A direct algorithm in some free boundary problems

C. M. Murea1 and D. Tiba2,3

1 Laboratoire de Mathématiques, Informatique et Applications,

Université de Haute Alsace,

4-6, rue des Frères Lumière, 68093 Mulhouse Cedex, France

cornel.murea@uha.fr2 Institute of Mathematics (Romanian Academy) and

Academy of Romanian Scientists, Bucharest, Romania3 Basque Center for Applied Mathematics,

Alameda Mazarredo, 14, E-48009 Bilbao, Basque Country, Spain

dan.tiba@imar.ro

Abstract

We study the wellknown obstacle problem which models applications fromelasticity [5], hydraulics [1], etc. Our approach enters the category of fixed domainmethods, uses just linear elliptic equations and approximates the coincidence setexplicitly.

MSC. 35R35, 35J86, 65N30keywords. obstacle problem, free boundary problems, penalization

1 Introduction

Free boundary problems are nonlinear partial differential equations modelling importantapplications from physics like change of phase phenomena, contact problems in elasticity,porous media propagation, etc. Besides the solution of the given differential equation,one has to find as well the so-called free boundary, which may be a surface or a wholeregion (mushy region) and which is in fact the main unknown of the problem. Thedifficulty of such problems is related to the geometric character of this unknown and tothe high nonlinearity of the corresponding equation. From this point of view, one maycompare free boundary problems with shape optimization problems [19, Ch. 5.1] and,in this paper, we shall essentially use techniques from optimal design.

It is to be noticed that from the very beginning, many of the proposed solutionmethods for free boundary problems are of fixed domain type, in the sense that the

1

unknown geometry is “embedded” via a certain procedure into the partial differentialequation: the Baiocchi transform in the dam problem [1], the enthalpy method for Stefanproblems [12], the regularization approach [2], [10] etc. The price to be paid is that theobtained partial differential equation is highly nonlinear and the free boundary has tobe recovered from certain properties of the solution (for instance as a level surface). Toovercome such difficulties, front tracking methods have been proposed [6], [15], but theirimplementation may be quite complex.

In this paper we propose a new algorithm for the elliptic obstacle problem whichis of fixed domain type in the sense that the finite element discretization is given inthe whole domain, independent of the position of the (unknown) free boundary. Ineach iteration a linear elliptic equation has to be solved in the whole domain and thecorresponding stiffness matrix is common for all iterations. This is a clear advantagefrom the point of view of the implementation and the approximating coincidence set isexplicitly computed in each iteration and it converges in the Hausdorff-Pompeiu sense[19] to the searched geometry. Moreover, we need just a scalar penalization parameterin our method.

Our approach is inspired from shape optimization techniques, but no shape optimiza-tion problem is used here although this is a known method in free boundary problems,[17], [8], [3]. One may compare the present approach to the recent works [18], [9]. Anefficient Lagrangian method together with a primal-dual active set strategy with reg-ularization is studied in [11]. We expect that our method may be applied as well toparabolic variational inequalities as in [2], [6], [23].

In the next section, we formulate the problem and the algorithm. Section 3 is devotedto the convergence analysis. The last section reports on some numerical experiments. Inthe examples, the algorithm has a very fast convergence, even in the case of nonsmoothobstacles. The initial iteration is arbitrary.

2 Formulation of the problem and the algorithms

Let D be a smooth domain in Rd, d ∈ N∗ and A : H10 (D) → H−1(D) be some ellipticoperator. In order to simplify the writing we shall take A = −∆ to be the Laplacianoperator, without loss of generality.

Let ψ : D → R denote the obstacle function. Here, we assume ψ ∈ H2(D) andψ|∂D ≤ 0 (to be consistent with the homogeneous Dirichlet conditions). Let f ∈ L2(D)be given. The classical formulation of the obstacle problem is:

−∆y = f in D+ = {x ∈ D; y(x) > ψ(x)} , (1)−∆y ≥ f in D \D+ = {x ∈ D; y(x) = ψ(x)} , (2)

y = ψ and∂y

∂n=∂ψ

∂non ∂D+ ∩D, (3)

y = 0 on ∂D. (4)

