# Theoretical Study of a Chain Sliding on  a Fixed Support

Authors: Jérôme Bastien; Claude-Henri Lamarque
Journal: Mathematical Problems in Engineering
 (2009)

---

## Abstract

A chain sliding on a fixed support, made out of some elementary rheological models (dry friction element and linear spring) can be covered by the existence and uniqueness theory for maximal monotone operators. Several behavior from quasistatic to dynamical are investigated. Moreover, classical results of numerical analysis allow to use a numerical implicit Euler scheme.


---

## Body

## 1. Introduction

This paper is the next step of a series of previous works dealing with modelling of discrete mechanical systems with finite number of degrees of freedom involving assemblies of classical smooth constitutive elements (in the mechanical point of view they correspond to linear or non linear springs, dashpots) and nonsmooth ones mainly based on St-Venant Elements. Let us cite basic rheological models [1], with different applications and developpements [2–7]. Delay or stochastic frame have also been investigated in [8–10].In this paper we examine a new model: it can be associated with motion of a discretized beam “sliding” on soil. We do not give more details on this discretization.This paper is organized as follows in Section2, the model is described. In Section 3, the general model is adapted to different dynamical, semi-dynamical or quasistatic cases. In Section 4, existence and uniqueness is addressed. In Section 5, numerical scheme is described and its convergence obtained.


## 2. Description of the Model

We refer to previous works for description of some rheological models (see for example [1, 6]).We consider the model of Figure2. (mi)1≤i≤n (with mi≥0) correspond to masses, (ki)0≤i≤n to stiffness, and (αi)1≤i≤n to St-Venant elements thresholds.The reader is referred to AppendixA.Letσ be the multivalued graph sign defined by (see Figure 1(a)).Two useful multivalued maximal monotone graphs.
(a)
The graphσ(b)
The graphβFigure 2
The studied model with mechanical components.(2.1)σ(x)={-1ifx<0,1ifx>0,[-1,1]ifx=0.
According to [11], this graph is maximal monotone. Therefore:(2.2)∀x∈ℝ,σ(x)=∂|x|.Let us assume (see Figure3) the followingFigure 3
The studied model with external forces(Fi)0≤i≤n, friction forces (gi)1≤i≤n, linear forces (fi)0≤i≤n, and with displacements (ui)0≤i≤n, (vi)1≤i≤n, x and ξ.(i)
This mechanical system is submitted to external forces(Fi)0≤i≤n+1: F0 is exerted on the spring with stiffness k0; For 1≤i≤n, Fi is exerted on material point of mass mi; Fn+1 is exerted on the spring with stiffness kn+1.(ii)
For1≤i≤n, gi is the friction force exerted by the support of the ith St-Venant.(iii)
For0≤i≤n, fi is elastic linear force exerted by the ith spring.(iv)
For0≤i≤n, ui is the displacement of the ith spring.(v)
For1≤i≤n, vi is the displacement of the ith St-Venant element.(vi)
ξ is the displacement of the spring with stiffness k0.(vii)
x is the displacement of the material point of mass mn.These two last notations are justified by the study of particular cases in the next sections.The different equations of the model are successively given by the fundamental Newton law:(2.3a)∀i∈{1,…,n},miv̈i=Fi+fi-1-fi+gi,
by the constitutive laws of linear springs:(2.3b)∀i∈{0,…,n},fi=-kiui,
by the constitutive of laws St-Venant elements:(2.3c)∀i∈{1,…,n},gi∈-αiσ(v̇i),
by the geometrical connexions:(2.3d)∀i∈{0,…,n-1},ξ+u0+u1+⋯+ui=vi+1,(2.3e)ξ+u0+u1+⋯+un=x,
and finally by the boundary conditions:(2.3f)F0=f0,(2.3g)Fn+1=fn.We can observe that (2.3d)–(2.3e) are equivalent to(2.4a)ξ+u0=v1,(2.4b)∀i∈{1,…,n-1},vi+1-vi=ui,(2.4c)x-vn=un.Now, we study systems (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), and (2.4a)–(2.4c).


## 3. Transformations of Equations

Now, as in [1, 6], we transform system (2.3a)-(2.3b)-(2.3c)-(2.3f)-(2.3g)-(2.4b)-(2.4c) to rewrite it under the usual form (A.7) according to different kinds of problem and of boundary conditions.Let us assume that the external forcingF1,…,Fn are known.


### 3.1. Dynamical Case

We assume in this section that(3.1)∀i∈{1,…,n},mi>0.Equations (2.3a)-(2.3b)-(2.3c)-(2.4a)–(2.4c) imply(3.2a)m1v̈1+α1σ(v̇1)-k0ξ+(k0+k1)v1-k1v2∋F1,(3.2b)∀i∈{2,…,n-1},miv̈i+αiσ(v̇i)-ki-1vi-1+(ki-1+ki)vi-kivi+1∋Fi,(3.2c)mnv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn-knx∋Fn.


#### 3.1.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions:(3.3a)ξ=0,x=0,(3.3b)andthereactionsF0andFn+1areunknown.We set, for allq∈ℕ*,(3.4)K(q)=(k0+k1-k100⋯000-k1k1+k2-k20⋯⋯000-k2k2+k3-k30⋯⋯0⋮⋮⋯⋯⋮⋮⋮00⋯00-kq-2kq-2+kq-1-kq-100⋯⋯⋯0-kq-1kq-1+kq)∈ℳq(ℝ).
Thus, by setting(3.5a)V=(v1,…,vn)t∈ℝn,(3.5b)F=(F1,…,Fn)t∈ℝn,(3.5c)K=K(n),(3.5d)ℳ=(m10⋯00m2⋯00⋯⋱00⋯⋯mn)
and defining the maximal monotone operator 𝒜 by(3.5e)𝒜(v1,…,vn)=α1σ(v1)×⋯×αnσ(vn),
equations (3.2a)–(3.2c) imply the system of equations(3.6)MV̈+𝒜V̇+KV∋F.
Reactions F0 and Fn+1 can be determined thanks to (2.3f)-(2.3g) which give(3.7a)F0=-k0v1,(3.7b)Fn+1=knvn.Set(3.8a)p=2n,(3.8b)M=(In00ℳ-1)∈ℳ2n(ℝ),
where In is the identity of ∈ℳn(ℝ) and for t∈ℝ, X=(V1,V2)∈ℝ2n, with V2=(V2,1,…,V2,n),(3.8c)𝒢(t,(V1,V2))=(V2ℳ-1F-ℳ-1KV1),(3.8d)ϕ(V1,V2)=∑i=1nαi|V2,i|.
Then, the system (3.6) is equivalent to (A.7) (see Appendix A).Reciprocally, if (3.6) and (3.7a)-(3.7b) hold, we define x, ξ, (ui)0≤i≤n, (fi)0≤i≤n, and (gi)1≤i≤n successivelly by(3.9)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b),∀i∈{1,…,n},gi=miv̈i-Fi-fi-1+fi.
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


