SIAM J. MATH. ANAL. c 2006 Society for Industrial and Applied Mathematics Vol. 38, No. 2, pp. 452–467

A UNIQUENESS CRITERION FOR THE SIGNORINI PROBLEM WITH COULOMB ∗

YVES RENARD†

Abstract. The purpose of this paper is to study the solutions to the Signorini problem with Coulomb friction (the so-called Coulomb problem). Some optimal a priori estimates are given, and a uniqueness criterion is exhibited. Recently, nonuniqueness examples have been presented in the continuous framework. It is proved, here, that if a solution satisfies a certain hypothesis on the tangential and if the friction coefficient is small enough, it is the unique solution to the problem. In particular, this result can be useful for the search of multisolutions to the Coulomb problem because it eliminates a lot of uniqueness situations.

Key words. unilateral contact, Coulomb friction, uniqueness of solution

AMS subject classifications. 35J85, 74M10

DOI. 10.1137/050635936

Introduction. The so-called Signorini problem with Coulomb friction (or simply the Coulomb problem) has been introduced by Duvaut and Lions [4]. It does not exactly represent the equilibrium of a solid which encounters an obstacle because when the equilibrium is reached (or any steady state solution) the friction condition is no longer an irregular law. The aim of this problem is in fact to be very close to a time semidiscretization of an evolutionary problem by an implicit scheme. The fact that several solutions could coexist in an implicit scheme (independently of the size of the time step) may be an indication that the evolutionary problem has a dynamical bifurcation. The first existence results for this problem were obtained by Neˇcas, Jaruˇsek, and Haslinger in [15] for a two-dimensional elastic strip, assuming that the coefficient of friction is small enough and using a shifting technique previously introduced by Fichera and later applied to more general domains by Jaruˇsek [11]. Eck and Jaruˇsek [5] give a different proof using a penalization method. We emphasize that most results on existence for frictional problems involve a condition of smallness for the friction

coefficient (and a compact support on ΓC ). Recently, examples of nonunique solutions have been given by Hild in [7] and [8] for a large friction coefficient. As far as we know, for a fixed geometry, it is still an open question whether or not there is uniqueness of the solution for a sufficiently small friction coefficient. In the finite element approximation framework, the presence of bifurcation has been studied in [9]. The present paper gives the first (partial) result of uniqueness of a solution to the Coulomb problem. The summary is the following. Section 1 introduces strong and weak formulations of the Coulomb problem. Section 2 gives optimal estimates on the solutions. In particular, a comparison is made with the solution to the frictionless contact problem. Section 3 gives an additional estimate for the Tresca problem, i.e., the problem with a given friction threshold. And finally, section 4 gives the partial

∗Received by the editors July 13, 2005; accepted for publication (in revised form) January 11, 2006; published electronically May 12, 2006. http://www.siam.org/journals/sima/38-2/63593.html †MIP, CNRS UMR 5640, INSAT, Complexe scientifique de Rangueil, 31077 Toulouse, France ([email protected]). 452

A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 453

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Fig. 1. Elastic body Ω in frictional contact. uniqueness result. It is proved in Proposition 5 for bidimensional problems and a friction coefficient less than one that there is no multisolution with one of the solutions having a tangential displacement with a constant . The major result is given by Proposition 6 using the notion of a multiplier in a pair of Sobolev spaces. 1. The Signorini problem with Coulomb friction. Let Ω ⊂ Rd (d = 2 or 3) be a bounded Lipschitz domain representing the reference configuration of a linearly elastic body. It is assumed that this body is submitted to a Neumann condition on a part of its boundary ΓN , to a Dirichlet condition on another part ΓD , and to a unilateral contact with static Coulomb friction condition on the rest of the boundary ΓC between the body and a flat rigid foundation (see Figure 1). This latter part ΓC is supposed to be of nonzero interior in the boundary ∂Ω of Ω. The problem consists in finding the displacement field u(t, x) satisfying

(1) − div σ(u)=f in Ω, (2) σ(u)=Aε(u)inΩ,

(3) σ(u)n = F on ΓN ,

(4) u = 0 on ΓD , where σ(u) is the tensor, ε(u) is the linearized strain tensor, n is the outward unit to Ω on ∂Ω, F and f are the given external loads, and A is the elastic coefficient tensor which satisfies classical conditions of symmetry and ellipticity.

On ΓC , it is usual to decompose the displacement and the stress vector in normal and tangential components as follows:

· − uN = u n,uT = u uN n, · − σN (u)=(σ(u)n) n,σT (u)=σ(u)n σN (u)n.

