Page 148 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

138 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
To solve this using Duhamel’s theorem, we first set f (t) = f (λ)with λ a timelike
constant.
Following the procedure outlined at the beginning of Example 8.6, we find
√
e −x s
U(x, s ) = f (λ)
s
The inverse transform is as follows:
x

u(x, t,λ) = f (λ)erfc √
2 t
Using Duhamel’s theorem,
t

∂ x
u(x, t) = f (λ)erfc √ dλ
∂t 2 t − λ
λ=0
which is a different form of the solution given in Example 8.6.

Problems
1. Show that the solutions given in Examples 8.6 and 8.9 are equivalent.
2. Use Duhamel’s theorem along with Laplace transforms to solve the following conduc-
tion problem on the half space:

u t = u xx
u(x, 0) = 0
u x (0, t) = f (t)

3. Solve the following problem first using separation of variables:
2
∂u ∂ u
= + sin(πx)
∂t ∂x 2
u(t, 0) = 0

u(t, 1) = 0
u(0, x) = 0

4. Consider now the problem

2
∂u ∂ u
= + sin(πx)te −t
∂t ∂x 2
with the same boundary conditions as Problem 7. Solve using Duhamel’s theorem.