Mathematical formulation of a dynamical system with dry friction subjected to external forces

In presence of dry friction, the physical description of the dynamic and static phases is as follows.

• •

  The phase of $x(t)$ at time $t$ is static when

  $$\dot{x}(t)=0\>\mbox{and}\>|b(x(t),t)|\leq f_{s}.$$

  It is important to emphasize that a static phase corresponds to $\dot{x}(t)=0$ on an time interval of positive Lebesgue measure and thus $\ddot{x}(t)=0$ on the same interval. In this case, the friction force takes the value

  $$\mathbb{F}(\dot{x}(t))=b(x(t),t)$$

  that is necessary for equilibrium.

• •

  Otherwise, the phase of $x(t)$ at time $t$ is dynamic when

  $$\dot{x}(t)=0\>\mbox{and}\>|b(x(t),t)|>f_{s}\>\mbox{or}\>\dot{x}(t)\neq 0.$$

  When $\dot{x}(t)\neq 0$, the friction force takes the value

  $$\mathbb{F}(\dot{x}(t))=\textup{sign}(\dot{x}(t))f_{d}.$$

  It is important to emphasize that $\dot{x}(t)=0$ and $|b(x(t),t)|>f_{s}$ happen on a negligible (isolated points) time set.

The transition from a static phase to a dynamic phase occurs as soon as

$$|b(x(t),t)|>f_{s}.$$

Conversely, from a dynamic phase the system enters into a static phase as soon as

$$\dot{x}(t)=0\>\mbox{and}\>|b(x(t),t)|\leq f_{s}.$$

Below, we make precise the mathematical formulation and obtain a solution which is $C^{1}$.

We want to define rigorously a function $x(t)$, which is $C^{1}$ and satisfies (1)-(2). To fix ideas, we consider initial conditions $\tau_{0}=0$, $x(\tau_{0})=x_{0}$, $\dot{x}(\tau_{0})=0$ in a static phase with

$$|b(x_{0},\tau_{0})|\leq f_{s}.$$

The first transition towards a dynamic phase occurs at time $\tau_{\frac{1}{2}}$ defined as

$$\tau_{\frac{1}{2}}\triangleq\inf\left\{t\geq\tau_{0},\>|b(x_{0},t)|>f_{s}\right\}$$

If $\tau_{0}<\tau_{\frac{1}{2}}$ then on the interval $[\tau_{0},\tau_{\frac{1}{2}})$ we define

$$x(t)\triangleq x_{0},\>\dot{x}(t)\triangleq 0.$$

If $|b(x_{0},t)|$ never goes beyond $f_{s}$ then $\tau_{\frac{1}{2}}=\infty$ and $x$ remains forever in a static phase. If $\tau_{0}=\tau_{\frac{1}{2}}$ there is no static (or stick) phase and thus the interval $[\tau_{0},\tau_{\frac{1}{2}})$ is void. Provided that $\tau_{\frac{1}{2}}$ is finite, we can define the time of return in static phase $\tau_{1}$ as

$$\tau_{1}\triangleq\inf\left\{t\>>\>\tau_{\frac{1}{2}},\>\dot{x}(t)=0\>\mbox{and}\>|b(x(t),t)|\leq f_{s}\right\}$$

where

$$\forall t>\tau_{\frac{1}{2}},\>\forall\varphi\in\mathbb{R},\>(b(x(t),t)-m\ddot{x}(t))(\varphi-\dot{x}(t))+f_{d}|\dot{x}(t)|\leq f_{d}|\varphi|$$

(6)

with the initial condition

$$x(\tau_{\frac{1}{2}})=x_{0},\>\quad\dot{x}(\tau_{\frac{1}{2}})=0.$$

This class of VI is standard and the theory tells that such a problem has one and only one solution in $C^{1}$ on the interval $[\tau_{\frac{1}{2}},\infty)$. See for instance [2] for a proof of this general result. Actually, we will see that $\tau_{1}>\tau_{\frac{1}{2}}$. We will discuss the properties of the solution in the next section. If $\tau_{1}=\infty$ then $x$ remains in a dynamic state forever, otherwise the same procedure can be repeated with initial condition $x(\tau_{1})$ that we denote $x_{1}$ at time $\tau_{1}$ where $|b(x_{1},\tau_{1})|\leq f_{s}$. By induction, we can define similarly the entrance times in static phase $\tau_{j}$ with corresponding states $x_{j}$ and the entrance times of dynamic phase $\tau_{j+\frac{1}{2}}$.

