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## independent variables

B

### 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 is a bounded C1,1 domain in Rn, n2, h : ×[0,) → [0,) and k : ×(0,) → [0,) are Carathéodory functions such thath(x,·)is nondecreasing, andk(x,·)is nonincreasing and singular at the origina.e.

xΩ. Additionally, k,s) and h,s) are allowed to be singular on ∂Ω for s > 0.

Under suitable additional hypothesis onhandk, we prove that the stated problem has a unique weak solution uH01(), and thatubelongs to C

. The behavior of the solution near∂Ωis also addressed.

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

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

### 1Introduction and statement of the main results

Let Ω be a bounded domain in Rn with C1,1 boundary, and consider a singular semilinear elliptic 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 any s ∈ (0,∞) and k(x,·) is continuous on (0,∞) a.e. x ∈ Ω), with k = k(x,s) allowed to be singular ats=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

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Problem (1.1) was studied, in the case where k(x,s) = a(x)sα, under different sets of assumptions onaandα, 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)uα(x) in Ω, u = 0 on ∂Ω, where a belongs to a suitable Lebesgue space. Among other results, they found a very weak solution inH10()(with test functions inCc1()) 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 withk(·,s)allowed to exhibit some kind of singularity onΩ, were studied in [1,24,28,30,31,37,38].

Diaz, Hernandez and Rakotoson [12] considered the problem





u=0 onΩ,

u>0 inΩ,

(1.2)

where d := dist(·,∂Ω), γ < 2, and a ∈ L() satisfies infa > 0. They studied the existence of solutionsu ∈ L1(Ω,d)(the d-weighted Lebesgue space) in the following very weak sense:

Z

u∆ϕ=

Z

such that ϕ=0 on ∂Ω. (1.3) Notice that the space of test functions involved in their notion of solution is strictly smaller than the corresponding space in the present paper (as given in Definition1.1below). In Theo- rem 2 they find, whenβ+γ<1, a very weak solution of problem (1.2), and prove that it be- longs toW01(Ω,k · kN(r),)∩Wloc2,q()for anyr ∈(0, 1)andq∈[1,), whereW01(Ω,k · kN(r),) is the space of the functions w : Ω → R such that w ∈ W1,1() and |∇w| belongs to the Lorentz spaceLN(r),(), withN(r):= nn1+r.

Regarding the case β+γ > 1, in theorem 1 they find a very weak solution of problem (1.2) that belongs toC Ω

∩Wloc2,q()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 H10() if, and only if, β+2γ<3. Additionally, in Theorem 4, they prove that, whenβ+γ>1, there exist positive constantsc1andc2such that the found solutionusatisfiesc1d

2γ 1+β

≤u≤c1d

2γ 1+β

inΩ. However, it is not obvious that, ifu∈ H01(), then uis a weak solution of problem (1.2), i.e., that (1.3) holds for any ϕ∈ H01().

The existence of classical solutionsu∈ C2()∩C Ω

of problem (1.2) was addressed by Mâagli [27] in the case when a ∈ Clocσ ()for some σ ∈ (0, 1), and dγa 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 Remark2.5below.

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

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

slim0+k(·,s) = a.e. inΩ, andk(·,s)is allowed to be singular on∂Ω, in the sense that

lim3xyk(x,s) = for any (y,s)∈Ω×(0,∞).

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Additionally, we allow the introduction of a second term, and consider the problem





u=k(·,u)−h(·,u) inΩ,

u=0 on ∂Ω,

u>0 inΩ,

(1.4)

where h=h(·,s)is allowed to be singular on∂Ωfor 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() for all ϕin H01(). We say thatu:Ω→Ris a weak solution of the problem

u=ψ inΩ, u=0 onΩ, (1.5)

ifu ∈H01()andR

h∇u,∇ϕi=R

ψϕfor all ϕin H01().

