# The Existence of Positive Solution for Semilinear Elliptic Equations with Multiple an Inverse Square Potential and Hardy-Sobolev Critical Exponents

Authors: M. Khiddi
Journal: Abstract and Applied Analysis
 (2019)

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## Abstract

Via the concentration compactness principle, delicate energy estimates, the strong maximum principle, and the Mountain Pass lemma, the existence of positive solutions for a nonlinear PDE with multi-singular inverse square potentials and critical Sobolev-Hardy exponent is proved. This result extends several recent results on the topic.


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## Body

## 1. Introduction and Main Result

LetΩ⊂RN be smooth open bounded with N>2. In this paper, we study the existence of solutions to the following nonlocal problem:(1)-Δu-∑1kμiux-ai2=∑1kKixu2∗si-2ux-aisiinΩ,u=0on∂Ω,for arbitrary k such that 1≤k<∞, 0<si<2, μi>0, ai∈Ω, ai≠aj if i≠j and 2si∗=2N-si/N-2, is the critical Sobolev-Hardy exponent (i=1,2,⋯,k).We suppose the following:H1  0<μi<μ¯,for everyi=1,2,⋯,kand∑i=1kμi<μ¯=N-2/22.H2 There is an i0, 1≤i0≤k, such that (2)min1≤i≤k⁡2-si2N-siSμi,siaiμi,siN-si/2-siKiai-N-2/2-si=2-si02N-si0Sμi0,si0ai0N-si0/2-si0Ki0ai0-N-2/2-si0,and(3)Ki0x=Ki0ai0+ox-ai03asx→ai0.H3  0<μi0≤μ¯-1 where i0 is given in (H2). The function Ki(x) is a positive bounded on Ω¯, for every (1≤i≤k). Furthermore,(4)0<k¯i=maxx∈Ω¯⁡Kx.The reason why we investigate (1) is the presence of the Hardy-Sobolev exponent and the so-called inverse square potential in the linear part, which cause the loss of compactness of embedding H01(Ω)↪L2∗(Ω), H01(Ω)↪L2(|x|-2,Ω) and H01(Ω)↪L2∗(s)(|x|-s,Ω)).Hence, we face a type of triple loss of compactness whose interacting with each other will result in some new difficulties. In last two decades, loss of compactness leads to many interesting existence and nonexistence phenomena for elliptic equations. Many important results on the singular problems with Hardy-Sobolev critical exponents (the case thats≠0 and μi=μ were obtained such as the existence and multiplicity of solutions in these works and these results give us very good insight into the problem; see, for example, [1–7] and references therein. In the present paper, we use a variational method to deal with problem (1) with general form and generalize the results in [8]. As k≥2 to our knowledge, there are no results on the existence of non-trivial solutions for (1). It is therefore significant for us to study the problem (1) deeply. However, because of the singularities caused by the terms |x-ai|-sp(i=1,2,…,k), our problem becomes more complicated to deal with than [8] and therefore we have to face more difficulties. Despite the multiple terms of hardy and the coefficients of the critical nonlinearity, but we will see how, they will play an important role in the search for the bubble whose energy is below the level of local compactness (PS). The existence result is obtained via constructing a minimax level within this range and the Mountain Pass Lemma due to A. Ambrosetti and P.H. Rabinowitz (see also[9]).Our main result is the following.Theorem 1.
Assume that conditions(H1), (H2), (H3) hold. Then problem (1) has at least one positive solution.The paper is organized as follows: in Section2, preliminary results about Palais-Smale condition for c in a suitable interval and construct some auxiliary functions and estimate their norms. In Section 3, fill the conditions of Mountain Pass Theorem and we establish our result.


