Proving this IVP has no solution

Question

Prove that:

$\dot{x} = f(x)$ where $f(x) = −1$ when $x ≥ 0$ and $f(x) = +1$ when $x < 0$

with $x(0)=0$

has no solution.

I can appreciate why it has no solution in plain English, but am struggling with how I should go about proving this mathematically.

Answer 1

Assuming $\dot x=\frac{dx}{dt}$

For $x\ge 0$, $x(t)=-t+A$

For $x<0$, $x(t)=t+B$ where $A$ and $B$ are constants of integration. Using the continuity of $x(t)$ at $t=0$, $x(0^-)=0=x(0^+)$, you can easily get $A=B=0$. Hence $x(t)= -t$ if $x\ge 0$ and $x(t)=t$ if $x<0$ is the solution.

Added

You can see that if we restrict $x(t)(>0)=-t$ by taking $t<0$ then $x(t)(<0)=t$ for $t>0$ doesn't hold.