Is $r(t) = \frac{1}{144}(T-t)^4\theta(T-t)$ a valid solution to $\ddot{r}=\sqrt{r},\,r(0)=\frac{T^4}{144}>0$? (with $\theta(t)$ the Heaviside step fn)

Question

Is $r(t) = \frac{1}{144}(T-t)^4\theta(T-t)$ a valid solution to $\ddot{r}=\sqrt{r},\,r(0)=\frac{T^4}{144}>0$? (with $\theta(t)$ the Heaviside unitary step function)

I am looking here for examples of solutions of finite duration to differential equations, and in this answer someone tells that the Norton's Dome could be an example, which is described by the differential equation $\ddot{r}=\sqrt{r}$, and as is better explained here and here, that leads to difficult interpretations of its solution with initial values $r(0)=0,\,r'(0)=0$ since its solution becomes $r(t)=\frac{1}{144}(t-T)^4\theta(t-T)$ so it could take any value of $T$ (since is already zero before time $T>0$ so the initial conditions don´t apply on $T$), which is weird as is discussed there, but I don´t wont to review that situation here.

As noted in this answer to this another question, the equation: $$\dot{x}=-\sqrt{x},\,x(0)=1$$ Has as solution $x(t)=\frac{1}{4}(2-t)^2\theta(2-t)$ for $x(0)=\frac{2^2}{4}>0$, since the rate of change is diminishing (minus sign), as a square root that is always positive, so after it reaches zero it will stay there forever (actually, Wolfram-Alpha gives a solution without the piecewise versions, so it should be mistaken from the explanation, but since the ODE is not Lipschitz, it falls under the non-uniqueness of solutions - I believe).

Now, following this answer to this other question, since these differential equations are behaving like piecewise differential equation for $t>T$ and $t<T$ (given there exist a point $r(T)=r'(T)=0$), and both stands the trivial zero solution, it should be possible to made solutions by "stitching" solutions combining the existing ones for $t>T$ and $t<T$.

With all these, I believe that the differential equation: $$\ddot{r}=\sqrt{r}$$ could stand finite duration solutions for an ending time $T>0$ determined by the initial condition $T = \sqrt[4]{144\,r(0)}$ for $r(0)>0$, leading to the solution: $$r(t) = \frac{1}{144}(T-t)^4\theta(T-t)$$ where is also interesting to note that the ending time condition $r(T)=0$ is somehow behaving like a boundary condition, since the solution to this second order differential equation get fully determined by only $r(0)$ without needing the initial value of the first derivative, which is also fully determined by the value of $r(0)$.

Since differently from $\dot{x}=-\sqrt{x}$, here in $\ddot{r}=\sqrt{r}$ the same argument about the positive squared root only says that the convexity/concavity is hold one or another, so I want to know if formally this piecewise solution solves the differential equation in the full time domain.

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Motivation

This part is not required to solve the question, if you want to skip it.

If the differential equation stands a finite-duration solution, there exist some "ending time" $T$ where $r(T)=\dot{r}(T)=0$ (see Finite Time Controllers by Vardia. T. Haimo).

So, if the differential equation has as trivial solution the zero function, I believe that If I can choose a finite duration solution $y(t) = r(t)\theta(T-t)$ with $\theta(t)$ the Heaviside unitary step function: $$\theta(t) = \begin{cases} 1,\quad t > 0 \\ 0,\quad t \leq 0 \end{cases}$$

I will have that: $$\begin{array}{r c l} \dot{y}(t) & = & \frac{d}{dt}\left(r(t)\theta(T-t) \right)\\ & = & \dot{r}(t)\theta(T-t) + r(t)\delta(T-t) \\ & = & \dot{r}(t)\theta(T-t) + \underbrace{r(T)\delta(T-t)}_{=\,0\,\text{if}\,r(T)\,=\,0} \\ & = & \dot{r}(t)\theta(T-t) \end{array}$$

And equivalently: $$\begin{array}{r c l} \ddot{y}(t) & = & \frac{d}{dt}\left(\dot{r}(t)\theta(T-t) \right)\\ & = & \ddot{r}(t)\theta(T-t) + \dot{r}(t)\delta(T-t) \\ & = & \ddot{r}(t)\theta(T-t) + \underbrace{\dot{r}(T)\delta(T-t)}_{=\,0\,\text{if}\,\dot{r}(T)\,=\,0} \\ & = & \ddot{r}(t)\theta(T-t) \end{array}$$

So, for the second order system with zero as trivial solution, if there exists a ending time $T$ such as $r(T)=\dot{r}(T)=0$ the finite duration solution could be made by any solution before time $T$ "stitched" with the trivial zero function solution after time $T$ (this because $y(t) \equiv r(t),\, t<T$).

