Page 26 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

16 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
The method would not have worked had the differential equation not been homoge-
neous. (Try it.) It also would not have worked if more than one boundary condition had been
nonhomogeneous. We will see how to get around these problems shortly.

Problems
1. Equation (2.9) could just as easily have been written as

τ ξξ 2
= =+λ

Show two reasons why this would reduce to the trivial solution or a solution for which
approaches infinity as τ approaches infinity, and that therefore the minus sign must
be chosen.
2. Solve the above problem with boundary conditions

U ξ (τ, 0) = 0 and U(τ, 1) = 0

using the steps given above.
Hint: cos(nπx) is an orthogonal set on (0, 1). The result will be a Fourier cosine series
representation of 1.

3. Plot U versus ξ for τ = 0.001, 0.01, and 0.1 in Eq. (2.18). Comment.

Example 2.2. A Steady Heat Transfer Problem in Two Dimensions
Heat is conducted in a region of height a and width b. Temperature is a function of two space
dimensions and independent of time. Three sides are at temperature u 0 and the fourth side is
at temperature u 1 . The formulation is as follows:

2
2
∂ u ∂ u
+ = 0 (2.19)
∂x 2 ∂y 2
with boundary conditions
u(0, x) = u(b, x) = u(y, a) = u 0

u(y, 0) = u 1 (2.20)

2.1.6 Scales and Dimensionless Variables
First note that there are two obvious length scales, a and b. We can choose either one of them
to nondimensionalize x and y. We define

ξ = x/a and η = y/b (2.21)

so that both dimensionless lengths are normalized.