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Please take a look also to the comments section, here, and in other people answers, since there are extended what are my apprehensions about the validity of the found answers.

I have found these two solutions to the following differential equations by playing on Wolfram-Alpha, and I would like to prove that they are formally solutions of each equation:

  • $$x'' = -\text{sgn}(x'),\quad \,x(0)=2,\,x'(0)=-2 \quad \Rightarrow \quad x(t) = \frac{1}{2}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^2\quad\text{(Eq. 1)}$$
  • $$y'' = \sqrt{|y'|},\quad \,y(0)=\frac{2}{3},\,y'(0)=-1 \quad \Rightarrow \quad y(t) = \frac{1}{12}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^3\quad\text{(Eq. 2)}$$

Notice that Wolfram-Alpha don't show close-form solutions for neither these equations as can be seen here and here.

The following notation is going to be used:

  • $\text{sgn}(t)$ is the Sign function, which fulfills that $\frac{\partial}{\partial t}\left(|t| \right) = \text{sgn}(t)$ and $\frac{\partial}{\partial t}\left(\text{sgn}(t) \right) = 2\ \delta(t)$
  • $\theta(t)$ is the Heaviside step function that fulfills $\frac{\partial}{\partial t}\left(\theta(t) \right) = \delta(t)$ and $\theta(t) = \frac{1}{2}\left(1+\text{sgn}(t) \right)$
  • $\delta(t)$ is the Dirac delta function that fulfills $\int_{-\infty}^\infty f(t)\delta(t)\,dt=f(0)$ with $\int_{-\infty}^\infty \delta(t)\,dt=1$

These definitions could be problematic, as it will reviewed later.

Motivation

Recently I figure out that no non-piecewise power series could have a finite extinction time due the Identity theorem, and found on this paper that a differential equation could have a solution that achieve a finite extinction time if and only if its nonlinear and have a singular solutions, so it must fail to fulfill uniqueness of solutions, but luckily the paper explains that within the initial conditions time and the finite extinction time uniqueness is still hold.

Since my intuition tells that classic mechanics system should achieve a finite extinction time where the movement due the system dynamics dies (as opposite of random thermal noise, which nature is external to the system dynamics as it were a random forcing force), so I started to look for some physics' examples, without finding many (I even tried to made them as in here and here), and those I found (here) were quite hard to understand (at least for me, I'm an electrician, nor a physicists neither a mathematician).

But a few days ago, someone in a Youtube comment named @siguc explains me the following:

"How about the motion of a brick on a horizontal surface with constant friction between the brick and the surface? Assuming the brick moves along the surface at $t=0$, it'll stop eventually. Newton's law: $x''=-k\ g\ \text{sgn}(x')$, where $g$ is $9.8\,\frac{m}{s^2}$ and $k$ is the friction coef.".

So I started to googling about this problem founding terms as Coulomb damping and the Stribeck curve, but the only place I found the same brick problem was in this Wiki page and no close-form solutions were shown.

Since the system were simple enough to be understood by me, I start by trying to see if I could find a solution to the simplest case by myself, so assuming here that $k\ g = 1$, I got the $\text{(Eq. 1)}$. Later I will explain why I choose those arbitrary initial conditions.

What I have done so far

On previous question I have found:

  • In this question another person (@KBS) proves on his answer that the solution I found could be formally a solution to $$u' = -\text{sgn}(u) \sqrt{|u|},\quad \,u(0)=1 \quad \Rightarrow \quad u(t) = \frac{1}{4}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^2\qquad\text{(Eq. 3)}$$
  • Later I found here that for a positive finite extinction time $T>0$ and a positive initial conditions $v(0)>0$ that determines $T(v(0))$, one can have that: $$\begin{array}{r c l} v' = -\text{sgn}(v) \sqrt[n]{|v|},\quad \,v(0)>0 \quad \Rightarrow \quad v(t) & = & \left[\frac{n-1}{n}\left(T-t\right)\right]^{\frac{n}{n-1}}\theta\left(T-t\right) \\ &\overset{!}{\equiv}& v(0)\cdot \left[\frac{1}{2}\left(1-\frac{t}{T}+\left|1-\frac{t}{T} \right|\right)\right]^{\frac{n}{n-1}}, \\ & & v(0) = \left[\frac{n-1}{n}\cdot T\right]^{\frac{n}{n-1}}\qquad\text{(Eq. 4)} \end{array}$$

