C 1 : POWER SERIES SOLUTIONS AND SPECIAL FUNCTIONSsdeuoc.ac.in/sites/default/files/sde_videos/DE - M Sc SLM(3).pdf · 6.6 charpit’s method 62 6.7 jacobi’s method 66 6.8 cauchy

Differential Equations 3

CONTENTS

CHAPTER 1 POWER SERIES SOLUTIONS 03

1.1 INTRODUCTION 03

1.2 POWER SERIES SOLUTIONS 04

1.3 REGULAR SINGULAR POINTS – 05

FROBENIUS SERIES SOLUTIONS

1.4 GAUSS’S HYPER GEOMETRIC EQUATION 07

1.5 THE POINT AT INFINITY 09

CHAPTER 2 SPECIAL FUNCTIONS 11

2.1 LEGENDRE POLYNOMIALS 11

2.2 BESSEL FUNCTIONS – GAMMA FUNCTION 15

CHEPTER 3 SYSTEMS OF FIRST ORDER EQUATIONS 20

3.1 LINEAR SYSTEMS 20

3.2 HOMOGENEOUS LINEAR SYSTEMS WITH 21

CONSTANT COEFFICIENTS

3.3 NON LINEAR SYSTEM 24

CHAPTER 4 NON LIEAR EQUATIONS 26

4.1 AUTONOMOUS SYSTEM 26

4.2 CRITICAL POINTS & STABILITY 28

4.3 LIAPUNOV’S DIRECT METHOD 31

4.4 SIMPLE CRITICAL POINTS -NON LINEAR SYSTEM 34

CHAPTER 5 FUNDAMENTAL THEOREMS 38

5.1 THE METHOD OF SUCCESSIVE APPROXIMATIONS 38

5.2 PICARD’S THEOREM 39


CHAPTER 6 FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 46

6.1 INTRODUCTION – REVIEW 46

6.2 FORMATION OF FIRST ORDER PDE 48

6.3 CLASSIFICATION OF INTEGRALS 50

6.4 LINEAR EQUATIONS 54

6.5 PFAFFIAN DIFFERENTIAL EQUATIONS 56

6.6 CHARPIT’S METHOD 62

6.7 JACOBI’S METHOD 66

6.8 CAUCHY PROBLEM 70

6.9 GEOMETRY OF SOLUTIONS 74

CHAPTER 7 SECOND ORDER PARTIAL DIFFERENTIAL EQUATIONS

7.1 CLASSIFICATION 78

7.2 ONE DIMENSIONAL WAVE EQUATION 81

7.3 RIEMANN’S METHOD 87

7.4 LAPLACE EQUATION 89

7.5 HEAT CONDUCTION PROBLEM 95 - 98


CHAPTER 1

POWER SERIES SOLUTIONS AND SPECIAL FUNCTIONS

1.1 Introduction

An algebraic function is a polynomial, a rational function, or any function that

satisfies a polynomial equation whose coefficients are polynomials. The elementary functions

consists of algebraic functions, the elementary transcendental functions or non algebraic

functions- the trigonometric functions and their inverses, exponential and logarithmic

functions and all others that can be constructed from these by adding or multiplying or taking

compositions.

Any other function is called a special function.

Consider the power series n

n

n xa

0

. The series has a radius of convergence R,

R0 such that the series converges for | x | < R and diverges for | x | > R.

We have, R = n

n

a

a

n

1lim

.

For the geometric series 1+ x + x 2 + … , R = 1 and for the exponential series

0 !n

n

n

x, R =

and the series

0

!n

nxn converges only for x = 0.

Suppose n

n

n xa

0

= f(x) for | x | < R . Then f(x) has derivatives of all orders and the series can

be differentiated term by term )(' xf 1

1

n

n

n xna , )('' xf 2

1

)1(

n

n

n xann and so on and

each series converges for | x | < R. In fact, we get nn

fa n

n ,!

)0(.

A function f(x) which can be expanded as a power series n

n

n xxa )( 0

0

, valid in some

neighborhood of x 0, is said to be analytic at x 0.

Polynomials, e x, sin x , cos x are analytic at all points, but 1/( 1+ x) is not at x = -1..


1.2. Power series solutions

It may be recalled that many differential equations can not be solved by the few

analytical methods developed and these methods can be employed only if the differential

equations are of a particular type. By applying the following method solutions can be

obtained as a power series and hence known as power series method.

Consider the equation .' yy

We may assume that this equation has a power series solution in the form y = n

n

n xa

0

that

converges for | x | < R, for some R.

Then ....32 2

321

' xaxaay . Since yy ' , by equating the coefficients of like powers

of x, we get a1=a0, 2a2=a1,3a3=a2,… which reduces to a1=a0,a2=a1/ 2 = a0/2!,a3=a0/3!,….

Thus we obtain, y = a0( 1+ x/1! + x 2 /2!+…..) = a0 ex, where a0 is left undetermined and

hence arbitrary.

Now let us consider the general second order homogeneous equation,

.0)()( ''' yxQyxPy (*).

If both P(x) and Q(x) are analytic at x =x0, we say x0 is an ordinary point of the equation

We may assume the solution of the equation (*) as a power series y = n

n

n xxa )( 0

0

valid for

|x-x0| < R, for some R. The various coefficients can be found in terms of a0 and a1, which is

left undetermined.

Consider .0'' yy Here P(x)=0 and Q(x) = 1, which are analytic at x = 0.

Assume y = n

n

n xa

0

. Then the equation gives the recurrence relation (n+1)(n+2) an+2+an=0 ,

for n=0,1,2,….. . .Substituting n =0,1,2,..successively and reducing we get a2n+1 = (-1)na1/

(2n+1)! and a2n=(-1)na0/(2n)!. Hence y = ....)!5!3

(...)!4!2

1(53

1

42

0 xx

xaxx

a

= a0 cos x + a1 sin x.

Consider the Legendre’s equation 0)1(21 '''2 yppxyyx , where p is a

constant.

Here P(x) = 21

2

x

x

and Q(x) =

21

)1(

x

pp

, which are analytic at x = 0.


Let y = n

n

n xa

0

. Then the equation gives the recurrence relation (n+1)(n+2)an+2-(n-1)an-

2nan+p(p+1) an= 0. Put n = 0,1,2,..which gives ,!3

)2)(1(,

!2

)1(1302 a

ppaa

ppa

,....!5

)4)(2)(3)(1(,

!4

)3)(1)(2(1504 a

ppppaa

ppppa

.

Thus y = a0

......

!4

)3)(1()2(

!2

)1(1 42 x

ppppx

pp

+a1

......

!5

)4)(2)(1)(3(

!3

)2)(1( 53 xpppp

xpp

x .

The radius of convergence for each of the series in the brackets is R = 1. The series in the first

bracket terminates for p = 0,2,4,6,.. and the series in the second bracket terminates for p =

1,3,5,….. The resulting polynomials are called Legendre polynomials whose properties will

be discussed later.

Ex. The equation 0)4121( 2'' yxpy , where p is a constant, has a power series

solution y = n

n

n xa

0

at x = 0. Show that the coefficients are related by the three term

recurrence relation 041)21()2)(1( 22 nnn aapann . If the dependent variable y

is replaced by y = w 4

2x

e

, show that the equation is transformed to 0''' pwxww and

its power series solution at x = 0, involves only a two term recurrence relation.

1.3. Regular singular points

x = x0 is a singular point of (*) if either P(x) or Q(x) is not analytic at x0. In this case the

power series solution may not exist in a neighborhood of x0. But the solutions near a singular

point is important in a physical context, and most of the cases they exist. Origin is a singular

point of 022

2

''' yx

yx

y and for x > 0, y = c1 x + c2 x-2 , is its general solution.

A singular point x0 of (*) is called regular singular if both ( x-x0)P(x) & (x-x0)2Q(x) are

analytic at x0.

Consider the Legendre’s equation 0)1(21 '''2 yppxyyx , for which x =1 ,

-1 are singular points but they are regular singular. For the Bessel equation of order p,


0)( 22'''2 ypxxyyx , where p is a non negative constant, x = 0 is a regular singular

point.

If x = x0 is regular singular point of (*), then by definition ( x-x0)P(x) & (x-x0)2Q(x) are

analytic at x0 and hence we may take ( x-x0)P(x) = n

n

n xxp )( 0

0

and

(x-x0)2Q(x) = n

n

n xxq )( 0

0

. A solution of the equation (*) as a Frobenius series

y = mxx 0n

n

n xxa )( 0

0

, where m is a real number and a0 is assumed non zero, can be

expected.

On substituting, y = mxx 0n

n

n xxa )( 0

0

in (*), and equating the coefficients, we get

the recursion formula 0])[(])()1)([(1

0

00

knkn

n

k

kn qpkmaqpnmnmnma

( ** ). Here )()(lim

0

0

0 xPxxxx

p

and )()(lim

2

0

0

0 xQxxxx

q

.

For n=0, ( ** )gives 0)1( 00 qmpmm ***, called the indicial equation, which

determines the values of m.

Substituting the values of m and taking n=1,2,3,.. in ( ** ) an’s can be determined in terms of

a0 and the respective solutions can be obtained .

Eg. Consider the equation 0)12(2 '''2 yyxxyx . x = 0 is a regular singular point of

the equation. Let us assume that the solution at x = 0, is y = mx n

n

n xa

0

.

we get the indicial equation, 02

1

2

1)1( mmm --(1). m = 1 , -1/2 . For m = 1 , -1/2

respectively we get the solutions on determining the an’s successively from the recurrence

relation ( ** ) as ...)35

4

5

2( 32

01 xxxay and ...)2

11( 221

02 xxxay which

are independent also and thereby the general solution is y = c1 y1 + c2 y2, where c1& c2 are

arbitrary constants.


Remark.

Let the roots of the indicial equation be real, say, m1& m2 with 21 mm .

Then the equation ( * ) has a Frobenius series solution corresponding to m1, the larger

exponent. If m2 = m1, there is no scope to get a second independent solution by the same

procedure and it may be found by some alternate method. If m1 - m2 is not a positive integer,

another independent solution corresponding to m2 can be obtained, and otherwise the method

may not be giving a second independent solution.

Ex.1.. Consider the equation 0)44(3 '''2 yxxyyx . Show that x = 0 is a regular

singular point and find the only one Frobenius series solution.

Ex.2.. Find the two independent solutions of 02 ''' xyyxy , at x = 0.

Ex.3. The Bessel equation of order p = ½ , namely 0)4

1( 2'''2 yxxyyx has x = 0 as a

regular singular point. The exponents m1 & m2 is such that m1 – m2 =1, but the method gives

two independent solutions, and determine them.

1.4. GAUSS’S HYPER GEOMETRIC EQUATION.

The equation 0])1([)1( ''' abyyxbacyxx , where a, b, c are constants – (A)

represents many classical equations and is known as Gauss’s Hyper geometric equation.

We have P(x) = )1(

)1(

xx

xbac

and Q(x) =

)1( xx

ab

.

The only singular points are, x = 0 & 1, and they are regular singular points.

We may proceed to investigate the solution at x = 0.

We get, p0 = c & q0 = 0, so the indicial equation is m(m-1)+mc = 0 which gives m1 = 0 &

m2 = 1-c. If 1-c is not a positive integer, i.e. if c is not zero or a negative integer, then (A)

has a solution of the form y = 0x n

n

n xa

0

. Substituting in (A) and equating to zero, the

coefficients of xn, we get the recursion formula ; nn ancn

nbnaa

))(1(

))((1

. With a0=1, we get

in succession all an’s and the solution, y = n

n

xncccn

nbbbnaaa

1 )1).....(1(!

)1).....(1()1)...(1(1 ,

called the hyper geometric function, denoted by, F(a ,b ,c ,x).

Since R = n

n

a

a

n

1lim

=

))(1(

))((lim

ncn

nbna

n

= 1, the series converges for |x| < 1.( Note

that the series reduces to a polynomial for a or b equal to zero or some negative integer.)


If 1-c is not zero or a negative integer a second independent solution can be obtained,

similarly. or by the substitution, y = x 1-cz, (A) becomes,

0)1)(1(}]1)]1)(1[()2{()1( ''' ycbcayxcbcaczxx --(B),

a hyper geometric equation with a, b, c replaced by (a-c+1), (b-c+1) and (2-c)

Hence the solution of ( B ), at x = 0 is,

z = F(a-c+1,b-c+1,2-c,x) or y = x 1-c F(a-c+1,b-c+1,2-c,x), when c is not a positive integer.

Thus if c is not an integer, then the general solution of (A), at x = 0 is, y = c1 F(a,b,c,x) +

c2 x 1-c F(a-c+1,b-c+1,2-c,x).

To find the solution at x = 1, we may take t = 1 - x, so that when x = 1 , t = 0.

(A) becomes 0])1()1[()1( ''' abyytbacbaytt . Hence the general solution

at x = 1 , when c-a-b is not an integer is, y = c1 F(a,b,a+b-c+1,1-x) +

c2 (1-x )c-a-bF(c-b,c-a,c-a-b+1,1-x).

Remark

The solution of the general hyper geometric equation

0)())(( ''' HyyDxCyBxxAx

Where A B is obtained through the map t = )(

)(

AB

Ax

which transforms the equation to

0][)1( .'.. HyyGtFyxx and x = A & x = B to t = 0 & t = 1 respectively.

Ex1. Show that ( 1+x )p= F(-p, b, b, -x), log(1+x) = x F(1, 1, 2, -x),

),23,21,21()(sin 21 xxFx

Ex2. Show that ),,,(lim

b

xabaF

be x

, )

4,

21,,(

limcos

2

2

a

xaaF

ax

Ex3. Consider the Chebychev’s equation, 0)1( 2'''2 ypxyyx , where p is a non

negative constant. Transform it into a hyper geometric equation by t= 2

1 x and show that its

general solution near x = 1 is, y =

)2

1,

23,

21,

21(

2

1)

2

1,

21,,(

2

1

21

xppF

xc

xppFc

Ex.4. Show that ),1,1,1(),,,(' xcbaFc

abxcbaF .


Ex.5. Show that the only solutions of the Chebychev’s equation, whose derivatives are

bounded near x = 1 are, y = )2

1,

21,,(1

xppFc

.

1.5. The point at infinity

It is of practical importance to study the solutions of a given differential equation, for large

values of x. By the transformation x = 1/t and taking t small this can be achieved.

Consider the Euler equation 024 '''2 yxyyx , which is transformed to

022 ...2 ytyyt , by the substitution x = 1/t. Since t = 0 is a regular singular point of the

transformed equation so is, x = for the original equation.

Consider the hyper geometric equation (A). By the map x=1/t , it is transformed to

0)2()1()1( ...2 abytytcbaytt . t = 0 is a regular singular point, with

exponents m = a , b. Hence x = is also a regular singular point with exponents a , b.

Confluent hyper geometric equation

Consider the hyper geometric equation 0])1([)1(2

2

abyds

dysbac

ds

ydss .

Changing s to x = bs, the equation becomes 0)1(

)()1( '''

ayy

b

xaxcy

b

xx ,

which has the regular singular points x = 0, b and . If we let b , then b will be merged

with , and this confluence of two regular singular points produce an irregular singular point

at for the limiting equation, 0)( ''' ayyxcxy , called the confluent hyper geometric

equation.


CHAPTER 2

SPECIAL FUNCTIONS – LEGENDRE POLYNOMIALS

2.1. Legendre Polynomials

For n, a non negative integer,

consider the Legendre’s equation 0)1(2)1( '''2 ynnxyyx --(L).

We are now proceeding to find the solutions of (L), bounded near x = 1 , a regular singular

point. Take, t = 2

1 x. Then x = 1 corresponds to t = 0 and the transformed equation is

0)1(]21[)1(2

2

ynndt

dyt

dt

ydtt , hyper geometric equation . t = 0 is regular singular

with indicial equation, m(m-1) + m = 0, giving the only exponent m = 0. The corresponding

Frobenius series solution is, y 1 = F( -n, n+1,1, t).

Let a second independent solution be y2=vy1. where

.....1

)1(

11

)1(

1111212

1

2

1

)1(

12

2

1

)(

2

1

'

taattyttty

ey

ey

vdt

tt

tdttp

, since y1 is a

polynomial with non zero constant term. Thus v = log t + a1 t+…… and y2 = y1 ( log t + a1t

+….). As t0, logt , y2 is unbounded at t = 0 i.e. at x = 1.

Thus the only bounded solutions of (L) bounded at x = 1 are constant multiples of

y1 = )2

1,1,,(

xnnF

- a polynomial of degree n , called the n th Legendre polynomial,

denoted by Pn(x). We may proceed to express the polynomial Pn(x) in the standard power

form and obtain a generating formula, known as the Rodrigue’s formula.

The power series solution, we have obtained earlier, at x = 0, reduces to a polynomial of

degree n, since p = n, a non negative integer, and there by a valid solution, bounded at x = 1,

also. Thus by the above observation about bounded solutions at x = 1, we get the earlier

solution as a constant multiple of Pn( x ).

On simplification,

Pn(x)=

n

nx

n

nx

nnnnx

nn)1(

2)!(

)!2(.....)1(

2!2

)2)(1)(1()1(

2!1

)1(1

2

2

222

--(1)

But Pn(x) is polynomial of degree n, which contains only odd or even powers of x according

as n is odd or even.

Hence, Pn(x)=anxn+an-2 x

n-2+….. --(2)


It is noted from (1) that Pn(1)=1 and using (2) Pn(-1)=(-1)n. Further from (1), we get

an= nn

n

2)!(

)!2(2

. Since a polynomial solution is valid everywhere, from the power series solution

we have obtained at x = 0, the recursion formula used in that context relates the coefficients of

Pn(x) in the form (2). Thus 2)1(

)1)(2(

kk a

kk

knkna and writing in the reverse order

with k = n, n-2, n-4,…..yields,

,)12(2

)1(2 nn a

n

nna

24

)32(4

)3)(2(

nn a

n

nna = na

nn

nnnn

)32)(12.(4.2

)3)(2)(1()1( 2

,……

Thus Pn(x) = nn

n

2)!(

)!2(2

....

)32)(12.(4.2

)3)(2)(1(

)12(2

)1( 42 nnn xnn

nnnnx

n

nnx --(3)

The coefficient of xn-2k in (3) can be simplified as )!2()!(!2

)!22()1(

knknk

knn

k

, and we obtain

Pn(x) =

2

0

n

k )!2()!(!2

)!22()1(

knknk

knn

k

xn-2k =

2

0

n

k

)()!(!2

)1( 22 kn

n

n

n

k

xdx

d

knk

= nn

n

n

kknn

kn

n

nx

dx

d

nx

knk

n

dx

d

n1

!2

1)1(

)!(!

!

!2

1 22

0

, called the Rodrigue’s formula,

which is used for computing the Legendre Polynomials directly.

We get, P0(x)=1, P1(x)=x, P2(x)=1/2(3x2-1), P3(x)=1/2(5x3-3x),…..

Ex.1. Assuming that n

n txPtxt

)(21

1

2

is true, show that Pn(1)=1, Pn(-1)=(-1)n,

P2n+1(0)=0 and P2n(0)=!2

)12...(3.1)1(

n

nn

n .

By differentiating both sides w.r.to t, and equating the coefficients of tn obtain the recursion

formula (n+1) Pn+1 )()()12()( 1 xnPxxPnx nn and use it to find P2(x) & P3(x) from

P1(x)=x and P0(x)=1.

Orthogonality of Legendre Polynomials

i.e. {Pn(x),n = 0,1,2,..} is a family of orthogonal functions in [-1,1]

Let f(x) be a function with at least n continuous derivatives in [-1,1] and consider the integral

I=

1

1

)()( dxxPxf n = dxxdx

dxf

n

n

n

n

n)1()(

!2

1 2

1

1

, by Rodrigue’s formula.

nmifn

nmif

dxxPxP nm

12

2

0

)()(

1

1


Applying integration by parts, I =

1

1

2

1

1

)1()(!2

1

n

n

n

nx

dx

dxf

n-

dxxdx

dxf

n

n

n

n

n)1()(

!2

1 2

1

1

1

1)1(

= - dxxdx

dxf

n

n

n

n

n)1()(

!2

1 2

1

1

1

1)1(

, since the expression in bracket vanishes at both the limits

Continuing to integrate by parts, we get I = dxxdx

dxf

n

n

nn

nnn

n

n

)1()(!2

)1( 2

1

1

)(

= dxxxfn

nn

n

n

)1()(!2

)1( 2

1

1

)(

.

Take f(x) = Pm(x), where m < n. Then f(n)(x) = 0, since Pm(x) is a polynomial of degree m.

Thus I = 0 .,0)()(

1

1

nmdxxPxP nm

Now let f(x) = Pn(x). Then f(n)(x)= !2

)!2(

n

nn

, and we get from above,

I =

dxxn

n n

n)1(

)!(2

)!2(1

1

2

22 2 dxx

n

n n

n)1(

)!(2

)!2(1

0

2

222

22 )!(2

)!2(

n

nn

dn

2

0

12cos , by the

substitution x = sin .

Thus I = 222 )!(2

)!2(

n

nn 3

2.....

12

22.

12

2

n

n

n

n =

12

2

n, on simplification.

Legendre series

Let f(x) be an arbitrary function, then )(0

xPa n

n

n

, where dxxPxfna nn )()(2

11

1

is called the Legendre series expansion of f(x). The expression of an’s are motivated by the

orthogonality properties of Legendre polynomials. Notice that if P(x) is a polynomial of

degree k, then P(x) = )(0

xPa n

k

n

n

.

Least square approximation

Let f(x) be a function defined in [-1,1] and consider the problem of finding a polynomial P(x)

of degree less than or equal to n, for a given n, such that the error estimate,

I = dxxPxf

1

1

2)]()([ is least. We will show that the approximation is uniquely fixed as


P(x) = )(0

xPa k

n

k

k

, where dxxPxfka kk )()(2

11

1

, and Pk(x) is the kth Legendre

polynomial.

