# Multiplicity of Solutions for a Class of Elliptic Problem ofp-Laplacian Type with a p-Gradient Term

Authors: Zakariya Chaouai; Soufiane Maatouk
Journal: International Journal of Mathematics and Mathematical Sciences
 (2019)

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## Abstract

We consider the following problem:-Δpu=c(x)|u|q-1u+μ|∇u|p+h(x)inΩ,u=0on∂Ω, where Ω is a bounded set in RN (N≥3) with a smooth boundary, 1<p<N, q>0, μ∈R⁎, and c and h belong to Lk(Ω) for some k>N/p. In this paper, we assume that c≩0 a.e. in Ω and h without sign condition and then we prove the existence of at least two bounded solutions under the condition that ck and hk are suitably small. For this purpose, we use the Mountain Pass theorem, on an equivalent problem to (P) with variational structure. Here, the main difficulty is that the nonlinearity term considered does not satisfy Ambrosetti and Rabinowitz condition. The key idea is to replace the former condition by thenonquadraticity condition at infinity.


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## Body

## 1. Introduction and Main Result

LetΩ be a bounded set in RN (N≥3) with a smooth boundary ∂Ω. In this paper, we are concerned with the following elliptic problem:P-Δpu=cxuq-1u+μ∇up+hxinΩ,u=0on∂Ω,where Δpu≔div(|∇u|p-2∇u) is the p-Laplacian operator, 1<p<N, q>0, μ∈R∗, and c and h belong to Lk(Ω) for some k>N/p.In the literature, there are many results concerning the existence, the uniqueness, and the multiplicity of solutions for models likeP under various assumptions on c and h. At first, it is important to mention that the sign of c plays a crucial role in the problem P regarding uniqueness, as well as existence, of bounded solutions. In this setting, we refer to ([1]) for more details. In the coercive case, that is, c(x)≤-α0 a.e. in Ω for some α0>0, Boccardo, Murat, and Puel ([2–4]) proved the existence of bounded solutions for more general divergence form problems with quadratic growth in the gradient by using the sub and supersolution method. Moreover, Barles and Murat ([5]) and Barles et at. ([6]) have treated the uniqueness question for similar problems. Notice that if we allow c(x)≤0 a.e. in Ω, then Ferone and Murat ([7, 8]) observed that finding solutions to P becomes rather complex without imposing some strong regularity conditions on the data. For the particular case c≡0, there had been many contributions ([9–11]). However, for c≤0 that may vanish only on some parts of Ω, the uniqueness of solutions was left open until the recent paper authored by Arcoya et at. ([12]). This last result was proved for p=2 and q=1 and under the following condition:(1)c,hbelongtoLkΩforsomek>N2,μ∈L∞ΩandmeasΩ∖Suppc>0,infu∈Wc,uH01Ω⁡∫Ω∇u2-μ+L∞Ωh+xu2>0,infu∈Wc,uH01Ω⁡∫Ω∇u2-μ-L∞Ωh-xu2>0.where Wc≔w∈H01(Ω):c(x)w(x)=0,a.e.inΩ. For a related uniqueness result see also Arcoya et at. ([13]).The case wherec(x)≩0 a.e. in Ω, the question of nonuniqueness has been being an open problem given by Sirakov ([14]) and it has received considerable attention by many authors. Moreover, it should be pointed out that the sign of h and whether μ is a function or a constant generate additional difficulties for solving P. In this setting, Jeanjean and Sirakov ([1]) showed the existence of two bounded solutions assuming that μ∈R∗, c, and h are in Lk(Ω) for some k>N/2 and satisfying(2)μh+LN/2Ω<CN,max⁡cLkΩ,μh-LkΩ<c¯,where c¯>0 depends only on N,k,meas(Ω),|μ|,[μh]+Lk(Ω), and CN is the optimal constant in Sobolev’s inequality. Here, h is allowed to change sign. Shortly after, this result was extended by Coster and Jeanjean ([15]) for μ is a bounded function such that μ(x)≥μ1>0 by using the degree topological method.Finally, in the case wherec is allowed to change sign and with c(x)≩0 a.e. in Ω, Jenajean and Quoirin ([16, Theorem 1.1]) showed the existence of two bounded positive solutions when h≩0; μ is a positive constant, and c+ and μh are suitably small.We would also like to mention that all the above quoted multiplicity results were restricted to the Laplacian operator with quadratic growth in the gradient, i.e.p=2, and for q=1. Moreover, it is interesting to mention that when c is allowed to change sign the solutions are positive.In this work, we prove the multiplicity of bounded solutions for the problemP by assuming the following assumption:Hc,hbelongstoLkΩforsomek>Np,hisallowedtochangesign,c≩0a.e.inΩ,q>0,μ∈R∗.Now, we give a brief exposition of the proof of our multiplicity result. At first, without loss of