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I have been studying the following problem of gradient systems. Given a system:

\begin{equation} \overset{\cdot}{x}=f(x), \ \ x=x(t) \in \mathbb{R}^3, \end{equation} where \begin{equation} f=-\nabla g, \ \ g\in C^1(\mathbb{R}^3), \end{equation} let's suppose that $x_0=0$ is an equilibrium point of this system. I have found references stating that when for example $x_0=0$ is also a local minimum of $g$, then this means that $0$ is a stable point of equilibrium. This is shown by means of the Lyapunov function $V=g(x)-g(0)$.

I have found no reference whatsoever about maxima. If we had to study the same system, but $x_0=0$ is now a local (or global) maximum of $g$, then is there a conclusion about the stability of $0$? I have tried defining the function $V=g(0)-g(x)$, but for this function we have that $\overset{\cdot}{V} \geqslant 0$ and not $\overset{\cdot}{V} \leqslant 0$. Is there even a way to define a Lyapunov function?

Thanks in advance.

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    $\begingroup$ Obviously, for a maximum the time evolution will make the solution cross the level sets of $g$ in an outward direction, never to come back. $\endgroup$ – LutzL Nov 7 '18 at 11:22
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If $g$ has a maximum, the equilibrium is unstable. (Think about the intuition: the system is pushing the point in the direction where $g$ decreases fastest!)

To show this rigorously, you can make the change of variables $s=-t$, which is equivalent to changing the sign of $g$ (so that the maximum becomes a minimum). Then apply the first result, to show that the system is stable when you use the backwards time variable $s$, hence unstable when the time $t$ runs in the usual direction.

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  • $\begingroup$ I tried considering $h=-g$ which has a minimum at $0$, but if i define $V(x)=h(x)-h(0)$ then $V$ is of class $C^1$, $V$ is positive but i cannot claim that the only point at which $V$ is equal to $0$ is $0$. I need that in order to explain that $V$ is a candidate for a Lyapunov function. And even if i did show that, then I would calculate \begin{equation} \overset{\cdot}{V} =\frac{dV}{dt}=\frac{dV}{dx} \cdot \frac{dx}{dt}=\nabla V \cdot \dot{x}=\nabla h \cdot \dot{x}= -\nabla g \cdot \dot{x}=f^2\geqslant 0 \end{equation} while I need it to be $\leqslant 0$. Any insight? $\endgroup$ – kleinmeinpouts Nov 8 '18 at 7:57
  • $\begingroup$ I'm not sure I understand the problem... In your original question, you're assuming that $g$ has a strict local maximum at $0$, aren't you? (Otherwise the statement isn't true.) So here you should assume that $h$ has a strict local maximum. Then with $s=-t$ you have $dx/ds = -dx/dt = - \nabla( -g ) = -\nabla h$ where $h?-g$ has a strict local minimum, so the theorem applies to the system $dx/ds=-\nabla h$. No need to try to redo the proof. $\endgroup$ – Hans Lundmark Nov 8 '18 at 9:50
  • $\begingroup$ I suppose that by strict local maximum you mean that there is a neighborhood of $0$ that for all points other than $0$, $g$ is strictly less than $g(0)$, right? What happens in the case i have a non-strict global maximum at $0$? It suffices to find a neighborhood $N$ of $0$ such that maximum is not attained at any other point of $N$, and $N$ will be the domain of the Lyapunov candidate i am trying to define. It is necessary for that definition to have $V$ being equal to $0$ only at $0$, that's why i need to clarify this; can i always choose such a neighborhood in the case of a global maximum? $\endgroup$ – kleinmeinpouts Nov 8 '18 at 23:49
  • $\begingroup$ Yes, that's the definition of strict local maximum. With a non-strict local minimum (to go back to the original setup), you have other equilibrium points arbitrarily close to the origin, so at least the origin can't be asymptotically stable. $\endgroup$ – Hans Lundmark Nov 9 '18 at 7:37
  • $\begingroup$ I guess that in the case where the number of local maxima or minima is finite, I can choose a ball centered at $0$ with sufficiently small radius so that no other maximum or minimum exists inside, then I define the Lyapunov function in that ball and I think I'm done. But what if there is an infinite number of maxima/minima in such a ball? Can I exclude that case using the compactness of the closure of the ball, or am I doomed in this case? $\endgroup$ – kleinmeinpouts Nov 10 '18 at 10:24

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