Stability of Differential Equations

Question

Pardon me if this is too basic of a question, but this is a fundamental concept I truly do not get. Given a matrix A in the system x'(t) = Ax(t), how do you determine the stability of the system? For instance:

Determine the stability of the following system: a'(t) = $\begin{bmatrix} 7&11&23&8 \\ 9&-7&5&-1 \\ -4&-2&3&4 \\ 5&-4&1&-3 \end{bmatrix}$a(t).

I know you have to do something with the eigenvalues, but besides that, I'm so, SO lost. Please be nice to me, as this is something I just cannot grasp. Any help is greatly appreciated.

Answer 1

This is a homogeneous linear system with constant coefficients, so any solution will be of the form $a=ve^{\lambda t}$, where $v$ is an eigenvector with eigenvalue $\lambda$. Notice $\lambda$ is in the exponent and it is a factor of $t$, so it will affect the behavior of $t$ as $t\rightarrow \infty$. Furthermore notice that if $a=0$ then $a'=0$, so $0$ is an equilibrium point. So now you just need to find the eigenvalues and eigenvectors of your system and look at what happens when you let $t \rightarrow\infty$, and that will give you the stability of your system.

Answer 2

First compute the eigenvalues. In Matlab,

a=[7 11 23 8; 9 -7 5 -1; -4 -2 3 4; 5 -4 1 -3];
eig(a)

The result is

-13.8129 + 0.0000i
7.6860 + 7.2588i
7.6860 - 7.2588i
-1.5591 + 0.0000i

Two eigenvalues have positive real part (i.e. $7.6860 \pm 7.2588i$), therefore, the system is unstable. I will leave how to compute the eigenvalues mathematically to you.