# Lyapunov function instead of linearization

Consider the following system of equations

$$\begin{cases} \dot x=y-x^2-x \\ \dot y=3x-x^2-y \\ \end{cases}$$ Then, the equilibriums are $(0,0)$ and $(1,2)$. Using linearization around $(1,2)$ one can obtain $$\begin{pmatrix} \dot x \\ \dot y \\ \end{pmatrix} = \begin{pmatrix} -3 & 1 \\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}$$ The matrix has two distinct eigenvalues and both of them have negative real values, implying that the equilibria $(1,2)$ is asymptotically stable.

My problem is with $(0,0)$, where the linearization fails, as we have eigenvalues with positive real values $$\begin{pmatrix} \dot x \\ \dot y \\ \end{pmatrix} = \begin{pmatrix} -1 & 1 \\ 3 & -1 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix}$$ So, we need to construct a Lyapunov function, but the "usual" $V(x,y)=ax^2+by^2$ doesn't seem to work in this case.

• is it really a problem to have two positive real eigenvalues? $(0,0)$ is simply unstable. – the_candyman Jul 3 '18 at 17:18
• The characteristic polynomial $0=q^2+2q-2=(q+1)^2-3$ has solutions $q=-1+\sqrt3>0$ and $q=-1-\sqrt3<0$ which characterizes $(0,0)$ as saddle point. – LutzL Jul 6 '18 at 16:44

## 2 Answers

Linear stability analysis comes to a useful conclusion as long as there are no eigenvalues with zero real part. If all eigenvalues have nonzero real part, the stability of the nonlinear system will match that of the linear system. Simply put, since you have two positive real eigenvalues, the fixed point is unstable in both systems.

The existence of a Lyapunov function implies stability, but it will be hard to show the existence of such a function, and even harder to show nonexistence, directly. Lyapunov functions should be used as a last resort, when linear stability analysis fails and you're pretty sure that the nonlinear system is stable, but you need to prove it.

• The problem is that (maybe I said it not so explicitly) one eigenvalue is real positive and the other one is real and negative. But, I understand that it still means instability since we have one eigenvalue that is positive. – user554578 Jul 3 '18 at 19:21
• Stable if all real parts are negative, inconclusive if any real part is 0, unstable otherwise. – AlexanderJ93 Jul 3 '18 at 19:23

Near the point $(0,0)$ the DE system can be approximated as

$$\dot x = y-x\\ \dot y = 3x-y$$

or

$$3 x\dot x = 3 x y - 3 x^2\\ y \dot y = 3 x y - y^2$$

now subtracting we have

$$\frac 12\frac{d}{dt}(y^2-3x^2)+(y^2-3x^2) = 0$$

so locally we have

$$\frac 12\frac{d}{dt}(3x^2-y^2)+3x^2-y^2 = 0$$

or calling $z = y^2-3x^2$ the equivalent system $\dot z + 2z = 0\;\;$ with solution $z = C e^{-2t}$ or

$$3x^2-y^2 = (\sqrt 3 x+ y)(\sqrt 3 x-y)=C e^{-2t}$$

as typical orbits for a saddle point which is unstable. I hope this helps.

• "I hope this helps." Not really, since $3x^2-y^2=Ce^{-t}$ does not contradict that $(0,0)$ is stable. – Did Jul 6 '18 at 11:48
• @Did Of course the point $(0,0)$ is unstable. My intention was to show the kind of orbits near that point as included now as a plot in the answer. – Cesareo Jul 6 '18 at 12:56
• Hmmm... yes the plot you added is something tangible, but on the side of the mathematics, what did you prove exactly? (Unrelated: $3x^2-y^2\propto e^{-2t}$, not $e^{-t}$.) – Did Jul 6 '18 at 15:06
• @Did As I conceive, Mathematics is also pleasure and beauty. Not only proof. – Cesareo Jul 6 '18 at 15:43
• @Did The correct formula is $3x^2-y^2 = C e^{-t}$ – Cesareo Jul 6 '18 at 16:13