EFFICIENT ALGORITHMS FOR PROBLEMS WITH FRICTION ∗ 1

EFFICIENT ALGORITHMS FOR PROBLEMS WITH FRICTION

∗

† , A. MATEI‡ , AND B.I. WOHLMUTH§ ¨ S. HUEBER

Abstract. In this paper, a nonconforming discretization method for the frictional contact between two bodies subjected to antiplane shear deformation is considered. The method is based on a mixed variational formulation where for the discretization of the Lagrange multiplier dual basis functions are used. Under some regularity assumptions on the solution, an optimal a priori error estimate is obtained. To solve the discrete nonlinear problem an inexact primal-dual active set strategy is introduced. Finally, numerical examples confirming the theoretical result for the a priori error estimate and illustrating the performance of the algorithm are presented. The results can easily be generalized to the case of Coulomb friction, and a numerical example is given. Key words. antiplane shear deformation, frictional contact, mixed formulation, dual Lagrange multipliers, nonconforming meshes, optimal a priori error estimates, inexact primal-dual active set strategy. AMS subject classifications. 65K10, 65N15, 65N30, 65N55, 74B10, 74M10, 74M15, 74S05

1. Introduction. The mechanical model used in this paper involve the particular type of deformation that a solid can undergo, the antiplane shear deformation. For a cylindrical body subject to antiplane shear, the displacement is parallel to the generators of the cylinder and is independent of the axial coordinate. The displacement field is described by a scalar function and therefore the governing equations and boundary conditions are quite simple. The antiplane shear (or longitudinal shear, generalized shear) may be viewed as complementary to the plane strain deformation, and represents the Mode III, fracture mode for crack problems. This model was considered by many authors, see for instance [BHP02, HMM02, Hor95, HS01, HS04, Kno76, Kno77, MMS01]. An excellent reference concerning the antiplane shear deformation in solid mechanics is the review article [Hor95], where modern developments for the antiplane shear model and its applications are described for both linear and nonlinear solid mechanics. Note that in [HMM02, MMS01] quasistatic antiplane contact problems were considered, and the study was made within the framework of variational inequalities. For the basics in the theory of variational inequalities we refer to [GLT81, Glo84]. In this paper, we study the antiplane shear deformation of two elastic bodies in frictional contact on their common boundary. To model the friction, we use Tresca’s law; a description of this law can be found in [DL76, Pan85] and very recently in [HS02, SST04]. Our study is based on a mixed variational formulation with dual Lagrange multipliers, the well-possednes of this weak formulation being guaranteed by arguments in the saddle point theory; details on the saddle point theory and its applications can be found, e.g., in [Bra97, BF91, ET76, Hcc96]. To discretize our frictional problem, we use a mortar technique on nonconforming meshes. For mortar ∗ This work was supported in part by the Deutsche Forschungsgemeinschaft, SFB 404, B8, and by the Research Project Nonlinear approximation and adaptivity: breaking complexity in numerical modeling and data representation carried out in the framework of the European Community Program Improving the Human Potential Network (HPRN-CT-2002-00286). † Institut f¨ ur Angewandte Analysis und Numerische Simulation (IANS), Universit¨ at Stuttgart, Pfaffenwaldring 57, D-70569 Stuttgart, Germany; [email protected]. ‡ Department of Mathematics, University of Craiova, A.I. Cuza street 13, 200585, Craiova, Romania; [email protected]. § Institut f¨ ur Angewandte Analysis und Numerische Simulation (IANS), Universit¨ at Stuttgart, Pfaffenwaldring 57, D-70569 Stuttgart, Germany; [email protected].

1

2 techniques with standard Lagrange multipliers, we refer to [BR03, BHL99, CHLS01, Hil00, HL02]. Mortar techniques with dual Lagrange multipliers were recently used in [HW05b, HW05a], within the framework of frictionless unilateral multibody contact problems. To solve the resulting nonlinear algebraic form of the discrete problem, we use a primal-dual active set strategy based on [Sta04]. The use of dual Lagrange multipliers allows us to combine this method with an optimal multigrid solver, see [WK03], and we obtain an inexact method. This approach results in an efficient iterative solver for the nonlinear problem with a negligible additional effort compared to solving a linear problem. The structure of this paper is the following one. In Section 2, we describe the frictional antiplane contact problem involving two elastic bodies, and we provide a week formulation. In Section 3, we propose a nonconforming discretization method based on dual Lagrange multipliers, see [Woh00]. We obtain an optimal a priori error estimate of order h1/2 + ν, 0 < ν ≤ 21 , for the solution and for the Lagrange multiplier if the solution is regular enough. In Section 4, the inexact primal-dual active set algorithm is presented. Finally, numerical examples confirming our theoretical results and illustrating the performance of our algorithm are given in Section 5. Section 6 contains the generalization of our algorithm to the case of Coulomb friction. 2. An antiplane frictional contact problem. Let us consider two deformable solids, B m , B s ⊂ R3 , that are cylinders, having the generators parallel to the x3 -axis, of a rectangular cartesian coordinate system Ox1 x2 x3 . The subscripts m and s are motivated by the presence of the mortar setting in the numerical analysis of our model. Everywhere below, we will use a superscript k to indicate that a quantity is related to the cylinder B k , k = m, s. We assume that the bodies are homogeneous, isotropic and elastic media; more precisely, we will use the constitutive law σ k = λk tr(ε(uk )) Id +2µk ε(uk )

in

Bk ,

(2.1)

where σ k = (σ kij ) denotes the stress field, ε(uk ) = (εij (uk )) the linearized strain P3 tensor, λk > 0 and µk > 0 are the Lam´e coefficients, tr(ε(uk )) = i=1 εii (uk ) and Id is the unit tensor in R3 . Moreover, we assume that the generators are sufficiently long so that end effects in the axial direction are negligible. Let us denote by Ω k a cross-section, which is a domain in R2 . Thus, B k = Ωk × (−∞, +∞). For each domain Ωk , we assume that its boundary Γk is Lipschitz continuous and is divided into three disjoint measurable parts Γk1 , Γk2 and Γk3 , with meas (Γk1 ) > 0. We assume that the bodies are clamped on Γk1 , body forces of density f k0 act on Ωk and surface tractions of density f k2 act on Γk2 . Moreover, we assume that the bodies in the initial configuration are in contact on their common boundary part Γ3 = Γ13 = Γ23 and that Γ3 is a compact k subset of ∂Ωk \Γ1 , k = m, s. We load the solid in a special way, as follows, f k0 = (0, 0, f0k ),

f0k = f0k (x1 , x2 ) : Ωk → IR,

(2.2)

f k2

f2k

(2.3)

=

(0, 0, f2k ),

=

f2k (x1 , x2 )

:

Γk2

→ IR.

The unit outward normal on Γk × (−∞, +∞) is denoted by nk , nk = (nk1 , nk2 , 0) and is defined a.e.. We note that on Γ3 , n := ns = −nm ,

σ s ns = −σ m nm ,

σ τ := σ sτ = −σ m τ .

For a vector v k , we denote by vnk and v kτ its normal and tangential parts on the boundary, given by vnk := v k · nk and v kτ := v k − vnk nk , respectively. Furthermore, for

3 a given stress field σ k , we denote by σnk and σ kτ the normal and the tangential parts on the boundary, that is σnk := (σ k nk ) · nk and σ kτ := σ k nk − σnk nk , respectively. The body forces (2.2) and the surface tractions (2.3) would be expected to give rise to a deformation of the elastic cylinder B k , such that the displacement uk is of the form uk = (0, 0, uk ),

uk = uk (x1 , x2 ) : Ωk → R.

