How to "formally" prove that $x(t)=\frac{1}{4}(2-t)^2\theta(2-t)$ solves $\dot{x}=-\text{sgn}(x)\sqrt{|x|},\,x(0)=1$? (with $\theta(t)$ the standard unitary step function).
I have found the equation studying these paper by V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers, and I believe the finite duration solution is given by:
$$x(t)=\frac{1}{4}(2-t)^2\theta(2-t)$$
Which I found from the comparison with the other equation $\dot{y}=-\sqrt{y},\,y(0)=1$ which solution is $y(t) = \frac{1}{4}(2-t)^2$.
Actually, the first solution I found was:
$$x(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2,$$
and so far, I have not been able to formally prove that $x(t)$ is the "true" solution, even when graphically is obvious (and theoretically, since ODEs stands Locally the Existence and Uniqueness of solutions, I believe it fits).
For every try, to made coincident both parts of the equation (right and left hand sides), I have to made an illegal operation like:
- assuming that is equivalent to use $\text{sgn}(x(t)\cdot\theta(2-t)) \cong \text{sgn}(x(t))\cdot\theta(2-t)$
- assuming that is equivalent to use $\theta(t) = \frac{1+\text{sgn}(t)}{2}$ or $\theta(t) = \frac{1}{2}\left(1+\frac{1}{\text{sgn}(t)}\right)$
- assuming that is equivalent to use $\text{sgn}(x(t)) \cong \frac{1}{\text{sgn}(x(t))}$
- assuming that is equivalent to use $\int \theta(t)\,dt = \int \theta^n(t)\,dt,\quad n\in \mathbb{Z}^+$
And other similar things which I beforehand know are mistaken (just by taking their derivatives I know it leads to different functions), but when displayed on graphs, they take the same values. So I know I am making mistakes (since this are distributions instead of functions and I don´t know how to formally work with them), but also because of been "vague" with the treatment have let me find the previous candidates to the solution, so I don´t know if I am finding something or it is everything wrong (maybe it is just the same as taking the solutions piecewise, but at least the differential equations at first doesn't look to be defined piecewise – at least for me, since I know beforehand that a finite duration solution will lead to definition issues on the diff. eq. since it becomes zero forever after the ending time, $t_F = 2$ on this example).
Hope you can elaborate on how to prove it explaining why the listed things are conceptually different (I am asking here for the "prove", not an explanation of how to find it, this because of my lack of background, it will be more useful to see how it is done and then try to search for the missing things I need to look for to understand the explanations). Beforehand thanks you very much.
Best Answer
Since $x(0)=1>0$, then one can see the solution is decreasing to zero and then stays there when it reaches this value. So, the differential equations simplifies to $\dot{x}=-\sqrt{x}$.
This can be rewritten as $$\dfrac{dx}{\sqrt{x}}=-ds$$ and integrating yields
$$\int_{x(0)}^{x(t)}\dfrac{dx}{\sqrt{x}}=-\int_0^tds$$ which yields
$$2(\sqrt{x(t)}-1)=-t.$$ Reorganizing and squaring then yields
$$x(t)=\dfrac{1}{4}(2-t)^2$$ but this is only valid for $t\in[0,2]$. For $t>2$, $x(t)=0$, so we multiply that expression by $\theta(2-t)$ to get the final result.
The expression you found with the absolute values is also valid and is equivalent to the one using the Heaviside function.