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MATHEMATICS-III
CONTENTS
Linear ODE with variable coefficients and series solutions
Special functions
Complex function – Differentiation and integration
Power series expansions of complex functions and contour integration
Conformal mapping
TEXT BOOKS:
•Advanced Engineering Mathematics by Kreyszig, John Wiley & Sons.
•Higher Engineering Mathematics by Dr. B.S. Grewal, Khanna Publishers.
REFERENCES:
1.Complex Variables Principles And Problem Sessions By A.K.Kapoor, World
Scientific Publishers
2.Engineering Mathematics-3 By T.K.V.Iyengar andB.Krishna Gandhi Etc
3.A Text Book Of Engineering Mathematics By N P Bali, Manesh Goyal
4.Mathematics for Engineers and Scientists, Alan Jeffrey, 6th Edit. 2013, Chapman &
Hall/CRC
5.Advanced Engineering Mathematics, Michael Greenberg, Second Edition. Person
Education
6.Mathematics For Engineers By K.B.Datta And M.A S.Srinivas,Cengage Publications
Power Series Method
The power series method is the standard method for solving linear ODEs with variable coefficients. It gives solutions in the form of power series. These series can be used for computing values, graphing curves, proving formulas, and exploring properties of solutions. In this section we begin by explaining the idea of the power series method.
From calculus we remember that a power series (in powers of x − x0) is an infinite series of the form
(1)
Here, x is a variable. a0, a1, a2, … are constants, called the coefficients of the series. x0 is a constant, called the center of the series. In particular, if x0 = 0, we obtain a power series in powers of x
(2)
We shall assume that all variables and constants are real.
2
0 0 1 0 2 00
( ) ( ) ( ) .m
mm
a x x a a x x a x x
2 3
0 1 2 30
.m
mm
a x a a x a x a x
Then we collect like powers of x and equate the sum of the coefficients of each occurring power of x to zero, starting with the constant terms, then taking the terms containing x, then the terms in x2, and so on. This gives equations from which we can determine the unknown coefficients of (3) successively.
Idea and Technique of the Power Series Method (continued)
The nth partial sum of (1) is
where n = 0, 1, … .If we omit the terms of sn from (1), the remaining expression is
This expression is called the remainder of (1) after the term an(x − x0)n.
Theory of the Power Series Method
2
0 1 0 2 0 0(6) ( ) ( ) ( ) ( )n
n ns x a a x x a x x a x x
+1 +2
+1 0 +2 0(7) R ( )=a ( - ) + ( - ) + .n n
n n nx x x a x x
In this way we have now associated with (1) the sequence of the partial sums s0(x), s1(x), s2(x), …. If for some x = x1 this sequence converges, say,
then the series (1) is called convergent at x = x1, the number s(x1) is called the value or sum of (1) at x1, and we write
Then we have for every n,(8)If that sequence diverges at x = x1, the series (1) is called divergent at x = x1.
Theory of the Power Series Method (continued)
1 1 ( ) ( ),lim n
n
s x s x
1 1 00
( ) ( ) .m
mm
s x a x x
1 1 1( ) ( ) ( ).
n ns x s x R x
Where does a power series converge? Now if we choose x = x0 in (1), the series reduces to the single term a0 because the other terms are zero. Hence the series converges at x0.In some cases this may be the only value of x for which (1) converges. If there are other values of x for which the series converges, these values form an interval, the convergence interval. This interval may be finite, as in Fig. 105, with midpoint x0. Then the series (1) converges for all x in the interior of the interval, that is, for all x for which(10) |x − x0| < Rand diverges for |x − x0| > R. The interval may also be infinite, that is, the series may converge for all x.
Theory of the Power Series Method (continued)
Legendre’s Equation.
Legendre Polynomials Pn(x)
Legendre’s differential equation(1) (1 − x2)y” − 2xy’ + n(n + 1)y = 0 (n constant)is one of the most important ODEs in physics. It arises in numerous problems, particularly in boundary value problems for spheres.
The equation involves a parameter n, whose value depends on the physical or engineering problem. So (1) is actually a whole family of ODEs. For n = 1 we solved it in Example 3 of Sec. 5.1 (look back at it). Any solution of (1) is called a Legendre function.The study of these and other “higher” functions not occurring in calculus is called the theory of special functions.
