10—Partial Differential Equations 249(IfA= 0everything is zero, so it has to beD.)
You can now write a general solution that satisfies three of the four boundary conditions. Combinethe coefficientsAandCinto one, and since it will be different for different values ofk, call itγn.
T(x,y) =
∑∞
n=1γnsin
nπx
a
sinhnπy
a
(10.25)
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=1γnsin
nπx
a
sinhnπb
a
=T 0
On this interval 0 < x < athese sine functions are orthogonal, so you take the scalar product of both
side with the sine.
∫a0dxsin
mπx
a
∑∞
n=1γnsin
nπx
a
sinhnπb
a
=
∫a0dxsin
mπx
a
T 0
a
2
γmsinh
mπb
a
=T 0
a
mπ
[
1 −(−1)m]
Just the oddmterms are present,m= 2`+ 1, so the result is
T(x,y) =
4
π
T 0
∑∞
`=01
2 `+ 1
sinh(
(2`+ 1)πy/a
)
sinh(
(2`+ 1)πb/a
)sin(2`+ 1)πx
a
(10.26)
You’re not done.
Does this make sense? The dimensions are clearly correct, but after that it takes some work.There’s really just one parameter that you have to play around with, and that’s the ratiob/a. If it’s
either very big or very small you may be able to check the result.
O
y
0
T 0
a b
a
x
O
0 0
a
b a
Ifa b, it looks almost like a one-dimensional problem. It is a thin slab with temperatureT 0
on one side and zero on the other. There’s little variation along thex-direction, so the equilibrium
equation is