Potential Analysis 9: 105–142, 1998. © 1998 Kluwer Academic Publishers. Printed in the Netherlands.
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A Proof of the Lewy–Stampacchia’s Inequality by a Penalization Method A. MOKRANE1 and F. MURAT2
1 Department of Mathematics, ÉcoleNormale Supérieure, Vieux Kouba, 16050 Algiers, Algeria 2 Laboratoire d’Analyse Numérique, Tour 55-65, Université Pierre et Marie Curie, 4, place Jussieu,
75252 Paris Cedex 05, France (Received: 8 November 1994; accepted: 7 September 1995) Abstract. In this paper we prove the Lewy–Stampacchia’s inequality for elliptic variational inequalities with obstacle involving fairly general Leray–Lions operators. The main novelty of the paper is the method of proof, which uses the natural penalization. One of the steps of the proof consists in 1,p proving, again thanks to the natural penalization, that the nonnegative cone of W0 () is dense in 0
the nonnegative cone of W −1,p ().
Mathematics Subject Classification (1991): 35J85, 35R05, 35A35, 49J40, 31C45. Key words: Variational inequalities, penalization, Lewy–Stampacchia’s inequality.
1. Introduction The prototype of the problems that we will consider here is the linear obstacle problem with obstacle 0. In this problem we are looking for a function u in H01 () and for a Radon measure µ which satisfy −1u − f = µ in , u = 0 on ∂, u > 0 in , µ > 0 in , hµ, ui = 0, or equivalently for the solution u of the variational inequality Z Du D(v − u) dx > hf, v − ui, ∀v ∈ K(0),
u ∈ K(0),
where K(0) = {v ∈ H01 (): v
> 0 a.e.
in }.
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When the right-hand side f is assumed to be in the ordered dual V2∗ of H01 (), where V2∗ = (H −1 ())+ − (H −1 ())+ , we will prove that the measure µ satisfies the Lewy–Stampacchia’s inequality, namely µ 6 f −.
(1.1)
The main novelty of the present paper is the method that we will use to prove the Lewy–Stampacchia’s inequality. We start from the penalized problem −1uε − 1 u− = f + − fˆε in D 0 (), ε ε u ∈ H 1 (), ε
0
where fˆε ∈ H01 (), fˆε > 0, is a smooth approximation of the negative part f − of f (we will prove in Section 5, again by a penalization method, that there exists such an approximation). We define 1 zε = fˆε − u− . ε ε Our main result consists to prove that zε− → 0 strongly in L2 (). Setting then 1 µε = u− , ε ε we have by definition of zε fˆε − µε = zε = zε+ − zε−
> −zε−,
which implies by passing to the limit that f − −µ > 0, i.e. the Lewy–Stampacchia’s inequality (1.1). In fact, we will prove that the Lewy–Stampacchia’s inequality, namely µ = A(u) − f
6 (f − A(ψ))− ,
(1.2)
holds in the general framework of nonlinear elliptic problems with obstacle of the type ( hA(u), v − ui > hf, v − ui, ∀v ∈ K(ψ), (1.3) u ∈ K(ψ),
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where 1,p
K(ψ) = {v ∈ W0 (): v
>ψ
a.e. in }. 1,p
Here the unknown is the function u ∈ W0 (). The data are the Leray–Lions 1,p pseudomonotone operator A(v) = −div(a(x, v, Dv)), which acts from W0 () 0 into W −1,p (), the obstacle ψ, which belongs to W 1,p () with ψ 6 0 on ∂, and the right-hand side f , which are assumed to be such that g = f − A(ψ) belongs to 0 0 the ordered dual Vp∗0 = (W −1,p ())+ − (W −1,p ())+ . On the function a(x, s, ξ ) we will make hypotheses of strong monotonicity in ξ , of local Lipschitz (or Hölder) continuity in ξ and of local Hölder continuity in s; note that these hypotheses do not imply the uniqueness of the solution of (1.3). The proof of (1.2) is then similar to the proof of the linear case, but is more technical. Setting g = f − A(ψ), (which is assumed to belongs to Vp∗0 ) we first consider (Subsection 4.1) the case where 1,p − ∞ −1,p 0 ()), g ∈ W0 () ∩ L () (and not only to W (1.4) ψ ∈ W 1,p () ∩ L∞ () (and not only to W 1,p ()), with ψ 6 0 on ∂. In such a case, we consider the penalized problem 1 A(uε ) − (uε − ψ)− = f = g + − g − + A(ψ). ε Setting 1 zε = g − − (uε − ψ)− , ε we prove similarly to the linear case that zε− → 0 strongly in L1 (). Defining 1 µε = (uε − ψ)− ε and passing to the limit in g − − µε = zε = zε+ − zε−
> −zε−,
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which follows from the definition of zε , we obtain that g − − µ > 0,
(1.5)
i.e. the Lewy–Stampacchia’s inequality (1.2) under the above assumptions (1.4) on g − and ψ. We then consider (Subsection 4.2) the general case where g − only belongs to −1,p 0 () and ψ only belongs to W 1,p () with ψ 6 0 on ∂. Approximating W 1,p − g by gˆn ∈ W0 () ∩ L∞ (), gˆn > 0 (existence of such approximations is proved, again by penalization, in Section 5), and ψ by ψn ∈ W 1,p () ∩ L∞ () with ψn 6 0 on ∂, we pass to the limit in (1.5), i.e. in gˆn − µn > 0, and deduce the Lewy–Stampacchia’s inequality in the general case. Actually the case where 1 < p 6 2 is more technical than the case p > 2 and requires special attention (see Subsection 4.1.5 and Sections 5 and 6). In short, our main goal in the present paper is to prove the Lewy–Stampacchia’s inequality (1.2), and the main interest of the paper is the method based on the natural penalization that we use to prove it. The hypotheses we make on the operator are moreover less restrictive than the hypotheses made by the previous authors who proved the Lewy–Stampacchia’s inequality, since the present method allows us to consider the case of pseudomonotone (and not only monotone) operators and require less regularity on the data. These hypotheses do not imply the uniqueness of the solution of the variational inequality (1.3), and therefore allow us to prove the Lewy–Stampacchia’s inequality in a framework where multiple solutions could exist. The existence of a solution of problem (1.3) is a classical result, which has been proved by various authors (see among other Hartman and Stampacchia [18] and Lions [20]). The Lewy–Stampacchia’s inequality was first proved by Lewy and Stampacchia [19] in the context of the superharmonic problem. It has then been extended by various authors to various cases of nonlinear elliptic operators, and became a powerful tool for proving existence and regularity results: see e.g. Mosco and Troianiello [23], Mosco [21], Bensoussan and Lions [1], Charrier and Troianiello [8], Troianiello [24], Vivaldi [25], Donati and Matzeu [13], Garroni and Troianiello [15], Hanouzet and Joly [17], Donati [12], Mosco and Frehse [22], Garroni and Menaldi [14], Boccardo and Gallouët [3], Garroni and Vivaldi [16] and Boccardo and Cirmi [2]. Penalization techniques have been widely used in the study of obstacle problems, see e.g. Brezis and Stampacchia [6], and Bensoussan and Lions [1]. The natural penalization has been used by Garroni and Menaldi [14] and Garroni and Vivaldi [16] to prove the Lewy–Stampacchia’s inequality in the context of integrodifferential equations which are not in divergence form. On the other hand, another proof of the Lewy–Stampacchia’s inequality, under hypotheses similar to ours but using the homographic penalization studied by Brauner and Nikolaenko [5], has recently been given by Boccardo and Cirmi [2].
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2. Statement of the Main Result Let be a bounded open subset of RN , with boundary ∂ (no smoothness is assumed on ∂). Let p and p 0 be fixed with 1 1 + 0 = 1, p p
1 < p, p 0 < ∞.
