ON THE MINIMIZATION OF SOME NONCONVEX DOUBLE OBSTACLE PROBLEMS

We consider a nonconvex variational problem for which the set of admis-sible functions consists of all Lipschitz functions located between two ﬁxed obstacles. It turns out that the value of the minimization problem at hand is equal to zero when the obstacles do not touch each other; other-wise, it might be positive. The results are obtained through the construction of suitable minimizing sequences. Interpolating these minimizing sequences in some discrete space, a numerical analysis is then carried out.


Introduction
Let Ω denote a bounded and convex domain in R n (n ≥ 2) of boundary ∂Ω and of closure Ω. Let ϕ : R n → R be a continuous function such that ϕ w i = 0 ∀i = 1, . . . , p, (1.1) ϕ(w) > 0 ∀w = w i , i = 1, . . . , p, (1.2) where w i 's are p elements of R n (p ≥ 2). For instance, in solid-solid phase transformations, the function ϕ could be some elastic energy that vanishes at wells w i 's. These wells stand for the stress-free states of a body represented by Ω. For further details about the physical background and also the mathematical modelling, we refer the reader to [2,3] and the references quoted therein. We assume that, for some physical reasons (e.g., some loads applied to the body Ω), the deformations are constrained not to exceed some fixed obstacles. To make precise the formulation of our problem, let α, β, and G be three Lipschitz continuous functions, that is, α, β, G ∈ W 1, ∞ (Ω) (1.3) such that Here W 1,∞ (Ω) is the usual Sobolev space of all weakly differentiable functions u : Ω → R such that u and |∇u| are essentially bounded. For a further discussion of Sobolev spaces, we refer the reader to [1]. We denote by K the following set: (1.5) Then we consider the following problem: More precisely, we consider the case when ∇G(x) ∈ Co w i a.e. in Ω, (1.7) where Co(w i ) denotes the convex hull of the wells w i 's. For the numerical purpose and in order to simplify, we will assume that Ω is polyhedral. Let (T h ) be a regular family of triangulations of Ω (see [11]). This means that where h K is the diameter of the N-simplex K and ρ K its roundness (i.e., the largest diameter of the balls that could fit in K). If P 1 (K) is the space of polynomials of degree 1 on K, we set

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A. Elfanni 537 where (1.10) Then we denote by I h the following approximate version of (1.6): The (double) obstacle problems were intensively studied by many authors in the case of convex variational problems, that is, the function ϕ in problem (1.6) is assumed to be convex. In this paper, we would like to investigate such problems in the presence of nonconvexity. Recall that the question of existence or nonexistence of minimizers for scalar noncovex variational problems with homogeneous boundary conditions is closed (see [10]). One can also see [9] for a contribution in this direction. The case of nonhomogeneous boundary conditions was first studied by Chipot (see [5]) and then other results were obtained in [6,7]. The main concern of our paper is to compute the value of I and obtain estimates for |I h − I| in terms of h. We denote by Ω the subset of Ω where the obstacles do not touch each other, that is, Moreover, if |Ω\Ω | = 0 (|Ω\Ω | is the Lebesgue measure of Ω\Ω ), one has, for h small enough, where C is a constant independent of h. The plan of our paper will be as follows. In Section 2, we present all the intermediate results that we will need to prove our main results which are exposed in Section 3. We will try as far as possible to have our paper self-contained.

Preliminary results
In this section, we collect all the ingredients which will be useful to prove our results. First we have the following lemma.
Lemma 2.1. If G is a Lipschitz continuous function satisfying (1.7), then there exists a Lipschitz functionG defined in R n such that

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where and denote, respectively, the infimum and supremum of functions and a · b is the scalar product of a, b ∈ R n .
Proof. Applying the mean value theorem after regularization (see [8] for details), one has Then letG We now prove that A. Elfanni 539 For every x, y ∈ R n , one has Moreover, for every x, y ∈ R n , one has Then the second inequality in (2.7) follows easily using the first inequality proved above. This completes the proof of the lemma. Now let (v 1 , . . . , v q ) be a basis of the space W spanned by the wells w i 's and denote by x z the points of the lattice of size h 1/2 spanned by the v i 's, that is, for any z = (z 1 , . . . , z q ) ∈ Z q , set Then we define the functions Λ h and V h by (2.14) By a unit cell of the lattice spanned by the h 1/2 v i 's we mean a set of the type Then one has the following lemma. (2.17) Proof. We give here an astute proof for the case when W is a one-dimensional space. This case was not treated in [6] where a technical proof for higher dimensions is given.
; the other case can be handled similarly. There exists λ ∈ [0, 1] such that Replacing the above infima by suprema and using the second inequality in Lemma 2.1, we obtain the result for the function V h .
We will also need the following lemma.
Proof. Let z ∈ Z q and x ∈ Ω. Using Lemma 2.1, one has

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Now we denote by x the component of x on P W (Ω), the orthogonal projection of Ω onto W. There exists z 0 such that x ∈ C z 0 , then x can be written as follows: Using the Lebesgue differentiation theorem, one obtains (1.7).

