Existence and regularity of positive solutions of quasilinear elliptic

arXiv:1703.08608v1 [math.AP] 24 Mar 2017

Existence and regularity of positive solutions of quasilinear elliptic problems with singular semilinear term José V. A. Goncalves Marcos L. M. Carvalho Carlos Alberto Santos∗

Abstract This paper deals with existence and regularity of positive solutions of singular elliptic problems on a smooth bounded domain with Dirichlet boundary conditions involving the Φ-Laplacian operator. The proof of existence is based on a variant of the generalized Galerkin method that we developed inspired on ideas by Browder [4] and a comparison principle. By using a kind of Moser iteration scheme we show L∞ (Ω)-regularity for positive solutions.

Contents 1. 2. 3. 4. 5. 6.

Introduction Main Results A family of Auxiliary Problems Applied Generalized Galerkin Method Comparison of Solutions and Estimates Proof of the Main Results 6.1. Pure Singular Problem - Existence of Solutions 6.2. Convex Singular Problem - Regularity of Solutions 7. Appendix - On Orlicz-Sobolev spaces

1

Introduction

This paper concerns existence and regularity of solutions to the singular elliptic problem − div(φ(|∇u|)∇u) =

a(x) in Ω, u > 0 in Ω, u = 0 on ∂Ω, uα

(1.1)

where Ω ⊂ RN , with N ≥ 2, is a bounded domain with smooth boundary ∂Ω, a is a non-negative function, 0 < α < ∞ and φ : (0, ∞) → (0, ∞) is of class C 1 and satisfies (φ1 ) ∗

(i) tφ(t) → 0 as t → 0, (ii) tφ(t) → ∞ as t → ∞,

Supported by CAPES/Brazil Proc. N o 2788/2015 − 02,

1

2 (φ2 ) tφ(t) is strictly increasing in (0, ∞), (φ3 ) there exist ℓ, m ∈ (1, N) such that ℓ−1 ≤

(tφ(t))′ ≤ m − 1, t > 0. φ(t)

We extend s 7→ sφ(s) to R as an odd function. It follows that the function Z t Φ(t) = sφ(s)ds, t ∈ R 0

is even and it is actually an N-function. Due to the nature of the operator ∆Φ u := div(φ(|∇u|)∇u) we shall work in the framework of Orlicz and Orlicz-Sobolev spaces namely LΦ (Ω), LΦe (Ω) and W01,Φ (Ω). We recall some basic notation on these spaces along with bibliographycal references in the Apendix. In the last years many research papers have been devoted to the study of singular problems like (1.1). In [21], Karlin & Nirenberg studied the singular integral equation Z 1 1 u(x) = G(x, y) dy, 0 ≤ x ≤ 1, u(y)α 0 where α > 0 and G(x, y) is a suitable potential. In [9], Crandall Rabinowitz & Tartar, addressed a class of singular problems which included as a special case, the model problem − ∆u =


(1.2)

where α > 0 and a : Ω → [0, ∞) is a suitable L1 -function. A broad literature on problems like (1.2) is available to date. We would like to mention [22, 34, 36] and their references. We would like to refer the reader to the very recent papers by Orsina & Petitta [29], Canino, Sciunzi & Trombetta [5] for the problem −∆u =

µ in Ω, u > 0 in Ω, u = 0 on ∂Ω. uα

In [29] µ is a nonnegative bounded Radon measure while in [5] µ is an L1 function. Other kinds of operators have been addressed and we mention Chu-Wenjie [7] and De Cave [11] for problems involving the p-Laplacian like −div(|∇u|p−2∇u) =

a(x) in Ω, u > 0 in Ω, u = 0 on ∂Ω; uα

Qihu Zhang [35] and Liu, Zhang & Zhao [26] for p(x)-Laplacian operator, −div(|∇u|p(x)−2∇u) =

a(x) in Ω, u > 0 in Ω, u = 0 on ∂Ω; uα

3 Boccardo & Orsina [3] and Bocardo & Casado-D´ıaz [2] for the problem −div(M(x)∇u) =


where M is a suitable matrix, Lazer & McKeena [24]; Goncalves & Santos [16], Hu & Wang [20] for problems involving the Monge-Ampére operator, e. g., det(D 2 u) =

a(x) in Ω, u < 0 in Ω, u = 0 on ∂Ω, (−u)γ

where a ∈ C ∞ (Ω), a > 0 and γ > 1. To the best of our knowledge singular problems like (1.1) in the presence of the operator ∆Φ were never studied and the main results of this paper (see Section 2) namely Theorems 2.1, 2.2 as well as Corollary 2.1 are new. Other problems which are special cases of (1.1) are −∆p u − ∆q u = a(x)u−α in Ω, u > 0 in Ω, u = 0 on ∂Ω,

(1.3)

where φ(t) = tp−2 + tq−2 with 1 < p < q < N, −

N X

∆pi u = a(x)u−α in Ω, u > 0 in Ω, u = 0 on ∂Ω.

(1.4)

i=1

where φ(t) =

PN

pj −2 , 1 < p1 < p2 < . . . < pN < ∞ and j=1 t

PN

j=1

1 > 1, pj

−div(a(|∇u|p )|u|p−2∇u) = a(x)u−α in Ω, u > 0 in Ω, u = 0 on ∂Ω.

(1.5)

where φ(t) = a(tp )tp−2 , 2 ≤ p < N and a : (0, ∞) → (0, ∞) is a suitable C 1 (R+ )-function. We also refer the reader to the paper [27], where the operator ∆Φ is employed. The operator ∆Φ appears in applied mathematics, for instance in Plasticity, see e.g. Fukagai and Narukawa [14] and references therein. We refer the reader to [31] for problems involving general operators.

2

Main Results

In this work, for each x ∈ Ω, we set d(x) = inf |x − y|. Our first result is. y∈∂Ω

Theorem 2.1 Assume that (φ1 ) − (φ3 ) and a ∈ L1 (Ω) hold. Then there is u such that u(α−1+ℓ)/ℓ ∈ W01,ℓ (Ω), u ≥ Cd a.e. in Ω, for some C > 0, and: (i) u ∈ W01,Φ (Ω), and Z

Ω

φ(|∇u|)∇u∇ϕdx =

Z

Ω

a(x) ϕdx, ϕ ∈ W01,Φ (Ω), α u

provided additionally that either ad−α ∈ LΦe (Ω) or 0 < α ≤ 1 and a ∈ Lℓ

(2.1) ∗ /(ℓ∗ +α−1)

(Ω),

4 1,Φ (ii) u ∈ Wloc (Ω), and

Z

φ(|∇u|)∇u∇ϕdx =

Ω

Z

Ω

a(x) ϕdx, ϕ ∈ C0∞ (Ω) α u

(2.2)

provided in addition that α ≥ 1. Next we will present some regularity results: Corollary 2.1 Under the conditions of the above Theorem, we have that: (i) u ∈ C(Ω) if a ∈ L∞ (Ω), ∗

