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Advances in Nonlinear Analysis

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On the fractional p-Laplacian equations with weight and general datum

Boumediene Abdellaoui
  • Corresponding author
  • Laboratoire d’Analyse Nonlinéaire et Mathématiques Appliquées, Département de Mathématiques, Université Abou Bakr Belkaïd, Tlemcen, Tlemcen 13000, Algeria
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/ Ahmed Attar
  • Laboratoire d’Analyse Nonlinéaire et Mathématiques Appliquées, Département de Mathématiques, Université Abou Bakr Belkaïd, Tlemcen, Tlemcen 13000, Algeria
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/ Rachid Bentifour
  • Laboratoire d’Analyse Nonlinéaire et Mathématiques Appliquées, Département de Mathématiques, Université Abou Bakr Belkaïd, Tlemcen, Tlemcen 13000, Algeria
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Published Online: 2016-12-02 | DOI: https://doi.org/10.1515/anona-2016-0072

Abstract

The aim of this paper is to study the following problem:

{(-Δ)p,βsu=f(x,u)in Ω,u=0in NΩ,

where Ω is a smooth bounded domain of N containing the origin,

(-Δ)p,βsu(x):=PVN|u(x)-u(y)|p-2(u(x)-u(y))|x-y|N+psdy|x|β|y|β

with 0β<N-ps2, 1<p<N, s(0,1), and ps<N. The main purpose of this work is to prove the existence of a weak solution under some hypotheses on f. In particular, we will consider two cases:

  • (i)

    f(x,σ)=f(x); in this case we prove the existence of a weak solution, that is, in a suitable weighted fractional Sobolev space for all fL1(Ω). In addition, if f0, we show that the problem above has a unique entropy positive solution.

  • (ii)

    f(x,σ)=λσq+g(x), σ0; in this case, according to the values of λ and q, we get the largest class of data g for which the problem above has a positive solution.

Keywords: Weighted fractional Sobolev spaces; nonlocal problems; entropy solution

MSC 2010: 49J35; 35A15; 35S15

1 Introduction and motivations

We consider the following problem:

{(-Δ)p,βsu=f(x,u)in Ω,u=0in NΩ,(1.1)

where

(-Δ)p,βsu(x):=PVN|u(x)-u(y)|p-2(u(x)-u(y))|x-y|N+psdy|x|β|y|β,

Ω is a smooth bounded domain containing the origin and f belongs to a suitable Lebesgue space.

This class of operators appear in a natural way when dealing with the improved Hardy inequality, namely, for all ϕ𝒞0(Ω) and for all q<p we have

Gs,p(ϕ)CΩΩ|v(x)-v(y)|p|x-y|N+qsw(x)p2w(y)p2𝑑x𝑑y,

where

Gs,p(ϕ)NN|ϕ(x)-ϕ(y)|p|x-y|N+ps𝑑x𝑑y-ΛN,p,sN|ϕ(x)|p|x|ps𝑑x,

ΛN,p,s is the optimal Hardy constant, w(x)=|x|-(N-ps)/p, and v(x)=ϕ(x)w(x). We refer to [18, 2, 4, 1] for a complete discussion about this fact.

In the same way, we can consider (-Δ)p,βs as an extension of the local operator -div(|x|-β|u|p-2u). This last one is strongly related to the classical Caffarelli–Khon–Nirenberg inequalities given in [11] and it was deeply analyzed in the literature. Notice that, as a consequence of the Caffarelli–Khon–Nirenberg inequalities, it is known that the weight |x|-β, with β<N-p, is an admissible weight in the sense that, if u is a weak positive supersolution to the problem

-div(|x|-β|u|p-2u)=0,

then it satisfies a weak Harnack inequality.

More precisely, there exists a positive constant κ>1 such that for all 0<q<κ(p-1) we have

(B2ρ(x0)uq(x)|x|-pβ𝑑x)1qCinfBρ(x0)u,

where B2ρ(x0)Ω and C>0 depends only on B.

We refer to [16, 19] and the references therein for a complete discussion and the proof of the Harnack inequality and its generalization of admissible weights.

Our objective in this work is to analyze the properties of the operator (-Δ)p,βs and to get the existence of a solution, in a suitable sense, to problem (1.1) for the largest class of the datum f.

The case of p-Laplacian equations is well known in the literature; we refer for example to [8, 7] where the authors proved the existence and the uniqueness of an entropy solution for L1 datum. The case of measure datum was treated in [13], where the existence of a renormalized solution is obtained.

The local case with weight was considered in [3]. The authors proved the existence and the uniqueness of an entropy solution for datum in L1.

For the operator (-Δ)p,βs the case p=2 and β=0 was analyzed in [22, 20]. Using a duality argument, in the sense of Stampacchia, the authors were able to prove the existence of a solution for any datum in L1. A more general semilinear problem was considered in [6], where the existence and the uniqueness of the solution is studied.

The case p2 and β=0, with regular data and variational structure, was treated in the year 2016 in [14, 12].

For general datum, based on some generalization of the Wolff potential theory, Kuusi, Mingione and Sire succeeded in [21] in obtaining the existence of a weak solution belonging to a suitable fractional Sobolev space.

In this paper, we will treat the case p2 and β>0. The argument considered in [21] seems to be too complicated to be adapted to our case.

Our approach is more simple and it is based on a suitable choice of a test function’s family and on some algebraic inequalities.

In the first part of the present paper, we will consider the case f(x,σ)=f(x). We prove the existence of a weak solution that is in an appropriate fractional Sobolev space. More precisely, we get the following existence result.

Theorem 1.1.

Assume that fL1(Ω). Then problem (1.1) has a weak solution u such that

ΩΩ|u(x)-u(y)|q|x-y|N+qs11|x|β|y|β𝑑y𝑑xMfor all q<N(p-1)N-s and for all s1<s,(1.2)

and Tk(u)Wβ,0s,p(Ω) for all k>0, where

Tk(a)={aif |a|k,ka|a|if |a|>k.

If p>2-sN, then uWβ,0s1,q(Ω) for all 1q<N(p-1)N-s and for all s1<s.

It is clear that for β=0, we reach the same existence and regularity result as was obtained in [21]. However, it seems that our approach is more simple and can be adapted for a large class of weighted nonlocal operators.

Next, assuming that f0, we show the existence of a positive entropy solution in the sense of Definition 2.9. The statement of our result is the following.

Theorem 1.2.

Assume that fL1(Ω) is such that f0. Then problem (1.1) has a unique entropy positive solution u in the sense of Definition 2.9 given below. Moreover, if un is the unique solution to the approximating problem

{(-Δ)p,βsun=fn(x)in Ω,un=0in NΩ

with fn=Tn(f), then Tk(un)Tk(u) strongly in Wβ,0s,p(Ω).

In the second part of the paper, we consider the case f(x,σ)=λσq+g(x). According to the values of q and λ, we prove the existence of an entropy solution for the largest class of the datum g.

The paper is organized as follows. In Section 2, we introduce some useful tools and preliminaries that we will use throughout the paper, like the weighted fractional Sobolev spaces and some related inequalities, a weak comparison principle and some algebraic inequalities. We also specify the sense in which the solutions to problem (1.1) are defined.

In Section 3, we begin by proving Theorem 1.1, namely, the case where f(x,σ)f(x). The main idea is to proceed by approximation and to pass to the limit using suitable test functions. In the second part of the section, we prove Theorem 1.2, more precisely, if f0, we are able to show that problem (1.1) has a unique positive entropy solution. In the same way, setting un as the solution of (1.1) with datum fnTn(f), we will prove that the sequence {Tk(un)}n converges to Tk(u) strongly in the corresponding weighted fractional Sobolev space.

In Section 4, we study the case where f(x,σ)=λσq+g(x) with λ>0 and g0. According to the values of q and λ, we get the largest class of the data g such that the problem (1.1) has a positive solution.

2 Functional setting and main tools

In this section, we give some functional settings that will be used below. We refer to [15, 23] for more details.

Let s(0,1), p1 and 0β<N-ps2. For simplicity of notation, we will set

dμ:=dx|x|2β and dν:=dxdy|x-y|N+ps|x|β|y|β.

Let ΩN; the weighted fractional Sobolev space Wβs,p(Ω) is defined by

Wβs,p(Ω){ϕLp(Ω,dμ):ΩΩ|ϕ(x)-ϕ(y)|p𝑑ν<+}.

Furthermore, Wβs,p(Ω) is a Banach space endowed with the norm

ϕWβs,p(Ω)=(Ω|ϕ(x)|p𝑑μ)1p+(ΩΩ|ϕ(x)-ϕ(y)|p𝑑ν)1p.

In the same way, we define the space Wβ,0s,p(Ω) as the completion of 𝒞0(Ω) with respect to the previous norm.

As in [5] (see also [15]) we can prove the following extension result.

Lemma 2.1.

Assume that ΩRN is a regular domain. Then for all wWβs,p(Ω) there exists w~Wβs,p(RN) such that w~|Ω=w and

w~Wβs,p(N)CwWβs,p(Ω),

where CC(N,s,p,Ω)>0.

The following weighted Sobolev inequality is obtained in [1] and will be used systematically in this paper.

Theorem 2.2 (Weighted fractional Sobolev inequality).

Assume that 0<s<1 and p>1 are such that ps<N. Let β<N-ps2. Then there exists a positive constant S(N,s,β) such that for all vC0(RN) we have

NN|v(x)-v(y)|p|x-y|N+psdx|x|βdy|y|βS(N,s,β)(N|v(x)|ps*|x|2βps*p)pps*,

where ps*=pNN-ps.

Moreover, if ΩRN is a bounded domain and β=N-ps2, then for all q<p there exists a positive constant C(Ω) such that

NN|v(x)-v(y)|p|x-y|N+psdx|x|βdy|y|βC(Ω)(N|v(x)|ps,q*|x|2βps,q*p)pps,q*

for all vC0(Ω), where ps,q*=pNN-qs.

