Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

Hint: Find a Lyapunov function $V$ of the form $$ V(x,y)=F(x)+y^2, $$ for some suitable nonnegative function $F$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.