It is possible to make equivalences among differential equations that share solutions?
I am thinking in two non-linear ODEs such they share the same solution of finite duration holding uniqueness among the initial time $t=0$ and until the finial extinction time $T=4$ which is a singular point, so, uniqueness on $t\in[0,\,4)$:
- $$y'(t)=-\operatorname{sgn}(y(t))\sqrt{|y(t)|},\quad y[0]=4 \quad \Rightarrow\quad y(t)=\frac{1}{4}(4-t)^2\theta(4-t) \tag{Eq. 1}\label{Eq. 1}$$
- $$2y(t)+(4-t)y'(t)=0,\quad y[0]=4 \quad \Rightarrow\quad y(t)=\frac{1}{4}(4-t)^2\theta(4-t)\tag{Eq. 2}\label{Eq. 2}$$
where $\theta(t)$ is the Heaviside step function. Here a plot in Demos showing the solution satisfies both differential equations with same initial condition.
Questions
It is possible to treat both equations as equivalent?, or at least, if It is possible to make some kind of equivalences among them? This because is easier to analyse (at least for me) the \eqref{Eq. 2} than \eqref{Eq. 1}, but I am interested in understand the behavior of \eqref{Eq. 1} through \eqref{Eq. 2}, as example... How could I have realize both equation share the same solution for some initial condition without knowing the solutions beforehand?
I am thinking in a general case by the way, no just this specific example.
Since the lack of answers I added later this comment:____________
As an example of what I am thinking/aiming to do: imagine I am working on a system modeled by something like \eqref{Eq. 1} but I don't know the answer yet, if I try to apply the Fourier Transform to find the solution $y(t)$ I would get stuck because of the nonlinear terms.
But if somehow I could demonstrate (I don't know how, maybe through Lyapunov's or Filippov's analysis), that the same solution of \eqref{Eq. 1} solves also \eqref{Eq. 2} (which I know it is possible because it actually happen in the example), I could try then to find the solution $y(t)$ by means of applying the Fourier Transform to \eqref{Eq. 2} (which I know it can be done - I reviewed it now at the 3nd added later)... If both systems shared solutions, Shouldn't also share the same Fourier Transform formula excepting for terms dependant on the initial conditions? Could I find the Fourier Transform of \eqref{Eq. 1} from analogy of the Fourier Transform of \eqref{Eq. 2} and matching terms?
Also, if somehow I am able to find the time $T$ when it stops moving and its final position $y[T]$, maybe I could made some approximation by using what I asked in this question and find the solution by Fourier or Laplace's means, es example, If I look for a system that decays oscillating (like this one as example), I could try to match it with the following frustrated 2nd order ODE and take the Fourier Transform between times $t=0$ and $t=T$:
$$h(t)=y(T)-(y(T)-y(t))\theta(T-t) \Rightarrow h''(t)+ah'(t)+bh(t)+c-(bh(T)+c)\theta(t-T)=0 $$
And find the original equation's solution by the Fourier transform of this replacement equation: Does this makes any sense?
5th Added later - more general match
Since I still have doubts in the comments, lets make an specific general example: as you could see in this previous question for a general initial condition \eqref{Eq. 1} got the following solution: $$y'(t)=-\operatorname{sgn}(y(t))\sqrt{|y(t)|},\quad y[0]=y_0 \quad \Rightarrow\quad y(t)=\frac{\operatorname{sgn}(y_0)}{4}\left(2\sqrt{|y_0|}-t\right)^2\theta\left(2\sqrt{|y_0|}-t\right) \tag{Eq. 4}\label{Eq. 4}$$
and the solution will stop moving at the finite extinction time $T=2\sqrt{|y_0|},\,\, 0\leq T<\infty$. Now, for $t<T$, you could check in Wolfram-Alpha I could made a replacement ODE like:
$$2y(t)+(2\sqrt{|y_0|}-t)y'(t)=0,\quad y[0]=y_0 \quad \Rightarrow\quad y(t)=\frac{\operatorname{sgn}(y_0)}{4}\left(2\sqrt{|y_0|}-t\right)^2\theta\left(2\sqrt{|y_0|}-t\right)\tag{Eq. 5}\label{Eq. 5}$$
which this time, differently from \eqref{Eq. 2} and its phase space depicted in the 2nd added later, in \eqref{Eq. 5} the singular point will move always just in the way the solutions of both ODEs matches in $0\leq t<T$.
Now, if I take the Fourier Transform of \eqref{Eq. 3} I will get: $$\mathbb{F} \left\{ \operatorname{sgn}(y(t))\sqrt{|y(t)|}\right\}=-iwY(iw) \tag{Eq. 6}\label{Eq. 6}$$
Then, doing the same on \eqref{Eq. 5} will give us: $$2Y(iw)-2iw\sqrt{|y_0|}Y(iw)-\frac{\partial}{\partial w}\left(iwY(iw)\right)=0$$ $$\Rightarrow -iwY(iw)=\frac{1}{2\sqrt{|y_0|}}\left[2Y(iw)-\frac{\partial}{\partial w}\left(iwY(iw)\right)\right]\tag{Eq. 7}\label{Eq. 7}$$
Does this mean that in general I could resolve as follows for any $Y(iw)$ on $0<t<T$? $$\boxed{\mathbb{F} \left\{ \operatorname{sgn}(y(t))\sqrt{|y(t)|}\right\}=\frac{1}{2\sqrt{|y_0|}}\left[2Y(iw)-\frac{\partial}{\partial w}\left(iwY(iw)\right)\right]}\tag{Eq. 8}\label{Eq. 8}$$
This as a general example, since I could made something similar for \eqref{Eq. 3} if it is valid, as example.
Going even further_____________
Now lets continuous the example: in the paper Finite Time Differential Equations is shown that the solution of the following differential equation do achieve a finite extinction time: $$y''(t)+\operatorname{sgn}(y(t))\sqrt{|y(t)|}+\operatorname{sgn}(y'(t))\sqrt{|y'(t)|}=0\tag{Eq. 9}\label{Eq. 9}$$
which I don't know any closed-form solution as you could see in Wolfram-Alpha. Now lets made a replacement ODE by:
$$y''(t)+\dfrac{2y(t)}{\left(2\sqrt{|y_0|}-t\right)} + \dfrac{2y'(t)}{\left(2\sqrt{|y'_0|}-t\right)} = 0\tag{Eq. 10}\label{Eq. 10}$$
Now, fortunately the solution of \eqref{Eq. 10} are found by Wolfram-Alpha when picking particular numbers, but unfortunately are some messy formulas, so I would pick as example some initial conditions that simplify as much as I was able to find the example of \eqref{Eq. 10}, namely $y[0]=\frac14$ and $y[0]=\frac14$, leading in Wolfram-Alpha to the following Sturm-Liouville equation:
$$\dfrac{\partial}{\partial t}\left(\dfrac{y'(t)}{(t-1)^2}\right)-\dfrac{2y(t)}{(t-1)^3} = 0,\quad y[0]=\frac14,\,y'[0]=\frac14 \tag{Eq. 11}\label{Eq. 11}$$
The result was a big mess of constants dependent of modified Bessel functions, If I didn't messed it up the answer could be reduced to:
$$y(t) = \dfrac{\sqrt{(t-1)^3}}{2\sqrt{2}}\biggr(\left(2J_2(2\sqrt{2})+\sqrt{2}J_3(2\sqrt{2})\right)K_3(2\sqrt{2-2t}) - \left(2K_2(2i\sqrt{2})+i\sqrt{2}K_3(2i\sqrt{2})\right)I_3(2\sqrt{2-2t})\biggr)\tag{Eq. 12}\label{Eq. 12}$$
Unfortunately, this time my idea of taking the Non-Lipschitz terms from one equation to make another equation looks like it didn't work, since the solution of \eqref{Eq. 12} is complex-valued as could be seen in Wolfram-Alpha. But it do keeps real-valued among time $0\leq t \leq 1$ so it is possible it would be solving solving the original ODE of \eqref{Eq. 9} so the replacement of \eqref{Eq. 10} work, but since has complex terms I was not able to test it on Demos this time, neither if is solving \eqref{Eq. 10} or \eqref{Eq. 11} so maybe the solution from \eqref{Eq. 12} is just wrong (I couldn't test it).
