Does finite-duration solutions to differential equation could be thought as the reciprocal of a system with finite-time blow-up?
Introduction
Recently I have found on the papers by Vardia T. Haimo Finite Time Differential Equations and Finite Time Controllers that there are finite-duration solutions to differential equations (which previously, I only know classic analytical solutions that at best vanishes at infinity - obviously not considering the zero function), and I have made some question here finite-duration trying to understand them through some examples (since I am not trained on Liapunov theory - I tried to research it but are not easy).
When searching for related information, I found on youtube a video of Terence Tao talking about finite-time blow-ups on the Navier-Stokes (NS) Equation:
- It was just coincidence because of the term finite-time, as in the first paper I mentioned.
- Looks like blow-ups on the NS equation are important since there is Millenium Prize question related (is a video from one of the most famous nowadays mathematicians), so must be something "interesting".
Obviously I don´t understand almost nothing, and this question is not related to the Millenium Prize neither to the NS Equation , but interestingly, he say something related to the blow ups for scalar differential equations, like they are not understand as much in the PDE cases (maybe is a translation mistake since I am not a native English speaker, so my apologizes in advance if I am mistaken).
By seeing the video just for curiosity, I search about these finite-time blow ups and I found this question where a simple example $\dot{y} = y^2,\,\,y(0)=1$ is given where its solution it is $y(t) = \frac{1}{1-t}$, which resembles the reciprocal of the solutions I have found so I believe they could be related.
Also, intuition says it should be the case, thinking as example in the Euler's Disk toy, where the wobble rate rises to infinity, but sounds abruptly stop indicating that the movement have indeed stop (instead like vanishing at infinity but keeping it under thermal noise-movement, argument I have saw for people discussing the solution of the pendulum with friction)... I believe that in this example the wobbling rate blow up is just an indication of its denominator going to zero as a finite duration solution.
So since I am looking for examples of finite duration solutions, maybe I could look for systems with finite time blow up behavior and taking its reciprocals.
Main text
On this question I am asking to formal validation about the differential equation: $$\ddot{r} = \sqrt{r}$$ for some initial conditions $r(0)>0$ if it could have as finite-duration solutions for an ending time $T>0$ the functions: $$r(t)=\frac{1}{144}(T-t)^4\theta(T-t)$$ with $\theta(t)$ the standard Heaviside unitary step function.
From now on, I will be making some things which could have some "definition" issues, why I still believe are not mistaken as is explained in the Motivation section here, so please keep with the question, an then later it could be argued if they are mistaken or not, or how to properly treat them.
Since the solution is a piecewise polynomial, and the solution to the finite-time blow-up equation is the inverse of a polynomial, I will try to see which equation is solved by the reciprocal of the finite duration solution: lets call $z(t)=\frac{1}{r(t)}$ and see in what the original equation becomes:
- $\displaystyle{\ddot{r} = \frac{d^2}{dt^2}\left(\frac{1}{z}\right)=\frac{2(\dot{z})^2-z\,\ddot{z}}{z^3}}$
- $\displaystyle{\sqrt{r}=\sqrt{\frac{1}{z}}=\frac{\sqrt{z}}{z}}$
So the differential equation $\ddot{r}=\sqrt{r}$ becomes: $$\ddot{z}+z\sqrt{z}-2\frac{(\dot{z})^2}{z}=0 \tag{Eq. A}$$ Where it is possible to check that: $$z(t) = \frac{144}{(T-t)^4}$$ solves Eq. A term by term: $$\frac{1}{(T-t)^6}\underbrace{\left\{2880+1728-4608\right\}}_{=\,0}=0$$
So at least in this case, indeed the the reciprocal of the solutions of a differential equation with finite duration solutions leads to a differential equation with finite-time blow ups, which is in the line with intuition: if something goes to zero its reciprocal should run to infinity.
But now I will become "illegal": thinking in the finite duration solution, its "direct reciprocal" will be: $$z_{dr}(t) = \frac{144}{(T-t)^4}\frac{1}{\theta(T-t)}$$ which indeed has a discontinuity at $t=T$ because the polynomial and because the $\frac{1}{\theta(T-t)}$, and is formally different from: $$z_1(t) = \frac{144}{(T-t)^4}\theta(T-t)$$ which also keep a discontinuity at $t=T$, but only because the polynomial this time.
