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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Throughout this article, we shall consider the following Dirichlet integro-differential problem
{(−Δ)su+|∇u|p=fin Ωu=0 in RN∖Ω | (1.1) |
for s∈(1/2,1), Ω⊂RN, 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 RN∖Ω |
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∗:=NN−2s+1, there is a unique non-negative distributional solution u∈W1,q0(Ω) of (1.1) for any 1≤q<p∗.
● Now, for x∈Ω, 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<N2s−1, then |∇u|δ1−s∈Lq(Ω) for all 1≤q<mNN−m(2s−1).
- If m=N2s−1, then |∇u|δ1−s∈Lq(Ω) for all 1≤q<∞.
- If m>N2s−1, 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 u∈C1,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 p≤2, then the existence of solution is obtained for all f∈L1(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 RN∖Ω,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∈[0,2s−1).
In [15,Thm. 1.1], they show that for ϵ=1 and under the integrability assumption
∫∞1g(s)s−1−p∗ds<∞, |
problem (1.2) admits a non-negative distributional solution u∈W1,q0(Ω), for all 1≤q<p∗,β where
p∗,β:=NN−2s+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 RN∖Ω, | (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 f∈Lm(Ω) with m≥1 and define v to be the unique solution to problem (1.3) with s>12. Then for all 1≤p<mNN−m(2s−1), there exists a positive constant C≡ˆC(Ω,N,s,p) such that
‖|∇v|δ1−s‖Lp(Ω)≤ˆC||f||Lm(Ω). | (1.4) |
Moreover,
1) If m=N2s−1, then |∇v|δ1−s∈Lp(Ω) for all 1≤p<∞.
2) If m>N2s−1, then v∈C1,σ(Ω) for some σ∈(0,1), and
‖|∇v|δ1−s‖L∞(Ω)≤C||f||Lm(Ω). |
In the case where f∈L1(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 f∈L1(Ω)∩Lmloc(Ω) with m>1. Let v the unique solution to problem (1.3). Suppose that m<N2s−1, then for any Ω1⊂⊂Ω′1⊂⊂Ω and for all 1≤p≤mNN−m(2s−1), there exists ˜C:=˜C(Ω,Ω1,Ω′1,N,s,p) such that
||∇v||Lp(Ω1)≤˜C(||f||L1(Ω)+||f||Lm(Ω′1)). | (1.5) |
Moreover,
1) If m=N2s−1, then |∇v|∈Lploc(Ω) for all 1≤p<∞.
2) If m>N2s−1, then v∈C1,σ(Ω) for some σ∈(0,1).
As a consequence we conclude that, if f∈Lm(Ω) with m>1, then
1) If m≥N2s−1, then ∫Ω|∇v|adx<∞ for all a<11−s.
2) If 1<m<N+2s2s−1, then ∫Ω|∇v|adx<∞ for all a<ˇP:=mNN(m(1−s)+1)−m(2s−1).
Remark 1.3. It is clear that a<a0=11−s 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 ϕ∈W1,p0(Ω), we have
C(Ω)∫Ω|ϕ|pδpdx≤∫Ω|∇ϕ|pdx<+∞. | (1.6) |
We prove now the optimality of a0. We argue by contradiction. Assume that, for 0≨f∈L∞(Ω), there exists a solution v to (1.3) such that v∈W1,p0(Ω) with p>11−s.
By using the classical Hardy inequality we obtain that
∫Ωvpδpdx≤∫Ω|∇v|pdx<+∞. |
By the results in [27] the solution behaves as v⋍δs, therefore, as a consequence, 1δp(1−s)∈L1(Ω), that is, p<11−s, 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 2s≤p<s1−s, 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 p≥2s. 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, p≥2s, 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, N≥1. 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), k≥1, 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 k−th order partial derivatives are locally Hölder continuous with exponent α in Ω.
For σ∈R, 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 u∈S(RN), the fractional Laplacian (−Δ)s is given by
(−Δ)su(x):=limϵ→0(−Δ)sϵu(x) |
where
(−Δ)sϵu(x):=∫RNu(x)−u(y)|x−y|N+2sχϵ(|x−y|)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 ϕ∈C(RN) belongs to Xs(Ω) if and only if the following holds
● supp(ϕ)⊂¯Ω.
