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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.

enter image description here

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.

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  • $\begingroup$ 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. $\endgroup$ – Evgeny Jul 6 '15 at 8:12
  • $\begingroup$ 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. $\endgroup$ – DaveNine Jul 6 '15 at 15:52
  • $\begingroup$ 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 :) $\endgroup$ – Evgeny Jul 6 '15 at 16:20
  • $\begingroup$ 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. $\endgroup$ – DaveNine Jul 8 '15 at 3:40
  • $\begingroup$ 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. $\endgroup$ – Evgeny Jul 8 '15 at 5:02
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I just want to point out that you can not use Lyapunov's second methods (Lyapunov functions) to show instability. The notion of Lyapunov's second method is not strong enough to do so. You can ONLY SHOW STABILITY.

In a less ambiguous way: if you can show stability its fine and you are save that the system is stable. If you can not show stability with Lyapunov's second method you can not conclude instability from this.

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    $\begingroup$ This makes a good Comment, but doesn't address the specific choice of a Lyapunov function (as proposed in the OP's second Edit). I think an Answer should at least contain that particular point. $\endgroup$ – hardmath Apr 11 '16 at 13:32

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