# The Solvability of Fractional Elliptic Equation with the Hardy Potential

Authors: Siyu Gao; Shuibo Huang; Qiaoyu Tian; Zhan-Ping Ma
Journal: Complexity
 (2020)

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## Abstract

In this paper, we study the existence and nonexistence of solutions to fractional elliptic equations with the Hardy potential−Δsu−λu/x2s=ur−1+δgu,inΩ,ux>0,inΩ,ux=0,inℝN∖Ω, where Ω⊂ℝN is a bounded Lipschitz domain with 0∈Ω, −Δs is a fractional Laplace operator, s∈0,1, N>2s, δ is a positive number, 2<r<rλ,s≡N+2s−2αλ/N−2s−2αλ+1, αλ∈0,N−2s/2 is a parameter depending on λ, 0<λ<ΛN,s, and ΛN,s=22sΓ2N+2s/4/Γ2N−2s/4 is the sharp constant of the Hardy–Sobolev inequality.


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## Body

## 1. Introduction

In this paper, we consider the solvability of the following fractional elliptic problem:(1)−Δsu−λux2s=ur−1+δgu,inΩ,u>0,inΩ,u=0,inℝN∖Ω,where Ω⊂ℝN is a bounded Lipschitz domain with 0∈Ω, s∈0,1, N>2s, 2<r<rλ,s≡N+2s−2αλ/N−2s−2αλ+1, 0<λ<ΛN,s, and ΛN,s=22sΓ2N+2s/4/Γ2N−2s/4 is the sharp constant of the Hardy–Sobolev inequality; the fractional Laplace operator −Δs is defined by(2)−Δsu=CN,sP.V.∫ℝNux−uyx−yN+2sdy,where P.V. stands for the Cauchy principal value and constant CN,s is a constant.Recently, much attention has been devoted to the study of fractional Laplacian equations. One of the reasons comes from the fact that the fractional Laplacian arises in various areas and different applications, such as phase transitions, finance, stratified materials, flame propagation, ultrarelativistic limits of quantum mechanics, and water waves. For more details, see [1–6] and references therein.For fractional elliptic problems with the Hardy potential, Abdellaoui et al. [7] obtained the existence and summability of solutions to a class of nonlocal elliptic problem:(3)−Δsu−λux2s=fx,u,inΩ,u>0,inΩ,u=0,inℝN∖Ω,with f∈LmΩ and 0<λ<ΛN,s. They mainly considered the summability of solutions to (3) with fx,u=fx and the existence and regularity of solutions to (3) with fx,u=hx/uσ. Mi et al. [8] obtained the combined influence of the Hardy potential and lower order terms on the existence and regularity of solutions to the problem:(4)−Δsu−λux2s+up=fx,inΩ,u>0,inΩ,u=0,inℝN∖Ω.Barrios et al. [9] discussed the existence and multiplicity of solutions to the following fractional elliptic equation:(5)−Δsu−λux2s=up+μuq,inΩ,u>0,inΩ,u=0,inℝN∖Ω,where 0<λ<ΛN,s, 0<q<1,(6)1<p<pλ,s=N+2s−2αλN−2s−2αλ,and αλ∈0,N−2s/2 is a parameter depending on λ. They shown that problem (5) has at least one solution if 1<p<pλ,s and problem (5) has no solution if p>pλ,s.Recently, Shang et al. [10] studied the existence and multiplicity of positive solutions to the following problem:(7)−Δsu−μux2s=λgxup+Kxu2s∗−1,inℝN,where s∈0,1, N>2s, 0<p<2s∗−1, and 0<μ<ΛN,s. Some other results of fractional elliptic equations with the Hardy potential, see [7, 9, 11–14] and references therein.The local version of quasilinear problem related to problem (8) has been considered by Boccardo et al. [15]. They analyzed the existence of nontrivial solutions to the following problem:(8)−Δpu=ur−2u+λgu,inΩ,u>0,inΩ,u=0,on∂Ω,where Ω⊂ℝN is a smooth bounded domain, 1<p<N, r>p, g:Ω×ℝ⟶ℝ is a Carathéodory function, and there exist constants c1>0 and q∈1,p such that gs≤c1sq−1 for any s>0.Motivated by the above works, the aim of this paper is to study the existence of solutions to problem (1) by the method of subsuper solutions and taking into advantage the combined effect of concave and convex nonlinearity.We make the following assumptions:(F1)(9)2<r<rλ,s≡N+2s−2αλN−2s−2αλ+1,whereαλ∈0,N−2s/2 is a parameter depending on λ.(F2) g:Ω×ℝ⟶ℝ is Carathéodory function, and there exist constants c1>0 and q∈1,2, such that, for any σ>0,(10)gσ≤c1σq−1.(F3) The function(11)u⟶ur−1+δguis nondecreasing.(F4) Define(12)M0=lim¯z⟶0+gzz,for allτ>0,(13)M1τ=inf0≤z≤τλ11/r−2gzz,whereλ1 is the first eigenvalue of −Δs in Ω.Now, we state our main result.Theorem 1.
