# Existence of a Generalized Solution for the Fractional Contact Problem

Authors: Leila Ait kaki; Nouria Arar; Mohammed S. Abdo; M. Daher Albalwi
Journal: Journal of Mathematics
 (2023)

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## Abstract

In this paper, we take into consideration the mathematical analysis of time-dependent quasistatic processes involving the contact between a solid body and an extremely rigid structure, referred to as a foundation. It is assumed that the constitutive law is fractional long-memory viscoelastic. The contact is considered to be bilateral and is modeled around Tresca’s law. We establish the existence of the generalized solution’s result. The proof is supported by the surjectivity of the multivalued maximum monotone operator, Rothe’s semidiscretization method, and arguments for evolutionary variational inequality.


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## Body

## 1. Introduction

It is well known that the empirical models containing the fractional derivative fit the experimental data more precisely than the model containing the integer derivative. The early application of the fractional derivative in viscoelasticity dates back to the beginning of this century. Germent and Baglet proposed the fractional Kelvin–Voigt model, and at the same time, Koeller obtained the fractional Maxwell model [1, 2]. The theoretical and numerical investigation has been done for fractional differential equations and inequalities in finite-dimensional spaces. On this subject, we cite, for instance [3–6]. The application of the variational framework, more precisely the variational inequalities, proves to be very useful in the study of engineering and mechanical problems. It was initiated by Panagiotopoulos; see monographs [7, 8], and several results have emerged; see [9–13]. In this paper, we are interested in studying the contact problem for a fractional viscoelastic law with long-term memory. The corresponding constitutive law is(1)σ=AεDδβ0Cw+Bεw+∫0δCt−μεwμdμ,where A,B, and C represent the fractional viscoelasticity, elasticity operator, and relaxation tensor, σ and w stand for the stress and the displacement fields, respectively, and Dδβ0C is related to Caputo fractional derivative of β order and 0<β<1.The long-memory viscoelastic constitutive law has been the subject of numerous papers; see [14–16]. The distinction of this study is the selection of contact with Tresca’s friction type, governed by a fractional constitutive law, and the use of Rothe’s approach and the maximum monotone operators’ surjectivity results to demonstrate the existence of a solution to a variational inequality. Actually, there are only a few contributions to the literature on the resolution of variational and hemivariational inequalities using Rothe’s approach; see, for example, [18–21].This article is structured according to the following outlines: in Section2, we state the equilibrium equations modeling the process of viscoelastic material coming into contact with a foundation. We give some of the notations and the assumptions on data. Then, we obtain the variational formulation of the problem and present the main results concerning the existence of a weak solution. In Section 3, we first prove the existence of the discrete generalized solution to the discrete problem, in the second, we establish the a priori estimates, and in the last, we pass to the limit to obtain the existence of solutions to the variational problem.


## 2. Setting of the Problem

Before setting up the physical model, we give some definitions and preliminary results.Definition 1.
For any positive integernandn−1≤s<n, the Caputo fractional derivative and fractional integral of orders of a given function w∈L10,b, are respectively defined as(2)Dδs0wδ=1Γn−s∫0δwnτdτδ−τs−n+1,∀δ∈0,b,Dδ−s0wδ=1Γs∫0δδ−μs−1wμdμ,∀δ∈0,b,where Γ⋅ is the standard Gamma function.Definition 2.
LetW be a real Banach space. An operator A:W⟶W′is said to be(a)
Aest hemicontinuous if the real function δ↦Au+δϑ,w is continuous from [0,1] in ℝ, for all u,ϑ,w∈W.(b)
bounded, ifA maps bounded sets of W into bounded sets of W′.(c)
A:W→2W′is monotone if, y−y0,x−x0≥0,∀y0∈Ax0,y∈Ax.(d)
A is maximal monotone if, A is monotone, and for all x∈DA,y∈2W′, such that y−z,x−x0≥0,∀x0∈W,z∈Ax0, then y∈Ax.(e)
The generalized gradient (subdifferential) of a convex functionℑatu is a subset of the dual space W′given by(3)∂ℑw=ϱ∈W′:ℑw−ℑϑ≤ϱ,w−ϑ,∀ϑ∈W.(f)
A sequencewn⊂W is weakly convergent to w∈W, noted wn⇀w, if for all m′∈W′, we have(4)limn⟶+∞wnw′W,W′=w,w′W,W′,where⋅,⋅W,W′ represents the duality between W and W′.Theorem 1 (see [22]).
LetF:W⟶−∞,+∞is convex, lower semicontinuous, andF≠+∞. Then,∂Fis maximal monotone.Theorem 2 (see [22]).
Let the following hypotheses holdH1: the operatorA:W→2W′is maximal monotone.H2: the operatorℬ:W→W′is monotone hemicontinuous and bounded.H3: B is A-coercive, meaning that a point u0∈DA and a number r>0 exist, that way(5)Bu,u−u0u⟶+∞whenu⟶+∞.Hence,A+B is surjective.


