Mathematical Tools for Physics

10—Partial Differential Equations 285

whereH~ is the heat flow vector, the power per area in the direction of the energy transport. H~.dA~=dP, the
power going across the areadA~. The total heat flowing into a volume is

ˆn

dQ

dt

=−

∮

dP=−

∮

H~.dA~

where the minus sign occurs because I want the heatin. For a small volume∆V, you now havem=ρ∆V and

mC

∂T

∂t

=ρ∆V C

∂T

∂t

=−

∮

H~.dA~

Divide by∆V and take the limit as∆V→ 0. The right hand side is the divergence

ρC

∂T

∂t

=− lim

∆V→ 0

1

∆V

∮

H~.dA~=−∇.H~ = +∇.κ∇T= +κ∇^2 T

Again, this assumes that the thermal conductivity,κ, is independent of position.

10.2 Separation of Variables
How do you solve these equations? I’ll start with the one-dimensional case and use the method ofseparation of
variables. The trick is to start by looking for a solution to the equation in the form of a product of a function
ofxand a function oft. T(x,t) =f(t)g(x). I do not assume that every solution to the equation will look
like this — that’s just not true. What will happen is that I’ll be able to express every solution as a sum of such
factored forms.
If you want to find out if you’ve got a solution, plug in:

∂T

∂t

=

κ

Cρ

∂^2 T

∂x^2

is

df

dt

g=

κ

Cρ

f

d^2 g

dx^2