Nonlocal Lazer-McKenna type problem perturbed by the Hardy's potential and its parabolic equivalence

In this paper, we study the effect of Hardy potential on the existence or non-existence of solutions to a fractional Laplacian problem involving a singular nonlinearity. Also, we mention a stability result.

Here 0 < s < 1, λ > 0, γ > 0, and Ω ⊂ R N (N > 2s) is a bounded smooth domain such that 0 ∈ Ω. Moreover, 0 ≤ µ, f ∈ L 1 (Ω). For 0 < λ ≤ Λ N,s , Λ N,s being the best constant in the fractional Hardy inequality, we find the necessary and sufficient condition for the existence of a positive weak solution to the above problem with respect to the data µ and f . Also, for a regular datum of f and with suitable assumptions, we have some existence and uniqueness results and calculate the rate of the growth of solutions. Moreover, we mention a non-existence and a complete blow-up result for the case λ > Λ N,s . Besides, we consider the parabolic equivalence of the above problem in the case µ ≡ 1, and some suitable f (x, t), i.e.
We will prove that for 0 < λ ≤ Λ N,s , Λ N,s = 4 s Γ 2 ( N+2s 4 ) Γ 2 ( N−2s 4 ) being the best constant in the fractional Hardy inequality, the above problem has a solution if and only if µ ∈ L 1 (Ω, δ s(1−γ) dx), δ(x) = dist(x, ∂Ω), and the datum of f satisfies the following integrability condition: where the constant β = β(N, s, λ) will be defined later in Lemma 2.2. In this lemma, we will see that any supersolution to (1) is unbounded near the origin and the nature of this unboundedness is like u(x) |x| −β in some open ball centered at the origin.
Also, we will see that there is no positive very weak (distributional) solution for the case λ > Λ N,s . This notion of the solution, which we consider for the non-existence result, is local in nature and we just ask the regularity needed to give distributional sense to the equation (similar to what is done in articles [1,2]). Moreover, this non-existence result is strong in the sense that a complete blow-up phenomenon occurs. By complete blow-up phenomenon, we mean that the solutions to the approximating problems (with the bounded weights (|x| 2s + ǫ) −1 and (u + ǫ) −γ instead of the terms |x| −2s and u −γ , respectively) tend to infinity for every x ∈ Ω, as 0 < ǫ ↓ 0.
In the above problem, (−∆) s stands for the fractional Laplacian operator, i.e. where P.V. is a commonly used abbreviation for the Cauchy principal value and is defined by the latter equation.
Here Γ denotes the Gamma function, and F u =û is the Fourier transform of u. By restricting the fractional Laplacian operator to act only on smooth functions that are zero outside Ω, we have the restricted fractional Laplacian (−∆ | Ω ) s . For this operator, the best alternative to the Dirichlet boundary condition is u ≡ 0 in R N \ Ω . For more details about fractional Laplacian, see [3,4,5]. Over the past decades, there has been much focus and also a vast literature about singular problems. Singularities appear in almost all fields of mathematics like differential geometry and partial differential equations. Singularities are the qualitative side of mathematics, and understanding of singularities always leads to a more detailed picture of the objects mathematics is dealing with, [6]. Many more details and references for the singular elliptic problems can be found in [7].
One famous type of singularities are the singularity of Hardy type, which is related to the inequality of the same name, and there are various generalizations of it. The well-known classical Hardy inequality is as follows: where Ω ⊂ R N , containing the origin, is a bounded domain and 1 ≤ p < N, [8,9]. The constant N−p p p is optimal and it is not attained in W 1,p 0 (Ω), meaning that the continuous embedding W 1,p 0 (Ω) ֒→ L p (Ω, |x| −p dx) is not compact. The intention of analyzing Hardy singularities has come from its widespread use in different branches of science. For details and references about the enormous literature for this topic, see the more recent book [10] and chapter 1 of [11]. Due to these motivations, over the past few decades, the study of general singularities has been considered.
In the pioneering works, [12,13], Baras and Goldstein studied the following singular Cauchy-Dirichlet heat problem in Ω = R N or else Ω to be a bounded smooth domain containing B 1 (0) = {x ∈ R N : x < 1}.
Authors assume that f and u 0 are non-negative and 0 ≤ V ∈ L ∞ (Ω \ B ǫ (0)), for each ǫ > 0, but V is singular at the origin. They say that V is too is the sharp constant in the following Hardy inequality: In the not too singular potential case, they found the necessary and sufficient condition for the existence of a nonnegative distributional solution to problem (2). Moreover, they obtained this solution as the limit of the solutions to the following approximate problem.
where V n (x) = min{V(x), n}. Also, for the too singular potential case, they showed that the problem has no solution even in the sense of distributions, and an instantaneous complete blow-up phenomenon occurs. Namely, u n (x, t) → +∞ for all (x, t) ∈ Ω × (0, T ) as n → ∞.