Assumptions on the operator 1,p Let A be a nonlinear operator of Leray–Lions type acting from W0 () into its 0 dual W −1,p (), which is defined by A(u) = −div(a(x, u, Du)). The function a: ×R ×R N → R N is assumed to be a strictly monotone Carathéodory function, i.e. ∀ (s, ξ ) ∈ R × R N , x → a(x, s, ξ ) is measurable, a.e. x ∈ , (s, ξ ) → a(x, s, ξ ) is continuous, (2.1) ∀ ξ ∈ R N , ∀η ∈ R N , ξ 6 = η, [a(x, s, ξ ) − a(x, s, η)][ξ − η] > 0. We also assume that there exist three constants α¯ > 0, β¯ > 0, γ¯ > 0, a function h¯ in L1 () and a function k¯ in Lp () such that, for a.e. x ∈ , for all s ∈ R and for all ξ ∈ RN one has ¯ > α¯ | ξ |p −[γ¯ |s|p−1 + |h(x)|], ¯ ¯ k(x)|+ |a(x, s, ξ )| 6 β[| | s | + | ξ |]p−1 . a(x, s, ξ )ξ
(2.2) (2.3)
We finally assume that for any m ∈ R + there exist four constants αm > 0, βm > 0, γm > 0, δm > 0 and three functions hm , km , lm in Lp () such that, for a.e. x ∈ , for all s ∈ R , t ∈ R with |s| 6 m, |t| 6 m, and |s − t| 6 εm for some small εm , and for all ξ ∈ R N , η ∈ R N one has αm | ξ − η |p if p > 2, | ξ − η |2 (2.4) [a(x, s, ξ ) − a(x, s, η)][ξ − η] > αm (|hm (x)|+ | ξ | + | η |)2−p if 1 < p 6 2, β [|k (x)| + |ξ | + |η|]p−2 |ξ − η| m m if p > 2, (2.5) |a(x, s, ξ ) − a(x, s, η)| 6 β |ξ − η|p−1 if 1 < p 6 2, m
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0 γm |s − t|(1+δm )/p [|lm (x)| + |ξ |]p−1 if p > 2, |a(x, s, ξ ) − a(x, t, ξ )| 6 γm |s − t|(1+δm )/2 [|lm (x)| + |ξ |]p−1 if 1 < p 6 2.
(2.6)
REMARK 2.1. Assumptions (2.1), (2.2) and (2.3) will be used in the proof of the existence result, of the a priori estimates and of the Lewy–Stampacchia’s inequality. Assumptions (2.4), (2.5) and (2.6) will be used only in the proof of the Lewy–Stampacchia’s inequality (and more specifically, in Subsections 4.1.4 and 2 4.1.5 below). REMARK 2.2. An example of an operator which satisfies all the above assumptions is the so-called p-Laplacian, defined for 1 < p < ∞ by −1p (u) = −div(|Du|p−2 Du). A more general example is the case where A(u) = −div(b(u)|Du|p−2 Du),
i.e. a(x, s, ξ ) = b(s)|ξ |p−2 ξ.
If b is a continuous function which satisfies α 6 b(s) 6 β, for some 0 < α 6 β < ∞, then hypotheses (2.1), (2.2), (2.3), (2.4) and (2.5) are satisfied; hypothesis (2.6) is satisfied if b is locally Hölder continuous with an Hölder exponent which 2 is strictly greater than 1/p 0 when p > 2 and than 1/2 when 1 < p 6 2. Assumptions on the obstacle and on the right-hand side Let ψ: → R be a given function (the obstacle) which is assumed to satisfy ψ ∈ W 1,p () with ψ
60
on ∂,
(2.7)
where the last assertion has to be understood in the usual weak sense (v − ψ)− ∈ W0 () ∀v ∈ W0 (). 1,p
1,p
Let K(ψ) be defined by 1,p
K(ψ) = {v ∈ W0 (): v
> ψ a.e.
in }.
(2.8)
Note that K(ψ) is non empty, since ψ + belongs to K(ψ). 0 Finally, let f ∈ W −1,p () be given and let g be defined by g = f − A(ψ) = f + div(a(x, ψ, Dψ)).
(2.9)
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We assume that g ∈ Vp∗0 ,
(2.10)
where Vp∗0 is the ordered dual space of W0 (), which is defined as the set of those 0 elements g of W −1,p () which are also elements of the space of Radon measures 0 M() and are such that g + and g − belong to W −1,p (), or equivalently as the set 0 of those elements g of W −1,p () which are such that there exist g p and g n (where the superscripts p and n stand for ‘positive’ and ‘negative’) such that ( g = gp − gn, 1,p
0
g p ∈ W −1,p (),
gp
> 0,
0
g n ∈ W −1,p (),
gn
> 0.
Our goal is to prove the following theorem. THEOREM 2.1. Under the above assumptions (2.1) to (2.10), there exists at least one function u, which is a solution of the variational inequality Z a(x, u, Du)(Dv − Du) dx > hf, v − ui, ∀v ∈ K(ψ), (2.11) u ∈ K(ψ), and which is such that the distribution µ defined by µ = −div(a(x, u, Du)) − f
(2.12)
satisfies the Lewy–Stampacchia’s inequality µ 6 g−.
(2.13)
REMARK 2.3. Theorem 2.1 states that there exists at least one solution u of the variational inequality (2.11) which also satisfies the Lewy–Stampacchia’s inequality (2.13). The existence of a solution u of the variational inequality (2.11) is well known (at least if p > 2, see Section 6 below). It is also well known that the existence of a solution of (2.11) is equivalent to the existence of a function u and of a Radon measure µ which satisfy −div(a(x, u, Du)) − f = µ in D 0 (), 1,p
u ∈ W0 (), u>ψ
a.e.
µ ∈ M(),
(2.14) (2.15)
in ,
(2.16)
µ > 0 in ,
(2.17)
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(u − ψ) = 0 µ − a.e.
in .
(2.18)
The Lewy–Stampacchia’s inequality thus states that µ satisfies (2.13), i.e. µ 6 g − = (f + div(a(x, ψ, Dψ))− .
2
REMARK 2.4. The hypotheses made in the present paper (and in particular (2.4), (2.5), (2.6)) do not imply the uniqueness of the solution of (2.11): there are indeed counterexamples to the uniqueness of the solution of the equation (which corresponds to ψ ≡ −∞) under hypotheses which are slightly more restrictive than (2.1), (2.2), (2.3), (2.4), (2.5), (2.6): see e.g. Carrillo and Chipot [7], Chipot and Michaille [9] or Boccardo et al. [4]. This is why Theorem 2.1 states the existence of at least one solution of (2.11) which satisfies the Lewy–Stampacchia’s inequality. On the other hand, it is fair to mention that uniqueness holds as soon as a strictly increasing zero-order term β(u) (the model of which is β(u) = λ|u|p−2 u, with λ > 0) is added to the pseudomonotone operator A(u) = − div(a(x, u, Du)). 2 0
REMARK 2.5. In view of (2.14), (2.15), (2.3) and (2.17), µ belongs to W −1,p ()∩ M(). Inequality (2.13) thus makes sense in the sense of Radon measures. Note also that according to a theorem of Deny [11] any function of W 1,p () is mea0 surable for the nonnegative measure µ which belongs to W −1,p (). This gives a 2 meaning to (2.18). REMARK 2.6. The proof of the Lewy–Stampacchia’s inequality that we will give below uses the natural penalization. This proof is direct and relatively simple (at least in the linear case, see Section 3 below). It does not require the assumption that A is a monotone operator, in contrast with Donati [12] and Mosco [21] who used the strict monotonicity of the operator A (remark that in general A is not monotone when a(x, s, ξ ) depends on s). Note also that the regularity assumptions (2.1) to (2.10) on the function a, on the right-hand side f and on the obstacle ψ are in some 2 sense minimal in the variational framework of functions with finite energy. REMARK 2.7. To prove the Lewy–Stampacchia’s inequality we will use a density 1,p lemma which asserts that the nonnegative cone of W0 () is dense in the nonnega0 tive cone of W −1,p (). This lemma was proved for example in Donati and Matzeu [13], and in Donati [12] by using the Lewy–Stampacchia’s inequality. We prove this density lemma in Section 5 below without using the Lewy–Stampacchia’s 2 inequality, but again the natural penalization method. 3. The Linear Case: A Simple Proof In order to better explain the ideas of the proof of Theorem 2.1, we will first present the proof in the model case where p = 2, A = −1 and ψ = 0. Specifically we
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will prove in this section the following theorem, which is a special case of Theorem 2.1. THEOREM 3.1 (Linear case with obstacle ψ = 0). Assume that f belongs to the ordered dual V2∗ of H01 (). Then there exists one function u ∈ H01 () and one measure µ ∈ H −1 () which satisfy −1u − f = µ in D 0 (), u ∈ H01 (), u > 0 a.e. in , (3.1) µ > 0 in D 0 (), −1 hµ, ui 1 = 0. H () H () 0
Moreover µ satisfies the Lewy–Stampacchia’s inequality µ 6 f −.
