10—Partial Differential Equations 253O
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. (10.30) 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. (10.24), 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. Now require that the derivative equals zero aty=b, so
Akcoskb= 0, or kb= (n+^12 )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 (10.31)
This is starting to get messy, so 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 is 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 the central axis of the box, aboutx=a/ 2. That it is even about this point
implies that the hyperbolic functions ofxshould be even aboutx=a/ 2. You can do this simply by
using acoshabout that point.
Asinky
(
Dcoshk(x−a 2 )
)
Put these together and you have a sum
∑∞n=0ansin
(
(n+^12 )πy
b
)
cosh(
(n+^12 )π(x−a 2 )
b
)
(10.32)
Each of these terms satisfies Laplace’s equation, satisfies the boundary conditions aty= 0andy=b,
and is even about the centerlinex=a/ 2. It is now a problem in Fourier series to match the conditions
atx= 0. They’re then automatically satisfied atx=a.
∑∞
n=0ansin
(
(n+^12 )πy
b
)
cosh