Similarly, for u ∈ H01(), we say that u is a weak supersolution (respectively a weak subsolution) of problem (1.5), and we write

(−∆uψ in Ω(resp. −∆uψinΩ), u=0 on ∂Ω,

to meanR

h∇u,∇ϕi ≥R

ψϕ(resp. R

h∇u,∇ϕi ≤R

ψϕ) for all nonnegative ϕin H01(). Definition 1.2. Let d : Ω → R be the distance function d := dist(·,∂Ω). For β ≥ 0 and γ∈ [0, 2)defineϑβ,γ :Ω→Rby:

ϑβ,γ:=dif β+γ<1,

ϑβ,γ := dln(ω0d1)if β+γ =1, where ω0 is an arbitrary number, which we fix from now on, such that ω0>diam(),

ϑβ,γ:=d

2γ 1+β

if β+γ>1.

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

Theorem 1.3. Let Ωbe 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∈;

k3) there existβ≥0,γ≥0,and B2 >0such that, for any s>0,k(·,s)≤B2dγsβ a.e.inΩ;

k4) there existδ>0and B1 >0such that, for any s∈(0,δ),k(·,s)≥ B1dγsβ a.e.inΩ.

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

h1) h is a Carathéodory function;

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

h3) h(·,s) ≤ B3dηsp a.e. inΩ for all s ∈ [0,∞), with B3 > 0, p > 1, 0 < η < γ+p+β if β+γ≤1,and0<η< γ+ (p+β)21+γ

β ifβ+γ>1.

Then:

i) If β+2γ < 3, then problem (1.4) has a unique weak solution u ∈ H01(). Moreover, u ∈ Wloc2,q()∩C Ω

for any q∈[1,∞),and u satisfies cϑβ,γ ≤ u≤c0ϑβ,γ inΩ,for some positive constants c and c0.

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

,then β+2γ<3.

Note that, in particular, Theorem1.3says that−∆u= dγ in Ω,u=0 on∂Ω,u> 0 inΩ, has a weak solution if, and only if,γ< 32.

The next theorem states that, when h is identically zero, the assertion i) of Theorem1.3 remains valid if the conditionk4)is replaced by the following milder condition:

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

Theorem 1.4. Let Ωbe a bounded domain in Rn with C1,1 boundary, and let k : Ω×(0,∞) → R satisfy the conditions k1)–k3) of Theorem1.3. Assume that β+γ < 32, and that the condition k5) holds. Then problem(1.1) has a unique weak solution u ∈ H10(),and u ∈ Wloc2,q()∩C Ω

for any q∈ [1,∞),and cd ≤u≤c0ϑβ,γ inΩ,for some positive constants c and c0.

Concerning the case whenhis nonidentically zero, our next result shows that the assertion i)of theorem1.3holds under a weaker condition thank4), at the expense of strengtheningh3).

Theorem 1.5. Let Ωbe a bounded domain in Rn with C1,1 boundary, and let k : Ω×(0,∞) → R satisfy the conditions k1)–k3) of Theorem1.3, withβandγsatisfyingβ>0,γ∈ [0, 2),andβ+γ<

3

2. Assume also the following condition:

k6) there existδ >0and B1>0such that, for any s∈(0,δ),k(·,s)≥B1sβ a.e.inΩ.

Let h: Ω×[0,∞)→Rsatisfy the conditions h1) and h2) of Theorem1.3, and the following

h4) h(·,s)≤ B3dηsp a.e.inΩfor all s∈[0,∞),with B3>0,p >1, 0<η< p−1ifβ+γ≤1 and0<η<(p−1)21+γ

β if1< β+γ< 32.

Then problem (1.4) has a unique weak solution u. Moreover, u∈Wloc2,q()∩C Ω

for any q∈ [1,∞), and there exist positive constants c and c0 such that c0ϑβ,0 ≤u≤cϑβ,γinΩ.

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 H01(). Indeed, Lazer and McKenna [23] considered the problem −u = auα in Ω, u = 0 on Ω, u > 0 in Ω, under the assumptions a ∈ Cγ

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

; and that u ∈ H01()if, and only if, α < 3. A clear-cut simple condition like that is elusive when the right hand side of the equation is not in the form auα; in [21] we addressed such a more general situation, but still did not consider the case when a spatial singularity is added. This latter situation is considered in the present work.