(3.2)
Proof. The proof of Theorem 3.1 that we will present in this section is slightly different of the proof of Theorem 2.1 that we will perform in Section 4 below. It is less technical and will be divided in four steps: 3.1. 3.2. 3.3. 3.4.
Penalization and a priori estimates Proof of the existence result Strong convergence of zε− in L2 () Proof of the Lewy–Stampacchia’s inequality.
3.1.
PENALIZATION AND A PRIORI ESTIMATES
Since f belongs to V2∗ , we have f = f + − f − where f + and f − are nonnegative elements of H −1 (). We approximate f − by a sequence fˆε which satisfies fˆε ∈ H01 (),
fˆε
> 0,
fˆε → f −
strongly in H −1 ().
(3.3)
Such an approximation exists, according to Lemma 5.1 which we state and prove in Section 5 below. We will assume that ε k fˆε k2H 1 () → 0 as ε → 0.
(3.4)
0
Such an hypothesis is licit since it only corresponds to relabel the subscripts in the sequence (3.3). We then consider the penalized problem −1u − 1 u− = f + − fˆ in D 0 (), ε ε ε ε (3.5) 1 uε ∈ H0 ().
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It is a classical result that problem (3.5) admits a solution. Using uε as test function in (3.5) we obtain 1 Z Z 1 2 + ˆ |Duε |2 dx + |u− ε | dx = H −1 () hf − fε , uε iH01 () ε
6
C k uε kH 1 () , 0
which implies k uε kH 1 () 6 C,
(3.6)
2 k u− ε kL2 () 6 Cε.
(3.7)
0
3.2.
PROOF OF THE EXISTENCE RESULT
Define 1 . µε = u− ε ε
(3.8)
In view of Equation (3.5), of hypothesis (3.3) on fˆε and of estimate (3.6) on uε , µε = −1uε − f + + fˆε is bounded in H −1 (). We can therefore extract a subsequence (still denoted by ε) such that (
uε * u weakly in H01 (), µε * µ weakly in H −1 ().
Let us prove, in the classical way, that (u, µ) is a solution of (3.1). Passing to the limit in Equation (3.5) yields −1u − µ = f + − f − = f
in D 0 ().
− We already know that u ∈ H01 (). From uε = u+ ε − uε and from (3.7), we deduce + that u = u , i.e. u > 0 a.e. in . The property µ > 0 follows from (3.8). To obtain the ‘complementary equality’ H −1 () hµ, uiH01 () = 0, we use (v − uε ) with v ∈ K(0) as test function in (3.5); since Z Z 1 1 − − u v dx 6 0 and − u− (−uε ) dx 6 0, ε ε ε ε 1 Here and in what follows, C denotes nonnegative constants which do not depend on ε, but can
vary from line to line.
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we deduce that Z Duε Dv dx
> hf + − fˆε , v − uε i +
115
Z |Duε |2 dx.
(3.9)
It is then easy to pass to the limit in (3.9) and to obtain that u solves the variational inequality Z DuD(v − u) dx > hf, v − ui, ∀v ∈ K(0), (3.10) u ∈ K(0), which, using v = 0 and v = 2u as test functions, yields Z DuDu dx = hf, ui,
i.e. hµ, ui = 0 since µ = −1u − f. 3.3. STRONG CONVERGENCE OF zε− IN L2 () We now define 1 . zε = fˆε − u− ε ε
(3.11)
The function zε belongs to H01 (), because fˆε has been chosen to belong to H01 (): this is the very reason for the approximation (3.3) of f − . Our goal is to prove that zε− converges strongly to zero in L2 (); this is the original part of the present paper. Equation (3.5) can be rewritten as −1uε + zε = f +
in D 0 ().
(3.12)
Using −zε− as test function in (3.12) we obtain Z Z − Duε D(−zε ) dx + |zε−|2 dx = hf + , −zε− i 6 0.
(3.13)
Define Eε = {x ∈ : zε (x) < 0}. Since fˆε
(3.14)
> 0, we have in view of the definitions (3.11) and (3.14) of zε and Eε
uε < 0
on Eε .
(3.15)
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Because of (3.14) and (3.15) we have Z − Du+ ε D(−zε ) dx = 0. − Since uε = u+ ε − uε , equation (3.13) reads as Z Z − − −Duε D(−zε ) dx + |zε− |2 dx 6 0.
− ˆ From the definition (3.11) of zε we have −u− ε = ε(zε − fε ); therefore −Duε = εD(zε − fˆε ) and we thus obtain Z Z − ˆ (3.16) ε D(zε − fε )D(−zε ) dx + |zε− |2 dx 6 0.
Using Young’s inequality in (3.16) gives Z Z Z − 2 − 2 ε |Dzε | dx + |zε | dx 6 −ε D fˆε Dzε− dx
6
ε 2
Z
|D fˆε |2 dx +
ε 2
Z
|Dzε−|2 dx.
In view of hypothesis (3.4) on fˆε this implies that k zε− k2L2 () 6
ε k D fˆε k2L2 () → 0. 2
We have proved that zε− → 0 strongly in L2 ().
(3.17)
3.4. PROOF OF THE LEWY–STAMPACCHIA’S INEQUALITY In view of definitions (3.11) and (3.8) of zε and µε , we have fˆε − µε = zε . Since zε > −zε− , this yields fˆε + zε−
> µε .
(3.18)
Passing to the limit in (3.18) thanks to the hypothesis (3.3) on fˆε and to the strong convergence (3.17) of zε− implies that f−
> µ,
i.e. the Lewy–Stampacchia’s inequality (3.2).
2
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4. Proof of Theorem 2.1 We will perform the proof of Theorem 2.1 without distinguishing between the cases p > 2 and 1 < p < 2, except for the second part of the proof of the convergence of zε− in L1 (), where some technicalities appear in the case 1 < p < 2. The proof will be divided in two parts: 4.1. The case where g − belongs to W0 () ∩ L∞ () and where ψ belongs to W 1,p () ∩ L∞ () 4.2. The general case: approximation of g and ψ and passage to the limit. 1,p
4.1. THE CASE WHERE g − BELONGS TO W0 () ∩ L∞ () AND WHERE ψ BELONGS TO W 1,p () ∩ L∞ () 1,p
In this case the proof will be performed in six steps: 4.1.1. Penalization and a priori estimates 4.1.2. Proof of the existence result 4.1.3. Strong convergence of zε− in L1 (): first part of the proof 4.1.4. Strong convergence of zε− in L1 (): second part of the proof when p > 2 4.1.5. Strong convergence of zε− in L1 (): second part of the proof when 1
ψ ∈ W 1,p () with ψ
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on ∂.
In this Subsection 4.1 we further assume that g = gp − gn, 0 g p ∈ W −1,p (), g p > 0, n 1,p g ∈ W0 () ∩ L∞ (), g n > 0,
(4.1)
where the superscripts p and n stand for ‘positive’ and ‘negative’ (note that we do not assume that g p = g + and g n = g − , although this could possibly hold true), and that ψ ∈ W 1,p () ∩ L∞ () with ψ
60
on ∂.
