The Levy Laplacian

CAMBRIDGE TRACTS IN MATHEMATICS General Editors b. bollobas, w. fulton, a. katok, f. kirwan, p. sarnak, b. simon, b. totaro

166 The L´evy Laplacian The L´evy Laplacian is an infinite-dimensional generalization of the well-known classical Laplacian. Its theory has been increasingly well-developed in recent years and this book is the first systematic treatment of it. The book describes the infinite-dimensional analogues of finite-dimensional results, and more especially those features that appear only in the generalized context. It develops a theory of operators generated by the L´evy Laplacian and the symmetrized L´evy Laplacian, as well as a theory of linear and nonlinear equations involving it. There are many problems leading to equations with L´evy Laplacians and to L´evy–Laplace operators, for example superconductivity theory, the theory of control systems, the Gauss random field theory, and the Yang–Mills equation. The book is complemented by exhaustive bibliographic notes and references. The result is a work that will be valued by those working in functional analysis, partial differential equations and probability theory.

Cambridge Tracts in Mathematics All the titles listed below can be obtained from good booksellers or from Cambridge University Press. For a complete series listing visit http://publishing.cambridge.org/stm/mathematics/ctm/ 142. Harmonic Maps between Rienmannian Polyhedra. By J. Eells and B. Fuglede 143. Analysis on Fractals. By J. Kigami 144. Torsors and Rational Points. By A. Skorobogatov 145. Isoperimetric Inequalities. By I. Chavel 146. Restricted Orbit Equivalence for Actions of Discrete Amenable Groups. By J. W. Kammeyer and D. J. Rudolph 147. Floer Homology Groups in Yang–Mills Theory. By S. K. Donaldson 148. Graph Directed Markov Systems. By D. Mauldin and M. Urbanski 149. Cohomology of Vector Bundles and Syzygies. By J. Weyman 150. Harmonic Maps, Conservation Laws and Moving Frames. By F. H´elein 151. Frobenius Manifolds and Moduli Spaces for Singularities. By C. Hertling 152. Permutation Group Algorithms. By A. Seress 153. Abelian Varieties, Theta Functions and the Fourier Transform. By Alexander Polishchuk 156. Harmonic Mappings in the Plane. By Peter Duren 157. Affine Hecke Algebras and Orthogonal Polynomials. By I. G. MacDonald 158. Quasi-Frobenius Rings. By W. K. Nicholson and M. F. Yousif 159. The Geometry of Total Curvature on Complete Open Surfaces. By Katsuhiro Shiohama, Takashi Shioya and Minoru Tanaka 160. Approximation by Algebraic Numbers. By Yann Bugeaud 161. Equivalence and Duality for Module Categories with Tilting and Cotilting for Rings. By R. R. Colby and K. R. Fuller 162. L´evy Processes in Lie Groups. By Ming Liao 163. Linear and Projective Representations of Symmetric Groups. By A. Kleshchev

The L´evy Laplacian M. N. FELLER

cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge cb2 2ru, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521846226 © CM. N. Feller 2005 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format isbn-13 isbn-10

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Contents

Introduction 1 1.1 1.2 1.3 2 2.1 2.2 2.3 3 3.1 3.2 3.3 4 4.1 4.2 4.3

page 1

The L´evy Laplacian Definition of the infinite-dimensional Laplacian Examples of Laplacians for functions on infinitedimensional spaces Gaussian measures

05 05

L´evy–Laplace operators Infinite-dimensional orthogonal polynomials The second-order differential operators generated by the L´evy Laplacian Differential operators of arbitrary order generated by the L´evy Laplacian

22 23

Symmetric L´evy–Laplace operator The symmetrized L´evy Laplacian on functions from the domain of definition of the L´evy–Laplace operator The L´evy Laplacian on functions from the domain of definition of the symmetrized L´evy–Laplace operator Self-adjointness of the non-symmetrized L´evy–Laplace operator Harmonic functions of infinitely many variables Arbitrary second-order derivatives Orthogonal and stochastically independent second-order derivatives Translationally non-positive case

v

09 13

30 33 40 40 44 48 53 54 59 64

vi

Contents

5 5.1

Linear elliptic and parabolic equations with L´evy Laplacians The Dirichlet problem for the L´evy–Laplace and L´evy–Poisson equations The Dirichlet problem for the L´evy–Schr¨odinger stationary equation The Riquier problem for the equation with iterated L´evy Laplacians The Cauchy problem for the heat equation

5.2 5.3 5.4 6 6.1 6.2 6.3 6.4 6.5 7 7.1 7.2 7.3 7.4

Quasilinear and nonlinear elliptic equations with L´evy Laplacians The Dirichlet problem for the equation
L U (x) = f (U (x)) The Dirichlet problem for the equation f (U (x),
L U (x)) = F(x) The Riquier problem for the equation
2L U (x) = f (U (x)) The Riquier problem for the equation f (U (x),
2L U (x)) =
L U (x) The Riquier problem for the equation f (U (x),
L U (x),
2L U (x)) = 0 Nonlinear parabolic equations with L´evy Laplacians The Cauchy problem for the equations ∂U (t, x)/∂t = f (
L U (t, x)) and ∂U (t, x)/∂t = f (t,
L U (t, x)) The Cauchy problem for the equation ∂U (t, x)/∂t = f (U (t, x),
L U (t, x)) The Cauchy problem for the equation ϕ(t, ∂U (t, x)/∂t) = f (F(x),
L U (t, x)) The Cauchy problem for the equation f (U (t, x), ∂U (t, x)/∂t,
L U (t, x)) = 0

Appendix. L´evy–Dirichlet forms and associated Markov processes A.1 The Dirichlet forms associated with the L´evy–Laplace operator A.2 The stochastic processes associated with the L´evy–Dirichlet forms Bibliographic notes References Index

68 68 84 86 88 92 92 94 96 99 103 108 108 115 121 126 133 133 137 142 144 152

Introduction

The Laplacian acting on functions of finitely many variables appeared in the works of Pierre Laplace (1749–1827) in 1782. After nearly a century and a half, the infinite-dimensional Laplacian was defined. In 1922 Paul L´evy (1886–1971) introduced the Laplacian for functions defined on infinite-dimensional spaces. The infinite-dimensional analysis inspired by the book of L´evy Lec¸ons d’analyse fonctionnelle [93] attracted the attention of many mathematicians. This attention was stimulated by the very interesting properties of the L´evy Laplacian (which often do not have finite-dimensional analogues) and its various applications. In a work [68] (published posthumously in 1919) Gˆateaux gave the definition of the mean value of the functional over a Hilbert sphere, obtained the formula for computation of the mean value for the integral functionals and formulated and solved (without explicit definition of the Laplacian) the Dirichlet problem for a sphere in a Hilbert space of functions. In this work he called harmonic those functionals which coincide with their mean values. In a note written in 1919 [92], which complements the work of Gˆateaux, L´evy gave the explicit definition of the Laplacian and described some of its characteristic properties for the functions defined on a Hilbert function space. In 1922, in his book [93] and in another publication [94] L´evy gave the definition of the Laplacian for functions defined on infinite-dimensional spaces and described its specific features. Moreover he developed the theory of mean values and using the mean value over the Hilbert sphere, solved the Dirichlet problem for Laplace and Poisson equations for domains in a space of sequences and in a space of functions, obtained the general solution of a quasilinear equation. We have mentioned here only a few of a great number of results given in L´evy’s book which is the classical work on infinite-dimensional analysis. The second half of the twentieth century and the beginning of twenty-first century follows a period of development of a number of trends originated 1

2

Introduction

in [93], and the infinite-dimensional Laplacian has become an object of systematic study. This was promoted by the appearance of its second edition Probl`emes concrets d’analyse fonctionnelle [95] in 1951 and the appearance, largely due to the initiative of Polishchuk, of its Russian translation (edited by Shilov) in 1967. During this period, there were published, among others, the works of: L´evy [96], Polishchuk [111–125], Feller [36–66], Shilov [132–135], Nemirovsky and Shilov [102], Nemirovsky [100, 101], Dorfman [28–33], Sikiryavyi [137–145], Averbukh, Smolyanov and Fomin [10], Kalinin [82], Sokolovsky [146–151], Bogdansky [13–22], Bogdansky and Dalecky [23], Naroditsky [99], Hida [75–78], Hida and Saito [79], Hida, Kuo, Potthoff and Streit [80], Yadrenko [158], Hasegawa [72–74], Kubo and Takenaka [85], Gromov and Milman [69], Milman [97, 98], Kuo [86–88], Kuo, Obata and Saito [89, 90], Saito [126–129], Saito and Tsoi [130], Obata [103–106], Accardi, Gibilisco and Volovich [4], Accardi, Roselli and Smolyanov [5], Accardi and Smolyanov [6], Accardi and Bogachev [1–3], Zhang [159], Koshkin [83, 84], Scarlatti [131], Arnaudon, Belopolskaya and Paycha [9], Chung, Ji and Saito [26], L´eandre and Volovich [91], Albeverio, Belopolskaya and Feller [8]. Many problems of modern science lead to equations with L´evy Laplacians and L´evy–Laplace type operators. They appear, for example, in superconductivity theory [24, 71, 152, 155], the theory of control systems [121, 122], Gauss random field theory [158] and the theory of gauge fields (the Yang–Mills equation) [4], [91]. L´evy introduced the infinite-dimensional Laplacian acting on a function U (x) by the formula M(x0 ,) U (x) − U (x0 ) →0 2


L U (x0 ) = 2 lim

(the L´evy Laplacian), where M(x0 ,) U (x) is the mean value of the function U (x) over the Hilbert sphere of radius  with centre at the point x0 . Given a function defined on the space of a countable number of variables we have n 1
∂ 2U
L U (x1 , . . . , xn , . . .) = lim , 2 n→∞ n k=1 ∂ x k while for functions defined on a functional space we have 1
L U (x(t)) = b−a

b a

δ 2 U (x) ds, δx(s)2

where δ 2 U (x)/δx(s)2 is the second-order variational derivative of U (x(t)).

Introduction

3

But already, in 1914, Volterra [154] had used different second-order differential expressions such as b
0 V (x(t)) = a

δ 2 V (x) ds δx(s)δx(s)

(the Volterra Laplacian), where δ 2 V (x)/δx(s)δx(τ ) is the second mixed variational derivative of V (x(t)). In 1966 Gross [70] and Dalecky [27] independently defined the infinite-dimensional elliptic operator of the second order which includes the Laplace operator
0 V (x(t)) = Tr V  (x), where V  (x) is the Hessian of the function V (x) at the point x. For a function V defined on a functional space,
0 V (x(t)) is the Volterra Laplacian, and for functions defined on the space of a countable number of variables, we have
0 V (x1 , . . . , xn , . . .) =

∞
∂2V k=1

∂ xk2

.

There exists a number of other examples of second-order infinite-dimensional differential expressions which considerably differ from the differential expressions of L´evy type. The corresponding references can be found in the bibliography to the monographs of Berezansky and Kondratiev [12] and Dalecky and Fomin [27]. The present book deals with the problems of the theory of equations with the L´evy Laplacians and L´evy–Laplace operators. It is based on the author’s papers [36–38, 40, 50–66] and the paper [8]. In Chapter 1 we give the definition of the L´evy Laplacian and describe some of its properties. In the foreword to his book [95], L´evy wrote: ‘In the theories which we mentioned, we essentially face the laws of great numbers similar to the laws of the theory of probabilities . . .’. The probabilistic treatment of the L´evy Laplacian in the second, third, and fourth chapters allows us to enlarge on a number of its interesting properties. Let us mention some of them. The L´evy Laplacian gives rise to operators of arbitrary order depending on the choice of the domain of definition of the operator. There is a huge number of harmonic functions of infinitely many variables connected with the L´evy Laplacian. The natural domain of definition of the L´evy Laplacian and that of the symmetrized L´evy Laplacian do not intersect. Starting from the non-symmetrized L´evy Laplacian, one can construct a symmetric and even a self-adjoint operator.

4

Introduction

Problems in the theory of equations with L´evy Laplacians are considered in Chapters 5–7. First, we concentrate our attention on the main classes of linear elliptic and parabolic equations with L´evy Laplacians. The equations which describe real physical processes are, as a rule, nonlinear. The theory of linear equations with the L´evy Laplacian is quite developed (see the bibliography). On the other hand, the theory of nonlinear equations with the L´evy Laplacian has only recently begun to be developed. The final two chapters deal with elliptic quasilinear and nonlinear and parabolic nonlinear equations with the L´evy Laplacian. We will see how striking is the difference (especially in the nonlinear case) between the theories of infinite-dimensional and n-dimensional partial differential equations. Finally in the Appendix we apply the results of Chapter 3 to the construction of Dirichlet forms associated with the L´evy–Laplace operator, and show the connection between these forms and Markov processes. There is no doubt that the reader of this book will see that the properties of the L´evy Laplacian, as a rule, have no analogues with the classical finitedimensional Laplacian. Moreover, the differences are so essential that one can call them pathological if the properties of the Laplace operator for functions of a finite number of variables are considered to be the norm. However, from another point of view the opposite statement is true as well. It should be emphasized that in this book we consider only the L´evy Laplacian. We do not consider here the problems of the theory of equations and operators of L´evy type (which naturally generalize the equations with L´evy Laplacians and L´evy–Laplace operators) considered in our papers [39, 41–49]. Unfortunately, a lot of the results concerning different trends originated in the book by L´evy are not included in this work although they undoubtedly deserve to be considered. In particular we do not discuss here the well-known approach to the L´evy Laplacian via white noise theory [80, 88]. I hope that this is compensated for to some extent by the large bibliography presented here. With great warmth I recollect numerous conversations on the topics discussed in this book with those who have departed: Yu. L. Dalecky (1926– 1997), O. A. Ladyzhenskaya (1922–2004), E. M. Polishchuk (1914–1987) and G. E. Shilov (1917–1975). During the preparation of this book for publication I was helped by Ya. I. Belopolskaya and I. I. Kovtun, and I am very grateful to them for their help.

1 The L´evy Laplacian

1.1 Definition of the infinite-dimensional Laplacian Let H be a countably-dimensional real Hilbert space. Consider a scalar function F(x) on H, where x ∈ H. L´evy introduced the infinite-dimensional differential Laplacian by M(x0 ,) F(x) − F(x0 ) . →0 2


L F(x0 ) = 2 lim

(1.1)

This definition assumes that F(x) has the mean value M(x0 ,) F(x), for  < 0 , and that the limit at the right-hand side of (1.1) exists. We define the mean value of the function F(x) over the Hilbert sphere x − x0 2H = 2 as the limit (if it exists) of the mean value, over the n-dimensional  sphere, of the function F( nk=1 xk f k ) = f (x1 , . . . , xn ), i.e., of the restriction of the function F(x) on the n-dimensional subspace with the basis { f k }n1 , xk = (x, f k ) H : M(x0 ,) F(x) = lim Mn F(x), n→∞  1 Mn F(x) = f (x1 , . . . , xn )dσn , sn n 

(xk −x0k )2 =2

k=1

where sn is the area, and dσn is the element of the n-dimensional sphere surface. In general, the mean value depends on the choice of the basis. It follows immediately from its definition that the mean value is additive and homogeneous: if there exists M(xo ,) Fk , k = 1, . . . , m, then there exists M(xo ,)

m 


m 
ck Fk = ck M(x0 ,) Fk .

k=1

k=1

5

6

The L´evy Laplacian

The mean value possesses the multiplicative property: if there exists M(xo ,) Fk , and the Fk are uniformly continuous in a bounded domain  ∈ H, which contains the sphere x − x0 2H = 2 , then there exists m m    M(x0 ,) Fk = M(x0 ,) Fk . k=1

k=1

This property follows from the following statement of L´evy. Let function F(x) be uniformly continuous on the sphere x − x0 2H = 2 , and let the average of the function F(x) exist (i.e., Mn → M, M = M(x0 ,) F ). Then for each δ > 0 we have 1 lim m n {x : | f (x1 , . . . , xn ) − M| > δ} = 0 n→∞ sn (here m n denotes the Lebesgue measure). Note that the definition of the Laplacian via mean values is valid for the finite-dimensional case as well. The definition (1.1) does not assume differentiability of the function F(x). However, if the function F(x) is twice strongly differentiable, then the following representation of the L´evy Laplacian holds. Lemma 1.1 Let the function F(x) be twice strongly differentiable in point x0 , and the Laplacian
L exist. Then n 1

L F(x0 ) = lim (F  (x0 ) f k , f k ) H , (1.2) n→∞ n k=1 where F  (x0 ) is the Hessian of the function F(x) in the point x0 , F  (x0 ) ∈ {H → H }, and { f k }∞ 1 is some chosen orthonormal basis in H. Indeed, it follows from the definition of the mean value that M(x0 ,) F(x) = MF(x0 + h), where M (h) is the mean value of the function over the sphere h2H = 1. Therefore, taking into account that for a ∈ H  M (a, h) H = 0, because s1n n h 2 =1 h k dσn = 0, we have k=1

k

1 1 {M(x0 ,) F(x) − F(x0 )} = 2 {MF(x0 + h) − F(x0 )} 2  

  1 1   = 2 M (F (x0 ), h) H + (F (x0 )h, h) H + r (x0 , h)  2  2

 1   = 2 lim Mn (F (x0 )h, h) H + r (x0 , h)  n→∞ 2 1 Mn r (x0 , h) ≥ lim Mn (F  (x0 )h, h) H limn→∞ 2 n→∞ 2

r (x0 , h) 2 and → 0 as h H → 0 ; h2H

1.1 Definition of the infinite-dimensional Laplacian

7

similarly we have 1 Mn r (x0 , h) 1 {M(x0 ,) F(x) − F(x0 )} ≤ limn→∞ Mn (F  (x0 )h, h) H + lim . n→∞ 2 2 2 From this we obtain 1 1 {M(x0 ,) F(x) − F(x0 )} − ε() ≤ limn→∞ Mn (F  (x0 )h, h) H 2 2 1 1 ≤ lim Mn (F  (x0 )h, h)) H ≤ 2 {M(x0 ,) F(x) − F(x0 )} + ε(), 2 n→∞  where ε() = 12 suph2H =1 |r (x0 , h)|, ε() → 0 as  → 0. Therefore,
L F(x0 ) = M(F  (x0 )h, h) H . Taking into consideration that, according to formula of Ostrogradsky,   ∂h k 1 1 Mn h 2k = h 2k dσn = dh 1 . . . dh n sn sn ∂h k n 

k=1

n 

h 2k =1

k=1

h 2k ≤1

n

π 2 / ( n2 + 1) vn 1 = = , n n sn n 2π 2 / ( 2 )  1 Mn h k h j = h k h j dσn = 0 for sn =

n 

k=1

j = k,

h 2k =1

(here h k = (h, f k ) H , vn is volume, sn is the area of surface of the sphere n 2 k=1 h k = 1, (s) the gamma function), we obtain that n 1
(F  (x0 ) f k , f k ) H . n→∞ n k=1


L F(x0 ) = M(F  (x0 )h, h) H = lim




If at the given point x0 the function F(x) is twice differentiable only with respect to the subspace Y of the space H (i.e., the second differential of the function F(x) at the point x0 does not exist for all increments h ∈ H, but d 2 F(x0 , y) = (FY (x0 )y, y) H exists for the increments y that form the subspace Y of the space H, and the second derivative of the function F(x) at the point x0 with respect to the subspace Y is the operator FY (x0 ) ∈ {Y → Y  }, where Y  is the space conjugate to Y ), then from (1.1) we deduce that n 1
(FY (x0 ) f k , f k ) H , n→∞ n k=1


L F(x0 ) = lim

(1.3)

provided that the basis { f k }∞ 1 is orthonormal in H and that f k ∈ Y. Now we give the formula for the infinite-dimensional Laplacian obtained by L´evy.

8

The L´evy Laplacian

Let there be a function F(x) = f (U1 (x), . . . , Um (x)), where f (u 1 , . . . , u m ) is a twice continuously differentiable function of m variables in the domain of values {U1 (x), . . . , Um (x)} in Rm , U j (x) are some twice strongly differentiable functions, and the
L U j (x) exist ( j = 1, . . . , m). Then
L F(x) exists, and m
∂ f 
L F(x) =
L U j (x). (1.4)  ∂u j u j =U j (x) j=1 Indeed, the second differential of the function F(x) at the point x for increment h ∈ H is m
∂ 2 f  d 2 F(x; h) = (F  (x)h, h) H = (U  (x), h) H (U j (x), h) H  ∂u i ∂u j ul =Ul (x) i i, j=1 m
∂ f  + (U j (x)h, h) H .  u j =U j (x) ∂u j j=1 According to (1.2),
L F(x) =

n ∂ 2 f  1
lim (Ui (x), f k ) H (U j (x), f k ) H  u l =Ul (x) n→∞ n ∂u ∂u i j i, j=1 k=1 m n
∂ f  1
+ lim (U j (x) f k , f k ) H .  u j =U j (x) n→∞ n ∂u j j=1 k=1 m


But n 1
(Ui (x), f k ) H (U j (x), f k ) H = 0, n→∞ n k=1

lim

(because (Ul (x), f k ) H → 0 as k → ∞), and n 1
(U j (x) f k , f k ) H =
L U j (x). n→∞ n k=1

lim

Therefore,
L F(x) =

m
∂ f 
L U j (x).  ∂u j u j =U j (x) j=1

A series of consequences follows from formula (1.4). 1. If the functions Uk (x) are harmonic in some domain , k = 1, . . . , m, then the function F(x) also is harmonic in .

1.2 Examples of Laplacians

9

2. The L´evy Laplacian is a ‘differentiation’. It is enough to set F(x) = U1 (x)U2 (x): then
L [U1 (x)U2 (x)] =
L U1 (x) · U2 (x) + U1 (x) ·
L U2 (x). 3. The Liouville theorem does not hold for harmonic functions of an infinite number of variables, i.e., there exists a function that is not equal to a constant which is harmonic and bounded in the whole space: it is sufficient to put F(x) = f (U (x)), where f (u) is a differentiable function in R1 bounded together with its derivative, U (x), which is a harmonic function in the whole of H . For example, F(x) = cos(α, x) H , α ∈ H.

1.2 Examples of Laplacians for functions on infinite-dimensional spaces For functions on a space of sequences, the L´evy Laplacian is an operator with an infinite number of partial derivatives, and for functions on spaces of functions of finitely many variables, the L´evy Laplacian is an operator in variational derivatives. Example 1.1 Let H = l2 be the space of sequences {x1 , . . . , xn , . . .} such that ∞ 2 k=1 x k < ∞. If the function F(x1 , . . . , xn , . . .) is twice strongly differentiable, and the Laplacian exists, then its Hessian is the matrix  ∂ 2 F(x) ∞     ∂ xi ∂ xk i,k=1 which induces a bounded operator in l2 : (F  (x)h, h)l2 =

∞
∂ 2 F(x) hi hk , ∂ xi ∂ xk i,k=1

and (1.2) yields that n 1
∂ 2 F(x) . n→∞ n ∂ xk2 k=1


L F(x1 , . . . , xn , . . .) = lim

Example 1.2 Let H = L 2 (0, 1) be the space of functions x(t), square integrable on [0, 1]. If the second differential of the twice differentiable function F(x(t)) has the form 1 2 1 1 δ F(x) 2 δ 2 F(x) 2 d F(x; h) = h(s)h(τ ) dsdτ, h (s) ds + δx(s)2 δx(s)δx(τ ) 0

0 0

10

The L´evy Laplacian

where the second variational derivative δ 2 F(x)/δx(s)2 and the second mixed variational derivative δ 2 F(x)/δx(s)δx(τ ) are continuous with respect to s and s, τ respectively (here h(t) ∈ L 2 (0, 1)), then one says that d 2 F(x; h) has normal form [95], and if 1 1 d F(x; h) = 2

0 0

δ 2 F(x) h(s)h(τ ) dsdτ, δx(s)δx(τ )

than one says that it has regular form [154]. We denote by B the set of all uniformly dense (according to the L´evy terminology) bases in L 2 (0, 1), i.e. orthonormal bases { f k }∞ 1 in L 2 (0, 1), such that lim (y, ϕn ) L 2 (0,1) = (y, 1) L 2 (0,1)

n→∞

for all y ∈ L 2 (0, 1),

 where ϕn (s) = n1 nk=1 f k2 (s). As has been shown by Polishchuk (in his comments to the Russian translation of [95]), all orthonormal bases which are the eigenfunctions of some Sturm– Liouville problem are uniformly dense. Let the function F(x) be twice strongly differentiable, and the second differential have normal form. Then 1
L F(x(t)) = 0

δ 2 F(x) ds δx(s)2

for arbitrary basis from B. Indeed, 1 1



(F (x)h, h) L 2 (0,1) =

δ(s − τ ) 0 0

δ 2 F(x) δ 2 F(x) ds + h(s)h(τ ) dsdτ δx(s)2 δx(s)δx(τ )

(δ(s − τ ) is the delta function), and, according to (1.2), 1 δ 2 F(x)  δ 2 F(x)
L F(x(t)) = lim ψn (s, τ ) dsdτ , ϕn (s) ds + 2 n→∞ δx(s) δx(s)δx(τ ) 0

where ψn (s, τ ) = But 1 1 0 0

1 n

n k=1

f k (s) f k (τ ).

δ 2 F(x) ψn (s, τ ) dsdτ → 0 δx(s)δx(τ )

as n → ∞,

1.2 Examples of Laplacians

11

because 1 1 0 0

≤

2 δ 2 F(x) ψn (s, τ ) dsdτ δx(s)δx(τ )

1 1  0 0

δ 2 F(x) 2 dsdτ δx(s)δx(τ )

1 1 ψn2 (s, τ ) dsdτ, 0 0

and 1 1 ψn2 (s, τ )dsdτ = 0 0

n 1
1 || f k ||4H = . 2 n k=1 n

Taking into account that { f k }∞ 1 ∈ B, we have 1
L F(x) = 0

δ 2 F(x) ds. δx(s)2

It is clear that if d 2 F(x; h) has regular form, then
L F(x(t)) = 0. Example 1.3 Let H = L 2;m (0, 1) be the space of vector functions x(t) = {x1 (t), . . . , xm (t)}, with components square integrable on [0, 1]. The second differential of the twice differentiable function F(x1 (t), . . ., xm (t)) is said to have normal form if d F(x; h) = 2

m 
i, j=1

1

0

δ 2 F(x) h i (s)h j (s) ds δxi (s)δx j (s) 1 1 + 0

0

 δ12 F(x) h i (s)h j (τ ) dsdτ , δxi (s)δx j (τ )

where the second partial variational derivatives δ 2 F(x)/δxi (s)δx j (s) and δ12 F(x)/δxi (s)δx j (τ ) of the function F(x) are continuous with respect to s and s, τ respectively (here h(t) ∈ L 2;m (0, 1)); and it is said to have regular form if d F(x; h) = 2

1 1 m  
i, j=1

0

0

δ12 F(x) h i (s)h j (τ ) dsdτ. δxi (s)δx j (τ )

12

The L´evy Laplacian

If F(x) is twice strongly differentiable, and the second differential has normal form, then m 
δ 2 F(x)
L F(x1 (t), . . . , xm (t)) = ds δx j (s)2 j=1 1

0

for an arbitrary uniformly dense basis in L 2;m (0, 1) (i.e., for the orthonormal ∞ ∞ basis { f k }∞ 1 = { f 1k , . . . , f mk }1 , such that { f jk }k=1 ∈ B, j = 1, . . . , m). Indeed, 

(F (x)h, h) L 2;m (0,1)

1 1 m   
δ(s − τ ) = i, j=1

0 0

+

δ 2 F(x) δxi (s)δx j (s)

δ12 F(x)  h i (s)h j (τ ) dsdτ δxi (s)δx j (τ )

and, according to (1.2), 1 1 m  

 δ 2 F(x) δ 2 F(x)
L F(x) = lim ψi jn (s, s) ds ϕ (s) ds + jn n→∞ δx j (s)2 δxi (s)δx j (s) i = j j=1 0

0

+

m
i, j=1

1 1 0 0

 δ12 F(x) ψi jn (s, τ ) dsdτ , δxi (s)δx j (τ )

where ϕ jn (s) =

n 1
f 2 (s), n k=1 jk

ψi jn (s, τ ) =

n 1
f ik (s) f jk (τ ). n k=1

But 1 1 0 0

1 0

δ12 F(x) ψi jn (s, τ ) dsdτ → 0, δxi (s)δx j (τ )

δ 2 F(x) ψi jn (s, s) ds → 0 (i = j) δxi (s)δx j (s)

as

n → ∞.

Taking into account that { f jk }∞ 1 ∈ B, j = 1, . . . , m, we have m 
δ 2 F(x)
L F(x) = ds. δx j (s)2 j=1 1

0

1.3 Gaussian measures

13

If d 2 F(x; h) has regular form, then
L F(x1 (t), . . . , xm (t)) = 0.

1.3 Gaussian measures The simplest measures in a finite-dimensional space which admit the extension to a Hilbert space are Gaussian measures. We consider a Gaussian measure in a Hilbert space which we need in the second, third and fourth chapters. We also consider a special Gaussian measure, the Wiener measure, which we need in the fifth chapter. First we define the Gaussian measure. The pair {H, A} where H is a Hilbert space and A is a σ -algebra of Borel sets from H is called a measurable Hilbert space. The measure µ whose characteristic functional has the form   1 χ(y) = exp − (K y, y) H + (a, y) H (y ∈ H ), 2 where a ∈ H, K is positive operator of a trace class in H , is called a Gaussian measure in Hilbert space {H, A}. Here a is the mean value, and K is the correlation operator of the measure µ. The measure is called centred if a = 0. In what follows we consider centred Gaussian measures. It follows from the above expression for the characteristic functional that all finite-dimensional projections µ P of the Gaussian measure µ are also Gaussian measures in a finite-dimensional space. In addition, the correlation operator of the measure µ P has the form K P = PKP (here P is an ortho-projector onto finite-dimensional (m-dimensional) subspace). The density of the measure µ P with respect to the Lebesgue measure in Rm is   1 1 −1 m . (x) = √ (K exp − x, x) R 2 P (2π)m det K P Finally, one should note the following important statement. The Gaussian measure µ is σ -additive if and only if K is a positive operator of trace class in H. The triple {H, A, µ} is called a measure space. Now we calculate some integrals with respect to a centred Gaussian measure which we need later. By direct computation we derive  µ(H ) = µ(d x) = 1. H

14

The L´evy Laplacian

Hence the triple {H, A, µ} is a probability space. Next we derive the expression for an integral of a function of a finite number of linear functionals with respect to a Gaussian measure. Let F(x) = f ((x, ϕ1 ) H , . . . , (x, ϕm ) H ), where f (u 1 , . . . , u m ) is a measurable function of m variables, and ϕ1 , . . . , ϕm are orthonormalized functions in H. As long as F(x) is a cylindrical function, we have m    − 12 τ jk u j u k 1 j,k=1 F(x) µ(d x) = √ f (u , . . . , u )e du 1 · · · du m , 1 m (2π)m det κ Rm

H

(K ϕ j , ϕk ) H mj,k=1 ,

where κ = and τ jk are the elements of the matrix inverse to κ. The existence of the integral at the left-hand side of this equality yields the existence of the integral on the right-hand side and vice versa. Now we compute the nth order moments   n m y1 ,...,yn = (x, yk ) H µ(d x) H

k=1

of the centred Gaussian measure µ. If n is odd, n = 2 p − 1, then m y1 ,...,yn = 0. If n is even, n = 2 p, then m y1 ,...,yn

∂n = (−1) ∂ε1 · · · ∂εn



p

H

 n   
 exp i(x, εk yk ) H µ(d x)  k=1

ε1 =···=εn =0

 n n    1  

∂  = (−1) p 1 exp − K εk yk ) , εk yk  H  ∂ε · · · ∂εn 2 k=1 k=1 n

ε1 =···=εn =0

=

p


(K ytk , yτk ) H

(y1 , . . . , yn ∈ H ),

{t,τ } k=1

 where {t,τ } is the sum over all possible partitions of a set of numbers 1, 2, . . . , n into p pairs (tk , tτk ), without taking into account the order of the pairs. In particular, m y1 y2 = (K y1 , y2 ) H , m y1 y2 y3 y4 = (K y1 , y2 ) H (K y3 , y4 ) H + (K y1 , y3 ) H (K y2 , y4 ) H + (K y1 , y4 ) H (K y2 , y3 ) H .