We have I = dxxPbxfn

k

kk

1

1

2

0

])()([ = dxxf

1

1

2)]([ +

n

k

kbk0

2

12

2- 2

1

10

)()( dxxPxfb k

n

k

k

= dxxf

1

1

2)]([ +

n

k

kbk0

2

12

2- 2

n

k

kk bak0 12

2=

dxxf

1

1

2)]([ +

n

k

kk abk0

2)(12

2-

n

k

kak0

2

12

2, which is least when b k = ak for k = 0 to n.

Hence the result.

Ex.1. If P(x) is a polynomial of degree n > 0 such that, 0)(

1

1

dxxPx k , for k = 0,1,..,n-1,

show that P(x) = c Pn(x), for some constant c.

Ex2. Show that among all the monic polynomials P(x) of degree n, )()!2(

)!(2 2

xPn

nn

n

is the

unique one so that dxxP 2

1

1

)]([

is least.

2.2. Bessel functions, The Gamma function

The differential equation 0)( 22'''2 ypxxyyx , where p is a non negative constant,

is known as the Bessel differential equation.

Note that x = 0 is a regular singular point of the equation with indicial equation m2-p2=0 and

exponents are m1 = p and m2 = -p. The equation has a solution in the form y = n

n

p xax

pn

n xa , where a0 0. The recurrence relation for an’s is, n(2p+n)an + an-2=0.

Since a-1=0, an = 0 for odd values of n. We get ))...(1(!2

1)1(

22nppn

an

n

n

.

Hence, we have,

y =

0

0

pxa))...(1(!2

)1(2

2

nppn

xn

nn

.

Taking a0 =1/2pp!, we get the solution,


Jp(x) =

0!2 p

xp

p

))...(1(!2)1(

2

2

nppn

xn

nn

=

)!(!

)2/()1(

2

0 npn

x Pnn

, called the

Bessel function of the first kind of order p.

Remark: In the above discussion we have used the notation p!, though p is a real number not

necessarily a non negative integer for which factorials are defined. We extend the definition

of factorial with the help of gamma function as follows.

For p > 0, we have dtetp tp

0

1)( .

The famous recurrence relation on gamma integral is obtained below.

Now

bdtetp tp

lim)1(

0

dtet t

b

p

0

1

=

dtetpetb

t

b

pbtp

0

1

0

lim=p

dtetb

t

b

p

0

1lim

=p p , since base

bb

p

0

Now .1)1(0

dte t Thus for any non negative integer n, )1( n =n )(n =n(n-1)

)1( n

=……= n(n-1)(n-2)…1 )1( = n!.

From the recurrence relation, presented as, )( p = p

p )1( -- ( I ), we can define )( p for

-1<p<0, since )1( p is available for p+1>0. For -2<p<-1, we again use ( I ) and the

extended definition , since -1<p+1<0. This process is continued to define )( p for all

negative real numbers, which are not integers.

Again from ( I ), we get, 0

lim

p)( p =

0

lim

p p

p )1( = . Hence, we can define,

)0( = and using ( I ) repeatedly define )( p = , for p any negative integer.

Now we have extended )( p for all values of p. We may now define p!= )1( p , for all

values of p or its reciprocal 1/p! = 1/ )1( p which vanishes by definition at any negative

integer p alone. Now with the above extension for factorial or its reciprocal Jp(x) is well

defined for all p > 0.

We have m1 - m2 = 2 p. There exists a Frobenius series solution corresponding to m2 = -p,

even when p =1/2, 3/2,..as a multiple of J-p(x) = )!(!

)2/()1(

2

0 npn

x pnn

.


The first term of this series is

px

p

2)!(

1 which is unbounded as x 0. Hence J-p(x) is

unbounded at x = 0 where as Jp(x) is bounded at x = 0, for p not an integer . Thus for p not an

integer, the general solution at x = 0 is y = c1 Jp(x) + c2 J-p(x).

For p = m a non negative integer J –m(x) =

)!(!

)2/()1(

2

0 nmn

x mnn

)!(!

)2/()1(

2

nmn

x mnn

mn

,

since 0)!(

1

nm for n = 0,1,…,m-1.

Thus J –m(x) =

)!()!(

)2/()1(

)(2

0 nmn

x mmnmn (-1)m

)!(!

)2/()1(

2

0 nmn

x mnn (-1)m Jm(x).

Hence Jm(x) & J-m(x) are not independent, when m = 0,1,2,….

Remark: The general solution is y = c1 Jp(x) + c2 Yp(x), where Yp(x)=

p

xJpxJ pp

sin

)(cos)( ,

for p not an integer and for m = 0,1,2,…Ym(x) = )(lim

xYp

p

.

Ex.1. Show that )2/1( .

We have 2)2/1(0

2

1

dtet t dse s2

0

, by the substitution t = s2.

dxe x

0

2 2

4)2/1( 40

2

dye y

0 0

)( 22

dxdye yx 4

0

2

0

2

drrde r ,

Changing to polar coordinates. Hence )2/1( .

Ex.2. When p = ½ , show that the general solution can be taken in equivalent forms

y = c1 J1/2 (x) + c2 Y-1/2(x) and y = xdxdx

sincos1

21 . Hence )(2

1 xJx = a cosx + b sinx

and )(2

1 xJx = c cos x + d sin x. Evaluate a,b,c,d and show that xx

xJ sin2

)(2

1

and

xx

xJ cos2

)(2

1

.

Properties of Bessel functions

We have Jp(x)= )!(!

)2/()1(

2

0 npn

x Pnn

.

Now dx

dxJx

dx

dp

p )()!(!2

)()1(

2

22

0 npn

xpn

Pnn

=)!1(!2

)()1(

12

122

0

npn

xpn

Pnn

=


)!1(!

)2/()1(

12

0

pnn

xx

pnnp = xp Jp-1(x).

i.e. )()( 1 xJxxJxdx

dp

p

p

p

…(1)

Similarly it can be shown that )()( 1 xJxxJxdx

dp

p

p

p

…..(2)

i.e. xp Jp’(x)+p xp-1 Jp(x) = xp Jp-1(x) …(1) & . x-p Jp

’(x) - p x-p-1Jp(x) = - x-p Jp+1(x)

..(2)

Now, (1)/xp Jp’(x) + (p/x) Jp(x) =Jp-1(x) ..(3) & (2)/x-p Jp

’(x) – (p/x) Jp(x)=-Jp+1(x)

..(4)

(3)+(4) 2 Jp’(x) = Jp-1(x)-Jp+1(x) ....(5) & (3)-(4) (2p/x) Jp(x) = Jp-1(x)+Jp+1(x)

…..(6)

The recurrence relation on Bessel functions is, (2p/x) Jp(x) = Jp-1(x)+Jp+1(x).

Orthogonality properties.

If n ’s are the positive zeroes of Jp(x), then

1

0

2

1 ,)(2

1

,0

)()(nmJ

nm

dxxJxxJnp

npmp

Let y = Jp(x) . Then 0)1(1

2

2''' y

x

py

xy . If a & b are distinct positive constants,

then u(x) = Jp(ax) & v(x) = Jp(bx) satisfy the equations, 0)(1

2

22''' u

x

pau

xu -- (1)

and 0)(1

2

22''' v

x

pbv

xv --(2).

(1)v – (2) u uvabuvvux

uvvudx

d)()(

1)( 22'''' --(3)

(3) x xuvabuvvuxdx

d)()]([ 22'' --(4)

Integrating from 0 to 1, we get, 1

0

1

0

''22 )( uvvuxxuvdxab = 0, if a & b are distinct

zeroes of Jp(x).

Let a = m & b = n . Then we have obtained,

1

0

,0)()( nmifdxxJxxJ npmp


(1)2x2u’ 02222 '2'22''''22

uupuuxaxuuux

or 222222'2 2])([2

xuaupxauxdx

d --(5).

Thus, dxxuaupxaux 2

1

0

21

0

2222'2 2])([2

..(6)

But u(x) = Jp(ax) and hence u’(1)=a Jp’(a). Thus, we get, from (6), with a replaced by n ,

that, 2'

1

0

2 )(2

1)( npnp JdxxxJ 2

1 )(2

1npJ .

Bessel series

Let f(x) be a function defined in [0,1] and n ’s be the positive zeroes of some fixed Bessel

function Jp(x), p 0. Then, )(1

xJa np

n

n

, where an =

0

2

1

)()()(

2dxxJxxf

Jnp

np

, is called

the Bessel series expansion of f(x).

The following theorem gives sufficient conditions for the expansion of a function as a Bessel

series.

Bessel Expansion Theorem : Assume that f(x) and f’(x) have at most a finite number of

jump discontinuities in [0,1]. If 0 < x < 1, then the Bessel series (B) converges to f(x), when x

is a point of continuity, and converges to ½ [f(x-)+f(x+)], when x is a point of discontinuity.

Ex.1. Prove that the positive zeroes of Jp(x) and Jp+1(x) occur alternately.

Ex.2. If

12/1,0

2/1,2/1

2/10,1

)(

x

x

x

xf , show that )()(

)2/()( 0

12

1

1 xJJ

Jxf n

nn

n

, where

n ’s are the positive zeroes of J0(x).

Ex.3. If F(x) = xp in [0,1), show that its Bessel series for a given p is

)()(

2

1 1

xJJ

x np

npn

p

. If g(x) is a well-behaved function in [0,1], then show that

dxxgx p )(

2

11

0

1 dxxJxxgJ

np

npn

)()()(

21

01 1

. By taking g(x) = xp, xp+1, deduce that

)1(4

112

pn

and )2()1(16

1124

ppn

. Taking p = ½ , derive that 6

1 2

2

n

and 90

1 4

4

n.


CHAPTER 3

SYSTEMS OF FIRST ORDER EQUATIONS

3.1. LINEAR SYSTEMS

Let x, y be variables depending on the independent variable t. Consider the following system

of first order differential equations

),,(

),,(

yxtGdt

dy

yxtFdt

dx

….(1).

The above system is called linear if the dependent variables x & y are appearing only in first

degree. Thus the corresponding linear system can be presented as,

)()()(

)()()(

222

111

tfytbxtadt

dy

tfytbxtadt

dx

….(2)

If f1(t) & f2(t) are identically zero, the system is called homogeneous. Thus the associated

homogeneous linear system is

ytbxtadt

dy

ytbxtadt

dx

)()(

)()(

22

11

……(3)

We assume that ai(t), bi(t), fi(t), i = 1, 2, are continuous in some interval [ a , b ].

The solution of (2) is a pair of functions, x = x(t) and y = y(t).

We require the support of the following theorems in our discussion.

Theorem 1. If t0 is any point in [ a , b ], and if x0 and y0 are given numbers, then (2) has a

unique solution x = x(t), y = y(t), valid in [ a , b ], such that x( t0 ) = x0 and y (t0 ) = y0.

( Proof is given later )

Theorem 2. If the linear homogeneous system (3) has two solutions, x = x1(t), y = y1(t),

and x = x2(t), y = y2(t), valid in [ a , b , then x = c1 x1(t) + c2 x2(t), y = c1 y1(t) + c2 y2(t) is also

a solution, for any two constants c1, c2.

Let W(t) = )()(

)()(

21

21

tyty

txtx. Then W(t) is called the Wronskian of the solutions (x1(t),y1(t)) and

(x2(t),y2(t)).


Theorem 3. If the two solutions (x1(t),y1(t)) and (x2(t),y2(t)) of the homogeneous system (3 )

has a Wronskian that does not vanish on [ a , b ], then x = c1 x1(t) + c2 x2(t), y = c1 y1(t) + c2

y2(t) , where c1 & c2 are arbitrary constants, is the general solution of (3 ) in [ a, b ].

Theorem 4. The Wronskian of two solutions of the homogeneous system, is either identically

zero or no where zero in [ a, b ].

Proof: We have dt

dW[ a1(t) + b2(t) ]W, which gives W(t) =

dttbtace )()( 21 , for some

constant c.

Then W(t) 0 , if c = 0 and W(t) 0 , for any t, if c 0 .

Remark: The two solutions x = x1(t), y = y1(t), and x = x2(t), y = y2(t),

valid in [ a , b ] of the homogeneous system are said to be linearly independent , if one is not

a constant multiple of the other which is equivalent to the condition that the Wronskian of the

solutions is not zero.

The following Theorem is a consequence of the above definition and Theorem 4.

Theorem 5. If the two solutions x = x1(t), y = y1(t), and x = x2(t), y = y2(t), are linearly

independent, then x = c1 x1(t) + c2 x2(t), y = c1 y1(t) + c2 y2(t) , where c1 & c2 are arbitrary

constants, is the general solution of (3 ) in [ a, b ].

Theorem 6. If the two solutions (x1(t),y1(t)) and (x2(t),y2(t)) of the homogeneous system are

linearly independent and x = xp(t), y = yp(t) is any particular solution of the corresponding

non- homogeneous system (2), then x = c1 x1(t) + c2 x2(t) + xp(t) , y = c1 y1(t) + c2 y2(t) + yp(t),

where c1 & c2 are arbitrary constants, is the general solution of (2 ) in [ a, b ].

Proof: Let (x(t),y(t) be a solution of (2). Then it can be easily shown that (x(t) - xp(t),y(t)-

yp(t)) is a solution of (3), and the result follows, by virtue of Theorem 5.

3.2. Homogeneous Linear System with constant coefficients.

Consider the system,

ybxadt

dy

ybxadt

dx

22

11

…(4) , where a1, a2,b1,b2 are constants. We may

assume that a solution of the system can be taken as

mt

mt

Bey

Aex …(5).

If we substitute (5) in equation (4),

we get, Amemt = a1 A emt + b1 B emt, bmemt = a2A emt + b2 B emt . Cancelling e mt throughout

gives the homogeneous linear algebraic system, (a1-m)A+b1B=0, a2A+(b2-m)B=0 ....(6).


It is clear that the system trivial solution A = 0, B = 0 yields the trivial solution

x = 0, y = 0 of (4). The system (6) has a non trivial solution iff .022

11

mba

bma

On expansion of the determinant, we get the quadratic equation

m2 – ( a1 + b2 ) m + ( a1b2 – a2b1) = 0 ….(7) , with roots, say, m = m1, m2.

For m = m1, the system (6) gives a non trivial solution, say, A1, B1. Then

tm

tm

eBy

eAx

1

1

11

11

We get the solution corresponding to m = m2, in a similar fashion as

tm

tm

eBy

eAx

2

2

22

22 .

The nature of the roots m1 & m2 are important whenever we try to write the general solution.

Case 1: Distinct Real roots.

If m1 and m2 are real and distinct, then x = c1 x1 + c2 x2, y = c1 y1 + c2 y2 is the general

solution.

Case 2. Complex roots

Let m = iba be the roots of (7). For m = a + ib, solve (6), to get A = A1 +iA2 , B = B1+iB2

Since we require real solutions alone, the general solution is a linear combination of

))sincos(),sincos(( 211211 btBbtBeybtAbtAex atat and

))cossin(),cossin(( 212212 btBbtBeybtAbtAex atat . These are obtained by

separating into real and imaginary parts, the solution,

mt

mt

Bey

Aex , obtained for m = a +ib .

Case 3: Two equal real roots

We get one solution as,

mt

mt

Bey

Aex. A second solution may be obtained in the form

mt

mt

etBBy

etAAx

)(

)(

21

21 and their linear combination gives the general solution.

Eg.1. Consider the system,

yxdt

dy

yxdt

dx

24

. Let x = A e mt , y = B e mt. Then after

cancellation of e mt, we get, the linear algebraic system (1- m ) A + B = 0 , 4 A + ( -2 – m )

B = 0.

For non trivial solution of the algebraic system, we have, m 2 + m – 6 = 0 i.e. m = -3 or 2.

With m = -3, the algebraic system becomes, 4 A + B = 0.


A non trivial is chosen as A = 1, B = -4 . Thus we have the solution, x = e -3t, y = -4 e -3t.

With m = 2, we get - A + B = 0. A non trivial solution is taken as, A = 1, B = 1. This gives

the solution, x = e 2t , y = e 2t. It may be noted that the solutions obtained are independent.

Hence the general solution is x = c1 e -3t + c2 e 2t , y = -4 c1 e -3t + c2 e 2t.


yxdt

dy

yxdt

dx43

. Let x = A e mt , y = B e mt. Then after

cancellation of e mt, we get, the linear algebraic system , ( 3 – m ) A – 4 B = 0, A + ( -1 – m )

B = 0 …( 1 )

For a non zero solution, ( 3 – m ) ( -1 – m ) + 4 = 0 i.e. m 2 – 2m + 1 = 0 or m = 1, 1.

With m = 1, ( 1 ) gives, A – 2 B = 0. Choose, A = 2, B = 1. Corresponding solution is,

x = 2 e t, y = e t .

A second solution linearly independent from the above is assumed to be, x = ( A1 + A2 t ) e t

and y = ( B1 + B2 t ) e t. Then we obtain, ( A 1 + A 2 t + A2 ) = 3 ( A1 + A2 t ) – 4 ( B1 + B2 t ) &

( B 1 + B 2 t + B2 ) = ( A1 + A2 t ) – ( B1 + B2 t ). Since these are identities in t, we get,

2 A2 – 4 B2 = 0, A2 – 2 B2 = 0, 2 A1 – A2 -4 B1 =0, A1 – 2 B1 – B2 = 0. A non zero solution is

taken as, A2 = 2, B2 = 1, A1 = 1, B1 = 0. Now we get another solution, x = ( 1 + 2t ) e t, y = e t.

The two solutions obtained are linearly independent. Hence, we get, x = 2 c1 et + c2 ( 1 + 2t ) et,

y = c1 e t + c2 t e t as the general solution.


yxdt

dy

yxdt

dx

25

24 . Let x = A e mt , y = B e mt. Then after

cancellation of e mt, we get, the linear algebraic system , ( m - 4 ) A + 2 B = 0, 5A +

( 2 – m ) B = 0 ( 1 )

For non trivial solution of (1), we have, m 2 – 6 m + 18 = 0 or m = i33 . Since the values of

m are complex, we are expecting complex values for A & B also.

Let A = A1 + i A2 , B = B1 + i B2 and substitute, m = i33 , in (1). We obtain,

( - 1 +3 i ) (A1 + i A2 ) + 2 (B1 + i B2 ) = 0, 5 (A1 + i A2) + ( -1 -3i )( B1 + i B2 ) = 0.

Equating the real and imaginary parts, - A1 – 3 A2 + 2 B1 = 0, 3 A1 – A2 + 2 B2 = 0, 5 A1 – B1

+ 3 B2 = 0, 5 A2 – 3 B1 – B2 = 0. Consider the coefficient matrix and reduce it to row echelon

form. A solution of the homogeneous algebraic system is, A1 = 2, A2 = 0, B1 = 1, B2 = -3.

The general solution is, x = e 3t( 2 c cos 3t + 2 d sin 3t ),

y = e 3t[c( cos 3t + 3 sin 3t) + d(sin 3t – 3 cos 3t)]


3.3 Non linear system – Volterra’s prey – predator equations.

Consider an island inhabited by foxes and rabbits. The foxes hunt the rabbits and rabbits feed

on carrots. We assume that there is abundant supply of carrots. As the rabbits become large in

number, foxes flourish as they hunt on rabbits and their population grows. As the foxes

become numerous and eat too many rabbits, the rabbit population declines. As a result the

foxes enter a period of famine and their population declines. As foxes decrease in number the

rabbits become more safe resulting in a population surge. As time goes on we can observe an

unending almost cyclic repetition of population growth and decline of either species.

We make a mathematical formulation of the above problem. Let x be the rabbit population

and y, the corresponding population of foxes at a given instant.

Since there is an unlimited supply of carrots, the rabbit population grows as in the case of a

first order reaction relative to the current population.

Thus in the absence of foxes, ,axdt

dx a > 0 . It is natural to assume that the number of

encounters between foxes and rabbits is jointly proportional to their populations. As these

encounters will enrich the fox population, but results in the decline of rabbit population, we

may correct the above equation as ,bxyaxdt

dx where a , b > 0. In a similar

manner, we obtain, ,dxycxdt

dy where c , d > 0. Thus, we have, the following non linear

system, describing the populations, ,bxyaxdt

dx

.dxycxdt

dy The above equations are called

Volterra’s prey – predator equations.

Eliminating t , we get x

dxdxc

y

dybya )()(

. The solution is , y a e –by = K x –c e dx, where

K = x0c y0

a e –dx0

– dy0 , for some initial solution, ( x0 , y0 ). Drawing the (x,y) graph is really

tough and Volterra has introduced an efficient approach in this regard as discussed below.

We note that , x , y being populations, are non negative. The plane is divided into 4 quadrants

and the bordering rays are used to represent the positive x, y, z , w directions. We may take,

z = y a e –by and w = K x –c e dx. Giving suitable values for x and y independently plot the

( y , z ) and ( x , w ) graphs in the respective quadrants and then obtain the ( x , y ) graph from

the ( z, w ) graph which is in fact the straight line z = w.


Note that 0dt

dx

dt

dy gives x = c/d and y = a/b, called the equilibrium populations.

Let x = X + c/d and y = Y + a/b. Then the system becomes,

dXYXb

ad

dT

dY

bXYYd

bc

dt

dX

.

Consider the linearised system,

Xb

ad

dT

dY

Yd

bc

dt

dX

. The solution of the linear system is a d 2 X 2

+ b 2 c Y 2 = L2, a family of ellipses concentric with the origin. The ( x , y ) graph turns out to

be an oval about the equilibrium point ( c/d , a/b )


CHAPTER 4

NON LINEAR EQUATIONS

4.1. Autonomous systems

Consider the system ),( yxFdt

dx , ),( yxG

dt

dy --( 1 ) .

