Systems of Differential Equations (Part 9. Complex Eigenvalues)
Posted by Suhaivi Hamdan on 5:54 PM with No comments
Complex Eigenvalues
In this section we will look at solutions to


where the eigenvalues of the matrix A are complex. With complex
eigenvalues we are going to have the same problem that we had back when we were
looking at second order differential equations.
We want our solutions to only have real numbers in them, however since
our solutions to systems are of the form,


we are going to have complex numbers come into our solution
from both the eigenvalue and the eigenvector.
Getting rid of the complex numbers here will be similar to how we did it
back in the second order differential equation
case, but will involve a little more work this time around. It’s easiest to see how to do this in an
example.
Example 1 Solve
the following IVP.
Solution
We first need the eigenvalues and eigenvectors for the
matrix.
So, now that we have the eigenvalues recall that we only
need to get the eigenvector for one of the eigenvalues since we can get the
second eigenvector for free from the first eigenvector.
We need to solve the following system.
Using the first equation we get,
So, the first eigenvector is,
When finding the eigenvectors in these cases make sure
that the complex number appears in the numerator of any fractions since we’ll
need it in the numerator later on.
Also try to clear out any fractions by appropriately picking the
constant. This will make our life
easier down the road.
Now, the second eigenvector is,
However, as we will see we won’t need this eigenvector.
The solution that we get from the first eigenvalue and
eigenvector is,
So, as we can see there are complex numbers in both the
exponential and vector that we will need to get rid of in order to use this
as a solution. Recall from the complex
roots section of the second order differential equation chapter that we can
use Euler’s formula to get the
complex number out of the exponential.
Doing this gives us,
The next step is to multiply the cosines and sines into
the vector.
Now combine the terms with an “i” in them and split these terms off from those terms that don’t
contain an “i”. Also factor the “i” out of this vector.
Now, it can be shown (we’ll leave the details to you) that
So, the general solution to a system with complex roots is
where
For our system then, the general solution is,
We now need to apply the initial condition to this to find
the constants.
This leads to the following system of equations to be
solved,
The actual solution is then,

As we did in the last section we’ll do the phase portraits
separately from the solution of the system in case phase portraits haven’t been
taught in your class.
Example 2 Sketch
the phase portrait for the system.
Solution
When the eigenvalues of a matrix A are purely complex, as they are in this case, the trajectories
of the solutions will be circles or ellipses that are centered at the
origin. The only thing that we really
need to concern ourselves with here are whether they are rotating in a
clockwise or counterclockwise direction.
This is easy enough to do.
Recall when we first looked at these phase portraits a couple of sections ago that if we pick a value of
Therefore at the point (1,0) in the phase plane the
trajectory will be pointing in a upwards direction. The only way that this can be is if the
trajectories are traveling in a counterclockwise direction.
Here is the sketch of some of the trajectories for this
problem.
The equilibrium solution in the case is called a center and is stable.

Note in this last example that the equilibrium solution is
stable and not asymptotically stable.
Asymptotically stable refers to the fact that the trajectories are
moving in toward the equilibrium solution as t increases. In this example
the trajectories are simply revolving around the equilibrium solution and not
moving in towards it. The trajectories
are also not moving away from the equilibrium solution and so they aren’t
unstable. Therefore we call the
equilibrium solution stable.
Not all complex eigenvalues will result in centers so let’s
take a look at an example where we get something different.
Example 3 Solve
the following IVP.
Solution
Let’s get the eigenvalues and eigenvectors for the matrix.
Now get the eigenvector for the first eigenvalue.
We need to solve the following system.
Using the second equation we get,
So, the first eigenvector is,
The solution corresponding the this eigenvalue and
eigenvector is
As with the first example multiply cosines and sines into
the vector and split it up. Don’t
forget about the exponential that is in the solution this time.
The general solution to this system then,
Now apply the initial condition and find the constants.
The actual solution is then,

Let’s take a look at the phase portrait for this problem.
Example 4 Sketch
the phase portrait for the system.
Solution
When the eigenvalues of a system are complex with a real
part the trajectories will spiral into or out of the origin. We can determine which one it will be by
looking at the real portion. Since the
real portion will end up being the exponent of an exponential function (as we
saw in the solution to this system) if the real part is positive the solution
will grow very large as t
increases. Likewise, if the real part
is negative the solution will die out as t
increases.
So, if the real part is positive the trajectories will
spiral out from the origin and if the real part is negative they will spiral
into the origin. We determine the
direction of rotation (clockwise vs. counterclockwise) in the same way that
we did for the center.
In our case the trajectories will spiral out from the
origin since the real part is positive and
will rotate in the counterclockwise direction as the last
example did.
Here is a sketch of some of the trajectories for this
system.
Here we call the equilibrium solution a spiral (oddly enough…) and in this
case it’s unstable since the trajectories move away from the origin.

If the real part of the eigenvalue is negative the
trajectories will spiral into the origin and in this case the equilibrium
solution will be asymptotically stable.
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