Linear and quasilinear complex equations of hyperbolic and mixed type

Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Type

Asian Mathematics Series A Series edited by Chung-Chun Yang Department of Mathematics, The Hong Kong University of Science and Technology, Hong Kong

Volume 1 Dynamics of transcendental functions Xin-Hou Hua and Chung-Chun Yang Volume 2 Approximate methods and numerical analysis for elliptic complex equations Guo Chun Wen Volume 3 Introduction to statistical methods in modern genetics Mark C.K. Yang Volume 4 Mathematical theory in periodic plane elasticity Hai-Tao Cai and Jian-ke Lu Volume 5 Gamma lines: On the geometry of real and complex functions Grigor A. Barsegian Volume 6 Linear and quasilinear complex equations of hyperbolic and mixed type Guo Chun Wen

Linear and Quasilinear Complex Equations of Hyperbolic and Mixed Type

Guo Chun Wen School of Mathematical Sciences, Peking University, Beijing, China

London and New York

First published 2002 by Taylor & Francis 11 New Fetter Lane, London EC4P 4EE Simultaneously published in the USA and Canada by Taylor & Francis Inc, 29 West 35th Street, New York, NY 10001 Taylor & Francis is an imprint of the Taylor & Francis Group © 2002 Guo Chun Wen This edition published in the Taylor & Francis e-Library, 2006.

“To purchase your own copy of this or any of Taylor & Francis or Routledge’s collection of thousands of eBooks please go to www.eBookstore.tandf.co.uk.” All rights reserved. No part of this book may be reprinted or reproduced or utilized in any form or by any electronic,mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers. Every effort has been made to ensure that the advice and information in this book is true and accurate at the time of going to press. However, neither the publisher nor the authors can accept any legal responsibility or liability for any errors or omissions that may be made. In the case of drug administration, any medical procedure or the use of technical equipment mentioned within this book, you are strongly advised to consult the manufacturer’s guidelines. British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging in Publication Data A catalog record for this book has been requested ISBN 0-203-16658-2 Master e-book ISBN

ISBN 0-203-26135-6 (Adobe eReader Format) ISBN 0–415–26971–7 (Print Edition)

Contents

Introduction to the series Preface Chapter I Hyperbolic complex equations of first order 1 2 3 4 5

3 4 5

1

Hyperbolic complex functions and hyperbolic pseudoregular functions 1 Complex forms of linear and nonlinear hyperbolic systems of first order equations 10 Boundary value problems of linear hyperbolic complex equations of first order 18 Boundary value problems of quasilinear hyperbolic complex equations of first order 25 Hyperbolic mappings and quasi-hyperbolic mappings 35

Chapter II Hyperbolic complex equations of second order 1 2

viii ix

Complex form of hyperbolic equations of second order 39 Oblique derivative problems for quasilinear hyperbolic equations of second order 43 Oblique derivative problems for general quasilinear hyperbolic equations of second order 50 Other oblique derivative problems for quasilinear hyperbolic equations of second order 59 Oblique derivative problems for degenerate hyperbolic equations of second order 66

39

vi

Contents

Chapter III Nonlinear elliptic complex equations of first and second order 1 2 3 4

Generalizations of Keldych–Sedov formula for analytic functions 79 Representation and existence of solutions for elliptic complex equations of first order 90 Discontinuous oblique derivative problems for quasilinear elliptic equations of second order 95 Boundary value problems for degenerate elliptic equations of second order in a simply connected domain 108

Chapter IV First order complex equations of mixed type 1 2 3 4 5

2 3 4 5

119

The Riemann–Hilbert problem for simplest first order complex equation of mixed type 119 The Riemann–Hilbert problem for first order linear complex equations of mixed type 126 The Riemann–Hilbert problem for first order quasilinear complex equations of mixed type 134 The Riemann–Hilbert problem for first order quasilinear equations of mixed type in general domains 138 The discontinuous Riemann–Hilbert problem for quasilinear mixed equations of first order 143

Chapter V Second order linear equations of mixed type 1

79

Oblique derivative problems for simplest second order equation of mixed type 157 Oblique derivative problems for second order linear equations of mixed type 162 Discontinuous oblique derivative problems for second order linear equations of mixed type 171 The Frankl boundary value problem for second order linear equations of mixed type 177 Oblique derivative problems for second order degenerate equations of mixed type 194

157

Contents Chapter VI Second order quasilinear equations of mixed type 1 2 3 4

vii

200

Oblique derivative problems for second order quasilinear equations of mixed type 200 Oblique derivative problems for second order equations of mixed type in general domains 209 Discontinuous oblique derivative problems for second order quasilinear equations of mixed type 218 Oblique derivative problems for quasilinear equations of mixed type in multiply connected domains 227 References Index

240 250

Introduction to the Series

The Asian Mathematics Series provides a forum to promote and reflect timely mathematical research and development from the Asian region, and to provide suitable and pertinent reference on text books for researchers, academics and graduate students in Asian universities and research institutes, as well as in the West. With the growing strength of Asian economic, scientific and technological development, there is a need more than ever before for teaching and research materials written by leading Asian researchers, or those who have worked in or visited the Asian region, particularly tailored to meet the growing demands of students and researchers in that region. Many leading mathematicians in Asia were themselves trained in the West, and their experience with Western methods will make these books suitable not only for an Asian audience but also for the international mathematics community. The Asian Mathematics Series is founded with the aim to present significant contributions from mathematicians, written with an Asian audience in mind, to the mathematics community. The series will cover all mathematical fields and their applications, with volumes contributed to by international experts who have taught or performed research in Asia. The material will be at graduate level or above. The book series will consist mainly of monographs and lecture notes but conference proceedings of meetings and workshops held in the Asian region will also be considered.

Preface

In this book, we mainly introduce first and second order complex equations of hyperbolic and mixed (elliptic-hyperbolic) type, in which various boundary value problems for first and second order linear and quasilinear complex equations of hyperbolic and mixed type are considered. In order to obtain the results on complex equations of mixed type, we need to first discuss some boundary value problems for elliptic and hyperbolic complex equations. In Chapters I and II, the hyperbolic pseudoregular functions and quasi-hyperbolic mappings are introduced, which are corresponding to pseudoanalytic functions and quasiconformal mappings in the theory of elliptic complex equations. On the basis of hyperbolic notations, the hyperbolic systems of first order equations and hyperbolic equations of second order with some conditions can be reduced to complex forms. In addition, several boundary value problems, mainly the Riemann– Hilbert problem, oblique derivative problems for some hyperbolic complex equations of first and second order are discussed in detail. In Chapter III, firstly the generalizations of the Keldych–Sedov formula for analytic functions are given. Moreover, discontinuous boundary value problems for nonlinear elliptic complex equations of first and second order are discussed. Besides some oblique derivative problems for degenerate elliptic equations of second order are also introduced. In Chapter IV, we mainly consider the discontinuous boundary value problems for first order linear and quasilinear complex equations of mixed type, which include the discontinuous Dirichlet problem and discontinuous Riemann–Hilbert problem. In the meantime we give some a priori estimates of solutions for the above boundary value problems. For the classical dynamical equation of mixed type due to S. A. Chaplygin [17], the first really deep results were published by F. Tricomi [77] 1). In Chapters V and VI, we consider oblique derivative boundary value problems for second order linear and quasilinear complex equations of mixed type by using a complex analytic method in a special domain and in general domains, which include the Dirichlet problem (Tricomi problem) as a special case. We mention that in the books [12] 1), 3), the author investigated the Dirichlet problem (Tricomi problem) for the simplest second order equation of mixed type, i.e. uxx +sgnyuyy =0 in

x

Preface

general domains by using the method of integral equations and a complicated functional relation. In the present book, we use the uniqueness and existence of solutions of discontinuous Riemann–Hilbert boundary value problem for elliptic complex equations and other methods to obtain the solvability result of oblique derivative problems for more general equations and domains, which includes the results in [12] 1), 3) as special cases. Similarly to the book [86] 1), the considered complex equations and boundary conditions in this volume are rather general, and several methods are used. There are two characteristics of this book: one is that mixed complex equations are included in the quasilinear case, and boundary value conditions are almost considered in the general oblique derivative case, especially multiply connected domains are considered. Another one is that complex analytic methods are used to investigate various problems about complex equations of hyperbolic and mixed type. We mention that some free boundary problems in gas dynamics and some problem in elasticity can be handled by using the results stated in this book. The great majority of the contents originates in investigations of the author and his cooperative colleagues, and many results are published here for the first time. After reading the book, it can be seen that many questions about complex equations of mixed type remain for further investigations. The preparation of this book was supported by the National Natural Science Foundation of China. The author would like to thank Prof. H. Begehr, Prof. W. Tutschke and Mr Pi Wen Yang, because they proposed some beneficial improving opinions to the manuscript of this book. Beijing August, 2001

Guo Chun Wen Peking University

CHAPTER I HYPERBOLIC COMPLEX EQUATIONS OF FIRST ORDER In this chapter, we first introduce hyperbolic numbers, hyperbolic regular functions and hyperbolic pseudoregular functions. Next we transform the linear and nonlinear hyperbolic systems of first order equations into complex forms. Moreover we discuss boundary value problems for some hyperbolic complex equations of first order. Finally, we introduce the so-called hyperbolic mappings and quasihyperbolic mappings.

1

1.1

Hyperbolic Complex Functions and Hyperbolic Pseudoregular Functions Hyperbolic numbers and hyperbolic regular functions

First of all, we introduce hyperbolic numbers and hyperbolic complex functions. The so-called hyperbolic number is z = x + jy, where x, y are two real numbers and j is called the hyperbolic unit such that j 2 = 1. Denote e2 = (1 − j)/2,

e1 = (1 + j)/2,

(1.1)

it is easy to see that


e1 + e2 = 1,

ek el =

ek , if k = l, 0, if k
= l,

k, l = 1, 2,

(1.2)

and (e1 , e2 ) will be called the hyperbolic element. Moreover, w = f (z) = u(x, y) + jv(x, y) is called a hyperbolic complex function, where u(x, y), v(x, y) are two real functions of two real variables x, y which are called the real part and imaginary part of w = f (z) and denote Re w = u(z) = u(x, y), Im w = v(z) = v(x, y). Obviously, z = x + jy = µe1 + νe2 ,

w = f (z) = u + jv = ξe1 + ηe2 ,

in which µ = x + y,

ν = x − y,

x = (µ + ν)/2,

y = (µ − ν)/2,

ξ = u + v,

η = u − v,

u = (ξ + η)/2,

v = (ξ − η)/2.

(1.3)

2

I. Hyperbolic Equations of First Order

z¯ = x−jywill be called the conjugate number of z. The absolute value of z is defined √ by |z| = |x2 − y 2 |, and the hyperbolic model of z is defined by z  = x2 + y 2 . The operations of addition, subtraction and multiplication are the same with the real numbers, but j 2 = 1. There exists the divisor of zero and denote by O = {z | x2 = y 2 } the set of divisors of zero and zero. It is clear that z ∈ O if and only if |z| = 0, and z has an inversion 1 z¯ 1 1 1 1 = = e1 + e2 = e1 + e2 , z z z¯ x + y x−y µ ν if and only if x + jy
∈ O, and if the hyperbolic numbers z1 = µ1 e1 + ν1 e2 , z2 = µ2 e1 + ν2 e2
∈ O, then 



z1 1 1 µ1 ν1 = (µ1 e1 + ν1 e2 ) e1 + e2 = e1 + e2 . z2 µ2 ν2 µ2 ν2 It is clear that |z1 z2 | = |z1 ||z2 |, but the triangle inequality is not true. As for the hyperbolic √ model of z, we have the triangle inequality:  z1 +z2  ≤  z1  +  z2 , and  z1 z2  ≤ 2  z1  z2  . In the following, the limits of the hyperbolic number are defined by the hyperbolic model. The derivatives of a hyperbolic complex function w = f (z) with respect to z and z¯ are defined by wz = (wx + jwy )/2,

wz¯ = (wx − jwy )/2,

(1.4)

respectively, and then we have wz¯ = (wx − jwy )/2 = [(ux − vy ) + j(vx − uy )]/2 = [(wx − wy )e1 +(wx + wy )e2 ]/2 = wν e1 + wµ e2 = [ξν e1 + ην e2 ]e1 +[ξµ e1 + ηµ e2 ]e2 = ξν e1 + ηµ e2 ,

(1.5)

wz = [(ux + vy ) + j(vx + uy )]/2 = wµ e1 + wν e2 = (ξe1 + ηe2 )µ e1 + (ξe1 + ηe2 )ν e2 = ξµ e1 + ην e2 . Let D be a domain in the (x, y)-plane. If u(x, y), v(x, y) are continuously differentiable in D, then we say that the function w = f (z) is continuously differentiable in D, and we have the following result. Theorem 1.1 Suppose that the hyperbolic complex function w = f (z) is continuously differentiable. Then the following three conditions are equivalent. (1) wz¯ = 0;

(1.6)

(2) ξν = 0, ηµ = 0;

(1.7)

(3) ux = vy , vx = uy .

(1.8)

Proof From (1.5), it is easy to see that the conditions (1),(2) and (3) in Theorem 1.1 are equivalent.

1. Hyperbolic Complex Functions

3

The system of equations (1.8) is the simplest hyperbolic system of first order equations, which corresponds to the Cauchy–Riemann system in the theory of elliptic equations. The continuously differentiable solution w = f (z) of the complex equation (1.6) in D is called a hyperbolic regular function in D. If the function w(z) is defined and continuous in the neighborhood of a point z0 , and the following limit exists and is finite: w(z)−w(z0 ) z−  z0  ξ(z)−ξ(z0 ) η(z)−η(z0 ) = µ→µlim e1 + e2 0 ,ν→ν0 µ − µ0 ν − ν0

w
(z0 ) = lim z→z

0

= [ξµ e1 + ην e2 ]|µ=µ0 ,ν=ν0 = wz (z0 ), then we say that w(z) possesses the derivative w
(z0 ) at z0 . From the above formula, we see that w(z) possesses a derivative at z0 if and only if ξ(z) = Re w(z) + Im w(z), η(z) = Re w(z) − Im w(z) possess derivatives at µ0 = x0 + y0 , ν0 = x0 − y0 respectively. Now, we can define some elementary hyperbolic regular functions according to series representations in their convergent domains as follows: z n = [µe1 + νe2 ]n = µn e1 + ν n e2 =

(x + y)n + (x − y)n (x + y)n −(x − y)n +j , 2 2

zn ex+y − ex−y z2 ex+y + ex−y + ··· + + · · · = eµ e1 + eν e2 = +j , 2! n! 2 2 ln(x + y) − ln(x − y) ln(x + y) + ln(x − y) +j , ln z = ln µ e1 + ln ν e2 = 2 2

ez = 1 + z +

sin z = z −

z3 z 2n+1 + · · · + (−1)n + · · · = sin µ e1 + sin ν e2 , 3! (2n + 1)!

cos z = 1 −

z 2n z2 + · · · + (−1)n + · · · = cos µ e1 + cos ν e2 , 2! (2n)! 

tgz =



1 1 sin z = (sin µe1 + sin νe2 ) e1 + e2 = tgµ e1 + tgν e2 , cos z cos µ cos ν 



cos z 1 1 ctgz = = (cos µe1 + cos νe2 ) e1 + e2 = ctgµ e1 +ctgν e2 , sin z sin µ sin ν (1+z)α = 1+αz+· · ·+

α(α−1) . . . (α−n+1) n z +· · · = (1+µ)α e1 +(1+ν)α e2 , n!

where n is a positive integer and α is a positive number. Moreover we can define the series expansion of hyperbolic regular functions and discuss its convergency.

4

I. Hyperbolic Equations of First Order

1.2

Hyperbolic continuous functions and their integrals

Suppose that w = f (z) = u(x, y) + jv(x, y) is any hyperbolic complex function in a domain D and possesses continuous partial derivatives of first order in D. Then for any point z0 ∈ D, we have ∆w = fx
(z0 )∆x + fy
(z0 )∆y + ε(∆z), where ∆w = f (z) − f (z0 ), fx
(z0 ) = ux (z0 ) + jvx (z0 ), fy
(z0 ) = uy (z0 ) + jvy (z0 ), and z = z0 + ∆z, ε is a function of ∆z and lim ε(∆z) → 0.

∆z→0

Suppose that C is a piecewise smooth curve in the domain D, and w = f (z) = u + jv = ξe1 + ηe2 is a continuous function in D. Then the integral of f (z) along C and D are defined by 



f (z)dz = C



C





udx + vdy + j[

vdx + udy] = C



f (z)dxdy =

udxdy + j

D

C



[ξdµe1 + ηdνe2 ],

vdxdy.

D

D

We easily obtain some properties of integrals of f (z) as follows. Theorem 1.2 (1) If f (z), g(z) are continuous functions in D and C is a piecewise smooth curve in D, then 



[f (z) + g(z)]dz = C



f (z)dz + C



[f (z) + g(z)]dxdy = D

g(z)dz, C





f (z)dxdy + D

g(z)dxdy. D

(2) Under the conditions in (1) and denoting M1 = maxz∈C  f (z) , M2 = supz∈D  f (z) , the length of C by l and the area of D by S, then  √ f (z)dz ≤ 2M1 l,  C



 D

f (z)dxdy ≤

√

2M2 S.

(3) If C is a piecewise smooth closed curve and G is the finite domain bounded by ¯ then we have Green’s formulas C, f (z) is continuously differentiable in G, 

j f (z)dz, 2 C G  j [f (z)]z dxdy = − f (z)d¯ z. 2 C G [f (z)]z¯dxdy =

1. Hyperbolic Complex Functions

5

(4) Under the conditions as in (3) of Theorem 1.1 and w = f (z) is a hyperbolic ¯ then regular function in G,  f (z)dz = 0. C

In the following, we introduce the definition of hyperbolic pseudoregular functions, and prove some properties of hyperbolic pseudoregular functions. 1.3

Hyperbolic pseudoregular functions and their properties

Let w(z), F (z), G(z) be continuous functions in a domain D and G(z), F (z) satisfy the conditions Im F (z)G(z)
= 0 in D. (1.9) Then for every point z0 ∈ D, we can obtain a unique pair of real numbers δ0 and γ0 , such that w(z0 ) = δ0 F (z0 ) + γ0 G(z0 ). (1.10) Setting W (z) = w(z) − δ0 F (z) − γ0 G(z),

(1.11)

W (z0 ) = 0.

(1.12)

it is easy to see that If the following limit exists and is finite: lim w(z ˙ 0 ) = z→z

0

w(z) − δ0 F (z) − γ0 G(z) W (z) − W (z0 ) = z→z lim , i.e. 0 z − z0 z − z0

(1.13)

w(z ˙ 0 ) = [Re W (z0 )+Im W (z0 )]µ e1 +[Re W (z0 )−Im W (z0 )]ν e2 = W
(z0 ), where µ = x + y, ν = x − y, then we say that w(z) ˙ is a (F, G)-derivative of w(z) at z0 . In order to express the existence of (1.13) by partial differential equations, we suppose again that Fz (z), Fz¯(z), Gz (z) and Gz¯(z) exist and are continuous

(1.14)

in a neighborhood of z0 . According to the definition of W (z), if Wz , Wz¯ exist, then Wz (z) = wz (z) − δ0 Fz (z) − γ0 Gz (z), Wz¯(z) = wz¯(z) − δ0 Fz¯(z) − γ0 Gz¯(z).

(1.15)

From (1.13), (1.15) and Theorem 1.1, we see that if w(z ˙ 0 ) exists, then Wz (z0 ) exists, Wz (z0 ) = w(z ˙ 0 ),

(1.16)

Wz¯(z0 ) = 0;

(1.17)

and

6

I. Hyperbolic Equations of First Order

and if wz (z), wz¯(z) are continuous in a neighborhood of z0 , and (1.17) holds, then we have (1.13) and (1.16). Since

W (z) =

   w(z) w(z ) w(z )   0 0       F (z) F (z0 ) F (z0 )       G(z) G(z0 ) G(z0 )    ,    F (z0 ) F (z0 )       G(z0 ) G(z0 ) 

(1.18)

(1.17) can be rewritten as   w (z )  z¯ 0    Fz¯(z0 )    Gz¯(z0 )



w(z0 ) w(z0 )   

F (z0 ) F (z0 )  = 0. G(z0 )

(1.19)

 G(z0 ) 

If (1.13) exists, then (1.16) can be written as

w(z ˙ 0) =

   w (z ) w(z ) w(z )   z 0 0 0       Fz (z0 ) F (z0 ) F (z0 )       Gz (z0 ) G(z0 ) G(z0 )    .    F (z0 ) F (z0 )       G(z0 ) G(z0 ) 

(1.20)

Unfolding (1.19) and (1.20) respectively, and arranging them, we obtain

where

¯ wz¯ = aw + bw,

(1.21)

¯ w˙ = wz − Aw − B w,

(1.22)

¯ F Gz¯ − Fz¯G F¯ Gz¯ − Fz¯G ¯ − F¯ G , b = F G ¯ − F¯ G , FG ¯ F Gz − Fz G F¯ Gz − Fz G ,B= A=− ¯ ¯ ¯ − F¯ G , FG − FG FG

(1.23)

a=−

here a(z), b(z), A(z) and B(z) are called the characteristic coefficients of the generating pair (F, G). Obviously F˙ = G˙ = 0, and ¯ Fz¯ = aF + bF¯ , Gz¯ = aG + bG, ¯ Fz = AF + B F¯ , Gz = AG + B G uniquely determine a, b, A and B. Denote them by a(F,G) , b(F,G) , A(F,G) and B(F,G) respectively.

1. Hyperbolic Complex Functions

7

From the above discussion, we see that if w(z ˙ 0 ) exists, then wz at z0 exists and (1.21), (1.22) are true. If wz and wz¯(z) exist and are continuous in a neighborhood z0 ∈ D, and (1.21) holds at z0 , then w(z ˙ 0 ) exists, and (1.22) is true. For any function w(z), if w(z) ˙ exists and is continuous in the domain D, then w(z) is called the first-class (F, G) hyperbolic pseudoregular function or hyperbolic pseudoregular function for short. It is clear that the following theorem holds. Theorem 1.3 w(z) is a hyperbolic pseudoregular function if and only if wz (z) and wz¯(z) exist and are continuous, and (1.21) holds. By (1.9), it is easy to see that every function w(z) has a unique expression w(z) = φ(z)F (z) + ψ(z)G(z),

(1.24)

where φ(z) and ψ(z) are two real-valued functions. Let K(z) = φ(z) + jψ(z).

(1.25)

Then we can give the following definition. If w(z) is the first-class (F, G) hyperbolic pseudoregular complex function, then K(z) = φ(z) + jψ(z) is called the second-class (F, G) hyperbolic pseudoregular function. Theorem 1.4 K(z) = φ(z)+jψ(z) is a second-class (F, G) hyperbolic pseudoregular function if and only if φ and ψ have continuous partial derivatives and F φz¯ + Gψz¯ = 0.

(1.26)

w(z) ˙ = F φz + Gψz

(1.27)

Under this condition, holds, where φz = (e1 φµ + e2 φν ) = [(φx + φy )e1 + (φx − φy )e2 ]/2, φz¯ = (e1 φν + e2 φµ ) = [(φx − φy )e1 + (φx + φy )e2 ]/2, ψz = (e1 ψµ + e2 ψν ) = [(ψx + ψy )e1 + (ψx − ψy )e2 ]/2, ψz¯ = (e1 ψν + e2 ψµ ) = [(ψx − ψy )e1 + (ψx + ψy )e2 ]/2. Proof From W (z) = [φ(z) − φ(z0 )]F (z) + [ψ(z) − ψ(z0 )]G(z), it follows that Wz (z0 ) = F (z0 )φz (z0 ) + G(z0 )ψz (z0 ), Wz¯(z0 ) = F (z0 )φz¯(z0 ) + G(z0 )ψz¯(z0 ).

(1.28)

8

I. Hyperbolic Equations of First Order

Thus the proof can be immediately obtained. Setting G = σ + jτ = (σ + τ )e1 + (σ − τ )e2 , F F ˜ + j τ˜ = (˜ σ + τ˜)e1 + (˜ − =σ σ − τ˜)e2 , G

−

(1.29)

where σ, τ, σ ˜ and τ˜ are real-valued functions, and τ
= 0, τ˜
= 0. Hence (1.26) is equivalent to the system of equations φx = σψx − τ ψy , φy = −τ ψx + σψy .

(1.30)

If φ and ψ have continuous partial derivatives up to second order, we find the derivatives with respect to x and y in (1.30), and then obtain ˜ x + γ˜ φy = 0, φxx − φyy + δφ

(1.31)

ψxx − ψyy + δψx + γψy = 0, where σ ˜y + τ˜x , δ˜ = τ˜ σy + τx δ= , τ

σ ˜x + τ˜y , τ˜ σx + τy . γ=− τ

γ˜ = −

(1.32)

In accordance with the following theorem, the elimination is reasonable. Theorem 1.5 Let δ, γ, δ˜ and γ˜ be determined by (1.32). Then the real-valued function φ (ψ) is the real (imaginary) part of a second-class hyperbolic pseudoregular function if and only if it has continuous partial derivatives up to second order, and satisfies the first (second) equation in (1.31). Proof

From the second formula in (1.31), we see that the function 

z

φ(z) = z0

[(σψx − τ ψy )dx + (−τ ψx + σψy )dy],

z0 , z ∈ D

is single-valued, and φ(z), ψ(z) satisfy system (1.30). The part of necessity can be derived from Theorem 2.3, Chapter II below. 1.4

Existence of a generating pair (F,G)

Theorem 1.6 Let a(z) and b(z) be two continuous functions in a bounded and closed domain D = {µ0 ≤ µ ≤ µ0 + R1 , ν0 ≤ ν ≤ ν0 + R2 }, where R1 , R2 are positive constants, and denote z0 = µ0 e1 + ν0 e2 . Then there exists a unique continuously differentiable hyperbolic pseudoregular function w(z) satisfying the complex equation wz¯ = a(z)w(z) + b(z)w(z),

(1.33)

1. Hyperbolic Complex Functions

9

and the boundary conditions w(z) = c1 (µ)e1 + c2 (ν0 )e2 ,

when

z ∈ L1 ,

w(z) = c1 (µ0 )e1 + c2 (ν)e2 ,

when

z ∈ L2 ,

(1.34)

where c1 (µ) and c2 (ν) are two real continuous functions on L1 , L2 respectively, L1 = {µ0 ≤ µ ≤ µ0 + R1 , ν = ν0 } and L2 = {µ = µ0 , ν0 ≤ ν ≤ ν0 + R2 }. The theorem is a special case of Theorems 3.3 and 3.4 below. Theorem 1.7 Let a(z) and b(z) be two continuous complex functions in the domain D as stated in Theorem 1.6. Then there exists a generating pair (F, G) in D, such that a = a(F,G) and b = b(F,G) . (1.35) Proof Denote by F (z) and G(z) two solutions of the complex equation (1.33) satisfying the boundary conditions w(z) = e1 + e2 = 1,

when z ∈ L1 ∪ L2

w(z) = e1 − e2 = j,

when z ∈ L1 ∪ L2 ,

and respectively. Then by Theorem 1.6, F (z) and G(z) have continuous partial derivatives, and ¯ Fz¯ = aF + bF¯ , and Gz¯ = aG + bG, F = 1, G = j,

when z ∈ L1 ∪ L2 .

Hence a = a(F,G) and b = b(F,G) . Whether Im F (z)G(z)
= 0 in D remains to be discussed. Theorem 1.8 Under the same conditions as in Theorem 1.7, and letting b = −a

z ∈ D,

or b = a,

(1.36)

there exists a generating pair (F, G) in D satisfying the complex equation (1.33), and F (z) = 1 or G(z) = j

in D.

(1.37)

Proof By the hypotheses in Theorem 1.7 and equation (1.36), there exists a unique generating pair (F, G) in D satisfying the conditions Fz¯ = a(F − F¯ ), F (z) = 1

¯ or Gz¯ = a(G + G),

or G(z) = j,

z ∈ D,

when z ∈ L1 ∪ L2 .

Hence we have (1.37). The above results are similar to those in [9]1) (see [89]).

10

2

I. Hyperbolic Equations of First Order

Complex Forms of Linear and Nonlinear Hyperbolic Systems of First Order Equations

In this section, we transform linear and nonlinear hyperbolic systems of first order equations into complex forms. 2.1

Complex forms of linear hyperbolic systems of first order equations

We consider the linear hyperbolic system of first order partial differential equations ⎧ ⎨ a11 ux ⎩

+ a12 uy + b11 vx + b12 vy = a1 u + b1 v + c1 ,

(2.1)

a21 ux + a22 uy + b21 vx + b22 vy = a2 u + b2 v + c2 ,

where the coefficients akl , bkl , ak , bk , ck (k, l = 1, 2) are known functions in D, in which D is a bounded domain. System (2.1) is called hyperbolic at a point in D, if at the point, the inequality I = (K2 + K3 )2 − 4K1 K4 > 0 (2.2) holds, in which  a

K1 = 



11

 a21

 a

b11  , b21 

K2 = 

11

 a21



b12  , b22 

 a

K3 = 

12

 a22



b11  , b21 

 a

K4 = 

12

 a22



b12  . b22 

If the inequality (2.2) at every point (x, y) in D holds, then (2.1) is called a hyperbolic system in D. We can verify that (2.2) can be rewritten as I = (K2 − K3 )2 − 4K5 K6 > 0, where K5 =

 a  11   a21



a12  , a22 

K6 =

 b  11   b21

(2.3)



b12  . b22 

If the coefficients akl , bkl (k, l = 1, 2) in D are bounded, and the condition I = (K2 + K3 )2 − 4K1 K4 = (K2 − K3 )2 − 4K5 K6 ≥ I0 > 0

(2.4)

holds, in which I0 is a positive constant, then (2.1) is called uniformly hyperbolic in D. In the following, we reduce system (2.1) to complex form. 1) If K2 , K3 are of same signs and K6
= 0 at the point (x, y) ∈ D, then we can solve (2.1) for vy , −vx and obtain the system of first order equations ⎧ ⎨ vy ⎩

= aux + buy + a0 u + b0 v + f0 ,

−vx = dux + cuy + c0 u + d0 v + g0 ,

(2.5)

where a, b, c, d are known functions of akl , bkl (k, l = 1, 2) and a0 , b0 , c0 , d0 , f0 , g0 are known functions of bkl , ak , bk , ck (k, l = 1, 2), and a = K1 /K6 ,

b = K3 /K6 ,

c = K4 /K6 ,

d = K2 /K6 .

2. Complex Forms of Hyperbolic Systems

11

The condition (2.2) of hyperbolic type is reduced to the condition ∆ = I/4K62 =

(b + d)2 − ac > 0. 4

(2.6)

There is no harm in assuming that a − c ≥ 0, because otherwise let y be replaced by −y, this requirement can be realized. If a, c are not of the same sign or one of them is equal to zero, then −ac ≥ 0, bd ≥ 0, and may be such that a ≥ 0, −c ≥ 0; or a, c are of same signs, then we may assume that a > 0, c > 0, because otherwise if v is replaced by −v, this requirement can be realized. Moreover, we can assume that 0 < c < 1, otherwise, setting v = h˜ v , herein h is a positive constant such that ˜ 4 = hK4 , K ˜ 6 = h2 K6 , and c˜ = K ˜ 4 /K ˜ 6 = c/h < 1, ˜bd˜ ≥ 0. h ≥ c + 1, we have K Multiply the first formula of (2.5) by −j and then subtract the second formula of (2.5). This gives vx − jvy = − j(aux + buy + a0 u + b0 v + f0 ) − dux − cuy − c0 u − d0 v − g0 . Noting z = x + jy, w = u + jv, and using the relations ⎧ ⎨ ux ⎩

= (wz¯ + w¯z + wz + w¯z¯)/2,

uy = j(−wz¯ + w¯z + wz − w¯z¯)/2,

vx = j(wz¯ − w¯z + wz − w¯z¯)/2,

vy = (−wz¯ − w¯z + wz + w¯z¯)/2,

we get j(wz¯ − w¯z¯) = −(aj + d)(wz¯ + w¯z + wz + w¯z¯)/2 −(c + bj)j(−wz¯ + w¯z + wz − w¯z¯)/2 +lower order terms, namely (1 + q1 )wz¯ + q2 w¯z = −q2 wz + (1 − q1 )w¯z¯ + lower order terms, in which q1 = [a − c + (d − b)j]/2,

q2 = [a + c + (d + b)j]/2.

Noting q0 = (1 + q1 )(1 + q¯1 ) − q2 q¯2 = [(2 + a − c)2 − (d − b)2 − (a + c)2 + (d + b)2 ]/4 = 1 + a − c − (d − b)2 /4 + (d + b)2 /4 − ac = 1 + a − c − (d − b)2 /4 + ∆ = 1 + a − c + σ = 1 + a − c + bd − ac = (1 + a)(1 − c) + bd > 0,

(2.7)

12

I. Hyperbolic Equations of First Order

where σ = ∆ − (b − d)2 /4 = bd − ac ≥ 0, thus we can solve (2.7) for wz¯, giving wz¯ − Q1 (z)wz − Q2 (z)w¯z¯ = A1 (z)w + A2 (z)w¯ + A3 (z), in which Q1 (z) =

−2q2 (z) , q0 (z)

Q2 (z) =

(2.8)

q1 + 1)] [q2 q¯2 − (q1 − 1)(¯ . q0

For the complex equation (2.8), if (a − c)2 − 4∆ ≥ 0, (1 + σ)2 − 4∆ ≥ 0, i.e. (K1 − K4 )2 − (K2 − K3 )2 + 4K5 K6 ≥ 0, (K62 + K2 K3 − K1 K4 )2 − (K2 − K3 )2 + 4K5 K6 ≥ 0,

(2.9)

then we can prove |Q1 | + |Q2 | = |Q1 Q1 |1/2 + |Q2 Q2 |1/2 < 1,

(2.10)

where |Q1 | = |Q1 Q1 |1/2 is the absolute value of Q1 . In fact, |2q2 | = |(a + c)2 − (d + b)2 |1/2 = |(a − c)2 − 4∆|1/2 , |q2 q2 − (q1 − 1)(q1 + 1)| = |(a − c)2 /4 − ∆ − [a − c + (b − d)j − 2] ×[a − c − (d − b)j + 2]/4| = |(1 + σ)2 − 4∆|1/2 , (1 + σ)2 + (a − c)2 + 2(1 + σ)(a − c) = [1 + σ + (a − c)]2 > 0, (1 + σ)2 (a − c)2 + (4∆)2 − 4∆(1 + σ)2 − 4∆(a − c)2 < (4∆)2 + (1 + σ)2 (a − c)2 + 8∆(1 + σ)(a − c), [(a − c)2 − 4∆][(1 + σ)2 − 4∆] < (4∆)2 + (1 + σ)2 (a − c)2 + 8∆(1 + σ)(a − c), 2{[(a − c)2 − 4∆][(1 + σ)2 − 4∆]}1/2 < 8∆ + 2(1 + σ)(a − c), (a − c)2 − 4∆ + (1 + σ)2 − 4∆ +2{[(a − c)2 − 4∆][(1 + σ)2 − 4∆]}1/2 < (1 + σ + a − c)2 , and then |(a − c)2 − 4∆|1/2 + |(1 + σ)2 − 4∆|1/2 < 1 + σ + a − c, thus, we can derive (2.10). 2) K2 , K3 at (x, y) ∈ D have same signs, K6 = 0, K5
= 0; by using similar methods, we can transform (2.1) into a complex equation in the form (2.8).

2. Complex Forms of Hyperbolic Systems

13

Now, we discuss the case: 3) K2 , K3 are not of same signs, K4
= 0 at the point (x, y) ∈ D, then we can solve (2.1) for uy , vy and obtain the system of first order equations ⎧ ⎨ vy ⎩

= aux + bvx + a0 u + b0 v + f0 ,

−uy = dux + cvx + c0 u + d0 v + g0 ,

(2.11)

where a, b, c, d are known functions of akl , bkl (k, l = 1, 2) and a0 , b0 , c0 , d0 , f0 , g0 are known functions of ak2 , bk2 , ak , bk , ck (k, l = 1, 2), and a = K5 /K4 ,

b = −K3 /K4 ,

c = K6 /K4 ,

d = K2 /K4 .

The condition (2.2) of hyperbolic type is reduced to the condition ∆ = I/4K42 = (b + d)2 /4 − ac > 0.

(2.12)

Similarly to (2.5), multiply the second formula of (2.11) by j, and then subtract the first formula of (2.11), we get −vy − juy = wz¯ − wz = −(a − d j)ux − (b − c j)vx + lower order terms = −(a − d j)(wz¯ + w¯z + wz + w¯z¯)/2 +(c − b j)(wz¯ − w¯z + wz − w¯z¯)/2 + lower order terms, namely (1 + q1 )wz¯ + q2 w¯z = (1 − q1 )wz − q2 w¯z¯ + lower order terms, where q1 =

(2.13)

[a − c − (d − b)j] [a + c − (d + b)j] , q2 = . 2 2

It is clear that q0 = (1 + q1 )(1 + q¯1 ) − q2 q¯2 =

[(2 + a − c)2 − (d − b)2 − (a + c)2 + (d + b)2 ] 4

= (1 + a)(1 − c) + bd > 0, thus we can solve (2.13) for wz¯, i.e. wz¯ − Q1 (z)wz − Q2 (z)w¯z¯ = A1 (z)w + A2 w¯ + A3 (z), in which Q1 (z) =

(2.14)

[(1 − q1 )(1 + q¯1 ) + |q2 |2 ] −2q2 (z) , Q2 = . q0 q0

4) K2 , K3 are not of same signs, K4 = 0, K1
= 0, by using similar methods as in 3), we can transform (2.1) into the complex equation of the form (2.14).

14

I. Hyperbolic Equations of First Order

2.2 Complex forms of nonlinear hyperbolic systems of first order equations Next, we consider the general nonlinear hyperbolic system of first order partial differential equations Fk (x, y, u, v, ux , uy , vx , vy ) = 0, k = 1, 2,

(2.15)

where the real functions Fk (k = 1, 2) are defined and continuous at every point (x, y) in D and possess continuous partial derivatives in ux , uy , vx , vy . For system (2.15), its condition of hyperbolic type can be defined by the inequality (2.2) or (2.3), but in which

K1 =

D(F1 , F2 ) , D(ux , vx )

D(F1 , F2 ) K4 = , D(uy , vy )

K2 =

D(F1 , F2 ) , D(ux , vy )

D(F1 , F2 ) K5 = , D(ux , uy )

K3 =

D(F1 , F2 ) , D(uy , vx )

(2.16)

D(F1 , F2 ) K6 = , D(vx , vy )

where Fkux , Fkuy , Fkvx , Fkvy (k = 1, 2) can be found as follows 

Fkux = 

Fkuy = 

Fkvx = 

Fkvy =

1

Fktux (x, y, u, v, tux , tuy , tvx , tvy )dt,

0 1

0

(2.17)

1

Fktvx (x, y, u, v, tux , tuy , tvx , tvy )dt,

0

0

Fktuy (x, y, u, v, tux , tuy , tvx , tvy )dt,

1

Fktvy (x, y, u, v, tux , tuy , tvx , tvy )dt.

By using the method in Subsection 2.1, for cases 1) K2 , K3 are of same signs, K5 or K6
= 0; 2) K2 , K3 are not of same signs, K1 or K4
= 0; then system (2.15) can be reduced to the complex form wz¯ − Q1 wz − Q2 w¯z¯ = A1 w + A2 w¯ + A3 ,

(2.18)

where z = x + jy, w = u + jv and Qk = Qk (z, w, wz , wz¯),

k = 1, 2,

Ak = Ak (z, w, wz , wz¯),

k = 1, 2, 3.

In particular, if (2.9) holds, from the condition of hyperbolic type in (2.2), it follows that (2.10) holds. Theorem 2.1 Let system (2.15) satisfy the condition of hyperbolic type as in (2.2) and the conditions from the existence theorem for implicit functions. Then (2.15) is solvable with respect to wz¯, and the corresponding hyperbolic complex equation of first order : (2.18) can be obtained.

2. Complex Forms of Hyperbolic Systems

15

As for the cases 3) K1 = K4 = 0, K2 , K3 are not of same signs, K5
= 0 or K6
= 0, and 4) K5 = K6 = 0, K2 , K3 are of same signs, K1
= 0 or K4
= 0, we can transform the quasilinear case of hyperbolic system: (2.15) into the complex forms by using a similar method in the next subsection. 2.3 Complex forms of quasilinear hyperbolic systems of first order equations Finally we discuss the quasilinear hyperbolic system of first order partial differential equations ⎧ ⎨ a11 ux + a12 uy + b11 vx + b12 vy = a1 u + b1 v + c1 , (2.19) ⎩ a21 ux + a22 uy + b21 vx + b22 vy = a2 u + b2 v + c2 , where the coefficients akl , bkl (k, l = 1, 2) are known functions in (x, y) ∈ D and ak , bk , ck (k = 1, 2) are known functions of (x, y) ∈ D and u, v ∈ IR. The hyperbolicity condition of (2.19) is the same as for system (2.1), i.e. for any point (x, y) ∈ D, the inequality I = (K2 + K3 )2 − 4K1 K4 = (K2 − K3 )2 − 4K5 K6 > 0 (2.20) holds, in which K1 = K4 =

 a  11   a21  a  12   a22



b11  , b21 

K2 =

b12  , b22 

K5 =



 a  11   a21  a  11   a21



b12  , b22  

a12  , a22 

 a K3 =  12  a22  b  K6 =  11  b21



b11  , b21  

b12  . b22 

We first consider the case: 1) K1 = K4 = 0, K2 , K3 are not of same signs, K6
= 0 at the point (x, y) ∈ D. From K1 = K4 = 0, there exist real constants λ, µ, such that a11 = λb11 ,

a21 = λb21 ,

a12 = µb12 ,

a22 = µb22 ,

thus K2 = λK6 , and then

K3 = −µK6 ,

K5 = λµK6 ,

I = (K2 − K3 )2 − 4K5 K6 = [(λ + µ)2 − 4λµ]K62 = (K2 + K3 )2 − 4K1 K4 = (λ − µ)2 K62 > 0.

It is easy to see that λ
= µ, i.e. K2
= −K3 , in this case, system (2.19) becomes the form ⎧ ⎨ b11 (λu + v)x + b12 (µu + v)y = a1 u + b1 v + c1 , (2.21) ⎩ b21 (λu + v)x + b22 (µu + v)y = a2 u + b2 v + c2 . Setting U = λu + v,

V = µu + v, and noting  U  u    Vu



Uv  Vv

  

=

 λ    µ



1   

1

= λ − µ
= 0,

16

I. Hyperbolic Equations of First Order

and

µU − λV U −V , v= , λ−µ µ−λ system (2.21) can be written in the form u=

⎧ ⎨ b11 Ux ⎩

+ b12 Vy = a
1 U + b
1 V + c1 ,

(2.22)

b21 Ux + b22 Vy = a
2 U + b
2 V + c2 ,

where a
1 = thus

⎧ ⎨ Ux ⎩

a1 − µb1 , λ−µ

b
1 =

−a1 + λb1 , λ−µ

a
2 =

a2 − µb2 , λ−µ

b
2 =

−a2 + λb2 , λ−µ

= [(a
1 b22 − a
2 b12 )U + (b
1 b22 − b
2 b12 )V + (c1 b22 − c2 b12 )]/K6 ,

Vy = [(a
2 b11 − a
1 b21 )U + (b
2 b11 − b
1 b21 )V + (c2 b11 − c1 b21 )]/K6 .

(2.23)

Subtracting the first equation from the second equation, the complex equation of W = U + jV Wz¯ + W z = A1 (z, W )W + A2 (z, W )W + A3 (z, W )

(2.24)

can be derived, where A1 , A2 , A3 are known functions of bkl , ak , bk , ck (k, l = 1, 2). Moreover, we consider system (2.19) with the condition 2) K5 = K6 = 0, K2 , K3 are of same signs, and K4
= 0 at the point (x, y) ∈ D. In this case, due to K5 = K6 = 0 at the point (x, y) ∈ D, there exist real constants λ, µ, such that a11 = λa12 ,

a21 = λa22 ,

b11 = µb12 ,

b21 = µb22 ,

thus K1 = λµK4 ,

K2 = λK4 ,

K3 = µK4 ,

and then I = (K2 + K3 )2 − 4K1 K4 = [(λ + µ)2 − 4λµ]K42 = (λ − µ)2 K42 > 0. It is clear that if λ
= µ, i.e. K2
= K3 , then system (2.19) can become the form ⎧ ⎨ a12 (λux ⎩

+ uy ) + b12 (µvx + vy ) = a1 u + b1 v + c1 ,

a22 (λux + uy ) + b22 (µvx + vy ) = a2 u + b2 v + c2 .

Let ξ=

x − µy , λ−µ

η=

−x + λy , λ−µ

it is easy to see that  ξ  x    ηx



ξy 

  ηy 

1 = (λ − µ)2

  1     −1



−µ  λ

  

=

1
= 0, λ−µ

(2.25)

2. Complex Forms of Hyperbolic Systems

17

and x = λξ + µη, y = ξ + η. Thus system (2.25) can be rewritten in the form ⎧ ⎨ a12 uξ ⎩

+ b12 vη = a1 u + b1 v + c1 ,

(2.26)

a22 uξ + b22 vη = a2 u + b2 v + c2 .

This system can be solved for uξ , vη , namely ⎧ ⎨ uξ ⎩

where

= a
1 u + b
1 v + c
1 ,

(2.27)

vη = a
2 u + b
2 v + c
2 ,

a
1 = (a1 b22 − a2 b12 )/K4 ,

a
2 = (a2 a12 − a1 a22 )/K4 ,

b
1 = (b1 b22 − b2 b12 )/K4 ,

b
2 = (b2 a12 − b1 a22 )/K4 ,

c
1 = (c1 b22 − c2 b12 )/K4 ,

c
2 = (c2 a12 − c1 a22 )/K4 .

Denoting ζ = ξ + jη, then system (2.27) can be written in the complex form wζ¯ + w¯ζ = A
1 (ζ, w)w + A
2 (ζ, w)w¯ + A
3 (ζ, w), in which

A
1 , A
2 , A
3

(2.28)

are known functions of ak2 , bk2 , ak , bk , ck (k = 1, 2).

For 3) K1 = K4 = 0, K2 , K3 are not of same signs, and K5
= 0, and 4) K5 = K6 = 0, K2 , K3 are of same signs, and K1
= 0, by using a similar method, we can transform (2.19) into the complex equations (2.24) and (2.28) respectively. We mention that it is possible that the case: b11 = λa11 ,

b21 = λa21 ,

b12 = µa12 ,

b22 = µa22

a22 = λa21 ,

b12 = µb11 ,

b22 = µb21

occurs for 1) and 2), and a12 = λa11 ,

occurs for 3) and 4), then we can similarly discuss. In addition, if λ(x, y), µ(x, y) are known functions of (x, y) in D, then in the left-hand sides of the two equations in (2.21) should be added b11 λx u + b12 µy u and b21 λx u + b22 µy u. It is sufficient to modify the coefficient of u and the system is still hyperbolic. For 2)–4), we can similarly handle. The complex equations as stated in (2.24) and (2.28) can be written in the form wz¯ + w¯z = A(z, w)w + B(z, w)w¯ + C(z, w),

(2.29)

which is equivalent to the system of first order equations ux = au + bv + f,

vy = cu + dv + g,

(2.30)

where z = x + jy, w = u + jv, A = (a + jb − c − jd)/2, B = (a − jb − c + jd)/2, C = f − g. Let W (z) = ve1 + ue2 , Z = xe1 + ye2 , (2.31) where e1 = (1+j)/2, e2 = (1−j)/2. From (2.30), we can obtain the complex equation WZ¯ = vy e1 + ux e2 = A1 W + A2 W + A3 = F (Z, W ),

(2.32)

in which A1 (Z, W ) = de1 + ae2 , A2 (Z, W ) = ce1 + be2 , A3 (Z, W ) = ge1 + f e2 [92].

18

3

I. Hyperbolic Equations of First Order

Boundary Value Problems of Linear Hyperbolic Complex Equations of First Order

In this section, we mainly discuss the Riemann–Hilbert boundary value problem for linear hyperbolic complex equations of first order in a simply connected domain. We first give a representation of solutions for the above boundary value problem, and then prove the uniqueness and existence of solutions for the above problem by using the successive iteration. 3.1 Formulation of the Riemann–Hilbert problem and uniqueness of its solutions for simplest hyperbolic complex equations Let D be a simply connected bounded domain in the x + jy-plane with the boundary Γ = L1 ∪ L2 ∪ L3 ∪ L4 , where L1 = {x = −y, 0 ≤ x ≤ R1 }, L2 = {x = y + 2R1 , R1 ≤ x ≤ R2 }, L3 = {x = −y − 2R1 + 2R2 , R = R2 − R1 ≤ x ≤ R2 }, L4 = {x = y, 0 ≤ x ≤ R2 − R1 }, and denote z0 = 0, z1 = (1 − j)R1 , z2 = R2 + j(R2 − 2R1 ), z3 = (1 + j)(R2 − R1 ) = (1 + j)R and L = L1 ∪ L4 , where j is the hyperbolic unit. For convenience we only discuss the case R2 ≥ 2R1 , the other case can be discussed by a similar method. We consider the simplest hyperbolic complex equation of first order: wz¯ = 0 in D.

(3.1)

The Riemann–Hilbert boundary value problem for the complex equation (3.1) may be formulated as follows: ¯ satisfying the boundary Problem A Find a continuous solution w(z) of (3.1) in D conditions Re [λ(z)w(z)] = r(z),

z ∈ L,

Im [λ(z0 )w(z0 )] = b1 ,

(3.2)

where λ(z) = a(z) + jb(z)
= 0, z ∈ L, and λ(z), r(z), b1 satisfy the conditions Cα [λ(z), L] = Cα [Re λ, L] + Cα [Im λ, L] ≤ k0 , Cα [r(z), L] ≤ k2 , |b1 | ≤ k2 ,

max z∈L1

1 , |a(z) − b(z)|

max z∈L4

1 ≤ k0 , |a(z) + b(z)|

(3.3) (3.4)

in which b1 is a real constant and α (0 < α < 1), k0 , k2 are non-negative constants. In particular, when a(z) = 1, b(z) = 0, i.e. λ(z) = 1, z ∈ L, Problem A is the Dirichlet problem (Problem D), whose boundary condition is Re [w(z)] = r(z),

z ∈ L,

Im [w(z0 )] = b1 .

(3.5)

Problem A with the conditions r(z) = 0, z ∈ L, b1 = 0 is called Problem A0 . On the basis of Theorem 1.1, it is clear that the complex equation (3.1) can be reduced to the form ξν = 0,

ηµ = 0,

(µ, ν) ∈ Q = {0 ≤ µ ≤ 2R, 0 ≤ ν ≤ 2R1 },

(3.6)

3. Linear Hyperbolic Equations

19

where µ = x + y, ν = x − y, ξ = u + v, η = u − v. Hence the general solution of system (3.6) can be expressed as ξ = u + v = f (µ) = f (x + y), u = [f (x + y) + g(x − y)]/2,

η = u − v = g(ν) = g(x − y),

i.e. (3.7)

v = [f (x + y) − g(x − y)]/2,

in which f (t), g(t) are two arbitrary real continuous functions on [0, 2R], [0, 2R1 ] respectively. From the boundary condition (3.2), we have a(z)u(z) + b(z)v(z) = r(z) on L,

λ(z0 )w(z0 ) = r(z0 ) + jb1 ,

i.e.

[a((1 − j)x) + b((1 − j)x)]f (0) + [a((1 − j)x) − b((1 − j)x)] × g(2x) = 2r((1 − j)x) on [0, R1 ], [a((1 + j)x) + b((1 + j)x)]f (2x) + [a((1 + j)x) − b((1 + j)x)]

(3.8)

× g(0) = 2r((1 + j)x) on [0, R], f (0) = u(0)+v(0) =

r(0)+b1 , a(0)+b(0)

g(0) = u(0)−v(0) =

r(0)−b1 . a(0)−b(0)

The above formulas can be rewritten as [a((1 − j)t/2)+b((1 − j)t/2)]f (0) + [a((1 − j)t/2) − b((1 − j)t/2)] × g(t) = 2r((1 − j)t/2), t ∈ [0, 2R1 ], [a((1 + j)t/2) + b((1 + j)t/2)]f (t) +[a((1 + j)t/2) − b((1 + j)t/2)] × g(0) = 2r((1 + j)t/2), t ∈ [0, 2R], i.e.

f (x + y) =

g(x − y) =

2r((1 + j)(x + y)/2) a((1 + j)(x + y)/2)+b((1 + j)(x + y)/2) [a((1+j)(x+y)/2) − b((1+j)(x+y)/2)]g(0) , − a((1+j)(x+y)/2) + b((1+j)(x+y)/2) 0 ≤ x+y ≤ 2R,

(3.9)

2r((1 − j)(x − y)/2) a((1 − j)(x − y)/2) − b((1 − j)(x − y)/2) [a((1−j)(x−y)/2)+b((1−j)(x−y)/2)]f (0) − , a((1−j)(x−y)/2)−b((1−j)(x−y)/2) 0 ≤ x−y ≤ 2R1 .

Thus the solution w(z) of (3.1) can be expressed as w(z) = f (x + y)e1 + g(x − y)e2 1 = {f (x + y) + g(x − y) + j[f (x + y) − g(x − y)]}, 2

(3.10)

20

I. Hyperbolic Equations of First Order

where f (x + y), g(x − y) are as stated in (3.9) and f (0), g(0) are as stated in (3.8). It is not difficult to see that w(z) satisfies the estimate ¯ ≤ M1 = M1 (α, k0 , k2 , D), Cα [w(z), D]

¯ ≤ M2 k2 = M2 (α, k0 , D)k2 , (3.11) Cα [w(z), D]

where M1 , M2 are two non-negative constants only dependent on α, k0 , k2 , D and α, k0 , D respectively. The above results can be written as a theorem. Theorem 3.1 Any solution w(z) of Problem A for the complex equation (3.1) possesses the representation (3.10), which satisfies the estimate (3.11).

3.2 Uniqueness of solutions of the Riemann–Hilbert problem for linear hyperbolic complex equations Now we discuss the linear case of the complex equation (2.32), namely wz¯ = A1 (z)w + A2 (z)w¯ + A3 (z),

(3.12)

and suppose that the complex equation (3.12) satisfies the following conditions: ¯ and satisfy Al (z) (l = 1, 2, 3) are continuous in D

Condition C

¯ = C[Re Al , D] ¯ + C[Im Al , D] ¯ ≤ k0 , C[Al , D]

l = 1, 2,

¯ ≤ k1 . C[A3 , D]

(3.13)

Due to w = u + jv = ξe1 + ηe2 , wz = ξµ e1 + ην e2 , wz¯ = ξν e1 + ηµ e2 , from the formulas in Section 1, equation (3.12) can be rewritten in the form ξν e1 + ηµ e2 = [A(z)ξ + B(z)η + E(z)]e1 ⎧ ⎨ ξν ⎩

+[C(z)ξ + D(z)η + F (z)]e2 , = A(z)ξ + B(z)η + E(z),

ηµ = C(z)ξ + D(z)η + F (z),

z ∈ D,

i.e. (3.14)

z ∈ D,

in which A = Re A1 + Im A1 ,

B = Re A2 + Im A2 ,

C = Re A2 − Im A2 ,

D = Re A1 − Im A1 ,

E = Re A3 + Im A3 ,

F = Re A3 − Im A3 .

The boundary condition (3.2) can be reduced to Re [λ(ξe1 + ηe2 )] = r(z),

Im [λ(ξe1 + ηe2 )]|z=z0 = b1 ,

(3.15)

where λ = (a + b)e1 + (a − b)e2 . Moreover, the domain D is transformed into Q = {0 ≤ µ ≤ 2R, 0 ≤ ν ≤ 2R1 , R = R2 − R1 }, which is a rectangle and A, B, C, D, E, F

3. Linear Hyperbolic Equations

21

are known functions of (µ, ν) ∈ Q. There is no harm in assuming that w(z0 ) = 0, otherwise through the transformation W (z) = w(z) − [a(z0 ) − jb(z0 )]

[r(z0 ) + jb1 ] , [a2 (z0 ) − b2 (z0 )]

(3.16)

the requirement can be realized. For convenience, sometimes we write z ∈ D or z ∈ Q and denote L1 = {µ = 0, 0 ≤ ν ≤ 2R1 }, L4 = {0 ≤ µ ≤ 2R, ν = 0}. Now we give a representation of solutions of Problem A for equation (3.12). Theorem 3.2 If equation (3.12) satisfies Condition C, then any solution w(z) of Problem A for (3.12) can be expressed as w(z) = w0 (z) + Φ(z) + Ψ(z)

in D,

w0 (z) = f (x + y)e1 + g(x − y)e2 , 

Ψ(z) =

x−y

0

Φ(z) = f˜(x + y)e1 + g˜(x − y)e2 ,

[Aξ + Bη + E]d(x − y)e1 +



(3.17)

x+y

[Cξ + Dη + F ]d(x + y)e2 ,

0

where f (x + y), g(x − y) are as stated in (3.9) and f˜(x + y), g˜(x − y) are similar to f (x + y), g(x − y) in (3.9), but Φ(z) satisfies the boundary condition Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

Im [λ(z0 )Φ(z0 )] = −Im [λ(z0 )Ψ(z0 )]. (3.18)

z ∈ L,

Proof It is not difficult to see that the functions w0 (z), Φ(z) are solutions of the ¯ which satisfy the boundary conditions (3.2) and (3.18) complex equation (3.1) in D, respectively, and Ψ(z) satisfies the complex equation [Ψ]z¯ = [Aξ + Bη + E]e1 + [Cξ + Dη + F ]e2 ,

(3.19)

and Φ(z) + Ψ(z) satisfies the boundary condition of Problem A0 . Hence w(z) = w0 (z)+Φ(z)+Ψ(z) satisfies the boundary condition (3.2) and is a solution of Problem A for (3.12). Theorem 3.3 Suppose that Condition C holds. Then Problem A for the complex equation (3.12) has at most one solution. Proof Let w1 (z), w2 (z) be any two solutions of Problem A for (3.12) and substitute them into equation (3.12) and the boundary condition (3.2). It is clear that w(z) = w1 (z) − w2 (z) satisfies the homogeneous complex equation and boundary conditions wz¯ = A1 w + A2 w Re [λ(z)w(z)] = 0,

in D,

if (x, y) ∈ L,

(3.20) w(z0 ) = 0.

(3.21)

On the basis of Theorem 3.2, the function w(z) can be expressed in the form w(z) = Φ(z) + Ψ(z), 

Ψ(z) =

0

x−y

[Aξ + Bη]e1 d(x − y) +

 0

(3.22)

x+y

[Cξ + Dη]e2 d(x + y).

22

I. Hyperbolic Equations of First Order

¯ of the point z0 = 0. We may Suppose w(z)
≡ 0 in the neighborhood (⊂ D) choose a sufficiently small positive number R0 < 1, such that 8M3 M R0 < 1, where M3 = max{C[A, Q0 ], C[B, Q0 ], C[C, Q0 ], C[D, Q0 ]}, M = 1+4k02 (1+k02 ) is a positive constant, and m = C[w(z), Q0 ] > 0, herein Q0 = {0 ≤ µ ≤ R0 } ∩ {0 ≤ ν ≤ R0 }. From (3.9),(3.10),(3.17),(3.18),(3.22) and Condition C, we have  Ψ(z) ≤ 8M3 mR0 ,

 Φ(z) ≤ 32M3 k02 (1 + k02 )mR0 ,

thus an absurd inequality m ≤ 8M3 M mR0 < m is derived. It shows w(z) = 0, (x, y) ∈ Q0 . Moreover, we extend along the positive directions of µ = x + y and ν = x − y successively, and finally obtain w(z) = 0 for (x, y) ∈ D, i.e. w1 (z) − w2 (z) = 0 in D. This proves the uniqueness of solutions of Problem A for (3.12). 3.3 Solvability of Problem A for linear hyperbolic complex equations of first order Theorem 3.4 If the complex equation (3.12) satisfies Condition C, then Problem A for (3.12) has a solution. Proof In order to find a solution w(z) of Problem A in D, we can express w(z) in the form (3.17). In the following, by using successive iteration we can find a solution of Problem A for the complex equation (3.12). First of all, substituting w0 (z) = ξ0 e1 + η0 e2 of Problem A for (3.1) into the position of w = ξe1 + ηe2 in the right-hand side of (3.12), the function w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z), 

Ψ1 (z) =

ν

0

[Aξ0 + Bη0 + E]e1 dν +

 0

(3.23)

µ

[Cξ0 + Dη0 + F ]e2 dµ,

¯ can be determined, where µ = x + y, ν = x − y, Φ1 (z) is a solution of (3.1) in D satisfying the boundary conditions Re [λ(z)Φ1 (z)] = −Re [λ(z)Ψ1 (z)],

z ∈ L,

(3.24)

Im [λ(z0 )Φ1 (z0 )] = −Im [λ(z0 )Ψ1 (z0 )]. Thus from (3.23), (3.24), we have  w1 (z) − w0 (z) = C[w1 (z) − w0 (z), D] ≤ 2M4 M (4m + 1)R
,


(3.25)

where M4 = maxz∈D¯ (|A|, |B|, |C|, |D|, |E|, |F |), m = w0 C(D) ¯ , R = max(2R1 , 2R), M = 1 + 4k02 (1 + k02 ) is a positive constant as in the proof of Theorem 3.3. Moreover, we substitute w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z) and the corresponding functions ξ1 (z) = Re w1 (z)+Im w1 (z), η1 (z) = Re w1 (z)−Im w1 (z) into the positions of w(z), ξ(z), η(z) in (3.17), and similarly to (3.23)–(3.25), we can find the corresponding functions Ψ2 (z), Φ2 (z) in D and the function w2 (z) = w0 (z) + Φ2 (z) + Ψ2 (z) in D.

3. Linear Hyperbolic Equations

23

It is clear that the function w2 (z) − w1 (z) satisfies the equality w2 (z) − w1 (z) = Φ2 (z) − Φ1 (z) + Ψ2 (z) − Ψ1 (z) = Φ2 (z) − Φ1 (z) + 

+

0

µ



ν

0

[A(ξ1 − ξ0 ) + B(η1 − η0 )]e1 dν

[C(ξ1 − ξ0 ) + D(η1 − η0 )]e2 dµ,

and  w2 − w1 ≤ [2M3 M (4m + 1)]2



R


0

R
dR
≤

[2M3 M (4m + 1)R
]2 , 2!

where M3 is a constant as stated in the proof of Theorem 3.3. Thus we can find a sequence of functions {wn (z)} satisfying wn (z) = w0 (z) + Φn (z) + Ψn (z), 

Ψn (z) = +

0



ν

[Aξn + Bηn + E]e1 dν +

(3.26)

µ

[Cξn + Dηn + F ]e2 dµ,

0

and wn (z) − wn−1 (z) satisfies wn (z)−wn−1 (z) = Φn (z)−Φn−1 (z) + Ψn (z)−Ψn−1 (z) 

= Φn (z)−Φn−1 (z) +

ν

0

[A(ξn−1 −ξn−2 )

+B(ηn−1 −ηn−2 )]e1 dν 

+

0

µ

[C(ξn−1 −ξn−2 )+D(ηn−1 −ηn−2 )]e2 dµ,

(3.27)

and then  wn − wn−1 ≤ [2M3 M (4m + 1)]n

 0

R


R
n−1 [2M3 M (4m + 1)R
]n dR
≤ . (3.28) (n − 1) ! n!

From the above inequality, it is seen that the sequence of functions {wn (z)}, i.e. wn (z) = w0 (z) + [w1 (z) − w0 (z)] + · · · + [wn (z) − wn−1 (z)](n = 1, 2, . . .)

(3.29)

uniformly converges to a function w∗ (z), and w∗ (z) satisfies the equality w∗ (z) = w0 (z) + Φ∗ (z) + Ψ∗ (z), 

Ψ∗ (z) =

0

ν

[Aξ∗ + Bη∗ + E]e1 dν +

 0

(3.30)

µ

[Cξ∗ + Dη∗ + F ]e2 dµ.

It is easy to see that w∗ (z) satisfies equation (3.12) and the boundary condition (3.2), hence it is just a solution of Problem A for the complex equation (3.12) in the closed ¯ ([87]2). domain D

24

I. Hyperbolic Equations of First Order

3.4 Another boundary value problem for linear hyperbolic complex equations of first order Now we introduce another boundary value problem for equation (3.12) in D with the boundary conditions Re [λ(z)w(z)] = r(z) on L1 ∪ L5 , Im [λ(z1 )w(z1 )] = b1 ,

(3.31)

where L1 = {y = −x,0 ≤ x ≤ R},L5 = {y = (R+R1 )[x/(R−R1 )−2R1 R/(R2 −R12 )], R1 ≤ x ≤ R = R2 −R1 },R2 ≥ 2R1 , λ(z) = a(z)+jb(z),z ∈ L1 , λ(z) = a(z) +jb(z) = 1+j,z ∈ L5 and λ(z), r(z), b1 satisfy the conditions Cα [λ(z), L1 ] ≤ k0 , Cα [r(z), L1∪L5 ] ≤ k2 , |b1 | ≤ k2 , max z∈L1

1 ≤ k0 , (3.32) |a(z)−b(z)|

in which b1 is a real constant and α (0 < α < 1), k0 , k2 are non-negative constants. The boundary value problem is called Problem A1 . On the basis of Theorem 1.1, it is clear that the complex equation (3.1) can be ¯ The general solution of system (3.6) can be expressed reduced to the form (3.6) in D. as w(z) = u(z) + jv(z) = [u(z) + v(z)]e1 + [u(z) − v(z)]e2 (3.33)

= f (x+y)e1 +g(x− y)e2

1 = {f (x+y)+g(x−y)+j[f (x+y)−g(x−y)]}, 2 where f (t) (0 ≤ t ≤ 2R), g(t) (0 ≤ t ≤ 2R1 ) are two arbitrary real continuous functions. Noting that the boundary condition (3.31), namely a(z)u(z) + b(z)v(z) = r(z) on L1 ∪ L5 , λ(z1 )w(z1 ) = r(z1 ) + jb1 , i.e. [a((1 − j)x) + b((1 − j)x)]f (0) + [a((1 − j)x) − b((1 − j)x)]g(2x) = 2r((1 − j)x) on [0, R1 ],

f (z1 ) = u(z1 )+v(z1 ) =

Re [λ(z)w(z)] = u(z)+v(z) = r



1+

r(z1 )+b1 a(z1 )+b(z1 )

R+R1 2RR1 j x−j R−R1 R−R1

(3.34)



on [R1 , R].

It is easy to see that the above formulas can be rewritten as [a((1−j)t/2)+b((1−j)t/2)]f (0)+[a((1−j)t/2)−b((1−j)t/2)] ×g(t) = 2r((1−j)t/2),

t ∈ [0, 2R1 ],

f (t) = r ((1 + j)R − (1 − j)R1 )



f (x+y) = f 

t + (1 − j)R1 , 2R



2R 2RR1 x− R−R1 R−R1

t ∈ [0, 2R],



4. Quasilinear Hyperbolic Equations

25

and then g(x − y) =

2r((1 − j)(x − y)/2) a((1 − j)(x − y)/2) − b((1 − j)(x − y)/2) −

[a((1−j)(x−y)/2)+b((1−j)(x−y)/2)]f (0) , a((1−j)(x−y)/2)−b((1−j)(x−y)/2)

(3.35)

0 ≤ x−y ≤ 2R1 , f (x+y) = r[((1+j)R−(1−j)R1 )

x+y +(1−j)R1 ], 2R

0 ≤ x+y ≤ 2R.

Substitute the above function f (x + y), g(x − y) into (3.33), the solution w(z) of (3.6) is obtained. We are not difficult to see that w(z) satisfies the estimate ¯ ≤ M1 , Cα [w(z), D]

¯ ≤ M2 k2 , Cα [w(z), D]

(3.36)

where M1 = M1 (α, k0 , k2 , D), M2 = M2 (α, k0 , D) are two non-negative constants. Next we consider Problem A1 for equation (3.12). Similarly to before, we can derive the representation of solutions w(z) of Problem A1 for (3.12) as stated in (3.17), where f (x + y), g(x − y) possess the form (3.35), and L = L1 ∪ L2 , z0 in the formula (3.18) should be replaced by L1 ∪ L5 , z1 . Moreover applying the successive iteration, the uniqueness and existence of solutions of Problem A1 for equation (3.12) can be proved, but L, z0 in the formulas (3.21) and (3.24) are replaced by L1 ∪ L5 , z1 . We write the results as a theorem. Theorem 3.5 Suppose that equation (3.12) satisfies Condition C. Then Problem A1 for (3.12) has a unique solution w(z), which can be expressed in the form (3.17), where f (x + y), g(x − y) possess the form (3.35).

4

Boundary Value Problems of Quasilinear Hyperbolic Complex Equations of First Order

In this section, we mainly discuss the Riemann–Hilbert boundary value problem for quasilinear hyperbolic complex equations of first order in a simply connected domain. We first prove the uniqueness of solutions for the above boundary value problem, and then give a priori estimates of solutions of the problem, moreover, by using the successive iteration, the existence of solutions for the above problem is proved. Finally, we also discuss the solvability of the above boundary value problem in general domains.

26

I. Hyperbolic Equations of First Order

4.1 Uniqueness of solutions of the Riemann–Hilbert problem for quasilinear hyperbolic complex equations In the subsection, we first discuss the quasilinear hyperbolic complex equation ¯ wz¯ = F (z, w), F = A1 (z, w)w + A2 (z, w)w + A3 (z, w) in D,

(4.1)

where D is a simply connected bounded domain in the x+jy-plane with the boundary Γ = L1 ∪ L2 ∪ L3 ∪ L4 as stated in Section 3. Suppose that the complex equation (4.1) satisfies the following conditions: Condition C ¯ for any continuous complex 1) Al (z, w) (l = 1, 2, 3) are continuous in z ∈ D function w(z) and satisfy ¯ = C[Re Al , D] ¯ + C[Im Al , D] ¯ ≤ k0 , C[Al , D]

¯ ≤ k1 . l = 1, 2, C[A3 , D]

(4.2)

¯ the equality 2) For any continuous complex functions w1 (z), w2 (z) in D, F (z, w1 )−F (z, w2 ) = A˜1 (z, w1 , w2 )(w1 −w2 )+ A˜2 (z, w1 , w2 )(w1 −w2 ) in D

(4.3)

¯ ≤ k0 , l = 1, 2 and k0 , k1 are non-negative constants. In particuholds, where C[A˜l , D] lar, when (4.1) is a linear equation, the condition (4.3) obviously holds. ¯ In order to give an a priori Cα (D)-estimate of solutions for Problem A, we need ¯ w1 , w2 , the above the following conditions: For any hyperbolic numbers z1 , z2 (∈ D), functions satisfy  A˜l (z1 , w1 ) − Al (z2 , w2 ) ≤ k0 [ z1 − z2 α +  w1 − w2 ], l = 1, 2,  A3 (z1 , w1 ) − A3 (z2 , w2 ) ≤ k2 [ z1 − z2 α +  w1 − w2 ],

(4.4)

in which α(0 < α < 1), k0 , k2 are non-negative constants. Similarly to (3.12) and (3.14), due to w = u + jv = ξe1 + ηe2 = ζ, wz = ξµ e1 + ην e2 , wz¯ = ξν e1 + ηµ e2 , the quasilinear hyperbolic complex equation (4.1) can be rewritten in the form ξν e1 + ηµ e2 = [A(z, ζ)ξ + B(z, ζ)η + E(z, ζ)]e1 ⎧ ⎨ ξν ⎩

+[C(z, ζ)ξ + D(z, ζ)η + F (z, ζ)]e2 , z ∈ D, = A(z, ζ)ξ + B(z, ζ)η + E(z, ζ),

ηµ = C(z, ζ)ξ + D(z, ζ)η + F (z, ζ),

i.e.

z ∈ D,

in which A = Re A1 + Im A1 , B = Re A2 + Im A2 , C = Re A2 − Im A2 , D = Re A1 − Im A1 , E = Re A3 + Im A3 , F = Re A3 − Im A3 .

(4.5)

4. Quasilinear Hyperbolic Equations

27

Obviously, any solution of Problem A for equation (4.1) possesses the same representation (3.17) as stated in Theorem 3.2. In the following, we prove the existence and uniqueness of solutions for Problem A for (4.1) with Condition C. Theorem 4.1 If Condition C holds, then Problem A for the quasilinear complex equation (4.1) has at most one solution. Proof Let w1 (z), w2 (z) be any two solutions of Problem A for (4.1) and substitute them into equation (4.1) and boundary condition (3.2). By Condition C, we see that w(z) = w1 (z) − w2 (z) satisfies the homogeneous complex equation and boundary conditions wz¯ = A˜1 w + A˜2 w¯ in D, (4.6) Re [λ(z)w(z)] = 0, z ∈ L, Im [λ(z0 )w(z0 )] = 0.

(4.7)

On the basis of Theorem 3.2, the function w(z) can be expressed in the form ˜ ˜ w(z) = Φ(z) + Ψ(z), ˜ Ψ(z) =



x−y

0

˜ + Bη]e ˜ 1 d(x − y) + [Aξ



x+y

0

˜ + Dη]e ˜ 2 d(x + y), [Cξ

(4.8)

˜ B, ˜ C, ˜ D ˜ and A˜1 , A˜2 is the same with that where the relation between the coefficients A, between A, B, C, D and A1 , A2 in (4.5). Suppose w(z)
≡ 0 in the neighborhood Q0 (⊂ ¯ of the point z0 = 0, we may choose a sufficiently small positive number R0 < 1, D) ˜ Q0 ], C[B, ˜ Q0 ], C[C, ˜ Q0 ], C[D, ˜ Q0 ]}. such that 8M5 M R0 < 1, where M5 = max{C[A, Similarly to the proof of Theorem 3.3, we can derive a contradiction. Hence w1 (z) = w2 (z) in D. 4.2 Solvability of Problem A for quasilinear hyperbolic complex equations Theorem 4.2 If the quasilinear complex equation (4.1) satisfies Condition C, then Problem A for (4.1) has a solution. Proof Similarly to the proof of Theorem 3.4, we use the successive iteration to find a solution of Problem A for the complex equation (4.1). Firstly, substituting w0 (z) = ξ0 e1 + η0 e2 of Problem A for (3.1) into the position of w = ξe1 + ηe2 in the right-hand side of (4.1), the function 

Ψ1 (z) =

0

w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z), 

ν

[Aξ0 + Bη0 + E]e1 dν +

0

(4.9)

µ

[Cξ0 + Dη0 + F ]e2 dµ,

¯ can be determined, where µ = x + y, ν = x − y, Φ1 (z) is a solution of (3.1) in D satisfying the boundary conditions Re [λ(z)Φ1 (z)] = −Re [λ(z)Ψ1 (z)],

z ∈ L,

Im [λ(z0 )Φ1 (z0 )] = −Im [λ(z0 )Ψ1 (z0 )].

(4.10)

28

I. Hyperbolic Equations of First Order

Moreover, we can find a sequence of functions {wn (z)} satisfying 

Ψn (z) =

0

wn (z) = w0 (z) + Φn (z) + Ψn (z), 

ν

[Aξn−1 + Bηn−1 + E]e1 dν +

0

(4.11)

µ

[Cξn−1 + Dηn−1 + F ]e2 dµ,

and wn (z) − wn−1 (z) satisfies wn (z)−wn−1 (z) = Φn (z)−Φn−1 (z)+Ψn (z)−Ψn−1 (z) = Φn (z) − Φn−1 (z) 

+

ν

 0µ

+

0

˜ n−1 −ξn−2 )+ B(η ˜ n−1 −ηn−2 )]e1 dν [A(ξ

(4.12)

˜ n−1 −ξn−2 )+ D(η ˜ n−1 −ηn−2 )]e2 dµ. [C(ξ

˜ |B|, ˜ |C|, ˜ |D|), ˜ we can obtain Denoting M5 = maxD¯ (|A|,  wn − wn−1 ≤ [2M5 M (4m + 1)]n

 0

R


[2M5 M (4m + 1)R
]n R
n−1 dR
≤ , (4.13) (n − 1) ! n!

in which m = w0 C(Q) , R
= max(2R1 , 2R), M = 1 + 4k02 (1 + k02 ) is a positive constant as in the proof of Theorem 3.3. The remained proof is identical with the proof of Theorem 3.4. 4.3 A priori estimates of solutions of the Riemann–Hilbert problem for hyperbolic complex equations We first give the boundedness estimate of solutions for Problem A. Theorem 4.3 If Condition C holds, then any solution u(z) of Problem A for the hyperbolic equation (4.1) satisfies the estimates ¯ ≤ M6 , C[w(z), D] ¯ ≤ M7 k, C[w(z), D]

(4.14)

in which M6 = M6 (α, k0 , k1 , k2 , D), k = k1 + k2 , M7 = M7 (α, k0 , D) are non-negative constants. Proof On the basis of Theorems 4.1 and 4.2, we see that under Condition C, Problem A for equation (4.1) has a unique solution w(z), which can be found by using ¯ are successive iteration. Due to the functions wn+1 (z) − wn (z) (n = 1, 2, . . .) in D ¯ is also continuous, continuous, the limit function w(z) of the sequence {wn (z)} in D and satisfies the estimate ¯ ≤ C[w(z), D]

∞ 

[2M5 M (4m + 1)R
]n
= e2M5 M (4m+1)R = M6 , n! n=0

(4.15)

where R
= max(2R1 , 2R). This is the first estimate in (4.14). As for the second estimate in (4.14), if k = k1 +k2 = 0, then it is true from Theorem 4.1. If k = k1 + k2 > 0,

4. Quasilinear Hyperbolic Equations

29

let the solution w(z) of Problem A be substituted into (4.1) and (3.2), and dividing them by k, we obtain the equation and boundary conditions for w(z) ˜ = w(z)/k: w˜z¯ = A1 w˜ + A˜2 w˜ + A3 /k,

¯ z ∈ D, ¯ z ∈ D,

Re [λ(z)w(z)] ˜ = r(z)/k,

(4.16) Im [λ(z0 )w(z ˜ 0 )] = b1 /k.

Noting that A3 /k, r/k, b1 /k satisfy the conditions ¯ ≤ 1, C[r/k, L] ≤ 1, |b1 /k| ≤ 1, C[A3 /k, D] ¯ in (4.15), we can obtain the estimate by using the method of deriving C[w(z), D] ¯ ≤ M7 = M7 (α, k0 , D). C[w(z), ˜ D] From the above estimate, the second estimate in (4.14) is immediately derived. Here we mention that in the proof of the estimate (4.14), we have not required that the coefficients λ(z), r(z) of (3.2) satisfy a H¨older (continuous) condition and only require that they are continuous on L. ¯ of solutions of Problem A for (4.1), and Next, we shall give the Cα (D)-estimates first discuss the linear hyperbolic complex equation (3.12) or (3.14). Theorem 4.4 Suppose that the linear complex equation (3.12) satisfies the conditions (3.13) and (4.4), i.e. the coefficents of (3.12) satisfies the conditions ¯ ≤ k0 , Cα [Al , D]

¯ ≤ k2 , Cα [A3 , D]

l = 1, 2,

(4.17)

in which α(0 < α < 1), k0 , k2 are non-negative constants. Then the solution w(z) = w0 (z) + Φ(z) + Ψ(z) satisfies the following estimates ¯ ≤ M8 , Cα [Ψ(z), D] ¯ ≤ M8 , Cα [w0 (z), D] ¯ ≤ M8 , Cα [w(z), D] ¯ ≤ M8 , Cα [Φ(z), D]

(4.18)

where w0 (z) is a solution of (3.1) as stated in (3.10), M8 = M8 (α, k0 , k1 , k2 , D) is a non-negative constant. Proof As stated before w0 (z) is the function as in (3.10), which satisfies the estimate (3.11), namely the first estimate in (4.18). In order to prove that Ψ(z) = Ψ1 (z) = Ψ11 (z)e1 + Ψ21 (z)e2 satisfies the second estimate in (4.18), from Ψ11 (z) =

 0

x−y

G1 (z)d(x − y),

Ψ21 (z) =

G1 (z) = A(z)ξ + B(z)η + E(z),

 0

x+y

G2 (z)d(x + y),

(4.19)

G2 (z) = C(z)ξ + D(z)η + F (z),

¯ with respect to and (4.17), we see that Ψ11 (z) = Ψ11 (µ, ν), Ψ21 (z) = Ψ21 (µ, ν) in D ν = x − y, µ = x + y satisfy the estimates ¯ ≤ M9 R
, Cα [Ψ2 (µ, ·), D] ¯ ≤ M9 R
, Cα [Ψ11 (·, ν), D] 1

(4.20)

30

I. Hyperbolic Equations of First Order

respectively, where R
= max(2R, 2R1 ), M9 = M9 (α, k0 , k1 , k2 , D) is a non-negative constant. If we substitute the solution w0 = w0 (z) = ξ0 e1 + η0 e2 of Problem A of (3.1) into the position of w = ξe1 + ηe2 in (4.19), and ξ0 = Re w0 + Im w0 , η0 = Re w0 − Im w0 , from (4.17) and (3.11), we obtain ¯ ≤ M10 , Cα [G2 (·, ν), D] ¯ ≤ M10 , Cα [G1 (µ, ·), D] ¯ ≤ M10 R
, Cα [Ψ2 (·, ν), D] ¯ ≤ M10 R
, Cα [Ψ11 (µ, ·), D] 1

(4.21)

in which M10 = M10 (α, k0 , k1 , k2 , D) is a non-negative constant. Due to Φ(z) = Φ1 (z) satisfies the complex equation (3.1) and boundary condition (3.18), and Φ1 (z) possesses a representation similar to that in (3.17), the estimate ¯ ≤ M11 R
= R
M11 (α, k0 , k1 , k2 , D) Cα [Φ1 (z), D]

(4.22)

can be derived. Thus setting w1 (z) = w0 (z)+Φ1 (z)+Ψ1 (z), w˜1 (z) = w1 (z)−w0 (z), it is clear that the functions w˜11 (z) = Re w˜1 (z) + Im w˜1 (z), w˜12 (z) = Re w˜1 (z) −Im w˜1 (z) satisfy as functions of µ = x + y, ν = x − y respectively the estimates: ¯ ≤ M12 R
, Cα [w˜ 1 (µ, ·), D] ¯ ≤ M12 R
, Cα [w˜11 (·, ν), D] 1 ¯ ≤ M12 R
, Cα [w˜ 2 (·, ν), D] ¯ ≤ M12 R
, Cα [w˜12 (µ, ·), D] 1

(4.23)

¯ M13 = where M12 = 2M13 M (4m + 1), M = 1 + 4k02 (1 + k0 ), m = Cα [w0 , D], maxD¯ [|A|, |B|, |C|, |D|, |E|, |F |]. By using successive iteration, we obtain the sequence of functions: wn (z) (n = 1, 2, . . .), and the corresponding functions w˜n1 = Re w˜n + Im w˜n , w˜n2 = Re w˜n − Im w˜n satisfying the estimates
n (M12 R
)n ¯ ≤ (M12 R ) , , Cα [w˜n1 (µ, ·), D] n! n!
n (M (M R ) R
)n 12 12 ¯ ≤ ¯ ≤ , Cα [w˜n2 (·, ν), D] , Cα [w˜n2 (µ, ·), D] n! n!

¯ ≤ Cα [w˜n1 (·, ν), D]

(4.24)

 ¯ the corresponding and denote by w(z) the limit function of wn (z) = nm=0 w˜n (z) in D, 1 2 functions w˜ = Re w(z) + Im w(z), w˜ = Re w(z) − Im w(z) satisfy the estimates





¯ ≤ eM12 R , Cα [w˜ 1 (µ, ·), D] ¯ ≤ eM12 R , Cα [w˜ 1 (·, ν), D] ¯ ≤ eM12 R
, Cα [w˜ 2 (·, ν), D] ¯ ≤ eM12 R
. Cα [w˜ 2 (µ, ·), D] Combining the first formula in (4.18), (4.20)–(4.24) and the above formulas, the last three estimates in (4.18) are derived. Theorem 4.5 Let the quasilinear complex equation (4.1) satisfy Condition C and (4.4). Then the solution w(z) of Problem A for (4.1) satisfies the following estimates ¯ ≤ M14 , Cα [w(z), D] ¯ ≤ M15 k, Cα [w(z), D]

(4.25)

where k = k1 + k2 , M14 = M14 (α, k0 , k1 , k2 , D), M15 = M15 (α, k0 , D) are non-negative constants.