C1 To give a clear sense to this decomposition, we assume ΓC to have the regular- ity. The unilateral contact condition is expressed by the following complementary condition:

≤ ≤ − (5) uN g, σN (u) 0, (uN g)σN (u)=0, where g is the normal gap between the elastic solid and the rigid foundation in refer- ence configuration (see Figure 2). 454 YVES RENARD

Ω g n

   Rigid foundation  

Fig. 2. Normal gap between the elastic solid Ω and the rigid foundation.

Denoting by F≥0 the friction coefficient, the static Coulomb friction condition reads as follows: | |≤−F (6) if uT =0, then σT (u) σN (u), u (7) if u =0 , then σ (u)=Fσ (u) T . T T N | | uT The friction satisfies the so-called maximum dissipation principle − · − · (8) σT (u) uT = sup ( μT uT ). μ ∈Rd−1 T |μ |≤−Fσ (u) T N 1.1. Classical weak formulation. We present here the classical weak formu- lation proposed by Duvaut [3] and Duvaut and Lions [4]. Let us introduce the Hilbert spaces { ∈ 1 Rd } V = v H (Ω; ),v = 0 on ΓD , { ∈ }⊂ 1/2 Rd X = v| : v V H (ΓC ; ), Γ C { ∈ } { ∈ } XN = vN| : v V ,XT = vT | : v V , Γ Γ C C     and their topological dual spaces V , X , XN , and XT . It is assumed that ΓC is suffi- ⊂ 1/2 ⊂ 1/2 Rd−1  ⊂ −1/2 ciently smooth such that XN H (ΓC ), XT H (ΓC ; ), XN H (ΓC ),  ⊂ −1/2 Rd−1 and XT H (ΓC ; ). 1/2 Classically, H (ΓC ) is the space of the restriction on ΓC of traces on ∂Ωof 1 −1/2 1/2 functions of H (Ω), and H (ΓC ) is the dual space of H00 (ΓC ), which is the space 1/2 of the restrictions on ΓC of functions of H (∂Ω) vanishing outside ΓC . We refer to [1] and [12] for a detailed presentation of trace operators. Now, the of admissible displacements is defined as { ∈ ≤ } (9) K = v V,vN g a.e. on ΓC . The maps a(u, v)= Aε(u):ε(v)dx, Ω l(v)= f ·vdx + F ·vdΓ,

Ω ΓN j(Fλ ,v )=−Fλ , |v | N T N T X ,X N N A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 455 represent the virtual work of elastic , the external load, and the “virtual work” of friction forces, respectively. Standard hypotheses are as follows: (10) a(·, ·) is a bilinear symmetric V-elliptic and continuous form on V × V : ∃ ∃ ≥ 2 ≤ ∀ ∈ α>0, M>0,a(u, u) α u V ,a(u, v) M u V v V u, v V, · ∃ | |≤ ∀ ∈ (11) l( ) is a linear continuous form on V ; i.e., L>0, l(v) L v V v V, ∈ (12) g XN , F∈ (13) MXN is a nonnegative multiplier in XN . F The latter condition ensures that j( λN ,vT ) is linear continuous on λN and  convex lower semicontinuous on vT when λN is a nonpositive element of XN (see, for instance, [2]). To satisfy condition (10), it is necessary that ΓD is of nonzero interior in the boundary of Ω and that the elastic coefficient tensor is uniformly elliptic (see [4]). We refer to Maz’ya and Shaposhnikova [14] for the theory of multipliers. The set

MXN denotes the space of multipliers from XN into XN , i.e., the space of function −→ R f :ΓC of finite

fvN X N f MX = sup . N v ∈X N N vN X v =0 N N −→ ∈ This is the norm of the linear mapping XN v (fv) XN . Of course, if F F F is a constant on ΓC , one has MX = . From the fact that Ω is supposed to N C1 be a bounded Lipschitz domain and ΓC is supposed to have the regularity, it is 1/2+ε possible to deduce that for d = 2 the space H (ΓC ) is continuously included in 1 ∩ ∞ MXN for any ε>0, and for d = 3 the space H (ΓC ) L (ΓC ) is included in MXN , continuously for the norm f + f ∞ (see [14]). In particular, the space H1(Γ ) L (Γ ) C C of Lipschitz continuous functions is continuously included in MXN . Condition (10) implies in particular that a(·, ·) is a scalar product on V and that the associated norm 1/2 v a =(a(v, v)) is equivalent to the usual norm of V : √ √ ≤ ≤ ∀ ∈ α v V v a M v V v V.

The continuity constant of l(·) can also be given with respect to: · a:

∃ La > 0, |l(v)|≤La v a ∀v ∈ V.

