Invertible point transformations and nonlinear differential equations

W.-H. Steeb was born in Pforzheim, Germany, in 1945 and was educated at the University of Kiel. He received his Ph.D. fr...

Author: Willi-Hans Steeb.

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W.-H. Steeb was born in Pforzheim, Germany, in 1945 and was educated at the University of Kiel. He received his Ph.D. from the University of Kiel in 1976 and his post-doctoral degree (Dr. habil.) from the University of Paderborn in 1981. Presently, he is Professor of Theoretical Physics and Applied Mathemat ics in the Department of Applied Mathematics and Nonlinear Studies, Rand Afrikaans University, Johannesburg, South Africa. His research interests are in the areas of statistical physics and nonlinear dynamics, in particular chaos with applications in electronics, quantum chaos and integrable systems. He is also involved in work on applications of computer algebra in science. He has more than 140 research papers and 17 books to his credit.

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WERTIILE POINT TRANSFORMATIONS AND NONLINEAR DIFFERENTIAL EQUATIONS Willi-Hans Steeb Rand Afrikaans University

World Scientific Singapore • •New Singapore NewJersey Jersey• L• London • Hong Kong

Published by World Scientific Publishing Co. Pte. Ltd. P O Box 128, Farrer Road, Singapore 9128 USA office: Suite IB, 1060 Main Street, River Edge, NJ 07661 UK office: 73 Lynton Mead, Totteridge, London N20 8DH

INVERTIBLE POINT TRANSFORMATIONS AND NONLINEAR DIFFERENTIAL EQUATIONS Copyright © 1993 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form orbyany means, electronic or mechanical, including photocopying, recording orany information storage and retrieval system now known or to be invented, without written permission from the Publisher.

ISBN 981-02-1355-7

Printed in Singapore.

Preface For nonlinear ordinary and partial differential equations the general solution usually cannot be given explicitly. It is desirable to have an approach by which it can be de termined whether a given nonlinear differential equation is integrable. Several methods have been employed for studying the existence of first integrals and the integrability of dynamical systems. A powerful tool to find integrable differential equations (both ordi nary or partial) is the Painleve test (Steeb and Euler 1988). For classical Hamiltonian systems we can apply the Ziglin analysis or the method of Noether symmetries (Eu ler and Steeb 1992). Other methods can also be applied to non-Hamiltonian systems: t h e direct method, the linear compatibility analysis, the use of Lax pairs, t h e method of Lie symmetries, t h e quasimonomial formalism, and the Carleman embedding proce dure (Kowalski and Steeb 1991). In this book we describe the method of t h e invertible point transformation. We also discuss the connection with other methods mentioned above. T h e invertible point transformation allows the construction of nonlinear differ ential equations from linear differential equations. In chapter 1 first-order ordinary differential equations and the invertible point transfor mation is studied. As an example we consider the Bernoulli equation. In chapter 2 we investigate second-order ordinary differential equations and t h e invertible point transformation. We also consider systems of second-order ordinary differential equations. A large number of examples illustrate the approach. T h e investigation of third-order differential equations is not very common in classical mechanics due to t h e specific form of the fundamental newtonian equations. These kind of equations appear in modelling of physical situations such as a radiating charged parti cle interacting with an external electromagnetic field. They appear also in the reduction procedure of nonlinear partial differential equations by similarity transformations (for example the Korteweg de Vries equation). In chapter 3 we discuss third-order ordinary differential equations and the invertible point transformation. T h e m e t h o d introduced by Lie considers t h e invariance of the form of the differential equation itself under invertible point transformations of one parameter. Lie himself showed t h a t for t h e one-dimensional free particle there are eight point transformations of one p a r a m e t e r t h a t maintain the invariance of t h e equation; the same situation occurs for a time-dependent oscillator. This is t h e maximum number of generators for a secondorder differential equation of the form cPu/dt2 = H(du/dt,u(t),t). Certain ordinary nonlinear differential equations also have eight symmetry generators. T h e free particle

v

equation d2U/dT2 = 0 has eight Lie symmetry vector fields. We generate a class of nonlinear second order differential equations with eight Lie symmetry vector fields by applying an invertible point transformation to the free-particle equation. We also show that this transformation permits us to obtain directly the symmetry generators for this class of equations by using the symmetry generators of the free particle. As examples we consider: the harmonic oscillator, the time-dependent oscillator, the Kepler problem, the particle in a constant magnetic field, and the chargemonopole interaction. In chapter 4 the connection between Lie point symmetries and the invertible point transformation is studied. In chapter 5 the connection between first integrals and the invertible point transforma tion is discussed. A particular problem is the identification of the classes of linearizable equations. For the second-order differential equations this amounts to finding the class of equations which are equivalent to the free-particle equation. The Cartan equivalence method provides the answer to this problem. We also consider third-order ordinary differential equations. In chapter 6 the Cartan method is introduced. In chapter 7 we study the invertible point transformation and the Painleve test. As an application we consider the anharmonic oscillator and the second Painleve transcen dents. For ordinary differential equations the Painleve analysis and the invertible point transformation can be used to construct integrable nonlinear equations or equations which are related to the Painleve transcendents. We compare both techniques for the second Painleve transcendents. First we give an introduction to the Painleve test and then consider its connection with the invertible point transformation. In chapter 8 we consider the Painleve test and partial differential equations. In chapter 9 partial differential equations and the invertible point transformation is discussed. We also study its connection with the Painleve test. In chapter 10 we consider difference equations and linearization. As an example the logistic equation is studied. Most of the calculations are checked with REDUCE. A collection of programs which are helpful in the study of the invertible point transformation is given in chapter 11. Chapter 12 gives a short introduction to the jet bundle formalism.

vi

Contents 1 First-order ordinary differential equation

1

2 Second-order ordinary differential equations 2.1 Second-order ordinary differential equation 2.2 Systems of second-order differential equations

7 7 27

3 Third-order differential equations

35

4 Lie point symmetries

45

5 First integrals and differential equation

61

6 Cartan equivalence method 6.1 Second-order differential equations

69 69

6.2 Third-order differential equations

76

7 Painleve test and linearization

87

8 Painleve test and partial differential equations

105

9 Partial differential equations

117

10 Difference equations

133

11 R E D U C E programs

141

12 Jet bundle formalism

171

References

175

Index

179

vii

Symbol Index

empty set the set of integers the set of positive integers: natural numbers the set of rational numbers the set of real numbers nonnegative real numbers the set of complex numbers the n-dimensional real linear space the n-dimensional complex linear space

0

z M Q 11 11+ C TV C" i %z AcB AnB AUB f°9 u,U t,T x,X u = U =

z II. II xy det / [,] Sjk

A x A d(.)

c

(u1,u2,...,un)T (UuU2,...,Unf

real part of the complex number z imaginary part of the complex number z the element x of Hn the subset A of the set B the intersection of the sets A and B the union of the sets A and B composition of mappings ( / o g)(x) = f(g{x)) dependent variables independent variables (time variables) independent variables (space variables) vector of dependent variables vector of dependent variables Lie symmetry vector field norm scalar product (inner product) determinant unit matrix, identity operator commutator Kronecker delta with Sjk — 1 for j = h and Sjk = 0 for j ■£ k eigenvalue cross product Grassmann product (exterior product, wedge product) exterior derivative Lagrangian function viii

Chapter 1 First-order ordinary differential equation An ordinary differential equation of first-order

4<**)-

(1)

may often be simplified or reduced to a standard form by introducing new variables, T, U by means of the transformation T(t) = G{t, u(t)),

U(T(t)) = F{t, u(f)).

(2)

We assume that H(t,u,u), G(t,u) and F(t,u) are smooth functions, i.e. C°°. All considerations are local. Geometrically, we regard (2) as a point transformation, for it transforms points (t,u) of the iu-plane to points (T,U) of the Ttf-plane. We assume that the Jacobian d(T U) J := a \ ' / - d6t a (*.«)

1 dG dG\ dt

du

9F

3F

^0

(3)

\ ~dJ to)

over a region R of the iu-plane. There is then no functional relation between u and U; for this would imply J = 0. (4) Moreover, if the point (Ti,Ui) corresponds to (t l ,«i) we can solve (2) uniquely for t,u in the neighbourhood of tuux. We thus obtain the inverse transformation t(T) = Q(T, U(T)),

u(t(T)) = P(T, U(T)).

(5)

Definition: We call the transformation given by (2) an invertible point transformation. 1

2

CHAPTER 1. FIRST ORDER ORDINARY DIFFERENTIAL

EQUATION

A curve in the iu-plane transforms into a curve in the T{/-plane. By means of (4) and (5) we can now transform (1) into («)

0.

H(T,U(T),§)=0. If (6) can be integrated to give

(7)

we obtain a solution of (1) on replacing T and U by their values (2) in terms of t and u. Here C is the constant of integration. From (2) we obtain dU _ dU_dT_ _dF__ dF_du dF_ H ~ dfdt ~ ~dt ~ du dt + ~dt' Since

dT _ dGdu dt ~ du dt+

we arrive at

d£du

du

dG dt

((8)

'

l(9)

'

d£

~du~ dt+ dt dT ~]>Gfa 5G' du dt dt

(10)

Let us now give several examples: Example 1: The Bernoulli equation dti

Po(t)-£ + Pi(t)u =

fl(t)u"

(n#l)

(11)

is transformed by the invertible point transformtion T(t) = t,

U(T(t)) = u*-»(t)

(12)

into the linear differential equation + (1 - n)P1(T)U = (1 - n)R(T).

Po(T)^

This can be seen as follows: Since

_ dT

(14)

Hi and

dU

dUdT

(13)

.,

-dT = Ir!t={1-n)u

,

_ndu

Tt

( 15 )

3 we have dU dll

„„du du

,

1T = ^~-n)u-*-.

a

(16)

Example 2: The homogeneous differential equation

du du

H

+ f

.(u\ fu\

{l)

(17)

=0

is transformed by

T(t) = t,

U(T(t)) =

^l

(18)

into the separable equation dU U + T— + f(U) = Q.

(19)

In examples 1 and 2 only the dependent variable is changed. D Example 3: The differential equation du dt

/

I at + bu + c \ at + 0V. + 7

(20)

may always be integrated by an invertible point transformation Here a, 6, c, a, ft, 7 are real constants. There are two cases to consider: Case 1:

D

(a

bx

-**{l u 0,) , o . "/,)**■

(21)

We make the invertible point transformation

t(T)=T + h,

u(t(T)) = U(T) + k

(22)

where h and k are constants. It follows that du dU ~dt ~~ dT

(23)

aT + bU + ah + bk + c aT + 0U + ah + (3k + 7

(24)

and at + bu + c at + j3u + f Since D ^ 0, the linear equations ah + bk + c = 0,

ah + /3k + 7 = 0

(25)

4

CHAPTER 1. FIRST ORDER ORDINARY DIFFERENTIAL

EQUATION

may be solved uniquely for h, k. With these values, (20) becomes dU dT

(aT + bU\ \aT + pU)'

J

(26)

This equation is homogeneous and may be integrated. Equations (22) may be interpreted as a translation of rectangular axes to the new origin (h, k), the point of intersection of the lines at + bu + c = 0,

at + 0u + 7 - 0.

(27)

When these lines are parallel, we have the second case: Case 2: D = det(*

>)=0.

(28)

We now put

T(t) = t,

U(T(t)) = at + bu® = !*±m.

(29)

These fractions are equal since D = 0. Now dU_ = 1+ ~ dT~ ' adt dU

(30)

and (20) is reduced to dU b (aU + =1+ dT = a \aU + in which the variables are separable. Example 4-' Let us consider du dt

□

i)

Zt — u — 5 -t + 3u + T

(31)

(32)

The lines 3t - u - 5 = 0,

-t + 3u + 7 = 0

(33)

meet in the point (1, —2). The change of variable t{T) = T + 1,

u(t(T)) = U(T) - 2

(34)

reduces (32) to the homogeneous equation dU _ 3T-U dT~ -T + 3U

(35)

5 or

(3T - U)dT + (T - 3U)dU = 0.

(36)

Since this admits the multiplier //. 1 T{3T -U) -- U) + U{T - == T(3T U(T --3*7) - W) = 3(T 2 - U2) the equation

3T- -u

T- -3U d,rr* » =n

^

d,T ++ ^

rpi _ ■Ul

(38)

J<2 .-U'

is exact. We obtain

(37)

(T + Uf{U - T) = C

(39)

where C is a constant. Since T{t) = t - l ,

U(T(t)) = u(t) + 2

(40)

we have (t + u(t) + l)2(u(t) -t + 3) = C.

□

(41)

Example 5: To solve du dt

2tu t2 +

u2-l

(42)

we write (42) into the form dt + du _ (t + u) 2 - 1 dt-du~ (t - u) 2 - 1'

(43)

This suggests the invertible point transformation

r(t) = t-u(t), tf(r(*)) = *+«(t). Thus we find

dU _ U2 - 1

(44) (45)

The integral of this separable equation is {T + 1)(U-1) C (T-l)(U + l)or

(46)

TU-l C +1 (47) U - T - C - l - K where C (or K) is a constant of integration. Hence, we obtain the general solution of (42) as t2 - u{tf - 1 = 2Ku(t). D (48)

6

CHAPTER 1. FIRST ORDER ORDINARY DIFFERENTIAL

Example 6: Consider the linear equation dlJ = A dT where A is a constant. The general solution is given by U(T) = AT + C

EQUATION

(49)

(50)

where C is the constant of integration. Applying the invertible point transformation to (49) we arrive at dudF d£_ (dGdu dG\ ,.. ( dt du+ dt ~ \du dt dt ) ' ' Consequently du (dF dF dG (52) "bT + A~dt dt \du or dF dG du ~^t+A~dt (53) dF dt AdG ■ du du If G does not depend on u we find

-£)dF

dG

~^t+A~m

du dt

(54)

dF du

If G(t, u) = t we arrive at dF

du dt

~m+A

Finally if F does not depend on t we have du _

A

dF du

■

A

(55)

(56)

du Thus the solution of (56) is given by F(u(t)) = At + C.

(57)

2

If, for example, F(u) = u we have u(t) = sjAt + C.

(58)

This means that (58) is a solution to du dt

A 2u

— = —.

D

(59)

Chapter 2 Second-order ordinary differential equations 2.1

Second-order ordinary differential equation

Second-order ordinary differential equations and invertible point transformations have been investigated by several authors (Duarte et al 1987, D u a r t e et al 1989, D u a r t e et al 1990, D u a r t e et al 1991, Euler et al 1991). T h e starting point is a linear second order differential equation or a second order differential equation where t h e exact solution is known. By applying an invertible point transformation a class of nonlinear ordinary differential equations is generated. We discuss t h e following cases: T h e free-particle

equation

2

dU

(1)

dT2=" the harmonic oscillator

S"-

the anharmonic oscillator

(2)

g+"-

(3)

the anharmonic system d2U

+ [/■" = 0, dT2 and the second Painleve transcendent

n

integer

tPll

^L = 2U3 + TU + a where o is a constant. 7

(4)

(5)

8

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL EQUATIONS

The most studied case is the free particle equation — T

(6)

- f l

dr2

Obviously, the general solution of (6) is given by U(T) dT++C2C2 w(r) ===CxT

(7) (7)

aation where d and C 2 are the constants of integration. The invertible point transformation is given by U(T(t)) = F(u(t), t), T(t) t) (8) T(t) = G(u{t), G(u(t),<) (8) J7(T(i)) = F(u(t)>*)> .1. We where F and G are smooth functions (C°°) of u and t. All considerations are local. We have to assume that dC dF 0 dGdF A (9) A: = (9) dt du

: ?rSr~£^

and

dGdu dGdu du dt

dG dG *o.„ dt

(10)

(10)

-du-Tt+lH**-

Applying the chain rule of differentiation we find that dT dG dGdu du dG dG =' + dt du dt lt -du-tt -m dt and

dU dt/

dUdT dt;dT

(dGdu dU dU (dGdu

,„ , (11a) (lla)

dG\ _dFdu dG\ dFdu

dF dF dt'

~dl =~-df~dl ~ IT l [to*++lt) ~dJ dT~dl-lr\'d^li ^)-^u'Ttdu+^-dt From (lib) we arrive at dr/ dU dT dT

—

dFdu 9F dF " a - J I + ~o7 du dt dt <7U a t ot dGdu dG dGdu d~G~' du dt dt du di dt

For the second derivative we obtain
/ti.v

vilc)

+ dJD = DDF f'Sf^ + fWl^W)) dU ,

[D{G))

where we introduced the notation dGdu dG D(G) '--du~ti+W =

D{G)'- ^dit

+

2 M/r,/,™ nd Gdu D(D(G));=27—-~

WW) ■= 2 ^

D(F):

{

dFdu

d^ D{F): ^t

d2G (du \2 +—. —

{ ))

dF

~'bmt+-dt =

,,,_,

(116) {nh)

+

dGcPu

%

d2G

j +~du-d*+-w

+ £ (£)' + g $ + |£

(12) (")

(13) (13)

(14a)

(14a)

2.1.

SECOND ORDER ORDINARY DIFFERENTIAL

and { {F )

DD )

_ d2F du

=2

: d~udtTt

+

d2 F

fdu\2

EQUATION

dF
w \Jt) + ^ ^

9

d2F

+ -dp-

{ub)

We prove now that ^ - = D(D(F))\u_u(t).

(15)

We have
~d¥

=

d2F (duV

d2F du

+

+

dFcPu

+

32F du

+

d2F

a ^ U d^diti fr;w frlditi m2SPF fdu\2 du2 \dtj

+

d2F du d£cPu a^F dudt dt + du dt2 + dt2 ■

(16)

From (13) we obtain D(D{F))

- SSPM™ du d IdFdu dt du \ du dt

(duV d2F

[dl)

du2

d2F ( du\2 du2 { dt)

+

+

dF\ dt)

d (dF du dF_ + dt \ du lt ~dt

du d2F

Ttd~u~dt

+

d2F du 2

d2F du

dFd2u d^dtTt ^l^ du dt2 +

dFd2u

d2F +W dt2

+

(17)

d2F

+ 2

+

dudt dtdu¥tH du dt &IW

lP

2

dF dt2 ' Inserting (lie) into (12) we find ^
(18)

Since D{G) / O w e find from condition (6) that D(G)D(D(F))-D(F)D(D(G))=0.

(19)

Equation (19) is implemented in a REDUCE program (see chapter 11). From (19) we obtain n

^

/ . \3

+ A3(u,t)l^\

/ , \2

,

+A 2 (tt,*)f^) +A 1 («,<)^ + Ao(«.0 = 0

(20)

10

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

where A3

:=

A2

:

Al

~

:=

(dGd2F

dFd2G\

°

. ,

A

l>a^-&T^J (dG
dn2dF2FdG dG dudi du du dudt

2 (dGd (dGd2FF

2 dF d2G ndF +2 2 dtdudtdt dudt

[dudt2

(dGd (dGd FF 22 \3t dt {-mm -

22

dFd dFd G\ G\

22 dFd G\G\ dF d 2 2 dt )) dt dudu

d22G du dudt dududt

dF ndF

2 F dG dd2F nJG

22

A :A 0 :=

EQUATIONS

dF d2G\ G\ dFd

-TuW)^

,t (21)

A_xi

. ._,. 1A .

dt dt2 )) A~ • Ul Ul ~ -dJ-dW)

Equations (21) are implemented in a REDUCE program (see chapter 11). Let us now consider special cases of (20). Example 1: Assume that G does not depend on u, i.e., G = G(t). Then we obtain A3 = 0 A2 A, 1 Ao

dGd22FF dGd dt du2

!,

2 ((dGd F _ dFdHA dFd2G\ 2dGcPF_ a 5 V dt* dudt "^ 5" V du ^2 dt2 /) 2 2 (dGd /3G3 FF

2 5 dFd F 92G\ G\

,

(22) ( 22 )

A _,,

A A° = i a T(-d7^--dJm « 5 - - a r « 5 -2J-)^"-

If G{t) = t, then (22) simplifies to A3 = 0 A2 A = 2

Ai = A!

5d22FF 2 du du2

AA -- 1l

32F i

\J

|

2-^-A-1 dudt A

(23) (23)

Ao = A° = dt2 a^A • Then (20) takes the form 2

2 2 (du\ 2 ,d „ A ,d Fdu du ,d2F ! d2F ,d2F xd F (du\ | + 2 A =0 2 +A 2 + + A + Adt du '[dtj dudt dt dt2

d22U u

W

A

a^ (*J

** S3* * ■ V = °

(24) (24)

2.1.

SECOND ORDER ORDINARY DIFFERENTIAL

EQUATION

11

where A=~ . du

a

(25)

Example 2: Here we consider the special case A3 = 0 and A! = 0. If we put A3 = 0 we obtain dG32F 3Fd2G . . =

tow

to aJ-

(26)

The solution of (26) is given by G(u,t) = f{t)F(u,t)

+ g(t)

(27)

where / and g are smooth functions. From the condition A] = 0, i.e. dGd2F dGd2F du dt2 + dt dudt

2dG^i_l)5Fd^G__dF^G_0 2

dt dudt

du dt

[(28)

~

'

we find two types of solutions for F and G, namely F(u,t)=a{t)u

F{U't)

=

+ b{t),

G(u,t) = C{a(t)u + b(t))+g{t)

dg

1

/

Jq dg dt

1

p(t)u + q(t)~~^+9 df + 9 dt dt

G(U <) =

p(t)u + q(t)-f> dt

(29a)

'

where C is constant. Inserting (29a) and (29b) into (21) yields A2 and A0.

(2%) □

Example 3: Let F(u,t) = eu,

G{u,t) = t.

(30)

Then we find A = e" and

(31)

Idus

+

^ UJ=°-

(32)

Since the general solution of (6) is given by (7) we find that the general solution of (32) is given by u(t) = HCit + C2). O (33) Example 4: Let k be an arbitrary constant. The differential equation d u

du

1 9q

-r- + ku— + -k'us dt' dt 9

= 0

(34)

12

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

is transformed into the free-particle equation if F(u, t) = - - \kt\ u 6

G(u, t) = - - \kt u 3

(35)

where

A- 1 .

(36)

uJ

The general solution t/(T) = CiT + C2 (37) for (6) permits us to find the general solution for (34). Inserting the invertible point transformation (35) into (37) we find t

«(0* >

2

=

= Cl

{W)~

It follows that

- \kt\

6(t-- d )

C ^=M>Xk kt - 1C tUc kt 2+ 6C u(t) =

+ C2.

2

x

where C\ and C? are constants of integration.

2

(38)

(39) ™

□

Example 5: Consider the differential equation for the free-falling body «PH

eft2

(40)

+
where g is a positive constant. Using the invertible point transformation F(u, t) = F(u,t) =u+ + -gt\ ^gt2, we obtain A = 1 and

0.

dT 2

G{u, G(u,t)t) = t

I I

(41)

(42)

A number of second-order ordinary differential equations
=

$dt2= ~~ *

H(t,:u(t),

( ' . - » ■
(43) <«>

can be derived from a Lagrangian function £(t,u, C{t,u,u).«).

(44) (44)

The Euler-Lagrange equation of motion is given by ddC
dC _ 0 3u

- — =

(45)

2.1. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATION

13

This equation follows from a variational principle. For example the free-particle equation

(Pu = 0 dt2

(46)

C{t,u,u)=l-u\

(47)

1

can be derived from Using the invertible point transformation we can find the Lagrangian function for linear or nonlinear ordinary differential equations which can be transformed into (46). For example a Lagrangian function for the free-falling body is given by C{t,u,u)=l-{u

(48)

+ gtf.

We find this Lagrangian function with the help of the invertible point transformation (41). Note that the equation of motion for the free falling body (40) can also be obtained from the Lagrangian function C(t,u,u)

=-u2

(49)

- gu.

Let us now consider the harmonic oscillator (50)

— + U = 0. dT2 The general solution of (50) is given by U(T) = d sin(T) + C2 cos(T)

(51)

where C\ and C2 are the constants of integration. Since (D(G))2^

= -^±D(D(G))

+ D(D(F))

(52)

we find that (50) takes the form -D{F)D(D(G))

+ D(G)D(D(F))

under the invertible point transformation (8). Let us now give two examples:

+ (D(G))3F = 0

(53)

14

CHAPTER

2. SECOND

ORDER ORDINARY

Example 1: Let

DIFFERENTIAL

EQUATIONS

u22 F(tl,t)=y. F(u,t)=j.

G(u,t) G(u,t) = = tiI,-

(54) (54)

Then we find that D(G) D(G) = = 1, 1,

D(D{G)) D(D(G)) = = 0, 0,

Thus we arrive at

du Dl.F) = —, D(F) =U u—, at

(duV D(D(F)) (' D(D(F)) = = l-^j it)

d2 u , „ . , + Uu—. d^"" ((55) ^

DD It) 2 wdt (-dl) \c T = °-

,_„-

(56) (56)

Example 2: Consider t h e differential equations
(57) \" /

and

T(t) = f -^rds 2 dS T^ = J Ip (s) p2{s)

(58) (58)

(59) (59)

(57) takes the form d?U „
(6

0)

lation where p satisfies (58). Applying the chain rule and the invertible point transformation (59) we find dUdT Idu 1 dp dU dU_dUdT_^du_ J_dp (61) u d ~~dt ' ' dT dt ~ pdt ~ ~f?~dl t~ dTit-'pdt > * dT

dr _ I Idu dU 1 u dp dT p2 p dt ' p2 dt'

Therefore

Theref°re Since

1

Hi ~7'

^ I

d fdU 1 \

T d< /'V ~ d/d!7U 1_\

dt\dTp2)

~

=

d2I/dTl 2^ dT dt p12 (PUdT

I^_JL^ H <

dUldp 2 3 ^ dTp dt d{7 1 dp

dT2 dt p2 dTp3dt " dTp dt ' p3 \dT p~~2dT~dl)

\ / (63) (63)

(64) (64)

2.1. SECOND ORDER ORDINARY DIFFERENTIAL

15

EQUATION

and d (\^}L (\du _ JL^E\ u dp\ — 2 dt \p dt pp*~dl) dt) dt\pdt~

-

1 (Pp 1 dudp i1 dpdu \
fo lp\ 2 (dp\ p p^Xdi) \c It)

u n u + 3

u d2p ~~fP~dl? p2 dt2 (65)

2 dpdu 2 ~~p "dl'dt ~'p2dt'di

+

ud2p 1\d?u d2u u d?p + o" u Idp\ (dpY 2 2+2 2 2 3 p'Ht ~ 2p~p^di 'p'dt p^\dt) dt p \dt)

it follows that
dUdp dUdp_

dpdu

Pu 2Pu

Pp Pp

u (dpV (dp\

pU +2 3 +p 2 +2 dT2 p '2dTH dT dt ~ ~' 2pPTtTt ~dt~dt+P Itw ' - 7* W J \-dt)[dt) ■ If^-p w-pu

Since cPu (Pu 22 / j N dt2 ~ -w (t)u, we have

dU dU

dp dp

du du

dT=P~di' -uTt>

(PP I Pp 1 ,, , -u2(t)p dt2 ~

(66)

,„„.

7'

u „ (PU1
<68>

From U = u/p vie arrive at 2

- II° -= gddTuip ^+ -U p 1

TT

2

or

dl* p p d2U SdT2 ^+ u

= 0= ■ U.

Thus the general solution of (57) is given by «(*) == Cip(t) sin(T{t)) + C,p{t) cos{T{t)) u(t) = Cip(t) sin(T(r)) + C2p(t) cos(T(t))

(69) (70) (71)

where p is any solution of (58). The solution of (58) is well-known. We have ab - (t) we can find p(t) (Lewis 1968). □ Let us now study the anharmonic oscillator tPU

gdT* ++U^ --==1).°. 3

(73)

This nonlinear equation can be solved in terms of elliptic functions. Let

:= U, fVi 1 :=U,

dU g. uU2 2~:= dT'

(74)

16

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

We obtain the autonomous system dUi_ Ui IT ~ '

dU2_ I T -

3

(75)

~Ul

with the first integral

i{ux,u2) = \ul + \ul

(76)

Then the solution of the initial value problem of system (75) can be expressed in Jacobi elliptic functions as _ (t/ 2 o/cj)sn(uT, i) + Uwcn(uT, i)dn(uT, i) l + (U10/V)2sn2{uT,i)

Ui(T) U2(T) =

U20cn(uT,i)dn(uT,i) [1 - {U10/V)2sn2(uT, [1 + (U10/V)2Sn2(wT,i)]2

(77a) i)}

2uUw(Uw/V)2sn{uT,i) [1 + sn2(uT,i)] 2 2 [l + (U10/V) sn (uT,i)]2 where

'1\1/2

w

= g)

V,

IP 2

V = {±Eyi\

E=^

IT* 4

+^

(776)

= I{UltUa)

(78)

with i = \/—l and sn, en, dn are the Jacobi elliptic sine, cosine, and delta functions, respectively. The initial values are U10 = U\{T = 0) and U20 = U2(T = 0). Now we apply the invertible point transformation (8) to (73). Since ,d2U__

(D(G)y dT2

D(F) D{D{G)) + D(D(F)) D(G)

(79)

we obtain + D(G)D(D{F))

-D(F)D(D(G))

+ (D(G))3F3

= 0.

(80)

Consequently

where

cPu

fdu\

^

UJ

+A3

2

_ (dGd F lI du du du du22

A3 =

A2 =

1

dG(P£ d2F dG 2+ dt du dudtdu

Idu\

+A2

w

2

dF8 G du du du du22

3F d2G dududt

, du

(81)

+Ai +Ao=0

^ \ du) I >{%)'>•)*•

OG (3GV dF d2G dt du2 + dt \du)

A"1

2.1. SECOND ORDER ORDINARY DIFFERENTIAL EQUATION (dGd22FF dG d2F _ (dGd 3G Ai1 — 1 2+ 2 \\du du dt dt dudt dudt

dF 3 d2G ndF

dFd G dFd22G du dt22 du

dt dudt dt dudt

_ jdGd fdGd22FF Ao \dt dt2

17 17

2 2 3G (3G\ dG(dG\ du\dt) ) du\dt

3\

,

)

dFd G (dGV J\ \ A, dFd22G — F3 A 1 dt dt2 ' [dt) I

(82)

Let us now give several examples: Example 1: If we make the ansatz F(u,t) = f(t)u, f(t)u,

G(u,t)=g(t) G{u,t)=g{t)

(83) (83)

we obtain from (82) that A22 = = 00 A3 = == A

(84a) (84a)

and

A A Al "J

(dgW Jl^\ \\
S 1

dt

It follows that

846 ((846) >

dt dt

cPu h(t)u + + f3{t)u f3{t)u33 = 0 ^ 2 + fi(t)-£ + fi{t)u *l"'~d~i dp

(85) (85)

where

/i(') -

dgdf dgdf
ffd

ftft

dgd22ff dgd

df
2 hit) = -dtw-TtlT ff tt Tt

fs(t) = -If

dg\* dt)

(86)

nd Here / and g are arbitrary smooth functions of t. We are now able to eliminate // aand g from system (86). We obtain g(t)^

-J )fJ/W = fWdsds

1/61 / 6

/(r) ==-Cf C3/ 3 /(*) h{t) = Mt)

7 (df3V 1 d2f3 ++ o73^' Tsr18/ ih~di3 dt ' " 3 6 ^ W 3^~W!{'dT)

+ f+

l'i l ' + h

~3dt 9

(t)explj f f l e xfl{s)dS p [ /\ / l ( ^ ]

+

(87a) (87a)

(876) (876) (87c) (87c)

18

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

Thus (73) is transformed into (85) under (8) where

<

G(u,t) = /t'^hs ti^-ds

F(u,t) = fcexp tt&®±) /s1/6(*)] - .

/

/(-)

with / 2 satisfying (87c) and / i and f3 are arbitrary smooth functions of t.

(88)

□

Remark: In chapter 7 we study this case in more detail in connection with the Painleve analysis. Let us now study special cases of example 1. Example 2: Let fi(t) = c\, fi^t) = c2, where c\ ^ 0. Then we find f{t)

=-t^l\

= -t-^l*.

(89)

G{u,t) = -e~Clt/3

(90)

g(t)

Consequently, the invertible point transformation F(u,t) =-uecit/3, transforms (73) into dPu du , — + C l - + c2 u + u = 0

(91)

with

2cJ = 9c 2 .

□

(92)

Example 3: Let fi(t) = 0, /a(i) = 1. Then from the invertible point transformation approach we find /2(<) = 0. (93) Thus we only find a special case.

□

Example 4: Let f2(t) = 0 and f3(t) = 1. Then

dh dt

+

2.2

3^=

= 0.

(94)

The general solution of (94) is given by

f

M=m+z-

D

(95)

2.1. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATION

19

Example 5: If we put h{t) = atn,

f3(t) = 0tm

(96)

we arrive at

/ 2 W = (6n +

^

m )

+

^ - ( l

+

=)g.

(97)

When we set a = 1/4, n = — 1 and m = —2 we find A(*) = £ .

/a(*)=g|?

/3(*) = | -

(98)

Examples 2 to 5 are special cases of example 1. Let us now consider the hodograph transformation. Example 6: Let F( u,t) Then Therefore

dF

dF

aT = 1' A3 =

G(u,£) = u.

=t,

aG

n

7T = 0 '

^ =

A2 = 0,

-t\

n '

Ai = 0,

Consequently, we obtain / I \ , (du \

dJ 2 ,u

0

(99) — -1 du

(100)

Ao = 0.

(101)

3

(102)

&-*{*) =°' ° The anharmonic system (4) has been studied in detail by Duarte et el (1991). Next we consider the second Painleve transcendent (see chapter 7)

f^ = 2U3 + TU + a (103) dl L and perform an invertible point transformation to find the anharmonic oscillator dr

-^

du

IL

+ f2(t)n + f3(t)u3

+ h(t)^

= 0.

(104)

This provides a condition on /i(t), /2M and fz{t) such that the anharmonic oscillator (104) can be transformed to the second Painleve transcendent (103). The invertible transformation is given by U(T(t)) = F(u(t), t)

T(t) = G(u(t),t),

(105)

where A

,

§

"

at du

§

^

0

du at

.

(106)

20

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL EQUATIONS

Since ( D ( G ) ) 2 g = -^D(D(G))

(107)

+ D(D(F))

we obtain from (103) that -D(F)D{D{G)) + D(G)D(D(F)) = 2(D{G))3F3 + (D(G))3GF + a{D{G)f.

(108)

Consequently cPu fa'.uY falu\ d?u + A 3 fdu\ ++ A. 2 fdu\u) ^ + A3(c

where

. du

U u)J M"
^

^

A3 = I_?£
2

2

du du du du2

du du2 du du

\ du du JJ

2

f 2dF^G__dFd^G _ (_ 9F52G n9Fd G flu 9u3i dududt dt du du22

A22 = | ~\ i

-■"(! 2 <9Fd2G G = (f 9F9 A1, = | ~ \^ du du dt dt22 '

9a 2FdG F9G du du22 dt

+

S)"-(£)>-«"(£)*

- " ( ! \du J 2 2

4+

2 2 dF d&*G G &£_dG F dG 2„d dt dt dudt dudt +' "dudt dudt dt

\du J J

2 3(dG\ dG

dFF 9G dG dudt dudt du

nd

„ / aG-\ d4

dG

GY

.,„„. (109)

. du + A ^ +Ao=0 +Ai + Ao = 0

n9F

+

\du)

dt

A-

2 d29FdG FdG dG (9GY (dG\2 33dG 2 dt du du -^-du-y-dT) du [dt )

.

-"£(£)■--£(£)>-* „„„9G fdGY dG fdGY\ A - 3 a — i -7T-3FG— \-z-) A du [dt) du {at) )

a ^{-ai^+ ff--(f)'-(f)'-"(f)>-^-^m-Hi ^(-ff <"»» Example 1: Let F(u,t) F(u, « )=« //(*)«, (*)«, With this special ansatz we find that

G(u,t) G(u,t) = g(t). g(t).

A3 = A2 = 0 and Ai Ax ==

(112a) 2

dy^ d£ 2a 2

L 9__ fd£i5

-tf-^

dt

(Ill)

*i

J

(1126)

2.1. SECOND ORDER ORDINARY DIFFERENTIAL

(

21 21

EQUATION

" /i7 (f)) U -

(
dfcPg , (dg\ \ , ( dg\ Idg\ - 2r/ 3 ( | ) 3 u3 3 - a ( | ) 3 " eftdi2 " A -- V -d¥Tt-Jtlfi-f9{-dt) da u~2f U J u -a(dl »o = - 1*J ^ — - •

(112c) (112c)

XI (X — \J \\i I v i i U W o X.'Ll. Q. L (JPXL du d2^ du 3 3 = 0 ■jTJ i W ^ + /«(*)" -jp + /h{t)-^ Mt)u + f/3(')" =0 3(t)u

where 2

(113) (113)

d d

L i_fliJ

(114a (114a)

df2

*»>- * VA " /iW =

.

d< (ft

)

7

dp dgcPf dfcPg dl
dt dt ' dtdt /2(f) == * g * j g m dgf dt.1

f9 (dg\

3

' [dl)

(1146) (1146)

W

A4f-

/3W = - 2 ( / f ) fs(t)

( I(114c) Hc)

•

nd Here / and g are arbitrary smooth functions of t. We are now able to eliminate: //a and g from system (114). We obtain h{s)ds\ f{i) Cf13/6 (i)(t)exp(j^^-\ exp f(t)--=-Cfl> (/ 3 / and

c2 eXP mdf (s)ds\ 9(t) = { A(t) 1 9(0 = ^^) CA1[lJ jA t) ' isfAt)

>(lh

(115a)

{U5b) (1156)

where

dfz 2 A(t) := ( - 6 / 3 ^ + 7 ( ^ ) 2 - 9f2 / : f/ 3 ^ -- 11 22 ^^ // || ++ 36/ 22 / 332 -- if* A . A(t)-,

-(-*£+'(*)'-

-stf/fV

(115c) (115c)

Inserting / and g into (114a) we find the condition

' ^ • - M # f ^ + I 8 ^ - ■»($),*+«^(S)'*

-^ A +-*« + 4f«-(*r + «(t)«

22

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

-(*)•*—(ty +»£*/,* + 7 2 § / 3 3 / 2 / 1 --1"4 f§ // 3l3// 1f3 --

/|/2 + 9 0 ^ ^ / 3 3 -

M*£J* dt2 h

54

EQUATIONS

.Jf3dfis3s ~ 21~dl~dVh' -hlTtTthh

9 0 ^ +18§/J 1 8 ^ + + 54™/}/, 5 4 ^ -- gofj}/, +

~(f)V«f A*-***+36 ( | ) ' j - _

3 6

f fah

8f*fi= =0.0. + 6 0 f &ft - - W&ftti 36/3V2/1 + "1-Stiff

(116) (116)

We also find this condition within the Painleve analysis of (104) (see chapter 7). For the case a ^ 0 one finds driven anharmonic oscillators. This case is not discussed here. Thus far we have considered only invertible point transformations to linearize nonlinear differential equations. There are also transformations including derivatives which can be used to linearize certain differential equations. Let us give several examples. Example 1: Consider the Riccati equation ^- = u2+t, u(t = 0) = l. (117) at This equation has no solution in terms of elementary functions. Bessel functions are needed to solve it. We assume a MacLaurin expansion for

u(t) = i + ±^le n=0

(118)(118)

n

-

and compute successive derivatives (du\ +2 d u =22(du\ +2 2

2

2

2

d3u

»S' ^£-SS+*£^ = "^ ^= ^ UJU) "^' ^ "^'
du dt

n -rz = 2u—, 2

J3 d*u

n

n

d"u

= 6dud u +2„

(119)

(119)

Therefore

u'(0) = 1,

u"(0) = 3,

u'"(0) = 8,

K('"»= 34,...

