1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

386 Chapter 6 Laplace Transform

1 + 21 iexp

[

iωt−

√

ω

2 (^1 +i)x

]

−α 2 iexp

[

−iωt−

√

ω

2 (^1 −i)x

]

= 1 +αexp

(

−

√

ω

2

x

)

sin

(

ωt−

√

ω

2

x

)

.

Example 3.
If a steel wire is exposed to a sinusoidal magnetic field, the boundary value–
initial value problem that describes its displacement is

∂^2 u

∂x^2 =

∂^2 u

∂t^2 −sin(ωt),^0 <x<^1 ,^0 <t,

u( 0 ,t)= 0 , u( 1 ,t)= 0 , 0 <t,

u(x, 0 )= 0 ,

∂u

∂t(x,^0 )=^0 ,^0 <x<^1.

The nonhomogeneity in the partial differential equation represents the effect
of the force due to the field. The transformed equation and its solution are

d^2 U

dx^2

=s^2 U− ω

s^2 +ω^2

, 0 <x< 1 ,

U( 0 ,s)= 0 , U( 1 ,s)= 0 ,

U(x,s)= ω

s^2 (s^2 +ω^2 )

cosh(^12 s)−cosh

(

s(^12 −x)

)

cosh(^12 s)

.

Several methods are available for the inverse transformation ofU.Anobvi-
ous one would be to compute

v(x,t)=L−^1

(cosh( 1

2 s)−cosh

(

s(^12 −x)

)

s^2 cosh(^12 s)

)

and writeu(x,t)as a convolution

u(x,t)=

∫t

0

sin

(

ω

(

t−t′

))

v

(

x,t′

)

dt′.

The details of this development are left as an exercise.
We could also use the Heaviside formula. The application is now rou-
tine, except in the interesting case where cosh(iω/ 2 )=0, that is, whereω=
( 2 n− 1 )π, one of the natural frequencies of the wire.
Let us supposeω=π,so

U(x,s)=s (^2) (s 2 π+π (^2) )
cosh(^12 s)−cosh

(

s(^12 −x)

)

cosh(^12 s)

.