Constants L and La can be chosen such that √ √ αLa ≤ L ≤ MLa. The classical weak formulation of problem (1)–(7) is given by ⎧ ⎨ Find u ∈ K satisfying (14) ⎩ − F − F ≥ − ∀ ∈ a(u, v u)+j( σN (u),vT ) j( σN (u),uT ) l(v u) v K. The major difficulty about (14) is due to the coupling between the friction thresh- old and the contact pressure σN (u). The consequence is that this problem does not represent a variational , in the sense that it cannot be derived from an optimization problem. 456 YVES RENARD

1.2. Neumann to Dirichlet operator. In this section, the Neumann to Dirich-

let operator on ΓC is introduced together with its basic properties. This will allow to

restrict the contact and friction problem to ΓC and obtain useful estimates. ∈  Let λ =(λN ,λT ) X ; then, under hypotheses (10) and (11), the solution u to ⎧ ⎨ Find u ∈ V satisfying (15) ⎩ a(u, v)=l(v)+λ, v ∀ v ∈ V X,X

is unique (see [4]). So it is possible to define the operator

E : X −→ X λ −→ u| . Γ C This operator is affine and continuous. Moreover, it is invertible and its inverse is − continuous. It is possible to express E 1 as follows: For w ∈ X, let u be the solution to the Dirichlet problem ⎧ ⎨ Find u ∈ V satisfying u| = w and Γ C (16) ⎩ a(u, v)=l(v) ∀ v ∈ V, v| =0; Γ C − then E 1(w) is equal to λ ∈ X defined by

  − ∀ ∈ λ, v X,X = a(u, v) l(v) v V. E−1 In a weak sense, one has the relation (u)=σ(u)n on ΓC . Now, under hypotheses (10) and (11) one has

2 1 2 C1 1 2 (17) E(λ ) − E(λ ) ≤ λ − λ  , X α X

where C1 is the continuity constant of the trace operator on ΓC and α is the coercivity constant of the bilinear form a(·, ·). One can verify it as follows. Let λ1 and λ2 be  1 2 given in XT and let u , u be the corresponding solutions to (15); then

1 − 2 2 ≤ 1 − 2 1 − 2  1 − 2 1 − 2 α u u a(u u ,u u )= λ λ ,u u  (18) V X ,X ≤ 1 − 2 1 − 2 C1 λ λ X u u V

and, consequently,

1 2 C1 1 2 (19) u − u ≤ λ − λ  . V α X Conversely, one has

E−1 1 − E−1 2 ≤ 2 1 − 2 (20) (u ) (u ) X MC2 u u X ,

where M is the continuity constant of a(·, ·) and C2 > 0 is the continuity constant of the homogeneous Dirichlet problem corresponding to (16) (i.e., with l(v) ≡ 0 and A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 457

w ∈ V ∀ ∈ C2 = sup v X v , where w| = v and a(w, z)=0 z V ). This latter estimate v=0 X Γ can be performed as follows: C

E−1 1 − E−1 2  − − (u ) (u ),v  E 1 1 − E 1 2 X ,X (u ) (u ) X = sup ∈ v X v X v=0 ⎛ ⎞ a(u1 − u2,w) = sup ⎝ inf ⎠ ∈ v X { ∈ } v X v=0 w V :w| =v Γ C ≤ 1 − 2 (21) Mγ u u V ,

w ∈ V where γ = sup v X inf v is the continuity constant of the homoge- v=0 {w∈V :w| =v} X Γ C ≤ neous Poisson problem with respect to a Dirichlet condition on ΓC . Using γ C2, this gives (20). · · It is also possible to define the following norms on ΓC relative to a( , ): v = inf w , a,ΓC a w∈V, w| =v Γ C λ, v λ, v X,X X,X λ − = sup = sup . a,ΓC v∈X v a,Γ v∈V v a v=0 C v=0

These are equivalent, respectively, to the norms in X and X: √ α √ v ≤ v ≤ Mγ v , X a,ΓC X C1

√ 1 √C1 λ  ≤ λ −a,Γ ≤ λ  . Mγ X C α X With these norms, the estimates are straightforward since the following lemma holds. Lemma 1. Let λ1 and λ2 be two elements of X and let u1 = E(λ1), u2 = E(λ2); then under hypotheses (10) and (11) one has

1 2 1 2 1 2 u − u = u − u = λ − λ − . a,ΓC a a,ΓC Proof. On the one hand, one has

u1 − u2 2 = inf w 2 = u1 − u2 2, a,ΓC w∈V a a w| =u1−u2 Γc 1 − 2 1 2 1 − 2 because u u is the minimum of 2 w a under the constraint w| = u u . This implies ΓC 1 − 2 2 1 − 2 1 − 2  1 − 2 1 − 2 u u = a(u u ,u u )= λ λ ,u u  a,ΓC X ,X 1 2 1 2 ≤ λ − λ − u − u , a,ΓC a,ΓC and finally