2

The boundary condition (4) may be replaced by other boundary conditions. Thedouble boundary condition (3) is due to the unknown character of the boundary ∂D+ andit determines it. The set D+ is called the noncoincidence set, while its complementaryis the coincidence set. The formulation (1)-(4) corresponds to the case of at least strongsolutions y ∈ H2(D) ∩ H10 (D). In general, the weak formulation of (1)-(4) is given inthe form of a variational inequality associated to the convex K ⊂ H10 (D):

∫D∇y · (∇y −∇v) dx ≤

∫Df (y − v) dx, y ∈ K, ∀v ∈ K, (5)

K = {v ∈ H10 (D); v ≥ ψ a.e. in D} . (6)

If the elliptic operator is symmetric (which is the case for the Laplace operator),then (5)-(6) is equivalent with the variational problem

minv∈K

{1

2

∫

D

|∇v|2 dx−∫

D

f v dx

}. (7)

In the case when f ∈ L2(D), ψ ∈ H2(D) with the compatibility condition ψ|∂D ≤ 0,it is known that the solution of (5)-(6) satisfies the regularity property y ∈ H2(D), [2,Thm. 2.5] and the formulation (1)-(4) may be used. Moreover, in this case, the obstacleproblem may be written as a multivalued equation

−∆y + β(y − ψ) ∋ f in D

where β ⊂ R × R is the maximal monotone graph given by

β(z) =

] −∞, 0], z = 0,0, z > 0,∅, z < 0.

(8)

Then β(y − ψ) ∈ L2(D) by the maximal regularity of the solution y ∈ H2(D).We state now our algorithm.

Algorithm 1

1) Choose n = 0, ǫ0 > 0, Ω0 ⊂ D open, ỹ−1 = 0;2) Compute yn ∈ H10 (D) as solution of the linear elliptic equation

−∆yn +1

ǫnχD\Ωn(yn − ψ) = f in D (9)

(here χD\Ωn is the characteristic function of D \Ωn, corresponding to the approximationof the coincidence set in iteration n);

3) yn = max{yn, ψ}, Ωn+1 = {x ∈ D; yn(x) > ψ(x)}, ǫn+1 = ǫn2 ;4) If ‖yn − yn−1‖L2(D) < tol then STOP else n=n+1 GO TO step 2.

3

Remark 1 Notice that by the regularity theory for elliptic equations, in Step 2 we havey ∈ H2(D)∩H10 (D). Then, by the Sobolev theorem in dimension d ≤ 3, we get yn ∈ C(D)and Ωn defined in Step 3 is open, for any n ∈ N. If f is in L∞(D), then yn is continuousin arbitrary dimension.

Remark 2 The algorithm uses just linear elliptic equations in the whole domainD. Thetype of penalization term from Step 2 may be compared with the approach developed inshape optimization problems in [18]. A classical penalization term for solving the obstacleproblem is 1

ǫ(yn − ψ)− where v−(x) = −v(x) if v(x) < 0 and v−(x) = 0 if v(x) ≥ 0,

see for example [7]. Another advantage of the Algorithm 1 is that it approximates thecoincidence set explicitly. We also underline its simplicity.

We denote now ψ1, ψ2 : D → R, ψ1 ≤ ψ2 in D, ψ1|∂D ≤ 0, ψ2|∂D ≥ 0 and

K̃ ={y ∈ H10(D); ψ1(x) ≤ y(x) ≤ ψ2(x) a.e. in D

}

which is a closed convex subset of H10 (D).

To K̃, the following variational inequality, similar to (5), may be associated:

∫

D

∇y · (∇y −∇v) dx ≤∫

D

f (y − v) dx, ∀v ∈ K̃ (10)

The existence of a unique solution y ∈ K̃ is wellknown.Clearly, one can define two coincidence sets, corresponding to the two obstacles:

D1 = {x ∈ D; y(x) = ψ1(x)}D2 = {x ∈ D; y(x) = ψ2(x)}

and associated to (10). Under regularity conditions, a description of (10) in classicalterms as (1)-(4) may be written.

We extend now Algorithm 1 to this situation.