(120) (120)

Consequently 3

4

17

31

101

^(i) l +tt + ^t* + ^i? + ^t* + ^t* + w * . . . .

(121)

(121)

Equation (117) can be linearized with the help of the transformation u{t) = - ~ | .

(1222 (1222

2.1. SECOND ORDER ORDINARY DIFFERENTIAL

23

EQUATION

We find the linear second-order differential equation d2y ^ f + f«/ = 0. D

(123)

Example 2: The nonlinear ordinary differential equation d2u du ^ +3 - ,

,«.,.. (124)

+ U3 =

o

occurs in the investigation of the modified Emden equation. Equation (124) can be linearized with the help of the transformations U(T) = u2{t(T)),

dT = u{t(T))dt{T)

u(t)~-

1 d

(125) (126)

~'W) ~dl'

Then we can find the general solution to (124). Let us first consider transformation (125): Since _ du dt dt

dU

dT=2u~dtdf and

J2udt_=2
fU dT

,.„„, (127)

ndu

'~ "dt

2

dt2 dT

udt2

we obtain u
2 ^

3udU +

YdT

„

TT

+uU

(128) '

{

= °-

With n / O w e find the linear differential equation
, (129) ,._.

(129)

(130)

Let us now consider transformation (126). Since

d_u = Lffi

dt

2\dt2V

uj,y\ \dt) j

and

** '(^-^m+ifw " dt dt ' I dt

W ~ J3 I dt* Y

r

2

K

'

032)

24

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

we obtain (with (j, ± 0)

EQUATIONS

g ==0.0. g

(133)

The general solution to (124) can now easily be found with the help of (133) and transformation (126). The general solution to (133) is given by <j>{t) = d < 2 + C2t + C3

(134)

where CUC2 and C3 are the constants of integration. Inserting solution (134) into (126) yields the general solution {) =

2t + (C2/C1)

(135)

t^ + (c2/c1)t + (c3/c1)

where & ? 0. If & = 0 and C2 jt 0, then,

u(t) =

1

i—--.

(136)

t + c3/c2-

If d = C2 = 0 and C 3 / 0, then u(t) = 0.

D

Examp/e 5: Consider the ordinary differential equation + A[^(<)]2m-2u = 0

~

(137)

where m = 1,2,..., and X is a real parameter. The smooth function <j> satisfies

(p

g

dt2

+ (08—1 = 0

(138)

with ^(0) = 1,

^ j p = 0.

(139)

Let z(t) = ((t))2m,

U{z(t)) = u(t).

(140)

We show that U satisfies the hypergeometric differential equation. First we notice that the integration of (138) yields

lf^V + J_^ = J_ 1 \dt J

1m

1m

(U1)

where we have taken into account the initial conditions (139). From (140) it follows that dU dt

dU dz dz dt

du dt

(142)

2.1.

SECOND ORDER ORDINARY DIFFERENTIAL

25 25

EQUATION

and dHJ_ _ ,zV dU(Pz
((143) H

'

From (140) we find that = 2 mm[ ^[( t^) ]r2 -- ^ . dt'

-|

(144)

Consequently, /'AA

4m 2 a

w)) Mmr-2*ii-- z ( l ---*)A '{ j t ) == 4m[0(i)] 1 dr\*

2

4m

\2

2

(145) (H5)

2

where we have used (141). From (144) we also obtain ^~

2m = 3m)z + 2m -— 1). = 2((t)) 2 ( ^ ) ) 2 m - 2-\(l ( ( l -- 3m)z

(146)

Again we have used (141). Inserting (144) through (146) into (143) yields

2(1 2(1-

,
,

+ 1)Z) p + l)z)-£~-*)^"+(7~ 2 ) S " + (7 -(a ~ (a++fi 1J ~- aa/3U ^U == 00

(147) (14? )

dU

where

„ — m —1 -AA aa ++ /3 ^ r ,, «P=~, <*P= , 0 == -~z2m ' 4m Equation (147) has three singular points at

_ 77 =

22m m— - 11 —. 2m '

, (148) (H8)

aa = 0, 0,11 and and oo. oo.

(149)

Remark: Equation (147) is called the hypergeometric differential equation.

□

Example J^: Let ++Mz) Pi(z)—+ P2(z)u = Q Tz+Mz)u~be a second order linear differential equation in the complex domain. Let z(z) == z, *(*) 2,

U(z(z)) === u(z)expr-J u(z)exp Pl(s)dS\.i(s)ds\ .

[ll'

(150) (150)

(151)

To find the differential equation for U(z) we set

Then

R(z) ::= = - jPl(s)ds. 2jPi(s)ds-

(152)

dU dU dz du dU R dU dU dz dll du D 1 — -— - — -= —-eR + - p i u eR 2T dz dz dz dz dz

.(153).

26

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

where we have used that (154)

dR_ 1 dz ~ 2 P l ' Since dz/dz = 1 we obtain from (153) that
cPu

1P= &*

du

B

+

*Tze

1 dpi

B

+

TiT*

B

1 i +

4^"e

R

/IPP\

(155)

Consequently §

+7(5)^ = 0

(156)

where

/(5)= MS) 4 ^ - ^ ( 2 ) -

o

(157)

2.2. SYSTEMS OF SECOND ORDER DIFFERENTIAL EQUATIONS

2.2

27

Systems of second-order differential equations

Here we consider systems of second order ordinary differential equations and the invertible point transformation. Several examples are given such as the Kepler problem and the motion of a particle in a magnetic field. We start from the n-dimensional free-particle equations (PUj

0,

j =

1,2,...,n.

(1)

The invertible point transformation is given by Ui(T(t)) = Fi(u(t),t),

T{t) = G{u{t),t)

(2a)

t(T) = Q(U(T), T)

(26)

with the inverse transformation Ui{t(T)) = Pi(U{T), T), where / dFx 5^ det det

"'

3Fj dFj 3^

; ^

k du, 5G dG

dF, \ dF1 \ 7^0 *0.

W

(3)

du„ dt dG dG dG

KdV, KdV, '"" d^

dt J) ~dl

We introduce the differential operator d(.) V"d(.)du * ( . ) * i i ,d(.)

D(.) Thus

=sH^-

D(G) := D(F,)

:= "dFtdu 4

ga^

(4)

dG

(5)

dF, 9F,

(6)

"~aT

Applying the invertible point transformation (2) to (1) we obtain D(G)D(D(Fi)) D(G)D(D(F))

D{Fi)D{D{G)) = - D(F,)D(D(G)) = 0,

l , 2 , . . . ,,n. n. i = 1,2,...

(7)

28

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

From (7) we obtain

t^^ttt^r-w^tt^^p^

fi,=0(8)

where » = 1 , 2 , . . . , n and - V* (<¥±dG _ dGdF dG_dF\\ {\du, dtn A, t :: -Y(——~ hi \duk du, duk du,) dt it

+

fri \duk du,

2 dG dd2F Fj{ du, dujdu dujduk k

3

r

dG d2F Fi{ _2ndG 2 duk duk dujdt dujdt

(

TiH 3 ■—

'

dFidG dGdFi dGdF dft dG { dt duk duk '~dTdu~k at dt duk

dFj dFi d2d G2G du, dujduk

2 dG d d2Fi 8G F{ 2nW dF dd22G G I dt dujdukk dt dujdu dukk dtduj du dtduj

2 dG d2F{ dGd dGd2Fj F{ JFi dFj d G d22G ndG + Vu ,J - 2 + 22 ~ dt dtduj duj dt dt dtduj dt duj dt dt dtduj

22 dFi{ dd GG dF dt dt dujduk dujduk

dFtd2G dFi 2 duj duj dt dt2

2 2 . _dG8 dGd F Fi dFi dFd2dG2G 2 ~ ~dt~di2~ dt dt '~ 'dTW

Li

v/ithi,j,k,l

=

,„. (9)

'^rw

l,...,n.

Remark: Equations (8) can also be obtained from a Lagrangian function which is constructed by starting from the Lagrangian function Co for the free particle 2 u (UD C {T,\J,XS')==lf2(\Y c00(T,v,u') !)I 2

(10)

2 ,=1

where dUJdT -> U[. The Euler-Lagrange equation is given by d dCo

dCo

»1 -=M1,2,. , . . - ,. .n, n

=

d^Wrm dTdUJ dUi " °'' where

dd

dd

""

d

+n

: 0T + hUl dUi df''~ 5r =ar + g ^ +

Wf5^ g^

n

(11)

d

(12)

2.2.

SYSTEMS

OF SECOND ORDER DIFFERENTIAL

29

EQUATIONS

Thus (8) can be obtained from the Lagrangian

A*,U)u) = i g ( m ) 2 z ) ( G ) .

<»>

If we start from other equivalent Lagrangians for the free-particle equation we also find equivalent Lagrangians for (8). This is a method for finding equivalent Lagrangians for system (8). Let us now give several examples. Example 1: Let Fi(t,u) = Y/Aik{t)uk,

(14)

G(t,u)=g(t).

Then A = - d g| ^ A , A„

A - =n 0, Ayw

u =

r ]k = 0, n T,

dg»
v

^dgdAij odgdAii dt dt

"~

AA..ii^9 A " d?

(PgJ^dA,,

(15)

it £ ~aWUk ~ w g ~irUk-

Consequently dg ^

It g

Aik

~w

dg"cPAik

g [ Tt~dT ~ A'kw) ~dT + Tt}-l ~dirUk ~^h

where i = 1,2,... ,n.

dAik

~^Uk

=

(16)

□

Example 2: This is a special case of example 1. We consider the re-dimensional isotropic oscillator. From the ansatz Aik(t) = See(ut)6ik,

g(t) = ^ M

(17)

we obtain the equations of motion of an n-dimensional harmonic isotropic oscillator cPui ■ iJui, dt2

D

i = \,2,...,n.

(18)

Example 3: This is also a special case of example 1. If we consider the two-dimensional motion of a charged particle in a plane perpendicular to the direction of a constant magnetic field, the equations of motion will be d2Ui

du2

-W = ^ ^

CPu2

-aW = -

2

dux

^

(19)

30

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

where w = e/2mc. System (19) can be obtained from the free-particle equations by

«<>=UU -T")-

tan(wt)

c

(20)

LJ

The Lagrangian function of (19) is given by

4)'

sec22(u;t). (uit). *»(«!«! + af) sec £(t, u,,u) u) = i(iij iiiu +Iw(u -I- u 2 ii 2 ) tan(urf) tan(w*) + -w(u i « ( « ?2 + i ( u 2 + u|) ii2) + w(«i« w(u,ii22 -- M C(t,u, I « 2»)) + 1u1 + (21) Let us consider system of second-order ordinary differential equations of the form1 U .„, ,/TT „ d2V ,.nnw r r ,TT ¥T — + h(U)L fe([/)L + g(I/)U q{U)V = 00 where

and

(22) (22)

([/1U,t/ U22,f/ ,U33))TT uU = (U [/ == l|U|| U ||U|| := := JU? ^/u[+u[+u] + U\ + US

(23a) (23o)

d\J- / , 1, dU3 U t dU 3 TT W T := U , 1x ^-=■ dU* - TT dU2 dU2 „ ddU * TT dU dU* * 2 TT n ddUUAA L = f/ 2 -=dT \ dT 3 dT ' 3 dT ' 1 dT ' * dT ' -U>dT)

to* (24)

(236)

where X denotes the cross product. >int of Two special cases of this equation are particularly important from a physical point view: The Kepler problem, where h(U) = 0,

q(U) =^^ l(u) =

(25)

q(U) ?([/) = 0.

(26)

and the charge-monopole problem where

»ro-£. h(U) = ~

We give some time-dependent generalizations of these systems. They are obtained ted by using invertible point transformations. We find the general expression for the system tern of differential equations which are equivalent to

n m + du d t / nL L + + Cc2 2 c/um Uu= =0i,0, -j^ + Ci,c,en CuCen SF under an invertible point transformation.

(27) (27)

2.2.

SYSTEMS

OF SECOND ORDER DIFFERENTIAL

EQUATIONS

31 31

Under the invertible point transformation U(T(*)) = f(t)u(t), /(<)u(t),

(28) (28)

T(t) T{t) = = g(t)

(27) will b e transformed t o d2u ,
I1 :=- uu xX rfll f

(30) (30)

"eft

and

d2g dt22 _

ld o 2ldf /

2fdt

^ _ f /» rn + l dtTt~ '= h

f

JTt~^f-

f

fu

f

dt dt

*** 1 rf2/
~ /■(*)•-*.

i%-Wil-<-

fit) -

~

fl/(n+3)(f,

h

{t}cXp

1

1

[(n

(29) (29)

u

(31)

[f(Ad~\

h{s)d

+ 3)J

"j

-£iVM*iH)d*

Hn+3) ) 9(t) --= - }fl / / 3 2 / ( " + 3W^pL^±^JMs (^)exp( g(t) 2)ds2)dSl

(32) (32)

;^ / ^) H^/H

+4,/( +3) A(*) == /f ,"i (t)ex p ! ( m

Mt) = J i ^ ( ^ (

(

2) A

and /2 satisfies 1)7x4/3 " ( (df ^ l3V\ \ 11 11^^/ /33. (n(n++l)/irf/ 11 ffid/i, ("(n++2)2)f a 2 3 (n + 33))2 J/3/23 I2 \A* jJ + (n (n + + 3)/ 3) /3 3
((n++4 4) ) 11

/ = Let

w-

=:: W~\t) W"1^) // (( '0) =

(34)

(35) (35)

32

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL

EQUATIONS

we find that (29) takes the form W - W^U

+ C^"(n+3)unl + C2W-^u"u

= 0.

(36)

For the case n = -3 the solution of (31) is given by />(<) = - r %

/»(*) =

Id 2 / fdt*

/(*) = /4 1/(m+4) (0/3 _2/lm+4) (<),

(37)

t

(LdA^o^f\idf

g(t) = J

\f3 dt ' fdtj fdt'

f3(s)f2(S)ds.

(38)

£xampZe i : Let m = - 3 , d = 0 in (27). Solving (31), with fx = 0, we arrive at

h{t) = Co,

f(t) = Mt),

where C 0 is a constant. Let

/2 = 1 ^ - 2 - 1 ^ )

(39)

ft(t) = W-l{t).

(40)

^cPu u 11
(41)

Then (29) is reduced to

under the invertible point transformation JJ{T(t)) = W-\t)u{t),

T{t) = yW-2(s)ds.

D

(42)

Example 2: The equation describing the pure charge-monopole interaction is a particular case of (27). Let C2 = 0 and n = - 3 . Then we find ^U L , T 2 . ^ = 0. Solving (31) for this case and choosing f3(t) = 1, f(t) = W-\t), that d2u 1 d2W C», „

^-w^"u

+

^

1=0

(43) we obtain from (29)

< 44 >

which describes a charge-monopole interaction plus a time-dependent linear force.

D

2.2. SYSTEMS

OF SECOND ORDER DIFFERENTIAL

EQUATIONS

33

Example 3: We consider the scattering of a charged particle. The magnetic field pro duced by a dipole, with magnetic moment M ::= = AAQ0k is

B ==--u[/- 55(3(M •U)U M) • U ) U ---UU22M)

(45)

and the equation for a charged particle in this field is given by d2\5 . 1 fdU ,rrt, ■ )Uu) U ^ + A ^ ^ x ( [ / * k - 3 -( 3(k k-U ) )= =0 0

(46) (46)

where A=

mc and the vector M is directed along the u3 axis. Applying the invertible point transfor mation (28) to (46) leads to _1 g~ ++ AW^f A(t)f ++ AW" AWu ++ XAAWu-|x^kf^u~^ x ("2t -- 3u3u) 3u^) ++ A/4(i)u W)"-1" u x* kk == 00 (47) (47)

where

4 df

m) =

m /sW

dd22gg

2

d2gdfdf d2g dt2 dt

f AW =~~\d J~d¥~ d_lf

(48a)

dtdtJ J

dt 1 dg

= ppTv ~ Tv

1 dfdg

m -= AW ~~p~di~di' J^JtJf

(486) (48fc)

These equations can be solved in terms of h and f3. We find /(4) = / » " 1 W « P I 1

-Jfi(s)ds\

/(4) = /»" (*)«p(-//i(')'fcJ

-)r,

g(t) = j/3-2(Sl2)exp \ A . )ds, («i)exp (-3Jh(s 1 -■ S 2/)ds A2W ^ 9(t)--

AW -

-AA

h{t) == + + < +2f AW ~ (jjdt dt \ / 3 dt *f7di / 3 dt dt ?) j■

(49) {49)

34

CHAPTER 2. SECOND ORDER ORDINARY DIFFERENTIAL EQUATIONS

Let f(t) =: W-\t).

(50)

Then (47) takes the form

S

- Wr^Z"

+ **«*%

>< <"* - 3«3U) - A ? u - u x k = 0. □

(51)

Chapter 3 Third-order differential equations The investigation of third-order ordinary differential equations d3U

(

du
V

It'

= H \t. »(*). dt ~~ 3

dt1)

(1)

is not very common in classical mechanics due to the specific form of the fundamental Newtonian equations. However, in some cases this kind of equation appears as modelling physical situations such as the radiating charged particle interacting with an external electromagnetic field. They also appear frequently in the reduction procedure of nonlin ear partial differential equations by similarity transformations. Consider, for example, the modified Korteweg de Vries equation

A similarity ansatz is

du „ 0,2du du cPu „ = 0. — _- 6u — + —— 3 at 6U ox + ox X

IH- Y

W=°-

U)(s)

u(x..*) == 3^/3' w(s)

5

(2)

X =

X 3«V3

where 5 is the similarity variable. Inserting the similarity ansatz into (2) yields cPw -,dw I dw\ „

—_ 6 w » _ _ ^

,„,

+a_j=0.

..

...

(4)

The famous Lorenz model which is an autonomous system of three first-order ordinary differential equations can be transformed into a third-order ordinary differential equa tion.

35

36

CHAPTER 3. THIRD-ORDER

DIFFERENTIAL

EQUATIONS

In this chapter we study the third-order ordinary equations which can be linearized by an invertible point transformation to the linear differential equation
(5)

C22TT ++ C C33 U(T) =^CiT d T 22 + C U(T):

(6)

The general solution of (5) is given by

where Cu C2 and C3 are the constants of integration. The invertible point transformation is given by U(T(t)) = F(u(t),t), F(u(t),t),

G(u(t),t). T(t) = G(u(t),t).

(7)

Q(U(T),T) t(T) = Q(U(T),T)

(8)

The inverse point transformation is given by u(t(T)) = P(U(T), P(U(T),T), T), where we assume that

dFdG

dFdG

du dt

dt du

A := d-f-ir ~ i i r # °A du dt ' ~dt~du~ * °'

(9) (9)

ative From (7) we obtain the following expressions for the derivatives of U. The first derivative is given by dU (10) dT

§ =.w§§

where and

dl

<">

V{Lr)

dG dGdu D(G) _ , , - ='- dGdu dG faTt + ~bT D(G) ~ -z--jr + ~ET dFdu dF D(F) = ~du~li dFdu + W dF

,l1a) „ , (lla) (Hi)

+

{m

D(F)=fa-dt W

For the second derivative we find that
(12) ( '

FinaUy the third derivative follows as ( n , ( nr m {D{Cj)) )

^^ >

^, U D(D(D(F)))(D(G)Y-A _ D{D(D{F))){D{G)f - 3[D(D(F))D(G) dT3 ~ (D(G)y (£(G))2 D(F)D(G)D(D(D(G))) {D(G)y (D(G))2

-

-

D(F)D(D(G))]D(D(G))

((13)

'

37 Thus the condition (5) yields D{D(D(F)))(D(G))2

- 3[D(D(F))D(G)

-

-D(F)D(G)D(D(D(G)))

D(F)D(D(G))}D(D(G)) (14)

= 0.

Remark: This equation is implemented in a REDUCE program in chapter 11. We now consider the special case that G does not depend on u. This case, too, is implemented as a REDUCE program in chapter 11. From (14) we obtain d*u d?u dud?u r. /,j sKdu
2 ^
33 .Jdu\ fdu\

,, ,. . Jd I' du\ .du . , u\2 „. ,, ,du j)+fs(t,u)-+f6(t,u) =0 +«'■">(*] +/.C«h»+^*'">Ss0 (15)

where d2F 2

u)--=.idu /i(t,u) 3-J£-,

h(t,

~6 dF ' du

22 I(dd FF G\ d22G\ 2 dt dudt = 33 &8L--WfHt,u)-2(t,u) = dF ' dG \ du dt I

I/ &F d3F t a ,,\ _= o3 du2dt /«(*,«) = M*. v)-s -Q-p dF \V du du

,

SPF &F du3 u) = hit, f3{t,u)=&£ dF du

2 2 d2Fd d2Fd G\ G\ 22 du du dt2 dFd_G_ dFdG du du dt dt I

(F (d G\ 2 2 2 2 2 2 (d dt G\2 dF dt F , ddtGd dt , d?F_ dt3 6 2 2 + /e(*,«) fd,A-1 = = 3 MA 9 dt dGdF dt . HF dF dG dF ' j6(t,u)-A p A QQQF dt du + Qp^ du \ -QQ dt ) ^du \~dt I du ~dt~du~ du

9 33 G G d 3 dt dt3 dG dG dt 3 dt d GdF 3 dtPGdF dt 3 dGdF dt dt dG_dF_ dt du dt du

(16) (0

>

under the invertible point transformation (7). Let us now consider several examples: Example 1: Assume that F(u,t) = f(t)u+h(t), F(u,t) = f(t)u+h(t),

G(t) = g(t). G(t) = g(t).

(17)

38

CHAPTER

3. THIRD-ORDER

DIFFERENTIAL

EQUATIONS

T h e n (16) becomes d2q\ ^ \ dt2 ' dg dtTt) 1

/ ( \df 3 fdt \\

= AW Mi) === AW /.w ===AW m =--=o, o, M*)AW-a —-■%■

fM-H

Mt) = 3

2 ((Pg\ (
d2gdf £gdl 2 1|P dt2dtdt , ,, l
I1 tlA / ^ - " dg dt3 dt

dg

dg \ dt I)

J dt dV

f

f6{tiU)= /e(*,i«) =

d3gdf\
9

f

f

dg f 1 J dtdt )I

' f

\I V dtdJ r V dtdt/ / 2 (d]gV cPgcPh d?i
^

(Pgdh (Pgdh H?dt dt3 dt

IF ^ isrr 7^ "izr

dg \ dt )

dg , dtJ

dt

dtJ

\ d* /

< ( m]\

dr

and (15) takes t h e form dru

$ ^

, , , d2u

W ++ /A 2W

, . du

. ,

,

$^ ++/.(*) A< *+/.<*,«)=<>. u + /s(M) = o. □

(19)

Let us now consider some special choices for / and g in (17). Example 2: Let f(t) = 1 a n d g(t) = t. T h e n df/dt = 0 and dg/dt (19) takes t h e form dd33uU d3/* „n r,
* + *=°- °

Example

3: T h e equation

d 3U u

du (fit

Example

(2°) ,

3 = 1) W +— dt3 +Tt=« di

is equivalent t o (5) under a n invertible point transformation.

= 1. We find that

.

W □

4: T h e nonlinear equation cPu 3 du d?u
(22)

39 is equivalent to (5). The invertible point transformation is given by

F{u,t) = u2,

(23)

G(t) = e*. a

Example 5: In 1879 Laguerre discussed the condition for a linear third-order differential equation d u uu du

~d¥ + a(t)lP+mti+c{t)u

(24)

+ d{t) = 0

to be reduced to (5). He obtained an invariant of the form .

l<Pa 1 da \db 2 3 1, + a + 6 ^ 3 ^ - 2 ^ 27a-36a

/ =

+ C

-

(25)

If this invariant is zero, then (24) can be transformed to (5) by an invertible point transformation. Let us consider some particular cases for which the Laguerre's invariant (25) is zero. If the time-dependent coefficients in (25) are constants the vanishing of / gives ha C

-

T

+

2 o ^ 3 = 0.

, (26)

If b = c = 0 in (25) the condition 1 = 0 leads to <Pa „ da 4 , „ - + 2 a - + - a * = 0.

.„_. (27)

This equation can be linearized (see chapter 2), i.e., it is equivalent to the free-particle equation (PU/dT2 = 0. The invertible point transformation is given by F(a,t)=--\t2, G{a,t) = - - \ t . a i a A Remark: There are linear equations, for example,

^

a

+u= 0

(28)

(29)

that cannot be reduced to (5). Example 6: Consider the Lorentz-Dirac equation o9u dt3

1 d2u — rdt2

1 , mr ku+—E(t) mr e

K

'

30 K

=0

.

'

where r := 2e 2 /3mc 3 . This equation is a special case of (25), with 6 = 0,

a=—, r

c=

, TUT

d=—K-L. rriT

(31)

40

CHAPTER 3. THIRD-ORDER DIFFERENTIAL

EQUATIONS

From the Laguerre's condition (I = 0) we find that if (32)

' - - & (30) will be reduced to (5).

By using the Cartan equivalence method (see chapter 6) we find a generalisation of the Laguerre's invariant for the general case of a nonlinear third-order differential equation . h

=

lcPdH IdHddH lddff 2 (BH\\ 18H6H + 6Wm-3Mdtm-2lt-to+27[-dj) 3"9i"9fi

+

8H & r (33)

where

(34)

± = l + il± + i±+u± dt

dt

du

du du

and u'= H(t,u,u,u). (35) The necessary conditions for a general third-order differential equation to be reduced to (9) are the vanishing of the following invariants

(36)

h =0 h 3

/3

=

ld3W 6~diF = 0 3

(37) 2

2

1 awd w 1 /3 w\ 1 a ^ + + ~6^^ 18"^""^ 36^J -°

(38)

If we apply the conditions (36)-(38) to an equation of the form (24) we find that both I2 and 73 vanish. Remark: With a REDUCE program (see chapter 11) we derive condition (25) by inserting (24) into (33). In another REDUCE program we check that (22) in fact satisfies h = 0. The condition on lx leads to the following restrictions on the functions /; given by (6) - 3 ^ / ^l + 9/3 = 0 du ^ 7 7 ^ - 2 / 2 + 6^ = 0 at du 2

9

d f: _ l 8 / l ^ + 2 7 ^ - 4 / ? + 36/1/, = 0 8u*

41

6

.d2h

6f

9fi

- mk ~ ^ 2

,9 /i 1Z-J1

Rf

dh

UJ1

-

d2f2

dfi

6 / i

dh , 0 , 3 / 3

30

l - 5 # -6f^+271

R e&h

-6f^

$°-Jl2 _ 1 8 / dh 2 ^ at dt

dh

7

~ 4 / ^ + ^ +W

18,*/«

dh

~ 6f*it+

¥ _ 5 4dhdu^ _

+ 212Il

3

=°

Jf>

ir - 9t -4/ i /*+12/^ = °

18/l/fi

_ 4 / | + 1 8 / a / s = 0.

(39)

at

Remark: In a REDUCE program we derive (39) (see chapter 11). If the coefficients /„• are constants, /, = &, we obtain from (39) the following conditions

ci

C3 = -±,

3C4 = CXC2

~2C\Ci + ZCxCt + 9C2C3 = 0,

-9dC6

- 2C| + 9C2C<, = 0.

(40)

Let us now consider another method to find linearizable third order nonlinear ordinary differential equations. Let

2-

and let

(41)

u{t) = dln
(a)

(42a)

M

^ ^ (426) dt' dMn* (42c) A3 Now we derive the differential equation for u for these three cases. We use the following „„t,t;„„ notation

W

U{t)

~

w

Ax(t) := j'u(t0)dt0

(43a)

A2{t) := / dh Ju(t0)dt0

(436)

t

t2

h

A3(t) := / dt2 J dt-, J u(t0)dt0.

(43c)

From (43a) through (43c) we find dAs dt

— iT-9

dA2 dt

—;— = Ai,

d^ dt

—.— = AQ = u

(44)

42

CHAPTER 3. THIRD-ORDER DIFFERENTIAL

EQUATIONS

and

deA" dt where n = 1,2,3. Prom (42a) we obtain

(45)

/i„_ie

u(t)»

\d

(46)

<j>dt

Therefore

4>{t)-- = e * .

(47)

d _ M e u ~dl ~

(48)

Taking the derivative we find

and ^

w-

e

- »M„2 , „AX

u +e

du

- „At (du

e

Tt= U

Consequently (42a) leads to the nonlinear equation

i + " 2 = 0-

, ,,2^

(49)

)' (5

du

°)

Thus (50) can be linearized. From (42b) we obtain 4(t) = eA\ The derivatives of (j> gives

and

(51)

I"*** d2<j>

il£l=^{A\

<«>

+ v).

(53)

Therefore A\ + u = 0.

(54)

du 2Aiu + — = 0 dt

(55)

Taking the derivative of (54) leads to

and

, 2u+2A

, du

>ltt

W=0-

(56)

43 To eliminate Ai we multiply (56) with u and insert (54). This leads to
(du\

"*i~U)

++

,

2u3 = 0.

,

(57)

Therefore (42) and (43b) yield (57). Thus (57) can be linearized. From (42c) we obtain

<j>{t) = tA\

(58)

= eA>(A22 + A1). dp

(59)

Consequently

It follows t h a t

Ai + A\ = 0

(60)

u + 2AXA2 = 0.

(61)

uA2 = 2A\.

(62)

du , — + 2uA2 + 2A\ = 0. at

(63)

S + 6A? = 0.

(64)

and From (60) and (61) we obtain From (61) it follows t h a t

Inserting (62) into (63) gives dt

Therefore
, . (65)

= 0

and
(66)

Multiplying (66) by u and inserting A\ from (65) leads to t h e nonlinear differential equation of third order

" ^ - ^ + 12U =°d3u

„ ,

du cPu

„

,„_. (6?)

Since t h e solution of (41) is given by 4>{t) = C1t + C2

(68)

we find with t h e help of (42c) a solution of (67), namely 2C3 "

(<) =

2

C?ta + 3C?* + 3CiC, 2 i + CS"

t69)

44

CHAPTER 3. THIRD-ORDER DIFFERENTIAL

EQUATIONS

Example: Consider (Pu
^ ^ ¥ UJ

U\

-J

4A

+ U* =

This equation can be linearized with the ansatz

„

°-0.

dy _

fdt =
,_„, (70) (70)

(71) (71)

Thus the general solution to (70) can be found. From ansatz (71) we find d2y dt*

du = JV-v dt

dy Idu f du ± ,A \ rfv = y —- + u + u-rdt {dt )

._„. (72)

2

fd u „ du d3y y 3 dt ' - [-d¥ + 3 K - + ' ) d*y = yH. dt* '

(73) (74)

Since 5 = 0 we find that d*y -n dt* The general solution of this linear equation is given by 33 22 y{t) Att + Ao A) y(t)--== AA3t3t + AA2t2t + Ait

(75)

(76)

where A 3 , A2, A\, AQ are the constants of integration. It follows that u(t) u =

3A3t2 ++ 2A22t ++ Aj Al A A33ti33 + A A22t2^ + A Ax1tf + A0

l(77) ;

and

2 3t At + B 3*2 + 22At £ u(t) = 3 2 U(t) - t«» + Al» + Bt B« + C + At where we assumed that A3 ^ 0. □

(78) (?8)

Chapter 4 Lie point symmetries In this chapter we describe the connection between the invertible point transformation and Lie point symmetries for ordinary differential equations. Calculations of Lie sym metry vector fields using the method of the Lie theory of extended groups have been described by various authors (Bluman and Cole 1974, Ovsiannikov 1982, Euler and Steeb 1992). To introduce the definition of a Lie symmetry vector field we consider a second-order ordinary differential equation

*($.£■<«>.')-■

m

The extension to higher order is straightfoward. In the study of the Lie theory of ex tended groups, relevant to Newtonian equations of motion, the one-parameter infinites imal point transformations act on solution curves in (u,t) space, transforming solution curves into solution curves. This is equivalent to the requirement that the Newtonian equation of motion be form invariant under these transformations. For such a transfor mation, the Lie point symmetry vector field (also called infinitesimal generator) is given by Z = ({u,t)-

+ r,{u,t)—

(2)

where £ and rj are smooth functions of u and t. To study the symmetry of the ordinary differential equation we have to evaluate the prolongation of the Lie symmetry vector field. The first prolonged Lie point symmetry vector field is given by ZW

= Z + VW-?r.

45

(3)

CHAPTER 4. LIE POINT

46

SYMMETRIES

The second prolonged Lie symmetry vector field takes the form

z<2> =:ZW where

,w

+K

drf

r,™

~ It''Udi>

(4)

OU

dt

-d( — u—rdt

(5)

The total time derivative is defined by ,„. (6)

d dd . d „d . d ;=—-(- u— 1- udu —. dt flu dt OT at du au From (1) we obtain the hyperplane AT(ii,u,u,i) = 0.

(7)

We assume that N is a smooth function (i.e. C°°) of u, tie, u and t. Definition: A vector field Z given by (2) is said to be a Lie point symmetry vector field for (1) if, whenever (1) is satisfied, Z^N(u,u,u,t)

(8)

=0

or equivalently ,dN

dN

MdN

mdN

„

^ir+'fc+'^+^M-0

(9)

where Z' 2 ' is the prolonged Lie symmetry vector field given by (4). □ The Lie symmetry vector fields of a given differential equation form a Lie algebra under the commutator. Thus if Z% and Z2 are Lie symmetry vector fields of a given differ ential equation, then \Z\,Z^ is also a Lie symmetry vector field, where [,] denotes the commutator. Definition: A vector space L over a field T (T = 1Z or C), with an operation Lx L —» L denoted by (z, y) —► [z, y] and called the commutator of z and y, is called a Lie algebra over J- if the following axioms are satisfied: (LI)

The commutator operation is bilinear.

(L2)

[z, z] = 0 for all z € L.

(L3)

[x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0

x,y,zeL.

47

We mention that Axiom (L3) is called the Jacobi identity. We notice that (LI) and (L2), applied to[x + y,x + y], imply anticommutativity. a Definition: We say that two Lie algebras L, U over F are isomorphic if there exists a vector space isomorphism

$([x,y])~[4(xU(y)} for all x, y £ L. Then
D

Example 1: Let N(u,u,u,t)=u == u++uu33 ==0.0. N(u,u,u,t)

(10)

We determine the Lie symmetry vector fields of (10). Since N(u,u,u,t) find that dN 5iv 22 = 3u,, ^^dt=_ 00 ,' -5— ^=3« f^ = 0, H = l. du au Thus condition (8) gives 77(2)(2) == 0 3u2n;? ++ 7?

33

= u + ui we we

(11) (11)

lu~ = °'

(12) (12)

where r/<2) is given by (5). Since dr; dn at dt and

drj dn at dt

. dn dr) au du'

(i) dn drj U V' — -z—h T; dt du

d£

d( d£

. d£ d£

dt

at dt

au du

,, „, (13)

.d(\ au/

.(dt — ^at

(14)

h- U - r -

- U

we obtain for (12) 2 2 2 „ drj22T) . dQ22n . d2i-l2 n ,dn .daV £ „ a ^f n . 2 d 2 £ -■ i £ 3d3d ^ 3 2A + o i 2 + 2n u ^ u + u 2 25 u 2 - - u 3 ^ - - u - | + 2 us 3 - ^ --2u2 u dt2 dt ^U dtdu ~ du2 au

0.

3d£ 3d£ uu

- n0 du~ (15)

where we have taken into account that 5u — = —u. -u3.3

(16)

Different powers of u are treated as linearly independent (as they are, since this: is a second-order equation). On equating the coefficients of separate powers of u to zero (bearing in mind that both £ and n are functions of u and t only), for the third and second powers we find 3

u3 and

:0 .

« 22

= 0 ^=*««,i) £(u,t) = f/,(t)tt 1(t)u + f/.W 2(t) 2 ad\ r> du2

2

„ad2 c;£ dudt

2

a d2*?n du2

dh d/t dt

(17) (17) , lfi ,

v(18) ;

CHAPTER 4. LIE POINT

48

SYMMETRIES

From (17) and (18) we obtain v(t,

M1..12 + 9i(t)u + g (t). u) = -j^u 2

(19)

Inserting (17) and (19) into (15) and separating out terms with u we find that ft(t) = 0. Therefore V(u,t) = gi(t)u + g2(t), Z{u,t) = f2{t) (20) and (15) takes the form

3{giu + 92y + u *£ + *» + » £ _ u>9l{t) _ **£ + 2 U 3 ^ dt

= 0. { 21)

Separating out terms with u, u3, u2, u and u°, respectively we find the following equations 2

djh

-dT

=

~dT

,

4ft 1 + ~n(« -="°.'