### 2.1. Physical Setting

Suppose that in the reference configuration, the domainD⊂ℝdd=2,3 which is considered bounded with a smooth boundary ∂D=Σ. Three open, disconnected sections make up this boundary: ΣD, Σℵ, and ΣC such that measΣD>0. Let 0,b time interval of interest, where b>0.The body is submitted to the action of body forces of densityf0. It also submitted to mechanical constraints on the boundary. The body is supposed squeezed into ΣD, thus the displacement field vanishes over ΣD. On Σℵ, there are the density of traction forces denoted fα. At some point, the body comes into contact over ΣC.Let denote byTd the space of the second order of symmetric tensors and by ⋅ and ⋅, respectively, the scalar product and the Euclidean norm in Td (resp in ℝd). We adopt the summation convention over repeated indices. All indices ι, j take values from 1 to d.(6)σ.h=σιjhιj,h=h.h1/2,∀σ,h∈Td,w.ϑ=wιϑι,w=w.w1/2,∀w,ϑ∈ℝd.LetD⊂ℝd, we shall use the notation(7)H=w=wι|wι∈L2D=L2Dd,H=σ=σιj|σιj=σjι∈L2D,H1=w=wι|εw∈H,H1=σ∈H|Divσ∈H,with ϵ:H⟶H and Div:H⟶H are, respectively, operators of deformation and divergence defined by(8)εw=ειjw,ειjw=12wιj+wjιand Divσ=σιj,j.The spaceH, H, H1, and H1 are Hilbert spaces endowed with the inner products given by(9)w,ϑH=∫Dwιϑιdx,σ,hℌ=∫Dσιjhιjdx,w,ϑH1=w,ϑH+εw,εϑℌ,σ,hℌ1=σ,hℌ+Divσ,DivhH.We use the notationϑ for the trace of ϑ on Σ. The normal and tangential components of ϑ on Σ are represented by the symbols ϑν and ϑτ, respectively, and are determined by the formulas ϑν=ϑ⋅ν, ϑτ=ϑ−ϑνν, σν=σν⋅ν, and στ=σν−σνν. The trace of the regular stress field σ applied to ν is noted σν.The physical model for the process is as follows:Problem 1.
Find a displacement fieldw:D×0,b⟶ℝd(10)σ=AεDδβ0Cw+Bεw+∫0δℭδ−μεwμdμinD×0,b,(11)Divσ+f0=0inD×0,b.
The boundary and initial conditions(12)w=0,onΣD×0,b,(13)σν=fℵ,onΣℵ×0,b,(14)σν=0,onΣC×0,b,(15)στ≤g,wτ≠0⇒στ=−gwτwτ,onΣC×0,b,(16)w0=w0,inD.For the sake of simplifying the notations in equations (10)–(16) and below, we do not show explicitly the dependence of all functions on, respectively, space and time variables x∈D∪Σ and δ∈0,b. The fractional viscoelastic constituency with memory law is given by (10), for 0<β<1, where A is linear viscosity function, B is linear elasticity, and C the linear relaxation function. The expression (11) is the equilibrium equation, we suppose the process is quasistatic. Here, the boundary conditions on displacement and traction are represented by (12) and (13). Conditions (14) and (15) are the bilateral contact and the friction conditions, respectively, it is modeled by Tresca’s law with an imposed friction bound g (hence g≥0 and the nonseparation condition uν=0). The equation given in (16) is the initial condition on displacement. To be able to give a variational formulation of the above problem, we give the following additional notations. We set(17)W=w∈H1∣w=0,onΣD,W˜=w∈W∣Dδβ0Cw∈W,equipped with the inner product w,ϑW=ϵw,ϵϑH1 for w, v∈W. In view of measΣD>0, there exist, respectively, CK>0 and cF>0, called Korn’s and Poincaré constants, such that(18)εϑH1≥cKvH1,∇ϱH≥cFϱH1,ϱ∈W.Let denote byV=L20,b;W the Bochner–Lebesgue function space. Since the space W is reflexive, it remains that V and its dual space V′=L20,b;W′ are reflexive too. To study the problem and present our main result, we make the following assumptions: the viscosity operator A, the elasticity one B, and the relaxation tensor C satisfy the conditions(19)aA:D×Td⟶Td,A=aιjkℓbAx,h=aιjkℓxhjk,∀h=hιj∈Td,a.e.x∈D.caιjkℓ∈L∞D,∀ι,j,k,ℓdthere existmA>0,such thatAx,ϱ⋅ϱ≥mAϱ2,for allϱ∈Td.(20)aB:D×Td→Td,B=bιjkℓ;bιjkℓ∈L∞D,∀ι,j,k,ℓ.bBx,h=bιjkℓxhjk,∀h=hιj∈Td,a.e.x∈D.(21)aℭ:D×0,b×Td→Td,ℭ=e˜ιjkℓ;e˜ιjkℓ∈L∞D,bℭx,δ,h=e˜ιjkℓx,δhjk,∀h=hιj∈Td,cℭis Lipschitz continuous with constantLℭ>0.The initial data and the density forces satisfy(22)w0∈W,f0∈L∞0,b;L2Dd,fℵ∈L∞0,b;L2Σℵd.