1.3 Gaussian measures

15

 Note that the computation of the integral H F(x) µ(d x) of the function F(x) admitting an approximation by a sequence of cylindrical functions FP (x) = F(P x) is reduced to the computation of the integral over the finite-dimensional subspace L P . If in addition some conditions hold which allow us to pass to the limit under the integral sign, we have    F(x) µ(d x) = lim FP (x)µ(d x) = lim FP (x)µ P (d x). P→E

H

P→E

H

LP

N

If P = PN , PN x = k=1 xk ek , xk = (x, ek ) H , where {ek }∞ 1 is a canonical basis in H, then µ P (d x) looks particularly simple: µ P (d x) = (2π)

−N /2

N 

λk e

−1/2

N  k=1

λ2k xk2

d x1 · · · d x N .

k=1

The canonical basis in H is the orthobasis which consists of the eigenvectors 1 of the operator T = K − 2 normalized in H ; T ek = λk ek , λk are eigenvalues of the operator T, k = 1, 2, . . . . Let F(x) = x2H . As long as the sequence of non-negative functions PN x2H converges to x2H monotonically, then one can go to the limit under the integral sign, and  ∞ N 

2 x H µ(d x) = lim (x, ek )2H µ(d x) = (K ek , ek ) H = Tr K . N →∞

H

k=1

k=1

H

Let F(x) = (Ax, x) H , where A is a bounded operator in H. Using the expressions for the second order moments, we obtain   N
(PN A PN x, x) H µ(d x) = (Ae j , ek ) H (x, e j ) H (x, ek ) H µ(d x) j,k=1

H

=

N


H

(Ae j , ek ) H (K e j , ek ) H .

j,k=1

We can go to a limit since (PN A PN x, x) H ≤ Ax2H . Therefore  ∞
(Ax, x) H µ(d x) = (AK ek , ek ) H = Tr AK . k=1

H

In the same way, using the expression for fourth order moments, we get   2 (Ax, x)2H µ(d x) = Tr AK + 2Tr (AK )2 . H

16

The L´evy Laplacian

Now consider the special Gaussian measure in Lˆ 2 (0, 1) with zero mean and correlation operator    1 1 1 1 1 K x(s) = min(t, s)x(s) ds − min(ξ, s)x(s) dsdξ 2 0 2 0 0 and show that this is the Wiener measure. Here Lˆ 2 (0, 1) is the space of real functions x(t), square integrable on [0, 1], satisfying the condition (x, 1) L 2 (0,1) = 0, where Lˆ 2 (0, 1) is a subspace of L 2 (0, 1). Let C0 (0, 1) be the space of real functions x(t), which are continuous on [0, 1] and satisfy the condition x(0) = 0. Wiener defined a measure in the space C0 (0, 1) as follows. Let 0 = t0 < t1 < t2 < · · · < tn = 1 be a partition of the interval [0, 1]. The set of functions y(t) ∈ C0 (0, 1) satisfying the condition ak < yk < bk , where yk = y(tk ), ak , bk are numbers (k = 1, 2, . . . , n) is called a quasi-interval. Consider quasi-intervals or so-called cylindrical sets in C0 (0, 1). Define the measure of a quasi-interval Q by the formula µW (Q) = b1

bn

n 
(yk − yk−1 )2  dy1 · · · dyn . · · · exp −    n 1/2 tk − tk−1 k=1 π n (tk − tk−1 ) a1 an

1

k=1

This measure admits a countably-additive continuation to a minimal σ -algebra in C0 (0, 1) which contains all cylindrical sets. The measure µW is called the Wiener measure. It is evident that the Wiener measure of the whole space C0 (0, 1) is equal to unity. The support of the Wiener measure is the set of functions which satisfy the H¨older conditions with index α < 1/2. Note that for a wide class of functionals (y) on the space C0 (0, 1) (in particular, for bounded and continuous functionals), Wiener’s integral can be calculated as follows. We replace the function y(t) by a broken line yn (t) with vertices at points (tk , y(tk )) and denote by n (y) the values of the functional for yn (t): n (y(t)) = (yn (t)) = ϕ(y1 , . . . , yn ), where ϕ(y1 , . . . , yn ) is a function of n variables, yk = y(tk ). Then  1 (y) µW (dy) = lim  1/2 n  n→∞ π n (tk − tk−1 ) C0  × Rn

k=1

n 
(yk − yk−1 )2  dy1 · · · dyn . ϕ(y1 , . . . , yn ) exp − tk − tk−1 k=1

1.3 Gaussian measures

17

In particular, it allows us to derive the formula the Wiener integral of the functional concentrated in m points. If (y) = f (y(t1 ), · · · , y(tm )), where t1 < t2 < · · · < tm , and f (y1 , . . . , ym ) is a function of m variables which is integrable in measure in Rm having the density n m   −1/2 
(yk − yk−1 )2  , (y1 , . . . , ym ) = π m (tk − tk−1 ) exp − tk − tk−1 k=1 k=1

then



 (y)µW (d x) =

f (y1 , . . . , ym )(y1 , . . . , ym ) dy1 · · · dym . Rm

C0 (0,1)

Consider in the space Lˆ 2 (0, 1) the Gaussian measure µ with zero mean and the correlation operator 1 K x(s) = 2

1

1 min(t, s)x(s) ds − 2

1 1

0

min(ξ, s)x(s) dsdξ ; 0

0

K −1 = T 2 is a closure in Lˆ 2 (0, 1) of the expression −2(d 2 x/dt 2 ) on the set of twice differentiable functions satisfying the condition x  (0) = x  (1) = 0. The operator K ∈ { Lˆ 2 (0, 1) → Lˆ 2 (0, 1)}. This is a trace class positive operator. Indeed, let us find the eigenvalues and eigenvectors of the operator K −1 . In other words, we find ρ such that the problem −2

d2x = ρ x, dt 2

x  (0) = x  (1) = 0

(x ∈ D K −1 )

has a non-trivial solution. The general solution of the differential equation 2

d2x +ρx =0 dt 2

has the form  x(t) = C1 cos

ρ t + C2 sin 2



ρ t. 2

It follows from x  (0) = x  (1) = 0 that  C2 = 0,

C1 sin

ρ t = 0, 2

18

The L´evy Laplacian

and for problem

√

ρ/2 = π k, k = 1, 2, . . ., there exist non-trivial solutions of the xk (t) = C1 cos πkt.

Therefore, the eigenvalues of the operator K −1 are ρk = 2π 2 k 2 . Its eigenvectors, normalized in Lˆ 2 (0, 1) (canonical basis in Lˆ 2 (0, 1)), are √ ek (t) = 2 cos πkt (k = 1, 2, . . .). The eigenvalues of K are νk = 1/(2π 2 k 2 ). So νk > 0, and Tr K =

∞ 1
1 1 = 2π 2 k=1 k 2 12

hence K is a positive operator of the trace class in H. Note that eigenvalues of the operator T = K −1/2 are √ λk = 2 πk. The Gaussian measure with such a correlation operator is a countably additive measure in Lˆ 2 (0, 1). With respect to the measure µ, almost all functions of the space Lˆ 2 (0, 1) are continuous and satisfy the H¨older condition with exponent α < 12 . Let Cˆ 0 (0, 1) be the set of all continuous functions on [0, 1] from Lˆ 2 (0, 1), i.e., the set of continuous functions orthogonal to unity: Cˆ 0 (0, 1) has the full measure in Lˆ 2 (0, 1). Calculate 

µ

1 x ∈ Cˆ 0 (0, 1) : α <

f (s) d x(s) < β

 ,

0

where f (t) is a bounded variation function on [0, 1] such that  f  L 2 (0,1) = 1. N Let P be an orthoprojector on R N, i.e. P x = k=1 xk ek , xk = (x, ek ) L 2 (0,1) . Then µ



1 x ∈ Cˆ 0 (0, 1) : α <

f (s) d x(s) < β



0

=

lim

N  
lim µ x ∈ Cˆ 0 (0, 1) : αl < ck x k < β l ,

αl →α,β l →β P→E

k=1

1.3 Gaussian measures

19

1 where ck = 0 f (s) dek (s), αl is a decreasing sequence converging to a number α, β l is an increasing sequence converging to the number β, and 

µ

x ∈ Cˆ 0 (0, 1) : αl <

N


ck xk < β l



k=1 N 

=

k=1

λk 

(2π )

 ···

N 2

αl <

N 

e

− 12



N  k=1

βl 2 λ2k xk2

ck xk <β l

d x1 · · · d x N =

1 π

1 2

 αl 2

k=1

N 

ck2 2 k=1 λk

− 12



N 

ck2 2 k=1 λk

e−η dη. 2

− 21

But ∞ ∞  2

√ ck2 2 = f (s) 2 sin πks ds =  f 2L 2 (0,1) , 2 λ k=1 k k=1 1

0

√ since 2 sin πkt, k = 1, 2, . . . , is a complete orthonormal system in L 2 (0, 1). Therefore 1 β   1 2 µ x ∈ Cˆ 0 (0, 1) : α < f (s) d x(s) < β = 1 e−η dη. π2 α

0

If f 1 (t), . . . , f n (t) are orthonormal functions in L 2 (0, 1) having bounded variation on [0, 1], then

µ



1 x ∈ Cˆ 0 (0, 1) : α j < ⎡

=

n 

⎢ 1 ⎣ 1 j=1 π 2

 f j (s) d x(s) < β j ( j = 1, 2, . . . , n)

0

β j

⎤

2 ⎥ e−η j dη j ⎦ .

αj

Choose f j (t) to be functions of this type. 1 Let 0 = t0 ≤ t1 ≤ · · · ≤ tn = 1 be a partition of [0, 1]. Put f j (t) = √t j −t j−1 for t ∈ [t j−1 , t j ], f j (t) = 0 for t ∈ / [t j−1 , t j ]; it is clear that ( f k , f j ) L 2 (0,1) = δk j .

20

The L´evy Laplacian

Then   µ x(t) ∈ Cˆ 0 (0, 1) : a j < x(t j ) − x(t j−1 ) < b j ( j = 1, 2, . . . , n)

=

√

n  

1

j=1

π2

bj

t j −t j−1 e

1

√

−η2j

n   dη j = j=1

aj t j −t j−1

b j

1

e

1

[π(t j − t j−1 )] 2

−t

ζ 2j j −t j−1

dζ j ,

aj

√ √ where a j = α j t j − t j−1 , b j = β j t j − t j−1 . There exists the one-to-one correspondence 1 y(t) = x(t) − x(0),

x(t) = y(t) −

y(s) ds 0

(y(t) ∈ C0 (0, 1),

x(t) ∈ Cˆ 0 (0, 1))

between the space C0 (0, 1) and the space Cˆ 0 (0, 1). The measure in the space Cˆ 0 (0, 1) is transferred by this correspondence into the space C0 (0, 1). In addition the set   x(t) ∈ Cˆ 0 (0, 1) : a j < x(t j ) − x(t j−1 ) < b j ( j = 1, 2, . . . , n) is mapped to the set  y(t) ∈ C0 (0, 1) : a j < y(t j ) − y(t j−1 ) < b j

 ( j = 1, 2, . . . , n) ,

and we have   µ y(t) ∈ C0 (0, 1) : a j < y(t j ) − y(t j−1 ) < b j ( j = 1, 2, . . . , n) =

b j

n 

1

j=1

[π (t j − t j−1 )] 2

e

1

−t

ζ 2j j −t j−1

dζ j .

aj

By the change of variables ζ j = z j − z j−1 ( j = 1, . . . , n), we obtain   µ y(t) ∈ C0 (0, 1) : a j < y(t j ) − y(t j−1 ) < b j ( j = 1, 2, . . . , n) 1

=  12 n  πn (t j − t j−1 )

b1

bn ···

a1

an

n 
(z j − z j−1 )2  dz 1 · · · dz n . exp − t j − t j−1 j=1

j=1

But this is the classical Wiener measure of a quasi-interval Q in the space C0 (0, 1). For this reason, one also calls the measure introduced in Lˆ 2 (0, 1) a Wiener measure, preserving the same notation µW .

1.3 Gaussian measures

21

To each functional F(x(t)) on Cˆ 0 (0, 1) integrable in the Wiener measure corresponds the functional 1 (y(t)) = F(y(t) − 



y(s) ds) 0

on C0 (0, 1), and Cˆ 0 (0,1) F(x)µW (d x) = C0 (0,1) (y) µW (dy). But Cˆ 0 (0, 1) has the full measure in Lˆ 2 (0, 1). If the functional (y) integrable in the Wiener measure on C0 (0, 1) corresponds to the functional F(x) on Lˆ 2 (0, 1) then   F(x)µW (d x) = (y) µW (dy). Lˆ 2 (0,1)

C0 (0,1)

2 L´evy–Laplace operators

Let L2 (H, µ) be the Hilbert space of functions F(x) on H square integrable in Gaussian measure µ with correlation operator K and zero mean value, and K be a positive operator of trace class such that ||x|| H ≤ ||K −1/2 x|| H , x ∈ D K −1/2 (here D K −1/2 denotes the domain of definition of the operator K −1/2 ), and  2 2 ||F||L2 (H,µ) = H F (x) µ(d x). The L´evy Laplacian is essentially infinite-dimensional: if F(x) is a cylindrical twice differentiable function F(x) = F(P x), P is the projection onto m-dimensional subspace, then its Hessian F  (x) is a finite-dimensional (mdimensional) operator, and m 1
(F  (x) f k , f k ) H = 0 n→∞ n k=1


L F(x) = lim

({ f k }∞ 1 is an orthonormal basis in H ). At the same time, the set C of cylindrical functions is everywhere dense in L2 (H, µ). If we now define the operator in L2 (H, µ) with everywhere dense domain of definition D L by LU =
L U, D L = C, then its closure L¯ = 0. In particular, we get L¯ = 0 if we define the operator on the well-known orthonormal system of Fourier–Hermite polynomials [25] m 1 ... m N (x) =

N 

Hm i ((K −1/2 x, f i ) H )

i=1

(N = 1, 2, . . . ; m i = 0, 1, 2, . . .), where {Hm (ξ )}∞ 0 are partially normalized ∞ Hermite polynomials, and { f i }1 is an orthonormal basis in H, f i ∈ H+ , because these polynomials are cylindrical functions. Hence it seems that the L´evy– Laplace operator is trivial. In fact, this is not true. There exist linear sets which are everywhere dense in L2 (H, µ) and on the functions from which the L´evy–Laplace operator is a non-trivial operator. 22

2.1 Infinite-dimensional orthogonal polynomials

23

For example, there exists a wide class of functions for which the L´evy Laplacian exists and is independent of the choice of the basis: namely, the class of twice differentiable functions F(x) such that the mean value of F(x) exists and its Hessian F  (x) is a thin operator: an operator is said to be thin if it has the form γ (x)E + S(x), where γ (x) is a scalar function, E is the identity operator, and S(x) is a compact operator. If F  (x) is a thin operator then
L F(x) = 2γ (x). The set of such functions includes, among others, the Shilov class [132, 134]. A set of functions of type F(x) = φ(x, ||x||2H ), where φ(x, ||x||2H ) = φ(P x, ||x||2H ), P is the projection on Rm , and φ(x, ξ ) are functions which are defined and twice differentiable on H × R1 is called the Shilov class. The Shilov class is dense in L2 (H, µ). If F(x) belongs to the Shilov class then
L F(x) = 2(∂φ/∂ξ )|ξ =||x||2H . Later in this chapter we construct a family of, complete in L2 (H, µ), orthogonal polynomial systems such that the L´evy Laplacian exists and does not depend on the choice of the basis. In addition, the L´evy Laplacian preserves the polynomial within the system. The L´evy Laplacian has some properties of both first and second order differential expressions, as well as other properties which are not true for either the second or the first order. The domain of definition consisting of orthogonal polynomial systems which do not contain C determines ‘the order’ of the operator, if one takes ‘the order’ to be the value by which the L´evy Laplacian lowers the degree of the polynomial. In the following we shall see a paradoxical property of the L´evy Laplacian: the same differential expression of L´evy–Laplace generates operators of any order depending on the choice of the domain of definition of the operator. If the domain of definition coincides with a complete orthonormal system of polynomials it is called natural (see, for example, Emch [34]).

2.1 Infinite-dimensional orthogonal polynomials Construct a, complete in L2 (H, µ), orthonormal system of polynomials such that the image of the Levy Laplacian belongs to the system. Let Hα ⊆ H0 ⊆ H−α , H+1 ≡ H+ , H0 ≡ H, H−1 ≡ H− , α > 0 be a chain of spaces from a Hilbert scale {Hβ }, −∞ < β < ∞, with the generating operator T = K −1/2 (its inverse, T −1 , is the Hilbert–Schmidt operator), (x, y) Hβ = (T β x, T β y) H (x, y ∈ Hβ ). We denote by H  ⊗ H  the tensor product of spaces H  and H  .

24

L´evy–Laplace operators

A linear combination of homogeneous forms of degrees not higher than m is called a polynomial of degree m on H. A real function m (x) = ϕ(x, . . . , x ), where ϕ(x, . . . , w) is a symmetric m-linear form (x, . . . , w ∈ H ),       m

m

m = 1, 2, . . . , is called a homogeneous form of degree m; the 0 are just numbers. ˆ m be the set of all measurable polynomials in L2 (H, µ) of degree Let  ˆ m is a subspace of L2 (H, µ), and  ˆ0 ⊂ ˆ1 less than or equal to m. Then  ˆ ˆ ⊂ · · · ⊂ m . We denote by m the orthogonal complement in m to m−1 , ˆ m = m ⊕  ˆ m−1 , m = 1, 2, . . . ; 0 =  ˆ 0. i.e.  ˆm = The subspaces 0 , 1 , . . . , m are mutually orthogonal,   ⊕ m  , and, as long as the set of measurable polynomials is dense k=0 k in L2 (H, µ), we have L2 (H, µ) = ⊕

∞


k .

k=0

ˆ m corresponds its projection onto subspace m , To each form m ∈  namely, the polynomial Pm ∈ m . Now we take, for m (x), the form γm (x) which represents the measurable extension to H of the continuous form (m!)−1/2 (γm , ⊗y m ) H m , where the kernel γm ∈ ⊗H−m is such that γm (x) ∈ L2 (H, µ): (γm , y ⊗ . . . ⊗ w)⊗H m is a symmetric m-linear continuous form (γm ∈ ⊗H−m , y, . . . , w ∈ H+ ). In the what follows we use the notation γm (x) = (m!)−1/2 (γm , ⊗x m )⊗H m for its measurable extension to H (although here x ∈ H, and not H+ ). The projection of such a form γm (x) onto m is a polynomial Pγm (x), which consists of γm (x) and the homogeneous forms of smaller degree ν (x, γm ), ν < m. In addition, (Pγm , Pγn )L2 (H,µ) = 0,

Pγm ∈ m ,

for all

for all

Pγn ∈ n ,

n < m.

To prove the existence of a measurable extension, we show that (Pγm , Pσm )L2 (H,µ) = (γm , σm )⊗H−m

for all γm ,

σm ∈ ⊗H−m .

(2.1)

To this end we calculate the integral of the nth differential in Gaussian measure. Theorem 2.1 Let F(x) ∈ L2 (H, µ), d i F(x; h 1 , . . . , h i ), i = 1, . . . , n, exist for arbitrary h 1 , . . . , h n ∈ H+2 , (i.e., F(x) is differentiable with respect to the subspace H+2 ) and there exists εi > 0 such that sup |d i F(x + εh i ; h 1 , . . . , h i )| ∀

|ε|≤εi

n 

j=i+1

[1 + (h j , x) H+ ] ∈ L(H, µ), i = 1, . . . , n.

2.1 Infinite-dimensional orthogonal polynomials

25

Then  d n F(x; h 1 , . . . , h n )µ(d x) H

=

[n/2]


(−1)

j




j 

 (h ik , h τk ) H+

{n, j; t,τ,s} k=1

j=0

F(x)

 (h si , x) H+ µ(d x) , (2.2)

i=1

H



n−2 j

where {n, j; t,τ,s} is the sum over all possible combinations of j pairs of numbers {t1 , τ1 }, . . . , {t j , τ j } chosen from numbers 1, 2, . . . , n (s1 , . . . , sn−2 j are the remaining numbers), without taking into account the order of the j pairs and of the remaining n − 2 j numbers. Proof. By the shift transformation, since we can pass to the limit, and differentiation under the integral sign for n = 1 we have    d  d F(x; h 1 )µ(d x) = F(x + εh 1 ) µ(d x) ε1 =0 dε1 H H   d − ε212 ||h 1 ||2H  + = e F(x)eε1 (h 1 ,x) H+ µ(d x) ε1 =0 dε1 H  = F(x)(h 1 , x) H+ µ(d x). (2.3) H

The rest we prove by induction. Assume that formula (2.2) is valid for n; we show that it is valid for n + 1 as well:  [n/2] j 

d n+1 F(x; h 1 , . . . , h n+1 )µ(d x) = (−1) j (h tk , h τk ) H+ {n, j;t,τ,s} k=1

j=0

H



×

n−2   j d F(x) (h si , x) H+ ; h 1 µ(d x) i=1

H

 −

F(x) H

n−2
j l=1

(h sl , h 1 ) H+

n−2 j

 (h si , x) H+ µ(d x) ,

i=1,i =l

where tk , τk , si = 2, 3, . . . , n + 1. It follows from (2.3) and the theorem’s conditions that 

[ n+1 2 ]

d H

n+1

F(x; h 1 , . . . , h n+1 )µ(d x) =




(−1) j

j 


 ×

F(x) H

(h tk , h τk ) H+

{n, j;t,τ,s} k=1

j=0

n−2 j i=1

 (h si , x) H+ µ(d x) .


26

L´evy–Laplace operators

ˆ ν are orthogonal if ν < m, we Let us show that (2.1) holds. Since m and  have !  m−1
(Pγm , Pσm )L2 (H,µ) = Pγm (x) σm (x) + ν (x) µ(d x) ν=1

H



=

Pγm (x) σm (x)µ(d x). H

It can be shown from Theorem 2.1 that  (Pγm , Pσm )L2 (H,µ) = σm (x)Pγm (x)µ(d x) H



∞


= H

(m!)− 2 (σm , f ν1 ⊗ . . . ⊗ f νm ) ⊗H m 1

ν1 ,...,νm =1

× (x, f ν1 ) H . . . (x, f νm ) H Pγm (x)µ(d x)  ∞
1 = (m!)− 2 (σm , f ν1 ⊗ . . . ⊗ f νm ) ⊗H m H

ν1 ,...,νm =1



× d m Pγm (x; T −2 f ν1 , . . . , T −2 f νm ) m

−

[2]


(−1) j

( f tk , f τk ) H−

{n, j;t,τ,s} k=1

j=1

×

j 


m−2 j

 ( f si , x) H Pγm (x) µ(d x).

i=1

Since m−2 j

ˆ m−2 j , ( f si , x) H ∈ 

Pγm ∈ 

i=1

we get  m−2 j H



( f si , x) H Pγm (x)µ(d x) = 0,

i=1

m j = 1, . . . , . 2

Therefore (Pγm , Pσm )L2 (H,µ) =

∞


(σm , f ν1 ⊗ . . . ⊗ f νm )⊗H m

ν1 ,...,νm =1

× (γm , T −2 f ν1 , . . . , T −2 f νm )⊗H m = (γm , σm )⊗H−m ; here { f k }∞ 1 is an orthonormal basis in H.

2.1 Infinite-dimensional orthogonal polynomials

27

Let us show the existence of the measurable extension. Let γm,k converge to γm in ⊗H−m , γm,k ∈ ⊗H m , γm ∈ ⊗H−m . Then (2.1) shows that the sequence Pγm,k (x) converges in L2 (H, µ). Choose a subsequence Pγm,ki (x) (the Pγm,ki (x) are continuous) which converges almost everywhere on H, and put Pγm (x) = lim Pγm,ki (x). Using (2.1) it can be shown that Pγm (x) is unique.

i→∞

Finally, as long as for all γm ∈ ⊗H−m there exists a sequence γm,k ∈ ⊗H m , which converges to γm in ⊗H−m , we have (Pγm , Pσm )L2 (H,µ) = (γm , σm )⊗H−m , for all γm , σm ∈ ⊗H−m . Wiener [157] constructed the system of orthogonal polynomials for the case of Wiener measure for the functionals defined on the space of continuous functions. We use this method to give the explicit form of polynomials by orthogonalizing the sequence of forms (more exactly, of the classes of forms) γm (x) = (m!)−1/2 (γm , ⊗x m )⊗H m ,

γm ∈ ⊗H−m .

Each polynomial Pγm (x) consists of γm (x) and of the forms of smaller degree ν (x, γm ), ν < m: P0 = 1,

Pγ1 (x) = γ1 (x), m

Pγm (x) = γm (x) +

[2]


m−2k (x, γm )

(m = 2, 3, . . .).

k=1

We construct m−2k (x, γm ) to obtain (Pγm , Pγm−2k )L2 (H,µ) = 0 given γm (x) (k = 1, . . . , [m/2]). Since the forms of even and odd degrees are orthogonal, we have (Pγ2 p , Pγ2q+1 )L2 (H,µ) = 0

( p, q = 0, 1, . . .).

For Pγ2 (x) = γ2 (x) + 0 (γ2 ), it follows from  (Pγ2 , P0 )L2 (H,µ) = [ γ2 (x) + 0 (γ2 )]µ(d x) = 0 H

that

 0 (γ2 ) =

γ2 (x)µ(d x). H

For Pγ3 (x) = γ3 (x) + 1 (x, γ3 ), and using the results from (Pγ3 , Pγ1 )L2 (H,µ) = 0 we have 1 (x, γ1 ) = (σ1 , x) H . According to Theorem 2.1, for n = 1 we have   d F(x; h)µ(d x) = F(x)(h, x) H+ µ(d x). H

H

28

L´evy–Laplace operators

∞ Therefore, taking into account that Pγ1 (x) = i=1 (γ , f i ) H (x, f i ) H , we have   (x, f i ) H Pγ3 (x)µ(d x) = d Pγ3 (x; T −2 f i )µ(d x) H

H



=

d γ3 (x; T −2 f i )µ(d x) + (σ1 , T −2 f i ) H = 0.

H

Hence (σ1 , T −2 f i ) H = −



d γ3 (y; T −2 f i )µ(dy)

H

3 = −√ 3! and 3 1 (x, γ3 ) = − √ 3!



(γ3 , T −2 f i ⊗ y ⊗ y)⊗H 3 µ(dy),

H

 (γ3 , x ⊗ y ⊗ y)⊗H 3 µ(dy). H

For Pγ4 (x) = γ4 (x) + 2 (x, γ4 ) + 0 (γ4 ), from the (Pγ4 , P0 )L2 (H,µ) = 0, we find that  0 (γ4 ) = − γ4 (x) + 2 (x, γ4 )]µ(d x).

condition

H

From the condition (Pγ4 , Pγ2 )L2 (H,µ) = 0 we find 2 (x, γ4 ) = (σ2 , x ⊗ x) H ⊗H . According to Theorem 2.1, for n = 2 we have   2 d F(x; h 1 , h 2 )µ(d x) = F(x)(x, h 1 ) H+ (x, h 2 ) H+ µ(d x) H

H

 −(h 1 , h 2 ) H+

F(x)µ(d x). H

Therefore, taking into account that (Pγ4 , P0 )L2 (H,µ) = 0 and that ∞
Pγ2 (x) = (γ2 , f j ⊗ f k ) H ⊗H (x, f j ) H (x, f k ) H , j,k=1

we have   (x, f i ) H (x, f k ) H Pγ4 (x)µ(d x) = d 2 Pγ4 (x; T −2 f j , T −2 f k )µ(d x) H



= H

H

d 2 γ4 (x; T −2 f j , T −2 f k )µ(d x) + 2(σ2 , T −2 f j ⊗ T −2 f k ) H ⊗H = 0.

2.1 Infinite-dimensional orthogonal polynomials

Hence

29

 1 d 2 γ4 (y; T −2 f j , T −2 f k )µ(dy) 2 H   3 =− (γ4 , T −2 f j ⊗ T −2 f k ⊗ y ⊗ y)⊗H 4 µ(dy) 2

(σ2 , T −2 f j ⊗ T −2 f k ) H ⊗H = −

H

and

  3 2 (x, γ4 ) = − (γ4 , x ⊗ x ⊗ y ⊗ y)⊗H 4 µ(dy). 2 H

Continuing in a similar way, we construct the following system of orthogonal polynomials: P0 (x) = 1, Pγ1 (x) = (γ1 , x) H ,  1 1 Pγ2 (x) = √ (γ2 , x ⊗ x) H ⊗H − √ (γ2 , x ⊗ x) H ⊗H µ(d x), 2! 2! H  1 3 Pγ3 (x) = √ (γ3 , ⊗x 3 )⊗H 3 − √ (γ2 , x ⊗ y 2 )⊗H 3 µ(dy), 3! 3! H  1 6 4 4 Pγ4 (x) = √ (γ4 , ⊗x )⊗H − √ (γ4 , ⊗x 2 ⊗ y 2 )⊗H 4 µ(dy) 4! 4! H   6 +√ (γ4 , ⊗x 2 ⊗ y 2 )⊗H 4 µ(dy)µ(d x) 4! H  1 −√ (γ4 , ⊗x 4 )⊗H 4 µ(d x), 4! H

etc. ∞ Let {γmq }q=1 be a complete orthonormal system in ⊗H−m . Lemma 2.1 The system of polynomials P0 = 1, Pγmq , with m, q = 1, 2, . . . , makes an orthonormal basis in L2 (H, µ). Indeed, (Pγmq , Pγnp )L2 (H,µ) = 0 for m = n, and, according to (2.1), (Pγmq , Pγmp )L2 (H,µ) = (γmq , γmp )⊗H−m = δ pq (here δ pq is the Kronecker delta); therefore (Pγmq , Pγnp )L2 (H,µ) = δmn δ pq . The completeness of the system P0 , Pγmq in L2 (H, µ) follows from the fact  ∞ that Pγmq ∈ m , L2 (H, µ) = ⊕ ∞ m=0 m , while the systems {γmq }q=1 are m complete in ⊗H− , m = 1, 2, . . . .


30

L´evy–Laplace operators

As usual it follows from Lemma 2.1 that given U ∈ L2 (H, µ) we have its expansion U (x) = U0 P0 +

∞


Umq Pγmq (x),

m,q=1

where

 U0 =

 U (x)µ(d x),

Umq =

H

U (x)Pγmq µ(d x). H

This series converges to U (x) in L2 (H, µ), and ||U ||2L2 (H,µ) = U02 +

∞


2 Umq .

m,q

2.2 The second-order differential operators generated by the L´evy Laplacian Consider operators of the second order generated by the differential expression of L´evy–Laplace. Choose kernels such that forms corresponding to them contain ||x||2lH . To this end we use the kernel corresponding to the identity operator E in H. Due to the theorem of Berezansky about a kernel [11] (E y, z) H = (δ, y ⊗ z) H− ⊗H− ,

δ ∈ H− ⊗ H− , y, z ∈ H+

and ||x||2lH = (⊗δl , ⊗x 2l )⊗H 2l . By (1.4), for m = 1 we have
L [||x||2H ]l = l[||x||2H ]l−1
L ||x||2H . Since ([||x||2H ] h, h) H = 2(Eh, h) H , by (1.2), we have
L ||x||2H = lim

n→∞

1 (2 + · · · + 2) = 2. n    n

Therefore
L [||x|| H ]2l = 2l||x||2(l−1) , H i.e., the L´evy Laplacian decreases the degree of the form (⊗δl , ⊗x 2l )⊗H 2l by 2.

2.2 The second-order differential operators

31

∞ Put γmq = amq , where {amq }q=1 is a complete orthonormal system of elem ments in ⊗H− such that

a2n,q =

n
˜ 2 p−2,q ] µ2n,q µ2n−2,q · · · µ2 p,q [⊗δ n− p+1 ⊗s

(n − p + 1)!

p=1

a2n−1,q =

+ s˜2n,q ,

n
˜ 2 p−1,q ] µ2n−1,q µ2n−3,q · · · µ2 p−1,q [⊗δ n− p ⊗s

(n − p)!

p=1

,

(n = 1, 2, . . .). (2.4)

∞ Here {smq }q=1 is a complete sequence of elements in ⊗H−m such that smq ∈ m ⊗H+ , and s˜ denotes that the symmetrization is applied to s. The numbers µmq are obtained in when bmq is orthogonalized (q = 1, 2, . . .). The system a2,q = µ2,q δ + s˜2,q is obtained when the sequence b2,q = δ + t2,q is orthogonal˜ + µ4,q δ ⊗s ˜ 2,q + s˜4,q is obtained when ized, the system a4,q = 12 µ4,q µ2,q δ ⊗δ ˜ + s2,q ⊗δ ˜ + t˜4,q is orthogonalized and so on, the sequence b4,q = µ2,q δ ⊗δ ∞ where {tmq }q=1 is a complete sequence in ⊗H−m , tm,q ∈ ⊗H+m . The complete∞ ness of the system {amq }q=1 in ⊗H−m follows from the completeness of the ∞ sequence {smq }q=1 . Denote Pamq ≡ Pmq . According to Lemma 2.1, the system of polynomials P0 , Pmq (x), for m, q = 1, 2, . . . , makes an orthonormal basis in L2 (H, µ). We denote by P the set of all possible linear combinations A0 P0 + N m,q=1 Amq Pmq , where Amq are arbitrary numbers, and N is a natural number.

Theorem 2.2 The L´evy Laplacian on P exists, does not depend on the choice of the basis, and is a second order operator. It decreases the degree of the polynomial Pmq (m ≥ 2) by 2:
L P0 = 0,


L P1q (x) = 0,


L Pmq (x) = 2[m(m − 1)]−1/2 µmq Pm−2,q (x)

(m = 2, 3, . . .).

Proof. According to (2.4), we have a2n,q (x) = (2n!)− 2 (a2n,q , ⊗x 2n )⊗H 2n 1

= (2n!)− 2 1

2n−2 p+2 n 
µ2n,q . . . µ2 p,q ||x|| (˜s2 p−2,q , ⊗x 2 p−2 )⊗H 2 p−2 H

(n − p + 1)! p=1  + (˜s2n,q , ⊗x 2n )⊗H 2n .

32

L´evy–Laplace operators

 4l  The 4lforms a2n,q (x) ∈ L2 (H, µ), since H ||x|| H µ(d x) < ∞. The integral H ||x|| H µ(d x) can be easily computed using the equality  2p (x, ek ) H µ(d x) =

λk √ 2π

∞ 2p

xk exp{−λ2k xk2 }d xk =

(2 p − 1)!!

−∞

H

2p

λk

,

where {ek }∞ 1 is a canonical basis in H . An orthogonal basis {ek }∞ 1 in H is called canonical if ek are eigenvectors of the operator T, normalized in H, i.e. T ek = λk ek and λk are eigenvalues of the operator T, k = 1, 2, . . . . For example, #2 "  ∞ ∞

1 1 4 ||x|| H µ(d x) = 2 + = 2TrK 2 + (TrK )2 . 4 2 λ λ k k k=1 k=1 H

For functions 1 ||x||2lH (˜sνq , ⊗x ν )⊗H ν , l! by Corollary 2 to (1.4), we obtain

1 1
L Hlν (x) =
L ||x||2lH · (˜sνq , ⊗x ν )⊗H ν + ||x||2lH ·
L (˜sνq , ⊗x ν )⊗H ν . l! l! Here

1 2 2l
L ||x|| H = ||x||2(l−1) , H l! (l − 1)! and by (1.2), n 1

L (˜sνq , ⊗x ν )⊗H ν = ν(ν − 1) lim (˜sνq , ⊗x ν−2 ⊗ f k ⊗ f k )⊗H ν = 0; n→∞ n k=1 Hlν (x) =

note (˜sνq , ⊗x ν−2 ⊗ f k ⊗ f k )⊗H ν → 0 as k → ∞, since s˜νq ∈ ⊗H+ν . Thus we have 2
L Hlν (x) = ||x||2(l−1) (˜sνq , ⊗x ν )⊗H ν H (l − 1)! for an arbitrary basis { f k }∞ 1 in H. Hence
L a2n,q (x) = 2µ2n,q [2n(2n − 1)]−1/2 a2n−2,q (x). If m = 2n + 1, then similarly we have
L a2n+1,q (x) = 2µ2n+1,q [(2n + 1)2n]−1/2 a2n−1,q (x). As the result we obtain
L Pmq (x) = 2µmq [m(m − 1)]−1/2 Pm−2,q (x).


2.3 Differential operators of arbitrary order

33

Define the operator
(2) L in L2 (H, µ) with the everywhere dense domain of definition D
(2) putting L


(2) L U =
L U, Theorem 2.3 The operator


(2) L

D
(2) = P. L

is closable.

Proof. Let Ul ∈ D
(2) , Ul → 0 and
(2) L → G ∈ L2 (H, µ) as l → ∞. L  N (l) Since Ul (x) = A0 (l)P0 + m,q=1 Amq (l)Pmq (x), we have lim {A20 (l)P0 l→∞  N (l) + m,q=1 A2mq (l)} = 0. Hence Amq = lim Amq (l) = 0 uniformly in m, q. l→∞

Since by Theorem 2.2, N (l)  1
 µ2,q A2,q (l) P0 √ 2 q=1


2L Ul (x) = 2

+

(l)−2 N (l) N



 µm+2,q [(m + 2)(m + 1)]−1/2 Am+2,q (l)Pmq (x) ,

q=1 m=1

we have G 0 = (G, P0 )L2 (H,µ) = lim

l→∞

G mq = =

N (l) √
 2 µ2,q A2,q (l) = 0, q=1

(G, Pmq )L2 (H,µ) = lim (
(2) L Ul , Pmq )L2 (H,µ) l→∞ lim 2µm+2,q [(m + 2)(m + 1)]−1/2 Am+2,q (l) l→∞

= 0,

which yields G = 0.




By Theorem 2.2 we compute
L U (x) for U ∈ P, and show that ∞ 2   1
D
¯ (2) = U ∈ L2 (H, µ) : √ µ2,q U2,q L 2 q=1

+

∞


 2 µ2m+2,q [(m + 2)(m + 1)]−1 Um+2,q <∞ .

m,q=1

2.3 Differential operators of arbitrary order generated by the L´evy Laplacian The L´evy–Laplace differential expression generates not only operators of the second order, but also operators of any order depending on the choice of the domain of definition of the operator.