3fli

2

From these equations we find that

fll

d29l d2g2 = u -—5-r-r- = 0. " ^ = 0, ' "55

92=0, »' ~-

(22)

must be a constant, say gi(t) = C-,. Thus

/»(<) = -ft* + ft-

(23)

Thus the Lie symmetry vector field is given by (24)

Z = (-C 1 f + ft)^ + C1W-|-. or au Consequently, we find two independent Lie symmetry vector fields Zx--

d ~~Jv Z2=t^

(25)

du'

For the commutator we find [Z1,Z2] = Z1.

(26)

Thus Zi and Z2 form a basis of a two-dimensional Lie algebra.

u

Example 2: Consider the free-particle equation
(27)

We find the eight Lie symmetry vector fields Zx-- = Zs = U-^,

d dT'

Z6 = T-^j,

Tirt =

d . dU'

r,

rr,9

Z3 = T—,

d d Z7 = T2-dJ, + TU-, dU

dT'

ZA4

d

= £/——

dU dU

ZS = U2—- +TU^-. dU dT'

D (28)

49

Example 3: As third example we consider the one-dimensional harmonic oscillator
(29)

The symmetry vector fields are given by Z2 = s i n ( t ) - ,

Zx = - ,

Z3 = cos(t)-,

Z4 = u -

sin(t) — + u2 cos(t)—, Z6 = u cosit)— - u2 sin(i) — at au at au Z7 = ucos(t)sm(t)— + sm2(t)—, Zs = usin2(t)—costt)sm(t)—. D (30) au ' at au at As in the case of the free-particle equation (27) we find eight Lie symmetry vector fields. Zh=u

Example 4: Consider the second Painleve transcendents ,

d?u at

.„,,

3 —2 = 2u + tu + a.

(31)

The second Painleve transcendent admits no Lie point symmetry vector fields.

□

Theorem: The number of Lie point symmetries which a second-order ordinary differential equation can possess is exactly one of 0, 1, 2, 3, or 8 (Mahomed and Leach 1990). All linear second-order ordinary differential equations are equivalent to the free parti cle equation (27), i.e. there is an invertible point transformation that transforms the equation into the free particle equation. Example 5: Consider the third-order differential equation

(32)

dT3 ' - 0.

The Lie symmetry vector fields are given by 7

d

Zi-gf,

z

*=Tw

7 z

z

-

d

7 -T

*-du>

° = T2ih>

9

7 -U

i

^- &f'

z T2

—

Zi u

~ du

^ i\+2TUw

D

(33)

Example 6: Consider the third-order linear differential equation — + u = 0. at6

(34)

50

CHAPTER 4. LIE POINT

SYMMETRIES

The Lie symmetry vector fields are given by a

Zl =

a

Z2 = K — , au

a7'

Z

,, , d

(35)

-^a~u

where / is a solution of (34). Since the general solution of (34) is given by u(t) ==d&e* u(t)-e - '* ' ++ C2 cos(v/3
(36) (36)

we find that the Lie symmetry vector fields of (34) are given by a Zo = u— du

* - » Q

Z3 = Zz = e-*-i-, e-*-£-, au ou

M

fi

a D

s i n ( V ^ t / 2 2)^-. e'/2^. Z5s = sm(V3t/2)e'' au

/ 2 Zt4 = cos(v cos(V3t/2)e'/ 3
(37)

£iamp/e 7: Consider the third-order differential equation du I
3*Tt\-dP)

z.

( (
(38) (38)

*V*V ={ [&))&'

This equation admits the Lie symmetry vector fields a

7 = 7T-, ° Z\ Zl =

at'

a° 7 = ST, Zl du

d° j.*6 a 7 = u——(Z3 i—, au dt'

Z2 =

dl' a , o^' a , 9 Z3 ZB = - 2 u t - r - + u2—•-*-£-, au dt at' z

> = -Ku+u2o\-4t'

7 Z4

=

9

i ^ 7°; — t-=— dt

du

i = uaH + tdl'a - U zn2' d=r - +uatn da~u( —. -2ur.—D z6 = dt au au T

z

< = -2^-u2t

(39)

2

+t

Tu

D

W

oint Remark: For first-order ordinary differential equations the connection with Lie point symmetries is as follows: If du /lit = H(t,u(t)) — = H(t,u(t)) It at ' admits the Lie symmetry vector field (2) then the differential one form H(t,u)dt H ^ d t r,{t,u) H(t,u)t(t,u) r,(t,u)-H(t,u){(t,u)

du d

is exact and the integration is known as quadrature. Theorem: Consider an n-th order ordinary differential equation n dd»u u _

=

J

du (1 ff^„(t),_.

d»-lu\

(40)

51

with n > 3. The maximum number of Lie symmetry vector fields is n + 4 (Mahomed and Leach 1990). Thus for n = 3 we find that the maximum number of Lie symmetry vector fields are given by 7. Next we give the transformation of the Lie symmetry vector fields d_ dU

d_ ST'

(41)

under the invertible point transformation U(T(t)) =

F(u(t),t),

u(t(T)) =

T(t) =

G(u(t),t),

t(T) =

(42a)

P(U(T),T)

(426)

Q(U(T),T)

with dGdF^_dG9F^,Q (43) ~~ dt du du dt Thus we have the transformation (u,t) —* (U,T). We assume that Q and P are C° functions of U and T. We find d_ dP_d_ dT ~* dTdu

+

d_ dP_d_ dU~* dUdu

dQ_d_ dTdV

+

dQ_d_ dUdf

(44)

In a more consistent notation we should write

w ~* %{u=F{u^T = G{u't)]iL+ w{u ^ - Z ( U

= F(u,t),T

= G(uJ))l

+

^(U

=

= F(uJ),T

nu i)>T=G(M))

-

= G(u,t})l

I (45)

In the following we use the notation given by (44). Let us now describe the connection between the Lie symmetry vector fields and the invertible point transformation. We start from the free-particle equation (PU dT2

0.

(46)

Under the invertible point transformation we find

$+*•<«.*> ( t t r + ^ ( ^ ) 2 + ^ t ) d i + A M = °

(4?)

52

CHAPTER 4. LIE POINT

where

2 (dGd (dGd2FF \\ du du du22

A3 =

A2s :~ Ai = 1

An0

dFd22G\ G\ dFd 9u du du du22/)

,

2 2 (dGd^ /ft?d2F dGd dGd FF dF dd22G G O 0F 2 + 1 1 du ftdu ^ftau2 a« ftft* 9u ftft* \ ft du du ftdu 2 2 (dGd2F F ndG 3 ndF d2 G (dGd^F 2 + 3G32F dF d G dt dtdu dtdu \\du du dt dt2 + dt dt dtdu dtdu " dt

(3Gd22FF (3Gd

dFd dFd22G\ G\

ft2 2

~-

A

ft ft22Jj A ft dt

\ft dt [dt

SYMMETRIES

d dFd i ^2G\ G\ ft du2 2j

ft ft* j dFd22G\ dFd G\2 du dt dt2 )) du

t,

(48)

(48) t

,, .

-1

The Lie symmetry vector fields of the free-particle equation (27) are given by (28). Using the invertible point transformation (42) we obtain the Lie symmetry vector fields for (47) dQd_ dQd dTdt dTdt

11

dPd_ dp d dTdu dTdu

dQdPd d z-rdQd+r +G 7
= 22

Zi 4F

z* =(GF9^ V

dT

+

z-F

+FdPd

- dUdidUdt + Fdud~u' dUdu dQd dPd Wd dPd Ze G= G +G Z6 -6 dUdidUdt + Gdudu~ dUdu d

4. (r2dP

W)dl+{GdT

-( df

dPd_ dp d dUdu dUdu

dQd Zi^F9Q9+FdPd

dp d

- dfdidTdt dfdH' dTdu dQd dPd zZI5 -F^-4+F FZ4*-FdTdi dTdu~ dTdt+ dTdu 2dQ dQ Z7 =- G(r +CF + GF)

dQd_ dQd dUdt dUdt

F^± dU) du

+

dP

++ rF GF \

d

du)dll

d (\ GFdT Q+F»mi dUJ dt

(49)

Example 1: The symmetry vector fields for the free-falling particle (Pu ^ 2+ +, 9--= = 0U dt

(50) (50)

can be found directly from (49) by using the transformation l 22 F{u,t) -gt, , F{u,t) = u u+ + -gt

G{u,t) G(u,t) = t.

a D

(51) (51)

Example 2: Consider the nonlinear equation i ,, , cPu , du du 1 (Pn 2 3 u _H? _ ++ fcuku_Tt ++_-fc . 9f e V = =0 0.

(52) (52)

53 53

It is transformed into the free-particle equation if 2 F(u,t) = -~ - - hIkt F(u, t) = t \,

G(u, t) = - - ---kt. -kt. G(u,t) uu 66 uu 33 The following symmetry vector fields for (52) are obtained from (8)

*-4*H+M£

Z2=u-

-

1,

(53)

3

3

u (n» *-« G--01-(>+*«*■-W£ *-G--Oi) 5 IN 9 z , = *i ((Il --- |ibtrf) |i«*)^| ++fcu fcu2*2<(—Jfctrf (ifcU<-"3) -1) £ fa ^4

V18

*Z5 «=*r*( (Iferf I * * -- l l)) | *+*(***-1 + td g f a r t - 1 - i * Vf c 2* »2 <)2 ■ ~ l8 " )

*-('-Hi^ ZT-

= -kt2 6

-OK

6

Zs8 = -kt ifci3 3 (l ( l - -yfcuf ijbuf")} I - +1 (hut O V 6 / O Tat \ 2 V2

^

:2 (-fcut V3

(-kut- -1 -- -k2u2t2

(-

^

- 1 - IkW 9

54

i8*3)

9u

+ - L f c V r 3 ) •£-. 108 108 / ou ou

P □

(54)

An n-th order (n > 3) ordinary differential equation has exactly one of n + 1, n + 2, or n + 4 (the maximum) Lie point symmetry vector fields. A necessary and sufficient condition (Mahomed and Leach 1990) for an n-th order ((nn >> 3) ordinary differential equation dnu

IF

„( , ,. du = H lt,u\ • < > • * • ■

1 dnd"~H\ u\ n l ' dt

-J

(55)

le nnto be linearizable via an invertible point transformation is that it must admit the dimensional abelian Lie algebra

nAi = At © Ax © . . . © At

(56)

where ffi denotes the direct sum and At is a one-dimensional Lie algebra. Suppose that (55) is linearizable to #'£/
dT

pj =a0

dT'

CHAPTER 4. LIE POINT

54

SYMMETRIES

by the invertible point transformation (42). Forming the commutator of any two, say Z« = 6 ( T , I 0 ^ + W ( r , f f ) ^ ,

* = 1.2

(58)

of the n + 1 Lie symmetry vector fields of (57) we find [Zi(T, 10, Za(r, U)] = (Zrf, - Z r f i ) ^ + (2i7» - 2 i « & ) ^ .

(59)

The commutator of these same operators in coordinates (t, u) is given by [Zx{t,ulZ2{t,u)]

= (^(Zrf) - Z1{Z1t))—

+ (Zi(Z 2 «) - ^ ( Z m ) ^

(60)

where, for example, Z2t = (2dt/dT + r)2dt/dU. The right hand sides of (59) and (60) are identical. The transformation technique employed here to find a class of one-dimensional equations with the maximum number of symmetry vector fields can be generalized to multidimen sional equations. This procedure can be useful to obtain directly the Lie symmetry vector fields of the equations. Now we consider systems of second-order differential equations (see also section 2.2). We start from the n-dimensional free-particle equations ^ T = 0,

j=l,2,...,n.

(61)

The invertible point transformation is given by T(t) = G(u(t), t)

(62a)

t(T) = Q(U(T),T).

(626)

Ui(T(t)) = Fi(u(t),t), with the inverse transformation Ui(t(T)) = Pi(V(T),T),

Applying the invertible point transformation we obtain the following class of equations "

» " »

dujdukdu,

« •

dUjduk

"

dUj

Aiih Tijk + Vij *- Aih~w+^^^ '~ti~dT~d7+^^ ~d7+Li j=ijt=n=i j=ik=i ~d7~dT ^ j=\

fc=i

a i

a i

a i

a i

a i

a i

ax

where A,*:

_y(dFLdG_ ^[\dukdu,

dGdFAdu, duk dui J dt

_ dG d2Fi Ay« = dui dujduk

dGdFi dG m dt duk duk

dFi d2G dui dujduk

dFidG dFj dG dt dt duk

= 0 (63)

55 d2F{

ndG

dG d2Fi

d2F; d2G

1 ijk = 2duj - — du ^ dt— + -^7~ s dt dujdui, k

dG d2Fj dt dtduj

,J

2 -du dtduj k

dGd2F{ duj dt

dF, d2G dFjd2G dt dtduj duj dt2

T _§GSiFi with i,j,k,l

dFt^G

•'~ at dt2

= 1 , . . . ,n.

dFt d2G dt dujduk

IRAS [

m dt*

'

The system of free-particle equations (61) has the following Lie symmetry vector fields 7 -

d T d Z72 =- T-^dT

a

d

dT' d -II— Z73i = u{—, u dT'

zZ&j

uU

d d

7 -

z

7

Zi

d -T— dU,

am' *- 'dT>a ^'-'du, n '"du,' a 2 T =T u Z U + Ui U u U (65) = T — + TTU — z7-Zsi U T

d 'dU3'

* == <w/

* ^+ % >w

^= ^< W

p'% 1 >auj ^ ^

where i,j = 1,2, . . . , n . The Lie algebra of these vector fields is sl(n + 2,7?.) whith dimension n 2 + 4n + 3. The transformation laws of the vector fields d/dT and d/dUi are as follows " dPj a

a

" ap3 a

a

aQ a

+ dT d duj dTdt' "*~*£J j ^ dT arat' Uj

dr dT

dQ d

[

dUi ~*^dUiduj dU.df dUi~* fri au, du: + amat

'

From (65) and (66) we find the Lie symmetry vector fields of (63) dQd " dPj d + + dT dt fr{ dT du,' ~" dTdt j^[dT duj'

Zi1

Z22

31

^ap,

d

dPj d ^dQAd + "+ Y ^A z7 4i- ' dU dt jr[dUidu

aqa »dPi d F^l^FX-^lA-

7 Zu--

dQd

&Tdt dTdt j^ST £[ dTduj dui

" ' dT dt *' £j ar aUj' * du, dt+ £ i 5 f / . ^ dQ d ^dP d Z5i Ze Z&j Zu -G di + 0fri duhdViduj' " ~ *•'dUjdt^ a^- a t + *' £l'^.dUjduk at/, au* ~ du{dUidt { du,' 'dTdt

+

jr[ dT du^

i

J

3

Z Z7= T =

+G

+

K «s<

-lrF Z78i = Zsi

G

+ +GG

( ar g ^ J 1a7at [*%*°£<>

dQ

GFi

+FrF?®-\°- 1

- [ dr

+F

F

+

G ££arWdu, aUj +G+hh th

GFT^—

+ FYYF^^-

Gt + dt + + GF' t i 9T duk + F*' & t ,

' | i ' a^ J at

' £[ dT duk

du 8du w* > *<

(67)

' 9U} duk (67)

' | j & ^ a^ &,t

56

CHAPTER

4. LIE POINT

SYMMETRIES

where i, j , k = 1 , 2 , . . . n. T h e Lie algebra of (63) is isomorphic to sl(n + 2, It) with the maximal dimension r? + 4n + 3. Example 1: If we consider the two-dimensional motion of a charged particle in a plane perpendicular to the direction of a constant magnetic field, t h e equations of motion is given by (Pllx du2 d2u2 = —2ui—r2 l(68) } dt dt ' ~dW dt* dt HF where u> = e/2mc. System (68) can be obtained from the free-particle equations by using the transformation 2

F,(u, 0 = £ Aik{t)uk, where A(<)

— tan(w<) \

~ { taa(«rf)

9(t) =

1 ) ' We find the Lie symmetry vector fields for (68) z Zx fi

Zi = U2-^—, au2

- d "di'

- d -~d^'

z Z2

ft

(69)

G(u, t) = g(t)

ta.n(uii)

(70)

z - d Z3 ~aV2

fi

fi

Z5 = ui\-u2-^—, aux du2

fi

Z6 = cos(2u;f)dux

sin(2cjt)-^— ou2

Z-, = sm(2ut)-T— + cos(2u>t)-^— dux du2 r\

Zs = -(cos(2ut)/2u)—

at

O

+ m sin(2u;t)- 5 — + OU-L

Q

r\

Ul

cos{2iJjt)T— ou2

Q

Z9 = (sin(2wi)/2u>) — at O

Ul

r\

cos(2u<)-^— + ux sin(2urf)-r— oui ou2 Q

Q

Z10 = (sin(2u;i)/2w)— + u2 sin(2wi)1- u2 cos(2wi)-— at oui ou2 Q

Zu = -(cos(2uit)/2oj)-£-

- u2 cos(2wt)-

Q

Q

1- u2 sin(2u><W—

at au\ u2 d 1 2 2 d d Zli = + U -^d-t + 2^-U^ ^-dV2 ut d d 1 2 2 d Zi3 = 2^^+UiU2^r+2(Ui-^^; a o Zl4 = (uj sin(2u><) + u2 cos(2wf))^- - w cos(2wf)(uJ + u\)-

au2

Q

us\n{2ut)(u\

+ u22)-^—

57 /-)

Zis = ( - u i cos(2wt) + u2 sin(2u;/)) — + u sin(2o>t)(uJ + u\)-r

rk

f\

\-u> cos(2urf)(uJ + t^)-^—. (71)

The procedure employed here to find the general class of n-dimensional equations with the maximum number of Lie symmetry vector fields, i.e., that can be reduced to the free equation, can be generalised to other classes of equations with a different symmetry group. For example, we can start from the equations of motion for the three-dimensional Kepler problem and find the class of equations that can be transformed, by (2), into these equations. We thus have a method for generating classes of integrable equations and for finding their Lie symmetry groups if we start from integrable equations with a known symmetry structure. Example 2: We consider the Lie symmetry vector fields for the equation ^

+ h(U)L + q(U)V = 0

(72)

U = (U1,U2,U3)T

(73)

L:=Uxf

(7,)

where and

where x denotes the cross product. Two special cases of this equation are particularly important from a physical point of view: the Kepler problem, where h(U) = 0,

q(U) = kU~3

(75)

and the charge-monopole problem, if h{U) = CU~3,

q(U) = 0.

(76)

We give the general expression for the system of differential equations which are equiv alent to the equation (P\J —-i + C1UnL + C2UmXJ = 0 (77) dl under an invertible point transformation. All the properties of the transformed system, its symmetry vector fields, invariants and solutions can be found from the knowledge of the corresponding properties of equation (77).

58

CHAPTER 4. LIE POINT

SYMMETRIES

The Lie algebra associated with the Lie symmetry vector fields of (77) is
7 -IT A--IT *2~U3du2

2

JL du3

(78)

Zt^K-L-K-jHf, Z^U^-U.Af aUz

Let

oU\

oUi 0U2

U(r(t)) = f(t)u, T(t)=g(t) be the invertible point transformation. Then (77) will be transformed to

+ A W ^ f + A(*) u + CiMtKl

$

(79)

+ C7f4(t)umu = 0

(80)

where l:=ux^ and

(81)

d2g

2

ji ~w=fu

fn+1

i=h

~di

m-tj7t=h'

r(¥)=u

(

82)

at Tt

The system (82) can be solved in terms of / i and / 3 , if in ^ - 3 , The system (82) can be solved in terms of h and / 3 , if n £ - 3 ,

/(*) = fit)

fV("+3)m

- /,

=

/3

/ (

1

1

/ f M ^ I

W+ 3ex,P^(B + 8)i//»W*"J

"

Wexp(^//

1

(^)

2 / (2 + 33)

g(t) -= dst - j/ / 3/ 3 "/ ^ () S(l S) el )xepx( p ( - ^ ± i i J / lf1((s22)ds 2\ g(t) (n + 3 ) / S ) < f S 2 ) d S l

(83)

m fi t) = //ji^m)+/ (^^+)3( 'f W - n 2)jh{s)d\ ) e xep x(& p ( ( -2n (-2+ 3-27/l(^j

{ /«(*) ==

and /2 satisfies h

~

(n + 4) 4) 1 /■dh\ d/ / 3 \ 2 2

(n + 3)V3 lJtj

1 1l ^

_ 1l dh ^ i , (n + 2 ) , 2

(n + 3)dt

F

(n + 3) 2 / l -

(84)

59 If we impose /,(*) = 0 in (83) we find

/-(*r'

Let /(f) =: W-\t).

(

-

)

*

"

(85)

■

Then (80) becomes

2 1 d d2W 1 W , ... m+4 m _ ((«+3) + ; »l + ,n C2V^-'+i) n c iw 777^7T U ++ C u»U u U= = 0. 0. dt ' -wdt* * W~ " Vtt l + C2W-l"' at' W at' For the case n = -3 the solution of (82) is 2

d
2 ~ ~JW

// i. = =

fi =

ld2/

(86) (86)

= /V(-«)^-V(*WJ f^^^m-^t // =

- /, a# 3§ ,

id/\ f/ 3 f + 2/dtj

id/ /dt'

t

= Jf3($)f(s)da. 9{t) =

(87)

If we impose the scaling Lie symmetry vector field ft

^

z"4+°§"4 z5 =

(88) (88)

ft

on (77) we find J2TT

+ CtlTL + C2t/2<"+1>U = 0.

(89)

Equation (89) has the Lie symmetry vector fields Zi,Z2,Z3,Zi symmetry symmetry vector vector field neia

and the additional Lie

—

i £ A rt/j a » i„.* « l -— *r -^ * , z*-5 = — 2^ r— oT

n + 1 £rj

(90)

ot/j

The Lie algebra associated with these Lie symmetry vector fields is a2© so(3). The time-dependent equation, with the same Lie symmetry structure but with transformed symmetry vector fields, obtained from (89) and the invertible point transformation (79) is ^

+ / i ^ + /au + <7x/,u"l + C2/u = 0.

(91)

The functions /,■ are given by (83) if n ± - 3 , and by (87) if n = - 3 . If we impose the fl.QciitionsJ 1J16 symmetry vector neici

ZS = T ^ on (89) we find

+

T^UlWi

(92)

60

CHAPTER 4. LIE POINT

- ^ + C1U-3L + C2U-4V

SYMMETRIES

= 0.

(93)

The Lie algebra associated with the Lie symmetry vector fields Z\, ..., Ze is sl(2,7l)©so(3) The transformed equation is

S + h ¥ + /2U + Clf3U~31 + C^u~4u = ° where the functions /,• are given by (87).

( 94

Chapter 5 First integrals and differential equation In this chapter we study the connection between the invertible point transformation, Lie symmetry vector fields and first integrals. First we introduce the definition of a first integral. Consider the n-th order ordinary differential equation dnu

IF

= 4 ««),£,. , , du

„(

We again assume that H(t, u,u,...)

dn~l1u\ u\ '' dtn-i)'

, . (1)

is a smooth function. A smooth function

d^iA r( , , du 1 ' ( * . " « . A ' " '' df"- J

(2)

is called a first integral (sometimes also called exact invariant or conserved quantity) y) of (l)if (3) (3)

P =0 whenever (1) is taken into account. Equation (3) can also be written as dd

9dl/ dldu dlcPu dldPu _|_ dl_du _|_ dI_cPu _|_ dI_
where un' stands for n dots and dnu/dtn

_|_

n dl _ ^ d_ u^ _ 0 = 0 nn du - dt dtnn ~ du

(4)

{

'

can be replaced by the right hand side of (1).

61

62

CHAPTER 5. FIRST INTEGRALS

AND DIFFERENTIAL

EQUATION

We now illustrate with several examples the concept of a first integral. Example 1: Consider the anharmonic oscillator

(.)

$ + «• = ». Multiplying the equation with du/dt we arrive at

ducPu du , „ * » + ** = *■ Then

is a first integral.

/

1 Idu\

du\

.„.

(6)

u* +

(7)

("'*) = 2 ( * J T

D

Example 2: Consider the damped anharmonic oscillator cPu

o

du

,„,

—+ c,-+c ! „ + „' = 0.

(8)

'HS=((^T)' + TM<.<)

,9)

Then

is a first integral of (8) if 2c\ = 9c2.

□

(10)

Example 3: Consider the time-dependent harmonic oscillator

^+^2W« = o

(ii)

and the nonlinear second order ordinary differential equation
1

«»(*)/, =

7'

From (11) and (12) we obtain d2p

cPu

(12)

, ,

u

(13)

0

w'-"*-?-

which, when multiplied by the integrating factor pdu/dt — udp/dt is immediately integrable to give the first integral

_ 1 f du I=

P

U

2\ Tt- i)

1 fu\

dp\ +

2-{-p) ■

,

v

W

63 In other words, 7 is a first integral of (11) if p satisfies (12).

□

Example 4- The classical equation of motion for a particle in a gravitational field moving in 1Z3 \ 0 is given by u = (u1)U2,U3)T

— + ^ - = 0,

(15)

where \i > 0 is the constant field strength. Each orbit is planar and so in any plane with polar coordinates (u, 9) we have the system of differential equations
2dud9 u dt dt

(16)

The three first integrals are: the energy

l((du\\h2\

.

£ =

2{[-dl)

+

(17a)

V2)

the magnitude of the angular momentum

»-m-rff

(176)

and the Runge-Lenz vector

R:=^x^. dt u

(17c)

M = K2 \ 0. □

(18)

The configuration space is

Example 5: Consider the so-called generalized Ermakov system dt2

+

(19a)

^*-i>($

^cPu2 + ^ ) u 2 = l dt2

5

p).

(196)

«2 U2 \ U l /

The first integral of (19) is given by

i=\(^u2-u^)

+

J\sM-s->9(s))ds.

a

(20)

Let 7 be a first integral of a given differential equation. Let U(T(t)) = F(u(t),t),

T(t) = G(u(t),t)

(21)

64

CHAPTER 5. FIRST INTEGRALS

AND DIFFERENTIAL

EQUATION

be the invertible point transformation which transforms the given differential equation into a new one. Now we can evaluate the first integral of this differential equation using the invertible point transformation. Example 1: Consider

(PV
where

(22) (22)

dU ~df'

(23) (23)

L L === UU xx ^ . Let

t

T(t)=g(t)= [f2(s)ds T(t)=g(t) = j f(s)ds

U(T(0) /{*)u, V(T(t)) = = f(t )u,

(24)

(24)

be the invertible point transformation. Then (22) is transformed into

(25) (25)

du 1l == uuXx It£

(26) (26)

f/ == W~\ W~\

(27) (27)

where and

Equation (25) describes a charge-monopole interaction plus a time-dependent linear near force. The first integrals of (22) are

JJ = =L L -- dU^U, CiU^V, ,

h

dU

= lT-

(u-

dU T) dT

1 fdvV E=\ ( ^ ) E=2(-df) 2 \dT) / 2 =■ ( » -

rfU,

rV

(28)

Then the first integrals for (25) can be found using the invertible point transformation ■tion (24). They are

j = l1- -C Cu-'u, lU-V

h II

£

dW \22 1E=\{w ' w d u d^-d-^uj =2

)-f(-

d 1 W™D\-W*D(^ = u. •^dt(£I\ fl + 2W ^-D) W*D (^V ¥ fa' ~ ^-D\ U* dt J \dtj dt \ dt dt J

= U

+

2

*-(-M"S-W

(29)

65 where t

-jw-

2 D(t)-D{t) = / W-2{s)ds. {s)ds.

(30) (30)

These time-dependent systems are three-dimensional Ermakov systems with a given symmetry structure. sived If a dynamical system admits a Lie symmetry vector field and the system can be derived Let from a Lagrangian function we can find a first integral from Noether's theorem.. Let us now describe a theorem which is helpful in finding first integrals from Lie symmetry netry vector fields of a certain class of differential equations (Hojman 1992). Consider cPu ( , s du\ -T^ = H(t,u(t),^-) u?2 H [t,u(t).v h dt)

dt

{'

where H satisfies

dtj

dH =0 .0. **= du = Oil

(31) (31) v

'

(32) (32)

If «««.«,*)#■ « , u du ou is a Lie symmetry vector field for (31), i.e. dH d d fd\ lflA^_^ ( = 0 ' du dr d~t{& ) -I—s dt \drj du du dc du with

dd dd d IT d .. d „d = -=- + u——h H— dt ' at dt du du

then

dt /= I = dt+ du

d ( d \ d (d\ — du —i \dr)

^ ^U*J 2. 1

=

(33) (33)

(34) (34) K

,„.-. (35) (36) (36 )

is a first integral for (31). The proof using REDUCE is given in chapter 11. The theorem can be extended to a system of second-order differential equations. ConCon sider (Pui d (37) ^ = H,(tMt), tf,(i,u(t),^), (37) = i), ii == 1.l , 2 , . . . , !n

IF

where the smooth functions H satisfy

1 E ^

OUi

0. =C

(38) (38)

'

66

CHAPTER 5. FIRST INTEGRALS

AND DIFFERENTIAL

EQUATION

If

££,.(*,u,u) A E6(*,u, ou

1•=i =1

(39) (39)

<

is a symmetry vector field for (37), i.e.,

4^4&l-E&|^-£?^4e; = 0, at \dt j

£rj OUJ

with

i = l,2,...,n

i = l,2,...,n

n 3 5o === a + n 5 ^ ^

dj

n

n a ++ S ii 77, 'a*

^ ^ S"^ g '^

then is a first integral for (37).

(40) (40)

£^[ ouj dt

'-£*♦£*«

(41) (41) (42)

The proof is straightforward (Hojman 1992). The theorem can be generalized as follows: Instead of (38) we assume that

gH + ^,,,0 +>/)-» u ^dHi

hid <

,«) (43)

where / is a smooth function. Then a first integral / is 1 " 9(m +1 " 7= 7 ■ 7=7£~^r / i=l dui J7§^(/^jv=i V d7j +

(44)

(44)

The proof is again straightforward (Hojman 1992). Example 1: Consider a two-dimensional harmonic oscillator, in polar coordinates (u,0), »,*),

g =-^+(f)2„ d20e *» d^ A symmetry vector field is given by

==

2 dude dude

;45&) (456)

"—udtTf «**■ 3, .

9

(46) (46)

U 0——. u e^-.

au

Since

dE-, BX ~diT

+

dH 3H2 2

>u

-M

u

=

45a)

(47)

67

we find that the function / takes the form

f(u,e) = u2.

(48)

The corresponding first integral is given by / = 6u2^. at

D

(49)

Let us now study the free-particle equation and the construction of Lie symmetry vec tor fields from first integrals. Conventionally the Lie symmetry vector fields of a onedimensional linear system are used to determine the invariants of first integrals associated with the system. Here the reverse procedure is adopted for the free-particle equation. The free particle has the: equation of motion
Then

dp

~di

(50)

du

(51)

lt du

= 0,

~dt

=

(52)

P

can be derived from the Hamiltonian . function H(u,P) = where

2

(53)

du _ dH dt ' dp

dp

dH ~dl ~~ du The two first integrals are given by

h = P,

1

2P

(54)

(55)

h = u --tp.

Their quotient

n-u~

■tp

(56) P is also a first integral (Mahomed and Leach 1985). We seek the set of one-parameter symmetry groups with which each of I1: h, h is associated. The functions £ and r/ of the Lie symmetry vector field A

Z = Z{u,t)jt+T,{u,t)—

si

r)

+ <;(u,p,t)~

(57)

68

CHAPTER 5. FIRST INTEGRALS

AND DIFFERENTIAL

EQUATION

is determined by the following equations: M dl dl „ + ,91 =0 i c^- + ^ du- + ^ Tt"= °

,M. (58) (58)

f

+ Tl

^dp2

„ 0 •"-<£^"fv-'s*"^" m

where

da22H# dudp

,a # „(1)

'

w

d??

,a2#

q

(59) (59)

dtdp

d£3#

(60)

- * - dt 9 p

' = §-If

«■»

by eliminating £ between (58) and (59) and insisting that £ and J; be independent of p. Thus for the first integral I\ = p, we d77 obtain ^ - PP ^ = 0 (61) df y dt Vdt ' where d_ (62) dt "= dt du Equation (61) yields a partial differential equation in which the terms are grouped to gether in powers of p. By equating coefficients of separate powers of p to zero we obtain

»w

(63)

P P 22 ::tt = >a{t) P p :

v =

da da, (<) ~dlu ++ 66(0

((64) 64)

and B, 6b ==: CC p° At + + B, (65) (65) P° : aa == At where a and 6 are determined using (64). Therefore the triplet of Lie symmetry vector fields with which Ii is associated is r,

&

°

at dt

du' du

5

Z\ + "U-r—, i = = t*— ;! + !-' z

In like manner, for J2, we obtain d

* - *W *-«!■ Z

Z^>2 — * =

2d

z

TT" 5 JT*>

at' dt

a z*3 == ir-du —.

z

(66) (66)

du

d

d zZe=t = t—.

(67) (67)

d

a . zZ9 = uu-^—

(68) (68)

tu dt *-t%o\+ hdu

° l-du

Similarly for 73, we find

0

,a

Z7 = tu— ++ u, 2-^-, We see that

* - * "at! - 5? du

u-z-, *Zg8 == "!• at'

Z = Z4 + Z Z9. Zil = 9.

* = Tudu

(69)

Consequently, we obtain the eight (independent) Lie symmetry vector fields of the freeparticle equation.

Chapter 6 Cartan equivalence method 6.1

Second-order differential equations

Consider the second-order ordinary differential equation
„ /

. . du\

-aV=H^li)

.„,

(1)

where H is a smooth function of t, u and li. We would like to find the condition on H so that (1) can be linearized. Cartan (1924) has proved that for an ordinary differential equation (1) of the secondorder in the plane, it is possible to define in the plane a projective connection having the integral curves of (1) as its geodesies. Naturally, the definition of this connection is invariant under the group of point transformations in the plane. We want to find the condition on H so that (1) can be transformed to the free particle equation under an invertible point transformation. We use the method given by Cartan (1924) (see also Gardner 1983, Kamran et al 1985, Hsu and Kamran 1988, Grissom et al 1989). We also study the question for third-order ordinary differential equations. We adopt the jet bundle formalism (see chapter 12 for an introduction). Our starting point is an exterior differential system on J'(7?., TV) whose solutions are in one-to-one correspondence with the solutions of (1). With coordinates t, u, p (li = p) on J1(7?., 11), we take the coframe given by dt,

du — pdt,

dp — H(t,u,p)dt.

69

(2)

70

CHAPTER 6. CARTAN EQUIVALENCE

METHOD

The solutions of (1) are the curves in J1(7Z.,7J) on which dt^O,

du-pdt

= 0,

dp-H(t,u,p)di

= 0.

(3)

The coframe we have chosen is not uniquely defined however and the solutions of (2) are the same as the solutions of Adt ± 0,

B(du - pdt) = 0,

BC(du - pdi) + D(dp - Hdt) = 0

(4)

where A, B and D are nowhere vanishing smooth functions and C is an arbitrary smooth function on ^("R^Tt). The prolongation of the invertible point transformation T(t) = G(t),

(5)

U(t,u) = F(t,u)

gives coordinate transformations on J 1 ^ , 7£) with T(t) = G(t),

p

U = F(t,u),

=

^

(6)

so that D = B/A under such changes of coordinates. Thus we are led to the problem of adapting the coframe, /

Wl

\

wa \u3

= j

/ A 0 0 B \ 0 BC

0 W dt \ 0 du - pdt \. B/A ) \ dp - H{t,u,p)dt j

(7)

That is, we consider the group G of matrices of the form S =

/ A 0

0 B

0 0

\

\ 0 BC B/A J

.

(8)

Then we try to use the natural G-action to provide a coframe of the forms (w1,w2,W3) which is adapted to our problem. First we compute dSS-1, which gives the 0th order principal components. Since 5-1 = and dS =

/ 1/A 0

0 1/B

0 0

\ (9)

V 0

-AC/B

A/B J

/ dA 0

0 dB

0 0

\

\ 0 d(BC) d{B/A) J

(10)

6.1.

SECOND-ORDER

DIFFERENTIAL

71

EQUATIONS

we find that dSS'1 has the form I a 0 dSS'1 = 0 6 \0 c

0 \ 0 b-aj

(11)

where ,

dA a:=

6 : =

~A'

dB "B'

C <=:=<*C+-dA.

(12)

Since u>! = Adt,

Ft OJ3 = BCdu - BCpdt + —dp

w2 = B d u - Bpdt,

R

A

A

Hdt

(13)

we find that daj2 = dB A du - pdB Adt-

dw! = dA A dt, dw3 = CdB Adu + BdC Adu-

CpdB Adt-

BpdC Adt-

Bdp A dt BCdp A dt + -dB A dp

B ,, , H ,„ , BH,A , BdH , , B dH , —— <M A
(14)

and u i A u ! = ABdt A du UJ1AUJ3 = ABCdt Adu + Bdt A dp

1 u>2 A u>3 = — - d u A dp A

B2 —Hdu Adt

B2 "TW^ A <^P-

A

A

Here A denotes t h e exterior product (also called wedge product or Grassmann

(15) product).

We notice that a^ A w,■ = 0 (for one-forms) and ddcj, = 0. We find dull = a Auii

du/3 = cAw2 Thus we have d

du>2 = b Aui2 + Dan u>i AUJ$ — ABCdt / A du D f)lj\ + (b — a) Au>3 -^—du A dt — ( BC + -r^r— dp A dt.

