Page 134 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

124 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
equation and boundary conditions as follows:

2
2
∂ y ∂ y
=
∂t 2 ∂x 2
y(x, 0) = y t (x, 0) = 0
y(0, t) = f (t)

y is bounded
Taking the Laplace transform with respect to time and applying the initial conditions
yields

2
d Y (x, s )
2
s Y (x, s ) =
dx 2
The solution may be written in terms of exponential functions
Y (x, s ) = Ae −sx + Be sx

In order for the solution to be bounded B = 0. Applying the condition at x = 0wefind

A = F(s )
where F(s ) is the Laplace transform of f (t).
Writing the solution in the form

e −sx
Y (x, s ) = sF(s )
s
and noting that the inverse transform of e −sx /s is the Heaviside step U x (t) where

U x (t) = 0 t < x
U x (t) = 1 t > x

and that the inverse transform of sF(s )is f (t), we find using convolution that
t

y(x, t) = f (t − µ)U x (µ)dµ = f (t − x) x < t
µ=0
= 0 x > t

For example, if f (t) = sin ω t

y(x, t) = sin ω(t − x) x < t

= 0 x > t