Remark 2.3.

As in the case β=0, if Ω is a bounded smooth domain of N, we can endow Wβ,0s,p(Ω) with the equivalent norm

|ϕ|Wβ,0s,p(Ω)=(ΩΩ|ϕ(x)-ϕ(y)|p|x-y|N+psdxdy|x|β|y|β)1p.

Now, for wWβs,p(N), we set

(-Δ)p,βsw(x)=PVN|w(x)-w(y)|p-2(w(x)-w(y))|x-y|N+psdy|x|β|y|β.

It is clear that for all w,vWβs,p(N) we have

(-Δ)p,βsw,v=12NN|w(x)-w(y)|p-2(w(x)-w(y))(v(x)-v(y))|x-y|N+psdxdy|x|β|y|β.

In the case where β=0, we denote (-Δ)p,βs by (-Δ)ps.

The following comparison principle extends the classical one obtained by Brezis–Kamin in [10]. See [22, 1] for the proof.

Lemma 2.4.

Let Ω be a bounded domain and let h be a non-negative continuous function such that h(x,σ)>0 if σ>0 and h(x,σ)σp-1 is decreasing. Let u,vWβ,0s,p(Ω) be such that u,v>0 in Ω and

{(-Δ)p,βsuh(x,u)in Ω,(-Δ)p,βsvh(x,v)in Ω.

Then uv in Ω.

The following algebraic inequalities can be proved using a suitable rescaling argument.

Lemma 2.5.

Assume that p1, a,bR+ and α>0. Then there exist positive constants c, c1, c2 such that

(a+b)αc1aα+c2bα(2.1)

and

|a-b|p-2(a-b)(aα-bα)c|ap+α-1p-bp+α-1p|p.(2.2)

In the case where α1, under the same conditions on a, b, p as above, we have

|a+b|α-1|a-b|pc|ap+α-1p-bp+α-1p|p.(2.3)

Since we are considering a solution with datum in L1, we need to use the concept of truncation. Recall that, for k>0 we have

Tk(a)={aif |a|k,ka|a|if |a|>k.

Define Gk(a)=a-Tk(a); if we take the above definition into consideration, it is not difficult to show the algebraic inequalities

|a-b|p-2(a-b)(Tk(a)-Tk(b))|Tk(a)-Tk(b)|p(2.4)

and

|a-b|p-2(a-b)(Gk(a)-Gk(b))|Gk(a)-Gk(b)|p,

where a,b and p1.

In the same way, we will use the classical weighted Marcinkiewicz spaces.

Definition 2.6.

For a measurable function u we set

Φu(k)=μ{xΩ:|u(x)|>k},

where dμ=|x|-2βdx.

We say that u is in the Marcinkiewicz space q(Ω,dμ) if Φu(k)Ck-q.

Since Ω is a bounded domain,

Lq(Ω,dμ)q(Ω,dμ)Lq-ε(Ω,dμ)

for all ε>0.

Since we are considering problems with general datum, we need to specify the concept of solution. We begin by the following definitions.

Definition 2.7.

Let u be a measurable function. We say that u𝒯β,01,p(Ω) if for all k>0 we have Tk(u)Wβ,0s,p(Ω).

Now, we are able to state what we mean by a solution to problem (1.1).

Definition 2.8.

Assume that fL1(Ω). We say that u is a weak solution to problem (1.1) if for all ϕ𝒞0(Ω) we have

12DΩ|u(x)-u(y)|p-2(u(x)-u(y))(ϕ(x)-ϕ(y))𝑑ν=Ωf(x)ϕ(x)𝑑x.

Following [6], we define the notion of entropy solution as follows.

Definition 2.9.

Consider fL1(Ω). We say that u𝒯0,β1,p(Ω) is an entropy solution to problem (1.1) if

Rh|u(x)-u(y)|p-1𝑑ν0as h,(2.5)

where

Rh={(x,y)N×N:h+1max{|u(x)|,|u(y)|} with min{|u(x)|,|u(y)|}h or u(x)u(y)<0},

and for all k>0 and ϕWβ,0s,p(Ω)L(Ω) we have

12DΩ|u(x)-u(y)|p-2(u(x)-u(y))[Tk(u(x)-ϕ(x))-Tk(u(y)-ϕ(y))]𝑑νΩf(x)Tk(u(x)-ϕ(x))𝑑x.

Remark 2.10.

Notice that for hk, choosing ϕ=Th-1(u), we obtain that

12DΩ|u(x)-u(y)|p-2(u(x)-u(y))[Tk(Gh-1(u(x)))-Tk(Gh-1(u(y)))]𝑑νΩf(x)Tk(Gh-1(u(x)))𝑑xk|u|>h-k-1|f(x)|𝑑x.

Since

|u(x)-u(y)|p-2(u(x)-u(y))[Tk(Gh-1(u(x)))-Tk(Gh-1(u(y)))]0

in DΩ, setting

R~h={(x,y)N×N:u(x)u(y)0 with |u(x)|h and h-k-1|u(y)|h}

and

R^h={(x,y)N×N:u(x)u(y)0 with |u(x)|h and h-k-1|u(x)|h},

we reach that

12R~h|u(x)-u(y)|p-1(h-u(y))𝑑νk|u|>h-k-1|f(x)|𝑑x

and

12R^h|u(x)-u(y)|p-1(h-u(x))𝑑νk|u|>h-k-1|f(x)|𝑑x.(2.6)

It is clear that

12{h-k-1u(y)<u(x)h}(u(x)-u(y))p𝑑νk|u|>h-k-1|f(x)|𝑑x

and

12{h-k-1u(x)<u(y)h}(u(y)-u(x))p𝑑νk|u|>h-k-1|f(x)|𝑑x.(2.7)

3 Existence results: Proofs of Theorems 1.1 and 1.2

In this section, we consider the problem

{(-Δ)p,βsu=fin Ω,u=0in NΩ,(3.1)

where fL1(Ω).

The main goal of this section is to show that problem (3.1) has a weak solution u in the sense of Definition 2.8. As in the local case, the main idea is to proceed by approximation and then pass to the limit by using suitable a priori estimates.

Before proving the main existence results, we need several lemmas.

Let {fn}nL(Ω) be such that fnf strongly in L1(Ω) and define un as the unique solution to the approximated problem

{(-Δ)p,βsun=fn(x)in Ω,un=0in NΩ.(3.2)

Notice that the existence and the uniqueness of un follows by using a classical variational argument in the space Wβ,0s,p(Ω).

The first a priori estimate is given by the following Lemma.

Lemma 3.1.

Let {un}n be defined as above. Then {un}n is bounded in the space Mp1(Ω,dμ) with p1=(p-1)NN-ps.

Proof.

Using Tk(un) as a test function in (3.2), we reach that

12DΩ|un(x)-un(y)|p-2(un(y)-un(x))[Tk(un(x))-Tk(un(y))]𝑑νkΩ|fn(x)|𝑑x.

Thus,

DΩ|un(x)-un(y)|p-2(un(y)-un(x))[Tk(un(x))-Tk(un(y))]dνCk.(3.3)

Recall that un=Tk(un)+Gk(un). Then by inequalities (2.4) and (3.3), we reach that

1kDΩ|Tk(un(x))-Tk(un(y))|p𝑑νMfor all k>0.

Now, using the weighted Sobolev inequality in Theorem 2.2, we get

S(N|Tk(un(x))|ps*|x|-2βps*p𝑑x)p/ps*DΩ|Tk(un(x))-Tk(un(y))|p𝑑νCk.

Since {|un|k}={|Tk(un)|=k}, we obtain that

μ{xΩ:|un|k}μ{xΩ:|Tk(un)|=k}Ω|Tk(un(x))|ps*kps*|x|-2βps*p𝑑x.

Hence,

μ{xΩ:|un|>k}CMps*pk-(ps*-ps*p).

Setting p1=ps*-ps*p=N(p-1)N-ps, we conclude that the sequence {un}n is bounded in the space p1(Ω,dμ) and the result follows. ∎

As a consequence we easily get that the sequence {|un|p-2un}n is bounded in the space Lσ(Ω,dμ) for all σ<NN-ps.

As in the local case, we prove now that the sequence {un}n is bounded in a suitable fractional Sobolev space. More precisely we have the following lemma.

Lemma 3.2.

Assume that {un}n is defined as in Lemma 3.1. Then

ΩΩ|un(x)-un(y)|q|x-y|N+qs11|x|β|y|β𝑑y𝑑xMfor all q<N(p-1)N-s and for all s1<s.(3.4)

Proof.

Let q<N(p-1)N-s be fixed. Since Ω is a bounded domain, it is sufficient to prove (3.4) for s1 very close to s. In particular, we fix s1 such that

pq(s-s1)p-q<β.

Define wn(x)=1-1(un+(x)+1)α, where α>0 is to be chosen later and un+(x)=max{un(x),0}.

In what follows, we denote by C1,C2, any positive constants that are independent of un and can change from one line to another.

Using wn as a test function in (3.2), we get

12DΩ|un(x)-un(y)|p-2(un(x)-un(y))(un+(x)+1)α-(un+(y)+1)α(un+(x)+1)α(un+(y)+1)α𝑑νΩfn(x)𝑑x.

Hence,

DΩ|un(x)-un(y)|p-2(un(x)-un(y))(un+(x)+1)α-(un+(y)+1)α(un+(x)+1)α(un+(y)+1)α𝑑νC1.(3.5)

Let vn(x)=un+(x)+1. Since

|un(x)-un(y)|p-2(un(x)-un(y))((un+(x)+1)α-(un+(y)+1)α)|un+(x)-un+(y)|p-2(un+(x)-un+(y))((un+(x)+1)α-(un+(y)+1)α)=|vn(x)-vn(y)|p-2(vn(x)-vn(y))(vnα(x)-vnα(y))

by (3.5), it follows that

DΩ|vn(x)-vn(y)|p-2(vn(x)-vn(y))vnα(x)-vnα(y)vnα(x)vnα(y)𝑑νC1.

