# Multiplicity Results for Weak Solutions of a Semilinear Dirichlet Elliptic Problem with a Parametric Nonlinearity

Authors: Ayékotan Messan Joseph Tchalla; Kokou Tcharie
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

This paper deals with the existence of weak solutions to a Dirichlet problem for a semilinear elliptic equation involving the difference of two main nonlinearities functions that depends on a real parameterλ. According to the values of λ, we give both nonexistence and multiplicity results by using variational methods. In particular, we first exhibit a critical positive value such that the problem admits at least a nontrivial non-negative weak solution if and only if λ is greater than or equal to this critical value. Furthermore, for λ greater than a second critical positive value, we show the existence of two independent nontrivial non-negative weak solutions to the problem.


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## Body

## 1. Introduction

In the last years, most works studied the existence, nonexistence, and multiplicity of nontrivial weak solutions of a semilinear Dirichlet problem of the form as follows:(1)−Δu=fλx,u,x∈Ω,u=0,x∈∂Ω,where Ω is a bounded domain in ℝN, λ is a real parameter, and fλ:Ω×ℝ⟶ℝ is a nonlinear function taking different forms. According to the values of λ, Ambrosetti et al. studied in [1], the existence and multiplicity of non-negative weak solutions of the problem (1) when fλx,u=λuq+up with 0<q<1<p. For example, by using variational method, they show the existence of infinitely many solutions of the problem as follows:(2)−Δu=λuq−1u+up−1u,x∈Ω,u=0,x∈∂Ω,for λ>0 and small. Later, Alama and Tarantello in [2] studied the semilinear Dirichlet problem (1) by searching non-negative solutions with(3)fλx,u=λu+ωxuq−1−hxur−1,where λ∈ℝ,Ω⊂ℝNN≥3 is a bounded domain with smooth boundary, ω and h are suitable functions, and 1<q<r. In this case also, the authors show the influence of values of λ on the existence and multiplicity of weak solutions of the problem. These different studies on nonexistence, existence, and multiplicity results for nontrivial weak solutions depending on a parameter for a Dirichlet problem for a semilinear elliptic equation were extensively investigated in the literature (see, for e.g., [3–6] and the references therein). Similar results, depending on a real parameter, are obtained in the case of quasilinear elliptic equations in bounded domains or in entire space ℝN. For example, we can mention the papers [7–9], which are devoted to the unbounded case. In [7], the authors deal with the nonexistence and existence of nontrivial weak solutions of the quasilinear problem:(4)−divax∇up−2∇u+uq−2u=λgxur−2uinΩ,ax∇up−2∂νu+bx∇up−2u=0,on∂Ω,where Ω is a smooth exterior domain in ℝN, ν is the unit vector of the outward normal on ∂Ω, 1<p<N,p<r<q<p∗=Np/N−p or p<q<r<p∗, and a,b, and g are suitable functions. They showed in different cases that the existence of weak solutions of the problem depends on the values of λ relative to the value of some critical value.In [8], Pucci and Radulescu studied the following problem in whole space:(5)−div∇up−2∇u+up−2u=λuq−2u−hxur−2uinℝN,u≥0,inℝN,where h>0 satisfies(6)0<∫ℝNhxq/q−r<∞,λ>0 is a parameter and 2≤p<q<minr,p∗ with p∗=Np/N−p if N>p and p∗=∞ if N≤p. They obtained that the nonexistence and multiplicity of nontrivial weak solutions of this quasilinear elliptic equation are corresponding to the smallness and the largeness of λ, respectively. In [10], Autuori and Pucci extended the results in [8] by solving a more general quasilinear elliptic equation with the same variational method. Motivated by these previous results, we are concerned in this paper with the existence, nonexistence, and multiplicity of nontrivial weak solutions of the following Dirichlet problem for a semilinear elliptic equation:(7)−Δu+u=λauu−buuinΩ,u=0on∂Ω,ελ,where Ω⊂ℝN, N≥3, is a bounded domain with smooth boundary, a, b are suitable non-negative functions, and λ is a real parameter. By taking inspiration on the method developed in [8, 10], we use variational arguments to study the existence and the multiplicity of nontrivial weak solutions of problem ελ according to the values of the parameter λ. To obtain our results in this work, we require inελ the following assumptions:A1a is a function continuous on ℝ+ and of C10,+∞,ℝ+ such that(8)at>0,att′>0for allt>0;,(9)at=0⟺t=0,(10)limt⟶+∞at=+∞.Let us set(11)At=∫0tsasdsfort≥0.A2 There exist p,q∈2,N such that p≤q<2∗=2N/N−2 and(12)p−2≤a′ttat≤q−2,for allt>0.ℬ1b is a function continuous on ℝ+ and of C10,+∞,ℝ+ such that(13)bt>0,btt′>0for allt>0,(14)bt=0⟺t=0,(15)limt⟶+∞bt=+∞.Let us set(16)Bt=∫0tsbsdsfort≥0.ℬ2 There exists r∈2,N such that q<r and(17)q−2<b′ttbt≤r−2,for allt>0.Some examples of functionsa and b in ελsatisfying the previous assumptions A1, A2ℬ1, and ℬ2:(1)
Foru≥0, we can have au=up−2 and bu=ur−2 in (ελ) with 2<p<r<N<2∗.(2)
Another example of functionsa and b is the following:(18)au=uα−2ln1+u,foru>0,0,foru=0,bu=uβ−2ln1+u,foru>0,0,foru=0,with3<α<α+1<β<N<2∗.In this case, for all t > 0, we have(19)α−3≤a′ttat≤α−2,α−2<β−3≤b′ttbt≤β−2.(3)
Letσ fixed in 1,+∞. For all u≥0,(20)au=uα−2lnσ+u,bu=uβ−2lnσ+u,with2+σσ+t0<α<α+σσ+t0<β<N<2∗,wheret0is the unique solution of the equationσlnσ+t−t=0on0,+∞. Hence, for all t > 0, we have(21)α−2−σσ+t0≤a′ttat≤α−2,α−2<β−2−σσ+t0≤b′ttbt≤β−2.Thus, it is clear that the functionsa and b of the Dirichlet problem ελ of our present work generalize the functions uq−2 and ur−22<q<r which appear in the main equation studied in [7, 8] or [10].The main goal of this paper is the proof of the following two theorems:Theorem 1.
By the fulfillment of assumptionsA1, A2ℬ1, and ℬ2, there exists a critical value λ^>0 such that the Dirichlet problem ελ admits at least a nontrivial non-negative weak solution if and only if λ≥λ^.Theorem 2.
Suppose that the assumptionsA1, A2ℬ1, and ℬ2 are fulfilled. Then, there exists a critical value λ¯ satisfying λ¯≥λ^ such that for all λ>λ¯, the Dirichlet problem ελ admits at least two nontrivial non-negative weak solutions.In Section2, we talk about Orlicz spaces, which we will use in our work. In Section 3, we give different imbeddings between the working spaces of this paper and prove the nonexistence of a nontrivial weak solution when λ in ελ is least than a positive number. The conditions for existence of weak solutions of ελ are established in Section 4. Section 5 has devoted to prove Theorem 1, and Section 6 deals with the proof of Theorem 2.