I have found these results in a laissez-faire way and not $100\%$ rigorously: I know that some definitions at the beginning have issues in the edge points, like the Heaviside function having or not a transition value of $1/2$ at $t=0$, which I have ignored, so I am using things like $\theta(t) = (\theta(t))^n$ which have some weird consequences, as in (Eq. 4) where the exclamation point is evidenced on the equivalence could be flawed. I have preferred the version with the absolute value function trying to be as far as possible of having derivatives of the Dirac Delta function $\delta'(t)$ which I don't know how to handle them.

Also has the consequence that if I evaluate the differential equation some issues happened at these edge points, like the rising of Dirac delta functions which broke the differential equation equality, but since it only happened on a zero-measure point, and the solutions doesn't have those problems, I believe they are valid as solutions.

With this laissez-faire approach I was able to just test some solutions of the form $x(t) = a\cdot(T-t)^b$ similar to the found for (Eq. 3) and (Eq. 4) and made them fit the differential equation, since from the mentioned paper I believe beforehand that there was an existent singular solution that could achieve a finite extinction time.

Since these special functions $\delta(t),\,\text{sgn}(t),\,\theta(t)$ are in reality distributions, which theory I know almost nothing, I would like to know if its possible to prove in a rigorous way they are indeed solutions of those differential equations.

BONUS TRACK: Do you believe it is possible to find a general solution for $x''=-k\ g\ \text{sgn}(x')$ using (Eq. 1)? At first I was concern about having an initial condition for the speed $x'(0)<0$, but I think now it makes sense since the system starts with an initial speed $|x'(0)|$ but through its dynamics it must start loosing speed immediately since friction is the only external force present in the sliding brick system.

PS: on other question like this people have attacked the intuition of having solutions that achieve a finite ending time, being valid opinions, please keep it out of the discussion here, since I am trying to explore these kind of solutions as alternative - instead, feel free to extend the discussion in the mentioned question which is more suitable for it.

Added later

About my worries, as example, in Wikipedia the Heaviside step function is defines as $\theta(0)=\frac{1}{2}$ which could generate issues since I have assumed as true that $\theta(t)=(\theta(t))^n$. Also, I am using that $$\frac{1}{2}\left(x +|x|\right) = \frac{1}{2}|x|\left(1+\frac{x}{|x|}\right) = \frac{1}{2}|x|\left(1+\text{sgn(x)}\right)=|x|\ \theta(x)$$, but if instead I take as factor the other term I get: $$\frac{1}{2}\left(x +|x|\right) = \frac{1}{2}x\left(1+\frac{|x|}{x}\right) = \frac{1}{2}x\left(1+\frac{1}{\text{sgn(x)}}\right)$$ And since in Wikipedia defined the sign function as having $\text{sgn}(0) = 0$ then the term $\frac{1}{\text{sgn(x)}}$ hidden a division by zero, which is obviously a big issue. These are examples of why I am worried about the validity, this sames issues also made struggles with the definitions at the beginning. Hope you could elaborate why is not a problem if it really is not.

