# Application of Lyapounov's Theorem

I have the following exercise to fullfill:

Given the system of differential equations $$x'=f(x)=-\nabla{g}$$ , where $$x(t)\in\Bbb{R^3}$$ and $$g$$ is $$C^1$$ and $$f(0)=0$$ and $$0$$ is a total maximum for $$g$$, decide about the stability of the point $$0$$.

My attempt: I consider the function $$V=g(0)-g(x)$$. Now, If the $$0$$ is an isolated point of maximum of $$g$$ and isolated equilibrium point of $$f$$ , we conclude that $$V$$ is a Lyapounov function with $$\nabla{V}\cdot{f}$$ $$=\nabla{g}\cdot\nabla{g}$$ , which is positive for $$x\neq0$$. So according to the Lyapunov theorem the $$0$$ is unstable.

My question is if we can solve the exercise given by not using the fact that $$0$$ is isolated, as stated in my proof. Thanks.

• Look up your definition of "total maximum" in contrast to "global maximum" etc. There the "isolated" property may already be included. – LutzL Oct 22 '18 at 17:56
• Thanks, for the feedback. I see also that if $f=0$ then as the solutions are constant, we see that $0$ is now stable...@LutzL – dmtri Oct 22 '18 at 18:24