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Research article Special Issues

A note on quasilinear equations with fractional diffusion

  • This contribution is part of the Special Issue: Partial Differential Equations from theory to applications—Dedicated to Alberto Farina, on the occasion of his 50th birthday Guest Editors: Serena Dipierro; Luca Lombardini
    Link: www.aimspress.com/mine/article/5752/special-articles
  • In this paper, we study the existence of distributional solutions of the following non-local elliptic problem {(Δ)su+|u|p=f in Ωu=0 in RNΩ,s(1/2,1). We are interested in the relation between the regularity of the source term f, and the regularity of the corresponding solution. If 1<p<2s, that is the natural growth, we are able to show the existence for all fL1(Ω). In the subcritical case, that is, for 1<p<p:=N/(N2s+1), we show that solutions are C1,α for fLm, with m large enough. In the general case, we achieve the same result under a condition on the size of the source. As an application, we may show that for regular sources, distributional solutions are viscosity solutions, and conversely.

    Citation: Boumediene Abdellaoui, Pablo Ochoa, Ireneo Peral. A note on quasilinear equations with fractional diffusion[J]. Mathematics in Engineering, 2021, 3(2): 1-28. doi: 10.3934/mine.2021018

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  • In this paper, we study the existence of distributional solutions of the following non-local elliptic problem {(Δ)su+|u|p=f in Ωu=0 in RNΩ,s(1/2,1). We are interested in the relation between the regularity of the source term f, and the regularity of the corresponding solution. If 1<p<2s, that is the natural growth, we are able to show the existence for all fL1(Ω). In the subcritical case, that is, for 1<p<p:=N/(N2s+1), we show that solutions are C1,α for fLm, with m large enough. In the general case, we achieve the same result under a condition on the size of the source. As an application, we may show that for regular sources, distributional solutions are viscosity solutions, and conversely.


    Throughout this article, we shall consider the following Dirichlet integro-differential problem

    {(Δ)su+|u|p=fin Ωu=0 in RNΩ (1.1)

    for s(1/2,1), ΩRN, p>1 and f a non-negative measurable function. When the nonlinear term appears in the righthand side the model (1.1) may be seen as a Kardar-Parisi-Zhang stationary problem driving by fractional diffusion (see [20] for the model in the local setting and [1] in the nonlocal case). The problem with the nonlinear term in the left hand side is the stationary counterpart of a Hamilton-Jacobi equation with a viscosity term, the principal nonlocal operator. See [30] and the references therein.

    The fractional Laplacian operator (Δ)s, and more general pseudo-differential operators, have been a classic topic in Harmonic Analysis and PDEs. Moreover, these is a renovated interest in these kind of operators. Non-local operators arise naturally in continuum mechanics, image processing, crystal dislocation, phase transition phenomena, population dynamics, optimal control and theory of games as pointed out in [6,10,11,12,18] and the references therein. For instance, the fractional heat equation may appear in probabilistic random-walk procedures and, in turn, the stationary case may do so in pay-off models (see [10] and the references therein). In the works [25] and [26] the description of anomalous diffusion via fractional dynamics is investigated and various fractional partial differential equations are derived from Lévy random walk models, extending Brownian walk models in a natural way. Fractional operators are also involved in financial mathematics, since Lèvy processes with jumps revealed as more appropriate models of stock pricing. The bounday condition

    u=0 in RNΩ

    which is given in the whole complement may be interpreted from the stochastic point of view as the fact that a Lèvy process can exit the domain Ω for the first time jumping to any point in its complement.

    Regarding the integro-differential problem that we discuss in the present manuscript, the main results of our research may be summarized as follows

    ● In the sub-critical scenario 1<p<p:=NN2s+1, there is a unique non-negative distributional solution uW1,q0(Ω) of (1.1) for any 1q<p.

    ● Now, for xΩ, setting δ(x)=dist(x,Ω)=dist(x,Ωc) (since Ω is a bounded regular domain), then if 1<p<p, with similar arguments to those in [1] and [15], we have

    - If m<N2s1, then |u|δ1sLq(Ω) for all 1q<mNNm(2s1).

    - If m=N2s1, then |u|δ1sLq(Ω) for all 1q<.

    - If m>N2s1, then |u|Cα(Ω) for some α(0,1).

    In the interval 1<p<p the result lies on the estimates for the Green function by Bogdan and Jakubowski in [8].

    ● For any 1<p<, u is C1,α provided the source is sufficiently small.

    ● Any solution uC1,α(Ω) with Hölder continuous source is a viscosity solution, and conversely.

    Notice that in the local case s=1, the main existing results can be summarized into two points: If p2, then the existence of solution is obtained for all fL1(Ω) using approximation arguments and suitable test function, see [7] and the references therein. However the truncating arguments are not applicable for p>2 including for L data. In the case of Lipschitz data, the author in [23] was able to get the existence and the uniqueness of a regular solution for all p. However this last argument is not applicable for Lm data including for p close to two.

    For the non local case, the first existence result was obtained in [15]. Indeed, they consider the problem

    {(Δ)su+ϵg(|u|)=ν in Ωu=0 in RNΩ,s(1/2,1), (1.2)

    with ϵ{1,1}, for a continuous and non-negative function g satisfying g(0)=0 and a non-negative Radon measure ν so that Ωδβdν< with β[0,2s1).

    In [15,Thm. 1.1], they show that for ϵ=1 and under the integrability assumption

    1g(s)s1pds<,

    problem (1.2) admits a non-negative distributional solution uW1,q0(Ω), for all 1q<p,β where

    p,β:=NN2s+1+β.

    In particular, this result implies that the Dirichlet problem (1.1) admits a solution u in W1,q0(Ω) for all q[1,p) and for 1<p<p. Moreover, for g Hölder continuous and bounded in R, solutions to (1.2) becomes strong for a Hölder continuous source.

    The regularity of solutions to (1.1) is strongly related to the corresponding issue for problems

    {(Δ)sv=f in Ωv=0 in RNΩ, (1.3)

    As a by-product of the results in [1,15,16], we have the following result which will be largely used throughout our paper.

    Theorem 1.1. Suppose that fLm(Ω) with m1 and define v to be the unique solution to problem (1.3) with s>12. Then for all 1p<mNNm(2s1), there exists a positive constant CˆC(Ω,N,s,p) such that

    |v|δ1sLp(Ω)ˆC||f||Lm(Ω). (1.4)

    Moreover,

    1) If m=N2s1, then |v|δ1sLp(Ω) for all 1p<.

    2) If m>N2s1, then vC1,σ(Ω) for some σ(0,1), and

    |v|δ1sL(Ω)C||f||Lm(Ω).

    In the case where fL1(Ω)Lmloc(Ω) where m>1, then as it was proved in [1], the above regularity results hold locally in Ω. More precisely we have

    Proposition 1.2. Assume that fL1(Ω)Lmloc(Ω) with m>1. Let v the unique solution to problem (1.3). Suppose that m<N2s1, then for any Ω1⊂⊂Ω1⊂⊂Ω and for all 1pmNNm(2s1), there exists ˜C:=˜C(Ω,Ω1,Ω1,N,s,p) such that

    ||v||Lp(Ω1)˜C(||f||L1(Ω)+||f||Lm(Ω1)). (1.5)

    Moreover,

    1) If m=N2s1, then |v|Lploc(Ω) for all 1p<.