generality, we solve the problemP by restricting it to the case μ is a positive constant. For μ is a negative constant, we replace u by -u in P, then we conclude. Next, we observe that the problems of type P do not have a variational formulation due to the presence of the p-gradient term. To overcome this difficulty, we perform the Kazdan-Kramer change of variable, that is, v=(eμu/p-1-1)/μ. Thus, we obtain the following equivalent problem P′:P′-Δpv=cxgv+hxfvinΩ,v=0on∂Ω,where(3)gs=p-1q-p+1μq1+μsp-1ln⁡1+μsq-1ln⁡1+μs,withs>-1μ,and(4)fs=1+μsp-1p-1p-1.We mean by bounded weak solutions ofP′ the functions v∈W01,p(Ω)∩L∞(Ω) satisfying(5)∫Ω∇vp-2∇v∇u=∫Ωcxgvu+∫Ωhxfvu,for any u∈W01,p(Ω)∩L∞(Ω). Obviously, if v>-1/μ is a solution of P′, then u=p-1/μln⁡1+μv is a solution of P. Hence, the solutions obtained here are not necessarily positive (compare with ([16])).One of the most fruitful ways to deal withP′ is the variational method, which takes into account that the weak solutions of P′ are critical points in W01,p(Ω) of the C1-functional(6)Iv=1p∫Ω∇vp-∫ΩcxGv-∫ΩhxFv,with G(s)=∫0sgtdt and F(s)=∫0sf(t)dt.In this work, to obtain the two critical points forI, we use the Mountain Pass Theorem to show one critical point and the standard lower semicontinuity argument to show the other. For the first one, according to the famous paper by Ambrosetti and Rabinowitz ([17]), the most important step is to show that I satisfies the Palais-Smale condition at the level c~ (see Definition 3). The fulfillment of this condition relies on the well-known Ambrosetti-Rabinowitz condition ((ARc) for short), namely,(7)thereexistθ>pands0>0suchthat0<θGs≤sgs,ass>s0.Unfortunately, this condition is somewhat restrictive and not being satisfied by many nonlinearitiesg. However, many researches have been made to drop the (ARc). We refer, for instance, to [18–22]. Notice that the nonlinearity g considered here does not satisfy (ARc). Moreover, since we do not assume any sign condition on h, the fulfillment of the Palais-Smale condition turns out more delicate (see, e.g., [16, 23]). To the best of our knowledge, only Jenajean and Quoirin ([16]), recently, proved the Palais-Smale condition under the assumptions c changes sign and h is positive, and without assuming (ARc). In their proof, for p=2 and q=1, the authors based one of the arguments on the positivity of h and the explicit determination of a function H;(8)Hs=gss-2Gs.In our situation, as h is allowed to change sign and the analog of their function H cannot be computed explicitly, due to our general consideration of p and q (1<p<N and q>0); hence, their arguments cannot be adapted.The key point to show the Palais-Smale condition in this paper is to prove thatg, among other conditions, satisfies the following (see Lemma 7):NQHs=gss-pGs→+∞,wheres→+∞.The conditionNQ is a variant of the well-knownnonquadraticity condition at infinity, which was introduced by Costa and Malgalhães ([18]) and is given as follows:CMthereexista>0,ν≥ν0>0suchthatliminfs→∞⁡Hssν≥a.Observe that sinceν>0, then NQ is weaker than CM. Moreover, it should be noted that NQ was considered by Furtado and Silva in their recent paper ([21]). Our result follows by using similar arguments.Concerning the existence of the second critical point handled by the standard lower semicontinuity argument, we look for a local minimum inW01,p(Ω) for the functional I. Indeed, we observe that I takes positive values in a large sphere, due to its geometrical structure (see Proposition 6), and I(0)=0.Now we state the main result of this paper.Theorem 1.
Assume thatH is satisfied. If ck and hk are suitably small, then the functional I has at least two critical points. Hence, the problem P has at least two bounded weak solutions.The paper is organized as follows. In Section2 we recall some preliminary results and show that the functional I has a geometrical structure. In Section 3 we prove our main result, Theorem 1.Notation. Through this paper, we use the following notations.(1)
The Lebesgue norm(∫Ωup)1/p in Lp(Ω) is denoted by .p for p∈[1,+∞[. The norm in L∞(Ω) is denoted by uL∞(Ω)≔esssupx∈Ωux. The Hölder conjugate of p is denoted by p′.(2)
The spacesW01,pΩ and W-1,p′Ω are equipped with Poincaré norm u≔(∫Ω|∇u|p)1/p and the dual norm ·∗≔∥·∥W-1,p′(Ω), respectively.(3)
We denote byB(0,R) the ball of radius R centered at 0 in W01,p(Ω) and ∂B(0,R) its boundary.(4)
We denote byCi,ci>0 any positive constants that are not essential in the arguments and that may vary from one line to another.


