Bar with both ends insulated
Now let us consider the situation where, instead of them being kept at
constant 0 degree temperature, the two ends of the bar are also sealed with
perfect insulation so that no heat could escape to the outside environment
(recall that the side of the bar is always perfectly insulated in the one-
dimensional assumption), or vice versa.
The new boundary conditions are
u
x
(0,
t
) = 0
and
u
x
(
L
,
t
) = 0
, reflecting the fact that there will be no heat
transferring, spatially, across the points
x
= 0 and
x
=
L
.
The heat
conduction problem becomes the initial-boundary value problem below.
(Heat conduction eq.)
α
2
u
xx
=
u
t
,
0 <
x
<
L
,
t
> 0,
(Boundary conditions)
u
x
(0,
t
) = 0, and
u
x
(
L
,
t
) = 0,
(Initial condition)
u
(
x
,
0) =
f
(
x
).
The first step is the separation of variables.
The equation is the same as
before.
Therefore, it will separate into the exact same two ordinary
differential equations as in the first heat conduction problem seen earlier.
The new boundary conditions separate into
u
x
(0,
t
) = 0
→
X
′
(0)
T
(
t
) = 0
→
X
′
(0) = 0
or
T
(
t
) = 0
u
x
(
L
,
t
) = 0
→
X
′
(
L
)
T
(
t
) = 0
→
X
′
(
L
) = 0
or
T
(
t
) = 0
As before, we cannot choose
T
(
t
) = 0.
Else we could only get the trivial
solution
u
(
x
,
t
) = 0, rather than the general solution.
Hence, the new
boundary conditions should be
X
′
(0) = 0 and
X
′
(
L
) = 0.
Again, we end up with a system of two simultaneous ordinary differential
equations.
Plus a set of two boundary conditions that goes with the spatial
independent variable
x
:
X
″ +
λX
= 0,
X
′
(0) = 0
and
X
′
(
L
) = 0
,
T
′ +
α
2
λ
T
= 0
.