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 nonexistence of solutions to the following fractional problem involving a singular nonlinearity: {(−Δ)su=λu|x|2s+μuγ+fin Ω,u>0in Ω,u=0in (RN∖Ω).\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\begin{aligned} \textstyle\begin{cases} (-\Delta )^{s} u = \lambda \frac{u}{ \vert x \vert ^{2s}} + \frac{\mu }{u^{\gamma }}+f & \text{in } \Omega, \\ u>0 & \text{in } \Omega, \\ u=0 & \text{in } (\mathbb{R}^{N} \setminus \Omega ). \end{cases}\displaystyle \end{aligned}$$ \end{document} Here 0<s<1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0 < s<1$\end{document}, λ>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\lambda >0$\end{document}, γ>0\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\gamma >0$\end{document}, and Ω⊂RN\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\Omega \subset \mathbb{R}^{N}$\end{document} (N>2s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$N > 2s$\end{document}) is a bounded smooth domain such that 0∈Ω\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0 \in \Omega $\end{document}. Moreover, 0≤μ,f∈L1(Ω)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0 \leq \mu,f \in L^{1}(\Omega )$\end{document}. For 0<λ≤ΛN,s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0< \lambda \leq \Lambda _{N,s}$\end{document}, ΛN,s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\Lambda _{N,s}$\end{document} being the best constant in the fractional Hardy inequality, we find a necessary and sufficient condition for the existence of a positive weak solution to the problem with respect to the data μ and f. Also, for a regular datum of f, under suitable assumptions, we obtain some existence and uniqueness results and calculate the rate of growth of solutions. Moreover, we mention a nonexistence and a complete blowup result for the case λ>ΛN,s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\lambda > \Lambda _{N,s}$\end{document}. Besides, we consider the parabolic equivalence of the above problem in the case μ≡1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mu \equiv 1$\end{document} and some suitable f(x,t)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$f(x,t)$\end{document}, that is, {ut+(−Δ)su=λu|x|2s+1uγ+f(x,t)in Ω×(0,T),u>0in Ω×(0,T),u=0in (RN∖Ω)×(0,T),u(x,0)=u0in RN,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document} $$\begin{aligned} \textstyle\begin{cases} u_{t}+(-\Delta )^{s} u = \lambda \frac{u}{ \vert x \vert ^{2s}} + \frac{1}{u^{\gamma }}+f(x,t) & \text{in } \Omega \times (0,T), \\ u>0 & \text{in } \Omega \times (0,T), \\ u =0 & \text{in } (\mathbb{R}^{N} \setminus \Omega ) \times (0,T), \\ u(x,0)=u_{0} & \text{in } \mathbb{R}^{N}, \end{cases}\displaystyle \end{aligned}$$ \end{document} where u0∈X0s(Ω)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$u_{0} \in X_{0}^{s}(\Omega )$\end{document} satisfies an appropriate cone condition. In the case 0<γ≤1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0<\gamma \leq 1$\end{document} or γ>1\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\gamma >1$\end{document} with 2s(γ−1)<(γ+1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$2s(\gamma -1)<(\gamma +1)$\end{document}, we show the existence of a unique solution for any 0<λ<ΛN,s\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$0< \lambda < \Lambda _{N,s}$\end{document} and prove a stabilization result for certain range of λ.

Here 0 < s < 1, λ > 0, γ > 0, and ⊂ R N (N > 2s) is a bounded smooth domain such that 0 ∈ . Moreover, 0 ≤ μ, f ∈ L 1 ( ). For 0 < λ ≤ N,s , N,s being the best constant in the fractional Hardy inequality, we find a necessary and sufficient condition for the existence of a positive weak solution to the problem with respect to the data μ and f . Also, for a regular datum of f , under suitable assumptions, we obtain some existence and uniqueness results and calculate the rate of growth of solutions. Moreover, we mention a nonexistence and a complete blowup result for the case λ > N,s . Besides, we consider the parabolic equivalence of the above problem in the case μ ≡ 1 and some suitable f (x, t), that is, where u 0 ∈ X s 0 ( ) satisfies an appropriate cone condition. In the case 0 < γ ≤ 1 or γ > 1 with 2s(γ -1) < (γ + 1), we show the existence of a unique solution for any 0 < λ < N,s and prove a stabilization result for certain range of λ.