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The paper is organized as follows: in Section 2, we collect some preliminary results.

Lemma2.4is an adaptation of Lemma 3.2 in [22] and states that, under suitable conditions, a solution in the sense of distributions of an elliptic problem, is also a weak solution in H01(). Remark2.5recalls a result, due to Mâagli [27], about existence, uniqueness, and behavior near the boundary, of positive classical solutions of problems of the form −∆u = adγuβ in Ω, u= 0 on∂Ω(for a suitable class of Hölder continuous functions a); and Remark 2.7recalls a sub-supersolution theorem for singular problems due to Loc and Schmitt [25]. In Section 3 we prove Theorems1.3,1.4and1.5, by combining the results of [27], [25], and Lemma2.4, jointly with some additional auxiliary results.

### 2Preliminaries

For w ∈ L1loc(), we write, as usual, w ∈ H01()0 to mean that wϕ ∈ L1() for any ϕ∈ H01(), and that the map ϕ→R

wϕis continuous onH01().

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

ϕ

d

L2() ≤ ck∇ϕkL2() for all ϕ∈ H01().

Remark 2.2. If ψ ∈ L1loc() and dψ ∈ L2(), then ψϕ ∈ L1() for any ϕ ∈ H01(). Moreover, ψ ∈ H01()0 and kψk(H1

0())0 ≤ ckdψk2 with c independent of ψ. Indeed, for ϕ ∈ H01(), from the Hölder and the Hardy inequalities, R

|ψϕ| ≤ kdψk2kd1ϕk2 ≤ ckdψk2kϕkH1

0(), wherecis the constant in the Hardy inequality of Remark2.1.

Remark 2.3. (See e.g., [10])λRis called a principal eigenvalue for−inΩ, with homoge- neous Dirichlet condition and weight function b∈ L(), if the problem−φ = λbφin Ω, φ=0 on ∂Ωhas a solutionϕ1 (called a principal eigenfunction) such thatϕ1 >0 inΩ. It is a well known fact that, for any C1,1 bounded domain Ω⊂ Rn, b∈ L(), and b+ 6≡0, there exists a unique positive principal eigenvalue λ1(b), and its eigenspace Vλ1 is a one dimen- sional subspace of C1

. Moreover, for each positive ϕ1 ∈ Vλ1, there are positive constants c1,c2 such thatc1dϕ1 ≤c2d in Ω. Consequently,|ln(ϕ1)| ∈L1(); and ϕ1t ∈ L1()if, and only if,t >−1.

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

Lemma 2.4. Let ψ ∈ Lloc()be such that |ψ| ∈ H10()0, and let u ∈ Wloc1,2()∩C() be a solution, in the sense of distributions, of the problem

∆u=ψ inΩ.

If there exist constants c >0 and r> 12 such that0≤ u ≤ cdr inΩ,then u ∈ H01()∩C1()∩ C Ω

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

Proof. Let ϕ∈ H01()such that supp(ϕ)⊂Ω. ThenR

h∇u,∇ϕi=R

ψϕ. Indeed, letδ>0 be such that supp(ϕ)⊂ δ, and let

ϕj jNbe a sequence inCc ()satisfying supp ϕj

⊂ Ωδ for allj, and such that

ϕj jNconverges toϕin H10(δ). Now,∇u|δ ∈ L2(δ,Rn), and so ζ → R

δh∇u,∇ζiis continuous on H01(δ). Also, R

∇u,∇ϕj

= R

ψϕj for all j. Then R

h∇u,∇ϕi=limjR

∇u,∇ϕj

=limjR

ψϕj =R

ψϕ.