An attempt for the bonus track If I used the assumption than $kg>0$, which make sense with the physics problem the equation is coming from, and I used the change of variable $x' = kg z$, the equation could become: $$\begin{array}{rcl} -\frac{x''}{kg} & = & \text{sgn}(x') \\ \iff -\frac{kgz'}{kg} & = & \text{sgn}(kg\ z) = \frac{kg\ z}{|kg\ z|} \overset{\text{since}\ kg>0}{=}\frac{kg\ z}{|kg|\ |z|} =\frac{z}{|z|} = \text{sgn}(z) \\ \Rightarrow z' & = & -\text{sgn}(z) \quad\text{which is equivalent to (Eq. 1)}\end{array}$$ With this, we have that the answer to: $$x''=-k\ g\ \text{sgn}(x'), \quad x(0) = 2kg,\,\,x'(0)=-2kg \Rightarrow x(t) = \frac{kg}{2}\left(1-\frac{t}{2}+\left|1-\frac{t}{2} \right|\right)^2$$ So for a general finite extinction time $T>0$ and constants such as $kg>0$ then an answer could be: $$ \begin{array}{l} x''=-k\ g\ \text{sgn}(x'), \qquad x(0) = T^2\frac{kg}{2}>0,\,\,x'(0)=-Tkg<0 \\ \Rightarrow x(t) = T^2\frac{kg}{8}\left(1-\frac{t}{T}+\left|1-\frac{t}{T} \right|\right)^2 = \frac{kg}{2}\left(T-t\right)^2\theta(T-t) \quad \text{(Eq. 6)} \end{array}$$

Does (Ec. 6) make sense-full as closed-form solution for the problem?

I have used that $s(t) = \int -(T-t)\ dt\cdot\theta(T-t) = \left[\frac{1}{2}(T-t)^2 + C_0\right]\theta(T-t)$ such as: $$s'(t) = -(T-t)\theta(T-t)\quad + \underbrace{\left[\frac{1}{2}(T-t)^2 + C_0\right]\delta(T-t)}_{C_0 \equiv 0,\,\text{so all the expression could be zero by}\,x\delta(x) = 0}$$

It is still a general solution?

Another example of the weird issues I have found, is that if in Wolfram Alpha, as example, I try to solve the equation (Eq. 1) with the founded solution of (Eq. 6) as: $$\frac{1}{kg}\frac{\partial^2}{\partial t^2}\left(\frac{kg}{2}\left(T-t\right)^2\theta(T-t)\right)+\text{sgn}\left(\frac{\partial}{\partial t}\left(\frac{kg}{2}\left(T-t\right)^2\theta(T-t)\right)\right)$$ I will find a mess as is shown here.

But if instead I solve the less rigorous equation: $$\left(\frac{1}{kg}\frac{\partial^2}{\partial t^2}\left(\frac{kg}{2}\left(T-t\right)^2\right)+\text{sgn}\left(\frac{\partial}{\partial t}\left(\frac{kg}{2}\left(T-t\right)^2\right)\right)\right)\cdot\theta(T-t)$$ it shows now that the proposed solution solves the equation (except in one isolated point $t=T$, which is also werid), and I find strange how and why the $\theta(t)$ function can cross through operations as it where a ghost (hope you understand what I am trying to say). Any source to some place where these properties are explain will be fantastic, Thanks.

