# Positive Solutions for Elliptic Problems with the Nonlinearity Containing Singularity and Hardy-Sobolev Exponents

Authors: Yong-Yi Lan; Xian Hu; Bi-Yun Tang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study multiplicity of positive solutions for a class of semilinear elliptic equations with the nonlinearity containing singularity and Hardy-Sobolev exponents. Using variational methods, we establish the existence and multiplicity of positive solutions for the problem.


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## Body

## 1. Introduction and Main Results

Consider the following semilinear elliptic equations with Dirichlet boundary value conditions:(1)−Δu−μux2=u2∗s−2xsu+λu−γ,inΩ,u>0,inΩ,u=0,in∂Ω,where Ω is a smooth bounded domain in RNN≥3, 0<s<2,2∗s=2N−s/N−2 is the Hardy-Sobolev critical exponent, 2∗=2∗0=2N/N−2 is the Sobolev critical exponent, μ<μ¯=ΔN−22/4, and γ∈0,1.The energy functional associated with problem (1) is defined by
(2)Iλu=12∫Ω∇u2−μu2x2dx−12∗s∫Ωu2∗sxsdx−λ1−γ∫Ωu1−γdx,for any u∈H01Ω. In general, a function u is called a weak solution of problem (1) if u∈H01Ω and ux>0 for all x∈Ω; it holds
(3)∫Ω∇u·∇v−μuvx2−u2∗s−1xsv−λu−γvdx=0,∀v∈H01Ω.The paper by Crandall et al. [1] is the starting point on semilinear problem with singular nonlinearity. There is a large literature on singular nonlinearities (see [2–14] and references therein). In particular, the following Dirichlet problem
(4)−Δu=uβ+μu−γ,inΩ,u>0,inΩ,u=0,in∂Ω,has been shown in [2], in which the authors proved that problem (4) possesses at least one solution for μ>0 small enough, and there exists no solution when μ is large. Chabrowski in [15] considered the Neumann boundary problems with singular superlinear nonlinearities by approximation and variational methods. When the superlinear term is subcritical, he obtained two solutions, a mountain-pass solution and a local minimizer solution. And, in the critical case, he obtained a local minimizer solution and proved that there is no moutain-pass solution.In recent years, people have paid much attention to the existence of solutions for problems with the Sobolev critical exponent (the case thats=0) (see [16–21] and the references therein); some authors also considered the singular problems with the Hardy-Sobolev critical exponent (the case that s≠0) (see [22–27] and the references therein).Up to our knowledge, the literature does not contain any result on the existence of positive solutions to the problem (1) with the nonlinearity containing singularity and Hardy-Sobolev exponents. Motivated by reasons above, the aim of this paper is to show the existence of positive solutions of problem (1). We study problem (1) and obtain at least two solutions via the Nehari method. It is well-known that the singular term leads to the nondifferentiability of the functional I on H01Ω, so I does not belong to C1H01Ω,R. In order to get the existence of multiple positive solutions of problem (1), we use the Nehari method and differentiate the two solutions by their different Nehari-type sets.The main result can be described as follows.Theorem 1.
Let 0<s<2,μ<μ¯=ΔN−22/4 and γ∈0,1. Then, there exists λ∗>0 small enough, such that problem 1 has at least two positive solutions for any λ∈0,λ∗.The paper is organized as follows: in Section2, we give some preliminaries; in Section 3, we prove Theorem 1. This idea is essentially introduced in 20]. Throughout this paper, we make use of the following notations:
(i)
The norm inH01Ω is denoted by
(5)∥u∥2=∫Ω∇u2−μu2x2dxBy Hardy inequality [28], we easily derive that the norm is equivalent to the usual norm:
(6)∥u∥02=∫Ω∇u2dx
(ii)
D1,2RNN≥3 denotes the space of the functions u∈L2∗RN such that ∣∇u∣∈L2RN, endowed with scalar product and norm, respectively, given by
(7)u,vRN=∫RN∇u·∇v−μuvx2dx,∥u∥RN2=∫RN∇u2−μu2x2dxthat coincides with the completion of C0∞RN with respect to the L2-norm of the gradient. By Hardy inequality [28], we easily derive that the norm is equivalent to the usual norm:
(8)∥u∥0,RN2=∫RN∇u2dxin D1,2RN.
(iii)
The norm inLpΩ is denoted by ·p
(iv)
C,C0,C1,C2,⋯ denote positive constants


