Solving $x’=-\text{sgn}(x)\sqrt{|x|}$: Uniqueness of solutions of finite duration

Question

Show that $$x(t) = \frac{\text{sgn}(x(0))}{4}\left(2\sqrt{|x(0)|}-t\right)^2\cdot\theta\!\left(2\sqrt{|x(0)|}-t\right)$$ is a solution to$$x'=-\text{sgn}(x)\sqrt{|x|}.$$
Is the solution unique?

This question was modified since I was able to find the answer to the first question.

Here, $\theta(t)$ is the Heaviside step function, and $\text{sgn}(x)$ is the Sign function.

This is a continuation of this question. By using the answer given by @KBS, I could find solutions with the same "structure" as the mentioned answer if $x(0)>0$, but for initial conditions $x(0)<0$, I struggle because the term $(2\sqrt{x(0)}-t)^2$ is, first, insensible to sign changes, and secondly, becomes complex, so I added an absolute value function to the initial condition, and later just pasted a term $\text{sgn}(x(0))$ in order to "make" the same solutions I find in the "slope-field" of wolfram-Alpha for negative initial values, leading to:

$$x(t) = \frac{\text{sgn}(x(0))}{4}\left(2\sqrt{|x(0)|}-t\right)^2\cdot\theta\!\left(2\sqrt{|x(0)|}-t\right)$$
which fulfill as I require:

1. $$\lim\limits_{t \to 0} x(t) \equiv x(0)$$ for any $x(0)\in \mathbb{R}$ (there is a discontinuity when arriving to $x(0^+)$ and $x(0^-)$ but since $x(0^+)\equiv x(0^-)\equiv 0$ I think there is no issues at all: checked here and then here.
2. It have a finite extinction time $T = 2\sqrt{|x(0)|}\geq 0$ where the system stop moving at zero $x(t) = 0,\ \forall t\geq T$.

But when I insert the function $x(t)$ into the equation: $$x'+\text{sgn}(x)\sqrt{|x|} = 0$$
I cannot make the terms to become zero: from the differentiation side I got some Dirac delta functions as shown here, and from the nonlinear term I cannot match the exponent in the sign functions outside and within the square root as shown here, so I don't know if the solution that we have found is indeed a solution to the differential equation "formally" speaking (at every point and real-valued initial condition).

• Is the $x(t)$ that we obtained indeed a formal solution to $x'+\text{sgn}(x)\sqrt{|x|} = 0$? (SOLVED)
• Is this solution unique? at least in the domain $t \in [0,\ T)$

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Added Later

Later I remembered this answer by @md2perpe where show that since $x\delta(x)=0$ in this situation where solution $x(t)=f(t)\theta(t)$ is such as:

1. $(f\theta)'=f'\theta+f\theta'=f'\theta+f\delta =f'\theta$ since $f(t)$ is a polynomial with the same argument as the Dirac's delta function $\delta(t)$, it happens to be $f\delta \equiv (c-t)^n\delta(c-t)=0$.
2. its equivalent $\text{sgn}(f\theta)\sqrt{|f\theta|} = \text{sgn}(f)\sqrt{|f|}\,\theta$

Then I do could match the left-hand-side $f'\theta$ with the right-hand-side $\text{sgn}(f)\sqrt{|f|}\,\theta$ at least while $t<T$, meaning that $x(t)$ is indeed a solution of finite duration (this, since the trivial solution $x(t)=0$ satisfies the differential equation after $t\geq T$).

Answer 1

IVPs for $x'(t)= -{\rm sgn}(x(t))\sqrt{|x(t)|}$ are uniquely solvable to the right, since $x \mapsto -{\rm sgn}(x)\sqrt{|x|}$ is monotone decreasing on $\mathbb{R}$. More general, let $f:\mathbb{R} \to\mathbb{R}$ be monotone decreasing, and let $x_1,x_2:[0,T) \to \mathbb{R}$ be solutions of $x'(t)=f(x(t))$, $x(0)=x_0$. Set $d(t):= (x_1(t)-x_2(t))^2$ $(t \in [0,T))$. Then $$ d'(t) = 2(x_1(t)-x_2(t))(f(x_1(t))- f(x_2(t))) \le 0 \quad (t \in [0,T)). $$ Hence $d$ is decreasing and $\ge 0$, and $d(0)=0$. Thus $d(t)=0$, that is $x_1(t)=x_2(t)$ $(t \in [0,T))$.