# Existence of Multiple Positive Solutions for Brezis–Nirenberg-Type Problems Involving Singular Nonlinearities

Authors: Mohammed El Mokhtar Ould El Mokhtar; Atika Matallah
Journal: Journal of Mathematics
 (2021)

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## Abstract

\This article is devoted to the existence and multiplicity to the following Brezis–Nirenberg-type problems involving singular nonlinearities:−Δu=up−1u+λu−1−β/xαuinΩu=0on∂Ω, where Ω is a smooth bounded domain in ℝNN≥3, 0∈Ω,λ>0, p=2∗−1 with 2∗≔2N/N−2 is the critical Sobolev exponent, 0≤α<Np+β/p+1, and 0<β<1. By using the Nehari manifold and maximum principle theorem, the existence of at least two distinct positive solutions is obtained.


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## Body

## 1. Introduction

The main purpose of this article is to investigate the existence of positive solutions of the following problem:(1)−Δu=up−1u+λu−1−βxαu,inΩ,u=0,on∂Ω,where Ω is a smooth bounded domain in ℝNN≥3, 0∈Ω,λ>0, p=2∗−1 with 2∗≔2N/N−2 is the critical Sobolev exponent, 0≤α<Np+β/p, and 0<β<1.In recent years, many authors have paid much attention to the following elliptic problem:(2)−Δu=up−1u+λfx,u,inΩ,u=0,on∂Ω,where Ω is a smooth bounded domain in ℝNN≥3, 0∈Ω,λ>0, 1<p≤2∗−1 with 2∗≔2N/N−2 is the critical Sobolev exponent, and f is a suitable function containing singularities on x; see [1–7] and references therein, after the work of Brezis and Nirenberg [8, 9]. Problem (2) becomes the well-known Brezis and Nirenberg problem and is studied extensively in [10]. Ding and Tang in [11] studied the existence of positive solutions with N≥3 and fx,u satisfying (AR) condition in the case λ=1.The novelty in this article is that the functionfx,u≔u/xαu−β with 0<β<1 presents a singular nonlinearity thing which will allow us to combine the perturbation with the variational methods. It should be noted that the problem studied in this work is not the fruit of the fertile imagination of a theorist; on the contrary, the problems dealt with in applied mathematics have their origins in different fields that we will cite as example: heterogeneous chemical catalysis, kinetic chemical catalysis, heat induction or electrical induction, non-Newtonian fluid theory, and viscous fluid theory. For further discussion on this subject, we refer the reader to [12].Before giving our main results, we state here some definitions, notations, and known results.We denote byℋ=ℋ01Ω\0 the closure of C0∞Ω\0 with respect to the norms:(3)u=∫Ω∇u2dx1/2.Since our approach is variational, we define the functionalJ on ℋ by(4)Ju≔12u2−1p+1∫Ωup+1dx−λ1−β∫Ωu1−βxαdx.A pointu∈ℋ is a weak solution of (1) if it satisfies(5)J′u,φ≔∫Ω∇u∇φdx−∫Ωup−1uφdx,−λ∫Ωu1−βxαφdx=0,for allφ∈ℋ.Here,.,. denotes the product in the duality ℋ′, ℋℋ′dual ofℋ.Let(6)S≔infu∈ℋ\0u2∫Ωup+1dx2/p+1.From [13], S is achieved.In our work, we research the critical points as the minimizers of the energy functional associated to problem (1) on the constraint defined by the Nehari manifold, which are solutions of our system.Letλ∗ and λ∗∗ be positive numbers such that(7)λ∗≔Sp+β/p−1p−1p+βA1+βp+β1+β/p−1,λ∗∗≔p−1p+βA1−β21/21+βp+β1/p−1S2p+1−p−11+β/p−1,where(8)A=2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p+β−α>0,with(9)0≤α<Np+1p+β.Now, we can state our main results.Theorem 1.
Assume thatN≥3, 0≤α<N/p+1p+β, 0<β<1, and λ verifying 0<λ<λ∗, then system (1) has at least one positive solution.Theorem 2.
In addition to the assumptions of Theorem1, there exists Λ∗∈0,λ∗∗ such that if λ satisfies 0<λ<Λ∗, then (1) has at least two positive solutions.This paper is organized as follows. In Section2, we give some preliminaries. Sections 3 and 4 are devoted to the proofs of Theorems 1 and 2.