## 2. Preliminaries

In this section, we will study the unperturbed problem(9)−Δu−μux2=u2∗s−2xsu,x∈RN,u∈Dr1,2RN,u>0,x∈RN.It is well-known that the nontrivial solutions of problem (9) are equivalent to the nonzero critical points of the energy functional
(10)I0u=12∫RN∇u2−μu2x2dx−12∗s∫RNu+2∗sxsdx,u∈Dr1,2RN.Obviously, the energy functionalI0u is well-defined and is of C2 with derivatives given by
(11)I0′u,v=∫RN∇u·∇v−μuvx2dx−∫RNu+2∗s−1xsvdx,u,v∈Dr1,2RN,(12)I0′′uv,w=∫RN∇v·∇w−μvwx2dx−∫RN2∗s−1u+2∗s−2xsvwdx,u,v,w∈Dr1,2RN.For allε>0, it is well-known that the function
(13)zεx=2ε2−sμ¯−μ/μ¯N−sμ¯−μμ¯μ¯/2−s/xμ¯−μ¯−με2−sμ¯−μ/μ¯+x2−sμ¯−μ/μ¯solves equation (9) and satisfies
(14)∫RN∇zε2−μzε2x2dx=∫RNzε2∗sxsdx.Moreoverzεx is the extremal function of the minimization problem
(15)S≔inf∫RN∇u2−μu2/x2dx∫RNu2∗s/xsdx2/2∗s:u∈D1,2RN,u≠0,SN−s/2−s=∫RN∇zε2−μzε2/x2dx=∫RNzε2∗s/xsdx.In view of [27, 29], we have the following exact local behavior of the solutions of (1).Lemma 2.
Let 0<s<2,μ<μ¯=ΔN−22/4. If u∈H01Ω is a positive solution of (1), then there exists r>0 small enough and some positive constants C1 and C2 such that
(16)C1x−μ¯−μ¯−μ≤∣ux∣≤C2x−μ¯−μ¯−μ,x∈B0,r\0.
Defineζ∈C01Ω such that ∣ζx∣≤1,∣∇ζx∣≤c for any x∈Ω, ζx≥0 for all ∈Ω, 0∈Ω, and
(17)ζx=1,ifx≤δ,0,ifx≥2δ.
Denotevεx=ζxzεx. Then, using an argument similar to [30], we have the following lemma.Lemma 3.
Let u∈H01Ω be a weak solution of problem (1). Then, for ε>0 small enough
(18)∫Ωu2∗s−1xsvεdx=OεN−2/4−2s,∫Ωuvε2∗s−1xsdx=OεN−2/4−2s,∫Ω∇vε2−μvε2x2dx=SN−s/2−s+OεN−s/2−s+OεN−2/2−s,∫Ωvε2∗sxsdx≥SN−s/2−s−OεN−2/2−s,∫Ωvεpdx=OεpN−2/4−2s.
Next, we define some Nehari-type sets, which are relevant in getting multiple positive solutions. Denote(19)Nλ=u∈H01Ω:∥u∥2=∫Ωu2∗sxs+λ∣u1−γdx,u≠0,Nλ+=u∈Nλ:1+γ∥u∥2>2∗s−1+γ∫Ωu2∗sxsdx,Nλ0=u∈Nλ:1+γ∥u∥2=2∗s−1+γ∫Ωu2∗sxsdx,Nλ−=u∈Nλ:1+γ∥u∥2<2∗s−1+γ∫Ωu2∗sxsdx.
Define the minimization problems(20)d+λ≔infu∈Nλ+Iλu.
Since(21)Iλu=12∫Ω∇u2−μu2x2dx−12∗s∫Ωu2∗sxsdx−λ1−γ∫Ωu1−γdx,it is easy to see that d+λ<0 for λ>0 and d+λ⟶0 as λ⟶0. Take λ3>0 such that d+λ+N−1SN−s/2−s>0 for any λ∈0,λ3. Denote
(22)λ4=2∗s−22∗s+γ−1S1+γN−s/4−2s+1−γ/21+γ2∗s−γ−11+γN−2/4−2sΩ1−2∗s−γ/2∗s,and set λ∗=minλ3,λ4.Lemma 4.
If λ∈0,λ∗, then Nλ0=∅. Moreover, for any u≠0, there exists a unique t+=t+u>0 such that t+uu∈Nλ− and
(23)t+>Tm≔∥u∥22∗s−1∫Ωu2∗s/xsdx1/2∗s−2,Iλt+u=maxt≥TmIλtu,and there exists a unique t−=t−u>0 such that t−uu∈Nλ+ and
(24)t−<TmandIλt−u=inf0≤t≤TmIλtu.Proof.
The proof is similar to [30]. We omit the details.