Since F and G are independent of t, the system is called autonomous. The solution of the

system is a pair of functions, ( x(t), y(t) ), describing a family of curves in the x-y plane,

called the phase plane. If t0 is any number and ( x0 , y0 ) is a given point in the phase plane,

there exists a unique curve ( x(t), y(t) ) passing through ( x0 , y0 ) and satisfying the system.

Such a curve is called a path in the phase plane and the plane with all these paths will be

called phase portrait of the system.

For a given path, we may use forward arrows to indicate the direction in which the path is

advancing as t . A point ( x0 , y0 ) at which both F and G vanish, is called a critical point

of the system. Since 0dt

dx and 0

dt

dy at a critical point ( x0 , y0 ), no path is passing

through a critical point and two different paths will not intersect, since there exists a unique

path through a given point.

Given an autonomous system, apart from its solution we are interested in the location of the

various critical points, arrangement of paths near critical points, stability of the critical points

and the phase portrait.

Stability : Let ( x0 , y0 ) be an isolated critical point and C = { (x(t) , y(t) ) | t } be a

path of (1). We say, C approaches ( x0 , y0 ) as t , if ),()(),((lim

00 yxtytxt

and

C enters ( x0 , y0 ) as t , if t

lim

xtx

yty

)(

)( 0 exists or or .

If a path C enters a critical point, then it approaches it in a definite direction.

Eg.1. Consider the system, xdt

dx , yx

dt

dy2 --(1). The origin is the only critical point,

and the general solution is obtained as, x = c1 e t, y = c1 e t + c2 e 2t --(2).

When c1 = 0, we get x = 0, y = c2 e 2t. In this case the path becomes the positive or negative

x – axis according as c2 > or < 0, and as t , the path approaches and enters the critical

point.


When c2 = 0, we get x = c1 e t and y = c1 e t. For c1 < 0, the path is the ray, y =x , x < 0 and for

c1 > 0, the path is the ray, y =x , x > 0, and both the paths enters the critical point, as t .

When 0, 21 cc , then the paths are ½ - parabolas y = x + (c2/c12) x2 . Each of these paths

enters ( 0 , 0 ) as t .

Eg.2. Consider the system, yxdt

dx43 , yx

dt

dy32 . The only critical point is ( 0 , 0 ).

We obtain the general solution as, x = 2 c1 e –t + c2 e t, y = c1 e -t + c2 e t .

When c1 = 0, we get x = c2 e t, y = c2 e t. In this case the path becomes the ½ – line y = x and

as t , the path approaches and enters the critical point.

When c2 = 0, we get x = 2 c1 e -t and y = c1 e -t. The path is the ½ - line y = ½ x , x < 0 and

both the paths enters the critical point, as t .

When 0, 21 cc , then the paths are distinct branches of hyperbolas ( x – y ) ( 2 y – x ) = C,

with asymptotes being y = x and y = ½ x and none of these paths approaches the critical point

( 0 , 0) as t or as t .

Eg.3. Consider the system, ydt

dx , x

dt

dy . The only critical point is ( 0 , 0 ).

We obtain the general solution as, x = - c1 sin t + c2 cos t, y = c1 cos t + c2 sin t , which are

circles, with common centre (0 , 0 ). All paths are closed ones and each of them encloses the

critical point and none of these paths approaches the critical point.

Eg.4. Consider the system, yxdt

dx , yx

dt

dy . The only critical point is ( 0 , 0 ).

By changing to polar coordinates, we get rd

dr

which gives the general solution as the

family of spirals r = c e . We have, 1

dt

d, so that as t , the spiral unwinds in the anti

clock wise fashion to infinity.

Stability and asymptotic stability .

Consider the autonomous system, ),( yxFdt

dx , ),( yxG

dt

dy --( 1 ) . For convenience,

assume that (0 , 0 ) is an isolated critical point of the system. This critical point is said to be

stable, if for each R > 0 given there exists Rr such that every path which is inside the

circle centered at ( 0 , 0 ) with radius r, for some t = t0, remains inside the circle centered


( 0 , 0 ) with radius R for all t > t0 , which is equivalent to say that paths which gets

sufficiently close to the critical point stay close to it in its due course i.e. as t .

The critical point is said to be asymptotically stable if there exists a circle with centre ( 0 , 0)

and radius r0 , such that a path which is inside this circle for some t = t0 approaches the centre

(0 , 0 ) as t .

4.2. Types of Critical points and stability of linear systems.

Consider the homogeneous linear system with constant coefficients,

ybxadt

dy

ybxadt

dx

22

11

---(1) which has evidently the origin as the only critical point by

assuming that a1 b2 – a2 b1 0.

Let x = A e mt, y = B e mt ---(2), be a non trivial solution,

when ever m 2 – ( a1 + b2 ) m + (a1 b2 – a2 b1) = 0 ---(3),

called the auxiliary equation of the system. Let m1 and m2 be the roots of (3).

We may distinguish the following 5 cases.

Major cases:

1. The roots m1 and m2 are real, distinct, and of the same signs.

2. The roots m1 and m2 are real, distinct, and of opposite signs

3. The roots m1 and m2 are complex conjugates, but not pure imaginary.

Border line cases :

4. The roots m1 and m2 are real, and equal.

5. The roots m1 and m2 are pure imaginary.

Case 1. : The critical point is called a node.

The general solution is

tmtm

tmtm

eBceBcy

eAceAcx

21

21

2211

2211 ---(4)

(a) The roots m1 and m2 are both negative.

Further assume, for precision, that m1 < m2 < 0.

When c1 = 0, we get, x = tm

eAc 2

22 , y = tm

eBc 2

22 --(5). If c2 > 0, then we get ½ of the line

y/x = B2/A2, which enters the critical point as t and for c2 < 0, we get the other ½ of

the same line which also enters ( 0 , 0 ) . as t


When c2 = 0, we get x = c1A1 e m1t and y = c1 B1 e m1

t –(6). For c1 < 0, the path is ½ of the

line, y/x = B1/A1, which enters the critical point as t and for c1 > 0, we get the other ½

of the same line which also enters ( 0 , 0 ) . as t .

When 0, 21 cc , then the paths are curves. Since m1 and m2 are both negative, these paths

also approaches ( 0 , ) as t . Considering the expression for y/x from (4), and since

m1 - m2 < 0, each of these paths enters ( 0 , 0 ) as t ( Note that y/x B2/A2 as

t .)

The critical point is referred as a NODE, and in this case it is asymptotically stable.

If m2 < m1 < 0, then the above conclusion holds good, with a change each curvilinear path

enters ( 0, 0 ) along the direction B1/A1. (b) The roots m1 and m2 are both positive.

The situation is exactly the same , but all the paths are approaching and enters ( 0 , 0 ) as

t .

2. Assume m1 < 0 < m2

The two ½ line paths represented by (5) enter ( 0 , 0 ) as t , the two ½ line paths

represented by (6) enter ( 0 , 0 ) as t .

But none of the curvilinear paths represented by (4), corresponding to 0, 21 cc ,

approaches ( 0 , 0 ) as t or t ; each of them is asymptotic to one of the ½ line

paths.

The critical point is called a SADDLE POINT which is always unstable.

3. Let 0,, 21 awhereibamibam .

The general solution is,

)]cossin()sincos([

)]cossin()sincos([

212211

212211

btBbtBcbtBbtBcey

btAbtAcbtAbtAcexat

at

--(8)

Suppose, a < 0. As t , all paths approach ( 0 , 0 ), but they do not enter it and they wind

around it in a spiral like manner. Changing to polar coordinates,

22

2

112

2

2

22

)(//

yx

ybxyabxa

yx

dtydxdtxdy

dt

d

-- (9)

dt

d > or < 0, if a2 > 0 or < 0.( Note that the discriminant D of the auxiliary equation is

negative, in the present context )

For y = 0, (9) gives dt

d=a2. Thus when a2 >0, then

dt

d> 0 which implies that as t , all

paths spiral about ( 0 , 0 ) in the anti clock wise sense. The sense will be clock wise, if a2 < 0.

The critical point is called a spiral, which is asymptotically stable.


If a > 0, the situation is the same, except that all paths approach ( 0 , 0 ) as t , and hence

it is an unstable spiral.

4. Let m1 = m2 = m, say.

Assume, m < 0.

(a) a1= b2 0 , a2 = b1 = 0. Let the common value be a. Then the system reduces to

yadt

dy

xadt

dx

and its general solution is x = c1 e mt , y = c2 e mt. The paths are ½ lines of

various slopes and since m < 0, each path enters ( 0 , 0 ) as t .

The critical point is an asymptotically stable border line node.

If m > 0, then all paths enter ( 0 , 0 ) as t . The critical point is an unstable border line

node.

(b) All other cases. Assume m < 0.

The general solution is,

tmtm

mtmt

etBBceBcy

etAAceAcx

)(

)(

12211

12211

1

.

When c2 = 0, we get the two ½ line paths lying on y/x = B/A.. Since, m < 0, both of them

enter ( 0 , 0 ) as t .

If c2 0 , the paths are curvilinear and all of them enter the critical point as , keep tangential

to y/x = B/A as they approach ( 0 , 0 ).

The critical point is again an asymptotically stable border line node.

If m > 0, then it is unstable.

5. We may refer case 3. with a = 0. Since the exponential factor is missing from the

solution, they reduce to periodic functions and each path is closed surrounding the origin. The

paths are actually ellipses.

The critical point is called a CENTRE, which is stable, but can not be asymptotically stable.

We may, summarise, some of the observations we have made in the sequence of the above

discussion about stability.

Theorem. The critical point ( 0 , 0 ) of the linear system (1) is stable iff both the roots of the

auxiliary equation have non positive real parts, and it is asymptotically stable ifi both roots

have negative real parts.

Taking, p = - ( m1 + m2) and q = m1 m2, we can reformulate the theorem as,

Theorem. The critical point ( 0 , 0 ) of the linear system (1) is asymptotically stable iff p and

q are both positive.


4.3. Liapunov’s direct method

In a physical system, if the total energy has a local minimum at certain equilibrium point, then

it is stable. This concept leads to a powerful method for studying stability problems.

Consider, the autonomous system, ),( yxFdt

dx , ),( yxG

dt

dy . Assume that ( 0 , 0 ) is an

isolated critical point of the system. Let C = [ x(t), y(t)] be a path. Let E(x,y) be a function

that is continuous and having continuous first partial derivatives in a region containing C. If

( x , y ) is point on C, then E(x,y) is a function of t alone, say, E(t). Its rate of change, as the

point moves along C is,

dt

dy

y

E

dt

dx

x

E

dt

dE

= G

y

EF

x

E

.

Let E(x,y) be a continuous function with continuous first partial derivatives in some region

containing the origin. If E( 0 , 0 ) = 0 , and then it is said to be positive definite if E (x,y) > 0,

for (x,y) 0 , and negative definite if E(x,y) < 0, for (x,y) 0 . Similarly, E is called positive

semi-definite if E(0,0) = 0 and E(x,y) 0, for (x,y) 0 and negative semi-definite if E(0,0)

= 0 and E(x,y) 0, for (x,y) 0 .

Functions of the form a x 2m + b y 2n , where m & n are positive integers and a & b are

positive constants, are positive definite. Note that E(x,y) is negative definite iff –E(x,y) is

positive definite; functions x 2m or y 2n are not positive definite.

Given the linear system (1), a positive definite function E(x,y) such that the derived function

H(x,y) = Gy

EF

x

E

is negative semi-definite is called a Liapunov function for (1). By the

earlier discussion, we get that, along a path C near the origin, dE/dt 0, and hence E is

decreasing along C as it advances.

Theorem. If there exists a Liapunov function E(x,y) for the system (1), then the critical point

( 0 , 0 ) is stable. Furthermore, if this function has the additional property that the derived

function H(x,y) is negative definite, then ( 0 , 0 ) is asymptotically stable.

Proof: Let C1 be a circle of radius R > 0 centered at the origin and it may be assumed that C1

is small enough that it is contained in the domain of definition of E. Since E(x,y) is

continuous and positive definite, it has a positive minimum m on C1. Since E(x,y) is

continuous at the origin and vanishes there, we can find 0 < r < R such that E(x,y) < m,

whenever (x,y) is inside the circle C2 of radius r nad centered at the origin. Let C be a path

which is inside C, for t = t0. Then E(t0 ) < m, and dE/dt 0 implies that E(t) E(t0) < m for


all t > t0. It follows that the path C can never reach the circle C1 for t > t0. Thus ( 0 , 0 ) is

stable.

Under the additional assumption, we claim further that E(t) 0 as t . This would

imply that the path C approaches ( 0 , 0 ) as t . Now along C, dE/dt < 0, E(t) is a

decreasing function. Since E(t) is bounded below by 0, E(t) L 0 ,say , as t . Then it

suffices to show that L = 0.

Suppose not. Choose 0 < r < r such that E(x,y) < L/2, whenever (x,y) is inside the circle C3

with radius r . Since H is negative definite, it has a negative maximum -k in the closed

annulus bounded by C1 and C3. Since this region contains C for t t0, E(t) = E( t0) +

dtdt

dEt

t

0

which gives E(t) E(t0 ) - k(t – t0 ). But since right side of the inequality becomes

negatively infinite as t , E(t) as t . This contradicts the fact that E(x,y) 0.

Thus L = 0, and the proof is complete.

Eg. Consider the equation of motion of a mass m attached to a spring,

02

2

kxdt

dxc

dt

xdm .

Here 0c is the viscosity of the medium through which the mass moves, and k> 0 is the

spring constant. The equivalent autonomous system is,

ym

cx

m

k

dt

dy

ydt

dx

The only critical point is ( 0 , 0 ). The kinetic energy of the mass is my 2/2, and the potential

energy due to the current elongation of the spring is 2

02

1kxkxdx

x

.

Thus the total energy of the mechanical system is E( x , y ) = ½ m y 2 + ½ k x 2.

Then E( x , y ) is positive definite and H( x , y ) = k x y + m y ( -k/m x-c/m y )= - c y 2 0 .

Thus E( x , y ) is a Liapunov function and the critical point is stable, by Theorem

Ex.1. Show that ( 0 , 0 ) is an asymptotically stable critical point of the system

35

3

2

3

yxdt

dy

yxdt

dx

.

Let E( x , y ) = a x 2m + b y 2n, where a , b > 0 and m , n are positive integers. E is positive

definite and H = 2ma x 2m-1( -3 x 3 - y ) + 2nb y 2n-1 ( x 5 – 2 y 3 ) = - 6 ma x 2m+2 – 2ma x 2m-1y


+ 2 nb x5 y 2n-1- 4 nb y 2n+2. Let m = 3, n = 1, a = 1 , b = 3. Then H = -18 x8 – 12 y 4 is

negative definite.

Now, E( x , y ) is a Liapunov function for the system with the derived function H(x,y),

negative definite. Thus the critical point ( 0 , 0 ) is asymptotically stable.

Ex.2. Show that ( 0 , 0 ) is an asymptotically stable critical point of the system

322

32

yyxdt

dy

xyxdt

dx

.

Take E( x , y ) = x 2 + y 2.

4.4. Simple critical points – Non linear system

Consider the autonomous system,

),(

),(

YxGdt

dy

yxFdt

dx

with an isolated critical point at ( 0 , 0 ).

Since F ( 0 , 0 ) = 0 = G( 0 , 0 ), assuming their Maclaurin’s series expansions about ( 0 , 0 )

and neglecting higher powers of x & y , for ( x , y ) close to ( 0 , 0 ), the system reduces to a

linear one.

More generally, we may take the system as,

),(

),(

22

11

yxgybxadt

dy

yxfybxadt

dx

It is assumed that a1 b2 – a2 b1 0 , so that the critical point will be isolated.

( 0 , 0 ) is called a ‘ simple critical point ’ of the system, if 0),(

)0,0(),(

lim

22

yx

yxf

yx

and 0),(

)0,0(),(

lim

22

yx

yxg

yx.

Theorem. Let ( 0 , 0 ) be a simple critical point of the non linear system,

),(

),(

22

11

yxgybxadt

dy

yxfybxadt

dx

--(1) .

If the critical point ( 0 , 0 ) of the associated linear system,

ybxadt

dy

ybxadt

dx

22

11

--(2)


falls under any of the three major cases ( Node, Saddle point, spiral ), then the critical point

( 0 , 0 ) of (1) is of the same type.

Remark: There will not be similarities among the paths in both the systems. In the non linear

case, paths will have more distortions.

If system (2) has the origin as a border line node ( centre ), then origin will be either a node or

spiral ( centre or spiral ) for the system (1).

Theorem. Let ( 0 , 0 ) be a simple critical point of the non linear system (1), and consider the

related linear system(2). If ( 0,0 ) is an asymptotically stable critical point of (2), then it is

asymptotically for (1).

Proof. We may construct a suitable Liapunov function for the system (1) to justify the claim.

The coefficients of the auxiliary equation of the linear system, namely, p & q will be positive,

by the assumption that ( 0 , 0 ) is asymptotically stable for (2).

Now define, E( x , y ) = ½ ( a x 2 + 2 b x y + c y 2 ), where a = pq

bababa )( 1221

2

2

2

2 ,

c = pq

bababa )( 1221

2

1

2

1 and b = -

pq

bbaa )( 1221 .

Note that p = - (a1 + b2 ) & q = ( a1 b2 – a2 b1 ). We have p ,q , a , b > 0 , and it can be

directly shown that ( ac – b2 ) > 0. Since b 2 – ac < 0 & a > 0, E(x,y) is positive definite.

It can also be easily obtained that,

H( x , y ) = (a x + b y ) ( a 1 x + b1 y ) + ( b x + c y ) ( a2 x + b2 y ) = -(x 2 + y 2 ), which is

negative definite.

Thus E( x , y ) is a Liapunov function for the linear system.

Now by using the continuity of f & g at ( 0 , 0 ), and shifting polar coordinates, it can be

shown that Gy

EF

x

E

, where F = ( a1 x + b1 y ) + f(x , y ) and G = ( a2 x + b2 y ) + g

( x , y ) is negative definite. Thus E ( x , y ) is a positive definite function with the derived

function related to the non linear system negative definite. Hence by Theorem, ( 0 , 0 ) is

asymptotically stable, for the system (1).

Eg. The equation of motion for damped vibrations of a simple pendulum is,

0sin)/()/(2

2

xagdt

dxmc

dt

xd ---(1) , where c > 0 and g is the acceleration due to

gravity.


The equivalent autonomous system is

ymcxagdt

dy

ydt

dx

)/(sin)/(

--(2)

(2) can be written as,

)sin)(/()/()/( xxagymcxagdt

dy

ydt

dx

--(3)

Thuis, f(x,y) = 0 and g(x,y) = (g/a) (x-sinx).

Since, 22

),(

)0,0(),(

lim

yx

yxg

yx0

sin

)0,0(),(

lim

22

yx

xx

yx ,

( 0 , 0 ) is a simple critical point of the non linear system (2). ( As ( x , y ) 0, for x 0,

22

|sin|

yx

xx0/sin1

||

|sin|

xx

x

xx ).

Now ( 0 , 0 ) is an isolated critical point of the associated linear system,

ymcxagdt

dy

ydt

dx

)/()/(

--(4) .

We have p = (c/m) > 0 , q = ( g / a ) > 0 , for (4)

Hence by Theorem, ( 0 , 0 ) is asymptotically stable. Thus by the last Theorem, ( 0 , 0 ) is an

asymptotically stable critical point of the original system (1).

Since x = 0 and y = dx/dt = 0 refers to the mean position and initial velocity, asymptotic

stability of ( 0 , 0 ) implies that the motion due to a slight disturbance of the simple pendulum

will die out with the passage of time.

We give a few more Theorems helpful in the investigation of an autonomous system

Theorem.1. A closed path of an autonomous system necessarily surrounds at least one

critical point.

Thus a system without critical points in a given region can not have closed paths in that

region.

Theorem.2. If y

G

x

F

is always positive or always negative in a certain region of the

phase plane, then the system can not have closed paths in that region.


Proof: Assume that the region contains a closed path C with interior R. Then by Green’s

Theorem. C R

GdxFdy )(y

G

x

F

dx dy 0. But along C dx = F dt & dy = G dt, so

that

C

GdxFdy )( = 0, a contradiction.

Theorem.3.( Poincare-Bendixson) Let R be bounded region of the phase plane together with

its boundary, and suppose R does not contain any critical points of the system. If C is a path

that enters R and remains in R in its further course, then C is either a closed path or it spirals

toward a closed path as t . Thus in either case the system has a closed path.

Theorem.4. ( Lienard) Let the functions f(x) and g(x) satisfy the following conditions. (1)

both are continuous and have continuous derivatives for all x (2) g(x) is an odd function such

that g(x) > 0 for x > 0, and f(x) is an even function (3) the odd function x

dxxfxF0

)()( has

exactly one positive zero at x = a, is negative for 0 < x < a, is positive and non decreasing for

x > a and F(x) as x . Then the Lienard’s equation 0)()(2

2

xgdt

dxxf

dt

xd has a

unique closed path surrounding the origin in the phase plane, and this path is approached

spirally by every other path as t .


CHAPTER 5

SOME FUNDAMENTAL THEOREMS

5.1. Method of successive approximations

Consider the initial value problem (IVP), y’ = f(x,y), y(x0 ) = y0, --( 1) , where f(x,y) is a

function continuous in some neighborhood of ( x0 , y0 ). A solution is geometrically a curve in

the x – plane that passes through ( x0 , y0 ) so that at each point on the curve the slope is

prescribed as f( x , y ).

