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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.

1 Answers 1

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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.

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    You should now be able to see the contradiction between assumptions and possible solutions. – 2017-01-21
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    @Lutzl...I dn't understand..Is there any mistake? – 2017-01-21
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    You got $x(t)=-t<0$ for $t>0$. The case study in contrast gave that $x(t)=+t+B$ for $x(t)<0$. Both can not be true at the same time. – 2017-01-21
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    Oh! Yes...I was thinking the sign of $t$ instead of $x$. See my edit. – 2017-01-22