(2.4)

Consequently, keeping in mind (2.1), the Cauchy stress vector on Γk × (−∞, +∞) is given by σ k nk = (0, 0, µk ∂n uk ), where, as usual, ∂n uk := ∇uk · nk . Moreover, we note that Div σ k = (0, 0, µk ∆uk ).

(2.5)

According to the physical setting, we have Div σ k + f k0 = 0 k

u =0 k

k

σ n =

f k2

in

Bk ,

(2.6)

on

Γk1 Γk2

× (−∞, +∞),

(2.7)

× (−∞, +∞).

(2.8)

on

Finally, we have to describe the frictional contact condition on Γ3 × (−∞, +∞). Everywhere in the sequel, we will use | · | to denote the Euclidean norm on IRd , d = 1 or 3. Since usn = um n = 0, on Γ3 × (−∞, +∞) the contact is bilateral. Let us model the friction using Tresca’s law,   |σ τ | ≤ g, |σ τ | < g ⇒ usτ − um on Γ3 × (−∞, +∞), (2.9) τ = 0,  ), β > 0, |σ τ | = g ⇒ σ τ = −β(usτ − um τ where g is a given function, called friction bound. We note that ukτ = (0, 0, uk ) and σ kτ = (0, 0, στk ) with στk = µk ∂n uk . In addition, we have ∂n u := µs ∂n us = −µm ∂n um

on

Γ3 .

(2.10)

Using (2.6)-(2.9) and keeping in mind (2.4)-(2.5), the study of our 3D mechanical model reduces to the study of the following mathematical problem. Problem 2.1. Find the displacement fields uk : Ω → R, k = m, s, such that µk ∆ uk + f0k = 0 k

u =0 k

k

µ ∂n u =

f2k

in

Ωk ,

(2.11)

on

Γk1 , Γk2 ,

(2.12)

on

and the friction law   |∂n u| ≤ g, |∂n u| < g ⇒ us − um = 0,  |∂n u| = g ⇒ ∂n u = −β(us − um ), holds.

(2.13)

on β > 0,

Γ3

(2.14)

4 In the study of Problem 2.1, we assume: f2k ∈ L2 (Γk2 ),

f0k ∈ L2 (Ωk ), 2

g ∈ L (Γ3 ),

(2.15)

g ≥ 0 a.e. on Γ3 .

(2.16)

In order to obtain a weak formulation for Problem 2.1, we introduce the Hilbert space o n V k := v k ∈ H 1 (Ωk ) | v k = 0 on Γk1 endowed with the inner product given by Z k k ∇uk · ∇v k dx, (u , v )V k :=

uk , v k ∈ V k ,

Ωk

and the associated norm, kv k kV k := k∇v k kL2 (Ωk )2 ,

vk ∈ V k ,

which is equivalent with k · k1,Ωk . Let us consider the product space V := V m × V s and let a : V × V → R be the bilinear form a(u, v) :=

2 X

µk (uk , v k )V k .

k=1

Clearly, this form is continuous with the continuity constant Ma = µm + µs and V -elliptic, with the V -ellipticity constant ma = min{µm , µs }. Defining f ∈ V such that ! Ã Z X Z k k k k f2 v ds , f0 v dx + (f, v)V = Γk 2

Ωk

k=m,s

due to Green’s formula, for regular functions um and us satisfying (2.11)-(2.13), we get Z (µm ∂n um v m + µs ∂n us v s ) ds, v = (v m , v s ) ∈ V. (2.17) a(u, v) = (f, v)V + Γ3

Keeping in mind (2.10), we define a Lagrange multiplier λ in M := (H 1/2 (Γ3 ))0 , Z ∂n u w ds, w ∈ H 1/2 (Γ3 ), (2.18) hλ, wiΓ3 := − Γ3

where h·, ·iΓ3 denotes the duality pairing. Furthermore, we introduce a nonempty closed convex set, Z o n g |w| ds, w ∈ H 1/2 (Γ3 ) . (2.19) Λ := ζ ∈ M : hζ, wiΓ3 ≤ Γ3

Clearly, λ ∈ Λ and we can rewrite (2.17) as a(u, v) = (f, v)V − hλ, [v]iΓ3 ,

v ∈ V,

(2.20)

5 where [v] denotes the jump defined on Γ3 by [v] := v s − v m ,

v = (v m , v s ) ∈ V.

Let us denote by b : V × M → R, the bilinear and continuous form b(v, ζ) := hζ, [v]iΓ3 .

(2.21)

Keeping in mind the properties of the trace operator, it can be shown that the form b(·, ·) satisfies the inf-sup property: there exists a constant α > 0 such that inf

sup

06=µ∈M 06=v∈V

b(v, ζ) ≥ α, kvkV kζkM

where k · kM denotes the natural norm of (H 1/2 (Γ3 ))0 . The friction law (2.14) leads to the identity Z Z g|[u]| ds. ∂n u [u] ds = − Γ3

Γ3

Taking into account the definition of λ, (2.18), the definition of b(·, ·), (2.21), and the definition of Λ, (2.19), we can write Z Z g|[u]| ds, ζ ∈ Λ. g|[u]| ds and b(u, ζ) ≤ b(u, λ) = Γ3

Γ3

Consequently, we get the inequality b(u, ζ − λ) ≤ 0,

ζ ∈ Λ.

Combining this, (2.20) and (2.21), we obtain the following weak formulation of Problem 2.1. Problem 2.2. Find u ∈ V and λ ∈ Λ such that a(u, v) + b(v, λ) = (f, v)V ,

v ∈ V,

b(u, ζ − λ)

ζ ∈ Λ.

≤ 0,

The well-possedness of Problem 2.2 is given by the following theorem. Theorem 2.1. Assume that (2.15) and (2.16) hold. Then, there exists a unique solution (u, λ) ∈ V × Λ of Problem 2.2. Moreover, if f1 , f2 are elements in V corresponding to the sets of data (f0 , f2 )1 , (f0 , f2 )2 , respectively, we have the stability result, ku1 − u2 kV + kλ1 − λ2 kM ≤

α + m a + Ma kf1 − f2 kV , α ma

(2.22)

where (u1 , λ1 ), (u2 , λ2 ) are the solutions of Problem 2.2 corresponding to f1 , f2 ∈ V, respectively. In order to prove the existence and the uniqueness of the solution of Problem 2.2, we can use Theorem 3.11 in [Hcc96]. The stability result can be obtained by standard techniques and using the inf-sup property of the bilinear form b(·, ·).

