Solution of Differential Equations by Means of One-Parameter Groups

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British Library Cataloguing in Publication Data

Hill, J. M. Solution of differential equations by means of one-parameter groups. 1. Differential equations I. Title 515.3'S 0A371 ISBN 0-273-08506-9

Library of Congress Cataloging in Publication Data

Hill, J. M. Solution of differential equations by means of one-parameter groups (Research notes in mathematics; 63) Bibliography; p. 1. Differential equations—Numerical solutions. 2. Groups, Theory of. I. T. II. Series. 0A371.H56.

515.3'5 82-621 ISBN 0-273-08506-9 AACR2.

Australian Cataloguing in Publication Data

Hill, J. M. (James M.) Solution of differential equations by means of one-parameter groups. Includes bibliographical references.

ISBN 0858968932. 1. Differential equations. 2. Groups, Theory of. I. Title. (Series: Research notes in mathematics; 63).

515.3'S

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JMHi11 University of Wollongong

Solution of differential equations by means of one -parameter groups

Pitman Advanced Publishing Program BOSTON LONDON. MELBOURNE

Preface

The trouble with solving differential equations is that whenever we are

successful we seldom stop to ask why.

The concept of one—parameter

transformation groups which leave the differential equation invariant provides the only unified understanding of all known special solution techniques.

In these notes I have attempted to present a fairly concise and

self—contained account of the use of one—parameter groups to solve differential equations.

The presentation is formal and is intended to appeal

to Applied Mathematicians and Engineers whose principal concern is obtaining solutions of differential equations.

I have included only the essentials of

the subject, sufficient to etiable the reader to attempt the group approach

when solving differential equations.

I have purposely not included all known

results since this would inevitably lead to unnecessarily reproducing large portions of existing accounts.

For example, for ordinary differential

equations, the account of the subject by L.E. Dickson, "Differential equations from the group standpoint", is still extremely readable and is recommended to the reader interested in pursuing the subject further.

For partial

differential equations the books by G.W. Bluman and J.D. Cole, "Similarity methods for differential equations", and by L.V. Ovsjannikov "Group properties of differential equations", contain several applications and examples which I have not reproduced here.

The first two chapters are introductory.

Chapter 1 gives a general

introduction with simple examples involving both ordinary and partial differential equations.

In Chapter 2 the concepts of one—parameter groups

and Lie series are introduced.

Just as ordinary methods of solving

differential equations often require a certain ingenuity so does the group approach.

In order to establish some familiarity with the group method I

have attempted to exploit our experience with linear equations. are aware that linear differential equations for the transformation

x1 =

f(x),

y1 =

g(x)y

devoted to implications of this result.

y(x)

Most of us

remain linear under

and Chapter 3 of these notes is

In Chapters 4 and 5 I have tried to

relate the usual theory for the group method with the results obtained in the third chapter.

In this respect these notes differ from most accounts of the

subject and I believe that a number of results given, especially in Chapter 3 are new.

The remaining two chapters are devoted to partial differential equations. For the most part the theory is illustrated with reference to diffusion related partial differential equations.

The theory for linear partial

differential equations is introduced in Chapter 6 for the classical diffusion or heat conduction equation and the Fokker—Planck equation. equations are treated in Chapter 7.

Non—linear

For partial differential equations the

group approach is less satisfactory since for boundary value problems both the equation and boundary conditions must remain invariant.

In these notes

we principally consider only the invariance of the equation and view the group method as a means of systematically deducing solution types of a given partial differential equation.

Although these notes appear as a research monograph they actually represent advanced teaching material and in fact form the basis of a post—graduate course given at the University of Wollongong for the past six years. therefore included numerous examples and exercises.

I have

In addition to the

exercises I have used the problems at the end of each chapter to conveniently

locate standard results for differential equations.

On occasions I have also

used these problems to include summaries of theory which is already adequately described in the literature. The existing theory of the solution of differential equations by means of one—parameter groups is by no means complete. the subject are highlighted in the text.

Many of the inadequacies of

When it does work it is very easy

and it is therefore an area of knowledge which every Applied Mathematician ought to be aware of.

Whatever the limitations of the group method may be,

it will always represent a profoundly interesting idea towards solving differential equations.

I hope these notes prove to be useful and complement

the existing literature.

James M. Hill, The University of Wollongong, Australia.

Acknowledgement

The author wishes to thank all of his students over the

past six years who have discovered various errors, spelling mistakes and omissions in a preliminary draft of these notes.

He is also grateful to Mrs.

Kerrie Gamble for her careful and thoroughly professional typing of the manuscript.

For

Desley,

and

Ruth.

Contents

1.

Introduction 1.1 1.2

2.

3.

First 4.1 4.2

4.3 4.4 4.5 4.6

5.

5.4

of standard linear equation8

First order equation y' + p(x)y = q(x) Second order homogeneous equation y" + p(x)y = 0 Third order equation y"+ p(x)y' + q(x)y = 0 Fourth order self—adjoint equation y" + [p(x)y']' + q(x)y = 0 Problems

order differential equation8 Infinitesimal versions of y' and y' = F(x,y) and the fundamental problem Integrating factors and canonical coordinates for y' = F(x,y) The alternative problem The fundamental problem and singular solutions of y' = F(x,y) Invariance of the associated first order partial differential equation Lie's problem and area preserving groups Problems

Second 5.1 5.2 5.3

groups and Lie series

One—parameter transformation groups Lie series and the commutation theorem Problems

Invariance 3.1 3.2 3.3 3.4

4.

Ordinary differential equations Partial differential equations Problems

One-parameter 2.1 2.2

1

and higher order differential equation8

Infinitesimal versions of y" and y" = F(x,y,y') Examples of the determination of and Determination of the most general differential equation invariant under a given one—parameter group Applications Problems

1

6 9

12 12 18

21

26 26 30 33 36 39

48

49 51

55 58 59 62 66

77 77 79

83 87 92

6.

Linear partial differential 6.1 6.2

equations

Formulae for partial derivatives

Classical groups for the diffusion equation 6.3 Simple examples for the diffusion equation

6.4 Moving boundary problems 6.5 Fokker-Planck equation 6.6 Examples for the Fokker—Planck equation 6.7 Non—classical groups for the diffusion equation Problems

7.

7.4

99

ioi 103 105 109 116 120 123

partial differential equations

135

Formulae for partial derivatives Classical groups for non—linear diffusion

136 140

Non-linear

7.1 7.2 7.3

97

Non—classical groups for non—linear diffusion Transformations of the non—linear diffusion equation Problems

References

146 148 150

159

1 Introduction

Although

a good deal of research over the past two centuries has been devoted

to differential equations our present understanding of them is far from complete.

These notes are concerned with obtaining solutions of differential

equations by means of one—parameter transformation groups which leave the equation invariant.

This subject was initiated by Sophus Lie [1] over a

hundred years ago.

Such an approach is not always successful in deriving

solutions.

However it does provide a framework in which existing special

methods of solution can be properly understood and also it is applicable to linear and non—linear equations alike.

In formulating differential equations

the Applied Mathematician inevitably makes certain assumptions.

Using group

theory these assumptions can be seen to hold the key to obtaining solutions of their equations.

The purpose of this chapter is to present a simple introduction to the subject for both ordinary and partial differential equations by means of simple familiar examples.

For ordinary differential equations comprehensive

accounts of the subject are given by Cohen [2], Dickson [3], Page [4] and more recently Bluman and Cole [5] and Chester [6].

For partial differential

equations the reader may consult Bluman and Cole [5] and Ovsjannikov [7] where additional references may also be found.

1.1

ORDINARY DIFFERENTIAL EQUATIONS

In order to illustrate some of the ideas developed in these notes we consider a simple example.

It is well known

differential equation

that

the 'homogeneous' first order

2

2

dx

(11)

xy

can be made separable by the substitution

u(x,y) =

y/x

and the resulting

solution is given by

=C, where

(1.2)

denotes an arbitrary constant.

C

We might well ask the following

questions:

Why does the substitution

Question I

equation for Question 2

u(x,y) =

y/x

lead to a separable

u ?

How do we interpret the degree of freedom embodied in the

arbitrary constant

C

in the solution?

Answers to these questions can be provided within the framework of transformations which leave the differential equation unaltered.

Consider

the following transformation,

x1ex,

y1ey,

C

where

C

is

C

(1.3)

an arbitrary constant.

We notice that (1.1) remains invariant

under (1.3) in the sense that the differential equation in the new variables x1

and

is identical to the original equation, namely

y1

2

2

—= x1 +y1 dy1

x1y1

dx1

(1.4)

.

Moreover we see that (1.3) satisfies the following: gives the identity transformation

(1)

c = 0

(ii)

—c

characterizes the inverse transformation

(iii)

if

x2 =

e6x1,

y2 = e6y1

x1 =

x,

y1 =

y, y =

x =

eCy1,

then the product transformation is also a

member of the set of transformations (1.3) and moreover is characterized by the parameter 2

c+5,

that

is

x2 = e

c+5

x, y2 = e

c+cS

y.

A transformation satisfying these three properties is said to be a one—

parameter

group

of transformatione.

We

observe that the usual associativity

law for groups follows from the property (iii).

With this terminology

established we might answer the above questions as follows: Answer 1 u

because

u(x1,y1) =

The substitution

y/x

leads to a separable equation for

is an invariant of (1.3) in the sense that

u(x,y)

u(x,y)

since, yl

u(x1,y1) =

u(x,y) =

= u(x,y)

—=

(1.5)

,

and it is this property which results in a simplification of (1.1).

In

general we shall see that if a differential equation is invariant under a one—parameter group of transformations then use of an invariant of the group results in a simplification of the differential equation.

If the

differential equation is of first order then it becomes separable while if the equation is of higher order then use of an invariant of the group permits a reduction in the order of the equation by one. Answer 2

From (1.2) and (1.3) we see that we have 2

y1

— =C xl

log x1 —

+ c

(1.6)

,

so that the degree of freedom in the solution (1.2) resulting from the arbitrary constant

C

is related to the invariance of the differential

equation (1.1) under the group of transformations (1.3) which is characterized That is, the transformation (1.3) permutes

by the arbitrary parameter

c.

the solution curves (1.2).

In general we shall see that for every one—

parameter group in two variables there are functions

u(x,y)

and

v(x,y)

such that the group becomes u(x1,y1) =

u(x,y)

,

v(x1,y1)

=

v(x,y)

+ c

.

(1.7) 3

Moreover if a first order differential equation is invariant under this group then in terms of these new variables =

4(u)

u

and

v

it takes the form,

(1.8)

,

and consequently has a solution of the form

v +

=

C

(1.9)

,

for appropriate functions

and

In order to give the reader some indication of the usefulness of the above we consider the following non—trivial equation, (1.10)

This is an Abel equation of the second kind (Murphy [8], page 25) which we see is not readily amenable to any of the standard devices.

However the

equation is clearly invariant under the group

x1ex, C

y1=e—Cy,

and therefore we choose

(1.11)

u(x,y) = xy

as the new dependent variable and the

differential equation (1.10) becomes, (1.12)

which can be readily integrated.

It is worthwhile emphasizing that not all

equations can be solved in such a simple manner. —

Consider for example,

(1.13)

,

which arises in finite elasticity (see Hill [9]).

This equation is again an

Abel equation of the second kind but in this case there is apparently no simple group such as (1. 11) which leaves the equation invariant.

In this general introduction it may be appropriate to mention here possible research areas for which group theory has not yet been applied.

4

The reader

might well like to bear these problems in mind with a view to developing results in these areas. Research area 1

Differential—difference equations.

It is well known that formal solutions of linear differential—difference equations, for example

-y(x—x0)

=

where

(1.14)

,

is a constant, can be expressed as

x0

r

y(x)

C.e

=

—W4X

(1.15)

,

j

are arbitrary constants and

where

denote the roots of

w =

If the equation is non—linear then there are no such general methods of Consider for example Hutchinson's equation which can be written as

solution.

4x)

=

y(x)[l

—

y(x—x0)]

(1.16)

.

This equation arises in theory of populations (see Hutchinson [10]).

What are

the implications of group theory, if any, for equations of this type?

(See

problems 19 and 20 of Chapter 4). Research area 2

Differential equations invariant under transformations which

cannot be characterized as one—parameter groups. A differential equation occurring in fluid dynamics is Tuck's equation (see Tuck [11]),

— 2

dt2

dx + dt

(5+3x)

+ 3x(1—x)

4x(1+x)(dtJ

It can be verified that if the usual way we let y

3x(1-.x)

dx

(1+x)

y =

+ 2y

x(t)

dx/dt

(117)

(1+x)

is a solution then so is

x(t)'.

If in

then (1.17) becomes

+ (5+3x) 4x(1+x)

2

'

which is again an Abel equation of the second kind.

(1 18)

From the invariance

5

property of (1.17) we can deduce that (1.18) remains invariant under the transfo rmat ion

(1.19) which clearly cannot be characterized as a one—parameter group.

Can we use

such invariance properties to determine solutions of differential equations? Research area 3

Abel equation of the second kind;

As we have already Indicated one of the most frequently occurring differential equations which is not always amenable to standard devices is the Abel equation of the second kind.

The general equation can be expressed

in the form (see Murphy [8], page 26) y

Equation

dx

=

a(x)

(1.20)

+ b(x)y .

(1.20) with arbitrary functions

a(x)

and

b(x)

would appear to be

a problem worthwhile studying.

1.2

PARTIAL DIFFERENTIAL EQUATIONS

Unlike ordinary differential equations the success of the group approach for partial differential equations depends to a considerable extent on the accompanying boundary conditions.

That is, the group approach is only

effective in the solution of boundary value problems if both the equation and boundary conditions are left unchanged by the one—parameter group.

For the

most part we confine our attention to specific differential equations rather than boundary value problems.

For any particular boundary value problem we

should always first look for any simple invariance properties.

These may be

more apparent from the physical hypothesis of the problem rather than its mathematical formulation.

If no such invariance

can be found and if the

problem merits a numerical solution then the group approach might still be

6

relevant as a means of checking the numerical technique with artificially imposed boundary conditions which permit an exact analytic solution. As an illustration we consider a boundary value problem for which both the partial differential equation and the boundary conditions are invariant under a simple one—parameter group.

Consider the problem of determining the source

solution for the one—dimensional diffusion or heat conduction equation for c(x,t), namely 2 3c

=

(t

—°°

>

x

<

< 00)

(1.21)

.

3x

The source solution of (1.21) is a solution which vanishes at infinity for all times and initially satisfies c(x,O) = where

c0ô(x)

(1.22)

,

is a constant specifying the strength of the source and

c0

ô(x)

is

We observe that both of (1.21) and (1.22)

the usual Dirac delta function.

are left unchanged by the transformation c

x1 = e x where

£

t1

,

denotes

2c

= e

t

c1 = e

,

—c

c

(1.23)

,

an arbitrary constant and we have made use of the

elementary property of delta functions, cS(Ax)

=

,

for any non—zero constant

(1.24)

.

Thus if

A.

and (1.22) then we have also

c1 =

is the solution of (1.21)

c =

Clearly this is the case if

has the functional form

4,(x,t)

for

=

t

_1

some function

(1.25)

,

ij

of the argument indicated.

Upon substituting (1.25)

into (1.21) we obtain the ordinary differential equation

+

+

= 0

,

(1.26) 7

where

E

and primes indicate differentiation with respect to

denotes

Equation (1 .26) can be reduced to the confluent hypergeometric equation (see

Murphy [8], page 321).

However the solution vanishing at infinity can

be readily verified to be simply, = Ae

A

where

(1.27)

,

denotes an arbitrary constant.

This constant is determined from

(1.22), namely

c(x,t)dx = c0

(1.28)

.

(1.27)

From this equation, (1.25) and

we find that the required solution of

the boundary value problem (1.21) and (1.22) becomes

c(x,t)

=

c0

e

—x2/4t

(4irt)

This

½

(t

>

solution is of course well known.

—°° < x

< co)

.

(1.29)

For our purposes it firstly serves as

a specific non—trivial boundary value problem for which the differential equation and boundary conditions are both invariant under a one—parameter group.

Secondly it serves to illustrate that knowledge of a one—parameter

group leaving the equation invariant enables, at least in the case of two independent variables, the partial differential equation to be reduced to an ordinary differential equation.

For more independent variables knowledge of

a group leaving the equation unchanged reduces the number of independent variables by one.

In these notes we give the general procedure for determining the group such as (1.23) whIch leaves a specific equation invariant.

We also give the

general technique for establishing the functional form of the solution such as that

8

given

by (1.25).

PROBLEMS

1.

Determine in each case the constants

a

and

8

such that the one—

parameter group ac x1 = e x

y1 = e

,

y

leaves the following differential equations invariant.

Use an invariant

of the group to integrate the equation.

(a) E =

3,12

+ By3

and

(A

B

are constants)

(b)

+ By = 0

(c)

x(A + xytl)

2.

Verify that,

(A, B

and

n

are constants)

y1=e —2c y,

x1=x+C,

is a one—parameter group of transformations and hence integrate the

differential equation (1 — 2x —

3.

logy)

2y = 0

Integrate the differential equation (x—y)2

=

(A

is a constant)

by observing that the equation admits the group

x1x+C, 4.

Given that

p(x)

y1=y+C. is a solution of the linear differential—difference

equation (1.14) show that p(x—x0) y(x)

p(x)

=

is a solution of the non—linear differential—difference equation dy(x)

=

y(x)[y(x)

—

y(x—x0)] 9

5.

Show that the transformation e

y(x)=

x

f(eXX0)

reduces equation (1.16) to the differential equation

f(t)

df(t) —

f(At)

dt

where 6.

t=e x-xO and X=e -xO

Show that with

w =

3(1-x)

=

dx

(1+x)

y/x

the differential equation (1.18) becomes

+

+

(1-x) (1+x)

Show further that the substitution 2

4

s =

(1—x)/(1+x)

2

(s —1)

w

2

dw SW — = 35 + 2w + — ds 4

and observe that the transformation (1.19) becomes Si

7.

yields

w1 =

—w

and

= —S.

Observe that the partial differential equation (1.21) remains invariant under the transformation c

x1 = e x

t1

,

= e

2€ t

,

c1 = c

so that the equation admits solutions of the form

Deduce the ordinary differential equation for

c(x,t) where 8.

A

and

Continuation.

4

c(x,t) =

and hence show that

2

"4dy + B

= A,j'

B

denote arbitrary constants.

For the non—linear diffusion equation

D(c)j—) where the diffusivity

D

is a function of

c

only, use the one—

parameter group and functional f6rm of the solution in the previous problem to deduce the ordinary differential equation 10

+ dD(4)

9.

Continuation.

+

For the case

= 0

D(c) = c

show that

the

ordinary

differential equation of the previous problem remains invariant under the group

41e 2c

c

=

and that with

the equation reduces to the Abel equation of

the second kind

+ p2 + p =

p = 10.

+

+

and

log

y =

Show that the singular solution

E.

corresponds to the solution

Continuation.

4(e)

constant.

Show that the transformation

p =

reduces the Abel

equation in the previous problem to q

+ fJq +

+

= 0

which has the same form as equation (1.20). 11.

By calculating the quantity, ac1

2 ac1

at1

show directly, using the chain rule for partial derivatives, that the classical diffusion equation (1.21) remains invariant under the following transformations, (i)

x1 = x + et,

t1 = t,

c1c

exp[_ —

2

1

(ii)

x1 = (1—ct)

=

(1—ct)

—

4(1—ct)

11

2 One-parameter groups and Lie series

In

this chapter we introduce the concepts of one—parameter groups and Lie For one—parameter groups there are two important results.

series.

the method of obtaining the global

Firstly

of the group from the infiniteBimal

Secondly the existence of canonical

coordinateB

for the group.

For

Lie series the important and remarkable result is the so—called Coninutation

These concepts are discussed below.

theorem.

2. 1

ONE—PARAMETER TRANSFORMATION GROUPS

In the (x,y) plane we say that the transformation =

is

f(x,y,c)

group of

a one-parameter

(i)

(identity) the value x =

(ii)

f(x,y,O)

(2.1)

,

tran8forn?ation8

y

=

the following properties hold:

characterizes the identity transformation,

c = 0

,

if

g(x,y,O)

(inverse) the parameter —c characterizes the inverse transformation, x =

(iii)

g(x,y,c)

y1 =

,

f(x1,y1,—c) x2 =

(closure) if the

,

y = g(x1,y1,—c)

f(x1,y1,cS)

two transformations

,

y2 = g(x1,y1,cS)

then the product of

is also a member of the set of transformations

(2.1) and moreover is characterized by the parameter x2 =

f(x,y,c+5)

,

c+tS

,

that

is

y2 =

Again we remark that the usual associativity law for groups follows from the closure property.

12

(a)

x1 = x ,

(b)

x1 = eCx

Some simple examples of one—parameter groups are: y1 = y + c ,

y1 = eCy

(translation group), (stretching group),

y sin c

x1 = x cos c

(c)

,

y1 = x sin c + y cos c

x1 = x

In order to show (c) forms a group we have inmiediately

when

=0

c

and

y = y1 cos c — x1 sin c

characterizes

the inverse and (ii) is satisfied.

x2 = x1 cos

— y1 sin d

(x

x2 =

and

cos c — y sin

=

x cos(c+5) — y

=

(x cos c — y sin c)sin

=

x sin(c+5) + y cos(c+tS)

y1 = y

so that —c

For (iii) we see that if

y2 = x1 sin d + y1 cos d

c)cos

and

group).

