A brick sliding in an horizontal plane after an initial push (under Coulomb's dry friction) - closed form solutions validation?


I am looking for simple mechanics models that could have closed-form solutions that achieves finite extinction times where it becomes zero for their own system dynamics and stays there forever after.

Recently I figure out that no solution to a linear differential equation, neither any solution described by a non-piecewise Power Series, could have a finite extinction time due the Identity theorem, and found on this paper that an ordinary differential equation (ODE) could indeed have a solution that achieve a finite extinction time, but if and only if its nonlinear and have a singular point in time (so it is Non-Lipschitz and stand singular solutions), so the ODE must have at least on point where it fail to fulfill uniqueness of solutions (luckily, within the initial conditions' time $t_0$ and the finite extinction time $T$ uniqueness is still hold I believe - see Filippov's differential equations). Be aware this is valid only for first and second order ODE's, I don't know if still stand for systems of ODEs, neither for PDE's (which I tried to review here).

Since my intuition tells that simple 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 by myself as in here and here), and those I found (here) have introduced non-linearities to model some specific behaviors, but not for modeling a finite duration, and were quite hard to understand (at least for me, I'm an electrician, nor a physicists neither a mathematician).

Looking for these kinds of finite duration systems, someone in a Youtube named @siguc told 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.".

where $\text{sgn}(x)$ is the is the Sign function.

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

Images from Wikipedia

In this section of the Wikipedia page is shown the differential equation that model the Newton's 2nd law and rules this system: $$ m\cdot x'' = -F - \left(\text{sgn}(x')\right)\cdot \mu_k\cdot m\cdot g$$

Now, assuming that the initial push have already happened, the brick is sliding with an initial velocity $x'(0)$, so the force that have produce this push could be already being considered as $F = 0$ (is absent now from the point of view of the brick), so the only force present is the friction that slows down the brick until it stop moving. If I am not mistaken, dividing from the mass $m$ and matching $k \equiv \mu_k >0$, I will have that the product $k\cdot g >0$ with $g = 9.8\,\frac{m}{s^2}$ the Earth's gravity acceleration constant recovering the differential equation mentioned by @siguc: $$x'' = -k\cdot g\cdot \text{sgn}(x') \tag{Eq. 1}\label{Eq. 1}$$

Main issue________________

As example, if I let $k\cdot g \equiv 1$ in \eqref{Eq. 1} and I review it in Wolfram-Alpha, it shows \eqref{Eq. 1} could have many solutions with different behaviors (as expected for a nonlinear differential equation):

Wolfram-Alpha solutions

So following my intentions of finding solutions of finite duration, I tried to solve the differential equation assuming there exist a time $T>0$ where the solution becomes $x(t) = 0,\,\forall\ t\geq T$, which I will called as Endiness constraint for now on (it is an "invented word" so far I know, so please keep it mind since I going to use it later).

With this, I found the following: If I use the change of variable $x' = k\cdot g \cdot z$, the equation \eqref{Eq. 1} could become: $$\begin{array}{rcl} -\frac{x''}{k\cdot g} & = & \text{sgn}(x') \\ \iff -\frac{k\cdot g\cdot z'}{k\cdot g} & = & \text{sgn}(k\cdot g\cdot z) \\ & = & \frac{k \cdot g\cdot z}{|k\cdot g\cdot z|} \\ & \overset{\text{since}\ k\cdot g>0}{=} & \frac{k\cdot g\cdot z}{|k\cdot g|\cdot |z|} \\ & = &\frac{z}{|z|} \\ &=& \text{sgn}(z) \\ \Rightarrow z' & = & -\text{sgn}(z) \qquad\text{(Eq. 2)}\tag{Eq. 2}\label{Eq. 2} \end{array}$$

Now for solving \eqref{Eq. 2}, thinking in the Endiness constraint I tried the following: since I want the solution to become by itself zero, and I know now that a non-piecewise power series doesn't going to work, I tried as Ansatz something of the form: $$z(t) = a \cdot (T-t)^n\cdot\theta(T-t)\tag{Eq. 3}\label{Eq. 3}$$ where $\theta(t)$ is the Heaviside step function and $a$ is a constant to be determined, so in this way, I will have that:

  • $z'(t) = a\cdot n\cdot (T-t)^{n-1}\cdot(-1)\cdot\theta(T-t) + \underbrace{\require{cancel}\cancel{a\cdot(T-t)^n\cdot\delta(T-t)}}_{=\,0\,\text{because}\,x^n\delta(x)\,=\,0} = -a\cdot n\cdot (T-t)^{n-1}\cdot\theta(T-t) \tag{Eq. 4}\label{Eq. 4}$ with $\delta(t)$ is the Dirac delta function, so I can move "freely" the $\theta(T-t)$ term in and out from the derivatives (as is explained here).
  • With the Ansatz $z(t)$ the solution will become zero at time $T$ continuously and stays there forever, and hopefully also its derivative will fulfill the same.

