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1

BASIC THEORY

Unit Structure :

1.1 Introduction.

1.2 Ordinary Differential Equations.

1.3 First Order ODE.

1.4 Existence and Uniqueness of Solutions (Scalar Case) .

1.5 Illustrative Examples .

1.6 Exercises.

1.1 I NTRODUCTIO N

As we already know, a differential equation - DE - is an equation

relating the following three items :

A function of one or more variables (the function being real valued or

vector values).

The independent variables of the function.

A finite number of derivatives of the function.

The highest order of derivatives of the function appearing in the

equation is the order of the differential equation.

Usually, the function in a differential equation is an unknown

function; it is an are observable quantity of a real process and therefore we

are interested in knowing the function. We use results and techniques of

mathematical analysis along with our geometric intuition and tease the

function out of the differential equation. We then speak of having solved

the differential equation.

Depending on the nature of the function (in which a differential

equation is set) we classify the differential equations in the following two

types :

I) A function nX:I

of a single real variable (say) t ranging in an

open interval I gives rise to the succession of derivatives : 02k2kdd X d X d XX X, , ,... .....dt dt dt dt

Now a differential equation in the function Xt is therefore an

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22,, , . . . . 0kkdX d X d XFtXdt dt dt

………………. (1)

Such an equation is said to be an ordinary differential equation. Thus

an ordinary differential equation is a differential equation in which the

constituent function :( )Xt X t

is a function of a single real variable.

We often use the acronym ODE in place of the full term : ordinary

differential equation.

II) On the other hand, there are differential equations in a function :ux ux

of a real multivariable 12.....

nxx x x which ranges in

an open subset

of

n. Such a function

uu xgives rise to mixed

partial derivatives :

122

12,1 ,1,:....niijnuinxuji j nxxDuuxx x

for various multi-indices 12....

n with 0,1, 2, ... ,ii

the

mixed partial derivative

Du having the order 12:. . .

n.

Now a differential equation in such a function

uu xof a multi -

variable 1.....

nxx x ranging in an open subset

of

nis an equation of

the type :

2,, , . . . . : 0

ii juuFx uDu mxx x ……… (2)

its order being m. Equation (2) is said to be a partial differential equation

in ()uxbecause it involves the mixed partial derivatives of u. We use the

acronym PDE for this type of differential equations.

There is more about the setting of a differential equation : In a

mathematical problem, a differential equation is accompanied by auxiliary

data. A solution of a differential equation is required to satisfy this auxiliary

data. To be more specific we are given a subset of the domain of a

prospective solution and some of its derivatives of the solution at the points

of this subset.

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In case of the ODE, the auxiliary data is said to consist of initial

conditions . An initial value problem consists of finding the solution of the

ODE which satisfies the accompanying initial conditions. Often, the pair

consisting of (a) an ordinary differential equation and (b) the initial

conditions is referred as the initial value problem -IVP-.

Often the initial conditions are given at the end points of an interval

which are then called the boundary conditions. Also, the resulting initial

value problem is called a boundary value problem .

In the case of a partial differential equation, the accompanying

auxiliary data is called the Cauchy data for PDE. The Cauchy problem

for a given PDE consists of finding the solution of the PDE which satisfies

the requirements of the given Cauchy data.

We will explain more about these terms initial conditions, Cauchy

data etc. - at later stages.

Partial differential equations, being more intricate mathematical

objects are studied by using the concepts and results of the ordinary

differential equations. Therefore, a basic course on differential equations

begins with a treatment of ordinary differential equations. In our treatment

of the subject also, we will develop enough theory of ODE and then apply it

to the partial differential equations.

Therefore, back to the theory of ODE.

1.2 ORDI NARY DIFFERE NTIAL EQUATIO NS

To begin with, we reorganize the form (1) of the ODE in the

following manner. Unraveling it, we separate the top order derivative and

express it as a function of the remain variable quantities, namely, 11,( ) , , . . . . . ,kkdX d XtXtdt dt

that is we form the equatio n 11,, ,kkkkdX d X d xftXdtdt dt

…………………… (3)

We regard the equation (3) as the standard form of an ODE (of

course, the ODE has order = k.) Note that the function :( )Xt X t

is a

vector valued function of the real variable t and as such it is a curve in n

.

Each ()Xt has n components : 12() () , () , . . . . , ()

nXt X t X t X t and

therefore all the derivative of it has n components :

12()() , () , . . . . . ()

ndXt d ddXt Xt Xtdt dt dtdt for 1

k.

Consequently, the function f appearing on the right hand side of (3) has n

components : 12,, . . . ,

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Consequently the DE (3) is actually the following system of ODE in the

functions : 12,, . . . . ,

tX t tX t

ntX t 1111122111, , ,...,, , ,...,, , ,...,kkkkkkkkkknnkkd X dX d Xft Xdtdt dtdXdX d Xft Xdtdt dtdXdX d Xft Xdtdt dt

……………… (4)

At this stage, we become more specific about the features of the

ODE (3) (or equivalently about the system (4).)

Let I be an open internal and let

denote an open subset of

n. We

consider the open sets

......nnI

(there being 1

k copies of

n in the above Cartesian product). This set is

being designated to accommodate the variable quantities :

11,( ) , ,kkdx d xtXt tdt dt

.

Clearly, the function f appearing on the right hand side of (3) must

have this set as its domain of definition.

Choosing 00,

tI x and 12 1,. . .

kww w in

n, we form the initial

condition 00 1 2 1,, , . . .

ktxw w w. Now, the initial value problem for the ODE

(3) is the following pair :

1100 11, , ,...., , ,......kkkkkdX d X d XftXdtdt dttxw w

……………. (5)

By a solution of the initial value problem (5) we mean an (at least) k

times continuously differentiable function (= curve in

)

:

XJ, J being an open interval with 0

tJ I, which satisfies

the following two items :

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The differential equation : 11() () ()...kkkkdXt d Xt d Xtftdt dt dt

for all

tJ

The initial conditions : 1000 0 111,, . . . . ,kkkdX tdXxt x t wwdtdt

Remarks :

(I) Though the independent variable t of the function X(t) in the ODE (1)

is stipulated to range in the interval I, we expect the solution ()tX t

of the initial value problems (4) to be defined only on a sub-internal J

of I (with 0t

J). Indeed, we come across concrete cases of the IVP in

which a solution exists only on a sub-inter J of I and therefore, we

grant this concession : a solution need be defined only on a sub-

interval J of I.

II) Often, the initial conditions are expressed more explicitly in terms of

the equations : 1000 0 111,, . . . . ,kkkdX tdXxt x t wwdtdt

.

An important special case of the ODE (3) is 2121,,kkkkdX d X dX d XfXdt dt dt dt

………………… (6)

in which the function f is independent of the variable :: . . . .

nnntf. In this case we say that the ODE (6) is

autonomous.

Returning to the initial value problem (5), there are two questions :

Does the initial value problem (5) admit a solution at all?

If it does is the solution unique?

Clearly, because f is the main ingredient of the ODE, the answer to

both of there questions naturally depends on the properties of f, especially

on its behaviour around the initial condition 00 11,, , . . . . . . .ktxw w

(i.e. how if

varies continuously, differentially etc. around 00 1 1,, , . . . . ,ktxw w

Following

two examples illustrate that answers to both the questions are (in general) in

the negative :

Let :

f be the function :

1010fx xx

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For this function we consider the (first order autonomous case of) the

initial value problem : () ,( ) 0dXfX x odt

.

We contend that this initial value problem has no solution.

For, if there was a solution :

XJ with 0

J, then (0) (0) 1dXfdt

implies that the solution ()tX t

is strictly

monotonic increasing in a neighborhood ,

of O.

On the other hand, 1( 0) ( )dXftdt

for all ,0

t implies

that ()tX t t

is strictly monotonic decreasing in ,0

. Thus, if the

solution of the above IVP exists then it is strictly monotonic increasing as

well as strictly monotonic decreasing . This prevents a solution!

Now let :

gbe the function

13000

gx xxx

For this g, we consider the autonomous initial value problem.

,0 0dXgX Xdt

Clearly, one solution of it is the function 3232120()300ttXtt

Another solution of the same IVP is the function 0() 0Xt

. Thus,

the IVP has at least two distinct solutions. (In fact it has an infinitude of

solutions. For each c > 0, the functions : 1()tc X t

is a solutions)

Of course, an IVP should admit a unique solution! In the following

sections we will concentrate our attention on first order ODE and for such

ODE we will introduce a condition - f being locally Lipschitz - which will

guarantees both - existence and uniqueness of the solution.

Above, we have been discussing ODE of arbitrary order :1 , 2 , 3 , . . . .n

and the IVP associated with them. But there is a

simplifying aspect of the ODE! Higher order ODE can be studied entirely in

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part of chapter 2). Therefore, for the time-being we will focus our interest

on first order ODE only.

1.3 FIRST ORDER ODE

We begin some more generalities related to first order ODE. As in

the preceding part, I denoted an open interval and

, an open subset of

n.

We consider a function :

:nfI

……………….. (7)

giving rise to the first order ODE :

(, )dXftXdt

……………………(8)

Note that for each tI

held fixed the map :

,: ; ,nftxf t x

is a vector field on

. Interpreting “t”

as the time variable we call the function (7) a time dependant vector field on

. And, often, we call a solution :XJ

of the ODE (8), an integral

curve of the vector field f.

An initial condition for the ODE (8) consists of a pair 00,tx

with 0tI

, 0x

and the associated initial value problem is :

00(, ) ,dXftX x t xdt

……………….. (9).

Finally, recall that a solution of (9) is an (at least) once continuously

differentiable curve

:XJ

(J being an open interval with 0tJ C I

) satisfying : ()(, () )dX tftXtdt

for all tI

and the initial condition 00Xt x

.

In the context of the IVP (9) we consider yet another equation the

following integral equation in an unknown function :XJ

: 00() (, () )ttXt x fsxs d s t J

…………………. (10).

Following result relates solutions of the IVP (9) and those of the

integral equation (10) :

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Proposition 1 :

A continuously differentiable curve :XJ

(J being an open subinterval of I with 0tJ

) is a solution of the IVP

(9) if and only if it satisfies the integral equation (10).

Proof : (I) - First suppose that the curve :XJ

satisfies the integral

equation (10). Putting 0tt

in (10) we get :

000000(, () )ttXt x fsXs d sxx

Thus Xsatisfies the initial condition.

Next, differentiating (10) we get

00( , ( ) )(, () )ttdX tdfsXs d sdt dtftXt

by fundamental theorem of integral calculus.

Above, we have verified that a solution tx t

of the integral

equation (1) is also a solution of the IVP (9). Conversely suppose, tx t

, tJ

is a solution of the IVP (9). Integrating the identify.

(( )dxtf t x t t Jdt

We get :

000(,ttttdx t x t X s dsdsf r x s ds

And therefore :

00(, )ttxt x f sX sd s

Thus we have :

00(,ttxt x f sX sd s

For all tJ

proving that a solution of the IVP (9) is also a solution of the

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1.4 EXISTE NCE A ND U NIQUE NESS OF SOLUTIO NS (SCALAR CASE)

In this section, we consider the scalar case, (i.e. a single differential

equation) of the initial value problem. Now

will be an open subset of

which, without loss of generality will be taken to be an open internal, we

write J for

. Thus we have on function

:fI J

Along with 0tI

, 0xJ

giving rise to the scalar case of the initial value

problem :

00(, ) ( )dXftx x t xdt

…………………… (11)

Following property of f ensures both, existence and uniqueness of

the solution of (11).

Definition 1 : f is locally lipschitz on J, uniformly in tI

if the following

two conditions are satisfied.

a) f is continuous on IJ

.

b) For each0tI

, 0xJ

there exist finite numbers 0, 0K

satisfying

the following :

i) 0000,, ,tt I xx J

and

ii) ,( , )ft x fty K x y

holds for all 00,tt t

and for all

pairs x, y in 00,xx

.

Remark : An autonomous ODE arises from a function :fJ

which is

independent of the time variable tI

. For such a function, the condition

(b) in the definition takes the following simpler from : for each 0xJ

,

there exist 0, 0K

satisfying :

i) 00,xx J

and

ii) ()fx fy K x y

for all x, y in J.

Also note that this condition implies continuity of f at every 0xJ

and therefore there is no separate mention of condition (a) in the definition

of local Lipschitz property of such a :fJ

.

Following proposition describes a broad class of functions with the

locally Lipschitz property :

Proposition 2 : If :fI J

is continuously differentiable on its

domain, then it has the locally Lipschitz property.

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Proof : Let 00,tx I J

be arbitrary. Using openness of IJ

, choose 0

such that 0000,,tt xx I J

.

Now, the function fx

is continuous on IJ

and therefore it is

bounded on the compact subset 0000,,tt xx

. We consider

any constant k > 0 with the property.

(, )ftx Kx

for all 00,tt t

and for all 00,xx x

.

Finally let 00,tt t

x, y in 00,xx

be arbitrary. By

the mean value theorem of differentia l calculus, we have :

,( , ) ( ) ,fft y ft x y x tx

for some z between x and y. Therefore,

,( , )(, )fft y ft x y x t z Ky xx

(Since 0000,,,tz t t x x

and therefore (,ftKx

. This proves the locally Lipschitz property of f.

Theorem 1 (Emil Picard) : If :fI J

is locally Lipschitz then for

any 0tI

, 0xJ

, the initial value problem.

00(, ) , ( )dfftx x t xdx

has a solution 00:,xt t J

for some 0

.

Proof : Choose 0

such that 0000,, ,tt I xx J

and

there exists 0K

for which ,( , )ft x fty K x y

holds for all 0000,,tt xx

. We choose M > 0 such that ,ft x M

for

all 00,tt t

, 00,xx x

.

Using the constants 0

, K > 0, M > 0, chosen above, we choose

one more constant

satisfying 10 min ,KM

.

We define a sequence of functions :

00:,kxt t k

recursively as follows : MATHEMATICS___________________________________________________________________________________________________________________________________________________

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00

0001002011( ) 0()() ( )() ( ( ) )(( ) )tttttktktxt xx t x f s x dsx t x f s x s dsx x f s x s ds

The sequence :kxk

of functions has the following two

properties :

a) 00() ,kxt x x

for each 00,tt t

b) 101() ()(1 ) !kkkkMK t txtx tk

Both these properties are derived using principle of mathematical

induction and the locally Lipschitz property of f. Using property (b) we

deduce that the sequenc e :kxk

is uniformly canchy on 00,tt

. For if 00,,,ttt t k p

, then 11() () () ()kpkp kjjjkxtx t x tx t

and therefore

111110() () () ()(1 ) !!kpkp kjjjkjjkpjkjjjkxtx t x tx tKt tMKjMKKj

!jjkKMKj

0

as k

.

This last observation is true because !jjkKj

is convergent,

converging 0kte

. Note that in the inequalities :

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() ()!jkp kjkKMxtx tKj

the right hand sides are independent

of t and therefore () ()kp kxtx t

uniformly on 00,tt

as k

, p being arbitrary. This completes the proof of our claim that the

sequence :tkxk

of functions is uniformly Cauchy on 00,tt

.

Using this last mentioned property of the sequence :kxk

we

define a function.

0000:,,xt t x x

by putting 00lim ( ), ,kkxt x t t t t

. The function x, thus defined,

is the uniform limit of the sequence :kxk

. Therefore we have :

000101lim ( )lim ( , ( )lim ( , ( )kktkkttkktxt x tx f s x s dsx f s x s ds

000101lim ( , ( )( , lim ( )tkkttkktx f s x s dsx f s x s ds

00(, ()ttxf s x s d s

All the above equations being valid because of the uniform

convergence of of 0kxtx

on 00,tt

.

Finally, the identity :

0000(, () , ,ttxt x f sxs d st t t

derived above has the

following two consequences.

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000()(, () )0( , ( ) )(, ()ttttdx t dx f s x s dsdt dtdf s x s dsdsftx t

by fundamental theorem of integral calculus.

2) 0000(, () )ttxt x f sxs d s

000xx

We have now verified that the function 0000:,,xt t x x J

is a solution of the given initial value

problem.

Remark : The above theorem there proves that the functions kxconstructed

above are approximate solutions of the initial value problem (11).

The sequence :tkxk

is called Picard’s scheme of approximate

solutions of the initial value problem.