Also note that under this construction, if happens to be true that $r(T)=\dot{r}(T)=0$, then also $r^{(n)}(T) = 0$ for all positive integer $n\geq 2$, indicating here that indeed the dynamics "die", so at least for modeling physics stuff, solutions like the discussed for the Norton's Dome shouldn't be happening in real life, since the position is zero and velocity is zero, kinetic energy is also zero, so position shouldn't be changing, so the system must be unperturbed in this value (I think this is true, but if experienced people are discussing this issue as is done on the already cited links, maybe there is something I am missing).

But looking this "formally", in my opinion, there are some definition issues: if I paste this solution on the differential equation I will have: $$\ddot{y}= \sqrt{y} \iff \ddot{r}\theta(T-t) = \sqrt{r}\cdot \sqrt{\theta(T-t)}$$ so even that conceptually $\theta(T-t) \equiv \sqrt{\theta(T-t)}$ (their plots are the same), I believe I am messing up with the formal treatment of this special functions (like for Dirac's delta under distribution theory), but I also believe the solutions still stands because of local existence and uniqueness of ODEs (in this case at least).

So somehow, being not-rigorous about the treatment of the the $\theta(t)$ function sometimes leads to garbage, but some other times, it will help to find the solutions, as example, using things like $\theta(t) = \frac{1+\text{sgn}(t)}{2}$ and other times using $\theta(t) = \frac{1}{2}\left(1+\frac{1}{\text{sgn}(t)}\right)$ or using $\text{sgn}(t) \equiv \frac{1}{\text{sgn}(t)}$ (the solution $x(t)=\frac{1}{4}(2-t)^2\theta(2-t)=\frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2$, as example when it works), things I am sure are mistaken (just seeing their derivatives which are different, even, there is a definition issue with the reciprocal versions), but sometimes if they suit, well... is part of what I am trying to understand.

Even so, when seeing the differential equation $\ddot{y} = \sqrt{y}$, there is nothing wrong after time $T$ since it will lead to $0 = 0$, some other people have noted that also I am having $\frac{\ddot{y}}{\sqrt{y}} = 1$ which lead to a division by zero after time $t\geq T$, so somehow it also could be wrong. Following the formalism of the cited paper Finite Time Controllers I believe there is no ambiguity, but for example this issue was importantly argued in this other question where I tried to built a finite duration solution from a smooth bump function $\in C_c^\infty$, and the argument also looks right (is completely valid refuse something if there rise a division by zero).

So I am looking to validate the presented approach somehow, formally speaking, because of these issues.

Answer 1

So, yes, the solution that you provide is the solution to the dynamical system you are are considering under the assumption that the the initial conditions of the system are $r(0)=T^4/144$ and $\dot{r}(0)=-T^3/36$. You can easily verify that with Wolfram or by successive differentiation. This solution is twice differentiable at all points $t\ge0$, so everything is fine. It also converges to zero finite-time.

However, for other initial conditions, the system may go to infinity such as when both $r(0)$ and $\dot r(0)$ are positive. This will also be the case when $r(0)>0$ and $\dot{r}(0)$ is not too negative (but not such that $r(0)=T^4/144$ and $\dot{r}(0)=-T^3/36$). In such case, the trajectory will be first decreasing and then quickly start increasing before $r(t)$ may hit the zero value and then go to infinity.

The system has no real solution when $r(0)<0$ as this would mean evaluating the square root of a negative number.

Finally, the solution may also have (real) solutions which are only defined on a compact set $[0,\tau]$. This is the case when $r(0)>0$ and $\dot{r}(0)$ is negative enough (but not such that $r(0)=T^4/144$ and $\dot{r}(0)=-T^3/36$). In that case, the solution will stop when $r(t)$ hits zero at some time $\tau$ that will depend on the values of the initial condition.