∗

(ii) u ∈ L∞ (Ω) if either a ∈ Lq (Ω) ∩ Lℓ /(ℓ +α−1) (Ω) and 0 < α ≤ 1 or a ∈ Lq (Ω) and α > 1, where N/ℓ < q ≤ q(α) with ∗ ℓ /s if 0 < s ≤ 1, (2.3) q(s) := (ℓ∗ + (α − 1)ℓ∗ /ℓ)/s if s > 1, (iii) there exists an only solution u ∈ W01,Φ (Ω) of Problem (1.1) in the sense of (2.1). We are going to take advantage of our techniques to show existence results to the singularconvex problem − div(φ(|∇u|)∇u) =

a(x) + b(x)uγ in Ω, u > 0 in Ω, u = 0 on ∂Ω, uα

(2.4)

where α, γ > 0. Theorem 2.2 Assume (φ1 ) − (φ3 ) and 0 ≤ γ < ℓ − 1. Assume in addition that ad−α ∈ LΦe (Ω) and 0 ≤ b ∈ Lσ (Ω) for some σ > ℓ/(ℓ − γ − 1). Then problem (2.4) admits a weak solution u ∈ W01,Φ (Ω) such that u ≥ Cd in Ω for some constant C > 0. Besides this, u ∈ L∞ (Ω) if ∗ ∗ b ∈ L∞ (Ω), and either a ∈ Lq (Ω) ∩ Lℓ /(ℓ +α−1) (Ω) with 0 < α ≤ 1 or a ∈ Lq (Ω) with α > 1, where N/ℓ < q ≤ q(α + γ) and q(s) was defined in (2.3). Remark 2.1 We note that: (a) solutions of both Theorems can be found by variational arguments in some particular cases, (b) if Ψ is an N-function such that Φ < Ψ 0 of the singular term in (2.4). Of course a regularized form of problem (1.1) corresponds to b = 0. Let us consider  aǫ (x)  + bǫ (x)uγ in Ω −∆Φ u = (3.1) (u + ǫ)α  u > 0 in Ω, u = 0 on ∂Ω for each ǫ > 0 given, where the L∞ (Ω)-functions are defined by

aǫ (x) = min{a(x), 1/ǫ}, bǫ (x) = min{b(x), 1/ǫ}, x ∈ Ω. Consider the map A := Aǫ : W01,Φ (Ω) × W01,Φ (Ω) −→ R, defined by Z h i aǫ (x)ϕ + γ − bǫ (x)(u ) ϕ dx, A(u, ϕ) := φ(|∇u|)∇u∇ϕdx − (|u| + ǫ)α Ω

(3.2)

Thus, finding a weak solution of (3.1) means to find u ∈ W01,Φ (Ω) such that A(u, ϕ) = 0 for each ϕ ∈ W01,Φ (Ω). Proposition 3.1 For each u ∈ W01,Φ (Ω), the particular, the operator T := Tǫ : W01,Φ (Ω) −→

(3.3)

functional A(u, .) is linear and continuous. In e W −1,Φ (Ω) defined by

hT (u), ϕi = A(u, ϕ), u, ϕ ∈ W01,Φ (Ω) is linear and continuous, and satisfies C kaǫ kΦe + Ckbǫ |u|γ kΦe . (3.4) ǫ Proof: Let u, ϕ ∈ W01,Φ (Ω). We shall use below the Hölder inequality and the embedding W01,Φ (Ω) ֒→ LΦ (Ω): Z aǫ (x)|ϕ| |A(u, ϕ)| ≤ φ(|∇u|)|∇u||∇ϕ| + + bǫ (x)(u+ )γ |ϕ| dx α ǫ Ω 2 ≤ 2kφ(|∇u|)∇ukΦe kϕk + α kaǫ kΦe kϕkΦ + 2kbǫ |u|γ kΦe kϕkΦ ǫ C ≤ (2kφ(|∇u|)∇ukΦe + α kaǫ kΦe + Ckbǫ |u|γ kΦe )kϕk. (3.5) ǫ It is enough to show that kbǫ |u|γ kΦe < ∞. Indeed, by using the embedding LΦ (Ω) ֒→ Lℓ (Ω) and γ ∈ (0, ℓ − 1) it follows by Lemma 7.2 that Z Z m ℓ ℓ−1 m−1 γ γ e e )dx Φ(bǫ (x)|u |)dx ≤ max{kbǫ k∞ , kbǫ k∞ } Φ(|u| Ω Ω Z Z γ e ≤ C + Φ(|u| )dx u≤1 u≥1 Z Z γℓ ℓ ≤ C |Ω| + |u| ℓ−1 dx ≤ C |Ω| + |u| dx u≥1 u≥1 Z ℓ ≤ C |Ω| + |u| dx ≤ C |Ω| + kukℓ , (3.6) kT (u)kW −1,Φe ≤ 2kφ(|∇u|)∇ukΦe +

Ω

6 where C = C(b, Φ, ǫ) > 0 is a constant. So A(u, .) is linear and continuous. The claims about T are now immediate. By proposition (3.1) the problem of finding a weak solution of (3.1) reduces to find u = uǫ ∈ W01,Φ (Ω) \ {0} such that T (uǫ ) = 0.

4

Applied Generalized Galerkin Method

In order to find u = uǫ ∈ W01,Φ (Ω) \ {0} such that T (uǫ ) = 0, we shall employ a Galerkin like method inspired in arguments found in Browder [4]. We are going to constrain the operator T to finite dimensional subspaces. As a first step take a ω ∈ W01,Φ (Ω) such that aω 6= 0 and aω ∈ L1 (Ω), (4.7) Let F ⊂ W01,Φ (Ω) be a finite dimensional subspace such that ω ∈ F . Now, consider the map TF : F → F ′ given by TF = IF′ ◦ T ◦ IF , where IF : (F, k.k) −→ (W01,Φ (Ω), k.k), IF (u) = u and let IF′ be the adjoint of IF . So, we have that TF = T F , because

hTF u, vi = hIF′ ◦ T ◦ IF u, vi = hT ◦ IF u, IF vi = hT u, vi, u, v ∈ F,

that is, Z h i aǫ (x)v + γ hTF (u), vi := φ(|∇u|)∇u∇v − − bǫ (x)(u ) v dx, u, v ∈ F. (|u| + ǫ)α Ω

(4.8)

The result below, which is a consequence of the Brouwer Fixed Point Theorem (see [25]), will play a central role in solving the finite dimensional equation TF (u) = 0. Proposition 4.1 Assume that S : Rs → Rs is a continuous map such that (S(η), η) > 0, |η| = r for some r > 0, where (·, ·) is the usual inner product in Rs and | · | is its corresponding norm. Then, there is η0 ∈ Br (0) such that S(η0 ) = 0. Proposition 4.2 The operator TF is continuous. Proof: Let (un ) ⊆ F be a sequence such that un → u in F . Since, the operator e ∆Φ : W01,Φ (Ω) → W −1,Φ (Ω) given by Z h−∆Φ u, vi := φ(|∇u|)∇u∇vdx, u, v ∈ W01,Φ (Ω), Ω

is continuous (see [14, Lemma 3.1]), we have that ∆Φ F is also continuous. To finish our proof, it remains to show that TF − ∆Φ F is continuous. By applying Lemma 7.3 and the embedding LΦ (Ω) ֒→ Lℓ (Ω), it follows, by eventually passing to a subsequence, that

7 (1) un → u a.e. in Ω; (2) there is h ∈ Lℓ (Ω) such that |un | ≤ h. Then for each v ∈ W01,Φ (Ω), aǫ (x)v aǫ (x)v −→ , α (|un | + ǫ) (|u| + ǫ)α

