# Autonomous ODE $\dot{x}=f(x)$: $\lim_{t\rightarrow\infty}x(t)=x^*\Rightarrow f(x^*)=0$

Let $$x : [0,\infty) \to \mathbb{R}^d$$ be a solution for the autonomous ODE $$\dot{x} = f(x)$$ where $$f : \mathbb{R}^d \to \mathbb{R}^d$$ is a Lipschitz continuous vector field. We know that $$\lim\limits_{t\to\infty}x(t)=x^*$$ where $$x^*\in\mathbb{R}^d$$. Show that $$f(x^*)=0$$.

I played around with just inserting the limit into $$f$$ as an argument and pulling it out (since $$f$$ is continuous) and I could also imagine where one would use the fact that $$f$$ is Lipschitz continuous. However, I have no idea where the fact that the ODE is autonomous and $$x$$ defined on $$[0,\infty)$$ is important. Hence I'm struggling with this proof.

• The fact that the system is autonomous tells you that the Lipschitz constant of $f$ is uniform in $t$, i.e. it doesn't depend on $t$, in contrast to a scenario where you have $f(x,t)$.
– TSF
May 26 '19 at 9:33
• Use the MVT to prove that if a function $x$ is differentiable and both $\lim_{t\rightarrow\infty} x(t)$ and $\lim_{t\rightarrow\infty} x'(t)$ exist, then $\lim_{t\rightarrow\infty} x'(t) = 0$. Then the above proposition follows trivially from the continuity of $f$. May 26 '19 at 9:51

Each component $$x_k$$ of $$x$$ is a real-valued function on $$[0, \infty)$$, so that we can apply the mean-value theorem: For $$n \in \Bbb N$$ $$x_k(n+1)-x_k(n) = \dot{x}_k(t_n) = f_k(x(t_n))$$ for some $$t_n \in (n, n+1)$$. For $$n \to \infty$$ the left-hand side has the limit zero. On the right-hand side $$t_n \to \infty \implies x(t_n) \to x^* \implies f_k(x(t_n)) \to f_k(x^*)$$ since $$f$$ is continuous.

It follows that $$f_k(x^*)=0$$ for each component of $$f$$, i.e. $$f(x^*) = 0$$.

Remark: The Lipschitz-continuity of $$f$$ guarantees the existence of a solution on $$[0, \infty)$$, but is not needed otherwise in the above proof.

The statement is simply untrue for the nonautomous case. Take for example $$f(x,t)=-x+e^{-t}$$

The solution tends towards zero but $$f(0,t)\neq0$$.

Assume that $$f(x^\ast)=V\neq 0$$ and $$f$$ is $$C$$-Lipschitz

For $$n$$ s.t. $$\frac{C}{n}< \frac{|V|}{2}$$, there is $$N$$ s.t. $$t\geq N$$ implies $$|x(t)-x^\ast|< \frac{1}{n}$$

Then $$|f(x(t))-f(x^\ast)|<\frac{C}{n}<\frac{|V|}{2}$$.

Hence $$f(x(t))=a(t)V +W(t),\ \frac{1}{2}.

Hence $$V\cdot \{ x(t)-x(N) \}= V\cdot \int^t_N\ f(x(t)) dt \geq \frac{1}{2}|V|^2 (t-N)$$

Hence $$x(t)$$ does not converge.