Lecture 14: Other Boundary Conditions for the Heat Equation
We have seen that the onedimensional heat equation can be used to model
the temperature in a rod which is insulated around its sides. The ﬁrst case we
solved assumed that both ends of the rod were held at 0
0
for all time
t >
0.
Other boundary conditions can be modelled, and we will treat two diﬀerent
cases.
Heat Equation with both ends insulated
If the ends of the rod are
insulated
, the boundary conditions on the partial
diﬀerential equation
u
t
=
α
2
u
xx
must be changed to
u
x
(0
,t
) =
u
x
(
L,t
) = 0 for
all
t >
0. This implies that no heat can ﬂow across either end of the rod. The
Method of Separation of Variables proceeds exactly as before, except that the
boundary conditions on the SturmLiouville problem
X
00
+
λX
= 0 must be
changed. With
u
(
x,t
) =
X
(
x
)
T
(
t
), the condition
u
x
(0
,t
)
≡
X
0
(0)
T
(
t
) = 0,
for all
t >
0 implies that
X
0
(0) = 0, and similarly the condition
u
x
(
L,t
) = 0
implies that
X
0
(
L
) = 0.
To solve the SturmLiouville problem
X
00
+
λX
= 0 with
X
0
(0) =
X
0
(
L
) =
0, we again need to treat the three cases
λ <
0
,λ
= 0
,λ >
0.
1. If
λ <
0, let
λ
=

K
2
. Then the general solution can be written as
X
(
x
) =
A
cosh(
Kx
) +
B
sinh(
Kx
)
,
X
0
(
x
) =
AK
sinh(
Kx
) +
BK
cosh(
Kx
)
.
Using the boundary conditions,
X
0
(0) =
AK
sinh(0) +
BK
cosh(0) = 0 +
BK
= 0
⇒
B
= 0
,
and
X
0
(
L
) =
AK
sinh(
KL
) = 0
⇒
A
= 0
,
since neither
K
nor
L
can be zero. This means that
X
(
x
)
≡
0 is the
only
solution, and therefore there are no negative eigenvalues.
2. If
λ
= 0, then the general solution is
X
(
x
) =
C
+
Dx
, and
X
0
(
x
) =
D
.
Both conditions
X
0
(0) =
X
0
(
L
) = 0 are satisﬁed if, and only if, the
constant
D
= 0. This means that
X
(
x
)
≡
C
, where
C
is a nonzero
constant, is a nonzero solution corresponding to the eigenvalue
λ
0
= 0.
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