In [17] Barrios, Bonis, Medina and Peral studied the solvability of the following superlinear problem: More precisely, for the case M = 0 and f ≥ 0, they proved the existence of a positive solution for every γ > 0 and λ > 0. Moreover, in the case M = 1 and f ≡ 1, they found a threshold Λ such that there exists a solution for every 0 < λ < Λ, and there does not for λ > Λ. Also in [28] authors considered the similar superlinear problem with the critical growth, namely when p = 2 * s − 1 = N+2s N−2s , and with a singular nonlinearity in the form u −q , q ∈ (0, 1). In the detailed article [29], Abdellaoui, Medina, Peral, and Primo studied the effect of the Hardy potential on the existence and summability of the solutions to a class of fractional Laplacian problems. We will use the essential tool introduced in this article, i.e., the weak Harnack's inequality, which they proved it by following the classical Moser and Krylov-Safonov idea. Also, we will take advantage of some of Calderón-Zygmund properties of solutions. See [29,Section 4] for the effect of the Hardy potential in some Calderón-Zygmund properties for the fractional Laplacian.
For the similar parabolic equivalence of (1), in [30], Giacomoni, Mukherjee and Sreenadh investigated the existence and stabilization results for the following parabolic equation involving the fractional Laplacian with singular nonlinearity: Under suitable assumptions on the parameters and datum, they studied the related stationary problem and then using the semi-discretization in time with the implicit Euler method, they proved the existence and uniqueness of the weak solution. It is worth noting that in [31,32], the authors have shown the same results for the local version of this problem for the general p-Laplacian case. Also for some of the recent papers on the optimal regularity results see [33,34]. The rest of the paper is as follows. In section 2, after introducing the functional setting we will outline our existence and non-existence theorems. Especially, we will have a theorem about the necessary and sufficient condition for the existence of a solution to problem (1) in the case λ ≤ Λ N,s , and a non-existence theorem in the case λ > Λ N,s . In section 3, we will provide proof of our existence theorems. In section 4, we will have some uniqueness results. Also, concerning uniqueness, with some regular assumptions on µ and f , we will show the existence and uniqueness of another notion of a solution so-called entropy solution for the case 0 < γ ≤ 1. Besides, we will mention a theorem about the rate of the growth of solutions to problem (1). Finally, in section 5, we will consider the parabolic version of problem (1) in the special case µ ≡ 1. Firstly, with the assumptions 0 < γ ≤ 1, or γ > 1, and 2s(γ − 1) < (γ + 1), we will show the existence of a unique solution for 0 < λ < Λ N,s and secondly, we will prove the stability for some range of λ. That is, we will find a positive constant λ * = λ * (N, s) < Λ N,s such that for any λ ∈ (0, λ * ), the solution to the parabolic problem converges to the unique solution of its stationary problem, as t → ∞.

Functional setting and existence, non-existence and blow-up results
Let 0 < s < 1, 1 ≤ p < ∞, and Ω be a bounded domain in R N . Also, let D Ω = R N × R N \ Ω c × Ω c , with Ω c = R N \ Ω. We define the following Banach space endowed with the norm: In the case p = 2, we denote by X s (Ω) the space X s,2 (Ω) which is a Hilbert space with the following inner product: Moreover, we define X s,p 0 (Ω) as the closure of C ∞ 0 (Ω) in X s,p (R N ). Equivalently, it can be shown that X s,p 0 (Ω) = u ∈ X s,p (R N ) : u = 0 a.e. in (R N \ Ω) . 4 It is easy to see that: This equality defines an equivalent norm for X s,p 0 (Ω) with (4). We denote it by It is worth noticing that, the continuous embedding of X s 2 0 (Ω) into X s 1 0 (Ω), holds for any s 1 < s 2 (see, e.g. [4, Proposition 2.1]). Besides, for the Hilbert space case, we have where C N,s is the normalization constant in the definition of (−∆) s . Thus Hardy inequality (3) also can be written as follows: For the proofs of the above facts see [35,Subsection 2.2] and [4]. Also, see [36,Section 2]. The following continuous embedding will be used in this paper.
where p * s = pN N−ps is the Sobolev critical exponent. Moreover, this embedding is compact for 1 ≤ q < p * s . See [4, Theorem 6.5 and Theorem 7.1].
Also we denote by X s,p loc (Ω), the set of all functions u such that uφ ∈ X s,p 0 (Ω) for any φ ∈ C ∞ c (Ω). When we say {u n } ⊂ X s,p loc (Ω) is bounded, we mean that {φu n } ⊂ X s,p 0 (Ω) is bounded for any fixed φ ∈ C ∞ c (Ω). Since we are dealing with the non-local operator (−∆) s , the following class of test functions will be used for defining the weak solution to problem (1).
Definition 2.2. We say that u ∈ X s 0 (Ω) is a weak energy supersolution (subsolution) to for all non-negative φ ∈ X s 0 (Ω). If u is a weak energy supersolution and subsolution, then we say that u is a weak energy solution.
We say that u is a weak solution to problem (1) if • u ∈ L 1 (Ω), and for every K ⋐ Ω, there exists C K > 0 such that u(x) ≥ C K a.e. in K and also u ≡ 0 in R N \ Ω ; and also together with these extra assumptions that the first and second terms on the right-hand side of the above equality be finite for any φ ∈ T (Ω). The well-posedness of the first and second terms on the right-hand side will be clear after the construction of solution.
As an analysis of the linear case with Hardy potential, firstly, we gather the following lemmas.