● The fractional Laplacian (−Δ)sϕ(x) exists for all x∈Ω and there is C>0 so that |(−Δ)sϕ(x)|≤C.
● There is φ∈L(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 Ω⊂IRN, then the fractional Sobolev Space Ws,p(Ω) is defined by
Ws,p(Ω)≡{ϕ∈Lp(Ω):∬Ω×Ω|ϕ(x)−ϕ(y)|pdν<+∞}, |
where dν=dxdy|x−y|N+ps.
Notice that Ws,p(Ω) is a Banach Space endowed with the norm
‖ϕ‖Ws,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 C∞0(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 v∈C∞0(IRN),
∬IR2N|v(x)−v(y)|p|x−y|N+psdxdy≥S(∫RN|v(x)|p∗sdx)pp∗s, |
where p∗s=pNN−ps and S≡S(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 RN∖Ω, | (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 u∈L1(Ω), and for all ϕ∈Xs, 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 ν∈M(Ω), be a Radon measure such that
∫Ωδβdν<∞,δ(x):=dist(x,Ωc), |
with β∈[0,2s−1). Then the problem (2.1) has a unique weak solution u in the sense of Definition 2.2 such that u∈W1,q0(Ω), for all 1≤q<p∗β where p∗,β:=NN−2s+1+β. Moreover
||u||W1,q0(Ω)≤C(N,q,Ω)∫Ωδβdν. | (2.2) |
For ν∈L1(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 f∈L1(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)|αdx≤Ckα−1||f||L1(Ω). |
We recall also the next comparison principle proved in [1]
Theorem 2.5. (Comparison Principle). Let g∈L1(Ω) and suppose that w1,w2∈W1,q0(Ω) for all 1≤q<NN−2s+1 are such that (−Δ)sw1,(−Δ)sw2∈L1(Ω) with
{(−Δ)sw1≤H1(x,w1,∇w1)+g in Ω,w1≤0 in RN∖Ω,{(−Δ)sw2≥H1(x,w2,∇w2)+g in Ω,w2≤0 in RN∖Ω, |
where H:Ω×IR×IRN→IR is a Carathéodoty function satisfying
1) H1(x,w1,∇w1),H1(x,w2,∇w2)∈L1(Ω),
2) for a.e. x∈Ω, we have
H1(x,w1,∇w1)−H1(x,w2,∇w2)=⟨B(x,w1,w2,∇w1,∇w2),∇(w1−w2)⟩+f(x,w1,w2) |
with B∈(La(Ω))N, a>N2s−1 and f∈L1(Ω) with f≤0 a.e. in Ω.
Then w1≤w2 in Ω.
Recall that we are considering problem (1.1), then we have the next definition.
Definition 2.6. A function u∈L1(Ω), with |∇u|p∈L1loc(Ω), is a distributional solution to problem (1.1) if for any ϕ∈Xs(Ω), there holds
∫Ωu(−Δ)sϕ+∫Ωϕ|∇u|p=∫Ωfϕ, |
and u=0 in RN∖Ω.
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:Ω→R is a strong solution to the equation
(−Δ)sw+|∇w|p=f |
in Ω if u∈C2s+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:RN→R is a viscosity subsolution to (1.1) in Ω, if u∈Lloc(RN), and for any open set U⊂Ω, any x0∈U and any ϕ∈C2(U) such that u(x0)=ϕ(x0) and ϕ≥u in U, if we let
v(x):={ϕ(x) in Uu(x) outside U, | (2.3) |
we have
(−Δ)sϕ(x0)+|∇ϕ(x0)|p≤f(x0), |
and v≤0 in RN∖Ω. On the other hand, a lower semicontinuous function u:RN→R is a viscosity supersolution to (1.1) in Ω if u∈Lloc(RN), and for any open set U⊂Ω, any x0∈U and any ψ∈C2(U) such that u(x0)=ψ(x0) and ϕ≤u 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)|p≥f(x0) |
and v≥0 in RN∖Ω. 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<11−s that we will be used later.