(4.2)
We will remove these regularity hypotheses on g n and ψ in Subsection 4.2. 4.1.1. Penalization and a priori estimates REMARK 4.1. Actually, we will not use hypotheses (4.1) and (4.2) in the first two subsections 4.1.1 and 4.1.2, in which we will only assume that the operator satisfies (2.1), (2.2), (2.3) and that the obstacle ψ: → R is a measurable function such that ∃v ? ∈ W0 () ∩ L∞ () with v ? 1,p
> ψ a.e.
in ,
(4.3)
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while the right-hand side is such that 0
f ∈ W −1,p ().
(4.4)
Note that the requirement v ? ∈ L∞ () of hypothesis (4.3) will be used only once in subsections 4.1.1 and 4.1.2; this requirement could be removed at the expense of a further passage to the limit, considering in a first step the obstacle ψn defined by ψn (x) = inf(n, ψ(x)), for which vn? = Tn (v ? ), where Tn is the truncation at height n, satisfies (4.3), and then passing to the limit in n. Note also that hypotheses (4.3) and (4.4) are satisfied when the hypotheses of 1,p Theorem 2.1 and (4.2) hold true, since then ψ + belongs to W0 () ∩ L∞ () and satisfies ψ + > ψ a.e. in . 2 Under hypotheses (2.1), (2.2), (2.3), (4.3) and (4.4), there exists at least (see Theorem 6.1 below) one solution uε of the penalized problem 1,p uε ∈ W0 (), (uε − ψ)− ∈ L2 (), Z Z 1 a(x, uε , Duε )Dv dx − (uε − ψ)− v dx ε = W −1,p0 () hf, viW 1,p () , 0 ∀ v ∈ W 1,p () ∩ L2 ().
(4.5)
0
Moreover uε satisfies (see Proposition 6.1 below) (uε − ψ)− uε ∈ L1 (), Z Z 1 a(x, uε , Duε )Duε dx − (uε − ψ)− uε dx ε = −1,p0 hf, uε i 1,p . W () W ()
(4.6)
0
Using in (4.5) the test function v ? and substracting from (4.6) we obtain Z Z 1 ? a(x, uε , Duε )D(uε − v ) dx − (uε − ψ)− (uε − v ? ) dx ε = hf, uε − v ? i.
(4.7)
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From (2.2) and (2.3) we get with the help of Young’s and Poincaré’s inequalities2 Z a(x, uε , Duε )Duε dx Z p ¯ > α¯ k Duε kLp () − [γ¯ |uε |p−1 + |h(x)|] dx > 3α¯ k Du kp ε Lp () − C, 4 Z Z ? a(x, uε , Duε )Dv dx 6 β¯ [|k(x)| ¯ + |uε | + |Duε |]p−1 |Dv ? | dx
6 α4¯ k Duε kpL () + C. p
For what concerns the second term of (4.7), we have v ? > ψ a.e. in , and thus Z Z 1 1 (uε − ψ)− (uε − v ? ) dx > |(uε − ψ)− |2 dx. (4.8) − ε ε Finally the right-hand side of (4.7) is estimated by |hf, uε − v ? i|
6 kf kW
−1,p0 ()
kuε kW 1,p () − hf, v ? i 0
6 α4 kDuε kpL () + C. p
From the above computation we deduce that k uε kW 1,p () 6 C,
(4.9)
0
k (uε − ψ)− k2L2 () 6 Cε.
(4.10)
4.1.2. Proof of the existence result Define 1 µε = (uε − ψ)− . ε
(4.11)
From Equation (4.5) we deduce that −div(a(x, uε , Duε )) − µε = f
in D 0 (),
(4.12)
2 Here and in what follows, C denotes nonnegative constants which do not depend on ε, but can
vary from line to line.
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A. MOKRANE AND F. MURAT 1,p
which in view of the growth condition (2.3) on a and of the W0 () estimate (4.9) 0 on uε implies that µε is bounded in W −1,p (). We can thus extract a subsequence (still denoted by ε) such that (
1,p
uε * u weakly in W0 (), 0
µε * µ weakly in W −1,p (). We now prove in the classical way that u is a solution of the variational inequality (2.11) and that (2.12) holds. We first observe that passing to the limit in (4.10) we obtain (u − ψ)− = 0, i.e. u ∈ K(ψ). Let now v be any element of K(ψ). The function Tk (v), where Tk is the trunca1,p tion at height k, belongs to W0 () ∩ L2 (). Using Tk (v) as test function in (4.5) and subtracting (4.6) we obtain Z Z 1 a(x, uε , Duε )(DTk (v) − Duε ) dx − (uε − ψ)− (Tk (v) − uε ) dx ε = hf, Tk (v) − uε i. When k satisfies k > kv ? kL∞ () (this is the only reason why we required v ? to belong to L∞ () in (4.3)), we have k > ψ(x) and therefore Tk (v) > ψ a.e. in , and thus Z 1 (uε − ψ)− (Tk (v) − uε ) dx 6 0. − ε Therefore hA(uε ), Tk (v) − uε i > hf, Tk (v) − uε i,
∀v ∈ K(ψ).
1,p
Since Tk (v) tends strongly to v in W0 () when k tends to infinity, this implies that hA(uε ), v − uε i > hf, v − uε i,
∀v ∈ K(ψ).
(4.13)
0
Since A(uε ) is bounded in W −1,p () we have, extracting if necessary a new subsequence, A(uε ) * ϕ
0
weakly in W −1,p ().
Taking v = u ∈ K(ψ) as test function in (4.13) we obtain lim suphA(uε ), uε i 6 hϕ, ui. ε→0
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Since the Leray–Lions operator A is pseudomonotone, this implies that hA(uε ), uε i → hA(u), ui,
ϕ = A(u) and
and passing to the limit in (4.13) we obtain that u solves the variational inequality (2.11). On the other hand, passing to the limit in (4.12) we obtain (2.12) with the help of ϕ = A(u). 4.1.3. Strong convergence of zε− in L1 (): first part of the proof This subsection and the two next ones are the main contributions of the present paper. From now on, we assume that hypotheses (4.1) and (4.2) hold true. Let us define 1 zε = g n − (uε − ψ)− . ε
(4.14) 1,p
The function zε belongs to W0 () because, in view of assumption (4.1) and of 1,p (2.7), g n and (uε − ψ)− belong to W0 (). We fix k > 0 (say e.g. k = 1) and we consider the function −Tk (zε− ), where Tk 1,p is the truncation at height k; this function belongs to W0 () ∩ L∞ (). Defining Eε = {x ∈ : − k < zε (x) < 0}, we have D(−Tk (zε− )) =
(
Dzε
in Eε ,
0
in \Eε .
On the other hand since by (4.1) g n that
(4.15)
> 0 on , the definition (4.14) of zε implies
uε − ψ < 0 on Eε .
(4.16)
In other terms Eε ⊂ {x ∈ : uε (x) − ψ(x) < 0}.
(4.17)
From the definition (4.14) of zε and from (4.17) we obtain 1 1 Dzε = Dg n − D(uε − ψ)− = Dg n + (Duε − Dψ) on Eε . ε ε
(4.18)
Similarly we deduce from (4.14) and (4.17) that 1 1 zε = g n − (uε − ψ)− = g n + (uε − ψ) on Eε , ε ε
(4.19)
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which implies that 1 −k < g n + (uε − ψ) < 0 on Eε . ε Using (4.16) and hypothesis (4.1) which implies that g n belongs here to L∞ (), this implies 0 > uε − ψ > −ε(g n + k) > −ε(kg n kL∞ () + k) on Eε . So we have |uε − ψ| 6 (kg n kL∞ () + k)ε
on Eε .
(4.20)
Since ψ belongs here to L∞ () by hypothesis (4.2), we deduce from (4.20) that setting m = kψkL∞ () + (kg n kL∞ () + k), one has when ε
61
|ψ| 6 m and
|uε | 6 m on Eε .
(4.21)
Using the definitions (2.9) and (4.14) of g and zε , as well as g = g p − g n (see (4.1)), the approximate equation (4.5) reads as Z Z Z a(x, uε , Duε )Dv dx − a(x, ψ, Dψ)Dv dx + zε v dx
(4.22)
= hg p , vi, 1,p
∀v ∈ W0 () ∩ L2 ().