#### 3.1.2. Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity and free at its right extremity so that we can write the boundary condition:(3.10a)ξ=0,(3.10b)reactionF0isunknown,(3.10c)displacementxisunknown,(3.10d)andexternalforcingFn+1isknown.As in Section3.1.1, by setting(3.11a)V=(v1,…,vn)t∈ℝn,(3.11b)F=(F1,…,Fn-1,Fn-Fn+1)t∈ℝn,(3.11c)K̃=(k0+k1-k100⋯000-k1k1+k2-k20⋯⋯000-k2k2+k3-k30⋯⋯0⋮⋮⋯⋯⋮⋮⋮00⋯00-kn-2kn-2+kn-1-kn-100⋯⋯⋯0-kn-1kn-1)∈ℳn(ℝ),(3.11d)ℳand𝒜asin(3.5d)-(3.5e),
we can prove that equations (3.2a)–(3.2c) imply the system of equations(3.12)MV̈+𝒜V̇+K̃V∋F.
Reactions F0 and displacement x can be determined thanks to(3.13a)F0=-k0v1,(3.13b)x=-Fn+1kn+vn.As in Section3.1.1, let us set(3.14a)p=2n,(3.14b)Mandϕdefinedby(3.8b)–(3.8d),
and for t∈ℝ, X=(V1,V2)∈ℝ2n, with V2=(V2,1,…,V2,n),(3.14c)𝒢(t,(V1,V2))=(V2ℳ-1F-ℳ-1K̃V1).
Then, system (3.12) is equivalent to (A.7).As in Section3.1.1, reciprocally, if (3.12) and (3.13a)-(3.13b) hold, we define x, ξ, (ui)0≤i≤n, (fi)0≤i≤n, and (gi)1≤i≤n successivelly by(3.15)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b),Lastequationof(3.9).
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


### 3.2. Semi-Dynamical Case

In this section, we assume that(3.16a)∀i∈{1,…,n-1},mi=0,(3.16b)mn=m>0.Equation (3.2b) implies(3.17a)∀i∈{2,…,n-1},αiσ(v̇i)+gi∋0,
with(3.17b)∀i∈{2,…,n-1},gi=-Fi-ki-1vi-1+(ki-1+ki)vi-kivi+1.
As in [6, 7], we introduce β, the inverse graph of σ (in the sens of [11], see Figure 1(b)):(3.18)β(x)={∅ifx∈]-∞,-1[⋃]1,+∞[,{0}ifx∈]-1,1[,ℝ-ifx=-1,ℝ+ifx=1.
We have(3.19)∀x∈ℝ,β(x)=∂ψ[-1,1](x),
where ∂ψ[-1,1] is the convex indicatrix function of the convex domain [-1,1]. Thus, (3.17a) is equivalent to(3.20)∀i∈{2,…,n-1},v̇i+∂ψ[-αi,αi](gi)∋0.
Similarly, (3.2a) gives(3.21a)v̇1+∂ψ[-α1,α1](g1)∋0,
with(3.21b)g1=-F1-k0ξ+(k0+k1)v1-k1v2.
and (3.2c) gives(3.22)mv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn-knx∋Fn.


#### 3.2.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions (3.3a)-(3.3b). As in [6, 7], let us set (3.23a)V=(v1,…,vn-1)t∈ℝn-1,(3.23b)G=(g1,…,gn-1)t∈ℝn-1,(3.23c)F=(F1,…,Fn-1)t∈ℝn-1,(3.23d)Z=(F1,…,Fn-2,Fn-1+kn-1vn)t∈ℝn-1,(3.23e)C=[α1,α1]×⋯×[-αn-1,αn-1]⊂ℝn-1,(3.23f)K̂=K(n-1)∈ℳn-1(ℝ), where K(q) is defined by (3.4). Thus, according to (3.17b)–(3.21b), we have(3.24)G=K̂V-Z,
and from (3.17a)–(3.21a) we can write(3.25)V̇+∂ψC(G)∋0,
Under the assumption(3.26)kn-1≥0∀i∈{1,…,n-1},ki>0,
the matrix K̂ is symmetric definite positive (see proof in Lemma B.1 of Appendix B), so that(3.27)V=K̂-1(G+Z),
and (3.25) gives(3.28)K̂-1(Ġ+Ż)+∂ψC(G)∋0,
which is equivalent to(3.29)Ġ+K̂∂ψC(G)∋-Ż.
For q integer and u vector of ℝm, we denote by(3.30)[u]q
the  qth  component  of  u. Equation (3.22) gives(3.31)mv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn∋Fn,
which can be rewritten under the following form:(3.32)v̈n+αnmσ(v̇n)-kn-1m[K̂-1(G+Z)]n-1+kn-1+knmvn∋Fnm.
Let u be the vector of ℝn-1 defined by(3.33)u=(0,…,0,1)t.
Note that(3.34)Z=F+kn-1vnu.
We set (3.35a)p=n+1,(3.35b)M=(K̂00I2)∈ℳn+1(ℝ),
and for all t∈ℝ, G∈ℝn-1, a,b∈ℝ, X=(G,a,b)t(3.35c)𝒢(t,X)=(-Ḟ-kn-1bubFnm+kn-1m[K̂-1(G+F+kn-1au)]n-1-kn-1+knma)(3.35d)ϕ(X)=ψ[α1,α1]×⋯×[-αn-1,αn-1]×{0}×{0}(X)+αnm|b|. Then, system (3.29)–(3.32) is equivalent to (A.7).ReactionsF0 and Fn+1 can be determined thanks to (3.36a)F0=-k0[K̂-1(G+F+kn-1au)]1,(3.36b)Fn+1=kna.Reciprocally, as in Section3.1.1, if (3.29)–(3.32) hold, we can determine G and Z thanks to(3.37)G=([X]1,…,[X]n-1)t,Z=kn-1[X]nu+F.
then we can calculate V thanks to (3.27). Successively, x, ξ, (ui)0≤i≤n, and (fi)0≤i≤n are defined by(3.38)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b).
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