Theorem 2.9. Assume that p>2s−11−sN+1, then for all 0≨f∈L∞(Ω), problem (1.1) has no weak solution u in the sense of Definition 2.2, such that u∈W1,p0(Ω).
Proof. Suppose by contradiction that problem (1.1) has a solution u with u∈W1,p0(Ω). It is clear that u solves the problem
(−Δ)su+⟨B(x),∇u⟩=f, |
where B(x)=|∇u|p−2∇u. Since p>2s−11−sN+1, then |B|∈Lσ(Ω) with σ>N2s−1 and then B∈KsN(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 ˆGs≃Gs, the Green function associated to the fractional laplacian. Hence
Gs(x,y)≃C(B)1|x−y|N−2s(δs(x)|x−y|s∧1)(δs(y)|x−y|s∧1). |
Using the fact that δs(x)|x−y|s≥C(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
δspδp≤Cupδp∈L1(Ω). |
Thus 1δp(1−s)∈L1(Ω). Since p(1−s)≥1, then we reach a contradiction.
Corollary 2.10. Let f be a Lipschitz function such that f≩0, then if p>11−s, problem (1.1) has no solution u such that u∈C1(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>1s−1 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 f∈Lm(Ω) with m≥1, and assume that 1<p<p∗. Then, the Dirichlet problem
{(−Δ)sw+|∇w|p=fin Ωw=0in RN∖Ω, |
has a unique distributional solution w verifying
● if m<N2s−1, then |∇w|∈Lqloc(Ω) for all 1≤q<mNN−m(2s−1);
● if m=N2s−1, then |∇w|∈Lqloc(Ω) for all 1≤q<∞;
● if m>N2s−1, then |∇w|∈Cα(Ω) for some α∈(0,1).
Moreover, if in addition f∈Cϵ(Ω), for some ϵ∈(0,2s−1), 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|p−1∈Lσ(Ω) with σ>N2s−1 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 f∈Cϵ(Ω), and let Ω′⋐Ω, open and let u be a distributional solution to problem (1.1). Since u∈L∞(RN) and f−|∇u|p∈L∞(Ω′), we apply Proposition 2.3 in [27] to derive
u∈Cβ(Ω″), for all β∈(0,2s),Ω″⋐Ω′. |
In particular, we have |∇u|∈Cβ−1(Ω″) for any β∈(1,2s). Consequently, f−|∇u|p∈Cϵ(Ω″). Appealing now to Corollary 2.4 in [27], we obtain u∈C2s+ϵ in a smaller subdomain of Ω″. Thus, u∈C2s+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 u∈Xs. Hence the integration by parts formula
∫Ωu(−Δ)sϕ=∫Ωϕ(−Δ)su |
holds for all ϕ∈Xs. For any ϕ∈C∞0(Ω) we hence obtain
∫Ωϕ(−Δ)su=∫Ωu(−Δ)sϕ=∫Ωfϕ−∫Ω|∇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 f∈C2ns+ϵ−n locally in Ω, for ϵ∈(0,2s−1) and n≥0, then u∈C2(n+1)s+ϵ−n locally in Ω.
In this subsection we will assume that p∗≤p<2s, then the first existence result for problem (1.1) is the following.
Theorem 3.3. Assume that p<2s, then for all f∈L1(Ω) with f≥0, the problem (1.1) has a maximal weak solution u such that u∈W1,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 n∈IN∗, the problem
{(−Δ)sun+|∇un|p1+1n|∇un|p=f in Ω,un=0 in RN∖Ω. | (3.1) |
has a unique solution un such that un∈W1,q0(Ω) for all 1≤q<NN−2s+1 and Tk(un)∈Hs0(Ω). To prove that, we proceed by approximation.
Let k∈IN∗ 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 RN∖Ω. | (3.2) |
where fk=Tk(f). We claim that the sequence {un,k}k is increasing in k, namely un,k≤un,k+1 for all k≥1 and n fixed. To see that, we have
(−Δ)sun,k+1+|∇un,k+1|p1+1n|∇un,k+1|p≥fk. |
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,k≤un,k+1 and then the claim follows. It is clear that un,k≤w for all n,k∈IN∗ where w is the unique solution to problem
{(−Δ)sw=f in Ω,w=0 in RN∖Ω. | (3.3) |
Notice that w∈W1,q0(Ω) for all 1≤q<NN−2s+1 and w∈Lr(Ω) for all 1≤r<NN−2s.
