3.3. Real Eigenvalues

Consider the linear homogeneous system

In order to find the eigenvalues, consider the characteristic polynomial

In this section we will consider the case of the quadratic equation above when it has two distinct real roots (that is, if ). The roots (eigenvalues) are

and

Here we know that the differential system has two linearly independent straight-line solutions

where (respectively ) is an eigenvector associated to the eigenvalue (respectively ). We also know that the general solution (which describes all of the solutions) to the system has the form

Keep in mind that and are two constant vectors.

Let us discuss the behavior of the solutions when (meaning the future) and when (meaning the past). Since the eigenvalues are distinct, one is bigger than the other one. Assume that we have

It is easy to see that we have

Behavior when

In this case we will consider the equation

Since

(because ) then it is clear that when , we have

Behavior when

In this case we will consider the equation

Since

(because ) then it is clear that when , we have

Remark:Since the two eigenvalues are real numbers, we have three cases to consider depending on their signs:

Case 1: Both are positive

In this case we have

meaning that the solutions emanate from the origin (if you go to the past, you will die at the origin). When , Y(t) explodes.

In this case the origin plays the role of a source. Clearly, the origin is the only equilibrium point.

Case 2: Both eigenvalues are negative

In this case we have

meaning that in the future the solutions die at the origin. When , Y(t) explodes.

In this case, the origin plays the role of a sink. Clearly, the origin is the only equilibrium point.

Case 3: The eigenvalues have different signs

In this case, the origin behaves like a saddle.

Remark: It is clear from the above discussions that one may decide about the signs of the eigenvalues just by looking at some solutions on the phase plane (depending whether we have a saddle, a sink or a source).

Example: Consider the three phase planes and decide about the sign-distribution of the associated eigenvalues.

Phase Plane I
Phase Plane II
Phase Plane III

Answer:

For the phase-plane I, the origin is a source. Therefore, the two eigenvalues are both positive.

For the phase-plane II, the origin is a saddle. Hence, the two eigenvalues are opposite signs.

For the phase-plane III, the origin is a sink. Hence, the two eigenvalues are negative.

Example: Consider the harmonic oscillator equation

Discuss the behavior of the spring-mass.

Answer: First, translate this equation to the system

where

The characteristic polynomial of this system is

The eigenvalues are

It is clear that both of them are negative. Hence, the origin is a sink. Meaning that, regardless of the initial condition, the mass will always tend to its equilibrium, or rest, position.
Note that if V is an eigenvector associated to the biggest eigenvalue , then all the solutions tend to the origin tangent to that vector V. In this case we have

Remark: The case when one of the two eigenvalues is zero will be discussed in another section separately.