# Boundary Element and Augmented Lagrangian Methods for Contact Problem with Coulomb Friction

Authors: Jia Wu; Shougui Zhang
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

In this paper, boundary element and augmented Lagrangian methods for Coulomb friction contact problems are presented. Based on the projection technique, both unilateral contact and Coulomb friction conditions are reformulated as fixed point problems. The original problem is deduced to a variational formulation with boundary integral operators. Then, we propose a new augmented Lagrangian method which can be dealt with the semismooth Newton method. Short theoretical results and the algorithm description are given. Numerical simulations show the performance of the method proposed.


---

## Body

## 1. Introduction

In many industrial applications or engineering problems, the contact between a deformable elastic body and a rigid obstacle plays an important role. Due to the nonlinear boundary condition, these problems are rather complicated from both the theoretical and numerical points of view. Therefore, reliable and efficient methods for the numerical simulation of friction contact problems are quite necessary for many areas of solid mechanics. For the existence and uniqueness of a solution, the weak formulation of the considered problem leads to a variational inequality which is a traditional method for the problem [1–4]. During the past decades, a vast majority of methods developed have been based on the variational inequalities. We note that there exist two types of approaches for the numerical solution of the contact problem. An option is to discrete the problem by the finite element method (FEM) [2–5] or the boundary element method (BEM) [6–10] and obtain a convex optimization problem in the finite dimensional space. Another option is to use the Lagrange multiplier which replaces the nonlinear problem with a sequence of linear problems in function spaces, and this idea has been introduced in [11–18]. Recently, some new methods for the numerical simulation of the friction contact problem have also been developed, and we mention the penalty method and Nitsche’s method [19–27].Fixed point methods such as projection techniques are also widely applied to complementary problems including contact problems in linear elasticity [12–14, 22, 25, 26]. The basic idea of these methods is to rewrite the inequality system as an equivalent equality system by using the projection. Although it is still nonlinear and not differentiable in the classical sense, semismooth Newton methods can be applied and easily implemented in terms of a primal-dual active set strategy. The main advantage of these methods is that the generalized derivatives of the nonlinear projection function are extremely easy to compute. This is quite important to deal with the nonlinearity of the problem. We refer to the series of recent papers by Stadler et al. [12, 13] and Chouly and Hild [22, 23], where the augmented Lagrangian method (ALM) and the semismooth Newton (SSN) methods have been applied to different contact problems.The BEM has been also employed to solve the contact problem in elasticity. The main benefit of BEM is the significant reduction of expense mesh generation because the formulation of the problem is reduced to the boundary of the domain. Consequently, the boundary variational formulation expresses in terms of boundary unknowns only. For contact problems, the key unknowns are displacement and stress on the contact boundary, which are considered as primary variables and can be obtained directly in the BEM [28–30]. Therefore, the BEM seems to be the natural way for these problems [6–10, 31–37].In this paper, we focus on the combination of augmented Lagrangian methods and boundary element methods for the Coulomb friction contact problem. To be precise, we use the projection technique to deal with the nonlinear boundary conditions and obtain the corresponding equivalent formulations. Then, the weak formulation of the original problem is deduced by using boundary integral operators. When compared with other methods, there is no inequality constraint, and a boundary integral equation is introduced, which is useful from both theoretical and numerical points of view. At each iteration step, our method only needs updating boundary values and solving a boundary variational problem, and we can apply the semismooth Newton method for the solution [11, 12, 15]. Using properties of projection and the Steklov–Poincaré operator (also known as the Dirichlet-to-Neumann mapping), we obtain the convergence in function spaces.The structure of this paper is as follows. In Section2, we introduce two equivalent formulations to replace the nonlinear condition, for the classical Coulomb frictional contact problems. A weak formulation with the boundary integral operators is presented in Section 3. Section 4 is devoted to an ALM with unconditional monotone convergence for all penalty parameters. In Section 5, we give two numerical examples to illustrate the behavior of the method, and some conclusions and perspectives are drawn in the last section.


