Electronic Journal of Qualitative Theory of Differential Equations2019, No. 54, 1–17; https://doi.org/10.14232/ejqtde.2019.1.54 www.math.u-szeged.hu/ejqtde/

Positive weak solutions of elliptic Dirichlet problemswith singularities in both the dependent and the

independent variables

Tomas GodoyB and Alfredo Guerin

FaMAF, Universidad Nacional de Cordoba, Ciudad Universitaria, Cordoba, 5000, Argentina

Received 29 April 2019, appeared 6 August 2019

Communicated by Maria Alessandra Ragusa

Abstract. We consider singular problems of the form −∆u = k (·, u) − h (·, u) in Ω,u = 0 on ∂Ω, u > 0 in Ω, where Ω is a bounded C1,1 domain in Rn, n ≥ 2, h :Ω × [0, ∞) → [0, ∞) and k : Ω × (0, ∞) → [0, ∞) are Carathéodory functions suchthat h (x, ·) is nondecreasing, and k (x, ·) is nonincreasing and singular at the origin a.e.x ∈ Ω. Additionally, k (·, s) and h (·, s) are allowed to be singular on ∂Ω for s > 0.Under suitable additional hypothesis on h and k, we prove that the stated problem hasa unique weak solution u ∈ H1

0 (Ω), and that u belongs to C(Ω). The behavior of the

solution near ∂Ω is also addressed.

Keywords: singular elliptic problems, variational problems, sub-supersolutionsmethod, finite energy solutions, positive solutions.

2010 Mathematics Subject Classification: 35J75, 35D30, 35J20.

1 Introduction and statement of the main results

Let Ω be a bounded domain in Rn with C1,1 boundary, and consider a singular semilinearelliptic problem of the form

−∆u = k (·, u) in Ω,

u = 0 on ∂Ω,

u > 0 in Ω,

(1.1)

where k : Ω× (0, ∞) → [0, ∞) is a Carathéodory function (i.e., k (·, s) is measurable for anys ∈ (0, ∞) and k (x, ·) is continuous on (0, ∞) a.e. x ∈ Ω), with k = k (x, s) allowed to besingular at s = 0.

Singular problems like (1.1) arise, for instance, in the study of chemical catalysts process,non-Newtonian fluids, the temperature of some electrical conductors whose resistance de-pends on the temperature, thin films, and micro electro-mechanical systems (see e.g., [4, 7, 13,15–17, 26, 33–35], and the references therein).

BCorresponding author. Email: godoy@mate.uncor.edu

2 T. Godoy and A. Guerin

Problem (1.1) was studied, in the case where k (x, s) = a (x) s−α, under different sets ofassumptions on a and α, in [2, 8, 11, 13, 16, 23, 36].

In [14] existence and nonexistence theorems were stated for Lane–Emden–Fowler equa-tions with convection and singular potential.

Recently, Chu, Gao and Gao [6], studied problems of the form −div (M (x)∇u) =

a (x) u−α(x) in Ω, u = 0 on ∂Ω, where a belongs to a suitable Lebesgue space. Amongother results, they found a very weak solution in H1

0 (Ω) (with test functions in C1c (Ω)) when

0 < α < 2 and 0 < α ∈ C(Ω)

.Problems of the form (1.1), with k = k (x, s) singular at s = 0, and with k (·, s) allowed to

exhibit some kind of singularity on ∂Ω, were studied in [1, 24, 28, 30, 31, 37, 38].Diaz, Hernandez and Rakotoson [12] considered the problem

−∆u = ad−γΩ u−β in Ω,

u = 0 on ∂Ω,

u > 0 in Ω,

(1.2)

where dΩ := dist (·, ∂Ω) , γ < 2, and a ∈ L∞ (Ω) satisfies infΩ a > 0. They studied theexistence of solutions u ∈ L1 (Ω, dΩ) (the dΩ-weighted Lebesgue space) in the following veryweak sense:

−∫

Ωu∆ϕ =

∫Ω

ad−γΩ ϕ for any ϕ ∈ C2 (Ω) such that ϕ = 0 on ∂Ω. (1.3)

Notice that the space of test functions involved in their notion of solution is strictly smallerthan the corresponding space in the present paper (as given in Definition 1.1 below). In Theo-rem 2 they find, when β + γ < 1, a very weak solution of problem (1.2), and prove that it be-longs to W1

0 (Ω, ‖ · ‖N(r),∞)∩W2,qloc (Ω) for any r ∈ (0, 1) and q ∈ [1, ∞), where W1

0 (Ω, ‖ · ‖N(r),∞)

is the space of the functions w : Ω → R such that w ∈ W1,1 (Ω) and |∇w| belongs to theLorentz space LN(r),∞ (Ω) , with N (r) := n

n−1+r .Regarding the case β + γ > 1, in theorem 1 they find a very weak solution of problem

(1.2) that belongs to C(Ω)∩W2,q

loc (Ω) for any q ∈ [1, ∞) ; and in Theorem 5, they prove that,when β + γ > 1 and γ < 2, the solution that they found belongs to H1

0 (Ω) if, and only if,β + 2γ < 3. Additionally, in Theorem 4, they prove that, when β + γ > 1, there exist positive

constants c1 and c2 such that the found solution u satisfies c1d2−γ1+β

Ω ≤ u ≤ c1d2−γ1+β

Ω in Ω. However,it is not obvious that, if u ∈ H1

0 (Ω) , then u is a weak solution of problem (1.2), i.e., that (1.3)holds for any ϕ ∈ H1

0 (Ω) .The existence of classical solutions u ∈ C2 (Ω) ∩ C

(Ω)

of problem (1.2) was addressed byMâagli [27] in the case when a ∈ Cσ

loc (Ω) for some σ ∈ (0, 1) , and dγΩa belongs to a suitable

class related to the notion of Karamata classes. Our results heavily depend on those found in[27], which are summarized in Remark 2.5 below.

The interested reader can find an updated panoramic view of the area in the researchbooks [19], [32], and in the survey article [18].

In this work we consider problem (1.1) when k : Ω× (0, ∞) → [0, ∞) is a Carathéodoryfunction, k = k (x, s) is allowed to be singular at s = 0, in the sense that

lims→0+

k (·, s) = ∞ a.e. in Ω,

and k (·, s) is allowed to be singular on ∂Ω, in the sense that

limΩ3x→y

k (x, s) = ∞ for any (y, s) ∈ ∂Ω× (0, ∞) .

Positive solutions to singular elliptic problems 3

Additionally, we allow the introduction of a second term, and consider the problem−∆u = k (·, u)− h (·, u) in Ω,

u = 0 on ∂Ω,

u > 0 in Ω,

(1.4)

where h = h (·, s) is allowed to be singular on ∂Ω for any s > 0. Under some further assump-tions on k and h, we prove existence and uniqueness results for weak solutions of problems(1.1) and (1.4).

The notion of weak solution that we use in this work is the usual one, given by the follow-ing definition.

Definition 1.1. Let ψ : Ω → R be a measurable function such that ψϕ ∈ L1 (Ω) for all ϕ inH1

0 (Ω). We say that u : Ω→ R is a weak solution of the problem

− ∆u = ψ in Ω, u = 0 on ∂Ω, (1.5)

if u ∈ H10 (Ω) and

∫Ω 〈∇u,∇ϕ〉 =

∫Ω ψϕ for all ϕ in H1

0 (Ω).Similarly, for u ∈ H1

0 (Ω) , we say that u is a weak supersolution (respectively a weaksubsolution) of problem (1.5), and we write

−∆u ≥ ψ in Ω (resp. − ∆u ≤ ψ in Ω ),

u = 0 on ∂Ω,

to mean∫

Ω 〈∇u,∇ϕ〉 ≥∫

Ω ψϕ (resp.∫

Ω 〈∇u,∇ϕ〉 ≤∫

Ω ψϕ) for all nonnegative ϕ in H10 (Ω).