Take v = −Tk (zε− ), which belongs to W0 () ∩ L∞(), as test function in (4.22). Using (4.15) we deduce that Z Z [a(x, uε , Duε ) − a(x, ψ, Dψ)]Dzε dx + zε− Tk (zε− ) dx Eε (4.23) = hg p , −Tk (zε− )i. 1,p
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Note that hg p , −Tk (zε− )i 6 0,
(4.24)
and write [a(x, uε , Duε ) − a(x, ψ, Dψ)] = [a(x, uε , Duε ) − a(x, uε , Dψ)] + [a(x, u , Dψ) − a(x, ψ, Dψ)].
(4.25)
ε
We deduce that from (4.23), (4.25), (4.18) and (4.24) that Z 1 [a(x, uε , Duε ) − a(x, uε , Dψ)][Duε − Dψ] dx ε Eε Z + zε− Tk (zε−) dx Z 6 − [a(x, uε , Duε ) − a(x, uε , Dψ)]Dgn dx Eε Z [a(x, uε , Dψ) − a(x, ψ, Dψ)]Dg n dx − E ε Z 1 − [a(x, uε , Dψ) − a(x, ψ, Dψ)][Duε − Dψ] dx. ε Eε 4.1.4. Strong convergence of zε− in L1 (): second part of the proof when p
(4.26)
>2
We will now estimate the various terms which enter in (4.26). In this subsection, we will assume that p > 2; the case where 1 < p < 2 will be treated in Subsection 4.1.5. Subsections 4.1.4 and 4.1.5 are the only parts of the proof of Theorem 2.1 where hypotheses (2.4), (2.5) and (2.6) are used. When p > 2 we deduce from the strong monotonicity (2.4) of a and from the L∞ () estimate (4.21) on uε that Z 1 [a(x, uε , Duε ) − a(x, uε , Dψ)][Duε − Dψ] dx ε Eε Z αm |Duε − Dψ|p dx. > ε Eε
(4.27)
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Similarly, the L∞ () estimate (4.21), the local Lipschitz continuity (2.5) of a with respect to ξ , and Young’s inequality together with the fact that (p−2)p 0 +p 0 = 1,p p and the W0 () estimate (4.9) on uε , yield3 Z n − [a(x, uε , Duε ) − a(x, uε , Dψ)]Dg dx Eε Z 6 βm [|km (x)| + |Duε | + |Dψ|]p−2|Duε − Dψ||Dgn| dx Eε Z αm 6 |Duε − Dψ|p dx 2ε E ε Z 0 0 p 0 /p [|km (x)| + |Duε | + |Dψ|](p−2)p |Dg n |p dx + Cm ε Eε Z αm 0 |Duε − Dψ|p dx + Cm ε p /p . 6 2ε Eε
(4.28)
Using now the local Hölder continuity (2.6) of a with respect to s, and the L∞ () estimates (4.21) and (4.20), we have for ε sufficiently small Z n − [a(x, uε , Dψ) − a(x, ψ, Dψ)]Dg dx Eε Z 0 6 γ |u − ψ|(1+δm /p ) [|l (x)| + |Dψ|]p−1 |Dg n | dx m
ε
m
Eε
6 [(kg kL n
Z
∞ ()
(4.29)
(1+δm /p 0 )
+ k)ε]
γm [|lm (x)| + |Dψ|]p−1 |Dg n | dx Eε
6 Cm ε(1+δ /p ) . m
0
Finally using Young’s inequality, and again (2.6), (4.21) and (4.20), we have for ε sufficiently small
3 Here and in what follows, C denotes nonnegative constants which do no depend on ε, but can m
depend on m and vary from line to line.
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Z 1 [a(x, u , Dψ) − a(x, ψ, Dψ)][Du − Dψ] dx − ε ε ε E ε Z αm 6 |Duε − Dψ|p dx 2ε Eε Z C 0 m + |a(x, uε , Dψ) − a(x, ψ, Dψ)|p dx ε Eε Z α 6 m |Duε − Dψ|p dx 2ε Eε Z Cm 0 0 + γmp |uε − ψ|1+δm [lm (x) + |Dψ|](p−1)p dx ε Eε Z αm |Duε − Dψ|p dx 6 2ε Eε Z Cm 0 n 1+δ m + γmp [lm (x) + |Dψ|]p dx [(kg kL∞ () + k)ε] ε Eε Z αm |Duε − Dψ|p dx + Cm ε δm . 6 2ε Eε
125
(4.30)
Summing up (4.26), (4.27), (4.28), (4.29) and (4.30), we have proved that for k fixed Z
zε− Tk (zε− ) dx → 0 if ε → 0.
This implies that zε− → 0 strongly in L1 (),
(4.31)
which is the desired result. 4.1.5. Strong convergence of zε− in L1 (): second part of the proof when 1
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Define the function Fε and the set Zε by
Fε = |hm (x)| + |Duε | + |Dψ|,
Zε = {x ∈ : Fε (x) 6 = 0}.
Note that Duε (x) = Dψ(x) = 0 in \Zε ; therefore we have in particular
Duε (x) − Dψ(x) = 0 in \Zε .
(4.32)
We deduce from (4.32), from the strong monotonicity (2.4) of a with respect to ξ and from the L∞ () estimate (4.21) on uε that Z 1 [a(x, uε , Duε ) − a(x, uε , Dψ)][Duε − Dψ] dx ε Eε Z 1 = [a(x, uε , Duε ) − a(x, uε , Dψ)][Duε − Dψ] dx ε Eε ∩Zε Z αm |Duε − Dψ|2 dx > ε Eε ∩Zε (|hm (x)| + |Duε | + |Dψ|)2−p Z αm |Duε − Dψ|2 = dx. ε Eε ∩Zε |Fε |2−p
(4.33)
Since the left-hand side of (4.33) is finite because of the growth condition (2.3) on a and because of the fact that uε and ψ belong to W 1,p (), the integral Z Eε ∩Zε
|Duε − Dψ|2 dx |Fε |2−p
is finite. Similarly the local Hölder continuity (2.5) of a with respect to ξ , the L∞ () estimate (4.21) on uε , (4.32), Young’s inequality for q defined by q = 2/(p − 1) (note that q > 1 when 1 < p < 2) and the facts that Fε is bounded in Lp () and that (2 − p)q 0 /q + q 0 = p, yield
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Z n [a(x, u , Du ) − a(x, u , Dψ)]Dg dx − ε ε ε Eε Z 6 β |Duε − Dψ|p−1 |Dg n | dx m Eε Z |Duε − Dψ|p−1 |Dg n | dx = βm Eε ∩Zε Z |Duε − Dψ|p−1 1/q ε |Fε |(2−p)/q |Dg n | dx = βm (2−p)/q 1/q |Fε | Eε ∩Zε ε Z αm |Duε − Dψ|(p−1)q 6 dx 2ε Eε ∩Zε |Fε |2−p Z 0 0 q 0 /q |Fε |(2−p)q /q |Dg n |q dx + Cm ε Eε ∩Zε Z αm |Duε − Dψ|2 0 dx + Cm ε q /q . 6 2−p 2ε Eε ∩Zε |Fε |
127
(4.34)
Using now the local Hölder continuity (2.6) of a with respect to s, the L∞ () estimates (4.20) and (4.21), we have for ε sufficiently small
Z n − [a(x, uε , Dψ) − a(x, ψ, Dψ)]Dg dx Eε Z 6 γm |uε − ψ|(1+δm )/2 [|lm (x)| + |Dψ|]p−1 |Dg n | dx
Eε
6 [(kg kL n
Z
∞ ()
+ k)ε]
γm [|lm (x)| + |Dψ|]p−1 |Dg n | dx Eε
(4.35)
(1+δm )/2
6 Cmε(1+δ )/2. m
For what concerns the last term of (4.26), we have, using (4.32) and Young’s inequality
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Z 1 − [a(x, u , Dψ) − a(x, ψ, Dψ)][Du − Dψ] dx ε ε ε Eε Z 1 [a(x, u , Dψ) − a(x, ψ, Dψ)][Du − Dψ] dx = − ε ε ε Eε ∩Zε Z 1 6 |a(x, uε , Dψ) − a(x, ψ, Dψ)| ε Eε ∩Zε
|Duε − Dψ| |Fε |(2−p)/2 dx |Fε |(2−p)/2 Z αm |Duε − Dψ|2 6 2ε dx |Fε |2−p Eε ∩Zε Z 1 + |a(x, uε , Dψ) − a(x, ψ, Dψ)|2 |Fε |2−p dx. 2αm ε Eε ∩Zε
(4.36)
In the last term of the right-hand side of (4.36) we use again (2.6), (4.20), (4.21) and the facts that Fε is bounded in Lp () and that 2(p − 1) + 2 − p = p; we obtain for ε sufficiently small Z 1 |a(x, uε , Dψ) − a(x, ψ, Dψ)|2 |Fε |2−p dx ε Eε ∩Zε Z 1 6 γm2 |uε − ψ|1+δm [|lm (x)| + |Dψ|]2(p−1) |Fε |2−p dx ε E ∩Z ε ε (4.37) 6 1ε [(kgn kL∞() + k)ε]1+δm Z γm2 [|lm (x)| + |Dψ|]2(p−1)|Fε |2−p dx Eε ∩Zε 6 C ε δm . m Summing up (4.26), (4.33), (4.34), (4.35), (4.36) and (4.37), we have proved that for k fixed Z zε− Tk (zε− ) dx → 0 if ε → 0.
