Mathematical Tools for Physics

10—Partial Differential Equations 298

O

0

b

y

0

F 0

a

0

x

−F 0 y/κ

O

b

y

0 a

−F 0 y/κ

x

If I can find a solution that equals−F 0 y/κon the left and right faces then it will cancel the+F 0 y/κthat
Eq. ( 27 ) provides. ButI can’t disturb the top and bottom boundary conditions. The way to do that is to find
functions that equal zero aty= 0and whose derivative equals zero aty=b. This is a familiar sort of condition
that showed up several times in chapter five on Fourier series. It is equivalent to saying that the top surface is
insulated so that heat can’t flow through it. You then use superposition to combine the solution with uniform
heat flow and the solution with an insulated boundary.
Instead of Eq. ( 21 ), use the opposite sign forα, so the solutions are of the form

(Asinky+Bcosky)(Csinhkx+Dcoshkx)

I require that this equals zero aty= 0, so that says

(0 +B)(Csinhkx+Dcoshkx) = 0

soB= 0. I require that the derivative equals zero aty=b, so

Akcoskb= 0, or kb= (n+^1 / 2 )π for n= 0, 1 , 2 ...

The value of the temperature is the same on the left that it is on the right, so

Csinhk0 +Dcoshk0 =Csinhka+Dcoshka =⇒ C=D(1−coshka)/sinhka (28)

This is starting to get messy, so I think it’s time to look around and see if I’ve missed anything that could
simplify the calculation. There’s no guarantee that there is any simpler way, but it’s always worth looking. The
fact that the system is the same on the left as on the right means that the temperature will be symmetric about