6 3. Nonconforming discretization and optimal a priori error estimates. In this section, we give the discretization of Problem 2.2, and we derive an optimal a priori error estimate for the discretization error. To do so, we introduce a suitable basis transformation of our nodal finite element basis. Let us assume that the bodies Ωk , k = m, s, are polygonal domains. To approximate V , we use lowest order finite elements on simplicial or quadrilateral triangulations. The finite element space associated with the shape regular triangulation Th,Ωk is denoted by S1 (Ωk , Th,Ωk ). The meshsize h is defined by the maximal diameter of the elements in Th,Ωm and Th,Ωs . For simplicity, we assume that Γk1 , k = m, s, and Γ3 can be written as union of edges, Γ3 both from the triangulation Th,Ωs of the slave side and from the triangulation Th,Ωm of the master side. Before introducing the discrete spaces, we decompose the set of all vertices into three disjoint sets S, M and N . By the first subset S, we denote all vertices on Γ3 of the triangulation Th,Ωs on the slave side, by M all vertices on Γ3 of the triangulation Th,Ωm on the master side. The set N contains all remaining ones. Then we have for the discrete spaces Vhk o n ¡ ¢ Vhk := vhk ∈ S1 Ωk , Th,Ωk : vhk = 0 on Γk1 ⊂ V k ,

and we define Vh := Vhm × Vhs . For the discretization of the Lagrange multiplier space we use dual shape functions, introduced in [Woh00]. In the case of linear or bilinear finite elements in 2D, the dual basis functions are associated with the vertices. We use discontinuous piecewise linear functions having value two at the associated vertex and value minus one at the two neighbor vertices as basis functions. We denote this discrete Lagrange multiplier space by Mh = span{ψp , p ∈ S}, where ψp denotes the basis function associated with the vertex p. Then the biorthogonality of the basis functions yields Z hψp , φq iΓ3 = δpq φq ds, p, q ∈ S, (3.1) Γ3

where φq are the standard nodal basis functions of S1 (Ωs , Th,Ωs ) associated with the vertex q. The finite element space Vh can be written in terms of the standard finite element basis φ as Vh = span{φp , p ∈ S ∪ M ∪ N }. Additionaly to the basis φ, ˆ see [WK03]. To introduce these we introduce the constrained finite element basis φ, basis functions, we define the entries of the coupling matrices D and M between the finite element basis functions φp and the basis functions for the Lagrange multiplier space ψp by D[p, q] := hψp , φq iΓ3 ,

p, q ∈ S,

M [p, q] := hψp , φq iΓ3 ,

p ∈ S, q ∈ M.

ˆ := D−1 M , Due to the biorthogonality (3.1), the matrix D is diagonal. In terms of M ˆ we obtain the constrained basis φ of Vh from the nodal basis φ of Vh by the transformation   ˆ   φN Id 0 0 φN ˆ := φˆM  :=  0 Id M ˆ >  φM  =: Qφ. φ (3.2) ˆ φ 0 0 Id S φS

We note that only basis functions associated with a node p ∈ M are changed, and that by definition b(φˆq , ψp ) = 0,

p ∈ S, q ∈ M.

7 For simplicity of notation, we use the same symbol for a function in Vh and Mh as for its algebraic representation with respect to the nodal basis. Let vh be the algebraic representation of an element vh ∈ Vh with respect to the basis φ and let vˆh be the ˆ Then corresponding algebraic representation with respect to the constrained basis φ. > we have the relation vh = Q vˆh . Now, after an easy computation, taking into account the biorthogonality (3.1), we get X X X X b(vh , µh ) = hµh , vp φ p − vq φq iΓ3 = hµh , vˆp φˆp iΓ3 = hµh , vˆp φp iΓ3 . p∈S

q∈M

p∈S

p∈S

Before introducing the discrete set Λh for the admissible Lagrange multiplier, we define for vh ∈ Vh the restriction to the slave side of the interface Γ3 by X vp φ p . vh,S := p∈S

Now we define the discrete convex set Λh by Z n Λh := µh ∈ Mh : hµh , vh,S iΓ3 ≤

g|vh,S |h ds, Γ3

o vh ∈ Vh ,

(3.3)

where the mesh dependent absolute value |vh,S |h of the function vh,S is given by X |vh,S |h := |vp |φp . p∈S

We remark that in general |vh,S |h 6= |vh,S |. Everywhere below in this section, we assume that g is a strictly positive constant. In this case, the convex set Λ h can be equivalently written as n o X Λh := µh ∈ Mh : µh = γp ψp , |γp | ≤ g, p ∈ S . (3.4) p∈S

The discrete formulation of Problem 2.2 is the following. Problem 3.1. Find uh ∈ Vh and λh ∈ Λh such that a(uh , vh ) + b(vh , λh ) = (f, vh )V , b(uh , ζh − λh ) ≤ 0,

vh ∈ V h , ζ h ∈ Λh .

Using the discrete inf-sup property for the spaces Mh and Vh , see, e.g., [Woh00], we get the existence and the uniqueness of the solution. In order to obtain an optimal a priori error estimate, several lemmas will be proved. We note that Λh 6⊂ Λ. Before presenting the first lemma, we have to consider for a function vh ∈ Vh the discrete jump vˆh,S on the interface Γ3 in the constrained basis and its mesh dependent absolute value by X X |ˆ vp |φp , vˆp φp , |ˆ vh,S |h = vˆh,S = p∈S

p∈S

respectively. Now we can prove the following lemma. Lemma 3.1. Let (u, λ) be the solution of Problem 2.2 and (uh , λh ) the solution of Problem 3.1. Then we have Z Z g|ˆ uh,S |h ds. g|[u]| ds, b(uh , λh ) = b(u, λ) = Γ3

Γ3

8

Proof. In order to prove the first equality, we define µ ∈ M by Z g sgn([u]) v ds, v ∈ H 1/2 (Γ3 ). hµ, viΓ3 := Γ3

Thus, we have µ ∈ Λ. Using the definition of Λ and taking into account that b(u, µ − λ) ≤ 0, we can write Z Z g |[u]| ds ≥ hλ, [u]iΓ3 ≥ hµ, [u]iΓ3 = Γ3

µ ∈ Λ,

g sgn([u]) [u] ds = Γ3

Z

g |[u]| ds. Γ3

Keeping in mind (2.21), we have the first relation. P up ) ψp . ObIn order to prove the second equality, we consider ζh = p∈S g sgn(ˆ viously, we have ζh ∈ Λh . Due to the relation b(uh , ζh − λh ) ≤ 0,

ζ h ∈ Λh ,

we deduce b(uh , λh ) ≥ h

X

g sgn(ˆ up )ψp , u ˆh,S iΓ3 .

p∈S

Using the biorthogonality relation of the basis function (3.1), we get Z Z X X X h g sgn(ˆ u p ) ψp , u ˆh,S iΓ3 = g sgn(ˆ up ) u ˆp φp ds = g |ˆ up | p∈S

Γ3

p∈S

=

Z

p∈S

φp ds Γ3

g |ˆ uh,S |h ds. Γ3

From this, we obtain b(uh , λh ) = hλh , u ˆh,S iΓ3 ≥

Z

g |ˆ uh,S |h ds. Γ3

Taking into account (2.21) and the definition of Λh , we have Z b(uh , λh ) ≤ g|ˆ uh,S |h ds. Γ3

Combining the last inequalities, we deduce the second relation. Furthermore, the following result holds. Lemma 3.2. Let (u, λ) ∈ V × Λ be the solution of Problem 2.2 and let (u h , λh ) ∈ Vh × Λh be the solution of the discrete Problem 3.1. Then there exists a positive constant C independent of the meshsize h, such that for all vh ∈ Vh , µh ∈ Mh , ª © ku − uh k2V + kλ − λh k2M ≤ C ku − vh k2V + kλ − µh k2M + b(u, λh − λ).

9 Proof. The proof follows directly from the arguments given in [HL02]. For the convenience of the reader, we adapt the proof to our situation and take into account the different definition of Λ and Λh . Evaluating a(u − uh , u − uh ), we find for vh ∈ Vh the relation a(u − uh , u − uh ) = a(u − uh , u − vh ) + a(u − uh , vh − uh ) = a(u − uh , u − vh ) − b(vh − uh , λ) + b(vh − uh , λh ) = a(u − uh , u − vh ) − b(vh − u, λ − λh ) − b(u − uh , λ − λh ). Due to the V -ellipticity of a(·, ·) and continuity of a(·, ·) and b(·, ·), we deduce from this evaluation, using a trace theorem, the following inequality, ku − uh k2V ≤ C (ku − uh kV + kλ − λh kM ) ku − vh kV − b(u − uh , λ − λh ).

