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Let $g: \mathbb{R} \to \mathbb{R}$ be a locally Lipschitz-continuous function with $g(0) = 0$ and $xg(x) < 0$ for all $x \neq 0$.

Consider the differential equation $\dot x = y, \dot y = g(x)$. I want to analyse the steady-state solutions for (Lyapnuov-) stability. Because of $g(0) = 0$ we have $(0,0)$ for a steady-state solution.

The methods I know for determining the stability of a differential equation require either differentiability or a Lyapunov function, and because $g$ is not necessarily differentiable, my idea was to construct a Lyapunov function.

I had much trouble constructing a function $L:\mathbb{R}^2 \to \mathbb{R}$ which fulfills $\langle \nabla L(x,y) , (y, g(x)) \rangle \leq 0$. Because of $xg(x) < 0$, $\partial_yL(x,y) = x$ and $\partial_x L(x,y) = -y$ would be great, but the Poincaré Lemma tells us, that there is no such function.

What other possibilities are there?

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  • $\begingroup$ The trajectories are cycles, and coincide with the level line of the functional $H(x,y)=y^2+G(x)$, for some suitable function $G$ depending on $g$. $\endgroup$
    – Did
    Jul 22, 2015 at 9:35
  • $\begingroup$ $G(x) = - \int_0^x g(s) ds$, but then $H(0,0)$ is not an isolated minimum. $\endgroup$
    – Cosmare
    Jul 22, 2015 at 9:39
  • $\begingroup$ I have deletad my answer. I misread the question. $\endgroup$ Jul 22, 2015 at 9:40
  • $\begingroup$ "H(0,0) is not an isolated minimum" Hmmm, why? Your $G$ is positive everywhere except at $G(0)=0$, no? (Minor point: multiply your $G$ by $2$.) $\endgroup$
    – Did
    Jul 22, 2015 at 9:45
  • $\begingroup$ Why is $G$ positive everywhere? We only have $xg(x) < 0$ and this property wasn't used yet. $\endgroup$
    – Cosmare
    Jul 22, 2015 at 9:46

1 Answer 1

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Following the comments by Did:

The trajectories passing close enough to $(0,0)$ are cycles, and coincide with the level lines of the functional $$H(x,y)=\frac{y^2}{2}+G(x),\qquad G(x) = \int_0^x g(t)\,dt$$

Indeed,

  1. $H$ is constant on each trajectory (this reflects the conservation of total energy: $y^2/2$ is kinetic energy and $G(x) $ potential energy )

  2. The point $(0,0)$ is a strict minimum of $H$ since $y^2$ has a strict minimum at $y=0$ and $G$ has a strict minimum at $x=0$ (considering the sign of its derivative).

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