1
Perturbation Techniques
In this series of lectures, we will like to be introduced to the basics of asymptotic
expansions and perturbation techniques. The most elementary application of perturbation
techniques is to algebraic equations which depend on a small parameter. We will study
three examples, each with increasing complexity. They will clearly demonstrate the ideas,
the problems that arise under certain conditions, the reasons for these difficulties
(nonuniformities in solutions), and the manner in which these nonuniformities are
removed.
Algebraic Equations:
Example 1
: Consider the quadratic equation (or, a polynomial of any degree)
2
(3
2 )
2
0,
0
1
x
x
ε
ε
ε

+
+
+
=
<
<<
(1)
When
ε
=0, the equation reduces to the solvable quadratic equation
2
3
2
(
2)(
1)
0.
x
x
x
x

+
=


=
(2)
Equation (2) is called the reduced equation
corresponding to the original problem in
equation (1). This equation has the zeros (or, roots) given by
1
2
1,
2.
x
x
=
=
(3)
Equation (1) is called the perturbed equation
, and it can be thought of as a function
( , )
f x
ε
whose zeros are being sought. Before we give a general result for the function
( , )
f x
ε
, let us assume that the zeros of the quadratic equation,
( ),
1,2,
i
i
x
x
i
ε
=
=
are
analytical functions of the small parameter
ε
. Thus, there exist a Taylor series expansion
in powers of
ε
which is convergent for every value of
ε
within some radius of
convergence
ε
0
. Therefore, let
°
°
°
°
°
4
2
0
3
2
3
0
1
2
3
1
sec
(
)
(
)
(
)
(
)
(
)
( )
. . . . .
i
i
i
first order
ond
zeroth
third
O
terms
O
order
order
order
O
O
O
x
x
x
x
x
x
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
∞
=
=
+
+
+
+
=
°
(4)
The aim is now to be able to find the coefficients
,
0,1,2,3,4,
......
i
x i
=
. Substituting
equation (4) into the equation (1) and expanding the resulting expression as an infinite
series in
ε
gives
2
3
2
2
3
0
1
2
3
0
1
2
3
(
...)
(3
2 )(
...)
2
0.
x
x
x
x
x
x
x
x
ε
ε
ε
ε
ε
ε
ε
ε
+
+
+
+

+
+
+
+
+
+
+
=
Now,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
2
3
2
2
2
3
2
3
2
0
1
2
3
0
0
1
2
3
1
2
3
2
2
2
3
0
0
1
0
2
1
0
3
1
2
(
...)
2
(
...)
(
...)
2
(2
)
(2
2
)
. . . . . .
x
x
x
x
x
x
x
x
x
x
x
x
x
x x
x x
x
x x
x x
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
ε
+
+
+
+
=
+
+
+
+
+
+
+
+
=
+
+
+
+
+
+
and
2
3
2
0
1
2
3
0
1
0
2
1
(3
2 )(
. . . )
3
(3
2
)
(3
2
)
. . .
x
x
x
x
x
x
x
x
x
ε
ε
ε
ε
ε
ε
+
+
+
+
+
=
+
+
+
+
+
Thus, equation (1) becomes
2
2
2
0
0
0
1
1
0
0
2
1
2
1
(
3
2)
(2
3
2
1)
(2
3
2
)
. . . .
0
x
x
x x
x
x
x x
x
x
x
ε
ε

+
+


+
+
+


+
=
(5)
Setting the coefficient of each power of
ε
to zero seperately gives
0
2
0
0
3
2
0
x
x
ε

+
=
(6)
1
0
1
1
0
2
3
2
1
0
x x
x
x
ε


+
=
(7)
2
2
0
2
1
2
1
2
3
2
0
x x
x
x
x
ε
+


=
(8)
Similar equations are obtained at higher orders in
ε
. Thus, we get an infinite sequence of
algebraic equations which need to be solved sequentially to determine the unknowns
,
0,1,2,3,4,
......
i
x i
=
.
Consider the equation (6). It has the two zeros (or, roots)
01
02
1,
2.
x
x
=
=
(9)
(Note that equation (6) is the same as the “reduced equation”). Now, consider equation
(7). It can be written as
1
0
0
(2
3)
2
1
x
x
x

=

(10)
This is a linear nonhomogeneous equation in the unknown variable
1
x
with the right hand
side being known due to the roots in equation (9). For each of the roots in equation (9),
equation (10) can be solved for
1
x
provided
0
(2
3)
0
x

≠
, which is clearly the case. Thus,
the two solutions of equation (10) are
11
01
12
02
1
1,
3
2.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 NA
 Power Series, Taylor Series, Quadratic equation, Elementary algebra

Click to edit the document details