Series Solutions to
Differential Equations
Before we get into finding series solutions to differential
equations we need to determine when we can find series solutions to
differential equations. So, let’s start
with the differential equation,
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(1)
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This time we really do mean nonconstant coefficients. To this point we’ve only dealt with constant
coefficients. However, with series
solutions we can now have nonconstant coefficient differential equations. Also, in order to make the problems a little
nicer we will be dealing only with polynomial coefficients.
Now, we say that x=x0
is an ordinary point if provided
both
are analytic at x=x0. That is to say that these two quantities have
Taylor series
around x=x0. We are going to be only dealing with
coefficients that are polynomials so this will be equivalent to saying that
for most of the problems.
If a point is not an ordinary point we call it a singular point.
The basic idea to finding a series solution to a
differential equation is to assume that we can write the solution as a power
series in the form,
and then try to determine what the an’s need to be.
We will only be able to do this if the point x=x0, is an ordinary point. We will usually say that (2)
is a series solution around x=x0.
Let’s start with a very basic example of this. In fact it will be so basic that we will have
constant coefficients. This will allow
us to check that we get the correct solution.
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Example 1 Determine
a series solution for the following differential equation about .
Solution
Notice that in this case p(x)=1 and so every point is an ordinary point. We will be looking for a solution in the
form,
We will need to plug this into our differential equation
so we’ll need to find a couple of derivatives.
Recall from the power series review section on power series that we can
start these at n=0 if we need to,
however it’s almost always best to start them where we have here. If it turns out that it would have been easier
to start them at n=0 we can easily
fix that up when the time comes around.
So, plug these into our differential equation. Doing this gives,
The next step is to combine everything into a single
series. To do this requires that we
get both series starting at the same point and that the exponent on the x be the same in both series.
We will always start this by getting the exponent on the x to be the same. It is usually best to get the exponent to
be an n. The second series already has the proper
exponent and the first series will need to be shifted down by 2 in order to
get the exponent up to an n. If you don’t recall how to do this take a
quick look at the first review section
where we did several of these types of problems.
Shifting the first power series gives us,
Notice that in the process of the shift we also got both
series starting at the same place.
This won’t always happen, but when it does we’ll take it. We can now add up the two series. This gives,
Now recalling the fact
from the power series review section we know that if we have a power series
that is zero for all x (as this is)
then all the coefficients must have been zero to start with. This gives us the following,
This is called the recurrence
relation and notice that we included the values of n for which it must be true.
We will always want to include the values of n for which the recurrence relation is true since they won’t
always start at n = 0 as it did in
this case.
Now let’s recall what we were after in the first
place. We wanted to find a series
solution to the differential equation.
In order to do this we needed to determine the values of the an’s. We are almost to the point where we can do
that. The recurrence relation has two
different an’s in it so
we can’t just solve this for an
and get a formula that will work for all n. We can however, use this to determine what
all but two of the an’s
are.
To do this we first solve the recurrence relation for the an that has the largest
subscript. Doing this gives,
Now, at this point we just need to start plugging in some
value of n and see what happens,
Notice that at each step we always plugged back in the
previous answer so that when the subscript was even we could always write the
an in terms of a0 and when the coefficient
was odd we could always write the an
in terms of a1. Also notice that, in this case, we were
able to find a general formula for an’s
with even coefficients and an’s
with odd coefficients. This won’t
always be possible to do.
There’s one more thing to notice here. The formulas that we developed were only
for k=1,2,… however, in this case
again, they will also work for k=0. Again, this is something that won’t always
work, but does here.
Do not get excited about the fact that we don’t know what a0 and a1 are. As you
will see, we actually need these to be in the problem to get the correct
solution.
Now that we’ve got formulas for the an’s let’s get a solution. The first thing that we’ll do is write out
the solution with a couple of the an’s
plugged in.
The next step is to collect all the terms with the same
coefficient in them and then factor out that coefficient.
In the last step we also used the fact that we knew what
the general formula was to write both portions as a power series. This is also our solution. We are done.
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Before working another problem let’s take a look at the
solution to the previous example. First,
we started out by saying that we wanted a series solution of the form,
and we didn’t get that.
We got a solution that contained two different power series. Also, each of the solutions had an unknown
constant in them. This is not a
problem. In fact, it’s what we want to
have happen. From our work with second order constant coefficient
differential equations we know that the solution to the differential equation
in the last example is,
Solutions to second order differential equations consist of
two separate functions each with an unknown constant in front of them that are
found by applying any initial conditions.
So, the form of our solution in the last example is exactly what we want
to get. Also recall that the following Taylor series,
Recalling these we very quickly see that what we got from
the series solution method was exactly the solution we got from first
principles, with the exception that the functions were the Taylor series for the actual functions
instead of the actual functions themselves.
Now let’s work an example with nonconstant coefficients
since that is where series solutions are most useful.