## 2. Notions on Orlicz Spaces (See Chapter 8 in [11])

Definition 1.
(definition of aN-function). Let ψ be a real-valued function defined on 0,∞ and having the following properties:(a)
ψ0=0,ψt>0ift>0,limt⟶+∞=+∞(b)
ψ is nondecreasing, that is, s>t implies ψs≥ψt(c)
ψ is right continuous, that is, if t≥0, then lims⟶t+ψs=ψt
Then, the real-valued functionΨ defined on 0,+∞ by(22)Ψt=∫0tψsds,is called an N-function.
Any such N-functionΨ has the following properties:(i)
Ψ is continuous on 0,∞(ii)
Ψ is strictly increasing(iii)
Ψ is convex(iv)
limt⟶0Ψt/t=0 and limt⟶+∞Ψt/t=+∞(iv)
The functiont⟼Ψt/t is strictly increasing on 0,∞
For any N-functionΨ=Ψt and an open set Ω⊂ℝN, the Orlicz space LΨΩ is defined. When Ψ satisfies Δ2-condition, i.e.,(23)Ψ2t≤kΨt,for allt≥0,for some constant k>0, then(24)LΨΩ=u:Ω⟶ℝmeasurable:∫ΩΨuxdx<∞.
Endowed with the norm(25)uΨ=infk>0:∫ΩΨuxkdx≤1,which is called the Luxembourg norm, the Orlicz space LΨΩ is a Banach space. It is known that if ∫ΩΨux/k0dx=1, then uΨ=k0 with k0>0.
The complement ofΨ is given by the Legendre transformation as follows:(26)Ψ˜s=maxt≥0st−Ψtfors≥0.
We say thatΨ and Ψ˜ are complementary N-functions of each other.
For allt>0, we have the inequality Ψ˜Ψt/t≤Ψt.
From Young’s inequality(27)st≤Ψt+Ψ˜s,a generalized version of Hölder’s inequality is obtained as follows:(28)∫Ωuxvxdx≤2uΨvΨ˜foru∈LΨΩ,v∈LΨ˜Ω.