On inverting we obtain

so (i) is satisfied.

x = x1 cos c + y1 sin c

(rotation

then we have

5 —

(x

d +

(x sin c + y cos c)cos d

sin c + y cos c) sin d

sin(c+tS)

and

and therefore (iii) is satisfied. The functions

=0

c

and

g(x,y,c)

are referred to as the global form

If for small values of the parameter

of the group. since

f(x,y,c)

we expand (2.1) then

c

gives the identity we have dx

x1 = x + c

+ 0(c2)

y1 = y + c

,

+ 0(c2)

where

If

indicates

0(c2)

(2.2)

,

c=O

c=O

terms involving only and

we introduce functions

and higher powers of

c2

c.

by

dy

dx c

=

= fl(x,y)

,

(2.3)

,

c=O

c=O

then we obtain x1 = x +

+ 0(c2) ,

y1 = y + cfl(x,y) + 0(c2) ,

and (2.4) is referred to as the infinitesimal form of the group. property

form of

of one-parameter tran8formatwn groups is the

group we can deduce

that

(2.4) The crucial

given the infinitesimal

the global form by integrating the following

13

of differential equations,

autonomous dx1

dy1 =

subject

=

,

(2.5)

,

to the initial conditions,

x1 = x

y1 = y

,

e = 0

when

(2.6)

.

A proof of this result can be found in Dickson [3].

Here we merely indicate

its validity by means of a simple example.

For the rotation group (c) we have dx1

dy1 ,

e = 0

and therefore on setting F(x,y) =

—y

=

,

we have from (2.3)

x

Thus in this case we need to integrate dy1

dx1 ,

subject

Introducing the complex variables

to the initial conditions (2.6).

z=x+iy and z1=x1+iy1 weobtain dz1

= iz1 and thus log z1 =

ic

+ log z

where we have used the initial conditions (2.6). z1 = e1Cz

Imaginary parts of rotation group (c). r =

(x

then we have

2

2½

+ y ) z =

On equating real and

we can readily deduce the global form of the

If we introduce polar coordinates

,

re 10

0 =

tan

and from

—1

(y/x)

z1 = e

ic

z

we see that the global form of

the rotation group (c) can be written alternatively as 01 = 0 + c. 14

That is, in terms of

defined by

(r, 0)

(r,0)

r1 =

r

and

coordinates the rotation group has

the appearance of the translation group.

This is a general property of one-'

parameter transformation groups. For

any given one-parameter transformation group (2.1) there extBtB

and v(x,y)

functwn8 u(x, y)

u(x1,y1) The function

u(x,y)

=

auch that the global form of the group becomes

v(x1,y1)

,

=

v(x,y)

(2.7)

+ c .

is said to be an invariant of the group while together

u(x,y)

are referred to as the canonical coordinateB of the group.

(u,v)

Methods for finding

—=

From (2.5) we obtain

dx1 dy1

(2.8)

n(x1,y1)

which we suppose integrates to yield

u(x1,y1) —

constant

so that from the

Initial conditions (2.6) we deduce the first equation of (2.7) and

Alternatively

known.

eliminating

c

u(x,y)

from

and (2.1)2.

We note that if

and suppose that from

u(x1,y1) =

u(x,y), namely a =

u(x,y)

we can deduce the explicit relation

Now for the purposes of integration in (2.5),

y1 =

is an

u(x,y)

In the integration of (2.8) let

v(x,y).

is

may be deduced directly from (2.1) simply by

invariant then so also Is any function of Method for finding

u(x,y)

a

is a

constant and from (2.5)i we have dx

(2.9)

.

=

If for some constant fX

x0

we define

by

dt

(2.10)

•

=

0

then from (2.6) and (2.9) we can deduce (2.7)2 where and hence

v(x,y)

v(x,y) =

Is known.

15

Example 1

Consider,

X1

(1+ex)

(1+cx)2y

=

'

(2.11)

.

The reader should verify that (i), (ii) and (iii) of the definition of a one— parameter are Indeed satisfied. x1 = x — cx2 + 0(c2) ,

so that from (2.4)

For small values of

and

fl(x,y) =

differentiatIng (2. 11) with respect to

-

(1+ex)

2

2xy.

Alternatively on

we have

c

dy1

2

= —x1

we have

y1 = y + 2cxy + 0(c2)

= —x2

dx1

c

-i-— = 2(1+cx)xy

,

and (2.5) confirms the given expressions for

=

2x1y1

E(x,y)

and

(2.12)

.

fl(x,y).

From

(2.12) we have X1

dx1 dy1

u(x,y) = x2y

which on integrating gives v(x,y) = x

we see that

-1

as an invariant while from (2.12)i

satisfies (2.7).

could be deduced directly from (2.11) by eliminating

Example

2

and

by the relations,

v

Show that

r

2

Alternatively the invariant

x y

C.

are related to the canonical coordinates

and

—

u

213

'

—

where the Jacobian is given by

On differentIating (2.7) with respect to dx1

r 16

dy1

+

dx1

=0

,

C

we obtain dy1

+

=1 -s----

,

(2.

14)

where

and

u1

denote

v1

and

u(x1,y1)

(2.5) and (2.14) we have on replacing

n=0

+

(x1,y1)

by

respectively.

From

(x,y),

Ti = 1

÷

,

v(x1,y1)

and (2.13) can be deduced immediately from these relations. A transformation in the

Example 3

(x,y)

plane is area preserving if

a(x1,y1) =

1

(2.15)

.

Show that (2.1) is area preserving if and only if —

u

only.

From (2.4) and (2.15) we can deduce on equating terms of order

c,

(2.16)

ax

From (2.13) and (2.16) we can deduce

1 a(x,y)

=o

and the required condition follows.

lit may be of Interest to note that since the set of area preserving transformations forms a group, the infinitesimal condition (2.16) is precisely the same as the global condition. respect to

c

That is, if we differentiate (2.15) with

we have dy1

dx1

y1

=0

and on multiplying this equation by

dx1

,y1,

we obtain

dy1

÷ ,

' —0

17

so that we have a

dx1

dy1

a

=0

+

(2.

.

17)

On using (2.5) we see that (2.17) is the same condition as (2.16)fl

2.2

LIE SERIES AND THE COMMUTATION THEOREM

Suppose we have the group (2.1) with infinitesimal version (2.4). the differential operator

We define

by

L

(2.18)

.

L =

Now for any function

$(x1,y1)

which does not depend explicitly on

c

we

have

dx

d41

de

dy1

de

ax1

denotes

where

L1

(x1,y1).

From (2.5) and (2.19) we obtain

(2.20)

,

denotes the differential operator

with

(x,y)

replaced by

d3

= L1 (L1($1))

,

—i-

= L1 [L1

(2. 21)

.

then by Maclaurin's expansion we have

•(c) =

=

4(O) +

c=O

de

and thus from (2.20) and (2.21) with

dc

c=O

c = 0

That is, we have

+

+

c=O

we obtain 3

2

=

18

L

Similarly we have

d2

If we let

(2.19)

de

4(x1,y1).

= L1(41) where

ay1

+

+ ...

(2.22)

,

= n=O

and we refer to such a series as a Lie series.

We notice that we can write

(2.22) as =

(2.23)

,

provided we interpret the differential operator

as the series operator,

)

in particular if we take

to be

4(x,y) e

x

cL

y

x

and

y

then from (2.23) we obtain,

(2.24)

.

On combining (2.23) and (2.24) we have the remarkable result, CL

x, e

CL

y) = e

CL

(2.25)

,

which is called the Commutation

theorem of Lie series (see

and Knapp

t121, page 17).

To illustrate (2.24) consider the rotation group (c). differential operator L = —y so that

a

L(x) =

In this case the

is given by

L

a

+ x

r

—y

and

L(y) = x

and the global form of the group can be

deduced from (2.24) using the expansions,

k2k

cos

£

= k=O

(—1) C (2k)!

k2k+l

(—1) C

sin c = k=O

(2k-i-1)i

It is worthwhile noting that using Lie series we can give a formal

Example 4

solution of any autonomous system of differential equations given initial values.

That is, consider =

F(X,Y)

,

= G(X,Y)

,

19

and

X

Y =

a,

at

t =

0.

The formal solution of this initial value

problem is x = etMa

y

,

F(a,8)

(2.26)

,

is defined by

M

where the operator N =

=

+

Consider two simple examples.

Firstly, the single differential equation,

dX

dt In this case we have M = -a2 —a-,

and

-a2,

M(a) =

M2(a) = 2a3

and in general

N(a)=(—1)n n!an+1 Hence

x =

etha =

and thus for

(a) = a

n0

lati <

n=O 1

we obtain the solution,

a 1-Fat

Secondly, consider (2.27)

A and

where

M= so that

20

B

are constants.

In this case we have

M(a) = (Aa + N2(cz) =

K2cz

M(8) = (Ba

,

,

N3(a) = K2(Aa

—

A8)

= K28

+ B8)

= K2(Bcz — A8)

,

M4(8)=K48, M5(cz)

= K4(Acz + B8)

M5(8) = K4(Ba — A8)

,

K = (A2 + B2)½. From these results and (2.26) we can

and so on, where

deduce the solutions, 2KX = [Ka + (Mi + B8) ]eKt + [lca - (Aa + B8) ]eKt 2KY = [K8 + (Ba —

A8)]eKt +

(Ba —

—

which of course could be established by more elementary methods (for example, differentiate (2.27)i with respect to [The

and make use of (2.27)2).

t

following problems which arise in continuum mechanics are useful

exercises in manipulating Lie series.] PROBLEMS

1.

tn cylindrical polar coordinates is

(r,O)

a transformation in the plane

area preserving if a(r1,01)

r r1

a(r,o)

Following the note at the end of Example 3, differentiate this equation with respect to

dr1 +

Hence

4

which

1

dr1 +

= 0

deduce there exists a function 4(r1,01,c) dc

tf

and show that,

c

r1 ao1

'

dc

r1 ar1

does not depend explicitly on r1 =

r,

01 = 0

when

such that

c=

0

C

is

the solution of this system for a one—parameter group with

4

as 21

an invariant, that is =

2.

4(r,0)

In the above problem obtain the one—parameter group given that,

where

Ar20 + BO

(a)

=

(b)

4(r,O) = Ar20 +

(c)

4(r,0) =

A

and

Br2log

sin

(0 +

r

0 cos 0)

denote arbitrary constants.

B

[Answers:

3.

(a)

r1 =

(b)

r1 =

(c)

r1 =

+ +

01 =

,

[r2

=

— Ac cos20]½,

r + Bc

01 = 0

Consider,

log(1 + cx)

x1 =

dx1

cx)y

dy1

and express in terms of

,

Calculate

y1 = (1 +

,

(x1,y1).

Hence or

otherwise deduce that this is not a one—parameter transformation group. 4.

Consider the one—parameter group, x1 =

f(x,y,c)

=

y1 =

g(x,y,c)

= y

x +

+

O(c)

+ cn(x,y) + 0c )

and introduce the operators,

P=L+w, Q=L-w, where

w

is defined by,

3x

Let

4(x,y)

notation

22

and

denote arbitrary functions and agree to use the for the Jacobian, that is

We consider the following Lie series,

e(x,y) = n=O

= n=O

= n=O

=

n0

Show that =

(f)

!

=

,

+

=

(ii)

[Hint: for (ii) start by considering 5.

If we suppose that

Continuation.

(e

Co

eL 4,

e

cL

r

=

fl

c

n=O show that,

(i)

k

= k=O

(ii)

;

=

Hence deduce,

= = n=O

[Hint: for (ii) start by considering P(;,1)

and use (i) and (ii) of

previous problem.] 6.

Continuation.

Observe that in particular,

(x1,y1) = (e

cL

x, e

eL

y) = e

1

and therefore, 3.

2

(x1,y1) =

1

+ ew +

P(w) +

P2(w) + ...

Verify 23

2

= Lw +

L(w) +

L (w) +

2

(,c1,y1)1 =

(ii)

7.

— Lw —

1

Q(w)

In cylindrical polar coordinates

—

(r,8)

r1 =

f(r,0,c)

=

+

+ 0(c2)

=

g(r,0,e)

= 0 +

+ 0(c2)

r

Q

(w) +

consider the one—parameter group,

and introduce operators,

P1=L+w1 , where

w1

w 1

and

w2

ar

ao'

are defined by w

2

r

Suppose that

£. L"(r)

r1 = =

A

r

=

=

n0

= n=O

a(r,0) =

=

Verify,

L(f1)

=

(i)

(ii)

A = 8.

eEl'2l

ContInuation.

Observe from Problem 6,

p=e cP1 1. Suppose that, =

24

r

a(r,0)

r

a(r,0)

P2=L+w2,

Verify,

(j)

(ii)

where

XklJk

n

o

k=O =

P3

! is given by

P3 = L +

+

Hence conclude, r1 a(r1,01) = r

=

[Hint: for (ii) start by considering

3 Invariance of standard linear equations

It

is well known

that

linear differential equations for

y(x)

remain linear

under transformations of the form,

x1 =

f(x,c)

,

y1 =

g(x,c)y

(3.1)

.

Throughout this chapter we consider only transformations (3.1) which we suppose form a one—parameter group such that infinitesimally we have x1 = x +

+ 0(62) ,

(3.2)

y1 = y + cfl(x)y + 0(62) .

We look for groups (3.1) which leave standard linear equations Invariant and deduce the form of the differential equation in terms of canonical coordinates (u,v) (see (2.7)).

initially the reader may well consider this

approach clumsy and irrelevant for such equations.

The object of the

exercise being to demonstrate the group approach in familiar situations with a view to the reader obtaining some insight into the relation between solutions and groups leaving the equation invariant.

Moreover even linear

equations are not always readily solved and the results obtained in sections 3.3 and 3.4 by this method are non—trivial and have not been given previously.

3.1

FIRST ORDER EQUATION

y' + p(x)y =

g(x)

For first order differential equations our primary objective is to introduce new variables such that the equation becomes separable.

For equations

invariant under a one—parameter group of transformations the appropriate new variables are the canonical coordinates (u,v) of the group. this procedure with the standard first

+ p(x)y = 26

q(x)

.

We illustrate

equation,

(3.3)

For convenience we Introduce the function

p (t)

by

s(x)

dt

s(x) = e where

(3.4)

Is some constant.

x0

With this definition the solution of (3.3) Is

known to be given by rx

s(x)y

—

s(t)q(t)dt

= C

(3.5)

,

J

"Co

where

Is a constant.

C

We now deduce (3.5) by finding a group (3.1) which

Invariant,

that Is

dy1

+ p(x1)y1 = q(x1)

Prom this equation and (3. 1) we deduce

+

+

f'(x)P(f)}

which becomes (3.3) provided p(x) =

+

g(x) q(f)

Y =

and

f(x)

f'(x)p(f)

are such that

g(x)

q(x)

=

,

,

g(x)

=

q(f)

From these equations and f(x) = x +

p(f) = we

cE(x) ÷ 0(c2)

+

p(x) +

0(c2)

obtain on equating terms of order

+

+

= 0

+

,

1

q(f)

,

(3.6)

+ cn(x) + 0(c2) , =

q(x) +

+

0(c2),

c

= 0

—

Hence we have n

where

+ C1

=

,

is a constant.

+

+

= C1

,

(3.7)

Thus

27

s(t)q(t)dt

= q(x)s(x) .IX

+

0

(3.8)

C1 —

where

is a further constant.

C2

[Of course in obtaining these results WE Now the

have had to solve an equation of the type (3.3), namely

global form of the group (3.1) is obtained by integrating (see (2.5) and (2.6) dx1

dy1 =

= n(x1)y1

,

subject to the initial conditions

(3.9)

,

x1 =

x,

y1 = y

when

c =

0.

From (3.8)2

and (3.9) we have dx1

dy1 1

= C1 —

—a-

and therefore, by integrating this equation we obtain, y1s(x1) = e

where

C1c

ys(x)

(3.10)

,

is defined by (3.4).

s(x)

From (3.8)i and

we have

q(x1)s(x1)dx1 = dc

(3.11)

,

+

and there are two cases to be considered. Firstly if

*.

C1 # 0

then (3.11) gives

1s(t)q(t)dt +

s(t)q(t)dt

+

and from this equation and (3. 10) we can deduce that our canonical

coordinates (u,v) (see (2.7)) are given by

28

+c

8(x) y

u(x,y) =

+

{c1

v(x,y)

log

=

+

In these coordinates it can be readily verified that the differential equation (3.3) becomes du —= dv

Cu 1

—

1

which is separable and integrates to give

log(1

—

— C1u) =

—

log

C3

is,

that

(1—C1u)e Civ where the

v

C3

C3

is a constant.

arbitrary

constant

if

Secondly

C1 = 0

C

This equation can be reconciled with (3.5) in (3.5) is

(C2 —

then from (3.10)

where

C3)/C1.

and (3.11) we have canonical

coordinates

u(x,y) — s(x)y ,

v(x,y)

—

J—

2jx0

s(t)q(t)dt

and in these coordinates (3.3) becomes du dv

C

Again our equation in canonical coordinates to

separable and can be integrated

give u — C2v =

C

which can also be reconciled with (3.5) where the constant both equations.

C

is the same in

We remark that (3.3) is invariant under other groups, in

addition to those considered here. 29

3.2

= 0

SECOND ORDER HOMOGENEOUS EQUATION y" +

For second order linear homogeneous equations we can without loss of generality (see problem 8) consider the normal form of the equation, namely 2

+ p(x)y = 0

(3.12)

.

We shall assume

and

are two linearly independent solutions of

(3.12) and for convenience we suppose their Wronskian is unity, that is —A

A

—

and

Our obj ective here is to relate

to a one—parameter group (3. 1)

For higher order linear differential equations

which leaves (3.12) invariant.

the use of canonical coordinates (u,v) simplifies the equation to one with constant coefficients. From (3.1) we have

+

= -.g-

f'dx

dx1

and

+

=

dx1

f'

dx

f'

+

-

14

X

f'

Clearly if (3.12) is to remain invariant there can be no term involving On equating the coefficient of

y'

to zero in (3.14) we obtain

y'.

f'(x)/g(x)2

constant, which must be unity if (3.1) is a one—parameter group and therefore we have f'(x) =

g(x)2

.

From (3.14) and (3.15) we find that the differential equation 2

d y1

+ p(x1)y1 = 0 dx1

becomes

30

(3.15)

invariant provided

and thus (3.12) remains

p(f)g4

+

2

—

=

p(x)

(3.16)

.

Equation (3.15) and (3.16) constitute two equations for the determination of From (3.6), (3.15) and (3.16) we find on equating terms of

the group (3.1). order

c, 'I,

=

2

+ p'E =

+2pE

The equation (3.17)2 for

E(x)

0

(3.17)

.

is a formally self—adjoint third order

differential equation (sometimes called anti self—adjoint, see Murphy [8], page 199) with first integral —

which

can

+ pE2 = constant ,

be verified by differentiation.

problem 20

or Murphy [8],

(3.18) It is well known

(see either

page 200) that the general solution of (3.17)2 is

given by = A$12

where is

A, B

+ 2B+1$2 + Cd,22

and

C

(3.19)

,

denote arbitrary constants.

the general solution consider for example =

+

'

=

+

[In order to see that (3.19) =

then we have

+

Observe that from the original differential equation we have =

and

—

substitution

gives zero.]

of this expression and those

for E'

and

into (3.17)2

Further from (3.12) and (3.19) we can deduce

31

+

=

+

+

=

+

+

,

— 2pE

and on substitution into (3.18) we find on using (3.13) that (3.18) becomes

(2EE"

—

= (AC — B2)

+

(3.20)

.

Now the global form of the one—parameter group (3.1) is obtained by solving the differential equations

dy1

dx1

E'(x1) 2

subject to the initial conditions

x1 =

x,

y1 = y

and

e =

0.

We find that

suitable canonical coordinates (u,v) are given by =

v(x,y)

u(x,y) =

E(x) ½ where

is some constant.

x0

f

In terms of (u,v) we find that the differential

equation (3.12) becomes

+

-

= 0

+ But from (3.20) we see that the differential equation finally becomes 2

+ (AC — B2)u = 0

(3.21)

.

dv

Thus for example, if

(AC — B2)

is positive the general solution of (3.12)

is given by

y(x) = E(x)½{C1 cos{K

where

C1, C2

(3.22)

+ C2 sin[K

are arbitrary constants and

K = (AC — B2)

.

We

have therefore

established the relationship between the general solution of (3.12) and the infinitesimal version of the one—parameter group of transformations leaving 32

In a sense, (3.22) is the 'inverse' of (3.19).

(3.12) invariant.

From

(3.19) we see that if we know a group leaving (3.12) unaltered then essentially we know a quadratic relation between the linearly independent solutions of (3.12).

We remark that solutions of (3.12) in the form of (3.22)

have been given previously although the function

appearing in (3.22)

has not been identified with the one—parameter group leaving the differential equation invariant (see for example Coppel [13], page 19). As an illustration of the above, consider the simple Euler equation, 2

i-i + dx

2

4x

2

= 0

In this case the equation is clearly invariant under the group and

y1 = y = x½

so that and

=

x.

we see from (3.19)

u = C1v + C2.

A = 1, B =

C

= 0

and

This expression confirms our

linearly independent solutions since in this case

3.3

= eCx

Since we have linearly independent solutions

q2(x) = x½ log x

therefore from (3.21) we have

x1

u =

and

v =

log

x.

THIRD ORDER HOMOGENEOUS EQUATION y" + p(x)y' + q(x)y =0

We see from problem 12 that for third order linear homogeneous differential equations we can without loss of generality consider the equation 3

+ p(x)

dx

X

+ q(x)y = 0

In this section we suppose that

(3.23)

.

41(x)

and

are linearly independent

solutions of the second order equation

+

y = 0

(3.24)

,

such that their Wronskian is unity.

We are concerned with finding one—

parameter groups of the form (3. 1) which leave (3.23) invariant.