Algorithm 2

1) Choose n = 0, ǫ0 > 0, Ω01 ⊂ D, Ω02 ⊂ D open subsets such that (D\Ω01)∩(D\Ω02) =

∅, ỹ−1 = 0;2) Compute yn ∈ H10 (D) as solution of the linear elliptic equation

−∆yn +1

ǫnχD\Ωn

1(yn − ψ1) +

1

ǫnχD\Ωn

2(yn − ψ2) = f in D (11)

3) Compute yn = min {ψ2, max{yn, ψ1}}, Ωn+11 = {x ∈ D; yn(x) > ψ1(x)}, Ωn+12 ={x ∈ D; yn(x) < ψ2(x)} ǫn+1 = ǫn2 ;

4) If ‖yn − yn−1‖L2(D) < tol then STOP else n=n+1 GO TO step 2.

4

Remark 3 In [17] an equation similar to (11) is used in a shape optimization problemobtained via the least squares method applied to a free boundary problem. The othersteps of the algorithm are different.

Remark 4 By the classical result of [4], the elastic-plastic torsion problem is equivalentwith a variational inequality of obstacle type and our algorithm may be applied as well.In [8], a related problem is studied by a comparable approach involving semi-smoothNewton methods.

Let βǫ : R → R denote the Yosida approximation of β given by (8). Assuming that{x ∈ D; yn(x) = ψ(x)} has zero measure, we can rewrite the step 2 of the Algorithm 1as

−∆yn +(β ′ǫn (yn−1 − ψ)

)(yn − ψ) = f. (12)

In (12), we take into account step 3 as well. Recall that the usual approximation byregularization of the variational inequality (5)-(6) is

−∆ỹn + βǫn (ỹn − ψ) = f in D, (13)

plus homogeneous boundary conditions on ∂D. Notice that βǫ(r) = β′ǫ(r)r, if r 6= 0,

which shows that (12) and (13) have very similar structure. Clearly, (13) is a nonlinearelliptic equation, while the decoupling operated in (12) allows to use linear ellipticequations. The form (9) puts into evidence the approximating coincidence set too.

Moreover, we have yn(x) ≤ ỹn(x) a.e. in D. To infer this, we use βǫ(r) = β ′ǫ(r)r, forr 6= 0, the concavity of βǫ(·) and the definition of the subdifferential of concave mapping:

(β ′ǫn (yn−1 − ψ)

)(yn − ψ) =

(β ′ǫn (yn−1 − ψ)

)(yn−1 − ψ)

+(β ′ǫn (yn−1 − ψ)

)(yn − ψ − yn−1 + ψ) ≥ βǫn (yn−1 − ψ)

+βǫn (yn − ψ) − βǫn (yn−1 − ψ) = βǫn (yn − ψ) .

We get−∆yn + βǫn (yn − ψ) ≤ f (14)

by the monotonicity of βǫ(·). This ends the argument by subtracting (13) from (14) andmultiplying by (yn − ỹn)+.

3 Convergence

The fixed domain character of Algorithm 1 is very helpful for the convergence analysisas well. We prove both the convergence of the solutions and of the geometry undergeneral assumptions. If certain regularity conditions are added, we show that the limitsatisfies (1)-(4).

The following two lemmas show that the general obstacle problem may be reducedto the case ψ|∂D = 0, in general.

5

Lemma 1 Denote by ŷ ∈ H2(D) ∩H10 (D) the solution of

−∆ŷ = f in D, ŷ = 0 on ∂D. (15)

Then y ≥ ŷ a. e. in D.

Proof. It is known that y, the solution of (1)-(4) satisfies y ∈ H2(D) ∩H10 (D) and(1)-(4) can be equivalently expressed in the language of multivalued equations as

−∆y + β(y − ψ) = f a.e in D, y = 0 on ∂D. (16)

where β ⊂ R × R is the maximal monotone graph (8).By regularity, the section of the multivalued function that occurs in (16) satisfied

β(y − ψ) ∈ L2(D). Clearly, (8) shows that β(y − ψ) ≤ 0 a.e. in D. Then a simplecomparison argument for (15), (16) ends the proof. 2

Denote ψ̂(x) = max{ψ(x), ŷ(x)} in D. Then ψ̂ ∈ H10 (D) and y(x) ≥ ψ̂(x) a.e. in D.

Lemma 2 The solution y of (1)-(4) satisfies the same problem with ψ replaced by ψ̂.

