Regularity for Obstacle Problems without Structure Conditions
aa r X i v : . [ m a t h . A P ] F e b REGULARITY FOR OBSTACLE PROBLEMSWITHOUT STRUCTURE CONDITIONS
GIACOMO BERTAZZONI, SAMUELE RICC `O
Abstract.
This paper deals with the Lipschitz regularity of minimizers for a class ofvariational obstacle problems with possible occurance of the Lavrentiev phenomenon. Inorder to overcome this problem, the availment of the notions of relaxed functional andLavrentiev gap are needed. The main tool used here is a ingenious Lemma which revealsto be crucial because it allows us to move from the variational obstacle problem to therelaxed-functional-related one. This is fundamental in order to find the solutions’ regularitythat we intended to study. We assume the same Sobolev regularity both for the gradient ofthe obstacle and for the coefficients.
February 26, 20211.
Introduction
In this manuscript we study the Lipschitz continuity of the solutions to variational obstacleproblems of the form min (cid:26)Z Ω f ( x, Dw ) : w ∈ K ψ (Ω) (cid:27) (1.1)in the case of p, q − growth condition, where Ω, f and K ψ (Ω) will be specified below andwhere we assume that Lavrentiev phenomenon may occur.The relationship between the ellipticity and the growth exponent we impose, namely (2.2),is the one considered for the first time in the series of papers [17], [18], [19], [20] and it issharp (in view of the well known counterexamples, see for instance [35]) also to obtain theLipschitz continuity of solutions to elliptic equations and systems and minimizers of relatedfunctionals with p, q − growth. Regularity results under non standard growth conditions, aresearch branch started after the pioneering papers by Marcellini [36], [37], [38], has recentlyattracted growing attention, see among the others [2], [6], [12], [15], [32], [39], [40], [41].The term Lavrentiev phenomenon refers to a surprising result first demonstrated in 1926by M. Lavrentiev in [33]. There it was shown that it is possible for the variational integralof a two-point Lagrange problem, which is sequentially weakly lower semicontinuous on theadmissible class of absolutely continuous functions, to possess an infimum on the dense sub-class of C admissible functions that is strictly greater than its minimum value on the fulladmissible class. Since the paper of Lavrentiev indeed, many contributions appeared in thisdirection, see [7], [8], [22], [23], [43].In our case, it is still possible to have occurrence of Lavrentiev phenomenon due to thenonstandard growth conditions required on the lagrangian. In this respect, under our as-sumptions, the Lavrentiev phenomenon can be reformulated in these terms:inf w ∈ ( W ,p ∩ { w ≥ ψ } ) Z Ω f ( x, Dw ) dx < inf w ∈ ( W ,q ∩ { w ≥ ψ } ) Z Ω f ( x, Dw ) dx (1.2) So, the aim of this work is to complement the results cointained in the paper [11], whereauthors obtain Lipschitz regularity results for obstacle problems with Sobolev regularity forthe coefficients and where the lagrangian f satisfies p, q − growth conditions without assumingthat Lavrentiev phenomenon may occur.This phenomenon is a clear obstruction to regularity, since (1.2) prevents minimizers tobelong to W ,q . Notice that (1.2) cannot happen if p = q or if f is autonomous (it not dependson variable x ) and convex. Moreover, as pointed out in Section 3 of [22], the appearanceof (1.2) has geometrical reasons and cannot be spotted in a direct way by standard ellipticregularity techniques. Therefore, the basic strategy in getting regularity results consistsin excluding the occurrence of (1.2) by imposing that the Lavrentiev gap functional L ( u ),defined in (2.10), vanishes on solutions.However, here in this manuscript we adopt a different viewpoint, following the lines of [20].We present a general Lipschitz regularity result by covering the case in which the Lavrentievphenomenon may occur. In this respect, a key role will be played by the relaxed functional.We therefore need to introduce the exact framework of relaxation in the case of obstacleproblems and then we will state our main result. The crucial step will be constituted byLemma 5.1 which is the natural counterpart of the necessary and sufficient condition to getthe absence of Lavrentiev phenomenon.We will state and prove the result with Sobolev dependence on both the obstacle and thepartial map x D ξ f ( x, ξ ). A model functional that is covered by our results is w Z Ω h | Dw | p + a ( x ) (1 + | Dw | ) q i dx with q > p > a ( · ) a bounded Sobolev coefficient.The plan of the paper is the following: in Section 2 we state our model problem and themain results of the paper, in Section 3 we present some preliminary results we need in thesequel; Section 4 is devoted to the presentation of our a-priori estimate and finally in Section5 we present the Lipschitz regularity results for solutions to the relaxed obstacle problem.2. Statement of the problem and of the main results
