Page 138 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

128 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
Taking the Laplace transform

sU − ς = U ςς
U(s, 0) = 0
U ς (s, 1) + HU(s, 1) = 0

The differential equation has a homogeneous solution
√ √
U h = A cosh( s ς) + B sinh( s ς)

and a particular solution
ς
U p =
s
so that
ς √ √
U = + A cosh( s ς) + B sinh( s ς)
s
Applying the boundary conditions, we find A = 0

1 + H
B =− √ √ √
s s cosh( s ) + H sinh( s )
The Laplace transform of the solution is as follows:

√
ς (1 + H)sinh( s ς)
U = − √ √ √
s s s cosh( s ) + H sinh( s )
The inverse transform of the first term is simply ζ. For the second term, we must first
find the poles. There is an isolated pole at s = 0. To obtain the residue of this pole note that
√ √
lim (1 + H)sinh ς s s τ lim (1 + H)(ς s + ··· )
− √ √ √ e = − √ √ =−ς
s → 0 s cosh s + H sinh s s → 0 s + H( s + ··· )
√
canceling the first residue. To find the remaining residues let s = x + iy.Then

(x + iy) [cosh x cos y + i sinh x sin y] + H [sinh x cos y + i cosh x sin y] = 0

Setting real and imaginary parts equal to 0 yields

x cosh x cos y − y sinh x sin y + H sinh x cos y = 0

and

y cosh x cos y + x sinh x sin y + H cosh x sin y = 0