A COUNTEREXAMPLE TO ANALYTICITY IN FRICTIONAL DYNAMICS ⋆

. We consider the motion of a particle acted on by dry friction and a force that is an analytic function of time. We give a counterexample to the claim that such motions are given by analytic functions of time. Several published arguments concerning existence and uniqueness in unilateral dynamics with friction rely on the analyticity of such motions. The counterexample invalidates those arguments for motions in three or more dimensions.


Introduction
Contact and impact are typically modelled with unilateral constraints: inequalities involving the coordinates of some bodies, which are satisfied when those bodies do not interpenetrate.Such constraints are essential components of models of robots [7,10] and other systems [6], and it is natural to enquire about the existence and uniqueness of solutions to initial value problems for such models [3].In the 1960s, it was discovered that the motion of a unilaterally constrained particle acted on by an external force is in general non-unique, even if the force is an infinitely differentiable function of time [1,4,14].However, in the 1990s, it was shown that such motions are unique if there is no friction and the force is an analytic function [1,13].(A function  is analytic if for every point  0 of its domain, the Taylor series of  at  0 converges to  () for all points  in some neighbourhood of  0 .)Several authors have explored how such results might be extended to models involving friction.Ballard and Basseville [2] presented arguments for the existence and uniqueness of solutions to initial value problems for a unilaterally constrained particle acted on by an analytic force and dry friction.Charles and Ballard [5] extended those arguments to finite collections of particles.A key step of those arguments is to derive a local solution given by a power series, and to claim that it corresponds to an analytic function.
In this paper, we present a simple counterexample to that claim.The counterexample invalidates the existence and uniqueness arguments presented in Ballard and Basseville [2] and Charles and Ballard [5] for unilaterally constrained particles in dimension  > 2, although those arguments are correct for  = 2 to the best of our knowledge.Consequently for  > 2, the only general existence result about unilateral dynamics with friction is that of Monteiro Marques [11], which only addresses situations with a single constraint and perfectly inelastic

Counterexample
Consider a particle in contact with a flat surface, acted on by a force and friction.The particle's position is an element of R  , but a normal force holds it in contact with the surface, so we describe its motion in R −1 using an orthogonal coordinate system tangent to that surface.The particle's tangential velocity v : [0,  ] → R −1 satisfies Newton's second law: where  is the particle's mass, the dot denotes the time derivative, F : [0,  ] → R −1 is the tangential force, and R : [0,  ] → R −1 is the tangential reaction.This reaction satisfies Coulomb's law of friction: and where  > 0 is the friction coefficient, and  () is the magnitude of the force normal to the surface.As discussed in Moreau [12], Ballard and Charles [3], we may write this compactly as where  denotes the subdifferential of a convex function  .For simplicity, we assume the particle has unit mass and that  () = 1.The velocity then satisfies the differential inclusion (This is equivalent to requiring that the differential equation v() = F() − (v()/‖v()‖) holds at times when v() ̸ = 0, and that the constraint ‖ v() − F()‖ ≤ 1 holds at times when v() = 0.) The following counterexample shows that even if the force F is an analytic function of time and the final time  > 0 is tiny, differential inclusion (1.1) need not have an analytic solution v.The force appearing in this counterexample is illustrated in Figure 1.The magnitude of this force initially equals the friction limit, but a component of the force orthogonal to its initial direction increases, causing the particle to slip.
Counterexample 1. Suppose the dimension is  > 2, the final time is  > 0, and the velocity v : [0,  ] → R −1 is an absolutely continuous function with v(0) = 0 that satisfies differential inclusion (1.1) for the force for almost all  ∈ [0,  ].Then v is not an analytic function.
To prove the velocity v() is not analytic, we derive its Taylor series at  = 0 and show that this series diverges for any  > 0.
Remark 1.1.It is not uncommon for differential equations to have divergent series solutions [9].However, the initial condition v(0) = 0 is necessary for a series solution of differential inclusion (1.1) to diverge for all  > 0, when F is analytic.Indeed, if v(0) ̸ = 0 then the tangential reaction is R() = −v()/‖v()‖ on a neighbourhood of  = 0, and this is an analytic function of v() on a neighbourhood of v(0), so the Cauchy-Kovalevskaya theorem [8] guarantees the existence of a unique analytic solution v() on a neighbourhood of  = 0. Remark 1.2.When working with unilateral constraints, the velocity must in general be formulated as a discontinuous function, so as to allow for impacts [2,5].Such formulations simplify to ours in the special case that a normal force holds the particle in contact with a surface.In this special case, the unilateral constraint is equivalent to a bilateral constraint: an equality constraint on the particle's coordinates.The existence and uniqueness of a solution to Counterexample 1 therefore follow from general results about bilaterally constrained problems with friction -specifically, Proposition 3.3 of [2] or Proposition 3.1 of [5].