## 2. Preliminary Results

Throughout this paper,C, Ci(i=1,2,3,⋯) represent all kinds of positive constants. We denote the standard norm of the Sobolev space H01(Ω) by (5)u=∫Ω∇u2dx1/2.Br(a) is a ball centered at a with radius r.O(εt) denotes Oεt<Cεt and o(εt) denotes oεt/εt→0 as ε→0. We will look for solutions of (1) by finding critical points of the C1-functional I: H01(Ω)→RN given by(6)Iu=12∫Ω∇u2dx-∑i=1kμi∫Ωu2x-ai2dx-∑i=1k12∗si∫ΩKixu2∗six-aisidxfor all H01(Ω). The function u∈H01(Ω) is said to be a solution of problem (1) if u satisfies (7)I′u,u=∫Ω∇u∇φdx-∑i=1kμi∫Ωuφx-ai2dx-∑i=1k∫ΩKixu2∗six-aisidxfor all φ∈H01(Ω).Problem (1) is well defined by the both inequalities, Sobolev-Hardy inequalities which is essentially due to Caffarelli, Kohn, and Nirenberg (see [10]):(8)∫Ωuqx-asdx2/q≤Cs,q∫Ω∇u2dx,∀u∈H01Ω,∀a∈Ω,where 2<q<2∗, and the Hardy inequality (see [11, 12]), that is a special case (q=s=2) of the above Sobolev-Hardy inequality.(9)∫Ωu2x-a2dx≤1μ¯∫Ω∇u2dx,∀u∈H01Ω,∀a∈Ω.By (8) and (9), for 0≤μ<μ¯, 0≤s<2, q=2∗(s) and a∈Ω we can define the best Sobolev-Hardy constant:(10)Sμ,sa=infu∈H01Ω∖0⁡∫Ω∇u2dx-μ∫Ωu2/x-a2dx∫Ωu2∗s/xsdx2/2∗s,In the case where s=0, then (2∗(0)=2∗); note Sμ,0a is the best constant in the Sobolev inequality, i.e.,(11)Sμ,0a=infu∈H01Ω∖0⁡∫Ω∇u2dx-μ∫Ωu2/x-a2dx∫Ωu2∗dx2/2∗.The best Sobolev-Hardy constant Sμ,sa is achieved only when Ω=RN by a family of functions:(12)Yμ,εax=2εN-sμ¯-μ/μ¯μ¯/2-sx-aaμε+x-a22-s/N-2μ¯-μN-2/2-sforRN∖0.Let (13)Cε=2εN-sμ¯-μμ¯μ¯/2-s,Uμ,εax=Yμ,εaxCε,where b(μ)=μ¯+μ¯-μ and a(μ)=μ¯-μ¯-μ, (see [13] for details). Moreover,(14)0≤aμ<N-22<bμ≤N-2.We consider ρ>0 such that B(a,2ρ)⊂Ω and define a cut function φ∈C0∞(Ω) such that 0≤φ≤1, |∇φ|≤C, φ=1 for |x-a|≤ρ and φ=0 for |x-a|>2ρ. Set(15)uμ,εax=φxUμ,εax,vμ,εax=uμ,εax∫Ωuμ,εax2∗s/x-asdx1/2∗s,so that (16)∫Ωvμ,εax2∗sx-asdx=1.Then we have the following estimates.Lemma 2.
For any0<μ<μ¯ and a∈Ω,(17)∫Ω∇vμ,εa2dx-μ∫Ωvμ,εax2x-a2dx=Sμ,sa+OεN-2/2-s,(18)∫Ωx-akvμ,εax2∗sx-asdx=OεkN-s/22-sμ¯-μ,(19)∫ΩKixvμ,εax2∗sx-asdx=Kia+Oε3N-s/22-sμ¯-μ.For a≠b, we have(20)∫Ωvμ,εax2x-b2dx=OεN-2/2-sμ¯-μifμ<μ¯-1,OεN-2/2-sln⁡εifμ=μ¯-1,Proof.
It is easy to get the following results (17),(18) (see [14].) We show (19) and (20) and for the proof (19). By using (18) and assumption (H3) we have(21)∫ΩKixvμ,εax2∗sx-asdx-∫ΩKiavμ,εax2∗sx-asdx≤∫ΩKix-Kiavμ,εax2∗sx-asdx≤C1∫Ωx-a3vμ,εax2∗sx-asdx=C2ε3N-s/22-sμ¯-μ.Now we show (20). Let a≠b and ε0=εN-2/22-sμ¯-μ. We have(22)∫Ωvμ,s,εax2x-b2dx=∫BρaWμ,s,εax2x-b2dx+∫Bρaφ2x-1Wμ,s,εax2x-b2dx=1a-b2∫BρaWμ,s,εax2dx+∫Bρaφ2x-1Wμ,s,εax2x-b2dx∫Bρa1x-b2-1a-b2Wμ,s,εax2dx=I1ε+I2ε+I3ε.