## 2. Preliminaries

Definition 1.
Letc∈ℝ, E a Banach space and I∈C1E,ℝ.(i)
unn is a Palais–Smale sequence at level c (in short PSc) in E for I if(10)Iun=c+on1,I′un=on1,whereon1 tends to 0 as n goes at infinity.(ii)
We say thatI satisfies the PSc condition if any PSc sequence in E for I has a convergent subsequence.


### 2.1. Nehari Manifold

It is well known thatJ is of class C1 in ℋ, and the solutions of (1) are the critical points of J which is not bounded below on ℋ. Consider the following Nehari manifold:(11)ℳ=u∈ℋ\0:J′u,u=0.Thus,u∈ℳ if and only if u∈ℋ\0 and(12)u2−∫Ωup+1dx−λ∫Ωu1−βxαdx=0.Note thatℳ contains every nontrivial solution of problem (1). In order to obtain the first positive solution, we give the following important lemmas.Lemma 1.
J is coercive and bounded from below on ℳ.Proof.
LetR0>0 such that Ω⊂B0,R0=x∈ℝN:x<R0. If u∈ℳ, then by (12) and the Hölder inequality, we obtain(13)∫Ωu1−βxαdx≤2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p+β−αup+1S−1−β/2,and we deduce that(14)Ju=p−12p+1u2+−λp+βp+11−β∫Ωu1−βxαdx≥p−12p+1u2+−λp+βp+11−βAup+1S−1−β/2,with(15)A=2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p+β−α>0,for 0≤α<N/p+1p+β.
Thus,J is coercive and bounded from below on ℳ.Define(16)ϕu=J′u,u.Then, foru∈ℳ,(17)ϕ′u,u=2u2−p+1∫Ωup+1dx−λ1−β∫Ωu1−βxαdx=βu2−p+β∫Ωup+1dx=λp+β∫Ωu1−βxαdx−p−1u2.Now, we splitℳ in three parts:(18)ℳ+=u∈ℳ:ϕ′u,u>0,ℳ0=u∈ℳ:ϕ′u,u=0,ℳ−=u∈ℳ:ϕ′u,u<0.We have the following results.Lemma 2.
Suppose thatu0 is a local minimizer for J on ℳ. If u0∉ℳ0, then u0 is a critical point of J.Proof.
Ifu0 is a local minimizer for J on ℳ, then u0 is a solution of the optimization problem:(19)minu∈ℋ\0/ϕu=0Ju.
Hence, there exists a Lagrange multipliersθ∈ℝ such that(20)J′u0=θϕ′u0inℋ′.
Thus,(21)J′u0,u0=θϕ′u0,u0.
However,ϕ′u0,u0≠0 since u0∉ℳ0. Hence, θ=0. This completes the proof. □Lemma 3.
There exists a positive numberλ∗ such that for all λ verifying(22)0<λ<λ∗,we have ℳ0=∅.Proof.
Let us reason by contradiction.
Suppose thatℳ0≠∅ such that 0<λ<λ∗. Then, by (17) and for u∈ℳ0, we have(23)1+βu2−p+β∫Ωup+1dx=0,λp+β∫Ωu1−βxαdx−p−1u2=0.
Moreover, by the Hölder inequality and the Sobolev embedding theorem, we obtain(24)u≥Sp+1/2p−11+βp+β1/p−1,and(25)u≤λp+βp−1A1/1+β.
From (24) and (25), we obtain λ≥λ∗, which contradicts our hypothesis. □Thus,ℳ=ℳ+∪ℳ−. Define(26)c≔infu∈ℳJu,c+≔infu∈ℳ+Ju,c−≔infu∈ℳ−Ju.For the sequel, we need the following lemma.Lemma 4.
(i)
For allλ such that 0<λ<λ∗, one has c≤c+<0.(ii)
There existsλ∗∗ defined in Theorem 2 such that for all λ such that 0<λ<λ∗∗, one has(27)c−>C0=C0λ,S,β.Proof.
(i)
Letu∈ℳ+. By (17), we have(28)1+βp+βu2>∫Ωup+1dx,and so(29)Ju=−1+β21−βu2+p+βp+11−β∫Ωup+1dx<1−p1+β2p+11−βu2<0,sincep+1>2 and 0<β<1. Then, we conclude that c≤c+<0.(ii)
Letu∈ℳ−. By (17), we get(30)1+βp+βu2<∫Ωup+1dx.
Moreover, by Sobolev embedding theorem, we have(31)∫Ωup+1dx≤S−p+1/2up+1.
This implies(32)u>Sp+1/2p−11+βp+β1/p−1,for allu∈ℳ−.
By (13), we get(33)Ju≥u2p−12p+1+−λu1−βp+βp+11−βAS1−β/2≥p−22p1+βp+β2/p−1Sp+1/p−1+−λ2p+β2p+1p−11−βA2S1−β/2.
Thus, for allλ such that(34)0<λ<λ∗∗=p−1p+βA1−β21/21+βp+β1/p−1S2p+1−p−11+β/p−1,we have Ju≥C0.Proposition 1.
(see [8, 9]).(i)
For allλ such that 0<λ<λ∗, there exists a PSc+ sequence in ℳ+.(ii)
For allλ such that 0<λ<λ∗∗, there exists aPSc− sequence in ℳ− and for each u∈ℋ\0, we write(35)tM≔tmaxu=1+βu2p+β∫Ωup+1dx1/p−1>0.Lemma 5.
Letλ be the real parameter such that 0<λ<λ∗. For each u∈ℋ\0, there exist unique t+ and t− such that 0<t+<tM<t−, t+u∈ℳ+, and t−u∈ℳ−:(36)Jt+u=infJtu,for0≤t≤tM,Jt−u=supJtu,fort≥0.Proof.
With minor modifications, we refer to [8, 9].