We now check that we can also define a sequence of sub phases of the dynamic phase, in which the solution satisfies a standard differential equation. This procedure is an alternative to the V.I. but is much less synthetic. The first dynamic sub-phase starts at $\tau_{\frac{1}{2}}\triangleq\inf\{t\geq\tau_{0},\>|b(x(t),t)|>f_{s}\}$ where we recall that $\tau_{0}\triangleq 0$. Define

$$x_{0}^{0}\triangleq x(0),\quad\tau_{0}^{0}\triangleq\tau_{\frac{1}{2}},$$

thus

$$|b(x_{0}^{0},\tau_{0}^{0})|=f_{s}.$$

We set $\epsilon_{0}^{0}\triangleq\textup{sign}(b(x_{0}^{0},\tau_{0}^{0}))$ and consider the differential equation

$$\begin{cases}&m\ddot{x}(t)=b(x(t),t)-\epsilon_{0}^{0}f_{d},\>t>\tau_{0}^{0},\\ &x(\tau_{0}^{0})=x_{0}^{0},\quad\dot{x}(\tau_{0}^{0})=0.\end{cases}$$

(8)

Then, we define

$$\tau_{0}^{1}\triangleq\inf\{t>\tau_{0}^{0},\>\dot{x}(t)=0\}\quad\mbox{and}\quad x_{0}^{1}\triangleq x(\tau_{0}^{1}).$$

• •

  If $|b(x_{0}^{1},\tau_{0}^{1})|\leq f_{s}$ then the dynamic phase ends here and a new static phase starts. We set

  $$\tau_{1}\triangleq\tau_{0}^{1}\quad\mbox{and}\quad x_{1}\triangleq x_{0}^{1}.$$

• •

  Otherwise $|b(x_{0}^{1},\tau_{0}^{1})|>f_{s}$ and we set

  $$\epsilon_{0}^{1}\triangleq\textup{sign}(b(x_{0}^{1},\tau_{0}^{1}))$$

  and again consider the differential equation

  $$\begin{cases}&m\ddot{x}(t)=b(x(t),t)-\epsilon_{0}^{1}f_{d},\>t>\tau_{0}^{1},\\ &x(\tau_{0}^{1})=x_{0}^{1},\quad\dot{x}(\tau_{0}^{1})=0\end{cases}$$

  (9)

  and introduce

  $$\tau_{0}^{2}\triangleq\inf\{t>\tau_{0}^{1},\>\dot{x}(t)=0\}.$$

  We can repeat the same procedure several times and thus define a sequence $\tau_{0}^{k},x_{0}^{k},\epsilon_{0}^{k}$ for $k\leq k(0)$ where

  $$k(0)\triangleq\inf\{k\geq 1,\>|b(x_{0}^{k},\tau_{0}^{k})|\leq f_{s}\}.$$

  If $k(0)=\infty$ then $x$ remains in a dynamic phase forever otherwise a new static phase starts at

  $$\tau_{1}\triangleq\tau_{0}^{k(0)}\quad\mbox{and}\quad x_{1}\triangleq x_{0}^{k(0)}.$$

  The dynamic phase on the interval $(\tau_{\frac{1}{2}},\tau_{1})$ is thus divided into dynamic sub-phases $(\tau_{0}^{k},\tau_{0}^{k+1})$ with

  $$\tau_{0}\leq\tau_{\frac{1}{2}}=\tau_{0}^{0}<\tau_{0}^{1}<\cdots<\tau_{0}^{k(0)}=\tau_{1}.$$

Similarly, if for any $j\geq 1$ we know $x_{j}$ and $\tau_{j}$ with $|b(x_{j},\tau_{j})|\leq f_{s}$. The first dynamic sub phase starts at

$$\tau_{j}^{0}\triangleq\tau_{j+\frac{1}{2}},\;x_{j}^{0}\triangleq x_{j},\>\mbox{where}\>|b(x_{j}^{0},\tau_{j}^{0})|=f_{s}.$$