The IVP is equivalent to the integral equation ( IE), y( x ) = y0 + x

x

dttytf

0

)](,[ --(2)

{ ( 1) is equivalent to (2) : Suppose y (x ) is a solution of ( 1 ). Then y ( x ) is indeed

continuous and if we integrate it from x0 to x, (2) is obtained. If y( x ) is a continuous

solution of (2), then y(x0) = y0 and by differentiating y’(x) = f(x,y(x)). }

We may suggest an iterative procedure to solve the IE (2). Start with the rough approximation

y0 ( x ) = y0. Substituting in the right side of (2), we get a new approximation as,

y1( x ) = y0 + x

x

dttytf

0

)](,[ 0 . Next use y1(x) in R.S. of (2), to get y2(x) = y0 + x

x

dttytf

0

)](,[ 1

This process can be continued to get yn( x ) = y0 +

x

x

n dttytf

0

)](,[ 1 .

The procedure is called Picard’s method of successive approximations.

Eg.1. Consider the IVP, y’ = y, y( 0 ) = 1.

Equivalent IE is, y( x ) = 1 + x

dtty0

)( . Then yn(x) = 1+

x

n dtty0

1 )( .

With y0( x ) = 1, y1(x) = 1+ x

dt0

1 .=1 + x, y2(x) = 1+

x

dtt0

)1( = 1 + x + x 2/2,

y3(x) = 1+

x

dttt0

2 )2/1( . = 3.22

132 xx

x . …..

It is clear that, yn ( x ) xexx

x ....3.22

132

.

Note that, y(x) = e x is a solution of the IVP.


Eg.2. Consider, y’ = x + y 2, y( 0 ) = 0.

We may take, y0 ( x ) = 0.

Then y1(x) = 0+ x

tdt0

= x2/2, y2(x) = dttt

x

)4/(0

4

= x 2/2 + x 5/20,

y3(x) =

x

dttttt0

7104 20/400/4/ = x 2/2 + x 5/20 + x 8/160+ x 11/4400,…….

Eg.3. Consider, y’ = x + y , y( 0 ) = 1.

It is not difficult to get the exact solution as, y(x) = 2 e x – x – 1.

With y0 ( x ) = 1, y1(x) = 1+

x

dtt0

)1( = !2

12x

x , y2(x) = 1+

x

dttt0

2 )!2/21( =

!31

32 x

xx , y3(x) = !43

143

2 xxxx ,…..

Note that yn ( x ) 1 + x + 2( e x – x – 1 ) = 2 e x – x – 1.

5.2. Picard’s Theorem

Theorem.1. Let f ( x , y ) and y

f

be continuous functions on a closed rectangular region R,

with sides parallel to the coordinate axes. If ( x0 , y0 ) is any interior point of R, then there

exists a number h > 0 with the property that the initial value problem,

y’ = f ( x , y ) , y( x0 ) = y0 ---(1),

has a unique solution y = y (x ) in [ x0 – h , x0 + h ].

Proof: We know that every solution of the IVP is also a continuous solution of the IE,

y( x ) = y0 + x

x

dttytf

0

)](,[ ---(2) and conversely.

We will show that (2) has a unique solution in [ x0 – h , x0 + h ] , for some h > 0.

We may a produce a sequence of functions following Picard’s method of successive

approximations.

Let y0 ( x ) = y0,

y1( x ) = y0 + x

x

dttytf

0

)](,[ 0 , y2(x) = y0 + x

x

dttytf

0

)](,[ 1 ,….,

yn( x ) = y0 +

x

x

n dttytf

0

)](,[ 1 ,……


Claim. The sequence < yn ( x ) > converges to a solution of the IVP, in [ x0 – h , x0 + h ] , for

some h > 0.

Since R is compact and f( x , y ) and y

f

are continuous in R, they are bounded.

Therefore there exists M , K > 0 such that Myxf ),( --(3) and

RyxKyxfy

),((),4(,),( .

Let ( x , y1 ) , ( x , y2 ) be distinct points in R.

Then by Mean value theorem, 21

*

21 ),(),(),( yyyxfy

yxfyxf

--(5) , for some y*

between y1 & y2. Then by (4), we get 2121 ),(),( yyKyxfyxf --(6).

Now choose h > 0 such that K h < 1 --(7) and the rectangular region R’ defined by

hxx 0 and Mhyy 0 is contained in R. Since ( x0, y0 ) is an interior to R, such an

h exists.

Note that yn(x) is the n th partial sum of the series,

1

10 )()()(n

nn xyxyxy --(8)

Thus to show that < yn( x ) > converges, it is sufficient to show that the series (*) converges.

(a) The graph of the functions y = yn ( x ), for hxx 0 , lies in R’ and hence in R, for

every n.

This is clear for y = y0(x) =y0.

Since ( t , y0(t) ) are in R’, we get Mhdttytfyxy

x

x

0

)](,[)( 001 . Thus graph of y = y1(x)

lies in R’. Now, it turns out that [t,y1(t)] are in R’ and Mhdttytfyxy

x

x

0

)](,[)( 102 .

Thus graph of y = y2 (x) lies in R’. Proceeding similarly we get the result(a).

Since y1( x ) is continuous in hxx 0 , there exists a constant a = max 01 )( yxy .

Since [t,y1(t)] , [t,y0(t)] are in R’, (6) KatytyKtytftytf )()())(,())(,( 0101 and

)()( 12 xyxy Kahdttytftytf

x

x

0

))](,()(,[( 01 =a(Kh).

Similarly, ahKKahKtytyKtytftytf 2

1212 )()()())(,())(,( so


)()( 23 xyxy 22

12 )(.))](,()(,[(

0

KhahahKdttytftytf

x

x

.

Continuing like this, we can show that, 1

1 )()()(

n

nn Khaxyxy .

Now each term of the series |)()(||)(|1

10

n

nn xyxyxy is dominated by the

corresponding term of the series, | y0 | + a + a (Kh) + a (Kh) 2 + … which converges being

essential a geometric series with common ration r = Kh numerically less than 1, by (7).

Thus, by comparison test, the series (8) converges uniformly in hxx 0 to a sum, say,

y(x) and hence < yn ( x ) > converges to y(x) uniformly in hxx 0 .

Since the graph of yn ( x ) lies in R’, the graph of the limit function y(x) also lies in R’.

Since each yn ( x ) is continuous , the uniform limit y( x ) is also continuous.

Now we proceed to prove that, y(x) is a solution of the IVP.

i.e. We have to show that, y( x ) - y0 - x

x

dttytf

0

)](,[ = 0 --(9)

But yn( x ) - y0 -

x

x

n dttytf

0

)](,[ 1 = 0. Now consider, | y( x ) - y0 - x

x

dttytf

0

)](,[ - 0 | =

| y( x ) - y0 - x

x

dttytf

0

)](,[ - [yn( x ) - y0 -

x

x

n dttytf

0

)](,[ 1 ] | =

| [y( x ) – yn ( x ) ] +

x

x

n dttytftytf

0

))](,())(,([ 1 | | [y( x ) – yn ( x ) ] | +

|

x

x

n dttytftytf

0

))](,())(,([ 1 | | [y( x ) – yn ( x ) ] | + K h max | [y( x ) – yn-1 ( x ) ] | , since

graph of y ( x ) lies in R’ and hence in R and by virtue of (6). The uniform convergence of

yn(x) to y ( x) will enable us to make the right side of the above inequality as small as we

please by taking n sufficiently large. Since the left side is independent of n, we get the

required result.

Now we settle the uniqueness part.

Let )(xy be another possible solution of the IVP in hxx 0 .

It is essential to show that the graph of )(xy also lies in R’. On the contrary assume, its graph

leaves R’. Then there exists some x1 in hxx 0 such that Mhyxy 01)( and the


continuity of )(xy at x = x0 will give Mhyxy 0)( if 0xx 01 xx . Then,

MhMhxx

Mh

xx

yxy

/

)(

0101

01

, where as by mean value theorem, there exists x*

between x0 & x1 such that, Mxyxfxyxx

yxy

*))(*,(*)(

)(

01

01

, since ( x*, *)(xy ) lies

in R’ . Hence, a contradiction.

Since both y(x) & )(xy are solutions of the IVP, )()( xyxy |

x

x

dttytftytf

0

))](,())(,([ |

Kh max )()( xyxy , since graph of both functions lie in R’.

So max )()( xyxy K h max )()( xyxy . But this will imply, max )()( xyxy = 0,

since K h < 1.

Thus, we have )(xy = y(x), for every x in the interval hxx 0 .

Remark. We notice that the continuity of y

f

is made of use of in the proof to the extent that

it implies (6). We can replace this requirement by a Lipschitz’s condition , namely, there

exists

K > 0 ,such that 2121 ),(),( yyKyxfyxf .

If we further drop this condition too, it is known that the IVP still has a solution, but the

uniqueness can not be ascertained.– Peano’s Theorem.

Consider the IVP, .0)0(,3' 2

1

yyy Let R be the rectangular region 1||,1|| yx . Here

f(x,y) = 3 y 1/2 is continuous in R. Two different solutions are, y= x 3 and y = 0.

Theorem.2. Let f( x , y ) be continuous and satisfy the Lipschitz’s condition,

2121 ),(),( yyKyxfyxf on a vertical strip . yandbxa If ( x0 , y0 )

is any point of the strip, then the initial value problem y’ = f( x , y ) , y ( x0 ) = y0 has a

unique solution in [ a , b ].

Proof.: The proof is similar to that of Theorem.1. and based on Picard’s method of successive

approximations.

Let M0 = | y0 |, M1 = max | y1(x) |, M = M0 + M1. It can be easily observed that, | y0( x ) |

M and | y1(x) – y0(x) | M .


Assume, bxx 0 . Then )()( 12 xyxy x

x

dttytftytf

0

))](,()(,[( 01

x

x0

dttytyKdttytftytf

x

x

)()())(,())(,( 0111

0

)( 0xxKM ,

)()( 23 xyxy x

x

dttytftytf

0

))](,()(,[( 12

x

x0

dttytyKdttytftytf

x

x

)()())(,())(,( 1212

0

2/)()( 2

0

2

0

2

0

xxMKdtxtMK

x

x

and in general, )()( 1 xyxy nn )!1/()( 1

0

1 nxxMK nn .

The same result holds for 0xxa , but ( x – x0 ) has to be replaced by | x – x 0 |.

Thus, we have, )()( 1 xyxy nn )!1/(|)(| 1

0

1 nxxMK nn )!1/()( 11 nabMK nn .

Now each term of the series |)()(||)(|1

10

n

nn xyxyxy ..(*) is dominated by the

corresponding term of the convergent numerical series,

........!2

)()(

22

abMKabKMMM and hence the series

1

10 )()()(n

nn xyxyxy converges uniformly in [ a , b ] to a limit function y(x).

The uniform convergence will readily imply that y(x) is a solution of the IVP in [ a , b ].

If possible, let )(xy be another solution to the IVP. We claim that, then yn ( x ) )(xy ,

also, so that )(xy =y(x).

Now, )(xy is continuous and )(xy = y0 + x

x

dttytf

0

)](,[ .

Let A = max 0)( yxy ,

Then for bxx 0 , )()( 1 xyxy x

x

dttytftytf

0

))](,()(,[( 0

x

x0

)()())(,())(,( 000

0

xxKAdtytyKdttytftytf

x

x

,

)()( 2 xyxy x

x

dttytftytf

0

))](,()(,[( 1


x

x0!2

)()()())(,())(,(

2

02

0

2

11

00

xxAKdtxtAKdtytyKdttytftytf

x

x

x

x

,…, and

in general, )()( xyxy n !

)( 0

n

xxAK

nn

.

A similar result is got for 0xxa .

Thus for any x in [ a , b ] , )()( xyxy n!

|)(| 0

n

xxAK

nn

!

)(

n

abAK

nn

. But from

exponential series, we get, for any r, r n / n! 0 as n . Thus the right side of the

above inequality tends to zero as n . Hence, we get, yn (x) )(xy also.

Remark. Picard’s method of successive approximations can be applied to systems of first

order equations, by starting with necessary number of initial conditions, by converting into a

system of integral equations. Picard’s theorem under suitable hypothesis holds good in this

context also.

Theorem. Let P(x), Q(x) and R(x) be continuous functions in[ a , b ]. If x0 is any point in

[ a , b ], and y0 , y0’ are any two numbers, then the initial value problem ,

)()()(2

2

xRyxQdx

dyxP

dx

yd , y(x0) = y0 and y’(x0) = y’0 , has a unique solution y = y (x)

in [ a, b ]


CHAPTER 6

FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS

6.1. Introduction - Review.

Partial differential equations arise from a context involving more than one independent

variable. For the analysis of a partial differential equation and its solution, geometrically, we

require good knowledge about representation of curves and surfaces, in 3 – dimension.

A curve C in 3 – dimension can be specified in parametric form as the collection of points

(x, y, z) satisfying the equations x = f1 ( t ), y = f2( t ), z = f3 ( t ), where the parameter t

varies in an interval I in R, and f1, f2, f3 are continuous functions on I. The standard

parameter is the arc length s from a fixed point A on C to a generic point P on C.

Equation to C, can also be presented in vector – parametric form as,

r = f1( t ) i + f2( t ) j + f3( t ) k. At any point P ( r ) on C, dt

rd gives a tangent direction

( If s is the arc length parameter, then ds

rd gives the unit tangent direction )

Eg. A straight line with direction ratios, l, m, n and passing through ( a , b , c ) can be written

in symmetric form as, l

xx 0 =

m

yy 0 =

n

zz 0.

Eg. A right circular helix on the cylinder x 2 + y 2 = a2, can be specified as, x = ta cos ,

y = ta sin , z = kt.

Equation to a surface is usually taken as F ( x , y , z ) = 0, where F is a continuously

differentiable function in R 3. Its equation can also be expressed in parametric form as,

x = F1( u , v ) , y = F2 ( u , v ) , z = F3 ( u , v ). If ,0),(

),( 21

vu

FF then u and v can be solved as

functions of x and y locally, say, u = ),( yx , v = ),( yx . Then from the last equation, we

get, z = F3 ( ),( yx , ),( yx ).

Suppose, the curve C : x = x( s ), y = y( s ), z = z( s ) lies on the surface S whose equation is,

F( x , y , z ) = 0.

Then F(x(s), y(s), z(s)) = 0, for every s.


On differentiating w. r. t. s,

we get, ds

dx

x

F

+

ds

dy

y

F

+

ds

dz

z

F

= 0. This implies that at the point P ( x , y , z) of the curve,

the direction ( x

F

,

y

F

,

z

F

) is perpendicular to the tangent direction (

ds

dx,

ds

dy,

ds

dz ) to C.

Since C lies entirely on S, the above is a tangent direction to the surface also.

Thus ( x

F

,

y

F

,

z

F

) is a normal direction to S at P.

Let u be a variable depending on the independent variable x, y, z,….. Then an equation of the

form, f ( x, y, z,…,u, x

u

,

y

u

,

z

u

,…..,

2

2

x

u

,

yx

u

2

,…. ) = 0, where f is a function with

continuous partial derivatives of order up to n, for some n, is called a partial differential

equation ( PDE ) of order n. The order of a PDE is the order of the highest order derivative

appearing in it.

A PDE is said to be quasi linear, if it is linear in its highest order derivatives; and semi –

linear, if it is quasi-linear and the coefficients of the highest order derivatives does not

contain the dependent variable or its derivatives. A PDE which is not quasi – linear is called

non-linear. A semi-linear PDE, which is linear in the dependent variable and its derivatives, is

called linear.

At the beginning, we may consider the case of only two independent variables, say, x and y

and one dependent variable, say, z. , depending on x and y.

We may use the notations, p = x

z

, q =

y

z

, r =

2

2

x

z

, s =

yx

z

2

, t = 2

2

y

z

,….

In this context, the first order PDE has the form, f( x , y, z , p , q ) = 0.

We require the following Theorem in Real Analysis, in many contexts which involves solving

one set of variables in terms of the others from a given functional equation.

Implicit Function Theorem. Let f be a continuously differentiable function from an open set

E of R n+m to R n. Let ( a, b ) E such that f ( a , b ) = 0. Let A = f’( a, b ).

If Ax is invertible then there exists W, a neighborhood of b in R m such that for each y in

W, there exists a unique x in R n , such that f ( x, y ) = 0.

The Theorem gives the logical support in finding x in terms of y, given f( x, y ) = 0


6.2.Formation of First Order PDE

Consider a family of surfaces of the form, F( u , v ) = 0, where F is an arbitrary function, and

u and v are given functions of x , y , z.

Differentiating, F ( u , v ) = 0, partially w .r. t. x , y, we get ,

u

F

puu zx +

v

F

pvv zx = 0 … ( 1 )

u

F

quu zy + v

F

qvv zy = 0 … ( 2 ).

On eliminating, u

F

and

v

F

, from ( 1 ) & ( 2 ), we get,

qvvquu

pvvpuu

zyzy

zxzx

= 0 ….( 3 ),

which can be simplified as, (uy vz – uz vy ) p + ( uz vx – ux vz ) q = ( ux vy – uy vx ) ….( 3 ), or,

in terms of Wronskians, ),(

),(

zy

vu

p +

),(

),(

xz

vu

q =

),(

),(

yx

vu

… ( 3 ), which is a quasi linear

equation.

Thus ),(

),(

zy

vu

p +

),(

),(

xz

vu

q =

),(

),(

yx

vu

is the PDE associated with the family of surfaces

F ( u , v ) = 0, where F is an arbitrary function and u & v are given functions of x, y, z.

Remark: Let u an v be functions of x and y. If v can be expressed as a function of u

alone, without involving x and y, then ),(

),(

yx

vu

= 0. Here, we say, u and v are functionally

dependent.

Let v = H ( u ), where H is some function. Then vx = H’(u) ux & vy = H’(u) uy.

Eliminating, H’(u), we get ),(

),(

yx

vu

= ( ux vy – uy vx ) = 0.

Now, consider, a two – parameter family of surfaces, z = F ( x, y, a, b ) …( 1 ),

where a and b are parameters.

Differentiating, partially w. r. t. x and y, we get, p = Fx ( x, y, a, b ) …( 2 )

and q = Fy ( x, y, a, b ) …( 3 ).


Suppose the matrix,

FybFF

FFF

xbb

yaxaa, is of rank 2. Then by Implicit function theorem, we

can solve for a and b from two of the above three equations, in terms x, y, z, p, q, and

substituting in the remaining equation, we get a PDE, f(x , y, z, p, q ) = 0.

Eg.1. Consider, z = x + a x 2 y 2 + b. ..( 1 ). Differentiating ( 1 ) w.r.t. x & y respectively,

p = 1 + 2a x y 2 …( 2 ) and q = 2 a x 2 y …( 3 ).

Eliminating a between ( 2 ) & ( 3 ), we get, x p – y q = x , a PDE.

Eg.2. Eliminate a and b to form a PDE, given z = a x + b y.

Differentiating partially w.r.t. x and y, we get, p = a & q = b . The PDE is obtained by

eliminating a and b, from the above equations.

Thus, z = x p + y q.

Eg.3. Eliminate the arbitrary function F, to form the PDE of the family of surfaces,

z = x + y + F ( x y ) …( 1 ).

On differentiation, p = 1 + F’ ( xy ) y …( 2 ) & q = 1 + F’ ( x, y ) x …( 3 ). Eliminating,

F’( x, y ) from equations ( 2 ) & ( 3 ), we get the PDE, x p – y q = x – y.

Eg.4. Eliminating F, form the PDE of the one – parameter family of surfaces,

F ( x + y , x - z ) = 0.

Let u = x + y, v = x - z .

On differentiation, w.r.t. x , y , Fu ( 1 + 0. p ) + Fv ( 1 – 1/ z p ) = 0 and

Fu ( 1 + 0. q ) + Fv ( -1/ z . q ) = 0. Eliminating, Fu and Fv, between the last two

equations,

We get the PDE, qz

pz

/11

/111

= 0. i.e. -1/ z q - 1 + 1/ z p = 0 , i.e. p – q = z .

Ex.1. Form the differential equation given, ( a ) z = y + F ( x 2 + y 2)

( b ) F ( x – z , y – z ) = 0 ( c ) z = F ( x / y )

Ex.2. Form the differential equation given, ( 1 ) z = ( x + a ) ( y + b )

( 2 ) 2 z = ( a x + y ) 2 + b ( 3 ) z 2 ( 1 + a 3 ) = ( x + a y + b ) 3

We have the classification of first order PDEs as given below.

( 1 ) Linear Equation

P ( x , y ) p + Q ( x , y ) q = R ( x , y ) z + S ( x , y )


(2) Semi-linear equation

P ( x , y ) p + Q ( x , y ) q = R ( x , y , z )

( 3 ) Quasi-linear equation

P( x , y , z ) p + Q ( x, y, z ) = R ( x , y , z )

( 4 ) Non-linear equation

f ( x, y, z, p, q ) = 0.

The solution of a first order PDE in x , y, z is a surface in 3 – dimension, called an integral

surface.

There are different classes of integrals for a given PDE.

6.3.Classification of Integrals

Consider the PDE, f ( x , y , z , p , q ) = 0.

(a ) Complete integral

A two-parameter family of solutions of the PDE f ( x , y , z , p , q ) = 0 … ( * ),

z = F ( x , y, a, b ) is called a complete integral of ( * ), if in the region of definition of the

PDE, the rank of the matrix

ybxbb

yaxaa

FFF

FFF is 2.

This condition implies that at least one of the 2 x 2 sub matrices,

ybb

xaa

FF

FF,

ybb

yaa

FF

FF,

ybxb

yaxa

FF

FFis non-singular i.e. invertible. It guarantees that we can solve for a and b from

two of the equations,

z = F ( x, y, a, b ) …( 1 ), p = Fx ( x, y, a, b ) …( 2 ) and q = Fy ( x, y, a, b ) …( 3 ).

in terms x, y, z, or p or q, and then elimination of a and b, by substituting in the remaining

equation, so that equation ( * ) is recovered or satisfied. This is a consequence of Implicit

Function Theorem.