(3.5)

Here and below, we denote by C a positive constant independent of the meshsize, whose value may change from place to place. Using Lemma 3.1, we can write for the last term in the previous estimate Z −b(u − uh , λ − λh ) = b(u, λh − λ) + b(uh , λ) − g|ˆ uh,S |h ds, Γ3

and from this, since b(uh , λ) ≤

R

Γ3

g|ˆ uh,S | ds and |ˆ uh,S | ≤ |ˆ uh,S |h , we deduce

−b(u − uh , λ − λh ) ≤ b(u, λh − λ).

(3.6)

On the other hand, taking into account the inf-sup property of the form b(·, ·), we find kµh − λh kM ≤ α =α

sup 06=wh ∈Vh

sup 06=wh ∈Vh

b(wh , µh − λh ) kwh kV b(wh , µh − λ) + a(uh − u, wh ) kwh kV

≤ C(kµh − λkM + kuh − ukV ), and, using the triangle inequality, we can write kλ − λh kM ≤ C(kµh − λkM + kuh − ukV ).

(3.7)

We conclude our proof using (3.5), (3.6) and (3.7). To estimate the first two terms in the right side of the inequality of Lemma 3.2, we use the approximation properties of the spaces Vh and Mh . In addition, we have to estimate the residual term b(u, λh − λ). To this end, we denote γsl := supp [u] ⊂ Γ3 , γst := Γ3 \γsl , and we introduce the sets W ∗ := ˚ γst ∩ γsl , n o W 0 := w ∈ Γ3 : [u](w) = 0 and ∃ r > 0 : [u](w − ε) [u](w + ε) < 0, 0 < ² < r , W := W ∗ ∪ W 0 .

Everywhere below we will use the following assumption. Assumption 3.1. The number of points in W is finite.

10 The minimum distance between the elements in W is denoted by a, i.e., a := inf{|wj − wk | : 1 ≤ j 6= k ≤ Nw }, where Nw denotes the number of points in W. By Assumption 3.1, Nw < ∞ and thus a > 0. For h < a2 =: h0 , we find between two neighbor points in W at least two vertices in S. Then the following lemma holds. Lemma 3.3. Let (u, λ) ∈ V × Λ be the solution of Problem 2.2 and let (u h , λh ) ∈ Vh × Λh be the solution of Problem 3.1. Under Assumption 3.1 and the regularity 3 assumption uk ∈ H 2 +ν (Ωk ), 0 < ν ≤ 12 , k = m, s, we then have the a priori error estimate X 1 b(u, λh − λ) ≤ Ch 2 +ν kλ − λh kM |uk | 32 +ν,Ωk k=m,s

for a positive constant C independent of h < h0 . Proof. For each element w ∈ W, we can choose an element in S denoted by pj(w) , such that |pj(w) − w| =

min

p∈S, [u](p)6=0

|w − p|,

pj(w) = w,

if

w ∈ W\S,

if

w ∈ S.

˜ Let us consider the set S, n o S˜ := p ∈ S : ∃ w ∈ W s.t. p = pj(w) .

For this set, we consider the following disjoint decomposition S˜ = S ∗ ∪ S l ∪ S r , where

n o S ∗ := p ∈ S : ∃ w ∈ W ∗ s.t. p = pj(w) , n o S l := p ∈ S : ∃ w ∈ W 0 s.t. p = pj(w) and pj(w) is on the left side of w , n o S r := p ∈ S : ∃ w ∈ W 0 s.t. p = pj(w) and pj(w) is on the right side of w ;

see Figure 3.1 for an example of an element in S ∗ and Figure 3.2 for an example of an element in S l . Let Ih be the standard interpolation operator on Γ3 X Ih [u] = [u](p)φp , p∈S

and let I˜h be a modified interpolation  [u](p)      [u](p sl j(w)+1 ) sl −1 ˜ (Ih [u])(p) = r   [u](pj(w)−1 ) srs−1    0

operator such that if

˜ p ∈ S\S,

if

p = pj(w) ∈ S l , sl :=

if if

p = pj(w) ∈ S r , sr := p = pj(w) ∈ S ∗ ,

|w−pj(w) | |pj(w)+1 −pj(w) | , |pj(w) −w| |pj(w) −pj(w)−1 | ,

11

[u] ≤h


Ih [u]





pj(w)−2

I˜h [u] pj(w) pj(w)+1



pj(w)−1

w





pj(w)+2

≥ a > 2h Fig. 3.1. w ∈ W ∗ , pj(w) ∈ S ∗ , [u], Ih [u], I˜h [u].

see, e.g., Figure 3.1 and Figure 3.2. In order to simplify the writing, everywhere below we will use the short notation pj±l instead of pj(w)±l , for l ∈ {0, 1, 2}. We note that 0 < sl , sr ≤ 21 . Moreover, the following identity takes place on Γ3 , [u] I˜h [u] ≥ 0.

(3.8)

Finally, we introduce the Lagrange interpolation operator Iˆh with respect to the nodes ³ ´ l r 0 S\(S ∪ S ) ∪ W . Due to the shape regularity of the grid given by the nodes in S,

and because of 0 < sl , sr ≤ 12 , the new grid is also shape regular. Let us evaluate b(u, λh − λ) using the interpolation operators introduced before. Firstly, using Ih and I˜h , we can write,

b(u, λh −λ) ≤ hλh −λ, [u]−Ih [u]iΓ3 +hλh −λ, Ih [u]− I˜h [u]iΓ3 +hλh −λ, I˜h [u]iΓ3 . (3.9) For the first term in the right side of the previous inequality, we get hλh − λ, [u] − Ih [u]iΓ3 ≤ kλh − λkM k[u] − Ih [u]k 12 ,Γ3 and from this, we obtain hλh − λ, [u] − Ih [u]iΓ3 ≤ Ckλh − λkM h1/2+ν

X

|uk | 32 +ν,Ωk .

(3.10)

k=m,s

In order to evaluate the second term, we use the inverse inequality to write kIh [u] − I˜h [u]k21 ,Γ3 ≤ 2

C kIh [u] − I˜h [u]k20,Γ3 h

and from this, we find kIh [u] − I˜h [u]k21 ,Γ3 ≤ 2

C h

pj

X

kIh [u] − I˜h [u]k20,[pj−1 ,pj+1 ] + C

∈S l ∪S r

We note that for each pj ∈ S ∗ we have 0 = (I˜h [u])(pj ) 6= (Ih [u])(pj ) = [u](pj ).

X ³

pj ∈S ∗

[u](pj )

´2

.

(3.11)

ag replacements 12

[u] 



pj(w)−2

pj(w)−1




pj(w)



w



pj(w)+1 pj(w)+2

I˜h [u] Iˆh [u]

Ih [u]

Fig. 3.2. w ∈ W 0 , pj(w) ∈ S l , [u], Ih [u], I˜h [u], Iˆh [u].

Let pj be an element in S l and let us evaluate kIh [u] − I˜h [u]k20,[pj−1 ,pj+1 ] . Firstly, we note that ³ ´2 kIh [u] − I˜h [u]k20,[pj−1 ,pj+1 ] ≤ Ch (Ih [u] − I˜h [u])(pj ) .