We will prove that for 0 < λ ≤ N,s , N,s = 4 s 2 ( N+2s 4 ) 2 ( N-2s 4 ) being the best constant in the fractional Hardy inequality, the problem has a solution if and only if μ ∈ L 1 ( , δ s(1-γ ) dx), δ(x) = dist(x, ∂ ), and the datum of f satisfies the following integrability condition: where the constant β = β(N, s, λ) will be defined later in Lemma 2.4. 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 > N,s . This notion of the solution, which we consider for the nonexistence result, is local in nature, and we just ask the regularity needed to give distributional sense to the equation (similar to that in [1,2]). Moreover, this nonexistence result is strong in the sense that a complete blowup phenomenon occurs. By complete blowup phenomenon we mean that the solutions to the approximating problems (with the bounded weights (|x| 2s + ) -1 and (u + ) -γ instead of the terms |x| -2s and u -γ , respectively) tend to infinity for every x ∈ as 0 < ↓ 0.
In the above problem, (-) s stands for the fractional Laplacian operator: (-) s u(x) = C N,s P.V. Here denotes the gamma function, and Fu =û is the Fourier transform of u. By restricting the fractional Laplacian operator to act only on smooth functions that are zero outside , we have the restricted fractional Laplacian (-| ) s . For this operator, the best alternative to the Dirichlet boundary condition is u ≡ 0 in (R N \ ). For more details about fractional Laplacian, see [3][4][5].
Over the past decades, there has been much focus and also a vast literature on 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 is the singularity of Hardy type, which is related to the inequality of the same name, and there are various its generalizations. The well-known classical Hardy inequality is as follows: where ⊂ R N is a bounded domain containing the origin, and 1 ≤ p < N [8,9]. The constant ( N-p p ) p is optimal, and it is not attained in W 1,p 0 ( ), meaning that the continuous embedding W 1,p 0 ( ) → L p ( , |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 Chap. 1 of [11]. Due to these motivations, over the past few decades, general singularities are widely studied.
In the pioneering works, Baras and Goldstein [12,13] studied the following singular Cauchy-Dirichlet heat problem in = R N or in a bounded smooth domain containing B 1 (0) = {x ∈ R N : x < 1}: The authors assume that f and u 0 are nonnegative and 0 ≤ V ∈ L ∞ ( \ B (0)) for each > 0, but V may be singular at the origin. They say that V is too singular if V (x) > C * (N) |x| 2 near x = 0, whereas V is not too singular if V (x) ≤ C * (N) |x| 2 near x = 0. Here C * (N) = (N-2) 2 4 is the sharp constant in the following Hardy inequality: In the not too singular potential case, they found necessary and sufficient conditions 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 blowup phenomenon occurs. Namely, u n (x, t) → +∞ for all (x, t) ∈ × (0, T) as n → ∞.
Barrios, Bonis, Medina, and Peral [17] 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 all γ > 0 and λ > 0. Moreover, in the case M = 1 and f ≡ 1, they found a threshold such that there exists a solution for every 0 < λ < and there does not for λ > . Also, in [28] the 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 of the form u -q , q ∈ (0, 1). In the detailed paper [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 paper, that is, the weak Harnack inequality, which they proved by following the classical Moser and Krylov-Safonov idea. Also, we will take advantage of some of Calderón-Zygmund properties of solutions. See [29,Sect. 4] for the effect of the Hardy potential in some Calderón-Zygmund properties for the fractional Laplacian.
For the similar parabolic equivalence of (1), Giacomoni, Mukherjee, and Sreenadh [30] investigated the existence and stabilization 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 a semidiscretization 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] authors have shown the same results for the local version of this problem for the general p-Laplacian case. Also, for some recent papers on the optimal regularity, see [33,34]. The rest of the paper is as follows. In Sect. 2, after introducing the functional setting, we outline our existence and nonexistence theorems. In particular, we provide a theorem about necessary and sufficient conditions for the existence of a solution to problem (1) in the case λ ≤ N,s and a nonexistence theorem in the case λ > N,s . In Sect. 3, we provide proofs of our existence theorems. In Sect. 4, we will present some uniqueness results. Also, concerning uniqueness, with some regularity assumptions on μ and f , we show the existence and uniqueness of the so-called entropy solution for the case 0 < γ ≤ 1. Besides, we mention a theorem about the rate of growth of solutions to problem (1). Finally, in Sect. 5, we consider the parabolic version of problem (1) in the particular case μ ≡ 1. Firstly, under the assumptions 0 < γ ≤ 1 or γ > 1 with 2s(γ -1) < (γ + 1), we show the existence of a unique solution for 0 < λ < N,s and secondly, we prove the stability for some range of λ; that is, we find a positive constant λ * = λ * (N, s) < 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, nonexistence, and blowup results
Let 0 < s < 1, 1 ≤ p < ∞, and let be a bounded domain in R N . Also, let D = R N × R N \ c × c with c = R N \ . We define the Banach space In the case p = 2, we denote by X s ( ) the space X s,2 ( ), which is a Hilbert space with the inner product Moreover, we define X s,p 0 ( ) as the closure of C ∞ 0 ( ) in X s,p (R N ). Equivalently, it can be shown that X s,p It is easy to see that This equality defines an equivalent norm for X s,p 0 ( ) with (4). We denote it by It is worth noting that the continuous embedding of X s 2 0 ( ) into X s 1 0 ( ) 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,Sect. 2.2] and [4]. Also, see [36,Sect. 2].
