A STUDY ON REGULAR PERTURBATION

PROBLEMS

By

Shareena . P. R

( M.Sc Mathematics)

CONTENT

TITLE PAGE NUMBER

PREFACE 1

CHAPTER 1 2

CHAPTER 2 5

CHAPTER 3 9

GOBLIOGRAPHY 26

1

PREFACE

The book “A study on regular perturbation problems” is intended for the PG

students in kerala university. In this book all the topic have been deal within a

simple and lucid manner. A sufficiently large number of problems have been

solved by studying this book , the student is expected to understand the concept of

regular perturbation, the fundamental ideas of perturbation. To do more problems

involving the regular perturbation and fundamental ideas of perturbation.

Suggestion for the further improvement of this book will be highly

appreciated.

Shareena . P.R.

2

CHAPTER 1

Defintion 1.1

Pertubation theory is the study of the effect of small distrurbance if the effect

are small, the distrurbance or perturbations are said to be regular, otherwise they

are said to be singular

Defintion 1.2

Asympotic Sequence

A set of function {∅n(휀)}n = 0,1,2…is an asymptotic as

휀 → 0푖푓, 푓표푟푒푎푐ℎ푖 > 0,∅nti(휀) = 0 (∅n(휀))as 휀 → 0, that is each subsequent

term gets smaller.

Examples

(a) {1, 휀, 휀2, 휀3,…}

(b) {1, 휀 / , 휀 / ,…}

(c) {1, 휀, 휀log 휀, 휀,휀2log휀,…}

Defintion 1.3

Asymptotic expansion

3

Define an asymptotic series,

y= yo+휀y1 +휀2y2+…,

where y1, y1, y2, … are sufficiently smooth functions.

The standard asymptotic sequence is {1,휀, 휀2, 휀3…} as 휀 0 and fn(x)

represents the members of asymptotic sequence then fn+1(휀) = 0(fn(휀)) as xa

that is → ( )( )

= 0. The general expression for asymptotic expansion of

function fn(휀) is the series of terms

f(x)=∑ 푎 푓푛(휀) + 푅 푎푠휀 → 0,푤ℎ푒푟푒푎 푎푟푒푐표푛푠푡푎푛푡푠푎푛푑

푅 = 푂(푓 (휀))푎푠휀 → 0푎푛푑 lim → RN=0.

Definition 1.4

The expression f(x) =∑ 푎 fn(휀) + RN ,where f(x;휀) depends on an

independent variable x and small parameter 휀. The coefficient of the gauge

function fn(휀) are functions of x and the remainder term after N terms is a function

of both x and 휀 is RN = O (fn+1(휀)) is said to be uniform asymptotic expansion, if

RN Cfn+1(휀), where c is the constant.

Example

f(x,휀)= = 1+휀sinx+휀2(sinx)2+…as 휀0.

The remainder term RN = 1+휀sinx+휀2(sinx)2+…-∑ 휀 (푠푖푛푥) ,

where → ( ) = (sinx)N +1

4

Defintion 1.5

The expression f(x) = ∑ an fn (휀) + RN, where f(x;휀) depends on an

independent variable x and small parameter ∈ is said to be non – uniform

asymptotic expansion, if there is no constants exists but the relation RN≤ Cfn+1 (휀)

satisfied is known as non-uniform asymptotic expansion

Example

f(x,휀)= = 1+휀x+휀2(x)2+…as 휀0.

The remainder term RN = 1+휀x+휀2(x)2+…-휀n(x)n,

Where → ( ) =(x)N+1. There is no fixed constant C exists such that

RN≤ 푐휀N+1.

5

CHAPTER 2

The Fundamental Ideas of Perturbation

2.1 Definition Of Regular And Singular Perturbations

Definition 2.1.1.

The problem which does not contain any small parameter is known as unperturbed

problem.

Example 2.1.1

(a) x2 + 3+ 1 = 0.

(b) +2 + y = 2x2 – 8x + 4, y(o) = 3, (0) = 3.

Definition 2.1.2. The problem which contains a small parameter is known as

perturbed problem.

Example 2.1.2.

(a) 휀x2 — x+휀 = 0.

(b) +y = 휀y2, y(0) = 1.

6

Depending upon the nature of perturbation, a perturbed problem can be divided

into two categories. They are,

1. Regularly perturbed

2. Singularly perturbed

Definition 2.1.3. The perturbation problem is said to be regular in nature, when the

order (degree) of the perturbed and the un-perturbed problem are same, when we

set 휀 = 0. Generally, the parameter presented at lower order terms. The following is

an example of regularly perturbed problem.

Example 2.1.3

(a) x2 -1 = 휀x

(b) ) + y =휀y2, y(0) = 1, (0) = -1

Definition 2.1.4. The perturbed problem is said to be singularly perturbed, when

the order (degree) of the problem is reduced when we set 휀 = 0. Generally, the

parameter presented at higher order terms and the lower order terms starts to

dominate. Sometime the above statement is considered as the definition of

singularly perturbation problem. The following is an example of singularly

perturbed problem.

Example 2.1.4

(a) 휀x2 -3x+8=0

(b) 휀 + = 2푥 + 1, y(0) = 0, y(1) = 4

7

2.2 The Fundamental Theorem Of Perturbation Theory

If A0, A1휀+…+AN휀N + O(휀N+1) = 0 for 휀 sufficiently small and if the coefficients A0,

A1… are independent of 휀 , then

A0 = A1 = …= AN = 0

2.3 Order Symbols

The letters O and o are order symbols. They are used to describe the rate at

which the function approaches to limit value.

If a function f(x) approaches to a limit value at the same rate of another function

g(x) at x x 0, then we can write f(x) = O(g(x)) as x —> xo. The functions are

said to be of same order as x x 0 . We can write it as,0

limxx

( )( )

= C where C is

finite. We can say here “f is big – oh of g”. If the expression f(x) = o(g(x)) as

the x x 0 means 0

limxx

( )( )

= 0. We can say here “ f is little – oh of g” x x 0 and

f(x) is smaller than g(x) as x x 0.

Example 2.3.1

(a) sin x = O(x) as x→ 0 since

0lim

xx = 1.

8

(b) = o(x) as x→ ∞.

(c)Sin x2 = o(x) as x →0 because 0

limxx

= 0.

(c) 3x+x3 = O(x) as x →0 since 0

limxx

= 3.

(e) e-x = o( ) as x→ ∞.

(f) sin(2x) = O(x) as x →0.

(g) x+e-x = O(x) as x →∞.

“Big-oh" notation and "Little-oh" notation are generally called Landau

symbols. The expression f(x)~ g(x) as x→x0 means 0

limxx

( )( )

= 1 is called "f is

asymptotically equal or approximately equal to g".

9

CHAPTER 3

Regular Perturbation problems

Very often, a mathematical problem cannot be solved exactly or, if the

exact solution is available, it exhibits such an intricate dependency in the

parameters that it is hard to use as such. It may be the case, however, that a

parameter can be identified, say 휀 , such that the solution is available and

reasonably simple for 휀 = 0. Then, one may wonder how this solution is altered for

non-zero but small 휀. Perturbation theory gives a systematic answer to this

question.

3.1 Solution Of Algebraic Equations.

Example 3.1.1

Consider the quadratic equation

x2-1 = 휀푥 (3.1)

The two roots of this equation are

x1 = + 1 + , 푥 = − 1 + (3.2)

For small , these roots are well approximated by the first few terms of their Taylor

series expansion (see figure 1)

푥 = 1 + + + 푂(휀 ), 푥2 = −1 + - + 푂(휀 ). (3.3)

10

Can we obtain (3.3) without prior knowledge of the exact solutions of (3.1)?.Yes,

using regular perturbation theory. The technique involves four steps.