#### 3.2.2. Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity and free at its right extremity so that we can write boundary condition (3.10a)–(3.10d).The calculus are similar to those of Section3.2.1; Equation (3.29) holds and (3.31) is replaced by(3.39)mv̈n+αnσ(v̇n)-kn-1vn-1+kn-1vn+Fn+1∋Fn.
Using notations (3.23a)–(3.23f), we obtain the system (A.7), where we set (3.40a)p=n+1,(3.40b)Mandϕaredefinedby(3.35b)and(3.35d),
and for all t∈ℝ, G∈ℝn-1, a,b∈ℝ, X=(G,a,b)t,(3.40c)𝒢(t,X)=(-Ḟ-kn-1bubFn-Fn+1m+kn-1m[K̂-1(G+F+kn-1au)]n-1-kn-1ma)∈ℳn+1(ℝ).
The reaction F0 and the displacement x can be determined thanks to (3.36a) and(3.41)x=-Fn+1kn+a.


### 3.3. Quasistatic Case

In this section, we assume that(3.42)∀i∈{1,…,n},mi=0.As it has been previously noticed, (3.17a)-(3.17b) and (3.21a)-(3.21b) are not modified, and (3.22) gives(3.43a)v̇n+∂ψ[-αn,αn](gn)∋0,
with(3.43b)gn=Fn-kn-1vn-1+(kn-1+kn)vn-knx.


#### 3.3.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions (3.3a)-(3.3b).As in Section3.2.1, following [6, 7], we set (3.44a)V=(v1,…,vn)t∈ℝn,(3.44b)G=(g1,…,gn)t∈ℝn,(3.44c)F=(F1,…,Fn)t∈ℝn,(3.44d)C=[α1,α1]×⋯×[-αn,αn]⊂ℝn,(3.44e)K=K(n)∈ℳn(ℝ),where K(q) is defined by (3.4).Thus, we have(3.45)G=KV-F,V̇+∂ψC(G)∋0
Under assumption(3.46)kn≥0,∀i∈{1,…,n-1},ki>0,
the matrix K is symmetric definite positive (see proof in Lemma B.1), so that(3.47)V=K-1(G+F),(3.48)Ġ+K∂ψC(G)∋-Ḟ.
We set (3.49a)p=n,(3.49b)M=K∈ℳn(ℝ),
and, for all t∈ℝ, for all X∈ℝn(3.49c)𝒢(t,X)=-Ḟ,(3.49d)ϕ(X)=ψC(X). Then, the system (3.48) is equivalent to (A.7).ReactionsF0 and Fn+1 can be determined thanks to (3.50a)F0=-k0[K-1(G+F)]1,(3.50b)Fn+1=kn[K-1(G+F)]n.


#### 3.3.2.  Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity so that we can write the boundary condition (3.10a) and (3.10b). Boundary conditions for its right extremity is given later.The calculus is similar to those of [6, 7].(i)  First Case: Displacement-Force Model
We assume that the displacementx is known and that the force Fn+1 is unknown.
We introduceV, G, C and matrix K defined by (3.44a), (3.44b), (3.44d), and (3.44e) and F defined by
(3.51)F=(F1,…,Fn-1,Fn+knx)t∈ℝn,
and we obtain, as in Section 3.3.1,
(3.52)G=KV-F,(3.53)Ġ+K∂ψC(G)∋-Ḟ.
By setting (3.54a)p=n,(3.54b)Mandϕaredefinedby(3.49b)and(3.49d),
and, for all t∈ℝ, for all X∈ℝn,
(3.54c)𝒢(t,X)=-Ḟ, we remark that system (3.53) is equivalent to (A.7).
ReactionsF0 and Fn+1 can be determined thanks to
(3.55a)Fn0=-k0[K-1(G+F)]1,(3.55b)Fn+1=-knx+kn[K-1(G+F)]n.(ii)  Second Case: Force-Displacement Model
We assume that external forcingFn+1 are known and displacement x is unknown.
The calculus are similar to the previous case.
Equation (3.43b) is replaced by
(3.56)gn=Fn+1-Fn-kn-1vn-1+kn-1vn.
Following the same method, we introduce V, G, and C defined by (3.44a)-(3.44b)-(3.44d), and matrix K̃ defined by (3.11c). Vector F is defined by
(3.57)F=(F1,…,Fn-1,Fn-Fn+1)t.
So, (3.52) is replaced by
(3.58)G=K̃V-F,
and (3.48) is replaced by
(3.59)Ġ+K̃∂ψC(G)∋-Ḟ.Remark 3.1.
As in [6], let us notice that matrix K̃ defined by (3.11c) for force-displacement model corresponds to matrix K(n) for displacement-force model defined by (3.4) with
(3.60)kn=0.According to previous remark, assumption(3.61)∀i∈{1,…,n-1},ki>0
and Lemma B.1 ensure that matrix K̃ is symmetric definite positive. Thus, like previously, the system is equivalent to(3.62)Ġ+K̃∂ψC(G)∋-Ḟ.
By giving p, ϕ defined by (3.49a)–(3.49d), 𝒢 defined by for all t∈ℝ, for all X∈ℝn,(3.63a)𝒢(t,X)=-Ḟ,
and M defined by(3.63b)M=K̃∈ℳn(ℝ),
we remark that system (3.62) is equivalent to (A.7). Reactions F0 and displacement x can be determined thanks to(3.64a)F0=-k0[K̃-1(G+F)]1,(3.64b)x=-Fn+1kn+[K̃-1(G+F)]n.