Now, using the fact that vn1 and by inequality (2.2), we get

DΩ|vnp+α-1p(x)-vnp+α-1p(y)|pvnα(x)vnα(y)𝑑νC2.(3.6)

Defining q1=qs1s<q and using the Hölder inequality, we find

ΩΩ|vn(x)-vn(y)|q|x-y|N+qs1dydx|x|β|y|β=ΩΩ|vn(x)-vn(y)|q|x-y|qs(vn(x)+vn(y))α-1(vn(x)vn(y))α(vn(x)vn(y))α(vn(x)+vn(y))α-1|x-y|(q-q1)sdydx|x|β|y|β|x-y|N(ΩΩ|vn(x)-vn(y)|p(vn(x)+vn(y))α-1|x-y|N+ps(v(x)v(y))α|x|β|y|β𝑑y𝑑x)qp   ×(ΩΩ(vn(x)+vn(y))α-1(v(x)v(y))α(vn(x)vn(y))αpp-q(vn(x)+vn(y))(α-1)pp-q|x-y|(q-q1)spp-qdydx|x-y|N|x|β|y|β)p-qq.(3.7)

Now, using the algebraic inequality (2.3), we have

|vn(x)-vn(y)|p(vn(x)+vn(y))α-1C|vn(x)p+α-1p-vn(y)p+α-1p|p.

Hence, taking into consideration that Ω×ΩDΩ and by (3.6), we get

(ΩΩ|vn(x)-vn(y)|p(vn(x)+vn(y))α-1|x-y|N+ps(v(x)v(y))α|x|β|y|β𝑑y𝑑x)qpC(DΩ|vn(x)p+α-1p-vn(y)p+α-1p|p|x-y|N+ps(vn(x)vn(y))α|x|β|y|β𝑑y𝑑x)qpC3.

So, going back to (3.7), we reach that

ΩΩ|vn(x)-vn(y)|q|x-y|N+qs1dydx|x|β|y|βC4(ΩΩ((vn(x)vn(y))α(vn(x)+vn(y))α)qp-q(vn(x)+vn(y))qp-q1|x-y|N-ps(q-q1)p-qdydx|x|β|y|β)p-qq.

By inequality (2.1), we have

(vn(x)+vn(y))(vn(x)vn(y)vn(x)+vn(y))αc(vn(x)+vn(y))α+1C5(vnα+1(x)+vnα+1(y)).

Therefore,

ΩΩ|vn(x)-vn(y)|q|x-y|N+qs1dxdy|x|β|y|βC5(ΩΩvn(α+1)qp-q(x)dxdy|x-y|N-ps(q-q1)p-q|x|β|y|β)p-qq+C5(ΩΩvn(α+1)qp-q(y)dxdy|x-y|N-ps(q-q1)p-q|x|β|y|β)p-qq.(3.8)

It is clear that

ΩΩvn(α+1)qp-q(x)dxdy|x-y|N-ps(q-q1)p-q|x|β|y|β=ΩΩvn(α+1)qp-q(y)dxdy|x-y|N-ps(q-q1)p-q|x|β|y|β,

hence we just have to estimate the first term. We have

ΩΩvn(α+1)qp-q(x)dxdy|x-y|N-ps(q-q1)p-q|x|β|y|β=Ωvn(α+1)qp-q(x)|x|β𝑑xΩdy|x-y|N-ps(q-q1)p-q|y|β.

Since Ω is a bounded domain, ΩBR(0). Thus

ΩΩvn(α+1)qp-q(x)dxdy|x-y|N-ps(q-q1)p-q|x|β|y|βBR(0)vn(α+1)qp-q(x)|x|β𝑑xBR(0)dy|x-y|N-ps(q-q1)p-q|y|β,

where vn=1 in BR(0)Ω. We set r=|x| and ρ=|y|. Then x=rx, y=ρy, where |x|=|y|=1.

Let

τ=(α+1)qp-qandθ=ps(q-q1)p-q.(3.9)

Then

ΩΩvnτ(x)dxdy|x-y|N-θ|x|β|y|βBR(0)vnτ(x)dx|x|β0RρN-1ρβrN-θ(|y|=1dHN-1(y)|x-ρry|N-θ)𝑑ρ.

We set σ=ρr; hence

ΩΩvnτ(x)dxdy|x-y|N-θ|x|β|y|βBR(0)vnτ(x)dx|x|2β-θ0RrσN-β-1(|y|=1dHN-1(y)|x-σy|N-θ)𝑑σ.

Defining

Kθ(σ)=|y|=1dHN-1(y)|x-σy|N-θ

as in [17], we find

Kθ(σ)=2πN-12β(N-12)0πsinN-2(ξ)(1-2σcos(ξ)+σ2)N-θ2𝑑ξ.

Notice that Kθ(σ)C|1-σ|-1+θ as σ1 and Kθ(σ)σθ-N as σ.

Therefore, there holds

ΩΩvnτ(x)dxdy|x-y|N-θ|x|β|y|βBR3(0)vnτ(x)dx|x|2β-θ0RrσN-β-1Kθ(σ)𝑑σ+BR(0)BR3(0)vnτ(x)dx|x|2β-θ0RrσN-β-1Kθ(σ)𝑑σ.(3.10)

Recall that r=|x|. Then if xBR(0)BR3(0), we have Rr<3. Hence taking into consideration that θ>0 and the behavior of Kθ near 1, we reach that

0RrσN-β-1Kθ(σ)𝑑σC103σN-β-1|1-σ|1-θ𝑑σ=C2<.

Now, if r|x|<R3, there holds

0RrσN-β-1Kθ(σ)𝑑σ=03σN-β-1Kθ(σ)𝑑σ+3RrσN-β-1Kθ(σ)𝑑σC2+(Rr)a3RrσN-β-a-1Kθ(σ)𝑑σ,

where a>0 is to be chosen later.

Since

3RrσN-β-a-1Kθ(σ)𝑑σ3σN-β-a-1Kθ(σ)𝑑σ,

using the fact that σN-β-a-1Kθ(σ)σ-1-β-a+θ as σ and choosing a>θ, it follows that

0σN-β-a-1K(σ)𝑑σC3<.

Now, going back to (3.10), there holds

ΩΩvnτ(x)dxdy|x-y|N-θ|x|β|y|βC1BR(0)vnτ(x)dx|x|2β-θ𝑑x+C2RaBR3(0)vnτ(x)dx|x|2β+a-θ𝑑xC(R)BR(0)vnτ(x)dx|x|2β+a-θ𝑑x.

Since q<(p-1)NN-s, we can choose α>0 in (3.9) such that τ<(p-1)NN-ps. By Lemma 3.1, choosing a>θ, very close to θ and using the Hölder inequality, we reach that

ΩΩvnτ(x)dydx|x-y|N-θ|x|β|y|βC6BR(0)vnτ(x)dx|x|2β+a-θC7for all n.(3.11)

Hence by (3.8) and (3.11), we conclude that

ΩΩ|un+(x)-un+(y)|q|x-y|N+qs1dydx|x|β|y|β=ΩΩ|vn(x)-vn(y)|q|x-y|N+qs1dxdy|x|β|y|βC8.

In the same way, by using 1-1/((un-(x)+1)α) as a test function in (3.2), we obtain that

ΩΩ|un-(x)-un-(y)|q|x-y|N+qs1dydx|x|β|y|βC9.

Combining the above estimates, we reach that

ΩΩ|un(x)-un(y)|q|x-y|N+qs1dydx|x|β|y|βC

and the result follows. ∎

Remark 3.3.

As a consequence we get the existence of a measurable function u such that Tk(u)Wβ,0s,p(Ω), |u|p-2uLσ(Ω,|x|-2βdx) for all σ<NN-ps, and Tk(un)Tk(u) weakly in Wβ,0s,p(Ω).

It is clear that unu a.e. in Ω. Since un=0 a.e. in NΩ, we have u=0 a.e. in NΩ.

Notice that by Lemma 3.1 we conclude that

|un|p-2un|u|p-2ustrongly in La(Ω,dμ) for all a<NN-ps.

Let

Un(x,y)=|un(x)-un(y)|p-2(un(x)-un(y))andU(x,y)=|u(x)-u(y)|p-2(u(x)-u(y)).

Since Ω is a bounded domain, by the result of Lemma 3.2 and using Vitali’s lemma, we obtain that

UnUstrongly in L1(Ω×Ω,dν).

We are now able to prove the first existence result.

Proof of Theorem 1.1.

It is clear that estimate (1.2) follows by using Lemma 3.2 and Fatou’s lemma.

Let ϕ𝒞0(Ω). Then, by using ϕ as a test function in (3.2), it follows that

12DΩ|un(x)-un(y)|p-2(un(x)-un(y))(ϕ(x)-ϕ(y))𝑑ν=Ωfn(x)ϕ(x)𝑑x.(3.12)

We set Φ(x,y)=ϕ(x)-ϕ(y). By (3.12), we have

12DΩU(x,y)Φ(x,y)𝑑ν+12DΩ(Un(x,y)-U(x,y))Φ(x,y)𝑑ν=Ωfn(x)ϕ(x)𝑑x.(3.13)

It is clear that

Ωfn(x)ϕ(x)𝑑xΩf(x)ϕ(x)𝑑xas n.

We claim that

DΩ(Un(x,y)-U(x,y))Φ(x,y)𝑑ν0as n.(3.14)

Since unu a.e. in Ω, it follows that

Un(x,y)Φ(x,y)|x-y|N+ps|x|β|y|βU(x,y)Φ(x,y)|x-y|N+ps|x|β|y|βa.e. in DΩ.

