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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.

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    $\begingroup$ 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 $f(x,t)$. $\endgroup$ – Tony May 26 at 9:33
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    $\begingroup$ 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 $f$. $\endgroup$ – Nicolas May 26 at 9:51
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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.

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

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

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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}<a(t) < \frac{3}{2},\ V\perp W(t)$.

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.

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