In the next chapter, we will generalize this result (the scalar case) so

as to become applicable to a system of first order ODE. We will also prove

that any two solutions of the initial value problem (11) agree on the overlap

of their domains.

1.5 ILLUSTRATIVE EXAMPLES

Example 1 : Obtain Picard’s scheme of approximate solutions of the initial

value problem : ,( 2 ) 3dxxxdt

and thereby obtain the solution of it.

Solution : This DE is an autonomous ODE with 0,, 2ftx x t

and 03x

. Therefore the approximate solutions are as follows :

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012222222() 3( ) 3 3 3 3 ( 2)3( 2)( ) 3 3 3 ( 2) 3 3 ( 2)2!333( ) 3 3 ( 2) ( 2) .. ( 2)1! 2!!333 ( 2) ( 2) ....1! 2!tttkkxtx t ds ttxts ds txtsss dsktt

13( 2)1ktk

Therefore, the solution of the IVP is

2(2 )0lim333lim 3 ( 2) ( 2) ( 2)1! 2!!33!kkkkktkxt x ttttktzek

Example 2 : Obtain approximate solution (to within 7t) of the initial value

problem :

2,( 0 ) 2dxxt t xdt

Solution : Here, 200,, 0 , 2ft x x t t t x

.

Therefore, 0() 2xt

2102343220() 2 ( 2 )223() 2 ( 2)4t

tx t s s dsttsx t s s s ds

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210022022220() 0 m a x , ( )2,22222 ( 2) 2ttt

txts x s dssds tssds ds tttr ds t

223222( 2) ( 2)2 2 ( 2) 2if 22 3.2tttttt

Again, note that 22if 2max , ( )( ) if 2tttx txt t

and consequently,

we get 2323422()2( 2) 2( 2) 2( 2) 2( 2)221! 2! 3! 4!ttxttt t tt

Using principle of mathematical induction, we get

221()222( 2) 2( 2) 2( 2) 2( 2)22...21! 2!!( 1 ) !kkktxt ttttttkk

Noting that 20!ktk

as k

for every t

we get :

222lim ( )222kkttif txt x te if t

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1.6 EXERCISES

Obtain solutions to within 5tof the following initial value problems :

1) 21(0) 0 1 12dxt x x for tdt

2) 2(0) 1dxxxdt

3) 21(0) 0dxxxdt

4) 3(1) 1dx xxdt t

5) 3, (0) 0dxtx x x xdt

6) , (0)dx txtxdt x

7) 3(0) 12dx xxxdt

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2

SYSTEMS OF FIRST ORDER ODE

Unit Structure :

2.1 Introduction.

2.2 Existence and Uniqueness of Solutions.

2.3 Uniqueness of a Solution.

2.4 The Autonomous ODE .

2.5 Solved Examples.

2.6 Higher Order ODE.

2.7 Exercises.

2.1 I NTRODUCTIO N

Basic concepts related to differential equations such as systems of

first order ordinary differential equations, the initial value problem

associated with such a system, a local solutions of the initial value problem

etc. were introduced in the first unit. At the end of the unit, we proved a

result regarding the solution of a single first order ODE.

We will extend the results of a single ODE to a system of first order

ODE and prove both, existence and uniqueness of solutions of an initial

value problem associated with a system of first order ODE. We will then

derive some simple results giving information about the nature of solution

of such an initial value problem.

We will conclude the chapter by explaining how a system of higher

order ODE can be reduced to a system of first order ODE. We can then

invoke the existence / uniqueness theorems and apply them to the first order

ODE and get some information about the solutions of the higher order

ODE.

2.2 EXISTE NCE A ND UNIQUE NESS OF SOLUTIO NS

We will use the same notations which were introduced in Unit 1.

Let

be an open subset ofn

, I an open interval and let :nfI

be a time dependent vector field having components: 12,, . . . . , :nff f I

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,dxft Xdt

… (1)

which when written in terms of its components becomes the following

system of first order ODE :

1112211( , ,......, )( , ,......, )( , ,......, )nnnnndxftX Xdtdxft X Xdtdxft X Xdt

... (2)

Between the expressions (1) and (2) the compact form (1) is more

convenient and therefore we will use it throughout this chapter, bearing in

mind that it is the same as the system (2).

Recall, given 00,tI x

we have the initial value problem :

00(, ) , ( )dxftX Xt xdt

… (3)

a solution of which is a continuously differentiable curve :XJ

satisfying ()(, ()dX tftXtdt

for all 00,( )tI X t x

. (J being an open

interval with 0tJ I

).

Now, towards the existence / uniqueness of solution of (3) we

introduce the locally Lipschitz property of f:

Definition :

The vector field f has the locally Lipschitz property if it satisfies

the following two conditions :

a) f is continuous on its dom ain and

b) For 00,tI x

, there exist two constants 0, 0K

such that

i) 000,, ,tt I B x

and

ii) ()ft x ft y Kx y

for all 00,tt t

and for all

x, y in 0,Bx

.

Remark :

We will also consider vector fields :nf

as a special case of :nfI

in which ,ft x is independent of :( , ) ( )tf t x f x

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such vector field give rise to the autonomo us ODE : ()dxfxdt

. Now, the

definition of locally Lipschitz property for such :nf

takes the

following simpler form : For any 0x

, there exist 0, 0K

such that 0,Bx

and ()fx fy Kx y

for all x, y, 0,Bx

.

Note that the above condition implies continuity of f at every 0x

and as such there is no separate mention of continuity on f.

Now we have the following result :

Proposition 1: If :nfI

is continuously differentiable on its

domain, then it has the locally Lipschitz property.

The proof of this proposition is on lines similar to that of Proposition 2 of

Unit 1.

Theorem 1: (Local Existence of solutions) : If :nfI

has the

locally Lipschitz property then the initial value problem :

00(, ) , ( )dxftx x t xdt

has a solution 00:,Xt t

.

Proof : We give here a sketchy proof. (To fill up all the details that are left

here, consult the proof of Theorem 1 in unit 1 )

For 00,tI x

choose b > 0 such that

000,, ,tb tbI B x b

and choose 0K

such that ()ft x ft y Kx y

holding for all 00,tt b t b

and for all x, y

in 0,Bxb

.

Using the fact that continuous functions are bounded on compact

subsets of their domains, we choose a constant M > 0 such that ft x M

holds for all 00,tt b t b

and for all 0,xB x b

.

We choose one more constant

which satisfying 10 min ,bkM

.

Now, we define the sequence of maps 00:,nkxt b t b

recursively

by putting : 00()xt x

0100() ( )ttxt x fs x d s

0201( ) ( ( ))ttxt x f s Xsd s

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001( ) ( ( ))tktkxt x f s X s d s

for each 00,tt b t b

.

About the sequence :kxk

we have the following :

a) 0() ,kxt B xb

for each 00,tt b t b

.

b) 1() ()(1 ) !kkkMKxtx tk

00,0tt b t b k

c) :kxk

is uniformly Cauchy on 00,tb tb

(verification of

these properties is left for the reader). We consider the uniform limit of the

sequence 000:,,xt t B x b

which is given by

00lim ( ),kkxt x t t t bt b

Thus, 001() l i m (, ()tkktxt x f sx s d s

00lim ( , ( )tkktxf s x s d s

00lim ( , ( )tkktxf s x s d s

0000(, l i m ()(, ()ttkkttxf s x s d s xf s x s d s

Thus the function ()tx t

satisfying the integral equation :

00() (, ()ttxt x f sxs d s

for all 00,tt b t b

.

Finally the validity of this integral equation has the following two

implications :

1) 0000() ( ,( )ttXt x fsxsd s

000xx

2) 00 ( , ( ))ttdx dfs x s d sdt dt

(, () )ftXt

This now proves that the curve 00:, ,Xt t

thus obtained

is a solution of the initial value problem.

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2.3 U NIQUE NESS OF A SOLUTIO N

We prove an inequality which will lead us to the uniqueness of the

solutions :

Proposition 2 : (Gronwall’s Inequality) : Let

:, 0fa b

:, 0ga b

be continuous functions and 0A

, a constant satisfying taft A fsgsd s

for all ,ta b

. Then .tagsd sft A e

for all ,ta b

.

Proof : First we assume A > 0 and put taht A f s gs d s

for all ,ta b

. Then 0ht

for all ,ta b

and () ()() ()ht f t g tht gt

that is, ()()htgtht

for all ,ta b

. Integrating this inequality over ,at

for ,ta b

we get ()log()athtgsd sha

.

Nothing that ha A

, we get the desired inequality in this case.

Now suppose A = 0. Then for each n

we have :

1taft fsg sd sn

, for all t [,)ab

Applying the above argument to1An

, we get 1agsd stft en

for every ,ta b

and for every n

. Holding t fixed and taking limit of

the last inequality as n

we get 1gsd sgsd stataft enAe

Now we prove the following essential uniqueness result of the solution :

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Proposition 3 : Let :, :xJ yJ

be two solutions of the initial

value problem : 00(, ) ( )dxftx x t xdt

.

then () ()xt yt

for all tJ J

.

Proof : Recall that both x and y, being solutions of the initial value

problem, satisfy the integral equations on their domain intervals :

0000( ) ( , ( ))( ) ( , ( ))ttttx t x f s x s dsy t x f s y s ds

Therefore 0() () , () , ()ttxt yt f sxs f s ys d s

which implies

: 0() () 0 , () , ()ttxt yt f sxs f s ys

00( ) ( )ttKxs ys

for all 0tt

.

Applying Gronwalls result with A = 0, we get

0( ) ( ) 0xt yt

for all 0tt

.

This gives the desired equality () ()xt yt

for all tJ J

.

Towards the uniqueness of the solution of the initial value problem

(3), we consider all the solutions of the initial value problem (3). Let the

totalit y of them be denoted by : ::xJ

the solutions x

being

thus indexed by a suitable indexing set

.

Above we have verified that any two solutions, say 1x

and 2x

are

equal on the overlap 12JJ

of their domains. Therefore, we patch

together all the solutions to get a maximal solution is the solution defined

on the largest open interval. It is obtained as follows.

Let :JU J clearly Jis an open sub internal of I with 0xJand all the solutions x patch up to get a solution :xJ of the

initial value problem (3).

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Because, we consider all the solutions of (3) we get that J is the

largest open internal on which the solution of (3) is defined. We summarize

all this discussion in the following theorem.

Theorem 2 : (Uniqueness of the solution) :

The initial value problem (3) has a unique (maximal) solution defined on

the largest open sub -interval J.

Clearly the solution is unique because it is defined on the largest and

hence unique internal J.

From now onwards we will consider this unique solution defined on

the maximal interval.

2.4 THE AUTO NOMOUS ODE

We note a few simple properties of the autonomous ODE :

()dxfxdt

determined by a locally Lipschitz and hence continuous vector field :nf

.

Now if :xJis a solution of this autonomous ODE then ()(( ) )dx tfx tdt tI and continuity of f and differentiability of x(t) and

hence its continuity together implies that ()dxtdtis continuous and thus the

curve :xJis continuously differential on I. This argument also gives

the following result. If :nf

is ktimes continuously differentiable

then the solution :xJ

is k + 1 times continuously differentiable on it s

domain interval.

Note that a solution of an autonomous ODE may not be defined for

all t

: We consider the initial value problem.

2,( 0 ) 1dxxxdt.

Clearly its solution is 1()1xtt which is defined on ,1only and

not on the whole of

.

It is a result that if the vector field f is compactly supported, then the

solution of the initial value problem (3) is defined for all t

. (We do not

prove this result here).

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2.5 SOLVED EXAMPLES

[Note : Recall, if :[ , ]nga b

is an integrable (vector valued) functions

with components12,, . . . ,ngg g then bagxd x 12() , () , . . . . , () ,bbbnaaagx d x gx d x gx d x . Equivalently written in the columnal

form, we have :

bagxd x12 bababnagx d xgx d xgx d x

We will use these notations in this article.

Example 1 : Obtain approximate solutions (upto 3t) of the following initial

value problem : 2 3 (0) 1 (0) 2dxxy xdtdyty ydt

Solution :

We have : 0() 1 ,xt 0() 2yt

1000() 1 2 . () 3 ()txt xs y s d s 018tds= 8t + 1

010() 2 ()tyt s ys d s

222 2 22tstsd s t

Thus 12181()()222txttytt.

Next 2110() 1 2 () 3 ()txt xs ysd s 20318 2 22tss d s

3211 8 13ttt 210() 2 ()tyt s ysd s 2023 22tssd s

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Thus, 32232211 8 1()3()32262tttxtytttt and so on.

Example 2 : Obtain approximate solution upto 5t :

2 (0) 1 (0) 2dxtx t y xdtdyxy t ydt

Solution : We have 00() 1 , () 2xt yt

232102102() 1 2 123() 2 2 2 22ttttxt s stytsd s t

Thus, 321212132()()222ttxtyttt

Next 22110() 1 () ()txt xs s ys d s 34052123tsssds

245521281 5ttt

2110() 2 () ()tyt xs ys s d s

54 3 2065 4 3219 722231 2 3219 72218 60 12 6tss s ssd stt t ttt and so on

Example 3 : Obtain approximate solutions (upto 3t) :

22 (1) 13 (1) 2 (1) 3dxyt xdtdyzt ydtdzxz zdt

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Solution : We have :000()1() 23()xtytzt

2111() 1 4 1 4 422ttxtsd s t 29422tt

3321112 2() 2 3 2 9 9 933 3 3tttyt s d s tt

11() 3 3 3 3 3 3tzt d s t t

Thus, 213119432()22() 933()3ttxttyt tztt

32124 4() 2 833tsxtss d s

424294 4 1 9 4 4262 36 2 394 4862 3tt ttt t

221() 2 9tyt s s d s

232399 1223 2 391 723 6tttt

2111() 3tzt xs zs d s

21423423934 ( 3 )2232 7 3 2 73442.4 4 2.8 432 7 4 1484 1 6tsss d stttttt

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Thus, 42232242239 44862 3()9 17()326()3 27 4148 4 16tt txtttytztttt

2.6 HIGHER ORDER ODE

As mentioned earlier, we associate a first order ODE with a order k

ODE and try to get information about the solutions of the higher order ODE

in terms of those of the associated first order ODE. In particular, we are

interested in a condition on the function 1(, , . . . )kkdx d xff t Xdt dt which will

ensure existence and uniqueness of the initial value problem for the higher

order ODE. In this article we explain the theory.

To begin with, we consider the ODE :

11(, , . . . )kkkkdx d x d xft xdt dt dt

… (4)

Together with the initial condition :

100 0 1011,. . .kkkdx d xxt x t w t wdt dt.

We introduce a new variable 12,, . . . . ,kyy y y where 1y ranges in and 2kyy

ranging in n

. Next we define :. . . .nnn kFI

by putting 1(, ) (, , . . . , . . . . )nFty Fty y23,, . . . ., ( , )Kyy y f t y

Now, the given initial value problem for the order k ODE give rise to

the following initial value problem in the first order ODE.

(, )dyFtydt

… (5)

the initial condition for it being 00 1 1,, . . . . ,kyt x w w.

Clearly the above first order ODE is actually the following system of

first order ODE :

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1223

12 (, )( , , ..... )kkkdyydtdyydtdyftydtfty y y

... (5*)

Note that taking 1yx the system (5) reduces to the given order k

ODE (4).

This shows that the solutions of the order k ODE (4) can be studied in terms

of the solutions of the first order ODE(5). Note that if f (of the ODE (4)) is

continuously differentiable on its do main, namely the set .....nnI

then F is continuously differentiable on its own domain, consequently the

existence and uniqueness result for the first order ODE applies to the initial

value problem : 00 11(, )( ) ( , .... )kdyFtydtyt x w w

The solution of which giving the solution of the above initial value problem

:

11,, . . .kkkdx d x d xft xdt dt dt

100 0 1011,. . .kkkdx d xxt x t w t wdt dt

.

We summarize this observation in the following :

Theorem 3 : If the function :. . . .nn nfI

is continuously

differ entiable on its domain of definition then the initial value problem.

11100 0 1011, ,.... ,, ,.... ,kkkkkkkdx d x d xft xdt dt dtdxdxxt x t w t wdtdt

has a unique solution.

Illustrative Examples :

These examples explain how we obtain a system of first order ODE from a

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1) The second order ODE : 22,dxkx kdt

being a given constant, gives rise

to the system :

xydyk xdt

Moreover the initial condition 00 0 0,dxxt x t ydt

gives the initial

condition 0000xtxyyt

for the (reduced) first order system.