Using the fact that u(x)=un(x)=ϕ(x)=0 for all xNΩ, we obtain

NΩNΩ(Un(x,y)-U(x,y))Φ(x,y)𝑑ν=0.

Thus,

DΩ(Un(x,y)-U(x,y))Φ(x,y)𝑑ν=Ω×Ω(Un(x,y)-U(x,y))Φ(x,y)𝑑ν+NΩΩ(Un(x,y)-U(x,y))Φ(x,y)𝑑ν   +ΩNΩ(Un(x,y)-U(x,y))Φ(x,y)𝑑ν=I1(n)+I2(n)+I3(n).

Using Lemma 3.2 and Remark 3.3, we easily find that

I1(n)0as n.

We now deal with I2(n). It is clear that for (x,y)Ω×BRΩ we have

|(Un(x,y)-U(x,y))Φ(x,y)|(|un(x)|p-1+|u(x)|p-1)|ϕ(x)|.

Since

sup{xSuppϕ,yBRΩ}1|x-y|N+psC,

we have

|(Un(x,y)-U(x,y))Φ(x,y)|x-y|N+ps|x|β|y|β|(|un(x)|p-1+|u(x)|p-1)|ϕ(x)||x|β|y|βQn(x,y).

Using Lemma 3.1 and Remark 3.3, we get QnQ strongly in L1(Ω×BRΩ) with

Q(x,y)=2|u(x)|p-1|ϕ(x)||x|-β|y|-β.

Thus by the dominated convergence theorem we arrive to I2(n)0 as n. In the same way, we obtain that I3(n)0 as n. Hence (3.14) follows and the claim is proved.

Therefore, passing to the limit in (3.13), we arrive at

12DΩU(x,y)Φ(x,y)𝑑ν=Ωf(x)ϕ(x)𝑑x.

Remark 3.4.

(i) It is clear that the same existence result holds if we replace f by a bounded Radon measure ν.

(ii) In the case where β=0, we get the same existence and regularity results as were obtained in [21].

3.1 The case of positive datum: Existence and uniqueness of the positive entropy solution

If f0, we choose fn=Tn(f), thus {un}n is an increasing sequence. In this case, we are able to prove that problem (3.1) has a unique entropy positive solution in the sense of Definition 2.9. Before beginning with the proof of Theorem 1.2, let us prove the following compactness result.

Lemma 3.5.

Let {un}n and u be defined as above. Then

Tk(un)Tk(u)strongly in Wβ,0s,p(Ω).

The proof of Lemma 3.5 will be a consequence of the following more general compactness result.

Lemma 3.6.

Let {un}nWβ,0s,p(Ω) be an increasing sequence such that un0 and (-Δ)p,βsun0. Assume further that {Tk(un)}n is bounded in Wβ,0s,p(Ω) for all k>0. Then there exists a measurable function u such that unu a.e. in Ω, Tk(u)Wβ,0s,p(Ω) for all k>0 and

Tk(un)Tk(u)strongly in Wβ,0s,p(Ω).

Proof.

Since {Tk(un)}n is bounded in Wβ,0s,p(Ω), using the monotony of the sequence {un}n, we get the existence of a measurable function u such that unu a.e. in Ω, Tk(u)Wβ,0s,p(Ω) and Tk(un)Tk(u) weakly in Wβ,0s,p(Ω). Since (-Δ)p,βsun0, we have

(-Δ)p,βsun,Tk(un)-Tk(u)0.

Thus,

DΩ|un(x)-un(y)|p-2(un(x)-un(y))(Tk(un(x))-Tk(un(y)))𝑑νDΩ|un(x)-un(y)|p-2(un(x)-un(y))(Tk(u(x))-Tk(u(y)))𝑑ν.(3.15)

Define

I1,nDΩ|un(x)-un(y)|p-2(un(x)-un(y))(Tk(un(x))-Tk(un(y)))𝑑ν

and

I2,nDΩ|un(x)-un(y)|p-2(un(x)-un(y))(Tk(u(x))-Tk(u(y)))𝑑ν.

For simplicity of notation, we set

Tn,k(x,y)|Tk(un(x))-Tk(un(y))|p-2(Tk(un(x))-Tk(un(y))).

We have

I1,n=DΩ|Tk(un(x))-Tk(un(y))|p𝑑ν+DΩ[Un(x,y)-Tn,k(x,y)](Tk(un(x))-Tk(un(y)))𝑑ν.

In the same way, using Young inequality, we obtain that

I2,n=DΩTn,k(x,y)(Tk(u(x))-Tk(u(y)))𝑑ν+DΩ[Un(x,y)-Tn,k(x,y)](Tk(u(x))-Tk(u(y)))𝑑νp-1pDΩ|Tk(un(x))-Tk(un(y))|p𝑑ν+1pDΩ|Tk(u(x))-Tk(u(y))|p𝑑ν+DΩ[Un(x,y)-Tn,k(x,y)](Tk(u(x))-Tk(u(y)))𝑑ν.

Combining the above estimates and going back to (3.15), we find

1pDΩ|Tk(un(x))-Tk(un(y))|p𝑑ν+DΩ[Un(x,y)-Tn,k(x,y)]×[(Tk(un(x))-Tk(u(x)))-(Tk(un(y))-Tk(u(y)))]dν   1pDΩ|Tk(u(x))-Tk(u(y))|pdν.(3.16)

Define

Kn(x,y)[Un(x,y)-Tn,k(x,y)][(Tk(un(x))-Tk(u(x)))-(Tk(un(y))-Tk(u(y)))].(3.17)

We claim that Kn(x,y)0 a.e. in DΩ. We set

D1={(x,y)DΩ:un(x)k,un(y)k},D2={(x,y)DΩ:un(x)k,un(y)k},D3={(x,y)DΩ:un(x)k,un(y)k},D4={(x,y)DΩ:un(x)k,un(y)k}.

Then DΩ=D1D2D3D4.

In D1 we have Un(x,y)-Tn,k(x,y)=0. Then Kn(x,y)=0. In the same way, if (x,y)D2, we have u(x)un(x)k and u(y)un(y)k. Then

[(Tk(un(x))-Tk(u(x)))-(Tk(un(y))-Tk(u(y)))]=0.

Thus, Kn(x,y)=0 in D2.

Assume that (x,y)D3. Then

Un(x,y)-Tn,k(x,y)=(un(x)-un(y))p-1-(k-un(y))p-10.

Since

[(Tk(un(x))-Tk(u(x)))-(Tk(un(y))-Tk(u(y)))]=-(Tk(un(y))-Tk(u(y)))0

by (3.17), it follows that Kn(x,y)0 in D3. In the same way, we can prove that Kn(x,y)0 in D4. Thus Kn(x,y)0 a.e. in DΩ and the claim follows.

Going back to (3.16), there results that

lim supnDΩ|Tk(un(x))-Tk(un(y))|p𝑑νDΩ|Tk(u(x))-Tk(u(y))|p𝑑ν.

Since Tk(un)Tk(u) weakly in Wβ,0s,p(Ω), we obtain Tk(un)Tk(u) strongly in Wβ,0s,p(Ω). ∎

Remark 3.7.

(i) As a consequence of the previous strong convergence we reach that

DΩKn(x,y)𝑑ν0as n.

(ii) Letting wn=1-11+un and w=1-11+u, and using wn as a test function in (3.2), we have

DΩ|un(x)-un(y)|p(1+un(x))(1+un(y))𝑑ν=Ωfn(x)wn(x)𝑑xΩf(x)w(x)𝑑xas n.

For k>0 fixed, we define the sets

An=DΩ{un(x)2k,un(y)k}andA=DΩ{u(x)2k,u(y)k}.

It is clear that for (x,y)An we have un(x)-un(y)12un(x). Thus,

DΩunp-1(x)χAn(x,y)𝑑νC(k)DΩ|un(x)-un(y)|p(1+un(x))(1+un(y))𝑑ν<C¯(k).(3.18)

Since unχ{An}(x,y)uχ{A} a.e. in DΩ, if p>2, we get

unχ{An}uχ{A}weakly in Lp-1(DΩ,dν).

(iii) From (3.18) we conclude that

ν{DΩAn}DΩAn𝑑νC~(k).

Hence by Fatou’s lemma, we reach that

ν{DΩA}DΩA𝑑νC~(k).

Now, we are in the position to prove the existence and the uniqueness of the entropy solution.

Proof of Theorem 1.2: Existence part.

It is clear that the existence of u follows by using Theorem 1.1, however the strong convergence of {Tk(un)}n in the space Wβ,0s,p(Ω) is a consequence of Lemma 3.5. To finish we just need to show that u is an entropy solution to problem (3.1) in the sense of Definition 2.9.

Let us begin by proving that (2.5) holds.

Since u,un0, the set Rh given in Definition 2.9 is reduced to

Rh={(x,y)N×N:h+1max{u(x),u(y)} with min{u(x),u(y)}h}.

Using T1(Gh(un)) as a test function in (3.2), we obtain

12DΩ|un(x)-un(y)|p-2(un(x)-un(y))[T1(Gh(un(x)))-T1(Gh(un(y)))]𝑑ν=Ωfn(x)T1(Gh(un(x)))𝑑xunhfn(x)𝑑x.(3.19)

It is not difficult to show that, for (x,y)Rh, we have

|un(x)-un(y)|p-2(un(x)-un(y))[T1(Gh(un(x)))-T1(Gh(un(y)))]0.

Thus, using Fatou’s lemma and by (3.19), we conclude that

12DΩ|u(x)-u(y)|p-2(u(x)-u(y))[T1(Gh(u(x)))-T1(Gh(u(y)))]𝑑νlim infn12DΩ|un(x)-un(y)|p-2(un(x)-un(y))[T1(Gh(un(x)))-T1(Gh(un(y)))]𝑑νΩf(x)T1(Gh(u(x)))𝑑xuhf(x)𝑑x.(3.20)

It is clear that for all (x,y)Rh we have

|u(x)-u(y)|p-2(u(x)-u(y))[T1(Gh(u(x)))-T1(Gh(u(y)))]|u(x)-u(y)|p-1.