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For eachj∈N, letµj :RRbe defined byµj(s):=0 ifs ≤ 1j,µj(s):=−3j2s3+14js2− 19s+ 8j if 1j < s< 2j, andµj(s):= sif 2j ≤s. Thenµj ∈C1(R),µj(s) =0 fors< 1j,µ0j(s)≥0 for 1j <s< 2j, andµ0j(s) =1 for 2j <s. Also, 0<µj(s)<sfor all s∈ 0,2j

. Letµj(u):=µj◦u. Then, for allj,∇ µj(u)= µ0j◦u

∇uin D0(). Sinceu∈Wloc1,2(), it follows thatµj(u)∈Wloc1,2(). Since supp µj(u)Ω, we haveµj(u)∈ H10(). Therefore, for allj,R

∇u,∇ µj(u)= R

ψµj(u), i.e., Z

{u>0}

µ0j◦u

|∇u|2 =

Z

ψµj(u). (2.1)

Now, µ0j◦u

|∇u|2 is nonnegative and limj µ0j◦u

|∇u|2 = |∇u|2 a.e. inΩ, and so, from (2.1) and Fatou’s lemma, we have

Z

|∇u|2 ≤limj Z

ψµj(u).

Let ϕ1 be the principal eigenfunction for − in Ω with homogeneous Dirichlet condition, and with weight function1, normalized by kϕ1k = 1. Since, for some positive constantc0, u≤ c0ϕr1in Ω, andϕr1 ∈ H01(), we have ψu∈ L1(). Now, limjψµj(u) = ψuinΩ, and, for anyj∈N,ψµj(u)≤ |ψu|inΩ. Then, Lebesgue’s dominated convergence theorem gives

jlim Z

ψµj(u) =

Z

ψu<∞.

ThusR

|∇u|2 <, and sou ∈H1(). As−u=ψin D0(),u∈ L()andψ∈ Lloc(), then the inner elliptic estimates (as stated e.g., in [5], Proposition 4.1.2, see also [20], Theorem 9.11) give that u ∈ C1(). From 0 ≤ u ≤ cdr in Ω, and u ∈ C(), we conclude that u∈C Ω

. Sinceu∈ H1(), u∈C Ω

, andu=0 onΩ, we getu∈ H01(). Asu∈ H01(), we have that ϕ→R

h∇u,∇ϕiis continuous on H10(). Therefore, since Cc ()is dense inH10(), and since, for anyϕ∈Cc(),

Z

h∇u,∇ϕi=

Z

ψϕ, (2.2)

we conclude that (2.2) holds for all ϕ∈ H01().

Remark 2.5. i) Letω0be as in Definition1.2,α<1, and ρ<2. Letz∈C([0,ω0])be such that z(0) = 0 and Rω0

0 t1ρLz(t)dt < , where Lz(t) := exp Rω0 t

z(s) s ds

. Letσ ∈ (0, 1), and let a∈ Clocσ ()satisfy, for some constantc>0,

1

cLz◦d ≤dρa≤cLz◦d inΩ. (2.3) Then, Theorem 1 in [27] says that the problem





∆u=auα inΩ,

u=0 on ∂Ω,

u>0 inΩ

(2.4)

has a unique classical solutionu ∈ C2()∩C Ω

; and that, for some positive constant c0, u satisfies,

c01

θρ◦d ≤u≤c0θρ◦d inΩ,

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where

θρ(t):= Z ω

0

0

Lz(s) s ds

11

α

if ρ =2, θρ(t):=t21αρ (Lz(t))11α if 1+α<ρ<2, θρ(t):=t

Z ω

0

t

Lz(s) s ds

11

α

ifρ=1+α, and

θρ(t):=t i f ρ <1+α.

ii) Let β≥ 0, and letγ< 2. If in i) we takeα:=−β, z :=0(thenLz =1), andρ :=γ, we get that the problem





v=dγvβ inΩ,

v=0 on∂Ω,

v>0 inΩ

(2.5)

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

; and that there exists positive constants c1 andc2 such that

c1ϑβ,γ ≤vβ,γ ≤c2ϑβ,γ inΩ, (2.6) whereϑβ,γ is as in Definition1.2.