## 2.1. Physical Setting

Suppose that in the reference configuration, the domainD⊂ℝdd=2,3 which is considered bounded with a smooth boundary ∂D=Σ. Three open, disconnected sections make up this boundary: ΣD, Σℵ, and ΣC such that measΣD>0. Let 0,b time interval of interest, where b>0.The body is submitted to the action of body forces of densityf0. It also submitted to mechanical constraints on the boundary. The body is supposed squeezed into ΣD, thus the displacement field vanishes over ΣD. On Σℵ, there are the density of traction forces denoted fα. At some point, the body comes into contact over ΣC.Let denote byTd the space of the second order of symmetric tensors and by ⋅ and ⋅, respectively, the scalar product and the Euclidean norm in Td (resp in ℝd). We adopt the summation convention over repeated indices. All indices ι, j take values from 1 to d.(6)σ.h=σιjhιj,h=h.h1/2,∀σ,h∈Td,w.ϑ=wιϑι,w=w.w1/2,∀w,ϑ∈ℝd.LetD⊂ℝd, we shall use the notation(7)H=w=wι|wι∈L2D=L2Dd,H=σ=σιj|σιj=σjι∈L2D,H1=w=wι|εw∈H,H1=σ∈H|Divσ∈H,with ϵ:H⟶H and Div:H⟶H are, respectively, operators of deformation and divergence defined by(8)εw=ειjw,ειjw=12wιj+wjιand Divσ=σιj,j.The spaceH, H, H1, and H1 are Hilbert spaces endowed with the inner products given by(9)w,ϑH=∫Dwιϑιdx,σ,hℌ=∫Dσιjhιjdx,w,ϑH1=w,ϑH+εw,εϑℌ,σ,hℌ1=σ,hℌ+Divσ,DivhH.We use the notationϑ for the trace of ϑ on Σ. The normal and tangential components of ϑ on Σ are represented by the symbols ϑν and ϑτ, respectively, and are determined by the formulas ϑν=ϑ⋅ν, ϑτ=ϑ−ϑνν, σν=σν⋅ν, and στ=σν−σνν. The trace of the regular stress field σ applied to ν is noted σν.The physical model for the process is as follows:Problem 1.
Find a displacement fieldw:D×0,b⟶ℝd(10)σ=AεDδβ0Cw+Bεw+∫0δℭδ−μεwμdμinD×0,b,(11)Divσ+f0=0inD×0,b.
The boundary and initial conditions(12)w=0,onΣD×0,b,(13)σν=fℵ,onΣℵ×0,b,(14)σν=0,onΣC×0,b,(15)στ≤g,wτ≠0⇒στ=−gwτwτ,onΣC×0,b,(16)w0=w0,inD.For the sake of simplifying the notations in equations (10)–(16) and below, we do not show explicitly the dependence of all functions on, respectively, space and time variables x∈D∪Σ and δ∈0,b. The fractional viscoelastic constituency with memory law is given by (10), for 0<β<1, where A is linear viscosity function, B is linear elasticity, and C the linear relaxation function. The expression (11) is the equilibrium equation, we suppose the process is quasistatic. Here, the boundary conditions on displacement and traction are represented by (12) and (13). Conditions (14) and (15) are the bilateral contact and the friction conditions, respectively, it is modeled by Tresca’s law with an imposed friction bound g (hence g≥0 and the nonseparation condition uν=0). The equation given in (16) is the initial condition on displacement. To be able to give a variational formulation of the above problem, we give the following additional notations. We set(17)W=w∈H1∣w=0,onΣD,W˜=w∈W∣Dδβ0Cw∈W,equipped with the inner product w,ϑW=ϵw,ϵϑH1 for w, v∈W. In view of measΣD>0, there exist, respectively, CK>0 and cF>0, called Korn’s and Poincaré constants, such that(18)εϑH1≥cKvH1,∇ϱH≥cFϱH1,ϱ∈W.Let denote byV=L20,b;W the Bochner–Lebesgue function space. Since the space W is reflexive, it remains that V and its dual space V′=L20,b;W′ are reflexive too. To study the problem and present our main result, we make the following assumptions: the viscosity operator A, the elasticity one B, and the relaxation tensor C satisfy the conditions(19)aA:D×Td⟶Td,A=aιjkℓbAx,h=aιjkℓxhjk,∀h=hιj∈Td,a.e.x∈D.caιjkℓ∈L∞D,∀ι,j,k,ℓdthere existmA>0,such thatAx,ϱ⋅ϱ≥mAϱ2,for allϱ∈Td.(20)aB:D×Td→Td,B=bιjkℓ;bιjkℓ∈L∞D,∀ι,j,k,ℓ.bBx,h=bιjkℓxhjk,∀h=hιj∈Td,a.e.x∈D.(21)aℭ:D×0,b×Td→Td,ℭ=e˜ιjkℓ;e˜ιjkℓ∈L∞D,bℭx,δ,h=e˜ιjkℓx,δhjk,∀h=hιj∈Td,cℭis Lipschitz continuous with constantLℭ>0.The initial data and the density forces satisfy(22)w0∈W,f0∈L∞0,b;L2Dd,fℵ∈L∞0,b;L2Σℵd.