A du

\

(16)

A dp J

/ w, \ / or 0 0 W «i \ / EitrjjWjAwi \ wj = 0 /J 0 «2 + EtflVyAu* V «3 / \ 0 7 0 - < * / \ "3 / V E i f c ^ i Au»» /

(17)

where / a 0 dSS" 1 = 0 / 5

I 0

7

0 0

\

fi-al

mod wi

(18)

CHAPTER 6. CARTAN EQUIVALENCE

72

METHOD

and all products of differential forms in (17) are wedge products. From (7) - (16) we find that after absorbtion of torsion by redefining a, /3 and 7, we have /f m \ UJ2

i\ \I

=

( a 0 0 0

O N /«!>\ 0

V0 7

W3 j

w2

j1

P-a) V w 3 /I

0

\

U>i A w 3

+

\<

o

(19)

;

Thus there are no torsion terms to normalize and we must prolong the problem. Consider the coframe on ^(11,11) x G given by (u>1,u>2,u>3,a,/?,7)T. The remaining freedom in modifying the 1-forms a, ,9,7 so as to leave (19) invariant determines a Lie group G'1' which acts on TZ6. The indeterminacy in our new coframe is given by /

G« =

Is Os \ 0 0 0 € GL(6,ft) 0 / 0 la

\0 g

f

(20)

J

where 1 3 is the 3 x 3 unit matrix. Since ddut = 0, we find that da

i = 1,2,3

= 2u>i A 7 + ao)i A«2 + &±>j A u>3

d/3 = 4d2 A 7 + /? A u> d-y =

fAa

(21)

+ pAu>3 + 6Au>2 — au>\ A u>3

where a and 6 are unabsorbable torsion terms and p, 8 give a basis for the right invariant 1-forms of G*1*. The G*1' action on o and b is calculated by computing dda and ddf. Since dda = 0 and dd-y = 0 we find da + 26 = 0,

db + 2p = 0 mod Wi,w2,u>3,a,7

(22)

so we can always translate a and 6 to 0. This yields the structure equations dmi — a A c^i 2 + <*>i A w 3

du>3 = 7 A w2 + (/? — a) A u>3 ! A 7 + /iii>2 A u 3

+ JJWJ A k^

i A«3 + J3W1 A u>2-

(23)

6.1. SECOND-ORDER DIFFERENTIAL EQUATIONS

73

Parametrically, the three basic invariants Jj, I2 and I3 take the form T,

J

2 -—

!■>

h3 =

„T v4 2 '

—

l1

~

2 1 /i3 (dd J_ ^2£H _ dd2H\ H\ 1 2

2A \d< 2A \dt dp2 2

, 1 (d d H 2 + ' 2.4 1A2BB [dtdpdu {dtdpdu

where

A d3H 2J322 dp a P33

d

8

dpdu)j dpdu

2

d HdH dp2 du du 3

d2H dH dpdu dpdu dp dp

- 2 du ^ 21)

[(24)

'

du2)

„d

(25) (25)

= +p +H 7t~~ di dt it ldu lp dp is the total time derivative. In terms of our original variables A, B, and C, a

dA (n„ („„ 1 dH\ = —;— 2C + - ^ - tax A \ A dp

a

li =

dB

„ T~ - "

C i

+

1 9d22H# 2B^U2

,_ A Cdff CdH „~ 2i \\ / 1 d822H # C d22H\ ,„ C<M CdA ,/f I1SSi? 2 Uh+ 2 77 =» «dC+K~A~ 7 ++ — + |(A^-d^ C \2ABdpdu ~ WW) ^ du ~A-dp~~ ' A dp ) -C Ju;,-|2B dp2 ) \2ABd~pdu'

Ui

d22H 1 <9 ff 2 ^' 2B W ~dpW U ^J ' (26)

If H = 0 in (1) we find from (24) that hh = = hh--==hh= 0.

(27)

Therefore the vanishing of all three invariants occurs if and only if (1) is transformable under the invertible point transformation (5) to the free-particle equation
(28)

2 ePu (duV ^ 2 =-W \dt)

(29) (29)

2 H(t,u,u,p) H(t, = -p P) -P2

(30)

hI1 = I,h == Ih3 = 0.

(31)

Example: Consider dt dt2 ~

\dt)

K

'

Then and Thus (29) can be linearized.

□

74

CHAPTER 6. CARTAN EQUIVALENCE

METHOD

In the case where all three invariants vanish the structure equations are the MaurerCartan equations for the symmetry group of (28). This is the six-dimensional subgroup of SL(3, IV) of matrices of the form 1 a-i ax 0 a2 \ h 62 k 63 63 , d e t 5 = 1l 61 V a 0 C2 / ci 0 c2 j which acts as fractional linear transformation in the plane by

(

s=

_ ait ait + + aa22 T = CiU c2' C1U ++ C2'

_ bbit1t + b62" + b22 2u + U = C\t Cit ++ C c22

(32)

,

.

If we let w ==5u) u> Su!

a» = (wi,(^2,013) u> (u>1,aj2,u>3)7',, and take

, t =

^Wi

u= u =fl

ao> i3

U>2

(34) (35)

W33 o)

3

then (33) gives
2 === e2 + +T T(2ti + u) e4) + TTh (2ei + e6

e*E7 === e£55 + + Tt33 + + U(e [/(«! + i)2e4) + + TUee dU TUe6 1+2e

where 1 dSS' d 5 5 - x ==

(36)

e2 \ I/' «iti 0 •. e3 Q e4 e55 «3 \v -- e< %e 0 -- (( eeii ++ fe.) £«) //

(37)

ddT == ddU ddT-ddU === 00

(38)

T h e equations are the structure equations (23) with Ii = 0, where a>i = e2, wi a = 2c! 2e1 + e 4 ,

^2 o) 2 = = £5, £5,

^3 = -63 ^3 = -63

0/9 = teix + 2e4 ,

7 = e<*■ 6.

(39)

These are also the structure equations for the affine group in the plane if we make the identification. (0i»«2,0ii,0i2,e w 1 , w3 ,3-, 0- /, 3- W ,-w (01,02,011,012,021,022) = ((wi,w 2 i,e 2 2 ) = i , 1 -, -77, ,aa- -/ /33 )) . Then (23) takes the form d0, = dOi = 0J0H 0 i 0 i i + 02012, 02012 J dd&n 6 u = -012021, -0r.!021,

d92 = 01021 + 02022 d6i12 d8 0 n 0 l 2 + 012022 2 = = --011012

(40)

6.1. SECOND-ORDER de21

DIFFERENTIAL

EQUATIONS

= - 0 2 1 0 1 2 - 022021,

<*022 = - 0 2 1 0 1 2 -

75 (41)

When I? and I3 both vanish, (23) can therefore be interpreted as the structure equations for a connection for the affine group on II2. The curvature 2-form is u

_ / - W 2 - { 0

0 -1,6,62

(42)

We followed in our derivation of the result the article of Kamran et al (1985). A slightly different derivation is described in the article of Grissom et al (1989).

76

6.2

CHAPTER 6. CARTAN EQUIVALENCE

METHOD

Third-order differential equations

Consider now third-order ordinary differential equations. Chern (1939) studied the ge ometry in the plane which arises from an ordinary differential equation of the third-order

<"

?"KS)

dt3

where invariance under the group of contact transformations is required. We find the condition on H for which (1) can be linearized to
dT3

(2)«

>■ '== 0.

We show that this is the case when a certain relative invariant / vanishes. The condition 1 = 0 has a simple geometrical meaning. By interpreting the integral curves of (1) as the points in a space of three dimensions, the connection so defined is a normal connection in the sense of Cartan. In the general case / / 0 it is also possible to define a generalized geometry in the plane with the elements of contact of the second order t,

u,

u,

u

as the elements of the space and with a certain five-parameter group as its fundamental group. We formulate the problem as a problem of equivalence of Pfaffian systems. We follow in our representation the article of Chern (1939). By regarding the u, u as independent variables, (1) may be written as a Pfaffian system du - H(t, u, ii, u)dt = 0,

dii — udt = 0,

du — iidt = 0.

(3)

We define the Pfaffian forms (differential one-forms) as Wj

=

a (du - Hdt + /3(du - iidt) + ^(du - udt))

U>2

=

\(du — udt)

LJ3

=

a (du — udt + v(du — iidt))

W4

(4)

:= Vi(dt + V2(du — iidt) + v3(dii — iidt))

in which the a, /?, 7, A, fi, v, V\, V2,1*3 are nine auxiliary variables. First we study the problem of equivalence of the Pfaffian system (4) and a new system du - Hdt = 0,

du- udt = 0,

du - udt = 0

(5)

6.2.

THIRD-ORDER

DIFFERENTIAL

77

EQUATIONS

under t h e group of contact transformations. If W»,

U!1,

U>3,

U>4

are four PfafHan forms formed from the system (5) with the new auxiliary variables a,

7,

P,

A,

p,,

P,

vu

v2,

v3

in t h e same way as wi, 1^2,^3 are formed from (3), we see t h a t a necessary and sufficient condition t h a t t h e two systems (3) and (5) be equivalent under t h e group of contact transformations is t h a t there be a transformation in the variables 1,

u,

u,

u;

a,

/?,

7,

A,

fi,

v,

vlz

v2,

v3

leaving t h e LJ'S invariant, i.e. u>i = ult

u>2 = W2,

1^3 = W3i

W4 = W4.

(6)

The method of equivalence developed by C a r t a n (1908, 1924) consists of reducing the number of auxiliary variables by imposing on t h e m relations of an invariant n a t u r e (i.e., invariant under t h e group of contact transformations). We apply the exterior derivatives of t h e Pfaffian forms a>;. As first conditions we suppose du>2 =

W3 A W4 mod u>2,

du>3 =

u^Au^

(7)

modw2,w3.

From (7) we find t h a t A = avr,

fi = ai»!.

Since t h e system u>i = 0,

u)2 = 0,

u3 = 0

is completely integrable and by taking account of (7) we see t h a t the exterior derivatives of uii are necessarily of t h e form du>x = dijj-i = du>3 =

Wn AUi

W32 Au>2 +

li>i2 AU2 + "U3 A u>3 W22 A " ! — ui3 Au>4 ^33 AU3 — U1AW4

dui

W42 A UJ2 +

">43 A LL>3 +

=

+

Wu

(8)

A UJ4

where t h e ID'S are new Pfaffian forms such that ion = — , a da

dvi r 2

w22 = a

wX2 = ~n{dP - vdf), v{ ,

vx

1 , w32 = —dv, Vi

w13 = —d~i fi da

dvi 1

W33 = a

, mod u>< i>!

CHAPTER 6. CARTAN EQUIVALENCE METHOD

78 1042 =

(dv2 — vdu>), 1043 = —dv3, IO44 = . (9) av\ a Vi T h e one-forms wn, ...,w44 are not completely determined by (8). We can add to t h e m suitable linear combinations of u>,- such t h a t t h e form of (8) is still preserved. In particular, t h e tflu, 1022, 1033, 1044 can be changed as follows V>ll =

lOll + <*lO>i + 020»2 + a3U3

W22 =

«*22 + ^2^2

(10) W33 = U>33 + C2W2 + C3W3 U>44 = ">44 + ^2^2 + C3UJ3 + e4u>4 where t h e o j , 02, 03, 62, C2, C3, e2, e3, e 4 are arbitrary functions of t h e variables. From (9) we have tun + 2u) 44 -1022 = 0 , m o d w;. «>11 + IO44 —™33 = 0 Consequently 1022 = «>11 + 21044, W33 = ">n + w4i (11) if we can choose a x a n d e 4 in (10) such t h a t t h e coefficients of Wx a n d w 4 in t h e two sides of (11) be equal. To see t h a t we can arrive at (11) it is necessary to find t h e coefficient of u>i A w 4 in dui, those of ux A w 2 , w2 A W4 in db>2, those of u>i A u;3, u>3 A W4 in 4 in <&i>4. From t h e relations (da \ (dH \ , ,. awi = I — A u>! I + a I -^- + 7 1 dt A au, m o d a>2, ai! A w 4 = — avtdt

A dii,

mod

tL>2,3

we see t h a t t h e coefficient of u>i A ui4 in dui^ is

1 /dtf

\

Vl Likewise we find: Coefficient of UJI A u>2 in du)2 = 0 Coefficient of CJ2 A W4 in du>2 = Coefficient of

u>i A 013 in du}3 = -

Coefficient of LJ3 A ui4 in du3 Coefficient of

_

= — ( 7 — 1/)

u>i A u i in duj^ =

. a

6.2. THIRD-ORDER

DIFFERENTIAL

EQUATIONS

79

Then (11) gives 2v3 °i = — . a

1 ( dH\ n e 4 = — \u-2~f-—r , Ui \ cm /

\dH i/ = 7 + - - — . 3 OH

Thus not only a^ and e4 but also i/ is determined. The auxiliary variables v and 7 being connected by the above equation, it is again possible to add any linear combination of U)i, u>2, o>3 to W13 without changing the forms of (8) and (11). We can thus arrive at u>13 = w22

(12)

if the coefficients of u>4 in both sides of the equation are equal. In order to have these coefficients it is necessary to calculate the coefficient of UJ3 A ui4 in dui and that of ui2 A u)4 in du>3. We find:

'■'"+,

)-(f

Coefficient of w3 A wA in duix =

— ( 7 ( -^r + 7 ) - [ -57- + , au / \ on

Coefficient of u>2 A w, in dw3 =

-r(z/ 2 — 71/ + /3).

Then (12) gives 2/3 = 7 ^~

2

+

2 977 _ M _ 1 / a ^ Y 37d£ 9u 9 \ 5u ^

+

1 d dH 33 eft 9u

where

~ » ( t , w ) : = _ +,,_+ « _ +* _

(13)

where $ is a smooth function of t, u, it, u. By determing 0 in this way, we see that wu is a linear combination of u>; and that the coefficient of u>2 A w4 m du>i is a n invariant. Its expression is of the form I/u3, where dH_ _ \dH_dH_ _2_ (dH\3 ~ ~~d^ ~ 3 du du ~ 27 \ du )

/=

+

\±dH_ 2 dt du

+

ldH_d_dH_ _ \&J>H_ 3 du dt du 6dt2 du'

( (14)

'

We implement the function I given by (14) in a REDUCE program. The function I is a relative invariant of (1). We divide our discussion into two cases, according to whether I = 0 or 7 ^ 0. First we consider the case 7 = 0. With a Httle change in notation we write (8) in the form du>\ = Wi A u>\ + w2 A UI3

du>2 = wi A u>2 + 2w3 A u;2 — u>3 A w 4

80

CHAPTER 6. CARTAN EQUIVALENCE

METHOD

du>3 = ui2 A u 2 + tui A u 3 + u) 3 A W3 - wi A w< du>i = u>4 A u>2 + ws A u>3 + w3 A u>4

(15)

where the u>j are five new Pfaffian forms linearly independent between them and inde pendent of u>i. The most general transformation on the iz\ which preserves (15) is of the form Wi = u>i, w2 = w2 + pu3, w3 = w3 + pu2 w4 = w4 + qu>2 + rui3 + pu4,

w 5 = u) 5 + rw 2 + su3

(16)

where the p, q, r, s are arbitrary smooth functions of the variables, which we take as new auxiliary variables. To find expressions for the exterior derivatives of W{ we apply the theorem of Poincare to (15). Putting TTi =

dwi — w2 A U>4 — 2w$ A u>i — w4 A u>3

=

dw2 — w2 A w3 — wi A u%

7T3 =

dw3 + w2 A u>4 + w$ A u>i

TT2

(17)

we obtain, from the identities ddu>i = 0,

ddu>2 = 0,

ddu>3 = 0

the relations 71"! A u>i

+

7T2 A ui3 = 0

TiAuj

+

2ir3 A u>2 = 0

(18)

K2 Ab>2 + 7Ti A U 3 -)- 7T3 A U)3 = 0. It follows t h a t fl"l A Wx A U>2 = 0,

5T! A a>2 A w 3 = 0,

7Tl A u>i A u>3 = 0

so t h a t ■Ki is of t h e form TTx = au>2 Au>3 + bu>3 AW] + cWi A u/2. It is possible to modify the Pfaffian forms w4 and u)5 to have a = b = c = 0. The most general transformation on the w; which preserves (15) and duii = w2 Au>4 + 2u>5 A u>i + w4 A ui3

(19)

is then wr = JBX,

u;2 = u) 2 + pu>3,

w3 = w3+ pu2,

w4 = w4 + pu4,

ws = w 5 .

(20)

6.2.

THIRD-ORDER

DIFFERENTIAL

81

EQUATIONS

where p is a new auxiliary variable. From (18) we then have 7T2 A UJ3 =

0

which gives 7r2 = 0,

mod u>3.

Since dw2 contains t h e t e r m dp A UJ3 we can choose a Pfaffian form w&, linearly indepen dent from t h e w, and from wi,..., w$, such t h a t we have (21)

dw2 = w2 A w3 -\- w4 A u>i + we A ui3

where t h e Pfafian form w6 is determined up to an additive term of u>3. T h e relations (18) then give 7T3 A U>2 = 0, w6 A u>3 A OJ2 + T 3 A w 3 = 0 which can be w r i t t e n as (TT3 — W6 A U>2) A LL>2 = 0,

(7T3 — u>6 A u>2) A OJ3 =

0.

This shows t h a t TT3 is of t h e form 7T3 = w6 A u>2 + au;2 A W3. We can choose w6 to make a — 0. T h e n we have i + we A ui2.

(22)

Equations (21) and (22) completely determine the Pfaffian form w6. It remains to find the expressions for t h e exterior derivatives of w4, iu 5 , u) 6 . For this purpose we apply the theorem of Poincare to t h e last equation of (15) and to (19), (21), (22). We define TT4

:=

dw4 + Wi A w4 + w2 A w$ + w3 A w4 — w6 A u)4

7r5

:=

dws + u>i A ws — w4 A u>4

7r6

:=

dw6 + u»i A we + w2 A w4 + 2w3 A we-

(23)

We obtain 27T5 A U)\ + 7T4 A LJ3

=

7T4 A U2 + 7T5 A W3 7r4 A u>i + we A UJ3 7T5 A Wi — 7T6 A OJ2

0

= 0 = 0 = 0.

, 24 >

It follows t h a t 7r5 A u>\ A u>2 = 7r5 A

UJ2

A u>3 =

TT5

A

UJ3

so t h a t 7T5 is of t h e form 7r5 = au2 A w3 + ^ 3 A a»! + ou»i A u>2.

A

CJJ

= 0

CHAPTER 6. CARTAN EQUIVALENCE METHOD

82

Relations (24) then show that TT4 and ire are of the form 1T4 =

ew2 A u>3 — cu>3 A U\ — 2awi A 012

7r6 =

fu>2 Au3 + aui3 A w j - ecji A u>2.

Consequently, we have the following Theorem: If the relative invariant / of (1) is zero, the problem of equivalence of the differential equation with respect to the group of contact transformations in the plane depends on the fundamental equations duii

=

Wi A a>i + w 2 A CJ3

dui2

=

*»i A wi + 2u>3 A u2 — W3 A ui4

dui3

=

» j A U2 + ti)i A 013 + M3 A U3 - u i A w 4

du>4 =

w4 A u>2 + ">5 A LJ3 + w3 A w 4

dw\

=

uij A u i + 2u»5 A U 1 + U I 4 A U 3

du>2

=

u;2 A t«3 + u)4 A u>! + u; 6 A w 3

dw3

=

—W2 A k)4 — ID5 A Wi + 11)6 A oi2

du; 4

=

u>! A tu4 — ty2 A 105 — tu3 A tu4 + ui 6 A <J4 + eu 2 A ui 3 — ca>3 A u>! — 2aw! A w2

dws

=

—wi A ws + W4 A 0J4 + au>2 A ui3 + 6w3 A u i + cu>i A w 2

dio 6

=

— wi A w6 — w2 A w4 — 2w3 Aw6 + fu2

A UJ3 + aw 3 A wi — ewi A UJ2.

(25) The complete system of invariants consits of the five fundamental invariants a, 6, c, e, / and of their covariant derivatives. If all the invariants a, b, c, e, / are zero, (1) is reducible

to

d3U

by a contact transformation.

^=°

„

By introducing six auxiliary variables, we have attached ten Pfaffian forms u>i,...w4, u>i, ..., t«6 to (1) in a way which is invariant under the group of contact transformations in the plane. When all the invariants a, 6, c, e, / are zero, as in the case of d3u/dt3 = 0 system (25) is the. equations of structure of a group in the parameters. This group is similar to the group Gio of transformations of contact in the plane which carry circles

83

6.2. THIRD-ORDER DIFFERENTIAL EQUATIONS

into circles. Hence in t h e general case system (25) is the equations of structure of a generalized space with Gio as its fundamental group and with U>1 = b}2 = k>3 =

0

as t h e equations of t h e elements of t h e space. T h e elements are thus the integral curves of(l).

Next we consider t h e case

I ?0. In the case / / O w e can choose vi as a function of t, u, u, u such t h a t the coefficient of u>2 A u>4 in du>i is equal to 1. T h e n w44 is a linear combination of uii and the coefficient of u>! A u>4 in u>4 is an invariant. It is easily found that this invariant is 1 dvi

v3

avi du

a

We can reduce it to zero by choosing » a = - ^ .

(26)

Of the auxiliary variables there remain only three arbitrary ones, which are a , 7, v2. Of the 42, u>3, u>4, WU, W13, W42 while t h e others are linear combinations of them. On taking account of t h e relations (11) and (12), we can write (8) in the form ! =

wn A aij + w13 A ui3 + CJ2 A LJ4 + aoij A u2 + 6ui2 A UI3

du>2

=

Wu A ui 2 — 0J3 A u>4 — 2aiu/2

A W3 — 2a2io2 A u>4

du>3

=

W13 A u»2 + i " n ^ 3 — ^ 1 A LJ4 + a.3U>2 A 0J3 — a2Lj3 A u>i

du>4

=

w42 f\u>2

(27)

+ b\u>\ A W3 + &2W2 A u>3 + 03^2 A Wi + 63(^3 A u>4

where the a's and t h e 6's are functions of t, u,u,u,
1 (

d

h =

i1 / avi \ avx

L^l\

1 dvx

v3dv1

dv3

Vi du

vi dt

dt

84

CHAPTERS.

CARTAN EQUIVALENCE

METHOD

These coefficients can be reduced to zero if we choose 7 =

v2

2 dH 33 du

1 dvj. vi dt

_ J_^i _ L^ai _ — [\Jt!i\

(28)

vi dii v\ dt dt \vi du ) ' Then both W\3 and W42 become linear combinations of w;. On writing w for wu we can write (27) in the form dui

=

w A u>! + u>2 A W4 + a^i A u;2 + iw2 A u>3 + CW2 A u 3 + eu>3 A w4

dw2 =

u) A W2 - ua A w4 + /ai 2 A u>3

0013 =

w A u>2 — wi A u>4 + owi A«2 + <7w2 A1J3 + «<;2 A w4

du>4 =

Awi A u;2 + io)! A u>3 + ju>2 A o)3 +fcu>2A ai4.

(29)

In these equations we regard a as an auxiliary variable and the Pfaffian form w is linearly independent of a>i. It is possible to choose v3 to make the coefficients a and 6 of (29) zero. The resulting equation are then of the form dui

=

1CAW1+U2AU4 + au>2 A u>3 4- bu>3 A u>4

4&J2

=

I B A U 2 - U 3 A U 4 + CU>3 A W4

duj3 = w A u>3 — u»i A ui3 + eaii Auj + /u>2 A w3 + 6u>2 A w4

(30)

ict»4 = ^W! A w2 4- ^Wj AU3 + iu>2 A u;3 + jo)2 A 0*4 where the a,b,...,j do not denote the same functions as those in (29). By the form of (30) the Pfaffian form w is completely determined. On applying the theorem of Poincare to (30) we can show that dw is of the form dw = fcwi A w2 + lui A u>3 + mu>i Au 4 + nw2 A o>3 + pu>2 A CJ4 + qu>3 A u>4 where we have the relations m = e = —-c, Consequently we have the following

p = a + -be.

(32)

(31)

6.2. THIRD-ORDER

DIFFERENTIAL

85

EQUATIONS

Theorem: If the relative invariant 7 ^ 0 , the complete system of invariants of (1) consists of the functions a, 6,... ,q in (30), (31) and of their covariant derivatives. When all the invariants a,b,...,q are zero, the equation (1) is reducible to
( 33 )

W*+U = » by a contact transformation.

When all the invariants a, b, ..., q are zero, (30) and (31) are the equations of structure of a five-parameter group Gs in the plane. The transformation group G5 is evidently the group which leaves the integral curves of (33) invariant. It is given by

T — T + e,

U-+eiU + t3e~T + eteiTcos (^-T\

+ t^Tsin

(^-T\

.

(34)

The corresponding Lie symmetry vector fields of (33) are given by Zx =

d w

d Z2 = UW,

Z3 = e-

y d —

* — (£•)•*& *—(T') "W

(35)

Hence, in the general case we have defined in the plane a generalized geometry with the elements of contact of the second order t, u, u, u as the elements of the space and with Gs as the fundamental group. Example: Consider * '3 ! B + ( £ ) ' + 3 ^ 2 + 1=0. dt \dt) dt dt This equation can be linearized to (33) with the invertible point transformation F{u,t) = eu,

G(u,t) = t.

When we insert H(t, u, u, u) = — (uf — 3iiu — 1 into (14) we find that J=-l.

(36)

86

CHAPTER 6. CARTAN EQUIVALENCE

Thus (36) cannot be linearized to (2). On the other hand cPu

lfi

fdu\ +

\dl)

can be linearized to (2). We have 7 = 0.

ducPu _ +

Q

!t~d¥

=

METHOD

Chapter 7 Painleve test and linearization The Painleve test is a powerful tool to investigate the integrability of nonlinear ordinary and partial differential equations (see Steeb and Euler 1988 and references therein). In some cases the Painleve test can lead to the linearization of the nonUnear differential equation. The nonlinear differential equation is considered in the complex domain. First we give a short introduction to ordinary differential equations and the Painleve test. Then we describe the connection between the Painleve property and the invertible point transformation. In chapter 8 we consider partial differential equations. Definition: An ordinary differential equation considered in the complex domain is said to have the Painleve property when every solution is single valued, except at the fixed singularities of the coefficients. That is, the Painleve property requires that the movable singularities are no worse than poles. A necessary condition that an n-th order ordinary differential equation of the form

— = H[zM*\-,1-^) where H is rational in w, ..., dn~1w/dzn~1 is that there is a Laurent expansion

(1)

and analytic in z, has the Painleve property

w(z) = (z-z1)kf:a:(z-ziy (*- -ziY

(2)

3=0

with n — 1 arbitrary expansion coefficients (besides the pole which is arbitrary). Notice that more than one branch point may arise. These subbranches may have less than n — 1 arbitrary expansion coefficients. 87

88

CHAPTER 7. PAINLEVE TEST AND

LINEARIZATION

A necessary condition that an autonomous system of k first-order ordinary differential equations (considered in the complex domain)

(3,

£ = V,w,

where V is rational in w has the Painleve property is that there is a Laurent expansion wi{z) = (z-z1)m''£aii(z-z1y,■ (*--*iY, i=o

i = l,2,...,fc

(4)

where k—1 expansion coefficients are arbitrary. More than one branch may arise. At least one branch must be a Laurent expansion with k — 1 arbitrary expansion coefficients. For the connection between the Painleve property and integrability we refer to the literature (see Steeb and Euler 1988 and references therein). If this necessary condition is satisfied we say that the system dvr/dz = V(w) passes the Painleve test. Remark: A remark is in order for applying the Painleve test for nonautonomous systems. The coefficients that depend on the independent variable must themselves be expanded in terms of t — t\, where
(5)

If nonautonomous terms enter the equation at lower order than the dominant balance the above mentioned expansion turns out to be unnecessary whereas if the nonautonomous terms are at dominant balance level they must be expanded with respect to t — t\. Example: The simplest example of a nonlinear equation which admits the Painleve property is given by the Riccati equation dw - W

, = 0.

(6)

We find a Laurent expansion of the form w(z) = (z-z1)-1f:wj(z-ziy.

(z-Zly. a

(7)

In order to find a linearization mapping we have to extend the ansatz (2) to vH(z) = 4r,(z)jrwii(z)
M^w

(8)

89 This ansatz comes from the Painleve test for partial differential equations (see chapter 8). We apply this extended ansatz now to four differential equations. Then we try to linearize them. We study the rate equation dw dz = — w — w the Lotka-Volterra models

dwi dz

= —Wi + WiW2

(10a)

W2 — W1W2

(106)

= —W-i + WiW2

(11a)

= —1^2 — VJ1W2

(116)

dwx dz

= C1W1 + C23W2W3

(12a)

dw2 dz

= C2W2 + C13WIW3

(126)

dw2 dz and

dwi dz du>2

dz and the system

(9)

dw3 (12c) = C3W3 + CX2W\W2 dz which arises in plasma physics. Equation (9) and system (11) have the Painleve property. System (10) does not have the Painleve property, and system (12) has the Painleve property if Ci = c2 = c3. Consider first (9). The Painleve test is now as follows: First we determine the dominant behaviour. Inserting the ansatz w(z) oc <j>n(z)w0(z)

(13)

into (9), we obtain n = — 1 and wo = <6'

(14)

. where ' := d/dz. Inserting the series ansatz CO

w(z)--

= f 'WE"',•(*)**<*) J=0

(15)

90

CHAPTER 7. PAINLEVE TEST AND

LINEARIZATION

into (9) and taking into account (14), we obtain j =

l : 4" = -<j>' -

i$Wx

j = 2 : w[ + w2 = — i«i — 2'w2 — w\ j = 3 : w2 + 2w3' = — w2 — 2'w3 — 2w\w2 and so on. Notice that if we put 4>(z) = z — z%, then we obtain the Laurent expansion of (9). If we require that Wj = 0 for j > 1, then we find the linear differential equation 4" + ' = 0

(16)

and the Backlund transformation w = f^'.

(17)

The general solution of (16) is given by
+ C3

(18)

where C\ and C2 are the constants of integration. Inserting this solution into (17) we find the general solution of (9). Another linearization procedure is as follows: Consider the Mdbius transformation v=

aw + +b ^— cw + +d

,._.

v 19;

where a, b, c, d € C and ad — --bcbe== 1.1. The system of all Mobius transformations forms a group with functional composition as group multiplication. The inverse transformation is given by w=

dv — -b b . —cv + 00' -cv

.„„. '

v(20)

Thus the Mobius transformation is a special invertible point transformation. Inserting (20) into (9) we find the linear ordinary differential equation dv —_ — = dz - ■

<21>,

ifa = 6 = — 1, c = 1, (i = 0. If we apply the inverse transformation V v

==

—w —11 w

(22)

91 to (21) then we obtain (9). We mention that this approach can be generaUzed to a system of n-coupled autonomous homogeneous Riccati equations of the projective type (Kowalski and Steeb 1991). Consider now system (10). Inserting the ansatz U)J OC 4>mW10,

W2 OC <j>nV)20

(23)

into (10), we find m = n = — 1 and u>io = ',

w?o = -'-

(24)

Inserting the series ansatz oo

w

oo

i = ~1 Yl Wiiftt

w

2 = 4>~X J2 wii
j=0

( 25 )

j=0

into (10), we obtain, for j = 1, 4" = —' + <j>'{w2i — w n ) -(/>" = -4> + 4>'{-w21

(26a)

+ tou)-

(266)

Equations (26a) and (26b) are not compatible. Therefore the series ansatz makes no sense. The result coincides with the fact that (10) does not pass the Painleve test. Logarithmic terms must be introduced. Ansatz (8) can also be extended in order to introduce logarithmic terms (so-called logarithmic psi series). Consider next the Lotka-Volterra model (11): With ansatz (8) we find m = n = — 1 and j

=

0 : w 1 0 = ',

j = 1: -4" j

=

2 :

w 2 0 = -4>'

" = -4/ + 4'(w21 = w'n+

wn)

4'-4'(w21-w11) W124'

= - I D u + ^11^21 + ^'(^22 -

w12)

">21 + U)22^' = —"^21 — ^ 1 1 ^ 2 1 + '(—w22 + ^ 1 2 )

and so on. To find a Backlund transformation we consider two possibilities. The simplest one is given by setting uiy = w2j — 0 for j > 1. Then we have 4" + ¥ = 0.

(27)

CHAPTER 7. PAINLEVE TEST AND LINEARIZATION

92

A Backlund transformation is given by w1 = -y(j>'

(28)

with wi = — w2. Since wi = —w2 it is obvious t h a t we find only a particular solution to (11) when we solve (27) and insert t h e solution into (28). An extension is possible. W h e n we p u t wXj = 0, w2j = 0 for j > 2, we find t h a t 4>, Wn, a n d w2i must satisfy the equations

4>" + 4>' = % , ! - % ) W'n

=

w'21

=

-WU

+

WnW21

-W2l-WUW2l.

This means t h a t u>n a n d w21 satisfy t h e Lotka-Volterra equation (11). A Backlund transformation is given by u>j = <£~V + wn (29a)

w2 = - « r y + w21.

(296)

Consider now system (12). For t h e sake of simplicity we p u t c 1 2 = C23 = c3i = 1. Moreover we set cx = c2 = c 3 = c. T h e series ansatz (i = 1,2,3) 00

w« = 4>ni Yl w'i<^ i=o

yields n\ = n2 = n3 = — 1. Furthermore we obtain W20W3Q j == 00 :: - W i o ^ ' == 1^20^30

-«w

(30a)

WIQW30 = 0 :: —w —w20204>' (j> == wxoiuao j =0

(306)

(j> = —w3030' = w-iow-ia U)10t«20 Jj == 00 :: —w

(30c)

w = cwio CW10 + + 1021^30 1^21^30 + W20W31 W20W31 j = 1 : u^io 'w =

(31a)

I I ^ M ++ «'io"'3i CW20 + +W W11W30 «'10"'31 ij == 11 :: wW202 0 == cu>2o

(316)

j == 11 :: u™30 ^ , == cu) CW30 + wuw WuW2020 + + wW W i 2l 30 + 102 10w

(31c)

and so on. A nontrivial solution t o system (30) is given by ">io = t"2o = 4>',

W30 = —'■

(32)

T h e simplest case for finding a Backlund transformation is given by setting Wij = w2j = W3J = 0 for j > 1. T h e n we find t h a t Wl

= w2 = <j>-l<j>\

w3 = - ^ " V

(33)

93 and satisfies the linear differential equation 4>" = c<j>'.

(34)

Consequently, solving (34) and inserting the solution into (33) we obtain a particular solution to (12). An extension is as follows: We put wij = w2j = w3j = 0 for j > 2. Then we obtain " = c4> — (j>'w2\ + (j>'w31 (35a) (j>" = c<j>' — 4>'wn + <j>'w31

(356)

" = c<j>' - 'wn + 'w2i

(35c)

and wn, w2i, and UJ31 satisfy (12) with ci = c2 = C3. Thus we have six equations for the four dependent variables 4>, win, w2\, u>3i. So we must prove that these equations are compatible. From (36) we find that u; n = w2\ = —w31. It follows that W\ = w2 = —w3 wt = 4>~x4>' - ""1 + \

(36)

and <j> fulfills the nonlinear ordinary differential equation

rr1 - \n>n = ~.

(37)

The left-hand side is the Schwarzian derivative of <j> with repect to z, which is denoted by {;z}. Hence, {^} = - y .

(38)

The Schwarzian derivative plays a central role for integrable differential equations (Steeb and Euler 1988). The extension of ansatz (19) to more than one dependent variable is as follows: We consider the transformation

v«£±§

(39)

(7w + D where

v=(v1,...,vk)T, w= (w1,...,wkf. Here A is a it x k matrix, B is h x 1, C is 1 X k and D is 1 x 1. The (k +1) X (k + 1) matrix ' A B ,C D is assumed to be invertible.

CHAPTER 7. PAINLEVE TEST AND

94

LINEARIZATION

Let us now describe the connection between the Painleve test and the invertible point transformation. First we consider the anharmonic oscillators

+ U3 = 0.

S

(40) (40)

Equation (40) has the Painleve property and therefore passes the Painleve test. It is well known that the anharmonic oscillator can be solved in terms of Jacobi elliptic functions. Let Ui = U and U% = dU/dT we obtain the autonomous system dU, IT U -dT=u *

dU2 dT ~ - f i u

w- *

-di = - '

(41)

^

with the first integral

m,Uu22))--= \vl + \ul Wi, 2 2 +&■

(42) (42)

Then the solution of system (40) can be expressed in Jacobi elliptic functions {U20/u)sn{uT, i) + + UiocajuT, U10ca{ujT,i)dn{u>T, i)dn(uT, i) „ m {VwMMuT, Ui{T) = ( ' l1 + {U(U w/V)hn*(u,T,i) m/V)ha>(uT,i) U2(T) == * '

2 U20 cn(uT,i)dn(uT,i) (U1 10 /V)2 sn [1 - (t/ sn 2(u>T,i)] (a;r,0] mai{uT,i)M^T,i) 0 /y) 2 [l /Vysn*(u>T,i)] [l + + (U(V /V)*sn*(u,T,i)f 10w 2 2uU10(U10/V) sn{uT, i) [1 + sn*(u>T,i)] /Vysn(uT,i) sn2(u>r, 01 22 [l + + (U(U /Vysn*(uT,i)} " [1 /V)*aS(uT,i)} 1010

l(43)

'

where 14 V ' V = 4(IE) = ( £)1/4

U

lJ V " =(\) = ( 2 )* ' y '

and

f/22

20 E E == u!f 2

u 4 J7 If*

++ w4!f~i(Ui,u = m,u2t))

with i = y—1 and sn, en, dn the Jacobi elliptic sine, cosine, and delta functions, respectively. The initial values are Uw = Ui(T = 0) and U20 = U2(T = 0). We find that (40) admits the two Lie symmetry vector fields Si-- =

a

df'

,S'a

,, d

d

-T—. = u-^-8U dT

(44)

Let us now discuss the Painleve test (Euler et al 1989) of the anharmonic oscillatorr dhi ^ + + Mt)^ ~dl?

*®i

3 3

+ f2(t)u ! ++ Mt)u /3(*)« === 0 +/«(*)«

(45)

95 where fu /2 and /3 are smooth functions of t. We assume that /3 ^ 0. For arbitrary functions fu /2, and /3 we find that (45) cannot explicity be solved. Obviously, we assume that fu /2 and /3 do not enter the Laurent expansion at dominant level. We apply the ansatz

„„) = * , - f > , W W .

