Stability of dynamical system

1. Dec 15, 2012

Suppose we have a dynamical system $x_{t+1} = Ax_{t}$ where $A$ is matrix, $x$ is vector. We suppose that $x$ always grow as time goes on.

If we treat equilibrium as the whole time evolution(path) of $x$ given $x_0 = a$ and no disturbance to the value of $x$ - that is $x$ follows from the initial condition, how would we be able to define stability of the system? What would be the equation?

2. Dec 15, 2012

pasmith

I think you mean that $\|x_t\|$ grows.

What do you mean by stability? Do you mean that a small change in the initial condition tends to 0 as $t \to \infty$? The formal expression of that is that the solution starting at $x_0$ is stable if and only if there exists some $\epsilon > 0$ such that for any solution $y_t$ starting at $y_0$, if $\|y_0 - x_0\| < \epsilon$ then $\|y_t - x_t\| \to 0$ as $t \to \infty$.

If so your system is unstable, since for $\|x_t\|$ to grow there must exist an eigenvalue $\lambda$ of A such that $|\lambda| > 1$. This eigenvalue has a corresponding eigenvector $v$, and we can assume that $\|v\| = 1$. Let $\epsilon > 0$, and $y_0 = x_0 + \frac12\epsilon v$. Then $\|y_0 - x_0\| = \frac12 \epsilon < \epsilon$, and
$$\|y_t - x_t\| = \left\|\frac12 \epsilon A^t v \right\| = \frac12 \epsilon |\lambda^t|$$
which tends to infinity as $t \to \infty$ for any strictly positive $\epsilon$.

Alternatively you could define stability to mean that the solution starting at $x_0$ is stable if and only if, for all finite $t$ and all $\epsilon > 0$, there exists $\delta > 0$ such that for any solution $y_t$ starting at $y_0$, if $\|x_0 - y_0\| < \delta$ then $\|x_t - y_t\| < \epsilon$. That's equivalent to requiring that $x_t - y_t$ is a continuous function of $x_0 - y_0$, which in this case it is.

Last edited: Dec 15, 2012