Let us present in details the setting of our problem.Here Ω is a bounded open set of R n , n ≥
2. We will deal with variational integral F ( u ) := Z Ω f ( x, Du ) dx (2.1)where f : Ω × R n → [0 , + ∞ ) is a Carath´eodory function which is convex and of class C with respect to the second variable. We consider p and q bounded by1 ≤ qp < r − nr n = 1 + 1 n − r (2.2)where r > n , so r < n and where we consider q > p ≥
2. We suppose that there exist twopositive constants ν , L and a function h : Ω → [0 , + ∞ ) such as h ( x ) ∈ L r loc (Ω) such that ν (1 + | ξ | ) p ≤ f ( x, ξ ) ≤ L (1 + | ξ | ) q (2.3) EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 3 ν (1 + | ξ | ) p − | λ | ≤ X i,j f ξ i ξ j ( x, ξ ) λ i λ j ≤ L (1 + | ξ | ) q − | λ | (2.4) | f xξ ( x, ξ ) | ≤ h ( x ) (1 + | ξ | ) q − (2.5)for all λ, ξ ∈ R n , λ = λ i , ξ = ξ i , i = 1 , , . . . , n , a.e. in Ω.Furthermore, we give meaning of our obstacle problem (1.1). Namely, the function ψ :Ω → [ −∞ , + ∞ ), called obstacle , belongs to the Sobolev space W ,p (Ω) and the class K ψ (Ω)is defined as follows K ψ (Ω) := (cid:8) w ∈ u + W ,p (Ω) : w ≥ ψ a.e. in Ω (cid:9) (2.6)where u is a fixed boundary value. In order to prove Theorem 2.3 we will need to assume u ∈ W ,q (Ω).To avoid trivialities, in what follows we shall assume that K ψ is not empty. We also assumethat a solution to (1.1) is such that f ( x, Du ) ∈ L (Ω). As it has been shown in [21], incase of non-standard growth condition (at least in the autonomous case), this turns to bethe right class of competitors. Remark 2.1.
Let us notice that, by replacing u by ˜ u = max { u , ψ } , we may assume thatthe boundary value function u satisfies u ≥ ψ in Ω . Indeed ˜ u = ( ψ − u ) + + u and since ≤ ( ψ − u ) + ≤ ( u − u ) + ∈ W , (Ω) , the function ( ψ − u ) + , and hence u − ˜ u , belongs to W , (Ω) . Moreover assumptions f ( x, Du ) ∈ L (Ω) and f ( x, Du ) ∈ L (Ω) imply f ( x, D ˜ u ) ∈ L (Ω) . Indeed we have Z Ω f ( x, D ˜ u ) dx = Z Ω ∩ { u ≥ ψ } f ( x, Du ) dx + Z Ω ∩ { u <ψ } f ( x, Dψ ) dx ≤ Z Ω [ f ( x, Du ) + f ( x, Dψ )] dx < + ∞ where we used that f ( x, ξ ) ≥ , by virtue of the left inequality in (2.3) . Throughout the paper we will denote by B ρ and B R balls of radii respectively ρ and R (with ρ < R ) compactly contained in Ω and with the same center, let us say x ∈ Ω. Moreover inthe sequel, constants will be denoted by C , regardless their actual value. Only the relevantdependencies will be highlighted.Our first main result is an a-priori estimate and reads as follows Theorem 2.2 (A priori estimate) . Let u ∈ K ψ (Ω) be a smooth solution to the obstacleproblem (1.1) under the assumptions (2.3) , (2.4) , (2.5) and (2.2) . If ψ ∈ W ,r loc (Ω) , then u ∈ W , ∞ loc (Ω) and the following estimate k Du k L ∞ ( B ρ ) ≤ C (cid:26)Z B R [1 + f ( x, Du )] dx (cid:27) β (2.7) G. BERTAZZONI, S. RICC `O holds for every < ρ < R and with positive constants C and β depending on n, r, p, q, ν, L, R, ρ and on the local bounds for k Dψ k W ,r and k h k L r . Now we want to present a meaningful definition of relaxation for problem (1.1) in the spiritof [1], [20], [22], [34]. Inspired by [14] and [30], we consider the set K ∗ ψ (Ω) := { w ∈ u + W ,q (Ω) : w ≥ ψ a.e. in Ω } Then, in spirit of [1] and [34], we introduce the relaxed functional F ( u ) := inf C ( u ) { lim inf j → + ∞ F ( u j ) } (2.8)where F is defined in (2.1) and where C ( u ) := (cid:8) { u j } j ∈ N ⊂ K ∗ ψ (Ω) : u j ⇀ u weakly in W ,p (Ω) (cid:9) (2.9)and consider the Lavrentiev gap L ( u ) := ( F ( u ) − F ( u ) if F ( u ) < ∞ F ( u ) = ∞ (2.10)Then, the Lipschitz regularity result we are able to prove is the following. Theorem 2.3.