Assume that the solution(s) of (3.1) can be Taylor expanded in varepsilon. Then

we have

x=X0+ 휀X1 + 휀2 X2 + O(휀3) (3.4)

for X0, X1, X2 to be determined.

Substitute (3.4) into (3.1) written as x2 - 1 - 휀 X = 0, and expand the left hand side

of the resulting equation in power series of 휀. Using

x2 = 푋 + 2휀X0X1 +휀2 (푋 + 2X0X2) + O(휀3), (3.5)

휀x = 휀X0 + 휀2 X1+O(휀3)

11

Figure 1: The root x1 plotted as a function of (solid line), compared with the

approximations by truncation of the Taylor series at O (2) , x1 = 1+ (dotted line),

and O (3), x1 = 1+ + (dashed line). Notice that even though the

approximations are a priori valid in the range <<1 only, the approximation

푥 = 1+ + is fairly good even up to = 2.

this gives

푋 – 1+ (2X0X1 –X0)+ 2 (푋 +2X0X2 ─X1) + O( 3)= 0 (3.6)

Equate to zero the successive terms of the series in the left hand

side of (3.6):

O (0) : 푋 ─1 = 0,

O (1) :2X0X1 – X0 = 0, (3.7)

O (2) : X2

1 + 2X0X1─X1 = 0,

O (3) : …

Successively solve the sequence of equations obtained in (3.7). Since X20-1=0 has

two roots, X0 = ±1, one obtains

X0 = 1, XI = , X2 = (3.8)

X0 = -1, XI = , X2 =

12

It can be checked that substituting (3.8) into (3.4) one recovers (3.3)From the

previous example it might not be clear what the advantage of regular perturbation

theory is, since one can obtain (3.3) more directly by Taylor expansion of the roots

in (3.2). To see the strength of regular perturbation theory, consider the following

equation

X2-1 = 푒 (3.9)

13

Figure 2: The solid line is the graph of two of the three solutions of (3.9) obtained

numerically and plotted as a function of (solid line). Also plotted are the

approximations by truncation of the Taylor series at O( 2), x1=1+∈ (dotted line),

and O( 3), x1= 1+ + (dashed line).

The solutions of this equation are not available; therefore the direct method

is inapplicable here. However, the Taylor series expansion of these solutions can

be obtained by perturbation theory. We introduce the expansion (3.4). We use

(recall that ez = 1+z+ + O (z3))

14

푒 = 휀푒 ( )

= 휀푒 푒 ( )

= 푒 + 푋 eXo+O(3) (3.10)

Substituting this expression in (3.9) written as x2 -1- 푒x= 0 and using (3.5), we

obtain

X20 – 1+ (2X0X1-eXo)+ 2(X2

1+2X0X1- X1eXo)+O( 3) = 0. (3.11)

Thus, the sequence of equations obtained is

O ( 0) : X20-1 = 0,

O ( 1) :2X0X1 – eX0 = 0

O ( 2) : X21 + 2X0X1-X1 eX0 = 0 (3.12)

O ( 3) : …

from which we obtain

Xo = 1, X1 = , X2 = (3.13)

X0 = -1, X1 = , X2=

or, equivalently,

15

x1 =1+ + +O(휀3)

x2 =1+ − +O(휀3)

The expression for x1 is compared to the numerical solution of (3.9) on figure 2.

Remark: In fact (3.9) has three solutions for 0 < < , with ≈ 0.43, and only

one for > .The solution which exists for all > 0 is the one with expansion

given in x2 in (3.14) ; the solution with the expansion given in x1 in (3.14)

disappears for > ; and the third solution (see figure 2: the solid line is the graph

of a two-valued function) cannot be obtained by regular perturbation.