Outline
First, we derive an algorithm to compute the Taylor series of v (Sect.2).We provide numerical evidence that this series diverges (Fig. 2) and explain the divergence intuitively.Then we discuss the specific results that our counterexample invalidates, and pinpoint the error in the arguments leading to those results (Sect.3).Appendix A gives a proof of divergence, and the supplementary material contains a Python implementation of our algorithm and Maxima code to verify our algebra.

Computing the Taylor series
In this section, we derive an algorithm for computing the Taylor series of the velocity in Counterexample 1, and provide evidence and an intuitive argument for the divergence of that series.We restrict attention to dimension  = 3. (We may obtain the velocity for  > 3 from that for  = 3, simply by setting the  − 3 additional components to zero.)We write the Taylor series at  = 0 of the velocity v in the form for some real coefficients   ,   with  0 ,  0 ̸ = 0, and some non-negative integers , .

Solving for the leading orders
First we show that  = 3 and  = 4.It follows from the initial condition v(0) = 0 that ,  ≥ 1.Also, for v() ̸ = 0 differential inclusion (1.1) reads Substituting the series expansions (2.1) and matching the coefficients of  0 in (2.3) gives , where 1 • is the indicator function.If  = , then this expression cannot be satisfied for  0 ̸ = 0. Thus, we have  ̸ = .Using this fact, matching the coefficients of  0 in (2.2) gives , then this expression cannot be satisfied.So, we must have  < , in which case this expression reads 1 =1  0 = 1 −  0 /| 0 |.As this cannot be satisfied for  = 1, it follows that  > 1 and  0 > 0. Matching the coefficients of  1 in (2.3), and using 1 <  <  (so that  ̸ = 2) gives As the terms involving  −1 and  2 have non-zero coefficients, they must be equal, and we conclude that

Algorithm
For the values of ,  in (2.4), functions ,  should satisfy the differential equations d d We write these equations in the form in terms of the intermediate functions By the Cauchy product formula, the coefficients of the formal power series for  ≥ 0. (In our notation, we sum over pairs of non-negative integers , , so that −   , and coefficients with negative indices vanish, so that  0 =  2 0 and  0 = 0.)Moreover, matching the coefficients of   and  −1 in (2.5) gives The idea of our algorithm is to treat  −2 ,  −1 ,   ,  −2 as unknowns, having already determined values for   ,   for  <  − 2. Equation (2.6) then provides expressions for   ,   ,   for  =  − 2,  − 1,  and hence for   ,   ,  −1 ,  −1 in terms of those unknowns.Substituting these expressions in (2.7) gives a linear equation for  −2 ,  −2 , which has a unique solution.(As the intermediate expressions for   ,  −1 involve  −1 ,   , it is interesting that the linear equation does not involve those variables.) We now apply the idea of the previous paragraph, treating  = 3, 4 and  ≥ 5 separately, as the formulas have "special cases" for small .(For instance, the sum ∑︀ +=     involves the term  2 −2 for  = 4 but the corresponding term is 2 2  −2 for  ≥ 5.) Full details of the following derivations are given in algebra.txt in the supplementary material.For  = 3 and 4, we get the linear equations with the unique solutions For  ≥ 5, we get where in which the terms   • are the parts of the sums in (2.6) that depend neither on the unknowns  −2 ,  −1 ,   ,  −2 , nor on the quantities   ,   for  =  − 2,  − 1,  that involve those unknowns: . (2.11) It turns out that   ,   ,   ,   ,   vanish for all odd .We show this by induction.In the base case, equation (2.8) gives  1 =  1 = 0, so (2.6) gives  1 =  1 =  1 = 0. Otherwise, suppose −2 is odd and that   ,   ,   ,   ,   vanish for all all odd  <  − 2. Consider any term of any of the sums • in (2.11).This term is of the form     for appropriate sequences ,  ∈ {, , , }, where one of ,  is odd and less than  − 2. It follows from the induction hypothesis that one of the factors   or   vanishes.Thus all the sums • vanish.It also follows from the induction hypothesis that  −4 = 0. Thus, equation (2.10) gives    =    = 0, so the linear equation (2.9) has the unique solution  −2 =  −2 = 0. Examining the sums giving  −2 ,  −2 ,  −2 in (2.6), we see that their terms are also of the form     , where one of the factors   or   vanishes.Therefore  −2 ,  −2 ,  −2 also vanish.This completes the induction.
In the light of the above discussion, Algorithm 1 computes the coefficients of the Taylor series of the velocity appearing in Counterexample 1.
Algorithm 1. Determining the coefficients of the Taylor series defined in (2.1).