γ + γ bǫ (x)(u+ n ) v −→ bǫ (x)(u ) v a.e. in Ω.

e is increasing, we obtain On the other hand, since Φ aǫ (x) a (x) a (x) a (x) ǫ ǫ ǫ e e ≤ Φ + Φ (|un | + ǫ)α − (|u| + ǫ)α (|un | + ǫ)α (|u| + ǫ)α 2a (x) ǫ e ≤ Φ ∈ L1 (Ω), α ǫ

(4.9)

because 0 ≤ aǫ ≤ 1/ǫ. So, by Lebesgue’s Theorem, Z aǫ (x) a (x) ǫ e Φ (|un | + ǫ)α − (|u| + ǫ)α dx → 0, Ω e ∈ ∆2 , we have and as a consequence of Φ

aǫ (x) a (x) ǫ

−

(|un | + ǫ)α (|u| + ǫ)α e → 0. Φ

By applying the Hölder inequality, we find that

Z

aǫ (x) aǫ (x) aǫ (x) aǫ (x)

kvkΦ → 0 − vdx ≤ 2 − α α α α (|u | + ǫ) (|u| + ǫ) (|u | + ǫ) (|u| + ǫ) e n n Ω Φ

for each v ∈ W01,Φ (Ω). Estimating as in (4.9), we have

γ + γ e bǫ |(u+ Φ n ) − (u ) |

for some C = C(a, Φ, ǫ) > 0. Arguing as above, we obtain Z

Ω

showing that TF is continuous.

γ + γ (u+ n ) + (u ) e ≤ Φ 2|bǫ |∞ 2 e + )γ ) + Φ((u e + )γ ) ≤ C Φ((u n ≤ C |u|ℓ + |h|ℓ + 2 ∈ L1 (Ω),

γ + γ bǫ (x)[(u+ n ) − (u ) ]vdx −→ 0

(4.10)

8 Proposition 4.3 There exists 0 6= u = uF = uǫ,F ∈ F such that TF (u) = 0 for each ǫ > 0 sufficiently small. Proof: Let s := dim F be the dimension of the subspace F , and set F = he1 , e2 , ..., es i. That is, each u ∈ F is uniquely expressed as u=

s X

ξj ej , ξ = (ξ1 , ξ2, · · · , ξs ) ∈ Rs .

j=1

Set |ξ| := kuk and consider the map i = iF : (Rs , |.|) → (F, k.k) given by i(ξ) = u. So, it follows by Proposition 4.2 and the fact that i is an isometry that the operator SF : Rs → Rs given by SF := i′ ◦ TF ◦ i (4.11) is continuous as well, where i′ is the adjoint of i. Besides this, by setting u := i(ξ) for ξ ∈ Rs , it follows from (φ3 ) and the embeddings W01,Φ (Ω) ֒→ LΦ (Ω) ֒→ Lℓ (Ω) ֒→ Lγ+1 (Ω) that (SF ξ, ξ) = (i′ ◦ TF ◦ i(ξ), ξ) = hTF (u), ui Z h i aǫ (x)|u| γ+1 ≥ φ(|∇u|)|∇u|2 − − b (x)|u| dx ǫ ǫα Ω Z 1 ≥ ℓ Φ(|∇u|)dx − α kaǫ kΦe kukΦ − |bǫ |∞ |u|γ+1 γ+1 ǫ Ω ≥ ℓ min{kukℓ , kukm} − C1 kuk − C2 kukγ+1 ,

(4.12)

for some positive constants C1 = C1 (ǫ) and C2 = C2 (ǫ). So, we can choose an r0 = r0 (ǫ) > 1 such that ℓr0ℓ − C1 r0 − C2 r0γ+1 > 0. More specifically, for each ξ such that |ξ| = r0 , we have (SF ξ, ξ) > 0. By the above, it follows from Proposition 4.1 that there exists a ξF ∈ B r0 (0) such that SF (ξF ) = 0, that is, letting u = uF = i(ξF ), it follows from (4.11), that hTF (u), vi = (SF (ξF ), η) = 0 for all v ∈ F, where v = i(η), and hence TF (u) = 0. As a consequence of this, we have Z h i aǫ (x)v + γ − bǫ (x)(u ) v dx = 0 for all v ∈ F. φ(|∇u|)∇u∇v − (|u| + ǫ)α Ω We claim that u = uǫ 6= 0 for enough small ǫ > 0. Indeed, otherwise by taking v = w and using Lebesgue’s Theorem, we obtain Z Z a(x)wdx = lim aǫ (x)wdx = 0, Ω

ǫ→0

Ω

but this is impossible by (4.7). This ends the proof of Proposition 4.3. The result below is a direct consequence of the Proposition proved just above.

9 Corollary 4.1 The number r0 > 0 and the function uF ∈ F found above satisfy: kuF k ≤ r0 , TF (uF ) = 0, and r0 > 0 does not depends on subspace F ⊂ W01,Φ (Ω) with 0 < dim F < ∞. ∗ ∗ Besides this, we can choose it independent of ǫ > 0 as well if 0 < α ≤ 1, a ∈ Lℓ /(ℓ +α−1) (Ω), and b ∈ Lσ (Ω) for some σ > ℓ/(ℓ − γ − 1). Proof: The first part of it was proved above. To show that r0 does not depends on ǫ > 0, we just redo the estimatives in (4.12) by using the hypotheses on a and b. Our aim below is to build a non-zero vector uǫ ∈ W01,Φ (Ω) such that T (uǫ ) = 0, where T was given by Proposition 3.1. This will provide us with some uǫ ∈ W01,Φ (Ω) such that Z h i aǫ (x)ϕ + γ φ(|∇u|)∇u∇ϕ − (4.13) − bǫ (x)(u ) ϕ dx = 0, ϕ ∈ W01,Φ (Ω). α (|u| + ǫ) Ω In this direction we have Lemma 4.1 There is a non-zero vector uǫ ∈ W01,Φ (Ω) such that T (uǫ ) = 0 or equivalently (4.13) holds true. Proof: Let w as in (4.7) and set n o 1,Φ 1,Φ A = F ⊂ W0 (Ω) | F is a finite dimensional subspace of W0 (Ω) and ω ∈ F , We assume that A is partially ordered by set inclusion. Take F0 ∈ A and set VF0 = uF ∈ F F ∈ A, F0 ⊂ F, TF (uF ) = 0 and kuF k ≤ r0 .