Furthermore we have

(Ih [u] − I˜h [u])(pj ) =

´ 1 ³ l l (1 − s )[u](p ) + s [u](p ) j j+1 1 − sl

and from this, we obtain the identity,

(Ih [u] − I˜h [u])(pj ) = Since 0 < sl ≤

1 2

Now we deduce

1 (Ih [u])(w). 1 − sl

(3.12)

and (3.12), we get

³ ´2 kIh [u] − I˜h [u]k20,[pj−1 ,pj+1 ] ≤ Ch (Ih [u] − Iˆh [u])(w) .

kIh [u] − I˜h [u]k20,[pj−1 ,pj+1 ] ≤ CkIh [u] − Iˆh [u]k20,[pj−1 ,pj+1 ] . Using the triangle inequality and the estimates for Lagrange interpolation operators, we find ³ ´2 kIh [u] − I˜h [u]k20,[pj−1 ,pj+1 ] ≤ Ch h1/2+ν |[u]|1+ν,[pj−1 ,pj+1 ] . (3.13)

In a similar way, for each pj in S r , we can also find the estimate (3.13). In order to estimate the last term in (3.11), let us introduce the following notation, © ª © ª S ∗ l := pj ∈ S ∗ : [u](pj+1 ) = 0 , S ∗ r := pj ∈ S ∗ : [u](pj−1 ) = 0 . (3.14)

We note that, for each pj ∈ S ∗ l , [u] 6= 0 on [pj , w] and [u] = 0 on (w, pj+2 ]. In addition, for each pj ∈ S ∗ r , [u] 6= 0 on [w, pj ] and [u] = 0 on [pj−2 , w). The regularity assumption on u yields [u] ∈ H 1+ν (Γ3 ) and, following [HW05a], for each point pj ∈ S ∗l ´2 ³ [u](pj ) ≤ Ch1+2ν |[u]|21+ν,[pj , pj+2 ] ,

13 and, for each point pj ∈ S ∗ r , ³ ´2 [u](pj ) ≤ Ch1+2ν |[u]|21+ν,[pj−2 , pj ] .

Using this last estimates for each pj ∈ S ∗ and (3.13) for each pj ∈ S r ∪S l , by summing and using a trace theorem, we get X 1 (3.15) |uk | 32 +ν,Ωk . kIh [u] − I˜h [u]k 12 ,Γ3 ≤ C h 2 +ν k=m,s

In a last step, we show hλh − λ, I˜h [u]iΓ3 ≤ 0. Firstly, we can find a constant δ > 0 such that ³ ´ sgn[u] = sgn [u] − δ(I˜h [u])

(3.16)

on

Γ3 ,

Using (3.8) we can write on Γ3

|[u]| = |δ(I˜h [u])| + |[u] − δ(I˜h [u])|. Taking into account Lemma 3.1, we get Z Z ˜ hλ, [u]iΓ3 = g |δ(Ih [u])| ds + Γ3

g |[u] − δ(I˜h [u])| ds.

(3.17)

Γ3

On the other hand, we have hλ, [u]iΓ3 = hλ, δ(I˜h [u])iΓ3 + hλ, [u] − δ(I˜h [u])iΓ3 ,

(3.18)

and due to the definition of Λ, we obtain hλ, δ(I˜h [u])iΓ3 ≤ hλ, [u] − δ(I˜h [u])iΓ3 ≤

Z

Z

g |δ(I˜h [u])| ds,

(3.19)

g |[u] − δ(I˜h [u])| ds.

(3.20)

Γ3

Γ3

Using (3.17)-(3.20), we deduce that in (3.19) and (3.20) we can write identities. Since δ > 0 we get Z g |I˜h [u]| ds. (3.21) hλ, I˜h [u]iΓ3 = Γ3

Now, let us write λh as a linear combination of the dual basis functions ψp and I˜h [u] as a linear combination of the standard basis functions φp , X X λh = γp ψp , I˜h [u] = αp φp , where αp = I˜h [u](p). p∈S

p∈S

Keeping in mind (3.1) and (3.21), we obtain Z ³ X ¯ hλh − λ, I˜h [u]iΓ3 = γ p α p φp − g ¯ Γ3

p∈(S\S ∗ )

X

p∈(S\S ∗ )

¯´ αp φp ¯ ds.

(3.22)

14 If we consider the subsets of Γ3 defined by n o Γ+ and 3 := x ∈ Γ3 : [u](x) > 0 taking into account (3.8), we deduce that

n o Γ− 3 := x ∈ Γ3 : [u](x) < 0 ,

∗ I˜h [u](p) > 0, p ∈ Γ+ 3 ∩ (S\S ), ∗ I˜h [u](p) < 0, p ∈ Γ− 3 ∩ (S\S ), supp I˜h [u] ⊂ supp[u].

From this, using (3.22), we can write Z Z X (γp − g)αp φp ds + hλh − λ, I˜h [u]iΓ3 = Γ+ 3 p∈(S\S ∗ )

(3.23)

X

(γp + g)αp φp ds.

Γ− 3 p∈(S\S ∗ )

Keeping in mind (3.4) and (3.23), we get (3.16), i.e., the last term in the right side of the inequality (3.9) is negative. Taking into account (3.9)-(3.16) we conclude Lemma 3.3. Based on the results obtained in Lemma 3.2 and Lemma 3.3 and using the well known approximation properties for the spaces Vh and Mh , by applying Young’s inequality, the following theorem holds. Theorem 3.4. Let (u, λ) ∈ V ×Λ be the solution of Problem 2.2 and let (u h , λh ) ∈ Vh × Λh be the solution of Problem 3.1. Under Assumption 3.1 and the regularity 3 assumption uk ∈ H 2 +ν (Ωk ), 0 < ν ≤ 12 , k = m, s, we then have the a priori error estimate X 1 |uk | 32 +ν,Ωk ku − uh kV + kλ − λh kM ≤ Ch 2 +ν k=m,s

for a positive constant C independent of the meshsize h < h0 . 4. Inexact primal-dual active set strategy. In this section, we present our algorithm to solve the discrete nonlinear Problem 3.1. We use an inexact primal-dual active set strategy. Primal-dual active set strategies for problems with friction were also considered in [Sta04]. In contrast to [Sta04], we do not use a penalty parameter and use a different nonlinear complementary function. Let uh ∈ Vh and λh ∈ Λh be the solution of the discrete Problem 3.1. For the discrete solution (uh , λh ) of Problem 3.1, we get due to λh ∈ Λh from the definition of the space Λh , see (3.3), the condition Z gφp ds, p ∈ S, (4.1) |λp,s | ≤ gp := Γ3

where the scaled value λp,s of λp is given by λp,s := λp D[p, p]. In terms of Lemma 3.1, the inequality condition of Problem 3.1 results for g > 0 in the algebraic system λp,s u ˆp = gp |ˆ up |,

p ∈ S.

(4.2)

We decompose S into two disjoint sets such that |λp,s | = gp for p ∈ A,

u ˆp = 0 for p ∈ I

(4.3)

15 and we call A active set and I inactive set. We observe that the nonlinear constraints (4.1) and (4.2) are the discrete constraints of (2.14). The constraints (4.1) and (4.2) can at each point with λp 6= 0 equivalently be written as ¡ ¢¡ ¢ |λp | ≤ gp , λp u ˆp |λp,s | − gp = 0, λp u ˆp ≥ 0, p ∈ S. (4.4) We use a nonlinear complementarity function C(ˆ up , λp ) defined by © ª C(ˆ up , λp ) := u ˆp λp − max 0, u ˆp λp + c (|λp,s | − gp ) , c > 0.

