1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

378 Chapter 6 Laplace Transform

from which we find the solution,

u(x,t)=

(

cos(πt)−^1

π

sin(πt)

)

sin(πx). 

Example 3.
Now we consider a problem that we know to have a more complicated solu-
tion:
∂^2 u
∂x^2 =

∂u

∂t,^0 <x<^1 ,^0 <t,

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

u(x, 0 )= 0 , 0 <x< 1.

The transformed problem is

d^2 U

dx^2

=sU, 0 <x< 1 ,

U( 0 ,s)=

1

s, U(^1 ,s)=

1

s.

The general solution of the differentialequationiswellknowntobeacombi-
nation of sinh(√sx)and cosh(√sx). Application of the boundary conditions
yields

U(x,s)=^1

s

cosh

(√

sx

)

+(^1 −cosh(

√

s))sinh(

√

sx)

ssinh(

√

s)

=sinh(

√

sx)+sinh(

√

s( 1 −x))

ssinh(√s).

This function rarely appears in a table of transforms. However, by extending
the Heaviside formula, we can compute an inverse transform.
WhenUis the ratio of two transcendental functions (not polynomials) ofs,
we wish to write

U(x,s)=

∑

An(x)s−^1 r

n

.

In this formula, the numbersrnare values ofsfor which the “denominator” of
Uis zero, or, rather, for which|U(x,s)|becomes infinite; theAnare functions
ofxbut nots.Fromthisformweexpecttodetermine

u(x,t)=

∑

An(x)exp(rnt).

This solution should be checked for convergence.
The hyperbolic sine (also the cosh, cos, sin, and exponential functions) is
not infinite for any finite value of its argument. ThusU(x,s)becomes infinite