Remark 2.6. Letβ≥0 andγ≥0 be such that β+2γ<3.

i) d1γϑβ,γβ ∈ L2(). Indeed, if β+γ < 1 then d1γϑβ,γβ = d1βγ ∈ L() ⊂ L2(). If β+γ=1, then d1γϑβ,γβ = ln ωd0

β

∈ L()⊂ L2(). Ifβ+γ >1, then d1γϑβ,γβ = d1γβ

2γ 1+β

=d

1

β+1(β+γ1)

and, sinceβ+2γ<3,− 2

β+1(β+γ−1)>−1, and so, again in this case,d1γϑβ,γβ ∈ L2()(because, forr∈R,dr ∈L2()whenever 2r+1>0).

ii) There exist positive constantscandτ> 12 such thatϑβ,γ ≤cdτinΩ. Indeed, if β+γ<1 thenϑβ,γ =d, if β+γ= 1 then ϑβ,γ = dln ωd0

and so, for any ε> 0,ϑβ,γ d1ε1

= dεln ωd0

∈ L(); and ifβ+γ>1, thenϑβ,γ =d

2γ 1+β

and, sinceβ+2γ<3, 21+γβ > 12. iii) Letvβ,γ be the solution of problem (2.5) given by Remark2.5. From i) and (2.6), it follows

thatd1γvβ,γβ ∈ L2().

iv) Letvβ,γ be as in iii). Then, by iii) and Remark 2.2,dγvβ,γβ ∈ H01()0.

v) Letvβ,γ be as in iii). Sincedγvβ,γβ ∈ H01()0 and since, by Remark2.5,vβ,γ ∈C2()∩ C Ω

, Lemma 2.4gives that vβ,γ ∈ H01(), and thatvβ,γ is a weak solution of problem (2.5). Moreover, since vβ,γ ∈ L() anddγvβ,γβ ∈ Lloc(), the inner elliptic estimates givevβ,γ∈Wloc2,q()for anyq∈[1,∞).

Remark 2.7. Let g: Ω×(0,∞)→Rbe a Carathéodory function. We say thatw ∈ L1loc()is a subsolution (supersolution) of the problem

z =g(·,z) inΩ (2.7)

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

h∇w,∇ϕi ≤ (≥)R

g(·,w)ϕ for all nonnegative ϕ∈Cc (). We say thatz∈ L1loc()is a solution, in the sense of distributions, of (2.7) if, and only if,z > 0 a.e. inΩ, and,R

h∇z,∇ϕi= R

g(·,z)ϕ for all ϕ∈ Cc().

According to Theorem 2.4 in [25], if (2.7) has a subsolutionzand a supersolution z(in the sense of distributions), both inLloc()∩Wloc1,2(), and such such that 0<z ≤z inΩ, and if there existsψ ∈ Lloc()such that |g(x,s)| ≤ ψ(x) a.e. x ∈ for all s ∈ [z(x),z(x)]; then (2.7) has a solutionzin the sense of distributions, which satisfies z≤z≤z inΩ.

### 3Proof of the main results

Remark 3.1. Let ϕ1 be a positive principal eigenfunction of − in Ω, with homogeneous Dirichlet boundary condition. If r > 12, then ϕr1 ∈ H01(). Indeed, ϕr1 ∈ L2(). Also, ϕ1r1 ∈ L2()and|∇ϕ1| ∈ L(), thus∇(ϕr1)∈ L2().

Lemma 3.2. Letηand p be as in the condition h3) of Theorem1.3. Then:

i) dγηϑβ,γp+β ∈ L().

ii) If, in addition,β+2γ<3,then d1ηϑβ,γp ∈L2()and d1γϑβ,γβ ∈ L2().

Proof. i) follows directly from the definition of ϑβ,γ and the facts that η < γ+p+β when β+γ ≤ 1, and that η < γ+ (p+β)21+γβ whenβ+γ > 1, and using, when β+γ = 1, that dεln ωd0

∈ L()for anyε>0.