For(23)I1ε=1a-b2∫BρaWμ,s,εax2dx=C1εN-2/2-s∫x≤ρ1ε2N-2/2-sx2aμ1+x22-s/N-2μ¯-μ2N-2/2-sdx=C1ε-N-2/2-s+NN-2/22-sμ¯-μ-N-2aμ/2-sμ¯-μ∫0ρ/ε0rN-1r2aμ1+r22-s/N-2μ¯-μ2N-2/2-sdr,we know that(24)rN-1r2aμ1+r22-s/N-2μ¯-μ2N-2/2-s~1r1-N+2aμ,intheneighbourhoodof0rN-1r2aμ1+r2-s/2N-2μ¯-μ2N-2/2-s~1r1-N+2bμ,intheneighbourhoodof+∞,and since a(μ)≤N-2/2<N/2 we deduct that(25)∫011r1-N+2aμdr<∞,forevery0<μ<μ¯(26)∫1ρ/ε01r1-N+2bμdr≤C1,if0<μ<μ¯-1,C2ln⁡ε0,ifμ=μ¯-1.By using (23) and (26), we obtain(27)I1ε≤C1εN-2/2-sμ¯-μ,if0<μ<μ¯-1,C2εN-2/2-sln⁡ε,ifμ=μ¯-1,
whereC1,C2>0 are constant.
For the second integral,(28)I2ε=∫Bρa1x-b2-1a-b2Wμ,s,εax2x-b2dx=C1ε-N-2/2-s∫x-a≤ρb-a2-x-b2x-b2a-b21ε02aμx-a/ε02aμ1+x-a/ε022-s/N-2μ¯-μ2N-2/2-sdx=C1ε-N-2/2-sε0N-2aμ∫y≤ρ/ε0-ε02y2-2ε0y,a-by2aμε0y+a-b2a-b21+y22-s/N-2μ¯-μ2N-2/2-sdy.For y≤ρ/ε0, we have –ε0|y|2-2〈y,(a-b)〉/ε0y+a-b|2a-b|2≤ρ+2a-b/a-b-ρ2|a-b|2.(29)I2ε≤C2ε-N-2/2-s-N-2aμ/2-sμ¯-μ+N-2N+1/22-sμ¯-μ∫y≤ρ/ε0yy2aμ1+y22-s/N-2μ¯-μ2N-2/2-sdy=C2ε3N-2/22-sμ¯-μ∫0ρ/ε0rNr2aμ1+r22-s/N-2μ¯-μ2N-2/2-sdr.Since 2a(μ)-N<1, this implies that (30)∫01rNr2aμ1+r22-s/N-2μ¯-μ2N-2/2-sdr<∞.Also, (31)rNr2aμ1+r2-s/2N-2μ¯-μ2N-2/2-s≤1r2bμ-N,and (32)2bμ-N>1⇔0≤μ<μ¯-94.Then, if 0≤μ<μ¯-9/4, we have(33)I2ε=∫Bρa1x-b2-1a-b2Wμ,s,εax2x-b2dx=Oε3N-2/22-sμ¯-μ.And, if μ=μ¯-9/4, we have(34)I2ε=∫Bρa1x-b2-1a-b2Wμ,s,εax2x-b2dx≤C1ε3N-2/22-sμ¯-μ∫011r2aμ-Ndr+∫1ρ/ε01rdr=C2ε3N-2/22-sμ¯-μ+C3ε2N-2/22-sμ¯-μln⁡ε=Oε3N-2/22-sμ¯-μln⁡ε.If μ¯-9/4<μ<μ¯, this implies that (35)2bμ-N<1,since 2a(μ)-N<1, we deduct that (36)∫011r2aμ-Ndr<∞(37)∫Bρa1x-b2-1a-b2Wμ,s,εax2x-b2dx≤C2ε3N-2/22-sμ¯-μ∫011r-N+2aμdr+∫1ρ/ε01r-N+2bμdr=C3ε3N-2/22-sμ¯-μ+C4εN-2/22-sμ¯-μ+N-2/2-s.We deduct that(38)I2ε≤Cε3N-2/22-sμ¯-μif0<μ<μ¯-94,Cε3N-2/22-sμ¯-μln⁡εifμ=μ¯-94,Cε3N-2/22-sμ¯-μ+CεN-2/22-sμ¯-μ+N-2/2-sifμ¯-94<μ<μ¯.So, if 0<μ<μ¯-1, we have(39)I2ε=∫Bρa1x-b2-1a-b2Wμ,s,εax2x-b2dx=oεN-2/2-sμ¯-μ.Let ρ<|a-b|, and ρ/2<|x-a|<ρ; we have |a-b|-ρ<|x-b|.(40)I3ε=∫Ω∖Bρaφ2x-1Wμ,s,εax2x-b2dx≤C1εN-2/2-sμ¯-μa-b-ρ2∫ρ/2ε0ρ/ε0rN-1r2aμ1+r22-s/N-2μ¯-μ2N-2/2-sdr≤C2εN-2/2-sμ¯-μ∫ρ/2ε0ρ/ε0rN-1-2aμ-4μ¯-μdr≤C2εN-2/2-sμ¯-μ∫ρ/2ε0ρ/ε0ε0-1+2μ¯-μdr≤C3εN-2/2-s.Then(41)I3ε=OεN-2/2-s.So, for 0<μ<μ¯-1 and taking (27),(39), and (41) in (22), we get (42)∫Ωvμ,s,εax2x-b2dx=OεN-2/2-sμ¯-μ+oεN-2/2-sμ¯-μ+OεN-2/2-s=OεN-2/2-sμ¯-μ.For the case μ=μ¯-1. Using (29) we have(43)I2ε≤C2εN-2/2-s∫y≤ρ/ε0-ε02y2-2ε0y,a-by2aμε0y+a-b2a-b21+y22-s/N-2μ¯-μ2N-2/2-sdy≤C3εN-2/2-s∫y≤ρ/ε0ε02y2+2ε0ya-by2aμ1+y22-s/N-2μ¯-μ2N-2/2-s≤C4εN-2/2-s∫0ρ/ε0ε02r2+2ε0ra-brN-1rN-41+r22-s/N-22N-2/2-sdr≤C5εN-2/2-s∫0ρ/ε0ε02r+2ε0a-bdr≤C6εN-2/2-s.