## 3. Variational Formulation of the Problem

In this section, we establish the existence of a weak solution to Problem1. First, we multiply the equilibrium equation by the test function, taking into account the initial and boundary conditions, and we obtain Problem 2 in the form of a variational inequality. Then, we reformulate the problem in operational form (Problem 3), which allows us to pass from a heavy scripture relating to tensorial products to a more convenient one. In order to apply the classical results of functional inclusions, we rewrite the last problem in the form of a nonlinear inclusion, thus giving Problem 4. Note that Problem 2, 3, and 4 are equivalent.To obtain the variational formulation of Problem1, we assume that the pair of functions u and σ are sufficiently smooth. We recall that Green’s formula holds:(23)σ,εϑH+Divσ,ϑH=∫Σσν⋅ϑ,∀ϑ∈H1.From the (11), the boundary conditions (12)–(15) and equality (23), we obtain for all ϑ∈W(24)σ,εϑH=∫Df0ϑ+∫Σℵfℵϑ+∫ΣCστϑ,=f,ϑ+∫ΣCστϑτ.Further,(25)σ,εϑH+∫ΣCgϑτ≥f,ϑ,∀ϑ∈W.Let defineJϑ=∫ΣCgϑτ, for all ϑ∈W, expression (25) leads to the following variational formulation:Problem 2.
Findw:0,b⟶W˜, such that(26)AεDδβ0Cwδ,εϑ−εwδH+Bεwδ,εϑ−εwδH+∫0δCδ−μεwμdμ,εϑ−εwδH+Jϑ−Jwδ≥fδ,εϑ−εwδW′,W,∀ϑ∈W,a.e.δ∈0,b.Let define the operatorsA, B, and C by the expressions(27)Awδ,ϑW′,W=Aεwδ,εϑℌ,ℬwδ,ϑW′,W=Bεwδ,εϑℌ,Cwδ,ϑW′,W=∫0δcδ,μwμdμ,ϑW′,W=∫0tℭt−sεwsds,εvℌ,where c is defined below. Note that since J is convex, we have(28)Jw−Jϑ≤ϱ,w−ϑ,∀ϱ∈∂Jw,∀ϑ∈W.The following problem can be established in light of all those considerations.Problem 3.
Findw∈L10,T;W˜, such thatx(29)ADδβ0Cwδ,ϑ−wδ+ℬwδ,ϑ−wδ+Cwδ,ϑ−wδ+Jϑ−Jwδ≥fδ,ϑ−wδW′,for allϑ∈Wanda.e.δ∈0,b.Let now suppose thatw∈AC0,b;W be a solution to Problem 3 and put u=Dδβ0Cw, for 0<β<1 where Dδβ0C is the Caputo fractional derivative. Note that when w∈AC0,b;W, then wδ=Dδ−β0uδ+w0 have a sense.So, we may rewrite our problem as a nonlinear inclusion, driven by a fractional integral operator, and it is given by the problem below.Problem 4.
Findu∈L10,b;W, such that(30)Auδ+BDδ−β0uδ+w0+CDδ−β0uδ+w0+∂JDδ−β0uδ+w0∋fδ,fora.e.δ∈0,b.Using data from Problems3 and 4, we reformulate the assumptions.HA : A∈LW,W′ is coercive, i.e., there exists a constant m˜A>0, such that(31)Aϑ,ϑ≥m˜Aϑ2,for allϑ∈W.HB : B∈LW,W′. Hc  c∈C0,b2,LW;W′ there exists Lc>0, such that(32)cμ,δ1−cμ,δ2W′≤Lcδ2−δ1,μ,δ1,δ2∈0,b, and m˜c=maxδ,μ∈0,b2cδ,μ. Hf  f∈L∞0,b;W′.Letℵ∈ℕ+ be fixed, h=b/ℵ, δk=kh, and for all f∈L10,b;W, ghk be defined by ghk=1/h∫δk−1δkgμdμ, for k=1,…,ℵ. Consider the so-called Rothe problem, i.e., the discretized problem corresponding to Problem 4, given by as follows:Problem 5.