For the proof of these lemmas see [29,Lemma 3.10], [29,Theorem 4.10] and [29,Theorem 3.4], respectively. In the next two theorems we have our existence results to problem (1). At first, we will prove that for 0 < λ < Λ N,s , and γ ≥ 1 the problem (1) admits a solution for the case µ ∈ L 1 (Ω), and f ∈ L 1 (Ω) ∩ X −s (Ω). It is crucial to indicate that our approach in the proof of Theorem 2.5, only works for the case γ ≥ 1. However if we further assume that µ ∈ L m (Ω), m = ( 2 * s 1−γ ) ′ (p ′ denotes the conjugate exponent of p) then the same approach works for γ < 1. For a result about the existence with less regularity assumption on µ, see [29,Theorem 5.3]. More precisely, the authors showed an existence result for the case µ ∈ L 1 (Ω, |x| −(1−γ)β dx).
In the following we denote the usual truncation operator and G n (σ) := σ − T n (σ).

If γ > 1, then there is a positive weak solution in X s
(Ω), then there is a positive weak solution in X s 0 (Ω) to problem (1). The next theorem gives a necessary and sufficient condition for the existence result to problem (1). Theorem 2.6 (A necessary and sufficient condition for the existence result). Let s ∈ (0, 1), 0 < λ ≤ Λ N,s , and γ > 0. Also assume that 0 ≤ f, µ ∈ L 1 (Ω). Then problem (1) has a positive weak solution if and only if Moreover, the solution u has the following regularity: • u ∈ X s 1 ,p 0 (Ω), for all s 1 < s and for all p < N N−s . Remark 2.7. A similar argument as in [41,Example 3.3] but with the fractional Laplacian instead of the Laplacian operator shows that problem (1) does not admit a solution for merely f ∈ L 1 (Ω).
The proof of these theorems will appear in the next section. In the following, we will have a non-existence and also a blow-up result for the case that λ > Λ N,s .
The following non-existence result is an immediate consequence of Lemma 2.2 and Lemma 2.3. More precisely, it is well known that the linear problem with Hardy potential has non positive supersolution if λ > Λ N,s . We only bring it here for completeness. Proof. We argue by contradiction. Let u be a positive very weak solution to problem (1). Therefore u satisfies . Then by using Lemma 2.3 and the positivity of g necessarily: for some B r (0) ⋐ Ω. On the other hand, by Lemma 2.2 we have for sufficiently small r, where β = N−2s 2 − α and α ∈ [0, N−2s 2 ) is given by the identity ) .
This non-existence result is strong in the sense that a complete blow-up phenomenon occurs. Namely, if u n is the solution to the following approximated problem with λ > Λ N,s , where the Hardy potential is substituted by the bounded weight (|x| 2s + 1 n ) −1 , and the singular nonlinearity is substituted by min{µ,n} In the same sprite of Theorem 2.8, the proof of this blow-up phenomenon can be obtained taking into consideration that any approximating sequence of non-negative supersolution to the linear problem blow-up in any point of Ω, if λ > Λ N,s , as it is proved in [29].

Proof of Theorem 2.5 and Theorem 2.6
First of all we prove Theorem 2.5. For this purpose let consider the following auxiliary problem: where g ∈ X −s (Ω). The function u ∈ X s 0 (Ω) is a weak energy solution to the above problem if u ≡ 0 in R N \ Ω and Here ·, · X −s (Ω),X s 0 (Ω) denotes the duality pairing between X −s (Ω) and X s 0 (Ω). The proof of the following Proposition about the existence result for (14), can be obtained by using the Hardy inequality and the classical variational methods. See for instance [43,Section 4.6]. Also, the uniqueness of the weak energy solution to (14) follows from the strict monotonicity of the operator (−∆) s u − λ u |x| 2s , for 0 ≤ λ < Λ N,s . The strict monotonicity of this operator is the direct consequence of the Hardy inequality.
Proposition 3.1. If g(x) ∈ L 2 (Ω), s ∈ (0, 1) and 0 < λ < Λ N,s , then there exists a unique positive weak energy solution to (14) in X s 0 (Ω). Before to continue, we need to define the set C as the set of functions v ∈ L 2 (Ω) such that there exist positive constants k 1 and k 2 such that where the constant β is given in Lemma 2.2, and δ(x) = dist(x, ∂Ω), x ∈ Ω, is the distance function from the boundary ∂Ω. Now, for every v ∈ C, define Φ(v) = w where w ∈ X s 0 (Ω) is the unique solution to the following problem for any fixed n: Here f n = T n ( f ), and µ n = T n (µ) are the truncations at level n. By Lemma 2.2, [29, Theorem 4.1] and a result of [37] it easily follows that w ∈ C. If we show that Φ : C → C has a fixed point w n , then w n ∈ C will be the weak solution to the following problem in X s 0 (Ω).
We apply the Schauder's fixed-point Theorem (see for example [43,Theorem 3.2.20]). We need to prove that Φ is continuous, compact and there exists a bounded convex subset of C ⊂ L 2 (Ω) which is invariant under Φ.
For continuity let v k → v in L 2 (Ω). It is obvious that for each n: Now, from the uniqueness of the weak solution to (14), For compactness, we argue as follows.