Then ifμ=μ¯-1 we get(44)I2ε=OεN-2/2-s.(45)I1ε=C2εN-2/2-sln⁡ε,ifμ=μ¯-1,(46)I3ε=OεN-2/2-s.
By (44), (45), and (46), we derive (20) for μ=μ¯-1.LetX be a Banach space and X-1 be the dual space of X. The functional I∈C1(X,R) is said to satisfy the Palais–Smale condition at level c ((PS)c in short), if any sequence un⊂X satisfying I(un)→c, I′(un)→0 strongly in X-1 as n→+∞ contains a subsequence converging in X to a critical point of the functional I.In our case,X=H01(Ω) and X-1=H-1(Ω).Lemma 3.
The functionalI satisfies (PS)c condition for any(47)c<min1≤i≤k⁡2-si2N-siSμi,siaiN-si/2-siKiaiN-2/2-si=c∗.Proof.
Supposeun is a (PS)c sequence for I with c<c∗. Then,(48)Iun→c,I′un→0asn→+∞.First, we show that un is bounded in H01(Ω). Let 2<θ<2∗(si) for all 1≤i≤k. So, 1/2-1/θ>0 and 1/θ-1/2∗(si)>0. By Hardy and Sobolev-Hardy inequality we have(49)c+un≥Iun-1θI′unun≥12-1θun2-∑i=1kμi∫Ωun2x-ai2dx+∑i=1k1θ-12∗si∫ΩKixun2∗six-aisidx≥12-1θ1-∑i=1kμiμ¯un2+∑i=1k1θ-12∗si∫ΩKixun2∗six-aisidx≥12-1θ1-∑i=1kμiμ¯un2.Therefore, up to a subsequence, we may assume that (50)un⇀uweaklyinH01Ω,un⇀uweaklyinL2Ω,x-ai-2dxfor1≤i≤k,un⇀uweaklyinL2∗siΩ,x-ai-sidxfor1≤i≤k,un→ua.e.onΩ.Then u∈H01(Ω) is a weak solution of problem (1). We may suppose that (51)∇un2⇀∇u2+ν,weak∗-senseofmeasures.Using the concentration-compactness principle due to Lions (cf. [[15], Lemma 1.2]), we obtain an atmost countable set Λ, a set of different points {xj}j∈Λ⊂Ω∖{a1,a1,⋯,ak}, real numbers νxj, νaj,τaj and σai for 1≤j≤k such that(52)ν≥∑j∈Λνxjδxj+∑j=1kνajδaj,and since 2<2∗(si)<2∗(1≤i≤k) we have (53)un2∗six-aisi⇀u2∗six-aisi+τaiδai,un2x-ai2⇀u2x-ai2+σaiδai,where δaj is the Dirac mass at aj∈RN.
Letε>0 such that for any j∈Λ, xj∉Bε(ai)(1≤i≤k). Choose a smooth cut-off function φi,ε centered at the point ai satisfying 0≤φi,ε≤1, φi,ε=1 for |x-ai|≤ε/2, φi,ε=0 for |x-ai|≥ε and |∇φi,ε|≤4/ε. Since unφi,ε is bounded, I′(un)unφi,ε=o(1), that is,(54)I′ununφi,ε=∫Ω∇un2φi,εdx+∫Ωun∇un∇φi,εdx-∑i=1kμi∫Ωun2x-ai2φi,εdx-∑i=1k∫ΩKixun2∗six-aisiφi,εdx.Moreover, we have(55)limε→0⁡limn→+∞⁡∫Ω∇un2φi,εdx≥limε→0⁡∫Ω∇u2φi,εdx+νai=νailimε→0⁡limn→+∞⁡∫Ωun2x-ai2φi,εdx=limε→0⁡∫Ωu2x-ai2φi,εdx+σai=σailimε→0⁡limn→+∞⁡∫ΩKixun2∗six-aisiφi,εdx=limε→0⁡∫ΩKixu2∗six-aisiφi,εdx+τai=τai.Arguing as in [3], we can prove that(56)limε→0⁡limn→+∞⁡∫Ωun∇un∇φi,εdx=0,limε→0⁡limn→+∞⁡∫Ωun2x-aj2φi,εdx=0,∀j≠i,limε→0⁡limn→+∞⁡∫ΩKixun2∗six-ajsiφi,εdx=0,∀j≠i.From (55)-(56), let n→+∞ and ε→0 in expression (54), we obtain(57)0≥νai-μiσai-Kiaiτai.By the definition of Sμi,siai, we deduce that(58)Sμi,siaiτai2/2∗si≤νai-μiσai.Combining (57) with (58), we get (59)Sμi,siaiτai2/2∗si≤Kiaiτai,which implies that (60)τai=0orτai≥Sμi,siaiN-si/2-siforevery1≤i≤k.Arguing by contradiction, let us suppose that there exist i0 such that(61)τai0≥Sμi0,si0ai0N-si0/2-si0.Thus,(62)c+o1=Iun-12I′unun=∑i=1k12-12∗si∫ΩKixun2∗six-aisidx=∑i=1k12-12∗si∫ΩKixu2∗six-aisidx+Kiaiτi.Letting n→+∞, we get (63)c≥∑i=1k12-12∗siKiaiτi,so, by (61), we obtain (64)c≥2-si02N-si0Sμi0,si0ai0N-si0/2-si0Ki0ai0N-2/2-si0,which contradicts the assumption that (65)c<min1≤i≤k⁡2-si2N-siSμi,siaiN-si/2-siKiaiN-2/2-si=c∗.Hence, up to a subsequence, we obtain that un→u strongly in H01(Ω).Lemma 4.
Under the assumptions of(H1), (H2), and (H3), there is a nonnegative function v0∈H01(Ω), such that supt≥0I(tv0)<c∗.Proof.
Let us prove only for the following case0<μi0<μ¯, for the other case the proof is the same. We consider the following functions on the interval [0,+∞((66)g¯t=t22∫Ω∇vμi0,εai02-μi0vμi0,εai02x-ai02-∑i≠i0μi∫Ωvμi0,εai02x-ai2dx-t2∗si02∗si0∫ΩKi0xvμi0,εai0x2∗si0x-ai0si0dx,and (67)Itvμi0,εai0=g¯t-∑i≠i0kt2∗si2∗si∫ΩKixvμi0,εai02∗six-aisidx.Using the following formula, (68)supt≥0⁡t22A-t2∗si02∗si0B=2-si02N-si0AN-si0/2-si0BN-2/si0-2,and using (17), (19), and (20), we have (69)supt≥0⁡g¯t=2-si02N-si0∫Ω∇vμi0,εai02-μi0vμi0,εai02/x-ai02dx-∑i≠i0μi∫Ωvμi0,εai02/x-ai2dxN-si0/2-si0∫ΩKi0xvμi0,εai0x2∗si0/x-ai0si0dxN-2/si0-2=2-si02N-si0Sμi0,si0ai0-C1∑i≠i0μi/ai-ai0εN-2/2-sμ¯-μ-oεN-2/2-sμ¯-μ+OεN-2/2-sN-si0/2-si0Ki0ai0+Oε3N-2/22-sμ¯-μN-2/si0-20=2-si02N-si0Sμi0,si0ai0N-si0/2-si0Ki0ai0N-2/2-si01-OεN-2/2-sμ¯-μ-oεN-2/2-sμ¯-μ+OεN-2/2-sN-si0/2-si01+Oε3N-2/22-sμ¯-μN-2/si0-2=2-si02N-si0Sμi0,si0ai0N-si0/2-si0Ki0ai0N-2/2-si0×1-OεN-2/2-sμ¯-μ-oεN-2/2-sμ¯-μ1-Oε3N-2/22-sμ¯-μ-oε3N-2/22-sμ¯-μ<2-si02N-si0Sμi0,si0ai0N-si0/2-si0Ki0ai0N-2/2-si0-OεN-2/2-sμ¯-μ=c∗,for 0<ε sufficiently small. And since, for all 1≤i≤k, the function Ki is a positive on Ω, we have (70)supt≥0⁡Itvμi0,εai0≤supt≥0⁡g¯t<c∗,for 0<ε sufficiently small.