## 3. Preliminaries and Nonexistence of Nontrivial Solution forλ Small

By conditionA1 and the definition of A and by condition ℬ1 and the definition of B, the functions A and B are N-functions, with ψs=sas or ψs=sbs, respectively.Lemma 1.
TheN-function A satisfies the Δ2-condition.Proof.
In fact, by (12),(29)for allt>0,p≤att2At≤q.
Thus, by Theorem 4.1 in [12], it follows that A satisfies the Δ2-condition.Lemma 2.
TheN-function B satisfies the Δ2-condition.Proof.
In fact, by (17)(30)for allt>0,q<btt2Bt≤r,and B satisfies the Δ2-condition by Theorem 4.1 in [12].
Let us consider the Sobolev spaceH01Ω, which is the completion of C0∞Ω with the norm(31)uH01Ω=∫Ω∇u21/2=∇u2.
LetX denote the completion of C0∞Ω with respect to the norm(32)uX=uH01Ω2+uB21/2,where(33)uB=infλ>0:∫ΩBuxλdx≤1,is the Luxembourg norm in the Orlicz space LBΩ. The space X is the space in which we will find our nontrivial weak solutions.Lemma 3.
The embeddingsX⟶H01Ω⟶L2∗Ω are continuous with uH01Ω≤uX,for allu∈X and u2∗≤C2∗uH01Ω,for allu∈H01Ω.Moreover,H01Ω⟶⟶LρΩ and X⟶⟶LρΩ are compact for all ρ such that 1≤ρ<2∗.Proof.
The first imbedding that i.e.,X⟶H01Ω is followed from the definition of the norm in X. The second imbedding is followed from Talenti’s work in [13]; C2∗ is the Talenti constant. The imbedding compact H01Ω⟶⟶LρΩ for ρ∈1,2∗ is obtained by Rellich’s Theorem. It follows that the mapping X⟶⟶LρΩ is compact for ρ∈1,2∗.Lemma 4.
Letζ0t=mintp,tq, ζ1t=maxtp,tq, ζ2t=mintq,tr, and ζ3t=maxtq,trfor allt≥0. Then,(34)ζ0ρAt≤Aρt≤ζ1ρAtforρ,t≥0,(35)ζ0uA≤∫ΩAuxdx≤ζ1uAforu∈LAΩ,(36)ζ2ρBt≤Bρt≤ζ3ρBtforρ,t≥0,(37)ζ2uB≤∫ΩBuxdx≤ζ3uBforu∈LBΩ.Proof.
The proof is given in [14] (see Lemma 2.1 of [14]). In fact by integrating the inequalities (29) and (30) respectively, we get inequalities (34) and (35). From (34) and (35) and the definition of Luxembourg norm we get respectively (35) and (37).Lemma 5.
The spaceLBΩ is imbedded continuously in LAΩ with uA≤ℋuB, where(38)ℋ=2kω,with k=max1,A1/B11/q and ω=max1,A1/BkA−11/2mesΩ.Proof.
This proof is based on the proof of Theorem 8.12 in [11]. Fix k, with k≥1, then(39)for allt≥1,AtBkt≤A1kqB1,by (34) and (36). Thus, by taking k=max1,A1/B11/q, we have(40)At≤Bkt,for allt≥1.
By the proof of (Theorem 8.12 [11]),(41)for allt≥A−112mesΩ,At≤ωBkt,whereω=max1,A1BkA−11/2mesΩ.
Letu∈LBΩ such that u≠0. Let us set Ω′u=x∈Ω:ux/2ωkuB≤A−11/2mesΩ and Ω″u=Ω−Ω′u.B being convex,we have(42)∫ΩAux2ωkuBdx=∫Ω′uAux2ωkuBdx+∫Ω″uAux2ωkuBdx≤12mesΩ∫Ω′udx+ω∫Ω″uBux2ωuBdx≤12+12∫ΩBuxuBdx≤1.
It follows thatuA≤2ωkuB and consequently LBΩ is imbedded continuously in LAΩ.Lemma 6.
TheN−functionA satisfies the following:(43)limt⟶+∞Aktt2∗=0for allk≥0.Proof.
The result is followed from property (34). In fact, by (34), for allk≥0,for allt≥1, we have(44)0≤Aktt2∗≤ζ1tAkt2∗=tq−2∗Ak.Lemma 7.
The imbeddingL2∗Ω⟶LAΩ is continuous with(45)uA≤CA∗u2∗,for allu∈L2∗Ω,where, CA∗=2max1;A−11/2mesΩ−2∗A11/2∗.Proof.
TheN-function A increases more slowly than t⟼t2∗ near infinity and by (34) it follows that for all t≥1,At≤A1t2∗. By the proof of (Theorem 8.12 [11]),(46)for allt≥A−112mesΩ,At≤ωA1t2∗,whereω=max1,A−112mesΩ−2∗.
Letu∈L2∗Ω such that u≠0. Let us set Ω′u=x∈Ω:ux/2ωA11/2∗u2∗≤A−11/2mesΩ and Ω′′u=Ω−Ω′u.(47)∫ΩAux2ωA11/2∗u2∗dx=∫Ω′uAux2ωA11/2∗u2∗dx+∫Ω″uAux2ωA11/2∗u2∗dx≤12mesΩ∫Ω′udx+ω∫Ω″uux2∗2ωu2∗2∗dx≤12+12∫Ωux2∗u2∗2∗dx≤1.
It follows thatuA≤2ωA11/2∗u2∗ and consequently L2∗Ω is imbedded continuously in LAΩ.Lemma 8.
The imbeddingH01Ω⟶LAΩ is continuous with(48)uA≤CAuH01Ωfor allu∈H01Ω,whereCA=C2∗CA∗.Moreover, the imbeddingsH01Ω⟶⟶LAΩ and X⟶⟶LAΩ are compacts.Proof.
The continuity of the imbeddingH01Ω⟶LAΩ is followed from Lemmas 3 and 7. By Lemma 6 and Theorem 8.36 in [11], it follows that the imbedding H01Ω⟶LAΩ is compact. It follows also that X⟶⟶LAΩ is compact because X⟶H01Ω is continuous by Lemma 3.Lemma 9.
Ifu is an element of X∖0 and λ is a real such that(49)∫Ω∇u2dx+∫Ωu2dx+∫Ωbuu2dx=λ∫Ωauu2dx,then λ>0 and there exists two positive constants k1 and k2 independent of u such that(50)k1f1λ≤uA≤k2f2λ,where f1 and f2 are two functions of λ.Proof.
Let us takeu∈X∖0 and λ∈ℝ such that (49) hold. Thus, we have 0<uH01Ω2≤λ∫Ωauu2dx. As a is a non-negative function in 0,+∞, it follows that λ>0. Let us show the second part of the Lemma. Firstly, by using respectively (29), (35), (48), and (49), we get(51)uA2≤qλCA2ζ1uA.
Thus, this last inequality yields(52)uA≥minqCA21/2−pλ1/2−p,qCA21/2−pλ1/2−p.
Secondly, by (29), (30), (48), and (49) we have(53)uA2≤CA2uH01Ω2≤CA2λ∫Ωauu2dx−∫Ωbuu2dx≤CA2q∫ΩλAux−Buxdx.
Let us setΩu=x∈Ω:ux≤1 and Ωu′=Ω−Ωu. Since λ>0, (34) and (36) imply(54)λAux−Bux≤λA1uxp−B1uxr,forx∈Ωu,λA1uxq−B1uxr,forx∈Ωu′.
Lets=p or s=q. By applying Young’s inequality ab≤aα/α+bβ/β for a,b>0 and α>1 and β>1, we have(55)λA1uxs≤srB1uxr+r−srλA1B1s/rr/r−s,where a=B1s/ruxs, b=λA1B1−s/r, α=r/s>1, and β=r/r−s>1. Inequality (55) yields(56)λA1uxs−B1uxr≤B1uxrsr−1+r−srλA1B1s/rr/r−s,(57)≤λr/r−sr−srA1B1s/rr/r−s,being s<r. Hence, (54) yields(58)λAux−Bux≤λr/r−pr−prA1B1p/rr/r−p,forx∈Ωu,λr/r−qr−qrA1B1q/rr/r−q,forx∈Ωu′.
Asr−p/r>r−q/r, it follows, from (58), that(59)λAux−Bux≤r−prmaxλr/r−pA1B1p/rr/r−p,λr/r−qA1B1q/rr/r−q,for allx∈Ω.
Thus, by (53) and (59), we get(60)uA≤CAqr−prmesΩ1/2maxλr/2r−pA1B1p/rr/2r−p,λr/2r−qA1B1q/rr/2r−q.
Finally, the last inequality and (52) yield the result (50).Definition 2.
An elementu of X is a weak solution of ελ if(61)∫Ω∇u∇vdx+∫Ωuvdx=λ∫Ωauuvdx−∫Ωbuuvdxfor allv∈X.
Consequently, the weak solutions ofελ are exactly the critical points of the energy functional Φλ:X⟶ℝ defined by(62)Φλu=12∫Ω∇u2dx+12∫Ωu2dx−λ∫ΩAudx+∫ΩBudx.Lemma 10.
Ifελ has a nontrivial weak solution u∈X, then λ≥λ0, where(63)λ0=minηκϖ2r−pp−2/pr−2,ηκϱ2r−qp−2/rp−2q,ηξϖ2r−pq−2/rq−2p,ηξϱ2r−qq−2/qr−2>0,with(64)η=CA−1qr−prmesΩ−1/2,κ=qCA21/p−2,ξ=qCA21/q−2,ϖ=A1B1p/rr/2r−p,ϱ=A1B1q/rr/2r−q.Proof.
Ifελ admits a nontrivial weak solution u∈X, then equality (49) is satisfied and λ>0 by Lemma 9. Let us now show that λ≥λ0. By inequalities (50) and (60) of the proof of Lemma 9, we get the result.
We claim that the setE=λ>0:εμadmits only the trivial solution for allμ<λ is not empty and bounded above. Indeed, by Lemma 9, for all λ<λ0, ελ admit only trivial solution. Thus, λ0∈E. Now suppose that for all M>0, there exists λM such that λM>M. Therefore, there exists a sequence λnn∈ℕ∗ of elements of E such that 0<λ0<λ1<λ2<⋯<λn<⋯ For all n∈ℕ, for all μ such that 0<μ<λn, εμ admits only trivial solution. By hypothesis, the sequence λnn∈ℕ tends to +∞. Hence, for all μ∈0,+∞, εμ admits only trivial solution. This contradicts the Lemma 9.
Let us define(65)λ^=supλ>0:εμadmits only the trivial solution for allμ<λ.
It is clear thatλ^≥λ0>0. In Section 5, we will prove that λ^ is the required critical value of the Theorem 1.


