Page 769 - Advanced_Engineering_Mathematics o'neil

22.4 Residues and the Inverse Laplace Transform 749

This means that
T 0
c = √ .
sI 0 ( sR)
The transform of the solution is therefore
√
T 0 I 0 ( sr)
U(r,s) = √ .
sI 0 ( sR)
We must invert this to obtain u(r,t). To use Theorem 22.7, we need the singularities of
√
T 0 I 0 ( zr)
tz
tz
e U(r, z) = e √ .
zI 0 ( zR)
tz
Singularities of e U(r, z) occur at zeros of the denominator. There is a simple pole at z = 0
because I 0 (0) = 1 = 0. Furthermore,
√ √
I 0 ( zR) = J 0 (i zR) = 0
√
if i zR is a zero of J 0 . These zeros are real, simple, and nonzero. If the positive zeros are labeled
√ √
j 1 , j 2 ,···, then all the zeros are ± j 1 ,± j 2 ,···. Therefore, I 0 ( zR) = 0if zR =±ij n for some
n. Then
2
2
z =− j /R .
n
2
2
tz
Therefore, e U(r, z) has simple poles at 0 and − j /R for n = 1,2,···. Inverting U(r,s) by
n
Theorem 22.7 yields the solution
∞

tz tz 2 2
u(r,t) = Res(e U(r, z), z = 0) + Res e F(z), z =− j /R .
n
n=1
There remains to compute these residues. First,
√
I 0 ( zr)
tz
Res(e U(r, z),0) = lim ze tz √
z→0 zI 0 ( zR)
√
I 0 ( zr)
tz
= lime √
z→0 I 0 ( zR)
I 0 (0)
= = 1.
I 0 (0)
For the residues at the other poles, use Corollary 22.1, since the poles are simple zeros of the
denominator of a function of the form g(z)/h(z) with
√
tz
e I 0 ( zr) √
g(z) = and h(z) = I 0 ( zR).
z
Then
2 2
Res g(z)/h(z),− j /R
n
2
e − j n t/R 2 I 0 (− j n ri/R) 1
= √
2
− j /R 2 d I 0 ( zR)
n dz − j n /R 2
2
√
2
e − j n t/R 2 I 0 (− j n ri/R) 2 z
= √

− j /R 2 RI ( zR) z=− j n /R 2
2
2
n
0
−2Ri − j n t/R 2 I 0 ( j n ri/R)
2
= e .

j n I ( j n i)
0
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October 14, 2010 15:37 THM/NEIL Page-749 27410_22_ch22_p729-750