1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

3.3 d’Alembert’s Solution 229

If we divide through the second equation bycand integrate, it becomes

ψ(x)−φ(x)=G(x)+A, 0 <x<a, (7)

whereG(x)stands for

G(x)=

1

c

∫x

0

g(y)dy (8)

andAis an arbitrary constant. Equations (6) and (7) can now be solved simul-
taneously to determine

ψ(x)=^12

(

f(x)+G(x)+A

)

, 0 <x<a,

φ(x)=^1

2

(

f(x)−G(x)−A

)

, 0 <x<a.

These equations giveψandφonly for values of the argument between 0
anda.Butx±ctmay take on any value whatsoever, so we must extend these
functions to define them for arbitrary values of their arguments. Let us desig-
nate them as

ψ(x)=^12

( ̃

f(x)+G ̃(x)+A

)

,

φ(x)=^1

2

( ̃

f(x)−G ̃(x)−A

)

,

wheref ̃andG ̃are some extensions offandG.(Thatisf ̃(x)=f(x)andG ̃(x)=
G(x)for 0<x<a.) However we choose these extensions, the wave equation
and the initial conditions are satisfied. Thus, they must be determined by the
boundary conditions,

u( 0 ,t)=ψ(ct)+φ(−ct)= 0 , t> 0 , (9)

u(a,t)=ψ(a+ct)+φ(a−ct)= 0 , t> 0. (10)

The first of these equations says that

̃f(ct)+G ̃(ct)+A+f ̃(−ct)−G ̃(−ct)−A= 0 ,

or
̃f(ct)+f ̃(−ct)+G ̃(ct)−G ̃(−ct)= 0.

As these equations must be true for arbitrary functionsfandG(because the
two functions are not interdependent), we must have individually

̃f(ct)=−f ̃(−ct), G ̃(ct)=G ̃(−ct). (11)

That is, ̃fis odd andG ̃is even.