Page 175 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Since p(x) > 0 or < 0 or every point a x b, according as p(a) > 0 or < 0, we can find a positive
solution p(x) from which, along with (x), we can obtain a solution y(x) = p(x) sin (x), not
identically zero, of the original equation (2).
Now for an integer n, (x) + 2 n is also a solution of the second equation of (8). Thus the
solutions y (x) and y (x) obtained from (x) and (x) + 2n are linearly dependent. So if the two
1 2
solutions y (x) and y (x) given by
1 2
y (x) = (x) sin (x)
1 1 1
y (x) = (x) sin (x)
2 2 2
are linearly dependent, then for some integer n
(x) = (x) + 2 .
1 2
Now, an initial condition for q(x),
(a) = ...(10)
gives a relation between y(a) and y (a) as follows
1
At x = a from (5) and (7) we have
z(a) = p(a) y (a) = (a) cos (a)
So p(a) y (a) sin (a) = (a) cos (a) sin (a)

or p(a) y (a) sin (a) = y(a) cos (a)
or p(a) y (a) sin (a) y(a) cos (a) = 0 ...(11)
In this section we shall be concerned with the problem of finding the solution y(x) corresponding
to the solution (x) satisfying the boundary conditions

(a) = , (b) =
at both ends of the interval a x b.
Condition (12) corresponds to the conditions
p(a) y (a) sin y(a) cos = 0
p(b) y (b) sin y(b) cos = 0 ...(13)
for y(x). It should be noted that the boundary value problem of finding the solution of (2)
satisfying the boundary conditions (13) between y and y is essentially different from the initial
value problem.

10.2 Green’s Function for One Dimensional Problem

Let us denote L (y), a differential operator
x
d dy
L (y) = p ( ) q ( )y ...(1)
x
x
x
dx dx
dy d dy
x
which is defined for every function y(x) such that and p ( ) are defined and continuous
dx dx dx
on the interval x b. Let us define Lagrange’s identity
d dz d dy
x
x
y L (z) z L (y) = p ( ) y z p ( )
x x dx dx dx dx

168 LOVELY PROFESSIONAL UNIVERSITY