From a further differentiation of (3.14) we obtain 33

= ___g__

dx13

-

+

f'3 dx3

f'3

-

ft

ft

+ 3gf"2

-

dx

f'4

3g'f"2 -

+

+

'f"

+

y

(3.25)

.

f'

ft

If (3.23) is to remain invariant we require the coefficient of

y"

in (3.25)

From this condition we deduce

to be zero.

f'(x) =

g(x)

(3.26)

,

and (3.25) becomes d3y

__! dx1

=

3

L

3

-

+

2

gdx

3

g

3

4dx g

2.&_. -

+ g

5

3

g

4

g

Using this equation and 3 dy1

dy1

i—

+ dx1

+ q(x1)y1

= 0

I

we obtain on multiplying by —

g2

the equation,

+

+

—

4gtg" + p(f)gg' + q(f)g3]y

=

o.

+

+

For invariance this equation must be identical with (3.23) and therefore f(x)

and

g(x)

as well as satisfying (3.26) must also satisfy

-

+ p(f)g2

=

p(x) (3.27)

+

-

+ p(f)ggt

From (3.6) and (3.26) we obtain 2E"t +

+

q(f)g3 =

=

q(x)

while from (3.27) we have

+ ptE = 0 (3.28)

F" + pE" +

+ q'E = 0

and these equations are only consistent if

34

=D

—

where then

(3.29)

,

Clearly If (3.23) is self—adjolnt (see problem 14)

is a constant.

D

p' = 2q

and (3.29) is trivially satisfied with the constant

However if (3.23) is not self—adjoint then as well as satisfying (3.29). general solution of

zero.

must be both a solution of

In terms of solutions of (3.24) the

is given by

+

=

Cx)

D

+

(3.30)

,

and we have —

where

A, B

= 4(AC — B2)

+ and

denote arbitrary constants and we have used the fact that

C

the Wronskian of for a given q(x) =

where

E(x)

I

is unity.

and

p(x)

(3.31)

,

we need to assume

+

If (3.23) is not self—adjoint then is given by

q(x)

D

(3.32)

,

X

is given by (3.30).

From the equations dx __.i

— r(

dy )

— r'( x1 )

we find that suitable canonical coordinates (u,v) are given by

v(x,y)

,

u(x,y) =

=

for some constant

x0.

f

In these coordinates the differential equation (3.23)

can be shown to become 3

+

—

+ V

+

+

+

— 0

which on using (3.28)i, (3.29) and (3.31) finally becomes

35

3

(3.33)

V

dv

Thus for third order differential equations (3.23) which are not self— adjoint, we can for a given function

p(x)

obtain a one—parameter group

(3. 1) which leaves the equation invariant provided the function

the form (3.32) for suitable constants

A, B, C

and

D.

q(x)

has

If this is the case

then the solution of (3.23) reduces to solving the third order linear equation (3.33)

the

with

If (3.23) happens to be seif—adjoint then

constant coefficients.

general solution can be obtained in the usual way for such equations from

the solutions of (3.24) (Murphy [8], page 200). For a simple illustration based on the example at the end of the previous section, suppose that

q(x)

from (3.32)

q(x)

=

x3{[A

for some constants

p(x) =

x2

then

41(x) = x½,

= x½ log x

and

must be given by + 2B log x + C(log x)2] 3D A, B, C

and

D.

ii

If this is the case (3.33) can be

solved in a straightforward manner and hence the solutions of (3.23) can be deduced.

3.4

FOURTH ORDER SELF-ADJOINT EQUATION y" +

[p(x)y']'

+ q(x)y = 0

From problems 16 and 17 we deduce that the general fourth order seif—adjoint equation

can be taken as +

In

+ q(x)y =

this section we suppose

(3.34)

0 .

41(x)

and

42(x)

are linearly independent

solutions of (3.35)

such that their Wronskian is unity.

36

On a further differentiation of (3.25) we find that the coefficient of y" Is zero provided, 2

g(x)3

f'(x) =

(3.36)

,

in which case we obtain, 4

(

1

g

dx14

g

3g2dx2

g2

(3.37)

From the equation 4 dy1

d

+

dy1

p(x1)

i—

+ q(x1)y1

= 0

and (3.37) we deduce that if the resulting equation is to be identical with (3.34) then we require

and

f(x)

to satisfy

g(x)

p(x) =

p(f)

gg'

[gig" -

(3.38)

+

+

g

g' g"'

3

g

2

-

32 g

+

g '2g" 3

g

3

g

From (3.6), (3.36) and (3.38) we have,

+

+

= 0 (3.39)

+ pE:']' +

and

=

+ 2q'E = 0

The two equations (3.39) are consistent only if

D

where

D

is a constant.

Thus in general for a given

(3.40)

p(x)

we need to 37

q(x)

assume

q(x) =

is given by

9p(x)2

D

(3.41)

,

dx

where

is a solution of

and has the general form (3.30) where

are linearly independent solutions of (3.35).

and

constants

A, B

and

are as in (3.30) then we have

C

+

—

Moreover if the

= (AC — B2)

(3.42)

.

The global form of the one—parameter group (3.1) can be deduced from dy1

dx1

= E(x1)

subject

=

,

to the initial conditions

x1 =

x,

y1 = y

when

c =

0.

Suitable

canonical coordinates (u,v) are given by =

v(x,y)

u(x,y) =

X0

where

x0

Making use of the result given in problem 18

is some constant.

3

(replacing

a(x)

and

with

B'(x)

and

E(x)1

respectively) we find

that in terms of (u,v) the differential equation (3.34) eventually becomes 4

2

+ 10(AC — B2) dv

+ [D + 9(AC — B2)2]u = 0

where we have made use of

(3.40) and (3.42).

is given by (3.41) then (3.34) constant coefficients. independent solutions

(3.43)

,

dv

Thus provided

q(x)

can be reduced to a linear equation with

Evidently the approach presupposes that the linearly and

+2

of the associated equation (3. 35) can be

readily obtained.

As an illustration suppose that solutions of (3.35) are

38

+1(x) =

1

p(x)

and

Is zero.

+2(x) =

x.

The linearly independent Hence

E(x)

is of the

form,

= A + 2Bx + Cx2 and

the

above approach is effective provided

q(x)

is given by

D

q(x)=

24'

(A + 2Bx + Cx )

for

some constants

A, B, C

D.

Consider for example

A =

0,

B

—½

C=D=1. Wehave

and

= x(x—1)

and

and

(3.43)

=

[x(1—x)f4

has the general solution

= (C1 + C2v)e

where

q(x)

,

C1, C2, C3

2

and

+ (C3 + C4v)e C4

are constants.

The solution of the original

equation can now be readily deduced.

[In the following problems

s(x)

is assumed to be defined by (3.4).

Also for

problems 4,5,6 and 7 a further arbitrary constant could be introduced into the condition restricting the coefficients.

In these problems we have assumed

that this constant has been absorbed into the constant defines

x0

in (3.4) which

s(x).]

PROBLEMS 1.

For Bernoulli's equation, + p(x)y =

q(x)y"

(n # 1)

show that =

q(x)s(x)

1-n

{(1_n)Ci

L0

+

C2}

= C1 —

39

2.

Continuation.

If the constant

C1

is non—zero deduce that suitable

canonical coordinates (u,v) are s(x)y

u(x,y)

=

+

v(x,y)

(1-n)C1

=

+ C2}

log

and therefore the differential equation becomes du

n-i

—C1).

Integrate this as a separable equation to obtain = C3

(1 —

and 3.

hence

deduce the solution of the original equation. If the constant

Continuation,

u(x,y)

s(x)y

=

,

v(x,y)

C1

is zero show that

= 1

2jx0 and that the differential equation becomes du dv

xi

2

Integrate to obtain u

and

with

40

1-n

—

C2(i—n)v =

C4

show that the same solution is obtained as in the previous problem C4 = (C2 — C3)/C1.

4.

Show that the generalised Riccati equation

E + p(x)y = q(x)

+ r(x)y2

remains invariant under (3.1) provided

r(x) =

q(x)s(x)2.

If this is the case show that —p(x)

1

—

ThX1

q(x)s(x)

q(x)s(i)

and that suitable canonical coordinates are, rx

u(x,y)

=

s(x)y

v(x,y)

,

s(t)q(t)dt

= 0

Hence show that the differential equation becomes du

2

and therefore the solution of the original equation is

y(x)

5.

=

+

(x)

c]

Show that the Abel equation of the first kind,

+

p(x)y

=

q(x)

+ r(x)y3

is invariant under the group of the previous problem provided r(x) =

q(x)s(x)3.

—1+u du dv

6.

Show that the differential equation becomes

3

Show that

+ p(x)y =

q(x)

+ r(x)log y

is invariant under (3.1) provided

q(x) =

r(x)log

s(x).

Show that, =

/

1

r(x)s(x)

= '

/

r(x)s(x)

'

41

u(x,y) = s(x)y

,

s(t)r(t)dt

v(x,y) =

and that the differential du

7.

=

log

0

equation becomes

u

Verify that the differential equation + p(x)y = q(x)ym +

admits

r(x)y"

the group (3.1) provided

r(x) =

q(x)s(x) n—rn

.

case deduce that, =

1

q(x)s(x)

fl(x)

1—rn

—p(x)

=

q(x)s(x) rx

u(x,y)

=

s(x)y

,

v(x,y)

I

=

I

s(t) 1-rnq(t)dt

JxO

and that the differential equation becomes du ——U dv

8.

rn

+U n

Show that the linear honxgeneous second order equation

can

be reduced to normal form either by,

(i) changing the dependent variable to rx -½1

y* where

a(t)dt

y=e

in which case we have

+ {b(x) or

42

(ii)

a(x)2

-

changing the independent variable to

= 0

x* where

If

this is the

(5 —I

fx

x*=j

a(t)dt

ds,

u

e

JxO

in

which case we have b(x)

dx*

9.

For the equation,

Continuation.

g

2

dx

0

dx* dx

3x 2

2

(1—x

(1—x )

2 )

show that the reductions to normal form given in the previous problem give rise to the following equations,

+

+

3

= 0

(')

dx

(ii) 10.

(1—x )

+

dx*

n(n+2)

(1—x

)

= 0

(1+x* )

With the notation of section 3.2 consider the non—homogeneous equation 2

+ p(x)y

q(x)

dx

If this equation is to remain invariant under the same group which leaves the homogeneous equation unaltered then show that

q(x)

must be

given by 3 2

q(x) = q0

where

q0

is a constant.

Hence show that the equation corresponding

to (3.21) becomes 2 du

2

+ (AC—B )u = q0

dv

43

11.

satisfies

If

,

where

K

satisfies

w(x) =

is a constant, show that

the non-

linear second order equation d2w

+ p(x)w

dx 12.

K2 =

The following three operations leave a linear differential equation linear,

y =

where

a(x)y*

changing the dependent variable to

(ii)

changing the independent variable to

(iii)

multiplication of the equation by a non—zero function

Show that by choosing

B(t) 3A(t)

=e

a(x)8'(x)

the

x*

where

x* = y(x)

such that

dt

0

,

y(x)

3

=

A(x)] —1

general linear third order equation + B(x)

A(x)

can

y

and

a,

fX

+ C(x)

+ D(x)y = 0

(*)

,

be reduced to an equation of the form, +

13.

y*

(i)

+ b(x)y =

a(x)

Continuation.

.

(**)

A second order linear equation is self—adjoint if it is

of the form

t[P(x) Show

+ Q(x)y =

that any second—order

self—adjoint by any previous problem.

44

0

linear differential equation can be made

one of the operations (i), (ii) and (iii) of the

14.

Continuation.

A third order equation is formally seif—adjoint (or

anti seif—adjoint) if it has the form (Murphy [8], page 199)

+

t{P(x)

+ Q(x)

+

=0

Show that the general equation (*) is self—adjoint if and only if

3dA

B(x)

ldf

D(x)

,

1dB

Make use of this result and the reduced equation (**) to show that no

combination of the operations (1), (ii) and (iii) of problem 12 can make a third order equation seif—adjoint unless it is originally self—adjoint. 15.

Continuation.

Show that if a third order equation is self—adjoint then

it remains seif—adjoint under (i), (ii) and (iii) of problem 12 provided

16.

is a constant multiple of

Cx(x)

Continuation.

y(x)8'(x)

Show using the operations of problem 12, that the general

fourth order equation

A(x) —f + B(x) can

+ E(x)y = 0

,

(+)

be reduced to one of the form, + a(x)

+ b(x)

dx

X

dx

by choosing

8

and

y fX

2

cz(x) 8'(x) 17.

+ D(x)

+ C(x)

Continuation.

3

= e

+

c(x)y

= 0

,

(++)

to be such that

B(t)

0

d ,

y(x) = [cz(x)8'(x)4A(x)] —1

A fourth order equation is self—adjoint if it has the

form

-44P(x)

+

t[Q(x)

+ R(x)y

= 0

Show that (+) is self—adjoint if and only if

45

B(x) = 2 18.

12 can

+

make (++) self—adjoint unless it is seif—adjoint

If (++)

originally.

—

Show that no combination of (i), (ii) and (iii) of

ContinuatIon.

problem

D(x) =

,

is self—adjoint show that these operations give

to another self—adjoint equation provided a(x)

rise

is a constant

multiple of y(x)13'(x).

[For the second part, if (++) the

is self—adjoint we have

b(x) =

a'(x)

and

equation becomes

+

dx*

(+4+)

+ R*y* = 0 ,

dx*

where, p*

= cx28'3

Q* =

cx28"+ 2cia'8" +

(ci" +

R* =

where

cz, 8,

a

and

acx"

c

+

(4cta"

Continuation.

2cx'2 + acx2)8'

a'ci'+ ccx)

are all functions of

differentiation with respect to

19.

—

x

x, primes denote

and we have taken

If in the previous problem the functions

are such that

iai

a

show that (4++) admits the factorization L2[A3L2y*] = 0

where

L2

is the second order operator defined by

L2y*

where

46

+ A2y*

dx*k A2

and

A3

are given by

y = a(x)

and

c(x)

Acx28' 20.

,

Verify by differentiation that the third order self—adjoint equation of problem 14 admits the first integral P(2yy" If

y'2)

+ P'yy' +

and

•1(x)

=

constant

are linearly independent solutions of the second

order equation + 2P'

and if

is given by

y(x)

A, B

where

and

P(2yy" — where

denote arbitrary constants then deduce that

C

y'2)

P'yy'

+

=

4(AC—B2)Pw2

+ Qy2

is the Wronskian of

w(x) U)

+ Qy = 0

d

namely

and

cp1

—

Hence conclude that this expression for

y

gives the general solution

to the seif—adjoint equation of problem 14. 21.

In the notation of section 3.3 deduce from (3.26) and (3.27) the equation,

dP(x))

[q(f) —! dP(f))g(x)3 = —

f

Hence deduce the condition (3.29). 22.

In the notation of section 3.4 deduce from (3.36) and (3.38) the equation,

-

—

{q(f)

—

.1.

=

— 9p(x)2 —

3

Hence deduce the condition (3.40).

47

4 First order differential equations

In

this chapter we discuss Lie'B fundamental problem (see Lie [1]) of finding

a one—parameter group which leaves a first order ordinary differential equation unaltered.

That is, for a given

F(x,y)

we wish to determine a one—

parameter group, x1 = x +

+ 0(c2)

(4.1)

y1 = y + cfl(x,y) + 0(c2)

,

such that the differential equation, =

F(x,y)

remains invariant.

(4.2)

,

This problem is by no means solved.

Much of the

literature is concerned with the alternative problem of finding differential equations which are left invariant by a given one—parameter group.

For this

aspect the reader should consult the standard tables of differential equations and their associated groups (see for example either Dickson [3], page 324 or Bluman and Cole [5], page 99).

We shall also consider the alternative

problem but with a view to situations not previously discussed.

For the

fundamental problem we highlight the role of singular and special solutions of (4.2) and we refer the reader to the discussion given by Page [4](page 113). Integral curves of (4.2)

z(x,y) =

constant,

evidently satisfy the first

order partial differential equation az

r + F(x,y)

= 0

.

(4.3)

We consider the invariance of (4.3) under a one—parameter group in the three variables

(x,y,z)

which we relate to integrating factors of (4.2).

sense this result provides a generalization of Lie's famous result for

48

In a

integrating factors (see problem 1).

This section deals briefly with the

group approach to partial differential equations and therefore the reader is perhaps best advised to avoid it until familiar with the material on partial differential equations described In

subsequent chapters. In the final

section of this chapter we attempt the solution of Lie's fundamental problem. Since the two functions

are not completely determined

and

by the single constraint (4.6) Lie's problem is rather to propose a second independent constraint on the group (4.1) which is in some sense compatible wIth (4.6) so as to simplify the subsequent analysis.

Here we propose that

the assumption that (4.1) Is area preserving may be such a constraint. Although the results obtained are by no means conclusive, different forms of Lie's problem are generated which at least convey some insight into the fundamental difficulties associated with the problem.

4.1

INFINITESIMAL VERSIONS OF PROBLEM

y'

AND

y' =

We calculate the infinitesimal version of

y'

F(x,y)

AND THE FUNDAMENTAL

as follows.

From (4.1) we

have,

dy 1

dx1

+0(c2) —

dx +

and on dividing through by d 1

—

dx1 —

dx 1

dy)

dx + dx

ay dx)

+

+

we obtain,

+

2

ay dx

Hence on using the binomial theorem for the denominator we have dy =

where

+ c'rr(x,y,y') + 0(c2)

rr(x,y,y')

,

(4.4)

is given by

49

dx

and this

is the infinitesimal version of

y'.

If (4.1) leaves (4.2) invariant then from (4.4), dy1 =

and

F(x1,y1) =

F(x,y) +

+

+ 0(c2)

we obtain

+ crr(x,y,y') =

F(x,y) +

cfr

and therefore from the terms of order

9x

+ 0(c2)

+ ii

c

we have

(4.6)

,

9y

where we have used (4.1) and (4.5).

differential equations determine two functions The functions

is

for

that E(x3y)

and

(4.6) is satisfied and

Lie 's fundamental problem for first order

a given

and n(x,y)

n(x,y)

can

F(x,y)

how can we systematically

such that

(4.6) is

satisfied.

be completely arbitrary provided

Equation (4.6) always admits the solution

#

However, this solution does not serve our purpose since in this case

=

when we come to deduce the global form of (4. 1) we need to solve dx1

dy1 =

=

,

and thus we are led back to our original problem (4.2). in the following section, it is also evident that

=

Further from (4.8)

is not an

acceptable solution of (4.6). If

and

fl(x,y)

are known functions then we show in the following

section that the condition (4.6) reduces to the existence of an integrating factor for the differential equation (4.2). 50

Moreover for given

E(x,y)

and

we may view (4.6) as a first order partial differential equation for the determination of

F(x,y).

Thus we may determine classes of differential

equations invariant under a known one—parameter group and this is the alternative problem which is discussed in the section thereafter.

4.2

INTEGRATING FACTORS AND CANONICAL COORDINATES FOR

If we introduce

by

A(x,y)

A =

y' =

F(x,y)

then (4.6) simplifies considerably

—

and we obtain

÷ F

afx

2

—O

4

a good deal simpler than (4.6), the interesting

aspect of (4.6) has been removed since (4.7) does not involve either If we introduce

directly. p(x,y)

by

p(x,y)

p =

A1

or

then we have

I

n(x,y)

=

F(x,y)E(x,y)

—

(4.8)

'

and (4.7) can be shown to become, (Fp) = 0

(4.9)

.

Hence if we write the original differential equation (4.2) as dy —

F(x,y)dx

= 0

(4.10)

,

then from (4.9) we see that

p(x,y)

is an integrating factor for (4.10).

This result is due originally to Lie [1] (see problem 1) and is generally given some prominence in the literature.

However, from the point of view of

actually solving differential equations the use of canonical coordinates is preferable.

Moreover as we have seen in the previous chapter canonical

coordinates can be used with higher order equations and therefore we will emphasize their use here.

From (4.8) and (4.9) we see that there exists a function

9z_ —

—F

9z '

z(x,y)

1

9y — (ri—FE)

.

such that

(4.11) 51

But we have

az

dx

dz =

dy—Fdx

dy

=0

=

Thus

where we have used (4. 10) and (4.11).

z(x,y) =

C

where

is a

C

constant represents the integral of (4.2) and we have using (4.11)

1

ay

ax

(4.12)

.

Thus if we introduce the operator a

L =

+

L

by

a ri

L(z) =

then (4.12) gives

and from the Commutation theorem (see (2.25)) we

1

have n

z(x1,y1) = e

r

z(x,y) =

C

n

L (z)

L

n0 and therefore z(x1,y1) = From

z(x,y)

+ c

(4.13)

.

thie equation and (2.7) we eee that if a fir8t order differential

equation iB invariant under a one-parameter group then the required integral

hae the form

z(x,y) (u, v)

where

=

v(x,y)

+

(4.14)

,

are the canonical coordinatee of the group and

i8

eome

function of u only. In

order to obtain (4.14) more directly we suppose that in terms of

canonical coordinates the differential equation (4.2) becomes dv

=

But clearly if this equation is invariant under then

4s

must be independent of

from the equation

52

v

u1 = u

and

v1 =

v + c

and the result (4.14) follows immediately

dv

=

$(u) Solve the differential equation,

Example 1

Y

dx

(x+x2+y2)

In this case we have — ax

—(1+2x)y

=

222' (x+x +y )

ay

(x+x2—y2)

222'

(x+x +y )

and from (4.6) we need to find =

(x+x2—y2)fl —

Unfortunately =

Try

1

and

(that is,

(x+x2+y2)

n

+y

and

n(x,y)

such that ,,2

—

(x+x2+y2)2

—

+

must now be determined by trial and

n(x,y)

constant)

then

F

error.

must satisfy

— (x+x2—y2)

=

Unfortunately even at this stage we cannot systematically solve this equation since the solution by Lagrange's method involves solving the original differential equation (see problem 3).

arrive

at the solution

=

x/y.

However, with some persistance we can

Thus the global form of the one—parameter

group is obtained by solving

dx1x1 dc

y1'

dy1 1

dc

subject to the initial conditions

x1 =

x,

y1 = y

when

£

= 0.

We obtain

y1=y+C, and the reader should verify that the given differential equation is indeed invariant under this group.

