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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 ?

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  • $\begingroup$ 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
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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 $x^2(t) \le x^2(0) = 0$.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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