## 3. Proof of Theorem1

In this section, we will prove Theorem1. The proof of Theorem 1 is based on solving the minimization problem (18) and the minimization problem
(25)d−λ≔infu∈N−Iλu.We divide the proof into two steps. In the first step, we prove that if there isw∈Nλ+ such that d+λ=Iλw and there is v∈Nλ− such that d−λ=Iλv, then w and v are two positive weak solutions of (1). In the second step, we prove that the minima d+λ in (18) and d−λ in (23) are achieved, respectively.Step 1.
Letw∈Nλ+ be such that d+λ=Iλw and v∈Nλ− such that d−λ=Iλv.Lemma 5.
For each φ∈H01Ω and φ≥0, we have the following:
There isη0 such that Iλw+ηφ≥Iλw for each 0≤η≤η0
ii tη−⟶1 as η⟶0+, where for each η≥0,tη− is the unique positive number satisfying tη−v+ηφ∈Nλ−Proof.
The proof follows exactly the scheme in the proof of Lemma3 in [31].Lemma 6.
For each φ∈H01Ω and φ≥0, we have w−γφ and v−γφ∈L1Ω. Moreover,
(26)∫Ω∇w∇φ−μwφx2−w2∗s−1xsφ−λw−γφdx≥0,∫Ω∇v∇φ−μvφx2−v2∗s−1xsφ−λv−γφdx≥0.
In particular,w,v>0 for all x∈Ω\0.Proof.
We only prove (24) since the proof of (25) is similar. Let φ≥0 and ε>0. By (i) of Lemma 5 and simple computations, we have that
(27)λ1−γ∫Ωw+εφ1−γ−w1−γεdx≤12ε∥w+εφ∥2−∥w∥2−12∗sε∫Ω∣w+εφ∣2∗sxs−w2∗sxsdx.Since the right hand side of the inequality has a finite limit value asε⟶0, by direct calculations, we conclude ε−1w+εφ1−γ−w1−γ increases monotonically as ε⟶0 and
(28)limε→0w+εφ1−γ−w1−γε=0,ifφx=0,1−γw−γφ,ifφx>0andwx>0,∞,ifφx>0andwx=0.The Fatou lemma yieldsw−γφ∈L1Ω and we get (24).Sincew,v≥0 and w,v≡0, by the strong maximum principle, it follows that
(29)w,v>0,∀x∈Ω\0.Lemma 7.
We have that w and v are positive weak solutions of (1).Proof.
For anyφ∈H01Ω and η>0, we define ψ=w+ηφ and ψ+=maxψ,0. Then, ψ+∈H01Ω. Since w∈Nλ, we obtain from (24) that
(30)0≤∫Ω∇w∇ψ+−μwψ+x2−w2∗s−1xsψ+−λ∣w−γψ+dx=∫x∈Ω,w+ηφ>0∇w∇ψ+−μwψ+x2−w2∗s−1xsψ+−λw−γψ+dx=∫Ω∇w∇ψ−μwψx2−w2∗s−1xsψ−λw−γψdx−∫x∈Ω,w+ηφ≤0∇w∇ψ−μwψx2−w2∗s−1xsψ−λw−γψdx≤η∫Ω∇w∇φ−μwφx2−w2∗s−1xsφ−λw−γφdx−η∫x∈Ω,w+ηφ≤0∇w∇φ−μuφx2dx.
Dividing byη and letting η⟶0, by w≥0,∀x∈Ω, we have the measure of x∈Ω,w+ηφ≤0 which tends to 0 as η⟶0, and we get that
(31)∫x∈Ω,w+ηφ≤0∇w∇φdx⟶0.
Therefore,(32)∫Ω∇w∇φ−μwφx2−w2∗s−1xsφ−λw−γφdx≥0.
Sinceφ is arbitrary, we get that w is a solution of (1). Similarly, we can prove that v is also a solution of (1).Step 2.
The minimad+λ and d−λ are achieved. The proof is exactly the same as [32]. We omit the details here.
We point out thatvε and the exact local behavior of w (see Lemma 2.) play essential roles. From Lemma 2., we have
(33)C1x−μ¯−μ¯−μ≤∣wx∣≤C2x−μ¯−μ¯−μ,x∈B0,r\0.
So there ism>0 such that wx≥m for x∈suppw\0.Lemma 8.
Under the assumptions of Theorem 1,
(34)d−λ<Iλw+2−s2N−sSN−s/2−s.Proof.
First, using an argument similar to the proofs in [31], we have η∗>0 such that w+η∗vε∈Nλ−. It remains to be proven that
(35)supIλw+ηvε:η>0<Iλw+2−s2N−sSN−s/2−s.Sincew is a solution, we obtain by direct computation that