,

and

the

v=y

Canonical coordinates (u,v) are given by

,

differential equation becomes

53

dv

—

1

1

— (1+u2) dyj

1x

Thus the solution

s

v + tan 1u =

C

or

y+tan

x

Alternatively if we write the differential equation as (4.10), namely

ydx

dy—

=0,

(x+x+y) then (4.8) gives the integrating factor

as

ji(x,y)

— 2

—

2

(x +y )

and we obtain dy

+ (x dy —ydx)

22

= 0

(x +y )

This integrates to give the previously obtained result. Example 2

ObtaIn a function

a(x)

or class of functions such that the

differential equation 2

remains invariant under a one—parameter group.

From (4.6) we have

_*(a2+2ay2+y4) This condition simplifies if

and

=

+

+ 2n(x)y2 =

=

—

and we obtain + y2]

From this equation we deduce on equating coefficients of powers of = Ax

provided 54

a(x)

+

B

,

fl(x)

=

takes the form,

—A

y,

(Ax+B)

where A, B

and

dx1 —a-

2' are

C

all constants.

From

dy1

= -Ay1

(Ax1 + B) ,

we deduce that suitable canonical coordinates u = (Ax+B)y and

v

,

(u,v)

are given by

log(Ax+B)

=

from the original differential equation we obtain

dv_

1

du

2

(u

+ Au + C)

This equation is separable and can be integrated for given values of the constants

4.3

A

and

C.

THE ALTERNATIVE PROBLEM

For a given

and

fl(x,y)

such that (4.6) is satisfied? differential equation in

F

can we obtain the most general

F(x,y)

We solve (4.6) as a first order partial (see problem 3).

The characteristic equations

are

= n(x,y)

,

,

(4.15)

,

(4.16)

and

dT

3x

(3y

3X)

ay

and in order to obtain the most general

independent

F(x,y)

integrals of (4.15) and (4.16).

we need to deduce two

In general (4.16) is a Riccati

equation which we solve using the known solution of (4.6), namely

=

Making the substitution (see problem 4) (4.17)

we obtain 55

di

418

ay'

ax

lay

which is linear and can be solved in the usual way. Obtain the most general first order differential equation

Example 3

invariant under a one—parameter group of the form,

y1=g(x)y

x1f(x) ,

Infinitesimally we have

+ 0(c2)

x1 = x +

+ 0(c2)

y1 = y +

,

and therefore the characteristic equations (4.15) and (4.16) become =

= n(x)y

,

(4.19)

,

and

=

n'(x)y

+ (r1(x)

(4.20)

.

From (4. 19) we have

dx

and therefore y s(x) = A where

A

(4.21)

,

is a constant and

s(x)

is defined by

s(x) = e

(4.22)

for some constant

4! +

From (4.19)i, (4.20) and (4.21) we obtain

x0.

—

dx

which integrates to give =

where

B

+

is a constant.

equation is obtained from

56

B

(4.23)

,

Hence our most general first order differential B =

that

is

•[s(x) y]

=

dx

In this case we can

verify

that suitable canonical coordinates

(u,v)

are

given by 1X

u(x,y) =

s(x)y

v(x,y)

,

=

dt

0

and that the differential equation becomes du

[Notice

=

this example generalizes problems 4, 5, 6 and 7 of Chapter 3.] Obtain the most general first order differential equation

Example 4

Invariant under the one-parameter group, =

where

k

ky

fl(x,y)

,

=

ky

is a constant.

In this case we have from (4.15)

dx and

therefore (4.24)

where

A

is a constant.

Making use of the result given in problem 6 we have

on performing the integration

(4.25)

where

B

is a constant and

s(x)

is defined by (4.22).

Since w =

we have from (4.17)

w =

and hence

—

n(x))

the required differential equation is

57

s(t)k )

—

dt +

= s(x)k

denotes an arbitrary function.

where

dt]

—

Suitable canonical coordinates are

—

dt

u(x,y) = y

v(x,y)

,

dt =

and on using du

du_ dx

-

1

dv

dv

1

dx

du

we see that the differential equation becomes ku du — e dv 4(u)

The remaining sections of this chapter are devoted to various aspects associated with Lie's fundamental problem and the condition (4.6).

4.4

THE FUNDAMENTAL PROBLEM AND SINGULAR SOLUTIONS OF z(x,y) =

Suppose the integral

C

F(x,y)

y' =

of (4.2) is solvable for

y

so that we

have y =

S(x,C)

(4.26)

.

But from (4.13) we see that if (4.2) is invariant under the one—parameter group (4.1) then y1 = and

S(x1,

therefore —

C-I-c)

on equating terms of order

C

we have

lasi +

where the partial derivatives in brackets refer to two arguments =

58

x

and —

C.

y

as a function of the

From this equation and (4.2) we deduce

F(x,y)E(x,y)

.

(4.27)

Now

A =

satisfies (4.7) and using (4.27) we see that (4.7) could be

—

deduced alternatively in the following two ways. Firstly, (4.7) follows from differentiating (4.2) partIally with respect to

In the bracket notation for the partial derivatives (4.2) becomes

C.

= F(x,y)

(4.28)

,

on partially differentiating with respect to

and

C

we obtain,

laxlaCJJ But we have, +

=

from which (4.7) can be deduced.

Secondly, (4.7) follows from the compatibil-

ity of the two equations (4.27) and (4.28) which the reader can

readily

verify.

We see from (4.27) that if and

E(x,y)

n(x,y)

fl(x,y0) =

y

y0(x)

Is a singular solution of (4.2) then

must be such that

F(x,y0)E(x,y0)

(4.29)

.

Hence, if as is often the case a singular solution of (4.2) is known, then (4.29) might well suggest the general nature of

and

n(x,y).

These

considerations indicate that singular solutions of first order differential equations perhaps play a more vital role than has been previously considered.

4.5

INVARIANCE OF THE ASSOCIATED FIRST ORDER PARTIAL DIFFERENTIAL EQUATION

In this section we consider the invariance of the associated first order partial differential equation (4.3).

We use the group approach for partial

differential equations which is described in detail in subsequent chapters. We look for a one—parameter group of transformations in three variables (x,y,z)

which leaves (4.3) invariant.

We use the convention that subscripts

59

denote partial differentiation with

x, y

and

as three independent

z

variables.

Suppose that the one—parameter group x1

+ 0c2

=x+

y1 = y + CT)(x,y,z) + 0(c2) z1

=z+

(4.30)

,

÷ 0(c2

— 3z1

leaves (4.3) unaltered. 3z

3z

3x

ax1

3y

c[c

3x1

—

+

=

and

as follows,

! .iL

÷

= 3x1

We calculate — and

+

therefore —

=

p+

c{c

+

+ 0c2

—

—

(4.31)

Similarly, az

=

3y1

_L

+

ax 3y1

__! ay

3y1

+

=

+

c +

+ s

+ 0(c2)

÷

—

and hence —

=

÷

+

—

13)2

+ —

(4.32) If

60

z =

is a solution of (4.3) then by invariance we have

=

therefore

z =

+ n(x,y,z)

=

and

also satisfies

C(x,y,z)

(4.33)

Now from,

+ F(x1,y1)

i—

=0

and (4.30), (4.31) and (4.32) we can deduce

+ FC

where

0

and we have used

=

—

+ F(ri-FF)

=

=

(4.34)

,

-F0.

From (4.3) and (4.33) we have

C =

and therefore (4.34) gives + —p--

(FO) = 0

(4.35)

.

Hence

0, that is

if

z

satisfies an equation of the type (4.33) as well as (4.3) then (4.35)

can

be deduced immediately since from (4.3) and (4.33) we have

is an integrating factor for (4.10).

-FC —

Clearly

C '

—

and (4.35) follows from the compatibility of these equations. In the above we have used the so—called non—classical approach for partial differential equations described in subsequent chapters.

We have shown that

the first order condition for invariance of (4.3) under (4.30) is equivalent to the existence of an integrating factor for (4.10).

Moreover this condition

conveys no more information than the condition for the compatibility of (4.3) and (4.33).

If we apply the claasical approach for partial differential

equations then on equating coefficients of 00 and deduce that

C =

$(z)

and that

derivatives as given in (4.7).

A =

—

0

to zero in (4.34) we

satisfies (4.7), with partial

Thus the classical approach. gives rise to the

61

well known result that if where

4.6

p

is an integrating factor then so also is

is the integral of the differential equation.

z

LIE'S PROBLEM AND AREA PRESERVING GROUPS

In this section for a given differential equation (4.2) we attempt to solve (4.6) assuming that the one—parameter group (4.1) is area preserving.

That

is, we assume there exists a sufficiently continuous and differentiable function

such that

G(x,y)

,

=

=

(4.36)

.

—

From (4.6) and (4.36) we obtain the second order partial differential equation for

G(x,y) 2

= ,(G,F) + F2 aG

+ 2F

(4 37)

2

which we require to solve for a prescribed function

F(x,y).

we can solve this equation by introducing two functions

In principle

A(G,F)

and

B(G,F)

such that

p = A(G,F)

= B(G,F)

,

The compatibility condition for

(4.38)

-

G(x,y)

together with (4.37) yields two

equations for the determination of the first order partial derivatives of F(x,y)

and the compatibility condition for this function gives the final

equation for

A(G,F)

and

Although the equation obtained is no more

B(G,F).

tractable than (4.37) the analysis does merit some simplifying features which would seem worthwhile reporting.

The following analysis should be contrasted

with other possible restrictions concerning the nature of the one—parameter group.

For example if

and

are assumed to be given as the

gradient of some function then this assumption appears to compound the subsequent analysis rather than simplify it.

62

The simplifying features

associated with (4.36) may not be due to the fact that the group happens to be area preserving but rather to the fact that (4.36) is embodied in the general expressions for G(x,y)

E(x,y)

(see (2.13)).

and

More precisely

is an invariant of the group and (4.36) results from (2.13) in the

case when the Jacobian In (2.13) Is a function of

u

only.

In order to solve (4.37) be means of (4.38) we need to assume that the Jacob Ian

—

a(G,F)

(4 39)

is non—zero and finite.

We also need the following elementary relations B

J

Writing the compatibility equation for

+ a(G,F)

G(x,y)

in the form

=0

we obtain

A + FB

F2

we see from (4.7) that (4.37) can be written as — 0 —

which on simplification yields, (4.42)

We note that it is in the derivation of (4.42) that the assumption (4.36) appears to significantly simplify the analysis. On solving (4.41) and (4.42) for

aF/ax

and

we obtain

63

aF 3x

aC

aC

I

Fac

IF3F

(4.43) 3F

where

3C

is given by

H(G,F)

H(G,F)

(444)

—

=

that

We note from (4. 38) and (4.43)

the given differential equation (4.2)

becomes

(4.45)

From the above equations we find after a long calculation that the

compatibility condition for

F(x,y)

becomes

(4.46) which can be written

as

(4.47) On comparing this equation with (4.7) we see that

(4.47)

is the statement

that (4.45) remains Invariant under the one—parameter group with infinitesimals

and =

—

B)

fl*(G,F)

given by

fl*(G,F) = —c

,

(4.48)

.

Thus an integrating factor for dF

is therefore

—

so that

64

(4.49)

— H(G,F)dG = 0 ,

the

.—

n*Y'.

n* = c2 3(A:B)

Now we can verify that —

compatibility condition for

(4.50)

F(x,y)

reduces to the statement

that the differential form

dG +

dF

— C

(.

2a(A,B)

)

a(G,F)

Problems 13 and 14 illustrate the above analysis

is an exact differential.

with two simple solutions of (4.47). expression for the Jacobian

For specific examples we need an

defined by (4.39).

J

From (4.38), (4.39),

(4.43) and (4.50) we find that —

j

=

C = A + FB

Using

(4.52)

.

dA +

(

we can simplify (4.51) to give — C dB —

— 0

c2 a(A,B) a(G,F)

Thus with

C1

=

dA +

the condition (4.47) is equivalent to the statement that dB = 0

AB

(4.53)

,

That is the compatibility condition for

is an exact differential.

F(x,y)

becomes

laB'

t'aA)

j

a(A,B)

a(A,B)

where the functions —

'

a(A,B)

are defined by

and —

1

(A+FB) A

4)

a(A,B)

—

F

455

- (A+FB)

particular method of solution of (4.54) is outlined in problems 15, 16, 17

and 18.

It is worthwhile noting that the differential forms (4.51) and (4.53) are consistent with that obtained from the requirement that

(A+FB)1

integrating factor for (4.9), provided we make use of the for

SF/ax

and

aF/ay.

Since from (4.40) and using

C = A + FB

must be an (4.43)

we have 65

dy—Fdx_ (A+FB)

C dF

ax JC

—

+

F

and (4.43), (4.44) and (4.52) yields precisely (4.51). PROBLEMS 1.

If the differential equation M(x,y)dx + N(x,y)dy = 0 is invariant under (4.1) show that

the

infinitesimal condition is

equivalent to the existence of an integrating factor

ji(x,y)

where

1

=

2.

is an integrating factor for both of the differential

If

equations,

M(x,y)dx ÷ e =

show that

3x

= 0

tan'(M/N) +

v20 =

3.

N(x,y)dy

and

N(x,y)dx —

M(x,y)dy

= 0

satisfies

= 0 ay

For the quasi—linear first order partial differential equation a(x,y,z)

÷ b(x,y,z)

= c(x,y,z)

(*)

,

show that the general solution is given by p =

where

is

•

an arbitrary function and

p(x,y,z)

and

o(x,y,z)

two independent integrals of the system of differential equations dx

[From

=

a(x,y,z)

p =

constant

dT 66

x

dT and

=

a =

b(x,y,z)

dz ,

= c(x,y,z)

constant we have

+bay +cOz =0,

are any

where subscripts denote partial differentiation with x, y three independent variables.

and

z

as

These two equations together with (*)

constitute three homogeneous equations for

a, b

and

c.

For non—trivial

solutions the determinant vanishes and this condition can be shown to become = o

from which the required condition fOllows.

derivatives are with = p

x

4.

If

+ p

as the independent variables, that is

y

=a+ a

,

etc.]

,

y

is a known

y0(x)

and

x

In the Jacobian partial

solution

q(x)

+ p(x)y =

of the Riccati equation

+ r(x)y2

show that the substitution

y = y0

+w1 gives rise to the linear

equation

+ 5.

—

p(x)]w

=

—r(x)

Show that the most general first order differential equation which admits the group =

is

=

,

-

-

=

where

s(x)

n(x)y +

-

s(t)dtj

is given by 1X

dt

s(x) = and

e

is an arbitrary function of the argument indicated.

67

6.

Using (4.15)2 and

(4.18),

make the substitution w =

and deduce the

equation, 1 —

7.

Show that the most general first order differential equation invariant under the group n-i

E(x,y) =

=

,

ri(x)y n

is -

dt + = s(x)

5.

is as defined in problem

Using canonical coordinates rx

u(x,y) =

s(x)y

,

v(x,y)

1

=

i

n1 s(x) n—i y

s(t) n—i dt

show that the differential equation becomes du dv 8.

u

n

For the Riccati equation given in problem 4, show that i y(x)=— r(x)z(x)

the

substitution

dz —, dx

gives rise to the linear equation,

+

+ —

q(x)r(x)z

= o

Deduce the normal form of this differential equation. 9.

Continuation.

E(x,y) =

Show that Riccati equation of problem 4, admits the group

E(x)

,

=

+

,

+

2rt

provided, (re)

68

and

-

+

=

satisfies

From these equations deduce that 1

+ 4{qr where

section

1

=

Can you reconcile this result with that of

a constant.

is

C

1

the equation,

3.2.

[The following three problems sununarize the three criteria given by

Dickson [3] (page 313) for the invariance of a differential equation under a one—parameter group.] 10.

Show that a

first order ordinary differential equation is invariant

under the group x1 = x +

if

+ 0c2

,

y1 = y + cfl(x,y) + 0(c2)

(**)

and only if Lz =

where

a

L =

and 11.

is the integral of the equation,

z(x,y)

+

is the operator

a

denotes an arbitrary function.

I

Continuation.

The differential operator associated with

(+)

N(x,y)dx + N(x,y)dy = 0 ,

is

L

given by P = N

The

ax

- M ay (LP)

is defined by

(LP) = LP — PL

Show that, (i)

(LP) =

—

.

69

the differential equation (+) is invariant under (**) if and only

(ii)

if the commutator (LP) is a constant multiple of the operator 12.

The first extension of the operator

Continuation.

L

is

L'

P.

where

1+71_L 3y' and where

is

ii

-

+

=

ii

given by

-

y'2

Show that the first order differential equation F(x,y,y') = 0

remains invariant under (**) if and only if L'F = 0

[The following six problems relate to section 4.6.] 13.

Assuming that

where

A =

f(G)

,

f

and

g

B =

g(G)F'

are functions of

C

only, show that equation (4.47)

simplifies to yield f"(l+f/g) +

f'(f/g)'

= 0

where primes denote differentiation with respect to equation and show that

J — ag

where

C.

Integrate this

is the integration constant.

a

Hence from the relations (4.40) deduce that fF

1

Co

where

and

C0

are further integration constants.

From these

results show that the original differential equation (4.2) in this case has the form —

for 70

(ay+B)h[ctx + log(ay+8)]

some arbitrary function

h

,

of the argument indicated.

Observe that

this equation can be solved by the substitution p = ax + log(ay+8) 14.

Assuming that

is identically zero show from (4.44) and (4.47)

H(G,F)

that

{Z'(F)[Z(F)G+m(F)]

B =

—

[L(F)m'(F)—m(F)&'(F)]}

C

where

+ n(F)

]½

2R(F)

[Z(F)G+m(F)]½

=

£, m

and

n

denote arbitrary functions of

denote differentiation with respect to

F.

F

Show that

and here primes J =

£(F)/2

and

from the relations (4.40) deduce

x =

+

£(F)

for some functions

and

p(F)

p(F),

'

+ q(F)

= £(F)

With s(F) = q(F) — Fp(F)

q(F).

show

that in this case the original differential equation (4.2) is the well known Clairaut's equation (Murphy [8], page 65) =

dx

+ (dxJ'

which has general solution 15.

y =

yx + s(y)

Assuming there exists some function

w(A,B)

for some constant such that

4s

and

y. iji

as

defined by (4.55) are given by

show that equation (4.54) becomes 2

a(A,B)

where the Laplacian

V2 =

V

2

a(A,B)

(**)

is given by

+

Show that (**) can be written alternatively as 71

________

v2 a(w,G) =

—

16.

(***)

a(A,B)

a(A,B)

a(A,B)

ContinuatIon.. From (4.55) and (*) conclude that

w(A,B)

satisfies the

first order partial differential equation aw

aw

BaA

1,

tan'(B/A)

+

AaB and

hence w =

denotes an arbitrary function of the argument indicated.

f

Introducing polar coordinates R = (A2+B2)½

tan'(B/A)

e =

,

show that F = where to 17.

[B—Ag(R)]/[A+Bg(R)]

g(R) =

Rf'(R)

and the prime denotes differentiation with respect

R

Continuation.

coordinates observe that (***) of problem 15

(R,e)

In

becomes 1 a(w,G) = 1 fa(w,V2G) — a(V2w,G)t, a(R,e) R a(R,e) R a(R,e) J

where

V

2

is given by

v2

R

R

On using w — e

+

f(R)

R aR

where 18.

72

g(R)

Continuation.

R

ae2

show that (****) simplifies to give

R2 ae2

RJaeJ

aR

is as defined in problem 16.

With

G = R2

and

h(G)

g(G½)

show that

G½[F+hGJ

G½[1_Fh(G)] A

B

,

{[1÷F2][1÷h(c)2])½

Hence show that

J =

(1+F2)/2

{[1+F2][1+h(G)2]}½

and that the relations (4.40) yield,

apart from arbitrary additive constants 2G½[F+h(G)]

2G½[ 1—Fh(G)]

X

{[1+F

2

2½' ][1+h(C) ]}

2

{[1+F

][1+h(G) ]}

Hence conclude that the original differential equation (4.2) in this case is

fy

=

-

xg[(x2+y2)½/2]j

+ yg[(x2+y2)½/2]J

dx

which Is solved using polar coordinates (see Murphy [8], page 67). 19.

Given the one—parameter group x1 = x +

y1 = y + cfl(x,y) + 0(C2)

+ 0(c2) ,

show that y1(x1—x0) = y(x—x0) +

Y(x_x0))]

—

(x—x0)

+ fl(x—x0, y(x_x0))} + 0(c2)

[This result can be verified by two distinct methods. (i)

Suppose that

y =

S(x,C)

y(x—x0) = S(x—x0,C) = S(x1—x0,C+c)

=S

and

and

,

,c-I-c) + 0(c2)

and

+

y(x-x0) +

where the partial derivative of x—x0

C

then

y1(x1—x0)

= S(x—x0,C) +

=

y1 = S(x1,C+c)

(x_xO) + S

+

+ 0(€2)

with respect to

C

has arguments

and is found from (4.27) to be given by 73

y(x—x0)) —

=

(x-x0)

y(x—x0))

from which the required result follows. Alternatively we have

(ii)

d

y(x—x0) = e

y(x)

and we require to find d

-x0

y1(x1—x0) = e

1 y1(x1)

x = x1

+ u(c )

From

A2

—

we have

+0ic2)

dx1{1

dx

and therefore

dx

dx1

Hence if we define the differential operators D

D

dx'

1

D1

and

D2

by

2

then we require to evaluate y1(x1—x0) = and type.

Knapp [12]

x = x

74

—

x0 ,

y = y(x—x0)

e

that

D2y(x+t) = In

(page 40) give formulae for operators of this

Observe that, e

and

+ cn(x,y)] +

-

(x,y)

(x+t)

order to calculate the order of

c

term arising from

we use the integral given in [12] (page 40).

=

y(x-x0)

We have

O[Dy(x+T)]*

+ U

dT + 0(c2)

where the 'star' in the integrand denotes that (x,y) in the square bracket becomes (x—x0—T, y(x—x0.-T)).

