Page 179 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes where every C is a function of . We shall determine the relations between C so that G(x, )
i i
satisfies the required properties for the Green’s function pertaining to the boundary condition
(2). Since G(x, ) is continuous at x = , we obtain
c y ( ) + c y ( ) = c y ( ) + c y ( ) ...(6)
1 1 2 2 3 3 4 4
By equation (12) of section (10.2), we obtain

1
c y ( ) + c y ( ) c y ( ) c y ( ) = ...(7)
1 1 2 2 3 3 4 4 p ( )
Finally from the boundary conditions (2) we obtain
c y (a) + c y (a) = (c y (b) + c y (b))
1 1 2 2 3 3 4 4
p(a) (c y (a) + c y (a) = p(b) (c y (b) + c y (b)) ...(8)
1 1 2 2 3 3 4 4
Also Green’s function should be symmetric i.e.
G(x, ) = G( , x) ...(8a)

Only the last relation of (8) must be changed according as the corresponding boundary conditions,
if we are concerned with Green’s function under the boundary conditions (3).

Example: Find the Green’s function for L y = 0 with the boundary conditions
x
y(0) = y(1), y (0) = y (1).
Solution:
The general solution of L y = 0 is of the form c x +c . Now taking as a fundamental system of the
x 1 2
solutions of y = 0, as
y (x) = (x), y (x) = 1, p(x) = 1, = 1
1 2
Let G(x, ) be given by the relation (5) where a = 0, b = 1 from the equations (6), (7) and (8) we have
c + c = c +c , c c = 1, c = (c + c ), c = c
1 2 3 4 1 3 2 3 4 1 3
Solving these equations, we obtain

1
2 c = 1, c = = c , (c c ) + c = c
1 1 2 3 1 3 2 4
1
c = c
2 2 4
1
2 c + = 0
2 2
1 1
c = /2, c = + /2
2 4 4 4
Therefore

1 1
x
G ( , ) x 1 for 0 x<
2 4 2
1 1
x 1 for x 1
2 4 2
1 1
x
or G ( , ) x G ( , ) .
x
2 4

172 LOVELY PROFESSIONAL UNIVERSITY