In this paper, we will use the following continuous embedding: where p * s = pN N-ps is the Sobolev critical exponent. Moreover, this embedding is compact for 1 ≤ q < p * s . See [4, Theorems 6.5 and 7.1]. Also, we denote by X s,p loc ( ) the set of all functions u such that uφ ∈ X s,p 0 ( ) for any φ ∈ C ∞ c ( ). When we say that {u n } ⊂ X s,p loc ( ) is bounded, we mean that {φu n } ⊂ X s,p 0 ( ) is bounded for any fixed φ ∈ C ∞ c ( ). Since we are dealing with the nonlocal operator (-) s , we will use the following class of test functions for defining the weak solution to problem (1): It can be shown that T ( ) ⊂ X s 0 ( ) ∩ L ∞ ( ) ∩ C 0,s ( ). See [29], where this class of test functions is used for dealing with problem (1). Moreover, every φ ∈ T ( ) is a strong solution to the equation (-) s φ = ϕ, and for every φ ∈ T ( ), there exists a constant β ∈ (0, 1) such that φ δ s ∈ C 0,β ( ); see [37]. It is easy to check that for u ∈ X s 0 ( ) and φ ∈ T ( ), One can show that (-) s : X s 0 ( ) → X -s ( ) is a continuous strictly monotone operator, where X -s ( ) is the dual space of X s 0 ( ).

Definition 2.1
We say that u ∈ L 1 ( ) is a very weak (distributional) supersolution (subsolution) to for all nonnegative φ ∈ T ( ). If u is a very weak (distributional) supersolution and subsolution, then we say that u is a very weak (distributional) solution.

Definition 2.2
We say that u ∈ X s 0 ( ) is a weak energy supersolution (subsolution) to for all nonnegative φ ∈ X s 0 ( ). If u is a weak energy supersolution and subsolution, then we say that u is a weak energy solution.

Definition 2.3
Let 0 ≤ μ, f ∈ L 1 ( ). We say that u is a weak solution to problem (1) if • u ∈ L 1 ( ), and for every K , there exists C K > 0 such that u(x) ≥ C K a.e. in K and also u ≡ 0 in (R N \ ); • Equation (1) is satisfied in the weak sense, that is, together with extra assumption that the first and second terms on the right-hand side of this equality are finite for any φ ∈ T ( ). 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, we first gather the following lemmas.
In the next two theorems, we present our existence results to problem (1). First, we will prove that for 0 < λ < N,s and γ ≥ 1, problem (1) admits a solution for the case μ ∈ L 1 ( ) and f ∈ L 1 ( ) ∩ X -s ( ). It is crucial to indicate that our approach in the proof of Theorem 2.7 only works in the case γ ≥ 1. However, if we further assume that μ ∈ L m ( ), m = ( 2 * s 1-γ ) (p denotes the conjugate exponent of p), then the same approach works for γ < 1. For a result on 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 ( , |x| -(1-γ )β dx).
In the following, we denote the usual truncation operator, and G n (σ ) := σ -T n (σ ).
The next theorem gives a necessary and sufficient condition for the existence of a solution to problem (1).

Theorem 2.8 (A necessary and sufficient condition for the existence of a solution)
Let s ∈ (0, 1), 0 < λ ≤ N,s , and γ > 0. Also, let 0 ≤ f , μ ∈ L 1 ( ). Then problem (1) has a positive weak solution if and only if Moreover, the solution u has the following regularity: Remark 2 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 ( ).
The proof of these theorems will appear in the next section. In the following, we will have a nonexistence and also a blowup result for the case λ > N,s .