34

L´evy–Laplace operators

The L´evy Laplacian of F(x) is the limit of a sequence of arithmetic means n 1
ξk (x), n k=1

where ξk (x) = (F  (x) f k , f k ) H . Any function ξk (x) on H which is measurable relative to a σ-algebra A is a random variable on the probability space {H, A, µ}. Let E ξk = H ξk (x)µ(d x), denote the mathematical expectation and Var ξk = ||ξk − E ξk ||2L2 (H,µ) , denote the variance of the random variable ξk . The convergence with probability 1 of a sequence of random variables corresponds to the convergence almost everywhere on H in measure µ of a sequence of measurable functions. If, in addition, the sequence ξk (x) satisfies the conditions which follow from the strong law of  large numbers, and lim n1 nk=1 E ξk = c, then for almost all x ∈ H we have n→∞

n 1
ξk (x) = c. n→∞ n k=1

lim

Therefore, it is possible to construct the forms F(x) of degree 2γ such that the L´evy Laplacian decreases the degree of these forms by 2γ , and finally to construct new polynomials by replacing the functions ||x||2lH in the polynomials Pmq (x) by the forms {F(x)}l . 2γ 2γ ∞ (2) Put γmq = amq (here amq ≡ amq ), where {amq }q=1 is the complete system 2γ

2γ

in ⊗H−m such that amq ∈ ⊗H m at m = 1, . . . , 2γ − 1, amq ∈ ⊗H−m at m = 2γ 2γ 2γ , 2γ + 1, . . . , and a1,q ≡ s˜1,q , . . . , a2γ −1,q ≡ s˜2γ −1,q . Then 2γ

a2γ n− j,q =

2γ 2γ n− p n µ2γ ˜
2γ n− j,q µ2γ n− j−2γ ,q . . . µ2γ p− j,q [⊗δ2γ ⊗s2γ p− j,q ]

(n − p)!

p=1

( j = 1, 2, . . . , 2γ − 1), 2γ a2γ ,q 2γ

=

a2γ n,q =

2γ µ2γ ,q δ2γ

+ s˜2γ ,q ,

2γ 2γ n− p+1 n µ2γ ˜ 2γ p−2γ ,q ]
⊗s 2γ n,q µ2γ n−2γ ,q . . . µ2γ p,q [⊗δ2γ

(n − p + 1)!

p=1

+ s˜2γ n,q where δ2γ =

∞

(n = 2, 3, . . .), γ −1 γ

(2.5) γ −1 γ

eν1 ⊗ . . . ⊗ T eν2γ , δ2 ≡ ν1 ,...,ν2γ =1 δν1 ν2 δν2 ν3 . . . δν2γ −1 ν2γ T 2γ ∞ ∞ δ, δ2γ ∈ ⊗H− , {ek }1 is a canonical basis in H , {smq }q=1 is a complete 2γ m m sequence of elements in ⊗H− such that smq ∈ ⊗H+ , and the numbers µmq 2γ are obtained in the process of orthogonalization of amq , q = 1, 2, . . . . 2γ

(2) 2γ ≡ Pmq , P Denote Pamq mq ≡ Pmq .

2.3 Differential operators of arbitrary order

35

2γ

According to Lemma 2.1, the system of polynomials P0 , Pmq , m, q = 1, 2, . . ., for each γ , γ = 1, 2, . . ., makes an orthonormal basis in L2 (H, µ). 2γ 2γ Denote by Pmq the set of all possible linear combinations A0 P0 + N 2γ (P(2) ≡ P). m,q=1 Amq Theorem 2.4 The L´evy Laplacian on P(2γ ) exists, is independent of the choice of basis, and is an operator of order 2γ . It decreases the polynomial degree 2γ Pmq , m ≥ 2γ , by 2γ :
L P0 = 0,


L P1,q (x) = 0,

2γ

...,


L P2γ −1,q (x) = 0,

2γ
L Pmq (x) = 2γ [m(m − 1) · · · (m − 2γ + 1)]−1/2 2γ

× [(2γ − 1)!!]µ2γ mq Pm−2γ ,q (x)

(m = 2γ , 2γ + 1, . . .)

for almost all x ∈ H. In particular, on P∞ the L´evy Laplacian is an infinite ∞ order operator
L P0 = 0,
L Pmq (x) = 0, m = 1, 2, . . . . Proof. According to (2.5), for arbitrary basis { f k }∞ 1 in H we have 2γ n,q (x) = (2γ n!)− 2 (a2γ n,q ⊗ x 2γ n )⊗H 2γ n = (2γ n!)− 2 1

2γ

×

1

2γ

n µ2γ 


2γ

2γ n,q

· · · µ2γ p,q [F(x)]n− p+1 (˜s2γ p−2γ ,q ⊗x 2γ p−2γ )⊗H 2γ p−2γ

p=1

+ (˜s2γ n,q F(x) =

∞


∞


(T

γ −1 γ

(n − p + 1)!  ⊗ x 2γ n )⊗H 2γ n ,

ek , f ν1 ) H · · · (T

γ −1 γ

ek , f ν2γ ) H

k=1 ν1 ,...,ν2γ =1

× (x, f ν1 ) H · · · (x, f ν2γ ) H =

∞


(T

γ −1 γ

2γ

x, ek ) H =

k=1

∞


2γ −2

λk

2γ

(x, ek ) H .

k=1

 2γ The forms 2γ n,q ∈ L2 (H, µ), because H F 2l (x)µ(d x) < ∞. One can easily compute integrals H F 2l (x)µ(d x). For example,  F 2 (x)µ(d x) = H

 
∞ H

4γ −4

λk

4γ

(x, ek ) H

k=1

+

∞


2γ −2

λj

2γ

2γ −2

(x, e j ) H λk

 2γ (x, ek ) H µ(d x)

j,k=1, j =k

= {(4γ − 1)!! − [(2γ − 1)!!]2 }Tr K 2 + [(2γ − 1)!!Tr K ]2 .

36

L´evy–Laplace operators For functions F l (x)(˜sνq , ⊗x ν )⊗Hν we have
L [F l (x)(sνq , ⊗x ν )⊗H ν =
L F l (x) · (sνq , ⊗x ν )⊗H ν +F l (x) ·
L (sνq , ⊗x ν )⊗H ν =
L F l (x) · (sνq , ⊗x ν )⊗H ν ,

since
L (sνq , ⊗x ν )⊗H ν = 0. Furthermore,
L F l (x) = l F l−1 (x)
L F(x). Let us compute the value
L F(x). For the above function F(x) we have ∞
2γ −2 (F  (x)h, h) H = 2γ (γ − 1) (T x, ei ) H (h, ei )2H . i=1

Let f i = ei . Then by (1.2) n 1
2γ −2 (T x, ek ) H , n→∞ n k=1


L F(x) = 2γ (γ − 1) lim 2γ −2

ξk (x) = (T x, ek ) H  E ξk =

is a sequence of independent random variables,

2γ −2 (T x, ek ) H µ(d x)

H

1 =√ 2π ∞

Var ξk = E ξk2 − [E ξk ]2 , E ξk2 =

∞

y2

y 2γ −2 e− 2 dy = (2γ − 3)!!,

−∞ 4γ −4

(T x, ek ) H

µ(d x) = (4γ − 5)!!,

−∞

∞

and, therefore, k=1 (Var ξk /k ) < ∞. According to the strong law of large numbers for independent random variables, with probability 1   lim n1 nk=1 (ξk − E ξk ) = 0. Since lim n1 nk=1 E ξk = (2γ − 3)!!, then 2

n→∞

n→∞

lim

n→∞

1 n

n


ξk = (2γ − 3)!!

k=1

almost surely, i.e.,
L F(x) = 2γ (2γ − 1)(2γ − 3)!! for almost all x ∈ H. For example, for γ = 2, n 1
(T x, ei )2H = 1, n→∞ n k=1

lim

and for γ = 3, n 1
(T x, ei )4H = 3 n→∞ n k=1

lim

almost surely.

2.3 Differential operators of arbitrary order

37

Let { f k }∞ 1 be an arbitrary basis in H. Then by (1.2) n
∞ 1
2γ −2 (T x, ei ) H ( f k , ei )2H . n→∞ n k=1 i=1


L F(x) = 2γ (2γ − 1) lim Let us represent sn (x) = sn (x) =

n

1 n

∞

2γ −2 ( f k , ei )2H i=1 (T x, ei ) H

k=1

in the form

n n ∞
1
1
2γ −2 (T x, ei )2γ −2 − (T x, ei ) H ( f k , ei )2H n i=1 n i=1 k=n+1

+

n
∞ 1
2γ −2 (T x, ei ) H ( f k , ei )2H = σn (x) − σˆ n + σ˘ n (x). n k=1 i=n+1

Since 2γ −2

ζi = (T x, ei ) H

∞


2γ −2 ζˆi = (T x, ei ) H

and

( f k , ei )2H

k=n+1

are sequences of independent random variables, E ζi = (2γ − 3)!!,

E ζˆi = (2γ − 3)!!

∞


( f k , ei )2 → 0 as i → ∞

k=n+1

∞

(because ( f k , ei )2H → 0 as i → ∞, k=n+1 ( f k , ei )2H → 0 as n → ∞) then by the strong law of large numbers lim σn (x) = (2γ − 3)!!, while lim σˆ n (x) = 0 n→∞ n→∞ almost surely. Since ∞


ζ˘k =

2γ −2

(T x, ei ) H

( f k , ei )2H

i=n+1

is a sequence of arbitrary random variables, E ζ˘k = (2γ − 3)!!

∞


( f k , ei )2H → 0

as k → ∞,

i=n+1

E|ζ˘k − E ζ˘k | ≤ c

∞


( f k , ei )2

i=n+1

and

∞
1 k=1

k

E |ζ˘k − E ζ˘k | < ∞,

then by the strong law of large numbers for arbitrary random variables we have lim σ˘ n (x) = 0 almost surely. n→∞ Thus, almost surely lim sn (x) = (2γ − 3)!!;

n→∞

38

L´evy–Laplace operators

that is,
L F(x) = 2γ (2γ − 1)[(2γ − 3)!!] for almost all x ∈ H. Since
L F l (x) = l F l−1 (x)
L F(x) = γ (2γ − 1)[(2γ − 3)!!]l F l−1 (x), l the L´evy Laplacian decreases the degree of the form (⊗δ2γ , ⊗x 2γ l )⊗H 2γ l by 2γ . Thus,


L [F l (x)(˜sνq , ⊗x ν )⊗H ν ] = 2γ (2γ − 1)[(2γ − 3)!!]l F l−1 (x)(˜sνq , ⊗x ν )⊗H ν for almost all x ∈ H. Hence
L 2γ n,q (x) = 2γ (2γ − 1)[2γ n(2γ n − 1) . . . (2γ n − 2γ + 1)]−1/2 2γ

2γ

2γ

× [(2γ − 3)!!]µ2γ n,q 2γ n−2γ ,q (x) for almost all x ∈ H. If m = 2γ (n − 1) + 1, . . . , m = 2γ n − 1, then similarly we have that almost everywhere on H 2γ


L 2γ (n−1)+1,q (x) = 2γ (2γ − 1)[(2γ (n − 1) + 1)(2γ (n − 1)) · · · × (2γ n − 4γ + 2)]−1/2 [(2γ − 3)!!]µ2γ (n−1)+1,q 2γ n−4γ +1,q (x), . . . , 2γ

2γ


L 2γ n−1,q (x) = 2γ (2γ − 1)[(2γ n − 1)(2γ n − 2) · · · (2γ n − 2γ )]−1/2 2γ

2γ

2γ

× [(2γ − 3)!!]µ2γ n−1,q 2γ n−2γ −1,q (x). This yields 2γ
L Pmq = 2γ [m(m − 1) · · · (m − 2γ + 1)]−1/2 [(2γ − 1)!!]µ2γ mq Pm−2γ ,q (x) 2γ

for almost all x ∈ H.


(2γ )

Define the operators
L , γ = 2, 3, . . ., in L2 (H, µ) with everywhere (2γ ) dense domains of definition D
(2γ ) putting
L U =
L U, D
(2γ ) = P2γ . L

Theorem 2.5 Operator

(2γ )
L

L

is closable.


Proof. Similar to that of Theorem 2.3. (2γ )
L

Applying Theorem 2.4, one can compute for U ∈ P2γ , and we show ) ¯ (2γ that the family of natural domains of definition of the operators
has the L

2.3 Differential operators of arbitrary order

39

form ∞ 2  
2γ 2γ D
¯ (2γ ) = U ∈ L2 (H, µ) : (2γ !)−1/2 µ2γ ,q U2γ ,q L

+

∞
m,q=1

$

q=1

%2 2γ 2γ µm+2γ ,q Um+2γ ,q

(m + 2γ )(m + 2γ − 1) · · · (m + 1)

 <∞ .

3 Symmetric L´evy–Laplace operator

In an infinite-dimensional space, there is no standard measure similar to the Lebesgue measure. If one considers a space with some other measure defined on it, for example, Gaussian measure, even the finite-dimensional Laplacian is not symmetric. Hence one has to symmetrize it. If F(x) is a cylindrical twice differentiable function, then its Hessian F  (x) is an m-dimensional operator, its gradient F  (x) ∈ Rm , and therefore the symmetrized L´evy Laplacian on such functions is equal to zero. An unexpected effect is that the symmetrized L´evy Laplacian can be equal to zero on functions for which the L´evy Laplacian itself is not equal to zero. For example, the symmetrized L´evy Laplacian is equal to zero on functions from the Shilov class. Moreover, it is equal to zero on functions from the natural domain of definition of the L´evy–Laplace operator. Nevertheless it is possible to construct the complete system of orthogonal polynomials such that the symmetrized L´evy Laplacian acts on such polynomials in a non-trivial way. In this case it appears that on these polynomials the L´evy Laplacian has no meaning. In the final section of this chapter, we again return to the L´evy Laplacian, and we see that it is not always necessary to symmetrize it. Given the L´evy Laplacian (not symmetrized), one can construct symmetric, and even self-adjoint, operators in the Hilbert space L2 (H, µ).

3.1 The symmetrized L´evy Laplacian on functions from the domain of definition of the L´evy–Laplace operator The symmetrized L´evy Laplacian, if it exists, can be determined by n   1
(F  (x) f k , f k ) H − (F  (x), f k ) H (x, f k ) H+ , n→∞ n k=1


C F(x) = lim

40

(3.1)

3.1 The symmetrized L´evy Laplacian

41

where F  (x) is the Hessian, F  (x) is the gradient of the function F(x) at the point x and { f k }∞ 1 is the chosen orthonormal basis in H, f k ∈ H+2 . If, in addition, the L´evy Laplacian also exists then n 1
(F  (x), f k ) H (x, f k ) H+ . n→∞ n k=1


C F(x) =
L F(x) − lim

Note that one can rewrite (3.1) in the form n   1
d 2 F(x; f k , f k ) − d F(x; f k )(x, f k ) H+ . n→∞ n k=1


C F(x) = lim

If we use Theorem 2.1 for n = 1:   d F(x; h)µ(d x) = F(x)(h, x) H+ µ(d x), H

H

then, assuming that U (x), V (x) satisfy the conditions of this theorem, we have   V (x)dU (x; h)µ(d x) = V (x)U (x)(h, x) H+ µ(d x) H

H

 −

U (x)d V (x; h)µ(d x). H

Applying this formula twice, we obtain   V (x)d 2 U (x; f k , f k )µ(d x) − V (x)dU (x; f k )(x, f k ) H+ µ(d x) H

H

 =−



d V (x; f k )(x, f k ) H+ U (x)µ(d x) + H

d 2 V (x; f k ; f k )U (x)µ(d x). H

If, in addition, one can pass to the limit under the integral sign, then   V (x) ·
C U (x)µ(d x) =
C V (x) · U (x)µ(d x). H

H

For the Shilov class of functions, F(x) = φ(x, ||x||2H ),  ∂φ 
C F(x) = 2  ∂ξ 

ξ =||x||2H

 ∂φ  −2  ∂ξ 

n 1
(x, f k ) H (x, f k ) H+ = 0 n→∞ n k=1

lim

ξ =||x||2H

42

Symmetric L´evy–Laplace operator

for almost all x ∈ H, because almost everywhere on H we have  lim n1 nk=1 (x, f k ) H (x, f k ) H+ = 1.

n→∞

Let us show that on polynomials Pmq (x), m, q = 1, 2, . . ., from the domain of definition of the L´evy–Laplace operator, the symmetrized L´evy Laplacian is equal to zero for almost all x ∈ H. The polynomials Pmq (x) consist of the forms amq (x), and they are from Hlν (x) =

1 ||x||2lH (˜sγ q , ⊗x)⊗H ν ; l!

see Section 2.2. For functions Hlν (x) we have
L Hlν (x) =

2 ||x||2(l−1) (˜sνq , ⊗x ν )⊗H ν . H (l − 1)!

Since (Hlν (x), h) H =

2 ||x||2(l−1) (x, h) H (˜sνq , ⊗x ν )⊗H ν H (l − 1)! 1 + ||x||2lH ν(˜sνq , ⊗x ν−1 ⊗ h)⊗H ν , l!

then, taking into account n 1
(˜sνq , ⊗x ν−1 ⊗ f k )⊗H ν (x, f k ) H+ = 0, n→∞ n k=1

lim

since s˜νq ∈ ⊗H+ν , we obtain
C Hlν (x) n   2 1
= ||x||2(l−1) 1 − (x, f k ) H (x, f k ) H+ (˜sνq , ⊗x ν )⊗H ν . lim H n→∞ n (l − 1)! k=1 We now show that sn (x) =

n   1
1 − (x, f k ) H (x, f k ) H+ n k=1

converges to zero almost everywhere on H . Let f k = ek . Then sn (x) =

n $ % 1
1 − (T x, ek )2H , n k=1

3.1 The symmetrized L´evy Laplacian

43

ζk = 1 − (T x, ek )2H is a sequence of independent random variables, 1 Eζk = 1 − √ 2π 1 Varζk = √ 2π

∞

∞

y 2 e− 2 y dy = 0, 1

2

−∞

(1 − y 2 )2 e− 2 y dy = 2 1

2

and

∞
Varζk k=1

−∞

k2

< ∞.

By the strong law of large numbers for independent random variables we deduce that lim sn (x) = 0 with probability equal to unity. This means that n→∞ 1 
C ||x||2lH = 0 l! for almost all x ∈ H. Let { f k }∞ 1 be an arbitrary basis in H, f k ∈ H+2 . Then n
∞ 1
sn = 1 − (x, f k ) H (T 2 x, ei ) H ( f k , ei ) H n k=1 i=1 n n ∞  1

1
= 1− (T 2 x, ei ) H (x, ei ) H − (x, f k ) H (T 2 x, ei ) H ( f k , ei ) H n i=1 n k=1 i=n+1 +

n ∞
1
(T 2 x, ei ) H (x, f k ) H ( f k , ei ) H = σn (x) − σˆ n (x) + σ˘ n (x). n i=1 k=n+1

Since ζi = [1 − (T 2 x, ei ) H (x, ei ) H ] is a sequence of independent random variables, then (as shown above) lim σn = 0 almost surely. n→∞ Since ∞
ζˆk = (x, f k ) H (T 2 x, ei ) H ( f k , ei ) H i=n+1

is a sequence of arbitrary random variables, ∞
Eζˆk = ( f k , ei )2H → 0 k=n+1

(because ( f k , ei ) H → 0 ∞), E|ζˆk − Eζˆk | ≤ c

as k → ∞, ∞


( f k , ei )2H

i=n+1

as k → ∞

∞

2 i=n+1 ( f k , ei ) H

and

∞
1 k=1

k

→ 0 as n →

E|ζˆk − Eζˆk | < ∞,

then by the strong law of large numbers for arbitrary random variables we have lim σˆ n = 0 almost surely. The law of large numbers also yields that n→∞ lim σ˘ n = 0 almost surely.

n→∞

44

Symmetric L´evy–Laplace operator

Therefore, almost surely lim sn = 0, which means that
C [ 1l ||x||2lH ] = 0 n→∞ for almost all x ∈ H. Thus,
C Hlν (x) = 0 for almost all x ∈ H. Hence, almost everywhere on H we have
C a2n,q (x) = 0. If m = 2n + 1, then we similarly have that almost everywhere on H
C a2n+1,q (x) = 0. Therefore
C Pmq (x) = 0 for almost all x ∈ H, m, q = 1, 2, . . . .

3.2 The L´evy Laplacian on functions from the domain of definition of the symmetrized L´evy–Laplace operator In Section 3.1 we showed that the symmetrized L´evy Laplacian is equal to zero on a wide class of functions, in particular, on polynomials from the natural domain of definition of the L´evy–Laplace operator. Do functions from L2 (H, µ) exist for which the symmetrized L´evy Laplacian is not equal to zero? The symmetrized L´evy Laplacian of G(x) is the limit of the sequence of arithmetic means n 1
ζk (x), ζk (x) = (G  (x) f k , f k ) H − (G  (x), f k ) H (x, f k ) H+ . n k=1  Even when H ζk (x)µ(d x) = 0 one would expect that
C G(x) = 0, if G(x) is such a function that the sequence ζk does not obey the law of large numbers. If in addition G(x) ∈ L2 (H, µ), then one can construct new polynomials by replacing the functions ||x||2lH in the Pmq (x) by the G(x)l . We construct in L2 (H, µ) a complete orthonormal system of polynomials Q 0 , Q mq (x), m, q = 1, 2, . . ., such that
C Q mq (x) = 0 and it belongs to L2 (H, µ). Assume that the correlation operator K of the measure µ is such that K 1/2 χ 1/2 (K −1/2 ) is a Hilbert–Schmidt operator, where ξ = χ 2 (η) is the inverse function of η = λ(ξ ), and λ(ξ ) is a certain monotonically increasing function √ that is continuous for ξ > 0 and such that λ(k) = λk ; χ(λk ) = k.

3.2 The symmetrized L´evy–Laplace operator

45

∞ We put γmq = αmq , where {αmq }q=1 is a complete orthonormal system of m elements in ⊗H− such that   ˜ 2 p−2,q n ν2n,q ν2n−2,q · · · ν2 p,q ⊗σ n− p+1 ⊗s
α2n,q = + s˜2n,q , (n − p + 1)! p=1   ˜ 2 p−1,q n ν2n−1,q ν2n−3,q · · · ν2 p−1,q ⊗σ n− p ⊗s
α2n−1,q = (n − p)! p=1

(n = 1, 2, . . .),

(3.2)

where σ =

∞


(χ(T )g j , gk ) H g j ⊗ gk ,

j,k=1 ∞ {gk }∞ 1 is a fixed orthonormal basis in H, gk ∈ H+2 , {smq }q=1 is a comm m plete sequence of elements in ⊗H− such that smq ∈ ⊗H+ , m = 1, 2, . . . , and the numbers νmq are obtained by the orthogonalization of αmq , m = 1, 2, . . . . Since by assumption, T −1 χ 1/2 (T ) is a Hilbert–Schmidt operator, the forms αmq (x) ∈ L2 (H, µ). Denote Pαmq ≡ Q mq . By Lemma 1.2 the system of polynomials Q 0 , Q mq , m = 1, 2, . . . , forms an orthonormal basis in L2 (H, µ).

Theorem 3.1
C Q mq (x) = 0,
C Q mq (x) ∈ L2 (H, µ). Proof. By (3.2), for an arbitrary basis { f k }∞ 1 in H, f k ∈ H+2 , we have α2n,q (x) = (2n!)− 2 (α2n,q , ⊗x 2n )⊗H 2n n 
ν2n,q · · · ν2 p,q {G(x)}n− p+1 (˜s2 p−2,q , ⊗x 2 p−2 )⊗H 2 p−2 1 = (2n!)− 2 (n − p + 1)! p=1  + (˜s2n,q , ⊗x 2n )⊗H 2n , 1

G(x) =

∞


(χ(T ) f j , f i ) H (x, f j ) H (x, f i ) H .

j,i=1

Let f i = ei . Then G(x) =

∞ √
k=1

k(x, ek )2H

46

Symmetric L´evy–Laplace operator

and, by (3.1), n √ 1
k[1 − λ2k (x, ek )2H ]. n→∞ n k=1


C G(x) = 2 lim

We show that the sequence pn (x) = in L2 (H, µ). For m < n we have  H

 H

n

√

k=1

k [1 − λ2k (x, ek )2 ] converges

m n m 2
4
2
k − k + k→ 0 n 2 k=1 nm k=1 m 2 k=1

[ pn (x) − pm (x)]2 µ(d x) =

as m, n → ∞. Now

1 n

pn2 (x)µ(d x) =

2 n2

n k=1

k=

n+1 , n

and

||
C G(x)||L2 (H,µ) = 2. Let { f k }∞ 1 be an arbitrary basis in H, f k ∈ H+2 . Then (G  (x), h) H = 2

∞


(χ(T ) f i , f j ) H (x, f j ) H (h, f i ) H ,

i, j=1

(G  (x)h, h) H = 2

∞


(χ(T ) f i , f j ) H (h, f j ) H (h, f i ) H ,

i, j=1

and by (3.1), n 1
[(χ(T ) f k , f k ) H − (χ(T )x, f k ) H (T 2 x, f k ) H ]. n→∞ n k=1


C G(x) = 2 lim

Similarly to the previous case, we can easily show that qn (x) =

n 1
[(χ(T ) f k , f k ) H − (χ(T )x, f k ) H (T 2 x, f k ) H ] n k=1

is a fundamental sequence. Taking into account that  

(x, α) H (x, β) H µ(d x) = (α, β) H− ,

H

(x, α) H (x, β) H (x, γ ) H (x, σ ) H µ(d x) = (α, β) H− (γ , σ ) H−

H

+ (α, γ ) H− (β, σ ) H− + (α, σ ) H− (β, γ ) H− ,

α, β, γ , σ ∈ H− ,

3.2 The symmetrized L´evy–Laplace operator

47

we obtain that  qn2 (x)µ(d x) H

1 = 2 n

 &
n   (χ(T ) f k , f k ) H − (χ(T )x, f k ) H (T 2 x, f k ) H H

j,k=1, j =k

  × (χ(T ) f j , f j ) H − (χ(T )x, f j ) H (T 2 x, f j ) H ' n  2
+ µ(d x) (χ(T ) f k , f k ) H − (χ(T )x, f k ) H (T 2 x, f k ) H = and lim

n→∞



k=1 n 


1 n2

 (χ(T ) f k , f j )2H + (χ(T ) f k , χ(T ) f j ) H− ( f k , f j ) H+ ,

j,k=1

qn2 (x)µ(d x) = 1. Therefore,

H

|
C G(x)||L2 (H,µ) = 2. The polynomials Q mq (x) consist of the forms αmq (x), and they are from Z lν (x) =

1 l G (x)(˜sνq , ⊗x ν )⊗H ν . l!

For functions Z lν we have
C Z lν (x) =

1 G l−1 (x)
C G(x)(˜sνq , ⊗x ν )⊗H ν (l − 1)!

(since s˜νq ∈ ⊗H+ν ). Therefore,
C α2n,q (x) ∈ L2 (H, µ) (since a2n ,q  (x) ∈ L2 (H, µ)). If m = 2n + 1, then similarly we have
C α2n+1,q (x) ∈ L2 (H, µ). Therefore,
C Q mq (x) ∈ L2 (H, µ).
We show that the L´evy Laplacian has no meaning on the polynomials Q mq , m = 2, 3, . . . ; q = 1, 2, . . . .√  k(x, ek )2H , and If f k = ek , then G(x) = ∞ k=1 n √ 1
k = ∞. n→∞ n k=1


L G(x) = 2 lim

If { f k }∞ 1 is an arbitrary basis in H, f k ∈ H+2 , then G(x) =

∞
j,k=1

(χ(T ) f j , f k ) H (x, f j ) H (x, f k ) H ,

48

Symmetric L´evy–Laplace operator

and n 1
(χ(T ) f k , f k ) H . n→∞ n k=1


L G(x) = 2 lim

Since n n n


(χ(T ) f k , f k ) H = (Pχ(T )P f k , f k ) H = (Pχ(T )Pϕk , ϕk ) H k=1

k=1

=

n


k=1

(χ(T )en k , en k ) H =

k=1

n


χ(λn k ),

k=1

where P is an orthogonal projection on the space with the basis { f k }n1 , ϕk are eigenvectors of operator Pχ(T )P, Pϕk = en k , then n 1

L G(x) = 2 lim χ(λn k ) = ∞, n→∞ n k=1 since χ (λn k ) → ∞ as k → ∞.

3.3 Self-adjointness of the non-symmetrized L´evy–Laplace operator Let us consider a statement which certainly can not be valid for the finitedimensional Laplacian. We show that, given a non-symmetrized L´evy Laplacian, one can construct a self-adjoint operator in the space of functions squareintegrable in Gaussian measure. We denote by T the set of all functions of the form V (x) = ϕ(Q(x))S(x), where Q(x) = x2H , S(x) are arbitrary twice strongly differentiable harmonic in H functions from L2 (H, µ), ϕ(ξ ) is an arbitrary fixed positive function defined and differentiable on [0, ∞) satisfying the (Lipschitz condition )with the Lipschitz constant c. Let, in addition, V (x) and ϕ  (x2H )/ϕ(x2H ) V (x) ∈ L2 (H, µ). The set T is linear. To show that it is everywhere dense in L2 (H, µ), it is sufficient to show that ˆ is everywhere dense, where T ˆ is the set of all functions the set T (x) = ϕ(x2H )(x), where (x) ∈ C. Here C is the set of cylindrical twice strongly differentiable functions. If  ∈ C, that is (x) = (P x), P is a projection on an

3.3 Self-adjointness of the non-symmetrized L´evy–Laplace operator 49 m-dimensional subspace, then its Hessian   (x) is a finite-dimensional (mdimensional) operator, and m 1
(  (x) f k , f k ) H = 0. n→∞ n k=1


L (x) = lim

ˆ ⊂ T. Therefore, (x) is a harmonic function in H, and T Let U ∈ L2 (H, µ). The set C is everywhere dense in L2 (H, µ), i.e., for all ε1 there exists 0 ∈ C such that U − 0 L2 (H,µ) ≤ ε1 . Since 0 ∈ C, we have 0 (x) = 0 (P x) for some projection on an mdimensional subspace with a basis { f k }m 1 . Choose 0 (x) = ϕ(x2H )0 (x)/ϕ(P x2H ); ˆ because 0 (x)/ϕ(P x2 ) ∈ C. Then 0 ∈ T, H U − 0 L2 (H,µ) ≤ U − 0 L2 (H,µ) + 0 − 0 L2 (H,µ) ≤ ε1 + 0 − 0 L2 (H,µ) . We have 0 −



0 2L2 (H,µ)

=

[0 (x) − ϕ(x2H )0 (x)/ϕ(P x2H )]2 µ(d x) H



=

[0 (x)/ϕ(P x2H )]2 [ϕ(P x2H ) − ϕ(x2H )]2 µ(d x) H



≤ c2

[0 (x)/ϕ(P x2H )]2

∞ 


2 (x, f k )2H µ(d x),

k=m+1

H

because the function ϕ(ξ ) satisfies the Lipschitz condition. Since 0 (x)/ϕ(P x2H ) depends only on (x, f 1 ) H , . . . , (x, f m ) H , and ∞ 2 k=m+1 (x, f k ) H depends on (x, f m+1 ) H , (x, f m+2 ) H , . . . , we have 0 − 0 2L2 (H,µ)   
∞ 2 ≤ c2 [0 (x)/ϕ(P x2H )]2 µ(d x) (x, f k )2H µ(d x) H

H

 = c2

k=m+1

[0 (x)/ϕ(P x2H )]2 µ(d x) H

×

∞ 
k=m+1

(K f k , f k ) H

2

+2

∞
k=m+1

 (K 2 f k , f k ) H .

50

Symmetric L´evy–Laplace operator

∞ Since 0 (x)/ϕ(P x2H ) ∈ L2 (H, µ), and k=1 (K f k , f k ) H = TrK < ∞, ∞ ∞ 2 2 (K f , f ) = TrK < ∞ for all { f } k k H k k=1 1 in H (K is a trace class operator), and we have  0 − 0 L2 (H,µ) ≤ ε2 . Therefore, U − 0 L2 (H,µ) ≤ ε1 + ε2 = ε ˆ and hence T is dense everywhere in L2 (H, µ). Lemma 3.1 The L´evy Laplacian on T exists, is independent of the choice of the basis, and is an operator of multiplication by the function 2ϕ  (x2H )/ϕ(x2H ): 2ϕ  (x2H ) V (x) ϕ(x2H ) Proof. By Corollary 2 to (1.4), we have
L V (x) =

(V ∈ T).


L V (x) =
L [ϕ(x2H )S(x)] =
L ϕ(x2H ) · S(x) + ϕ(x2H ) ·
L S(x). By (1.4) for m = 1, we obtain
L ϕ(x2H ) = ϕ  (x2H )
L x2H = 2ϕ  (x2H )

(since
L x2H = 2).

In addition, since S(x) is harmonic,
L S(x) = 0. Therefore, 2ϕ  (x2H )
L V (x) = 2ϕ  (x2H )S(x) = V (x). ϕ(x2H )


Lemma 3.2 If
L V (x) = (2ϕ  (x2H )/ϕ(x2H ))V (x), then the function V (x) has the form V (x) = ϕ(x2H )S(x), where S(x) is an arbitrary harmonic function. ( ) Proof. It follows from
L V (x) = 2ϕ  (x2H )/ϕ(x2H ) V (x), for V (x) = 0, that
L V (x)
L ϕ(x2H ) = . V (x) ϕ(x2H ) Hence,
L {ln |V (x)|} =
L {ln ϕ(x2H )}, i.e.,   |V (x)|
L ln = 0. ϕ(x2H ) Therefore |V (x)| ln = G(x), ϕ(x2H )

3.3 Self-adjointness of the non-symmetrized L´evy–Laplace operator 51

where G(x) is an arbitrary harmonic function. Hence, V (x) = ϕ(x2H e G(x) = ϕ(x2H )S(x),


where S(x) is an arbitrary harmonic function.

We define an operator L in L2 (H, µ) with everywhere dense domain of definition D L , putting LU =
L U,

D L = T.

Theorem 3.2 The operator L is essentially self-adjoint. Proof. The operator L is symmetric, since D L is dense in L2 (H, µ) and by Lemma 3.1  2ϕ  (x2H ) (L V1 , V2 )L2 (H,µ) = V1 (x)V2 (x)µ(d x) ϕ(x2H ) H

= (V1 , L V2 )L2 (H,µ)

V1 , V2 ∈ D L .

for all

Let us show that L is essentially self-adjoint. Consider the operator L ∗ adjoint to L in L2 (H, µ). Let Z ∈ D L ∗ . Then for any V ∈ D L we have (L V, Z )L2 (H,µ) = (V, L ∗ Z )L2 (H,µ) ; that is,

 H

2ϕ  (x2H ) V (x)Z (x)µ(d x) = ϕ(x2H )



V (x)(L ∗ Z )(x)µ(d x)

H

ϕ(x2H )SV (x),

or, taking onto account that V (x) = we get   2ϕ  (x2 )  H ∗ ϕ(x2H )SV (x) Z )(x) µ(d x) = 0. Z (x) − (L ϕ(x2H ) H

This equality holds for functions having the form ϕ(x2H )S(x) where S(x) ∈ C (the set of cylindrical functions) and ϕ(ξ ) > 0 (for example, it holds if S(x) belongs to the complete orthonormal Hermite–Fourier polynomial system). Hence almost everywhere on H we have L∗ Z =

2ϕ  (x2H ) Z. ϕ(x2H )

Thus for all Z ∈ D L ∗ 2ϕ  (x2H ) Z ∈ L2 (H, µ) ϕ(x2H )

and

L∗ Z =

2ϕ  (x2H ) Z. ϕ(x2H )

52

Symmetric L´evy–Laplace operator

By the equality 



   2ϕ (x2H ) Z (x) 2ϕ (x2H ) V (x) U (x) Z (x)µ(d x) = µ(d x) ϕ(x2H ) ϕ(x2H ) H H it for any ) Z ∈ L2 (H, µ), L ∗ Z = (  is 2 easy 2to ) see ( that 2  2ϕ (x H )/ϕ(x H ) Z and 2ϕ (x H )/ϕ(x2H ) Z ∈ L2 (H, µ). This shows that   2ϕ  (x2H ) D L ∗ = Z ∈ L2 (H, µ) : (H, µ) . Z ∈ L 2 ϕ(x2H ) domain D L ∗ is )dense in L2 (H, µ) since D L ∗ ⊃ D L . The operator L ∗ Z = ( The 2ϕ  (x2H )/ϕ(x2H ) Z , Z ∈ D L ∗ , is a self-adjoint operator. By the self-adjointness of L ∗ we deduce that L is essentially self-adjoint.