2) The initial value problem :

322532224301 , 02 , 03dx dx d xtt x tdt dt dtdx d xxdt dt

reduces to the following system of first order ODE along with the initial

conditions.

25,0 1,0243 , ( 0 ) 3dxyxdtdyzydtdzzt y t x t zdt

3) The third order system of ODE :33.xx ydyx ydt

is equivalent to the following system of first order ODE in 123456,,,,,zz z z z z z

3124354651262.izzzzzzdzzdtzzzzzz

which is obtained by putting

2212 345622,,dx dy d x d yzx z y z z z zdt dt dt dt

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2.7 EXERCISES

1) Prove : A continuously differentiable 22:f

has the locally

Lipschitz property.

2) Given a continuous, 22

matrix valued function 2:AM

let

the time dependent vector field 22:f

be given by ,( )ft x A tx

for all 2,tx

prove that f is locally Lipschitz.

3) Give an example of a 22:f

which is continuous but not locally

Lipschtiz.

4) Obtain approximate solutions of the following initial value problems.

a) 212dxxy x xdt

213dyxy y ydt

b) 220 2dxxy xdt

34 0 3dyyx ydt

c) 1,0 2dxxdt y

103dyydt x

4) 34, 0 1dx dyxxdt dt

2,0 2dy dxyydt dt

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3

LINEAR SYSTEMS OF ODE (I)

Unit Structure :

3.1 Introduction

3.2 The Exponential of a Linear Endomorphism

3.3 Properties o f the Exponential

3.4 Exercise

3.1 I NTRODUCTIO N

We now consider following system of first order ordinary differential

equa tions in the function :( )xt x t

12,( ) , ( ) ,. . . , ( ) :nxt x t x t

111 1 12 21221 1 22 2211 2 2............nnnnnnnnn ndxax ax axdtdxax ax axdtdxax ax axdt

… (1)

where the coefficients ijaappearing on the right hand sides of the system (1)

are all constant real numbers. (The case in which () ,ij ijaa t t

will be

discussed in the next chapter). Writing Xt in the columnal form : 12()()()nxtxtXtxt

and collecting the coefficients ijain a matrix A, that is 1,ijAa i j n

, we

rewrite the system (1) in the matrix form : .dXAXdt

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For a given vector nw

with 12nwwww

we form the initial condition 0Xw

. Thus, we now have the initial value problem :

., ( 0 )dXAX X wdt

… (2)

Later on, we will consider an arbitrary 0t

and the initial value problem :

0., ( )dXAX X t wdt

… (2

)

obtained by bringing 0tin place of t = 0. The solution of this more general 2IVP

is easily obtained from the solution of (2). Therefore, we treat the

particular case (2) in detail first.

Here we are taking .fX A X

in our model differential equation dXfXdt

(the first order, autonomous case) treating the matrix ijAa

as

a linear transformation (= linear endomorphism) ofn

, its action on a vector nx

being given by

11 12 1121 22 2 212, ,...., ,........................, ,....nnnnn n naa axaa a xAxxaa a

11 1 12 2 121 1 22 2 211 2 2.............................................nnnnnnnnn nax ax axax ax axax ax ax

.

where 12nnxxXx

.

Thus in our treatment, the symbol A is made to play a double rule : (i) A as the nn

matrix and (ii) A as a linear transformation (= linear transformation) of n

.

Note that when n = 1 , the system (1) reduces to the single differential

equation dxaxdt

.

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The solution of the initial value problem ,dxaxdt

(0)X

ww

in this

one-dimensional case (i.e. n = 1) is the familiar function : ,tatw e t

.

Recall :

22 3311! 2! 3!tata t a t ae

The comparison between the on - dimensional initial value problem and

its n-dimensional case suggests that we expect the solution of the IVP(2) to be

a curve of the form

:, ( ) . . . . . . . ( * * )ntAXt X t e w

where tAe is an nn

matrix

which has the power series expansion :

22 33........1! 2! 3!tA t A t AI

suggested by (*) above. To carry forward the analogy, we call Ae, the

exponential of A. We will define the new quantity Ae first. (Replacing A by tA

for t

, will then yield tAe). Once this is accomplished, we will verify that

the curve :( )tAXt X t ew

is indeed the solution of the initial value problem

(2).

3.2 THE EXPO NENTIAL OF A LI NEAR E NDOMORPHISM

Let :nnA

be a linear endomorphism having its matrix ija

, we

choose a finite constant C satisfying the inequality : ()Ax Cx

for all nx

(e.g. 32max :1 ,ijCn a i jn

will do the job). Note that

the above inequality implies :

kkAx Cx

and kkkktA x t C x

for every nx

, every t

and every k

. In particular for the vector field :nnf

given

by .,nfx A x x

, we have :

()()fx fy A x A yAx yCx y

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for any x, y, in n

. In other words the vector field f(x)=Ax has the Lipschitz

property. Consequently the initial value problem (2) has a unique solution.

Next, we define a map :nnB

as follows : Let nx

be

arbitrary. Then we have : 23() ()().......1! 2! 3!Ax AxAxx

2.......1! 2!CCx C xxex

This shows that the infinite sum :

23() () ()1! 2! 3!Ax A x A xx

converges absolutely. We put :

2() ()()1! 2!Ax A xBx x

Thus the map B; e xpressed in terms of A, is given by 231! 2! 3!AA ABI

Note that each power kA is a linear transformation of n

and this

implies the linearlity of B. in fact, for any a, b in

, any x, y in n

, we have

222( ) ( ) ( ) ( ) .....1!2!( ) ( ) ( ) ( ) ....1! 2! 2! 2!AABa x b y a x b y a x b y a x b yab a bax by A x A y A x A y

22() ()() ()................1! 2!1! 2!Ax A xAy A yaxby

() ()aB x bB y

We adapt the notation Ae for B. occasionally we use the notation exp(A) for Ae. Thus 2() ()exp( ) ( ) ( ).....1! 2!AAx A xAx ex x

for every nx

.

Thus, each linear endomorphism A of n

gives rise to the linear

endomorphism Ae. Moreover, if t is any real number then tA is also a linear

endomorphism and it gives rise to the exponentia l tAe given by : MATHEMATICS______________________________________________________________________________________

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22 33.....1! 2! 3!tAtA t A t AeI

it being a linear endomorphism of n

where 22 33() () ().....1! 2! 3!tAtA x t A x t A xex x

for every nx

.

Having defined the exponential Ae, we obtain the solution of the IVP(2) in

terms of the exponentiation. Consider the map :nnX

given by ()tAxt e w

22() ()....1! 2!tA w t A wwt

Because the infinite series defining each X(t) converge s absolutely we

get that ()tX t

is differentiable and the derivative ()dXtdt is obtained by

termwise differentiation of the infinite series defining X(t). Thus we have 2232233() ( ) ( ) ( )0.......1! 1! 2!() () ().....1! 2! 3!()dX t A w tA w t A wdttA w t A w t A wAwAxt

Thus, we have : ()()dX tAxtdt

for every t

. Moreover we have : (0) 0 0.....Xww

.

This completes the proof that the map :nX

given by ()taxt e w

is a

solution of the IVP(2). We summar ize this observation in the following.

Theorem 1 : The map :nX

given by ()taxt e w

is the solution of the

initial value problem ., ( 0 )dXAX X wdt

. The proof of the following is self

evident :

Corollary : The curve :nX

given by 0()tt Axt e w

for all t

is

the solution of the initial value problem : 0.,( )dxAX xt wdt

.

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3.3 PROPERTIES OF THE EXPO NENTIAL

Following few properties help u s sum the infinite series defining Ae and get

the matrix of it.

I) If A is a diagonal matrix, say 12nA

then we have 12kkkknA

and therefore

211222211111!2!1Anne

211222231....21....1! 2!1....1! 2! 3!nnn

12neee

II) If abAba

thencos , sinsin cosAabbeebb

.

Proof : Let ai b

so that Re( ), ( )mab I

and

MATHEMATICS______________________________________________________________________________________

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Re ( )() R emmIAI

Moreover, 2Re ( ) Re ( )() R e () R emmmmIIAII

2222Re Re ( ) ( ), 2Re ( )2Re ( ), Re Re ( ) ( )Re ( )( ) RemmmmmmmmIIIIIIII

In general Re ( )( ) RekkmkkkmIAI

holding for every k

.

Consequently :

22222....1! 2!Re ( )Re ( )1011....( ) Re011!2!( ) ReAmmmmAAeIIIII

22322ReRe()Re 1....]]1! 2!1! 2! 3!()11.......] Re 1 Re Re ]1! 2!1! 2!mmmmmIIIII

2222ReRe()Re 1....] 0.......]1! 2!1! 2!()0.......] Re 1 Re Re .......]1! 2!1! 2!mmmmIIII

2222ReReRe 1....] 1...]12 !1! 2!()()(1)...] Re 1 Re 1 Re 1 ...]1! 2!1! 2!mm mmm mII III I

_____________________________________________________________________________________Chapter 3 : Linear Systems of ODE (I)

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2222Re(1...) 1...)1! 2!1! 2!1...) Re 1...)1! 2!1! 2!mmII

ReRecos sinsin coscos sinsin cosmmaaaaaeI eIe eebe beb e bbbebb

III) If A and B are linear endomorphisms of n

with the property AB BA

, then AB A B B Aee e e e

.

Proof :

The class ical binomial theorem applies to the powers ;kAB k

: 0KKjkjjJAB k A B

.

Therefore 01!mABmeABm

0001!!! () !km kmkmBmk m k

00!( ) !km kmkABkm k

0!!kmk mABke

0000!! ! !kkkkABAB A Bke k eee

It can be proved on similar lines that AB B Aee e

IV) Let Aa I B

where a is a real number and B is a strictly upper

triangular nn

matrix.

MATHEMATICS______________________________________________________________________________________

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12123211000nnnnbbbbb

So that 0nB

andaI B B aI

. Then we have

21...1! 2! ( 1)!AanBB Bee In

.

V) Let :nnA

be any linear transformation and let :nnB

be

an invertible linear transformation. Then 11BABAeB e B

.

Proof : For each k

we have 11kkBAB B A B

and therefore :

1211112121111....1!2!11.....1!2!....1! 2!BAB

Ae I B A B B ABB I B B AB B A BAABIBBe B

VI) We recollect here a few elementary facts of linear algebra culminating

in a formula relating two sets of coordinates on n

. These results will be used

in a conjunction with (IV) above to solve systems of linear ODE.

Let a linear transformation :nnA

have all real and distinct eigen-

values 12,. . . .n

with respective eigen-vectors 12,. . . .:1nii iff f A f f in

. Now, i

are all distinct implies that the

set 12,. . . .nff f

is a vector basis of n

. Thus we have two vector bases of n

now :

i) The standard vector basis 12,. . . .nee e

with 0, ....0,1, 0....0ie

and

ii) The 12,. . . .nff f

consisting of the eigen -vectors of A.

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Let the linear transformation :nnB

be given by 1iiBe f i n

. Clearly B is invertible. Putting 1nj ij iifb e

we get the

matrix ijb

it is the matrix of the linear map B (with respect to the standard

basis 1,....nee

of n

).

Now we note that the linear transformation 1BA B

has the set 1,....nee

as its eigen vectors with the respective eigen-values, ,....in

.

Consequently the matrix of 1BA B

with respect to the standard basis 1,....nee

is diagonalized.

121nBA B

.

Let 12,. . . .nyy y

be the coordinates on n

determined by the vector

basis 12,. . . .nff f

. As usual, 12,. . . .nxx x

are the Cartesian coordinates of n

- they are the coordinates determined by the standard basis 1,....nee

n

.

Now we have :

111nnyxByx

.

Where 1B

is the matrix 1ijb

Examples :

In this section we use the theory developed in the preceding sections to

solve linear systems of differential equations.

Example 1 :

a) Solve the following IVP. MATHEMATICS______________________________________________________________________________________

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2(0) 12 2 (0) 22(0) 3dxxyz xdtdyyz ydtdzzzdt

b) The same system of differential equations but take the initial conditions

: 11 , ( 1 )2 , ( 1 ) 3xyz

.

Solution :

a) We rewrite the IVP in the form :

2 1 1 (0) 1022(0) 20 0 2 (0) 3xxxdyyydtzzz

We have : 211 011022 2 002002 000I

Next, note that 2011 002002 000000 000

and

011 000002 000000 000k

for all 3k

.

Therefore, we have

22211011 002exp 0 2 2002 0002002000 000ttteIt

22101 200 1ttttet

According to the Theorem 1 we have :

_____________________________________________________________________________________Chapter 3 : Linear Systems of ODE (I)

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222211 5 3101 2 2 2 600 133ttxttttttyt etetzt

That is 2215 3txt e t t

2226ttyt e tzt e

To get the solution of (b) we apply the corollary to the Theorem 1 w hich

suggests that the variable t in above is to be replaced by t – 1 this gives the

solution of (b).

22121215 1 3 131ttxt e t tet t

212122126 164tttyt e tetzt e

Example 2 : Solve

5 2 , (0) 22 5 , (0) 3dxxy xdtdyxy ydt

Solution : We have :

52 ( 0 ) 225(0) 3xx xdyy ydt

According to property (II) of Section 3.3, we have

52 2exp253xttyt

5cos2 , sin 2 2sin 2 , cos23tttett

Therefore 52cos2 3sin2txt e t t

52sin2 3cos2tyt e t t

Example 3 : Solve MATHEMATICS______________________________________________________________________________________

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3 , (0) 14 , (0) 23 , (0)dxxz xdtdyyydtdzzz zdt

Solution : Because the middle equation is independent of x, z, we solve it

(taking into account the initial condition on it). This gives 42tyt e

.

Next, we deal with the coupled pair of the remaining equation and the

initial conditions on them :

13 ( 0 ) 131(0) 4xx xdzz zdt

This gives the sol ution :

cos( 3 ) sin( 3 ) 1esin ( 3 ) cos( 3 ) 4txtttttyt

cos3 , sin3 1sin3 , cos34cos3 4 sin3sin3 4 cos3ttttettttett

That is cos3 4 sin 3txt e t t

4cos3 sin3tzt e t t

.

Putting together all of them, we get

cos3 4 sin 3txt e t t

4tyt e

4cos3 sin3tzt e t t

3.4 EXERCISES

1) Compute the exponential of each of the matrixes.

(i) (ii) (iii) _____________________________________________________________________________________Chapter 3 : Linear Systems of ODE (I)

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300020005

100042004

300130013

2) Obtain th e matrix for ,tbet

for the B given below.

(i) (ii)

30 0 003 0 000 2 400 42B

200320102B

3) Solve the following initial value problems.

i) 34 1 2dxxy xdt

43 1 3dyxy ydt

31 4dzyz zdt

ii) 432 1 2dxxyz xdt

41 5dyyz ydt

41 6dzzzdt

iii) 32 1 4dxxy xdt

23 1 3dyxy ydt

4) Solve the initial value problem :

111222333444(0) 122 0 0(0) 122 0 0(0) 100 02(0) 100 1 2xxxxxxdxxxdtxxx

MATHEMATICS______________________________________________________________________________________

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4

LINEAR DIFFERE NTIAL EQUATIO NS

Unit structure :

4.1 Objectives

4.2 Introduction

4.3 The Second Order Homogeneous Equation

4.3.1 Homogeneous Equations with Constant Coefficients

4.3.2 Initial Value Problem for Second Order Equations

4.4 Linear Dependence and Independence of Solutions

4.4.1 Wronskian, a formula for the Wronskian

4.4.2 Abel’s Identity

4.5 The Second Order Nonhomogeneous Equations

4.6 The Homogeneous Equations of order 𝑛

4.7 Initial Value Problem for 𝑛௧ Order Equations

4.8 The Nonhomogeneous Equations of Order 𝑛

4.9 Exercise

4.10 Summary

4.1 OBJECTIVES

The learner will be able to:

solve homogeneous and non-homogeneous second order differential

equations.

check linear dependence and linear independence of set of functions.

learn definition of Wronskian and apply Wronskian to check linear

dependence of set of functions.

prove Abel’s Identity and solve homogeneous and non-homogeneous

𝑛௧ order differential equations.

4.2 I NTRODUCTIO N

A second order linear differential equation has the general form

𝑃(𝑥)𝑑ଶ𝑦

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦

𝑑𝑥+𝑅(𝑥)𝑦=𝐺(𝑥)−−−−(4.2.1)

where 𝑃,𝑄,𝑅 and 𝐺 are continuous functions .

