1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

3.6 Wave Equation in Unbounded Regions 241

solution of the wave equation can come to our aid again here. We know that
the solution of Eq. (1) has the form

u(x,t)=ψ(x+ct)+φ(x−ct).

The two initial conditions boil down to

ψ(x)+φ(x)=f(x), 0 <x,

ψ(x)−φ(x)=G(x)+A, 0 <x.

As in the finite case, we have defined

G(x)=^1

c

∫x

0

g(y)dy

andAis any constant.
From the two initial conditions we obtain

ψ(x)=^12

(

f(x)+G(x)+A

)

, x> 0 ,

φ(x)=^1

2

(

f(x)−G(x)−A

)

, x> 0.

BothfandGare known forx>0. Thus

ψ(x+ct)=^12

(

f(x+ct)+G(x+ct)+A

)

is defined for allx>0andt≥0. Butφ(x−ct)is not yet defined forx−ct<0.
That means that we must extend the functionsfandGin such a way as to
defineφfor negative arguments and also satisfy the boundary condition. The
sole boundary condition is Eq. (4), which becomes

u( 0 ,t)= 0 =ψ(ct)+φ(−ct).

In terms of ̃fandG ̃, extensions offandG,thisis

0 =f(ct)+G(ct)+A+f ̃(−ct)−G ̃(−ct)−A.

SincefandGare not dependent on each other, we must have individually

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

That is, ̃fisfo, the odd extension off,andG ̃isGe, the even extension ofG.
Finally, we arrive at a formula for the solution:

u(x,t)=^1

2

[

fo(x+ct)+Ge(x+ct)

]

+^1

2

[

fo(x−ct)−Ge(x−ct)

]

. (6)