Definition 1.2. Let dΩ : Ω → R be the distance function dΩ := dist (·, ∂Ω). For β ≥ 0 andγ ∈ [0, 2) define ϑβ,γ : Ω→ R by:

• ϑβ,γ := dΩ if β + γ < 1,

• ϑβ,γ := dΩ ln(ω0d−1Ω ) if β + γ = 1, where ω0 is an arbitrary number, which we fix from

now on, such that ω0 > diam (Ω) ,

• ϑβ,γ := d2−γ1+β

Ω if β + γ > 1.

We assume, from now on, n ≥ 2. Let us state our main results.

Theorem 1.3. Let Ω be a bounded domain in Rn with C1,1 boundary, and let k : Ω× (0, ∞) → R

satisfy the following conditions:

k1) k is a nonnegative Carathéodory function;

k2) s→ k (·, s) is nonincreasing on (0, ∞) a.e. x ∈ Ω;

k3) there exist β ≥ 0, γ ≥ 0, and B2 > 0 such that, for any s > 0, k (·, s) ≤ B2d−γΩ s−β a.e. in Ω;

k4) there exist δ > 0 and B1 > 0 such that, for any s ∈ (0, δ) , k (·, s) ≥ B1d−γΩ s−β a.e. in Ω.

Let h : Ω× [0, ∞)→ R satisfy the following conditions:

h1) h is a Carathéodory function;

4 T. Godoy and A. Guerin

h2) h (x, ·) is nondecreasing on [0, ∞) , and h (·, 0) = 0 a.e. in Ω;

h3) h (·, s) ≤ B3d−ηΩ sp a.e. in Ω for all s ∈ [0, ∞) , with B3 > 0, p > 1, 0 < η < γ + p + β if

β + γ ≤ 1, and 0 < η < γ + (p + β) 2−γ1+β if β + γ > 1.

Then:

i) If β + 2γ < 3, then problem (1.4) has a unique weak solution u ∈ H10 (Ω). Moreover, u ∈

W2,qloc (Ω) ∩ C

(Ω)

for any q ∈ [1, ∞) , and u satisfies cϑβ,γ ≤ u ≤ c′ϑβ,γ in Ω, for some positiveconstants c and c′.

ii) If problem (1.4) has a weak solution u ∈ H10 (Ω) ∩ C

(Ω)

, then β + 2γ < 3.

Note that, in particular, Theorem 1.3 says that −∆u = d−γΩ in Ω, u = 0 on ∂Ω, u > 0 in Ω,

has a weak solution if, and only if, γ < 32 .

The next theorem states that, when h is identically zero, the assertion i) of Theorem 1.3remains valid if the condition k4) is replaced by the following milder condition:

k5) there exist δ > 0 and a measurable set E ⊂ Ω such that |E| > 0 and infE×(0,δ) k > 0,where inf stands for the essential infimum, and |E| denotes the Lebesgue measure of E.

Theorem 1.4. Let Ω be a bounded domain in Rn with C1,1 boundary, and let k : Ω× (0, ∞) → R

satisfy the conditions k1)–k3) of Theorem 1.3. Assume that β + γ < 32 , and that the condition k5)

holds. Then problem (1.1) has a unique weak solution u ∈ H10 (Ω) , and u ∈ W2,q

loc (Ω) ∩ C(Ω)

forany q ∈ [1, ∞) , and cdΩ ≤ u ≤ c′ϑβ,γ in Ω, for some positive constants c and c′.

Concerning the case when h is nonidentically zero, our next result shows that the assertioni) of theorem 1.3 holds under a weaker condition than k4), at the expense of strengthening h3).

Theorem 1.5. Let Ω be a bounded domain in Rn with C1,1 boundary, and let k : Ω× (0, ∞) → R

satisfy the conditions k1)–k3) of Theorem 1.3, with β and γ satisfying β > 0, γ ∈ [0, 2) , and β + γ <32 . Assume also the following condition:

k6) there exist δ > 0 and B1 > 0 such that, for any s ∈ (0, δ) , k (·, s) ≥ B1s−β a.e. in Ω.

Let h : Ω× [0, ∞)→ R satisfy the conditions h1) and h2) of Theorem 1.3, and the following

h4) h (·, s) ≤ B3d−ηΩ sp a.e. in Ω for all s ∈ [0, ∞) , with B3 > 0, p > 1, 0 < η < p− 1 if β+ γ ≤ 1

and 0 < η < (p− 1) 2−γ1+β if 1 < β + γ < 3

2 .

Then problem (1.4) has a unique weak solution u. Moreover, u ∈W2,qloc (Ω)∩C

(Ω)

for any q ∈ [1, ∞) ,and there exist positive constants c and c′ such that c′ϑβ,0 ≤ u ≤ cϑβ,γ in Ω.

Remark 1.6. Let us stress that the strength of the singularity, which is the theme in the back-ground of the present work, needs to be limited if one expects weak solutions in H1

0 (Ω).Indeed, Lazer and McKenna [23] considered the problem −∆u = au−α in Ω, u = 0 on ∂Ω,u > 0 in Ω, under the assumptions a ∈ Cγ

(Ω)

, minΩ a > 0, α > 0, and Ω a bounded reg-ular domain. They proved that there exists a unique solution u ∈ C2 (Ω) ∩ C

(Ω)

; and thatu ∈ H1

0 (Ω) if, and only if, α < 3. A clear-cut simple condition like that is elusive when theright hand side of the equation is not in the form au−α; in [21] we addressed such a moregeneral situation, but still did not consider the case when a spatial singularity is added. Thislatter situation is considered in the present work.

Positive solutions to singular elliptic problems 5

The paper is organized as follows: in Section 2, we collect some preliminary results.Lemma 2.4 is an adaptation of Lemma 3.2 in [22] and states that, under suitable conditions, asolution in the sense of distributions of an elliptic problem, is also a weak solution in H1

0 (Ω).Remark 2.5 recalls a result, due to Mâagli [27], about existence, uniqueness, and behavior nearthe boundary, of positive classical solutions of problems of the form −∆u = ad−γ

Ω u−β in Ω,u = 0 on ∂Ω (for a suitable class of Hölder continuous functions a); and Remark 2.7 recalls asub-supersolution theorem for singular problems due to Loc and Schmitt [25]. In Section 3 weprove Theorems 1.3, 1.4 and 1.5, by combining the results of [27], [25], and Lemma 2.4, jointlywith some additional auxiliary results.

2 Preliminaries

For w ∈ L1loc (Ω) , we write, as usual, w ∈

(H1

0 (Ω))′ to mean that wϕ ∈ L1 (Ω) for any

ϕ ∈ H10 (Ω) , and that the map ϕ→

∫Ω wϕ is continuous on H1

0 (Ω) .

Remark 2.1. Let us recall the Hardy inequality (as stated, e.g., in [29, Theorem 1.10.15], seealso [3, p. 313]): There exists a positive constant c such that

∥∥ ϕdΩ

∥∥L2(Ω)

≤ c ‖∇ϕ‖L2(Ω) for all

ϕ ∈ H10 (Ω) .