If in the integral we make the

substitution p

then

x0

x

T

we have

=

y(x—x0)

(x-x0)dp + 0c2

(p,y(p))

+

and thus = y(x—x0) ÷ c

(x-x0)

E(x.-x0, Y(x.-x0))]

+ 0c2 The result now follows since =

crl(x-x0,

y(x.-x0)) ÷

0(c2

Note that we can check the validity of this second method by using the integral given in [12] to evaluate x1 — x0 =

+ cE(x,y)] + 0(c2)

Proceeding as above we obtain x1

x0 = x — x0 + cE(x,y) + 0(c2)

which of course is the desired result.] 20.

Continuation. x1 = x +

Obtain a one—parameter group

+ 0c2

,

y1 = y + cTl(x)y + 0c2

which leaves the following differential—difference equations invariant, (a)

dx

(x) = —y(x—x0)

75

(b)

(x) =

y(x)[1

(e)

(x) =

y(x)[y(x)

y(x—x0)] —

y(x—x0)]

Can we use these groups to simplify or integrate the equation? [Answers:

(a)

x1=x+ac,

y1=e8cy,

(b)

x1=x+C,

y1=y,

(c)

x1e

where

a

and

-1

$

(1—c

—aE:

),

cic

are arbitrary constants. d

y(x—x0) = e

y1=e y,

dx

y(x) =

r L

n=0

(—x ) 0 n!

Notice that since,

n y

(x)

differential—difference equations are really 'infinite' order differential equations.]

76

5 Second and higher order differential

equations First

order differential equations can be Invariant under an infinite number

of one—parameter groups.

Second and higher order equations differ in that

they are invariant under at most a finite number of groups. equations are invariant under at most 8 while for differential equations are invariant under at most Dickson [3], page 353).

n >

2,

Second order nth order

n + 4

groups (see

Higher order equations also differ from first order

ones in that if there exists a one—parameter group leaving the equation invariant then this group can be systematically determined.

Much of the

literature is concerned with obtaining the most general second order differential equation invariant under a given group and again we advise the reader to consult standard tables of such differential equations (see for In the first section of this chapter we

example Dickson [3], page 349).

deduce the condition (5.8) for a second order differential equation to be invariant under a one—parameter group.

In the next section we give four

examples making use of this condition.

In the section thereafter we give

examples of the determination of the most general second and higher order differential equations invariant under a given one—parameter group.

In the

final section we give three applications from the nuclear industry due to Axford [14], [15], [16] and [17].

5.1

We

INFINITESIMAL VERSIONS OF

y"

AND

y"

=

consider the general second order differential equation 2

t I

dx

2

y, dxJ

'51 77

and look for a one—parameter group, =

x +

+ 0c2

,

which leaves (5.1) invariant.

y1 = y + cfl(x,y) + 0(c2) ,

(5.2)

Throughout this chapter we shall use the

notation (5.3)

80

that

from results given in the previous chapter we have

(5.4)

= z + cvr(x,y,z) + 0(c2)

where

ir(x,y,z)

is given by

ay

ax

In

order to calculate the infinitesimal version of

y"

we proceed as

follows, 2

d y1 2

dx1

d

dx

dy1

dx1

dx dx1

and thus we have

ax

dx2

ay

az dx2

lax

(5.6)

ay

)dx2J

If (5.1) is left invariant by (5.2) then on using F(x1,y1,z1) =

F(x,y,z)

+

+

+

+ 0(c2)

and (5.1) and (5.6) we deduce the condition that (5.2) leaves (5.1) invariant, namely

tax

78

ay

az

tax

ay

ax

ay

az)

(57)

If

involves powers of

F

then generally we can determine

z

from (5.7) by equating coefficients of the powers of

and

On using (5.5) we

z.

find that (5.7) becomes

(ay

axJ

(5.8)

axjazj

ay

ax2 aXaY

ax az

ax

ay azj

=0. In the following section we illustrate with examples how solutions and

of (5.8) can be deduced.

5.2

EXAMPLES OF THE DETERMINATION OF

Example 1 n(x,y)

Show that if

AND

ri(x,y)

is independent of

F(x,y,y')

y'

and

then

take the following forms,

p(x)y

E(x,y) = If

((x,y)

+

,

n(x,y)

is independent of

F(x,y,z)

z

=

p'(x)y2

+

+

.

(5.9)

then (5.8) becomes

(5.10)

ax2 + ILI1

-

2 -

2

3

ay2

lay2

=0, and from the coefficients of

axay

'

and

we have

ay2 79

We notice that from the coefficients of

and (5.9) follows inmiediately.

we also have

z°

and

z

3p"(x)y +

3p(x)F

—

[p'(x)y2

[p(x)y + =

Hence either

p(x)

(5.11)

+

[p"(x)y2 + rf'(x)y + t"(x)J

+

—

+

is identically zero and

and

=

F(x,y)

is

a solution of the partial differential equation

+ or

p(x)

t(x)J

+

=

—

is non—zero in which case

F(x,y)

+

+

t"(x)J

must be a linear function of

y

(see problems 1 and 2). Example 2

y" = 0

Show that

is invariant under precisely 8 one—parameter

groups.

From (5.8) with

identically zero we have

F

2

2

2

(5.12)

2

and as in the previous example we deduce that of the form (5.9).

From the coefficients of

must be

and z

and

z°

in (5.12) and (5.9)

we deduce =

and

2fl'(x)

,

p"(x)

=

rf'(x)

=

t"(x)

hence

p(x) =

C1x

+ C2 ,

T)(x) = C3x + C6 , where

C1, C2, ..., C8

= C3x2

t(x)

+ C4x + C5

= C7x + C8

denote arbitrary constants.

gives rise to a one—parameter group leaving example the group generated by

80

= 0

y" = 0

C3, that is take

Each of these constants invariant.

C3 =

1

Consider for

and assume all the

other constants are zero.

dx1

From

dy1

2

X1)'1

dc

x1 = x

and

y1 = y

,

c = 0

when

we find that the global form of the group

is

x We

x (1—cx)

= 1

y 1 = (1—cx)

have +

= (1—ex)

so that clearly y" = Show

Example 3

=

is

cy

0

,

=

remains

(1-cx)3

invariant

under this group.

that the differential equation

Xf +

,

not invariant under any one—parameter group. With

F = xy + e

+

z

we see that (5.8) becomes

3

z2 -

- 2

tay2

+ ez

From the term involving

,

so that C3

ax

ay

= C1x + C2

we deduce

ax' and

denote arbitrary constants.

involving

eZ

n(x,y) = C1(y—x) + C3 From the coefficient of

where z°

C1, C2

and

in the term not

we have

—C1xy = y(C1x + C2) + x[C1(y—x) + C3] 81

which is clearly only satisfied if

C1 = C2 = C3 = 0

.

Hence there is no one-

parameter group leaving the given differential equation invariant. Obtain the most general invariant one—parameter group for the

Example 4

second order differential equation 2

+ p(x)y = 0

(5.13)

.

dx

As in example 1 with

and

y)

F(x,y) = —p(x)y

are given by (5.9) and from (5.11)

y)

we deduce

p"(x) + p(x)p(x) = 0 =

p(x)C(x)

C"(x) +

fl"(x) +

,

= 0

+

(5.14)

,

= 0

and hence the given differential equation is invariant under 8 distinct one— parameter groups.

We notice that the group arising from

that considered in section 3.2.

The group arising from

the invariance of (5.13) under the addition to We consider

p(x)

in more detail.

y

and

ri(x)

merely reflects

of any solution of (5.13).

For this group we find that suitable

canonical coordinates are

v(x,y)

,

u(x,y) =

=

and we have du d(uv)

=

dx —

Hence on using (5.13) and (5.14)1 we obtain

= p(x)

-

p"(x)y

= 0

,

dx2

and therefore we have u = C1uv + C2

where

82

C1

and

C2

are constants.

is

From this equation we readily deduce

1X

y = C1p(x)

dt

+ C2p(x)

p(t) 2

I

which is a well known result for the general solution of (5.13). It is also worthwhile noting that embodied in (5.14)2 =

i)

(ax) = 0,

is the group C

=x,

y1 = e y

which reflects the invariance of all linear homogeneous equations under stretchings of

In this case suitable canonical coordinates are

y.

v(x,y)

u(x,y) = x , and

with

w(u)

w(u)

the

=

=

log

y

defined by

dv

differential equation (5. 13) becomes the first order Riccati equation

(Murphy [8], page 15), dw 2 1-+w +p(u)0.

5.3

DETERMINATION OF THE MOST GENERAL DIFFERENTIAL EQUATION INVARIANT UNDER A GIVEN GROUP

In the notation of section 5.1 suppose have

deduced two independent

invariants

and

for given A(x,y)

and

B(x,y,z)

n(x,y)

we

(say) of the

characteristic equations dx

where

=

dT

Tr(x,y,z)

=

dz ,

is defined by (5.5).

= ir(x,y,z)

,

(5.15)

The basic result for the determination

of the most general second order equation invariant under this group is that this equation is given by =

where

•(A,B)

,

is an arbitrary function of the arguments indicated.

(5.16) Clearly (5.16) 83

is of second order and is invariant under the given group.

In order to see

that there can be no more general equation than (5.16) we refer the reader to the comment following problem 5.

We remark also that the most general third order differential equation invariant under the given group is given by

B,

=

where

'V

(5.17)

,

denotes an arbitrary function. Obtain the most general second order differential equation

Example 5

invariant under the group x1

y1 =

f(x)

g(x)y

and hence deduce the most general linear invariant second order equation. From example 3 of the previous chapter we have already obtained

s(x)y

A =

where

B =

—

-

—=

(5.18)

,

are as previously defined.

and

s(x),

fl(x)y]

2ri(x))

Now

-

X

,

dA

(5.19)

and on incorporating the denominator of (5.19) into the arbitrary function of (5.16) we deduce the required second order equation

+ fE'(x)

fl(x)2

— 2

(5 20)

E(x)

dx2

where

=

—

denotes an arbitrary function and

A

and

B

are given by (5.18).

The most general linear homogeneous second order differential equation invariant under the given group is obtained by taking

to be given

by = aA

84

+

,

(5.21)

where

cx

and

are

B

We find from (5.18), (5.20) and (5.21) that

constants.

this equation becomes + b(x)y = 0

+ a(x)

where a(x) =

2

—a +

{fl(x)2 —

and on eliminating

between these two later equations we obtain —

+

1

da —

a(x)2)

=

a

— B2

Hence this result is consistent with that obtained in section 3.2 (see also problem 8 part (i) of Chapter 3).

Find the most general second order differential equation

Example 6

invariant under the group =

where

k

ky

ky

=

,

is a constant.

From example 4 of Chapter 4 we can deduce

A—y+logs(x) B =

-

n(x))

with the usual definition for dB = dA

-

s(x).

ok

dt}

(5.22)

,

On differentiating we can show

+

-

k

n'(x)1

—

ldxJ

—

J

and thus the required differential equation is

+

k

=

+

— (5.23)

85

denotes an arbitrary function and

where

A

and

are given by (5.22).

B

Hence a differential equation of the form (5.23) can be reduced to the first order equation

+ kB +

— 0

Obtain the most general linear third order equation of the form

Example 7 3

+

p(x)

dx

+ q(x)y = 0

X

(5.24)

,

which is invariant under the group given in example 5. In the notation of example 5 we have from (5.18) and (5.19) =

B

s(x)

+

—

+

(x))y

—

On differentiating this equation with respect to

result by

+ (dB)2}

+

+

=

+

+ 3n(x)2 —

-

-

+

the

and multiplying the

we obtain

B

B{B

Hence

A

—

31i(x)E'(x)

—

-

most general third order equation invariant under the given group

is obtained by equating the expression on the right—hand side of this equation to

1P1(A,B,dB/dA) = ciA

B,

where fl(x) =

86

a,

and —

y/3

y

+

.

The most general linear equation arises from + yB

are constants. and we find

In order to obtain (5.24) we require

—+

/

(x) =

3

-

q(x) =

+

+

+

+

and thus

which agrees with the result obtained in section 3.3.

APPLICATIONS

5.4

In this section we consider three specific second order differential equations which arise from various problems in the nuclear industry.

These applications

are due to Axford [14], [15], [16] and [17] and we refer the reader to these papers for the full motivation and detailed analysis.

For additional

applications the reader should consult Bluman and Cole [5] (page 116).

For

our purposes the problems considered illustrate the scope and limitations of the group approach for problems arising from a practical context. Example 8.

Reactor core optimization

In Axford [14] for the problem of

determining the appropriate fuel distribution which minimizes the ratio of the critical mass of the core to the reactor power when the power output is prescribed, the following differential equation is obtained

- y'2

+ where

y(x)

—0

+

denotes the non—dimensional thermal flux, primes denote

differentiation with respect to two cases considered are geometry while For

cx =

(5.25)

,

0

a =

1

ci

x

and

ci

and

are known

constants.

The

= 0 which corresponds to assuming a slab

corresponds to a cylindrical geometry.

equation (5.25) becomes 87

(5.26)

and it is instructive to solve this equation first by standard devices and then by the group approach. we let

y'

z =

Since (5.26) does not depend explicitly on

x

and we obtain in the usual way the first order differential

equation

y

dy

(5.27)

z

We recognise this as an equation of the Bernoulli type and therefore set w = z2

and deduce w = C1y2 - 28y3

where

C1

denotes an integration constant. dy

½

On integrating

= dx

we obtain in a straightforward manner C'

(5.28)

y(x) = + C2)]

B[1 + where

C2

constant

denotes a further integration constant and we are assuming the C1

is positive.

Alternatively we may deduce the general solution

(5.28) from the group approach in the following way.

We observe that (5.26)

remains invariant under the two one—parameter groups,

x,=x-i-c,

y,=y,

(5.29)

x,ex,

y,e —2cy.

(5.30)

and

The group (5.29) is merely the formal statement that (5.26) does not depend explicitly on

x

again set

y'

88

z =

and

therefore since

and obtain (5.27).

y'

is an invariant of this group we

However the group (5.30) means that

(5.27) is invariant under

y1=e —2c y,

z1=e —3c z,

and therefore we select

u* = zy

3

2

as the new dependent variable and (5.27)

becomes the separable first order differential equation du* —

(u*2÷213)

—

yu*

dy —

This equation readily integrates to yield

C1/y

=

- 2B

from which the solution obtained previously can be deduced. For

cx

non—zero equation (5.25) is still invariant under (5.30) and we

therefore select

u = yx2

substitution, t =

log

as the new dependent variable. and

x

p =

du/dt

With this

equation (5.25) becomes the Abel

equation of the second kind (Murphy [8], page 25) p

=

2(cx-1)u

- Bu2 -

(cx-1)p +

This equation can be reduced to standard from (see equation (1.20)) by the p = qu

substitution q

which for for

= 2(cx—1)

cx =

1

find

We

.

—

— (cz—1)

(5.31)

,

can be integrated to finally obtain the following solution

y(x), namely 2C 1

y(x) =

+ where again

C1

other values of

and cx

(5.32)

+ C2

denote integration constants.

We note however for

the solution of (5.31) is by no means apparent.

The next two examples discussed in detail in [15], [16] and [17] arise from the steady state heat conduction equation with non—linear thermal

89

conductivity

and non—linear source terms

k(T)

divlk(T)grad T] + S(T) = 0 where

T

S(T), that is

(5.33)

,

denotes the temperature.

These problems arise in the context of

thermal instability phenomena in rods and plates in the sense that if the rate of energy produced by the heat source exceeds the rate at which energy can be transferred out across the boundary then a steady state temperature distribution cannot exist (see Axford [15]).

The particular thermal

conductivity and source terms considered are, k(T) =

k0(T/T0)1

(5.34)

,

S(T) = S0(TIT0)6 , where

y, 6, k0,

S(T)

and

= S0 exp(T/T0)

(5.35)

,

denote constants.

T0

Power law conductivity and source term

Example 9.

In Axford [15] and [16]

the following differential equation in non—dimensional variables is deduced from (5.33), (5.34) and y1 {y" +

where

a

and

+ yy '

a, y

a =

1

and

(5.

,

are constants and again

slab geometry while three constants

= 0

+

a = 0

corresponds to the plate or

corresponds to a rod or cylindrical geometry. 6

36)

The

encompass a wide variety of physical behaviour

and the full analysis of (5.36) involves consideration of a number of special cases which are detailed in [15] and [16].

Here we restrict our attention to

results which can be deduced rapidly and simply from the group approach.

For

any practical problem we first examine simple groups leaving the equation invariant before using the theory given in the first section of this chapter and equation (5.8).

For the examples of this section, this aspect is

summarized in problems 9, 10 and 11.

90

If we look for a simple stretching group

y1=eacy,

c x1=ex,

(5.37)

leaving (5.36) invariant then we fin.d 2

(5.38)

a =

provided take

S #

1

u =

+ y.

Assuming for the time being that this is the case we

as the new dependent variable so that equation (5.36) becomes

+

+ (cx+2a)xu' + a(cx+a—1)u} +

and with the substitutions

t =

log

and

x

p =

du/dt

this equation becomes

+ (a+2a+2ay—1)pu + a(a+a+ya—l)u2 + yp2 +

up

= 0

= 0

In general this is again an Abel equation of the second kind. cases give rise to standard equations. is homogeneous while if

a

1

and

cS =

For example if

S =

(1+y)(a+3)/(cx—1)

(5.39)

.

However special

y + 3

the equation

the equation is of

the Bernoulli type. If

cS =

1

+ y

then (5.36) becomes (5.40)

which is invariant under the one—parameter group x1 = x

C

y1 = e y

,

Thus we take

w =

y'/y

(5.41)

.

as the new dependent variable (see example 4) and

(5.40) becomes

+

w + (1+y)w2 +

= 0

(5.42)

,

which is a Ricatti equation (Murphy [8], page 15). are special cases which can be readily solved.

Clearly

a = 0

or

y =

—1

For the final example we

consider the equation arising from (5.33) for the case of constant thermal

91

conductivity and exponential source term. Example 10.

Constant coliductivity and

exponential

source term

With non—

dimensional variables Axford [15] and [171 obtains the equation

y" + For

a

=

(5.43)

.

zero this equation can be readily integrated by means of the

standard substitution

z =

dy/dx.

a

For

non—zero we look for a one—

parameter group leaving (5.43) invariant of the form £

x1 = e x We find

a =

—2

y1 = y + ac

,

(5.44)

.

and therefore on eliminating

as an invariant of the group.

u =

With

c

from (5.44) we obtain

u

as the dependent variable

(5.43) becomes

x2(uu"—u'2) + axuu' + 2(1—a)u2 =

and the usual substitutions = p2 + (1—a)up —

up

For

a =

1

t

=

log

and

x

2(1—a)u2 —

p =

du/dt

yield,

(5.45)

.

equation (5.45) is the same as equation (5.27) and therefore the

solution can be deduced from (5.28).

However for

a #

1

(5.45) must be

solved as an Abel equation of the second kind.

The examples of this section illustrate how simple groups leaving the equation invariant may be utilized to reduce the order of the differential equation.

The resulting differential

standard type

equations may or may

not

be of a

with a simple solution.

PROBLEMS 1.

In the notation of example 1 of section 5.2 show that if zero then F(x,y) =

92

G(x)y

+ H(x)

p(x)

is non-

where

p"(x) CC) X — —

and deduce that

-

for constants

C1

and

Continuation.

If

p(x)

that

F(x,y)

— —

—

3p(x) must be such that

and

p(x),

+

2.

H( X )

p(x)

P(X)

dt

= C1

PX

+

p(t)

C2.

is identically zero and

=

E'(x)

is given by 1X

+

F(x,y) = ½

Jx0

E(x) where 3.

If

4

p(x)

show

E(t)

denotes an arbitrary function of the argument Indicated. is non—zero but arbitrary show that the differential equation

2

+ p(x)y2 = 0 dx

is invariant under at most 6 one—parameter groups.

Show that

2

+

=0

dx

can be reduced to a first order Abel equation of the second kind (Murphy [8], page 25). 4.

Find the most general second order differential equations which are invariant under the one—parameter groups given in problems 5 and 7 of Chapter 4.

5.

Show that the second order differential equation F(x,y,y',y") = 0 is invariant under the one-parameter group x1 = x + cE(x,y) + 0(c2)

,

y1 = y +

+ 0(c2

,

(*)

93

if and only if L" F = 0

where

L"

is the second extension of the operator

where

a

a

L

and

ir

a

a

are the infinitesimal versions of

a

and

y'

and

y"

(Note that if

respectively and are defined by (5.5) and (5.6). B(x,y,y')

L, namely

A(x,y),

are three independent integrals of the

C(x,y,y',y")

equations =

1(x,y)

=

,

= 7T(x,y,y')

,

=

,

then the most general second order equation invariant under (*) takes = 0

the form

or

C =

[For a detailed discussion of the following problems the reader should consult Dickson [3], page 358.] 6.

Given two one—parameter groups with operators L1 =

11(x,y)

a

+

a ri1(x,y) -b-—

L9 =

show that the first extension of the commutator

E2(x,y)

a

+ ii2(x,y)

(L1L2)

a

is identical to

the commutator of their respective first extensions, that is

(L'1L'2).

(See problems 11 and 12 of Chapter 4.) 7.

Continuation. y" =

F(x,y,y')

operators

L1

Show that if the second order differential equation, is invariant under two one—parameter groups with and

group with operator 8.

Continuation.

If

then it is also invariant under the one—parameter

L2

(L1L2). L1

and

L2

that there exists an operator invariant and is such that (L1L3) = aL1 + bL3 94

leave L3

y" =

F(x,y,y')

invariant show

which also leaves the equation

for some constants 9.

a

and

b.