Assume that f satisfies the hypothesis (2.3) , (2.4) , (2.5) and (2.2) . TheDirichlet problem min {F ( u ) : u ∈ K ψ (Ω) } with F defined as in (2.8) and u ∈ W ,q (Ω) , has at least one locally Lipschitz continuoussolution. Preliminary results
First of all, we state the following lemma which has important application in the so calledhole-filling method. Its proof can be found for example in [31, Lemma 6.1].
Lemma 3.1.
Let h : [ ρ , R ] → R be a non-negative bounded function and < ϑ < , A, B ≥ and β > . Assume that h ( s ) ≤ ϑ h ( t ) + A ( t − s ) β + B for all ρ ≤ s < t ≤ R . Then h ( r ) ≤ c A ( R − ρ ) β + c B where c = c ( ϑ, β ) > . We now present the higher differentiability result we need in the sequel. Our hypothesisimplies the ones needed for this theorem to be true. The proof can be found in [29].
Theorem 3.2.
Let u ∈ K ψ (Ω) be a solution to the obstacle problem Z Ω D ξ f ( x, Du ) · D ( ϕ − u ) dx ≥ EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 5 under the assumptions (2.3) , (2.4) , (2.5) for the exponents ≤ p < q < n < r such that (2.2) holds true. Then the following estimate Z B R/ | D ( V p ( Du ( x ))) | dx ≤ C (cid:16) k Dψ k L r ( B R/ ) + k D ψ k L r ( B R/ ) + k Du k L p ( B R/ ) (cid:17) α (3.2) holds for all balls B R/ ⊂ B R/ ⊂ B R ⋐ Ω , with C = C ( ν, L, q, p, r, n, R ) and α = α ( p, q, r, n ) ,where we defined V p ( ξ ) := ξ (cid:0) | ξ | (cid:1) p − (3.3) Remark 3.3.
One can easily check that, for p ≥ , there exists an absolute constant c suchthat | ξ | p ≤ c | V p ( ξ ) | (3.4) Remark 3.4.
Thanks to the fact that our first main result Theorem 2.2 is an a prioriestimate, we can use indifferently the variational inequality (4.1) and the obstacle problem (1.1) in the following.
Once estimate (3.2) is estabilished, by applying a suitable approximation procedure (see [29]for the details) we get the following corollary.
Corollary 3.5.
Let u and ψ be as above. Then the following implication holds Dψ ∈ W ,r loc (Ω) = ⇒ (cid:0) | Du | (cid:1) p − Du ∈ W , (Ω) (3.5)4. A priori estimate
The linearization procedure.
The linearization procedure is a process which goesback to [25] and later was refined in [16], see also [26], [27], [28]. We will follow the lines of [3].We consider a smooth function h ε : (0 , ∞ ) → [0 ,
1] such that h ′ ε ( s ) ≤ s ∈ (0 , ∞ ) and h ε ( s ) = ( s ≤ ε s ≥ ε Consider the function ϕ = u + t · η · h ε ( u − ψ )with η ∈ C (Ω), η ≥ < t << Z Ω D ξ f ( x, Du ) · D ( ϕ − u ) dx ≥ ϕ ∈ W ,q loc (Ω), ϕ ≥ ψ . We have Z Ω D ξ f ( x, Du ) · D ( η h ε ( u − ψ )) dx ≥ ∀ η ∈ C (Ω)Since η L ( η ) = Z Ω D ξ f ( x, Du ) · D ( η h ε ( u − ψ )) dx G. BERTAZZONI, S. RICC `O is a bounded positive linear functional, by Riesz Representation Theorem there exists anon-negative measure λ ε such that Z Ω D ξ f ( x, Du ) · D ( η h ε ( u − ψ )) dx = Z Ω η dλ ε ∀ η ∈ C (Ω)It is not difficult to prove that the measure λ ε is independent to ε . Therefore we can write Z Ω D ξ f ( x, Du ) · D ( η h ε ( u − ψ )) dx = Z Ω η dλ ∀ η ∈ C (Ω)By Theorem 3.2, we have that V p ( Du ) := (1 + | Du | ) p − Du ∈ W , (Ω) (4.2)Now, in order to identify the measure λ , we may pass to the limit as ε ↓ Z Ω − div( D ξ f ( x, Du )) χ [ u = ψ ] η dx = Z Ω η dλ ∀ η ∈ C (Ω) (4.3)By introducing g := − div( D ξ f ( x, Du )) χ [ u = ψ ] (4.4)and combining our results we obtain Z Ω D ξ f ( x, Du ) · Dη dx = Z Ω g η dx ∀ η ∈ C (Ω) (4.5)We are left to obtain an L r estimate for g : since Du = Dψ a.e. on the contact set, by (2.4)and (2.5) and the assumption Dψ ∈ W ,r loc (Ω; R n ), we have | g | = (cid:12)(cid:12) div( D ξ f ( x, Du )) χ [ u = ψ ] (cid:12)(cid:12) = | div( D ξ f ( x, Dψ )) |≤ n X k =1 | f ξ k x k ( x, Dψ ) | + n X k,i =1 | f ξ k ξ i ( x, Dψ ) ψ x k x i |≤ h ( x ) (1 + | Dψ | ) q − + L (1 + | Dψ | ) q − | D ψ | that is g ∈ L r loc (Ω).4.2. A priori estimate.