Remark 2.2. If ψ ∈ L1loc (Ω) and dΩψ ∈ L2 (Ω) , then ψϕ ∈ L1 (Ω) for any ϕ ∈ H1

0 (Ω).Moreover, ψ ∈

(H1

0 (Ω))′ and ‖ψ‖(H1

0 (Ω))′ ≤ c ‖dΩψ‖2 with c independent of ψ. Indeed, for

ϕ ∈ H10 (Ω) , from the Hölder and the Hardy inequalities,

∫Ω |ψϕ| ≤ ‖dΩψ‖2 ‖d

−1Ω ϕ‖2 ≤

c‖dΩψ‖2‖ϕ‖H10 (Ω), where c is the constant in the Hardy inequality of Remark 2.1.

Remark 2.3. (See e.g., [10]) λ ∈ R is called a principal eigenvalue for −∆ in Ω, with homoge-neous Dirichlet condition and weight function b ∈ L∞ (Ω) , if the problem −∆φ = λbφ in Ω,φ = 0 on ∂Ω has a solution ϕ1 (called a principal eigenfunction) such that ϕ1 > 0 in Ω. It is awell known fact that, for any C1,1 bounded domain Ω ⊂ Rn, b ∈ L∞ (Ω) , and b+ 6≡ 0, thereexists a unique positive principal eigenvalue λ1 (b) , and its eigenspace Vλ1 is a one dimen-sional subspace of C1 (Ω). Moreover, for each positive ϕ1 ∈ Vλ1 , there are positive constantsc1, c2 such that c1dΩ ≤ ϕ1 ≤ c2dΩ in Ω. Consequently, |ln (ϕ1)| ∈ L1 (Ω) ; and ϕt

1 ∈ L1 (Ω) if,and only if, t > −1.

The following lemma is an adaptation of Lemma 3.2 in [22]

Lemma 2.4. Let ψ ∈ L∞loc (Ω) be such that |ψ| ∈

(H1

0 (Ω))′ , and let u ∈ W1,2

loc (Ω) ∩ C (Ω) be asolution, in the sense of distributions, of the problem

−∆u = ψ in Ω.

If there exist constants c > 0 and r > 12 such that 0 ≤ u ≤ cdr

Ω in Ω, then u ∈ H10 (Ω) ∩ C1 (Ω) ∩

C(Ω)

, and u is a weak solution of −∆u = ψ in Ω, u = 0 on ∂Ω.

Proof. Let ϕ ∈ H10 (Ω) such that supp (ϕ) ⊂ Ω. Then

∫Ω 〈∇u,∇ϕ〉 =

∫Ω ψϕ. Indeed, let δ > 0

be such that supp (ϕ) ⊂ Ωδ, and let

ϕj

j∈Nbe a sequence in C∞

c (Ω) satisfying supp(

ϕj)⊂

Ωδ for all j, and such that

ϕj

j∈Nconverges to ϕ in H1

0 (Ωδ). Now, ∇u|Ωδ∈ L2 (Ωδ, Rn) , and

so ζ →∫

Ωδ〈∇u,∇ζ〉 is continuous on H1

0 (Ωδ). Also,∫

Ω

⟨∇u,∇ϕj

⟩=∫

Ω ψϕj for all j. Then∫Ω 〈∇u,∇ϕ〉 = limj→∞

∫Ω

⟨∇u,∇ϕj

⟩= limj→∞

∫Ω ψϕj =

∫Ω ψϕ.

6 T. Godoy and A. Guerin

For each j ∈N, let µj : R→ R be defined by µj (s) := 0 if s ≤ 1j , µj (s) := −3j2s3 + 14js2−

19s + 8j if 1

j < s < 2j , and µj (s) := s if 2

j ≤ s. Then µj ∈ C1 (R) , µj (s) = 0 for s < 1j , µ′j (s) ≥ 0

for 1j < s < 2

j , and µ′j (s) = 1 for 2j < s. Also, 0 < µj (s) < s for all s ∈

(0, 2

j

).

Let µj (u) := µj u. Then, for all j, ∇(µj (u)

)=(µ′j u

)∇u in D′ (Ω). Since u ∈W1,2

loc (Ω) ,

it follows that µj (u) ∈W1,2loc (Ω). Since supp

(µj (u)

)⊂ Ω, we have µj (u) ∈ H1

0 (Ω). Therefore,for all j,

∫Ω

⟨∇u,∇

(µj (u)

)⟩=∫

Ω ψµj (u), i.e.,∫u>0

(µ′j u

)|∇u|2 =

∫Ω

ψµj (u) . (2.1)

Now,(µ′j u

)|∇u|2 is nonnegative and limj→∞

(µ′j u

)|∇u|2 = |∇u|2 a.e. in Ω, and so, from

(2.1) and Fatou’s lemma, we have∫Ω|∇u|2 ≤ limj→∞

∫Ω

ψµj (u) .

Let ϕ1 be the principal eigenfunction for −∆ in Ω with homogeneous Dirichlet condition,and with weight function 1, normalized by ‖ϕ1‖∞ = 1. Since, for some positive constant c′,u ≤ c′ϕr

1 in Ω, and ϕr1 ∈ H1

0 (Ω) , we have ψu ∈ L1 (Ω). Now, limj→∞ ψµj (u) = ψu in Ω, and,for any j ∈N,

∣∣ψµj (u)∣∣ ≤ |ψu| in Ω. Then, Lebesgue’s dominated convergence theorem gives

limj→∞

∫Ω

ψµj (u) =∫

Ωψu < ∞.

Thus∫

Ω |∇u|2 < ∞, and so u ∈ H1 (Ω). As −∆u = ψ in D′ (Ω), u ∈ L∞ (Ω) and ψ ∈ L∞loc (Ω),

then the inner elliptic estimates (as stated e.g., in [5], Proposition 4.1.2, see also [20], Theorem9.11) give that u ∈ C1 (Ω). From 0 ≤ u ≤ cdr

Ω in Ω, and u ∈ C (Ω) , we conclude thatu ∈ C

(Ω). Since u ∈ H1 (Ω) , u ∈ C

(Ω)

, and u = 0 on ∂Ω, we get u ∈ H10 (Ω) .

As u ∈ H10 (Ω) , we have that ϕ →

∫Ω 〈∇u,∇ϕ〉 is continuous on H1

0 (Ω). Therefore, sinceC∞

c (Ω) is dense in H10 (Ω) , and since, for any ϕ ∈ C∞

c (Ω) ,∫Ω〈∇u,∇ϕ〉 =

∫Ω

ψϕ, (2.2)

we conclude that (2.2) holds for all ϕ ∈ H10 (Ω) .