( b ) General integral

Let z = F ( x, y, a, b ) …( 1 ), be the complete integral of ( * ), where a and b are arbitrary

constants, referred as parameters in the geometrical context that ( 1 ) represents a two-

parameter family of surfaces in 3 – dimension.


Let us assume that a and b are functionally related, so that b = ( a ), where is an

arbitrary function.

Correspondingly, we get a one-parameter subfamily, z = F ( x, y, a, ( a ) ) of the two-

parameter family of surfaces represented by ( 1 ).

The envelope of this family, if it exists, is also a solution of the PDE ( * ) , called the General

Integral.

The envelope is obtained by eliminating the parameter between the equations,

z = F ( x, y, a, ( a ) ) …( 2 ) and 0 = )(' aFF ba … ( 3 ), obtained by differentiating

( 2 ) partially w .r. t. the parameter a.

The elimination will give G ( x , y , z ) = 0, a surface in 3-dimension.

If instead of an arbitrary function , we use a definite relation between a and b, like b = a

or b = a + 2 a 2 or b = sin a, etc , and proceeding to find the envelope of the corresponding

sub-family of ( 1 ), then the resulting envelope, if it exists, is a solution of ( * ), called a

particular integral.

c ) Singular Integral

The envelope of the two-parameter family of surfaces z = F ( x, y, a, b ) …( 1 ), if it exists,

is also a solution of the PDE ( * ), called the singular integral.

The envelope can be obtained by eliminating the parameters a and b from the equations,

z = F ( x, y, a, b ) …( 1 ), 0 = Fa ( x, y, a, b ) … ( 2 ) and 0 = Fb ( x, y, a, b) …( 3 ).

( d ) Special Integral

In certain cases, we can obtain solutions, which are not falling under the above classes, called

Special Integrals.

For the PDE, p – q = z , z = 0 is solution, which is not belonging to the three class of

solutions mentioned above.

Theorem.1. The envelope of a 1-parameter family, z = F ( x, y, a ) of solutions of the PDE

f ( x, y, z, p , q ) = 0, if it exists, is also a solution of the PDE.

Proof: The envelope is obtained by eliminating the parameter a between

z = F ( x, y, a ) ..( 1 ) and 0 = Fa ( x, y, a ) …( 2 ).

Thus the envelope is z = G ( x, y ) = F ( x, y, a(x, y )), where a( x, y ) is obtained from ( 2 ),

by solving for a in terms of x and y.

For points on the envelope, Gx = Fx + Fa ax = Fx and Gy =Fy + Fa ay = Fy , since on the

envelope Fa = 0. Thus, the envelope has the same partial derivatives as those of a member


of the family at a given point. Since the PDE at a point is a relation to be satisfied by the

coordinates of the point and the partial derivatives at that point, we get that the envelope is

also a solution of the PDE.

Theorem.2. The Singular integral is a solution.

Proof: Let z = F ( x, y, a, b ) …( 1 ) be the complete integral of f( x, y, z, p , q ) = 0 …( * )

The singular integral of ( * ) is obtained by eliminating a and b between,

z = F ( x, y, a, b ) …( 1 ) , 0 = Fa ( x, y, a, b ) …..( 2 ) 0 = Fb ( x, y, a, b ) …(3).

Hence the envelope is z = G ( x, y ) = F ( x, y, a ( x, y ) , b ( x, y )), where a( x, y ) & b ( x, y)

are obtained from ( 2 ) & ( 3 ) by solving for a & b in terms of x & y.

For the envelope, Gx = Fx + Fa ax + Fb bx = Fx and Gy = Fy + Fa ay + Fb by = Fy, since Fa = 0,

Fb = 0 on the envelope. Thus at any point on the envelope the partial derivatives will be the

same as a member of the family. Hence the envelope is also a solution of ( * ).

Remark : Recall that an envelope of a family at given point on it, touches a member of the

family.

Remark: The singular integral can also be determined directly from the given PDE ( * ) by

the following procedure.

The singular solution is obtained by eliminating, p and q from the equations,

f( x, y, x, p , q ) = 0 ..( * ), fp ( x, y, z, p, q ) = 0 ..( ** ), fq ( x, y, z, p, q ) = 0 …( *** ),

treating p & q as parameters.

Let z = F ( x, y, a, b ) be the complete integral of ( * ).

Then, f( x, y, F(x, y, a, b ), Fx ( x, y, a, b), Fy( ( x, y, a, b )) = 0, which holds for every a & b.

It can be differentiated partially w.r.t. a & b , so that,

fz Fa + fp Fxa + fq Fya = 0 and fz Fb + fp Fxb + fq Fyb = 0 .

Since on the singular integral, Fa = 0 & Fb = 0, the above equations will simplify to,

fp Fxa + fq Fya = 0 and fp Fxb + fq Fyb = 0 …( # ).

Since the matrix

ybxbb

yaxaa

FFF

FFF is 2, and Fa = 0 & Fb = 0, we get

ybxb

yaxa

FF

FF, non-

singular. Hence ( # ) gives, fp = 0 & fq = 0. Hence, the result.

Eg. It can be shown that z = F ( x, y, a, b ) = a x + b y + a 2 + b2 …( 1 )is a complete integral

of the PDE, f( x, y, z, p, q ) = z – px- qy – p2 – q2 = 0. …( 2 )

From ( 1 ) by differentiating partially w.r.t. x & y, we get,p = a & q = b.


Then equation ( 2 ) is satisfied by ( 1 ) i.e. ( 1) is a solution of ( 2), for any two arbitrary

constants, a & b.

Further,

ybxbb

yaxaa

FFF

FFF =

102

012

by

ax is of rank 2.

Thus, ( 1 ) is a complete integral of ( 2 ).

We can find particular integrals by relating a & b, and finding the envelope of the

corresponding sub-family.

Let b = a.

Then we get the family, z = a( x + y ) + 2 a 2. Differentiating w. r. t. a, 0 = ( x + y ) + 4 a.

On elimination of a , we get the envelope as, z = - ( x + y ) / 4 ( x + y ) + 2 ( - ( x + y )/4 ) 2

= - ( x + y ) 2/ 8 or ( x + y ) 2 + 8 z = 0, a particular integral.

The singular integral is obtained as follows.

Eliminate a & b from, z = ax + by + a 2 + b 2 , 0 = x + 2 a, 0 = y + 2 b .

The singular integral is, ( x 2 + y 2 ) + 4 z = 0.

Remark: A PDE can have more than one complete integral, so that the term ‘ Complete’ may

not be misinterpreted, and the particular integrals or the singular integral are not members of

the family z = F ( x, y, a, b ), for some values of a & b.

6.4.Linear equations

The following Theorem provides a method for finding the General integral of a given quasi

linear equation.

Theorem. The general integral of the quasi linear equation,

P( x , y , z ) p + Q ( x, y, z ) = R ( x , y , z ) … ( 1 ) , where P, Q, R are continuously

differentiable functions of x, y, and z is F ( u, v ) = 0 … ( 2 ) ,where F is an arbitrary

function and u and v are functions such that u ( x, y, z ) = c1 and v( x, y, z ) = c2 are two

independent solutions of the system of ordinary differential equations,

P

dx

Q

dy

R

dz --- ( 3 )

Proof: Since u ( x, y, z ) = c1 is a solution of ( 3 ), du = 0 u x dx + uy dy + uz dz = 0,

and hence ux P + uy Q + uz R = 0 ……( 4 ).

Similarly, we get, vx P + vy Q + vz R = 0 ……( 5 ). Thus, from equations ( 4 ) & ( 5 ),

),(),( zyvu

P

=

),(),( xzvu

Q

=

),(),( yxvu

R

…( 6 ) ( Here, we use the assumption

that u and v are independent )


But we have seen earlier that F ( u, v ) = 0 produces the PDE,

),(

),(

zy

vu

p +

),(

),(

xz

vu

q =

),(

),(

yx

vu

….( 7 ).

Now, from ( 6 ) & ( 7 ), it is obtained that, P p + Q q = R. Thus, F ( u, v ) = 0 is a solution of

the given PDE and it is general integral, by the presence of the arbitrary function.

Remark: It may be noted that the general integral is a family of surfaces and a member may

be fixed by assigning a specific form for the function F.

For given constants c1 & c2, the solutions u ( x, y, z ) = c1 and v( x, y, z ) = c2 of the

auxiliary equation, P

dx

Q

dy

R

dz, determines the curve of intersection of the surfaces

u ( x, y, z ) = c1 and v( x, y, z ) = c2 .

Eg.1. Find the general integral of z ( x p – y q ) = y 2 - x 2.

Consider, the auxiliary equation, P

dx

Q

dy

R

dz i.e.

zx

dx zy

dy

22 xy

dz

.

i.e. zx

dx zy

dy

)( yxz

dydx

22 xy

dz

.i.e. x

dx

y

dy

and

)( yxz

dydx

22 xy

dz

.

The solutions are, x y = c1 and ( x + y ) 2 + z 2 = c2.

The general integral is , F ( x y , ( x + y ) 2 + z 2 ) = 0, where F is an arbitrary function.

Eg.2. Find the general integral of z ( x p + y q ) = x + y

Consider, the auxiliary equation, P

dx

Q

dy

R

dz i.e.

zx

dx

zy

dy

yx

dz

.

i.e. x

dx

y

dy and

)( yxz

dydx

yx

dz

or dx + dy = z dz.

The solutions are, x/y = c and 2 (x + y ) - z 2 = d.

The general integral is, F ( x/y, 2 ( x + y ) – z 2 ) = 0, where F is arbitrary.

Ex. Determine the general integral of (a) ( y + 1 ) p + ( x + 1 ) q = z

( b ) x ( y – z ) p + y ( z – x ) q = z ( x – y ) ( c ) x p + y q = z

6.5.Pfaffian Differential Equations

A differential equation of the form,

n

i ini dxxxF1 1 0),......,( … ( 1 ), where Fi’ s

continuous functions, is called a Pfaffian differential equation in the n-variables x1,…..,xn.


The Pfaffian differential form

n

i ini dxxxF1 1 ),......,( is said to be exact, if there exists a

continuously differentiable function u( x1,…..,xn ) such that du =

n

i ini dxxxF1 1 ),......,( ,

and is called integrable, if there exists a non-zero differentiable function ( x1,…..,xn ) such

that

n

i ini dxxxF1 1 ),......,( is exact. Here, is called an integrating factor.

If the Pfaffian differentiable form is exact as described above, then the solution of the

equation ( 1 ) is u( x1,…..,xn ) = c, a constant. (The term ‘ exact’ or ‘integrable’ is also

attributed to the corresponding equations also.)

Theorem.1. The Pfaffian differential equation, P ( x, y ) dx + Q ( x, y ) dy = 0, in two

variables x & y, is always integrable.

Proof. If Q ( x, y ) = 0, then equation reduces to dx = 0, and it is exact for u ( x, y ) = x.

Other wise, we get Q

P

dx

dy , first order ordinary differential equation. By Picard’s

Theorem, the above equation has got a solution.

Theorem.2. Let u( x, y ) and v ( x, y ) be two functions such that 0

y

v. If, further

),(

),(

yx

vu

=0,

then there exists a functional relation F ( u, v ) = 0, between u & v without involving x & y

directly.

Proof. Since 0

y

v, we can eliminate y between the u = u ( x, y ) & v = v( x, y ) to obtain a

relation, F( u, v, x ) = 0….( * )

We claim that F is independent of x. On differentiating ( * ) w. r. t. x & y, we get,

x

u

u

F

x

F

+

x

v

v

F

= 0 ….( 1 ) and

y

u

u

F

+

y

v

v

F

= 0 ..(2)

Case.1. 0

x

v.

Then ( 1 ) y

v

- ( 2 )

x

v

),(

),(

yx

vu

u

F

y

v

x

F

= 0.

But then ),(

),(

yx

vu

=0, which gives

y

v

x

F

= 0

x

F

= 0 , since 0

y

v.


Case 2. x

v

= 0.

Then ( 1 ) reduces to x

u

u

F

x

F

= 0. But

),(

),(

yx

vu

= 0, with

x

v

= 0 & 0

y

v

x

u

= 0, and thus

x

F

=0.

Thus, in either case, x

F

=0, which implies that F is independent of x, or F is a function of u &

v only.

Theorem.3. Let X = ( P, Q, R ), where P, Q, R are continuously differentiable functions of x,

y, z and be a differentiable function of x, y, z. If X . Curl X = 0,

then X . Curl X = 0.

Proof: Consider, X . Curl X =

zyx z

Q

y

RP

,,

)()( = 2

zyx z

Q

y

RP

,,

)()( +

zyx zPR

yPQ

,,

)()( = 2

zyx z

Q

y

RP

,,

)()(= 2 ( X . Curl X ).

Since X . Curl X = 0, we get from above that, X . Curl X =0, for any .

Conversely, if X . Curl X =0 and 0 , then from the above identity, X . Curl

X = 0.

Remark: A Pfaffian differential form in 3 variable x, y, z can be taken as,

P( x, y, z ) dx + Q ( x, y, z ) dy + R ( x, y, z ) dz = X . rd , where X = ( P, Q, R ) and

),,( zyxr

Theorem.4. A necessary and sufficient condition for the Pfaffian differential equation,

X . rd = P( x, y, z ) dx + Q ( x, y, z ) dy + R ( x, y, z ) dz = 0 …( * ) to be integrable is

X . Curl X = 0.

Proof: Suppose, ( * ) is integrable. Then there exist differentiable functions (x, y, z ) 0

and u( x, y, z ) such that du = (x, y, z ) (P( x, y, z ) dx + Q ( x, y, z ) dy + R ( x, y, z ) dz ).

But du = x

u

dx +

y

u

dy +

z

u

dz. On comparison, P =

x

u

, Q =

y

u

, R =

z

u

.

i.e. X = u . But then curl X = curl u =0. Thus X . Curl X =0 .

Since (x, y, z ) 0 , from Theorem.4. we get X . Curl X = 0.


Conversely, assume X . Curl X = 0.

Treating, z as a constant, ( * ) becomes P( x, y, z ) dx + Q ( x, y, z ) dy = 0…( ** ), a

Pfaffian differential equation in the two variables x & y, which is always integrable, by

Theorem 1.

Then there exists (x, y, z ) 0 and U (x, y, z ) such that dU = P dx + Q dy.

Then P = x

U

, Q =

y

U

( # ).

Now using ( # ) in ( * ) x

U

dx +

y

U

dy +

z

U

dz + ( R -

z

U

) dz = 0.

i.e. dU + K dz = 0…( *** ). , where K = ( R - z

U

) .

Now we claim that K is a function of U and z alone, so that ( *** ) is a Pfaffian

differential equation in two variable U & z, and thereby integrable.

Since it is assumed that X . Curl X = 0, by Theorem.2. X . Curl X = 0.

But X = ( P , Q, R ) = (x

U

,

y

U

,

z

U

+ K ) = (

x

U

,

y

U

,

z

U

) + ( 0, 0, K )

= U + ( 0 , 0, K ). Then curl X = curl U + curl ( 0, 0, K ) = 0 + (

,

y

K)0,

x

K

Thus 0 = X . Curl X = (x

U

,

y

U

,

z

U

+ K ) . (

,

y

K)0,

x

K

=

x

U

y

K

y

U

x

K

=

),(

),(

yx

KU

= 0. Hence K can be expressed as a function of U and z, with out involving x or y.

Thus ( *** ) becomes, . dU + K( U, z ) dz = 0, which is integrable.

Let the solution be ( U, z ) = c. The solution of ( * ) is thus u ( x, y, z ) = c, by substituting

for U = U ( x, y, z ).

Theorem.5. The Pfaffian differential equation,

X . rd = P( x, y, z ) dx + Q ( x, y, z ) dy + R ( x, y, z ) dz = 0 …( * )

is exact iff Curl X = 0.

Proof: We have, X = u iff X = u and du = u . rd .

Thus X . rd = 0 is exact iff there exists u ( x, y, z ) such that X . rd = du, by definition,

i.e iff X . rd = u . rd iff . X = u iff curl X = 0.


Eg.1. Show that yz dx + zx dy + xy dz = 0 is exact and solve it.

Here X = ( yz, xz, xy ). Then curl X = 0. Hence by Theorem 5., equation is exact.

Solution is, yzx = c, a constant

Eg.2. Solve ( 6x + yz ) dx + ( xz – 2y ) dy + ( xy + 2z ) dz = 0.

Here X = ( 6x + yz, xz – 2y, xy + 2z ). Then curl X =0. Hence by Theorem 5., equation is

exact. The solution is obtained by direct integration as 3 x2 + yzx – y 2 + z 2 = c.

Eg.3. Solve ( y 2 + yz ) dx + ( xz + z 2 ) dy + ( y 2 – xy ) dz = 0.

Here X = ( y 2 + yz, xz + z 2, y 2 - xy ). Then X . curl X =0.

Hence by Theorem 4., equation is integrable. ( Note that equation is not exact since X 0 )

Treat z as a constant. Then equation becomes, ( y 2 + yz ) dx + ( xz + z 2 ) dy = 0.

i.e. y ( y + z ) dx + z ( x + z ) dy = 0,i.e. 1/(x + z ) dx + z/ y(y + z ) dy = 0.

i.e. 1/(x + z ) dx + [1/y - 1/(y + z ) ] dy = 0. log ( x + z ) + log y – log ( y + z ) = log c

i.e. U = ( x + z ) y/( y + z ).

Let be the integrating factor.

Then P = Ux y ( y + z ) = y/( y + z )

or = 1/ ( y + z ) 2.

Thus K = R – Uz = ( y 2 – xy )/( y + z ) 2 - y( y – x )/( y + z ) 2 = 0.

The original equation becomes, d U = 0, whose solution is U = c i.e. ( x + z ) y/( y + z ) = c.

Eg.4. Show that yz dx + 2 xz dy – 3 xy dz = 0 is integrable and solve it.

Here X = ( yz, 2 xz, -3 xy ). Then X . curl X =0.

Hence by Theorem 4., equation is integrable. ( Note that equation is not exact, since X 0 )

Treat z as a constant. Then equation becomes, y z dx + 2 xz dy = 0 or y d x + 2 x dy = 0,

i.e. dx / x+ 2 dy/ y = 0, whose solution is U = x y 2 = c.

If is the integrating factor,

then . yz = y 2 = y/z. Now K = y/z .( -3 xy ) – 0 = - 3 x y 2/z = - 3 U/z.

The solution of the original equation is dU + - 3 U/z dz = 0 U / z 3 = c i.e. x y 2 / z 3 = c.

Remark: The variables in the Pfaffian differential equations are independent and the

equation is written in terms of the differentials of these variables, and no dependent variable is

appearing in the equation. Pfaffian differential equations are used in finding the complete

integral of a given non-linear PDE.


Ex.1. Solve ( 1 + yz ) dx + x ( z – x ) dy - ( 1 + xy ) dz = 0

Ex.2. Test for integrability and solve, z( z – y ) dx + z ( x + z ) dy + x ( x + y ) dz = 0..

6.6. Charpit’s Method

The partial differential equations, f( x, y, z, p, q ) = 0 ….( 1 ) and

g( x, y, z, p, q ) = 0 ….( 2 ) are said to be compatible in a domain D in 3-dimension, if

( a ) J = 0),(

),(

qp

gf, on D and ( b ) for p = ),,( zyx and ),,( zyxq obtained by

solving ( 1 ) & ( 2 ) algebraically, the Pfaffian differential equation,

dz = ),,( zyx dx + ),,( zyx dy ..( 3 ) is integrable.

The condition J = 0),(

),(

qp

gf, guarantees the solvability of p & q in terms of the remaining

variables x, y, z from ( 1 ) & ( 2 ), by virtue of the Implicit Function Theorem.

Theorem.1. A necessary and sufficient condition for the integrability of the Pfaffian

differential equation dz = ),,( zyx dx + ),,( zyx dy , where ),,( zyx and ),,( zyx are the

expressions for p & q obtained from f( x, y, z, p, q ) = 0 ….( 1 ) and

g ( x, y, z, p, q ) = 0 ….( 2 ) is that, [ f, g ] = ),(

),(

px

gf

+ p

),(

),(

pz

gf

+

),(

),(

qy

gf

+ q

),(

),(

qz

gf

= 0.

Proof: Let )1,,( X . Then the Pfaffian differential equation,

dz = ),,( zyx dx + ),,( zyx dy i.e. ),,( zyx dx + ),,( zyx dy - dz = 0, becomes rdX . = 0

Hence by the necessary and sufficient condition for the integrability of a Pfaffian differential

equation, we get .X curl .X = 0.

i.e. )1,,( . )(,,( yxzz = 0, i.e. zyzx …( 3 ).

By substituting for & for p & q respectively, wherever necessary and feasible, and by

differentiating f( x, y, z, p, q ) = 0 w . r. t. x & z,

we get fx + fp x + fq x = 0 …( 4 ) and fz + fp z + fq z = 0 ..( 5 )

( 5 ) + ( 4 ) ( fx + fz ) + fp ( x + z ) + fq ( x + z ) = 0 .. ( 6 )

Similarly from ( 2 ), we get ( gx + gz ) + gp ( x + z ) + gq ( x +

z ) = 0 .. ( 7 )

( 6 ) gp – ( 7 ) fp ( x + z ) = J

1

),(

),(

),(

),(

pz

gf

px

gf …( 8 ).

By differentiating equations ( 1 ) & ( 2 ) w. r. t. y & z, and proceeding as above, we get,


)( zy = - J

1

),(

),(

),(

),(

qz

gf

qy

gf …( 9 ).

Now from equations, ( 3 ), ( 8 ) & ( 9 ), we get [ f, g ] = 0.

