A nonlinear problem witha weight and a nonvanishing boundary datum
aa r X i v : . [ m a t h . A P ] N ov A NONLINEAR PROBLEM WITH A WEIGHTAND A NONVANISHING BOUNDARY DATUM
REJEB HADIJI
Abstract.
We consider the problem:inf u ∈ H g (Ω) , k u k q =1 Z Ω p ( x ) |∇ u ( x ) | dx − λ Z Ω | u ( x ) | dx where Ω is a bounded domain in IR n , n ≥ p : ¯Ω −→ IR is a givenpositive weight such that p ∈ H (Ω) ∩ C ( ¯Ω), 0 < c ≤ p ( x ) ≤ c , λ is a real constant and q = nn − and g a given positive boundary data.The goal of this present paper is to show that minimizers do exist. Wedistinguish two cases, the first is solved by a convex argument while thesecond is not so straightforward and will be treated using the behaviorof the weight near its minimum and the fact that the boundary datumis not zero. Introduction
Let Ω be a bounded domain in IR n of class C , n ≥
3. Let us considerthe minimization problem S ( p, g ) = inf u ∈ H g (Ω) , k u k q =1 Z Ω p ( x ) |∇ u ( x ) | dx (1.1)where H g (Ω) = { u ∈ H (Ω) s.t. u = g on ∂ Ω } ,g ∈ H ( ∂ Ω) ∩ C ( ∂ Ω) is a given boundary datum and q = nn − is the criticalSobolev exponent.Note that it is well known that H (Ω) ֒ → L r (Ω) is continuous for any1 ≤ r ≤ nn − . Moreover this embedding is compact for 1 ≤ r < nn − .We suppose that the weight p : Ω → IR is a smooth function such that0 < c ≤ p ( x ) ≤ c ∀ x ∈ ¯Ω and p is in H (Ω) ∩ C (Ω).In this paper, we ask the question whenever the problem (1.1) has aminimizer. Note that if the infimum (1.1) is achieved by some u then wehave − div( p ( x ) ∇ u ) = Λ u q − in Ω, u > u = g on ∂ Ω,(1.2)
Mathematics Subject Classification.
Key words and phrases.
Critical Sobolev exponent, Sobolev inequality, boundary data.Convex problem. where Λ ∈ IR is the Lagrange multiplier associated to the problem (1.1).These kind of problems, which are known to bear features of noncompact-ness are studied by many authors. First existence results for the problemwith a linear perturbation are due to Brezis-Nirenberg. Set S λ ( p, g ) = inf u ∈ H g (Ω) , k u k q =1 Z Ω p ( x ) |∇ u ( x ) | dx − λ Z Ω | u ( x ) | dx (1.3)They showed that if g = 0 and p = 1, then S λ (1 ,
0) is attainted as soonas S λ (1 , < S and this is the case if n ≥
4, 0 < λ < λ , or n = 3 and0 < λ ∗ < λ < λ where λ is the first eigenvalue of − ∆ and λ ∗ depends onthe domain, (see [6]). They showed also that if g λ = 0 and p = 1 thenthe infimum in (1.1) is achieved, (see [7]). Our approach uses their method.In the case of p = 1 and g = 0, Coron, Bahri and Coron exploited thetopology of the domain. They proved that equation − ∆ u = u q − in Ω, u > u = 0 on ∂ Ω,has a solution provided that the domain has nontrivial topology, (see [8] and[3]).We refer to [13], [14] for the study of existence and multiplicity solutionsof problem (1.1) with the presence of a smooth and positive weight andwith homogeneous Dirichlet boundary condition. Nevertheless, in [12], it isshown that if p is discontinuous then a solution of S ( p,
0) still exists.In [10], the authors studied the minimization problem on compact manifoldsin the case λ = 0 with many variants.For more general weights, depending on x and on u , in a recent paper writ-ten with Vigneron, we showed that in the case of homogeneous Dirichletboundary condition and in the presence of a linear perturbation the corre-sponding minimizing problem possesses a solution. The model of the weightis p ( x, u ) = α + | x | β | u | k with positive parameters α , β and k . Note that inthis case natural scalings appear and the answer depends on the ratio βk .For more details, we refer to [2] and [15].To motivate our problem, we briefly recall that it is inspired by the studyof the classical Yamabe problem which has been the source of a large liter-ature, (see for example [1], [3], [6], [8], [10] and [16]), we refer to [15] andthe references therein for many recent developments in quasi-linear ellipticequations.In this paper, we will assume that if g p has a global minimum a ∈ Ω such that satisfies: p ( x ) ≤ p + γ | x − a | α ∀ x ∈ B ( a, R ) ⊂ Ω , (1.4)for constants α > γ > R > NONLINEAR PROBLEM WITH A WEIGHT 3
The following auxiliary linear Dirichlet problem will play an importantrole in this paper:(1.5) (cid:26) − div( p ∇ v ) = 0 in Ω ,v = g on ∂ Ω . Statement of the main result.
