ODE $f:\text{<0 if$tx > 0$},\text{>0 if$tx<0$}$; show that $x(t)\equiv0$ is the only solution to $\dot{x}=f(t,x)\hspace{0,3cm}, x(0)=0$

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 ?

• Does x=0 even solve it? – M.B. May 26 at 11:35
• Well I assume that any function that respects these two constraints is a viable option – 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 $$x^2(t) \le x^2(0) = 0$$.
• How does it come, that we now have $t$ instead of $xt$ in the if conditions to the right ? – Christian Singer May 26 at 14:04
• @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. – Martin R May 26 at 14:10
• 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) – Christian Singer May 26 at 14:13
• @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. – Martin R May 26 at 14:14
• 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. – Christian Singer Jun 1 at 18:04