# Show that the solution $u \equiv 0$ of $u''+g(u)=0$ where $g(0)=0,g'(0)>0$ is stable but not asymptotically stable

The original question states

Let $g$ be a continuously differentiable function such that $g(0)=0$. Show that the solution $u \equiv 0$ of $u''+g(u)=0$ is not stable when $g'(0)<0$, and stable but not asymptotically stable when $g'(0)>0$

The first part is easy enough with Lyapunov linearization. However I am not able to prove the second part. I tried using Lyapunov functions but didn't find one.

I do think that the solutions whose value at 0 is close enough to 0 are all periodic. for example, $u''+u=0$ yield $\sin(x), \cos(x)$, while $u''+e^u-1=0$ and $u''+\sin(u)=0$ lead these and these solutions respectively. If this is true, it would solve the exercise, but I'm not sure if I can prove it.

The Lyapunov function is $$V(u,u')= \int_0^u g(u)\, du+\frac12 (u')^2.$$ Indeed, $g(0)=0$, $g'(0)>0$ implies that for sufficiently small $u>0$, $g(u)$ is positive, and for sufficiently small $u<0$, $g(u)<0$, thus, in some neighborhood of $0$, $\int_0^u g(u)\, du>0$. Hence, $V(u,u')$ is positive definite. Obviously, $V(0,0)=0$ and $V$ is continuously differentiable.
The derivative $$V'= \frac{\partial V}{\partial u} u'+\frac{\partial V}{\partial u'}u''= g(u)u'+u'(-g(u))=0,$$ thus, the origin is stable. Since $V(u,u')$ is the first integral of the system, it is not asymptotically stable.