Our main result is the following:
Theorem 1.1.
Let us assume that the dimension n ≥ and g ∈ H ( ∂ Ω) ∩ C ( ∂ Ω) is a given boundary datum. Let v be the unique solution of (1.5).We have (1) Let || v || q < and let assume that g and having a constant sign.Assume that p has a global minimum a ∈ Ω that satisfies (1.4). Thenfor every n ∈ [3 , α + 2[ the infimum S ( p, g ) is achieved in H g (Ω) . (2) If || v || q ≥ then for every n ≥ the infimum S ( p, g ) is achieved in H g (Ω) . The next proposition tell us that one has Σ g = { u ∈ H g (Ω) , k u k q = 1 } 6 = ∅ which ensures that S ( p, g ) is well defined: Proposition 1.2.
Let g ∈ H ( ∂ Ω) ∩ C ( ∂ Ω) be given boundary datum and v be the unique solution of (1.5), we have • If || v || q < ,then there is a bijection between Σ and Σ g . • If || v || q ≥ , then Σ g = ∅ . Our problem depends on || v || q . More precisely, we will use a convexargument to show that if || v || q ≥ || v || q < p near its minimum and the fact that g has a constantsign. We will argue by contradiction, supposing that minimizing sequenceconverges weakly to some limit u . The fact that the boundary datum isnot 0 will give us that u is not identically 0. Then, by using a suitable testfunctions, we will show equality (4.2) below which is due to term of order 0.After precise computations, we get strict inequality in (4.29) which is dueto the next term in the same expansion, which is lead to a contradiction.Since the nonlinearity of the problem is as stronger as n is low, it is rathersurprising that the infimum is achieved for lower dimensions n ∈ [3 , α + 2[.Note that the presence of p is more significative if α > n ∈ [3 , α + 2[. Remark that if α = 0 then infimum of p = p + γ is not p .For general boundary data g , we do not have control over the normal deriva-tive of a solution of (1.2) on the boundary of Ω and then, standard Pohozaevidentity cannot be used.1.2. Structure of the paper.
The paper is structured as follows: In sec-tion 2 we give the notations and some preliminary results.In the next section, we state two results related to our main result namely,Theorem 3.1 which gives the sign of the Lagrange-multiplier associated to
REJEB HADIJI minimizers of S ( p, g ) given by Theorem 1.1 and Theorem 3.2 which gener-alizes our main result in case of the presence of a linear perturbation.In section 4, we will focus on the proof of Theorem 1.1, which is the mainresult of this paper, it will be proved by a contradiction argument that spansthe whole of this section.In section 5, we give the proof of Theorem 3.1.The last section is dedicated to the problem of existence of minimizer in thepresence of a linear perturbation and the proof of Theorem 3.2.2. Notations and preliminary results
Sobolev inequality says that there exist
M > Z Ω p ( x ) |∇ φ | dx ≥ M (cid:18)Z Ω | φ | q dx (cid:19) q for all φ ∈ H (Ω) . The best constant is defined by S ( p,
0) = inf u ∈ H (Ω) , || u || q = 1 Z Ω p ( x ) |∇ u | dx. Set S = S (1 ,
0) = inf u ∈ H (Ω) , || u || q = 1 Z Ω |∇ u | dx. We know that when the domain is IR n , the constant S (1 ,
0) is achieved bythe functions: U x , ε ( x ) = (cid:18) εε + | x − x | (cid:19) n − , x ∈ IR n where x ∈ IR n and ε >
0, (see [1], [6], [16]). Let us denote by(2.1) u x ,ε ( x ) = U x , ε ( x ) ψ ( x )where ψ ∈ C ∞ ( IR n ), ψ ≡ B ( x , r ) ψ ≡ B ( x , r ) ⊂ Ω, r >
0. Wehave(2.2) Z Ω p ( x ) |∇ u x ,ε | dx = p ( x ) K + O ( ε n − ) , (2.3) Z Ω | u x ,ε | q dx = K + O ( ε n ) , where K and K are positive constants with K K q = S .We have also u x , ε ⇀ H (Ω) . − ∆ U x , ε = c n U q − x , ε in IR n . It is well known that S in never achieved for bounded domain, (see [6]). NONLINEAR PROBLEM WITH A WEIGHT 5
In the the presence of the weight p we have Proposition 2.1.