4 Harmonic functions of infinitely many variables

It is well known that the only harmonic function of a single variable is a linear function. It is also known that in a multi-dimensional case the number of harmonic functions grows. This number increases rapidly if we are interested in infinite-dimensional harmonic functions associated with the L´evy Laplacian. Let us give several examples (which have no finite-dimensional counterparts) of harmonic functions defined on a Hilbert space H : 1. Cylindrical twice differentiable functions. Let F(x) be a cylindrical twice differentiable function F(x) = F(P x), and P be a projection on m-dimensional subspace. Then the Hessian F  (x) is a finite-dimensional (m-dimensional) operator and (1.2) leads to m 1
(F  (x) f k , f k ) H = 0; n→∞ n k=1


L F(x) = lim

here { f k }∞ 1 is an orthonormal basis in H . 2. Twice differentiable functions F(x), whose Hessians F  (x) are compact operators in H. Indeed, since Hessian F  (x) is a compact self-adjoint operator in H, we have (F  (x) f k , f k ) H → 0 as k → ∞ and, according to (1.2), n 1
(F  (x) f k , f k ) H = 0 n→∞ n k=1


L F(x) = lim

for arbitrary orthonormal basis { f k }∞ 1 in H. For example, if H = L 2 (a, b), and the second differential of a function F(x) has the regular form (Example 1.2), then 

b

F (x)h(s) = a

δ 2 F(x) h(s) ds δx(s)δx(τ )

53

54

Harmonic functions of infinitely many variables

(δ 2 F(x)/δx(s)δx(τ ) is a continuous function with respect to s, τ, δ 2 F(x)/ δx(s)δx(τ ) = δ 2 F(x)/δx(τ )δx(s)), and F  (x) is a compact operator in L 2 (a, b).  m Another example of this kind is given by H = l2 , and F(x) = ∞ k=1 x k  for integer numbers m ≥ 3. In this case the Hessian F (x) is a matrix ∞ m(m − 1)δik xkm−2 i,k=1 (here δik is the Kronecker symbol), xkm−2 → 0 as  k → ∞, and F (x) is a compact operator in l2 . 3. Functions of the form F(x) = f (U1 (x), . . . , Um (x)), where f (u 1 , . . . , u m ) is a twice continuously differentiable function of m variables over the set {U1 (x), . . . , Um (x)} in Rm , and functions U j (x) are harmonic in the whole space H, j = 1, . . . , m. Indeed, since
L U j (x) = 0, j = 1, . . . , m, we derive, using (1.4),
L F(x) = 4.

H = l 2,

F(x) =

∞


m
∂ F 
L U j (x) = 0.  ∂u j u j =U j (x) j=1

αk λ2m−2 xk2m k

(αk = O(1/ln1+ε k),

ε > 0),

k=1

F(x) =

∞  k


λ j x j λ2m−1 xk2m , k

k=1 j=1

F(x) =

∞ 
k=1

F(x) =

∞


λ2m−1 xk2m+1 , k

k=1

2m+2 λ2m x2  k xk − k , (2m − 1)!!(2m + 2)(2m + 1) 2

m ≥ 1 is an integer, xk = (x, ek )l2 , ek = (0, . . . , 1, 0, . . .), K −1/2 ek = λk ek ,    k = 1, 2, . . . . k As a result of the strong law of large numbers, all these functions are harmonic µ–almost everywhere on l2 .

4.1 Arbitrary second-order derivatives Theorem 4.1 Let F(x) be a twice differentiable function with respect to a subspace Hα for some α > 0,  |(FHα (x) f k , f k ) H | p µ(d x) < ∞ H

for some p, 2/3 < p ≤ 1, and { f k } be an orthonormal basis in H , f k ∈ Hα . If p n 
n  p |(FHα (x) f k , f k ) H | µ(d x) = O , (4.1) ψε (n) k=1 H

4.1 Arbitrary second-order derivatives

55

where ψε (n) is one of functions (ln n)1+ε ,

ln n(ln ln n)1+ε ,

...,

ln n · · · ln · · · ln n (ln · · ln n)1+ε  · m−1

m

for all ε > 0, then
L F(x) = 0

for almost all x ∈ H.

(4.2)

On the other hand, there exists a function F(x) such that (4.1) is satisfied for ε = 0, but (4.2) is not valid. Estimate (4.1) (and, therefore, the harmonicity of F(x)) does not depend on the choice of the basis from a class of square close bases. Proof. The L´evy Laplacian (1.3) is the limit of the sequence of arithmetic mean  values n1 nk=1 ξk (x), where ξk (x) = (FHα (x) f k , f k ) H . Any function (x) on H that is measurable with respect to a σ -algebra A, is a random variable on the probability space {H, A, µ}. Its mean value is given by  E = (x) µ(d x), H

and the convergence of a sequence of random variables with probability 1 corresponds to the convergence of the sequence of measurable functions almost everywhere on H with respect to µ. Therefore, {ξk (x)}∞ 1 is a sequence of arbitrary (dependent) random variables,  np  for all ε > 0. ψε (n) k=1  This follows from the theorem of Petrov [108] that n1 nk=1 ξk (x) → 0 almost surely, i.e., E|ξk | p < ∞, and

n


E |ξk | p = O

n 1
(FHα (x) f k , f k ) H = 0 n→∞ n k=1


L F(x) = lim

for almost all x ∈ H. Let us show that there exists a function such that the estimates of the theorem are fulfilled for ε = 0 , but the function ceases to be harmonic. Consider F(x) =

x yk ∞ k
k=1

− 12

− 12

ϕk (λk z k )dz k dyk

56

Harmonic functions of infinitely many variables

on l2 , where ϕk (ζ ) are defined on R1 and ϕk (ζ ) = (2π )1/2 keζ

2

/2

ϕk (ζ ) = −(2π)1/2 keζ

for

2

/2

for

k+1  1 , , 2ψ0 (k) 2kψ0 (k)  k+1 1  ,− ζ ∈ − , 2kψ0 (k) 2ψ0 (k)

ζ ∈



and ϕk (ζ ) = 0 outside these intervals if k > k1 ; if k ≤ k1 then ϕk (ζ ) = 1 for ζ > 0; and ϕk (ζ ) = −1 for ζ < 0. Choose the number k1 to satisfy the condition (k1 ψ0 (k1 )/(k1 + 1)) > 1; xk = (x, ek )l2 ; {ek }∞ 1 is a canonical basis in l 2 . In this case n 1

L F(x) = lim ϕk (λk xk ). n→∞ n k=1 The random variables ξk (x) = ϕk (λk xk ) are independent. For all k, we have E ξk = 0. If k ≤ k1 , then E |ξk | = 1, and for k > k1 we have E |ξk | = 1/ψ0 (k). Therefore, if p = 1, then

n
n E |ξk | = O ψ0 (n) k=1 and condition (4.1) is satisfied for ε = 0. For k > k1 , the probability P(|ξk | ≥ k) = µ{x ∈ l2 : |ξk (x)| ≥ k} k+1/2kψo(k)  2 1 2 = √ , e−ζ /2 dζ ≥ √ 2π 2π e1/8 kψ0 (k) 1/2ψ0 (k)

because for k > k1 we have (kψ0 (k)/(k + 1)) > 1, and for  1 k+1  ζ ∈ , 2ψ0 (k) 2kψ0 (k) 2 − (k+1)  ζ2 2 2 we have e− 2 ≥ e 8k ψ0 (k) ≥ e−1/8 . The series ∞ k=k1 1/(kψ0 (k) diverges, there∞ fore k=1 P(|ξk | ≥ k) = ∞, and, according to the lemma of Borel–Cantelli, P(|ξk | ≥ k i.m.t.) = 1, where i.m.t. means ‘infinitely many times’.  Therefore, the relation n1 nk=1 ξk (x) → 0 almost surely (i.e.
L F(x) = 0 for almost all x ∈ l2 ) is not valid. Indeed, if (4.2) holds, then ξk (x)/k → 0 almost surely, which contradicts the fact that P(|ξk | ≥ k i.m.t.) = 1. To prove the final part of the theorem, choose the orthonormal basis {gk }∞ 1 ∞ in H, gk ∈ Hα , to be different from { f k }∞ 1 and square close to { f k }1 , i.e.,

∞
k=1

gk − f k 2H < ∞.

4.1 Arbitrary second-order derivatives

57

Then for p ≤ 1 n  n  

  |(FHα (x)gk , gk ) H | p µ(d x) − |(FHα (x) f k , f k ) H | p µ(d x)  k=1

k=1

H

≤

n 
k=1

=

n 
k=1

H

|(FHα (x)gk , gk ) H − (FHα (x) f k , f k ) H | p µ(d x)

H

|(FHα (x)(gk − f k ), gk ) H + (FHα (x) f k , gk − f k ) H | p µ(d x)

H

≤ 2c p

n


p

gk − f k  H ,

k=1

since F(x) is differentiable with respect to Hα , and hence the operator FHα (x) ∈ {Hα → H−α } : FHα (x)h H−α ≤ C(x)h Hα is bounded, we get |(FHα (x)h, g) H | ≤ C(x)h Hα g Hα , and



|(FHα (x)h, g) H |µ(d x) ≤ ch H g H ,

H

 c=

C(x)µ(d x),

h, g ∈ Hα

H

But p ∞
gk − f k  H k=1

[kψε (k)]

2− p 2

≤

∞ 


gk −

f k 2H

" ∞  p/2


k=1

k=1

1 kψε (k)

# 2−2 p ,

by the H¨older inequality, and ∞


gk − f k 2H < ∞,

k=1

∞
k=1

1 <∞ kψε (k)

(ε > 0),

so, by Kronecker’s lemma, n


  2− p p gk − f k  H = o [nψε (n)] 2 .

k=1 2− p

Hence, taking into account that [nψε (n)] 2 = o(n p /ψε (n)) for p > 23 n  p p and we obtain k=1 H |(FHα (x) f k , f k ) H | µ(d x) = O(n /ψε (n)),

58

Harmonic functions of infinitely many variables 

n

|(FHα (x)gk , gk ) H | p µ(d x) = O(n p /ψε (n)). According to the first statement of this theorem, n 1

L F(x) = lim (FHα (x)gk , gk ) H = 0 n→∞ n k=1 k=1

H

for almost all x ∈ H.




Condition (4.1) of the theorem is in any case not necessary. One can see this by choosing the function ∞ 1
3 F(x) = λk xk λk+1 xk+1 6 k=1 on l2 . For this function, n 1
λk xk λk+1 xk+1 . n→∞ n k=1


L F(x) = lim

The random variables ξk (x) = λk xk λk+1 xk+1 are orthogonal, E ξk = 0, E ξk2 =  1, nk=1 E ξk2 = n ≤ c(n 2 /(ψε (n) ln2 n)), and it follows from Theorem 4.2 to be presented below that
L F(x) = 0 for almost all x ∈ l2 . At the same time, condition (4.1) is not satisfied, because   1 ∞ 2 2 2 E |ξk | = |λk xk λk+1 xk+1 | µ(d x) = √ |y|e−y /2 dy = , π 2π n

−∞

l2

and k=1 E |ξk | = π2 n. A similar phenomenon can be seen in the case ∞ 1
F(x) = λk xk3 6 k=1 on l2 . For this function, n 1
λk x k . n→∞ n k=1


L F(x) = lim

The random variables ξk (x) = λk xk are independent, E ξk = 0, E ξk2 = 1, n 2 2 k=1 E ξk = n ≤ c(n /ψε (n)), and it follows from Theorem 4.3, also presented below, that
L F(x) = 0 for almost all x ∈ l2 . At the same time, condition (4.1) is not satisfied, because  ∞  2 1 −y 2 /2 E |ξk | = |λk xk | µ(d x) = √ , |y|e dy = π 2π −∞

l2

and

n k=1

E |ξk | =

*

2 π

n.

4.2 Orthogonal and stochastically independent derivatives

59

The reason for this phenomenon is the fact that, for given sums of orthogonal and independent random variables, there exist more precise estimates than for a sum of arbitrary (dependent) random variables. However, in this case the function F(x) has to satisfy additional restrictions.

4.2 Orthogonal and stochastically independent second order derivatives Theorem 4.2 Let F(x) be a function twice differentiable with respect to the subspace Hα for some α > 0, and ξk (x) = (FHα f k , f k ) ∈ L2 (H, µ), { f k }∞ 1 be an orthonormal basis in H, f k ∈ Hα . Let, in addition, functions ξk (x) satisfy the conditions (ξk , 1)L2 (H,µ) = 0, (ξ j , ξk )L2 (H,µ) = 0 for j = k, j, k = 1, 2, . . . . If n


ξk (x)2L2 (H,µ) = O



k=1

 n2 , ψε (n) ln2 n

(4.3)

with ψε (n) determined in Theorem 4.1, then
L F(x) = 0

for almost all x ∈ H.

(4.4)

On the other hand, there exists a function F(x) such that (4.3) is satisfied for ε = 0 but (4.4) does not hold. Estimate (4.3) (and, therefore, the harmonicity of F(x)) does not depend on the choice of basis from the class of square close bases. Proof. The L´evy Laplacian (1.3) is the limit of the sequence of arithmetic means 1 n ∞ k=1 ξk (x). It follows from the conditions of the theorem that {ξk (x)}1 is a n sequence of orthogonal random variables, with expectation E ξk = (ξk , 1)L2 (H,µ) = 0, E ξk ξ j = 0 ( j = k), and variance Var ξk = ξk 2L2 (H,µ) < ∞, and

n
k=1

Var ξk = O



 n2 . ψε (n) ln2 n

According to Petrov’s theorem [109], if {ζk }∞ of or1 is the sequence   thogonal random variables, Bn = nk=1 Var ζk → ∞, then n1 nk=1 ζk = √ o( Bn f (Bn ) ln n) almost surely, where f (τ ) is a positive function on [τ0 , ∞)  such that ∞ f (n) = ψε (n), ε > 0). It folk=k0 (1/k f (k)) < ∞ (for example, n lows from this theorem that if E ζk = 0, k=1 Var ζk = O(n 2 /ψε (n) ln2 n),  then n1 nk=1 ζk → 0 almost surely. Therefore, according to the conditions

60

Harmonic functions of infinitely many variables

of Theorem 4.2,

1 n

n

k=1 ξk

→ 0 almost surely, i.e., n 1
(FHα (x) f k , f k ) H = 0 n→∞ n k=1


L F(x) = lim

for almost all x ∈ H. Now let us show that there exists a function F(x) such that although the conditions of Theorem 4.2 are fulfilled, for ε = 0, F(x) fails to be harmonic. In the space l2 , consider a function F(x) defined on l2 , by + ∞ k−1  1
k F(x) = λ j x j λk xk3 2 6 k=k0 ψ0 (k) ln k j=1 where xk = (xk , ek )l2 , {ek }∞ 1 is a canonical basis in l 2 and k0 is chosen so as to ensure that ln · · · ln k0 > 0. m

Then n 1

L F(x) = lim n→∞ n k=k0

The random variables

+

ξk (x) =

+

k  k λjxj. ψ0 (k) ln2 k j=1

k  k λjxj ψ0 (k) ln2 k j=1

are orthogonal: E ξk = 0, Var ξk = Therefore n
k=1

Var ξk = O

k . ψ0 (k) ln2 k

n2 ψ0 (n) ln2 n

,

and condition (4.3) is satisfied for ε = 0.  At the same time, the relation n1 nk=1 ξk (x) → 0 for almost all x ∈ l2 does not hold, because n n n−1 k

1
ξk (x)
1 − ξk (x) = ξ j (x) n k=k0 k k(k + 1) j=1 k=k0 k=k0

 (Abel’s transformation), the series ∞ k=k0 ξk (x)/k diverges, whereas the series ∞ k k=k0 1/(k(k + 1)) j=1 ξ j (x) converges almost everywhere on l 2 . Indeed, ∞ ∞

ξk (x) = ck ξˆk (x), k k=k0 k=k0

4.2 Orthogonal and stochastically independent derivatives

61

where ξˆk (x) =

ξk (x) ξk (x)L2 (H,µ)

is a orthonormal system, 1 ck = √ kψ0 (k) ln k is a positive monotone decreasing numerical sequence, ∞


ck2 ln2 k =

k=k0

∞
k=k0

1 = ∞, kψ0 (k)

and, by the Men’shov–Rademacher theorem, it follows that this series diverges. Due to the orthogonality of ξk (x) and condition (4.3) at ε = 0, we have  
∞ k ∞ k  2 1/2  


1 1   ξ j (x)µ(d x) ≤ ξ j (x) µ(d x)  k(k + 1) k(k + 1) k=k0 k=k0 j=1 j=1 l2

∞


=

k=k0

1 k(k + 1)

and the series

k  


∞ k=k0

j=1

l2

∞ 1/2
ξ 2j (x)µ(d x) ≤A k=k0

l2

1/(k(k + 1))

k j=1

1 < ∞, √ (k + 1) ψ0 (k) ln k

ξ j (x) converges for almost all x ∈ l2 .

Proof of the last statement of the theorem. ∞ Let {gk }∞ 1 be an orthonormal basis in H, gk ∈ Hα , distinct from { f k }1 and ∞ square close to { f k }1 . Without loss of generality, assume that (FHα (x)gn , gn ) H n

→0

 (a necessary condition for n1 nk=1 (FHα (x)gk , gk ) H → 0). Then n  n  

  (FHα (x)gk , gk )2H µ(d x) − (FHα (x) f k , f k )2H µ(d x)  k=1

k=1

H

H

n   
  (FHα (x)gk , gk ) H − (FHα (x) f k , f k ) H = k=1

H

    × (FHα (x)gk , gk ) H + (FHα (x) f k , f k ) H µ(d x) n   
  (FHα (x)(gk − f k ), gk ) H + (FHα (x) f k , gk − f k ) H = k=1

H

n   
 × (FHα (x)gk , gk ) H + (FHα (x) f k , f k ) H µ(d x) ≤ 4c kgk − f k  H , k=1

62

Harmonic functions of infinitely many variables

because it follows from the condition of differentiability of the function F(x) with respect to subspace Hα that  |(FHα (x)h, g) H |µ(d x) ≤ ch H g H , h, g ∈ Hα . H

But ∞ ∞ ∞ 1/2 


kgk − f k  H 1 1/2 , ≤ gk − f k 2H , kψε (k) k 3 ψε (k) k=1 k=1 k=1

and ∞


∞


1 < ∞ (ε > 0), kψ ε (k) k=1 k=1 ,  so, according to the Kronecker lemma, nk=1 kgk − f k  H = o( n 3 ψε (n)). Hence, taking into account that gk − f k 2H < ∞,

n


(FHα (x) f k , f k ) H 2L2 (H,µ) = O



k=1

 n2 , ψε (n) ln2 n

we have n


(FHα (x)gk , gk ) H 2L2 (H,µ) = O



k=1

 n2 . ψε (n) ln2 n

According to the first statement of the theorem, n 1
(FHα (x)gk , gk ) H = 0 n→∞ n k=1


L F(x) = lim for almost all x ∈ H.




Theorem 4.3 Let function F(x) be twice differentiable with respect to the subspace Hα for some α > 0, and ξk (x) = (FHα f k , f k ) H ∈ L2 (H, µ), where { f k }∞ 1 is an orthonormal basis in H, f k ∈ Hα . Let also the functions ξk (x) satisfy the conditions   m  m  g pk (ξ pk (x))µ(d x) = g pk (ξ pk (x))µ(d x) (4.5) H

k=1

k=1

H

for any finite number of functions ξ p1 (x), . . . , ξ pm (x) and for any bounded and continuous functions g p1 (τ ), . . . , g pm (τ ), τ ∈ R1 . If n
k=1

ηk (x)2L2 (H,µ) = O

 n2  , ψε (n)

(4.6)

4.2 Orthogonal and stochastically independent derivatives

where ηk (x) = ξk (x) − Theorem 4.1, and

 H

63

ξk (x)µ(d x), the function ψε (n) is determined in

n  1
ξk (x)µ(d x) = 0; n→∞ n k=1

lim

H

then
L F(x) = 0 for almost all x ∈ H.

(4.7)

On the other hand, there exists a function F(x) for which at ε = 0 (4.6) is satisfied, but (4.7) does not hold. Estimate (4.6) (and, therefore, the harmonicity of F(x)) does not depend on the choice of the basis from the class of square close bases. Proof. It follows from the conditions of the theorem that {ξk (x)}∞ 1 is a sequence of independent random variables, {ηk (x)}∞ is the sequence given by 1 ηk (x) = ξk (x) − E ξk (x), and the variances satisfy Var ξk = ηk (x)2L2 (H,µ) < ∞, and

n


  Var ξk = O n 2 /ψε (n) .

k=1

 According to Theorem 25 from Petrov [110], n1 nk=1 ηk → 0 almost surely.  But, by the above-stated condition of our theorem, n1 nk=1 E ξk → 0 almost  everywhere on H n1 nk=1 ξk → 0, i.e., n 1
(FHα (x) f k , f k ) H = 0 n→∞ n k=1


L F(x) = lim for almost all x ∈ H.




Let us show that there exists a function F(x) such that at ε = 0 (4.6) holds but F(x) is not harmonic. Consider the function xk yk ∞
F(x) = ϕk (λk z k ) dz k dyk k=k1 − 2k

1 1 ψ0 (k1 )

− 2k

1 1 ψ0 (k1 )

on l2 , where ϕk (ζ ) is a function on R1 such that, for k > k1 we have   1 2 ϕk (ζ ) = (2π)1/4 keζ /4 for ζ ∈ 0, , 2kψ0 (k)   1 2 ϕk (ζ ) = −(2π)1/4 keζ /4 for ζ ∈ − ,0 , 2kψ0 (k)

64

Harmonic functions of infinitely many variables

and ϕk (ζ ) = 0 outside of these intervals; xk = (x, ek )l2 , {ek }∞ 1 is a canonical basis in l2 ; the number k1 is chosen to be such that ln · · · ln k 1    > 0. m

The L´evy Laplacian of this function has the form n 1
ϕk (λk xk ). n→∞ n k=k1


L F(x) = lim

The random variables ξk (x) = ϕk (λk xk ) are independent, E ξk = 0,

Var ξk =

k . ψ0 (k)

Therefore n


Var ξk = O

k=1

 n2  , ψ0 (n)

and condition (4.6) is satisfied for ε = 0. For k > k1 , the probability P(|ξk | ≥ k) = µ{x ∈ l2 : |ξk | ≥ k} 1 = √ 2π

1 2kψ0 (k)



− 2kψ1 (k)

e−1/8k1 ψ0 (k1 ) dζ ≥ √ , 2π kψ0 (k) 2

e−ζ

2

/2

2

0

because for ζ ∈ (−1/(2kψ0 (k)), 1/(2kψ0 (k))) we have e−ζ /2 ≥ e−1/8k1 ψ0 (k1 ) .  ∞ The series ∞ k=k1 1/(kψ0 (k)) diverges, therefore k=k1 P(|ξk | ≥ k) = ∞, and, according to the Borel–Cantelli lemma P(|ξk | ≥ k for i.m.t.) = 1.  Therefore, the relation n1 nk=k1 ξk → 0 almost surely is not satisfied (i.e.,
L F(x) = 0 for almost all x ∈ l2 ). Indeed, if (4.7) did hold, then it would be necessary that ξk (x)/k → 0 almost surely, which contradicts the equality P(|ξk | ≥ k for i.m.t.) = 1. The proof of the final part of Theorem 4.3 is similar to the final part of the proof of the Theorem 4.2. 2

2

2

4.3 Translationally non-positive case One can find functions for which conditions of Theorems 4.1–4.3 are not satisfied but which, nevertheless, are harmonic. To this end one more condition can be useful and will be given below. Notice that this condition does not include the requirement of orthogonality or stochastic independence of second derivatives, and so it is a certain complement to Theorem 4.1.

4.3 Translationally non-positive case

65

Let us construct an example of a function for which the conditions of harmonicity of Theorems 4.1–4.3 are not satisfied, but it is harmonic µ-almost everywhere. Consider the function ∞
1 F(x) = (e y2k−1 +y4k−2 + e−2y2k−1 + e−2y4k−2 ) ln 2kλ 2k−1 λ4k−2 k=1 on l2 , where yk = λk xk , xk = (x, ek ) H , {ek }∞ 1 is a canonical basis in l 2 . The L´evy Laplacian of this function is n 1

L F(x) = lim ξk (x), n→∞ n k=1 where ξ pk = 0

for pk = 2k − 1, 4k − 2, λ2k−1 ξ2k−1 (x) = (e y2k−1 +y4k−2 + 4e−2y2k−1 ), ln 2kλ4k−2 λ4k−2 ξ4k−2 (x) = (e y2k−1 +y4k−2 + 4e−2y4k−2 ), k = 1, 2, . . . . ln 2kλ2k−1

(4.8)

The condition of harmonicity (4.1) in Theorem 4.1 is not satisfied, because  cλ2k−1 |ξ2k−1 (x)| p µ(d x) = , ln 2kλ4k−2 l2  cλ4k−2 |ξ4k−2 (x)| p µ(d x) = . ln 2kλ2k−1 l2

Both the conditions (ξk , ξ j )L2 (H,µ) = 0 ( j = k) in Theorem 4.2 and (4.5) of Theorem 4.3 are not satisfied. Nevertheless, the function F(x) is µ–harmonic almost everywhere. It satisfies the conditions of Theorem 4.4 given below, and hence is harmonic. Theorem 4.4 Let function F(x) be twice differentiable with respect to the subspace Hα for some α > 0, and ξk (x) = (FHα f k , f k ) H ∈ L2 (H, µ), where { f k }∞ 1 is some orthonormal basis in H, f k ∈ Hα . If the functions ξk (x) satisfy the conditions 

 ξ j (x)ξk (x)µ(d x) ≤

H



ξk (x) ≥ 0,  ξ j (x)µ(d x) ξk (x)µ(d x) for j = k,

H

ξk (x)µ(d x) ≤ ∞,

sup k≥1 H

H ∞ η 2
k L (H,µ) 2

k=1

k2

< ∞,

66

Harmonic functions of infinitely many variables

where

 ηk (x) = ξk (x) −

ξk (x)µ(d x), H

and if

n  1
ξk (x)µ(d x) = 0, n→∞ n k=1

lim

H

then
L F(x) = 0

for almost all x ∈ H.

Proof. It follows from the conditions of the theorem that {ξk }∞ 1 is the sequence of arbitrary (dependent) random variables, and ξk (x) ≥ 0, E ξ j ξk ≤ E ξ j E ξk for j = k,  ∞
Var ξk < ∞, sup ξk (x) µ(d x) < ∞. k2 k≥1 k=1 H

Note that the Gramm matrix (ηˆ j , ηˆ k )L2 (H,µ) ∞ ˆ k (x)}∞ j,k=1 of functions {η 1 , where ηˆ k (x) = ηk (x)/ηk (x)L2 (H,µ) , i.e., the matrix of the coefficients of correlation of the random variables {ξk (x)}∞ 1 , is a translationally non-positive matrix1 , since (ηˆ j , ηˆ k )L2 (H,µ) = 1, and by the condition of the theorem    ξ j (x)ξk (x) µ(d x) ≤ ξ j (x) µ(d x) ξk (x) µ(d x) for j = k H

H

H

so (ηˆ j , ηˆ k )L2 (H,µ) ≤ 0 for j = k, j, k = 1, 2, . . . .  According to the Etemadi theorem [35], n1 nk=1 ηk (x) → 0 almost surely.  But, by the assumptions of our theorem, n1 nk=1 E ξk (x) → 0, therefore 1 n k=1 ξk (x) → 0 almost surely, i.e., n n 1
(FHα (x) f k , f k ) H = 0 n→∞ n k=1


L F(x) = lim for almost all x ∈ H.




Let us come back to the above example. The functions ξk (x), k = 1, 2, . . . , given by (4.8), satisfy the conditions of Theorem 4.4: ξ2k−1 (x) > 0, ξ pk (x) = 0 1

for

ξ4k−2 (x) > 0,

pk = 2k − 1, 4k − 2,

The matrix a jk ∞ j,k=1 is called translationally non-positive, if for some ν > 0, a jk − νδ jk ≤ 0.

4.3 Translationally non-positive case

67

so we have ξk (x) ≥ 0;  5e4 + 8e ξ2k−1 (x)ξ4k−2 (x) µ(d x) = ln2 2k l2





(4e2 + e)2 , ln2 2k l2 l2    ξ2i−1 (x)ξ4l−2 (x) µ(d x) = ξ2i−1 (x) µ(d x) ξ4l−2 (x) µ(d x) for i = l;

≤

ξ2k−1 (x) µ(d x)

ξ4k−2 (x) µ(d x) =

l2

l2

therefore



l2

 ξ j (x)ξk (x) µ(d x) ≤

l2

 ξ j (x) µ(d x)

l2

η2k−1 2L2 (H,µ) =

ξk (x) µ(d x), l2

bλ22k−1 ln2 2kλ24k−2

, η4k−2 2L(H,µ) =

bλ24k−2 ln2 2kλ22k−1

,

hence ∞ η 2
k L (H,µ) 2



k2

k=1

cλ2k−1 ξ2k−1 (x) µ(d x) = , ln 2kλ4k−2

l2

(b, c are constants), so

< ∞,

 ξ4k−2 (x) µ(d x) =

cλ4k−2 ln 2kλ2k−1

l2 n  1
lim ξk (x) µ(d x) = 0. n→∞ n k=1 l2

According to Theorem 4.4,
L F(x) = 0 for almost all x ∈ l2 .

5 Linear elliptic and parabolic equations with L´evy Laplacians

The theory of linear equations with L´evy Laplacians is an essential part of infinite-dimensional analysis. Numerous authors have obtained solutions of various kinds of the problems in a variety of functional classes.

5.1 The Dirichlet problem for the L´evy–Laplace and L´evy–Poisson equations Consider the Dirichlet problem for the L´evy–Laplace equation
L U (x) = 0, and for the L´evy–Poisson equation
L U (x) = (x). Let  be a bounded open set in a Hilbert space H (i.e. a domain in H ), and ¯  =  ∪ be a domain in H with boundary . The Dirichlet problem for the equation
L U (x) = (x) is to find the function U (x), which satisfies the equation
L U (x) = (x) in the domain  ⊂ H and is equal to some given function G(x) on the boundary of this domain. We define in H the domain  = {x ∈ H : 0 ≤ Q(x) < 1} with the boundary = {x ∈ H : Q(x) = 1} where Q(x) is a twice differentiable function such that is a convex bounded surface in H, and
L Q(x) is a constant positive non-zero number. We will call such a domain fundamental. As simple examples, we mention the ball ¯ = {x ∈ H : x2H ≤ 1}  68

5.1 The Dirichlet problem for L´evy–Laplace equations

69

and the ellipsoid ¯ = {x ∈ H : (Bx, x) H ≤ 1},  where B is a thin operator (i.e., B = γ E + S(x), E is the identity operator, and S(x) is a compact operator in H ), and γ > 0. Consider the Shilov class of functions, i.e. the set of functions of the form (x) = φ(x, x2H ), where φ(x, ξ ) is a function defined and twice differentiable on H × R1 , such that φ(x, x2H ) = φ(P x, x2H ), P is a projection on an m-dimensional subspace. In this case, ∂φ 
L (x) = 2  . ∂ξ ξ =x2H This class contains cylindrical functions (x), i.e. functions of the form (x) = (P x). Theorem 5.1 (G. E. Shilov) Let G(x) = g(x, x2H ) be a function from the ¯ be a fundamental domain for which the function Q(x) has Shilov class and  the form Q(x) = x2H − (x) + 1, where (x) is a cylindrical positive, twice differentiable function. Then in this class there exists a unique solution of the problem
L U (x) = 0

in ,

U (x) = G(x) on ,

given by U (x) = g(x, (x)).

(5.1)

Proof. The function g(x, (x)) is cylindrical and, therefore, harmonic. On the boundary , the function Q(x) = 1, therefore U (x) = g(x, x2H ) = G(x) there. There exists a unique solution to the problem in the Shilov class. Since the function U (x) is cylindrical for some projector P on Rm we have U (x) = U (P x) = u((x, a1 ) H , . . . , (x, am ) H ),

a1 , . . . , am ∈ H,

where u(ξ1 , . . . , ξm ) is a function of m variables. If U (x0 ) = c at the point x0 ∈ H then there exists a hyperplane α(x0 ) which goes through the point x0 such that U (x) remains constant. Note that a hyperplane in H is a manifold of finite co-dimension α(x) = {x ∈ H : (b1 , x) H = c1 , . . . , (bm , x) = cm },

b1 , . . . , bm ∈ H.

70

Linear equations with L´evy Laplacians

On the intersection of α(x0 ) and we also have U (z) = c, z ∈ . Therefore, for any harmonic function U (x) in  we have |U (x)|x∈¯ ≤ sup |U (x)|. It yields x∈

the uniqueness of the solution of the Dirichlet problem for the Levy–Laplace equation.
Theorem 5.2 (G. E. Shilov) Let (x) = φ(x, x2H ) be a function from the Shilov class. Under conditions of Theorem 5.1 there exists a unique solution to the problem
L U (x) = (x) in ,

U (x) = G(x) on ,

which is equal to 1 U (x) = g(x, (x)) − 2

(x) 

φ(x, τ )dτ.

(5.2)

x2H

Proof. From (5.2) by (1.4), we obtain 1
L U (x) =
L g(x, (x)) − φ(x, (x))
L (x) 2 1 + φ(x, x2H )
L x2H = φ(x, x2H ), 2 since (x) and g(x, (x)) are cylindrical and, therefore, harmonic, and by (1.2) we have
L x2H = 2. On the boundary we have Q(x) = 1, therefore we have here U (x) = g(x, x2H ) = G(x). The uniqueness of the solution to the Dirichlet problem for the Levy–Poisson equation follows from the uniqueness of the solution to the Dirichlet problem for the L´evy–Laplace equation.
One can extend (5.1) and (5.2) to general fundamental domains ¯ = {x ∈ H : 

0 ≤ Q(x) ≤ 1,
L Q(x) = c > 0}

and to the class of functions of the form (x) = φ(x, (Ax, x) H ), where A = γ A E + S A , E is the identity operator, and S A is a compact operator in H, φ(x, ξ ) is a function defined and twice differentiable on H × R1 , such that φ(x, (Ax, x) H ) = φ(P x, (Ax, x) H ), and P is a projection on an m-dimensional subspace. In this case, ∂φ 
L (x) = 2γ A  . ∂ξ ξ =(Ax,x)

5.1 The Dirichlet problem for L´evy–Laplace equations

71

We introduce a function T (x) =

1 − Q(x) ,
L Q(x)

which has the following properties: 1 0 < T (x) ≤ , T (x) = 0 at x ∈ ,
L T (x) = −1 at x ∈ .
L Q(x) Theorem 5.3 1. Let G(x) = g(x, (Ax, x) H ), A = γ A E + S A , and the domain ¯ be fundamental. Then the function  U (x) = g(x, 2γ A T (x) + (Ax, x) H ))

(5.3)

is the solution to the problem
L U (x) = 0

in

,

U (x) = G(x) on .