Hence, we get the existence of un such that un,k↑un strongly in Lσ(Ω) for all 1≤σ<NN−2s.
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,k→un strongly in W1,α0(Ω) for all α<NN−2s+1. Since the sequence {uk,n}k is increasing in k, then the limit un is unique. Thus, up to a further subsequence, ∇un,k→∇un 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(un−vn)=ˆH(|∇un|)−ˆH(|∇vn|), |
where ˆH(|ξ|)=−|ξ|p1+1n|ξ|p. Since |ˆH(|ξ|)|≤n, then we obtain that un−vn∈L∞(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 un≤w 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 RN∖Ω. | (3.4) |
Thus
(−Δ)sun+|∇un|p1+1n+1|∇un|p≥f. |
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+1≤un≤w for all n.
Hence, there exists u such that un↓u strongly in Lσ(Ω) for all 1≤σ<NN−2s.
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|x−y|N+2sdxdy+∫Ωgn(|∇un|)Tj(un)dx≤Cj. |
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,ε≤1, then taking into consideration that
(un(x)−un(y))(vn,ε(x)−vn,ε(y))≥0, |
it follows that
∫Ωgn(|∇un|)vn,ε(x)≤∫Ωfdx≤C. |
Letting ε→0, we reach that ∫Ωgn(|∇un|)dx≤C an the claim follows. Define hn=f−gn, then ||hn||L1(Ω)≤C. As a consequence and by the compactness result in Theorem 2.3, we reach that, up to a subsequence, un→u strongly in W1,α0(Ω) for all 1≤α<NN−2s+1 and then, up to an other subsequence, ∇un→∇u a.e in Ω. Hence gn→g 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 gn→g strongly in L1(Ω).
Notice that, using T1(Gj(un)) as a test function in (3.4) it holds that
∫un≥j+1gndx≤∫un≥jfdx→0 as j→∞. |
Let ε>0 and consider E⊂Ω to be a measurable set, then
∫Egndx=∫{E∩{un<j+1}}gndx+∫{E∩{un≥j+1}}gndx≤∫{E∩{un<j+1}}|∇Tj+1(un)|pdx+∫{un≥j+1}fdx. |
By (3.5), letting n→∞, we can chose |E| small enough such that
lim supn→∞∫{E∩{un<j+1}}|∇Tj+1(un)|pdx≤ε2. |
In the same way and since f∈L1(Ω), we reach that
lim supn→∞∫{un≥j+1}fdx≤ε2. |
Hence, for |E| small enough, we have
lim supn→∞∫Egndx≤ε. |
Thus by Vitali lemma we obtain that gn→g 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|p≤f. |
Hence ˆu≤un and then ˆu≤u.
Remark 3.4. 1) The existence of a unique solution to the approximating problem (3.4) holds for all p≥1.
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<NN−1 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,g≩0, the problem
{(−Δ)su+|∇u|p=g(x)u1+au+f in Ω,u=0 in RN∖Ω. | (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 p≥p∗, 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 K⊂Ω, we define
Capσ,p(K)=inf{‖ψ‖Wσ,p0(Ω):ψ∈Wσ,p0(Ω),0≤ψ≤1 and ψ≥χK a.e. inΩ}. | (3.7) |
Now, if U⊂Ω is an open set, then
Capσ,p(U)=sup{Capσ,p(K):K⊂U compact of Ω with K⊂U}. |
For any borel subset B⊂Ω, the definition is extended by setting:
Capσ,p(B)=inf{Capσ,p(U), U open subset of Ω, B⊂U}. |
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 x0∈Ω, then the problem
{(−Δ)su+|∇u|p=δx0in Ω,u=0 in RN∖Ω, | (3.8) |
has non solution u such that u∈W1,p0(Ω).
