Okay, it is finally time to completely solve a partial
differential equation. In the previous
section we applied separation of variables to several partial differential
equations and reduced the problem down to needing to solve two ordinary
differential equations. In this section
we will now solve those ordinary differential equations and use the results to
get a solution to the partial differential equation. We will be concentrating on the heat equation
in this section and will do the wave equation and Laplace’s equation in later
sections.
The first problem that we’re going to look at will be the
temperature distribution in a bar with zero temperature boundaries. We are going to do the work in a couple of
steps so we can take our time and see how everything works.
The first thing that we need to do is find a solution that
will satisfy the partial differential equation and the boundary
conditions. At this point we will not
worry about the initial condition. The
solution we’ll get first will not satisfy the vast majority of initial
conditions but as we’ll see it can be used to find a solution that will satisfy
a sufficiently nice initial condition.
|
Example 1 Find
a solution to the following partial differential equation that will also satisfy
the boundary conditions.
Solution
Okay the first thing we technically need to do here is
apply separation of variables. Even though we did that in the previous
section let’s recap here what we did.
First, we assume that the solution will take the form,
and we plug
this into the partial differential equation and boundary conditions. We separate the equation to get a function
of only t on one side and a
function of only x on the other
side and then introduce a separation constant. This leaves us with two ordinary
differential equations.
We did all of
this in Example 1
of the previous section and the two ordinary differential equations are,
The time
dependent equation can really be solved at any time, but since we don’t know
what is yet let’s hold off on that one. Also note that in many problems only the
boundary value problem can be solved at this point so don’t always expect to
be able to solve either one at this point.
The spatial
equation is a boundary value problem and we know from our work in the
previous chapter that it will only have non-trivial solutions (which we want)
for certain values of ,
which we’ll recall are called eigenvalues.
Once we have those we can determine the non-trivial solutions for each
,
i.e. eigenfunctions.
Now, we
actually solved the spatial problem,
in Example
1 of the Eigenvalues and Eigenfunctions section of the previous chapter
for . So, because we’ve solved this once for a
specific L and the work is not all
that much different for a general L
we’re not going to be putting in a lot of explanation here and if you need a
reminder on how something works or why we did something go back to Example 1
from the Eigenvalues and Eigenfunctions section for a reminder.
We’ve got three
cases to deal with so let’s get going.
In this case we
know the solution to the differential equation is,
Applying the
first boundary condition gives,
Now applying
the second boundary condition, and using the above result of course, gives,
Now, we are
after non-trivial solutions and so this means we must have,
The positive
eigenvalues and their corresponding eigenfunctions of this boundary value
problem are then,
Note that we
don’t need the in the eigenfunction as it will just get
absorbed into another constant that we’ll be picking up later on.
The solution to
the differential equation in this case is,
Applying the
boundary conditions gives,
So, in this
case the only solution is the trivial solution and so is not an eigenvalue for this boundary value
problem.
Here the
solution to the differential equation is,
Applying the
first boundary condition gives,
and applying
the second gives,
So, we are
assuming and so and this means . We therefore we must have and so we can only get the trivial solution
in this case.
Therefore,
there will be no negative eigenvalues for this boundary value problem.
The complete
list of eigenvalues and eigenfunctions for this problem are then,
Now let’s solve
the time differential equation,
and note that
even though we now know we’re not going to plug it in quite yet to
keep the mess to a minimum. We will
however now use to remind us that we actually have an
infinite number of possible values here.
This is a
simple linear (and separable for that matter) 1st order
differential equation and so we’ll let you verify that the solution is,
Okay, now that
we’ve gotten both of the ordinary differential equations solved we can
finally write down a solution. Note
however that we have in fact found infinitely many solutions since there are
infinitely many solutions (i.e.
eigenfunctions) to the spatial problem.
Our product
solution are then,
We’ve denoted
the product solution to acknowledge that each value of n will yield a different
solution. Also note that we’ve changed
the c in the solution to the time
problem to to denote the fact that it will probably be
different for each value of n as
well and because had we kept the with the eigenfunction we’d have absorbed it
into the c to get a single constant
in our solution.
|
So, there we have it.
The function above will satisfy the heat equation and the boundary
condition of zero temperature on the ends of the bar.
The problem with this solution is that it simply will not
satisfy almost every possible initial condition we could possibly want to
use. That does not mean however, that
there aren’t at least a few that it will satisfy as the next example
illustrates.
So, we’ve seen that our solution from the first example will
satisfy at least a small number of highly specific initial conditions.
Now, let’s extend the idea out that we used in the second
part of the previous example a little to see how we can get a solution that
will satisfy any sufficiently nice initial condition. The Principle of Superposition is, of course,
not restricted to only two solutions.
For instance the following is also a solution to the partial
differential equation.
and notice that this solution will not only satisfy the
boundary conditions but it will also satisfy the initial condition,
Let’s extend this out even
further and take the limit as . Doing this our solution now becomes,
This solution will satisfy any initial condition that can be
written in the form,
This may still seem to be very restrictive, but the series
on the right should look awful familiar to you after the previous chapter. The series on the left is exactly the Fourier sine series we looked at in that
chapter. Also recall that when we can
write down the Fourier sine series for any piecewise smooth function on .
So, provided our initial condition is piecewise smooth after
applying the initial condition to our solution we can determine the as if we were finding the Fourier sine series
of initial condition. So we can either
proceed as we did in that section and use the orthogonality of the sines to
derive them or we can acknowledge that we’ve already done that work and know
that coefficients are given by,
So, we finally can completely solve a partial differential
equation.
|
Example 3 Solve
the following BVP.
Solution
There isn’t really all that much to do here as we’ve done
most of it in the examples and discussion above.
First, the solution is,
The
coefficients are given by,
If we plug these in we get the solution,
|
That almost seems anti-climactic. This was a very short problem. Of course some of that came about because we
had a really simple constant initial condition and so the integral was very
simple. However, don’t forget all the
work that we had to put into discussing Fourier sine series, solving boundary
value problems, applying separation of variables and then putting all of that
together to reach this point.
While the example itself was very simple, it was only simple
because of all the work that we had to put into developing the ideas that even
allowed us to do this. Because of how
“simple” it will often be to actually get these solutions we’re not actually
going to do anymore with specific initial conditions. We will instead concentrate on simply developing
the formulas that we’d be required to evaluate in order to get an actual
solution.
So, having said that let’s move onto the next example. In this case we’re going to again look at the
temperature distribution in a bar with perfectly insulated
boundaries. We are also no longer going
to go in steps. We will do the full
solution as a single example and end up with a solution that will satisfy any
piecewise smooth initial condition.
|
Example 4 Find
a solution to the following partial differential equation.
Solution
We applied separation of variables to this problem in Example 2 of the
previous section. So, after assuming
that our solution is in the form,
and applying
separation of variables we get the following two ordinary differential
equations that we need to solve.
We solved the
boundary value problem in Example
2 of the Eigenvalues and Eigenfunctions section of the previous chapter
for so as with the first example in this section
we’re not going to put a lot of explanation into the work here. If you need a reminder on how this works go
back to the previous chapter and review the example we worked there. Let’s get going on the three cases we’ve
got to work for this problem.
The solution to
the differential equation is,
Applying the
first boundary condition gives,
The second
boundary condition gives,
Recall that and
so we will only get non-trivial solutions if we require that,
The positive
eigenvalues and their corresponding eigenfunctions of this boundary value
problem are then,
The general
solution is,
Applying the
first boundary condition gives,
Using this the
general solution is then,
and note that
this will trivially satisfy the second boundary condition. Therefore is an eigenvalue for this BVP and the
eigenfunctions corresponding to this eigenvalue is,
The general
solution here is,
Applying the
first boundary condition gives,
The second
boundary condition gives,
We know that and so . Therefore we must have and so, this boundary value problem will
have no negative eigenvalues.
So, the
complete list of eigenvalues and eigenfunctions for this problem is then,
and notice that
we get the eigenvalue and its eigenfunction if we allow
in the first set and so we’ll use the
following as our set of eigenvalues and eigenfunctions.
The time
problem here is identical to the first problem we looked at so,
Our product
solutions will then be,
and the
solution to this partial differential equation is,
If we apply the
initial condition to this we get,
and we can see
that this is nothing more than the Fourier
cosine series for on and so again we could use the orthogonality
of the cosines to derive the coefficients or we could recall that we’ve
already done that in the previous chapter and know that the coefficients are
given by,
|
The last example that we’re going to work in this section is
a little different from the first two.