Our starting point is now (4.5). As long as we are proving an apriori estimate, we make use of the fact that u ∈ W , ∞ loc (Ω), which is needed in order to let(4.5) to be satisfied. By this further requirement and Theorem 3.2, the “second variation”system holds Z Ω n X i,j =1 f ξ i ξ j ( x, Du ) u x j x s D x i ϕ + n X i =1 f ξ i x s ( x, Du ) D x i ϕ ! dx = Z Ω g D x s ϕ dx (4.6)for all s = 1 , . . . , n and for all ϕ ∈ W , (Ω). We fix 0 < ρ < R with B R compactly containedin Ω and we choose η ∈ C (Ω) such that 0 ≤ η ≤ η ≡ B ρ , η ≡ B R and | Dη | ≤ CR − ρ . We test (4.6) with ϕ = η (1 + | Du | ) γ u x s EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 7 for some γ ≥ D x i ϕ = 2 η η x i (1 + | Du | ) γ u x s +2 η γ (1 + | Du | ) γ − | Du | D x i ( | Du | ) u x s + η (1 + | Du | ) γ u x s x i Inserting in (4.6) we get:0 = Z Ω n X i,j =1 f ξ i ξ j ( x, Du ) u x j x s η η x i (1 + | Du | ) γ u x s dx + Z Ω n X i,j =1 f ξ i ξ j ( x, Du ) u x j x s η (1 + | Du | ) γ u x s x i dx + Z Ω n X i,j =1 f ξ i ξ j ( x, Du ) u x j x s η γ (1 + | Du | ) γ − | Du | D x i ( | Du | ) u x s dx + Z Ω n X i =1 f ξ i x s ( x, Du ) 2 η η x i (1 + | Du | ) γ u x s dx + Z Ω n X i =1 f ξ i x s ( x, Du ) η (1 + | Du | ) γ u x s x i dx + Z Ω n X i =1 f ξ i x s ( x, Du ) 2 η γ (1 + | Du | ) γ − | Du | D x i ( | Du | ) u x s dx − Z Ω g η η x s (1 + | Du | ) γ u x s dx − Z Ω g η γ (1 + | Du | ) γ − | Du | D x s ( | Du | ) u x s dx − Z Ω g η (1 + | Du | ) γ u x s x s dx := I ,s + I ,s + I ,s + I ,s + I ,s + I ,s + I ,s + I ,s + I ,s We sum in the previous equation all terms with respect to s from 1 to n , and we denote by I − I the corresponding integrals. We can estimate them following the lines of [4].Summing up the nine terms and using (2.4), we obtain Z Ω η (1 + | Du | ) p − + γ | D u | dx ≤ C Θ (1 + γ ) (cid:20)Z Ω ( η m + | Dη | m ) (1 + | Du | )( q − p + γ ) m dx (cid:21) m (4.7)where the constant C depends on ν, L, n, p, q but is independent of γ and where we setΘ = 1 + k g k L r (Ω) + k h k L r (Ω) G. BERTAZZONI, S. RICC `O and m = rr − p ≥
2, we have (cid:20)Z Ω η ∗ (1 + | Du | )( γ + p ) ∗ dx (cid:21) ∗ = (cid:20)Z Ω η ∗ (1 + | Du | )( γ + p ) ∗ dx (cid:21) ∗ ≤ C Z Ω (cid:12)(cid:12)(cid:12) D h η (1 + | Du | ) γ + p i(cid:12)(cid:12)(cid:12) dx ≤ C Z Ω | Dη | (1 + | Du | ) γ + p dx + C (1 + γ ) Z Ω η h (1 + | Du | ) γ + p − | Du | | D u | i dx = C Z Ω | Dη | (1 + | Du | ) γ + p dx + C (1 + γ ) Z Ω η (1 + | Du | ) γ + p − | Du | | D u | dx = C Z Ω | Dη | (1 + | Du | ) γ + p dx + C (1 + γ ) Z Ω η (1 + | Du | ) γ + p − − | Du | | D u | dx ≤ C Z Ω | Dη | (1 + | Du | ) γ + p dx + C (1 + γ ) Z Ω η (1 + | Du | ) γ + p − | D u | dx where we set 2 ∗ = nn − n ≥ n = 2and we can observe that 2 ∗ >