Note that by Proposition 4.3 and Corolary 4.1, VF0 6= ∅. σ σ Since VF0 ⊂ Br0 (0), then V F0 ⊂ Br0 (0), where V F0 denotes the weak closure of VF0 . As a matter σ of this fact, V F0 is weakly compact. Consider the family σ B := V F | F ∈ A . Claim. B has the finite intersection property. σ σ σ Indeed, let {V F1 , V F2 , ..., V Fp } be a finite subfamily of B and set F := span{F1 , F2 , ..., Fp }. σ

By the very definition of VFi , we have that uF ∈ V Fi , i = 1, 2, ..., p, that is p \

σ

V Fi 6= ∅.

i=1

This ends the proof of the Claim. Since B r0 is weakly compact, it follows that (cf. [28, Thm. 26.9]) \ σ W := V F 6= ∅. F ∈A

10 σ

Let uǫ ∈ W . Then uǫ ∈ V F for each F ∈ A. σ

Take F0 ∈ A such that span{ω, uǫ} ⊂ F0 . Since uǫ ∈ V F0 , it follows by [12, Thm. 1.5] and the definition of VF0 that there are sequences (un ) = (un,ǫ ) ⊂ VF0 and (Fn ) = (Fn,ǫ ) ⊂ A such that un ⇀ uǫ in W01,Φ (Ω), un ∈ Fn , kun k ≤ r0 , F0 ⊂ Fn , and Z

φ(|∇un |)∇un ∇vdx =

Ω

Z Ω

aǫ (x) + γ + bǫ (x)(un ) vdx (|un | + ǫ)α

(4.14)

for each v ∈ Fn . comp

Now, by eventually taking subsequences and using W01,Φ (Ω) ֒→ LΦ (Ω), we obtain that un → uǫ in LΦ (Ω), un → uǫ a.e. in Ω and (|un |) is bounded away by some function in LΦ (Ω). Set vn = un − uǫ and note that vn ∈ Fn , because un ∈ Fn and uǫ ∈ F0 ⊂ Fn in (4.14). Then Z

aǫ (x) + γ limh−∆Φ (un ), un − uǫ i = lim + bǫ (x)(un ) (un − uǫ )dx (|un | + ǫ)α Ω Z aǫ (x) γ ≤ lim + bǫ (x)|un | |un − uǫ |dx. ǫα Ω

(4.15)

comp

As W01,Φ (Ω) ֒→ LΦ (Ω), we have Z aǫ (x) ≤ 1 kaǫ k e kun − uǫ kΦ → 0. (u − u )dx n 0 Φ ǫα α Ω ǫ

Recalling that γ < ℓ − 1, W01,Φ (Ω) ֒→ Lℓ (Ω) and (un ) is bounded in Lℓ (Ω), we get Z

γ

bǫ (x)|un | |un − uǫ |dx ≤ |bǫ |∞

Ω

Z

|un |

γℓ ℓ−1

Ω

ℓ−1 ℓ |un − uǫ |ℓ dx

ℓ−1 Z ℓ ℓ ≤ |bǫ |∞ |Ω| + |un | dx |un − uǫ |ℓ → 0. Ω

Now, by using the facts above, it follows from (4.14) that limh−∆Φ (un ), un − uǫ i ≤ 0, and a consequence of this, we have that un → uǫ in W01,Φ (Ω), because −∆Φ satisfies the condition (S+ ) (see [6, Prop. A.2]). So, passing to a subsequence if necessary, we have (1) ∇un → ∇uǫ a.e. in Ω, (2) there is h ∈ LΦ (Ω) such that |∇un | ≤ h.

11 Since ϕ ∈ W01,Φ (Ω), it follows of the fact that tφ(t) is nondecreasing in [0, ∞) and (2), that |φ(|∇un|)∇un ∇ϕ| ≤ φ(|∇un |)|∇un ||∇ϕ| ≤ φ(h)h|∇ϕ| e ≤ Φ(φ(h)h) + Φ(|∇ϕ|) ≤ Φ(2h) + Φ(|∇ϕ|) ∈ L1 (Ω),

that is, it follows by the Lebesgue Theorem, that Z Z φ(|∇un |)∇un ∇ϕdx −→ φ(|∇uǫ|)∇uǫ ∇ϕdx. Ω

Ω

Now, by passing to the limit in (4.14) and using the above informations, we get that uǫ satisfies (4.13), that is, Tǫ (uǫ ) = T (uǫ ) = 0 for each ǫ > 0, since ϕ ∈ W01,Φ (Ω) was taken arbitrarily. By arguments as in the proof of Proposition 4.3 we infer that uǫ 6≡ 0. Lemma 4.2 The function uǫ ∈ C 1,αǫ (Ω), for some 0 < αǫ ≤ 1, and it is a solution of (3.1). Proof: By Lemma 4.1, it remains to show that uǫ > 0. Set −u− ǫ as a test function in (4.13). So, it follows by Remark 7.1 (see Appendix), that Z Z − − 2 ℓ Φ(|∇uǫ |)dx ≤ φ(|∇u− ǫ |)|∇uǫ | dx Ω Ω Z aǫ (x) u− dx, (4.16) = − − | + ǫ)α ǫ (|u Ω ǫ which implies that u− ǫ ≡ 0. So, uǫ satisfies Z Z Z aǫ (x) φ(|∇uǫ |)∇uǫ ∇ϕ = ϕdx+ bǫ (x)uγǫ ϕdx, ϕ ∈ W01,Φ (Ω). α Ω Ω (uǫ + ǫ) Ω

(4.17)

Finally, for each p ∈ (m, ℓ∗ ), it follows that |f (x, t)| :=

aǫ (x) tp + γ p−1 + b (x)(t ) ≤ C (1 + |t| ) and lim =0 ǫ ǫ t→∞ Φ∗ (λt) (|t| + ǫ)α

for each ǫ > 0 given. So by [33, Corollary 3.1], uǫ ∈ C 1,αǫ (Ω) for some 0 < αǫ ≤ 1. Now, by summing up the term uǫ φ(uǫ ) to both sides of (4.17) and applying [6, Proposition 5.2] we infer that uǫ > 0. In conclusion, uǫ is a solution of (3.1).

5

Comparison of Solutions and Estimates

Let n ≥ 1 be an integer and take ǫ = 1/n. Let un ∈ W01,Φ (Ω) ∩ C 1,αn (Ω), for some αn ∈ (0, 1], denotes the solution of (3.1), both for b = 0 and b ≥ 0 not identically null, given by Lemma 4.2, that is, − ∆Φ un =

an (x) + bn (x)uγ in Ω, un > 0 in Ω, un = 0 on ∂Ω (un + 1/n)α

We have the following result on comparison of solutions.

(5.18)

12 Lemma 5.1 The following inequalities hold: (i) un + 1/n ≥ u1 for each integer n ≥ 1 (ii) u1 ≥ Cd in Ω for some C > 0 which independs of n. Proof. First, we consider b = 0 in (5.18), that is, − ∆Φ un =

an (x) in Ω, un > 0 in Ω, un = 0 on ∂Ω. (un + 1/n)α

(5.19)

So, by (5.19) we have div(φ(|∇u1|)∇u1 ) −

a1 (x) ≥ 0 in Ω, (u1 + 1)α

(5.20)

in the weak sense. On the other hand, since an (x) a1 (x) ≥ in Ω. (wn + 1/n)α ((wn + 1/n) + 1)α we get by (5.19) that div(φ(|∇(un + 1/n)|)∇(un + 1/n)) −

a1 (x) ≤ 0 in Ω, ((un + 1/n) + 1)α

(5.21)

in the weak sense, (test finctions are taken non-negative). By applying Theorem 2.4.1 in [30] to (5.20) and (5.21), we obtain un + 1/n ≥ u1 . Now, since ∂Ω is smooth, it follows by [15, Lemma 14.16] that the distance function x 7→ d(x) satisfies ∂d < 0 on Ω \ Ωδ , d ∈ C 2 (Ω), d > 0 on Ωδ and ∂η where Ωδ = {x ∈ Ω | d(x) > δ} for some δ > 0, and η stands for the exterior unit normal to ∂Ω. Now, since u1 ∈ W01,Φ (Ω) ∩ C 1,α1 (Ω) is a solution of − ∆Φ u =

a1 (x) in Ω, u > 0 in Ω, u = 0 on ∂Ω, (u + 1)α

(5.22)

it follows by [33, Lemma 4.2] that ∂u1 < 0 on Ω \ Ωδ . ∂η So there is a constant C > 0 such that ∂u1 ∂d ≤C on Ω \ Ωδ , ∂η ∂η and as a consequence Cd(x) ≤ u1 (x) for x ∈ Ω.