Proof. For simplify of tipping we assume that x0=0∈Ω 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 u∈W1,p0(Ω). Then u∈Wσ,p0(Ω) for all σ<1. We claim that (−Δ)su∈W−σ,p(Ω), the dual space of Wσ,p0(Ω), for all σ∈(2s−1,2s). To see that, we consider ϕ∈C∞0(Ω), then
|∫Ω(−Δ)suϕdx|≤∬IR2N|u(x)−u(y)||ϕ(x)−ϕ(y)||x−y|N+2sdxdy≤(∬IR2N|u(x)−u(y)|p|x−y|N+p(2s−σ)dxdy)1p(∬IR2N|ϕ(x)−ϕ(y)|p′|x−y|N+p′σdxdy)1p′. |
Since 2s−σ∈(0,1), then (∬IR2N|u(x)−u(y)|p|x−y|N+p(2s−σ)dxdy)1p≤C(σ,s,N,Ω)||u||W1,p0(Ω). Thus
|∫Ω(−Δ)suϕdx|≤C||u||W1,p0(Ω)||ϕ||Wσ,p′0(Ω), |
and then the claim follows. Hence going back to problem (3.8), we deduce that δ∈L1(Ω)+W−σ,p(Ω).
As in [7], let us now show that if ν∈W−σ,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 A⊂A0 and Capσ,p′(A0)=0. Let K⊂A0 be a compact set, then there exists a sequence {ψn}n∈C∞0(Ω) such that 0≤ψn≤1, ψn≥χK and ||ψn||p′Wσ,p′0(Ω)→0 as n→∞. It is clear that ψn→χK a.e in Ω, as n→∞. Hence
ν(K)=limn→∞∫ψndν=limn→∞⟨ψn,ν⟩Wσ,p′0(Ω),W−σ,p0(Ω). |
Thus
|ν(K)|≤lim supn→∞|⟨ψn,ν⟩Wσ,p′0(Ω),W−σ,p0(Ω)|≤lim supn→∞||ν||W−σ,p0(Ω)||ψn||Wσ,p′0(Ω)=0. |
Therefore, we conclude that for any compact set K⊂A0, we have |ν(K)|=0. Hence |ν(A0)|=0 and the result follows.
Notice that if h∈L1(Ω), then |h|<<Capσ,p′. As a conclusion, we deduce that δ<<Capσ,p′ for all σ∈(2s−1,2s).
Since p>p∗, we can choose σ0∈(2s−1,2s) such that p′σ0<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|α−1)+ with 0<α<N−σ0p′p′, we obtain that w∈Wσ,p′0(Ω). Notice that, for all v∈Wσ,p′0(Ω), we know that
Capσ,p′{|v|≥k}≤Ck||v||Wσ,p′0(Ω). |
Since w(0)=∞, then {0}⊂{|w|≥k} for all k>0. Thus
Capσ,p′{0}≤Ck||w||Wσ,p′0(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∈(2s−1,2s).
Let consider now the next problem
{(−Δ)su+|∇u|p=λg(x)u+f in Ω,u=0 in RN∖Ω. | (3.9) |
with g≩0 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 g∈L1(Ω). We say that g is an admissible weight if
C(g,p)=infϕ∈W1,p0(Ω)∖{0}(∫Ω|∇ϕ|pdx)1p∫Ωg|ϕ|dx>0. | (3.10) |
∙ If g∈LpNN(p−1)+p(Ω) with g≩0, 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 f∈L1(Ω) with f≥0 and for all λ>0, the problem (3.9) has a solution u such that u∈W1,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 RN∖Ω. | (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)dx≤kλ∫Ωg(x)undx+k‖f‖L1(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)σ−C2(k), |
Thus using (3.10) for Hk(un) it holds that
∫Ω|∇Hk(un)|pdx≥C(g,p)(∫ΩgHk(un)dx)p≥C(g,p)[C1(k)(∫Ωgundx)p−C2(k)] |
where C1(k),C2(k)>0 are independent of n.
Therefore, going back to (3.12), we conclude that
C1(k)(∫Ωgundx)pdx≤kC(g,p)[λ∫Ωg(x)undx+‖f‖L1(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−(2s−1). Thus, in this case and for all λ>0, problem (3.9) has a solution u such that u∈W1,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 λ≤ΛN,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 2s≤p<s1−s and in the same way as above we can show the next existence result.