There are two types of second order linear differential equation:

(a) Homogeneous linear equations: If in (4.2.1), 𝐺(𝑥)=0 then the

differential equation is called Homogeneous linear differential equation.

(b) Non-homogeneous linear equations: If in (4.2.1), 𝐺(𝑥)≠0 then the

differential equation is called Non-homogeneous linear differential

equation.

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4.3 THE SECO ND ORDER HOMOGE NEOUS EQUATIO N

Consider the homogeneous equation of the form

𝑃(𝑥)𝑑ଶ𝑦

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦

𝑑𝑥+𝑅(𝑥)𝑦=0−−−− ( 4 . 3 )

We say that 𝑦∗(𝑥) is a solution of (4.3) if it satisfies (4.3) that is

𝑃(𝑥)𝑑ଶ𝑦∗(𝑥)

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦∗(𝑥)

𝑑𝑥+𝑅(𝑥)𝑦∗(𝑥) = 0

Theorem 4.3 : If 𝑦ଵ(𝑥) and 𝑦ଶ(𝑥) are two solutions of the linear

homogeneous equation (4.3) and if 𝑐ଵ and 𝑐ଶ are any constants, then

𝑦∗(𝑥)=𝑐ଵ𝑦ଵ(𝑥)+𝑐ଶ𝑦ଶ(𝑥)

is also a solution of (4.3).

Proof. Since 𝑦ଵ(𝑥) and 𝑦ଶ(𝑥) are two solutions of (4.3), we have

𝑃(𝑥)𝑑ଶ𝑦ଵ(𝑥)

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦ଵ(𝑥)

𝑑𝑥+𝑅(𝑥)𝑦ଵ(𝑥) = 0

and

𝑃(𝑥)𝑑ଶ𝑦ଶ(𝑥)

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦ଶ(𝑥)

𝑑𝑥+𝑅(𝑥)𝑦ଶ(𝑥) = 0

Consider

𝑃(𝑥)𝑑ଶ𝑦∗(𝑥)

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦∗(𝑥)

𝑑𝑥+𝑅(𝑥)𝑦∗(𝑥)

=𝑃 ( 𝑥 )𝑑ଶ(𝑐ଵ𝑦ଵ(𝑥)+𝑐ଶ𝑦ଶ(𝑥))

𝑑𝑥ଶ

+𝑄(𝑥)𝑑(𝑐ଵ𝑦ଵ(𝑥)+𝑐ଶ𝑦ଶ(𝑥))

𝑑𝑥+𝑅(𝑥)(𝑐ଵ𝑦ଵ(𝑥)+𝑐ଶ𝑦ଶ(𝑥))

=𝑃(𝑥)ቈ𝑐ଵ𝑑ଶ𝑦ଵ(𝑥)

𝑑𝑥ଶ+𝑐ଵ𝑑ଶ𝑦ଶ(𝑥)

𝑑𝑥ଶ

+𝑄(𝑥)ቈ𝑐ଵ𝑑𝑦ଵ(𝑥)

𝑑𝑥+𝑐ଵ𝑑𝑦ଶ(𝑥)

𝑑𝑥+𝑅(𝑥)[𝑐ଵ𝑦ଵ(𝑥)+𝑐ଶ𝑦ଶ(𝑥)]

= 𝑐 ଵቈ𝑃(𝑥)𝑑ଶ𝑦ଵ(𝑥)

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦ଵ(𝑥)

𝑑𝑥+𝑅(𝑥)𝑦ଵ(𝑥)

+𝑐ଶቈ𝑃(𝑥)𝑑ଶ𝑦ଵ(𝑥)

𝑑𝑥ଶ+𝑄(𝑥)𝑑𝑦ଵ(𝑥)

𝑑𝑥+𝑅(𝑥)𝑦ଵ(𝑥)

=0

Thus 𝑐ଵ𝑦ଵ(𝑥)+𝑐ଶ𝑦ଶ(𝑥) is also a solution of (4.3).

Remark 4.3 : If 𝑦ଵ(𝑥) and 𝑦ଶ(𝑥) are two linearly independent solutions of

(4.3) and 𝑃(𝑥) is never 0 then the general solution is given by 𝑐ଵ𝑦ଵ(𝑥)+

𝑐ଶ𝑦ଶ(𝑥) where 𝑐ଵ and 𝑐ଶ are arbitary constants.

4.3.1 HOMOGE NEOUS EQUATIO NS WITH CO NSTA NT COEFFICIE NTS

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## Page 47

𝐴𝑑ଶ𝑦

𝑑𝑥ଶ+𝐵𝑑𝑦

𝑑𝑥+𝐶𝑦=0−−−− ( 4 . 3 . 1 )

where 𝐴,𝐵 and 𝐶 are constants and 𝐴≠0 is called homogeneous equations

with constant coefficients.

Let 𝑦=𝑒௫ then substituting 𝑦 in (4.3.1) we get

𝑒௫(𝑎𝑘ଶ+𝑏𝑘+𝑐)=0

As 𝑒௫≠0 we get that 𝑦=𝑒௫ is the solution of (4.3.1) if 𝑘 is the root of

𝑎𝑘ଶ+𝑏𝑘+𝑐=0−−−−(4.3.2)

The equation (4.3.2) is called the auxiliary equation (or characteristic

equation) of (4.3.1). The equation (4.3.2) is an algebraic equation in 𝑘

whose solution can be found using the formula .

𝑘ଵ=−𝑏+√𝑏ଶ−4𝑎𝑐

2𝑎, 𝑘ଶ=−𝑏−√𝑏ଶ−4𝑎𝑐

2𝑎.

Depending on the sign of √𝑏ଶ−4𝑎𝑐 we have the following cases:

Case (i): 𝑏ଶ−4𝑎𝑐>0

In this case the equation (4.3.2) has two real and distinct roots 𝑘ଵ and 𝑘ଶ

and the general solution of (4.3.1) is given by 𝑦=𝑐ଵ𝑒భ௫+𝑐ଵ𝑒మ௫.

Case (ii): 𝑏ଶ−4𝑎𝑐=0

In this case the equation (4.3.2) has only one distinct real root say 𝑘 and the

general solution of (4.3.1) is given by 𝑦=𝑐ଵ𝑒௫+𝑐ଵ𝑥𝑒௫.

Case (iii): 𝑏ଶ−4𝑎𝑐<0

In this case the equation (4.3.2) has two complex roots say 𝑘ଵ=𝑎+𝑖𝑏 and

𝑘ଶ=𝑎−𝑖𝑏 and the general solution of (4.3.1) is given by 𝑦=

𝑒௫(𝑐ଵcos𝑏𝑥+𝑐ଶsin𝑏𝑥).

Example 4.3.1.1. Solve the equation

𝑦ᇱᇱ+𝑦ᇱ−6𝑦=0

Solution . Here the auxiliary equation is given as 𝑘ଶ+𝑘−6=0

Thus, we have two distinct real roots 𝑘ଵ=2 ,𝑘ଶ=− 3. Hence the general

solution of the given equation is

𝑦=𝑐ଵ𝑒ଶ௫+𝑐ଶ𝑒ିଷ௫

Example 4.3.1.2. Solve the equation

4𝑦ᇱᇱ+12𝑦ᇱ+9𝑦=0

Solution . Here the auxiliary equation is given as 4𝑘ଶ+12𝑘+9=0

Thus, we have (2𝑘 +3)ଶ=0 which gives only one real root 𝑘=−ଷ

ଶ.

Hence the general solution of the given equation is

𝑦=𝑐ଵ𝑒ିଷ

ଶ௫+𝑐ଶ𝑥𝑒ିଷ

ଶ௫

Example 4.3.1.3. Solve the equation

𝑦ᇱᇱ−6𝑦ᇱ+13𝑦=0

Solution . Here the auxiliary equation is given as 𝑘ଶ−6𝑘+13=0 which

has no real roots. The complex roots of the auxiliary equation are given by

𝑘ଵ=3+2 𝑖 , 𝑘 ଶ=3−2 𝑖

and the general solution is giv en by

𝑦=𝑒ଷ௫(𝑐ଵcos2𝑥 +𝑐 ଶsin2𝑥).

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4.3.2 I NITIAL VALUE PROBLEM FOR SECO ND ORDER EQUATIO NS

An initial value problem for second order equation (4.2.1) is to find a

solution of the equation which also satisfies the initial condition 𝑦(𝑥)=𝑦

and 𝑦ᇱ(𝑥)=𝑦ଵ where 𝑦 and 𝑦ଵ are given constants.

For example, to find the solution of the initial value problem 𝑦ᇱᇱ+𝑦ᇱ−

6𝑦 = 0 with 𝑦(0)=1 and 𝑦ᇱ(0)=0, we first find the root of the auxiliary

equation 𝑘ଶ+𝑘−6=0 . From example (4.3.1.1) we know that the general

solution is given by

𝑦=𝑐ଵ𝑒ଶ௫+𝑐ଶ𝑒ିଷ௫

To satisfy the initial condition, we need

𝑦(0)=𝑐ଵ+𝑐ଶ=0

and

𝑦ᇱ(0)=2 𝑐ଵ−3𝑐ଶ=0

which gives 𝑐ଵ=ଷ

ହ and 𝑐ଶ=ଶ

ହ. Hence the solution of the initial value

problem is given by

𝑦=3

5𝑒ଶ௫+2

5𝑒ିଷ௫

Theorem 4.3.2[Uniqueness Theorem]: If 𝑃(𝑥) and 𝑄(𝑥) are continuous

functions on an open interval 𝐼 containing 𝑥, then the equation

𝑦ᇱᇱ+𝑃(𝑥)𝑦ᇱ+𝑄(𝑥)𝑦=𝑅 ( 𝑥 )

with initial condition 𝑦(𝑥)=𝑦 and 𝑦ᇱ(𝑥)=𝑦ଵ has a unique solution .

Proof. Let 𝑦 and 𝑦∗ be any two solutions of the equation 𝑦ᇱᇱ+𝑃(𝑥)𝑦ᇱ+

𝑄(𝑥)𝑦=𝑅 ( 𝑥 ) with initial conditions 𝑦(𝑥)=𝑦 and 𝑦ᇱ(𝑥)=𝑦ଵ. Let

𝑚=𝑦−𝑦∗. Then

𝑚ᇱᇱ+𝑃(𝑥)𝑚ᇱ+𝑄(𝑥)𝑚=0

and 𝑚(𝑥)=𝑚ᇱ(𝑥)=0

Consider, 𝐸(𝑥)=𝑚ଶ(𝑥)+൫𝑚ᇱ(𝑥)൯ଶ. Then 𝐸(𝑥)≥0 and 𝐸(𝑥)=0.

Differentiating 𝐸(𝑥) we get

𝐸ᇱ(𝑥)=2 𝑚(𝑥)𝑚ᇱ(𝑥)+2𝑚ᇱ(𝑥)𝑚ᇱᇱ(𝑥)

=2 𝑚ᇱ(𝑥)[𝑚(𝑥)+𝑚ᇱᇱ(𝑥)]

=2 𝑚ᇱ(𝑥)[𝑚(𝑥)−𝑃(𝑥)𝑚ᇱ(𝑥)−𝑄(𝑥)𝑚(𝑥)]

=− 2 𝑃(𝑥)ቀ൫𝑚ᇱ(𝑥)൯ଶቁ+2𝑚(𝑥)𝑚ᇱ(𝑥)[1−𝑄(𝑥)]

By Cauchy-Schwartz inequality,

𝑚(𝑥)𝑚ᇱ(𝑥)[1−𝑄(𝑥)]≤(1+|𝑄(𝑥)|)ቀ𝑚ଶ(𝑥)+൫𝑚ᇱ(𝑥)൯ଶቁ

Thus,

𝐸ᇱ(𝑥)≤− 2 𝑃(𝑥)ቀ൫𝑚ᇱ(𝑥)൯ଶቁ+2(1+|𝑄(𝑥)|)ቀ𝑚ଶ(𝑥)+൫𝑚ᇱ(𝑥)൯ଶቁ

=(1+|𝑄(𝑥)|)𝑚ଶ(𝑥)+(1+|𝑄(𝑥)|+2|𝑃(𝑥)|)൫𝑚ᇱ(𝑥)൯ଶ

Let 𝐾 ≥ 1+max ௫∈ூ{|𝑄(𝑥)|+2|𝑃(𝑥)|} . Then

𝐸ᇱ(𝑥)≤𝐾𝐸(𝑥)

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## Page 49

Suppose, there exist some 𝑥ଵ∈𝐼 such that 𝐸(𝑥ଵ)>0 with 𝑥ଵ>𝑥. Then

𝑑

𝑑𝑥൫𝑒ି௫𝐸(𝑥)൯=𝑒ି௫൫𝐸ᇱ(𝑥)−𝐾𝐸(𝑥)൯≤0

Thus, 𝑒ି௫𝐸(𝑥) is a decreasing function and hence 𝑒ି௫భ𝐸(𝑥ଵ)≤

𝑒ି௫బ𝐸(𝑥)=0. This implies that 𝐸(𝑥ଵ)≤0 which is a contradiction.

Thus, we get 𝐸(𝑥)=0 for all 𝑥∈𝐼 which implies that 𝑚(𝑥)=0 .

4.4 LI NEAR DEPE NDENCE A ND I NDEPE NDENCE OF SOLUTIO NS

A set of solutions {𝑓ଵ,𝑓ଶ,⋯,𝑓} of a differential equation is linearly

independent on an interval 𝐼 if and only if the only values of the scalars

𝑐ଵ,𝑐ଶ,⋯,𝑐 such that

𝑐ଵ𝑓ଵ(𝑥)+𝑐ଶ𝑓ଶ(𝑥)+⋯+𝑐 𝑓(𝑥)=0 , ∀ 𝑥∈𝐼

are 𝑐ଵ=𝑐ଶ=⋯=𝑐 =0 .

A set which is not linearly ind ependent is called a linearly dependent set.

4.4.1 WRO NSKIA N, A FORMULA FOR THE WRO NSKIA N

Let 𝑓ଵ,𝑓ଶ,⋯,𝑓 be functions in 𝐶ିଵ(𝐼) where 𝐼 is an interval. Then

Wronskian of these functions is defined by

𝑊[𝑓ଵ,𝑓ଶ,⋯,𝑓](𝑥)=ተተ𝑓ଵ(𝑥) 𝑓 ଶ(𝑥)⋯𝑓 (𝑥)

𝑓ଵᇱ(𝑥) 𝑓ଶᇱ(𝑥)⋯𝑓 ᇱ(𝑥)

⋮⋮ ⋮ ⋮

𝑓ଵ(ିଵ)(𝑥) 𝑓ଶ(ିଵ)(𝑥) ⋯ 𝑓(ିଵ)(𝑥)ተተ

For example, if 𝑓ଵ(𝑥)= sin𝑥 and 𝑓ଶ(𝑥)=c o s𝑥 on (−∞,∞) then

𝑊[𝑓ଵ,𝑓ଶ](𝑥)=ቚsin𝑥 cos𝑥

cos𝑥 −sin𝑥ቚ=− 1

Theorem 4.4.1 : Let 𝑓ଵ,𝑓ଶ,⋯,𝑓 be functions in 𝐶ିଵ(𝐼) where 𝐼 is an

interval. If 𝑊[𝑓ଵ,𝑓ଶ,⋯,𝑓] is non-zero at some point 𝑥∈𝐼 then

{𝑓ଵ,𝑓ଶ,⋯,𝑓} is linearly independent on 𝐼.

Proof. Consider 𝑐ଵ,𝑐ଶ,⋯,𝑐 such that

𝑐ଵ𝑓ଵ(𝑥)+𝑐ଶ𝑓ଶ(𝑥)+⋯+𝑐 𝑓(𝑥)=0

for all 𝑥∈𝐼.

Differentiating 𝑛−1 times, we get the system of equation

𝑐ଵ𝑓ଵ(𝑥)+𝑐ଶ𝑓ଶ(𝑥)+⋯+𝑐 𝑓(𝑥)=0 ,

𝑐ଵ𝑓ଵᇱ(𝑥)+𝑐ଶ𝑓ଶᇱ(𝑥)+⋯+𝑐 𝑓ᇱ(𝑥)=0 ,

⋮

𝑐ଵ𝑓ଵିଵ(𝑥)+𝑐ଶ𝑓ଶିଵ(𝑥)+⋯+𝑐 𝑓ିଵ(𝑥)=0

where the unknowns are 𝑐ଵ,𝑐ଶ,⋯,𝑐. The determinant of the coefficients of

this system is the Wronskian of the functions 𝑓ଵ,𝑓ଶ,⋯,𝑓. We know that i f

the determinant of the coefficients of a homogeneous system is non-zero,

then the system has a unique solution.