4. Quasilinear Hyperbolic Equations

31

Proof According to the proof of Theorem 4.3, from the first formula in (4.25), the second formula in (4.25) is easily derived. Hence we only prove the first formula in (4.25). Similarly to the proof of Theorem 4.4, we see that the function Ψ1 (z) = Ψ11 (µ, ν)e1 +Ψ21 (µ, ν)e2 still possesses the estimate (4.20). Noting that the coefficients ¯ and w, and applying the condition (4.4), we can derive are the functions of z ∈ D similar estimates as in (4.21). Hence we also obtain estimates similar to (4.22) and (4.23), and the constant M12 in (4.23) can be chosen as M12 = 2M13 M (4m + 1), m = ¯ Thus the first estimate in (4.25) can be derived. Cα [w0 (z), D]. Moreover, according to the above methods, we can obtain estimates for [Re w + Im w]ν , [Re w − Im w]µ analogous to those in (4.14) and (4.25). 4.4 The boundary value problem for quasilinear hyperbolic equations of first order in general domains In this subsection, we shall generalize the domain D to general cases. 1. The boundary L1 of the domain D is replaced by a curve L
1 , and the boundary of the domain D
is L
1 ∪ L
2 ∪ L3 ∪ L4 , where the parameter equations of the curves L
1 , L
2 are as follows: L
1 = {γ1 (x) + y = 0, 0 ≤ x ≤ l}, L
2 = {x − y = 2R1 , l ≤ x ≤ R2 },

(4.26)

in which γ1 (x) on 0 ≤ x ≤ l = 2R1 − γ1 (l) is continuous and γ1 (0) = 0, γ1 (x) > 0 on 0 < x ≤ l, and γ1 (x) is differentiable except at isolated points on 0 ≤ x ≤ l and 1 + γ1
(x) > 0. By this condition, the inverse function x = σ(ν) of x + γ1 (x) = ν can be found, and σ
(ν) = 1/[1 + γ1
(x)], hence the curve L
1 can be expressed by x = σ(ν) = (µ + ν)/2, i.e. µ = 2σ(ν) − ν, 0 ≤ ν ≤ 2R1 . We make a transformation µ ˜=

2R[µ − 2σ(ν) + ν] , 2R − 2σ(ν) + ν

ν˜ = ν,

2σ(ν) − ν ≤ µ ≤ 2R, 0 ≤ ν ≤ 2R1 ,

(4.27)

its inverse transformation is µ=

1 [2R − 2σ(ν) + ν]˜ µ + 2σ(ν) − ν, 2R

ν = ν˜,

0≤µ ˜ ≤ 2R, 0 ≤ ν˜ ≤ 2R1 . (4.28)

The transformation (4.27) can be expressed by ⎧ ⎪ ⎪ ˜ ⎪x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

1 µ + ν˜) = (˜ 2 2R(x + y) + 2R(x − y) − (2R + x − y)[2σ(x + γ1 (x)) − x − γ1 (x)] , = 4R − 4σ(x + γ1 (x)) + 2x + 2γ1 (x)

⎪ ⎪ ⎪ ⎪ y˜ = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ =

1 (˜ µ − ν˜) 2 2R(x + y) − 2R(x − y) − (2R − x + y)[2σ(x + γ1 (x)) − x − γ1 (x)] , 4R − 4σ(x + γ1 (x)) + 2x + 2γ1 (x)

(4.29)

32

I. Hyperbolic Equations of First Order

where γ1 (x) = −y, and its inverse transformation is ⎧ ⎪ ⎪ ⎪ x ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ y ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

[2R − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ x + y˜) 1 = (µ + ν) = 2 4R x + γ1 (x) − x˜ + y˜ , +σ(x + γ1 (x)) − 2 [2R − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ 1 x + y˜) = (µ − ν) = 2 4R x + γ1 (x) + x˜ − y˜ . +σ(x + γ1 (x)) − 2

(4.30)

z ) the transformation (4.29) and the Denote by z˜ = x˜ + j y˜ = f (z), z = x + jy = f −1 (˜ inverse transformation (4.30) respectively. In this case, the system of equations and boundary conditions are ξν = Aξ + Bη + E, Re [λ(z)w(z)] = r(z), z0


ηµ = Cξ + Dη + F, z ∈ L
1 ∪ L
2 , L
1

z ∈ D
,

Im [λ(z0 )w(z0 )] = b1 ,

(4.31) (4.32)

L
2

= l − jγ1 (l), λ(z), r(z), b1 on ∪ satisfy the conditions (3.3),(3.4). in which ¯
satisfies Condition C, through the transformation (4.28) Suppose system (4.31) in D and ξν˜ = ξν , ηµ˜ = [2R − 2σ(ν) + ν]ηµ /2R, system (4.31) is reduced to ξν˜ = Aξ + Bη + E,

ηµ˜ = [2R − 2σ(ν) + ν]

[Cξ + Dη + F ] . 2R

(4.33)

z ), the boundary condition Moreover, through the transformation (4.30), i.e. z = f −1 (˜ (4.32) is reduced to Re [λ(f −1 (˜ z ))w(f −1 (˜ z ))] = r(f −1 (˜ z )), z˜ ∈ L1 ∪ L2 , z0 ))w(f −1 (˜ z0 )] = b1 , Im [λ(f −1 (˜

(4.34)

in which z˜0 = f (z0 ) = 0. Therefore the boundary value problem (4.31),(4.32) is transformed into the boundary value problem (4.33),(4.34). On the basis of Theorems 4.1 and 4.2, we see that the boundary value problem (4.33), (4.34) has a unique solution w(˜ z ), and then w[f (z)] is just a solution of the boundary value problem (4.31),(4.32) in D
. Theorem 4.6 If the complex equation (4.1) satisfies Condition C in the domain D
with the boundary L
1 ∪ L
2 ∪ L3 ∪ L4 , where L
1 , L
2 are as stated in (4.26), then Problem A with the boundary conditions Re [λ(z)w(z)] = r(z), z ∈ L
1 ∪ L
2 , Im [λ(z0 )w(z0 )] = b1 has a unique solution w(z).

4. Quasilinear Hyperbolic Equations

33

2. The boundary L1 , L4 of the domain D are replaced by the two curves L

1 , L

4 respectively, and the boundary of the domain D

is L

1 ∪ L

2 ∪ L

3 ∪ L

4 , where the parameter equations of the curves L

1 , L

2 , L

3 , L

4 are as follows: L

1 = {γ1 (x) + y = 0, 0 ≤ x ≤ l1 },

L

2 = {x − y = 2R1 , l1 ≤ x ≤ R2 },

L

3 = {x + y = 2R, l2 ≤ x ≤ R2 },

L

4 = {−γ4 (x) + y = 0, 0 ≤ x ≤ l2 },

(4.35)

in which and γ1 (0) = 0, γ4 (R2 ) = 2R − R2 ; γ1 (x) > 0, 0 ≤ x ≤ l1 ; γ4 (x) > 0, 0 ≤ x ≤ l2 ; γ1 (x) on 0 ≤ x ≤ l1 , γ4 (x) on 0 ≤ x ≤ l2 are continuous, and γ1 (x), γ4 (x) possess derivatives except at finite points on 0 ≤ x ≤ l1 , 0 ≤ x ≤ l2 respectively, and 1 + γ1
(x) > 0, 1 + γ4
(x) > 0, z1

= x − jγ1 (l1 ) ∈ L2 , z3

= x + jγ2 (l2 ) ∈ L3 . By the conditions, the inverse functions x = σ(ν), x = τ (µ) of x + γ1 (x) = ν, x + γ4 (x) = µ can be found respectively, namely µ = 2σ(ν) − ν, 0 ≤ ν ≤ l1 + γ1 (l1 ),

ν = 2τ (µ) − µ, l2 + γ4 (l2 ) ≤ µ ≤ 2R1 . (4.36)

We make a transformation µ ˜ = µ,

ν˜ =

2R1 (ν − 2τ (µ) + µ) , 2R1 − 2τ (µ) + µ

0 ≤ µ ≤ 2R, 2τ (µ) − µ ≤ µ ≤ 2R1 ,

(4.37)

its inverse transformation is µ=µ ˜,

ν=

2R1 − 2τ (µ) + µ ν˜ + 2τ (µ) − ν, 2R1

0≤µ ˜ ≤ 2R,

0 ≤ ν˜ ≤ 2R1 . (4.38)

Hence we have 4R1 x − [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 + x + y) 1 µ + ν˜) = , x˜ = (˜ 2 4R1 − 4τ (x + γ1 (x)) + 2x + 2γ1 (x) 1 2R1 y + [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 − x − y) , y˜ = (˜ µ − ν˜) = 2 4R1 − 4τ (x + γ1 (x)) + 2x + 2γ1 (x)

(4.39)

and its inverse transformation is 4R1 x˜ + [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 − x˜ + y˜) 1 x = (µ + ν) = , 2 4R1 4R1 y˜ − [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 + x˜ − y˜) 1 . y = (µ − ν) = 2 4R1

(4.40)

Denote by z˜ = x˜ + j y˜ = g(z), z = x + jy = g −1 (˜ z ) the transformation (4.39) and its inverse transformation (4.40) respectively. Through the transformation (4.38), we have 2R1 − 2τ (µ) + µ (u + v)ν˜ = (u + v)ν , (u − v)µ˜ = (u − v)µ , 2R1 system (4.31) in D

is reduced to ξν˜ =

2R1 − 2τ (µ) + µ [Aξ + Bη + E], 2R1

ηµ˜ = Cξ + Dη + F,

z ∈ D
,

(4.41)

34

I. Hyperbolic Equations of First Order

where D
is a bounded domain with boundary L
1 ∪ L
2 ∪ L3 ∪ L4 and L
1 = L

1 . Moreover, through the transformation (4.40), the boundary condition (4.32) on L

1 ∪ L

4 is reduced to z ))w(g −1 (˜ Re [λ(g −1 (˜ z ))] = r(g −1 (˜ z )), z˜ ∈ L
1 ∪ L4 , Im [λ(g −1 (˜ z0 ))w(g −1 (˜ z0 )] = b1 ,

(4.42)

in which z˜0 = g(z0 ) = 0. Therefore the boundary value problem (4.31),(4.32) in D

is transformed into the boundary value problem (4.41),(4.42). On the basis of Theorem 4.6, we see that the boundary value problem (4.41),(4.42) has a unique solution w(˜ z ), and then w[g(z)] is just a solution of the boundary value problem (4.31),(4.32). Theorem 4.7 If the complex equation (4.1) satisfies Condition C in the domain D

with the boundary L

1 ∪ L

2 ∪ L

3 ∪ L

4 , then Problem A with the boundary conditions Re [λ(z)w(z)] = r(z), z ∈ L

1 ∪ L

4 , Im [λ(z0 )w(z0 )] = b1 has a unique solution w(z), where z1

= x − jγ1 (l1 ) ∈ L2 , z3

= x + jγ4 (l2 ) ∈ L3 . Now, we give an example to illustrate the above results. When R2 = 2R1 , the boundary of the domain D is L1 = {x = −y, 0 ≤ x ≤ R1 }, L2 = {x = y + 2R1 , R1 ≤ x ≤ R1 }, L3 = {x = −y + 2R1 , R1 ≤ x ≤ 2R1 }, L4 = {x = y, 0 ≤ x ≤ R1 }. We replace L1 ∪ L4 by a left semi-circumference L

1 ∪ L

4 with the center R1 and the radius R1 , namely 

L

1 = {x − y = ν, y = −γ1 (x) = − R12 − (x − R1 )2 , 0 ≤ x ≤ R1 }, L

4 = {x + y = µ, y = γ4 (x) =



R12 − (x − R1 )2 , 0 ≤ x ≤ R1 },

where 1 + γ1
(x) > 0, 0 < x ≤ R1 , 1 + γ4
(x) > 0, 0 ≤ x < R1 , and x = σ(ν) = x = τ (µ) =

R1 + ν − R1 + µ −



R12 + 2R1 ν − ν 2 , 2



R12 + 2R1 µ − µ2 . 2

It is clear that according to the above method, the domain D can be generalized to a general domain D

, namely its boundary consists of the general curves L

1 , L

2 , L

3 , L

4 with some conditions, which includes the circumference L = {|z − R1 | = R1 }. Finally, we mention that some boundary value problems for equations (3.12) and (4.1) with one of the boundary conditions Re [λ(z)w(z)] = r(z), z ∈ L1 ∪ L2 , Im [λ(z1 )w(z1 )] = b1 , Re [λ(z)w(z)] = r(z), z ∈ L3 ∪ L4 , Im [λ(z3 )w(z3 )] = b1 , Re [λ(z)w(z)] = r(z), z ∈ L1 ∪ L5 , Im [λ(z0 )w(z0 )] = b1 , Re [λ(z)w(z)] = r(z), z ∈ L1 ∪ L6 , Im [λ(z0 )w(z0 )] = b1 ,

5. Quasi-hyperbolic Mappings

35

can be discussed, where λ(z), r(z), b1 satisfy the conditions similar to those in (3.3),(3.4),(3.31) and Lj (j = 1, . . . , 5), zj (j = 0, 1, 3) are as stated in Section 3, R2 − 2R1 , 0 ≤ x ≤ R2 }, and λ(z) = 1 + j, z ∈ L5 , λ(z) = 1 − j, z ∈ L6 . L6 = {y = R2 x For corresponding boundary value problems of hyperbolic systems of first order complex equations, whether there are similar results as before? The problem needs to be investigated.

5

Hyperbolic Mappings and Quasi-hyperbolic Mappings

Now, we introduce the definitions of hyperbolic mappings and quasi-hyperbolic mappings, and prove some properties of quasi-hyperbolic mappings.

5.1

Hyperbolic mappings

A so-called hyperbolic mapping in a domain D is a univalent mapping given by a hyperbolic continuously differentiable function w = f (z) = u + jv satisfying the simplest hyperbolic system of first order equations ux = v y ,

vx = uy ,

(5.1)

which maps D onto a domain G in the w-plane. By Theorem 1.1, system (5.1) is equivalent to the system ξν = 0, ηµ = 0, (5.2) where ξ = u + v, η = u − v, µ = x + y, ν = x − y. Noting   µx    µy



νx   νy 

 1

= 

1



1 

 −1 









 ξu ηu  1 1     = −2,  =  = −2,  ξv ηv   1 −1 

(5.3)

if we find a homeomorphic solution of (5.2), then the solution w = u + jv = ξ[(µ + ν)/2, (µ − ν)/2]e1 + η[(µ + ν)/2, (µ − ν)/2]e2

(5.4)

of the corresponding system (5.1) is also homeomorphic. In fact, ξ = ξ(µ),

η = η(ν)

(5.5)

is a homeomorphic solution of (5.2), if ξ(µ) and η(ν) are strictly monotonous continuous functions of µ(µ0 ≤ µ ≤ µ1 ) and ν(ν0 ≤ ν ≤ ν1 ) respectively. When [ξ(µ), η(ν)] is univalent continuous in ∆ = {µ0 ≤ µ ≤ µ1 , ν0 ≤ ν ≤ ν1 } and D(∆) is the closed domain in the z = x + jy-plane corresponding to ∆, it is easy to see that [u(x, y), v(x, y)] is a homeomorphic solution of (5.1) in D(∆).

36

I. Hyperbolic Equations of First Order

5.2

Quasi-hyperbolic mappings

In this subsection, we first discuss the uniformly hyperbolic system in the complex form wz¯ = Q(z)wz , Q(z) = a(z) + jb(z), (5.6) where Q(z) is a continuous function satisfying the condition |Q(z)| ≤ q0 < 1, here q0 is a non-negative constant. On the basis of the representation wz¯ = ξν e1 + ηµ e2 , wz = ξµ e1 + ην e2 , Q = q1 e1 + q2 e2 , from (5.6), it follows that ξν = q1 ξµ ,

ηµ = q 2 ην .

(5.7) 2

¯ = Due to Q = a + jb = q1 e1 + q2 e2 , here q1 = a + b, q2 = a − b, thus |Q| = |QQ| 2 2 2 |a − b | = |q1 q2 | ≤ q0 < 1, and a representation theorem of solutions for (5.6) can be obtained. Theorem 5.1 Let χ(z) be a homeomorphic solution of (5.6) in a domain D, and w(z) be a solution of (5.6) in D. Then w(z) can be expressed as w(z) = Φ[χ(z)],

(5.8)

where Φ(χ) is a hyperbolic regular function in the domain G = χ(D). Proof Suppose that z(χ) is the inverse function of χ(z), we can find {w[z(χ)]}χ¯ = wz zχ¯ + wz¯z¯χ¯ = wz [zχ¯ + Q¯ zχ¯ ], and zχ ], χχ = 1 = χz zχ + χz¯z¯χ = χz [zχ + Q¯ zχ¯ ]. χχ¯ = 0 = χz zχ¯ + χz¯z¯χ¯ = χz [zχ¯ + Q¯ zχ¯ = 0, consequently Φ(χ) = w[z(χ)] From the above equalities, we see χz
= 0, zχ¯ + Q¯ satisfies [Φ(χ)]χ¯ = 0, χ ∈ G = χ(D). This shows that Φ(χ) is a hyperbolic regular function in G = χ(D), therefore the representation (5.8) holds. Next we prove the existence of a homeomorphic solution of equation (5.6) with some conditions for the coefficient of (5.6). From (5.7), we see that the complex equation (5.6) can be written in the form ξν = (a + b)ξµ ,

ηµ = (a − b)ην .

(5.9)

Let ∆ = {µ0 ≤ µ ≤ µ0 + R1 , ν0 ≤ ν ≤ ν0 + R2 }, in which µ0 , ν0 are two real numbers and R1 , R2 are two positive numbers, if a, b possess continuously differentiable

5. Quasi-hyperbolic Mappings

37

derivatives with respect to µ, ν in ∆, then the solution w = ξe1 + ηe2 of (5.9) in ∆ is a homeomorphism, provided that one of the following sets of conditions b > 0, −b < a < b,

(5.10)

b < 0, b < a < −b,

(5.11)

holds and ξ and η are strictly monotonous continuous functions of µ (µ0 ≤ µ ≤ µ0 + R1 ) and ν (ν0 ≤ ν ≤ ν0 + R2 ) respectively. Thus we have the following theorem. Theorem 5.2 Denote by D the corresponding domain of ∆ in the (x + jy)-plane and let w(z) be a continuous solution of (5.6) in D. If (5.10) or (5.11) in D holds, ξµ (ξν ) > 0 and ηµ (ην ) > 0 except some possible isolated points on µ (µ0 ≤ µ ≤ µ0 + R1 ) (ν (ν0 ≤ ν ≤ ν0 + R2 )), then the solution w(z) in D is a homeomorphism. In particular, if the coefficient Q(z) = a + jb of the complex equation (5.6) is a hyperbolic constant, which satisfies the condition |Q(z)|2 = |QQ| = |a2 − b2 | = |q1 q2 | ≤ q02 < 1,

z ∈ D,

we make a nonsingular transformation µ = −(a + b)σ + τ,

ν = σ − (a − b)τ.

(5.12)

Thus system (5.9) can be transformed into the system ξσ = 0,

ητ = 0,

(σ, τ ) ∈ G,

(5.13)

where the domain G is the corresponding domain of ∆ under the transformation (5.12). According to the discussion of hyperbolic mappings, we see that system (5.13) in G possesses a homeomorphic solution, hence system (5.9) in ∆ has a homeomorphic solution and then the complex equation (5.6) in D has a homeomorphic solution. The above result can be written as a theorem. Theorem 5.3 Suppose that Q(z) = a + jb is a hyperbolic constant and |Q(z)| < 1. Then the complex equation (5.6) in D has a homeomorphic solution. 5.3

Other Quasi-hyperbolic mappings

Now, we consider the hyperbolic complex equation w¯z = Q(z)wz ,

(5.14)

where Q(z) is a continuous function in D = {0 ≤ x ≤ R1 , 0 ≤ y ≤ R2 } satisfying the condition |Q(z)| ≤ q0 < 1. If Q(z) in D is a hyperbolic regular function of z¯, we introduce a transformation of functions W = w¯ − Qw,

i.e. w =

W + Q(z)W . 1 − |Q(z)|2

(5.15)

38

I. Hyperbolic Equations of First Order

Then (5.14) is reduced to the complex equation Wz = 0.

(5.16)

The solution W (z) of (5.16) in D is a hyperbolic regular function of z¯. As stated before, the complex equation (5.16) possesses a homeomorphic solution, which realizes a hyperbolic mapping in D. Moreover, if Q(z) is a hyperbolic regular function in D, we find the partial derivative with respect to z¯ in (5.14), and obtain w¯zz¯ = Q(z)wzz¯,

i.e. wzz¯ = Q(z)w¯zz¯,

(5.17)

from |Q(z)| < 1, it follows that (1 − |Q(z)|2 )wzz¯ = 0,

i.e. wzz¯ = 0,

(5.18)

the solution of the above complex equation (5.18) is called a hyperbolic harmonic complex function. A hyperbolic harmonic complex function w(z) can be expressed as w(z) = u(z) + jv(z) = φ(z) + φ(z) + ψ(z) − ψ(z) = φ(z) + ψ(z) + φ(z) − ψ(z) = f (z) + g(z), in which φ(z), ψ(z) are hyperbolic regular functions, hence f (z) = φ(z)+ψ(z), g(z) = φ(z) −ψ(z) are hyperbolic regular functions. This is a representation of hyperbolic harmonic functions through hyperbolic regular functions. Hence in order to find a hyperbolic harmonic function, it is sufficient to find solutions of the following two boundary value problems with the boundary conditions Re f (x) = Re φ0 (x), Re f (jy) = Re φ1 (y), and Im g(x) = Im φ0 (x), Im g(jy) = Im φ1 (y), respectively, where φ0 (x), φ1 (y) are given hyperbolic complex functions on {0 ≤ x ≤ R1 }, {0 ≤ y ≤ R2 } respectively, and R1 , R2 are two positive constants. At last we mention that the notations of hyperbolic numbers and hyperbolic complex functions are mainly used in this and next chapters. From Chapter III to Chapter VI except in Section 5, Chapter V, we do not use them. The references for this chapter are [5],[9],[12],[19],[26],[29],[32],[34],[38],[44],[51], [59],[68],[74],[80],[83],[85],[87],[89],[92],[97].

CHAPTER II HYPERBOLIC COMPLEX EQUATIONS OF SECOND ORDER In this chapter, we mainly discuss oblique derivative boundary value problems for linear and quasilinear hyperbolic equations of second order in a simply connected domain. Firstly, we transform some linear and nonlinear uniformly hyperbolic equations of second order with certain conditions into complex forms, give the uniqueness theorem of solutions for the above boundary value problems. Moreover by using the successive iteration, the existence of solutions for several oblique derivative problems is proved. Finally we introduce some boundary value problems for degenerate hyperbolic equations of second order with certain conditions.

1

Complex Form of Hyperbolic Equations of Second Order

This section deals with hyperbolic equations of second order in the plane domains, we first transform some linear and nonlinear uniformly hyperbolic equations of second order with certain conditions into complex forms, and then we state the conditions of some hyperbolic complex equations of second order.

1.1 Reduction of linear and nonlinear hyperbolic equations of second order Let D be a bounded domain, we consider the linear hyperbolic partial differential equation of second order auxx + 2buxy + cuyy + dux + euy + f u = g,

(1.1)

where the coefficients a, b, c, d, e, f, g are known continuous functions of (x, y) ∈ D, in which D is a bounded domain. The condition of hyperbolic type for (1.1) is that for any point (x, y) in D, the inequality I = ac − b2 < 0,

a > 0,

(1.2)

holds. If a, b, c are bounded in D and I = ac − b2 ≤ I0 < 0,

a>0

(1.3)

40

II. Hyperbolic Equations of Second Order

in D, where I0 is a negative constant, then equation (1.1) is called uniformly hyperbolic in D. Introduce the notations as follows ( )z = ( )zz =

( )x + j( )y ( )x − j( )y ( )xx − ( )yy , ( )z¯ = , ( )zz¯ = , 2 2 4 ( )xx + ( )yy + 2j( )xy ( )xx + ( )yy − 2j( )xy , ( )z¯z¯ = , 4 4

(1.4)

( )x = ( )z + ( )z¯, ( )y = j[( )z − ( )z¯], ( )xy = j[( )zz − ( )z¯z¯], ( )xx = ( )zz + ( )z¯z¯ + 2( )zz¯, ( )yy = ( )zz + ( )z¯z¯ − 2( )zz¯, equation (1.1) can be written in the form 2(a − c)uzz¯ + (a + c + 2bj)uzz + (a + c − 2bj)uz¯z¯ + (d + ej)uz + (d − ej)uz¯ + f u = g in D.

(1.5)

If a
= c in D, then equation (1.5) can be reduced to the complex form uzz¯ − Re [Q(z)uzz + A1 (z)uz ] − A2 (z)u = A3 (z) in D,

(1.6)

in which Q=

d + ej f g a + c + 2bj . , A1 = , A2 = , A3 = a−c a−c 2(a − c) 2(a − c)

If (a + c)2 ≥ 4b2 , then the conditions of hyperbolic type and uniformly hyperbolic are transformed into |Q(z)| < 1 in D, (1.7) and |Q(z)| ≤ q0 < 1 in D,

(1.8)

respectively. For the nonlinear hyperbolic equation of second order Φ(x, y, u, ux , uy , uxx , uxy , uyy ) = 0 in D,

(1.9)

from (1.4) we have Φ = F (z, u, uz , uzz , uzz¯). Under certain conditions, equation (1.9) can be reduced to the real form auxx + 2buxy + cuyy + dux + euy + f u = g in D,

(1.10)

and its complex form is as follows a0 uzz¯ − Re [quzz + a1 uz ] − a2 u = a3 in D,

(1.11)

1. Complex Form of Hyperbolic Equations

41

in which 

a=

1

Φτ uxx (x, y, u, ux , uy , τ uxx , τ uxy , τ uyy )dτ = a(x, y, u, ux , uy , uxx , uxy , uyy ),

0



2b =

0



c= d=



Φτ uyy (x, y, u, ux , uy , τ uxx , τ uxy , τ uyy )dτ = c(x, y, u, ux , uy , uxx , uxy , uyy ), 1

Φτ ux (x, y, u, τ ux , τ uy , 0, 0, 0)dτ = d(x, y, u, ux , uy ),

0 1

Φτ uy (x, y, u, τ ux , τ uy , 0, 0, 0)dτ = e(x, y, u, ux , uy ),

0



f=

Φτ uxy (x, y, u, ux , uy , τ uxx , τ uxy , τ uyy )dτ = 2b(x, y, u, ux , uy , uxx , uxy , uyy ),

1

0



e=

1

0

1

Φτ u (x, y, τ u, 0, 0, 0, 0, 0)dτ = f (x, y, u),

g = −Φ(x, y, 0, 0, 0, 0, 0, 0) = g(x, y), and a0 = 2(a − c) =



1

0

Fτ uzz¯ (z, u, uz , τ uzz , τ uzz¯)dτ = a0 (z, u, uz , uzz , uzz¯),

q = 2(a+c+2bj) = −2 a1 = 2(d + ej) = −2 a2 = f = −

 0





1

0

0

Fτ uzz (z, u, uz , τ uzz , τ uzz¯)dτ = q(z, u, uz , uzz , uzz¯),

1

Fτ uz (z, u, τ uz , 0, 0, 0)dτ = a1 (z, u, uz ),

(1.12)

1

Fτ u (z, τ u, 0, 0, 0, 0)dτ = a2 (z, u),

a3 = −F (z, 0, 0, 0, 0) = a3 (z). The condition of uniformly hyperbolic type for equation (1.10) is the same with (1.3). If a
= c in D, the complex equation (1.11) can be rewritten in the form uzz¯ − Re [Quzz + A1 uz ] − A2 u = A3 in D,

(1.13)

where Q = q/a0 ,

A1 = a1 /a0 ,

A2 = a2 /a0 ,

A3 = a3 /a0

are functions of z ∈ D, u, uz , uzz , uzz¯, the condition of uniformly hyperbolic type for (1.13) is as stated in the form (1.8). As stated in [12] 3), for the linear hyperbolic equation (1.1) or its complex form (1.6), if the coefficients a, b, c are sufficiently smooth, through a nonsingular transformation of z, equation (1.1) can be reduced to the standard form uxx − uyy + dux + euy + f u = g

(1.14)

uzz¯ − Re [A1 (z)uz ] − A2 (z)u = A3 (z).

(1.15)

or its complex form

42

II. Hyperbolic Equations of Second Order

1.2

Conditions of some hyperbolic equations of second order

Let D be a simply connected bounded domain with the boundary Γ = L1 ∪ L2 ∪ L3 ∪ L4 as stated in Chapter I, where L1 = {x = −y, 0 ≤ x ≤ R1 }, L2 = {x = y + 2R1 , R1 ≤ x ≤ R2 }, L3 = {x = −y − 2R1 + 2R2 , R2 − R1 ≤ x ≤ R2 }, L4 = {x = y, 0 ≤ x ≤ R2 − R1 }, and denote z0 = 0, z1 = (1 − j)R1 , z2 = R2 + j(R2 − 2R1 ), z3 = (1 + j)(R2 − R1 ), L = L3 ∪ L4 , here there is no harm in assuming that R2 ≥ 2R1 . In the following, we mainly consider second order quasilinear hyperbolic equation in the form uzz¯ − Re [A1 (z, u, uz )uz ] − A2 (z, u, uz )u = A3 (z, u, uz ), (1.16) whose coefficients satisfy the following conditions: Condition C ¯ for all continuously differentiable 1) Al (z, u, uz )(l = 1, 2, 3) are continuous in z ∈ D ¯ functions u(z) in D and satisfy ¯ = C[Re Al , D] ¯ + C[Im Al , D] ¯ ≤ k0 , l = 1, 2, C[A3 , D] ¯ ≤ k1 . C[Al , D]

(1.17)

¯ the equality 2) For any continuously differentiable functions u1 (z), u2 (z) in D, ¯ F (z, u1 , u1z ) − F (z, u2 , u2z ) = Re [A˜1 (u1 − u2 )z ] + A˜2 (u1 − u2 ) in D

(1.18)

holds, where ¯ ≤ k0 , l = 1, 2, C[A˜l (z, u1 , u2 ), D]

(1.19)

in (1.17),(1.19), k0 , k1 are non-negative constants. In particular, when (1.16) is a linear equation, from (1.17) it follows that the conditions (1.18), (1.19) hold. ¯ of solutions for some boundary value In order to give a priori estimates in Cα (D) problems, we need to add the following conditions: For any two real numbers u1 , u2 ¯ w1 , w2 , the above functions satisfy and hyperbolic numbers z1 , z2 ∈ D,  Al (z1 , u1 , w1 ) − Al (z2 , u2 , w2 )  ≤ k0 [ z1 − z2 α +  u1 − u2 α +  w1 − w2 ], l = 1, 2,  A3 (z1 , u1 , w1 ) − A3 (z2 , u2 , w2 ) 

(1.20)

≤ k1 [ z1 − z2 α +  u1 − u2 α +  w1 − w2 ], where α (0 < α < 1), k0 , k1 are non-negative constants. It is clear that (1.16) is the complex form of the following real equation of second order uxx − uyy = aux + buy + cu + d in D, (1.21) in which a, b, c, d are functions of (x, y)(∈ D), u, ux , uy (∈ IR), and A1 =

a + jb c d , A2 = , A3 = in D. 2 4 4

2. Quasilinear Hyperbolic Equations

43

Due to z = x + jy = µe1 + νe2 , w = uz = ξe1 + ηe2 , and wz =

wx + jwy wx − jwy = ξµ e1 + ην e2 , wz¯ = = ξν e1 + ηµ e2 , 2 2

the quasilinear hyperbolic equation (1.16) can be rewritten in the form ξν e1 + ηµ e2 = [A(z, u, w)ξ + B(z, u, w)η + C(z, u, w)u + D(z, u, w)]e1 ⎧ ⎨ ξν ⎩

+[A(z, u, w)ξ + B(z, u, w)η + C(z, u, w)u + D(z, u, w)]e2 , i.e.

(1.22)

= A(z, u, w)ξ + B(z, u, w)η + C(z, u, w)u + D(z, u, w) in D,

ηµ = A(z, u, w)ξ + A(z, u, w)η + C(z, u, w)u + D(z, u, w)

in which A=

a+b a−b c d ,B= ,C= ,D= . 4 4 4 4

In the following, we mainly discuss the oblique derivative problem for linear hyperbolic equation (1.6) and quasilinear hyperbolic equation (1.16) in the simply connected domain. We first prove that there exists a unique solution of the boundary value problem and give a priori estimates of their solutions, and then prove the solvability of the boundary value problem for general hyperbolic equations.

2

Oblique Derivative Problems for Quasilinear Hyperbolic Equations of Second Order

Here we first introduce the oblique derivative problem for quasilinear hyperbolic equations of second order in a simply connected domain, and give the representation theorem of solutions for hyperbolic equations of second order.

2.1 Formulation of the oblique derivative problem and the representation of solutions for hyperbolic equations The oblique derivative problem for equation (1.16) may be formulated as follows: ¯ satisfying Problem P Find a continuously differentiable solution u(z) of (1.16) in D the boundary conditions 1 ∂u = Re [λ(z)uz ] = r(z), 2 ∂l u(0) = b0 ,

z ∈ L = L3 ∪ L 4 ,

Im [λ(z)uz ]|z=z3 = b1 ,

(2.1)

44

II. Hyperbolic Equations of Second Order

where l is a given vector at every point on L, λ(z) = a(z) + jb(z) = cos(l, x) + j cos(l, y), z ∈ L, b0 , b1 are real constants, and λ(z), r(z), b0 , b1 satisfy the conditions Cα [λ(z), L] ≤ k0 , Cα [r(z), L] ≤ k2 , max z∈L3

|b0 |, |b1 | ≤ k2 ,

1 1 , max ≤ k0 , |a(z) − b(z)| z∈L4 |a(z) + b(z)|

(2.2)

in which α(0 < α < 1), k0 , k2 are non-negative constants. The above boundary value problem for (1.16) with A3 (z, u, w) = 0, z ∈ D, u ∈ IR, w ∈ C I and r(z) = b0 = b1 = 0, z ∈ L will be called Problem P0 . By z = x + jy = µe1 + νe2 , w = uz = ξe1 + ηe2 , the boundary condition (2.1) can be reduced to Re [λ(z)(ξe1 + ηe2 )] = r(z), u(0) = b0 , Im [λ(z)(ξe1 + ηe2 )]|z=z3 = b1 ,

(2.3)

where λ(z) = (a + b)e1 + (a − b)e2 . Moreover, the domain D is transformed into Q = {0 ≤ µ ≤ 2R, 0 ≤ ν ≤ 2R1 }, R = R2 − R1 , which is a rectangle and A, B, C, D are known functions of (µ, ν) and unknown continuous functions u, w, and they satisfy the condition ⎧ λ(z3 )w(z3 ) = λ(z)[ξe1 + ηe2 ]|z=z3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Re [λ(z)w(2Re1 + νe2 )] = r(z), ⎪ ⎪ ⎪ ⎪ ⎨

= r(z3 ) + jb1 , u(0) = b0 ,

if (x, y) ∈ L = {µ = 2R, 0 ≤ ν ≤ 2R1 },

3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Re [λ(z)w(µe1 + 0e2 )] ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(2.4)

= r(z),

if (x, y) ∈ L4 = {0 ≤ µ ≤ 2R, ν = 0},

where λ(z), r(z), b0 , b1 are as stated in (2.1). We can assume that w(z3 ) = 0, otherwise through the transformation W (z) = {w(z) − [a(z3 ) − jb(z3 )][r(z3 ) + jb1 ]} /[a2 (z3 ) − b2 (z3 )], the requirement can be realized. It is not difficult to see that the oblique derivative boundary value problem (Problem P ) includes the Dirichlet boundary value problem (Problem D) as a special case. In fact, the boundary condition of Dirichlet problem (Problem D) for equation (1.21) is as follows u(z) = φ(z) on L = L3 ∪ L4 .

(2.5) √ We find the derivative with respect to the tangent direction s = (x∓jy)/ 2 for (2.5), in which ∓ are determined by L3 and L4 respectively, it is clear that the following equalities hold: Re [λ(z)uz ] = r(z),

z ∈ L,

Im [λ(z)uz ]|z=z3 = b1 ,

(2.6)

2. Quasilinear Hyperbolic Equations

45

in which ⎧ 1−j ⎪ ⎪ √ ⎪ ⎪ ⎨ 2

λ(z) = a + jb = ⎪

⎪ 1+j ⎪ ⎪ ⎩ √



b+ 1 = Im

2

on L3 , on L4 ,

φx r(z) = √ on L = L3 ∪ L4 , 2



√ 1−j φ +φ √ uz (z3 ) = − x √ x |z=z3 −0 = − 2φx |z=z3 −0 , 2 2 2



(2.7)



√ 1+j √ uz (z3 ) = 2φx |z=z3 +0 , b0 = φ(0), 2 √ √ √ √ in which a = 1/ 2
= b = −1/ 2 on L3 and a = 1/ 2
= −b = −1/ 2 on L4 . b− 1 = Im

Noting that Problem P for (1.16) is equivalent to the Riemann–Hilbert problem (Problem A) for the complex equation of first order and boundary conditions: wz¯ = Re [A1 w] + A2 u + A3 in D, z ∈ L,

Re [λ(z)w(z)] = r(z),



and the relation u(z) = 2Re

Im [λ(z)w(z)]|z=z3 = b1 ,

(2.8) (2.9)

z

0

w(z)dz + b0 in D,

(2.10)

from Theorem 1.2 (3), Chapter I, and equation (2.8), we see that 

2Re

Γ



w(z)dz =

Γ



w(z)dz + 

= 2j D

Γ

w(z)d¯ z

[wz¯ − w¯z ]dxdy = 4

 D

Im [wz¯]dxdy = 0,

the above equality for any subdomain in D is also true, hence the function determined ¯ In this case we by the integral in (2.10) is independent of integral paths in D. may choose that the integral path is along two family of characteristic lines, namely first along one of characteristic lines: x + y = µ (0 ≤ µ ≤ 2R) and then along one of characteristic lines: x − y = ν (0 ≤ ν ≤ 2R1 ), for instance, the value of u(z ∗ )(z ∗ = x∗ + jy ∗ ∈ D, y ∗ ≤ 0) can be obtained by the integral u(z ∗ ) = 2Re





w(z)dz + s1

s2



w(z)dz + b0 ,

where s1 = {x + y = 0, 0 ≤ x ≤ (x∗ − y ∗ )/2}, s2 = {x − y = x∗ − y ∗ , (x∗ − y ∗ )/2 ≤ x ≤ x∗ −y ∗ }, in which x∗ −y ∗ is the intersection of the characteristic line: {x−y = x∗ −y ∗ } through the point z ∗ and real axis. In particular, when Aj = 0, j = 1, 2, 3, equation (1.16) becomes the simplest hyperbolic complex equation uzz¯ = 0.

(2.11)

46

II. Hyperbolic Equations of Second Order

Problem P for (2.11) is equivalent to Problem A for the simplest hyperbolic complex equation of first order wz¯ = 0 in D (2.12) with the boundary condition (2.9) and the relation (2.10). Hence similarly to Theorem 3.1, Chapter I, we can derive the representation and existence theorem of solutions of Problem A for the simplest equation (2.12), namely Theorem 2.1 Any solution u(z) of Problem P for the hyperbolic equation (2.11) can be expressed as (2.10), where w(z) is as follows w(z) = f (x + y)e1 + g(x − y)e2 1 = {f (x + y) + g(x − y) + j[f (x + y) − g(x − y)]}, 2 r((1 + j)R) + b1 , f (2R) = u(z3 ) + v(z3 ) = a((1 + j)R) + b((1 + j)R) g(0) = u(z3 ) − v(z3 ) =

(2.13)

r((1 + j)R) − b1 , a((1 + j)R) − b((1 + j)R)

here f (x + y), g(x − y) possess the forms 2r((1 − j)(x − y)/2 + (1 + j)R) a((1 − j)(x − y)/2 + (1 + j)R) − b((1 − j)(x − y)/2 + (1 + j)R) [a((1−j)(x−y)/2+(1 + j)R)+b((1 − j)(x − y)/2 + (1 + j)R)]f (2R) − , a((1 − j)(x − y)/2 + (1 + j)R)−b((1 − j)(x − y)/2 + (1 + j)R) 0 ≤ x − y ≤ 2R1 , (2.14) 2r((1+j)(x+y)/2)−[a((1 + j)(x + y)/2)−b((1 + j)(x + y)/2)]g(0) f (x + y)= , a((1 + j)(x + y)/2) + b((1 + j)(x + y)/2) 0 ≤ x + y ≤ 2R. g(x − y)=

Moreover u(z) satisfies the estimate ¯ ≤ M1 = M1 (α, k0 , k2 , D), C 1 [u(z), D] ¯ ≤ M2 k2 = M2 (α, k0 , D)k2 , Cα1 [u(z), D] α

(2.15)

where M1 , M2 are two non-negative constants only dependent on α, k0 , k2 , D and α, k0 , D respectively. Proof Let the general solution w(z) = uz = 12 {f (x + y) + g(x − y) + j[f (x + y) − g(x − y)]} of (2.12) be substituted in the boundary condition (2.1), we obtain a(z)u(z) + b(z)v(z) = r(z) on L, λ(z3 )w(z3 ) = r(z3 ) + jb1 ,

i.e.

[a((1 − j)x + 2jR) + b((1 − j)x + 2jR)]f (2R)+[a((1 − j)x + 2jR) −b((1 − j)x + 2jR)]g(2x − 2R) = 2r((1 − j)x + 2jR) on L3 ,

2. Quasilinear Hyperbolic Equations

47

[a((1 + j)x) + b((1 + j)x)]f (2x) + [a((1 + j)x) −b((1 + j)x)]g(0) = 2r((1 + j)x) on L4 , f (2R) = u(z3 )+v(z3 ) =

r((1 + j)R)+b1 , a((1 + j)R)+b((1 + j)R)

r((1 + j)R)−b1 , a((1 + j)R)−b((1 + j)R) and the above formulas can be rewritten as g(0) = u(z3 )−v(z3 ) =

[a((1 − j)t/2 + (1 + j)R) + b((1 − j)t/2 + (1 + j)R)]f (2R) + [a((1 − j)t/2 + (1 + j)R) − b((1 − j)t/2 + (1 + j)R)]g(t) = 2r((1 − j)t/2 + (1 + j)R),

t ∈ [0, 2R1 ],

[a((1 + j)t/2) + b((1 + j)t/2)]f (t) + [a((1 + j)t/2) −b((1 + j)t/2)]g(0) = 2r((1 + j)t/2),

t ∈ [0, 2R],

thus the solution w(z) can be expressed as (2.13),(2.14). From the condition (2.2) and the relation (2.10), we see that the estimate (2.15) of u(z) for (2.11) is obviously true. Next we give the representation of Problem P for the quasilinear equation (1.16). Theorem 2.2 Under Condition C, any solution u(z) of Problem P for the hyperbolic equation (1.16) can be expressed as 

u(z) = 2Re

0

z

w(z)dz + b0 in D,

w(z) = W (z) + Φ(z) + Ψ(z) in D, W (z) = f (µ)e1 + g(ν)e2 , Φ(z) = f˜(µ)e1 + g˜(ν)e2 , 

Ψ(z) =

0



ν

[Aξ +Bη+Cu + D]e1 dν +

(2.16)

µ

2R

[Aξ +Bη+Cu+D]e2 dµ,

where f (µ), g(ν) are as stated in (2.14) and f˜(µ), g˜(ν) are similar to f (µ), g(ν) in (2.14), but r(z), b1 are replaced by −Re [λ(z)Ψ(z)], −Im [λ(z3 )Ψ(z3 )], namely Re [λ(z)Φ(z)] = −Re [λ(x)Ψ(x)],

z ∈ L,

Im [λ(z3 )Φ(z3 )] = −Im [λ(z3 )Ψ(z3 )]. (2.17)

Proof Let the solution u(z) of Problem P be substituted into the coefficients of equation (1.16). Then the equation in this case can be seen as a linear hyperbolic equation (1.15). Due to Problem P is equivalent to the Problem A for the complex equation (2.8) with the relation (2.10), from Theorem 3.2, Chapter I, it is not difficult to see that the function Ψ(z) satisfies the complex equation [Ψ]z¯ = [Aξ + Bη + Cu + D]e1 + [Aξ + Bη + Cu + D]e2 in D,

(2.18)

48

II. Hyperbolic Equations of Second Order

and Φ(z) = w(z) − W (z) − Ψ(z) satisfies the complex equation and the boundary conditions ξν e1 + ηµ e2 = 0, (2.19) Re [λ(z)(ξe1 + ηe2 )] = −Re [λ(z)Ψ(z)] on L,

(2.20)

Im [λ(z)(ξe1 + ηe2 )]|z=z3 = −Im [λ(z3 )Ψ(z3 )].

By the representation of solutions of Problem A for (1.16) as stated in (3.17), Chapter I, we can obtain the representation (2.16) of Problem P for (1.16). 2.2 Existence and uniqueness of solutions of Problem P for hyperbolic equations of second order Theorem 2.3 If the complex equation (1.16) satisfies Condition C, then Problem P for (1.16) has a solution. Proof We consider the expression of u(z) in the form (2.16). In the following, by using successive iteration we shall find a solution of Problem P for equation (1.16). Firstly, substitute 

u0 (z) = 2Re

0

z

w0 (z)dz + b0 , w0 (z) = W (z) = ξ0 e1 + η0 e2 ,

(2.21)

into the position of u, w = ξe1 + ηe2 in the right-hand side of (1.16), where w0 (z) is the same function with W (z) in (2.16) and satisfies the estimate (2.15). Moreover, we have 

u1 (z) = 2Re 

Ψ1 (z) =

0

+

0

x−y



z

w1 (z)dz + b0 , w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z),

[Aξ0 + Bη0 + Cu0 + D]e1 d(x − y)

(2.22)

x+y

2R

[Aξ0 + Bη0 + Cu0 + D]e2 d(x + y),

from the first equality in (2.22), the estimate ¯ ≤ M3 C[w0 (z), D] ¯ + k2 , C 1 [u0 (z), D]

(2.23)

can be derived, where M3 = M3 (D). From the third and second equalities in (2.22), we can obtain ¯ ≤ 2M4 [(4 + M3 )m + k2 + 1]R
, C[Ψ1 (z), D] ¯ ≤ 8M4 k 2 (1 + k 2 )[(4 + M3 )m + k2 + 1]R
, C[Φ1 (z), D] 0 0


¯ ≤ 2M4 M [(4 + M3 )m + k2 + 1]R , C[w1 (z) − w0 (z), D]

(2.24)

2. Quasilinear Hyperbolic Equations

49

where M4 = maxD¯ (|A|, |B|, |C|, |D|), R
= max(2R1 , 2R), m = w0 (z) C(D) ¯ , M = 1 + 4k02 (1 + k02 ) is a positive constant. Thus we can find a sequence of functions {wn (z)} satisfying 

un+1 (z) = 2Re

z

0

wn+1 (z)dz + b0 , 

wn+1 (z) = w0 (z) + Φn (z) + 

+

ν

[Aξn + Bηn + Cun + D]e1 dν

0

(2.25)

µ

2R

[Bηn + Aηn + Cun + D]e2 dµ,

and then  wn − wn−1  ≤ {2M4 M [(4 + M3 )m + 1]}n ×



R


0

R
n−1 dR
(n − 1) !

{2M4 M [(4 + M3 )m + 1)]R
}n . ≤ n!

(2.26)

From the above inequality, it is seen that the sequence of functions {wn (z)}, i.e. wn (z) = w0 (z) + [w1 (z) − w0 (z)] + · · · + [wn (z) − wn−1 (z)](n = 1, 2, . . .)

(2.27)

¯ uniformly converges a function w∗ (z), and w∗ (z) satisfies the equality in D w∗ (z) = ξ∗ e1 + η∗ e2



= w0 (z) + Φ∗ (z) + 

+

ν

[Aξ∗ + Bη∗ + Cu∗ + D]e1 dν

0

(2.28)

µ

2R

[Aξ∗ + Bη∗ + Cu∗ + D]e2 dµ, 

and the function u∗ (z) = 2Re

z

0

w∗ (z)dz + b0 ,

(2.29)

¯ is just a solution of Problem P for equation (1.16) in the closed domain D. Theorem 2.4 Suppose that Condition C holds. Then Problem P for the complex ¯ equation (1.16) has at most one solution in D. Proof Let u1 (z), u2 (z) be any two solutions of Problem P for (1.16), we see that u(z) = u1 (z) − u2 (z) and w(z) = u1z (z) − u2z (z) satisfies the homogeneous complex equation and boundary conditions wz¯ = Re [A˜1 w] + A˜2 u in D,

(2.30)

Re [λ(z)w(z)] = 0 on L, Im [λ(z3 )w(z3 )] = 0,

(2.31)



and the relation u(z) = 2Re

0

z

¯ w(z)dz, z ∈ D.

(2.32)

50

II. Hyperbolic Equations of Second Order

From Theorem 2.2, we see that the function w(z) can be expressed in the form w(z) = Φ(z) + Ψ(z), 

Ψ(z) =

0

ν

˜ + Bη ˜ + Cu]e ˜ 1 dν + [Aξ



µ

2R

˜ + Bη ˜ + Cu]e ˜ 2 dµ, [Aξ

(2.33)

moreover from (2.32), ¯ ≤ M3 C[w(z), D] ¯ C1 [u(z), D]

(2.34)

can be obtained, in which M3 = M3 (D) is a non-negative constant. By using the process of iteration similar to the proof of Theorem 2.3, we can get  w(z) = w1 − w2 ≤

{2M5 M [(4 + M3 )m + 1]R
}n , n!

˜ |B|, ˜ |C|). ˜ Let n → ∞, it can be seen w(z) = w1 (z) − w2 (z) = where M5 = maxD¯ (|A|, ¯ This proves the uniqueness of solutions of Problem P for 0, Ψ(z) = Φ(z) = 0 in D. (1.16).

3

Oblique Derivative Problems for General Quasilinear Hyperbolic Equations of Second Order

¯ of solutions for the oblique In this section, we first give a priori estimates in C 1 (D) derivative problem, moreover by using the estimates of solutions, the existence of solutions for the above problem for general quasilinear equation is proved. Finally we discuss the oblique derivative problem for hyperbolic equations of second order in general domains. 3.1 A priori estimates of solutions of Problem P for hyperbolic equations of second order From Theorems 2.3 and 2.4, we see that under Condition C, Problem P for equation (1.16) has a unique solution u(z), which can be found by using successive iteration. Noting that wn+1 (z)−wn (z) satisfy the estimate (2.26), the limit w(z) of the sequence of functions {wn (z)} satisfies the estimate ¯ ≤ e2M5 M [(4+M3 )m+1]R
= M6 , max  w(z) = C[w(z), D] ¯ z∈D

(3.1)

and the solution u(z) of Problem P is as stated in (2.10), which satisfies the estimate ¯ ≤ R ∗ M6 + k2 = M 7 , C 1 [u(z), D] where R∗ = 2R1 + 2R. Thus we have

(3.2)

3. General Hyperbolic Equations

51

Theorem 3.1 If Condition C holds, then any solution u(z) of Problem P for the hyperbolic equation (1.16) satisfies the estimates ¯ ≤ M7 , C 1 [u, D] ¯ ≤ M8 k, C 1 [u, D]

(3.3)

in which M7 = M7 (α, k0 , k1 , k2 , D), k = k1 + k2 , M8 = M8 (α, k0 , D) are non-negative constants. ¯ In the following, we give the Cα1 (D)-estimates of solution u(z) for Problem P for (1.16). Theorem 3.2 If Condition C and (1.20) hold, then any solution u(z) of Problem P for the hyperbolic equation (1.16) satisfies the estimates ¯ ≤ M9 , C 1 [u, D] ¯ ≤ M10 , C 1 [u, D] ¯ ≤ M11 k, Cα [uz , D] α α

(3.4)

in which k = k1 + k2 , Mj = Mj (α, k0 , k1 , k2 , D), j = 9, 10, M11 = M11 (α, k0 , D) are non-negative constants. Proof Similarly to Theorem 4.3, Chapter I, it suffices to prove the first estimate in (3.4). Due to the solution u(z) of Problem P for (1.16) is found by the successive iteration through the integral expression (2.16), we first choose the solution 

u0 (z) = 2Re

z

0

w0 (z)dz + b0 , w0 (z) = W (z) = ξ0 e1 + η0 e2 ,

(3.5)

of Problem P for the equation uzz¯ = 0 in D,

(3.6)

and substitute u0 , w0 into the position of u, w = ξe1 + ηe2 on the right-hand side of (1.16), where w0 (z) is the same function with W (z) in (2.16) and w0 (z), u0 (z) satisfy the first estimates ¯ = Cα [Re w0 D] ¯ ≤ M12 k2 , C 1 [u0 , D] ¯ ≤ M13 k2 = M14 , (3.7) ¯ + Cα [Im w0 , D] Cα [w0 , D] α where Mj = Mj (α, k0 , D), j = 12, 13, and then we have 

u1 (z) = 2Re

0

z

w1 (z)dz + b0 , w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z),

Ψ1 (z) = Ψ11 (z)e1 + Ψ21 (z)e2 , Ψ11 (z) = Ψ21 (z) =



 0

ν

G1 (z)dν,

(3.8)

µ

2R

G2 (z)dµ, G1 (z) = G2 (z) = Aξ0 + Bη0 + Cu0 + D.

From (3.7) and the last three equalities in (3.8), it is not difficult to see that Ψ11 (z) = Ψ11 (µ, ν), Ψ21 (z) = Ψ21 (µ, ν) satisfy the H¨older continuous estimates about ν, µ respectively, namely ¯ ≤ M15 R
, Cα [Ψ2 (µ, ·), D] ¯ ≤ M15 R
, Cα [Ψ11 (·, ν), D] 1

(3.9)

52

II. Hyperbolic Equations of Second Order

in which M15 = M15 (α, k0 , k1 , k2 , D), there is no harm assuming that R
= max(2R1 , 2R) ≥ 1. By Condition C and (1.20), we can see that G1 (z), Ψ11 (z) G2 (z), Ψ21 (z) about µ, ν satisfy the the H¨older conditions respectively, i.e. ¯ ≤ M16 R
, Cα [Ψ1 (µ, ·), D] ¯ ≤ M16 R
, Cα [G1 (µ, ·), D] 1 (3.10) ¯ ≤ M16 R
, Cα [Ψ2 (·, ν), D] ¯ ≤ M16 R
, Cα [G2 (·, ν), D] 1 where M16 = M16 (α, k0 , k1 , k2 , D). Moreover we can obtain the estimates of Ψ1 (z), Φ1 (z) as follows ¯ ≤ M17 R
, Cα [Φ1 (z), D] ¯ ≤ M17 R
, Cα [Ψ1 (z), D]

(3.11)

in which M17 = M17 (α, k0 , k1 , k2 , D). Setting w˜11 (z) = Re w˜1 (z) + Im w ˜1 (z), w˜21 (z) = Re w˜1 (z) − Im w˜1 (z), w˜1 (z) = w1 (z) − w0 (z) and u˜1 (z) = u1 (z) − u0 (z), from (3.8)– (3.11), it follows that ¯ ≤ M18 R
, Cα [w˜ 2 (z), D] ¯ ≤ M18 R
, Cα [w˜11 (z), D] 1
¯ ≤ M18 R
, C 1 [˜ ¯ Cα [w˜1 (z), D] α u1 (z), D] ≤ M18 R ,

(3.12)

where M18 = M18 (α, k0 , k1 , k2 , D). According to the successive iteration, the estimates of functions w˜n1 (z) = Re w˜n (z) + Im w˜n (z), w˜n2 (z) = Re w˜n (z) − Im w˜n (z), w˜n (z) = wn (z) − wn−1 (z) and the corresponding function u˜n (z) = un (z) − un−1 (z) can be obtained, namely
n
n ¯ ≤ (M18 R ) , Cα [w˜ 2 (z), D] ¯ ≤ (M18 R ) , Cα [w˜n1 (z), D] n n! n!
n
n R ) (M (M 18 18 R ) ¯ ≤ ¯ ≤ , Cα1 [˜ . Cα [w˜n (z), D] un (z), D] n! n! Therefore the sequences of functions

wn (z) = w0 (z) +

n 

w˜m (z), un (z) = u0 (z) +

m=1

n 

(3.13)

u˜m (z) (n = 1, 2, . . .)

m=1

¯ respectively, and w(z), u(z) satisfy the estiuniformly converge to w(z), u(z) in D mates ¯ ≤ M9 = eM18 R
, C 1 [u(z), D] ¯ ≤ M10 , Cα [w(z), D] (3.14) α this is just the first estimate in (3.4). 3.2 The existence of solutions of Problem P for general hyperbolic equations of second order Now, we consider the general quasilinear equation of second order uzz = F (z, u, uz ) + G(z, u, uz ), F = Re [A1 uz ] + A2 u + A3 , G = A4  uz  +A5 | u | , σ

τ

(3.15) z ∈ D,

3. General Hyperbolic Equations

53

where F (z, u, uz ) satisfies Condition C, σ, τ are positive constants, and Aj (z, u, uz ) (j = 4, 5) satisfy the conditions in Condition C, i.e. ¯ ≤ k0 , C[Aj (z, u, uz ), D]

j = 4, 5,

and denote by Condition C
the above conditions. Theorem 3.3

Let the complex equation (3.15) satisfy Condition C
.

¯ (1) When 0 < max(σ, τ ) < 1, Problem P for (3.15) has a solution u(z) ∈ C11 (D). ¯ (2) When min(σ, τ ) > 1, Problem P for (3.15) has a solution u(z) ∈ C11 (D), provided that M19 = k1 + k2 + |b0 | + |b1 | (3.16) is sufficiently small. (3) In general, the above solution of Problem P is not unique, if 0 < max(σ, τ ) < 1. Proof

(1) Consider the algebraic equation for t : M8 {k1 + k2 + 2k0 tσ + 2k0 tτ + |b1 | + |b0 |} = t,

(3.17)

where M8 is a non-negative constant in (3.3), it is not difficult to see that equation (3.17) has a unique solution t = M20 ≥ 0. Now, we introduce a closed and convex ¯ whose elements are the functions u(z) satisfying subset B ∗ in the Banach space C 1 (D), the conditions ¯ C 1 [u(z), D] ¯ ≤ M20 . u(z) ∈ C 1 (D), (3.18) We arbitrarily choose a function u0 (z) ∈ B ∗ for instance u0 (z) = 0 and substitute it into the position of u in coefficients of (3.15) and G(z, u, uz ), from Theorems 2.3 and 2.4, it is clear that problem P for uzz¯ −Re [A1 (z, u0 , u0z )uz ]−A2 (z, u0 , u0z )u−A3 (z, u0 , u0z ) = G(z, u0 , u0z ),

(3.19)

has a unique solution u1 (z) ∈ B ∗ . By Theorem 3.1, we see that the solution u1 (z) satisfies the estimate in (3.18). By using the successive iteration, we obtain a sequence of solutions um (z)(m = 1, 2, ...) ∈ B ∗ of Problem P, which satisfy the equations um+1zz¯ − Re [A1 (z, um , umz )um+1z ] − A2 (z, um , umz )um+1 ¯ −A3 (z, um , umz ) = G(z, um , umz ) in D,

(3.20)

m = 1, 2, . . .

and um+1 (z) ∈ B ∗ . From (3.20), we see that u˜m+1 (z) = um+1 (z) − um (z) satisfies the equations and boundary conditions u˜m+1zz¯ − Re[A˜1 u˜m+1z ] − A˜2 u˜m+1 ¯ = G(z, um , umz ) − G(z, um−1 , um−1z ) in D,

m = 1, 2, . . . ,

Re [λ(z)(˜ um+1 (z)] = 0 on L, Im [λ(z3 )(˜ um+1 (z3 )] = 0.

(3.21)

54

II. Hyperbolic Equations of Second Order

¯ ≤ 2k0 M20 , M20 is a solution of Noting that C[G(z, um , umz ) − G(z, um−1 , um−1z ), D] the algebraic equation (3.17) and according to Theorem 3.1, ¯ ≤ M20 um+1 , D]  u˜m+1 = C 1 [˜

(3.22)

can be obtained. Due to u˜m+1 can be expressed as 

u˜m+1 (z) = 2Re 

Ψm+1 (z) =

x−y

0

+

z

0



wm+1 (z)dz, wm+1 (z) = w0 (z) + Φm+1 (z) + Ψm+1 (z),

˜ m + Bη ˜ m + Cu ˜ m + G]e ˜ 1 d(x − y) [Aξ

x+y

2R

(3.23)

˜ m + Bη ˜ m + Cu ˜ m + G]e ˜ 2 d(x + y), [Aξ

˜ and A, ˜ B, ˜ C, ˜ G ˜ is the same as that of in which the relation between A˜1 , A˜2 , G ˜ = G(z, um , umz ) − G(z, um−1 , um−1z ). A1 , A2 , A3 and A, B, C, D in Section 1 and G By using the method in the proof of Theorem 2.3, we can obtain
n ¯ ≤ (M21 R ) ,  um+1 − um = C 1 [˜ um+1 , D] n!

where M21 = 4M22 M (M3 + 2)(2m0 + 1), R
= max(2R1 , 2R), m0 = w0 (z) C(D) ¯ , ˜ Q], C[B, ˜ Q], C[C, ˜ Q], C[G, ˜ Q]}, M = 1 + 4k 2 (1 + k 2 ). From herein M22 = max{C[A, 0 0 the above inequality, it is seen that the sequences of functions: {um (z)}, {wm (z)}, i.e. um (z) = u0 (z)+[u1 (z)−u0 (z)]+· · ·+[um (z)−um−1 (z)](m = 1, 2, . . .), wm (z) = w0 (z)+[w1 (z)−w0 (z)]+· · ·+[wm (z)−wm−1 (z)](m = 1, 2, . . .),

(3.24)

uniformly converge to the functions u∗ (z), w∗ (z) respectively, and w∗ (z) satisfies the equality w∗ (z) = w0 (z) + Φ∗ (z) 

+

0



+

x−y

[Aσ∗ + Bη∗ + Cu∗ + D]e1 d(x − y)

(3.25)

x+y

2R

[Bσ∗ + Aη∗ + Cu∗ + D]e2 d(x + y), 

and the function u∗ (z) = 2Re

0

z

w∗ (z)dz + b0 ,

(3.26)

is just a solution of Problem P for the nonlinear equation (3.15) in the closed ¯ domain D. (2) Consider the algebraic equation for t : M8 {k1 + k2 + 2k0 tσ + 2k0 tτ + |b0 | + |b1 |} = t,

(3.27)

3. General Hyperbolic Equations

55

it is not difficult to see that equation (3.27) has a solution t = M20 ≥ 0, provided that M19 in (3.16) is small enough. If there exist two solutions, then we choose the minimum of both as M20 . Now, we introduce a closed and convex subset B∗ ¯ whose elements are of the functions u(z) satisfying the of the Banach space C 1 (D), conditions ¯ C 1 [u(z), D] ¯ ≤ M20 . u(z) ∈ C 1 (D), (3.28) By using the same method in (1), we can find a solution u(z) ∈ B∗ of Problem P for equation (3.15) with min(σ, τ ) > 1. (3) We can give an example to explain that there exist two solutions for equation (3.15) with σ = 0, τ = 1/2, namely the equation uxx − uyy = Au1/2 ,

A = 8sgn(x2 − y 2 ) in D

(3.29)

has two solutions u1 (x, y) = 0 and u2 (x, y) = (x2 − y 2 )2 /4, and they satisfy the boundary conditions Re [λ(z)uz ] = r(z), where

z ∈ L,

λ(z) = 1 − i,

u(0) = b0 , Im [λ(z)uz ]|z=0 = b1 ,

z ∈ L1 ,

b0 = 0, r(z) = 0,

λ(x) = 1 + i,

z ∈ L = L1 ∪ L4 ,

(3.30)

z ∈ L4 , b1 = 0.

3.3 The existence of solutions of Problem P for hyperbolic equations of second order in general domains In this subsection, we shall generalize the domain D to general cases. 1. The boundaries L3 , L4 of the domain D are replaced by the curves L
3 , L
4 , and the boundary of the domain D
is L1 ∪ L2 ∪ L
3 ∪ L
4 , where the parameter equations of the curves L
3 , L
4 are as follows: L
3 = {x + y = 2R, l ≤ x ≤ R2 }, L
4 = {x + y = µ, y = γ1 (x), 0 ≤ x ≤ l},

(3.31)

in which γ1 (x) on 0 ≤ x ≤ l = γ1 (l) is continuous and γ1 (0) = 0, γ1 (x) > 0 on 0 < x ≤ l, and γ1 (x) possesses the derivatives on 0 ≤ x ≤ l except some isolated points and 1 + γ1
(x) > 0. By the condition, we can find the inverse function x = τ (µ) = (µ + ν)/2 of x + γ1 (x) = µ, and then ν = 2τ (µ) − µ, 0 ≤ µ ≤ 2R. We make a transformation µ ˜ = µ, ν˜ =

2R1 (ν − 2τ (µ) + µ) , 2R1 − 2τ (µ) + µ

0 ≤ µ ≤ 2R,

2τ (µ) − µ ≤ ν ≤ 2R1 ,

(3.32)

its inverse transformation is µ=µ ˜, ν =

2R1 − 2τ (µ) + µ ν˜ + 2τ (µ) − µ, 2R1

0≤µ ˜ ≤ 2R,

0 ≤ ν˜ ≤ 2R1 . (3.33)

56

II. Hyperbolic Equations of Second Order

Hence we have 4R1 x − [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 + x + y) 1 µ + ν˜) = , x˜ = (˜ 2 4R1 − 4τ (x + γ1 (x)) + 2x + 2γ1 (x) 4R1 y + [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 − x − y) 1 µ − ν˜) = , y˜ = (˜ 2 4R1 − 4τ (x + γ1 (x)) + 2x + 2γ1 (x)

(3.34)

and 4R1 x˜ + [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 − x˜ + y˜) 1 x = (µ + ν) = , 2 4R1 4R1 y˜ − [2τ (x + γ1 (x)) − x − γ1 (x)](2R1 + x˜ − y˜) 1 y = (µ − ν) = . 2 4R1

(3.35)

Denote by z˜ = x˜ + j y˜ = f (z), z = x + jy = f −1 (˜ z ) the transformations (3.34) and (3.35) respectively. In this case, setting w(z) = uz , equation (1.16) in D
and boundary condition (2.1) on L
3 ∪ L
4 can be rewritten in the form ξν = Aξ + Bη + Cu + D,

ηµ = Aξ + Bη + Cu + D,

¯
, z∈D

Re [λ(z)w(z)] = r(z), z ∈ L
3 ∪ L
4 , u(0) = b0 , Im [λ(z3 )w(z3 )] = b1 , in which z3 = l + jγ1 (l), λ(z), r(z), b1 on and w(z) satisfy the relation 

u(z) = 2Re

0

z

L
3 ∪ L
4

(3.36) (3.37)

satisfy the condition (2.2), and u(z)

w(z)dz + b0 in D
.

(3.38)

¯
satisfies Condition C, through the transformation Suppose equation (1.16) in D (3.33), we have ξν˜ = [2R1 − 2τ (µ) + µ]ξν /2R1 , ηµ˜ = ηµ , system (3.36) is reduced to ξν˜ =

[2R1 − 2τ (µ) + µ][Aξ + Bη + Cu + D] , ηµ˜ = Aξ + Bη + Cu + D. 2R1

(3.39)

Moreover through the transformation (3.35), the boundary condition (3.37) on L
3 ∪L
4 is reduced to Re [λ(f −1 (˜ z ))w(f −1 (˜ z ))] = r(f −1 (˜ z )), z˜ ∈ L3 ∪ L4 , z3 ))w(f −1 (˜ z3 )] = b1 , Im [λ(f −1 (˜

(3.40)

in which z˜3 = f (z3 ). Therefore the boundary value problem (3.36),(3.37) is transformed into the boundary value problem (3.39),(3.40). On the basis of Theorems 2.2 and 2.3, we see that the boundary value problem (3.39),(3.40) has a unique solution w(˜ z ), and then w[f (z)] is just a solution of the boundary value problem (3.36),(3.37). Theorem 3.4 If equation (1.16) satisfies Condition C in the domain D
with the boundary L1 ∪L2 ∪L
3 ∪L
4 , then Problem P with the boundary condition (3.37)(w = uz ) has a unique solution u(z) as stated in (3.38).

3. General Hyperbolic Equations

57

2. The boundaries L3 , L4 of the domain D are replaced by two curves L

3 , L

4 respectively, and the boundary of the domain D

is L1 ∪ L2 ∪ L

3 ∪ L

4 , where the parameter equations of the curves L

3 , L

4 are as follows: L

3 = {x − y = ν, y = γ2 (x),

l ≤ x ≤ R2 },

L

4

0 ≤ x ≤ l},

= {x + y = µ, y = γ1 (x),

(3.41)

in which γ1 (0) = 0, γ2 (R2 ) = R2 − 2R1 ; γ1 (x) > 0, 0 ≤ x ≤ l; γ2 (x) > 0, l ≤ x ≤ R2 ; γ1 (x) on 0 ≤ x ≤ l, γ2 (x) on l ≤ x ≤ R2 are continuous, and γ1 (x), γ2 (x) possess the derivatives on 0 ≤ x ≤ l, l ≤ x ≤ R2 except isolated points respectively, and 1+γ1
(x) > 0, 1−γ2
(x) > 0, z3
= l+jγ1 (l) = l+jγ2 (l) ∈ L3 (or L4 ). By the conditions, we can find the inverse functions x = τ (µ), x = σ(ν) of x + γ1 (x) = µ, x − γ2 (x) = ν respectively, namely ν = 2τ (µ) − µ,

0 ≤ µ ≤ 2R,

µ = 2σ(ν) − ν,

0 ≤ ν ≤ l − γ2 (l).

(3.42)

0 ≤ µ ≤ 2σ(ν) − ν,

0 ≤ ν ≤ 2R1 ,

(3.43)

We first make a transformation µ ˜=

2Rµ , 2σ(ν) − ν

ν˜ = ν,

its inverse transformation is µ=

(2σ(ν) − ν)˜ µ , 2R

ν = ν˜,

0 ≤ µ ≤ 2R,

0 ≤ ν ≤ 2R1 .

(3.44)

The above transformation can be expressed by 2R(x + y) + (x − y)[2σ(x − γ2 (x)) − x + γ2 (x)] 1 µ + ν˜) = , x˜ = (˜ 2 2[2σ(x − γ2 (x)) − x + γ2 (x)] 2R(x + y) − (x − y)[2σ(x − γ2 (x)) − x + γ2 (x)] 1 µ − ν˜) = , y˜ = (˜ 2 2[2σ(x − γ2 (x)) − x + γ2 (x)]

(3.45)

and its inverse transformation is 1 [2σ(x − γ2 (x)) − x + γ2 (x)](˜ x + y˜) + 2R(˜ x − y˜) x = (µ + ν) = , 2 4R [2σ(x − γ2 (x)) − x + γ2 (x)](˜ x + y˜) − 2R(˜ x − y˜) 1 . y = (µ − ν) = 2 4R

(3.46)

Denote by z˜ = x˜ + j y˜ = g(z), z = x + jy = g −1 (˜ z ) the transformation (3.45) and the inverse transformation (3.46) respectively. Through the transformation (3.44), we have 2σ(ν) − ν (u + v)ν˜ = (u + v)ν , (u − v)µ˜ = (u − v)µ , (3.47) 2R system (3.36) in D

is reduced to ξν˜ = Aξ + Bη + Cu + D,

ηµ˜ =

2σ(ν) − ν [Aξ + Bη + Cu + D] in D
. 2R

(3.48)

58

II. Hyperbolic Equations of Second Order

Moreover through the transformation (3.46), the boundary condition (3.37) on L

3 ∪L

4 is reduced to Re [λ(g −1 (˜ z ))w(g −1 (˜ z ))] = r(g −1 (˜ z )), z˜ ∈ L
3 ∪ L
4 ,

(3.49)

z3 ))w(g −1 (˜ z3 )] = b1 , Im [λ(g −1 (˜

in which z˜3 = g(z3
). Besides the relation (3.38) is valid. Therefore the boundary ¯

is transformed into the boundary value problem value problem (3.36),(3.37) in D (3.48),(3.49). On the basis of Theorem 3.4, we see that the boundary value problem (3.48),(3.49) has a unique solution w(˜ z ), and then w[g(z)] is just a solution of the boundary value problem (3.36),(3.37) in D

, but we mention that the conditions of curve L
3 , L
4 through the transformation z = g −1 (˜ z ) must satisfy the conditions of the curves in (3.31). For instance, if z3 ∈ L3 , γ1 (x) ≥ x + 2l − 2R on 2R − 2l ≤ x ≤ l, then the above condition holds. If z3 ∈ L4 , γ2 (x) ≥ 2l − x on l ≤ x ≤ 2l, then we can similarly discuss. For other case, it can be discussed by using the method as stated in Section 2, Chapter VI below. Theorem 3.5 If equation (1.16) satisfies Condition C in the domain D

with the boundary L1 ∪ L2 ∪ L

3 ∪ L

4 , then Problem P with the boundary conditions Re[λ(z)uz ] = r(z),

z ∈ L

3 ∪ L

4 , u(0) = b0 ,

Im[λ(z3

)uz (z3

)] = b1

(3.50)

has a unique solution u(z) as stated in (3.38) in D

. By using the above method, we can generalize the domain D to more general domain including the disk D

= {||z − R1 || < R1 }. For the domain D

, we choose R2 = 2R1 , the boundary L

of the domain D

consists of L

1 , L

2 , L

3 , L

4 , namely 







L

1 = y = −γ1 (x) = − R12 − (x − R1 )2 , L

2 = y = −γ2 (x) = − R12 − (x − R1 )2 , 







L

3 = y = γ3 (x) = L

4 = y = γ4 (x) =



0 ≤ x ≤ R1 , 

R1 ≤ x ≤ 2R1 , 

R12 − (x − R1 )2 ,

R1 ≤ x ≤ 2R1 ,

R12 − (x − R1 )2 ,

0 ≤ x ≤ R1 ,



where 1 + γ1
(x) > 0, 1 + γ4
(x) > 0 on 0 < x ≤ R1 , 1 − γ2
(x) > 0, 1 − γ3
(x) > 0 on R1 ≤ x < 2R1 . The above curves can be rewritten as L

1 L

2

⎧ ⎨

= ⎩x = σ1 (ν) = ⎧ ⎨

= ⎩x = τ1 (µ) =

R1 + ν − R1 + µ −





⎫

R12 + 2R1 ν − ν 2 ⎬ , ⎭ 2 ⎫

R12 + 2R1 µ − µ2 ⎬ , ⎭ 2

4. Other Oblique Derivative Problems

L

3 L

4

⎧ ⎨

= ⎩x = σ2 (ν) = ⎧ ⎨

= ⎩x = τ2 (µ) =

R1 + ν − R1 + µ −





59 ⎫

R12 + 2R1 ν − ν 2 ⎬ , ⎭ 2 ⎫

R12 + 2R1 µ − µ2 ⎬ , ⎭ 2

where σ1 (ν), σ2 (ν) are the inverse functions of x + γ1 (x) = ν, x − γ3 (x) = ν on 0 ≤ ν ≤ 2R1 , and x = τ1 (µ), x = τ2 (µ) are the inverse functions of x − γ2 (x) = µ, x + γ4 (x) = µ on 0 ≤ µ ≤ 2R1 respectively. Through a translation, we can discuss the unique solvability of corresponding boundary value problem for equation (1.16) in any disk ||z − z0 || < R, where z0 is a hyperbolic number and R is a positive number.

4

Other Oblique Derivative Problems for Quasilinear Hyperbolic Equations of Second Order

In this section, we discuss other oblique derivative problems for quasilinear hyperbolic equations. Firstly the representation theorem of solutions for the above boundary value problems is given, moreover the uniqueness and existence of solutions for the above problem are proved. The results obtained include the corresponding result of the Dirichlet boundary value problem or the Darboux problem([12]3) as a special case.

4.1 Formulation of other oblique derivative problems for quasilinear hyperbolic equations We first state four other oblique derivative problems for equation (1.16), here the domain D is the same as that in Section 1, but R2 = 2R1 . ¯ satisfyProblem P1 Find a continuously differentiable solution u(z) of (1.16) in D ing the boundary conditions Re [λ(z)uz ] = r(z), u(0) = b0 ,

z ∈ L1 ∪ L2 ,

(4.1)

Im [λ(z)uz ]|z=z1 = b1 ,

where b0 , b1 are real constants, λ(z) = a(z) + jb(z), z ∈ L1 ∪ L2 , and λ(z), r(z), b0 , b1 satisfy the conditions Cα [λ(z), L1 ∪ L2 ] ≤ k0 , |b0 |, |b1 | ≤ k2 ,

max z∈L1

Cα [r(z), L1 ∪ L2 ] ≤ k2 ,

1 , |a(z) − b(z)|

max z∈L2

1 ≤ k0 , |a(z) + b(z)|

in which α(0 < α < 1), k0 , k2 are non-negative constants.

(4.2)

60

II. Hyperbolic Equations of Second Order If the boundary condition (4.1) is replaced by Re [λ(z)uz ] = r(z),

z ∈ L1 ,

Re [Λ(x)uz (x)] = R(x),

u(0) = b0 ,

Im [λ(z)uz ]|z=z1 = b1 ,

x ∈ L0 = (0, R2 ),

(4.3)

where b0 , b1 are real constants, λ(z) = a(z) + jb(z), z ∈ L1 , Λ(x) = 1 + j, x ∈ L0 , and λ(z), r(z), R(x), b0 , b1 satisfy the conditions Cα [λ(z), L1 ] ≤ k0 ,

Cα [r(z), L1 ] ≤ k2 ,

Cα [R(x), L0 ] ≤ k2 ,

max z∈L1

|b0 |, |b1 | ≤ k2 ,

1 ≤ k0 , |a(z) − b(z)|

(4.4)

in which α(0 < α < 1), k0 , k2 are non-negative constants, then the boundary value problem for (1.16) will be called Problem P2 . If the boundary condition in (4.1) is replaced by Re [λ(z)uz ] = r(z),

z ∈ L2 ,

Re [Λ(x)uz (x)] = R(x),

u(0) = b0 ,

Im [λ(z)uz ]|z=z1 = b1 ,

x ∈ L0 = (0, R2 ),

(4.5)

where b0 , b1 are real constants, λ(z) = a(z) + jb(z), z ∈ L2 , Λ(x) = 1 − j, x ∈ L0 , and λ(z), r(z), R(x), b0 , b1 satisfy the conditions Cα [λ(z), L2 ] ≤ k0 ,

Cα [r(z), L2 ] ≤ k2 ,

Cα [R(x), L0 ] ≤ k2 ,

max z∈L2

|b0 |, |b1 | ≤ k2 ,

1 ≤ k0 , |a(z) + b(z)|

(4.6)

then the boundary value problem for (1.16) is called Problem P3 . For Problem P2 and Problem P3 , there is no harm in assuming that w(z1 ) = 0, otherwise through the transformation W (z) = {w(z) − [a(z1 ) − jb(z1 )][r(z1 ) + jb1 ]}/[a2 (z1 ) − b2 (z1 )], the requirement can be realized. If the boundary condition in (4.1) is replaced by u(x) = s(x),

uy = R(x),

x ∈ L0 = (0, R2 ),

(4.7)

where s(z), R(x) satisfy the conditions Cα1 [s(x), L0 ] ≤ k2 ,

Cα [R(x), L0 ] ≤ k2 ,

(4.8)

then the boundary value problem for (1.16) is called Problem P4 . In the following, we first discuss Problem P2 and Problem P3 for equation (2.11).

4. Other Oblique Derivative Problems

61

4.2 Representations of solutions and unique solvability of Problem P2 and Problem P3 for quasilinear hyperbolic equations Similarly to Theorem 2.1, we can give the representation of solutions of Problem P2 and Problem P3 for equation (2.11), namely Theorem 4.1 Any solution u(z) of Problem P2 for the hyperbolic equation (2.11) can be expressed as  z (4.9) u(z) = 2Re w(z)dz + b0 in D, 0

where w(z) is as follows w(z) = f (x + y)e1 + g(x − y)e2 1 = {f (x + y) + g(x − y) + j[f (x + y) − g(x − y)]}, 2

(4.10)

here f (x + y), g(x − y) possess the forms f (x + y) = Re [(1 + j)uz (x + y)] = R(x + y), 0 ≤ x + y ≤ 2R, g(x−y) =

2r((1−j)(x−y)/2)−[a((1−j)(x−y)/2)+b((1−j)(x−y)/2)]f (0) , a((1 − j)(x − y)/2) − b((1 − j)(x − y)/2) 0 ≤ x − y ≤ 2R1 ,

f (0) = u(z1 ) + v(z1 ) =

(4.11)

r((1 − j)R1 ) + b1 . a((1 − j)R1 ) + b((1 − j)R1 )

As for Problem P3 for (2.11), its solution can be expressed as the forms (4.9), (4.10), but where f (x + y), g(x − y) possess the forms 2r((1+j)(x+y)/2+(1−j)R1 ) a((1+j)(x+y)/2+(1−j)R1 )−b((1+j)(x+y)/2+(1−j)R1 ) [a((1+j)(x+y)/2+(1−j)R1 )−b((1+j)(x+y)/2+(1−j)R1 )]g(2R1 ) , − a((1 + j)(x + y)/2+(1 − j)R1 )+b((1 + j)(x + y)/2+(1 − j)R1 ) 0 ≤ x + y ≤ 2R, (4.12) r((1 − j)R1 ) − b1 g(2R1 ) = u(z1 ) − v(z1 ) = , a((1 − j)R1 ) − b((1 − j)R1 ) g(x − y) = Re [(1 − j)uz (x − y)] = R(x − y), 0 ≤ x − y ≤ 2R1 .

f (x + y) =

Moreover the solution u(z) of Problem P2 and Problem P3 satisfies the estimate ¯ ≤ M23 , Cα1 [u(z), D]

¯ ≤ M24 k2 , Cα1 [u(z), D]

(4.13)

where M23 = M23 (α, k0 , k2 , D), M24 = M24 (α, k0 , D) are two non-negative constants. Next we give the representation of solutions of Problem P2 and Problem P3 for the quasilinear hyperbolic equation (1.16).

62

II. Hyperbolic Equations of Second Order

Theorem 4.2 Under Condition C, any solution u(z) of Problem P2 for the hyperbolic equation (1.16) can be expressed as 

u(z) = 2Re

0

z

w(z)dz + b0 ,

W (z) = f (µ)e1 + g(ν)e2 , 

Ψ(z) =

w(z) = W (z) + Φ(z) + Ψ(z) in D, Φ(z) = f˜(µ)e1 + g˜(ν)e2 , 

ν

2R1

[Aξ + Bη + Cu + D]e1 dν +

0

(4.14)

µ

[Aξ + Bη + Cu + D]e2 dµ,

where f (µ), g(ν) are as stated in (4.11) and f˜(µ), g˜(ν) are similar to f (µ), g(ν) in (4.11), but the functions r(z), R(x), b1 are replaced by the corresponding functions −Re [λ(z)Ψ(z)], −Re [Λ(x)Ψ(x)], −Im [λ(z1 )Ψ(z1 )], namely Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L1 ,

Re [Λ(x)Φ(x)] = −Re [Λ(x)Ψ(x)],

x ∈ L0 ,

(4.15)

Im [λ(z1 )Φ(z1 )] = −Im [λ(z1 )Ψ(z1 )]. As to Problem P3 for (1.16), its solution u(z) possesses the expression (4.14), where W (z) is a solution of Problem P3 for (2.11) and Φ(z) is also a solution of (2.11) satisfying the boundary conditions Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L2 ,

Re [Λ(x)Φ(x)] = −Re [Λ(x)Ψ(x)],

x ∈ L0 ,

(4.16)

Im [λ(z1 )Φ(z1 )] = −Im [λ(z1 )Ψ(z1 )]. Proof Let the solution u(z) of Problem P2 be substituted into the coefficients of equation (1.16). Due to Problem P2 is equivalent to the Problem A2 for the complex equation (2.8) with the relation (2.10), where the boundary conditions are as follows Re [λ(z)w(z)] = r(z) on L1 ,

Re [Λ(x)w(x)] = R(x) on L0 ,

(4.17)

Im [λ(z1 )w(z1 )] = b1 , According to Theorem 3.2 in Chapter I, it is not difficult to see that the function Ψ(z) satisfies the complex equation [Ψ]z¯ = [Aξ + Bη + Cu + D]e1 + [Aξ + Bη + Cu + D]e2 in D,

(4.18)

and noting that W (z) is the solution of Problem A2 for the complex equation (2.12), hence Φ(z) = w(z) − W (z) − Ψ(z) = ξe1 + ηe2 satisfies the complex equation and the boundary conditions ξµ e1 + ηλ e2 = 0, (4.19)

4. Other Oblique Derivative Problems

63

Re [λ(z)(ξe1 + ηe2 )] = −Re [λ(z)Ψ(z)] on L1 , Re [Λ(x)(ξe1 + ηe2 )] = −Re [Λ(x)Ψ(x)] on L0 ,

(4.20)

Im [λ(z)(ξe1 + ηe2 )]|z=z1 = −Im [λ(z1 )Ψ(z1 )]. The representation of solutions of Problem A2 for (1.16) is similar to (3.17) in Chapter I, we can obtain the representation (4.14) of Problem P2 for (1.16). Similarly, we can verify that the solution of Problem P3 for (1.16) possesses the representation (4.14) with the boundary condition (4.16). By using the same method as stated in the proofs of Theorems 2.3 and 2.4, we can prove the following theorem. Theorem 4.3 If the complex equation (1.16) satisfies Condition C, then Problem P2 (Problem P3 ) for (1.16) has a unique solution. 4.3 Representations of solutions and unique solvability of Problem P4 for quasilinear hyperbolic equations It is clear that the boundary condition (4.7) is equivalent to the following boundary condition ux = s
(x),

uy = R(x),

Re [(1 + j)uz (x)] = σ(x),

x ∈ L0 = (0, R2 ),

u(0) = s(0),

Re [(1 − j)uz (x)] = τ (x),

i.e. (4.21)

u(0) = r(0),

in which

s
(x) − R(x) s
(x) + R(x) , τ (x) = . (4.22) 2 2 Similarly to Theorem 4.1, we can give the representation of solutions of Problem P4 for equation (2.11). σ(x) =

Theorem 4.4 Any solution u(z) of Problem P4 for the hyperbolic equation (2.11) can be expressed as (4.9), (4.10), where b0 = s(0), f (x + y), g(x − y) possess the forms s
(x + y) + R(x + y) , 0 ≤ x + y ≤ 2R, 2 s
(x − y) − R(x − y) , 0 ≤ x − y ≤ 2R1 , g(x − y) = τ (x − y) = 2 f (x + y) = σ(x + y) =

(4.23)

and

s
(2R1 ) − R(2R1 ) s
(0) + R(0) , g(2R1 ) = . 2 2 Moreover u(z) of Problem P4 satisfies the estimate (4.13). f (0) =

(4.24)

Next we give the representation of Problem P4 for the quasilinear hyperbolic equation (1.16).