    2) If m>N2s1, then vC1,σ(Ω) for some σ(0,1).

    As a consequence we conclude that, if fLm(Ω) with m>1, then

    1) If mN2s1, then Ω|v|adx< for all a<11s.

    2) If 1<m<N+2s2s1, then Ω|v|adx< for all a<ˇP:=mNN(m(1s)+1)m(2s1).

    Remark 1.3. It is clear that a<a0=11s is optimal. Before proving the optimality of a0, let us recall the next Hardy inequality that will be used systematically in what follows.

    Proposition 1.4. (Hardy inequality) Assume that Ω is a bounded regular domain of IRN and 1<p<N. Then there exists a positive constant C(Ω) such that for all ϕW1,p0(Ω), we have

    C(Ω)Ω|ϕ|pδpdxΩ|ϕ|pdx<+. (1.6)

    We prove now the optimality of a0. We argue by contradiction. Assume that, for 0fL(Ω), there exists a solution v to (1.3) such that vW1,p0(Ω) with p>11s.

    By using the classical Hardy inequality we obtain that

    ΩvpδpdxΩ|v|pdx<+.

    By the results in [27] the solution behaves as vδs, therefore, as a consequence, 1δp(1s)L1(Ω), that is, p<11s, a contradiction.

    Hence, the bound for the exponent of the gradient seems to be natural if we impose that the solution lies in the Sobolev space W1,p0(Ω) for the problem with reaction gradient term.

    In the case of absorption gradient term, this affirmation seems to be difficult to prove, however, in Theorem 2.9, we will show that the non existence result holds, at least, for large values of p and for all bounded non negative data.

    In the case of gradient reaction term and for 2sp<s1s, the authors in [1] proved the existence of a solution u with |u|Lploc(Ω) using a fixed point argument. In the present paper we will use the same approach to get the existence of a solution for p2s. However, in addition to the regularity condition of f, smallness condition on the source term ||f||Lm(Ω) is also needed.

    The paper is organized as follows. In Section 2, we introduce the functional setting and we precise the notion of solutions that we will use throughout this work as the weak sense and the viscosity sense. We give also some useful estimates for weak solutions and the general comparison principle. A non existence result is proved using suitable estimates on the Green function for the fractional Laplacian with drift term.

    The existence of a solution is proved in Section 3. In the Subsection 3.1 we treat the case of natural growth behavior in the gradient term, namely the case 1<p<2s. In this case existence of a solution is obtained for all L1 datum. As a complement of the result proved in [15], we prove that if p>p, the existence of a solution for general measure data ν is not true and additional hypotheses on ν related to a fractional capacity are needed.

    Problem with a linear zero order reaction term is also analyzed. In such a case we are able to show existence for data in L1 and then a breaking of resonance occurs under natural hypotheses on the zero order term and p.

    Some additional regularity results are obtained in the subcritical case 1<p<p.

    The general case, p2s, is treated in Subsection 3.3. Here and since we will use fixed point theorem, we need to impose some additional condition on the regularity and the size of f. The existence result is obtained in a suitable weighted Sobolev space under additional hypotheses on p. The above existence result holds trivially for the case s=1 and then can be seen as an extension of the existence result obtained in [23] in the framework of Lm datum.

    The analysis of the viscosity solution is done is Section 4 where it is also proved that weak solution is a viscosity solution and viceversa if the data f is sufficiently regular and s is close to 1, compare with [28].

    Some related open problems are given in the last section.

    In what follows, Ω will denote a bounded, open and C2 domain in RN with bounded boundary, N1. We introduce some functional-space notation. By USC(Ω), LSC(Ω) and C(Ω), we denote the spaces of upper semi-continuous, lower semi continuous and continuous real-valued functions in Ω, respectively. Moreover, the space Ck(Ω), k1, is defined as the set of functions which derivatives of orders k are continuous in Ω. Also, the Hölder space Ck,α(Ω) is the set of Ck(Ω) whose kth order partial derivatives are locally Hölder continuous with exponent α in Ω.

    For σR, we define the truncation operator as follows

    Tk(σ):=max(k,min(k,σ)).

    Finally, for any u, we denote by

    u+=max{0,u} and u=max{0,u}.

    In order to introduce the notion of distributional solutions, we give some definitions. For s(12,1) and uS(RN), the fractional Laplacian (Δ)s is given by

    (Δ)su(x):=limϵ0(Δ)sϵu(x)

    where

    (Δ)sϵu(x):=RNu(x)u(y)|xy|N+2sχϵ(|xy|)dy

    with:

    χt(|x|):={0,|x|<t1,|x|t.

    For larger class of functions the fractional Laplacian can be defined by density. See [17] or [29] for instance.

    Definition 2.1. We say that a function ϕC(RN) belongs to Xs(Ω) if and only if the following holds

    supp(ϕ)¯Ω.

    The fractional Laplacian (Δ)sϕ(x) exists for all xΩ and there is C>0 so that |(Δ)sϕ(x)|C.

    There is φL(Ω,δsdx) and ϵ0>0 so that

    |(Δ)sϵϕ(x)|φ(x),

    a.e. in Ω and for all ϵ(0,ϵ0).

    Before staring the sense for which solutions are defined, let us recall the definition of the fractional Sobolev space and some of its properties.

    Assume that s(0,1) and p>1. Let ΩIRN, then the fractional Sobolev Space Ws,p(Ω) is defined by

    Ws,p(Ω){ϕLp(Ω):Ω×Ω|ϕ(x)ϕ(y)|pdν<+},

    where dν=dxdy|xy|N+ps.

    Notice that Ws,p(Ω) is a Banach Space endowed with the norm

    ϕWs,p(Ω)=(Ω|ϕ(x)|pdx)1p+(Ω×Ω|ϕ(x)ϕ(y)|pdν)1p.

    The space Ws,p0(Ω) is defined as the completion of C0(Ω) with respect to the previous norm.

    If Ω is a bounded regular domain, we can endow Ws,p0(Ω) with the equivalent norm

    ||ϕ||Ws,p0(Ω)=(Ω×Ω|ϕ(x)ϕ(y)|pdν)1p.

    Notice that if ps<N, then we have the next Sobolev inequality, for all vC0(IRN),

    IR2N|v(x)v(y)|p|xy|N+psdxdyS(RN|v(x)|psdx)pps,

    where ps=pNNps and SS(N,s,p).

    In the following definition, we introduce the class of distributional solutions.

    Assume that ν is a bounded Radon measure and consider the problem

    {(Δ)sv=ν in Ω,v=0 in RNΩ, (2.1)

    Let us begin by precising the sense in which solutions are defined for general class of data.

    Definition 2.2. We say that u is a weak solution to problem (2.1) if uL1(Ω), and for all ϕXs, we have

    Ωu(Δ)sϕdx=Ωϕdν,

    where Xs is given in Definition 2.1.

    As a consequence of the properties of the Green function, the authors in [16] obtain the following regularity result.