Example 3.1.2

Consider the cubic equation

x3 –x+ = 0 (3.15)

We look for a solution of the form

x= x0+ x1+ 2x2 + O(휀3) (3.16)

Using this expansion in the equation, expanding, and equating coefficients of

to zero, we get

x30 – x0 = 0

3x20

x1 - xl + 1 = 0

3x0 x2 – x2 + 3x0

x21 = 0

16

Note that we obtain a nonlinear equation for the leading order solution x0, and

nonhomogeneous linearized equations for the higher order corrections x1, x2,..This

structure is typical of many perturbation problems.

Solving the leading-order perturbation equation, we obtain the three roots

X0 = 0, 1.

Solving the first-order perturbation equation, we find that

x1 =

The corresponding solutions are

x = 휀 +O(휀 2), x = 1- 휀 + O (휀2)

Continuing in this way, we can obtain a convergent power series expansion about

휀 = 0 for each of the three distinct roots of (3.15). This result is typical of regular

perturbation problems. .

An alternative but equivalent method to obtain the perturbation series is to

use the Taylor expansion

x(휀) = x(0) + x(0)휀+ ! x (O) (휀2 )+…

where the dot denotes a derivative with respect to휀. To compute the coefficients,

we repeatedly differentiate the equation with respect to 휀 and set 휀 = 0 in the

result. For example, setting 휀= 0 in (3.15), and solving the resulting equation for

x(0), we get x(0) = 0, 1. Differentiating (3.15) with respect to 휀, we get

3x2 x - x+1 = 0.

Setting 휀 = 0 and solving for x(0), we get the same answer as before.

17

Example 3.1.3

Consider the quadratic equation

(1-휀)x2 – 2x+1 = 0 (3.18)

Suppose we look for a straight forward power series expansion of the form

x= x0 +휀x1 + O(휀2)

We find that

x20 – 2x0 + 1 =0,

2(x0 – 1) x1 = 푥

Solving the first equation, we get x0 = 1. The second equation then becomes 0= 1.

It follows that there is no solution of the assumed form.

This difficulty arises because x= 1 is a repeated root of the unperturbed

problem. As a result, the solution

x= ±

does not have a power series expansion in 휀, but depends on 휀

An expansion

x= x0 + 휀 x1 + 휀x2 + O(휀 )

leads to the equations x0= 1, x21 = 1, or

x= 1 휀 + O(휀)

18

in agreement with the exact solution.

3.2 Solution Of First Order Differential Equitions

Example 3.2.1

Consider the differential equation

y′ + y2 = 0 (3.19)

which has been disturbed by a small effect, so that (3.19) has to be modified to

read

y′ + y2 = 휀x, y(1) =1 (3.20)

where 휀 is small. If then become necessary to determine by how much the solution

of (3.19) has been altered because of the presence of the disturbing function 휀x.

We refer to this change in the solutions as a perturbation.

A precise perturbation theory is extremely difficult. In this example, we shall aim

to give a rough outline of a method by which this problem can be handled. Call

y0(x) a solution of satisfying y(1) = 1, and denote the solution of (3.20) by

y(x) = y0(x) + p(x) (3.21)

where p(x) is the perturbation. We next expand y(x) in a series in powers of 휀, so

that

y(x) = y0(x) + 휀y1(x) + 휀2y2(x)+… (3.22)

Comparing (3.21) with 3.22), we see that

p(x) = 휀y1(x) + 휀2y2(x)+… (3.23)

19

The term 휀y1(x) is called the first order perturbation; the second term 휀2y2(x) is

called the second order perturbation, and so on.