Dividing (1) by 1 − x2, we obtain the standard form needed in Theorem 1 of Sec. 5.1 and we see that the coefficients −2x/(1 − x2) and n(n + 1)/(1 − x2) of the new equation are analytic at x = 0, so that we may apply the power series method. Substituting
(2)
and its derivatives into (1), and denoting the constant n(n + 1) simply by k, we obtain
By writing the first expression as two separate series we have the equation
0
( ) m
mm
y x a x
2 2 1
2 1 0
(1 ) ( 1) 2 0.m m m
m m mm m m
x m m a x x ma x k a x
2
2 2 1 0
( 1) ( 1) 2 0.m m m m
m m m mm m m m
m m a x m m a x ma x ka x
The reduction of power series to polynomials is a great advantage because then we have solutions for all x, without convergence restrictions. For special functions arising as solutions of ODEs this happens quite frequently, leading to various important families of polynomials. For Legendre’s equation this happens when the parameter n is a nonnegative integer because then the right side of (4) is zero for s = n, so that an+2 = 0, an+4 = 0, an+6 = 0, …. Hence if n is even, y1(x) reduces to a polynomial of degree n. If n is odd, the same is true for y2(x). These polynomials, multiplied by some constants, are called Legendre polynomials and are denoted by Pn(x).
Polynomial Solutions. Legendre Polynomials Pn(x)
Series Solutions
Near a Regular Singular Point, Part I
We now consider solving the general second order linear
equation in the neighborhood of a regular singular point x0.
For convenience, will will take x0 = 0.
Recall that the point x0 = 0 is a regular singular point of
iff
iff
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
0at analytic are )()(
)(and)(
)(
)( 22 xxqxxP
xRxxxp
xP
xQx
on convergent ,)( and )(0
2
0
n
n
n
n
n
n xxqxqxxpxxp
Series Solutions
Near a Regular Singular Point, Part I
We now consider solving the general second order linear
equation in the neighborhood of a regular singular point x0.
For convenience, will will take x0 = 0.
Recall that the point x0 = 0 is a regular singular point of
iff
iff
0)()()(2
2
yxRdx
dyxQ
dx
ydxP
0at analytic are )()(
)(and)(
)(
)( 22 xxqxxP
xRxxxp
xP
xQx
on convergent ,)( and )(0
2
0
n
n
n
n
n
n xxqxqxxpxxp
Transforming Differential Equation
Our differential equation has the form
Dividing by P(x) and multiplying by x2, we obtain
Substituting in the power series representations of p and q,
we obtain
0)()()( yxRyxQyxP
0
2
0
,)(,)(n
n
n
n
n
n xqxqxxpxxp
0)()( 22 yxqxyxxpxyx
02
210
2
210
2 yxqxqqyxpxppxyx
Comparison with Euler Equations
Our differential equation now has the form
Note that if
then our differential equation reduces to the Euler Equation
In any case, our equation is similar to an Euler Equation but
with power series coefficients.
Thus our solution method: assume solutions have the form
0,0for ,)(0
0
2
210
n
nr
n
r xaxaxaxaaxxy
02
210
2
210
2 yxqxqqyxpxppxyx
02121 qqpp
000
2 yqyxpyx
Example 1: Regular Singular Point (1 of 13)
Consider the differential equation
This equation can be rewritten as
Since the coefficients are polynomials, it follows that x = 0 is
a regular singular point, since both limits below are finite:
012 2 yxyxyx
02
1
2
2
yx
yx
yx
2
1
2
1limand
2
1
2lim
2
2
020 x
xx
x
xx
xx
Example 1: Euler Equation (2 of 13)
Now xp(x) = -1/2 and x2q(x) = (1 + x )/2, and thus for
it follows that
Thus the corresponding Euler Equation is
As in Section 5.5, we obtain
We will refer to this result later.
0,2/1,2/1,2/1 3221100 qqppqqp
0
2
0
,)(,)(n
n
n
n
n
n xqxqxxpxxp
020 2
00
2 yyxyxyqyxpyx
2/1,1011201)1(2 rrrrrrrxr
012 2 yxyxyx
Example 1: Differential Equation (3 of 13)
For our differential equation, we assume a solution of the form
By substitution, our differential equation becomes
or
012 2 yxyxyx
0
2
0 0
1
1)(
,)(,)(
n
nr
n
n n
nr
n
nr
n
xnrnraxy
xnraxyxaxy
0120
1
000
n
nr
n
n
nr
n
n
nr
n
n
nr
n xaxaxnraxnrnra
0121
1
000
n
nr
n
n
nr
n
n
nr
n
n
nr
n xaxaxnraxnrnra
Example 1: Combining Series (4 of 13)
Our equation
can next be written as
It follows that
and
0121
1
000
n
nr
n
n
nr
n
n
nr
n
n
nr
n xaxaxnraxnrnra
01)(121)1(21
10
n
nr
nn
r xanrnrnraxrrra
01)1(20 rrra
,2,1,01)(12 1 nanrnrnra nn
Example 1: Indicial Equation (5 of 13)
From the previous slide, we have
The equation
is called the indicial equation, and was obtained earlier when
we examined the corresponding Euler Equation.
