# Multiplicity of Positive Solutions for Semilinear Elliptic Systems

Authors: Tsing-San Hsu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the effect of the coefficienth(x) of the critical nonlinearity on the number of positive solutions for semilinear elliptic systems. Under suitable assumptions for f(x), g(x), and h(x), we should prove that for sufficiently small λ,μ>0, there are at least k+1 positive solutions of the semilinear elliptic systems -Δu=λf(x)|u|q-2u+(α/(α+β))h(x)|u|α-2u|v|β, -Δv=μg(x)|v|q-2v+(β/(α+β))h(x)|u|α|v|β-2v, where 0∈Ω⊂ℝN is a bounded domain, α>1, β>1, and N/(N-2)<q<2<α+β=2* for N>4.


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## Body

## 1. Introduction and Main Results

ForN≥3, α>1, β>1, and 1≤q<2<α+β=2*=2N/(N-2), consider the semilinear elliptic systems
(Pλ,μ){-Δu=λf(x)|u|q-2u+αα+βh(x)|u|α-2u|v|βinΩ,-Δv=μg(x)|v|q-2v+βα+βh(x)|u|α|v|β-2vinΩ,u=v=0on∂Ω,
where λ,μ>0, Ω⊂ℝN is a bounded domain with smooth boundary ∂Ω.Letf, g, and h satisfy the following conditions.(H1)
f, g, and h are positive continuous functions in Ω¯ and maxx∈Ω¯h(x)=1.(H2)
There existk points a1,a2,…,ak∈Ω and some σ≥N-2 such that h(ai) are strict maxima and satisfy
(1)h(ai)=maxx∈Ω¯h(x)=1∀1≤i≤kand h(x)=h(ai)+O(|x-ai|σ) as x→ai uniformly in i.Recent studies [1–10] have investigated the elliptic systems with subcritical or critical exponents and have proved the existence of a ground state solution or the existence of at least two positive solutions for these problems. For the case of N>4, α>1, β>1, and 2<q<α+β=2*=2N/(N-2), Lin [11] constructs the k compact Palais-Smale sequences that are suitably localized in correspondence of k maximum points of h. Under assumptions (H1)-(H2), she has showed that there are at least k positive solutions of the problem (Pλ,μ) for sufficiently small λ,μ>0. In this paper, we study the problem (Pλ,μ) and complement the results of [11] to the case 1≤q<2. Under assumptions (H1)-(H2), we should prove that there exist at least k+1 positive solutions of the problem (Pλ,μ) for sufficiently small λ,μ>0.LetE=H01(Ω)×H01(Ω) be the space with the standard norm
(2)∥(u,v)∥E=(∫Ω‍(|∇u|2+|∇v|2)dx)1/2.
Associated with the problem (Pλ,μ), we consider the C1-functionalIλ,μ, for (u,v)∈E,
(3)Iλ,μ(u,v)=12∥(u,v)∥E2-1q∫Ω‍(λf(x)|u|q+μg(x)|v|q)dx-12*∫Ω‍h(x)|u|α|v|βdx.
The weak solution (u,v)∈E of the problem (Pλ,μ) is the critical point of the functional Iλ,μ; that is, (u,v)∈E satisfies
(4)∫Ω‍(∇u∇φ1+∇v∇φ2)dx-λ∫Ω‍f(x)|u|q-2uφ1dx-μ∫Ω‍g(x)|v|q-2vφ2dx-α2*∫Ω‍h(x)|u|α-2u|v|βφ1dx-β2*∫Ω‍h(x)|u|α|v|β-2vφ2dx=0
for any (φ1,φ2)∈E.LetD1,2(ℝN)={u∈L2*(ℝN)∣∇u∈(L2(ℝN))N} with the norm ∥u∥2=∫ℝN‍|∇u|2dx, and let S be the best Sobolev constant defined by
(5)S=infu∈D1,2(ℝN)∖{0}∫ℝN‍|∇u|2dx(∫ℝN‍|u|2*dx)2/2*S(=infu∈H01(Ω)∖{0}∫Ω‍|∇u|2dx(∫Ω‍|u|2*dx)2/2*)>0,
and let
(6)Sα,β=infu,v∈H01(Ω)∖{0}∫Ω‍(|∇u|2+|∇v|2)dx(∫Ω‍|u|α|v|βdx)2/(α+β);
then, by [1, Theorem 5], we have
(7)Sα,β=((αβ)β/(α+β)+(βα)α/(α+β))S,
where α+β=2*.Set(8)Λ1=(2-q2*-q)2/(2*-2)((2*-q)γ∞2*-2|Ω|(2*-q)/2*)-2/(2-q)×SN/2+q/(2-q)>0,
where γ∞=max{|f|L∞(Ω),|g|L∞(Ω)}.The main results of this paper are given as follows.Theorem 1.
Assume that(H1) holds. If λ,μ>0 satisfy 0<λ2/(2-q)+μ2/(2-q)<Λ1, then there exists at least one positive ground state solution of the problem (Pλ,μ).Theorem 2.
Under the assumptions(H1)-(H2), and N/(N-2)<q<2 and N>4, there exists a positive number Λ*∈(0,Λ1) such that for λ,μ>0 and λ2/(2-q)+μ2/(2-q)<Λ*, the problem (Pλ,μ) has k+1 positive solutions.This paper is organized as follows. In Section2, we consider the Nehari manifold
(9)𝒩λ,μ={(u,v)∈E∖{0}∣〈Iλ,μ′(u,v),(u,v)〉=0},
where
(10)〈Iλ,μ′(u,v),(u,v)〉=∥(u,v)∥E2-∫Ω‍(λf(x)|u|q+μg(x)|v|q)dx-∫Ω‍h(x)|u|α|v|βdx.
Note that 𝒩λ,μ contains all nontrivial weak solution of the problem (Pλ,μ). Using the argument of Tarantello [12, 13], we split 𝒩λ,μ into two parts 𝒩λ,μ+ and 𝒩λ,μ- for 0<λ2/(2-q)+μ2/(2-q)<Λ1. In Section 3, we prove Theorem 1. In Section 4, since Iλ,μ satisfies the (PS)γ-condition for γ∈(-∞,(1/N)(Sα,β)N/2-C0(λ2/(2-q)+μ2/(2-q))), for sufficiently small λ, μ, and some restriction on q and N, we construct the k compact Palais-Smale sequences which are suitably localized in correspondence with the k maximum points of h and which converge to distinct solutions of the problem (Pλ,μ) belonging to 𝒩λ,μ-. Hence, we prove Theorem 2 (one is the ground state solution belonging to 𝒩λ,μ+ and the others are in 𝒩λ,μ-).