Finduhkk=1ℵ⊂W, such that uh0=0, and(33)Auhk+Bwhk+Cwhk+∂Jwhk∋fhk,where whk and Cwhk for δ∈0,δk, and k=1,2,…,ℵ, are defined by(34)whk=w0+hβΓβ+1∑j=1kuhjk−j+1β−k−jβ,(35)Cwhk=∑j=1k∫δj−1δjcδk,μwhjdμ.First, we shall show the existence of solution for a Problem5.Theorem 3.
Suppose that conditionsHA, HB, Hc, andHfare verified. Hence, there existsh0>0, such that for allh∈0,h0, there is at least one solution to Problem 5.Proof.
Givenuh0,uh1,…,uhn−1, we claim that there exist uhn∈W, such that (33) hold. From equality (34), we obtain elements wh0,wh1,…,whn−1. We have to show that there exists elements uhn∈W and such that inclusion (33) holds. Let denote by(36)ϑn−1=w0+hβΓβ+1∑j=1n−1uhjn−j+1β−n−jβ,c^=hβΓβ+1.
To accomplish this, we will demonstrate that the multivalued operator(37)W∋ϑ↦Aϑ+B+Cϑn−1+c^ϑ+∂Jϑn−1+c^ϑis surjective.
SuppositionsHA, HB, and Hc imply the linearity and boundedness of the operator(38)W∋ϑ↦Aϑ+B+Cϑn−1+c^ϑ∈W∗,so, it is hemicontinuous, and satisfies the condition(39)Aϑ+Bϑn−1+c^ϑ+Cϑn−1+c^ϑ,ϑ≥m˜A−c^B+bm˜cϑ2−B+bm˜cϑn−1ϑ.
Thus, under the conditionm˜A−c^B+bm˜c>0, the operator(40)Aϑ+B+Cϑn−1+c^ϑ,is coercive and monotone for allϑ∈W.
For the mappingx↦Jx, which it is, continuous and convex for all x∈W, then by Theorem 1, we obtain that ∂Jϑn−1+c^ϑ⊂2W′ is maximal monotone for all ϑ∈W. The coercivity is so fulfilled if Γβ+1m˜A/B+bm˜c1/β>h. Therefore, for all h∈0,h0, where h0 is given by(41)h0=m˜AΓ1+β2B+bm˜c1/β.
We may now use Theorem2 to guarantee the surjectivity of the operator ϑ↦Aϑ+B+Cϑn−1+c^ϑ+∂Jϑn−1+c^ϑ, for all 0<h<h0. Thus, we conclude that there exists elements uhn∈W, satisfying the (33). This completes the proof of the theorem.
Note thatuhn∈W is solution of (33), also means that there exists ςhn∈∂Jwhk, such that(42)Auhk+Bwhk+Cwhk+ςhn=fhk.It is noteworthy that hereafter, we useCte to denote a generic positive constant independent of h. The sequence of solutions to Rothe Problem 5 is estimated in the finding that follows:Lemma 1.
Let’s say that all four hypothesesHA, HB, Hc, andHf,are fulfilled, then there existsh1>0independent ofh, such that for allh∈0,h1, the solutions of Problem 5k=nsatisfy(43)maxn=1,2,…,ℵuhn≤Cte,(44)maxn=1,2,…,ℵwhn≤Cte,where the constantCtehere is independent ofhandn.Proof.
