10—Partial Differential Equations 296
flowingintothis area has the reversed sign,
+κ∂T/∂y (26)
The total power flow is the integral of this over the area of the top face.
LetLbe the length of this long rectangular rod, its extent in thez-direction. The element of area along
the surface aty=bis thendA=Ldx, and the power flow into this face is
∫a
0
Ldxκ
∂T
∂y
∣
∣
∣
∣
y=b
The temperature function is the solution Eq. ( 23 ), so differentiate that equation with respect toy.
∫a
0
Ldxκ
4
π
T 0
∑∞
`=0
[(2`+ 1)π/a]
2 `+ 1
cosh
(
(2`+ 1)πy/a
)
sinh
(
(2`+ 1)πb/a
)sin
(2`+ 1)πx
a
at y=b
=
4 LκT 0
a
∫a
0
dx
∑∞
`=0
sin
(2`+ 1)πx
a
and this sum does not converge.I’m going to push ahead anyway, temporarily pretending that I didn’t notice this
minor difficulty with the series. Just go ahead and integrate the series term by term and hope for the best.
=
4 LκT 0
a
∑∞
`=0
a
π(2`+ 1)
[
−cos
(
(2`+ 1)π
)
+ 1
]
=
4 LκT 0
π
∑∞
`=0
2
2 `+ 1
=∞
This infinite series for the total power entering the top face is infinite. The series doesn’t converge (use the
integral test).
This innocuous-seeming problem is suddenly pathological because it would take an infinite power source
to maintain this temperature difference. Why should that be? Look at the corners. You’re trying to maintain
a non-zero temperature difference (T 0 − 0 ) between two walls that are touching. This can’t happen, and the