Repeated Eigenvalues
Consider the linear homogeneous system
In order to find the eigenvalues consider the Characteristic polynomial
In this section, we consider the case when the above quadratic equation has double real root (that is if
) the double root (eigenvalue) is
In this case, we know that the differential system has the straight-line solution
where
is an eigenvector associated to the eigenvalue
. We also know that the general solution (which describes
all the solutions) of the system will be
where
is another solution of the system which is linearly independent from the straight-line solution
. Therefore, the problem in this case is to find
.
Search for a second solution.
Let us use the vector notation. The system will be written as
where A is the matrix coefficient of the system. Write
The idea behind finding a second solution
where
, linearly independent from
, is to look for it as
is some vector yet to be found. Since
and
(where we used
Simplifying, we obtain
), then (because
is a solution of the system) we must have
or
This equation will help us find the vector
. Note that the vector
(why?). This will help establish the linear independence of
from
will automatically be linearly independent from
.
Example. Find two linearly independent solutions to the linear system
Answer. The matrix coefficient of the system is
In order to find the eigenvalues consider the Characteristic polynomial
Since
eigenvector
, we have a repeated eigenvalue equal to 2. Let us find the associated
. Set
Then we must have
which translates into
This reduces to y=0. Hence we may take
Next we look for the second vector
. The equation giving this vector is
which translates into
the algebraic system
where
Clearly we have y=1 and x may be chosen to be any number. So we take x=0 for example to get
Therefore the two independent solutions are
The general solution will then be
Qualitative Analysis of Systems with Repeated Eigenvalues
Recall that the general solution in this case has the form
where is the double eigenvalue and
when
If
is the associated eigenvector. Let us focus on the behavior of the solutions
(meaning the future). We have two cases
, then clearly we have
In this case, the equilibrium point (0,0) is a sink. On the other hand, when t is large, we have
So the solutions tend to the equilibrium point tangent to the straight-line solution. Note that is
, then the
solution is the straight-line solution which still tends to the equilibrium point.
If
, then Y(t) tends to infinity as
are moving along the straight-line solution.
, except of course the constant solution. Note again that if
Another example of the repeated eigenvalue's case is given by harmonic oscillators.
, then we