1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

240 Chapter 3 The Wave Equation

Consider the problem

∂^2 u

∂x^2 =

1

c^2

∂^2 u

∂t^2 ,^0 <t,^0 <x, (1)

u(x, 0 )=f(x), 0 <x, (2)

∂u

∂t(x,^0 )=g(x),^0 <x, (3)

u( 0 ,t)= 0 , 0 <t. (4)

We require in addition that the solutionu(x,t)be bounded asx→∞.
On separating variables, we makeu(x,t)=φ(x)T(t)and find that the fac-
tors satisfy

T′′+λ^2 c^2 T= 0 , 0 <t,

φ′′+λ^2 φ= 0 , 0 <x,

φ( 0 )= 0 , |φ(x)| bounded.

The solutions are easily found to be

φ(x;λ)=sin(λx), T(t;λ)=Acos(λct)+Bsin(λct),

and we combine the productsφ(x;λ)T(t;λ)in a Fourier integral

u(x,t)=

∫∞

0

(

A(λ)cos(λct)+B(λ)sin(λct)

)

sin(λx)dλ. (5)

The initial conditions become

u(x, 0 )=f(x)=

∫∞

0

A(λ)sin(λx)dλ, 0 <x,

∂u

∂t

(x, 0 )=g(x)=

∫∞

0

λcB(λ)sin(λx)dλ, 0 <x.

Both of these equations are Fourier integrals. Thus the coefficient functions
are given by

A(λ)=

2

π

∫∞

0

f(x)sin(λx)dx, B(λ)=

2

πλc

∫∞

0

g(x)sin(λx)dx.

It is sufficient to demand that

∫∞

0 |f(x)|dxand

∫∞

0 |g(x)|dxboth be finite in
order to guarantee the existence ofAandB.
The deficiency of the Fourier integral form of the solution given in Eq. (5)
is that the formula gives no idea of whatu(x,t)looks like. The d’Alembert