For v ∈ C, let w be the solution to (15). If λ s 1 (Ω) is the first eigenvalue of (−∆) s in X s 0 (Ω), [38,Proposition 9], then we have where in the last inequality we have used the Hardy inequality. Testing (15) For the first term on the right-hand side of the above equality we have the following estimate: where in the last inequality we have used the Hölder inequality. Once more using Hölder inequality gives Ω f n w dx ≤ C 2 Ω |w| 2 dx 1 2 for some C 2 > 0. Thus combining this inequality with (17), (18), and (19) we obtain which implies that Φ(L 2 (Ω)) is contained in a ball of finite radius in L 2 (Ω). Therefore the intersection of this ball with 4 , which means that Φ(L 2 (Ω)) is relatively compact in L 2 (Ω) by the compactness of the embedding (6).
Proposition 3.2. For every K ⋐ Ω, there exists C K > 0 such that {w n }, the solution to (16), satisfies w n (x) ≥ C K a.e. in K, for each n.
Proof. Let us consider the following problem: Existence of the weak solution v n follows from a similar proof to problem (16). In the same way of [17,Lemma 3.2] we can show that v n ≤ v n+1 a.e. in Ω. Also for each K ⋐ Ω, there exists C K > 0 such that v 1 (x) ≥ C K a.e. in K. Now subtracting the weak formulation of (20) from the weak formulation of (16) and using (w n − v n ) − as a test function (see [44,Theorem 20]) we conclude that w n ≥ v n a.e. in Ω. Therefore, for every K ⋐ Ω, there exists C K such that For the left-hand side, by using (7) and the following elementary inequality we get For the first term on the right-hand side we have For the second term on the right-hand side of (21), note that Also for the last term: Thus from (21), (23), (24), (25) and (26) we obtain If we show that the term is uniformly bounded in n, then (27) gives T . For proving the boundedness of (28), it is enough to consider φ = G k (w n ) as a test function in (16) as follows, where G k (σ) := σ − T k (σ).
Note that for the left-hand side we have used [44,Proposition 3]. In order to estimate the terms on the right-hand side of this equality for uniformly in n, we have the following. For the second term on the right-hand side of (29) we have the following estimate uniformly in n: For Ω f n G k (w n ) dx, we have the following estimate: . For the first term on the right-hand side of (29) we can write: For the last term in (30), by using the Hölder inequality with exponents a = 2 * s and b = 2N N+2s < N 2s , noting that the integration can be over Ω, because of w n ≡ 0 in R N \ Ω, and the embedding (6) we obtain . Combining the above estimates, from (29) we get Now Hardy inequality shows the boundedness of the term, R N |G k (w n )| 2 |x| 2s dx, and therefore we obtain the boundedness of (28) by using the fact that w 2 n ≤ 2(T 2 Moreover, we get the boundedness of G k (w n ) X s 0 (Ω) uniformly in n. Now we show that {T k (w n )} is bounded in X s loc (Ω). For this purpose first note that by Proposition 3.2, for any compact set K ⋐ Ω, there exists C(K) > 0 such that a.e. in K. Therefore For (x, y) ∈ K × K, define α n := T k (w n )(x) C , and β n := T k (w n )(y) C . Since α n , β n ≥ 1, we have the following estimate by applying an elementary inequality Now by the definition of α n and β n , we obtain Thus we get the boundedness of {T k (w n )} ∞ n=1 in X s loc (Ω) by using (5) and the boundedness of {T Since the rest of the proof can be obtained proceeding as in the case γ = 1, for the sake of brevity it is left to the reader.
Now we are ready to proof Theorem 2.5.
Proof of Theorem 2.5. There exists u ∈ X s loc (Ω) (u ∈ X s 0 (Ω) in the case γ ≤ 1) such that up to a subsequence • w n → u weakly in X s loc (Ω) (weakly in X s 0 (Ω) in the case γ ≤ 1).
k (u) weakly in X s 0 (Ω). Also, by using the embedding (6), up to a subsequence we may have • w n → u in L r (Ω), for any r ∈ [1, 2 * s ).
Now for every fixed φ ∈ T (Ω), by the estimates above, we could pass to the limit and obtain Also, for every φ ∈ T (Ω), we have Since for every K ⋐ Ω, there exists C K > 0 such that w n (x) ≥ C K a.e. in K and also w n ≡ 0 in R N \ Ω and because of w n (x) → u(x) a.e. in Ω, thus u is a weak solution to problem (1). Finally note that if we take γ such that 4γ (γ+1) 2 > λ Λ N,s , then by testing (16) with w γ n , and using the inequality (22) together with Hardy inequality, it easily follows that u γ+1 2 ∈ X s 0 (Ω).

13
By now, in Theorem 2.8 we have shown that for λ > Λ N,s there is no positive solution to problem (1). Also, in Theorem 2.5 we have proved the existence of a positive solution for λ < Λ N,s . The following remark for λ = Λ N,s may be interesting.