Substituting (3.22) in (3.20), we obtain

(yo′+휀y1′+휀2y2′+ …+(yo+휀y1+휀2y2+…)2 = 휀x (3.24)

Carrying out the indicated multiplication, then collecting coefficient of like powers

of 휀, we have

(yo′ + y0

2) + (y1′+ 2y0y1)휀+ (y2′+ 2y0 + y2 + y12)휀2+ …= 휀x (3.25)

Next we take like powers of 휀. There results

yo ′+ y0′= 0,

(y1′+ 2y0y1) = x,

(y2′+ 2y0y2 + y12) = 0 (3.26)

…………………………

By solving each equation (3.26) is succession, we can thus determine the functions

y1(x), y2(x), y3(x)… in (3.22). Each of these functions, however, must satisfies an

initial condition. Since the initial condition associated with the original equation

(3.19) is y(1) = 1 and since y0 is a solution of (3.19) so that yo(1) = 1, this is initial

condition will be satisfied if in (3.22), we assume

yo(1) = 1, y1(1) = 0, y2(1) = 0,…

We illustrate the details of the above method in the next example

In practice the first and second order perturbations terms of (3.23) are usually

sufficient

20

Example 3.2.2

Find the first and second order perturbation terms in the solution of

y′ + y2 = 0, for which y(1) = 1 (3.28)

due to the precense of a disturbing function 휀x, where 휀 is small Solution. Because

of the disturbing function 휀x, (3.28) must be modified to read

y′ + y2 = 휀x (3.29)

For which y(1) = 1. Following the procedure outlined above we let, see (3.22) and

(3.27)

y(x) = y0(x) + 휀y1(x) + 휀2y2(x) +… (3.30)

with initial conditions

y0(1) = 1, y1(1) = 0, y2(1) = 0… (3.31)

Substituting (3.30) in (3.31), we obtain,

(y0′+ y02) + (y1′+ 2y0y1) 휀 + (y2′+ 2y0y2 + y1

2)휀2+…= 휀x (3.32)

Equating coefficients of like powers of 휀0, 휀, 휀2, we obtain from (3.32) the system

of equations

y0′+ y02 = 0

(y1′+ 2y0y1) = x (3.33)

(y2′+ 2y0y2 + y12) = 0

A solution of the first equation of (3.33), satisfying the initial condition y0(1)=1 of

(3.31), is

21

y0 = (3.34)

Substituting (3.34) in the second equation of (3.33)

y1′ + y1 = x (3.35)

A solution of (3.35) satisfying y1(1) = 0 of (3.31) is

y1 = (푥 − ) (3.36)

Substituting (3.34) and (3.36) in the third equation of (3.33), we obtain

y2′+ 푦 = −(푥 − 2 + )

A solution of (3.37) satisfying y2(1) = 0 of (3.31), is

y2 = − ( − − )− (3.38)

Substituting (3.34), (3.36), (3.38) and (3.30), we obtain

y= + 푥 − − 3푥 − 14푥 − + (3.39)

The solution of (3.28) satisfies y(1) = 1, that is, its solution if there were no

disturbing function 휀x present, is . Because of the disturbing function 휀x, the first

and second order perturbation terms are, respectively, the second and third term in

(3.39).

22

3.3 Eigenvalue problems

Spectral perturbation theory studies how the spectrum of an operator is perturbed

when the operator is perturbed. In general this question is a difficult one, and

subtle phenomena may occur, especially in connection with the behavior of the

continuous spectrum of the operators. Here, we consider the simplest case of the

perturbation in an eugenvalue.

Letℋ be a Hilbert space with inner product <.,.>, and

퐴 :퐷(퐴 ) ⊂ ℋ → ℋ a linear operator in ℋ, with domain D(A), depending

smoothly on a real parameter 휀. We assume that :

(a) 퐴 is self adjoint, so that

(x, 퐴 y) = (퐴 x,y) for all x, y ∈ D(퐴 )

(b) 퐴 has a smooth branch of simple eigenvalues ⋋ ∈ ℝ with eigenvectors

푥 ∈ ℋ, meaning that

퐴 푥 =⋋ 푥 . (3.40)

We will compute the perturbation in the eigenvalue from its value at 휀 = 0

when 휀 is small but nonzero

A concrete example is the perturbation in the eigenvalues of a symmetric

matrix. In that case, we have ℋ=ℝn with the Euclidean inner product

<x,y> = xTy,

and 퐴 : ℝn→ ℝn is a linear transformation with and n× n symmetric

matrix (aij). The perturbation in the eigenvalues of a Hermitian matrix

corresponds to ℋ = ℂn with inner product <x,y> = x-Ty. A we illustrate

below with the schrodinger equation of quantum mechanics, spectral

problems for differential equations can be formulated in terms of unbounded

operators acting in infinite – dimensional Hilbert spaces.