For the second order differential equation

(+)

x dx

dx2

y

functions

that for all

y

the only one—parameter groups leaving (+) invariant

f(y)

take the form

g(x)

i,(x,y) =

n(x,y) =

where

denotes an arbitrary constant and

A

functions of

=

Continuation.

g(x)

and

h(x)

are

such that

x

+

10.

+

+ A}

-

If

—

2g'(x)}

f(y) —

nf'(y)

show that if

f(y) =

6

y +

1

the only one—

parameter group leaving (+) invariant is given by =

If

iS =

y + 1

x

n(x,y)

,

= (1+y—6)

show that the differential equation (+) Is invariant under

the group (++) where

and

g(x)

h(x)

satisfy the following differential

equations

g" +

=0

+ a(2—a) N—

h" + 11.

h' +

= 0

If

f(y) = Be3'

ContInuation.

and

y = 0

show that for

a

1

the

only one—parameter group leaving (+) invariant is given by (-H-) with g(x)

and

h(x)

given by

95

4A

2A

where g(x)

A

and

is

the arbitrary constant in problem 9.

12.

B

Show that

ac_

h(x) = 4A

log x + 2(2A-B)

is a further arbitrary constant.

the

classical diffusion equation 2 ac 2'

ax

admits travelling wave solutions of the form —

B(x)]

is a constant and

A(x)

c(x,t) =

where

w

A(x)sin[wt

A" = AB'2

where

and

B(x)

satisfy

AB" + 2A'B' + k2A = 0

,

k = (w/D)½.

Observe that

+ k2A2 — 0

remains invariant under the one—parameter group

x1=x,

C A1=eA,

and deduce the second order differential equation

w" + 6w' + 4w3 + 2k2(w' + w =

96

a =

are given by

h(x)

g(x) = —2Ax log x + Bx ,

where

For

A'/A.

1

show that

6 Linear partial differential equations

For partial differential equations the calculations involved in the determination of a one—parameter group leaving the equation invariant are generally fairly lengthy.

In order to keep these calculations to a minimum

we first consider a restricted class of one—parameter transformation groups applicable to linear partial differential equations. considered in the following chapter. variable

c

Non—linear equations are

Specifically for a single dependent

and two independent variables

x

and

t

we consider

transformations of the form x1 =

f(x,t,c)

= g(x,t,c) c1 =

h(x,t,c)c

where the functions

= x

=

t

+

+ 0(c2)

+ cn(x,t) + 0(c2)

= c +

f, g

(6.1)

,

+ 0(c2) and

h

do not depend explicitly on

c.

If the

transformation (6.1) leaves a given partial differential equation invariant and if

then from

c =

c1 =

4(x1,t1)

on equating terms of order

c

we have

+

fl(x,t)

For known functions

=

E(x,t), ri(x,t)

(6.2)

.

and

t(x,t), equation (6.2) when solved

as a first order partial differential equation, yields the functional form of the similarity solution in terms of an arbitrary function.

This arbitrary

function is determined by substitution of the functional form of the solution into the given partial differential equation.

In the case of two independent

variables the resulting equation is an ordinary differential equation.

For 97

more than two independent variables the procedure reduces the number of independent variables by one.

In the following section we give the formulae for the infinitesimal versions of the partial derivatives

ac/ax, Dc/Dt, D2c/Dx2, D2c/axat

and

Although we make no use of the last two partial derivatives, they

D2c/3t2.

are included for completeness.

For the remainder of the chapter we

principally consider groups of the form (6.1) and the corresponding solutions of diffusion type equations.

In particular we consider the classical

diffusion equation 2

(6.3) Dx

and

the

Fokker—Planck equation which we assume given in the form,

DC =

where

p(x)

D

p(x) and

DC1

q(x)

D

+ j—(q(x)c)

(6.4)

,

are functions of

x

only.

In the determination of

groups leaving an equation invariant there are two methods, termed and non—classical.

The classical approach equates the infinitesimal version

of the given partial differential equation to zero without making use of equation (6.2).

The non—classical procedure which is considerably more

complicated makes use of (6.2) and includes the classical groups as special cases.

For the most part we obtain results from the classical procedure.

However in the final section we discuss the non—classical approach with reference to equation (6.3).

The results given in this chapter for (6.3) are

due to Bluman and Cole [18] and Bluman [19J(see also Bluman and Cole 15], page 206) while the general equation (6.4) was first studied by Nariboli [20 1. [20] several special examples are analysed in detail. present some new results for equation (6.4). how the most general function 98

q(x)

In section 6.5 we

We show for arbitrary

p(x)

can be found such that (6.4) admits a

In

classical group of transformations leaving the equation invariant.

6.1

FORMULAE FOR PARTIAL DERiVATIVES

For the one—parameter group of transformations (6.1) we assume that the Jacobian, a(x1,t1)

a(X,t)

=

ax1

(6.5)

=

is non—zero and finite. =

1

+

+

From (6.1) and (6.5) we have

+

(6.6)

.

Now for the partial derivative ac1/ax1 a(c1,t1)

ac1

= 3(x1,t1)

we have

a(C1,t1) = J

(6.7)

a(X,t)

and on substituting (6.1)2, (6.1)3 and (6.6) into (6.7) we obtain

=

3x1

3x

+

+

—

a(x,t)

+

+ 0c2

which simplifies to give (6.8)

Similarly from a(c1,x1) 1

at1

= — a(x1,t1) = — J

(6.9)

a(x,t)

we obtain

L=ft÷

c{c*+

If we introduce

+0c2

(c

and

ff2

.

(6.10)

by

99

f

IL

z

ie

31

(89)

pUP

=

fxe

(o19) +

=

+

+

+

(zr9)

•

iapio

' '3ZC —

I

—

z

ç

Ix UIo1J

(99)

'(zr9) e

=

no

(ii

+

+

-

2ujsn 1(ir9)

XC

xe

xe

+

—

1

+

woij

——

13e

xe —

1OJ

001

+

+

—

—

ç

— —

(11''x)e

'

I

—

c

(E19) suaflr?nba

xe

xe

xe

xc

xe

xe

+

-

Z9

-

+

NOISfkIdIQ NOLLVflöa

U0T439S

UI

9) (01

I

dnoi2

a3flpap 9q4

(E9)

30 aq4 UOTST13TP

(ci

9)

+ xe

'Xe

xe

xe

+

•

asn 30

31

PUT3

9q4

l9pUfl

UOT

UOTSfl33TP

(r9)

9q4

SUOT43Un3

ipns

xe

z

+

xe

xc

+

(cr9)

'

(cr9)

ST

30 dnoi2

30

STU.L ST

1Cldurrs

30

10I

derivatives to zero.

we deduce that

From the coefficient of

the coefficient of =

11

(t)x

n(t) while

from

we have

ac/st

+ p(t)

n =

(6.16)

,

where the prime here denotes differentiation with respect to denotes an arbitrary function of

t.

t

and

p(t)

On equating the coefficient of

to zero and making use of (6.16) we deduce that

[flutX2

=

denotes

+ P'(t)x) + o(t)

(6.17)

,

where

a(t)

of

in equation (6.15) and using (6.17) we obtain

c

—

1

[n" (t)x2

a further arbitrary function of

+

phI(t)x)

+

a'

(t)

+

n"(t)

t.

From the coefficient

=

from which it is apparent that we require = 0

p"(t)

,

= 0

a'(t)

,

=

—

From these equations it is now a simple matter to deduce the classical group of the diffusion equation, namely = K + 5t + Bx + yxt

where

T1(x,t) =

a +

C(x,t) =

-y

a,

y,

x

2

5, A

Sx

ti

+

-

and

T+ K

A

denote six arbitrary constants and for

comparison purposes we have adopted the same notation used in Bluman and Cole [5] and [18].

Some of these constants give rise to standard or even

trivial solutions of (6.3).

However it is instructive for the reader to

deduce the global form of the one—parameter group and the resulting similarity solutions of the diffusion equation. 102

The constants

K,

a

and

A

represent

respectively the invariance of (6.3) under translations of stretching of

c

The constants

(see problem 1).

B,

y

x

and

and 6

t

and

are

considered in the examples of the following section.

6.3

SIMPLE EXAMPLES FOR THE DIFFUSiON EQUATION

The general classical similarity solution of (6.3) is obtained from (6.2) and (6.18) with all the constants in (6.18) non—zero.

For purposes of illustration

it is useful to consider the solutions arising from one non—zero constant with the others taken to be zero. Example

= 1,

B

I

a = y = 6 =

=

K

0.

=

In this case the global form

of the one—parameter group is obtained by solving

dt1

dx1

dc1

— x1

2t1

—

— 0

subject to the initial conditions x1 = x when

c =

t1

,

0.

=

t

c1 = c

,

(6.19)

,

In this case we find

x1=ex,

c1c,

t1=e2ct,

so that clearly the constant simultaneous stretchings of

B

x

reflects and

the invariance of (6.3) under From (6.2) we obtain the partial

t.

differential equation

+ 2t

x

=0

which on solving gives rise to the functional form previously considered in Chapter 1 (page 7). 6 = 1,

Example 2

a = B=

=A

= K =

0.

In order to deduce the global

form of this group we require to solve dx

dt

=

t1

dc

= 0

x

=

— T c1 103

with initial conditions (6.19). We find = x + ct,

t1 =

t

c1 = c exp[_

,

—

Further

from (6.2) the functional

t

x

=

—

form of

the solution is obtained by solving

C

which yields c(x,t) =

denotes

where

(6.20)

,

an arbitrary function of

t.

On substituting (6.20) into

(6.3) we readily deduce the ordinary differential equation

+

= 0

and therefore

— denotes

where

From this equation and (6.20) we

an arbitrary constant.

see that the constant

also

iS

gives rise to the well known source solution

(1.29).

y = 1,

Example 3 dx1

cx =

=A

8=

dt

=

x1t1

,

= K =

dc

=

,

0. x2

=

together with the initial conditions (6. 19).

t

+

c1

,

(6.21)

From (6.21)2 we have (6.22)

=

and

In this case we have

therefore

(6.21)i becomes

dx1

x1t

dc

(1—€t)

which on integration yields X1 104

= (ict)

(6.23)

Using (6.22) and (6.23) in (6.21)3 and integrating the resulting equation we find

2) c1 =

(6.24)

exp 4(l—ct)J

In order to determine the functional form of the corresponding similarity solution we have from (6.2)

(6.25)

.

On solving this equation we find that

c(x,t) =

e

—x2/4t

(6.26)

, —

t

where

r

denotes an arbitrary function of the argument Indicated.

On

substitution of (6.26) in (6.3) we find that we have simply =

so the constant

c(x,t) =

gives rise to the solution

y e

—x2/4t

+

'

x —

t

which again includes the source solution (1.29) as well as the solution of (6.3) which is the derivative of the source solution with respect to

x.

Thus although no new solutions are obtained by considering separately the constants in (6.18), these simple examples illustrate the basic procedure in simple terms.

In order to obtain non—trivial results we need to consider

the full group (6.18).

This is done in the following section with reference

to moving boundary problems (see also problems 6, 7, 8 and 9).

6.4

MOVING BOUNDARY PROBLEMS

Problems involving the classical diffusion equation (6.3) and an unknown moving boundary

x =

X(t)

occur in many areas of science, engineering and

105

industry (see Ockendon and Hodgkins [21] and Wilson, Solomon and Boggs [22]). The literature on these problems is scattered throughout many diverse disciplines and it is not possible here to consider the subject in detail. The purpose of this section is to identify the moving boundaries which remain invariant under the classical group (6.18).

x =

X(t)

These boundaries

relate to most of the exact analytic results which are available for such problems and therefore might provide a useful guide to the solution of other problems with unknown boundaries.

Typically a moving boundary problem takes the form

O<x<X(t) (6.27)

c(X(t),t) = 0 together with either prescribed on

(X(t),t)

,

—k(t)

(or a linear combination of these)

or

c

=

and prescribed initial data for

x = 0

c

and

X.

We note

that the dot denotes differentiation with respect to time and that for some problems the initial condition on

c

may

not

Such problems are

be present.

However it is instructive to observe the precise nature

clearly non—linear.

of this non—linearity for those problems which can be transformed to fixed Assuming that both

boundary value problems.

zero and that the prescribed data for explicitly involve

'r =

,

=

and with

t

c(x,t) =

=

then

(l,'r)

and

on

or

5C(t)

x = 0

are never does not

we can make the transformation.

X(t)

C(p,'r)

c

X(t)

(6.28)

,

the moving boundary problem (6.27) becomes

,

—

0 < p < 1

,

(6.29) C(1,T) = 0

106

,

(1,T)

=

Thus in principle we have transformed (6.27) to a fixed boundary value problem except that now the equation to be solved is non—linear, although not of the usual type of non—linearity with which we are familiar. for prescribed data

c(0,t) = c0

and

X(0) = a

where

c0

For example,

and

a

are

constants and no initial condition we supplement (6.29) with

T(0)

C(0,i) = c0 , In this

case

X(t)

C =

/2d0}

4(p).

We can readily deduce

(6.31)

,

is a root of

b =

b = c0e

and

(6.30)

.

a solution exists in the form

= c0

where

= a

—b/2

—hcJ2/2

tie

i(t)) is

(or

+ 2bt)2

X(t) = (a2

do1

(6.32)

,

given by

(6.33)

.

We observe that (6.33) actually includes the well known moving boundary X(t) =

as a special case (see Crank [231, page 99).

Moreover

Langford [24] has generated general solutions of the diffusion equation with a moving boundary of the form (6.33).

These solutions have been further

generalized by Bluman [19] and 'I'ait [25].

moving boundaries

page 235) which

X(t)

Here we simply consider the general

due to Bluman 119] (see also Bluman and Cole [51,

remain invariant under (6.18).

Applications of the resulting

similarity solutions involve complicated eigenfunction expansions and we refer the reader to the original papers [19], [24] and [25] for these details as well as for related references. When

deducing

the functional form of the solution

c(x,t)

from (6.2) we

require two independent integrals of the system of differential equations,

107

= E(x,t)

= ?(x,t)c

= n(x,t)

,

(6.34)

.

,

If we suppose

that

dx =

6 35

dt

admits the integral

w(x,t) =

constant then w

is the similarity variable

and from dc = c dt fl(x,t)

with

x

(6 36)

replaced by

x =

x(w,t)

we can deduce (on treating

w

as a

x

and

constant) that the solution takes the functional form, c(x,t) =

w

If we from

in both

and

being regarded as a function of

now consider the boundaries

x1 =

X(t1)

x =

X(t)

t.

left invariant by (6.1) then

we have

dX =

(6

dt

so that the similarity variable w

38)

defines the invariant boundaries from the

equation

w(X(t),t) = where

(6.39)

,

denotes an arbitrary constant.

In particular for the classical

group we can deduce from (6. 18)1, (6.18)2 and (6. 38)

df

K+ót

X

3. )

(6.40)

J

In the integration of (6.40) there are four cases which must be considered separately. Case (i)

ay.

In this case we find that the invariant boundary takes

the form X(t) = At + B + 108

,

(6.41)

where the constants A =

A

and

,

B

are defined by

B

K8—&Z

=

(6.42)

czy-8 We

observe that (6.41) contains

as a special case and that since it

contains six arbitrary constants

8,

y,

6, K

and

it may perhaps be

utilized as an approximate expression for an unknown moving boundary. 82 =

Case (ii)

ay,

Case (iii)

8

=y

X(t) = Kt +

=

0,

t

#

0.

6

(6.43)

.

In this case we find

62 +

=8=y

Case (iv)

w =

For this case we have

+ w0(t+B) —

X(t) =

simply

y # 0.

(6.44)

.

=

0,

6 #

0.

For this case the similarity variable is

and the invariant boundaries are therefore

t =

constant.

In Bluman [19] (and Bluman and Cole [5], page 235) the above moving boundaries are exploited for an inverse moving boundary problem in the sense that the heat input on the moving boundary is not prescribed but rather

determined so as to be consistent with the assumed special form of moving boundary.

The resulting solutions of (6.3) corresponding to the above four

cases are outlined in problems 6, 7, 8 and 9.

6.5

FOKKER-PLANCK EQUATION

In this section we consider classical groups of the form (6.1) which leave the Fokker—Planck equation (6.4) invariant.

Nariboli [20] considers in some

detail a number of special cases of (6.4) and also gives extensive references to physical and biological applications of (6.4).

Bluman [26] gives a

detailed analysis of a boundary value problem for the special case of (6.4) with

p(x)

equal to a constant (see also, Bluman and Cole [5], page 258).

109

Here we first obtain the general form of the group for arbitrary This analysis has not been given previously.

q(x).

results for particular functions introduce the functions = JX

dY

1(x)

and

1(x)

,

J(x)

and

p(x)

J(x)

p(x)

We then illustrate these It is convenient here to

q(x).

which we define by

+ g(x)

=

(6.45)

p(y)

in the interval under

p(x)

Further throughout this section primes denote differentiation

consideration.

with respect to the argument indicated.

From the formulae for the transformed partial derivatives (6.8), (6.10)

we find that the

and (6.13) and making use of (6.4) to eliminate condition for invariance of (6.4) becomes

- lulic. -

+

C

Ic.

+

—2

—

=

+

2

+

is a function of

t

matter to deduce that =

ri'(t)

+ çq'(x))

(6.46)

.

=

n(t),

that

it is a simple

From the coefficient of

only.

is given by

P(X)½ 1(x) + p(t)P(X)½

denotes an arbitrary function of

integral defined by (6.45)i.

110

c(Eq"(x)

we have immediately that

From the coefficient of

obtain

q'(x)c}

—

k

+ (p"(x) +

p(t)

+

2

2

+ (p'(x)

where

(p'(x)

(6.47)

t

and

From the coefficient of

1(x)

is the indefinite in (6.46) we

where

p'(t)

—

—

=

—

—

denotes a further arbitrary function of

o(t)

defined by (6.45)2.

+ a(t)

.acji). t

and

On substituting the above expressions for

into the equation obtained by equating the coefficient of

c

(6.48)

, J(x) E,

is

and

in (6.46) to

zero we obtain the equation,

n"(t) 1(x)2 + p"(t) 1(x) —

n'(t)

=

where the

+

function

o'(t)

—

p(x)½

+

,

(6.49)

is defined by

2p(X)½ J'(x) + J(x)2 -

=

n(t)

4q'(x)

.

(6.50)

In the analysis of (6.49) there are two distinct cases to consider. Case (1)

=

where p(t)

In this case

C11(x)2 + 2C21(x) + C3 ,

C1, C2 and

non—zero.

p(t)

denote arbitrary constants and the functions

C3

and

n(t),

are obtained by solving the following equations,

o(t)

4C1r1'(t)

—

(6.51)

= 0 3C

p"(t) —

o'(t) —

Case (ii)

C1p(t) —

n"(t) 4

p(t)

=

,

—

zero.

n'(t)

34

(6.52)

—

In this case we have

+

=

2C11(x)2 + C3

which upon integration gives

=

C11(x)

2

C4

+ C3 + 1(x)

2

(6.53)

111

where

C4

and

ri(t)

denotes a further arbitrary constant.

a(t) are

determined by solving the following equations,

0

n"(t) — 4C1T1'(t)

In

(6.54)

"I

——

For this case the functions

n

p(x), we obtain the most general

both cases f or given

q(x)

such that

(6.4) admits a classical one—parameter group of transformations leaving the equation invariant, by solving

4q'(x)

+ J(x)2 —

where for case (1)

f(I)

= C112

f(I)

=

f(I)

(6.55)

,

is given by

(6.56)

+ 2C21 + C3 ,

while for case (ii) we have C4

f(I)=CI 2

(6.57)

1

We solve (6.55) by taking defined by (6.45)1.

I

as the independent variable where

1(x)

is

From (6.55) we have

(6.58)

,

and on using J =

(log

+

equation (6.58) simplifies to yield the Ricatti equation f(I)

where

u

u =

If

112

for

(6.59)

,

is defined by

(log

the time

being

—

(6.60)

.

we assume that

u =i1(I)

is the solution of the

Ricattl equation (6.59) then from (6.60) we see that for a given function p(x)

the function

q(x), such that (6.4) admits a classical group, is given

by

q(x) =

=

p'(x)

-

,

1(x)

and we observe that in general Clearly,with

f(I)

(6.61)

,

J

2

contains four arbitrary constants.

given by either (6.56) or (6.57), equation (6.59) has

a number of simple solutions for special values of the constants and

C4.

C1, C2, C3

For solutions of the Ricatti equation we refer the reader to

Murphy [81 (page 15).

Here we give the general solution of (6.59) in terms

of confluent hypergeometric functions, that is solutions of the second order different ial equation

zw"(z) + (c—z)w'(z) — aw(z) = 0 where

a

and

c

(6.62)

,

are constants (see Murphy [8], page 331).

If

c

is non—

integer then (6.62) has linearly independent solutions,

1F1(a,c;

w1(z) =

z)

(6.63)

w2(z) = where

2—c; z)

is defined by

1F1(a,c; z)

k

(a)

1F1(a,c;

k

k=0

where the symbol (a)o =

(6.64)

,

a) =

Is defined by

(a)k

1

(6.65)

(a)k

=

a(a+1)(a+2)

We observe that if

a

(a+k—1)

(k

1)

is a negative integer then the series (6.64) reduces.

to a polynomial expression.

113

(6.59)

In order to reduce we make

the

to a second order linear differential equation

transformation

u(l)

(6.66)

and (6.59) becomes

(6.67)

We consider the two cases separately. Case (i)

non—zero and

p(t)

f(I)

given by (6.56).

In this case we let

2

C

,

v(I)

z)w'(z)

—

z =

=

(6.68)

,

+ and (6.67) becomes

zw"(z) +

—

where the constant

a =

Case (ii)

z =

1

= 0

(6.69)

,

is given by

a

—ci

fcc 1

+

aw(z)

(6.70)

3/2 8C1

zero and

p(t)

—i- i2

v(I)

,

given by (6.57).

f(I)

=

In this case we let

(6.71)

,

and (6.67) gives 2m —

zw"(z) + where the constants

+

a =

and

a

in

z]w'(z)

,

m =

in

aw(z) = 0

—

,

(6.72)

are given by

—

(1+C4)½)

.