2. Let (x) = φ(x, (Bx, x) H ), B = γ B E + S B . Then the function 1 U (x) = g(x, 2γ A T (x) + (Ax, x) H ) H − 2γ B

2γ B T (x)+(Bx,x) H 

φ(x, τ )dτ (5.4) (Bx,x) H

is a solution to the problem
L U (x) = (x) in

,

U (x) = G(x) on .

The proof obviously follows from (1.4) and from the fact that
L T (x) = −1, and, hence
L [2γ A T (x) + (Ax, x) H ] = 0,
L [2γ B T (x) + (Bx, x) H ] = 0 in  In addition, T  = 0, and, therefore, on we have U (x) = G(x).

.


Consider equations in variational derivatives and functionals F(x) on the Hilbert space of functions L 2 (0, 1). Lemma 5.1 If a twice strongly differentiable functional V (x) attains its minimum (resp. maximum) on x0 (t), and
L V (x0 ) exists, then
L V (x0 ) ≥ 0

(resp.
L V (x0 ) ≤ 0).

The proof obviously follows from the fact that d 2 V (x0 ; h) ≥ 0 (resp. d 2 V (x0 ; h) ≤ 0) for all h ∈ L 2 (0, 1), at the point of minimum (resp. maximum) x0 , and, therefore by Lemma 1.1,
L V (x0 ) = M{d 2 V (x0 ; h)} ≥ 0

(resp.
L V (x0 ) ≤ 0).


72

Linear equations with L´evy Laplacians

Lemma 5.2 Let W (x) be twice strongly differentiable harmonic in , contin¯ functional. Then uous in  sup W (x) = sup W (x), ¯ x∈

inf W (x) = inf W (x).

¯ x∈

x∈

x∈

Indeed, given x(t) ∈ L 2 (0, 1) define 1 xη (t) = 2η

t+η x(s)ds,

x(t) = 0

outside of

[0, 1].

t−η

The function xη (t) is called a Steklov mean of a function x(t). η ¯η = Let L 2 (0, 1) be the image of L 2 (0, 1) under this mapping, and  η η ¯  ∩ L 2 (0, 1), η = ∩ L 2 (0, 1). It is known that xη (t) ∈ L 2 (0, 1), xη → x in ¯ η , η are closed in L 2 (0, 1) and compact for each L 2 (0, 1) as η → 0, and  fixed η. ¯ η , it attains its maxSince W (x) is harmonic in η and continuous on  imum value on η . Let, on the contrary, sup W (x) = W (x0 ) = Mη > m η = ¯ x∈

sup W (x). We set x∈ η

Mη − m η V (x) = W (x) + 2d

1 x 2 (s)ds, 0

where 1 d = sup x∈ η

x 2 (s)ds, 0

and show that it attains its maximum at a certain point x0 ∈ η . Let sup V (x) = x∈ η

V (x ), x ∈ η . Then we have Mη − m η V (x ) = W (x ) + 2d

1 x 2 (s)ds ≤ m η +

Mη − m η 2

0

Mη − m η = Mη − < Mη . 2 ¯ η , V (x) ≥ W (x), therefore sup V (x) ≥ sup W (x) = Mη . But for all x ∈  ¯η x∈

¯η x∈

This means that x0 ∈ η . By Lemma 5.1,
L V (x0 ) ≤ 0. Substituting V (x) here and taking into account that
L W (x0 ) = 0 (because x0 ∈ η ), we obtain that Mη ≤ m η , which is a contradiction. Thus, sup W (x) = sup W (x). ¯η x∈

x∈ η

5.1 The Dirichlet problem for L´evy–Laplace equations

73

Since limη→0 sup W (x) = sup W (x), ¯η x∈

limη→0 sup W (x) = sup W (x),

¯ x∈

x∈ η

x∈

we have sup W (x) = sup W (x). ¯ x∈

x∈

Similarly we obtain inf W (x) = inf W (x). ¯ x∈




x∈

Theorem 5.4 The solution to the Dirichlet problem for the equation
L U (x) = 0 is unique in the class of functionals U (x) that are twice strongly differentiable ¯ and such that
L U (x) exists. in , continuous in  Proof. Suppose that under the conditions of the theorem    
L U1 (x) = 0,
L U2 (x) = 0 in , and U1  = U2  = G(x).



Then for the functional W (x) = U1 (x) − U2 (x) we have  
L W (x) = 0 in , W  = 0.

By Lemma 5.2, W (x) = 0, and so U1 (x) ≡ U2 (x).




Theorem 5.5 The solution to the Dirichlet problem for the equation
L U (x) = (x) is unique in the class of functionals U (x) that are twice strongly differen¯ and such that
L U (x) exists. tiable in , continuous in  Proof. Suppose that under the conditions of the theorem
L U1 (x) = (x),


L U2 (x) = (x)

in ,

    and U1  = U2  = 0.



Then for the functional W (x) = U1 (x) − U2 (x) we have  
L W (x) = 0 in , W  = 0.

By Lemma 5.2, W (x) = 0, and so U1 (x) ≡ U2 (x).




Consider the class of Gˆateaux functionals; that is, the set of functionals of the form 1 F(x) =

1 ···

0

f (x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N , 0

where f (ξ1 , . . . , ξ N ; t1 , . . . , t N ) is a continuous function of 2N variables

74

Linear equations with L´evy Laplacians

(−∞ < ξk < ∞, 0 ≤ tk ≤ 1, k = 1, 2, . . . , N ), and # " N

L Q f (ξ1 , . . . , ξ N ; t1 , . . . , t N ) = O b . ξk2 , 0 < b < 4 k=1 Theorem 5.6 (E. M. Polishchuk) Assume that 1 G(x) =

1 ···

0

g(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N , 0

¯ is fundamental and dQ(x; h). belongs to the Gˆateaux class, the domain  2 d Q(x, h) have normal form. Then the functional 1 U (x) = (4π T (x)) N /2

1

1 ···

0

∞ ×

dt1 · · · dt N 0

∞ ···

−∞

g(ζ1 , . . . , ζ N ; t1 , . . . , t N )e

−

N  k=1

(x(tk )−ζk )2 4T (x)

dζ1 · · · dζ N , (5.5)

−∞

where T (x) = (1 − Q(x))/(
L Q(x)), solves the problem
L U (x) = 0

in , U (x) = G(x) on .

Proof. Rewrite (5.5) in the form 1 U (x) =

1 ···

0

u(x(t1 ), . . . , u(t N ); t1 , . . . , t N , T (x))dt1 · · · dt N ,

(5.6)

0

where u(ξ1 , . . . , ξ N ; t1 , . . . , t N ; τ ) =

∞ ∞ 1 · · · g(ζ1 , . . . , ζ N ; t1 , . . . , t N ) (4π τ ) N /2 −∞ −∞ & ' N
(ξk − ζk )2 × exp − dζ1 · · · dζ N 4τ k=1

is a solution to the Cauchy problem

  u

N
∂u ∂ 2u = 2 ∂τ k=1 ∂ξk

τ =0

(0 < τ ≤

1
L Q

= g(ξ1 , . . . , ξ N ; t1 , . . . , t N ).

, −∞ < ξk < ∞), (5.7)

5.1 The Dirichlet problem for L´evy–Laplace equations

75

Let us derive the expression for the second differential of (5.6). Since we can differentiate under the integral sign we obtain 1 d U (x; h) =

···

2

1 &
N

0

∂ 2u h(t j )h(tk )dt j dtk ∂ξ j ∂ξk j,k=1

0

N
∂ 2u h(tk )dT (x; h) ∂ξk ∂τ k=1  2 ∂u ∂ 2u  2 + d dT (x; h) + T (x; h) dt1 · · · dt N . ∂τ 2 ∂τ

+2

By virtue of Lemma 1.1, taking into account that the differentials of T (x) have normal form (because according to the condition of the theorem, the differentials of Q(x) have normal form) we obtain ' 1  1 &
N ∂ 2u ∂u
L U (x) = · · ·
L T (x) dt1 · · · dt N . + 2 ∂τ k=1 ∂ξk 0

0

But
L T (x) = −1, therefore 1
L U (x) =

··· 0

 1 &
N 0

k=1

∂ 2u ∂u − ∂τ ∂ξk2

' dt1 · · · dt N .

Since u(ξ1 , . . . , ξ N ; t1 , . . . , t N ; τ ) is a solution to the Cauchy problem (5.7) then
L U (x) = 0 in the domain . On the boundary , we have T (x) = 0. Therefore, on this boundary 1 U (x) =

1 ···

u(x(t1 ), . . . , x(t N ); t1 , . . . , t N , 0)dt1 · · · dt N

0

0

1

1

=

··· 0

g(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N = G(x). 0




Theorem 5.7 (E. M. Polishchuk) Let 1 (x) =

1 ···

0

ϕ(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N 0

be a functional from the Gˆateaux class. Under the conditions of Theorem 5.6,

76

Linear equations with L´evy Laplacians

the functional 1 U (x) = (4π T (x)) N /2

1

1 ···

dt1 · · · dt N

0

0

∞ ×

∞ ···

−∞

−

T (x)1 0

···

∞ dt1 . . . dt N

N  k=1

∞ ···

−∞

0 −

−

N  k=1

(x(tk )−ζk )2 4T (x)

dζ1 · · · dζ N

−∞

1

0

×e

g(ζ1 , . . . , ζ N ; t1 , . . . , t N )e

(x(tk )−ζk )2 4(T (x)−t)

−∞

ϕ(ζ1 , . . . , ζ N ; t1 , . . . , t N ) [4π(T (x) − t)] N /2

 dζ1 · · · dζ N dt

(5.8)

is a solution to the problem
L U (x) = (x) in , U (x) = G(x) on . Proof. Rewrite (5.8) in the form 1 U (x) =

1  u(x(t1 ), . . . , x(t N ); t1 , . . . , t N ; T (x)) ···

0

0

 + v(x(t1 ), . . . , x(t N ); t1 , . . . , t N ; T (x)) dt1 · · · dt N ,

where u(ξ1 , . . . , ξ N ; t1 , . . . , t N ; τ ) is a solution to the Cauchy problem (5.7), and v(ξ1 , . . . , ξ N ; t1 , . . . , t N ; τ ) τ  ∞ ∞ ϕ(ζ1 , . . . , ζ N ; t1 , . . . , t N ) =− ··· [4π(τ − t)] N /2 0

×e

−

−∞

N  k=1

−∞

(ξk −ζk )2 4(τ −t)

 dζ1 · · · dζ N dt

is a solution to the Cauchy problem N
∂ 2v 2 k=1 ∂ξk

∂v = ϕ(ξ1 , . . . , ξ N ; t1 , . . . , t N ) ∂τ

−

  v

τ =0

(0 < τ ≤

1 , −∞ < ξk < ∞),
L Q

= 0.

The final part of the proof is similar to the proof of Theorem 5.6.




5.1 The Dirichlet problem for L´evy–Laplace equations

77

Note that solutions (5.5) and (5.8) belong to the uniqueness class mentioned in Theorems 5.4 and 5.5. Let us continue our study of the class of Gˆateaux functionals. The Dirichlet problem for the L´evy–Laplace and L´evy–Poisson equations in the class of Gˆateaux functionals is connected with the finite-dimensional heat equation (Theorems 5.6 and 5.7). At the same time, the fundamental solution of the heat equation gives rise to the Wiener measure. This lets us obtain the solution of the Dirichlet problem using the integral with respect to the Wiener measure. Let Lˆ 2 (0, 1) be the space of functions x(t), square integrable on [0, 1] and satisfying the condition (x, 1) L 2 (0,1) = 0; Lˆ 2 (0, 1) is a subspace of L 2 (0, 1). In Lˆ 2 (0, 1), one can define the Wiener measure – the Gaussian measure with the zero mean value and the correlational operator 1 K x(s) = 2

1

1 min(t, s)x(s)ds − 2

0

11 min(ξ, s)x(s)dsdξ 0 0

(see Section 1.3). Theorem 5.8 Let 1 G(x) =

1 ···

0

g(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N 0

be a functional from the Gˆateaux class. Let g(ξ1 , . . . , ξ N ; t1 , . . . , t N ) be twice continuously differentiable in ξ1 , . . . , ξ N and, together with its derivatives, inteN 2 ξi N N /2 − 12 i=1 grable in R with respect to the measure having density (1/(2π ) )e . 2 ˆ Let in addition the domain  be fundamental, and d Q(x; h) and d Q(x; h) have normal form. Then there exists a unique solution to the problem
L U (x) = 0 in

, U (x) = G(x) on ,

which is equal to  1 U (x) =

1 ···

Lˆ 2 (0,1) 0

, g(x(t1 ) + 2 T (x) Y (t1 ), . . . , x(t N )

0

, + 2 T (x) Y (t N ); t1 , . . . , t N )dt1 · · · dt N µw (dy),

(5.9)

where T (x) =

1 − Q(x) ,
L Q(x)

Y (tk ) =

y(σtk ) − y(σtk−1 ) ; √ tk

σtk =

k
i=1

ti ,

σt0 = 0.

78

Linear equations with L´evy Laplacians

Proof. Let Cˆ 0 (0, 1) denote the set of all continuous functions from Lˆ 2 (0, 1). There exists a one-to-one correspondence between C0 (0, 1) and Cˆ 0 (0, 1) given by 1 z(t) = y(t) − y(0),

y(t) = z(t) −

z(s)ds 0

(y(t) ∈ C0 (0, 1),

z(t) ∈ Cˆ 0 (0, 1)).

If a functional F(y) is integrable in the Wiener measure on Lˆ 2 (0, 1) and (z) = 1 F(z − 0 z(s)ds) is integrable with respect to the classical Wiener measure on C0 (0, 1), then   F(y)µw (dy) = (z)µw (dz). C0

Cˆ 0

The space Cˆ 0 (0, 1) has full measure in Lˆ 2 (0, 1) hence   F(y)µw (dy) = (z)µw (dz). Lˆ 2 (0,1)

C0 (0,1)

Since

⎡

y(σtk ) − y(σtk−1 ) = ⎣z(σtk ) −

1

⎤ z(s)ds ⎦ − [z(σtk−1 ) −

0

1 z(s)ds] 0

= z(σtk ) − z(σtk−1 ), we can rewrite (5.9) in the form 

1

U (x) =

1 ···

C0 (0,1) 0

, g(x(t1 ) + 2 T (x) Z (t1 ), . . . , x(t N )

0

, + 2 T (x) Z (t N ); t1 , . . . , t N )dt1 · · · dt N µw (dz), √ where Z (tk ) = (z(σtk ) − z(σtk−1 )/ tk )(z(t) ∈ C0 (0, 1)). Let us find the second differential of the functional U (x). By differentiating under the integral sign we obtain  1 d U (x; h)) =

···

2

C0 (0,1) 0

1 
N 0

∂ 2 g(x (t1 ), . . . , x (t N ); t1 , . . . , t N ) ∂ξ j ∂ξk j,k=1

 δT (x; h)  δT (x; h)  × h(t j ) + Z (t j ) √ h(tk ) + Z (tk ) √ T (x) T (x)

5.1 The Dirichlet problem for L´evy–Laplace equations

N
∂g(x (t1 ), . . . , x (t N ); t1 , . . . , t N )

+

∂ξk

k=1

79

Z (tk )

 δ 2 T (x; h) [δT (x; h)]2  − dt1 · · · dt N µw (dz), √ 2T 3/2 (x) T (x) √ where x (tk ) = x(tk ) + 2 T Z (tk ). By virtue of Lemma 1.1 taking into account that the differentials of T (x) have normal form (because by the condition of the theorem the differentials of Q(x) have normal form) we obtain ×

 1
L U (x) =

··· C0 (0,1) 0

1
N  k=1

0

∂ 2 g(x (t1 ), . . . , x (t N ); t1 , . . . , t N ) ∂ξk2


L T (x) ∂g(x (t1 ), . . . , x (t N ); t1 , . . . , t N )  Z (tk ) dt1 · · · dt N µw (dz). + √ ∂ξk T (x) But
L T (x) = −1, therefore  1
L U (x) =

··· C0 (0,1) 0

−√

1
N  0

k=1

∂ 2 g(x (t1 ), . . . , x (t N ); t1 , . . . , t N ) ∂ξk2

1 ∂g(x (t1 ), . . . , x (t N ); t1 , . . . , t N )  dt1 · · · dt N µw (dz). Z (tk ) ∂ξk T (x)

By changing the order of integration, we obtain
L U (x) =

1 N 
j,k=1

0

1  ··· 0 C0 (0,1)

 ∂ 2 g( (t ), . . . ,  (t ); t , . . . , t ) x 1 x N 1 N ∂ξk2

Z (tk ) ∂g(x (t1 ), . . . , x (t N ); t1 , . . . , t N )  dt1 · · · dt N µw (dz) −√ ∂ξk T (x) 1 1 N 
= · · · Ik (x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N , j,k=1

0

0

where  Ik (x(t1 ), . . . x(t N ); t1 , . . . , t N ) = C0 (0,1)

 ∂ 2 g( (t ), . . . ,  (t ); t , . . . , t ) x 1 x N 1 N ∂ξk2

Z (tk ) ∂g(x (t1 ), . . . , x (t N ); t1 , . . . , t N )  dt1 · · · dt N µw (dz). −√ ∂ξk T (x)

80

Linear equations with L´evy Laplacians

Let us compute the Wiener integral of the functional concentrated at points σt0 , σt1 , . . . σt N . If the functional F(x) = f (z(τ1 ), . . . , z(τm )), where f (ξ1 , . . . , ξm ) is a function on Rm , and 0 < τ1 < τ2 < · · · < τm ≤ 1, then  f (z(τ1 ), . . . , z(τm ))µw (dz) C0 (0,1)

= [

m 



π −m/2 (τi − τi−1

f (z 1 , . . . , z m )e )]1/2

−

m  i=1

(z i −z i−1 )2 τi −τi−1

dz 1 · · · dz m .

Rm

i=1

Since σt0 < σt1 < σt2 < · · · < σt N N t2 , . . . , σt N = i=1 ti ), we have

(here

σt0 = 0, σt1 = t1 , σt2 = t1 +

Ik (x(t1 ), . . . x(t N ); t1 , . . . , t N ) ∞ ∞  2 ˆ ˆ x (t N ); t1 , . . . , t N ) 1 ∂ g(x (t1 ), . . . ,  = · · · N  ∂ξk2 (π N ti )1/2 −∞ −∞ i=1

−√ ×e

ˆ x (t1 ), . . . ,  ˆ x (t N ); t1 , . . . , t N )  1 z k − z k−1 ∂g( √ tk ∂ξk T (x)

−

N 

(z i −z i−1 )2 ti

dz 1 · · · dz N , √ ˆ x (tk ) = x(tk ) + 2 T ((z k − z k−1 )/√tk ). where  √ √ Let us change z k to ζk via (z k − z k−1 )/ tk = ζk / 2; we obtain i=1

Ik (x(t1 ), . . . x(t N ); t1 , . . . , t N ) ∞ ∞  2 ˘ ˘ x (t N ); t1 , . . . , t N ) 1 ∂ g(x (t1 ), . . . ,  = · · · (2π) N /2 ∂ξk2 −∞

−∞

˘ x (t N ); t1 , . . . , t N )  − 12  ζi2 ˘ x (t1 ), . . . ,  ζk ∂g( e i=1 dζ1 · · · dζ N ∂ξk 2T (x) N ∞ ∞ 2 ˘ ˘ x (t N ); t1 , . . . , t N ) − 12  ζi2 ∂ g(x (t1 ), . . . ,  1 i=1 = ··· e dζ1 · · · dζ N (2π) N /2 ∂ξk2 N

−√

−∞

1 + (2π) N /2

∞ ···

−∞ N 

×

−∞

∞

−∞

√

˘ x (t N ); t1 , . . . , t N ) ˘ x (t1 ), . . . ,  1 ∂g( ∂ξk 2T (x)

∂ − 12 ζi2 e i=1 dζ1 · · · dζ N , ∂ζk

5.1 The Dirichlet problem for L´evy–Laplace equations

81

√ ˘ x (tk ) = x(tk ) + 2T ζk . where  Integrating by parts (in the second summand) we obtain that Ik = 0. Therefore,
L U (x) = 0 in the domain . On the boundary , the functional U (x) is equal to G(x), because on , we have T (x) = 0, and the Wiener measure of the whole space Lˆ 2 (0, 1) is equal to 1. In addition, U (x) lies in the uniqueness class indicated in Theorem 5.4.
Theorem 5.9 Let 1 (x) =

1 ···

0

ϕ(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N 0

be a functional from the Gˆateaux class. Let ϕ(ξ (t1 ), . . . , ξ (t N ); t1 , . . . , t N ) be twice continuously differentiable in ξ1 , . . . , ξ N and, together with its derivatives, integrable in R N with respect to the measure having the denN 2 ξi N /2 − 12 i=1 sity (1/(2π) )e . Then under the conditions of Theorem 5.8. there exists the unique solution to the problem
L U (x) = (x) in , U (x) = G(x) on , which is equal to  1 U (x) =

1 , · · · g(x(t1 ) + 2 T (x)Y (t1 ), . . . , x(t N )

Lˆ 2 (0,1) 0

0

, + 2 T (x)Y (t N ); t1 , . . . , t N )dt1 · · · dt N µw (dy) T (x)  1 1 , − · · · ϕ(x(t1 ) + 2 T (x) − ξ Y (t1 ), . . . , x(t N ) 0

Lˆ 2 (0,1) 0

0

, + 2 T (x) − ξ Y (t N ); t1 , . . . , t N )dt1 . . . dt N µw (dy)dξ. Proof. If one has the solution to the Dirichlet problem for the L´evy–Laplace equation (Theorem 5.8), it is sufficient to show that the solution of the Dirichlet problem with homogeneous boundary conditions for the L´evy–Poisson equation
L V (x) = (x) in , V (x) = 0 on , is equal to T (x) 

1

V (x) = −

1 ···

0 Lˆ 2 (0,1) 0

0

, ϕ(x(t1 ) + 2 T (x) − ξ Y (t1 ), . . . , x(t N )

, + 2 T (x) − ξ Y (t N ); t1 , . . . , t N )dt1 · · · dt N µw (dy)dξ.

82

Linear equations with L´evy Laplacians

Computing the second differential and the L´evy Laplacian of the functional V (x) (the scheme of computation does not differ from that given in the proof of Theorem 5.8), we obtain 1
L V (x) = −

1 ···

0

ϕ(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N
L T (x) 0

T (x) 

1

−

··· 0 Lˆ 2 (0,1) 0

1
N  k=1

0

∂ 2 ϕ(x,ξ (t1 ), . . . , x,ξ (t N ); t1 , . . . , t N ) ∂ξk2


L T (x) +√ Y (tk ) T (x) − ξ ∂ϕ(x,ξ (t1 ), . . . , x,ξ (t N ); t1 , . . . , t N )  × dt1 · · · dt N µw (dy)dξ, ∂ξk √ where x,ξ (tk ) = x(tk ) + 2 T − ξ Y (tk ). But
L T (x) = −1, and the second summand is equal to zero (see the proof of Theorem 5.8). Therefore
L V (x) = (x) in the domain . On the boundary the functional V (x) = 0, because on we have T (x) = 0, and the Wiener measure of the whole space Lˆ 2 (0, 1) is equal to 1. In addition, V (x) lies in the uniqueness class indicated in Theorem 5.5.
Consider the class of functionals which can be represented in the form of a series in terms of orthogonal polynomials P0 , Pmq (x), with m, q = 1, 2, . . . , introduced in Section 2.2. Theorem 5.10 Let G(x), be a functional defined and continuous on L 2 (0, 1), and which admits the representation in the form of a series in the polynomials ¯ Let P0 , Pmq (x), m, q = 1, 2, . . . absolutely and uniformly converging on . 2 ¯ the domain  be fundamental, and d Q(x; h) and d Q(x; h) have normal form. Then there exists a unique solution of the problem
L U (x) = 0 in

, U (x) = G(x) on ,

which is equal to ∞


U (x) = G 0 P0 +

G mq Pˆ mq (x),

m,q=1

where



G0 =

 G(x)µw (d x),

Lˆ 2 (0,1)

G mq =

G(x)Pmq µw (d x), Lˆ 2 (0,1)

5.1 The Dirichlet problem for L´evy–Laplace equations

ˆ lν,q (x) = 1 H l!

l  

 Lˆ 2 (0,1)

1 ×

, [x(tk ) + 2 T (x) Y (tk )]2 dtk

k=1 0

1 ...

0

1

s(tl+1 , . . . , tl+q )

l+q 

, [x(t j ) + 2 T (x) Y (t j )]dt j µw (dy),

j=l+1

0

l 1 Hlν,q (x) = l! k=1

83

1

1 x (tk )dtk

0

1 ...

2

0

l+q 

s(tl+1 , . . . , tl+q )

x(t j )dt j ,

j=l+1

0

2l + ν = m, are the forms according to which the polynomials Pmq (x) are constructed (see (2.4)), T (x) =

1 − Q(x) ,
L Q(x)

Y (tk ) =

y(σtk ) − y(σtk−1 ) , √ tk

σ tk =

k


ti ;

σt0 = 0.

i=1

Proof. The forms Hlν,q (x), and, therefore, the polynomials Pmq (x), are Gˆateaux functionals satisfying the conditions of Theorem 5.8. Therefore,   ˆ ˆ
L Pmq (x) = 0 in , Pmq  = Pmq (x). By Lemma 5.2 we have inf Pˆ mq (x) = inf Pˆ mq (x) ≤ Pˆ mq (x) ≤ x∈

ˆ x∈

sup Pˆ mq (x) = sup Pˆ mq (x) for a functional Pˆ mq (x) harmonic in  and conx∈ ˆ x∈   ¯ tinuous in . Since Pˆ mq  = Pmq (x), we have |Pˆ mq (x)| ≤ sup |Pmq (x)|.

x∈

ˆ of the series By the absolute and uniform convergence on  ∞ ∞ ˆ G 0 + m,q G mq Pmq (x), the series G 0 + m,q G mq Pmq (x) converges uni ∞ 2 ˆ ˆ formly on . The series ∞ m,q G mq d Pmq (x; h), m,q G mq d Pmq (x; h), h ∈ ∞ L 2 (0, 1), and m,q G mq
L Pˆ mq (x) converge uniformly on  as well.   This along with the equality
L Pˆ mq (x) = 0 in , Pˆ mq (x) =   Pmq (x), m, q = 1, 2, . . . means that
L U (x) = 0 in  U  = G(x), and that U (x) lies in the uniqueness class indicated in Theorem 5.4.
Theorem 5.11 Let (x) be a functional defined and continuous on L 2 (0, 1), and which admits the representation as the series in the polynomials ¯ Then P0 , Pmq (x), m, q = 1, 2, . . . , absolutely and uniformly converging on . under the conditions of Theorem 5.10 there exists the unique solution of the problem
L U (x) = (x) in , U (x) = G(x) on ,

84

Linear equations with L´evy Laplacians

which is equal to U (x) = G 0 P0 +

" # ∞
ˆ ˘ G mq Pmq (x) − 0 P0 + mq Pmq (x) ,

∞
m,q=1

where

m,q=1





0 =

(x)µw (d x),

mq (x) =

Lˆ 0 (0,1)

˘ lν,q (x) = 1 H l!

Lˆ 0 (0,1)

T (x)  0

1

, [x(tk ) + 2 T (x) − ξ Y (tk )]2 dtk

k=1 0

1 ···

0

l  

Lˆ 0 (0,1)

1 ×

(x)Pmq (x)µw (d x),

s(tl+1 ), . . . , s(tl+q )

l+q 

[x(t j )

j=l+1

0

, + 2 T (x) − ξ Y (t j )]dt j µw (dy)dξ,

2l + ν = m.

Proof. Given the solution of the Dirichlet problem for the L´evy–Laplace equation (Theorem 5.10), it is sufficient to show that the solution of the Dirichlet problem with homogeneous boundary conditions for the L´evy–Poisson equation
L V (x) = (x) in , V (x) = 0 on is equal to " # ∞
V (x) = − 0 P0 + mq P˘ mq (x) . m,q=1

The forms Hlν,q(x) , and, therefore the polynomials P˘ mq (x) as well, are Gˆateaux functionals which satisfy the conditions of Theorem 5.9. Therefore 
L P˘ mq (x) = −Pmq (x) in , P˘ mq  = 0. The final part of the proof does not differ from that given in Theorem 5.10.


5.2 The Dirichlet problem for the L´evy–Schr¨odinger stationary equation We now consider the Dirichlet problem for the L´evy–Schr¨odinger equation
L U (x) + P(x)U (x) = 0,

(5.10)

where U (x) is a function to be determined and P(x) is a given function on H.

5.2 The L´evy–Schr¨odinger stationary equation

85

Theorem 5.12 Let G(x) be a function defined and continuous in a bounded domain  ∪ of the space H. If V (x) is a solution of the equation
L V (x) = −P(x), then the solution of (5.10) is equal to U (x) = (x)e V (x) , where (x) is an arbitrary harmonic function on H. The solution of the Dirichlet problem
L U (x) + P(x)U (x) = 0 in

, U (x) = G(x) on



has the form U (x) = (x)e V (x) , where V (x) is the solution of the Dirichlet problem for the L´evy–Poisson equation  
L V (x) = −P(x) in , V  = 0,

and (x) is the solution of the Dirichlet problem for the L´evy–Laplace equation  
L (x) = 0 in ,   = G(x).

The solution exists and is unique in the same functional classes where there exists the unique solution of the Dirichlet problem both for the L´evy–Laplace equation and the L´evy–Poisson equation. Proof. By formula (1.4) for m = 1, we derive   U (x) 
U (x) L  
L ln  = (x) U (x) and
L V (x) = −P(x),    (x)  where V (x) = ln U for U (x) = 0 and for arbitrary harmonic function (x) , (x) = 0. V (x) Hence, U (x) = (x)e .       The conditions V  = 0, and   = G(x) ensure V  = G(x). The final statement of the theorem is clear.
Note that, in contrast with the finite-dimensional case, the condition P(x) ≤ 0 is not required for the uniqueness of the solution of equation (5.10). This, in particular, follows from the uniqueness theorems for the solutions of the L´evy–Laplace and L´evy–Poisson equations given in Section 5.1.

86

Linear equations with L´evy Laplacians

5.3 The Riquier problem for the equation with iterated L´evy Laplacians The Riquier problem for the equation with iterated L´evy Laplacians F(x, U (x),
L U (x), . . . ,
rL U (x)) = (x), where the function F(x, u 1 , . . . , u r ) is defined on H × Rr , consists in finding a function U (x) satisfying this equation in a bounded domain  ⊂ H and such that U (x),
L U (x), . . . ,
rL−1 U (x), are equal to given functions G 0 (x), G 1 (x), . . . , G r −1 (x) on the boundary of this domain. It is especially simple to solve the Riquier problem for the polyharmonic equation
rL U (x) = 0. Given functions G 0 (x), G 1 (x), . . . , G r −1 (x) defined and continuous in a bounded domain  ∪ of the space H, one can reduce the Riquier problem
rL U (x) = 0

in , U (x) = G 0 (x),


L U (x) = G 1 (x), . . . ,
rL−1 U (x) = G r =1 (x)

on

to the Dirichlet problem for the L´evy–Laplace equation and to (r − 1) Dirichlet problems for the L´evy–Poisson equation. Indeed, the Riquier problem for the polyharmonic equation can be reduced to the following system:  
L Ur −1 (x) = 0 in , Ur −1  = G r −1 (x),  
L Ur −2 (x) = Ur −1 (x) in , Ur −2  = G r −2 (x), . . . ,  
L U1 (x) = U2 (x) in , U1  = G 1 (x),  
L U (x) = U1 (x) in , U  = G 0 (x).

It is clear that there exists a unique solution to the Riquier problem for the polyharmonic equation in the same functional classes in which there exists a unique solution of the Dirichlet problem both to the L´evy–Laplace and L´evy– Poisson equations. Consider now the linear equation for iterated L´evy Laplacians with harmonic coefficients
rL U (x) + H1 (x)
rL−1 U (x) + · · · + Hr (x)U (x) = 0, where H j (x) are harmonic function on H, j = 1, 2, . . . , r.

(5.11)

5.3 The equation with iterated L´evy Laplacians

87

The equation with constant coefficients
rL U (x) + a1
rL−1 U (x) + · · · + ar U (x) = 0 is a special case of equation (5.11). Theorem 5.13 (E. M. Polishchuk) Let G 0 (x), G 1 (x), . . . , G r −1 (x) be functions defined and continuous in a bounded domain  ∪ of the space H. If all roots 1 (x), . . . , r (x) of the equation r (x) + H1 (x)r −1 (x) + · · · + Hr (x) = 0

(5.12)

are different, then the solution of equation (5.11) is equal to U (x) =

r


 j (x) exp

1 2

j=1

  j (x)||x||2H ,

where  j (x) are arbitrary harmonic functions. ¯  j (x) = i (x), j = i, then the solution of If, in addition, for all x ∈ , the Riquer problem
rL U (x) + H1 (x)
rL−1 U (x) + · · · + Hr (x)U (x) = 0 U (x) = G 0 (x),
L U (x) =

G 1 (x), . . . ,
rL−1 U (x)

in

,

= G r −1 (x) on

has the form U (x) =

r


 j (x) exp

1 2

j=1

  j (x)||x||2H ,

(5.13)

where  j (x) satisfy r Dirichlet problems for the L´evy–Laplace equation:  
L  j (x) = 0 in ,  j (x) =  j (x), j = 1, . . . , r,

and  j (x) solve the system of linear equations r


 j (x)mj (x) exp

j=1

1 2

  j (x)||x||2H = G m (x), m = 0, . . . , r − 1.

(5.14)

Proof. By (1.4),  j (x) are harmonic functions. Since the  j (x) are harmonic functions as well, and
L ||x||2H = 2, then by (1.4),
mL U (x) =

r
j=1

 j (x)mj (x) exp

1 2

  j (x)||x||2H ,

m = 1, . . . , r.

Substituting these expressions into (5.11), and taking into account (5.12), we obtain an identity.

88

Linear equations with L´evy Laplacians

The determinant of the system (5.14) is equal to exp

r 1


2

  j (x)||x||2H [1 (x) − 2 (x)] · · · [1 (x) − r (x)]

j=1

× [2 (x) − 3 (x)] · · · [2 (x) − r (x)] × · · · × [r −1 (x) − r (x)]. It is not equal to zero, because, according to the theorem condition, we have ¯ i = j. Therefore, we have reduced the Riquier probi (x) =  j (x) for x ∈ , lem to r Dirichlet problems for L´evy–Laplace equations.
Note that solution (5.13) exists and is unique in the same functional classes where there exists a unique solution of the Dirichlet problem for the L´evy– Laplace equation. In a similar way one can write the solution of the Riquier problem in the case of multiple roots of equation (5.12). However, one should take into account that if the root l (x) has multiplicity kl , l = 1, . . . , p, then p 
||x||2H + ··· l1 (x) + l2 (x) 2 l=1   ||x||2 kl−1   1 H + lkl (x) exp l (x)||x||2H , 2 2 p where lkl (x) are harmonic functions (l = 1, . . . , p), l=1 kl = r.

