Let $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ continuous fulfilling $$ \begin{cases} f(t,x)<0, & \text{if $xt>0$}.\\ f(t,x)>0, & \text{if $xt<0$}. \end{cases} $$ Show that, $x(t)\equiv0$ is the only solution to the initial value problem $$\dot{x}=f(t,x)\hspace{1cm} x(0)=0$$

So obviously $x(t)\equiv0$ solves the problem on $\mathbb{R}^2$ in order to show that it's the only solution I would try to use the Picard-Lindelöf theorem. In order to be able to use it I would furthermore have to prove that $f$ is locally Lipschitz in $x$. But how can I do this ? I know of this theorem that says that if $f$ is locally differentiable w.r.t $x$ it's locally Lipschitz. Would that help ?

  • $\begingroup$ Does x=0 even solve it? $\endgroup$ – M.B. May 26 at 11:35
  • $\begingroup$ Well I assume that any function that respects these two constraints is a viable option $\endgroup$ – Christian Singer May 26 at 11:37

The continuity of $f$ implies that $f(t, 0) = 0$ for all $t \in \Bbb R$, so that $x(t) \equiv 0$ is a solution.

For any solution $x$ which is defined in an interval $I$ containing $t=0$ $$ \frac{d}{dt}(x^2(t)) = 2 x(t) \dot x(t) = 2 x(t)f(t, x(t)) \begin{cases} \le 0 & \text{ if } t > 0 \\ \ge 0 & \text{ if } t < 0 \end{cases} $$ because if $x(t) \ne 0$ then $x(t)$ and $f(t, x(t))$ have the opposite sign for $t> 0$, and the same sign for $t < 0$.

It follows that $x^2(t) \le x^2(0) = 0$.

  • $\begingroup$ How does it come, that we now have $t$ instead of $xt$ in the if conditions to the right ? $\endgroup$ – Christian Singer May 26 at 14:04
  • 1
    $\begingroup$ @ChristianSinger: You have to consider all possible combinations. For example, if $t > 0$ and $x(t) < 0$ then $f(t, x(t)) > 0$ so that the product $x(t) \cdot f(t, x(t))$ is negative, and so on. And if $t=0$ or $x(t)=0$ then the product is zero. $\endgroup$ – Martin R May 26 at 14:10
  • $\begingroup$ Ah! I see! You could take the coordinate systems with axes $x$ and $t$ and get that in quadrant II and IV $f$ is always positive and in I and III always negative. (except for the axes themselves) $\endgroup$ – Christian Singer May 26 at 14:13
  • 1
    $\begingroup$ @ChristianSinger: Exactly. Or put it like this: The product is zero if $t=0$ or $x(t) = 0$. Otherwise $x(t)$ and $f(t,x(t))$ have the opposite/same sign if $t$ is positive/negative. $\endgroup$ – Martin R May 26 at 14:14
  • $\begingroup$ This might be coming a bit late but upon reading this answer again, I wonder how the continuity of $f$ implies that $x(t)\equiv 0$ is a solution. $\endgroup$ – Christian Singer Jun 1 at 18:04

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.