Page 653 - Advanced_Engineering_Mathematics o'neil

17.4 Laplace Transform Techniques 633

This can be inverted using the geometric series, as we did for the problem with the wave equation
on [0, L] in Section 16.5. Recall that the geometric series is
∞
1
n
n
= (−1) u for |u| < 1.
1 + u
n=0
In the following, the third line is obtained from the second by multiplying numerator and
√
denominator by e − s/kx :
√
√ √
cosh s/kx e s/kx + e − s/kx
√
= √ √
cosh s/kL e s/kL + e − s/kL
√ √ √ √
e s/kx − s/kL + e − s/kx − s/kL
e
e
= √
1 + e −2 s/kL

√ √ 1
− s/k(L−x) − s/k(L+x)
= e + e √
1 + e −2 s/kL
∞
√ √ √
− s/k(L−x) − s/k(L+x)
n −2n s/kL
= e + e (−1) e
n=0
∞
√
√
n − s/k((2n+1)L−x) − s/k((2n+1)L+x)
= (−1) e + e .
n=0
Then
T 0
U(x,s) =
s
∞
√ √
T 0 n − s/k((2n+1)L−x) − s/k((2n+1)L+x)

− (−1) e + e .
s
n=0
Taking the inverse transform term by term, we have the solution
u(x,t) = T 0
∞
n (2n + 1)L − x (2n + 1)L + x
(−1) erfc √ + erfc √ .
− T 0
2 kt 2 kt
n=0
Temperature Distribution in a Semi-Infinite Bar
We will solve the boundary value problem
2
∂u ∂ u
= k for x > 0,t > 0
∂t ∂x 2
and
u(x,0) = 0,u(0,t) = f (t).
This models diffusion in a thin homogeneous bar lying along the x-axis with a given initial
temperature function. Because there is no bound on x, we also impose the condition
lim u(x,t) = 0for t > 0.
x→∞
Take the transform in the time variable to obtain
2
∂ U
sU(x,s) = k
∂x 2

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October 14, 2010 15:25 THM/NEIL Page-633 27410_17_ch17_p611-640