10—Partial Differential Equations 248Use the same method used before for heat flow in one dimension: separation of variables. Assumea 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. (10.8), when you divide byfgthe resulting equation is separated into a term involving
xonly 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 (10.23)
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= 0
andx=a. This suggests looking for solutions that vanish there, and that in turn says you should
work with sines ofx. 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= 0andx=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
The accompanying equation forgis now
d^2 g(y)
dy^2
= +k^2 g =⇒ g(y) =Csinhky+Dcoshky
(Or exponentials if you prefer.) The combined, separated solution to∇^2 T= 0is
(Asinkx+Bcoskx)(Csinhky+Dcoshky) (10.24)
The general solution will be a sum of these, summed over various values ofk. This is where you have
to apply the boundary conditions to determine the allowedk’s.
left: T(0,y) = 0 =B(Csinhky+Dcoshky), so B= 0
(This holds for allyin 0 < y < b, so the second factor can’t vanish unless bothCandDvanish. If
that is the case theneverything vanishes.)