(46,

j=0

The condition that (45) passes the Painleve test is given by the differential equation

#l£/f*l - M ^ f fi^fif^ 9

3 (f| )\, /3/1+e (f)' f *-« W (f f )',,,,+» f |/;V/SA I2 f(f)"**

^jte^M-m ^iA,, h^IUi.m

+90 /33 - 2 7 f ^ ff -

-m

2

^J hdht2t (#) V/ | +^1 9#2 ^ (f /!/, {%)) V?s#f h'

3

- 57§f /I/, + 72f to - 14f/33/3 -

2 d d _ 9 0 f nh I ++ 1l88*g£Lj -/ ,+<M yj/, +60f -90fy3/ + 5 , ^^/ , ++336 ( f ) /|-K 4fth+w 4tifitfj■(§)*- - 36f -36^/2/?

+8 ^

= 0.

at

at

(47)

We solve constraint (47) in such a way that an expression yielding / 2 as a function of /j and /3 wiU be given. For any set of functions {/(*), /2{t), f3{t)} satisfying the integrability condition we show that (45) can be written in the form (Euler et al 1991, Estevez et al 1991) —

-(AT

+ B)U + U3 = 0

(48)

where A and B are constants. This is the form of the second Painleve transcendent whose solutions are known to be uniform functions of T and free of movable branch points. Only the particular case A = 0 can be solved in terms of Jacobi elliptic functions. Let us first give the form of the solved constraint. We define the function M as follows

M(t) = (36/JAV U ^

~ 1 ( f ) 2 + 2A/ 3 f + 8/|/= + 12/|f - 36/|/2) (49)

where the smooth function A(i) is given by

m = /,■,-^0-pf-i/A«*l

(so)

96

CHAPTER 7. PAINLEVE TEST AND

LINEARIZATION

After some manipulations one can show that condition (47) can be written in terms of Mas
-dT+wA ~^)~dt

'

Let us now perform the uniform change of variables in (51)

(52a)

dT = $(t)dt

#(*) Hi) ==-fA^expl-ljMs)^] fl/3(t)exp I-1-} f1(s)ds)

.

(526)

With this change, (51) takes the form

(53)

The general solution of (53) is given by M(T) = AT + B, where A and B aree the tain constants of integration. Using (50) - (53) in this last expression for M(T) we obtain

*-^K-- T (5) ! 1/3

+2/1/>f+12/

/3 (, 1 )expf-i// -IB- (B +hA/^ Ajdst1f^M^P (-\JMs2)ds 1 ( S22\\)^jj

.f+8/,v,)

/3 U (/32/3exp(-lJMs)ds\\ exp f-lJfi(s)ds\\

(54)

which yields the solution of the constraint.

owing The result follows easily from the above analysis. Let us perform in (49) the following change of variables dT = 9{t)dt, u(t) = X(t)U(T(t)) (55) where A(t) and #(*) are given by (50) and (52b), respectively. Then (45) takes theiform form g _ M ( T ) [ / + t/ 3 = 0.

(56)

Since M(T) = AT + B, we finally obtain —

-(AT

+ B)U + U3 = 0.

(57)

For A = 0 this is a nonlinear differential equation whose general solution can be given ren inin terms of elliptic functions. For general A and B this is the second Painleve transcendent.

97 Let us now discuss special cases of (47) (Euler et al 1989). We recall these cases here, because we discuss them in connection with the invertible point transformation. Case I. Let fi{t) = c l t / 2 (t) = c2, and f3(t) = c3, where cj, c2, and c3 are constants (c3 ^ 0). Then we obtain from condition (47) that c|c?(2c? - 9c2) = 0.

(58)

Case II. Let /i(t) = 0 and / 3 (r) = 1. Then we find
-^

(59)

This is a special case of the second Painleve transcendent. The solutions have no branch points, and are therefore uniform functions in t (see Ince 1956, Davis 1962). Case III. Let f2(t) = 0 and / 3 (i) = 1. Then (47) takes the form

-W + 3 ^ A

+2

(it)

+

YHfl + 9fl = °-

(60)

This equation admits the particular solutions fi(t) = j ,

fi{t) = | .

(61)

Thus (60) admits more than one branch in the Painleve analysis. Equation (60) does not pass the Painleve test, because it admits non-integer resonances (rational resonances). However, (60) passes the so-called weak Painleve test (see Steeb and Euler 1988 and references therein). Case IV. A case where flt / 2 and f3 are non-constant and satisfy (45) is given by

Mt) = lt,

m=±

Mt) = ~-

(62)

Equation (45) together with the functions given by (62) arises in the Painleve analysis of external driven anharmonic oscillators (Fournier et al 1988). This differential equation can be integrated in terms of Jacobi elliptic functions. Case V. Equation (45) together with /i(*) = ^ .

/»(*) = ^ i ,

/»(«) = - ^

(63)

98

CHAPTER 7. PAINLEVE TEST AND

LINEARIZATION

occurs in the Painleve analysis of the Lorenz model (Tabor and Weiss 1981). The functions fx, f?, and f3 satisfy (47). Then (45) together with the functions given by (63) can be solved in terms of elliptic functions as follows: Applying the transformation u(t) = t^gp/*)

(64)

to (45) where fi, fi and /3 are given by (63), yields

g = 2,3

(65)

with a = < I/4 . Case VI. The equation d?u

2du

„

,„„.

^ ^ f =°

™

+

1 ,

+

arises in the group theoretical reduction of a nonlinear wave equation. Consequently, we have fi(t) = 2/t, f2(t) = 0 and f3(t) = 1/t. These functions satisfy (47). Therefore, (66) passes the Painleve test. Now we ask whether the equation derived above can be found from (40) with the help of the invertible point transformation. The invertible transformation is given by T(t) = G(u(t),t), where

U(T(t)) = F(u(t), t)

(67)

dGdF dGdF dGdF dGdF .„ ' du dt ^ dt du

A

(68)

We obtain from (40) that

d?u d?u ^

A +

A

3 (du\ Idu\ 3

^ j

where

2

Idu\ 'u\ , du k fa + A , ^ j + A 1 ^ + A0 == 00 2

d G _ (dGd F _dFd A -L?G (02)* j*) A 33 == [8G&L 22 22 +

I du du du du

2 _ (dGd F (dG
2 F dG nd&F_dG

2 (dGd fdGPFF ~ ^\ du du dt22

dGd2F ndGPF_

\du j

2

+

22

(dGd FF (dGd

j

A

dG (dG\2 dt \duJ

2

22

dFd dFd GG

J

-i^m)^y(dG\*\

dG (dG\2 du\dt J A

.

+ F A ~ [dt dt2 ' ' dt dt2 {-dl) =\lH-d*-lH-dW '♦ (£) V j) A --

ka A0:

3\

-££♦•£(£)"'•)*-

_ ndF d£FG_ d G _ dF&G dt dudt dt dudt dudt du dt2

=

1

KSW*--i

_ ndF dF
= 2

du du

,„_. (69) (69)

3\

t

I

( 7(70) °)

99 We are not able to handle this general case. We make a particular choice for F and G, namely F(u,t) = f(t)u, G(u(t),t) = g(t). (71) With this special ansatz we find that A3 == A2 == 00

(72)

dj_dg__ 2dldg_

«P d*sS di dt dt J dt2 dg , Wf dt1

A =_ Ai

2 ((Pfdg (d fdg df
A„

*

(

Ao — = ^ AQ

*

('dg\3

)

>

■

%-U — dtJ f dtJ

It follows that

,

J --v (§)'«' [li) u u 2f

^ — .

(73)

3 3 = 0 ^T + l(0^ + + /h(t)~ + /2(*)" f2(t)u + + f/3W" 3(t)u ^0

(74)

where

dg rfjL J 2

dt dt dt J^MZIM.

h/fl =

~

2

2

dfdg

d d 2 L l_

dg df1Jf dt

'

fh-

3

d d d 3 dgd / df
^dg 1,

_ W

'

o(M f_ h/ 3 = ~ V~dj) ■

-(4)'

dt

(75) Here / and g are arbitrary functions of t. We are now able to eliminate / and g from system (75). We obtain t J fl/2,

N

^ - -J /TW4

(76)

^ d s

g(t)--

= c/ 1/6w

(76)

f {s)d

1 /M = = C /3 3 ^ eex xp p ( / ^ ) f(t) (/ 3 : )

l#i

/»(*) = / J W -" 33 J"5T i

2 2 +

r9 J l

/l +

#3

18/ i 8 / 3 dt*

7 1 /d/3y 2 V
(77) 11^/3 2 66 //,3 Ad 'i ' '

(78)

(?8)

To summarize: Equation (40) is transformed into (45) under (71) where

(cex ^// ( )^j/ 3 1/6f^(t))(<)j Uu,,

F(u,t) == f c e x pP (\)fi(s)da) 1 S F(u,r)

G(t) G(t) "=- /

fl%)f-^-ds d /is)"

with / . satisfying (78) and /j and / 3 are arbitrary smooth functions of t.

(79)

100

CHAPTER 7. PAINLEVE TEST AND

LINEARIZATION

Let us now compare the two approaches. When we insert (28c) into (8) we find that (8) is satisfied identically. Thus the invertible point transformation with the special ansatz (23) is a special case of the Painleve approach. Let us now look at the special cases which we discussed for the Painleve approach. We find that some cases given in the Painleve analysis cannot be found with the invertible point transformation. Case I. Let fi(t) = c%, f2(t) = c 2 . Then we find

f(t) = -ec^3

g(t) = -e-Clt/3.

(80)

Consequently, the invertible point transformation F(u,t) = = -ueCit'3,

3 = cit-e-^l G(t) =-e~ '3

(81)

transforms (40) into (Pu d?u __ ++

du du

_ + C2U C C ll _ + C2U + +

u3 u

, ■> =0 0 =

,„.. ( 8 2 )

with 2c\ = 9c2. Case II. Let f\{t) = 0, /3(f) = 1. Then from the invertible point transformation approach we find / 2 (i) = 0. Thus we only find a special case. By calculating the Lie symmetry vector fields of (59) we find that if A, B ^ 0 the equation admits no Lie symmetry vector fields. If B = 0 and A ^ 0 we find that (59) admits the Lie symmetry vector field Z = d/dt. Therefore (59) cannot be obtained from (40) by applying any invertible point transformation. It can be shown that the equations related by an invertible point transformation have the same number of Lie symmetry vector fields (see chapter 4). Therefore all the equations equivalent to (40) by an invertible point transformation must have two symmetry generators with the same Lie algebra, i.e. [Z\, Z2] = —Z\. Case III. Let f2(t) = 0 and f3(t) - 1. Then

frf/i + 3f/?==0.0. dt +

A

(83)

The general solution is given by 1

hit) ~«*=i^W " 2
AW ="

3 t

0*)

101 of (60) has only one Lie symmetry vector field, namely

z

d d -t— + u — dt du = at du

and therefore cannot be obtained from (40) by an invertible point transformation. Case IV and V. If we put h(t) = at",

f3(t) = fit™

we arrive at

r ,.\ ,„ .at"-1 2 , ,„ / * ( t ) = (6» + ™ ) _ + 5 «V-(l + When we set a = 1/4, n = — 1 and m = —2 we find

/i(*) = ^

/»(*) = ^

T

m\ m ) - .

/»(*) = §•

, (85)

(86)

We find / 3 of (62) by setting £ = 1/32 and f3 of (63) by setting 0 = - 1 / 8 . Case VI: The functions

/i(0 = f,

/>(*) = 0,

M0 = J

(87)

are not solutions of (78). We could conjecture that this is related to the fact that we have used the special ansatz (71). That this is not so can be demonstrated by determing the Lie symmetry vector fields of (45), with fit / 2 and / 3 satisfying (87). We only find the Lie symmetry vector field

z

d

=-24t+l- du

(88)

We conclude that this case cannot be obtained from (40) with an invertible point transformation. Three remarks are in order: Equation (40) can be considered in the complex plane. Then the singularties are poles of order one. The singularities form a rectangular lattice. Applying the invertible point transformation and using the solution (43) we can now study the pattern of the singularities of the transformed equation. The fifty ordinary differential equations of second order of Painleve type are just representatives of equivalence classes. The group under which the classification is done is given by

T(t)--

=m

U(T(t)) =

M*Wt) + M*) M*Ht) + M*)

(89

where tpi, ip2, fa, V"4 a n d 4> a r e analytic functions of t. We mention that for certain choices of the parameters, the Painleve transcendents II - V admit one-parameter families of

CHAPTER 7. PAINLEVE TEST AND

102

LINEARIZATION

solutions expressible in terms of classical transcendental functions, such as Airy, Bessel, Weber-Hermite, and Whittaker, respectively (Gromak 1978).

Next we consider the second Painleve transcendent

±±L = 2U3 + TU + a

(90)

and perform an invertible point transformation to find the anharmonic oscillator

-^ + fi(t)-£ + f2(t)u + f3u3 = 0.

(91)

This provides a condition on fi(t), f2(t) and fs{t) such that the anharmonic oscillator (91) can be transformed to the second Painleve transcendent (90). The Painleve test for (91) have be discussed above. The condition on /j(<), f?{t) and f3(t) is given by (47). We show that the conditions are the same for the two approaches. Now we ask whether the equation derived above can be found from (90) with the help of the invertible point transformation. The invertible transformation is given by T(t) = G(u(t),t), where .

dGdF dGdF dt du

U(T(t)) = F(u(t),t)

(92)

dGdF dGdF

(93)

t n

' du dt * °

We obtain from (90) that (^U
/
(du\ (du\

UJ __ ^ + AA3 U^-jJ + AA2 (^-J +

3

+

2

+Ai, du+Ao

^

„

,„,.

(94)

=U + A 1 ~ + A0 = 0

(94)

where

-my

*•- (-££ + S i - - (£)'-« (£)"-(£)>du du2

\

au2 du

\ouJ

\ouJ

_ /(

dF d22G ndFd

2 dFd dFd2G G

d22FdG FdG

I

dudt du a^at

at du au at

at ' "aauuat at du a« du dt

A-> "•22 — 1 1- 2

22 1

a^22

-»" (£)'£-

-3a(

1

dn2dF2FdG dG

v a u> /

2 2

dG 3(dG\ dG ,r,3l 'dG\ - 6F3 dt au J at I\ du)

a*/

2 2 2 2 a dG t'dG\ G_ — dF ™ d2G FdG r ^dG nd F -- /f ' _ dFd Z^ o—— dG +1d*dFdG (dGY F3 22 -22 +- 2 1 6F 22 dt dudt dudt dudt dudt dtdt dtdt du du 'du \,dt){it) dt du ' '~ 1\{ du dt ,

Ai M

i

103 o r ^ < 9 G dG\ -3.FG— —

du

du \ dt J

2

2

(m\*\

i

[mj)

1

n dG - 3 a — I -*-)

.

i

A" 1

3

(dG\3\

'-(-^-^-»-(f)'-"©'-(f)>-™ \dt) [dt) j1 dFd G d FdG « 1' A 0 = — ^ — ^ r2 + -TTT-^ 2 1^

dt dt

dt dt

nr,,(dG\

„„(dG\ - FG\

2F3 —

\dt)

- a

A" 1

(95)

We now make a particular choice for .F and G, namely F(u, t) = f(t)u, F(u,t) f(t)u,

(96) (96)

G(u,t) = g(t). g(t).

With this special ansatz we find that (97a) (97a)

A22 = =0 A3 = == A and

dgdf

2dg_df_ 2 f A A

•A1 1l

"

Ttdir dt dt W dt2 ~dg ' dg f 1 * dt dt

-(*r

(dgd2/ (^fl_ dfd2d g\Lfl\ (dg \ \333 ^ (^l)3 (k u+ f u a 2 + \ dt dt [did* dtdt2 {dtJ {dt A A„ (976) A AQ ° ~dg • (976) fdg 1f dt

For a = 0 it follows that 3

^

+ fi(t)~ ^+A ( * ) ^ + /+, (h(t)u * ) « ++/Mt)u ,(ty where

d dd 2 l L_

l L_

fx{t) =

AM =

m

hit) ==

= = 0

(98) (98)

ffl fdll

dt dt J dP_ dt2 d±dt_ fdg 1f ~dl dt

df-dtdt>->°{dt) dt dt2 ' dtdt2 fdg 1 dt

(dgV (dg\3 \dt)

/aW = =- 2 ( / | ) 2 . hit)

<>%)

( 9(99a) 9a)

{ m )

(996)

(99c) (99c)

We are now able to eliminate / and g from system (99). We obtain /e

(J/A(-)-). ■

(t)exp (\jh(s)dA f{t) =--Cfl Cfl/6(t)exp /(<) =

(100) (100)

104

CHAPTER 7. PAINLEVE TEST AND

LINEARIZATION

By inserting (100) and (99c) into (99b) we find
m

(f//iM*)

(|i)l)

where

M{t)

6/3 +7

2/l/3

12Mi+36/2/l 8/i2/l

=w* (- ^ (f)' - f -

- )•

18/1

(102) Inserting the derived / and g in (99a) we find condition (47). To summarize: The condition (47) that the anharmonic oscillator (45) passess the Painleve test is identical to the condition that the anharmonic oscillator (45) can be transformed to the second Painleve transcendent. The Painleve property is not preserved in general under an invertible point transforma tion. For example, the equation

(Pu
1 (du\ 'u\ -(a u \c11/

. a a +u5 + tu + u

, (103)

where a is a constant, has movable branch points, since it has solutions of the form

u(t) = -£uj(t-tly-1'2

(104)

3=0

with Uj, j = 0,1,2,...

constants. If we make the invertible point transformation F(u,t) = u2,

G{u,t) = t

we find that U(T) satisfies the second Painleve transcendent.

(105)

Chapter 8 Painleve test and partial differential equations The Painleve property for ordinary differential equations, namley that all movable sin gularities in the solutions are poles, has been known since the famous work of Sonya Kovalevskaya, to be related to integrability. In recent years, it has been generalized so as to apply to (nonlinear) partial differential equations. The conjecture is that a partial differential equation is integrable (by the inverse scattering transform or Backhand trans form, or some other method), if and only if it possesses the Painleve property possibly after a change of variables (see Steeb and Euler 1988). Let us now consider partial differential equations (Ward 1984, Weiss 1987, Steeb and Euler 1988). The major difference between analytic functions of one complex variable and several complex variables is that, in general, the singularities of a function of several complex variables cannot be isolated. If / = f(zi,...,zn) is a meromorphic function of n complex variables (2n real variables), the singularities of / occur along analytic manifolds of (real) dimension 2n — 2. These manifolds are determined by conditions of the form ^(z1,...,zn) = 0 (1) where is an analytic function of ( z 1 ; . . . , z„) in a neighbourhood of the manifold. Definition: We say that a partial differential equation has the Painleve property when the solutions of the partial differential equation are single valued about the movable, singularity manifolds. For partial differential equations we require that the solution be a single-valued functional of the data, i.e. arbitrary functions. This is a formal property and not a restriction on the data itself (Weiss 1987).

105

W6CHAPTER

8. PAINLEVE TEST AND PARTIAL DIFFERENTIAL

EQUATIONS

To verify if a partial differential equation has the Painleve property we introduce a method for expanding a solution of a nonlinear partial differential equation about a movable, singular manifold. Let u = u(zi, . . . , 2 „ ) b e a solution of the partial differential equation and assume that oo

u^E"/3=0 .tf

(2)

Uj = Uj = Uj(zi,...,z Uj(zU .n). ■ , 2 » )

(3)

u ■

i=o

where (j> and are analytic functions of (z\,..., zn) in a neighbourhood of the manifold (1). Substitution of the ansatz (2) into the partial differential equation determines the possible values of a and defines the recursion relations for Uj,j = 0,1,2, When a is an integer and (2) is a valid and general expansion about the manifold (1), then the solution has a single valued representation about (1). If this representation is valid for all allowed movable singularity manifolds then the partial differential equation has the Painleve property. For a specific partial differential equation it is necessary to identify all possible values for a and then find what the form of the resulting Psi series is. The Psi series for nonlinear partial differential equations contain a lot of information about the solutions of the partial differential equations. For equations which have the Painleve property one can find the Lax pairs and Backlund transformations. For equa tions that do not have the Painleve property it is still possible to obtain single valued ex pansions by specializing the arbitrary functions that appear in the Psi series expansions. The specialization leads to a system of partial differential equations for the formally arbitrary data. For specific systems, and we conjecture in general, these equations are integrable. The form of the resulting reduction enables the identification of integrable reductions of the original system. If the Painleve test is passed, it is proposed as a sufficient condition for integrability. The Painleve property is a statement about how the solutions behave as functionals of the data in the neighbourhood of a singularity manifold and not a statement about the data itself. Expansions about characteristic manifolds are required to be singlevalued. Essential singularities are found to be determined by certain Psi series involving nonconstant leading orders and resonances. To illustrate the nature of the Painleve property we consider examples of equations with and without the Painleve property. In some of these examples we find a linearization of the nonlinear partial differential equation. Example 1: A simple case of an equation with the Painleve property is Burgers equation du dt

du dx

d2u dx2'

(4)

107 We find t h e series oo

u(*,0M_1£«^'-

(5)

Examination of t h e recursion relations for the Uj yields a system of the form

o-w+» (£)'*-*(*■

*g.|....).

m

Evaluation of t h e recursions yields ,

=

0:

u0 =

,

d<j> cty dt

3 == 1 :

;

- 2 / ox

d d _ dd22<j> dx dx2

(7)

=2: ^U + u , ^-^J = 0 -

The relation (compatibility condition) at j = 2 is satisfied identically and t h e expansion is valid with arbitrary functions <j> and u2. Truncation at constant level leads to

d*

d

-dI

d'

+ Ul

dx- = w

(8)

and Ui satisfies the Burgers equation. W h e n we insert the trivial solution of the Burgers equation U\ = 0 we have 2

d<j> d 4, dt ~~' dx2

0)

and

u = -2r1JEquation (9) is t h e linear diffusion equation.

(io)

□

Example 2: Another example where we can Unearize the partial differential equations with t h e Painleve test is the system dux

dui

-dT-Cl^

=

-UlU2

du2 du2 -^r - c2—-= at ox

uxu2.

, (llQ)

,..,. (116)

Here we find an expansion of the form

u1 = r1Eu^ 1=0

(12a)

108CHAPTER 8. PAINLEVE TEST AND PARTIAL DIFFERENTIAL

EQUATIONS

OO

(126)

l uui2 = 4>"2j<^ *- YJu2j
where "10

-£♦•£

(13a)

-S"*K-

(136)

=

«20

:

The truncation at constant level leads to Ui1 "

1 rV n =
" 2 u»

= ^f It«u» tM + + «una =

(14a)

,

(146)

where u u and u21 satisfy system (11). If we insert the trivial solution (15)

"" 1n1 = "21 "21 = 0 into system (14) we find that <j> satisfies the linear partial differential equation d2<j> . .d2 , a2<* n (ci + C 2 ) + ClC2 dt* ' 3x^ a^-°

S-( ci+c ')S +cic >§=°

(i6) (16)

u1=l-^ c2—hn4> "i = 1 - ^ -+Hc2—Jln<£

(17a)

U "2=

(176) (176)

and

I T T - C I - T - )ln^. 2 =\at U " ox) -

Thus we have linearized system (11).

*

&

)

*

*

■

D

Example 3: The inviscid Burgers equation is given by

I--S-0-

(18)

It is well known that the inviscid Burgers equation admits the implicit solution u(x,t) = f(x / ( I•++tu(x,t)) «(x,t) = iu( a: .0)

(19)

where / is a smooth function. From (15) we have

J-K^V 1+, M.«:)r=,v £-( £)''-rV

(20a)

<"•» (206)

109 whence (18) follows. Obviously we would call (18) integrable. The solution exists as long as (21) l-i/'/0. (21) Hence we can expect that for some x there will be a first t value such that 1-tf = = 0. 0, At this point there will be a shock. In transport processes this corresponds to a critical itral length. Equation (18) does not have the Painleve property. We find an expansion of the form orm u = u0 + u1r+u22"

(22) (22)

+ ...

where n = 1/2. For t h e expansion coefficients uj we obtain the relations collecting cting terms with t h e same power of <j>. We find an anfinite coupled system fof t h e expansion ision coefficients us. For t h e first two expansion coefficients we have

d±_d± r4>1 /1/2: 2 : uud± =d± dx

(23a) 23a)

dt

du0d , +u du0 + 1 2dd

x-^ + b' /x 2U'd-x=0-If we require that u 0 satisfies the inviscid Burgers equatton ,o

236) (236)

■ *"■■

du0 = at ~dt

du00 OX ox

(24)

then we find from the infinite coupled system for the expansion coefficients that Ul

= u2 = ... = 0.

(25)

Inserting (25) into (23) gives

f ^ y ^ + f —(d4>Vd V —4, - 2dd ^ 0 V d 4>=Q' (dVd24> [dxl \dx)

22

dt dt

2

2

n

2 2

\dt) \dt)

dxdx '

It follows t h a t

Q

dx dt dt dtdx dtdx dx

(26) (26)

d<j> d
M. u =. M u =

(27) (27)

■ d(f> dx dx where <j> satisfies (26) a n a u satisfies (18). Thuh we have a Backlund transformation. tion. ation Equation (26) also arises in many non-Painleve equations at the resonance. Equation (26) can be linearized applying the Legendre transformation t =

H d

dx'

X =

dW dw dt

,„ . 28a)

110CHAPTER 8. PAINLEVE TEST AND PARTIAL DIFFERENTIAL 94 t) = -gTl

EQUATIONS ,Mt. (*ob)

dW t = -jj—

(29)

(x,t) + W(e,r,) = xe + tV. It follows that

fi2W

lFPW

with the general solution

w

( 30)

82W

-^-^m w=Q

(31)

W(e,r]) = G(eJr,) + T,H(eM. Under the invertible point transformation T(x,t) = t,

X(x,t)

= u(x,t),

U(X(x,t),T(x,t))

(32)

=x

we find that (18) takes the form

dU -X. &T dT~

( '

The general solution of (33) is given by (34)

U(X,T) = g(X) - XT where g is an arbitrary smooth function. Inserting the transformation (32) gives

(35)

tu(x, t) + x = g(u(x,<)). We find that (18) admits the Lie symmetry vector fields (Euler and Steeb 1992) d 3 d d 3 d Z\ = TT:, Zt = — , Z3 = t— + x—, Z4 = x— + u—

at

ex

m

ZS = f ; T - I T '

z

xt

d

e

d

+

ex

a*

au

Z6 = X— + U2 —

a

a

aU

* = x2a-x + xtm + {x + tu)ua-u

z

* = Tx + Jt ~<« * > * '

(36)

The group theoretical reduction of (18) applying the Lie symmetry vector field Za yields the particular solution u(x,t) = - - . (37) The group theoretical reduction of (18) applying the Lie symmetry vector field Z4 gives the particular solution

■"•■>- -14/FF

where C is a constant of integration.

D

<->

Ill Example 4: The two dimensional case of the inviscid Burgers equation is given by du

du du

du du

m=urx+udy-

%

(39)

■

W

We find the implicit solution u(x,y,t): tu{x,y,t),y u(x,y,t) = = f(x f{x + tu(x,y,t),y

tu(x,y,t)). + tu{x,y,t)).

(40) (40)

Again we find an expansion of the form U2 u = uu00 + + + uUl1<4 4>1/2 + u2<j> ^ + +. . ... .

where

(41) (41)

84, (42)

-whf

u0 ■

dx-+dy~

and du0

du du 0 0

duQ 0 I du 1 2d 2d4>

I1 2idt d

nn

U o Uo u U 0 5 ^^y-~2 " ~aa~dy" '2U'dy"°'dT~ ^~2Ul~2 dy''d~x" ~bT' Uo" l^~

( 4(43. 3

.

Again we can set «j = u2 = ... = 0. Then satisfies tht equation

d2 dV dt2\dx^dy) dy)

(£♦

and (43) takes the form

fdi + df\ dtdx dtdxdtdt \dx

dy) dy)

d24> d didy dt \dx didy \dx "•" dy) dy)

+ &)'(&+&+&)+(2)"(!M>C)"

£--£—£-*

(44)

(45)

leve To summarize: We find that the inviscid Burgers equation (18) does not pass the Painleve sion test, whereas the Burgers equation (4) with a ^ 0 passes the test. We find an expansion of the form oo OO

uu=f'E^ = j—0

6 4 4 (46) )

for the Burgers equation, where and u2 are arbitrary. Although the inviscid urgBurg t dd d ers equation does not pass the Painleve test (only the weak kainleve eest), we find tion Backlund transformation for the inviscid Burgers equation. For the Burgers squation ans(4) the Painleve test provides an auto-Backlund transformation and a Backlund trans D formation (so-called Cole-Hopf transformation) to the linear diffusion equation. □

U2CHAPTER

8. PAINLEVE TEST AND PARTIAL DIFFERENTIAL

EQUATIONS

Example 5: The Korteweg-de Vries equation du

du

m+uTx

SPu

+ ,7

„

.,„. (47)

-^ = 0

possesses the Painleve property. The expansion about the singular manifold has the form oo

2

= r E u ^-

u

(48)

3=0

We find that the resonances occur at i = -1,4,6.

(49)

Resonances are those values of j at which it is possible to introduce arbitrary functions into the expansion (48). For each nontrivial resonance there occurs a compatibility condition that must be satisfied if the solution is to have a single-valued expansion. The resonance at j = — 1 corresponds to the arbitrary function defining the singular manifold (1). The compatabiUty conditions at j = 4 and 6 are satisfied identically. From the recursion relations, we find: /n >
(50) (50)

~^{i

== l1::UUlj ji ~■ „

d*di dxdt

(8 \otej

2

d
(51) (51)

=

'-'■^(£)'^ SS-(S)'3

92^ j = 3: dzdi

+

Ui + 4(T

d^u dx2

2

dx~dx->-3ff[w)

/ ,9^V U3 +
9^j

=°

n

^=°

« (52)

(53)

j = 4 : compatibility condition:

9V +a 2 ^ U2

U3 fc (&st *&* a^ - "( {&) » " • ) j -= °" dx \dxdt

+

(54) (54)

Owing to (53) the compatibility condition (54) is satisfied identically. The compatability condition at j = 6 is also satisfied identically. Thus the Korteweg-de Vries equation possesses the Painleve property. We now specialize (48) by setting the resonance functions u" 4 = "u6e = = 0.

(55) (55)

113 Furthermore, by requiring u3 = 0

(56)

we find that Uj = 0,

; > 3

(57)

if du2

+u

-dT

du2

^

+
diu2

(58)

^ = °-

We thus obtain the following Backlund transformation d2 u = \2cr—\n + u2

(59)

where u and u2 satisfy (47) and d4>d (d\ (d\22 dd dx dt

^A ddP deP „ (d24>\2

-(S)'-

d'<j>

dxdt

+

„

= 0

d*<j>

W

(60)

d^U2 + ad^'~ - 0.

(61)

g-»>

(62)

The substitution

reduces (60) and (61) to the Lax pair (in v) for the Korteweg-de Vries equation.

□

Example 6: The Kadomtsev-Petviashvili equation d2u

d Idu

+

d3u\

du

+u

W d-x{m rx

+

=0

w)

„

,„„.

(63)

possesses the Painleve property. The expansion about the singular manifold = 0 is oo

u = (j>-2J2uj3

(64)

j = -1,4,5,6.

(65)

with resonances at Therefore, subject to the noncharacteristic condition dd> Tr-yLO ox

when

= 0

(j>,U4,us,ue are arbitrary functions of z, j/,< in the expansion (64). The Backlund trans formation for (63) is u = u0(f>~2 + U i ^ _ 1 + u2 (66)

114 CHAPTER 8. PAINLEVE TEST AND PARTIAL DIFFERENTIAL

EQ UATIONS

which obtains

(o+Y „„.--121012UJ (g)' u0 =

d52A ^

19 " 1 = 12n M

>=

d &4> dj_ dP£ „ 4- dt
~dj

3

a

W

dx

dx dx

2

di

dx 2

/d \ 2 2 (&±\

32

2

dx

d +

(?l\ (—) dy =0 d

\\dx) dx I

\\dx) dx /

2 d2,P+ P* dL64> ,0*4, d2<j> 0*4> Ui ++ dx4 + dy2 ++ U2 dxdt d^i w dy w W ==0- 0

(67)

(67)

iction. The system (67) is not overdetermined since 114,1*5,1*6 may vanish without restriction. From (66) and (67) it follows that d2 u == Vl1 2 ^ 12 nl
uU2

d4>

(d24>\ dx2 4 3 d4> Wdx \dx )

8P4>

+4 3 *+f %_ + -d% w~ +

dx

dx (d4) d dy dy d4> \dx/ \dx)

d dx

2

1

++

\ dx )

dx

and

(68) (—) dy

i\dx) ==°0 d4>

\dxl

/d4>\

(dt

% + <**} +5

2

\

dy d<j>

\dx) \dx)

\\dx 5x

(69) (69)

=0

(70)

j)

where 2 (d24>\ (d24>\ 2 2 d dx i _ i1 ML. dx /{4>;x} < *. x -i. = A & Xf x ~ 2 d4> dx d4> ' '- dx dj_ 2 \ W ' dx ) \dx 1

V dx /

n\\ (71)

l

;

V dx /

ivative Equation (70) is the modified equation formulated in terms of the Schwarzian derivative ariant (71) and (69) is a Miura transformation from (70) to (63). Equation (70) is invariant under the Moebius group aip + b (72) 4> =

* = °P^ Y

where ad — be ^ 0.

□

aj} + d

aj> + d

( 72 ) K

'

115 Example

7: T h e Schwarzian

Korteweg

de Vries

equation

dip dt + {fa x} == A ■§^ {fax} \

(73) (73)

di> dx

where

93V>

{fax} : _ c V

3 "2

(P*\

2

dx*

di\> dx <9i V dx 1 is the Schwarzian derivative, has singularities of the form

== --4>-J

I

/

(75)

3=0

with arbitrary ,u0,Ul if t h e non-characteristic condition

/o0

dx a^-al*

(76) (76)

is verified. If 9 ^ / 8 a : = 0 the expansion about the characteristic manifold has the form orm

Hx,t) == m #*,*)

+^ l ^

(77) (77)

j=0

ition where <j>(x,t) = x - x0(t) and fa = fa(t). T h e Schwarzian Korteweg de Vries equation olds. has single valued expansions about the characteristic and noncharacteristic manifolds. The Painleve property requires all movable singularity manifolds to be single valued, lued, whether characteristic or not. D Example

8: A derivative

Schwarzian

equation

oy

# f ++ lifa*}fc{**> == 00

w

(78) (78)

dx dx

has non-characteristic singularities of the form

fa[x,t) -*..)-*-£**<

m

(79) i=o i=o ding where , fa, fa, fa are arbitrary. Therefore, (78) has a single valued expansion depending tion. on t h e m a x i m u m number of arbitrary functions allowed for the order of the equation. However, about t h e characteristic manifold where dfa/dx = 0 CO

CO

fax,t) fax, t)= =f(t) f(t) + »¥4ak j=0 j=0 /t=0 k=0

(so)

U6CHAPTER

8. PAINLEVE TEST AND PARTIAL DIFFERENTIAL

EQUATIONS

where

.-J+irfi and = x — x0(t).

(81)

The expansions (80) are highly multiple valued as functionals of <j>.

a Example 9: An example of an equation with a non-constant resonance is the Rand equation u

(duY 2&u dx-*=3(m)

(82)

It has the leading order u = u0<j>a + ■ • ■ where

/d+\ 3 (a-l)(a-2) = 3

dt

d<j>

tj U .

(83)

\dx' This quadratic equation for a determines the leading order. Of course a is a non-constant functional of d/dt and d/dx. The resonance condition u = u0a + u 1 ^ a + r + ■ ■ ■

(84)

determines the resonances r = - 1 , 0, 4 - 3a.

(85)

It is the case here that one resonance, 4 — 3a, is a functional of the singular manifold, <j>.

Chapter 9 Partial differential equations In this chapter we consider partial differential equations in one space x and one time variable t. The dependent variable is u. All considerations are local. All maps are C°°. The invertible point transformation takes the form T(t,x)

= G(u{x,t),x,t), G(u(x,t),x,t),

X(t,x)

=

H(u(x,t),x,t) H{u{x,t),x,t) (la) (la)

U{X(x U(X(x,t),T(x,t)) ,t),T(x ,*)) = = = F(u(x,t),x,t) F{u{x ,t),x J) where

t,u(x ,0)) d(H{x, t,u{x,t))),G{x, d(H(x,t,u(^))),G(x,tMx,t))) /O. d(x,t) d(x,t)

d{H,G ,F) d(H,G,F) #0, d(x,t, d(x,t,u)u) '

(16)

Example: An example of such a transformation is the hodograph transformation = u(x,t), = u(x,t),

U(X{x,t),T{x,t)) U(X(x,t),T(x,t))

dX__dH_du 8H_ dX dH du dH dt' dt ~ du du dt+ dt

dGdu dT dT_dGdu du dt dt dt '~ du

[x,t)=t, T(x,t)=t,

X(x,t) X(x,t)

= x. x. =

(2) (2)

a a

From (1) we find that

dX _ dH_du dX_ dH du dx du dx du We introduce the notation DT(G) :

dG du dG ~ ~du~dt+ dt '

dH dH_ dx' ' dx

dT dT_ dGdu ~aZ dx ' du dx

dH du du dt

DT{H):

dH dt

+

dG dt

l(3a) a

dG dx

(36)

DT{F) :

>

dF du du dt

dF dt

(4a)

+ JX{F) : du dx di (46) DX{H) : DX(G) ■-- du dx du dx dx ' «m-££^(«) dx' «w-fre+& «<°)-£s s. dG du

mce

dG

dU__dU_dX dU dX dU L dt dt ~ ~ dX dX~dT

dH du

+

dH

dF du 8F dF dUdT_dFdu dU dT + dT dt ~ ~du~ du dtdt dtdt 1117 1 7

dFdu

dl

(5)

CHAPTER 9. PARTIAL DIFFERENTIAL

118

EQUATIONS

it follows that dU IdHdu . 8H\ d>( + dX V du dt dt )

dU (dGdu 8G\ _ dFdu d_F dT [du dt + dt) ~ du dt + dt '

+

(

'

Therefore dU

DT(F)-^DT(H)

dT Since

dU dUdX dx ~ dX dx

W(5 find

(7)

DT{G) +

dF dx

dUdT dFdu dT dx ~ du dx

(8)

that BU IdHdu dX \ du dx

dU (dGdu 8G\ 8H\ + + dx) ' dT { du dx dx )

dFdu

dF dx

~ du dx

(9)

Ciansequentl;y

Dxm-%.

dU

DX{G) (10)

dX DX{H) For higher derivatives it is convenient to introduce the operators d2G(du\2 d2Gdu dGd2u d2G nrTUrvrfm DT(DT(G)) = — ( - ] + n2 — - + - +— d2

DxmTic))-——— {

(

2

dG

°^

+

+

d2 du

—

d2

°

+

(11a)

°

+

'' ~ Bu dx dt dudx dt du dxdt dudt dx dtdx d2G (du\2 „ d2G du dGd2u 82G DX(DX(G)) = — ^ j + 2 — - + _ +—

(

(life) '

(lie)

where DX(DT(G))

= DT{DX{G)).