## 2. Preliminaries and Geometry of the FunctionalI

In this section, we recall the standard definitions of Palais-Smale sequence at the levelc~ and Palais-Smale condition at the level c~ for I and we prove that the functional I defined in (6) has a geometrical structure.Let us define the level atc~ as follows:(9)c~=infγ∈Γ⁡maxt∈0,1⁡Iγt,where Γ=γ∈C0,1,W01,pΩ:γ0=0,γ1=v0 is the set of continuous paths joining 0 and v0, where v0∈W01,p(Ω) is defined in Proposition 6 below.Definition 2.
LetE be a Banach space with dual space E∗ and un is a sequence in E. We say that un is a Palais-Smale sequence at the level c~ for I if(10)Iun→c~,I′unE∗→0.Definition 3.
We say thatI satisfies the Palais-Smale condition at the level c~ if any Palais-Smale sequence at the level c~ for I possesses a convergent subsequence.In order to prove thatI has a geometrical structure, we need some properties of g, which we gather in the following lemma without proof.Lemma 4.
(1)
g ( s ) / | s | p - 2 s → c as s→0, where c=0 if q>p-1 and c=1 if q=p-1. (2)
g ( s ) / | s | q - 1 s → ( p - 1 ) q - p + 1 as s→0, for all q>0. (3)
g ( s ) / s p - 1 → + ∞ and G(s)/sp→+∞ as s→+∞, for all q>0.Lemma 5.
(1)
Ifq≥p-1, then we have (11)gs≤c0sr+c1sp-1, for alls>-1/μ, and for all r∈(p-1,p). (2)
If0<q<p-1, then we have (12)gs≤c1sr+c2sq, for alls>-1/μ, and for all r∈(p-1,p).Proof.
By using Lemma4(1), there exists η>0 such that for all |s|<η we have(13)gs≤c1sp-1.Let δ∈(0,1). If s≥η, then we have(14)gs≤c2η,μ,δsp-1+δ.Moreover, simple calculation yields(15)g′s=p-1q-p+1μq-11+μsp-2ln⁡1+μsq-1p-1ln⁡1+μs+q.Now, if -1/μ<s≤-η, then we have |g(s)|≤|g(T)|, where T=(e-q/p-1-1)/μ. Hence,(16)gs≤c3η,μ,δsp-1+δ.By combining (14) and (16), (1) holds. To prove the property (2), we use Lemma 4(2) and the same previous argument.Proposition 6.
Assume thatH holds. If ck and hk are suitably small, then the functional I has a geometrical structure; that is, I satisfies the following properties:(i)
There existsρ>0 such that for all v in ∂B(0,ρ), I(v)≥β, where β>0.(ii)
There existsv0∈W01,p(Ω) such that v0>ρ and I(v0)≤0.Proof.