We multiply equation (33) by taking k=n and ϑ=whn−uhn∈W to get(45)Auhn,uhn+Bwhn,uhn+Cwhn,uhn+Jwhn−Jwhn−uhn≤fhn,uhn.
By the expression (34) of whn and hypotheses HB and Hc, we have(46)Bwhn,uhn=Bw0+c^∑j=1nuhjn−j+1β−n−jβ,uhn≥−Bu0W′uhn−Buhjuhnc^∑j=1nn−j+1β−n−jβ−c^Buhn2,(47)Cwhn,uhn=∑j=1n∫δj−1δjcδn,μu0+c^∑ℓ=1juhℓn−ℓ+1β−n−ℓβdμ,uhn=ϱhn,uhn+∫δn−1δncδn,μw0+hβΓβ+1uhndμ,uhn≥−2bm˜cw0+c^∑ℓ=1n−1uhℓn−ℓ+1β−n−ℓβuhn−bm˜chβΓβ+1uhn2,where(48)ϱhn=∑j=1n−1∫δj−1δjcδn,μw0+hβΓβ+1∑ℓ=1juhℓn−ℓ+1β−n−ℓβdμ+∫δn−1δncδn,μu0+hβΓβ+1∑ℓ=1n−1uhℓn−ℓ+1β−n−ℓβdμ.
According to the definition ofJ, we have(49)Jwhk−Jwhk−uhn≤g∫ΣCwτhk−wτhk−uτhkda≤gcFuhj,where cF is Poincaré’s constant. This implies that(50)ςhn,uhn≥Jwhk−Jwhk−uhn≥−gcFuhn.
Now, we conclude that because of the coercivity of the operatorA and expressions (46), (47), and (50),(51)fhn,uhn≥Auhn,uhn+Bwhn,uhn+Cwhn,uhn+Jwhk−Jwhk−uhn≥mAuhn2−c^Buhn2−Bw0W′+gcpuhn−c^B∑j=1n−1uhjn−j+1β−n−jβuhn−2bm˜cw0+c^∑ℓ=1n−1uhℓn−ℓ+1β−n−ℓβuhn−bm˜cc^uhn2.
Thus, gives(52)fznW′+c^B+2bm˜c+w0∑j=1n−1uzjn−j+1β−n−jβ+Bw0W′+gcp≥m˜Å−c^B+bm˜cuzn.
If we chooseh1=m˜AΓ1+β/2B+bm˜c1/β, we deduce that(53)m˜A−c^B+bm˜c≥m˜A2for allh∈0,h1,(note that we have h0>h1). Subsequently,(54)2fhnW′m˜A+2Bw0W′+gcpm˜A+2c^B+2bm˜c+u0m˜A∑j=1n−1uhjn−j+1β−n−jβ≥uhn.
From hypothesisHf, fhnW′≤supδ∈0,bfδW′, for all h>0; n=1,ℵ¯. Next, thanks to the generalized discrete Gronwall inequality, see that for(55)c0=2fhnW′m˜A+2Bw0W′+gcpm˜Auhn≤c0exp2m˜Ac^B+2bm˜c+w0∑j=1n−1n−j+1β−n−jβ≤Cte.
So, the estimate (43) is verified.
Next, by the definition of theuhn, which is written in terms of whn, we find the following estimate:(56)whn≤w0+1Γβ+1∑ℓ=1nuhℓδn−ℓ+1β−δn−ℓβ≤w0+δnβCteΓβ+1≤w0+bβCteΓβ+1=Cte.
Thus concludes the proof of Lemma1.It is now necessary to demonstrate our major result, that a solution to Problem3 exists. We define the interpolate functions u¯h, w¯h:0,b⟶W by(57)u¯hδ=uhn,w¯hδ=whn,andςδ=ςhn,δ∈δn−1,δn,n=1,…,ℵ,as h⟶0, where uhn∈L1/κ0,b;W is a solution to Problem 4.Lemma 2.
Assume thatHA,HB,Hc, andHf, hold. Letκ∈0,η, whereη=max1/2,β, andhnbe a sequence such thathn⟶0, asn⟶∞, to make simple, let it denoted byh. Then, we have(58)u¯h⇀uweakly inL1/κ0,b;W,(59)w¯hδ⇀Dδ−β0uδ+w0weakly inW,ash⟶0.Proof.
Let0<κ<η=max1/2,β, then by applying estimate (43), we have(60)u¯hL1/κ0,b;W1/κ=∫0bu¯hμ1/κdμ=∑ι=1ℵ∫δι−1διuhι1/κdμ=ℓ11/κ∑ι=1ℵ∫δι−1διdμ≤Cte.
Hence, it follows thatu¯h is bounded in L1/κ0,b;W and consequently, we deduce that there exists u∈L1/κ0,b;W, such that(61)u¯h⟶uweakly inL1/κ0,b;W,ash⟶0.