Theorem 3.9. Suppose that f∈Lm(Ω) with m>N/[p′(2s−1)]. Then there is λ∗>0 such that if ||f||Lm(Ω)≤λ∗, problem (1.1) admits a solution uδ1−s∈W1,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 N≥2. 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={v∈W1,10(Ω):vδ1−s∈W1,pm0(Ω) and (∫Ω|∇(vδ1−s)|pmdx)1pm≤l12s}, | (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 v∈E, then |∇v|pmδpm(1−s)∈L1(Ω) and
(∫Ω|∇v|pmδpm(1−s)dx)1pm≤ˆC0(Ω)l1p. |
Define now the operator
T:E→W1,10(Ω)v→T(v)=u |
where u is the unique solution to problem
{(−Δ)su=−|∇v|p+f in Ω,u=0 in RN∖Ω,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<2s−1 such that |f−|∇v|p|δβ∈L1(Ω). To do that we have just to show that |∇v|pδβ∈L1(Ω).
It is cleat that |∇v|p∈L1loc(Ω), moreover, we have
∫Ω|∇v|pδβdx=∫Ω|∇v|pδp(1−s)δβ−p(1−s)dx≤(∫Ω|∇v|pmδpm(1−s)dx)1m(∫Ωδ(β−p(1−s))m′dx)1m′. |
If p(1−s)<2s−1, we can chose β<2s−1 such that p(1−s)<β. Hence ∫Ωδ(β−p(1−s))m′dx<∞.
Assume that p(1−s)≥2s−1, then s∈(12,p+2p+1]. Notice that, since p<s1−s, then p(1−s)−(2s−1)<1−s. Since m>Np′(2s−1)>1s, then (p(1−s)−(2s−1))m′<1. Hence we get the existence of β<2s−1 such that (p(1−s)−β)m′<1 and then we conclude.
Then using the fact that v∈E, we reach that |∇v|pδβ+f∈L1(Ω). Therefore the existence of u is a consequence of Theorems 2.3 and 1.1. Moreover, |∇u|∈Lα(Ω) for all α<NN−2s+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))dy−∫ΩGs(x,y)|∇v(y)|2sdy, |
then
∇u(x)=∫Ω∇xGs(x,y)f(y))dy−∫Ω∇xGs(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)|δ1−s≤C2(Ω,N,s)(I1(x)+I2(x)+J1(x)+J2(x)), |
where
I1(x)=δ1−s(x)∫{|x−y|<δ(x)}Gs(x,y)|x−y||∇v(y)|2sdy, |
I2(x)=1δs(x)∫{|x−y|≥δ(x)}Gs(x,y)|∇v(y)|2sdy, |
J1(x)=δ1−s(x)∫{|x−y|<δ(x)}Gs(x,y)|x−y|f(y)dy, |
and
J2(x)=1δs(x)∫{|x−y|≥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,Ji∈Lpm(Ω) for i=1,2 and
||Ii||Lpm(Ω)+||Ji||Lpm(Ω)≤˘C(|||∇v|δ1−s||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|δ1−s||Lpm(Ω)≤˘C(Ω,N,s,m,p)C2(Ω,N,s,m,p)(|||∇v|δ1−s||Lpm(Ω)+||f||Lm(Ω))≤ˉC(Ω,N,s,m,p)(l1p+||f||Lm(Ω))=l1p. |
Thus u∈E 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 u∈E such that T(u)=u. Thus, u∈W1,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(2s−1)1−s and p<s(2s−1)N(1−s)−1. | (3.15) |
If f∈Cϵ(Ω), for some ϵ∈(0,2s−1), 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|p−1∈Lσloc(Ω) for some σ>N2s−1 which is the key point in order to get the desired regularity.
In the case where f≩0, we can prove also that u≩0, more precisely, we have
Corollary 3.12. Assume that the above conditions hold. Let f∈Cϵ(Ω)∩L∞(¯Ω), for some ϵ∈(0,2s−1). If f(x)≥0 for all x∈Ω, then the solution from Theorem 3.9 is non-negative. Moreover, if f1≤f2 and u1 and u2 are the corresponding strong solutions to f1 and f2 from Corollary 3.11, respectively, then u1≤u2.