## 3.1. Dynamical Case

We assume in this section that(3.1)∀i∈{1,…,n},mi>0.Equations (2.3a)-(2.3b)-(2.3c)-(2.4a)–(2.4c) imply(3.2a)m1v̈1+α1σ(v̇1)-k0ξ+(k0+k1)v1-k1v2∋F1,(3.2b)∀i∈{2,…,n-1},miv̈i+αiσ(v̇i)-ki-1vi-1+(ki-1+ki)vi-kivi+1∋Fi,(3.2c)mnv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn-knx∋Fn.


### 3.1.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions:(3.3a)ξ=0,x=0,(3.3b)andthereactionsF0andFn+1areunknown.We set, for allq∈ℕ*,(3.4)K(q)=(k0+k1-k100⋯000-k1k1+k2-k20⋯⋯000-k2k2+k3-k30⋯⋯0⋮⋮⋯⋯⋮⋮⋮00⋯00-kq-2kq-2+kq-1-kq-100⋯⋯⋯0-kq-1kq-1+kq)∈ℳq(ℝ).
Thus, by setting(3.5a)V=(v1,…,vn)t∈ℝn,(3.5b)F=(F1,…,Fn)t∈ℝn,(3.5c)K=K(n),(3.5d)ℳ=(m10⋯00m2⋯00⋯⋱00⋯⋯mn)
and defining the maximal monotone operator 𝒜 by(3.5e)𝒜(v1,…,vn)=α1σ(v1)×⋯×αnσ(vn),
equations (3.2a)–(3.2c) imply the system of equations(3.6)MV̈+𝒜V̇+KV∋F.
Reactions F0 and Fn+1 can be determined thanks to (2.3f)-(2.3g) which give(3.7a)F0=-k0v1,(3.7b)Fn+1=knvn.Set(3.8a)p=2n,(3.8b)M=(In00ℳ-1)∈ℳ2n(ℝ),
where In is the identity of ∈ℳn(ℝ) and for t∈ℝ, X=(V1,V2)∈ℝ2n, with V2=(V2,1,…,V2,n),(3.8c)𝒢(t,(V1,V2))=(V2ℳ-1F-ℳ-1KV1),(3.8d)ϕ(V1,V2)=∑i=1nαi|V2,i|.
Then, the system (3.6) is equivalent to (A.7) (see Appendix A).Reciprocally, if (3.6) and (3.7a)-(3.7b) hold, we define x, ξ, (ui)0≤i≤n, (fi)0≤i≤n, and (gi)1≤i≤n successivelly by(3.9)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b),∀i∈{1,…,n},gi=miv̈i-Fi-fi-1+fi.
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


### 3.1.2. Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity and free at its right extremity so that we can write the boundary condition:(3.10a)ξ=0,(3.10b)reactionF0isunknown,(3.10c)displacementxisunknown,(3.10d)andexternalforcingFn+1isknown.As in Section3.1.1, by setting(3.11a)V=(v1,…,vn)t∈ℝn,(3.11b)F=(F1,…,Fn-1,Fn-Fn+1)t∈ℝn,(3.11c)K̃=(k0+k1-k100⋯000-k1k1+k2-k20⋯⋯000-k2k2+k3-k30⋯⋯0⋮⋮⋯⋯⋮⋮⋮00⋯00-kn-2kn-2+kn-1-kn-100⋯⋯⋯0-kn-1kn-1)∈ℳn(ℝ),(3.11d)ℳand𝒜asin(3.5d)-(3.5e),
we can prove that equations (3.2a)–(3.2c) imply the system of equations(3.12)MV̈+𝒜V̇+K̃V∋F.
Reactions F0 and displacement x can be determined thanks to(3.13a)F0=-k0v1,(3.13b)x=-Fn+1kn+vn.As in Section3.1.1, let us set(3.14a)p=2n,(3.14b)Mandϕdefinedby(3.8b)–(3.8d),
and for t∈ℝ, X=(V1,V2)∈ℝ2n, with V2=(V2,1,…,V2,n),(3.14c)𝒢(t,(V1,V2))=(V2ℳ-1F-ℳ-1K̃V1).
Then, system (3.12) is equivalent to (A.7).As in Section3.1.1, reciprocally, if (3.12) and (3.13a)-(3.13b) hold, we define x, ξ, (ui)0≤i≤n, (fi)0≤i≤n, and (gi)1≤i≤n successivelly by(3.15)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b),Lastequationof(3.9).
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


## 3.1.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions:(3.3a)ξ=0,x=0,(3.3b)andthereactionsF0andFn+1areunknown.We set, for allq∈ℕ*,(3.4)K(q)=(k0+k1-k100⋯000-k1k1+k2-k20⋯⋯000-k2k2+k3-k30⋯⋯0⋮⋮⋯⋯⋮⋮⋮00⋯00-kq-2kq-2+kq-1-kq-100⋯⋯⋯0-kq-1kq-1+kq)∈ℳq(ℝ).
Thus, by setting(3.5a)V=(v1,…,vn)t∈ℝn,(3.5b)F=(F1,…,Fn)t∈ℝn,(3.5c)K=K(n),(3.5d)ℳ=(m10⋯00m2⋯00⋯⋱00⋯⋯mn)
and defining the maximal monotone operator 𝒜 by(3.5e)𝒜(v1,…,vn)=α1σ(v1)×⋯×αnσ(vn),
equations (3.2a)–(3.2c) imply the system of equations(3.6)MV̈+𝒜V̇+KV∋F.
Reactions F0 and Fn+1 can be determined thanks to (2.3f)-(2.3g) which give(3.7a)F0=-k0v1,(3.7b)Fn+1=knvn.Set(3.8a)p=2n,(3.8b)M=(In00ℳ-1)∈ℳ2n(ℝ),
where In is the identity of ∈ℳn(ℝ) and for t∈ℝ, X=(V1,V2)∈ℝ2n, with V2=(V2,1,…,V2,n),(3.8c)𝒢(t,(V1,V2))=(V2ℳ-1F-ℳ-1KV1),(3.8d)ϕ(V1,V2)=∑i=1nαi|V2,i|.
Then, the system (3.6) is equivalent to (A.7) (see Appendix A).Reciprocally, if (3.6) and (3.7a)-(3.7b) hold, we define x, ξ, (ui)0≤i≤n, (fi)0≤i≤n, and (gi)1≤i≤n successivelly by(3.9)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b),∀i∈{1,…,n},gi=miv̈i-Fi-fi-1+fi.
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


