Mathematical Tools for Physics

10—Partial Differential Equations 295

As compared to the previous approximation whereab, you can’t as easily tell whether this is plausible or not.
You can however learn from it.
At the very top, wherey=bthis reduces to the constantT 0 that you’re supposed to have at that position.
Recall again the Fourier series for a constant on 0 < x < a.
Move down fromy=bby the distancea, so thatb−y=a. That’s a distance from the top equal to the
width of the rectangle. It’s still rather close to the end, but look at the series for that position.

x=0

y=b

y=b−a

x=a

T(x,b−a)≈

4

π

T 0

∑∞

`=0

1

2 `+ 1

e−(2`+1)πsin

(2`+ 1)πx

a

For= 0, the exponential factor ise−π= 0. 043 and for= 1this factor ise−^3 π= 0. 00008. This means that
measured from theT 0 end, within the very short distance equal to the width, the temperature has dropped 95%
of the way down to its limiting value of zero. The temperature in the rod is quite uniform until you’re very close
to the heated end.

The Heat Flow into the Box
All the preceding analysis and discussion was intended to make this problem and its solution sound oh-so-plausible.
There’s more, and it isn’t pretty.
The temperature on one of the four sides was given as different from the temperatures on the other three
sides. What will the heat flow into the region be? That is, what power must you supply to maintain the
temperatureT 0 on the single wall?
At the beginning of this chapter, Eq. ( 1 ), you have the equation for the power through an areaA, but that
equation assumed that the temperature gradient∂T/∂xis the same all over the areaA. If it isn’t, you simply
turn it into a density.

∆P=−κ∆A

∂T

∂x

, and then

∆P

∆A

→

dP

dA

=−κ

∂T

∂x

(25)

Equivalently, just use the vector form from Eq. ( 6 ),H~ =−κ∇T. In Eq. ( 19 ) the temperature isT 0 alongy=b,
and the power density (energy/(time.area)) flowing in the+ydirection is−κ∂T/∂y, so the power density