Proof. Suppose that there is a point x0∈Ω 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 0≤f(x1)−0=(−Δ)su(x1)<0.
We next prove the last statement in the Corollary 3.12. Let f1≤f2. Let u1 and u2 be the corresponding strong solutions from Corollary 3.11, and assume that
minΩ(u2−u1)=u2(x0)−u1(x0)<0. |
Hence ∇(u1−u2)(x0)=0 and (−Δ)s(u2−u1)(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 f∈Cϵ(Ω) and that
p<p∗ |
or
N<s(2s−1)1−s,p∗≤p<s(2s−1)N(1−s)−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)⊂U⋐Ω and let u∈C2(U). Then:
|P.V.∫Bϵ(x)u(x)−u(y)|x−y|N+2sdy|≤cϵ |
where cϵ is independent of x and cϵ→0 as ϵ→0.
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 ϕ∈C2(U), U⋐Ω, touching u from below at x0∈U. Then (−Δ)su(x0) is finite and moreover:
(−Δ)su(x0)+|∇ϕ(x0)|p≥f(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
∫Br∖Bρu(0)−u(y)|y|N+2sdy=∫Br∖Bρϕ(y)−u(y)|y|N+2sdy−∫Br∖Bρϕ(y)|y|N+2sdy≤−∫Br∖Bρϕ(y)|y|N+2sdy, |
where we have used that ϕ touches u from below. As ρ→0, the last integral converges since ϕ∈C2(Br). Hence
limρ→0∫Br∖Bρu(0)−u(y)|y|N+2sdy∈[−∞,M], | (4.2) |
where
M:=limρ→0(−∫Br∖Bρϕ(y)|y|N+2sdy). |
Also, from the fact that u is a supersolution, we have u≥0 in RN∖Ω. Thus
∫RN∖Bru(0)−u(y)|y|N+2sdy≤∫¯Ω∖Br−u(y)|y|N+2sdy. | (4.3) |
Since u∈LSC(¯Ω), there is a constant m so that
u(y)≥m, for all y∈¯Ω∖Br. |
Hence from (4.3), it follows
∫RN∖Bru(0)−u(y)|y|N+2sdy≤−m∫RN∖Br1|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
∫RN∖Bru(0)−u(y)|y|N+2sdy=∫RN∖Brϕr(0)−ϕr(y)|y|N+2sdy=(−Δ)sϕr(0)−P.V.∫Br−ϕ(y)|y|N+2sdy≥−|∇ϕ(0)|p+f(0)+ρ. |
By letting r→0, and then ρ→0, we derive (4.1).
We now give the main result of this section.
Theorem 4.3 (Comparison principle for viscosity solutions). Assume that f∈C(Ω). Let v∈USC(¯Ω) be a subsolution and u∈LSC(¯Ω) be a supersolution, respectively, of (1.1). Then v≤u in Ω.
Proof. We argue by contradiction. Assume that there is x0∈Ω so that:
σ:=supΩ(v−u)=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|x−y|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
limϵ→0|xϵ−yϵ|2=0, |
hence ¯x=¯y. Moreover
Ψϵ(xϵ,yϵ)≥Ψϵ(¯x,¯x) |
implies that:
limϵ→01ϵ|xϵ−yϵ|2=0. |
As a consequence, by letting ϵ→0 in (4.4) and using the semicontinuity of u and v, we obtain
σ=limϵ→0(v(xϵ)−u(yϵ)). | (4.5) |
Also, observe that ¯x∈Ω, because otherwise there is a contraction with v≤u in RN∖Ω.
