Mathematical Tools for Physics

10—Partial Differential Equations 291

O

0

b

y

0

T 0

a

0

x

Look at this problem from several different angles, tear it apart, look at a lot of special cases, and see what
can go wrong. In the process you’ll see different techniques and especially a lot of applications of Fourier series.
This single problem will illustrate many of the methods used to understand boundary value problems.
Use the same method that I used before for heat flow in one dimension: separation of variables. Assume a
solution to be the product of a function ofxand a function ofy, then plug into the equation.

T(x,y) =f(x)g(y), then ∇^2 T=

d^2 f(x)

dx^2

g(y) +f(x)

d^2 g(y)

dy^2

= 0

Just as in Eq. ( 7 ), when you divide byfgthe resulting equation is separated into a term involvingxonly and
one involvingyonly.
1
f

d^2 f(x)

dx^2

+

1

g

d^2 g(y)

dy^2

= 0

Becausexandycan be varied independently, these must be constants adding to zero.

1

f

d^2 f(x)

dx^2

=α, and

1

g

d^2 g(y)

dy^2

=−α (20)

As before, the separation constant can be any real or complex number until you start applying boundary conditions.
You recognize that the solutions to these equations can be sines or cosines or exponentials or hyperbolic functions
or linear functions, depending on whatαis.
The boundary conditions state that the surface temperature is held at zero on the surfacesx= 0and
x=a. This suggests looking for solutions that vanish there, and that in turn says you should work with sines of
x. In the other direction the surface temperature vanishes on only one side so you don’t need sines in that case.
Theα= 0case gives linear functions isxand iny, and the fact that the temperature vanishes onx= 0and
x=akills these terms. (It does doesn’t it?) Pickαto be a negative real number: call itα=−k^2.

d^2 f(x)

dx^2

=−k^2 f =⇒ f(x) =Asinkx+Bcoskx