64

II. Hyperbolic Equations of Second Order

Theorem 4.5 Under Condition C, any solution u(z) of Problem P4 for the hyperbolic equation (1.16) can be expressed as (4.14), where b0 = s(0), W (z) is a solution of Problem A4 for (2.12) satisfying the boundary condition (4.21) (W = uz ), and Φ(z) is also a solution of (2.12) satisfying the boundary conditions Re [(1 + j)Φ(x)] = −Re [(1 + j)Ψ(x)],

Re [(1 − j)Φ(x)] = −Re [(1 − j)Ψ(x)]. (4.25)

Proof Let the solution u(z) of Problem P4 be substituted into the coefficients of equation (1.16). Due to Problem P4 is equivalent to the Problem A4 for the complex equation (2.8) with the relation (2.10) and the boundary conditions Re [(1 + j)w(x)] = σ(z),

Re [(1 − j)w(x)] = τ (x),

x ∈ L0 ,

(4.26)

according to Theorem 3.2 in Chapter I, it can be seen that the function Ψ(z) satisfies the complex equation (4.18) and noting that W (z) is the solutions of Problem A4 for the complex equation (2.12), i.e. (4.19), hence Φ(z) = w(z) − W (z) − Ψ(z) satisfies equation (4.19) and boundary conditions Re [(1 + j)(ξe1 + ηe2 )] = −Re [(1 + j)Ψ(x)], Re [(1 − j)(ξe1 + ηe2 )] = −Re [(1 − j)Ψ(x)],

x ∈ L0 .

(4.27)

Similarly to Theorem 4.2, the representation of solutions of Problem A4 for (2.8) is similar to (3.17) in Chapter I, we can obtain the representation (4.14) of Problem P4 for (1.16). By using the same method as stated in the proofs of Theorems 2.3 and 2.4, we can prove the following theorem. Theorem 4.6 If the complex equation (1.16) satisfies Condition C, then Problem P4 for (1.16) has a unique solution. Besides we can discuss the unique solvability of Problem P2 , Problem P3 and Problem P4 for general quasilinear hyperbolic equation (3.15), and generalize the above results to the general domains D
with the conditions (3.31) and (3.41) respectively. Similarly to Problem P as in Section 2, we can discuss Problem P1 for equation (2.11), here the solution w(z) of equation (2.12) is as follows w(z) = f (x + y)e1 + g(x − y)e2 1 = {f (x + y) + g(x − y) + j[f (x + y) − g(x − y)]}, 2 r((1 − j)R1 ) + b1 , f (0) = u(z1 ) + v(z1 ) = a((1 − j)R1 ) + b((1 − j)R1 ) g(2R1 ) = u(z1 ) − v(z1 ) =

r((1 − j)R1 ) − b1 , a((1 − j)R1 ) − b((1 − j)R1 )

(4.28)

4. Other Oblique Derivative Problems

65

here f (x + y), g(x − y) possess the forms 2r((1−j)(x−y)/2)−[a((1−j)(x−y)/2)+b((1−j)(x−y)/2)]f (0) , a((1 − j)(x − y)/2) − b((1 − j)(x − y)/2) (4.29) 0 ≤ x − y ≤ 2R1 , 2r((1+j)(x+y)/2+(1−j)R1 ) f (x+y) = a((1+j)(x+y)/2+(1−j)R1 )−b((1+j)(x+y)/2+(1−j)R1 ) [a((1+j)(x+y)/2+(1−j)R1 )−b((1+j)(x+y)/2+(1−j)R1 )]g(2R1 ) , − a((1+j)(x+y)/2+(1−j)R1 )+b((1+j)(x+y)/2+(1−j)R1 ) 0 ≤ x + y ≤ 2R1 .

g(x − y) =

Moreover when we prove that Problem P1 of equation (1.16) has a unique solution u(z), the integrals in (2.16),(2.22),(2.25),(2.28) and (2.33) possess the similar forms, and the integral path in (2.21) can be chosen, for instance the integral Ψ(z) in (2.16) is replaced by 

Ψ(z) =



ν

2R1

[Aξ + Bη + Cu + D]e1 dν +

0

µ

[Aξ + Bη + Cu + D]e1 dµ.

(4.30)

Now we explain that the Darboux problem is a special case of Problem P1 . The so-called Darboux problem is to find a solution u(z) for (1.16), such that u(z) satisfies the boundary conditions u(x) = s(x),

x ∈ L0 ,

u(z) = φ(x),

z = x + jy ∈ L1 ,

(4.31)

where s(x), φ(x) satisfy the conditions C 1 [s(x), L0 ] ≤ k2 ,

C 1 [φ(x), L1 ] ≤ k2 ,

(4.32)

herein k2 is a non-negative constant [12]3). From (4.31), we have u(x) = s(x),

x ∈ L0 ,

Re [(1−j)uz ] = φx (x) = φ
(x),


u(0) = s(0),

Im [(1−j)uz ]|z=z1 −0 = −φ (1),

u(x) = s(x),

x ∈ L0 ,

u(0) = b0 , where we choose √ λ(z) = (1−j)/ 2,

Re [λ(z)uz ] = r(z),

z ∈ L1 ,

i.e. z ∈ L1 ,

(4.33)

Im [λ(z)uz ]|z=z1 = b1 , √ r(z) = φ
(x)/ 2,

z ∈ L1 ,

b0 = s(0),

√ b1 = −φ
(1)/ 2, (4.34)

it is easy to see that the boundary conditions (4.33),(4.34) possess the form of the boundary condition (4.1). This shows that the above Darboux problem is a special case of Problem P2 . For more general hyperbolic equations of second order, the corresponding boundary value problems remain to be discussed.

66

5

II. Hyperbolic Equations of Second Order

Oblique Derivative Problems for Degenerate Hyperbolic Equations of Second Order

This section deals with the oblique derivative problem for the degenerate hyperbolic equation in a simply connected domain. We first give the representation theorem of solutions of the oblique derivative problem for the hyperbolic equation, and then by using the method of successive iteration, the existence and uniqueness of solutions for the above oblique derivative problem are proved. 5.1 Formulation of the oblique derivative problem for degenerate hyperbolic equations It is known that the Chaplygin equation in the hyperbolic domain D possesses the form K(y)uxx + uyy = 0 in D, (5.1) where K(y) possesses the first order continuous derivative K
(y), and K
(y) > 0 on y1 < y < 0, K(0) = 0, and the domain D is a simply connected domain with the boundary L = L0 ∪ L1 ∪ L2 , herein L0 = (0, 2), 



 y

L1 = x+

0



−K(t)dt = 0, x ∈ (0, 1) ,

L2 = x−

 y 0



−K(t)dt = 2, x ∈ (1, 2)

are two characteristic lines, and z1 = x1 + jy1 = 1 + jy1 is the intersection point of L1 and L2 . In particular, if K(y) = −|y|m , m is a positive constant, then 

y

0



−K(t)dt =

 0

y

|t|m/2 dt = −

 0

|y|

d

2 2 |t|(m+2)/2 = − |y|(m+2)/2 . m+2 m+2

In this case, the two characteristic lines L1 , L2 are as follows: 2 2 |y|(m+2)/2 = 0, L2 : x + |y|(m+2)/2 = 2, i.e. m+2 m+2 m + 2 2/(m+2) m+2 x) (2 − x)]2/(m+2) . , L2 : y = −[ L1 : y = −( 2 2

L1 : x −

In this section we mainly consider the general Chaplygin equation of second order K(y)uxx + uyy = dux + euy + f u + g in D,

(5.2)

where D, K(y) are as stated in (5.1), its complex form is the following equation of second order uzz¯ = Re [Quzz + A1 uz ] + A2 u + A3 , z ∈ D, (5.3) where Q=

K(y) + 1 , K(y) − 1

A1 =

d + je , K(y) − 1

A2 =

f , 2(K(y) − 1)

A3 =

g , 2(K(y) − 1)

5. Degenerate Hyperbolic Equations

67

and assume that the coefficients Aj (z)(j = 1, 2, 3) satisfy Condition C. It is clear that equation (5.2) is a degenerate hyperbolic equation. The oblique derivative boundary value problem for equation (5.2) may be formulated as follows: ¯ which satisfies the Problem P1 Find a continuous solution u(z) of (5.3) in D, boundary conditions 1 ∂u = Re [λ(z)uz˜] = r(z), z ∈ L = L1 ∪ L2 , 2 ∂l (5.4) u(0) = b0 , Im [λ(z)uz˜]|z=z1 = b1 , 

where l is a given vector at every point z ∈ L, uz˜ = [ −K(y)ux + juy ]/2, uz¯˜ = 

[ −K(y)ux − juy ]/2, b0 , b1 are real constants, λ(z) = a(x) + jb(x) = cos(l, x) + j cos(l, y), z ∈ L, and λ(z), r(z), b0 , b1 satisfy the conditions Cα1 [λ(z), Lj ] ≤ k0 , |b0 |, |b1 | ≤ k2 ,

max z∈L1

Cα1 [r(z), Lj ] ≤ k2 ,

1 , |a(z) − b(z)|

max z∈L2

j = 1, 2,

1 ≤ k0 , |a(z) + b(z)|

(5.5)

in which α (1/2 < α < 1), k0 , k2 are non-negative constants. For convenience, we can assume that uz˜(z1 ) = 0, i.e. r(z1 ) = 0, b1 = 0, otherwise we make a transformation uz˜ − [a(z1 ) − jb(z1 )][r(z1 ) + jb1 ]/[a2 (z1 ) − b2 (z1 )], the requirement can be realized. ¯ r(z) = 0, z ∈ L and b0 = b1 = 0 Problem P1 with the conditions A3 (z) = 0, z ∈ D, will be called Problem P0 . For the Dirichlet problem (Tricomi problem D) with the boundary condition: 

u(x) = φ(x) on L1 = AC = x = − 

L2 = BC = x = 2 +

 0

y





0

y





−K(t)dt, 0 ≤ x ≤ 1 , 

−K(t)dt, 1 ≤ x ≤ 2 ,

we find the derivative for (5.6) according to s = x on L = L1 ∪ L2 , and obtain us = ux + uy yx = ux − (−K(y))−1/2 uy = φ
(x) on L1 , us = ux + uy yx = ux + (−K(y))−1/2 uy = φ
(x) on L2 , i.e. (−K(y))1/2 U + V = (−K(y))1/2 φ
(x)/2 = r(x) on L1 , (−K(y))1/2 U − V = (−K(y))1/2 φ
(x)/2 = r(x) on L2 , i.e. Re [(1 − j)(U˜ + j V˜ )] = U˜ − V˜ = r(x) on L1 , Im [(1 − j)(U˜ + j V˜ )] = [−U˜ + V˜ ]|z=z1 −0 = −r(1 − 0), Re [(1 + j)(U˜ + j V˜ )] = U˜ + V˜ = r(x) on L2 , Im [(1 + j)(U˜ + j V˜ )] = [U˜ + V˜ ]|z=z1 +0 = r(1 + 0),

(5.6)

68

II. Hyperbolic Equations of Second Order

where



U=

−K(y)ux /2 = U˜ ,

V = −uy /2 = −V˜ ,

a + jb = 1 − j,

a = 1
= b = −1 on L1 ,

a + jb = 1 + j,

a = 1
= −b = −1 on L2 .

From the above formulas, we can write the complex forms of boundary conditions of U˜ + j V˜ : z ∈ Lj (j = 1, 2),

Re [λ(z)(U˜ + j V˜ )] = r(z), ⎧ ⎨1 − j

⎧ ⎨(−K(y))1/2 φ
(x)/2

on L1 , = a + jb, λ(z) = ⎩ r(x) = ⎩ 1 + j = a − jb, (−K(y))1/2 φ
(x)/2 on L2 , 

and u(z) = 2Re

z

0

(U − jV )dz + φ(0) in D.

(5.7)

Hence Problem D is a special case of Problem P1 . 5.2 Unique solvability of Problem P for Chaplygin equation (5.1) in the hyperbolic domain D In the subsection, we discuss the Chaplygin equation (5.1) in the hyperbolic domain D, where the arcs L1 = AC, L2 = BC are the characteristics of (5.1), i.e. 

x+

y

0



−K(t)dt = 0,

Setting that

0 ≤ x ≤ 1, 

µ=x+

y



0

x−

−K(t)dt,

and then µ + ν = 2x,

µ−ν =2



y



0

−K(t)dt = 2,

ν =x− 



(µ − ν)y = 2 −K(y),

y



(5.8)

−K(t)dt,

(5.9)

−K(y) = (µ − ν)y /2,

(5.10)

y

0



−K(t)dt,

0





1 ≤ x ≤ 2.



yµ = 1/2 −K(y) = −yν ,

xµ = 1/2 = xν , hence we have



Ux = Uµ + Uν ,

Vy = 

Vx = Vµ + Vν ,

Uy =

−K(y)(Vµ − Vν ), (5.11)

−K(y)(Uµ − Uν ),

and K(y)Ux − Vy = K(y)(Uµ + Uν ) − 

Vx + Uy = Vµ + Vν +



−K(y)(Vµ − Vν ) = 0, (5.12)

−K(y)(Uµ − Uν ) = 0 in D,

5. Degenerate Hyperbolic Equations 

where U = ux /2, V = −uy /2 and U˜ =





−K(y)

µ

−K(y)U, V˜ = −V . Noting that

1 K
= − (−K)−1/2 K
(y)yµ = , 2 4K

we have

69





−K(y)

ν

= −K
/4K,



K
U , 4K  K
U (U˜ + V˜ )ν = −K(y)Uν − Vν − . 4K Moreover, by (5.11) and (5.14) we obtain (U˜ − V˜ )µ =

(5.13)

−K(y)Uµ + Vµ +

(5.14)



−2 −K(y)(U˜ + V˜ )ν 

=

−K(y)[(U˜ + V˜ )µ − (U˜ + V˜ )ν ] −



−K(y)[(U˜ + V˜ )µ + (U˜ + V˜ )ν ]



= − −K(y)(U˜ + V˜ )x + (U˜ + V˜ )y 1 K
(y) ˜  = − (−K(y))−1/2 K
(y)U = U , 2 −K(y)(U˜ − V˜ )µ 2 2K(y) 

=

(5.15)

−K(y)[(U˜ − V˜ )µ + (U˜ − V˜ )ν ]



+ −K(y)[(U˜ − V˜ )µ − (U˜ − V˜ )ν )] 

=

−K(y)(U˜ − V˜ )x + (U˜ − V˜ )y

1 K
(y) ˜ = − (−K(y))−1/2 K
(y)U = U. 2 2K(y) Thus equation (5.15) can be written as a system of equations (U˜ + V˜ )ν = −(−K)−1/2

K
(y) ˜ U, 4K(y)

(U˜ − V˜ )µ = (−K)−1/2

K
(y) ˜ U, 4K(y)

i.e. (5.16)

˜ y ] = A1 W ˜ z¯˜ = 1 [(−K(y))1/2 W ˜ x −j W ˜ (z)+A2 W ˜ (z) in ∆, W 2 ˜ = U˜ + j V˜ , A1 (z) = A2 (z) = −jK
/8K in ∆ = {0 ≤ µ ≤ 2, 0 ≤ ν ≤ 2}, in which W and  z u(z) = 2Re (U − jV )dz + b0 in D, (5.17) 0

−1/2

where U − jV = (−K)

U˜ + j V˜ .

In the following, we first give the representation of solutions for the oblique derivative problem (Problem P1 ) for system (5.16) in D. For this, we first discuss the system of equations (U˜ + V˜ )ν = 0, (U˜ − V˜ )µ = 0 in D (5.18)

70

II. Hyperbolic Equations of Second Order

with the boundary condition 1 ∂u = Re [λ(z)(U˜ + j V˜ )] = r(z), z ∈ L, 2 ∂l u(0) = b0 , Im [λ(z)(U˜ + iV˜ )]|z=z1 = b1 ,

(5.19)

in which λ(z) = a(z) + jb(z) on L1 ∪ L2 . Similarly to Chapter I, the solution of Problem P1 for (5.18) can be expressed as ξ = U˜ + V˜ = f (µ), η = U˜ − V˜ = g(ν), f (µ) − g(ν) f (µ) + g(ν) , V˜ (x, y) = , U˜ (x, y) = (5.20) 2 2 ˜ (z) = (1 + j)f (µ) + (1 − j)g(ν) , i.e. W 2 in which f (t), g(t) are two arbitrary real continuous functions on [0, 2]. For convenience, denote by the functions a(x), b(x), r(x) of x the functions a(z), b(z), r(z) of z in (5.19), thus the first formula in (5.19) can be rewritten as a(x)U˜ (x, y) + b(x)V˜ (x, y) = r(x) on L, i.e. [a(x)−b(x)]f (x+y)+[a(x)+b(x)]g(x−y) = 2r(x) on L, [a(t/2) + b(t/2)]f (0) + [a(t/2) − b(t/2)]g(t) = 2r(t/2),

i.e.

t ∈ [0, 2],

[a(t/2+1)−b(t/2+1)]f (t)+[a(t/2+1)+b(t/2+1)]g(2) = 2r(t/2+1),

t ∈ [0, 2],

where f (0) = U˜ (0) + V˜ (0) =

r(1) + b1 , a(1) + b(1)

g(2) = U˜ (2) − V˜ (2) =

r(1) − b1 . a(1) − b(1)

Noting that the boundary conditions in (5.19), we can derive









1 2r(ν/2) − (a(ν/2) + b(ν/2))f (0) U˜ = f (µ) + , 2 a(ν/2) − b(ν/2) 2r(ν/2) − (a(ν/2) + b(ν/2))f (0) 1 f (µ) − , or V˜ = 2 a(ν/2) − b(ν/2)




2r(µ/2 + 1) − (a(µ/2 + 1) − b(µ/2 + 1))g(2) 1 g(ν) + , U˜ = 2 a(µ/2 + 1) + b(µ/2 + 1)


(5.21)



2r(µ/2 + 1) − (a(µ/2 + 1) − b(µ/2 + 1))g(2) 1 −g(ν) + , V˜ = 2 a(µ/2 + 1) + b(µ/2 + 1) if a(x) − b(x)
= 0 on [0,1] and a(x) + b(x)
= 0 on [1,2] respectively. From the above formulas, it follows that ˜ (x)] = U˜ + V˜ = 2r(x/2+1)−(a(x/2+1)+b(x/2+1))f (0) , Re[(1+j)W a(x/2+1)−b(x/2+1) (5.22) 2r(x/2)−(a(x/2)−b(x/2))g(2) ˜ ˜ ˜ , x ∈ [0, 2], Re[(1−j)W (x)] = U − V = a(x/2) + b(x/2)

5. Degenerate Hyperbolic Equations

71

if a(x) − b(x)
= 0 on [0,1] and a(x) + b(x)
= 0 on [1,2] respectively. From (5.22),

˜ (z) = W

⎧  2r(µ/2 + 1) − (a(µ/2 + 1) − b(µ/2 + 1))g(2) 1 ⎪ ⎪ ⎪ (1 + j) ⎪ ⎪ ⎨2 a(µ/2 + 1) + b(µ/2 + 1)  ⎪ ⎪ 2r(ν/2) − (a(ν/2) + b(ν/2))f (0) ⎪ ⎪ ⎪ +(1 − j) ⎩

(5.23)

a(ν/2) − b(ν/2)

can be derived. Next, we find the solution of Problem P1 for system (5.16). From (5.16), we have U˜ + V˜ = −



ν

2

(−K)−1/2

˜ = U˜ +j V˜ = − 1+j W 2



K
(y) ˜ U dν, 4K(y)

ν

2

(−K)−1/2

U˜ − V˜ =

 0

µ

(−K)−1/2

K
(y) ˜ U dµ, 4K(y)

K (y) ˜ K
(y) ˜ 1−j  µ (−K)−1/2 U dν + U dµ 4K(y) 2 0 4K(y)


1 − j  µ K
(y) 1 + j  ν K
(y) U dν + U dµ, =− 2 2 2 4K(y) 0 4K(y) the above last two integrals are along two characteristic lines s2 and s1 respectively. But according to the method in [66]1), if we denote by s1 the member of the family  of characteristic lines dx/dy = − −K(y), and by s2 the member of the family of 

characteristic lines dx/dy = 

ds1 = 

ds2 =

−K(y) passing through the point P ∈ D, and



(dx)2 + (dy)2 =



(dx)2

+

(dy)2

=

1+

dy dx

dy 1+ dx



2

dx = 

2

dx =

√ 1−K dx = − 1 − Kdy, −K √ 1−K dx = 1 − Kdy, −K

(5.24)

then system (5.1) can be rewritten in the form (U˜ + V˜ )s1 = (U˜ + V˜ )x xs1 +(U˜ + V˜ )y ys1 =√

  1 −K(y)(U˜ + V˜ )x −(U˜ + V˜ )y 1−K

1 K
(y) ˜ 1 = √ (−K(y))−1/2 K
(y)U = − √ U, 2 1−K 2 1−K K(y) (U˜ − V˜ )s2 = (U˜ − V˜ )x xs2 +(U˜ − V˜ )y ys2 =√

  1 −K(y)(U˜ − V˜ )x +(U˜ − V˜ )y 1−K

1 1 K
(y) ˜ (−K(y))−1/2 K
(y)U = √ =− √ U, 2 1−K 2 1−K K(y)

(5.25)

72

II. Hyperbolic Equations of Second Order

thus we obtain the system of integral equations ξ = U˜ + V˜ = − η = U˜ − V˜ =

 0

 0 s2

s1

 s1 1 K
(y) ˜ K
(y) ˜ U ds1 = U dy, 0 2 K(y) 2 (1 − K) K(y) 

1

 s2 1 K
(y) ˜ K
(y) ˜  U ds2 = U dy, 0 2 K(y) 2 (1 − K) K(y)

(5.26)

1

where the integrals are along the two families of characteristics s1 and s2 respectively. Similarly to next subsection, the solution U˜ + V˜ , U˜ − V˜ can be obtained by the method of successive iteration, which can be expressed as ˜ (z) = U˜ + j V˜ = W (z) + Φ(z) + Ψ(z), W U˜ + V˜ = −



s1

0

K
(y) ˜ U ds1 , 2 (1−K) K(y) 

1

U˜ − V˜ =



s2

0

(5.27) K
(y) ˜ U ds2 , 2 (1−K) K(y) 

1

where W (z), Φ(z) are the solutions of system (5.18) satisfying the boundary condition (5.19) respectively, but the function r(z), b1 in the boundary condition of Φ(z) should be replaced by −Re [λ(z)Ψ(z)] on L1 ∪L2 and −Im [λ(z1 )Ψ(z1 )], and then the function 

u(z) = 2Re

0

z

(U − jV )dz + b0 = 2Re

 0

z

[(−K(y))−1/2 U˜ + j V˜ ]dz + b0 in D, (5.28)

is just the solution of Problem P1 for (5.16) and the solution is unique. Here we mention that firstly it suffices to consider the case of y1 ≤ y ≤ −δ, where δ is a sufficiently small positive number, and when we find out the solution of Problem P1 for equation (5.1) with the condition y1 ≤ y ≤ −δ, and then let δ → 0. Theorem 5.1 If the Chaplygin equation (5.1) in D satisfies the above conditions, then Problem P1 for (5.1) in D has a unique solution. Finally we mention that the boundary condition of Problem P1 : z ∈ L1 ∪ L2

Re [λ(z)(U˜ + j V˜ )] = r(z), can be replaced by 



2U˜ (x) = S(x) = −K[y(x)]ux ,

i.e. u(x) =

0

x



S(x)dx −K[y(x)]

2V˜ (x) = R(x) = uy on AB = L0 = (0, 2), where Cα [R(x), AB] ≤ k2 < ∞, Cα1 [s(x), AB] ≤ k2 .

+u(0) = s(x), (5.29)

5. Degenerate Hyperbolic Equations

73

5.3 Unique solvability of the oblique derivative problem for degenerate hyperbolic equations In this subsection, we prove the uniqueness and existence of solutions of Problem P4 for the degenerate hyperbolic equation (5.2), the boundary condition of Problem P4 is as follows u(x) = s(x), uy (x) = R(x) on L0 = (0, 2), (5.30) where s(x), R(x) satisfy the condition Cα2 [s(x), L0 ], Cα2 [R(x), L0 ] ≤ k2 , the above boundary value problem is also called the Cauchy problem for (5.2). Making a transformation of function v(z) = u(z) − yR(x) − s(x) in D,

(5.31)

equation (5.2) and boundary condition (5.30) are reduced to the form K(y)vxx + vyy = dvx + evy + f v + G, G = g + f (yR + s) + eR + d[yR
(x) + s
(x)]



(5.32)





−K(y)[yR (x) + s (x)] in D, v(x) = 0,

vy (x) = 0 on L0 .

(5.33)

Hence we may only discuss Cauchy problem (5.32),(5.33) and denote it by Problem P4 again. According to Subsection 4.3, Problem P4 for (5.32) is equivalent to the boundary value problem A for the hyperbolic system of first order equations, the relation and the boundary conditions √   2 −K 1 −d 1 K
(y) √ ξs1 = √ ξν = √ −e− ξ 2 K(y) 1−K 2 1−K −K 

+



√   −d 2 −K 1 1 K
(y) √ ξ ηµ = √ −e+ ηs 2 = √ 2 K(y) 2 1−K 1−K −K 

+



−d 1 K
(y) √ +e− η − fv − G , 2 K(y) −K



(5.34) 

1 K
(y) −d √ +e+ η − fv − G , 2 K(y) −K

ξ = U˜ + V˜ ,

η = U˜ − V˜ ,

vy = ξ − η,

v(x) = 0.

In particular, if K(y) = −|y|m h(y), m is a positive constant, then m|y|m−1 h(y) |y|m hy m hy K
(y) = − = + . K(y) K(y) K(y) y h

74

II. Hyperbolic Equations of Second Order

Integrating the hyperbolic system in (5.34) along the characteristics s1 , s2 , we obtain the system of integral equations as follows 

v(z) =

y

0

ξ(z) = − 

η(z) =

(ξ − η)dy in D,

 0

y

[A1 ξ + B1 η + C1 (ξ + η) + Dv + E]dy,

y

0

[A2 ξ + B2 η + C2 (ξ + η) + Dv + E]dy,

z ∈ s1 ,

(5.35)

z ∈ s2 ,

in this case e hy e hy e hy e hy A1 = − − , B1 = − , A2 = − + , B2 = + , 2 4h 2 4h 2 4h 2 4h f G 1 d 1 d m m C1 = − √ − , C2 = − √ + , D=− , E=− . 2 −K 4y 2 −K 4y 2 2 In the following we may only discuss the case of K(y) = −|y|m h(y), because otherwise it can be similarly discussed. In order to find a solution of the system of integral equations (5.35), we need to add the condition lim

y→0

|y|d(x, y) = 0, |y|m/2

i.e. d(x, y) ≈ ε(y)|y|m/2−1 ,

(5.36)

where ε(y) → 0 as y → 0. It is clear that for two characteristics s01 : x = x1 (y, z0 ), s02 : x = x2 (y, z0 ) passing through P0 = z0 = x0 + jy0 ∈ D, we have 

|x1 − x2 | ≤ 2|

0

y

√

−Kdy| ≤ M |y|m/2+1 for y
< y < 0,

(5.37)



for any z1 = x1 + jy ∈ s01 , z2 = x2 + jy ∈ s02 , where M (> max[4 h(y)/(m + 2), 1]) is a positive constant. Suppose that the coefficients of (5.35) possess continuously differentiable with respect to x ∈ L0 and satisfy the condition |Aj |, |Ex |,

|Ajx |, |Bj |, |Bjx |, |yCj |, |yCjx |, |D|, √ ¯ |1/ h|, |hy /h| ≤ M, z ∈ D, j = 1, 2.

|Dx |,

|E|,

(5.38)

According to the proof of Theorem 5.1, it is sufficient to find a solution of Problem P4 for arbitrary segment −δ ≤ y ≤ 0, where δ is a sufficiently small positive number, and choose a positive constant γ(< 1) close to 1, such that the following inequalities hold: 3M δ [ε(y)M + m/2]δ m/2 + < γ, 2 m+2 (5.39) 2ε(y)M + m 6δ 2 M 2 8δM + < γ. + m+6 m+2 m+2 Similar to [66]1), a solution of Problem P4 for (5.35) on −δ < y < 0 can be found. Firstly, let y0 ∈ (−δ, 0) and D0 be a domain bounded by the boundary y = 0, s01 , s02 ,

5. Degenerate Hyperbolic Equations

75

we choose v0 = 0, ξ0 = 0, η0 = 0 and substitute them into the corresponding positions of v, ξ, η in the right-hand sides of (5.35), and obtain ξ1 (z) = − 

η1 (z) =

0



y

[A1 ξ0 +B1 η0 +C1 (ξ0 +η0 )+Dv0 + E]dy = −

0



y

[A2 ξ0 +B2 η0 +C2 (ξ0 +η0 )+Dv0 +E]dy =

v1 (z) = Re



y

0



y

Edy,

0

y

0

Edy,

z ∈ s01 ,

z ∈ s02 ,

(5.40)

(ξ0 − η0 )dy = 0 in D0 .

By the successive iteration, we find the sequences of functions {vk }, {ξk }, {ηk }, which satisfy the relations ξk+1 (z) = − 

ηk+1 (z) = 

vk+1 (z) =



[A1 ξk + B1 ηk + C1 (ξk + ηk ) + Dvk + E]dy,

y

[A2 ξk + B2 ηk + C2 (ξk + ηk ) + Dvk + E]dy,

0

0

y

0

y

(ξk − ηk )dy in D0 ,

z ∈ s01 , z ∈ s02 ,

(5.41)

k = 0, 1, 2, . . . .

¯ 0 satisfy the estimates We can prove that {vk , }, {ξk }, {ηk } in D |vk (z)|,

|ξk (z)|,

|ηk (z)| ≤ M

k 

γ j |y|,

|ξk (z) + ηk (z)|,

j=0

|vk (z1 )−vk (z2 )|,

|ξk (z1 )−ξk (z2 )|,

|ηk (z1 )−ηk (z2 )| ≤M

k 

γ j |y|m/2+1 ,

j=0

|vk+1 (z) − vk (z)|,

|ξk+1 (z) − ξk (z)|,

|ξk+1 (z) + ηk+1 (z) − ξk (z) − ηk (z)|, −vk (z1 ) − vk (z2 )|,

(5.42)

|ηk+1 (z) − ηk (z)| ≤ M γ k |y|, |vk+1 (z1 ) − vk+1 (z2 )

|ξk+1 (z1 ) − ξk+1 (z2 ) − ξk (z1 ) + ξk (z2 )|,

|ηk+1 (z1 ) − ηk+1 (z2 ) − ηk (z1 ) + ηk (z2 )| ≤ M γ k |y|m/2+1 . In fact, from (5.40), it follows that the first formula with k = 1 holds, namely |v1 (z)| = 0 ≤ M |y|,

|ξ1 (z)| ≤ M |y|,

|η1 (z)| ≤ M |y| = Mγ 0 |y| ≤ M

1  j=0

Moreover we get |v1 (z1 )−v1 (z2 )| = 0, |ξ1 (z)+η1 (z)| ≤ | ≤



0

y



[E(z1 )−E(z2 )]dy| ≤ 2| 1 

0

y

Ex [x1 −x2 ]dy|

4 M 2 |y|m/2+2 ≤ M γ|y|m/2+1 ≤ M γ j |y|m/2+1 , m+4 j=0

γ j |y|.

76

II. Hyperbolic Equations of Second Order 

|ξ1 (z1 )−ξ1 (z2 )| ≤ | ≤|



y

0

≤ M|

y

0

[E(xj (t, z1 )+jt)−E(xj (t, z2 )+jt)]dt|

|Ex ||xj (t, z1 ) − xj (t, z2 )|dy| ≤ M |



y

0



y

0

M |y|m/2+1 dy| ≤ M γ|y|m/2+1 ≤ M 

|η1 (z1 )−η1 (z2 )| = |

γ j |y|m/2+1 ,

j=0

y

0

|x1 − x2 |dy| 1 

[E(xj (t, z1 )+jt)−E(xj (t, z2 )+jt)]dt| ≤ M

1 

γ j |y|m/2+1 ,

j=0

|v1 (z) − v0 (z)| = |v1 (z)| ≤ M γ|y|, |ξ1 (z) − ξ0 (z)| = |ξ1 (z)| ≤ M γ|y|, |η1 (z)−η0 (z)| = |η1 (z)| ≤ M γ|y|, |v1 (z1 )−v1 (z2 )−v0 (z1 )−v0 (z2 )| ≤ M γ|y|m/2+1 , |ξ1 (z)+η1 (z)−ξ0 (z)−η0 (z)| = |ξ1 (z)+η1 (z)| ≤ γ|y|m/2+1 ≤ M γ|y|m/2+1 , |ξ1 (z1 )−ξ1 (z2 )−ξ0 (z1 )+ξ0 (z2 )| = |ξ1 (z1 )−ξ1 (z2 )| ≤ M γ|y|m/2+1 , |η1 (z1 )−η1 (z2 )−η0 (z1 )+η0 (z2 )| = |η1 (z1 )−η1 (z2 )| ≤ M γ|y|m/2+1 . In addition, we use the inductive method, namely suppose the estimates in (5.42) for k = n are valid, then they are also valid for k = n + 1. In the following, we only give the estimates of |vn+1 (z)|, |ξn+1 (z)|, |ξn+1 + ηn+1 (z)|, the other estimates can be similarly given. From (5.41), we have 

|vn+1 (z)| ≤ |

[ξn −ηn ]dy| ≤ 2M |

0



|ξn+1 (z)| ≤ |

 y n

y

⎡ y

0

≤ M|

0 j=0

⎣(|A1 |+|B1 |+|D|)M ⎡



y

0



n 

γ j ydy| ≤ M

n 

n 

γ j |y|2 ≤ M

j=0

γ j |y|+|C1 |M

j=0



n+1 

γ j |y|,

j=0 n 

⎤

γ j |y|m/2+1 +|E|⎦ dy|

j=0

⎤

n ε(y) m  ⎣3M √ + γ j |y| + γ j |y|m/2+1 + 1⎦ dy| 4|y| 2|y| h j=0 j=0

⎧ ⎨ 3

≤ M |y| ⎩

2



M |y| + |ε(y)|M +

⎫

 n n+1 ⎬  m |y|m/2  γ j + 1⎭ ≤ M γ j |y|, 2 m + 2 j=0 j=0

and  y
 2

|ξn+1 (z)+ηn+1 (z)| ≤ |

0

[A2 (z2 )ξn (z2 )−A1 (z1 )ξn (z1 )+B2 (z2 )ηn (z2 )

j=1

−B1 (z1 )ηn (z1 )] + C2 (z2 )(ξn (z2 ) + ηn (z2 )) −C1 (z1 )(ξn (z1 ) + ηn (z1 ))



+D(z2 )vn (z2 ) − D(z1 )vn (z1 ) + E(z2 ) − E(z1 )] dy|.

5. Degenerate Hyperbolic Equations

77

Noting that |D(z2 )vn (z2 ) − D(z1 )vn (z1 )| = |[D(z2 ) − D(z1 )]vn (z2 ) + D(z1 ) ×[vn (z2 ) − vn (z1 )]| ≤

n 

γ j |y||D(z2 ) − D(z1 )| + M 2

j=0

n 

γ j |y|m/2+1

j=0

≤ M2

n 

γ j |y||x2 −x1 |+M 2

j=0

n 

γ j |y|m/2+1

j=0

≤ (M |y|+1)M 2

n 

γ j |y|m/2+1 ,

j=0

and |A2 (z2 )ξn (z2 ) − A1 (z1 )ξn (z1 ) + B2 (z2 )ηn (z2 ) − B1 (z1 )ηn (z1 )| ≤ |[A2 (z2 )−A2 (z1 )]ξn (z2 )+[A2 (z1 )−A1 (z1 )]ξn (z2 )+A1 (z1 )[ξn (z2 )−ξn (z1 )] +[B2 (z2 )−B2 (z1 )]ηn (z2 )+B1 (z1 )[ηn (z2 )−ηn (z1 )]| +[B2 (z1 )−B1 (z1 )]ηn (z2 ) ≤ 2M |y|[M |x1 − x2 | + |y|m/2 ]

n 

  h   y  |ξ (z ) − ηn (z2 )|  2h  n 2

γj + 

j=0

≤ (2M |y| + 3)M 2

n 

γ j |y|m/2+1 ,

j=0

we get |ξn+1 (z) + ηn+1 (z)| ≤ |



y

0

[(3M |y| + 4)M 2

n 

γ j |y|m/2+1

j=0

+(|C1 | + |C2 |)M 

n  j=0

γ j |y|m/2+1 + M 2 |y|m/2+1 ]dy|

6M 2 y 2 8M |y| m + + |ε(y)|M + m+6 m + 2⎤ 2 n 2  2M j |y|⎦ × γ + m + 2 j=0 m+4

≤ M |y|m/2+1

≤M

n+1 



γ j |y|m/2+1 .

j=0

On the basis of the estimates (5.42), we can derive that {vk , }, {ξk }, {ηk } in D0 uniformly converge to v∗ , ξ∗ , η∗ satisfying the system of integral equations ξ∗ (z) = −



0

y

[A1 ξ∗ + B1 η∗ + C1 (ξ∗ + η∗ ) + Dv∗ + E]dy,

z ∈ s1 ,

78

II. Hyperbolic Equations of Second Order 

η∗ (z) =

y

0y

v∗ (z) =

0

[A2 ξ∗ + B2 η∗ + C2 (ξ∗ + η∗ ) + Dv∗ + E]dy,

z ∈ s2 ,

(ξ∗ − η∗ )dy in D0 ,

and the function v∗ (z) satisfies equation (5.32) and boundary condition (5.33), hence u∗ (z) = v ∗ (z) + yR(x) + s(x) is a solution of Problem P4 for (5.2). From the above discussion, we can see that the solution of Problem P4 for (5.2) in D is unique. Theorem 5.2 If the equation (5.2) in D satisfies the above conditions, then Problem P4 for (5.2) in D has a unique solution. Now we mention that if we denote ˜ (z) = U˜ +j V˜ = |y|m/2 U −jV = 1 [|y|m/2 ux +juy ], W 2 m/2 ˜ (z) = U˜ −j V˜ = |y| U +jV = 1 [|y|m/2 ux −juy ], W 2 ˜ (z) = |y|m/2 U−jV is a solution of the first order hyperbolic complex equation then W ˜ + A2 (z)W ˜ + A3 (z)u + A4 (z) in D, ˜ z¯˜ = A1 (z)W W 







e m d − A1 = − +j , 4|y|m/2 8|y| 4 A2 = −

e m d + +j , 4|y|m/2 8|y| 4

f A3 = − , 4

(5.43)

g A4 = − , 4

and 

u(z) = 2Re

0



z

uz dz = 2Re

0

z

(U −jV )d(x+jy) = 2Re

 0

z

(U +jV )d(x−jy)

is a solution of equation (5.2) with K(y) = −|y|m . By using the similar method, we can prove the solvability of Problem P1 , Problem P2 and Problem P3 for equation (5.2). Moreover for general domain D
with noncharacteristics boundary, we can also discuss the solvability of Problem P1 , Problem P2 , Problem P3 and Problem P4 for equation (5.2). Besides we can discuss the solvability of corresponding boundary value problems for the hyperbolic equation in the form uxx + K(y)uyy = dux + euy + f u + g in D (5.44) under certain conditions, where K(y) is as stated in (5.2). The references for the chapter are [2],[7],[12],[13],[24],[25],[34],[41],[44],[47],[54], [60],[66],[70],[79],[85],[87],[89],[95].

CHAPTER III NONLINEAR ELLIPTIC COMPLEX EQUATIONS OF FIRST AND SECOND ORDER In this chapter, we discuss the representation and existence of solutions of discontinuous boundary value problems for nonlinear elliptic complex equations of first and second order, which will be used in latter chapters.

1

Generalizations of Keldych–Sedov Formula for Analytic Functions

It is known that the Keldych–Sedov formula gives the representation of solutions of the mixed boundary value problem for analytic functions in the upper half-plane (see [53]). But for many problems in mechanics and physics, one needs a more general formulas of solutions of the discontinuous Riemann–Hilbert boundary value problem for analytic functions in the upper half-plane and other special domains. In this section, we shall establish the representations of solutions of the general discontinuous boundary value problem for analytic functions in the upper half-plane and upper halfunit disk. In the following sections and chapters we shall give applications to some nonlinear elliptic complex equations and quasilinear equations of mixed type.

1.1 General discontinuous boundary value problem for analytic functions in the upper half-plane Let D be the upper half-plane and a(x), b(x), c(x) be known real functions on L = {−∞ < x < ∞, y = 0}, where a(x), b(x) possess discontinuities of first kind at m distinct points xj (j = 1, . . . , m, −∞ < x1 < · · · < xm < ∞), m is a positive integer, and c(x) = O(|x−xj |−βj ) in the neighborhood of xj (j = 1, 2, . . . , m) on L, herein βj (< 1, j = 1, 2, . . . , m) are non-negative constants, such that βj +γj < 1, γj (j = 1, . . . , m) are as stated in (1.3) below. Denote λ(x) = a(x) − ib(x) and |a(x)| + |b(x)| =
0, there is no harm in assuming that |λ(x)| = 1, x ∈ L∗ = L\{x1 , . . . , xm }. Suppose that λ(x), c(x) satisfy the conditions λ(x) ∈ Cα (Lj ), |x − xj |βj c(x) ∈ Cα (Lj ), j = 1, 2, . . . , m,

(1.1)

80

III. Elliptic Complex Equations

where Lj is the line segment from the point xj−1 to xj on L, x0 = xm , Lj (j = 1, 2, . . . , m) do not include the end points, L1 = {x < x1 } ∪ {x > xm }, α(0 < α < 1) is a constant, and the function λ(x) ∈ Cα (L∞ )(L∞ is a neighborhood of the point ∞) is indicated as λ(1/x) ∈ Cα (L∗ ), here L∗ (⊂ L) is a neighborhood of the point x = 0. The discontinuous Riemann–Hilbert boundary value problem for analytic functions in D may be formulated as follows. Problem A Find an analytic function Φ(z) = u(z)+iv(z) in D, which is continuous ¯ in D∗ = D\{x 1 , x2 , . . . , xm } satisfying the boundary condition z ∈ L∗ .

Re [λ(x)Φ(x)] = au − bv = c(x),

(1.2)

Problem A with the condition c(x) = 0 on L∗ is called Problem A0 . Denote by λ(xj − 0) and λ(xj + 0) the left limit and right limit of λ(x) as x → xj (j = 1, 2, . . . , m) on L, and eiφj =

λ(xj −0) , λ(xj +0)





γj =



φj Kj = + Jj , π



1 φj λ(xj −0) ln = −Kj , πi λ(xj +0) π

Jj = 0 or 1,

(1.3)

j = 1, . . . , m,

in which 0 ≤ γj < 1 when Jj = 0 and −1 < γj < 0 when Jj = 1, j = 1, . . . , m, and 

m φj 1 1 − γj K = (K1 + · · · + Km ) = 2 2 j=1 π



(1.4)

is called the index of Problem A and Problem A0 . If λ(x) on L is continuous, then K = ∆Γ arg λ(x)/2π is a unique integer. If the function λ(x) on L is not continuous, we can choose Jj = 0 or 1, hence the index K is not unique. We can require that the solution Φ(z) satisfy the condition −δ

Φ(z) = O(|z − xj | ),

⎧ ⎨ βj

+ τ, for γj ≥ 0, and γj < 0, βj > |γj |, δ=⎩ |γj | + τ, for γj < 0, βj ≤ |γj |, j = 1, . . . , m,

(1.5)

in the neighborhood (⊂ D) of xj , where τ (< α) is an arbitrary small positive number. In order to find the solution of Problem A for analytic functions, we first consider Problem A0 . Making a transformation Ψ(z) =

Φ(z) , Π(z)

Π(z) =

m  z j=1

− xj z +i

γ j

,

(1.6)

in which γj (j = 1, . . . , m) are as stated in (1.3), the boundary condition Re [λ(x)Φ0 (x)] = 0,

x ∈ L∗ ,

(1.7)

1. Generalizations of Keldych–Sedov Formula

81

of Problem A0 for analytic functions Φ0 (z) is reduced to the boundary condition Re [Λ(x)Ψ0 (x)] = 0,

Λ(x) = λ(x)

Π(x) , |Π(x)|

x ∈ L∗ ,

(1.8)

of Problem A∗0 for analytic functions Ψ0 (z) = Φ0 (z)/Π(z). Noting that 

λ(xj − 0) Π(xj − 0) Λ(xj − 0) = Λ(xj + 0) λ(xj + 0) Π(xj + 0)



λ(xj − 0) −iπγj e = ±1, λ(xj + 0)

=

(1.9)

the index of Λ(x) on L is 



m m φj 1 1 1 ∆L arg Λ(x) = − γj = Kj , K= 2π 2 j=1 π 2 j=1

(1.10)

which is the same as the index of λ(x) on L. If 2K is even, provided that we change the signs of Λ(x) on some line segments of Lj (j = 1, . . . , m), then the new function Λ∗ (x) on L is continuous, its index is K too. When 2K is odd, we rewrite the boundary condition (1.8) in the form



Re Λ(x)

x − x0 x + i Ψ0 (x) = 0, x ∈ L∗ , x + i x − x0

(1.11)

where x0 (∈ L) is a real number and x0
∈ {x1 , . . . , xm }, thus similarly to before we change the signs of Λ(x)(x − x0 )|x + i|/(x + i)|x − x0 | on some line segments of L, such that the new function Λ∗ (x) on L is continuous, its index is K ∗ = K − 1/2. Next we find an analytic function Ψ∗ (z) = i

z−i z+i

[K]



z − x0 iS(z) e in D, z+i

(1.12)

which satisfies the homogeneous boundary condition Re [Λ∗ (x)Ψ∗ (x)] = 0,

x ∈ L∗ ,

(1.13)

where [K] is the integer part of K, S(z) is an analytic function in D satisfying the boundary condition ⎡

Re [S(x)] = arg ⎣Λ∗ (x)

x−i x+i

[K]



⎤

x − x0 ⎦ , x+i

x ∈ L,

Im [S(i)] = 0.

(1.14)

Hence Problem A∗0 for analytic functions possesses the solution ⎧ ⎨ Ψ∗ (z),

Ψ0 (z) = ⎩

when 2K is even,

(z − x0 )Ψ∗ (z)/(z + i),

(1.15) when 2K is odd,

82

III. Elliptic Complex Equations

and then Problem A0 for analytic functions has a non-trivial solution in the form

X(z) = Π(z)Ψ0 (z) =

⎧ z ⎪ ⎪ ⎪ ⎪i ⎨ z

−i +i

K

Π(z)eiS(z) ,

when 2K is even,

 ⎪ ⎪ z−i [K] z−x0 ⎪ ⎪ ⎩i Π(z)eiS(z) ,

z+i

z+i

(1.16) when 2K is odd.

Take into account that X(z) has a zero of order [K] at the point z = i for K ≥ 0 and a pole of order |[K]| at the point z = i for K < 0, and a zero of order 1 at the point z = x0 when 2K is an odd integer, moreover, X(z) satisfies the homogeneous boundary condition (1.7), it is clear that iλ(x)X(x) is a real-valued function on L. Let us divide the nonhomogeneous boundary condition (1.2) by iλ(x)X(x), and obtain 

Re



Φ(x) c(x) λ(x)c(x) , = = iX(x) iX(x) iλ(x)X(x)

x ∈ L∗ .

(1.17)

By using the Schwarz formula, we get   1  ∞ λ(t)c(t) Q(z) Φ(z) = dt + , iX(z) πi −∞ (t − z)iX(t) i

  X(z)  ∞ λ(t)c(t) dt + Q(z) . Φ(z) = πi −∞ (t − z)X(t)

(1.18)

If K ≥ 0, the function Q(z) possesses the form

Q(z) = i

[K]  j=0



cj

z−i z+i

j

+ cj

z−i z+i

−j 

⎧ ⎪ ⎪ ⎨0,

when 2K is even, (1.19) + ⎪ x0 + z ⎪ , when 2K is odd, ⎩ic∗ x0 − z

where c∗ , c0 are arbitrary real constants, and cj (j = 1, . . . , [K]) are arbitrary complex constants, from this we can see that the general solution Φ(z) includes 2K+1 arbitrary real constants. If 2K is odd, we note (z − x0 )/[(t − z)(t − x0 )] = 1/(t − z) − 1/(t − x0 ), then the integral in (1.18) is understood as the difference of two integrals of Cauchy type. If K < 0, we have to take ⎧ ⎪ ⎪ ⎨ ic∗

Q(z) =

= 0,

x0 + z ⎪ ⎪ , ⎩ ic∗ x0 − z

when 2K is even, (1.20) when 2K is odd,

and require that the function in the square bracket of (1.18) has at least a zero point

1. Generalizations of Keldych–Sedov Formula

83

of order |[K]| at z = i. From 

∞

−∞



∞

λ(t)c(t) 1+(z−i)/(x0 +i) x0 +i dt+ic∗ (1−(z−i)/(t−i))(t−i)X(t) 1−(z−i)/(x −∞ 0 −i) x0 −i

= =

λ(t)c(t) x0 + z dt + ic∗ (t − z)X(t) x0 − z

∞  

∞

j=0 −∞





∞ λ(t)c(t)(z − i)j z−i z − i x0 + i  dt + ic∗ 1 + j+1 (t − i) X(t) x0 + i x0 − i j=0 x0 − i

j

(1.21)

  ∞  ∞ 2ic∗ x0 λ(t)c(t) x0 +i  λ(t)c(t) dt+ic∗ + dt+ = (z−i)j x0 −i j=1 −∞ (t−i)j+1 X(t) (x0 −i)j+1 −∞ (t−i)X(t) 

∞

in the neighborhood of z = i, this shows that 

∞

−∞



∞

−∞

λ(t)c(t) dt = 0, j = 1, . . . , −K(= |[K]|), X(t)(t − i)j

when 2K is even,

λ(t)c(t) 2ic∗ x0 dt+ = 0, j = 2, . . . , [−K]+1(= |[K]|), X(t)(t − i)j (x0 − i)j

(1.22)

when 2K is odd, then the function in the square bracket of (1.18) has a zero point of order |[K]| at z = i, hence the function Φ(z) is analytic at z = i. Besides when 2K is odd, c∗ = i

x0 − i  ∞ λ(t)c(t) dt x0 + i −∞ X(t)(t − i)

(1.23)

is a determined constant. Therefore when K < 0, Problem A has −2K − 1 solvability conditions. Thus we have the following theorem. Theorem 1.1 Problem A for analytic functions in D = {Im z > 0} has the following solvability result. (1) If the index K ≥ 0, the general solution Φ(z) of Problem A possesses the form (1.18), (1.19), which includes 2K + 1 arbitrary real constants. (2) If the index K < 0, Problem A has −2K − 1 solvability conditions as stated in (1.22). When the conditions hold, the solution of Problem A is given by the second formula in (1.18), in particular   X(z)(z − i)|[K]|  ∞ 2ic∗ x0 λ(t)c(t) Φ(z) = dt+ (1.24) πi (x0 − z)(x0 − i)|[K]| −∞ (t − z)(t − i)|[K]| X(t)

in |z − i| < 1, where the constant c∗ is determined as stated in (1.23). Finally, we mention that if x1 , . . . , xm are first kind of discontinuities of c(x) and if γj > 0, j = 1, 2, . . . , m, then the solution Φ(z) of Problem A is bounded in

84

III. Elliptic Complex Equations

D∗ = D\{x1 , . . . , xm }. In general, if γj ≤ 0 (1 ≤ j ≤ m), the solution Φ(z) of Problem A may not be bounded in the neighborhood of xj in D∗ = D\{x1 , . . . , xm }. We have ⎧ ⎨ O(|z − xj |−γj ), if γj < 0, Jj = 1, (1.25) Φ(z) = ⎩ O(ln |z − xj |), if γj = 0, Jj = 0, in the neighborhood of xj on D∗ , but the integral 

z

Φ(z)dz in D i

is bounded. In particular, if m = 2n and ⎧ ⎨ 1,

λ(x) =

⎩

i,

x ∈ (x2j−1 , x2j ), x ∈ (x2j , x2j+1 ),

j = 1, . . . , n,

and xj (j = 1, . . . , m = 2n) are first kind of discontinuous points of c(x), we can choose γ2j−1 = 1/2, K2j−1 = 0, γ2j = −1/2, K2j = 0, j = 1, . . . , n, and then the index of themixed boundary value problem is K = 0. In this case one can choose Π(z) = Πnj=1 (z − x2j−1 )/(z − x2j ). From the formula (1.18) with K = 0, the Keldych–Sedov formula of the mixed boundary value problem for analytic functions in the upper half-plane is derived [53]. If we chose γ2j−1 = −1/2, K2j−1 = 1, γ2j = −1/2, K2j = 0, j = 1, . . . , n, and the index of the mixed boundary value problem is K = n = m/2, then the representation of solutions of the mixed boundary value problem for analytic functions can be written from (1.18) with K = n, which includes 2K + 1 = m + 1 arbitrary real constants, where the function  Π(z) = 1/ Πnj=1 (z − x2j−1 )(z − x2j ).

1.2 The general discontinuous boundary value problem for analytic functions in the upper half-disk Now we first introduce the general discontinuous Riemann–Hilbert problem (Problem B) for analytic functions in the unit disk D = {|z| < 1} with the boundary conditions Re [λ(z)Φ(z)] = au − bv = c(z),

Γ = {|z| = 1},

(1.26)

where λ(z) = a(z) − ib(z), |λ(z)| = 1 on Γ, and Z = {z1 , z2 , . . . , zm } are first kind of discontinuous points of λ(z) on Γ, and λ(z), c(z) satisfy the conditions λ(z) ∈ Cα (Γj ),

|z − zj |βj c(z) ∈ Cα (Γj ),

j = 1, 2, . . . , m,

(1.27)

herein Γj is the arc from the point zj−1 to zj on Γ and z0 = zm , and Γj (j = 1, 2, . . . , m) does not include the end points, α(0 < α < 1) is a constant.

1. Generalizations of Keldych–Sedov Formula

85

Denote by λ(zj − 0) and λ(zj + 0) the left limit and right limit of λ(z) as z → zj (j = 1, 2, . . . , m) on Γ, and eiφj =



λ(zj − 0) , λ(zj + 0)

γj =

φj + Jj , π

Jj = 0 or 1,



Kj =





λ(zj − 0) 1 φj ln − Kj , = πi λ(zj + 0) π

(1.28)

j = 1, . . . , m,

in which 0 ≤ γj < 1 when Jj = 0 and −1 < γj < 0 when Jj = 1, j = 1, . . . , m. The index K of Problem B is defined by (1.4). Let βj + γj < 1, j = 1, . . . , m, we require that the solution Φ(z) possesses the property −δ

Φ(z) = O(|z − zj | ),

⎧ ⎨ βj

+ τ, for γj ≥ 0, and γj < 0, βj > |γj |, δ=⎩ |γj | + τ, for γj < 0, βj ≤ |γj |, j = 1, . . . , m,

(1.29)

in the neighborhood (⊂ D∗ ) of zj , where τ (< α) is an arbitrary small positive number. By using a similar method as stated in Subsection 1, we can obtain the formula for solutions of the boundary value problem. Theorem 1.2 Problem B for analytic functions in D = {|z| < 1} has the following solvability result. (1) If the index K ≥ 0, the general solution Φ(z) of Problem B possesses the form Φ(z) =

  X(z)  (t + z)λ(t)c(t) dt + Q(z) , 2πi Γ (t − z)tX(t) ⎧ ⎪ ⎪ ⎨ 0,

with Q(z) = i

[K]  j=0

(cj z j + cj z −j ) + ⎪

⎪ ⎩ ic∗

(1.30)

when 2K is even, z0 + z , z0 − z

(1.31) when 2K is odd,

where the constant c∗ , c0 are arbitrary real constants, and cj (j = 1, . . . , [K]) are arbitrary complex constants, which includes 2K + 1 arbitrary real constants. (2) If the index K < 0, Problem B has −2K − 1 solvability conditions given by  Γ

 Γ

λ(t)c(t) dt = 0, X(t)tj

j = 1, . . . , −K(= |[K]|),

λ(t)c(t) dt+ic∗ z0−j+1 = 0, X(t)tj

when 2K is even,

j = 1, . . . , [−K]+1(= |[K]|),

when 2K is odd. (1.32)

When the conditions hold, the solution of Problem B possesses the form Φ(z) =

  X(z)z [K]  ic∗ λ(t)c(t) dt + , |[K]|−1 πi Γ (t − z)X(t)t|[K]| (z0 − z)z0

(1.33)

86

III. Elliptic Complex Equations

where the constant c∗ is determined via (1.32) as 

c∗ = i

Γ

λ(t)c(t) dt. X(t)t

In the above formula, X(z) is a non-trivial solution of the homogeneous boundary value problem (Problem B0 ) for analytic functions in the form ⎧ ⎨ iz K Π(z)eiS(z) ,

X(z) = ⎩

iz

[K]

when 2K is even,

(z − z0 )Π(z)e

iS(z)

Π(z) = ,

when 2K is odd,

m 

(z − zj )γj ,

(1.34)

j=1

in which S(z) is an analytic function in D satisfying the boundary conditions z [K] ], Re [S(z)] = arg[Λ∗ (z)¯

z ∈ Γ,

Im [S(0)] = 0,

the function Λ∗ (z) is similar to that in (1.13) [85]11),[86]1). In addition, through the conformal mapping from the upper half-unit disk D = {|z| < 1, Im z > 0} onto the unit disk G = {|ζ| < 1}, namely ζ(z) = −i

z 2 + 2iz + 1 , z 2 − 2iz + 1



z(ζ) =



 1 1 + iζ − 2(1 − ζ 2 ) , ζ +i

we can obtain the result of the general discontinuous Riemann–Hilbert problem (Problem C) for analytic functions in the upper half-unit disk D = {|z| < 1, Im z > 0}, namely w(z) = Φ[ζ(z)] in D = {|z| < 1, Im z > 0} (1.35) is the solution of Problem C for analytic functions. In order to the requirement in latter chapters, we give a well posed version (Problem B
) of Problem B for analytic functions in D = {|z| < 1}, namely we find ¯ an analytic function Φ(z), which is continuous in D\Z and satisfies the boundary condition (1.26) and the point conditions Im [λ(zj
)Φ(zj
)] = bj ,

j = 1, . . . , m,

(1.36)


z1
, . . . , zm

where (
∈ Z) are distinct points on Γ and bj (j = 1, . . . , m) are real constants, and we choose the index K = (m − 1)/2 of λ(z) on Γ = {|z| = 1}. The homogeneous problem of Problem B
with the conditions c(z) = 0 on Γ and bj = 0 (j = 1, . . . , m) will be called Problem B0
. Theorem 1.3 solution.

Problem B
for analytic functions in D = {|z| < 1} has a unique

Proof First of all, we verify the uniqueness of solutions of Problem B
. Let Φ1 (z), Φ2 (z) be two solutions of Problem B
for analytic functions. Then the function Φ(z) = Φ1 (z) − Φ2 (z) is a solution of Problem B0
with the homogeneous boundary conditions Re [λ(z)Φ(z)] = 0 on Γ = {|z| = 1},

Im [λ(zj
)Φ(zj
)] = 0, j = 1, . . . , m.

(1.37)

1. Generalizations of Keldych–Sedov Formula

87

According to the method in the proof of Theorem 1.1 or [8]2),[80]1) and [85]11), we see that if Φ(z)
≡ 0 in D, then m = 2K + 1 ≤ 2ND + NΓ ≤ 2K,

(1.38)

where ND , NΓ are numbers of zero points in D and Γ∗ = Γ\Z respectively. This contradiction proves that Φ(z) ≡ 0, i.e. Φ1 (z) = Φ2 (z) in D. Now we prove the existence of solutions of Problem B
for analytic functions. By the representation (1.30) of the general solution of Problem B for analytic functions, it is easy to see that the general solution Φ(z) can be written as Φ(z) = Φ0 (z) +

m 

dj Φj (z),

(1.39)

j=1

where Φ0 (z) is a special solution of Problem B
, and Φj (z) (j = 1, . . . , m) are a complete system of linearly independent solutions of Problem B0
and dj (j = 1, . . . , m) are arbitrary real constants. In the following, we prove that there exists a unique system of real constants d
j (j = 1, . . . , m) such that |d
1 | + · · · + |d
m | =
0 satisfying the equalities m 

d
j Φj (zj
) = λ(zj
)[c(zj
) + ibj ] − Φ0 (zj
),

j = 1, . . . , m.

(1.40)

j=1




Then the analytic function Φ(z) = Φ0 (z) + m j=1 dj Φj (z) satisfies the boundary conditions (1.26) and (1.36), and thus is a solution of Problem B
. According to the algebraic theory, it suffices to verify that the homogeneous system of algebraic system of equations: (1.40), i.e.

Φ∗ (zj
) =

m 

d
j Φj (zj
) = 0,

j = 1, . . . , m

(1.41)

j=1




has no non-trivial solution. Noting that the analytic function Φ∗ (z) = m j=1 dj Φj (z)

is a solution of Problem B0 , from the uniqueness of solutions of Problem B , we see that Φ∗ (z) = 0. This proves the existence of solutions of Problem B
for analytic functions.

Next, we consider that D is the upper half-unit disk, a(z), b(z) possess discontinuities of first kind at m distinct points z1 , . . . , zm ∈ Γ ∪ L0 = Γ
= ∂D, which are arranged according to the positive direction of ∂D. Here Γ = {|z| = 1, Im z > 0}, L0 = {−1 ≤ x ≤ 1, y = 0} and z1 , . . . , zn−1 ∈ Γ = {|z| = 1, Im z > 0}, xn = −1, . . . , xm = x0 = 1 ∈ L0 , where n (< m), m are positive integers, and c(z) = O(|z − zj |−βj ) in the neighborhood of zj (j = 1, 2, . . . , m) on Γ, in which βj (< 1, j = 1, 2, . . . , m) are non-negative constants, such that βj + γj < 1, γj (j = 1, . . . , m) are as stated in (1.29). Denote λ(z) = a(z) − ib(z) and |a(z)| + |b(z)| =
0, there is no harm in assuming that |λ(z)| = 1, z ∈ Γ
= Γ ∪ L0 . Suppose that λ(z), c(z) satisfy conditions again (1.27).

88

III. Elliptic Complex Equations

Problem C Find an analytic function Φ(z) = u(z)+iv(z) in D, which is continuous ¯ on D∗ = D\Z satisfying the boundary condition z ∈ Γ∗ = Γ
\Z,

Re [λ(z)Φ(z)] = au − bv = c(z),

(1.42)

here Z = {z1 , . . . , zm }. Problem C with the condition r(z) = 0 on Γ∗ is called Problem C0 . The index K of Problem C is the same as stated in (1.4). We can require that the solution Φ(z) satisfies the condition (1.29). In order to find the solution of Problem C for analytic functions, it suffices to choose a conformal mapping from the upper half-unit disk onto the upper half plane or the unit disk. In the following we shall use the other method, namely first find a solution of Problem A for analytic functions in D+ = {Im z > 0} with the boundary condition Re [λ(x)Φ(x)] = r(x) on L = (−∞, ∞), ⎧ ⎨r(x)

on L0 = (−1, 1), r(x) = ⎩ c(x) on L1 = (−∞ < x < −1) ∪ (1 < x < ∞),

(1.43)

in which λ(x), c(x) on L2 = (−∞ < x ≤ −1) ∪ (1 ≤ x < ∞) are appropriate functions, such that λ(x), |x − xj |βj c(x) are piecewise H¨older continuous functions and continuous at the points x = −1, 1, and the index of λ(x) on L is K = 0. For instance, setting ⎧ ⎪ λ(−1 + 0) on (−∞, −1], ⎪ ⎪ ⎪ ⎨

λ(x) =

λ(x) on (−1, 1),

⎪ ⎪ ⎪ ⎪ ⎩ λ(16/(x + 3) − 3)

(1.44) on [1, ∞),

and denoting x2m−j = 16/(xj + 3) − 3, j = n + 1, . . . , m − 1, we can determine that the index of above function λ(x) on L is K = 0. On the basis of Theorem 1.1, the solution Ψ(z) of Problem A can be expressed in the form (1.18),(1.19) with K = 0 ˜ and λ(z), c(z) are as stated in (1.27). Thus the function Φ(z) = Φ(z) − Ψ(z) is analytic in D and satisfies the boundary condition ˜ Re [λ(z)Φ(z)] = r˜(z) =

⎧ ⎨ r(z) − Re [λ(z)Ψ(z)], ⎩

0,

z ∈ Γ,

z ∈ L0 .

(1.45)

Next, similarly to Section 1, we make a transformation ˜ Φ(z) ˜ Ψ(z) = , Π(z)

Π(z) =

2m−n−1  j=n+1,j=m

z − xj z +i

γ j

,

(1.46)

in which γj (j = n + 1, . . . , m − 1, m + 1, . . . , 2m − n − 1) are similar to those in (1.28), the boundary condition ˜ Re [λ(z)Φ(z)] = 0, z ∈ L, (1.47)

1. Generalizations of Keldych–Sedov Formula

89

˜ of Problem C˜0 for the analytic function Φ(z) is reduced to the boundary condition ˜ = 0, Re [Λ(z)Ψ(z)]

z ∈ L,

(1.48)

λ(zj − 0) −iπγj e = ±1, λ(zj + 0)

(1.49)

Λ(z) = λ(z)Π(z)/|Π(z)|,

˜ ˜ for the analytic function Ψ(z) = Φ(z)/Π(z). Noting that 

λ(zj − 0) Π(zj − 0) Λ(zj − 0) = Λ(zj + 0) λ(zj + 0) Π(zj + 0)



=

j = n + 1, . . . , m − 1, m + 1, . . . , 2m − n − 1, the index of Λ(z) on L is the same as the index of λ(z) on L. Due to 2K = 0 is even, provided that we change the sign of Λ(z) on some arcs Lj = (xj−1 , xj ) (j = n + 1, . . . , 2m − n − 1), Ln+1 = (−∞, xn+1 ) ∪ (x2m−n−1 , ∞), then the new function Λ∗ (z) on Γ is continuous, the index of λ(z) on L has not been changed. Moreover we find a solution of Problem A for analytic functions in Im z > 0 with the boundary conditions Re [S(z)] = arg Λ∗ (z) on L = (−∞, ∞),

Im S(i) = 0,

(1.50)

−iS(z) ˆ ˜ . and denote Ψ(z) = Ψ(z)e

ˆ Now we extend the analytic function Ψ(z) as follows ˆ Φ(z) =

⎧ ˆ ⎪ ⎨ Ψ(z)

in D = {|z| < 1,

⎪ ⎩ −Ψ(¯ ˆ z)

˜ = {|z| < 1, in D

Im z > 0}, (1.51) Im z < 0}.

ˆ It can be seen that the analytic function Φ(z) in {|z| < 1} satisfies the boundary condition ˆ Re [Λ(z)Φ(z)] = R(z) on |z| = 1, (1.52) where ⎧ ⎨λ(z),

⎧ ⎨r˜(z)eIm S(z)

on Γ0 = {|z| = 1, Im z > 0}, Λ(z) = ⎩ R(z) = ⎩ ˜ 0 = {|z| = 1, Im z < 0}. −˜ r(¯ z )eIm S(¯z) on Γ λ(¯ z ),

(1.53)

We can find the index K
of Λ(z) on |z| = 1 and by Theorem 1.2, the analytic function ˆ Φ(z) in D with the boundary condition (1.52) can be found, i.e. 

ˆ ˆ i1 + ζ Φ(z) =Φ 1−ζ



in D,

(1.54)

ˆ where Φ(z) is an analytic function as the function Φ(z) in (1.18), but in which λ(z), c(z), K are replaced by λ[i(1 + ζ)/(1 − ζ)], R[i(1 + ζ)/(1 − ζ)], K
respectively, ˆ herein λ(z), R(z) are as stated in (1.53). It is clear that Φ(z) includes 2K
+ 1 arbitrary real constants when K
≥ 0 and −2K
− 1 solvability conditions when K
< 0.

90

III. Elliptic Complex Equations

Thus the solution of Problem C for analytic functions in the upper half-unit disk D is obtained, i.e. iS(z) ˆ w(z) = Ψ(z) + Φ(z)Π(z)e in D. (1.55) Theorem 1.4 When the index K ≥ 0, Problem C for analytic functions in D has a solution in the form (1.55) including 2K + 1 arbitrary real constants, and when K < 0, under −2K − 1 conditions, Problem C for analytic functions possesses the solution as stated in (1.55). Moreover, the above solution of Problem C for analytic functions can be expressed by (1.35). The Keldych–Sedov formula for analytic functions in the upper half-plane possesses important applications to the Tricomi problem for some equations of mixed type (see [12]1),3)). But more general boundary value problems for equations of mixed type cannot be solved by this formula. Due to we have Theorems 1.1–1.4, such that the above general problems can be solved. In addition, we can give the representation of solutions to the discontinuous Riemann–Hilbert boundary value problem for analytic functions in the zone domain D = {0 < Im z < 1}, which can be used to solve some boundary value problems for nonlinear problems in mechanics.

2

Representation and Existence of Solutions for Elliptic Complex Equations of First Order

In this section, we shall establish the representations for solutions of the general discontinuous boundary value problem for elliptic complex equations of first order in the upper half-unit disk. Moreover, we shall prove the existence of solutions for nonlinear elliptic complex equations of first order. 2.1 Representation of solutions of the discontinuous Riemann–Hilbert problem for elliptic complex equations in the upper half-unit disk Let D be an upper half-unit disk with the boundary Γ
= Γ ∪ L0 as stated in Section 1. We consider the nonlinear uniformly elliptic systems of first order equations Fj (x, y, u, v, ux , vx , uy , vy ) = 0 in D,

j = 1, 2.

Under certain conditions, the system can be transformed into the complex form wz¯ = F (z, w, wz ),

F = Q1 wz + Q2 w¯z¯ + A1 w + A2 w¯ + A3 ,

z∈D

(2.1)

(see [86]1)), in which F (z, w, U ) satisfy the following conditions. Condition C 1) Qj (z, w, U ), Aj (z, w) (j = 1, 2), A3 (z) are measurable in z ∈ D for all continu¯ ous functions w(z) in D∗ = D\Z and all measurable functions U (z) ∈ Lp0 (D∗ ), and

2. Elliptic Equations of First Order

91

satisfy ¯ ≤ k0 , Lp [Aj , D]

j = 1, 2,

¯ ≤ k1 , Lp [A3 , D]

(2.2)

where Z = {z1 , . . . , zm }, D∗ is any closed subset in D, p0 , p (2 < p0 ≤ p), k0 , k1 are non-negative constants. 2) The above functions are continuous in w ∈ C I for almost every point z ∈ D, U ∈C I, and Qj = 0 (j = 1, 2), Aj = 0 (j = 1, 2, 3) for z
∈ D. 3) The complex equation (2.1) satisfies the uniform ellipticity condition |F (z, w, U1 ) − F (z, w, U2 )| ≤ q0 |U1 − U2 |,

(2.3)

for almost every point z ∈ D, in which w, U1 , U2 ∈ C I and q0 (< 1) is a non-negative constant. Problem A The discontinuous Riemann–Hilbert boundary value problem for (2.1) is to find a continuous solution w(z) in D∗ satisfying the boundary condition: Re [λ(z)w(z)] = c(z),

z ∈ Γ∗ = Γ
\Z,

(2.4)

where λ(z), c(z) are as stated in Section 1 satisfying Cα [λ(z), Γj ] ≤ k0 ,

Cα [|z − zj |βj c(z), Γj ] ≤ k2 ,

j = 1, . . . , m,

(2.5)

herein α (1/2 < α < 1), k0 , k2 are non-negative constants. Assume that (βj + γj )/β < 1, β = min(α, 1 − 2/p0 )/2, γj , βj (j = 1, . . . , m) are as stated in (1.28), (1.29). Problem A with A3 (z) = 0 in D, c(z) = 0 on Γ∗ is called Problem A0 . The index K of Problem A and Problem A0 is defined as in (1.4). In order to prove the solvability of Problem A for the complex equation (2.1), we need to give a representation theorem for Problem A. Theorem 2.1 Suppose that the complex equation (2.1) satisfies Condition C, and w(z) is a solution of Problem A for (2.1). Then w(z) is representable by w(z) = Φ[ζ(z)]eφ(z) + ψ(z),

(2.6)

¯ which quasiconformally maps D onto the unit where ζ(z) is a homeomorphism in D, disk G= {|ζ| < 1} with boundary L = {|ζ| = 1}, where ζ(−1) = −1, ζ(i) = i, ζ(1) = 1, Φ(ζ) is an analytic function in G, ψ(z), φ(z), ζ(z) and its inverse function z(ζ) satisfy the estimates ¯ ≤ k3 , Cβ [ψ, D]

¯ ≤ k3 , Cβ [φ, D]

¯ ≤ k3 , Cβ [ζ(z), D]

¯ ≤ k3 , Cβ [z(ζ), G]

¯ ≤ k3 , Lp0 [|φz¯| + |φz |, D] ¯ ≤ k3 , Lp0 [|ψz¯| + |ψz |, D] ¯ ≤ k3 , Lp0 [|χz¯| + |χz |, D] ¯ ≤ k4 , Cβ [z(ζ), G]

(2.7) (2.8) (2.9)

in which χ(z) is as stated in (2.14) below, β = min(α, 1 − 2/p0 )/2, p0 (2 < p0 ≤ p), kj = kj (q0 , p0 , k0 , k1 , D) (j = 3, 4) are non-negative constants.Moreover, if the

92

III. Elliptic Complex Equations

coefficients Qj (z) = 0 (j = 1, 2) of the complex equation (2.1) in D, then the representation (2.6) becomes the form w(z) = Φ(z)eφ(z) + ψ(z),

(2.10)

and if K < 0, Φ(z) satisfies the estimate ¯ ≤ M1 = M1 (p0 , β, δ, k, D), Cδ [X(z)Φ(z), D] in which m 

X(z) =




|z−zj |ηj |z−zn |2ηn |z−zm |2ηm ,

j=1,j=n,m

ηj =

|γj |+τ, γj < 0, βj ≤ |γj |, |βj |+τ, other case.

(2.11)

(2.12)

Here γj (j = 1, . . . , m) are real constants as stated in (1.28) and δ, τ (0 < δ < min(β, τ )) are sufficiently small positive constants, and M1 is a non-negative constant. Proof We substitute the solution w(z) of Problem A into the coefficients of equation (2.1) and consider the following system


ψz¯ =

Qψz + A1 ψ + A2 ψ¯ + A3 ,

φz¯ = Qφz + A, Wz¯ = QWz ,

⎧ ⎨ A1

A=⎩

Q=

Q1 + Q2 wz /wz for wz
= 0, 0 for wz = 0 or z
∈ D,

+ A2 w/w ¯ for w(z)
= 0,

(2.13)

0 for w(z) = 0 or z
∈ D,

W (z) = Φ[ζ(z)].

By using the continuity method and the principle of contracting mappings, we can find the solution 1   f (ζ) dσζ , ψ(z) = T f = − π D ζ −z (2.14) φ(z) = T g, ζ(z) = Ψ[χ(z)], χ(z) = z + T h ¯ 2 < p0 ≤ p, χ(z) is a homeomorphism in of (2.13), where f (z), g(z), h(z) ∈ Lp0 ,2 (D), ¯ Ψ(χ) is a univalent analytic function, which conformally maps E = χ(D) onto the D, unit disk G(see [85]11)), and Φ(ζ) is an analytic function in G. We can verify that ψ(z), φ(z), ζ(z) satisfy the estimates (2.7) and (2.8). It remains to prove that z = z(ζ) satisfies the estimate (2.9). In fact, we can find a homeomorphic solution of the last ¯ equation in (2.13) in the form χ(z) = z + T h such that [χ(z)]z , [χ(z)]z¯ ∈ Lp0 (D) [80]1),[85]9). Next, we find a univalent analytic function ζ = Ψ(χ), which maps χ(D) onto G, hence ζ = ζ(z) = Ψ[χ(z)]. By the result on conformal mappings, applying the method of Lemma 2.1, Chapter II in [86]1), we can prove that (2.9) is true. When Qj (z) = 0 in D, j = 1, 2, then we can choose χ(z) = z in (2.14), in this case Φ[ζ(z)] can be replaced by the analytic function Φ(z), herein ζ(z), Ψ(z) are as stated in (2.14), it is clear that the representation (2.6) becomes the form (2.10). Thus the analytic function Φ(z) satisfies the boundary conditions Re [λ(z)eφ(z) Φ(z)] = c(z) − Re [λ(z)ψ(z)],

z ∈ Γ∗ .

(2.15)

On the basis of Theorem 1.2 and the estimate (2.7), Φ(z) satisfies the estimate (2.11).

2. Elliptic Equations of First Order

93

2.2 Existence of solutions of the discontinuous Riemann–Hilbert problem for nonlinear complex equations in the upper half-unit disk Theorem 2.2 ments hold.

Under the same conditions as in Theorem 2.1, the following state-

(1) If the index K ≥ 0, then Problem A for (2.1) is solvable, and the general solution includes 2K + 1 arbitrary real constants. (2) If K < 0, then Problem A has −2K − 1 solvability conditions. Proof Let us introduce a closed, convex and bounded subset B1 in the Banach ¯ × Lp0 (D) ¯ × Lp0 (D), ¯ whose elements are systems of functions space B = Lp0 (D) ¯ + Lp0 (f, D) ¯ + Lp0 (g, D), ¯ which q = [Q(z), f (z), g(z)] with norms  q = Lp0 (Q, D) satisfy the condition |Q(z)| ≤ q0 < 1 (z ∈ D),

¯ ≤ k3 , Lp0 [f (z), D]

¯ ≤ k3 , Lp0 [g(z), D]

(2.16)

where q0 , k3 are non-negative constants as stated in (2.3) and (2.7). Moreover introduce a closed and bounded subset B2 in B, the elements of which are systems of functions ω = [f (z), g(z), h(z)] satisfying the condition ¯ ≤ k4 , Lp0 [f (z), D] where Πh =

− π1



D [h(ζ)/(ζ

¯ ≤ k4 , Lp0 [g(z), D]

|h(z)| ≤ q0 |1 + Πh|,

(2.17)

2

− z) ]dσζ .

We arbitrarily select q = [Q(z), f (z), g(z)] ∈ B1 , and using the principle of con¯ of the integral equation tracting mappings, a unique solution h(z) ∈ Lp0 (D) h(z) = Q(z)[1 + Πh]

(2.18)

can be found, which satisfies the third inequality in (2.17). Moreover, χ(z) = z + T h ¯ Now, we find a univalent analytic function ζ = Φ(χ), is a homeomorphism in D. which maps χ(D) onto the unit disk G as stated in Theorem 2.1. Moreover, we find an analytic function Ψ(ζ) in G satisfying the boundary condition in the form Re [Λ(ζ)Φ(ζ)] = R(ζ),

ζ ∈ L = ζ(Γ),

(2.19)

in which ζ(z) = Ψ[χ(z)], z(ζ) is its inverse function, ψ(z) = T f, φ(z) = T g, Λ(ζ) = λ[z(ζ)] exp[φ(z(ζ))], R(ζ) = r[z(ζ)] − Re [λ[z(ζ)]ψ(z(ζ))], where Λ(ζ), R(ζ) on L satisfy conditions similar to λ(z), c(z) in (2.5) and the index of Λ(ζ) on L is K. In the following, we first consider the case of K ≥ 0. On the basis of Theorem 1.2, we can find the analytic function Φ(ζ) in the form (1.30), here 2K +1 arbitrary real constants can be chosen. Thus the function w(z) = Φ[ζ(z)]eφ(z) + ψ(z) is determined. Afterwards, we find out the solution [f ∗ (z), g ∗ (z), h∗ (z), Q∗ (z)] of the system of integral equations f ∗ (z) = F (z, w, Πf ∗ )−F (z, w, 0)+A1 (z, w)T f ∗ +A2 (z, w)T f ∗ +A3 (z, w),

(2.20)

W g ∗ (z) = F (z, w, W Πg ∗ +Πf ∗ )−F (z, w, Πf ∗ )+A1 (z, w)W +A2 (z, w)W ,

(2.21)

94

III. Elliptic Complex Equations S
(χ)h∗ (z)eφ(z) = F [z, w, S
(χ)(1 + Πh∗ )eφ(z) + W Πg ∗ + Πf ∗ ] (2.22)

− F (z, w, W Πg ∗ + Πf ∗ ),

h∗ (z) , S
(χ) = [Φ(Ψ(χ))]χ , (2.23) [1 + Πh∗ ] and denote by q ∗ = E(q) the mapping from q = (Q, f, g) to q ∗ = (Q∗ , f ∗ , g ∗ ). According to Theorem 2.1 from Chapter IV in [86]1), we can prove that q ∗ = E(q) continuously maps B1 onto a compact subset in B1 . On the basis of the Schauder fixed-point theorem, there exists a system q = (Q, f, g) ∈ B1 , such that q = E(q). Applying the above method, from q = (Q, f, g), we can construct a function w(z) = Φ[ζ(z)]eφ(z) + ψ(z), which is just a solution of Problem A for (2.1). As for the case of K < 0, it can be similarly discussed, but we first permit that the function Φ(ζ) satisfying the boundary condition (2.15) has a pole of order |[K]| at ζ = 0, and find the solution of the nonlinear complex equation (2.1) in the form w(z) = Φ[ζ(z)]eφ(z) + ψ(z). From the representation, we can derive the −2K − 1 solvability conditions of Problem A for (2.1). Q∗ (z) =

Besides, we can discuss the solvability of the discontinuous Riemann–Hilbert boundary value problem for the complex equation (2.1) in the upper half-plane and the zone domain. For some problems in nonlinear mechanics as stated in [90], it can be solved by the results in Theorem 2.2. 2.3 The discontinuous Riemann–Hilbert problem for nonlinear complex equations in general domains In this subsection, let D
be a general simply connected domain with the boundary Γ
= Γ
1 ∪ Γ
2 , herein Γ
1 , Γ
2 ∈ Cα1 (0 < α < 1) and their intersection points z
, z

with the inner angles α1 π, α2 π(0 < α1 , α2 < 1) respectively. We discuss the nonlinear uniformly elliptic complex equation wz¯ = F (z, w, wz ),

F = Q1 wz + Q2 w¯z¯ + A1 w + A2 w¯ + A3 ,

z ∈ D
,

(2.24)




in which F (z, w, U ) satisfies Condition C in D . There exist m point Z = {z1 = z
, . . . , zn−1 , zn = z

, . . . , zm } on Γ
arranged according to the positive direction successively. Denote by Γ
j the curve on Γ
from zj−1 to zj , j = 1, 2, . . . , m, and Γ
j (j = 1, . . . , m) does not include the end points. Problem A
The discontinuous Riemann–Hilbert boundary value problem for (2.1) is to find a continuous solution w(z) in D∗ = D
\Z satisfying the boundary condition: Re [λ(z)w(z)] = c(z),

x ∈ Γ∗ = Γ
\Z,

Im [λ(zj
)w(zj
)] = bj ,

j = 1, . . . , m,

(2.25)

where zj
, bj (j = 1, . . . , m) are similar to those in (1.36), λ(z), c(z), bj (j = 1, . . . , m) are given functions satisfying Cα [λ(z), Γj ] ≤ k0 ,

Cα [|z − zj |βj c(z), Γj ] ≤ k2 ,

|bj | ≤ k2 ,

j = 1, . . . , m, (2.26)

3. Elliptic Equations of Second Order

95

herein α (1/2 < α < 1), k0 , k2 are non-negative constants, and assume that (βj + γj )/β < 1, β = min(α, 1 − 2/p0 )/α0 , βj (j = 1, . . . , m) are similar to those in (1.29), α0 = max(1/α1 , 1/α2 , 1). Problem A with A3 (z) = 0 in D, r(z) = 0 on Γ
, and bj = 0 (j = 1, . . . , m) is called Problem A
0 . The index K = (m − 1)/2 of Problem A and Problem A0 is defined as in (1.4). In order to give the unique result of solutions of Problem A
for equation (2.24), we need to add one condition: For any complex functions wj (z) ∈ C(D∗ ), Uj (z) ∈ Lp0 (D
)(j = 1, 2, 2 < p0 ≤ p), the following equality holds: F (z, w1 , U1 ) − F (z, w1 , U2 ) = Q(U1 − U2 ) + A(w1 − w2 ) in D
,

(2.27)

in which |Q(z, w1 , w2 , U1 , U2 )| ≤ q0 , A(z, w1 , w2 , U1 , U2 ) ∈ Lp0 (D). Especially, if (2.24) is a linear equation, then the condition (2.27) obviously is true. Applying a similar method as before, we can prove the following theorem. Theorem 2.3 If the complex equation (2.24) in D
satisfies Condition C, then Problem A
for (2.24) is solvable. If Condition C and the condition (2.27) hold, then the solution of Problem A
is unique. Moreover the solution w(z) can be expressed as (2.6)–(2.9), where β = min(α, 1 − 2/p0 )/α0 . If Qj (z) = 0 in D, j = 1, 2 in (2.24), then the representation (2.6) becomes the form w(z) = Φ(z)eφ(z) + ψ(z),

(2.28)

and w(z) satisfies the estimate ¯ ≤ M1 = M1 (p0 , β, δ, k, D), Cδ [X(z)w(z), D]

(2.29)

in which m 

X(z) =

|z − zj |ηj |z − z1 |max(1/α1 ,1)η1 |z − zn |max(1/α2 ,1)ηn ,

j=2,j=1,n




ηj =

|γj | + τ,

if γj < 0,

|βj | + τ,

if γj ≥ 0,

βj ≤ |γj |,

(2.30)

and γj < 0, βj < |γj |,

here γj (j = 1, . . . , m) are real constants as stated in (1.28), δ, τ (0 < δ < min(β, τ )) are sufficiently small positive constants, and M1 is a non-negative constant.

