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17.4 Laplace Transform Techniques 635

in which the “constants” c 1 and c 2 may depend on s. We will require that U(x,s)→0as x →∞,
so choose c 2 = 0. This leaves
√ A
U(x,s) = c 1 e s/kx + .
s
To obtain c 1 , take the Laplace transform of the boundary condition u(0,t) = B[1 − H(t − t 0 )] to
obtain

B e −t 0 s
U(0,s) = L[B]− L[B(t − t 0 )]= − B .
s s
Then
B e −t 0 s A
U(0,s) = − B = c 1 + .
s s s
Solve for c 1 to obtain

B − A B
c 1 = − e −t 0 s .
s s
This gives us

B − A B √ A
U(x,s) = − e −t 0 s e − s/kx + .
s s s
Invert this expression to obtain the solution in terms of the error function and the complementary
error function:

x x
u(x,t) = Aerf √ + Berfc √ (1 − H(t − t 0 ))
2 kt 2 kt
x x

+ Aerf √ + Berfc √ H(t − t 0 )
2 kt 2 kt

x
− Berfc √ H(t − t 0 ).
2 k(t − t 0 )

SECTION 17.4 PROBLEMS

2
∂u ∂ u
1. Solve 3. = k for x > 0,t > 0,
∂t ∂x 2
2
∂u ∂ u u(x,0)= e ,u(0,t) = 0, lim u(x,t) = 0
−x
= k for 0 < x < L,t > 0,
∂t ∂x 2 x→∞
4. Apply the Laplace transform with respect to t to the
u(x,0) = 0,u(0,t) = 0,u(L,t) = T 0 = constant.
problem
2
2. Solve ∂u = k ∂ u for 0 < x < L,t > 0,
∂t ∂x 2
2
∂u ∂ u
= k for x > 0,t > 0 u(x,0) = 1,u(0,t) = u(L,t) = 0.
∂t ∂x 2
2
u(x,0) = 0,u(0,t) = t , lim u(x,t) = 0 Then use the transform with respect to x to solve the
x→∞ resulting problem for U(x,s).
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October 14, 2010 15:25 THM/NEIL Page-635 27410_17_ch17_p611-640