Confirmation: It didn't work, user @Nasser verified it in Mathematica in the following answer.
Anyway, I think it should be clear now what kind if things I am trying to understand. Even if it didn't work for different ODEs, the replacement equation do work for ODEs that share the same solution, if there is a way to new this beforehand knowing the solutions it could be useful for solving Non-Lipschitz ODEs.
2nd added later - answer to @DWade64 comment - not mandatory reading
Their fields look very different indeed in Wolfram-Alpha, only similar on the neighborhood of the solution.
3nd added later - not mandatory reading
Here I just were trying my hypothesis for \eqref{Eq. 2}:
$$\begin{array}{c} 2y(t)+(4-t)y'(t)=0\qquad \Biggr/\mathbb{F}\{\cdot\}\\ \Rightarrow 2\hat{y}-4iw\hat{y}-i\frac{\partial}{\partial w}\left(iw\hat{y}\right) = 0\\ \Rightarrow 2\hat{y}-4iw\hat{y} -i\left[i\hat{y}+iw\hat{y}'\right] = 0\\ \Rightarrow \left(3-4iw\right)\hat{y}+w\hat{y}'=0 \end{array}$$ Here I got an ODE for the Fourier Transform $\hat{y}$ which Wolfram-Alpha solves as: $$\hat{y}(iw) = c_1\frac{e^{4iw}}{w^3}$$ and taking the inverse Fourier Transform on Wolfram-Alpha (here I match constants since I am used to work with the electrician definition) I get: $$y(x)=c_1\frac{i}{4}(t-4)|t-4|$$ and matching the initial value $y[0]=4$ I find $c_1=i$ so I recover for $0<t<4$: $$y(x)=\frac{1}{4}(4-t)|4-t|,\quad \hat{y}(iw) = i\frac{e^{4iw}}{w^3}$$
So, at least in principle, If it is possible to find somehow that the solution of \eqref{Eq. 1} matches the solution of \eqref{Eq. 2}, it could have been possible to find the solution of \eqref{Eq. 1} by solving \eqref{Eq. 2} knowing beforehand it will hold only between $t\in\{0,\,T=4\}$.
Keep in sight that it is not exactly the same solution so it is not also exactly the same spectrum 1 spectrum 2:
$$\int\limits_{-\infty}^{\infty} \frac{1}{4}(4-t)^2\theta(4-t)e^{-iwt}\,dt = 4\pi\delta(w)- i\frac{e^{4iw}}{2w^3}$$
$$\int\limits_{0}^{\infty} \frac{1}{4}(4-t)^2\theta(4-t)e^{-iwt}\,dt = \frac{i+w(4-8iw)}{2w^3}-i\frac{e^{4iw}}{2w^3}\equiv \int\limits_{0}^{4} \frac{1}{4}(4-t)^2\theta(4-t)e^{-iwt}\,dt$$
when you could notice the the upper bound of the integral don't change the result since the solutions do have stopped moving at time $t=4$.
4th Added later - not mandatory reading
Just as curiosity, the same solution of \eqref{Eq. 1} and \eqref{Eq. 2} solves for $0\leq t<4$: $$y''(t)=-\frac{1}{2}\operatorname{sgn}(y'(t)),\quad y[0]=4,\,y'[0]=-2 \quad \Rightarrow\quad y(t)=\frac{1}{4}(4-t)^2\theta(4-t) \tag{Eq. 3}\label{Eq. 3}$$
Here added in Desmos.