## 4. Basic Results for Existence of Nontrivial Solution

The results in the previous section require us to work from now on withλ>0.Lemma 11.
The energy functionalΦλ in coercive on X.Proof.
Letu∈X.(66)Φλu=12∫Ω∇u2dx+12∫Ωu2dx−λ∫ΩAudx+∫ΩBudx,≥12uH01Ω2−λ∫ΩAudx−12∫ΩBudx+12∫ΩBudx.
By using Young’s inequality, as we use it in (55), we have(67)λAux−12Bux≤2q/r−qr−prcλ,for allx∈Ω,where cλ=maxλr/r−pA1/B1p/rr/r−p,λr/r−qA1/B1q/rr/r−q. Therefore, we have(68)Φλu≥12uH01Ω2+12∫ΩBudx−2q/r−qr−prcλmesΩ,≥12uH01Ω2+12ζ2uB−2q/r−qr−prcλmesΩ,≥12uH01Ω2+12uB2−1−2q/r−qr−prcλmesΩ,≥12uX2−2q/r−qr−prcλmesΩ−12.
In conclusion,Φλ is coercive in X.Lemma 12.
LetA˜ and B˜ respectively be the complements of N-functions A and B. Then, we have(69)pp−1≤A˜′ssA˜s≤qq−1fors>0,(70)qq−1≤B˜′ssB˜s≤rr−1fors>0,where A˜′s and B˜′s are respectively the derivatives of A˜s and B˜s.Proof.
The results are obtained by using (29) and (30) and the proof of the point (2.7) of Lemma 2.5 in [14].Proposition 1.
The spacesLAΩ and LBΩ are reflexives and separables Banach spaces.Proof.
The N-functionsA and B satisfy Δ2-condition respectively by Lemmas 1 and 2. Moreover, the inequalities (69) and (70) of Lemma 12 imply that the N-functions A˜ and B˜ satisfy the Δ2-condition. Thus, by (Theorem 8.20 and Remark 8.22 in [11]), the spaces LAΩ and LBΩ are reflexives and separables Banach spaces.Proposition 2.
X,.X is a separable reflexive Banach space.Proof.
LetY=H01Ω×LBΩ, which we endow with the norm uY=uH01Ω+uB. Since H01Ω and LBΩ are reflexives Banach spaces, then Y,.Y is a separable and reflexive Banach space by Theorem 1.23 in [11]. Let us consider the operator Γ:X,.X⟶Y,.Y defined by Γu=u,u. Γ is well defined, linear, and isometric. Therefore, ΓX is closed subspace of Y and so ΓX is separable and reflexive by Theorem 1.22 in [11]. Consequently, X,.Y is a separable reflexive Banach space, being isomorphic to a separable, reflexive space. Finally, we conclude that X,.X is a separable reflexive Banach space because reflexivity and separability are preserved under equivalent norms.Lemma 13.
Letun be a sequence in X such that Φλunn is bounded. Then, un admits a weakly convergent subsequent in X.Proof.
The proof comes from the coercivity ofΦλ in X and the reflexivity of the space X.Lemma 14.
The functionalΦΔu=1/2∫Ω∇u2dx is convex, of class C1 and is particularly sequentially weakly lower semicontinuous in X.Proof.
The convexity of functionalΦΔu=1/2∫Ω∇u2dx is followed from the convexity of function t⟼t2 in ℝ.
Let us show the continuity ofΦΔu on X. Let unn be a sequence of which elements are in X and let u∈X such that un⟶u in X. Let unkk be an arbitrary subsequence of unn. The subsequence unk⟶u in X and hence, by Lemma 3, ∇unk⟶∇u in L2ΩN. By Theorem 4.9 in [15], there exists a subsequence unkjj of unk and a function ψ∈L1Ω such that ∇unkj⟶∇u a.e in Ω as j⟶∞ and ∇unkj≤ψ a.e in Ω for all j∈ℕ. So ∇unkj2⟶∇u2 a.e in Ω as j⟶∞ and ∇unkj2≤ψ2∈L1Ω a.e in Ω for all j∈ℕ. The dominated convergence theorem implies that ∇unkj2⟶∇u2 in L1Ω as j⟶∞. The subsequence unkk being arbitrary, we deduce that ∇un2⟶∇u2 in L1Ω as n⟶∞. Then, we get the continuity of ΦΔ on X and also, ΦΔ is sequentially weakly lower semicontinuous in X by Corollary 3.9 in [15]. Moreover, ΦΔ is Gateaux-differentiable in X and for all u,φ∈X, we have(71)ΦΔ′u,φ=∫Ω∇u∇φdx.
Let us finish the proof by showing thatΦΔ′ is continuous. Let unn be a sequence of functions in X and u∈X such that un⟶u in X when n⟶∞. By a simple calculus we have(72)ΦΔ′un−ΦΔ′uX′≤un−uH01Ω≤un−uX.
This inequality yields the continuity ofΦΔ′.Lemma 15.
The functionalΦ2u=1/2∫Ωu2dx is convex, of class C1 and sequentially weakly lower semicontinuous in X. Furthermore, if un is a sequence of elements of X and u belongs to X such that un converge weakly to u in X, then Φ2′un⟶Φ2′u in X′.Proof.
The convexity of the functionalΦ2 is obvious. The continuity of Φ2 in X follows from Lemma 3. Thus, Φ2 is sequentially weakly lower semicontinuous in X by Corollary 3.9 in [15]. To complete the proof of the theorem, it suffices to show the last part of the theorem. Therefore, let us take un⊂X and u∈X such that un⇀u weakly in X. Since X⟶⟶L2Ω is compact by lemma 3, we have Φ2′un−Φ2′uX′≤un−u2 and then Φ2′un⟶Φ2′u in X′.Proposition 3.