SupposeF1−F4 hold. Then, there exists a positive constant δ0, such that, for all δ∈λ1/M0,δ0, problem (1) has at least a nonnegative solution if M0>λ1/δ0, where M0 is defined by (12).Remark 1.
In order to prove the above theorem, we study directly to the pseudodifferential operator, without the harmonic extension to an extra dimension by transforming the nonlocal problem into a local problem due to Caffarelli and Silvestre [16].Remark 2.
To establish the upper bound forr (see (9)), we consider a radial solution w=Ax2s−N/2+β with constant A>0 to the problem:(14)−Δsw−λwx2s=wr−1.
We obtain(15)Aγβx−2s+2s−N/2+β−λAx−2s+2s−N/2+β=Ar−1x2s−N/2+βr−1,where(16)γβ≔π2sΓN+2s+2β/4ΓN+2s−2β/4ΓN−2s−2β/4ΓN−2s+2β/4.
In order to have homogeneity, we have(17)2s−N2+β=−2sr−2.
Thus, we deduce thatγβ−λ=Ar−2. Since A>0, we conclude that γβ−λ>0. Note that the map γ:0,N−2s/2⟼0,ΛN,s is decreasing about β, see [17, 18]. Hence, there is a unique element αλ such that γαλ=λ. Thus, we have αλ>β, that is,(18)αλ>−2sr−2+N−2s2,which implies that(19)r<N+2s−2αλN−2s−2αλ+1≔rλ,s.
Therefore, we can construct a supersolution to problem (1) for r<rλ,s, just modifying the w found above. Thus, rλ,s is the threshold for the existence to problem (1).Now, we consider the nonexistence of solution to problem (1).Theorem 2.
SupposeF1−F4 hold. Then, problem (1) has no solution in H0s if for some τ>1, M1τ>0, and δ>λ1/M1τ, where M0 and M1 are defined by (12) and (13), respectively.The following two examples also appeared in [15].Remark 3.
An example of functiongσ≡σq with 0<q<1, which satisfies conditions (10) and (11) for any δ∈0,δ0, such that problem (1) has at least one positive solution. In this condition M0=∞, by (52), we have ηr−qc1−1C12−qC1−C1r≥δ. Define(20)ΦC1=ηr−qc1−1C12−qC1−C1r.
It is easy to see that(21)ddC1ΦC1=0⟺C1=C1,0≔3−q2−q+r1/r−1.
We have to prove thatδ is smaller than the minimum of ΦC1. Therefore, we have(22)δ0=ΦC1,0=ηr−qc1−13−q2−q+r1/r−13−q2−q+r2−q/r−1−3−q2−q+r1−q+r/r−1.
Moreover, by (10) and (13), we have for any τ>1,(23)M1τδ0≤M11δ0≤c1δ0inf0<z≤λ11/r−2zq−2≡c1δ0λ1q−2/r−2.
Thus, for anyτ>1, M1τ≤c1λ1q−2/r−2. Hence, problem (1) has no solution at least δ>c1λ1r−q/r−2. Therefore, the result of the above theorem is more general than [9].Remark 4.
We consider the function(24)gσ=σσθ−11+σσα−1,for 0<α<θ<1. We easily deduce that conditions (10) and (11) are fulfilled and M0=∞. On the contrary,(25)M1τ=inf0≤z≤τλ11/r−2zθ−11+zα.
Ifα>θ−1, the function zθ−1/1+zα is monotone decreasing for z≥0. Then,(26)M1τ=τλ1θ−1/r−21+τλ1α/r−2.
Similarly, in this case, problem (1) has no solution provided(27)δ>λ1r−θ−1/r−21+λ1α/r−2.Remark 5.
The functionτ⟶M1τ is nonincreasing. Hence, if δ>λ1/M1τ0, for some τ0>1, the results of Theorem 1 will be true for any τ>τ0.The paper is organized as follows. In Section2, we present some definitions and preliminary tools, which will be used in the Proof of Theorems 1 and 2. The Proof of Theorems 1 and 2 are given in Section 3 and Section 4, respectively.


