Page 181 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes 2
d
Example: Find generalized Green’s function for L = 2 , with the boundary conditions
x dx
y (0) = y (1) = 0.
Solution:
The general solution of y (x) = 0 is a polynomial of degree 1. Hence there exists a non-trivial
solution y (x) = 1 of the boundary value problem. So from the condition (2) we have
0
L G(x, ) = 1, that is, G (x, ) = 1.
x xx
Hence we have

x 2
G(x, ) = A + A x + x
1 2 2

x 2
= B + B x + x >
1 2 2
By the boundary conditions G (0, x) = 0, G (1, ) = 0, we obtain
x x
A = 0, B = 1. So the condition
2 2
G x ( 0, ) G x ( 0, ) 1
holds automatically. By the continuity at x = , that is G( + 0, ) G( 0, ) = 0, we obtain
B A = 0. Hence we obtain
1 1
x 2
G(x, ) = A + x
1 2
x 2
= A + x + x >
1 2
Finally, from the relation

1
G ( , ) ( )d = 0,
x
y
0
0
we obtain A = 0. Thus the generalized Green’s function is given by
1
x 2
G(x, ) = x
2

x 2
= x + x > .
2

Self Assessment

d 2
2. Find the generalized Green’s function for L = 2 , with the boundary conditions
x dx
1
y( 1) = y(1), y ( 1) = y (1). (Hint: take y (x) = )
0 2

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