## 2. Nehari Manifold

Throughout this paper,(H1) will be assumed. First, we give some notations.Notations. We make use of the following notations.Lp(Ω), 1≤p≤∞, denote Lebesgue spaces; the norm Lp is denoted by |·|Lp(Ω) for 1≤p≤∞.E=[H01(Ω)]2, endowed with norm ∥z∥E2=∥(u,v)∥E2=|∇u|22+|∇v|22.The dual space of a Banach spaceE will be denoted by E-1.|z|=|(u,v)|=(|u|,|v|) and tz=t(u,v)=(tu,tv) for all z∈E and t∈ℝ.z=(u,v) is said to be nonnegative in Ω if u≥0 and v≥0 in Ω.z=(u,v) is said to be positive in Ω if u>0 and v>0 in Ω.|Ω| is the Lebesgue measure of Ω.Br(a)={x∈ℝN∣|x-a|<r} is a ball in ℝN.O(εt) denotes |O(εt)|/εt≤C as ε→0 for t≥0.O1(εt) means that there exist the constants C1,C2>0 such that C1εt≤O1(εt)≤C2εt as ε is small.on(1) denotes on(1)→0 as n→∞.γ∞=max{|f|L∞(Ω),|g|L∞(Ω)}.C, Ci will denote various positive constants, the exact values of which are not important.LetKλ,μ:E→ℝ be the functional defined by
(11)Kλ,μ(z)=∫Ω‍(λf(x)|u|q+μg(x)|v|q)dx∀z=(u,v)∈E.We know thatIλ,μ is not bounded below on E. From the following lemma, we have that Iλ,μ is bounded from below on the Nehari manifold 𝒩λ,μ defined in (9).Lemma 3.
The energy functionalIλ,μ is coercive and bounded below on 𝒩λ,μ.Proof.
Ifz=(u,v)∈𝒩λ,μ, then by (10), the Hölder inequality, and the Sobolev embedding theorem, we get
(12)Iλ,μ(z)=2*-22*2∥z∥E2-2*-q2*qKλ,μ(z)(13)≥1N∥z∥E2-2*-q2*qγ∞S-q/2|Ω|(2*-q)/2*×(λ2/(2-q)+μ2/(2-q))(2-q)/2∥z∥Eq.
Hence, we have that Iλ,μ is coercive and bounded below on 𝒩λ,μ.Define(14)Φλ,μ(z)=〈Iλ,μ′(z),z〉.
Then, for z∈𝒩λ,μ,
(15)〈Φλ,μ′(z),z〉=2∥z∥E2-qKλ,μ(z)-2*∫Ω‍h(x)|u|αvβdx(16)=(2-q)∥z∥E2-(2*-q)∫Ω‍h(x)|u|αvβdx(17)=(2*-q)Kλ,μ(z)-(2*-2)∥z∥E2.
We apply the method in [12]; let
(18)𝒩λ,μ+={z∈𝒩λ,μ∣〈Φλ,μ′(z),z〉>0},𝒩λ,μ0={z∈𝒩λ,μ∣〈Φλ,μ′(z),z〉=0},𝒩λ,μ-={z∈𝒩λ,μ∣〈Φλ,μ′(z),z〉<0}.By using equality (17), we get that Kλ,μ(z)>0 for z∈𝒩λ,μ+. Moreover, we have the following results.Lemma 4.
LetΛ1 be a constant defined as in (8). If 0<λ2/(2-q)+μ2/(2-q)<Λ1, then 𝒩λ,μ0=∅.Proof.
Assuming the contrary, there existλ,μ>0 with 0<λ2/(2-q)+μ2/(2-q)<Λ1 such that 𝒩λ,μ0≠∅. Then, by (16) and (17), for u∈𝒩λ,μ0, we have
(19)∥z∥E2=2*-q2-q∫Ω‍h(x)|u|α|v|βdx=2*-q2*-2Kλ,μ(z).
Using (H1) and both the Hölder and the Sobolev inequalities, we get
(20)∥z∥E≥(2-q2*-qS2*/2)1/(2*-2),(21)∥z∥E≤(2*-q2*-2S-q/2|Ω|(2*-q)/2*γ∞)1/(2-q)×(λ2/(2-q)+μ2/(2-q))1/2.
This implies
(22)λ2/(2-q)+μ2/(2-q)≥(2-q2*-q)2/(2*-2)((2*-q)γ∞2*-2|Ω|(2*-q)/2*)-2/(2-q)×SN/2+q/(2-q)=Λ1,
which is a contradiction.For eachz∈E with ∫Ω‍h(x)|u|α|v|βdx>0, we write
(23)tmax=((2-q)∥z∥E2(2*-q)∫Ω‍h(x)|u|α|v|βdx)1/(2*-2)>0.
Then, the following lemma holds.Lemma 5.
Suppose that0<λ2/(2-q)+μ2/(2-q)<Λ1, and z∈E with ∫Ω‍h(x)|u|α|v|βdx>0. Then, there exist unique 0<t+<tmax<t- such that t+z∈𝒩λ,μ+, t-z∈𝒩λ,μ- and
(24)Iλ,μ(t+z)=inf0≤t≤tmaxIλ,μ(tz);Iλ,μ(t-z)=supt≥0Iλ,μ(tz).Proof.
This is similar to the proof of Hsu [14, Lemma 2.7].Applying Lemma4 (𝒩λ,μ0=∅ for 0<λ2/(2-q)+μ2/(2-q)<Λ1), we write 𝒩λ,μ=𝒩λ,μ+∪𝒩λ,μ- and define
(25)θλ,μ=infz∈𝒩λ,μIλ,μ(z);θλ,μ+=infz∈𝒩λ,μ+Iλ,μ(z);θλ,μ-=infz∈𝒩λ,μ-Iλ,μ(z).The following lemma shows that the minimizers on𝒩λ,μ are usual critical points for Iλ,μ.Lemma 6.
For the case whenλ∈(0,Λ1), if z0 is a local minimizer for Iλ,μ on 𝒩λ,μ, then Iλ,μ′(z0)=0 in E-1.Proof.
See Brown and Zhang [15, theorem 2.3].Lemma 7.