## 3. Proof of Main Result 1

We verify that the functionalI satisfies the mountain pass geometry. To this end, we consider the energy level.(71)c1=infγ∈Γ⁡supt∈0,1⁡Iγt,where (72)Γ=γ∈C0,1;H01Ω:γ0=0,Iγ1<0. For any u∈H01(Ω)∖0, by Hardy and Sobolev-Hardy inequality, (8) and (9) (take q=2∗(si)), we get that (73)Iu=12∫Ω∇u2-∑i=1kμi∫Ωu2x-ai2dx-12∗si∫ΩKixu2∗six-aisidx≥121-1μ¯∑i=1kμi∫Ω∇u2dx-∑i=1kk¯iCsi,2∗si-2∗si/22∗si∫Ω∇u2dx2∗si/2≥121-1μ¯∑i=1kμiu2-∑i=1kk¯iCsi,2∗si-2∗si/22∗siu2∗si.Hence, there exists ρ0>0 small enough such that (74)r=infu=ρ0⁡Iu>0.Then (75)Iu≥r,for all u∈H01(Ω) with u=ρ0. Let v∈H01(Ω) given in Lemma 4. Since limt→+∞I(tv)=-∞, hence there exists t0>0 such that t0v≥ρ0 and I(t0v)<0, by Lemma 4, we obtain (76)c1≤supt∈0,1⁡Itt0v≤supt≥0⁡Itv<c∗.Moreover, by the Mountain Pass Theorem [9] and Lemma 3, we obtain that c1 is critical value of I at point u and thus is a solution of problem (1). Then the rest of the proof follows exactly the same lines as that in [3]. In order to find the positive solution of (1), we replace I(u) with I+(u) defined as follows: (77)I+u=12∫Ω∇u2dx-∑i=1kμi∫Ωu2x-ai2dx-∑i=1k12∗si∫ΩKixu+2∗six-aisidxwhere u+=maxu,0. Repeating the above arguments, we find a critical point of I+ and by applying the maximum principle we obtain a positive solution. So, the proof of Theorem 1 is therefore completed.



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