Proving that these solutions are formally solving these differential equations: $x'' = -\text{sgn}(x')$ and $y'' = \sqrt{|y'|}$ Please take a look also to the comments section, here, and in other people answers, since there are extended what are my apprehensions about the validity of the found answers.
I have found these two solutions to the following differential equations by playing on Wolfram-Alpha, and I would like to prove that they are formally solutions of each equation:

*

*$$x'' = -\text{sgn}(x'),\quad \,x(0)=2,\,x'(0)=-2 \quad \Rightarrow \quad x(t) = \frac{1}{2}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^2\quad\text{(Eq. 1)}$$

*$$y'' = \sqrt{|y'|},\quad \,y(0)=\frac{2}{3},\,y'(0)=-1 \quad \Rightarrow \quad y(t) = \frac{1}{12}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^3\quad\text{(Eq. 2)}$$
Notice that Wolfram-Alpha don't show close-form solutions for neither these equations as can be seen here and here.
The following notation is going to be used:

*

*$\text{sgn}(t)$ is the Sign function, which fulfills that $\frac{\partial}{\partial t}\left(|t| \right) = \text{sgn}(t)$ and $\frac{\partial}{\partial t}\left(\text{sgn}(t) \right) = 2\ \delta(t)$

*$\theta(t)$ is the Heaviside step function that fulfills $\frac{\partial}{\partial t}\left(\theta(t) \right) = \delta(t)$ and $\theta(t) = \frac{1}{2}\left(1+\text{sgn}(t) \right)$

*$\delta(t)$ is the Dirac delta function that fulfills $\int_{-\infty}^\infty f(t)\delta(t)\,dt=f(0)$ with $\int_{-\infty}^\infty \delta(t)\,dt=1$
These definitions could be problematic, as it will reviewed later.

Motivation
Recently I figure out that no non-piecewise power series could have a finite extinction time due the Identity theorem, and found on this paper that a differential equation could have a solution  that achieve a finite extinction time if and only if its nonlinear and have a singular solutions, so it must fail to fulfill uniqueness of solutions, but luckily the paper explains that within the initial conditions time and the finite extinction time uniqueness is still hold.
Since my intuition tells that classic mechanics system should achieve a finite extinction time where the movement due the system dynamics dies (as opposite of random thermal noise, which nature is external to the system dynamics as it were a random forcing force), so I started to look for some physics' examples, without finding many (I even tried to made them as in here and here), and those I found (here) were quite hard to understand (at least for me, I'm an electrician, nor a physicists neither a mathematician).
But a few days ago, someone in a Youtube comment named @siguc explains me the following:

"How about the motion of a brick on a horizontal surface with constant friction between the brick and the surface? Assuming the brick moves along the surface at $t=0$, it'll stop eventually. Newton's law: $x''=-k\ g\ \text{sgn}(x')$, where $g$ is $9.8\,\frac{m}{s^2}$ and $k$ is the friction coef.".

So I started to googling about this problem founding terms as Coulomb damping and the Stribeck curve, but the only place I found the same brick problem was in this Wiki page and no close-form solutions were shown.
Since the system were simple enough to be understood by me, I start by trying to see if I could find a solution to the simplest case by myself, so assuming here that $k\ g = 1$, I got the $\text{(Eq. 1)}$. Later I will explain why I choose those arbitrary initial conditions.

What I have done so far
On previous question I have found:

*

*In this question another person (@KBS) proves on his answer that the solution I found could be formally a solution to $$u' = -\text{sgn}(u) \sqrt{|u|},\quad \,u(0)=1 \quad \Rightarrow \quad u(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^2\qquad\text{(Eq. 3)}$$

