This document discusses regular perturbation theory and its application to solving algebraic equations. It begins by defining regular and singular perturbations. For regular perturbations, the order of the perturbed and unperturbed problems are the same when the perturbation parameter is set to zero. The document then shows how regular perturbation theory can be used to solve algebraic equations. Specifically, it demonstrates obtaining the series solutions for the quadratic equation x^2-1=ε and the cubic equation x^3-x+ε=0 by assuming power series solutions in ε and solving the equations order-by-order in ε. This yields convergent power series expansions for the roots as functions of the small perturbation parameter ε.

1. A STUDY ON REGULAR PERTURBATION
PROBLEMS
By
Shareena . P. R
( M.Sc Mathematics)

2. CONTENT
TITLE PAGE NUMBER
PREFACE 1
CHAPTER 1 2
CHAPTER 2 5
CHAPTER 3 9
GOBLIOGRAPHY 26

3. 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.

4. 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( )}n = 0,1,2…is an asymptotic as
→ 0 , ℎ > 0, ∅nti( ) = 0 (∅n( )) as → 0, that is each subsequent
term gets smaller.
Examples
(a) {1, , 2
, 3
,…}
(b){1, /
, /
,…}
(c) {1, , log , , 2
log ,…}
Defintion 1.3
Asymptotic expansion

5. 3
Define an asymptotic series,
y= yo+ y1 + 2
y2+…,
where y1, y1, y2, … are sufficiently smooth functions.
The standard asymptotic sequence is {1, , 2, 3
…} as 0 and fn(x)
represents the members of asymptotic sequence then fn+1( ) = 0(fn
( )) as xa
that is →
( )
( )
= 0. The general expression for asymptotic expansion of
function fn( ) is the series of terms
f(x)=∑ ( ) + → 0, ℎ
= ( ( )) → 0 lim → RN=0.
Definition 1.4
The expression f(x) =∑ fn( ) + RN ,where f(x; ) depends on an
independent variable x and small parameter . The coefficient of the gauge
function fn( ) are functions of x and the remainder term after N terms is a function
of both x and is RN = O (fn+1( )) is said to be uniform asymptotic expansion, if
RN  Cfn+1( ), where c is the constant.
Example
f(x, )= = 1+ sinx+ 2
(sinx)2
+…as 0.
The remainder term RN = 1+ sinx+ 2
(sinx)2
+…-∑ ( ) ,
where → ( ) = (sinx)N
+1

6. 4
Defintion 1.5
The expression f(x) = ∑ an fn ( ) + RN, where f(x; ) 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 ( )
satisfied is known as non-uniform asymptotic expansion
Example
f(x, )= = 1+ x+ 2
(x)2
+…as 0.
The remainder term RN = 1+ x+ 2
(x)2
+…- n
(x)n
,
Where → ( ) =(x)N
+1. There is no fixed constant C exists such that
RN≤ N+1.

7. 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) x2
— x+ = 0.
(b) +y = y2
, y(0) = 1.

8. 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 = 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 = x
(b) )
+ y = y2
, 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 = 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) x2
-3x+8=0
(b)
+ = 2 + 1, y(0) = 0, y(1) = 4

9. 7
2.2 The Fundamental Theorem Of Perturbation Theory
If A0, A1 +…+AN
N
+ O( N+1
) = 0 for sufficiently small and if the coefficients A0,
A1… are independent of , 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
lim
xx
( )
( )
= 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
lim
xx
( )
( )
= 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
0
lim
xx
= 1.

10. 8
(b) = o(x) as x→ ∞.
(c)Sin x2
= o(x) as x →0 because
0
lim
xx
= 0.
(c) 3x+x3
= O(x) as x →0 since
0
lim
xx
= 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
lim
xx
( )
( )
= 1 is called "f is
asymptotically equal or approximately equal to g".

11. 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 , such that the solution is available and
reasonably simple for = 0. Then, one may wonder how this solution is altered for
non-zero but small . 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 = (3.1)
The two roots of this equation are
x1 = + 1 + , = − 1 + (3.2)
For small , these roots are well approximated by the first few terms of their Taylor
series expansion (see figure 1)
= 1 + + + ( ), 2 = −1 + - + ( ). (3.3)

12. 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+ X1 + 2
X2 + O( 3
) (3.4)
for X0, X1, X2 to be determined.
Substitute (3.4) into (3.1) written as x2
- 1 - X = 0, and expand the left hand side
of the resulting equation in power series of . Using
x2
= + 2 X0X1 + 2
( + 2X0X2) + O( 3
), (3.5)
x = X0 + 2
X1+O( 3
)

13. 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
= 1+

+

is fairly good even up to  = 2.
this gives
– 1+ (2X0X1 –X0)+ 
2
( +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
) : ─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 X2
0-1=0 has
two roots, X0 = ±1, one obtains
X0 = 1, XI = , X2 = (3.8)
X0 = -1, XI = , X2 =

14. 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 =  (3.9)

15. 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
))

