Page 609 - Advanced_Engineering_Mathematics o'neil

16.5 Laplace Transform Techniques 589

Now
s s A

Y (x,s) = k 1 e sx/c − − k 1 e −sx/c .
c c s 3
Since e sx/c →∞ and x →∞, we must choose k 1 = 0. This gives us
A A
Y(x,s) = e −sx/c − .
s 3 s 3
−1
The solution is y(x,t) = L [Y(x,s)].Now
A A

−1 2
L = t
s 3 2
and, from the shifting theorems in Chapter 3,
A A x x
2
−1 −sx/c
L e = t − H t −
s 3 2 c c
in which H is the Heaviside function. The solution is therefore
A x 2 x A
2
y(x,t) = t − H t − − t .
2 c c 2
Since

x x 0 if x > ct,
2
t − H t − =
c c (t − x/c) 2 if x ≤ ct,
then
2
At
− for x > ct,
y(x,t) = 2 2
A x A
2 c 2 − xt for x ≤ ct.
c
A Problem on [0, L] Next consider the boundary value problem:
2
2
∂ y ∂ y
= c 2 for 0 < x < L,t > 0,
∂t 2 ∂x 2
∂y
y(x,0) = (x,0) = 0,
∂t
and
∂y
y(0,t) = 0, E (L,t) = f (t).
∂t
This models vibrations in an elastic bar of constant density ρ, uniform cross section and length
2
L. f (t) is a force per unit length acting parallel to the bar at the right end x = L.Here c = E/ρ,
where E is Young’s modulus for the material of the bar. The bar is initially at rest and lying flat
along the interval [0, L].
Apply the Laplace transform L, with respect t, using the boundary conditions and the oper-
ational rule for taking the transform of derivatives. This is similar to the problem just solved and
we obtain
s 2

Y (x,s) − Y(x,s) = 0,
c 2
in which Y (x,s) = ∂Y/∂x and s is carried along as a parameter. This is a second order linear

homogeneous constant coefficient differential equation. The characteristic equation is
2
s
2
λ − = 0,
c

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October 14, 2010 15:23 THM/NEIL Page-589 27410_16_ch16_p563-610