(i) Ifun converge strongly to u in LAΩ, then aunun converge to auu in LA˜Ω. (ii) If un converge strongly to u in LBΩ, then bunun converge to buu in LB˜Ω.Proof.
(i) Suppose thatun converge strongly to u in LAΩ. Let vnkk be a fixed subsequence of the sequence n⟼aunun. For all k, we have vnk=aunkunk, where unkk is a subsequence of the sequence un, which converge to u in LAΩ. Thus, ∫ΩAunk−udx⟶0, as k⟶∞. It follows that, there exists a subsequence unkjj of unkk and a positive function ψ∈L1Ω such that Aunkj−u⟶0 a.e in Ω and 0≤Aunkj−u≤ψx a.e in Ω,for allj∈ℕ. As A is continuous and strictly increasing in ℝ+, we have unkj⟶ua.e in Ω. Hence, vnkj⟶v=auua.e in Ω and A˜aunkjunkj−auu⟶0a.e in Ω. On other hand, there exists h∈L1Ω such thatA˜aunkjunkj−auu≤hxa.e in Ω. Indeed, by the fact that Aunkj−u<ψxa.e in Ω and the function t⟼att is continuous and increases strictly in ℝ+, we have(73)A˜aunkjunkj−auu≤A˜2aA−1ψx+uxA−1ψx+ux,forx∈Ωa.e.
By (13) and since A˜ satisfy Δ2-condition, we have(74)A˜aunkjunkj−auu≤δA˜AA−1ψx+uxA−1ψx+ux,forx∈Ωa.e,≤δAA−1ψx+ux,forx∈Ωa.e,≤δ′AmaxA−1ψx,uxforx∈Ωa.e,where δ and δ′ are constants. Thus, hx=δ′AmaxA−1ψx,ux. In consequence, it yields that A˜vnkj−v=A˜aunkjunkj−auu⟶0in L1Ω, as j⟶∞. Due to arbitrariness of vnkk, we deduce that A˜vn−v=A˜aunun−auu⟶0 in L1Ω, as n⟶∞. Hence, aunun⟶auu in LA˜Ω.
(ii) The proof is analogous to (i).Lemma 16.
The functionalΦAu=∫ΩAudx is convex, of class C1 and sequentially weakly lower semicontinuous in X. Furthermore, if unn is a sequence of elements in X and u belongs to X such that un converge weakly to u∈X, then ΦA′un⟶ΦA′u in X′.Proof.
The convexity ofΦA follows from the convexity of the positive function t⟼At defined on ℝ. The function t⟼At is also continuous and vanishes in zero. By Lemma 3 in [16], for all k>1, for ε>0, such that, 0<ε<1/k, we have(75)Aa+b−Aa≤εφεa+ψεb,for alla,b∈ℝ,where φεt=Akt−kAt, ψεt=2ACεt, for allt∈ℝ with 1/Cε=εk−1.
Letunn be a sequence of X and u an element of X such that un⟶u in X. u belongs to LAΩ by Lemma 8 and we have ∫ΩAudx<∞.un⟶u in X implies un⟶u in H01Ω. So, un⟶u a.e in Ω. By (35), properties of function ζ1 and Lemma 8, we have(76)0≤∫Ωφεun−udx=∫ΩAkun−u−kAun−udx≤ζ1kun−uA≤kqζ1un−uA≤kqζ1un−uX.
Asun⟶u in X, there exists M>0 such that for all integer n,(77)0≤∫Ωφεun−udx≤kqM.
Sinceu belongs to LAΩ, Aθu∈L1Ω for all θ∈ℝ and thus,(78)0≤∫Ωψεudx<∞,for allε>0.
Hence, by Theorem 2 in [16], it follows that ΦAun⟶ΦAu, as n⟶∞. This assures the continuity of the functional ΦA. It then follows that ΦA is sequentially weakly lower semicontinuous in X by Corollary 3.9 in [15]. Furthermore, ΦA is Gateaux-differentiable in X and for all u,φ∈X,(79)ΦA′u,φ=∫Ωauxuxφxdx.
Letun⊂X and u∈X such that un⇀u in X as n⟶∞. Let us show that ΦA′un⟶ΦA′u in X′. By Hölder’s inequality, we have(80)ΦA′un−ΦA′uX′=supφX=1φ∈X∫Ωaunun−auuφdx≤2supφX=1φ∈Xaunun−auuA˜φA≤2aunun−auuA˜.
Sinceun⇀u in X, by Lemma 8 and Proposition 3, it follows that aunun−auuA˜⟶0 and consequently, ΦA′un⟶ΦA′u in X′. In particular, this shows that ΦA is class C1 in X and the proof of the Lemma is finished.Lemma 17.
The functionalΦBu=∫ΩBudx is convex, of class C1 and sequentially weakly lower semicontinuous in X. Moreover, if unn is a sequence of elements in X and u∈X such that un⇀u∈X as n⟶∞, then ΦB′un⇀∗ΦB′u in X′.Proof.
The convexity ofΦB follows from the convexity of the positive function t⟼Bt defined on ℝ. The proof of continuity of ΦB in X is analogous to the proof of continuity of ΦA in X in the previous Lemma. In fact, by Lemma 3 in [16], for all k>1, for ε>0, such that, 0<ε<1/k, we have(81)Bx+y−Bx≤εφεx+ψεy,for allx,y∈ℝ,where φεt=Bkt−kBt, ψεt=2BCεt, forallt∈ℝ with 1/Cε=εk−1.
Letunn⊂X and u∈X such that un⟶u in X. By the definition of the space X, we have X⟶LBΩ, ∫ΩBudx<∞, and un⟶u in LBΩ. un⟶u in X implies un⟶u in H01Ω. So, un⟶ua.e in Ω. By (35) and properties of function ζ2, we have(82)0≤∫Ωφεun−udx≤krζ2un−uX.
Asun⟶u in X, there exists M>0 such that for all integer n, 0≤∫Ωφεun−udx≤krM. Since u belongs to LBΩ, Bθu∈L1Ω for all θ∈ℝ and thus,(83)0≤∫Ωψεudx<∞,for allε>0.
