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