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17.4 Laplace Transform Techniques 631

11. Derive equation (17.13). Hint: This can be done using subject to the initial condition that
complex function theory and a contour integral. Here
is another way. Let ∞ √ π
∞ 2
F(0) = e −ζ 2 dζ = .
F(x) = e −ζ 2 cos(xζ)dζ. 0
0
Compute F (x) by differentiating under the integral This integral for F(0) is familiar from statistics and is

sign and show that F (x)=−xF(x)/2. Solve for F(x) assumed to be known. Finally, let x = α/β.

17.4 Laplace Transform Techniques

In this section, we make use of the Laplace transform to solve diffusion problems. As we did with
the wave equation, we will look at two typical problems. First, however, we need two functions
and a transform formula that occur frequently when dealing with diffusion problems.
The error function is defined by
2 t 2
erf(t) = √ e −u du
π 0
and the complementary error function by
2 ∞ 2
erfc(t) = √ e −u du.
π t
These are also used in probability and statistics. If the standard result that
√
∞ 2 π

e −u du =
0 2
is used, it is routine to check that
erfc(t) = 1 − erf(t).

The transform formula that we will need is
k 1 √

L erfc √ (s) = e −k s .
2 t s
Armed with these tools, we will look at two problems.

Temperatures in a Homogeneous Slab

We will solve the following:
2
∂u ∂ u
= k for 0 < x < L,t > 0,
∂t ∂x 2
u(x,0) = T 0 = constant,
and
∂u
u(L,t) = (0,t) = 0for t > 0.
∂x
This problem models the temperature distribution in a homogeneous solid slab or bar bounded
by the planes x =0 and x = L with the left side insulated (no flow of heat energy across this face)
and the right end kept at temperature zero. The initial temperature in the slab is constant.
Apply the Laplace transform with respect to t to the heat equation. As in Section 16.5, where
the Laplace transform was used to analyze wave motion, the resulting differential equation will

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October 14, 2010 15:25 THM/NEIL Page-631 27410_17_ch17_p611-640