Remark 3.6. In the borderline case λ = Λ N,s , by invoking the improved version of Hardy inequality, [45], one can define the space H(Ω) as the completion of C ∞ 0 (Ω) with respect to the norm: It can be proved that X s 0 (Ω) H(Ω) X s,q 0 (Ω), for all q < 2. By invoking the classical variational methods in the space H(Ω) and the same techniques used above, a similar existence result can be obtained in this new function space. See [45,Remark 1] and also [40] for the details. Proof of Theorem 2.6. Let consider u as a weak solution to problem (1) and φ n ∈ T (Ω) as the weak energy solutions to the following problems: where the iteration starts with The Comparison Principle for fractional Laplacian operator implies that φ 0 ≤ φ 1 ≤ · · · ≤ φ n−1 ≤ φ n ≤ φ, where φ := lim n→∞ φ n is the weak energy solution to Using φ n as a test function in (1) implies that On the other hand, by the definition φ n , we have Combining (32) and (33) and noticing that φ n−1 Therefore, the sequence { f φ n } is uniformly bounded in L 1 (Ω). Also, since { f φ n } is increasing, applying the Monotone Convergence Theorem and invoking Lemma 2.2 we obtain 14 Also, from Remark 2.1, it follows that Ω µ δ s(γ−1) dx < +∞. Now assume that for some r and B r (0) ⋐ Ω, and Ω µ δ s(γ−1) dx < +∞.
Let u n ∈ X s 0 (Ω) be the weak energy solutions to the problems where Here f n = T n ( f ). Again we have u 0 ≤ u 1 ≤ · · · ≤ u n−1 ≤ u n in R N . Using φ ∈ X s 0 (Ω), the solution to (31), as a test function in (36) we obtain On the other hand, using u n as a test function in the weak formulation of (31), we get From (37) and (38) and using Lemma 2.2 together with (34), and (35) we obtain Notice that in the last inequality, we have used u 0 ≥ cδ s , and φ ∼ c 1 δ s near the boundary, ∂Ω, for some c 1 > 0, since φ is the solution to (31). This follows by a result of [37] together with the Comparison Principle for the fractional Laplacian.
Since u n is increasing and also uniformly bounded in L 1 (Ω), by the Monotone Convergence Theorem we conclude that u := lim n→∞ u n is a function in L 1 (Ω). We want to show that u is a weak solution to problem (1). For this purpose let ψ ∈ X s 0 (Ω) ∩ L ∞ (Ω) be the unique positive weak energy solution to Using ψ as a test function in (36) and noting that ψ ∼ δ s , from (39) we get Thus by applying the Monotone Convergence Theorem we get by the Dominated Convergence Theorem we have Therefore, u satisfies the equation (1) in the following weak sense: Tesing T k (u n ) in (36), and using (35), we can show that T k (u n ) → T k (u) weakly in X s 0 (Ω) (similar to the arguments in the proof of Proposition 3.3). Moreover, since λ u n−1 |x| 2s + 1 Noting that since we have N > 2s, therefore N N−s < 2. Now, by invoking Theorem 5 and Proposition 10 in chapter 5 of the reference book [48], we get that u ∈ X s 1 ,p 0 (Ω), for all s 1 < s and for all p < N N−s . (In [48], X s,p 0 (Ω) reads as Λ

Some uniqueness results and the rate of the growth of solutions
In this section, we have some uniqueness results. Also, with some summability assumptions on the data of µ and f , we find the rate of the growth of solutions.
At first for the special case µ ≡ 1, by studying the behaviour of solutions near the boundary we discuss the uniqueness of solutions to problem (1).

Proposition 4.1. If µ ≡ 1 then the solution obtained to problem (1) in Theorem 2.5 behaves as:
for any x ∈ Ω, and some k 1 > 0, where r > diam(Ω). Here β is as defined in Lemma 2.2.
Proof. First of all notice that by Lemma 2.2, there exist a constant C 1 > 0 such that Now let w be the weak energy solution to the following problem.

Remark 4.2. Notice that by using the estimates in Proposition 4.1 and applying the Hölder inequality and the fractional Hardy-Sobolev inequality (and convexity of Ω only for
, where in the last inequality, we used the continuous embedding of X s 2 0 (Ω) into X s 1 0 (Ω), for any s 1 < s 2 .
• If γ = 1: • If γ > 1: If in addition we assume 2s(γ − 1) < γ + 1, then For general domains with some boundary regularity, the fractional Hardy-Sobolev inequality is proved for s ∈ [ 1 2 , 1). See [51,52,53]. But in [50], the authors proved the fractional Hardy-Sobolev inequality for any s ∈ (0, 1), by using the fact that the domain is a convex set and its distance from the boundary is a superharmonic function.
Let u 1 and u 2 be two solutions in X s loc (Ω) to problem (1) and define w = u 1 − u 2 . Then we have The fractional Hardy-Sobolev inequality and a density argument, shows that the equality (43) holds for all φ ∈ X s 0 (Ω), see remark 4.2. This means that w ∈ X s 0 (Ω). Now by using w − as a test function in (43) and applying Hardy inequality we deduce that w − ≡ 0. So we reach at the conclusion that u 1 ≥ u 2 . Similar argument shows that u 1 ≤ u 2 . Therefore u 1 = u 2 , and the uniqueness follows. The assumption µ ≡ 1 is taken for the purpose of simplification. However, we can assume any µ ≥ m, for some positive constant m, such that

and the above argument works. For a further discussion see [30, Theorem 5.2] which is about a Brezis-Oswald type result concerning uniqueness.
