Mathematical Tools for Physics

10—Partial Differential Equations 292

The accompanying equation forgis now

d^2 g(y)

dy^2

= +k^2 g =⇒ g(y) =Csinhky+Dcoshky

(Or exponentials if you prefer.) The combined, separated solution to∇^2 T= 0is

(Asinkx+Bcoskx)(Csinhky+Dcoshky) (21)

The general solution will be a sum of these, summed over various values ofk. This is where you have to apply
the boundary conditions to determine the allowedk’s.

left: T(0,y) = 0 =B(Csinhky+Dcoshky), so B= 0

(This holds for allyin 0 < y < b, so the second factor can’t vanish unless bothCandDvanish. If that is the
case theneverythingvanishes.)

right: T(a,y) = 0 =Asinka(Csinhky+Dcoshky), so sinka= 0

(Again, the factor withycan’t vanish or everything vanishes. IfA= 0then everything vanishes. All that’s left
issinka= 0.)
bottom: T(x,0) = 0 =AsinkxD, so D= 0

(IfA= 0everything is zero, so it’s got to beD.)
You can now write a general solution that satisfies three of the four boundary conditions. Combine the
coefficientsAandCinto one, and since it will be different for different values ofk, call itγn.

T(x,y) =

∑∞

n=1

γnsin

nπx

a

sinh

nπy

a

(22)

Thenπ/aappears becausesinka= 0, and the limits onnomit the negativenbecause they are redundant.
Now to find all the unknown constantsγn, and as before that’s where Fourier techniques come in. The
fourth side, aty=b, has temperatureT 0 and that implies

∑∞

n=1

γnsin

nπx

a

sinh

nπb

a

=T 0