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:
- $$\lim\limits_{t \to 0} x(t) \equiv x(0)$$ 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.
- 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 $t \in [0,\ T)$
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:
- $(f\theta)'=f'\theta+f\theta'=f'\theta+f\delta =f'\theta$ since $f(t)$ is a polynomial with the same argument as the Dirac's delta function$\delta(t)$, it happens to be $f\delta \equiv (c-t)^n\delta(c-t)=0$.
- 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 $t\geq T$). You could see it's numerical verification on Desmos.
Update 2
@CalvinKhor have sugest me this comment by @ThomasRichard in MathOverflow suggesting a real life physical model that fits the differential equation: "It might be useful to mention that the ODE $y′=−\sqrt{y^{+}}$, for which non uniqueness holds, actually models a real physical system. $y(t)$ represents the water height in a pierced cylindrical bucket. And the non uniqueness of the Cauchy problem with initial condition $y(0)=0$ is natural in this model: if you know the bucket is empty at time $0$, it is hard to tell if there was water in it before and when it got empty."
And later, @CalvinKhor also shared the explanation in this other comment, which I like to share since I found it really interesting and suits here:
