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It is $x(t) = e^{c_1-t}\cdot\theta(c_1-t)$ a solution to $\dot{x} = -|x|$ with $\theta(t)$ the unitary step function?

I am trying to understand solutions of finite duration to differential equations. I have made recently a question here where in the comments of the answer, to understand the analysis, the counterexample $\dot{y} = - y$ is given as a ODE which solution "vanishes at infinity" (actually $y(t) = e^{c_1-t}$).

From there, I am trying to understand the effect of introducing an "absolute value" component similar to the other question diff. eq., but without lowering the order of the exponent of the right-side-hand polynomial.

Now, after trying to solve the equation without success, and making many illegal things, I found that when plotted, the function: $$x(t) = e^{c_1-t}\cdot\theta(c_1-t)$$ behave as fulfilling $x'+|x| = 0$ for times $t<c_1$, for times $t>c_1$ every component is zero so the equation is also fulfilled (but with definitions problems since you can do the fraction $\dot{x}/|x|$ without dividing by zero - but this will be share with every possible finite duration solution, as also happen in the other question example).

But int the point $t=c_1$ there is a singularity where the equation is not fulfilled, but since it happen only in a zero-measure point, I want to know it it still been possible to consider $x(t)$ as a solution to $\dot{x}=-|x|$.

Also, since the solution $x(t)$ is not achieving zero continuously at $t=c_1$, this solution is different from the problems analyzed by V. T. Haimo: Finite Time Differential Equations and Finite Time Controllers, so I don´t think the solution is a "proper" finite-duration solution, so the theory explained there won´t apply, and I believe is more suitable/similar to the differential equations with piecewise continuous arguments EPCAs theory studied in Generalized Solutions Of Functional Differential Equations by Joseph Wiener, but unfortunately I don´t have enough background to fully understand any of these references (I only read the more easy concepts).

Joako
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1 Answers1

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Your solution has a discontinuity at $t=c_1$, so it is not differentiable there. This means that $x(t) = e^{c_1-t}\cdot\theta(c_1-t)$ does not solve the differential equation in the usual sense because this function is not everywhere differentiable.

There is an easy way to find the solutions to your dynamical system. We have three options.

  1. If $x(0)>0$, then $\dot{x}(t)=-x(t)$ for all $t\ge0$, and we have that $x(t)=e^{-t}x(0)$, $t\ge0$.
  2. If $x(0)=0$, then $x(t)=x(0)$, $t\ge0$.
  3. If $x(0)<0$, then $\dot{x}(t)=x(t)$ for all $t\ge0$, and we have that $x(t)=e^{t}x(0)$, $t\ge0$.
KBS
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  • Thanks you very much for answering. The same answer will stand for $\dot{x}=-|x|-1$ and $x(t) = \left(e^{c_1-t}-1\right)\theta(c_1-t)$, isn't?? (now $x(t=c_1) = 0$ so $x(c_1-t)\delta(c_1-t) = 0$ so the singularity could be avoided - or I am mistaken and it is indeed a different situation and $x(t)$ is a solution?) – Joako Mar 21 '22 at 20:34
  • @Joako I am not sure what you mean by "the same answer" but the fix you suggest does not fix all the issues as the derivative at $t=c_1$ is not continuous. It is indeed equal to -1 from the right and 0 from the left. – KBS Mar 21 '22 at 20:39
  • Does then $x(t)=x(0)e^{-\text{sgn}(x(0))t}$ solve the differential equation? – Joako Nov 13 '23 at 02:28