Mathematical Tools for Physics

10—Partial Differential Equations 293

On this interval 0 < x < athese sine functions are orthogonal, so you take the scalar product of both side with
the sine. ∫
a

0

dxsin

mπx

a

∑∞

n=1

γnsin

nπx

a

sinh

nπb

a

=

∫a

0

dxsin

mπx

a

T 0

a

2

γmsinh

mπb

a

=T 0

a

mπ

[

1 −(−1)m

]

Only the oddmterms are present,m= 2`+ 1, so the result is

T(x,y) =

4

π

T 0

∑∞

`=0

1

2 `+ 1

sinh

(

(2`+ 1)πy/a

)

sinh

(

(2`+ 1)πb/a

)sin

(2`+ 1)πx

a

(23)

You’re not done.
Does this make sense? The dimensions are clearly correct, but after that it takes some work. There’s really
only 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

ab

a

x

O

0 0

a
ba
Ifab, 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

∇^2 T= 0 =

∂^2 T

∂x^2

+

∂^2 T

∂y^2

≈

∂^2 T

∂y^2

This simply says that the second derivative with respect toyvanishes, so the answer is the straight lineT=A+By,
and with the condition that you know the temperature aty= 0and aty=byou easily find

T(x,y)≈T 0 y/b