Thus if 𝑊[𝑓ଵ,𝑓ଶ,⋯,𝑓](𝑥)≠0 then the homogeneous system has a

unique solution which is 𝑐ଵ=𝑐ଶ=⋯=𝑐 =0 .

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4.4.2 ABEL’S IDENTITY

Let 𝑦ଵ(𝑥) and 𝑦ଶ(𝑥) be two solutions of 𝑦ᇱᇱ+𝑃(𝑥)𝑦ᇱ+𝑄(𝑥)𝑦=0. Then

the Wronskian is given by 𝑦ଵ(𝑥)𝑦ଶᇱ(𝑥)−𝑦ଵᇱ(𝑥)𝑦ଶ(𝑥). Then

𝑊ᇱ(𝑥)=𝑦ଵ(𝑥)𝑦ଶᇱᇱ(𝑥)−𝑦ଵᇱᇱ(𝑥)𝑦ଶ(𝑥)

Since 𝑦ଵ and 𝑦ଶ are solutions of the given equation, we have

𝑦ଵᇱᇱ(𝑥)=− 𝑃(𝑥)𝑦ଵᇱ(𝑥)−𝑄(𝑥)𝑦ଵ(𝑥)

and

𝑦ଶᇱᇱ(𝑥)=− 𝑃(𝑥)𝑦ଶᇱ(𝑥)−𝑄(𝑥)𝑦ଶ(𝑥)

Substituting this in 𝑊ᇱ(𝑥) we get

𝑊ᇱ(𝑥)=− 𝑃(𝑥)𝑊(𝑥)

That is the Wronskian satisfies the first order linear equation

𝑊ᇱ(𝑥)+𝑃(𝑥)𝑊(𝑥)=0

which on solving gives

𝑊(𝑥)=𝑊(𝑥)𝑒∫(௧)ௗ௧ೣ

ೣబ

This formula is known as th e Abel’s Identity.

4.5 THE SECO ND ORDER NONHOMOGE NEOUS EQUATIO NS

Theorem 4.5.1: If 𝑦(𝑥) is any solution of 𝑃(𝑥)𝑦ᇱᇱ+𝑄(𝑥)𝑦ᇱ+𝑅(𝑥)𝑦=

𝑆(𝑥). Then 𝑦∗(𝑥)=𝑦(𝑥)+𝑦(𝑥) is the general solution of 𝑃(𝑥)𝑦ᇱᇱ+

𝑄(𝑥)𝑦ᇱ+𝑅(𝑥)𝑦=𝑆(𝑥) where 𝑦(𝑥) is the general solution of 𝑃(𝑥)𝑦ᇱᇱ+

𝑄(𝑥)𝑦ᇱ+𝑅(𝑥)𝑦=0 .

Proof. Consider

𝑃(𝑥)𝑦∗ᇱᇱ+𝑄(𝑥)𝑦∗ᇱ+𝑅(𝑥)𝑦∗

=𝑃 ( 𝑥)ቀ𝑦ᇱᇱ(𝑥)+𝑦ᇱᇱ(𝑥)ቁ+𝑄(𝑥)ቀ𝑦ᇱ(𝑥)+𝑦ᇱ(𝑥)ቁ

+𝑅(𝑥)ቀ𝑦(𝑥)+𝑦(𝑥)ቁ

=൫𝑃(𝑥)𝑦ᇱᇱ(𝑥)+𝑄(𝑥)𝑦 ᇱ(𝑥)+𝑅(𝑥)𝑦 (𝑥)൯

+ቀ𝑃(𝑥)𝑦ᇱᇱ(𝑥)+𝑄(𝑥)𝑦 ᇱ(𝑥)+𝑅(𝑥)𝑦 (𝑥)ቁ

=0+𝑆 ( 𝑥)

=𝑆 ( 𝑥)

Thus, 𝑦∗(𝑥) is the solution of 𝑃(𝑥)𝑦ᇱᇱ+𝑄(𝑥)𝑦ᇱ+𝑅(𝑥)𝑦=0. Here 𝑦(𝑥)

is called the complementary function.

Remark: Thus, to solve a non-homogeneous differential equation, we just

need to find the particular solution 𝑦(𝑥).

The following table gives a list of particular solutions for some familiar

𝑆(𝑥). 𝑆(𝑥) 𝑦(𝑥) 𝑘𝑒௫ 𝐶𝑒௫ 𝑘+𝑘ଵ𝑥+⋯+𝑘𝑥 𝐶+𝐶ଵ𝑥+⋯+𝐶𝑥 𝑘cos𝑎𝑥 𝐶cos𝑎𝑥+𝐷sin𝑎𝑥 𝑘sin𝑎𝑥 𝐶cos𝑎𝑥+𝐷sin𝑎𝑥 _________________________________________________________________________________ 50MATHEMATICS_________________________________________________________________________________________munotes.in

## Page 51

Example 4.5.1.1: Solve the equation

𝑦ᇱᇱ+3𝑦ᇱ+2𝑦=4𝑥ଶ+1

Solution. We first find 𝑦(𝑥). Consider the auxiliary equation

𝑘ଶ+3𝑘+2=0

(𝑘+2)(𝑘+1)=0

Hence the complementary function is 𝑦=𝐶ଵ𝑒ିଶ௫+𝐶ଶ𝑒ି௫.

Now, from the above table the particular solution is 𝑦(𝑥)=𝐴 𝑥ଶ+𝐵𝑥+𝐶

where the constants 𝐴,𝐵 and 𝐶 are to be determined.

Consider

𝑦ᇱᇱ+3𝑦ᇱ+2𝑦=2 𝐴 𝑥ଶ+(6𝐴+2𝐵)𝑥+(2𝐴+3𝐵 +2𝐶 )=4 𝑥ଶ+1

Then comparing the coefficients of powers of 𝑥 we get

2𝐴 = 4

(6𝐴+2𝐵)=0

(2𝐴+3𝐵 +2𝐶 )=1

which on solving gives 𝐴=2 ,𝐵=− 6 ,𝐶=ଵହ

ଶ. Hence the general solution

of the given equation is

𝑦=𝐶ଵ𝑒ିଶ௫+𝐶ଶ𝑒ି௫+൬2𝑥ଶ−6𝑥+15

2൰

4.6 THE HOMOGE NEOUS EQUATIO NS OF ORDER 𝒏

A linear differential equation of order 𝑛 is an equation of the form

𝑃(𝑥)𝑦()+𝑃ିଵ(𝑥)𝑦(ିଵ)+⋯+𝑃 ଵ(𝑥)𝑦ᇱ+𝑃(𝑥)𝑦=0−−−( 4 . 6 )

is called a Homogeneous Equation of order 𝑛.

Remarks:

1. If 𝑓 is a solution to (𝐴) then 𝐶𝑓 is also a solution to (4.6).

2. If 𝑓 and 𝑔 are the solutio ns to (𝐴) then 𝐶ଵ𝑓+𝐶ଶ𝑓 is also a solution

to (4.6).

3. If 𝑓ଵ,𝑓ଶ,⋯,𝑓 are the solutions to (4.6) then 𝐶ଵ𝑓ଵ+𝐶ଶ𝑓ଶ+⋯+

𝐶𝑓 is also a solution to (4.6).

Example 4.6.1: Find th e general solution of 𝑦ᇱᇱᇱ+4𝑦ᇱᇱ−7𝑦ᇱ−10𝑦=0 .

Solution. Here the auxiliary equation is given by

𝑘ଷ+4𝑘ଶ−7𝑘−10=0

whose roots are given by 𝑘ଵ=− 1 ,𝑘 ଶ=2 ,𝑘ଷ=− 5. Thus, the general

solution is given by

𝑦=𝐶ଵ𝑒ି௫+𝐶ଶ𝑒ଶ௫+𝐶ଷ𝑒ିହ௫

where 𝐶ଵ,𝐶ଶ,𝐶ଷ are arbitrary constants.

4.7 I NITIAL VALUE PROBLEM FOR 𝒏𝒕𝒉 ORDER EQUATIO NS

An initial value problem for 𝑛௧ order equation (4.6) is to find a solution of

the equation which also satisfies the initial condition 𝑦(𝑥)=𝑦, 𝑦ᇱ(𝑥)=

𝑦ଵ,⋯,𝑦(ିଵ)(𝑥)=𝑦ିଵ where 𝑦,𝑦ଵ,⋯,𝑦ିଵ are given constants. _________________________________________________________________________________ 51_________________________________________________________________________________________Chapter 4 : Linear Differential Equationmunotes.in

## Page 52

Example 4.7.1: Find the solution of the initial value problem 𝑦ᇱᇱᇱ+4𝑦ᇱᇱ−

7𝑦ᇱ−10𝑦=0 with 𝑦(0)=− 3 ,𝑦ᇱ(0)=12 and 𝑦ᇱᇱ(0)=−36.

Solution. Here, the general solution is given by

𝑦=𝐶ଵ𝑒ି௫+𝐶ଶ𝑒ଶ௫+𝐶ଷ𝑒ିହ௫

Now, we use the initial conditions to determine the constants 𝐶ଵ,𝐶ଶ,𝐶ଷ.

Thus we get

𝐶ଵ+𝐶ଶ+𝐶ଷ=− 3

−𝐶ଵ+2𝐶ଶ−5𝐶ଷ=12

𝐶ଵ+4𝐶ଶ+25𝐶ଷ=−36

which on solving gives 𝐶ଵ=−ହ

ଶ,𝐶ଶ=1 ,𝐶ଷ=−ଷ

ଶ. Hence the solution to

the initial value problem is

𝑦=−5

2𝑒ି௫+𝑒ଶ௫−3

2𝑒ିହ௫

4.8 THE NONHOMOGE NEOUS EQUATIO NS OF ORDER 𝒏

General form of the 𝑛௧ order non-homogeneous equation is given by

𝑃(𝑥)𝑦()+𝑃ିଵ(𝑥)𝑦(ିଵ)+⋯+𝑃 ଵ(𝑥)𝑦ᇱ+𝑃(𝑥)𝑦=𝑓 ( 𝑥 )−−−( 4 . 8 )

where 𝑓(𝑥) is a given function.

Similar to second order non-homogeneous, the general solutions of (4.8) is

also given as

𝑦=𝑦+𝑦

where 𝑦 is the general solution of corresponding homogeneous equation

and 𝑦 is the particular solution of (4.8).

Example 4.8.1: Solve 𝑦ᇱᇱᇱ+4𝑦ᇱᇱ−7𝑦ᇱ−10𝑦=100𝑥ଶ−64𝑒ଷ௫

Solution. Here the general solution of 𝑦ᇱᇱᇱ+4𝑦ᇱᇱ−7𝑦ᇱ−10𝑦=0 is

given by

𝑦=𝐶ଵ𝑒ି௫+𝐶ଶ𝑒ଶ௫+𝐶ଷ𝑒ିହ௫

Now for 𝑓(𝑥)=100𝑥ଶ−64𝑒ଷ௫ we consider a trial function similar to the

structure of 𝑓(𝑥) as

𝑦=𝐴+𝐴ଵ𝑥+𝐴ଶ𝑥ଶ+𝐵𝑒ଷ௫

Substituting this to the given non -homogeneous equation, we get

(−10𝐴−7𝐴ଵ+8𝐴ଶ)+(−10𝐴ଵ−14𝐴ଶ)𝑥−10𝐴ଶ𝑥ଶ+32𝐵𝑒ଷ௫

=100𝑥ଶ−64𝑒ଷ௫

On comparing the coefficients, we get

(−10𝐴−7𝐴ଵ+8𝐴ଶ)=0

(−10𝐴ଵ−14𝐴ଶ)=0

−10𝐴ଶ=100

32𝐵=−64

which on solving gives

𝐴=−89

5,𝐴ଵ=14,𝐴ଶ=−10,𝐵=−2

Hence the general solution is

𝑦=𝐶ଵ𝑒ି௫+𝐶ଶ𝑒ଶ௫+𝐶ଷ𝑒ିହ௫+൬−89

5+14𝑥−10𝑥ଶ−2𝑒ଷ௫൰ _________________________________________________________________________________ 52MATHEMATICS_________________________________________________________________________________________munotes.in

## Page 53

4.9 EXERCISE

1. Solve 𝑦ᇱᇱᇱ+4𝑦ᇱᇱ−𝑦ᇱ−4𝑦=0

2. Solve 𝑦ᇱᇱᇱ+4𝑦ᇱᇱ−3𝑦ᇱ−18𝑦=0

3. Solve 𝑦ᇱᇱᇱ+6𝑦ᇱᇱ+12𝑦ᇱ+8𝑦=0

4. Solve 𝑦ᇱᇱᇱ−5𝑦ᇱᇱ−𝑦ᇱ+5𝑦=10𝑥−63𝑒ିଶ௫+29sin2𝑥

5. Solve 𝑦(ସ)+8𝑦ᇱᇱ+16𝑦=64𝑥sin2𝑥

6. Solve the initial value problem: 𝑦(ସ)+2𝑦ᇱᇱᇱ−2𝑦ᇱ−𝑦=24𝑥𝑒ି௫+

24𝑒௫−8sin𝑥,with 𝑦 (0)=− 2 ,𝑦ᇱ(0)=0 ,𝑦ᇱᇱ(0)=6 ,𝑦ᇱᇱᇱ(0)=

10

4.10 SUMMARY In this unit, solving homogeneous and non-homogeneous second order

differential equations are discussed. Linear dependence and linear

independence of set of functions is discussed. Definition of Wronskian and

applications of Wronskian to check linear depen dence of set of functions is

discussed. Abel’s Identity is discussed and methods to solve homogeneous

and non-homogeneous 𝑛௧ order differential equations are also discussed in

this unit.

References :

1. Differential Equations with Applications and Historical Notes by G.F.

Simmons.

2. Advanced Differential Equations by M.D. Raisinghania.

3. ODE and PDE by M.D. Raisinghania.

4. Schaum's outline of Differential Equations by Richard Bronson.

*****

_________________________________________________________________________________ 53_________________________________________________________________________________________Chapter 4 : Linear Differential Equationmunotes.in

## Page 54

/,1($56<67(062)2'(,,

8QLW6WUXFWXUH

5.1 Introduction

5.2 The Initial Value Problem

5.3 The solution of the Homogeneous Equation

5.4 The Inhomogeneous Equation

,1752'8&7,21

We consider a generalization of the type of systems studied in the

preceding chapter. The new systems to be studied will be inhomogeneous,

linear, first order systems with time dependant coefficients.

Throughout the chapter, I denotes an open interval. We consider a

family of continuous functions :

:, 1 ,:, 1aIij nijuIini

This family gives rise to the following system of non-homogeneous

ODE :

111 111221 12211( ) ...... ( ) ( )( ) ...... ( ) ( )( ) ...... ( ) ( )nnnnnnnn n ndXat X at X u tdtdXat X at X u tdtdXat X at X u tdt

… (1)

We also consider the same system but without the () :iut

_____________________________________________________________________________________ 54____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 55

111 11221 1211( ) ...... ( )( ) ...... ( )( ) ...... ( )nnnnnnnn ndXat X at XdtdXat X at XdtdXat X at Xdt

… (2)

We call (2) the homogeneous part of the system (1).

Our method of obtaining solutions of (1) consists of obtaining (i) a

particular solution of the inhomogeneous system (1), then (ii) obtain the space

of all solutions of the homogeneous system (2) and then combine (i) and (ii) to

get all the solutions of the system (1).

We use the following abridged notations to which the operations of

linear algebra will be applicable.

For each ,( )tI A t

is the nn

matrix (), ()ijat u t

denotes the column 1()()nutut

and as usual ()Xt is the column 1()()nxtxt

.

