Mathematical Tools for Physics

10—Partial Differential Equations 307

be represented by a sum of sines inx, and the same statement will hold for theycoordinate. This leads to the
form of the sum

V(x,y,z) =

1

2

V 0 +

∑∞

n=1

∑∞

m=1

αnmsin

(nπx

a

)

sin

(mπy

b

)

e−knmz

whereknmis thek 3 of the preceding equation. What happened to the other term inz, the one with the positive
exponent? Did I say that I’m looking for solutions in the domainz > 0?
Atz= 0this must match the boundary conditions stated, and as before, the orthogonality of the sines on
the two domains allows you to determine the coefficients. You simply have to do two integrals instead of one.
See problem 19.

V(x,y,z >0) =

1

2

V 0 +

8 V 0

π^2

∑∞

oddn

∑∞

oddm

1

nm

sin

(nπx

a

)

sin

(mπy

b

)

e−knmz (45)