(5.23)

13 This ends the proof of Lemma 5.1 for b = 0. If b is not identically null, we redo the above proof by considering (5.20) and obtaining (5.21) as a consequence of b be non-negative. This ends the proof. We have the following estimates. Lemma 5.2 Let un ∈ C 1,αn (Ω) be a solution of (5.19). Then there is a constant C > 0 such that k[(un + 1/n)(α+ℓ−1)/ℓ − (1/n)(α+ℓ−1)/ℓ ]k1,ℓ ≤ C, for all integer n ≥ 1. where k · k1,ℓ above is the norm of W01,ℓ . Proof. At first notice that un , [(un + 1/n)α − (1/n)α ] ∈ W01,Φ (Ω) ∩ C 1,αn (Ω) ⊂ W01,ℓ (Ω). By estimating, we get ℓαΦ(1) ℓαΦ(1)

Z

Z

|∇un |ℓ (un +

|∇un |≥1

1 α−1 ) dx ≤ n

Z 1 1 α−1 |∇un |ℓ (un + )α−1 dx ≤ |∇un | (un + ) dx + n n |∇un |≥1 |∇un | 1, we have Z

|∇un |ℓ (un + 1/n)α−1 dx ≤ |∇un |≤1 Z i h ℓαΦ(1) |Ω| + (un + 1/n)α−1 dx.

ℓαΦ(1)

(5.26)

un >1

Summing up (5.24) and (5.26), we obtain a positive constant C such that Z Z α−1+ℓ ℓ ∇ (un + 1/n) ℓ dx ≤ C 1 + Ω

un >1

(un + 1/n)

α−1

dx .

(5.27)

Now, by picking ǫ such that 0 < ǫ < ℓ − ℓ(α − 1)/(α + ℓ − 1), it follows from (5.27), using un > 1 and of the embbeding W01,ℓ (Ω) ֒→ Lℓ (Ω) ֒→ Lℓ−ǫ (Ω), that Z

α+ℓ−1 ℓ−ǫ α−1+ℓ ℓ

(un + 1/n) ℓ dx

∇ (un + 1/n) ℓ

≤ C 1+ ℓ un >1

α+ℓ−1 ℓ−ǫ

, ≤ C 1 + ∇ (un + 1/n) ℓ

ℓ

for some C > 0. That is, [(un + 1/n)(α+ℓ−1)/ℓ − (1/n)(α+ℓ−1)/ℓ ] is bounded in W01,ℓ (Ω) as well. This ends the proof of Lemma 5.2.

6

Proof of The Main Results

We begin proving Theorem 2.1 that treats about existence of positive solution to the pure singular problem 1.1.

15

6.1

Pure Singular Problem - Existence of Solutions

Proof of (i) of Theorem 2.1 Assume first that ad−α ∈ LΦ˜ (Ω). Since un ∈ W01,Φ (Ω) satisfies (5.19), it follows from Remark 7.1, Lemma 7.1, (5.23) and Hölder inequality, that Z Z ℓζ0 (k∇un kΦ ) ≤ ℓ Φ(|∇un |)dx ≤ φ(|∇un |)|∇un|2 dx Ω Z Ω Z an (x) a(x) = |un |dx 1 α un dx ≤ C α Ω (un + n ) Ω d Z Z a(x) = C + |un |dx (6.1) dα Ω/Ωδ Ωδ Z Z a(x) |un |dx ≤ C |un |dx + C dα (x) Ω

a Ω

≤ Ckun kΦ + 2C α kun kΦ , d Φe cpt

where we used an ≤ a just above. It follows from our assumptions and from W01,Φ (Ω) ֒→ LΦ (Ω), ∗ ∗ that (un ) ⊂ W01,Φ (Ω) is bounded. If 0 < α ≤ 1 and a ∈ Lℓ /(ℓ +α−1) (Ω), then the boundedness of (un ) in W01,Φ (Ω) is a consequence of Corollary 4.1. So, in both cases, up to subsequences, there exist u ∈ W01,Φ (Ω) and θ ∈ LΦ (Ω) such that (1) un ⇀ u in W01,Φ (Ω), (2) un → u in LΦ (Ω), (3) un → u a.e. in Ω, (4) 0 ≤ un ≤ θ.

As a first consequence of these facts, it follows from Lemma 5.1 and (3) that u ≥ Cd a.e. in Ω. Now, by using un − u as a test function in (5.19) and following similar arguments as in (4.15), we get Z an (x) h−∆Φ un , un − ui ≤ (un − u)dx α (u + 1/n) h Ω n a i

(6.2) ≤ C + 2 α kun − ukΦ d Φe

for some C > 0 independent of n. Since, the operator −∆Φ is of the type S+ , it follows from (2) and (6.2) that un → u in W01,Φ (Ω). To finish our proof, given ϕ ∈ W01,Φ (Ω), it follows from Lemma 5.1, that

α an a d a ≤ ϕ |ϕ| ∈ L1 (Ω), |ϕ| ≤ C α α α (un + 1/n) d un + 1/n d

that is, by passing to the limit in (5.19), we obtain that u is a solution of (1.1). This ends our proof. We were not able to employ the above arguments in the proof of (ii) of Theorem 2.1, because in such case we do not know if a/dα belongs to LΦe (Ω), that is, the sequence (un ) likely is not

16 1,Φ (Ω). This bounded in W01,Φ (Ω). Instead, it was possible to show that (un ) is bounded in Wloc was done by applying Lemma 5.2.

Proof of (ii) of Theorem 2.1. Given U ⊂⊂ Ω, let δU = min{d(x) / x ∈ U} > 0. So, it follows from Lemma 5.1, that un + 1/n ≥ CδU := CU > 0 in U, that is, for n > 1 enough big, we can take (un + 1/n − CU )+ as a test function in (5.19), to obtain Z