For anyg∈W′ and δ∈0,b, note that function δ−μβ−1χ0,tμg for μ∈0,b, is in L1/κ′0,b;W′, where κ′=1−κ, indeed(62)δ−⋅β−1χ0,δgL1/κ′0,b;W′κ′=∫0δgW′1/κ′δ−μβ−1/κ′dμ≤1−κβ−κgW′1/κ′δ−μβ−κ/1−κ|0δ<∞.
Hence, this leads(63)g,1Γβ∫0δδ−μβ−1u¯hμdμ−1Γβ∫0δδ−μβ−1uμdμ≤1Γβ∫0δδ−μβ−1g,u¯hμ−uμdμ≤1Γβδ−⋅β−1χ0,δgL1/κ′0,b;W′u¯h−uL1/κ0,b;W⟶0,ash⟶0.
Thus, we deduce that(64)Dδ−β0u¯hδ⟶Dδ−β0uδweakly inW,ash⟶0,for all δ∈0,b. Moreover, remark that we have(65)Dδ−β0u¯hδ=c^∑j=1nuhjn−j+1β−n−jβ=1Γβ∫0δnδn−μβ−1u¯hμdμ1Γβ∫0δδn−μβ−1u¯hμdμ−∫δδnδn−μβ−1u¯hμdμ,so, it entails(66)I=Dδ−β0u¯hδ−w¯hδ+w0≤1Γβ∫δδnδn−μβ−1u¯hμdμ+1Γβ∫0δδn−μβ−1−δ−μβ−1u¯hμdμ.
Using estimate (43) again, one has(67)I≤ℓ1Γβ∫δδnδn−μβ−1dμ+∫0δδ−μβ−1−δn−μβ−1dμ≤ℓ1Γβ+1δn−δβ−δnβ+δn−δβ+δnβ,for δ∈δn−1,δn. Therefore, estimate (67), ensure that, for all δ∈0,b(68)w¯hδ−w0−Dδ−β0u¯hδ⟶0strongly inW,ash⟶0.
Employing the latter and taking convergence (64) into account, we get(69)w¯hδ⇀Dδ−β0uδ+w0weakly inW,ash⟶0.Lemma 3.
Letuhn∈Wandςhn∈∂Jwhkare solutions of equation (42). Then, we have(70)ςhn⟶ςweakly inW′,whereςδ∈∂Jwδandwδis given by the convergence (59).Proof.
Let’s remind thatJw¯hδ=∫ΣCgγw¯hδdμ, where γw¯hδ=w¯τhδ; furthermore, using the compactness of trace map γ:W⟶L2ΣC, it follows from the weak convergence w¯hδ⇀wδ=Dδ−β0uδ+w0, in W that w¯hδ⟶wδ, strongly in L2ΣC. Subsequently, we deduce that(71)Jw¯hδ⟶Jw¯δ.
For allδ∈δn−1,δn, we have first(72)ςhn,ϑ≤Jwhk−ϑ−Jwhk,and second(73)ςhn,ϑ≤Jwhk−ϑ−Jwhk≤g∫ΣCγw¯hδ−γϑ−γw¯hδda,≤g∫ΣCγϑda≤gcFϑ.
This leads to(74)ςhnW′≤gcF.Then, there exists a subsequence noted ςhn, such that ςhn⟶ς weakly in W′ (W′ is reflexive space). We claim that ςδ∈∂Jwδ. Indeed, we have that the subdifferential ∂Jwδ is identical to γ∗∂ℑγwδ, where ℑ:L2ΣC⟶ℝ. It is sufficient to prove then that(75)ℑγwδ−ℑγϑ≤ς˜,γwδ−γϑL2ΣC=γ∗ς˜,wδ−ϑW′,for allϑ∈W.
Otherwise, let us show thatγ∗ς˜δ=ςδ, with ς˜δ∈∂ℑγwδ. Combining the strong convergence(76)γwhδ⟶γwδinL2ΣC.with the weak convergence of a subsequence of ς˜hδ to ς˜δ, we pass to the limit as h⟶0, we have(77)limℑγw¯hδ−ℑγϑ≤limς˜hδ,γw¯hδ−γϑL2ΣC=ς˜δ,γwδ−ϑL2ΣC,for allϑ∈W.
Sinceγ∗ς˜h=ςh, and γ∗ is linear, it follows that(78)ςhδ=γ∗ς˜hδ⇀γ∗ς˜δweakly inW′.
The uniqueness of limit guarantees thatγ∗ς˜δ=ςδ. Now by definition of the subdifferential of J, we have(79)Jwδ−Jϑ=ℑγwδ−ℑγϑ≤γ∗ς˜δ,wδ−ϑW′,for allϑ∈W.
This entailsςδ is in ∂Jwδ. Thus, complete the proof of the lemma.Theorem 4.