This implies that zε− tends to zero strongly in L1 (), i.e. (4.31), which is the desired result. 4.1.6. Proof of the Lewy–Stampacchia’s inequality From now on we return to the general case 1 < p < ∞.
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Coming back to the definition (4.14) of zε , we have 1 zε = zε+ − zε− = g n − (uε − ψ)− , ε which together with the definition (4.11) of µε yields g n + zε− = zε+ + µε
> µε .
Passing to the limit in this inequality thanks to the strong convergence (4.31) of zε− to zero, we deduce that gn
> µ.
(4.38)
This is the Lewy–Stampacchia’s inequality in the case where g n = g − . This completes the proof of Theorem 2.1 in the case where g n = g − and where g − and ψ satisfy the regularity assumptions (4.1) and (4.2). 4.2. THE GENERAL CASE; APPROXIMATION OF g AND ψ AND PASSING TO THE LIMIT
Let now g − be in W −1,p () and let ψ be in W 1,p () with ψ 6 0 on ∂ as assumed in hypotheses (2.10) and (2.7). By Lemma 5.1, that we will state and prove in Section 5 below, there exists gˆn such that ( 1,p gˆn ∈ W0 (), gˆn > 0 for each n, (4.39) 0 gˆn → g − strongly in W −1,p (). 0
Using the truncation Tk at height k, which has the property that for any h ∈ 1,p W0 () 1,p
Tk (h) → h strongly in W0 () if k → ∞,
(4.40)
we can assume by a further approximation that each of those functions gˆn also belongs to L∞ (). Similarly we approximate ψ by ψn = Tn (ψ), which has in particular the following properties ( ψn ∈ W 1,p () ∩ L∞ (), ψn 6 0 on ∂, (4.41) ψn → ψ strongly in W 1,p (). Note that in (4.39) as well as in the whole of Subsection 4.2, n is used as a subscript to denote the index of a sequence; this has to be contrasted with Subsection 4.1, where the superscript n was used to denote the ‘negative’ part of g.
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Define now fn = g + − gˆn + A(ψn ).
(4.42)
Then g = g + − gˆn and ψ = ψn satisfy hypotheses (4.1) and (4.2); for n fixed, we have proved in Subsection 4.1 that there exists at least one solution un of the variational inequality (
hA(un ), v − un i > hfn , v − un i,
∀v ∈ K(ψn ),
un ∈ K(ψn ),
(4.43)
such that the distribution µn defined by µn = A(un ) − fn ,
(4.44)
satisfies (4.38), i.e. the inequality µn
6 gˆn .
(4.45)
Using in (4.43) the test function v = ψ + , which belongs to K(ψn ), we deduce 1,p as in the proof of estimate (4.9) that un is bounded in W0 (). It follows from the 0 growth condition (2.3) on a that A(un ) is bounded in W −1,p (), and from (4.44), 0 (4.42), (4.39) and (4.41) that µn is bounded in W −1,p (). We can thus extract a subsequence (still denoted by n) such that 1,p
un * u weakly in W0 (), A(un ) * ϕ
0
weakly in W −1,p (), 0
µn * µ weakly in W −1,p (). We now prove in the classical way that u is a solution of the variational inequality (2.11). From un ∈ K(ψn ) and from (4.41) we first deduce that u ∈ K(ψ). For any v ∈ K(ψ), the function Tn (v) belongs to K(ψn ) and strongly converges to v in 1,p W0 () when n tends to infinity. Since fn strongly converges to g + −g − +A(ψ) = 0 f in W −1,p (), using Tn (u) as test function in (4.43) implies that lim suphA(un ), un i 6 hϕ, ui. n
Since the operator A is pseudomonotone, this implies that ϕ = A(u) and
hA(un ), un i → hA(u), ui.
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Using Tn (v) as test function in (4.43) therefore proves that u is a solution of (
hA(u), v − ui > hf, v − ui,
∀v ∈ K(ψ),
u ∈ K(ψ), i.e. (2.11). Finally, passing to the limit in (4.44) and (4.45) proves (2.12) and (2.13) and 2 completes the proof of Theorem 2.1. REMARK 4.2. The proof that we used at the end of Subsection 4.2 to prove that u is a solution of (2.11) did not use the Lewy–Stampacchia’s inequality. If we choose to use this inequality, an alternative way of proof is the following one: since gˆn 0 tends to g − strongly in W −1,p (), and since we have 0 6 µn 6 gˆ n by (4.45), the 0 compactness of the intervals of W −1,p () (see Lemma 2.8 of [10]) implies that µn 0 is relatively compact in the strong topology of W −1,p (). Therefore, extracting a 0 subsequence, A(un ) = µn + fn converges strongly in W −1,p () and we can pass 2 to the limit in (4.43) and (4.44) like in the case of an equation.
0
5. A Density Lemma for the Nonnegative Cone of W−1,p () The goal of this section is to prove a density lemma. This lemma was proved for example in Donati and Matzeu [13] and Donati [12] by using the Lewy–Stampacchia’s inequality. We prove it here by using the natural penalization. 1,p
LEMMA 5.1. The nonnegative cone of W0 () is dense in the nonnegative cone 0 of W −1,p (). In other terms, let f be such that 0
f ∈ W −1,p (),
f
> 0.