## 3.1.2. Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity and free at its right extremity so that we can write the boundary condition:(3.10a)ξ=0,(3.10b)reactionF0isunknown,(3.10c)displacementxisunknown,(3.10d)andexternalforcingFn+1isknown.As in Section3.1.1, by setting(3.11a)V=(v1,…,vn)t∈ℝn,(3.11b)F=(F1,…,Fn-1,Fn-Fn+1)t∈ℝn,(3.11c)K̃=(k0+k1-k100⋯000-k1k1+k2-k20⋯⋯000-k2k2+k3-k30⋯⋯0⋮⋮⋯⋯⋮⋮⋮00⋯00-kn-2kn-2+kn-1-kn-100⋯⋯⋯0-kn-1kn-1)∈ℳn(ℝ),(3.11d)ℳand𝒜asin(3.5d)-(3.5e),
we can prove that equations (3.2a)–(3.2c) imply the system of equations(3.12)MV̈+𝒜V̇+K̃V∋F.
Reactions F0 and displacement x can be determined thanks to(3.13a)F0=-k0v1,(3.13b)x=-Fn+1kn+vn.As in Section3.1.1, let us set(3.14a)p=2n,(3.14b)Mandϕdefinedby(3.8b)–(3.8d),
and for t∈ℝ, X=(V1,V2)∈ℝ2n, with V2=(V2,1,…,V2,n),(3.14c)𝒢(t,(V1,V2))=(V2ℳ-1F-ℳ-1K̃V1).
Then, system (3.12) is equivalent to (A.7).As in Section3.1.1, reciprocally, if (3.12) and (3.13a)-(3.13b) hold, we define x, ξ, (ui)0≤i≤n, (fi)0≤i≤n, and (gi)1≤i≤n successivelly by(3.15)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b),Lastequationof(3.9).
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


## 3.2. Semi-Dynamical Case

In this section, we assume that(3.16a)∀i∈{1,…,n-1},mi=0,(3.16b)mn=m>0.Equation (3.2b) implies(3.17a)∀i∈{2,…,n-1},αiσ(v̇i)+gi∋0,
with(3.17b)∀i∈{2,…,n-1},gi=-Fi-ki-1vi-1+(ki-1+ki)vi-kivi+1.
As in [6, 7], we introduce β, the inverse graph of σ (in the sens of [11], see Figure 1(b)):(3.18)β(x)={∅ifx∈]-∞,-1[⋃]1,+∞[,{0}ifx∈]-1,1[,ℝ-ifx=-1,ℝ+ifx=1.
We have(3.19)∀x∈ℝ,β(x)=∂ψ[-1,1](x),
where ∂ψ[-1,1] is the convex indicatrix function of the convex domain [-1,1]. Thus, (3.17a) is equivalent to(3.20)∀i∈{2,…,n-1},v̇i+∂ψ[-αi,αi](gi)∋0.
Similarly, (3.2a) gives(3.21a)v̇1+∂ψ[-α1,α1](g1)∋0,
with(3.21b)g1=-F1-k0ξ+(k0+k1)v1-k1v2.
and (3.2c) gives(3.22)mv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn-knx∋Fn.


### 3.2.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions (3.3a)-(3.3b). As in [6, 7], let us set (3.23a)V=(v1,…,vn-1)t∈ℝn-1,(3.23b)G=(g1,…,gn-1)t∈ℝn-1,(3.23c)F=(F1,…,Fn-1)t∈ℝn-1,(3.23d)Z=(F1,…,Fn-2,Fn-1+kn-1vn)t∈ℝn-1,(3.23e)C=[α1,α1]×⋯×[-αn-1,αn-1]⊂ℝn-1,(3.23f)K̂=K(n-1)∈ℳn-1(ℝ), where K(q) is defined by (3.4). Thus, according to (3.17b)–(3.21b), we have(3.24)G=K̂V-Z,
and from (3.17a)–(3.21a) we can write(3.25)V̇+∂ψC(G)∋0,
Under the assumption(3.26)kn-1≥0∀i∈{1,…,n-1},ki>0,
the matrix K̂ is symmetric definite positive (see proof in Lemma B.1 of Appendix B), so that(3.27)V=K̂-1(G+Z),
and (3.25) gives(3.28)K̂-1(Ġ+Ż)+∂ψC(G)∋0,
which is equivalent to(3.29)Ġ+K̂∂ψC(G)∋-Ż.
For q integer and u vector of ℝm, we denote by(3.30)[u]q
the  qth  component  of  u. Equation (3.22) gives(3.31)mv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn∋Fn,
which can be rewritten under the following form:(3.32)v̈n+αnmσ(v̇n)-kn-1m[K̂-1(G+Z)]n-1+kn-1+knmvn∋Fnm.
Let u be the vector of ℝn-1 defined by(3.33)u=(0,…,0,1)t.
Note that(3.34)Z=F+kn-1vnu.
We set (3.35a)p=n+1,(3.35b)M=(K̂00I2)∈ℳn+1(ℝ),
and for all t∈ℝ, G∈ℝn-1, a,b∈ℝ, X=(G,a,b)t(3.35c)𝒢(t,X)=(-Ḟ-kn-1bubFnm+kn-1m[K̂-1(G+F+kn-1au)]n-1-kn-1+knma)(3.35d)ϕ(X)=ψ[α1,α1]×⋯×[-αn-1,αn-1]×{0}×{0}(X)+αnm|b|. Then, system (3.29)–(3.32) is equivalent to (A.7).ReactionsF0 and Fn+1 can be determined thanks to (3.36a)F0=-k0[K̂-1(G+F+kn-1au)]1,(3.36b)Fn+1=kna.Reciprocally, as in Section3.1.1, if (3.29)–(3.32) hold, we can determine G and Z thanks to(3.37)G=([X]1,…,[X]n-1)t,Z=kn-1[X]nu+F.
then we can calculate V thanks to (3.27). Successively, x, ξ, (ui)0≤i≤n, and (fi)0≤i≤n are defined by(3.38)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b).
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