(i) If0<λ2/(2-q)+μ2/(2-q)<Λ1 and z=(u,v)∈𝒩λ,μ+, then one has
(26)Kλ,μ(z)>0,Iλ,μ(z)<0.
In particular, θλ,μ≤θλ,μ+<0.
(ii) If0<λ2/(2-q)+μ2/(2-q)<(q/2)2/(2-q)Λ1 and z=(u,v)∈𝒩λ,μ-, then one has u≢0, v≢0 in Ω,
(27)∥z∥E>(2-q2*-q)1/(2*-2)SN/4,
and θλ,μ->d0 for some positive constant d0=d0(λ,μ,q,N,S,γ∞,|Ω|).Proof.
(i) Letz=(u,v)∈𝒩λ,μ+. By (16) and (17), we have
(28)Kλ,μ(z)>2*-22*-q∥z∥E2>0,2-q2*-q∥z∥E2>∫Ω‍h(x)|u|α|v|βdx.
Then,
(29)Iλ,μ(z)=(12-1q)∥z∥E2+(1q-12*)∫Ω‍h(x)|u|α|v|βdx<[(12-1q)+(1q-12*)2-q2*-q]∥z∥E2=-2-qqN∥z∥E2<0.
By the definition of θλ,μ, θλ,μ+, we deduce that θλ,μ≤θλ,μ+<0.
(ii) Letz∈𝒩λ,μ-; by (16) and the Hölder and the Sobolev inequalities, we get
(30)2-q2*-q∥z∥E2<∫Ω‍h(x)|u|α|v|βdx≤S-2*/2∥z∥E2*.
This implies
(31)∫Ω‍h(x)|u|α|v|βdx>2*-q2-q∥z∥E>(2-q2*-q)12*-2SN4∀z∈𝒩λ,μ-.
By (13) and (31), we obtain that u≢0, v≢0 in Ω, and
(32)Iλ,μ(z)≥∥z∥Eq[1N∥z∥E2-q-(2*-q2*q)γ∞S-q/2|Ω|(2*-q)/2*m×(λ2/(2-q)+μ2/(2-q))(2-q)/2(2*-q2*q)]>(2-q2*-q)q/(2*-2)SqN/4×[1N(2-q2*-q)(2-q)/(2*-2)S(2-q)N/4-(2*-q2*q)γ∞S-q/2|Ω|(2*-q)/2*×(λ2/(2-q)+μ2/(2-q))(2-q)/21N(2-q2*-q)(2-q)/2*-2].
Thus, if 0<λ2/(2-q)+μ2/(2-q)<(q/2)2/(2-q)Λ1, for all z∈𝒩λ,μ-, then
(33)Iλ,μ(z)≥d0(λ,μ,q,N,S,γ∞,|Ω|)>0.


## 3. Existence of a Ground State Solution

First of all, we define the Palais-Smale (denote by (PS)) sequences and (PS)-condition inE for Iλ,μ as follows.Definition 8.
(i) Forγ∈ℝ, a sequence {zn} is a (PS)γ-sequence in E for Iλ,μ if Iλ,μ(zn)=γ+on(1) and Iλ,μ′(zn)=on(1) strongly in E-1 as n→∞.
(ii)Iλ,μ satisfies the (PS)γ-condition in E if any (PS)γ-sequence {zn} in E for Iλ,μ contains a convergent subsequence.Proof of Theorem1.
Using the same argument as in Wu [16, Proposition 9] or Hsu [14, Proposition 3.3], there exists a minimizing sequence {zn} for Iλ,μ on 𝒩λ,μ such that
(34)Iλ,μ(zn)=θλ,μ+on(1),Iλ,μ′(zn)=on(1)inE-1.
Since Iλ,μ is coercive on 𝒩λ,μ (see Lemma 3), we get that {zn} is bounded in E. Then, there exist a subsequence {zn=(un,vn)} and zλ,μ1=(uλ,μ1,vλ,μ1)∈E such that
(35)un⇀uλ,μ1,vn⇀vλ,μ1weaklyinH01(Ω),un⟶uλ,μ1,vn⟶vλ,μ1almosteverywhereinΩ,un⟶uλ,μ1,vn⟶vλ,μ1stronglyinLs(Ω)∀1≤s<2*.
This implies
(36)Kλ,μ(zn)=Kλ,μ(zλ,μ1)+on(1)asn⟶∞.
First, we claim that zλ,μ1 is a nontrivial solution of (Pλ,μ). By (34) and (35), it is easy to verify that zλ,μ1 is a weak solution of (Pλ,μ). From zn∈𝒩λ,μ and (12), we deduce that
(37)Kλ,μ(zn)=q(2*-2)2(2*-q)∥zn∥E2-2*q2*-qIλ,μ(zn).
Let n→∞ in (37); by (34), (36), and θλ,μ<0, we get
(38)Kλ,μ(zλ,μ1)≥-2*q2*-qθλ,μ>0.
Thus, zλ,μ1∈𝒩λ,μ is a nontrivial solution of (Pλ,μ). Now, we prove that zn→zλ,μ1 strongly in E and Iλ,μ(zλ,μ1)=θλ,μ. By (37), if z∈𝒩λ,μ, then
(39)Iλ,μ(z)=1N∥z∥E2-2*-q2*qKλ,μ(z).
In order to prove that Iλ,μ(zλ,μ1)=θλ,μ, it suffices to recall that zλ,μ1∈𝒩λ,μ, by (39) and applying Fatou's lemma to get
(40)θλ,μ≤Iλ,μ(zλ,μ1)=1N∥zλ,μ1∥E2-2*-q2*qKλ,μ(zλ,μ1)≤liminfn→∞(1N∥zn∥E2-2*-q2*qKλ,μ(zn))≤liminfn→∞Iλ,μ(zn)=θλ,μ.
This implies that Iλ,μ(zλ,μ1)=θλ,μ and limn→∞∥zn∥E2=∥zλ,μ1∥E2. Let z~n=zn-zλ,μ1; then Brézis-Lieb lemma [17] implies
(41)∥z~n∥E2=∥zn∥E2-∥zλ,μ1∥E2.
Therefore, zn→zλ,μ1 strongly in E. Since Iλ,μ(zλ,μ1)=Iλ,μ(|zλ,μ1|)=θλ,μ and |zλ,μ1|∈𝒩λ,μ+, by Lemma 6 we may assume that zλ,μ1 is a nontrivial nonnegative solution of (Pλ,μ). By an argument of Hsu [18, Lemma 4.2], we can deduce that uλ,μ1≢0 and vλ,μ1≢0 in Ω. Finally, from the maximum principle [19], we deduce that zλ,μ1 is positive in Ω.Remark 9.
z
λ
,
μ
1
∈
𝒩
λ
,
μ
+ and Iλ,μ(zλ,μ1)=θλ,μ=θλ,μ+.Proof.