But similarly to the analysis presented here for the solutions of the form $y(t)=r(t)\theta(T-t)$, if I force $z_1(t)=z(t)\theta(T-t)$ on Eq. A, I can built the equation in the form: $$\left(\ddot{z}+z\sqrt{z}-2\frac{(\dot{z})^2}{z}\right)\theta(T-t)=0$$ but this time, I don´t have the cancellation of the terms $z(T)\delta(T-t) \neq 0$ neither $\dot{z}(T)\delta(T-t) \neq 0$ (with $\delta(t)$ the Dirac's Delta function), so in principle it shouldn't work, but somehow (don't knowing If because the singularity solves the problem or it inherited from the finite duration solutions), at least "it looks" like $z_1(t)$ is indeed a solution to Eq. A, like a piecewise blow-up solution instead of a finite duration solution.
The questions
- Does every differential equation who stand finite-duration solutions lead to a differential equation with solutions that behave as having finite-time blow-ups if I take the reciprocal of their solutions?
Added later
I am not 100% sure about the following, but is the reason I added a last sub-question
I believe now that there is at least a family of finite duration function defined by $\dot{y}=−\sqrt[n]{y}$ which can be solved by $y(t)=\left[\frac{n−1}{n}(T−t)\right]^{\frac{n}{n−1}}\theta(T−t)$, as I explained better on this answer, and they also fulfill that the change of variable $z(t)=\frac{1}{y(t)}$ leads to the differential equation $\dot{z}=−z^{\frac{2n−1}{n}}$ which stand solutions $z(t)=\left[\frac{n}{n−1}\frac{1}{T−t}\right]^{\frac{n}{n−1}}$ so they are related with finite time blow ups as its reciprocals.
So, with this, I believe now that at least the reciprocal of a finite-duration solution will be always related to a finite time blow up, as intuition says that something that is going to zero in finite time (let say at time $t=T$), it should have its reciprocal rising to infinity at the same time $t=T$.
2nd Added later
On the comments, @CalvinKhor showed that question 1 is false: the example $\dot{z}=z^2,\,z(0)=1$ which has solution $z(t)=\frac{1}{1-t}$ showing a finite-time blow up behavior at $t=1$, but taking its reciprocal $r=\frac{1}{z}$ and assuming that ${r(t)}^2\neq 0\,\forall\,t$, then it will lead to the differential equation $\dot{r}=-1,\,r(0)=1$ which has solution $r(t)=1-t$: here, since $\dot{r}=-1$ is a Lipschitz differential equation, finite duration solutions should not be supported (if I have properly understood Vardia T. Haimo papers, this is actually the case for this scalar ODE, but I am not 100% sure because of the constant Right-Hand value it could be non-autonomous).
This made a really interesting tool: Non Lipschitz ODEs, since have non-unique solutions are considered "problematic", like in the Norton's Dome example given above: the solutions to $\ddot{r}=\sqrt{r}$ under the initial conditions $r(0)=r'(0)=0$ are problematic since the solution $r(t)=\frac{1}{144}(T-t)^4$ will rise from zero at some constant arbitrary $T$ which is not dependent on the initial conditions, since the solution is indeed behaving as $r(t)=\frac{1}{144}(t-T)^4\theta(t-T)$ so its been already zero for any value of $0\leq t<T$ (see details here and here).
But if is true what it is said on the papers by Vardia T. Haimo cited above, when the scalar autonomous Non-Lipzchitz ODE $\ddot{x}=F(x,\dot{x})$ achieves the time where $F(0,0)$ the dynamics of the system "dies", so the "rise from zero" of the solutions of the Norton's Dome in reality cannot happen.
Keeping this in mind, and since not every finite-time blow up could be mapped to a finite duration solution through the reciprocal, but I believe that the converse is true at least for scalar variables (question 2 must be proved true first), maybe, this analysis could be used to discard singular solutions of differential equations like in the NS eq., is a long shoot, but made this question interesting for analysis... or conversely, the singularities that match a finite duration solution could be visualized just as an artifact because a variable in the denominator is achieving zero in a finite time (or an indication of this is happening in real life).
It looks like scalar real-valued finite duration solutions to initial value problems of autonomous ODEs with positive final times, are like the "well behaved twins" of this non-unique solutions, since they are fully determined by the initial conditions and have a theory with tools that could be apply to other kind of problems.
In this question I explained what I believe is a real life application, if you are interested.
3rd added later
Working into this another question, due other users comments and answers I found about Regular singular points, where a similar analysis is done for differential equation that could be solved through the Frobenius method. With this, maybe depending of how behave the Singular solution the main question will be true or not, but at least I think there are going to be scenarios where it is going to be true.