## 2. Setting of the Problem

We consider the classical Coulomb frictional contact problem with a rigid foundation. LetΩ⊂ℝ2 be an open and bounded domain with a Lipschitz boundary Γ=∂Ω. The boundary Γ consists of three mutually disjoint parts ΓD, ΓN, and ΓC≠∅, where Dirichlet, Neumann, and frictional contact conditions are prescribed. Suppose that there are no volume forces acting on the body. The case of nonvanishing volume forces can be treated using Newton potentials. Let n and t be the outward normal and tangential vectors of Γ, respectively. For a given boundary traction t^∈L2ΓN2 and obstacle g∈H1/2ΓC, the problem consists in finding the displacement u satisfying(1)−divσu=0inΩ,(2)σu=AεuinΩ,(3)u=0onΓD,(4)σun=t^onΓN,where σu is the stress tensor, εu=∇u+∇uT/2 is the linearized strain tensor, and A is the fourth-order elasticity tensor which satisfies usual conditions of symmetry, coercivity, and boundedness.OnΓ, we use decomposition of the displacement and the stress vector fields in normal and tangential as follows:(5)un=u⋅n,ut=u−unn,un=u⋅n,ut=u−unn,σnu=σnun⋅n,σtu=σun−σnun.For a given obstacleg∈H1/2ΓC, the unilateral contact condition is given by(6)un≤g,σnu≤0,σnuun−g=0onΓC,and the Coulomb friction condition reads as(7)ifut=0,thenσtu≤sσnuonΓC,(8)ifut≠0,thenσtu=−sσnuututonΓC,where s is the friction coefficient.Let us introduce the following Hilbert spaces:(9)V≔v∈H1Ω2,v=0onΓD,K≔v∈V,un−g≤0onΓC,and notations:(10)au,v≔∫Ωσu:εvdx,jv≔∫ΓCsvtdsx,Lv≔∫ΓNt^x⋅vxdsx.In the above definition, for anyA=aij∈R2×2,B=aij∈R2×2, the product is defined as A:B=∑i,jaijbij.Problems (1)–(8) are then equivalent to the following variational inequality:(11)findu∈Ksatisfyingau,v−u+ju−jv≥Lv−u,∀v∈K,or a convex minimization problem:(12)findu∈K,such thatJu=minv∈KJv≔12av,v−Lv+jv.It has been proved that problem (11) or equivalently (12) has a unique solution in the theory of variational inequalities [1, 13, 23].For this problem, the major difficulty is due to the coupling between the contact condition and the friction condition. Moreover, the transition between contact and noncontact is characterized by a change in the type of the boundary condition with constraint. In this paper, we rewrite the friction contact conditions (6)–(8) as equivalent projection formulas. Let us introduce the projection notation a+ for a∈ℝ:(13)a+≔a,ifa>0,0,otherwise,and the projection notation xα for x∈ℝ2, α>0:(14)xα≔x,ifx≤α,αxx,otherwise.As a result, we obtain the following results for the contact and friction conditions onΓC.Lemma 1.
For allρ1>0, the contact conditions (6) on ΓC are equivalent to(15)σnu+ρ1un−g−σnu+=0onΓC.The proof can be found in [22, 35–37]. For the friction condition, we also have the following result according to [23, 37].Lemma 2.
For allρ2>0 and s>0, the contact conditions (7) and (8) on ΓC are equivalent to(16)σtu+ρ2ut−σtusσnu=0onΓC.Proof.
Firstly, let us assume that (7) and (8) hold. Let ut=0, and then from (7), we have(17)ρ2ut−σtusσnu=−σtus|σnu=−σtu.
For the caseut≠0, from (8), we have(18)ρ2ut−σtu=ρ2+sσnuutut.
It results thatρ2ut−σtu=ρ2ut+sσnu≥sσnu, which means that(19)ρ2ut−σtusσnu=sσnuutut.
Note thatρ2ut−σtu and ut have the same orientation. Then, using (8) yields(20)σtu=−sσnuutut=−ρ2ut−σtusσnu.
On the other hand, letut and σtu in (16) hold. We first assume that ρ2ut−σtu≤sσnu, that is, ρ2ut−σtusσnu=ρ2ut−σtu. Thus, (16) is(21)σtu+ρ2ut−σtu=0,so that ut=0 and σtu=ρ2ut−σtu≤sσnu. Then, (7) is satisfied. Finally, we consider the case ρ2ut−σtu>sσnu, which means that ut2+σtu2≠0 and there exists ω∈0,1 such that ρ2ut−σtusσnu=ωρ2ut−σtu. Then, from (16), we have(22)σtu=ρ2ωω−1ut,σtu=sσnu,so that in this case,(23)sσnu=ρ2ω1−ωut.
This implies, due toρ2ω/ω−1<0 and relationship (16), that ut≠0, σtu≠0, and σtu=−sσnuut/ut. Hence, the friction conditions (7) and (8) hold.


## 3. Boundary Weak Formulation of the Frictional Contact Problem

In order to give a boundary variational formulation to the frictional contact problem, we apply Green’s formula to (1)–(8) and obtain the following variational formulation:(24)∫Ωσu:∈vdx=∫ΓCσun⋅vds+Lv,∀v∈V.LetGx,y be the fundamental solution of the two-dimensional Lamé equation:(25)Gx,y=λ+3μ4πμλ+2μlog1x−yI+λ+μλ+3μx−yx−yTx−y2,where Lamé constants λ and μ are given by Young’s modulus E>0 and Poisson ratio v∈0,1/2 as(26)λ=Ev1+v1−2v,μ=E21+v,and I is the 2×2 unit matrix. We then introduce boundary integral operators: the single layer potential V, the double layer potential K, the adjoint double layer potential K′, and the hypersingular integral operator W:(27)Vϕx=∫ΓGx,yϕydsy,V:H−1/2Γ2⟶H1/2Γ2,Kϕx=∫ΓTnyGx,yϕydsy,K:H1/2Γ2⟶H1/2Γ2,K′ϕx=∫ΓTnxGx,yϕydsy,K′:H−1/2Γ2⟶H−1/2Γ2,Wϕx=−Tnx∫ΓTnyGx,yϕydsy,W:H1/2Γ2⟶H−1/2Γ2,where Tn is the boundary traction operator defined by Tnu≔σuΓn. Now, we can define the Steklov–Poincaré operator S by(28)S=W+12I+K′V−112I+K.Properties of the boundary integral operators yield that the Steklov–Poincaré operatorS is linear, symmetric, and positive definite on H01/2Γ,ΓD=v∈H1/2Γ2,v=0onΓD [30–32], i.e., there exists a positive constant α>0 such that for any v∈H01/2Γ,ΓD,(29)Sv,vΓ≥αvH01/2Γ,ΓD2.Furthermore, we note the Dirichlet-to-Neumann mapping onΓ:(30)S:H1/2Γ2⟶H−1/2Γ2uΓ⟼σuΓ,and SuΓ=σuΓ, and then we get(31)∫Ωσu:∈vdx=∫ΓSuΓ⋅vds,∀v∈V.Let us define(32)Su,vΓ≔∫ΓSux⋅vxdsx,σu,vΓC≔∫ΓCσux⋅vxdsx.By substituting (31) into (24), we use σtu=σun−σnun and obtain the boundary weak formulation of (1)–(4):(33)Su,vΓ−σnu,vnΓC−σtu,vtΓC=Lv,∀v∈H01/2Γ,ΓD.Consequently, we obtain boundary weak formulation (33) and projection fixed point problems (15) and (16), which are equivalent to the original problems (1)–(8) and useful for the numerical and theoretical analysis.


