9Partial differential equations (PDEs)
IA Differential Equations
9.3 The diffusion equation
Heat conduction in a solid in one dimension is modelled by the diffusion equation
∂T
∂t
= κ
∂
2
T
∂x
2
This is known as a parabolic PDE (parabolic because it resembles y = ax
2
).
Here T (x, t) is the temperature and the constant κ is the diffusivity.
Example. Consider an infinitely long bar heated at one end (
x
= 0). Note
in general the “velocity”
∂T/∂t
is proportional to curvature
∂
2
T/∂x
2
, while in
the wave equation, it is the “acceleration” that is proportional to curvature.
In this case, instead of oscillatory, the diffusion equation is dissipative and all
unevenness simply decays away.
Suppose
T
(
x,
0) = 0,
T
(0
, t
) =
H
(
t
) =
(
0 t < 0
1 t > 0
, and
T
(
x, t
)
→
0 as
x → ∞
. In words, this says that the rod is initially cool (at temperature 0), and
then the end is heated up on one end after t = 0.
There is a similarity solution of the diffusion equation valid on an infinite
domain (or our semi-infinite domain) in which
T
(
x, t
) =
θ
(
η
), where
η
=
x
2
√
κt
.
Applying the chain rule, we have
∂T
∂x
=
dθ
dη
∂η
∂x
=
1
2
√
κt
θ
0
(η)
∂
2
T
∂x
2
=
1
2
√
κt
dθ
0
dη
∂η
∂x
=
1
4κt
θ
00
(η)
∂T
∂t
=
dθ
dη
∂η
∂t
= −
1
2
x
2
√
κ
1
t
3/2
θ
0
(η)
= −
η
2t
θ
0
(η)
Putting this into the diffusion equation yields
−
η
2t
θ
0
= κ
1
4κt
θ
00
θ
00
+ 2ηθ
0
= 0
This is an ordinary differential equation for
θ
(
η
). This can be seen as a first-
order equation for
θ
0
with non-constant coefficients. Use the integrating factor
µ = exp(
R
2η dη) = e
η
2
. So
(e
η
2
θ
0
)
0
= 0
θ
0
= Ae
−η
2
θ = A
Z
η
0
e
−u
2
du + B
= α erf(η) + B
where erf(η) =
2
√
π
R
η
0
e
−u
2
du from statistics, and erf(η) → 1 as η → ∞.
Now look at the boundary and initial conditions, (recall
η
=
x/
(2
√
κt
)) and
express them in terms of η. As x → 0, we have η → 0. So θ = 1 at η = 0.
Also, if x → ∞, t → 0
+
, then η → ∞. So θ → 0 as η → ∞.
So
θ
(0) = 1
⇒ B
= 1. Colloquially,
θ
(
∞
) = 0 gives
α
=
−
1. So
θ
= 1
−erf
(
η
).
This is also sometimes written as
erfc
(
η
), the error function complement of
η
. So
T = erfc
x
2
√
κt
In general, at any particular fixed time t
0
, T (x) looks like
x
T
with decay length
O
(
√
κt
). So if we actually have a finite bar of length
L
, we
can treat it as infinite if
√
κt L, or t L
2
/κ