How Do We Define Stability in a Dynamically Growing System?

[nacadaryo]

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?

 

[pasmith]

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.

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

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?

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.