(5.1)
We will construct fˆε such that (
1,p fˆε ∈ W0 (), fˆε
fˆε → f
>0
0
strongly in W −1,p (). 0
Proof. For f ∈ W −1,p (), f (
−div(|Dv|p−2 Dv) = f 1,p
for each ε > 0,
v ∈ W0 (),
(5.2)
> 0, we define v, vε and fˆε by in D 0 (), (5.3)
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Z Z 1 p−2 |Dvε | Dvε Dw dx + (vε − v)w dx = 0, ε
(5.4)
1,p
∀w ∈ W0 () ∩ L2 (), 1,p
vε ∈ W0 (),
vε − v ∈ L2 (),
1 fˆε = − (vε − v) = −div(|Dvε |p−2 Dvε ). (5.5) ε REMARK 5.1. The existence of vε is ensured by Theorem 6.2 below. In the case where p > 2, or more exactly where p
, > N2N +2 0
this result follows from the facts that W0 () ⊂ L2 () ⊂ W −1,p () and that 1,p
1 w → −div(|Dw|p−2Dw) + w ε 1,p
is a monotone, coercive, continuous and bounded operator from W0 () into 0 W −1,p (). When p > 1 is close to 1, it is unclear whether the problem 1 −div(|Dvε |p−2 Dvε ) + (vε − v) = 0 in D 0 (), ε (5.6) 1,p vε ∈ W0 (), has a solution or not. Indeed the operator w → −div(|Dw|p−2 Dw) + (1/ε)w 0 1,p is not defined from W0 () into W −1,p () (because (1/ε)v does not in gen0 0 1,p eral belong to W −1,p ()) since W0 () is not embedded in W −1,p () when 0 1,p p < 2N/(N + 2): indeed the embedding W0 () ⊂ W −1,p () would imply 1,p kϕkW −1,p0 () 6 CkϕkW 1,p () , thus hϕ, ϕi 6 Ckϕk2 1,p and therefore W0 () ⊂ W0 ()
0
L2 (), a fact which is false when p < 2N/(N + 2). On the other hand, the operator w → − div (|Dw|p−2 Dw) + (1/ε)w is well defined, continuous, bounded, 1,p coercive and monotone on W0 () ∩ L2 (), but there is no reason for (1/ε)v to 0 belongs to W −1,p () + L2 () when 1 < p < 2N/(N + 2). This is the reason 2 why we replaced (5.6) by (5.4) and used Theorem 6.2. Using (vε − v)+ , which belongs to W0 () ∩ L2 (), as test function in the difference of (5.4) and (5.3), we obtain, since f > 0 Z Z [|Dvε |p−2 Dvε − |Dv|p−2 Dv]D(vε − v)+ dx + 1 |(vε − v)+ |2 dx ε (5.7) = −hf, (v − v)+ i 6 0, 1,p
ε
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which implies in particular that vε − v
60
a.e. in .
We thus have constructed fˆε such that 1 1,p fˆε = − (vε − v) ∈ W0 (), ε
fˆε
> 0.
0 It remains to prove that fˆε strongly converges to f in W −1,p (). Using in (5.4) the test function (vε − v) we obtain Z Z 1 p−2 |Dv | Dv Dv dx + |vε − v|2 dx ε ε ε ε (5.8) Z p−2 = |Dvε | Dvε Dv dx,
which implies that 1 p kDv ε kLp () + kv ε − vk2L2 () ε
6 C.
(5.9)
Therefore vε * v
1,p
weakly in W0 ().
(5.10)
Rewriting now (5.8) as Z Z 1 p−2 p−2 2 [|Dvε | Dvε − |Dv| Dv][Dvε − Dv] dx + ε |vε − v| dx Z = − |Dv|p−2 Dv[Dvε − Dv] dx,
we deduce from (5.10) that Z [|Dvε |p−2 Dvε − |Dv|p−2 Dv][Dvε − Dv] dx → 0.
When p
(5.11)
> 2, we have for some αp > 0
Z
|Dvε − Dv|p dx
αp
6
Z
[|Dvε |p−2 Dvε − |Dv|p−2 Dv][Dv ε − Dv] dx, 1,p
which implies, with the help of (5.11), that v ε strongly converges to v in W0 (). When 1 < p < 2, we have for some αp > 0 R |Dvε −Dv|2 αp (|Dvε |+|Dv|)2−p dx Z (5.12) 6 [|Dvε |p−2 Dvε − |Dv|p−2 Dv][Dv ε − Dv] dx.
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Using then Hölder’s inequality with q = 2/p > 1 we obtain, since (2−p)pq 0 /2 = p Z |Dvε − Dv|p dx Z |Dv ε − Dv|p = (|Dv ε | + |Dv|)(2−p)p/2 dx ε | + |Dv|)(2−p)p/2 (|Dv p/2 Z 1/q 0 Z |Dvε − Dv|2 p 6 dx (|Dv | + |Dv|) dx , ε ε 2−p (|Dv | + |Dv|) which implies, with the help of (5.12), (5.11) and (5.9), that v ε strongly converges 1,p to v in W0 (). For any 1 < p < ∞, we have thus proved that vε → v
1,p
strongly in W0 ().
(5.13)
Therefore fˆε = −div(|Dvε |p−2 Dvε ) → −div(|Dv|p−2 Dv) = f 0
strongly in W −1,p () as required. This completes the proof of Lemma 5.1.
2
6. Appendix: Proof of Two Existence Results The goal of this Appendix is to prove two existence results (Theorems 6.1 and 6.2). Although these results could be considered as well known, their statements and proofs deserve some attention (see Remark 5.1 above) and so we preferred to present complete proofs here. When 1 < p < 2 and p is close to 1, it is not clear whether the penalized problem 1,p u ∈ W0 (), ε Z Z 1 a(x, uε , Duε )Dv dx − (uε − ψ)− v dx (6.1) ε 1,p = hf, vi, ∀v ∈ W0 (), has a solution or not, even if ψ belongs to W 1,p (). Indeed there is no reason 1,p for (uε − ψ)− v to belong to L1 () when v only belongs to W0 (), since the 1,p embedding W0 () ⊂ L2 () only holds when p
, > N2N +2
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i.e. when p is not too close to 1. For this reason we replace (6.1) by (6.3) in Theorem 6.1 below. THEOREM 6.1. Let 1 < p < ∞, be a bounded open set of R N , f ∈ W −1,p (), and a be a Carathéodory function which satisfies (2.1), (2.2) and (2.3). Finally let ψ: → R be a measurable function such that 0
1,p
∃v ? ∈ W0 () with v ?
>ψ
a.e. in .
(6.2)
Then for each ε > 0 there exists at least one uε such that 1,p uε ∈ W0 (), (uε − ψ)− ∈ L2 (), Z Z 1 a(x, uε , Duε )Dv dx − (uε − ψ)− v dx = hf, vi, ε 1,p 2 ∀ v ∈ W0 () ∩ L (). PROPOSITION 6.1. Every solution of (6.3) satisfies − 1 (uε − ψ) uε ∈ L (), Z Z 1 a(x, uε , Duε )Duε dx − (uε − ψ)− uε dx = hf, uε i. ε
(6.3)
(6.4)
Proof. Let uε be any solution of (6.3). Since the function v = Tk (uε )+ , where 1,p Tk is the truncation at height k, belongs to W0 () ∩ L2 (), we have Z Z 1 a(x, uε , Duε )DTk (uε )+ dx − (uε − ψ)− Tk (uε )+ dx ε (6.5) + = hf, Tk (uε ) i. When k tends to infinity, Tk (uε )+ tends strongly to u+ ε in W0 (), and thus Z 1 (uε − ψ)− Tk (uε )+ dx is bounded independently of k. ε 1,p
1 The monotone convergence theorem then implies that (uε − ψ)− u+ ε ∈ L () and that Z Z 1 1 (uε − ψ)− Tk (uε )+ dx → (uε − ψ)− u+ when k → ∞. ε dx ε ε
Similarly, using Tk (uε )− as test function in (6.3), we prove that (uε − ψ)− u− ε ∈ 1 L () and that Z Z 1 1 − − (uε − ψ) Tk (uε ) dx → (uε − ψ)− u− when k → ∞. ε dx ε ε
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Passing to the limit in (6.5) and in its analogue for v = −Tk (uε )− completes the 2 proof of Proposition 6.1. Proof of Theorem 6.1 In this proof ε > 0 is fixed; we will thus remove the subscript ε when it is not essential. First step: approximation and a priori estimates Consider the problem −div(a(x, un , Dun )) − 1 Tn (un − ψ)− = f ε n 1,p u ∈ W0 (),
in D 0 (),
(6.6)
where Tn is the truncation at height n. In contrast with (6.1), this problem has a solution for any p > 1. Indeed even if |ψ| is infinite on a set of positive measure, 1,p Tn (w − ψ)− belongs to L∞ () for any w ∈ W0 (), and the operator 1 w → −div(a(x, w, Dw)) − Tn (w − ψ)− ε 0
is well defined from W0 () in W −1,p (), and is a pseudomonotone, continuous, 0 1,p bounded and coercive operator from W0 () in W −1,p (). The existence of a solution of (6.6) then follows from classical theorems (see e.g. [20]). Using in (6.6) the test function (un − v ? ), where v ? is given by (6.2), we have Z Z 1 n n n ? a(x, u , Du )D(u − v ) dx − Tn (un − ψ)− (un − v ? ) dx ε (6.7) n ? = hf, u − v i. 1,p
We now observe that Tn (un − ψ)− vanishes except on the set E n = {x ∈ : un (x) < ψ(x)}. Since ψ(x) 6 v ? (x) for a.e. x ∈ , |ψ| is finite a.e. on E n ; every term in the following computation is thus finite a.e. on and we have 1 n − n ? − Tn (u − ψ) (u − v ) ε 1 1 n − n n − ? = − ε Tn (u − ψ) (u − ψ) − ε Tn (u − ψ) (ψ − v ) (6.8) 1 n − n > − ε Tn (u − ψ) (u − ψ) > 1 |Tn (un − ψ)− |2 > 0 a.e. on . ε
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137
From (6.7), (6.8) and the coercivity and growth conditions (2.2), (2.3) we deduce that ||un ||W 1,p () 0
6 C, () 6 C,
||A(un )||W −1,p0 Z 1 |Tn (un − ψ)− |2 dx ε
6 C,
where the constant C does not depend on n. Extracting a subsequence (still denoted with the superscript n), there exist uε , ϕ and h such that, when n tends to infinity (recall that ε is fixed) un * uε
1,p
weakly in W0 () and a.e. in ,
A(un ) * ϕ
(6.9)
0
weakly in W −1,p (),
(6.10)
Tn (un − ψ)− * h weakly in L2 ().