## 2.1. Nehari Manifold

It is well known thatJ is of class C1 in ℋ, and the solutions of (1) are the critical points of J which is not bounded below on ℋ. Consider the following Nehari manifold:(11)ℳ=u∈ℋ\0:J′u,u=0.Thus,u∈ℳ if and only if u∈ℋ\0 and(12)u2−∫Ωup+1dx−λ∫Ωu1−βxαdx=0.Note thatℳ contains every nontrivial solution of problem (1). In order to obtain the first positive solution, we give the following important lemmas.Lemma 1.
J is coercive and bounded from below on ℳ.Proof.
LetR0>0 such that Ω⊂B0,R0=x∈ℝN:x<R0. If u∈ℳ, then by (12) and the Hölder inequality, we obtain(13)∫Ωu1−βxαdx≤2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p+β−αup+1S−1−β/2,and we deduce that(14)Ju=p−12p+1u2+−λp+βp+11−β∫Ωu1−βxαdx≥p−12p+1u2+−λp+βp+11−βAup+1S−1−β/2,with(15)A=2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p+β−α>0,for 0≤α<N/p+1p+β.
Thus,J is coercive and bounded from below on ℳ.Define(16)ϕu=J′u,u.Then, foru∈ℳ,(17)ϕ′u,u=2u2−p+1∫Ωup+1dx−λ1−β∫Ωu1−βxαdx=βu2−p+β∫Ωup+1dx=λp+β∫Ωu1−βxαdx−p−1u2.Now, we splitℳ in three parts:(18)ℳ+=u∈ℳ:ϕ′u,u>0,ℳ0=u∈ℳ:ϕ′u,u=0,ℳ−=u∈ℳ:ϕ′u,u<0.We have the following results.Lemma 2.
Suppose thatu0 is a local minimizer for J on ℳ. If u0∉ℳ0, then u0 is a critical point of J.Proof.
Ifu0 is a local minimizer for J on ℳ, then u0 is a solution of the optimization problem:(19)minu∈ℋ\0/ϕu=0Ju.
Hence, there exists a Lagrange multipliersθ∈ℝ such that(20)J′u0=θϕ′u0inℋ′.
Thus,(21)J′u0,u0=θϕ′u0,u0.
However,ϕ′u0,u0≠0 since u0∉ℳ0. Hence, θ=0. This completes the proof. □Lemma 3.
There exists a positive numberλ∗ such that for all λ verifying(22)0<λ<λ∗,we have ℳ0=∅.Proof.
Let us reason by contradiction.
Suppose thatℳ0≠∅ such that 0<λ<λ∗. Then, by (17) and for u∈ℳ0, we have(23)1+βu2−p+β∫Ωup+1dx=0,λp+β∫Ωu1−βxαdx−p−1u2=0.
Moreover, by the Hölder inequality and the Sobolev embedding theorem, we obtain(24)u≥Sp+1/2p−11+βp+β1/p−1,and(25)u≤λp+βp−1A1/1+β.
From (24) and (25), we obtain λ≥λ∗, which contradicts our hypothesis. □Thus,ℳ=ℳ+∪ℳ−. Define(26)c≔infu∈ℳJu,c+≔infu∈ℳ+Ju,c−≔infu∈ℳ−Ju.For the sequel, we need the following lemma.Lemma 4.
(i)
For allλ such that 0<λ<λ∗, one has c≤c+<0.(ii)
There existsλ∗∗ defined in Theorem 2 such that for all λ such that 0<λ<λ∗∗, one has(27)c−>C0=C0λ,S,β.Proof.
(i)
Letu∈ℳ+. By (17), we have(28)1+βp+βu2>∫Ωup+1dx,and so(29)Ju=−1+β21−βu2+p+βp+11−β∫Ωup+1dx<1−p1+β2p+11−βu2<0,sincep+1>2 and 0<β<1. Then, we conclude that c≤c+<0.(ii)
Letu∈ℳ−. By (17), we get(30)1+βp+βu2<∫Ωup+1dx.
Moreover, by Sobolev embedding theorem, we have(31)∫Ωup+1dx≤S−p+1/2up+1.
This implies(32)u>Sp+1/2p−11+βp+β1/p−1,for allu∈ℳ−.
By (13), we get(33)Ju≥u2p−12p+1+−λu1−βp+βp+11−βAS1−β/2≥p−22p1+βp+β2/p−1Sp+1/p−1+−λ2p+β2p+1p−11−βA2S1−β/2.
Thus, for allλ such that(34)0<λ<λ∗∗=p−1p+βA1−β21/21+βp+β1/p−1S2p+1−p−11+β/p−1,we have Ju≥C0.Proposition 1.
(see [8, 9]).(i)
For allλ such that 0<λ<λ∗, there exists a PSc+ sequence in ℳ+.(ii)
For allλ such that 0<λ<λ∗∗, there exists aPSc− sequence in ℳ− and for each u∈ℋ\0, we write(35)tM≔tmaxu=1+βu2p+β∫Ωup+1dx1/p−1>0.Lemma 5.
Letλ be the real parameter such that 0<λ<λ∗. For each u∈ℋ\0, there exist unique t+ and t− such that 0<t+<tM<t−, t+u∈ℳ+, and t−u∈ℳ−:(36)Jt+u=infJtu,for0≤t≤tM,Jt−u=supJtu,fort≥0.Proof.
With minor modifications, we refer to [8, 9].