Remark 2.5. i) Let ω0 be as in Definition 1.2, α < 1, and ρ < 2. Let z ∈ C ([0, ω0]) be such thatz (0) = 0 and

∫ ω00 t1−ρLz (t) dt < ∞, where Lz (t) := exp

( ∫ ω0t

z(s)s ds

). Let σ ∈ (0, 1) , and let

a ∈ Cσloc (Ω) satisfy, for some constant c > 0,

1c

Lz dΩ ≤ dρΩa ≤ cLz dΩ in Ω. (2.3)

Then, Theorem 1 in [27] says that the problem−∆u = auα in Ω,

u = 0 on ∂Ω,

u > 0 in Ω

(2.4)

has a unique classical solution u ∈ C2 (Ω) ∩ C(Ω); and that, for some positive constant c′, u

satisfies, (c′)−1

θρ dΩ ≤ u ≤ c′θρ dΩ in Ω,

Positive solutions to singular elliptic problems 7

where

θρ (t) :=(∫ ω0

0

Lz (s)s

ds) 1

1−α

if ρ = 2,

θρ (t) := t2−ρ1−α (Lz (t))

11−α if 1 + α < ρ < 2,

θρ (t) := t(∫ ω0

t

Lz (s)s

ds) 1

1−α

if ρ = 1 + α,

and

θρ (t) := t i f ρ < 1 + α.

ii) Let β ≥ 0, and let γ < 2. If in i) we take α := −β, z := 0 (then Lz = 1), and ρ := γ, we getthat the problem

−∆v = d−γΩ v−β in Ω,

v = 0 on ∂Ω,

v > 0 in Ω

(2.5)

has a unique classical solution vβ,γ ∈ C2 (Ω) ∩ C(Ω)

; and that there exists positive constantsc1 and c2 such that

c1ϑβ,γ ≤ vβ,γ ≤ c2ϑβ,γ in Ω, (2.6)

where ϑβ,γ is as in Definition 1.2.

Remark 2.6. Let β ≥ 0 and γ ≥ 0 be such that β + 2γ < 3.

i) d1−γΩ ϑ

−ββ,γ ∈ L2 (Ω). Indeed, if β + γ < 1 then d1−γ

Ω ϑ−ββ,γ = d1−β−γ

Ω ∈ L∞ (Ω) ⊂ L2 (Ω). If

β + γ = 1, then d1−γΩ ϑ

−ββ,γ =

(ln(ω0

dΩ

))β ∈ L∞ (Ω) ⊂ L2 (Ω). If β + γ > 1, then d1−γΩ ϑ

−ββ,γ =

d1−γ−β 2−γ

1+β

Ω = d− 1

β+1 (β+γ−1)Ω and, since β + 2γ < 3, − 2

β+1 (β + γ− 1) > −1, and so, again in

this case, d1−γΩ ϑ

−ββ,γ ∈ L2 (Ω) (because, for r ∈ R, dr

Ω ∈ L2 (Ω) whenever 2r + 1 > 0).

ii) There exist positive constants c and τ > 12 such that ϑβ,γ ≤ cdτ

Ω in Ω. Indeed, if β + γ < 1

then ϑβ,γ = dΩ, if β + γ = 1 then ϑβ,γ = dΩ ln(ω0

dΩ

)and so, for any ε > 0, ϑβ,γ

(d1−ε

Ω

)−1=

dεΩ ln

(ω0dΩ

)∈ L∞ (Ω) ; and if β + γ > 1, then ϑβ,γ = d

2−γ1+β

Ω and, since β + 2γ < 3, 2−γ1+β > 1

2 .

iii) Let vβ,γ be the solution of problem (2.5) given by Remark 2.5. From i) and (2.6), it follows

that d1−γΩ v−β

β,γ ∈ L2 (Ω) .

iv) Let vβ,γ be as in iii). Then, by iii) and Remark 2.2, d−γΩ v−β

β,γ ∈(

H10 (Ω)

)′ .v) Let vβ,γ be as in iii). Since d−γ

Ω v−ββ,γ ∈

(H1

0 (Ω))′ and since, by Remark 2.5, vβ,γ ∈ C2 (Ω) ∩

C(Ω)

, Lemma 2.4 gives that vβ,γ ∈ H10 (Ω) , and that vβ,γ is a weak solution of problem

(2.5). Moreover, since vβ,γ ∈ L∞ (Ω) and d−γΩ v−β

β,γ ∈ L∞loc (Ω) , the inner elliptic estimates

give vβ,γ ∈W2,qloc (Ω) for any q ∈ [1, ∞) .

Remark 2.7. Let g : Ω× (0, ∞) → R be a Carathéodory function. We say that w ∈ L1loc (Ω) is

a subsolution (supersolution) of the problem

− ∆z = g (·, z) in Ω (2.7)

8 T. Godoy and A. Guerin

in the sense of distributions, if, and only if: w > 0 in Ω, g (·, w) ∈ L1loc (Ω) , and

∫Ω 〈∇w,∇ϕ〉 ≤

(≥)∫

Ω g (·, w) ϕ for all nonnegative ϕ ∈ C∞c (Ω). We say that z ∈ L1

loc (Ω) is a solution, in thesense of distributions, of (2.7) if, and only if, z > 0 a.e. in Ω, and,

∫Ω 〈∇z,∇ϕ〉 =

∫Ω g (·, z) ϕ

for all ϕ ∈ C∞c (Ω) .

According to Theorem 2.4 in [25], if (2.7) has a subsolution z and a supersolution z (in thesense of distributions), both in L∞

loc (Ω) ∩W1,2loc (Ω) , and such such that 0 < z ≤ z in Ω, and if

there exists ψ ∈ L∞loc (Ω) such that |g (x, s) | ≤ ψ (x) a.e. x ∈ Ω for all s ∈ [z (x) , z (x)] ; then

(2.7) has a solution z in the sense of distributions, which satisfies z ≤ z ≤ z in Ω.

3 Proof of the main results

Remark 3.1. Let ϕ1 be a positive principal eigenfunction of −∆ in Ω, with homogeneousDirichlet boundary condition. If r > 1

2 , then ϕr1 ∈ H1

0 (Ω). Indeed, ϕr1 ∈ L2 (Ω). Also,

ϕr−11 ∈ L2 (Ω) and |∇ϕ1| ∈ L∞ (Ω) , thus ∇ (ϕr

1) ∈ L2 (Ω).

Lemma 3.2. Let η and p be as in the condition h3) of Theorem 1.3. Then:

i) dγ−ηΩ ϑ

p+ββ,γ ∈ L∞ (Ω) .

ii) If, in addition, β + 2γ < 3, then d1−ηΩ ϑ

pβ,γ ∈ L2 (Ω) and d1−γ

Ω ϑ−ββ,γ ∈ L2 (Ω) .

Proof. i) follows directly from the definition of ϑβ,γ and the facts that η < γ + p + β whenβ + γ ≤ 1, and that η < γ + (p + β) 2−γ

1+β when β + γ > 1, and using, when β + γ = 1, thatdε

Ω ln(ω0

dΩ

)∈ L∞(Ω) for any ε > 0.

To see the first assertion of ii) note that, by h3), 2 (1− η + p) > 0 when β + γ ≤ 1. Now,(d1−η

Ω ϑpβ,γ

)2= d2(1−η+p)

Ω when β + γ < 1, and (d1−ηΩ ϑ

pβ,γ)

2 = d2(1−η+p)Ω

(ln(ω0

dΩ

))2p when β +

γ = 1. Thus, in both cases, d1−ηΩ ϑ

pβ,γ ∈ L∞ (Ω) ⊂ L2 (Ω). If β + γ > 1, then d1−η

Ω ϑpβ,γ =

d1−η+p 2−γ

1+β

Ω and, by h3),

2(

1− η + p2− γ

1 + β

)+ 1 > 2

(1−

(γ + (p + β)

2− γ

1 + β

)+ p

2− γ

1 + β

)+ 1

=3− β− 2γ

β + 1> 0.

Thus, again in this case, d1−ηΩ ϑ

pβ,γ ∈ L2 (Ω).