(i) To prove this lemma we distinguish two cases onq. Firstly, if 0<q<p-1, then by using Lemma 5(2) and Hölder’s inequality, we get(17)∫ΩcxGv≤c1ckvr+1k′+c2ckvq+1k′.We choose r>p-1 with r close to p-1 such that (r+1)k′<pN/N-p, which exists due to the assumption k>N/p. Obviously, (q+1)k′<pN/N-p. Thus, by using Sobolev’s embedding we get(18)∫ΩcxGv≤C1ckvr+1+C2ckvq+1.Moreover, from the definition of the function f in (4), we have(19)fv≤c1+vp-1,forsomec>0.Using Sobolev’s embedding, we get(20)∫ΩhxFv≤C3hk+C4hkvp.By the definition of I in (6), we deduce that(21)Iv≥1pvp-C1ckvr+1-C2ckvq+1-C3hk-C4hkvp.Now, let v in ∂B0,ρ. Then, we have(22)Iv≥1pρp-ckC1ρr+1+C2ρq+1-hkC3+C4ρp.We take ρ sufficiently large, such that ck≤ρ-r-2+p and hk≤ρ-1 (which are sufficiently small by hypothesis), then(23)Iv≥1pρp-Cρp-1≥ρp-11pρ-C=β1.Secondly, q≥p-1, we choose again r as above such that pk′<(r+1)k′<pN/N-p. Then, by using Lemma 5(1) and Sobolev’s embedding, we get(24)∫ΩcxGv≤c1ckvr+1+c2ckvp.Now, as the first case, we get(25)Iv≥1pρp-C′ρp-1≥ρp-11pρ-C′=β2.Finally, we summarize the two cases and get(26)Iv≥β,whereβ=min⁡β1,β2.(ii) To prove the second property, we show that I(tv)→-∞ as t→+∞. For this, let v∈C0∞(Ω) be a positive function such that cv≩0. By the definition of I in (6), we have (27)Itv=tpp∫Ω∇vp-∫ΩcxGtv-∫ΩhxFtv=tp1p∫Ω∇vp-∫ΩcxGtvtpvpvp-∫ΩhxFtvtpvpvp.From inequality (19), we get(28)∫ΩhxFtvtpvpvp≤cast→+∞.Moerever, by Lemma 4(3), we get (29)∫ΩcxGtvtpvpvp→+∞ast→+∞.Thus, we deduce the desired result.Finally, we stress that sinceI has a geometrical structure, then the existence of a Palais-Smale sequence at the level c~ for I is ensured. This can be observed directly from the proof given in ([17]), or alternatively using Ekeland’s variational principle ([24]).


## 3. Proof of Theorem1

Recall from introduction that the proof of our main result is divided into two steps as follows. In the first step, we show the existence of the first critical point for theC1-functional I by using the Mountain Pass Theorem due to Ambrosetti-Rabinowitz ([17]). Precisely, we show that the functional I satisfies the Palais-Smale condition at the level c~. In the second step, we show the existence of the second critical point of I on B(0,ρ) (which is a local minimum) by using the lower semicontinuity argument. Moreover, we are going to see that these critical points are not the same. Finally, we show that any solution of problem P is bounded.