Hence, by Theorem 2 in [16], it follows that ΦBun⟶ΦBu as n⟶∞. This assures the continuity of the functional ΦB in X. It then follows that ΦB is sequentially weakly lower semicontinuous in X by Corollary 3.9 of [15]. Furthermore, ΦB is Gateaux-differentiable in X and for all u,φ∈X,(84)ΦB′u,φ=∫Ωbuxuxφxdx.
Letun⊂X and u∈X such that un⟶u in X as n⟶∞. Let us show that ΦB′un⟶ΦB′u in X′. By Hölder’s inequality, we have(85)ΦB′un−ΦB′uX′=supφX=1φ∈X∫Ωbunun−buuφdx≤2bunun−buuB˜.
According to Proposition3 (ii), it follows that bunun−buuB˜⟶0 and consequently, ΦB′un⟶ΦB′u in X′. Thus, ΦB is class C1 in X.
Let us show the last part of this Lemma. Letun⊂X and u∈X such that un⇀u in X. We shall proof that ΦB′un⇀∗ΦB′u in X′ as n⟶∞. Let φ∈X. For any integer n, let us set vn=bunun and consider an arbitrary subsequence vnkk of the sequence vn. We have vnk=bnkunk, where unkk is a subsequence of un. Then, unk⇀u weakly in X when k⟶∞. By Lemma 3, it follows that unk⟶u in L2Ω. Thus, there exists a subsequence unkjj of unkk such that unkj⟶u a.e in Ω and unkj≤g a.e in Ω, where g∈L1Ω. In consequence, for all φ∈X, vnkjφ⟶buuφ a.e in Ω and there exists h∈L1Ω such that bunkjunkjφ−buuφ≤h a.e in Ω. In fact, by (30) and (36) we have(86)bunkjunkj−buuφ≤bunkjunkj+buuφ,≤rB1φmaxunkjq−1,unkjr−1+maxuq−1,ur−1a.einΩ,≤rB1φmaxgq−1,gr−1+maxuq−1,ur−1a.einΩ.
Thus,h=rB1φmaxgq−1,gr−1+maxuq−1,ur−1∈L1Ω. Hence,(87)∫Ωvnkj−buuφdx=∫Ωbunkjunkj−buuφdx⟶0.
In conclusion,ΦB′un⇀∗ΦB′u in X′, as n⟶∞, since vnkk is an arbitrary subsequence of the sequence vn.
For anyx,u∈Ω×ℝ, let us set(88)fx,u=λauu−buu,(89)Fx,u=∫0ufx,vxdx=λAu−Bu.Lemma 18.
For any fixedu∈X, the functional ℱu:X⟶ℝ,v⟼∫Ωfx,uxvxdx is in X′ (X′ is the Banach dual space of X. In particular, if vn⇀vinX, then ℱuvn⟶ℱuv.Proof.
Let us takeu∈X. It is easy to show that ℱu is linear. Moreover, for all v∈X, by using Hölder inequality and imbedding properties of X in LAΩ and X in LBΩ, we have(90)ℱuv≤λ∫Ωauuvdx+∫Ωbuuvdx≤2λauuA˜vA+2buuB˜vB≤2CAλauuA˜+2buuB˜vX.
Thus, we conclude thatℱu is continuous on X. Then, it follows that, vn⇀vinX yields ℱuvn⟶ℱuv.Lemma 19.
The functionalΦλ is of class C1 and sequentially weakly lower semicontinuous in X, that is if un⇀u in X, then(91)Φλu≤liminfn⟶∞Φλun.Proof.
Lemmas14–17 imply Φλ∈C1X. Let unn⊂X and u∈X such that un⇀u in X. The definition of Φλ and (89) allow us to write(92)Φλu−Φλun=12uH01Ω2−unH01Ω2+12u22−un22+∫ΩFx,unx−Fx,ux.
Sinceun⇀u in X, Lemmas 14 and 15 yield respectively that(93)uH01Ω2≤liminfn⟶∞unH01Ω2,(94)u22≤liminfn⟶∞un22.
Hence, by (92), we get(95)limsupn⟶∞Φλu−Φλun≤limsupn⟶∞∫ΩFx,unx−Fx,ux.
By (88) and (89), for all s∈0,1,(96)Fux,u+sun−u=∂F∂ux,u+sun−u=fx,u+sun−u=fx,u+un−u∫0sfux,u+tun−udt.where fux,u=∂f/∂ux,u=λua′u+au−ub′u+bu, forallu∈X.
Multiplying (96) by un−u and integrating the result over [0,1] with respect to s, we obtain(97)Fx,un−Fx,u=fx,uun−u+un−u2∫01∫0sfux,u+tun−udtds.
According to the assumptionsA2 and ℬ2, we have(98)for allu∈X,fux,u≤q−1λau−bu.
Afterwards, (29) and (30) imply(99)for allu∈X∖0,fux,u≤qq−1λAu−Buu−2.
By (24), for all u∈X∖0, we have(100)λAux−Buxu−2≤λA1uxp−2−B1uxr−2,forx∈Ωu,λA1uxq−2−B1uxr−2,forx∈Ωu′.
Lets=p−2 or s=q−2. By applying Young’s inequality ab≤aα/α+bβ/β for a,b>0 and α>1 and β>1, we have(101)λA1uxs≤sr−2B1uxr−2+r−2−sr−2λA1B1s/r−2r−2/r−2−s,where we take a=B1s/r−2uxs, b=λA1B1−s/r−2, α=r−2/s>1, and β=r−2/r−2−s>1. Inequality (101) yields(102)λA1uxs−B1uxr−2≤B1uxr−2sr−2−1+r−2−sr−2λA1B1s/r−2r−2/r−2−s,(103)≤λr−2/r−2−sr−2−sr−2A1B1s/r−2r−2/r−2−s,being s<r−2. Hence, (100) yields(104)λAux−Buxux−2≤λr−2/r−pr−pr−2A1B1p−2/r−2r−2/r−p,forx∈Ωu,λr−2/r−qr−qr−2A1B1q−2/r−2r−2/r−q,forx∈Ωu′.
As>r−p/r−2r−q/r−2, it follows, from (104), that(105)λAux−Buxux−2≤r−pr−2cλ′,for allx∈Ω,where(106)cλ′=maxλr−2/r−pA1B1p−2/r−2r−2/r−p,λr−2/r−qA1B1p−2/r−2r−2/r−q.
From (104) and (105), we obtain(107)for allu∈X,fux,u≤qq−1r−pr−2cλ′.
Consequently, (97) yields(108)∫ΩFx,un−Fx,udx≤∫Ωfx,uun−udx+qq−1r−pcλ′/2r−2∫Ωun−u2dx≤ℱuun−u+qq−1r−pcλ′2r−2un−u22.
Sinceun⇀u in X, Lemmas 3 and 18 yield respectively ℱuun−u⟶0 and un−u2⟶0. Thus, limsupn⟶∞∫ΩFx,unx−Fx,uxdx≤0 and then, by (95), we get the claim (91).