### 3.2.2. Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity and free at its right extremity so that we can write boundary condition (3.10a)–(3.10d).The calculus are similar to those of Section3.2.1; Equation (3.29) holds and (3.31) is replaced by(3.39)mv̈n+αnσ(v̇n)-kn-1vn-1+kn-1vn+Fn+1∋Fn.
Using notations (3.23a)–(3.23f), we obtain the system (A.7), where we set (3.40a)p=n+1,(3.40b)Mandϕaredefinedby(3.35b)and(3.35d),
and for all t∈ℝ, G∈ℝn-1, a,b∈ℝ, X=(G,a,b)t,(3.40c)𝒢(t,X)=(-Ḟ-kn-1bubFn-Fn+1m+kn-1m[K̂-1(G+F+kn-1au)]n-1-kn-1ma)∈ℳn+1(ℝ).
The reaction F0 and the displacement x can be determined thanks to (3.36a) and(3.41)x=-Fn+1kn+a.


## 3.2.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions (3.3a)-(3.3b). As in [6, 7], let us set (3.23a)V=(v1,…,vn-1)t∈ℝn-1,(3.23b)G=(g1,…,gn-1)t∈ℝn-1,(3.23c)F=(F1,…,Fn-1)t∈ℝn-1,(3.23d)Z=(F1,…,Fn-2,Fn-1+kn-1vn)t∈ℝn-1,(3.23e)C=[α1,α1]×⋯×[-αn-1,αn-1]⊂ℝn-1,(3.23f)K̂=K(n-1)∈ℳn-1(ℝ), where K(q) is defined by (3.4). Thus, according to (3.17b)–(3.21b), we have(3.24)G=K̂V-Z,
and from (3.17a)–(3.21a) we can write(3.25)V̇+∂ψC(G)∋0,
Under the assumption(3.26)kn-1≥0∀i∈{1,…,n-1},ki>0,
the matrix K̂ is symmetric definite positive (see proof in Lemma B.1 of Appendix B), so that(3.27)V=K̂-1(G+Z),
and (3.25) gives(3.28)K̂-1(Ġ+Ż)+∂ψC(G)∋0,
which is equivalent to(3.29)Ġ+K̂∂ψC(G)∋-Ż.
For q integer and u vector of ℝm, we denote by(3.30)[u]q
the  qth  component  of  u. Equation (3.22) gives(3.31)mv̈n+αnσ(v̇n)-kn-1vn-1+(kn-1+kn)vn∋Fn,
which can be rewritten under the following form:(3.32)v̈n+αnmσ(v̇n)-kn-1m[K̂-1(G+Z)]n-1+kn-1+knmvn∋Fnm.
Let u be the vector of ℝn-1 defined by(3.33)u=(0,…,0,1)t.
Note that(3.34)Z=F+kn-1vnu.
We set (3.35a)p=n+1,(3.35b)M=(K̂00I2)∈ℳn+1(ℝ),
and for all t∈ℝ, G∈ℝn-1, a,b∈ℝ, X=(G,a,b)t(3.35c)𝒢(t,X)=(-Ḟ-kn-1bubFnm+kn-1m[K̂-1(G+F+kn-1au)]n-1-kn-1+knma)(3.35d)ϕ(X)=ψ[α1,α1]×⋯×[-αn-1,αn-1]×{0}×{0}(X)+αnm|b|. Then, system (3.29)–(3.32) is equivalent to (A.7).ReactionsF0 and Fn+1 can be determined thanks to (3.36a)F0=-k0[K̂-1(G+F+kn-1au)]1,(3.36b)Fn+1=kna.Reciprocally, as in Section3.1.1, if (3.29)–(3.32) hold, we can determine G and Z thanks to(3.37)G=([X]1,…,[X]n-1)t,Z=kn-1[X]nu+F.
then we can calculate V thanks to (3.27). Successively, x, ξ, (ui)0≤i≤n, and (fi)0≤i≤n are defined by(3.38)(2.3b),(2.4a)–(2.4c),(3.3a)-(3.3b).
Then, we can deduce (2.3a), (2.3b), (2.3c), (2.3f), (2.3g), (2.4b), and (2.4c).


## 3.2.2. Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity and free at its right extremity so that we can write boundary condition (3.10a)–(3.10d).The calculus are similar to those of Section3.2.1; Equation (3.29) holds and (3.31) is replaced by(3.39)mv̈n+αnσ(v̇n)-kn-1vn-1+kn-1vn+Fn+1∋Fn.
Using notations (3.23a)–(3.23f), we obtain the system (A.7), where we set (3.40a)p=n+1,(3.40b)Mandϕaredefinedby(3.35b)and(3.35d),
and for all t∈ℝ, G∈ℝn-1, a,b∈ℝ, X=(G,a,b)t,(3.40c)𝒢(t,X)=(-Ḟ-kn-1bubFn-Fn+1m+kn-1m[K̂-1(G+F+kn-1au)]n-1-kn-1ma)∈ℳn+1(ℝ).
The reaction F0 and the displacement x can be determined thanks to (3.36a) and(3.41)x=-Fn+1kn+a.


## 3.3. Quasistatic Case

In this section, we assume that(3.42)∀i∈{1,…,n},mi=0.As it has been previously noticed, (3.17a)-(3.17b) and (3.21a)-(3.21b) are not modified, and (3.22) gives(3.43a)v̇n+∂ψ[-αn,αn](gn)∋0,
with(3.43b)gn=Fn-kn-1vn-1+(kn-1+kn)vn-knx.