By replacing \eqref{Eq. 3} and \eqref{Eq. 4} into \eqref{Eq. 2}, I could realize the following: $$-a\cdot n\cdot (T-t)^{n-1}\cdot\theta(T-t) \overset{???}{=} -\text{sgn}\left(a \cdot (T-t)^n\cdot\theta(T-t)\right)\tag{Eq. 5}\label{Eq. 5}$$ where from inspection I could determine the following:

  1. Since in the Right-Hand-Side of \eqref{Eq. 5} I have a Sign function, which can only have three possible values $\{-1;\,0;\,1\}$, and in the Left-Hand-Side of \eqref{Eq. 5} I have a polynomial in the variable $t$, the only possible way to have an equality is by letting $n \equiv 1$.
  2. Now, with $n=1$ \eqref{Eq. 5} becomes $-a\cdot\theta(T-t) = -\text{sgn}\biggr(a\cdot\underbrace{(T-t)}_{>0\,\text{when}\,t<T}\cdot\theta(T-t) \biggr)$, which by the same argument fixes $a \in \{-1;\,0;\,1\}$, which could be checked by inspection.

So with this, I can use as solution of \eqref{Eq. 2} $z(t) = \pm (T-t)\cdot\theta(T-t) \tag{Eq. 6}\label{Eq. 6}$

Now, turning back to the original variable and integrating with respect time (there is a step I am not $\mathit{100\%}$ sure is right so I will show it with an exclamation symbol "$!$"): $$\begin{array}{rcl} x'(t) & = & \pm \,k\cdot g\cdot (T-t)\cdot\theta(T-t)\qquad \Biggr/\,\int\,dt\\ \Rightarrow x(t) & = & \pm \,k\cdot g\cdot \displaystyle{\int} (T-t)\cdot\theta(T-t)\,dt \\ & \overset{!}{=} & \pm \,k\cdot g\cdot \displaystyle{\int}(T-t)\,dt \cdot\theta(T-t) \\ \Rightarrow x(t) & = & \pm \,k\cdot g\cdot \left[\frac{1}{2}\left(T-t\right)^2 + C_0 \right]\cdot\theta(T-t) \qquad\text{(Eq. 7)}\tag{Eq. 7}\label{Eq. 7} \end{array}$$
where now there is an integration constant $C_0$. But again, by derivation of $x(t)$ show in \eqref{Eq. 7} and using again the Endiness constraint I will have that: $$x'(t) = \pm \,k\cdot g\cdot \left[(-1)\cdot\frac{\require{cancel}\cancel{2}}{\require{cancel}\cancel{2}}\cdot(T-t)+\require{cancel}\cancel{0}\right]\cdot\theta(T-t) + \underbrace{\require{cancel}\cancel{(\pm)\,k\cdot g\cdot\left[\frac{1}{2}(T-t)^2 + C_0\right]\delta(T-t)}}_{C_0\,\equiv\,0,\,\text{so all the expression could be zero by}\,x\delta(x)\,=\,0} = \text{(Eq. 7)}$$ which imply I have now the closed-form "particular solutions" to \eqref{Eq. 1} (they are not general solutions): $$x(t) = \pm \,\frac{k\cdot g}{2}\cdot\left(T-t\right)^2\cdot\theta(T-t) \equiv \pm\,T^2\cdot\frac{k \cdot g}{8}\cdot\left(1-\frac{t}{T}+\left|1-\frac{t}{T}\right|\right)^2\tag{Eq. 8}\label{Eq. 8} $$

calling now $x_+(t)$ and $x_-(t)$ when choosing the positive and negative sign versions respectively.

Now I have a solution $x(t)$ that fulfill the Endiness constraint as intended: it becomes softly zero at time $t=T$, meaning here that both the solution as its first derivative continuously becomes zero at this finite extinction time $T$.

