1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

380 Chapter 6 Laplace Transform

whereqandpare the obvious choices. We take

√

rn=+inπin all calculations:

p′(s)=sinh

(√

s

)

+s^1

2

√

s

cosh

(√

s

)

,

p′(rn)=^1

2

inπcosh(inπ)=^1

2

inπcos(nπ),

q(rn)=sinh(inπx)+sinh

(

inπ( 1 −x)

)

=i

[

sin(nπx)+sin

(

nπ( 1 −x)

)]

.

Hence the portion ofu(x,t)that arises from eachrnis

An(x)exp(rnt)= 2 sin(nπx)+sin(nπ(^1 −x))

nπcos(nπ)

exp

(

−n^2 π^2 t

)

.

Part c. On assembling the various pieces of the solution, we get

u(x,t)= 1 +

2

π

∑∞

1

sin(nπx)+sin(nπ( 1 −x))

ncos(nπ) exp

(

−n^2 π^2 t

)

.

The same solution would be found by separation of variables but in a slightly
different form.

Example 4.
Now consider the wave problem

∂^2 u

∂x^2 =

∂^2 u

∂t^2 ,^0 <x<^1 ,^0 <t,

u( 0 ,t)= 0 , ∂u(∂^1 x,t)= 0 , 0 <t,

u(x, 0 )= 0 , ∂u(∂xt,^0 )=x, 0 <x< 1.

The transformed problem is

d^2 U

dx^2

=s^2 U−x, 0 <x< 1 ,

U( 0 ,s)= 0 , U′( 1 ,s)= 0 ,

and its solution (by undetermined coefficients or otherwise) gives

U(x,s)=

sxcosh(s)−sinh(sx)
s^3 cosh(s).
The numerator of this function is never infinite. The denominator is zero at
s=0ands=±i( 2 n− 1 )π/2(n= 1 , 2 ,...). We shall again use the Heaviside
formula to determine the inverse transform ofU.