Finally, d1−γΩ ϑ

−ββ,γ = d1−β−γ

Ω ∈ L∞(Ω) ⊂ L2(Ω) when β + γ < 1, and d1−γΩ ϑ

−ββ,γ =(

ln(ω0

dΩ

))−β ∈ L∞(Ω) ⊂ L2(Ω) when β + γ = 1. If β + γ > 1, then d1−γΩ ϑ

−ββ,γ = d

1−γ−β 2−γ1+β

Ω and,

since 2(1− γ− β 2−γ

1+β

)+ 1 = 3−β−2γ

β+1 > 0, we have, again in this case, d1−γΩ ϑ

−ββ,γ ∈ L2 (Ω) .

Remark 3.3. Assume the conditions k1), k3), h1), and h3) of Theorem 1.3. Assume also thatβ + 2γ < 3. Then, for any ε > 0, k(·, εvβ,γ) and h

(·, εvβ,γ

)belong to

(H1

0 (Ω))′. Indeed, by k1)

and h1), k(·, εvβ,γ

)and h

(·, εvβ,γ

)are measurable functions, and by k3) and h3),

dΩk(·, εvβ,γ

)≤ ε−βB2d1−γ

Ω v−ββ,γ ≤ ε−βB2c−β

2 d1−γΩ ϑ

−ββ,γ a.e. in Ω,

dΩh(·, εvβ,γ

)≤ εpB3d1−η

Ω vpβ,γ ≤ εpB3cp

3 d1−ηΩ ϑ

pβ,γ a.e. in Ω.

Positive solutions to singular elliptic problems 9

Then, by Lemma 3.2, dΩk(·, εvβ,γ

)and dΩh

(·, εvβ,γ

)belong to L2 (Ω) , and so, by Remark 2.2,

k(·, εvβ,γ

)and h

(·, εvβ,γ

)belong to

(H1

0 (Ω))′ .

Lemma 3.4. Assume the conditions k1), k3), k4), h1), and h3), of Theorem 1.3, and let vβ,γ be thesolution, given by Remark 2.5, of problem (2.5). Then, for any ε positive and small enough, εvβ,γ is asubsolution, in the sense of distributions, of problem (1.4) and, if in addition, β + 2γ < 3, then εvβ,γ

is a weak subsolution of (1.4).

Proof. By Lemma 3.2 i), there exists a positive constant c1 such that d−ηΩ ϑ

pβ,γ ≤ c1d−γ

Ω ϑ−ββ,γ in Ω,

and by Remark 2.5, there exist positive constants c2 and c3 such that c2ϑβ,γ ≤ vβ,γ ≤ c3ϑβ,γ

in Ω. Thus, for some positive constant c4, d−ηΩ vp

β,γ ≤ c1d−γΩ v−β

β,γ in Ω. Then, for any ε positive

and small enough, 12 ε−βB1d−γ

Ω v−ββ,γ ≥ εpB3d−η

Ω vpβ,γ in Ω. By diminishing ε if necessary, we can

assume that, in addition, ε < min

1, δ‖vβ,γ‖∞

, 12 B1

. By Remark 2.5, vβ,γ satisfies, in the senseof distributions,

− ∆vβ,γ = d−γΩ v−β

β,γ in Ω. (3.1)

Then, in the sense of distributions,

−∆(εvβ,γ

)= εd−γ

Ω v−ββ,γ ≤

12

ε−βB1d−γΩ v−β

β,γ (3.2)

≤ ε−βB1d−γΩ v−β

β,γ − εpB3d−ηΩ vp

β,γ ≤ k(·, εvβ,γ

)− h

(·, εvβ,γ

)in Ω;

where, in the last inequality, we have used that, since ε < δ‖vβ,γ‖∞

we have εvβ,γ ≤ δ in Ω, and

then, by k4), k(·, εvβ,γ

)≥ ε−βB1d−γ

Ω v−ββ,γ a.e. in Ω. Thus, for any nonnegative ϕ ∈ C∞

c (Ω) ,

∫Ω

⟨∇(εvβ,γ

),∇ϕ

⟩≤∫

Ω

(k(·, εvβ,γ

)− h

(·, εvβ,γ

))ϕ. (3.3)

and so εvβ,γ is a subsolution, in the sense of distributions, of problem (1.4). Now supposeβ + 2γ < 3. By Remark 2.6 v), εvβ,γ ∈ H1

0 (Ω) , and by Remark 3.3, k(·, εvβ,γ

)and h

(·, εvβ,γ

)belong to

(H1

0 (Ω))′. Thus

(k(·, εvβ,γ

)− h

(·, εvβ,γ

))ϕ ∈ L1 (Ω) for any ϕ ∈ H1

0 (Ω) and, sinceC∞

c (Ω) is dense in H10 (Ω) , it follows that (3.3) holds for any nonnegative ϕ ∈ H1

0 (Ω) .

Remark 3.5. Let us recall the following well known result: Let g : Ω × (0, ∞) → R be aCarathéodory function such that s→ g (x, s) is nonincreasing for a.e. x ∈ Ω, and consider theproblem

−∆u = g (·, u) in Ω,

u = 0 on ∂Ω,

u > 0 in Ω.

(3.4)

Let u ∈ H10 (Ω) be a weak subsolution of problem (3.4) and let u ∈ H1

0 (Ω) be a weak su-persolution of the same problem. Then u ≤ u a.e. in Ω. Indeed, we have, in weak sense,−∆ (u− u) ≤ g (·, u)− g (·, u) , u− u = 0 on ∂Ω. Taking (u− u)+ as test function, and notingthat (g (·, u)− g (·, u)) (u− u)+ ≤ 0 a.e. in Ω, we conclude that u ≤ u.

Lemma 3.6. Assume the conditions k1), k4), h1), and h3) of Theorem 1.3. If problem (1.4) has a weaksolution u ∈ H1

0 (Ω) ∩ C(Ω)

, then γ ≤ 32 .

10 T. Godoy and A. Guerin

Proof. Let u ∈ H10 (Ω) ∩ C

(Ω)

be a weak solution of problem (1.4). For ρ > 0, let Aρ :=x ∈ Ω : dΩ (x) ≤ ρ. Since u ∈ C

(Ω)

and u = 0 on ∂Ω, there exists ρ > 0 such that u ≤ δ inAρ. Then, by k4),

k (·, u) ≥ B1d−γΩ u−β a.e. in Aρ. (3.5)

Let ϕ1 be a positive principal eigenfunction for −∆ in Ω, with homogeneous Dirichlet condi-

tion and weight function 1. Let ε > 0, and let ϕ := ϕ12+ε

1 . Then ϕ ∈ H10 (Ω) ∩ L∞ (Ω). Note

that, by k1) and h1), k (·, u) ϕ and h (·, u) ϕ are nonnegative measurable functions, and that,by h3),

h (·, u) ϕ = dΩh (·, u) d−1Ω ϕ ≤ B3d2−η

Ω ‖u‖p−1∞

∣∣∣d−1Ω u

∣∣∣ ∣∣∣d−1Ω ϕ

∣∣∣≤ B3 ‖dΩ‖2−η

∞ ‖u‖p−1∞

∣∣∣d−1Ω u

∣∣∣ ∣∣∣d−1Ω ϕ

∣∣∣ a.e. in Ω;

and so, by the Hölder and the Hardy inequalities, h (·, u) ϕ ∈ L1 (Ω). Now, taking into account(3.5) and that k is nonnegative,

B1 ‖u‖−β∞

∫Aρ

d−γΩ ϕ ≤ B1

∫Aρ

d−γΩ u−β ϕ ≤

∫Aρ

k (·, u) ϕ (3.6)

≤∫

Ωk (·, u) ϕ =

∫Ω〈∇u,∇ϕ〉+

∫Ω

h (·, u) ϕ < ∞.

