Mathematical Tools for Physics

10—Partial Differential Equations 302

The separated solutions are then

f(x)g(z) =enπix/L

(

Aenπz/L+Be−nπz/L

)

(35)

The solution forz > 0 is therefore the sum

V(x,z) =

∑∞

n=−∞

enπix/L

(

Anenπz/L+Bne−nπz/L

)

(36)

The coefficientsAnandBnare to be determined by Fourier techniques. First however, look at thez-behavior.
As you move away from the plane toward positivez, the potential should not increase without bound. Terms
such aseπz/Lhowever increase withz. This means that the coefficients of the terms that increase exponentially
inzcannot be there.

An= 0forn > 0 , and Bn= 0forn < 0

V(x,z) =A 0 +B 0 +

∑∞

n=1

enπix/LBne−nπz/L+

∑−^1

n=−∞

enπix/LAnenπz/L (37)

The combined constantA 0 +B 0 is really one constant; you can call itC 0 if you like. Now use the usual Fourier
techniques given that you know the potential atz= 0.

V(x,0) =C 0 +

∑∞

n=1

Bnenπix/L+

∑−^1

n=−∞

Anenπix/L

The scalar product ofemπix/Lwith this equation is

〈

emπix/L,V(x,0)

〉

=







2 LC 0 (m= 0)

2 LBm (m > 0 )

2 LAm (m < 0 )

(38)