### 3.3.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions (3.3a)-(3.3b).As in Section3.2.1, following [6, 7], we set (3.44a)V=(v1,…,vn)t∈ℝn,(3.44b)G=(g1,…,gn)t∈ℝn,(3.44c)F=(F1,…,Fn)t∈ℝn,(3.44d)C=[α1,α1]×⋯×[-αn,αn]⊂ℝn,(3.44e)K=K(n)∈ℳn(ℝ),where K(q) is defined by (3.4).Thus, we have(3.45)G=KV-F,V̇+∂ψC(G)∋0
Under assumption(3.46)kn≥0,∀i∈{1,…,n-1},ki>0,
the matrix K is symmetric definite positive (see proof in Lemma B.1), so that(3.47)V=K-1(G+F),(3.48)Ġ+K∂ψC(G)∋-Ḟ.
We set (3.49a)p=n,(3.49b)M=K∈ℳn(ℝ),
and, for all t∈ℝ, for all X∈ℝn(3.49c)𝒢(t,X)=-Ḟ,(3.49d)ϕ(X)=ψC(X). Then, the system (3.48) is equivalent to (A.7).ReactionsF0 and Fn+1 can be determined thanks to (3.50a)F0=-k0[K-1(G+F)]1,(3.50b)Fn+1=kn[K-1(G+F)]n.


### 3.3.2.  Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity so that we can write the boundary condition (3.10a) and (3.10b). Boundary conditions for its right extremity is given later.The calculus is similar to those of [6, 7].(i)  First Case: Displacement-Force Model
We assume that the displacementx is known and that the force Fn+1 is unknown.
We introduceV, G, C and matrix K defined by (3.44a), (3.44b), (3.44d), and (3.44e) and F defined by
(3.51)F=(F1,…,Fn-1,Fn+knx)t∈ℝn,
and we obtain, as in Section 3.3.1,
(3.52)G=KV-F,(3.53)Ġ+K∂ψC(G)∋-Ḟ.
By setting (3.54a)p=n,(3.54b)Mandϕaredefinedby(3.49b)and(3.49d),
and, for all t∈ℝ, for all X∈ℝn,
(3.54c)𝒢(t,X)=-Ḟ, we remark that system (3.53) is equivalent to (A.7).
ReactionsF0 and Fn+1 can be determined thanks to
(3.55a)Fn0=-k0[K-1(G+F)]1,(3.55b)Fn+1=-knx+kn[K-1(G+F)]n.(ii)  Second Case: Force-Displacement Model
We assume that external forcingFn+1 are known and displacement x is unknown.
The calculus are similar to the previous case.
Equation (3.43b) is replaced by
(3.56)gn=Fn+1-Fn-kn-1vn-1+kn-1vn.
Following the same method, we introduce V, G, and C defined by (3.44a)-(3.44b)-(3.44d), and matrix K̃ defined by (3.11c). Vector F is defined by
(3.57)F=(F1,…,Fn-1,Fn-Fn+1)t.
So, (3.52) is replaced by
(3.58)G=K̃V-F,
and (3.48) is replaced by
(3.59)Ġ+K̃∂ψC(G)∋-Ḟ.Remark 3.1.
As in [6], let us notice that matrix K̃ defined by (3.11c) for force-displacement model corresponds to matrix K(n) for displacement-force model defined by (3.4) with
(3.60)kn=0.According to previous remark, assumption(3.61)∀i∈{1,…,n-1},ki>0
and Lemma B.1 ensure that matrix K̃ is symmetric definite positive. Thus, like previously, the system is equivalent to(3.62)Ġ+K̃∂ψC(G)∋-Ḟ.
By giving p, ϕ defined by (3.49a)–(3.49d), 𝒢 defined by for all t∈ℝ, for all X∈ℝn,(3.63a)𝒢(t,X)=-Ḟ,
and M defined by(3.63b)M=K̃∈ℳn(ℝ),
we remark that system (3.62) is equivalent to (A.7). Reactions F0 and displacement x can be determined thanks to(3.64a)F0=-k0[K̃-1(G+F)]1,(3.64b)x=-Fn+1kn+[K̃-1(G+F)]n.


## 3.3.1. Clamped Mechanical System

We assume that our mechanical system is clamped at its two extremities so that we can write the boundary conditions (3.3a)-(3.3b).As in Section3.2.1, following [6, 7], we set (3.44a)V=(v1,…,vn)t∈ℝn,(3.44b)G=(g1,…,gn)t∈ℝn,(3.44c)F=(F1,…,Fn)t∈ℝn,(3.44d)C=[α1,α1]×⋯×[-αn,αn]⊂ℝn,(3.44e)K=K(n)∈ℳn(ℝ),where K(q) is defined by (3.4).Thus, we have(3.45)G=KV-F,V̇+∂ψC(G)∋0
Under assumption(3.46)kn≥0,∀i∈{1,…,n-1},ki>0,
the matrix K is symmetric definite positive (see proof in Lemma B.1), so that(3.47)V=K-1(G+F),(3.48)Ġ+K∂ψC(G)∋-Ḟ.
We set (3.49a)p=n,(3.49b)M=K∈ℳn(ℝ),
and, for all t∈ℝ, for all X∈ℝn(3.49c)𝒢(t,X)=-Ḟ,(3.49d)ϕ(X)=ψC(X). Then, the system (3.48) is equivalent to (A.7).ReactionsF0 and Fn+1 can be determined thanks to (3.50a)F0=-k0[K-1(G+F)]1,(3.50b)Fn+1=kn[K-1(G+F)]n.