Once again, because of the interest in uniqueness, we have another definition to solutions of (1). In fact, we would like to consider the entropy solution. The motivation of the definition comes from the works [54,55]. In what follows, we would consider 0 < γ ≤ 1.
Definition 4.1. Assume 0 ≤ µ, f ∈ L 1 (Ω), and 0 < γ ≤ 1. We say that u is an entropy solution to (1) if • for every K ⋐ Ω, there exists C K > 0 such that u(x) ≥ C K in K and also u ≡ 0 in R N \ Ω ; • T k (u) ∈ X s 0 (Ω), for every k, and u satisfies the following family of inequalities: for any k and any φ ∈ X s 0 (Ω) ∩ L ∞ (Ω), and also together with this extra assumption that the second term on the right-hand side of the above inequality be finite for any φ ∈ X s 0 (Ω) ∩ L ∞ (Ω). The well-posedness of this term will be clear after the construction of entropy solution.
Let u and v be two entropy solution. Testing u with φ = T h (v) and v with φ = T h (u) in the weak formulation of entropy inequalities, we have and Adding up the left-hand sides of (44) and (45) and restricting them to we have the following estimate by using Hardy inequality Also, summing the right-hand sides of (44) and (45) when restricted to A h 0 gives Now, consider the set A h When restricted to A h 1 , we have the following for the left-hand side of (44): On the other hand, when restricted to A h 1 , the right-hand side of (44) is which goes to zero as h → ∞. Finally on the remaining set A h 2 = {x ∈ Ω : |u − T h (v)| < k, |v| < h, |u| ≥ h}, the left-hand side of (44) is as follows which goes to zero as h → ∞. The right-hand side of (44), when restricted to A h 2 , is as follows which also goes to zero as h → ∞. Similarly, we can estimate the left-hand side of (45) on the sets B h and On the other hand for the right-hand side of (45) on the sets B h 1 = {x ∈ Ω : |v − T h (u)| < k, |u| ≥ h} and B h 2 = {x ∈ Ω : |v − T h (u)| < k, |u| < h, |v| ≥ h}, we have: 19 Putting all the estimates (46), (47), (48), (49), (50), (51), (52), (53), (54), and (55) together we obtain Therefore u ≡ v, and the uniqueness is proved. Now, we construct an entropy solution for the case 0 < γ ≤ 1, µ ∈ L 2 * s 1−γ ′ (Ω) ∩ L 2 (Ω) and a datum of f ∈ L 1 (Ω) such that satisfies the integrability condition (10). Let consider the following approximating problems: Here µ n = T n (µ) and f n = T n ( f ). The increasing behaviour of µ n (u n + 1 n ) −γ + f n , and the monotonicity of the operator (−∆) s u − λ u |x| 2s will ensure the existence of an increasing sequence of solutions to problems (56). Testing n=1 is a bounded sequence in X s 0 (Ω) for each fixed k and each fixed φ ∈ X s 0 (Ω) ∩ L ∞ (Ω). Therefore, up to a subsequence is an increasing sequence of non-negative functions, once more the strict monotonicity of (−∆) s implies that T k (u n − φ) → T k (u − φ) strongly in X s 0 (Ω) (see for example [29,Lemma 2.18] for this compactness result). Now, using T k (u n − φ) as a test function in (56), and noting that (because of u 1 ∼ cδ s , near the boundary, and applying the Hölder and the fractional Hardy-Sobolev inequalities) we may pass to the limit and find an entropy solution even with the equalities instead of the inequalities in Definition 4.1.
Notice that from the above estimate and by Fatou's Lemma we deduce for any φ ∈ X s 0 (Ω) ∩ L ∞ (Ω), and any k > 0. We end this section by a Calderón-Zygmund type property to solutions of problem (1). See [41] for this property in the local case without the presence of singular nonlinearity and [15] for the case without the Hardy potential.
As mentioned before in Lemma 2.2, any supersolution to (1) is unbounded, i.e., u(x) |x| −β in a neighborhood of the origin. Now we have the following result, which says this rate is precisely the rate of the growth of u for the regular data of µ and f . Proof. We follow [29,Theorem 4.1]. Also see [15,Lemma 3.3]. Let k ≥ 1. By the change of variable v(x) := |x| β u(x), it can be checked that v solves: where the operator L β is as follows: See [29,Section 2] for the properties of this operator and the associated weighted fractional Sobolev space.
Using G k (v) as a test function in (57), and following the proof of [29, Theorem 4.1] we obtain where A k := {x ∈ Ω : v(x) ≥ k}. Applying the weighted Sobolev inequality [29,Proposition 2.11] in the left-hand side of (58), and noting that |x| βγ ≤ C 2 , in Ω, gives For the first term in the right-hand side of the above inequality, by using the Hölder inequality we get Similarly, for the second term Putting the results together, we obtain On the other hand, since Ω is bounded, there exists a constant C 4 > 0 such that Moreover, for any z > k, we have that A z ⊂ A k and G k (v)χ A z ≥ (z − k). Thus from (59) and (60) we have 2s , we obtain that there exists k 0 such that ψ(k) ≡ 0, for any k ≥ k 0 . Thus v(x) ≤ k 0 , a.e. in Ω. This means that u(x) ≤ k 0 |x| −β , a.e. in Ω.