3

Discontinuous Oblique Derivative Problems for Quasi linear Elliptic Equations of Second Order

This section deals with the oblique derivative boundary value problems for quasilinear elliptic equations of second order. We first give the extremum principle and representation of solutions for the above boundary value problem, and then obtain a priori estimates of solutions of the above problem, finally we prove the uniqueness and existence of solutions of the above problem.

96

III. Elliptic Complex Equations

3.1 Formulation of the discontinuous oblique derivative problem for elliptic equations of second order Let D be the upper half-unit disk as stated in Section 1 and Γ
= Γ ∪ L0 of D be the boundary, where Γ = {|z| = 1, Im z ≥ 0} and L0 = (−1, 1). We consider the quasilinear uniformly elliptic equation of second order auxx + 2buxy + cuyy + dux + euy + f u = g in D,

(3.1)

¯ and u, ux , uy ∈ IR. Under where a, b, c, d, e, f, g are given functions of (x, y) ∈ D certain conditions, equation (3.1) can be reduced to the the complex form uzz¯ = F (z, u, uz , uzz ),

F = Re [Quzz +A1 uz ]+A2 u+A3 in D,

(3.2)

where Q = Q(z, u, uz ), Aj = Aj (z, u, uz ) and 1 uz = [ux − iuy ], uz¯ = 2 −a + c − 2bi −d − ei , A1 (z) = , Q(z) = a+c a+c z = x + iy,

1 [ux + iuy ], uzz¯ = 2 −f A2 (z) = , 2(a + c)

1 [uxx + uyy ], 4 g A3 (z) = . 2(a + c)

Suppose that equation (3.2) satisfies the following conditions. Condition C I for almost 1) Q(z, u, w), Aj (z, u, w) (j = 1, 2, 3) are continuous in u ∈ IR, w ∈ C every point z ∈ D, u ∈ IR, w ∈ C I, and Q = 0, Aj = 0 (j = 1, 2, 3) for z
∈ D. 2) The above functions are measurable in z ∈ D for all continuous functions ¯ u(z), w(z) on D∗ = D\Z, and satisfy ¯ ≤ k0 , j = 1, 2, Lp [Aj (z, u, w), D]

¯ ≤ k1 , A2 (z, u, w) ≥ 0 in D, (3.3) Lp [A3 (z, u, w), D]

in which p0 , p (2 < p0 ≤ p), k0 , k1 are non-negative constants, Z = {−1, 1}. 3) Equation (3.2) satisfies the uniform ellipticity condition, namely for any number u ∈ IR, w ∈ C I, the inequality |Q(z, u, w)| ≤ q0 < 1

(3.4)

for almost every point z ∈ D holds, where q0 is a non-negative constant. The discontinuous oblique derivative boundary value problem for equation (3.2) may be formulated as follows: ¯ Problem P Find a continuously differentiable solution u(z) of (3.2) in D∗ = D\Z, which is continuous in D and satisfies the boundary conditions 1 ∂u = Re [λ(z)uz ] = r(z), 2 ∂ν

z ∈ Γ∗ = Γ
\Z,

u(−1) = b0 ,

u(1) = b1 ,

(3.5)

3. Elliptic Equations of Second Order

97

where Z = {−1, 1} is the set of discontinuous points of λ(z) on Γ∗ , ν is a given vector at every point on Γ∗ , λ(z) = a(x) + ib(x) = cos(ν, x) − i cos(ν, y), cos(ν, n) ≥ 0 on Γ∗ . If cos(ν, n) ≡ 0 on Γ∗ = Γ
\Z, then the condition u(1) = b1 can be canceled. Here n is the outward normal vector at every point on Γ∗ , δ0 (< 1) is a constant, b0 , b1 are real constants, and λ(z), r(z), b0 , b1 satisfy the conditions Cα [λ(z), Γj ] ≤ k0 ,

Cα [|z − zj |βj r(z), Γj ] ≤ k2 ,

j = 1, 2,

|b0 |, |b1 | ≤ k2 .

(3.6)

Herein α (1/2 < α < 1), k0 , k2 are non-negative constants. We assume that (βj + γj )/β < 1, β = min(α, 1 − 2/p0 )/2, βj (j = 1, . . . , m) are as stated in (1.27). Problem P with A3 (z) = 0 in D, r(z) = 0 on Γ, b0 = b1 = 0 is called Problem P0 . The index of Problem P is K, where K is defined as in (1.4), here we choose K = 0, and K = −1/2 if cos(ν, n) ≡ 0 on Γ∗ . If A2 (z) = 0 in D, the last point condition in (3.5) can be replaced by Im [λ(z)uz ]|z=0 = b2 , (3.7) and we do not need the assumption cos(ν, n) ≥ 0 on Γ, where b2 is a real constant satisfying the condition |b2 | ≤ k2 . Then the boundary value problem for (3.2) will be called Problem Q. In the following, we only discuss the case of K = 0, and the case of K = −1/2 can be similarly discussed. 3.2

The representation theorem of Problem P for equation (3.2)

We first introduce a theorem. Theorem 3.1 Suppose that equation (3.2) satisfies Condition C. Then there exist two solutions ψ(z), Ψ(z) of the Dirichlet problem (Problem D) of (3.2) and its related homogeneous equation uzz¯ − Re [Q(z, u, uz )uzz + A1 (z, u, uz )uz ] − A2 (z, u, uz )u = 0 in D,

(3.8)

satisfying the boundary conditions ψ(z) = 0,

Ψ(z) = 1 on Γ

(3.9)

respectively, and ψ(z), Ψ(z) satisfy the estimates ¯ ≤ M2 , C 1 [Ψ(z), D] ¯ ≤ M2 , Cβ1 [ψ(z), D] β ¯ ≤ M3 , Lp0 [ψzz¯, D]

¯ ≤ M3 , Lp0 [Ψzz¯, D]

Ψ ≥ M4 > 0 in D,

(3.10)

where β (0 < β ≤ α), Mj = Mj (q0 , p0 , β, k0 , k1 , D) (j = 2, 3, 4) are non-negative constants. Proof We first assume that the coefficients Q = Aj = 0 (j = 1, 2, 3) of (3.2) in the ε-neighborhood of z = −1, 1, i.e. Dε = {|z ± 1| ≤ ε, Im z ≥ 0}, ε > 0, where ε = 1/m (m is a positive integer). Introduce the transformation and its inversion ζ(z) = −i

z 2 + 2iz + 1 , z 2 − 2iz + 1



z(ζ) =



 1 1 + iζ − 2(1 − ζ 2 ) . ζ +i

(3.11)

98

III. Elliptic Complex Equations

The function ζ(z) maps D onto G = {|ζ| < 1}, such that the boundary points −1, 0, 1 are mapped onto the points −1, −i, 1 respectively. Through the transformation, equation (3.2) is reduced to the equation uζ ζ¯ = |z
(ζ)|2 {Re [Quζζ /(z
(ζ))2 +(A1 /z
(ζ)−Qz

(ζ)/(z
(ζ))3 )uζ ]+A2 u+A3 } (3.12) ¯ satisfies conditions similar to Condition C. in G. It is clear that equation (3.12) in G Hence equation (3.12) and its related homogeneous equation uζ ζ¯ = |z
(ζ)|2 {Re [Quζζ /(z
(ζ))2 +(A1 /z
(ζ)−Qz

(ζ)/(z
(ζ))3 )uζ ]+A2 u} in G (3.13) possess the solutions ψ(ζ), Ψ(ζ) satisfying the boundary conditions ψ(ζ) = 0,

Ψ(ζ) = 1 on L = ζ(Γ),

and ψ[ζ(z)], Ψ[ζ(z)] in D are the solutions of Problem D of (3.2),(3.8) satisfying the boundary condition (3.9) respectively, and ψ(z), Ψ(z) satisfy the estimate (3.10), but the constants Mj = Mj (q0 , p0 , β, k0 , k1 , D, ε) (j = 2, 3, 4). Now we consider ⎧ ⎨ ψ(z)

˜ ψ(z) =⎩

in D, (3.14)

˜ = {|z| < 1, Im z < 0}. −ψ(¯ z ) in D

˜ in ∆ = {|z| < 1} is a solution of the elliptic equation It is not difficult to see that ψ(z) ˜ zz + A˜1 uz ] − A˜2 u = A˜3 in ∆, uzz¯ − Re [Qu

(3.15)

where the coefficients ⎧ ⎨Q(z),

⎧ ⎨A1 (z),

⎧ ⎨A2 (z),

⎧ ⎨A3 (z)

⎧ ⎫ ⎨D ⎬

˜= Q A˜1 = ⎩ A˜3 = ⎩ in ⎩ ⎭ , A˜2 = ⎩ ⎩Q(¯ ˜ z ), z) A2 (¯ −A3 (¯ D z) A1 (¯ z ), ˜ is the symmetrical domian of D with respect to the real axis. It is clear that where D the coefficients in ∆ satisfy conditions similar to those from Condition C. Obviously ˜ the solution ψ(z) satisfies the boundary condition ψ(z) = 0 on ∂∆ = {|z| = 1}. ˜ Denote by ψm (z) the solution of equation (3.2) with Q = Aj = 0(j = 1, 2, 3) in the ε = 1/m-neighborhood of z = −1, 1, we can derive that the function ψ˜m (z) in ∆ satisfies estimates similar to ψ(z) in (3.10), where the constants Mj (j = 2, 3) are independent of ε = 1/m. Thus we can choose a subsequence of {ψ˜m (z)}, which uniformly converges to ψ∗ (z), and ψ∗ (z) is just a solution of Problem D for the original equation (3.2) in D. Noting that the solution Ψ(z) = ψ(z) + 1 of Problem D for equation (3.8) is equivalent to the solution ψ(z) of Problem D for the equation uzz¯ − Re [Quzz + A1 uz ] − A2 u = A2 in D

(3.16)

with the boundary condition ψ(z) = 0 on Γ, by using the same method, we can prove that there exists a solution Ψ(z) of Problem D for (3.8) with the boundary condition Ψ(z) = 1 on Γ, and the solution satisfies the estimates in (3.10).

3. Elliptic Equations of Second Order

99

Theorem 3.2 Suppose that equation (3.2) satisfies Condition C, and u(z) is a solution of Problem P for (3.2). Then u(z) can be expressed as 

u(z) = U (z)Ψ(z)+ψ(z), U (z) = 2Re

0

z

w(z)dz+b0 , w(z) = Φ[ζ(z)]eφ(z) ,

(3.17)

where ψ(z), Ψ(z) are as stated in Theorem 3.1 satisfying the estimate (3.10), ζ(z) ¯ which quasiconformally maps D onto the unit disk G= is a homeomorphism in D, {|ζ| < 1} with boundary L, where ζ(−1) = −1, ζ(1) = 1, ζ(i) = i, Φ(ζ) is an analytic function in G, φ(z), ζ(z) and its inverse function z(ζ) satisfy the estimates ¯ ≤ k3 , Cβ [φ(z), D]

¯ ≤ k3 , Cβ [ζ(z), D]

¯ ≤ k3 , Cβ [z(ζ), G]

¯ ≤ k3 , Lp0 [|φz¯| + |φz |, D]

¯ ≤ k3 , Cβ [z(ζ), G]

¯ ≤ k4 , Lp0 [|χz¯| + |χz |, D]

(3.18)

in which χ(z) is as stated in (2.14), β = min(α, 1 − 2/p0 )/2, p0 (2 < p0 ≤ p), kj = kj (q0 , p0 , k0 , k1 , D)(j = 3, 4) are non-negative constants. Proof We substitute the solution u(z) of Problem P into the coefficients of equation (3.2). It is clear that (3.2) in this case can be seen as a linear equation. Firstly, on the basis of Theorem 3.1 there exist two solutions ψ(z), Ψ(z) of Problem D of (3.2) and its homogeneous equation (3.8) satisfying the estimate (3.10). Thus the function U (z) =

u(z) − ψ(z) in D, Ψ(z)

(3.19)

is a solution of the equation Uzz¯ − Re [QUzz + AUz ] = 0,

A = A1 − 2(ln ψ)z¯ + 2Q(ln Ψ)z in D,

(3.20)

and w(z) = Uz is a solution of the first order equation 1 wz¯ = [Qwz + Qw¯z¯ + Aw + Aw] ¯ in D 2

(3.21)

satisfying the boundary condition 1 ∂U [ + (ln Ψ)ν U ] = r(z) − Re [λ(z)ψz ] on Γ∗ , i.e. 2 ∂ν

(3.22)

Re [λ(z)Uz + (ln Ψ)ν U/2] = r(z) − Re [λ(z)ψz ] on Γ∗ . By the following Lemma 3.3, we see that (ln Ψ)ν > 0 on Γ∗ , and similarly to Theorem 2.1, the last formula in (3.17) can be derived, and φ(z), ζ(z) and its inverse function z(ζ), χ(z) satisfy the estimates (2.7)–(2.9). Now we consider the linear homogeneous equation uzz¯ − Re [Quzz + A1 (z)uz ] − A2 (z)u = 0 in D, and give a lemma.

(3.23)

100

III. Elliptic Complex Equations

Lemma 3.3 Let the equation (3.23) in D satisfy Condition C, and u(z) be a continuously differentiable solution of (3.23) in D. If M = maxz∈D u(z) ≥ 0, then there exists a point z0 ∈ ∂D, such that u(z0 ) = M . If z0 = x0 ∈ (−1, 1), and u(z) < u(z0 ) ¯ in D\{z 0 }, then ∂u u(z0 ) − u(z) > 0, (3.24) = lim z(∈l)→z ∂l |z − z0 | 0 where z (∈ D) approaches z0 along a direction l, such that cos(l, y) > 0. Proof From the result in Section 2, Chapter III, [86]1), we see that the solution u(z) in D attains its non-negative maximum M at a point z0 ∈ ∂D. There is no harm in assuming that z0 is a boundary point of ∆ = {|z| < R}, because we can choose a subdomain(∈ D) with smooth boundary and the boundary point z0 , and then make a conformal mapping. Thus this requirement can be realized. By Theorem 3.1, we find a continuously differentiable solution Ψ(z) of (3.23) in ∆ satisfying the boundary condition: Ψ(z) = 1, z ∈ ∂∆ = {|z| = R}, and can derive that 0 < Ψ(z) ≤ 1, z ∈ ∆. Due to V (z) = u(z)/Ψ(z) is a solution of the following equation LV = Vzz¯ − Re [A(z)Vz ] = 0,

A(z) = −2(ln Ψ)z + A1 (z) in ∆,

(3.25)

it is clear that V (z) < V (z0 ), z ∈ ∆, and V (z) attains the maximum at the point ˜ = z0 . Afterwards, we find a continuously differentiable solution V˜ (z) of (3.25) in ∆ {R/2 ≤ |z| ≤ R} satisfying the boundary condition V˜ (z) = 0,

z ∈ ∂∆;

V˜ (z) = 1,

|z| =

R . 2

˜ and It is easy to see that ∂ V˜ /∂s = 2Re [iz V˜z ], z ∈ ∂ ∆, z V˜z ∂ V˜ = 2Re , ∂n R

z ∈ ∂∆,

∂ V˜ z V˜z = −4Re , ∂n R

|z| =

R , 2

˜ where s, n are the tangent vector and outward normal vector on the boundary ∂ ∆. Noting that W (z) = V˜z satisfies the equation Wz¯ − Re [A(z)W ] = 0,

˜ z ∈ ∆,

˜ and the index of iz on the and the boundary condition Re [izW (z)] = 0, z ∈ ∂ ∆, ˜ equals to 0, hence W (z) has no zero point on ∂∆, thus ∂ V˜ /∂n = boundary ∂ ∆ 2Re [zW (z)/R] < 0, z ∈ ∂∆. The auxiliary function Vˆ (z) = V (z) − V (z0 ) + εV˜ (z),

˜ z ∈ ∆,

by selecting a sufficiently small positive number ε, such that Vˆ (z) < 0 on |z| = R/2, ˜ on the basis of the obviously satisfies Vˆ (z) ≤ 0, z ∈ ∂∆. Due to LVˆ = 0, z ∈ ∆, maximum principle, we have Vˆ (z) ≤ 0, z ∈ ∂∆,

i.e. V (z0 ) − V (z) ≥ −ε[V (z0 ) − V (z)],

˜ z ∈ ∆.

3. Elliptic Equations of Second Order

101

Thus at the point z = z0 we have ∂ V˜ ∂V ≥ −ε > 0, ∂n ∂n

∂u ∂V ∂Ψ ∂ V˜ ∂Ψ =Ψ +V ≥ −ε +V > 0. ∂n ∂n ∂n ∂n ∂n

Moreover, noting the condition cos(l, n) > 0, cos(l, s) > 0, ∂U /∂s = 0 at the point z0 , where s is the tangent vector at z0 , it follows the inequality ∂u ∂u ∂u = cos(l, n) + cos(l, s) > 0. ∂l ∂n ∂s

(3.26)

Theorem 3.4 If equation (3.2) satisfies Condition C and for any uj (z) ∈ C 1 (D∗ ), j = 1, 2, uzz ∈ C I, the following equality holds: ˜ zz + A˜1 uz ]− A˜2 u, F (z, u1 , u1z , u1zz ) − F (z, u2 , u2z , u2zz ) = −Re [Qu ¯ < ∞, j = 1, 2, then the solution u(z) of Problem P is unique. where Lp [A˜j , D] Proof Suppose that there exist two solutions u1 (z), u2 (z) of Problem P for (3.2), it can be seen that u(z) = u1 (z) − u2 (z) satisfies the homogeneous equation and boundary conditions ˜ zz + A˜1 uz ] + A˜2 u in D, uzz¯ = Re [Qu 1 ∂u = 0, 2 ∂ν

z ∈ Γ∗ ,

(3.27) u(−1) = 0,

u(1) = 0.

If the maximum M = maxD¯ u(z) > 0, it is clear that the maximum point z ∗
= −1 and 1. On the basis of Lemma 3.3, the maximum of u(z) cannot attain on (−1, 1), hence its maximum M attains at a point z ∗ ∈ Γ∗ . If cos(ν, n) > 0 at z ∗ , from Lemma 3.3, we get ∂u/∂ν > 0 at z ∗ , this contradicts the boundary condition in (3.27); if cos(ν, n) = 0 at z ∗ , denote by Γ
the longest curve of Γ including the point z ∗ , so that cos(ν, n) = 0 and u(z) = M on Γ
, then there exists a point z
∈ Γ\Γ
, such that at z
, cos(ν, n) > 0, ∂u/∂n > 0, cos(ν, s) > 0 (< 0), ∂u/∂s ≥ 0 (≤ 0), hence (3.26) at z
holds, it is impossible. This shows z ∗
∈ Γ. Hence maxD+ u(z) = 0. By the similar method, we can prove minD+ u(z) = 0. Therefore u(z) = 0, u1 (z) = u2 (z) in D. Theorem 3.5 Suppose that equation (3.2) satisfies Condition C, then the solution u(z) of Problem P for (3.2) satisfies the estimates 1/β ¯ = Cβ [u(z), D]+C ¯ ¯ ≤ M5 , C˜δ1 [u(z), D] uz , D] δ [|X(z)|

¯ ≤ M6 (k1 + k2 ), C˜δ1 [u(z), D]

(3.28)

in which β = min(α, 1 − 2/p0 ), X(z) = |z + 1|2η1 |z − 1|2η2 , M5 = M5 (p0 , β, δ, k, D), M6 = M6 (p0 , β, δ, k0 , D) are two non-negative constants. Proof We first verify that any solution u(z) of Problem P for (3.2) satisfies the estimate ¯ ≤ M7 = M7 (p0 , α, k, D). ¯ + C[|X(z)|1/β uz , D] S(u) = C[u(z), D]

(3.29)

102

III. Elliptic Complex Equations

Otherwise, if the above inequality is not true, there exist sequences of coefficients: m m m {Qm }, {Am j }(j = 1, 2, 3), {λ }, {r }, {bj }(j = 1, 2) satisfying the same conditions of Q, Aj (j = 1, 2, 3), λ, r, bj (j = 0, 1), and {Qm }, {Am j }(j = 1, 2, 3) weakly converge in D to Q0 , A0j (j = 1, 2, 3), and {λm }, {rm }, {bm }(j = 0, 1) uniformly converge j on Γ∗ to λ0 , r0 , b0j (j = 0, 1) respectively. Let um is a solution of Problem P for m m m (3.2) corresponding to {Qm }, {Am j } (j = 1, 2, 3), {λ }, {r }, {bj }(j = 0, 1), but maxD¯ |um (z)| = Hm → ∞ as m → ∞. There is no harm in assuming that Hm ≥ 1. Let U m = um /Hm . It is clear that U m (z) is a solution of the boundary value problem m m m m Uzmz¯ − Re [Qm Uzz + Am = 1 Uz ] − A2 U

∂U m rm (z) = , z ∈ Γ∗ , ∂νm Hm

U m (−1) =

Am 3 , Hm

bm bm 0 , U m (1) = 1 . Hm Hm

From the conditions in the theorem, we have m Lp [Am 2 U +





Cα |z − zj |βj

Am 3 ¯ ≤ M7 , C[λj , Γj ] ≤ M7 , , D] Hm

rm (z) ∗ , Γ ≤ M7 , j = 1, . . . , m, Hm

 m  b   j     Hm 

≤ M7 , j = 0, 1,

where M7 = M7 (q0 , p0 , α, K, D) is a non-negative constant. According to the method in the proof of Theorem 2.3, we denote 

wm = Uzm , U m (z) = 2Re

z

−1

wm (z)dz +

bm 0 , Hm

and can obtain that Um (z) satisfies the estimate ¯ + Cδ [|X(z)|1/β U m , D] ¯ ≤ M8 , Cβ [U m (z), D] z

(3.30)

in which M8 = M8 (q0 , p0 , δ, α, K, D), δ (> 0) are non-negative constants. Hence from {U m (z)} and {|X(z)|1/β Uzm }, we can choose subsequences {U mk (z)} and ¯ respect{|X(z)|1/β Uzmk }, which uniformly converge to U 0 (z) and |X(z)|1/β Uz0 in D ively, and U 0 (z) is a solution of the following boundary value problem 0 Uz0z¯ = Re [Q0 Uzz + A01 uz ] + A02 U 0 = 0 in D,

∂U 0 = 0 on Γ∗ , ∂ν

U 0 (−1) = 0,

U 0 (1) = 0.

By the result as stated before, we see that the solution U 0 (z) = 0. However, from S(U m ) = 1, the inequality S(U 0 ) > 0 can be derived. Hence the estimate (3.29) is true. Moreover, by using the method from S(U m ) = 1 to (3.30), we can prove the first estimate in (3.28). The second estimate in (3.28) can be derived from the first one.

3. Elliptic Equations of Second Order

103

3.3 Existence of solutions of the discontinuous oblique derivative problem for elliptic equations in the upper half-unit disk Theorem 3.6 solvable.

If equation (3.2) satisfies Condition C, then Problem P for (3.2) is

Proof Noting that the index K = 0, we introduce the boundary value problem Pt for the linear elliptic equation with a parameter t(0 ≤ t ≤ 1): Lu = uzz¯ − Re [Quzz + A1 (z)uz ] = G(z, u),

G = tA2 (z)u + A(z)

(3.31)

¯ and the boundary condition (3.5). It is evident that when for any A(z) ∈ Lp0 (D) t = 1, A(z) = A3 (z), Problem Pt is just Problem P. When t = 0, the equation in (3.31) is Lu = uzz¯ −Re [Quzz +A1 uz ] = A(z),

i.e. wz¯ −Re [Qwz +A1 w] = A(z),

(3.32)

where w = uz . By Theorem 3.7 below, we see that Problem P for the first equation in (3.32) has a unique solution u0 (z), which is just a solution of Problem P for equation (3.31) with t = 0. Suppose that when t = t0 (0 ≤ t0 < 1), Problem Pt0 is solvable, ¯ i.e. Problem Pt for (3.31) has a unique solution u(z) such that |X(z)|1/β uz ∈ Cδ (D). We can find a neighborhood Tε = {|t − t0 | < ε, 0 ≤ t ≤ 1, ε > 0} of t0 , such that for every t ∈ Tε , Problem Pt is solvable. In fact, Problem Pt can be written in the form Lu−t0 [G(z, u)−G(z, 0)] = (t−t0 )[G(z, u)−G(z, 0)]+A(z),

z∈D

(3.33)

and (3.5). Replacing u(z) in the right-hand side of (3.33) by a function u0 (z) with ¯ especially, by u0 (z) = 0, it is obvious that the the condition |X(z)|1/β u0z ∈ Cδ (D), boundary value problem (3.33),(3.5) then has a unique solution u1 (z) satisfying the ¯ Using successive iteration, we obtain a sequence of conditions |X(z)|1/β u1z ∈ Cδ (D). ¯ solutions: {un (z)} satisfying the conditions |X(z)|1/β unz ∈ Cδ (D)(n = 1, 2, . . .) and Lun+1 −t0 [G(z, un+1 )−G(z, 0)] = (t−t0 )[G(z, un )−G(z, 0)]+A(z), Re [λ(z)un+1z ] = r(z),

z ∈ Γ,

un+1 (−1) = b0 ,

un+1 (1) = b1 ,

z ∈ D,

n = 1, 2, . . . .

From the above formulas, it follows that L(un+1 − un )z¯ − t0 [G(z, un+1 ) − G(z, un )] = (t − t0 )[G(z, un ) − G(z, un−1 )], Re [λ(z)(un+1z − unz )] = 0, un+1 (−1) − un (−1) = 0,

z ∈ D,

z ∈ Γ,

(3.34)

un+1 (1) − un (1) = 0.

Noting that ¯ ≤ |t − t0 |k0 C˜ 1 [un − un−1 , D], ¯ Lp [(t − t0 )(G(z, un ) − G(z, un−1 )), D] δ

(3.35)

104

III. Elliptic Complex Equations

¯ = Cβ [un − un−1 , D] ¯ + Cδ [|X(z)|1/β (unz − un−1z ), D], ¯ and where C˜δ1 [un − un−1 , D] applying Theorem 3.5, we get ¯ ≤ |t − t0 |M6 C˜ 1 [un − un−1 , D]. ¯ C˜δ1 [un+1 − un , D] δ

(3.36)

Choosing the constant ε so small that 2εM6 < 1, it follows that ¯ ¯ ≤ C˜ 1 un − un−1 , D , C˜δ1 [un+1 − un , D] δ 2

(3.37)

and when n, m ≥ N0 + 1 (N0 is a positive integer), ¯ ≤ 2−N0 C˜δ1 [un+1 −un , D]

∞ 

¯ ≤ 2−N0 +1 C˜ 1 [u1 −u0 , D]. ¯ 2−j C˜δ1 [u1 −u0 , D] δ

(3.38)

j=0

Hence {un (z)} is a Cauchy sequence. According to the completeness of the Banach ¯ there exists a function u∗ (z) ∈ C˜ 1 (D), ¯ so that C˜ 1 [un − u∗ , D] ¯ → 0 for space C˜δ1 (D), δ δ n → ∞. From (3.38), we can see that u∗ (z) is a solution of Problem Pt for every t ∈ Tε = {|t − t0 | ≤ ε}. Because the constant ε is independent of t0 (0 ≤ t0 < 1), therefore from the solvability of Problem Pt when t = 0, we can derive the solvability of Problem Pt when t = ε, 2ε, . . . , [1/ε] ε, 1. In particular, when t = 1 and A(z) = A3 (z), Problem P1 for the linear case of equation (3.2) is solvable. Next, we discuss the quasilinear equation (3.2) satisfying Condition C, but we first ¯ dist(z, Γ) < assume that the coefficients Q = 0, Aj (j = 1, 2, 3) = 0 in Dm = {z ∈ D, 1/m}, here m(≥ 2) is a positive integer, namely consider m m uzz¯ = Re [Qm uzz + Am 1 uz ] + A2 u + A3 in D,

(3.39)

where m

⎧ ⎨Q(z, u, uz ),

Q =⎩ 0,

Am j

⎧ ⎨Aj (z, u, uz )

=⎩ 0

in

⎧ ˜m ⎨D ⎩

⎫

= D\Dm⎬ Dm

⎭

, j = 1, 2, 3.

Now, we introduce a bounded, closed and convex set BM in the Banach space B = ¯ any element of which satisfies the inequality C˜δ1 (D), ¯ ≤ M5 , C˜δ1 [u(z), D]

(3.40)

where M5 is a non-negative constant as stated in (3.28). We are free to choose an arbitrary function U (z) ∈ BM and insert it into the coefficients of equation (3.39). It is clear that the equation can be seen as a linear equation, hence there exists a unique solution u(z) of Problem P , and by Theorem 3.5, we see u(z) ∈ BM . Denote by u(z) = S[U (z)] the mapping from U (z) ∈ BM to u(z), obviously u(z) = S[U (z)] maps BM onto a compact subset of itself. It remains to verify that u(z) = S[U (z)] continuously maps the set BM onto a compact subset. In fact, we arbitrarily select ¯ → 0 as n → ∞. a sequence of functions: {Un (z)}, such that C˜δ1 [Un (z) − U0 (z), D]

3. Elliptic Equations of Second Order

105

Setting un (z) = S[Un (z)], and subtracting u0 (z) = S[U0 (z)] from un (z) = S[Un (z)], we obtain the equation for u˜n = un (z) − u0 (z): u˜n¯z −Re [Qm (z, Un , Unz )˜ unzz +Am unz ]−Am un = Cn , 1 (z, Un , Unz )˜ 2 (z, Un , Unz )˜ ˜m ˜m ˜m Cn = Cn (z, Un , U0 , u0 ) = A˜m 3 − Re [Q u0zz + A1 u0z ] − A2 u0 ,

(3.41)

˜ m = Qm (z, Un , Unz )−Qm (z, U0 , U0z ), A˜m = Am (z, Un , Unz )−Am (z, U0 , U0z ), in which Q j j j j = 1, 2, 3, and the solution u˜n (z) satisfies the homogeneous boundary conditions Re [λ(z)uz ] = 0,

z ∈ Γ∗ = Γ\Z,

u(−1) = 0,

u(1) = 0.

(3.42)

Noting that the function Cn = 0 in Dm , according to the method in the formula (2.43), Chapter II, [86]1), we can prove that ¯ → 0 as n → ∞. Lp [Cn , D] On the basis of the second estimate in (3.28), we obtain ¯ ≤ M6 Lp [Cn , D], ¯ C˜δ1 [un (z) − u0 (z), D]

(3.43)

¯ → 0 as n → ∞. This shows that u(z) = S[U (z)] in the thus C˜δ1 [un (z) − u0 (z), D] set BM is a continuous mapping. Hence by the Schauder fixed-point theorem, there exists a function u(z) ∈ BM , such that u(z) = S[u(z)], and the function u(z) is just a solution of Problem P for the quasilinear equation (3.39). Finally we cancel the conditions: the coefficients Q = 0, Aj (j = 1, 2, 3) = 0 in Dm = {z, dist(z, Γ) < 1/m}. Denote by um (z) a solution of Problem P for equation (3.39). By Theorem 3.5, we see that the solution satisfies the estimate (3.28). Hence from the sequence of solutions: um (z), m = 2, 3, . . ., we can choose a subsequence {umk (z)}, for convenience denote {umk (z)} by {um (z)} again, which uniformly con¯ and u0 (z) satisfies the boundary condition (3.5) of verges to a function u0 (z) in D, Problem P . At last, we need to verify that the function u0 (z) is a solution of equation (3.2). Construct a twice continuously differentiable function gn (z) as follows gn (z) =

⎧ ⎨ 1,

˜ n = D\D ¯ n, z∈D

⎩ 0,

z ∈ D2n ,

0 ≤ gn (z) ≤ 1 in Dn \D2n ,

(3.44)

where n( ≥ 2) is a positive integer. It is not difficult to see that the function um n (z) = gn (z)um (z) is a solution of the following Dirichlet boundary value problem m m m um nz z¯ − Re [Q unzz ] = Cn in D,

(3.45)

um n (z) = 0 on Γ,

(3.46)

where m m m m m m m m Cnm = gn [Re (Am 1 uz )+A2 u ]+u [gnz z¯−Re (Q gnzz )]+2Re [gnz uz¯ −Q gnz uz ]. (3.47)

106

III. Elliptic Complex Equations

By using the method from the proof of Theorem 3.5, we can obtain the estimates of m ˜ um n (z, t) = u (z, t) in Dn , namely ˜ Cβ1 [um n , Dn ] ≤ M 9 ,

 um n Wp2

0

˜ n)≤ (D

M10 ,

(3.48)

where β = min(α, 1 − 2/p0 ), 2 < p0 ≤ p, Mj = Mj (q0 , p0 , α, k0 , k1 , Mn
, gn , Dn ), j = ˜ 2n ]. Hence from {um (z)}, we can choose a sub9, 10, here Mn
= max1≤m<∞ C 1,0 [um , D n sequence {unm (z)}, such that {unm (z)}, {unmz (z)} uniformly converge to u0 (z), u0z (z) ˜ n , respectively. and {unmzz (z)}, {unmzz¯(z)} weakly converge to u0zz (z), u0zz¯(z) in D m ˜ 2 , {um (z)} has a subsequence For instance, we take n = 2, um (z) = u (z) in D 2 2 ˜ 2 , the limit function of which is u0 (z) in D ˜ 2 . Next, we take n = 3, {um2 (z)} in D ˜ from {um 3 (z)} we can select a subsequence {um3 (z)} in D3 , the limit function is u0 (z) ˜ 3 . Similarly, from {um (z)}(n > 3), we can choose a subsequence {umn (z)} in in D n ˜ n . Finally from {umn (z)} in D ˜ n , we choose ˜ n and the limit of which is u0 (z) in D D the diagonal sequence {umm (z)} (m = 2, 3, 4, . . .), such that {umm (z)}, {ummz (z)} uniformly converge to u0 (z), u0z (z) and {ummzz (z)}, {ummzz¯(z)} weakly converge to u0zz (z), u0zz¯(z) in any closed subset of D respectively, the limit function u(z) = u0 (z) is just a solution of equation (3.2) in D. This completes the proof. Theorem 3.7 If equation (3.2) with A2 (z) = 0 satisfies Condition C, then Problem Q for (3.2) has a unique solution. Proof By Theorem 2.3, we choose D
= D, n = m = 2, z1 = −1, z2 = 1 and K = 0, the second linear equation in (3.32) with A(z) = A3 (z) has a unique solution w0 (z), and the function  z w0 (z)dz + b0 (3.49) u0 (z) = 2 Re −1

is a solution of Problem Q for the first linear equation in (3.32). If u0 (1) = b
= b1 , then the solution is just a solution of Problem P for the linear equation (3.2) with A2 (z) = 0. Otherwise, u0 (1) = b

= b1 , we find a solution u1 (z) of Problem Q with the boundary conditions Re [λ(z)u1z ] = 0 on Γ,

Im [λ(z)u1z ]|z=0 = 1,

u1 (−1) = 0.

On the basis of Theorem 3.4, it is clear that u1 (1)
= 0, hence there exists a real constant d
= 0, such that b1 = b
+ du1 (1), thus u(z) = u0 (z) + du1 (z) is just a solution of Problem P for the linear equation (3.2) with A2 (z) = 0. As for the quasilinear equation (3.2) with A2 = 0, the existence of solutions of Problem Q and Problem P can be proved by the method as stated in the proof of last theorem.

3.4 The discontinuous oblique derivative problem for elliptic equations in general domains In this subsection, let D
be a general simply connected domain, whose boundary Γ
= Γ
1 ∪ Γ
2 , herein Γ
1 , Γ
2 ∈ Cα2 (1/2 < α < 1) have two intersection points z
, z

with

3. Elliptic Equations of Second Order

107

the inner angles α
π, α

π (0 < α
, α

< 1), respectively. We discuss the quasilinear uniformly elliptic equation uzz¯ = F (z, u, uz , uzz ),

z ∈ D
,

F = Re [Quzz + A1 uz ] + A1 u + A3 ,

(3.50)

in which F (z, u, uz , uzz ) satisfy Condition C in D
. There are m points Z = {z1 = z
. . . , zn−1 , zn = z

, . . . , zm } on Γ
arranged according to the positive direction successively. Denote by Γ
j the curve on Γ
from zj−1 to zj , j = 1, 2, . . . , m, z0 = zm , and Γ
j (j = 1, 2, . . . , m) does not include the end points. Problem P
The discontinuous oblique derivative boundary value problem for (3.50) is to find a continuous solution w(z) in D∗ = D
\Z satisfying the boundary condition: 1 ∂u = Re [λ(z)uz ] = c(z), 2 ∂ν

z ∈ Γ∗ = Γ
\Z,

u(zj ) = bj ,

j = 1, . . . , m,

(3.51)

where cos(ν, n) ≥ 0, λ(z), c(z) are given functions satisfying Cα [λ(z), Γj ] ≤ k0 ,

Cα [|z − zj |βj c(z), Γj ] ≤ k2 ,

|bj | ≤ k2 ,

j = 1, . . . , m, (3.52)

herein α (1/2 < α < 1), k0 , k2 are non-negative constants, and assume that (βj + γj )/β < 1, β = min(α, 1 − 2/p0 )/α0 , βj (j = 1, . . . , m) are as stated in (1.27), α0 = max(1/α1 , 1/α2 , 1). Problem P
with A3 (z) = 0 in D, r(z) = 0 on Γ
is called Problem P0
. If cos(ν, n)
≡ 0 on each of Γj (j = 1, . . . , m), we choose the index K = m/2 − 1 of Problem P
, which is defined as that in Subsection 2.3. If A2 = 0 in D, the last point conditions in (3.51) can be replaced by u(zn ) = bn ,

Im [λ(z)uz ]|zj
= bj ,

j = 1, . . . , n − 1, n + 1, . . . , m.

(3.53)

Here zj
(
∈ Z, j = 1, . . . , n − 1, n + 1, . . . , m) ∈ Γ
are distinct points and the condition cos(ν, n) ≥ 0 on Γ
can be canceled. This boundary value problem is called Problem Q
. Applying a similar method as before, we can prove the following theorem. Theorem 3.8 Let equation (3.50) in D
satisfy Condition C similar to before. Then Problem P
and Problem Q
for (3.50) are solvable, and the solution u(z) can be expressed by (3.17), but where β = min(α, 1 − 2/p0 )/α0 . Moreover, if Q(z) = 0 in D, then the solution u(z) of equation (3.50) possesses the form in (3.17), where w(z) = Φ(z)eφ(z) + ψ(z) and u(z) satisfies the estimate ¯ = Cδ [u(z), D] ¯ + Cδ [X(z)w(z), D] ¯ ≤ M11 = M11 (p0 , β, δ, k, D), C˜δ1 [u, D]

(3.54)

in which X(z) is given as X(z) =

m  j=2,j=n








|z − zj |ηj |z − z1 |η1 /α |z − zn |ηn /α ,

(3.55)

108

III. Elliptic Complex Equations

where ηj (j = 1, . . . , m) are as stated in (2.30). Besides the solution of Problem P
and Problem Q
for (3.50) are unique, if the following condition holds: For any real functions uj (z) ∈ C 1 (D∗ ), Vj (z) ∈ Lp0 (D)(j = 1, 2), the equality ˜ 1 −V2 )+ A˜1 (u1 −u2 )z ]+ A˜2 (u1 −u2 ) in D
, F (z, u1 , u1z , V1 )−F (z, u1 , u1z , V2 ) = Re [Q(V ˜ ≤ q0 in D
, A˜1 , A˜2 ∈ Lp0 (D
). holds, where |Q| Finally we mention that the above results can be generalized to the case of second order nonlinear elliptic equation in the form uzz¯ = F (z, u, uz , uzz ), F = Re [Q(z, u, uz , uzz ) + A1 (z, u, uz )uz ] + A2 (z, u, uz )u + A3 (z, u, uz ) in D satisfying the conditions similar to Condition C, which is the complex form of the second order nonlinear elliptic equation Φ(x, y, u, ux , uy , uxx , uxy , uyy ) = 0 in D with certain conditions (see [86]1)).

4

Boundary Value Problems for Degenerate Elliptic Equations of Second Order in a Simply Connected Domain

This section deals with the oblique derivative problem for the degenerate elliptic equation of second order in a simply connected domain. We first give a boundedness estimate of solutions of the oblique derivative problem for the equation, and then by using the principle of compactness, the existence of solutions for the above oblique derivative problem is proved.

4.1 Formulation of boundary value problems for degenerate elliptic equations Let D be a simply connected domain with the boundary Γ
= Γ ∪ L0 , where Γ ∈ Cα2 (0 < α < 1) in the upper half plane with the end points −1, 1 and L0 = [−1, 1], and the inner angles of D at −1, 1 equal α1 π, α2 π, herein 0 < α1 , α2 < 1. We consider the elliptic equation of second order Lu = y m uxx + uyy + a(x, y)ux + b(x, y)uy + c(x, y)u = d(x, y) in D,

(4.1)

here m is a positive number. Its complex form is as follows uzz¯ − Re [Q(z)uzz + A1 (z)uz ] + A2 u = A3 in D,

(4.2)

4. Degenerate Elliptic Equations

109

where Q(z) =

1 − ym , 1 + ym

A1 (z) = −

a + bi , 1 + ym

A2 (z) = −

c , 2(1 + y m )

A3 (z) =

d . 2(1 + y m )

Suppose that equation (4.2) satisfies the following conditions. Condition C ¯ and satisfy The coefficients Aj (z)(j = 1, 2) are continuously differentiable in D ¯ ≤ k0 , Cα1 [Aj (z), D]

j = 1, 2, 3,

A2 = −

c c ≥ − ≥ 0 on D, 2(1 + y m ) 4

(4.3)

in which α(0 < α < 1), k0 are non-negative constants. The oblique derivative boundary value problem is as follows. Problem P In the domain D, find a solution u(z) of equation (4.1), which is ¯ and satisfies the boundary condition continuously differentiable in D, lu =

1 ∂u +σ(z)u = φ(z), 2 ∂ν

z ∈ Γ,

∂u = ψ(x), x ∈ L0 , ∂y

(4.4)

u(−1) = b0 , u(1) = b1 , where ν is any unit vector at every point on Γ, cos(ν, n) ≥ 0, σ(z) ≥ σ0 > 0, n is the unit outer normal at every point on Γ, λ(z) = cos(ν, x) − i cos(ν, y), and λ(z), φ(z), ψ(x) are known functions and b0 , b1 are known constants satisfying the conditions Cα1 [η, Γ] ≤ k0 , η = λ, σ,

Cα1 [φ, L0 ] ≤ k0 ,

Cα1 [ψ, L0 ] ≤ k0 ,

|b0 |, |b1 | ≤ k0 ,

(4.5)

in which α (1/2 < α < 1), k0 , σ0 are non-negative constants. Problem P with the conditions: A3 (z) = 0 in D, φ(z) = 0 on Γ, ψ(z) = 0 on L0 and b0 = b1 = 0 is called Problem P0 . If cos(ν, n) = 1, here n is a outward normal vector on Γ, then Problem P is the Neumann boundary value problem (Problem N ), and if cos(ν, n) > 0, σ(z) = 0 on Γ, then Problem P is the regular oblique derivative problem, i.e. third boundary value problem (Problem O), in this case we choose σ(z) > 0 on Γ. If cos(ν, n) = 0 and σ(z) = 0 on Γ, then from (4.4), we can derive 

u(z) = 2 Re



z

−1

uz dz + b0 = r(z) on Γ, u(1) = b1 = 2 Re

1

−1

uz dz + b0 .

(4.6)

In this case, Problem P is called Problem D. In the following, there is no harm in assuming d(z) = 0 in (4.1). 4.2

A priori estimates of solutions for Problem P for (4.1)

First of all, we give a lemma and then give a priori estimate of boundedness of solutions of Problem P for (4.1).

110

III. Elliptic Complex Equations

Lemma 4.1 Suppose that equation (4.1) or (4.2) satisfies Condition C and Lu ≥ ¯ of (4.1) attains its positive 0 (or Lu ≤ 0) in D, if the solution u(z) ∈ C 2 (D) ∩ C(D) maximum (or negative minimum ) at a point x0 ∈ (−1, 1), and maxΓ u(z) < u(x0 ) (or minΓ u(z) > u(x0 )) on Γ, then lim

y→0

∂u(x0 , y) ∂u(x0 , y) < 0 (or lim > 0), y→0 ∂y ∂y

(4.7)

if the limit exists. Proof

Assume that the first inequality is not true, namely lim

y→0

∂u(x0 , y) = M
≥ 0. ∂y

(4.8)

Obviously M
= 0. Denote M = u(x0 ), B = maxD¯ |b(z)| and by d the diameter of D. Thus there exists a small positive constant ε < M such that maxΓ u(z) ≤ M − ε. Making a function εu(z) , v(z) = (M eBd − εeBy ) we have v(z) ≤

ε(M − ε) εM on Γ, < M eBd −εeBd M eBd −ε

v(x) ≤ v(x0 ) =

εM on L0 . M eBd −ε

(4.9)

Noting that Lu ≥ 0, the function v(x, y) satisfies the inequality y m vxx + vyy + a(x, y)vx + ˜b(x, y)vy + c˜(x, y)v ≥ 0 in D, where ˜b = b − 2εBeBy /(M eBd − εeBy ), c˜(x, y) = c − ε(B + b)BeBy /(M eBd − εeBy ) ≤ 0 in D. According to the above assumption, we get ε2 BM ∂v(x0 , y) = > 0. y→0 ∂y (M eBd − ε)2 lim

Hence v(x, y) attains its maximum in D, but from (4.9), it is impossible. This proves the first inequality in (4.7). Similarly we can prove the second inequality in (4.7). Now we choose a positive constant η < 1, and consider the equation Lη u = (y+η)m uxx +uyy +a(x, y)ux +b(x, y)uy +c(x, y)u = d in D.

(4.10)

It is easy to see that (4.10) is a uniformly elliptic equation in D. From Theorem 3.6, we can derive that for every one of η = 1/n > 0 (n = 2, 3, . . .), there exists a solution un (z) of Problem D for equation (4.10). In the following, we shall give some estimates of the solution un (z). Lemma 4.2 If Condition C holds, then any solution un (z) of Problem P for (4.10) with d = 0 satisfies the estimate ¯ ≤ M12 = M12 (α, k0 , D), C[un (z), D]

(4.11)

4. Degenerate Elliptic Equations

111

where M12 is a non-negative constant. Proof that

We first discuss Problem D and choose two positive constants c1 , c2 such c1 ≥ c2 +max |r(z)|+max ec2 y ,

c2 > max |ψ(x)|+max |b|+2 max |d|+1,

¯ D

Γ

¯ D

L0

¯ D

and make a transformation of function v(z) = c1 − ec2 y ± un (z), thus we have Lη v ≤ −c2 (c2 + b)ec2 y + c(c1 − ec2 y ) + 2 max |d| < 0, ¯ D

v > 0 on Γ,

vy =

z = x + iy ∈ D,

∂v = −c2 ec2 y ± ψ(z) < 0 on L0 , ∂y

(4.12)

by the extremum principle for elliptic equations, the function v(z) cannot take the negative minimum in D, hence v(z) = c1 − ec2 y ± un (z) ≥ 0,

¯ i.e. c1 ≥ ec2 y ∓ un (z) in D,

(4.13)

hence |un (z)| ≤ c1 − ec2 y ≤ c1 = M12 . For other case, we introduce an auxiliary function v(z) = c1 − ec2 y ± un (z), where c1 , c2 are two positive constants satisfying the conditions c2 > max |b(z)| + max |ψ(x)| + max ec2 y + 2 max |d|, ¯ D

L0

c1 > c2 + max ec2 y ¯ D



¯ D

¯ D

|φ(z)| c2 1+ +max . Γ σ0 σ0

(4.14)

We can verify that the function v(z) satisfies the conditions Lη v < 0 in D,

lv > 0 on Γ,

vy < 0 on L0 ,

(4.15)

hence v(z) cannot attain the negative minimum in D. Thus |un (z)| ≤ c1 − ec2 y ≤ c1 = M12 . This completes the proof. Secondly from the sequence of solutions: {un (z)} of Problem P for equation (4.10), we can choose a subsequence {unk (z)}, which uniformly converges to a solution u∗ (z) ¯ of (4.1) in any closed subset of D\L. In fact by Condition C and the estimate (4.11), we can derive the estimate of the solution un (z) as follows Cβ1 [un (z), D] ≤ M13 = M13 (β, k0 , D, η),

(4.16)

where η = 1/n > 0 and β (0 < β ≤ α) is a constant. Lemma 4.3 If Condition C holds, then any solution un (z) of Problem P for (4.10) satisfies the estimate (4.16). From the above lemma, we can derive that the limit function u∗ (z) of {unk (z)} satisfies the first boundary condition in (4.4). In order to prove that u∗ (z) satisfies

112

III. Elliptic Complex Equations

the second boundary condition in (4.4), we write the similar results in [24]1) as a lemma. Lemma 4.4

Suppose that Condition C holds and 0 < m < 2, or m ≥ 2, a(x, y) = O(y m/2−1+ε ),

ay = O(y m/2−2+ε ),

(4.17)

where ε is sufficiently small positive number. Then any solution um (z) of Problem P for (4.10) with d = 0 satisfies the estimate |uny |, |((y + η)m+ε − η m+ε )u2nx | ≤ M14 = M14 (α, k0 , D) in Rn,2δ0 ,

(4.18)

where Rn,δ = {|x − x0 | < ρ − δ, 0 < y < δl }(Rn,δ ⊂ D), x0 ∈ (−1, 1), δ0 , δ1 , δ, ρ (0 < δ ≤ 2δ0 < ρ, δ1 < 1/n) are small positive constants, and M14 is a non-negative constant. Proof

(1) First of all, we prove the estimate (uy )2 ≤ [(y + η)m+ε1 − η m+ε1 ](ux )2 + M15 in Rn,δ0 = D∗ ,

(4.19)

in which ε1 (< ε), M15 are non-negative constants. f = f (x) = X 4 = [ρ2 − (x − x0 )2 ]4 , g = g(x) = X 2 = [ρ2 − (x − x0 )2 ]2 are functions of x and F = η m+ε1 − Y m+ε1 , G = 1 − Y ε1 , H = −Y ε1 are functions of Y = y + η, and introducing an auxiliary function v(z) = f [F (ux )2 + G(uy )2 ] + gu2 + H in D∗ ,

(4.20)

if v(z) attains a positive maximum value at a point z ∗ ∈ D∗ , then ˜ η (v) = Lη (v) + cv ≤ 0 at z ∗ . v(z) > 0, vx = vy = 0, L

(4.21)

From (4.20), we get vx = 2f [F ux uxx + Guy uxy ] + 2guux + f
[F (ux )2 + G(uy )2 ] + g
u2 = 0, vy = 2f [F ux uxy + Guy uyy ] + 2guuy + f [F
(ux )2 + G
(uy )2 ] + H
= 0, ˜ η (v) = 2f [F ux Lη (ux )+Guy Lη (uy )]+2guLη (u)+2f F [Y m (uxx )2 +(uxy )2 ] L m

2

2

m

2

2

(4.22)

m


+2f G[Y (uxy ) +(uyy ) ]+2g[Y (ux ) +(uy ) ]+4Y f [F ux uxx +Guy uxy ] +4f [F
ux uxy +G
uy uyy ]+Y m f

[F (ux )2 +G(uy )2 ]+f [F

(ux )2 +G

(uy )2 ] +4Y m g
uux +Y m g

u2 +H

+af
[F (ux )2 +G(uy )2 ] +ag
u2 +bf [F
(ux )2 +G
(uy )2 ]+bH
+2cH, in which Y = y + η, and from (4.10), we obtain Lη (ux ) = −(ax ux + bx uy + cx u), Lη (uy ) = −(mY m−1 uxx + ay ux + by uy + cy u), m

2

2f F Y (uxx ) = 2f F Y

−m

(4.23) 2

(uyy + aux + buy + cu) ,

4. Degenerate Elliptic Equations

113

and then we have m 2mf F [2guuy + f [F
(ux )2 + G
(uy )2 ] + H
] − (4.24) Y Y 2mf G uy (aux + buy + cu) − 2f Guy (ay ux + by uy + cy u). ×ux uxy + Y

2f Guy Lη (uy ) = −

˜ η (v), it is not difficult to derive Substituting (4.23),(4.24) into L 1˜ mF )ux uxy Lη (v) = 2(F Y −m + G)(Y m (uxy )2 + (uyy )2 ) + 2(2F
− f Y +4[G
uy + F Y −m (aux + buy + cu)]uyy − 2F ux (ax ux + bx uy + cx u) 2mG uy (aux + buy + cu) −2Guy (ay ux + by uy + cy u) + Y    f
2 f

f
−2 2 +a +2F Y −m (aux + buy + cu)2 + Y m (4.25) f f f

 m ×[F (ux )2 + G(uy )2 ] + − + b [F
(ux )2 + G
(uy )2 ] + F

(ux )2 Y  
f
g g mg

2 m 2 2 m g − uuy +G (uy ) +2 [Y (ux ) +(uy ) ]+4Y uux −2 f f f2 fY     f
g
g

g
2 H

+ (−m/Y + b)H
+ 2cH −2 2 +a . + Ym u + f f f f Moreover by (4.11),(4.17),(4.20) and (4.25), we obtain


1˜ Lη (v) = (2+o(1))[Y m (uxy )2 +(uyy )2 ]+O Y m+ε1 −1 |ux uxy | f +Y m−2+2ε+ε1 |ux |2 +(Y m/2−1+ε+ε1 |ux |+Y ε1 −1 |uy |+Y ε1 )|uyy | +Y m/2−2+ε |ux uy |+Y −1 (uy )2 +Y m/2−1+ε+ε1 |ux |+Y −1 |uy |+Y ε1 





Y m Y m/2−1+ε + + [Y m+ε1 (ux )2 +(uy )2 ] X2 X +(1−ε1 )(m+ε1 )2 (1+o(1))Y m−2+ε1 (ux )2 +ε1 (m+1−ε1 )(1+o(1))Y ε1 −2 (uy )2 

(4.26)



Ym Ym (uy )2 Y −1 +2 2 (ux )2 +2 2 +O |u |+ |uy | x X X X3 X2 ε1 (m+1−ε1 )(1+o(1)) ε1 −2 + Y . X4

When 0 < ε1 < min(ε,1), it is easy to see that the right-hand side of (4.26) is positive, which contradicts (4.21), hence v(z) cannot have a positive maximum in D∗ . On the basis of the estimate (4.11), we see that v(z) on the upper boundary {|x− 2 x0 | < ρ,y = 1/n} of D∗ is bounded, and v(x) ≤ f Gψ 2 +gM12 on the lower boundary {|x−x0 | < ρ,y = 0} of D∗ , moreover v(x) < 0 on the left-hand side and right-hand side {|x−x0 | = ρ,0 < y < 1/n} of D∗ . Thus the estimate (4.19) is derived.

114

III. Elliptic Complex Equations

(2) Now we give the estimate |((y +η)m+ε −η m+ε )u2nx | ≤ M16 ∈ Rn,δ0 = D∗ ,

(4.27)

in which M16 is independent of η. In fact, we introduce the auxiliary function (4.20), where we choose that f = f (x) = X 4 = [(ρ−δ0 )2 −(x−x0 )2 ]4 ,g = g(x) = X 2 = [(ρ−δ0 )2 −(x−x0 )2 ]2 , and F = Y m+ε1 −η m+ε1 , G = Y ε2 , H = Y ε3 , herein Y = y +η, ε2 ,ε3 are positive constants satisfying 0 < 2ε3 < ε2 ≤ ε1 /2. If v(z) attains a positive maximum value at a point z ∗ ∈ D∗ , then we have (4.21). Substituting (4.10),(4.20),(4.23) into (4.22), we get ˜ η (v) L = 2F [Y m (uxx )2 +(uxy )2 ]+2G[Y m (uxy )2 f g +(uyy )2 ]+2 [Y m (ux )2 +(uy )2 ]+Σ, f

(4.28)

in which



mG uy uyy −2F ux (ax ux +bx uy +cx u) Y mG uy (aux +buy +cu) −2Guy (ay ux +by uy +cy u)+2 Y

Σ = 4F
ux uxy +2 2G
+





+ Ym





f
2 f
f

−2 2 +a [F (ux )2 +G(uy )2 ]+b[F
(ux )2 f f f 

+G
(uy )2 ]+F

(ux )2 +G

(uy )2 +4Y m 



+ Ym



(4.29)



g
f
g uux − f f2



f
g
g

g
H

+bH
+2cH −2 2 +a u2 + . f f f f

From (4.11),(4.17),(4.20) and (4.29), it follows that


Σ = O Y m+ε1 −1 |ux uxy |+Y ε2 −1 |uy uyy |+Y m+ε1 −2 |ux |2 +Y m/2−2+ε+ε2 

×|ux uy |+Y ε2 −2 |uy |2 +Y m+ε1 |ux |+Y ε2 −1 |uy |+

Y m Y m/2−1+ε + X2 X





×(Y

m+ε1

Ym Y ε3 −2 |ux | +Y |uy | )+ 3 |ux |+ ≥−G(Y m |uxy |2 +|uyy |2 ) X X4 2

ε2

(4.30)

2





g Y ε3 −2 −2 (Y m |ux |2 +|uy |2 )+O Y m+ε1 −2 |ux |2 +Y ε2 −2 |uy |2 + . f X4 By (4.28),(4.30), if we can verify the following inequality  m

2

2

G[Y (uxy ) +(uyy ) ]+O Y



m+ε1 −2

2

|ux | +Y

ε2 −2

Y ε3 −2 |uy | + > 0, X4 2

(4.31)

4. Degenerate Elliptic Equations

115

˜ η (v) > 0. Noting that F
,G
,H
are positive, and from (4.11), then the inequality L (4.20), (4.22), we have g H
2|F ux uxy +Guy uyy | ≥ F
|ux |2 +G
|uy |2 −2 |uuy |+ f f 1 u2 1+o(1)
H. ≥ F
|ux |2 + (H
−
2 ) ≥ F
|ux |2 + f G f Hence

(F 2 |ux |2 +Y m G2 |uy |2 )(Y m |uxy |2 +|uyy |2 ) = Y m (F ux uxy +Guy uyy )2 +(Y m Guy uxy −F ux uyy )2

(4.32)

Ym
1+o(1)
2 (F |ux |2 + H). ≥ 4 f By (4.19),(4.20),(4.32), we obtain 

(F 2 |ux |2 +Y m G2 |uy |2 ) Y m |uxy |2 +|uyy |2 

1 Y ε3 −2 + O Y m+ε1 −2 |ux |2 +Y ε2 −2 |uy |2 + G X4



(4.33)

Ym ε3 +o(1) ε3 −1 2 (mY m+ε1 −1 |ux |2 + ≥ Y ) 4 X4



+Y

m+ε2

(Y

m+ε1 +ε2

2

|ux | +Y M15 )Y ε2

−ε2

O Y



m+ε1 −2

Y ε2 −2 |ux | + > 0. X4 2

From (4.22), we see that ux ,uy cannot simultaneously be zero. By (4.33), we have ˜ η (v) > 0 holds. This contradicts (4.21). Therefore v(z) cannot (4.31), such that L attain a positive maximum in D∗ . On the basis of (4.11),(4.20) and the boundary condition (4.4), we see that v ≤ 2 f Gψ 2 +gM12 +H on the lower boundary of D∗ is uniformly bounded. Moreover the function v(z) is uniformly bounded on the upper, left-hand and right-hand boundaries of D∗ . Thus the estimates in (4.18) are derived. Now we prove a theorem as follows. ¯ Then any Theorem 4.5 Suppose that Condition C, (4.17) hold and ax +c ≤ 0 in D. solution un (z) of Problem P for (4.10) satisfies the estimate |ux | ≤ M17 = M17 (α,k0 ,D) in D,

(4.34)

where we assume that the inner angles αj π(j = 1,2) of D at z = −1,1 satisfy the conditions 0 < αj (
= 1/2) < 1,j = 1,2 and M17 is a non-negative constant.

116

III. Elliptic Complex Equations

Proof

We find the derivative with respect to x to equation (4.10), and obtain (y+η)m uxxx +uxyy +a(x,y)uxx +b(x,y)uxy +[ax +c(x,y)]ux =F (x,y), F = f (x,y)−bx uy −cx u in D.

(4.35)

On the basis of Lemmas 4.1 and 4.4, we have |F (x,y)| = |f (x,y)−bx uy −cx u| ≤ M18 < ∞ in D, and equation (4.35) can be seen as a elliptic equation of ux , and the solution ux satisfies the boundary conditions ∂u ∂r(z) = cos(s,x)ux +cos(s,y)uy = on Γ, ∂s ∂s

(ux )y = ψ
(x) on L0 ,

(4.36)

in which s is the tangent vector at every point Γ. Noting that the angles αj π(j = 1,2) satisfy the conditions 0 < αj (
= 1/2) < 1,j = 1,2, it is easy to see that cos(s,x)
= 0 at z = −1,1. Thus the first boundary condition in (4.6) can be rewritten in the form ux = R(z) = −

∂r(z) cos(s,y) 1 uy + on Γ, cos(s,x) cos(s,x) ∂s

(4.37)

here R(z) is a bounded function in the neighborhood (⊂ Γ) of z = −1,1, hence by the method in the proof of Lemma 4.2, we can prove that the estimate (4.34) holds. As for cos(s,x) = 0 at z = −1 or z = 1, the problem remains to be solved? Theorem 4.6 Suppose that Condition C and (4.17) hold. Then Problem P for (4.1) or (4.2) has a unique solution. Proof As stated before, for a sequence of positive numbers: η = 1/n, n = 2,3,..., we have a sequence of solutions: {un (z)} of the corresponding equations (4.10) with η = 1/n(n = 2,3,...), which satisfy the estimate (4.16), hence from {un (z)}, we can choose a subsequence {unk (z)}, which converges to a solution u0 (z) of (4.2) in D ∪Γ satisfying the first boundary condition in (4.4). It remains to prove that u0 (z) satisfies the other boundary condition in (4.4). For convenience, we denote unk (z) by u(z),x0 is any point in −1 < x0 < 1, and give a small positive number β, there exists a sufficiently small positive number δ, such that |ψ(x)−ψ(x0 )| < β when |x−x0 | < δ. Moreover we consider an auxiliary function v(z) = F (ux )2 ±uy +G+f, ⎧ m+1+ε 2 ⎨Y −η m+1+ε2 ,

F =⎩

G = −Cy ε2 −τ ∓ψ(x0 ),

f = −C(x−x0 )2 ,

0 < m < 1,

Y m+ε1 −(m+ε1 )η m−1+ε1 Y +(m+ε1 −1)η m+ε1 ,

(4.38) m ≥ 1,

where Y = y +η,η = 1/n,ε2 (0 < ε2 ≤ ε1 /3) are positive constants and C is an undetermined positive constants. We first prove that v(z) cannot attain its positive maximum in D∗ = {|x−x0 |2 +y 2 < σ 2 ,y > 0}. Otherwise there exists a point z ∗ , such that

4. Degenerate Elliptic Equations

117

v(z ∗ ) = maxD∗ v(z) > 0, and then vx = 2F ux uxx ±uxy +f
= 0,

vy = 2F ux uxy ±uyy +F
(ux )2 +G
= 0,

˜ η (v)=2F ux Lη (ux )±Lη (uy )+2F [Y m (uxx )2 +(uxy )2 ]+4F
ux uxy L +[F

+bF
+cF ](ux )2 ±cuy +Y m f



(4.39)

+af
+2cf +G

+bG
+2cG, and from (4.23) and (4.39), we obtain m [2F ux uxy +F
(ux )2 +G
] Y m ± (aux +buy +cu)∓(ay ux +by uy +cy u). Y

±Lη (uy ) = −

(4.40)

Moreover by (4.23) and (4.39)–(4.40), we have ˜ η (v) = −2F ux (ax ux +bx uy +cx u)− m [2F ux uxy +F
(ux )2 +G
L Y ∓(aux +buy +cu)]∓(ay ux +by uy +cy u) +2F [Y m (uxx )2 +(uxy )2 ]∓4F
ux (2F ux uxx +f
) +[F

+bF
+cF ](ux )2 ±cuy +Y m f

+af
+2cf +G

+bG
+2cG

2 m (4.41) ux = 2F Y m [uxx ∓2F
Y −m (ux )2 ]2 +2F uxy − 2Y m




2 −m 2 2 +[F − F +bF +cF −8F Y F (ux ) ](ux ) Y  ma m + −2F (bx uy +cx u)± ∓ay ∓4F
f
ux +G

− G
+bG
Y   Y m2 m 2 (bu +2cG− +2a F (u ) ± +cu)∓(b u x x y y y +cy u) 2Y 2 Y ±cuy +Y m f

+af
+2cf. Choosing a sufficiently small positive number σ, such that the domain D∗ = {|x− x0 |2 +y 2 < σ 2 }∩{y > 0} ⊂ D, σ < min[ρ−2δ0 ,δ1 ], where δ0 ,δ1 are constants as stated in Lemma 4.4, and |ψ(z)−ψ(x0 )| < τ , we can obtain



˜ η (v) ≥ ε2 (m+1+ε2 )(1+o(1))Y m+ε2 −1 (ux )2 +O 1 |ux | L Y

(4.42)

+Cε2 (m+1−ε2 )(1+o(1))y ε2 −2 > 0 if 0 < m < 1, and ˜ η (v) ≥ (m+ε1 )(m+ε1 −1)Y m+ε1 −2 (ux )2 +O(Y m/2+ε−2 )|ux | L +Cε2 (m+1−ε2 )(1+o(1))Y ε2 −2 > 0 if m ≥ 1,

ε = ε1 ,

in which we use Lemmas 4.2 and 4.4, the conditions (4.17), (4.38), (4.41) and F (ux )2 = O(Y 2ε2 ), F
(ux )2 = O(Y −1+2ε2 ) if m ≥ 1.

(4.43)

118

III. Elliptic Complex Equations

It is clear that (4.42),(4.43) contradict (4.21), hence v(z) cannot attain a positive maximum in D∗ . From (4.38), we get v(z) = F (ux )2 ±[uy −ψ(x0 )]−τ −C[(x−x0 )2 +y ε2 ].

(4.44)

Moreover it is easy to see that v(z) < 0 on the boundary of D∗ , provided that the constant C is large enough. Therefore v(z) ≤ 0 in D∗ . From F ≥ 0 and (4.42)–(4.44), the inequality ±[uy −ψ(x0 )]−τ −C[(x−x0 )2 +Y ε2 ] ≤ 0, i.e. (4.45) |uy −ψ(x0 )| ≤ τ +C[(x−x0 )2 +Y ε2 ] in D∗ is derived. Firstly let η → 0 and then let z → x0 , τ → 0, we obtain lim∂u/∂y → ψ(x0 ). Similarly we can verify limz(∈D∗ )→x0 uy = ψ(x0 ) when x0 = −1,1. Besides we can also prove that u(z) → u(x0 ) as z(∈ D∗ ) → x0 , when x0 = −1,1. This shows that the limit function u(z) of {un (z)} is a solution of Problem P for (4.1). Now we prove the uniqueness of solutions of Problem P for (4.1), it suffices to verify that Problem P0 has no non-trivial solution. Let u(z) be a solution of Problem P0 for (4.1) with d = 0 and u(z)
≡ 0 in D. Similarly to the proof of Theorem 3.4, we see that its maximum and minimum cannot attain in D ∪Γ. Moreover by using Lemma 4.1, we can prove that the maximum and minimum cannot attain at a point in (−1,1). Hence u(z) ≡ 0 in D. Finally we mention that for the degenerate elliptic equation K(y)uxx +uyy = 0,

K(0) = 0,

¯ K
(y) > 0 in D,

(4.46)

which is similar to equation (4.1) satisfying Condition C and other conditions as before, hence any solution u(z) of Problem P0 for (4.46) satisfies the estimates (4.11),(4.18) and (4.34) in D, provided that the inner angles αj π(j = 1,2) of D at z = −1,1 satisfy the conditions 0 < αj (
= 1/2) < 1,j = 1,2. Equation (4.46) is the Chaplygin equation in elliptic domain. Besides, oblique derivative problems for the degenerate elliptic equations of second order uxx +y m uyy +a(x,y)ux +b(x,y)uy +c(x,y)u = f (x,y) in D, y m1 uxx +y m2 uyy +a(x,y)ux +b(x,y)uy +c(x,y)u = f (x,y) in D needs to be considered, where m,m1 ,m2 are non-negative constants. The references for this chapter are [3],[6],[11],[15],[18],[23],[24],[30],[33],[38],[39], [40],[46],[48],[50],[53],[58],[60],[65],[67],[76],[78],[80],[81],[82], [85],[86],[94],[96],[99].

CHAPTER IV FIRST ORDER COMPLEX EQUATIONS OF MIXED TYPE In this chapter, we introduce the Riemann–Hilbert boundary value problem for first order complex equation of mixed (elliptic-hyperbolic) type in a simply connected domain. We first prove uniqueness and existence of solutions for the above boundary value problem, and then give a priori estimates of solutions for the problem, finally discuss the solvability of the above problem in general domains. The results in this chapter will be used in the following chapters.

1

The Riemann–Hilbert Problem for Simplest First Order Complex Equation of Mixed Type

In this section, we discuss the Riemann–Hilbert boundary value problem for the simplest mixed complex equation of first order in a simply connected domain. Firstly, we verify a unique theorem of solutions for the above boundary value problem. Moreover, the existence of solutions for the above problem is proved.

1.1 Formulation of the Riemann–Hilbert problem for the simplest complex equation of mixed type Let D be a simply connected bounded domain in the complex plane C I with the boundary ∂D = Γ ∪ L, where Γ(⊂ {y > 0}) ∈ Cα (0 < α < 1) with the end points z = 0, 2 and L = L1 ∪L2 , L1 = {x = −y, 0 ≤ x ≤ 1}, L2 = {x = y + 2, 1 ≤ x ≤ 2}. Denote D+ = D ∩ {y > 0}, D− = D ∩ {y < 0} and z1 = 1 − i. Without loss of generality, we may assume that Γ = {|z − 1| = 1, y ≥ 0}, otherwise through a conformal mapping, this requirement can be realized. We discuss the mixed system of first order equations ux −vy = 0, vx +sgny uy = 0 in D.

(1.1)

120

IV. First Order Mixed Complex Equations

Its complex form is the following complex equation of first order


wz¯ wz∗






= 0 in

D+ D−



,

(1.2)

where w = u + iv,

z = x + iy,

1 wz = [wx + iwy ], 2

1 wz∗ = [wx − iw¯y ]. 2

The Riemann–Hilbert boundary value problem for the complex equation (1.2) may be formulated as follows: ¯ Problem A Find a continuous solution w(z) of (1.2) in D∗ = D\({0, 2} ∪ {x ± y = ∗ ¯ 2, Im z ≤ 0}) or D = D\({0, 2} ∪ {x ± y = 0, Im z ≤ 0}) satisfying the boundary conditions (1.3) Re [λ(z)w(z)] = r(z), z ∈ Γ, Re [λ(z)w(z)] = r(z),

z ∈ Lj (j = 1 or 2),

Im [λ(z1 )w(z1 )] = b1 ,

(1.4)

where λ(z) = a(z) + ib(z), |λ(z)| = 1, z ∈ Γ ∪ Lj (j = 1 or 2), b1 is a real constant, and λ(z), r(z), b1 satisfy the conditions Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

|b1 | ≤ k2 ,

Cα [λ(z), Lj ] ≤ k0 ,

Cα [r(z), Lj ] ≤ k2 ,

j = 1 or 2,

a(z) − b(z)
= 0 on L1 ,

(1.5)

or a(z) + b(z)
= 0 on L2 ,

in which α(0 < α < 1), k0 , k2 are non-negative constants. For convenience, we may assume that w(z1 ) = 0, otherwise through a transformation of the function w(z) − λ(z1 )[r(z1 ) + ib1 ], the requirement can be realized. This Riemann–Hilbert problem (Problem A) for (1.2) with r(z) = 0, z ∈ Γ ∪ L1 (or L2 ) and b1 = 0 will be called Problem A0 . The number 1 K = (K1 + K2 ), 2

(1.6)

is called the index of Problem A and Problem A0 , where 

Kj =



φj + Jj , Jj = 0 or 1, π

eiφj =

λ(tj − 0) , λ(tj + 0)

γj =

φj − Kj , π

j = 1, 2, (1.7)

√ √ in which t1 = 2, t2 = 0, λ(t) = (1 − i)/ 2 on L0 = [0, 2] or λ(t) = (1 + i)/ 2 on L0 = [0, 2] and λ(t1 − 0) = λ(t2 + 0) = exp(7πi/4) or exp(πi/4). Here we only discuss the case of K = (K1 +K2 )/2 = −1/2 on the boundary ∂D+ of D+ . In order to ensure that the solution w(z) of Problem A is continuous in the neighborhood(⊂ D− ) of the point z = 0 or z = 2, we need to choose γ1 > 0 or γ2 > 0 respectively.

1. Simplest Mixed Complex Equation

121

1.2 Uniqueness of solutions of the Riemann–Hilbert problem for the simplest complex equation of mixed type Problem A for (1.2) has at most one solution.

Theorem 1.1

Proof Let w1 (z), w2 (z) be any two solutions of Problem A for (1.2). It is clear that w(z) = w1 (z) − w2 (z) is a solution of Problem A0 for (1.2) with boundary conditions z ∈ Γ,

Re [λ(z)w(z)] = 0,

z ∈ Lj (j = 1 or 2),

Re [λ(z)w(z)] = 0,

(1.8) Im [λ(z1 )w(z1 )] = 0.

(1.9)

Due to the complex equation (1.2) in D− can be reduced to the form ηµ = 0 in D− ,

ξν = 0,

(1.10)

where µ = x + y, ν = x − y, ξ = u + v, η = u − v, the general solution of system (1.10) can be expressed as ξ = u + v = f (µ) = f (x + y),

η = u − v = g(ν) = g(x − y),

i.e.

(1.11) f (x + y) + g(x − y) f (x + y) − g(x − y) , v(z) = , 2 2 in which f (t), g(t) are two arbitrary real continuously differentiable functions on [0, 2]. Noting the boundary condition (1.9), we have u(z) =

au+bv = 0 on L1 or L2 ,

[av − bu]|z=z1 = 0, i.e.

[a((1−i)x)+b((1−i)x)]f (0) +[a((1−i)x)−b((1−i)x)]g(2x) = 0 on [0, 1],

or (1.12)

[a((1+i)x−2i)+b((1+i)x−2i)]f (2x−2) +[a((1+i)x−2i)−b((1+i)x−2i)]g(2) = 0 on [1, 2], w(z1 ) = 0,

[u+v]|z1= f (0) = 0,

or [u−v]|z1= g(2) = 0.

The second formula in (1.12) can be rewritten as [a((1−i)t/2)+b((1−i)t/2)]f (0) +[a((1−i)t/2) − b((1 − i)t/2)]g(t) = 0, f (0) = g(t) = 0, or [a((1+i)t/2+1−i)+b((1−i)t/2+1−i)]f (t) (1.13) +[a((1−i)t/2+1−i)−b((1−i)t/2+1−i)]g(2) = 0, g(t) = f (t) = 0,

t ∈ [0, 2].

Thus the solution (1.11) becomes 1 u(z) = v(z) = f (x + y), g(x − y) = 0, or 2 1 u(z) = −v(z) = g(x − y), f (x + y) = 0, 2

(1.14)

122

IV. First Order Mixed Complex Equations

if a(z) − b(z)
= 0 on L1 or a(z) + b(z)
= 0 on L2 respectively. In particular, we have 1 u(x) = v(x) = f (x), x ∈ [0, 2], or 2 1 u(x) = −v(x) = g(x), x ∈ [0, 2]. 2

(1.15)

Next, due to f (0) = 0, or g(2) = 0, from (1.15), we can derive that u(x) − v(x) = 0,

i.e. Re [(1 + i)w(x)] = 0,

x ∈ [0, 2], or

u(x) + v(x) = 0,

i.e. Re [(1 − i)w(x)] = 0,

x ∈ [0, 2].

(1.16)

Noting the index K = −1/2 of Problem A for (1.2) in D+ and according to the result in Section 1, Chapter III, and [85]11),[86]1), we know that w(z) = 0 in D+ . Thus u(z) + v(z) = Re [(1 − i)w(z)] = f (x + y) = 0,

g(x − y) = 0, or

u(z) − v(z) = Re [(1 + i)w(z)] = g(x − y) = 0,

f (x + y) = 0,

(1.17)

obviously w(z) = u(z) + iv(z) = w1 (z) − w2 (z) = 0 on D− .

(1.18)

This proves the uniqueness of solutions of Problem A for (1.2).

1.3 Existence of solutions of the Riemann–Hilbert problem for the simplest complex equation of mixed type Now, we prove the existence of solutions of the Riemann–Hilbert problem (Problem A) for (1.2). Theorem 1.2

Problem A for (1.2) has a solution.

Proof As stated before, the general solution of (1.2) in D− can be expressed as f (x + y) − g(x − y) f (x + y) + g(x − y) , v(z) = , 2 2 (1 + i)f (x + y) + (1 − i)g(x − y) , i.e. w(z) = 2

u(z) =

(1.19)

in which f (t), g(t) are two arbitrary real continuously differentiable functions on [0, 2]. Taking into account the boundary condition (1.4), we have

1. Simplest Mixed Complex Equation

123

au + bv = r(x) on L1 or L2 , i.e. [a((1−i)x)+b((1−i)x)]f (0) + [a((1−i)x)−b((1−i)x)]g(2x) = 2r((1−i)x) on [0, 1], f (0) = [a(z1 ) + b(z1 )]r(z1 ) + [a(z1 ) − b(z1 )]b1 , or

(1.20)

[a((1+i)x−2i)+b((1+i)x−2i)]f (2x−2) +[a((1+i)x−2i)−b((1+i)x−2i)]g(2) = 2r((1+i)x−2i) on [1, 2], g(2) = [a(z1 ) − b(z1 )]r(z1 ) − [a(z1 ) + b(z1 )]b1 . The second and third formulas in (1.20) can be rewritten as [a((1 − i)t/2) − b((1 − i)t/2)]g(t) = 2r((1 − i)t/2) − [a((1 − i)t/2) + b((1 − i)t/2)]f (0),

t ∈ [0, 2], or

[a((1+i)t/2+1−i)+b((1+i)t/2+1−i)]f (t)

(1.21)

= 2r((1+i)t/2+1−i) −[a((1+i)t/2+1−i)−b((1+i)t/2+1−i)]g(2),

t ∈ [0, 2],

thus the solution (1.19) possesses the form 1 1 u(z) = {f (x + y) + g(x − y)}, v(z) = {f (x + y) − g(x − y)}, 2 2 2r((1−i)(x−y)/2)−[a((1−i)(x−y)/2)+b((1−i)(x−y)/2)]f (0) g(x−y) = , a((1−i)(x−y)/2)−b((1−i)(x−y)/2) 1 1 (1.22) u(z) = {g(x−y)+f (x + y)}, v(z) = {−g(x−y)+f (x + y)}, 2 2 2r((1 + i)(x + y)/2 + 1 − i) f (x + y) = a((1 + i)(x + y)/2 + 1 − i) + b((1 + i)(x + y)/2 + 1 − i) [a((1 + i)(x + y)/2 + 1 − i) − b((1 + i)(x + y)/2 + 1 − i)]g(2) , − a((1 + i)(x + y)/2 + 1 − i) + b((1 + i)(x + y)/2 + 1 − i) if a(z) − b(z)
= 0 on L1 , or a(z) + b(z)
= 0 on L2 respectively. In particular, we get 1 1 u(x) = {f (x) + g(x)}, v(x) = {f (x) − g(x)}, 2 2


g(x) =



2r((1 − i)x/2) − [a((1 − i)x/2) + b((1 − i)x/2)]f (0) , a((1 − i)x/2) − b((1 − i)x/2)

1 1 u(x) = {g(x) + f (x)}, v(x) = {−g(x) + f (x)}, 2 2

124

IV. First Order Mixed Complex Equations

f (x) =

2r((1+i)x/2+1−i) a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i)

[a((1+i)x/2+1−i)−b((1+i)x/2+1−i)]g(2) , − a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) From the above formulas, it follows that

(1.23) x ∈ [0, 2].

u(x)−v(x)=

2r((1−i)x/2)−[a((1−i)x/2)+b((1−i)x/2)]f (0) , or a((1−i)x/2)−b((1−i)x/2)

u(x)+v(x) =

2r((1+i)x/2+1−i) a((1+i)x/2+1−i)+b((1+i)x/2+1−i) −

[a((1+i)x/2+1−i)−b((1+i)x/2+1−i)]g(2) , a((1+i)x/2+1−i)+b((1+i)x/2+1−i)

(1.24) x ∈ [0,2]

f (0) = [a(1−i)+b(1−i)]r(1−i)+[a(1−i)−b(1−i)]b1 , or g(2) = [a(1−i)−b(1−i)]r(1−i)−[a(1−i)+b(1−i)]b1 , i.e. Re [(1+i)w(x)] = s(x), s(x) =

2r((1−i)x/2)−[a((1−i)x/2)+b((1−i)x/2)]f (0) , a((1 − i)x/2) − b((1 − i)x/2) x ∈ [0, 2], or

Re [(1−i)w(x)] = s(x) =

2r((1+i)x/2+1−i) a((1+i)x/2+1 − i)−b((1+i)x/2+1 − i) −

(1.25)

[a((1+i)x/2+1−i)−b((1+i)x/2+1−i)]g(2) , a((1+i)x/2+1 − i)−b((1+i)x/2+1 − i) x ∈ [0, 2],

if a((1 − i)x) − b((1 − i)x)
= 0 on [0,1] or a((1 + i)x − 2i) + b((1 + i)x − 2i)
= 0 on [1,2] respectively. We introduce a conformal mapping ζ = ζ(z) from the domain D+ onto the upper half-plane G = {Im ζ > 0}, such that the three points z = 0, 1, 2 map to ζ = −1, 0, 1 respectively, it is not difficult to derive that the conformal mapping and its inverse mapping can be expressed by the elementary functions, namely ζ(z) =

 5 5(z − 1) , z(ζ) = 1 + (1 − 1 − 16ζ 2 /25). 2ζ (z − 1)2 + 4

Denoting W (ζ) = w[z(ζ)] and

⎧ ⎪ λ[z(ζ)], ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨1−i

Λ(ζ) = ⎪

ζ ∈ Γ1 = ζ(Γ),

√ , ζ ∈ Γ2 = {−1 ≤ Re ζ ≤ 1, Im ζ = 0}, or 2

⎪ ⎪ ⎪ ⎪ 1+i ⎪ ⎪ ⎪ √ , ⎩

2

ζ ∈ Γ2 = {−1 ≤ Re ζ ≤ 1, Im ζ = 0},

(1.26)

1. Simplest Mixed Complex Equation

125

in which the points ζ1 = 1, ζ2 = −1 are the discontinuous points of Λ(ζ) on ∂G = {Im ζ = 0}, from (1.6),(1.7), it can be seen that the index of Λ(ζ) on ∂G = {Im ζ = 0} is K = −1/2. Hence, according to the result of Theorem 1.1, Chapter III, we know that the discontinuous Riemann–Hilbert boundary value problem for analytic functions W (ζ) in G with the boundary condition ⎧ ⎨ r[z(ζ)],

Re [Λ(z)W (ζ)] = R(ζ) =

⎩

ζ ∈ Γ1 , √ s[z(ζ)]/ 2, ζ ∈ Γ2 ,

(1.27)

has a unique solution W (ζ) in G as follows: W (ζ) = and X(ζ) = i

  X(ζ)  ∞ Λ(t)R(t) 2+ζ in G, dt + ic∗ πi 2−ζ −∞ (t − ζ)X(t)

ζ −2 Π(ζ)eiS(ζ) , Π(ζ) = ζ −i c∗ =



ζ −1 ζ +i

γ1 

ζ +1 ζ +i

(1.28)

γ2

,

2i + 1  ∞ Λ(t)R(t) dt, 2 + i −∞ X(t)(t − i)

and S(ζ) is an analytic function in Im ζ > 0 with the boundary condition

Re [S(t)] = arg[Λ1 (t)



t−2 ] on Im t = 0, Im [S(i)] = 0, t+i

(1.29)

where γj (j = 1, 2) are as stated in (1.7) and Λ1 (t) = λ(t)Π(t)(t − 2)|x + i|/[|Π(t)| ×|t − 2|(x + i)], and the boundedness of w(z) or boundedness of integral of the solution w(z) in the neighborhood ⊂ D\{0, 2} of t1 = 2 and t2 = 0 is determined by Jj = 0, γj > 0, or Jj = 0, γj = 0 and Jj = 1(j = 1, 2) respectively. Hence Problem A for (1.2) has a solution w(z) in the form ⎧ ⎪ W [ζ(z)], z ∈ D+ \{0, 2}, ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ 2 (1 + i)f (x + y) + (1 − i) ⎪ ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ 2r((1−i)(x−y)/2)−[a((1−i)(x−y)/2)+b ((1−i)(x−y)/2)]f (0) ⎪ ⎪ ⎪× , ⎪ ⎪ ⎨ a((1 − i)(x − y)/2) − b ((1 − i)(x − y)/2) w(z) = ⎪
⎪ 2(1+i)r((1+i)(x+y)/2+1−i) ⎪ ⎪ 1 (1−i)g(x−y)+ ⎪ ⎪ ⎪ ⎪ 2 a((1+i)(x+y)/2+1−i)+b ((1+i)(x+y)/2+1−i) ⎪ ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ [a((1 + i)(x + y)/2 + 1 − i) + b ((1 + i)(x + y)/2 + 1 − i)]g(2) ⎪ ⎪ ⎪ , − ⎪ ⎪ ⎪ a((1+i)(x+y)/2+1−i)+b ((1+i)(x+y)/2+1−i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ z = x + iy ∈ D− \{0, 2},

or

(1.30)

126

IV. First Order Mixed Complex Equations

in which f (0), g(2) are as stated in (1.24), W (ζ) in D+ \{0, 2} is as stated in (1.28), and from (1.23), we derive that f (x + y) = u(x + y) + v(x + y) = Re [(1 − i)W (ζ(x + y))], g(x − y) = u(x − y) − v(x − y) = Re [(1 + i)W (ζ(x − y))],

(1.31)

where W [ζ(x + y)] and W [ζ(x − y)] are the values of W [ζ(z)] on 0 ≤ z = x + y ≤ 2 and 0 ≤ z = x − y ≤ 2 respectively. From the foregoing representation of the solution w(z) of Problem A for (1.2) and the mapping ζ(z), we can derive that w(z) satisfies the estimate Cβ [w(z)X(z), D+ ] + Cβ [w± (z)Y ± (z), D− ] ≤ M1 ,

(1.32)

in which X(z) = Π2j=1 |z − tj |2|γj |+δ , Y ± (z) = |x ± y − tj |2|γj |+δ , w± (z) = Re w ± Im w, β(0 < β < δ), δ are sufficiently small positive constants, and M1 = M1 (β, k0 , k2 , D) is a non-negative constant [85]15). Finally, we mention that if the index K is an arbitrary even integer or 2K is an arbitrary odd integer, the above Riemann–Hilbert problem for (1.2) can be considered, but in general the boundary value problem for K ≤ −1 have some solvability conditions or its solution for K ≥ 0 is not unique.