Therefore, using the fact that

uhf(x)𝑑x0as h

and by (3.20), we conclude that

Rh|u(x)-u(y)|p-1𝑑ν0as h.

Hence (2.5) holds.

Recall that

Un(x,y)=|un(x)-un(y)|p-2(un(x)-un(y))andU(x,y)=|u(x)-u(y)|p-2(u(x)-u(y)).

Let vWβ,0s,p(Ω)L(Ω). Taking Tk(un-v) as a test function in (3.2), we reach that

12DΩUn(x,y)[Tk(un(x)-v(x))-Tk(un(y)-v(y))]𝑑ν=Ωfn(x)Tk(un(x)-v(x))𝑑x.

One immediately sees that

Ωfn(x)Tk(un(x)-v(x))𝑑xΩf(x)Tk(u(x)-v(x))𝑑xas n.

We now deal with the first term. We have

Un(x,y)[Tk(un(x)-v(x))-Tk(un(y)-v(y))]=:K1,n(x,y)+K2,n(x,y),(3.21)

where

K1,n(x,y)=|(un(x)-v(x))-(un(y)-v(y))|p-2((un(x)-v(x))-(un(y)-v(y)))×[Tk(un(x)-v(x))-Tk(un(y)-v(y))]

and

K2,n(x,y)=[Un(x,y)-|(un(x)-v(x))-(un(y)-v(y))|p-2((un(x)-v(x))-(un(y)-v(y)))]×[Tk(un(x)-v(x))-Tk(un(y)-v(y))].

It is clear that K1,n(x,y)0 a.e. in DΩ. Since

K1,n(x,y)|(u(x)-v(x))-(u(y)-v(y))|p-2((u(x)-v(x))-(u(y)-v(y)))×[Tk(u(x)-v(x))-Tk(u(y)-v(y))]a.e. in DΩ

as n, using Fatou’s lemma, we obtain that

DΩK1,n(x,y)𝑑νDΩ[|(u(x)-v(x))-(u(y)-v(y))|p-2((u(x)-v(x))-(u(y)-v(y)))]×[Tk(u(x)-v(x))-Tk(u(y)-v(y))]dν.(3.22)

We now deal with K2,n.

We set

wn=un-v,w=u-v,σ1(x,y)=un(x)-un(y),σ2(x,y)=wn(x)-wn(y).

Then

K2,n(x,y)=[|σ1(x,y)|p-2σ1(x,y)-|σ2(x,y)|p-2σ2(x,y)]×[Tk(un(x)-v(x))-Tk(un(y)-v(y))].

We claim that

DΩK2,n(x,y)𝑑νDΩ[U(x,y)-|(u(x)-v(x))-(u(y)-v(y))|p-2((u(x)-v(x))-(u(y)-v(y)))]×[Tk(u(x)-v(x))-Tk(u(y)-v(y))]dνas n.(3.23)

We divide the proof of the claim into two cases according to the value of p.

The singular case: p(1,2].

In this case, we have

||σ1(x,y)|p-2σ1(x,y)-|σ2(x,y)|p-2σ2(x,y)|C|σ1(x,y)-σ2(x,y)|p-1=C|v(x)-v(y)|p-1.

Thus,

|K2,n(x,y)|C|v(x)-v(y)|p-1|Tk(un(x)-v(x))-Tk(un(y)-v(y))|K~2,n(x,y).(3.24)

Since Tk(un)Tk(u) strongly in Wβ,0s,p(Ω), we get

K~2,nC|v(x)-v(y)|p-1|Tk(u(x)-v(x))-Tk(u(y)-v(y))|strongly in L1(DΩ,dν)

as vWβ,0s,p(Ω)L(Ω). Using the dominated convergence theorem, we reach that

DΩK2,n(x,y)𝑑νDΩ[U(x,y)-|(u(x)-v(x))-(u(y)-v(y))|p-2((u(x)-v(x))-(u(y)-v(y)))]×[Tk(u(x)-v(x))-Tk(u(y)-v(y))]dν

as n and then (3.23) follows in this case.

The degenerate case: p>2. This case is more relevant. As in the previous case, we have

||σ1(x,y)|p-2σ1(x,y)-|σ2(x,y)|p-2σ2(x,y)|C1|σ1(x,y)-σ2(x,y)|p-1+C2|σ2(x,y)|p-2|σ1(x,y)-σ2(x,y)|C1|v(x)-v(y)|p-1+C2|v(x)-v(y)||wn(x)-wn(y)|p-2C1|v(x)-v(y)|p-1+C2|v(x)-v(y)||un(x)-un(y)|p-2.

Thus,

|K2,n(x,y)|C1|v(x)-v(y)|p-1|Tk(un(x)-v(x))-Tk(un(y)-v(y))|+C2|v(x)-v(y)||un(x)-un(y)|p-2|Tk(wn(x))-Tk(wn(y))|K¯2,n(x,y)+Kˇ2,n(x,y).

The term K¯2,n(x,y) can be treated as K~2,n defined in (3.24). Hence it remains to deal with Kˇ2,n(x,y).

We define

D1={(x,y)DΩ:un(x)k~,un(y)k~},

where k~k+v is a large constant. Using the fact that Tk~(un)Tk~(u) strongly in Wβ,0s,p(Ω), we obtain that

Kˇ2,n(x,y)χ{D1}C2|v(x)-v(y)||u(x)-u(y)|p-2|Tk(w(x))-Tk(w(y))|χ{u(x)k~,u(y)k~}

strongly in L1(DΩ,dν).

Now, consider the set

D2={(x,y)DΩ:un(x)k1,un(y)k1},

where k1>k+v. Then Kˇ2,n(x,y)χ{D2}=0.

Hence we just have to deal with sets of the form

D3={(x,y)DΩ:un(x)2k,un(y)k}

or

D4={(x,y)DΩ:un(y)2k,un(x)k}.

We will use Remark 3.7 and a duality argument.

It is clear that for (x,y)D3 we have

Kˇ2,n(x,y)χ{D3}(x,y)C(k)|v(x)-v(y)||Tk(wn(x))-Tk(wn(y))|unp-2(x)χ{D3}(x,y).

From Remark 3.7, we know that

unp-2χ{D3}up-2χ{u(x)2k,u(y)k}weakly in Lp-1p-2(DΩ,dν).(3.25)

Notice that

[|v(x)-v(y)||Tk(wn(x))-Tk(wn(y))|]p-1p-1p|Tk(wn(x))-Tk(wn(y))|p+1p|v(x)-v(y)|p(p-1)p-1p|Tk(wn(x))-Tk(wn(y))|p+1p(2v)p(p-2)|v(x)-v(y)|p=:Ln(x,y).

It is clear that LnL strongly in L1(DΩ,dν) with

L(x,y)=p-1p|Tk(w(x))-Tk(w(y))|p+1p(2v)p(p-2)|v(x)-v(y)|p.(3.26)

Thus by (3.25), (3.26) and using a duality argument, we find that

Kˇ2,nχ{D3}C2|v(x)-v(y)||u(x)-u(y)|p-2|Tk(w(x))-Tk(w(y))|χ{u(x)2k,u(y)k}

strongly in L1(DΩ,dν).

We can treat the set D4 in the same way.

Therefore, combining the above estimates and using the dominated convergence theorem, we conclude that

DΩK2,n(x,y)𝑑νDΩ[U(x,y)-|(u(x)-v(x))-(u(y)-v(y))|p-2((u(x)-v(x))-(u(y)-v(y)))]×[Tk(u(x)-v(x))-Tk(u(y)-v(y))]dνas n,

and the claim follows.

Hence by (3.21)–(3.23), we conclude that

12DΩU(x,y)[Tk(u(x)-v(x))-Tk(u(y)-v(y))]𝑑νΩf(x)Tk(u(x)-v(x))𝑑x(3.27)

and the result follows at once. ∎

It is clear that if u is an entropy solution of (3.1), then for all wC0(Ω) we have

12DΩU(x,y)(w(x)-w(y))𝑑ν=Ωf(x)w(x)𝑑x,(3.28)

where U(x,y)=|u(x)-u(y)|p-2(u(x)-u(y)).

Moreover, we can prove that (3.28) holds for all wWβ,0s,p(Ω)L(Ω) such that w0 in the set {u>k} for some k>0. More precisely we have the following lemma.

Lemma 3.8.

Assume that u is an entropy solution to (3.1) with f0. Then for all wWβ,0s,p(Ω)L(Ω) such that for some k>0 we have w0 in the set {u>k}, we obtain

12DΩU(x,y)(w(x)-w(y))𝑑ν=Ωf(x)w(x)𝑑x.(3.29)

Proof.

Let wWβ,0s,p(Ω)L(Ω) be such that w0 in the set {u>k0} for some k0>0. Define vh=Th(u-w) with hk0+w+1.

Since u is an entropy solution to (3.1), by (3.27), for k fixed such that kmax{k0,w}, we have

12DΩU(x,y)[Tk(u(x)-vh(x))-Tk(u(y)-vh(y))]𝑑νΩf(x)Tk(u(x)-vh(x))𝑑x.(3.30)

It is clear that

Ωf(x)Tk(u(x)-vh(x))𝑑xΩf(x)w𝑑xas h.

Notice that for hw we have {uw-h}=, thus for h as above there results that

{|u(x)-w(x)|h}{(u(x)-w(x))h}.

Define

Ah{(x,y)DΩ:|u(x)-w(x)|<h,|u(y)-w(y)|<h},Bh{(x,y)DΩ:|u(x)-w(x)|h,|u(y)-w(y)|h}={(x,y)DΩ:(u(x)-w(x))h,(u(y)-w(y))h},Eh{(x,y)DΩ:(u(x)-w(x))h,|u(y)-w(y)|h},Fh{(x,y)DΩ:|u(x)-w(x)|<h,(u(y)-w(y))>h}.

