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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.

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    $\begingroup$ Look up your definition of "total maximum" in contrast to "global maximum" etc. There the "isolated" property may already be included. $\endgroup$ – LutzL Oct 22 '18 at 17:56
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    $\begingroup$ 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 $\endgroup$ – dmtri Oct 22 '18 at 18:24

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