(4.5)

The equivalence between the constraints (4.4) and the condition C(ˆ u p , λp ) = 0 can be easily shown by a straightforward computation. We remark that in the nonlinear complementary function we use the nodal value λp and the scaled nodal value λp,s . To guarantee equivalence between C(ˆ up , λp ) = 0, p ∈ S and (4.1) and (4.2), we have to make sure that if C(ˆ up , λp ) = 0 and λp = 0 we also have u ˆp = 0. To determine the sets A and I we use an iterative scheme based on the nonlinear complementary function (4.5). Let k be the corresponding iteration index. The sign of λkp,s for p ∈ Ak is defined by the previous step. With the decomposition of the vertices introduced in Section 3, we write for the solution vector uh = (uN , uM , uS )> and for the Lagrange multiplier λh = λS . For the first line of Problem 3.1, the algebraic system in the standard basis φ reads as Ah uh + Bh λh = fh with B = (0, −M > , D)> . The transformation ˆ introduced in (3.2), leads to the of this algebraic system to the constrained basis φ, ˆh λh = fˆh , where we have the relations Aˆh = QAh Q> and fˆh = Qfh . system Aˆh u ˆh + B ˆ has the form B ˆh in the constrained basis φ ˆh = (0, 0, D)> . The coupling matrix B Combining this system with (4.3) and eliminating the Lagrange multiplier leads to 0 ˆ AN N B AˆMN B @ AˆAN 0

AˆN M AˆMM AˆAM 0

AˆN A AˆMA AˆAA 0

AˆN I AˆMI AˆAI IdII

1 0 fˆN u ˆN CB u C B ˆM ˆ f CB M C = B A@ u ˆA A @ fˆA + gA u ˆI 0 10

1

C C, A

(4.6)

where Aˆk,l , k, l ∈ {N , M, A, I}, are the block stiffness matrices of Aˆh associated with the corresponding basis functions. The entries of the vectors gA are the values ±gp with p ∈ A. We remark that the sign of the entry gp of gA at the point p ∈ A will be specified by the sign of λp in the previous iteration step. The Lagrange multiplier can be computed by the residual, i.e., ¢ ¡ (4.7) λh = D−1 fˆS − AˆSN u ˆN − AˆSM u ˆM − AˆSS u ˆS . To solve this linear problem, we use an optimal multigrid method. Due to the use of a multigrid method as an iterative solver, we can formulate the inexact primal-dual active set algorithm. Based on the condition C(ˆ up , λp ) = 0, we introduce the primaldual active set strategy as an iterative scheme. We note that primal-dual active set methods in combination with exact solvers can be analyzed as a Newton method, see, e.g., [HIK03, Sta04]. In combination with a multigrid method it can be interpreted as a nonlinear multigrid method.

16

Algorithm 1: Inexact primal-dual active set strategy (0) Set k = 1, choose c > 0 and m ∈ N. Set A1 = ∅ and I1 = S. Initialize u ˆ1,0 h as an initial solution for the multigrid method. (1) For i = 1, . . . , m, compute , Ak , Ik ). uk,i−1 u ˆk,i h h = M G(ˆ (2) If kˆ uk,0 ˆk,m uk,m h −u h k / kˆ h k small enough, then stop. (3) Set Ak+1 and Ik+1 to © ª k,m Ak+1 := p ∈ S : u ˆk,m + c(|λk,m p λp p,s | − gp ) > 0 , © ª k,m Ik+1 := p ∈ S : u ˆk,m + c(|λk,m p λp p,s | − gp ) ≤ 0 .

(4.8)

=u ˆk,m (4) Set u ˆk+1,0 h , set h

(gA )p = sign(λk,m p )gp

for

p ∈ Ak+1 ,

set k = k + 1 and go to step (1). In this algorithm we denote by u ˆk,i uk,i−1 , Ak , Ik ) the iterate after one h = M G(ˆ h multigrid step for the linear system (4.6) with A = Ak and I = Ik . The values are computed by (4.7). We remark that for m = ∞, we solve the linear system λk,m p (4.6) exactly, and we are in the case of the exact primal-dual active set strategy. To motivate the choice of the active and inactive sets as presented in (4.8), we refer to Table 4.1. This Table contains for any primal-dual pair (ˆ ukh , λkh ) satisfying u ˆkp = 0 for k p ∈ Ik and |λp,s | = gp for p ∈ Ak the choice of the active and inactive sets Ak+1 and Ik+1 . We point out that for each node p ∈ Ak+1 we have λk,m 6= 0. Thus the vector p = 0, gA in (4.6) is always well defined. The choice of Ik+1 guarantees that from λk,m p = 0. it follows u ˆk+1,m p Table 4.1 Choice for the active set Ak+1 and the inactive set Ik+1 for a primal-dual pair (ˆ ukh , λkh ) satisfying u ˆkp = 0 for p ∈ Ik and |λkp,s | = gp for p ∈ Ak .

u ˆkp = 0 |λkp,s | = gp

|λkp,s | ≤ gp |λkp,s | > gp u ˆkp λkp > 0 u ˆkp λkp ≤ 0

p∈

u ˆkp λkp + c(|λkp,s | − gp )

choice

Ik Ik Ak Ak

≤0 >0 >0 50

Ik+1 Ak+1 Ak+1 Ik+1

We mention that the value of the positive constant c has no influence on the choice of the active and inactive sets A and I when we solve the linear system exactly. It only influences the choice in the inexact approach. Our algorithm guarantees that for a not admissible Lagrange multiplier λk,m the sign is preserved in the next step, p where the node p is active.

17 5. Numerical Examples. In this last section, we confirm our theoretical results for the discretization error and show the performance of our inexact primal-dual active set Algorithm 1. The implementation of this algorithm is based on the finite element toolbox UG; see [BBJ+ 97]. As material, we use for all examples granite with Young’s modulus E = 106 and Poisson’s ratio ν = 0.25. For Lam´e’s parameters, we then have E Eν and µ = 2(1+ν) . We focus on the the plain strain relation given by λ = (1+ν)(1−2ν) 2 1 discretization errors of the solution u in the L - and the H -norm. The error of the Lagrange multiplier is considered in a weighted L2 -norm k · k− 12 ,h,Γ3 given by kλk2− 1 ,h,Γ3 := 2

X

he kλk20,e ,

e∈Γs3

where we denote by e the edges of the discretization of Γs3 and by he the length of the edge e. We remark that using an inverse estimate it is also possible to establish a priori error estimates for the Lagrange multiplier in the weighted L2 -norm. Since there is in general no analytical solution available, we use for computing the discretization errors a reference solution uref and λref on a fine mesh satisfying href ≤ 1/4 h. Using href = 1/2 h as reference meshsize does not yield reliable values for the discretization PSfrag replacements errors. Stress and Jump (g=0.6)

Γs1

0.6

Ωs

0.5 0.4 0.3

r φ P1 P2 P3

Γ3

0.2

Ωm

λ

h

0.1

Γm 1

−1

s

m

150*(uh−uh )

0 −0.5

0 0.5 x1−coordinate

1

Fig. 5.1. Left: problem definition and grid on level 0; middle: contour lines of the solution; right: Lagrange multiplier λh and amplified jump [uh ] on Γ3 .