To see the first assertion of ii)note that, byh3), 2(1−η+p) > 0 when β+γ ≤ 1. Now, d1ηϑβ,γp 2

= d2(1η+p) when β+γ < 1, and (d1ηϑβ,γp )2 = d2(1η+p) ln ωd0

2p

when β+ γ = 1. Thus, in both cases, d1ηϑβ,γp ∈ L() ⊂ L2(). If β+γ > 1, then d1ηϑpβ,γ = d1η+p

2γ 1+β

and, byh3), 2

1−η+p2−γ 1+β

+1>2

1−

γ+ (p+β)2γ 1+β

+p2−γ 1+β

+1

= 3β−2γ β+1 >0.

Thus, again in this case,d1ηϑpβ,γ ∈ L2().

Finally, d1γϑβ,γβ = d1βγ ∈ L() ⊂ L2() when β+γ < 1, and d1γϑβ,γβ = ln ωd0

β

∈ L()⊂ L2()whenβ+γ=1. Ifβ+γ>1, thend1γϑβ,γβ =d1γβ

2γ 1+β

and, since 2 1−γβ21+γβ

+1= 3β

β+1 >0, we have, again in this case,d1γϑβ,γβ ∈ L2(). Remark 3.3. Assume the conditions k1), k3), h1), and h3) of Theorem1.3. Assume also that β+2γ<3. Then, for anyε>0,k(·,εvβ,γ)andh ·,εvβ,γ

belong to H01()0. Indeed, by k1) and h1),k ·,εvβ,γ

andh ·,εvβ,γ

are measurable functions, and by k3) and h3), dk ·,εvβ,γ

εβB2d1γvβ,γβεβB2c2βd1γϑβ,γβ a.e. inΩ, dh ·,εvβ,γ

εpB3d1ηvpβ,γεpB3c3pd1ηϑβ,γp a.e. in Ω.

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Then, by Lemma3.2,dk ·,εvβ,γ

anddh ·,εvβ,γ

belong to L2(), and so, by Remark2.2, k ·,εvβ,γ

andh ·,εvβ,γ

belong to H01()0.

Lemma 3.4. Assume the conditions k1), k3), k4), h1), and h3), of Theorem 1.3, and let vβ,γ be the solution, given by Remark2.5, of problem(2.5). Then, for anyεpositive and small enough, εvβ,γ is a subsolution, in the sense of distributions, of problem (1.4) and, if in addition, β+2γ < 3,thenεvβ,γ is a weak subsolution of (1.4).

Proof. By Lemma3.2 i),there exists a positive constantc1 such thatdηϑβ,γp ≤c1dγϑβ,γβ 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 constantc4,dηvβ,γp ≤ c1dγvβ,γβ in Ω. Then, for anyε positive and small enough, 12εβB1dγvβ,γβεpB3dηvpβ,γ in Ω. By diminishingε if necessary, we can assume that, in addition, ε < min

1,kv δ

β,γk,12B1 . By Remark 2.5, vβ,γ satisfies, in the sense of distributions,

∆vβ,γ = dγvβ,γβ inΩ. (3.1) Then, in the sense of distributions,

εvβ,γ

=εdγvβ,γβ1

2εβB1dγvβ,γβ (3.2)

εβB1dγvβ,γβεpB3dηvpβ,γ ≤ k ·,εvβ,γ

−h ·,εvβ,γ

inΩ;

where, in the last inequality, we have used that, since ε< k δ

vβ,γk we have εvβ,γδin Ω, and then, byk4),k ·,εvβ,γ

εβB1dγvβ,γβ a.e. inΩ. Thus, for any nonnegativeϕ∈Cc (), Z

εvβ,γ ,∇ϕ

Z

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 suppose β+2γ < 3. By Remark2.6 v),εvβ,γ ∈ H01(), and by Remark 3.3, k ·,εvβ,γ

andh ·,εvβ,γ belong to H01()0. Thus k ·,εvβ,γ

−h ·,εvβ,γ

ϕ∈L1()for any ϕ∈ H01()and, since Cc()is dense inH01(), it follows that (3.3) holds for any nonnegative ϕ∈ H01(). Remark 3.5. Let us recall the following well known result: Let g : Ω×(0,∞) → R be a Carathéodory function such thats→ g(x,s)is nonincreasing for a.e. x∈Ω, and consider the problem





∆u= g(·,u) inΩ,

u=0 on∂Ω,

u>0 inΩ.