(6.73)

+ We note that when

114

C4

is zero the soluticn for this case coincides with that

for case (1) when the constant

is zero.

C2

As a simple Illustration of the above, consider case (I) when the constants and

C1, C2

are such that

C3

a, as given by (6.70)

For case (I) we have from (6.66), — 1)

u(I) =

and when

a = —(1/2)

value —(1/2).

and (6.68)

(6.74)

,

the linearly Independent solutions of (6.69) obtained

from (6.63) are essentially (that Is, apart from an arbitrary multiplicative constant)

*

w1(z) = e

z

eY

I

— z

—i-j

dy

y

(6.75) =

From (6.74) and (6.75) we deduce that for case (I) with

a = —(1/2)

the

solution of the Ricatti equation (6.59) becomes -

(1_z){C5

u(I)

dY) - z½ezl

—

(6.76) z z

C5

denotes a further arbitrary constant and

z

as a function of

I

is defined by (6.68)i.

In the following section we consider a number of special cases of (6.4).

The first example is due to Bluman [261 who gives a similar analysis relating to the confluent hypergeometric functions but with reference only to the special case

p(x) =

1

and

q(x)

arbitrary.

The remaining examples are due

to Nariboli [201.

115

6.6

EXAMPLES FOR tHE FOKKER—PLANCK EQUATION

Example

1(x)

= x

J(x)

,

arbitrary. In this case we have from (6.45)

q(x)

p(x) = 1,

4

q(x)

=

while from (6.50) we obtain 2

4(x) =q(x)

—2q (x)

Further (6.49) becomes ri"(t)

X

8

2

+

p"(t) 2

X

—

—

ri'(t)

=

a

+

÷

q'(x)

As before there are two cases to consider. Case (1)

In this case

non—zero.

p(t)

q(x)

must satisfy the Ricatti

equation q(x)2

— 2q'(x)

p(t) and

and Case

(ii)

p(t)

are obtained from (6.52).

a(t)

zero.

In this case

+ 4,(x) = 2C1x2 + C3

$' (x) so

= C1x2 + 2C2x + C3

that

q(x) 2

Bluman

+ C3 +4

= C1x2

a(t) are obtained from (6.54).

and

and

[See

C

2q'(x)

—

[26]

or Bluman

and Cole [5] (page 258)

for a full discussion and

application of this example.] Example 5 from case (i)

p(x) = 1,

of the

C1=b2, 116

q(x)

=

bx

where

b

previous example we have

C2=0,

C3=—2b,

is a constant.

In

this

case

and therefore

p(t)

n(t),

4b2ri'(t)

—

b2p(t)

p"(t) —

are

a(t)

determined from the equations

= 0

= 0

' \/ +

——

4

/

and

b

2

From these equations we readily deduce n(t) =

a +

+

bt +Ke—bt

p(t) = óe

o(t)=ybe —2bt where

a, 8,

y,

+A,

6, K

and

denote six arbitrary constants.

A

Altogether

using (6.47) and (6.48) we obtain E(x,t) =

bx(8e2bt

ri(x,t)=a+8e C(x,t) = A ÷

2bt

+ Kebt)

+

—

+ye —2bt

bye—2bt

bóxe

—

bt

2 2

— b x

2bt

which is in agreement with the result given by Nariboli [20].

As a simple illustration consider the case constants zero.

8

=1

and the remaining

The global form of the one—parameter group is obtained by

solving dx1 =

dt1

2bt1 ,

=e

dc1

2bt1 ,

2 2 2bt1 = —b x1e c1

subject to the usual initial conditions (6.19). X1

x

=

(l—2cbe

I

1

—

log(l—2cbe

2bt )

)

222bt cbxe (1—2cbe

Further

=

2bt ½

We find

2b )

the partial differential equation (6.2) becomes

117

bx

ac

-b22 xc

ac

+

which has similarity variable

w = xe

—bt

and the functional form of the

solution is given by

c(x,t) = e On

—bx2/2

—bt )

substituting this functional form into (6.4) with

p(x) =

I

and

q(x) = bx

we obtain simply = 0

,

so that

+

c(x,t) = where

The more general solution

denote arbitrary constants.

and

types for this example are summarized in problem 12. Consider the equation,

Example 6 ac

2

a —i

=a

(xc) + b

a

(xc)

ax

where

a

and

denote arbitrary constants.

b

In this case we have the

following results:

q(x)=a+bx,

p(x)=ax, ½

J Cx) = b

,

1(x) = 2 2

b —x

½

+ 3[a) ½

3a

From these results we find from (6.49) that a(t)

are determined from —

b2ri'(t)

=—

Altogether 118

we find,

= 0

,

p(t)

is zero and

ri(t)

and

=

fl(x,t) =

—

where

bx[8e1)t

—

+ ye

b[Bebt

and

3, y

x,

bt

a + A

—

—bt

ye_bt)

—

—

denote four

.A

arbitrary constants.

The various

solution types for this example are summarized in problem 13. ConsIder the equation

Example 7

2 a2 = —i (x c)

+ b

(xc)

ax

where

b

p(x)=x2, 1(x) = and

(6.49)

In this case we have

is again an arbitrary constant.

log

q(x)=(b+2)x, x

J(x)

,

=

(b+3)

,

4(x)

=

(b+1)2

becomes,

(log x)2 +

p"(t)

log

—

n"(t)

a'(t)

n'(t) (b+1)2

Thus we obtain, =

x

+ yt2

+

p(t) = K + cSt

a(t) = where

a,

+

-

5, K

13, y,

-

and

A

+ A

denote six arbitrary constants.

From (6.47)

and (6.48) we find x + (K-h5t)x T)(x,t) =

=

a + —

—

+ (log x)2 —

(b-i-i)2

{S +

(a+2$t+yt2) —

log x

—

(b+3)(K+ót) +

A

and the various solution types are simunarized in problem 14. 119

Finally in this section we remark that Nariboli [20] also details the analysis for the following two equations., 2

ac =

a

r

i(1—x ) Ct

ax

ac

1

a

at

m

where

ma rm

1x

4ax For these equations we have

denotes an arbitrary constant.

respectively in the notation of (6.4) p(x) = (1—x

p(x) We

x

22 )

q(x)

,

= —4x(1—x

mi-I

mx

q(x)

,

2 )

m

refer the reader to [20] for the subsequent analysis of these equations.

6.7

NON-CLASSICAL GROUPS FOR THE DIFFUSION EQUATION

In this section for the diffusion equation (6.3) we make use of equation (6.2) so that the left—hand side of (6.15) depends on ac/ax.

We then equate the coefficients of

c

c

and

only through ac/ax

c

and

to zero and obtain

two equations for the determination of the non—classical groups of (6.3). For the one—parameter group (6.1) we introduce

A(x,t)

and

B(x,t)

defined

by

B(x,t)

,

A(x,t) =

(6.77)

,

=

so that (6.2) becomes

—Ac—B-—. ax at

(6.78)

On differentiating (6.78) partially with respect to

x

and making use of

(6.3) it is a simple matter to deduce 2

axat

r

(

ax

jax

(6.79)

On substituting (6.78) and (6.79) into (6.15) and then equating to zero the 120

______

coefficients of

and

c

then remarkably the resulting equations

ac/ax

simplify to give,

aAa2A ax

2A3B ax'

2

(6.80)

2 ax

ax

2

3x

These equations determine the non-classical group of (6.3).

Although equations

(6.80) are non—linear and clearly a good deal more complicated than the original problem, we observe that any special solution of (6.80) can be employed to reduce the diffusion equation to an ordinary differential equation. If we assume that

for some function

B =

integrating (6.80)2 with respect to of

I'(x,t)

then on

and neglecting an arbitrary function

x

t, we have a2

(6.81)

ax

On substituting these expressions for

A

and

namely

single equation for the determination of aci,

2

atax

ax

2

49x

into (6.80)i we obtain a

B

2

a3ci

8

ax

ax

2

ax

2

2

+

8

ax

2 ax)

+

ax axat

= 0

We simply note that the classical

This equation is clearly complicated.

group (6.18) arises from the case A(x,t) = 0 ,

where

h(x,t)

B(x,t)

=

—

(6.82)

.

lxxx = 0

and that if

C(x,t)

is zero then

(6.83)

,

denotes any solution of (6.3).

The resulting solution

c(x,t)

satisfies ac ax

h 121

Example 8

•(x,t) = ci log x

If

a = —(1/2)

a =

or

A(x,t) = 0

—(3/2).

a = —(1/2) then

If

B(x,t)

,

then from (6.82) we have either

—

=

so that (6.78) becomes ac at

which

19c_

has general solution

c(x,t)

2

=

j- +

=

On substituting this functional

t

form

into (6.3)

= 0 ,

and

therefore +

c(x,t) =

where If

denote arbitrary constants.

and

a = —(3/2) A(x,t) =

+

—

then

B(x,t)

,

=

—

I

and (6.78) becomes ac

3ac

3

x

2

c(x,t)

=

x4(w) ,

w =

j— +

3t

and from (6.3) we deduce c(x,t) =

where

122

and

+

3t)

+ q1x

denote arbitrary constants.

we find that

PROBLEMS 1.

Show for the diffusion equation that the one—parameter group arising from the constants

and

K

A

in (6.18) (that is, with

=

y

= 6 =

0)

becomes

t1t+ac,

X1=x+KC,

c1=ecc,

and that the functional form of the solution of (6.3) is c(x,t) = e

Ax/K

4(czx—Kt)

and relate this

Hence deduce the ordinary differential equation for result to problem 12 of Chapter 5. 2.

For the diffusion equation with the constants

and

6

cx =

y

= K =

A

= 0

and

8

non—zero in (6.18), show that the global form of the one—

parameter group becomes, + t

=

,

t1

c1 = c exp

=

+4

Show also that the functional form of the solution is

c(x,t)

tJ

48

and obtain the resulting ordinary differential equation for 3.

With

where

c(x,t) =

y =

(x/X(t))2

4.

deduce from the diffusion

equation (6.3) X(t)2

+ =

Hence conclude that if

X(t)

+ YX(t)X(t))}

takes the form

X(t) = (a+28t)½

where

a

and

8

are

arbitrary constants then (6.3) admits separable

solutions of the form 123

c(x,t) =

and

satisfies

4)(y)

y4"(y) +

—

denotes a further arbitrary constant.

a

Show by a simple change

of independent variable that this equation reduces to the confluent hypergeometric equation (see Langford [241). 4.

With

where

c(x,t) = dT =

X

=

1

X(t) =

(x+2Bt+yt2)1

(*)

and 1

=

2• xXt

exp

4X(t)

½

X(t)

verify that the diffusion equation (6.3) simplifies to give

aT

2

ap

2

where

5.

4

½ With

Continuation.

where

c(x,t) =

defined by equation (*) of the previous problem and

exp {-

=

show that the multi—dimensional diffusion equation, ac

—

a2c

2+

ax

(2v+1) ac x

ax'

becomes

ap

where

124

5

-

+ (2v+1)

=

p

ap

4

is as previously defined (see Taft [25]).

p,

T

and

X

are as

is given by

6.

if

82

82 > cxy show from (6.18) that the equation

and

corresponding to (6.37) becomes

T

I8+yt—(8

(a+28t+yt

I

2 ¼

A

—cry)

2

[frf-yt+(8 -cry')

)

½

(A2+yw2) +

times exp {_

where

2

is given by (6.42)1

and

w

and

(cx+28t+yt2)½} , p

are defined by

x—(At+B) L (cz+28t+yt )2

=

U

[62+A2(cry 82)])

21

1

2(8

where

B

is given by (6.42)2.

By substituting (**) into the diffusion

equation (6.3) deduce the following ordinary differential equation for

+

+

where the constants D = If

E

82 < cry

= 0

and

D

E

are given by

[A2(82—ory')—62] — A

=

show that in place of the square bracket in (**) we have

+

exp

[The solutions for

4(w)

]) tan1

+

can be expressed in terms of confluent

hypergeometric functions (see Bluman and Cole [5], page 215).] 7.

If

82 = cry

and

y # 0

deduce from (6.18) and (6.36) the following

functional form

c(x,t) =

where

w,

L

(t+8)½ and

exp

-

w2(t+e)} ,

+

(***)

-

M are given by

125

=

(4

I

1

+

L = ½(K—58)

Further

1

1

+ 12J (t+8)f (t+8) M

,

(6.3)

show from

¼(62+28+4A)

=

satisfies,

and (***) that = 0

—

[The solutions of this equation can be expressed in terms of Airy functions (see [5], page 217).] 8.

If

8

=y

and

= 0

a

show from (6.18) that (6.37) becomes

0

÷ At -

c(x,t) =

wt}

,

{-

where

w

is given by 2

(4

c5t

=x

—i--

—

Kt

Deduce from (****) and (6.3) that

=0

+ K4'(w) +

[This

satisfies

equation also has solutions expressible in terms of Airy functions

(see [5], page 218).] 9.

If

a =

8=y

= 0

and

CS #

0

show from (6.18) that the source solution

results, namely

I

c(x,t)

exp

(t+Ic)½

=

r

4(t+K)

denotes an arbitrary constant.

where 10.

(

i

Show that the equation

I can be transformed to the classical diffusion equation — 2

126

'

by means of the transformation, c(x,t) =

a(x)C(y,t)

if and only if

takes the form,

p(x)

p(x) = (C1x + C2) where 11.

and

C1

y =

,

4/3

denote arbitrary constants.

C2

Observe from example 5 that the similarity variable

w

for the classical

group of ac =

2 ac

a

+ b

(xc)

ax

Is

obtained by integrating,

dj

bt —bi (&e+Ke

x

)

dt —

y,

where the constants

and

K

are the same as those used In

By making the transformation

example 5.

p= e

&

2br

show that the IntegratIon of (+) can be effected by considering separately four distinct cases In a completely analogous manner to the corresponding integration for the classical diffusion equation.

Deduce

In each case the similarity variable. 12.

WIth reference to example 5 establish the following solution types for which

0

as

t

0

and

c(x, t)

is given by

c(x,t) = (I)

= —y = 1,

=

= XT + ó* + K* —

=

0.

6*(12_1)½

T(T2_1)A '2exp

bo*[w(T2_l)½

+

127

= —a = 1,

(ii)

y

0

(T2_1)½

+ K*

2 T—1

T

times

—i---

=

=

=0.

x+

=

bA*4(w)

—

•"(w) +

T —

exp {_

205*+2K*)WT(T2_1)cj}

=0

— bw'(w) —

(iii) B=—y=1, a=0. TX

+ tS* + K*T2 , (T4_1)1 T2_1I

T

—

{

=

=0

—

(iv)

B

=y =

—(x/2)

= 1.

+

(13=

(T2—1)

(T2—1)2 2 2

=

(T21)½ -

{ LT2_1) b {(1+2X*) + 4(tS*+K*)bw — bw2}

(v)

=

TX

03=

=

1,

a =

+ tS* + y1(T2—y2) 2

22½'

[(T —1)(T —ji )]

128

(tS*+K*)21fl

+

exp

= 0

*(w,t) =

2(1—i.i

2

T

k[(w2+12)12

exp {

1

22

times

b{(1+2A*)(1+V2)

—

2(2K*

2

)

—

q)"(w) —

—

V

2

X*

+ (1+112)6*)

—

22

'

= 0

—

(1+112)

=

2(1-jA

)

In each of the cases

Te

2 )

is defined by

bt

6*, K*

and the constants based on

I

1

'

2

and

K

6,

A

A*

and

are appropriately redefined constants

respectively and do not necessarily refer to

the same constant in each case. 13.

WIth reference to example 6 establish the following solution types where is given by

c(x,t)

c(x,t) = (1)

XT (A)

= T(T—1)

= (T—1)

—

+

(ii)

w =

A*—1

(t—

=

(i—i)

= 0

exp {—

+ (2a—bw)4'(w) — b(A*+1)4(w) = 0

bix 2

(iii)

=

,

= T

(121)

w =

XT

(11)2

exp

a(t —1)J

(T+1)

aw4"(w) + 2a4'(w) —

(iv)

(i—i)

ij,(w,t)

=0

2bA*

=

T

(T—1)

2

exp J—

bt2x a(T—1)

+

(T_1)1 J

129

+

(v)

2

U) = (r—1)(wr—1)

{-

(T-1)(UT-1)

=

ac$"(w) ÷ In

=0

+

+

each of the above cases

is

T

—

4

= 0

defined by

T=ebt and 14.

A*

denotes an arbitrary constant.

With reference to example 7 establish the following solution types where c(x,t)

is given by c(x,t) =

(i)

a=B=O, y=1. K

2t —

*(w,t) =

exp

{x +

(w)

K

—

(ii)

a—I = 0,

*(w,t)

—

—

130

t-

—

÷

(w)

+

—

2(b+3)t]}

=0

= (1/2).

tAexp

ir

+ (b+3)cS

4'(w) — A4(w) = 0

+ (b+1)

jt

½ —

(2c5+b+3))'

(iii)

I =

B = 0,

W =

log

ci =

1.

x — Kt —

(St2

+ (b+3)

= exp

(K-I-b+3)t2

-

+

(iv)

+

B = 0,

cx =

K +

(k?)]t (S2t3

(b+3) 4[A

+

K+

2

(b+3)2

+T14)(W)0.

y =1.

(S+Kt+logx =

P(t)A

—

exp

t2+u2)¼

—

(b+3_K)(t2-I.p2)½}

(22 (4

4"(w) +

15.

—

For the boundary value problem, ac

I

c(x,O)

c(x,t)

where

c0

(t>O,

a

c05(x—x0) ac ,

and

ox

x0

(x,t)

+ 0

as

x +

and

provided

the functions

satisfy = 0

Hence with

±00

denote arbitrary constants, show that the initial

condition remains invariant under (6.1) fl(x,t)

—°°<x<°°)

,

fl(x0,O)

10 E 1(x0)

= 0

,

1(x0,O)

=

show that the functions

(x

p(t)

and

a(t)

131

in (6.47) and (6.48) satisfy

p(O) = —

= 0 ,

ri"(O)

a'o'

16.

r2 + 0

8

10

p'(O) 2

—

0

2

For case (I) of section 6.5 show that one group leaving

Continuation.

the boundary value problem of the previous question invariant is, C(x,t) =

2p(x)1

82 = C1.

c(x,t)

where

,

fl(x,t)

= 0

+ (C2/B)(1—cosh(Bt)) — 81 cosh(Bt) — J sinh(8t),

=

where

sinh(Bt)

Hence show that the solution takes the form

4(t) (I)

exp

"½

=

812

1

—h--

coth(8t) +

denotes an arbitrary function of

28 sinh(8t)

3

From the partial

t.

differential equation deduce that =

c(x,t)

+

C)2

exp

[p(x)sinh(8t)]

B

+ + C2

c21

+

where

a = [c1c3

—

/

4C1

1

Blo

28 sinh(8t)

denotes an arbitrary constant.

and

Showthatas t÷O,

c(x,t)

[p(x)Bt]½

exp

I '.

t

and observe that for given 1(x) the constant the condition J

132

c(x,t)dx

— c0 .

is determined from

17.

For case (ii) of section 6.5 show that the functions

Continuation. n(t)

and

o(t)

are given by

.2 (st) sinh

fl(t) =

822 o(t) =

C

= C1.

where again

sinh

—

—

2

(st)

Deduce from (6.2) that the functional form of the

solution of the boundary value problem of question 15 becomes Yt

c(x,t) = exp {—

where $(w)

is the similarity variable,

w =

y = C3/4

and

satisfies the differential equation C4

÷

—

—i

4(w)

= 0

4w With

and

ci

defined by 4(w)

=

=

so that —

+

where

n = (1+C4)½/2.

correct behaviour as

where

0

,

Show further that the solution 0

t

previous question when

=

=0

[i(ci)

is given by

1(d))

is a constant and

function of the first kind.

giving the

and agreeing with the result of the

C2 = C4 = 0

+

$(w)

I

n

denotes the usual modified Bessel

Verify that as

t + 0

4t

133

18.

Obtain the solutions of the boundary value problem of

Continuation.

question 15 for the following three equations: 2

2

=a

(ii)

(iii)

where

(xc) + b

and

(xc)

(x2c) + b

=

a

(xc)

b

denote arbitrary constants.

Answers:

a'

(i)

c(x,t)

(ii)

c(x,t) =

where

b

C01

m(t)

m(t) =

1½

—bt ,

b(x—x0e

c(x,t) =

1

L

are defined by

y(t)

4Ort)'ixI

134

exp

)

2(l=e2bt)

=

b

a(1—e

(iii)

—bt2

exp

_2btJ

y(t)

and

1

Ut)

+ (b+1)t]2

1

—

4

t

7 Non-linear partial differential equations

For non—linear partial differential equations we need to consider more general transformations which leave the given equation invariant.

chapter for a single dependent variable x

and

In this

and for two independent variables

c

we consider one—parameter groups of the form

t

f(x,t,c,t)

=

x+

+

0(c2)

= g(x,t,c,c)

=

t + cn(x,t,c)

+

0(c2)

+

0(c2)

x1 =

c1 =

h(x,t,c,c)

= c

+

For known functions

fl(x,t,c)

and

(7.1)

,

?(x,t,c)

the similarity

variable and functional form of the solution are obtained by solving the first order partial differential equation,

+

fl(x,t,c)

= ?(x,t,c)

(7.2)

.

In the first section we give formulae for the infinitesimal versions of the first and second order partial derivatives of completeness we also give formulae for make no use of these results.

c(x,t). and

Again for

a2c/at2

although we

In the second section we deduce the classical

groups of the non—linear diffusion equation, namely ac =

where

D(c)

a

I

1D(c)

ad

rj

(7.3)

'

denotes an arbitrary function of

c.