(6.11)
From the almost everywhere convergence of un to uε we deduce that Tn (un − ψ)− → (uε − ψ)−
a.e. in ,
(6.12)
which together with (6.11) implies h = (uε − ψ)−
and
(uε − ψ)− ∈ L2 ().
Passing to the limit in (6.6) yields Z 1 (uε − ψ)− v dx = hf, vi, hϕ, vi − ε Second step: end of the proof We will now prove that Z 1 lim sup Tn (un − ψ)− (un − uε ) dx ε n→∞
(6.13)
1,p
∀v ∈ W0 () ∩ L2 ().
6 0.
(6.14)
(6.15)
Indeed, as said before, |ψ| is finite on the set E n = {x ∈ : un (x) < ψ(x)} and Tn (un − ψ)− vanishes except on E n ; every term in the following computation is thus finite a.e. on and we have Tn (un − ψ)− (un − uε ) = Tn (un − ψ)− [(un − ψ) − (uε − ψ)] = Tn (un − ψ)− [(un − ψ)+ − (un − ψ)− (6.16) − (uε − ψ)+ + (uε − ψ)− ] 6 −Tn(un − ψ)− (un − ψ)− + Tn(un − ψ)− (uε − ψ)−.
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In view of (6.11) and (6.13) we have Z Z n − − Tn (u − ψ) (uε − ψ) dx → |(uε − ψ)− |2 dx,
while using (6.9) and (6.12) we have, by Fatou’s Lemma Z Z n − n − |(uε − ψ)− |2 dx. lim inf Tn (u − ψ) (u − ψ) dx > n→∞
(6.17)
(6.18)
Inequality (6.15) therefore follows from (6.16), (6.17) and (6.18). Using (un − uε ) as test function in (6.6) we have Z 1 Tn (un − ψ)− (un − uε ) dx = hf, un − uε i, hA(un ), un − uε i − ε
(6.19)
which using (6.9), (6.10) and (6.15) yields lim suphA(un ), un i 6 hϕ, uε i.
(6.20)
n→∞
Since the operator A is pseudomonotone, this implies that ϕ = A(uε ). From (6.14) we then deduce that the equation in (6.3) holds. This completes the proof 2 of Theorem 6.1. A result similar to Theorem 6.1 is the following one, which we used in Section 5 (see Remark 5.1). THEOREM 6.2. Let 1 < p < ∞, be a bounded open set of R N , f ∈ W −1,p (), 1,p v ∈ W0 (), and a be a Carathéodory function which satisfies (2.1), (2.2) and (2.3). Then for each ε > 0 there exists at least one vε such that 1,p vε ∈ W0 (), vε − v ∈ L2 (), Z Z 1 a(x, vε , Dvε )Dw dx + (vε − v)w dx = hf, wi, (6.21) ε 1,p ∀w ∈ W0 () ∩ L2 (). 0
Proof. The proof of Theorem 6.2 is very similar to the proof of Theorem 6.1; we thus only sketch it. Here again, ε > 0 is fixed; we will thus remove the subscript ε when it is not essential. First step: approximation and a priori estimate Consider the problem −div(a(x, v n , Dv n )) + 1 Tn (v n − v) = f ε n 1,p v ∈ W0 (),
in D 0 (),
(6.22)
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139
which has a solution since the operator w → −div(a(x, w, Dw))−(1/ε)Tn (w−v) 1,p is a pseudomonotone, continuous, bounded and coercive operator from W0 () 0 into W −1,p (). Using the test function (v n − v) in (6.22), we deduce with the help of the coercivity and growth conditions (2.2) and (2.3) that n kv kW 1,p () 6 C, 0 n kA(v )kW −1,p0 () 6 C, (6.23) Z Z 1 1 |Tn (v n − v)|2 dx 6 Tn (v n − v)(v n − v) dx 6 C, ε ε where the constant C does not depend on n. Extracting a subsequence (still denoted with the superscript n), there exist vε , ϕ and h such that, when n tends to infinity (recall that ε is fixed) 1,p
v n * vε weakly in W0 () and a.e. in ,
(6.24)
0
weakly in W −1,p (),
(6.25)
Tn (v n − v) * h weakly in L2 ().
(6.26)
A(v n ) * ϕ
From the almost everywhere convergence of v n to vε we deduce that Tn (v n − v) → (vε − v) a.e. in ,
(6.27)
which together with (6.26) implies h = vε − v
and
(vε − v) ∈ L2 ().
Passing to the limit in (6.22) yields Z 1 hϕ, wi + (vε − v)w dx = hf, wi, ε Second step: end of the proof We will now prove that Z 1 Tn (v n − v)(v n − vε ) dx lim sup − ε n→∞
(6.28)
1,p
∀w ∈ W0 () ∩ L2 ().
6 0.
(6.29)
(6.30)
Indeed Z Tn (v n − v)(v n − vε ) dx
Z =
Z Tn (v − v)(v − v) dx + n
Tn (v n − v)(v − vε ) dx.
n
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In view of (6.26) and (6.28) we have Z Z Tn (v n − v)(v − vε ) dx → (vε − v)(v − vε ) dx,
while using (6.24) and (6.27), we have, by Fatou’s Lemma Z Z n n |vε − v|2 dx. lim inf Tn (v − v)(v − v) dx > n→∞
(6.31)
(6.32)
Inequality (6.30) therefore follows from (6.31) and (6.32). Using (v n − vε ) as test function in (6.22) we have Z 1 n n hA(v ), v − vε i + Tn (v n − v)(v n − vε ) dx = hf, v n − v ε i, ε
(6.33)
which using (6.24), (6.25) and (6.30) yields lim suphA(v n ), v n i 6 hϕ, vε i.
(6.34)
n→∞
Since the operator A is pseudomonotone, this implies that ϕ = A(vε ). From (6.29) we deduce that the equation in (6.21) holds. This completes the proof of 2 Theorem 6.2. Note Added in Proof In a forthcoming paper ‘Proving the Lewy–Stampacchia’s inequality by penalization’, to appear in Atti Sem. Mat. Fis. Univ. Modena 46, Suppl. dedicato a C. Vinti, 1998, 315–334, we improve the results of the present paper by proving that every solution of the variational inequality (2.11) satisfies the Lewy–Stampacchia’s inequality. Acknowledgements This work was done during the visits of the first author to the Laboratoire d’Analyse Numérique of Paris 6 University within the framework of the French-Algerian cooperation grant ‘Accord programme 92 MDU 219’, the support of which is gratefully acknowledged. References 1. 2.
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