1 2 1 2 u − u ≤ λ − λ − . a,ΓC a,ΓC 458 YVES RENARD

On the other hand, one has

λ1 − λ2,v 1 − 2 1 2 X,X a(u u ,w) λ − λ − = sup = sup inf , a,ΓC ∈ v∈X v a,Γ v∈X w V v a,Γ v=0 C v=0 C w| =v ΓC u1 − u2 w ≤ sup inf a a = u1 − u2 = u1 − u2 , ∈ a a,ΓC v∈X w V v a,Γ v=0 C w| =v ΓC which ends the proof of the lemma. 1.3. Direct weak inclusion formulation. Let

{ ∈ ≤ } KN = vN XN : vN 0 a.e. on ΓC

be the (translated) set of admissible normal displacements on ΓC . The normal cone  ∈ in XN to KN at vN XN is defined as  NK (v )={μ ∈ X : μ ,w − v  ≤ 0 ∀w ∈ K }. N N N N N N N X ,X N N N N

In particular, N (v )=∅ if v ∈/ K . The subgradient of j(Fλ ,u ) with respect KN N N N N T to the second variable is given by F { ∈  F ∂2j( λN ,uT )= μT XT : j( λN ,vT ) ≥ j(Fλ ,u )+μ ,v − u  ∀v ∈ X }. N T T T T X ,X T T T T

With this notation, problem (14) is equivalent to the following problem: ⎧ ⎪ Find u ∈ V, λ ∈ X , and λ ∈ X satisfying ⎪ N N T T ⎪ ⎪ ⎨⎪ E (uN ,uT )= (λN ,λT ), (22) ⎪  ⎪ −λ ∈ NK (u − g)inX , ⎪ N N N N ⎪ ⎩⎪ − ∈ F  λT ∂2j( λN ,uT )inXT .

More details on this equivalence can be found in [13]. Remark 1. Inclusion −λ ∈ N (u − g) is equivalent to the complementarity N KN N relations

u ≤ g, λ ,v  ≥ 0 ∀v ∈ K , λ ,u − g =0, N N N X ,X N N N N X ,X N N N N which is the weak formulation of the strong complementarity relations (5) for the − ∈ F contact conditions. Similarly, the second inclusion λT ∂2j( λN ,uT ) represents the friction condition. 1.4. Hybrid weak inclusion formulation. We will now consider the sets of admissible stresses. The set of admissible normal stresses on ΓC can be defined as

Λ = {λ ∈ X : λ ,v  ≥ 0 ∀v ∈ K }. N N N N N X ,X N N N N A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 459

∗ This is the opposite of KN , the polar cone to KN . The set of admissible tangential stresses on ΓC can be defined as Λ (Fλ )={λ ∈ X : −λ ,w  + Fλ , |w | ≤ 0 ∀w ∈ X }. T N T T T T X ,X N T X ,X T T T T N N With this, problem (14) is equivalent to the following problem: ⎧ ⎪ ∈ ∈  ∈  ⎪ Find u V, λN X , and λT X satisfying ⎪ N T ⎪ ⎪ E ⎨ (uN ,uT )= (λN ,λT ), (23) ⎪ ⎪ −(u − g) ∈ NΛ (λ )inX , ⎪ N N N N ⎪ ⎩⎪ −u ∈ N F (λ )inX , T ΛT ( λN ) T T where the two inclusions can be replaced by inequalities as follows: ⎧ Find u ∈ V, λ ∈ X , and λ ∈ X satisfying ⎪ N N T T ⎪ ⎪ ⎪ E ⎨⎪ (uN ,uT )= (λN ,λT ), (24) ⎪ λ ∈ Λ , μ − λ ,u − g ≥ 0 ∀μ ∈ Λ , ⎪ N N N N N X ,X N N ⎪ N N ⎪ ⎪ ⎩ λ ∈ Λ (Fλ ), μ − λ ,u  ≥ 0 ∀μ ∈ Λ (Fλ ). T T N T T T X ,X T T N T T

Remark 2. The inclusion −u ∈ N F (λ ) implies the complementarity T ΛT ( λN ) T relation

λ ,u  = Fλ , |u | T T X ,X N T X ,X T T N N and the weak maximum dissipation principle

