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634 CHAPTER 17 The Heat Equation

and

U(0,s) = f (s), lim U(x,t) = 0.
x→∞
This differential equation has general solution
√ √
U(x,s) = c 1 e s/kx + c 2 e − s/kx .
Since u(x,t) → 0as x →∞, then U(x,s) → 0as x →∞, requiring that c 1 = 0. Furthermore,
u(0,t) = f (t) implies that U(0,s) = F(s) = c 2 . Note here that c 2 may depend on s, which is a
parameter in the transform with respect to t. Therefore,
√
− s/kx
U(x,s) = F(s)e .
The solution is
√

−1 −1 − s/kx
u(x,t) = L [u(x,s)](t) = L F(s)e .
We can also write this solution as a convolution
u(x,t) = f (t) ∗ g(t),
where
√
−1 − s/kx
g(t) = L e (t).
A Semi-Infinite Bar with Discontinuous Temperature at the Left End
We will solve the boundary value problem
2
∂u ∂ u
= k for x > 0,t > 0,
∂t ∂x 2
u(x,0) = A for x > 0,
and

B for 0 ≤ t ≤ t 0
u(0,t) =
0 for t > t 0 .
Here t 0 , A, and B are positive constants.
This problem models the temperature distribution in a thin, homogeneous bar extending
along the nonnegative x-axis with a constant initial temperature A and a discontinuous temper-
ature function at the left end where x = 0. The Laplace transform is a natural approach for this
problem, because this transform is well suited to treating piecewise continuous functions. Begin
by writing

u(0,t) = B[1 − H(t − t 0 )]
in which H is the Heaviside function. Apply the Laplace transform with respect to t in the heat
equation using the condition that u(x,0) = A to obtain
∂ 2 s A
U(x,s) − U(x,s) =− .
∂x 2 k k
As usual, think of this as a differential equation in x. The general solution is
√ √ A
U(x,s) = c 1 e s/kx + c 2 e − s/kx + ,
s

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October 14, 2010 15:25 THM/NEIL Page-634 27410_17_ch17_p611-640