U (x) =

5.4 The Cauchy problem for the heat equation Consider the heat equation ∂U (t, x) =
L U (t, x), ∂t where U (t, x) is a function on [0, T ] × H. It follows from the work of Polishchuk [118], that the Cauchy problem for the heat equation ∂U (t, x) =
L U (t, x), ∂t

U (0, x) = G(x)

corresponds to its dual problem, i.e., the Dirichlet problem for the L´evy–Laplace equation
L U (x) = 0

in , U = G(x)

on .

5.4 The Cauchy problem for the heat equation

89

The part of the time variable t is played by the function T (x) =

1 − Q(x) ;
L Q(x)

T (x) > 0 at x ∈ , T (x) = 0 at x ∈ ,

which is defined on the fundamental domain  ∪ = {x ∈ H : 0 ≤ Q(x) ≤ 1,


L Q(x) = const.}.

Based on theorems from Section 5.1 we can formulate the following theorems. For the class of functions of the form (x) = φ(x, (Ax, x) H ), where A = γ A E + S A , φ(x, ξ ) is a function on H × R1 , such that φ(x, (Ax, x) H ) = φ(P x, (Ax, x) H ), then P is a projection on an m-dimensional subspace: Theorem 5.14 Let G(x) = g(x, (Ax, x) H ), A = γ A E + S A . Then U (x) = g(x, 2γ A t + (Ax, x) H ) is the solution of the problem ∂U (t, x) =
L U (t, x), ∂t For the class of Gˆateaux functionals:

U (0, x) = G(x).

Theorem 5.15 Let 1 G(x) =

1 ···

g(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N ,

0

0

be a functional from the Gˆateaux class. Then 1 U (x) = (4π t) N /2

1

1 ···

0

dt1 · · · dt N 0

∞ ∞ N
(x(tk ) − ζk )2 dζ1 · · · dζ N × · · · g(ζ1 , . . . , ζ N ; t1 , . . . , t N ) exp − 4t k=1 −∞

−∞

is the unique solution of the problem ∂U (t, x) =
L U (t, x), ∂t Theorem 5.16 Let 1 G(x) =

1 ···

0

U (0, x) = G(x).

g(x(t1 ), . . . , x(t N ); t1 , . . . , t N )dt1 · · · dt N , 0

90

Linear equations with L´evy Laplacians

be a functional from the Gˆateaux class. Let g(ξ1 , . . . , ξ N ; t1 , . . . , t N ) be twice continuously differentiable in ξ1 , . . . , ξ N and, together with its derivatives, integrable in R N with respect to the measure having density N 2 (1/(2π) N /2 ) exp(− 12 i=1 ξi ). Then there exists a unique solution of the problem ∂U (t, x) =
L U (t, x), ∂t

U (0, x) = G(x),

which is equal to  1 U (x) =

1 ···

Lˆ 2 (0,1) 0

√ g(x(t1 ) + 2 tY (t1 ), . . . , x(t N )

0

√ + 2 tY (t N ); t1 , . . . , t N )dt1 · · · dt N µw (dy),

where Y (tk ) =

y(σtk ) − y(σtk−1 ) ; √ tk

σ tk =

k


ti ,

σt0 = 0.

i=1

For the class of functionals which admit the representation series in orthopolynomials: Theorem 5.17 Let the functional G(x) be defined and continuous on L 2 (0, 1) and be represented in the form of an absolutely and uniformly converging series in the polynomials P0 , Pmq (x), m, q = 1, . . . . Then there exists the unique solution of the problem ∂U (t, x) =
L U (t, x), ∂t

U (0, x) = G(x),

which is equal to U (x) = G 0 P0 +

∞


G mq Pˆ mq (x),

m,q=1

where





G0 =

G(x)µw (d x), Lˆ 2 (0,1)

ˆ lν,q (x) = 1 H l!



Lˆ 2 (0,1)

l  1  k=1

0

G mq =

G(x)Pmq µw (d x), Lˆ 2 (0,1)

√ [x(tk ) + 2 tY (tk )]2 dtk

5.4 The Cauchy problem for the heat equation

1 ×

1 ···

0

s(tl+1 , . . . , tl+q )

l+q 

91

√ [x(t j ) + 2 tY (t j )]dt j µw (dy),

j=l+1

0

2l + ν = m, Hlν,q (x) =

1 l  

1 l! k=1

0

1

1 ···

x 2 (tk )dtk 0

s(tl+1 , . . . , tl+q ) 0

l+q 

x(t j )dt j ,

j=l+1

2l + ν = m are the forms according to which the polynomials Pmq (x) are constructed (see k √ (2.4)), Y (tk ) = (y(σtk ) − y(σtk−1 ))/ tk ; σtk = i=1 ti , σt0 = 0. Proofs of Theorems 5.14, 5.15, 5.16, and 5.17 coincide with proofs of Theorems 5.3, 5.6, 5.8, and 5.10, respectively.

6 Quasilinear and nonlinear elliptic equations with L´evy Laplacians

Solutions of Dirichlet and Riquier problems for quasilinear and nonlinear elliptic equations are constructed in the same functional classes in which the solution of the Dirichlet problem for the L´evy–Laplace equation exists (the reduction of the problem). This allows us to cover various functional classes, since the Dirichlet problem for linear equations with L´evy Laplacians has been solved in numerous works for a wide variety of classes.

6.1 The Dirichlet problem for the equation ∆ L U (x) = f (U (x)) It was shown by L´evy that a general solution of the equation solved with respect to the L´evy Laplacian (quasilinear)
L U (x) = f (U (x)),

(6.1)

where U (x) is a function on H, and f (ξ ) is a given continuous function of a single variable in the range of {U (x)} in R1 , is given by the formula 1 ϕ(U (x)) − ||x||2H = (x), 2 where

 ϕ(ξ ) =

(6.2)

dξ ; f (ξ )

(x) is a harmonic function on H. The solution of the Dirichlet problem  
L U (x) = f (U (x)) in , U  = G(x),

where  is a bounded domain in H with a boundary and G(x) is a given function, is reduced to the solution of the Dirichlet problem for the L´evy–Laplace 92

6.1 The Dirichlet problem for the equation
L U (x) = f (U (x))

93

equation

  1   in ,   = ϕ(G(x)) − ||x||2H  . 2 Indeed, from (6.2) by (1.4) for m = 1, we obtain
L (x) = 0

1 ϕξ (U (x))
L U (x) −
L ||x||2H =
L (x). 2 But ϕξ (ξ ) = 1/ f (ξ ); by (1.2),
L ||x||2H = 2, and
L (x) = 0; therefore
L U (x) = f (U (x)). It is also clear that, since (6.2) is solved with respect to (x), the Dirichlet problem for equation (6.1) is reduced to the Dirichlet problem for the L´evy– Laplace equation. Example 6.1 We solve the Dirichlet problem in a unit ball in the space H for the equation
L U (x) = U 2 (x),   U  2 = (T (x − x0 ), x − x0 ) H , ||x|| H =1

where T is a positive compact operator in H, x0 ∈ H, and x0  H > 1. For this equation, f (ξ ) = ξ 2 ,

1 ϕ(ξ ) = − , ξ

and, by (6.2), its solution has the form 1 . 2 x H + (x) 2

U (x) = − 1

It allows us to find (x) on the surface x2H = 1. The solution of the Dirichlet problem in a ball of the radius 1 for the L´evy–Laplace equation  2 + (T (x − x0 ), x − x0 ) H 
L (x) = 0,   2 = − x H =1 2(T (x − x0 ), x − x0 ) H has the form (see Section 5.1) (x) = −

2 + (T (x − x0 ), x − x0 ) H . 2(T (x − x0 ), x − x0 ) H

Substituting (x) into the solution of the equation, we obtain the solution of the problem: U (x) =

2(T (x − x0 ), x − x0 ) H . (1 − x2H )(T (x − x0 ), x − x0 ) H + 2

94

Nonlinear elliptic equations with L´evy Laplacians

6.2 The Dirichlet problem for the equation f (U (x), ∆ L U (x)) = F(x) Consider the nonlinear equation which is not solved with respect to the L´evy Laplacian f (U (x),
L U (x)) = F(x),

(6.3)

where U (x) is a function on H, and F(x) is a given function defined on H, f (ξ, ζ ) is a given function defined on R 2 . Theorem 6.1 Let f (ξ, ζ ) be a twice continuously differentiable function of two variables in the range of {U (x),
L U (x)} in R2 , and the function F(x) satisfy the conditions
L F(x) = γ = 0, where γ is a constant. Then the solution of equation (6.3) (written implicitly) has the form f (U (x), ω(U (x), (x))) − F(x) = 0,

(6.4)

where ζ = ω(ξ, c) is a solution of the (nonlinear) ordinary differential equation f ζ (ξ, ζ )

dζ γ + f ξ (ξ, ζ ) = , dξ ζ

(6.5)

and (x) is an arbitrary harmonic function on H. If (6.4) is solvable with respect to (x), (x) = ϕ(U (x), F(x)) (here ϕ(ξ, β) is some function on R2 ), then the Dirichlet problem   f (U (x),
L U (x)) = F(x) in , U  = G(x),

is reduced to the Dirichlet problem for the L´evy–Laplace equation  
L (x) = 0 in ,   = ϕ(G(x), F(x)).

If, in addition, (6.4) is uniquely solvable with respect to (x), and if the Dirichlet problem for the L´evy–Laplace equation has a unique solution in some functional class, then the solution of the Dirichlet problem for equation (6.3) is unique in this class. Proof. Applying (1.4) twice with m = 2, we obtain from (6.4) that f ξ (U, ω(U, ))
L U (x) + f ζ (U, ω(U, ))

∂ω ∂ω ×
L U (x) +
L (x) −
L F(x) = 0. ∂ξ ∂c

6.2 The Dirichlet problem for the equation f (U (x),
L U (x)) = F(x) 95 Since
L (x) = 0,
L F(x) = γ , we have

∂ω   f ξ (U, ω(U, )) + f ζ (U, ω(U, ))
L U (x) = γ . ∂ξ But, according to the condition of the theorem, ω(ξ, c) satisfies the equation (6.5), i.e., f ξ (U, ω) + f ζ (U, ω)

γ dω = ; dξ ω

therefore, ωγ
L U = γ . Hence
L U = ω(U, ). Substituting this into (6.3) and taking into account (6.4), we obtain the identity. The final conclusions of Theorem 6.1 are obvious.
Example 6.2 We solve the Dirichlet problem in a unit ball of the space l2 for the equation ||x||l22 [
L U (x)]2 − 4U (x)
L U (x) + ||x||l22 = 0, ∞ 
1  U  2 = cosh xkm , m ≥ 3. ||x||l =1 2 2 k=1 Rewrite it in the form 4U (x)
L U (x) = xl22 . [
L U (x)]2 + 1 For this equation we have f (ξ, ζ ) =

4ξ ζ , (ξ 2 + 1)

γ = 2,

and (6.5) takes the form 2ξ ζ dζ = 1. (ξ 2 + 1) dξ

√ Its solution is ζ = ± 2ξ/c − 1. √ Substituting the expression ω(U, ) = ± (2U (x)/(x)) − 1, into (6.4), we obtain the solution of the equation: U (x) =

1 xl42 4

+  2 (x)

2(x)

.

It allows us to find (x) on the surface xl22 = 1 which corresponds to the boundary condition problem. The solution of the Dirichlet problem in a unit

96

Nonlinear elliptic equations with L´evy Laplacians

ball for the L´evy–Laplace equation  
L (x) = 0,  

' & ∞
1 m = exp ± xk xl2 =1 2 2 k=1

has the form (see Section 5.1)

' & ∞
1 m (x) = exp ± xk . 2 k=1

Substituting (x) into the solution of the equation, we obtain the solution of the problem: ' & '! & ∞ ∞

1 U (x) = . ||x||l42 exp ∓ xkm + exp ± xkm 4 k=1 k=1

6.3 The Riquier problem for the equation ∆2L U (x) = f (U (x)) Consider the equation solved with respect to the iterated L´evy Laplacian (quasilinear)
2L U (x) = f (U (x)),

(6.6)

where U (x) is a function on H, and f (ξ ) is a given function on R1 . Theorem 6.2 Let f (ξ ) be a continuous function of a single variable in the range of {U (x)} in R1 . Then the solution of (6.6) (written implicitly) has the form 1 0 (U (x), 0 (x), 1 (x)) = ||x||2H , (6.7) 2 where  dξ 0 (ξ, c0 , c1 ) = + c0 , 1 (ξ, c1 ) +  1 (ξ, c1 ) = ± 2 f (ξ ) dξ + c1 ; 0 (x), 1 (x) are arbitrary harmonic functions on H. In addition,
L U (x) = 1 (U (x), 1 (x)). If (6.7), (6.8) are solvable with respect to 0 (x), 1 (x): ( ) 0 (x) = ϕ0 ||x||2H , U (x),
L U (x) , 1 (x) = ϕ1 (U (x),
L U (x))

(6.8)

6.3 The Riquier problem for the equation
2L U (x) = f (U (x))

97

(here ϕ0 (α, ξ, ζ ) is a function on R3 , ϕ1 (ξ, ζ ) is a function on R2 ), then the Riquier problem    
2L U (x) = f (U (x)) in , U  = G 0 (x),
L U (x) = G 1 (x),



where G 0 (x), G 1 (x) are some given functions, is reduced to a couple of Dirichlet problems for the L´evy–Laplace equations: the solution to the Riquier problem has the form of (6.7), where 0 (x), 1 (x) are solutions of the problems      
L 0 (x) = 0 in , 0  = ϕ0 ||x||2H  , G 0 (x), G 1 (x) ,  
L 1 (x) = 0 in , 1  = ϕ1 (G 0 (x), G 1 (x)).

If, in addition, (6.7), (6.8) are uniquely solvable with respect to 0 (x), 1 (x), and if the Dirichlet problem for the L´evy–Laplace equation has a unique solution in some functional class, then the solution of the Riquer problem for equation (6.6) is unique in the same class. Proof. From (6.7) by (1.4) for m = 3, we obtain ∂ 0 (U, 0 , 1 ) ∂ 0 (U, 0 , 1 )
L U (x) +
L 0 (x) ∂ξ ∂c0 ∂ 0 (U, 0 , 1 ) 1 +
L 1 (x) =
L ||x||2H . ∂c1 2 In so far as ∂ 0 (ξ, c0 , c1 ) 1 = , ∂ξ 1 (ξ, c1 )


L 0 (x) =
L 1 (x) = 0,

and
L ||x||2H = 2, we have
L U (x) = 1 (U (x), 1 (x)) (using formula (6.8)). By (1.4) for m = 2, we obtain
2L U (x) =

∂ 1 (U, 1 ) ∂ 1 (U, 1 )
L U (x) +
L 1 (x). ∂ξ ∂c1

In so far as ∂ 1 (ξ, c1 ) f (ξ ) f (ξ ) = *  , = ∂ξ 1 (ξ, c1 ) ± 2 f (ξ ) dξ + c1 and
L 1 (x) = 0, we have
2L U (x) =

f (U (x))
L U (x). 1 (U (x), 1 (x))

98

Nonlinear elliptic equations with L´evy Laplacians But
L U (x) = 1 (U (x), 1 (x)); therefore
2L U (x) = f (U (x)).

The solution U (x) (given implicitly by (6.7)) satisfies equation (6.6) in  if
L 0 (x) =  0 in , and 
L 1 (x) = 0 in . By  the assumptions of the theorem    we get 0  = ϕ0 (x2H  , G 0 (x), G 1 (x)), 1  = ϕ1 (G 0 (x), G 1 (x)). Hence the solution to the Riquier problem has the form of (6.7), where 0 (x), 1 (x) are solutions of the Dirichlet problems      
L 0 (x) = 0 in , 0  = ϕ0 ||x||2H  , G 0 (x), G 1 (x) ,  
L 1 (x) = 0 in , 1  = ϕ1 (G 0 (x), G 1 (x)).




The final conclusion of Theorem 6.2 are clear.

Example 6.3 We solve the Riquier problem in a unit ball of L 2 (0, 1) for the equation
2L U (x) = eU (x) , 1   U 2 = cos x(s) ds, ||x|| L

 
L U (x)

2 (0,1)

=1

||x||2L (0,1) =1 2

0

=

√

3 exp

 1 1 2

 cos x(s) ds .

0

For this equation, we have f (ξ ) = eξ , therefore

, 1 (ξ, c1 ) = ± 2eξ + c1 , , √ 2eξ + c1 − c1 1 0 (ξ, c0 , c1 ) = ± √ ln , √ + c0 , c1 2eξ + c1 + c1

c1 > 0.

According to (6.7) we obtain the solution of the equation 1 1, 1  U (x) = ln 1 (x)sh −2 1 (x) x2L 2 (0,1) − 0 (x) . 2 2 2 In addition, ,
L U (x) = ± 2eU (x) − 1 (x). It allows us to find 0 (x) and 1 (x) on the surface x2L 2 (0,1) = 1 corresponding to the boundary conditions. The solutions of the Dirichlet problems

( ) 6.4 The equation f U (x),
2L U (x) =
L U (x)

99

in a unit ball of L 2 (0, 1) for the L´evy–Laplace equation
L 0 (x) = 0,

  0 

  1 1 √ 1 = cos x(s) ds ln(2 ± 3) − exp − x2L (0,1) =1 2 2 2 0

and
L 1 (x) = 0,

  1 

x2L

2 (0,1)

=1

 1  = exp cos x(s) ds 0

have the form (see Section 5.1) 1  1  1  1 0 (x) = − exp − exp − 1 − x 2 (s) ds 2 2 2 0

1 ×

 √ cos x(s) ds ln(2 ± 3),

0

1  1 1  2 1 (x) = exp exp − 1 − x (s) ds cos x(s) ds . 2 

0

0

Substituting 0 (x) and 1 (x) into the solution of the equation, we obtain the solution of the problem: 1 1    1 2 U (x) = ln exp exp − 1 − x (s) ds cos x(s) ds 2 2 1

0

× sinh−2

0

1

1  1 exp exp − 1 − 4 2 2

1 x 2 (s) ds 0

1

 cos x(s)) ds

0

√ ) 1 ( × x2L 2 (0,1) − 1 + ln 2 ± 3 . 2 (

)

6.4 The Riquier problem for the equation f (U (x), ∆2L U (x)) = ∆ L U (x) Consider the nonlinear equation not solved with respect to the iterated L´evy Laplacian, but solved with respect to the L´evy Laplacian f (U (x),
2L U (x)) =
L U (x), where U (x) is a function on H, and f (ξ, ζ ) is a given function on R2 .

(6.9)

100

Nonlinear elliptic equations with L´evy Laplacians

Theorem 6.3 Let f (ξ, ζ ) be a twice continuously differentiable function of two variables in the range of {U (x),
2L U (x)} in R2 . If ζ = ω(ξ, c0 ) is the solution of the (nonlinear) ordinary differential equation f ζ (ξ, ζ )

dζ ζ + f ξ (ξ, ζ ) = , dξ f (ξ, ζ )

(6.10)

then the solution of (6.9) (written implicitly) has the form (U (x), 0 (x), 1 (x)) = where

 (ξ, c0 , c1 ) =

1 ||x||2H , 2

(6.11)

dξ + c1 ; f (ξ, ω(ξ, c0 ))

0 (x), 1 (x) are the arbitrary harmonic functions on H. In addition,
L U (x) = f (U (x), ω(U (x), 0 (x))).

(6.12)

If (6.11) and (6.12) are solvable with respect to 0 (x), 1 (x): 0 (x) = ϕ0 (U (x),
L U (x)), ( ) 1 (x) = ϕ1 ||x||2H , U (x),
L U (x) (here ϕ0 (ξ, ζ ) is a function on R2 , ϕ1 (α, ξ, ζ ) is a function on R3 ), then the Riquier problem ( ) f U (x),
2L U (x) =
L U (x) in ,     U  = G 0 (x),
L U (x) = G 1 (x),



where G 0 (x), G 1 (x) are some given functions, is reduced to a couple of Dirichlet problems for the L´evy–Laplace equations: the solution to the Riquier problem has the form (6.11), where 0 (x), 1 (x) are solutions of the problems  
L 0 (x) = 0 in , 0  = ϕ0 (G 0 (x), G 1 (x)),      
L 1 (x) = 0 in , 1  = ϕ1 ||x||2H  , G 0 (x), G 1 (x) .



If, in addition, (6.11), (6.12) are uniquely solvable with respect to 0 (x), 1 (x) and if the Dirichlet problem for the L´evy–Laplace equation has a unique solution in some functional class, then the solution of the Riquer problem for equation (6.9) is unique in the same class.

( ) 6.4 The equation f U (x),
2L U (x) =
L U (x)

101

Proof. From (6.11) by (1.4) for m = 3, we obtain ξ (U, 0 , 1 )
L U (x) + c0 (U, 0 , 1 )
L 0 (x) 1 + c1 (U, 0 , 1 )
L 1 (x) =
L ||x||2H . 2 In so far as 1 , f (ξ, ω(ξ, c0 )) and
L ||x||2H = 2, we have ξ (ξ, c0 , c1 ) =


L 0 (x) =
L 1 (x) = 0,


L U (x) = f (U (x), ω(U (x), 0 (x))) (formula (6.12)). Applying (1.4) for m = 2, twice we obtain
2L U (x) = f ξ (U, ω(U, 0 ))
L U (x) + f ζ (U, ω(U, 0 ))
L ω(U, 0 ) = f ξ (U, ω(U, 0 ))
L U (x)  ∂ω(U,  )  ∂ω(U, 0 ) 0 + f ζ (U, ω(U, 0 ))
L U (x) +
L 0 (x)) . ∂ξ ∂c0 In so far as
L U (x) = f (U (x), ω(U (x), 0 (x))), and
L 0 (x) = 0, we have  ∂ω(U, 0 ) 
2L U (x) = f ξ (U, ω(U, 0 ) + f ζ (U, ω(U, 0 )) f (U, ω(U, 0 )). ∂ξ But, by virtue of the condition of the theorem, ω(ξ, c0 ) satisfies (6.10), i.e., f ξ (ξ, ω(ξ, c0 )) + f ζ (ξ, ω(ξ, c0 ))

ω(ξ, c0 ) ∂ω(ξ, c0 ) = ; ∂ξ f (ξ, ω(ξ, c0 ))

therefore
2L U (x) = ω(U (x), 0 (x)).

(6.13)

Substituting (6.13) into (6.9) and taking into account (6.12), we obtain the identity. The solution U (x) (given implicitly by (6.11)) satisfies the equation (6.9) in  if
L 0(x) = 0 in ,
L 1 (x) = 0 in . By of the theorem   the assumptions   2  we get 0  = ϕ0 (G 0 (x), G 1 (x)), 1  = ϕ1 x H  , G 0 (x), G 1 (x) . Hence the solution to the Riquier problem has the form of (6.11), where 0 (x), 1 (x) are solutions of the Dirichlet problems  
L 0 (x) = 0 in , 0  = ϕ0 (G 0 (x), G 1 (x)),      
L 1 (x) = 0 in , 1  = ϕ1 ||x||2H  , G 0 (x), G 1 (x) .

The final conclusion of Theorem 6.3 is clear.






102

Nonlinear elliptic equations with L´evy Laplacians

Example 6.4 Let us solve the equation $ 2 %n
L U (x) − nU n+1 (x)
2 U (x) + nU n (x)[
L U (x)]2 = 0.

(6.14)

Rewrite this equation in the form & $ U (x)
2L U (x)

%n '1/2
2L U (x) − =
L U (x). nU n (x)

For this equation, we have

+ f (ξ, ζ ) =

ξζ −

ζn , nξ n

and (6.10) takes the form [ξ n+2 − ξ ζ n−1 ] Its solution is ζ = c0 ξ. Therefore (ξ, c0 , c1 ) =

dζ − ξ n+1 ζ + ζ n = 0. dξ

√ nξ 1 arc cosh (n−1)/2 + c1 c0 c0

According to (6.11) we obtain the solution of the equation (6.14): + 1  , 0n−1 (x) U (x) = cosh 0 (x) x2H − 1 (x) , n 2

(6.15)

where 0 (x), 1 (x) are arbitrary harmonic functions. Example 6.5 Let us solve the Riquier problem in a unit ball of the space H for equation (6.14) with n = 3 $ 2 %3
L U (x) − 3U 4 (x)
2L U (x) + 3U 3 (x)[
L U (x)]2 = 0,   U  2 = (T (x − x0 ), x − x0 ) H , ||x|| H =1   3 2 
L U  2 = (T (x − x0 ), x − x0 ) H2 , ||x|| H =1 3 where T is a positive compact operator in H, x0 ∈ H, and x0  H > 1. According to (6.15) we have  x2  , 0 (x) H U (x) = √ cosh 0 (x) − 1 (x) . 2 3 In addition,


L U (x) =

1 0 (x)U 2 (x) − 03 (x). 3

6.5 The equation f (U (x),
L U (x),
2L U (x)) = 0

103

It allows us to find 0 (x) and 1 (x) on the surface x2H = 1 corresponding to the boundary conditions. The solutions of the Dirichlet problems in a unit ball of H for the L´evy–Laplace equation  
L 0 (x) = 0, 0  2 = (T (x − x0 ), x − x0 ) H , ||x|| H =1 √  Arch 3 1 
L 1 (x) = 0, 1  2 = − , ||x|| H =1 2 (T (x − x0 ), x − x0 )1/2 H have the form (see Section 5.1) 0 (x) = (T (x − x0 ), x − x0 ) H , √ 1 arc cosh 3 1 (x) = − . 2 (T (x − x0 ), x − x0 )1/2 H Substituting the above expressions for 0 (x) and 1 (x) into the solution of the equation, we obtain the solution of the problem:  1 ) 1/2 ( U (x) = (T (x − x0 ), x − x0 ) H cosh (T (x − x0 ), x − x0 ) H x2H − 1 2   ) 1 2 1/2 ( sinh (T (x − x0 ), x − x0 ) H x2H − 1 . + 3 2

6.5 The Riquier problem for the equation f (U (x), ∆ L U (x), ∆2L U (x)) = 0 Consider the nonlinear equation ( ) f U (x),
L U (x),
2L U (x) = 0,

(6.16)

where U (x) is a function on H, and f (ξ, η, ζ ) is a given function on R3 . Theorem 6.4 Let f (ξ, η, ζ ) be a twice continuously differentiable function of three variables in the range of {U (x),
L U (x),
2L U (x)} in R3 . Let in addition the equation f (ξ, η, ρη) = 0

(6.17)

have a solution η = φ(ξ, ρ) (although the equation f (ξ, η, ζ ) = 0, generally speaking is not solvable with respect to η). Then the solution of (6.16) (written implicitly) has the form (U (x), 0 (x), 1 (x)) =

1 ||x||2H , 2

(6.18)

104

Nonlinear elliptic equations with L´evy Laplacians

where

 (ξ, c0 , c1 ) =

dξ + c1 , φ(ξ, ω(ξ, c0 ))

ρ = ω(ξ, c0 ) is the solution of the (nonlinear) ordinary differential equation φρ (ξ, ρ)

dρ + φξ (ξ, ρ) = ρ; dξ

(6.19)

0 (x), 1 (x) are the arbitrary harmonic functions on H. In addition,
L U (x) = φ(U (x), ω(U (x), 0 (x))).

(6.20)

If (6.18), (6.20) are solvable with respect to 0 (x), 1 (x): 0 (x) = ϕ0 (U (x),
L U (x)), ( ) 1 (x) = ϕ1 ||x||2H , U (x),
L U (x) (here ϕ0 (ξ, η) is a function on R2 , ϕ1 (α, ξ, η) is a function on R3 ), then the Riquier problem ( ) f U (x),
L U (x),
2L U (x) = 0 in ,     U  = G 0 (x),
L U (x) = G 1 (x),



where G 0 (x), G 1 (x) are some given functions, is reduced to a couple of Dirichlet problems for the L´evy–Laplace equations: the solution to the Riquier problem has the form of (6.18), where 0 (x), 1 (x) are solutions of the problems  
L 0 (x) = 0 in , 0  = ϕ0 (G 0 (x), G 1 (x)),      
L 1 (x) = 0 in , 1  = ϕ1 ||x||2H  , G 0 (x), G 1 (x) .



If, in addition, (6.18) and (6.20) are uniquely solvable with respect to 0 (x), 1 (x), the equation (6.17) has a unique non-trivial solution η, and if the Dirichlet problem for the L´evy–Laplace equation has a unique solution in some functional class, then the solution of the Riquer problem for equation (6.16) is unique in the same class. Proof. From (6.18) by (1.4) for m = 3, we obtain ξ (U, 0 , 1 )
L U (x) + c0 (U, 0 , 1 )
L 0 (x) 1 + c1 (U, 0 , 1 )
L 1 (x) =
L ||x||2H . 2

6.5 The equation f (U (x),
L U (x),
2L U (x)) = 0

105

In so far as ξ (ξ, c0 , c1 ) =

1 , φ(ξ, ω(ξ, c0 ))


L 0 (x) =
L 1 (x) = 0,

and
L ||x||2H = 2, we have
L U (x) = φ(U (x), ω(U (x), 0 (x))) (formula (6.20)). Applying (1.4) for m = 2 twice, we obtain
2L U (x) = φξ (U, ω(U, 0 ))
L U (x) + φρ (U, ω(U, 0 ))
L ω(U, 0 ) = φξ (U, ω(U, 0 ))
L U (x)  ∂ω(U,  )  ∂ω(U, 0 ) 0 + φρ (U, ω(U, 0 ))
L U (x) +
L 0 (x)) . ∂ξ ∂c0 In so far as
L U (x) = φ(U (x), ω(U (x), 0 (x))), and
L 0 (x) = 0, we have  ∂ω(U, 0 ) 
2L U (x) = φξ (U, ω(U, 0 ) + φρ (U, ω(U, 0 )) φ(U, ω(U, 0 )). ∂ξ But, by virtue of the condition of the theorem, ω(ξ, c0 ) satisfies (6.19), i.e., φρ (ξ, ω(ξ, c0 ))

dω(ξ, c0 ) + φξ (ξ, ω(ξ, c0 )) = ω(ξ, c0 ); dξ

therefore
2L U (x) = ω(U (x), 0 (x))φ(U (x), ω(U (x), 0 (x))).

(6.21)

Substituting (6.20) and (6.21) into (6.16) we obtain the identity f (U, φ(U, ω(U, 0 )), ω(U, 0 )φ(U, ω(U, 0 ))) = 0, since by the condition of the theorem the equality η = φ(ξ, ρ) follows from the equality f (ξ, η, ρη) = 0. The solution U (x) (given implicitly by (6.18)) satisfies the equation (6.16) in  if
L 0 (x) = 0 in , and
L 1 (x) = 0 in . By the conditions of the     theorem we get 0  = ϕ0 (G 0 (x), G 1 (x)), 1  = ϕ1 x2H  , G 0 (x), G 1 (x) . Hence the solution to the Riquier problem has the form of (6.18), where 0 (x), 1 (x) are solutions of the Dirichlet problems  
L 0 (x) = 0 in , 0  = ϕ0 (G 0 (x), G 1 (x)),      
L 1 (x) = 0 in , 1  = ϕ1 ||x||2H  , G 0 (x), G 1 (x) .

The final conclusion of Theorem 6.4 is clear.






106

Nonlinear elliptic equations with L´evy Laplacians

Example 6.6 Consider the Riquier problem in an ellipsoid (Bx, x) H < 1 in the space H for the equation $ 2 %2
L U (x) − U (x)
L U (x)
2L U (x) 1 + U 2 (x)[
L U (x)]2 − [
L U (x)]3 = 0, (6.22)  2  2 2   U = 2ex H ,
L U (x) = e2x H . (6.23) (Bx,x) H =1

(Bx,x) H =1

Here B = E + S, where E is the identity operator and S is a self-adjoint compact operator in H. For this equation we have 1 f (ξ, η, ζ ) = ζ 2 − ξ ηζ + ξ 2 η2 − η3 = 0 2 and (6.17) has the form 1 ρ 2 η2 − ρξ η2 + ξ 2 η2 − η3 = 0. 2 This equation has the non-trivial solution η = φ(ξ, ρ) = ρ 2 − ρξ +

ξ2 , 2

and (6.19) takes the form (2ρ − ξ )

dρ + (ξ − ρ) = ρ. dξ

Its solution is ρ = ω(ξ, c0 ) = ξ + c0 . Therefore φ(ξ, ω(ξ, c0 )) = and

 (ξ, c0 , c1 ) =

ξ 2 /2

ξ2 + c0 ξ + c02 , 2

2 ξ + c0 dξ + c1 = arctan + c1 . 2 c0 c0 + c 0 ξ + c0

According to (6.18) we obtain the solution of the equation (6.22)  1    U (x) = tan x2H − 21 (x) 0 (x) −1 0 (x). 4 In addition,

(6.24)

1 2 U (x) + 0 (x)U (x) + 02 (x). 2 It allows us to find 0 (x) and 1 (x) on the surface (Bx, x) H = 1 corresponding to the boundary conditions. The solutions of the Dirichlet problems
L U (x) =

6.5 The equation f (U (x),
L U (x),
2L U (x)) = 0

107

in a ellipsoid of H for the L´evy–Laplace equation   2 
L 0 (x) = 0, 0  = −ex H ,  (Bx,x) H =1   1$ 2 % 
L 1 (x) = 0, 1  = x2H − π e−x H  2 (Bx,x) H =1

have the form (see Section 5.1) 1 [1 − (Sx, x) H − πe−1+(Sx,x) H ]. 2 Substituting 0 (x) and 1 (x) into the solution (6.24) of the equation we obtain the solution of the problem (6.22), (6.23): 0 (x) = −e1−(Sx,x) H ,

U (x) =

1 (x) =

2e1−(Sx,x) H $ %) . (1 1 + tan 4 e1−(Sx,x) H 1 − (Bx, x) H

7 Nonlinear parabolic equations with L´evy Laplacians

Solutions of the Cauchy problem for nonlinear parabolic equations are constructed in the same functional classes in which the solution of the Cauchy problem for the ‘heat equation’ exists (the reduction of the problem). This allows one to cover various functional classes, since the Cauchy problem for linear parabolic equations with L´evy Laplacians has been solved in numerous works for a wide variety of classes. Each theorem of this chapter includes both a construction of the solution for the problem without initial data and a construction of the solution for the Cauchy problem. The solution to the Cauchy problem has the same form as the solution to the problem without initial data, but the arbitrary harmonic functions (x) and (x) in the representation of the solution to the problem without initial data are changed for the specially chosen functions (t, x) and (t, x).