## 2. Preliminaries and Function Setting

In this section, we recall some known results for reader’s convenience.Denote the space(28)ℒs≔u:ℝN⟶ℝis measurable:∫ℝNux1+xN+2sdx<∞,equipped with the norm(29)uℒs≔∫ℝNux1+xN+2sdx.LetΩ be an open subset of ℝN. Given u∈ℒs and φ in the Schwartz class, the distribution −Δsu in D′Ω is defined as(30)−Δsu,φ=∫ℝNu−Δsφdx,for anyφ∈Cc∞Ω.We give some useful facts for the fractional Sobolev space.Definition 1.
Lets∈0,1, and define the fractional Sobolev space:(31)HsℝN=u∈L2ℝN:ξsu^∈L2ℝN.
We need to consider the spaceX0sΩ, which is defined as(32)X0sΩ=u∈HsℝN,u=0a.e.inℝN\Ω,with the norm(33)uX0sΩ=∫Qux−uy2x−yN+2sdxdy1/2,where Q=ℝ2N\Ωc×Ωc. The pair X0sΩ,.X0sΩ yields a Hilibert space (see Lemma 7 in [19]).We have to use the classical Sobolev theorem.Theorem 3 (see [20], Theorem 6.5).
Lets∈0,1, then there exists a positive constant S=SN,s, such that, for any measurable and compactly supported function u:ℝN⟶ℝ, we have(34)uL2s∗ℝN2≤S∫ℝN∫ℝNux−uy2x−yN+2sdxdy,where 2s∗ is the so-called Sobolev critical exponent.In this paper, we consider the existences of energy solution to problem (1) with the critical and subcritical cases.Definition 2.
We say thatu∈X0sΩ is an energy solution to problem (1), if for any φ∈X0sΩ,(35)∫Ωur−2uφdx<∞,∫Ωuφx2sdx<∞,CN,s2∫Qux−uyφx−φyx−yN+2sdxdy−λ∫Ωuφx2sdx=∫Ωur−1φdx+δ∫Ωguφdx.We also need to consider the weak solution to problem (1).Definition 3.
We say thatu∈L1Ω is a weak solution to problem (1), if u≥0 a.e. in Ω, u=0 in ℝN∖Ω,(36)∫Ωλux2s+ur−1+δguδsdx<∞,and for all φ∈C2s+βΩ∩CsΩ¯, β>0,(37)∫Ωu−Δsφdx=∫Ωλux2s+ur−1+δguφdx,where φ=0 in ℝN∖Ω and δx≔distx,∂Ω.Definition 4.
Ifu satisfies(38)−Δsu−λux2s≥ur−1+δgu,inΩ,u≥0,inℝN∖Ω,in the weak sense, we say that u is a supersolution to problem (1).
Ifu satisfies(39)−Δsu−λux2s≤ur−1+δgu,inΩ,u≤0,inℝN∖Ω,in the weak sense, we say that u is a subsolution to problem (1).Now, we recall the comparison lemma.Lemma 1 (see [9]).
Letu∈HsℝN and v∈HsℝN be solutions, respectively, to(40)−Δsu=f1,inΩ,u=g1,inℝN∖Ω,−Δsv=f2,inΩ,v=g2,inℝN∖Ω.
Then,ux≤vx for all x∈ℝN if f1≤f2 and g1≤g2.For the supercritical case, we need a prior regularity result, see [9], Lemma 2.2.Lemma 2.
Givenf∈L1Ω,δsxdx, where δx=distx,∂Ω. There exists a unique weak solution vx∈L1Ω to(41)−Δsv=f,inΩ,v=0,inℝN∖Ω,in the sense that(42)∫Ωv−Δsϑ=∫Ωfϑ,for all ϑ∈C2Ω¯ with ϑ=0 in ℝN∖Ω.Moreover,vL1Ω≤CfL1Ω,δsxdx, for some constant C independent of f. In addition, if f≥0(43)−Δsv=f,inΩ,v≥0,inℝN∖Ω.Then,v≥0a.e.inΩ.


## 3. The Existence Result

We are now ready to prove Theorem1 by employing the idea contained in [9, 15], whose proof will be split into several steps.Proof of Theorem 1.
Step 1: subsolution to problem (1). We first consider the eigenvalue problem:(44)−Δsφ1=λ1φ1,inΩ,φ1=0,inℝN∖Ω.Note that the eigenfunctionφ1≥0 belongs to X0s∩L∞Ω.SupposeδM0>λ1, where M0 is given in (12), by F4, for all δ∈λ1/M0,δ0, taking t small enough, we have(45)gtφ1tφ1>λ1δ.Therefore, forx∈Ω,(46)−Δstφ1=λ1tφ1<δgtφ1≤δgtφ1+tφ1r−1+λtφ1x2s.Therefore,(47)−Δstφ1≤δgtφ1+tφ1r−1+λtφ1x2s,inΩ,tφ1=0,inℝN∖Ω.Thus,u¯≔tφ1 is a subsolution to problem (1).Next, we consider supersolution to problem (1) in the subcritical and supercritical case, respectively.Step 2: supersolution for subcritical and critical case:2<r≤2s∗. We look for a supersolution of the form wx≔Ax−β with A≥0 and β>0 as real parameters and verify(48)β<N−2s2.Sincer≤2s∗, we obtain(49)r−1β<β+2s,(50)βr<N.By (49), we deduce that(51)−Δsw−λwx2s≥wr−1,inΩ,for the appropriate choice ofA.Letη≔infΩw>0. Taking u¯=C1w with 0<C1<1, which is a suitable constant such that, for δ small enough, and by (10) we have(52)ηr−p≥δc1C11−qC1−C1r,whereδ appears in (1).By (52), we obtain(53)−Δsu¯−λu¯x2s≥u¯r−1+δgu¯,inΩ,u¯≥0,inℝN∖Ω.Thus, we have concluded thatC1w is a supersoution to (1) for 2<r≤2s∗. Moreover, by (48) and (50), we obtain that(54)u¯∈LrΩ,u¯2x2s∈L1Ω.Definewj in L1ℝN is the weak solution to(55)−Δsωj+1=wjr−1+λwjx2s+δgwj,inΩ,ωj+1=0,inℝN∖Ω,forj≥1 and w0≔u¯. We now check that this definition makes sense and wj are monotone and satisfy(56)0≤u¯≤w1≤w2≤⋯≤wj≤wj+1≤⋅⋅⋅≤u¯a.e.Ω.Foru¯, there is nothing to prove. Suppose the result is true up to order j. Then,(57)−Δswj+1=wjr−1+λwjx2s+δgwj,≤u¯r−1+λu¯x2s+δgu¯,inΩ,ωj+1=0,inℝN∖Ω.Sowj is well-defined by (54) and Lemma 2. By the induction hypothesis, for x∈Ω,(58)−Δswj+1−wj=wjr−1−wj−1r−1+λwj−wj−1x|2s+δgwj−gwj−1≥0,andwj+1−wj=0,inℝN∖Ω. Then, by Lemma 1, we obtain wj+1≥wjinΩ.Similarly, for anyx∈Ω,(59)−Δsu¯−wj≥u¯r−1−wjr−1+λu¯−wjx2s+δgu¯−gwj≥0,andu¯−wj+1≥0inℝN∖Ω. Then, wj+1≤u¯ a.e. in Ω. We conclude that (56) holds.We can defineuδ≔limj⟶∞wj in L1Ω. Moreover, by (9), (54), and (56),(60)−Δs/2wjL2ℝN=λ∫Ωwjwj−1x2sdx+∫Ωwjwj−1r−1dx+δ∫Ωwjgwj−1dx≤λ∫Ωu¯2x2sdx+∫Ωu¯rdx+δc1∫Ωu¯qdx≤C.Hence, up to a subsequence, we know thatwj⇀uδ in X0sΩ. By monotony, the whole sequence weakly converges. Therefore, we can pass to the limit in (55) and conclude that uδ≥0 is a minimal energy solution of (1).Step 3: supersolution for supercritical case:2s∗<r<rλ,s. If r<rλ,s, where rλ,s is given in (9). For constant A≥0, there exists a radial function vx≔Ax−2s/r−2 such that(61)−Δsv−λvx2s=vr−1,inℝN.Sincer>2s∗>2N−2s/N−2s, then(62)v∈Llocr−1ℝN,vx2s∈Lloc1ℝN.Takingu¯=C1v, where the constant C1>0 is given by (54), we obtain(63)−Δsu¯−λu¯x2s≥u¯r−1+δgu¯,inΩ,u¯>0,inℝN∖Ω.Moreover by (62), u¯≥0 satisfies (36). By Lemma 2, define wj to be the weak solutions of (55). Moreover, similarly, by the induction hypothesis, we can conclude that(64)0≤u¯≤w1≤w2≤⋅⋅⋅≤wj≤wj+1≤⋅⋅⋅≤u¯a.e.Ω.Note thatF2 and F4 hold, for all δ∈λ1/M0,δ0, define w1 as the solution of(65)−Δsw1=u¯r−1+λu¯x2s+δgu¯,inΩ,w1>0,inΩ,w1=0,inℝN∖Ω.Set(66)Fu¯=u¯r−1+λu¯x2s+δgu¯=tφ1r−1+λtφ1x2s+δgtφ1.Then,w1∈W01,p∩L∞Ω. By (44), we obtain that(67)−Δsw1=Fu¯≥−Δsu¯,inΩ,w1=0,inℝN∖Ω.We deduce from the comparison principle thatu¯≤w1 in Ω.On the contrary, by (11), the function F is nondecreasing. Therefore,(68)−Δsu¯≥Fu¯≥Fu¯=−Δsw1,inΩ,u¯≥w1,inℝN∖Ω.By the comparison principle we deduce thatw1≤u¯ in Ω.In particular, for allx∈Ω, wj is a nondecreasing sequence which is bounded. Therefore, wj monotone converges in L1ℝN to a weak nonnegative solution uδ to (1) for 2s∗<r<rλ,s.Therefore, forδ small enough, we have built a minimal solution in both subcritical and supercritical case. Let(69)M=supδ>0:problem1has a solution,that is, we show thatM>0.Step 4:M<∞, for 2<r<rλ,s. We consider the following eigenvalue problem with the Hardy potential:(70)−Δsψ1−λψ1x2s=λ1ψ1,inΩ,ψ1=0,inℝN∖Ω.Since0<λ<ΛN,s, problem (70) is well defined. Taking ψ1 as a test function in problem (1), we obtain that(71)CN,s2∫Qψ1x−ψ1yux−uyx−yN+2sdxdy−λ∫Ωψ1ux|2sdx=∫Ωur−1ψ1dx+δ∫Ωguψ1dx.Sinceψ1 is a solution to (70), it follows that(72)∫Ωur−1+δguψ1dx=λ1∫Ωuψ1dx.If2s∗<r<rλ,s. Taking φ1 as a test function in (1), where φ1≥0 is solution to problem (44), we have(73)∫Ωu−Δsφ1dx=∫Ωλux2s+ur−1+δguφ1dx≥∫Ωur−1+δguφ1dx.Moreover,φ1 is also a classical solution (see Remark 2.1 in [21]). From (72), we immediately deduce that(74)λ1∫Ωuφ1dx≥∫Ωur−1+δguφ1dx.Since there exist structural positive constantsb0 and b1 such that tr−2t+δgt>b0δb1t, for any t>0. From (70) and (73), we obtain that b0δb1<λ1. This implies that M<∞ for r<rλ,s.We complete the Proof of Theorem1.


## 4. Nonexistence Result

In this section, we consider the nonexistence of solution to problem (1) in H0s.Proof of Theorem 2.
Suppose that problem (1) has a solution u∈H0s under the conditions of Theorem 2. Then, there exists a constant δ>0 such that δφ1≤uinΩ, where φ1 is the first eigenfunction of −Δs, that is, φ1 satisfies (44).
Letμ∈λ1,λ1+ε, where ε>0 is a small constant. Denote ψ=δφ1. Then, we have(75)−Δsψ=λ1ψ≤μψ<λ1+εu,inΩ,ψ=0,inℝN∖Ω.
Furthermore, for anyτ>1, δ<λ1/M1τ; then, for ε small enough, we deduce that(76)λ1+εu≤ur−1+δgu+λux2s≡−Δsu,inΩ,u=0,inℝN∖Ω.
Thus, according to (75) and (76), we have(77)−Δsψ≤μψ,inΩ,−Δsu≥μu,inΩ,u≥ψ,inΩ,u=ψ=0,inℝN∖Ω.
Hence, it is possible to construct the subsolution and supersolution to the problem:(78)−Δsu=μu,inΩ,u=0,inℝN∖Ω,with μ∈λ1,λ1+ε. However, this is impossible.



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