Then

DΩU(x,y)[Tk(u(x)-vh(x))-Tk(u(y)-vh(y))]𝑑ν=Ah+Bh+Eh+Fh=IAh+IBh+IEh+IFh.

It is clear that

IAh=AhU(x,y)[Tk(w(x))-Tk(w(y))]𝑑ν=AhU(x,y)[w(x)-w(y)]𝑑ν=Ah{u(x)<k0,u(y)<k0}U(x,y)[w(x)-w(y)]𝑑ν+Ah{u(x)>k0,u(y)>k0}U(x,y)[w(x)-w(y)]𝑑ν+Ah{u(x)>k0,u(y)k0}U(x,y)[w(x)-w(y)]𝑑ν+Ah{u(x)k0,u(y)>k0}U(x,y)[w(x)-w(y)]𝑑ν=I1(h)+I2(h)+I3(h)+I4(h).

Since Tk(u)Wβ,0s,p(Ω), we have

I1(h){u(x)<k0,u(y)<k0}U(x,y)[w(x)-w(y)]𝑑νas h.

Using the properties of w, we have I2(h)=0. Let us consider now I3(h). We have

I3(h)=Ah{k0<u(x)<2k0,u(y)k0}U(x,y)[w(x)-w(y)]𝑑ν+Ah{u(x)>2k0,u(y)k0}U(x,y)[w(x)-w(y)]𝑑ν=J1(h)+J2(h).

As above, since Tk(u)Wβ,0s,p(Ω), we obtain

J1(h){k0<u(x)<2k0,u(y)<k0}U(x,y)[w(x)-w(y)]𝑑νas h.(3.31)

For J2(h) we have

|U(x,y)[w(x)-w(y)]|w|U(x,y)|=w|u(x)-u(y)|p-1.

Using the fact that

{u(x)>2k0,u(y)k0}|U(x,y)|𝑑ν<,

by the dominated convergence theorem, we conclude that

J2(h){u(x)2k0,u(y)<k0}U(x,y)[w(x)-w(y)]𝑑νas h.(3.32)

Hence by (3.31) and (3.32), we obtain that

I3(h){k0<u(x)<2k0,u(y)<k0}U(x,y)[w(x)-w(y)]𝑑ν+{u(x)2k0,u(y)<k0}U(x,y)[w(x)-w(y)]𝑑ν

as h.

We can treat I4(h) in the same way. Hence,

IAhDΩU(x,y)[w(x)-w(y)]𝑑νas h.

We now deal with IBh. It is clear that if (x,y)Bh, then vh(x)=vh(y)=h, hence

IBh=BhU(x,y)[Tk(u(x)-h)-Tk(u(y)-h)]𝑑ν0.

Now, for (x,y)Eh, we have u(x)h-w>k0, thus w(x)=0. Hence,

EhEh{u(y)<h-w-1}Eh{hu(x),u(y)h-w-1}E1(h)E2(h).

It is clear that for (x,y)E2(h) we have w(x)=w(y)=0. Then

U(x,y)[Tk(u(x)-v(x))-Tk(u(y)-vh(y))]=U(x,y)[Tk(Gh(u(x)))-Tk(Gh(u(y)))]0.

Thus,

E2(h)U(x,y)[Tk(u(x)-v(x))-Tk(u(y)-vh(y))]𝑑ν0.

Therefore, we conclude that

IEhE1(h)U(x,y)[Tk(Gh(u(x)))-Tk(w(y))]𝑑ν-2kE1(h)|U(x,y)|𝑑ν.

Let h1=h-w-1. Then by (2.5) we reach that

E1(h)|U(x,y)|𝑑νu(x)>h1,u(y)<h1-1|U(x,y)|𝑑ν0as h.

Thus, IEho(h).

In the same way, we can prove that IFho(h). Therefore, we reach that

lim infh12DΩU(x,y)[Tk(u(x)-v(x))-Tk(u(y)-vh(y))]𝑑ν12DΩU(x,y)(w(x)-w(y))𝑑ν.

As a conclusion, and going back to (3.30), we have proved that

12DΩU(x,y)(w(x)-w(y))𝑑νΩf(x)w(x)𝑑x.

Substituting w by -w in the above inequality, we obtain that

12DΩU(x,y)(w(x)-w(y))𝑑ν=Ωf(x)w(x)𝑑x,

which is the desired result. ∎

Now we are in a position to prove the uniqueness result in Theorem 1.2.

Proof of Theorem 1.2: Uniqueness part.

Let u be the entropy positive solution defined in Theorem 1.1. Recall that u=lim supun, where un is the unique solution to the approximated problem (3.2).

Assume that v is another entropy positive solution to problem (3.1). We claim that unv for all n. To prove the claim, we fix n and define wn=(un-v)+. Then wn=(un-Tk(v))+, where kun. Hence wnWβ,0s,p(Ω)L(Ω) and wn0 in the set {v>un}. Therefore, using wn as a test function in (3.2) and taking into consideration the identity (3.29) in Lemma 3.8, we reach that

12DΩUn(x,y)(wn(x)-wn(y))𝑑ν=Ωfn(x)wn(x)𝑑xΩf(x)wn(x)𝑑x=12DΩV(x,y)(wn(x)-wn(y))𝑑ν,

where

Un(x,y)=|un(x)-un(y)|p-2(un(x)-un(y))andV(x,y)=|v(x)-v(y)|p-2(v(x)-v(y)).

Thus,

12DΩ(Un(x,y)-V(x,y))(wn(x)-wn(y))𝑑ν0.

Using the fact that

(Un(x,y)-V(x,y))(wn(x)-wn(y))C|wn(x)-wn(y)|p,

it follows that wn0, hence unv for all n and the claim follows. As a consequence we obtain that uv.

Let us now prove that vu. To this end, we will follow closely the argument used in [7].

Since u and v are entropy solutions to (3.1), for hk, we have

12DΩU(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑νΩf(x)Tk(u(x)-Th(v(x)))𝑑x(3.33)

and

12DΩV(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]𝑑νΩf(x)Tk(v(x)-Th(u(x)))𝑑x.(3.34)

It is clear that

Ωf(x)Tk(u(x)-Th(v(x)))𝑑x+Ωf(x)Tk(v(x)-Th(u(x)))𝑑x0as h.

Thus from (3.33) and (3.34), we have

I(h)12DΩU(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν+12DΩV(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]𝑑ν=P(h)+Q(h)o(h).(3.35)

Let

DΩ1(h){(x,y)DΩ:u(x)<h and u(y)<h}

and

DΩ2(h){(x,y)DΩ:v(x)<h and v(y)<h}.

Then

P(h)=DΩ1(h)U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν+DΩDΩ1(h)U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν=P1(h)+P2(h)

and

Q(h)=DΩ2(h)V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]𝑑ν+DΩDΩ2(h)V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]𝑑ν=Q1(h)+Q2(h).

We claim that P2(h)o(h) and Q2(h)o(h).

Let us begin by proving that P2(h)o(h). Recall that uv. Then

DΩDΩ1(h)={(x,y)DΩ:u(x)h}{(x,y)DΩ:u(y)h}.

If u(x)h and u(y)h, then v(x)h and v(y)h. Thus,

U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]=U(x,y)[Tk(u(x)-h)-Tk(u(y)-h)]0.

On the other hand, by (2.5), we find

{u(x)>h,u(y)<h-1}|U(x,y)|𝑑ν=o(h).

Hence,

{u(x)h}U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν{u(x)h,h-1u(y)h}U(x,y)[Tk(u(x)-h)-Tk(u(y)-Th(v(y)))]𝑑ν+o(h).(3.36)

Notice that for (x,y){u(x)h,h-1u(y)h} we have

U(x,y)[Tk(u(x)-h)-Tk(u(y)-Th(v(y)))]=U(x,y)[Tk(u(x)-h)+Tk(Th(v(y))-u(y))]0.

Then by (3.36) there results that

{u(x)h}U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑νo(h).(3.37)

In the same way, we can prove that

{u(y)h}U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑νo(h).(3.38)

Thus combining (3.37) and (3.38), we arrive at P2(h)o(h) as claimed.

We now deal with Q2(h). Recall that

Q2(h)=DΩDΩ2(h)V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]𝑑ν.

As above, we have

DΩDΩ2(h)={(x,y)DΩ:v(x)h}{(x,y)DΩ:v(y)h}M1(h)M2(h)M3(h),

where

M1(h)={(x,y)DΩ:v(x)h and v(y)h},M2(h)={(x,y)DΩ:v(x)h and v(y)<h},

and

M3(h)={(x,y)DΩ:v(x)<h and v(y)h}.

Let

Z1(h)={(x,y)DΩ:v(x)-Th(u(x))k}andZ2(h)={(x,y)DΩ:v(y)-Th(u(y))k}.

If (x,y)Z1(h)Z2(h), we have

V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]=0.

Hence we can assume that

(x,y)(Z1(h)Z2(h))(Z2(h)Z1(h))Y1(h)Y2(h).

Therefore, we conclude that

Q2(h)=M1(h)Y1(h)+M2(h)Y1(h)+M3(h)Y1(h)+M1(h)Y2(h)+M2(h)Y2(h)+M3(h)Y2(h)=J1(h)+J2(h)+J3(h)+T1(h)+T2(h)+T3(h).(3.39)

Let us begin by proving that J1(h)o(h). Notice that

M1(h)Y1(h)={(x,y)DΩ:v(x)h,v(y)h and v(x)-Th(u(x))k,v(y)-Th(u(y))<k}.