7.1 The Cauchy problem for the equations (∂U (t, x)/∂t) = f (∆ L U (t, x)) and (∂U (t, x)/∂t) = f (t, ∆ L U (t, x)) Consider the nonlinear equation ∂U (t, x) = f (
L U (t, x)), ∂t

(7.1)

where U (t, x) is a function on [0, T ] × H, and f (ζ ) is a given function of a single variable. Theorem 7.1 Let f (ζ ) be a twice continuously differentiable function of a single variable in the range of {
L U (t, x)} in R1 . 108

7.1 The equation

∂U (t,x) ∂t

= f (
L U (t, x))

109

1. Then the solution of equation (7.1) has the form U (t, x) = f ((x))t + (x)

||x||2H + (x), 2

where (x), (x) are arbitrary harmonic functions on H. 2. Let the equation

∂ V (τ, x)  ||x||2H f ζ − X = 0, t+ 2  ||x|| H τ =X − 2 ∂τ 2

(7.2)

(7.3)

where V (τ, x) is the solution of the Cauchy problem for the ‘heat equation’ ∂ V (τ, x) =
L V (τ, x), ∂τ

V (0, x) = U0 (x)

(and (∂ 2 V (τ, x)/∂τ 2 ) = 0), is solvable with respect to X = χ (t, x) χ (0, x) = (1/2)||x||2H ). Then the solution of the Cauchy problem ∂U (t, x) = f (
L U (x)), ∂t

(7.4) (and

U (0, x) = U0 (x)

for the nonlinear equation (7.1) has the form U (t, x) = f ((t, x))t + (t, x) where

||x||2H + (t, x), 2

(7.5)

∂ V (τ, x)   ||x||2 , τ =χ (t,x)− 2 H ∂τ  x2H  (t, x) = V χ(t, x) − , x −(t, x)χ(t, x); 2

(t, x) =

V (τ, x) is the solution of the Cauchy problem (7.4). If, in addition, equation (7.3) is uniquely solvable with respect to χ(t, x) and if the Cauchy problem for the ‘heat equation’ has a unique solution in some functional class, then the solution of the Cauchy problem for equation (7.1) is unique in the same class. Proof. 1. From (7.2) by (1.4), we obtain ∂U (t, x) = f ((x)), ∂t
L U (t, x) = f ζ ((x))t
L (x) +

||x||2H
L (x) 2

1 + (x)
L ||x||2H +
L (x) = (x), 2

110

Nonlinear parabolic equations with L´evy Laplacians

since by harmonicity
L (x) =
L (x) = 0, and by (1.2),
L ||x||2H = 2. Substituting these expressions into (7.1), we obtain the identity. 2. From (7.5) by (1.4), we obtain ∂U (t, x) ∂(t, x) = f ((t, x)) + f ζ ((t, x))t ∂t ∂t ||x||2H ∂(t, x) ∂ (t, x) + + , 2 ∂t ∂t

(7.6)


L U (t, x) = f ζ ((t, x))t
L (t, x) +

||x||2H
L (t, x) + (t, x) +
L (t, x). 2

(7.7)

By direct computation we find the expressions for ∂ (t, x)/∂t and
L (t, x):   x2 ∂ V χ(t, x) − 2 H , x ∂ (t, x) ∂(t, x) ∂χ(t, x) = − χ(t, x) − (t, x) , ∂t ∂t ∂t ∂t  x2H 
L (t, x) =
L V χ(t, x) − , x −
L (t, x)χ(t, x) 2 −(t, x)
L χ(t, x). In so far as

 ∂ V χ(t, x) − ∂t

x2H 2

,x



∂χ(t, x) ∂ V (τ, x)   ||x||2 τ =χ (t,x)− 2 H ∂τ ∂t ∂χ(t, x) = (t, x) , ∂t

=

we have ∂ (t, x) ∂(t, x) = −χ(t, x) . ∂t ∂t In so far as  ||x||2H 
L V χ(t, x) − ,x 2  ∂ V (τ, x)  =  ||x||2 [
L χ(t, x) − 1] τ =χ (t,x)− 2 H ∂τ   +
L V (τ, x) ||x||2 = (t, x)
L χ(t, x) τ =χ (t,x)− 2 H

 ∂ V (τ, x)  − −
L V (τ, x)  ||x||2 τ =χ (t,x)− 2 H ∂t = (t, x)
L χ(τ, x),

(7.8)

7.1 The equation

∂U (t,x) ∂t

= f (
L U (t, x))

111

since, by condition (7.4) of the theorem (∂ V (τ, x)/∂t) =
L V (τ, x), we have
L (t, x) = −χ (t, x)
L (t, x).

(7.9)

Substituting (7.8) into (7.6), and (7.9) into (7.7) and taking into account the condition (7.3) of the theorem we derive ∂U (t, x) = f (t, (t, x)) ∂t

∂(t, x) ||x||2H + f ζ ((t, x))t + − χ(t, x) = f ((t, x)), 2 ∂t
L U (t, x) = (t, x))

||x||2H + f ζ ((t, x))t + − χ(t, x)
L (t, x) = (t, x). 2 Substituting these expressions into equation (7.1), we obtain the identity. Putting t = 0 in (7.5) and taking into account that χ(0, x) = ||x||2H /2 and hence, (0, x) = V (χ(0, x) − x2H /2, x) − (0, x) and ||x||2H /2 = V (0, x) − (0, x)||x||2H /2, we obtain U (0, x) = (0, x)

||x||2H + (0, x) = V (0, x) = U0 (x). 2


The final statement of the theorem is obvious. Example 7.1 Let us solve the Cauchy problem ∂U (t, x) = ln
L U (t, x) ∂t

(
L U (t, x) > 0),

U (0, x) = (Ax, x)2H ,

where A = E + S, E is the identity operator, and S is some self-adjoint compact operator in H. For this equation, f (ζ ) = ln ζ, and the solution of the Cauchy problem ∂ V (τ, x) =
L V (τ, x), ∂τ, x

V (0, x) = (Ax, x)2H

has the form (see Section 5.4) V (τ, x) = [2τ + (Ax, x) H ]2 . Therefore, (7.3) takes the form X2 −

 1 ||x||2H 1  ||x||2H t − (Sx, x) H X − (Sx, x) H − = 0, 2 2 4 2 8

112

Nonlinear parabolic equations with L´evy Laplacians

and its solution is given by

* 1 ||x||2H χ(t, x) = − (Sx, x) H + (Ax, x)2H + 2t . 4 2 From (7.5) we derive the solution of the problem:

* 2 U (t, x) = t ln 2 (Ax, x) H + (Ax, x) H + 2t −

2 * 1 (Ax, x) H − (Ax, x)2H + 2t + (Ax, x)2H . 4

Let us consider a nonlinear equation ∂U (t, x) = f (t,
L U (t, x)), ∂t

(7.10)

where U (t, x) is a function on [0, T ] × H, and f (t, ζ ) is a given function of two variables. Theorem 7.2 Let f (t, ζ ) be some function of two variables twice continuously differentiable with respect to ζ and continuous in the domain [0, T ] × {
L U (t, x)} in R2 ; {
L U (t, x)} is a range in R1 . 1. Then the solution of equation (7.10) has the form t U (t, x) =

f (s, (x)) ds + (x)

||x||2H + (x), 2

(7.11)

0

where (x), (x) are arbitrary harmonic functions on H. 2. Let the equation t 0

f ζ



∂ V (τ, x)  ||x||2H − X = 0, s, ds +  ||x||2H τ =X − 2 ∂τ 2

(7.12)

where V (τ, x) is the solution of the Cauchy problem for the ‘heat equation’ ∂ V (τ, x) =
L V (τ, x), ∂τ

V (0, x) = U0 (x)

(7.13)

for some given function U0 (x) (and ∂ 2 V (τ, x)/∂τ 2 = 0), be solvable with respect to X = χ(t, x) (and χ(0, x) = (||x||2H /2)). Then the solution of the Cauchy problem ∂U (t, x) = f (t,
L U (x)), ∂t

U (0, x) = U0 (x)

7.1 The equation

∂U (t,x) ∂t

= f (
L U (t, x))

113

for the nonlinear equation (7.10) can be determined according to the formula t U (t, x) =

f (s, (t, x)) ds + (t, x)

||x||2H + (t, x), 2

(7.14)

0

where

∂ V (τ, x)   ||x||2 , τ =χ (t,x)− 2 H ∂τ ||x||2H (t, x) = V (χ(t, x) − , x) − (t, x)χ(t, x); 2

(t, x) =

V (τ, x) is the solution of the Cauchy problem (7.13). If, in addition, equation (7.12) is uniquely solvable with respect to χ(t, x) and if the Cauchy problem for the ‘heat equation’ has a unique solution in some functional class, then the solution of the Cauchy problem for equation (7.10) is unique in the same class. Proof. 1. From (7.11), by (1.4) and by harmonicity of (x) and (x), we obtain ∂U (t, x) = f (t, (x)), ∂t t ||x||2H
L U (t, x) =
L (x) f ζ (s, (x)) ds
L (x) + 2 0

+ (x) +
L (x) = (x). Substituting these expressions into (7.10), we obtain the identity. 2. From (7.14), by (1.4), we obtain ∂U (t, x) = f (t, (t, x)) + ∂t

t

f ζ (s, (t, x)) ds

0

∂(t, x) ∂t

||x||2H ∂(t, x) ∂ (t, x) + + , 2 ∂t ∂t t
L U (t, x) = f ζ ((t, x)) ds
L (t, x)

(7.15)

0

+

||x||2H
L (t, x) + (t, x) +
L (t, x). 2

(7.16)

Substituting the expressions for ∂ (t, x)/∂t and
L (t, x) into (7.15), (7.16) (formulae (7.8), (7.9)) and taking into account condition (7.12) of the theorem,

114

Nonlinear parabolic equations with L´evy Laplacians

we obtain ∂U (t, x) = f (t, (t, x)) ∂t  t  ∂(t, x) ||x||2H + − χ(t, x) f ζ (s, (t, x)) ds + 2 ∂t 0

= f (t, (t, x)),
L U (t, x) = (t, x) +

 t

f ζ (s, (t, x)) ds +

 ||x||2H − χ(t, x)
L (t, x) 2

0

= (t, x). Substituting these expressions into equation (7.10), we obtain an identity. Putting t = 0 in (7.14) and taking into account that χ (0, x) = ||x||2H /2, we obtain

x2H ||x||2H U (0, x) = (0, x) + (0, x) = V χ(0, x) − ,x 2 2 = V (0, x) = U0 (x). The final statement of the theorem is obvious. Example 7.2 Let us solve the Cauchy problem ∂U (t, x) 1 = 4(t + 1)e− 2
L U (t,x) , ∂t  ||x||2H  U (0, x) = ||x||2H 1 − ln . 2 Here f (t, ζ ) = 4(t + 1)e− 2 ζ , 1

and the solution of the Cauchy problem ∂ V (τ, x) =
L V (τ, x), ∂t has the form (see Section 5.4)

 ||x||2H  V (0, x) = ||x||2H 1 − ln 2

  ||x||2H  V (τ, x) = (2τ + ||x||2H ) 1 − ln τ + . 2 Therefore, (7.12) takes the form (1 + t)2 X −

||x||2H = 0, 2




7.2 The equation

∂U (t,x) ∂t

= f (U (t, x),
L U (t, x))

115

and its solution is given by χ(t, x) =

||x||2H . 2(1 + t)2

From (7.14) we derive the solution of the problem:  ||x||2H  U (t, x) = ||x||2H 1 − ln . 2(1 + t)2

7.2 The Cauchy problem for the equation (∂U (t, x)/∂t) = f (U (t, x), ∆ L U (t, x)) Consider the nonlinear equation ∂U (t, x) = f (U (t, x),
L U (t, x)), ∂t

(7.17)

where U (t, x) is a function on [0, T ] × H, and f (ξ, ζ ) is a given function of two variables. Theorem 7.3 Let f (ξ, ζ ) be a twice continuously differentiable function of two variables in the range of {U (t, x),
L U (t, x)} in R2 . Let the equation η = f (ξ, cη) be solvable with respect to η, η = φ(ξ, c), and the variables ξ and c be separable: φ(ξ, c) = α(c)β(ξ ) (here α(c), β(ξ ) are functions on R1 , β(ξ ) = 0). 1. Then the solution of equation (7.17) (written implicitly) has the form

||x||2H ϕ(U (t, x)) = α((x)) t + (x) + (x), (7.18) 2  where ϕ(ξ ) = (dξ /β(ξ )), (x), (x) are arbitrary harmonic functions on H. 2. Let the equation    ∂ V (τ, x)/∂τ  αc t  ||x||2H f (V (τ, x), ∂ V (τ, x)/∂τ ) τ =X − 2

   ||x||2 ∂ V (τ, x)/∂τ  H  + γc − X = 0, (7.19) 2  f (V (τ, x), ∂ V (τ, x)/∂τ ) τ =X − ||x||2 H 2 where γ (c) = c α(c), V (τ, x), is the solution of the Cauchy problem for the ‘heat equation’ ∂ V (τ, x) =
L V (τ, x), ∂τ

V (0, x) = U0 (x),

(7.20)

116

Nonlinear parabolic equations with L´evy Laplacians

( 2 ) ∂ V (τ, x)/∂τ 2 = 0 , is solvable with respect to X = χ (t, x) (and χ(0, x) = ||x||2H /2). Let the function ϕ exist and have an inverse ϕ −1 . Then the solution of the Cauchy problem ∂U (t, x) = f (U (t, x),
L U (x)), ∂t

U (0, x) = U0 (x)

for the nonlinear equation (7.17) can be determined by

||x||2H ϕ(U (t, x)) = α((t, x)) t + (t, x) + (t, x), 2

(7.21)

where ∂ V (τ, x)    ∂τ  (t, x) =  ∂ V (τ, x)  f V (t, x), ∂τ

, τ =χ (t,x)− 12 ||x||2H

  ||x||2H  (t, x) = ϕ V χ(t, x) − , x −γ ((t, x))χ (t, x); 2 V (t, x) is solution of the Cauchy problem (7.20). If, in addition, the equations η = f (ξ, cη) and (7.19) are uniquely solvable, respectively, with respect to η and χ(t, x) and if the Cauchy problem for the ‘heat equation’ has a unique solution in some functional class, then the solution of the Cauchy problem for equation (7.17) is unique in the same class. Proof. 1. From (7.18), by (1.4), we derive ϕξ (U (t, x))

∂U (t, x) = α((x)), ∂t



||x||2H ϕξ (U (t, x))
L U (t, x) = αc ((x))
L (x) t + (x) 2

2 ||x|| H 1 2 + α((x))
L (x) + (x)
L ||x|| H 2 2 +
L (x) = (x)α((x)) (since
L (x) =
L (x) = 0,
L ||x||2H = 2). Hence ∂U (t, x) = α((x))β(U (t, x)), ∂t
L U (t, x) = (x)α((x))β(U (t, x)) (since ϕξ (ξ ) = 1/β(ξ )). Substituting these expressions into (7.17), we obtain the identity α((x))β(U (t, x)) = f (U (t, x), (x)α((x))β(U (t, x))),

7.2 The equation

∂U (t,x) ∂t

= f (U (t, x),
L U (t, x))

117

since by the conditions of the theorem the equality η = α(c)β(ξ ) follows from η = f (ξ, cη). 2. From (7.21), by (1.4), we derive ∂U (t, x) ||x||2H ∂(t, x) = α((t, x)) + α((t, x)) ∂t 2 ∂t 2

||x|| ∂ (t, x) ∂(t, x) H + αc ((t, x)) t + (t, x) + , (7.22) ∂t 2 ∂t  ||x||2H  ϕξ (U (t, x))
L U (t, x) = αc ((t, x))
L (t, x) t + (t, x) 2 ||x||2H + α((t, x))
L (t, x) + (t, x)α((t, x)) +
L (t, x). 2 (7.23)

ϕξ (U (t, x))

By direct computation we derive expressions for ∂ (t, x)/∂t and
L (t, x):   2 /2, x ∂ V χ(t, x) − x H ∂ (t, x) 1  =   ∂t ∂t β V χ(t, x) − x2 /2, x H

∂χ(t, x) ∂(t, x) −γc ((t, x)) χ(t, x) − γ ((t, x)) , ∂t ∂t  1 x2H  
L V χ(t, x) −
L (t, x) =   ,x 2 x 2 β V χ(t, x) − H , x 2

−γc ((t, x))
L (t, x)χ(t, x)

− γ ((t, x))
L χ(t, x).

In addition,  ∂ V χ(t, x) −

x2H 2

∂t

,x



 ∂ V (τ, x)  =  ∂τ 

τ =χ (t,x)− 12 ||x||2H

It follows from the condition of the theorem that  ∂ V (τ,x)   ∂τ  (t, x) =  ∂ V (τ,x)  f V (t, x), ∂τ

∂χ(t, x) . ∂t

,

τ =χ (t,x)− 12 ||x||2H

and hence  ∂ V (τ, x)   ∂τ 

τ =χ (t,x)− 12 ||x||2H

  x2H  , x (t, x)α((t, x)), = β V χ(t, x) − 2

118

Nonlinear parabolic equations with L´evy Laplacians

(since from η = f (ξ, cη) it follows that η = α(c)β(ξ )) . Hence,   x2   ∂ V χ(t, x) − 2 H , x x2H  ∂χ(t, x) = γ ((t, x))β V χ(t, x) − ,x . ∂t 2 ∂t Therefore ∂ (t, x) ∂(t, x) = −γc ((t, x))χ(t, x) . (7.24) ∂t ∂t Furthermore, 1
L V (χ (t, x) − ||x||2H , x) 2  ∂ V (τ, x)   = χ(t, x) − 1] +
V (τ, x) [
  L L τ =χ (t,x)− 12 ||x||2H τ =χ(t,x)− 12 ||x||2H ∂τ  ∂ V (τ, x)  =
L χ(t, x)  ∂τ  1 2 τ =χ (t,x)− 2 ||x|| H   ∂ V (τ, x)   − −
L V (τ, x)   ∂τ τ =χ (t,x)− 12 ||x||2H   ∂ V (τ, x)  =  ∂τ  1 2 τ =χ (t,x)− 2 ||x|| H

  x2H  , x (t, x)α((t, x))
L χ(t, x), ×
L χ(t, x) = β V χ (t, x) − 2 which yields
L (t, x) = −γc ((t, x))χ(t, x)
L (t, x).

(7.25)

Substituting (7.24) into (7.22), and (7.25) into (7.23), we obtain, by condition (7.19) of the theorem,  1 ∂U (t, x) = α((t, x)) + αc ((t, x))t β(U (t, x)) ∂t  ||x||2  ∂(t, x) H + γc ((t, x)) − χ(t, x) 2 ∂t = α((t, x)), 1
L U (t, x) = γ ((t, x)) β(U (t, x))   ||x||2  H + αc ((t, x))t + γc ((t, x)) − χ (t, x) 2 ×
L ((t, x) = γ (t, x). Hence ∂U (t, x) = α((t, x))β(U (t, x)), ∂t


L U (t, x)) = γ ((t, x))β(U (t, x)).

7.2 The equation

∂U (t,x) ∂t

= f (U (t, x),
L U (t, x))

119

Substituting these expressions into equation (7.17) and taking into account that by the conditions of the theorem the equality η = f (ξ, cη) yields η = α(c)β(ξ ), we obtain the identity α((t, x))β(U (t, x)) = f (U (t, x), γ ((t, x))β(U (t, x))). Putting t = 0 in (7.21) and taking into account that χ(0, x) = 1/2||x||2H , and, therefore, (0, x) = ϕ(V (χ(0, x) − x2H /2, x)) − γ ((0, x))||x||2H /2 = ϕ(V (0, x)) − γ ((0, x))x2H /2, we obtain ||x||2H + (0, x) 2 = ϕ(V (0, x)) = ϕ(U0 (x))

ϕ(U (0, x)) = (0, x)α((0, x))

and U (0, x) = U0 (x). The final statement of the theorem is obvious.




Example 7.3 Let us solve the Cauchy problem ∂U (t, x) [
L U (t, x)]2 = ∂t U (t, x)

(U (t, x) = 0),

U (0, x) = ||x||4H .

Here f (ξ, ζ ) = ζ 2 /ξ,

α(c) = 1/c2 ,

β(ξ ) = ξ,

ϕ(ξ ) = ln |ξ |,

and the solution of the Cauchy problem ∂ V (τ, x) =
L V (τ, x), ∂τ has the form (see Section 5.4)



V (0, x) = ||x||4H

2 ||x||2H V (τ, x) = 4 τ + . 2 Therefore, (7.19) takes the form 1 X 2 − ||x||2H X − 4t = 0, 2 and its solution is given by  1 1 1 2 χ (t, x) = ||x|| H + ||x||4H + 16t. 4 2 4 From (7.21), we derive the expression for the solution of the problem + !2 ||x||2H ||x||4H U (t, x) = + + 16t 2 4 ' & 16t * × exp −  2 2 . ||x|| H ||x||4H + + 16t 2 4

120

Nonlinear parabolic equations with L´evy Laplacians

The solution of the Cauchy problem for the quasilinear equation ∂U (t, x) =
L U (t, x) + f 0 (U (t, x)), (7.26) ∂t where f 0 (ξ ) is a given function of one variable has an especially simple form. By Theorem 7.3, we have Corollary. The solution of the Cauchy problem ∂U (t, x) =
L U (t, x) + f 0 (U (t, x)), ∂t for the quasilinear equation (7.26) has the form 

U (0, x) = U0 (x)

ϕ(U (t, x)) = t + ϕ(V (t, x)),

(7.27)

where ϕ(ξ ) = (dξ / f 0 (ξ )), and V (t, x) is the solution of the Cauchy problem for the ‘heat equation’ ∂ V (t, x) =
L V (t, x), V (0, x) = U0 (x). ∂t Indeed, since f (ξ, ζ ) = ζ + f 0 (ξ ), in (7.26), it follows that α(c) = 1/(1 − c), β(ξ ) = f 0 (ξ ). In so far as we have αc (c) = γc = 1/(1 − c)2 , (7.19) takes the form 1 t + ||x||2H − X = 0, 2 and hence 1 χ(t, x) = t + ||x||2H . 2 By (7.21) we have



1 2 ϕ(U (t, x)) = α((t, x))t + γ ((t, x)) ||x|| H − χ (t, x) 2

1 2 +ϕ V χ (t, x) − ||x|| H , x . 2

In so far as 1 (t, x) , γ ((t, x) = , 1 − (t, x) 1 − (t, x)

1 1 χ(t, x) = t + ||x||2H , V χ(t, x) − ||x||2H , x = V (t, x), 2 2 α((t, x)) =

we have ϕ(U (t, x)) = t + ϕ(V (t, x)).

7.3 The equation ϕ(t, ∂U∂t(t,x) ) = f (F(x),
L U (t, x))

121


Example 7.4 Let us solve the Cauchy problem ∂U (t, x) =
L U (t, x) − U m (x), ∂t U (x, 0) = (Ax, x)2H ,

m > 1,

where A = E + S, E is the identity operator, and S is a self-adjoint compact operator in H. Here f 0 (ξ ) = −ξ m , therefore ϕ(ξ ) =

1 . (m − 1)ξ m−1

The solution of the Cauchy problem ∂ V (τ, x) =
L V (τ, x), ∂τ has the form (see Section 5.4)

V (x, 0) = (Ax, x)2H

V (τ, x) = [2τ + (Ax, x) H ]2 . By (7.27) we derive the solution of the Cauchy problem: U (t, x) =

[2t + (Ax, x) H ]2 1

{(m − 1)t[2t + (Ax, x) H ]2m−2 + 1} m−1

.

7.3 The Cauchy problem for the equation ϕ(t, ∂U (t, x)/∂t) = f (F(x), ∆ L U (t, x)) Consider the nonlinear equation

∂U (t, x) ϕ t, = f (F(x),
L U (t, x)), ∂t

(7.28)

where U (t, x) defined on [0, T ] × H ; ϕ(t, ξ ), and f (α, ζ ) are given functions of two variables, and F(x) is a given function on H. Theorem 7.4 Let ϕ(t, ξ ) and f (α, ζ ) be functions of two variables continuous, respectively, in the domain [0, T ] × {∂U (t, x)/∂t} (here {∂U (t, x)/∂t} is the range of ∂U (t, x)/∂t in R1 ) and in the range of {F(x),
L U (t, x)} in R2 . Let in addition, the equations ϕ(t, ξ ) = a, f (cα, ζ ) = a be solvable, respectively,

122

Nonlinear parabolic equations with L´evy Laplacians

with respect to ξ and ζ, i.e., ξ = h(t, a) and ζ = g(α, a). Assume that h(t, a) and g(α, a) and the partial derivatives h a (t, a) and ga (α, a) are continuous in their domains of definition, and F(x) is such that
L F(x) = c = 0, where c is a constant. 1. Then the solution of equation (7.28) has the form  t  1 F(x) c U (t, x) = h(s, (x))ds + g(α, (x))dα + (x), 0

(7.29)

0

where (x), (x) are arbitrary harmonic functions on H. 2. Let the equation  t   ∂ V (τ, x)    ds h a s, f X,  τ =X − 1c F(x) ∂τ 0  1 F(x)   ∂ V (τ, x)   c  + dα = 0, ga α, f X,  1 τ =X − c F(x) ∂τ X

(7.30)

where V (τ, x) is the solution of the Cauchy problem for the ‘heat equation’ ∂ V (τ, x) =
L V (τ, x), ∂τ

V (0, x) = U0 (x)

(7.31)

be solvable with respect to X = χ(t, x), with χ(0, x) = 1c F(x). Then the solution of the Cauchy problem

∂U (t, x) ϕ t, = f (F(x),
L U (t, x)), U (0, x) = U0 (x) ∂t for the nonlinear equation (7.28) has the form  t  1 F(x) c U (t, x) = h(s, (t, x))ds + g(α, (t, x))dα + (t, x), (7.32) 0

where

χ (t,x)

  ∂ V (τ, x)  (t, x) = f χ(t, x), ,  τ =χ (t,x)− 1c F(x) ∂τ ( ) 1 (t, x) = V χ(t, x) − F(x), x ; c

V (t, x) is the solution of Cauchy problem (7.31). If, in addition, the equations ϕ(t, ξ ) = α, f (cα, ζ ) = α and (7.30) are uniquely solvable, respectively, with respect to ξ, ζ and χ(t, x) and if the Cauchy problem for the ‘heat equation’ has a unique solution in some functional class, then the solution of the Cauchy problem for equation (7.28) is unique in the same class.

7.3 The equation ϕ(t, ∂U∂t(t,x) ) = f (F(x),
L U (t, x))

123

Proof. 1. From (7.29) by (1.4), we obtain ∂U (t, x) = h(t, (x)), ∂t  t
L U (t, x) = h a (s, (x)) ds
L (x) 0



1 c

+ 0

+g

F(x)

ga (α, (x))dα
L (x)

1

1 F(x), (x)
L F(x) +
L (x). c c

But by harmonicity
L (x) =
L (x) = 0, and by the conditions of the theorem,
L F(x) = c; therefore 1 
L U (t, x) = g F(x), (x) . c Substituting the expressions ∂U (t, x)/∂t and
L U (t, x) into (7.28), we obtain the identity

1 ϕ(t, h(t, (x))) = f F(x), g F(x), (x) , c since by the condition of the theorem, ϕ(t, ξ ) = f (cα, ζ ). 2. From (7.32) by (1.4), we obtain  t ∂U (t, x) ∂(t, x) = h(t, (t, x)) + h a (s, (t, x)) ds ∂t ∂t 0  1 F(x) c ∂(t, x) + ga (α, (t, x))dα ∂t χ (t,x) ∂χ(t, x) ∂ (t, x) − g(χ(t, x), (t, x)) + , ∂t ∂t  t
L U (t, x) = h a (s, (t, x)) ds
L (t, x) 0



+

1 c

F(x)

χ (t,x)

ga (α, (t, x))dα
L (t, x)

1

1 F(x), (t, x)
L F(x) c c − g(χ(t, x), (t, x))
L χ(t, x) +
L (t, x). +g

(7.33)

Let us compute ∂ (t, x)/∂t:

 ∂ (t, x) ∂χ (t, x) ∂ V (τ, x)  = .  ∂t ∂τ ∂t τ =χ (t,x)− 1 F(x) c

(7.34)

124

Nonlinear parabolic equations with L´evy Laplacians

But the relation g(α, f (α, ζ )) = ζ in the condition of the theorem yields    ∂ V (τ, x)  ∂ V (τ, x)  = g χ(t, x), f χ(t, x),   1 1 τ =χ (t,x)− c F(x) τ =χ(t,x)− c F(x) ∂τ ∂τ = g(χ(t, x), (t, x)). Therefore ∂ (t, x) ∂χ(t, x) = g(χ (t, x), (t, x)) . (7.35) ∂t ∂t Let us compute
L (t, x)   ∂ V (τ, x)  1
L χ(t, x) −
L F(x)
L (t, x) =  1 τ =χ (t,x)− c F(x) ∂τ c   +
L V (τ, x) . 1 τ =χ (t,x)− c F(x)

But from g(α, f (α, ζ )) = ζ ∂ V (τ, x)  =  τ =χ (t,x)− 1c F(x) ∂τ =

it follows that    ∂ V (τ, x)  g χ(t, x), f χ(t, x),  τ =χ(t,x)− 1c F(x) ∂τ g(χ(t, x), (t, x)).

Therefore
L (t, x) = g(χ(t, x), (t, x))
L χ(t, x)  ∂ V (τ, x)   − −
L V (τ, x)  . τ =χ (t,x)− 1c F(x) ∂τ Taking into account that ∂ V (τ, x)/∂τ =
L V (τ, x), by (7.31) we have
L (t, x) = g(χ(t, x), (t, x))
L χ(t, x).

(7.36)

Substituting (7.35) into (7.33), and (7.36) into (7.34), we obtain, taking into account (7.30),  t ∂U (t, x) = h(t, (t, x)) + h a (s, (t, x)) ds ∂t 0  1 F(x)  ∂(t, x) c + ga (α, (t, x))dα ∂t χ (t,x) = h(t, (t, x)),  1 F(x)  t  c 
L U (t, x) = h a (s, (t, x)) ds + ga (α, (t, x))dα 0

1

χ (t,x)

 ×
L (t, x) + g F(x), (t, x) c 1  = g F(x), (t, x) . c

7.3 The equation ϕ(t, ∂U∂t(t,x) ) = f (F(x),
L U (t, x))

125

Substituting these expressions into equation (7.28), we obtain the identity 1  ϕ(t, h(t, (t, x))) = f F(x), (t, x) . c This follows from the condition ϕ(t, ξ ) = f (cα, ζ ). Putting t = 0, in (7.32) and taking into account that χ(0, x) = 1c F(x), we obtain 1 U (0, x) = (0, x) = V (χ(0, x) − F(x), x) = V (0, x) = U0 (x). c


The final statement of the theorem is obvious.

Example 7.5 Let us solve the Cauchy problem   ∂U (t, x) exp = (t + 1)[
L U (t, x) + (Ax, x) H ], ∂t 1 U (0, x) = (Bx, x)2H , 4 where A = E + S, B = E − S, E is the identity operator, and S is a selfadjoint compact operator in H. Here F(x) = (Ax, x) H = ||x||2H + (Sx, x) H ,
L F(x) = 2, exp{ξ } ϕ(t, ξ ) = , f (cα, ζ ) = 2α + ζ, t +1 and hence h(t, a) = ln(t + 1)a,

g(α, a) = −2α + a.

The solution of the Cauchy problem ∂ V (τ, x) =
L V (t, x), ∂τ

V (0, x) =

1 (Bx, x)2H 4

has the form (see Section 5.4)  2 1 V (τ, x) = τ + (Bx, x) H . 2 Therefore, (7.30) takes the form  2 1   1 1 1 X − (Sx, x) H − ||x||2H X − (Sx, x) H − t = 0, 2 2 2 4 and its solution is  1 1 1 1 2 χ(t, x) = ||x|| H + ||x||4 + t + (Sx, x) H . 4 2 4 2

126

Nonlinear parabolic equations with L´evy Laplacians

By (7.32), we obtain the solution of the problem:    t  3 1 t+1 2 U (t, x) = ln (t + 1) ||x|| H + 2 ||x||4H + t − t 4 2  1 1 1 1 + ||x||2H ||x||4H + t − ||x||2H (Sx, x) H + (Sx, x)2H . 2 4 2 4

7.4 The Cauchy problem for the equation f (U (t, x), ∂U (t, x)/∂t, ∆ L U (t, x)) = 0 Consider the nonlinear equation ( ) ∂U (t, x) f U (t, x), (7.37) ,
L U (t, x) = 0 ∂t where U (t, x) is function on [0, T ] × H, and f (ξ, η, ζ ) is a function of three variables. Theorem 7.5 Let the function f (ξ, η, ζ ) satisfy the following conditions: (1) The function f (ξ, η, ζ ) is a twice continuously differentiable function of three variables in the range of {U (t, x), ∂U (t, x)/∂t,
L U (t, x)} in R3 . (2) The equation f (ξ, η, aη) = 0

(7.38)

is solvable with respect to η, η = φ(ξ, a) (although the equation f (ξ, η, ζ ) = 0, generally speaking can not be solved with respect to η), and assume that the variables ξ and a can be separated, i.e. φ(ξ, a) = α(a)β(ξ ) where α(a), β(ξ ) are functions on R1 , β(ξ ) = 0. 1. Then the solution of equation (7.37) (written implicitly) has the form ϕ(U (t, x)) = α((x))t + γ ((x)) where

 ϕ(ξ ) =

dξ , β(ξ )

x2H + (x), 2

(7.39)

γ (a) = aα(a)

and (x), (x) are arbitrary harmonic functions on H. 2. Let the function f (ξ, η, ζ ) satisfy in addition the following conditions. (3) The equation  1 ∂ V (τ, x) ∂ V (τ, x)  f V (τ, x), , =0 a ∂τ ∂τ

(7.40)

7.4 The equation f (U (t, x), ∂U∂t(t,x) ,
L U (t, x)) = 0 can be solved with respect to a, a = σ (V (τ, x), ∂ V (τ, x)/∂τ ). Assume that the equation    ∂ V (τ, x)  αa σ V (τ, x), t 2  x τ =X − 2 H ∂τ     x2 ∂ V (τ, x)  H + γa σ V (τ, x), − X = 0,  x2H τ =X − 2 ∂τ 2

127

(7.41)

can be solved with respect to X = χ(t, x), with χ(0, x) = x2H /2. Here V (τ, x) is the solution of the Cauchy problem for the ‘heat equation’ ∂ V (τ, x) =
L V (τ, x), V (0, x) = U0 (x). (7.42) ∂τ Let the function ϕ exist and have an inverse ϕ −1 . Then the solution of the Cauchy problem ( ) ∂U (t, x) f U (t, x), ,
L U (t, x) = 0, U (0, x) = U0 (x) ∂t for the nonlinear equation (7.37) has the form ϕ(U (t, x)) = α((t, x))t + γ ((t, x)) where

x2H + (t, x), 2

(7.43)



dξ , γ (a) = aα(a), β(ξ )   ∂ V (τ, x)  (t, x) = σ V (τ, x),   ∂τ ϕ(ξ ) =

τ =χ (t,x)−

x2H 2

,

  x2H  (t, x) = ϕ V χ(t, x) − , x −γ ((t, x))χ (t, x); 2 V (t, x) is solution of the Cauchy problem (7.42). If in addition equations (7.38), (7.40), and (7.41) have respectively unique solutions η, a and χ(t, x) and the Cauchy problem for the ‘heat equation’ has a unique solution in a certain functional class, then the solution to the Cauchy problem for equation (7.37) is unique in the same class. Proof. 1. From (7.39), by (1.4) we derive
L ϕ(U (t, x)) = ϕξ (U (t, x))
L U (t, x) = αa ((x))
L (x)t + γa ((x))
L (x) 1 + γ ((x))
L x2H +
L (x) 2 = (x)α((x))

x2H 2

128

Nonlinear parabolic equations with L´evy Laplacians

(since
L (x) =
L (x) = 0,
L x2H = 2). Moreover, ∂ϕ(U (t, x)) ∂U (t, x) = ϕξ (U (t, x)) = α((x)). ∂t ∂t From this we deduce that ∂U (t, x)) = α((x))β(U (t, x)) = φ(U (t, x), (x)), ∂t
L U (t, x) = (x)α((x))β(U (t, x)) = (x)φ(U (t, x), (x)) since ϕξ (ξ ) = 1/β(ξ ). Substituting these relations into (7.37) we derive ( ) f U (t, x), φ(U (t, x), (x)), (x)φ(U (t, x), (x)) = 0. The last equality holds identically as, due to condition (2), φ(U (t, x), (x)) = φ(U (t, x), (x)). 2. Let us rewrite (7.43) in the form  x2  H ϕ(U (t, x)) = α((t, x)t + γ ((t, x)) − χ (t, x) 2   x2H  + ϕ V χ (t, x) − ,x . (7.44) 2 From (7.44) ∂ϕ(U (t, x)) ∂U (t, x) = ϕξ (U (t, x)) ∂t ∂t = α((t, x))   x2  ∂(t, x) H + αa ((t, x))t + γa ((t, x)) − χ (t, x) 2 ∂t    ∂χ(t, x) ∂ V (τ, x)  − γ ((t, x)) − ϕξ (V (τ, x)) .  x2 τ =χ(t,x)− 2 H ∂τ ∂t (7.45) Since (t, x) = σ (V, (∂ V /∂τ ))|τ =χ (t,x)+(x2H /2) and χ(t, x) solve (7.41), which yields αa (σ )t + γa (σ )[(x2H /2) − χ(t, x)] = 0, we deduce from (7.45) ∂U (t, x) = α((t, x)) ∂t    ∂ V (τ, x)  − γ ((t, x)) − ϕξ (V (τ, x))  ∂τ 

ϕξ (U (t, x)

 ∂χ(t, x) τ =χ (t,x)−

x2H 2

∂t

. (7.46)

Furthermore,  = σ (V, ∂ V /∂τ ), and hence by condition (3) satisfies the equation  1 ∂V ∂V  f V, , = 0.  ∂τ ∂τ

7.4 The equation f (U (t, x), ∂U∂t(t,x) ,
L U (t, x)) = 0

129

Thus, since f (V, (1/)(∂ V /∂τ ), (1/)(∂ V /∂τ )) = 0 and by condition (2) we get 1 ∂V = φ(V, ).  ∂τ Taking into account that γ () = α() and ϕξ (ξ ) = 1/β(ξ ) we derive γ () − and

α()β(V ) − 1 ∂V · = β(V ) ∂τ β(V )

1 

·

∂V ∂τ

=

 1 ∂ V (τ, x)  γ ((t, x)) = ·  β(V (τ, x)) ∂τ 

φ(V, ) − 1 · β(V )

τ =χ (t,x)−

x2H 2

∂V ∂t

.