The following nonexistence result is an immediate consequence of Lemmas 2.4 and 2.5. More precisely, it is well known that the linear problem with Hardy potential has a nonpositive supersolution if λ > N,s . We only bring it here for completeness. Page 10 of 42 Theorem 2.9 Let s ∈ (0, 1), λ > N,s , and γ > 0. Then there is no positive very weak solution to problem (1).
Proof We argue by contradiction. Let u be a positive very weak solution to problem (1).
. Then by using Lemma 2.5 and the positivity of g we have for some B r (0) . On the other hand, by Lemma 2.4 we have for sufficiently small r, where β = N-2s 2α, and α ∈ [0, N-2s 2 ) is given by the identity ) .
This nonexistence result is strong in the sense that a complete blowup phenomenon occurs. Namely, let u n be the solution to the following approximated problem with 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} Then u n (x 0 ) → ∞ for any x 0 ∈ as n → ∞.
In the same spirit of Theorem 2.9, we can prove this blowup phenomenon by taking into consideration that any approximating sequence of nonnegative supersolution to the linear problem blows up at any point of if λ > N,s , as it is proved in [29].

Proof of Theorem 2.7 and Theorem 2.8
We first prove Theorem 2.7. For this purpose, let us consider the following auxiliary problem: where g ∈ X -s ( ). The function u ∈ X s 0 ( ) is a weak energy solution to this problem if Here ·, · X -s ( ),X s 0 ( ) denotes the duality pairing between X -s ( ) and X s 0 ( ). The proof of the following proposition on the existence for (14) can be obtained by using the Hardy inequality and classical variational methods. See, for instance, [43,Sect. 4.6]. Also, the uniqueness of the weak energy solution to (14) follows from the strict monotonicity of the operator (-) s uλ u |x| 2s for 0 ≤ λ < N,s . The strict monotonicity of this operator is a direct consequence of the Hardy inequality.
Before we continue, we need to define the set C of functions v ∈ L 2 ( ) such that there exist positive constants k 1 and k 2 such that where the constant β is given in Lemma 2.4, and δ(x) = dist(x, ∂ ), x ∈ , is the distance function from the boundary ∂ . Now for v ∈ C, define (v) = w, where w ∈ X s 0 ( ) 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 2.4, [29,Theorem 4.1], and a result of [37], it easily follows that w ∈ C. If we show that : 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 ( ): We apply the Schauder fixed-point theorem (see, e.g., [43,Theorem 3.2.20]). We need to prove that is continuous and compact and that there exists a bounded convex subset of It is obvious that for each n, Now from the uniqueness of the weak solution to (14) we For compactness, we argue as follows.
For v ∈ C, let w be a solution to (15). If λ s 1 ( ) is the first eigenvalue of (-) s in X s 0 ( ) [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 this equality, we have the estimate where in the last inequality, we have used the Hölder inequality. Once more using the Hölder inequality gives f n w dx ≤ C 2 ( |w| 2 dx) 1 2 for some C 2 > 0. Thus combining this inequality with (17), (18), and (19), we obtain which implies that (L 2 ( )) is contained in a ball of finite radius in L 2 ( ). Therefore the intersection of this ball with C is invariant under . Moreover, we have 4 , which means that (L 2 ( )) is relatively compact in L 2 ( ) by the compactness of embedding (6).

Proposition 3.2 For every K
, there exists C K > 0 such that {w n }, the solutions to (16), satisfy w n (x) ≥ C K a.e. in K for each n.
Proof Let us consider the following problem: The existence of a 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 . Also, for each K , 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 nv n )as a test function (see [44,Theorem 20]), we conclude that w n ≥ v n a.e. in . Therefore, for every K , there exists n=1 be the sequence of solutions to (16).
Proof We will follow the proof of [29, For the left-hand side, by using (7) and the 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 For proving the boundedness of (28), it suffices to consider φ = G k (w n ) as a test function in (16) Note that for the left-hand side, we have used [44,Proposition 3]. To estimate the terms on the right-hand side of this equality 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 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), using the Hölder inequality with exponents a = 2 * s and b = 2N N+2s < N 2s , noting that the integration can be over because of w n ≡ 0 in R N \ , and embedding (6), we obtain Combining the above estimates, from (29) we get Now the 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 k (w n ) + G 2 k (w n )), that is, Moreover, we get the boundedness of G k (w n ) X s 0 ( ) uniformly in n. Now we show that {T k (w n )} is bounded in X s loc ( ). For this purpose, first note that by Proposition 3.2, for any compact set K , there exists C(K) > 0 such that Therefore T k (w n ) ≥ T k (w 1 ) ≥C := min k, C(K) .
