Let $x\in \mathbb{R}^n$ and $f:\mathbb{R}^n \to \mathbb{R}^n$ be smooth. Suppose $x \cdot f(x) < 0$ for all nonzero $x$. We want to show that all solutions to $x'(t)=f(x(t))$ tend to $0$.
Usually these are pretty straightforward. Indeed, for such solutions $x$, we have $$\frac{d}{dt} ||x|| = \frac{x \cdot x'}{||x||} =\frac{x\cdot f(x)}{||x||} <0$$

and so the norm of each solution is decreasing in $t$. But given this info, what stops there from being a solution which is bounded away from $0$? I know that for any initial data $(t_0,x_0)$, $||x(t)||\le ||x_0||$, but I don't think that helps too much. I might be missing something easy, but any help is appreciated.

  • $\begingroup$ For some intuition, suppose the norm converges to something nonzero and moreover the trajectory even converges to something nonzero (the latter is not guaranteed of course). Then what happens if you restart the process at the limit? On the one hand, it should be frozen, because the derivative went to zero. On the other hand it can't be frozen by the initial hypothesis. In the "real" problem you have to tackle this directionality issue as well. $\endgroup$ – Ian Jul 10 '17 at 22:07

Hint: Use continuity of solutions w.r.t. initial conditions and a compactness argument. Some details:

1) Assume $|x(t)|\rightarrow r>0$ as $t$ goes to infinity

2) By compactness you may find a sequence of times $t_n\rightarrow \infty$ and $y_0$ so that $x(t_n)\rightarrow y_0$. Let $y(t)$ be the solution starting at $y_0$

3) Show that for $s\geq 0$ you have pointwise convergence $x(t_n+s)\rightarrow y(s)$.

4) Conclude that $|y(s)|=r$ for all $s\geq 0$ and get a contradiction.


If $X(t)$ is any solution

$$ \dfrac{d}{dt} \|X\|^2 = 2 X \cdot X' = 2 X \cdot f(X)$$ Now $x \cdot f(x)$ is strictly negative on $\mathbb R^n \backslash \{0\}$, and is continuous, so it is bounded away from $0$ on compact sets. Suppose there was $\epsilon > 0$ and a solution $X(t)$ with $\|X(t_0)\| \ge \|X(t)\| \ge \epsilon $ for all $t > t_0$, There is $\delta > 0$ such that $x \cdot f(x) \le -\delta$ for all $x$ with $\|X(t_0)\| \ge \|x\| \ge \epsilon$. And then

$$\|X(t)\|^2 \le \|X(t_0)\|^2 - 2 \delta (t - t_0)$$ which is impossible for $t > t_0 + \|X(t_0)\|^2/(2\delta)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.