We claim thatzλ,μ1∈𝒩λ,μ+. On the contrary, assume that zλ,μ1∈𝒩λ,μ- (𝒩λ,μ0=∅ for λ2/(2-q)+μ2/(2-q)∈(0,Λ1)); then by Lemma 5, there exist unique t1+ and t1- such that t1+zλ,μ1∈𝒩λ,μ+ and t1-zλ,μ1∈𝒩λ,μ-. In particular, we have t1+<t1-=1. Since
(42)ddtIλ,μ(t1+zλ,μ1)=0,d2dt2Iλ,μ(t1+zλ,μ1)>0,
there exists t1+<t-≤t1- such that Iλ,μ(t1+zλ,μ1)<Iλ,μ(t-zλ,μ1). By Lemma 5,
(43)Iλ,μ(t1+zλ,μ1)<Iλ,μ(t-zλ,μ1)≤Iλ,μ(t1-zλ,μ1)=Iλ,μ(zλ,μ1),
which is a contradiction. Hence, zλ,μ1∈𝒩λ,μ+ and Iλ,μ(zλ,μ1)=θλ,μ=θλ,μ+.


## 4. Existence ofk+1 Solutions

Throughout this section,(H1)-(H2) will be assumed. First of all, we want to show that Iλ,μ satisfies the (PS)γ-condition in E for γ∈(-∞,(1/N)(Sα,β)N/2-C0(λ2/(2-q)+μ2/(2-q))), where C0 is defined in the following lemma.Lemma 10.
If{zn}⊂E is a (PS)γ-sequence for Iλ,μ with zn⇀z weakly in E, then Iλ,μ′(z)=0 and there exists a constant C0=C0(q,N,S,γ∞,|Ω|)>0 such that Iλ,μ(z)≥-C0(λ2/(2-q)+μ2/(2-q)).Proof.
Letzn=(un,vn) and z=(u,v). If {zn} is a (PS)γ-sequence for Iλ,μ with zn⇀z weakly in E, it is easy to check that Iλ,μ′(z)=0 in E-1. Then, we get 〈Iλ,μ′(z),z〉=0; that is, ∫Ω‍h(x)|u|α|v|βdx=∥z∥E2-Kλ,μ(z). Thus, by (13), the Hölder, the Young, and the Sobolev inequalities, we have
(44)Iλ,μ(z)≥1N∥z∥E2-2*-q2*qγ∞S-q/2|Ω|(2*-q)/2*×(λ2/(2-q)+μ2/(2-q))(2-q)/2∥z∥Eq≥1N∥z∥E2-1N∥z∥E2-C0(λ2/(2-q)+μ2/(2-q))=-C0(λ2/(2-q)+μ2/(2-q)),
where C0=C0(q,N,S,γ∞,|Ω|)>0.Lemma 11.
If{zn}⊂E is a (PS)γ-sequence for Iλ,μ, then {zn} is bounded in E.Proof.
See Hsu and Lin [8, Lemma 2.3].Recall that(45)Sα,β=infu,v∈H01(Ω)∖{0}∥(u,v)∥E2(∫Ω‍|u|α|v|βdx)2/(α+β),
and let
(46)c*=1N(Sα,β)N/2-C0(λ2/(2-q)+μ2/(2-q)),
where C0>0 is given in Lemma 10.Lemma 12.
I
λ
,
μ satisfies the (PS)γ-condition in E for γ∈(-∞,c*).Proof.
Let{zn}⊂E be a (PS)γ-sequence for Iλ,μ with γ∈(-∞,c*). Write zn=(un,vn). We know from Lemma 11 that {zn} is bounded in E, and then zn⇀z=(u,v) weakly up to a subsequence; z is a critical point of Iλ,μ. Furthermore, we may assume that un⇀u, vn⇀v weakly in H01(Ω) and un→u, vn→v strongly in Ls(Ω) for all 1≤s<2*, and un→u, vn→v a.e. on Ω. Hence, we have that Iλ,μ′(z)=0 and
(47)Kλ,μ(zn)=Kλ,μ(z)+on(1).
Let u~n=un-u,v~n=vn-v and z~n=(u~n,v~n). Then, we obtain
(48)∥z~n∥E2=∥zn∥E2-∥z∥E2+on(1),
and by an argument of Han [20, Lemma 2.1],
(49)∫Ω‍h(x)|u~n|α|v~n|βdx=∫Ω‍h(x)|un|α|vn|βdx-∫Ω‍h(x)|u|α|v|βdx+on(1).
Since Iλ,μ(zn)=γ+on(1), Iλ,μ′(zn)=on(1) in E-1 and (47)–(49), we deduce that
(50)12∥z~n∥E2-12*∫Ω‍h(x)|u~n|α|v~n|βdx=γ-Iλ,μ(z)+on(1),(51)∥z~n∥E2-∫Ω‍h(x)|u~n|α|v~n|βdx=on(1).
Hence, we may assume that
(52)∥z~n∥E2⟶l,∫Ω‍h(x)|u~n|α|v~n|βdx⟶l.
Assume that l≠0; by the definition of Sα,β, |h|L∞(Ω)=1 and (52), we obtain
(53)Sα,βl2/2*=Sα,βlimn→∞(∫Ω‍h(x)|u~n|α|v~n|βdx)2/2*≤|h|L∞(Ω)2/2*limn→∞∥z~n∥2=l,
which implies that l≥(Sα,β)N/2. In addition, from Lemma 10, (50), and (52), we get
(54)γ=(12-12*)l+Iλ,μ(z)≥1N(Sα,β)N/2-C0(λ2/(2-q)+μ2/(2-q)),
which is a contradiction. Hence, l=0; that is, zn→z strongly in E.From assumption(H2), we can choose r0∈(0,1) such that
(55)Br0(ai)¯∩Br0(aj)¯=∅fori≠j,1≤i,j≤k,
and ∪i=1kBr0(ai)¯⊂Ω, where Br0(ai)¯={x∈ℝN∣|x-ai|≤r0} and h(ai)=|h|∞=1 for 1≤i≤k.Define(56)Qi(z)=∫Ω‍ψi(x)(|∇u|2+|∇v|2)dx∫Ω‍(|∇u|2+|∇v|2)dx,z=(u,v)∈E∖{0},
where ψi(x)=min{1,|x-ai|}, i=1,2…,k.Then, we have the following separation result.Lemma 13.
