# Stability of zero

Determine stability of zero in \begin{cases} x'=y \\ y'=-f(x) \end{cases}

Here $f: \mathbb{R} \rightarrow \mathbb{R}$ is class $\mathcal{C}^1, f(0)=0$ and $xf(x)>0$ for $x \neq 0$.

Could you help me solve this problem?

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Hint: Find a Lyapunov function $V$ of the form $$V(x,y)=F(x)+y^2,$$ for some suitable nonnegative function $F$.
Thank you, but shouldn't $V$ go from $\mathbb{R}^2$ to $\mathbb{R}$? I know I should check that $V^{-1}(0)=(0,0),$ and that $<\text{grad} V(x,y), g(x,y)> \le 0$ for $(x,y) \neq 0$. Here $g(x.y) = (y, -f(x))$ – Bruce May 29 '14 at 20:21
So... did you find some $F$ such that $\frac{\mathrm d}{\mathrm dt}V(x(t),y(t))=0$ for every $t$? – Did May 30 '14 at 7:25
No, $V(x,y)=-x^2+y^2$ is NOT invariant by the dynamics (and these considerations do not show that $(0,0)$ is unstable). – Did May 30 '14 at 12:29
Ok, you are right. $\frac{\mathrm d}{\mathrm dt}V(x(t),y(t))=(F(x(t))+y(t)^2)'=F'(x(t)x'(t)+2y(y)y'(t)=x'(t)(F'(x(t))-f(x))$. Is that correct? – Bruce May 30 '14 at 12:52
Make it $F(x)=\int_0^xf(s)ds$ and we can (finally!) close this exchange. Good job. (Here is a suggestion; next time, start to think before the 15th comment...) – Did Jun 1 '14 at 9:39