I am given the following system of odes \begin{gather} \begin{pmatrix} \dot{x}\\ \dot{y} \end{pmatrix} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} - b(x^2 + y^2) \begin{pmatrix} x\\ y \end{pmatrix} + (x^2 + y^2)^2 \begin{pmatrix} x\\ y \end{pmatrix} \end{gather} I want to study the stability of the $(0, 0)$ solution in terms of $b$. I cannot understand the argument they are using. It goes as follows: \begin{equation} V(x, y) = \frac{1}{2}(x^2+y^2) \end{equation} is a Lyapunov function for $(0, 0)$ under certain conditions. Exactly,
- $V(0, 0) = 0$
- $\dot{V}(x, y) = DV(x,y) f(x, y) = \\ = x(y-bx(x^2+y^2) + x(x^2+y^2)^2) + y(-x -by(x^2+y^2)+y(x^2+y^2)^2) = \\ = -b (x^2+y^2)^2 + (x^2+y^2)^3$
Then, if $b>0$ for all $(x,y)$ such that $x^2+y^2 < \sqrt{b}$ we have that $\dot{V} < 0$, therefore $V$ is a strict Lyapunov function and the solution $(0,0)$ is asymptotically stable.
If $b \leq 0$ then $\dot{V} > 0$ for all $(x, y) \neq (0, 0)$, therefore it is inestable (because if time goes the other way it would be asymptotically stable).
The parts I am not understanding are in bold.
First of all, when we talk about stability I thought solutions only needed to be defined for positive time. Therefore, by time "going the other way", what do we mean? Are we considering $-t$? Also, I don't see why this would mean instable.
Thanks in advance. Any help would be appreciated.
When looking at the limit of the states as time, denoted by $t$, goes to $-\infty$. This is equivalent to first defining a new "time-variable" $\tau = -t$ and look at the limit at $\tau \to \infty$, with
$$ \frac{d\,V(x,y)}{d\tau} = \frac{d\,V(x,y)}{dt} \frac{d\,t}{d\tau} = -\frac{d\,V(x,y)}{dt}. $$
From this it is hopefully clear that for $b \leq 0$ it follows that as $\tau \to \infty$ and thus $t \to -\infty$, then the state goes to the origin. So as time $t$ becomes more negative the states get closer to the origin. Therefore, as time $t$ becomes more positive the states get further from the origin, which is equivalent to the origin of the system being unstable.
It can be noted that this is not a necessary condition for instability. Namely, an equilibrium point of a system is also considered unstable when only some of its state diverge from the considered equilibrium point. Instead one could use Chetaev's theorem to more generally determine instability.