(36)Iλw+ηvε−Iλw=η22∥vε∥2+η∫Ω∇w∇vε−μwvεx2dx−∫Ωw+ηvε2∗s−w2∗s2∗sxsdx−λ∫Ωw+ηvε1−γ1−γ−w1−γ1−γdx=η22∥vε∥2−∫Ωw+ηvε2∗s−w2∗s2∗sxs−w2∗s−1ηvεxsdx−λ∫Ωw+ηvε1−γ1−γ−w1−γ1−γ−w−γηvεdx.Note that the following inequality (see [31]) holds: there is α>0 and 0<δ<N−s/N−2 such that λr+t1−γ/1−γ−r1−γ/1−γ−r−γt≥−αtδ for each r≥m and t≥0.By direct calculations, we can get that(37)Iλw+ηvε−Iλw≤η22∥vε∥2−η2∗s2∗s∫Ωvε2∗sxsdx−η2∗s−1∫Ωwvε2∗s−1xsdx+αηδ∫Ωvεδdx.SoIλw+ηvε−Iλw⟶0 as η⟶0 and Iλw+ηvε−Iλw⟶−∞ as η⟶+∞. Hence, we only consider the right-hand side of the above inequality in the case of η∈η0,η1 for some 0<η0<η1<+∞. Hence, we obtain from Lemma 3 that
(38)supη>0Iλw+ηvε−Iλw≤2−s2N−s∥vε∥22∗s/2∗s−2−∫Ωvε2∗sxsdx−2/2∗s−2−OεN−2/4−2s+OεδN−2/4−2s=2−s2N−sSN−s/2−s−OεN−s/4−2s−OεN−2/4−2s+OεδN−2/4−2s<2−s2N−sSN−s/2−s,forε>0sufficientlysmall.The proof is complete.Lemma 9.
The minimum d−λ in (23) is achieved by v∈Nλ− with Iλv=d−λ.Proof.
Letvn be any sequence in Nλ− such that Iλvn⟶d−λ. It is easy to see that vn is bounded in H01Ω. Then, there exists a v∈H01Ω and a subsequence of vn, still denoted by vn such that
(39)vn⇀v,inH01Ω.
Setun=vn−v and assume that
(40)∥un∥2⟶a2,∫Ωun2∗sxsdx⟶b2∗s.
Sincevn∈Nλ−, according to Bre’zis-Lieb’s Lemma (see Lemma 1.32 in MW [33]) and the Sobolev embedding theorem, one gets
(41)a2+∥v∥2=λ∫Ωv1−γdx+b2∗s+∫Ωv2∗sxsdx.
We claim thatv≠0. Indeed, if v=0, then a2=b2∗≥SN−s/2−s≠0 (since for any u∈Nλ−, ∥u∥ is bounded away from zero) and this means that
(42)d−λ=limn⟶∞Iλvn=Iλ0+12a2−12∗sb2∗s≥2−s2N−sSN−s/2−s,which contradicts the previous lemma.
From the assumptionλ∈0,λ∗, by Lemma 4, we have 0<t+<Tm<t− such that t+v∈Nλ+ and t−v∈Nλ−. For t>0, we define
(43)δt=12a2t2−12∗sb2∗st2∗s,gt=Iλtv+δt.
Now, we consider the cases(i)
t−<1
(ii)
t−≥1 and b>0
(iii)
t−≥1 and b=0
Case (i). Fromt−<1,
(44)g′t−=It−v,v+δ′t−=a2t−−b2∗st−2∗s−1.
Defineht≐a2t−b2∗st2∗s−1, then ht is increasing as t<Tm,; ht is decreasing as t>Tm, so ht has maximum value hTm. As t−>Tm, we have g′t−≥hTm>0. g′1=0 and g′t−>0, so we can see that g is increasing on t−,1. Then, we have
(45)d−λ=g1>gt−≥Iλt−v=Iλ0+t−22a2−b2∗s>Iλt−v≥d−λ,which is a contradiction.
Case (ii). We sett0=a2/b2∗sN−2/22−s. We know that δ attains the unique maximum at t0 and δt0≥2−s/2N−sSN−s/2−s. Moreover, δ′t>0 for 0<t<t0 and δ′t<0 for t>t0.
From the assumption onλ∈0,λ∗, we know g1≥gt0. If t0≤1, we have
(46)d−λ=g1≥gt0=Iλt0v+δt0≥Iλt0v+2−s2N−sSN−s/2−s,which contradicts the previous lemma. Thus, we have t0>1. By virtue of g′t≤0 for t≥1, we obtain ∂/∂tIλtv≤−δ′t≤0 for 1≤t≤t0 and
(47)d−λ=g1=Iλv+12a2−12∗sb2∗s≥Iλv+2−s2N−sSN−s/2−s,which also contradicts the previous lemma.
Case (iii). Ifa≠0, then we obtain from the fact that vn∈Nλ− by some computations that ∂/∂tIλtvt=1<0 and ∂2/∂2tIλtvt=1<0, from the definition of Nλ−, which contradicts t−≥1. So
(48)a=0,vn⟶vinH01Ω.
Hence, we havev∈Nλ− with Iλv=d−λ.
The proof is complete.



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