2. In the case n = 2 we assume that 2 ∗ > m . Thanks to theleft hand side of (2.2), we know that 1 ≤ qpp ≤ qp ≤ q − p γ + p ≤ q − p γ EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 9 so we can write (cid:20)Z Ω η ∗ (1 + | Du | )( γ + p ) ∗ dx (cid:21) ∗ ≤ C Z Ω | Dη | (1 + | Du | ) q − p + γ dx + C (1 + γ ) Z Ω η (1 + | Du | ) γ + p − | D u | dx Thanks to (4.7), we finally get (cid:20)Z Ω η ∗ (1 + | Du | )( γ + p ) ∗ dx (cid:21) ∗ ≤ C Θ (1 + γ ) (cid:20)Z Ω ( η m + | Dη | m ) (1 + | Du | )( q − p + γ ) m dx (cid:21) m from which we deduce "Z B ρ (1 + | Du | )( γ + p ) ∗ dx ∗ ≤ C Θ (1 + γ )( R − ρ ) (cid:20)Z B R (1 + | Du | )( q − p + γ ) m dx (cid:21) m for any 0 < ρ < R .At this point we introduce the quantity σ defined as σ = q − p − p m (4.9)where we observe that σ > m >
1, in fact σ = q − p − p m ≥ p − p − p m = p − p m > q − p σ + p m Therefore "Z B ρ (1 + | Du | )( γ + p ) ∗ dx ∗ ≤ C Θ (1 + γ )( R − ρ ) (cid:20)Z B R (1 + | Du | )( q − p + γ ) m dx (cid:21) m = C Θ (1 + γ )( R − ρ ) (cid:20)Z B R (1 + | Du | )( σ + p m + γ ) m dx (cid:21) m which allow us to say that "Z B ρ (1 + | Du | )( γ + p ) ∗ dx ∗ ≤ C Θ (1 + γ )( R − ρ ) k | Du | k σL ∞ ( B R ) (cid:20)Z B R (1 + | Du | ) γ m + p dx (cid:21) m (4.10)We now inductively define the exponents γ := 0 γ k +1 := 1 m (cid:20)(cid:16) γ k + p (cid:17) ∗ − p (cid:21) (4.11) α k := m γ k + p k ≥
1. It follows that α k +1 = (cid:16) γ k + p (cid:17) ∗ χ α k + τ (4.12)where we have set χ := 2 ∗ m and τ := 2 ∗ α r = 2 ∗ p r (4.13)By induction we can prove that α k +1 = α χ k + τ k − X i =0 χ i = p χ k + τ k − X i =0 χ i (4.14)and γ k +1 = α m ( χ k −
1) + τm k − X i =0 χ i = p m ( χ k −
1) + τm k − X i =0 χ i (4.15)Now we consider 0 < ρ < R and set R k = ρ + R − ρ k ∀ k ≥ R k +1 ≤ R k for all k ≥ R k − R k +1 = R − ρ k +1 . We rewrite (4.10) with ρ = R k +1 and R = R k . We obtain "Z B Rk +1 (1 + | Du | )( γ k + p ) ∗ dx ∗ ≤ C Θ (1 + γ k )( R k − R k +1 ) k | Du | k σL ∞ ( B Rk ) "Z B Rk (1 + | Du | ) m γ k + p dx m from which we deduce "Z B Rk +1 (1 + | Du | )( γ k + p ) ∗ dx ∗ ≤ C k +1 Θ (1 + γ k )( R − ρ ) k | Du | k σL ∞ ( B Rk ) "Z B Rk (1 + | Du | ) m γ k + p dx m and we can write Z B Rk +1 (1 + | Du | )( γ k + p ) ∗ dx (4.16) EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 11 ≤ (cid:20) C k +1 Θ (1 + γ k )( R − ρ ) (cid:21) ∗ k | Du | k ∗ σ L ∞ ( B Rk ) "Z B Rk (1 + | Du | ) m γ k + p dx χ For each k ∈ N , we define: A k := "Z B Rk (1 + | Du | ) α k dx αk (4.17)where we can use the definitions (4.11) and (4.12) for α k and α k +1 . So we can rewrite (4.16) as A k +1 ≤ (cid:20) C k +1 Θ (1 + γ k )( R − ρ ) (cid:21) ∗ αk +1 (cid:20) k | Du | k ∗ σ L ∞ ( B Rk ) (cid:21) αk +1 A αk χαk +1 k Iterating this inequality we obtain A k +1 ≤ k Y i =1 (cid:20) C i +1 Θ (1 + γ i )( R − ρ ) (cid:21) ∗ χk − i αk +1 (cid:20) k | Du | k ∗ σ L ∞ ( B R ) (cid:21) αk +1 P k − i =0 χ i A χk α αk +1 (4.18)Thanks to (4.14), it’s easy to see thatlim k → + ∞ α k +1 k − X i =0 χ i = 1 α ( χ −
1) + τ (4.19)and that lim k → + ∞ χ k α α k +1 = α ( χ − α ( χ −