16. 14
 = ( )
= ( )
=  + eXo
+O(
3
) (3.10)
Substituting this expression in (3.9) written as x2
-1-
x
= 0 and using (3.5), we
obtain
X2
0 – 1+ (2X0X1-eXo
)+ 
2
(X2
1+2X0X1- X1eXo
)+O(
3
) = 0. (3.11)
Thus, the sequence of equations obtained is
O (
0
) : X2
0-1 = 0,
O (
1
) :2X0X1 – eX0
= 0
O (
2
) : X2
1 + 2X0X1-X1 eX0
= 0 (3.12)
O (
3
) : …
from which we obtain
Xo = 1, X1 = , X2 = (3.13)
X0 = -1, X1 = , X2=
or, equivalently,

17. 15
x1 =1+

+ +O( 3
)
x2 =1+

− +O( 3
)
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+
2
x2 + O( 3
) (3.16)
Using this expansion in the equation, expanding, and equating coefficients of 
to zero, we get
x3
0 – x0 = 0
3x2
0 x1 - xl + 1 = 0
3x0 x2 – x2 + 3x0 x2
1 = 0

18. 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 = +O( 2
), x =  1- + O ( 2
)
Continuing in this way, we can obtain a convergent power series expansion about
= 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( ) = x(0) + x(0) +
!
x (O) ( 2
)+…
where the dot denotes a derivative with respect to . To compute the coefficients,
we repeatedly differentiate the equation with respect to and set = 0 in the
result. For example, setting = 0 in (3.15), and solving the resulting equation for
x(0), we get x(0) = 0,  1. Differentiating (3.15) with respect to , we get
3x2
x - x +1 = 0.
Setting = 0 and solving for x(0), we get the same answer as before.

19. 17
Example 3.1.3
Consider the quadratic equation
(1- )x2
– 2x+1 = 0 (3.18)
Suppose we look for a straight forward power series expansion of the form
x= x0 + x1 + O( 2
)
We find that
x2
0 – 2x0 + 1 =0,
2(x0 – 1) x1 =
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 , but depends on
An expansion
x= x0 + x1 + x2 + O( )
leads to the equations x0= 1, x2
1 = 1, or
x= 1  + O( )

20. 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
= x, y(1) =1 (3.20)
where 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 x.
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 , so
that
y(x) = y0(x) + y1(x) + 2
y2(x)+… (3.22)
Comparing (3.21) with 3.22), we see that
p(x) = y1(x) + 2
y2(x)+… (3.23)

21. 19
The term y1(x) is called the first order perturbation; the second term 2
y2(x) is
called the second order perturbation, and so on.
Substituting (3.22) in (3.20), we obtain
(yo
′
+ y1′+ 2
y2′+ …+(yo+ y1+ 2
y2+…)2
= x (3.24)
Carrying out the indicated multiplication, then collecting coefficient of like powers
of , we have
(yo
′
+ y0
2
) + (y1′+ 2y0y1) + (y2′+ 2y0 + y2 + y1
2
) 2
+ …= x (3.25)
Next we take like powers of . 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

22. 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 x, where is small Solution. Because
of the disturbing function x, (3.28) must be modified to read
y′ + y2
= x (3.29)
For which y(1) = 1. Following the procedure outlined above we let, see (3.22) and
(3.27)
y(x) = y0(x) + y1(x) + 2
y2(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′+ y0
2
) + (y1′+ 2y0y1) + (y2′+ 2y0y2 + y1
2
) 2
+…= x (3.32)
Equating coefficients of like powers of 0
, , 2
, we obtain from (3.32) the system
of equations
y0′+ y0
2
= 0
(y1′+ 2y0y1) = x (3.33)
(y2′+ 2y0y2 + y1
2
) = 0
A solution of the first equation of (3.33), satisfying the initial condition y0(1)=1 of
(3.31), is

23. 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 = ( − ) (3.36)
Substituting (3.34) and (3.36) in the third equation of (3.33), we obtain
y2′+ = −
( − 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= + − − 3 − 14 − + (3.39)
The solution of (3.28) satisfies y(1) = 1, that is, its solution if there were no
disturbing function x present, is . Because of the disturbing function x, the first
and second order perturbation terms are, respectively, the second and third term in
(3.39).

24. 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.
Let ℋ be a Hilbert space with inner product <.,.>, and
: ( ) ⊂ ℋ → ℋ a linear operator in ℋ, with domain D(A), depending
smoothly on a real parameter . We assume that :
(a) is self adjoint, so that
(x,
y) = (
x,y) for all x, y ∈ D( )
(b)
has a smooth branch of simple eigenvalues ⋋ ∈ ℝ with eigenvectors
∈ ℋ, meaning that
=⋋ . (3.40)
We will compute the perturbation in the eigenvalue from its value at = 0
when 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> = xT
y,
and
: 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-T
y. 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.

25. 23
We use the expansions.
= A0 + A1+…+ An+…
= x0 + x1+…+ xn+…
⋋ = ⋋0 + ⋋1+…+ ⋋n+…
in the eigenvalue problem (3.40), equate coefficients of , 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-⋋ , we find that
<x,z> = (x,(A-⋋ )y)
= <(A-⋋ ) , >
= 0

26. 24
In many cases, the necessary solvability condition in this proposition is also
sufficient, and then we say trhat A - ⋋ 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
⋋=
〈 , 〉
〈 , 〉
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 is an eigenvector of , then is also a
solution for any scalar . If
= 1+ c1+O( 2
)
then
= x0 + (x1+c1x0) + O( 2
).
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

27. 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.

28. 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.