For $x' = y$, and $y' = -x - y$, Find all equilibrium points and decide whether they are stable, asymptotically stable, or unstable.

For $x' = y$, and $y' = -x - y$, Find all equilibrium points and decide whether they are stable, asymptotically stable, or unstable.

I found that the equilibrium points are (0,0). Then I try to find the eigenvalues. The characteristic equation I found is $\lambda + \lambda^2+1=0$. I realize that they are complex eigenvalues...The real part of this complex eigenvalues are negative. Does this mean this is stable ? Is this also asymptotically stable? What are the differences between asymptotically stable and stable?

Your given system is a linear time-invariant system. It can be written as

$$\dot{\boldsymbol{x}}=\begin{bmatrix}0 &1\\ -1 & -1\\ \end{bmatrix}\boldsymbol{x}.$$

In order to determine the stability of the system, you need to determine the eigenvalues of the coefficient matrix

$$\boldsymbol{A}=\begin{bmatrix}0 &1\\ -1 & -1\\ \end{bmatrix}.$$

Now, there are three cases (for real systems with real coefficients) that need to be distinguished:

Case 1 (Asymptotic stability = Stable + attractive): All eigenvalues have a strictly negative real part. This system matrix is said to be hurwitz. For an asymptotically stable equilibrium point, we know that trajectories starting close enough (for linear systems it implies that any trajectory will converge to the origin) to the origin will converge towards the equilibrium point for $t\to \infty$.

Case 2 (Instability): There is at least one eigenvalue which has a strictly positive real part.

Case 3 (Stability): All eigenvalues have a real part that is $\leq 0$. And we have at least one eigenvalue with real part equal to zero. As long as the multiplicity of these eigenvalues is $\leq 1$ we have a stable system. This is sometimes called marginally stable. In contrast to asymptotic stability, we do not have attractivity in this case. That means that trajectories that start close to the origin will stay close to the origin but they will not converge to the origin for $t\to \infty$.

• +1 Old question but much better than the other answer @oks – Dylan Nov 28 '17 at 23:01

You are right.

All you need to do next is solve the characteristic equation. If the real part of the solutions of the characteristic equation is positive, then the equilibrium is unstable. http://www.math24.net/equilibrium-points-of-linear-autonomous-systems.html

• I found that the real part of this complex eigenvalues are negative. Does this mean this is stable ? Is this also asymptotically stable? What are the differences between asymptotically stable and stable? – afsdf dfsaf Feb 22 '14 at 17:37
• Yes. negative real part means asymptotically stable. tutorial.math.lamar.edu/Classes/DE/EquilibriumSolutions.aspx. I don't know what the difference between stable and asymptotically stable is. – oks Feb 22 '14 at 17:49