Thus∫

Ω d−γΩ ϕ < ∞, therefore

∫Ω d−γ+ 1

2+ε

Ω < ∞. Then −γ + 12 + ε > −1, i.e., γ < 3

2 + ε. Sincethis holds for any ε > 0, the lemma follows.

Remark 3.7. Let us mention that, if the following condition k7) holds:

k7) There exists B1 > 0 such that, for any s > 0, k (·, s) ≥ B1d−γΩ s−β a.e. in Ω,

then the conclusion of Lemma 3.6 remains valid when the assumption u ∈ H10 (Ω) ∩ C

(Ω)

is weakened to u ∈ H10 (Ω) ∩ L∞ (Ω). Indeed, define ϕ as in the proof of Lemma 3.6, and

observe that (3.6) holds with Aρ replaced by Ω.

Proof of Theorem 1.3.. We first prove i). Let vβ,γ be the solution, given by Remark 2.5, of prob-lem (2.5). By Lemma 3.4, for ε positive and small enough, z := εvβ,γ is a weak subsolution

of (1.4). Let z := B1

1+β

2 vβ,γ. By Remark 3.3, k (·, z) − h (·, z) ∈(

H10 (Ω)

)′ and so, taking into

account Remark 2.6 v), z := B1

1+β

2 vβ,γ is a weak supersolution of problem (1.4). In particularz and z are a subsolution and a supersolution, respectively, in the sense of distributions, ofthe problem −∆u = k (·, u)− h (·, u) in Ω. By diminishing ε if necessary we can assume thatz ≤ z in Ω. Moreover, by (2.6), there exist positive constants c1 and c2 such that z ≥ c1ϑβ,γ andz ≤ c2ϑβ,γ in Ω. Thus, for a.e. x ∈ Ω and for s ∈ [z (x) , z (x)] ,

|k (x, s)− h (x, s)| ≤ k (x, z (x)) + h (x, z (x))

≤ B2d−γΩ (x) c−β

1 ϑ−ββ,γ (x) + B3d−η

Ω (x) cp2 ϑ

pβ,γ (x) .

Also, d−γΩ ϑ

−ββ,γ and d−η

Ω ϑpβ,γ belong to L∞

loc (Ω). Then Remark 2.7 gives a solution u, in the senseof distributions, of the problem −∆u = k (·, u)− h (·, u) in Ω, that satisfies z ≤ u ≤ z a.e. in Ω.Consequently, for some positive constants c1 and c2,

c1ϑβ,γ ≤ u ≤ c2ϑβ,γ a.e. in Ω. (3.7)

Positive solutions to singular elliptic problems 11

Therefore, |k (·, u)− h (·, u)| ≤ B2d−γΩ u−β + B3d−η

Ω up ≤ B2c−β1 d−γ

Ω ϑ−ββ,γ + B3cp

2 d−ηΩ ϑ

pβ,γ a.e. in Ω.

By Remark 2.6, d1−γΩ ϑ

−ββ,γ ∈ L2 (Ω) and by Lemma 3.2, d1−η

Ω ϑpβ,γ ∈ L2 (Ω) , then, by Remark 2.2,

|k (·, u)− h (·, u)| ∈(

H10 (Ω)

)′. Also, from (3.7), k (·, u)− h (·, u) ∈ L∞loc (Ω) and u ∈ L∞ (Ω).

Then, by the inner elliptic estimates, u ∈W2,qloc (Ω) for any q ∈ [1, ∞). Therefore u ∈ C (Ω). By

(3.7), u is also continuous on ∂Ω. Then u ∈ C(Ω). Notice that (from the definition of ϑβ,γ)

there exist positive constants c1, c2 and τ, such that c1dΩ ≤ ϑβ,γ ≤ c2dτΩ in Ω, and so, by (3.7),

an estimate of the same kind holds for u. Then, by Lemma 2.4, u ∈ H10 (Ω) , and u is a weak

solution of problem (1.4). The uniqueness assertion of the theorem follows from Remark 3.5.Now we prove ii). Suppose that u ∈ H1

0 (Ω) ∩ C(Ω)

is a weak solution of problem (1.4).As u ∈ C

(Ω)

, u > 0 in Ω, and 0 ≤ k (·, u) ≤ B2d−γΩ u−β, we have k (·, u) ∈ L∞

loc (Ω). Alsoh (·, u) ∈ L∞

loc (Ω). Therefore, by the inner elliptic estimates,

u ∈ C(Ω)∩W2,q

loc (Ω) for any q ∈ [1, ∞) . (3.8)

Note that, by Lemma 3.6, γ < 2. By Lemma 3.4, for ε positive and small enough, εvβ,γ is asubsolution, in the sense of distributions, of (1.4). For s > 0, let g (·, s) := k (·, s) − h (·, s).Then

− ∆(u− εvβ,γ

)≥ g (·, u)− g

(·, εvβ,γ

)in D′ (Ω) ; (3.9)

and, by (3.8) and Remarks 2.5 and 2.6, u− εvβ,γ ∈ C(Ω)∩W2,q

loc (Ω) for any q ∈ [1, ∞). Then,from (3.9), using a suitable family of mollifiers ϕεε>0 , we get

− ∆(u− εvβ,γ

)≥ g (·, u)− g

(·, εvβ,γ

)a.e. in Ω. (3.10)

Let us see that the open set V :=

x ∈ Ω : u (x) < εvβ,γ (x)

is empty. By way of contradiction,suppose V 6= ∅. From h2) and k2), s→ g (·, s) is nonincreasing a.e. in Ω, and then (3.10) givesthat −∆

(u− εvβ,γ

)≥ 0 a.e. in V. Moreover, from u− εvβ,γ ∈ C

(Ω)

and u− εvβ,γ = 0 on ∂Ω,it follows easily that u− εvβ,γ = 0 on ∂V. Then, by the Aleksandrov maximum principle, (asstated, e.g., in [20], Theorem 9.1), u ≥ εvβ,γ in V, which is impossible. Therefore, as stated,V = ∅, and so

u ≥ εvβ,γ in Ω. (3.11)

On the other hand, from the definition of vβ,γ, k3), and h2),

−∆(

B1

1+β

2 vβ,γ

)= B2d−γ

Ω

(B

11+β

2 vβ,γ

)−β

≥ k(·, B

11+β

2 vβ,γ

)≥ k

(·, B

11+β

2 vβ,γ

)− h

(·, B

11+β

2 vβ,γ

)= g

(·, B

11+β

2 vβ,γ

)in D′ (Ω)

and so −∆(

B1

1+β

2 vβ,γ − u)≥ g

(·, B

11+β

2 vβ,γ)− g (·, u) in D′ (Ω) , which gives

−∆(

B1

1+β

2 vβ,γ − u)≥ g

(·, B

11+β

2 vβ,γ

)− g (·, u) a.e. in Ω.