1) + τ = δ (4.20)We define: M k := k Y i =1 (cid:2) C i +1 (1 + γ i ) (cid:3) ∗ χk − i αk +1 = exp ( ∗ α k +1 k X i =1 χ k − i log (cid:2) C i +1 (1 + γ i ) (cid:3)) and it isn’t difficult to see, thanks to (4.14), thatlim k → + ∞ M k ≤ M < + ∞ (4.21)Last but not least, as proved in [4], we can say that X := lim k → + ∞ k X i =1 ∗ χ k − i α k +1 < ∞ (4.22) Thanks to (4.19), (4.20), (4.21) and (4.22), letting k → + ∞ , noticing that R > R , we canrewrite (4.18) as follows k | Du | k L ∞ ( B ρ ) ≤ M Θ X ( R − ρ ) X k | Du | k ∗ σ α χ − τ "Z B R (1 + | Du | ) p dx δp Now we have that E := 2 ∗ σ α ( χ −
1) + 2 τ = ∗ σ p ( χ −
1) + ∗ α r = ∗ (cid:2) q − p (cid:0) + m (cid:1)(cid:3) p (cid:0) ∗ m − (cid:1) + ∗ p r = 2 ∗ (cid:2) q − p (cid:0) + m (cid:1)(cid:3) p (cid:2) ∗ m − ∗ r (cid:3) (4.23)and we have to prove that E <
1. Now, if n ≥ ∗ (cid:20) q − p (cid:18)
12 + 12 m (cid:19)(cid:21) < p (cid:20) ∗ m − ∗ r (cid:21) ∗ (cid:20) q − p (cid:18)
12 + 12 m (cid:19)(cid:21) < ∗ p (cid:20) m − ∗ + 1 r (cid:21) q < p (cid:20)
12 + 12 m + 12 m − ∗ + 1 r (cid:21) but using the equality 12 m − ∗ = 1 n − r (4.24)we know that 12 + 12 m + 12 m − ∗ + 1 r = 12 + 22 ∗ + 2 n − r − ∗ + 1 r = (cid:20)
12 + 12 ∗ + 2 n (cid:21) − r = (cid:20)
12 + n − n + 2 n (cid:21) − r = n + n − n − r = 1 + 1 n − r EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 13 and, thanks to (2.2), the thesis is proved. On the other hand, if n = 2, then passing to thelimit 2 ∗ → ∞ in (4.23) we deduce, thanks to (4.8) and (4.9), that σp (cid:0) m + r (cid:1) < q − p − p m < p (cid:18) m + 1 r (cid:19) q < p (cid:20)
12 + 1 m + 1 r (cid:21) q < p (cid:20)
12 + r − r + 1 r (cid:21) q < p (cid:20) − r (cid:21) which is nothing but (2.2) with the choice n = 2. Thus we can use the Young’s inequalitywith exponents 2 α ( χ −
1) + 2 τ ∗ σ and 2 α ( χ −
1) + 2 τ [2 α ( χ −
1) + 2 τ ] − ∗ σ to get: k | Du | k L ∞ ( B ρ ) ≤ k | Du | k L ∞ ( B R ) + (cid:20) M Θ X ( R − ρ ) X (cid:21) ϑ "Z B R (1 + | Du | ) p dx δ ϑp for an exponent ϑ = ϑ ( n, r, p, q ) > ρ < ρ < R < R , once more by Lemma 3.1we finally get k | Du | k L ∞ ( B ρ ) ≤ (cid:20) M Θ X ( R − ρ ) X (cid:21) ϑ (cid:20)Z B R (1 + | Du | ) p dx (cid:21) δ ϑp so we can write, thanks to (2.3) k Du k L ∞ ( B ρ ) ≤ (cid:20) M Θ X ( R − ρ ) X (cid:21) ϑ (cid:20)Z B R (1 + | Du | ) p dx (cid:21) δ ϑp ≤ (cid:20) M Θ X ( R − ρ ) X (cid:21) ϑ (cid:20)Z B R ν f ( x, Du ) dx (cid:21) δ ϑp ≤ C (cid:26)Z B R [1 + f ( x, Du )] dx (cid:27) β Remark.
We can rewrite the equation (4) in a different way as follows k Du k L ∞ ( B ρ ) ≤ C (cid:2) k h k L r ( B R ) (cid:3) α γ (cid:26)Z B R [1 + f ( x, Du )] dx (cid:27) γp (4.25) Approximation in case of Occurrence of Lavrentiev Phenomenon andproof of Theorem 2.3
Approximation procedure.
We start by proving a lemma which will be necessary inthe following. The proof of this lemma follows similarly as in [5], see also [20], where thesame result has been achieved for functionals. Here the novelty is the application to obstacleproblems.
Lemma 5.1.
For each u ∈ K ψ (Ω) , there exists a sequence u k ∈ K ∗ ψ (Ω) such that u k ⇀ u weakly in W ,p (Ω) and F ( u ) = lim k → + ∞ F ( u k ) Proof.