−λ ,u  = sup −μ ,u  , T T X ,X T T X ,X T T ∈ F T T μT ΛT ( λN ) which is the weak formulation of (8). 2. Optimal a priori estimates on the solutions to the Coulomb problem. For the sake of simplicity, a vanishing contact gap (g ≡ 0) will be considered in the following. Remark 3. In the case of a nonvanishing gap, it is possible to find ug ∈ V such that ug| = gn, and then w = u − ug is solution to the problem Γ C⎧ ⎪ ∈ ∈  ∈  ⎪ Find w V, λN X , and λT X satisfying ⎪ N T ⎪ ⎪ ⎪ a(w, v)=l(v) − a(ug,v)+λ ,w  + λ ,w  , ⎨ N N X ,X T T X ,X N N T T (25) ⎪ ⎪ ⎪ −w ∈ NΛ (λ )inX , ⎪ N N N N ⎪ ⎩⎪ −w ∈ N F (λ )inX , T ΛT ( λN ) T T i.e., a contact problem without gap but with a modified source term. 460 YVES RENARD

Following Remarks 1 and 2, a solution (u, λ) to problem (22) (i.e., a solution u to problem (14)) satisfies the complementarity relations

λ ,u  =0, N N X ,X N N λ ,u  = Fλ , |u | . T T X ,X N T X ,X T T N N This implies

λ, u ≤ 0, X,X which expresses the dissipativity of contact and friction conditions. The first conse- quence of this is that solutions to problem (14) can be bounded independently of the friction coefficient. Proposition 1. Assuming hypotheses (10), (11), and (13) are satisfied and g ≡ 0, let (u, λ) be a solution to problem (22), which means that u is a solution to problem (14); then

u ≤ L , λ − ≤ L , a a a,ΓC a L M u ≤ , λ  ≤ Lγ . V α X α Proof. One has

2 u = a(u, u)=l(u)+λ, u ≤ La u a, a X,X which states the first estimates. The estimate on λ − can be performed using a,ΓC the intermediary solution uN to the following problem with a homogeneous Neumann condition on ΓC : (26) a(uN ,v)=l(v) ∀v ∈ V.

N Since u a ≤ La for the same reason as for u, and using Lemma 1, one has − 2 − N − N  − N  ≤− N  ≤ λ 0 − = a(u u ,u u )= λ, u u  λ, u  La λ −a,Γ . a,ΓC X ,X X ,X C The two last estimates can be stated thanks to equivalence of norms introduced in section 1.1. c It is possible to compare u a to the corresponding norm of the solution u to the Signorini problem without friction defined as follows: ⎧ ⎨ Find uc ∈ K satisfying (27) ⎩ a(uc,v− uc) ≥ l(v − uc) ∀ v ∈ K.

It is well known that under hypotheses (10) and (11), this problem has a unique solution (see [12]). Proposition 2. Assuming hypotheses (10), (11), and (13) are satisfied and g ≡ 0, let u be a solution to problem (14), let uc be the unique solution to problem (27), and let uN be the solution to problem (26); then

c N u a ≤ u a ≤ u a. A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 461

Proof. One has

a(uN ,uN )=l(uN ),a(uc,uc)=l(uc),a(u, u)=l(u)+λ ,u  . T T X ,X T T Since uc is the solution to the Signorini problem without friction, it minimizes over K 1 − N the energy 2 a(v, v) l(v). The solution u minimizes this energy functional over V . Thus, since u ∈ K, one has 1 1 1 a(uN ,uN ) − l(uN ) ≤ a(uc,uc) − l(uc) ≤ a(u, u) − l(u), 2 2 2 and the following relations allow one to conclude that

a(uc,uc) − a(uN ,uN )=l(uc − uN ),

a(u, u) − a(uc,uc)=l(u − uc)+λ ,u  T T X ,X T T because then 1 1 a(uc,uc) − a(uN ,uN ) ≤ 0, 2 2 1 1 1 a(u, u) ≤ a(uc,uc)+Fλ , |u | ≤ a(uc,uc). N T X ,X 2 2 N N 2 It is also possible to estimate how far from uc is a solution u to problem (14). Let ∈ us introduce the following norms on ΓC .Forv X let us define

vT a,Γ = inf w a = inf z a,Γ , C w∈V z∈X C w =v z =v T T T T

vN a,Γ = inf w a = inf v a,Γ . C w∈V z∈X C w =v z =v N N N N One has

v ≤ v , v ≤ v . T a,ΓC a,ΓC N a,ΓC a,ΓC Now, for λ ∈ X, let us define λ ,v  λ ,v  T T X ,X T T X ,X T T λ − = sup T = sup T , T a,ΓC v ∈X v a,Γ v∈X v a,Γ T T T C  C v =0 v=0 T λ ,v  λ ,v  N N X ,X N N X ,X N N λ − = sup N = sup N . N a,ΓC v ∈X v a,Γ v∈X v a,Γ N N N C  C v =0 v=0 N Then, the following equivalences of norms are immediate:

√ α √ v ≤ v ≤ γ M v , N X N a,ΓC N X C1 N N √ α √ v ≤ v ≤ γ M v , T X T a,ΓC T X C1 T T 462 YVES RENARD