In terms of these notations the systems (1) and (2) take the following

compact forms :

() ()dxAt X utdt

… (1)

()dxAt Xdt

, the homogenous part of the above … (2)

7+(,1,7,$/9$/8(352%/(0

Given 00,ntI x

, we consider the IVP :

00() () . . . ( )dxAt X ut X t xdt

… (3)

Note that the vector field :nnfI

given by (, ) () ()ftx A tx y t

is locally Lipschitz : _____________________________________________________________________________________ 55MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 56

-XVWLILFDWLRQ Let 00,ntI x

be arbitrary. Choose 0

such that 00,tt I

. Now the map :nAI

being continuous on its

domain I, is bounded one the compact interval 00,tt t

and for any x,

y in 0,Bx

we get.

,( , ) ( ) ( ) ( ) ( )() ( )ft x fty A tx u t A ty u tAt x y

and therefore

,,().ft x fty A t x yAt x yKx y

For all 00,tt t

and for all x, y in 0,Bx

.

Therefore, the basic existence and uniqueness results are applicable. The

IVP (3) has a unique solution defined on the largest open sub-interval ,

of

I. We prove that ,

= I.

3URSRVLWLRQ

The solution of the IVP (3) is defined on the whole of I.

3URRI

(Sketchy, by contradiction method). Assume the contradictory : ,CI

, say,

right ha nd end point of I, so that 0,tI

.

Now, being the solution of the IVP (3) the curve :,nX

satisfies the integral equation :

000() () () ()ttttXt x A s Xsd s usd s

Using continuity of the maps :, :nnAI M uI

we get a finite

constant M such that () , ()As M us M

for all 0,st

and therefore,

we have :

000()() ()ttttXt x A s Xs d s us d s

_____________________________________________________________________________________ 56____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 57

00()ttxM X s d s M

.

By Gronwall’s lemma, we get :

0()Mxt x M e

for all 0,tt

. Thus, the set 0() : ,Xt t t

is a bounded subset of n

and therefore, the limit lim ( )tXt

exists. We call it ny

.

Having arrived at the po int y in n

, we consider the initial value

problem :

() ()dxAt X ut X ydt

.

Let :,0nX

be a solution of this IVP. Clearly the

two solutions :

:, ,: ,nnXX

agree on the overlap and

theref ore, they patch up to give a solution : :,nX

which

contradicts the assumed maximality of the interval ,

. Therefore, we must

have :

right hand end point of I. Similar reasoning leades us to

left

hand end point of I and therefor e ,I

.

Thus, every solution of (2) whatever be the initial condition is defined

on the whole of

.

7+(62/87,212)7+(+202*(1(286(48$7,21

We consider the set of all the solutions of the homogeneous equation

(2). Let the set be denoted by V.

3URSRVLWLRQ The set V has the structure of a n dimensional vector space.

3URRILet a, b in

, X, Y, in V be arbitrary. We prove that aX bY

also is

in V :

()ddX dYaX bY a bdtdt dtaAt X bAt Y

()At a X b Y

because ()Atis linear. _____________________________________________________________________________________ 57MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 58

Thus, ()daX bY A t aX bYdt

i.e. aX bY V

.

This shows that V is a real vector space. Actually V is isomorphic with n

, the brief explanation of which is as follows.

Choose 0tI

arbitrarily and hold it fix. For each nx

we consider

the unique solution of the initial value problem :

0() , ( )dxAt X X t xdt

.

We denote the unique solution of it by xX where we have attached the

suffix x to the solution xX to indicate the dependence of the solution on the

initial condition.

Now, we have an association rule xxX

associating the unique xXX

with each nx

. In other words, we have the map :

;nxVx X

… (4)

(Which associates each nx

, the element xX of V). It is easy to show that

this map is an isomorphism. First the linearity of the map : Let a, b in

, x, y

in n

be arbitrary. We consider the two curves :

:nxyaX bX I

and :nax byXI

.

It is clear that both are solutions of the IVP with the same initial

condition ax + by and therefore by the uniqueness of the solution, we get the

desired equality.

Clearly 0xX

implies 0x

. The implies that the linear map (4) is

injective. Finally, let X be any element o f V. Let 00Xt x

. Then 0xXX

showing that the map (4) is a sur jective map.

We have explained now that the map (3) is linear, it is injective and

surjective as well. Therefore (4) is a linear isomorphism between V and n

,

i.e. V is indeed n -dimensional real vector space.

We consider a vector basis 12.....nXX X

of the solution space V. We

call it a IXQGDPHQWDOV\VWHPRIVROXWLRQV of the homogeneous ODE (2). _____________________________________________________________________________________ 58____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 59

Clearly for each tI

, the vectors 12() , () , . . . . , () ,nXt X t X t

are linearly

independent vectors of n

. Putting them along the columns of a nn

maxtrix,

we denote the resulting nn

matrix by ()Wt thus :

12() () () ()nWt X t X t X t

or if the vector ()jXt has the

coordinates : 12() () , () , . . . . . . , ()jj j n jXt x tx t xt

then () (); 1 ,ijWt x t i j n

.

We call the resulting map :nWI M

a IXQGDPHQWDOPDWUL[RI

VROXWLRQV of the homogeneous part (2). Note that for each ,( )tI W t

is an

invertible matrix. Here is a simple example :

We consider the 2 dimensional case in which the 22

matrix ()At is

the constant matrix 23()32At

for all t

. It gives rise to the system of

homogeneous ODE :

2332dxxydtdxxydt

… (*)

Putting 2212cos3sin3(),( ),sin3cos3ttttXt e X t e ttt

, we get the

fundamental system 12,XX of solution space of (*) and the resulting

fundamental matrix 2:( )WM

given by 2cos3 , sin3()sin3 , cos2tttWt ett

f o r

all t

.

7+(,1+202*(1(286(48$7,21

We now consider the inhomogeneous ODE (1) and its solution space.

To begin with, we have the following result relating the solutions of the two

equations (1) and (2).

3URSRVLWLRQ Let :nYI

be a solution of the inhomogeneous system (1). _____________________________________________________________________________________ 59MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 60

a) If :nXI

is a solution of the homogeneous system (2) then XY

is a

solution of the inhomogeneous system (2).

b) Then the solution of the (inhomogeneous) initial value problem : 00() () , ( )dxAt X ut X t xdt

is given by 01() () () () ()ttXt Yt Wt Ws usd s

for all tI

.

The proof of the theorem is a straight forward application of the

fundamental theorem of integral calculus (applied to integration of vector

valued functions).

3URRI(a) We have 0100 0()ttXt Yt Wt Ws u sd s

00000 since * 0ttxW tdsx

(b) First note that

1212() () ()()() ()nnddWt X t X t X tdt dtdX tdX t dX tdt dtdt

1212() () () () () ()() () () ()() ()nnAt X t At X t At X tAt X t X t X tAt W t

Now, we have :

_____________________________________________________________________________________ 60____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 61

10() () () () ()ttddXt Y t W t W s u s d sdt dt

1100() () () () () () ()ttttdddYt Wt Ws usd s Wt Ws usdt dtdt

1110( ) ( ) ( ) ( ). ( ) ( ) ( ) ( ) ( )() () () () () ()tttAt Y t At W t W s us d s W t W t utAt Y t W t W s usd s ut

() () () .At xt ut

Thus ()() () ()dX tat x t utdt

all tI

and 00() .Xt x

,OOXVWUDWLYH([DPSOH

(We do not solve the example completely we indicate only a few steps leaving further

details for the reader to settle.)

Solve :

3 2 (0) 13 2 (0) 2dxxy xdtdyxy t ydt

6ROXWLRQ,QFRPSOHWH

We have : 2cos3 sin3()sin3 cos3tttWt ett

And 2cos3 sin3 1()sin3 cos3 2tttYt ett

2cos3 2 sin3.sin3 2 cos3tttett

There fore 2220cos3 sin 3 cos3 sin3 cos3 sin3 2().sin3 cos3 sin3 cos3 sin3 cos3ttstt t tssXt eeedstt t tss s

_____________________________________________________________________________________ 61MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 62

2220cos3 2 sin3 cos3 sin3 cos3 sin3 2sin 3 2cos 3 sin3 cos3 sin3 cos3tttstt ttsseeedstt ttss s

22cos3 2 sin3 cos3 sin3 *sin3 2cos3 sin3 cos3 **tttttteetttt

where in the last column, * 20(2cos3 sin3 )tses s s d s

and ** 020( cos3 2sin3 )tses s s d s

+,*+(525'(52'(

As usual, I denotes an open interval, for a natural number n, we consider a single

ODE.

121212() () . . . () ()nnnnnnndx d dat atat x b tdt dt dt

… (5)

in an unknown function :xI

the coefficients 12,,naa a

, b being smooth

functions on I. Equation (5) is OLQHDU because the left hand side of it is a linear

combination of ,, . . . .nndx d xxdtdt Again, if b 0

then we say that the equation (5) is

homogeneous.

Recall the initial valve problem for (5) is the following. Given to 01,,. . .nIx x x

all constant real numbers, find a n times continuously differentiable

function :,xJ

J being an open interval with 0tJ I

such that the

following two requirements are satisfied :

(i) 111()()() . . . . () () ()nnnnndx t d x tatat x t b tdtdt

for all tJ

and (ii) 01011.100()() , () . . .nnndx tdxxt x t xxdtdt

We will reduce the ODE (5) to a linear system of first order ODE and get

information of solutions of the former in terms of those of the reduced system.

Towards this aim, we consider the following object.

_____________________________________________________________________________________ 62____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 63

1,xnfndxdtYdxdt

(ii) 111 () 1()()nAtaatt

(iii) 0()()utbt

In the matrix A( t) in (ii) their being zeros at all the vacant places, including

the main diagonal, and the + 1 entries being just above the main diagonal and parallel

to it. Now, we consider the system

() ()dxAt x utdt

… (5

)

Along with its homogeneous part : ()dxAt xdt

… (6)

Note that the given (order n) ODE (5) is equivalent to the first order system

(5') while the homogeneous part of (5) is equivalent to (6). We recall the results of

the preceding sections obtained for the linear systems, now applicable to (5') which

we transcribe them so as to become applicable to the equation (5).

Thus we consider a fundamental system 1,nYY

of the solution space of

(6). This system yields functions 12,, . . . . :nxx x I

such that 12121211112111,.........nnnnnnnnnnxxxdxdx dxdtdtdtYYYdx dxdxdtdtdt

Now, we have the following important facts :

(1) 1,nYY

are linearly independent solutions of (6) implies 1...nxx are

solutions of the homogeneous part of (5). Moreover any solution x of the _____________________________________________________________________________________ 63MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 64

homogeneous part of (5) is expressible as a linear combination of the functions 12... .nxx x

11 2 2.nnxc x c x c x

(2) The solutions 1...nxx (of the homogeneous part of (5) are linearly independent

over I.

11 2 2... 0nndx dx d x

implies 12... 0.ndd d

This proves the following important :

3URSRVLWLRQ : The solution space of the homogeneous part of (5) is a n-dimensional

real vector space.

Now, given any set 1... :nff I

of n times continuously differentiable functions,

we associate with it the function :

1... ) :nWW f f I

given by : ( ) det ( )jjidfWttdt

for all tI

The function W is called the Wronskian of the family 1{. . . } .nff

Note that when the functions 1....nxx I

form a vector berries of the

solution space the matrix :

1(. . . ) :nWW x x I

given by W( t) = det ()jijdxtdt

for all t I

is the fundamental matrix of the

homogenous part (6) of (5')

$62/87,212)7+(121+202*(1(286(48$7,21

Suppose, a fundamental system 1{. . . . . . }nxx of solutions of the homogeneous

equation (6) is found. We discuss a method – attributed to Lagrange-which yields a

solution of the non- homogeneous ODE (5).

Recall, W( t) stands for the fundamental matrix with its (ij)th entry ()jijdxtdt.

For each i, 1in

, we consider the n

n matrix denoted by W j(t) obtained from

W(t) by replacing its jth column by the column _____________________________________________________________________________________ 64____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 65

()00tb

We adopt the notations : D(t) for det(W(t)) and Di(t) for det(Wi(t)).

Now, to obtain the desired solution we consider a function :xI

which is in the

form 11 2 2() () () () () . () ()nnxt v tx t v tx t t v t x t

= 1() ()njjjvt xt

,

where :( 1 )jvI jn

are unknown functions which are required to satisfy

a number of identities.

These identities will determine the functions jv which in turn specify the x which

will be the desired solution. Now differentiating x(t). We get

11jjjnnjjjdx dvdxvxdt dt dt

The first requirement on th e jv is :

10jjnjdvxdt

… (i)

So that we are left with

1njjidxdxvdt dt

… (*)

Now differentiating (*) above, we get

221122jnnjjjjjdx d v d xdxvdt dtdt dt

The second requirement on the jv is

10jnjjdv dxdt dt

… (ii)

Leaving us with 21222njjjdxdxvdt dt

… (**)

Continuing this procedure we get analogous ident ities, the requirement on jv at the

last stage being :

111.nnjnjjdv d xbdtdt

_____________________________________________________________________________________ 65MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 66

Thus, we have the following two strings of identities : 1111()njjjnjjjnnnjxjnnjnnjnjjxv xdxdxvdt dtIdxdvdt dtdv d xdtdt

1110()()njjjnjjjnnjjnjdvxdtdv dxdt dtIIdv d xbtdtdt

Multiplying the equations in (I) by an, an – 1,…., 1a, 1 and adding we get

1111()() () ()nnnnnndx d xatat x a t b tdt dt

thus showing that the function x(t) is a solution of the inhomogeneous equation.

On the other hand, using the simultaneous equations in (ii), we get :

()()jjdv D tdt D t

and therefore 0()()()tjjtDsvt d s j nDs

This leads us to the desired solution

1() () () .njjjxt v t x t t I

(;(5&,6(6

(1) Prove that each solution of the inhomogeneous equation (5) is defined on I.

(2) Prove : If a solution :xI

of the homogeneous equation (6) vanishes at same

t0 I

, then x

(3) Solve : 53( 0 ) 1dxxxdt

32 ( 0 ) 2dyxy t ydt

(4) Solve : 34( 0 ) 1dxxyt xdt

_____________________________________________________________________________________ 66____________________________________________________________________________________________________Chapter 5 : Linear Systems of ODE (II)munotes.in

## Page 67

34 ( 0 ) 1dyxy ydt

(5) Same D. E as above but x(1) = 1 , y(1) = 1.

_____________________________________________________________________________________ 67MATHEMATICS____________________________________________________________________________________________________munotes.in

## Page 68

0(7+2'2)32:(56(5,(6

8QLW6WUXFWXUH

6.1 Introduction.

6.2 Power Series. (A Quick Review) .

6.3 Method Of Power Series.

6.4 I llustrative Examples.

6.5 Legendre Equation, Lege ndre Polynomials.

6.6 Frobaneous Method.

Exerc ises.

,1752'8&7,21In this chapter, we study a type of second order ODE (scalar case) which

gives solutions in the form of absolutely convergent power series. These ODE

contain in their form, functions (e.g. the coefficient functions) which are analytic

in sense that they a dmit absolutely convergent power series expansions. Naturally

the method of solving such DE makes use of techniques and properties of

absolutely convergent power series. Therefore we call this method the PHWKRGRI

SRZHUVHULHV .

The reader will realize that this method applies not only to second order

linear ODE, it actually is applicable to a wider class of ODE of any order.

Recall, at the elementary level we could solve simple DE in terms of

elementary function such as the polynomials, the logarithm function, the

exponential functions, the trigonometric functions and so on. But soon we find

that things start going the opposite way : Differential equations generate new

functions as their solutions. Such functions are called VSHFLDOIXQFWLRQV . Most of

these functions are in the form of power series and as such are obtained by the

methods of power series. There is a more powerful method which is called

)UREDQHRXVPHWKRG . We discuss briefly this method also. Using this method we

introduce two special functions : (i) the Legendre polynomials and (ii) the Bessel

functions. We derive some of their properties.

We begin our treatment of special functions by recalling basic facts of

power series. 68________________________________________________________________________________________________________________________________________________________________________________________Chapter 6 : Method of Power Seriesmunotes.in

## Page 69

32:(56(5,(6$48,&.5(9,(: A SRZHUVHULHV is an infinite sum of the type:

a0 + a1(t – t0) + a 2(t – t0)2 + ….. …..+ a k(t – t0)k + ……….

