Consider $f_1,f_2,V\in \scr{C}^1(\Bbb{R}^2;\Bbb{R})$. Suppose that
- (H1) $\langle (f_1,f_2),\nabla V\rangle\leq 0$ in $\Bbb{R}^2$;
- (H2) $\nabla V(x)\neq 0$ for all $x\in \Bbb{R}^2$.
Suppose that $c\in \mathrm{Image}(V)$. Show that the subset $\Sigma=\{x\in \Bbb{R}^2; V(x)\leq c\}$ is invariant by the flow of \begin{align*} p'(s)=(f_1,f_2)^t(p(s))\tag{*} \end{align*} i.e., every maximal solution $p(s)$ of $(*)$ that starts in $\Sigma$ remains in $\Sigma$. Can we weaken the hypothesis (H2)?
I was able to easily prove $\Sigma$ invariation, which follows straightfoward from the fact \begin{align*} 0&\geq \langle (f_1(x(s),y(s)),f_2(x(s),y(s)),(\partial_x V(x(s),y(s)),\partial_yV(x(s),y(s)))\rangle\\ &=\langle (x'(s),y'(s)),(\partial_x V(x(s),y(s)),\partial_yV(x(s),y(s)))\rangle\\ &=\partial_x V(x(s),y(s))\cdot x'(s)+\partial_yV(x(s),y(s)))\cdot y'(s)\\ &=\frac{d\, V}{ds}(x(s),y(s)). \end{align*}
But, can we weaken (H2)? I insist to review the proof, but each time I read again, it seems I didn't use (H2) at all! Furthermore, all "counterexamples" I can think (functions for $V$ which have critic points, with graphs being saddles or something, and respectively functions $f_1$ and $f_2$) do not work to yield any contradiction!