1 C1 √ λ  ≤ λ − ≤ √ λ  , N X N a,ΓC N X γ M N α N

1 C1 √ λ  ≤ λ − ≤ √ λ  . T X T a,ΓC T X γ M T α T And the following result can be easily deduced.  Lemma 2. There exists C3 > 0 such that for all λ ∈ X

λ − ≤ C λ − , λ − ≤ C λ − . T a,ΓC 3 a,ΓC N a,ΓC 3 a,ΓC F∈ This also allows us to define an equivalent norm on MXN given for MXN by

Fv N a,ΓC F a = sup , v ∈X v N N N a,ΓC v =0 N which satisfies √ √ √α F ≤ F ≤ C1√γ M F a MX a. N C1γ M α With these definitions, the following result holds. Lemma 3. There exists C4 > 0 such that

F|v | ≤ C F v ∀v ∈ X . T a,ΓC 4 a T a,ΓC T T Proof. One has

F|v | ≤ F |v | . T a,ΓC a T a,ΓC · Moreover, it is known (see [1]) that the norm X is equivalent to the norm N |v (x) − v (y)|2 v 2 = v 2 + N N dxdy, N 1/2,Γ N L2(Γ ) d C C |x − y| ΓC ΓC and it is easy to verify that |v | ≤ v for any v ∈ X . Thus, the T 1/2,ΓC T 1/2,ΓC T T result can be deduced from the previously presented equivalences of norms. c Of course the tangential stress on ΓC corresponding to u is vanishing. The ∈ F tangential stress corresponding to u can be estimated as follows. As λT ΛT ( λN ), one has λ ,v  −Fλ , |v | T T X ,X N T X ,X T N λ − = sup T ≤ sup N T a,ΓC v ∈X v v ∈X v T T T a,ΓC T T T a,ΓC v =0 v =0 T T ≤ C F λ − . 4 a N a,ΓC Now, with the result of Proposition 1 this means that

(28) λ − ≤ L C C F , T a,ΓC a 3 4 a and the following result holds. A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 463

. u a uc a    

0 F a

.

Fig. 3. Admissibility zone for ua.

Proposition 3. Assuming hypotheses (10), (11), and (13) are satisfied and g ≡ 0, let u be a solution to problem (14) and let uc be the solution to problem (27); then c u − u a ≤ LaC3C4 F a, LC C uc − u ≤ 3 4 F . V α a ∈  c ∈  Proof. With λ X and λ X the corresponding stresses on ΓC , because −λc ∈ N (uc ) and −λ ∈ N (u − g) and because of the fact that N is a N KN N N KN N KN monotone set-valued map, one has  c − c −  ≤ λ λN ,u uN 0. N N X ,X N N c Now, u − u a can be estimated as follows: c 2 c c c c c u − u = a(u − u, u − u)=λ − λ, u − u ≤ λ −a,Γ u − u a, a X,X T C which gives the result taking into account (28). The latter result implies that if problem (14) has several solutions, then they are c 1 2 in a ball of radius LaC3C4 F a centered around u . In particular, if u and u are 1 2 two solutions to problem (14), one has u −u a ≤ 2LaC3C4 F a. This is illustrated by Figure 3. 4. F Remark For a friction coefficient constant on ΓC , the graph in Figure 3 can be more precise for F = F a small, since, from the proof of Proposition 2 and 2 ≤ c 2 the continuity result given by the latter proposition, one can deduce u a u a + Fλc , |uc | at least if λc , |uc | < 0. Of course, if λc , |uc | =0, N T X ,X N T X ,X N T X ,X N N N N N N the solution uc to the Signorini problem without friction is also a solution to the Coulomb problem for any friction coefficient. 3. Elementary estimates on the Tresca problem. What is usually called ∈  the Tresca problem is the friction problem with a given friction threshold. Let θ XN be given. Then it can be formulated as follows: ⎧ ⎨ Find u ∈ K satisfying (29) ⎩ − − ≥ − ∀ ∈ a(u, v u)+j(θ, vT ) j(θ, uT ) l(v u) v K. 464 YVES RENARD

It is well known that under standard hypotheses (10), (11), and (13), this problem 1 − has a unique solution (see [12]) which minimizes the functional 2 a(u, u)+j(θ, u) l(u). In fact, it is not difficult to verify that all the estimates given in the previous section for the solutions to the Coulomb problem are still valid for the solution to the Tresca problem. Moreover, the solution to the Tresca problem continuously depends on the friction threshold θ. This result is stated in the following lemma. Lemma 4. Assuming hypotheses (10), (11), and (13) are satisfied, if u1, u2 are the 1 ∈ 2 ∈ solutions to problem (29) for a friction threshold θ ΛN and θ ΛN , respectively, 1 2 then there exists a constant C5 > 0 independent of θ and θ such that the following estimate holds:

1 2 2 1 2 u − u ≤ C θ − θ − . a 5 a,ΓC Proof. One has

a(u1,u2 − u1) − l(u2 − u1)+j(θ1,u2) − j(θ1,u1) ≥ 0,

a(u2,u1 − u2) − l(u1 − u2)+j(θ2,u1) − j(θ2,u2) ≥ 0,

which implies

1 2 2 1 2 1 2 u − u ≤θ − θ , |u |−|u |  , a T T X ,X N N

which gives the estimate using Proposition 1 (in fact, C5 ≤ 2C4La). Remark 5. It does not seem possible to establish a Lipschitz continuity with respect to the friction threshold θ. Such a result would automatically imply the uniqueness of the solution to the Coulomb problem for a sufficiently small friction coefficient. 4. A uniqueness criterion. Hild in [7, 8] exhibits some multisolutions for the Coulomb problem on triangular domains. These solutions have been obtained for a large friction coefficient (F > 1) and for a tangential displacement having a constant sign. For the moment, it seems that no multisolution has been exhibited for an arbitrary small friction coefficient in the continuous case, although such a result exists for finite element approximation in [6], albeit for a variable geometry. As far as we know, no uniqueness result has been proved even for a sufficiently small friction coefficient. The result presented here is a partial uniqueness result, which determines some cases where it is possible to say that a particular solution of the Coulomb problem is in fact the unique solution. A contrario, this result can be used to search multisolutions for an arbitrary small friction coefficient, by the fact that it eliminates a lot of situations. The partial uniqueness results we present in this section are deduced from the estimate given by the following lemma. Lemma 5. Assuming hypotheses (10), (11), and (13) are satisfied and g ≡ 0,ifu1 and u2 are two solutions to problem (14) and λ1 and λ2 are the corresponding contact

stresses on ΓC , then one has the following estimate: 1 − 2 2 1 − 2 2 ≤− 2 1 − 2  ∀ ∈− F 1 2 u u a = λ λ −a,Γ ζ λ ,u u ζ ∂2j( λ ,u ). C T T T X ,X N T T T Proof. One has

1− 2 2 1− 2 2  1 − 2 1 − 2   1 − 2 1 − 2  u u a = λ λ −a,Γ = λ λ ,u u + λ λ ,u u . C N N N N X ,X T T T T X ,X N N T T A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 465

1 2 1 2 Because NK is a monotone set-valued map, one has λ − λ ,u − u  ≤ 0. N N N N N X ,X N N Thus

1 − 2 2 ≤ 1 − − 2 1 − 2  ∀ ∈− F 1 2 u u a (λ ζ)+(ζ λ ),u u ζ ∂2j( λ ,u ). T T T T X ,X N T T T F But ∂2j( λN ,uT ) is also a monotone set-valued map with respect to its second vari- 1 2 2 − ≤ 1 2 − able, which implies the result (and also the fact that u u a infζ∈−∂2j(Fλ ,u ) ζ N T 2 λ − ). T a,ΓC An immediate consequence of this lemma is the following result for a vanishing tangential displacement. Proposition 4. Assuming hypotheses (10), (11), and (13) are satisfied and g ≡ F 0,ifu is a solution to problem (14) such that uT =0a.e. on ΓC and if C3C4 a < 1, then u is the unique solution to problem (14). Proof. Let us assume that u is another solution to problem (14). Then from Lemma 5 one has

2 u − u ≤ζ − λ ,u − u  ∀ζ ∈−∂2j(Fλ , u ), a T T T X ,X N T T T

but, because u = 0 and due to the complementarity relations λ , u  = T T T X ,X T T Fλ , |u | and ζ,u  = Fλ , |u | , it implies using Lemma 3 N T X ,X T X ,X N T X ,X N N T T N N that

− 2 ≤F − | | ≤ F − − 2 u u (λN λN ), uT  C3C4 a λ λ −a,Γ uT uT a X ,X C a,ΓC N N ≤ F − 2 C3C4 a u u a,

which concludes the proof. In the case d = 2, it is possible to give a result to a solution having a tangential displacement with a constant sign on ΓC . We will say that a tangential displacement u ∈ X is strictly positive if μ ,u  > 0 for all μ ∈ X such that μ ≥ 0 T T T T X ,X T T T T T (i.e., μ ,v  ≥ 0 for all v ∈ X , v ≥ 0, a.e. on Γ ) and μ =0. T T X ,X T T T C T T T Proposition 5. Assuming hypotheses (10), (11), and (13) are satisfied, g ≡ 0, F and d =2,ifu is a solution to problem (14) such that uT > 0 and C3 a < 1, then F u is the unique solution to problem (14) (when is constant over ΓC , the condition reduces to C3F < 1).