## 3. Proof of Theorem 1

Now, taking as a starting point the work of Tarantello [14], we establish the existence of a local minimum for J on ℳ+.Proposition 2.
For allλ such that 0<λ<λ∗, the functional J has a minimizer u0+∈ℳ+ and it satisfies(i)
Ju0+=c=c+(ii)
u0+ is a nontrivial solution of (1)Proof.
If0<λ<λ∗, then by Proposition 1 (i) there exists a unPSc+ sequence in B¯R⊂ℳ+; thus, it bounded by Lemma 1. Then, there exists u0+∈ℋ and we can extract a subsequence which will denote by un such that(37)un⟶u0+weakly inℋ,un⟶u0+strongly inL1−βΩ,x−α,un⟶u0+a.e inΩ.
By (13) and (37), we have(38)limn⟶∞∫Ωun1−βxαdx=∫Ωu0+1−βxαdx+o1.
Thus, by (37), u0+ is a weak nontrivial solution of (1). Now, we show that un converges to u0+ strongly in ℋ. Suppose otherwise. By the lower semicontinuity of the norm and then either u0+<liminfn⟶∞un, we obtain(39)c≤Ju0+=p−1/2p+1u0+2+−λp+βp+11−β∫Ωu0+1−βxαdx<liminfn⟶∞Jun=c.
We get a contradiction. Therefore,un converge to u0+ strongly in ℋ. Moreover, we have u0+∈ℳ+. If not, then by Lemma 5, there are two numbers t0+ and t0−, uniquely defined so that t0+u0+∈ℳ+ and t−u0+∈ℳ−. In particular, we have t0−<t0+=1. Since(40)ddtJtu0+⇃t=t0+=0,d2dt2Jtu0+⇃t=t0+>0,there exists t0−<t−≤t0+ such that Jt0−u0+<Jt+u0+. By Lemma 5, we get(41)Jt0−u0+<Jt−u0+<Jt0+u0+=Ju0+,which contradicts the fact that Ju0+=c+. Since Ju0+=Ju0+ and u0+∈ℳ+, then by Lemma 2, we may assume that u0+ is a nontrivial nonnegative solution of (1). By the Harnack inequality, we conclude that u0+>0 (see, for example, [15]).


