Page 133 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS WITH LAPLACE TRANSFORMS 123
Thus for each of the sets of poles we have

4g(−1) n (2n − 1)πς (2n − 1)πτi
R(s = s n ) = sin exp
2
π (2n − 1) 2 2 2
Now adding the residues corresponding to each pole and its conjugate we find that the
final solution is as follows:

∞
8 (−1) n (2n − 1)πς (2n − 1)πτ
y(ς, τ) = g ς + sin cos
π 2 (2n − 1) 2 2 2
n=1
Suppose that instead of a constant force at ζ = 1, we allow g to be a function of τ.In
this case, the Laplace transform of y(ζ, τ) takes the form

G(s )sinh(ςs )
Y (ς, s ) =
s cosh s

The simple pole with residue gζ is not present. However, the other poles are still at the
same s n values. The residues at each of the conjugate poles of the function

sinh(ς s )
F(s ) =
s cosh s
are

2(−1) n (2n − 1)πς (2n − 1)πτ
sin sin = f (ς, τ)
π(2n − 1) 2 2

According to the convolution theorem

τ

y(ς, τ) = y(τ − τ )g(τ )dτ

τ =0
τ
∞ n
4 (−1) (2n − 1)πς (2n − 1)πτ

y(ς, τ) = sin g(τ − τ )sin dτ .
π (2n − 1) 2 2
n=0
τ
In thecasethat g = constant, integration recovers the previous equation.
Example 8.2. An infinitely long string is initially at rest when the end at x = 0 undergoes
a transverse displacement y(0, t) = f (t). The displacement is described by the differential