Proof. Let us assume that u is another solution to problem (14), with λN and λT

the corresponding contact stresses on ΓC . Then from Lemma 5 one has

2 u − u ≤ζ − λ , u − u  ∀ζ ∈−∂2j(Fλ ,u ). a T T T X ,X N T T T F − F F Because uT > 0, one has λT = λN and ∂2j( λN ,uT ) contains λN . Thus, taking F ζ = λN , one obtains

2 2 u−u ≤F(λ − λ ), u − u  ≤ λ−λ −a,Γ F(u−u) a ≤ F a u−u , a N N T T X ,X C a T T

which implies u = u when F a < 1.

Of course, the same reasoning is valid for uT < 0. 466 YVES RENARD

. ξ +1

0

-1 Γ C u T 0

Γ C

.

Fig. 4 u ξ . Example of tangential displacement T and a possible corresponding multiplier for d =2.

→ Let us now define the space of multipliers M(XT XN ) of the functions ξ : → Rd · ΓC such that ξ n = 0 a.e. on ΓC and such that the following two equivalent norms are finite: · · ξ vT X ξ v a,Γ N T C ξ M(X →X ) = sup and ξ a = sup . T N v ∈X v v ∈X v T T T X T T T a,ΓC v =0 T v =0 T T C1 → Because ΓC is assumed to have the regularity, M(XT XN ) is isomorphic to d−1 (MXN ) . F It is possible to give a more general result assuming that λT = λN ξ, with ∈ → | |≤ ξ M(XT XN ). It is easy to see that this implies that ξ 1 a.e. on the support ∈ of λN and, more precisely, that ξ DirT (uT ) a.e. on the support of λN , where DirT (.) d is the subderivative of the convex map R x −→ | xT |. This means that it is ∈ reasonable to assume that ξ DirT (uT ) a.e. on ΓC . Proposition 6. Assuming hypotheses (10), (11), and (13) are satisfied and g ≡ F ∈ → 0,ifu is a solution to problem (14) such that λT = λN ξ, with ξ M(XT XN ), ∈ F ξ DirT (uT ) a.e. on ΓC , and C3 a ξ a < 1, then u is the unique solution to problem (14).

Proof. Let us assume that u is another solution to problem (14), with λN and λT the corresponding contact stresses on ΓC . Then from Lemma 5 one has 2 u − u ≤ζ − λ , u − u  ∀ζ ∈−∂2j(Fλ ,u ). a T T T X ,X N T T T F F ≤ Then, a possible choice is ζ = λN ξ, which, together with the fact that ξ a F a ξ a, gives − 2 ≤F − −  ≤ F − − u u ξ(λN λN ), uT uT  C3 a ξ a λ λ −a,Γ u u a a X ,X C T T ≤ F − 2 C3 a ξ a u u a, which implies u = u when C3 F a ξ a < 1. Remark 6. Using equivalences of norms, one can deduce that a more restrictive√ F F α√ condition than C3 a ξ a < 1 is the condition MX ξ M(X →X ) < . N T N C1C3γ M As illustrated in Figure 4, for d = 2, the multiplier ξ has to vary from −1to+1 each time the sign of the tangential displacement changes from negative to positive. A UNIQUENESS CRITERION FOR THE COULOMB PROBLEM 467

→ The set M(XT XN ) does not contain any multiplier having a discontinuity of the first kind. This implies that in order to satisfy the assumptions of Proposition 6 the tangential displacement of the solution u cannot pass from a negative value to a positive value, being zero on only a single point of ΓC . Perspectives. As far as we know, the result given by Propositions 4, 5, and 6 are the first results dealing with the uniqueness of the solution to the Coulomb problem without considering a regularization of the contact or the friction law. In the future, it may be interesting to investigate the following open problems. Is it possible to prove that, for a sufficiently regular domain and a sufficiently F regular loading, a solution of the Coulomb problem is necessarily such that λT = λN ξ ∈ → with ξ M(XT XN )? This could be a way to prove a uniqueness result for a sufficiently small friction coefficient and regular loadings.

The more the tangential displacement uT oscillates around 0 (i.e., the more uT changes its sign for d = 2), the more the multiplier ξ varies and thus the greater

ξ M(X →X ) is. Does it mean that a multisolution for an arbitrary small friction T N coefficient and a fixed geometry has to be searched with very oscillating tangential displacement (necessarily for all the solutions)? Finally, the convergence of finite element methods in the uniqueness framework given by Proposition 6 will be presented in [10].

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