Proof. We denote K̂ ={v ∈ H10 (D); v(x) ≥ ψ̂(x) a.e. in D

}. By Lemma 1, y ∈ K̂

and ∆y + f ≤ 0 a.e. in D, by (16).For any v ∈ K̂, we compute

∫

D

(∆y + f)(v − y)dx =∫

D

(∆y + f)(ψ̂ − y)dx

+

∫

D

(∆y + f)(v − ψ̂)dx ≤∫

D

(∆y + f)(ψ̂ − y)dx = 0

The last equality is a consequence of (1)-(2) and the fact that y(x) = ψ(x) means

ŷ ≤ ψ(x) and, consequently, y(x) = ψ̂(x).Integrating by parts in the first term of the above inequality, we get

∫

D

∇y (∇y −∇v) dx ≤∫

D

f(y − v)dx, ∀v ∈ K̂.

This ends the proof. 2

Theorem 1 i) On a subsequence, yn → y̌ weakly in H10 (D) and Ωn → Ω in the com-plementary Hausdorff-Pompeiu topology.

ii) If y̌ ∈ C1(D) and {Ωn} are uniformly of class C, then y̌ is the solution of (1)-(4)with D+ = Ω. The convergence is valid without taking subsequences.

6

Proof. i) By Lemma 1 and 2, we may assume ψ = 0 on ∂D. Then yn − ψ is inH10 (D) and may be used as test function in relation (9). We obtain

∫

D

|∇(yn − ψ)|2 dx+1

ǫn

∫

D\Ωn

(yn−ψ)2dx =∫

D

f(yn−ψ) dx−∫

D

∇ψ·∇(yn−ψ)dx. (17)

Then {yn} is bounded in H10 (D) and

1

ǫn

∫

D\Ωn

(yn − ψ)2dx ≤ Const, ∀n ∈ N. (18)

We denote by y̌ ∈ H10 (D) the weak limit of yn in H10 (D) on a subsequence and byΩ the limit of Ωn (again on a subsequence) in the complementary Hausdorff-Pompeiutopology.

ii) If we assume that {Ωn} is a family of open sets satisfying the uniform C property(i.e. the boundaries ∂Ωn can be locally represented as graphs of uniformly continuousmappings), then it is known that D \ Ωn → D \ Ω in the Hausdorff-Pompeiu comple-mentary metric too, [19] Appendix 3.

Let C ⊂ D \Ω be any compact subdomain. Then C ⊂ D \Ωn, for n ≥ nC , by the Γproperty of the Hausdorff-Pompeiu complementary convergence, [19]. By (18), we get

1

ǫn

∫

C

(yn − ψ)2dx ≤ Const, ∀n ≥ nC .

Then y̌|K = ψ a.e. in C. By letting C → D \Ω, we obtain y̌|D\Ω = ψ a.e. in D \Ω (herewe also use that the boundary of an open set of class C has zero Lebesgue measure,[19]).

Take now any ρ ∈ C∞0 (Ω). Again by the Γ property, we have that supp ρ ⊂ Ωn forn ≥ nρ. Consequently, ρ may be used as test function on (9), for n ≥ nρ and we maylet n→ ∞. We obtain that

∫

D

∇y̌ · ∇ρ dx =∫

D

f ρ dx.

As supp ρ ⊂ Ω the integrals are in fact defined on Ω and we obtain that

−∆y̌ = f in Ω (19)

in the sense of distributions.We also notice that y̌−ψ ∈ H10 (Ω) since y̌−ψ = 0 a.e. in D\Ω and domains of class

C have the Keldysh-Hedberg stability property, [19]. We underline that for domains ofclass C, the usual trace theorems cannot be applied and the above relation on (∂Ω)∩Dhas to be carefully interpreted [19, Ch. 2].

7

Finally, let Q ⊂ Ω be any compact subdomain. Again by the Γ-property of theHausdorff-Pompeiu complementary convergence for the open sets, we get Q ⊂ Ωn, forn ≥ nQ. Then, yn(x) > ψ(x) for x ∈ Q, n ≥ nQ. By yn → y̌ strongly in L2(D), we infery̌(x) ≥ ψ(x) a.e. in Q. Consequently, y̌(x) ≥ ψ(x) a.e. in Ω. The points on ∂Ω\∂D, areminimum points for y̌ − ψ. Under C1 regularity assumptions, we obtain ∇ (y̌ − ψ) = 0on ∂Ω \ ∂D, that is (3) is satisfied for D+ = Ω.