Then for (x,y)M1(h)Y1(h) we have

V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]=V(x,y)[k-(v(y)-Th(u(y)))]

If v(x)v(y), then V(x,y)[k-(v(y)-Th(u(y)))]0. Therefore, we just have to consider the case where (x,y)M1(h)Y1(h) with v(x)<v(y). Thus,

|V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|=(v(y)-v(x))p-1[k-(v(y)-Th(u(y)))].

Now taking into consideration that (x,y)M1(h)Y1(h), we get

0(v(y)-v(x))Th(u(y))+k-(Th(u(x))+k)Th(u(y))-Th(u(x))u(y)-u(x).

Therefore, we conclude that

|V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|2k(u(y)-u(x))p-1.

If u(x)h, then u(y)h, hence

V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]=V(x,y)[Tk(v(x))-Tk(v(y))]0.

It then remains to consider the case u(x)<h. We distinguish the following three cases:

  • (i)

    If u(y)>(h+1), by (2.5), we reach that

    {u(y)>h+1,u(x)<h}|U(x,y)|𝑑ν=o(h).

  • (ii)

    If h<u(y)(h+1), then 0k-(v(y)-Th(u(y)))u(y)-u(x), thus

    |V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|(u(y)-u(x))p.

    Now, by (2.7), we get

    {h<u(y)h+1,u(x)<h}(u(y)-u(x))p𝑑ν=o(h).

  • (iii)

    We now deal with the set u(y)h. Since (x,y)M1(h)Y1(h), we have u(y)(h-k). Therefore, if u(x)<(h-k-1), using again (2.6), we reach that

    {u(y)>(h-k),u(x)<(h-k-1)}|U(x,y)|𝑑ν=o(h).

    Let us assume that (h-k-1)<u(x)u(y)<h. In this case, we have

    [k-(v(y)-Th(u(y)))]=u(y)-(v(y)-k)u(y)-(v(x)-k)u(y)-u(x).

    So for (x,y)M1(h)Y1(h) with h-k-1<u(x)u(y)<h we get

    |V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|=(v(y)-v(x))p-1[k-(v(y)-Th(u(y)))](u(y)-u(x))p.

    Now, by using again (2.7), it follows that

    {h-k-1<u(x)u(y)<h}|V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|𝑑ν={h-k-1<u(x)u(y)<h}(u(y)-u(x))p𝑑ν=o(h).

Therefore, combining the above estimates, we obtain J1(h)o(h).

For J2(h) we have v(y)h<v(x) and

Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))k-(v(y)-Th(u(x)))0.

Thus,

V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]0for all (x,y)M2(h)Y1(h).

Hence, J2(h)0.

We now deal with J3(h). We have v(x)h<v(y) and for all (x,y)M3(h)Y1(h) we have

|V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|=(v(y)-v(x))p-1[k-(v(y)-Th(u(y)))].

If v(x)(h-1), then by (2.6) we have

{v(y)>h,v(x)<h-1}|V(x,y)[Tk(v(x)-Th(u(x)))-Tk(v(y)-Th(u(y)))]|𝑑ν2k{v(y)>h,v(x)<h-1}|V(x,y)|𝑑ν=o(h).(3.40)

Now, assume (h-1)<v(x)h. Since (x,y)M3(h)Y1(h), we obtain v(y)k+Th(u(y))k+u(y). Thus, u(y)>(h-k). It is clear that

0v(y)-v(x)Th(u(y))-Th(u(x))u(y)-u(x).(3.41)

Hence following the same discussion as in case (iii) in the analysis of J1(h) and using (3.40) and (3.41), we obtain that J3(h)0.

Notice that in a symmetric way we can prove that T1(h)+T2(h)+T3(h)o(h).

Going back to (3.39), it holds that Q2(h)o(h) and then the claim follows.

Therefore, going back to the definition of I(h) given in (3.35) and taking into consideration that u<h in the set {v<h}, we get

I(h)12DΩ1(h)U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν+12DΩ2(h)V(x,y)[Tk(v(x)-u(x))-Tk(v(y)-u(y))]𝑑ν+o(h)12DΩ2(h)(V(x,y)-U(x,y))[Tk(v(x)-u(x))-Tk(v(y)-u(y))]𝑑ν+12DΩ1(h)DΩ2(h)U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν+o(h)I1(h)+I2(h)+o(h).

It is clear that

I1(h)CDΩ2(h)|Tk(v(x)-u(x))-Tk(v(y)-u(y))|p𝑑ν.

We claim that I2(h)o(h).

Notice that

DΩ1(h)DΩ2(h)=N1(h)N2(h)N3(h),

where

N1(h){(x,y)DΩ:u(x)h,u(y)h,v(x)>h,v(y)>h},N2(h){(x,y)DΩ:u(x)h,u(y)h,v(x)>h,v(y)h},N3(h){(x,y)DΩ:u(x)h,u(y)h,v(x)h,v(y)>h}.

Since

12N1(h)U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]𝑑ν0,

we conclude that

I2(h)12N2(h)+12N3(h)=I21(h)+I22(h).

For (x,y)N2(h) we will consider the following three main cases:

(I) If h-u(x)v(y)-u(y), then 0h-v(y)u(x)-u(y). Hence,

U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]=U(x,y)[Tk(v(y)-u(y))-Tk(h-u(x))]0.

(II) If u(x)-u(y)0h-v(y), then u(x)-u(y)0 and h-u(x)v(y)-u(y). Thus,

U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]=U(x,y)[Tk(v(y)-u(y))-Tk(h-u(x)]0.

(III) Now consider the case where 0u(x)-u(y)h-v(y). It is clear that 0u(x)-u(y)v(x)-v(y). Hence,

|U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]|=(u(x)-u(y))p-1[Tk(h-u(x))-Tk(v(y)-u(y))]2k(v(x)-v(y))p-1.

If v(y)h-1 or v(x)h+1, by (2.5), we get

N2(h){{v(x)>h,v(y)<h-1}{v(x)>h+1,v(y)<h}}|V(x,y)|𝑑ν=o(h).

Thus, we deal with the set {h-1<v(y)h and v(x)h+1}.

It is clear that if v(y)-u(y)k, then h-u(x)k. Thus,

|U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]|=0.

Assume that h-u(x)k. Then v(y)-u(y)k, hence

|U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]|(v(x)-v(y))p-1[(h-u(x))-(v(y)-u(y))](v(x)-v(y))p.

Therefore, using (2.7), we obtain

N2(h){h-1<v(y)hv(x)h+1}{h-ku(x)}|U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]|𝑑νN2(h){h-1<v(y)hv(x)h+1}(v(x)-v(y))p𝑑ν=o(h).

We now consider the set {v(y)-u(y)<k<h-u(x)}. Then u(x)<h-k, and thus u(y)<h-k. As above, we have

|U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]|(v(x)-v(y))p-1[(k-(v(y)-u(y))](v(x)-v(y))p

Thus using again (2.7), we obtain

N2(h){h-1<v(y)hv(x)h+1}{u(x)<h-k}|U(x,y)[Tk(u(x)-Th(v(x)))-Tk(u(y)-Th(v(y)))]|𝑑νN2(h){h-1<v(y)hv(x)h+1}(v(x)-v(y))p𝑑ν=o(h).

Therefore, we conclude that I21(h)o(h). In the same way and using a symmetric argument, we can prove that I22(h)o(h).

Hence I2(h)o(h) and the claim follows.

In conclusion, we have proved that

CDΩ2(h)|Tk(v(x)-u(x))-Tk(v(y)-u(y))|p𝑑νo(h).

Let h. There results that

DΩ|Tk(v(x)-u(x))-Tk(v(y)-u(y))|p𝑑ν=0.

Thus Tk(u)=Tk(v) for all k. Then u=v. ∎

4 Problem with reaction term and general datum

In this section, we consider the problem

{(-Δ)p,βsu=λuq+g(x)in Ω,u0in Ω,u=0in NΩ,(4.1)

where λ,q>0 and g0. According to the values of q and λ, we will prove that problem (4.1) has an entropy solution in the sense of Definition 2.9.

Let us begin with the case q<p-1. We have the following existence result.

Theorem 4.1.

Assume that q<p-1. Then for all gL1(Ω) and for all λ>0 problem (4.1) has a positive entropy solution.

Proof.

Without loss of generality we can assume that λ=1. We set gn=Tn(g). Then gn0 and gng strongly in L1(Ω). Define un to be the unique solution to the approximated problem

{(-Δ)p,βsun=unq+gnin Ω,un0in Ω,un=0in NΩ.

Notice that the existence of un can be obtained as a critical point of the functional

J(u)=12pDΩ|u(x)-u(y)|p𝑑ν-1q+1Ωu+q+1𝑑x-Ωgnu𝑑x.

However, the uniqueness follows by using the comparison result in Lemma 2.4. It is clear that by the same comparison principle we obtain that unun+1.

We claim that {unp-1}n is uniformly bounded in L1(Ω). To prove the claim we argue by contradiction. Assume that Cnunp-1L1(Ω) as n. We set

vn=unCn1p-1.

Then vnp-1L1(Ω)=1 and vn solves the problem

{(-Δ)p,βsvn=Cnq-p+1p-1vnq+Cn-1gnin Ω,vn0in Ω,vn=0in NΩ.(4.2)

We set

GnCnq-p+1p-1vnq+Cn-1gn.

Then GnL1(Ω)0 as n. Taking into consideration the results of Lemmas 3.1 and 3.2, we get the existence of a measurable function v such that Tk(v)Wβ,0s,p(Ω), vp-1Lσ(Ω,|x|-2βdx) for all σ<NN-ps, and Tk(vn)Tk(v) weakly in Wβ,0s,p(Ω).

Since σ>1, using Vitali’s lemma, we can prove that vnp-1vp-1 strongly in L1(Ω). Thus, vp-1L1(Ω)=1.