=

……………….. (1)

where a 0, a1, a2, ………., a k, ……… are real constants . It is absolutely convergent

if there exists a r > 0 such that

The lub of all r > 0 satisfyi ng the above inequality is the UDGLXVRIFRQYHUJHQFH of

the power series (1), we denote it by R. If (1) is absolutely convergent with R as

its radius of convergence, it follows that for each

t (t0 – R, t 0 + R) the infinite sum

converges, giving rise the function :

f :(t0 – R, t 0 + R) —ℝ……………….. (2)

where

for each t (t0 – R, t 0 + R).

The function (2) is the VXPIXQFWLRQ of the (absolutely convergent) power

series (1). It is a basic result that the sum function (2) is infinitely differentiable on

its domain and the kth derivative ( kℤ+) is obtained by differentiating the infinite

series termwise . In particular we have :

Consequently the power series (1) becomes

f(t) =

………. (3)

An important implication of (3) is the following result :

If the functions 69____________________________________________________________________________________________MATHEMATICS____________________________________________________________________________________________munotes.in

## Page 70

f :(t0 – R, t 0 + R) —ℝ

g :(t0 – R, t 0 + R) —ℝ

admit the power series expansions :

then we have the following basic fact :

f(t) g(t)

if and only if a k = b k for all k ℤ+ that is, if and only if

=

holds for all k ℤ+

We make use of this basic result in what is to follow in this chapter.

0(7+2'2)32:(56(5,(6 We consider a second order ODE of the type :

………. (4)

where the coefficient functions P(t) and Q(t) admit absolutely convergent power

series expansions on an interval (-R, R) :

Recall, according to the theory of linear second order homogeneous ODE

discussed in Unit4, the solution space of the O DE (4) is a two dimensional vector

space.

Now, because the functions P(t), Q(t) admit absolutely convergent power

series expansions - we call such functions DQDO\WLF – we expect a solution of (4)

also to be analytic :

x(t) =

………. (6)

We prove below the following two results : 70________________________________________________________________________________________________________________________________________________________________________________________Chapter 6 : Method of Power Seriesmunotes.in

## Page 71

(a) Indeed, a solution t ı— x (t) of (4) has a power series expansion (6)

and obtain the constants c k, k ≥ 2 in terms of the constants a k, bk, k, l in

ℤ+ (The constants c 0, c1, will play the role of the arbitrary constants in

the solution of the second order ODE (4).)

(b) The infinite series (6) is absolutely convergent in the interval (-

R, R).

We proceed to prove these two claims.

Differentiating the power series (6) for the solution x(t), we get

for all t in (-R, R). Substituting these power series expansions along with

those for P(t), Q(t) in the given DE we get :

In above the coefficient of the power tn-2 for n ≥ 2 is :

n(n-1)c n + a 0(n-1)c n-1 + a 1(n-2)c n-2 + ………. + a n-2c1 + b0cn-2+ b 1cn-3 +

………. + b n-2c0

Equating it with zero we get the following succession of equations :

n(n-1)c n = -[a0(n-1)c n-1 + a1(n-2)c n-2 + ….. + a n-2c1]

- [b0cn-2 + b 1cn-3 + ………. + b n-2c0] ………. (*)

for n ≥ 2. These equations show that the constants (c n : n ≥ 2) can be

obtained recursively in terms of arbitrary constants c0, c1 and the given

constants a 0, a1, a2, ………., b 0, b1, b2, ………. Thus the solution (6) is

formally obtained. It remains only to prove that the formal series (6) is

absolutely convergent for |t| < R and hence determines a function x: (-R,

R) —ℝ which then becomes the solution of (2). Towards the

justification of this claim, we have :

n(n-1)|c n| ≤ |a0|(n-1)|c n-1| + |a 1|(n-2)|c n-2| + ………. + |a n-2||c1|

+ |b 0||cn-2| + |b 1||cn-3| + ………. + |b n-2||c1|

≤ (n-1) [|a 0||cn-1| + |a 1||cn-2| + ………. + |a n-2||c1| + …]

+ [|b0||cn-2| + |b 1||cn-3| + ………. + |b n-2||c0|] ………. (6) 71____________________________________________________________________________________________MATHEMATICS____________________________________________________________________________________________munotes.in

## Page 72

Now, let a number r satisfying 0 < r < R be arbitrary chosen. Also,

choose one more constant say S with 0 < r < S < R. By the absolute

convergence of the two series in (5) in (-R, R) and by the choice S < r, we

have

We choose a D > 0 such that

Consequently, we have |an| ≤

, |bn| ≤

for all n ℤ+. Next, we consider

an arbitrary m ℕ (to be fixed later) and for this m, another constant M,

(again, larger enough but finite) so that the following inequalities hold for

0 ≤ k ≤ m -1 :

|ck| ≤

………. (8)

Substituting the estimates (7), (8) in the inequality (6) we get :

m(m-1)|c m| ≤ (m-1)

+

= (m – 1)

+

≤ (m-1)

+

=

Therefore, we have

|cm| ≤

………. (**)

72________________________________________________________________________________________________________________________________________________________________________________________Chapter 6 : Method of Power Seriesmunotes.in

## Page 73

At this stage we fix m. It should be so large that the expansion in

the above last inequality (**) is ≤ 1. With this choice of m, we have |c m| ≤

. This inequality together with the inequalities (*) imply :

|ck| ≤

for0 ≤ k ≤ m. Now, application of principle of mathematical induction and

the inequality (6) together imply that, the inequalities (8) are true for all k

ℤ+. This ensures that the series (6) defining the function t ı—x(t) is

absolutely convergent for all t with |t| < r. Again, this is true for all r with 0

< r < R and therefore the series in (6) is absolutely convergent for all t (-

R, R). This leads us to the following :

Theorem 1 :The function

x : (-R, R) —ℝ

given by

wherec 0, c1 are arbitrary constants and the c k, k > 2 satisfying (*).

Remark :Of all the constants

in (6) the constants c 2,

c3, ………., c k, ………. are expressed in terms of the constants a k, bk, k

ℤ+,the last constants namely c 0, c1 remaining unspecified. They are the

two arbitrary constants of the second order ODE (4).

,/8675$7,9((;$03/(6(I) The DE :

. Here, P(t) 0 and Q(t) 10.

Let x(t) = c 0 + c 1t + c 2t2 + ………. be a solution of the equation. Then

we get

= 2.1c 2 + 3.2c 3t + 4.3c 4t2 + ………. and therefore 2.1c 2 +

3.2c 3t + 4.3c 4t2 + ………. + 10(c 0 + c1t + c 2t2 + ……) = 0 , that is :

(10c 0 + 2.1c 2) + (10c 1 + 3.2c 3)t + (10c 2 + 4.3c 4)t2 + …………… +

(10c k + (k+2)(k+1)c k+2)tk + ………. = 0

Therefore we get :

10c k + (k+2)(k+1)c k+2 = 0 for k = 2, 3, ………. This gives :

ck+2 =

for all k ℤ+.

This recurrence relation gives the following succession :

c2 =

c3 =

c 4 =

=

c5 =

=

, c6 =

=

, c 7 =

=

………… 73____________________________________________________________________________________________MATHEMATICS____________________________________________________________________________________________munotes.in

## Page 74

c2k =

………. c2k+1 =

……….

Therefore,

x(t) = c 0 + c1t + c 2t2 + c3t3 + c4t4 + ……….

= (c 0+ c2t2 + c4t4 + ……….) + (c 1t + c 3t3 + c5t5……….)

= c0

+ c 1

= c0

+

=

=

Where A = c 0, B =

are arbitrary constants.

(II) Solve

Solution : Let x(t) = c 0 + c 1t + c 2t2 + ……….

Now, we have :

= 2.1c 2 + 3.2c 3t + 4.3c 4t2+ ..... + (k + 2)(k + 1)c k+2tk + .....

and therefore,

2c1t + 2.2c 2t2 + ………. + 2kc ktk + ……….

4x(t) = 4c 0 + 4c 1t + 4c 2t2 + ………. + 4c ktk + ……….

Therefore,

(12c 2 + 4c 0) + (6c 3 + 6c 1)t + (4.3c 4 + 8c 2)t2

+ ………. + [(k + 2)(k + 1)c k+2 + 2(k + 2)c k]t4

+ ………. = 0

Equating the coefficients of the powers of t with zero, we get

c2=

, c3 = -c1, c4 =

, ………., c k+2 =

, ………. 74________________________________________________________________________________________________________________________________________________________________________________________Chapter 6 : Method of Power Seriesmunotes.in

## Page 75

This gives the solution :

x(t) = c 0 + c1t -

……….

(III) We consider here a first order O DE, the solution of which is to be

obtained following a similar procedure :

= 5x.

Assuming the solution to be the power series :

x(t) = c 0 + c1t + c2t2 + ………. + c ktk……….

We get :

= c1 + 2c 2t + 3c 3t2 + ………. + (k + 1)c k+1tk + ……….

Substituting these power series in the given differential equations, we

get :

c1 + 2c 2t + 3c 3t2 + ………. + (k + 1)c k+1tk + ……….

= 5c 0 + 5c 1t + 5c 2t2 + ………. + 5cktk + ……….

Equating the coefficients, we get :

c1 = 5c 0, 2c 2 = 5c 1, 3c 3 = 5c 2, ……….,kc k = 5c k-1, for all k ≥ 1 and

therefore c k =

for all k ≥ 1 .

This gives :

x(t) =

……….

= c0e5t.

/(*(1'5((48$7,216/(*(1'5(32/<120,$/6 For an arbitrary real number , we consider the differential equation

(1 – t2)

+ 2t

+ ( + 1) x = 0 ………. (9)

for |t| < 1. We rewrite it in the form

= 0 ………. (9')

The two equivalent forms (9) and (9') of the differential equation are called the

/HJHQGUHHTXDWLRQ involvingthe parameter . It is a particular case of the ODE

(4) in which P(t) =

, Q(t) =

both admitting power series solution in (-1,

1) i.e. R = 1. According to Theorem 1 the Legendre equation has a solution given

by an infinite power series converging absolutely in the interval (-1, 1). The

resulting function (which depends on the parameter ) is called the /HJHQGUH

IXQFWLRQ.

It can be proved that the coefficients c n, n ℤ+ in the expansion

75____________________________________________________________________________________________MATHEMATICS____________________________________________________________________________________________munotes.in

## Page 76

of the solution of (9) satisfy the recurrence relations :

cn =

for n ≥ 2. In particular, if the parameter takes an integral value say = m, then

(n - 1)(n - 2) - ( + 1) = 0 for n = m + 1 and consequently, c m + 1 = 0. This further

implies that one solution of the Legendre equation is a polynomial. Because the

Legendre equation is a homogeneous linear differential equation, the polynomial

solution of it is determined to within a multiplicative constant. A particular

polynomial solution of it denoted by P m(t) is the polynomial :

Pm(t) =

Pm(t) is the /HJHQGUHSRO\QRPLDORIGHJUHHP .

7+()52%$1(2860(7+2' We consider a homogeneous linear second order O DE of the type :

………. (10)

Like the ODE (9) it is more general than (4) because of the coefficients t2 of

and t of

. Again, the functions P(t) and Q(t) have the power series expansions :

P(t) = a 0 + a1t + a 2t2 + ……….

Q(t) =b 0 + b 1t + b 2t2 + ……….

both the power series being absolutely convergent in an interval (-R, R).

It turns out that the solution is in the form of power series with t = 0 as a

singular point of the solution. The method of getting a solution of (10) is called

the )UREDQHRXVPHWKRG . It is explained below.

We expect the solution of (10) to be a function of type :

x(t) = ts(c0 + c1t + c 2t2 + ……….) ………. (*)

wheres is a real number and c 0 is non-zero. We have therefore to find s, c 0, c1,

……….

Assuming the series (*) to be absolutely convergent in (-R, R) we consider

its derivatives :

………. (**) 76________________________________________________________________________________________________________________________________________________________________________________________Chapter 6 : Method of Power Seriesmunotes.in

## Page 77

………. (***)

Substituting these power series expansions for x(t),

,

in equation

(10) we get :

0

Because the factor ts ≠ 0, we get that the expansion within the bracket must

be identically zero. Therefore the coefficient of each power of t in above must be

zero. This gives the following succession of equations :

………. (****)

We solve these equations to get the values of c 0, c1, c2, ……….

To begin with, we consider the first equation in the set (****). Since c 0 ≠

0, we get

s(s – 1) + a 0s + b 0 = 0

This equation is called the LQGLFLDOHTXDWLRQ . This equation, which is a

quadratic equation in s, when solved it gives two values for s to be substituted in

the solution (*).

We then consider arbitrary c 0 ≠ 0 and using the succession of equations in

(****) we obtain c 0, c1, c2, ……….

The procedure described in above is applied in the next section where we

obtain a family of special functions called %HVVHOIXQFWLRQV .

%(66(/)81&7,216 For any p ℤ+we consider the ODE :

theODE being called the %HVVHO

VHTXDWLRQ . 77____________________________________________________________________________________________MATHEMATICS____________________________________________________________________________________________munotes.in

## Page 78

Clearly, the indicial equation of this D.E. is :

s(s – 1) + s – p2 = 0

It gives s ± p. Therefore, one solution of the Bessel's equation, denoted by J p(t) is

of the form :

Jp(t) = t p(c0 + c1t + c 2t2 + ……….)

where c 0 is an arbitrary constant. Taking c 0 =

we get

Jp(t) = tp

Substituting the power series expansion of J p(t),

,

, in

the differential equation (11) we get

s(s – 1)c 0 + (s + 1)sc 1t + (s + 2)(s + 1)c 2t2 + ……….

+ sc 0 + (s + 1)c 1t + (s + 2)c 2t2 + ……….

- p2c0–p2c1t – p2c2t2 - p2c3t3- ……….

= 0

Equating coefficients of powers of t gives :

[s(s – 1) + s – p2] c0 = 0

[(s + 1)s + s + 1 – p2]c1 = 0

[(s + n)(s + n - 1) + (s + n) – p2]cn + cn+2 = 0 n ≥ 2

The first of these equations in the indicial equation giving s = ±p, the second gives

c1 = 0 and the last equation gives c n=

. Therefore the Bessel functions are

given by

Jp(t) =

When s = -p, we get the relation :

cn =

for n ≥ 2.

EXERCISES : 78________________________________________________________________________________________________________________________________________________________________________________________Chapter 6 : Method of Power Seriesmunotes.in

## Page 79

Obtain solutions in the form of power series of the following D.E.

(i)

(ii)

(= constant)

(iii)

(iv)

(v)

(vi)

*****

79____________________________________________________________________________________________MATHEMATICS____________________________________________________________________________________________munotes.in

## Page 80

67850±/,289,//(7+(25<

8QLW6WUXFWXUH

.1 Introduction :

.2 The S turmian Bou ndary Value Problem :

.3 Vibrations of an Elastic String

.4 Unit End Exercises

,QWURGXFWLRQ

Recall, the initial value problem for an ordinary differentia l equation:

11(, , , . . . )kkkkdX d X d XftXdtdtdt

is to obtain a solution :XI

of it when the values :

10() , ()oodXXt X t vdt

,….,1011kkkdtvdt

of the solution and its

derivatives are prescribed at a single point to of its domain internal. A

%RXQGDU\ 9DOXH 3UREOHP (BVP), is another fundamental problem in the

theory of ODE in which the solution of an ODE is required to satisfy a number

of conditions at two points of its domain (the two points actually being the

boundary points of the domain interval.)

In this chapter, we will study an important type of boundary value

problems associated with a certain type of linear second order ODE; we call

the BVP the ³Sturm Liouille eigenvalue problem.´ The resulting theory is very

vast and makes use of results from functional analysis. Therefore, we only

outline the theory introducing the concepts and stating the results without

proof. We will illustrate the scope of the theory by using it to solve the

vibrating string problem.

7KHVWXUPDLQ%RXQGDU\9DOXH3UREOHP

In the following, I stands for the interval > a,b@. All the functions ()tX t

, ()tY t

, ()tZ t

etc appearing in the disc ussion are assumed to be

defined on internals containing I. _____________________________________________________________________________________ 80_______________________________________________________________________________________________Chapter 7 : Sturm-Liouville Theorymunotes.in

## Page 81

We use the following notations :

C(I) is the vector space of all continuous functions 1:, ( )XI R CI

is

the subspace of C(I) which are continuously differentiable on I while 2()CI

consist of those :XI

in C(I) which are twice continuously differentiable

on I.

To introduce the type of boundary value problems we want to discuss, we

consider the following data:

(I) The functions p, q, r : I

with the followi ng properties :

(i) p is continuously differentiable on I i.e. 1()pC I

(ii) q, r are continuous on I.

