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?

• 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. – LutzL Nov 7 '18 at 11:22

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.
• 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? – kleinmeinpouts Nov 8 '18 at 7:57
• 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. – Hans Lundmark Nov 8 '18 at 9:50
• 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? – kleinmeinpouts Nov 8 '18 at 23:49
• 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? – kleinmeinpouts Nov 10 '18 at 10:24