2

The Riemann–Hilbert Problem for First Order Linear Complex Equations of Mixed Type

In this section we discuss the Riemann–Hilbert boundary value problem for first order linear complex equations of mixed (elliptic-hyperbolic) type in a simply connected domain. Firstly, we give the representation theorem and prove the uniqueness of solutions for the above boundary value problem, secondly by using the method of successive iteration, the existence of solutions for the above problem is proved.

2.1 Formulation of Riemann–Hilbert problem of first order complex equations of mixed type Let D be a simply connected bounded domain in the complex plane C I with the boundary ∂D = Γ ∪ L, where Γ, L = L1 ∪ L2 , D+ = D ∩ {y > 0}, D− = D ∩ {y < 0} and z1 = 1 − i are as stated in Section 1. We discuss the first order linear system of mixed (elliptic-hyperbolic) type equations ⎧ ⎨ ux − vy = au + bv + f, z = x + iy ∈ D, (2.1) ⎩ vx + sgny uy = cu + dv + g,

2. Linear Mixed Complex Equations

127

in which a, b, c, d, f, g are functions of (x, y)(∈ D), its complex form is the following complex equation of first order


wz wz∗






= F (z, w), F = A1 (z)w + A2 (z)w + A3 (z) in

D+ D−



,

(2.2)

where w = u + iv, A1 =

1 1 wz = [wx + iwy ], wz∗ = [wx − iw¯y ], 2 2 a + ib + ic − d f + ig A2 = , A3 = . 4 2

z = x + iy,

a − ib + ic + d , 4

Suppose that the complex equation (2.2) satisfies the following conditions. Condition C Aj (z) (j = 1, 2, 3) are measurable in z ∈ D+ and continuous in D− in D∗ = ¯ ¯ 2} ∪ {x ± y = 0, Im z ≤ 0}), and D\({0, 2} ∪ {x ± y = 2, Im z ≤ 0}) or D∗ = D\({0, satisfy (2.3) Lp [Aj , D+ ] ≤ k0 , j = 1, 2, Lp [A3 , D+ ] ≤ k1 , C[Aj , D− ] ≤ k0 ,

j = 1, 2,

C[A3 , D− ] ≤ k1 .

(2.4)

where p (> 2), k0 , k1 are non-negative constants. 2.2 The representation and uniqueness of solutions of the Riemann– Hilbert problem for mixed complex equations We first introduce a lemma, which is a special case of Theorem 2.1, Chapter III. Lemma 2.1 Suppose that the complex equation (2.2) satisfies Condition C. Then any solution of Problem A for (2.2) in D+ with the boundary conditions (1.3) and Re [λ(x)w(x)] = s(x),

λ(x) = 1 − i or 1 + i,

x ∈ L0 , Cα [s(x), L0 ] ≤ k3 ,

(2.5)

can be expressed as w(z) = Φ(z)eφ(z) + ψ(z), z ∈ D+ ,

(2.6)

where Im [φ(z)] = 0, z ∈ L0 = (0, 2), and φ(z), ψ(z) satisfies the estimates Cβ [φ, D+ ] + Lp0 [φz¯, D+ ] ≤ M2 , Cβ [ψ, D+ ] + Lp0 [ψz¯, D+ ] ≤ M2 ,

(2.7)

in which k3 , β (0 < β ≤ α), p0 (2 < p0 ≤ 2), M2 = M2 (p0 , β, k, D+ ) are non-negative constants, k = (k0 , k1 , k2 , k3 ), Φ(z) is analytic in D+ and w(z) satisfies the estimate Cβ [w(z)X(z), D+ ] ≤ M3 (k1 + k2 + k3 ),

(2.8)

128

IV. First Order Mixed Complex Equations

in which


X(z) = |z − t1 |η1 |z − t2 |η2 , ηj =

2|γj | + δ, if γj < 0, δ, γj ≥ 0,

j = 1, 2,

(2.9)

here γj (j = 1, 2) are real constants as stated in (1.7) and δ is a sufficiently small positive constant, and M3 = M3 (p0 , β, k0 , D+ ) is a non-negative constant. Theorem 2.2 If the complex equation (2.2) satisfies Condition C in D, then any solution of Problem A with the boundary conditions (1.3), (1.4) for (2.2) can be expressed as w(z) = w0 (z) + W (z), (2.10) where w0 (z) is a solution of Problem A for the complex equation (1.2) and W (z) possesses the form ˜ φ(z) ˜ ˜ w(z) = Φ(z)e + ψ(z) in D+ ,

W (z) = w(z) − w0 (z),

1 g(ζ) ˜ dσζ , φ(z) = φ˜0 (z) + T g = φ˜0 (z) − π D+ ζ − z 

W (z) = Φ(z) + Ψ(z),

Ψ(z) =

2

˜ ψ(z) = T f in D+ ,



ν

g1 (z)dνe1 +

µ

0

(2.11)

g2 (z)dµe2 in z ∈ D− ,

˜ in which φ(z) = 0 on L0 , e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic function in D+ , and


g(z) =

A1 + A2 w/w, 0, w(z) = 0,

g1 (z) = Aξ + Bη + E,

w(z)
= 0,

f = A1 ψ˜ + A2 ψ˜ + A3 in D+ ,

(2.12)

−

g2 (z) = Cξ + Dη + F in D ,

where ξ = Re w+Im w, η = Re w−Im w, A = Re A1 +Im A1 , B = Re A2 +Im A2 , C = Re A2 − Im A2 , D = Re A1 − Im A1 , E = Re A3 + Im A3 , F = Re A3 − Im A3 , and ˜ ˜ φ(z), ψ(z) satisfy the estimates ˜ ˜ Cβ [φ(z), D+ ] + Lp0 [φ˜z¯, D+ ] ≤ M4 , Cβ [ψ(z), D+ ] + Lp0 [ψ˜z¯, D+ ] ≤ M4 ,

(2.13)

˜ where M4 = M4 (p0 , β, k, D+ ) is a non-negative constant, Φ(z) is analytic in D+ and − Φ(z) is a solution of equation (1.2) in D satisfying the boundary conditions ˜

z ∈ Γ,

˜ ˜ Re [λ(z)(eφ(z) Φ(z) + ψ(z))] = r(z), ˜ φ(x) ˜ ˜ + ψ(x))] = s(x), Re [λ(x)(Φ(x)e

x ∈ L0 ,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L0 ,

z ∈ L1 or L2 ,

Im [λ(z1 )Φ(z1 )] = −Im [λ(z1 )Ψ(z1 )].

(2.14)

2. Linear Mixed Complex Equations

129

Moreover the solution w0 (z) of Problem A for (1.2) satisfies the estimate (1.32), namely Cβ [w0 (z)X(z), D+ ] + Cβ [w0± (z)Y ± (z), D− ] ≤ M5 (k1 + k2 )

(2.15)

where w0± (z) = Re w0 (z) ± Im w0 (z), Y ± (z) = 2j=1 |x ± y − tj |ηj , j = 1, 2, X(z), ηj = 2|γj | + δ (j = 1, 2), β are as stated in (1.32), and M5 = M5 (p0 , β, k0 , D) is a nonnegative constant. Proof Let the solution w(z) be substituted in the position of w in the complex equation (2.2) and (2.12), thus the functions g1 (z), g2 (z) and Ψ(z) in D− in (2.11),(2.12) can be determined. Moreover we can find the solution Φ(z) of (1.2) with the boundary condition (2.14), where ⎧ ⎪ 2r((1 − i)x/2) − 2R(1 − i)x/2) ⎪ ⎪ +Re [λ(x)Ψ(x)], ⎪ ⎪ ⎨ a((1 − i)x/2) − b((1 − i)x/2)

s(x) = ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

x ∈ L0 , or

2r((1+i)x/2+1−i)−2R((1+i)x/2+1−i) + Re [λ(x)Ψ(x)], a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i)

x ∈ L0 , (2.16)

here and later R(z) = Re [λ(z)Ψ(z)] on L1 or L2 , thus ⎧ ˜ φ(z) ˜ ⎨ Φ(z)e

w(z) = w0 (z) + W (z) = ⎩

˜ + ψ(z) in D+ ,

w0 (z) + Φ(z) + Ψ(z) in D− ,

(2.17)

is the solution of Problem A for the complex equation


wz w¯z∗

⎧ ⎫ ⎨ D+ ⎬



= A1 w + A2 w¯ + A3 in ⎩ , D− ⎭

(2.18)

which can be expressed as in (2.10) and (2.11).

2.3 The unique solvability of the Riemann–Hilbert problem for first order complex equations of mixed type Theorem 2.3 Let the mixed complex equation (2.2) satisfy Condition C. Then Problem A for (2.2) has a solution in D. Proof In order to find a solution w(z) of Problem A in D, we express w(z) in the form (2.10)–(2.12). In the following, we shall find a solution of Problem A by using the successive iteration. First of all, denoting the solution w0 (z) = (ξ0 e1 + η0 e2 ) of Problem A for (1.2), and substituting them into the positions of w = (ξe1 + ηe2 ) in the right-hand side of (2.2), similarly to (2.10)–(2.12), we have the corresponding functions g0 (z), f0 (z) in D+ and the functions

130

IV. First Order Mixed Complex Equations

W1 (z) = w1 (z) − w0 (z),

˜

˜ 1 (z)eφ1 (z) + ψ˜1 (z), w1 (z) = Φ

1   g0 (ζ) dσζ , φ˜1 (z) = φ˜0 (z) − π D+ ζ − z w1 (z) = w0 (z) + W1 (z), 

Ψ1 (z) =

2

ψ˜1 (z) = T f0 in D+ , (2.19)

W1 (z) = Φ1 (z) + Ψ1 (z),

ν



[Aξ0 +Bη0 +E]e1 dν +

0

µ

[Cξ0 +Dη0 +F ]e2 dµ in D−

can be determined, where µ = x + y, ν = x − y, and the solution w0 (z) satisfies the estimate (2.15), i.e. Cβ [w0 (z)X(z), D+ ] + Cβ [w0± (z)Y ± (z), D− ] ≤ M6 = M5 (k1 + k2 ),

(2.20)

where β, X(z), Y ± (z) are as stated in (2.15). Moreover, we find an analytic function ˜ 1 (z) in D+ and a solution Φ1 (z) of (1.2) in D− satisfying the boundary conditions Φ ˜

˜ 1 (z) + ψ˜1 (z))] = r(z), Re [λ(z)(eφ1 (z) Φ

z ∈ Γ,

˜ 1 (x)eφ˜1 (x) + ψ˜1 (x))] = s1 (x), Re [λ(x)(Φ

x ∈ L0 ,

Re [λ(x)Φ1 (x)] = −Re [λ(x)Ψ1 (x)],

z ∈ L0 ,

Re [λ(z)Φ1 (z)] = −Re [λ(z)Ψ1 (z)],

z ∈ L1 or L2 ,

(2.21)

Im [λ(z1 )Φ1 (z1 )] = −Im [λ(z1 )Ψ1 (z1 )], in which ⎧ ⎪ 2r((1−i)x/2)−2R1 ((1−i)x/2) ⎪ ⎪ +Re [λ(x)Ψ1 (x)], ⎪ ⎪ ⎨ a((1−i)x/2)−b((1−i)x/2)

s1 (x) = ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

x ∈ L0 , or

2r((1+i)x/2+1−i)−2R1 ((1+i)x/2+1−i) +Re [λ(x)Ψ1 (x)], a((1+i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i)

x ∈ L0 ,

here and later R1 (z) = Re [λ(z)Ψ1 (z)] on L1 or L2 , and w1 (z) = w0 (z) + W1 (z) =

⎧ ˜ 1 (z)eφ˜1 (z) ⎨Φ ⎩

+ ψ˜1 (z) in D+ ,

w0 (z) + Φ1 (z) + Ψ1 (z) in D−

(2.22)

satisfies the estimate Cβ [w1 (z)X(z), D+ ] + C[w1± (z)Y ± (z), D− ] ≤ M7 = M7 (p0 , β, k, D),

(2.23)

˜ 1 (z) are similar to the functions in Theorem 2.2. Furthermore we where φ˜1 (z), ψ˜1 (z), Φ substitute w1 (z) = w0 (z) + W1 (z) and the corresponding functions w1+ (z) = ξ1 (z) = Re w1 (z)+Im w(z), w1− (z) = η1 (z) = Re w1 (z)−Im w(z) into the positions of w, ξ, η in (2.11),(2.12), and similarly to (2.19)–(2.22), we can find the corresponding functions

2. Linear Mixed Complex Equations

131

˜ 2 (z) in D+ and Ψ2 (z), Φ2 (z) and W2 (z) = Φ2 (z) + Ψ2 (z) in D− , and φ˜2 (z), ψ˜2 (z), Φ the function ⎧ ˜ 2 (z)eφ˜2 (z) + ψ˜2 (z) in D+ , ⎨Φ w2 (z) = w0 (z) + W2 (z) = ⎩ (2.24) w0 (z) + Φ2 (z) + Ψ2 (z) in D− satisfies a similar estimate of the form (2.23). Thus there exists a sequence of functions {wn (z)} as follows

wn (z) = w0 (z) + Wn (z) =

⎧ ˜ n (z)eφ˜n (z) + ψ˜n (z) in D+ , ⎪ Φ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ w0 (z) + Φn (z) + Ψn (z) in D − , ⎪ ⎪ ⎨  ν ⎪ Ψ (z) = [Aξn−1 +Bηn−1 +E]e1 dν ⎪ n ⎪ ⎪ 2 ⎪ ⎪ ⎪  ⎪ µ ⎪ ⎪ ⎪ ⎩ + [Cξn−1 +Dηn−1 +F ]e2 dµ 0

(2.25) in D− ,

and then

√ + ν ± [Aξ0 +Bη0 +E]e1 dν| |[w1± (z)−w0± (z)]Y ± (z)| ≤ |Φ± 1 (z)Y (z)|+ 2 |Y (z) 

−

+|Y (z)

µ

0

2



[Cξ0 + Dη0 + F ]e2 dµ| ≤ 2M8 M (4m + 1)R
in D− ,

(2.26) where m = max{C[w0+ (z)Y + (z), D− ] + C[w0− (z)Y − (z), D− ]}, M8 = maxz∈D− (|A|, |B|, |C|, |D|, |E|, |F |), R
= 2, M = 1 + 4k02 (1 + k02 ), M5 is a constant as stated in (2.20). It is clear that wn (z) − wn−1 (z) satisfies 

wn (z)−wn−1 (z) = Φn (z)−Φn−1 (z)+ 

+

0

µ

2

ν

[A(ξn −ξn−1 )+B(ηn −ηn−1 )]e1 dν (2.27)

[C(ξn − ξn−1 ) + D(ηn − ηn−1 )]e2 dµ in D− ,

where n = 1, 2, . . .. From the above equality, we can obtain ± ]Y ± (z)| ≤ [2M8 M (4m + 1)]n |[wn± − wn−1

×



R


0

R
n−1 [2M8 M (4m + 1)R
]n dR
≤ in D− , (n − 1) ! n!

and then we can see that the sequence of functions {wn± (z)Y ± (z)},

(2.28) i.e.

± wn± (z)Y ± (z) = {w0± (z)+[w1± (z)−w0± (z)]+· · ·+[wn± (z)−wn−1 (z)]}Y ± (z)

(n = 1, 2,. . .) in the equality

D−

uniformly converge to functions

w∗± (z)Y ± (z),

(2.29)

and w∗ (z) satisfies

w∗ (z) = w0 (z) + Φ∗ (z) + Ψ∗ (z), 

Ψ∗ (z) =

2

+

ν

[Aξ∗ + Bη∗ + E]e1 dν  0

µ

[Cξ∗ + Dη∗ + F ]e2 dµ in D− ,

(2.30)

132

IV. First Order Mixed Complex Equations

where ξ ∗ = Re w∗ + Im w∗ , η = Re w∗ − Im w∗ , and w∗ (z) satisfies the estimate


C[w∗± (z)Y ± (z), D− ] ≤ e2M8 M (4m+1)R .

(2.31)

˜ n (z)eφ˜n (z) + ψ˜n (z)) Moreover, we can find a sequence of functions {wn (z)}(wn (z) = Φ + + ˜ in D and Φn (z) is an analytic function in D satisfying the boundary conditions ˜

˜ n (z)eφn (z) + ψ˜n (z))] = r(z), Re [λ(z)(Φ ˜

˜ n (x)eφn (x) + ψ˜n (x))] = s(x), Re [λ(x)(Φ

z ∈ Γ, x ∈ L0 ,

(2.32)

in which ⎧ ⎪ 2r((1−i)x/2)−2Rn ((1−i)x/2) ⎪ ⎪ +Re [λ(x)Ψn (x))], ⎪ ⎪ ⎨ a((1−i)x/2)−b((1−i)x/2)

sn (x) =

x ∈ L0 , or

⎪ ⎪ 2r((1+i)x/2+1−i)−2Rn ((1−i)x/2 +1−i) ⎪ ⎪ ⎪ +Re [λ(x)Ψn (x)], ⎩

a((1+i)x/2+1−i)+b((1+i)x/2+1−i)

x ∈ L0 ,

(2.33) here and later Rn (z) = Re [λ(z)Ψn (z)] on L1 or L2 . From (2.31), it follows that Cβ [sn (x)X(x), L0 ] ≤ 2k2 k0 +

[2M8 M (4m + 1)R
]n = M9 , n!

(2.34)

and then the estimate Cβ [wn (z)X(z), D+ ] ≤ M3 (k1 + k2 + M9 ),

(2.35)

thus from {wn (z)X(x)}, we can choose a subsequence which uniformly converge a function w∗ (z)X(z) in D+ . Combining (2.31) and (2.35), it is obvious that the ¯ satisfies the estimate solution w∗ (z) of Problem A for (2.2) in D Cβ [w∗ (z)X(z), D+ ] + C[w∗± (z)Y ± (z), D− ] ≤ M10 = M10 (p0 , β, k, D),

(2.36)

where M10 is a non-negative constant. Theorem 2.4 Suppose that the complex equation (2.2) satisfies Condition C. Then Problem A for (2.2) has at most one solution in D. Proof Let w1 (z), w2 (z) be any two solutions of Problem A for (2.2). By Condition C, we see that w(z) = w1 (z) − w2 (z) satisfies the homogeneous complex equation and boundary conditions Lw = A˜1 w + A˜2 w¯ in D, (2.37) Re [λ(z)w(z)] = 0,

z ∈ Γ,

Re [λ(z)w(z)] = 0,

z ∈ L1 or L2 ,

Re [λ(x)w(x)] = s(x),

x ∈ L0 ,

Re [λ(z1 )w(z1 )] = 0.

(2.38)

2. Linear Mixed Complex Equations

133

From Theorem 2.2, the solution w(z) can be expressed in the form

w(z) =

⎧ ˜ φ(z) ˜ ˜ ⎪ , φ(z) = T˜g in D+ , ⎪ ⎪ Φ(z)e ⎪ ⎪ ⎪ ⎧ ⎪ ⎪ ⎪ ⎨ A1 + A2 w/w, ¯ w(z)
= 0, z ∈ D+ , ⎪ ⎪ ⎪ g(z) = ⎪ ⎨ ⎩ 0, w(z) = 0, z ∈ D+ , ⎪ ⎪ ⎪ ⎪ ⎪ Φ(z) + Ψ(z), ⎪ ⎪ ⎪ ⎪  ν  µ ⎪ ⎪ ⎪ ⎪ ⎩ Ψ(z) = [Aξ + Bη]e1 dν + [Cξ + Dη]e2 dµ 2

0

(2.39)

in D− ,

˜ where Φ(z) is analytic in D+ and Φ(z) is a solution of (1.2) in D− satisfying the ˜ boundary condition (2.14), but ψ(z) = 0, z ∈ D+ , r(z) = 0, z ∈ Γ, and ⎧ ⎪ −2R((1−i)x/2) ⎪ ⎪ +Re [λ(x)Ψ(x)], ⎪ ⎪ ⎨ a((1−i)x/2) − b((1−i)x/2)

s(x) = ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

x ∈ L0 ,

or

−2R((1 − i)x/2 + 1 − i) +Re [λ(x)Ψ(x)], a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i)

x ∈ L0 .

By using the method in the proof of Theorem 2.3, we can derive that |w± (z)Y ± (z)| ≤

[2M8 M (4m + 1)R
]n in D− . n!

(2.40)

Let n → ∞, we get w± (z) = 0, i.e. w(z) = w1 (z) − w2 (z) = 0, Ψ(z) = Φ(z) = 0 in ˜ φ(z) ˜ D− . Noting that w(z) = Φ(z)e satisfies the boundary conditions in (2.38), we see ˜ that the analytic function Φ(z) in D+ satisfies the boundary conditions ˜

˜ Re [λ(z)eφ(z) Φ(z)] = 0,

z ∈ Γ,

˜ = 0, Re [λ(x)Φ(x)]

x ∈ L0 ,

(2.41)

˜ and the index of the boundary value problem (2.41) is K = −1/2, hence Φ(z) = 0 in ˜ φ(z) ˜ D+ , and then w(z) = Φ(z)e = 0 in D+ , namely w(z) = w1 (z) − w2 (z) = 0 in D+ . This proves the uniqueness of solutions of Problem A for (2.2). From Theorems 2.3 and 2.4, we see that under Condition C, Problem A for equation (2.2) has a unique solution w(z), which can be found by using successive iteration, and w(z) of Problem A satisfies the estimates Cβ [w(z)X(z), D+ ] ≤ M11 , C[w± (z)Y ± (z), D− ] ≤ M12 ,

(2.42)

where w± (z) = Re w(z) ± Im w(z), X(z), Y ± (z) are as stated in (1.32), and β(0 < β < δ), Mj = Mj (p0 , β, k, D) (j = 11, 12) are non-negative constants, k = (k0 , k1 , k2 ). Moreover, we can derive the following theorem. Theorem 2.5 Suppose that equation (2.2) satisfies Condition C. Then any solution w(z) of Problem A for (2.2) satisfies the estimates Cβ [w(z)X(z), D+ ] ≤ M13 (k1 + k2 ), C[w± (z)Y ± (z), D− ] ≤ M14 (k1 + k2 ),

(2.43)

134

IV. First Order Mixed Complex Equations

in which Mj = Mj (p0 , β, k0 , D)(j = 13, 14) are non-negative constants. Proof When k1 + k2 = 0, from Theorem 2.3, it is easy to see that (2.43) holds. If k1 + k2 > 0, then it is seen that the function W (z) = w(z)/(k1 + k2 ) is a solution of the homogeneous boundary value problem Lw = F (z, w)/k, F/k = A1 W + A2 W + A3 /k in D, Re [λ(z)W (z)] = r(z)/k,

z ∈ Γ,

Re [λ(z)W (z)] = r(z)/k,

z ∈ Lj ,

Im [λ(z1 )W (z1 )] = b1 /k, j = 1 or 2,

where Lp [A3 /k, D+ ] ≤ 1, C[A3 /k, D− ] ≤ 1, Cα [r(z)/k, Γ] ≤ 1, Cα [r(z)/k, Lj ] ≤ 1, j = 1 or 2, |b1 /k| ≤ 1. On the basic of the estimate (2.42), we can obtain the estimates Cβ [w(z)X(z), D+ ] ≤ M13 , C[w± (z)Y ± (z), D− ] ≤ M14 , (2.44) where Mj = Mj (p0 , β, k0 , D) (j = 13, 14) are non-negative constants. From (2.44), it follows the estimate (2.43). From the estimates (2.43),(2.44), we can see the regularity of solutions of Problem A for (2.2). In the next section, we shall give the H¨older estimate of solutions of Problem A for first order quasilinear complex equation of mixed type with the more restrictive conditions than Condition C, which includes the linear complex equation (2.2) as a special case.

3

The Riemann–Hilbert Problem for First Order Quasilinear Complex Equations of Mixed Type

In this section we discuss the Riemann–Hilbert boundary value problem for first order quasilinear complex equations of mixed (elliptic-hyperbolic) type in a simply connected domain. We first give the representation theorem and prove the uniqueness of solutions for the above boundary value problem, and then by using the successive iteration, the existence of solutions for the above problem is proved.

3.1 Representation and uniqueness of solutions of Riemann–Hilbert problem for first order quasilinear complex equations of mixed type Let D be a simply connected bounded domain as stated in Subsection 2.1. We discuss the quasilinear mixed (elliptic-hyperbolic) system of first order equations ⎧ ⎨ ux ⎩

− vy = au + bv + f,

vx + sgny uy = cu + dv + g,

z = x + iy ∈ D,

(3.1)

3. Quasilinear Mixed Complex Equations

135

in which a, b, c, d, f, g are functions of (x, y) (∈ D), u, v (∈ IR), its complex form is the following complex equation of first order




wz wz∗




= F (z, w),

F = A1 w + A2 w + A3 in

D+ D−



,

(3.2)

where Aj = Aj (z, w), j = 1, 2, 3, and the relations between Aj (j = 1, 2, 3) and a, b, c, d, f, g are the same as those in (2.2). Suppose that the complex equation (3.2) satisfies the following conditions. Condition C I for almost every point z ∈ D+ , 1) Aj (z, w) (j = 1, 2, 3) are continuous in w ∈ C and are measurable in z ∈ D+ and continuous on D− for all continuous functions ¯ ¯ w(z) in D∗ = D\({0, 2}∪{x±y = 2, y ≤ 0}) or D∗ = D\({0, 2}∪{x±y = 0, y ≤ 0}), and satisfy Lp [Aj , D+ ] ≤ k0 , j = 1, 2, Lp [A3 , D+ ] ≤ k1 , (3.3) C[Aj , D− ] ≤ k0 , j = 1, 2, C[A3 , D− ] ≤ k1 . 2) For any continuous functions w1 (z), w2 (z) on D∗ , the following equality holds: F (z, w1 ) − F (z, w2 ) = A˜1 (z, w1 , w2 )(w1 − w2 ) + A˜2 (z, w1 , w2 )(w1 − w2 ) in D, (3.4) where Lp [A˜j , D+ ] ≤ k0 ,

C[A˜j , D− ] ≤ k0 ,

j = 1, 2

(3.5)

in (3.3),(3.5), p (> 2), k0 , k1 are non-negative constants. In particular, when (3.2) is a linear equation (2.2), the condition (3.4) is obviously valid. The boundary conditions of Riemann–Hilbert problem for the complex equation (3.2) are as stated in (1.3),(1.4). Let the solution w(z) of Problem A be substituted in the coefficients of (3.2). Then the equation can be viewed as a linear equation (2.2). Hence we have the same representation theorems as Lemma 2.1 and Theorem 2.2. Theorem 3.1 Suppose that the quasilinear complex equation (3.2) satisfies Condition C. Then Problem A for (3.2) has a unique solution in D. Proof We first prove the uniqueness of the solution of Problem A for (3.2). Let w1 (z), w2 (z) be any two solutions of Problem A for (3.2). By Condition C, we see that w(z) = w1 (z)−w2 (z) satisfies the homogeneous complex equation and boundary conditions wz¯ = A˜1 w + A˜2 w¯ in D, (3.6) z ∈ L1 or L2 , Re [λ(z1 )w(z1 )] = 0, (3.7) ˜ where the conditions on the coefficients Aj (j = 1, 2) are the same as in the proof of Theorem 2.4 for the linear equation (2.2). Besides the remaining proof is the same in the proof of Theorems 2.3 and 2.4. Re [λ(z)w(z)] = 0,

136

IV. First Order Mixed Complex Equations

Next noting the conditions (3.3),(3.4), by using the same method, the existence of solutions of Problem A for (3.2) can be proved, and any solution w(z) of Problem A for (3.2) satisfies the estimate (2.43). In order to give the H¨older estimate of solutions for (3.2), we need to add the following condition. ¯ w1 , w2 , the above functions satisfy 3) For any complex numbers z1 , z2 (∈ D), |Aj (z1 , w1 ) − Aj (z2 , w2 )| ≤ k0 [|z1 − z2 |α + |w1 − w2 |],

j = 1, 2,

|A3 (z1 , w1 ) − A3 (z2 , w2 )| ≤ k1 [|z1 − z2 |α + |w1 − w2 |],

z ∈ D− ,

(3.8)

in which α(1/2 < α < 1), k0 , k1 are non-negative constants. On the basis of the results of Theorem 4.4 in Chapter I and Theorem 2.3 in Chapter III, we can derive the following theorem. Theorem 3.2 Let the quasilinear complex equation (3.2) satisfy Condition C and (3.8). Then any solution w(z) of Problem A for (3.2) satisfies the following estimates Cδ [X(z)w(z), D+ ] ≤ M15 ,

Cδ [Y ± (z)w± (z), D− ] ≤ M16 ,

(3.9)

in which w± (z) = Re w(z) ± Im w(z) and X(z) =

2 

|z − tj |ηj ,

Y ± (x) =

j=1

2 




|x ± y − tj |ηj ,

ηj =

2|γj |+2δ,

γj ≥ 0,

2δ,

j=1

if γj < 0,

(3.10)

here γj (j = 1, 2) are real constants as stated in (1.7) and δ is a sufficiently small positive constant, and Mj = Mj (p0 , β, k, D) (j = 15, 16) are non-negative constants, k = (k0 , k1 , k2 ). 3.2 Existence of solutions of Problem A for general first order complex equations of mixed type Now, we consider the general quasilinear mixed complex equation of first order


Lw =

wz¯ wz¯∗






= F (z, wz ) + G(z, w),

F = A1 w + A2 w¯ + A3 ,

z∈

G = A4 | w | , σ

D+



D−

, (3.11)

z ∈ D,

in which F (z, w) satisfies Condition C, σ is a positive constant, and A4 (z, w) satisfies the same conditions as Aj (j = 1, 2), where the main condition is ¯ ≤ k0 , C[A4 (z, w), D] and denote the above conditions by Condition C
.

(3.12)

3. Quasilinear Mixed Complex Equations Theorem 3.3

137

Let the mixed complex equation (3.11) satisfy Condition C
.

(1) When 0 < σ < 1, Problem A for (3.11) has a solution w(z). (2) When σ > 1, Problem A for (3.11) has a solution w(z), provided that M17 = k1 + k2 + |b1 |

(3.13)

is sufficiently small. Proof

(1) Consider the algebraic equation for t : (M13 + M14 ){k1 + k2 + 2k0 tσ + |b1 |} = t,

(3.14)

in which M13 , M14 are constants stated in (2.43). It is not difficult to see that the equation (3.14) has a unique solution t = M18 ≥ 0. Now, we introduce a closed and ¯ whose elements are the function w(z) convex subset B ∗ of the Banach space C(D), satisfying the condition C[w(z)X(z), D+ ] + C[w± (z)Y ± (z), D− ] ≤ M18 .

(3.15)

∗

We arbitrarily choose a function w0 (z) ∈ B for instance w0 (z) = 0 and substitute it into the position of w in the coefficients of (3.11) and G(z, w). From Theorem 3.1, it is clear that problem A for Lw − A1 (z, w0 )w − A2 (z, w0 )w¯ − A3 (z, w0 ) = G(z, w0 ),

(3.16)

has a unique solution w1 (z). By (2.43), we see that the solution w1 (z) satisfies the estimate in (3.15). By using successive iteration, we obtain a sequence of solutions wm (z)(m = 1, 2, ...) of Problem A, which satisfy the equations Lwm+1 − A1 (z, wm )wm+1z − A2 (z, wm )w¯m+1 ¯ +A3 (z, wm ) = G(z, wm ) in D,

(3.17)

m = 1, 2, . . .

and wm+1 (z)X(z) ∈ B ∗ , m = 1, 2, . . .. From (3.17), we see that w˜m+1 (z) = wm+1 (z) −wm (z) satisfies the complex equation and boundary conditions ˜ G(z) ˜ = G(z, wm )−G(z, wm−1 ) in D, ¯ Lw˜m+1 − A˜1 w˜m+1 − A˜2 w¯m+1 = G, Re [λ(z)w˜m+1 (z)] = 0 on Γ ∪ Lj ,

j = 1 or 2,

(3.18)

Im [λ(z1 )w˜m+1 (z1 )] = 0,

˜ ¯ ≤ 2k0 M18 , M18 is a solution of the where m = 1, 2, . . . . Noting that C[X(z)G(z), D] algebraic equation (3.14) and according to the proof of Theorem 2.3, ± C[w˜m+1 X(z), D+ ] + C[w˜m+1 (z)Y ± (z), D− ] ≤ M18

(3.19)

can be obtained. The function w˜m+1 can be expressed as wm+1 (z) = w0 (z) + Φm+1 (z) + Ψm+1 (z), 

Ψm+1 (z) =

2

+

x−y

 0

˜ m+1 + Bη ˜ m+1 + E]e ˜ 1 d(x − y) [Aξ

x+y

˜ m+1 + F˜ ]e2 d(x + y) in D− , [C˜ ξ˜m+1 + Dη

(3.20)

138

IV. First Order Mixed Complex Equations

˜ B, ˜ C, ˜ D, ˜ E, ˜ F˜ and A˜1 , A˜2 , G ˜ is the same as that of in which the relation between A, A, B, C, D, E, F and A1 , A2 , A3 in (2.12). By using the method from the proof of Theorem 2.5, we can obtain ± C[w˜m+1 X(z), D+ ] + C[w˜m+1 (z)Y ± (z), D− ] ≤

(M20 R
)m , m!

¯ herein M19 = where M20 = 2M19 M (M5 + 1)(4m + 1), R
= 2, m = C[w0± (z)Y ± (z), D], ˜ ˜ ˜ ˜ ˜ ˜ max{C[A,Q], C[B,Q], C[C,Q], C[D,Q], C[E,Q], C[F ,Q]}, M = 1 + 4k02 (1 + k02 ). From the above inequality, it is seen that the sequence of functions {wm (z)X(z)}, i.e. ± ± ± wm (z)Y ± (z) = {w0± (z)+[w1± (z)−w0± (z)]+· · ·+[wm (z)−wm−1 (z)]}Y ± (z)

(3.21)

(m = 1, 2, . . .) uniformly converge to w∗± (z)Y ± (z), and similarly to (2.30), the corresponding function w∗ (z) satisfies the equality w∗ (z) = w0 (z) + Φ∗ (z) + Ψ∗ (z), 

Ψ∗ (z) =

2

+

x−y

 0

[Aξ∗ + Bη∗ + E]e1 d(x − y)

x+y

(3.22)

[Cξ∗ + Dη∗ + F ]e2 d(x + y) in D− ,

and the function w∗ (z) is just a solution of Problem A for the quasilinear equation (3.11) in the closure of the domain D. (2) Consider the algebraic equation (M13 + M14 ){k1 + k2 + 2k0 tσ + |b1 |} = t

(3.23)

for t. It is not difficult to see that equation (3.23) has a solution t = M18 ≥ 0, provided that M17 in (3.13) is small enough. Now, we introduce a closed and convex ¯ whose elements are the functions w(z) satisfying subset B∗ of the Banach space C(D), the conditions C[w(z)X(z), D+ ] + C[w± (z)Y ± (z), D− ] ≤ M18 .

(3.24)

By using the same method as in (1), we can find a solution u(z) ∈ B∗ of Problem A for equation (3.11) with σ > 1.

4

The Riemann–Hilbert Problem for First Order Quasilinear Equations of Mixed type in General Domains

This section deals with the Riemann–Hilbert boundary value problem for quasilinear first order equations of mixed (elliptic-hyperbolic) type in general domains.

4. Mixed Equations in General Domains

139

4.1 Formulation of the oblique derivative problem for second order equations of mixed type in general domains Let D be a simply connected bounded domain D in the complex plane C I with the boundary ∂D = Γ ∪ L, where Γ, L are as stated in Section 1. Now, we consider the domain D
with the boundary Γ ∪ L
1 ∪ L
2 , where the parameter equations of the curves L
1 , L
2 are as follows: L
1 = {γ1 (x) + y = 0, 0 ≤ x ≤ l},

L
2 = {x − y = 2, l ≤ x ≤ 2},

(4.1)

in which γ1 (x) on 0 ≤ x ≤ l = γ1 (l) + 2 is continuous and γ1 (0) = 0, γ1 (x) > 0 on 0 ≤ x ≤ l, and γ1 (x) is differentiable on 0 ≤ x ≤ l except finitely many points and 1 + γ1
(x) > 0. Denote D
+ = D
∩ {y > 0} = D+ , D
− = D
∩ {y < 0} and z1
= l − iγ1 (l). Here we mention that in [12]1),3), the author assumes that the derivative of γ(x) satisfies γ1
(x) > 0 on 0 ≤ x ≤ l and other conditions. We consider the first order quasilinear complex equation of mixed type as stated in (3.2) in D
, and assume that (3.2) satisfies Condition C in D
. The oblique derivative boundary value problem for equation (3.2) may be formulated as follows: Problem A
Find a continuous solution w(z) ¯ of (3.2) in D∗ = D\{0, L
2 }, which satisfies the boundary conditions Re [λ(z)w(z)] = r(z),

z ∈ Γ,

Re [λ(z)w(z)] = r(z),

z ∈ L
1 ,

(4.2) (4.3)

Im [λ(z)uz¯]|z=z1
= b1 , where λ(z) = a(x) + ib(x) and |λ(z)| = 1 on Γ ∪ L
1 , and b0 , b1 are real constants, and λ(z), r(z), b0 , b1 satisfy the conditions Cα [λ(z), Γ] ≤ k0 , Cα [λ(z), L
1 ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

Cα [r(z), L
1 ] ≤ k2 ,

max
z∈L1

|b1 | ≤ k2 , 1 ≤ k0 , |a(x) − b(x)|

(4.4)

in which α (1/2 < α < 1), k0 , k2 are non-negative constants. The boundary value problem for equation (3.2) with A3 (z, u, uz ) = 0, z ∈ D, u ∈ IR, uz ∈ C I, r(z) = 0, z ∈ Γ ∪ L
1 and b0 = b1 = 0 will be called Problem A
0 . The number 1 (4.5) K = (K1 + K2 ) 2 is called the index of Problem A
and Problem A
0 as stated in Section 1. Similarly we only discuss the case of K = −1/2 on ∂D+ , because in this case the solution of Problem A
is unique. Besides we choose γ1 > 0. In the following, we first discuss the domain D
and then discuss another general domain D

.

140

IV. First Order Mixed Complex Equations

4.2 The existence of solutions of Problem A for first order equations of mixed type in general domains 1. By the conditions in (4.1), the inverse function x = σ(ν) of x + γ1 (x) = ν = x − y can be found and σ
(ν) = 1/[1 + γ1
(x)]. Hence the curve L
1 can be expressed by x = σ(ν) = (µ + ν)/2, i.e. µ = 2σ(ν) − ν, 0 ≤ ν ≤ l + γ1 (l). We make a transformation   µ−2σ(ν)+ν µ ˜=2 , ν˜ = ν, 2σ(ν)−ν ≤ µ ≤ 2, 0 ≤ ν ≤ 2, (4.6) 2−2σ(ν)+ν where µ, ν are real variables, its inverse transformation is µ = [2 − 2σ(ν) + ν]˜ µ/2 + 2σ(ν) − ν,

ν = ν˜,

0≤µ ˜ ≤ 2,

0 ≤ ν˜ ≤ 2.

(4.7)
−

It is not difficult to see that the transformation in (4.6) maps the domain D onto D− . The transformation (4.6) and its inverse transformation (4.7) can be rewritten as ⎧ ⎪ 1 4x − (2 + x − y)[2σ(x + γ1 (x)) − x − γ1 (x)] ⎪ ⎪ , x˜ = (˜ µ + ν˜) = ⎪ ⎪ ⎨ 2 4 − 4σ(x + γ1 (x)) + 2x + 2γ1 (x) (4.8) ⎪ ⎪ 4y − (2 − x + y)[2σ(x + γ1 (x)) − x − γ1 (x)] 1 ⎪ ⎪ ⎪ µ − ν˜) = , ⎩ y˜ = (˜ 2 4 − 4σ(x + γ1 (x)) + 2x + 2γ1 (x) and ⎧ x + y˜) ⎪ ⎪ x = 1 (µ + ν) = [2 − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ ⎪ ⎪ ⎪ ⎪ 2 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x + γ1 (x) − x˜ + y˜ ⎪ ⎪ +σ(x + γ1 (x)) − , ⎪ ⎨ 2 (4.9) ⎪ ⎪ x + y˜) ⎪ ⎪ y = 1 (µ − ν) = [2 − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ ⎪ ⎪ ⎪ 2 4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ x + γ1 (x) + x˜ − y˜ ⎪ ⎪ ⎩ . +σ(x + γ1 (x)) − 2 −1 Denote by z˜ = x˜ + j y˜ = f (z), z = x + jy = f (˜ z ) the transformation (4.8) and the inverse transformation (4.9) respectively. In this case, the system of equations is ξν = Aξ + Bη + E,

ηµ = Cξ + Dη + F,
−

z ∈ D
− ,

(4.10)




which is another form of (3.2) in D . Suppose that (3.2) in D satisfies Condition C, through the transformation (4.7), we obtain ξν˜ = ξν , ηµ˜ = [2 − 2σ(ν) + ν]ηµ /2 in D
− , and then ξν˜ = Aξ + Bη + E − (4.11) [2 − 2σ(ν) + ν][Cξ + Dη + F ] in D , ηµ˜ = 2 and through the transformation (4.8), the boundary condition (4.3) is reduced to Re [λ(f −1 (˜ z ))w(f −1 (˜ z ))] = r(f −1 (˜ z )), Im [λ(f

−1

(˜ z1 ))w(f

−1

(˜ z1 )] = b1 ,

z˜ ∈ L1 ,

(4.12)

4. Mixed Equations in General Domains

141

in which z˜1 = f (z1
). Therefore the boundary value problem (4.10),(4.3) is transformed into the boundary value problem (4.11),(4.12), i.e. the corresponding Problem A in D. On the basis of Theorem 3.1, we see that the boundary value problem (3.2)(in D+ ),(4.11),(4.2),(4.12) has a unique solution w(˜ z ), and w(z) is just a solution of Problem A for (3.2) in D
with the boundary conditions (4.2),(4.3). Theorem 4.1 If the mixed equation (3.2) in D
satisfies Condition C in the domain D
with the boundary Γ ∪ L
1 ∪ L
2 , where L
1 , L
2 are as stated in (4.1), then Problem A
for (3.2) with the boundary conditions (4.2), (4.3) has a unique solution w(z). 2. Next let the domain D

be a simply connected domain with the boundary Γ ∪ L

1 ∪ L

2 , where Γ is as stated before and L

1 = {γ1 (x) + y = 0, 0 ≤ x ≤ l},

L

2 = {γ2 (x) + y = 0, l ≤ x ≤ 2},

(4.13)

in which γ1 (0) = 0, γ2 (2) = 0, γ1 (x) > 0, 0 ≤ x ≤ l, γ2 (x) > 0, l ≤ x ≤ 2, γ1 (x) on 0 ≤ x ≤ l, γ2 (x) on l ≤ x ≤ 2 are continuous, and differentiable except at isolated points, and 1 + γ1
(x) > 0, 1 − γ2
(x) > 0. Denote D

+ = D

∩ {y > 0} = D+ and D

− = D

∩ {y < 0} and z1

= l − iγ1 (l) = l − iγ2 (l). We consider the Riemann– Hilbert problem (Problem A
) for equation (3.2) in D

with the boundary conditions (4.2) and Re [λ(z)w(z)] = r(z), z ∈ L

2 , Im [λ(z1

)w(z1

)] = b1 , (4.14) where z1

= l − iγ1 (l) = l − iγ2 (l), and λ(z), r(z) satisfy the corresponding condition Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

Cα [r(z), L

2 ] ≤ k2 ,

max

z∈L1

|b1 | ≤ k2 ,

1 , |a(x) − b(x)|

max

z∈L2

Cα [λ(z), L

2 ] ≤ k0 , 1 ≤ k0 , |a(x) + b(x)|

(4.15)

in which α (1/2 < α < 1), k0 , k2 are non-negative constants. By the conditions in (4.13), the inverse function x = τ (µ) of x − γ2 (x) = µ can be found, namely ν = 2τ (µ) − µ,

0 ≤ µ ≤ 2.

(4.16)

We make a transformation µ ˜ = µ,

ν˜ =

2ν , 2τ (µ) − µ

0 ≤ µ ≤ 2,

0 ≤ ν ≤ 2τ (µ) − µ,

(4.17)

where µ, ν are real variables, its inverse transformation is µ=µ ˜ = x˜ + y˜, [2τ (µ) − µ]˜ ν 2 [2τ (x − γ2 (x)) − x + γ2 (x)](˜ x − y˜) , = 2

ν=

(4.18) 0≤µ ˜ ≤ 2, 0 ≤ ν˜ ≤ 2.

142

IV. First Order Mixed Complex Equations

Hence we have 1 2(x − y) + (x + y)[2τ (x − γ2 (x)) − x + γ2 (x)] x˜ = (˜ µ + ν˜) = , 2 2[2τ (x − γ2 (x)) − x + γ2 (x)] 1 −2(x − y) + (x + y)[2τ (x − γ2 (x)) − x + γ2 (x)] y˜ = (˜ µ − ν˜) = , 2 2[2τ (x − γ2 (x)) − x + γ2 (x)] 1 x = (µ + ν) = 2 1 y = (µ − ν) = 2

(4.19)

1 [(2τ (x − γ2 (x)) − x + γ2 (x))(˜ x − y˜) + 2(˜ x + y˜)], 4 1 [(−2τ (x − γ2 (x)) + x − γ2 (x))(˜ x − y˜) + 2(˜ x + y˜)]. 4

Denote by z˜ = x˜ + j y˜ = g(z), z = x + jy = g −1 (˜ z ) the transformation (4.18) and its inverse transformation in (4.19) respectively. Through the transformation (4.18), we obtain (u + v)ν˜ = [τ (µ) − µ/2](u + v)ν ,

(u − v)µ˜ = (u − v)µ in D
− .

(4.20)

System (4.10) in D

− is reduced to ξν˜ = [τ (µ) − µ/2][Aξ + Bη + E]

in D
− .

(4.21)

ηµ˜ = Cξ + Dη + F Moreover, through the transformation (4.19), the boundary condition (4.14) on L

2 is reduced to Re [λ(g −1 (˜ z ))w(g −1 (˜ z ))] = r[g −1 (˜ z )], z ∈ L
2 , Im [λ(g −1 (z1
))w(g −1 (z1
)] = b1 ,

(4.22)

in which z1
= z˜1

= g(z1

). Therefore the boundary value problem (4.10),(4.14) is transformed into the boundary value problem (4.21),(4.22). According to the method in the proof of Theorem 4.1, we can see that the boundary value problem (3.2) (in D+ ), (4.21), (4.2), (4.22) has a unique solution w(˜ z ), and then w(z) is a solution of the boundary value problem (3.2),(4.2),(4.14). But we mention that through the transformation (4.17) or (4.19), the boundaries L

1 , L

2 are reduced to L
1 , L
2 respectively, such that L
1 , L
2 satisfy the condition as stated in (4.1). In fact, if the intersection z1

of L

1 and L

2 belongs to L2 and γ1 (x) ≥ 2(1 − l) + x, 2 − 2l ≤ x ≤ l, then the above requirement can be satisfied. If z1

∈ L1 = {x + y = 0}, γ2 (x) ≥ 2l − x, l ≤ x ≤ 2l, then we can proceed similarly. Theorem 4.2 If the mixed equation (3.2) satisfies Condition C in the domain D

with the boundary Γ ∪ L

1 ∪ L

2 , where L

1 , L

2 are as stated in (4.13), then Problem A
for (3.2), (4.2), (4.14) in D

has a unique solution w(z).

5. Discontinuous Riemann–Hilbert Problem

5

143

The Discontinuous Riemann–Hilbert Problem for Quasilinear Mixed Equations of First Order

This section deals with the discontinuous Riemann–Hilbert problem for quasilinear mixed (elliptic-hyperbolic) complex equations of first order in a simply connected domain. Firstly, we give the representation theorem and prove the uniqueness of solutions for the above boundary value problem. Afterwards by using the method of successive iteration, the existence of solutions for the above problem is proved. 5.1 Formulation of the discontinuous Riemann–Hilbert problem for complex equations of mixed type Let D be a simply connected domain with the boundary Γ ∪ L1 ∪ L2 as stated as before, where D+ = {|z − 1| < 1, Im z > 0}. We discuss the first order quasilinear complex equations of mixed type as stated in (3.2) with Condition C. In order to introduce the discontinuous Riemann-Hilbert boundary value problem for the complex equation (3.2), let the functions a(z), b(z) possess discontinuities of first kind at m − 1 distinct points z1 , z2 , . . . , zm−1 ∈ Γ, which are arranged according to the positive direction of Γ, and Z = {z0 = 2, z1 , . . . , zm = 0} ∪ {x ± y = 0, x ± y = 2, y ≤ 0}, where m is a positive integer, and r(z) = O(|z −zj |−βj ) in the neighborhood of zj (z = 0, 1, . . . , m) on Γ, in which βj (j = 0, 1, . . . , m) are sufficiently small positive numbers. Denote λ(z) = a(z) + ib(z) and |a(z)| + |b(z)| =
0, there is no harm in assuming that |λ(z)| = 1, z ∈ Γ∗ = Γ\Z. Suppose that λ(z), r(z) satisfy the conditions λ(z) ∈ Cα (Γj ),

|z − zj |βj r(z) ∈ Cα (Γj ),

j = 0, 1, . . . , m,

(5.1)

herein Γj is an arc from the point zj−1 to zj on Γ and Γj (j = 1, . . . , m) does not include the end points, α(0 < α < 1) is a constant. ¯ and w(z) on Problem A∗ Find a continuous solution w(z) of (3.2) in D∗ = D\Z Z maybe become infinite of an order lower than unity, which satisfies the boundary conditions Re [λ(z)w(z)] = r(z), z ∈ Γ, (5.2) Re [λ(z)w(z)] = r(z),

z ∈ Lj (j = 1 or 2),

Im [λ(z)w(z)]|z=z1 = b1 ,

(5.3)

where b1 are real constants, λ(z) = a(x) + ib(x)(|λ(z)| = 1), z ∈ Γ ∪ Lj (j = 1 or 2), and λ(z), r(z), b1 satisfy the conditions Cα [λ(z), Γj ] ≤ k0 ,

Cα [r(z), Γj ] ≤ k2 ,

|b1 | ≤ k2 ,

Cα [λ(z), Lj ] ≤ k0 ,

Cα [r(z), Lj ] ≤ k2 ,

j = 1 or 2,

1 ≤ k0 , or max z∈L1 |a(z) − b(z)|

1 ≤ k0 , max z∈L2 |a(z) + b(z)|

(5.4)

144

IV. First Order Mixed Complex Equations

in which α (1/2 < α < 1), k0 , k2 are non-negative constants. The above discontinuous Riemann–Hilbert boundary value problem for (3.2) is called Problem A∗ . Denote by λ(zj −0) and λ(zj +0) the left limit and right limit of λ(z) as z → zj (j = 0, 1, . . . , m) on Γ, and iφj

e

λ(zj − 0) , = λ(zj + 0) 







1 φj λ(zj − 0) γj = ln − Kj , = πi λ(zj + 0) π

φj Kj = + Jj , Jj = 0 or 1, π

(5.5)

j = 0, 1, . . . , m,

in which zm = 0, z0 = 2, λ(z) = exp(−iπ/4) on L0 = (0, 2) and λ(z0 − 0) = λ(zn + 0) = exp(−iπ/4), or λ(z) = exp(iπ/4) on L0 and λ(z0 − 0) = λ(zn + 0) = exp(iπ/4), and 0 ≤ γj < 1 when Jj = 0, and −1 < Jj < 0 when Jj = 1, j = 0, 1, . . . , m, and 

m  γj 1 φj − K = (K0 + K2 + · · · + Km ) = 2 2π 2 j=0



(5.6)

is called the index of Problem A∗ . Now the function λ(z) on Γ ∪ L0 is not continuous, we can choose Jj = 0 or 1 (0 ≤ j ≤ m), hence the index K is not unique. Here we choose the index K = −1/2. Let βj + γj < 1, j = 0, 1, . . . , m. We can require that the solution u(z) satisfy the conditions −δ

⎧ ⎨ βj +τ,

uz = O(|z−zj | ), δ = ⎩

for γj ≥ 0,

|γj |+τ, for γj < 0,

and γj < 0, βj ≤ |γj |,

βj > |γj |,

(5.7)

j = 1, . . . , m,

in the neighborhood of zj in D∗ , where τ (< α) is a small positive number. For Problem A∗ of the quasilinear complex equation (3.2), we can prove that there exists a unique solution by using a similar method as stated in the last section. Next we discuss the more general discontinuous Riemann–Hilbert problem. As stated before, denote L = L1 ∪ L2 , L1 = {x = −y, 0 ≤ x ≤ 1}, L2 = {x = y + 2, 1 ≤ x ≤ 2}, and D+ = D ∩ {y > 0}, D− = D ∩ {y < 0}. Here, there are n points E1 = a1 , E2 = a2 , . . . , En = an on the segment AB = (0, 2) = L0 , where a0 = 0 < a1 < a2 < · · · < an < an+1 = 2, and denote by A = A0 = 0, A1 = (1 − i)a1 /2, A2 = (1−i)a2 /2, . . . , An = (1−i)an /2, An+1 = C = 1−i and B1 = 1−i+(1+i)a1 /2, B2 = 1 − i + (1 + i)a2 /2, . . . , Bn = 1 − i + (1 + i)an /2, B = Bn+1 = 2, on the segments [n/2] AC, CB respectively. Moreover, we denote D1− = D− ∩ {∪j=0 (a2j ≤ x − y ≤ a2j+1 )}, [(n+1)/2] − − ˜ 2j+1 = D− ∩{a2j ≤ x−y ≤ a2j+1 }, (a2j−1 ≤ x+y ≤ a2j )} and D D2 = D− ∩{∪j=1 − ˜ 2j j = 0, 1, . . . , [n/2], D = D− ∩ {a2j−1 ≤ x + y ≤ a2j }, j = 1, . . . , [(n + 1)/2], and + − D∗− = D− \Z, Z = {∪n+1 j=0 (x ± y = aj , y ≤ 0)}, D∗ = D ∪ D∗ . ¯ where Problem A∗ Find a continuous solution w(z) of (3.2) in D∗ = D\Z, Z = {z0 , z1 , . . . , zm , a1 , . . . , an } ∪ {x ± y = aj , y ≤ 0, j = 1, . . . , n}, and the above solution w(z) satisfies the boundary conditions (5.2) and

5. Discontinuous Riemann–Hilbert Problem

145

Re [λ(z)w(z)] = r(z), z ∈ L3 =

[n/2] j=0

A2j A2j+1 ,

Re [λ(z)w(z)] = r(z), z ∈ L4 =

[(n+1)/2] j=1

B2j−1 B2j ,

(5.8)

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 , j = 0, 1, . . . , [n/2], Im [λ(z)w(z)]|z=B2j−1 = c2j , j = 1, . . . , [(n + 1)/2], where cj (j = 1, . . . , n + 1) are real constants, λ(z) = a(x) + ib(x), |λ(z)| = 1, z ∈ Γ, and λ(z), r(z), cj (j = 1, . . . , n + 1) satisfy the conditions Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

|cj | ≤ k2 ,

Cα [λ(z), Lj ] ≤ k0 ,

Cα [r(z), Lj ] ≤ k2 ,

1 , max z∈L3 |a(x) − b(x)|

1 and max ≤ k0 , z∈L4 |a(x) + b(x)|

j = 0, 1, . . . , n + 1,

j = 3, 4,

(5.9)

where α (1/2 < α < 1), k0 , k2 are non-negative constants. The above discontinuous Riemann–Hilbert boundary value problem for (3.2) is called Problem A∗ . Denote by λ(tj − 0) and λ(tj + 0) the left limit and right limit of λ(z) as z → tj = zj (j = 0, 1, . . . , m, zm+k = ak , k = 1, . . . , n, zn+m+1 = 2) on Γ ∪ L0 (L0 = (0, 2)), and eiφj =

γj =

φj + Jj , π

Jj = 0 or 1,



Kj =



λ(tj − 0) , λ(tj + 0) 

1 λ(tj − 0) ln πi λ(tj + 0)



=

φj − Kj , π (5.10)

j = 0, 1, . . . , m + n,

in which [a] is the largest integer not exceeding the real number a, λ(z) = exp(−iπ/4) on L
1 = AB ∩ D1− and λ(a2j + 0) = λ(a2j+1 − 0) = exp(−iπ/4), j = 0, 1, . . . , [n/2], and λ(z) = exp(iπ/4) on L
2 = AB ∩ D2− and λ(a2j−1 + 0) = λ(a2j − 0) = exp(iπ/4), j = 1, . . . , [(n + 1)/2], and 0 ≤ γj < 1 when Jj = 0, and −1 < γj < 0 when Jj = 1, j = 0, 1, . . . , m + n, and 

m+n  φj γj 1 − K = (K0 + K1 + · · · + Km+n ) = 2 2π 2 j=0



(5.11)

is called the index of Problem A∗ and Problem A∗0 . We can require that the solution w(z) in D+ satisfy the conditions

146

IV. First Order Mixed Complex Equations

w(z) = O(|z − zj |−τ ),

τ = γj
+ δ,

j = 0, 1, . . . , m + n, +

(5.12)

γj


in the neighborhood of zj (0 ≤ j ≤ m + n) in D , where = max(0, −γj ) (j =
1, . . . , m − 1, m + 1, . . . , m + n), γm = max(0, −2γm ), γ0
= max(0, −2γ0 ) and γj (j = 0, 1, . . . , m + n) are real constants in (5.10), δ is a sufficiently small positive number, and choose the index K = −1/2. Now we explain that in the closed domain D− , the functions u + v, u − v corresponding to the solution w(z) in the neighborhoods of the 2n + 2 characteristic lines Z0 = {x + y = 0, x − y = 2, x ± y = aj (j = m + 1, . . . , m + n), y ≤ 0} may be not bounded if γj ≤ 0(j = m, . . . , m + n + 1). Hence if we require that u + v, u − v in D− \Z0 is bounded, then it needs to choose γj > 0 (j = 0, 1, . . . , m + n + 1). 5.2 Representation of solutions for the discontinuous Riemann–Hilbert problem We first introduce a lemma. Lemma 5.1 Suppose that the complex equation (3.2) satisfies Condition C. Then there exists a solution of Problem A∗ for (3.2) in D+ with the boundary conditions (5.2) and Re [λ(z)w(z)]|z=x = s(x),

⎧ ⎨1 − i

λ(x) = ⎩

Cβ [s(x), L
j ] ≤ k3 , on L
1 = D1− ∩ AB,

j = 1, 2, (5.13)

1 + i on L
2 = D2− ∩ AB,

and w(z) satisfies the estimate Cβ [w(z)X(z), D+ ] ≤ M21 (k1 + k2 + k3 ),

(5.14)

in which k3 is a non-negative constant, s(x) is as stated in the form (5.25) below, γj
+δ X(z) = Πm+n , herein γj
= max(0, −γj )(j = 1, . . . , m − 1, m + 1, m + j=0 |z − zj |

n), γ0 = max(0, −2γ0 ), γm = max(0, −2γm ) and γj (j = 0, 1, . . . , m + n) are real constants in (5.10), β (0 < β < δ), δ are sufficiently small positive numbers, and M21 = M21 (p0 , β, k0 , D+ ) is a non-negative constant. By using the method as in the proofs of Theorems 2.1–2.3, Chapter III, Theorem 1.2 and Lemma 2.1, we can prove the lemma. Theorem 5.2

Problem A∗ for equation (1.2) in D has a unique solution w(z).

Proof First of all, similarly to Theorem 1.2, the solution w(z) = u(z) + iv(z) of equation (1.2) in D− can be expressed as (1.19). According to the proof of Theorem 1.2, we can obtain f (x + y) on L
1 = D1− ∩ AB and g(x − y) on L
2 = D2− ∩ AB in the form

5. Discontinuous Riemann–Hilbert Problem

g(x) = k(x) =

147

2r((1−i)x/2)−[a((1−i)x/2)+b ((1−i)x/2)]h2j a((1 − i)x/2) − b ((1 − i)x/2) − ˜ 2j+1 = D ˜ 2j+1 on L ∩ AB,

f (x) = h(x) =

2r((1 + i)x/2 + 1 − i) a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) −

(5.15)

[a((1+i)x/2+1−i)−b ((1+i)x/2+1−i)]k2j−1 a((1+i)x/2+1−i)+b ((1+i)x/2+1−i) − ˜ 2j = D2j ∩ AB, on L

˜ j− (j = 1, 2, . . . , 2n + 1) are as stated in Subsection 5.1, and where D h2j = Re [λ(A2j+1 )(r(A2j+1 ) + ic2j+1 )] + Im [λ(A2j+1 )(r(A2j+1 ) + ic2j+1 )] ˜ 2j+1 , on L

j = 0, 1, . . . , [n/2],

k2j−1 = Re [λ(B2j−1 )(r(B2j−1 ) + ic2j )] − Im [λ(B2j−1 )(r(B2j−1 ) + ic2j )] ˜ 2j , on L

j = 1, . . . , [(n + 1)/2],

− − ˜ 2j+1 = D ˜ 2j+1 ˜ 2j = D ˜ 2j where L ∩ AB, j = 0, 1, . . . , [n/2], L ∩ AB, j = 1, . . . , [(n + 1)/2]. By using Theorem 2.2, Chapter III, choosing an appropriate index K = −1/2, there exists a unique solution w(z) of Problem A∗ in D+ with the boundary conditions (5.2) and

⎧ ⎨ k(x),

Re [λ(x)w(x)] = ⎩

λ(x) = h(x),

and denote

⎧ ⎨ h(x)

Re [λ(x)w(x)] =

⎩

⎧ ⎪ ⎨1 − i

on L
1 = D1− ∩ AB,

⎪ ⎩1 + i

on L
2 = D2− ∩ AB,

on L
1 ,

k(x) on L
2 ,

Cβ [X(x)k(x), L
1 ] ≤ k2 ,

(5.16)

Cβ [X(x)h(x), L
2 ] ≤ k2 ,

herein β(0 < β ≤ α < 1), k2 are non-negative constants. Next we find a solution w(z) of Problem A∗ for (1.2) in D− with the boundary conditions Re [λ(z)w(z)] = r(z) on L3 ∪ L4 , Im [λ(z)w(z)]|z=A2j+1 = c2j+1 , Im [λ(z)w(z)]|z=B2j−1 = c2j ,

j = 0, 1, . . . , [n/2],

(5.17)

j = 1, . . . , [(n + 1)/2],

and (5.16), where cj (j = 1, . . . , n + 1) are as stated in (5.8). By the result and the ˜ 1− method in Chapters I and II, we can find the solution of Problem A∗ for (1.2) in D in the form

148

IV. First Order Mixed Complex Equations

w(z) = w(z) ˜ + λ(A2j+1 )[r(A2j+1 ) + ic2j+1 ], 1 w(z) ˜ = [(1 + i)f2j+1 (x + y) + (1 − i)g2j+1 (x − y)], 2 f2j+1 (x + y) = Re [λ(x + y)w(x + y)], g2j+1 (x − y) =

2r((1 − i)(x − y)/2) ˜− , in D 2j+1 a((1 − i)(x − y)/2) − b((1 − i)(x − y)/2) j = 0, 1, . . . , [n/2],

w(z) = w(z) ˜ + λ(B2j−1 )[r(B2j−1 ) + ic2j ], 1 w(z) ˜ = [(1 + i)f2j (x + y)+(1 − i)g2j (x − y)], f2j (x + y) 2 2r((1+i)(x+y)/2+1−i) , = h(x + y) = a((1+i)(x+y)/2+1−i)+b ((1+i)(x+y)/2+1−i) ˜−, g2j (x − y) = Re [λ(x − y)w(x − y)] in D 2j



j = 1, . . . ,



n+1 . 2

Furthermore, from the above solution we can find the solution of Problem A∗ for (1.2) in D− \{D1− ∪ D2− }, and the solution w(z) of Problem A∗ for (1.2) in D− possesses the form 1 w(z) = [(1 + i)f (x + y) + (1 − i)g(x − y)], 2 f (x + y) = Re [(1 − i)w(x + y)] in D− \D2− , (5.18)

g(x − y) = k(x − y) in D1− , f (x + y) = h(x + y) in D2− , g(x − y) = Re [(1 + i)w(x − y)] in D− \D1− ,

where k(x), h(x) are as stated in (5.15), and w(z) is the solution of Problem A∗ for (1.2) with the boundary conditions (5.2),(5.16). Theorem 5.3 Let the complex equation (3.2) satisfy Condition C. Then any solution of Problem A∗ for (3.2) can be expressed as w(z) = w0 (z) + W (z) in D,

(5.19)

where w0 (z) is a solution of Problem A∗ for equation (1.2), and W (z) possesses the form W (z) = w(z) − w0 (z) in D,

˜

φ(z) ˜ ˜ w(z) = Φ(z)e + ψ(z),

g(ζ) 1 ˜ dσζ , φ(z) = φ˜0 (z) + T g = φ˜0 (z) − + π ζ −z D

˜ ψ(z) = T f in D+ ,

5. Discontinuous Riemann–Hilbert Problem

149

W (z) = Φ(z) + Ψ(z) in D− ,

Ψ(z) =

⎧ ν  µ ⎪ ⎪ g 1 (z)dνe1 + g 2 (z)dµe2 ⎪ ⎪ ⎨ a2j+1 0  µ ⎪ ν 1 ⎪ ⎪ ⎪ g (z)dνe1 + g 2 (z)dµe2 ⎩ 2

a2j−1

− in D2j+1 , − in D2j ,

j = 0, 1, . . . , [n/2],

(5.20)

j = 1, . . . , [(n+1)/2],

in which e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic ˜ function in D+ , Im [φ(z)] = 0 on L0 , and ⎧ ⎨ A1

g(z) = ⎩

+ A2 w/(w),

0, w(z) = 0,

w(z)
= 0,

z ∈ D+ ,

(5.21)

f = A1 T f + A2 T f + A3 in D+ , g 1 (z) = Aξ + Bη + E, g 2 (z) = Cξ + Dη + F in D− ,

˜ ˜ where ξ = Re w + Im w, η = Re w − Im w, and φ(z), ψ(z) satisfy the estimates ˜ Cβ [φ(z), D+ ] + Lp0 [φ˜z¯, D+ ] ≤ M22 , ˜ Cβ [ψ(z), D+ ] + Lp0 [ψ˜z¯, D+ ] ≤ M22 ,

(5.22)

˜ is where p0 (0 < p0 ≤ p), M22 = M22 (p0 , α, k, D+ ) are non-negative constants, Φ(z) analytic in D+ and Φ(z) is a solution of equation (1.2) in D− satisfying the boundary conditions ˜ ˜ ˜ Re [λ(z)eφ(z) Φ(z)] = r(z) − Re [λ(z)ψ(z)], ˜ φ(x)

˜ Re [λ(x)(Φ(x)e

˜ + ψ(x))] = s(x),

x ∈ L0 ,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ Γ,

z ∈ L0 ,

z ∈ L3 ∪ L4 ,

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0,

j = 0, 1, . . . , [n/2],

Im [λ(z)(Φ(z) + Ψ(z))]|z=B2j−1 = 0,

j = 1, . . . , [(n + 1)/2].

(5.23)

Moreover the solution w0 (z) of Problem A∗ for (1.2) satisfies the estimate in the form Cβ [X(z)w0 (z), D+ ] + C[w0± (µ, ν)Y ± (µ, ν), D− ] ≤ M23 (k1 + k2 ) γj
+δ

(5.24)


γj +δ in which X(z) = Πm+n , Y ± (z) = Y ± (µ, ν) = Πm+n , j=0 |z − zj | j=0 |x ± y − zj | ± w0 (µ, ν) = Re w0 (z)± Im w0 (z), w0 (z) = w0 (µ, ν), µ = x + y, ν = x − y, γj
(j = 0, 1, ..., m + n) are as stated in (5.14), M23 = M23 (p0 , β, k0 , D) is a nonnegative constant.

150

IV. First Order Mixed Complex Equations

Proof Let the solution w(z) of Problem A∗ be substituted into the complex equation (3.2) and the solution w0 (z) = ξ0 e1 + η0 e2 of Problem A∗ for equation (1.2) be substituted in the position of w in (5.21). Thus the functions f (z), g(z) in D+ and g1 (z), g2 (z) and Ψ(z) in D− in (5.20),(5.21) can be determined. Moreover, by ˜ Theorem 5.2, we can find an analytic function Φ(z) in D+ and a solution Φ(z) of (1.2) in D− with the boundary conditions (5.23), where ⎧ ⎪ 2r((1 − i)x/2) − 2R((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ a((1 − i)x/2) − b((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1 − i)x/2) + b((1 − i)x/2)]h2j ⎪ ⎪ ⎪− + Re [λ(x)Ψ(x)], ⎪ ⎪ a((1 − i)x/2) − b((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ x ∈ (a2j , a2j+1 ), j = 0, 1, . . . , [n/2],

s(x) = ⎪

⎪ 2r((1+i)x/2+1−i)−2R((1+i)x/2+1−i) ⎪ ⎪ ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1 + i)x/2 + 1 − i) − b((1 + i)x/2 + 1 − i)]k2j−1 ⎪ ⎪ − ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ +Re [λ(x)Ψ(x)], x ∈ (a , a ), j = 1, . . . , [(n + 1)/2], 2j−1

(5.25)

2j

in which the real constants h2j (j = 0, 1, ..., [n/2]), k2j−1 (j = 1, ..., [(n + 1)/2]) are as stated in (5.15), thus ⎧ ˜ ˜ φ(z) ⎨ Φ(z)

w(z) = w0 (z) + W (z) = ⎩

˜ + ψ(z) in D+ ,

w0 (z) + Φ(z) + Ψ(z) in D− ,

is the solution of Problem A∗ for the complex equation


wz w¯z∗



⎧ ⎫ ⎨D + ⎬

= A1 w + A2 w¯ + A3 in ⎩ ⎭ , D−

(5.26)

and the solution w0 (z) of Problem A∗ for (1.2) satisfies the estimate (5.24). 5.3 Existence and uniqueness of solutions of the Riemann–Hilbert problem for (3.2) Theorem 5.4 Suppose that the complex equation (3.2) satisfies Condition C. Then Problem A∗ for (3.2) is solvable. Proof In order to find a solution w(z) of Problem A∗ in D, we express w(z) in the form (5.19)–(5.21). On the basis of Theorem 5.2, we see that Problem A∗ for (1.2) has a unique solution w0 (z)(= ξ0 e1 + η0 e2 ), and substitute it into the position of w = ξe1 + ηe2 in the right-hand side of (3.2). Similarly to (2.19), from (5.19)–(5.21),

5. Discontinuous Riemann–Hilbert Problem

151

we obtain the corresponding functions g0 (z), f0 (z) in D+ , g01 (z), g02 (z) in D− and the functions 

1 φ˜1 (z) = φ˜0 (z)− π

D+

g0 (ζ) dσζ , ζ −z

˜ = T f0 in D+ , ψ(z)

⎧ ν  µ ⎪ − ⎪ g01 (z)dνe1 + g02 (z)dµe2 in D2j+1 , ⎪ ⎪ ⎨ a2j+1 0 Ψ1 (z) = ⎪ ν  µ ⎪ − ⎪ ⎪ g01 (z)dνe1 + g02 (z)dµe2 in D2j , ⎩ 2

j = 0, 1, . . . , [n/2],

(5.27)

j = 1, . . . , [(n+1)/2],

a2j−1

can be determined, where µ = x + y, ν = x − y, and the solution w0 (z) satisfies the estimate (5.24), i.e. Cβ [w0 (z)X(z), D+ ] + Cβ [w0± (z)Y ± (z), D− ] ≤ M23 (k1 + k2 ).

(5.28)

˜ Moreover, by Theorem 5.2, we can find an analytic function Φ(z) in D+ and a solution − Φ1 (z) of (1.2) in D satisfying the boundary conditions ˜ 1 (z)eφ˜1 (z) + ψ˜1 (z))] = r(z), Re [λ(z)(Φ ˜ 1 (x) + ψ˜1 (x))] = s(z), Re [λ(x)(Φ

z ∈ Γ,

x ∈ L0 ,

Re [λ(x)Φ1 (x)] = Re [λ(x)(W1 (x) − Ψ1 (x))], Re [λ(z)Φ1 (z)] = −Re [λ(z)Ψ1 (z)],

z ∈ L0 ,

z ∈ L3 ∪ L 4 ,

Im [λ(z)(Φ1 (z) + Ψ1 (z))]|z=A2j+1 = 0,

j = 0, 1, . . . , [n/2],

Im [λ(z)(Φ1 (z) + Ψ1 (z))]|z=B2j−1 = 0,

j = 1, . . . , [(n + 1)/2],

(5.29)

where ⎧ ⎪ 2r((1 − i)x/2) − 2R1 ((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ a((1 − i)x/2) − b ((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1 − i)x/2) + b ((1 − i)x/2)]h2j ⎪ ⎪ + Re [λ(x)Ψ1 (x)], − ⎪ ⎪ a((1 − i)x/2) − b ((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ x ∈ (a2j , a2j+1 ), j = 0, 1, . . . , [n/2],

s1 (x) = ⎪

⎪ 2r((1+i)x/2+1−i)−2R1 ((1+i)x/2+1−i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b ((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1 + i)x/2 + 1 − i) − b ((1 + i)x/2 + 1 − i)]k2j−1 ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b ((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ +Re [λ(x)Ψ (x)], x ∈ (a , a ), j = 1, . . . , [(n + 1)/2], 1

2j−1

2j

in which the real constants h2j , k2j−1 are as stated in (5.15), and w1 (z) = w0 (z) + W1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z) in D−

(5.30)

152

IV. First Order Mixed Complex Equations

satisfies the estimate Cβ [w1 (z)X(z), D+ ]+C[w1± (µ, ν)Y ± (µ, ν), D− ] ≤ M24 = M24 (p0 , β, k, D− ),

(5.31)

here w1± (µ, ν) = Re w1 (µ, ν) ± Im w1 (µ, ν), Y ± (µ, ν), X(z), Y ± (z) are as stated in (5.24). Furthermore we substitute w1 (z) = w0 (z)+W1 (z) and corresponding functions w1 (z), ξ1 (z) = Re w1 (z) + Im w1 (z), η1 (z) = Re w1 (z) − Im w1 (z) into the positions w(z), ξ(z), η(z) in (5.20), (5.21), and similarly to (5.27)–(5.30), we can find the ˜ 2 (z) in D+ and Ψ2 (z), Φ2 (z) and W2 (z) = corresponding functions φ˜2 (z), ψ˜2 (z), Φ Φ2 (z) + Ψ2 (z) in D− . The function ⎧ ˜ 2 (z)eφ˜2 (z) ⎨Φ

w2 (z) = w0 (z) + W2 (z) = ⎩

+ ψ˜2 (z) in D+ ,

w0 (z) + Φ2 (z) + Ψ2 (z) in D−

(5.32)

satisfies the similar estimate in the form (5.31). Thus there exists a sequence of functions {wn (z)} as follows ⎧ ˜ n (z)eφ˜n (z) ⎨Φ

wn (z) = w0 (z) + Wn (z) = ⎩

+ ψ˜n (z) in D+ ,

w0 (z) + Φn (z) + Ψn (z) in D− ,

⎧ ν  µ ⎪ 1 2 ⎪ gn−1 (z)e1 dν + gn−1 (z)e2 dµ ⎪ ⎪ ⎨ a2j+1 0 Ψn (z) =⎪ ν  µ ⎪ 1 2 ⎪ ⎪ gn−1 (z)e1 dν + gn−1 (z)e2 dµ ⎩ 2

a2j−1

1 gn−1 (z) = Aξn−1 +Bηn−1 +E,

− in D2j+1 , − in D2j ,

j = 0, 1, . . . , [n/2],

(5.33)

j = 1, . . . , [(n+1)/2],

2 gn−1 (z) = Cξn−1 +Dηn−1 + F in D− ,

and then ± |[w1± (µ, ν) − w0± (µ, ν)]Y ± (µ, ν)| ≤ |Φ± 1 (µ, ν)Y (µ, ν)|




√ + 2 |Y − (µ, ν)| +

 

+|Y (µ, ν)| |

0



µ

max |



1≤j≤n+1

a2j+1

g01 (z)e2 dµ|


ν

g01 (z)e1 dν| + |

+ max | 1≤j≤n+1



µ

a2j−1

 2

ν



g02 (z)e1 dν| 

(5.34)

g02 (z)e2 dµ|

−

≤ 2M25 M (4m + 1)R in D , where m = C[w0+ (µ, ν)Y + (µ, ν), D− ]+C[w0− (µ, ν)Y − (µ, ν), D− ], M = 1+ 4k02 (1+k02 ),

5. Discontinuous Riemann–Hilbert Problem

153

˜ |B|, ˜ |C|, ˜ |D|). ˜ It is clear that wn (z) − wn−1 (z) satisfies R
= 2, M25 = maxz∈D− (|A|, wn (z) − wn−1 (z)

=

 µ ⎧ ν 1 1 2 2 − ⎪ ⎪ [g −g ]e dν + [gn−1 −gn−2 ]e2 dµ in D2j+1 , 1 ⎪ n−1 n−2 ⎪ ⎪ a2j+1 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ j=0, 1,. . ., [n/2], Φn (z) − Φn−1 (z) +⎪  ν  µ ⎪ 1 1 2 2 − ⎪ ⎪ [g −g ]e dν + [gn−1 −gn−2 ]e2 dµ in D2j , 1 ⎪ n−1 n−2 ⎪ ⎪ a2j−1 ⎪ 2 ⎪ ⎪ ⎪ ⎩

(5.35)

j=1,. . ., [(n+1)/2].

Moreover, we can find the solution w(z) of Problem A∗ for (3.2) in the set [n/2] ˜ − [(n+1)/2] ˜ − D− \{(∪j=0 D D2j )} = D− \{D1− ∪ D2− }. From the above result, 2j+1 ) ∪ (∪j=1 ± |[wn± − wn−1 ]Y ± | ≤ [2M25 M (4m + 1)]n



R


0
n

[2M25 M (4m + 1)R ] ≤ n!

R
n−1 dR
(n − 1) !

(5.36)

−

in D ,

can be obtained, and then we see that the sequence of functions {wn± (µ, ν)Y ± (µ, ν)}, i.e. ± wn± (µ, ν)Y ± (µ, ν) = {w0± +[w1± −w0± ]+. . .+[wn± −wn−1 ]}Y ± (µ, ν)(n = 1, 2, . . .) (5.37)

in D− uniformly converge to w∗± (µ, ν)Y ± (µ, ν), and w∗ (z) = [w+ (µ, ν) + w− (µ, ν) −i(w+ (µ, ν) − w− (µ, ν))]/2 satisfies the equality w∗ (z) = w0 (z)+Φ∗ (z)+Ψ∗ (z),

⎧ ν  µ ⎪ − ⎪ g∗1 (z)e1 dν + g∗2 (z)e2 dµ in D2j+1 , j = 0,1,...,[n/2], ⎪ ⎪ ⎨ a2j+1 0 Ψ∗ (z) = ⎪  ν  µ ⎪ − ⎪ ⎪ g∗1 (z)e1 dν + g∗2 (z)e2 dµ in D2j , j = 1,...,[(n+1)/2], ⎩ 2

(5.38)

a2j−1

g∗1 (z) = Aξ∗ +Bη∗ +E, g∗2 (z) = Cξ∗ +Dη∗ +F in D− , and the corresponding function u∗ (z) is just a solution of Problem A∗ for equation (3.2) in the domain D− and w∗ (z) satisfies the estimate


C[w∗± (µ, ν)Y ± (µ, ν), D− ] ≤ M26 = e4M25 M (2m+1)R .

(5.39) φ˜n (z)

˜ n (z)e In addition, we can find a sequence of functions {wn (z)}(wn (z) = Φ + ψ˜n (z)) ˜ n (z) is an analytic function in D+ satisfying the boundary conditions in D+ and Φ ˜

˜ n (z)eφn (z) + ψ˜n (z))] = r(z), z ∈ Γ, Re [λ(z)(Φ ˜ n (x)eφ˜n (x) + ψ˜n (x))] = s(x), x ∈ L0 , Re [λ(x)(Φ

(5.40)

154

IV. First Order Mixed Complex Equations

in which

⎧ 2r((1 − i)x/2) − 2Rn ((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ a((1 − i)x/2) − b((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1 − i)x/2) + b((1 − i)x/2)]h2j ⎪ ⎪ + Re [λ(x)Ψn (x)], ⎪ −√ ⎪ ⎪ 2[a((1 − i)x/2) − b((1 − i)x/2)] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ x ∈ (a2j , a2j+1 ), j = 0, 1, . . . , [n/2],

sn (x) = ⎪

⎪ 2r((1+i)x/2+1−i)−2Rn ((1+i)x/2+1−i) ⎪ ⎪ ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1 + i)x/2 + 1 − i) − b((1 + i)x/2 + 1 − i)]k2j−1 ⎪ ⎪ − ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

+Re [λ(x)Ψn (x)],

x ∈ (a2j−1 , a2j ),

(5.41)

j = 1, . . . , [(n + 1)/2],

in which the real constants h2j , k2j−1 are as stated in (5.15). From (5.31),(5.39), it follows that Cβ [X(x)sn (x), L0 ] ≤ M27 = M27 (p0 , β, k, D), (5.42) and then the estimate Cβ [wn (z)X(z), D+ ] ≤ M21 (k1 + k2 + M27 ).

(5.43)

Thus from {wn (z)X(x)}, we can choose a subsequence which uniformly converges to a function w∗ (z)X(z) in D+ . Combining (5.43) and (5.39), it is obvious that the ¯ satisfies the estimate solution w∗ (z) of Problem A∗ for (3.2) in D Cβ [w∗ (z)X(z), D+ ] + C[w∗± (z)Y ± (z), D− ] ≤ M28 = M28 (p0 , β, k, D),

(5.44)

where M28 is a non-negative constant. Theorem 5.5 If the complex equation (3.2) satisfies Condition C, then Problem A∗ for (3.2) has at most one solution in D. Proof Let w1 (z), w2 (z) be any two solutions of Problem A∗ for (3.2). By Condition C, we see that w(z) = w1 (z) − w2 (z) satisfies the homogeneous complex equation


wz w¯z∗

⎧ ⎫ ⎨ D+ ⎬



= A1 w + A2 w¯ in ⎩ D− ⎭

(5.45)

and boundary conditions Re [λ(z)w(z)] = 0,

z ∈ Γ,

Re [λ(x)w(x)] = s(x), Re [λ(z)w(z)] = 0,

x ∈ L0 ,

z ∈ L3 ∪ L4 ,

Im [λ(z)w(z)]|z=A2j+1 = 0,

j = 0, 1, . . . , [n/2],

Im [λ(z)w(z)]|z=B2j−1 = 0,

j = 1, . . . , [(n + 1)/2],

(5.46)

5. Discontinuous Riemann–Hilbert Problem in which

s(x) =

⎧ ⎪ 2r((1 − i)x/2) − 2R((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ a((1 − i)x/2) − b((1 − i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ +Re [λ(x)Ψ(x)], x ∈ (a2j , a2j+1 ),

155

j = 0, 1, . . . , [n/2], (5.47)

⎪ ⎪ 2r((1+i)x/2+1−i)−2R((1+i)x/2+1−i) ⎪ ⎪ ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

+Re [λ(x)Ψ(x)],

x ∈ (a2j−1 , a2j ),

j = 1, . . . , [(n + 1)/2],

From Theorem 5.3, the solution w(z) can be expressed in the form

w(z) =

⎧ ˜ φ(z) ˜ ˜ ⎪ , φ(z) = T˜g in D+ , Φ(z)e ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ⎪ ⎪ ⎪ ⎨ A1 + A2 w/w, ¯ w(z)
= 0, z ∈ D+ , ⎪ ⎪ ⎪ g(z) = ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ 0, w(z) = 0, z ∈ D+ , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Φ(z) + Ψ(z), ⎪ ⎪ ⎨  µ ⎧ ν ⎪ ˜ + Bη]e ˜ 1 dν + ˜ + Dη]e ˜ 2 dµ ⎪ [ Aξ [Cξ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ a2j+1 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ ⎨ in D2j+1 , j = 0, 1, . . . , [n/2], ⎪ ⎪ ⎪ ⎪ ⎪ Ψ(z) = ⎪  ν  µ ⎪ ⎪ ⎪ ⎪ ⎪ ˜ + Bη]e ˜ 1 dν + ˜ + Dη]e ˜ 2 dµ ⎪ ⎪ [Aξ [Cξ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2 a2j−1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ −

(5.48)

in D2j , j = 1, . . . , [(n + 1)/2].