Suppose that a ∈ Ω be a global minimum of p . Set p = p ( a ) . If g = 0 , we have S ( p, is never achieved and S ( p,
0) = p S (1 ,
0) = p S. Proof.
When g = 0, the functions u a,ε k u a,ε k q are admissible test functions for S ( p,
0) and we have as ε → p S ≤ S ( p, ≤ Z Ω p ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ∇ u a,ε k u a,ε k q (cid:12)(cid:12)(cid:12)(cid:12) dx = p S + Z Ω ( p ( x ) − p ) (cid:12)(cid:12)(cid:12)(cid:12) ∇ u a,ε k u a,ε k q (cid:12)(cid:12)(cid:12)(cid:12) dx + o (1)= p S + o (1) . Passing to the limit ε → S ( p,
0) = p S. This implies that S ( p,
0) is not achieved. Indeed, let us suppose that S ( p, u . Using the fact that S is never achived in boundeddomains, we obtain p S < p Z Ω |∇ u | dx ≤ Z Ω p ( x ) |∇ u | dx = p S. This leads to a contradiction.2.1.
The auxiliary Dirichlet problem.
The linear Dirichlet problem (1.5)has a unique solution which solves the following problemmin v ∈ H g (Ω) Z Ω p ( x ) |∇ v ( x ) | dx. (2.4)Let us give now the proof of Proposition 1.2: Recall thatΣ g = { u ∈ H g (Ω) , k u k q = 1 } . In the first case we can construct a bijection between Σ and Σ g . Indeed,let us define, for t in IR and u ∈ Σ the function(2.5) f ( t ) = Z Ω | tu + v | q since f is smooth, f ′′ ( t ) = q ( q − R Ω | tu + v | q − u , f (0) < t →∞ f ( t ) = ∞ , using the intermediate value theorem and the convexity of f , we ob-tain, for every u in Σ , the existence of a unique t ( u ) > || t ( u ) u + v || q = 1.Let us denote by ϕ : Σ → Σ g the function defined by ϕ ( u ) = t ( u ) u + v . Let u and u in Σ such that ϕ ( u ) = t ( u ) u + v = ϕ ( u ) = t ( u ) u + v , wehave nesseceraly || t ( u ) u || q = || t ( u ) u || q , this implies that t ( u ) = t ( u )and u = u . Therefore we have that ϕ is one to one function. Let w ∈ Σ g , REJEB HADIJI w = v , set u = w − v || w − v || q , we have t ( u ) = || w − v || q and ϕ ( w − v || w − v || q ) = w . Thus, ϕ is a bijection.Suppose || v || q ≥
1, let ζ ∈ C ∞ c (Ω) is such that || v − ζv || q <
1. Observe that v − ζv = g on boundary. The same argument as above gives t > || v − tζv || q = 1. 3. Statement of further results
The sign of the Euler-Lagrange.