## 4. Boundary Element and Augmented Lagrangian Methods for the Contact Problem

With the above results, we present our boundary element and augmented Lagrangian methods for the Coulomb frictional contact problem as follows.We will detail the above algorithm in Section5 using the semismooth Newton method.Letu∗ and σ∗ be the unique solution of the problems (1)–(8) and corresponding stress tensor on the boundary Γ, respectively. From Lemma 1 and Lemma 2, u∗ and σ∗ satisfy (15) and (16). In order to analyze the convergence of Algorithm 1, we introduce the following notation:(34)Bρun,σn≔Bun,σnu;g,ρ≔ρun−g−σnu,and will use the following projection properties on ΓC [12, 13, 35].Algorithm 1: Augmented Lagrangian method (ALM).
Step 0: choose initial functionσn0,σt0∈L2ΓC and large enough ρ1,ρ2∈ℝ+, and set k≔0.Step 1: finduk+1 and σk+1 such that Suk+1,vΓ−σnk+1,vnΓC−σtk+1,vtΓC=Lv,∀v∈H01/2Γ,ΓD, with σnk+1+ρ1unk+1−g−σnk+=0onΓC,σtk+1+ρ2utk+1−σtkμσnk=0onΓC.Step 2: check some stopping criterion, and if it is satisfied then stop, else updatek≔k+1 and go to Step 1.Lemma 3.
For allunk,σnk∈L2ΓC generated by the first condition in Step 1 in Algorithm 1, it holds that(35)σnk+1−σn∗,Bρ1un∗,σn∗−Bρ1unk+1,σnkΓC≥σnk+1−σn∗ΓC2.Lemma 4.
For allσtk,utk∈L2ΓC2 generated by the second condition in Step 1 in Algorithm 1, it holds that(36)σtk+1−σt∗,ρ2ut∗−σt∗−ρ2utk+1−σtkΓC≥σtk+1−σt∗ΓC2.Proof.
Let us separateΓC into four subparts ΓC1, ΓC2, ΓC3, and ΓC4, where(37)ρ2utk+1−σtk≥sσnu,ρ2ut∗−σt∗≥sσnuonΓC1,ρ2utk+1−σtk≥sσnu,ρ2ut∗−σt∗<sσnuonΓC2,ρ2utk+1−σtk<sσnu,ρ2ut∗−σt∗≥sσnuonΓC3,ρ2utk+1−σtk<sσnu,ρ2ut∗−σt∗<sσnuonΓC4.
From (16) and the second condition in Step 1 in Algorithm 1, we have(38)σtk+1−σt∗≤ρ2ut∗−σt∗−ρ2utk+1−σtkonΓC1,σtk+1−σt∗<ρ2ut∗−σt∗−ρ2utk+1−σtkonΓC2,σtk+1−σt∗<ρ2ut∗−σt∗−ρ2utk+1−σtkonΓC3,σtk+1−σt∗=ρ2ut∗−σt∗−ρ2utk+1−σtkonΓC4.
Sinceσtk+1−σt∗ and ρ2ut∗−σt∗−ρ2utk+1−σtk have the same orientation, it follows that(39)σtk+1−σt∗,ρ2ut∗−σt∗−ρ2utk+1−σtkΓC≥σtk+1−σt∗ΓC2.Theorem 1.
Letuk be the sequence generated by Algorithm 1; then, uk converges to u∗ in H01/2Γ,ΓD as k⟶∞.Proof.
Letδuk≔uk−u∗, δunk≔unk−un∗, and δutk≔utk−ut∗, then δuk∈H01/2Γ,ΓD, δσtk,δutk∈L2ΓC2, and δunk,δσnk∈L2ΓC. We take u=u∗, σ=σ∗, and v=δuk+1 in (33). This yields(40)Su∗,δuk+1Γ−σn∗,δunk+1ΓC−σt∗,δutk+1ΓC=Lδuk+1.
Letv=δuk+1 in the equation in Step 1 in Algorithm 1, then we have(41)Suk+1,δuk+1Γ−σnk+1,δunk+1ΓC−σtk+1,δutk+1ΓC=Lδuk+1.
Subtracting (40) from (41) results in(42)Sδuk+1,δuk+1Γ=δσnk+1,δunk+1ΓC+δσtk+1,δutk+1ΓC.
Using Lemmas3 and 4 and Young’s inequality, we obtain(43)δσnk+1,δunk+1ΓC=ρ1−1δσnk+1,Bρ1unk+1,σnk−Bρ1un∗,σn∗ΓC+ρ1−1δσnk+1,δσnkΓC≤ρ1−1δσnk,δσnk+1ΓC−ρ1−1δσnk+1ΓC2≤2ρ1−1δσnkΓC2−2ρ1−1δσnk+1ΓC2,δσtk+1,δutk+1ΓC=ρ2−1δσtk+1,ρ2utk+1−σtk−ρ2ut∗−σt∗ΓC+ρ2−1δσtk+1,δσtkΓC≤ρ2−1δσtk+1,δσtkΓC−ρ2−1σtk+1−σt∗ΓC2≤2ρ2−1δσtkΓC2−2ρ2−1δσtk+1ΓC2.
From (31) and (42), it follows that(44)Sδuk+1,δuk+1Γ≤2ρ1−1δσnkΓC2−2ρ1−1δσnk+1ΓC2+2ρ2−1δσtkΓC2−2ρ2−1δσtk+1ΓC2,(45)Sδuk+1,δuk+1Γ≥αδuk+1H01/2Γ,ΓD2.
Combining (44) and (45) yields(46)αδuk+1H01/2Γ,ΓD2≤2ρ1−1δσnkΓC2−2ρ1−1δσnk+1ΓC2+2ρ2−1δσtkΓC2−2ρ2−1δσtk+1ΓC2.
Summing up (46) with respect to k, we obtain(47)∑k=0∞αδuk+1H01/2Γ,ΓD2≤2ρ1−1δσn0ΓC2+2ρ2−1δσt0ΓC2<∞,this shows that(48)limk⟶∞δuk+1H01/2Γ,ΓD2=0.
Consequently,uk converges to u∗ in H01/2Γ,ΓD.
Note that in (46), we can observe that the larger value of parameters ρ1 and ρ2 results in faster convergence to the algorithm. Therefore, this method is reliable and efficient for friction contact problems.