Remark: In the proof of the Theorem, x, y, z are taken as independent variable and p & q as

variables depending on them, since we are discussing the integrability of a Pfaffian

differential equation in the variables x, y, z. The basic fact that x & y are independent and p &

q are partial derivatives of z( x, y ) w.r.t. x, y is not used explicitly, anywhere.

Remark: If the partial differential equations, f( x, y, z, p, q ) = 0 ….( 1 ) and

g( x, y, z, p, q ) = 0 ….( 2 ) are compatible then they have a one-parameter family of

common solutions given by the associated Pfaffian differential equation.

Charpit’s method : Consider the partial differential equation, f ( x, y, z, p, q ) = 0 ….( 1 ) .

The above theorem provides a tool for finding a complete integral of ( 1 ).

A PDE, g( x, y, z, p, q, a ) = 0 ….( 2 ) compatible with ( 1 ) , for each constant value of a,

can be obtained as follows.

Assuming the compatibility of ( 1 ) & ( 2 ), we get the necessary & sufficient condition as,

[ f, g ] = ),(

),(

px

gf

+ p

),(

),(

pz

gf

+

),(

),(

qy

gf

+ q

),(

),(

qz

gf

= 0 … ( 3 ).

Expanding the Wronskians and interpreting ( 3 ) in the context of the system of equations

( 1 ) & ( 2 ) which has x, y, z, p, q as independent variables and f, g as variables depending on

them, we have a quasi linear differential equation, in the independent variables x, y, z, p, q

and the dependent variable g and its first order partial derivatives gx, gy, gz, gp, gq , since f is

given.

In this context, the above condition ( 3 ) may be presented as,

fp gx + fq gy + ( p fp + q fq ) gz – ( fx + p fz ) gp – ( fy + q fz ) gq = 0 ..( 3 )

Since ( 3 ) is a quasi linear partial differential equation, its solution g can be obtained through

the auxiliary equation,

pf

dx =

qf

dy =

)( qp qfpf

dz

= -

)( zx pff

dp

= -

)( zy qff

dq

=

0

dg …( 4 )

Any solution of ( 4 ) containing at least p or q can be taken as g( x, y, z, p, q, a ) = 0 ….( 2 ),

where the arbitrary constant a will arise naturally.


Since by the assumption, the given PDE ( 1 ) and ( 2 ) which can be obtained as in the

above discussion are compatible, the Pfaffian differential equation,

dz = ),,,( azyx dx + ),,,( azyx dy , where ),,,( azyx & ),,,( azyx are the

expressions for p & q obtained algebraically from ( 1 ) & ( 2 ), is integrable.

The solution of this Pfaffian differential obtained as, z = F ( x, y, z, a, b ), where F is a known

function and b is another arbitrary constant, is readily a solution of the original PDE ( 1 ), in

the form of a complete integral.

Remark: We can drop the last expression in the auxiliary equation, since it will give only the

trivial solution g = a constant function, which is not desirable

Remark: The PDEs f( x, y, z, p, q ) = 0 ….( 1 ) and g( x, y, z, p, q ) = 0 ….( 2 ) are

compatible amounts to the assumption that they share a one-parameter family of common

solutions, and does not mean that the equations are equivalent which assumes that their

solutions agree completely.

Eg.1. Consider the PDE, f = p 2 x 2 + q 2 y 2 – 4 = 0 …( 1 ).

The auxiliary equation to find a PDE, g( x, y, z, p, q, a ) = 0 ….( 2 ), compatible with ( 1 ) is,

pf

dx =

qf

dy =

)( qp qfpf

dz

= -

)( zx pff

dp

= -

)( zy qff

dq

…( * )

i.e. px

dx22

= qy

dy22

= )2.2.( 22 qyqpxp

dz

= -

)2( 2xp

dp= -

)2( 2yq

dq …( * )

We may consider, qy

dy22

= -)2( 2yq

dq i.e.

y

dy= -

q

dq g = q y – a = 0 …( 2 )

Now from ( 1 ) & ( 2 ), q = a /y , p = xa /4 2 .

Consider, dz = xa /4 2 dx + a /y dy, whose solution is z = 24 a

log x + a log y + b.

Thus, we have got the complete integral of ( 1 ) as, z = 24 a log x + a log y + b.

Eg.2. Find a complete integral of f = p + q - pq = 0 …( 1 ), by Charpit’s method.

The auxiliary equation to find a PDE, g( x, y, z, p, q, a ) = 0 ….( 2 ), compatible with ( 1 ) is,

pf

dx =

qf

dy =

)( qp qfpf

dz

= -

)( zx pff

dp

= -

)( zy qff

dq

…( * )

i.e. dx/( 1 – q ) = dy/( 1 – p ) = dz/ p ( 1-q)+q(1-p)) = - dp/0 = - dq/0

A solution is q = a. Then ( 1 ) p = a/( a – 1 ).


Now consider, dz = p dx + q dy = a/(a-1) dx + a dy, whose solution is z = ax /(a-1) + a y + b,

and hence a complete integral of ( 1).

Eg.3. Consider, ( p2 + q 2 ) – q z = 0 …( 1 )

The auxiliary equation is, dx/ 2py = dy/( 2qy-z) = dz/[2(p2+q2)y-qz] = - dp/-pq = - dq/ p2

Consider, dp/pq = - dq/p2 or p dp + q dq = 0 p 2 + q 2 = a 2, say.

Using, ( 1 ) q = a 2y/z , p = a 222 yaz /z.

Consider, dz = a 222 yaz /z dx + a 2y/z dy i.e. ( z dz – a 2 y dy )/ 222 yaz = a dx.

On solving, we get a complete integral as, 222 yaz = ax + b.

Ex.1. Find a complete integral of p 2 q 2 + x2 y2 = x 2 q2 ( x 2 + y 2 )

Ex.2. Show that x 2 p + q – xz = 0 and xp – yq – x = 0 are compatible and hence find a

1- parameter family of common solutions.

Ex.3. Find a complete integral of ( a ) p 2 + q 2 – 1 = 0 ( b ) ( p 2 + q 2 ) x – pz = 0,

by Charpit’s method.

Ex.4. Find a complete integral of p 2y = 2 ( z + xp + yq )

6.7. Jacobi’s method

Now let us consider a first order PDE, f ( x, y, z, ux, uy, uz ) = 0 ..( 1 ), where u is a variable

depending on the independent variables x, y, z. ( It may be noted that the depended variable

is not explicitly appearing in the equation )

A solution u = F ( x, y, z, a, b, c ) of ( 1 ), where a, b, c are arbitrary constants & F is a

known function, is said to be a complete integral , if

zcycxcc

zbybxbb

zayaxaa

FFFF

FFFF

FFFF

is of rank 3.

For a PDE of the above type, a method will be developed to find the complete integral, similar

to the development of the Charpit’s method.

Two one-parameter families of partial differential equations,

h1 ( x, y, z, ux, uy, uz , a ) = 0 ..( 2 ) and h2 ( x, y, z, ux, uy, uz , b ) = 0 ..( 3 ) are said to be

compatible with f ( x, y, z, ux, uy, uz ) = 0 ..( 1 ), if 0),,(

),,( 21

zyx uuu

hhf and

the Pfaffian differential equation,


du = ux( x, y, z, a, b ) dx + uy( x, y, z, a, b ) dy + uz( x, y, z, a, b ) dz … ( 4 ), is integrable,

where ux( x, y, z, a, b ), uy( x, y, z, a, b ) and uz( x, y, z, a, b ) are obtained from ( 1 ) , ( 2 )

& ( 3 ), algebraically, for all values of a & b.

Since, u is expected to be a function of x, y, z, the differential equation is integrable iff it is

exact. (Since, we have du = ux dx + uy dy + uz dz )

The equation ( 4 ) is exact iff the mixed derivatives of order 2 agree,

i.e. uxy = uyx, uyz = uzy, uxz= uzx.

Theorem. If the one-parameter families of partial differential equations,

h1 ( x, y, z, ux, uy, uz , a ) = 0 ..( 2 ) and h2 ( x, y, z, ux, uy, uz , b ) = 0 ..( 3 ) are

compatible with f ( x, y, z, ux, uy, uz ) = 0 .. ( 1 ), then ),(

),(

xux

hf

+

),(

),(

yuy

hf

+

),(

),(

zuz

hf

= 0,

where

h = h 1 or h2 .

Proof: Differentiating, ( 1 ) partially w.r.t. x, y, z , we get,

x

f

+ xx

x

uu

f

+ xy

y

uu

f

+ xz

z

uu

f

= 0 …( 4 )

y

f

+ yx

x

uu

f

+ yy

y

uu

f

+ yz

z

uu

f

= 0 …( 5 )

z

f

+ zx

x

uu

f

+ zy

y

uu

f

+ zz

z

uu

f

= 0 …( 6 )

Similarly, ( 2 ) & ( 3 )

x

h

+ xx

x

uu

h

+ xy

y

uu

h

+ xz

z

uu

h

= 0 …( 7 )

y

h

+ yx

x

uu

h

+ yy

y

uu

h

+ yz

z

uu

h

= 0 …( 8 )

z

h

+ zx

x

uu

h

+ zy

y

uu

h

+ zz

z

uu

h

= 0 …( 9 ), where h = h1, h2.

(4) xu

h

- ( 7 )

xu

f

),(

),(

xux

hf

+

),(

),(

xy uu

hf

uxy +

),(

),(

xz uu

hf

uzx = 0 …( 10 ), since uxy = uyx

& uxz = uzx.

Similarly from ( 5 ) & ( 8 ), ),(

),(

yuy

hf

+

),(

),(

yx uu

hf

uxy +

),(

),(

yz uu

hf

uzy = 0 …( 11 ),


since uxy = uyx & uyz = uzy .

Similarly from ( 6 ) & ( 9 ), ),(

),(

zuz

hf

+

),(

),(

zx uu

hf

uxz +

),(

),(

zy uu

hf

uzy = 0 …( 12 ),

since uxz = uzx & uyz = uzy .

Now, ( 10 ) + ( 11 ) + ( 12 ) ),(

),(

xux

hf

+

),(

),(

yuy

hf

+

),(

),(

zuz

hf

= 0, which is the required

condition.

Remark: Given the PDE, f ( x, y, z, ux, uy, uz ) = 0 .. ( 1 ), we may find two families of

partial differential equations of the same type, h1 ( x, y, z, ux, uy, uz , a ) = 0 ..( 2 ) and h2

( x, y, z, ux, uy, uz , b ) = 0 ..( 3 ) , so that they are compatible. These families of PDEs are

obtained using ( 1 ), by means of the compatibility condition established in the above

Theorem, namely,

),(

),(

xux

hf

+

),(

),(

yuy

hf

+

),(

),(

zuz

hf

= 0 -…( * ). The Wronskians may be expanded to get the

quasi linear PDE, x

h

u

f

x

+

y

h

u

f

y

+

z

h

u

f

z

-

x

f

u

h

x

-

y

f

u

h

y

-

z

f

u

h

z

= 0 … ( * ),

in the independent variables x, y, z, ux, uy, uz and the dependent variable h.

Two independent solutions h1 = 0 & h2 = 0 of the associated auxiliary equation,

xuf

dx

=

yuf

dy

=

zuf

dz

= -

xf

du x

= -

yf

du y

= -

zf

du z

…( ** ) are obtained.

Then from equations ( 1 ), ( 2 ), ( 3 ) ux, uy, uz are obtained algebraically in terms of x, y, z

and the two arbitrary constants.

Substitute these expressions in du = ux dx + uy dy + uz dz , which will be an exact

differential equation and the solution obtained as,

u = F ( x, y, z, a, b, c ) is a complete integral of the original PDE .

The above procedure is known as the Jacobi’s method.

Eg.1. Consider, z 2 + z uz = ux 2 + uy

2.

The auxiliary equation is dx/ -2ux = dy/-2uy = dz/z = dux/0 = duy/0 = duz/-2z-uz

Thus dux = 0 = duy ux = a, uy = b. Then uz = (a 2 + b 2 – z 2)/z.

Then du = a dx + b dy + (a 2 + b 2 – z 2)/z dz u = ax + by + ( a 2 + b 2 ) log z – z 2/2 + c.

Eg.2. Consider, z + 2 uz = ( ux + uy ) 2


The auxiliary equation is,

dx/-2( ux + uy ) = dy/ -2( ux + uy ) = dz/2 = - dux/0 = - duy/0 = -duz/1.

Thus dux = 0 = duy ux = a, uy = b. Then uz = [( a + b ) 2 – z ]/2.

Then du = a dx + b dy + [( a + b ) 2 – z ]/2 dz u = ax + by + ( a + b ) 2 z/2 - z 2/4 + c

Eg.3. Consider, x ux + y uy = uz 2

Auxiliary equation is, dx/x = dy/ y = dz/- 2 uz = - dux/ux = - duy/uy = -duz/0

Then dux/ux + dx/x = 0 and dy/y + duy/uy = 0. xux = a, yuy = b.

Then uz = ba .

Then du = a/x dx + b/y dy + ba dz u = a logx + b logy + ba z + c

Eg.4. Consider, f( ux , uy, uz ) = 0, where f is a given function.

The auxiliary equation is, dx/( ) = dy/ ( ) = dz/( ) = - dux/0 = - duy/0 = -duz/0.

Thus dux = duz = duy = 0 ux = a, uy = b, uz = c, where f( a, b, c ) = 0

Then du = a dx + b dy + c dz u = ax + by + cz + d, f( a, b, c ) = 0.

Ex. Find the complete integral by Jacobi’s method, given ux 2 + uy

2 + uz = 1.

Remark: We can apply Jacobi’s method to find a complete integral of f ( x, y, z, p, q ) = 0.

Let the solution be u( x, y, z ) = c , a constant. But, here z = z(x, y ).

Differentiating partially u( x, y, z(x,y) ) = c w. r. t. x, we get, ux + uz p = 0.

Thus p = - ux / uz and similarly, q = - uy / uz .

Now substituting in f ( x, y, z, p, q ) = 0, the above expressions for p & q , we get

G( x, y, z, ux, uy, uz ) = 0.

By employing Jacobi’s method, a solution is obtained as u = u ( x, y, z, a, b ) + C.

But, we have already taken u = c, in the beginning, and we are expecting only 2 arbitrary

constants in the complete integral .

The situation can be overcome with C = c, so that the complete integral of f ( x, y, z, p, q ) = 0

is obtained as u( x, y, z, a, b ) = 0.

Eg. Consider, p 2 x + q 2 y = z.

Assume the solution as u( x, y, z ) = c. Then p = - ux / uz & q = - uy / uz .

On substitution, we get, x ux 2 + y uy

2 – z uz 2 = 0.

The auxiliary equation is,

dx/2xux = dy/ 2yuy = dz/- 2zuz = - dux/ux2 = - duy/uy

2 = -duz/-uz2

The 2 independent solutions are, xux2 = a, yuy

2 = b. Then z uz 2 = a + b


Then ux = 2

1

x

a, uy =

21

y

b, uz =

21

z

ba

On integrating, du = 2

1

x

adx +

21

y

bdy +

21

z

badz, we get

u = 2 ( ax) ½ +2 ( by ) ½ + 2 ( (a + b)z ) ½ + c.

Thus we have a complete integral of the given PDE as,

( ax) ½ + ( by ) ½ + ( (a + b)z ) ½ = 0.

Ex. Find the complete integral of ( a ) ( p 2 + q 2 ) y = qz ( b ) pqxy = z 3

6.8. The Cauchy Problem

The problem of determining the integral surface of a given PDE, containing a given curve is

called a Cauchy problem.

Different approaches are made for solving the problem according to the type of the equation

given.

Quasi linear equation

Let C be a given curve with parametric equation, x = x0 ( s ), y = y0 ( s ), z = z0 ( s ), where s

is the parameter.

The problem is to determine the integral surface of the quasi linear PDE,

P ( x, y, z ) p + Q ( x, y, z ) q = R ( x, y, z ) …( 1 ) containing the given curve C.

Let F ( u, v ) = 0 …( 2 ) be the general integral of ( 1 ), where u ( x, y, z ) = c1

& v ( x, y, z ) = c2 are two independent solutions of the auxiliary equation,

dx / P = dy/Q = dz/R …( 3 ). Here we have to fix the function F, so that the surface

F ( u, v ) = 0, contains C.( i.e. we have to relate the arbitrary constants c1 & c2.)

This can be done as follows.

Substitute x = x0 ( s ), y = y0 ( s ), z = z0 ( s ) in the equations, u ( x, y, z ) = c1

& v ( x, y, z ) = c2 , and then eliminate s between the resulting equations, so that the required

relation between i.e. the form of the function F will be obtained.

Eg.1. Consider the PDE, x 3 P + y ( 3 x 2 + y ) q = z ( 2 x 2 + y ) …( 1 ), and the curve

C : x0 = 1, y0 = s, z0 = s ( 1 + s ) ….( 2 ). We have to find the integral surface of ( 1 )

containing C.

First, we have to find two independent solutions of the auxiliary equation,

)2()3( 223 yxz

dz

yxy

dy

x

dx

0

///

)2()3( 223

zdzydyxdx

yxz

dz

yxy

dy

x

dx

&


)3( 23 yxy

dy

x

dx

. Now, -dx/x + dy/y – dz/z = 0 &

yx

dydxyx

y

dy

x

dxyx

3

2

3

2 )3()3(

xyyx

xdydydxyx

y

dy

3

2 )3(.

We get, u = y/xz = c1 & v =

2

3

cy

xyyx

. Substituting, x = 1, y = s, z = s ( 1 + s )

in u = c1 & v = c2, 11

1c

s

& 2

21c

s

s

. Eliminating s, we get, c1 c2 – c1 – c2 + 2 = 0.

Thus the required surface is, uv – u – v + 2 = 0

i.e. xz

y.

y

xyyx 3

- xz

y-

y

xyyx 3

+ 2 = 0.

Ex.1. Find the general integral of ( x – y ) y 2 p + ( y – x ) x 2 q = ( x 2 + y 2 ) z and the

integral surface containing the curve xz = a 2, y = 0.

Ex.2. Find the integral surface of the PDE, ( x – y ) p + ( y – x – z ) q = z passing through the

circle, z = 1, x 2 + y 2 = 1.

Non linear Equation

Let us consider the Cauchy problem for non linear PDE. We are assuming that the solution is

a particular integral.

Let F ( x, y, z, a, b ) = 0 ..( 1 ) be the Complete integral of f ( x, y, z, p, q ) = 0 …..( 2 )

To fix a particular integral, we have to isolate a one parameter subfamily S of ( 1 ). Let E

be the envelope of this sub family which contains the given curve C. Since E is the envelope

of the sub family, it will touch each member of the family S. As C lies entirely on E, C will

be tangential to each member of S, for some parameter s.

This will then imply that, F ( x0(s), y0(s), z0(s), a, b ) = 0 ….( 3 )

and also that s

F

( x0(s), y0(s), z0(s), a, b ) = 0 ….( 4 ) , for some s.

Now, eliminate s between ( 3 ) & ( 4 ) to get a relation between a & b, and there by the sub

family S. Then E is found as the envelope of S.

It can be observed that the Cauchy problem for a non linear equation may have more than one

solution, unlike the situation with the quasi linear equations.

Eg.1.. Find the integral surface of ( p 2 + q2 ) x = p z …( 1 ) containing the curve

C: x = 0, y = s 2, z = 2s.


The auxiliary equation is, pf

dx =

qf

dy =

)( qp qfpf

dz

= -

)( zx pff

dp

= -

)( zy qff

dq

…( * )

i.e. zpx

dx

2 =

qx

dy

2 =

)( pz

dz = -

2q

dp

= -

pq

dq …( * )

- 2q

dp

= -

pq

dq p 2 + q 2 = a 2 …( 2) .

Using ( 1 ), we get p = z

xa 2

& q = z

xaza

222 .

Now consider, dz = z

xa 2

dx + z

xaza

222 dy.

On integration, we obtain the complete integral as, z 2 = a2 x 2 + ( a y + b ) 2. ….( 3 )

Substituting, x = 0, y = s 2 , z= 2s , we get, 4 s 2 = ( a s 2 + b ) 2 ….( 4 )

Integrating ( 4 ) partially www. R. t. s, 8 s = ( a s 2 + b ) 4 a s or 2 = a ( a s 2 + b ) …( 5 )

Eliminating, s between ( 4 ) & ( 5 ), ab = 1 or b = 1/a.

The corresponding one parameter subfamily S of ( 3 ) is, z 2 = a2 x 2 + ( a y + 1/a ) 2. ….( 6 )

Or a 4 ( x 2 + y 2 ) + a 2 ( 2y – z 2 ) + 1 = 0 …( 7 ).To find the envelope of S, differentiate (7 )

partially w. r. t. a, to get 4 a 3 ( x 2 + y 2 ) + 2a ( 2y – z 2 ) = 0 ….( 8 ).

Eliminating a between ( 7 ) & ( 8 ), the envelope is obtained as, z 2 = 222 yxy ,

which is the required solution.

Eg.2. Find the complete integral and the integral surface of p 2 x + q y – z = 0, containing

the curve C: x + z = 0, y = 1.

The auxiliary equation is, pf

dx =

qf

dy =

)( qp qfpf

dz

= -

)( zx pff

dp

= -

)( zy qff

dq

…( * )

i.e. px

dx

2 =

y

dy =

qyxp

dz

22 = -

)1( pp

dp

= -

0

dq …( * ) q = a. Using the given PDE,

p = x

ayz . Consider the Pfaffian differential equation, dz =

x

ayz dx + a dy

whose solution is, bxayz or on squaring to remove the ambiguity,

( a y – z + x + b ) 2 = 4 b x. …( * )

The curve can be parametrised as, x = s, y = 1, z = -s.

Substituting in ( * ), ( a + b + 2s ) 2 = 4 b s …( ** ).