## 5. Existence of Weak Solutions ofελ for Large Values of λ

Let(109)λ¯=infΦAu=1u∈X12∫Ω∇u2+u2+∫ΩBuxdxwithΦAu=∫ΩAuxdx.We remark thatλ¯>0. Indeed, by the fact that ΦAu=1⇒uA=1, (48) implies that(110)∫Ω∇u2+u2dx≥uH01Ω2≥1CA2,for allu∈XwithΦAu=1.By the property (37), Lemma 5 and the monotony of function ζ2, we have(111)∫ΩBuxdx≥ζ2uB≥ζ21ℋ,for allu∈XwithΦAu=1.Thus,(112)λ¯≥12CA2+ζ21ℋ>0.Lemma 20.
For allλ>λ¯, the functional Φλ admits in X a global nontrivial non-negative minimizer m with negative energy, that is Φλm<0.Proof.
For eachλ>0, Lemmas 11 and 19 and Corollary 3.23 in [15] assure to Φλ the existence of a global minimizer m∈X, that is(113)Φλm=infv∈XΦλv.
So,m is a solution of ελ. Let us prove that m is nontrivial as soon as λ>λ¯. For that we shall just prove that infv∈XΦλv<0. Let λ>λ¯. By the definition of λ¯, there exists a function φ∈X, with ∫ΩAuxdx=1, such that(114)λ∫ΩAuxdx=λ>12∫Ω∇φ2+φx2+∫ΩBφxdx≥λ¯.
Hence,(115)12∫Ω∇φ2+φx2−λ∫ΩAuxdx+∫ΩBφxdx<0and then(116)Φλφ<0.
Therefore,(117)Φλm=infv∈XΦλv≤Φλφ<0.
Hence, it follows that, forλ>λ¯, equation ελ has a nontrivial weak solution m∈X with Φλm<0. To complete the proof, we can assume m≥0 a.e in Ω because m∈X and Φλm=Φλm.
Let us define(118)λ˜=infλ>0:ελhas an on trivial weak solution.
By Lemma20, it is clear that this definition is meaningful. Furthermore, it follows easily that λ¯>λ˜.Theorem 3.
For anyλ>λ˜, the problem ελ admits a nontrivial non-negative weak solution uλ∈X.Proof.
Fixλ>λ˜. By definition of λ˜, there exists h∈λ˜,λ such that Φh has a nontrivial critical point uh∈X. Without loss of generality, we assume that uh≥0a.e in Ω, since uh is also a solution of εh. We can easily see that uh is a subsolution for ελ. Let us consider the following minimization problem:(119)infv∈CΦλvwhereC=v∈X:forallx∈Ω,vx≥uhx.
We remark easily thatC is a closed and convex set. Thus, C is weakly closed. Moreover, being Φλ is coercive in X by Lemma 11, it follows that it is coercive in C. Finally, Φλ is sequentially weakly lower semicontinuous in X and so in C. Hence, Corollary 3.23 of [15] implies that there exists uλ∈C such that(120)Φλuλ=infv∈CΦλv.
Now, we shall prove thatuλ is a solution of ελ. Let ϑ∈C0∞Ω and ϵ>0. Let us set ϑϵ=max0,uh−uλ−ϵϑ and ωϵ=uλ+ϵϑ+ϑϵ. It is clear that ωϵ∈C and(121)0≤Φλ′uλ,ωε−uλ=εΦλ′uλ,ϑ+Φλ′uλ,ϑε.
Hence(122)Φλ′uλ,ϑ≥−1εΦλ′uλ,ϑε.
Let us set(123)Ωε=x∈Ω:uλx+εϑx≤uhx<uλx.
It is obvious thatΩϵ⊂suppϑ. Since, uh is a subsolution of ελ and ϑϵ≥0, it turns out that Φλ′uλ,ϑϵ≤0. Consequently, as(124)Φλ′uλ,ϑε=Φλ′uh,ϑε+Φλ′uλ−Φλ′uh,ϑε,fux=λauxux−buxux,
We have(125)Φλ′uλ,ϑε≤Φλ′uλ−Φλ′uh,ϑε≤∫Ωε∇uλ−∇uh⋅∇uh−uλ−εϑdx+∫Ωεuλ−uhuh−uλ−εϑdx−∫Ωεfuλ−fuh⋅uh−uλ−εϑdx≤∫Ωε∇uλ−∇uh⋅∇−εϑdx+∫Ωεuλ−uhuh−uλ−εϑdx−∫Ωεfuλ−fuh⋅uh−uλ−εϑdx.(i)
ForΩε=∅, it is clear that Φλ′uλ,ϑϵ≤0. Hence by (122), we have Φλ′uλ,ϑ≥0.(ii)
Also, whenΩε≠∅, we have 0≤uh−uλ−εϑ=uh−uλ+εϑ<εϑ in Ωε because in this case ϑ≤0 in Ωϵ. Then, (125) yields(126)Φλ′uλ,ϑε≤ε∫Ωε∇uh−∇uλ⋅∇ϑdx+∫Ωεuλ−uhϑdx+∫Ωεfuλ−fuh⋅ϑdx≤ε∫Ωεψxdx,where ψx=∇uh−∇uλ.∇ϑ+uλ−uh+fuλ−fuhϑ. We claim that ψ∈L1suppϑ. Indeed, ∇uh and ∇uλ are in L2ΩN and uλ and uh are in Lloc1Ω since X⊂H01Ω. Moreover, by A1 and ℬ1, the functions t⟼att and t⟼btt are continuous in ℝ+. Thus, the functions auλuλ, auhuh, buλuλ, and buhuh are continuous in Ω. Consequently, auλuλ, auhuh, buλuλ, and buhuh belong to Lloc1Ω. Finally, fuλ−fuh∈Lloc1Ω since fuλ−fuh≤λauλuλ+auhuh+buλuλ+buhuh. Therefore, the claim is obtained. Thus,(127)limε⟶0+∫Ωεψxdx=0,since Ωϵ⟶0 as ϵ⟶0+. Then, (126) implies that Φλ′uλ,ϑε≤°ε as ϵ⟶0+. So, by (122), it follows that Φλ′uλ,ϑ≥°1 as ϵ⟶0+.
We deduce that,Φλ′uλ,ϑ≥0 for all ϑ∈C0∞Ω, that is Φλ′uλ,ϑ=0 for all ϑ∈C0∞Ω. Thus, uλ is a weak solution of ελ in X because X is the completion of C0∞Ω with respect to the norm .X. Finally, uλ is nontrivial and non-negative, since uλ>uh.Lemma 21.
λ˜=λ^.Proof.
This proof is the same to that one did in the step 5 of the proof of (Theorem 1.1 in [7]). In fact, by Theorem 3, we have λ˜≥λ^. Indeed, for all λ such that ελ a weak nontrivial solution, we have λ>λ^ and thus, λ˜≥λ^. Let us show that λ˜≤λ^. Suppose that λ˜>λ^. For λ<λ˜, the problem ελ cannot admit a nontrivial solution u∈X since this would contradict the minimality of λ˜. Thus, for all λ∈λ^,λ˜ the unique solution of ελ is u≡0. This assertion is still again impossible because it would contradict the maximality of λ^. Hence, λ˜=λ^.Theorem 4.
The problemελ^ admits a nontrivial non-negative weak solution in X.Proof.
Letλnn∈ℕ∗ be a strictly decreasing sequence converging to λ^ and un∈X be a nontrivial non-negative weak solution of ελnforalln∈ℕ∗. For all n, we have(128)∫Ω∇un∇vdx=−∫Ωunvdx+λn∫Ωaununvdx−∫Ωbununvdx,for allv∈X.
By (29), (30), (49), and Lemma 4 we have(129)uH01Ω+qζ2unB≤qλnζ1unA.
Furthermore, by inequality (60) and the monotony of ζ1 there exists a constant KA such that(130)unH01Ω+qζ2unB≤KAqCλn,for alln∈ℕ∗,where(131)Cλn=maxλn1+pr/2r−pA1B1p/rpr/2r−p,λn1+qr/2r−pA1B1p/rqr/2r−p,λn1+pr/2r−qA1B1q/rpr/2r−q,λn1+qr/2r−qA1B1q/rqr/2r−q,for alln∈ℕ∗.
Thus, by monotony of sequenceλnn∈ℕ∗, we get(132)unH01Ω+qζ2unB≤KAqCλ1,for alln∈ℕ∗.
Therefore, the sequencesunH01Ωn and unBn are bounded, and it yields that the sequence unXn is bounded. According to Theorem 3.18 in [15], Propositions 1 and 2 and Lemma 8 allow to extract, from the sequence unn, a subsequence still relabeled unn and satisfying(133)un⇀uinX,un⟶uinLAΩ,un⇀uinLBΩ,un⟶ua.e inΩ,∇un⇀∇uinL2ΩN,for some u∈X. We claim that u, which is clearly non-negative by (133), is the solution we are looking for. In fact, for all v∈X,(134)∫Ω∇un∇vdx⟶∫Ω∇u∇vdx,as n⟶∞, since ∇un⇀∇uinL2ΩN by (133). Since un⇀u in X, Lemma 15 yields in particular that for all v∈X,(135)∫Ωunvdx⟶∫Ωuvdx,as n⟶∞. Moreover, Lemmas 16 and 17 imply that for all v∈X(136)∫Ωaununvdx⟶∫Ωauuvdx,∫Ωbununvdx⟶∫Ωbuuvdx,as n⟶∞. By passing to the limit in (128) as n⟶∞, we get by (134)–(136)(137)∫Ω∇u∇vdx+∫Ωuvdx=λ^∫Ωauuvdx−∫ΩbuuvdxΩ,for all v∈X. Hence, u is a weak non-negative solution of ελ^. It remains to show that the solution u is nontrivial. Since un⇀u in X by (133), Lemma 8 yields in particular that uA=limunA. Moreover, (52) applied to each un≡0, implies that(138)unA≥minqCA21/2−pλn1/2−p,qCA21/2−qλn1/2−q.
Thus, by passing to the limit asn⟶∞, we get(139)uA≥minqCA21/2−pλ^1/2−p,qCA21/2−qλ^1/2−q,since λn↘λ^ and λ^>0. Consequently, u is nontrivial and non-negative by (133).
Proof of Theorem1. Section 3, Lemma 20, and Theorems 3 and 4 show the existence of λ^>0 such that for all λ≥λ^, ελ admits at least a nontrivial non-negative weak solution in X.