The parabolic case and a stabilization result
In this section, we study on the following evolution problem in Ω × (0, T ), where u 0 ∈ X s 0 (Ω) satisfies an appropriate cone condition which will be precised later. In what follows, we will mention an existence and uniqueness and also a stabilization result to problem (61).
First of all, we define a notion of a weak solution. Before it, we need the following class of test functions.
Notice that Aubin-Lions-Simon Lemma, see [56], implies that the following embedding is compact.
Definition 5.1. Assume u 0 ∈ L 2 (Ω), and f ∈ L 2 (Ω × (0, T )). We say that u ∈ A(Ω T ) is a weak supersolution (subsolution) to problem (61) if and also together with this extra assumption that the second term on the right-hand side of the above inequality be finite for any φ ∈ A(Ω T ). The well-posedness of the second term on the right-hand side will be clear after the construction of solution.
If u is a weak supersolution and subsolution then we say that u is a weak solution. Notice that by the embedding (62), the initial condition u(x, 0) = u 0 make sense.
Before outlining our theorems, we need to define the following sets: • Let U Sing γ be the set of all functions in L 2 (Ω) such that there exists k 1 > 0 such that where r > diam(Ω). •

22
Also, we need to the following definition.
Proof. We follow the proof of [30,Theorem 2.4]. For any ǫ > 0, let consider the following approximating problem: The existence of a unique energy solution easily follows by the classical variational methods. Indeed let X s 0 (Ω) + := {u ∈ X s 0 (Ω) | u ≥ 0}, and consider the corresponding energy functional to problem (64) as follows: Notice that the last term is well-defined since g ∈ L ∞ (Ω, |x| β dx) ⊂ L 2 (Ω). Using Hardy inequality, one can show that this functional I ǫ,θ : X s 0 (Ω) + → R is weakly lower semi-continuous, coercive and strictly convex. Since X s 0 (Ω) + is a closed subspace of the reflexive space X s 0 (Ω) + , therefore the existence of a unique minimizer is obvious by the classical theory (for instance see [57,Chapter 1]). Therefore, as a consequence, we get the existence of a unique energy solution to problem (64).
Let 0 < ǫ 1 ≤ ǫ 2 . We want to show that u ǫ 2 ,θ ≤ u ǫ 1 ,θ a.e. in Ω. This easily follows by subtracting the weak formulations of u ǫ i ,θ , i = 1, 2, and using (u ǫ 2 ,θ − u ǫ 1 ,θ ) + as a test function which together with the Hardy inequality implies (u ǫ 2 ,θ − u ǫ 1 ,θ ) + ≡ 0, a.e. in Ω. Now let w ∈ X s 0 (Ω) ∩ U Sing γ be the unique energy solution to Notice that for the general γ > 1, we only know that w ∈ X s loc (Ω). But since 2s(γ − 1) < γ + 1, thanks to Remark 4.2, we get w ∈ X s 0 (Ω) too. Now define u := Mw, for some M > 1. Because of the same singular behaviour of w and g near the origin, and noting that g is bounded, near the boundary, ∂Ω, and w behaves as cδ s , near the boundary, we can choose M large enough (independent of ǫ) such that in Ω.
for any φ ∈ T (Ω). But in fact, we want to show that u θ is an energy solution. For this purpose let u := mw, for some m > 0. If we choose m small enough such that in Ω, (which is possible by taking into consideration the behavior of w and g near the origin and the boundary, ∂Ω) then u will be a subsolution to problem (63) and with the similar arguments as in above we obtain u ≤ u θ a.e. in Ω. Thus u ≤ u θ ≤ u, which implies that u θ ∈ U Sing γ . On the other hand, by invoking the Hardy inequality and also because of the restrictions 0 < γ ≤ 1, or γ > 1 with 2s(γ − 1) < γ + 1, a density argument shows that (65) holds for all φ ∈ X s 0 (Ω) (see Remark 4.2). This means that u θ ∈ X s 0 (Ω) is the unique energy solution to problem (63). Now, let g ∈ L m (Ω), m > N 2s , which is possible if mβ < N, or equivalently α > N−2s 2 − N m . Since λ = λ(α), given by (9), is a continuous decreasing function for α ∈ [0, N−2s 2 ), this recent condition is equivalent to 0 < λ < λ * , for some λ * < Λ N,s . Thus Comparison Principle for the fractional Laplacian operator together with Theorem 4.4 gives u(x) ≤ C|x| −β a.e. in R N . Now, the interior regularity theory for the fractional Laplacian, that follows from [37, Proposition 1.1], implies that u ∈ C(Ω \ B ǫ (0)), for anyΩ ⋐ Ω and any ǫ > 0 small enough. Moreover, by following the proof of [49,Theorem 1.4] we obtain the continuity of u up to the boundary of Ω. This completes the proof.