IfQi(z)≤r0/3 and Qj(z)≤r0/3 for z∈E∖{0}, then i=j.Proof.
For anyz∈E∖{0} satisfying Qi(z)≤r0/3(1≤i≤k), we get
(57)r03∥z∥E2≥∫Ω‍ψi(x)(|∇u|2+|∇v|2)dx≥∫Ω∖Br0(ai)‍ψi(x)(|∇u|2+|∇v|2)dx≥r0∫Ω∖Br0(ai)‍(|∇u|2+|∇v|2)dx,
which implies that
(58)∥z∥E2≥3∫Ω∖Br0(ai)‍(|∇u|2+|∇v|2)dx,1≤i≤k.
Hence, from (58), we obtain
(59)2∥z∥E2≥3(∫Ω∖Br0(ai)‍(|∇u|2+|∇v|2)dx+∫Ω∖Br0(aj)‍(|∇u|2+|∇v|2)dx)≥3∥z∥E2ifi≠j,
which is a contradiction.Fori=1,2,…,k, we set
(60)𝒩λ,μi={u∈𝒩λ,μ-∣Qi(z)<r03},∂𝒩λ,μi={u∈𝒩λ,μ-∣Qi(z)=r03},
and define
(61)θλ,μi=inf𝒩λ,μiIλ,μ(z),θ~λ,μi=inf∂𝒩λ,μiIλ,μ(z).Recall that the best Sobolev constantS is defined as
(62)S=infu∈D1,2(ℝN)∖{0}∫ℝN‍|∇u|2dx(∫ℝN‍|u|2*dx)2/2*.
It is well known that
(63)U(x)=[N(N-2)](N-2)/4[1+|x|2](N-2)/2
is a minimizer of S, and |∇U|L2(ℝN)2=|U|L2*(ℝN)2*=SN/2. Fix a maximum point ai of h (1≤i≤k). Let ηi∈C0∞(Ω) be a cut-off function such that 0≤ηi≤1, |∇ηi|≤C, and ηi(x)=1 for |x-ai|<r0/2, ηi(x)=0 for |x-ai|>r0. We define
(64)uεi(x)=ε(2-N)/2ηi(x)U(x-aiε)=c1ε(N-2)/2ηi(x)[ε2+|x-ai|2](N-2)/2,
where c1=[N(N-2)](N-2)/4 and ε>0.From now on, we assume thatN/(N-2)<q<2 and N>4.Lemma 14.
There existε0>0, Λ2∈(0,(q/2)2/(2-q)Λ1), such that for ε∈(0,ε0) and λ2/(2-q)+μ2/(2-q)∈(0,Λ2), one has
(65)supt≥0Iλ,μ(tαuεi,tβuεi)<c*uniformlyini,
where c* is the positive constant given in Lemma 12.
In particular,0<θλ,μ-≤θλ,μi<c* for all 1≤i≤k.Proof.
It is well known that (or see Brézis and Nirenberg [21], Cheng and Ma [22, Lemma 3.2], Struwe [23], and Willem [24, Lemma 1.46]) as ε→0+,
(66)|uεi|L2*(Ω)2=|U|L2*(ℝN)2+O(εN-2),(67)|∇uεi|L2(Ω)2|∇U|L2(ℝN)2+O(εN-2).
For N/(N-2)<q<2, N>4 and ε<r0/2,
(68)|uεi|Lq(Ω)q=∫Br0/2(ai)‍[ε(2-N)/2U(x-aiε)]qdx+O(εN-2)≥Cεθ+O(εN-2),whereθ=N-(N-2)q2.
Setz¯εi=(αuεi,βuεi). By Lemma 5, there exists tεi>0 such that zεi=tεiz¯εi∈𝒩λ,μ- for 0<λ2/(2-q)+μ2/(2-q)<Λ1. Furthermore,
(69)Qi(zεi)=∫Ω‍ψi(x)|∇uεi|2dx∫Ω‍|∇uεi|2dx=∫(Ω-ai)/ε‍ψi(ai+εy)|∇(ηi(ai+εy)U(y))|2dy∫(Ω-ai)/ε‍|∇(ηi(ai+εy)U(y))|2dy⟶ψi(ai)=0asε⟶0.
Hence, there exists ε¯0>0 for any
(70)ε∈(0,ε¯0),Qi(zεi)<r03,
which implies
(71)zεi=tεiz¯εi∈𝒩λ,μiforε∈(0,ε¯0),
and then
(72)θλ,μ-≤θλ,μi≤Iλ,μ(zεi)≤supt≥0Iλ,μ(ttεiz¯εi)=supt≥0Iλ,μ(tz¯εi).
First, we consider the functional I0,0:E→ℝ defined by
(73)I0,0(u,v)=12∥(u,v)∥E2-12*∫Ω‍h(x)|u|α|v|βdx.
Step I. Show that supt≥0I0,0(z¯εi)≤(1/N)(Sα,β)N/2+O(εN-2).
According to condition(H2), we conclude that
(74)|∫Ω‍h(x)|uεi(x)|2*dx-∫Ω‍h(ai)|uεi(x)|2*dx|≤∫Ω‍|h(x)-h(ai)||uεi(x)|2*dx=O(∫Br0(ai)‍|x-ai|σ|uεi(x)|2*dx)=O(εσ).
From (66), (74), h(ai)=1, and σ≥N-2, we can deduce that
(75)(∫Ω‍h(x)|uεi(x)|2*dx)2/2^*=(|uεi|L2*(Ω)2*+O(εσ))2/2^*=|uεi|L2*(Ω)2+O(εσ)=|U|L2*(ℝN)2+O(εN-2).
Using (67) and (75), then
(76)|∇uεi|L2(Ω)2(∫Ω‍h(x)|uεi(x)|2*dx)2/2^*=|∇U|L2(ℝN)2+O(εN-2)|U|L2*(ℝN)2+O(εN-2)=S+O(εN-2).
Since
(77)supt≥0(A2t2-B2*t2*)=1N(AB2/2*)N/2,foranyA>0,B>0,
by (7) and (76), we conclude that
(78)supt≥0I0,0(tz¯εi)=1N((α+β)|∇uεi|L2(Ω)2(αα/2ββ/2∫Ω‍h(x)|uεi(x)|2*dx)2/2*)N/2≤1N(Sα,β)N/2+O(εN-2).