## 5. Numerical Examples

In this section, we present two test examples for the algorithm proposed in Section4 for the solution of friction contact problems. In order to simplify the numerical process, we use constant boundary elements to approximate u and σ and solve the corresponding linear systems. Here, we choose uk+1−uk∞,ΓC≤10−8uk+1∞,ΓC as the stopping criterion, and all computations are performed in Matlab codes. To deal with the nonlinearity of Algorithm 1, we apply the semismooth Newton method to Step 1 [12, 13] as given in Algorithm 2.Algorithm 2: Semismooth Newton (SSN) method.
Step 0: chooseu0,σ0∈H01/2Γ,ΓD×L2ΓC2 and large enough ρ1,ρ2∈ℝ+, and set k=0.Step 1: determineAnk+1=x∈ΓC:ρ1unk−g−σnk>0,ℐnk+1=ΓC\Ank+1,Atk+1=x∈ΓC:ρ2utk−σtk>sσnk,ℐtk+1=ΓC\Atk+1.Step 2: solveSuk+1,vΓ−σnk+1,vnΓC−σtk+1,vtΓC=Lv,∀v∈H01/2Γ,ΓD, with the boundary conditions σnk+1=0onℐnk+1,σnk−1unk+1−gonAnk+1;,σtk+1=σtk−ρ2utk+1onℐtk+1,sσnksignρ2utk−σtkonAtk+1; and obtain uk+1 and σk+1 on ΓC.Step 3: updatek=k+1 and go to Step 1.


### 5.1. First Example

We consider an elastic square bodyΩ=0,1×0,1 with a rigid foundation and ΓC=0,1×0. The Neumann boundary condition is given by t^=0,−1⊤ on 0.5,1×1, t^=1,0⊤ on 0×0.5,1, and t^=0 on the rest of ΓN. And, the symmetry conditions (i.e., un=0,σtu=0) are applied on ΓS≔1×0,1. Young’s modulus and Poisson’s ratio are E = 10000 and v=0.2, respectively. This problem is not V-elliptic but only K-elliptic [38].LetN denote the number of boundary elements on Γ. First, we apply our method to this problem with N = 320, ρ1 = 10000, and ρ2 = 10000, and the initial and deformed configurations of the body are given in Figure 1. In addition, the surface displacement and traction are depicted in Figures 2 and 3, respectively. We observe that ΓC is divided into two parts: the right part where the body remains in contact with the axis y = 0 and the left part of ΓC where it separates from this axis with a separation point near (0.26, 0), and the contact part is divided into a slip part and a stick part with a transition point from the slip to the stick near (0.47, 0). It can be seen that our results are in a good agreement with those in ref. [38].Figure 1
Initial and deformed configurations.Figure 2
Normal and tangential displacements onΓC.Figure 3
Normal and tangential tractions onΓC.Next, we solve the problem by choosing different parametersρ1 and ρ2 to investigate the convergence behavior of our method. Table 1 gives the number of iterations for N = 320. We note that numerical results converge quickly when the parameters ρ1 and ρ2 are sufficiently large.Table 1
Number of iterations to differentN and ρ.
ρ2=102ρ2=103ρ2=104ρ2=105ρ2=106ρ1=1022525252525ρ1=1032513131313ρ1=1042513131313ρ1=1052513131313ρ1=1062513131313


### 5.2. Second Example

In this example, we consider another model problem of the elastic square bodyΩ=0,1×0,1 in contact with a rigid foundation on the bottom side. In addition to the Neumann data t^=0,−x21−x2⊤ applied on the top of the body, we set symmetric Neumann data t^=0,2y21−y2⊤ on the left side and t^=0,−2y21−y2⊤ on the right side, respectively. Similarly, the symmetry conditions (i.e., un=0,σtu=0) are applied on ΓS≔0.5×0,1 for this example [38]. We use Young’s modulus E = 10000 and Poisson’s ratio v=0.3 for the test.We chooseN = 320, ρ1 = 10000, and ρ2 = 10000 again and apply our method to this problem. Figure 4 depicts the initial and deformed configurations of the half of the body (because of the symmetry). Figures 5 and 6 give the numerical results of the surface displacement and traction. One can observe that there is a transition point between contact and separation near (0.08, 0) on ΓC, which matches well with the numerical results in ref. [38]. In Table 2, we also list the number of iterations for the convergence with respect to different parameters ρ1 and ρ2. As it can be seen from our tests, our method also converges quickly when ρ1 and ρ2 are sufficiently large.Figure 4
Initial and deformed configurations.Figure 5
Normal and tangential displacements onΓC.Figure 6
Normal and tangential tractions onΓC.Table 2
Number of iterations to differentN and ρ.
ρ2=102ρ2=103ρ2=104ρ2=105ρ2=106ρ1=1023938383838ρ1=1032912121212ρ1=1042910888ρ1=1052910888ρ1=1062910888