So far so good, until I checked the initial conditions: at initial conditions at time $t_0=0$, I will have that $x(0) = \pm\,T^2\cdot\frac{k\cdot g}{2}$ and $x'(0)=\mp\,T\cdot k\cdot g$, and noting that since $k\cdot g\cdot T>0$ I have both initial conditions have opposite signs, which have confuse me. Thinking in a brick starting at a general position $x_+(0)=T^2\cdot\frac{k\cdot g}{2}>0$ makes sense since distances are positive quantities, but now its initial speed is $x_+'(0)=-T\cdot k\cdot g<0$ meaning the brick is going in a different direction it should be going after the initial push, and similarly, if I choose $x_-'(0)=T\cdot k\cdot g>0$ I will got the speed in the right sign/direction, but now the initial position $x_-(0)=-T^2\cdot\frac{k\cdot g}{2}<0$ will be negative, which is "weird" for a distance, and trying to introduce some complex time $T\in\mathbb{C}$ trying to fit both the initial conditions make no sense in my opinion.

What I believe have happened is that with the solutions I have found, tacitly I have fixed the spatial coordinates origin at $x(t)=0$ since $x(T)=0$, so I have to think about the problem as it "things were coming towards the final position", but I am not completely sure about this, which is in part why I making this question.


I am looking for theoretical and experimental validation of the solutions I have found on \eqref{Eq. 8}, explaining how it could be visualized.

Points I would like to see in the answer:

  1. Do you know any source where this problem \eqref{Eq. 1} is solved through closed-form solutions? Wanting to know if the treatment done here is right, and if there are other solutions (any book, paper, educational website or video - serious sources please).
  2. Could you give this problem and solutions a basic explanation? Like if you were making a high-school class, with free-body diagrams, assumptions made, and drawings explaining what is happening in the solutions (this due my confusion about how to fit the figure of the problem with the solutions I have found).
  3. How well/bad this solution of \eqref{Eq. 8} fit the real life and traditional ways of solving this problem?, here I wanting to know if this solutions Do match solutions through "energy analysis"?, Also, If the solutions fits the experimental results at "classroom scale"? like an inclined plane experiments with a non-rolling object, but continued after the falling object reach the horizontal table, it have now an initial speed and will move for a while over the table.
  4. Now I have 2 formulas for determining the final time $T$ (finite extinction time): $T = \sqrt{\frac{2\cdot|x(0)|}{k\cdot g}}$ and $T = \frac{|x'(0)|}{k\cdot g}$: Does they always coincide? What are the implications of having a "predetermined" final time?
  5. Note that both previous formula implies that the distance the brick slides is related to the inital speed as $d_{\text{max}}\equiv|x(0)|=\frac{|x'(0)|^2}{2\cdot k\cdot g}\equiv \frac{m\cdot|x'(0)|^2}{2\cdot \mu_k \cdot g}$, which kind of resembles the kinetic energy formula; Does this distance fits experimental results like when a fast car slams its brakes?

Added later

Unfortunately a grumpy moderator closed the question, but luckily I have figure out how to answer questions $(2)$ to $(5)$:

First noting that if I make the change of variable $\tau = T-t$ the \eqref{Eq. 8} becomes: $$x(T-\tau) = \pm\,\frac{k\cdot g}{2}\tau^2\theta(\tau) \tag{Eq. 9}\label{Eq. 9}$$ so the solution is indeed a displaced version of a solution referenced from the point $x(T)=0$, which solves my doubts about how to imaging the solution within the figure of the sliding brick shown on Wikipedia.

But also \eqref{Eq. 9} solves the other questions: since the solution is quadratic and only dependent from the $\tau^2$ term, means the solution it is of constant acceleration, matching the solution with the traditional way to solve it through the uniform acceleration laws like the Torricelli's equation $v_f^2-v_0^2 = 2a\Delta x$ which implies the formula of question $(5)$ as the equation $v_f = v_0+at$ directly implies the second formula of question $(4)$ (and both together implies the first one), so matching $$|x''|\equiv a \equiv |-k\cdot g\cdot \text{sgn}(x')|= k\cdot g\tag{Eq. 10}\label{Eq. 10}$$ from \eqref{Eq. 1}, it makes a lot of sense that the solution of this problem under Coulomb damping fit the solution of an uniform accelerated system for times $t<T$, which at least for me validates the closed-form solutions founded as the application of the Endiness constraint as it were detailed above, in this, the simplest problem I could think off. The graphical description of this problem as a uniform accelerated system could be found in the following Youtube video, from where I figure out the resemblance.

I hope its focused enough now. If is kept closed, please comment the sources.

  • 1
    $\begingroup$ Hi Joako. Welcome to Phys.SE. Note that res. recom. qs. are restricted on Phys.SE. For the record users (not moderators) closed the question. $\endgroup$
    – Qmechanic
    Jan 27 at 2:16