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I've been working on a problem where I ended up with the following non linear ODE:

$$\frac{d^2x(t)}{dt^2}=A~\text{sign}(x(t)),$$

where $A$ is just some constant.

I realize that for a specific sign of the function $x(t)$ the solution is a parabola but I need to know beforehand the sign of the function for a specific $t$... Is there a way to solve this equation?

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    $\begingroup$ I'm unsure of how to handle situations when $x(t) = 0$, but one approach is as follows: We can note that $\text{sign}(x(t)) = \frac{x(t)}{\sqrt{(x(t))^2}}$ (again, for non-zero $x(t)$). Thus, your differential equation becomes: $$\frac{d^2}{dt^2}x(t)= A \frac{x(t)}{\sqrt{(x(t))^2}}$$ Wolfram Alpha can solve this (with step-by-step solution), and it doesn't seem to difficult to follow. $\endgroup$ – apnorton Apr 26 '13 at 21:22
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For now I will assume $A$ to be strictly positive. Let $x(t)$ be a solution on an interval $I$. On an interval where $x(t)$ is strictly positive, $x(t)$ is a parabol pointing up. On an interval where $x(t)$ is strictly negative, $x(t)$ is a parabol pointing down. On an interval where $x(t)$ is zero, well... $x(t)$ is zero. (you seem to have already proved this)

If $x(t)$ never reaches 0, then $x(t)$ is constant-sign, because it is continuous (it has to be because it must be two times differentiable). So $x(t)$ is a parabol and resolution is over.

If $x(t)$ reaches zero, let $K$ be the subset of $I$ where $x(t)=0$. K is a closed subset of $I$ so it's the intersection of $I$ and a reunion of closed intervals and singletons. Let $[b,c]$ be one of them (we can have $b=c$).

Suppose that we have room in $I$ at the left of $b$. Then we have a small iterval $]a,b[$ where $x(t) \neq 0$. On that interval $x(t)$ is a parabol. But then $x(t)$ can't be differentiated two times at $b$ unless $x(t)$ is $0$ on $]a,b[$, which is absurd.

Same goes at the right of $c$. We have proved that if $x(t)$ is zero somewhere, $x(t)$ is zero everywhere.

So the solutions are : positive upward parabol, negative downward parabol, and zero. All these solutions are maximal when $I=\mathbb{R}$

Just a thought : This is a special case of DE, since the function that gives the second derivative is not continous. In particular, we don't have the usual theorems of existence.

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  • $\begingroup$ Thank you very much for the answer. I need to look at it with a little more attention since the solutions were not what I would expect for the problem. Thank you once again. $\endgroup$ – PML Apr 26 '13 at 22:14
  • $\begingroup$ The basic idea is that we can solve the DE very easily if x stays positive, stays negative, or stays zero. The problem comes when x changes sign : can we then create a solution with cut-and-paste ? The answer is no. That's why my answer is a big wall of text with no calculus. $\endgroup$ – justt Apr 26 '13 at 22:25
  • $\begingroup$ After reading your proof with more attention I got it. Thank you again for your time $\endgroup$ – PML Apr 26 '13 at 22:36

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