Let $p\in C^1 (\mathbb R)$ and consider the differential equation $$\dot x=p(x)x.$$
a) Show: If $p(0)>0$, then $0$ is unstable.
b) Let $s>0$. Show: $s$ is asymptotically stable iff there is a $r>0$ such that $p$ satisfies the following conditions:
- $p(x)>0$ for $s-r<x<s$
- $p(x)<0$ for $s<x<s+r$
What I've got for a):
First of all we see that $0$ is a stationary point. Let $x_0$ be an initial point. Then, for $x_0>0$ any solution $\varphi$ will only take positive values (since otherwise there would be a $t'$ such that $\varphi (t')=0$ by the intermediate value theorem and then $\varphi'\equiv 0$ by Picard-Lindelöf since $p(x)x$ is continuous and therefor the solution unique, but $\varphi(0)=x_0>0$). Now, if we have $p(0)>0$, then on any neighbourhood $U_\varepsilon (0)$ we have $\dot\varphi >0$ and $0$ can't be asymptotically stable. Also it can't be lyapunov-stable since we can choose a neighbourhood $U_\epsilon(0)$ where $p(x)>0$ and thus $\varphi$ leaves the neighbourhood $U$.
For b):
It's quite visual what happens here: If we start at $0<x_0<s$ any solution $\varphi$ always stays in between $[0,s]$ for the reason mentioned above and if $s$ is asymptotically stable there is a neighbourhood $U_\epsilon(s)$ such that $\lim_{t\to\infty}\varphi(t)=s$ on that neighbourhood. The problem though is that this doesn't necessarily mean $\dot\varphi>0$ on that neighbourhood so I could choose $r=\epsilon$. Or does it? How do I write that down mathematically accurate?
Edit: I think the converse of b) can be proven easily: If there is an $r>0$ such that $p(x)>0$ on $(s-r, r)$ and we start at some $x_0\in (s-r,r)$, then $\dot\varphi >0$ since $p(x)x>0$. So $\varphi$ is strictly monotonous, but also bounded because of $0<\varphi <s$ so we get $\lim_{t\to\infty}\varphi(t)=s$ and thus $s$ is asymptotically stable. Note that $\varphi$ always stays in $[x_0,s)$, so $s$ also must be liapunov-stable.