## 5.1. First Example

We consider an elastic square bodyΩ=0,1×0,1 with a rigid foundation and ΓC=0,1×0. The Neumann boundary condition is given by t^=0,−1⊤ on 0.5,1×1, t^=1,0⊤ on 0×0.5,1, and t^=0 on the rest of ΓN. And, the symmetry conditions (i.e., un=0,σtu=0) are applied on ΓS≔1×0,1. Young’s modulus and Poisson’s ratio are E = 10000 and v=0.2, respectively. This problem is not V-elliptic but only K-elliptic [38].LetN denote the number of boundary elements on Γ. First, we apply our method to this problem with N = 320, ρ1 = 10000, and ρ2 = 10000, and the initial and deformed configurations of the body are given in Figure 1. In addition, the surface displacement and traction are depicted in Figures 2 and 3, respectively. We observe that ΓC is divided into two parts: the right part where the body remains in contact with the axis y = 0 and the left part of ΓC where it separates from this axis with a separation point near (0.26, 0), and the contact part is divided into a slip part and a stick part with a transition point from the slip to the stick near (0.47, 0). It can be seen that our results are in a good agreement with those in ref. [38].Figure 1
Initial and deformed configurations.Figure 2
Normal and tangential displacements onΓC.Figure 3
Normal and tangential tractions onΓC.Next, we solve the problem by choosing different parametersρ1 and ρ2 to investigate the convergence behavior of our method. Table 1 gives the number of iterations for N = 320. We note that numerical results converge quickly when the parameters ρ1 and ρ2 are sufficiently large.Table 1
Number of iterations to differentN and ρ.
ρ2=102ρ2=103ρ2=104ρ2=105ρ2=106ρ1=1022525252525ρ1=1032513131313ρ1=1042513131313ρ1=1052513131313ρ1=1062513131313


## 5.2. Second Example

In this example, we consider another model problem of the elastic square bodyΩ=0,1×0,1 in contact with a rigid foundation on the bottom side. In addition to the Neumann data t^=0,−x21−x2⊤ applied on the top of the body, we set symmetric Neumann data t^=0,2y21−y2⊤ on the left side and t^=0,−2y21−y2⊤ on the right side, respectively. Similarly, the symmetry conditions (i.e., un=0,σtu=0) are applied on ΓS≔0.5×0,1 for this example [38]. We use Young’s modulus E = 10000 and Poisson’s ratio v=0.3 for the test.We chooseN = 320, ρ1 = 10000, and ρ2 = 10000 again and apply our method to this problem. Figure 4 depicts the initial and deformed configurations of the half of the body (because of the symmetry). Figures 5 and 6 give the numerical results of the surface displacement and traction. One can observe that there is a transition point between contact and separation near (0.08, 0) on ΓC, which matches well with the numerical results in ref. [38]. In Table 2, we also list the number of iterations for the convergence with respect to different parameters ρ1 and ρ2. As it can be seen from our tests, our method also converges quickly when ρ1 and ρ2 are sufficiently large.Figure 4
Initial and deformed configurations.Figure 5
Normal and tangential displacements onΓC.Figure 6
Normal and tangential tractions onΓC.Table 2
Number of iterations to differentN and ρ.
ρ2=102ρ2=103ρ2=104ρ2=105ρ2=106ρ1=1023938383838ρ1=1032912121212ρ1=1042910888ρ1=1052910888ρ1=1062910888


## 6. Conclusion

In this paper, we have studied a new method for the solution of contact and friction problems and its convergence analysis. The method is developed by the ALM and the BEM. In particular, we have shown that our method shows unconditional convergence and converges quickly when the parametersρ1 and ρ2 are sufficiently large. Although this method requires solving a nonlinear elasticity problem at each iteration, it can be easily dealt with by using the SSN. The numerical examples demonstrate the remarkable efficiency and reliability of the method.



---

*Source: 7490736-2020-07-17.xml*