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 ( ) by using (5)  Remark 4 For the case 0 < γ < 1, if furthermore we assume μ ∈ L ( 2 * s 1-γ ) ( ), then the sequence {w n } ∞ n=1 is bounded in X s 0 ( ). Indeed, you just have to keep in mind that Since the rest of the proof can be obtained proceeding as in the case γ = 1, for brevity, we leave it left to the reader. Now we are ready to prove Theorem 2.7.
Proof of Theorem 2.7 There exists u ∈ X s loc ( ) (u ∈ X s 0 ( ) in the case γ ≤ 1) such that, up to a subsequence, • w n → u weakly in X s loc ( ) (weakly in X s 0 ( ) in the case γ ≤ 1); weakly in X s 0 ( ). Also, using embedding (6), up to a subsequence, we have • w n → u in L r ( ) for any r ∈ [1, 2 * s ); • w n (x) → u(x) pointwise a.e. in . Now for every fixed φ ∈ T ( ), by the estimates above, we can pass to the limit and obtain Also, for every φ ∈ T ( ), we have Since for every K , there exists C K > 0 such that w n (x) ≥ C K a.e. in K and also w n ≡ 0 in (R N \ ), and because of w n (x) → u(x) a.e. in , u is a weak solution to problem (1).
By now, in Theorem 2.9, we have shown that for λ > N,s , there is no positive solution to problem (1). Also, in Theorem 2.7, we have proved the existence of a positive solution for λ < N,s . The following remark for λ = N,s may be interesting.
Remark 5 In the borderline case λ = N,s , by invoking the improved version of Hardy inequality [45] we can define the space H( ) as the completion of C ∞ 0 ( ) with respect to the norm It is known that X s 0 ( ) H( ) X s,q 0 ( ) for all q < 2. By invoking the classical variational methods in the space H( ) and the same techniques used above we can obtain a similar existence result in this new function space. See [45,Remark 1] and also [40] for the details. Now, in the spirit of [29,Theorem 4.10], we prove Theorem 2.8, which gives a necessary and sufficient condition for the existence of a solution to (1).
Proof of Theorem 2.8 Let u be a weak solution to problem (1), and let φ n ∈ T ( ) be weak energy solutions to the following problems: where the iteration starts with The comparison principle for fractional Laplacian operator implies that Using φ n as a test function in (1) yields 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 ( ). Also, since {f φ n } is increasing, applying the monotone convergence theorem and invoking Lemma 2.4, we obtain Also, from Remark 1 it follows that μ δ s(γ -1) dx < +∞.
Let u n ∈ X s 0 ( ) be 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 ( ), 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), using Lemma 2.4 together with (34) and (35), we obtain Note that in the last inequality, we have used u 0 ≥ cδ s and φ ∼ c 1 δ s near the boundary ∂ for some c 1 > 0, since φ is a 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 ( ), by the monotone convergence theorem we conclude that u := lim n→∞ u n is a function in L 1 ( ). We want to show that u is a weak solution to problem (1). For this purpose, let ψ ∈ X s 0 ( ) ∩ L ∞ ( ) 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 equation (1) in the following weak sense: Testing 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 ( ) (similarly to the arguments in the proof of Proposition 3.3). Moreover, since λ u n-1 |x| 2s + 1 n + μ (u n-1 + 1 n ) γ + f n converges strongly in L 1 ( , δ s dx), by mimicking the proofs of [

Some uniqueness results and the rate of the growth of solutions
In this section, we give some uniqueness results. Also, under some summability assumptions on the data of μ and f , we find the growth rate of solutions.
First, for the particular case μ ≡ 1, by studying the behavior 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 2.7 behaves as follows:
for any x ∈ and some k 1 > 0, where r > diam( ). Here β is as defined in Lemma 2.4.
Proof First, notice that by Lemma 2.4 there exists a constant C 1 > 0 such that Now let w be a weak energy solution to the problem for some k 1 , k 2 > 0, r > diam( ), and any x ∈ . By the comparison principle for the fractional Laplacian operator (see, e.g., [5, Proposition 2.17]) we obtain u(x) ≥ w(x), which, together with (41) and (42), gives (40).
Remark 6 Note that by using the estimates in Proposition 4.1 and applying the Hölder and fractional Hardy-Sobolev inequalities (and convexity of only for 0 < s < 1 2 ) [50, Theorem 1.1] we get where in the last inequality, we used the continuous embedding of X s 2 0 ( ) into X s 1 0 ( ) for s 1 < s 2 , • For γ = 1, If, in addition, 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]. However, in [50] the authors proved the fractional Hardy-Sobolev inequality for any s ∈ (0, 1) by using the facts that the domain is a convex set and its distance from the boundary is a superharmonic function.