(12)

Analogously, we define the operators DT{DT{H)),...,

DX{DX{F)).

The derivative of (6) with respect to t yields ^L(DX(H))2 +^(DT(G))2

+ 2-^L.DT(G)DT(H) + ^DT(DT(G))

+

^DT(DT(H))

= DT(DT(F)).

The derivative of (9) with respect to t yields ^DX(H)DT(H)

+ -^~DT{G)DX{H)

+

^DT(DX(H))

(13a)

119 Q2 r r

+

fir/

o2 rr

dXdfDT{H)DX{G)

+

gflDT(G)DX(G)

+ —(DT{DX(G))

=

DT(DX(F)). (136)

The derivative of (6) with respect to i yields (13b). The derivative of (9) with respect to x yields ^{DX{H)f

+ 2-^fDX(G)DX(H)

+ ^(DX(G))2

+

+ ^(DX(DX(G))

^DX(DX(H))

= DX(DX(F)).

(13c)

Let us now give several examples: Example 1: Consider the inviscid Burgers equation du dt

du dx

(14)

Let T{x,t) = t,

X{x,t) = u{x,t),

U{X(x,t),T(x,t))

=x

(15)

be the invertible point transformation. Then G(u,t,x)

H{u,t,x)

= t,

= u,

F(u,t,x)

= X.

(16)

Therefore DT{G) = 1, DX{G) = 0,

vrm-% arm-*

DT{F) = 0

(17a)

DX{F) = 1.

(176)

Inserting (17) into (7) and (10) gives dudU dU dT ~~ ~~dtdX' Thus we arrive at

dU 1 du' dX ~~

du df~~x-

(18a)

(186)

The general solution of (18b) is given by U(X,T) = g(X) - XT

(19)

CHAPTER 9. PARTIAL DIFFERENTIAL

120

EQUATIONS

where g is an arbitrary smooth function. Inserting the invertible point transformation (15) into (19) yields the general solution of (14) (20)

tu(x, t) + x = g(u(x, £)). Example 2: Let T(x, t) = = t,t, T(x,t)

U(X(x,t),T(x,t)) = x U(X(x, t), T(x, ()) =

X(x,t) X{x, t) = u(x, u(x,t), t),

(21)

and

du dd22uu du du du dx' dt dx2 1- u—. dx We derive the partial differential equation for U(X,T). dT dT_ , dt ~ ' and

dT n <^_ dx

From (21) we obtain dX _ du dx ~ dx'

dX dX__dudu

— dx =~ 0,'

dt ~ ~W dt' ~dt

dU dU dX dV__dVdX_ dx ~= dX dx

Therefore

(22)

+

dU du dX dx

dUd£&T_ dT = dT dx ~- 1 '

(23) (24)

1.

(25)

Analogously, dUdXL dU__dU_dX du dt ~ dX dt and therefore

dUdu dUdu dX dt+

+

dUdT dU9T_Q — II dT dt dt ~

dU dU_ — dT ~

(26) (27)

II

From (25) we obtain

It follows that

(d22U dX dX 2 [dX dx \ax

+

+

d (dU du\ = 0. dx [dXdx) dx [dXdx)~~

(28)

d2U 8T\ dT^ dXdTdXj dXdTdXJ

(29)

, du du dU_d^u dUd2u _ 1— + dx dX dX dx* dx* ~— 0. dx

Consequently d2U (S>u\2 dUd2u + 22 + 2 2 dX [dx) dXdx dXdx ~ dX [c'xj

(30)

To summarize:

"

v -_ A\ '

dU du _ Qrp dT dt dV m~~dU_' dX

dx

du du 11 = dU> dx 5x" W ' dX

dx

2 dd2uU

dx* dx^

=

d22U dX22 dX 3 ~ (du\ fdU\3'

[ax) [ax)

(31)

121 Inserting (31) into (22) gives (dU_\2dU

(dU\2

_ SHJ_ _

(32) (32)

[ax) dT - ax* \dx) ■ Example 3: Let T(x,t) = t,

X(x,t)

= J—jds,

U(X(x,t),T(x,t))

= u(x,t).

(33)

Let

22 du du „d , , 2d uu (34) dt dx2' We show that U(X, T) satisfies the linear diffusion equation. From (33) we obtain

dT

,

dt = 1, dX dt

dT „ „ - = 0, dx

dX 1 -dx S - : u(x, t)

-j™»-

J u2(s,t)

du dx

(35)

From (33) we find dU__dU_dX_

dx ~ dX dx Therefore

+

dU_dT_du

((36)

dflh ~ dx'

'

dU du dX=Udx-

(37) (37)

Analogously, dU__9U_dX_ dt ~ dX dt and therefore

+

dU_dT_du dT dt ~ dt

((38)

'

dU du dU du dX dx + dT' = ~df

(39)

From (37) we obtain

(duY 1

d fdU\

d2u

(40)

U

dx [dXj ~ \dx) + oV2' It follows that

(d2UdX d2U dT\ 2 \dX dx ' dXdTdXj

(du\2

[dxj

+U

d2u

dx2

(41)

Since ^

OX

=

0

(42)

CHAPTER 9. PARTIAL DIFFERENTIAL

122

EQUATIONS

and

(43)

** = I ox

we obtain d2U dX2

u 2

( du\

+u

qd

u

....

(44)

—(&} w-

Inserting (33), (37), (39) and (44) into (34) gives

(45)

*L = *L.

dT dX2 Equation (45) is the linear diffusion equation. □

K

'

Example ^: We consider the Legendre transformation and linearization. Consider the nonlinear partial differential equation (du\ \dt)

d2u (9u\ dx2~+\dx')

d2u ~d¥~

dudu d2u _ o7i~diarx~dt~

'

. . '

(

We show that (46) can be linearized with the following transformation du e(a»,«) = dx

iU:,t)-x(e.,v)

(47a)

du

(476)

dW

= de

t(e, ») =

(47c)

dW

(474)

drj

u(x, t) + W(e{x, t), t)(x, 0 ) = xe(x, t) + Uj(x, t).

(47e)

This transformation is called Legendre transformation. Thus (47) is not an invertible point transformation, since it involves derivatives. For the sake of completeness we study this transformation here. Finally we give the solution to the linearized equation. Since e = ^ ( * ( e , ,),*(*,,))

(48)

we obtain taking the derivative with respect to t d2udx 1 ~= dx2 de

+

d2u dt d2ud2W dt ~ dx22 de dxdt de dt22

+

d2u d2W dxdt dedn dxdtdcdr)

( (49)

'

123 where we have used (47c) and (47d). Analogously, we find d2u d2W d^cPW

dd22uud d2W

(50a) (50a)

0 = dxdt dt2 + dt2 dedn dxTt-w Wdefr, 2 22 2 _dd2ud ud W W d2u d
(506) '

{

(50c)

If we set p

d2ud2u (d2uV ^dT W-\d-x-d-t)

(51) (5 1)

2

we obtain PW_(PW_ (d2W\2 de2 dn2 \dedn)

_1 p'

(52)

Inserting these expressions into (46) we obtain the linear partial differential equation 2

2d

£

W

„ d2W

+2

^ ^

+i?

,d2W

=0

„

(53)

V -

Since (53) can be written in the form 2 a\ ( a ad\ \ (( a d d\2 Id e— + TI-Z- U - ^ + ' / T ^ )W = 0 \ \ dt %) \ de 'dr)J /

(54)

we find that the general solution of (53) is given by

^,,) = G0+^^) where G and H are two arbitrary smooth functions.

(55) (55)

□

Example 5: The nonlinear partial differential equation ^

-

^

+ fi(x + t)f2(x-t)Smu

=0

(56) (56)

ooth is an extended one-dimensional sine-Gordon equation, where / j and / 2 are two smooth functions. Let X(x,t):= j h(s)ds

57a) (57a)

x-t

T(x,t):=

j f2(s)ds

;576) (57b)

124

CHAPTER 9. PARTIAL DIFFERENTIAL

EQUATIONS (57c) (57c)

U(X(x, t), T(x, 0 ) := u{x, t). We show that U(X, T) satisfies the sine-Gordon equation 4

au-

dTdX

(58)

= sin I/.

From (57a) and (57b) we obtain

ax fc =/!(« + «),

ax

ar

,,

™=fi(x = / . ( . ++ti), ), l r

Since

dU dX dU dU__WdX_ dx ~ 8X dX dx

we obtain using (59)

Analogously,

,,

* = ' • < - -*)

(59a)

dT ^ = -f -f22(x(x-t). -1).

(596) (596)

du dUdU_dT_du dT dT dx dx ~ dx

((59c) C)

~di~

+

dU , , dU,, ay/i(* + 0 + ^ / . ( « -

au - * >

=

*

(60)

■

at/ at/ ax at/ ar au dU__dU_dX_ dUdT_du + at " ' dx dt ' ar at " " a* at ~ ax at dT at ~ at

and

l(61)

'

du dU r, at/,, ayA(» + *)--gfM*EF<W+O-»M-O-£-

62 (62)

aa /at/ , a dUt /r, , \A a d2iuu / a t / , .. ^(^/^ + t) + - / 2 ( x - t ) ) = ^ .

(63) (63)

-

*

>

=

*

<>

■

From (60) we find

Therefore d2U , 2Jl 2 g-jph+

ax

d2U Jlj2 dXdT fihi

^ axdT

2 a2u2f2 2j2 2 + at/ d/i ++ at/ <*/2 a u 2 ^ ar ^ ax ax ds <{« arards* ~ dx ax' " 2

v ; (64)

Analogously, from (62) we obtain

a /ait,,

N /l(l + i )

at/,,

65 (65)

2 (aT -ar - *V a* ' Ba 7(§«■+•>-£«■-*)-£•

<>

/2(l

Therefore 2 agg^ t/, 2 2Jl 2 1

dX a* '

d2U W flfi

"dXdT dXdTJlj2

d2u 2 a^4fi dudh , dudf gt/ M/4f»2 2h

' dT* dT Ji ^ dX ds+

^a2uu2

dT dt ' " at2"

,(66) fifn w

125 Consequently, we obtain

d2U - S[- sin u U = 0. 1 1 dTdX » dfdx= °- °

, , (67) 67

4A

< >

Example 6: Consider the linear partial differential equation du &U = ' dt3 dt '~b¥ m and the invertible point transformation T(x,t) X(x,t) u(x,t), T(x,t)=t,= t, X{x,t) = = u(x,t),

(68) (68)

U(X(x,t),T(x,t)) U(X(x,t),T{x,t))

= x. = x.

(69) (69)

Then we obtain dT dT_ dt~' dt and

8T ^L-Q dx ~

dX dX__dudu ~~5t dt ~ dt' dt'

n

'

dX _ du du dX dx ~ dx

dU dU dX dU dT _ dx dU_ _ dU_dX_ dU_dT dx _ 1. dx dx ~ dX dx +' dTdx~ dT dx ' dx~ '

Therefore

dU du dUdu dXdx dxYx 8U dU dT 8U_dU8T ~ dT dT dt dt ++ dt dt ~

It follows that

dU ^ dT dT

and

+ +

,

,„, ria),

~l =

(71a)

dU dX dU_<»£_ dX dt ~~ dX dt ~

dU du ^ = 0 dX dXdt~ dt

dU du du _ _ _QJ~ 221 dt~= - Wmr' a* dX dX

(70) '

(

[

716)

' (72) '

x

du _ _L_ 1 = ~dx~ ~dU' dx~ du_dX dX

{(73)

'

From (70) we find fd22UdX dUd (d U8X d22U U dT\du 8T\ du dU d22u _ =0 + 22 + [dX dx ' dXdT [dX dXdT dx) dx dX dX dx' dx2

((74)

2 2 dd22Uu (du\ dud (duV dUd2u u 2 2 + 2 2=Q + dX \dx) dXdx d-x \Tx) dxdx- ~ - '

(75) (75)

. '

From (71a) we find 22

dd uu _ dx2 2 "

dx ~

2 2 d&U(du\ U (du\2 22 dX [dx)

dd22UV

dx \dx) _ _jdxl_ dx2 dU (duV ou /du\3 dx dX

w) (dx)

m) (76) {

°>

CHAPTER 9. PARTIAL DIFFERENTIAL

126 and

S

en,

")"£ + S = °.

KdX)

dx2

EQUATIONS

dX2

Taking the dervative of (77) with respect to x we arrive at

(auV (d2U dX 2

(dXJ

(dX dx

(dU\3
d2U dT\ d2u dXdT dx / dx2

■+

^U_dX^ d3U dT av3 3 a — ~ avavr dX dx dX2dT a_ dx

\ dX I dx3

\

'

Consequently n

3

(dUY

(d2Udu\

d2u

2

(dUY&u +

8PUdu +

n

(ax) \d-x Tx) a ? (ox) 5? mai = °

( 79a >

or

ax 2 ; Therefore

+

+ (ftTJ vax; ftv 9i3 ^9x 3-9x = °-

_&U_dU_ of&uV
(796)

(80)

From (68) we obtain

_avm (vuV of _ dx*ax+ \ax2) au

'W

T^rY

(£)

ax

or

au

(81)

#u

(M\

QX3

AdJp)

[ax)

[ax)

(82)

Definition: Partial differential equations that can be solved either by an appropriate inverse scattering transform scheme or by a transformation to a linear partial differential equation are said to be linearizable.

127

T h e most well-known linearizable partial differential equations are of the form du

dnu

+/

, /

dn~1u\

du

«-^ rfe'-"'ftF0'

Definition: (83).

n 2

--

A partial differential equation is said to be semilinear

,

,

(83; (83)

if it is of the forrr form

There also exist linearizable equations of the form du

m

9"u , ,,9"u

=9{u)

d^

, //

+ f u

dn~ dn1-u\xu\ 'dx*-1)'

du du

[ 'd-x'--

71 > 2

(84;

where dg/du ^ 0. Definition: A partial differential equation is said to be quasilinear if it is of the form form (84). Example 1: A quasilinear linearizable equation is du __ dd f( 1 du\ du du\

1 du du

~dt dx '\u2 dx) +au^dx' u2 dx ~di-dx~\^dx~) where a is an arbitrary constant.

(85) (85)

Q

Example 2: The Harry-Dym equation du

~di~' is linearizable.

dx3 Wl2)

(86)

□

The most general equation of the form du du

~di

2 u ,, ,d ^d'u

. (1 du\ du\

._. (87)

which is linearizable is the equivalent to (85), which via an extended hodograph trans formation is mapped to the Burgers' equation. Definition: An extended hodographic transformation comprises of the change of the vari able u —> dv/dx = (u(x, t)) followed by a pure hodograph transformation (Clarkson et al 1989).

128

CHAPTER 9. PARTIAL DIFFERENTIAL

EQUATIONS

The Harry-Dym equation (86) can be transformed either into the Korteweg de Vries equation, or to the modified Korteweg de Vries equation. Definition: Two partial differential equations are equivalent if one can be obtained from the other by a transformation involving the dependent variables

(88) (88)

u -= = (v) 6(v)

and/or the introduction of a potential variable dv

u == — ,

ddx?

oorr

du du

. . (89) m

7— =v = v. Yx dx -

ransExample: The Burgers' equation is equivalent to the heat equation. We find the transformation from the Painleve analysis of the Burgers equation (see chapter 8). OI

The only third-order semilinear partial differential equations that are linearizablee are equivalent to the folowing six equations du

(Pu dPu

.„„.

du du

((90) 90)

Tt=dx^^d-x dt dx3 dx

,_,,

du dru (Pu du du du du ~dl~~= dV^ dx3 uYx+1dx Vt d-x dx du idx- fix du / 33 du d33uu 1 fdu\ 'du\ , ,. „ „.3u .) — du du du ■(aerfv + 3 = + fif 'dx+ + ^Tx dx) dt ~ dx ~ 8 I « 5 ? - 8 ( f c J +< ^ ^ f a T f a

nm -H(£)

33

2 22 du 3du fd2u\ u\ (i du dduu ■idu(d fduVY1 1 + ~di~lh?~ ' Idx \dx2)

2

3n,J[du\

»-5?-2s(a?J ( (^jj -2^((&J d3uu 33 (d2u\2 ^ 3 ~di~~ dx ' ' 2 \dx2) du 3u

=

where

(du\1 fduY \dx)

(duY 'du\ 1

•(

,__.

( (92) 92)

. .

(93) 93

( ) du + du ,„,. du (94) + A1\du (94) + 1

-^-

j ^ ^dx

)d-x

3„, Jdu\2*u\

■ - ^ . , ( S?j

(dP\2 2 = 4P3 = 4 P 3 --SP -£ -t \~du~)

(S)

(91) (9 1)

du +1

~dx~

(95)

(96) W

and a, /?,7,5, and e are arbitrary constants. Equation (90) is a linear partial differential equation sometimes referred to as the Airy equation in moving coordinates. Equation (91) is the Korteweg de Vries equation. Equation (92) is the modified Korteweg de Vries equation. Equation (93) is the Calogero-Degasperis-Fokas equation and (94) is unnamed and involves the Weierstrass elliptic function P(u).

12£ 129 The Calogero-Degasperis-Fokas equation can be put into rational form. Let U = e u/2 .

(97) (97;

Then

/du\ (dU\ dU &U cPU 33 d9 ^ 7 dU 3 dt m ~= dx a? '-' 22dx& \fj

2 2 WUi«u f. + (Q[/ ++ ?u- + ^ + l)?> &x ■ ,

rr2

nrr

2

^ U

(9s; (98)

If we apply a pure hodograph transformation to a partial differential equation in potential form (that is, an equation that does not depend explicitly on the dependent variable) that also does not depend explicitly on the independent variables, then the resulting equation is also in potential form with no explicit dependence on the independent variables. We now consider second-order quasilinear partial differential equations. Proposition: The most general second-order, quasilinear partial differential equation of the form du ,. N,d922uu ,J/ du\ du\ du ..„. (99) ~dt

» =*«>fei+ /(«.&]

(")

ation with dg/du ^ 0, which may be transformed via an extended hodograph transformation to a semilinear partial differential equation of the form

dS

=■

d22SS

„(. „(„ dS\ dS\

,lnn,

(100) (100)

\S, -r— dx^2 +hGG (s'dx)

df dx lif~

\ ' dxj

is given by . ,8 .d2u

8u du =9{u) g{u)

2

m m= Wdx-

2 .8u | +h{u) + ,H£ [dxj {Y [f-2){c-*) ) a(gg"

+

2 (du\ g'\(du\

Vl

X

,im,

(101) (l01)

X

where ' = d/du, and g(u) and b(u) are arbitrary functions that are twice and once differentiable, respectively. Furthermore, (100) is equivalent to

(dv\iPv~2d%v^u(Hdv(dv\ dv d1=(dv_Y \ — h r dt \dxj to22 h n \dx)

di~ \dx)

dx

((102) }

\dx)

^

'

which is transformed via a pure hodograph transformation to 1 l dU d2=Uvu_du H(BUY ) ). d_u ((*LY H 2

dT dT ~ dX dX2 ' -dx dX \ \\dXJ

[dx) j

For the proof we refer to Clarkson et al (1989).

)

((103) 103) v

'

CHAPTER 9. PARTIAL DIFFERENTIAL

130

EQUATIONS

Finally let us now consider third-order quasilinear partial differential equations. Proposition: The most general third-order, quasilinear partial differential equation of the form 2 du ,a33uu , (/ du du d2du\ u\ du ,. .d dg ,. „0 ,,„„. (104) 104 du T ~al B!

» 'Wfl?+/(-.fc'wj'

i^°

(

)

ilinear that may be transformed via an extended hodograph transformation to a semilinear partial differential equation of the form

80 au

=

22 a o\ a&u3o3 + „((T1 do du d o\ u,

+G '' dx

„„,

(105) (105)

{ dx'dx?)

wm ^wm) dT

is given by

du\ du „ / du\ d u +_„B // du\ du + ( du\ d u di =9{u)^fe 3+ B «(«'fc]fe u +*B y dx) dx Yt = aT / " \du\8u > d-x)Yx (gg"<" ("' Yx) a* g'\ dud u du du

2 2

, .Pu , .EPu

2

2

♦J ( f y

W\„( 8u\du (gg» g>\dud*u ~ 3 ~ J ~dx~!h? 3 5 ; 3{U'dx-)dx-+(T

(106)

prime denotes derivative with respect to u, and g(u) and B(u, ux) are arbitrary functions. :tions. Furthermore, (105) is equivalent to (dv\~3d~33vd3v

dv

(1)

dx*

+

(dv TJ(dv

d2dv\2v\

„„„, (107)

[dx'dx2)

that is transformed via a pure hodograph transformation to

2

(aoy1 m ((soy1 a22o (aoy3\ (duy ax \ax) ))• dT~ dx*~ \dx ) \ax) [ax) ~axH{{ax) ' ^ax^{dx) 3

2

do (a 2o\ du_ad*0u 3 3 (d u\ 6 dT = ~~ dx ' ' \ax )2

•§*((§r

(108) (108)

Kingston and Sophocleous (1991) classified all invertible point transformation between tween generalized Burgers equation of the form du

m

2

+

du ,adu2u + b(x dx '<>*? = 0.

(109)

The invertible point transformation relating (109) and dO TrdO — + U— +

2 /vrrnd 0 a(X,T)— = 0

(110)

must be of the form X X === cC3C5I + ClC!Cji c\t ++ cc22,, 3csx +

TT == clt c\t++ cc44, ,

TT = —3 " + , U Cl C

1/ =

C55 c

h Ci

(111) (111)

131 or xc3ce - d ^ = ^ 2 — — + c2,

1 T = c5--5—-,

C3C4U (112) [ / = c3C6(ut - z) - —— + c j1 (112)

where Ci,..., eg = const.. In both cases 6 a(X,X) = —^—.

(113)

For invariance it is necessary for the function a to satisfy the functional equation 0 a(x,t) { >T)-^~-

(114) (1U )

a X

Winternitz and Gazeau (1992) classified variable coefficient Korteweg-de Vries equations [uations

|| ++ //(( M ) ugg ++ ff((M)g g ==0,0, M ) u

ff M)

/ ##00, ,

, g^O *0

(115)

he clasinto equivalence classes under an invertible point transformation The aim of the clas [uations sification is to identify all classes of variable coefficient Korteweg de Vries equations lations, that have nontrivial symmetry groups, i.e. groups of local Lie point transformations, transforming solutions amongst each other. quation Calogero and De Lillo (1992) studied the Eckhaus nonlinear partial differential equation

i®¥. + ^!* 4. [ 2 ( JL|¥|» ) + |$|« - Vlx) ) * = 0. at

ox2

\

\dx

j

(116) (116)

)

They showed that (116) is C-integrable, namely it can be linearized, and solved,I, by an appropriate change of dependent variable. The linearizing transformation reads:: <^(x,i) = C(i)*(i,«)exp ^(x,i) C(i)*(i,r)exp

(117a)

j/ d ds\q{s,t)\A s|tf(M)|2

\ -1/2 / 2 2 2 J ds\(s,t)\ \ *(*,t) === {x,t) ds\(s,t)A #(x,t) <£(x,t) \clc(t)(t) + +2 2j ■W

1 dC(t) _i_^) = da(t) mm
■(«(<),

1 dC(t) dt '

C(t)

Jawmam)r

(1176) (1176)

/

0)

^3(*(a(t),t)^(a(i),f]

(117c) (inc) (117d)

132

CHAPTER 9. PARTIAL DIFFERENTIAL

EQUATIONS

and satisfies the linear partial differential equation

£ + g-?(.»-•.

(1.8)

Here the real function a(t) is arbitrary. In the special case in which ^{x,t) and <j>{x,t) vanish (sufficiently fast) as x —► —oo it is convenient to set a = — oo, so that the transformations (117) read simply 4(x,t) = CV(x,t)exp

I J ds\V{s,t)\2 j

/

*(*,*) = *(x,t)

»

|C| 2 + 2 |

(119a)

v -i/2

ds\{s,t)\2\

(1196)

with C a constant. The Unearizing transformation is independent of the external poten tial V(x) (provided it is real, as we assuming). The potential V(x) only enters in the linear Schrodinger equation (118).

Chapter 10 Difference equations We consider systems of autonomous ordinary difference equations defined on some domain U i n a Euclidean space TV. Let u = (Ul,u2,...,un)T.

(1)

The system of difference equations is given by u.i+i = ,(u,),

i = l,2,...,n

(2)

where & ara essumed do ob smooth functions snd d takes the ealues 0 , 0 , 1 , 2 , . . . h i h set of difference equations defines a m a p $ which maps u ( to u ( + 1 . We also use $ to define t h e system of difference equations (2) itself. A class of nonlinear difference equations can b e linearized with the help of invertible point transformations. As an example, consider ut + 1 = ut/(l+u(). Then the invertible point transformation ut = l/yt yields yt+1 = yt + 1. In the following we describe a linearization technique using Lie point symmetries. T h e Lie symmetry vector fields then provide the invertible point transformation which linearizes the nonlinear difference equation. Let us introduce t h e Lie point symmetry vector fields of difference equations (Maeda 1987). A continuous group acting on U is called a symmetry group of a difference equation $ if a n d only if each element commutes with $ . Any vector field of a symmetry group is called a symmetry operator. Let

d

Z = P'{u)lk'~fa,

(3)

be a smooth differential operator defined on U. Then, Z is a Lie symmetry vector field (2) if and only if *.Z = Z (4) holds, where # . denotes t h e differential of $ . In other words, the functional equations hold true 133

CHAPTER 10. DIFFERENCE

134

6(*(u)) = E & ( # ( « ) ,

i = l,2,...,n.

EQUATIONS

(5)

Equation (5) is coordinate-free. Let (j/,-) be another local coordinate in U. Suppose that $ and Z are expressed in terms of these coordinates as Ittt+i = ^i(yt))

» = 1,2,... ,n

(6a)

=S ffly) s-

Z--

(66)

When (5) holds, then ' and £' satisfy (5) automatically, and the converse is also true. We describe now a technique for determining the new coordinates by means of Lie sym metry vector fields. We also show that the dimension of the difference equations is reduced. When the difference equation is one-dimensional, it is shown that the derived difference equation becomes linear, and a summation procedure may be applied to ob tain the analytic expression of the solution. Let $ denote an n-dimensional difference equation given by (2). Throughout this section, we suppose that the new coordinate system introduced below covers a sufficiently large domain that $, suitably restricted, makes sense. We begin by stating the main theorem (Maeda 1987). Theorem: Suppose that a system of difference equations $ admits s Lie symmetry vector fields Za, such that for any integer I, (I < I < s — 1), [Za,Zp] = 0,

modZ1,...tZ,

(7a)

holds for arbitrary integers a, ft (1 < a, /3 < I + 1), Ta,nk(Z1,...,Zs)

=s

(76)

where [ , ] denotes the commutator. Then there is a local coordinate system (j/,-) in which the difference equation is expressed as Vit+i = Uit+gi{Vi+it,Ui+2t,---,ynt), Vat+i = 9a(y/3t),

i = l,...,s

a,/3 = s + l,s + 2,...,n.

(8a) (86)

Remark: The Lie symmetry vector fields of a system of difference equations form a Lie algebra. If Z\ and Zi are Lie symmetry vector fields of a system of difference equation, then obviously [Z%, Z2] is also a Lie symmetry vector field. Proof. It follows by induction with respect to a that Za are expressed as the following forms

Z =

^h

z

iM

^w/t}Mh h

Q=2 s

'-' -

(9)

135 From (5) it follows t h a t each cj>' must take the form (8). D The coordinates J/, are obtained by solving the differential equations Zayi = S,a,

a - l,...,mra(i,s),

i = l,...,n.

(10)

where <5,a denotes t h e Kronecker delta. T h e condition (7) means t h a t {Za} forms a completely integrable system and we choose a specific local coordinate system with respect to t h e foliation. Though the condition does not require t h a t {Za} spans a Lie algebra, it contains t h e case t h a t they form a solvable or a nilpotent Lie algebra. In particular, when s Lie symmetry vector fields commute with one another, t h a t is, form a commutative Lie algebra, the difference equation is expressed in a simpler form Vit+i = Vit + gi(yat), ya+i = 9c(ypt),

«= l,2,...,a

a,0 = s + l,s + 2,...,n.

(11) (12)

The coordinates are obtained by solving the equations Zayi = Sia.

(13)

It can be seen from (8) t h a t t h e original difference equation is composed of an (n — s)dimensional difference equation and s remaining equations (8a) in which a variable yu is contained as a forward difference. Therefore, if a solution of (8b) is obtained in some way, we can construct t h e whole solution step by step by way of a summation procedure. Let us now consider one-dimensional difference equations $ : ut+1 = (ut).

(14)

When 4> admits a Lie symmetry vector field, then (14) becomes a linear difference equa tion. In t h e one-dimensional case, (5) takes the form a{u)) = i{u)^(u). au

(15)

If we find a nontrivial solution (,, the new coordinate y is obtained by solving (10), that is

**-lwr

(16)

This is t h e invertible point transformation. Then, t h e theorem shows t h a t the difference equation is expressed with the coordinate y as a simple linear difference equation 2/i+i = yt + A

(17a)

136

CHAPTER 10. DIFFERENCE

EQUATIONS

where A is a constant. The general solution of (17a) is given by yt = At + C

(176)

where C is an arbitrary constant. A similar discussion is possible by starting from an equation slightly different from (15). Assume that f (*(«)) = Bi{u)^{u) (18) where <j> is the function appearing in (14) and B is a constant not equal to 1 or 0. With the use of a nontrivial solution £ of (18), we again introduce a new variable y by (16). Then, since (18) is equivalent to 1 dyt+i =: dyt fidyt (19) B i • (T where d denoting exterior differentiation, the difference equation is expressed in terms of y as yt+i = ^Vt + A

(20)

where A is a constant. The general solution of (20) is given by Vt

1\*

AB

-(M°+£x

(21)

where C is the constant of integration. In this case, also, we can write down the analytic expression of the solution of the original difference equation. Now we give some examples. We consider two nonlinear difference equations. Example 1: The so-called Riccati difference equation is given by

where a, 6, c, d € % with c ^ 0 and

aut + b *m = ^ cut + d cut + d

(22)

ad-bcjt 0.

(23)

In order to apply the similarity method, it is necessary to obtain a nonzero function £ which satisfies fau + b\ ad-be ,„,.

H^J=e(u)(^MF

(24)

We find that a quadratic polynomial

£(u) = cu2 + (d-a)u-b

(25)

137 satisfies (24). According to (16), i.e.,

y{u)=j

ds cs2 + (d — a)s — 6

(26)

the new variable y is obtained as follows (i) {a - d)2 + 46c > 0: y(u) =

1 a(uj — u 2 )

In

u — Ui

(27)

{ ( « i ) = £ ( u , ) = 0.

u — Ui

(ii) (a - d) 2 + 46c = 0: 1

y(u) =

(28)

{(«i)=0.

a(u — Ui)'

(iii) (a - df + 46c < 0: 1 J/(U) = — t a n aui

f 6

, /l /
"1

c

la — d 2c

1/2 (29)

In the respective cases, the difference equation is transformed to a linear equation of the form (20). T h e Riccati equation is transformed to a first-order linear equation. □ Remark: It is well known how to transform the Riccati difference equation (22) into a second-order linear difference equation. Let ut =

vt Wf+l

a 1- - . c

(30)

Then (22) transforms into _ c2vt + c(a + d)vt+x : : c6 — ad

Vtj.1 "4+2 —

•

L-l

(31)

Let (q,p) be a Cartesian coordinate system in 7£ 2 , and put Mi = 1Z2 \ 0 and M = RP1. There is a n a t u r a l projection 7r from Mi onto the one-dimensional real projective space M. T h e Riccati equation is a difference equation on the projective space M. If we define a mapping * on Mi by (q,p)^>(cp + dq,ap + bq) (32)

CHAPTER 10. DIFFERENCE

138

EQUATIONS

a mapping $ subject to o \P = $ o 7T

TT

(33)

is introduced on M where $ corresponds to the Riccati difference equation. A vector field Zi = (ap + fig)—+ (cp+ <*
(34)

on Aft is a symmetry operator of \P, it follows that the flow of Z = 7r.Zi

(35)

commutes with $. Example 2: Consider the difference equation $ defined on the unit interval / = [0,1] ut+i = aut(l — ut)

(36)

where a E [1,4] and u0 6 [0,1]. This difference equation (the so-called logistic equation) can show chaotic behaviour in the range a 6 [3.57,..., 4]. We now seek a £ which satisfies (18), where <j> = au(\ — u). We look for f among functions of the form (a fc-th order polynomial)1/*, Then the solution of (18) can be found in two cases: a = 2 and a = 4. (i) For a = 2, we have « u ) =1--u,

B =i

(37)

Since £ has a zero point | , we restrict the domain to I0 = [0, | ] and rewrite (36) on I0 formally. Using (16), the new variable is obtained as 3/(u) = - l n ( i - u )

(38)

and the difference equation becomes yt+1 = 2Vt-U2.

(39)

The general solution of (39) is given by yt = C2t-ln2

(40)

where C is the constant of integration. In terms of u we find that the solution of the initial value problem of (36) is given by ut = \-\{l-2u0f. Equation (41) satisfies (36) (a = 2) for every initial value in / .

(41)

139

(ii) For a = 4, we cannot find a single-valued function which satisfies (36) throughout /. Since

((u) = (u(l - u)Y"

(42)

and B = 1/2 satisfies (18) in To, we have the invertible point transformation y(u) = cos _1 (l - 2 a )

(43)

u = -(l-cosy)

(44)

and, accordingly,

by applying the procedure formally. Then (36) is transformed to cost/, + i = cos2y,.

(45)

When u is sufficiently small, the $ is indeed expressed as a linear equation 2/<+i = 2yt.

(46)

The general solution of (46) is given by y, = C2*

(47)

where C is the constant of integration. Therefore the solution of the initial value problem is given by ut = - - 5 cos(cos- 1 (l - 2«0)21).

(48)

Owing to the periodicity of the cosine function, this satisfies (36) for every initial value in /, and is the solution of the initial value problem. □ We can derive an interesting property from the above expressions. Let C be a complex plane and $ be a mapping on C defined by z - > z2.

(49)

Then $ has two important invariant curves, namely, the real axis and the unit circle whose centre is the origin. We define a mapping 7r from C onto the real axis such that z = q + ip^x=

-{l-g).

(50)

When we restrict ^ on each invariant curve, a mapping $ is induced by 7T 0 * = $ 0 7T

(51)

140

CHAPTER 10. DIFFERENCE

EQUATIONS

which corresponds to each difference equation treated above. If we put z = e1 in the latter case, then a flow given by 6 —> 0ec on the unit circle commutes with \t, and the motion projected from it on the real axis appears as a symmetry of $ . Finally let us give some other nonlinear difference equation which can be linearized. The difference equation u ( + 1 = u ( (3 - 4u 2 ),

woe [-1,1],

* = 0,1,2,...

(52)

is linearizable with the solution of the initial value problem ut = sin(3'sin _1 (u 0 )).

(53)

The difference equation u!+1 = 1 6 u ( ( l - u t ) ( l - 2 u t ) 2 ,

uo g [0,1],

t = 0,1,2,...

(54)

admits the solution u t = sin2(4tsin-1(^/iIo"))

(55)

of the initial value problem. The difference equation

1 - - Put)

ut+i = *Uti\~"fl~fUt\

1- -

jkvr

«o€[0,l],

fc2e[0,l],

t = 0 , 1 , 2 , . . . (56)

can be solved in terms of Jacobi elliptic functions ut = sn2(2<sn"1(v/u0", it), it).

(57)

In all these examples we also have to restrict ourselves to subdomains in the linearization. It seems probable that a symmetry group can always be associated with some difference equations which can be solved to obtain analytic expressions of solutions, even though the equations in the above examples are solved in a heuristic way. The similarity method seems to be a useful and systematic method for obtaining analytic expressions for difference equations.