## 3.3.2.  Clamped-Free Mechanical System

We assume that our mechanical system is clamped at its left extremity so that we can write the boundary condition (3.10a) and (3.10b). Boundary conditions for its right extremity is given later.The calculus is similar to those of [6, 7].(i)  First Case: Displacement-Force Model
We assume that the displacementx is known and that the force Fn+1 is unknown.
We introduceV, G, C and matrix K defined by (3.44a), (3.44b), (3.44d), and (3.44e) and F defined by
(3.51)F=(F1,…,Fn-1,Fn+knx)t∈ℝn,
and we obtain, as in Section 3.3.1,
(3.52)G=KV-F,(3.53)Ġ+K∂ψC(G)∋-Ḟ.
By setting (3.54a)p=n,(3.54b)Mandϕaredefinedby(3.49b)and(3.49d),
and, for all t∈ℝ, for all X∈ℝn,
(3.54c)𝒢(t,X)=-Ḟ, we remark that system (3.53) is equivalent to (A.7).
ReactionsF0 and Fn+1 can be determined thanks to
(3.55a)Fn0=-k0[K-1(G+F)]1,(3.55b)Fn+1=-knx+kn[K-1(G+F)]n.(ii)  Second Case: Force-Displacement Model
We assume that external forcingFn+1 are known and displacement x is unknown.
The calculus are similar to the previous case.
Equation (3.43b) is replaced by
(3.56)gn=Fn+1-Fn-kn-1vn-1+kn-1vn.
Following the same method, we introduce V, G, and C defined by (3.44a)-(3.44b)-(3.44d), and matrix K̃ defined by (3.11c). Vector F is defined by
(3.57)F=(F1,…,Fn-1,Fn-Fn+1)t.
So, (3.52) is replaced by
(3.58)G=K̃V-F,
and (3.48) is replaced by
(3.59)Ġ+K̃∂ψC(G)∋-Ḟ.Remark 3.1.
As in [6], let us notice that matrix K̃ defined by (3.11c) for force-displacement model corresponds to matrix K(n) for displacement-force model defined by (3.4) with
(3.60)kn=0.According to previous remark, assumption(3.61)∀i∈{1,…,n-1},ki>0
and Lemma B.1 ensure that matrix K̃ is symmetric definite positive. Thus, like previously, the system is equivalent to(3.62)Ġ+K̃∂ψC(G)∋-Ḟ.
By giving p, ϕ defined by (3.49a)–(3.49d), 𝒢 defined by for all t∈ℝ, for all X∈ℝn,(3.63a)𝒢(t,X)=-Ḟ,
and M defined by(3.63b)M=K̃∈ℳn(ℝ),
we remark that system (3.62) is equivalent to (A.7). Reactions F0 and displacement x can be determined thanks to(3.64a)F0=-k0[K̃-1(G+F)]1,(3.64b)x=-Fn+1kn+[K̃-1(G+F)]n.


## 4. Existence of Uniqueness Results

Thus, as proved in [1], all the systems of Section 3 can be written under the form (A.7) and, according to Proposition A.1 (see Appendix A), have a unique solution. For all systems, Table 1 provides the corresponding integer p, function ϕ, and matrix M. It is easy to prove that ϕ is convex proper and lower semi-continuous function on ℝp and that M is symmetric positive definite.Table 1
The dimension of the system, the convex function and the symmetric positive definite matrix used for the above described mechanical models.
SystempfunctionϕmatrixM(3.6)2nϕ(V1,V2)=∑i=1nαi|V2,i|(In00ℳ-1)(3.12)2nϕ(V1,V2)=∑i=1nαi|V2,i|(In00ℳ-1)(3.29)–(3.32)n+1ϕ(t,(G,a,b))=ψ[-α1,α1]×⋯×[-αn-1,αn-1]×{0}×{0}(G,a,b)+αnm|b|(K̂00I2)(3.29)–(3.39)n+1ϕ(t,(G,a,b))=ψ[-α1,α1]×⋯×[-αn-1,αn-1]×{0}×{0}(G,a,b)+αnm|b|(K̂00I2)(3.48)nϕ(X)=ψ[-α1,α1]×⋯×[-αn,αn](X)K(3.53)nϕ(X)=ψ[-α1,α1]×⋯×[-αn,αn](X)K(3.62)nϕ(X)=ψ[-α1,α1]×⋯×[-αn,αn](X)K̃


## 5. Convergence of Numerical Scheme

All the models examined here can be written under the form (A.7). Based on [1, 12], general writing of the implicit Euler scheme corresponds to(5.1)∀n∈{0,…,N-1},Xn+1-Xnh+M∂ϕ(Xn)∋𝒢(tn,Xn),X0=ξ.
with time step h, discretized time tn=hn, and approximations X0,…,XN of the exact solution provided by the numerical scheme. Previous studies [12] ensure that this numerical scheme is convergent with order 1/2 (systems (3.6), (3.12), (3.29)–(3.32), and (3.29)–(3.39)) or 1 (systems (3.48), (3.53), and (3.62)).In practice for computation of solutions, three cases can be distinguished, based on further expression ofXn+1:(5.2)Xn+1=[I+hM∂ϕ]-1(Xn+h𝒢(tn,Xn)),
where I is the identity and [I+hM∂ϕ]-1 is the inverse of the graph I+hM∂ϕ (see [11]). According to [11], [I+hM∂ϕ]-1 is a monovalued operator, providing a unique solution Xn+1∈ℝp. In the first case, effective computations of Xn+1 associated with diagonal matrix M is explicit: this situation corresponds to systems (3.6) and (3.12). In the second case, ϕ is defined as the indicatrix function of a closed convex set: this situation corresponds to systems (3.48), (3.53), and (3.62). Effective computation of Xn1 is given by the projection of a given vector on a closed convex set (see [6]). In the third case (for systems (3.29)–(3.32) and (3.29)–(3.39)), ϕ is involving indicatrix function of a closed convex set and a norm function. In such case, computation of Xn+1 leads to the following problem: according to [11], Xn+1 is the solution of minimization problem: considering ∥·∥M the norm define by the inner product given by (A.4)(5.3a)∥x∥M=xtM-1x,(5.3b)Zn=Xn+h𝒢(tn,Xn)
solve(5.3c)minx∈D(∂ϕ)ϕ(x)+12h∥x-Zn∥M2
and such problem can be solved in practice following efficient algorithms [13].


## 6. Conclusion

In this paper, a mechanical system involving finite degrees of freedom and nonsmooth terms have been investigated from the mechanical point of view. Dynamical, semi-dynamical, and quasistatic modeling have been established. The main results are theoretical ones:(i)
all the problems are well posed;(ii)
it has been explained how a numerical approximation of solutions can be effectively computed.All the mechanical systems have been considered in a deterministic frame. Theoretical results and corresponding effective computations could be extended to the stochastic frame.



---

*Source: 361296-2010-03-22.xml*