Page 170 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 9: Adjoint and Self-Adjoint Equations

so that Notes

( 1) n 1
x
y ( ) 2 3 sin nx
n
n 1
This type of series is known as a Fourier series.
This example is, of course, rather artificial, and we could have integrated (14) directly. There are,
however, many boundary value problems for which this eigenfunction expansion method is the
only way to proceed analytically.

Example 2: Consider the inhomogeneous equation
2
(1 x )y 2xy + 2y = f(x) on 1 < x < 1, ...(15)
with f C[ 1, 1], subject to the condition that y should be bounded on [ 1, 1]. We begin by noting
that there is a solubility condition associated with this problem. If u(x) is a solution of the
homogeneous problem, then, after multiplying through by u and integrating over [ 1, 1], we
find that

1 1 1
u (1 x 2 )y u (1 x 2 )y u ( ) ( )dx
x
f
x
1 1 1
If u and y are bounded on [ 1, 1], the left hand side of this equation vanishes, so that
1
u ( ) ( )dx 0. Since the Legendre polynomial, u = P (x) = x, is the bounded solution of the
f
x
x
1 1
homogeneous problem, we have
1
f
x
P 1 ( ) ( )dx 0
x
1
Now, to solve the boundary value problem, we first construct the eigenfunction solutions by
solving Ly = y, which is
2
(1 x )y 2xy + (2 )y = 0
The choice 2 = n(n + 1), with n a positive integer, gives us Legendre’s equation of integer
order, which has bounded solutions y (x) = P (x). These Legendre polynomials are orthogonal
n n
over [ 1, 1]. If we now write
x
f ( ) A P ( ),
x
m m
m 0
x
where A = 0 by the solubility condition, and then expand y(x) = B P ( )
m m
1 m 0
we find that
{2 m(m + 1)}B = A for m 0
m m
The required solution is therefore

1 A
y ( ) A 0 B P ( ) m P m ( )
x
x
x
1 1
2 2 m (m 1)
m 2
with B an arbitrary constant.
1
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