Intuitive argument and numerical evidence for divergence
If   does not vanish for all large enough , and each of the sums in (2.11) satisfies • / −4 = constant + (1/) for even  as  → ∞, then (2.9) gives For such a sequence, there is no  > 0 such that ∑︀ ∞ =0     converges.We implemented the above algorithm using Python's fractions module, which provides exact arithmetic for rational numbers, and provide the resulting implementation as series.py in the supplementary material.We find the following initial terms of the series: as  → 0. Figure 2 plots the ratio   / −2 , which does indeed decrease nearly linearly, as suggested by (2.12).

Relation to previous work
Ballard and Basseville [2] argued for the analyticity of the motion of a single particle, under the action of an analytic force and Coulomb friction, and Charles and Ballard [5] extended those arguments to multiple particles.Both papers apply Lemma 3.4 of [2] to draw this conclusion.This lemma involves a function  that is hypothesized to be analytic in a neighbourhood of the origin.While both papers claim that the function to which they apply this lemma ( G in [2], p. 70 and g in [5], p. 13) is analytic in a neighbourhood of the origin, this claim is false in dimension  > 2, as illustrated in the example below.Moreover, Counterexample 1 shows that it is not possible to circumvent this error: the counterexample directly contradicts the results about the existence of a local analytic solution presented in Proposition 3.5 and Theorem 4.1 of [2] and Proposition 4.1 of [5].As these existence results are central to the uniqueness arguments presented in Theorem 4.2 of [2] and Theorem 4.2 of [5], the counterexample also invalidates those arguments in dimension  > 2.
Example 3.1.For a single particle  that is initially at rest but slides immediately thereafter, which is acted on by a tangential force F() ∈ R −1 and normal force  (), the function in Charles and Ballard [5], p. 13 is g (, ũ, ṽ where S 0+2 () = ∑︀ 0+2 =0 u    is the truncated Taylor series for the tangential position of the particle for suitable coefficients u  ∈ R −1 , and  0 is the integer such that Ṡ0+2 () has a single non-zero term.Now, if the force satisfies F() = (2, 0) ⊤ and | ()| = 1, then the solution for the tangential velocity is v() = (, 0) ⊤ , so the truncated Taylor series satisfies Ṡ0+2 () = (, 0) ⊤ with  0 = 0.It follows that g (, ũ, ṽ) = 1 which has a simple pole at  = 0 whenever the second component of ṽ does not vanish.Thus g is not analytic on any neighbourhood of the origin.

Appendix A. Proof of Counterexample 1
The proof is based on Algorithm 1.It is straightforward, but laborious as the algorithm involves many intermediate sums, each of which must be bounded.The file algebra.txt in the supplementary material contains computer algebra code for all calculations involved. Let Lemma A.2. Suppose  ≥ 182 is even and   holds.Then   / −2 ≤ /5.9.
Proof.As   holds, we have
Step 1: Base case ( ≤200 ).Our implementation of Algorithm 1, which uses exact arithmetic for rational numbers from Python's fractions module, gives max in which each maximum is taken only over even values of .Therefore  ≤200 holds.(While one might find a proof involving a "smaller" base case, for instance  ≤50 , we do not pursue this as we believe it would still require computer algebra.) Step 2: Showing that  ≤−4 ⇒   −2 .Suppose that  ≤ holds where  :=  − 4 ≥ 200.As   holds, we have |  / −2 + 1| ≤ / 2 , so Lemma A.2 gives Similarly, we have () As   and  −2 hold, Lemma A.1 gives and the penultimate steps of (A.6) give Applying Lemma A.3 to the sums in the following expressions, we obtain for any  ̸ = 0.By d'Alembert's ratio test, it follows that the series ∑︀ ∞ =0,  even     diverges for all non-zero .Therefore this series is not the Taylor series at  = 0 of an analytic function.This completes the proof.

Figure 1 .
Figure 1.The tangential force and reaction acting on the particle in Counterexample 1.