I am given the ODE $$\dot{x}=-f(x)$$ ( $\dot{x}=\frac{dx}{dt}$ with $x$ being a function of $t$)
with $$f(0)=0$$ $x \in \mathbb{R}$
such that $f$ is locally Lipschitz on $(-a,a)$, $a >0$
and $xf(x) >0$ for all $x \in (-a,a)- \{0\}$
Prove that $0$ is an asymptotically stable equilibrium point of the ODE.
I tried the following $$\dot{x} x =- f(x)x $$ $$\frac{x^2}{2} = -\int_0^x f(y)ydy$$ and take $$V(x)=\frac{x^2}{2} + \int_0^x f(y)ydy$$ as a Lyapunov function. Now $f(x)x >0$ so $V(x) >0$ Also it's clear that $V(0)=0$ be So $V(x)$ is positive defined.
And this is where I am stuck, for 0 to asymptotically stable then $V^*(x) := \frac{dV}{dt}$ must be negative defined, but I can't see why this is. How can I prove this?
The Lyapunov function is $V(x)=x^2/2$. Its derivative
$$ \dot V= x\dot x= -xf(x) $$ is negative definite.