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To illustrate the method we consider the heat equation
 |
(2.48) |
with the boundary conditions
 |
(2.49) |
for all time and the initial condition, at
, is
 |
(2.50) |
where
is a given function of
.
The temperature,
, is assumed seperable in
and
and we
write
 |
(2.51) |
so that (2.48) becomes
 |
(2.52) |
or, on dividing by
,
 |
(2.53) |
where
is the separation constant. In fact, we expect
to be
negative as can be seen from the time equation. Here we have
with solution
Thus,
will increase in time if
is positive and decrease in
time only if
is negative. It is obviously unphysical for the
temperature to increase in time without any additional heating
mechanism and so we must assume that
is negative. To force this we
set
 |
(2.54) |
The spatial equation is then
 |
(2.55) |
which is the simple harmonic motion equation with trigonometric
solutions. Thus, the solution to (2.55) is
 |
(2.56) |
Now, applying the boundary conditions (2.49), we find that
so that, in (2.56),
and
 |
(2.57) |
As for the wave equation, we take the most general solution by adding
together all the possible solutions, satisfying the boundary
conditions, to obtain
 |
(2.58) |
The final step is to apply the initial conditions, namely
 |
(2.59) |
Again we must invert the Fourier series and we do this by multiplying
the equation by
and integrating between
and
.
Thus, we obtain,
 |
(2.60) |
for all positive integers,
.
Note that because every term in the solution for
has a
negative exponential in it, the temperature must decrease in time and
the final solution will tend to
This is different from the wave equation where the oscillations simply
continued for all time. This trivial solution,
, is a
consequence of the particular boundary conditions chosen here.
Example 2. .28Solve
subject to the boudary conditions
and the initial condition
The solution is given by (2.58) with, from
(2.60),
 |
(2.61) |
Integration by parts yields, for even
,
and
for odd
.
For
problems involving heat flow there is no reason why the temperature
should have the same value at each end of the rod. The next few
examples will show the effect of different boundary conditions.
Example 2. .29Solve
subject to the boudary conditions
and the initial condition
In this example
and
are not equal. To proceed we
consider the final steady state after a long period of time.
Eventually we expect all temperature variations to be smoothed out in
time to leave a solution that is independent of time. This solution
satisfies the equation
and has solution
where the constants
and
are chosen to satisfy the boundary
conditions in
. Thus,
However, this does not satisfy the initial condition. Thus, we add on
a solution,
, that satisfies the heat equation and
the boundary conditions
Thus, the solution
satisfies the heat equation and the boundary conditions for the full
problem. From (2.58) we have
The initial condition on
can be written as
Thus, we have
Hence, from (2.60) we have
Thus,
Example 2. .30Solve
subject to
Thus,
and
Integration by parts gives
Hence,
Hence, the solution is
Instead of specifying the value of the temperature at the ends of the
rod we could fix
instead. This corresponds to
fixing the heat flux that enters or leaves the system. For example,
if
, then no heat enters the system and the
ends are said to be insulated. The following example illustrates the
case when one end is insulated and the other has a fixed temperature.
Example 2. .31Solve the heat equation subject to the boundary conditions
and the initial condition
In this case the steady state solution must satisfy the boundary
conditions and is simply
Again we look for a seperable solution of the form
and the solution reduces to
Applying the boundary condition at
gives
and the condition at
gives
so that, in this case
must be an odd integer times
. Thus,
Hence, the solution is
Finally, the initial condition is given by inverting the Fourier
series at
. Thus,
Again multiply by
and integrate
between
and
to get
Next: About this document ...
Up: Partial Differential Equations of
Previous: Modelling: Derivation of the
Prof. Alan Hood
2000-10-30