Step II. Let C0 be the positive constant given in Lemma 10. We can choose δ1>0 such that for all 0<λ2/(2-q)+μ2/(2-q)<δ1, we have
(79)c*=1N(Sα,β)N/2-C0(λ2/(2-q)+μ2/(2-q))>0.
Since Iλ,μ is continuous in E, Iλ,μ(0)=0, and {z¯εi} is uniformly bounded in E for any 0<ε<min{ε¯0,r0/2} (see (67)), then there exists t0>0 (independent of ε) such that for any 0<ε<min{ε¯0,r0/2},
(80)sup0≤t≤t0Iλ,μ(tz¯εi)<c*,uniformlyini,mlylinni,∀0<λ2/(2-q)+μ2/(2-q)<δ1.
According to condition (H1), fmin=minx∈Ω¯f(x)>0 and gmin=minx∈Ω¯g(x)>0. Applying the results of Step I and (68), we have that for N/(N-2)<q<2 and N>4,
(81)supt≥t0Iλ,μ(tz¯εi)=supt≥t0(I0,0(tz¯εi)-tqqKλ,μ(tz¯εi))≤1N(Sα,β)N/2+O(εN-2)-t0qqm(λ+μ)×∫Br0/2(ai)‍|uεi|qdx≤1N(Sα,β)N/2+O(εN-2)-(λ+μ)O1(εθ),
where m=min{αq/2fmin,βq/2gmin} and θ=N-((N-2)q)/2.
Therefore, we can chooseλ=O1(ετ1) and μ=O1(ετ2) such that
(82)2-qqθ<τ1,τ2<(N-2)-θ.
This implies that(83)min{τ1,τ2}+θ<22-qmin(τ1,τ2),min{τ1,τ2}+θ<N-2,(λ+μ)O1(εθ)=O1(εmin{τ1,τ2}+θ),λ2/(2-q)+μ2/(2-q)=O1(ε2/(2-q)min{τ1,τ2}).
There exist δ2>0, ε0∈(0,min{ε¯0,r0/2}) such that for all 0<λ2/(2-q)+μ2/(2-q)<δ2 and 0<ε<ε0, we have
(84)O(εN-2)-(λ+μ)O1(εθ)<-C0(λ2/(2-q)+μ2/(2-q)).
Thus, we can choose Λ2=min{(q/2)2/(2-q)Λ1,δ1,δ2}>0. Then, for all λ2/(2-q)+μ2/(2-q)∈(0,Λ2), there holds
(85)supt≥0Iλ,μ(tz¯εi)<c*uniformlyini.
Step III. For 0<λ2/(2-q)+μ2/(2-q)<Λ2 and 0<ε<ε0, by Lemma 7, (72), and (85), we get
(86)0<θλ,μ-≤θλ,μi≤Iλ,μ(tz¯εi)<c*∀1≤i≤k.To proceed, we need to quote the concentration-compactness principle (see [24, 25]) about the case of systems.Lemma 15.
Let{un,vn}⊂H01(Ω)×H01(Ω) be a sequence such that
(87)un⇀u,vn⇀vweaklyinH01(Ω);un⟶u,vn⟶va.e.onΩ,l|∇(un-u)|2+|∇(vn-v)|2⇀μ~weaklyinthesenseofmeasures,measures.|un-u|α|vn-v|β⇀ν~weaklyinthesenseofmeasures.
Then, it follows that
(88)limsupn→∞∫ℝN‍(|∇un|2+|∇v|n2)dx=∫ℝN‍(|∇u|2+|∇v|2)dx+∥μ~∥,limsupn→∞∫ℝN‍|un|α|vn|βdx=∫ℝN‍|u|α|v|βdx+∥ν~∥,∥ν~∥2/(α+β)≤Sα,β-1∥μ~∥.
Moreover, if u≡v≡0 and ∥ν~∥2/(α+β)=Sα,β-1∥μ~∥, then μ~ and ν~ concentrate at a single point.Proof.
See Han [20, Lemma 2.2].Lemma 16.
For anyi∈{1,2,…,k}, there exist Λ~i>0 such that
(89)θ~λ,μi>1N(Sα,β)N/2∀0<λ2/(2-q)+μ2/(2-q)<Λ~i.Proof.
Fixi∈{1,2,…,k}. Assume the contrary. There then exists a sequence {(λn,μn)} with (λn,μn)→(0,0) as n→∞ such that θ~λn,μni→c≤(1/N)(Sα,β)N/2 as n→∞. Consequently, there exists a sequence {zn=(un,vn)}⊂∂𝒩λn,μni such that as n→∞,
(90)∫Ω‍(|∇un|2+|∇vn|2)dx=∫Ω‍(λnf(x)|un|q+μng(x)|vn|q)dx+∫Ω‍h(x)|un|α|vn|βdx,(91)Iλn,μn(zn)⟶c≤1N(Sα,β)N/2asn⟶∞.
It then follows easily that {zn} is uniformly bounded in E, and since f and g are continuous on Ω¯, we obtain
(92)Kλn,μn(zn)=∫Ω‍(λnf(x)|un|q+μng(x)|vn|q)dx=on(1)asn⟶∞.
From (90), and by the Hölder and the Sobolev inequalities, we can fix m0>0 such that
(93)∫Ω‍(|∇un|2+|∇vn|2)dx≥m0,∫Ω‍h(x)|un|α|vn|βdx≥m0.
Thus, up to a subsequence, we infer that
(94)limn→∞∫Ω‍(|∇un|2+|∇vn|2)dx=limn→∞∫Ω‍h(x)|un|α|vn|βdx=l>0.
Furthermore, by |h|L∞(Ω)=1, we deduce
(95)l=limn→∞∫Ω‍h(x)|un|α|vn|βdx≤|h|L∞(Ω)limn→∞∫Ω‍|un|α|vn|βdx≤Sα,β-2*/2limn→∞(∫Ω‍(|∇un|2+|∇vn|2)dx)2*/2≤Sα,β-2*/2l2*/2,
which implies
(96)l≥(Sα,β)N/2.
On the other hand, we have, as n→∞,
(97)1Nl=12∥zn∥2dx-12*∫Ω‍h(x)|un|α|vn|βdx-1qKλn,μn(zn)+on(1)=Iλn,μn(zn)+on(1)≤1N(Sα,β)N/2.