LetHA,HB, Hc, andHfbe satisfied. Assume thatκ∈0,η, whereη=max1/2,β. Then, there exists a solutionw∈L10,b;WandDδ0Cw∈L1/κ0,b;W, satisfying Problem 3.Proof.
Consider nowA˜, B˜, and C˜, the Nemytskii operators that correspond to A, B, and C, respectively, which are defined by A˜uδ=Auδ,B˜uδ=Bw0+Dδ−β0uδ, and C˜uδ=Cw0+Dδ−β0uδ. Since A∈LW,W′ and u¯h⟶u weakly in L1/κ0,b;W, as h⟶0, then we have(80)A˜u¯h⇀A˜uweakly inL1/κ¯0,b;W′.
Note that1/κ>2, then L1/κ0,b;W′⊂V. Now by hypothesis that B∈LW,W′, one has(81)Bw0+Dδ−β0u¯hδ,ϑδ≤Bw0+ℓ1bβΓβ+1Bϑδ.
Furthermore, we use the convergence(82)Bw0+Dδ−β0u¯hδ,ϑδ⟶Bw0+Dδ−β0uδ,ϑδa.e. in0,b.
Applying the Lebesgue dominated convergence theorem, we deduce that(83)ℬ˜u¯h⇀ℬ˜uweakly inV′.
Now by hypothesis onc, C∈ℒW,W′,c∈C0,b;ℒW,W′, for δ∈δn−1,δn(84)Cwhn−∫0δcδ,μwμdμ=∑j=1n∫δj−1δjcδn,μwhjdμ−∫δj−1δcδ,μwμdμ≤∫0δcδn,μwhjdμ−cδ,μwμdμ+∑j=1n∫δδjcδn,μwhjdμ.
Using the boundedness ofuhj, j=1,…,n and the Lipschitz continuity of c⋅,μ, this implies(85)Cuhn−∫0δcδ,μuμdμ≤hCte.
That is to conclude(86)Cwhn⟶∫0δcδ,μwμdμ,strongly inW,ash⟶0.
Sincec∈C0,b;ℒW,W′, it is obvious that ∫0δcδ,μwμdμ belongs to V. That leads to(87)∫0bC˜w¯h,ϑδdδ⟶∫0b∫0δcδ,μwμdμ,ϑδdδ,∀ϑ∈V.
It is clear thatfh⟶f strongly in V′, as h⟶0.
The above convergences entail for allv∈V(88)A˜uh,ϑV′,V⟶A˜u,ϑV′,V,B˜uh,ϑV′,V⟶B˜u,ϑV′,V,C˜wh,ϑV′,V⟶C˜w,ϑV′,V,ςh,ϑV′,V⟶ς,ϑV′,V,for ϑ∈W and h⟶0. By the strong convergence of fh to f in W′ and the fact that fh and f belong to L∞0,b;W′, we have(89)fh,whV′,V⟶f,wV′,V,ash⟶0.
All above convergences imply that(90)0≤limh⟶0supA˜uh+B˜uh+C˜wh+ςh,ϑV′,V−limh⟶0inffh,ϑV,V′.
We obtain the following inequality, for allϑ∈V(91)ADδβ0Cw+Bw+Cw+ς,ϑV′,V≥f,ϑV′,V.
Finally, taking into account the formulation (25), we deduce that w∈L10,b;W and Dδ0Cw in L1/κ0,b;W is a solution of Problem 3. This completes the proof of Theorem 4.


## 4. Conclusion

In this paper, we have already considered the problem of the deformation of a body that comes into contact with a rigid foundation. The novelty here was that we took into account the combination of a fractional Kelvin–Voigt’s constitutive law and a frictional Tresca’s contact [17] law. Also, we have established the existence of weak solutions by going through a series of problems that allows us to find the first discrete solutions by means of Rothe’s method and some results on nonlinear inclusions. Then, by passing to the limit, we have found a weak solution to the problem, satisfying a certain regularity.



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