Table 5.1 First Example: Relative L2 (Ω)-error and relative H 1 (Ω)-error of uh with respect to uref , weighted L2 (Γ3 )-error of λh with respect to λref and the numerical convergence rates.

level 0 1 2 3 4 5 6

kuh − uref k0,Ω /kuref k0,Ω 1.210198e + 00 1.762545e − 01 4.855714e − 02 1.342324e − 02 3.395710e − 03 8.468612e − 04 2.062731e − 04

− 2.78 1.86 1.85 1.98 2.00 2.04

|uh − uref |1,Ω /|uref |1,Ω 8.679373e − 01 5.503170e − 01 3.744576e − 01 1.969682e − 01 9.866146e − 02 4.907663e − 02 2.395136e − 02

− 0.66 0.56 0.93 1.00 1.01 1.03

kλh − λref k− 12 ,h,Γ3 1.525284e − 01 6.264323e − 02 7.003110e − 03 1.907398e − 03 7.817159e − 04 2.646597e − 04 8.864475e − 05

− 1.28 3.16 1.88 1.29 1.56 1.58

5.1. First Example. In this chapter, we consider the example shown in the left picture of Figure 5.1. We set the lower rectangle Ωm = [−1, 1] × [0, 1] and the

18 m upper as Ωs = [−1, 1] × [1, 2]. On Γm 1 , being the bottom of the lower rectangle Ω , m we set the Dirichlet values equal to u = 0. We assume the top of the upper body Ωs with given Dirichlet values us = 0.22 exp(−60x21 ). On all the remaining boundary segments, we set homogeneous Neumann boundary conditions. For the friction bound we use g = 0.6. The contour lines of the numerical solution uh are presented in the picture in the middle of Figure 5.1. The right picture in this figure shows the Lagrange multiplier λh and the jump uh amplified with the factor 150 at the interface Γ3 . The discretization errors and the convergence rates are presented in Table 5.1. Here we used the finite element solution on Level 8 as reference solution uref . The grid on Level 0 is shown in the left picture of Figure 5.1. On Level 8, we have 525 825 nodes on Ωs and 394 497 on Ωm . The convergence rates in the L2 -norm are asymptotically 2.0 and in the H 1 -norm, we get 1.0. As also observed in the linear case, we get better convergence rates for the Lagrange multiplier. The best approximation error in the weighted L2 (Γ3 )-norm is of order h1.5 . PSfrag replacements

Ω

φ

Ωs P2

Γm 1

P1

r

m

Γ3 Γs1

P3

Fig. 5.2. Left: problem definition; middle: grid on level 1; right: contour lines of the solution.

5.2. Second Example. Now we consider the problem depicted in the left picture of Figure 5.2. Here the interface is a curved one. By means of this example, we illustrate the performance and flexibility of our algorithm. To fix the geometry, we set the three points P1 , P2 and P3 equal to P1 = (0, 1.5), P2 = (−1, 0) and P3 = (1, 0). The radius r of the upper half disc is set to be r = 1 and for the angle φ we choose φ = π/4. The lower body plays the role of the slave side Ωs an the upper one the role of the master side Ωm . We fix the domain at the bottom of the master side Ωm . At the top of the upper body Ωs , we set the Dirichlet value to us = 0.055x1 . As done in the first example, we set the friction bound equal to g = 0.6. Stress and Jump (g=0.6)

Displacement (g=0.6) 0.015

0.6

λ

0.4

50*(u −u )

h

s h

m h

0.01

0.2

0.005

0

0

−0.2

−0.005

−0.4

−0.01

−0.6 −0.5

0 angle φ

0.5

−0.015

s h m u h

u

−0.5

0 angle φ

0.5

Fig. 5.3. Left: Lagrange multiplier λh and amplified jump [uh ] for g = 0.6; right: displacement ush on slave side and um h on master side for g = 0.6.

The right picture in Figure 5.2 shows the contour lines of the solution. Figure 5.3

19 Table 5.2 Second example: Relative L2 (Ω)-error and relative H 1 (Ω)-error of uh with respect to uref , weighted L2 (Γ3 )-error of λh with respect to λref and the numerical convergence orders.

level 0 1 2 3 4 5 6

kuh − uref k0,Ω /kuref k0,Ω

|uh − uref |1,Ω /|uref |1,Ω

8.934322e − 02 3.096022e − 02 8.651862e − 03 2.427230e − 03 6.665729e − 04 1.791767e − 04 4.652240e − 05

3.231679e − 01 1.893279e − 01 1.034272e − 01 5.643122e − 02 3.041368e − 02 1.613158e − 02 8.311285e − 03

− 1.53 1.84 1.83 1.86 1.90 1.95

− 0.77 0.87 0.87 0.89 0.91 0.96

kλh − λref k− 12 ,h,Γ3 7.718450e − 02 2.216194e − 02 8.133636e − 03 3.781621e − 03 1.303146e − 03 3.521447e − 04 1.671239e − 04

− 1.80 1.45 1.10 1.54 1.89 1.08

shows the Lagrange multiplier λh , the jump [uh ] amplified with the factor 50 and the s solution of the master and slave side um h , uh on the interface Γ3 . The discretization errors and the corresponding convergence rates are listed in Table 5.2. Again we use as reference solution the solution on Level 8. The grid on Level 1 is presented in the picture in the middle of Figure 5.2. To illustrate the performance of the inexact primal-dual active set Algorithm 1, we use as multigrid solver for the linear problems a W-cycle with 3 pre- and post-smoothing steps of the symmetric Gauß–Seidel iteration. For the constant c in Algorithm 1, we use c = 0.01. We observe that the value of c has to be small enough to get convergence of the active sets. This value depends on the relation between the stresses λp and the displacements u ˆp and therefore on the material data. On each level, we start the multigrid iteration with u ˆ1,0 h = 0. We compare the inexact approach (m = 1), where we update the active set A after each multigrid step, with the exact approach (m = ∞), where we solve for each choice of the set A the resulting linear system. For each of these two approaches, we compare the case where we start on each level from A1 = ∅ with the nested approach. Here we inherit the active set A from level l − 1 to l and use the results as A1 . Table 5.3 Comparison between exact and inexact active set strategy for the second example given in Figure 5.2 for the two approaches A1+ = A1− = ∅ and the nested approach.

level 0 1 2 3 4 5 6 7 8

A1 = ∅ exact inexact Kl Ml 2 2 3 3 3 3 4 5 5 5 6 6 7 7 8 8 9 9

Kl 2 2 1 1 2 2 1 2 2

nested approach exact inexact |Ak | Ml |Ak | 0 2 2 0 2 2 4 2 2 4 6 1 6 10 3 10 12 10 18 20 2 18 20 38 40 2 38 40 78 3 78 80 78 154 156 3 154 158 156 310 312 4 310 314 310

312

20 For the exact strategy, we denote by Kl the step of the active set iteration on level l in which the correct active sets are found. For the inexact approach Ml denotes the number of the multigrid iterations after which the correct active sets are found and kept. Table 5.3 shows Kl and Ml for the two choices of A1 . Here we denote by |Ak | the number of active nodes in the k-th. step. For the case A1 = ∅, we observe that almost on each level Kl = Ml . Due to that there is no need to solve the resulting linear system exactly. Furthermore we remark that the required number of steps depends linearly on the level. For the nested approach, we observe K l ≤ Ml and the number of necessary steps is bounded independently of the refinement level. For the exact strategy, we get Kl ≤ 2 and for the inexact strategy Ml ≤ 4. Although Ml ≥ Kl , the inexact strategy is much more efficient because each step requires only one multigrid iteration. Table 5.4 Necessary number of multigrid iterations for the four considered cases for the second example given in Figure 5.2 for the two approaches.