φ(|∇un|)|∇un |

2

≤

U

≤ ≤

Z

φ(|∇un |)|∇un |2 dx

Zun +1/n≥CU un +1/n≥C Z U

1

CUα−1

a(x) dx (un + 1/n)α−1

(6.3)

adx < ∞,

Ω

because a ∈ L1 (Ω), and α ≥ 1. So, it follows from Remark 7.1 and Lemma 7.1, that (un ) ⊂ W 1,Φ (U) is bounded. That is, there exist (uUn1 ), uU ∈ W 1,Φ (U) such that uUn1 ⇀ uU in W 1,Φ (U), uUn1 → uU in LΦ (U), uUn1 (x) → uU (x) a.e. in U. In particular, it follows from Lemma 5.1 and of the pointwise convergence that u ≥ Cd a.e. in U. Hence, by using a Cantor diagonalization argument applied to an exhaustion Uk of Ω with 1,Φ 1,Φ Uk ⊂⊂ Uk+1 ⊂⊂ Ω, we show that there is u ∈ Wloc (Ω) such that uk → u in Wloc (Ω) and u ≥ Cd a.e. in Ω. Now, we are going to show that this u satisfies the equation in (1.1). Given ϕ ∈ C0∞ (Ω), let Θ ⊂⊂ Ω be the support of ϕ. So, by very above informations, we have that (a) un ⇀ u in W 1,Φ (Θ), (b) un → u in LΦ (Θ), (c) un (x) → u(x) a.e. in Θ and there exists θ ∈ LΦ (Θ) such that un ≤ θ in Θ. So, by using ϕ(un − u) as a test function in (5.19), LΦ (Θ) ֒→ L1 (Θ), and (b) above, we obtain Z Z 1 an |ϕ(un − u)|dx (6.4) φ(|∇un |)∇un ∇(ϕ(un − u)) dx ≤ α cd Θ Θ ≤ Cϕ kakLΦ(Θ) kun − ukLΦ (Θ) −→ 0, e where Θ ⊂⊂ Ω is the support of ϕ. That is, Z Z φ(|∇un |)∇un ∇(un − u)ϕdx = φ(|∇un|)∇un ∇ϕ(un − u)dx + on (1). Θ

(6.5)

Θ

e Besides this, it follows from Holder’s inequality, (b) above and the property Φ(φ(t)t) ≤ Φ(2t)

17 for t > 0, that Z Z φ(|∇un|)∇un ∇ϕ(un − u) dx ≤ Cϕ φ(|∇un |)|∇un ||un − u|dx Θ

Θ

≤ Cϕ kφ(|∇un |)|∇un |kLΦe (Θ) kun − ukLΦ (Θ) → 0 ≤ Cϕ kun − ukLΦ(Θ) → 0,

and using this information in (6.5), we obtain that Z φ(|∇un|)∇un ∇(un − u)ϕdx = on (1).

(6.6)

Ω

Besides this, we note that Z Z φ(|∇u|)∇u∇(un − u)ϕdx ≤ φ(|∇u|)∇u∇[ϕ(un − u)]ϕdx Θ ZΘ + φ(|∇u|)∇u∇ϕ(un − u)dx , Θ

and the first integral on the right side goes to zero, due to (a) above, and the second one converges to zero due to (b) above. That is, Z φ(|∇u|)∇u∇(un − u)ϕdx → 0. (6.7) Θ

So, it follows from (4.13) and (4.15), that Z φ(|∇un |)∇un − φ(|∇u|)∇u, ∇un − ∇u ϕdx → 0,

(6.8)

Θ

and a consequence of this togheter with the Lemma 6 in [10], we have that ∇un (x) → ∇u(x) a.e. in Θ, that is, φ(|∇un (x)|)∇un (x) → φ(|∇u(x)|)∇u(x) a.e. in Θ. In addition, since (φ(|∇un |)∇un ) ⊂ (LΦe (Θ))N is bounded, it follows from Lemma 2 in [17] that φ(|∇un |)∇un ⇀ φ(|∇u|)∇u in (W 1,Φ (Θ))N . 1,Φ Now, passing to limit in (5.19), we obtain that u ∈ Wloc (Ω) satisfies Z Z a(x) φ(|∇u|)∇u∇ϕdx = ϕdx. α Ω Ω u

Besides this, it follows from Lemma 5.2, that α−1−ℓ ℓ

un

⇀ v in W01,ℓ (Ω),

that is, u(α−1−ℓ)/ℓ ∈ W01,ℓ (Ω) as well. This ends our proof. Below, we take advantage of the former arguments to show existence of solutions to Problem 2.4. The greatest effort is done to show L∞ -regularity of its solutions.

18

6.2

Convex Singular Problem - Regularity of Solutions

Proof of Theorem 2.2: Since 0 < γ < ℓ − 1 and 0 ≤ a ∈ Lq (Ω) for some q > ℓ/(ℓ − γ − 1), it follows by arguments similar to those used in the proof of Theorem 2.1 that there exist both a sequence of aproximating solutions still denoted by (un ) and a corresponding solution u ∈ W01,Φ (Ω) to problem (2.4) such that u ≥ Cd in Ω for some constant C > 0. Claim. u ∈ L∞ (Ω). The proof of this Claim uses arguments driven by a Moser Iteration Scheme. Parts of our argument were motivated by reading [19]. However our proof in the present paper is selfcontained. In order to show the Claim, set β1 := (ℓ + α − 1)q ′ > 0, βk∗ := βk + β1 , βk+1 :=

ℓ∗ ∗ β , δ := ℓ∗ /(q ′ ℓ), ℓq ′ k

where 1/q ′ + 1/q = 1. We point out that δ > 1 because q > N/ℓ. In addition, 2δ k − δ k−1 − 1 β1 , βk∗ = 2δ k−1 + δ k−2 + ... + 1 β1 = δ−1 βk =

(6.9)

2δ k − δ k−1 − δ β1 , δ−1

(6.10)

and since δ > 1, βk ր ∞. β /(q ′ +α)

Now, taking k0 such that βk0 , βk0 + q ′ (α − 1) > 1, we have that unk each k ≥ k0 and using it in (4.13), we obtain βk q′

Z

βk q′

φ(|∇un |)|∇un|2 un

+α−1

Ω

is a test function for

βk Z ′ +α βk +α+γ an unq q′ dx ≤ dx + bun α Ω (un + 1/n) Z βk βk ′ +α+γ q′ ≤ aun + bunq dx

(6.11)

Ω

βk ′

βk ′

kq . ≤ kakq kun kβqk + kbk∞ kun kβqk kuα+γ n We claim that kuα+γ kq is bounded. Indeed, n if (α + γ)q ≤ 1, it follows that α ≤ 1, because q > N/ℓ > 1. In this case, it follows from Corollary 4.1 that un is bounded in W01,Φ (Ω). In particular, there exists θ0 ∈ L1 (Ω) such that un ≤ θ0 , that is, 1 kuα+γ kq ≤ (|Ω| + kθ0 k1 ) q . n If (α + γ)q > 1 we distinguish between the two cases: α > 1 and α ≤ 1. In the case α > 1, we find by using that ((un + 1/n)(ℓ+α−1)/ℓ ) is bounded in W 1,ℓ (Ω) and ∗ W 1,ℓ (Ω) ֒→ Lℓ (Ω) that ℓ1∗ Z ∗ (α−1) ℓ∗ +(α−1) ℓℓ 1+ ℓ dx = kun(ℓ+α−1)/ℓ )kℓ∗ ≤ C, un kun kℓ∗ +(α−1) ℓ∗ = ℓ

Ω

19 that is, by using our assumption q ≤ q(α + γ), it follows from its definition (see (2.3) for this) that (α + γ)q ≤ ℓ∗ + (α − 1)ℓ∗ /ℓ. So, kun k(α+γ)q ≤ C, ∗

(6.12)

∗

because Lℓ +(α−1)ℓ /ℓ (Ω) ֒→ L(α+γ)q (Ω). In the case α ≤ 1, again we have that un is bounded in W01,Φ (Ω). W01,Φ (Ω) ֒→ L(γ+α)q (Ω), see (2.3) again, that