    Theorem 2.3. Suppose that s(12,1) and let νM(Ω), be a Radon measure such that

    Ωδβdν<,δ(x):=dist(x,Ωc),

    with β[0,2s1). Then the problem (2.1) has a unique weak solution u in the sense of Definition 2.2 such that uW1,q0(Ω), for all 1q<pβ where p,β:=NN2s+1+β. Moreover

    ||u||W1,q0(Ω)C(N,q,Ω)Ωδβdν. (2.2)

    For νL1(Ω), setting T:L1(Ω)W1,θ0(Ω), with T(f)=u, then T is a compact operator.

    Related to Tk(u) and for s>12, we have the next regularity result obtained in [1].

    Theorem 2.4. Assume that fL1(Ω) and define u to be the unique weak solution to problem (2.1), then Tk(u)W1,α0(Ω)Hs0(Ω) for any α<2s, moreover

    Ω|Tk(u)|αdxCkα1||f||L1(Ω).

    We recall also the next comparison principle proved in [1]

    Theorem 2.5. (Comparison Principle). Let gL1(Ω) and suppose that w1,w2W1,q0(Ω) for all 1q<NN2s+1 are such that (Δ)sw1,(Δ)sw2L1(Ω) with

    {(Δ)sw1H1(x,w1,w1)+g in Ω,w10 in RNΩ,{(Δ)sw2H1(x,w2,w2)+g in Ω,w20 in RNΩ,

    where H:Ω×IR×IRNIR is a Carathéodoty function satisfying

    1) H1(x,w1,w1),H1(x,w2,w2)L1(Ω),

    2) for a.e. xΩ, we have

    H1(x,w1,w1)H1(x,w2,w2)=B(x,w1,w2,w1,w2),(w1w2)+f(x,w1,w2)

    with B(La(Ω))N, a>N2s1 and fL1(Ω) with f0 a.e. in Ω.

    Then w1w2 in Ω.

    Recall that we are considering problem (1.1), then we have the next definition.

    Definition 2.6. A function uL1(Ω), with |u|pL1loc(Ω), is a distributional solution to problem (1.1) if for any ϕXs(Ω), there holds

    Ωu(Δ)sϕ+Ωϕ|u|p=Ωfϕ,

    and u=0 in RNΩ.

    We denote by Gs the Green kernel of (Δ)s in Ω and by Gs[] the associated Green operator defined by

    Gs[f](x):=ΩGs(x,y)df(y).

    See [8] and [14] for the estimates of the Green function.

    Definition 2.7. A function u:ΩR is a strong solution to the equation

    (Δ)sw+|w|p=f

    in Ω if uC2s+α(Ω), for some α>0 and

    (Δ)su(x)+|u(x)|p=f(x)

    for every x in Ω.

    The other class of solutions that we shall consider is the class of viscosity solutions. Unlike the distributional scenario, the notion of viscosity solutions requires the punctual evaluation of the equation using appropriate test functions that touch the solution from above or below. We refer to [5] and [13] for more details.

    Definition 2.8. An upper semicontinuous function u:RNR is a viscosity subsolution to (1.1) in Ω, if uLloc(RN), and for any open set UΩ, any x0U and any ϕC2(U) such that u(x0)=ϕ(x0) and ϕu in U, if we let

    v(x):={ϕ(x) in Uu(x) outside U, (2.3)

    we have

    (Δ)sϕ(x0)+|ϕ(x0)|pf(x0),

    and v0 in RNΩ. On the other hand, a lower semicontinuous function u:RNR is a viscosity supersolution to (1.1) in Ω if uLloc(RN), and for any open set UΩ, any x0U and any ψC2(U) such that u(x0)=ψ(x0) and ϕu in U, if we define v as

    v(x):={ψ(x) in Uu(x) outside U, (2.4)

    there holds

    (Δ)sψ(x0)+|ψ(x0)|pf(x0)

    and v0 in RNΩ. Finally, a viscosity solution to (1.1) is a continuous function which is both a subsolution and a supersolution to (1.1).

    To end this section, we prove the next non existence result that justifies in some way the condition p<11s that we will be used later.

    Theorem 2.9. Assume that p>2s11sN+1, then for all 0fL(Ω), problem (1.1) has no weak solution u in the sense of Definition 2.2, such that uW1,p0(Ω).

    Proof. Suppose by contradiction that problem (1.1) has a solution u with uW1,p0(Ω). It is clear that u solves the problem

    (Δ)su+B(x),u=f,

    where B(x)=|u|p2u. Since p>2s11sN+1, then |B|Lσ(Ω) with σ>N2s1 and then BKsN(Ω) the Kato class of function defined by formula (30) in [8]. Thus

    u(x)=ΩˆGs(x,y)f(y)dy,

    where ˆGs is the Green function associated to the operator (Δ)s+B(x). From the result of [8], we know that ˆGsGs, the Green function associated to the fractional laplacian. Hence

    Gs(x,y)C(B)1|xy|N2s(δs(x)|xy|s1)(δs(y)|xy|s1).

    Using the fact that δs(x)|xy|sC(Ω)δs(x), we reach that

    u(x)C(B)δs(x)Ωf(y)δ(y)dy.

    Therefore, using the Hardy inequality we deduce that

    δspδpCupδpL1(Ω).

    Thus 1δp(1s)L1(Ω). Since p(1s)1, then we reach a contradiction.

    Corollary 2.10. Let f be a Lipschitz function such that f0, then if p>11s, problem (1.1) has no solution u such that uC1(Ω) with |u|Lp(Ω).

    Remark 2.11. It is clear that the above result makes a significative difference with the local case and the general existence result proved in [23] for Lipschitz function. We conjecture that the non existence result holds at least for all p>1s1 as in the case of gradient reaction term.

    In this section we consider the case of natural growth in the gradient, namely we will assume that p<2s. Then using truncating arguments, we are able to show the existence of a solution to problem (1.1) for a large class of data. We also treat the case where a linear reaction term appears in (1.1).

    In the case where p<p, then for more regular data f, we can show that the solution is in effect a classical solution.

    Theorem 3.1. Let fLm(Ω) with m1, and assume that 1<p<p. Then, the Dirichlet problem

    {(Δ)sw+|w|p=fin Ωw=0in RNΩ,

    has a unique distributional solution w verifying

    if m<N2s1, then |w|Lqloc(Ω) for all 1q<mNNm(2s1);

    if m=N2s1, then |w|Lqloc(Ω) for all 1q<;

    if m>N2s1, then |w|Cα(Ω) for some α(0,1).

    Moreover, if in addition fCϵ(Ω), for some ϵ(0,2s1), then the C1,α distributional solution is a strong solution.

    Proof. It is clear that the existence and the uniqueness follow using [15,1], however, the regularity in the local Sobolev space follows using Proposition 1.2. Notice that, in this case |u|p1Lσ(Ω) with σ>N2s1 and then we can iterate the local regularity result in Proposition 1.2 to deduce that |u|Lθloc(Ω) for all θ>0. Hence |u|Ca(Ω) for some a<1.

    Now, assume that fCϵ(Ω), and let ΩΩ, open and let u be a distributional solution to problem (1.1). Since uL(RN) and f|u|pL(Ω), we apply Proposition 2.3 in [27] to derive

    uCβ(Ω), for all β(0,2s),ΩΩ.