level 1 2 3 4 5 6 7 8

A1 = ∅ exact inexact 13 7 16 7 26 11 33 11 36 10 42 11 49 12 54 12

nested approach exact inexact 10 6 6 6 8 8 14 8 12 7 6 6 12 7 12 7

lin. problem 6 6 8 9 7 6 7 7

To compare the computational cost to solve the nonlinear problem for the four cases we consider the necessary numbers of multigrid iterations. These numbers are shown in Table 5.4. The last column contains the necessary number of multigrid iterations to solve the nonlinear problem with the exact active and inactive sets A and I given. Then the nonlinear problem reduces to a linear one. It turns out that the number of necessary multigrid iterations for the inexact approach with the nested strategy is almost the same as for solving the linear problem. So, the additional effort for solving the nonlinear problem instead of solving the linear one is negligible. For the exact approach starting with A1 = ∅ on each level, the number depends linearly on the level. We obtain that the inexact strategy is cheaper compared to the exact one. In combination with the nested approach, the inexact strategy yields a very efficient iterative solver for our nonlinear problem. Finally, we mention that our inexact primal-dual active set Algorithm 1 can also handle nonconstant friction bounds g. If we use g = (x21 − 1) + 1.2 instead of g = 0.6 in the second example, we find the results shown in Figure 5.4. 6. Generalization of Algorithm 1 to Coulomb friction. In this last section, we want to generalize our algorithm to a multibody contact problem with Coulomb friction. We use linear elasticity as material law as given in (2.1). In this case, we have to replace in the conditions for the tangential direction (2.9) the constant friction bound g by a solution dependent one. We have to replace it by F|σ n |, where σ n := (σn) · n and F denotes the friction coefficient. The contact conditions in

21 2 1

2 1

Stress and Jump (g=(x −1)+1.2)

Displacement (g=(x −1)+1.2) 0.015

1

+g −g λh

0.5

0.01

s h um h

u

0.005

50*(ush−um ) h

0 0 −0.005 −0.01

−0.5 −0.5

0 angle φ

−0.015

0.5

−0.5

0 angle φ

0.5

2 Fig. 5.4. Left: Lagrange multiplier λh and amplified jump ush − um h for g = (x1 − 1) + 1.2; s m 2 right: displacement uh on slave side and uh on master side for g = (x1 − 1) + 1.2.

normal direction are the nonpenetration condition of the two bodies. More details on the nonlinear contact conditions can be found, e.g., in [KO88]. Now we have to find two active sets. The active set of nodes on the contact interface being in contact in normal direction with the second body and the active set for the Lagrange multiplier in tangential direction. For the first one we use the Algorithm given in [HW05b]. For the second one we use Algorithm 1. We remark that here we have to update the values for the friction gp after each step. Combining the inexact versions of these two algorithms leads to an efficient iterative solver for the complex nonlinear problem of Coulomb friction. We mention that after each multigrid step we update both; the two active sets and the solution dependent friction bound. PSfrag replacements Γs2

H

Γm 1

Ωm

φ Γm 1

Ωs L

ri

M

ra Γm 1

Fig. 6.1. Problem Definition in 3D and 2D; initial grid and distorted grid on level 0; effective von Mises stress on level 2.

We consider an example in the three-dimensional case. The geometry is shown in the left picture of Figure 6.1. On the top of the inner cylinder, we apply some Neumann boundary condition. Due to the axisymmetry the problem can be reduced to a twodimensional setting. The resulting geometry can be seen in the second picture of Figure 6.1. For the geometry we choose L = 5, H = 15, ri = 27, ra = 30, M = (24.5, 7.0)> and φ = 13. For the softer inner body Ωs , playing the role of the slave side, we use for the material data λs = 57.6923 and µs = 38.4615. For the outer body Ωm , playing the role of the master side, we set λm = 576.923 and µm = 384.615. At the top of the inner cylinder, we apply a Neumann data f2s = 17.0. Due to the axisymmetry, we have on the left side of Ωs , where r = 0 holds, homogeneous Dirichlet data in radial direction. We fix Ωm at the three segments Γm 1 , as shown in the second picture of Figure 6.1. For the friction coefficient F we choose F = 0.7. We remark that the two bodies penetrate in there reference configuration.

22 For the Lam´e operator in cylinder coordinates, we refer to [JL01]. We denote by ur the displacement in the radial direction, by uz the displacement in the axial direction and by uθ the displacement in circumferential direction. Since we are in the axisymmetric case, we have uθ = 0 and therefore we get the relations εrθ = εθr = εzθ = εθz = 0. For the other entries of the strain tensor ε we get µ ¶ µ ¶ 1 ∂ur ∂uz 1 ∂uθ ∂uz ∂ur . , εzz = , εθθ = + ur , εrz = εzr = + εrr = ∂r ∂z r ∂θ 2 ∂z ∂r Now we can write for the stress strain relation given by Hooke’s law      λ + 2µ λ λ 0 εrr σ rr   σ zz   λ λ + 2µ λ 0     εzz  .  σ θθ  =  λ λ λ + 2µ 0   εθθ  2εrz σ rz 0 0 0 µ

Figure 6.2 shows the normal and the tangential parts of the discrete Lagrange multiplier on the contact interface. We remark, that the two bodies are in contact, where we have λh · n 6= 0. In tangential direction, we get two active zones where we have |λh · t| = F|λh · n|. Between these two sets the two bodies are glued together in tangential direction, i.e., (ush − um h ) · t = 0. The tangential part of the displacements on the slave and the master side is shown in the right picture of Figure 6.2. Contact Stress (F=0.7)

Displacement (F=0.7)

10

λh⋅ n

8

λh⋅ t

0.25

ush⋅ t

0.2

u ⋅t

m h

0.15 6 0.1

4

0.05

2 0 0

0 5 10 z−coordinate

15

0

5 10 z−coordinate

15

Fig. 6.2. Left: normal part λh · n and tangential part λh · t of the Lagrange multiplier λh ; right: tangential part ush · t of the displacement on the slave side and tangential part um h · t of the displacement on the master side.

REFERENCES [BBJ+ 97]

[BF91] [BHL99]

[BHP02]

[BR03]

P. Bastian, K. Birken, K. Johannsen, S. Lang, N. Neuß, H. Rentz–Reichert, and C. Wieners. UG – a flexible software toolbox for solving partial differential equations. Computing and Visualization in Science, 1:27–40, 1997. F. Brezzi and M. Fortin. Mixed and hybrid finite element methods. Springer-Verlag, New York, 1991. F. Ben Belgacem, P. Hild, and P. Laborde. Extension of the mortar finite element method to a variational inequality modeling unilateral contact. Mathematical Models and Methods in Applied Sciences, 9:287–303, 1999. A. Borrelli, C. Horgan, and C. Patria. Saint-Venant’s principle for antiplane shear deformations of linear piezoelectric materials. SIAM Journal on Applied Mathematics, 62(6):2027–2044, 2002. F. Ben Belgacem and Y. Renard. Hybrid finite element methods for the Signorini problem. Mathematics of Computation, 72(243):1117–1145, 2003.

23 [Bra97] [CHLS01]

[DL76] [ET76] [Glo84] [GLT81] [Hcc96]

[HIK03] [Hil00]

[HL02] [HMM02]

[Hor95] [HS01]

[HS02]

[HS04]

[HW05a] [HW05b]

[JL01] [Kno76]

[Kno77] [KO88]

[MMS01]

[Pan85] [SST04] [Sta04] [WK03] [Woh00]

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