Page 389 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Differential and Integral Equation

Notes Now from solution (b) using boundary conditions
X(0) = A + B =0
and X(L) = Ae x Be x 0
Giving A B =0, so that X(x) = 0 therefore 0 which is absurd.

Hence (a) and (b) are not the solutions of wave equation (i). The third solution (c) is periodic (in
time). Therefore the solution is u x ,t A cos x B sin x C cos ct D sin ct 0. Using the
boundary conditions (i) and (ii), we have

u 0,t A C cos ct D sin ct 0.
Hence A = 0

and u , L t B sin L C cos ct D sin ct 0.

this gives sin L = 0
or L n

n
or
L
where n = 1, 2, 3......, (i.e. a + ive integer).
Hence the solution of equation (i) satisfying boundary conditions is
n ct n ct n x
u n , x t C n cos D n sin sin ...(vii)
L L L
Now using initial conditions (iii) and (iv), we have

n x
u x ,0 C n sin f x
n
L
y n c n ct n c n ct sinn x
and C n sin D n cos
t L L L L L
t 0
n c n x
= D n sin g x .
L L
Clearly these will not be satisfied if we take only a single term as our solution. The equation
(i) is a linear and homogeneous therefore the sum of different solutions will still be a solution.
This instead of (vii), the solution may be taken as

n ct n ct n x
u , x t C cos D sin sin
n n ...(viii)
n 1 L L L
Therefore using initial conditions

n x
u x ,0 C n sin f x
n 1 L
y n c n x
and D n sin g x
t t 0 n 1 L L

382 LOVELY PROFESSIONAL UNIVERSITY