Differentiating ( ** ) w. r. t. s, ( a + b + 2s ) = b. … ( *** )

Eliminating s between ( ** ) & ( *** ) , b ( b + 2 a ) = 0 i.e. b = 0 or b = -2a.


It cab easily seen that b = 0 leads to no solution.

Take, b = - 2a. Then ( * ) ( a y – z + x - 2a ) 2 = - 8a x. ( $ )

Differentiating partially w.r.t. a, ( a y – z + x - 2a ) ( y – 2 ) = - 4 x. ( # ).

The envelope is obtained by eliminating a between ( $ ) & ( # ) as x y = z ( y-2 ).

Ex.1. Find the integral surface of p 2 x + pq y – 2 p z – x = 0 containing x = z, y = 1.

Ex.2. Determine the integral surface of p q = z containing x = 0, z = y 2.

Ex.3. Find the particular solution of ( x – y ) y 2 p + ( y – x ) x 2 q = ( x 2 + y 2 ) z containing

the curve xz = a 2, y = 0.

6.9 Geometry of Solutions

We may start our discussion with the Semi linear PDE,

P ( x, y ) p + Q ( x , y ) q = R ( x, y, z ) … ( 1 )

Here it is assumed that P, Q, R are continuously differentiable functions.

Consider the equation, ),(

),(

yxP

yxQ

dx

dy … ( 2 ), whose solution is a one parameter family of

curves in the x-y plane.

( 2 ) can also be written as, dt

dx = P ( x, y ) .. (3.1 ),

dt

dy= Q ( x, y ) …( 3.2 )

Along these curves ( x = x ( t ), y = y ( t )), z ( x, y ) will satisfy the ordinary differential

equation, ),(

),,(

),(

),(),(

yxP

zyxR

yxP

zyxQzyxP

dx

dyzz

dx

dz yx

yx

or

dt

dz=

dx

dz.

dt

dx= R (x, y, z ) ….( 3.3. ).

The one parameter family of curves in the x – y plane determined by ( 2 ) are called the

Characteristic curves of ( 1 ).

Let ( x0, y0 ) be a point in the x-y plane. Then by Picard’s theorem the initial value problem,

dt

dx = P ( x, y ) .. (3.1 ),

dt

dy= Q ( x, y ) …( 3.2 ) , x ( 0 ) = x0, y ( 0 ) = y0 has a unique

solution, x( t ) = x ( x0, y0, t ), y = y ( x0, y0, t ) , which is the unique characteristic curve

passing through ( x0, y0 ). At ( x0, y0 ) on this curve, prescribe the z – value as z0.

Then along this curve ( 3.3 ) will have a unique solution as,

z = z ( x0, y0, t ) such that z ( 0 ) = z0.

Let be a given curve in the x-y plane such that intersects each of the characteristic curves.

If is equipped with the data, being the value of z at each point ( x0, y0 ), we can


completely determine z( x, y ) for the region in the x-y plane containing the characteristic

curves. Here is called an initial data curve.

It can be seen that can not be chosen arbitrarily, but has to satisfy the ‘admissibility

criterion’.

Eg. Solve the semi linear equation, x q – y p = z with the initial condition that

z ( x, o ) = f ( x ), 0x , where f ( x ) is a given function.

Here the initial data curve is the positive x – axis and it carries the information about the

solution, namely, at ( x , o ) the solution is z = f ( x ).

The Characteristic curves are determined by, y

x

dx

dy x 2 + y 2 = c 2.

Thus the Characteristic curves are the concentric circles at the origin. Note that meets each

of the characteristic curves at a unique point.

Along a Characteristic curve x 2 + y 2 = c 2, 22 xc

z

y

z

dx

dz

i.e. 22 xc

dx

dz

dz

z = k( c ) )/(sin 1 cxe

=

22

1sin

22 yx

x

eyxk .

By the initial condition, f ( x ) = z ( x, 0 ) = k(x ) 2

e or k ( x ) = f ( x ) 2

e .

Thus we have the solution, z ( x, y ) =

22

1sin222 yx

x

eyxf

.

We now consider, a quasi linear equation, P ( x, y, z ) p + Q ( x, y, z ) q = R ( x, y, z ) …(1 )

Any solution defines a surface z = z ( x, y ) in 3- dimension, with normal direction at a point

( x, y, z ) on it being ( p, q, - 1). Hence from the given PDE we can make the following

observation. “ Any surface z = z ( x, y ) is a solution of the PDE ( 1 ) iff the tangent plane at

each of its points contains the directions ( P, Q, R )”.. Here ( P, Q, R ) is called a characteristic

direction and the integral curves of the resulting vector field or direction field are called the

characteristic curves. The Characteristic curves are determined by the system of ordinary

differential equations, P

dx =

Q

dy =

R

dz , the auxiliary equation, we have used earlier for

finding the general integral.

The equations may be rewritten as,

dt

dx = P ( x, y, z ) ,

dt

dy = Q ( x, y, z ),

dt

dz = R ( x, y, z ) .. ( 2 ).


Given a point ( x0, y0, z0 ) in space , there exists a unique solution for the above system, say,

x = x ( t ), y = y ( t ), z = z ( t ) such that x( 0 ) = x0, y ( 0 ) = y0, z ( 0 ) = z0, which is

geometrically the Characteristic curve passing through ( x0, y0, z0 ).

Let be a curve in space with equation x = x0 ( s ), y = y0 ( s ) , z = z0 ( s ). Then there

exists a unique characteristic curve passing through each point of , and these curves taken

together will determine an integral surface.

The curve of intersection of two integral surfaces is a characteristic curve and if a

characteristic curve meets an integral surface then it should lie entirely on that surface.

The system ( 2 ) produces a two parameter family of curves in space and any one parameter

subfamily will generate an integral surface.

Eg.1. Consider the initial value problem - determine the integral surface of

z p + q = 1 containing the curve C : x = s, y = s , z = s/2.

The given equation is quasi linear and the characteristic equations are,

dt

dx = z ,

dt

dy = 1,

dt

dz = 1 ..( * ) . We may solve it under the initial conditions x ( s, 0 ) = s,

y ( s, 0 ) = s, z ( s, 0 ) = s/2, so that we can determine all the characteristic curves through each

point of the given curve C. The surface generated by these curves will be the integral surface

containing C.

( * ) z ( s, t ) = t + a , y ( s, t ) = t + b, initially. Under the initial conditions, a = s/2, b = s.

Now, we get, x ( s, t ) = ½ t 2 + ½ st + c. From x( s , o ) = s, we get c = s.

Thus, we have the solution as, x = ½ t 2 + ½ st + s , y = t + s, z = t + s/2 .

We get the required surface by eliminating s & t from the above three equations.

Solving for s & t in terms of x & y, s = 2/1

2

2

y

yx

, t = 2/1 y

xy

. Substituting in the last

equation, z = )2(2

24 2

y

yxy

.

Eg.2. Solve the Cauchy problem, 2 p + y q = z , x = s, y = s 2, z = s.

The characteristic equations are, dt

dx = 2 ,

dt

dy = y,

dt

dz = z ..( * ).

Initial conditions are x ( s, 0 ) = s , y ( s , 0 ) = s 2, z ( s, 0 ) = s.

We get the solution, x = s + 2t, y = s 2 e t , z = s e t .

Eliminating s & t , the solution to the Cauchy problem is obtained as,

zyxz

yez 22

.


Ex.1. Solve p + q = z 2 under the condition z ( x, 0 ) = sin x

Ex.2. Find the integral surface of xz p – yz q = y 2 – x 2 , passing through the straight line

x/2 = y/1=z/1

Finally, we consider a non linear equation, f ( x, y, z, p , q ) = 0 …. ( * ).

Let ( x0, y0, z0 ) be a point in space, and z = z ( x, y ) be an integral surface

through ( x0, y0, z0 ).

Then the equation to the tangent plane at ( x0, y0, z0 ) to this surface is,

z – z0 = p ( x – x0 ) + q ( y – y0 ) ……..( 1)

Assuming, 0qf , we can solve for q from ( * ) as q = q ( x, y, z, p ).

Thus p & q are not independent at ( x0, y0, z0 ) and ( 1 ) describes a one parameter p –

family of planes through ( x0, y0, z0 ), and its envelope is called the Monge cone at ( x0, y0, z0 )

of the PDE.

Thus a surface z = z ( x, y ) is an integral surface iff it must be tangential to the Monge cone

at each point on it.

The equation to the Monge cone at ( x0, y0, z0 ) can be determined by eliminating the

parameter p between ( 1 ) and 0 = ( x – x0 ) + ( y – y0 ) dq/dp,

where q = q ( x0, y0, z0, p ), obtained from ( * ).

But from ( * ), we get fp + fq ( dq/dp) = 0 . Hence, on substitution, ( x – x0) fq = ( y – y0)

fp or

qp f

yy

f

xx 00

., which can be extended using ( 1 ) as

qp f

yy

f

xx 00

=

qp qfpf

zz

0 ..( 2 )

Thus the Monge cone at ( x0, y0, z0 ).can be determined by eliminating, p & q from

f ( x, y, z, p , q ) = 0 …. ( * ), z – z0 = p ( x – x0 ) + q ( y – y0 ) ……..( 1), and

qp f

yy

f

xx 00

=

qp qfpf

zz

0 ..( 2 )

Here ( 2 ) represents the generators of the Monge cone.


CHAPTER 7

SECOND ORDER PARTIAL DIFFERENTIAL EQUATIONS

Let x , y be independent variables and u be a variable depending on x & y.

A partial differential equation which contains second order partial derivatives of u as well as

the basic variables x, y, z will be called a second order partial differential equation. We will

be discussing in some detail certain classical second order equations arising from physical

contexts alone. They belong to the class of semi linear equations.

A second order PDE in the form,

R(x, y ) uxx +s ( x, y ) uxy + T ( x, y ) uyy + g ( x, y, u, ux, uy ) = 0 ….( 1 ),

where R, S, T are continuous functions of x and y, is called a semi linear equation.

7.1. Classification

Consider the second order semi linear PDE,

R(x, y ) uxx +S ( x, y ) uxy + T ( x, y ) uyy + g ( x, y, u, ux, uy ) = 0 ….( 1 )

It may be also assumed that R, S , T have continuous partial derivatives w. r. t. x & y.

The above equation can be reduced to certain canonical forms according to the type of the

equation, and their solutions can be obtained.

Consider S 2 - 4 R T . The equation ( 1 ) is hyperbolic, parabolic, elliptic according as

S 2 - 4 R T >, = ,< 0, respectively.

We may transform the independent variables x & y to new variables and as , =

(x, y )

& = ( x, y ).

Then xxx uuu , yyy uuu and

xxxxxxxxxxxx uuuuuuu 22,……

Then R(x, y ) uxx + S ( x, y ) uxy + T ( x, y ) uyy = )( 22

yyxx TSRu

+ )( 22

yyxx TSRu + )2)(2( yyyxyxxx TSRu + ),,,,( uuuG

Taking the forms,

A ( u, v ) = R u 2 + S u v + T v 2, B ( u1, v1, u2, v2 ) = R u1 u2 + (S/2)(u1 v2 + u2 v1) + T v1 v2,

it can be shown by direct computation that,

),,,(),(),( 2

yxyxyxyx BAA ( 4 R T – S 2 ) 4/2

yxyx ….( * )

The PDE ( 1 ) is transformed to,


),( yxA u + 2 ),,,( yxyxB u + ),( yxA u + ),,,,( uuuH = 0 …( ** )

We may choose and so that the equation ( 1 ) will be assuming simpler forms according

to its types.

Hyperbolic Equation ( S 2 – 4 R T > 0 )

Consider 02 TSR .. ( 2 ).

This equation has two distinct real roots ),(&),( yxyx .

Consider the equations, dx

dy + ),( yx = 0 and

dx

dy + ),( yx = 0 .

Let the solutions be f ( x, y ) = c & g ( x, y ) = d. Then choose = f(x, y ) and = g

( x, y )

Theses choices make 0),( yxA and ),( yxA = 0.

Hence the equation ( 1 ) is transformed to ),,,,( uuuu , which is called the

canonical form of the hyperbolic equation.

Parabolic Equation ( S 2 – 4 R T = 0 )

In this case, the roots of the equation ( 2 ) will coincide, say, ),( yx .

Take = f(x, y ) as in the above case. Take as function of x & y so that it is independent

from . Since these functions are independent, 0 yxyx .

Now by the choice of , 0),( yxA . Thus ( * ) ),,,( yxyxB = 0. Since is

chosen independent of , ),( yxA 0 . Thus we get the transformed equation as,

),,,,( uuuu , called the Canonical form of the parabolic equation.

Elliptic Equation ( S 2 – 4 R T < 0 )

In this case the roots of ( 2 ) are complex, and proceeding as in the case of hyperbolic type

determine the functions, = f(x, y ) and = g ( x, y ), which will be complex conjugates. We

make a further transformation,

2

and

i2

. Then the equation will finally

reduces to ),,,,( uuuuu , in terms of the real variables and , which is

called the canonical form of the elliptic equation.

Eg.1. We may reduce to canonical form, uxx – x 2 uyy = 0.

Here R = 1, S = 0, T = - x 2, so that S2 – 4 R T = 4 x 2 > 0. The equation is hyperbolic type.

Consider 02 TSR i.e. 022 x x .


Now the equations, dx

dy + ),( yx = 0 and

dx

dy + ),( yx = 0 becomes

dx

dy= -x , x

y + x 2/2 = c, y – x 2/2 = d . Now take, = y + x 2/2 and = y – x 2/2.

We get u x = u x- u y , uy = u + u ,

u xx = u x 2 - 2 x 2 u + x 2 u + u - u ,

uyy = u + 2 u + u .

Thus the equation is transformed to, u =

)(4

uu.

Eg.2. Consider y 2 uxx – 2 x y uxy + x 2 uyy - yx uy

xu

x

y 22

= 0.

Here S 2 – 4 R T = 0; equation is parabolic. The only root of 02 TSR is x/y.

dx

dy+

y

x = 0 x 2 + y 2 = c. Take = x 2 + y 2.

Choose = x 2 - y 2 so that these functions are independent.

The equation reduces to u = 0.

Eg.3. We may reduce to canonical form, uxx + x 2 uyy = 0.

Here S 2 – 4 R T = - 4 x 2 < 0 ; equation is elliptic.

02 TSR becomes 022 x ix .

We get = iy + x 2/2 and = -iy + x 2/2. Further

2

= x 2/2 and

i2

= y.

The equation is transformed to

2

uuu

.

Ex.1. Transform ( n – 1 ) 2 uxx - y 2n uyy = n y 2n-1 uy into canonical form, where n is an

integer.

Ex.2. Transform uxx – 4 x 2 uyy = 1/x ux to canonical form.

7.2. ONE DIMENSIONAL WAVE EQUATION

Let y = y ( x, t ) be the transverse displacement of a string at the position x at the instant t

from the mean position, being the x- axis. We may consider a small segment of the string of

length s between two neighbouring points P & Q. The forces acting on this portion of the

string are the tensions T1 & T2 respectively at P & Q along the tangential directions.


Resolving the forces along the x- direction and the y – direction,

T2 2cos = T1 1cos = T, say, and

( s ) ytt = T2 2sin - T1 1sin = T (2tan -

1tan ) , where is the linear density

and 1 and .

2 the inclination of the tangents at P & Q .

Here 2tan = ( yx )|Q = ( yx )|P + ( yxx )|P x , approximately and

1tan = ( yx )|P

Hence ( s ) ytt = T ( yxx )|P x , approximately. Taking the limit as Q P, we get

ytt = 21 x

xx

y

yT

. Assuming small displacements, yx is negligible, and there by we get,

yxx = 2

1

c ytt , for some constant c, which is the one dimensional wave equation.

Let us proceed to find the solution of the following initial value problem.

Consider an infinite string placed along the x – axis and undergoing vibrations about it, so

that at the position x and at the instant t, its vertical displacement y is given by the

equation,

yxx = 2

1

c ytt , x and t > 0 : with the initial conditions y ( x, 0 ) = f ( x ) &

yt ( x, 0 ) = g ( x ), x .

The wave equation is hyperbolic. Using the transformation, = x – ct, = x + ct, the wave

equation can be reduced to y = 0.

The solution is y = F ( ) + G( ), where F & G are arbitrary functions.

In terms of the original variables x & t, y ( x, t ) = F ( x – c t ) + G ( x + c t ) …( * )

By the initial condition y ( x, 0 ) = f ( x ), F ( x ) + G ( x ) = f( x ) …( 1 )

Since yt ( x , t ) = F’ ( x – c t ) .-c + G’ ( x + c t ) . c , using the condition yt ( x, 0 ) = g ( x ),

-c F’ ( x ) + c G’ ( x ) = g ( x ) .

Then for a suitable x0, - c F ( x ) + c G ( x ) = x

x

dssg

0

)( . …( 2 )

Thus F ( x ) =

x

x

dssgxcfc

0

)()(2

1 and G ( x ) =

x

x

dssgxcfc

0

)()(2

1.

Thus the solution of the problem is,

y ( x , t ) =

ctx

ctx

dssgc

ctxfctxf)(

2

1

2

)()(, called the d’ Alembert’s solution.


Remark: The straight lines x – ct = a constant & x + ct = a constant in x-t plane are called

characteristic curves.

It can be shown that a given pair of characteristics of different types will fix the solution if the

data is supplied on both of them.

Suppose the data is given on the characteristics, = 0 & = 0 i.e. assume y ( 0, ) =

g( ) and y ( , 0 ) = f ( ), for some given functions g & f.

From the solution, y ( , ) = F ( ) + G( ) ,

g( ) = F ( 0 ) + G ( ) and f ( ) = F ( ) + G ( 0 ) so that y ( , ) = F ( ) + G( )

=

f ( ) + g( ) - f ( 0 ) , since f( 0 ) = g ( 0 ).

But the solution can not be uniquely fixed if the data is given only on one characteristic.

Domain of dependence and range of influence

Let P ( x1, t1 ) be any point with t1 > 0.

Then we have, y ( x1 , t1 ) =

11

11

)(2

1

2

)()( 1111

ctx

tcx

dssgc

ctxfctxf, so that ,

y ( P ) =

B

A

dssgc

BfAf)(

2

1

2

)()(, where A ( x1 – c t1, 0 ) & B ( x1 + c t1, 0 ) are the

points at which the characteristics x – c t = x1 – c t1 & x + c t = x1 + c t1 through P meets the

x – axis.

Here y ( P ) depends on the data given on the line segment AB, which is called the domain

of dependence for P. The data at A ( x1 , 0 ) on x – axis will influence the solution y ( x , t )

at any point P ( x, t ) lying in the angular region bounded by the characteristics through A.

Hence this region is called the range of influence of A.

Vibrations of a semi- infinite string.

Consider the motion of an infinite string placed along the positive x- axis and tied at the end

x = 0, and undergoing transverse vibrations about the mean position, namely, the x-axis.

Then we have the following problem:-

yxx = 2

1

c ytt , x0 and t > 0 , with the initial conditions y ( x, 0 ) = u ( x ) &

yt ( x, 0 ) = v ( x ), x0 , and the boundary conditions y ( 0, t ) = 0 = yt ( 0, t ).

The d’ Alembert’s solution obtained earlier in the case of infinite string may not suit the

situation, since at present the data ( i.e. u & v ) is available only for x > 0, but even for x ,


t > 0 the above mentioned solution requires informations at x – ct which can assume

negative values.

To overcome the situation we may give odd extensions to u & v , by defining,

U ( x ) =

0,)(

0,)(

xxu

xxu and V ( x ) =

0,)(

0,)(

xxv

xxv .

We have gone for odd extensions to take care of the homogeneous boundary condition given

at x = 0.

We claim that the solution to the current problem is,

y ( x , t ) =

ctx

ctx

dssVc

ctxUctxU)(

2

1

2

)()( …( & )

Put x = 0. Then y( 0 , t ) =

ct

ct

dssVc

ctUctU)(

2

1

2

)()( = 0, since U & V are odd

functions.

Put t = 0. Then y ( x, 0 ) =

x

x

dssVc

xUxU)(

2

1

2

)()( = u ( x ), for x > 0.

Similarly, we can show that yt ( x , 0 ) = v ( x ) & yt ( 0 , t ) = 0, by taking

csGdssV )()( in ( & ) , then differentiating w.r.t. t and finally substituting t = 0.

Vibrations of a finite string

Consider a string of length l placed along the x – axis and tied at both ends at x = 0 & x = l,

making transverse vibrations about the x- axis.

We have,

yxx = 2

1

c ytt , 10 x and t > 0 , with the initial conditions y ( x, 0 ) = u ( x ) &

yt ( x, 0 ) = v ( x ), 10 x , and the boundary conditions y ( 0, t ) = 0 = y ( l, t ) &

yt ( 0, t ) = 0 = yt ( l, t )..

The d’ Alembert’s solution obtained earlier in the case of infinite string may not serve the

purpose, since at present the data ( i.e. u & v ) is available only for 10 x , but even for

these x & t > 0 the above mentioned solution requires information at x – ct & x + ct,

which can assume values outside [ 0 , l ]. We may extent the data functions to cope up with

the situation. First, u & v will be given odd extensions to [ -l, l] and then periodic extensions

with period 2 l to cover x .

Define, U ( x ) =

0,)(

0,)(

xlxu

lxxu and V ( x ) =

0,)(

0,)(

xlxv

lxxv , and then


U ( x + r. 2l ) = U ( x ) , V ( x + r. 2l ) = V ( x ), for lxl , r = ,...2,1

Assuming U ( x ) & V ( x ) can be expanded as Fourier sine series, we have,

U ( x ) =

l

xmu

m

m

sin

1

, where dsl

smsu

lu

l

m

0

sin)(2

and

V ( x ) =

l

xmv

m

m

sin

1

, where dsl

smsv

lv

l

m

0

sin)(2

.