Let u 1 and u 2 be solutions in X s loc ( ) to problem (1) in the particular case where μ ≡ 1 and either 0 < γ ≤ 1 or γ > 1 with 2s(γ -1) < γ + 1 and define w = u 1u 2 . Then we have The fractional Hardy-Sobolev inequality and a density argument show that equality (43) holds for all φ ∈ X s 0 ( ); see Remark 6. This means that w ∈ X s 0 ( ). Now by using was a test function in (43) and applying the ardy inequality we deduce that w -≡ 0. So we arrive at the conclusion that u 1 ≥ u 2 . A similar argument shows that u 1 ≤ u 2 . Therefore u 1 = u 2 , and the uniqueness follows.

Remark 7
The assumption μ ≡ 1 is taken for simplification. However, the above argument works for any μ ≥ m for some positive constant m such that Once again, because of the interest in uniqueness, we have another definition of solutions to (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 consider 0 < γ ≤ 1.

Definition 4.2
Let 0 ≤ μ, f ∈ L 1 ( ) and 0 < γ ≤ 1. We say that u is an entropy solution to (1) if • for every K , there exists C K > 0 such that u(x) ≥ C K in K and also u ≡ 0 in R N \ ; • T k (u) ∈ X s 0 ( ) for every k, and u satisfies the following family of inequalities: for all k and all φ ∈ X s 0 ( ) ∩ L ∞ ( ), together with the assumption that the second term on the right-hand side of the inequality is finite for all φ ∈ X s 0 ( ) ∩ L ∞ ( ). The well-posedness of this term will be clear after the construction of entropy solution.
Let u and v be entropy solutions. Testing u with φ = T h (v) and v with φ = T h (u) in the weak formulation of entropy inequalities, we have and Summing the left-hand sides of (44) and (45) and restricting them to we have the following estimate by using the 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 1 = {x ∈ : |u -T h (v)| < k, |v| ≥ h}. When restricted to A h 1 , for the left-hand side of (44), we have: 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 which goes to zero as h → ∞.
The right-hand side of (44), when restricted to A h 2 , is which also goes to zero as h → ∞. Similarly, we can estimate the left-hand side of (45) on the sets B h On the other hand, for the right-hand side of (45) on the sets B h and Putting all 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 where 0 < γ ≤ 1, μ ∈ L ( 2 * s 1-γ ) ( ) ∩ L 2 ( ), and a datum of f ∈ L 1 ( ) satisfying the integrability condition (10). Let consider the following approximating problems: Here μ n = T n (μ) and f n = T n (f ). The increasing behavior of μ n (u n + 1 n ) -γ +f n and the monotonicity of the operator (-) s uλ u |x| 2s ensures the existence of an increasing sequence of solutions to problems (56). Testing (56) n=1 is a bounded sequence in X s 0 ( ) for each fixed k and each fixed φ ∈ X s 0 ( ) ∩ L ∞ ( ). Therefore, up to a subsequence, is an increasing sequence of nonnegative functions, the strict monotonicity of (-) s again implies that T k (u nφ) → T k (uφ) strongly in X s 0 ( ) (see, e.g., [29,Lemma 2.18] for this compactness result). Now, using T k (u nφ) as a test function in (56) and noting that (because u 1 ∼ cδ s near the boundary and applying the Hölder and 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 4.2. Note that from the above estimate by Fatou's lemma we deduce for all φ ∈ X s 0 ( ) ∩ L ∞ ( ) and k > 0. We end this section by a Calderón-Zygmund-type property of solutions to 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 2.4, any supersolution to (1) is unbounded, that is, u(x) |x| -β in a neighborhood of the origin. Now we have the following result, which says that this rate is precisely the rate of 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) we can check that v solves where the operator L β is as follows: See [29,Sect. 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 ∈ : v(x) ≥ k}. Applying the weighted Sobolev inequality [29, Proposition 2.11] to the left-hand side of (58) and noting that |x| βγ ≤ C 2 in , we have For the first term on the right-hand side of this inequality, by using the Hölder inequality we get Similarly, for the second term, Putting these results together, we obtain On the other hand, since 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 ≥ (zk). 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 . This means that u(x) ≤ k 0 |x| -β a.e. in .