    In particular, we have |u|Cβ1(Ω) for any β(1,2s). Consequently, f|u|pCϵ(Ω). Appealing now to Corollary 2.4 in [27], we obtain uC2s+ϵ in a smaller subdomain of Ω. Thus, uC2s+ϵ locally in Ω.

    We prove that u is a strong solution. Since the term f|u|p is Cϵ in Ω, and then, by appropriate extension, in ¯Ω, we deduce from [16,Lemma 2.1(ⅱ)] that uXs. Hence the integration by parts formula

    Ωu(Δ)sϕ=Ωϕ(Δ)su

    holds for all ϕXs. For any ϕC0(Ω) we hence obtain

    Ωϕ(Δ)su=Ωu(Δ)sϕ=ΩfϕΩ|u|pϕ.

    Therefore

    (Δ)su(x)=f(x)|u(x)|p

    for almost everywhere x in Ω. By continuity, it holds in the full set Ω.

    Remark 3.2. Observe that the reasoning employed to prove the above result gives the precise way in which the function f transfers its regularity to a solution u. Indeed, if fC2ns+ϵn locally in Ω, for ϵ(0,2s1) and n0, then uC2(n+1)s+ϵn locally in Ω.

    In this subsection we will assume that pp<2s, then the first existence result for problem (1.1) is the following.

    Theorem 3.3. Assume that p<2s, then for all fL1(Ω) with f0, the problem (1.1) has a maximal weak solution u such that uW1,p0(Ω) and Tk(u)W1,α0(Ω)Hs0(Ω) for any 1<α<2s and for all k>0.

    Proof. We divide the proof into two steps.

    The first step: We show for a fixed positive integer nIN, the problem

    {(Δ)sun+|un|p1+1n|un|p=f in Ω,un=0 in RNΩ. (3.1)

    has a unique solution un such that unW1,q0(Ω) for all 1q<NN2s+1 and Tk(un)Hs0(Ω). To prove that, we proceed by approximation.

    Let kIN and define un,k to be the unique solution to the approximating problem

    {(Δ)sun,k+|un,k|p1+1n|un,k|p=fk in Ω,un,k=0 in RNΩ. (3.2)

    where fk=Tk(f). We claim that the sequence {un,k}k is increasing in k, namely un,kun,k+1 for all k1 and n fixed. To see that, we have

    (Δ)sun,k+1+|un,k+1|p1+1n|un,k+1|pfk.

    Thus un,k+1 is a supersolution the problem solved by un,k. Setting H(x,s,ξ)=|ξ|p1+1n|ξ|p, then by the comparison principle in Theorem 2.5, it follows that un,kun,k+1 and then the claim follows. It is clear that un,kw for all n,kIN where w is the unique solution to problem

    {(Δ)sw=f in Ω,w=0 in RNΩ. (3.3)

    Notice that wW1,q0(Ω) for all 1q<NN2s+1 and wLr(Ω) for all 1r<NN2s.

    Hence, we get the existence of un such that un,kun strongly in Lσ(Ω) for all 1σ<NN2s.

    For n fixed, we set hn,k:=fk|un,k|p1+1n|un,k|p, then |hn,k|f+n. Thus ||hn,k||L1(Ω)||f||L1(Ω)+n|Ω|. Hence using the compactness result in Theorem 2.3 we deduce that up to a subsequence, un,kun strongly in W1,α0(Ω) for all α<NN2s+1. Since the sequence {uk,n}k is increasing in k, then the limit un is unique. Thus, up to a further subsequence, un,kun a.e. in Ω. Hence using the dominated convergence Theorem it holds that

    |un,k|p1+1n|un,k|p|un|p1+1n|un|p strongly in La(Ω) for all a<.

    Hence un solves the problem (3.1). To proof the uniqueness of un, we assume that vn is another solution to problem (3.1), then

    (Δ)s(unvn)=ˆH(|un|)ˆH(|vn|),

    where ˆH(|ξ|)=|ξ|p1+1n|ξ|p. Since |ˆH(|ξ|)|n, then we obtain that unvnL(Ω). Finally using the comparison principle in Theorem 2.5 it follows that un=vn and then we conclude.

    Second step: Consider the sequence {un}n obtained in the first step, then we know that unw for all n. We claim that un is decreasing in n. Recall that un is the unique solution to the problem

    {(Δ)sun+|un|p1+1n|un|p=f in Ω,un=0 in RNΩ. (3.4)

    Thus

    (Δ)sun+|un|p1+1n+1|un|pf.

    Hence un is a supersolution to the problem solved by un+1. As a consequence and using the comparison principle in Theorem 2.5, it follows that un+1unw for all n.

    Hence, there exists u such that unu strongly in Lσ(Ω) for all 1σ<NN2s.

    We set gn(|un|)=|un|p1+1n|un|p, and let j>0, using Tj(un) as a test function in (3.4) it follows that

    DΩ(Tj(un(x))Tj(un(y)))2|xy|N+2sdxdy+Ωgn(|un|)Tj(un)dxCj.

    Hence {Tj(un)}n is bounded in Hs0(Ω) for all j>0 and then, up to a subsequence, we have Tj(u)Tj(u) weakly in Hs0(Ω). We claim that {gn}n is bounded in L1(Ω). To see that, we fix ε>0 and we use vn,ε=unε+un as a test function in (3.4). It is clear that vn,ε1, then taking into consideration that

    (un(x)un(y))(vn,ε(x)vn,ε(y))0,

    it follows that

    Ωgn(|un|)vn,ε(x)ΩfdxC.

    Letting ε0, we reach that Ωgn(|un|)dxC an the claim follows. Define hn=fgn, then ||hn||L1(Ω)C. As a consequence and by the compactness result in Theorem 2.3, we reach that, up to a subsequence, unu strongly in W1,α0(Ω) for all 1α<NN2s+1 and then, up to an other subsequence, unu a.e in Ω. Hence gng a.e. in Ω where g(x)=|u|p. Since p<2s, then by Theorem 2.4 and using Vitali Lemma we conclude that

    Tk(un)Tk(u) strongly in W1,σ0(Ω) for all σ<2s.

    In particular

    Tk(un)Tk(u) strongly in W1,p0(Ω). (3.5)

    Hence to get the existence result we have just to show that gng strongly in L1(Ω).

    Notice that, using T1(Gj(un)) as a test function in (3.4) it holds that

    unj+1gndxunjfdx0 as j.

    Let ε>0 and consider EΩ to be a measurable set, then

    Egndx={E{un<j+1}}gndx+{E{unj+1}}gndx{E{un<j+1}}|Tj+1(un)|pdx+{unj+1}fdx.

    By (3.5), letting n, we can chose |E| small enough such that

    lim supn{E{un<j+1}}|Tj+1(un)|pdxε2.

    In the same way and since fL1(Ω), we reach that

    lim supn{unj+1}fdxε2.

    Hence, for |E| small enough, we have

    lim supnEgndxε.

    Thus by Vitali lemma we obtain that gng strongly in L1(Ω). Therefore we conclude that u is a solution to problem (1.1).