Joako
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    For the bonus track: \begin{align} x''=-kg\ \mathrm{sgn}(x'),\quad \iff \quad \mathrm{sgn}(x')x''=-kg,\quad \iff\quad \mathrm{sgn}(x')x'=-kgt+C \end{align} Defining the parameter $\xi$ such that $x'=\xi$ we then arrive at the parametric solution \begin{align} x=C_2-\frac{1}{2kg}\mathrm{sgn}(\xi)\xi^2,\quad t=C-\frac{1}{kg}\mathrm{sgn}(\xi)\xi. \end{align} With your initial conditions $C=2/kg$ and $C_2=2(1-1/kg)$. The parameter $\xi$ must be restricted for a finite, positive domain. – Eli Bartlett Nov 23 '22 at 07:56
  • @EliBartlett thanks for commenting. You have use the same kind of things I am using, which conceptually I don't know if they are formally valid, since they could be mistaken within distribution theory analysis, like using $\frac{1}{\text{sgn}(x)}\equiv\text{sgn}(x)$. I still trying to understand the parametric part, but there is something I don't get in the third line: Since you have used that $|x'|=\text{sgn}(x'),x'$, you could have used the first line making $-\frac{x',x''}{kg}=-kgt+C$ which solutions $x(t)=\int \sqrt{(kgt)^2/2+c_1t+c_2}\ dt$ could not be fulfilling the first ODE. (...) – Joako Nov 27 '22 at 00:42
  • @EliBartlett (...) Because of these kind of issues, is why I am worried about the answers of the main question being valid answers or not. Maybe you could elaborate in an answer. – Joako Nov 27 '22 at 00:44
  • @EliBartlett Also the third line means that $|x'| = -kgt+C$ which it would be only valid in the reals if $-kgt+C>0$ for every $0<t\leq T$ which maybe fix $C(T)$... but with these issues, the only thing I understand is that these kind of ODE with special functions required special caution. I think that any function with a $f'(T)=0$ could be candidate by making $g(t)=f(t)\theta(T-t)$ and then testing $g(t)$ in the differential equation, but I am not sure about what else it will need. – Joako Nov 27 '22 at 02:19
  • @EliBartlett Hi Eli, I have incorporated what I think could be a closed-form solution to the bonus track for the assumption $kg>0$ and final extinction time $0<T<\infty$... Does it fit with the solutions you give on the first comment? (I still struggle to understand the parametric part you have used). Hope you could comment. – Joako Nov 30 '22 at 11:37
  • @EliBartlett I have corrected some constants on the general answer of Eq. 6, but I think is quite general for solutions that achieve a finite ending time, Have you ever see those solutions to the brick sliding in the horizontal plane before? – Joako Dec 07 '22 at 16:39
  • Hi @EliBartlett !, I have incorporated a detailed question about the closed-form solutions of the brick sliding in an horizontal plane after an initial push here. Hope you could take a look. Best regards. – Joako Jan 21 '23 at 23:25

1 Answers1

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Perhaps I don't understand the problem, but I don't think that it's difficult to prove within distribution theory that those are solutions to the differential equations (I skip the initial conditions since I'm sure that you can check them yourself).

The first solution can be written $x(t) = 2(1-t/2)^2 \, \theta(1-t/2).$ If $f$ is a smooth function (here meaning $C^\infty$) and $u$ is a distribution then the product rule $(fu)'=f'u+fu'$ is valid. Also, if $g$ is affine, then the chain rule $(u\circ g)'=g'\cdot(u'\circ g)$ is valid. Furthermore, $\theta'=\delta$ and $f\delta=f(0)\delta.$ Thus, $$ x'(t) = 2(-1/2)\cdot 2(1-t/2)\,\theta(1-t/2) + 2(1-t/2)^2 \, (-1/2)\delta(1-t/2) = -2(1-t/2)\,\theta(1-t/2), $$ and $$ x''(t) = -2(-1/2)\,\theta(1-t/2) - 2(1-t/2)\,(-1/2)\delta(1-t/2) = \theta(1-t/2). $$ Now note that $$ \operatorname{sgn}(x'(t)) = \operatorname{sgn}(-2(1-t/2)\,\theta(1-t/2)) = -\operatorname{sgn}(1-t/2)\,\theta(1-t/2) = -\theta(1-t/2) = -x''(t). $$

Likewise for the other differential equation and its solution.