Let u be a minimizer for the prob-lem (1.1), then, it satisfies the following Euler-Lagrange equation − div( p ( x ) ∇ u ) = Λ u q − in Ω, u > u = g on ∂ Ω, || u || q = 1 , (3.1)where Λ ∈ IR is the Lagrange multiplier associated to the problem (1.1), let v be defined by (1.5). The sign of Λ is given by the following: Theorem 3.1.
The sign of Λ is as the following: If || v || q < then Λ > ,if || v || q > then Λ < and if || v || q = 1 then Λ = 0 . Presence of a linear perturbation.
Over the course of the proof ofTheorem 1.1, one also reaps the following compactness result.
Theorem 3.2.
We assume that p , g and v satisfy the same conditions as inTheorem 1.1. Assume that || v || q < . Let us denote by λ the first eigenvalueof the operator − div( p ∇ . ) with homogeneous Dirichlet boundary condition.Then for λ < λ we have the infimum in S λ ( p, g ) is achieved in the followingcases: (1) λ > , α > and n ≥ , (2) λ > , α ≤ and n ∈ [3 , α + 2[ . (3) λ < , n = 3 or with α > and n = 5 with α > . In the presence of a linear perturbation, we will highlight a competitionbetween three quantities, the dimension n , the exponent α in (1.4) and theterm of the linear perturbation. As we will see and as in Theorem 1.1 thebehavior of p near its minimum plays an important role. The exponent α = 2 is critical in the case λ = 0.4. Proof of Theorem 1.1.
Let us start by proving the first part of Theorem 1.1. Suppose that || v || q <
1. Since the function u is a solution of S ( p, − g ) if and only if − u issolution of S ( p, g ), it suffices to consider the case g ≥ u j ) be a minimizing sequence for S ( p, g ), that is, Z Ω p ( x ) |∇ u j ( x ) | dx = S ( p, g ) + o (1) NONLINEAR PROBLEM WITH A WEIGHT 7 and || u j || q = 1 , u j = g in ∂ Ω . Since g ≥
0, we may always assume that u j ≥
0, indeed, ( | u j | ) is alsoa minimizing sequence. Since ( u j ) is bounded in H we may extract asubsequence still denoted by ( u j ) such that ( u j ) converges weakly in H to a function u ≥ a.e. , ( u j ) converges strongly to u in L (Ω), and ( u j )converges to u a.e. on Ω with u = g on ∂ Ω.Using a standard lower semicontinuity argument, we infer that || u || q ≤ || u || q = 1.Arguing by contradiction, let us assume that || u || q < . We will prove that this is not possible with the assistance of several lem-mas. We start by giving the first-order term of the energy R Ω p ( x ) |∇ u ( x ) | dx ,next, we show that u satisfies some kind Euler-Lagrange equation and thenit is smooth. Finally, we compute the second-order term and highlight acontradiction.4.1. The first-order term.Lemma 4.1.
For every w ∈ H g (Ω) such that || w || q < , we have S ( p, g ) − Z Ω p ( x ) |∇ w ( x ) | dx ≤ p S (cid:18) − Z Ω | w | q (cid:19) q , (4.1) For the weak limit u , we have equality: S ( p, g ) − Z Ω p ( x ) |∇ u ( x ) | dx = p S (cid:18) − Z Ω | u | q (cid:19) q . (4.2) Proof.