So, it follows from

α+γ kuα+γ kq = kun kα+γ ≤ C, n (α+γ)q ≤ κkun k

for some κ, C > 0. Thus, in both cases, it follows from (6.11) and the estimates just above that there exists a constant c0 > 0 such that Z βk βk βk q′ 2 q′ +α−1 (6.13) dx ≤ (kak + kbk c )ku k φ(|∇u |)|∇u | u n q ∞ 0 n βk . n n q′ Ω On the other hand, it follows by Lemma 7.1 that Z Z βk βk βk ℓΦ(1) 2 q′ +α−1 ℓ q′ +α−1 dx ≥ φ(|∇u |)|∇u | u β |∇u | u n n n n k n q′ Ω q′ |∇un |≥1

(6.14)

and so it follows from (6.13) and (6.14), that Z Z βk βk ℓΦ(1) ℓΦ(1) ℓ q′ +α−1 ℓ q′ +α−1 β |∇u | u β |∇u | u dx ≤ dx n n k n k n q′ q′ Ω |∇un | 0. To do this, we are going to consider two cases again: α ≤ 1 and α > 1. βk

+α−1

If α ≤ 1 notice that Lβk (Ω) ֒→ L q′ (Ω). Hence Z βk ′ +α−1 1− 1 + 1−α β /q ′ +α−1 β /q ′ unq dx = kun kβkk /q′ +α−1 ≤ |Ω| q′ βk kun kβkk kun kα−1 βk .

(6.17)

Ω

On the other hand, since u1 ≤ un , we have ku1kβk ≤ kun kβk ,

(6.18)

20 and by the embedding Lβk (Ω) ֒→ L1 (Ω) we get ku1 k1 ≤ |Ω|

1− β1

k

ku1 kβk .

(6.19)

Combining (6.18) and (6.19) we have kun kα−1 βk ≤ |Ω|

(1−α)(1− β1 ) k

ku1kα−1 . 1

(6.20)

So by (6.17) and (6.20) we infer that Z βk ′ +α−1 2−α− q1′ β /q ′ unq dx ≤ |Ω| ku1 kα−1 kun kβkk . 1

(6.21)

Ω

Now by applying (6.21) in (6.15), we get ℓΦ(1) βk q′

Z

ℓ

βk q′

|∇un | un Ω

+α−1

βk ℓΦ(1) 2−α− q1′ ′ α−1 |Ω| ku1k1 βk kun kβqk dx ≤ ′ q βk q′

+ (kakq + kbk∞ c0 )kun kβk .

(6.22)

Let α > 1. By Hölder Inequality, (α − 1)q < (α + γ)q and (6.12), we have Z q1 Z βk βk +α−1 q′ q′ (α−1)q un dx dx ≤ kun kβk un Ω

βk q′

≤ kun kβk

βk q′

≤ kun kβk

Ω

|Ω| +

|Ω| + 1

Z

[un ≥1]

(α+γ)q kun k(α+γ)q βk ′

q1

un(α−1)q dx q1

≤ (|Ω| + C) q kun kβqk .

(6.23)

Now by applying (6.23) in (6.15), we get Z βk βk 1 ℓΦ(1) ℓΦ(1) ′ ℓ q′ +α−1 q βk |∇un | un βk (|Ω| + C) kun kβqk dx ≤ ′ ′ q q Ω βk q′

+ (kakq + kbk∞ c0 )kun kβk .

(6.24)

So, it follows from (6.22) (the case α ≤ 1) and (6.24) (the case α > 1) that the inequality (6.16) is true for B > 0 defined by  ′ 2−α− q1′ ℓΦ(1) α−1  q ku1 k1 + kakq + kbk∞ c0 , 0 < α ≤ 1, ′ |Ω| ℓΦ(1) q B := 1 ′ ℓΦ(1)  q q + kak + kbk c , α > 1. (|Ω| + C) q ∞ 0 ′ ℓΦ(1) q This shows the inequality (6.16). Now since ℓ Z βk +β1 ℓ Z βk +q ′ (α−1) ℓq ′ ′ ℓ ∇ un ℓq′ dx = dx, |∇un | un q βk + β1 Ω Ω

21 it follows from (6.16) and W01,ℓ (Ω) ֒→ Lℓ (Ω), that

∗ ℓ ∗ ℓ ∗ βk βk

βk βk q′ q′ ℓ ′

ℓq kun kβk+1 = u ≤ µ B ku k n βk , ′ ℓq ℓ∗ ∗

(6.25)

for some µ > 0. Set Fk+1 := βk+1 ln(kun kβk+1 ). So, it follows from the last inequality, that ∗ βk+1 q ′ βk βk Fk+1 ≤ + ln B + ′ ln(kun kβk ) ℓ ln µ + ℓ ln βk∗ ℓq ′ q ∗ ℓ ≤ ℓ∗ ln (µBβk∗ ) + ′ Fk qℓ = λk + δFk ,

(6.26)

where λk := ℓ∗ ln (µBβk∗) . Now, by using (6.9) and (6.10), we can infer that

λk = b + ℓ∗ ln 2δ k−1 + δ k−2 + ... + 1 ,

where b := ℓ∗ ln(µBβ1 ), that is,

Fk ≤ δ k−1 F1 + λk−1 + δλk−2 + ... + δ k−2 λ1 . So, δ k−1 F1 + λk−1 + δλk−2 + ... + δ k−2 λ1 Fk ≤ 2δk −δk−1 −δ βk β1 δ−1 F1 +

= Since

λk−1 k−2 + λδk−2 + ... δk−1 k−1 2δ−1−1/δ β1 δ−1

+

λ1 δ

.

(6.27)

n λn b ℓ∗ 2δ − δ n−1 − 1 = n + n ln δn δ δ δ−1 n ∗ ℓ 2δ b , ≤ n + n ln δ δ δ−1

it follows from (6.27), that Fk ≤ βk ≤ ≤ ≤

F1 + b

F1 +

1 δk−1

b δ−1

k−1 1 + ... 1δ + ℓ∗ δk−1 ln 2δδ−1 + ... + 1δ ln

+ℓ

∗

2δ−1−1/δk−1 β1 δ−1

1 δk−1

F1 +

b δ−1

+ ℓ∗ ln

F1 +

b δ−1

+ ℓ∗

ln

2δk−1 δ−1

1 δ

+ ... + ln

2δ−1−1/δk−1 β1 δ−1 2 1 + ... 1δ δ−1 δk−1 2δ−1−1/δk−1 β1 δ−1

1 2 ln δ−1 + δ−1 k−1 2δ−1−1/δ β1 δ−1

ln δ

k−1 δk−1

+ ln δ

P∞

n n=1 δn

2δ δ−1

2δ δ−1

+ ... 1δ

→ d0 .

