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So I found that $(\pm 1,0)$ and $(0,0)$ are steady states and its trace of the linear system is always $-1$. This implies all three points are sinks (fixed points).

Is the question for (a) implying I need a Lyapunov function for EACH point?

I know for $(0,0)$, I would do $L_0 = y^2/2 - \int_{0}^{t} t - t^3 dt = y^2/2 -x^2/2 + x^4/4$

and for $(\pm 1,0)$, $L_{\pm 1} = y^2/2 - \int_{\pm 1}^{t} t - t^3 dt$

For $L_0$, I find that $L_0' < 0$ for all $(x,y) \neq (0,0)$. But I also find that $L_0(\pm 1, 0) < 0$, so any open set works for my basin of attraction?

Lemon
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1 Answers1

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In general you need a Lyapunov function for each point. Any neighborhood about a point is sufficient for local stability. The question asks for "reasonably large" which is not really well-defined, but I think it wants you to find as large a region as you can so that you can draw a goodphase portrait in part C.