Page 140 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

130 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
Note that as a partial check on this solution, we can evaluate the result when H →∞ as

∞ ∞ n+1
−2 2(−1)
2
2
2
−y τ −n π τ
u(ς, τ) = sin ς y n e n = sin(nπς)e
y n cos y n nπ
n=1 n=1
in agreement with the separation of variables solution. Also, letting H → 0wefind
∞
2 sin(y n ς)
2
y τ
u(ς, τ) = e n
y 2 sin(y n )
n=1 n
2n−1
with y n = π again in agreement with the separation of variables solution.
2
Example 8.5. Next we consider a conduction (diffusion) problem with a transient source q(τ).
(Nondimensionalization and normalization are left as an exercise.)
u τ = u ςς + q(τ)
u(ς, 0) = 0 = u ς (0,τ)

u(1,τ) = 1

Obtaining the Laplace transform of the equation and boundary conditions we find

sU = U ςς + Q(s )
U ς (0, s ) = 0
1
U(1, s ) =
s
A particular solution is

Q(s )
U P =
s
and the homogeneous solution is

√ √
U H = A sinh(ς s ) + B cosh(ς s )

Hence the general solution is

Q √ √
U = + A sinh(ς s ) + B cosh(ς, s )
s