Let w ∈ H g (Ω) such that || w || q <
1. Therefore we can find aconstant c ε,a > || w + c ε,a u ε,a || q = 1 . Using Brezis-Lieb Lemma (see [4]), we obtain c qε,a = 1 K (cid:18) − Z Ω | w | q (cid:19) + o (1)(4.3) REJEB HADIJI where K is defined in (2.3). Careful expansion as ε → n ≥ Z Ω p ( x ) |∇ u a,ε ( x ) | dx ≤ p K + O ( ε n − ) if (cid:26) n ≥ n − < α,p K + A ε α + o ( ε α ) if (cid:26) n ≥ n − > α,p K + A ε n − | log ε | + o ( ε n − | log ε | ) if (cid:26) n ≥ α = n − , with K = ( n − Z IR n | y | (1 + | y | ) n dy and where A , A and A are positive constants depending only on n , γ and α , and for n = 3 and for α > ε → Z p ( x ) |∇ u a,ε ( x ) | dx = p K + [ ω Z R ( p + γr α ) | ψ ′ ( r ) | dr + ω kα Z R | ψ | r α − dr ] ε + o ( ε ) . where ψ is defined as in (2.1). Therefore for n = 3 and α > Z p ( x ) |∇ u a,ε ( x ) | dx = p K + A ε + o ( ε ) . where A is a positive constant.Remark that regardless of dimension n as long as n ≥ α > Z Ω p ( x ) |∇ u a,ε ( x ) | dx ≤ p K + o (1) . (4.6)Using w ε = w + c ε,a u ε,a as testing function in S ( p, g ) we obtain S ( p, g ) ≤ Z Ω p ( x ) |∇ w ( x ) | dx + c ε,a Z Ω p ( x ) |∇ u a,ε ( x ) | + o (1)Using (4.3), the fact that K K q = S and taking into account (4.6) we get thefirst assertion of the Lemma 4.1.For the second part, thanks to (4.1), it suffices to prove one inequality for u . S ( p, g ) − Z Ω p ( x ) |∇ u ( x ) | dx ≥ p S (cid:18) − Z Ω | u | q (cid:19) q . (4.7)Set v j = u j − u so that v j = 0 in ∂ Ω and ( v j ) converges weakly to 0 in H and a.e. We have by Sobolev inequality
NONLINEAR PROBLEM WITH A WEIGHT 9 Z Ω p ( x ) |∇ v j | ≥ p S || v j || q . (4.8)On the other hand, we have (see [4])1 = Z Ω | v j | q + Z Ω | u | q + o (1) . (4.9)Since ( u j ) is a minimizing sequence we have S ( p, g ) = Z Ω p ( x ) |∇ v j | + Z Ω p ( x ) |∇ u | + o (1) , (4.10)hence, combining (4.8), (4.9) and (4.10) we obtain the desired conclusion.We will now use the fact that g is not identically zero. A consequence ofthe above lemma is the following: Lemma 4.2.
The function u satisfies ( − div( p ∇ u ) = p S (cid:0) − R Ω | u | q (cid:1) − qq | u | q − u in Ω u = g on ∂ Ω(4.11)
Moreover, u is smooth, u ∈ L ∞ (Ω) and u > in Ω . Proof.
Applying (4.1) to w = u + tϕ , ϕ ∈ C ∞ (Ω) and | t | small enough, wehave S ( p, g ) ≤ Z Ω p ( x ) |∇ u | − t Z Ω p ( x ) ∇ u ∇ ϕ + o ( t )+ p S (cid:18) − Z Ω | u | q − qt Z Ω | u | q − uϕ + o ( t ) (cid:19) q , thus S ( p, g ) ≤ Z Ω p ( x ) |∇ u | − t Z Ω p ( x ) ∇ u ∇ ϕ + p S (cid:18) − Z Ω | u | q (cid:19) q (cid:18) − t R Ω | u | q − uϕ − R Ω | u | q + o ( t ) (cid:19) . Hence, by using (4.7) we obtain for every ϕ ∈ C ∞ (Ω) − Z Ω p ( x ) ∇ u ∇ ϕ − (cid:18) − Z Ω | u | q (cid:19) − qq Z Ω | u | q − uϕ = 0 . (4.12)Since u = g on ∂ Ω we obtain (4.11).For proving the regularity of u , it suffices, in view of the standard ellipticregularity theory to show that u is in L t (Ω) for all t < ∞ . To see this, weshall apply Lemma A1 of [5], then, u is as smooth as the regularity of p and g permits. By using the strong maximum principle, and the fact that g ≥ g u > . The second-order term.
Now, we need a refined version of (4.1).Similarly as in the proof of (4.1), let c ǫ,a be defined by 1 = R Ω | u + c ǫ,a u ǫ,a | q .We can write c ε,a = c (1 − δ ( ε ))(4.14)with c q = 1 K (cid:18) − Z Ω | u | q (cid:19) and lim ε → δ ( ε ) = 0 . (4.15) Lemma 4.3.