(3.4)

Let u ∈ H01() be a weak subsolution of problem (3.4) and let u ∈ H01() 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 noting that (g(·,u)−g(·,u)) (u−u)+≤0 a.e. inΩ, we conclude thatu≤u.

Lemma 3.6. Assume the conditions k1), k4), h1), and h3) of Theorem1.3. If problem(1.4)has a weak solution u∈ H01()∩C Ω

,thenγ32.

(10)

Proof. Let u ∈ H01()∩C Ω

be a weak solution of problem (1.4). For ρ > 0, let Aρ := {x∈ :d(x)≤ρ}. Sinceu∈ C Ω

andu=0 onΩ, there existsρ> 0 such thatu≤δ in Aρ. Then, byk4),

k(·,u)≥B1dγuβ a.e. in Aρ. (3.5) Let ϕ1be a positive principal eigenfunction for−in Ω, with homogeneous Dirichlet condi- tion and weight function 1. Letε > 0, and let ϕ := ϕ

1 2+ε

1 . Then ϕ ∈ H01()∩L(). Note that, byk1) and h1), k(·,u)ϕ and h(·,u)ϕ are nonnegative measurable functions, and that, byh3),

h(·,u)ϕ=dh(·,u)d1ϕ≤ B3d2ηkukp1d1u d1ϕ

≤B3kdk2ηkukp1d1u d1ϕ

a.e. inΩ;

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

B1kukβ

Z

Aρdγϕ≤B1 Z

Aρdγuβϕ

Z

Aρk(·,u)ϕ (3.6)

Z

k(·,u)ϕ=

Z

h∇u,∇ϕi+

Z

h(·,u)ϕ<∞.

ThusR

dγϕ <∞, thereforeR

dγ+12+ε < ∞. Thenγ+ 12+ε > −1, i.e., γ < 32+ε. Since this holds for anyε>0, the lemma follows.

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

k7) There existsB1 >0 such that, for anys>0,k(·,s)≥ B1dγsβ a.e. inΩ,

then the conclusion of Lemma 3.6 remains valid when the assumption u ∈ H01()∩C Ω is weakened to u ∈ H01()∩L(). Indeed, define ϕ as in the proof of Lemma 3.6, and observe that (3.6) holds with Aρreplaced byΩ.

Proof of Theorem1.3.. We first provei). Letvβ,γ be the solution, given by Remark2.5, 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 := B

1+1β

2 vβ,γ. By Remark 3.3, k(·,z)−h(·,z) ∈ H01()0 and so, taking into account Remark2.6 v), z := B

1 1+β

2 vβ,γ is a weak supersolution of problem (1.4). In particular z and z are a subsolution and a supersolution, respectively, in the sense of distributions, of the problem−∆u = k(·,u)−h(·,u)in Ω. By diminishing εif necessary we can assume that z≤zinΩ. Moreover, by (2.6), there exist positive constantsc1andc2such thatz≥ c1ϑβ,γand z≤c2ϑβ,γ inΩ. Thus, fora.e.x∈ and for s∈[z(x),z(x)],

|k(x,s)−h(x,s)| ≤k(x,z(x)) +h(x,z(x))

≤ B2dγ(x)c1βϑβ,γβ(x) +B3dη(x)c2pϑβ,γp (x).

Also,dγϑβ,γβ anddηϑpβ,γ belong toLloc(). Then Remark2.7gives a solutionu, in the sense of distributions, of the problem−∆u=k(·,u)−h(·,u)inΩ, that satisfiesz≤ u≤z a.e. inΩ.

Consequently, for some positive constantsc1 andc2,

c1ϑβ,γ ≤ u≤c2ϑβ,γ a.e. inΩ. (3.7)

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