Unstable equilibrium point

Question

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$. Suppose there is $V: W \rightarrow \mathbb{R} \in C^{1}$ in a neighboorhood of an equilibrium point $x_0$ (that is, $f(x_0) = 0$) such that $V(x_0) = 0$ and $\frac{d}{dt}V(\phi_t(x)) > 0, \forall x \neq x_0$. If there is $(x_n)_{n \in \mathbb{N}}$ such that $x_n \rightarrow x_0$ and $V(x_n) > 0$ then $x_0$ is not stable.

My attempt:

First of all, note that if $x \in W$ and $x \neq x_0$, then $x$ is regular.

If $f(x) = 0$, then $\phi_t(x) = x$, $\forall t$. Then, $V(\phi_t(x)) = V(x)$ and that can't happen, because $\frac{d}{dt}V(\phi_t(x)) > 0$.

I think I must use the Tubular Flow Theorem.

Can someone help me? I would appreciate =)

Answer 1

Suppose, for the sake of a contradiction, that $x_0$ is stable. Pick $\epsilon >0$ such that $\overline{B(x_0,\epsilon)} \subset W$. From the stability assumption we can then pick $\delta>0$ such that $|x-x_0| < \delta$ implies that $|\varphi(t,x) -x_0| < \epsilon$ for all $t \ge 0$.

Since $V$ is continuous on the compact set $\overline{B(x_0,\epsilon)}$ we know that $V$ achieves its maximum on this set. Call the maximum $M$. Since $V(x_n) >0$ and $x_n \to x_0$ as $n \to \infty$ we know that $M >0$.

Since $x_n \to x_0$ as $n \to \infty$ we can pick $x:=x_n \in B(x_0,\delta) \backslash \{x_0\}$ for some fixed $n$ large enough. The above shows that $\varphi(t,x) \in B(x_0,\epsilon)$ for all $t \ge 0$. Thus $t \mapsto V(\varphi(t,x))$ is increasing (by assumption) and bounded above. This allows us to conclude that $$ \lim_{t\to \infty} V(\varphi(t,x)) \in (0,M]. $$

Set $m = V(x) = V(x_n) >0$. The above tells us that $m < M$ and that $m \le V(\varphi(t,x)) \le M$ for all $t \ge 0$. Now consider the set $$ K = \{ y \in \overline{B(x_0,\epsilon)} : m \le V(y) \le M \}, $$ which is nonempty since $\varphi(t,x) \in K$ for $t \ge 0$ and compact since $V$ is continuous. Since $V$ is $C^1$ and $f$ is continuous we may then define $$ L = \min\{\nabla V(y) \cdot f(y) : y \in K \}. $$ Note that for any $y \in K$ can compute $$ \nabla V(y) \cdot f(y) = \nabla V(\varphi(t,y)) \cdot f(\varphi(t,y)) \Big\vert_{t=0} =\frac{d}{dt} V(\varphi(t,y)) \Big\vert_{t=0} >0 $$ by assumption, so we find that actually $L >0$.

Now we compute, for any $t>0$, $$ M \ge V(\varphi(t,x)) = V(\varphi(0,x)) + \int_0^t \frac{d}{dt} V(\varphi(s,x)) ds \\ =V(x) + \int_0^t \nabla V(\varphi(s,x))\cdot f(\varphi(s,x)) ds \ge m + tL, $$ which yields a contradiction for $t$ large enough. Thus $x_0$ is not a stable equilibrium.