We have δ ( ε ) K c q ≥ p ε n − (cid:18) c Z Ω u q − ψ | x − a | n − c q ( q − Du ( a ) (cid:19) (4.16) + q − c q K δ ( ε ) + o ( δ ( ε )) + o ( ε n − ) . where D is a positive constant. Proof.
First case q ≥
3. We need the following inequality, for all a ≥ b ≥ a + b ) q ≥ a q + qa q − b + qab q − + b q (4.17)which follows from t q + qt q − + qt + 1(1 + t ) q ≤ t such that t = ba if a = 0.Using (4.17) and the fact that u > Z Ω | u + c ε,a u ε,a | q ≥ Z Ω u q + qc q − ε,a Z Ω uu q − ε,a + qc ε,a Z Ω u q − u ε,a + c qε,a Z Ω u qε,a . and thus 1 ≥ Z Ω u q + qc q − ε,a Z Ω uu q − ε,a + qc (1 − δ ( ε )) Z Ω u q − u ε,a (4.18) + qc q (cid:18) − qδ ( ε ) + q ( q − δ ( ε ) + o ( δ ( ε ) (cid:19) Z Ω u qε,a . NONLINEAR PROBLEM WITH A WEIGHT 11
On the other hand we have Z Ω uu q − ε,a = ε n − Du ( a ) + o ( ε n − )(4.19)where D is a positive constant, and Z Ω u q − u ε,a = ε n − Z Ω u q − ψ | x − a | n − + o ( ε n − ) . (4.20)Combining (2.3), (4.18), (4.19) and (4.20) we obtain (4.16).Second case 2 < q <
3. In what follows C denote a positive constantindependent of ε . The keys are the two following inequalities, we have forall a ≥ b ≥ | ( a + b ) q − ( a q + qa q − b + qab q − + b q ) | ≤ Ca q − b if a ≤ b (4.21)and | ( a + b ) q − ( a q + qa q − b + qab q − + b q ) | ≤ Cab q − if a ≥ b (4.22)which follows respectively from | (1 + t ) q − ( t q + qt q − + qt + 1) | t ≤ C (4.23)for t ≥ | (1 + t ) q − ( t q + qt q − + qt + 1) | t q − ≤ C (4.24)for t ≤ t such that t = ba if a = 0.Using (4.21) and (4.22) we get1 = Z Ω | u + c ε,a u ε,a | q (4.25) = Z Ω u q + qc q − ε,a Z Ω uu q − ε,a + qc ε,a Z Ω u q − u ε,a + c qε,a Z Ω u qε,a + R (1) ε + R (2) ε . where R (1) ε ≤ C Z { x,u ≥ c ε,a ψU a,ε } u | ψU a,ε | q − and R (2) ε ≤ C Z { x,u
Now, using (4.16) and the fact that δ ( ε ) = o (1) we infer S ( p, g ) ≤ R Ω p |∇ u | + p K c − c ε n − R Ω div( p ∇ u ) ψ | x − a | n − (4.31) − c " ε n − K c q (cid:18) c Z Ω u q − ψ | x − a | n − + c q − Du ( a ) (cid:19) + q − δ ( ε ) + o ( δ ( ε )) Ω p ( x ) |∇ u a,ε | dx + c δ ( ε ) Z Ω p ( x ) |∇ u a,ε | dx + o ( ε n − ) . Since R Ω p ( x ) |∇ u a,ε | dx = K + o (1) we obtain S ( p, g ) ≤ R Ω p |∇ u | + c R Ω p ( x ) |∇ u a,ε | dx − ( q − c δ ( ε ) + o ( δ ( ε )) − c "Z Ω div( p ∇ u ) ψ | x − a | n − + c − q K Z Ω u q − ψ | x − a | n − + DK u ( a ) ! ( K + o (1)) ε n − + o ( ε n − ) . This leads to S ( p, g ) ≤ Z Ω p |∇ u | + c Z Ω p ( x ) |∇ u a,ε | dx − ( q − c K δ ( ε ) + o ( δ ( ε )) − c DK K u ( a ) ε n − + o ( ε n − ) . We know that δ ( ε ) = o (1) thus S ( p, g ) ≤ Z Ω p |∇ u | + c Z Ω p ( x ) |∇ u a,ε | dx (4.32) − c DK K u ( a ) ε n − + o ( ε n − ) . We are now able to give a precise asymptotic behavior of the RHS of (4.30).This will possible thanks to the fact that u ( a ) = 0, namely u ( a ) >