As the solution of (1)-(4) is unique, the above convergences are valid without takingsubsequences. 2

Remark 5 If f ∈ L∞(D), then y ∈ W 2,p(D) for any p ≥ 1. By the Sobolev theorem,we get y ∈ C1(D) in arbitrary dimension. Regularity results for the solution of thevariational inequality and for the free boundary may be found in [14] and [16]. Thisjustifies the regularity assumption.

Remark 6 The above argument shows that the sequences constructed in Algorithm 1have accumulation points. Under regularity assumptions, the limit point is unique andgives the solution of (1)-(4). The algorithm has a constructive character both for thefunctions and the geometry.

We close this section with a stability result in L2(D) for the Algorithm 2 appliedto the bilateral obstacle problem. We assume that the sets {x ∈ D; yn(x) = ψ1(x)},{x ∈ D; yn(x) = ψ2(x)} have zero measure.

Theorem 2 The sequence {yn} is bounded in L2(D). Moreover, there is C > 0, inde-pendent of n, such that:

∫

D\Ωn2

(yn − ψ2)2+dx+∫

D\Ωn1

(yn − ψ1)2−dx ≤ Cǫn. (20)

Proof. We use transformations similar to (14), in the equation (11). We get

−∆yn + βǫn(yn − ψ1) +1

ǫnχD\Ωn

2(yn − ψ2) ≤ f, (21)

where β is given by (8) and βǫn is its regularization.We multiply (21) by (yn − ψ2)+ and we use that

βǫn(yn − ψ1)(yn − ψ2)+ ≥ 0. (22)

While βǫn(yn−ψ1) may take negative values, this happens for yn ≤ ψ1, that is yn−ψ2 ≤ 0(since ψ1 ≤ ψ2). Then (yn − ψ2)+ = 0 and (22) follows. We infer∫

D

∇(yn − ψ2)2+ +1

ǫn

∫

D\Ωn2

(yn − ψ2)2+ ≤∫

D

f(yn − ψ2)+ +∫

D

∇ψ2 · ∇(yn −ψ2)+. (23)

8

By the conditions ψ2 |∂D ≥ 0 and yn |∂D = 0, we have (yn − ψ2)+ = 0 on ∂D and thePoincaré inequality shows that {(yn − ψ2)+} is bounded in H10 (D), by (23).

We introduce now the maximal monotone graph γ ⊂ R × R by

γ(r) =

0, r < 0,[0,+∞[ , r = 0,

∅, r > 0.(24)

and denote by γǫ, ǫ > 0, its Yosida approximation. Notice that γǫ is a convex functionin R.

Equation (11) may be rewritten in the form

−∆yn +1

ǫnχD\Ωn

1(yn − ψ1) + γ′ǫn(yn−1 − ψ2)(yn − ψ2) = f (25)

under our assumption on the level sets of yn − ψ1, yn − ψ2.We compute

γ′ǫn(yn−1 − ψ2)(yn − ψ2) = γ′ǫn(yn−1 − ψ2)(yn−1 − ψ2) (26)+γ′ǫn(yn−1 − ψ2)(yn − ψ2 − yn−1 + ψ2) = γǫn(yn−1 − ψ2)+γ′ǫn(yn−1 − ψ2)(yn − ψ2 − yn−1 + ψ2) ≤ γǫn(yn−1 − ψ2)

+γǫn(yn − ψ2) − γǫn(yn−1 − ψ2) = γǫn(yn − ψ2)

using the subdifferential of convex mappings.By (25), (26), we obtain

−∆yn +1

ǫnχD\Ωn

1(yn − ψ1) + γǫn(yn − ψ2) ≥ f. (27)

Multiply (27) by −(yn − ψ1)− ∈ H10 (D), due to yn = 0 on ∂D, ψ1 ≤ 0 on ∂D:∫

D

∇(yn − ψ1)2− +1

ǫn

∫

D\Ωn1

(yn − ψ1)2− − γǫn(yn − ψ2)(yn − ψ1)− (28)

≤ −∫

D

f(yn − ψ1)− +∫

D

∇ψ1 · ∇(yn − ψ1)−.