(6.28)

22 Now, going back to the definition of Fk , we obtain Fk

|un (x)| ≤ kun k∞ = lim sup kun kβk ≤ lim sup e βk ≤ ed0 for all x ∈ Ω, k→∞

k→∞

and |u(x)| = lim |un (x)| ≤ ed0 a.e x ∈ Ω, n→∞

because un (x) → u(x) a.e. in Ω, that is, u ∈ L∞ (Ω). This ends our proof. Proof of Corollary 1.1: First, let u ∈ W01,Φ (Ω) be a solutionpof (1.1). Take ϕ ∈ W01,Φ (Ω), and ϕn ∈ C0∞ (Ω) such that ϕn → ϕ in W01,Φ (Ω). So, by taking θ ǫθ + |ϕn − ϕk |θ − ǫ, for some θ ∈ N, as a test function, we obtain Z a(x) p θ ǫθ + |ϕn − ϕk |θ − ǫ dx (6.29) 0 ≤ α Ω u Z |ϕn − ϕk |θ−1 ≤ φ(|∇u|)|∇u| (θ−1)/θ |∇ϕn − ∇ϕk |dx Ω ǫθ + |ϕn − ϕk |θ ≤ kφ(|∇u|)|∇u|kLΦ˜ (Ω) k∇ϕn − ϕk kLΦ (Ω) , for every ǫ > 0 given. Making ǫ → 0, we find that aϕ n is a Cauchy sequence in L1 (Ω), α u so that (aϕn )/uα → υ ∈ L1 (Ω). Since, ϕn (x) → ϕ(x) a.e. in Ω, we have that υ = (aϕ)/uα . By hypothesis, u ∈ W01,Φ (Ω) satisfies (see (2.2)) Z Z a(x) φ(|∇u|)∇u∇ϕndx = ϕn dx, α Ω Ω u and, passing to the limit, it follows that Z Z a(x) φ(|∇u|)∇u∇ϕdx = ϕdx for all ϕ ∈ W01,Φ (Ω). α Ω Ω u

(6.30)

To complete the proof of the uniqueness, let v ∈ W01,Φ (Ω) be another solution of (1.1). Now assuming that u, v ∈ W01,Φ (Ω), u 6= v satisfy (6.30), setting ϕ = u − v and using the fact that ∆Φ is stricly monotone, we obtain Z 0 ≤ (φ(|∇u|)∇u − φ(|∇v|)∇v)(∇u − ∇v)dx (6.31) Ω Z 1 1 = a(x) α − α (u − v)dx < 0, u v Ω impossible. Now, we proceed to the regularity. First (i). In this case, we have an = a for n large enough. So, as a consequence of the Comparison Principle, like at the end of the proof in Lemma 5.1, that un+1 ≥ un . Besides this, if we assume that n 1 1o Ω0 := x ∈ Ω / un+1(x) + ⊂⊂ Ω, > un (x) + n+1 n

23 is not empty, then we would obtain −∆Φ (un+1 + 1/(n + 1)) ≤ −∆Φ (un + 1/n) in Ω0 , that is, 1 un+1 (x) + n+1 ≤ un (x) + n1 in Ω0 . This is impossible. So, we have 0 ≤ un − uk ≤

1 1 − in Ω. k n

Since (un ) ⊂ C 1 (Ω), we obtain that un converges uniformilly to u, that is, u ∈ C(Ω). Proof of (ii). It just follows from the same arguments that we used to proof Theorem 2.2, by taking b = 0. This ends our proof.

7

Appendix - On Orlicz-Sobolev spaces

In this section we present for, the reader’s convenience, several results/notation used in the paper. The reader is referred to [1, 32] regarding basics on Orlicz-Sobolev spaces. The usual norm on LΦ (Ω) is, (Luxemburg norm), Z u(x) kukΦ = inf λ > 0 | Φ dx ≤ 1 , λ Ω while the Orlicz-Sobolev norm of W 1,Φ (Ω) is kuk1,Φ

N X

∂u

= kukΦ +

∂xi . i=1

Φ

We denote by W01,Φ (Ω) the closure of C0∞ (Ω) with respect to the Orlicz-Sobolev norm of W 1,Φ (Ω). It remind that e = max{ts − Φ(s)}, t ≥ 0. Φ(t) s≥0

e are N-functions satisfying the ∆2 -condition, (cf. [32, p 22]). In It turns out that Φ and Φ 1,Φ addition, LΦ (Ω) and W (Ω) are reflexive and Banach spaces. Remark 7.1 It is well known that (φ3 ) implies that the condition

(φ3 )

′

φ(t)t2 ℓ≤ ≤ m, t > 0, Φ(t)

e ∈ ∆2 . is verified. Furthermore, under this condition, Φ, Φ

By the Poincaré Inequality (see e.g. [17]), that is, the inequality Z Z Φ(u)dx ≤ Φ(2dΩ |∇u|)dx, Ω

Ω

where dΩ = diam(Ω), it follows that kukΦ ≤ 2dΩ k∇ukΦ for all u ∈ W01,Φ (Ω).

24 As a consequence of this, we have that kuk := k∇ukΦ defines a norm in W01,Φ (Ω) that is equivalent to k · k1,Φ . Let Φ∗ be the inverse of the function Z t −1 Φ (s) t ∈ (0, ∞) 7→ N+1 ds s N 0 which can be extended to R by Φ∗ (t) = Φ∗ (−t) for t ≤ 0. We say that an N-function Ψ grows essentially more slowly (grows more slowly) than Υ, denoted by Ψ 0 t→∞ Φ∗ (t) lim

(Ψ(t) ≤ Υ(kt) for all t ≥ t0 for some k, t0 > 0). The imbeddings below (cf. [1]) were used in this paper. First, we have cpt

W01,Φ (Ω) ֒→ LΨ (Ω) if Φ < Ψ 0, e Φ(t)

e e e ζ2 (t)Φ(ρ) ≤ Φ(ρt) ≤ ζ3 (t)Φ(ρ), ρ, t > 0, Z e ζ2 (kukΦe ) ≤ Φ(u)dx ≤ ζ3 (kukΦe ), u ∈ LΦe (Ω). Ω

25 Lemma 7.3 Let Let Φ be an N-function satisfying ∆2 . Let (un ) ⊂ LΦ (Ω) be a sequence such that un → u in LΦ (Ω). Then there is a subsequence (unk ) ⊆ (un ) such that (i) unk (x) → u(x) a.e. x ∈ Ω, (ii) there is h ∈ LΦ (Ω) such that |unk | ≤ h a.e. in Ω. R Proof:(Sketch) We have that Ω Φ(un − u)dx → 0. By [1] LΦ (Ω) ֒→ L1 (Ω). So there are a subsequence, we keep the notation, and e h ∈ L1 (Ω) such that un → u a.e. in Ω and Φ(un − u) ≤ e h a.e. in Ω. Since Φ is convex, increasing and satisfies ∆2 , we have |un − u| + |u| C Φ(|un |) ≤ CΦ ≤ [Φ(|un − u|) + Φ(|u|)] 2 2 C e [h + Φ(|u|)], ≤ 2

that is,

−1

|un | ≤ Φ

C e (h + Φ(|u|)) 2

:= h ∈ LΦ (Ω),

because e h ∈ L1 (Ω), Φ(|u|) ∈ L1 (Ω), and Z Z K e −1 Φ(h)dx = Φ Φ (h + Φ(|u|)) dx 2 Ω Ω =

Z Ω

showing that h ∈ LΦ (Ω).

K e (h + Φ(|u|)) dx < ∞, 2

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28 Marcos L. M. Carvalho Jos´ e V.A. Gon¸calves Universidade Federal de Goi´ as Instituto de Matem´ atica e Estat´ıstica 74001-970 Goiˆ ania, GO - Brasil emails: [email protected] [email protected] Carlos Alberto Santos Universidade de Bras´ılia Departamento de Matem´ atica Bras´ılia, DF 70910–900, Brazil email: [email protected]