Solve a nonlinear ODE 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?
 A: 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.