Notice that−γǫn(yn − ψ2)(yn − ψ1)− ≥ 0. (29)

since −γǫn(yn − ψ2) may take negative values just for yn ≥ ψ2 ≥ ψ1 and in this case(yn − ψ1)− = 0. By (28), (29) we obtain:

∫

D

∇(yn − ψ1)2− +1

ǫn

∫

D\Ωn1

(yn − ψ1)2− (30)

≤ −∫

D

f(yn − ψ1)− +∫

D

∇ψ1 · ∇(yn − ψ1)−.

9

Relation (30) shows that {(yn−ψ1)−} is bounded in H10 (D), by the Poincaré inequality.We use the inequality

(x− b)+ ≤ (x− a)+ + (a− b)+

and we have

(yn − ψ1)+ ≤ (yn − ψ2)+ + (ψ2 − ψ1)+ = (yn − ψ2)+ + ψ2 − ψ1.

Relation (23) shows that {(yn − ψ1)+} is bounded in L2(D). In combination with (30),it yields {yn} bounded in L2(D).

Relation (20) follows by adding (23) and (30) and using the already establishedboundedness of all the terms except the penalization term. This ends the proof. 2

Remark 7 In fact, the above proof shows that {yn} bounded in Lp(D), p > 2 dependingon the dimension of D. Relation (20) says that the sequence {yn} does not overpassthe obstacles ψ1, ψ2 in the limit. The proof also provides partial information on {∇yn},but it is unclear whether {yn} is bounded in H10 (D). This would ensure convergenceproperties, similar to Theorem 1. In the numerical examples from the next section, suchconvergence properties appear to be true.

4 Numerical tests

We have used the software FreeFem++ v 3.12, http://www.freefem.org. Exceptingthe Test 4, the solutions satisfy homogeneous boundary conditions on ∂D.

Test 1.

This is just a simple example with explicit solution, to test the software and Algorithm1 for solving (5)-(6). We set D = {(x1, x2);

√x21 + x

22 < 1}. The obstacle ψ ∈ H2(D)

is given by the formula

ψ(x1, x2) =

12,

√x21 + x

22 ≤ 14

8(√

x21 + x22

)3− 13 (x21 + x22) + 5

√x21 + x

22 − 116 , 14 <

√x21 + x

22 <

34

12−

√x21 + x

22,

34≤

√x21 + x

22 ≤ 1

For f ∈ L2(D) given by

f(x1, x2) =

−1,√x21 + x

22 ≤ 14

−36√x21 + x

22 + 26 − 52√x2

1+x2

2

, 14<

√x21 + x

22 <

34

1

2√

x21+x2

2

, 34≤

√x21 + x

22 ≤ 1

10

the corresponding solution y ∈ H2(D) ∩H10 (D) of (5)-(6) has the form

y(x1, x2) =

12,

√x21 + x

22 ≤ 14

4(√

x21 + x22

)3− 13

2(x21 + x

22) +

52

√x21 + x

22 +

732, 1

4<

√x21 + x

22 <

34

12− 1

2

√x21 + x

22,

34≤

√x21 + x

22 ≤ 1

We have applied the Algorithm 1 with a mesh of 10850 triangles and 5554 verticesand a fixed penalization parameter ǫn = 10

−3, for all n ∈ N. The algorithm ends after8 iteration when we have started from D \ Ω0 = {(x1, x2);

√(x1 − 0.5)2 + x22 ≤ 0.05}

and it ends after 6 iterations starting from D \ Ω0 = ∅. The computed coincidence setcorresponds to the theoretical coincidence set {(x1, x2);

√x21 + x

22 ≤ 0.25}.

We use the notion of convergence order following [20]. In order to estimate theconvergence order, we have computed

‖yn+1 − y‖L2(D)‖yn − y‖L2(D)

, n = 0, . . . , 4

and we have obtained 0.73, 0.27, 0.33, 0.38, 0.83 which indicates in this example thatthe algorithm converges linearly.

Test 2.

We use Algorithm 1 for solving (5)-(6). We set D = {(x1, x2);√x21 + x

22 < 1}. The

obstacle ψ is given by the formula

ψ(x1, x2) =

12, 0.5 ≤ x1 ≤ 0.7 and − 0.1 ≤ x2 ≤ 0.1

−12.5 ((x1 + 0.4)2 + x22) + 0.5, (x1 + 0.4)2 + x22 ≤ 0.08−1

2, otherwise.