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Example 2 Find
a series solution around for the following differential equation.
Solution
As with the first example p(x)=1 and so again for this differential equation every point is
an ordinary point. Now we’ll start
this one out just as we did the first example. Let’s write down the form of the solution
and get its derivatives.
Plugging into the differential equation gives,
Unlike the first example we first need to get all the
coefficients moved into the series.
Now we will need to shift the first series down by 2 and
the second series up by 1 to get both of the series in terms of xn.
Next we need to get the two series starting at the same
value of n. The only way to do that for this problem is
to strip out the n=0 term.
We now need to set all the coefficients equal to
zero. We will need to be careful with
this however. The n=0 coefficient is in front of the series and the n=1,2,3… are all in the series. So, setting coefficient equal to zero gives,
Solving the first as well as the recurrence relation
gives,
Now we need to start plugging in values of n.
There are a couple of things to note about these
coefficients. First, every third
coefficient is zero. Next, the
formulas here are somewhat unpleasant and not all that easy to see the first
time around. Finally, these formulas
will not work for k=0 unlike the
first example.
Now, get the solution,
Again, collect up the terms that contain the same
coefficient, factor the coefficient out and write the results as a new series,
We couldn’t start our series at k=0 this time since the general term doesn’t hold for k=0.
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Now, we need to work an example in which we use a point
other that x=0. In fact, let’s just take the previous example
and rework it for a different value of x0. We’re also going to need to change up the
instructions a little for this example.
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Example 3 Find
the first four terms in each portion of the series solution around for the following differential equation.
Solution
Unfortunately for us there is nothing from the first
example that can be reused here.
Changing to completely changes the problem. In this case our solution will be,
The derivatives of the solution are,
Plug these into the differential equation.
We now run into our first real difference between this
example and the previous example. In
this case we can’t just multiply the x
into the second series since in order to combine with the series it must be x+2.
Therefore we will first need to modify the coefficient of the second
series before multiplying it into the series.
We now have three series to work with. This will often occur in these kinds of
problems. Now we will need to shift
the first series down by 2 and the second series up by 1 the get common
exponents in all the series.
In order to combine the series we will need to strip out
the n=0 terms from both the first
and third series.
Setting coefficients equal to zero gives,
We now need to solve both of these. In the first case there are two options, we
can solve for a2 or we
can solve for a0. Out of habit I’ll solve for a0. In the recurrence relation we’ll solve for
the term with the largest subscript as in previous examples.
Notice that in this example we won’t be having every third
term drop out as we did in the previous example.
At this point we’ll also acknowledge that the instructions
for this problem are different as well.
We aren’t going to get a general formula for the an’s this time so we’ll have to be satisfied with just
getting the first couple of terms for each portion of the solution. This is often the case for series
solutions. Getting general formulas
for the an’s is the
exception rather than the rule in these kinds of problems.
To get the first four terms we’ll just start plugging in
terms until we’ve got the required number of terms. Note that we will already be starting with an
a0 and an a1 from the first two terms
of the solution so all we will need are three more terms with an a0 in them and three more
terms with an a1 in
them.
We’ve got two a0’s
and one a1.
We’ve got three a0’s
and two a1’s.
We’ve got four a0’s
and three a1’s. We’ve got all the a0’s that we need, but we still need one more a1’. So, we’ll need to do one more term it looks
like.
We’ve got five a0’s
and four a1’s. We’ve got all the terms that we need.
Now, all that we need to do is plug into our solution.
Finally collect all the terms up with the same coefficient
and factor out the coefficient to get,
That’s the solution for this problem as far as we’re
concerned. Notice that this solution
looks nothing like the solution to the previous example. It’s the same differential equation, but
changing x0 completely
changed the solution.
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Let’s work one final problem.
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Example 4 Find
the first four terms in each portion of the series solution around for the following differential equation.
Solution
We finally have a differential equation that doesn’t have
a constant coefficient for the second derivative.
So is an ordinary point for this differential
equation. We first need the solution
and its derivatives,
Plug these into the differential equation.
Now, break up the first term into two so we can multiply
the coefficient into the series and multiply the coefficients of the second
and third series in as well.
We will only need to shift the second series down by two
to get all the exponents the same in all the series.
At this point we could strip out some terms to get all the
series starting at n=2, but that’s
actually more work than is needed.
Let’s instead note that we could start the third series at n=0 if we wanted to because that term
is just zero. Likewise the terms in
the first series are zero for both n=1
and n=0 and so we could start that
series at n=0. If we do this all the series will now start
at n=0 and we can add them up
without stripping terms out of any series.
Now set coefficients equal to zero.
Solving this gives,
Now, we plug in values of n.
Now, from this point on all the coefficients are
zero. In this case both of the series
in the solution will terminate. This
won’t always happen, and often only one of them will terminate.
The solution in this case is,
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