md2perpe
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  • Thanks for the answer. Since I don't know distribution theory (as electrical engineer I were taught to work with $\delta(x)$ as a function, but knowing some operations are forbidden), I have my doubts about the handle and validity of the solutions I found. As example, as in your answer I have used that $$\begin{array}{r c l} 1/2[1-t/2+|1-t/2|]^2 & = & 1/2[|1-t/2|\ (1+(1-t/2)/|1-t/2|)]^2 \ & =&1/2[|1+t/2|\ (1+sgn(1-t/2))]^2 \ & =&1/2[2\ |1-t/2|\ \theta(1-t/2)]^2 \ & = &1/2[|2-t|\ \theta(2-t)]^2 \ &=&1/2 (2-t)^2\ \theta(2-t) \end{array}$$ using that $\theta(x)\equiv (\theta(x))^n$, (...) – Joako Nov 30 '22 at 10:13
  • (...) but as example, in Wiki the Heaviside step function is defined as $\theta(0) = 1/2$ so it imply $1/2 \equiv [1/2]^n$ which is false in general, this kind of things is what I am worried about. As other example, if instead of factorizing by $|1-t/2|$ I were used $(1-t/2)$ in the first line of my previous comment eqn., I have that $1/2[(1-t/2)\ (1+1/sgn(1-t/2))]^2$ and since the sign function in Wikipedia is defined as $sgn(0) = 0$ I have a division by zero. (...) – Joako Nov 30 '22 at 10:17
  • Could you elaborate (hopefully in a simplified way), Why these mentioned things aren't problems when working with distributions?. Also, I have get really interested in the reason behind the properties you mentioned about the product and chain rules: Do you have any source to those demonstration, hopefully with counterexamples?. Again, thanks you very much. – Joako Nov 30 '22 at 10:24
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    When an ordinary function is treated as a distribution, values at isolated points do not matter. That's because those do not affect the value of the integral. – md2perpe Nov 30 '22 at 19:12
  • excuse my ignorance, but I don't know what integral you are referring to. I hope you could elaborate or share a source to the definitions you are thinking of. – Joako Nov 30 '22 at 23:59
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    https://en.wikipedia.org/wiki/Distribution_(mathematics) – md2perpe Dec 01 '22 at 06:53
  • I have see it before but doesn't understood the meaning with the integral, so, if the pseudo-function behave well against a smooth bump function, then I could ignore problematic points if they have zero-measure, is interesting. The test function must be smooth? I tried to find one with closed-form in the reals as in the freq. domain unsuccesfully (here), since I don't get answers I tried thinks like $\left(1-x^2\right)^4$ which fulfill both, but it isn't smooth (and had other issues like this – Joako Dec 02 '22 at 01:45
  • Also, did you check the last thing I added?... how $\theta(t)$ can go in and out of derivatives in demostrations it also related to be treated as a distribution? – Joako Dec 02 '22 at 01:48
  • @Joako. Distributions are defined as linear functionals on a space of smooth functions. Functions are distributions through integration against the smooth function. Operations on distributions are often defined by moving them to the smooth function through how it can be done in the case of functions, e.g. since $\int f'(x) , \phi(x) , dx = -\int f(x) , \phi'(x) , dx$ if $\phi$ vanishes outside a bounded set, then if $u$ is a distribution we defined $u'$ by $\langle u', \phi \rangle = -\langle u, \phi'\rangle,$ where $\langle u, \phi \rangle$ denotes the action of $u$ on $\phi.$ – md2perpe Dec 03 '22 at 09:09
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    @Joako. I'm not sure what you mean with $\theta$ going in and out of derivatives, but if $f(0)=0$ then $(f\theta)' = f'\theta + f\theta' = f'\theta + f(0)\delta = f'\theta$ since $\theta'=\delta$ and $f\delta=f(0)\delta.$ – md2perpe Dec 03 '22 at 09:14
  • Nice! I get it know I think. Since I am looking for a function $f(t)$ with a finite extinction time $0<T<\infty$ so it is equivalent to say $f(t)=f(t)\theta(T-t)$, and it happens I am testing with functions $f(t)=a(T-t)^m$ it ends happening that $f'\equiv (f\theta)'=f'\theta$ since the missing part made things of the form $q(t)(T-t)\delta(T-t)$ such as $q(t)$ allows to have the property $(T-t)\delta(T-t)=0$. Is quite interesting since allows me to test the solution avoiding things that Wolfram-Alpha got stacked when solving, like terms $\text{Abs}'(r(t))$ which are not simplified. – Joako Dec 04 '22 at 08:34