Page 143 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS WITH LAPLACE TRANSFORMS 133
Thus we have

√
U(x, s ) = F(s )e −x s
Multiplying and dividing by s gives

√
e −x s
U(x, s ) = sF(s )
s
√
The inverse transform of e −x s /s is
√
e −x s x
L −1 = erfc √
s 2 t
and we have seen that

L{ f }= sF(s ) − f (0)
Thus, making use of convolution, we find

t

x x

u(x, t) = f (0)erfc √ + f (t − µ)erfc √ dµ
2 t 2 µ
µ=0
Example 8.7. Now consider a problem in cylindrical coordinates. An infinite cylinder is
initially at dimensionless temperature u(r, 0) = 1 and dimensionless temperature at the surface
u(1, t) = 0. We have
∂u 1 ∂ ∂u
= r
∂t r ∂r ∂r
u(1, t) = 0

u(r, 0) = 1
u bounded

The Laplace transform with respect to time yields

1 d dU
sU(r, s ) − 1 = r
r dr dr

with
1
U(1, s ) =
s