Chapter 11 REDUCE programs Computer algebra is a powerful tool in the study of a wide class of problems in mathematics, physics, and engineering. A large number of problems in physics, mathematics and engineering involve the manipulation of large algebraic expressions. Here computer algebra is very useful. One of the most important general-purpose computer algebra systems available today is REDUCE (Hearn 1991, Steeb and Lewien 1992). REDUCE is mainly a software package for algebraic computations, but can also be used for numerical analysis. In particular we can combine algebraic computations with numerical analysis. REDUCE is based on LISP. It has a large number of capabilities. For example, we can integrate and differentiate functions with symbolic entries. REDUCE also provides the user with a complete programming language. REDUCE allows rule oriented programming. For example the derivatives of the Jacobi elliptic functions can be denned and then used in differentiations. We give a collection of REDUCE programs which are helpful in the study of invertible point transformations. Thus a large number of results in the book are checked using REDUCE. Notice that REDUCE does not differentiate uppercase from lowercase letters, i.e., ts = TS etc.. We use the following notation: us stands for u, uc for U, ts for t and tc for T. Sometimes we also use u instead of us. In the programs we mainly use differentiation. For example the REDUCE program depend us, ts; us := ts*ts; result := df(us,ts);

assumes that us depends on ts. We then set us = {ts)2. Thereafter we differentiate us with respect to ts. The output is obviously 2 * ts. The command df(us,ts,2); evaluates the second derivative of us with respect to ts. 141

142

CHAPTER 11. REDUCE

PROGRAMS

C h a p t e r 1: The first program chapll.our shows that (41) is the general solution of (32). '/.chapll.our; depend u s , t s ; A := ( ( t s + us + l ) " 2 ) * ( u s - t s + 3 ) ; B := d f ( A , t s ) ; d f ( u s . t s ) := (3*ts - us - 5 ) / ( - t s + 3*us + 7 ) ; B; In the program chapl2.our we show that (48) is the general solution of (42). '/,chapl2.our; depend u s , t s ; A := t s * t s - us*us - 1 - 2*k*us; B := d f ( A . t s ) ; d f ( u s . t s ) := ( 2 * t s * u s ) / ( t s * t s + us*us - 1); B; 2*k*us := t s * t s - us*us - 1; B; In the program chapl3.our we evaluate the right hand side of (53) for given functions F and G. '/chapl3.our; depend F, us, ts; depend G, us, ts; '/.depend us, ts; resl := (-df(F,ts)+A*di(G,ts))/(df(F,us)-A*df(G,us)); '/.Example; F := us"2; G := ts; resl;

143 Chapter 2: The expression on the left hand side of (19) of chapter 2 D(D(F))D(G)

- D(F)D{D{G))

= 0

is implemented in the REDUCE program chap21.our. We also separate out the dif ferent coefficients with the command coeffn(...). We then obtain (20)-(21) of chapter 2. '/.chap21.our; operator D; depend H, US, TS; depend F, US, TS; depend G, US, TS; depend US, TS; let df(US.US) = 1; for all H let D(H) = df(H,US)*df(US,TS)+df(H,TS); result := D(D(F))*D(G) - D(F)*D(D(G)); rl r2 r3 r4 r5

:= coeffn(result,df(US,TS,2),1); := coeffn(result,df(US,TS),3); := coeffn(result,df(US,TS),2); := coeffn(result,df(US,TS),l); := coeffn(coeffn(result,df(US,TS),0),df(US,TS,2),0) ;

The output is: RESULT := 2*DF(F,US,TS)*DF(G,US)*DF(US,TS)**2 + 2*DF(F,US,TS) *DF(G,TS)*DF(US,TS) + DF(F,US,2)*DF(G,US)*DF(US,TS)**3 + DF(F ,US,2)*DF(G,TS)*DF(US,TS)**2 - 2*DF(F,US)*DF(G,US,TS)*DF( US,TS)**2 - DF(F,US)*DF(G,US,2)*DF(US,TS)**3 - DF(F,US)*DF (G,TS,2)*DF(US,TS) + DF(F,US)*DF(G,TS)*DF(US,TS,2) + DF(F,TS, 2)*DF(G,US)*DF(US,TS) + DF(F,TS,2)*DF(G,TS) - 2*DF(F,TS)* DF(G,US,TS)*DF(US,TS) - DF(F,TS)*DF(G,US,2)*DF(US,TS)**2 - DF (F,TS)*DF(G,US)*DF(US,TS,2) - DF(F,TS)*DF(G,TS,2)$ Rl := DF(F,US)*DF(G,TS) - DF(F,TS)*DF(G,US)$

144

CHAPTER 11. REDUCE PROGRAMS

R2 := DF(F,US,2)*DF(G,US) - DF(F,US)*DF(G,US,2)$ R3 := 2*DF(F,US,TS)*DF(G,US) + DF(F,US,2)*DF(G,TS) - 2*DF(F, US)*DF(G,US,TS) - DF(F,TS)*DF(G,US,2)$ R4 := 2*DF(F,US,TS)*DF(G,TS) - DF(F,US)*DF(G,TS,2) + DF(F,TS, 2)*DF(G,US) - 2*DF(F,TS)*DF(G,US,TS)$ R5 := DF(F,TS,2)*DF(G,TS) - DF(F,TS)*DF(G,TS,2)$

In the program chap22.our we evaluate the functions A3, A2, Ai and A0 given by (21) for the functions F and G given by (35). '/,chap22.our; depend F, us, ts; depend G, us, ts; F := ts/us - k*ts*ts/6; G := 1/us - k*ts/3; delta := df(F,us)*df(G.ts)-df(G,us)*df(F,ts); lam3 := (df(G,us)*df(F,us,2)-df(F,us)*df(G,us,2))/delta; lam2 := (df(G,ts)*df(F,us,2)+2*df(F,us,ts)*df(G,us) -2*df(F,us)*df(G.us.ts)-df(F,ts)*df(G,us,2))/delta; laml := (df(G,us)*df(F,ts,2)+2*df(G,ts)*df(F.us.ts) -2*df(F,ts)*df(G,us,ts)-df(F,us)*df(G,ts,2))/delta; lamO := (df(G,ts)*df(F,ts,2)-df(F,ts)*df(G,ts,2))/delta;

The output of chap22.our is: DELTA := 1/US**3$ LAM3 := 0$ LAM2 := 0$ LAM1 := K*US$ LAMO := (K**2*US**3)/9$

The next program chap23.our shows that (33) is the general solution of (32).

145 '/,chap23.our; depend u s , t s ; us := l o g ( c l * t s + c 2 ) ; r e s u l t := d f ( u s , t s , 2 )

+ (df(us,ts))**2;

In c h a p 2 4 . o u r we evaluate the inverse transformation of (35). 51chap24. our; solve({UC = TS/US - k*TS*TS/6, TC = 1/US - k * T S / 3 } , { T S , U S } ) ; The o u t p u t is: {{TS=( - SQRT(6*K*UC + 9*TC**2) - 3*TC)/K, US=( - SQRT(6*K*UC + 9*TO*2))/(2*K*UC + 3 * T O * 2 ) > , {TS=(SQRT(6*K*UC + 9*TO*2) - 3*TC)/K, US=SQRT(6*K*UC + 9*TC**2)/(2*K*UC + 3*TC**2)}}$ Thus we have two solutions. The following program c h a p 2 5 . o u r shows that (20) can also be obtained from the Lagrangian function

f_HD(F)r 2

D(G)

since (46) can be derived from the Lagrangian function (47). '/,chap25. o u r ; depend L, ts, u, ud; depend F, ts, u; depend G, ts, u; L := (l/2)*(df(F,u)*ud+di(F,ts))-2/(df(G,u)*ud+df(G,ts)); LUD := df(L,ud); resl := df(LUD.ts) + ud*df(LUD,u) + udd*df(LUD.ud) - df(L,u); result := num(resl); rl := coeffn(result,udd,l); r2 r3 r4 r5

:= := := :=

coeffn(result,ud,3); coeffn(result,ud,2); coeffn(result,ud,1); coeffn(coeffn(result,ud,0),udd,0);

146

CHAPTER

11. REDUCE

The output is: RES1 := (2*DF(F,U,TS)*DF(F,U)*DF(G,U)*DF(G,TS)*UD**2 + 2*DF(F ,U,TS)*DF(F,U)*DF(G,TS)**2*UD - 2*DF(F,U,TS)*DF(F ,TS)*DF(G,U)**2*UD**2 - 2*DF(F,U,TS)*DF(F,TS)*DF( G,U)*DF(G,TS)*UD + DF(F,U,2)*DF(F,U)*DF(G,U)*DF(G ,TS)*UD**3 + DF(F,U,2)*DF(F,U)*DF(G,TS)**2*UD**2 - DF(F,U,2)*DF(F,TS)*DF(G,U)**2*UD**3 - DF(F,U,2)*DF (F,TS)*DF(G,U)*DF(G,TS)*UD**2 - 2*DF(F,U)**2*DF(G,U, TS)*DF(G,TS)*UD**2 - DF(F,U)**2*DF(G,U,2)*DF(G,TS )*UD**3 - DF(F,U)**2*DF(G,TS,2)*DF(G,TS)*UD + DF(F,U )** 2 *DF(G,TS)**2*UDD + DF(F,U)*DF(F,TS,2)*DF(G,U)*DF (G,TS)*UD + DF(F,U)*DF(F,TS,2)*DF(G,TS)**2 + 2*DF(F, U)*DF(F,TS)*DF(G,U,TS)*DF(G,U)*UD**2 - 2*DF(F,U)* DF(F,TS)*DF(G,U,TS)*DF(G,TS)*UD + DF(F,U)*DF(F,TS)* DF(G,U,2)*DF(G,U)*UD**3 - DF(F,U)*DF(F,TS)*DF(G,U,2) *DF(G,TS)*UD**2 + DF(F,U)*DF(F,TS)*DF(G,U)*DF(G,TS,2 )*UD - 2*DF(F,U)*DF(F,TS)*DF(G,U)*DF(G,TS)*UDD - DF( F,U)*DF(F,TS)*DF(G,TS,2)*DF(G,TS) - DF(F,TS,2)*DF (F,TS)*DF(G,U)**2*UD - DF(F,TS,2)*DF(F,TS)*DF(G,U)* DF(G,TS) + 2*DF(F,TS)**2*DF(G,U ) TS)*DF(G,U)*UD + DF( F,TS)**2*DF(G,U,2)*DF(G,U)*UD**2 + DF(F,TS)**2*DF (G,U)**2*UDD + DF(F,TS)**2*DF(G,U)*DF(G,TS,2))/(DF(G ,U)**3*UD**3 + 3*DF(G,U)**2*DF(G,TS)*UD**2 + 3*DF(G,U)* DF(G,TS)**2*UD + DF(G,TS)**3)$ RESULT := 2*DF(F,U,TS)*DF(F,U)*DF(G,U)*DF(G,TS)*UD**2 + 2*DF( F,U,TS)*DF(F,U)*DF(G,TS)**2*UD - 2*DF(F,U,TS)*DF(F,TS)*DF( G,U)**2*UD**2 - 2*DF(F,U,TS)*DF(F,TS)*DF(G,U)*DF(G,TS)*UD + DF(F,U J 2)*DF(F,U)*DF(G,U)*DF(G,TS)*UD**3 + DF(F,U,2)*DF(F,U )*DF(G,TS)**2*UD**2 - DF(F,U,2)*DF(F,TS)*DF(G,U)**2*UD**3 DF(F J U,2)*DF(F,TS)*DF(G,U)*DF(G,TS)*UD**2 - 2*DF(F,U)**2*DF(G ,U,TS)*DF(G,TS)*UD**2 - DF(F,U)**2*DF(G,U,2)*DF(G,TS)*UD** 3 - DF(F,U)**2*DF(G,TS,2)*DF(G,TS)*UD + DF(F,U)**2*DF(G,TS)** 2*UDD + DF(F,U)*DF(F,TS,2)*DF(G,U)*DF(G,TS)*UD + DF(F,U)*DF(F ,TS,2)*DF(G,TS)**2 + 2*DF(F,U)*DF(F,TS)*DF(G ) U,TS)*DF(G,U) *UD**2 - 2*DF(F,U)*DF(F,TS)*DF(G,U,TS)*DF(G,TS)*UD + DF(F,U)* DF(F,TS)*DF(G,U,2)*DF(G,U)*UD**3 - DF(F,U)*DF(F,TS)*DF(G,U,2) *DF(G,TS)*UD**2 + DF(F,U)*DF(F,TS)*DF(G,U)*DF(G,TS ) 2)*UD - 2* DF(F,U)*DF(F,TS)*DF(G,U)*DF(G,TS)*UDD - DF(F,U)*DF(F,TS)*DF(G ,TS,2)*DF(G,TS) - DF(F,TS,2)*DF(F,TS)*DF(G,U)**2*UD - DF(F ,TS,2)*DF(F,TS)*DF(G,U)*DF(G,TS) + 2*DF(F ) TS)**2*DF(G,U,TS

PROGRAMS

147 )*DF(G,U)*UD + DF(F,TS)**2*DF(G,U,2)*DF(G,U)*UD**2 + DF(F,TS) **2*DF(G,U)**2*UDD + DF(F,TS)**2*DF(G,U)*DF(G,TS,2)$ Rl := DF(F,U)**2*DF(G,TS)**2 - 2*DF(F,U)*DF(F,TS)*DF(G,U)*DF( G,TS) + DF(F,TS)**2*DF(G,U)**2$ R2 := DF(F,U,2)*DF(F,U)*DF(G,U)*DF(G,TS) - DF(F,U,2)*DF(F,TS) *DF(G,U)**2 - DF(F,U)**2*DF(G,U,2)*DF(G,TS) + DF(F,U)*DF(F,TS )*DF(G,U,2)*DF(G,U)$ R3 := 2*DF(F,U,TS)*DF(F,U)*DF(G,U)*DF(G,TS) - 2*DF(F,U,TS)*DF (F,TS)*DF(G,U)**2 + DF(F,U,2)*DF(F,U)*DF(G,TS)**2 - DF(F,U,2) *DF(F,TS)*DF(G,U)*DF(G,TS) - 2*DF(F,U)**2*DF(G,U,TS)*DF(G,TS) + 2*DF(F,U)*DF(F,TS)*DF(G,U,TS)*DF(G,U) - DF(F,U)*DF(F,TS)* DF(G,U,2)*DF(G,TS) + DF(F,TS)**2*DF(G,U,2)*DF(G,U)$ R4 := 2*DF(F,U,TS)*DF(F,U)*DF(G,TS)**2 - 2*DF(F,U,TS)*DF(F,TS )*DF(G,U)*DF(G,TS) - DF(F 1 U)**2*DF(G,TS,2)*DF(G,TS) + DF(F,U) *DF(F,TS,2)*DF(G,U)*DF(G,TS) - 2*DF(F,U)*DF(F,TS)*DF(G,U,TS)* DF(G,TS) + DF(F,U)*DF(F,TS)*DF(G,U)*DF(G,TS,2) - DF(F,TS,2)+ DF(F,TS)*DF(G,U)**2 + 2*DF(F,TS)**2*DF(G,U,TS)*DF(G,U)$ R5 := DF(F,U)*DF(F,TS,2)*DF(G,TS)**2 - DF(F,U)*DF(F,TS)*DF(G, TS,2)*DF(G,TS) - DF(F,TS,2)*DF(F,TS)*DF(G,U)*DF(G,TS) + DF (F,TS)**2*DF(G,U)*DF(G,TS,2)$ Note t h a t we have to take into account t h a t

A=

dGd£_dGd£ dGdF dGdF dt du du &t dt du du dt

In program c h a p 2 6 . o u r we evaluate the left hand side of (53). '/.chap26.our; operator D; depend H, US, TS; depend F, US, TS; depend G, US, TS; depend US, TS; let df(US.US) = 1;

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148

f o r a l l H l e t D(H) = df(H,US)*df(US,TS)+df(H.TS); result

:= D(D(F))*D(G) - D(F)*D(D(G)) + (D(G)*D(G)*D(G))*F;

The output of the program is: RESULT := 2*DF(F,US,TS)*DF(G,US)*DF(US,TS)**2 + 2*DF(F,US,TS) *DF(G,TS)*DF(US,TS) + DF(F,US,2)*DF(G,US)*DF(US,TS)**3 + DF(F ,US,2)*DF(G,TS)*DF(US,TS)**2 - 2*DF(F,US)*DF(G,US,TS)*DF( US,TS)**2 - DF(F,US)*DF(G,US,2)*DF(US,TS)**3 - DF(F,US)*DF (G,TS,2)*DF(US,TS) + DF(F,US)*DF(G,TS)*DF(US,TS,2) + DF(F,TS, 2)*DF(G,US)*DF(US,TS) + DF(F,TS,2)*DF(G,TS) - 2*DF(F,TS)* DF(G,US,TS)*DF(US,TS) - DF(F,TS)*DF(G,US,2)*DF(US,TS)**2 - DF (F,TS)*DF(G,US)*DF(US,TS,2) - DF(F,TS)*DF(G,TS,2) + DF(G,US) **3*DF(US,TS)**3*F + 3*DF(G,US)**2*DF(G,TS)*DF(US,TS)**2*F + 3*DF(G,US)*DF(G,TS)**2*DF(US,TS)*F + DF(G,TS)**3*F$ In program c h a p 2 7 . o u r we use the result from c h a p 2 6 . o u r to evaluate (56). To find the correct result we have to put df(us,ts) —> usd and df(us,ts,2) —> usdd. 5£chap27.our; depend F, US, TS; depend G, US, TS; F := (US*US)/2; G := TS; RESULT := 2*DF(F,US,TS)*DF(G,US)*usd**2 + 2*DF(F,US,TS) *DF(G,TS)*usd + DF(F,US,2)*DF(G,US)*usd**3 + DF(F ,US,2)*DF(G,TS)*usd**2 - 2*DF(F,US)*DF(G,US,TS)*DF( US,TS)**2 - DF(F,US)*DF(G,US,2)*usd**3 - DF(F,US)*DF (G,TS,2)*usd + DF(F,US)*DF(G,TS)*usdd + DF(F,TS, 2)*DF(G,US)*usd + DF(F,TS,2)*DF(G,TS) - 2*DF(F,TS)* DF(G,US,TS)*usd - DF(F,TS)*DF(G,US,2)*usd**2 - DF (F,TS)*DF(G,US)*usdd - DF(F,TS)*DF(G,TS,2) + DF(G,US) **3*usd**3*F + 3*DF(G,US)**2*DF(G,TS)*usd**2*F + 3*DF(G,US)*DF(G,TS)**2*usd*F + DF(G,TS)**3*F; The output of the program is: RESULT := (US**2 + 2*US*USDD + 2*USD**2)/2$

In c h a p 2 8 . o u r we derive (77b) from (77a).

149 '/.chap28. o u r ; o p e r a t o r s n , e n , dn; f o r a l l x l e t d f ( s n ( x ) , x ) = (cn(x))*(dn(x)); f o r a l l x l e t d f ( c n ( x ) , x ) = -(sn(x))*(dn(x)); f o r a l l x l e t d f ( d n ( x ) , x ) = -(k**2)*(sn(x))*(cn(x)); depend f, t c ; k := i ; f := ( ( U 2 0 / o m ) * s n ( o m * t c ) + U 1 0 * c n ( o m * t c ) * d n ( o m * t c ) ) / (1 + ( U 1 0 / V ) - 2 * ( s n ( o m * t c ) ) - 2 ) ; df(f,tc); for a l l y l e t (cn(y))*2 = 1 - ( s n ( y ) ) " 2 ; df(f,tc); The ouput is: (V**2*( - DN(TC*0M)**2*SN(TC*0M)**3*0M*U10**3 - 2*DN(TC*0M)** 2*SN(TC*OM)*CN(TC*OM)**2*OM*U10**3 - DN(TC*0M)**2*SN( TC*0M)*V**2*0M*U10 - DN(TC*0M)*SN(TC*0M)**2*CN(TC* 0M)*U20*U10**2 + DN(TC*0M)*CN(TC*0M)*V**2*U20 + SN( TC*0M)**3*CN(TC*0H)**2*QM*U10**3 + SN(TC*OM)*CN(TC* OM)**2*V**2*OM*U10))/(SN(TC*OM)**4*U10**4 + 2*SN(TC *0H)**2*V**2*U10**2 + V**4)$ (V**2*(DN(TC*0M)**2*SN(TC*0M)**3*0M*U10**3 - DN(TC*0M)**2*SN( TC*0M)*V**2*0M*U10 - 2*DN(TC*0M)**2*SN(TC*0M)*0M* U10**3 - DN(TC*0M)*SN(TC*0M)**2*CN(TC*0M)*U20*U10**2 + DN(TC*0M)*CN(TO0M)*V**2*U20 - SN(TC*0M)**5*0M*U10**3 - SN(TC*0M)**3*V**2*0M*U10 + SN(TO0M)**3*0M*U10**3 + SN(TC*0M)*V**2*0M*U10))/(SN(TC*0M)**4*U10**4 + 2*SN( TC*0M)**2*V**2*U10**2 + V**4)$

Obviously, (80) and (108) can also be easily implemented by slightly modifying program chap26.our. In c h a p 2 9 . o u r we find t h e expansion coefficients of (121). '/.chap29.our; operator a; for n:=l step 1 until 8 sum (x**(n))*a(n);

CHAPTER 11. REDUCE PROGRAMS

150

u := 1 + ws; b := df(u,x) - u*u - x; dO := coeiin(b,x,0); l i s t := solve({dO},{a(l)}); first(list); a(l) := part(ws,2); dl := coeffn(b,x,l); solve({dl},{a(2)}); part(ws.l); a(2) := part(ws,2); d2 := coeffn(b,x,2); solve «d2},{a(3)}); part(ws.l); a(3) := part(ws,2); d3 := coeffn(b,x,3); solve«d3},{a(4)}); part(ws,l); a(4) := part(ws,2); The output is as follows: A(l) A(2) A(3) A(4)

:= := := :=

1$ 3/2$ 4/3$ 17/12$

In chap210.our we show that the linearization ansatz (122) leads to (123). */.chap210.our; operator u, y; depend u, ts; depend y, ts; A := df(u,ts) - u"2 - ts; u := - (l/y)*df(y,ts); solve(A,di(y,ts,2));

The output is as follows: {DF(Y,TS,2)= - Y*TS}$

151 Tresse (1896) showed that the compatibility of (21) imposes the annulment of the fol lowing expression 02A, 0d*A2 d2A3 9A0 d\3 ^ - T - 2 ^ - ^ - + 3 - 5 ^ - + 6A 3 ^—+3A0-5

ou' d2A,

ouot d2Ax

at2 „32A0

SA, 9A3 5A, 9A2 3 A 3 - 5 - - 3 A 1 - 5 - - A 2 - 5 — + 2A 2 -r-- = 0

du au at . . 5A3 . . 5A0 „. 3A2

at d\0

ou 9A2

at SA,

-^ + 2 i^- 3 ^^ + 6 A o -ar + 3 A 3 ^r- 3 A o ^r- 3 A 2 ^7- A i ^r + 2 A i ^7 = °In program Tresse.our we implement these two expressions. Xtresse.our; depend F, u s , t s ; depend G, u s , t s ; d e l t a :=

df(F,us)*df(G.ts)-df(G,us)*df(F.ts);

13 := 12 :=

(df(G,us)*df(F,us,2)-df(F,us)*df(G,us,2))/delta; (df(G,ts)*df(F,us,2)+2*df(F,us,ts)*df(G,us) -2*df(F,us)*df(G,us,ts)-df(F,ts)*df(G,us,2))/delta; 11 := ( d f ( G , u s ) * d f ( F , t s , 2 ) + 2 * d f ( G , t s ) * d f ( F , u s , t s ) -2*df(F,ts)*df(G,us,ts)-df(F,us)*df(G,ts,2))/delta; 10 := ( d f ( G , t s ) * d f ( F , t s , 2 ) - d f ( F , t s ) * d f ( G , t s , 2 ) ) / d e l t a ;

r e s u l t l := d f ( l l , u s , 2 ) - 2*df(12,us,ts) + 3*df(13,ts,2) + 6*13*df(10,us) + 3*10*df(13,us) - 3*13*df(11,ts) 3 * l l * d f ( 1 3 , t s ) - 1 2 * d f ( l l , u s ) + 2*12*df(12,ts);

-

r e s u l t 2 := - d f ( 1 2 , t s , 2 ) + 2*df(11,us,ts) - 3*df(10,us,2) + 6*10*df(13,ts) + 3*13*df(10,ts) - 3*10*df(12,us) 3*12*df(10,us) - l l * d f ( 1 2 , t s ) + 2 * l l * d f ( 1 1 , u s ) ;

-

In program chap211.our we evaluate (7) for n = 2 (chapter 2, section 2.2). Xchap211.our; operator D; depend H, US, TS; depend F l , US1, US2, TS; depend F2, US1, US2, TS; depend G, US1, US2, TS;

152

CHAPTER 11. REDUCE PROGRAMS

depend US1, TS; depend US2, TS; l e t df(USl.USl) = 1; l e t df(US2,US2) = 1; for a l l H l e t D(H) = df(H,USl)*df(US1,TS)+ df(H,US2)*df(US2,TS)+df(H.TS); r e s u l t l := D(D(F1))*D(G) - D(F1)*D(D(G)); r l l := rl2 := rl3 := r l 4 := rl5 := rl6 := rl7 := rl8 := rl9 := rllO :=

coeffn(resultl,df(USl,TS,2),l); coeffn(resultl,df(US2,TS,2),l); coeffn(resultl,df(USl,TS),3); coeffn(coeffn(reaultl,df(USl,TS),2),df(US2,TS),l); coeffn(coeffn(resultl,df(USl,TS),l),df(US2,TS),2); coeffn(resultl,df(US2,TS),3); coeffn(coeffn(resultl,df(USl,TS),2),df(US2,TS),0); coeffn(coeffn(resultl,df(USl,TS),l),df(US2,TS),l); coeffn(coeffn(resultl,df(US2.TS),2),df(US1.TS),0); coeffn(coeffn(coeffn( resultl,df(US1.TS),1),df (US2.TS),0),df(US2.TS.2),0); r i l l := coeffn(coeffn(coeffn( resultl,df(US2 ) TS),l),df(USl,TS),0),df(USl J TS > 2),0); r l l 2 := coeffn(coeffn(coeffn(coeffn (resultl,df(USl,TS),0),df(USl,TS,2),0),df(US2.TS.2),0), df(US2,TS),0); result2 := D(D(F2))*D(G) - D(F2)*D(D(G)); s l l := s l 2 := s l 3 := s l 4 := sl5 := sl6 := sl7 := s l 8 := sl9 := sllO :=

coeffn(result2,df(USl,TS,2),l); coeffn(result2,df(US2,TS,2),l); coeffn(result2,df(USl,TS),3); coeffn(coeffn(result2,df(US1.TS),2),df(US2.TS),1); coeffn(coeffn(result2,df(USl,TS),l),df(US2,TS),2); coeffn(result2,df(US2,TS),3); coeffn(coeffn(result2,df(USl,TS),2),df(US2,TS),0); coeffn(coeffn(result2,df(USl,TS),l),df(US2,TS),l); coeffn(coeffn(result2,df(US2,TS),2),df(USl,TS),0); coeffn(coeffn(coeffn( result2,df(USl,TS),l),df(US2,TS),0),df(US2,TS,2),0); s i l l := coeffn(coeffn(coeffn( result2,df(US2,TS),l),df(USl,TS),0),df(USl,TS,2),0); s l l 2 := coeffn(coeffn(coeffn(coeffn

153 (resuU2,df(USl,TS),0),df(USl,TS,2),0),df (US2,TS,2) ,0), df(US2,TS),0); The output is:

Rll := DF(G,TS)*DF(F1,US1) - DF(G,US1)*DF(F1,TS) - DF(G,US1)* DF(F1,US2)*DF(US2,TS) + DF(G,US2)*DF(F1,US1)*DF(US2,TS)$ R12 := DF(G,TS)*DF(F1,US2) + DF(G,US1)*DF(F1,US2)*DF(US1,TS) - DF(G,US2)*DF(F1,TS) - DF(G,US2)*DF(F1,US1)*DF(US1,TS)$ R13 :=

- DF(G,US1,2)*DF(F1,US1) + DF(G,US1)*DF(F1,US1,2)$

R14 := - 2*DF(G,US1,US2)*DF(F1,US1) - DF(G,US1,2)*DF(F1,US2) + 2*DF(G,US1)*DF(F1,US1,US2) + DF(G,US2)*DF(F1,US1,2)$ R15 := - 2*DF(G,US1,US2)*DF(F1,US2) + DF(G,US1)*DF(F1,US2,2) - DF(G,US2,2)*DF(F1,US1) + 2*DF(G,US2)*DF(F1,US1,US2)$ R16 :=

- DF(G,US2,2)*DF(F1,US2) + DF(G,US2)*DF(F1,US2,2)$

R17 := - 2*DF(G,TS,US1)*DF(F1,US1) + DF(G,TS)*DF(F1,US1,2) DF(G,US1,2)*DF(F1,TS) + 2*DF(G,US1)*DF(F1,TS,US1)$ R18 := 2*( - DF(G,TS,US1)*DF(F1,US2) - DF(G,TS,US2)*DF(F1,US1 ) + DF(G,TS)*DF(F1,US1,US2) - DF(G,US1,US2)*DF(F1, TS) + DF(G,US1)*DF(F1,TS,US2) + DF(G,US2)*DF(F1,TS,US1))$ R19 := - 2*DF(G,TS,US2)*DF(F1,US2) + DF(G,TS)*DF(F1,US2,2) DF(G,US2,2)*DF(F1,TS) + 2*DF(G,US2)*DF(F1,TS,US2)$ R110 := - 2*DF(G,TS,US1)*DF(F1,TS) - DF(G,TS,2)*DF(F1,US1) + 2*DF(G,TS)*DF(F1,TS,US1) + DF(G,US1)*DF(F1,TS,2)$ Rill := - 2*DF(G,TS,US2)*DF(F1,TS) - DF(G,TS,2)*DF(F1,US2) + 2*DF(G,TS)*DF(F1,TS,US2) + DF(G,US2)*DF(F1,TS,2)$ R112 :=

- DF(G,TS,2)*DF(F1,TS) + DF(G,TS)*DF(F1,TS,2)$

Sll := DF(G,TS)*DF(F2,US1) - DF(G,US1)*DF(US2,TS)*DF(F2,US2) - DF(G,US1)*DF(F2,TS) + DF(G,US2)*DF(US2,TS)*DF(F2,US1)$

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154

512 := DF(G,TS)*DF(F2,US2) + DF(G,US1)*DF(US1,TS)*DF(F2,US2) - DF(G,US2)*DF(US1,TS)*DF(F2,US1) - DF(G,US2)*DF(F2,TS)$ 513 :=

- DF(G,US1,2)*DF(F2,US1) + DF(G,US1)*DF(F2,US1,2)$

514 := - 2*DF(G,US1,US2)*DF(F2,US1) - DF(G,US1,2)*DF(F2,US2) + 2*DF(G,US1)*DF(F2,US1,US2) + DF(G,US2)*DF(F2,US1,2)$ 515 := - 2*DF(G,US1,US2)*DF(F2,US2) + DF(G,US1)*DF(F2,US2,2) - DF(G,US2,2)*DF(F2,US1) + 2*DF(G,US2)*DF(F2,US1,US2)$ 516 :=

- DF(G,US2,2)*DF(F2,US2) + DF(G,US2)*DF(F2,US2,2)$

517 := - 2*DF(G,TS,US1)*DF(F2,US1) + DF(G,TS)*DF(F2,US1,2) DF(G,US1,2)*DF(F2,TS) + 2*DF(G,US1)*DF(F2,TS,US1)$

-

518 := 2*( - DF(G,TS,US1)*DF(F2,US2) - DF(G,TS,US2)*DF(F2,US1 ) + DF(G,TS)*DF(F2,US1,US2) - DF(G,US1,US2)*DF(F2, TS) + DF(G,US1)*DF(F2,TS,US2) + DF(G,US2)*DF(F2, TS,US1))$ 519 := - 2*DF(G,TS,US2)*DF(F2,US2) + DF(G,TS)*DF(F2,US2,2) DF(G,US2,2)*DF(F2,TS) + 2*DF(G,US2)*DF(F2,TS,US2)$

-

S110 := - 2*DF(G,TS,US1)*DF(F2 ) TS) - DF(G,TS,2)*DF(F2,US1) + 2*DF(G,TS)*DF(F2,TS,US1) + DF(G,US1)*DF(F2,TS,2)$ S i l l := - 2*DF(G,TS,US2)*DF(F2,TS) - DF(G,TS,2)*DF(F2,US2) + 2*DF(G,TS)*DF(F2,TS,US2) + DF(G,US2)*DF(F2,TS,2)$ S112 :=

- DF(G,TS,2)*DF(F2,TS) + DF(G,TS)*DF(F2,TS,2)$

Now we derive (19) from the Lagrangian function (21) (program c h a p 2 1 2 . o u r ) . '/.chap212.our; depend L, t s , u l , u 2 , u l d , u 2 d ; L := ( l / 2 ) * ( u l d " 2 + u2d"2) + o m * ( u l * u 2 d - u l d * u 2 ) +om*(ul*uld+u2*u2d)*tan(om*ts) +om"2*(ul"2 + u 2 " 2 ) * ( s e c ( o m * t s ) ) " 2 / 2 ; LUD1 := d f ( L . u l d ) ; r e s l := d f ( L U D l . t s )

+ uld*df(LUDl.ul) +

155 uldd*df(LUDl,uld)

-

df(L,ul);

LUD2 := d f ( L , u 2 d ) ; r e s 2 := d f ( L U D 2 , t s ) + u2d*df(LUD2,u2) + u2dd*df(LUD2,u2d) - d f ( L , u 2 ) ; for a l l q l e t

(sec(q))"2 = 1 + (tan(q))~2;

resl; res2; The output is: - U2D*0M + U1DD$ U1D*0M + U2DD$

chapter 3: We now consider the third derivative. We derive (9)-(10). In t h e program c h a p 3 1 . o u r we implement t h e left hand side of (16) and separate out the different coefficients. 5ichap31.our; o p e r a t o r D; depend H, U, TS depend F , U, TS depend G, U, TS depend U, TS; l e t df(U,U) = 1; f o r a l l H l e t D(H) = d f ( H , U ) * d f ( U , T S ) + d f ( H . T S ) ; r e s := D ( D ( D ( F ) ) ) * ( D ( G ) ) " 2 - 3*D(D(F))*D(D(G))*D(G) + 3*(D(D(G)))"2*D(F) - D(D(D(G)))*D(F)*D(G); rl r2 r3 r4 r5 r6 r7 r8

= coeffn(coeffn(res,df(U,TS,3),l),di(U,TS),l); = coeffn(coeffn(res,df(U,TS,3),1),df(U,TS),0); = coeffn(res,df(U,TS,2),2); = coeffn(res,df(U,TS),5); = coeffn(res,df(U,TS),4); = coefin(res,df(U,TS),3); = coeffn(coeffn(res,df(U,TS,2),l),df(U,TS)>2); = coeffn(coeffn(res,df(U,TS),2),df(U,TS,2),0) ;

CHAPTER 11. REDUCE PROGRAMS

156

r9 := c o e f f n ( c o e f f n ( r e s , d f ( U , T S , 2 ) , l ) , d f ( U , T S ) , l ) ; rlO = coeffn(coeffn(res,df(U,TS,2),1),df(U,TS),0); coeffn(coeffn(coeffn(res,df(U,TS),l),df(U,TS,2),0),dl(U,TS,3),0); rll = rl2 = coeffn(coeffn(coeffn(res,df(U,TS,3),0),dl(U,TS,2),0),df(U,TS),0); In program chap32.our we consider the special case of (14) where the function G does not depend on u. We find (15) as well as (16). '/.chap32.our; operator D; depend H, U, TS; depend F, U, TS; depend G, TS; depend U, TS; let df(U,U) = 1; for all H let D(H) = df(H,U)*df(U,TS)+df(H,TS); res := D(D(D(F)))*(D(G))"2 - 3*D(D(F))*D(D(G))*D(G) + 3*(D(D(G)))"2*D(F) - D(D(D(G)))*D(F)*D(G); rl r2 r3 r4 r5 r6 r7

= = = =

coeffn(res,df(U,TS,3),l); coeffn(coeffn(res,df(U,TS,2),l).clf(U,TS),l); coeffn(coeffn(res,df(U,TS,2),1),df(U.TS),0); coeffn(res,df(U,TS),3);

= coeffn(res,df(U,TS),2); = coeffn(coeffn(res,df(U,TS),l),df(U,TS,2),0); = coeffn(coeffn(coeffn(res,df(U,TS) 1 0),df(U,TS,3),0) J df(U,TS,2),0);

The output is: RES := 3*DF(F,U,TS,2)*DF(G,TS)**2*DF(U,TS) - 6*DF(F,U, TS)*DF(G,TS,2)*DF(G,TS)*DF(U,TS) + 3*DF(F,U,TS)*DF(G,TS) **2*DF(U,TS,2) + DF(F,U,3)*DF(G,TS)**2*DF(U,TS)**3 + 3*DF( F,U,2,TS)*DF(G,TS)**2*DF(U,TS)**2 - 3*DF(F,U,2)*DF(G,TS ,2)*DF(G,TS)*DF(U,TS)**2 + 3*DF(F,U,2)*DF(G,TS)**2*DF(U ,TS,2)*DF(U,TS) - DF(F,U)*DF(G,TS,3)*DF(G,TS)*DF(U,TS) + 3*DF(F,U)*DF(G,TS,2)**2*DF(U,TS) - 3*DF(F,U)*DF(G,TS,2)* DF(G,TS)*DF(U,TS,2) + DF(F,U)*DF(G,TS)**2*DF(U,TS,3) + DF( F,TS,3)*DF(G,TS)**2 - 3*DF(F.TS,2)*DF(G,TS,2)*DF(G,TS) DF(F,TS)*DF(G,TS,3)*DF(G,TS) + 3*DF(F,TS)*DF(G,TS,2)**2$

157 Bl := DF(F,U)*DF(G,TS)**2$ R2 := 3*DF(F,U,2)*DF(G,TS)**2$ R3 := 3*DF(G,TS)*(DF(F,U,TS)*DF(G,TS)

- DF(F,U)*DF(G,TS,2))$

R4 := DF(F,U,3)*DF(G,TS)**2$ R5 := 3*DF(G,TS)*(DF(F,U,2,TS)*DF(G,TS)

-

DF(F,U,2)*DF(G,TS,2))$

R6 := 3*DF(F,U,TS,2)*DF(G,TS)**2 - 6*DF(F,U,TS)*DF(G,TS,2)* DF(G,TS) - DF(F,U)*DF(G,TS,3)*DF(G,TS) + 3*DF(F,U)*DF(G,TS,2)**2$ R7 := DF(F,TS,3)*DF(G,TS)**2 - 3*DF(F,TS,2)*DF(G,TS,2)*DF(G, TS) - DF(F,TS)*DF(G,TS,3)*DF(G,TS) + 3*DF(F,TS)*DF(G I TS,2)**2$

In c h a p 3 3 . o u r we implement (33). Then we show t h a t (22) can be linearized. y.chap33.our; o p e r a t o r D; depend Q, t s , u , u d , u d d ; depend H, t s , u , u d , u d d ; f o r a l l Q l e t D(Q) = d f ( Q , t s )

+ ud*df(Q,u) + udd*df(q,ud)

+ H*df(Q,udd);

H := - 3 * u d * u d d / u + 3*udd + 3 * u d " 2 / u - 2*ud; result

:= ( l / 6 ) * D ( D ( d f ( H . u d d ) ) ) - ( l / 3 ) * d f ( H , u d d ) * D ( d f ( H , u d d ) ) -(l/2)*D(df(H,ud)) + (2/27)*(df(H,udd))"3 +(l/3)*df(H,ud)*df(H.udd) + df(H,u);

The o u t p u t is: RESULT := 0$

In c h a p 3 4 . o u r we derive (39) from (33). '/,chap34. our; operator D; depend Q, ts, u, ud, udd; depend H, ts, u, ud, udd;

CHAPTER 11. REDUCE PROGRAMS

158 depend f l , t s , u; depend f 2 , t s , u; depend f 3 , t s , u; depend f 4 , t s , u; depend f 5 , t s , u; depend f 6 , t s , u;

for a l l Q l e t D(Q) = d f ( Q . t s ) + ud*df(Q,u) + udd*df(Q,ud) + H*df(q,udd); H := -fl*ud*udd - f2*udd - f3*ud"3 - f4*ud~2 - f5*ud - f6; result := (l/6)*D(D(df(H.udd))) - (l/3)*df(H,udd)*D(df(H.udd)) -(l/2)*D(df(H,ud)) + (2/27)*(df(H,udd))-3 +(l/3)*df(H,ud)*df(H,udd) + df(H,u); rl r2 r3 r4 r5 r6

= = = = = =

coeffn(coeffn(result,udd.l),ud,l); coeffn(coeffn(result,udd,l),ud,0); coeffn(result,ud,3); coeffn(result,ud,2); coeffn(coeffn(result,ud,l),udd,0); coeffn(coeffn(result,ud,0),udd,0);

The output is: RESULT := ( - 18*DF(F1,U,TS)*UD**2 - 9*DF(F1,U,2)*UD**3 - 18* DF(F1,U)*UD**3*F1 - 18*DF(F1,U)*UD**2*F2 - 54*DF( F1,U)*UD*UDD - 9*DF(F1,TS,2)*UD - 18*DF(F1,TS)* UD**2*F1 - 18*DF(F1,TS)*UD*F2 + 9*DF(F1,TS)*UDD 18*DF(F2,U,TS)*UD - 9*DF(F2,U,2)*UD**2 - 18*DF(F2, U)*UD**2*F1 - 18*DF(F2,U)*UD*F2 - 63*DF(F2,U)* UDD - 9*DF(F2,TS,2) - 18*DF(F2,TS)*UD*F1 - 18*DF( F2,TS)*F2 + 27*DF(F3,U)*UD**3 + 81*DF(F3,TS)*UD **2 + 54*DF(F4,TS)*UD - 27*DF(F5,U)*UD + 27*DF(F5, TS) - 54*DF(F6,U) - 4*UD**3*F1**3 + 36*UD**3*F1 *F3 - 12*UD**2*F1**2*F2 + 18*UD**2*F1*F4 + 54*UD** 2*F2*F3 - 18*UD*UDD*F1**2 + 162*UD*UDD*F3 - 12*UD* F1*F2**2 + 36*UD*F2*F4 - 18*UDD*F1*F2 + 54*UDD*F4 - 18*F1*F6 - 4*F2**3 + 18*F2*F5)/54$ Rl := ( - 3*DF(F1,U) - Fl**2 + 9*F3)/3$ R2 := (DF(Fl.TS) - 7*DF(F2,U) - 2*F1*F2 + 6*F4)/6$ R3 := ( - 9*DF(F1,U,2) - 18*DF(F1,U)*F1 + 27*DF(F3,U) - 4*F1 **3 + 36*Fl*F3)/54$

159 R4 := ( - 6*DF(F1,U,TS) - 6*DF(F1,U)*F2 - 6*DF(F1,TS)*F1 - 3 * DF(F2,U,2) - 6*DF(F2,U)*F1 + 27*DF(F3,TS) - 4*F1**2*F2 + 6*F1*F4 + 18*F2*F3)/18$ R5 := ( - 3*DF(F1,TS,2) - 6*DF(F1,TS)*F2 - 6*DF(F2,U,TS) - 6* DF(F2,U)*F2 - 6*DF(F2,TS)*F1 + 18*DF(F4,TS) - 9*DF(F5, U) - 4*F1*F2**2 + 12*F2*F4)/18$ R6 := ( - 9*DF(F2,TS,2) - 18*DF(F2,TS)*F2 + 27*DF(F5,TS) - 54 *DF(F6,U) - 18*F1*F6 - 4*F2**3 + 18*F2*F5)/54$ Equation (75) is derived in program c h a p 3 5 . o u r . We insert ansatz (71) into (70). Then we obtain (75). '/.chap35. o u r ; operator u, y; depend u, ts; depend y, ts; A := di(u,ts,3) + 4*u*df(u,ts,2) + 6*u*u*df(u,ts) + 3*(df(u,ts))"2 + u"4; u := (l/y)*df(y,ts); A; The output is: DF(Y,TS,4)/Y$ Since t / / 0 we find t h a t

A

„

— - = 0.