Now taking Tk(vn) as a test function in (4.2) and using the fact that GnL1(Ω)0, we conclude that

Tk(vn)Wβ,0s,p(Ω)0

as n. Hence Tk(v)=0 for all k. Then v0. Thus we reach a contradiction with the fact that vp-1L1(Ω)=1.

Therefore, unp-1L1(Ω)C for all n and the claim follows.

Since q<p-1, we conclude that the sequence {unq+gn}n is bounded in L1(Ω), and thus we get the existence of a measurable function u such that unquq, up-1Lσ(Ω,|x|-2βdx) for all σ<NN-ps and Tk(un)Tk(u) weakly in Wβ,0s,p(Ω).

Since {unq+fn}n is an increasing sequence, using Lemma 3.6, we conclude that Tk(un)Tk(u) strongly in Wβ,0s,p(Ω). Now, by Theorem 1.2 we obtain that u is an entropy solution to problem (4.1) in the sense of Definition 2.9.

We now prove that u is the minimal solution of (4.1).

Let u¯ be another entropy positive solution to problem (4.1). Recall that u=limnun, so to finish we have to show that unu¯ for all n. Fix n and consider the sequence {wn,i}i defined by wn,0=0 with wn,i+1 being the unique solution to the problem

{(-Δ)p,βswn,i+1=wn,iq+gnin Ω,wn,i+10in Ω,wn,i+1=0in NΩ.

It is clear that the sequence {wn,i}i is increasing in i with wn,iun for all i. Hence wn,iw¯n is a solution to problem (4.4). Now, by the comparison principle in Lemma 2.4 we conclude that w¯n=un, and by an iteration argument we can prove that wn,iu¯ for all i. Hence unu¯ and the result follows. ∎

In the case where q=p-1, the problem is related to the first eigenvalue of the operator (-Δ)p,βs. More precisely, we set

λ1=infϕWβ,0s,p(Ω),ϕ012DΩ|ϕ(x)-ϕ(y)|p𝑑νΩ|ϕ|p𝑑x.

As in the case β=0, it is not difficult to show that λ1>0 and that λ1 is attained.

Now, we can formulate our existence result.

Theorem 4.2.

Assume that q=p-1. If λ<λ1, then for all gL1(Ω) problem (4.1) has a minimal entropy positive solution.

To prove Theorem 4.2, we need the following classical regularity result.

Lemma 4.3.

Let u be the unique solution to the problem

{(-Δ)p,βsu=fin Ω,u=0in NΩ,(4.3)

where |f||x|ps*βLm(Ω,|x|-ps*βdx) for some m>Nps. Then uL(Ω).

Proof.

We follow closely the Stampacchia argument given in [24]. Using Gk(u(x)), with k>0, as a test function (4.3), and taking into consideration that

U(x,y)(Gk(u(x))-Gk(u(y)))|Gk(u(x))-Gk(u(y))|p,

where U(x,y)=|u(x)-u(y)|p-2(u(x)-u(y)), we reach that

12DΩ|Gk(u(x))-Gk(u(y))|p|x-y|N+psdx|x|βdy|y|βΩ|f||Gk(u(x))|𝑑x.

By the Weighted Fractional Sobolev Inequality in Theorem 2.2, it follows that

SGk(u)Lps*(Ω,|x|-ps*βdx)pAk|f||Gk(u(x))|𝑑x,

where Ak={xΩ:|u(x)|k}. We set dω=dx|x|ps*β. Then

Ak|f||Gk(u(x))|𝑑x=Ak(|f||x|ps*β)|Gk(u(x))|𝑑ωGk(u)Lps*(Ω,dω)p(|f||x|ps*β)Lm(Ω,dω)|Ak|dω1-1m-1ps*.

Thus,

CGk(u)Lps*(Ω,dω)p-1ps*(|f||x|ps*β)Lm(Ω,dω)|Ak|dω1-1m-1ps*.

Let h>k. Since AhAk, there results that

(h-k)|Ah|dωp-1ps*(|f||x|ps*β)Lm(Ω,dω)|Ak|dω1-1m-1ps*.

Hence,

|Ah|dωC(|f||x|ps*β)Lm(Ω,dω)ps*p-1|Ak|dωps*p-1(1-1m-1ps*)(h-k)ps*p-1.

We set Φ(k)=|Ah|dω. Then

Φ(h)CΦps*p-1(1-1m-1ps*)(k)(h-k)ps*p-1.

Since m>Nps, we have

ps*p-1(1-1m-1ps*)>1.

By the classical result of Stampacchia (see [24]) we get the existence of k0>0 such that Φ(h)=0 for all hk0, hence we conclude the proof. ∎

Proof of Theorem 4.2.

We follow closely the argument used in the proof of Theorem 4.1.

Define un to be the unique solution to the approximated problem

{(-Δ)p,βsun=λunp-1+gnin Ω,un0in Ω,un=0in NΩ.(4.4)

Notice that, since λ<λ1, the existence of un can be obtained as a critical point of the functional

J(un)=12pDΩ|u(x)-u(y)|p𝑑ν-λpΩ|u|p𝑑x-Ωgnu𝑑x.

However, the uniqueness follows by using the comparison result in Lemma 2.4. It is clear that by using the same comparison principle we obtain that unun+1.

We claim that {unp-1}n is uniformly bounded in L1(Ω). We argue by contradiction. Assume that

Cnunp-1L1(Ω)

as n. We set

vn=unCn1p-1.

Then vnp-1L1(Ω)=1 and vn solves the problem

{(-Δ)p,βsvn=λvnp-1+gnCnin Ω,vn0in Ω,vn=0in NΩ.

We set Gnvnp-1+gnCn, thus GnL1(Ω)C. Taking into consideration the results of Lemmas 3.1 and 3.2, we get the existence of a measurable function v such that Tk(v)Wβ,0s,p(Ω), vp-1Lσ(Ω,|x|-2βdx) for all σ<NN-ps and Tk(vn)Tk(v) weakly in Wβ,0s,p(Ω).

Since σ>1, using Vitali’s lemma, we can prove that vnp-1vp-1 strongly in L1(Ω). Thus, vp-1L1(Ω)=1. It is clear that Gnλvp-1 strongly in L1(Ω). Thus v solves

{(-Δ)p,βsv=λvp-1in Ω,v0in Ω,v=0in NΩ.(4.5)

We claim that vL(Ω). From the previous discussion we know that vp-1Lσ(Ω,|x|-2βdx) for all σ<NN-ps. Thus setting a1=(p-1)NN-ps-(p-1)-ε, with ε very small, and using an approximation argument, we can take va1 as a test function in (4.5) to conclude that

DΩ|v(x)-v(y)|p-2(v(x)-v(y))(va1(x)-va1(y))𝑑νC.

Hence by using inequality (2.2), it follows that

DΩ|va1+p-1p(x)-va1+p-1p(y)|p𝑑νC.

Using the weighted Sobolev inequality in Theorem 2.2, we reach that

Ω|v(x)|(a1+p-1)ps*p|x|ps*β𝑑x<.

Now we set a2=(a1+p-1)ps*p-(p-1). Then using va2 as a test function in (4.5) and following the same argument as above, we conclude that

Ω|v(x)|(a2+p-1)ps*p|x|ps*β𝑑x<.

Now consider the sequence an+1=(an+p-1)ps*p-(p-1). It is clear that an and by an induction argument we can prove that Ωvan𝑑x< for all n. Thus using Theorem 4.3, we conclude that v is an energy solution to problem (4.5) and that vL(Ω). Now, by using v as a test function in (4.5) and taking into consideration that λ<λ1, it follows that vWβ,0s,p(Ω)=0, which is a contradiction to the fact that vp-1L1(Ω)=1. Therefore, the claim follows. The rest of the proof follows exactly the same argument as in the proof of Theorem 4.1. ∎

Let us now consider the case q>p-1. We follow closely the argument used in [9]. It is clear that in this case additional conditions on g are needed in order to guarantee the existence of a positive solution.

More precisely, if gL1(Ω), we define w to be the unique positive solution to the problem

{(-Δ)p,βsw=gin Ω,w=0in NΩ.

We are able to prove the following result.

Theorem 4.4.

Assume that gL1(Ω) verifies wq(x)g(x) a.e. in Ω. Then there exists a positive constant λ¯ such that for all λ<λ¯ problem (4.1) has a minimal entropy positive solution.

Proof.

Recall that by the results of Lemmas 3.1 and 3.2 we know that wp-1Lσ(Ω,|x|-2βdx) for all σ<NN-ps and Tk(w)Wβ,0s,p(Ω).

Let v be the minimal solution to the problem

{(-Δ)p,βsv=g+wqin Ω,v=0in NΩ.

It is not difficult to show that v2p-1w, hence by using the hypothesis on g, it follows that

(-Δ)p,βsv=g+wqg+2q1-pvq.

Then v is a supersolution to (4.1) for λλ¯=2q1-p. Fixing λ as above and defining the sequence {un}n by u0=0, we have that un+1 is the unique solution to the following problem:

{(-Δ)p,βsun+1=unq+gn+1in Ω,un+1=0in NΩ.

By an induction argument, we can prove that unv for all n and that the sequence {un}n is increasing in n. Thus {unq+gn}n is increasing and bounded in L1(Ω). Now, using the same compactness argument as in the proofs of Theorems 4.1 and 4.2, we get the existence result. ∎

Acknowledgements

The authors would like to express their gratitude to the anonymous referee for their comments and suggestions that improved the last version of the manuscript.

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About the article

Received: 2016-03-29

Revised: 2016-09-13

Accepted: 2016-10-07

Published Online: 2016-12-02


Funding Source: Ministerio de Economía y Competitividad

Award identifier / Grant number: MTM2013–40846-P

Work partially supported by Project MTM2013–40846-P, MINECO, Spain.


Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 144–174, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2016-0072.

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© 2019 Walter de Gruyter GmbH, Berlin/Boston. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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