The parabolic case and a stabilization result
In this section, we study on the following evolution problem: where u 0 ∈ X s 0 ( ) satisfies an appropriate cone condition, which will be specified later. In what follows, we consider the existence and uniqueness and also a stabilization for problem (61).
First of all, we define the notion of a weak solution. We need the following class of test functions: Note that the Aubin-Lions-Simon lemma (see [56]) implies that the following embedding is compact: Definition 5.1 Let u 0 ∈ L 2 ( ) and f ∈ L 2 ( × (0, T)). We say that u ∈ A( T ) is a weak supersolution (subsolution) to problem (61) if • for every K × (0, T), there exists C K > 0 such that u(x, t) ≥ C K a.e. in K and also and also that the second term on the right-hand side of the inequality is finite for any φ ∈ A( T ) (the well-posedness of the second term on the right-hand side will be clear after the construction of a solution); • u(x, 0) ≥ (≤)u 0 (x) a.e. in . If u is a weak supersolution and subsolution, then we say that u is a weak solution. Note that by embedding (62) the initial condition u(x, 0) = u 0 makes sense.
Before outlining our theorems, we need to define the following sets: • U Sing γ , the set of all functions in L 2 ( ) such that there exists k 1 > 0 such that Also, we need the following definition. for any 0 < γ ≤ 1 or γ > 1 with 2s(γ -1) < γ + 1: Moreover, there exists a positive constant λ * < N,s such that for any λ ∈ (0, λ * ), this unique solution also belongs to W ( ).
Proof We follow the proof of [30,Theorem 2.4]. For any > 0, consider the following approximating problem: The existence of a unique energy solution easily follows by the classical variational methods. Indeed, let X s 0 ( ) + := {u ∈ X s 0 ( )|u ≥ 0}, and consider the corresponding energy func-tional to problem (64): Note that the last term is well-defined since g ∈ L ∞ ( , |x| β dx) ⊂ L 2 ( ). Using the Hardy inequality, we can show that this functional I ,θ : X s 0 ( ) + → R is weakly lower semicontinuous, coercive, and strictly convex. Since X s 0 ( ) + is a closed subspace of the reflexive space X s 0 ( ) + , the existence of a unique minimizer is obvious by the classical theory (e.g., see [57,Chap. 1]). Therefore, as a consequence, we get the existence of a unique energy solution to problem (64).
Let 0 < 1 ≤ 2 . We want to show that u 2 ,θ ≤ u 1 ,θ a.e. in . This easily follows by subtracting the weak formulations of u i ,θ , i = 1, 2, and using (u 2 ,θu 1 ,θ ) + as a test function, which, together with the Hardy inequality, implies (u 2 ,θu 1 ,θ ) + ≡ 0 a.e. in . Now let w ∈ X s 0 ( ) ∩ U Sing γ be the unique energy solution to Note that for general γ > 1, we only know that w ∈ X s loc ( ). However, since 2s(γ -1) < γ +1, thanks to Remark 6, we also get w ∈ X s 0 ( ). Now define u := Mw for some M > 1. Because of the same singular behavior of w and g near the origin, noting that g is bounded near the boundary ∂ and w behaves as cδ s near the boundary, we can choose M large enough (independent of ) such that is a strictly monotone operator for 0 < λ < N,s (this strict monotonicity is an easy consequence of [30, Lemma 3.1] and the Hardy inequality), and thus u ,θ ≤ u. Therefore u θ ≤ u, where u θ := lim →0 + u ,θ . This implies that u θ is a very weak (distributional) solution to problem (63), that is, for any φ ∈ T ( ). 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 (which is possible by taking into consideration the behavior of w and g near the origin and the boundary ∂ ), then u will be a subsolution to problem (63), and by similar arguments as before we obtain u ≤ u θ a.e. in . Thus u ≤ u θ ≤ u, which implies that u θ ∈ U Sing γ . On the other hand, by invoking the Hardy inequality, 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 ( ) (see Remark 6). This means that u θ ∈ X s 0 ( ) is the unique energy solution to problem (63). Now let g ∈ L m ( ), 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 * < N,s . Thus the comparison principle for the fractional Laplacian operator, together with Theorem 4.3, gives u(x) ≤ C|x| -β a.e. in R N . Now the interior regularity theory for the fractional Laplacian, which follows from [37, Proposition 1.1], implies that u ∈ C(˜ \ B (0)) for any˜ and any > 0 small enough. Moreover, by following the proof of [49,Theorem 1.4] we obtain the continuity of u up to the boundary of . This completes the proof.