Hence, together with (96), we get
(98)l=(Sα,β)N/2,
and then from (95), we also have
(99)limn→∞∫Ω‍h(x)|un|α|vn|βdx=limn→∞∫Ω‍|un|α|vn|βdx=l.
Therefore,
(100)limn→∞∫Ω‍|un|α|vn|βdx=l.
Set z~n=(u~n,v~n)=zn/∥zn∥; then, we have ∥z~n∥=1. Moreover, by (94),(98), and (100), we get
(101)limn→∞∫Ω‍|u~n|α|v~n|βdx=limn→∞∫Ω‍|un|α|vn|βdx∥zn∥2*=Sα,β-N/(N-2).
Thus, up to a subsequence, we may assume that
(102)u~n⇀u,v~n⇀vweaklyinH01(Ω);u~n⟶u,v~n⟶va.e.onΩ,|∇(u~n-u)|2+|∇(v~n-v)|2⇀μ~weaklyinthesenseofmeasures,measures.|u~n-u|α|v~n-v|β⇀ν~weaklyinthesenseofmeasures.
Since Ω is bounded, from (101) and Lemma 15, we deduce that
(103)1=∫Ω‍(|∇u|2+|∇v|2)dx+∥μ~∥,(104)Sα,β-N/(N-2)=∫Ω‍|u|α|v|βdx+∥ν~∥,(105)∥ν~∥2/(α+β)≤Sα,β-1∥μ~∥.
If ∫Ω‍(|∇u|2+|∇v|2)dx≠0 and ∥μ~∥≠0, we deduce that
(106)1=(∫Ω‍(|∇u|2+|∇v|2)dx+∥μ~∥)(α+β)/2>(∫Ω‍(|∇u|2+|∇v|2)dx)(α+β)/2+∥μ~∥(α+β)/2≥Sα,β(α+β)/2∫Ω‍|u|α|v|βdx+Sα,β(α+β)/2∥ν~∥=Sα,β(α+β)/2·Sα,β-N/(N-2)=1,
which is a contradiction.
Thus,∫Ω‍(|∇u|2+|∇v|2)dx=0 or ∥μ~∥=0. If ∥μ~∥=0, from (103)–(105), we get ∫Ω‍(|∇u|2+|∇v|2)dx=1 and ∫Ω‍|u|α|v|βdx=Sα,β-N/(N-2). Then,
(107)∫Ω‍(|∇u|2+|∇v|2)dx(∫Ω‍|u|α|v|βdx)2/(α+β)=Sα,β,
which means that Sα,β is achieved by (u,v). It is impossible since Sα,β cannot be achieved on any bounded domain Ω. Hence,
(108)∫Ω‍(|∇u|2+|∇v|2)dx=0,∥μ~∥=1.
Then, u≡v≡0 on Ω, and from (103), (104), we easily have ∥ν~∥2/(α+β)=Sα,β-1=Sα,β-1∥μ~∥. By Lemma 15, we conclude that x0∈Ω¯ such that
(109)ofmeasures,|∇u~n|2+|∇v~|2⇀δx0weaklyinthesenseofmeasures,resress..|u~n|α|v~n|β⇀Sα,β-N/(N-2)δx0weaklyinthesenseofmeasures.
Observe that Qi(z~n)=Qi(zn)=r0/3;
(110)r03=limn→∞Qi(zn)=limn→∞∫Ω‍ψi(x)(|∇u~n|2+|∇v~n|2)dx∫Ω‍(|∇u~n|2+|∇v~n|2)dx=ψi(x0),
which implies that x0≠ai by the definition of ψi(x). On the other hand, from (95) and (101), we get
(111)Sα,β-N/(N-2)h(x0)=limn→∞∫Ω‍h(x)|u~n|α|v~n|βdx=limn→∞∫Ω‍h(x)|un|α|vn|βdx∥zn∥2*=limn→∞∫Ω‍|un|α|vn|βdx∥zn∥2*=limn→∞∫Ω‍|u~n|α|v~n|βdx=Sα,β-N/(N-2),
which is impossible, because h(x) is not a constant function by condition (H2).Throughout this section, takeΛ*=min{Λ2,min1≤i≤kΛ~i}; Λ2 and Λ~i are as in Lemmas 14 and 16. Using the idea of Tarantello [12], we have the following results. For z=(u,v), φ=(φ1,φ2)∈E, we define
(112)z-φ=(u-φ1,v-φ2),〈z,φ〉=∫Ω‍(∇u∇φ1+∇v∇φ2)dx,Gλ,μ(z,φ)=∫Ω‍(λf(x)|u|q-2uφ1+μg(x)|v|q-2vφ2)dx,H(z,φ)=αα+β∫Ω‍h(x)|u|α-2u|v|βφ1dx+βα+β∫Ω‍h(x)|u|α|v|β-2vφ2dx.Lemma 17.
For each0<λ2/(2-q)+μ2/(2-q)<Λ* and z=(u,v)∈𝒩λ,μi(1≤i≤k), there exist ϵ>0 and a differentiable function ξ:Bϵ(0)⊂E→ℝ+ such that ξ(0)=1, ξ(ϕ)(z-ϕ)∈𝒩λ,μi for all ϕ∈Bϵ(0) and
(113)〈ξ′(0),φ〉=2〈z,φ〉-qGλ,μ(z,φ)-2*H(z,φ)(2-q)∥z∥E2-(2*-q)H(z,z),
for all φ=(φ1,φ2)∈E.Proof.
Forz∈𝒩λ,μi, define a function Fz:ℝ×E→ℝ by
(114)Fz(ξ,ϕ)=〈Iλ,μ′(ξ(z-ϕ)),ξ(z-ϕ)〉=ξ2∥z-ϕ∥2-ξqGλ,μ(z-ϕ,z-ϕ)-ξα+βH(z-ϕ,z-ϕ).
Then, Fu(1,0)=〈Iλ,μ′(z),z〉=0 and
(115)ddξFz(1,0)=2∥z∥E2-qGλ,μ(z,z)-(α+β)H(z,z)=(2-q)∥z∥E2-(2*-q)H(z,z)<0.