=0

(7.47)

Substituting (7.47) into (7.46) we obtain ∂U (t, x) = φ(U (t, x), (t, x)). ∂t From (7.44), by (1.4) we derive

(7.48)


L ϕ(U (t, x)) = ϕξ (U (t, x))
L U (t, x) = γ ((t, x))    x2 H + αa ((t, x))t + γa ((t, x)) − χ (t, x)
L (t, x) 2 − γ ((t, x))
L χ(t, x)   x2H   + ϕξ (V (τ, x)) ,x . x2H
L V (χ(t, x) − τ =χ (t,x)− 2 2 (7.49) Since (t, x) = σ (V, (∂ V /∂τ ))|τ =χ (t,x)− x2 , and χ(t, x) solves (7.41) we 2 get from (7.49) that ϕξ (U (t, x))
L U (t, x) = γ ((t, x)) − γ ((t, x))
L χ(t, x)   x2H   + ϕξ (V (τ, x)) ,x . x2H
L V χ(t, x) − τ =χ (t,x)− 2 2 But ( x2H )
L V χ(t, x) − ,x  2 ∂ V (τ, x)  =  ∂τ  x2H τ =χ (t,x)−

2     1  2 ×
L χ(t, x) −
L x H +
L V (τ, x)  2

τ =χ (t,x)−

x2H 2

(7.50)

130

Nonlinear parabolic equations with L´evy Laplacians    ∂ V (τ, x)   = [
L χ(t, x) − 1] +
L V (τ, x)   ∂τ  x2H x2 τ =χ (t,x)− 2 τ =χ (t,x)− 2 H  ∂ V (τ, x)  =
L χ(t, x)  ∂τ  x2 τ =χ (t,x)− 2 H   ∂ V (τ, x)   − −
L V (τ, x)   ∂τ x2 τ =χ (t,x)− 2 H  ∂ V (τ, x)  =
L χ(t, x)  ∂τ  x2H τ =χ (t,x)−

2

since, by the condition of the theorem, ∂ V (τ, x)/∂τ =
L V (τ, x). If we substitute the value of
L V (χ(t, x) − (x2H /2), x) in (7.50) we obtain ϕξ (U (t, x))
L U (t, x)   ∂ V (τ, x)   = γ ((t, x)) − γ ((t, x)) − ϕξ (V (τ, x))  ∂τ 

τ =χ (t,x)−

x2H 2


L χ(t, x).

Now keeping in mind (7.47) we derive ϕξ (U (t, x))
L U (t, x) = γ ((t, x)) and
L U (t, x) = (t, x)φ(U (t, x), (t, x)).

(7.51)

Substituting (7.48) and (7.51) into (7.37) we get ( ) f U (t, x), φ(U (t, x), (t, x)), (t, x)φ(U (t, x), (t, x)) = 0. By condition (2) of the theorem this equality holds identically. If we put t = 0 in (7.43), take into account that χ (0, x) = (1/2)x2H and consequently (0, x) = ϕ(V (0, x)) − γ ((0, x))x2H /2, we obtain ϕ(U (0, x)) = γ ((0, x)) = ϕ(U0 (x))

x2H x2H + ϕ(V (0, x)) − γ ((0, x)) 2 2

which yields U (0, x) = U0 (x). The final assertion of the theorem is obvious.




Example 7.6 Let us solve the Cauchy problem  ∂U (t, x) 3  ∂U (t, x) 2  ∂U (t, x)  − U (t, x) +
L U (t, x) ∂t ∂t ∂t = U (t, x)
L U (t, x), U (0, x) = x4H . (7.52)

7.4 The equation f (U (t, x), ∂U∂t(t,x) ,
L U (t, x)) = 0

131

In this case we have f (ξ, η, ζ ) = η3 − ξ η2 + ηζ = ξ ζ and (7.38) has the form η3 − ξ η2 + aη2 = aξ η. This equation has the non-trivial solutions η = −a and η = ξ. Consider first the solution to (7.38) of the form η = φ(ξ, a) = −a. Here α(a) = −a, γ (a) = −a 2 , β(ξ ) = 1, ϕ(ξ ) = ξ. Rewriting (7.40) in the form ( ∂∂τV )3 ( ∂∂τV )2 ( ∂∂τV )2 ∂V =V − V + a3 a2 a ∂τ we get  ∂V  ∂V a = σ V, = V. ∂τ ∂τ The solution to the Cauchy problem ∂ V (τ, x) =
L V (τ, x), ∂τ

V (0, x) = x4H

is given by (see Section 5.4)  x2H 2 V (τ, x) = 4 τ + 2 and hence  ∂ V (τ, x)  2 σ V (τ, x), .  x2H = τ =X − 2 ∂τ X Equation (7.41) is now of the form (t − 4)X + 2x2H = 0 and has the solution X = χ(t, x) = −

2x2H . t −4

From (7.43) we derive the solution of the problem (7.52): U (t, x) =

t(t − 4)3 + 32x6H . 2(t − 4)2 x2H

This solution is defined for all points (t, x) ∈ [0, T ] × H except (4, x) or (t, 0).

132

Nonlinear parabolic equations with L´evy Laplacians

Consider now another solution η = φ(ξ, a) = ξ to (7.38). Here α(a) = 1, γ (a) = a, β(ξ ) = ξ, ϕ(ξ ) = ln ξ. This time (7.41) has the form X−

x2H = 0, 2

and its solution is given by x2H . 2 From (7.43) we derive one more solution of the problem (7.52): X = χ(t, x) =

U (t, x) = x4H et . This solution is defined for all points t ∈ [0, T ], x ∈ H . Remark. In the particular case when (7.37) can be solved with respect to ∂U (t, x)/∂t and has the form ∂U (t, x) = f 0 (U (t, x),
L U (t, x)), ∂t where f 0 (ξ, ζ ) is a function of two arguments, the function σ (V, ∂ V /∂τ ) from (7.41) is determined by the data of the problem. Indeed, in this case (7.40) is written in the form  ∂V  1 ∂V = f 0 V, a ∂τ ∂τ which yields ∂V  ∂V  ∂τ ). a = σ V, = ∂τ f 0 (V, ∂∂τV

Note that the Cauchy problem for the equation ∂U (t, x)/∂t = f 0 (U (t, x),
L U (t, x)) was considered in Section 7.2.

Appendix L´evy–Dirichlet forms and associated Markov processes

Dirichlet forms appear in various problems of modern mathematical physics, stochastic analysis, quantum mechanics and many other areas. There is a close connection between these forms and Markov processes (see, e.g. [7]). We construct a Dirichlet form associated with the infinite-dimensional L´evy–Laplace operator and a Markov process generated by this Dirichlet form. To this end we use the self-adjoint operator in L2 (H, µ) generated by the nonsymmetrized L´evy Laplacian from Section 3.3.

A.1 The Dirichlet forms associated with the L´evy–Laplace operator Recall that a Dirichlet form is by definition a Markovian closed symmetric bilinear form. A symmetric form E(U, V ) on L2 (H, µ) is called Markovian if the following property holds [67]: for each ε > 0, there exists a real function φε (t), t ∈ R1 , such that φε (t) = t

for all t ∈ [0, 1],

−ε ≤ φε (t) ≤ 1 + ε

0 ≤ φε (s) − φε (t) ≤ s − t

whenever

for all t ∈ R1 ,

t < s,

and for any U ∈ DE , we have that φε (U ) ∈ DE

and E(φε (U ), φε (U )) ≤ E(U, U ).

−1/2

introduced in Section 2.1 (K is the correlation operator Recall the operator T = K of the Gaussian measure µ). Require in addition that T −1/2 is a Hilbert–Schmidt operator. ,  −2 f , f ) < ∞. For example, if { f }∞ coincides with the Assume and that ∞ (T k k H k 1 k=1 canonical basis {ek }∞ 1 in H, then ∞ ,
k=1

(T −2 ek , ek ) H =

∞
1 = Tr T −1 < ∞ λ k k=1

since T −1 is a trace operator. We consider the L´evy–Laplace operator LU =
L U, D L = T introduced in Section 3.3, where the set T consists of functions of the form V (x) = ϕ(Q(x))S(x).

133

134 Appendix. L´evy–Dirichlet forms and associated Markov processes

2 Moreover the function Q(x) is chosen to be of the form Q(x) = x H + φ(x), with φ(x) = ∞ (x, f ) . Such a choice of Q(x) makes D ∩ D = , where E is the k H L E k=1 Dirichlet form. Denote by T the set of all functions of the form

V (x) = ϕ(Q(x))S(x) such that V (x), [ϕ  (Q(x))/ϕ(Q(x))]2 V (x) ∈ L2 (H, µ). Here S(x) is an arbitrary, twice strongly differentiable harmonic in H functions from L2 (H, µ), Q(x) =

∞


ζk (x) = x2H + φ(x),

(A.1)

k=1

ζk (x) = (x, f k )2H + (x, f k ) H , φ(x) =

∞
(x, f k ) H ( f k ∈ H+ , x ∈ H ), k=1

and ϕ(ξ ) is a fixed function on R . We choose ϕ(ξ ) to be a solution to the equation 1

ϕ(ξ )ϕ  (ξ ) + 2ϕ(ξ )ϕ  (ξ ) + [ϕ  (ξ )]2 = 0.

(A.2)



By the substitution κ(ξ ) = ϕ (ξ )/ϕ(ξ ), for ξ such that ϕ(ξ ) = 0, (A.2) can be reduced to the Riccati equation κ  (ξ ) + 2κ 2 (ξ ) + 2κ(ξ ) = 0.

∞

(A.3)

The series k=1 ζk (x), converges µ-almost everywhere on H, since φ(x) < ∞, for µ-almost all x ∈ H. Actually  √  ∞ 2 (K f k , f k ) H 2 1 2 (K f k , f k ) H . |(x, f k ) H |µ(d x) = ye− 2 y dy = √ π 2π H

0

∞ 

∞ Hence k=1 H |(x, f k ) H |µ(d x) < ∞ and by Levi’s theorem k=1 (x, f k ) H < ∞, µ-almost everywhere on H . The function V  (x) = ϕ(Q(x))S(x) is twice differentiable along the subspace H+ 2 since both φ(x) = ∞ k=1 (x, f k ) H and x H and hence the above Q(x) given by (A.1) possess this property. To check that φ(x) is twice differentiable let us compute dφ(x; h) =

∞ ∞

1 (h, f k ) H ≤ [(T h, T h) H (T −1 f k , T −1 f k ) H ] 2 k=1

k=1

∞ ,
= h H+ (T −2 f k , f k ) H < ∞,

d 2 φ(x; h) = 0,

(h ∈ H+ ).

k=1

It is obvious that the set T is linear and is everywhere dense in L2 (H, µ). Define on L2 (H, µ) the bilinear form E(U, V ) by E(U, V ) = lim En (U, V ), n→∞

where

 En (U, V ) = H

n 1
(U  (x), f k ) H (VH + (x), f k ) H µ(d x). n k=1 H+

A.1 The Dirichlet forms associated with the L´evy–Laplace operator 135

Lemma A.1 The form E(U, V ) on T exists and E(U, V ) = (κ 2 (Q)U, V )L2 (H,µ) where κ(ξ ) is a positive solution of the Riccati equation (A.3). The form E(U, V ) is symmetric and densely defined. Proof. For U, V ∈ T we have U (x) = ϕ(Q(x))SU (x), V (x) = ϕ(Q(x))SV (x). Therefore dU (x; h) = (U H + (x), h) H = ϕ  (Q(x))(Q  (x), h) H SU (x) + ϕ(Q(x))(SU (x), h) H , d V (x; h) = (VH + (x), h) H = ϕ  (Q(x))(Q  (x), h) H SV (x) + ϕ(Q(x))(SV (x), h) H

(h ∈ H+ ),

and n 1
(U  (x), f k ) H (VH + (x), f k ) H n k=1 H+

= [ϕ  (Q(x))]2 SU (x)SV (x) + ϕ  (Q(x))ϕ(Q(x))

n 1
(Q  (x), f k )2H n k=1

m 1
(Q  (x), f k ) H [SU (x)(SV (x), f k ) H n k=1

+ SV (x)(SU (x), f k ) H ] + ϕ 2 (Q(x))

m 1
(S  (x), f k ) H (SV (x), f k ) H . n k=1 U

Since d Q(x; h) = (Q  (x), h) H =

∞


2(x, f k ) H ( f k , h) H + (h, f k ) H ,

k=1

we obtain (Q  (x), f k ) H = 2(x, f k ) H + ( f k , f k ) H . Furthermore lim

n→∞

n 1
(Q  (x), f k )2H = 1 n k=1

due to the convergence (x, f k ) H → 0 as k → ∞, µ-almost all x ∈ H . Finally we have n 1
(U H + (x), f k ) H (VH + (x), f k ) H = [ϕ  (Q(x))]2 SU (x)SV (x) n→∞ n k=1

 ϕ (Q(x)) 2 U (x)V (x) = κ 2 (Q(x))U (x)V (x) = ϕ(Q(x))

lim

and E(U, V ) = (κ 2 (Q)U, V )L2 (H,µ)

U, V ∈ T.


136 Appendix. L´evy–Dirichlet forms and associated Markov processes

Lemma A.2 The L´evy Laplacian
L exists on T and acts as the operator of multiplication by the function −κ 2 (Q(x))
L V (x) = −κ 2 (Q(x))V (x), where κ(ξ ) is a positive solution of the Riccati equation (A.3). Proof. For V (x) ∈ T, we have V (x) = ϕ(Q(x))SV (x). Therefore d V (x; h) = (VH + (x), h) H

= ϕ  (Q(x))(Q  (x), h) H SV (x) + ϕ(Q(x))(SV (x), h) H ,

d 2 V (x; h) = (VH+ (x)h, h) H = ϕ  (Q(x))(Q  (x), h)2H SV (x) + ϕ  (Q(x))(Q  (x)h, h) H SV (x) + 2ϕ  (Q(x))(Q  (x), h) H (SV (x), h) H + ϕ(Q(x))(SV (x)h, h) H . From the formula (1.3) n 1
(Q  (x, f k )2H n→∞ n k=1


L V (x) = ϕ  (Q(x))SV (x) lim

+ ϕ  (Q(x))SV (x) lim

n→∞

+ 2ϕ  (Q(x)) lim

n→∞

+ ϕ(Q(x)) lim

n→∞

n 1
Q  (x) f k , f k ) H n k=1

n 1
(Q  (x), f k ) H (SV (x), f k ) H n k=1

n 1
(S  (x) f k , f k ) H n k=1 V m 1
(Q  (x), f k )2H n→∞ n k=1

= ϕ  (Q(x))SV (x) lim

+ ϕ  (Q(x))SV (x) lim

n→∞

n 1
(Q  (x) f k , f k ) H , n k=1

since SV is a harmonic function. In the proof of Lemma A.1 we have shown that for µ-almost all x ∈ H n 1
(Q  (x), f k )2H = 1. n→∞ n k=1

lim

In addition since d 2 Q(x; h) = (Q  (x)h, h) H = 2h2H , we have (Q  (x) f k , f k ) H = 2, and hence
L Q(x) = lim

n→∞

n 1
(Q  (x) f k , f k ) = 2. n k=1

Hence
L V (x) = ϕ  (Q(x))SV (x) + 2ϕ  (Q(x))SV (x) =

ϕ  (Q(x)) + 2ϕ  (Q(x)) V (x). ϕ(Q(x))

A.2 The processes associated with the L´evy–Dirichlet forms

Taking into account that ϕ(ξ ) is governed by (A.2) we get 

ϕ (Q(x)) 2 V (x) = −κ 2 (Q(x))V (x).
L V (x) = − ϕ(Q(x))

137




Define the operator L in L2 (H, µ) with everywhere dense domain of definition D L putting LU =
L U,

D L = T.

Theorem A.1 The operator L is an essentially self-adjoint operator and −L is positive. The closure − L¯ of −L is a self-adjoint positive operator. Proof. The proof of essential self-adjointness of L is similar to the proof in Theorem 3.2. Furthermore, since a general solution κ(ξ ) to the Riccati equation (A.3) has the form κ(ξ ) = −

1 , 1 + ce2ξ

where c is a constant and κ 2 (ξ ) > 0, for − ∞ < ξ < ∞ we obtain (−LU, U )L2 (H,µ) = (κ 2 (Q)U, U )L2 (H,µ) > 0,

for all U ∈ D L .


The final conclusions of the theorem are evident. Consider the form  , , ¯ − LU, − L¯ V E(U, V ) =

L2 (H,µ)

U, V ∈ DE ,

DE = D√− L¯ .

(A.4)

Theorem A.2 The form E(U, V ) given by (A.4) is symmetric, densely defined, positive and closed. It is a Dirichlet form in L2 (H, µ): that is E(U, V ) is a symmetric, closed, bilinear Markov form. Proof. The first three assertions result from Lemma √A.1, Lemma A.2 and Theorem A.1. ¯ √ Furthermore the form E(U, V ) is closed since − L is a closed operator (recall that − L¯ is positive and self-adjoint). To prove that E(U, V ) is a Dirichlet form we use the sufficient condition from [67]. Let U ∈ DE ,

V = (0 ∨ U ) ∧ 1.

Then V ∈ DE and estimate E(V, V ) ≤ E(U, U ) is obvious since E(U, V ) = (|κ(Q)|U, |κ(Q)|V )L2 (H,µ) .
Due to Theorem A.2 it is natural to call E(U, V ) the L´evy–Dirichlet form.

A.2 The stochastic processes associated with the L´evy–Dirichlet forms Theorem A.3 A Markov process ξx (t) on H, associated with the L´evy–Dirichlet form ¯ can be constructed as the limit ( for n → ∞) generated by the L´evy–Laplace operator L,

138 Appendix. L´evy–Dirichlet forms and associated Markov processes

of a family of the diffusion processes ξx,n (t) associated with the forms , , En (U, V ) = ( l¯n U, l¯n V )L2 (H,µ) ,

(A.5)

where ln U (x) = −

n $  % 1
(U H+ (x) f k , f k ) H − (U H + (x), f k ) H (x, f k ) H+ . n k=1

(A.6)

Proof. The forms En (U, V ) are symmetric and densely defined. Applying the integration by parts formula (2.2) for n = 1 and for F = VdU, we rewrite (A.5) in the form  n 1
En (U, V ) = (U  (x), f k ) H (VH + (x), f k ) H µ(d x) U, V ∈ DE n k=1 H+ H

(see Lemma A.1). Let f k = ek , where {ek }∞ 1 is the canonical basis in H. Since U H + (x) =

∞
∂U ek , ∂ xk k=1

xk = (x, ek ) H ,

U H + (x)e j =

∞
∂ 2U ek , ∂ x j ∂ xk k=1

(x, ek ) H+ = λ2k xk ,

we deduce from (A.6) that ln = −

n 2 ∂ ∂ 1
− λ2k xk . 2 n k=1 ∂ xk ∂ xk

(A.7)

For each n ∈ N the symmetrized n-dimensional Laplacian ln is positive and essentially self-adjoint on the dense set T in L2 (H, µ). It generates a contractive positivity preserving semigroup Tn (t) = e−ln t ¯

(t ≥ 0)

on L2 (H, µ) such that Tn (t) · 1 = 1. Hence Tn (t) is a Markov semigroup and thus the corresponding form En (U, V ) is a Markov form. There is one-to-one correspondence between the semigroup Tn (t) and the transition probability Pn (t, x, B) = P{ξx,n (t) ∈ B|ξx,n (0) = x} of a diffusion process ξx,n (t) defined on the probability space (, F, P); B is a Borel subset of H . For any bounded measurable function f one has  (Tn (t) f )(x) = f (y)P(t, x, dy) = E( f (ξx,n (t))) (A.8) H

for µ-almost all x ∈ H and t ≥ 0. For any bounded continuous function f on H there exists a unique solution to the Cauchy problem ∂U (t, x)/∂t + ln U (t, x) = 0, for t > 0, with U (0, x) = f (x) considering U (t, x) as a function of x1 , . . . , xn , the variables xn+1 , xn+2 , . . . of U (t, x) being considered as parameters. We prove that Tn (t) converges strongly in L2 (H, µ) to T (t), that is for all f ∈ L2 (H, µ) Tn (t) f − T (t) f L2 (H,µ) → 0 as n → ∞.

A.2 The processes associated with the L´evy–Dirichlet forms

139

Choose f ∈ DE . Then n ¯

n¯

e−t lm f − e−t ln f L2 (H,µ) ≤ e−t lm f − e−t m ln f L2 (H,µ) + e−t m ln f − e−t ln f L2 (H,µ) . ¯

¯

¯

¯

For m ≥ n we have lm ≥ (n/m)ln , since

 m ∂f 2 1
µ(d x) (lm f, f )L2 (H,µ) = m k=1 ∂ xk H

n n 1
∂f 2 µ(d x) = (ln f, f )L2 (H,µ) . m k=1 ∂ xk m

 ≥ H

Hence, n

e−tlm ≤ e−t m ln and n ¯

*

*($ % ) n ¯ n ¯ e−t m ln − e−t l¯m  e−t m ln − e−t l¯m f, f L (H,µ) 2 * ($ % ) n l¯ ¯m −t m −t l n ≤ 2 e −e f, f L (H,µ) ,

e−t m ln f − e−t lm f L2 (H,µ) ≤ ¯

2

since Tn (t) ≤ 1. By Duhamel’s formula we have e

n l¯ −t m n

f −e

−t l¯m

t f =

 n  ¯ n¯ e−(t−s) m ln l¯m − l¯n e−slm f ds. m

(A.9)

0 n

It is easy to check by direct computation that the operators e−(t−s) m ln and e−slm commute, since (n/m)ln and lm commute. To prove the latter property it is sufficient to recall (A.7). In fact, setting qk = (∂ 2 /∂ xk2 ) − λ2k xk (∂/∂ xk ) we see immediately that q j qk = qk q j for all k, j = 1, . . . n which yields ln lm = lm ln and thus (n/m)ln and lm commute. From this we get from (A.9) t  n  n¯ n ¯ ¯ ¯ e−t m ln f − e−t lm f = e−(t−s) m ln −slm l¯m − l¯n f ds m 0

and

 n   n ¯ ¯ ¯ ¯ e−t m ln f − e−t lm f 2L2 (H,µ) ≤ 2 e−t m ln − e−t lm f, f L2 (H,µ) ⎛ t ⎞    n n¯ ¯ = 2 ⎝ e−(t−s) m ln −slm l¯m − l¯n f ds, f ⎠ m L2 (H,µ)

0

 t   n  =2 ds l¯m − l¯n f, G st f m L2 (H,µ) 0

  t    n  n  ≤2 l¯m − l¯n f, f l¯m − l¯n G st f, G st f ds, m m L2 (H,µ) L2 (H,µ) 0

140 Appendix. L´evy–Dirichlet forms and associated Markov processes

n ¯

where G st f = e−(t−s) m ln −lm s f. But ¯

2  t   n  ds l¯m − l¯n G st f, G st f m L2 (H,µ) 0

≤t

 t   n  ds l¯m − l¯n G st f, G st f m L2 (H,µ) 0

 t =t

e 0

n ¯ [l¯m − l¯n ]e−2slm f ds, f m

 ¯ − e−2t lm f, f

n l¯ −2(t−s) m n

L2 (H,µ)

 1 −2t n l¯n e m =t 2 L2 (H,µ) t −2t n l¯n −2t l¯m 2 m ≤ e −e  f L2 (H,µ) ≤ t f 2L2 (H,µ) , 2

since Tn (t) ≤ 1. By Lemma A.1 we know that En ( f, f ) is a Cauchy sequence which implies   1/4  n  n¯ ¯ 1/2  f L2 (H,µ) e−t m ln f − e−t lm f L2 (H,µ) ≤ 4t l¯m − l¯n f, f m L2 (H,µ)   1/4 n 1/2 = 4t Em ( f, f ) − En ( f, f )  f L2 (H,µ) m → 0 as m, n → ∞, n

In addition for m ≥ n we have e−tln ≤ e−t m ln from which we get   1/4  n  n ¯ ¯ 1/2  f L2 (H,µ) e−t m ln f − e−t ln f L2 (H,µ) ≤ 4t l¯n − l¯n f, f m L2 (H,µ)   1/4 n 1/2 En ( f, f ) = 4t 1 −  f L2 (H,µ) m → 0 as m, n → ∞. Thus, lim

e−t lm f − e−t ln f L2 (H,µ) = 0 ¯

m>n,n→∞

¯

for all f ∈ DE ,

(A.10)

for any t from a finite interval. ¯ The sequence e−t ln f is a Cauchy sequence for any t > 0, f ∈ DE . Since DE is dense ¯ in L2 (H, µ), and the family e−t ln is uniformly bounded we conclude that (A.10) holds for all f ∈ L2 (H, µ). Hence lim Tn (t) f = T (t) f

n→∞

for all f ∈ L2 (H, µ).

(A.11)

The semigroup T (t) is a contraction in C2 (H, µ) since in the strong limit this property of Tn (t) is inherited. Moreover T 1 = 1, since Tn 1 = 1 for all n and T is positivity preserving, since the Tn are positivity preserving. Hence T is a Markov semigroup

A.2 The processes associated with the L´evy–Dirichlet forms

141

in L2 (H, µ). By the Kolmogorov–Ionescu Tulcea construction there exists a Markov process ξx (t) such that E( f (ξx (t))) = T (t) f (x) for µ-almost all x ∈ H, for any bounded measurable function f defined on H . From (A.8) and (A.11) we deduce that (T (t) f )(x) = lim E( f (ξx,n (t))) = E( f (ξx (t))) n→∞

for µ-almost all x ∈ H.




Bibliographic notes

Chapter 1 The Laplacian for functions on a Hilbert space was introduced by L´evy [92–95]. The representation of the Laplacian (1.2) (Lemma 1.1) for the case of functionals defined on functional spaces was obtained by Polishchuk [121]. Formula (1.4) is due to L´evy [92– 95]. It was used by many authors. Corollary 3 from this formula is due to Polishchuk in [116]. The theory of Gaussian measures in a Hilbert space is presented in many works (see, e.g., [27, 153]). Measure in the space of continuous functions was defined as early as 1923 by Wiener [156]. In a Hilbert space of functions which are orthogonal to unity, the Wiener measure is introduced in the book by Shilov and Phan Dich Tinh [136].

Chapter 2 This chapter is the extended presentation of the article [50] by Feller (see also [52]). The complete orthonormalized systems of polynomials for the Wiener measure were constructed by Cameron and Martin [25], by Ito [81], by Wiener [157], and for general Gaussian measures – by Vershik [153]. Theorem 2.1 for the case of the Wiener integral of multiple variations was proved by Owchar in [107]. Other theorems of this chapter are due to the author of this book.

Chapter 3 The results of Sections 3.1 and 3.2 were published in the article by Feller [51] (see also [52]), and the results of Section 3.3 were published in the article by Feller [53].

Chapter 4 This chapter presents the results of Feller [54–57, 59].

142

Bibliographic notes

143

Chapter 5 5.1. The Dirichlet problem for the L´evy–Laplace and L´evy–Poisson equations was studied by L´evy [92–95], Polishchuk [116, 119], Feller [36, 37], Shilov [132, 134], Dorfman [28], Sikiryavyi [137, 140], Kalinin [82] and Bogdansky [19]. The notion of a fundamental domain in a Hilbert space was introduced by Polishchuk in [116]. The uniqueness theorems were proved by Feller [36, 37]. 5.2. The Dirichlet problem for the stationary L´evy–Schr¨odinger equation in various functional classes was considered by Feller [38] and Shilov [134]. 5.3. The Riquier problem for a linear equation in iterated L´evy Laplacians was studied by Polishchuk [117], Shilov [134], and, for a polyharmonic equation, by Feller [40]. 5.4. The Cauchy problem for the ‘heat equation’ in different functional classes was considered in the works of Dorfman [29], Sokolovsky [148], Bogdansky [14, 15, 17, 18, 20] and Bogdansky and Dalecky [23]. The correspondence between the Cauchy problem for the ‘heat equation’ and the Dirichlet problem for the L´evy–Laplace equation was revealed by Polishchuk in [118]. Theorems 5.1 and 5.2 are due to Shilov [132], Theorems 5.6, 5.7 and 5.13 are due to Polishchuk [116, 117]. Other theorems of this chapter are due to Feller.

Chapter 6 Elliptic quasilinear equations with L´evy Laplacians were studied by L´evy [92, 95], Shilov [134], Sikiryavyi [140], Sokolovsky [146, 150], Feller [61]. Elliptic nonlinear equations with L´evy Laplacians were studied by Feller [60, 62–64]. The presentation of the material in Section 6.1 follows the studies of L´evy [92, 95], and in Sections 6.2– 6.4 we follow articles by Feller [60–64].

Chapter 7 Parabolic nonlinear equations with L´evy Laplacians appear in the works of Shilov [134] (a mixed problem for a nonlinear equation with iterated L´evy Laplacians), Feller [65, 66] (the Cauchy problem for nonlinear parabolic equations). Quasilinear equations appear in the work of Sokolovsky [151]. The presentation of the material in this chapter follows articles by Feller [65, 66].

Appendix The stochastic processes associated with the L´evy Laplacian were considered by Accardi, Roselli and Smolyanov [5], Accardi and Smolyanov [6], Accardi and Bogachev [1–3], Kuo, Obata and Saito [90] and Albeverio, Belopolskaya and Feller [8]. This Appendix follows the articles by Feller [53] and by Albeverio, Belopolskaya and Feller [8].

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Index

Canonical basis in H 15, 32 Cauchy problem for equation heat 88 nonlinear 108, 112, 115, 121, 126 quasilinear 120 Diffusion processes 138 Dirichlet forms 133, 137 Dirichlet problem for equation L´evy–Laplace 69, 71, 74, 77, 82 L´evy–Poisson 70, 71, 75, 81, 83 L´evy–Schrodinger 84 nonlinear 94 quasilinear 92 Formula Duhamel 139 L´evy 7 Ostrogradsky 7 Fourier–Hermite polynomials 22 Functionals Gateaux class 73, 74, 77 in the form of series 82 Functions cylindrical 22, 40, 48, 53 harmonic 8, 48, 53, 134 Shilov class 23, 41, 69, 70 Fundamental domain in H 68 Gradient of a function 6, 41 Hessian of a function 6, 41, 53 L´evy–Dirichlet forms 133, 137 L´evy Laplacian 5, 6

L´evy Laplacian for functions on a space l2 9 L 2 (0, 1), 10 L 2 ; m (0, 1), 12 Lemma Borel–Cantelli 56, 64 Kronecker 57, 62 Markov processes 133, 137 semigroup 138, 140 Mathematical expectation 34, 43, 55, 59 Matrix Gramm 66 translationally non-positive 67 Mean value of function 5 Gaussian measure 13, 22 random variable 34, 43, 55, 59 Measure centred 13 Gaussian 13, 22 Lebesgue 6, 13 Wiener 16, 77 Mixed variational derivative 10 Moments of the measure 14 Operator compact 23, 63 correlation 13, 16, 22, 133 Hilbert–Schmidt 23, 44, 133 generating 23 thin 23, 69 trace class 13, 17, 22, 50 Operator L´evy–Laplace arbitrary order 33

152

Index

Operator L´evy–Laplace (cont.) essentially self-adjoint 51, 137 multiplication by the function 50, 136 second order 30 symmetrized 40 Orthonormal basis in H 6, 22, 41 L2 (H, µ) 29, 31, 35, 45 Orthonormal system polynomials 23

L 2 (0, 1) 9 Lˆ 2 (0, 1) 16, 77 L 2 ; m(0, 1) 11 L2 (H, µ) 22 Steklov mean 72 Stochastic processes 137 Strong law of large numbers 34, 43, 54 Sturm–Liouville problem 10 Surface in H 68

Partial variational derivative 11 Probability space 14

Theorem Berezansky 30 Etemadi 66 Levi B. 134 Liouville 9 Men’schov–Rademacher 61 Petrov 55, 59, 63 Polischuk 74, 75, 87 Shilov 69, 70 Tensor product 23

Riccati equation 134 Riquier problem for equation linear 87 nonlinear 99 polyharmonic 86 quasilinear 96 Space C0 (0, 1) 16 H5 H+ 23 H− 23 Hβ 23 l2 9

Uniformly dense basis in L 2 (0, 1) 10 L 2 ; m(0, 1) 12 Variance 34, 43, 59 Variational derivative 10

153