Proceeding exactly as in the above proof of V = ∅ (with u and εvβ,γ(x) replaced by B1

1+β

2 vβ,γ

and u, respectively) we get that V :=

x ∈ Ω : B1

1+β

2 vβ,γ < u

is empty, and consequently,

u ≤ B1

1+β

2 vβ,γ in Ω. (3.12)

12 T. Godoy and A. Guerin

Thus, taking into account (3.11) and (3.12), by Remark 2.5, there exist positive constants c1 andc2 such that

c1ϑβ,γ ≤ u ≤ c2ϑβ,γ in Ω. (3.13)

In order to conclude the proof, we take u as test function in problem (1.1), to get∫

Ω B2d−γΩ u1−β≤∫

Ω k (·, u) u =∫

Ω |∇ (u)|2 +∫

Ω h (·, u) u < ∞. Let us consider the three possible cases.

Case β + γ ≤ 1. Since γ ≤ 32 we have β + 2γ = β + γ + γ ≤ 1 + γ < 3.

Case β + γ > 1 and β ≤ 1. In this case, by (3.13), d−γΩ u1−β ≥ c1−β

1 d−γΩ ϑ

1−ββ,γ = c1−β

1 d−γΩ d

(1−β) 2−γ1+β

Ω

in Ω. Since∫

Ω d−γΩ u1−β < ∞ we get that −γ + (1− β) 2−γ

1+β > −1, i.e., β + 2γ < 3.

Case β + γ > 1 and β > 1. Here we have, by (3.13), d−γΩ u1−β ≥ (c′′)1−β d−γ

Ω ϑ1−ββ,γ in Ω and so,

as in the second case, we get β + 2γ < 3.

Proof of Theorem 1.4. Assume β + γ < 32 . Let δ and E be given by k5), let m0 := infE×(0,δ), and

let λE and φE be the principal eigenvalue and the positive principal eigenfunction respectively,for −∆ in Ω, with homogeneous Dirichlet condition and weight function m0χE, with φE nor-malized by ‖φE‖∞ = 1. Let ε ∈ (0, minδ, λ−1

E ), and let z := εφE. Then z ∈ W2,q (Ω) for anyq ∈ [1, ∞) and thus, in particular, z ∈ C

(Ω). Also,

− ∆z = ελEm0χEφE ≤ m0χE ≤ k (·, z) in D′ (Ω) . (3.14)

By the properties of the principal eigenfunctions (see Remark 2.3) there exist positive constantsc1 and c2 such that c1dΩ ≤ z ≤ c2dΩ in Ω. Let vβ,γ be the solution, given by Remark 2.5, of

problem (2.5), and let z := B1

1+β

2 vβ,γ. By diminishing ε if necessary, we can assume z ≤ z inΩ. As in the proof of Theorem 1.3, Remark 2.7 applies to obtain a solution u, in the sense ofdistributions, of the problem −∆u = k (·, u) in Ω, such that z ≤ u ≤ z a.e. in Ω. Then, for somepositive constants c and c′, cdΩ ≤ u ≤ c′ϑβ,γ a.e. in Ω. As in the proof of Theorem 1.3, we have

u ∈W2,qloc (Ω)∩C

(Ω)

for any q ∈ [1, ∞). Note also that, by k3), dΩk (·, z) ≤ B2c−β1 d1−γ−β

Ω a.e. inΩ, and so, since d1−γ−β

Ω ∈ L2 (Ω) (because β + γ < 32 ), Remark 2.2 gives k (·, z) ∈

(H1

0 (Ω))′ ,

and so, as in the proof of Theorem 1.3, using Lemma 2.4, we get u ∈ H10 (Ω) , and that u is a

weak solution of problem (1.1). Finally, the uniqueness of the weak solution u is proved, as inTheorem 1.3, using Remark 3.5 .

Remark 3.8. Assume the hypothesis of Theorem 1.5, and let η and p be as in the condition h4)there. Then, η < p− 1 if β ≤ 1, and η < 2(p−1)

1+β if β > 1. Indeed, if β + γ ≤ 1, then η < p− 1.

If β ≤ 1 and β + γ > 1, then η < (p− 1) 2−γ1+β < p− 1 (because 2−γ

1+β < 1 when β + γ > 1). If

β > 1, then β + γ > 1, and so η < (p− 1) 2−γ1+β ≤

2(p−1)1+β .

Lemma 3.9. Assume the hypothesis of Theorem 1.5, and let η and p be as in the condition h4) there.Then:

i) d1−ηΩ ϑ

pβ,0 ∈ L2 (Ω) .

ii) dΩϑ−ββ,0 ∈ L2 (Ω) .

iii) ϑp+ββ,0 d−η

Ω ∈ L∞ (Ω) .

Positive solutions to singular elliptic problems 13

Proof. To see i), note that, by h4), 1− η + p > 0 when β ≤ 1. If β < 1 then d1−ηΩ ϑ

pβ,0 = d1−η+p

Ω ∈L∞ (Ω) ⊂ L2 (Ω). If β = 1 then d1−η

Ω ϑpβ,0 = d1−η+p

Ω

(ln(ω0

dΩ

))p ∈ L∞(Ω) ⊂ L2 (Ω). If β > 1 then

d1−ηΩ ϑ

pβ,0 = d

1−η+ 2p1+β

Ω and, by h4), 1− η + 2p1+β > 1−

( 2(p−1)1+β

)+ 2p

1+β > 0. Therefore, again in

this case, d1−ηΩ ϑ

pβ,0 ∈ L∞ (Ω) ⊂ L2 (Ω).

To prove ii), observe that dΩϑ−ββ,0 = d1−β

Ω ∈ L∞ (Ω) when β < 1, and that dΩϑ−ββ,0 =(

ln(ω0

dΩ

))−1 ∈ L∞ (Ω) when β = 1. Note also that, if β > 1, then dΩϑ−ββ,0 = d

1−β 21+β

Ω and

2(1− β 2

1+β

)+ 1 = 3−β

β+1 > 0. Then dΩϑ−ββ,0 ∈ L∞(Ω) ⊂ L2(Ω) when β > 1.

To see iii), note that ϑp+ββ,0 d−η

Ω = dp+β−ηΩ when β < 1, and ϑ

p+ββ,0 d−η

Ω = d−η+p+1Ω

(ln(ω0

dΩ

))p+1

when β = 1. Since, in both cases, p + β − η > β + 1 > 0, we obtain ϑp+ββ,0 d−η

Ω ∈ L∞ (Ω) ⊂

L2 (Ω) when β ≤ 1. If β > 1, then ϑp+ββ,0 d−η

Ω = d−η+(p+β) 2

1+β

Ω and, by h4), −η + (p + β) 21+β >

−( 2(p−1)

1+β

)+ (p + β) 2

1+β = 2 > 0. Thus, again in this case, ϑp+ββ,0 d−η

Ω ∈ L∞ (Ω) ⊂ L2 (Ω) .

Lemma 3.10. Assume the hypothesis of Theorem 1.5 and let η and p be as in the condition h4) there.Let B and D be positive constants, let F : Ω× (0, ∞)→ R be defined by

F (·, s) := Ds−β − Bd−ηΩ sp, (3.15)

and let vβ,0 be the solution, provided by Remark 2.5, of problem (2.5) taking there γ = 0. Then, for ε

positive and small enough, εvβ,0 is a weak subsolution of the problem−∆u = F (·, u) in Ω,

u = 0 on ∂Ω,

u > 0 in Ω.

