# How to prove a fixed point is stable?

\begin{align*} \dot{x} &= 2 x - \frac{8}{5} x^2 - xy\\ \dot{y} &= \frac{5}{2} y - y^2 - 2 xy \end{align*}

So I have this dynamical system. I linearized it and found that the fixed point (at $x = 1.25$, $y = 0$) is the boundary case (https://en.wikipedia.org/wiki/Linear_dynamical_system#/media/File:LinDynSysTraceDet.jpg). $\tau$ = -2 and $\bigtriangleup$ = 0, which means the linearization says that it is a line of fixed point, which doesn't make sense since there is only 1 fixed point. My professor says that for non-linear system there is disturbance for the boundary case, so 3 possible outcomes are possible. It can be either saddle node, line of fixed points, or stable point. I tried pplane software (you can download it online for free too), and when I zoomed it for x = [1.2, 1.3], y = [-.05, .05]. It does look like there is a line of fixed point slightly above the fixed point (1.25, 0). I'm kind of stuck here. How do I prove the stability of this fixed point now? I also found the Lyapunov function for the system: $$V = \ln(x) - \frac{4}{5}\ln(y).$$

The two eigenvalues are -2 & 0 with eigenvectors (1,0) & (5, -8) respectively for fixed point (1.25, 0). This problem is just very weird. I have no idea what eigenvalue of 0 means. I also graphed out all the eigenvectors of the other fixed points too. Basically, I can't tell if the fixed point (1.25, 0) is stable or not. Please help!!

SOME IDEAS: Let's say I have a region D bounded by four points: (1.24, .01), (1.26, .01), (1.24, 0), (1.26, 0). It's easy to see that the Lyapunov inside this region is always positive, and $\dot{V}$ is always negative when x, y > 0. This proves that the fixed point (1.25, 0) is stable at least around this neighborhood. And that means it has to be stable! Does this approach mathematically sound? Thank you!

• @Moo, yes I'm fully aware of that, but my question tailors specifically to the fixed point I mentioned. – Phu Nguyen Mar 10 '18 at 2:00
• How is your $V$ supposed to be a Lyapunov function when it is not defined for $y=0$? In general you do not say that you have a Lyapunov function for the system. You would say that you have a Lyapunov function for asymptotic stability/stability/instability for a given equilibrium point. – MrYouMath Mar 10 '18 at 7:59
• @MrYouMath you are correct! – Phu Nguyen Mar 10 '18 at 10:14

The system \begin{align*} \dot{x} &= 2 x - \frac{8}{5} x^2 - xy\\ \dot{y} &= \frac{5}{2} y - y^2 - 2 xy \end{align*}

has other equilibrium points as well. For example $(0,0)$ is an equilibrium point. The linearized system does not tell you the whole story unless you have a structurally stable equilibrium point such as a saddle point.

• I know it has 3 three fixed points, but the only fixed point I care about is the one I mentioned above. – Phu Nguyen Mar 10 '18 at 1:58
• Did you find the eigenvalues and eignvectors of your linearizes system at this equilibrium point? – Mohammad Riazi-Kermani Mar 10 '18 at 2:04
• The two eigenvalues are -2 & 0 with eigenvectors (1,0) & (5, -8) respectively. I'm not sure how to interpret this though. Under normal circumstances, I would be able to tell if this is stable or unstable based on the eigenvalues and eigenvectors, but I can't tell from this problem. Please help! – Phu Nguyen Mar 10 '18 at 6:23
• any idea on what to do next? Thank you! – Phu Nguyen Mar 11 '18 at 1:32
• The zero eigenvalue is not predictable. I do not know what to do next. – Mohammad Riazi-Kermani Mar 11 '18 at 1:58

Let $$(x, y) = (-5 \xi/8 + \zeta + 5/4, \xi)$$. The system in the new coordinates is $$\dot \xi = \frac 1 4 \xi^2 - 2 \xi \zeta, \\ \dot \zeta = -2 \zeta + \frac 5 {32} \xi^2 - \frac 1 4 \xi \zeta - \frac 8 5 \zeta^2.$$ The linearized $$(\xi, \zeta)$$ system has one zero and one negative eigenvalue. Then the center manifold of the form $$\zeta = h(\xi)$$ exists (an approximation to it can be found by constructing a power series solution to $$\dot \zeta = h'(\xi) \dot \xi$$).

Since $$h(0) = h'(0) = 0$$, we necessarily have $$h(\xi) = O(\xi^2)$$. Therefore $$\dot {\xi_c} = \xi_c^2/4 + O(\xi_c^3)$$ for the vector field restricted to the center manifold. The fact that $$\xi_c = 0$$ is unstable implies that $$(\xi, \zeta) = (0, 0)$$ is also unstable. In the original coordinates, the trajectories starting at points close to $$(5/4, 0)$$ converge to $$(5/4, 0)$$ if $$y < 0$$ and to $$(0, 5/2)$$ if $$y > 0$$.