Thanks to the Hardy inequality, following the idea of [30,Theorem 4.1], that is. applying the semidiscretization in time with implicit Euler method and also invoking the result of Theorem 5.3, we obtain the following existence result to problem (61).
Remark 8 By invoking [42,Proposition 5.3] it is straightforward to obtain that if λ > N,s , then problem (61) has no solution. Moreover, a similar complete blowup phenomenon occurs as in the stationary case.
Finally, the following theorem is on stabilization to problem (61). By stabilization we mean that ifû(x) is the unique solution to a stationary problem with the datum of f (x), then u(x, t), the solution to the parabolic problem, converges toû(x) as t → ∞. Theorem 5.5 Let s ∈ (0, 1), let either 0 < γ ≤ 1 or γ > 1 with 2s(γ -1) < γ + 1, and let 0 < λ < λ * . Also assume that u 0 ∈ D(L) Then if u(x, t) is the unique positive weak solution to problem (61), then whereû is the unique weak solution to (1) with μ ≡ 1.
Since proofs of the theorems in this section are essentially the same as those of the corresponding ones in [30], we give them in the Appendix. The functions u η t andũ η t satisfy 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 , summing from k = 1 to n ≤ n, and using Young's inequality, (67), and embedding (6), we get, for a constant C > 0, For the first term in the left-hand side of (70), similarly to (2.7) in the proof of [31, Theorem 0.9], we have the equality n k=1 (u ku k-1 )u k dx = 1 2 n k=1 |u ku k-1 | 2 dx and define u = mw, m > 0, and u = Mw, M > 0. By a direct computation we have Since w behaves as c 1 |x| -β near the origin and behaves as c 2 δ s near the boundary ∂ , we can choose m > 0 small enough and M > 0 large enough such that Since u 0 ∈ U Sing γ , we can choose u and u such that it satisfies the above inequalities and u ≤ u 0 ≤ u. By the monotonicity of the operator (-) s u-λ u |x| 2s -u -γ , applying it iteratively, we get u ≤ u k ≤ u for all k. This implies for a.e. (x, t) ∈ [0, T] × , Thus Sing γ uniformly for t ∈ [0, T]. Now we can estimate the singular term in (70) as follows: T u 1-γ dx < +∞, γ > 1 with 2s(γ -1) < γ + 1.
By the definition of u η t andũ η t , noting that u k ∈ L ∞ ( , |x| β dx) for all k, we obtain that On the other hand, for t ∈ [t k-1 , t k ), we have Integrating both sides of (70) over (t k-1 , t k ) and using the above estimates, the Hardy Inequality, and (71), we get that u η t ,ũ η t are bounded in L 2 [0, T]; X s 0 ( ) .
Now we want to obtain another a priori estimate.
Putting the results together in (94), as r → 0 + , we obtain Therefore (95) and (92) give the equality. Since the maps t → u 1-γ (x, t) dt and t → . We will use the m-accretive operator theory. Let u 0 ∈
Since we can easily check that (L(u) -L(v))w dx ≥ 0, we have w ≡ 0 a.e. in or, equivalently, |x| β (uv) ≤ f 1f 2 L ∞ ( ,|x| β dx) . Reversing the roles of u and v gives This proves that L is m-accretive in W ( ). Now the rest of the proof van be obtained by invoking [58,Theorem 4.2], as explained in [31, Proposition 0.1].
Let u, u ∈ D(L) be the sub-and supersolutions, respectively, to (1) with μ ≡ 1 such that u ≤ u 0 ≤ u, which is possible because of u 0 ∈ D(L) . Let u be a weak solution of (61), and let v 1 and v 2 be the unique solutions to (61) with initial conditions u and u, respectively. Since λ ∈ (0, λ * ) and u, u ∈ D(L) L ∞ ( ,|x| β dx) , Theorem 5.4 gives v 1 , v 2 ∈ C([0, T]; W ( )). Taking u 0 = u (respectively, u 0 = u), we consider the sequence {u k } (respectively, {u k }) which is nondecreasing (respectively, nonincreasing) as solutions to the iteration given by (68). Moreover, we consider the sequence {u k } obtained in iteration (68) and starting with the initial condition u 0 . Then by the choice of η t we have Now consider the maps t → v 1 (x, t) and t → v 2 (x, t), which are nondecreasing and nonincreasing, respectively (by similar reasoning as that in [59,Lemma 10.6]  Similarly, we get Finally, using (96), we conclude that u(t) →û in L ∞ ( , |x| β dx) as t → ∞.