Then the solution becomes,

y( x , t ) =

l

xmu

m

m

sin

1

l

ctmcos +

c

l

l

ctm

l

xm

m

v

m

m sinsin

1

.

Remark: The solution of the problem can be obtained by an alternate method, called the

method of separation of variables.

Here we assume that the solution can be written as y ( x, t ) = X ( x ) T ( t ).

Then the equation becomes, Tc

T

X

X2

''''

. Here the right side is a function of t alone, where

as the left side is a function of x alone. Hence each of them must be a constant, say, .

Therefore, X’’ - X = …( 1 ) and T’’ – c 2 T = 0 …( 2 )

From 0 = y ( 0, t ) = X ( 0 ) T ( t ) , we infer X ( 0 ) = 0 and similarly from y ( l , t ) = 0, we

get X ( l ) = 0. ( For otherwise, we will be reaching only the trivial solution only )

Case 1. > 0.

The solution of ( 1 ) is, X ( x ) = A xx Bee , where A & B are arbitrary constants.

The conditions X ( 0 ) = 0 = X ( l ) A = 0 & B = 0. This leads to the trivial solution.

Case 2. = 0.

Now ( 1 ) X ( x ) = A + B x. Again the conditions X ( 0 ) = 0 = X ( l ) will imply A = 0

= B, and hence only the trivial solution.

Case 3. < 0.

Now from ( 1 ) we get X ( x ) = xBxA sincos . X ( 0 ) = 0 A = 0.

But X ( l ) = 0 B lsin = 0. Thus for non trivial solution we consider the possibility

lsin = 0 which gives nl , n = 1,2,3, . Taking 2

22

l

nn

, called the eigen

values of the equation ( 1 ), we get the solutions,

l

xnBX nn

sin , n = 1, 2,3, and

correspondingly


( 2 )

l

ctnD

l

ctnCtT nnn

sincos)( , n= 1, 2, ….

Hence, for n = 1, 2, …, we have the solutions,

yn (x, t ) =

l

xn

l

ctnb

l

ctna nn

sinsincos .

Since the boundary conditions are homogeneous by the method of super imposition, we get a

solution as y ( x, t ) =

1

),( txyn . Substituting yn ( x , t ) and applying the initial conditions,

we get dsl

smsu

la

l

m

0

sin)(2

and dsl

smsv

cmb

l

m

0

sin)(2

.

Note that we get the same solution obtained earlier.

Remark: The solution to the problem of vibrations of a finite string is unique – a

consequence of the following Theorem,

Theorem. The solution of the problem, ytt - c 2 yxx = F( x, t ), 10 x and t > 0 , with

the initial conditions y ( x, 0 ) = u ( x ) & yt ( x, 0 ) = v ( x ), 10 x ,

and the boundary conditions y ( 0, t ) = 0 = y ( l, t ) , if it exists, is unique.

Proof: Let there be two solutions, say, u1 & u2.

Let W = u1 - u2 .

Then W satisfies the problem, Wtt - c 2 Wxx = 0, 10 x and t > 0 , with the initial

conditions W ( x, 0 ) = 0 & Wt ( x, 0 ) = 0, 10 x ,

and the boundary conditions W ( 0, t ) = 0 = W ( l, t ) .

We will show that W = 0.

Consider E ( t ) = dxWWc

l

tx 0

222 . Here E ( t ) is s a differentiable function and W is twice

differentiable.

Therefore

l l

xxt

l

txttt dxWWcWWcdxWWdt

dE

0 0

2

0

22 =

l

xxttt dxWcWW0

2 )(2 = 0, since

for every t, W ( 0, t ) = 0 = W ( l, t ) Wt ( 0, t ) = 0 = Wt ( l, t ).

Thus E = a constant. But W ( x, 0 ) = 0 , 10 x Wx ( x, 0 ) = 0 & Wt ( x, 0 ) = 0,

given. Thus E ( 0 ) = 0 E = 0. Hence Wx = 0 = Wt , 10 x and t > 0.

This implies that W ( x, t ) = a constant and hence W = 0, since W ( x , 0 ) = 0.


7.3. Riemann’s Method.

This method can be employed for solving linear, second order, hyperbolic equations, in

canonical form.

Let L [ u ] = uxy + a ( x, y ) ux + b ( x, y ) uy + c ( x, y ) u = f ( x, y ) ……( 1 ), where a, b, c, f

are continuously differentiable functions of x & y. Being a hyperbolic equation in canonical

form, the characteristics are x = a constant , y = a constant. A solution of ( 1 ) is a function

with continuous second order partial derivatives.

Let v ( x, y ) be a function with continuous second order partial derivatives.

Then v uxy – u vxy = ( v ux )y – ( u vy )x, avux = ( av u )x – u ( av )x, bv uy = ( b v u )y – u

( b v )y

so that v L [ u ] – u M [ v ] = Ux + Vy, where U = a u v – u vy, V = b u v + v ux and

M [ v ] = vxy – ( a v )x – ( b v )y + c v.

Here L , M are differential operators and M is called the adjoint of L.

We require the following - Green’s Theorem – Let C be closed curve bounding the region D

and U and V be differentiable functions in D and continuous on C.

Then CD

yx VdxUdydxdyVU .

We will discuss the following Cauchy problem

Let be a smooth initial curve such that it is nowhere parallel to the x or y axes.

Assume that u and ux ( or uy ) are prescribed along . We want a solution of ( 1 ) in some

neighborhood of .

Let P ( , ) be a point at which the solution to the above Cauchy problem is required.

Let the characteristics through P intersect the initial data curve at Q and R ( so that PQ is

horizontal and PR is vertical)

Let D be the region bounded by the closed contour C = PQRP.

Then by the application of Green’s Theorem ,

P

R

Q

P

R

QD

VdxUdyVdxUdydxdyvuMuvL )(][][ ---( * )

It can be shown that

Q

P

x

Q

P

Q

P

dxvbvuuvVdx )( ….( ** )

Substituting in ( * ),

[ u v ]P = [ u v ]Q +

Q

P

x dxvbvu )( +

R

P

y dyvavu )( -


R

Q

VdxUdy )( + dxdyvuMuvLD

][][ …...(&)

Choose v(x, y; , ) so that M [ v ] = 0, vx = bv on y = , vy = a v, on x = and v = 1

at P( , ). Such a function is called a Riemann function for the problem.

Now ( & ) [ u ]P = [ u v ]Q -

R

Q

bdxadyuv )( -

)( dxvudyuv x

R

Q

y + dxdyvfD

…...( I )

( I ) gives u at P when u and ux are given along the curve .

Since, [ u v ]R – [ u v ]Q =

R

Q

yx dyuvdxuv ])()[( , from ( I ) , we get,

[ u ]P = [ u v ]R -

R

Q

bdxadyuv )( -

)( dyvudxuv y

R

Q

x + dxdyvfD

…...( II )

( II ) can be used to find u at P, when u and uy are given along the curve .

On adding ( I ) & ( II ),

[ u ]P = { [ u v ]Q + [ u v ]R }/2 + ½ )( dyudxuv y

R

Q

x -

½ )( dyvdxvu y

R

Q

x -

R

Q

bdxadyuv )( …( III ),

which can be used for finding u at P, when u , ux and uy are given along the curve .

Remark: Consider the wave equation, y = 0 …( 1 ).

Take the Riemann function as ,;,(v ) = 1 , and using formula ( III ) under Riemann’s

method,

[ u ]P =

QR

duduRuQu

2

1

2

)()(. …( 2 )

But x = 2/ and t = c2/ .

( 1 ) becomes, uxx = (1/c2 ) utt and the solution ( 2 ) will reduce to the d’ Alembert’s

solution obtained earlier.


7.3. Laplace’s Equation

The Laplace’s equation in two dimension is yyxx uuu 2 and a solution of the equation is

known as a Harmonic function.

There are many boundary value problems associated with harmonic functions.

Let D be the interior of a simple closed smooth curve B and f , h be continuous functions on

the boundary B.

Dirichlet Problem.

To find a function u( x, y ) harmonic in D and agrees with f on the boundary B.

Neumann Problem.

To find a function u( x, y ) harmonic in D and satisfies un = f on B, where n is the unit,

outward normal to B.

Robin Problem.

To find u ( x, y ) harmonic in D and satisfies un + h u = 0 on B, where h is non egative.

We need the following theorem in complex analysis.

Maximum & Minimum Principles

Suppose u ( x, y ) is harmonic in a bounded domain D and continuous on BDD .

Then u attains its maximum as well as minimum at some point on B.

Proof: Let the maximum of u on B be M. Suppose the maximum of u on D is not attained at

any point on B. Then it will be attained at some point in D, say, P( x0, y0 ). Let M0 = u(x0, y0).

Then M0 > M.

Consider v ( x, y ) = u ( x, y ) + 2

0

2

02

0 )()(4

yyxxR

MM

…. ( 1 ), for each point in

D where R is the radius of a circle with center P containing D.

Then v ( x, y ) is continuous on D and v( x0, y0 ) = u(x0, y0) = M0.

On B, we have 00

4),( M

MMMyxv

.

Thus v( x , y ) also attains its maximum at some point Q in D itself.

Then at Q , 0,0 yyxx vv .i.e. 0 yyxx vv , at Q.

However, in D vxx + vyy = uxx + uyy + 2

0

R

MM =

2

0

R

MM > 0.

Thus we reach a contradiction.

Hence the maximum of u in D is attained at some point on the boundary B.

The minimum value of u in D is attained at some point on the boundary B also.

Apply the above discussion to –u instead of u.


Theorem. The solution of the Dirichlet problem, if it exists, is unique.

Proof: Let u1 & u2 be two solutions of the problem. Then v = u1 – u2 is also harmonic in D and

on B, v = 0. Then by Maximum & minimum principles applied to v gives v = 0 in D.

i.e. u1 = u2 in D .

Green’s identity: If U( x, y ) & V ( x, y ) are functions on the boundary B of a closed region

D, then by Green’s Theorem D B

yx VdxUdydSVU )()( .

Let U = x & V = y .

Then dsn

dSBD

yyyyxxxx

….( % ),

where n is the unit out ward normal to B.

Interchanging & and subtracting the equations, we get,

dsnn

dSD B

22 …( $ )

Theorem. Let u ( x , y ) be a solution of the Neumann problem. Then B

dssf .0)(

Proof: Let = 1 and u in Green’s identity …( $ ).

Theorem. The solution to the Neumann problem, is unique upto an additive constant.

Proof: Let u1 & u2 be two solutions of the problem. Let v = u1 – u2.

Then 02 v on D and 0

n

v on B.

Take = u in Green’s identity …( % ). Then 02

dSvD

, which implies

0v , since it is continuous. Thus v = a constant.

Dirichlet Problem for the upper ½ plane

Consider the problem, 0,,0 yxuu yyxx ….( 1 ),

and u ( x , 0 ) = f ( x ) , x with the assumption that u is bounded as y and

u and ux vanishes as | x | .

Solution is obtained by Fourier Transform Method.

Let ),( yU be the Fourier transform of u ( x, y ) w.r.t. x. i.e. ),( yU =

dxeyxu xi

),(

2

1.


Then by taking the Fourier transform of ( 1 ) , 02 UU yy ….( 2 )

Solution of ( 2 ) is, yy eBeAU )()( .

Since u is bounded as y , U is also bounded as y .

Hence for > 0, A ( ) = 0, and for < 0 , B ( ) = 0

Thus ),( yU = yeU ||)0,( .

But .),()]([)]0,([)0,( sayKxfFxuFU

Thus ),( yU = yeU ||)0,( = yeK ||)( .

We can compute directly

22

||1 2)(

xy

yeF y

.

Thus by Convolution Theorem, u ( x, y ) = f(x) *

22

2

xy

y

=

dxy

yf

22 )(

2)(

2

1 =

y

d

xy

f

22 )(

)(.

Neumann Problem for the upper ½ plane

Consider 0,,0 yxuu yyxx and uy ( x , 0 ) = g ( x ) , x ,

with the assumption that u is bounded as y and u and ux vanishes as | x | .

and dxxg

)( = 0.

Solution is found by converting it to a Dirichlet problem. Let v ( x, y ) = uy ( x, y )

Then u ( x, y ) = y

a

dxv ),( .

In terms of v, the problem becomes,

0,,0

yxuu

yuuvv yyxxyyyxxyyyxx and v ( x , 0 ) = uy ( x, 0 ) = g ( x ).

Thus from the above solution of the Dirchlet problem, v(x, y ) =

y

d

xy

g

22 )(

)(.

Hence, u ( x , y ) =

1y

a

dd

xy

g

22 )(

)( =

1

d

ax

yxg

22

22

log)( .

Dirichlet Problem for the interior of a circle.

Consider a circle of radius a, centered at the origin.

Consider the problem, ,011

2

2 ur

ur

uu rrr r < a …( 1 ), subject to the

boundary condition u ( a , ) = f ( ) …..( 2 ).


Since the equation is linear and homogeneous, we assume that the solution is in the separated

form, i.e. u ( r , ) = R ( r ) H ( ) ….( 3 )

Then ( 1 ) H

H

R

Rr

R

Rr ''''2

, say.

Here, is a constant. Hence we get, r2 R’’ + r R’ - R = 0 …( 4 ) & H” + H = 0 …( 5 )

But H is a periodic function with period 2 . Hence < 0 will not supply a feasible

solution.

When = 0, then ( 4 ) & ( 5 ) gives R = A + B log r, H = C + D …( 6 )

Since u is bounded inside the circle , but log r - as r 0, we get B = 0 , and hence

R = A. Further since H is periodic we get C = 0.

Thus under this case we get, u = a constant.

Let < 0. Assume = 2 .

Then H = A cos + B sin . Then the periodicity of H will fix as 1,2,3,….

Correspondingly, ( 4 ) R ( r ) = C r n + D r –n. Since u has to be bounded and r –n.

as r 0, we get D = 0.

Combining ( 6 ) and the solutions corresponding to n = 1, 2, 3, … by super position the

solution is,

u ( r, ) = nbnaa

rann

n

sincos2 1

0

, …( 7 ) , for some constants an & bn.

By the given boundary conditions, u ( a , ) = f ( ) .

Then ( 7 ) an =

2

0

cos)(1

dnf and bn =

2

0

sin)(1

dnf . Substituting these

coefficients in the series solution for u ( r, ), with = r/a we can obtain the solution in

the form of an integral formula,

dfu )()cos(21

1

2

1),(

2

0

2

2

, known as the Poisson integral formula.

Dirichlet Problem for the exterior of a circle.

Similar to the above problem we have to determine a harmonic function u in the region r > a,

if u is given at points on r = a. Since the region is unbounded we may impose the further

condition that u is bounded as r .

Proceeding as above, the solution is obtained as,

dfu )()cos(21

1

2

1),(

2

0

2

2

.


Neumann Problem for the interior of a circle

The problem is to solve ,011

2

2 ur

ur

uu rrr r < a, subject to the boundary condition

that )(fr

u

on r = a.

As in Dirichlet’s problem we get, u ( r, ) = nbnaa

rann

n

sincos2 1

0

, …..( 1 )

where an’s & bn’ s are constants to be fixed based on the boundary conditions.

Differentiating ( 1 ),

r

au ),( )(sincos

1

fnbnaa

nnn

an =

2

0

cos)( dnfn

a and bn =

2

0

sin)( dnfn

a.

On substitution and simplification,

we get u ( r, ) =

dfraraaa

)()cos(2log22

2

0

220

Remark: The solution to the corresponding problem for the exterior of the circle r = a is,

u ( r, ) =

dfraraaa

)()cos(2log22

2

0

220

Dirichlet’s problem for a rectangle.

Consider the Problem -: uxx + uyy = 0, 0 < x < a , 0 < y < b; u ( x , 0 ) = f ( x ) , ax 0

u( x , b ) = 0 , ax 0 and u ( 0 , y ) = 0, u ( a , y ) = 0 , by 0

Solution is found by the method of separation of variables.

Assume that u ( x , y ) = X ( x ) Y ( y ).

Then we get X’’ - X = 0 & Y’’ + Y = 0 . These equations are solved using the

conditions X ( 0 ) = 0 = X ( a ), Y ( b ) = 0.

We will be getting non zero solutions corresponding to the case = -

2

2

an

, n = 1,2,3,..

as Xn = Bn sin

a

xn and Yn = En sinh

a

byn )(

By the method of super imposition we may assume the solution as,

U ( x, y ) =

1

nnYX

1

na sin

a

xn sinh

a

byn )(. Now using the boundary

condition u ( x , 0 ) = f ( x ), we get an = dxa

xnxf

a

bna

a

0

sin)(

sinh

2

.


7.5. Heat Conduction Problem

Consider a homogeneous, isotropic solid. Let V be an arbitrary volume inside the solid

bounded by the surface S. If V is an volume element then the heat energy stored in it is

Vuc , where c is the specific heat, u is the temperature as function of its position and time,

and is the density.

Thus the total Heat energy stored in V = VucV

… ( 1 )

For an element S of the of the bounding surface, heat flow across it is, nuk . S, where

k is the thermal conductivity factor and n is the unit, outward drawn normal.

Thus the total flux across S is S

nuk . S = VukV

. , by Gauss Divergence theorem.

Since no heat energy is created or destroyed in V,

the rate of change of Heat energy in V = Flux across S.

Thus, dt

dVuc

V

= VukV

. . i.e. 0)).()((

Vukuc

tV

.

Since V is arbitrary, we get, 0).(

uk

t

uc . Assuming the k is a constant throughout

the body, we have the heat conduction equation, uLt

u 2

, where L is a constant.

The one dimensional ( Solid body is a straight rod ) heat conduction equation is, 2

2

x

uL

t

u

Heat Conduction – Infinite rod

Consider an infinite homogeneous rod placed along the x- axis, sufficiently thin so that heat is

uniformly distributed over any cross section and insulated to prevent any loss or gain from

external sources. Let u ( x , t ) be the temperature at the position x, at the instant t.

The problem is to solve, ut = k uxx, x , t > 0, …( 1)

under the initial condition, u ( x, 0 ) = f ( x ), x .

We may use the Fourier transform method .

Let ),()],([ tUtxuF dxetxu xi

),(

2

1.

Then ( 1 ) Ut + k 2 U = 0.

Its solution is U ( , t ) = A ( ) kte2 , where A ( ) is an arbitrary function, which can

be fixed by the initial condition, since A ( ) = U ( , 0 ) .


We have, U ( , 0 ) = F[u(x,0)] = dxeoxu xi

),(

2

1= dxexf xi

)(

2

1= F( ), say.

Hence, U ( , t ) = F ( ) kte2 .

Then by convolution theorem, u ( x , t ) = f ( x ) * F -1( kte2 ) =

defkt

ktx

4

2

)(2

1

Note that, F -1( kte2 ) =

kt

x

kt 4exp

2

1 2

.

Remark: Convolution Theorem : - F ( f * g ) = F ( f ) . F ( g ), where F stands for the Fourier

Transform, and f * g is the convolution product defined as,

( f * g) ( x ) =

dgxf

)()(2

1.

Heat conduction – Finite rod

Consider the heat conduction problem in a finite rod of length l, placed along the x-axis

extending from x = 0 to x = l, under the additional homogeneous boundary conditions that

U ( 0 , t ) = 0 = u ( l, t ), t > 0.

We may use the method of separation of variables, by taking u ( x, t ) = X ( x ) T ( t )

The equation gives, ,''''

kT

T

X

X a constant.

We can notice that for non zero solution, must be negative. Let = - 2 .

Then we have, X ‘’ + 2 X = 0 & T’ +

2 kT = 0.

The boundary conditions will become, X ( 0 ) = 0 = X ( l ).

X ‘’ + 2 X = 0 X ( x ) = xBxA sincos . Now X ( 0 ) = 0 gives A = 0.

Further X ( l ) = 0 gives 0sin lB . To get non zero solution, we consider 0sin l i.e.

,...3,2,1, nnl . Hence Xn ( x ) = Bn

l

xnsin , n=1, 2, 3,..

Correspondingly, T’ + 2 kT = 0 Tn ( t ) = Cn

2

22

expl

ktn .

Thus, un ( x , t ) = an

l

xnsin

2

22

expl

ktn .

By the principle of superposition, u ( x , t ) =

1

an

l

xnsin

2

22

expl

ktn .


Now u ( x , 0 ) = f ( x ) implies

1

an

l

xnsin = f ( x ) , lx 0 .

Then an =

0

2

lf(x)

l

xnsin dx.

Thus we have the solution,

u ( x , t ) =

1

an

l

xnsin

2

22

expl

ktn , where an =

0

2

lf(x)

l

xnsin dx.

Theorem. The solution of the problem, ut – k uxx = F ( x , t ), 0 < x < l, t > 0 satisfying the

initial condition u ( x, 0 ) = f ( x ), lx 0 and boundary conditions

u ( 0 , t ) = 0 = u ( l, t ), 0t , if exists, is unique.

Proof: Let u1 & u2 be two solutions. Take v = u1 – u2.

Then v satisfies, vt – k vxx = 0, 0 < x < l, t > 0, v ( x, 0 ) = 0, lx 0 and

v ( 0 , t ) = 0 = v ( l, t ), 0t .

Let E ( t ) = dxtxvk

l

),(2

1

0

2

. Note that E ( t ) 0 .

Then dt

dE dxvv

kt

l

0

1 lxxx

l

vvdxvv0

0

- dxvx

l2

0

- 02

0

dxvx

l

, by v (0, t ) = 0

= v (l, t )

Thus E ( t ) is a decreasing function. But from v ( x , 0 ) = 0, we get E ( 0 ) = 0.

Thus E ( t ) 0. But E ( t ) 0 . Thus E = 0 v ( x , t ) = 0, lx 0 , 0t . i.e. v = 0.

Remark: The solution to the Heat conduction problem in a finite rod is unique, by the above

Theorem.

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