We are going to consider the temperature distribution in a thin circular
ring. We will consider the lateral
surfaces to be perfectly insulated and we are also going to assume that the
ring is thin enough so that the temperature does not vary with distance from
the center of the ring.
So, what does that leave us with? Let’s set as shown below and then let x be the arc length of the ring as
measured from this point.
We will measure x
as positive if we move to the right and negative if we move to the left of . This means that at the top of the ring we’ll
meet where (if we move to the right) and (if we move to the left). By doing this we can consider this ring to be
a bar of length 2L and the heat
equation that we developed
earlier in this chapter will still hold.
At the point of the ring we consider the two “ends” to be in
perfect thermal contact. This means that at the two ends both the
temperature and the heat flux must be equal.
In other words we must have,
If you recall from the section
in which we derived the heat equation we called these periodic boundary
conditions. So, the problem we need to
solve to get the temperature distribution in this case is,
|
Example 5 Find
a solution to the following partial differential equation.
Solution
We applied separation of variables to this problem in Example 3 of the
previous section. So, if we assume the
solution is in the form,
we get the
following two ordinary differential equations that we need to solve.
As we’ve seen
with the previous two problems we’ve already solved a boundary value problem
like this one back in the Eigenvalues and Eigenfunctions section of the
previous chapter, Example 3 to
be exact with . So, if you need a little more explanation
of what’s going on here go back to this example and you can see a little more
explanation.
We again have
three cases to deal with here.
The general solution
to the differential equation is,
Applying the
first boundary condition and recalling that cosine is an even function and
sine is an odd function gives us,
At this stage
we can’t really say anything as either or sine could be zero. So, let’s apply the second boundary
condition and see what we get.
We get
something similar. However notice that
if then we would be forced to have and this would give us the trivial solution
which we don’t want.
This means
therefore that we must have which in turn means (from work in our
previous examples) that the positive eigenvalues for this problem are,
Now, there is
no reason to believe that or . All we know is that they both can’t be zero
and so that means that we in fact have two sets of eigenfunctions for this
problem corresponding to positive eigenvalues. They are,
The general
solution in this case is,
Applying the
first boundary condition gives,
The general
solution is then,
and this will
trivially satisfy the second boundary condition. Therefore is an eigenvalue for this BVP and the
eigenfunctions corresponding to this eigenvalue is,
For this final
case the general solution here is,
Applying the
first boundary condition and using the fact that hyperbolic cosine is even
and hyperbolic sine is odd gives,
Now, in this
case we are assuming that and so . This turn tells us that . We therefore must have .
Let’s now apply
the second boundary condition to get,
By our
assumption on we again have no choice here but to have and so for this boundary value problem there
are no negative eigenvalues.
Summarizing up
then we have the following sets of eigenvalues and eigenfunctions and note
that we’ve merged the case into the cosine case since it can be here
to simplify things up a little.
The time
problem is again identical to the two we’ve already worked here and so we
have,
Now, this
example is a little different from the previous two heat problems that we’ve
looked at. In this case we actually
have two different possible product solutions that will satisfy the partial
differential equation and the boundary conditions. They are,
The Principle
of Superposition is still valid however and so a sum of any of these will
also be a solution and so the solution to this partial differential equation
is,
If we apply the
initial condition to this we get,
and just as we
saw in the previous two examples we get a Fourier series. The difference this time is that we get the
full Fourier series for a piecewise smooth
initial condition on . As noted for the previous two examples we
could either rederive formulas for the coefficients using the orthogonality
of the sines and cosines or we can recall the work we’ve already done. There’s really no reason at this point to
redo work already done so the coefficients are given by,
Note that this
is the reason for setting up x as
we did at the start of this problem. A
full Fourier series needs an interval of whereas the Fourier sine and cosines series
we saw in the first two problems need .
|
Okay, we’ve now seen three heat equation problems solved and
so we’ll leave this section. You might
want to go through and do the two cases where we have a zero temperature on one
boundary and a perfectly insulated boundary on the other to see if you’ve got
this process down.