By using the method in the proof of Theorem 5.4, we can get that |w± (z)Y ± (z)| ≤

[2M25 M (4m + 1)R
]n in D− . n!

(5.49)

Let n → ∞, we get w± (z) = 0, i.e. w(z) = w1 (z) − w2 (z) = 0, and then Φ(z) = ˜ φ(z) ˜ Ψ(z) = 0 in D− , thus s(x) = 0 on L0 . Noting that w(z) = Φ(z)e satisfies the ˜ boundary conditions in (5.46), we see that the analytic function Φ(z) in D+ satisfies the boundary conditions ˜

˜ Re [λ(z)eφ(z) Φ(z)] = 0,

z ∈ Γ,

˜ = 0, Re [λ(x)Φ(x)]

x ∈ L0 ,

(5.50)

˜ and the index of the boundary value problem (5.50) is K = −1/2, hence Φ(z) = 0 in ˜ φ(z) ˜ D+ , and then w(z) = Φ(z)e = 0 in D+ , namely w(z) = w1 (z) − w2 (z) = 0 in D+ . This proves the uniqueness of solutions of Problem A∗ for (3.2). From Theorems 5.4 and 5.5, we see that under Condition C, Problem A∗ for equation (3.2) has a unique solution w(z), which can be found by using successive iteration, and the solution w(z) satisfies the estimate (5.44), i.e. Cβ [w(z)X(z), D+ ] + C[w± (µ, ν)Y ± (µ, ν), D− ] ≤ M29 ,

(5.51)

156

IV. First Order Mixed Complex Equations

where k = (k0 , k1 , k2 ), M29 = M29 (p0 , β, k, D) is a non-negative constant. Moreover we have Theorem 5.6 Suppose that equation (3.2) satisfies Condition C. Then any solution w(z) of Problem A∗ for (3.2) satisfies the estimates (5.44) and Cβ [w(z)X(z), D+ ] + C[w(z), D− ] ≤ M30 (k1 + k2 ),

(5.52)

where X(z), Y (z) are as stated in (5.24) respectively, and M30 = M30 (p0 , β, δ, k0 , D) is a non-negative constant. From the estimates (5.51) and (5.52), we can see the regularity of solutions of Problem A∗ for (3.2). Finally, we mention that if the index K is an arbitrary even integer or 2K is an arbitrary odd integer, the above Problem A∗ for (3.2) can be considered. But in general Problem A∗ for (3.2) with K ≤ −1 has −2K − 1 solvability conditions or when K ≥ 0 its general solution includes 2K + 1 arbitrary real conditions. For more general first order complex equations of mixed type, the discontinuous Riemann–Hilbert boundary value problem remains to be discussed. The references for this chapter are [3],[8],[12],[16],[20],[25],[35],[36],[42],[44],[52], [55],[60],[63],[73],[75],[83],[85],[95],[98].

CHAPTER V SECOND ORDER LINEAR EQUATIONS OF MIXED TYPE In this chapter, we discuss the oblique derivative boundary value problem for second order linear equation of mixed (elliptic-hyperbolic) type in a simply connected domain. We first prove uniqueness and existence of solutions for the above boundary value problem, and then give a priori estimates of solutions for the problem, finally discuss the existence of solutions for the above problem in general domains. In books [12]1),3), the author investigated the Dirichlet problem (Tricomi problem) for the mixed equation of second order, i.e. uxx + sgny uyy = 0. In [69], the author discussed the Tricomi problem for the generalized Lavrent
ev-Bitsadze equation uxx + sgny uyy + Aux + Buy + Cu = 0, i.e. uξη + auξ + buη + cu = 0 with the conditions: a ≥ 0, aξ + ab − c ≥ 0, c ≥ 0 in the hyperbolic domain. In this section, we cancel the above assumption in [69] and obtain the solvability result on the discontinuous Poincar´e problem, which includes the corresponding results in [12]1),3),[69] as special cases.

1

Oblique Derivative Problems for Simplest Second Order Equation of Mixed Type

In this section, we introduce the oblique derivative boundary value problem for simplest mixed equation of second order in a simply connected domain, and verify the uniqueness and existence of solutions for the above boundary value problem. Let D be a simply connected bounded domain in the complex plane C I with the boundary ∂D = Γ ∪ L, where Γ(⊂ {y > 0}) ∈ Cα2 (0 < α < 1) with the end points z = 0, 2 and L = L1 ∪ L2 , L1 = {x = −y, 0 ≤ x ≤ 1}, L2 = {x = y + 2, 1 ≤ x ≤ 2}, and denote D+ = D ∩ {y > 0}, D− = D ∩ {y < 0} and z1 = 1 − i. We may assume that Γ = {|z − 1| = 1, y ≥ 0}, otherwise through a conformal mapping, the requirement can be realized. 1.1 The oblique derivative problem for simplest second order equation of mixed type In A. V. Bitsadze’s books [12]1),3), the author discussed the solvability of several boundary value problems including the Dirichlet problem or Tricomi problem

158

V. Second Order Linear Mixed Equations

(Problem D or Problem T ) for the second order equation of mixed type uxx + sgny uyy = 0 in D,

(1.1)




the equation is so–called Lavrent ev-Bitsadze equation, its complex form is as follows:




uz¯z uz ∗ z ∗

where




= 0 in

D+



D−

,

(1.2)

1 uz∗ = uz¯, wz∗ = [wx − iw¯y ]. 2

Now we formulate the oblique derivative boundary value problem as follows: Problem P Find a continuously differentiable solution u(z) of (1.2) in D∗ = ¯ ¯ and sat¯ + y = 0, 2}, which is continuous in D D\{0, x − y = 2} or D∗ = D\{x isfies the boundary conditions 1 ∂u = Re [λ(z)uz ] = r(z), 2 ∂l

z ∈ Γ,

u(0) = b0 ,

u(2) = b2 ,

(1.3)

1 ∂u = Re [λ(z)uz¯] = r(z), z ∈ Lj (j = 1 or 2), Im [λ(z)uz¯]|z=z1 = b1 , (1.4) 2 ∂l where l is a given vector at every point on Γ ∪ Lj (j = 1 or 2), λ(z) = a(x) + ib(x) = cos(l, x) − i cos(l, y), if z ∈ Γ, and λ(z) = a(z) + ib(z) = cos(l, x) + i cos(l, y), if z ∈ Lj (j = 1 or 2), b0 , b1 are real constants, and λ(z), r(z), b0 , b1 , b2 satisfy the conditions Cα [λ(z),Γ] ≤ k0 , cos(l, n) ≥ 0 on Γ,

Cα [r(z),Γ] ≤ k2 , |bj | ≤ k2 ,

Cα [λ(z),Lj ] ≤ k0 ,

j = 0, 1, 2,

max z∈L1

Cα [r(z),Lj ] ≤ k2 ,

j=1 or 2,

1 1 or max ≤ k0 , z∈L2 |a(z)+b(z)| |a(z)−b(z)| (1.5)

in which n is the outward normal vector at every point on Γ, α (1/2 < α < 1), k0 , k2 are non-negative constants. For convenience, we may assume that uz (z1 ) = 0, otherwise through a transformation of function Uz¯(z) = uz¯(z) − λ(z1 )[r(z1 ) + ib1 ], the requirement can be realized. The boundary value problem for (1.2) with r(z) = 0, z ∈ Γ ∪ Lj (j = 1 or 2) and b0 = b1 = b2 = 0 will be called Problem P0 . The number 1 K = (K1 + K2 ), 2

(1.6)

is called the index of Problem P and Problem P0 , where 



φj Kj = + Jj , π

Jj = 0 or 1,

eiφj =

λ(tj − 0) , λ(tj + 0)

γj =

φj − Kj , π

j = 1, 2, (1.7)

1. Simplest Mixed Equation of Second Order

159

in which t1 = 2, t2 = 0, λ(t) = eiπ/4 on L0 = (0, 2) and λ(t1 − 0) = λ(t2 + 0) = exp(iπ/4), or λ(t) = ei7π/4 on L0 and λ(t1 − 0) = λ(t2 + 0) = exp(i7π/4). Here we choose K = 0, or K = −1/2 on the boundary ∂D+ of D+ if cos(ν, n) ≡ 0 on Γ and the condition u(2) = b2 can be canceled. In this case the solution of Problem P for (1.2) is unique. In order to ensure that the solution u(z) of Problem P is continuously differentiable in D∗ , we need to choose γ1 > 0. If we require that the ¯ is only continuous, it is suffices to choose −2γ1 < 1, −2γ2 < 1. solution u(z) in D Problem P in this case still includes the Dirichlet problem as a special case. If the boundary condition Re [λ(z)uz¯] = r(z) on Lj (j = 1 or 2) in (1.4) is replaced by Re [λ(z)uz ] = r(z) on Lj (j = 1 or 2), then Problem P does not include the above Dirichlet problem (Problem D) as a special case. Setting that uz = w(z), it is clear that Problem P for (1.2) is equivalent to the Riemann–Hilbert boundary value problem (Problem A) for the first order complex equation of mixed type

+ wz¯ D = 0 in , (1.8) w¯z∗ D− with the boundary conditions Re [λ(z)w(z)] = r(z), Re [λ(z)w(z)] = r(z),

z ∈ Γ,

z ∈ Lj (j = 1 or 2), 

and the relation u(z) = 2Re

0

u(2) = b2 ,

(1.9)

Im[λ(z1 )w(z1 )] = b1 ,

z

w(z)dz + b0 in D,

(1.10)

in which the integral path in D− is as stated in Chapter II. On the basis of the result in Section 1, Chapter IV, we can find a solution w(z) of Problem A for the mixed complex equation (1.8) as stated in (1.30), Chapter IV, but the function λ(x) in the integral formula in D+ should be replaced by λ(x) on L0 = (0, 2), the function w(z) in D− should be replaced by w(z) in the second formula in (1.30), Chapter IV. Hence we have the following theorem. Theorem 1.1 Problem P for the mixed equation (1.2) has a unique solution in the form (1.10), where w(z) = w(z) ˜ + λ(z1 )[r(z1 ) − ib1 ], ⎧ ⎪ ⎪ ⎨ W [ζ(z)],

z ∈ D+ \{0, 2},

w(z) ˜ =⎪1 1 ⎪ ⎩ (1 − i)f (x + y) + (1 + i)g(x − y), 2 2 2r((1 − i)(x − y)/2) g(x − y) = a((1 − i)(x − y)/2) − b((1 − i)(x − y)/2) −

[a((1−i)(x−y)/2)+b((1−i)(x−y)/2)]f (0) , or a((1 − i)(x − y)/2) − b((1 − i)(x − y)/2)

160

V. Second Order Linear Mixed Equations 1 1 w(z) ˜ = (1 + i)g(x − y) + (1 − i)f (x + y), 2 2 f (x + y) =

2r((1 + i)(x + y)/2 + 1 − i) a((1 + i)(x + y)/2 + 1 − i) + b((1 + i)(x+y)/2+1 −i)

(1.11)

[a((1+i)(x+y)/2+1−i)−b((1+i)(x+y)/2+1−i)]g(2) , − a((1+i)(x+y)/2+1−i)+b((1+i)(x+y)/2+1−i) z = x + iy ∈ D− \{0, 2}, in which f (0) = [a(z1 ) + b(z1 )]r(z1 ) + [a(z1 ) − b(z1 )]b1 , g(2) = [a(z1 ) − b(z1 )]r(z1 ) − [a(z1 ) + b(z1 )]b1 , W (ζ) on D+ \{0, 2} is as stated in (1.28), Chapter IV, but where the function λ(z) on L0 is as stated in (1.7), and λ(z), r(z), b1 are as stated in (1.3), (1.4). Moreover the functions f (x + y) = U (x + y, 0) − V (x + y, 0) = Re [(1 + i)W (ζ(x + y))], g(x − y) = U (x − y, 0) + V (x − y, 0) = Re [(1 − i)W (ζ(x − y))],

(1.12)

where U = ux /2, V = −uy /2, W [ζ(x + y)] and W [ζ(x − y)] are the values of W [ζ(z)] on 0 ≤ z = x + y ≤ 2 and 0 ≤ z = x − y ≤ 2 respectively. From the above representation of the solution u(z) of Problem P for (1.2), we can derive that u(z) satisfies the estimate ¯ + Cβ [w(z)X(z), D+ ] + C[w± (z)Y ± (z), D− ] ≤ M1 k2 , (1.13) Cβ [u(z), D] in which k2 are as stated in (1.5), w± (z) = Re w ∓ Im w and X(z) =

2 

|z − tj |ηj ,

Y ± (z) =

j=1

|x ± y − tj |ηj ,

j=1




ηj =

2 

2|γj | + δ, if γj < 0, δ, γj ≥ 0,

(1.14)

j = 1, 2,

here γj (j = 1, 2) are real constants as stated in (1.7) and β, δ (β < δ) are sufficiently small positive constant, and M1 = M1 (p0 , β, k0 , D+ ) is a non-negative constant. From the estimate (1.13), we can see the regularity of solutions of Problem P for (1.2). Finally, we mention that if the index K is an arbitrary even integer or 2K is an arbitrary odd integer, the above oblique derivative problem for (1.1) or (1.2) can be considered, but in general these boundary value problems for K ≤ −1 have some solvability conditions or their solutions for K ≥ 0 are not unique. 1.2 The Dirichlet boundary value problem for simplest second order equation of mixed type The Dirichlet problem (Problem D or Problem T ) for (1.2) is to find a solution of (1.2) with the boundary conditions u(z) = φ(z) on Γ ∪ Lj (j = 1 or 2)

(1.15)

1. Simplest Mixed Equation of Second Order

161

where φ(z) satisfies the condition C 1 [φ(z), Γ ∪ Lj ] ≤ k2 ,

j = 1 or 2,

(1.16)

In the following we shall explain that Problem D is a special case of Problem P . In fact, denote w = uz in D, Problem D for the mixed equation (1.2) is equivalent to Problem A for the mixed equation (1.8) with the boundary condition (1.9) and the relation (1.10), in which ⎧ ⎪ i(z ⎪ ⎨

λ(z) = a + ib = ⎧ φ ⎪ ⎪ ⎨ θ

− 1),

1−i ⎪ ⎪ ⎩ √ 2

θ = arg(z − 1) on Γ,

1+i on L1 , or √ on L2 , 2

on Γ,

r(z) = ⎪ φx φx ⎪ ⎩ √ on L1 , or √ on L2 , 2 2 

(1.17)



1+i φx − φx b1 = Im √ uz¯(z1 − 0) = √ |z=z1 −0 = 0, or 2 2 

b1 = Im



1−i √ uz¯(z1 + 0) = 0; b0 = φ(0), 2

√ √ √ √ in which a = 1/ 2
= b = −1/ 2 on L1 or a = 1/ 2
= −b = −1/ 2 on L2 . As for the index K = −1/2 of Problem D on ∂D+ , we can derive as follows: According to (1.17), the boundary conditions of Problem D in D+ possess the form Re [i(z − 1)w(z)] = r(z) = φθ , 

Re 






on Γ,

1−i φ ((1 − i)x/2) √ w(x) = s(x) = √ , 2 2

x ∈ [0, 2], or



1+i φ
((1 + i)x/2 + 1 − i) √ Re √ w(x) = s(x) = , 2 2

x ∈ [0, 2],

it is clear that the possible discontinuous points of λ(z) on ∂D+ are t1 = 2, t2 = 0, and λ(t1 + 0) = e3πi/2 , λ(t2 − 0) = eπi/2 , λ(t1 − 0) = λ(t2 + 0) = eπi/4 , λ(t1 − 0) = e−5πi/4 = eiφ1 , λ(t1 + 0) λ(t1 − 0) = eπi/4 = eiφ1 , λ(t1 + 0)

or λ(t1 − 0) = λ(t2 + 0) = e7πi/4 , λ(t2 − 0) = eπi/4 = eiφ2 or λ(t2 + 0)

λ(t2 − 0) = e−5πi/4 = eiφ2 . λ(t2 + 0)

162

V. Second Order Linear Mixed Equations

In order to insure the uniqueness of solutions of Problem D, we choose that 1 φ1 5 − K1 = − − (−1) = − < 0, π 4 4 φ2 1 0 ≤ γ2 = − K2 = < 1, or π 4 φ1 1 0 ≤ γ1 = − K1 = < 1, π 4 1 φ2 5 −1 < γ2 = − K2 = − − (−1) = − < 0, π 4 4

−1 < γ1 =

thus

1 K 1 + K2 = − , or 2 2 1 K 1 + K2 K1 = 0, K2 = −1, K = =− . 2 2 ¯ In this case, the unique solution w(z) is continuous in D∗ = D\{0, x − y = 2} or ¯ D∗ = D\{x + y = 0, 2}; for the first case, w(z) in the neighborhood of t2 = 0 is bounded, and w(z) in the neighborhood of t1 = 2 possesses the singularity in the form |z − 2|−1/2 and its integral (1.10) is bounded; for the second case, w(z) in the neighborhood of t1 = 2 is bounded, and w(z) in the neighborhood of t2 = 0 possesses the singularity of |z|−1/2 and its integral is bounded. If we require that the solution w(z) = uz is bounded in D+ \{0, 2}, then it suffices to choose the index K = −1, in this case the problem has one solvability condition. K1 = −1,

K2 = 0,

K=

From Theorem 1.1, it follows that the following theorem holds. Theorem 1.2 Problem D for the mixed equation (1.2) has a unique continuous solution in D as stated in (1.10), where λ(z), r(z), b1 are as stated in (1.17) and √ W (ζ) in D+ \{0, √ 2} is as stated in (1.28), Chapter IV, but in which λ(x) = (1 + i)/ 2 or (1 − i)/ 2 on L0 and f (x + y), g(x − y) are as stated in (1.12). [85]15)

2

Oblique Derivative Problems for Second Order Linear Equations of Mixed Type

In this section, we mainly discuss the uniqueness and existence of solutions for second order linear equations of mixed type.

2.1 Formulation of the oblique derivative problem for mixed equations of second order Let D be a simply connected bounded domain D in the complex plane C I with the boundary ∂D = Γ ∪ L as stated in Section 1. We consider the linear mixed equation

2. Oblique Derivative Problems

163

of second order uxx + sgny uyy = aux + buy + cu + d in D,

(2.1)

where a, b, c, d are functions of z(∈ D), its complex form is the following equation of second order ⎧ ⎨ uzz = Re [A1 (z)uz ] + A2 (z)u + A3 (z) in D + , (2.2) ⎩ uz∗ z∗ = Re [A1 (z)uz ] + A2 (z)u + A3 (z) in D− , where 1 uz = [ux − iuy ], 2

z = x + iy,

1 uz∗ = [ux + iuy ] = uz¯, 2 A1 =

a + ib , 2

uz¯z∗

c A2 = , 4

1 uz¯ = [ux + iuy ], 2 1 = [(uz¯)x − i(uz¯)y ] = 2

A3 =

1 uzz¯ = [uxx + uyy ], 4 1 [uxx − uyy ], 4

d in D. 4

Suppose that equation (2.2) satisfies the following conditions: Condition C The coefficients Aj (z) (j = 1, 2, 3) in (2.2) are measurable in z ∈ D+ and continuous in D− and satisfy Lp [Aj , D+ ] ≤ k0 ,

Lp [A3 , D+ ] ≤ k1 ,

j = 1, 2,

C[Aj , D− ] ≤ k0 ,

j = 1, 2,

A2 ≥ 0 in D+ ,

C[A3 , D− ] ≤ k1 .

(2.3) (2.4)

where p (> 2), k0 , k1 are non-negative constants. If the condition (2.4) is replaced by Cα [Aj , D− ] ≤ k0 ,

j = 1, 2,

Cα [A3 , D− ] ≤ k1 ,

in which α(0 < α < 1) is a real constant, then the conditions will be called Condition C
. The oblique derivative boundary value problem (Problem P ) for equation (2.2) is ¯ to find a continuously differentiable solution u(z) of (2.2) in D∗ = D\{0, x − y = 2} ¯ ¯ and satisfies the boundary or D∗ = D\{x + y = 0, 2}, which is continuous in D conditions (1.3) and (1.4), in which b0 , b2 is a real constant satisfying the condition |b0 |, |b2 | ≤ k2 . The index K is defined as stated in Section 1, now we discuss the case 1 K = (K1 + K2 ) = 0, 2

(2.5)

where 

Kj =



φj + Jj , Jj = 0 or 1, π

eiφj =

λ(tj − 0) , λ(tj + 0)

γj =

φj − Kj , π

j = 1, 2, (2.6)

in which t1 = 2, t2 = 0, λ(t) = eiπ/4 on L0 = (0, 2) and λ(t1 − 0) = λ(t2 + 0) = exp(iπ/4), or λ(t) = ei7π/4 on L0 and λ(t1 − 0) = λ(t2 + 0) = exp(i7π/4). We mention

164

V. Second Order Linear Mixed Equations

that if A2 = 0 in D, or cos(l, n) ≡ 0 on Γ, then we do not need the point condition u(2) = b2 in (1.3) and only choose the index K = −1/2. Because, if cos(l, n) ≡ 0 on Γ, from the boundary condition (1.3), we can determine the value u(2) by the value u(0), namely 

u(2) = 2Re

0



2

uz dz+u(0) = 2

2

0

Re [i(z − 1)uz ]dθ+b0 = 2



0

π

r(z)dθ+b0 ,

in which λ(z) = i(z − 1), θ = arg(z − 1) on Γ. In brief, we choose that ⎧ ⎪ ⎪ ⎨ 0,

⎧ ⎫ ⎨ u(0) = b0 , u(2) = b2 ⎬

K =⎪

1 ⎪ ⎩− ,

the point conditions are ⎩

2

u(0) = b0

⎭

⎧ ⎨ cos(l, n)
≡ 0

, if ⎩ on Γ. cos(l, n) ≡ 0

In order to ensure that the solution u(z) of Problem P is continuously differentiable in D∗ , we need to choose γ1 > 0 or γ2 > 0. If we only require that the solution is continuous, it suffices to choose −2γ2 < 1, −2γ1 < 1 respectively. In the following, we shall only discuss the case: K = 0, and the case: K = −1/2 can be similarly discussed. 2.2 The representation and uniqueness of solutions for the oblique derivative problem for (2.2) Now we give the representation theorems of solutions for equation (2.2). Theorem 2.1 Let equation (2.2) satisfy Condition C in D+ , u(z) be a continuous solution of (2.2) in D+ and continuously differentiable in D∗+ = D+ \{0, 2}. Then u(z) can be expressed as u(z) = U (z)Ψ(z) + ψ(z) in D+ , 

U (z) = 2Re

0

z

w(z)dz + b0 , w(z) = Φ(z)eφ(z) in D+ ,

(2.7)

where ψ(z), Ψ(z) are the solutions of equation (2.2) in D+ and uzz¯ − Re [A1 uz ] − A2 u = 0 in D+

(2.8)

respectively and satisfy the boundary conditions ψ(z) = 0, Ψ(z) = 1 on Γ ∪ L0 ,

(2.9)

where ψ(z), Ψ(z) satisfies the estimates Cβ1 [ψ, D+ ] ≤ M2 , Cβ1 [Ψ, D+ ] ≤ M3 ,

 ψ Wp20 (D+ ) ≤ M2 ,

 Ψ Wp20 (D+ ) ≤ M3 , Ψ(z) ≥ M4 > 0,

(2.10) z ∈ D+ ,

(2.11)

2. Oblique Derivative Problems

165

in which β (0 < β ≤ α), p0 (2 < p0 ≤ p), Mj = Mj (p0 , β, k, D) (j = 2, 3, 4) are non-negative constants, k = (k0 , k1 , k2 ). Moreover U (z) is a solution of the equation A = −2(ln Ψ)z¯ + A1 in D+ ,

Uzz¯ − Re [AUz ] = 0,

(2.12)

where Im [φ(z)] = 0, z ∈ L0 = (0, 2) and φ(z) satisfies the estimate Cβ [φ, D+ ] + Lp0 [φz¯, D+ ] ≤ M5 ,

(2.13)

in which β(0 < β ≤ α), M5 = M5 (p0 , β, k0 , D) are two non-negative constants, Φ(z) is analytic in D+ . If u(z) is a solution of (2.2) in D+ satisfying the boundary conditions (1.3) and Re [λ(z)uz ]|z=x = s(x),

λ(x) = 1 + i or 1 − i,

x ∈ L0 , Cβ [s(x), L0 ] ≤ k3 , (2.14)

then the following estimate holds: Cβ [u(z), D+ ] + Cβ [uz X(z), D+ ] ≤ M6 (k1 + k2 + k3 ),

(2.15)

in which k3 is a non-negative constant, s(x) can be seen as stated in the form (2.23) below, X(z) is as stated in (1.14), and M6 = M6 (p0 , β, k0 , D+ ) is a non-negative constant. Proof According to the method in the proof of Theorem 3.1, Chapter III, the equations (2.2),(2.8) in D+ have the solutions ψ(z), Ψ(z) respectively, which satisfy the boundary condition (2.9) and the estimates (2.10),(2.11). Setting that U (z) =

u(z) − ψ(z) , Ψ(z)

(2.16)

it is clear that U (z) is a solution of equation (2.12), which can be expressed the second formula in (2.7), where φ(z) satisfies the estimate (2.13) and Φ(z) is an analytic function in D+ . If s(x) in (2.14) is a known function, then the boundary value problem (2.2),(1.3),(2.14) has a unique solution u(z) as stated in the form (2.7), which satisfies the estimate (2.15). Theorem 2.2 Suppose that the equation (2.2) satisfies Condition C. Then any solution of Problem P for (2.2) can be expressed as 

u(z) = 2Re

0

z

w(z)dz + b0 ,

w(z) = w0 (z) + W (z),

(2.17)

where w0 (z) is a solution of Problem A for the complex equation (1.8) with the boundary conditions (1.3), (1.4)(w0 (z) = u0z ), and W (z) possesses the form W (z) = w(z) − w0 (z) in D,

˜

φ(z) ˜ ˜ w(z) = Φ(z)e + ψ(z) in D+ ,

g(ζ) 1 ˜ ˜ dσζ , ψ(z) φ(z) = φ˜0 (z) + T g = φ˜0 (z) − = T f in D+ , + π D ζ −z 

W (z) = Φ(z) + Ψ(z), Ψ(z) =

2

ν

g 1 (z)dνe1 +



0

µ

g 2 (z)dµe2 in D− ,

(2.18)

166

V. Second Order Linear Mixed Equations

in which e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, l = x − y, φ˜0 (z) is an analytic ˜ function in D+ , such that Im [φ(x)] = 0 on L0 , and


g(z)=

A1 /2+A1 w/(2w), ¯ 0,

w(z) = 0,

w(z)
= 0,

z ∈ D+ ,

g 1 (z) = g 2 (z) = Aξ +Bη +Cu+D, A=

ReA1 +ImA1 , 2

B=

f (z)=Re[A1 φ˜z ]+A2 u+A3 in D− ,

ξ = Rew +Imw,

ReA1 −ImA1 , 2

η = Rew −Imw,

(2.19)

C = A2 , D = A3 in D− ,

˜ where Φ(z) and Φ(z) are the solutions of equation (1.8) in D+ and D− respectively satisfying the boundary conditions ˜ φ(z) ˜ ˜ Re [λ(z)(Φ(z)e + ψ(z))] = r(z), ˜ φ(x)

˜ Re [λ(x)(Φ(x)e

z ∈ Γ,

˜ + ψ(x))] = s(x),

x ∈ L0 ,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L0 ,

(2.20)

z ∈ L1 or L2 ,

Im [λ(z1 )Φ(z1 )] = −Im [λ(z1 )Ψ(z1 )], where λ(x) = 1 + i or 1 − i, x ∈ L0 . Moreover by Theorem 1.2, Chapter IV, the solution w0 (z) of Problem A for (1.8) and u0 (z) satisfy the estimate in the form Cβ [u0 (z), D]+Cβ [w0 (z)X(z), D+ ]+Cβ [w0± (z)Y ± (z), D− ] ≤ M7 (k1 +k2 ),

(2.21)

in which w± (z) = Re w(z) ∓ Im w(z), X(z), Y ± (z) are as stated in (1.14), 

u0 (z) = 2Re

0

z

w0 (z)dz + b0 ,

(2.22)

and M7 = M7 (p0 , β, k0 , D) is a non-negative constant. From (2.22), it follows that ¯ ≤ M8 {Cβ [w0 (z)X(z), D+ ] + Cβ [w± (z)Y ± (z), D− } + k2 , Cβ [u0 (z), D] 0 where M8 = M8 (D) is a non-negative constant. Proof Let u(z) be a solution of Problem P for equation (2.2), and w(z) = uz , u(z) be substituted in the positions of w, u in (2.19), thus the functions g(z), ˜ ˜ f (z), g1 (z), g2 (z), and ψ(z), φ(z) in D+ and Ψ(z) in D− in (2.18),(2.19) can be ˜ determined. Moreover we can find the solution Φ(z) in D+ and Φ(z) in D− of (1.8) with the boundary conditions (2.20), where ⎧ ⎪ 2r((1−i)x/2)−2R((1−i)x/2) ⎪ ⎪ +Re [λ(x)Ψ(x)] ⎪ ⎪ ⎪ ⎪ ⎪ a((1 − i)x/2) − b((1 − i)x/2) ⎪ ⎨

s(x) =

2r((1+i)x/2+1−i)−2R((1+i)x/2+1−i)

⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ +Re [λ(x)Ψ(x)] on L , 0

or (2.23)

2. Oblique Derivative Problems

167

here and later R(z) = Re [λ(z)Ψ(z)] on L1 or L2 , thus ⎧ ˜ ⎪ ˜ φ(z) ⎨ Φ(z)

w(z) = w0 (z) + W (z) =

⎪ ⎩w

0 (z)

˜ + ψ(z) in D+ ,

+ Φ(z) + Ψ(z) in D− ,

is the solution of Problem A for the complex equation


wz

⎧ ⎫ ⎨ D+ ⎬



= Re [A1 w] + A2 u + A3 in ⎩ , D− ⎭

w¯z∗

(2.24)

which can be expressed as the second formula in (2.17), and u(z) is a solution of Problem P for (2.2) as stated in the first formula in (2.17). Theorem 2.3 If equation (2.2) satisfies Condition C, then Problem P for (2.2) has at most one solution in D. Proof Let u1 (z), u2 (z) be any two solutions of Problem P for (2.2). By Condition C, we see that u(z) = u1 (z)−u2 (z) and w(z) = uz satisfies the homogeneous equation and boundary condition


wz



w¯z∗




= Re [A1 w] + A2 u in z ∈ Γ,

Re [λ(z)w(z)] = 0, Re [λ(z)w(z)] = 0,

D+ D−

u(0) = 0,

z ∈ Lj (j = 1 or 2),



,

(2.25)

u(2) = 0, (2.26)

Im [λ(z1 )w(z1 )] = 0.

From Theorem 2.2, the solution w(z) can be expressed in the form

w(z) =

⎧ ˜ φ(z) ˜ ⎪ Φ(z)e ⎪ ⎪ ⎪ ⎪ ⎨

˜ + ψ(z),

⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Ψ(z)

ν

w0 (z) + Φ(z) + Ψ(z), 

=

2

˜ φ(z) = φ˜0 (z) + T˜g in D+ ,

˜ ψ(z) = T f,

[Aξ +Bη+Cu]e1 dν +

 0

(2.27) µ

−

[Aξ +Bη+Cu]e2 dµ in D ,

˜ where g(z) is as stated in (2.19), Φ(z) in D+ is an analytic function and Φ(z) is a ˜ ˜ solution of (1.8) in D− satisfying the boundary condition (2.20), φ(z), ψ(z) possess ˜ the similar properties as φ(z), ψz (z) in Theorem 2.1. If A2 = 0 in D+ , then ψ(z) = 0. ˜ Besides the functions Φ(z), Φ(z) satisfy the boundary conditions ⎧ ˜ ⎨ Re [λ(x)Φ(x)] ⎩

= s(x),

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))]

on L0 ,

(2.28)

where s(x) is as stated in (2.23). From (2.17) with b0 = 0, we can obtain ¯ ≤ M8 {C[w(z)X(z), D+ ] + C[w± (z)Y ± (z), D− }. C[u(z), D] 0

(2.29)

168

V. Second Order Linear Mixed Equations

By using the method of iteration, the estimate C[w(z), D− ] ≤

[2M9 M (4m + 1)R
]n n!

(2.30)

can be derived, where M9 = max{C[A, D− ], C[B, D− ], C[C, D− ]}, M = 1 + 4k02 (1 + k02 ), and m = C[w(z), D− ] > 0. Let n → ∞, from (2.29), it follows that w(z) = 0 in D− , and Ψ(z) = 0, Φ(z) = 0, z ∈ D− . Thus the solution u(z) = 2 Re 0z w(z)dz is the solution of equation (2.8) with the boundary conditions Re [λ(z)uz ] = 0 on Γ,

u(2) = 0, (2.31) in which λ(x) = 1 + i, or 1 − i, x ∈ L0 . Similarly to the proof of Theorem 3.4, Chapter III, we can obtain u(z) = 0 on D+ . This shows the uniqueness of solutions of Problem P for (2.2). 2.3

Re [λ(x)uz (x)] = 0 on L0 = (0, 2),

u(0) = 0,

The solvability of the oblique derivative problem for (2.2)

Theorem 2.4 Suppose that the mixed equation (2.2) satisfies Condition C. Then Problem P for (2.2) has a solution in D. Proof It is clear that Problem P for (2.2) is equivalent to Problem A for the complex equation of first order and boundary conditions:


wz w¯z∗

⎧ ⎫ ⎨ D+ ⎬



= F,

F = Re [A1 w] + A2 u + A3 in ⎩ , D− ⎭

Re [λ(z)w(z)] = r(z), Re [λ(z)w(z)] = r(z),

z ∈ Γ,

z ∈ Lj (j = 1 or 2),

(2.32)

(2.33)

Im [λ(z1 )w(z1 )] = b1 ,

and the relation (2.17). From (2.17), it follows that ¯ ≤ M8 [C(w(z)X(z), D+ ) + C(w± (z)Y ± (z), D− )] + k2 , C[u(z), D] ±

(2.34)

±

where X(z), Y (z), w (z) are as stated in (1.14) respectively, M8 = M8 (D) is a non-negative constant. In the following, by using successive iteration, we shall find a solution of Problem A for the complex equation (2.32) in D. Firstly denoting the solution w0 (z)(= ξ0 e1 + η0 e2 ) of Problem A for (1.8) and u0 (z) in (2.17), and substituting them into the position of w = (ξe1 + ηe2 ), u(z) in the right-hand side of (2.32), similarly to (2.18),(2.19), we have the corresponding functions f1 (z), g1 (z), g21 (z), g21 (z) and ˜ 1 (z)eφ˜1 (z) + ψ˜1 (z) in D+ , w1 (z) = Φ 1 φ˜1 (z) = φ˜0 (z)+T g1 = φ˜0 (z)− π 

W1 (z) = Φ(z)+Ψ(z), Ψ(z) =

2

 D+ ν

g1 (ζ) dσζ , ψ˜1 (z) = T f1 in D+ , ζ −z

g11 (z)dνe1 +



0

µ

g12 (z)dµe2 in D− ,

(2.35)

2. Oblique Derivative Problems

169

where µ = x + y, ν = x − y, where Φ1 (z) is a solution of (1.8) in D− satisfying the boundary conditions Re [λ(x)Φ1 (x)] = Re [λ(x)(W1 (z) − Ψ1 (x))], Re [λ(z)Φ1 (z)] = −Re [λ(z)Ψ1 (z)],

z ∈ L0 ,

z ∈ L1 or L2 ,

(2.36)

Im [λ(z1 )Φ1 (z1 )] = −Im [λ(z1 )Ψ1 (z1 )], and w1 (z) = w0 (z) + W1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z) in D−

(2.37)

satisfies the estimate Cβ [w(z)X(z), D+ ] + C[w1± (z)Y ± (z), D− ] ≤ M10 = M10 (p0 , β, k, D− ).

(2.38)

Furthermore we substitute w1 (z) = w0 (z)+W1 (z) and corresponding functions w1 (z), ξ1 (z) = w+ (z) = Re w1 (z)−Im w1 (z), η1 (z) = w− (z) = Re w1 (z)+Im w1 (z), u1 (z) into the positions w(z), ξ(z), η(z), u(z) in (2.18),(2.19), and similarly to (2.35)–(2.37), ˜ 2 (z) in D+ , Ψ2 (z), Φ2 (z) and we can find the corresponding functions ψ˜2 (z), φ˜2 (z), Φ W2 (z) = Φ2 (z) + Ψ2 (z) in D− , and the function ˜ 2 (z)eφ˜2 (z) + ψ˜2 (z) in D+ , w2 (z) = Φ w2 (z) = w0 (z) + W2 (z) = w0 (z) + Φ2 (z) + Ψ2 (z) in D−

(2.39)

satisfies the similar estimate in the form (2.38). Thus there exists a sequence of functions: {wn (z)} and ˜ n (z)eφ˜n (z) + ψ˜n (z) in D+ , wn (z) = Φ wn (z) = w0 (z) + Wn (z) = w0 (z) + Φn (z) + Ψn (z), 

Ψn (z) =

2

ν

gn1 (z)e1 dν +



0

µ

(2.40)

gn2 (z)e2 dµ in D− ,

and then ± |[w1± (z) − w0± (z)]Y ± (z)| ≤ |Φ± 1 (z)Y (z)|  ν √ + 2[|Y + (z) [Aξ0 + Bη0 + Cu0 + D]e1 dν|

+|Y − (z)

 0

2

µ

(2.41)

[Aξ0 + Bη0 + Cu0 + D]e2 dµ|] ≤ 2M11 M (4m + 1)R
in D− ,

where M11 = maxz∈D− (|A|, |B|, |C|, |D|), m = C[w0 (z)X(z), D− ], R
= 2, M = 1 + 4k02 (1 + k02 ). It is clear that wn (z) − wn−1 (z) satisfies wn (z) − wn−1 (z) = Φn (z) − Φn−1 (z) + 

+

0

µ

 2

ν

[A(ξn − ξn−1 ) + B(ηn − ηn−1 ) + C(un − un−1 )]e1 dν (2.42)

[A(ξn − ξn−1 ) + B(ηn − ηn−1 ) + C(un − un−1 )]e2 dµ in D− ,

170

V. Second Order Linear Mixed Equations

where n = 1, 2, . . .. From the above equality, the estimate ± |[wn± − wn−1 ]Y ± (z)| ≤ [2M11 M (4m + 1)]n ×



R


0


n

[2M11 M (4m + 1)R ] ≤ n!

in

R
n−1 dR
(n − 1) !

(2.43)

D−

can be obtained, and then we can see that the sequence of functions: {wn± (z)Y ± (z)}, i.e. ± wn± (z)Y ± (z) = {w0± (z) + [w1± (z) − w0± (z)] + · · · + [wn± (z) − wn−1 (z)]}Y ± (z) (2.44)

(n = 1, 2, . . .) in D− uniformly converge to w∗± (z)Y ± (z), and w∗ (z) = [w∗+ (z) + w∗− (z) − i(w∗+ (z) − w∗− (z))]/2 satisfies the equality 

Ψ∗ (z) =

2

w∗ (z) = w0 (z) + Φ∗ (z) + Ψ∗ (z), 

ν

[Aξ∗ +Bη∗ +Cu∗ +D]e1 dν +

0

µ

[Aξ∗ +Bη∗ +Cu∗ +D]e2 dµ in D− ,

(2.45)

and the corresponding function u∗ (z) is just a solution of Problem P for equation (2.2) in the domain D− , and w∗ (z) satisfies the estimate


C[w∗± (z)Y ± (z), D− ] ≤ e2M11 M (4m+1)R .

(2.46)

In the meantime we can obtain the estimate Cβ [wn (z)X(z), D+ ] ≤ M12 = M12 (p0 , β, k, D),

(2.47)

hence from the sequence {wn (z)}, we can choose a subsequence, which uniformly converges to w∗ (z) in D+ , and w∗ (z) satisfies the same estimate (2.47). Combining (2.46) and (2.47), it is obvious that the solution w∗ (z) = uz of Problem A for (2.2) ¯ satisfies the estimate in D Cβ [w∗ (z)X(z), D+ ] + C[w∗± (z)Y ± (z), D− ] ≤ M13 = M13 (p0 , β, k, D), where M13 is a non-negative constant. Moreover the function u(z) in (2.17) is a solution of Problem P for (2.2), where w(z) = w∗ (z). From Theorems 2.3 and 2.4, we see that under Condition C, Problem A for equation (2.32) has a unique solution w(z), which can be found by using successive iteration and the corresponding solution u(z) of Problem P satisfies the estimates Cβ [u(z), D+ ] + Cβ [uz X(z), D+ ] ≤ M14 , ¯ + C[u± Y ± (z), D] ¯ ≤ M15 , C[u(z), D] z

(2.48)

where X(z), Y ± (z) is as stated in (1.14), and Mj = Mj (p0 , β, k, D) (j = 14, 15) are non-negative constants, k = (k0 , k1 , k2 ). Moreover we can derive the following theorem.

3. Discontinuous Oblique Derivative Problems

171

Theorem 2.5 Suppose that equation (2.2) satisfies Condition C. Then any solution u(z) of Problem P for (2.2) satisfies the estimates Cβ [u(z), D+ ] + Cβ [uz X(z), D+ ] ≤ M16 (k1 + k2 ), ± − C[u(z), D− ] + C[u± z Y (z), D ] ≤ M17 (k1 + k2 ),

(2.49)

in which Mj = Mj (p0 , β, k0 , D) (j = 16, 17) are non-negative constants. From the estimates (2.48),(2.49), we can see that the regularity of solutions of Problem P for (2.2) (see [85]15)).

3

Discontinuous Oblique Derivative Problems for Second Order Linear Equations of Mixed Type

This section deals with an application of method of integral equations to second order equations of mixed type. We mainly discuss the discontinuous Poincar´e boundary value problem for second order linear equation of mixed (elliptic-hyperbolic) type, i.e. the generalized Lavrent
ev-Bitsadze equation with weak conditions by the method of integral equations. We first give the representation of solutions for the above boundary value problem, and then give the solvability conditions of the above problem by the Fredholm theorem for integral equations. 3.1 Formulation of the discontinuous Poincar´ e problem for mixed equations of second order Let D be a simply connected bounded domain D in the complex plane C I with the boundary ∂D = Γ ∪ L as stated in Section 1. We consider the second order linear equation of mixed type (2.1) and its complex form (2.2) with Condition C
. In order to introduce the discontinuous Poincar´e boundary value problem for equation (2.2), let the functions a(z), b(z) possess the discontinuities of first kind at m + 2 distinct points z0 = 2, z1 , . . . , zm+1 = 0 ∈ Γ and Z = {z0 , z1 , . . . , zm+1 }, which are arranged according to the positive direction of Γ, where m is a positive integer, and r(z) = O(|z − zj |−βj ) in the neighborhood of zj (j = 0, 1, . . . , m + 1) on Γ, in which βj (j = 0, 1, . . . , m + 1) are small positive numbers. Denote λ(z) = a(x) + ib(x) and |a(x)| + |b(x)| =
0, there is no harm in assuming that |λ(z)| = 1, z ∈ Γ∗ = Γ\Z. Suppose that λ(z), r(z) satisfy the conditions λ(z) ∈ Cα (Γj ),

|z − zj |βj r(z) ∈ Cα (Γj ),

j = 0, 1, . . . , m + 1,

(3.1)

herein Γj is an arc from the point zj−1 to zj on Γ and zm+1 = 0, and Γj (j = 0, 1, . . . , m + 1) does not include the end points, and α (0 < α < 1) is a constant. Problem Q Find a continuously differentiable solution u(z) of (2.2) in D∗ = ¯ Z( ˜ Z˜ = {0, x − y = 2, y ≤ 0} or Z˜ = {x + y = 0, y ≤ 0, 2}), which is continuous in D\

172

V. Second Order Linear Mixed Equations

¯ and satisfies the boundary conditions D 1 ∂u +εσ(z)u = Re [λ(z)uz ]+εσ(z)u = r(z)+Y (z)h(z), 2 ∂l 1 ∂u = Re [λ(z)uz¯] = r(z), 2 ∂l

z ∈ L1 or L2 ,

z ∈ Γ,

u(0) = b0 ,

Im [λ(z)uz¯]|z=z1 = b1 ,

(3.2) (3.3)

where l is a vector at every point on Γ ∪ Lj (j = 1 or 2), z 1 = 1 − i, b0 , b1 are real constants, λ(z) = a(x)+ib(x) = cos(l, x)−i cos(l, y), z ∈ Γ, and λ(z) = a(x)+ib(x) = cos(l, x) + i cos(l, y), z ∈ Lj (j = 1 or 2), and λ(z), r(z), b0 , b1 satisfy the conditions Cα [λ(z), Γ] ≤ k0 ,

Cα [σ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

Cα [λ(z), Lj ] ≤ k0 ,

Cα [σ(z), Lj ] ≤ k0 ,

1 , max z∈L1 |a(x) − b(x)|

1 or max ≤ k0 , z∈L2 |a(x) + b(x)|

|b0 |, |b1 | ≤ k2 ,

Cα [r(z), Lj ] ≤ k2 ,

j = 1 or 2,

(3.4)

in which α (1/2 < α < 1), k0 , k2 are non-negative constants, ε is a real parameter. Besides, the functions Y (z), h(z) are as follows Y (z) = η

m+1 

|z − zj | |z−z∗ | , γj

l

∗

z∈Γ ,

j=0

⎧ ⎨ 0,z ∈Γ, if K ≥ −1/2,

h(z) = ⎩

hj ηj (z), z ∈Γj , if K <−1/2,

(3.5)

in which Γj (j = 0, 1, . . . , m) are arcs on Γ∗ = Γ\Z and Γj ∩Γk = φ, j
= k, hj ∈ J (J = φ if K ≥ −1/2; J = 1, . . . , 2K
− 1 if K < −1/2; K
= [|K| + 1/2]) are unknown real constants to be determined appropriately, herein h1 = 0, l = 1 if 2K is odd, z∗ (
∈ Z) ∈ Γ∗ is any fixed point, and l = 0 if 2K is even, Γj (j = 1, . . . , 2K
− 1) are non-degenerate, mutually disjointed arcs on Γ, and Γj ∩ Z = φ, j = 1, . . . , 2K
− 1, ηj (z) is a positive continuous function on the interior point set of Γj , such that ηj (z) = 0 on Γ\Γj and Cα [ηj (z), Γ] ≤ k0 ,

j = 1, . . . , 2K
− 1,

(3.6)

and η = 1 or −1 on Γj (0 ≤ j ≤ m + 1, Γm+1 = (0, 2)) as stated in [93]. The above discontinuous Poincar´e boundary value problem for (2.2) is called Problem Q. ¯ r(z) = 0, z ∈ Γ ∪ Lj (j = 1 or 2) and Problem Q for (2.2) with A3 (z) = 0, z ∈ D, b0 = b1 = 0 will be called Problem Q0 . Denote by λ(zj − 0) and λ(zj + 0) the left limit and right limit of λ(z) as z → zj (0 ≤ j ≤ m + 1) on Γ ∪ L0 , and eiφj =

λ(zj − 0) , λ(zj + 0)





φj Kj = + Jj , π

γj =

λ(zj − 0) φj 1 ln = − Kj , πi λ(zj + 0) π

Jj = 0 or 1,

j = 0, 1, . . . , m + 1,

(3.7)

3. Discontinuous Oblique Derivative Problems

173

in which zm+1 = 0, z0 = 2, λ(z) = eiπ/4 on L0 = (0, 2) and λ(z0 − 0) = λ(zm+1 + 0) = exp(iπ/4), or λ(z) = e−iπ/4 on L0 and λ(z0 − 0) = λ(zm+1 + 0) = exp(−iπ/4), and 0 ≤ γj < 1 when Jj = 0 and −1 < Jj < 0 when Jj = 1, j = 0, 1, . . . , m + 1, and 

m+1  φj γj 1 − K = (K0 + K2 + · · · + Km+1 ) = 2 2π 2 j=0



(3.8)

is called the index of Problem Q and Problem Q0 . Let βj +γj < 1, j = 0, 1, . . . , m+1, we can require that the solution u(z) satisfy the condition: uz = O(|z − zj |−δj ) in the neighborhood of zj (j = 0, 1, . . . , m + 1) in D∗ , where τj =

⎧ ⎨ βj +τ, ⎩

for γj ≥ 0,

|γj |+τ,

and γj < 0, βj > |γj |, βj ≤ |γj |,

for γj < 0,

⎧ ⎨ 2τj ,

δj = ⎩

τj ,

j = 0, m+1, (3.9) j = 1, . . . , m,

and τ, δ(< τ ) are small positive numbers. In order to ensure that the solution u(z) of Problem Q is continuously differentiable in D∗ , we need to choose γ1 > 0 or γ2 > 0 respectively.

3.2 The representation and solvability of the oblique derivative problem for (2.2) Now we write a representation theorem of solutions for equation (2.2), which is similar to Theorem 2.2. Theorem 3.1 If equation (2.2) satisfies Condition C
and ε = 0, A2 ≥ 0 in D+ , then any solution of Problem Q for (2.2) can be expressed as 

u(z) = 2Re

z

0

w(z)dz + c0 ,

w(z) = w0 (z) + W (z),

(3.10)

where w0 (z) is a solution of Problem A for equation (1.8) with the boundary conditions z ∈ Γ,

Re [λ(z)w(z)] = r(z) + Y (z)h(z), Re [λ(z)w(z)] = r(z),

z ∈ Lj (j = 1 or 2),

(3.11) Im [λ(z)w(z)]|z=z1 = b1 ,

and W (z) possesses the form W (z) = w(z) − w0 (z),

˜

φ(z) ˜ ˜ W (z) = Φ(z)e + ψ(z),

g(ζ) 1 ˜ ˜ dσζ , ψ(z) φ(z) = φ˜0 (z) + T g = φ˜0 (z) − = T f in D+ , + π D ζ −z 

W (z) = Φ(z) + Ψ(z),

Ψ(z) =

2

ν

g 1 (z)dνe1 +



0

µ

g 2 (z)dµe2 in D− ,

(3.12)

174

V. Second Order Linear Mixed Equations

in which e1 =

1+i 1−i , e2 = , 2 2




g(z) =

A1 /2+A1 w/(2w),

0, w(z) = 0,

µ = x + y,

w(z)
= 0,

A=

Re A1 +Im A1 , 2

f (z) = Re [A1 φ˜z ]+A2 u+A3 in D+ ,

z ∈ D+ ,

g 1 (z) = g 2 (z) = Aξ +Bη+Cu+D, B=

ν = x − y, and

ξ = Re w−Im w,

Re A1 −Im A1 , 2

C = A2 ,

η = Re w+Im w,

(3.13)

D = A3 in D− ,

˜ = 0 on L0 = (0, 2), and where φ˜0 (z) is an analytic function in D+ such that Im [φ(x)] ˜ Φ(z), Φ(z) are the solutions of the equation (1.8) in D+ , D− respectively satisfying the boundary conditions ˜

˜ ˜ Re [λ(z)eφ(z) Φ(z)] = r(z) − Re [λ(z)ψ(z)], ˜ φ(x) ˜ ˜ Re [λ(x)(Φ(x)e + ψ(x))] = s(x),

x ∈ L0 ,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ Γ,

z ∈ L0 ,

(3.14)

z ∈ L1 or L2 ,

Im [λ(z 1 )Φ(z 1 )] = −Im [λ(z 1 )Ψ(z 1 )], where

⎧ ⎪ 2r((1−i)x/2)−2R((1−i)x/2) ⎪ ⎪ +Re [λ(x)Ψ(x)], ⎪ ⎪ ⎪ ⎪ ⎪ a((1 − i)x/2) − b((1 − i)x/2) ⎪ ⎨

or

s(x) = ⎪ 2r((1+i)x/2+1−i)−2R((1+i)x/2+1−i) ⎪ ⎪ ⎪ ⎪ a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) ⎪ ⎪ ⎪ ⎪ ⎩ +Re [λ(x)Ψ(x)]

(3.15)

on L0 ,

in which s(x) can be written similar to (2.9). Moreover from Theorem 1.1, if the index K ≤ −1/2, the solution u0 (z)(w0 (z) = u0z (z)) of Problem Q for (1.2) satisfies the estimate in the form ¯ + Cδ [w0 (z)X(z), D+ ] + C[w± (z)Y ± (z), D− ] ≤ M18 (k1 + k2 ) Cδ [u0 (z), D] 0

(3.16)

in which δ is a small positive constant, p0 (2 < p0 ≤ p), M18 = M18 (p0 , δ, k0 , D) are two non-negative constants, ηj ± w0± (z) = Re w0 (z) ∓ Im w0 (z), X(z) = Πm+1 j=0 |z−tj | , Y (z) =

2 j=1

|x ± y−tj |ηj ,

ηj = 2|γj | + δ, j = 0, m + 1, ηj = |γj | + δ, j = 1, . . . , m, and



u0 (z) = 2Re

0

z

w0 (z)dz + c0 .

(3.17)

3. Discontinuous Oblique Derivative Problems

175

In order to prove the solvability of Problem Q for (2.2), denote w = uz and consider the equivalent boundary value problem (Problem B) for the mixed complex equation ⎧ ⎪ ⎪ ⎪ wz ⎪ ⎪ ⎪ ⎨

− Re [A1 (z)w] = εA2 (z)u + A3 (z),

w¯z∗ − Re [A1 (z)w] = A3 (z),

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ u(z)



= 2Re

0

z ∈ D+ ,

z ∈ D− ,

(3.18)

z

w(z)dz + b0

with the boundary conditions Re [λ(z)w] = r(z) − εσ(z)u + Y (z)h(z), Re [λ(z)uz¯] = r(z),

z ∈ Lj (j = 1 or 2),

z ∈ Γ,

(3.19)

Im [λ(z)uz¯]|z=z1 = b1 ,

where b0 , b1 are real constants are as stated in (3.2),(3.3). According to the method in Section 5, Chapter IV, we can find the general solution of Problem B1 for the mixed complex equation ⎧ ⎨ wz ⎩

− Re [A1 (z)w] = A3 (z),

w¯z∗ − Re [A1 (z)w] = A3 (z),

z ∈ D+ , z ∈ D− ,

(3.20)

with the boundary conditions Re [λ(z)w(z)] = r(z) + Y (z)h(z), Re [λ(z)w(z)] = r(z),

z ∈ Γ,

z ∈ Lj (j = 1 or 2),

(3.21) Im [λ(z)w(z)]|z=z1 = b1 ,

which can be expressed as w(z) ˜ = w0 (z) +

2K+1 

ck wk(z)

(3.22)

k=1

in which w0 (z) is a special solution of Problem B1 and wk (z)(k = 1, . . . , 2K + 1, K ≥ 0) is the complete system of linear independent solutions for the homogeneous problem of Problem B1 . Moreover, denote by H2 u the solution of Problem B2 for the complex equation ⎧ ⎨ wz ⎩

− Re [A1 (z)w] = A2 (z)u,

w¯z∗ − Re [A1 (z)w] = A2 (z)u,

z ∈ D+ , z ∈ D− ,

(3.23)

with the boundary conditions Re [λ(z)w(z)] = −σ(z)u + Y (z)h(z), Re [λ(z)w(z)] = 0,

z ∈ Γ,

z ∈ Lj (j = 1 or 2),

Im [λ(z)w(z)]|z=z1 = 0,

(3.24)

176

V. Second Order Linear Mixed Equations

and the point conditions ⎧ ⎨ 1, . . . , 2K

Im [λ(aj )w(aj )] = 0,

j∈J =⎩ φ,

+ 1,

K ≥ 0,

(3.25)

K < 0,

where aj ∈ Γ\Z are the distinct points. It is easy to see that H2 is a bounded ± ± + − ¯ (i.e. C(u, D)+C(X(z)u ¯ operator from u(z) ∈ C˜ 1 (D) z , D ) + C(Y (z)uz , D ) < ∞) ± ± + − ˜ ¯ ¯ to w(z) ∈ Cδ (D) (i.e. Cδ (u, D) + Cδ (X(z)w(z), D ) + Cδ (Y (z)w (z), D ) < ∞), herein X(z), Y ± (z) are functions as stated in (3.16). Furthermore denote 

u(z) = H1 w + c0 = 2Re

z

0

w(z)dz + c0 ,

(3.26)

where c0 is arbitrary real constant. It is clear that H1 is a bounded operator from ¯ to u(z) ∈ C˜ 1 (D). ¯ On the basis of Theorem 3.1, the function w(z) X(z)w(z) ∈ C˜δ (D) can be expressed as an integral. From (3.26) and w(z) = w(z) ˜ + εH2 u, we can obtain a nonhomogeneous integral equation (K ≥ 0): u − εH1 H2 u = H1 w(z) + c0 +

2K+1 

ck H1 wk (z).

(3.27)

k=1

¯ we can use the Fredholm Due to H1 H2 is a completely continuous operator in C˜ 1 (D), theorem for the integral equation (3.27). Denote by εj (j = 1, 2, . . .) : 0 < |ε1 | ≤ |ε2 | ≤ · · · ≤ |εn | ≤ |εn+1 | ≤ · · ·

(3.28)

are the discrete eigenvalues for the homogeneous integral equation u − εH1 H2 u = 0.

(3.29)

Noting that Problem Q for the complex equation (2.2) with ε = 0 is solvable, hence |ε1 | > 0. In the following, we first discuss the case of K ≥ 0. If ε
= εj (j = 1, 2, . . .), i.e. it is not an eigenvalue of the homogeneous integral equation (3.29), then the nonhomogeneous integral equation (3.27) has a solution u(z) and the general solution of Problem Q includes 2K + 2 arbitrary real constants. If ε is an eigenvalue of rank q as stated in (3.28), applying the Fredholm theorem, we obtain the solvability conditions for nonhomogeneous integral equation (3.27), there is a system of q algebraic equations to determine the 2K + 2 arbitrary real constants, setting that s is the rank of the corresponding coefficients matrix and s ≤ min(q, 2K + 2), we can determine s equalities in the q algebraic equations, hence Problem Q for (2.2) has q − s solvability conditions. When these conditions hold, then the general solution of Problem Q includes 2K + 2 + q − s arbitrary real constants. As for the case of K < 0, it can be similarly discussed. Thus we can write the above result as in the following theorem. Theorem 3.2 Suppose that the linear mixed equation (2.2) satisfies Condition C
. If ε
= εj (j = 1, 2, . . .), where εj (j = 1, 2, . . .) are the eigenvalues of the homogeneous integral equation (3.29). Then

4. Frankl Boundary Value problem

177

(1) When K ≥ 0, Problem Q for (2.2) is solvable, and the general solution u(z) of Problem Q for (2.2) includes 2K + 2 arbitrary real constants. (2) When K < 0, Problem Q for (2.2) has −2K−1−s solvability conditions, s ≤ 1. If ε is an eigenvalue of homogeneous integral equation (3.29) with the rank q. (3) When K ≥ 0, Problem Q for (2.2) has q − s solvability conditions and s ≤ min (q, 2K + 2). (4) When K < 0, Problem Q for (2.2) has −2K − 1 + q − s solvability conditions and s ≤ min (−2K − 1 + q, 1 + q). Moreover we can derive the solvability result of Problem P for equation (2.2) with the boundary condition (3.2), in which h(z) = 0.

4

The Frankl Boundary Value Problem for Second Order Linear Equations of Mixed Type

This section deals with the Frankl boundary value problem for linear second order equations of mixed (elliptic-hyperbolic) type, i.e. for generalized Lavrent
ev-Bitsadze equations. We first give representation formula and prove uniqueness of solutions for the above boundary value problem, moreover we obtain a priori estimates of solutions, finally by the method of parameter extension, the existence of solutions is proved. In the books [12]1),3), the Frankl problem was discussed for the special mixed equations of second order: uxx +sgny uyy = 0. In the book [73], the Frankl problem was discussed for the mixed equation with parabolic degeneracy sgny|y|m uxx + uyy = 0, which is a mathematical model of problem of gas dynamics. There the existence of solutions of Frankl problem was proved by using the method of integral equations. In this section, we will not use this method. We are proving the solvability of the Frankl problem for generalized linear Lavrent
ev-Bitsadze equations, generalizing the corresponding result from [12]1),3).

4.1 Formulation of the Frankl problem for second order equations of mixed type Let D be a simply connected bounded domain in the complex plane C I with the boundary ∂D = Γ ∪ A
A ∪ A
C ∪ CB, where Γ(⊂ {x > 0, y > 0}) ∈ Cµ2 (0 < µ < 1) with the end points A = i and B = a, A
A = {x = 0, −1 ≤ y ≤ 1}, A
C = {x − y = 1, x > 0, y < 0} is the characteristic line and CB = {1 ≤ x ≤ a, y = 0}, and denote D+ = D ∩ {y > 0}, D− = D ∩ {y < 0}. Without loss of generality, we may assume that Γ = {x2 /a2 + y 2 = 1, x > 0, y > 0}, otherwise, through a conformal mapping from D+ onto the domain D
+ = {x2 /a2 + y 2 < 1, x > 0, y > 0}, such that three boundary points i, 0, 1 are not changed, then the above requirement can be realized.

178

V. Second Order Linear Mixed Equations

Frankl Problem Find a continuously differentiable solution u(z) of equation ¯ and satisfies the (2.2) in D∗ = D\{1, a, i, −i, x + y = 0}, which is continuous in D boundary conditions u = ψ1 (s) on Γ,

(4.1)

u = ψ2 (x) on CB,

(4.2)

∂u = 0 on A
A, ∂x u(iy) − u(−iy) = φ(y), −1 ≤ y ≤ 1.

(4.3) (4.4)

Here ψ1 (s), ψ2 (x), φ(y) are given real-valued functions satisfying the conditions Cα1 [ψ1 (s),S]≤k2 ,

Cα1 [ψ2 (x),CB]≤k2 ,

Cα1 [φ(y),A
A]≤k2 ,

(4.5)

ψ1 (0)=ψ2 (a),

in which S = {0 ≤ s ≤ l}, s is the arc length parameter on Γ normalized such that s = 0 at the point B, l is the length of Γ, and α (0 < α < 1), k2 are nonnegative constants. The above boundary value problem is called Problem F and the corresponding homogeneous problem is called Problem F0 . Let

1 1 U = ux , V = − uy , W = U + iV in D, 2 2 then equation (2.2) can be written as the complex equation


Wz

⎧ ⎫ ⎨ D+ ⎬



= Re [A1 W ] + A2 u + A3 in ⎩ , D− ⎭

W z∗

(4.6)



(4.7)

z

u(z) = 2Re a

W (z)dz + ψ1 (0).

If A1 = A2 = A3 = 0 in D, then it is clear that 1 1 U (x, y) = ux = [f (x + y) + g(x − y)], 2 2 1 1 −V (x, y) = uy = [f (x + y) − g(x − y)], 2 2

(4.8)

in D− . From the boundary conditions (4.1)–(4.3), it follows that 1 1 1 −V (0, y) = [u(0, y)]y = [u(0, −y)]y + φ
(y) 2 2 2 1 1 = V (0, −y) + φ
(y) = −F (y) + φ
(y), 2 2

U (0, y) =

1 ∂u = 0, 2 ∂x

F (y) = −V (0, −y),

−1 ≤ y ≤ 0,

(4.9)

4. Frankl Boundary Value problem

179

and then 1 U (0, y) = [f (y) + g(−y)] = 0, −1 ≤ y ≤ 0, 2 1 1 −V (0, y) = [f (y) − g(−y)] = −F (y) + φ
(y), −1 ≤ y ≤ 0, 2 2 1 U (0, y) + V (0, y) = g(−y) = F (y) − φ
(y), −1 ≤ y ≤ 0, 2 1 U (0, y) − V (0, y) = f (y) = −F (y) + φ
(y), −1 ≤ y ≤ 0, 2 f (y) = −g(−y), f (y) = g(−y) − 2F (y) + φ
(y),

−1 ≤ y ≤ 0, i.e.

(4.10)

f (y − x) = −g(x − y), f (y − x) = g(x − y) − 2F (y − x) + φ
(y − x), 1 U (x, y) + V (x, y) = g(x − y) = F (y − x) − φ
(y − x), 2

0 ≤ x − y ≤ 1,

U (x, y) − V (x, y) = f (x + y) = −g(−x − y) 1 = −F (x + y) + φ
(x + y), 2

0 ≤ −x − y ≤ 1.

Hence 1 U (x, y) = [f (x + y) − f (y − x)], 0 ≤ x − y ≤ 1, 2 1 1 −V (x, y) = [f (x+y)−f (y−x)]−F (y − x)+ φ
(y−x), 0 ≤ x−y ≤ 1, 2 2 1
U (x, 0) + V (x, 0) = g(x) = F (−x) − φ (−x), 0 ≤ x ≤ 1. 2

(4.11)

In particular we have 1 1 1 U (x, 0) = [f (x) − f (−x)] = [f (x) + F (−x) − φ
(−x)], 2 2 2 1 −V (x, 0) = [f (x) − f (−x)] − F (−x) + 2 1 1 = [f (x) − F (−x) + φ
(−x)] 2 2 The boundary conditions of the Frankl problem are

1
φ (−x) 2

(4.12)

on OC.

∂u = 2Re [λ(z)W (z)] = r(z), z ∈ Γ ∪ CB, u(a) = b0 = ψ1 (0), ∂l 1 ∂u = r(0, y) = Re [λ(iy)W (iy)] = 0, −1 ≤ y ≤ 1, U (0, y) = 2 ∂x

(4.13)

1 1 Re [λ(x)W (x)] = r(x) = √ [F (−x)− φ
(−x)], 2 2

(4.14)

x ∈ L0 = (0, 1),

180

V. Second Order Linear Mixed Equations

in which l is the tangent vector on the boundary Γ, and ⎧ ⎪ cos(l, x) − i cos(l, y), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1, ⎪ ⎨

λ(z) =

⎧ ⎪ ψ1
(s) on Γ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 0 on AO, ⎪ ⎨

= BA,

r(z) = ⎪ 1 1 ⎪ √ [F (−x) − φ
(−x)] on OC, ⎪ ⎪ ⎪ 2 2 ⎪ ⎪

1+i ⎪ ⎪ √ , ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎩

⎪ ⎪ ⎩ ψ
(x)

1,

2

on CB.

We shall prove the solvability of the Frankl problem for equation (2.2) by using the methods of parameter extension and symmetry extension. We can choose the index K = −1/2 of λ(z) on the boundary ∂D+ of D+ . In fact, due to the boundary condition 1 Re [λ(z)W (z)] = Re [λ(z)(ux − iuy )] = r(z) on ∂D+ = AO ∪ OB ∪ BA, 2

(4.15)

and λ(z) = 1 on AO ∪ CB, λ(z) = exp(iπ/4) on OC, λ(z) = cos(l, x) − i cos(l, y) on Γ, denote t1 = 0, t2 = 1, t3 = a, t4 = i, it is seen λ(a + 0) = exp(i3π/2) and λ(i − 0) = exp(iπ), we have 



λ(tj −0) , λ(tj +0)

Kj =

φj +Jj , π

eiφ1 =

λ(t1 −0) ei0 = iπ/4 = e−iπ/4 , λ(t1 +0) e

0 < γ1 =

eiφ2 =

λ(t2 −0) eiπ/4 = i0 = eiπ/4 , λ(t2 +0) e

−1 < γ2 =

Jj = 0 or 1,

eiφ3 =

ei0 λ(t3 − 0) = i3π/2 , λ(t3 + 0) e

eiφ4 =

eiπ λ(t4 − 0) = i0 = eiπ , λ(t4 + 0) e

eiφj =

0 ≤ γ3 =

γj =

φj −Kj , π

j = 1,. . .,4,

φ1 1 3 −K1 = − −K1 < < 1, π 4 4 φ2 1 3 −K2 = −K2 = − < 0, π 4 4

(4.16)

φ3 3 1 − K3 = − − K3 = < 1, π 2 2

0 ≤ γ4 =

φ4 − K4 = 1 − K4 = 0 < 1, π

here [b] is the largest integer not exceeding the real number b, we choose K1 = −1, K3 = −2, K2 = K4 = 1. Under these conditions, the index K of λ(z) on the boundary ∂D+ of D+ is just as follows: 1 1 K = (K1 + K2 + K3 + K4 ) = − . 2 2

(4.17)

Noting that U (0, y) = 0 on A
A, we can extend W (z) onto the reflected domain ˜ of D about the segment A
A. In fact, we introduce the function D ⎧ ⎨ W (z)

˜ (z) = W ⎩

in D,

˜ −W (−¯ z ) on D,

(4.18)

4. Frankl Boundary Value problem

181

˜ (z) is a solution of the equation this function W ⎧ ⎫ ˜z ⎬ ⎨W

⎧

⎫

⎨D⎬ ˜ ] + A˜2 u˜ + A˜3 in = Re [A˜1 W ⎩ ˜ ⎭ ⎩D ˜⎭ W z∗ 

u˜(z) = 2Re

z

1

(4.19)

˜ (z)dz + ψ1 (0), W

with the boundary conditions ˜ W ˜ (z)]=˜ r(z), 2Re[λ(z)

! ˜ BC, z ∈ Γ∪CB ∪ Γ∪

˜ W ˜ (x)]=˜ r(x), Re[λ(x)

u(a)=b0 =ψ1 (0)=u(−a),

(4.20)

˜ 2 =(0,1)∪(−1,0), x∈ L

in which ⎧

⎧

⎧

⎨ A1 (z), ⎨ A2 (z), ⎨ A3 (z) in D, A˜1 = ⎩ A˜3 = ⎩ A˜2 = ⎩ ˜+ ∪ D ˜ −, z ), A2 (−¯ A3 (−¯ z ) in D z ), −A1 (−¯

and

⎧ ⎨ λ(z),

˜ λ(z) =⎩

⎧ ⎨ r(z)

λ(−¯ z ),

⎧ 1+i ⎪ ⎪ ⎪ √ , ⎨

r˜(z) = ⎩

(4.21)

Γ ∪ CB,

˜ ∪ BC, ˜ −r(−¯ z) Γ (4.22)

⎧

⎨ r(z) on OC = (0, 1), 2 ˜ r˜(z) = ⎩ λ(z) =⎪ 1−i ⎪ ˜ = (−1, 0), ⎪ √ , −r(−¯ z ) on CO ⎩ 2

˜ BC ˜ = (−a, −1), CO ˜ and AB ˜ are the reflected curves of Γ, CB, OC and BA herein Γ, ˜ about the imaginary axis, respectively. We choose the index of the function λ(z) on ˜ + ∪AO) of the elliptic domain D+ ∪ D ˜ + ∪AO as K = −1/2. In the boundary ∂(D+ ∪ D ˜ ˜ λ(z) = exp(iπ/4) on OC, λ(z) fact, noting that λ(z) = 1 on CB ∪ BC, = exp(−iπ/4) ˜ we denote t1 = 0, t2 = 1, t3 = a, t4 = i, t5 = −a, t6 = −1, it is seen on CO, λ(a + 0) = exp(i3π/2), λ(i − 0) = λ(i + 0) = exp(iπ), λ(−a − 0) = exp(iπ/2), hence we have 



Kj =

φj +Jj , π

eiφ1 =

˜ 1 −0) e−iπ/4 λ(t = = e−iπ/2 , ˜ 1 +0) eiπ/4 λ(t

eiφ2 =

˜ 2 −0) eiπ/4 λ(t = i0 = eiπ/4 , ˜ 2 +0) e λ(t

Jj = 0 or 1,

ejφj =

˜ j −0) λ(t , ˜ j +0) λ(t

0 < γ1 = −1 < γ2 =

γj =

φj −Kj , π

j = 1,. . ., 6,

φ1 1 1 −K1 = − −K1 = < 1, π 2 2

φ2 1 3 −K2 = −K2 = − < 0, π 4 4

182

V. Second Order Linear Mixed Equations

eiφ3 = eiφ4 = iφ5

e

˜ 3 −0) λ(t ei0 = i3π/2 = e−i3π/2 , ˜ 3 +0) e λ(t ˜ 4 − 0) λ(t eiπ = iπ = ei0 , ˜ e λ(t4 + 0)

˜ 5 − 0) eiπ/2 λ(t = = i0 = eiπ/2 , ˜ 5 + 0) e λ(t

eiφ6 =

˜ 6 −0) λ(t ei0 = −iπ/4 = eiπ/4 , ˜ λ(t6 +0) e

0 < γ3 =

0 ≤ γ4 =

φ3 3 1 −K3 = − −K3 = < 1, π 2 2

φ4 − K4 = 0 − K4 = 0 < 1, π (4.23)

φ5 1 1 − K5 = − K5 = < 1, 0 < γ5 = π 2 2 −1 < γ6 =

φ6 1 3 −K6 = −K6 = − < 0. π 4 4

˜ If we choose K1 = −1, K2 = K6 = 1, K3 = −2, K4 = K5 = 0, the index K of λ(z) is just 1 1 K = (K1 + K2 + · · · + K6 ) = − . (4.24) 2 2 We can discuss the solvability of the corresponding boundary value problem (4.19), (4.20), and then derive the existence of solutions of the Frankl problem for equation (2.2). 4.2 Representation and a priori estimates of solutions to the Frankl problem for (2.2) First of all, similarly to Lemma 2.1, we can prove the following theorem. Theorem 4.1 Let equation (2.2) satisfy Condition C in D+ , and u(z) be a continuous solution of (2.2) in D∗+ = D+ \{0, 1, a, i}. Then u(z) can be expressed as u(z) = U (z)Ψ(z) + ψ(z) in D+ , 

z

U (z) = 2Re a

w(z)dz + b0 , w(z) = Φ(z)eφ(z) in D+ .

(4.25)

Here ψ(z), Ψ(z) are the solutions of equation (2.2) in D+ and uzz¯ − Re [A1 uz ] − A2 u = 0 in D+ ,

(4.26)

respectively and satisfy the boundary conditions ψ(z) = 0, Ψ(z) = 1 on Γ ∪ L, ∂Ψ(z) ∂ψ(z) = 0, = 0 on AO, ∂x ∂x

(4.27)

where L = (0, a). They satisfy the estimates Cγ1 [X(z)ψ(z), D+ ] ≤ M19 ,  X(z)ψ(z) Wp20 (D+ ) ≤ M19 ,

(4.28)

4. Frankl Boundary Value problem

183

Cγ1 [X(z)Ψ(z), D+ ] ≤ M20 , X(z)Ψ(z) Wp20 (D+ )≤ M20 ,Ψ(z) ≥ M21 > 0, z ∈ D+ , (4.29) in which X(z) = |x + y − t1 |η1 4j=2 |z − tj |ηj , ηj = max{−γj + δ, δ}, j = 1, 2, 3, 4, herein tj , γj (j = 1, 2, 3, 4) are as stated in (4.16), δ, γ(γ < δ) are small positive constants, p0 (2 < p0 ≤ p), Mj = Mj (p0 , γ, k, D) (j = 19, 20, 21) are non-negative constants, k = (k0 , k1 , k2 ). Moreover, U (z) is a solution of the equation Uzz¯ − Re [AUz ] = 0, A = −2(ln Ψ)z¯ + A1 in D+ ,

(4.30)

where Im [φ(z)] = 0, z ∈ ∂D+ , Re [φ(0)] = 0 and φ(z) satisfies the estimate Cβ [φ(z), D+ ] + Lp0 [φz¯, D+ ] ≤ M22 ,

(4.31)

in which β (0 < β ≤ α), M22 = M22 (p0 , α, k0 , D+ ) are two non-negative constants, Φ(z) is analytic in D+ . If u(z) is a solution of Problem F, then W (z) = uz satisfies the boundary conditions Re [λ(z)W (z)] = r(z) on Γ ∪ AO, u(a) = b0 = ψ1 (0), ⎧ 1+i ⎪ ⎨ √

Re [λ(x)W (x)] = r(x), λ(x) = ⎪ ⎩

on L0 = (0, 1), 2 1, on L1 = (1, a).

(4.32)

(4.33)

Theorem 4.2 Suppose that equation (2.2) satisfies Condition C. Then any solution of the Frankl problem for (2.2) can be expressed as 

z

u(z) = 2Re a

W (z)dz + b0 , b0 = ψ1 (0).

(4.34)

Here W (z) is a solution of the equation


Wz¯ ¯ z∗ W






= Re [A1 W ] + A2 u + A3 in

D+ D−



,

(4.35)

satisfying the boundary conditions (4.13) − (4.14) (W (z) = uz ), and W (z) possesses the form ˜

φ(z) ˜ ˜ W (z) = Φ(z)e + ψ(z),

g(ζ) 1  ˜ dσζ , φ(z) = φ˜0 (z) + T g = φ˜0 (z) − + π D ζ −z 

W (z) = Φ(z) + Ψ(z),

Ψ(z) =

1

ν

g1 (z)dνe1 +

˜ ψ(z) = T f in D+ ,  0

µ

g2 (z)dµe2 in D− ,

(4.36)

184

V. Second Order Linear Mixed Equations

in which e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic ˜ function in D+ , such that Im φ(x) = 0 on Γ ∪ AO ∪ L, and


W (z)
= 0

A1 /2+A1 W /(2W ),

g(z) =

0,

W (z) = 0

f (z) = Re [A1 φ˜z ] + A2 u + A3 in D+ ,

f (z) = Re [A1 uz ] + A2 u + A3 ,

g 1 (z) = g 2 (z) = Aξ +Bη+Cu+D, A=

Re A1 +Im A1 , 2

B=

in D+

ξ = Re W +Im W,

Re A1 −Im A1 , 2

(4.37)

η = Re W −Im W, D = A3 in D− ,

C = A2 ,

˜ where Φ(z) in D+ and Φ(z) in D− are solutions of the equation


Wz¯ W z∗






= 0 in

D+ D−



,

(4.38)

satisfying the boundary conditions ˜ ˜ ˜ = r(z) − Re [λ(z)ψ(z)], Re [λ(z)eφ(z) Φ(z)] ˜ φ(z) ˜

Re [e

˜ Φ(z)] = −Re [ψ(z)],

z ∈ Γ,

z = iy ∈ AO,

˜

φ(x) ˜ ˜ ] = r(x) − Re [λ(x)ψ(x)], Re [λ(x)Φ(x)e

x ∈ L0 = (0, 1),

Re [λ(x)(Φ(x) + Ψ(x))] = r(x) = Re [λ(x)W (x)], Re [Φ(x)] = −Re [Ψ(x)],




x ∈ OA ,

(4.39)

x ∈ L0 = (0, 1),

u(a) = b0 = ψ1 (0),

where λ(x) on L = (0, a) is as stated in (4.15). Proof Let u(z) be a solution of the Frankl problem for equation (2.2), and W (z) = uz , u(z) be substituted in the positions of w, u in (4.37). Thus the functions ˜ ˜ g(z), f (z), g1 (z), g2 (z), and ψ(z), φ(z) in D+ and Ψ(z) in D− in (4.36),(4.37) are ˜ determined. Moreover, we can find the solution Φ(z) in D+ and Φ(z) in D− of (4.38) with the boundary condition (4.39), where r(z) as stated in (4.15), namely r(z) = H(F, φ),

z ∈ Γ ∪ AO ∪ L,

⎧ ˜ φ(z) ⎪ ˜ ⎨ Φ(z)e

thus W (z) =

(4.40)

˜ + ψ(z) in D+ ,

⎪ ⎩ Φ(z) + Ψ(z)

in D− ,

(4.41)

is the solution of Problem A for the complex equation (4.35) with the boundary conditions (4.13),(4.14), which can be expressed as in (4.36), and u(z) is a solution of the Frankl problem for (2.2) as stated in (4.34). Next, we discuss the uniqueness of solutions of the Frankl problem for (2.2).

4. Frankl Boundary Value problem

185

Theorem 4.3 Suppose that the mixed equation (2.2) satisfies Condition C. Then ¯ ∩ C 1 (D). the Frankl problem for (2.2) has at most one solution u(z) ∈ C(D) Proof We consider equation (2.2) in D+ . As stated before, if u1 (z), u2 (z) are two solutions of the Frankl problem for (2.2), then u(z) = u1 (z) − u2 (z) is a solution of the homogeneous equation ⎧ ⎫ ⎨uz z¯ ⎬ ⎩u

z¯z ∗

⎭

⎧ ⎫ ⎨D + ⎬

= Re [A1 uz ]+A2 u in ⎩ −⎭ , D 

z

u(z) = 2 Re

W (z)dz, a

˜ u(z) = U (z)Ψ(z),

W (z) = uz in D,

(4.42)

˜ φ(z) ˜ Uz = Φ(z)e in D+ ,

W (z) = Φ0 (z) + Φ(z) + Ψ(z) in D− , in which Ψ(z) =

 1

ν

g 1 (z)dνe1 +



µ

0

g 2 (z)dµe2 ,

g 1 (z) = g 2 (z) = Aξ + Bη + Cu in D− ,

˜ ˜ ˜ ˜ and Ψ(z), Φ(z), φ(z), Φ(z) are similar to those in Theorem 4.2, and Φ(z), Φ(z), Φ0 (z) are solutions of equation (4.38) in D+ and D− respectively satisfying the conditions ˜ = r(z) = 0 on Γ∪AO∪CB, 2Re [λ(z)Φ(z)]

U (a) = 0,

1 ˜ = [f (x) + F (−x)] on L0 = (0, 1), Re [λ(x)Φ(x)] 2 ˜ Φ(x) = Φ(x), Φ0 (x) = uz (x) − Ψ(x) − Uz (x)e−φ(x) on L0 , Re [Φ0 (z)] = Re [Ψ(z)],

(4.43)

Im [Φ0 (z)] = Im [uz −Ψ(z)−Uz (z)e−φ(z) ] on A
O.

 ˜ According to Theorem 4.1, the solution U˜ (z) = 2Re az Φ(z)dz of equation (1.2) satisfies the boundary conditions

U˜ (z) = S(z) = 0 on Γ ∪ CB, U˜ (x) = 2



x

˜ Φ(x)dx =2

a

= S(x) =

 0



x

a

˜

Uz (x)e−φ(x) dx =



x

0

˜ Φ(x)dx −

 0

a

˜ Φ(x)dx

(4.44)

x

1 [f (x)+F (−x)]dx = g(x)+ U˜ (ix) on (0, 1), 2

  where g(x) = 0x f (x)dx, 0x F (−x)dx = U˜ (ix)/2. Besides the harmonic function U˜ (z) in D+ satisfies the boundary condition

∂ U˜ (z) = 0 on AO. ∂x

(4.45)

Moreover, there exists a conjugate harmonic function V˜ (z) in D+ suchthat V˜ (0) = 0.  From the above last formula, we can derive that V˜ (iy) = 0y V˜y dy = 0y U˜x dy = 0 on

186

V. Second Order Linear Mixed Equations

AO. By the Cauchy theorem, we have  ∂D+

=−

[U˜ (z) + iV˜ (z)]2 dz = 0 

a

1



+i

1

[V˜ (x)]2 dx −

0





[U˜ (iy)]2 dy +

Γ

[V˜ (z)]2



1

0



dy dx +i ds ds ds

(4.46)

[U˜ 2 (x) − V˜ 2 (x) + 2i U˜ (x)V˜ (x)]dx.

   Due to the continuity of U˜y on (0,1), V˜ (x) = 0x V˜x dx = − 0x U˜y dx = 2 0x V˜ (x)dx = x ˜ − 0 [f (x) − F (−x)]dx = −g(x) + U (ix)/2 is obtained. From the imaginary part in (4.46) and the above formula, it is clear that  Γ

[V˜ (z)]2

 1  1 ∂y 1 ds + [U˜ (iy)]2 dy + 2 {[g(x)]2 − [U˜ (ix)]2 }dx = 0. ∂s 4 0 0

(4.47)

Hence, we get U˜ (iy) = 0 on AO,

g(x) = 0 on OC,




and then f (x) = g (x) = 0, F (−x) = [U˜ (ix)]x /2 = 0 on OC. Due to the function ˜ r(z) = S(z) = 0 on ∂D+ in (4.43),(4.44) and the index K = −1/2, hence Φ(z) = 0 in + D , and then the solution u(z) of the homogeneous Frankl problem for (4.42) in D+ satisfies u(z) = u1 (z) − u2 (z) = 0. Moreover, we can derive u(z) = u1 (z) − u2 (z) = 0 in D− . This proves the uniqueness of solutions for the Frankl problem for (2.2) in D. Finally we give an a priori estimate of solutions to the Frankl problem for equation (2.2). From the estimate we can see the singular behavior of uz at the discontinuity set Z = {1, a, i, −i, x + y = 0}. It becomes infinity of an order not exceeding 3/4 at z = 1, infinite of order not exceeding a small positive number δ at the points {i, −i} and uz is bounded at the point set {a, x + y = 0}. In fact, we can prove that z = i, −i are removable singular points. In [12]3), the author pointed out that uz can become infinity of an order less than 1. ¯ and the funcTheorem 4.4 Suppose that equation (2.2) satisfies Condition C in D tion r(z) in (4.14) is H(F, φ), especially 1 1 r(x) = H(F, φ) = √ [F (−x) − φ
(−x)], 2 2

x ∈ L0 = (0, 1).