    If ˆu is an other solution to (1.1), then

    (Δ)sˆu+|ˆu|p1+1n|ˆu|pf.

    Hence ˆuun and then ˆuu.

    Remark 3.4. 1) The existence of a unique solution to the approximating problem (3.4) holds for all p1.

    2) Problem of uniqueness of solution to problem (1.1) is an interesting open problem including for the local case s=1 where partial results are known in the case 1<p<NN1 or p=2.

    3) As a consequence of the previous result and following closely the same argument we can prove that for all p<2s, for all a>0 and for all (f,g)L1(Ω)×L1(Ω) with f,g0, the problem

    {(Δ)su+|u|p=g(x)u1+au+f in Ω,u=0 in RNΩ. (3.6)

    has a positive solution u.

    In the case where the datum f is substituted by a Radon measure ν, existence of solutions holds for all 1<p<p as it was proved in [15]. However, if pp, then the situation changes completely as in the local case, and, additional hypotheses on ν related to a fractional capacity Capσ,p are needed, with σ<1.

    The fractional capacity Capσ,p is defined as follow.

    For a compact set KΩ, we define

    Capσ,p(K)=inf{ψWσ,p0(Ω):ψWσ,p0(Ω),0ψ1 and  ψχK a.e. inΩ}. (3.7)

    Now, if UΩ is an open set, then

    Capσ,p(U)=sup{Capσ,p(K):KU compact of Ω with KU}.

    For any borel subset BΩ, the definition is extended by setting:

    Capσ,p(B)=inf{Capσ,p(U), U open subset of ΩBU}.

    Notice that, using Sobolev inequality, we obtain that if Capσ,p(A)=0 for some set A⊂⊂Ω, then |A|=0. We refer to [24] and [31] for the main properties of this capacity.

    To show that the situation changes for the set of general Radon measure, we prove the next non existence result.

    Theorem 3.5. Assume that p>p, 12<s<1 and let x0Ω, then the problem

    {(Δ)su+|u|p=δx0in Ω,u=0 in RNΩ, (3.8)

    has non solution u such that uW1,p0(Ω).

    Proof. For simplify of tipping we assume that x0=0Ω and we write δ for δ0. We follow closely the argument used in [4]. Assume by contradiction that for some p>p, problem (3.8) has a solution uW1,p0(Ω). Then uWσ,p0(Ω) for all σ<1. We claim that (Δ)suWσ,p(Ω), the dual space of Wσ,p0(Ω), for all σ(2s1,2s). To see that, we consider ϕC0(Ω), then

    |Ω(Δ)suϕdx|IR2N|u(x)u(y)||ϕ(x)ϕ(y)||xy|N+2sdxdy(IR2N|u(x)u(y)|p|xy|N+p(2sσ)dxdy)1p(IR2N|ϕ(x)ϕ(y)|p|xy|N+pσdxdy)1p.

    Since 2sσ(0,1), then (IR2N|u(x)u(y)|p|xy|N+p(2sσ)dxdy)1pC(σ,s,N,Ω)||u||W1,p0(Ω). Thus

    |Ω(Δ)suϕdx|C||u||W1,p0(Ω)||ϕ||Wσ,p0(Ω),

    and then the claim follows. Hence going back to problem (3.8), we deduce that δL1(Ω)+Wσ,p(Ω).

    As in [7], let us now show that if νWσ,p(Ω), then ν<<Capσ,p. Notice that, if in addition, ν is nonnegative, then we can prove that

    ν(A)C(Capσ,p(A))1p,

    and we deduce easily that ν<<Capσ,p. Here we give the proof without the positivity assumption on ν.

    Let A⊂⊂Ω be such that Capσ,p(A)=0, then there exists a Borel set A0 such that AA0 and Capσ,p(A0)=0. Let KA0 be a compact set, then there exists a sequence {ψn}nC0(Ω) such that 0ψn1, ψnχK and ||ψn||pWσ,p0(Ω)0 as n. It is clear that ψnχK a.e in Ω, as n. Hence

    ν(K)=limnψndν=limnψn,νWσ,p0(Ω),Wσ,p0(Ω).

    Thus

    |ν(K)|lim supn|ψn,νWσ,p0(Ω),Wσ,p0(Ω)|lim supn||ν||Wσ,p0(Ω)||ψn||Wσ,p0(Ω)=0.

    Therefore, we conclude that for any compact set KA0, we have |ν(K)|=0. Hence |ν(A0)|=0 and the result follows.

    Notice that if hL1(Ω), then |h|<<Capσ,p. As a conclusion, we deduce that δ<<Capσ,p for all σ(2s1,2s).

    Since p>p, we can choose σ0(2s1,2s) such that pσ0<N. To end the proof, we have just to show that Capσ0,p{0}=0. Without loss of generality, we can assume that Ω=B1(0). Since σp<N, setting w(x)=(1|x|α1)+ with 0<α<Nσ0pp, we obtain that wWσ,p0(Ω). Notice that, for all vWσ,p0(Ω), we know that

    Capσ,p{|v|k}Ck||v||Wσ,p0(Ω).

    Since w(0)=, then {0}{|w|k} for all k>0. Thus

    Capσ,p{0}Ck||w||Wσ,p0(Ω) for all k.

    Letting k, it holds that Capσ,p{0}=0 and the result follows.

    As a direct consequence of the above Theorem we obtain that for p>p, to get the existence of a solution to problem (1.1) with measure data ν, then necessarily ν is continuous with respect to the capacity Capσ,p for all σ(2s1,2s).

    Let consider now the next problem

    {(Δ)su+|u|p=λg(x)u+f in Ω,u=0 in RNΩ. (3.9)

    with g0 and λ>0. As in local case studied in [3], we can show that under natural conditions on q and g, the problem (3.9) has a solution for all λ>0. Moreover, the gradient term |u|q produces a strong regularizing effect on the problem and kills any effect of the linear term λgu.

    Before stating the main existence result for problem (3.9), let us begin by the next definition.

    Definition 3.6 Let g be a nonnegative measurable function such that gL1(Ω). We say that g is an admissible weight if

    C(g,p)=infϕW1,p0(Ω){0}(Ω|ϕ|pdx)1pΩg|ϕ|dx>0. (3.10)

    If gLpNN(p1)+p(Ω) with g0, then using the Sobolev inequality in the space W1,p0(Ω), it holds that g satisfies (3.10).

    If p<N and g(x)=1|x|σ with σ<1+Np, then using the Hardy-Sobolev inequality in the space W1,p0(Ω), we deduce that g satisfies (3.10).

    Now, we are able to state the next result.

    Theorem 3.7. Assume that 1<p<2s and suppose that g is an admissible weight in the sense given in (3.10). Then for all fL1(Ω) with f0 and for all λ>0, the problem (3.9) has a solution u such that uW1,p0(Ω) and Tk(u)W1,α0(Ω)Hs0(Ω) for any 1<α<2s and for all k>0.