*Later I found here that for a positive finite extinction time $T>0$ and a positive initial conditions $v(0)>0$ that determines $T(v(0))$, one can have that: $$\begin{array}{r c l} v' = -\text{sgn}(v) \sqrt[n]{|v|},\quad \,v(0)>0 \quad \Rightarrow \quad v(t) & = & \left[\frac{n-1}{n}\left(T-t\right)\right]^{\frac{n}{n-1}}\theta\left(T-t\right) \\ &\overset{!}{\equiv}& v(0)\cdot \left[\frac{1}{2}\left(1-\frac{t}{T}+\left|1-\frac{t}{T} \right|\right)\right]^{\frac{n}{n-1}}, \\ & & v(0) =  \left[\frac{n-1}{n}\cdot T\right]^{\frac{n}{n-1}}\qquad\text{(Eq. 4)} \end{array}$$
I have found these results in a laissez-faire way and not $100\%$ rigorously: I know that some definitions at the beginning have issues in the edge points, like the Heaviside function having or not a transition value of $1/2$ at $t=0$, which I have ignored, so I am using things like $\theta(t) = (\theta(t))^n$ which have some weird consequences, as in (Eq. 4) where the exclamation point is evidenced on the equivalence could be flawed. I have preferred the version with the absolute value function trying to be as far as possible of having derivatives of the Dirac Delta function $\delta'(t)$ which I don't know how to handle them.
Also has the consequence that if I evaluate the differential equation some issues happened at these edge points, like the rising of Dirac delta functions which broke the differential equation equality, but since it only happened on a zero-measure point, and the solutions doesn't have those problems, I believe they are valid as solutions.
With this laissez-faire approach I was able to just test some solutions of the form $x(t) = a\cdot(T-t)^b$ similar to the found for (Eq. 3) and (Eq. 4) and made them fit the differential equation, since from the mentioned paper I believe beforehand that there was an existent singular solution that could achieve a finite extinction time.
Since these special functions $\delta(t),\,\text{sgn}(t),\,\theta(t)$ are in reality distributions, which theory I know almost nothing, I would like to know if its possible to prove in a rigorous way they are indeed solutions of those differential equations.

BONUS TRACK: Do you believe it is possible to find a general solution for $x''=-k\ g\ \text{sgn}(x')$ using (Eq. 1)? At first I was concern about having an initial condition for the speed $x'(0)<0$, but I think now it makes sense since the system starts with an initial speed $|x'(0)|$ but through its dynamics it must start loosing speed immediately since friction is the only external force present in the sliding brick system.
PS: on other question like this people have attacked the intuition of having solutions that achieve a finite ending time, being valid opinions, please keep it out of the discussion here, since I am trying to explore these kind of solutions as alternative - instead, feel free to extend the discussion in the mentioned question which is more suitable for it.

Added later
About my worries, as example, in Wikipedia the Heaviside step function is defines as $\theta(0)=\frac{1}{2}$ which could generate issues since I have assumed as true that $\theta(t)=(\theta(t))^n$. Also, I am using that $$\frac{1}{2}\left(x +|x|\right) = \frac{1}{2}|x|\left(1+\frac{x}{|x|}\right) = \frac{1}{2}|x|\left(1+\text{sgn(x)}\right)=|x|\ \theta(x)$$, but if instead I take as factor the other term I get:
$$\frac{1}{2}\left(x +|x|\right) = \frac{1}{2}x\left(1+\frac{|x|}{x}\right) = \frac{1}{2}x\left(1+\frac{1}{\text{sgn(x)}}\right)$$
And since in Wikipedia defined the sign function as having $\text{sgn}(0) = 0$ then the term $\frac{1}{\text{sgn(x)}}$ hidden a division by zero, which is obviously a big issue.
These are examples of why I am worried about the validity, this sames issues also made struggles with the definitions at the beginning. Hope you could elaborate why is not a problem if it really is not.