Statement and proof of the main results
First we assume that the obstacles do not touch each other, that is, Thus the following constants are positive:

2)
where Ω 1 and Ω 2 denote the following sets: , Then one has the following theorem.

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where C * = 2q max i w i max i v i .
Proof. We consider the following Lipschitz functions: where dist(x, ∂Ω i ) denotes the distance from x to the boundary ∂Ω i of Ω i , i = 1, 2. The functions Λ h and V h are defined by (2.14). Then one has We prove (3.7); (3.8) can be proved the same way. According to Lemma 2.3, one has On the other hand, one has where M 1 is defined by (3.2). Since 0 < h < (M 1 /C * ) 2 , one gets It is clear by (2.25) that the functions u h i 's coincide with G at the boundaries of Ω i 's, respectively. Since ∂Ω ⊂ ∂Ω 1 ∪ ∂Ω 2 , the function (3.14) coincides with G at the boundary of Ω. Moreover, u h is a Lipschitz function. Indeed, let x ∈ Ω 1 and y ∈ Ω 2 . There exists z ∈ ∂Ω 1 ∩ ∂Ω 2 such that

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Thus where C is a constant independent of x, y, and h (in the sequel, we will denote by C every constant which does not depend on h). Then the function u h belongs to K. Now we can prove (3.4). Due to (2.25), one has Therefore, ∇u h = w i except in a neighborhood N h of the boundaries ∂Ω i of measure less than Ch 1/2 . Using (1.1) and the fact that ∇u h is uniformly bounded, one obtains for every 0 < h < inf((M 1 /C * ) 2 , (M 2 /C * ) 2 ), which obviously implies (3.4). Due to (2.25) again, one has Letû h denote the interpolate of u h . Clearly,û h ∈ K h and one has We prove (3.22). Let x ∈ Ω. There exists an N-simplex K ∈ T h such that x ∈ K. Let y be any vertex of the N-simplex K. One has u h (y) =û h (y). (3.23) Therefore, one has

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Using the mean value theorem, one obtains Since ∇u h is uniformly bounded, ∇û h is also uniformly bounded (see [4]). Moreover, x − y ≤ h K ≤ h. Thus there exists a constant C so that (3.22) holds. Therefore, using the triangle inequality (recall that h < 1), one gets Notice that except maybe on the set S composed of simplices where interpolation occurred. Since on this set ∇û h remains bounded, one has where |S| is the Lebesgue measure of S. When So, in the set where dist(x, ∂Ω 1 ∪ ∂Ω 2 ) ≥ C * h 1/2 + h, one has thatû h equals the interpolate of Λ h or V h . We denote by S 1 the set First we have To estimate |S ∩ S 1 |, one sees that interpolation occurs on an h-neighborhood of the set where Λ h and V h have discontinuity in their gradients.

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Clearly, Λ h and V h have a jump in their gradients on a unit cell of the lattice spanned by h 1/2 v i 's when one of the functions is equal to another. These two functions are equal on a set of (q − 1)dimensional measure bounded by Ch (q−1)/2 . Since in Lemma 2.2 the supremum and infimum are taken on a finite number of functions, it is clear that (3.35) where N(h) is the number of cells of size h 1/2 included in P W (Ω). Clearly, N(h)h q/2 is less than or equal to the q-dimensional measure of P W (Ω). Therefore, Combining (3.28), (3.33), and (3.36), one obtains This completes the proof of the theorem.

Remark 3.2.
We chose the cells C z 's of size h 1/2 among those of size h α , α ∈ (0, 1), since they provide the best estimate (see [6] for details).  There exists ν ∈ R n such that w i · ν = 0 ∀i = 1, 2, . . . , p. We end this paper by considering the case where the two obstacles touch each other. We denote by Ω the following open set: (3.43) One has, for every u ∈ K, (3.48) Now let u ∈ K(Ω ). We defineũ ∈ K(Ω) as follows: