I have the following exercise:
Let $V : \Re^{L} \rightarrow \Re$ be twice continuously differentiable. Let $x^{*}$ be an isolated minimum of $V$. Show that $V$ is a strict Lyapunov function for the system $$ \dot{x} = -\nabla V(x) = (-\frac{\partial{V}}{\partial{x_{1}}}, …,-\frac{\partial{V}}{\partial{x_{L}}}) $$ and deduce that $x^{*}$ is asymptotically stable.
I understand that I need to show that $V(x^{*})= 0$, $V(x)>0 \; \text{if} \; x \in \Re^{L} \text{ \ {$x^{*}$}}$, and $\dot{V} < 0 $ in $\Re^{L}$.
I am able to get the following:
$$ \dot{V} = <\nabla V, \dot{x}> = \frac{\partial{V}}{\partial{x_{1}}}(-\frac{\partial{V}}{\partial{x_{1}}}) + … + \frac{\partial{V}}{\partial{x_{L}}}(-\frac{\partial{V}}{\partial{x_{L}}}) = \sum_{i = 1}^{L}-(\frac{\partial{V}}{\partial{x_{i}}})^{2}$$ $$ \Rightarrow \dot{V} < 0 $$.
I am unable to show that $V(x^{*}) = 0 $ and that $V(x)>0 \; \text{if} \; x \in \Re^{L} \text{ \ {$x^{*}$}}$. I understand the argument going from Lyapunov function to these facts but not the other way around or how to show this here. Any help/tips are welcome.
Thanks in advance!
Best Answer
In order to show asymptotically stability you do not need that $V(x^*)=0$. Namely if $V(x^*)=c$, with $c$ some positive or negative constant, one could also define a new Lyapunov function $V'(x)=V(x)-c$. This new function does satisfies $V'(x^*)=0$ and its gradient is the same as that of $V(x)$, so the rest this holds. In your case it is given that $x^*$ is an isolated minimum, so in the neighborhood of $x=x^*$ one has $V(x)>V(x^*)\ \forall\,x\neq x^*$. Performing the shift $V'(x)=V(x)-V(x^*)$ would therefore show what you wanted to proof, which is thus equivalent to showing that $V(x)>V(x^*)\ \forall\,x\neq x^*$.