CHAPTER 11. REDUCE

160

PROGRAMS

chapter 4: In chap41.our we evaluate (9) for (10), i.e. we find the determining equation for r\ and I for (10). '/.chap42. our; operator D; depend Q, ts, u, ud; depend H, ts, u, ud; depend eta, ts, u; depend xi, ts, u; for all q let D(q) = df(q.ts) + ud*df(q,u) + udd*df(q,ud); etal := D(eta) - ud*D(xi); eta2 := D(etal) - udd*D(xi);

H := udd + u*u*u; r l := xi*df(H,ts) + eta*df(H,u) + etal*df(H,ud) + eta2*df(H.udd); udd := - u*u*u; rl; The output is: Rl := 2*DF(ETA,U,TS)*UD + DF(ETA,U,2)*UD**2 + DF(ETA,U)*UDD + DF(ETA,TS,2) - 2*DF(XI,U,TS)*UD**2 - DF(XI,U,2)*UD**3 - 3*DF (XI,U)*UD*UDD - DF(XI,TS,2)*UD - 2*DF(XI,TS)*UDD + 3*U**2*ETA$ UDD :=

- U**3$

rl; 2*DF(ETA,U,TS)*UD + DF(ETA,U,2)*UD**2 - DF(ETA,U)*U**3 + DF( ETA.TS.2) - 2*DF(XI,U,TS)*UD**2 - DF(XI,U,2)*UD**3 + 3*DF( XI,U)*U**3*UD - DF(XI,TS,2)*UD + 2*DF(XI,TS)*U**3 + 3*U**2*ETA$ Next we show (program chap42.our) that Zi given by (30) is in fact a Lie symmetry vector field of (29).

161 '/,chap42.our; operator D; depend Q, ts, u, ud; depend N, ts, u, ud; depend eta, ts, u; depend xi, ts, u;

for a l l q l e t D(q) = d f ( q , t s ) + ud*df(q,u) + udd*df(Q,ud); e t a l := D(eta) - ud*D(xi); eta2 := D(etal) - udd*D(xi); N := udd + u; eta := u * c o s ( t s ) * s i n ( t s ) ; xi := ( s i n ( t s ) ) * ( s i n ( t s ) ) ; r l := xi*df(N,ts) + eta*df(N,u) + etal*df(N,ud) + eta2*df(N.udd); udd := - u; rl; The output is: Rl := UDD :=

- (3*SIN(TS)*C0S(TS))*(U + UDD)$ - U$

r l ; 0$ chapter 5: First we show (chap51.our) that (9) is a first integral of (8) if condition (10) is satisfied. '/,chap51. our; operator D; depend Q, t s , u, ud; depend F I , t s , u, ud;

162

CHAPTER 11. REDUCE PROGRAMS

f o r a l l q l e t D(q) = d f ( q , t s )

+ ud*df(q,u) + H*df(q,ud);

FI := ((ud + c l * u / 3 ) * 2 + ( u * 4 ) / 2 ) * e x p ( 4 * c l * t s / 3 ) ; r e s u l t 1 := D ( F I ) ; H := - c l * u d - c2*u - u"3; r e s u l t 1; result2

:= r e s u l t 1 / ( e x p ( 4 * c l * t s / 3 ) ) ;

r e s u l t 3 := c o e f f n ( r e s u l t 2 , u , 2 ) ; r e s u l t 4 := s o l v e ( r e s u l t 3 = 0 , c 2 ) ; result5 result6

:= c o e f f n ( c o e f f n ( r e s u l t 2 , u d , l ) , u , l ) ; := s o l v e ( r e s u l t 5 = 0 , c 2 ) ;

The output is: RESULT1 := (2*E**((4*TS*C1)/3)*(9*H*U*C1 + 27*H*UD + 9*U**4* Cl + 27*U**3*UD + 2*U**2*C1**3 + 15*U*UD*C1**2 + 27*UD**2*Cl))/27$ resultl; (2*E**((4*TS*C1)/3)*U*(2*U*C1**3 - 9*U*C1*C2 + 6*UD*C1**2 27*UD*C2))/27$ RESULT2 := (2*U*(2*U*C1**3 - 9*U*C1*C2 + 6*UD*C1**2 - 27*UD* C2))/27$ RESULT3 := (2*C1*(2*C1**2 -

9*C2))/27$

RESULT4 := { C 2 = ( 2 * C l * * 2 ) / 9 } $ RESULT5 := (2*(2*C1**2 -

9*C2))/9$

RESULT6 := { C 2 = ( 2 * C l * * 2 ) / 9 } $

Next we show ( c h a p 5 2 . o u r ) that /j given by (28) is a first integral of (22).

163 '/,chap52.our; o p e r a t o r D; depend Q, t c , u l , u 2 , u 3 , u d l , u d 2 , u d 3 ; depend F I , t c , u l , u 2 , u 3 , u d l , u d 2 , u d 3 ; f o r a l l Q l e t D(Q) = d f ( q , t c ) + u d l * d f ( Q , u l ) + H l * d f ( Q , u d l ) + u d 2 * d f ( Q , u 2 ) + H2*df(Q,ud2) + ud3*df(Q,u3) + H3*df(Q,ud3); FI := u d l * ( u l - t c * u d l ) + ud2*(u2 - t c * u d 2 ) + u d 3 * ( u 3 - t c * u d 3 ) ; result

:= D ( F I ) ;

HI := - C l * ( u 2 * u d 3 H2 := - C l * ( u 3 * u d l H3 := - C l * ( u l * u d 2

U3*ud2)/((sqrt(ul*ul+u2*u2+u3*u3))~3) Ul*ud3)/((sqrt(ul*ul+u2*u2+u3*u3))"3) U2*udl)/((sqrt(ul*ul+u2*u2+u3*u3))"3)

result; The output is: RESULT := - 2*TC*UD1*H1 - 2*TOUD2*H2 - 2*TC*UD3*H3 + U1*H1 + U2*H2 + U3*H3$ result;

0$

In program c h a p 5 3 . o u r we demonstrate that (36) is a first integral of (31). '/.chap53.our; operator D; depend Q, ts, u, ud; depend H, ts, u; depend xi, ts, u, ud; depend FI, ts, u, ud; for all Q let D(q) = df(Q,ts) + ud*df(Q,u) + H*df(q,ud); FI := df(xi,u) + df(D(xi),ud); resultl := D(FI);

164

CHAPTER 11. REDUCE

PROGRAMS

r e s u l t 2 := D(D(xi)) - xi*df(H,u) - df(H,ud)*D(xi); r e s u l t 3 := d f ( r e s u l t 2 , u d ) ; r e s u l t 4 := r e s u l t 1 - r e s u l t 3 ; The output is RESULT1 := DF(H,U)*DF(XI,UD,2)*UD + DF(H,TS)*DF(XI,UD,2) + 2* DF(XI,U,TS,UD)*UD + 2*DF(XI,U,TS) + 2*DF(XI,U,UD,2)*H*UD + 3* DF(XI,U,UD)*H + DF(XI,U,2,UD)*UD**2 + 2*DF(XI,U,2)*UD + 2*DF( XI,TS,UD,2)*H + DF(XI,TS,2,UD) + DF(XI,UD,3)*H**2$ RESULT2 := DF(H,U)*DF(XI,UD)*UD - DF(H,U)*XI + DF(H,TS)*DF(XI ,UD) + 2*DF(XI,U,TS)*UD + 2*DF(XI,U,UD)*H*UD + DF(XI,U,2)*UD**2 + DF(XI,U)*H + 2*DF(XI,TS,UD)*H + DF(XI,TS,2) + DF(XI,UD,2)*H**2$ RESULT3 := DF(H,U)*DF(XI,UD,2)*UD + DF(H,TS)*DF(XI,UD,2) + 2* DF(XI,U,TS,UD)*UD + 2*DF(XI,U,TS) + 2*DF(XI,U,UD,2)*H*UD + 3* DF(XI,U,UD)*H + DF(XI,U,2,UD)*UD**2 + 2*DF(XI,U,2)*UD + 2*DF( XI,TS,UD,2)*H + DF(XI,TS,2,UD) + DF(XI,UD,3)*H**2$ RESULT4 := 0$ chapter 6: REDUCE provides a user contributed package EXCALChy E. Schrufer for the manipulation of differential forms (see REDUCE user manual). After going into REDUCE one has to load the package with the command load.package excalc;

In chap61.our we show that the third equation of (16) is satisfied. 7.chap61; pform oml=l, om2=l, om3=l, al=l, bl=l, cl=l, rl=2, r2=2, r3=2; pform ts=0, u=0, p=0, h=0, a=0, b=0, c=0; fdomain h=h(ts,u,p); oml := a*d(ts); om2 := b*d(u) - b*p*d(ts); om3 := b*c*d(u) - b*c*p*d(ts) + (b/a)*d(p) - (b/a)*h*d(ts);

165

a l := ( l / a ) * d ( a ) ; b l rl

:= ( l / b ) * d ( b ) ; c l

:= d ( c ) + ( c / a ) * d ( a ) ;

:= d ( o m 3 ) ;

r2 := cl-om2 + ( b l - a l ) ~ o m 3 - ( b / a ) * ( < B ( h , u ) ) * ( d ( u ) - d ( t s ) ) (b*c + ( b / a ) * ( < B ( h , p ) ) ) * ( d ( p ) - d ( t s ) ) ;

-

r 3 := r l - r 2 ; The output is: Rl := ( - d ( A ) * d ( P ) * B + d(A)"d(TS)*B*H + d(B)"d(P)*A + d(B)~d (U)*A**2*C - d(B)"d(TS)*A**2*C*P - d(B)"d(TS)*A*H + d ( C)*d(U)*A**2*B - d(C)-d(TS)*A**2*B*P - d(P)~d(TS)*
chapter 7: We show that (40) passes the Painleve test. The expansion coefficient a4 and the pole position are arbitrary (constants of integration). operator a; for n:=0 step 1 until 8 sum (x**(n-l))*a(n); u := ws; b := df(u,x,2) - u*u*u; c := num(b); dO := coeffn(c,x,0); second(solve({dO},{a(0)})); a(0) := part(ws,2); dl := coeffn(c,x,l); solve({dl},{a(l)}); part(ws.l);

CHAPTER 11. REDUCE PROGRAMS

166 a(l) := part(ws,2); d2 := coeffn(c,x,2); solve«d2},{a(2)}); part(ws.l) ; a(2) := part(ws,2); d3 := coeffn(c,x,3); solve({d3},{a(3)}); part(ws.l); a(3) := part(ws,2); d4 := coeffn(c,x,4);

solve({d4},{a(4)}); part(ws,l); a(4) := part(ws,2); The output is:

DO := A(0)*( - A(0)**2 + 2)$ A(0) := SQRT(2)$ Dl := - 6*A(1)$ A(l) := 0$ D2 := - 6*A(2)$ A(2) := 0$ D3 := - 4*A(3)$ A(3) := 0$ D4 := 0$

A(4) := ARBC0MPLEX(1)$ chap8: We show that Burgers equation (4) passes the Painleve test. The function u2 is arbitrary. '/Burgers. our; depend depend depend depend depend depend

phi, x, ts; u, x, ts; uO, x, ts; ul, x, ts; u2, x, ts; u3, x, ys;

u := uO/phi + ul + u2*phi + u3*(phi)"2 + u4*(phi)"3;

167 A := d f ( u , t s ) + u * d f ( u , x ) B := A * ( p h i " 3 ) ; cO := c o e f f n ( B , p h i , 0 ) ; first(solve(cO,uO)); uO := p a r t ( w s , 2 ) ; c l := c o e f f n ( B , p h i , l ) ; c2 := c o e f f n ( B , p h i , 2 ) ; c3 := c o e f f n ( B , p h i , 3 ) ;

-

df(u,x,2);

first(solve(cl,ul)) ; u l := p a r t ( w s , 2 ) ; The output is A := ( - 3*DF(PHI,X,2)*PHI**5*U4 - 2*DF(PHI,X,2)*PHI**4*U3 DF(PHI,X,2)*PHI**3*U2 + DF(PHI,X,2)*PHI*U0 - 6*DF(PHI,X )**2*PHI**4*U4 - 2*DF(PHI,X)**2*PHI**3*U3 - 2*DF(PHI,X) **2*U0 + 2*DF(PHI,X)*DF(U0,X)*PHI - 2*DF(PHI,X)*DF(U2,X )*PHI**3 - 4*DF(PHI,X)*DF(U3,X)*PHI**4 + 3*DF(PHI,X)* PHI**8*U4**2 + 5*DF(PHI,X)*PHI**7*U3*U4 + 4*DF(PHI,X)* PHI**6*U2*U4 + 2*DF(PHI,X)*PHI**6*U3**2 + 3*DF(PHI,X)* PHI**5*U1*U4 + 3*DF(PHI,X)*PHI**5*U2*U3 + 2*DF(PHI,X)* PHI**4*U0*U4 + 2*DF(PHI,X)*PHI**4*U1*U3 + DF(PHI,X)*PHI **4*U2**2 + DF(PHI,X)*PHI**3*U0*U3 + DF(PHI,X)*PHI**3* U1*U2 - DF(PHI,X)*PHI*UO*Ul - DF(PHI,X)*U0**2 + 3*DF( PHI,TS)*PHI**5*U4 + 2*DF(PHI,TS)*PHI**4*U3 + DF(PHI, TS)*PHI**3*U2 - DF(PHI,TS)*PHI*UO - DF(U0,X,2)*PHI** 2 + DF(U0,X)*PHI**5*U4 + DF(U0,X)*PHI**4*U3 + DF(UO,X)* PHI**3*U2 + DF(U0,X)*PHI**2*U1 + DF(U0,X)*PHI*U0 + DF( U0,TS)*PHI**2 - DF(U1,X,2)*PHI**3 + DF(U1,X)*PHI**6* U4 + DF(U1,X)*PHI**5*U3 + DF(U1,X)*PHI**4*U2 + DF(U1,X) *PHI**3*U1 + DF(U1,X)*PHI**2*U0 + DF(U1,TS)*PHI**3 - DF (U2,X,2)*PHI**4 + DF(U2,X)*PHI**7*U4 + DF(U2,X)*PHI**6* U3 + DF(U2,X)*PHI**5*U2 + DF(U2,X)*PHI**4*U1 + DF(U2,X) *PHI**3*U0 + DF(U2,TS)*PHI**4 - DF(U3,X,2)*PHI**5 + DF( U3,X)*PHI**8*U4 + DF(U3,X)*PHI**7*U3 + DF(U3,X)*PHI **6*U2 + DF(U3,X)*PHI**5*U1 + DF(U3,X)*PHI**4*U0)/PHI**3$ B := - 3*DF(PHI,X,2)*PHI**5*U4 - 2*DF(PHI,X,2)*PHI**4*U3 DF(PHI,X,2)*PHI**3*U2 + DF(PHI,X,2)*PHI*U0 - 6*DF(PHI,X)**2* PHI**4*U4 - 2*DF(PHI,X)**2*PHI**3*U3 - 2*DF(PHI,X)**2*U0 + 2 DF(PHI,X)*DF(UO,X)*PHI - 2*DF(PHI,X)*DF(U2,X)*PHI**3 - 4*DF( PHI,X)*DF(U3,X)*PHI**4 + 3*DF(PHI,X)*PHI**8*U4**2 + 5*DF(

CHAPTER 11. REDUCE PROGRAMS

168

PHI,X)*PHI**7*U3*U4 + 4*DF(PHI,X)*PHI**6*U2*U4 + 2*DF(PHI, X)*PHI**6*U3**2 + 3*DF(PHI,X)*PHI**5*U1*U4 + 3*DF(PHI,X)* PHI**5*U2*U3 + 2*DF(PHI,X)*PHI**4*U0*U4 + 2*DF(PHI,X)*PHI**4* U1*U3 + DF(PHI,X)*PHI**4*U2**2 + DF(PHI,X)*PHI**3*U0*U3 + DF( PHI,X)*PHI**3*U1*U2 - DF(PHI,X)*PHI*U0*U1 - DF(PHI,X)*UO** 2 + 3*DF(PHI,TS)*PHI**5*U4 + 2*DF(PHI,TS)*PHI**4*U3 + DF(PHI, TS)*PHI**3*U2 - DF(PHI,TS)*PHI*UO - DF(U0,X,2)*PHI**2 + DF (U0,X)*PHI**5*U4 + DF(U0,X)*PHI**4*U3 + DF(U0,X)*PHI**3*U2 + DF(U0,X)*PHI**2*U1 + DF(UO,X)*PHI*UO + DF(U0,TS)*PHI**2 - DF( U1,X,2)*PHI**3 + DF(U1,X)*PHI**6*U4 + DF(U1,X)*PHI**5*U3 + DF(U1,X)*PHI**4*U2 + DF(U1,X)*PHI**3*U1 + DF(U1,X)*PHI**2*U0 + DF(U1,TS)*PHI**3 - DF(U2,X,2)*PHI**4 + DF(U2,X)*PHI**7*U4 + DF(U2,X)*PHI**6*U3 + DF(U2,X)*PHI**5*U2 + DF(U2,X)*PHI**4* Ul + DF(U2,X)*PHI**3*U0 + DF(U2,TS)*PHI**4 - DF(U3,X,2)*PHI** 5 + DF(U3,X)*PHI**8*U4 + DF(U3,X)*PHI**7*U3 + DF(U3,X)*PHI**6 *U2 + DF(U3,X)*PHI**5*U1 + DF(U3,X)*PHI**4*U0$ CO := DF(PHI,X)*U0*( - 2*DF(PHI,X) - U0)$ UO :=

- 2*DF(PHI,X)$

Cl := 2*DF(PHI,X)*( - DF(PHI,X,2) + DF(PHI,X)*U1 + DF(PHI,TS))$ C2 := 2*( - DF(PHI,X,TS) + DF(PHI,X,3) - DF(PHI,X,2)*U1 - DF( PHI,X)*DF(U1,X))$ C3 := - 3*DF(PHI,X,2)*U2 - 4*DF(PHI,X)**2*U3 - 4*DF(PHI,X)* DF(U2,X) + DF(PHI,X)*U1*U2 + DF(PHI,TS)*U2 - DF(U1,X,2) + DF( U1,X)*U1 + DF(U1,TS)$ Ul := (DF(PHI,X,2) - DF(PHI,TS))/DF(PHI,X)$

Since

g*° .2" we find that ((> satisfies the linear partial differential equation (heat equation)

d d24 dt =~= dx2'

169 chapter 9: In program c h a p 9 1 . o u r we evaluate DX(DX(H)),

DX(DT{H))

and

DT{DT(H)).

'/,chap91.our operator DX; operator DT; depend K, US, depend H, US, depend F, US, depend G, US,

XS, XS, XS, XS,

TS; TS; TS; TS;

depend US, XS, TS; let df(US.US) = 1; f o r a l l K l e t DX(K) = df(K,US)*df(US,XS)+df(K,XS); f o r a l l L l e t DT(L) = d f ( L , U S ) * d f ( U S , T S ) + d f ( L , T S ) ; r e s u l t 1 := DX(DT(H)); r e s u l t 2 := DX(DX(H)); r e s u l t 3 := DT(DT(H)); The output is RESULT1 := DF(H,US,XS)*DF(US,TS) + DF(H,US,TS)*DF(US,XS) + DF (H,US,2)*DF(US,XS)*DF(US,TS) + DF(H,US)*DF(US,XS,TS) + DF(H,XS,TS)$ RESULT2 := 2*DF(H,US,XS)*DF(US,XS) + DF(H,US,2)*DF(US,XS)**2 + DF(H,US)*DF(US,XS,2) + DF(H,XS,2)$ RESULT3 := 2*DF(H,US,TS)*DF(US,TS) + DF(H,US,2)*DF(US,TS)**2 + DF(H,US)*DF(US,TS,2) + DF(H,TS,2)$

c h a p t e r 10 In program c h a p l O l . o u r we show t h a t (22) can be linearized with t h e help of (30).

170

CHAPTER 11. RED UCE PROGRAMS

'/.chaplOl .our; operator u, v; R := u ( t + l ) * ( c * u ( t ) + d ) - (a*u(t)+b); u ( t ) := v ( t ) / v ( t + l ) + a / c ; u ( t + l ) := v ( t + l ) / v ( t + 2 ) + a / c ; solve(R,v(t+2)); The output is: {V(T + 2)=( - C*(V(T + 1)*A + V(T + 1)*D + V(T)*C))/(A*D - B*C)}$

Chapter 12 Jet bundle formalism In this chapter we give a survey of t h e jet bundle formalism. In chapter 6 we used the jet bundle formalism for the C a r t a n equivalence method. First of all, let us introduce t h e notation used. We study the system of partial differ ential equations within the jet bundle formalism. This obviously includes the ordinary differential equations as a special case. A triple (N,K,M)

is called a fibred manifold, if M and N are differentiable manifolds and n : N —> M is a surjective submersion. For example, let N = Km+n

= { ( * ! , . . . , xm,uu

M = Km = { ( a * , . . . , x m ) }

.. . , « „ ) } ,

and T-(xi,...,xm,uu...,u„)

=

(xu...,xm).

The so-called base manifold M represents the independent variables. In most cases in physics M = %i or an open subset of H4. T h e manifold N represents b o t h the dependent variables (the fields) and t h e independent variables. In most cases in physics TV will b e an open subset of the Euclidean space 7l 4 x 71". Now let d i m M = m and dimiV = n + m a n d let (x,-,Uj) (1 < i < m , 1 < j < n) denote the coordinate function defined by a fibre chart. Sections of N are defined as smooth maps s:M

-> N

such that 7T o s =

171

1M

CHAPTER 12. JET BUNDLE

172

FORMALISM

where 1 M is the identity map of M. We call the functions (x{,Uj) the fibre coordinates on N. The r-jet bundle Jr(N) is given by the equivalence classes of sections of N having rth order contact. The coordinate functions on JT(N) are denoted by (X,, UJt U,{, . . . , U,)iij 2 ,..t r /

where i, i 1 ( . . . , »'r € { 1 , . . . , m}, j € { 1 , . . . , n} and 1 < i2 < i2 < • ■ ■ < ir < m. u^...,,, corresponds to the partial derivative of Uj with respect to *{, ...£,„. The infinite jet bundle is denoted by J{N). Within the jet bundle formalism a system of partial differ ential equations of order r is defined to be a submanifold of Jr(N). Consider a system of partial differential equations of order r

d'u,

*"* (*W*). §*:>••' dx^dx ... i2

\ i=0 dxir)

(1)

where k = 1 , . . . , q. Within the jet bundle formalism we consider the submanifold Fk (xi, Uj, Uji,..., u^ ...,r) = 0

(2)

and the contact forms Oj = duj — Ujidii,

...

,0, tl ,...; r = d"jii...;,._, — Uj,1...,rjt
(3)

Throughout we use summation convention. Here d() denotes the exterior derivative of (). Consider now the vector field D{ defined on J(N)

D,

9 d dii

dd '' duj

■ ■ +

d duih.,

(4)

Uji^.. AT

l'r

the summation on the right hand side is restricted to 1 < ij < i2 < ■ • • < iT < m. D{ is sometimes called the operator of total differentiation. Together with (2) we consider all differential consequences DiFk = 0,

...,

DhDh...DirFk

= Q.

(5)

Let fi = dxi A dx2 A . . . A dxm

(6)

be the volume form on M. The quantity xm will play the role of the time coordinate. Definition. The (m — 1) differential-form u = fa(xi,ui,uji...)(—

[—

jJ fni )j

(7)

\dxa

defined on J(N) is called a conservation law of (2) if

o-roM = o

(8)

173 whenever s : M —* N is a solution to (2). js is the jet extension of s up to infinite order and 9/3x a _lfi denotes the contraction. We recall that we use summation convention. Remark: Another possibility for defining conserved currents is: The (m — 1) differential form u> given above is called a conservation law if a\v G J, where J is the differential ideal generated by Fk, DiFk,..., and the contact forms. We notice that the first definition is the more general one. For deriving conserved currents we consider the vector field Z = aj^- + b]^axj ouj

(9)

where a, and bj depend upon (i,, u,, Uj,,...). The prolongation up to infinite order of the vector field Z is given by

z = z + ca— + • • • + c^.iOUji

(io)

+• ■ •

OUJIL..,,

where Cji = Di(bj)

;

-

;

ujkDi(ak)

;

(n)

cj,,...,-, = Dir{eih.,.<,_,) - Uj, 1 ...i r _ 1 rA r (a,). Definition: The system of partial differential equations (1) which is described within the jet bundle formalism by (2) and the contact forms is called invariant under the vector field Z if L2Fk=0, fc = l,2 q (12) where = stands for the restriction to solutions of (1). Again we can give a definition which is not so general, but frequently used. Here the system of partial differential equations is called invariant if L^r7^ 6 J, Lt^i £ J> ■ ■ ■ i where J is the differential ideal generated by Fk, DiFk, and the contact forms. If the vector field Z leaves invariant the system of partial differential equations (1), then the vector field V = V~

+ Di{Vi)-£-

+ ■■■ + Dn...tr(U3)^—

+ ■■■

(13)

does so, where Uj = 6j - aiUji.

(14)

CHAPTER 12. JET BUNDLE

174

FORMALIS

In the following we consider the vector field V instead of Z for investigating the syi metries. The vector field (13) is called the vertical vector field. Assume that the vector field V is integrable to the corresponding group action u —+ exp(£V)u.

(1

Then, owing to invariance, a solution s : M —> N is carried into a new solution exp(eV] Theorem: Assume that the system of partial differential equations (1) is invariant und the vector field V. Let w be a conservation law of (1). Then Lyu) is also a conservatii law of (1). Proof. Let u be a conservation law. Since solutions s : M —* N are mapped in solutions by the transformation group which is generated by V we obtain (j exp(eV)s)*du> = 0.

(1

Owing to the identity jjj

exp( £ V»*u; = (j exp(ey) 5 )*Li^

(1

we find 0 = -yU exp(eV)s)'du> = (j exp(eV)s)'d{Lvuj). (1 de In the last step we have used the fact that the Lie derivative and the exterior derivati commute, i.e., dLvQ = Lvd{). (1 Setting e = 0, it follows that (js)'d{Lvuj) = 0 (2 which completes the proof.

References Aguirre M and Krause J 1984 J. Math. Phys. 25 210 Bluman G W and Cole J 1974 Similarity Berlin

Method for Differential

Equations

Springer,

Berkovich L M and Rozov N K 1972 Diff. Eq. 8 1609 Calogero F and DeLillo S 1992 J. Phys. A: Math. Cartan E 1908 Ann.

Gen. 25 L287

Ec. Normale 2 5 57

Cartan E 1924 Bull. Soc. Math.

France 52 205

Chern S-S 1939 Am. J. Math. 6 1 97 Clarkson P A, Fokas A S and Ablowitz M J 1989 SI AM J. Appl.

Math.

Cohen A 1931 An Introduction New York

Groups Stechert,

Davis H T 1962 Introduction New York

to the Lie Theory of One-Parameter

to Nonlinear

Differential

4 9 1188

and Integral Equations

Dover,

Duarte L G S, D u a r t e S E S and Moreira I C 1987 J. Phys. L701

A: Math.

Gen.

20

Duarte L G S, D u a r t e S E S and Moreira I C 1989 J. Phys. L201

A: Math.

Gen.

22

Duarte L G S, Euler N, Moreira I C and Steeb W-H 1990 J. Phys. 23 1457

175

A: Math.

Gen.

Duarte L G S, Moreira I C, Euler N and Steeb W-H 1991 Physica Scripta 43 449 Estevez P G 1991 J. Phys. A: Math. Gen. 24 1153 Euler N and Steeb W-H 1992 Continuous symmetries, Lie algebras and differential eq tions, Bibliographisches Institut, Mannheim Euler N, Steeb W-H and Cyrus K 1989 / . Phys. A: Math. Gen. 22 L195 Euler N, Steeb W-H, Duarte L G S and Moreira I C 1991 Int. J. Theor. 1267

Phys.

Fournier J D, Levine G and Tabor M 1988 J. Phys. A: Math. Gen. 21 33 Grissom C, Thompson G and Wilkens G 1989 / . of Differential Equations 77 1 Gromak V I 1978 Diff. Urav. 14 2131 Hojman S A 1992 J. Phys. A: Math. Gen. 25 L291 Hsu L and Kamran K 1988 Letters in Mathematical Physics 15 91 Ince E L 1956 Ordinary Differential Equations Dover, New York Jackiw R 1980 Ann. Phys. NY 129 183 Jung C and Scholz H-J 1988 Phys. A: Math. Gen. 21 2301 Junker G and Inomata A 1985 Phys. Lett. 110A 195 Kamran N, Lamb K G and Shadewick W F 1985 J. Differential Geometry 22 139 Katzin G H and Levine J 1983 J. Math. Phys. 24 1761 Kingston J G and Sophocleous C 1991 Phys. Lett. A 155 15 Kowalski K and Steeb W-H 1990 Carleman linearization World Scientific Publishir Singapore Leach P G L 1985 J. Math. Phys. 26 2510 Leach P G L and Mahomed F M 1985 Quaest. Math. 8 241 176

Leach P G L and Gorringe V M 1990 J. Phys. A: Math. Gen. 23 2765 Lewis H R 1968 J. Math. Phys. 9 1976 Lie S 1891 Vorlesunger uber Differentialgleichunger Teubner, Leipzig Maeda S. 1987 IMA J. Appl. Math. 38 129 Mahomed F M and Leach P G L 1985 Quaestiones Math. 8 241 Mahomed F M and Leach P G L 1990 J. Math. Analysis and Appl. 151 80 Moreira I C 1983 Suppl. Cie. Cult 35 252 Moreira I C 1984 Hadronic J. 8 475 Moreira I C 1990 Europhys. Lett. 12 391 Moreira I C and Ritter 0 M 1991 J. Phys. A: Math. Gen. 24 3181 Moreira I C, Ritter 0 M and Santos F C 1985 J. Phys. A: Math.

Gen. 18 L427

Ovsiannikov L V 1962 Gruppovye Svoystva Differentsialny Uravneni USSR Academy of Science, Novosibirsk Prince G E and Eliezer C J 1980 J. Phys. A: Math. Gen. 13 815 Prince G E and Eliezer C J 1981 J. Phys. A: Math. Gen. 14 587 Sarlet W, Mahomed F M and Leach P G L 1987 /. Phys. A: Math. Gen. 20 277

Steeb W-H and Euler N 1988 Nonlinear Evolution Equations and Painleve Test World Scientific Publishing, Singapore Steeb W-H Problems in Mathematical Physics, Volume I, Bibliographisches Institut, 1990, Mannheim Steeb W-H and Lewien D 1992 Algorithms and Computation with REDUCE, Bibli ographisches Institut, Mannheim 177

Tabor M and Weiss J 1981 Phys. Rev. A 24 2157 Tresse M A Determination des Invariants Ponctuels de I'Equation dinaire du Second Ordre y" = u(x,y,y'), S. Hirzel, Leipzig

Differentielle

Ward R S 1988 Nonlinearity 1, 671 Ward R S 1984 Phys. Lett. 102A 279 Weiss J 1984 J. Math. Phys. 25 2226 Weiss J 1986 J. Math. Phys. 27, 1293 Winternitz P and Gazeau J P 1992 Phys. Lett. A 167 246 Wulfman C E and Wybourne B G 1976 J. Phys. A: Math. Gen. 9 507

178

Index Anharmonic oscillator Bernoulli equation Bessel function Burgers equation Cartan method Charge monopole problem Commutator Conserved quantity Elliptic functions Ermakov system Euler-Lagrange equation Exact invariant Exterior product First integral Free-falling body Free-particle equation Grassmann product Harmonic oscillator Harry-Dym equation Hodograph transformation Hypergeometric differential equation Infinitesimal generator Invertible point transformation Inviscid Burgers transformation Jacobi identity Jet bundle formalism Kadomtsev-Petviashvili equation Kepler problem Korteweg-de Vries equation Lagrangian function Laguerre invariant Legendre transformation Lie algebra Lie symmetry vector field Logistic equation Lorenz-Dirac equation Lotka-Volterra model Maurer-Cartan equation Miura transformation Modified Emden equation

179

7, 15, 62 2 22 106 69 30 46 61 16 63 12 61 71 61 12 7 71 7, 13 127 28, 117 25 45 1 108 47 171 113 30 112 12 39 109, 122 46 45 138 39 89 74 114 23

35 90 87 87 116 141 136 22 63 7,49 93, 114 115 123 71 30

Modified Korteweg-de Vries equation Moebius transformation Painleve property Painleve test Rand equation REDUCE Riccati difference equation Riccati equation Runge-Lenz vector Second Painleve transcendents Schwarzian derivative Schwarzian Korteweg-de Vries equation Sine-Gordon equation Wedge product Vector product

180

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