## 4. Proof of Theorem 2

Next, we establish the existence of a local minimum forJ on ℳ−. For this, we require the following lemmas.Lemma 6.
Letun be a PSc sequence for J for some c∈ℝ with un⇀u in ℋ.
Then,J′u=0 and Ju≥−λ2/βCp,β,A,S, with Cp,β,A,S>0.Proof.
It easy to prove thatJ′u=0, which implies that J′u,u=0, and(42)∫Ωup+1dx=u2−λ∫Ωu1−βxαdx.
Therefore,(43)Ju=p−12p+1u2+−λp+βp+11−β∫Ωu1−βxαdx.
From (13) and considering u small enough, we get(44)∫Ωu1−βxαdx≤2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p+β−αu1−βS−1−β/2,which implies that(45)Ju≥p−12p+1u2−λp+βp+11−βS−1−β/2Au1−β,with(46)A=2πN/2p+βNΓN/2p+β−αp+1p+β/p+1R0N/p+1p−1+β−α.
Using (44) and function ft=Dt2−λEt1−β, we obtain that(47)ft≥−λ2/1+βCp,β,A,S,for allt>0small enough,where(48)Cp,β,A,S=D1−βE2D2/1+ββ1−β,with(49)D=p−12p+1,E=p+βp+11−βS−1−β/2A.
We haveCp,β,A,S>0 since 0<β<1.
Then, we conclude that(50)Ju≥−λ2/1+βCp,β,A,S.Lemma 7.
Letλ∈0,λ∗∗, then the functional J satisfies the PSc condition in ℋ with c<c∗, where(51)c∗=p−1p+β21+βp+1Sp+1/p−1−λ2/1+βCp,β,A,S.Proof.
If0<λ<λ∗∗, then by Proposition 1 (ii) there exists a un, PSc sequence in ℳ; thus, it bounded by Lemma 1. Then, there exists u∈ℋ and we can extract a subsequence which will denoted by un such that(52)un⇀uweakly inℋ,un⇀uweakly inLp+1Ω,un⟶ua.e inΩ.
Then,u is a weak solution of (1). Let vn=un−u, then by Brezis–Lieb [16], we obtain(53)vn2=un2−u2+on1,and(54)∫Ωvnp+1dx=∫Ωunp+1−dx∫Ωup+1dx+on1.
Since(55)Jun=c+on1,J′un=on1,and by (53) and (54), we deduce that(56)12vn2−1p+1∫Ωvnp+1dx=c−Ju+on1,vn2−∫Ωvnp+1dx=on1.
Hence, we may assume that(57)vn2⟶l,∫Ωvnp+1dx⟶l.
Moreover, by the Sobolev inequality, we have(58)vn2≥S∫Ωvnp+1dx.
Combining (58) and (57), we obtain(59)l≥l2/p+1S.
Either(60)l=0orl≥Sp+1/p−1.
Then, from (56), (57), and Lemmas 6 and 7, we obtain(61)c≥p−12p+1l+Jun≥c∗,which is a contradiction. Therefore, l=0, and we conclude that un converges to u strongly in ℋ.
Thus,(62)Junconverges toJu=c,asntends to+∞.Lemma 8.
There existv∈ℋ and Λ∗>0 such that for λ∈0,Λ∗, one has(63)supt≥0Jtv<c∗.