Equation (7.3) is well

known in the literature (see for example, Knight and Philip [27], Munier et al [28] and Shampine [29]). D(c) = cm

In particular the power law diffusivities

have received a good deal of attention (see Babu and Genuchten

[30], Munier et al [28] and Crundy [31]).

The results of section 7.2 are due 135

to Ovsjannilcov [7] and Bluman and Cole [5] (page 295).

In section 7.3 we

briefly consider the non—classical approach for the non—linear diffusion equation (7.3)

Although we present no new results in this section the

governing equations are summarized for the reader interested in pursuing the matter further.

In section 7.4 we give two results for equation (7.3) due to

Munier et al [28].

Although not directly related to group methods these

results involve important transformations of non—linear diffusion equations. The first result shows that every equation of the form (7.3) can by a sequence of transformations be reduced to an equation of the form D(x)

ac

1

(7.4)

,

so that the power law diffusivity

D(c) =

c2

plays an important role.

The

second result due to Munier et al [28] is that the most general inhomogeneous and non—linear diffusion equation with diffusivity

D(x,c)

which can be

reduced by transformations to the classical linear diffusion equation (6.3) is given by equation (7.53).

7.1

FORMULAE FOR PARTIAL DERIVATIVES

In this section we deduce the infinitesimal versions of the partial derivatives

ac/ax,

ac/at,

a2c/ax2,

a2c/axat

and

a2c/3t2.

We again use

the convention that subscripts denote partial differentiation with c

as three independent variables.

ax

cax'

x

at

t

x, t

Thus for example we have,

catS

Now either directly from (7.1) or from the definition of a one—parameter group we have

136

x = x1 —

+ O(C2)

t = t1

+ 0(c2)

—

and

and therefore ax

=

up to order

we obtain ax

+

lx +

—

1

c

=

÷ 0(c2)

+

(7.5) at L

First

=—

cii +ri —

for

at = 1

+

—

+

+

we have

ac/ax

3c

x

1

at

and using (7.1) and (7.5) we obtain aC1

=

+ c1c

+

+

11

+

I

x

I.

+

+ s

ft)}

+ 0(c2)

÷

which simplifies to give

I

3C = —

+

3c

c

÷

3c ac} —

—

+ 0

—

2

)

(7.6) Similarly for

we have

ac/at

aC

a

at so that from (7.1) and (7.5) we obtain ac1 =

+ c

3t1

x

+

+

c

I.

+ 01c2)

which

becomes aC1

= 3c

÷ c{

ac +

ac

ac

—

c

÷

0c2 (7.7) 137

Again for convenience we introduce

+

=

-

and

-

-

—

—

112

such that

ac

—

(7.8)

ac

ac

+

112 =

'

so that we have simply

=

2

+ 0 c)

÷

(7.9)

.

We observe that (7.8) can be written alternatively as

axax (7.10) —

at

For the infinitesimal version of 2 a

c1 =

=

T

1

ax

+

1

which using (7.5) and (7.9) gives

Cl x

+

+

C(11

nx +n

+

c

ad1

On simplifying this result we obtain a2 c1

2

=

+

I

+

C

From the first equation of (7.8) we have

138

+ 0(c2)

.

(7.11)

= XT

XX

+ )

xe

3X

—

Z

xe

ie ulolJ (11'L)

z

=

b

(zr'L)

+

+

3X

xe

'Xe +

(

33

—

—

—

xe

Ixe) -

+

ei't

xe

-

33U

-

+

J

U

xo

XQ)

Ixeiie +

(crL)

ioj =

uioij e

(c'L)

z

=

pirn

Z(8.1)

z =

+

ei't

JLe)

+

—

+ 3

1

x

+ U

+

ezrnpep

2UTSn z

=

Uj

(Xe lexe

JLeJ

+ 3

—

+

J

+

+ U

(cri)

ptw (cl'L)

woia

i)

(8

eq

jo

L)

uz.toqs

('iT

eq eqi ewes 2uisn

10 L) (cT

eqi

z

=

+

'ie'xe

+

ix

+

— —

33

eJ

—

ie

-

+

U

X3

—

xc

-

-

xc

—

ie

-

—

+

J

lie xc z

-

UZ

+

+

J

(91L) U93

B

z J

+

lie

ie

ie

1

—

(ielxe

-

+

ie

-

-

+

+

{iexe1ie

(LIL) (LIL) puB

pue

ZL

'WDISSVID

aq

LUoiJ

(EVL)

iCq

x

U

MOISIkMIcI

(EL) z =

xc aiaqz.i Os?'

inoq2noiqi

[Xe)

(8rL)

'

qipt i3adsal oi

indicated.

Now using

D'(c1)

÷ EçD'(c) + 0(c2)

D(c)

D(c1) = =

D'(c)

+ C1D"(c) + 0(c2)

we find from (7.6), (7.7) and

(7. 13)

that (7. 18) remains invariant under (7.1)

provided

+

= D(c)

—

—

+

—

nxx

ac

ac

ac

-

+

+

x

Ô(c)

(7.19) where the last term,

involving the two curly brackets, arises from

eliminating

by means of equation (7.18).

On equating the

coefficients of the various partial derivatives in (7.19) to zero we obtain the following equations: 2

c

2

nx=o' cc

—

+ D'(c)fl

c

= 0

0

141

2

in

,

c

ac ac t

C

t

x

0

XC

+ D'(c)fl X = 0

—

xx

+2D(c)C xc +2D'(c)C =

—D(c)?

,

C

+ D(c)T)

—D(c)ri

c

=0,

D'(cfl

+

t

x

x

xx

xx

+

D'(c)

D(c)

=0

0,

(7.20) (7.21)

,

=0.

(7.22)

From the first seven of these equations we can readily deduce =

n = n(t)

,

+ D(c)

c

—

=

cc

=

0.

fD(c) 1

—

ax so that 2

t.

But from (7.21) and

(7.21)

From

n'(t)'

we obtain

,

(7.25)

)

either ax

=

(7.26)

,

or the diffusivity '

ID'(c)J

D(c)

is such that

,,

=0

is

D(c) = a(c+8)m

142

and

+ 4(x,t)

n'(t) — 2

and therefore =

x

deduce

(7.24) we can

where

(7.24)

,

denotes an arbitrary function of

where

that

(7.23)

,

a, B

and

,

m denote arbitrary constants.

(7.27) If (7.26) holds then from

(7.20) and (7.25) we can readily deduce

x + K

=

fl(x,t,c)

= 6 +

yt

,

A

denote arbitrary constants.

(7.28)

= 0

where

and

y, 6, K

(7.28) is applicable to all functions Alternatively if

—

D(c).

has the form (7.27) then from (7.25) we have

D(c)

(c+B)[2

=

We note that the group

r)'(t))

(7.29)

,

and on substituting this expression into (7.20) we find

(7.30)

,

_D(c)(3

while substitution of (7.29) into (7.22) and making use of (7.30) gives

(7.31)

.

=

ax We see that (7.30) and (7.31) give rise to two

=

while for

a2

we have,

fl"(t)

=

K + Ax

fl(x,t,c) = 6 +

in

= 0

we have

in

F(x,t,c) =

while for

namely for all constants

= n"(t) = 0

m = —4/3

Thus for all

cases,

yt

,

=

(c+8)(2X—y)

= —4/3

we have

(7.32)

143

E(x,t,c) = K + Ax + px2

where

rl(x,t,c) =

tS

C(x,t,c) =

-

+ yt

+ 2A

5, K, A

y,

(7.33)

,

denote

and

-

y) The resulting

arbitrary constants.

ordinary differential equations corresponding to (7.28), (7.32) and (7.33) are given in problems 1, 2 and 3.

Source solution of (7.3) with a power law diffusivity.

Example 1

(7.34)

,

such that

vanishes at infinity while initially satisfies

c(x,t)

c(x,O) = c0

where

c0

and

delta function.

Consider

tS(x)

(7.35)

,

denote arbitrary constants and

m

cS(x)

is the usual Dirac

Noting the elementary property (1.24) of delta functions we

see directly that (7.34) and (7.35) remain invariant under the one—parameter group

x1ex, c

that

t1=e (m+2)c t

,

c1=e —c c

is,

E(x,t,c) = x

,

ri(x,t,c)

=

(m+2)t

This equation corresponds to (7.32) with y = (m+2).

with

B

= —c

,

B =

=K

We note that the more general case with

=

0, D(c)

(7.36)

.

A =

1

and

given by (7.27)

non—zero appears not to admit a simple group leaving (7.35)

From (7.2) and (7.36) we see that the similarity variable and functional form

of the solution are obtained by solving x

r + (m+2)t

We find that

144

=

-c

where

i

,

c(x,t) =

(7.37)

,

On substituting (7.37) into (7.34) we obtain

n =

+

= 0

vanishes

and since

at infinity, the constant of integration is zero

and we obtain

+

= 0

A further integration gives

2)m

I

=

where

'

2(m+2),)'

—

(7.38)

denotes an arbitrary constant.

C

With

C

given by

2

C

(7.39)

= 2(m+2)

zero for

we take

c(x,0)dx

>

so that the condition

= c0

J

becomes

I m

1

2

w1 —

u

2i

= c0

j

1

Thus with

w =

(cos0)

I

J

where

sin

0

and using the formula

d0= 12

r(x)

constant

C

m

denotes the usual gamma function, we can readily deduce that the in (7.38) is given by

145

In

C

2

13

-

in

—

2(2-I-in)

0

(2-fin)

in

r[i

(7.40

.

+

Thus altogether from (7.37) arid (7.38) we have that the source solution of (7.34) is given by 1

1

c(x,t) =

1

t

c(x,t) = 0

where

n =

I

mx

—

2 I

- 2nt

2(m+2)t ,

(m+2)1,

wi

ii)

'

<

(7.41)

J

>

w =

C

,

Is given by (7.40) and

is defined

by equation (7.39).

7.3

NON—CLASSICAL GROUPS FOR NON-LINEAR DIFFIJSI ON

Although there are no known non—classical groups of (7.3) we derive here the governing equations in order to illustrate the non—classical approach in a non—linear context.

With

A(x,t,c) ,

A(x,t,c) =

and

B(x,t,c)

B(x,t,c)

defined by

(7.42) =

we have from (7.2)

(7.43)

On differentiating (7.43) partially with respect to (7.43) to eliminate

and

ac/st

x

and using (7.18) and

respectively it is a simple matter

to deduce

=

2

+

-s-fl-) +

(7.44)

— Bx

+

—

We remind the reader that the subscripts refer to partial derivatives of functions of three Independent variables

146

x, t

and

c.

Writing

(7.45)

and substituting (7.43) and (7.44) Into (7.19) we obtain the following cubic expression in

0, namely —

+

—

—

+

=

—

—

2D(c)

[A

+

—

+ 2D'(c)O

+

—

-

+ D"(cfl02

+

+

[B

(A—B0—D'(c)02)

+

(7.46)

.

On equating to zero the coefficients of following equations for the determination of D(c)Bcc —

o2}

— Bc)

-

—

—

+

[Ac — B

D'(c)B

0

00

and

A(x,t,c)

and

we obtain the B(x,t,c):

= 0

2BB

o2,

(7.47) 0

+ 2[D(c)A ]

B

,

xc

t

00

,

A

t

=

D(c)A

We observe that with

xx

—

— 2AB

D'(c) 2AB = D(c)B — 2BB + AB xx x c D(c)

x

+ A2 D'(c)

D(c) = 1,

D(c)

A =

a(x,t)c

and

B =

b(x,t)

(7.47) reduce precisely to (6.80) for the two functions

a(x,t)

equations and

b(x,t).

Equations (7.47) are recorded for purposes of illustration and we make no attempt here to obtain special solutions.

147

TRANSFORMATIONS OF THE NON-LINEAR DIFFUSION EQUATiON

7.4

The most widely known tranformation of a non—linear partial differential equation is for Burgers' equation (see for example [32], [33] and [34]) 2

3u

u

the classical diffusion equation, assuming that

(7.48)

D

is a

In this section we give two Important results for the non—linear

constant.

diffusion equation (7.3) due to Munier et al [28].

A number of related

transformations can be found in the literature (see for example, Knight and Philip [27] and Storm [35]).

The first result due to Munier et al [28] is that every non—linear

diffusion equation of the form (7.3) can be transformed to the following equation with a simpler non—linearity, namely

—i

v where

v(c,t)

(7.50)

,

is essentially the flux associated with equation (7.3).

see this we define

(7.51)

.

On multiplying (7.3) by

D(c)

partially with respect to

=

I1D(c)

We now introduce 3v

by

u(x,t)

u(x,t) = D(c)

j—j

v(c,t)

x

and differentiating the resulting equation we find

(7.52)

.

u(x,t)

so that (7.52) becomes

I '

148

To

which on usIng (7.3) and (7.51) simplifIes to give (7.50).

We note that the

equivalence of (7.4) and (7.50) is readily seen. The second result due to Munier et al [28] is that the most general inhomogeneous and non—linear diffusion equation with diffusivity

D(x,c)

which can be transformed to the classical diffusion equation (6.3) takes the form

at

A.

(7

ax

and

y

6

denote arbitrary constants.

53)

In order to see that

(7.53) can be reduced to the classical diffusion equation we can without loss of generality consider the equation, ac

a

,

at

ax

llcJ

Instead

ac ax

of working with (7.54) with independent variables

(x,t)

we consider

the same equation

for

aw

a

at

ac

(7.55)

acf

ltwJ

v(c,t). Making the transformation w(c,t)

=

(7.56)

v(c,t)

it is a simple matter to show that (7.55) becomes av

2av 2

=

v

(7.57)

.

ac

This

is clearly the same equation as (7.50) with

by introducing

x

D(c)

unity and therefore

such that

v(c,t) E u(x,t)

,

(7.58)

equation (7.57) is equivalent to the classical diffusion equation (6.3) for c(x,t).

149

PROBLEMS

1.

For the non—linear diffusion equation (7.3) show that the similarity variable and functional form of the solution corresponding to the group x + K,

E(x,t,c)

rl(x,t,c) =

2(t+5),

0

are respectively x+K (t+5)

Hence show that the resulting ordinary differential equation is = 0

D(4)4"(w) + D'(4)q'(w)2 2.

For the non—linear diffusion equation with m

D(c) =

show that the similarity variable and functional form of the solution corresponding to the group =

(1+A)x + =

K

,

rl(x,t,c)

=

2(t+6)

2A —

are given respectively by

=

Ix +

(t+6)

K ,

-

c =

2

Show that the resulting ordinary differential equation is

+

+

(i-i-A)

u4'(w) —

= 0

Show that this equation can be reduced to a first order ordinary differential equation by observing that the above equation remains invariant under the one—parameter group of transformations c

'so

2dm

3.

For the non—linear diffusion equation with D(c) =

the special case of the group

consider

jix2 ÷ (l+A)x ÷ K,

=

C(x,t,c) =

=

2(t+6),

(c+8)(2px+A)

—

for which the constants

fl(x,t,c)

K,

A,

and

ji

m satisfy

(A÷l)m = 4jiK

In this case show that the similarity variable and functional form of solution are =

c =

(t+6) -½

1

exp

-2

÷ (A÷I))

-8 ÷

exp

Show that the resulting ordinary differential equation is — —

÷ IL

= 0

which can be reduced to a first order ordinary differential equation by observing that the above equation remains invariant under the group

41e—3c/2

e

4.

Show that the non—classical approach applied to a

F(x)

Ii

gives rise to the following four equations for the one—parameter group (7.1), B

B

cc

C

151

A +

+ c2[F(x)—3JBB

= 0

(AB)] + F'(x)B2 + {2[1]

-

—

BA}

-

2

2

F(x) [A + AA I + t c

where

A(x,t,c)

and

F' (x)AB

—

AA

c

=

2 C

- 2AB

x

2 c

are defined by (7.42) and subscripts

B(x,t,c)

denote partial derivatives with respect to the three independent variables 5.

x, t

and

c.

The non—linear axially symmetric diffusion or heat conduction equation in cylindrical and spherical regions can be transformed into an equation of the form

kac

a2c

ac

(*)

ax

where

k =

and

1

k

regions respectively.

2

corresponds to cylindrical and spherical

By considering the classical invariance of (*)

under the one—parameter group

x + cE(x) + 0c2)

x1

t + cri(t) + 0c2 =c

+ CC(x,c) + 0(c2)

show that for both values of constant

k

a group exists if either

f(c)

or if

f(c) = where

a, B

situation 152

and

m

denote arbitrary constants. In the latter

show that for both values of

k,

is

= x, where

y, ó

n(t)

and

= (nry+2)t +

denote further

X

For the special case of

k =

?(x,c)

cS,

arbitrary

=

yc +

X

(**)

,

constants.

show that a more general group than

1

(**) exists provided f(c) =

and deduce for example that (*) admits groups of the form,

where again 6.

=

(log

y

and

Continuation.

x +

(Wy)]

x +

y)(c+$)

cS

n(t)

,

=

2t +

S

denote further arbitrary constants.

Deduce the similarity variables, functional forms of the

solutions and the resulting ordinary differential equations for the groups given in the previous problem. 7.

Observe that 2

ax

2

xax

remains invariant under the group, c

x1 = e x,

t1 = e

(2+mn)c

c1 = e

t,

ne

c

Use this group to deduce the source solutions of (***) for k = 2

n =

given that for these values of

k

the group with

k =

n =

—2

1

and and

respectively leaves the appropriate initial condition invariant

—3

as well.

[This is because with rectangular cartesian coordinates appropriate initial condition for

k =

1

(X,Y,Z)

the

is

c(X,Y,O) = c0S(X)ó(Y)

while for

k = 2

we have

c(X,Y,Z,O) = c0cS(X)S(Y)S(Z)

,

153

where as usual

denotes a constant specifying the strength of the

c0

source.]

8.

Show that 2 ôc

2 ac

13cac

at'

ax2

remains invariant under x1 = x +

+

0(c2)

y1 = y +

+

0(c2)

=

t+

+

0(c2)

2 y--1—-1--- +0(c),

provided

satisfy the Cauchy—Riemann equations,

and

E1(x,y)

namely

and 9.

x, (3, y

and

ô

ô = 0

For

Continuation. =

x

denote arbitrary constants. and

and

2

2

given by

= 2xy

—

deduce the following similarity variables and functional form of the solution,

22 (x+y)

I

yx

,

2

2

y

2

2

(3(x+y) Hence from (+) deduce the following partial differential equation for

aw)

154

aT

B

10.

For

Continuation.

= e

and

= 0

nx = e sin fly

nx

cos fly

given by

and

,

deduce the following similarity variable

and functional form of the

solution, = e

-nx

sin ny

,

T

exp ft

t

=

4(w,t) —

c(x,y,t) =

e

-nx cos

log(sin ny)

fly —

Hence from (-I-) deduce the following partial differential equation for 4(w,t)

a 2

n

3T1T 3TJ

2

(nw)

2

e

2

[Theabove six problems are due to Nariboli 11.

Show that the source solution for Burgers' equation, namely au

+ u 3u = D

2 3u

3x

u(x,0)

= u0 5(x)

x-'±°°,

as

where

u0

is a constant, remains invariant under the one—parameter

group x1

= e cx

t1

,

= e

2c

t

,

u1 = e

—c

u

Hence deduce that u(x,t) =

and that -2De [c

—w

2

/4D

+

]55

where w =

and

is a constant..

C

condition that the constant —

C

is given by

C

sinh(Du0)

—

12.

Deduce from the initial

Show that the classical groups of the non-linear wave equation, 2 ac

2

=

f(c) 2ac —i

where

f(c)

(-H-)

,

ax

at

is non—constant, are summarized by the following three cases:

f(c) arbitrary =

yx + 6

rI(x,t,c) =

yt: + K

= 0

(ii)

f(c) =

= yx

yt

=

C(x,t,c) (iii)

+ 6 + Xmx

+ K

=

f(c) = cx(c+8)2

F(x,t,c) =

yx +

T1(x,t,c) =

yt

t(x,t,c)

6

+ 2Xx + jix2

+ K

+

In each case deduce the similarity variables, functional forms of the solution and the resulting ordinary differential equations. 13.

Continuation.

The fundamental solution of the non—linear wave equation

(+1-) of the previous problem satisfies the initial data

(x,0)

c(x,0) = 0 ,

156

=

6(x)

.

(-H-I-)

Show that for all wave speeds

the fundamental solution remains

f(c)

invariant under the one—parameter group C

x1 = e x

t1 =

,

e

C

c1 = c

,

t

and hence takes the form c(x,t) =

4(xt1)

Deduce from (++) that

satisfies

4)

f(4))24)"(w)

w24)"(w) + 2w4)' (w) =

where

14.

the ordinary differential equation

xt'.

w=

For the linear case

Continuation.

deduce that

f(c) =

f0

where

f0

is a constant

is given by

4)'(w)

C 2

where

denotes an arbitrary constant.

C

Hence show that the

fundamental solution in this case is given by

x+f0t

c(x,t)

—

0

Can you 15.

ft 0

determine the constant

Continuation.

C

from the initial data (4-f-f)?

f(c) = c

For the case

show that the ordinary

differential equation of problem 13 remains invariant under the one—

group

parameter

w1=ew,

C

C

Hence with,

±

Tlogw,

deduce the Abel equation of the second kind, 2

P

'p

+

+ ' (1—'p

)

= 0

.

('—'p ) 157

___

16.

Obtain the classical groups and resulting solutions of the following partial differential equations: (1)

the telegrapher's equation, 2

a

where (ii)

2

a

and

D

where

5

and

D

2 ac

2' are constants.

the non—linear Burgers' equation, aU

where (iv)

are constants.

the diffusion equation with convection,

ax_D

(ill)

2'

U

2

+ u

D

au

= D

2 au

is a constant.

Barenblatt's equation (see Barenblatt et al [37]) for flow in fissured rocks, 2

3 2

ax

where

158

a

and

$

are constants.

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