Thanks to Hardy inequality and following the idea of [30, Theorem 4.1], i.e. applying the semi-discretization in time with implicit Euler method, and also invoking the result of Theorem 5.1, we will obtain the following existence result to problem (61).

Remark 5.3. By invoking
Finally, the following theorem is about a stabilization result to problem (61). By stabilization, we mean that if u(x) is the unique solution to the stationary problem with the datum of f (x), then u(x, t), the solution to the parabolic problem, converges toû(x), as t → ∞.
Since proofs of the theorems in this section are essentially the same as proofs of the corresponding ones in [30], we will give them in the appendix.

Appendix
Here we give the proofs of Theorem 5. Let η t = T n and for 0 ≤ k ≤ n, define t k = kη t and Also, define Clearly we have f η t (·, t) ∈ L ∞ (Ω, |x| β dx) ⊂ L 2 (Ω), t ∈ [0, T ], and for 1 < p < +∞, Now, let θ = η t , and g = η t f k + u k−1 ∈ L ∞ (Ω, |x| β dx) in problem (63). Then, Theorem 5.1 implies the existence of u k ∈ X s 0 (Ω) ∩ U Sing γ as a solution to the following problem: where the above iteration starts from the initial condition of problem (61), i.e. u 0 (x). Now, for 1 ≤ k ≤ n, and t ∈ [t k−1 , t k ), inspired by the implicit Euler method, we define The funtions u η t andũ η t satisfies Now, in what follows, we establish some uniform estimates in η t for u η t andũ η t .
Multiplying (68) by η t u k , integrating over R N and summing from k = 1 to n ′ ≤ n, using Young's inequality, (67) and the embedding (6) we get for a constant C > 0 For the first term in the left-hand side of (70), similar to (2.7) in the proof of [31, Theorem 0.9], we have the following equality Now, let w ∈ X s 0 (Ω) ∩ U Since w behaves as c 1 |x| −β near the origin and behaves as c 2 δ s , near the boundary, ∂Ω, we can choose m > 0 small enough, and M > 0 large enough, such that in Ω, in Ω, Since u 0 ∈ U Sing γ , we can choose u and u such that it satisfies the above inequalities and u ≤ u 0 ≤ u. From the monotonicity of the operator (−∆) s u − λ u |x| 2s − u −γ , and applying it iteratively we get u ≤ u k ≤ u, for all k. This implies for a.e. (x, t) ∈ [0, T ] × Ω, Thus u η t ,ũ η t ∈ U Sing γ uniformly for each t ∈ [0, T ]. Now, for the singular term in (70), we can estimate as follows: T Ω u 1−γ dx < +∞, 0 < γ ≤ 1, T Ω u 1−γ dx < +∞, γ > 1, with 2s(γ − 1) < γ + 1.
Now we want to obtain another a priori estimate. Multiplying (68) by u k − u k−1 , integrating over R N and summing from k = 1 to n ′ ≤ n, using Young's inequality we get which implies η t 2 By using the convexity of the term − 1 1−γ Ω u 1−γ dx, we get Notice that the existence and uniqueness is guaranteed by Theorem 5.1. Subtracting the weak formulations of these two equations and using w := |x| β (u − v) − f 1 − f 2 L ∞ (Ω,|x| β dx) + as a test function, we obtain Ω w 2 |x| β dx + θ Ω L(u) − L(v) w dx ≤ 0.
Since we can easily check that Ω L(u) − L(v) w dx ≥ 0, thus w ≡ 0 a.e. in Ω, or equivalently |x| β (u − v) ≤ f 1 − f 2 L ∞ (Ω,|x| β dx) . Reversing the roles of u and v gives This proves that L is m-accretive in W(Ω Let u, u ∈ D(L) L ∞ (Ω,|x| β dx) be the sub and supersolution respectively to (1) with µ ≡ 1 such that u ≤ u 0 ≤ u, which is possible because of u 0 ∈ D(L) L ∞ (Ω,|x| β dx) . Let u denotes the weak solution of (61) and v 1 and v 2 be the unique solutions to (61) with the initial conditions u and u, respectively. Since λ ∈ (0, λ * ), and u, u ∈ D(L) L ∞ (Ω,|x| β dx) , thus Theorem 5.2 gives v 1 , v 2 ∈ C([0, T ]; W(Ω)). Taking u 0 = u (respectively u 0 = u), we consider the sequence {u k } (respectively {u k }) which is non-decreasing (respectively non-increasing) as solutions to the iteration given by (68). Moreover, we consider the sequence {u k } as the one that is obtained in the iteration (68), and starts with the initial condition u 0 . Then by the choice of η t we may have Now consider the maps t → v 1 (x, t) and t → v 2 (x, t), which are non-decreasing and non-increasing, respectively (by similar reasoning as the one in [ Similarly, we getṽ 2 = S (t)ṽ 2 . Thusṽ 1 andṽ 2 are the stationary solutions to (61) i.e. solves (1) with µ ≡ 1. On the other hand, by the uniqueness of solutions to the stationary problem,ṽ 1 =ṽ 2 =û. Now, applying the Dini's Theorem (see [60,Theorem 7.13]) gives in L ∞ (Ω, |x| β dx) as t → ∞.