In particular,c<c∗ for all λ∈0,Λ∗.Proof.
Letφεx satisfy (6). Then, we have(64)λ1−β∫Ωφε1−βxαdx>0.
We consider the two functions:(65)ft≔Jtφε,gt=t22φε2−tp+1p+1∫Ωφεp+1dx.
Then, for allλ∈0,λ∗∗,(66)f0=0<c∗.
By the continuity off, there exists t0>0 such that(67)ft<c∗,∀t∈0,t0.
On the other hand, we have(68)maxt≥0gt=p−1p+β21+βp+1Sp+1/p−1.
Then, we obtain(69)supt≥0Jtφε<p−1p+β21+βp+1Sp+1/p−1−λ2/1+βCp,β,A,S.
Now, takingλ>0 such that(70)−λt01−β1−β∫Ωφε1−βxαdx<−λ2/1+βCp,β,A,S,we obtain(71)0<λ<t01+β1−βCp,β,A,S1+β/1−β∫Ωφε1−βxαdx1+β/1−β=Λ1.
Set(72)Λ∗=minλ∗∗,Λ1.
We deduce thatc−<c∗ for all λ∈0,Λ∗, then there exists tn>0 such that tnwn∈ℳ− with wn satisfying (6) and for all λ∈0,Λ∗,(73)c−≤Jtnwn≤supt≥0Jtwn<c∗.Lemma 9.
For allλ such that 0<λ<Λ∗=minλ∗∗,Λ1, the functional J has a minimizer u0− in ℳ− and it satisfies(i)
Ju0−=c−>0(ii)
u0− is a nontrivial solution of (1) in ℋProof.
By Proposition1 (ii), there exists a PSc− sequence un for J, in ℳ− for all λ∈0,λ∗∗. From Lemmas 7, 8, and 4 (ii), for λ∈0,Λ1, J satisfies PSc− condition and c−>0. Then, we get that un is bounded in ℋ. Therefore, there exist a subsequence of un still denoted by un and u0−∈ℳ− such that un converges to u0− strongly in ℋ and Ju0−=c−>0 for all λ∈0,Λ∗.
Finally, by using the same arguments as in the proof of Proposition2, for all λ∈0,λ∗, we have that u0− is a solution of (1).Now, we complete the proof of Theorem2. By Proposition 2 and Lemma 9, we obtain that (1) has two positive solutions u0+∈ℳ+ and u0−∈ℳ−. Since ℳ+∩ℳ−=∅, then u0+ and u0− are distinct.Conclusion 1.
In our work, we have searched the critical points as the minimizers of the energy functional associated to the problem on the constraint defined by the Nehari manifoldℳ, which are solutions of our problem. Under some sufficient conditions on coefficients of equation of (1) such that N≥3, 0≤α<Np+β/p+1, and β∈0,1, we split ℳ in two disjoint subsets ℳ+ and ℳ−; thus, we consider the minimization problems on ℳ+ and ℳ−, respectively.



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