0. Oneneeds to distinguish between dimensions and the parameter α . Four casesfollow from (4.4) and (4.32): • The case when n ≥ n < α + 2. We have S ( p, g ) ≤ Z Ω p |∇ u | + c (cid:0) p K + o ( ε n − ) (cid:1) − c DK K u ( a ) ε n − + o ( ε n − ) . Consequently, we have(4.33) S ( p, g ) ≤ Z Ω p |∇ u | + p c K − c DK K u ( a ) ε n − + o ( ε n − ) . • The case when n ≥ n > α + 2. We have S ( p, g ) ≤ Z Ω p |∇ u | + c ( p K + A ε α + o ( ε α )) − c DK K u ( a ) ε n − + o ( ε n − ) . Therefore, we have S ( p, g ) ≤ Z Ω p |∇ u | + p c K − c DK K u ( a ) ε n − + A c ε α + o ( ε α ) + o ( ε n − ) . Hence, if n < α + 2 then S ( p, g ) ≤ R Ω p |∇ u | − c DK K u ( a ) ε n − + o ( ε n − ) . (4.34) • The case when n ≥ α = n −
2. We have S ( p, g ) ≤ Z Ω p |∇ u | + c (cid:0) p K + A ε n − | log ε | + o ( ε n − | log ε | ) (cid:1) − c DK K u ( a ) ε n − + o ( ε n − ) . thus we get(4.35) S ( p, g ) ≤ Z Ω p |∇ u | + p c K − c DK K u ( a ) ε n − + o ( ε n − ) . • n = 3 and α >
1. We have S ( p, g ) ≤ Z Ω p |∇ u | + c [ p K + A ε + o ( ε )] − c DK K u ( a ) ε + o ( ε ) . Hence we have(4.36) S ( p, g ) ≤ Z Ω p |∇ u | + p c K − c DK K u ( a ) ε + o ( ε ) . Now, thanks to (4.33), (4.34), (4.35), (4.36) and the fact that u ( a ) > The case || v || q ≥ . For the proof of the second part of Theorem 1.1we set α := inf u ∈ H g (Ω) , k u k q =1 Z Ω p ( x ) |∇ u ( x ) | dx and β := inf u ∈ H g (Ω) , k u k q ≤ Z Ω p ( x ) |∇ u ( x ) | dx. Indeed using the convexity of the problem β , it is clear that the infimum in β is achieved by some function w ∈ H g (Ω) satisfying || w || q ≤
1. Necessarilywe have equality. Let us reason by contradiction, if we had || w || q <
1, let
NONLINEAR PROBLEM WITH A WEIGHT 15 ζ ∈ ∞ c (Ω), for t real and small such that we have || w + tζ || q <
1, using w + tζ as test function in β we obtain that w would be the unique solution of thefollowing Euler-Lagrange equation:(4.37) (cid:26) − div( p ∇ w ) = 0 in Ω ,w = g on ∂ Ω . that is mean w and v coincide, this leads to a contradiction since || v || q ≥ α is achieved.Since || w || q = 1 we have R Ω p ( x ) |∇ w ( x ) | dx = β ≤ α ≤ R Ω p ( x ) |∇ w ( x ) | dx .Thus α = β .5. The sign of the Euler-Langange multiplier. Proof ofTheorem 3.1
We follow an idea of [11]. Let u be a minimizer for the problem (1.1) and v be defined by (1.5), using the fact that problem (1.5) has a unique solutionwhich minimizes (2.4), we remark that we have || v || q = 1 if and only if wehave Λ = 0.Using (3.1) and (1.5) we obtain (cid:26) − div( p ( x ) ∇ ( u − v )) = Λ u q − in Ω, u − v = 0 on ∂ Ω.(5.1)First, suppose that || v || q <
1. Multiplying (5.1) by u − v and integrating weobtain(5.2) Λ( || u || qq − Z Ω | u | q − v ) = Z Ω p ( x ) |∇ ( u − v ) | . From H¨older inequality and the fact that || u || q = 1 we obtain(5.3) || u || qq − Z Ω | u | q − v ≥ − || v || q > . Putting together (5.2) and (5.3) and using the fact that u = v we see thatΛ >
0. Suppose now that || v || q >
1. For t ∈ IR , let us define the function f by f ( t ) = Z Ω | tu + (1 − t ) v | q dx. Note that the function f is smooth and convex since f ′′ ( t ) = q ( q − R Ω tu +(1 − t ) v | q − ( u − v ) ≥ f (0) = || v || qq > f (1) = || u || qq = 1 . We may use the following:
Lemma 5.1.