    Proof. Fix λ>0 and define {un}n to be a sequence of positive solutions to problem

    {(Δ)sun+|un|p=λg(x)un1+1nun+f in Ω,un=0 in RNΩ. (3.11)

    To reach the desired result we have just to show that the sequence {g(x)un1+1nnu}n is uniformly bounded in L1(Ω). To do that, we use Tk(un) as a test function in (3.11), hence

    ||Tk(un)||2Hs0(Ω)+Ω|un|pTk(un)dxkλΩg(x)undx+kfL1(Ω). (3.12)

    It is clear that

    Ω|un|pTk(un)=Ω|Hk(un)|pdx

    where Hk(σ)=σ0(Tk(t))1pdt. By a direct computation we obtain that

    Hk(σ)C1(k)σC2(k),

    Thus using (3.10) for Hk(un) it holds that

    Ω|Hk(un)|pdxC(g,p)(ΩgHk(un)dx)pC(g,p)[C1(k)(Ωgundx)pC2(k)]

    where C1(k),C2(k)>0 are independent of n.

    Therefore, going back to (3.12), we conclude that

    C1(k)(Ωgundx)pdxkC(g,p)[λΩg(x)undx+fL1(Ω)]+C2(k).

    Since p>1, then by Young inequality we reach that {gun}n is uniformly bounded in L1(Ω). The rest of the proof follows exactly the same compactness arguments as in the proof of Theorem 3.3.

    Remark 3.8. In the case where g(x)=1|x|2s, the Hardy potential, the condition (3.10) holds if p>NN(2s1). Thus, in this case and for all λ>0, problem (3.9) has a solution u such that uW1,p0(Ω) and Tk(u)W1,α0(Ω)Hs0(Ω) for all α<2s.

    Notice that, in this case, without the absorption term |u|p, the existence of solution holds under the restriction λΛN,s, where ΛN,s is the Hardy constant, and with integral condition on the datum f near the origin. We refer to [2] for more details.

    For 2sp<s1s and in the same way as above we can show the next existence result.

    Theorem 3.9. Suppose that fLm(Ω) with m>N/[p(2s1)]. Then there is λ>0 such that if ||f||Lm(Ω)λ, problem (1.1) admits a solution uδ1sW1,p0(Ω).

    Proof. The proof follows closely the argument used in [19] and [1], however, for the reader convenience we include here some details.

    Without loss of generality we can assume that N2. Fix λ>0 such that if ||f||Lm(Ω)λ, then there exists l>0 satisfies

    ˉC(Ω,N,s,m,p)(l+||f||Lm(Ω))=l1p,

    where ˉC(Ω,N,s,m,p) is a positive constant which only depends on the data, it is independent of f and its will be specified below.

    Define now the set

    E={vW1,10(Ω):vδ1sW1,pm0(Ω) and (Ω|(vδ1s)|pmdx)1pml12s}, (3.13)

    It is clear that E is a closed convex set of W1,10(Ω). Using Hardy inequality in (1.6), we deduce that if vE, then |v|pmδpm(1s)L1(Ω) and

    (Ω|v|pmδpm(1s)dx)1pmˆC0(Ω)l1p.

    Define now the operator

    T:EW1,10(Ω)vT(v)=u

    where u is the unique solution to problem

    {(Δ)su=|v|p+f in Ω,u=0 in RNΩ,u>0 in Ω. (3.14)

    To prove that T is well defined we will use Theorem 2.3, namely we show the existence of β<2s1 such that |f|v|p|δβL1(Ω). To do that we have just to show that |v|pδβL1(Ω).

    It is cleat that |v|pL1loc(Ω), moreover, we have

    Ω|v|pδβdx=Ω|v|pδp(1s)δβp(1s)dx(Ω|v|pmδpm(1s)dx)1m(Ωδ(βp(1s))mdx)1m.

    If p(1s)<2s1, we can chose β<2s1 such that p(1s)<β. Hence Ωδ(βp(1s))mdx<.

    Assume that p(1s)2s1, then s(12,p+2p+1]. Notice that, since p<s1s, then p(1s)(2s1)<1s. Since m>Np(2s1)>1s, then (p(1s)(2s1))m<1. Hence we get the existence of β<2s1 such that (p(1s)β)m<1 and then we conclude.

    Then using the fact that vE, we reach that |v|pδβ+fL1(Ω). Therefore the existence of u is a consequence of Theorems 2.3 and 1.1. Moreover, |u|Lα(Ω) for all α<NN2s+1+β. Hence T is well defined.

    Now following the argument used in [1], for l defined as above and using the regularity result in Theorem 1.1 where we choose the constant ˉC strongly related to the constant ˆC defined in formula (1.4), we can prove that T is continuous and compact on E and that T(E)E. For the reader convenience we included some details

    We have

    u(x)=ΩGs(x,y)f(y))dyΩGs(x,y)|v(y)|2sdy,

    then

    u(x)=ΩxGs(x,y)f(y))dyΩxGs(x,y)|v(y)|pdy.

    Thus

    |u(x)|Ω|xGs(x,y)|Gs(x,y)Gs(x,y)(|v(y)|pdy+f(y))dy.

    Taking into consideration the properties of the Green function, it holds that

    |u(x)|δ1sC2(Ω,N,s)(I1(x)+I2(x)+J1(x)+J2(x)),

    where

    I1(x)=δ1s(x){|xy|<δ(x)}Gs(x,y)|xy||v(y)|2sdy,
    I2(x)=1δs(x){|xy|δ(x)}Gs(x,y)|v(y)|2sdy,
    J1(x)=δ1s(x){|xy|<δ(x)}Gs(x,y)|xy|f(y)dy,

    and

    J2(x)=1δs(x){|xy|δ(x)}Gs(x,y)f(y)dy.

    Following the arguments used in [1] and using the regularity result in Theorem 2.3, we get the existence of a positive constant ˘C:=˘C(Ω,N,s,m) such that Ii,JiLpm(Ω) for i=1,2 and

    ||Ii||Lpm(Ω)+||Ji||Lpm(Ω)˘C(|||v|δ1s||Lpm(Ω)+||f||Lm(Ω)).

    Hence assuming that ˉC(Ω,N,s,m,p)=˘C(Ω,N,s,m,p)C2(Ω,N,s,m,p), it follows that

    |||v|δ1s||Lpm(Ω)˘C(Ω,N,s,m,p)C2(Ω,N,s,m,p)(|||v|δ1s||Lpm(Ω)+||f||Lm(Ω))ˉC(Ω,N,s,m,p)(l1p+||f||Lm(Ω))=l1p.

    Thus uE and then T(E)E. In the same way we can prove that T is compact.

    Therefore by the Schauder Fixed Point Theorem, there exists uE such that T(u)=u. Thus, uW1,pmloc(Ω) solves (1.1), at least in the sense of distribution.

    Remark 3.10. 1) It is clear that the above argument does not take advantage of the fact that the gradient term appears as an absorption term.

    2) The existence of a solution can be also proved independently of the sign of f.

    As in Theorem 3.1, if in addition we suppose that f is more regular, then under suitable hypothesis on s and p, we get the following analogous result of Theorem 3.1.

    Corollary 3.11. Assume that the conditions of Theorem 3.9 hold. Assume in addition that

    N<s(2s1)1s and p<s(2s1)N(1s)1. (3.15)

    If fCϵ(Ω), for some ϵ(0,2s1), then the C1,α distributional solutions from Theorem 3.9 is a strong solution.