We notice that ψ is not continuous, but the method still works.Let f ∈ L2(D) given by

f(x1, x2) =

{3.5, x21 + x

22 ≤ 0.16

−0.001, otherwise.

The mesh has 109898 triangles and 55350 vertices and we use a fixed penalizationparameter ǫn = 10

−4, for all n ∈ N. Starting from D \ Ω0 = ∅, the history of error ofcomputed solution between two consecutive iterations is presented in Table 1.

After 11 iterations the relative error no longer change. The computed solution isshown in Figure 2 and the computed coincidence set in Figure 3.

The values of‖yn+1 − yn‖L2(D)‖yn − yn−1‖L2(D)

, n = 1, . . . , 9.

11

Figure 1: The obstacle for the Test 2.

are: 0.26, 0.31, 0.63, 0.58, 0.49, 0.38, 0.29, 0.16, 0.13 and 4.37363e-13, so, in this example,the algorithm converges superlinearly.

We have tested the stability of the algorithm when f , the right-hand side in (5),is perturbed by δf : D → R defined by δf(x1, x2) = k100 + k10 sin(10k x1) cos(10k x2)for k ∈ {1, 2, 3, 4, 5}. We denote the computed solution of (5)-(6) with the right-handside f and f + δf by yh and yh + δyh, respectively. The notation doesn’t take intoaccount the (nonlinear) dependence on k, for simplicity of writing. We observe thatthe second column of Table 2 is bounded, which shows that the convergence propertyproved in Theorem 1 is satisfied in this example. Similar comment for the third columnin Table 2. Moreover, the coincidence set associated to the perturbed solutions cannotbe distinguished graphically from the unperturbed one represented in Figure 3.

12

iteration ‖yn − yn−1‖L2(D)0 0.3208391 0.0849192 0.0266143 0.0168364 0.0098345 0.0048456 0.0018867 0.0005608 9.32461e-059 1.22327e-0510 5.35013e-18

Table 1: History of the error for Test 2.

k ‖δyh‖L2(D) / ‖δf‖L2(D) ‖δyh‖L2(D) / ‖yh‖L2(D) ‖δyh‖L2(D)1 0.0167631 0.00351167 0.001508322 0.0166544 0.00699263 0.003003453 0.0164874 0.01031800 0.004431764 0.0166252 0.01380200 0.005928175 0.0167180 0.01719910 0.00738730

Table 2: Test2. Stability when f is perturbed.

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Figure 2: The computed solution and the obstacle for the Test 2.

Figure 3: The computed coincidence set (blue) for the Test 2.

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Test 3.

We consider the torsion of an elastic-plastic prism studied in [7, p. 133] and [21]. Thecross-section of the prism is D = [0, 1] × [0, 1]. We solve the problem (5)-(6) withf(x) = −8 and ψ(x) = −dist(x, ∂D), for all x ∈ D. We use a mesh of 39216 triangles,19865 vertices and size h = 1

128. The penalization parameter is ǫn = 0.003. The

coincidence set presented in Figure 4 is similar to the above references.

Figure 4: The computed coincidence set (blue) for the Test 3.

The Algorithm 1 stops after 6 iterations. The algorithm presented in [21] stops after7 iterations, while the over-relaxation algorithm with projection presented in [7, p. 133]stops after 93 iterations. The same problem was solved in [7, p. 206] by the followingpenalization method: find uǫ ∈W 1,40 (Ω) such that

a(uǫ, v) +1

ǫ

∫

Ω

(1 − |∇uǫ|2

)− ∇uǫ · ∇v dx =∫

Ω

f v dx, ∀v ∈W 1,40 (Ω).

For ǫ = 0.625×10−3, f(x) = −10, the algorithm stops after 186 iterations. We concludethat the number of iterations of Algorithm 1 in order to obtain the numerical solutionis comparable with the algorithm from [21] and it is significantly reduced by rapport ofthe algorithms from [7, p. 133, p. 206]. We also underline the simplicity of Algorithm1.

Test 4.

We consider the second example from [3]. The domain is

D ={(x1, x2); (x1 − 0.5)2 + (x2 − 0.5)2 ≤ 0.52, x2 > 0.5

}

∪ ([0, 1] × [0.2, 0.5]) \ ([0.3, 0.45] × [0.2, 0.45] ∪ [0.6, 0.75] × [0.2, 0.35]) .

