Page 102 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

92 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS

In the special case where f (x) = u 0
√
x/2 t

2u 0 2 x
u(x, t) = √ exp(−σ )dσ = u 0 erf √ (5.106)
π 2 t
0
where erf(p) is the Gauss error function defined as
p
2 2
erf(p) = √ exp(−σ )dσ (5.107)
π
0
Example 5.9 (Steady conduction in a quadrant). Next we consider steady conduction in the
region x ≥ 0, y ≥ 0inwhich thefaceat x = 0iskeptatzerotemperature andthe face at
y = 0 is a function of x: u = f (x). The solution is also assumed to be bounded.

u xx + u yy = 0 (5.108)

u(x, 0) = f (x) (5.109)
u(0, y) = 0 (5.110)

Since u(0, y) = 0 the solution should take the form e −αy sin α x, which is, according to our
2
experience with separation of variables, a solution of the equation ∇ u = 0. We therefore
assume a solution of the form
∞

u(x, y) = B(α)e −α y sin α xdα (5.111)
0
with
∞
2

B(α) = f (ς)sinας dς (5.112)
π
0
The solution can then be written as
∞ ∞
2
u(x, y) = f (ς) e −α y sin α x sin ας d α d ς (5.113)
π
ς=0 α=0
Using the trigonometric identity for 2 sin ax sin aς = cos a(ς − x) − cos a(ς + x) and noting
that
∞
y
e −α y cos aβ d α = (5.114)
2
β + y 2
0