    Notice that the condition (3.15) is used in order to show that |u|p1Lσloc(Ω) for some σ>N2s1 which is the key point in order to get the desired regularity.

    In the case where f0, we can prove also that u0, more precisely, we have

    Corollary 3.12. Assume that the above conditions hold. Let fCϵ(Ω)L(¯Ω), for some ϵ(0,2s1). If f(x)0 for all xΩ, then the solution from Theorem 3.9 is non-negative. Moreover, if f1f2 and u1 and u2 are the corresponding strong solutions to f1 and f2 from Corollary 3.11, respectively, then u1u2.

    Proof. Suppose that there is a point x0Ω so that u(x0)<0. Since u is continuous in RN (see Proposition 1.1 in [27]), we have u attains its negative minimum at an interior point x1 of Ω. Hence

    u(x1)=0,(Δ)su(x1)<0.

    But hence we obtain the contradiction 0f(x1)0=(Δ)su(x1)<0.

    We next prove the last statement in the Corollary 3.12. Let f1f2. Let u1 and u2 be the corresponding strong solutions from Corollary 3.11, and assume that

    minΩ(u2u1)=u2(x0)u1(x0)<0.

    Hence (u1u2)(x0)=0 and (Δ)s(u2u1)(x0)<0, so we have the contradiction

    f1(x0)=(Δ)su1(x0)+|u1(x0)|p>(Δ)su2(x0)+|u2(x0)|p=f2(x0).

    In this section, we investigate the relation between distributional solutions and viscosity solutions. Let us recall that according to Theorem 3.1 and Corollary 3.11, to obtain strong solutions to (1.1) it is sufficient that fCϵ(Ω) and that

    p<p

    or

    N<s(2s1)1s,pp<s(2s1)N(1s)1 and ||f||Lm(Ω)λ,

    for λ defined in Theorem 3.9. In this section we show that strong solutions to (1.1) are viscosity solutions. The converse is also true provided a comparison principle for viscosity solutions. We prove it in the next subsection.

    We prove a comparison result for viscosity solutions of problem (1.1). This result requires a continuous source term f.

    In order to state the result, we shall need some technical lemmas that could have interest by themselves. For related results see [22].

    We start with a usual property for the fractional Laplacian of smooth functions. See [21,Lemma 2.6] for the proof.

    Lemma 4.1. Let Bϵ(x)UΩ and let uC2(U). Then:

    |P.V.Bϵ(x)u(x)u(y)|xy|N+2sdy|cϵ

    where cϵ is independent of x and cϵ0 as ϵ0.

    Observe that in the definition of viscosity solutions, we do not evaluate the given equation in the solution u. However, the following lemma states an extra information when u is touched from below or above by C2-test functions.

    Lemma 4.2. Let u be a viscosity supersolution to (1.1) and suppose that there exists ϕC2(U), UΩ, touching u from below at x0U. Then (Δ)su(x0) is finite and moreover:

    (Δ)su(x0)+|ϕ(x0)|pf(x0). (4.1)

    A similar result holds for subsolutions.

    Proof. We assume that x0=0 and u(0)=0. For r>0 so that Br:=B(0,r)U, define:

    ϕr(x):={ϕ(x), in Bru(x), outside Br,

    Hence for all 0<ρ<r

    BrBρu(0)u(y)|y|N+2sdy=BrBρϕ(y)u(y)|y|N+2sdyBrBρϕ(y)|y|N+2sdyBrBρϕ(y)|y|N+2sdy,

    where we have used that ϕ touches u from below. As ρ0, the last integral converges since ϕC2(Br). Hence

    limρ0BrBρu(0)u(y)|y|N+2sdy[,M], (4.2)

    where

    M:=limρ0(BrBρϕ(y)|y|N+2sdy).

    Also, from the fact that u is a supersolution, we have u0 in RNΩ. Thus

    RNBru(0)u(y)|y|N+2sdy¯ΩBru(y)|y|N+2sdy. (4.3)

    Since uLSC(¯Ω), there is a constant m so that

    u(y)m, for all y¯ΩBr.

    Hence from (4.3), it follows

    RNBru(0)u(y)|y|N+2sdymRNBr1|y|N+2sdy<.

    This fact, together with (4.2), imply that (Δ)su(0)[,).

    We now prove the estimate (4.1), and consequently that (Δ)su(0) is finite. For ρ>0, we have by Lemma 4.1 that

    |P.V.Brϕ(y)|y|N+2sdy|ρ,

    choosing r small enough. Hence

    RNBru(0)u(y)|y|N+2sdy=RNBrϕr(0)ϕr(y)|y|N+2sdy=(Δ)sϕr(0)P.V.Brϕ(y)|y|N+2sdy|ϕ(0)|p+f(0)+ρ.

    By letting r0, and then ρ0, we derive (4.1).

    We now give the main result of this section.

    Theorem 4.3 (Comparison principle for viscosity solutions). Assume that fC(Ω). Let vUSC(¯Ω) be a subsolution and uLSC(¯Ω) be a supersolution, respectively, of (1.1). Then vu in Ω.

    Proof. We argue by contradiction. Assume that there is x0Ω so that:

    σ:=supΩ(vu)=v(x0)u(x0)>0.

    As usual, we double the variables and consider for ϵ>0 the function

    Ψϵ(x,y):=v(x)u(y)1ϵ|xy|2.

    By the upper semi continuity of v and u, there exist xϵ and yϵ in ¯Ω so that

    Mϵ:=sup¯ΩׯΩΨϵ=Ψϵ(xϵ,yϵ).

    By compactness, xϵ¯x and yϵ¯y, up to subsequence that we do not re-label. From

    Ψϵ(xϵ,yϵ)Ψϵ(x0,x0) (4.4)

    and the upper boundedness of v and u in ¯Ω, we derive

    limϵ0|xϵyϵ|2=0,

    hence ¯x=¯y. Moreover

    Ψϵ(xϵ,yϵ)Ψϵ(¯x,¯x)

    implies that:

    limϵ01ϵ|xϵyϵ|2=0.

    As a consequence, by letting ϵ0 in (4.4) and using the semicontinuity of u and v, we obtain

    σ=limϵ0(v(xϵ)u(yϵ)). (4.5)

    Also, observe that ¯xΩ, because otherwise there is a contraction with vu in RNΩ.

    Define the C2 test functions

    ϕϵ(x):=v(xϵ)1ϵ|xϵyϵ|2+1ϵ|xyϵ|2,
    ψϵ(y):=u(yϵ)1ϵ|xϵyϵ|2+1ϵ|xϵy|2.

