1

How to "formally" prove that $x(t)=\frac{1}{4}(2-t)^2\theta(2-t)$ solves $\dot{x}=-\text{sgn}(x)\sqrt{|x|},\,x(0)=1$? (with $\theta(t)$ the standard unitary step function).

I have found the equation studying these paper by V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers, and I believe the finite duration solution is given by: $$x(t)=\frac{1}{4}(2-t)^2\theta(2-t)$$

Which I found from the comparison with the other equation $\dot{y}=-\sqrt{y},\,y(0)=1$ which solution is $y(t) = \frac{1}{4}(2-t)^2$.

Actually, the first solution I found was: $$x(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2}\right|\right)^2,$$ and so far, I have not been able to formally prove that $x(t)$ is the "true" solution, even when graphically is obvious (and theoretically, since ODEs stands Locally the Existence and Uniqueness of solutions, I believe it fits).

For every try, to made coincident both parts of the equation (right and left hand sides), I have to made an illegal operation like:

  • assuming that is equivalent to use $\text{sgn}(x(t)\cdot\theta(2-t)) \cong \text{sgn}(x(t))\cdot\theta(2-t)$
  • assuming that is equivalent to use $\theta(t) = \frac{1+\text{sgn}(t)}{2}$ or $\theta(t) = \frac{1}{2}\left(1+\frac{1}{\text{sgn}(t)}\right)$
  • assuming that is equivalent to use $\text{sgn}(x(t)) \cong \frac{1}{\text{sgn}(x(t))}$
  • assuming that is equivalent to use $\int \theta(t)\,dt = \int \theta^n(t)\,dt,\quad n\in \mathbb{Z}^+$

And other similar things which I beforehand know are mistaken (just by taking their derivatives I know it leads to different functions), but when displayed on graphs, they take the same values. So I know I am making mistakes (since this are distributions instead of functions and I don´t know how to formally work with them), but also because of been "vague" with the treatment have let me find the previous candidates to the solution, so I don´t know if I am finding something or it is everything wrong (maybe it is just the same as taking the solutions piecewise, but at least the differential equations at first doesn't look to be defined piecewise - at least for me, since I know beforehand that a finite duration solution will lead to definition issues on the diff. eq. since it becomes zero forever after the ending time, $t_F = 2$ on this example).

Hope you can elaborate on how to prove it explaining why the listed things are conceptually different (I am asking here for the "prove", not an explanation of how to find it, this because of my lack of background, it will be more useful to see how it is done and then try to search for the missing things I need to look for to understand the explanations). Beforehand thanks you very much.

Joako
  • 1,380

1 Answers1

2

Since $x(0)=1>0$, then one can see the solution is decreasing to zero and then stays there when it reaches this value. So, the differential equations simplifies to $\dot{x}=-\sqrt{x}$.

This can be rewritten as $$\dfrac{dx}{\sqrt{x}}=-ds$$ and integrating yields

$$\int_{x(0)}^{x(t)}\dfrac{dx}{\sqrt{x}}=-\int_0^tds$$ which yields

$$2(\sqrt{x(t)}-1)=-t.$$ Reorganizing and squaring then yields

$$x(t)=\dfrac{1}{4}(2-t)^2$$ but this is only valid for $t\in[0,2]$. For $t>2$, $x(t)=0$, so we multiply that expression by $\theta(2-t)$ to get the final result.

The expression you found with the absolute values is also valid and is equivalent to the one using the Heaviside function.

KBS
  • 7,114
  • Thanks for answering. How do you know from Eq. 1 that it will stay at zero forever after $t=2$?? You use this in your argument (as I did), but I am looking to figure out Why is that right... is obvious after you know the solution, but it is not obvious at all for me How from Eq. 1 you can figure out you must take a multiplication by the step function. Please elaborate into that. – Joako Mar 20 '22 at 16:58
  • 1
    I guess by equation 1, you mean the differential equation. I have explained that in the first sentence of my answer. If you start from $x(0)=1$, then $x(t)$ is decreasing. So, either it is decreasing without reaching zero (like the solution to $\dot{x}=-x$ would do) or it reaches zero in finite-time, say at time $T$. When this is the case, then we have that $\dot{x}(T)=0$ since $x(T)=0$ and so $x(t)=0$ for all $t\ge T$. From the explicit solution, we see that it reaches zero in finite-time with $T=2$. After that, we have that $x(t)=0$ for all $t\ge T=2$. – KBS Mar 20 '22 at 17:11
  • And How do you know it will never rise from zero after time $T=2$, thinking here in what happens, for example, in the Norton's Dome Example? here, here – Joako Mar 20 '22 at 19:57
  • 1
    Because $0$ is an equilibrium point of the system. Just look at the differential equation. It is immediate to see that once the trajectory has reached the zero value, it just stays there. To convince you, just assume that the initial condition is $x(0)=0$, what happens next? – KBS Mar 20 '22 at 20:00
  • I am not familiar with equilibrium points... I have read know that it is when $\dot{x} = 0$, but to be a stable point other criteria must be being fulfilled (I get lost here), so for me is not obvious from the equation to see what you are saying (sorry about my lack of background).. Could you elaborate, please, about how the diff. eq. tells you that if reach $x(0)=0$ it will remain there forever? – Joako Mar 20 '22 at 20:15
  • 1
    Then you will need to read some material on dynamical systems because without that, it will be difficult to understand them. If you start at $x(0)=0$ then the derivative $\dot{x}(0)=0$. If the derivative is zero at $t=0$, then the function is locally constant at $t=0$. But then, this argument extends forward in time. So, if $x(0)=0$, then $x(t)=0$ for all $t\ge0$. The same thing happens when the trajectory starting at $x(0)=1$ reaches zero at $t=2$. – KBS Mar 20 '22 at 20:26
  • Actually in the mentioned paper they use the same argument about how it "extends forward in time", but I don´t get it (yet), of How it happen (I am still searching for a reference, so if you have any, is welcome)... in the paper, also a second order system is investigated, where there are solutions that aren't of finite duration coexisting with finite duration ones, depending on the parameters values. Do you believe that the same argument could be use to prove this other related question I have made? – Joako Mar 20 '22 at 20:39
  • 1
    Read first the book by Khalil, for instance, to get a better understanding of dynamical systems and their stability. The book by Strogatz is also a good starting point. Only then, you can go back to those papers. You will not be able to understand them if you do not have at least a basic knowledge of dynamical systems. Regarding your other question, I do not not know. – KBS Mar 20 '22 at 20:45
  • I have read Nonlinear Dynamics and Chaos by Strogatz a decade ago when studying engineering (just because of curiosity, it wasn't on the syllabus), and I don´t remember to see on any place the existence of finite-duration-solutions of ODEs, neither the Liapunov analysis used on the paper – Joako Mar 20 '22 at 21:10
  • 1
    I am not talking about finite duration solutions but basic results on dynamical systems which you seem to ignore. Without them, there is no chance you will understand what you are trying to learn. I am sorry but not knowing about equilibrium points is a pretty strong signal that you need to learn more about dynamical systems. – KBS Mar 20 '22 at 21:18
  • I have read pages 18 and 19 of the Strogatz and I remember why an equilibrium point is stable now. Thanks a lot. – Joako Mar 20 '22 at 21:30
  • No problem. Just refresh your dynamical systems theory and then get back to the papers by Haimo. – KBS Mar 20 '22 at 21:33