show that $\dfrac{dV}{dt}<0$

Question

suppose that $V(x,y)$ is negative definite with a max at $(x_0,y_0)$. show that V(x,y)=-S(x,y) is a lyapunov function for the system: \begin{equation} \dfrac{dx}{dt}=\dfrac{\partial S}{\partial x}\\ \dfrac{dy}{dt}=\dfrac{\partial S}{\partial y} \end{equation}

What I did u=is direct substitution to get: $\dfrac{dV}{dt} = -\left[\left(\dfrac{\partial S}{\partial x}\right)^2+\left(\dfrac{\partial S}{\partial y}\right)^2\right]\leq 0$

How can I show that $\dfrac{dV}{dt}<0$?

Answer 1

$\frac{dV}{dt}$ can't be $<0$ everywhere. It's $0$ at the maximum. Because by definition $V'<0$ everywhere except at a fixed point $(x_0,y_0)$ in which case $V'(x_0,y_0)=0$. Now show that the max $(x_0,y_0)$ is a fixed point: Since $(x_0,y_0)$ is a max, you have $0=\frac{\partial V}{\partial x}|_{x_0}=-\frac{\partial S}{\partial x}|_{x_0}=\frac{dx}{dt}$ and $0=\frac{\partial V}{\partial y}|_{y_0}=-\frac{\partial S}{\partial y}|_{y_0}=\frac{dy}{dt}$. Thus, the maximum is also a fixed point, and so by definition, $\frac{dV}{dt}|_{(x_0,y_0)}=0$