Page 88 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

78 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS

The length and time scales are clearly the same as the problem in Chapter 2. Hence, τ =
2
tα/L and ξ = x/L. If we choose U = (u − u 0 )/(u 1 − u 0 ) we make the boundary condition
at x = 0 homogeneous but the condition at x = L is not. We have the same situation that we
had in Section 2.3 of Chapter 2. The differential equation, one boundary condition, and the
initial condition are homogeneous. Proceeding, we find

U τ = U ξξ
U(τ, 0) = U(0,ξ) = 0 (5.20)

U ξ (τ, 1) + B[U(τ, 1) − 1] = 0

where B = hL/k. It is useful to relocate the nonhomogeneous condition as the initial condition.
As in the previous problem we assume U(τ, ξ) = V (τ, ξ) + W(ξ).

V τ = V ξξ + W ξξ
W(0) = 0
W ξ (1) + B[W(1) − 1] = 0
(5.21)
V (τ, 0) = 0
V ξ (τ, 1) + BV (τ, 1) = 0

V (0,ξ) =−W(ξ)

Set W ξξ = 0. Integrating twice and using the two boundary conditions on W,

Bξ
W(ξ) = (5.22)
B + 1
The initial condition on V becomes

V (0,ξ) =−Bξ/(B + 1) . (5.23)

Assume V (τ, ξ) = P(τ)Q(ξ), substitute into the partial differential equation for V , and divide
by PQ as usual.

P Q 2
= =±λ (5.24)
P Q
We must choose the minus sign for the solution to be bounded. Hence,

2
P = Ae −λ τ
(5.25)
Q = C 1 sin(λξ) + C 2 cos(λξ)