The roots r1 = 1, r2 = ½, of the indicial equation are called the
exponents of the singularity, for regular singular point x = 0.
The exponents of the singularity determine the qualitative
behavior of solution in neighborhood of regular singular point.
0)1)(12(13201)1(2 20
0
0
rrrrrrraa
01)(121)1(21
10
n
nr
nn
r xanrnrnraxrrra
Example 1: Recursion Relation (6 of 13)
Recall that
We now work with the coefficient on xr+n :
It follows that
01)(121)1(21
10
n
nr
nn
r xanrnrnraxrrra
01)(12 1 nn anrnrnra
1,
112
1)(32
1)(12
1
2
1
1
nnrnr
a
nrnr
a
nrnrnr
aa
n
n
nn
Example 1: First Root (7 of 13)
We have
Starting with r1 = 1, this recursion becomes
Thus
2/1 and 1 ,1for ,
11211
1
rrnnrnr
aa n
n
1,
1211112
11
nnn
a
nn
aa nn
n
215325
13
012
01
aaa
aa
1,
!12753
)1(
etc,32175337
0
023
nnn
aa
aaa
n
n
Example 1: First Solution (8 of 13)
Thus we have an expression for the n-th term:
Hence for x > 0, one solution to our differential equation is
1,
!12753
)1( 0
n
nn
aa
n
n
1
0
1
1
00
0
1
!12753
)1(1
!12753
)1(
)(
n
nn
n
nn
rn
n
n
nn
xxa
nn
xaxa
xaxy
Example 1: Radius of Convergence for
First Solution (9 of 13)
Thus if we omit a0, one solution of our differential equation is
To determine the radius of convergence, use the ratio test:
Thus the radius of convergence is infinite, and hence the series
converges for all x.
0,
!12753
)1(1)(
1
1
xnn
xxxy
n
nn
10
132lim
)1(!13212753
)1(!12753limlim
111
1
nn
x
xnnn
xnn
xa
xa
n
nn
nn
nn
n
n
n
n
Example 1: Second Root (10 of 13)
Recall that
When r1 = 1/2, this recursion becomes
Thus
2/1 and 1 ,1for ,
11211
1
rrnnrnr
aa n
n
1,
122/1212/112/12
111
nnn
a
nn
a
nn
aa nnn
n
312132
11
012
01
aaa
aa
1,
!12531
)1(
etc,53132153
0
023
nnn
aa
aaa
n
n
Example 1: Second Solution (11 of 13)
Thus we have an expression for the n-th term:
Hence for x > 0, a second solution to our equation is
1,
!12531
)1( 0
n
nn
aa
n
n
1
2/1
0
1
2/1
02/1
0
0
2
!12531
)1(1
!12531
)1(
)(
n
nn
n
nn
rn
n
n
nn
xxa
nn
xaxa
xaxy
Example 1: Radius of Convergence for
Second Solution (12 of 13)
Thus if we omit a0, the second solution is
To determine the radius of convergence for this series, we can
use the ratio test:
Thus the radius of convergence is infinite, and hence the series
converges for all x.
10
12lim
)1(!11212531
)1(!12531limlim
111
1
nn
x
xnnn
xnn
xa
xa
n
nn
nn
nn
n
n
n
n
1
2/1
2!12531
)1(1)(
n
nn
nn
xxxy
Example 1: General Solution (13 of 13)
The two solutions to our differential equation are
Since the leading terms of y1 and y2 are x and x1/2, respectively,
it follows that y1 and y2 are linearly independent, and hence
form a fundamental set of solutions for differential equation.
Therefore the general solution of the differential equation is
where y1 and y2 are as given above.
1
2/1
2
1
1
!12531
)1(1)(
!12753
)1(1)(
n
nn
n
nn
nn
xxxy
nn
xxxy
,0),()()( 2211 xxycxycxy
Shifted Expansions & Discussion
For the analysis given in this section, we focused on x = 0 as
the regular singular point. In the more general case of a
singular point at x = x0, our series solution will have the form
If the roots r1, r2 of the indicial equation are equal or differ by
an integer, then the second solution y2 normally has a more
complicated structure. These cases are discussed in Section 5.7.
If the roots of the indicial equation are complex, then there are
always two solutions with the above form. These solutions are
complex valued, but we can obtain real-valued solutions from
the real and imaginary parts of the complex solutions.
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