According to the implicit function theorem, there exist ϵ>0 and a differentiable function ξ:Bϵ(0)⊂E→ℝ such that ξ(0)=1;
(116)〈ξ′(0),φ〉=2〈z,φ〉-qGλ,μ(z,φ)-2*H(z,φ)(2-q)∥z∥E2-(2*-q)H(z,z),Fz(ξ(φ),φ)=0∀φ∈Bϵ(0),
which is equivalent to
(117)〈Iλ,μ′(ξ(φ)(z-φ)),ξ(φ)(z-φ)〉=0∀φ∈B(0;ϵ);
that is, ξ(φ)(z-φ)∈𝒩λ,μ for all φ∈Bϵ(0). Furthermore, by the continuity of the functions ξ and Qi, we have that
(118)(2-q)∥ξ(φ)(z-φ)∥2-(2*-q)H(ξ(φ)(z-φ),ξ(φ)(z-φ))<0,Qi(ξ(φ)(z-φ))<r03
still holds if ϵ is sufficiently small. This implies that ξ(φ)(z-φ)∈𝒩λ,μi.Proposition 18.
If0<λ2/(2-q)+μ2/(2-q)<Λ*, then there exists a (PS)θλ,μi-sequence {zni}⊂𝒩λ,μi in E for Iλ,μ.Proof.
If𝒩λ,μi¯ denotes the closure of 𝒩λ,μi, at first we note that 𝒩λ,μi¯=𝒩λ,μi∪∂𝒩λ,μi for all i=1,2,…,k. It then follows from Lemmas 14 and 16, that
(119)θλ,μi<θ~λ,μifori=1,2,…,k,0<λ2/(2-q)+μ2/(2-q)<Λ*.
Hence,
(120)θλ,μi=inf{Iλ,μ(z)∣z∈𝒩λ,μi¯}fori=1,2,…,k.
Now, we fix i∈{1,2,…,k}. Applying the Ekeland variational principle [26], there exists a minimizing sequence {zni}⊂𝒩λ,μi¯ such that
(121)Iλ,μ(zni)<θλ,μi+1n,Iλ,μ(zni)≤Iλ,μ(φ)+1n∥φ-zni∥Eforeachφ∈𝒩λ,μi¯.
Using (119), we may assume that zni∈𝒩λ,μi for n sufficiently large. Applying Lemma 17 with z=zni, we obtain the function ξn:Bϵn(0)→ℝ for some ϵn>0 such that ξn(φ)(zni-φ)∈𝒩λ,μi for all φ∈Bϵn(0). Let 0<δ<ϵn and z∈E∖{0}; we set
(122)φδ=δz∥z∥E
and zδ=ξn(φδ)(zni-φδ). Since zδ∈𝒩λ,μi, we deduce from (121) that
(123)Iλ,μ(zδ)-Iλ,μ(zni)≥-1n∥zδ-zni∥E.
By the mean-value theorem, we obtain
(124)〈Iλ,μ′(zni),(zδ-zni)〉+o(∥zδ-zni∥E)≥-1n∥zδ-zni∥E.
Therefore,
(125)〈Iλ,μ′(zni),-φδ〉+(ξn(φδ)-1)〈Iλ,μ′(zni),(zni-φδ)〉≥-1n∥zδ-zni∥E+o(∥zδ-zni∥E).
Now, we observe that ξn(φδ)(zni-φδ)∈𝒩λ,μi, and consequently we get from (125) that
(126)-δ〈Iλ,μ′(zni),z∥z∥E〉+(ξn(φδ)-1)ξn(φδ)〈Iλ,μ′(zδ),ξn(φδ)(zni-φδ)〉+(ξn(φδ)-1)〈Iλ,μ′(zni)-Iλ,μ′(zδ),(zni-φδ)〉≥-1n∥zδ-zni∥E+o(∥zδ-zni∥E).
Then, we write the pervious inequality in the following form:
(127)〈Iλ,μ′(zni),z∥z∥E〉≤∥zδ-zni∥Eδn+o(∥zδ-zni∥E)δ+(ξn(φδ)-1)δ〈Iλ,μ′(zni)-Iλ,μ′(zδ),(zni-φδ)〉.
We can find a constant C>0 independent of δ such that
(128)∥zδ-zni∥≤δ+C(|ξn(φδ)-1|),limδ→0|ξn(φδ)-1|δ≤∥ξn′(0)∥≤C.
For a fixed n, let δ→0 in (127). Using the fact that
(129)limδ→0∥zδ-zni∥E=0,
we obtain
(130)〈Iλ,μ′(zni),z∥z∥E〉≤Cn.
This implies
(131)Iλ,μ(zni)=θλ,μi+on(1),Iλ,μ′(zni)=on(1)inE-1.Now, we complete the proof of Theorem2. By Lemmas 12, 14 and Proposition 18, for all λ2/(2-q)+μ2/(2-q)∈(0,Λ*), there exists a sequence {zni}⊂𝒩λ,μi and z0i=(u0i,v0i)∈E, 1≤i≤k, such that
(132)Iλ,μ(zni)=θλ,μi+on(1),Iλ,μ′(zni)=on(1)inE-1,zni⟶z0istronglyinE.
Moreover, {zni}⊂𝒩λ,μ-, and by Lemma 7 (ii), we get z0i∈𝒩λ,μ-, u0i≢0, v0i≢0 in Ω,
(133)∥z0i∥E>(2-q2*-q)1/(2*-2)SN/4,θλ,μi≥θλ,μ->0fori=1,2,…,k.
Thus, z0i is a nontrivial solution of the problem (Pλ,μ) and Iλ,μ(z0i)=θλ,μi for i=1,2…,k. Set u+=max{u,0} and v+=max{v,0}. Replace the terms ∫Ω‍h(x)|u|α|v|βdx and ∫Ω‍(λf(x)|u|q+μg(x)|v|q)dx of the functional Iλ,μ by ∫Ω‍h(x)u+αv+βdx and ∫Ω‍(λf(x)u+q+μg(x)v+q)dx, respectively. It then follows that z0i is a nonnegative solution of the problem (Pλ,μ). Applying the maximum principle [19], z0i is a positive solution of the problem (Pλ,μ). Since Qi(z0i)<r0/3,
(134)zλ,μ1∈𝒩λ,μ+,z0i∈𝒩λ,μi⊂𝒩λ,μ-fori=1,2,…,k,
where zλ,μ1 is a positive solution of equation (Pλ,μ) as in Theorem 1. From Lemma 13, we conclude that 𝒩λ,μi are disjoint for i=1,2…,k. This implies that z0i(1≤i≤k) and zλ,μ1 are distinct positive solutions of the problem (Pλ,μ).



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*Source: 746380-2013-04-11.xml*