Define the C2 test functions
ϕϵ(x):=v(xϵ)−1ϵ|xϵ−yϵ|2+1ϵ|x−yϵ|2, |
ψϵ(y):=u(yϵ)−1ϵ|xϵ−yϵ|2+1ϵ|xϵ−y|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)|p≤f(xϵ) |
and
(−Δ)su(yϵ)+|∇ψϵ(yϵ)|p≥f(yϵ) |
Therefore:
(−Δ)sv(xϵ)−(−Δ)su(yϵ)≤f(xϵ)−f(yϵ)+|∇ψϵ(yϵ)|p−|∇ϕϵ(xϵ)|p. | (4.6) |
Since f∈C(Ω) and
∇yψϵ(yϵ)=−∇xϕϵ(xϵ), |
we have that the right hand side in (4.6) tends to 0 as ϵ→0. Thus, we obtain
lim infϵ→0∫RNv(xϵ)−v(xϵ+z)−u(yϵ)+u(yϵ+z)|z|N+2sdz=lim infϵ→0((−Δ)sv(xϵ)−(−Δ)su(yϵ))≤0. | (4.7) |
Let A1,ϵ:={z∈RN:xϵ+z,yϵ+z∈Ω}. Hence for z∈A1,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:=RN∖A1,ϵ. We will justify that we are allowed to use Fatou's Theorem in
lim infϵ→0∫A2,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,yϵ∈Br(¯x) for all ϵ<ϵ0. Take z∈A2,ϵ. We show now that |z|≥2r. Indeed, to reach a contradiction, assume that |z|<2r. Since z∉A1,ϵ, it follows that xϵ+z or yϵ+z does not belong to Ω. Without loss of generality, assume xϵ+z∉Ω. Hence
|xϵ+z−¯x|<3r, |
and so xϵ+z∈B3r(¯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 z∉B2r when z∈A2,ϵ, we have
∫A2,ϵ|v(xϵ)||z|N+2sdz≤C∫RN∖B2r1|z|N+2sdz<∞. |
On the other hand
∫A2,ϵ|v+(z+xϵ)||z|N+2sdz≤∫RN∖B2r|v+(z+xϵ)||z|N+2sdz=∫RN∖B2r(xϵ)|v+(y)||y−xϵ|N+2sdy. |
Since v is a subsolution, we have v≤0 in RN∖Ω. Hence
∫A2,ϵ|v+(z+xϵ)||z|N+2sdz≤∫¯Ω∖B2r(xϵ)|v+(y)||y−xϵ|N+2sdy≤∫¯Ω∖Br(¯x)|v+(y)||y−xϵ|N+2sdy≤1rN+2s∫¯Ω∖Br(¯x)v+(y)dy. |
Observe that the last integral is finite since v∈L1loc(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 infϵ→0∫A2,ϵv(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sdz≥∫RNlim infϵ→0v(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sχA2,ϵ(z)dz≥∫RN∖A¯xσ+u(¯x+z)−v(¯x+z)|z|N+2sdz. | (4.11) |
Here A¯x:={z∈RN:¯x+z∈Ω} 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 infϵ→0[v(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)]≥σ−v(¯x+z)+u(¯x+z) |
for a.e. z∈RN. Moreover, the inequality u≥v in RN∖Ω implies that the last integral in (4.11) is non-negative. Then
lim infϵ→0∫A2,ϵv(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sdz≥0. | (4.12) |
Therefore by Fatou Lemma, (4.12) and (4.7), we deduce
∫RNlim infϵ→0v(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sχA1,ϵdz≤lim infϵ→0∫A1,ϵv(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sdz≤lim infϵ→0∫A1,ϵv(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sdz+lim infϵ→0∫A2,ϵv(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sdz≤lim infϵ→0∫RNv(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2sdz≤0. |
Hence
lim infϵ→0v(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)|z|N+2s≤0 |
almost everywhere in A1,ϵ. In particular for z∈A¯x. We then have by the lower semicontinuity of −v and u in ¯Ω and (4.5), that
0≥lim infϵ→0[v(xϵ)−u(yϵ)−v(xϵ+z)+u(yϵ+z)]≥σ+u(¯x+z)−v(¯x+z). |
Since z∈A¯x is arbitrary, we conclude σ≤v(x)−u(x) for a.e. in ¯Ω, which implies for x∈∂Ω
0≥v(x)−u(x)≥lim supy→x,y∈Ω(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 u∈C1,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 u∈C1,α(Ω) be such that
(−Δ)su(x)+|∇u(x)|p=f(x),for all x∈Ω. |
Let U⊂Ω be open, take x0∈U and let ϕ∈C2(U) be such that u(x0)=ϕ(x0) and ϕ≥u 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 f∈Cϵ(Ω)∩Lm(Ω), for some ϵ>0 and m>N2s−1. 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 p≤2s and for all f∈L1(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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