Let u ∈ K ψ (Ω) such that F ( u ) < ∞ . Then, for all k , there exists u ( k ) h ∈ K ∗ ψ (Ω) suchthat u ( k ) h ⇀ u , as h → + ∞ , weakly in W ,p (Ω) and F ( u ) ≤ lim h → + ∞ Z Ω f ( x, Du ( k ) h ) dx ≤ F ( u ) + 1 k Moreover, by the weak convergence of u ( k ) h in W ,p (Ω) we getlim h → + ∞ k u ( k ) h − u k L p (Ω) = 0and, for h sufficiently large, Z Ω | Du ( k ) h | p dx ≤ Z Ω f ( x, Du ( k ) h ) dx ≤ F ( u ) + 1Then for all k there exists h k such that for all h ≥ h k k u ( k ) h − u k L p (Ω) < k and for h = h k , by denoting w k = u ( k ) h k , we have k w k − u k L p (Ω) < k and Z Ω | Dw k | p dx ≤ C then w k ⇀ u , as k → + ∞ , in the weak topology of W ,p (Ω) and F ( u ) ≤ Z Ω f ( x, Dw k ) dx ≤ F ( u ) + 1 k i.e. lim k → + ∞ Z Ω f ( x, Dw k ) dx = F ( u ) (cid:3) For the approximation we are gonna now consider the situation where the Lavrentiev phe-nomenon may occur. We are gonna use the following Theorem 5.2 in order to rewrite f through a suitable sequence of regular functions: that, together with Theorem 2.3, will per-mit us to prove that the supplementary assumption u ∈ W , ∞ loc (Ω) we supposed in Section 4can be removed. EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 15
Theorem 5.2.
Let f be satisfying the growth conditions (2.3) , f ξξ and f ξx be two Carath´eodoryfunctions, satisfying (2.4) and (2.5) and f strictly convex at infinity. Then there exists asequence of C − functions f lk : Ω × R n → [0 , + ∞ ) , f lk convex in the last variable and strictlyconvex at infinity, such that f lk converges to f as l → ∞ and k → ∞ for a.e. x ∈ Ω ,for all ξ ∈ R n and uniformly in Ω × K , where Ω ⋐ Ω and K being a compact set of R n .Moreover the functions f lk satisfy the hypothesis (2.3) , (2.4) , (2.5) with constants which areindependent on k and satisfy the additional hypothesis necessaries to conclude our proof withconstants which are dependent only on k .Proof. We argue as in [18], [20] and [24]. For the sake of completeness, we give a sketch ofthe arguments of the proof.Let B be the unit ball of R n centered in the origin and consider a positive decreasing sequence ε l →
0. We introduce f l ( x, ξ ) = Z B × B ρ ( y ) ρ ( η ) f ( x + ε l y, ξ + ε l η ) dη dy where ρ is a positive symmetric mollifier and f lk ( x, ξ ) = f l ( x, ξ ) + 1 k (1 + | ξ | ) q (5.1)It is easy to check that the sequence f lk satisfies the hypothesis (2.3), (2.4), (2.5) withconstants which are independent on k and satisfies the additional hypothesis necessaries toconclude our proof with constants which are dependent only on k . (cid:3) Proof of Theorem 2.3.
Now we are ready to give the proof of Theorem 2.3.
Proof.
For u ∈ W ,q (Ω), let us consider the variational problemsinf (cid:26)Z Ω f lk ( x, Du ) dx : u ∈ K ∗ ψ (Ω) (cid:27) (5.2)where f lk are defined in (5.1). By semicontinuity arguments and direct methods of theCalculus of Variations, there exists u lk ∈ K ∗ ψ (Ω), a solution to (5.2). By the growth conditionsand the minimality of u lk , remembering that u ∈ K ∗ ψ (Ω) because it is not restrictive toassume u ≥ ψ as observed in Remark 2.1, we get Z Ω | Du lk | p dx ≤ Z Ω f lk ( x, Du lk ) dx ≤ Z Ω f lk ( x, Du ) dx = Z Ω f l ( x, Du ) dx + 1 k Z Ω (1 + | Du | ) q dx Moreover, the properties of the convolutions imply that f l ( x, Du ) l →∞ −→ f ( x, Du ) a.e. in Ωand since Z Ω f l ( x, Du ) dx ≤ C Z Ω (1 + | Du | ) q dx where is fundamental the hypothesis that u ∈ W ,q (Ω). By