How to prove a fixed point is stable?

Question

\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!

Answer 1

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.

Answer 2

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 $y > 0$.