(4.48)

Then any solution u(z) of the Frankl problem for equation (2.2) in D+ satisfies the estimate C˜γ1 [u, D+ ] = Cγ [u(z), D+ ] + C[uz X(z), D+ ] ≤ M23 (k1 + k2 ),

(4.49)

where X(z) is as stated in (4.29), i.e. X(z) = |x + y − t1 |η1

4  j=2

|z − tj |ηj ,

ηj = max{−γj + δ, δ},

j = 1, 2, 3, 4, (4.50)

4. Frankl Boundary Value problem

187

and M23 = M23 (p0 , γ, δ, k0 , D) is a non-negative constant. Proof On the basis of the uniqueness of solutions of the Frankl problem for (2.2) in Theorem 4.3 and the results in [12]3) by using reductio ad absurdum we can derive the estimate (4.49). In fact, from (4.83),(4.84) in the proof of Theorem 4.6 ˜ n+1 , u˜n+1 ] (W ˜ n+1 (z) = Wn+1 (z)−Wn (z), u˜n+1 (z) = below, we see that the function [W un+1 (z) − un (z)) is a solution of the boundary value problem ˜ n+1 ) = (t − t0 )G(z, u˜n , W ˜ n ), [Wn+1 ]z¯ − t0 G(z, u˜n+1 , W

z ∈ D+ ,

z ∈ Γ ∪ AO ∪ CB,

˜ n+1 (z)] = 0, Re [λ(z)W

˜ n+1 (z)] − t0 H(F˜n+1 , 0) = (t − t0 )H(F˜n , 0)] on L0 = (0, 1), Re [λ(z)W 

u˜n+1 (z) =

z

a

(4.51)

z ∈ D+ ,

˜ n+1 (z)dz, W

˜ u˜n , w˜n ) = G(z, un+1 , wn+1 ) − G(z, where G(z, u, W ) = Re [A1 W ] + A2 u + A3 , G(z, ˜ n+1 of the un , wn ), F˜n+1 = Fn+1 − Fn . On the basis of Theorem 4.1, the solution W boundary value problem (4.51) can be expressed as ˜ n+1 (z) + ψ˜n+1 (z) in D+ , u˜n+1 (z) = Un+1 (z)Ψ 

U˜n+1 (z) = 2 Re

z

˜ n+1 (z)eφ˜n+1 (z) in D+ , ˜ n+1 (z)dz, Un+1z = Φ Φ

1

(4.52)

˜ n+1 (z) = Φ0 (z) + Φn+1 (z) + Ψn+1 (z) in D− , W n+1 where Φn+1 (z), Φ0n+1 (z) are the solutions of equation (4.38) in D− , Ψn+1 (z) is a ˜ n+1 (z) are the solutions of solution of the equation in (4.51) in D− , ψ˜n+1 (z), Ψ the equation in (4.51) and its homogeneous complex equation in D+ satisfying the boundary conditions ˜ n+1 (z) = 1 on Γ ∪ L, ψ˜n+1 (z) = 1, Ψ ˜ n+1 (z) ∂Ψ ∂ ψ˜n+1 (z) = 0, = 0 on AO. ∂x ∂x

(4.53)

According to the proof of Theorem 4.3, we see that the function U˜n+1 (z) satisfies the boundary conditions U˜n+1 (z) = S(z) = 0 on Γ ∪ CB, U˜n+1 (x) = Re 

S(x) =

0

x



x

a

U˜n+1 (a) = 0,

˜ n+1x dx = S(x), Φ

[f (x)+ F˜n+1 (−x)]dx −

1  x ˜
φ (−x)dx 2 0 n+1

1 1 = g(x) + U˜n+1 (ix) + φ˜n+1 (−x) on (0, 1), 2 2

(4.54)

188

V. Second Order Linear Mixed Equations

where 

g(x) =

0



x

f (x)dx,



U˜n+1 (ix) , F˜n+1 (−x)dx = 2

x

0

U˜n+1 (−x) = −

x

0

φ˜
n+1 (−x)dx.

Besides we can see that the harmonic function U˜n+1 (z) in D+ satisfies the boundary condition ∂ U˜n+1 (z) = 0 on AO. (4.55) ∂x Moreover there exists a conjugate harmonic function V˜n+1 (z) in D+ such that V˜n+1 (0) = 0. We shall verify that |X(x)U˜n+1x | = 0, lim max + n→∞ D

Suppose that limn→∞

1 0

+2

0

1



n→∞ 0

1

[U˜n+1 (iy)]2 dy = 0.

(4.56)

|U˜n+1 (iy)|dy = C > 0, due to 





lim

Γ

[V˜n+1 (z)]2

 1 ∂y ds + [U˜n+1 (iy)]2 dy ∂s 0



1 [g(x)] − [t0 U˜n+1 (ix) + (t − t0 )U˜n (ix)]2 dx = 0, 4 2

(4.57)

 provided that |t − t0 | is sufficently small, such that |t − t0 |2 01 |U˜n (iy)|2 dy ≤ 1 2 ˜ 0 |Un+1 (iy)| dy/2 for n = nk → ∞, then similarly to (4.47), from (4.57) we can derive that (4.58) U˜n+1 (iy) = 0 on AO, g(x) = 0 on OC.

  This contradiction proves that 01 [U˜n+1 (ix)]2 dx = 0, 01 [U˜n+1 (ix)]2 dx = 0 and U˜n+1 = + Un+1 (z) − Un (z) = 0 in D for n ≥ N0 , where N0 is a sufficiently large positive number. Hence u˜n+1 = un+1 (z) − un (z) = ψn+1 (z) − ψn (z) in D+ for n ≥ N0 . Similarly to the proof of the first estimate in (4.28), we can obtain

Cγ1 [X(z)˜ un+1 (z), D+ ] ≤ M24 |t − t0 |Cγ1 [X(z)˜ un (z), D+ ],

(4.59)

in which M24 = M24 (p, γ, δ, k0 , D+ ) is a non-negative constant. Choosing the constant ε so small that εM24 ≤ 1/2 and |t − t0 | ≤ ε, it follows that 1 un+1 , D+ ] ≤ εM24 |t − t0 |Cγ1 [˜ un , D+ ] ≤ Cγ1 [˜ un , D+ ], C˜γ1 [˜ 2 un+1 , D+ ] ≤ 2−n+N0 C˜γ1 [˜

∞ 

2−j C˜γ1 [u1 − u0 , D+ ] ≤ 2−n+N0 +1 C˜γ1 [u1 −u0 , D+ ]

j=N0

for n > N0 . Therefore there exists a continuous function u∗ (z) on D+ , such that u∗ (z) =

∞  j=0

u˜n+1 =

∞  j=0

[un+1 − un (z)].

4. Frankl Boundary Value problem From the estimate of

n

j=0 [uj+1 (z)

189

− uj (z)] = un+1 (z) − u0 (z) in D+ , the estimate

¯ = Cγ [un+1 , D+ ] + C[un+1z X, D+ ] ≤ M25 C˜γ1 [un+1 , D]

(4.60)

can be derived, where M25 = M25 (p0 , γ, δ, k0 , D+ ) is a non-negative constant. Moreover we can derive a similar estimate of u∗ (z) in D+ and D− , which gives the estimate (4.49).

4.3

The solvability of the Frankl problem for (2.2)

Theorem 4.5 Suppose that the mixed equation (2.2) satisfies Condition C and ¯ i.e. A1 (z) = A2 (z) = 0 in D, ⎧ ⎨ uzz ⎩

= A3 (z),

uz¯z∗ = A3 (z),

z ∈ D+ ,

(4.61)

z ∈ D− .

Then the Frankl problem for (4.61) has a solution in D. Proof It is clear that the Frankl problem for (4.61) is equivalent to the following Problem A for the complex equation of first order and boundary conditions:


Wz W z∗

⎧ ⎫ ⎨ D+ ⎬



= A3 (z) in ⎩ , D− ⎭

Re [λ(z)W (z)] = r(z) = H(F, φ), Re [λ(z)W (z)] = r(z) = 0, and the relation



(4.62)

z ∈ Γ ∪ AO ∪ L,

λ(z) = 1,

z ∈ OA
,

(4.63)

z

u(z) = 2Re a

W (z)dz + b0 in D,

(4.64)

in which λ(z), r(z) are as stated in (4.15) and (4.40). In order to find a solution W (z) of Problem A in D, we can express W (z) in the form (4.34)–(4.37). In the following, by using the method of parameter extension, we shall find a solution of Problem A for the complex equation (4.62). We consider equation (4.62) and the boundary conditions with the parameter t ∈ [0, 1]: Re [λ(z)W (z)] = tH(F, φ) + R(z) on ∂D+ = Γ ∪ AO ∪ L,

(4.65)

in which H(F, φ) on ∂D+ = Γ ∪ AO ∪ L is as stated in (4.48), and R(z)X(z) ∈ Cγ (∂D+ ), this problem is called Problem Ft . When t = 0, the unique solution of Problem F0 for the complex equation (4.61) can be found by a method given in Section 1, and its solution [W0 (z), u0 (z)] can be

190

V. Second Order Linear Mixed Equations

expressed as u0 (z) = 2Re



z

a

˜ (z) in D, W0 (z) = W

W0 (z)dz + b0 ,

˜ ˜ (z) = Φ(z) ˜ W + ψ(z),

1 ˜ ψ(z) = T A3 = − π 

W (z) = Φ(z)+Ψ(z),

Ψ(z) =

1



ν

A3 (z)e1 dν +

D+



µ

0

b0 = ψ1 (0),

A3 (ζ) dσζ in D+ , ζ −z

(4.66)

A3 (z)e2 dµ in D− ,

˜ where Φ(z) is an analytic function in D+ and Φ(z) is a solution of (4.38) in D− satisfying the boundary conditions ˜ (z)] = R(z), Re [λ(z)W

z ∈ Γ ∪ L,

˜ ˜ + Ψ(z))] = R(z), Re [λ(z)(Φ(z)

λ(z) = 1,

z = iy ∈ AO,

Re [λ(z)(Φ(z) + Ψ(z))] = R(z),

λ(z) = 1,

z = iy ∈ OA
,

˜ (x)], Re [λ(x)(Φ(x)+Ψ(x))] = R(x) = Re [λ(x)W

z = x ∈ OC,

(4.67)

u(a) = b0 .

Suppose that when t = t0 (0 ≤ t0 < 1), Problem Ft0 is solvable, i.e. Problem Ft0 ¯ We can find a neighborhood for (4.62) has a solution [W0 (z), u0 (z)] (u0 (z) ∈ C˜γ1 (D)). Tε = {|t − t0 | ≤ ε, 0 ≤ t ≤ 1}(0 < ε < 1) of t0 such that for every t ∈ Tε , Problem Ft is solvable. In fact, Problem Ft can be written in the form


Wz ¯ z∗ W

⎧ ⎫ ⎨ D+ ⎬



= A3 (z) in ⎩ , D− ⎭

(4.68) +

Re [λ(z)W (z)] − t0 H(F, φ) = (t − t0 )H(F, φ) + R(z) on ∂D . Replacing W (z), u(z) in the right-hand sides of (4.68) by a function W0 (z) ∈ Cγ (∂D+ ) ¯ especially, by the and the corresponding function u0 (z) in (4.66), i.e. u0 (z) ∈ C˜γ1 (D), solution [W0 (z), u0 (z)] of Problem F0 , it is obvious that the boundary value problem for such an equation in (4.68) then has a solution [W1 (z), u1 (z)], u1 (z) ∈ C˜γ1 (∂D). Using successive iteration, we obtain a sequence of solutions [Wn (z), un (z)], un (z) ∈ ¯ n = 1, 2, . . . , which satisfy the equations and boundary conditions C˜γ1 (D),


Wn+1z ¯ n+1z∗ W

⎧ ⎫ ⎨ D+ ⎬



= A3 (z) in ⎩ , D− ⎭

Re [λ(z)Wn+1 (z)] − t0 H(Fn+1 , φ) = (t − t0 )H(Fn , φ) + R(z) on ∂D+ , Re [λ(z)Wn+1 (z)] = 0,

(4.69)

z ∈ OA
.

From the above formulas, it follows that [Wn+1 − Wn ]z¯ = 0,

z ∈ D,

Re [λ(z)(Wn+1 (z)−Wn (z))]−t0 [H(Fn+1 −Fn , 0)] = (t−t0 )[H(Fn −Fn−1 ,0)].

(4.70)

4. Frankl Boundary Value problem

191

Noting that |t − t0 |Cγ [XH(Φn − Φn−1 , 0), L0 ] ≤ |t − t0 |Cγ [X(Φn − Φn−1 ), L0 ],

(4.71)

and applying Theorem 4.4, we have C˜γ1 [un+1 − un , D+ ] ≤ M26 C˜γ1 [Φn − Φn−1 , D+ ],

(4.72)

where M26 = M26 (p0 , γ, δ, k0 , D+ ). Choosing the constant ε so small that εM24 ≤ 1/2 and |t − t0 | ≤ ε, it follows that 1 C˜γ1 [un+1 − un , D+ ] ≤ εM26 C˜γ1 [un − un−1 , D+ ] ≤ Cγ1 [un − un−1 , D+ ], 2

(4.73)

and when n, m ≥ N0 + 1 (N0 is a positive integer), C˜γ1 [un+1 − un , D+ ] ≤ 2−N0

∞ 

2−j C˜γ1 [u1 − u0 , D+ ] ≤ 2−N0 +1 C˜γ1 [u1 − u0 , D+ ]. (4.74)

j=0

Hence {un (z)} is a Cauchy sequence. According to the completeness of the Banach space C˜γ1 (D+ ), there exists a function u∗ (z) ∈ Cγ1 (D+ ), and W∗ (z) = u∗z (z), so that C˜γ1 [un − u∗ , D+ ] = Cγ [un − u∗ , D+ ] + C[X(Wn − W∗ ), D+ ] → 0 as n → ∞. We can see that [W∗ (z), u∗ (z)] is a solution of Problem Ft for every t ∈ Tε = {|t − t0 | ≤ ε}. Because the constant ε is independent of t0 (0 ≤ t0 < 1), therefore from the solvability of Problem F0 when t0 = 0, we can derive the solvability of Problem Ft when t = ε, 2ε, . . . , [1/ε]ε, 1. In particular, when t = 1 and R(z) = 0, Problem F1 i.e. the Frankl problem for (4.61) in D+ is solvable. As for the solution [W (z), u(z)] in D− , it can be obtained by (4.10),(4.11) and the method in Chapters I and II, namely 

z

u(z) = 2Re a

W (z)dz + b0 on D− ,

W (z) = Φ(z) + Ψ(z),



Ψ(z) =

1

b0 = ψ1 (0), 

ν

A3 (z)e1 dν +

0

µ

A3 (z)e2 dµ,

1 Φ(z) = [(1 + i)f (x + y) + (1 − i)g(x − y)], 2

(4.75)

f (x + y) = Re [(1 − i)(W (x + y) − Ψ(x + y))], g(x − y) = Re [(1 + i)(W (x − y) − Ψ(x − y))],

z ∈ D− ∩ {x + y ≥ 0},

where W (x + y), W (x − y) are the values on 0 ≤ z = x + y ≤ 1, 0 ≤ x − y ≤ 1 of the solution W (z) of Problem F for (4.61) in D+ and Ψ(x + y), Ψ(x − y) are the values on 0 ≤ z = x + y ≤ 1, 0 ≤ x − y ≤ 1 of Ψ(z) respectively. Moreover, the function W (z) in D− ∩ {x + y ≤ 0} can be obtained by (4.75),(4.18). In fact, from (4.75) we have found the function W (z) on OC
= {x + y = 0, 0 ≤ x ≤ 1/2}, by (4.18), we ˜ (z) = −W (−¯ obtain the function W z ) on OC

= {x − y = 0, −1/2 ≤ x ≤ 0}, and

192

V. Second Order Linear Mixed Equations

˜ ˜ denote σ(x) = Re [(1 − i)Ψ(z)] on OC
, τ (x) = Re [(1 + i)Ψ(z)] on OC

. Hence the − solution u(z) in D ∩ {x + y ≤ 0} is as follows: 

u(z) = 2Re

z

W (z)dz + u(0),

0

˜ Ψ(z) =



ν

0

A˜3 (z)e1 dν +



µ

0

A˜3 (z)e2 dµ,

1 ˜ W (z) = [(1 − i)f (x + y) + (1 + i)g(x − y)] + Ψ(z), 2 f (x + y) = τ ((x + y)/2) + Re W (0) + Im W (0), g(x − y) = σ((x − y)/2) + Re W (0) − Im W (0),

(4.76)

z ∈ D− ∩ {x + y ≤ 0},

in which Φ(z) and Ψ(z) are the functions from (4.75). Furthermore, we can prove that the solution u(z) satisfies the boundary conditions (4.1)–(4.4). This completes the proof. Theorem 4.6 Suppose that the mixed equation (2.2) satisfies Condition C. Then the Frankl problem for (2.2) has a solution in D. Proof Similarly to the proof of Theorem 4.5, we see that the Frankl problem for (2.2) is equivalent to Problem A for first order complex equation and boundary conditions:


Wz ¯ z∗ W

⎧ ⎫ ⎨ D+ ⎬



= G,

G = G(z, u, W ) = Re [A1 W ] + A2 u + A3 in ⎩ , D− ⎭

Re [λ(z)W (z)] = r(z) = H(F, φ),

z ∈ Γ ∪ AO ∪ L,

and the relation (4.64), in which r(z) = H(F, φ) on z ∈ ∂D stated in (4.63).

+

(4.77) (4.78)

= Γ ∪ AO ∪ L is as

In order to find a solution W (z) of Problem A in D, we can express W (z) in the form (4.34)–(4.37). In the following, by using the method of parameter extension, a solution of Problem A for the complex equation (4.77) will be found. We consider the equation and boundary conditions with the parameter t ∈ [0, 1]: Wz = tG + K(z),

G = G(z, u, W ) = Re [A1 W ] + A2 u + A3 in D+ ,

Re [λ(z)W (z)] = tH(F, φ) + R(z) on ∂D+ = Γ ∪ AO ∪ L, where K(z) ∈ Lp Ft .

(D+ )

(4.79) (4.80)

+

and R(z)X(z) ∈ Cγ (∂D ). This problem is called Problem

When t = 0, the complex equation (4.79) becomes the equation Wz = K(z),

in D+ .

(4.81)

From Theorem 4.5, we can find the unique solution of Problem F0 for (4.79). Suppose that when t = t0 (0 ≤ t0 < 1), Problem Ft0 is solvable, i.e. Problem Ft0 for (4.79) has ¯ We can find a neighborhood Tε = {|t − t0 | ≤ a solution [W0 (z), u0 (z)] (u0 ∈ C˜γ1 (D)).

4. Frankl Boundary Value problem

193

ε, 0 ≤ t ≤ 1}(0 < ε < 1) of t0 such that for every t ∈ Tε , Problem Ft is solvable. In fact, Problem Ft can be written in the form Wz − t0 G(z, u, W ) = (t − t0 )G(z, u, W ) + K(z) in D+ , Re [λ(z)W (z)] − t0 H(F, φ) = (t − t0 )H(F, φ) + R(z) on ∂D+ .

(4.82)

Replacing W (z), u(z) in the right-hand sides of (4.82) by a function W0 (z) ∈ Cγ (∂D+ ) ¯ especially, by the and the corresponding function u0 (z) in (4.66), i.e. u0 (z) ∈ C˜γ1 (D), solution [W0 (z), u0 (z)] of Problem F0 , it is obvious that the boundary value problem for such equation in (4.82) then has a solution [W1 (z), u1 (z)], u(z) ∈ C˜γ1 (∂D). Using ¯ successive iteration, we obtain a sequence of solutions [Wn (z), un (z)], un (z) ∈ C˜γ1 (D), n = 1, 2, . . . , which satisfy the equations and boundary conditions in D+ ,

(4.83)

Re [λ(z)Wn+1 (z)]−t0 H(Fn+1 , φ) = (t − t0 )H(Fn , φ)+R(z) on ∂D+ .

(4.84)

Wn+1z −t0 G(z, un+1 , Wn+1 ) = (t − t0 )G(z, un , Wn ) + K(z),

From the above formulas, it follows that [Wn+1 − Wn ]z¯ − t0 [G(z, un+1 , Wn+1 ) − G(z, un , Wn )] = (t − t0 )[G(z, un , Wn ) − G(z, un−1 , Wn−1 )], z ∈ D+ , Re [λ(z)(Wn+1 (z) − Wn (z))] − t0 [H(Fn+1 − Fn , 0)] = (t − t0 )[H(Fn − Fn−1 , 0)],

(4.85)

z ∈ L0 .

Noting that Lp [(t−t0 )(G(z, un , Wn )−G(z, un−1 , Wn−1 )), D+ ] ≤ 2k0 |t−t0 |C˜γ1 [un −un−1 , D+ ], |t − t0 |Cγ [XH(Fn − Fn−1 , 0), L0 ] ≤ |t − t0 |Cγ [X(Fn − Fn−1 ), L0 ], (4.86) and according to the method in the proof of Theorem 4.4, we can obtain C˜γ1 [un+1 − un , D+ ] ≤ M27 [2k0 + 1]C˜γ1 [un − un−1 , D+ ],

(4.87)

where M27 = M27 (p0 , γ, δ, k0 , D+ ). Choosing the constant ε so small that εM27 (2k0 + 1) ≤ 1/2 and |t − t0 | ≤ ε, it follows that 1 C˜γ1 [un+1 − un , D+ ] ≤ εM27 (2k0 + 1)C˜γ1 [un − un−1 , D+ ] ≤ C˜γ1 [un − un−1 , D+ ], (4.88) 2 and when n, m ≥ N0 + 1 (N0 is a positive integer), C˜γ1 [un+1 − un , D+ ] ≤ 2−N0

∞ 

2−j C˜γ1 [u1 − u0 , D+ ] ≤ 2−N0 +1 C˜γ1 [u1 − u0 , D+ ].

j=0

Hence {un (z)} is a Cauchy sequence. According to the completeness of the Banach space C˜γ1 (D+ ), there exists a function u∗ (z) ∈ Cγ1 (D+ ), and W∗ (z) = u∗z (z), so that

194

V. Second Order Linear Mixed Equations

C˜γ1 [un − u∗ , D+ ] = Cγ [un − u∗ , D+ ] + C[X(Wn − W∗ ), D+ ] → 0 as n → ∞. We can see that [W∗ (z), u∗ (z)] is a solution of Problem Ft for every t ∈ Tε = {|t − t0 | ≤ ε}. Because the constant ε is independent of t0 (0 ≤ t0 < 1), therefore from the solvability of Problem Ft0 when t0 = 0, we can derive the solvability of Problem Ft when t = ε, 2ε, . . . , [1/ε] ε, 1. In particular, when t = 1 and K(z) = 0, R(z) = 0, Problem F1 , i.e. the Frankl problem for (2.2) in D+ is solvable. The existence of the solution [W (z), u(z)] of Problem F for (2.2) in D− can be obtained by the method in Chapters I and II.

5

Oblique Derivative Problems for Second Order Degenerate Equations of Mixed Type

In this section we discuss the oblique derivative problem for second order degenerate equations of mixed type in a simply connected domain. We first give the representation of solutions of the boundary value problem for the equations, and then prove the uniqueness of solutions for the problem. Moreover we introduce the possibility to prove the existence of the above oblique derivative problem.

5.1 Formulation of oblique derivative problems for degenerate equations of mixed type Let D be a simply connected bounded domain in the complex plane C I with the boundary ∂D = Γ ∪ L, where Γ(⊂ {y > 0})  ∈ Cα2 (0 < α < 1) with the end  points z = 0, 2 and L = L1 ∪ L2 , L1 = {x + 0y −K(t)dt = 0, x ∈ (0, 1)}, L2 =  

{x − 0y −K(t)dt = 2, x ∈ (1, 2)}, and z1 = x1 + jy1 = 1 + jy1 is the intersection point of L1 and L2 . In this section, we use the hyperbolic numbers. Denote D+ = D ∩ {y > 0}, D− = D ∩ {y < 0}. We may assume that Γ = {|z − 1| = 1, y ≥ 0}, and consider the linear degenerate mixed equation of second order Lu = K(y)uxx + uyy = dux + euy + f u + g in D,

(5.1)

where K(y) possesses the first order continuous derivatives K
(y), and K
(y) > 0 on y
= 0, K(0) = 0. The following degenerate mixed equation is a special case Lu = sgny|y|m uxx + uyy = dux + euy + f u + g in D,

(5.2)

where m is a positive constant, d, e, f, g are functions of z(∈ D). Similarly to (5.43), ˜ (z) = U˜ −iV˜ = y m/2 U +iV = [y m/2 ux −iuy ]/2, W ˜ z¯˜ = [y m/2 W ˜ x+ Chapter II, we denote W ˜ y ]/2 in D+ and W ˜ (z) = U+j ˜ V˜ = |y|m/2 U−jV = [|y|m/2 ux+juy ]/2, W ˜ z¯˜ = [|y|m/2 W ˜ x− iW ˜ y ]/2 in D− , then equation (5.2) in D can be reduced to the form jW

5. Degenerate Mixed Equations

195

⎧ ⎫ ˜¯⎬ ⎨W z˜

⎩W ˜¯

z˜




˜ + A2 (z)W ˜ + A3 (z)u + A4 (z) in = A1 (z)W ⎭ ⎧ im ⎪ ⎪ ⎪ ⎪ ⎨ 8y



D+



D−

,



d d m e ie ⎧ +⎫ + + = m/2 + i , ⎨D ⎬ 4y m/2 4 4y 8y 4   in A1 = ⎪ ⎩ −⎭ −d −d je m e jm ⎪ D ⎪ ⎪ + − − = +j ⎩ m/2 m/2 8|y| 4|y| 4 4|y| 8|y| 4

A2

+

  ⎧ d m e ⎪ ⎪ ⎪ − + i , ⎪ ⎨ 4y m/2 8y 4   =⎪ m e −d ⎪ ⎪ ⎪ + +j , ⎩ m/2

4|y|

and

8|y|

⎧ f ⎪ ⎪ ⎪ ⎨ ,

⎧g ⎪ ⎪ ⎨ ,

⎧

(5.3)

⎫

⎨D + ⎬ 4 A3 = ⎪ 4 A4 = ⎪ g in ⎩ ⎭ f ⎪ ⎪ D− ⎩− ⎪ ⎩− , 4 4

4

 z ⎧ ⎪ ⎪ uz dz+u(0) in D+ , ⎨ 2 Re 0z u(z) = ⎪ ⎪ ⎩ 2 Re (U −jV )d(x+jy)+u(0) 0

in D−

is a solution of equation (5.2). Suppose that equation (5.2) satisfies the following conditions: Condition C The coefficients Aj (z) (j = 1, 2, 3) in (5.2) are continuous in D+ and continuous in D− and satisfy C[Aj , D+ ] ≤ k0 ,

j = 1, 2,

C[A3 , D+ ] ≤ k1 ,

C[Aj , D− ] ≤ k0 ,

j = 1, 2,

C[A3 , D− ] ≤ k1 .

A2 ≥ 0 in D+ ,

(5.4)

where p (> 2), k0 , k1 are non-negative constants. If the above conditions is replaced by Cα1 [Aj , D± ] ≤ k0 ,

j = 1, 2,

Cα1 [A3 , D± ] ≤ k1 ,

(5.5)

in which α (0 < α < 1) is a real constant, then the conditions will be called Condition C
. Now we formulate the oblique derivative boundary value problem as follows: Problem P Find a continuously differentiable solution u(z) of (5.2) in D∗ = ¯ and satisfies the boundary conditions ¯ D\{0, L2 }, which is continuous in D lu =

∂u = 2Re [λ(z)uz ] = r(z), ∂l

z ∈ Γ,

u(0) = b0 ,

u(2) = b2 ,

(5.6)

Re [λ(z)uz¯˜] = r(z), z ∈ L1 , Im [λ(z)uz¯˜]|z=z1 = b1 , (5.7) √ where uz˜ = [ −Kux + iuy ]/2, λ(z) = a(x) + ib(x) = cos(l, x) − i cos(l, y), if z ∈ Γ and λ(z) = a(z) + jb(z), if z ∈ L1 , b0 , b1 , b2 are real constants, and λ(z)(|λ(z)| = 1), r(z), b0 , b1 , b2 satisfy the conditions

196

V. Second Order Linear Mixed Equations Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

cos(l, n) ≥ 0 on Γ,

Cα [λ(z),L1 ] ≤ k0 ,

|b0 |, |b1 |, |b2 | ≤ k2 ,

Cα [r(z),L1 ] ≤ k2 ,

1 ≤ k0 , max z∈L1 |a(z) − b(z)|

(5.8)

in which n is the outward normal vector at every point on Γ, α (1/2 < α < 1), k0 , k2 are non-negative constants. For convenience, we may assume that uz˜(z1 ) = 0, otherwise through a transformation of function Uz˜(z) = uz˜(z)−λ(z1 ) [r(z1 )+jb1 ]/[a2 (z1 )+ b2 (z1 )], the requirement can be realized. If cos(l, n) = 0 on Γ, where n is the outward normal vector on Γ, then the problem is called Problem D, in which  u(z) = 2Re 0z uz dz + b0 = φ(z) on Γ. ¯ r(z) = 0, z ∈ Γ ∪ Lj (j = 1 or 2) and Problem P for (5.2) with A3 (z) = 0, z ∈ D, b0 = b1 = 0 will be called Problem P0 . The number 1 K = (K1 + K2 ) 2 is called the index of Problem P and Problem P0 , where 

Kj =



φj + Jj , π

eiφj =

Jj = 0 or 1,

λ(tj − 0) , λ(tj + 0)

γj =

φj − Kj , π

j = 1, 2, (5.9)

in which t1 = 2, t2 = 0, λ(t) = eiπ/4 on L0 = (0, 2) and λ(t1 − 0) = λ(t2 + 0) = exp(iπ/4). Here we choose K = 0, or K = −1/2 on the boundary ∂D+ of D+ if cos(ν, n) ≡ 0 on Γ and the condition u(2) = b2 can be canceled. In fact, if cos(l, n) ≡ 0 on Γ, from the boundary condition (5.6), we can determine the value u(2) by the value u(0), namely 

u(2) = 2Re



2

0

uz dz+u(0) = 2

0



2

Re [i(z−1)uz ]dθ+b0 = 2

0

π

r(z)dθ+b0 ,

(5.10)

in which λ(z) = i(z −1), θ = arg(z −1) on Γ. In order to ensure that the solution u(z) of Problem P is continuously differentiable in D∗ , we need to choose γ1 > 0. If we require that the solution is only continuous, it suffices to choose −2γ2 < 1, −2γ1 < 1 respectively. In the following, we shall only discuss the case: K = 0, and the case: K = −1/2 can be similarly discussed. Problem P in this case still includes the Dirichlet problem (Problem D) as a special case. 5.2 Representation and uniqueness of solutions of oblique derivative problem for degenerate equations of mixed type Now we give the representation theorem of solutions for equation (5.2). Theorem 5.1 Suppose that the equation (5.2) satisfies Condition C
. Then any solution of Problem P for (5.2) can be expressed as 

u(z) = 2Re

0

z

w(z)dz + b0 , w(z) = w0 (z) + W (z),

(5.11)

5. Degenerate Mixed Equations

197

where w0 (z) is a solution of Problem A for the complex equation Wz¯˜ = 0 in D

(5.12)

with the boundary conditions (5.6), (5.7) (w0 (z) = u0z on Γ, w0 (z) = u0˜z on L1 ), and W (z) in D− possesses the form 

W (z) = Φ(z) + Ψ(z),

Ψ(z) =



ν

g1 (z)dνe1 +

2

0

µ

g2 (z)dµe2 in D− ,

(5.13)

in which e1 = (1 + j)/2, e2 = (1 − j)/2, µ = x − 2|y|m/2+1 /(m + 2), ν = x + 2|y|m/2+1 /(m + 2), ˜1 η+ Cu+ ˜ D, ˜ g1 (z) = A˜1 ξ + B

ξ = Re w+Im w,

˜2 η+ Cu+ ˜ D, ˜ g2 (z) = A˜2 ξ + B

C˜ = −



f , 4|y|m/2



˜=− D

g , 4|y|m/2



A˜1 =

d m 1 − −e , 4|y|m/2 2y |y|m/2

A˜2 =

d m 1 − −e , 4|y|m/2 2|y| |y|m/2



η = Re w−Im w,

˜1 = B





d m 1 − +e , 4|y|m/2 2y |y|m/2 

˜2 = B

(5.14) 

d m 1 − +e in D− , 4|y|m/2 2|y| |y|m/2

Φ(z) is the solutions of equation (5.2), and w(z) in D+ and Φ(z) in D− satisfy the boundary conditions Re [λ(z)w(z)] = r(z),

z ∈ Γ,

Re [λ(x)w(x)] = s(x),

Re [λ(x)Φ(x)] = Re [λ(x)(W (z) − Φ(x))], Re [λ(z)(Φ(z) + Ψ(z))] = 0,

z ∈ L1 ,

x ∈ L0 ,

x ∈ L0 = (0, 2),

(5.15)

Im [λ(z1 )(Φ(z1 ) + Ψ(z1 ))] = 0,

where λ(x) = 1 + i, x ∈ L0 . Moreover by Section 5, Chapter II, we see that w0 (z) is a solution of Problem A for equation (5.12), and 

u0 (z) = 2Re

0

z

w0 (z)dz + b0 in D.

(5.16)

Proof Let u(z) be a solution of Problem P for equation (5.2), and w(z) = uz , u(z) be substituted in the positions of w, u in (5.13), (5.14), thus the functions g1 (z), g2 (z), and Ψ(z) in D− in (5.13),(5.14) can be determined. Moreover we can find the solution Φ(z) in D− of (5.12) with the boundary condition (5.15), where s(x) on L0 is a function of λ(z), r(z), Ψ(z), thus w(z) = w0 (z) + Φ(z) + Ψ(z) in D− −

(5.17)

is the solution of Problem A in D for equation (5.2), which can be expressed as the second formula in (5.11), and u(z) is a solution of Problem P for (5.2) as stated in the first formula in (5.11).

198

V. Second Order Linear Mixed Equations

Theorem 5.2 Suppose that equation (5.2) satisfies Condition C
. Then Problem P for (5.2) has at most one solution in D. Proof Let u1 (z), u2 (z) be any two solutions of Problem P for (5.2). It is easy to see that u(z) = u1 (z) − u2 (z) and W (z) = uz˜ satisfy the homogeneous equation and boundary conditions


Wz˜ Wz˜






= A1 W + A2 W + A3 u in

Re [λ(z)W (z)] = 0,

z ∈ Γ,

Re [λ(z)W (z)] = 0,

z ∈ L1 ,

D+ D−

u(0) = 0,



,

(5.18)

u(2) = 0, (5.19)

Im [λ(z1 )W (z1 )] = 0,

where W (z) = uz in D+ . According to the method as stated in Section 5, Chapter II, the solution W (z) in the hyperbolic domain D− can be expressed in the form W (z) = Φ(z) + Ψ(z), 

Ψ(z) =

2

ν

˜ + Bη+ ˜ Cu]e ˜ 1 dν + [Aξ

 0

µ

˜ + Bη+ ˜ Cu]e ˜ 2 dµ in D− , [Aξ

(5.20)

where Φ(z) is a solution of (5.12) in D− satisfying the boundary condition (5.15). Similarly to the method in Section 5, Chapter II, Ψ(z) = 0, Φ(z) = 0, W (z) = 0, z ∈ D− can be derived. Thus the solution u(z) = 2 Re 0z w(z)dz is the solution of the homogeneous equation of (5.2) with homogeneous boundary conditions of (5.6) and (5.7): 2Re [λ(z)uz ] = 0 on Γ, in which λ(x) = 1 + i,

Re [λ(x)uz˜(x)] = 0 on L0 ,

u(0) = 0,

u(2) = 0,

(5.21)

x ∈ L0 = (0, 2).

Now we verify that the above solution u(z) ≡ 0 in D+ . If u(z)
≡ 0 in D+ , noting that u(z) satisfies the boundary condition (5.21), and similarly to the proof of Theorem 3.4, Chapter III, we see that its maximum and minimum cannot attain in D+ ∪ Γ. Hence u(z) attains its maximum and minimum at a point z ∗ = x∗ ∈ L0 = (0, 2). By using Lemma 4.1, Chapter III, we can derive that ux (x∗ ) = 0, uy (x∗ )
= 0, and then  1  Re [λ(x)uz˜(x∗ )] = −K(y) ux (x∗ ) + uy (x∗ )
= 0, 2 this contradicts the second equality in (5.21). Thus u(z) ≡ 0 in D+ . This completes the proof. 5.3 Solvabilty problem of oblique derivative problems for degenerate equations of mixed type From the above discussion, we see in order to prove the existence of solutions of Problem P for equation (5.2), the main problem is to find a solution of the oblique

5. Degenerate Mixed Equations

199

derivative problem for the degenerate elliptic equation of second order, i.e. equation (5.2) in elliptic domain D+ , and the oblique derivative boundary conditions is (5.6) and Re [λ(x)uz˜(x)] = s(x) on L0 = (0, 2), i.e.  (5.22) 1  −K(y) ux + uy = s(x) on L0 , 2 which is more general than the case as stated in Section 4, Chapter III. We try to solve the problem by using the method of integral equations or the method of auxiliary functions, which will be discussed in detail in our other publishers. The references for this chapter are [1],[10],[12],[17],[21],[22],[28],[37],[43],[47],[49], [57],[62],[66],[69],[70],[73],[77],[85],[91],[93].

CHAPTER VI SECOND ORDER QUASILINEAR EQUATIONS OF MIXED TYPE This chapter deals with several oblique derivative boundary value problems for second order quasilinear equations of mixed (elliptic-hyperbolic) type. We shall discuss oblique derivative boundary value problems and discontinuous oblique derivative boundary value problems for second order quasilinear equations of mixed (elliptichyperbolic) type. Moreover we shall discuss oblique derivative boundary value problems for general second order quasilinear equations of mixed (elliptic-hyperbolic) type and the boundary value problems in multiply connected domains. In the meantime, we shall give a priori estimates of solutions for above oblique derivative boundary value problems.

1

Oblique Derivative Problems for Second Order Quasilinear Equations of Mixed Type

In this section, we first give the representation of solutions for the oblique derivative boundary value problem, and then prove the uniqueness and existence of solutions of the problem and give a priori estimates of solutions of the above problem. Finally we prove the solvability of oblique derivative problems for general quasilinear second order equations of mixed type.

1.1 Formulation of the oblique derivative problem for second order equations of mixed type Let D be a simply connected bounded domain D in the complex plane C I as stated in Chapter V. We consider the second order quasilinear equation of mixed type uxx + sgny uyy = aux + buy + cu + d in D,

(1.1)

where a, b, c, d are functions of z(∈ D), u, ux , uy (∈ IR), its complex form is the following complex equation of second order Luz =


uzz uz¯z∗


= F (z, u, uz ),

F = Re [A1 uz ]+A2 u+A3 in

D+ D−

 ,

(1.2)

1. Oblique Derivative Problems

201

where Aj = Aj (z, u, uz ), j = 1, 2, 3, and 1 uzz¯ = [uxx + uyy ], 4 a + ib A1 = , A2 = 2

1 1 uz¯z∗ = [(uz¯)x − i(uz¯)y ] = [uxx − uyy ], 2 4 c d , A3 = in D. 4 4

Suppose that the equation (1.2) satisfies the following conditions, namely Condition C I for almost every 1) Aj (z, u, uz ) (j = 1, 2, 3) are continuous in u ∈ IR, uz ∈ C point z ∈ D+ , and measurable in z ∈ D+ and continuous in D− for all continuously ¯ ¯ differentiable functions u(z) in D∗ = D\{0, x − y = 2} or D∗ = D\{x + y = 0, 2} and satisfy Lp [Aj , D+ ] ≤ k0 ,

A2 ≥ 0 in D+ ,

j = 1, 2, Lp [A3 , D+ ] ≤ k1 ,

C[Aj , D− ] ≤ k0 ,

j = 1, 2,

C[A3 , D− ] ≤ k1 .

(1.3)

2) For any continuously differentiable functions u1 (z), u2 (z) in D∗ , the equality F (z, u1 , u1z ) − F (z, u2 , u2z ) = Re [A˜1 (u1 − u2 )z ] + A˜2 (u1 − u2 ) in D

(1.4)

holds, where A˜j = A˜j (z, u1 , u2 ) (j = 1, 2) satisfy the conditions Lp [A˜j , D+ ] ≤ k0 , C[A˜j , D− ] ≤ k0 ,

j = 1, 2

(1.5)

in (1.3),(1.5), p (> 2), k0 , k1 are non-negative constants. In particular, when (1.2) is a linear equation, the condition (1.4) obviously holds. Problem P The oblique derivative boundary value problem for equation (1.2) is to find a continuously differentiable solution u(z) of (1.2) in D∗ = D\{0, x − y = 2}, which is continuous in D and satisfies the boundary conditions 1 ∂u = Re [λ(z)uz ] = r(z), 2 ∂l

z ∈ Γ,

Im [λ(z)uz¯]z=z1 = b1 ,

1 ∂u = Re [λ(z)uz¯] = r(z), 2 ∂l u(0) = b0 ,

z ∈ L1 ,

(1.6)

u(2) = b2 ,

where l is a given vector at every point on Γ ∪ L1 , λ(z) = a(x) + ib(x) = cos(l, x) ∓i cos(l, y), and ∓ are determined by z ∈ Γ and z ∈ L1 respectively, b0 , b1 , b2 are real constants, and λ(z), r(z), b0 , b1 , b2 satisfy the conditions Cα [λ(z),Γ] ≤ k0 , Cα [λ(z),L1 ] ≤ k0 ,

Cα [r(z),Γ] ≤ k2 , Cα [r(z),L1 ] ≤ k2 ,

cos(l,n) ≥ 0 on Γ,

|b0 |,|b1 |,|b2 | ≤ k2 ,

maxz∈L1 [1/|a(x)−b(x)|] ≤ k0 ,

(1.7)

in which n is the outward normal vector at every point on Γ, α (1/2 < α < 1), k0 , k2 are non-negative constants. For convenience, we may assume that w(z1 ) = 0,

202

VI. Second Order Quasilinear Mixed Equations

otherwise through a transformation of function W (z) = w(z) − λ(z1 )[r(z1 ) − ib1 ], the requirement can be realized. Here we mention that if A2 (z) = 0 in D1 , we can cancel the assumption cos(l, n) ≥ 0 on Γ, and if the boundary condition Re [λ(z)uz¯] = r(z), z ∈ L1 is replaced by Re [λ(z)uz ] = r(z), z ∈ L1 , then Problem P does not include the Dirichlet problem (Tricomi problem) as a special case. The boundary value problem for (1.2) with A3 (z, u, uz ) = 0, z ∈ D, u ∈ IR, uz ∈ C I, r(z) = 0, z ∈ ∂D and b0 = b2 = b1 = 0 will be called Problem P0 . The number 1 K = (K1 + K2 ) 2 is called the index of Problem P and Problem P0 , where  λ(tj − 0) φj φj , γj = − Kj , + Jj , Jj = 0 or 1, eiφj = Kj = π λ(tj + 0) π

(1.8)

j = 1, 2, (1.9)

in which [a] is the largest integer not exceeding the real number a, t1 = 2, t2 = 0, λ(t) = exp(iπ/4) on L0 and λ(t1 −0) = λ(t2 +0) = exp(iπ/4), here we only discuss the case of K = 0 on ∂D+ if cos(l, n)
≡ 0 on Γ, or K = −1/2 if cos(l, n) ≡ 0 on Γ, because in this case the last point condition in (1.6) can be eliminated, and the solution of Problem P is unique. In order to ensure that the solution u(z) of Problem P in D∗ is continuously differentiable, we need to choose γ1 > 0. If we require that the solution ¯ is only continuous, it suffices to choose −2γ1 < 1, −2γ2 < 1. of Problem P in D Besides, if A2 = 0 in D, the last condition in (1.6) is replaced by Im [λ(z)uz¯]|z=z2 = b2 ,

(1.10)

where the integral path is along two family of characteristic lines similar to that in (2.10), Chapter II, z2 (
= 0, 2) ∈ Γ, and b2 is a real constant with the condition |b2 | ≤ k2 and here the condition cos(l, n) ≥ 0 is canceled, then the boundary value problem for (1.2) will be called Problem Q. 1.2 The existence and uniqueness of solutions for the oblique derivative problem for (1.2) Similarly to Section 2, Chapter V, we can prove the following results. Lemma 1.1 Let equation (1.2) satisfy Condition C. Then any solution of Problem P for (1.2) can be expressed as  z u(z) = 2Re w(z)dz + b0 , w(z) = w0 (z) + W (z) in D, (1.11) 0

where the integral path in D− is the same as in Chapter II, and w0 (z) is a solution of Problem A for the equation 

D+ wz¯ = 0 in , (1.12) Lw = wz∗ D−

1. Oblique Derivative Problems

203

with the boundary condition (1.6) (w0 (z) = u0z ), and W (z) possesses the form ˜ φ(z) ˜ ˜ W (z) = w(z) − w0 (z) in D, w(z) = Φ(z)e + ψ(z) in D+ ,  g(ζ) 1 ˜ ˜ dσζ , ψ(z) φ(z) = φ˜0 (z) + T g = φ˜0 (z) − = T f in D+ , (1.13) π D+ ζ − z  ν  µ W (z) = Φ(z) + Ψ(z), Ψ(z) = g1 (z)dνe1 + g2 (z)dµe2 in D−, 2

0

˜ in which Im [φ(z)] = 0 on L0 = (0, 2), e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic function in D+ and continuous in D+ , such that Im [φ(x)] = 0 on L0 , and
¯ w(z)
= 0, A1 /2+A1 w/(2w), f (z)=Re[A1 φ˜z ]+A2 u+A3 in D+ , g(z)= + 0, w(z)=0, z ∈ D , g1 (z) = g2 (z) = Aξ +Bη +Cu+D,

ξ = Rew +Imw,

η = Rew −Imw,

(1.14)

ReA1 +ImA1 ReA1 −ImA1 , B= , C = A2 , D = A3 in D−, 2 2 ˜ where Φ(z) is an analytic function in D+ and Φ(z) is a solution of equation (1.12) in D− satisfying the boundary conditions ˜ φ(z) ˜ ˜ + ψ(z))] = r(z), z ∈ Γ, Re [λ(z)(Φ(z)e A=

˜ φ(x) ˜ ˜ Re [λ(x)(Φ(x)e + ψ(x))] = s(x),

x ∈ L0 ,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L0 ,

(1.15)

z ∈ L1 ,

Im [λ(z1 )Φ(z1 )] = −Im [λ(z1 )Ψ(z1 )], in which λ(x) = 1 + i, x ∈ L0 = (0, 2) and s(x) is as stated in (2.23), Chapter V. Moreover by Theorem 1.1, Chapter V, the solution w0 (z) of Problem A for (1.12) and u0 (z) satisfy the estimate in the form Cβ [u0 (z), D]+Cβ [w0 (z)X(z), D+ ]+Cβ [w0± (µ, ν)Y ± (µ, ν), D− ] ≤ M1 (k1 +k2 ), (1.16) where X(z) =

2 

|z − tj |ηj ,

j=1


ηj =

Y ± (z) = Y ± (µ, ν) = [|ν − 2||µ − 2|]ηj , 2|γj | + δ, γj < 0, δ, γj ≥ 0,

(1.17) j = 1, 2,

w0± (µ, ν)

herein = Re w0 (z) ∓ Im w0 (z), w0 (z) = w0 (µ, ν), µ = x + y, ν = x − y, and γ1 , γ2 are the real constants in (1.9), β(< δ), δ are sufficiently small positive constants, and  z ¯ u0 (z) = 2Re w0 (z)dz + b0 in D (1.18) 0

where p0 (2 < p0 ≤ p), M1 = M1 (p0 , β, k0 , D) are non-negative constants.

204

VI. Second Order Quasilinear Mixed Equations

Theorem 1.2 Suppose that equation (1.2) satisfies Condition C. Then Problem P for (1.2) has a unique solution u(z) in D. Theorem 1.3 Suppose that the equation (1.2) satisfies Condition C. Then any solution u(z) of Problem P for (1.2) satisfies the estimates C˜β1 [u, D+ ] = Cβ [u(z), D+ ] + Cβ [uz X(z), D+ ] ≤ M2 , ± − C˜ 1 [u, D− ] = Cβ [u(z), D− ] + C[u± z (µ, ν)Y (µ, ν), D ] ≤ M3 ,

C˜β1 [u, D+ ] ≤ M4 (k1 + k2 ),

(1.19)

C˜ 1 [u, D− ] ≤ M4 (k1 + k2 ),

where X(z), Y ± (µ, ν) are stated in (1.17), and Mj = Mj (p0 , β, k0 , D) (j = 2, 3, 4) are non-negative constants. ¯ 1.3 Cα1 (D)-estimate of solutions of Problem P for second order equations of mixed type ¯ Now, we give the Cα1 (D)-estimate of solutions u(z) for Problem P for (1.2), but it needs to assume the following conditions: For any real numbers u1 , u2 and complex numbers w1 , w2 , we have |Aj (z1 , u1 , w1 )−Aj (z2 , u2 , w2 )| ≤ k0 [|z1 −z2 |α +|u1 −u2 |α +|w1 −w2 |], |A3 (z1 , u1 , w1 )−A3 (z2 , u2 , w2 )| ≤k1 [|z1 −z2 | +|u1 −u2 | +|w1 −w2 |], α

α

j = 1, 2, z1 ,z2 ∈D− , (1.20)

where α (0 < α < 1), k0 , k1 are non-negative constants. Theorem 1.4 If Condition C and (1.20) hold, then any solution u(z) of Problem P for equation (1.2) in D− satisfies the estimates ± − C˜β1 [u, D− ] = Cβ [u, D− ]+Cβ [u± z (µ, ν)Y (µ, ν), D ]

≤ M5 ,

C˜β1 [u, D− ] ≤ M6 (k1 + k2 ),

(1.21)

in which u± z (µ, ν) = Re uz ∓ Im uz , β (0 < β ≤ α), M5 = M5 (p0 , β, k, D), M6 = M6 (p0 , β, k0 , D) are non-negative constants, k = (k0 , k1 .k2 ). Proof Similarly to Theorem 1.3, it suffices to prove the first estimate in (1.21). Due to the solution u(z) of Problem P for (1.2) is found by the successive iteration through the integral expressions (1.11), (1.13) and (1.14), we first choose the solution of Problem A of (1.12) in the form (1.18), i.e.  z u0 (z) = 2Re w0 (z)dz + b0 , w0 (z) = ξ0 (z)e1 + η0 (z)e2 in D, (1.22) 0

and substitute them into the positions of u0 , w0 in the right-hand side of (1.14), we can obtain Ψ1 (z), w1 (z), u1 (z) as stated in (1.11)–(1.14). Denote

1. Oblique Derivative Problems  u1 (z) = 2Re Ψ11 (z) = Ψ21 (z) =



0

205

z

w1 (z)dz + b0 ,

w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z),

ν

2



G1 (z)dν,

G1 (z) = Aξ0 + Bη0 + Cu0 + D,

G2 (z)dµ,

G2 (z) = Aξ0 + Bη0 + Cu0 + D,

(1.23)

µ

0

from the last two equalities in (1.23), it is not difficult to see that G1 (z) = G1 (µ, ν), Ψ11 (z) = Ψ11 (µ, ν) and G2 (z) = G2 (µ, ν), Ψ21 (z) = Ψ21 (µ, ν) satisfy the H¨older estimates about ν, µ respectively, namely Cβ [G1 (·, ν), D− ] ≤ M7 ,

Cβ [Ψ11 (·, ν), D− ] ≤ M7 R,

Cβ [G2 (µ, ·), D− ] ≤ M8 ,

Cβ [Ψ21 (µ, ·), D− ] ≤ M8 R,

(1.24)

where Mj = Mj (p0 , β, k, D) (j = 7, 8) and R = 2. Moreover, from (1.23), we can derive that Ψ11 (µ, ν), Ψ21 (µ, ν) about µ, ν satisfy the H¨older conditions respectively, namely Cβ [Ψ11 (µ, ·), D− ] ≤ M9 R, Cβ [Ψ21 (·, ν), D− ] ≤ M9 R, (1.25) where M9 = M9 (p0 , β, k, D). Besides we can obtain the estimate of Φ1 (z), i.e. Cβ [Φ1 (z), D− ] ≤ M10 R = M10 (p0 , β, k, D)R,

(1.26)

in which Φ1 (z) satisfies equation (1.12) and boundary condition of Problem P , but in which the function Ψ(z) is replaced by Ψ1 (z). Setting w1 (z) = w0 (z) + Φ1 (z) + Ψ1 (z) and by the first formula in (1.23), we can find the function u1 (z) from w1 (z). Furthermore from (1.25),(1.26), we can derive that the functions w˜1± (z) = w˜1± (µ, ν) = Re w˜1 (z) ∓ Im w˜1 (z) (w˜1 (z) = w1 (z) − w0 (z)) and u˜1 (z) = u1 (z) − u0 (z) satisfy the estimates Cβ [w˜1± (µ, ν)Y ± (µ, ν), D− ] ≤ M11 R, Cβ [˜ u1 (z), D− ] ≤ M11 R,

(1.27)

where M11 = M11 (p0 , β, k, D). Thus, according to the successive iteration, we can obtain the estimates of functions w˜n± (z) = w˜n± (µ, ν) = Re w˜n (z) ∓ Im w˜n (z) (w˜n (z) = wn (z) − wn−1 (z)) and the corresponding function u˜n (z) = un (z) − un−1 (z) satisfy the estimates Cβ [w˜n± (µ, ν)Y ± (µ, ν), D− ] ≤

(M11 R)n , n!

(M11 R)n . n!

(1.28)

u˜m (z) + u0 (z), n = 1, 2, . . . ,

(1.29)

Cβ [˜ un (z), D− ] ≤

Therefore the sequences of functions wn (z) =

n  m=1

w˜m (z) + w0 (z), un (z) =

n  m=1

uniformly converge to w(z), u(z) in any close subset of D∗ respectively, and w(z), u(z) satisfy the estimates

206

VI. Second Order Quasilinear Mixed Equations Cβ [w± (µ, ν)Y ± (µ, ν), D− ] ≤ eM11 R ,

Cβ [u(z), D− ] ≤ M5 ,

(1.30)

this is just the first estimate in (1.21). From the estimates (1.19) and (1.21), we can see the regularity of solutions of Problem P for (1.2). Moreover it is easy to see that the derivatives [w+ (µ, ν)]ν , [w− (µ, ν)]µ satisfy the estimates similar to those in (1.30). As for Problem Q for (1.2), we can similarly discuss its unique solvability.

1.4 The solvability for the oblique derivative problem for general second order quasilinear equations of mixed type Now, we consider the general quasilinear equation of second order  

uzz D+ = F (z, u, uz ) + G(z, u, uz ), z ∈ , Luz = uz¯z∗ D− F = Re [A1 uz ] + A2 u + A3 ,

G = A4 |uz | + A5 |u| , σ

τ

(1.31)

z ∈ D,

where F (z, u, uz ) satisfies Condition C, σ, τ are positive constants, and Aj (z, u, uz ) (j = 4, 5) satisfy the conditions in Condition C, the main conditions of which are Lp [Aj (z, u, uz ), D+ ] ≤ k0 ,

C[Aj (z, u, uz ), D− ] ≤ k0 ,

j = 4, 5,

and denote the above conditions by Condition C
. Theorem 1.5

Let the complex equation (1.31) satisfy Condition C
.

¯ (1) When 0 < max(σ, τ ) < 1, Problem P for (1.31) has a solution u(z) ∈ C(D). ¯ (2) When min(σ, τ ) > 1, Problem P for (1.31) has a solution u(z) ∈ C(D), provided that (1.32) M12 = k1 + k2 + |b0 | + |b1 | is sufficiently small. (3) When min(σ, τ ) > 1, Problem P for the equation 
uzz = F (z, u, uz ) + εG(z, u, uz ), Luz = uz¯z∗


z∈

D+ D−

 ,

(1.33)

¯ provided that the positive number ε in (1.33) is approhas a solution u(z) ∈ C(D), priately small, where the functions F (z, u, uz ), G(z, u, uz ) are as stated in (1.31). Proof

(1) Consider the algebraic equation M4 {k1 + k2 + 2k0 tσ + 2k0 tτ + |b0 | + |b1 |} = t

(1.34)

1. Oblique Derivative Problems

207

for t, where M4 is the constant stated in (1.19). It is not difficult to see that equation (1.34) has a unique solution t = M13 ≥ 0. Now, we introduce a bounded, closed and ¯ whose elements are the functions convex subset B ∗ of the Banach space B = C˜ 1 (D), u(z) satisfying the condition ± − C˜ 1 [u(z), D] = Cβ [u(z), D]+Cβ [uz X(z), D+ ]+C[u± z (µ, ν)Y (µ, ν), D ] ≤ M13 . (1.35)

We arbitrarily choose a function u0 (z) ∈ B for instance u0 (z) = 0 and substitute it into the position of u in the coefficients of (1.31) and G(z, u, uz ). From Theorem 1.2, it is clear that problem P for Luz − Re [A1 (z, u0 , u0z )uz ] − A2 (z, u0 , u0z )u − A3 (z, u0 , u0z ) = G(z, u0 , u0z ), (1.36) has a unique solution u1 (z). From Theorem 1.4, we see that the solution u1 (z) satisfies the estimate in (1.35). By using successive iteration, we obtain a sequence of solutions um (z) (m = 1, 2, ...) ∈ B ∗ of Problem P, which satisfy the equations Lum+1z − Re [A1 (z, um , umz )um+1z ] − A2 (z, um , umz )um+1 +A3 (z, um , umz ) = G(z, um , umz ) in D,

(1.37)

m = 1, 2, . . .

and um+1 (z) ∈ B ∗ . From (1.37), we see that and u˜m+1 (z) = um+1 (z) − um (z) satisfies the equations and boundary conditions L˜ um+1z −Re[A˜1 u˜m+1z ]−A˜2 u˜m+1 =G(z,um ,umz )−G(z,um−1 ,um−1z ) in D,

(1.38)

Re[λ(z)˜ um+1z ] = 0 on Γ, Re[λ(z)˜ um+1¯z ] = 0 on L1 , Im[λ(z)˜ um+1¯z ]|z=z1 =0, ¯ ≤ in which m = 1, 2, . . . . Noting that C[G(z, um , umz ) − G(z, um−1 , um−1z ), D] 2k0 M13 , M13 is a solution of the algebraic equation (1.34), and according to Theorem 1.3, the estimate ¯ ≤ M14 = M14 (p0 , β, k0 , D)  u˜m+1 = C˜ 1 [˜ um+1 , D]

(1.39)

can be obtained. Due to u˜m+1 can be expressed as  z wm+1 (z)dz, in D, wm+1 (z) = Φm+1 (z) + Ψm+1 (z), u˜m+1 (z) = 2Re  Ψm+1 (z) =

2

0

x−y



+ 0

˜ m+1 + Bη ˜ m+1 + Cu ˜ m+1 + D]e ˜ 1 d(x − y) [Aξ

x+y

(1.40)

˜ m+1 + Bη ˜ m+1 + Cu ˜ m+1 + D]e ˜ 2 d(x + y) in D− , [Aξ

in which Φm+1 (z), Ψm+1 (z) are similar to the functions Φ(z), Ψ(z) in (1.13), the rela˜ and A, ˜ B, ˜ C, ˜ D ˜ is the same as that of A1 , A2 , A3 and A, B, C, D tion between A˜1 , A˜2 , G ˜ = G(z, um , umz ) − G(z, um−1 , um−1z ). By using the method from the in (1.14) and G proof of Theorem 1.3, we can obtain ¯ ≤  um+1 − um = C˜ 1 [˜ um+1 , D]

(M14 R
)m , m!

208

VI. Second Order Quasilinear Mixed Equations

where M14 = 2M4 (M15 + 1)M R(4m0 + 1), R = 2, m0 = w0 (z)X(z) C(D) ¯ , herein ˜ Q], C[B, ˜ Q], C[C, ˜ Q]}, C[D, ˜ Q]}, M = 1 + 4k 2 (1 + k 2 ). From the M15 = max{C[A, 0 0 above inequality, we see that the sequence of functions: {um (z)}, i.e. um (z) = u0 (z) + [u1 (z) − u0 (z)] + · · · + [um (z) − um−1 (z)],

m = 1, 2, . . .

(1.41)

uniformly converges to a function u∗ (z), and w∗ (z) = u∗z satisfies the equality w∗ (z) = w0 (z) + Φ∗ (z) + Ψ∗ (z) in D−,  x−y Ψ∗ (z) = + [Aξ∗ + Bη∗ + Cu∗ + D]e1 d(x − y) 2

 +

0

x+y

(1.42)

[Aξ∗ + Bη∗ + Cu∗ + D]e2 d(x + y) in D−, 

and the function u∗ (z) = 2Re

0

z

¯ w∗ (z)dz + b0 in D

(1.43)

¯ is just a solution of Problem P for the general quasilinear equation (1.31) in D. (2) Consider the algebraic equation M4 {k1 + k2 + 2k0 tσ + 2k0 tτ + |b0 | + |b1 |} = t

(1.44)

for t. It is not difficult to see that the equation (1.44) has a solution t = M13 ≥ 0, provided that the positive constant M12 in (1.32) is small enough. Now, we introduce ¯ whose elements a bounded, closed and convex subset B∗ of the Banach space C˜ 1 (D), are the functions u(z) satisfying the condition ± − C˜ 1 [u(z), D] = Cβ [u(z), D]+Cβ [uz X(z), D+ ]+C[u± z (µ, ν)Y (µ, ν), D ] ≤ M13 . (1.45)

By using the same method as in (1), we can find a solution u(z) ∈ B∗ of Problem P for equation (1.31) with min(σ, τ ) > 1. (3) There is no harm in assuming that k1 , k2 in (1.3),(1.7) are positive constants, ¯ we introduce a bounded, closed and convex subset B
of the Banach space C˜ 1 (D), whose elements are the functions u(z) satisfying the condition C˜ 1 [u(z), D] ≤ (M4 + 1)(2k1 + k2 )

(1.46)

where M4 is a constant as stated in (1.19) and we can choose an appropriately small ¯ ≤ k1 . Moreover, we are free to choose positive number ε such that C[εG(z, u, uz ), D] a function u0 (z) ∈ B
for instance u0 (z) = 0 and substitute it into the position of u in the coefficients of (1.33) and G(z, u, uz ). From Theorem 1.2, it is seen that there exists a unique solution of Problem P for Lu − Re [A1 (z, u0 , u0z )uz ] − A2 (z, u0 , u0z )u − A3 (z, u0 , u0z ) = G(z, u0 , u0z ), and u1 (z) ∈ B
. Thus similarly to the proof in (1), by the successive iteration, a solution of Problem P for equation (1.33) can be obtained.

2. Mixed Equations in General Domains

209

By using a similar method as before, we can discuss the solvability of Problem ˜ for equation (1.2) or (1.31) with the boundary P˜ and the corresponding Problem Q conditions Re [λ(z)uz ] = r(z), z ∈ Γ, u(0) = b0 , u(2) = b2 , Re [λ(z)uz¯] = r(z),

z ∈ L2 ,

Im [λ(z)uz¯]|z=z1 = b1

in which the coefficients λ(z), r(z), b0 , b1 , b2 satisfy the condition (1.7), but where the conditions Cα [λ(z), L1 ] ≤ k0 , Cα [r(z), L1 ] ≤ k2 , maxz∈L1 [1/|a(x) − b(x)|] ≤ k0 are replaced by Cα [λ(z), L2 ] ≤ k0 , Cα [r(z), L2 ] ≤ k2 , maxz∈L2 [1/|a(x) + b(x)|] ≤ k0 and in (1.9) the condition λ(t) = eiπ/4 on L0 = (0, 2) and λ(t1 −0) = λ(t2 +0) = exp(iπ/4) is replaced by λ(t) = e−iπ/4 on L0 = (0, 2) and λ(t1 − 0) = λ(t2 + 0) = exp(−iπ/4). ¯ ¯ Besides the set D∗ = D\{0, x−y = 2} in Condition C is replaced by D∗ = D\{x+y = 0, 2}, if the constant γ2 > 0 in (1.9).

2

Oblique Derivative Problems for Second Order Equations of Mixed Type in General Domains

This section deals with oblique derivative boundary value problem for second order quasilinear equations of mixed (elliptic-hyperbolic) type in general domains. We prove the uniqueness and existence of solutions of the above problem. In refs. [12]1),3), the author discussed the Dirichlet problem (Tricomi problem) for second order equations of mixed type: uxx + sgny uyy = 0 by using the method of integral equations and a complicated functional relation. In the present section, by using a new method, the solvability result of oblique derivative problem for more general domains is obtained. 2.1 Oblique derivative problem for second order equations of mixed type in another domain. Let D be a simply connected bounded domain D in the complex plane C I with the boundary ∂D = Γ ∪ L, where Γ(⊂ {y > 0}) ∈ Cµ2 (0 < µ < 1) with the end points z = 0, 2 and L = L1 ∪ L2 ∪ L3 ∪ L4 , and L1 = {x + y = 0, 0 ≤ x ≤ a/2},

L2 = {x − y = a, a/2 ≤ x ≤ a},

L3 = {x + y = a, a ≤ x ≤ 1 + a/2},

L4 = {x − y = 2, 1 + a/2 ≤ x ≤ 2},

(2.1)

where a (0 < a < 2) is a constant. Denote D+ = D ∩ {y > 0} and D− = D ∩ {y < 0}, D1− = D− ∩{x−y < a} and D2− = D− ∩{x+y > a}. Without loss of generality, we may assume that Γ = {|z − 1| = 1, y ≥ 0}, otherwise through a conformal mapping, this requirement can be realized. We consider the quasilinear second order mixed equation (1.2) and assume that (1.2) satisfies Condition C in D, here D− is as stated before. Problem P for(1.2)

210

VI. Second Order Quasilinear Mixed Equations

in D is to find a continuously differentiable solution of (1.2) in D∗ = D\{0, a, 2} satisfying the boundary conditions 1 ∂u = Re[λ(z)uz ] = r(z), z ∈ Γ, 2 ∂l u(0) = b0 , u(a) = b1 , u(2) = b2 , (2.2) 1 ∂u = Re[λ(z)uz¯] = r(z), z ∈ L1 ∪L4 , 2 ∂l Im[λ(z)uz¯]|z=zj = bj+2 , j = 1, 2, where l is a given vector at every point on Γ ∪ L1 ∪ L4 , z1 = (1 − i)a/2, z2 = (1 + a/2) + i(1 − a/2), λ(z) = a(x) + ib(x) = cos(l, x) − i cos(l, y), z ∈ Γ, and λ(z) = a(x) + ib(x) = cos(l, x) + i cos(l, y), z ∈ L1 ∪ L4 , bj (j = 0, 1, . . . , 4) are real constants, and λ(z), r(z), bj (j = 0, 1, . . . , 4) satisfy the conditions Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

Cα [r(z), Lj ] ≤ k2 ,

j = 1, 4,

cos(l, n) ≥ 0 on Γ,

1 , max z∈L1 |a(x)−b(x)|

Cα [λ(z), Lj ] ≤ k0 ,

|bj | ≤ k2 ,

j = 0, 1, . . . , 4,

(2.3)

1 max ≤ k0 , z∈L4 |a(x)+b(x)|

in which n is the outward normal vector at every point on Γ, α (1/2 < α < 1), k0 , k2 are non-negative constants. The number 1 K = (K1 + K2 + K3 ), 2 is called the index of Problem P on the boundary ∂D+ of D+ , where  φj λ(tj −0) φj +Jj , Jj = 0 or 1, eiφj = , γj = −Kj , j = 1,2,3, Kj = π λ(tj +0) π

(2.4)

(2.5)

in which [b] is the largest integer not exceeding the real number b, t1 = 2, t2 = 0, t3 = a, λ(t) = eiπ/4 on (0, a) and λ(t3 − 0) = λ(t2 + 0) = exp(iπ/4), and λ(t) = e−iπ/4 on (a, 2) and λ(t1 − 0) = λ(t3 + 0) = exp(−iπ/4). Here we only discuss the case K = 1/2, or K = 0 if cos(l, n) = 0 on Γ, because in this case the solution of Problem P is unique and includes the Dirichlet problem (Tricomi problem) as a special case. We mention that if the boundary condition Re [λ(z)uz¯] = r(z), z ∈ Lj (j = 1, 4) is replaced by Re [λ(z)uz ] = r(z), z ∈ Lj (j = 1, 4), then Problem P does not include the Dirichlet problem (Tricomi problem) as a special case. In order to ensure that the solution u(z) of Problem P is continuously differentiable in D∗ , we need to choose γ1 > 0, γ2 > 0 and can select γ3 = 1/2. If we only require that the solution u(z) in D is continuous, it is sufficient to choose −2γ1 < 1, −2γ2 < 1, −γ3 < 1.

2. Mixed Equations in General Domains

211

Besides, we consider the oblique derivative problem (Problem Q) for equation (1.2) with A2 = 0 and the boundary condition (2.2), but the last point conditions u(a) = b1 , u(2) = b2 in (2.2) is replaced by Im [λ(z)uz ]|zj
= cj , zj
(j

j = 1, 2,

(2.6)

∗

in which = 1, 2) ∈ Γ = Γ\{0, 2} are two points and c1 , c2 are real constants and |c1 |, |c2 | ≤ k2 . Similarly to Section 1, we can give a representation theorem of solutions of Problem P for equation (1.2), in which the functions Ψ(z) in (1.13), λ(x), s(x) on L0 in (1.15), X(z), Y ± (µ, ν) in (1.17) are replaced by  µ ⎧ ν ⎪ g (z)dνe + g2 (z)dµe2 , z ∈ D1− , ⎪ 1 1 ⎨ a 0 Ψ(z) =  ν (2.7)  µ ⎪ ⎪ ⎩ g1 (z)dνe1 + g2 (z)dµe2 , z ∈ D2− , 2

λ(x) =




a

1 + i,

x ∈ L
0 = (0, a),

1 − i,

x ∈ L

0 = (a, 2),

⎧ ⎪ 2r((1−i)x/2)−2Re[λ((1−i)x/2)Ψ((1−i)x/2) ⎪ ⎪ +Re[λ(x)Ψ(x)] ⎪ ⎪ a((1−i)x/2)−b((1−i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ [a((1−i)x/2)+b((1−i)x/2)]f (0) ⎪ ⎪ , x ∈ (0,a), − ⎪ ⎪ ⎪ a((1−i)x/2)−b((1−i)x/2) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2r((1+i)x/2+1−i)−2Re[λ((1+j)x/2+1−i)Φ((1+j)x/2+1−i)] ⎪ ⎪ ⎪ ⎪ a((1+i)x/2+1−i)+b((1+i)x/2+1−i) ⎪ ⎨ s(x)= a((1+i)x/2+1−i)−b((1+i)x/2+1−i)g(2)−h(x) ⎪ − , ⎪ ⎪ ⎪ a((1+i)x/2+1−i)+b((1+i)x/2+1−i) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ h(x) = Re[λ(x)Ψ(x)] ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ×[a((1+i)x/2+1−i)+b((1+i)x/2+1−i)]/2, x ∈ (a,2), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ f (0) = [a(z1 )+b(z1 )]r(z1 )+[a(z1 )−b(z1 )]b3 , ⎪ ⎪ ⎪ ⎪ ⎩ g(2) = [a(z2 )−b(z2 )]r(z2 )−[a(z2 )+b(z2 )]b4 , 3 3   X(z) = |z − tj |ηj , Y ± (z) = Y ± (µ, ν) = [|µ − tj ||ν − tj |]ηj , j=1


ηj =

j=1

2 max(−γj , 0) + δ, max(−γj , 0) + δ,

(2.8)

(2.9)

(2.10)

j = 1, 2, j = 3,

respectively, δ is a sufficiently small positive constant, besides L1 and the point z1 in (1.15) should be replaced by L1 ∪ L4 and z1 = (1 − i)a/2, z2 = 1 + a/2 + (1 − a/2)i. Now we first prove the unique solvability of Problem Q for equation (1.2).

212

VI. Second Order Quasilinear Mixed Equations

Theorem 2.1 If the mixed equation (1.2) in the domain D satisfies Condition C, then Problem Q for (1.2) has a unique solution u(z) as stated in the form  z u(z) = 2Re w(z)dz + b0 in D, (2.11) 0

where ˜ φ(z) ˜ ˜ w(z) = Φ(z)e + ψ(z) in D+ ,  g(ζ) 1 ˜ dσζ , φ(z) = φ˜0 (z) + T g = φ˜0 (z) − π ζ −z + D w(z) = w0 (z) + Φ(z) + Ψ(z) in D− , ⎧ ν  µ ⎪ ⎪ g (z)dνe + g2 (z)dµe2 , ⎪ 1 1 ⎨ Ψ(z) =

0

a

⎪ ⎪ ⎪ ⎩





ν

2

g1 (z)dνe1 +

µ

g2 (z)dµe2 , a

˜ ψ(z) = T f in D+ , (2.12)

z ∈ D1− , z ∈ D2− ,

˜ where Im [φ(z)] = 0 on L0 = (0, 2), e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic function in D+ and continuous in D+ , such that ˜ Im [φ(z)] = 0 on L0 , and f (z), g(z), g1 (z), g2 (z) are as stated in (1.14), and Φ(z) is + an analytic function in D and Φ(z) is a solution of equation (1.12) satisfying the boundary conditions: the first conditions and Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L1 ∪ L4 ,

Im [λ(zj )Φ(zj )] = −Im [λ(zj )Ψ(zj )], Im [λ(zj
)Φ(zj
)]

= cj ,

j = 1, 2,

(2.13)

j = 1, 2,

in which λ(x), s(x) are as stated in (2.8),(2.9). Proof By using a similar method as stated in Section 2, Chapter V, we can prove Theorem 2.1, provided that L1 or L2 in the boundary conditions, Section 2, Chapter V is replaced by L1 ∪ L4 ; the point conditions Im [λ(z1 )w(z1 )] = b1 in Section 2, Chapter V is replaced by Im [λ(zj )w(zj )] = bj , j = 1, 2 and so on; the formula (2.14), Chapter V is replaced by
λ(x) =

Re [λ(z)uz ] = Re [λ(z)w(z)] = s(x), 1 + i, x ∈ L
0 = (0, a), 1 − i, x ∈ L

0 = (a, 2),

Cβ [s(x), L
0 ] + Cβ [s(x), L

0 ] ≤ k3 .

Theorem 2.2 Suppose that the equation (1.2) satisfies Condition C. Then Problem P for (1.2) has a unique solution u(z), and the solution u(z) satisfies the estimates C˜ 1 [u(z), D+ ] = Cβ [u(z), D+ ] + Cβ [uz X(z), D+ ] ≤ M15 , β

± − Cˆ 1 [u(z), D− ] = Cβ [u(z), D− ] + C[u± z Y (z), D ] ≤ M15 ,

C˜β1 [u(z), D+ ] ≤ M16 (k1 +k2 ),

Cˆ 1 [u(z), D− ] ≤ M16 (k1 +k2 ),

(2.14)

2. Mixed Equations in General Domains

213

where 3  X(z)= |z −tj |ηj ,




3  Y (z)= |x±y −tj |ηj , ±

j=1

ηj =

j=1

2max(−γj ,0)+δ, max(−γj ,0)+δ,

j =1,2, j =3,

(2.15) herein t1 = 2, t2 = 0, t3 = a, γ1 , γ2 , γ3 are real constants in (2.5), β(< δ), δ are sufficiently small positive constants, and M15 = M15 (p0 , β, k, D), M16 = M16 (p0 , β, k0 , D) are non-negative constants, k = (k0 , k1 , k2 ). Proof First of all, we prove the uniqueness of solutions of Problem P for (1.2). Suppose that there exist two solutions u1 (z), u2 (z) of Problem P for (1.2). By Condition C, we can see that u(z) = u1 (z) − u2 (z) and w(z) = uz satisfies the homogeneous equation and boundary conditions


wz




= f,

w¯z∗

f = Re [A1 w] + A2 u in

Re [λ(z)w(z)] = r(z), u(2) = 0,

z ∈ Γ,

u(0) = 0,

Re [λ(z)w(z)] = 0,

D+ D−

 ,

(2.16)

u(a) = 0,

z ∈ L1 ∪ L4 ,

(2.17)

and (2.11). By using the method of proofs in Theorems 2.3 and 2.4, Chapter V, w(z) = uz = 0, u(z) = 0 in D− can be derived. Thus we have   1+i 1−i ∂u ∂u √ uz = = 0 on (0, a), 2Re √ uz = = 0 on (a, 2), (2.18) ∂l ∂l 2 2 √ it is clear √ that (1 − i)/ 2 = cos(l, x) − i cos(l, y) = exp(−iπ/4) on (0, a) and (1 + i)/ 2 = cos(l, x) + i cos(l, y) = exp(iπ/4) on (a, 2). On the basis of the maximum principle of solutions for (1.2) with A3 = 0 in D+ , if maxD+ u(z) > 0, then its maximum M attains at a point z ∗ ∈ Γ ∪ L0 , obviously z ∗
= 0, a and 2, and we can prove z ∗
∈ Γ by the method as stated in the proof of Theorem 2.3, Chapter V. Moreover, it is not difficult to prove that if z ∗ ∈ L0 then ∂u/∂l
= 0 at z ∗ . This contradicts (2.17). Thus maxD+ u(z) = 0. By the similar method, we can prove minD+ u(z) = 0. Hence u(z) = 0, u1 (z) = u2 (z) in D+ .

2Re

Secondly, we first prove the existence of solutions of Problem P for the linear equation (1.2) with A2 = 0, i.e.


uzz = Re [A1 uz ] + A3 , uz¯z∗ = Re [A1 uz ] + A3 ,

z ∈ D+ , z ∈ D− ,

(2.19)

By Theorem 2.1, we can prove the solvability of Problem P for (2.19). In fact if u0 (a) = b1 , u0 (2) = b2 , then the solution u0 (z) is just a solution of Problem P for (1.2). Otherwise, u0 (a) = c
1
= b1 , or u0 (2) = c
2
= b2 , we find a solution u2 (z) of

214

VI. Second Order Quasilinear Mixed Equations

Problem P for (2.19) with the boundary conditions Re [λ(z)ukz ] = 0,

z ∈ Γ,

Re [λ(z)uk¯z ] = 0,

z ∈ L 1 ∪ L4 ,

Im [λ(z)uk¯z ]|z=zj
= δkj ,

uk (0) = 0, k = 1, 2,

(2.20)

k, j = 1, 2,

in which δ11 = δ22 = 1, δ12 = δ21 = 0. It is clear that    u (a) u (a)  2  1  J = 
= 0.  u1 (2) u2 (2) 

(2.21)


0), such that Because otherwise there exist two real constants d1 , d2 (|d1 | + |d2 | = d1 u1 (z) + d2 u2 (z)
≡ 0 in D and d1 u1 (a) + d2 u2 (a) = 0, d1 u1 (2) + d2 u2 (2) = 0.

(2.22)

According to the proof of uniqueness as before, we can derive d1 u1 (z) + d2 u2 (z) ≡ 0 in D, the contradiction proves J
= 0. Hence there exist two real constants d1 , d2 , such that d1 u1 (a) + d2 u2 (a) = c
1 − b1 , d1 u1 (2) + d2 u2 (2) = c
2 − b2 ,

(2.23)

thus the function u(z) = u0 (z) − d1 u1 (z) − d2 u2 (z) in D

(2.24)

is just a solution of Problem P for equation (1.2) in the linear case. Moreover, we can obtain that the solution u(z) of Problem P for (1.2) satisfies the estimates in (2.14), we can rewrite in the form Cˆ 1 [u, D] = Cβ [u(z), D] + Cβ [uz X(z), D+ ] ± − +C[u± z (µ, ν)Y (µ, ν), D ] ≤ M17 ,

(2.25)

Cˆ 1 [u, D] ≤ M18 (k1 + k2 ), where X(z), Y ± (µ, ν) are as stated in (2.10), and M17 = M17 (p0 , β, k, D), M18 = M18 (p0 , β, k0 , D) are non-negative constants, k = (k0 , k1 , k2 ). By using the estimate and method of parameter extension, the existence of solutions of Problem P for quasilinear equation (1.2) can be proved. 2.2 Oblique derivative problem for second order equations of mixed type in general domains Now, we consider the domain D
with the boundary ∂D
= Γ ∪ L
, where L
= L
1 ∪ L
2 ∪ L
3 ∪ L
4 and the parameter equations of the four curves L
1 , L
2 , L
3 , L
4 are L
1 = {γ1 (x) + y = 0, 0 ≤ x ≤ l1 }, L
3 = {x + y = a, a ≤ x ≤ l2 },

L
2 = {x − y = a, l1 ≤ x ≤ a},

L
4 = {γ2 (x) + y = 0, l2 ≤ x ≤ 2},

(2.26)

2. Mixed Equations in General Domains

215

in which γ1 (0) = 0, γ2 (2) = 0; γ1 (x) > 0 on 0 ≤ x ≤ l1 = γ(l1 ) + a; γ2 (x) > 0 on l2 = −γ(l2 ) + a ≤ x ≤ 2; γ1 (x) on 0 ≤ x ≤ l1 , γ2 (x) on l2 ≤ x ≤ 2 are continuous, and γ1 (x), γ2 (x) are differentiable on 0 ≤ x ≤ l1 , l2 ≤ x ≤ 2 except some isolated points, and 1 + γ1
(x) > 0 on 0 ≤ x ≤ l1 , 1 − γ2
(x) > 0 on l2 ≤ x ≤ 2. Denote D
+ = D
∩ {y > 0} = D+ , D
− = D
∩ {y < 0}, D1
− = D
− ∩ {x < a} and D2
− = D
− ∩ {x > a}. Here we mention that in [12]1),3), the author assumed 0 < −γ1
(x) < 1 on 0 ≤ x ≤ l1 and some other conditions. We consider the quasilinear second order equation of mixed (elliptic-hyperbolic) type: (1.2) in D
. Assume that equation (1.2) satisfies Condition C, but the hyperbolic domain D− is replaced by D
− . Problem P
The oblique derivative problem for equation (1.2) is to find a continuously differentiable solution of (1.2) in D∗ = D
\{0, a, 2} for (1.2) satisfying the boundary conditions

1 ∂u = Re [λ(z)uz ] = r(z), z ∈ Γ, u(0) = b0 , u(a) = b1 , u(2) = b2 , 2 ∂l 1 ∂u = Re [λ(z)uz¯] = r(z), z ∈ L
1 ∪L
4 , Im [λ(z)uz¯]|z=zj = bj+2 , j = 1,2. 2 ∂l

(2.27)

Here l is a given vector at every point on Γ ∪ L
1 ∪ L
4 , z1 = l1 − iγ1 (l1 ), z2 = l2 − iγ2 (l2 ), λ(z) = a(x) + ib(x) = cos(l, x) − i cos(l, y), z ∈ Γ, and λ(z) = a(x) + ib(x) = cos(l, x) + i cos(l, y), z ∈ L
1 ∪ L
4 , bj (j = 0, 1, . . . , 4) are real constants, and λ(z), r(z), bj (j = 0, 1, . . . , 4) satisfy the conditions

Cα [λ(z), Γ] ≤ k0 , Cα [λ(z), L
j ] ≤ k0 , cos(l, n) ≥ 0 on Γ, max
z∈L1

1 , |a(x) − b(x)|

Cα [r(z), Γ] ≤ k2 , Cα [r(z), L
j ] ≤ k2 , |bj | ≤ k2 , max
z∈L4

j = 1, 4,

0 ≤ j ≤ 4,

(2.28)

1 ≤ k0 , |a(x) + b(x)|

in which n is the outward normal vector at every point on Γ, α (1/2 < α < 1), k0 , k2 are non-negative constants. In particular, if L
j = Lj (j = 1, 2, 3, 4), then Problem P
in this case is called Problem P . In the following, we discuss the domain D
with the boundary Γ∪L
1 ∪L
2 ∪L
3 ∪L
4 , where L
1 , L
2 , L
3 , L
4 are as stated in (2.26), and γ1 (x), γ2 (x) satisfy the conditions 1 + γ1
(x) > 0 on 0 ≤ x ≤ l1 and 1 − γ2
(x) > 0 on l2 ≤ x ≤ 2. By the conditions, the

216

VI. Second Order Quasilinear Mixed Equations

inverse functions x = σ(ν), x = τ (µ) of x + γ1 (x) = ν = x − y, x − γ2 (x) = µ = x + y can be found respectively, namely µ = 2σ(ν) − ν,

0 ≤ ν ≤ a,

ν = 2τ (µ) − µ,

a ≤ µ ≤ 2.

(2.29)

They are other expressions for the curves L
1 , L
4 . Now, we make a transformation in D
− : µ ˜=

a[µ − 2σ(ν) + ν] , a − 2σ(ν) + ν

µ ˜ = µ,

ν˜ =

ν˜ = ν,

2σ(ν) − ν ≤ µ ≤ a,

a[2τ (µ)−µ−2]+(2−a)ν , 2τ (µ) − µ − a

0 ≤ ν ≤ a, (2.30)

a ≤ µ ≤ 2,

a ≤ ν ≤ 2τ (µ)−µ,

in which µ, ν are variables. If (µ, ν) ∈ L
1 , L
2 , L
3 , L
4 , then (˜ µ, ν˜) ∈ L1 , L2 , L3 , L4 respectively. The inverse transformation of (2.30) is 1 [a − 2σ(ν) + ν]˜ µ + 2σ(ν) − ν a 1 = [a − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ x + y˜)+2σ(x+γ1 (x))−x−γ1 (x), a

µ=

ν = ν˜ = x˜ − y˜, ν= =

0≤µ ˜ ≤ a, 0 ≤ ν˜ ≤ a, µ = µ ˜ = x˜ + y˜,

1 [(2τ (µ) − µ)(˜ ν − a) − a(˜ ν − 2)] 2−a

(2.31)

1 x − y˜−a)−a(˜ x − y˜−2)], [(2τ (x−γ2 (x))−x+γ2 (x))(˜ 2−a a≤µ ˜ ≤ 2,

a ≤ ν˜ ≤ 2.