For all t ∈ [0 , we have f ( t ) > . Proof.
Arguing by contradiction, since f is continuous, by the intermediatevalue theorem there exists t ∈ [0 ,
1[ such that f ( t ) = 1. Using t u + (1 − t ) v ∈ Σ g as testing function in S ( , p, g ) we have(5.5) S ( p, g ) = Z Ω p ( x ) |∇ u | ≤ Z Ω p ( x ) |∇ ( t u + (1 − t ) v | Multiplying (1.5) by u − v and integrating we obtain(5.6) Z Ω p |∇ v | = Z Ω p ∇ u ∇ v Using (5.5), (5.6) and the fact that t < Z Ω p |∇ u | ≤ Z Ω p |∇ v | Since v is the unique solution of (1.5) we obtain that u = v which clearlycontradicts (5.4). This complete the proof of Lemma 5.1.By the convexity of f and Lemma 5.1 we deduce that f ′ (1) ≤
0. But f ′ (1) = q R Ω | u | q − ( u − v ) and then by (5.1) we have f ′ (1) = q Λ R Ω p ( x ) |∇ ( u − v ) | .We conclude that Λ < Existence of minimizer in the presence of a linearperturbation: Proof of Theorem 3.2
First, we claim that if problem (1.3) has a solution then λ < λ . Indeed,let u be a solution of (1.1) and v satisfying (1.5), we have − div( p ( x ) ∇ ( u − v ) = Λ( λ, u ) u q − + λu in Ω, u > u − v = 0 on ∂ Ω.(6.1)where Λ( λ, u ) is a Euler-Lagrange multiplier. Since || v || q <
1, using section5, we find that Λ( λ, u ) >
0. Let ϕ be the eigenfunction of the operator − div( p ∇ . ) with homogeneous Dirichlet boundary condition correspondingto λ . Multiplying (6.1) by ϕ and integrating we obtain − Z Ω div( p ( x ) ∇ ( u − v )) ϕ = λ Z Ω ( u − v ) ϕ = Λ( λ, u ) Z Ω u q − ϕ + λ Z Ω uϕ . Then we get ( λ − λ ) Z Ω ( u − v ) ϕ ≥ λ Z Ω vϕ and thus λ < λ .The proof of Theorem 1.1 is similar to the one of Theorem 1.1 so that webriefly outline it. We need only to take into account the linear perturbation NONLINEAR PROBLEM WITH A WEIGHT 17 term. We will then follow exactly all the steps in the proof of Theorem 1.1untill (4.32), we just need to account the linear perturbation. We get S λ ( p, g ) ≤ R Ω p |∇ u | + c (cid:0)R Ω p ( x ) |∇ u a,ε | dx − λ R Ω | u a,ε | dx (cid:1) (6.2) − c DK K u ( a ) ε n − + o ( ε n − ) . From [6] we have k u a,ε k = K ε + O ( ε n − ) if n ≥ ,C ε | log ε | + O ( ε ) if n = 4 ,C ε + O ( ε ) if n = 3(6.3)where C , C and C are positive constants. Using (4.4), (6.2) and (6.3) andthe fact that u ( a ) > References [1] T. Aubin,
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