We set f(x) = −1 and ψ(x) = 0, for all x ∈ D. We solve the problem (5), but weimpose nonhomogenous boundary conditions

K ={v ∈ H1(D); v ≥ ψ a.e. in D, v = g on ∂D

}

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where

g(x) =

{0, x2 ≤ 0.5c exp(−200(x1 − 0.5)2), otherwise.

Also, at the Step 2, Algorithm 1, the solution of the linear elliptic equation has nonho-mogenous boundary conditions.

We use a mesh of 4874 triangles, 2555 vertices and the penalization parameter isǫn = 0.003. The Algorithm 1 stops after 4 iterations if c = 3.

Figure 5: Test 4. Coincidence sets (blue) for c=3.

In [3], a level set method with artificial time variable is introduced which convergesto the steady obstacle problem.

Also, we have solved the third example from [3]. The number of time advancingiterations for the level set method is greater than 60, see Figure 3, (f) from [3], whilethe Algorithm 1 stops after 6 iterations.

Test 5.

Now we test the Algorithm 2 for the bilateral elastic-plastic torsion problem discussedin [13] and [22]. We can put this problem in the form (10). Let D = [0, 1] × [0, 1],ψ1(x, y) = −dist ((x, y), ∂D), ψ2(x, y) = 0.2 for all (x, y) ∈ D and set

g(x) =

6x, 0 < x ≤ 1/6,2(1 − 3x), 1/6 < x ≤ 1/3,6(x− 1/3), 1/3 < x ≤ 1/2,

2 (1 − 3(x− 1/3)) , 1/2 < x ≤ 2/3,6(x− 2/3), 2/3 < x ≤ 5/6,

2 (1 − 3(x− 2/3)) , 5/6 < x ≤ 1

and

f(x, y) =

300, (x, y) ∈ S = {(x, y) ∈ D; |x− y| ≤ 0.1 and x ≤ 0.3} ,−70 exp(y)g(x), x ≤ 1 − y and (x, y) /∈ S,15 exp(y)g(x), x > 1 − y and (x, y) /∈ S.

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We use a mesh of 9662 triangles, 4960 vertices, the size h = 164

and the penalizationparameter is ǫn = 0.003. The computed solution after 15 iterations is presented inFigure 6 and the corresponding coincidence sets in Figure 7.

Figure 6: Test 5. Computed solution

Figure 7: Test 5. Coincidence sets (blue) for the bottom obstacle (left) and for the topobstacle (right)

In [13], an augmented lagrangian active set strategy is employed. At each iteration,a reduced linear system associated with the inactive set is solved. Without multigridnested iterations, the algorithm stops after 30 iterations.

In [22], at each iteration, linear systems associated to the complementary of thecoincidence sets are solved. There are no information about the number of iterations,only on the CPU time; for example the total CPU time for the “up down” algorithm is

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916 s and the compute time of a linear system is 46 s. For instance, the CPU time is15.12 s on a PC with Intel i5 2.53 GHz and 4 Go RAM in our algorithm.

Test 6.

We solve (10), where the computational domain is D = {(x1, x2);√x21 + x

22 < 1},

the two obstacles are defined by ψ1(x1, x2) = −0.15 and ψ2(x1, x2) = 0.241, for all(x1, x2) ∈ D and the right side member is

f(x1, x2) =

{5, x21 + x

22 ≤ 0.25

−5, otherwise.

We have applied the Algorithm 2 with a mesh of 11320 triangles and 5789 vertices anda fixed penalization parameter ǫn = 10

−7, for all n ∈ N. Starting from D \ Ω01 = ∅ andD \ Ω02 = ∅, after 6 iterations we have obtained the relative error ‖yn − yn−1‖L2(D) =0.004868. The computed solution after 6 iterations is presented in Figure 8 and thecorresponding coincidence sets in Figure 9.

After iteration 6 there are small oscillations of the relative error which becomesidentically 0 after iteration 29 and the algorithm stops.

Figure 8: The computed solution after 6 iterations for the Test 6.

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Figure 9: Test 6: the computed coincidence set (blue) after 6 iterations for the bottomobstacle (left) and for the top obstacle (right)

Figure 10: The computed solution after 29 iterations for the Test 6.

Acknowledgments

The second author was supported by Grant 145/2011 CNCS, Romania.

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