It is not difficult to see that the transformations in (2.31) map the domains D1
− , D2
− onto the domains D1− , D2− respectively. Moreover, we have 2ax − (a + x − y)[2σ(x + γ1 (x)) − x − γ1 (x)] 1 µ + ν˜) = , x˜ = (˜ 2 2a − 4σ(x + γ1 (x)) + 2x + 2γ1 (x) 1 2ay − (a − x + y)[2σ(x + γ1 (x)) − x − γ1 (x)] , y˜ = (˜ µ − ν˜) = 2 2a − 4σ(x + γ1 (x)) + 2x + 2γ1 (x) 1 1 x = (µ + ν) = [a − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ x + y˜) 2 2a 1 +σ(x + γ1 (x)) − (x + γ1 (x) − x˜ + y˜), 2 1 1 [a − 2σ(x + γ1 (x)) + x + γ1 (x)](˜ y = (µ − ν) = x + y˜) 2 2a 1 +σ(x + γ1 (x)) − (x + γ1 (x) + x˜ − y˜), 2

(2.32)

2. Mixed Equations in General Domains

217

and 1 [2τ (x−γ2 (x))−x+γ2 (x)](x+y+a)+2(x−y)−2a(1+x) x˜ = (˜ µ+˜ ν) = , 2 2[2τ (x−γ2 (x))−x+γ2 (x)−a] 1 [2τ (x−γ2 (x))−x+γ2 (x)](x+y−a)−2(x−y)+2a(1−y) y˜= (˜ µ−˜ ν) = , 2 2[2τ (x−γ2 (x))−x+γ2 (x)−a] 1 1 [2τ (x−γ2 (x))−x+γ2 (x))(˜ x − y˜ −a) x = (µ+ν) = 2 2(2−a)

(2.33)

+2(˜ x + y˜ +a)−2a˜ x], 1 1 x + y˜ +a) y = (µ−ν) = [2τ (x−γ2 (x))−x+γ2 (x))(−˜ 2 2(2−a) +2(˜ x + y˜ −a)−2a˜ y ]. Denote by z˜ = x˜ + i˜ y = f (z), z˜ = x˜ + i˜ y = g(z) and z = x + iy = f −1 (˜ z ), z = x+iy = g −1 (˜ z ) the transformations and their inverse transformations in (2.32), (2.33) respectively. Through the transformation (2.30), we have 1 (U −V )µ˜ = [a−2σ(ν)+ν](U −V )µ in D1
− , a 2τ (µ) − µ − a (U + V )ν , (U − V )µ˜ = (U − V )µ in D2
− . (U + V )ν˜ = 2−a (U +V )ν˜ = (U +V )ν ,

(2.34)

Equation (1.2) in D
− can be rewritten in the form ξν = Aξ + Bη + Cu + D,

ηµ = Aξ + Bη + Cu + D,

z ∈ D
− ,

(2.35)

where ξ = U + V = (ux − uy )/2, η = U − V = (ux + uy )/2, under the transformation (2.34), it is clear that system (2.35) in D
− is reduced to ξν˜ = Aξ +Bη+ Cu+D, ξν˜ =

1 ηµ˜ = [a−2σ(ν)+ν][Aξ +Bη+Cu+D] in D1− , a

2τ (µ)−µ−a [Aξ +Bη+Cu+D], 2−a

(2.36)

ηµ˜ = Aξ +Bη+Cu+D in D2− .

Moreover, through the transformations (2.32),(2.33), the boundary condition (2.23) on L
1 ∪ L
4 is reduced to z ))w(f −1 (˜ z ))] = r[f −1 (˜ z )], Re [λ(f −1 (˜

z˜ ∈ L1 , Im [λ(f −1 (˜ z3 ))w(f −1 (˜ z3 ))] = b1 ,

z ))w(g −1 (˜ z ))] = r[g −1 (˜ z )], Re [λ(g −1 (˜

z˜ ∈ L4 ,

Im [λ(g −1 (˜ z4 ))w(g −1 (˜ z4 ))] = b2 ,

(2.37)

in which z˜3 = f (z3 ), z˜4 = g(z4 ). Therefore the boundary value problem (1.2) (in D+ ), (2.35), (2.27), (2.6) is transformed into the boundary value problem (1.2), (2.36), (2.27), (2.37). According to the proof of Theorem 2.1, we see that the boundary value

218

VI. Second Order Quasilinear Mixed Equations

problem (1.2), (2.36), (2.27), (2.37) has a unique solution w(˜ z ), and then w[˜ z (z)] is a solution of the boundary value problem (1.2),(2.2) (w = uz ) in D
− , and the function ⎧  z ⎪ ⎪ w(z)dz + b0 in D+ , ⎪ 2Re ⎪ ⎪ 0 ⎪ ⎪ ⎪  z ⎨ w[f (z)]dz + u(a) in D1
− , u(z) = 2Re (2.38) ⎪ a ⎪ ⎪ ⎪  z ⎪ ⎪ ⎪ ⎪ w[g(z)]dz + u(2) in D2
− , ⎩ 2Re 2

is just a solution of Problem P for (1.2) in D
, where u(a) = b1 , u(2) = b2 . Theorem 2.3 If the mixed equation (1.2) in the domain D
satisfies Condition C, then Problem P
for (1.2) with the boundary condition (2.2) has a unique solution u(z) as stated in (2.38), where z1 = l1 − iγ1 (l1 ), z2 = l2 − iγ2 (l2 ). By using the above method and the method in Section 4, Chapter IV, we can discuss the unique solvability of Problem P
for equation (1.2) in some more general domains D

including the domain D

= {|z − 1| < 1, Im z ≥ 0} ∪ {|z − a/2| < a2 /4, Im z < 0} ∪ {|z − 1 − a/2| < (2 − a)2 /4, Im z < 0}.

3

Discontinuous Oblique Derivative Problems for Second Order Quasilinear Equations of Mixed Type

This section deals with discontinuous oblique derivative problems for quasilinear second order equations of mixed (elliptic-hyperbolic) type in a simply connected domain. Firstly, we give a representation theorem and prove the uniqueness of the solution for the above boundary value problem, and then by using the method of successive iteration, the existence of solutions for the above problem is proved.

3.1 Formulation of discontinuous oblique derivative problems for second order equations of mixed type Let D be a simply connected domain with the boundary Γ ∪ L1 ∪ L2 as stated before, where D+ = {|z −1| < 1, Im z > 0}. We discuss the second order quasilinear complex equations of mixed type as stated in (1.2) with Condition C. In order to introduce the discontinuous oblique derivative boundary value problem for equation (1.2), let the functions a(z), b(z) possess discontinuities of first kind at m − 1 distinct points z1 , z2 , . . . , zm−1 ∈ Γ, which are arranged according to the positive direction of Γ, and Z = {z0 = 2, z1 , . . . , zm = 0} ∪ {x + y = 0, x − y = 2, Im z ≤ 0}, where m is a positive integer, and r(z) = O(|z − zj |−βj ) in the neighborhood of zj (j = 0, 1, . . . , m) on Γ, in which βj (j = 0, 1, . . . , m) are sufficiently small positive numbers. Denote λ(z) = a(z) + ib(z) and |a(z)| + |b(z)| =
0. There is no harm in assuming that

3. Discontinuous Oblique Derivative Problems

219

|λ(z)| = 1, z ∈ Γ∗ = Γ\Z. Suppose that λ(z), r(z) satisfy the conditions λ(z) ∈ Cα (Γj ),

|z − zj |βj r(z) ∈ Cα (Γj ),

j = 1, . . . , m,

(3.1)

herein Γj is the arc from the point zj−1 to zj on Γ and z0 = 2, and Γj (j = 1, . . . , m) does not include the end points, α (0 < α < 1) is a constant. Besides, there exist n points E1 = a1 , E2 = a2 , . . . , En = an on the segment AB = L0 = (0, 2) and E0 = 0, En+1 = 2, where a0 = 0 < a1 < a2 < · · · < an < an+1 = 2. Denote by A = A0 = 0, A1 = (1 − i)a1 /2, A2 = (1 − i)a2 /2, . . . , An = (1 − i)an /2, An+1 = C = 1 − i and B1 = 1 − i + (1 + i)a1 /2, B2 = 1 − i + (1 + i)a2 /2, . . . , Bn = 1 − i + (1 + i)an /2, Bn+1 = B = 2 on the segments AC, CB respectively. Moreover, we denote [n/2] [(n+1)/2] D1− = D− ∩{∪j=0 (a2j ≤ x−y ≤ a2j+1 )}, D2− = D− ∩{∪j=1 (a2j−1 ≤ x+y ≤ a2j )} − − ˜ ˜ − = D− ∩ {a2j−1 ≤ and D2j+1 = D ∩ {a2j ≤ x − y ≤ a2j+1 }, j = 0, 1, . . . , [n/2], D 2j n+1 − − x + y ≤ a2j }, j = 1, . . . , [(n + 1)/2], and D∗ = D \{∪j=0 (x ± y = aj , y ≤ 0)}, D∗ = D+ ∪ D∗− . The discontinuous oblique derivative boundary value problem for equation (1.2) may be formulated as follows: ¯ which is continuously Problem P
Find a continuous solution u(z) of (1.2) in D, + − differentiable in D∗ = D ∪ D∗ and satisfies the boundary conditions 1 ∂u = Re [λ(z)uz ] = r(z), 2 ∂l

z ∈ Γ, [n/2]



1 ∂u = 2Re [λ(z)uz¯] = r(z), 2 ∂l

z ∈ L3 =

1 ∂u = 2Re [λ(z)uz¯] = r(z), 2 ∂l

z ∈ L4 =

Im [λ(z)uz¯]|z=A2j+1 = c2j+1 ,

j = 0, 1, . . . , [n/2],

Im [λ(z)uz¯]|z=B2j−1 = c2j ,

A2j A2j+1 ,

j=0

(3.2)

[(n+1)/2]



B2j−1 B2j ,

j=1

j = 1, . . . , [(n + 1)/2],

u(zj ) = bj , j = 0, 1, . . . , m, u(aj ) = bm+j ,

(3.3)

j = 1, . . . , n,

where l is a vector at every point on Γ ∪ L3 ∪ L4 , bj (j = 0, 1, . . . , m + n), cj (j = 0, 1, . . . , n + 1, c0 = b0 ) are real constants, λ(z) = a(x) + ib(x) = cos(l, x) − i cos(l, y), z ∈ Γ, λ(z) = a(x) + ib(x) = cos(l, x) + i cos(l, y), z ∈ L3 ∪ L4 , and λ(z), r(z), cj (j = 0, 1, . . . , n + 1) satisfy the conditions Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

|cj | ≤ k2 ,

Cα [λ(z), Lj ] ≤ k0 ,

Cα [r(z), Lj ] ≤ k2 ,

j = 3,4,

cos(l, n) ≥ 0 on Γ,

1 ≤ k0 , max z∈L3 |a(x) − b(x)|

j = 0, 1, . . . , n + 1, |bj | ≤ k2 , j = 0, 1, . . . , m+n,

1 ≤ k0 , max z∈L4 |a(x) + b(x)|

(3.4)

220

VI. Second Order Quasilinear Mixed Equations

where n is the outward normal vector at every point on Γ, α (1/2 < α < 1), k0 , k2 are non-negative constants. The above discontinuous oblique derivative boundary value problem for (1.2) is called Problem P
. Problem P
for (1.2) with A3 (z, u, uz ) = 0, ¯ r(z) = 0, z ∈ Γ ∪ L3 ∪ L4 , bj = 0(j = 0, 1, . . . , m + n) and cj = 0 z ∈ D, (j = 0, 1, . . . , n + 1) will be called Problem P0
. Moreover we give the same definitions as in (5.10), (5.11), Chapter IV, but choose K = (m + n − 1)/2, or K = (m + n)/2 − 1 if cos(ν, n) ≡ 0 on Γ, and the condition u(zm ) = bm can be canceled. Besides we require that the solution u(z) in D+ satisfies the conditions
−δ

uz = O(|z − zj | ), δ =

βj + τ,

for γj ≥ 0,

|γj | + τ,

and γj < 0,

for γj < 0,

βj > |γj |,

βj ≤ |γj |,

j = 0, 1, . . . , m + n, (3.5) in the neighborhood of zj (0 ≤ j ≤ m), aj (1 ≤ j ≤ n) in D+ , where γj
=
max (0, −γj ) (j = 0, 1, . . . , m + n), γ0
= max (0, −2γ0 ), γm = max (0, −2γm ) and γj (j = 0, 1, . . . , m + n) are real constants as stated in (5.10), Chapter IV, δ is a sufficiently small positive number. Now we explain that in the closed domain D− , the derivatives ux + uy , ux − uy of the solution u(z) in the neighborhoods of the 2n + 2 characteristic lines Z
= {x + y = 0, x − y = 2, x ± y = aj (j = 1, . . . , n), y ≤ 0} may be not bounded if γj ≤ 0(j = 0, 1, . . . , n + 1). Hence if we require that the derivative uz of u(z) in D− \Z
is bounded, then we need to choose γj > 0 (j = 0, 1, . . . , n + 1). If we only require that the solution u(z) is continuous in D, it suffices to choose −2γ0 < 1, −2γm < 1, −γj < 1 (j = 1, . . . , m − 1, m + 1, . . . , m + n). Furthermore, we need to introduce another oblique derivative boundary value problem. Problem Q
If A2 (z) = 0 in D, we find a continuously differentiable solution u(z) of ¯ and satisfies the boundary conditions (3.2),(3.3), (1.2) in D∗ , which is continuous in D but the point conditions in (3.3) are replaced by u(2) = b0 = d0 ,

Im [λ(z)uz ]|z=zj
= dj ,

j = 1, . . . , m + n,

(3.6)

where zj
(
∈ Z) ∈ Γ(j = 0, 1, . . . , m + n) are distinct points, dj (j = 0, 1, . . . , m + n) are real constants satisfying the conditions |dj | ≤ k2 , j = 0, 1, . . . , m + n, but we do not assume cos(ν, n) ≥ 0 on each Γj (j = 1, . . . , m).

3.2 Representations of solutions for the oblique derivative problem for (1.2) First of all, we give the representation of solutions of Problem Q
for the equation


uz¯z uz¯z∗




= 0 in

D+ D−

 .

(3.7)

3. Discontinuous Oblique Derivative Problems

221

It is clear that Problem Q
for (3.7) is equivalent to the following boundary value problem (Problem A
) for the first order complex equation
Lw =

wz¯



w¯z∗


= 0 in

D+



D−

,

(3.8)

with the boundary conditions Re [λ(z)w(z)] = r(z) on Γ, Re [λ(z)w(z)] = r(z) on L3 ∪ L4 , Im [λ(z)w(z)]|z=zj
= bj ,

j = 1, . . . , m + n,

Im [λ(z)w(z)]|z=A2j+1 = c2j+1 , Im [λ(z)w(z)]|z=B2j−1 = c2j , and the relation

 u(z) = 2 Re 2

(3.9)

j = 0, 1, . . . , [n/2], j = 1, . . . , [(n + 1)/2],

z

w(z)dz + b0 ,

(3.10)

where the integral path is appropriately chosen. Thus from Theorem 5.2, Chapter IV, we can derive the following theorem. Theorem 3.1 The boundary value problem Q
for (3.7) in D has a unique continuous solution u(z) as stated in (3.10), where the solution w(z) of Problem A
for (3.8) in D− possesses the form 1 w(z) = [(1 − i)f (x + y) + (1 + i)g(x − y)], 2 f (x + y) = Re [(1 + i)w(x + y)] in D− \{D2− }, g(x − y) = k(x − y) in D1− ,

(3.11)

f (x + y) = h(x + y) in D2− , g(x − y) = Re [(1 − i)w(x − y)] in D− \{D1− }, herein w(x + y)(0 ≤ x + y ≤ 2), w(x − y)(0 ≤ x − y ≤ 2) are values of the solution w(z) of Problem A
for (3.8) in D+ with the first boundary condition in (3.9) and the boundary condition

Re [λ(x)w(x)] =

⎧ ⎨ k(x) on L
1 = D1− ∩ AB, ⎩ h(x) on L
= D− ∩ AB, 2 2

(3.12)

222

VI. Second Order Quasilinear Mixed Equations

in which k(x), h(x) can be expressed as k(x) =

2r((1−i)x/2)−[a((1−i)x/2)+b((1−i)x/2)]h2j a((1 − i)x/2) − b((1 − i)x/2) ˜ 2j+1 = D ˜ − ∩ AB, on L 2j+1

h(x) =

2r((1 + i)x/2 + 1 − i) a((1 + i)x/2 + 1 − i) + b((1 + i)x/2 + 1 − i) −

(3.13)

[a((1+i)x/2+1−i)−b((1+i)x/2+1−i)]k2j−1 a((1+i)x/2+1−i)+b((1+i)x/2+1−i) ˜ 2j = D− ∩ AB, on L 2j

˜ − (j = 1, 2, . . . , 2n + 1) are as stated before, and where D j h2j = Re [λ(A2j+1 )(r(A2j+1 ) + ic2j+1 )] + Im [λ(A2j+1 )(r(A2j+1 ) + ic2j+1 )] ˜ 2j+1 , on L

j = 0, 1, . . . , [n/2],

k2j−1 = Re [λ(B2j−1 )(r(B2j−1 ) + ic2j )] − Im [λ(B2j−1 )(r(B2j−1 ) + ic2j )] ˜ 2j , on L

j = 1, . . . , [(n + 1)/2],

˜ 2j+1 = D ˜ − ∩ AB, where L 2j+1 [(n + 1)/2].

j = 0, 1, . . . , [n/2],

˜ 2j = D ˜ − ∩ AB, L 2j

j = 1, . . . ,

Next we give the representation theorem of solutions of Problem Q
for equation (1.2). Theorem 3.2 Suppose that equation (1.2) satisfies Condition C. Then any solution of Problem Q
for (1.2) can be expressed as  z u(z) = 2Re w(z)dz + c0 , w(z) = w0 (z) + W (z) in D, (3.14) 2

where w0 (z) is a solution of Problem A
for the complex equation (3.8) with the boundary condition (3.2), (3.6) (w0 (z) = u0z ), and w(z) possesses the form ˜ φ(z) ˜ ˜ w(z) = Φ(z)e + ψ(z),   1 g(ζ) ˜ ˜ dσζ , ψ(z)=T f in D+ , φ(z)= φ˜0 (z)+T g= φ˜0 (z)− π ζ −z + D

w(z) = W (z)+w0 (z) in D,

W (z) = Φ(z)+Ψ(z) in D− ,  µ ⎧ ν − ⎪ g (z)dνe + g2 (z)dµe2 in D2j+1 , j = 0,1,...,[n/2], ⎪ 1 1 ⎪ ⎨ a2j+1 0 Ψ(z) =  ν  µ ⎪ ⎪ − ⎪ g1 (z)dνe1 + g2 (z)dµe2 in D2j , j = 1,...,[(n+1)/2], ⎩ 2

a2j−1

(3.15)

3. Discontinuous Oblique Derivative Problems

223

in which e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic ˜ function in D+ , Im [φ(z)] = 0 on L
= (0, 2), and
¯ w(z)
= 0, A1 /2+A1 w/(2w), f (z) = Re[A1 φ˜z ]+A2 u+A3 , g(z) = 0, w(z) = 0, z ∈ D+ , (3.16) g1 (z) = g2 (z) = Aξ +Bη +Cu+D, ξ = Rew +Imw, η = Rew −Imw, ReA1 +ImA1 ReA1 −ImA1 , B= , C = A2 , D = A3 in D+ , 2 2 ˜ where Φ(z) is an analytic function in D+ and Φ(z) is a solution of the equation (3.8) in D− satisfying the boundary conditions ˜ φ(z) ˜ ˜ Re [λ(z)(Φ(z)e + ψ(z))] = r(z), z ∈ Γ, A=

˜ φ(x) ˜ ˜ Re [λ(x)(Φ(x)e + ψ(x))] = s(x), ˜ φ(z)

˜ Im [λ(z)(Φ(z)e

x ∈ L0 = (0, 2),

˜ + ψ(z))| z=zj
= bj ,

j = 1, . . . , m + n,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L0 ,

(3.17)

z ∈ L3 ∪ L4 ,

Im [λ(z)(Φ(z) + Ψ(z))]|z=A2j+1 = 0,

j = 0, 1, . . . , [n/2],

Im [λ(z)(Φ(z) + Ψ(z))]|z=B2j−1 = 0,

j = 1, . . . , [(n + 1)/2],

where s(x) can be written as in (3.19) below. Moreover, the solution u0 (z) of Problem Q
for (3.7) the estimate C˜ 1 [u0 (z), D− ] = Cβ [u0 (z), D− ] + C[w± (µ, ν)Y ± (µ, ν), D− ] ≤ M19 (k1 + k2 ) (3.18) 0




γj +δ in which Y ± (z) = Y ± (µ, ν) = Πn+1 , w0± (µ, ν) = Re w0 (z) ∓ Im w0 (z), j=0 |x ± y − aj | w0 (z) = w0 (µ, ν), µ = x+y, ν = x−y are as stated in (5.24), Chapter IV and u0 (z) is as stated in the first formula of (3.14), where w(z) = w0 (z), M19 = M19 (p0 , β, k0 , D) is a non-negative constant.

Proof Let u(z) be a solution of Problem Q
for equation (1.2), and w(z) = uz , u(z) be substituted in the positions of w, u in (3.16). Thus the functions g(z), f (z), ˜ ˜ ψ(z), φ(z) in D+ and g1 (z), g2 (z), Ψ(z) in D− in (3.15),(3.16) can be determined. ˜ Moreover, we can find the solution Φ(z) in D+ and Φ(z) in D− of (3.8) with the boundary condition (3.17), where s(x)=

2r((1−i)x/2)−[a((1−i)x/2)+b((1−i)x/2)]h2j , a((1−i)x/2)−b((1−i)x/2) x ∈ (a2j ,a2j+1 ),

s(x)=

j = 0,1,...,[n/2], (3.19)

2r((1+i)x/2+1−i)−[a((1+i)x/2+1−i)−b((1+i)x/2+1−i)]k2j−1 , a((1+i)x/2+1−i)+b((1+i)x/2+1−i) x ∈ (a2j−1 ,a2j ),

j = 1,...,[(n+1)/2],

224

VI. Second Order Quasilinear Mixed Equations

in which the real constants hj (j = 0, 1, . . . , n) are of the form h2j = Re [λ(A2j+1 )(r(A2j+1 ) + ic2j+1 ) + Im [λ(A2j+1 )(r(A2j+1 ) + ic2j+1 )] ˜ 2j+1 , on L

j = 0, 1, . . . , [n/2],

k2j−1 = Re [λ(B2j−1 )(r(B2j−1 ) + ic2j ) − Im [λ(B2j−1 )(r(B2j−1 ) + ic2j )] ˜ 2j , on L

j = 1, . . . , [(n + 1)/2],

˜ 2j+1 = D ˜ − ∩AB, j = 0, 1, . . . , [n/2], L ˜ 2j = D ˜ − ∩AB, j = 1, . . . , [(n+1)/2]. where L 2j+1 2j Thus ⎧ ˜ ˜ ˜ φ(z) ⎨ Φ(z) + ψ(z) in D+ , w(z) = w0 (z) + W (z) = ⎩ w (z) + Φ(z) + Ψ(z) in D− , 0 is the solution of Problem A
for the complex equation
+ 
D wz = Re [A1 w] + A2 u + A3 in , wz∗ D−

(3.20)

and u(z) is a solution of Problem Q
as stated in (3.14). 3.3 Unique solvability for the discontinuous oblique derivative problem for (1.2) Theorem 3.3 Suppose that equation (1.2) satisfies Condition C. Then Problem Q
for (1.2) has a unique solution in D. Proof The proof is similar to the proof of Theorems 2.3 and 2.4, Chapter V, but the boundary condition on Lj (j = 1 or 2) and the point condition in which are modified. For instance the boundary condition on Lj (j = 1 or 2) and the point condition in (1.4), Chapter V are replaced by that on L3 ∪ L4 and Im [λ(z)w(z)]|z=A2j+1 = 0, j = 0, 1, . . . , [n/2], Im [λ(z)w(z)]|z=B2j−1 = 0, j = 1, . . . , [(n + 1)/2] respectively; the integral in (2.18), Chapter V is replaced by  µ ⎧ ν − ⎪ ⎪ g0 (z)e1 dν + g0 (z)e2 dµ in D2j+1 , j = 0, 1, . . . , [n/2], ⎪ ⎨ a2j+1 0 Ψ1 (z) =  ν  µ ⎪ ⎪ (3.21) − ⎪ g0 (z)e2 dµ in D2j , j = 1, . . . , [(n+1)/2], ⎩ g0 (z)e1 dν + 2

a2j−1

g0 (z) = Aξ0 + Bη0 + Cu0 + D in D− , and so on; and the characteristic lines through the points z1 = 1 − i are replaced by the characteristic lines through the points A2j+1 (j = 0, 1, . . . , [n/2]), B2j−1 (j = 1, . . . , [(n + 1)/2]. Moreover we can obtain the estimates of solutions of Problem Q
for (1.2).

3. Discontinuous Oblique Derivative Problems

225

Theorem 3.4 Suppose that equation (1.2) satisfies Condition C. Then any solution u(z) of Problem Q
for (1.2) satisfies the estimates C˜ 1 [u(z), D− ] = Cβ [u(z), D− ] + C[w± (µ, ν)Y ± (µ, ν), D− ] ≤ M20 , C˜β1 [u(z), D+ ] = Cβ [u(z), D+ ] + Cβ [uz X(z), D+ ] ≤ M21 (k1 + k2 ),

(3.22)

˜1

C [u(z), D− ] ≤ M21 (k1 + k2 ) = M22 , where X(z) =

m+n 




|z − aj |γj +δ ,

Y ± (z) = Y ± (µ, ν) =

j=0

n+1 




|x ± y − aj |γj +δ ,

j=0

w0± (µ, ν) = Re w0 (z)∓Im w0 (z),

w0 (z) = w0 (µ, ν),

µ = x+y,

(3.23)

ν = x−y,


= in which γj
= max (0, −γj ) (j = 1, . . . , m − 1, m + 1, m + n), γ0
= max (0, −2γ0 ), γm max (0, −2γm ) and γj (j = 0, 1, . . . , m + n) are real constants as stated before, β (0 < β < δ), δ are sufficiently small positive numbers, and k = (k0 , k1 , k2 ), M20 = M20 (p0 , β, k, D), M21 = M21 (p0 , β, δ, k0 , D) are two non-negative constants.

From the estimate (3.22), we can see the regularity of solutions of Problem Q
for (1.2). Next, we consider the oblique derivative problem(Problem P
) for the equation (1.2). Theorem 3.5 Suppose that the mixed equation (1.2) satisfies Condition C. Then Problem P
for (1.2) has a solution in D. Proof First of all, we prove the uniqueness of solutions of Problem P
for (1.2). Suppose that there exist two solutions of Problem P
for (1.2). By Condition C, it can be seen that u(z) = u1 (z) − u2 (z) and w(z) = uz satisfy the homogeneous equation and boundary conditions
 
wz D+ = f, f = Re [A1 w] + A2 u in , w¯z∗ D− u(zj ) = 0,

j = 0, 1, . . . , m,

u(aj ) = 0,

j = 1, . . . , n,

z ∈ Γ,

Re [λ(z)w(z)] = 0,

Im [λ(z)w(z)]|z=A2j+1 = 0,

j = 0, 1, . . . , [n/2],

Im [λ(z)w(z)]|z=B2j−1 = 0,

j = 1, . . . , [(n + 1)/2].

Re [λ(z)w(z)] = 0,

z ∈ L3 ∪L4 ,

(3.24)

By using the method of proof in Theorem 3.3, w(z) = uz = 0, u(z) = 0 in D− can be verified. Thus we have 2Re [λ(x)uz ] =

∂u = 0 on L
= (0, 2), ∂l

226

VI. Second Order Quasilinear Mixed Equations

it is clear that λ(x) = cos(l, x)−i cos(l, y) = exp(−iπ/4) on L
1 and λ(x) = cos(l, x)− i cos(l, y) = exp(iπ/4) on L
2 . On the basis of the maximum principle of solutions for the equation uzz = Re [A1 uz ] + A2 u, z ∈ D+ , (3.25) if maxD+ u(z) > 0, then its maximum attains at a point z ∗ ∈ Γ ∪ L
, obviously z ∗
= zj (j = 0, 1, . . . , m), aj (j = 1, . . . , n), and we can prove z ∗
∈ Γ by the method as stated in the proof of Theorem 3.4, Chapter III. Moreover, it is not difficult to prove that if z ∗ ∈ L
, then ∂u/∂l
= 0 at z ∗ . Hence maxD+ u(z) = 0. By the similar method, we can prove minD+ u(z) = 0. Hence u(z) = 0, u1 (z) = u2 (z) in D. Secondly, we first prove the existence of solutions of Problem P
for equation (1.2) in the linear case. From Theorem 3.3, it can be seen that Problem Q
for (1.2) has a solution u∗ (z) in D, if u∗ (zj ) = bj , j = 0, 1, . . . , m, u∗ (aj ) = bm+j , j = 1, . . . , n, then the solution u∗ (z) is just a solution of Problem P
for (1.2). Otherwise, [u∗ (a
1 ), . . . , u∗ (a
m+n )] = [d∗1 , . . . , d∗n+m ], in which a
j = zj , j = 1, . . . , m, a
j = aj−m , j = m + 1, . . . , m + n, we find m + n solutions u1 (z), . . . , um+n (z) of Problem Q
for (3.25) with the boundary conditions Re [λ(z)ukz ] = 0, uk (2) = 0,

z ∈ Γ,

Re [λ(z)ukz ] = 0,

Im [λ(z)uk¯z ]|z=zj
= δjk ,

Im [λ(z)uk¯z ]|z=A2j+1 = 0,

z ∈ L 3 ∪ L4 ,

j, k = 1, . . . , m + n, (3.26)

j = 0, 1, . . . , [n/2],

Im [λ(z)uk¯z ]|z=B2j−1 = 0, j = 1, . . . , [(n + 1)/2].  ¯ It is obvious that U (z) = m+n k=1 uk (z)
≡ 0 in D, moreover we can verify that  
 u1 (a1 ) · · · um+n (a
1 )        . . . . .. . J = 
= 0, . .      u1 (a
) · · · um+n (a
)  m+n m+n thus there exist m + n real constants c
1 , c
2 , · · · , c
m+n , which are not equal to zero, such that c
1 u1 (a
k ) + c
2 u2 (a
k ) + · · · + c
m+n um+n (a
k ) = d∗k − bk , k = 1, · · · , m + n, thus the function u(z) = u∗ (z) −

m+n 

¯ c
k uk (z) in D

k=1

is just a solution of Problem P
for the linear equation (1.2) with A2 = 0 in D. In addition we can derive that the solution u(z) of Problem P
for (1.2) satisfies the estimates similar to (3.22). Afterwards we consider the equation with the parameter t ∈ [0, 1]: 

 uzz D+ Luz = == Re [A1 uz ]+ t[A2 u+A3 ] + A(z) in , (3.27) uz¯z∗ D−

4. Problems in Multiply Connected Domains

227

where A(z) is any function in D satisfying the condition C[A(z)X(z), D+ ] + C[A± (µ, ν)Y ± (µ, ν), D− ] < ∞. By using the method of parameter extension, namely when t = 0, we see that Problem P
for such equation has a unique solution by the above discussion. Moreover assuming that when t = t0 ∈ (0, 1], Problem P
for equation (3.27) has a solution, then we can prove that there exists a small positive constant ε, such that for any t ∈ {|t−t0 | ≤ ε, t ∈ [0, 1]}, Problem P
for such equation (3.27) has a solution. Thus we can derive that there exists a solution u(z) of Problem P
for equation (3.27) with t = 1, especially when A(z) = 0 in D, i.e. Problem P
for equation (1.2) has a solution u(z). This completes the proof.

4

Oblique Derivative Problems for Quasilinear Equations of Mixed Type in Multiply Connected Domains

In this section we discuss the oblique derivative boundary value problems for quasilinear second order equations of mixed (elliptic-hyperbolic) type in multiply connected domains. We first give a representation of solutions for the above boundary value problem, and then prove the uniqueness and existence of solutions of the above problem and give a priori estimates of solutions of the above problem. In the book [9]2), the author proposed the Dirichlet boundary value problem (Tricomi problem) for second order equations of mixed type in multiply connected domains. In [12] 1),3) the author only discussed the Dirichlet problem (Problem T2 ) for the Lavrent
ev-Bitsadze equation of mixed (elliptic-hyperbolic) type: uxx + sgny uyy = 0 in a special doubly connected domain. Up to now we have not seen that other authors have solved it in multiply connected domains. In this section, we try to discuss the oblique derivative problem for quasilinear equations of mixed type in multiply connected domains, which includes the Dirichlet problem (Problem T2 ) as a special case.

4.1 Formulation of the oblique derivative problem for second order equations of mixed type Let D be an N -connected bounded  domain D in the complex plane C I with the N 2 boundary ∂D = Γ ∪ L, where Γ = j=1 Γj ∈ Cα (0 < α < 1) in {y > 0} with the end points z = a1 = 0, b1 , a2 , b2 , . . . , aN , bN = 2, and L = ∪2N j=1 Lj , L1 = {x = −y, 0 ≤ x ≤ 1}, L2 = {x = −y + b1 , b1 ≤ x ≤ b1 + (a2 − b1 )/2}, L3 = {x = y + a2 , b1 + (a2 − b1 )/2 ≤ x ≤ a2 }, L4 = {x = −y + b2 , b2 ≤ x ≤ b2 + (a3 − b2 )/2}, . . . , L2N −1 = {x = y + aN , bN −1 + (aN − bN −1 )/2 ≤ x ≤ aN }, L2N = {x = y + 2, 1 ≤ x ≤ 2}, in which a1 = 0 < b1 < a2 < b2 < · · · < aN < bN = 2, and denote D+ = D ∩ {y > 0}, D− = D ∩ {y < 0}, D1− = D− ∩{x + y < b1 }, D2− = D− ∩ {b1 < − − x + y < a2 }, D3− = D− ∩ {a2 < x + y < b2 }, . . . , D2N −2 = D ∩ {bN −1 < x + y < − − aN }, D2N −1 = D ∩{aN < x+y}, and z1 = 1−i, z2 = b1 +(a2 −b1 )(1−i)/2, . . . , zN = bN −1 + (aN − bN −1 )(1 − i)/2. We assume that the inner angles π/α2j−1 , π/α2j of D+ at the points z = aj , bj (j = 1, . . . , N ) are greater than zero and less than π.

228

VI. Second Order Quasilinear Mixed Equations

We consider the quasilinear second order equation of mixed type: (1.1) and its complex form (1.2) with Condition C. The oblique derivative boundary value problem for equation (1.2) may be formulated as follows: Problem P

Find a continuous solution ¯ which is conu(z) of equation (1.2) in D, ¯ tinuously differentiable in D∗ = D\Z and satisfies the boundary conditions 1 ∂u 1 ∂u = Re [λ(z)uz ] = r(z), z ∈ Γ, (4.1) = Re [λ(z)uz¯] = r(z), z ∈ L
, 2 ∂l 2 ∂l Im [λ(z)uz¯]|z=zj = cj , j = 1, . . . ,N, u(aj ) = dj , u(bj ) = dN +j , j = 1, . . . ,N, (4.2) where Z = {x ± y = aj , x ± y = bj , j = 1, . . . , N, y ≤ 0}, L
= ∪N j=1 L2j−1 , l is a vector at every point on Γ ∪ L
, λ(z) = a(x) + ib(x) = cos(l, x) ∓ i cos(l, y), z ∈ Γ ∪ L
, ∓ are determined by z ∈ Γ and L
respectively, cj , dj , dN +j (j = 1, . . . , N ) are real constants, and λ(z), r(z), cj , dj , dN +j (j = 1, . . . , N ) satisfy the conditions Cα [λ(z), Γ] ≤ k0 ,

Cα [r(z), Γ] ≤ k2 ,

cos(l, n) ≥ 0 on Γ,

Cα [λ(z), L
] ≤ k0 ,

|cj |, |dj |, |dN +j | ≤ k2 ,

1 max , z∈L1 |a(x) − b(x)|

Cα [r(z), L
] ≤ k2 ,

j = 1, . . . , N,

(4.3)

1 max

≤ k0 , z∈L |a(x) + b(x)|

in which n is the outward normal vector on Γ, L

= ∪N j=2 L2j−1 , α (1/2 < α < 1), k0 , k2 are non-negative constants. Here we mention that if A2 = 0 in D+ , then we can cancel the condition cos(l, n) ≥ 0 on Γ, and if the boundary condition Re [λ(z)uz¯] = r(z), z ∈ L
is replaced by Re [λ(z)uz ] = r(z), z ∈ L
, then Problem P

does not include the Dirichlet problem (Tricomi problem). The boundary value problem for (1.2) with A3 (z, u, uz ) = 0, z ∈ D, u ∈ IR, uz ∈ C I, r(z) = 0, z ∈ Γ ∪ L
and cj = 0 (j = 0, 1, . . . , N ) and dj = 0 (j = 1, . . . , 2N ) will be called Problem P0

. The number 1 K = (K1 + K2 + · · · + K2N ), 2

(4.4)

is called the index of Problem P

and Problem P0

on the boundary ∂D+ of D+ , where  λ(tj −0) φj φj Kj = , γj = −Kj , j = 1,...,N, (4.5) +Jj , Jj = 0 or 1, ejφj = π λ(tj +0) π in which [a] is the largest integer not exceeding the real number a, and t1 = a1 = 0, t2 = b1 , t3 = a2 , t4 = b2 , . . . , t2N −1 = aN , t2N = bN = 2, and λ(t) = eiπ/4 on lj = (aj , bj ),

λ(t2j−1 + 0) = λ(t2j − 0) = eiπ/4 ,

j = 1, . . . , N.

4. Problems in Multiply Connected Domains

229

If cos(l, n)
≡ 0 on each of Γj (j = 1, . . . , N ), then we select the index K = N − 1 on ∂D+ . If cos(l, n) ≡ 0 on Γj (j = 1, . . . , N ), then we select the index K = N/2 − 1 on ∂D+ , and the last N point conditions in (4.2) can be eliminated. In this case, Problem P includes the Dirichlet problem (Tricomi problem) as a special case. Now we explain that in the closed domain D− , the derivative ux ± uy of the solution u(z) in the neighborhoods of 4N characteristic lines x ± y = aj , x ± y = bj (j = 1, . . . , N ) may not be bounded if γj αj ≤ 0(j = 1, . . . , 2N ). Hence if we require that the solution u(z) in D− \Z is bounded, where Z = {x + y = aj , x + y = bj , x − y = aj , x − y = bj , y ≤ 0 (j = 1, . . . , N )}, then it needs to choose γj > 0 (j = 1, . . . , 2N ), herein γj (j = 1, . . . , 2N ) are as stated in (4.5). If we require that solution u(z) is only continuous in D, it suffices to choose −γj αj < 1 (j = 1, . . . , 2N ). Moreover, we need to introduce another oblique derivative boundary value problem. Problem Q

If A2 = 0 in D, one has to find a continuously differentiable solution ¯ and satisfies the boundary conditions u(z) of (1.2) in D∗ , which is continuous in D (4.1),(4.2), but the last N point conditions are replaced by Im [λ(z)uz¯]|z=zj
= d
j ,

j = 1, . . . , N,

(4.6)

where zj
(j = 1, . . . , N − 1) are distinct points, such that zj
∈ Γ, zj

∈ Z (j = 1, . . . , N − 1) and d
j (j = 1, . . . , N ) are real constants satisfying the conditions |d
j | ≤ k2 , j = 1, . . . , N . In the case the condition cos(l, n) ≥ 0 on Γ can be canceled and we choose the index K = N − 1.

4.2 Representation and uniqueness of solutions for the oblique derivative problem for (1.2) Now we give representation theorems of solutions for equation (1.2). Theorem 4.1 Suppose that equation (1.2) satisfies Condition C. Then any solution of Problem P

for (1.2) can be expressed as 

z

u(z) = 2 Re 0

w(z)dz + d1 , w(z) = w0 (z) + W (z),

(4.7)

where w0 (z) is a solution of Problem A for the equation


wz¯ wz∗




= 0 in

D+ D−

 ,

(4.8)

230

VI. Second Order Quasilinear Mixed Equations

with the boundary conditions (4.1) and (4.2)(w0 (z) = u0z ), and W (z) possesses the form ˜

φ(z) ˜ ˜ W (z) = w(z) − w0 (z), w(z) = Φ(z)e + ψ(z),   1 g(ζ) ˜ ˜ dσζ , ψ(z) φ(z) = φ˜0 (z) + T g = φ˜0 (z) − = T f in D+ , π D+ ζ − z

W (z) = Φ(z) + Ψ(z) in D− ,  ν ⎧ µ − ⎪ ⎪ g (z)dµe + g1 (z)dνe1 in D2j−1 , j = 1, 2, . . . , N, 2 2 ⎨ 0 2  ν Ψ(z) =  µ ⎪ − ⎪ g2 (z)dµe2 + g1 (z)dνe1 in D2j , j = 1, 2, . . . , N − 1, ⎩ 0

(4.9)

aj+1

˜ in which Im [φ(z)] = 0 on L0 = ∪N j=1 lj , lj = (aj , bj ), j = 1, . . . , N, e1 = (1 + i)/2, e2 = (1 − i)/2, µ = x + y, ν = x − y, φ˜0 (z) is an analytic function in D+ , and
¯ w(z)
= 0, A1 /2+A1 w/(2w), g(z) = f (z) = Re [A1 φ˜z ]+A2 u+A3 in D+ , 0, w(z) = 0, z ∈ D+ , g1 (z) = g2 (z) = Aξ + Bη + Cu + D, A=

Re A1 + Im A1 , 2

Re A1 − Im A1 , 2

B=

η = Re w − Im w,

ξ = Re w + Im w, C = A2 ,

D = A3 in D− ,

(4.10) ˜ where Φ(z) is analytic in D+ and Φ(z) is a solution of equation (4.8) in D− satisfying the boundary conditions ˜ φ(z) ˜ ˜ Re [λ(z)(Φ(z)e + ψ(z))] = r(z), ˜ φ(x)

˜ Re [λ(x)(Φ(x)e

z ∈ Γ,

˜ + ψ(x))] = s(x),

x ∈ L0 ,

Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], Re [λ(z)Φ(z)] = −Re [λ(z)Ψ(z)],

z ∈ L0 ,

(4.11)

z ∈ L
,

Im [λ(zj )Φ(zj )] = −Im [λ(zj )Ψ(zj )],

j = 1, . . . , N,

where λ(x) = 1+i on L0 , and s(x) is as stated in (4.14) below. Moreover, the solution u0 (z) of Problem P

for (4.8) in D− satisfies the estimate in the form Cβ [u0 (z), D] + Cβ [X(z)w(z), D] + C[w0± (µ, ν)Y ± (µ, ν), D− ] ≤ M23 (k1 + k2 ) (4.12) |γj |αj +δ |γj |αj +δ , Y ± (z) = Y ± (µ, ν) = Π2N , in which X(z) = Π2N j=1 [|z − tj | j=1 [|µ − tj ||ν − tj |] δ, β (0 < β < δ), γj (j = 1, . . . , 2N ) are as stated in (4.5), w0± (µ, ν) = Re w0 (z) ∓ Im w0 (z), w0 (z) = w0 (µ, ν), µ = x + y, ν = x − y, and  z u0 (z) = 2 Re w0 (z)dz + d1 (4.13) 0

4. Problems in Multiply Connected Domains

231

and M23 = M23 (p0 , β, k0 , D) is a non-negative constant. Proof Let u(z) be a solution of Problem P

for equation (1.2), and w(z) = uz , u(z) be substituted in the positions of w, u in (4.10), thus the functions g(z), f (z), ˜ ˜ ψ(z), φ(z) in D+ and g1 (z), g2 (z), Ψ(z) in D− in (4.9),(4.10) can be determined. ˜ Moreover, we can find the solution Φ(z) in D+ and Φ(z) in D− of (4.8) with the boundary conditions (4.11), where s(x) =

2r((1−i)x/2)−2Re [λ((1−i)x/2)Ψ((1−i)x/2)]−h(x) +Re [λ(x)Ψ(x)] (4.14) a((1 − i)x/2) − b((1 − i)x/2)

on L0 , in which   (1−i)x (1−i)x +b [Re (λ(z1 )(r(z1 )+ic1 ))+Im (λ(z1 )(r(z1 )+ic1 ))], h(x) = a 2 2 and

Ψ(x) =

⎧ ν  ν ⎪ ⎪ g (z)dνe = g1 ((1 − i)ν/2)dνe1 , z = x + iy = (1 − i)x ∈ L1 , ⎪ 1 1 ⎪ ⎪ 2 2 ⎪ ⎪ ⎪ µ  µ ⎪ ⎪ ⎪ ⎨ g2 (z)dµe2 = g2 ((1 + i)µ/2 + (1 − i)aj /2))dµe2 , 0

0

⎪ ⎪ ⎪ z = x + iy = (1 + i)x − aj i ∈ L2j−1 , j = 2, . . . , N, ⎪ ⎪ ⎪ ⎪ µ  µ ⎪ ⎪ ⎪ ⎪ g2 (z)dµe2 = g2 ((1+i)µ/2+1−i))dµe2 , z = (1+i)x−2i ∈ L2N . ⎩ 0

0

⎧ ˜ ˜ ˜ φ(z) ⎨ Φ(z) + ψ(z) in D+ ,

Thus w(z) = w0 (z) + W (z) =

⎩ w (z) + Φ(z) + Ψ(z) in D− , 0

is the solution of Problem A for the complex equation 
+
D wz = Re [A1 w] + A2 u + A3 in , w¯z∗ D−

(4.15)

and u(z) is a solution of Problem P for (1.2) as stated in the first formula in (4.7). Theorem 4.2 Suppose that equation (1.2) satisfies Condition C. Then Problem P

for (1.2) has at most one solution in D. Proof Let u1 (z), u2 (z) be any two solutions of Problem P

for (1.2). It is clear that u(z) = u1 (z) − u2 (z) and w(z) = uz satisfies the homogeneous equation and boundary conditions

+ D wz = Re [A1 w] + A2 u in , w¯z∗ D− (4.16) Re [λ(z)w(z)] = 0, z ∈ Γ, Re [λ(x)w(x)] = s(x), x ∈ L0 , u(0) = 0, Re [λ(z)w(z)] = 0,

z ∈ L
,

Im [λ(zj )w(zj )] = 0,

j = 1, . . . , N,

232

VI. Second Order Quasilinear Mixed Equations

in which s(x) =

−2Re [λ((1 − i)x/2)Ψ((1 − i)x/2)] + Re [λ(x)Ψ(x)] on L0 . a((1 − i)x/2) − b((1 − i)x/2)

(4.17)

From Theorem 4.1, the solution w(z) can be expressed in the form ⎧ ˜ ˜ ˜ = T f, φ(z) ˜ = φ˜0 (z)+ T˜g in D+ , ˜ φ(z) ⎪ Φe + ψ(z), ψ(z) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Φ(z) + Ψ(z) in D− , ⎪ ⎪ ⎪ ⎧ µ  ν ⎪ ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ [Aξ +Bη+Cu]edµe + [Aξ +Bη+Cu]dνe1 in D2j−1 , ⎪ ⎨ 2 ⎪ ⎪ 0 2 ⎪ ⎪ w(z) = (4.18) ⎪ ⎪ ⎪ ⎨ j = 1, 2, . . . , N, ⎪ ⎪ ⎪ ⎪  ν Ψ(z) =  µ ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ [Aξ +Bη+Cu]edµe + [Aξ +Bη+Cu]dνe1 in D2j , ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ 0 a ⎪ j+1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎩ j = 1, 2, . . . , N − 1, ˜ where g(z) is as stated in (4.10), Φ(z) in D+ is an analytic function and Φ(z) is a − solution of (4.8) in D satisfying the boundary condition (4.11), in which W (z) = ˜ ˜ w(z). If A2 = 0 in D+ , then ψ(z) = 0, besides the functions Φ(z), Φ(z) satisfy the boundary conditions ⎧ ˜ ⎨ Re [λ(x)Φ(x)] = s(x), x ∈ L0 , (4.19) ⎩ Re [λ(x)Φ(x)] = Re [λ(x)(W (x) − Ψ(x))], where s(x) is as stated before. Noting that C[u(z), D− ] ≤ M24 {C[X(z)w(z), D+ ] + C[w± (z)Y ± (z), D− ]} and applying the method of iteration, we can get |w± (z)Y ± (z)| ≤

[2N M25 M ((4 + M24 )m + 1)R
]n in D− , n!

where M24 = M24 (D), M25 = maxD− [|A|, |B|, |C|], M = 1 + 4k02 (1 + k02 ), m = C[w(z), D− ], R
= 2. Let n → ∞, we can derive w± (z) = 0, i.e. w(z) = w1 (z) − w2 (z) = 0, u(z) = 0, Ψ(z) = Φ(z) = 0 in D− , and s(x) = 0 on L0 . Besides, noting that the solution u(z) of the equation uzz¯ = Re [A1 uz ] + A2 u in D+ ,

(4.20)

1 ∂u = Re[λ(z)uz ] = 0, z ∈ Γ∪L0 , u(aj ) = 0, u(bj ) = 0, j = 1,...,N, (4.21) 2 ∂l and the index of the above boundary value problem is K = N − 1, on the basis of Theorem 3.7, Chapter III, we see that u(z) = 0 in D+ . This proves the uniqueness of solutions of Problem P

for (1.2) in D. As for the general equation (1.2), we can prove the uniqueness of solutions of Problem P

by the extremum principle for elliptic equations of second order by the method in the proof of Theorem 3.4, Chapter III.

4. Problems in Multiply Connected Domains 4.3

233

The solvability for the oblique derivative problem for (1.2)

First of all, we prove the existence of solutions of Problem P

for equation (3.7) in D. It is obvious that Problem P

for equation (3.7) is equivalent to the following boundary value problem (Problem A

) for (3.8) with the boundary conditions Re [λ(z)w(z)] = r(z) on L
,

Re [λ(z)w(z)] = r(z) on Γ, Im [λ(z)w(z)]|z=zj = cj ,

j = 1, . . . , N,

u(aj ) = dj , u(bj ) = dN +j ,

(4.22)

j = 1, . . . ,N,

and the relation  u(z) = 2 Re 0

z

w(z)dz + d1 .

(4.23)

Similarly to the method in the proof of Theorem 3.1, we can get the following theorem. Theorem 4.3

Problem P

for (3.7) in D has a unique continuous solution u(z).

Proof From the second and third boundary conditions in (4.22), we can obtain the following conditions: Re [λ(x)w(x)] = k(x) on L0 , k(x) =

2r((1 − i)x/2) − [a((1−i)x/2)+b((1−i)x/2)]h on L0 , a((1 − i)x/2) − b((1 − i)x/2)

(4.24)

h = [Re (λ(z1 )(r(z1 )+ic1 ))+Im (λ(z1 )(r(z1 )+ic1 ))]. According to the method in the proof of Theorem 2.2, Chapter III, we can find a solution w(z) of (3.8) in D+ with the first boundary condition in (4.22) and (4.24). Thus we can find the solution of Problem P

for (3.7) in D as stated in (4.23), and the solution w(z) of Problem A

for (3.8) in D1− possesses the form w(z) = w(z) ˜ + λ(z1 )[r(z1 ) − ic1 ], 1 w(z) ˜ = [(1 − i)f (x + y) + (1 + i)g(x − y)], 2 f (x + y) = Re [(1 + i)w(x + y)], g(x − y) =

2r((1 − i)(x − y)/2) in D1− \{0, b1 }. a((1−i)(x−y)/2) − b((1−i)(x−y)/2)

234

VI. Second Order Quasilinear Mixed Equations

Similarly we can write the solution of Problem P

in Dj− (j = 2, 3, . . . , 2N − 1) as w(z) = w(z)+λ(z ˜ j )[r(zj )−icj ], 1 w(z) ˜ = [(1−i)fj (x+y)+(1+i)gj (x−y)] in Dj− , 2

j = 2,...,N,

f2j (x+y)=

2r((1+i)(x+y)/2+(1−i)aj+1 /2) , a((1+i)(x+y)+(1−i)aj+1 /2)−b((1+i)(x+y)/2+(1−i)aj+1 /2)

g2j (x−y)=

2r((1−i)(x−y)/2) − in D2j , a((1−i)(x−y)/2)+b((1−i)(x−y)/2)

j =1,2,...,N −1,

f2j−1 (x+y) = Re[(1−i)w(x+y)], g2j−1 (x−y) =

2r((1−i)(x−y)/2) − in D2j−1 , a((1−i)(x−y)/2)−b((1−i)(x−y)/2)

j = 2,...,N, (4.25)

in which Dj− (j = 2, 3, . . . , 2N − 1) are as stated before. Theorem 4.4 Suppose that the mixed equation (1.2) satisfies Condition C. Then Problem P

for (1.2) in D has a solution. Proof It is clear that Problem P

for (1.2) is equivalent to Problem A

for the complex equation of first order and boundary conditions:

+ D wz = Re [A1 w] + A2 u + A3 in , (4.26) w¯z∗ D− Re [λ(z)w(z)] = r(z),

z ∈ Γ,

Im [λ(zj )w(zj )] = cj ,

u(aj ) = dj ,

Re [λ(z)w(z)] = r(z), u(bj ) = dN +j ,

z ∈ L
,

j = 1, . . . ,N,

(4.27)

and the relation (4.23). In order to find a solution w(z) of Problem A

for (4.26) in D, we express w(z) in the form (4.9)–(4.10) and use the successive iteration. First of all, denoting the solution w0 (z) of Problem A

for (4.26), and substituting w0 (z)(= ξ0 e1 + η0 e2 ) and the corresponding function u0 (z) into the positions of w(z) (= ξe1 + ηe2 ), u(z) in the right hand side of (4.26),(4.9) and (4.10), thus the corresponding functions g0 (z), f0 (z) and the functions ˜ 1 (z)eφ˜1 (z) + ψ˜1 (z), W1 (z) = w1 (z) − w0 (z), w1 (z) = Φ  g0 (ζ) 1 dσζ , ψ˜1 (z) = T f0 in D+ , φ˜1 (z) = φ˜0 (z) − π ζ −z + D w1 (z) = w0 (z) + W1 (z), W1 (z) = Φ1 (z) + Ψ1 (z), (4.28) g0 (z) = Aξ0 + Bη0 + Cu0 + D in D− , ⎧ µ  ν ⎪ − ⎪ g0 (z)dµe2 + g0 (z)dνe1 in D2j−1 , j = 1, 2, . . . , N, ⎨ 0µ 2ν   Ψ1 (z) = − ⎪ ⎪ g0 (z)dµe2 + g0 (z)dνe1 in D2j , j = 1, 2, . . . , N − 1, ⎩ 0

aj+1

4. Problems in Multiply Connected Domains

235

can be determined, where µ = x + y, ν = x − y, and the solution w0 (z) = u0z , u0 (z) ˜ 1 (z), Φ1 (z) of (4.8) satissatisfies the estimate (4.12). Moreover, we find a solution Φ fying the boundary conditions ˜ 1 (z)eφ˜1 (z) + ψ˜1 (z))] = r(z), z ∈ Γ, Re [λ(z)(Φ ˜ 1 (x)eφ˜1 (x) + ψ˜1 (x))] = s1 (x), Re [λ(x)(Φ

x ∈ L0 ,

Re [λ(x)Φ1 (x)] = Re [λ(x)(W1 (x) − Ψ1 (x))], Re [λ(z)Φ1 (z)] = −Re [λ(z)Ψ1 (z)],

z ∈ L0 ,

(4.29)

z ∈ L
,

Im [λ(zj )Φ1 (zj )] = −Im [λ(zj )Ψ1 (zj )],

j = 1, . . . , N,

where λ(x) = 1 + i on L0 and the function s1 (x) =

2r((1−i)x/2)−2Re [λ((1−i)x/2)Ψ1 ((1−i)x/2)]−h(x) a((1−i)x/2)−b((1−i)x/2)

+ Re [λ(x)Ψ1 (x)], 

 (1 − i)x (1 − i)x +b [Re (λ(z1 )(r(z1 ) + ic1 )) h(x) = a 2 2

(4.30)

+Im (λ(z1 )(r(z1 ) + ic1 ))] on L0 , in which w1 (z) satisfies the estimate Cβ [u1 (z), D] + Cβ [X(z)w1 (z), D+ ] + C[w1± (µ, ν)Y ± (µ, ν), D− ] ≤ M26 , w1± (µ, ν)

(4.31)

±

here = Re w1 (µ, ν) ∓ Im w1 (µ, ν), X(z), Y (µ, ν) is as stated in (4.12), M26 = M26 (p0 , β, k, D) is a non-negative constant. Thus we can obtain a sequence of functions: {wn (z)} and wn (z) = w0 (z) + Wn (z) = w0 (z) + Φn (z) + Ψn (z) in D− ,  ν ⎧ µ − ⎪ ⎪ g (z)dµe + gn−1 (z)dνe1 in D2j−1 , j = 1, 2, . . . , N, n−1 2 ⎨ 0 2  ν Ψn (z) =  µ ⎪ − ⎪ gn−1 (z)dµe2 + gn−1 (z)dνe1 in D2j , j = 1, 2, . . . , N − 1, ⎩ 0

aj+1

gn−1 (z) = Bξn−1 + Aηn−1 + Cun−1 + D in D− , (4.32) and then

|[w1± (µ, ν) − w0± (µ, ν)]Y ± (µ, ν)| ± ≤ |Φ± 1 (µ, ν)Y (µ, ν)| +

+|Y − (µ, ν)|[|

 2

√

2{|Y + (µ, ν)| |

ν

g0 (z)e1 dν| +

N −1  j=1

 |



µ

0

g0 (z)e2 dµ| (4.33)

µ

g0 (z)e1 dν|] aj+1

≤ 2N M27 M ((4 + M24 )m + 1)R
in D− ,

236

VI. Second Order Quasilinear Mixed Equations

where m = C[w0+ (µ, ν)X ± (µ, ν), D− ]+C[w0− (µ, ν)Y ± (µ, ν), D− ], M = 1+4k02 (1+k02 ), ˜ |B|, ˜ |C|, ˜ |D|). ˜ It is clear that wn (z) − wn−1 (z) satisfies R
= 2, M27 = maxz∈D− (|A|, wn (z)−wn−1 (z) = Φn (z)−Φn−1 (z)+Ψn (z)−Ψn−1 (z) in D− , ⎧ µ  ν ⎪ − ⎪ [g (z)−g (z)]dµe + [gn (z)−gn−1 (z)]dνe1 in D2j−1 , ⎪ n n−1 2 ⎪ ⎪ 0 2 ⎪ ⎪ ⎪ ⎪ ⎨ j = 1,2,...,N,  ν Ψn (z)−Ψn−1 (z)=  µ ⎪ − ⎪ [g (z)−g (z)]dµe + [gn (z)−gn−1 (z)]dνe1 in D2j , ⎪ n n−1 2 ⎪ ⎪ 0 aj+1 ⎪ ⎪ ⎪ ⎪ ⎩ j = 1,2,...,N −1, (4.34) where n = 1, 2, . . .. From the above equality, ± (µ, ν)]Y ± (µ, ν)| |[wn± (µ, ν) − wn−1



≤ [2N M27 M ((4 + M24 )m + 1)]n ≤

0

R


R
n−1 dR
(n − 1) !

(4.35)

[2N M27 M ((4 + M24 )m + 1)R
]n in D− n!

can be obtained, and we can see that the sequences of functions {wn± (µ, ν)Y ± (µ, ν)}, i.e. ± wn± (µ, ν)Y ± (µ, ν) = {w0± + [w1± − w0± ] + · · · + [wn± − wn−1 ]}Y ± (µ, ν)

(4.36)

(n = 1, 2, . . .) in D− uniformly converge to w∗± (µ, ν)X ± (µ, ν), and w∗ (z) = [w+ (µ, ν)+ w− (µ, ν) −i(w+ (µ, ν) − w− (µ, ν))]/2 satisfies the equality w∗ (z) = w0 (z)+Φ∗ (z)+Ψ∗ (z) in D− ,  ν ⎧ µ − ⎪ ⎪ g (z)dµe + g∗ (z)dνe1 in D2j−1 , j = 1, 2, . . . , N, ∗ 2 ⎪ ⎨ Ψ∗ (z) =

0

⎪ ⎪ ⎪ ⎩



0

2



µ

ν

g∗ (z)dµe2 + aj+1

(4.37) − g∗ (z)dνe1 in D2j ,

j = 1, 2, . . . , N − 1,

g∗ (z) = Bξ∗ + Aη∗ + Cu∗ + D in D− , and the corresponding function u∗ (z) is just a solution of Problem P

for equation (1.2) in the domain D− and w∗ (z) satisfies the estimate


C[w∗± (µ, ν)Y ± (µ, ν), D− ] ≤ M28 = e2N M27 M ((4+M24 )m+1)R .

(4.38)

4. Problems in Multiply Connected Domains

237

Besides we see that the solution w∗ (z) = uz of Problem A

for (4.26) and the cor¯ satisfy the estimate responding function u∗ (z) in D Cβ [u∗ (z), D] + Cβ [w∗ (z)X(z), D+ ] + C[w∗± (µ, ν)Y ± (µ, ν), D− ] ≤ M29 ,

(4.39)

where M29 = M29 (p0 , α, k, D) is a non-negative constant. Moreover the function u(z) in (4.23) is a solution of Problem P

for (1.2). From the proof of Theorem 4.4, we can obtain the estimates of any solution u(z) of Problem P

and the corresponding function w(z) = uz . Theorem 4.5 Suppose that equation (1.2) satisfies Condition C. Then any solution u(z) of Problem P

for (1.2) satisfies the estimates C˜ 1 [u(z), D] = Cβ [u(z), D] + Cβ [X(z)w(z), D+ ] +C[w± (µ, ν)Y ± (µ, ν), D− ] ≤ M30 , C˜β1 [u(z), D] ≤ M31 (k1 + k2 ),

(4.40)

where X(z) =

2N 

|z − tj ||γj |αj +δ , Y ± (z) = Y ± (µ, ν) =

j=1

2N  [|µ − tj ||ν − tj |]|γj |αj +δ ,

(4.41)

j=1

herein γj (j = 1, . . . , 2N ) are real constants in (4.5), β(0 < β < δ), δ are sufficiently small positive constants, and k = (k0 , k1 , k2 ), M30 = M30 (p0 , β, k, D), M31 = M31 (p0 , β, δ, k0 , D) are two non-negative constants. Next, we consider the oblique derivative problem (Problem Q

) for the equation (1.2). Theorem 4.6 Suppose that the mixed equation (1.2) with A2 = 0 satisfies Condition C. Then its Problem Q

has a solution in D. Proof It is clear that Problem Q

is equivalent to the following boundary value problem 
+
D wz¯ = Re [A1 w(z)] + A3 in , (4.42) w¯z∗ D− Re [λ(z)w(z)] = r(z),

z ∈ Γ,

Re [λ(z)w(z)] = r(z),

Im [λ(zj
)w(zj
)] = d
j ,

z ∈ L
,

(4.43)

j = 1, . . . , N.

Noting that w(z) satisfies the second boundary condition in (4.11), namely Re [λ(x)w(z)] = s(x), s(x) = Re [λ(x)Ψ(x)]

(4.44)

2r((1−i)x/2)−2Re [λ((1− i)x/2)Ψ((1−i)x/2)]−h(x) on L0 , + a((1−i)x/2)−b((1−i)x/2)

238

VI. Second Order Quasilinear Mixed Equations

and the index K = N − 1, similarly to the proof of Theorems 2.3 and 2.4, Chapter V, we can find a unique solution w(z) of the boundary value problem (4.42)–(4.44), and the function u(z) in (4.23) is just a solution of Problem Q

in D. Finally, we mention that the above result includes the Dirichlet problem (Tricomi ˜j | = Rj }, problem) as a special case. In fact, if Γ1 = {|z − 1| = 1}, Γj = {|z − a a ˜j = bj−1 +(aj − bj−1 )/2, Rj = (aj − bj−1 )/2, j = 2, . . . , N, R1 = 1, the boundary condition of the Dirichlet problem is u(z) = φ(x) on Γ ∪ L
,

(4.45)

which can be rewritten as Re [λ(z)w(z)] = r(z) on Γ,

Im [λ(zj )w(zj )] = cj , j = 1, . . . , N,  z w(z)dz + d0 in D, Re [λ(z)w(z)] = r(z) on L
, u(z) = 2Re

(4.46)

0

in which d0 = φ(0), ⎧ ⎪ i(z − 1), θ = arg(z − 1) on Γ1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ i(z−˜ aj )/Rj , θ = arg(z−˜ aj ) on Γj , √ λ(z) = a + ib = ⎪ (1 + i)/ 2 on L1 , ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (1 − i)/√2 on L

= ∪N L , j=2

j = 2,. . .,N, (4.47)

2j−1

⎧ ⎪ ⎨ φθ /Rj on Γj , j = 1, . . . , N, r(z) = φx
N ⎪ ⎩ √ on L = ∪j=1 L2j−1 , 2  (4.48) φx − φx 1+i c1 = Im √ uz¯(z1 − 0) = √ |z=z1 −0 = 0, 2 2  1−i cj = Im √ uz¯(zj + 0) = 0, j = 2, . . . , N, 2 √ √ √ √ in which a = 1/ 2
= b = −1/ 2 on L1 and a = 1/ 2
= −b = −1/ 2 on L

.

and

If the index of Problem D on ∂D+ is K = N/2 − 1, we can argue as follows: According to (4.46),(4.47), the boundary conditions of Problem D in D+ possess the form Re [i(z − 1)w(z)] = r(z) = φθ , z ∈ Γ,  φ
((1 − i)x/2) 1−i √ , Re √ w(x) = s(x) = 2 2 x ∈ lj = (aj , bj ),

j = 1, . . . , N.

4. Problems in Multiply Connected Domains

239

It is clear that the possible points of discontinuity of λ(z) on ∂D+ are t1 = a1 = 0, t2 = b1 , t3 = a2 , t4 = b2 , . . . , t2N −1 = aN , t2N = bN = 2, and λ(a1 − 0) = λ(bj + 0) = eiπ/2 , 3iπ/2

λ(bN + 0) = λ(aj − 0) = e λ(aj + 0) = λ(bj − 0) = e

iπ/4

λ(t1 − 0) = eiπ/4 = eiφ1 , λ(t1 + 0)

,

j = 1, 2, . . . , N − 1, ,

j = 2, . . . , N, j = 1, . . . , N,

0 < γ1 =

1 φ1 1 − K1 = − 0 = < 1, π 4 4

λ(t2N − 0) φ2N 5 1 = e−5iπ/4 = eiφ2N , −1 < γ2N = −K2N = − −(−1) = − < 0, λ(t2N + 0) π 4 4 1 φj 5 λ(tj − 0) = e5iπ/4 = eiφj , 0 < γj = − Kj = −1 = < 1, j = 3, 5, . . . , 2N −1, λ(tj + 0) π 4 4 λ(tj − 0) φj 1 1 −iπ/4 iφj = e , −1 < γj = − Kj = − = − < 0, j = 2, 4, . . . , 2N −2. =e λ(tj + 0) π 4 4 Thus K1 = K2 = K4 = · · · = K2N −2 = 0, K3 = K5 = · · · = K2N −1 = 1, K2N = −1, in the case, where the index of Problem D on ∂D+ is 1 N − 1. K = (K1 + K2 + · · · + K2N ) = 2 2 Hence Problem D for (1.2) has a unique continuous solution u(z) in D. If we require that the derivative w(z) = uz of the solution u(z) is bounded in D\Z, it suffices to replace K2 = K4 = · · · = K2N −2 = −1, K2N = −2 and then the index K = −1. In this case the problem has one solvability condition. The oblique derivative problems for the Chaplygin equation of second order K(y)uxx + uyy = 0,

K(0) = 0,

K
(y) > 0 in D,

where D is a multiply connected domain, was proposed by L. Bers in [9]2), but the problem was still not solved. For more general second order quasilinear degenerate equations of mixed type and second order mixed equations in higher dimensional domains, many boundary value problems need to be investigated and solved. The references for this chapter are [4],[6],[12],[14],[27],[31],[37],[39],[45],[56],[61], [64],[71],[72],[73],[85],[88],[90],[98].

.

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Index algebraic equation 53, 54, 87, 137, 138, 176, 206, 207, 208 algebraic theory 87 analytic function 79–93, 99, 125, 128, 130, 132, 133, 149, 150, 151, 153, 155, 165–167, 174, 184, 190, 202, 212, 223, 228, 230, 232 a priori estimate 25, 28, 42, 43, 50, 109, 119, 157, 177, 182, 186, 200, 227 auxiliary function 100, 111, 112, 114, 116, 199 A.V. Bitsadze 157

Banach space 53, 55, 93, 104, 137, 138, 191, 193, 207, 208 boundary condition 9, 18–24, 27, 29, 30, 32, 34, 38, 43–46, 48, 49, 53, 55–59, 60–65, 67, 68, 70, 72, 73, 78, 80–82, 84, 86, 88, 89, 91–100, 101, 103, 105–107, 109, 111, 112, 115, 116, 120, 121, 122, 125, 127–130, 132, 133, 135, 137, 139, 140, 141–144, 146–148, 149, 150, 151, 153–155, 158, 159, 160, 161, 163–169, 172–175, 177–185, 187, 189, 190, 192, 193, 195–199, 201–203, 205, 207, 209–215, 217–220, 223, 224–226, 228–235, 237, 238 boundary value problem 1, 18, 24–25, 31, 32, 34, 35, 38, 39, 42–44, 56, 58–60, 65, 67, 72, 73, 78, 84, 90, 97, 102, 103, 107, 108, 119, 126, 133, 134, 139, 141–143, 155, 157, 158, 160, 165, 171, 175, 177, 178, 182, 187, 190, 193, 194, 195, 200, 201,

217, 219, 220, 227, 228, 232, 237, 238, 239 boundedness 28, 108, 109, 125

¯ 26, 29, 42 Cα (D)-estimate ¯ 50 C 1 (D)-estimate ¯ 51, 204 Cα1 (D)-estimate Cauchy problem 73 Cauchy–Riemann system 3 Cauchy sequence 104, 191, 193 Cauchy theorem 186 Chaplygin equation 66, 68, 72, 118, 239 characteristic coefficient 6 characteristic lines 45, 66, 71, 177, 202, 224, 229 characteristics 68, 72, 74, 78 closed and convex set 53, 55, 93, 104, 137, 138, 207, 208 coefficients matrix 176 compact subset 94, 104 completely continuous operator 176 completeness 104, 191, 193 complete system of linearly independent solutions 87, 175 complex equation 1, 3, 8, 9, 12, 13, 16–24, 27, 29, 32, 34, 36–38, 45, 47–49, 53, 62–64, 90, 91, 94, 95, 120, 121, 127–129, 132, 135, 137, 143, 146, 148, 150, 154, 165, 167, 168, 175, 176, 178, 184, 187, 189, 192, 197, 200, 206, 218, 221, 222, 234 complex equation of mixed (elliptic-hyperbolic) type 119, 126, 129, 134, 136, 156, 159

Index Condition C 21, 22, 25–28, 30, 32, 34, 42, 47–53, 56, 58, 62–64, 67, 90, 91, 94–101, 104, 106–112, 116, 118, 127–129, 132–136, 139–143, 146, 148, 150, 154, 155, 156, 163–165, 167, 168, 170, 171, 182, 183, 185, 186, 189, 192, 195, 201, 202, 204, 206, 209, 212, 215, 218, 222, 224, 225, 227, 228, 229, 231, 234, 237 Condition C
53, 163, 173, 176, 195, 196, 198, 206 condition of hyperbolic type 10, 11, 13, 14, 39, 40 condition of uniformly hyperbolic type 41 conformal mapping 86, 92, 100, 124, 157, 177, 209 conjugate harmonic function 185, 188 continuity method 92 continuous function 70, 172, 188 continuously differentiable 2–4, 36, 74, 108, 109, 159, 164, 173, 196, 202, 210, 219, 226 continuously differentiable function 8, 42, 105, 121, 122, 201 continuously differentiable solution 3, 43, 59, 96, 158, 163, 171, 177, 195, 201, 220, 229 continuous mapping 105 continuous solution 37, 107, 120, 139, 144, 162, 219, 228 convergency 3 convergent domain 3 Darboux problem 59, 65 degenerate elliptic equation 108, 118, 199 degenerate equation of mixed type 194, 196, 199 degenerate hyperbolic equation 39, 66, 73 degenerate mixed equation 194 diagonal sequence 106

251 Dirichlet (boundary value) problem 18, 44, 59, 67, 97, 105, 157, 159, 160, 196, 202, 209, 210, 227, 229, 238 discontinuities of first kind 79, 83, 143, 171, 218 discontinuity 186 discontinuous boundary value problem 79 discontinuous oblique derivative (boundary value) problem 95, 96, 103, 106, 107, 171, 200, 218, 219, 224 discontinuous Poincar´e (boundary value) problem 157, 171, 172 discontinuous point 84, 97, 125, 161 discontinuous Riemann–Hilbert problem 79, 80, 90, 91, 93, 94, 125, 143, 144, 145, 146, 156 discrete eigenvalue 176 divisor of zero 2 doubly connected domain 227 eigenvalue 176, 177 elementary function 124 elliptic (complex) equation 3, 10, 96, 98, 103, 106, 108, 111, 116, 232 elliptic domain 118, 181, 199 equation of mixed type 89, 90, 139, 140, 158, 171, 200, 204, 209, 214, 218, 227, 228 estimates of functions 52, 205 estimates of solutions 28, 108, 110, 111, 136, 224, 235, 237 existence and uniqueness of solutions 27, 48, 66, 150, 202 existence of solutions 25, 39, 52, 55, 79, 87, 90, 93, 103, 106, 108, 119, 122, 126, 134, 143, 157, 177, 182, 193, 198, 213, 218, 226, 233 expression of curve 216 expression of solutions 48, 50, 62 expression theorem 46 extremum principle 95, 232

252 (F, G)-derivative 5 (F, G)-hyperbolic pseudoregular complex function 7 finite kind of discontinuous points 84 Frankl (boundary value) problem 177, 179, 180, 182–184, 186, 188, 189, 191, 193, 194 Fredholm theorem 171, 176 gas dynamics 177 general boundary value problem 90 general Chaplygin equation 66 general discontinuous boundary value problem 79, 84, 90 general discontinuous Riemann–Hilbert problem 84, 86, 144 general domain 31, 34, 50, 55, 58, 64, 78, 94, 106, 119, 138–140, 157, 209, 214, 218 general hyperbolic equation 43, 52, 65 general Lavrent
ev–Bitsadze equation 157, 171, 177 general nonlinear hyperbolic system 14 general quasilinear equation of mixed type 200, 206, 208 general quasilinear hyperbolic equation 50, 52, 64 general quasilinear mixed equation 136 general solution 24, 46, 82, 83, 85, 87, 93, 121, 122, 156, 175, 176 generating pair 6, 8, 9 Green’s formula 4 harmonic function 185, 188 higher dimensional domain 238 H¨older continuous condition 29, 51, 52, 205 H¨older continuous estimate 51, 204 H¨older continuous function 88 H¨older estimate 134, 136 H¨older function 88 homeomorphic (solution) 35–37, 92

Index homeomorphism 37, 91–93, 99 homogeneous boundary condition 81, 82, 86, 101, 105 homogeneous boundary value problem 86, 134, 178, 198 homogeneous (complex) equations 21, 27, 49, 97, 101, 132, 154, 187, 195 homogeneous equation 98, 99, 185, 198, 213, 225, 231 homogeneous Frankl problem 186 homogeneous integral equation 176, 177 homogeneous problem 175, 198 homogeneous system 87 hyperbolic 10, 17 hyperbolic (complex) equation 1, 14, 28, 37, 39, 55, 63 hyperbolic complex functions 1–3, 38 hyperbolic constant 37 hyperbolic continuous function 4 hyperbolic continuously differentiable function 35 hyperbolic domain 66, 68, 157, 198, 212 hyperbolic element 1 hyperbolic equation 28, 39, 42, 43, 46–48, 50, 51, 61, 62, 65, 78 hyperbolic harmonic complex function 38 hyperbolicity condition 15 hyperbolic mapping 1, 35, 37 hyperbolic model 2 hyperbolic number 1, 2, 26, 38, 42, 59 hyperbolic pseudoregular functions 1, 5, 7, 8 hyperbolic regular functions 1, 3, 5, 36–38 hyperbolic system 1, 10, 14, 35, 73 hyperbolic unit 1 implicit function 14 index 80, 81, 84, 88, 90, 93, 95, 97, 100, 103, 107, 120, 125, 126, 139, 144, 145, 156, 158, 160–164, 180, 181,

Index 186, 196, 202, 228, 229, 232, 237, 238, 239 integral equation 93, 171, 176 integral expression 51, 204 integral formula 159 integral of Cauchy type 82 integral path 45, 65, 159, 202 inverse function 31, 32, 36, 55, 57, 59, 91, 93, 99, 140, 141, 216 inverse mapping 124 inverse transformation 31–33, 55–57, 140, 141, 216 inversion 97 Keldych–Sedov formula 79, 84, 90 Lavrent
ev–Bitsadze equation 157, 227 L. Bers 239 linear and nonlinear hyperbolic complex equation 10, 20, 22, 39 linear and quasilinear hyperbolic complex equation 39 linear and quasilinear hyperbolic complex system 9 linear complex equation 29, 134 linear complex equation of mixed type 126 linear degenerate mixed equation 194 linear elliptic equation 103 linear equation 42, 95, 99, 104, 106, 135, 200, 226 linear equation of mixed type 157, 162, 171, 172 linear homogeneous equation 99 linear hyperbolic complex equation 18, 25, 29, 39, 41, 43, 47 linear hyperbolic system 10 linear mixed equation 162, 176 linear system of mixed type 126 mathematical model 177 maximum 100, 101, 110, 118, 198, 213, 226 maximum point 101 maximum principle 100, 211, 213, 226 mechanics 79

253 method of integral equation 171, 177, 209 method of iteration 168, 193, 232 method of parameter extension 177, 180, 189, 192, 214, 227 method of successive iteration 66, 72, 143 minimum 55, 118, 198 mixed boundary value problem 79, 84 mixed (complex) equation 127, 129, 137, 141, 142, 162, 165, 167, 168, 171, 175, 176, 177, 184, 189, 192, 212, 218, 224, 237, 239 mixed equation with parabolic degeneracy 177 mixed system 119 monotonous continuous function 37 multiply connected domain 200, 227, 238, 239 negative minimum 110, 111 Neumann boundary value problem 109 non-degenerate, mutually disjointed arcs 172 nonhomogeneous integral equation 176 nonlinear boundary condition 82 nonlinear complex equation 94 nonlinear elliptic (complex) equation 79, 90, 108 nonlinear equation 54 nonlinear hyperbolic equation 40 nonlinear hyperbolic system 14 nonlinear mechanics 90, 94 nonlinear uniformly elliptic system 90 non-singular transformation 37, 41 non-trivial solution 82, 86, 87, 118 oblique derivative (boundary value) problem 39, 43, 44, 50, 59, 66, 67, 69, 73, 95, 108, 109, 139, 157, 158, 160, 162–164, 168, 173, 194–196, 198–202, 206, 209, 211, 214, 215, 219, 224, 225, 228, 233, 237, 239

254 parameter equation 31, 33, 55, 57, 139, 214 partial differential equation 5, 10, 14, 15, 39 piecewise smooth curve 4 physics 79 point condition 86, 97, 107, 164, 176, 202, 211, 212, 220, 224, 229 positive maximum 110, 113, 115, 118 principle of compactness 108 principle of contracting mapping 92, 93 principle of extremum 111 Problem A 18–23, 26–30, 32, 34, 45, 46–48, 80, 83, 88, 89, 91–95, 120–122, 126, 127, 129, 132–138, 140, 141, 159, 165–168, 170, 173, 184, 189, 192, 197, 202, 203, 204, 229, 231, 233 Problem A0 18, 21, 80, 82, 120, 121 Problem A1 24, 25 Problem A2 62 Problem A4 64 Problem A
94, 95, 139–142, 219–222 Problem A
0 95, 139 Problem A

233, 234, 237 Problem A∗ 144–150, 153–156 Problem A∗0 81, 91, 95, 145 Problem A∗ 143, 144 Problem B 84–86, 175 Problem B0 86 Problem B1 175 Problem B2 175 Problem B
86, 87 Problem B0
86, 87 Problem C 86, 88, 90 Problem C0 88 Problem C
136, 137, 139, 171 Problem D 18, 44, 67, 68, 97–99, 109, 110, 111, 158, 159, 160, 161, 162, 196, 238, 239 Problem F 178, 183, 191, 193 Problem F0 178, 189, 190, 192 Problem F1 194

Index Problem Ft 189, 190, 192, 194 Problem Ft0 189, 190, 192, 193 Problem N 109 Problem O 109 Problem P 43–56, 58, 64, 68, 96, 97, 99, 101–106, 109–111, 116, 118, 158–161, 163, 165–168, 170, 171, 195–198, 201–204, 205, 207–215, 218, 229, 230, 231 Problem P0 44, 67, 97, 109, 118, 158, 196, 197, 202 Problem P1 59, 64–72, 78, 104 Problem P2 60–63, 64, 65, 78 Problem P3 60–63, 64, 78 Problem P4 60, 63, 64, 73, 74, 78 Problem Pt 103, 104 Problem Pt0 103 Problem P˜ 209 Problem P
107, 204, 215, 218, 219, 225 Problem P0
107, 220, 223–225 Problem P

228–229, 231–234, 236, 237 Problem P0

228 Problem Q 97, 106, 171–177, 202, 206, 211, 212 Problem Q0 172, 173 ˜ 209 Problem Q Problem Q
107, 108, 220–226 Problem Q

229, 237 Problem T 158 Problem T2 227 Process of iteration 50 quasi-hyperbolic mapping 1, 35, 37 quasilinear (complex) equation 27, 30, 34, 135, 138, 139, 144 quasilinear (complex) equation of mixed type 79, 134, 143, 199, 200 209, 215, 218, 227, 228 quasilinear degenerate equation of mixed type 239 quasilinear elliptic equation 95 quasilinear equation 104–106, 214

Index quasilinear hyperbolic equation 31, 42, 43, 47, 59–61, 63 quasilinear hyperbolic system 15 quasilinear mixed (complex) equation 136, 143, 209 quasilinear mixed system 134 quasilinear uniformly elliptic equation 96, 107 rank 176, 177 reductio ad absurdum 187 reflected domain 180 regularity of solution 134, 156, 160, 171, 206, 225 representation 20, 25, 30, 35, 36, 38, 43, 46–48, 63, 64, 79, 87, 90–92, 94, 134, 160, 182, 198, 229 representation of solutions 18, 20, 25, 48, 61, 63, 64, 69, 79, 84, 90, 95, 146, 171, 200, 220, 227 representation theorem 36, 43, 59, 66, 91, 97, 126, 134, 135, 143, 164, 173, 196, 211, 218, 222 Riemann–Hilbert (boundary value) problem 18, 20, 25, 28, 45, 119–122, 125, 126, 127, 129, 134, 135, 138, 141, 150, 159 removable singular point 185 Schauder fixed-point theorem 94, 105 Schwarz formula 82 sequence 28, 106, 116 sequence of coefficients 102 sequence of functions 23, 28, 30, 49, 50, 52, 54, 75, 104, 131, 132, 138, 152, 153, 169, 170, 205, 207, 208, 235, 236 sequence of solutions 53, 103, 105, 111, 116, 137, 190, 193 series expansion 3 simplest complex equation of mixed type 119, 121, 122 simplest equation of mixed type 157, 160

255 simplest hyperbolic (complex) equation 18, 45, 46 simplest hyperbolic system 3, 35 simplest mixed equation 157 simply connected domain 18, 25, 42, 43, 66, 106, 108, 119, 126, 134, 139, 141, 143, 157, 194, , 200, 208, 218 singularity 162 solvability 22, 25, 27, 43, 78, 83, 91, 94, 104, 119, 157, 160, 168, 173, 175, 177, 180, 182, 189, 191, 194, 200, 206, 209, 211, 213, 231 solvability condition 83, 89, 94, 126, 156, 160, 162, 171, 176, 177, 239 subsequence 102, 105, 106, 111, 116, 132, 154, 170 successive iteration 18, 22, 25, 27, 28, 30, 39, 48, 50–53, 75, 103, 126, 129, 133, 134, 137, 150, 155, 168, 170, 190, 193, 204, 205, 207, 208, 218, 234 symmetry extension 180 system (of equations) 32, 35, 37, 69, 71, 72, 140 system of integral equations 72, 74, 77, 93 third boundary value problem 109 transformation 21, 31–34, 37, 44, 55–58, 60, 67, 80, 88, 97, 111, 120, 140, 141, 142, 158, 196, 202, 216, 217 triangle inequality 2 Tricomi problem 67, 90, 157, 202, 209, 210, 213, 227, 229, 236 uniformly bounded 115 uniformly converge 23, 49, 52, 54, 77, 98, 102, 106, 111, 131, 132, 138, 153, 154, 170, 205, 208, 236 uniformly elliptic system 90 uniformly elliptic equation 110 uniformly hyperbolic 10, 40 uniformly hyperbolic system 36

256 unique continuous solution 162, 221, 233, 239 uniqueness and existence of solutions 18, 25, 59, 72, 95, 119, 157, 162, 200, 209, 227 uniqueness of solutions 20, 22, 25, 87, 118, 121, 122, 126, 127, 133–134, 143, 150, 154, 162, 164, 168, 177, 184, 186, 187, 194, 196, 202, 206, 213, 225, 229, 233 uniqueness theorem 39, 119, 232 unique solution 25, 28, 32, 34, 43, 50, 53, 56, 58, 63, 64, 65, 72, 78, 86, 93, 103, 104, 106, 116, 125, 133, 135, 137, 141, 142, 144, 147, 150, 155, 159, 162, 165, 189, 192, 204, 207, 208, 212, 217, 224, 227, 238

Index unique solvability 59, 61, 63, 64, 68, 73, 129, 206, 211, 218, 224 unique system 87 unit disk 84, 86, 92, 99 univalent analytic function 92, 93 univalent continuous 35 univalent mapping 35 upper half-plane 79, 84, 90, 94, 124 upper half-(unit) disk 79, 84, 86, 88, 90, 93, 96, 103

weakly converge 102, 106 well posed version 86

zone domain 90, 94