For each $k\geq 0$, to define the dynamic sub phase starting at $\tau_{j}^{k}$, we set

$$\epsilon_{j}^{k}\triangleq\textup{sign}(b(x_{j}^{k},\tau_{j}^{k}))$$

and consider the differential equation

$$\begin{cases}&m\ddot{x}(t)=b(x(t),t)-\epsilon_{j}^{k}f_{d},\>t>\tau_{j}^{k},\\ &x(\tau_{j}^{k})=x_{j}^{k},\quad\dot{x}(\tau_{j}^{k})=0.\end{cases}$$

(10)

Then, we uniquely define $\tau_{j}^{k+1}$ by

$$\tau_{j}^{k+1}\triangleq\inf\{t>\tau_{j}^{k},\dot{x}(t)=0\}.$$

This procedure defines the dynamic sub phase $(\tau_{j}^{k},\tau_{j}^{k+1})$. On this subinterval $\epsilon_{j}^{k}=\textup{sign}(\dot{x}(t))$, so (10) is also the equation

$$m\ddot{x}(t)=b(x(t),t)-\textup{sign}(\dot{x}(t))f_{d}.$$

(11)

We set $x_{j}^{k+1}\triangleq x(\tau_{j}^{k+1})$ and we proceed as follows:

• •

  if $|b(x_{j}^{k+1},\tau_{j}^{k+1})|\leq f_{s}$ then we start a new static phase and we set

  $$\tau_{j+1}\triangleq\tau_{j}^{k+1}\>\mbox{and}\>x_{j+1}\triangleq x_{j}^{k+1}.$$

• •

  on the other hand, if $|b(x_{j}^{k+1},\tau_{j}^{k+1})|>f_{s}$ then we start a new dynamic sub phase on the interval $(\tau_{j}^{k+1},\tau_{j}^{k+2})$.

We can define

$$k(j)\triangleq\inf\{k\geq 1,\>|b(x_{j}^{k},\tau_{j}^{k})|\leq f_{s}\},$$

thus the dynamic phase on the interval $(\tau_{j+\frac{1}{2}},\tau_{j+1})$ is divided into dynamic sub-phases $(\tau_{j}^{k},\tau_{j}^{k+1})$ with

$$\tau_{j}\leq\tau_{j+\frac{1}{2}}=\tau_{j}^{0}<\tau_{j}^{1}<\cdots<\tau_{j}^{k(j)}=\tau_{j+1}.$$

This procedure gives explicitly the solution of the V.I.. We will see in the next section that, when considering the case $b(x,t)=K(\beta T(t)-x)$, we have the additional advantage that the solution is explicit in a dynamic sub phase.

When $b(x,t)=K(\beta T(t)-x)$, the differential equation (10) has an explicit solution. For $t\in[\tau_{j}^{k},\tau_{j}^{k+1})$,

$$x(t)=x_{j}^{k}\cos\omega\left(t-\tau_{j}^{k}\right)+\int_{\tau_{j}^{k}}^{t}\sin\omega(t-s)\dfrac{K\beta T(s)-\epsilon_{j}^{k}f_{d}}{m\omega}\textup{d}s$$

(12)

and $\tau_{j}^{k+1}$ is defined by the equation

$$x_{j}^{k}\sin\omega\left(\tau_{j}^{k+1}-\tau_{j}^{k}\right)=\int_{\tau_{j}^{k}}^{\tau_{j}^{k+1}}\cos\omega(\tau_{j}^{k+1}-s)\dfrac{K\beta T(s)-\epsilon_{j}^{k}f_{d}}{m\omega}\textup{d}s.$$

(13)

We then set

$$x_{j}^{k+1}\triangleq x(\tau_{j}^{k+1})=x_{j}^{k}\cos\omega\left(\tau_{j}^{k+1}-\tau_{j}^{k}\right)+\int_{\tau_{j}^{k}}^{\tau_{j}^{k+1}}\sin\omega(\tau_{j}^{k+1}-s)\dfrac{K\beta T(s)-\epsilon_{j}^{k}f_{d}}{m\omega}\textup{d}s.$$

(14)

The above discussion allows to define an algorithm to construct the trajectory $x(t)$ of the initial problem (1) and (2) with $b(x,t)$ given in (3). We construct the sequence $\{x_{j},\tau_{j},\>j\geq 0\}$ where

$$|b(x_{j},\tau_{j})|\leq f_{s}.$$

When $j=0$, it corresponds to the initial condition $x_{0}\triangleq x(0)$ and $\tau_{0}\triangleq 0$. For each $j\geq 0$, we first compute

$$\tau_{j+\frac{1}{2}}\triangleq\inf\{t\geq\tau_{j},\>|b(x_{j},t)|>f_{s}\}$$

and then we define a subsequence with respect to $k$.

• •

  Define

  $$\tau_{j}^{0}\triangleq\tau_{j+\frac{1}{2}},\quad x_{j}^{0}\triangleq x_{j},\quad\epsilon_{j}^{0}\triangleq\textup{sign}(b(x_{j}^{0},\tau_{j}^{0})).$$

• •

  For each $k\geq 0$, using $\epsilon_{j}^{k}$, define $\tau_{j}^{k+1}$ using (13) and $x_{j}^{k+1}$ using (14).

• •

  If

  $$|b(x_{j}^{k+1},\tau_{j}^{k+1})|\leq f_{s}$$

  then

  $$\tau_{j+1}\triangleq\tau_{j}^{k}\>\mbox{and}\>x_{j+1}\triangleq x_{j}^{k+1}$$

  otherwise we define

  $$\epsilon_{j}^{k+1}\triangleq\textup{sign}(b(x_{j}^{k+1},\tau_{j}^{k+1}))$$

  and repeat the procedure with the definition of $\tau_{j}^{k+2},x_{j}^{k+2}$.

The sub interval $(\tau_{j},\tau_{j+\frac{1}{2}})$ is a static or stick subphase, and the sub interval $(\tau_{j+\frac{1}{2}},\tau_{j+1})$ is a dynamic or slip phase, itself subdivided into dynamic subphases. The trajectory $x(t)$ is completely defined by this cascade of phases and subphases.

Since Equation (13) is transcendental, we need also an algorithm to solve it. We are going to state an approximation which simplifies considerably the calculations. This approximation is called quasistatic which means that the excitation variation is slow enough to be neglected during any dynamic phase. From then on, an interesting consequence of this approximation is that there are no subphases in the dynamic phases. We will see below that, with the definition of sub phases of the Section 2.4 in mind, one has $x_{j}^{1}$ and $\tau_{j}^{1}$ satisfy

$$|K(\beta T(\tau_{j}^{1})-x_{j}^{1})|\leq f_{s}.$$

So there is only one sequence $\tau_{j},x_{j}$ and once $\tau_{j+\frac{1}{2}}$ is defined, the equation for $\tau_{j}^{1}$ becomes simply an equation for $\tau_{j+1}$ which is

$$x_{j}\sin\omega\left(\tau_{j+1}-\tau_{j}\right)=\int_{\tau_{j}}^{\tau_{j+1}}\cos\omega(\tau_{j+1}-s)\dfrac{K\beta T(s)-\epsilon_{j}f_{d}}{m\omega}\textup{d}s.$$

(15)

with

$$\epsilon_{j}\triangleq\textup{sign}(\beta T(\tau_{j+\frac{1}{2}})-x_{j}).$$

Next finding $x_{j+1}$ uses Equation (14) as follows:

$$x_{j+1}\triangleq x_{j}\cos\omega\left(\tau_{j+1}-\tau_{j+\frac{1}{2}}\right)+\int_{\tau_{j+\frac{1}{2}}}^{\tau_{j+1}}\sin\omega(\tau_{j+1}-s)\dfrac{K\beta T(s)-\epsilon_{j}f_{d}}{m\omega}\textup{d}s.$$

(16)

The quasistatic approximation consists in making the approximation

$$\forall s\in(\tau_{j+\frac{1}{2}},\tau_{j+1}),\quad T(s)=T(\tau_{j+\frac{1}{2}}).$$