23

We use the expansions.

퐴 = A0 + 휀A1+…+휀 An+…

푥 = x0 + 휀x1+…+휀 xn+…

⋋ = ⋋0 + 휀 ⋋1+…+휀 ⋋n+…

in the eigenvalue problem (3.40), equate coefficients of 휀 , and rearrange

the result. We find that

(A0 - ⋋0I)x0 = 0, (3.41)

(A0 - ⋋0I)x1 = -A1x0 +⋋1x0, (3.42)

(A0 - ⋋0I)xn = ∑ {-Aixn-i+ ⋋ixn-i}. (3.43)

Assuming that x0 ≠ 0, equation (3.41) that ⋋0 is an eigenvalue of A0 and x0

is an eigenvector. Equation (3.42) is then a singular equation for x1. The

following proposition gives a simple, but fundamental, solvability condition

for this equation

Proposition 3.3.1

Suppose that A is a self – adjoint operator acting in a Hilbert space ℋ and

⋋ ∈ ℝ. If z ∈ ℋ, a necessary condition for the existence of a solution by

the y ∈ ℋ of the equation

(A-⋋I)y = z (3.44)

is that

(x, z) = 0,

For every eigenvector x of A with eigenvalue ⋋

Proof:-Suppose z ∈ ℋ and y ia a solution of (3.44). If x ia an eigenvector

of A, then using (3.44) and the self – adjointness of A-⋋ 퐼 , we find that

<x,z> = (x,(A-⋋ 퐼)y)

= <(A-⋋ 퐼)푥, 푦 >

= 0

24

In many cases, the necessary solvability condition in this proposition is also

sufficient, and then we say trhat A - ⋋ 퐼 satisfies the fredholm alternative; for

example, this is true in the finite dimensional case, or when A is an elliptic partial

differential operator.

Since A0 is self – adjoint and ⋋0 is a simple eigenvalue with eigenvector x0,

equation (3.44) it is solvable for x1 only if the right hand side is orthogonal to x0,

which implies that

⋋=⟨푥 ,퐴 푥 ⟩⟨푥 , 푥 ⟩

This equation gives the leading order perturbation in the eigenvalue, and is the

most important result of the expansion.

Assuming that the necessary solvability condition in the proposition is

sufficient, we can then solve (3.42) for x1. A solution for x1 is not unique, since

we can add to it an arbitrary scalar multiple of x0. This nonuniqueness is a

consequence of the fact that if 푥 is an eigenvector of 퐴 , then 푐 푥 is also a

solution for any scalar 푐 . If

푐 = 1+휀c1+O(휀2)

then

푐 푥 = x0 +휀 (x1+c1x0) + O(휀2).

Thus, the addition of c1x0 to x1 corresponds to arescaling of the eigenvector by a

factor that is close to one.

This expansion can be continued to any order. The solvability condition for

(3.43) determines ⋋n, and the erquation may then be solved for xn, up to an

25

arbitrary vector cnx0. The appearance of singular problems, and the need to impose

solvability conditions at each order which determine parameters in the expansion

and allow for the solution of higher order corrections, is a typical structure of many

perturbation problems.

26

Bibliography

[1] John.K.Hunter., Asymptotic Analysis And Singular Perturbation Theory,

University of California,2004. .

[2] James.G.Sinunonds.James.E.Nlann Jr, A First Look at Perturbation Theroy,

second edition.

[3] J.Kevorkian, J.D.Cole, Perturbation Methods In Applied Math-ematics,1981.

[4] Ravi.P.agarwal, Donal.O.Regan, Ordinary And Partial Differential Equations.