(3.16)

Proof. Note that, for x ∈ Ω and s > 0, F (x, s) = Ds−β − Bd−ηΩ (x) sp ≥ 1

2 Ds−β wheneversp+β ≤ 1

2 B−1DdηΩ (x). Let Q := 1

2 B−1D. Then

F (x, s) ≥ 12

Ds−β for x ∈ Ω and s ∈(

0, Q1

p+β dη

p+β

Ω (x))

. (3.17)

By Remark 2.5, vβ,0 ≤ c1ϑβ,0, with c1 a positive constant. By Lemma 3.9, there exists a positive

constant c2 such that ϑp+ββ,0 d−η

Ω ≤ c2 in Ω. Then, for ε > 0,(εvβ,0

)p+β ≤ (εc1)p+β c2dη

Ω in Ω. (3.18)

Thus, from (3.17) and (3.18), we have, for ε positive and small enough,

F(·, εvβ,0

)≥ 1

2D(εvβ,0

)−β in Ω. (3.19)

We have also, for some positive constant c,∣∣F (·, εvβ,0)∣∣ ≤ Dε−βv−β

β,0 + Bεpd−ηΩ vp

β,0 ≤ c(

ϑ−ββ,0 + d−η

Ω ϑpβ,0

)in Ω,

the last inequality by Remark 2.5. By Lemma 3.9, dΩϑ−ββ,0 ∈ L2 (Ω) and d1−η

Ω ϑpβ,0 ∈ L2 (Ω).

Then, by Remark 2.2,F(·, εvβ,0

)∈ (H1

0(Ω))′. (3.20)

14 T. Godoy and A. Guerin

By diminishing ε, if necessary, we can assume that ε ≤ 12 Dε−β Then (3.19) and (3.20) give that

εvβ,0 satisfies, in weak sense,[c]c− ∆

(εvβ,0

)= εv−β

β,0 ≤12 D(εvβ,0

)−β ≤ F(·, εvβ,0

)in Ω,

εvβ,0 = 0 on ∂Ω.

Lemma 3.11. Assume the hypothesis of Theorem 1.5 and let η and p be as in the condition h4) there.Then:

i) d1−γΩ ϑ

−ββ,0 ∈ L2 (Ω) .

ii) d1−ηΩ ϑ

pβ,γ ∈ L2 (Ω) .

Proof. To prove i), observe that d1−γΩ ϑ

−ββ,0 = d1−β−γ

Ω when β < 1, and that 2 (1− β− γ) + 1 > 0

(because β + γ < 32 ). Then d1−γ

Ω ϑ−ββ,0 ∈ L2(Ω) when β < 1. If β = 1, then d1−γ

Ω ϑ−ββ,0 =

d−γΩ

(ln(ω0

dΩ

))−1 and γ < 32 − β = 1

2 . Therefore, since(

ln(ω0

dΩ

))−1 is bounded, we obtain that

d1−γΩ ϑ

−ββ,0 ∈ L2 (Ω) when β = 1. If β > 1 then d1−γ

Ω ϑ−ββ,0 = d

1−γ−β 21+β

Ω and 2(1− γ− β 2

1+β

)+ 1 >

2(1−

( 32 − β

)− β 2

1+β

)+ 1 = 2β

β−1β+1 > 0. Thus d1−γ

Ω ϑ−ββ,0 ∈ L2 (Ω) when β > 1.

To prove ii), note that, by h4), 2(1− η + p)+ 1 > 2(1− (p− 1)+ p)+ 1 > 0 when β+γ ≤ 1.Then, taking into account that d1−η

Ω ϑpβ,γ = d1−η+p

Ω when β + γ < 1, and that d1−ηΩ ϑ

pβ,γ =

d1−η+pΩ

(ln(ω0

dΩ

))p when β + γ = 1, we conclude that d1−ηΩ ϑ

pβ,γ ∈ L2 (Ω) when β + γ ≤ 1. If

β + γ > 1 then d1−ηΩ ϑ

pβ,γ = d

1−η+p 2−γ1+β

Ω and, by h4),

2(

1− η + p2− γ

1 + β

)+ 1 > 2

(1− (p− 1)

2− γ

β + 1+ p

2− γ

1 + β

)+ 1 =

3β− 2γ + 7β + 1

>1

β + 1

(3β− 2

(32− β

)+ 7)=

5β + 4β + 1

> 0

and so, again in this case, d1−ηΩ ϑ

pβ,γ ∈ L2 (Ω) .

Proof of Theorem 1.5. Let B1, B3, η, and p, be given by k6) and h4). Let vβ,γ be the uniqueclassical solution (given by Remark 2.5) of problem (2.5). Let F be defined by (3.15), takingthere D such that 0 < D < B1 infΩ

(d−γ

Ω

)and B = B3; and let ε > 0 be small enough such

that εvβ,0 is a weak subsolution of problem (3.16). Then u := εvβ,0 is a weak subsolution ofproblem (1.4). Since β + 2γ < β + 2

( 32 − β

)= 3− β ≤ 3, by Remark 2.6, vβ,γ ∈ H1

0 (Ω) , and

vβ,γ is a weak solution of problem (2.5). Note that u := B1

1+β

2 vβ,γ is a weak supersolution ofproblem (1.4), thus, by Remark 3.5, u ≤ u in Ω. Also, by (2.6), there exist positive constants c1

and c2 such that u ≥ c1ϑβ,0 and u ≤ c2ϑβ,γ in Ω. Then, for a.e. x ∈ Ω and s ∈ [u (x) , u (x)] ,

|k (x, s)− h (x, s)| ≤ k (x, u (x)) + h (x, u (x))

≤ B2d−γΩ (x) c−β

1 ϑ−ββ,0 (x) + B3d−γ

Ω (x) cp2 ϑ

pβ,γ (x) .

Since d−γΩ ϑ

−ββ,γ and d−γ

Ω ϑpβ,γ belong to L∞

loc (Ω) , Remark 2.7 gives a solution u, in the sense ofdistributions, of the problem −∆u = k (·, u)− h (·, u) in Ω, that satisfies u ≤ u ≤ u in Ω. Then,for some positive constants c1 and c2,

c1ϑβ,0 ≤ u ≤ c2ϑβ,γ a.e. in Ω. (3.21)

Positive solutions to singular elliptic problems 15

Therefore, |k (·, u)− h (·, u)| ≤ B2d−γΩ u−β + B3d−Ωup ≤ B2c−β

1 d−γΩ ϑ

−ββ,0 + B3cp

2 d−ηΩ ϑ

pβ,γ a.e. in Ω.

By Lemma 3.11, d1−γΩ ϑ

−ββ,0 ∈ L2 (Ω) and d1−η

Ω ϑpβ,γ ∈ L2 (Ω) ; then, by Remark 2.2,

|k (·, u)− h (·, u)| ∈(

H10 (Ω)

)′. Also, from (3.21), we get k (·, u) − h (·, u) ∈ L∞loc (Ω) and

u ∈ L∞ (Ω). Then, by the inner elliptic estimates, u ∈ W2,qloc (Ω) for any q ∈ [1, ∞). There-

fore u ∈ C (Ω). By (3.21), u is also continuous on ∂Ω. Then u ∈ C(Ω). Notice that, by the

definition of ϑβ,γ, there exist positive constants c2 and τ, such that ϑβ,γ ≤ c2dτΩ in Ω, and so,

by (3.21), u ≤ c′dτΩ in Ω for some positive constant c′. Then, by Lemma 2.4, u ∈ H1

0 (Ω) , andu is a weak solution of problem (1.4). Finally, the uniqueness assertion of the theorem followsfrom Remark 3.5.

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