    Then ϕϵ touches v from above at xϵ and ψϵ touches u from below at yϵ. By Lemma 4.2, we have

    (Δ)sv(xϵ)+|ϕϵ(xϵ)|pf(xϵ)

    and

    (Δ)su(yϵ)+|ψϵ(yϵ)|pf(yϵ)

    Therefore:

    (Δ)sv(xϵ)(Δ)su(yϵ)f(xϵ)f(yϵ)+|ψϵ(yϵ)|p|ϕϵ(xϵ)|p. (4.6)

    Since fC(Ω) and

    yψϵ(yϵ)=xϕϵ(xϵ),

    we have that the right hand side in (4.6) tends to 0 as ϵ0. Thus, we obtain

    lim infϵ0RNv(xϵ)v(xϵ+z)u(yϵ)+u(yϵ+z)|z|N+2sdz=lim infϵ0((Δ)sv(xϵ)(Δ)su(yϵ))0. (4.7)

    Let A1,ϵ:={zRN:xϵ+z,yϵ+zΩ}. Hence for zA1,ϵ, we have from the inequality

    Ψϵ(xϵ,yϵ)Ψϵ(xϵ+z,yϵ+z)

    that

    v(xϵ)v(xϵ+z)u(yϵ)+u(yϵ+z)0. (4.8)

    Define A2,ϵ:=RNA1,ϵ. We will justify that we are allowed to use Fatou's Theorem in

    lim infϵ0A2,ϵv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdz (4.9)

    by showing that the integrand is bounded from below by an L1 function. Firstly, let r>0 so that B3r(¯x)Ω and take ϵ0 small enough such that xϵ,yϵBr(¯x) for all ϵ<ϵ0. Take zA2,ϵ. We show now that |z|2r. Indeed, to reach a contradiction, assume that |z|<2r. Since zA1,ϵ, it follows that xϵ+z or yϵ+z does not belong to Ω. Without loss of generality, assume xϵ+zΩ. Hence

    |xϵ+z¯x|<3r,

    and so xϵ+zB3r(¯x)Ω which is a contradiction. Next, notice that

    v(xϵ)v(xϵ+z)|z|N+2s|v(xϵ)||z|N+2s|v+(xϵ+z)||z|N+2s. (4.10)

    Hence, using that zB2r when zA2,ϵ, we have

    A2,ϵ|v(xϵ)||z|N+2sdzCRNB2r1|z|N+2sdz<.

    On the other hand

    A2,ϵ|v+(z+xϵ)||z|N+2sdzRNB2r|v+(z+xϵ)||z|N+2sdz=RNB2r(xϵ)|v+(y)||yxϵ|N+2sdy.

    Since v is a subsolution, we have v0 in RNΩ. Hence

    A2,ϵ|v+(z+xϵ)||z|N+2sdz¯ΩB2r(xϵ)|v+(y)||yxϵ|N+2sdy¯ΩBr(¯x)|v+(y)||yxϵ|N+2sdy1rN+2s¯ΩBr(¯x)v+(y)dy.

    Observe that the last integral is finite since vL1loc(RN) by definition. In this way, recalling (4.10), the term

    v(xϵ)v(xϵ+z)|z|N+2s

    is bounded from below by an L1-integrable function. A similar result follows for

    u(yϵ+z)u(yϵ)|z|N+2s.

    Hence, we may use Fatou Lemma in (4.9) and derive

    lim infϵ0A2,ϵv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdzRNlim infϵ0v(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sχA2,ϵ(z)dzRNA¯xσ+u(¯x+z)v(¯x+z)|z|N+2sdz. (4.11)

    Here A¯x:={zRN:¯x+zΩ} and we have used the a.e. pointwise convergence of χA2,ϵ to χA¯x, [9,Lemma 4.3] together with a diagonal argument to conclude for a subsequence

    lim infϵ0[v(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)]σv(¯x+z)+u(¯x+z)

    for a.e. zRN. Moreover, the inequality uv in RNΩ implies that the last integral in (4.11) is non-negative. Then

    lim infϵ0A2,ϵv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdz0. (4.12)

    Therefore by Fatou Lemma, (4.12) and (4.7), we deduce

    RNlim infϵ0v(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sχA1,ϵdzlim infϵ0A1,ϵv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdzlim infϵ0A1,ϵv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdz+lim infϵ0A2,ϵv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdzlim infϵ0RNv(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2sdz0.

    Hence

    lim infϵ0v(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)|z|N+2s0

    almost everywhere in A1,ϵ. In particular for zA¯x. We then have by the lower semicontinuity of v and u in ¯Ω and (4.5), that

    0lim infϵ0[v(xϵ)u(yϵ)v(xϵ+z)+u(yϵ+z)]σ+u(¯x+z)v(¯x+z).

    Since zA¯x is arbitrary, we conclude σv(x)u(x) for a.e. in ¯Ω, which implies for xΩ

    0v(x)u(x)lim supyx,yΩ(v(y)u(y))σ.

    A contradiction with the hypothesis.

    In this subsection we prove that strong and viscosity solutions coincide.

    Theorem 4.4. Any strong solution uC1,α(Ω) to problem (1.1) is a viscosity solution as well.

    Remark 4.5. For conditions to ensure the existence of strong solutions to problem (1.1) see Theorem 3.1, Theorem 3.9 and Corollary 3.11.

    Proof. The proof is straightforward, we give it by completeness. Let uC1,α(Ω) be such that

    (Δ)su(x)+|u(x)|p=f(x),for all xΩ.

    Let UΩ be open, take x0U and let ϕC2(U) be such that u(x0)=ϕ(x0) and ϕu in U. Define

    v(x):={ϕ(x), in Uu(x), outside U. (4.13)

    Hence, since u is C1, u(x0)=ϕ(x0) and then we have that

    (Δ)sϕ(x0)+|ϕ(x0)|p=(Δ)sϕ(x0)+|u(x0)|p.

    By the assumption on ϕ, we have that (Δ)sϕ(x0)(Δ)su(x0) and so the u is a viscosity sub-solution. In a similar way, u is a super-solution and the conclusion follows.

    Theorem 4.6. Assume that the condition (3.15) holds that fCϵ(Ω)Lm(Ω), for some ϵ>0 and m>N2s1. We suppose that ||f||Lm(Ω)λ defined in Theorem 3.9. Then any viscosity solution is a strong solution.

    Proof. To prove the converse, assume that u is a viscosity solution to problem (1.1). In view of Theorem 3.9 and Corollary 3.11, there exists a distributional solution v (which is also strong in view of the assumptions on f). Since any strong solution is of viscosity, we consequently infer from the Comparison Theorem 4.3 that u=v. This ends the proof of the theorem.

    (1) For the existence of solution using approximating argument, the limitation p<2s seems to be technical, we hope that the existence of a solution holds for all p2s and for all fL1(Ω). For p>2s, this is an interesting open question, even for the Laplacian, with Lm data. Notice that this is not the framework of the paper [23].

    (2) For p>2s, it seems to be interesting to eliminate the smallness condition ||f||Lm(Ω) and to treat more general set of p without the condition (3.15).

    (3) In order to understand a bigger class of linear integro-differential operators, is seems necessary to obtain alternative techniques independent of the representation formula.

    The first and the third authors are partially supported by Project MTM2016-80474-P, MINECO, Spain. The first author also is partially supported by DGRSDT, Algeria and an Erasmus grant from Autonoma University of Madrid. The second author is partially supported by CONICET and grant PICT 2015-1701 AGENCIA.

    The authors would like to thank the anonymous reviewer for his/her careful reading of the paper and his/her many insightful comments and suggestions.

    The authors declare no conflict of interest.



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