Lebesgue Dominated CovergenceTheorem we deduce thereforelim l → + ∞ Z Ω | Du lk | p dx ≤ lim l → + ∞ Z Ω f l ( x, Du ) dx + 1 k Z Ω (1 + | Du | ) q dx = Z Ω f ( x, Du ) dx + 1 k Z Ω (1 + | Du | ) q dx By Theorem 5.2, the functions f lk satisfy (2.3), (2.4) and (2.5), so we can apply the a-prioriestimate (4.25) to u lk and obtain, by standard covering arguments for all B ⋐ Ω, k Du lk k L ∞ ( B ) ≤ C [ k h l k L r (Ω) ] α γ (cid:26)Z Ω [1 + f lk ( x, Du lk )] dx (cid:27) γp Since k h l k L r (Ω) = k (1 + h ) l k L r (Ω) ≤ k h k L r (Ω) , we obtain k Du lk k L ∞ ( B ) ≤ C [ k h k L r (Ω) ] α γ (cid:26)Z Ω [1 + f lk ( x, Du lk )] dx (cid:27) γp ≤ C [ k h k L r (Ω) ] α γ (cid:26)Z Ω [1 + f l ( x, Du ) + 1 k (1 + | Du | ) q ] dx (cid:27) γp ≤ C [ k h k L r (Ω) ] α γ (cid:26)Z Ω [1 + f ( x, Du ) + 1 k (1 + | Du | ) q ] dx (cid:27) γp where C, γ, β depend on n, r, p, q, L, ν, B,
Ω and on the local bound for k Dψ k W ,r but areindependent of l, k . Therefore we conclude that there exist u k ∈ K ψ (Ω), for all k ∈ N , suchthat u lk l →∞ −→ u k weakly in W ,p (Ω) u lk l →∞ −→ u k weakly star in W , ∞ loc (Ω)and by the previous estimates k Du k k L p (Ω) ≤ lim inf l →∞ k Du lk k L p (Ω) ≤ Z Ω f ( x, Du ) dx + Z Ω (1 + | Du | ) q dx and k Du k k L ∞ ( B ) ≤ lim inf l →∞ k Du lk k L ∞ ( B ) ≤ C [ k h k L r (Ω) ] α γ (cid:26)Z Ω [1 + f ( x, Du ) + 1 k (1 + | Du | ) q ] dx (cid:27) γp Thus we can deduce that there exists, up to subsequences, u ∈ K ψ (Ω), thanks to the factthat K ψ (Ω) is convex and close, such that u k → u weakly in W ,p (Ω) u k → u weakly star in W , ∞ loc (Ω) EGULARITY FOR OBSTACLE PROBLEMS WITHOUT STRUCTURE CONDITIONS 17
Now, we have that the functional is lower semicontinuous in W , (Ω) and u k → u weakly in W ,p (Ω), so u k → u weakly in W , (Ω) too.Hence, thanks to (3.3), by compact embedding and noticing that Z B | D ( V p ( Du ε ( x ))) | dx ≤ C (cid:18)Z Ω (1 + | Du ε ( x ) | ) p dx (cid:19) β with C and β depending on n, r, p, q, ν, L, R, ρ , we infer V p ( Du k ) ⇀ w weakly in W , loc (Ω) V p ( Du k ) → w strongly in L loc (Ω)from which we deduce, together with inequality (3.4), Du k → w strongly in L ploc (Ω)We thus have the strong convergence u k → u in W ,p (Ω) + u and the limit function u still belongs to K ψ (Ω), since this set is closed.For any fixed k ∈ N , using the uniform convergence of f l to f in Ω × K (for any K compactsubset of R n ) and the minimality of u lk , we get for all w ∈ K ∗ ψ (Ω) Z Ω f ( x, Du k ) dx ≤ lim inf l →∞ Z Ω f ( x, Du lk ) dx = lim inf l →∞ Z Ω f l ( x, Du lk ) dx ≤ lim inf l →∞ Z Ω f l ( x, Du lk ) dx + 1 k Z Ω (1 + | Du lk | ) q dx ≤ lim inf l →∞ Z Ω f l ( x, Du lk ) dx + 1 k Z Ω (1 + | Du lk | ) q dx ≤ lim inf l →∞ Z Ω f l ( x, Dw ) dx + 1 k Z Ω (1 + | Dw | ) q dx Then, for Ω → Ω, using Dominated Convergence Theorem, we have Z Ω f ( x, Du k ) dx ≤ Z Ω f ( x, Dw ) dx + 1 k Z Ω (1 + | Dw | ) q dx By definition (2.8), we have F ( u ) ≤ lim inf k →∞ Z Ω f ( x, Du k ) dx ≤ Z Ω f ( x, Dw ) dx ∀ w ∈ K ∗ ψ (Ω) (5.3)Let v ∈ K ψ (Ω). By Lemma 5.1, there exists u k ∈ K ∗ ψ (Ω) such that u k ⇀ v weakly in W ,p (Ω)and lim k →∞ Z Ω f ( x, Du k ) dx = F ( v )By (5.3), F ( u ) ≤ Z Ω f ( x, Du k ) dx and we can conclude that F ( u ) ≤ lim k →∞ Z Ω f ( x, Du k ) dx = F ( v ) ∀ v ∈ K ψ (Ω)Then u ∈ W , ∞ loc (Ω) is a solution to the problem min {F ( u ) : u ∈ K ψ (Ω) } . (cid:3) References [1]
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