## 6. Second Solution for Large Values of the Parameter

By variational methods, we prove, in this section, that the Dirichlet problemελ admits at least two nontrivial weak solutions if λ is sufficiently large.Lemma 22.
For anym∈X∖0 and λ>0 there exist ϱ∈0,mH01Ω and α=αϱ>0 such that Φλu≥α for all u∈X, with uH01Ω=ϱ.Proof.
Letm∈X∖0 and λ>0. Let u be in X. By (35) and (48) and monotony of the function ζ1,(140)Φλu≥12uH01Ω2−λ∫ΩAudx≥12uH01Ω2−λζ1uA≥12uH01Ω2−λζ1CAuH01Ω≥12−λmaxCApuH01Ωp−2,CAquH01Ωq−2uH01Ω2.
So, to find the result of the Lemma, it is enough to takeϱ such that(141)0<ϱ<min2λCAp1/2−p,2λCAq1/2−q,mH01Ω, (142)α=αϱ=12−λmaxCApϱp−2,CAqϱq−2ϱ2>0.
In Lemma20, we have shown that for all λ>λ¯, the problem ελ admits a nontrivial non-negative weak solution m∈X which is above all a global minimizer of Φλ in X, with Φλm<0. In this section, we are looking for a second nontrivial weak solution of ελ when λ>λ¯.Lemma 23.
There existsκ^>0 such that(143)for allx,y∈ℝ2,bxx−byyx−y≥κ^Bx−y≥0.Proof.
(i)
Ifx=y, the inequality is verified for all κ^>0.(ii)
Ifx≠0 and y=0 (or x=0 and y≠0), the inequality is obtained by (30) and κ^=q>0.(iii)
Suppose now thatx≠y with x≠0 and x≠0. By hypothesis ℬ1, the function t⟼btt is strictly increased in 0;+∞ and consequently it is strictly increased in −∞;+∞ because it is an odd function. Thus, for x≠y with x≠0 and x≠0, bxx−byyx−y>0 and(144)φx,y=bxx−byyx−yBx−y>0.
Let(145)κ=infx≠yx,y∈ℝ∗2φx,y>0.
In conclusion, takingκ^=maxq,κ, we get the result of this Lemma.Theorem 5 (see Theorem A.3 in [10]).
LetX,⋅ and X,.E be two Banach spaces such that X⟶E. Let Φ:X⟶ℝ be a C1 functional with Φ0=0. Suppose that there exist ϱ,α>0 and m∈X such that mE>ϱ, Φm<α, and Φu≥α for all u∈X with uE=ϱ. Then, there exists a sequence unn∈X such that for all n(146)c≤Φun≤c+1n2andΦ′unX′≤2n,where(147)c=infγ∈Γmaxt∈0,1Φγt,Γ=γ∈C0,1;X:γ0=0,γ1=m,Proof.
One can find the proof of this theorem in the section (Appendix A in [10]).
Proof of Theorem2. Let λ be a strictly positive fixed number such that λ>λ¯. Then, let m∈X be the global minimizer of Φλ given by Lemma 20. Thanks to Lemma 22 and the fact that Φλm<0, the assumptions of Theorem 5 which is a variant of Ekeland’s variational principle are fulfilled for the energy functional Φλ of ελ. Hence, there exists a sequence unn⊂X such that(148)Φλun⟶c,Φ′unX′⟶0,where(149)c=infγ∈Γmaxt∈0,1Φγt,Γ=γ∈C0,1;X:γ0=0,γ1=m.
In the sequel, we shall prove that the sequenceunn strongly converges to some u in X and that u is a second nontrivial non-negative weak solution of ελ.Step 1.
By Lemma11 (coercivity of Φλ on X), the sequence unn is bounded in X and consequently, unn is bounded in H01Ω. For the sequel, let us use the argument of the proof of Theorem 4. Hence, by Propositions 1 and 2 and Lemma 8 we can extract from the sequence unn, a subsequence still relabeled unn and satisfying(150)un⇀uinX,un⟶uinLAΩ,un⇀uinLBΩ,un⟶ua.e inΩ,∇un⇀∇uinL2ΩN,for some u∈X. It remains to show that u is a nontrivial solution of (ελ), with u≠m. We have(151)Φλ′un,v=∫Ω∇un∇vdx−∫Ω−un+λaunun−bununvdx,for alln,for allv∈X.
By the same argument used in the proof Theorem4, we have, for all v∈X,(152)∫Ω∇un∇vdx⟶∫Ω∇u∇vdx,∫Ωunvdx⟶∫Ωuvdx,∫Ωaununvdx⟶∫Ωauuvdx,∫Ωbununvdx⟶∫Ωbuuvdx,as n⟶∞. Hence, passing to the limit as n⟶∞ in (151), using and the fact that Φλ′un,v⟶0 as n⟶∞ for all v∈X, we get(153)∫Ω∇un∇vdx+∫Ωuvdx=λ∫Ωauuvdx−∫Ωbuuvdx,forallv∈X,so u is a weak solution of (ελ).Step 2.
We claim that(154)ℐan=∫Ωaunun−auuun−udx⟶0,as n⟶∞. By hypothesis A1, the function t⟼att is strictly increased in 0;+∞ and consequently, it is strictly increased in −∞;+∞ because it is an odd function. Thus, we have(155)ℐan≥0.
By (150), un⟶u in LAΩ. Thus, aunun⟶auu in LA˜Ω by Proposition 3. Therefore, by Hölder inequality, we get(156)0≤∫Ωaunun−auuun−udx≤2aunun−auuA˜un−uA⟶0,as n⟶∞. This gives the proof of the claim.Step 3.
In this step, we show thatun−uX⟶0 as n⟶∞. Let us set(157)ℐbn=∫Ωbunun−buuun−udx.
For the same arguments, as forℐan, we have(158)ℐbn≥0.
Note thatΦλ′un−Φλ′u,un−u⟶0 as n⟶∞, since un⟶u in X and Φλ′un⟶0 in X′ as n⟶∞. Hence, by (154)(159)un−uH01Ω2+un−u22+ℐbn=Φλ′un−Φλ′u,un−u+λℐan⟶0,as n⟶∞. Furthermore, we have(160)un−uH01Ω2⟶0,as n⟶∞. By (143) and (159), it follows that(161)∫ΩBun−udx≤1κ^ℐbn⟶0,as n⟶∞. Thus, by section 8.13 in [11],(162)un−uB⟶0asn⟶∞,since the N-function B satisfied the Δ2-condition. Finally,(163)un−uX⟶0asn⟶∞,by (160) and (162).Step 4.
Sinceun⟶u in X and Φλ∈C1X, we get Φλu=c=limn⟶∞Φλun. So, u is a second independent nontrivial weak solution of (ελ), with Φλu=c>0>Φλm. We can assume u≥0 a.e. in Ω, since u is also a solution of (ελ) due to Φλu=Φλu. This concludes the proof.


## 7. Conclusion

In this paper we studied the nonexistence, the existence, and multiplicity results for nontrivial weak solutions of the semilinear elliptic Dirichlet problem with (ελ) involving a positive parameter λ. Our main results are obtained in the Theorems 1 and 2 by using variational methods.



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