6.4 More Difficult Examples 383
6.4 More Difficult Examples
The technique of separation of variables, once mastered, seems more straight-
forward than the Laplace transform. However, when time-dependent bound-
ary conditions or inhomogeneities are present, the Laplace transform offers
a distinct advantage. Following are some examples that display the power of
transform methods.
Example 1.
A uniform insulated rod is attached at one end to an insulated container of
fluid. The fluid is circulated so well that its temperature is uniform and equal
to that at the end of the rod. The other end of the rod is maintained at a con-
stant temperature. A dimensionless initial value–boundary value problem that
describes the temperature in the rod is
∂^2 u
∂x^2 =∂u
∂t,^0 <x<^1 ,^0 <t,
∂u( 0 ,t)
∂x =γ∂u( 0 ,t)
∂t , u(^1 ,t)=^1 ,^0 <t,
u(x, 0 )= 0 , 0 <x< 1.The transformed problem and its solution are
d^2 U
dx^2=sU, 0 <x< 1 ,dU
dx(^0 ,s)=sγU(^0 ,s), U(^1 ,s)=1
s,U(x,s)=cosh(√
sx)+√
sγsinh(√
sx)
s(cosh(√
s)+√
sγsinh(√
s))=q(s)
p(s).
Aside froms=0, the denominator has no real zeros. Thus we again search
for complex zeros by employing
√
s=ξ+iη. The real and imaginary parts of
the denominator are to be computed by using the addition formulas for cosh
and sinh. The requirement that both real and imaginary parts be zero leads to
the equations
(
cosh(ξ )+ξγsinh(ξ )
)
cos(η)−ηγcosh(ξ )sin(η)= 0 , (1)
ηγsinh(ξ )cos(η)+(
sinh(ξ )+ξγcosh(ξ ))
sin(η)= 0. (2)We may think of these as simultaneous equations in sin(η)and cos(η).Because
sin^2 (η)+cos^2 (η)=1, the system has a solution only when its determinant is
zero. Thus after some algebra, we arrive at the condition
(
1 +ξ^2 γ^2 +η^2 γ^2
)
sinh(ξ )cosh(ξ )+ξγ(
sinh^2 (ξ )+cosh^2 (ξ )