We will use the notation $T_{j+\frac{1}{2}}\triangleq T(\tau_{j+\frac{1}{2}})$. From (16), we obtain

$$\sin(\omega(\tau_{j+1}-\tau_{j+\frac{1}{2}}))=0$$

which implies

$$\tau_{j+1}=\tau_{j+\frac{1}{2}}+\frac{\pi}{\omega}.$$

We turn to (16) which simplifies considerably using

$$|\beta T_{j+\frac{1}{2}}-x_{j}|=\frac{f_{s}}{K},$$

(17)

and we obtain

$$x_{j+1}=x_{j}+2\epsilon_{j}\frac{f_{s}-f_{d}}{K}.$$

(18)

Of course, we need to find

$$\tau_{j+\frac{1}{2}}=\inf\{t>\tau_{j},\>|\beta T(t)-x_{j}|>\frac{f_{s}}{K}\}$$

and then define

$$\tau_{j+1}=\tau_{j+\frac{1}{2}}+\frac{\pi}{\omega}.$$

Also

$$\epsilon_{j+1}\triangleq\textup{sign}(\beta T_{j+\frac{3}{2}}-x_{j+1}).$$

When we are in a static phase at the value $x_{j}$, starting at $\tau_{j}$, we can interpret the condition (17) as a condition that the future temperature must satisfy to initiate a new dynamic phase.

When $f\triangleq f_{s}=f_{d}$, we do not need to define the static phases as in Section 2.2 above. Indeed, we have seen in the proof of Proposition 1, that if $\dot{x}(t)=0$ on a set of positive measure, then necessarily we have $|\beta T(t)-x(t)|\leq\dfrac{f}{K}$. This means that during a static phase, the V.I. is also satisfied. Therefore, the problem is simply

$$\forall\varphi\in\mathbb{R},\;(m\ddot{x}(t)-K(\beta T(t)-x(t)))(\dot{x}(t)-\varphi)+f|\dot{x}(t)|\leq f|\varphi|$$

(19)

with

$$x(0)=x_{0},\>\dot{x}(0)=0.$$

In the case $f_{s}=f_{d},$ we also see from the approximation formula (18) that $x_{j}$ is constant. So the system enters in the static mode at the same point. Since in our assumption $\dot{x}(0)=0$ and $x(0)=x_{0},$ we have $x_{j}=x_{0}$. This is only an approximation. The alternance of static and dynamic modes follows the sequence of times, $\tau_{j},$$\tau_{j+\frac{1}{2}}$, such that

$$\tau_{j+\frac{1}{2}}=\inf\left\{t>\tau_{j}\>|\;T(t)\notin\left[\frac{x_{0}}{\beta}-\frac{f}{\beta K},\frac{x_{0}}{\beta}+\frac{f}{\beta K}\right]\right\}$$

(20)

$$\tau_{j+1}-\tau_{j+\frac{1}{2}}=\dfrac{\pi}{\omega}$$

with $\tau_{0}=0$. What is not an approximation is the following discussion. In a static mode $\tau_{j}\leq t\leq$$\tau_{j+\frac{1}{2}},$ we have

$$|\beta T(t)-x(t)|\leq\dfrac{f}{K}$$

(21)

and in a dynamic mode,$\tau_{j+\frac{1}{2}}\leq t\leq\tau_{j+1}$ we have the equation

$$m\ddot{x}(t)=K(\beta T(t)-x(t))-\epsilon_{j}f_{s}$$

(22)

and $x(t)$ is $C^{1}$, with $\ddot{x}(t)$ locally bounded.

Let $T>0,N\in\mathbb{N}^{\star}$ and $\{t_{n}\}_{n=0}^{N}$ a family of time which discretizes $[0,T]$ such that $t_{n}\triangleq nh$ where $h\triangleq\frac{T}{N}$. We will construct a sequence $\{(X_{t_{n}}^{h},\dot{X}_{t_{n}}^{h})\}_{n=0}^{N}$ where for each $n$, $(X_{t_{n}}^{h},\dot{X}_{t_{n}}^{h})$ is meant to approximate $(x(t_{n}),\dot{x}(t_{n}))$.