642:527
FALL 2007
THE HEAT EQUATION IN A DISK
Periodic and singular Sturm-Liouville problems
In these notes we study the two-dimensional heat equation in a disk of radius
a
:
α
2
∇
2
u
(
x,y,t
) =
∂
∂t
u
(
x,y,t
)
,
x
2
+
y
2
≤
a
2
.
Here
∇
2
, also written as
△
, is the two-dimensional Laplacian, that is,
∇
2
u
=
u
xx
+
u
yy
.
We will impose a homogeneous Dirichlet boundary condition at the boundary of the disk,
i.e., require that
u
(
x,y,t
) = 0 if
x
2
+
y
2
=
a
2
; this could easily be replaced by a Neumann
or Robin boundary condition.
We will also impose an initial condition specifying the
temperature distribution at time
t
= 0.
Since the domain and the PDE have rotational symmetry it is natural to work in polar
coordinates (
r,θ
), where
x
=
r
cos
θ
and
y
=
r
sin
θ
. We must then rewrite the
x
and
y
derivatives contained in
∇
2
as derivatives with respect to
r
and
θ
. The result is derived
in Section 16.7 of our text, with the final result appearing in equation (18); that equation
refers to cylindrical coordinates, but to pass to polar coordinates one simply neglects all
derivatives with respect to
z
.
The
r
derivatives appearing in (18) can conveniently be
rewritten as
∂
2
u
∂r
2
+
1
r
∂u
∂r
≡
1
r
∂
∂r
parenleftbigg
r
∂u
∂r
parenrightbigg
.
Thus our problem becomes
1
r
∂
∂r
parenleftbigg
r
∂u
∂r
parenrightbigg
+
1
r
2
∂
2
u
∂θ
2
=
1
α
2
∂u
∂t
,
0
≤
r<a,
0
≤
θ<
2
π, t>
0;
(1)
u
(
a,θ,t
) = 0
,
0
≤
θ<
2
π,t>
0;
(2)
u
(
r,θ,
0) =
f
(
r,θ
)
,
0
≤
r<a,
0
≤
θ<
2
π.
(3)
To attack this problem we will first find, by separation of variables, solutions of the PDE
(1) that also satisfy the boundary conditions (2); then we will form a solution that also
satisfies the initial condition (3) as a superposition of these product solutions.
To apply the separation of variables method we look for solutions of (1) in the form
u
(
r,θ,t
) =
R
(
r
)Θ(
θ
)
T
(
t
)
.
Inserting this proposed solution into (1) leads to
Θ
T
r
d
dr
parenleftbigg
r
dR
dr
parenrightbigg
+
RT
r
2
d
2
Θ
dθ
2
=
R
Θ
α
2
dT
dt
.
We divide this equation by
R
Θ
T
to obtain
1
r
(
rR
′
)
′
R
+
1
r
2
Θ
′′
Θ
=
1
α
2
T
′
T
.
(4)
1