(iii) p(t) > 0 for all .tI

(II) Constants 12 1 2,, ,

with 2212O

and 2221O

(Note that 2212O

is equivalent to the property that of the pair 12

, at least one

is non -zero. The other inequality also has similar interpretation).

(III) Arbitra ry constants 12,.

Using the functions in (I) we construct the linear, second order ordina ry

differential equation :

()pX q X r

… (1)

i.e. ,.dd Xpt t qtX t rt t Idt dt

in the unknown function tX t

The equation : PX q X

… (2)

is the homogeneous part of it.

We require the solution function ()tX t

of (1) to satisfy the boundary

conditions :

121122()Xa X aXb X b

… (3)

Taking together the D E(1) and the boundary conditions (3) we get the pair

pX qX r

_____________________________________________________________________________________ 81MATHEMATICS_______________________________________________________________________________________________munotes.in

## Page 82

121122(),Xa X a Xb X b

… (4)

The pair (4) is said to constitute the 6WXUPLDQERXQGDU\YDOXHSUREOHP .

In above, taking 120, 0r

we get the KRPRJHQHRXV 6WXUPLDQ

ERXQGDU\YDOXHSUREOHP

PX qX

12120Xa X a Xb X b

… (5)

In above for every 2() ,XC I

we put

() ( ).LX p X q XpX p X qX

Note that 2()XC I

implies that L(X) is conti nuous. Thus we get the map

2:( ) ( )LC I CI

of the indicated vector spaces. Clearly L is linear.

Now for any 2,i n ( )XY C I we have ..LX tYt Xt L Y t

..dX tddYpt Y t X t tdt dtdt

for all .tI

We refer to this equality as the .Lagrange Identity Integrating

this identity over the inter val ,,ab we get .( ) .baLX t Yt Xt L Y t d t

..pb X bY b X b Y b

pa X aYa X aY a

… (*)

Moreover, if both the functions X, Y satisfy the boundary conditions in (5) then

it follows that the R.H. S. of (*) is zero and consequently we have :

.bbaaLx yYtd t XtLx td t

… (**)

We will explain more about this equality (**) at a later stage.

Here is a short list of properties of the spaces of solutions of the

boundary value problem (4) and its homogeneous part (5).

(1) A finite linear combination :

11 2 2...nnCX CX C X

of solution s 12,, ,nXX X

of the B.V.P. (5) is also a solution of the BVP (5).

_____________________________________________________________________________________ 82_______________________________________________________________________________________________Chapter 7 : Sturm-Liouville Theorymunotes.in

## Page 83

(2) If X and Y are solutions of the in homogeneous BVP (5) then the

difference X – Y is a solution of the homogeneous boundary value problem (5).

(3) If :XI R

is a solution solution of the BVP (5) and :YI

a

solution of the non -homogeneous B.V.P. (4) then X + Y is a solution of the

BVP (4).

(4) Finally, let :YI

be a fixed solution of the inhomogeneous BVP (4).

Then every other solution :YI

of (4) can be expressed in the form YXY

for a unique solution X of the homogeneous BVP (5).

At this stage we describe a condition which ensures a unique solution of

the boundary value problem (4). Towards this aim, recall that the solution

space of the second order, linear homogeneous ODE (2) is a 2 dimensional

vector space and we call a basis of this vector space a IXQGDPHQWDOV\VWHP of

the ODE.

We choose a fundamental system 12,XX of the ODE (2). Next, using

it and the constants1212,

, of the boundary conditions of (5) we form

the quantity.

11 1 1 2 112 1 1 2 121 1 2 2 222 1 2 2 2RX a X aRX b X bRX a X bRX b X b

and we consider the determinant 11 1221 22detRRRR

which we denote by W. Now

we have the following result.

7KHRUHP The boundary value problem (4) has a unique solution if and only

if 0.W

3URRI

Let 12,XX be a fundamental system of solutions of the DE(2). Using 12XX and the variation of constants formula we choose a particular

solution *:YI

of the differential equation (1). Now a general solution of

(1) has the form:

*11 2 2,Yt Y t CY t CY t t I

… (*)

C1 , C2 being some constants.

_____________________________________________________________________________________ 83MATHEMATICS_______________________________________________________________________________________________munotes.in

## Page 84

Now we consider (*) to be a solution of the BVP (6). Clearly, (*) is a solution

of BVP (6) if and only if the constants 12,CCsatisfy the following

simultaneous equations:

*11 1 1 1 2 1 2*22 1 2 12 2 2RY CR cRRC RC R

…(**)

where******11222,,RY Y a Y a RY Y b Y b

(here, of course *Y

being he derivative *dYdt).

Clearly the equations (**) are satisfied if and only if the matrix of the

coefficients of 12,CCin (**) is non-singular, that is, if and only if:

11 1221 22detRRWORR

Here is a simple illustrative case.

The boundary value problem :

2210(0) (0)dXXtdtXX X

,

being some constants.

We claim that the BVP has a unique solution.

In fact, here, we have 121

, 10,

21

, 12sin , cosXt t Xt t

is a fundamental system of solutions of the

homogeneous part. 220.dXXdt

Therefore, we get : 11 12 21 22,0 w h i l e 1RR R R

. This gives 10 .W

giving existence

and uniqueness of the solution. In fact 121c o s s i nXt C t C t

is a

general solution of the ODE in the BVP above Now 12(0) (0) .XX C C

and 2XC

give 21andCC

Therefore the unique solution of the above B.V.P. is 1sin .Xtcost t

Next, let the functions ,: ,pqI

the constants 12 1 2,,

and the

differential operator 2:,LC I CI

all be as in the pre ceding section. In

addition let :rI

be a continuous function with 0rt

holding for all .tI

For a real number

we consider the linear homogeneous ODE. _____________________________________________________________________________________ 84_______________________________________________________________________________________________Chapter 7 : Sturm-Liouville Theorymunotes.in

## Page 85

()LX r X

… (6)

along with the boundary conditions :

121200Xa X aXb X b

Thus we have the linear homogeneous boundary value problem :

121200pX pX q r XXa X a Xb X b

… ()

which involves the real parameter

.

A Value

of the parameter for which a non-zero solution XX

of ()

exists is called an eigen value

of the boundary value problem ().

Sturm – Liouville eigen-value problem consists of getting the get E

consisting of all the eigen values

, the corresponding eigen-functions X

and studying the function space C(I) in terms of the eigen functions. :XE

. We will state the main theorem, proving only a part of it, and use

it to solve the vibrating st ring problem in the next section.

Before going further, we consider the following three conce pts:

(1) The inner product ,:0,CI CI

given by ,,,batt t d t

in C(I)

(2) The norm :,CI o

given by ,f o r a l lCI

(3) The uniform norm ,CI o

given by lub _ : .ttI

Note that if .baCr t d t

, then we have C

……. ()

holds for all .CI

Now, for

, we consider the following two subspaces of C(I) :

(1) V

is the solution space of the (second order, linear, homogeneous)

ODE (6). _____________________________________________________________________________________ 85MATHEMATICS_______________________________________________________________________________________________munotes.in

## Page 86

(2) W

is the subspace of V

consisting of all the solutions of the

boundary value problem (). If 0W

then we call W

the eigen - space of the B.V.P () with

as its eigen - value..

We prove the following three properties of the spaces W

,

(I) Each W

is a SURSHU subspace of V

(Thus if 0W

then it

is a 1-dimensional subspace of V

(II) If 12,

are in ,

with 12,

and if 12,t h e n ,WW

with respect to the inner

product , !.

(III) There is a countable subset :kk

of real numbers such that 0kW

for each k

and 0W

if k

for any

.

We prove property (I) : choose a fundamental system 12,XXof

solutions of the equation (6). Their l inear dependence implies

det 1212Xt XtXtXt

for all tI

.

In particular we have

det 12120Xa XaXaXa

Now, if both 12XX were in W(

), then they would satisfy the boundary

conditions. In particular, 11 2 20Xa X a

and 12 220Xa X a

But 2212o

implies det 12120Xa XaXaXa

which contradicts the

above stipulation of linear independence. This proves that W(

) is a proper

subspace of V

(i.e. it is either the {0} subspace or it is one dimensional

subspace of V

.)

Property (II) follows by the property () , , ( )LX Y X L Y

for all X, Y

in C (I).

Property (III) is a consequence of the fact that ( C (I), ) is a separable

metric space.

Now we state without proof the main theorem. _____________________________________________________________________________________ 86_______________________________________________________________________________________________Chapter 7 : Sturm-Liouville Theorymunotes.in

## Page 87

7KHRUHP6WXUP± /LRQYLOOH

(1) The boundary value problem () has a non-zero solution only for a

countable (finite or denumerable) set of

i.e. the solution spaces W

are non -trivial only for a countable collection of real numbers

.

L et 12kE

be the subset of

consisting of those

such that 0i f a n d o n l y i fkW

for some .kE

(k

are the

eigen values of the Sturm-Liouville problem and kW

are the eigen

spaces).

(2) The set E has no limit point.

(3) Each kW

is a 1 dimensional subspace of 2CI. For each kE

we

choose kkXW

with 1.kX

Now we have ..kkWX X

(4) If ,t h e n,0 .ke k eke X X i eXX

(5) If 2XCI

then,kkkXt Xt X t X t

for all tI

where the convergence of to infinite series to X(t) is uniform in tI

.

(6) If () i s s u c h t h a t , 0kXC IXX

holds for all kXthen X Ł 0.

The proof of this theorem makes use of the properties of a compact

operator on a separable Hilbert Space and as such it is to be studied from a

suitable advanced text -book on ODE (which usually refer to text-books of

functional analysis).

We state here two more results without proof :

7KHRUHP6WXUP/LRXYLOOH6HSDUDWLRQ7KHRUHP

The zeros of two linearly independent solutions of L(X) = 0 separate each

other.

Thus if 12Xa n d X are two independent solutions of L(X) = 0, their

between two consecutive zeros of 1Xis a zero of 2X and between two

consecutive zeros of 2X is a zero of 1X.

7KHRUHP7KH&RPSDULVRQ7KHRUHP

Consider two eigen-value problems of Sturm Liouville with the respective date ,,,,pqr

and *** **,,,,pqr

over the same interv al I = >a, b@. _____________________________________________________________________________________ 87MATHEMATICS_______________________________________________________________________________________________munotes.in

## Page 88

If *,pp

*,qq

*rr

in I and *0,

*0

hold with strict inequality in at least one place, then the corresponding

eigenvalues satisfy *nn

for all n.

In the next section, we study the dynamics of an elastic vibrating string in

which we will use the results of the Sturm -Liouville Theorem.

9,%5$7,2162)$1(/$67,&675,1*

The vibrating motion of a stretched elastic string is governed by a partial

differential equation called the (one-dimensional) wave equation . Wave

equation is solved by a method called method of separation of variables. The

resulting analysis makes use of the Sturm Liouville theory. We will therefore

study the problem of the vibrating elastic string as an application of the Sturm-

Liouville Theory.

We first explain the PDE, the wave equation of the vibrating string.

A string of natural length L is held horizontally along the X-axis of a vertical

XOY -plane. Its ends, A, B remain tied to the points (0, 0) and ( L, 0)

respectively . The string is plucked slightly and then is set in motion in such a

way that each point C of the string vibrates vertically. We study the vibrating

motion of the string as the collective vertically oscillating motion of each point

C of the string.

Therefore we consider an arbitrary point C of the string Let ,lA C x

(The real number x and the point C determine each other and therefore we may

take of ³the point x´ instead of ³the point C´.)

Let at an instant (, )tY t x

be the instantaneous y-coordinate of the

point C (Since the oscillatory motion of C is only in the vertical direction, x-

coordinate of C remains constant.) Therefore the (oscillatory) motion of the

point C is described by the function (, )tY t x

and the motion of the whole

string is given by the function ,,1tx Ytx t x

… ()

Now, the basic equations of motion enable us to derive the equation

222220,YYct x Ltx

…(10)

satisfied by the function Y(t, x). Equation (10) is the wave equation which

is satisfied by the vibrating string. In (10) c is a constant determin ed by the

mass of the string its elastic properties and the gravitational constant.)

_____________________________________________________________________________________ 88_______________________________________________________________________________________________Chapter 7 : Sturm-Liouville Theorymunotes.in

## Page 89

Suppose the string was plucked slighlty and released with initial velocity

(= initial velocity of each point C). so that the string executes the vibration

motion as described above. We consider two continuous functions :

:0 , , :0 ,fL g L

describing initial position and initial

velocity of the string, that i s :

0, , 0, , 0 .YYx f x xg x x Lt

Now, we have the initial boundary value problem for the function ,, ;tx Ytx

222220, 0YYct x Ltx

…(10)

,0 0 , 0,Yt YtL t

0,0, , 0 .YYx f x xg x x Lt

We want to find out the fun ction ,tx Yt

To begin with, we consider all the solutions of the wave equation which are of

the type X(x). T(t):

,.Ytx XxTt

Now, 22....22,,,YYtx Xx Tt X x Tttx

(the dots indicating

differentiation (twice) with respect to the appropriate variable)

Now the equation takes the form

....2Xx Tt cXx Tt

Assuming 0, 0 0, 0 ,Xx Tt f o r t x L

we get

.. ..21Tt XxTt Xxc

…(11)

This shows that the common value in (11) is independent of t, x i.e. it must be a

constant say d.

....21Tt XxdTt Xxc

…(12)

Now if 0d

we would get ..22with 0.Tt cdTt cd

If 0d

, we

would get T(t) = At + B for some constants a

, B (A is not zero because,

otherwise Tt B

which gives Y(t, x) = B. X(x) implying that the motion of _____________________________________________________________________________________ 89MATHEMATICS_______________________________________________________________________________________________munotes.in

## Page 90

the string is independent of t ie the string is station any). On the other hand, if

d ! o, we would get

..dc t dc tTt A e B e t

In either case (i.e d ! 0, or d = 0) the factor T(t) becomes unbounded as t

ranges in 0,

. This renders Y(t, x) also unbounded (ie the string stretching

limit less, another physical impossibility). Therefore we are left with the

possibility d 0. We put 2ford

. Then the above ODE (12) take

the form :

....222,Tt c Tt Xx Xx

.. (13)

We consider the second equation :

..2,0 .Xx Xx x L

Its general solution is :

cos sinXx x x x

,

being constants. This gives: ,c o s s i nYtx Tt x x

with the condition 0Tt

. Now 0, 0 .Yt x Tt

implies 0

and therefore ,. s i n w i t h 0 ,Yt x Tt x Tt

0.

But we have Y(t, L) = 0 and therefore sin 0L

which implies that Lkf o r k

. Therefore the parameter

can take the values ,,kkkL

.

This shows that :

.c o ssinkkkcLcLTt T tttkk

and ,kLX x X x Sin x kk

.

Thus, we get a sequence of solutions :

.c o s .s i n s i nkkkcLcL LYt xttxkkK

for k

, X,kk

being arbitrang constants. Now the general solution Y(t, x)

will be a linear combination of all of them.

,. c o s . s i n s i nkkkcLcL LYt xttxkkk

Differentiating the infinite series partially with respect to t gives : ,.s i ncos sinkkkYCLCLCL Ltxttxtkkkk

_____________________________________________________________________________________ 90_______________________________________________________________________________________________Chapter 7 : Sturm-Liouville Theorymunotes.in

## Page 91

In particular, we have

0, sin0, sinkkkkLfx Y xxkYLgx xxtk

These are nothing but the Fourier expansions of the given functions f(x), g(x), ,kk

being their Fourier coefficients which are calculated using the standard

trigonometric identities. Now we have the solution of the vibrating string

problem:

,. c o s.s i n s i nkkkCLCL LYt xttxkkk

8QLW(QG([HUFLVHV

Find the eigen values and the eigen functions of the following boundary value

problems :

(1) ..00 0 , 1XX X X

(2) ..00 0 ,00 .XX X X L f o r L

(3) ..0, 0 0 0 0.XX X X Lf o r L

(4) .1() 0 1 0 0tX t X X e

(Hint : Try ,rter

)

(5) ..2210 , 0 0tteXe X X X

(Hint: TaketXe u

)

(6) ..1310 , 1 0 , 0 .tX tX X X e

_____________________________________________________________________________________ 91MATHEMATICS_______________________________________________________________________________________________munotes.in