An attempt for the bonus track
If I used the assumption than $kg>0$, which make sense with the physics problem the equation is coming from, and I used the change of variable $x' = kg z$, the equation could become:
$$\begin{array}{rcl}
-\frac{x''}{kg} & = & \text{sgn}(x') \\
\iff -\frac{kgz'}{kg} & = & \text{sgn}(kg\ z) = \frac{kg\ z}{|kg\ z|} \overset{\text{since}\ kg>0}{=}\frac{kg\ z}{|kg|\ |z|} =\frac{z}{|z|} = \text{sgn}(z) \\
\Rightarrow z' & = & -\text{sgn}(z) \quad\text{which is equivalent to (Eq. 1)}\end{array}$$
With this, we have that the answer to:
$$x''=-k\ g\ \text{sgn}(x'), \quad x(0) = 2kg,\,\,x'(0)=-2kg \Rightarrow x(t) = \frac{kg}{2}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^2$$
So for a general finite extinction time $T>0$ and constants such as $kg>0$ then an answer could be:
$$ \begin{array}{l} x''=-k\ g\ \text{sgn}(x'), \qquad x(0) = T^2\frac{kg}{2}>0,\,\,x'(0)=-Tkg<0 \\ \Rightarrow x(t) = T^2\frac{kg}{8}\left(1-\frac{t}{T}+\left|1-\frac{t}{T} \right|\right)^2 = \frac{kg}{2}\left(T-t\right)^2\theta(T-t) \quad \text{(Eq. 6)} \end{array}$$
Does (Ec. 6) make sense-full as closed-form solution for the problem?
I have used that $s(t) = \int -(T-t)\ dt\cdot\theta(T-t) = \left[\frac{1}{2}(T-t)^2 + C_0\right]\theta(T-t)$ such as:
$$s'(t) = -(T-t)\theta(T-t)\quad + \underbrace{\left[\frac{1}{2}(T-t)^2 + C_0\right]\delta(T-t)}_{C_0 \equiv 0,\,\text{so all the expression could be zero by}\,x\delta(x) = 0}$$
It is still a general solution?

Another example of the weird issues I have found, is that if in Wolfram Alpha, as example, I try to solve the equation (Eq. 1) with the founded solution of (Eq. 6) as:
$$\frac{1}{kg}\frac{\partial^2}{\partial t^2}\left(\frac{kg}{2}\left(T-t\right)^2\theta(T-t)\right)+\text{sgn}\left(\frac{\partial}{\partial t}\left(\frac{kg}{2}\left(T-t\right)^2\theta(T-t)\right)\right)$$
I will find a mess as is shown here.
But if instead I solve the less rigorous equation:
$$\left(\frac{1}{kg}\frac{\partial^2}{\partial t^2}\left(\frac{kg}{2}\left(T-t\right)^2\right)+\text{sgn}\left(\frac{\partial}{\partial t}\left(\frac{kg}{2}\left(T-t\right)^2\right)\right)\right)\cdot\theta(T-t)$$
it shows now that the proposed solution solves the equation (except in one isolated point $t=T$, which is also werid), and I find strange how and why the $\theta(t)$ function can cross through operations as it where a ghost (hope you understand what I am trying to say). Any source to some place where these properties are explain will be fantastic, Thanks.
 A: Perhaps I don't understand the problem, but I don't think that it's difficult to prove within distribution theory that those are solutions to the differential equations (I skip the initial conditions since I'm sure that you can check them yourself).
The first solution can be written $x(t) = 2(1-t/2)^2 \, \theta(1-t/2).$ If $f$ is a smooth function (here meaning $C^\infty$) and $u$ is a distribution then the product rule $(fu)'=f'u+fu'$ is valid. Also, if $g$ is affine, then the chain rule $(u\circ g)'=g'\cdot(u'\circ g)$ is valid. Furthermore, $\theta'=\delta$ and $f\delta=f(0)\delta.$ Thus,
$$
x'(t) 
= 2(-1/2)\cdot 2(1-t/2)\,\theta(1-t/2) + 2(1-t/2)^2 \, (-1/2)\delta(1-t/2) 
= -2(1-t/2)\,\theta(1-t/2),
$$
and
$$
x''(t)
= -2(-1/2)\,\theta(1-t/2) - 2(1-t/2)\,(-1/2)\delta(1-t/2)
= \theta(1-t/2).
$$
Now note that
$$
\operatorname{sgn}(x'(t))
= \operatorname{sgn}(-2(1-t/2)\,\theta(1-t/2))
= -\operatorname{sgn}(1-t/2)\,\theta(1-t/2)
= -\theta(1-t/2)
= -x''(t).
$$
Likewise for the other differential equation and its solution.
