# Finding a Lyapunov Function for a system involving a trigonometric function

I'm dealing with determining if $(0,0)$ is stable or not for the following system via constructing a Lyapunov function. The system is

$$\begin{cases} x'(t)=(1-x)y+x^2\sin{(x)}& \\ y'(t)=-(1-x)x+y^2\sin{(y)}& \end{cases}$$

My initial guess was to choose $V(x,y)=\frac{1}{2}x^2+\frac{1}{2}y^2$, however this leads to $\dot{V}=x^3 \sin{x}+y^3 \sin{y}$, which, for small $x$ and $y$, is positive. However, this does not agree with numerical evidence and also looking at the linearization method, which shows that the origin is indeed a stable node.

Might someone have a suggestion for a Lyapunov function, as well as the domain to choose for it? I suppose I might as well ask if it would be valid to approximate the sine terms by its argument since we would be looking at a small neighborhood around the origin.

EDIT: Here is a streamplot of the system in a neighborhood of the origin,it appears that the origin is unstable, so I guess I was wrong with my initial assumption. I guess this agrees with my choosing of a Lyapunov function because the function is positive for all x and y(except at the origin), implying instability.

EDIT2: After thinking for a little bit, the Lyapunov function $V(x,y)=\frac{1}{2}x^2+\frac{1}{2}y^2$ work, with the domain $\Omega = \{ (x,y)\in \mathbb{R}^2 \vert -\pi < x < \pi$ and $-\pi < y <\pi \}$ so that $\dot{V}(x,y)>0$ in $\Omega$. This establishes instability of the origin.

• I don't understand your reasoning. Linearization yields a system $x' = y$, $y' = -x$ and it's a center for linearized system (so nothing particular can be said about original non-linear equilibrium). Could you add your numerical evidences to your question? Because what you've found is a Lyapunov function for time-reversed system of equations: so, in backward time the origin has a Lyapunov function that shows it's stable equilibrium $\Rightarrow$ original system has an origin as repelling equilibrium. – Evgeny Jul 6 '15 at 8:12
• Was I missing something with the lnearization? I found the same system as you did, however I just went on finding the eigenvalues to be imaginary, implying stability. I'm editing to include numerical evidence that I found. – DaveNine Jul 6 '15 at 15:52
• I think that you've made a typo in your calculations. For quick check I use the fact that If we have system $x' = y$, $y' = -x$, it's equivalent to $x'' = - x$ which is an equation of harmonic oscillator. So, for linearized system the origin is a center. However, it's impossible to say something about stability of original equilibrium basing only on linearization. Your choice of Lyapunov function was correct from the beginning :) – Evgeny Jul 6 '15 at 16:20
• Thanks! This is a good example then for which linearization does not work to find information about stability. I also realized that $\dot{V}$ is only positive in a certain domain, but that's easily fixable. – DaveNine Jul 8 '15 at 3:40
• You're welcome! There are also few classic examples of complex foci that illustrate the idea of linearization-telling-nothing-in-nonhyperbolic-cases. Also, you don't need to have $\dot{V}$ to be positive on all plane -- it's enough to have it positive in small neighbourhoods of origin. – Evgeny Jul 8 '15 at 5:02