### 3.1. First Critical Point: Palais-Smale Condition

In this subsection, we prove thatI satisfies the Palais-Smale condition at the level c~. Precisely, we show that any Palais-Smale sequence at the level c~ for I is bounded in W01,p(Ω), and, then, it has a strongly convergent subsequence.They key point to prove the boundedness of the Palais-Smale sequence at the levelc~ in W01,p(Ω) is to show that g verifies the nonquadraticity condition at infinity NQ. Indeed, we have the following lemma.Lemma 7.
The functiong defined in (3) verifies the nonquadraticity condition at infinity NQ.Proof.
To proveNQ, we show that H is increasing and unbounded for s sufficiently large. We recall that H(s)=g(s)s-pG(s). Then, by simple calculations, we get(30)H′s=Cμs1+μsp-2ln⁡1+μsq-11-pln⁡1+μsμs+q,where C=(p-1)q-p+1/μq. Thus, H is increasing for s large enough. Moreover, H is unbounded. Indeed, by contradiction, if H is bounded, then there exists a positive constant M such that(31)Hs≤M,forslargeenough.In addition, from the definition of H and using integration by parts on G, we get(32)Hs=-C1μln⁡1+μsq1+μsp-1+qC∫0s1+μtp-1ln⁡1+μtq-1dt.By choosing δ∈p-1,p, we obtain(33)Hssδ=-1μln⁡1+μsq1+μsp-1sδ+qC∫0s1+μtp-1ln⁡1+μtq-1sδ≤Msδ.When s→+∞, we obtain H(s)/sδ→+∞ and M/sδ→0. Hence, we have a contradiction. As a conclusion, the function g verifies NQ.Lemma 8.
Letun be a Palais-Smale sequence at the level c~ for I in W01,p(Ω). Then, un is bounded in W01,p(Ω).Proof.
Letun be a Palais-Smale sequence at the level c~ for I in W01,p(Ω). We prove by contradiction that un is bounded in W01,p(Ω). We assume that un is unbounded in W01,p(Ω), that is, un→+∞.
For all integern≥0, we define(34)Izn≔max0≤t≤1⁡Itun,wherezn=tnun,tn∈0,1.We are going to prove that I(zn)→+∞ and also Izn is bounded, which is the desired contradiction.
(a) Showing that I ( z n ) → + ∞: We set vn≔un/un; then (vn) is bounded in W01,p(Ω). Hence, there exists a subsequence denoted again vn such that vn converges weakly and strongly to v in W01,p(Ω) and in Ls(Ω) for some 1≤s<p∗, respectively. Moreover, vn also converges to v almost everywhere in Ω. Recall that p∗≔Np/N-p is Sobolev conjugate.
Now, we claim by contradiction that v ≡ 0 a.e. in Ω.
Sinceun is Palais-Smale type sequence, then we have(35)Iun→c~,I′un∗→0.Hence,(36)∫Ω∇unp-2∇un∇φ-∫Ωcxgunφ-∫Ωhxfunφ=ϵn,for all φ∈W01,p(Ω) and for some ϵn→0 as n→+∞. We divide both sides of (36) by unp-1, to obtain(37)∫Ωcxgununp-1φ=ϵnunp-1+∫Ω∇vnp-2∇vn∇φ+∫Ωhxfununp-1φ.On the one hand, since vn converges weakly to v in W01,p(Ω) and by the inequality (19), then for n large enough the second and the third terms of the right-hand side of (37) are bounded.
On the other hand, ifv≢0 in Ω, then cv≢0 in Ω. Now, we choose φ∈W01,p(Ω) such that cvφ>0 in Ωφ and cvφ≡0 in Ω∖Ωφ, with |Ωφ|>0. Since vnun=un in Ω, then by using Lemma 4(3), we obtain(38)liminfcxgununp-1φ=liminfcxvnp-1gvnunvnunp-1φ=+∞inΩφ.Hence, by using Fatou’s lemma in (37) we obtain the unbounded term in the left-hand side of (37). Hence, the claim (i.ev≡0 a.e. in Ω.)
Sinceun→+∞, then there exists M>0 such that un>M, for n large enough. Moreover, we have(39)Izn≥IMunun=IMvn=Mpp-∫ΩcxGMvn-∫ΩhxFMvn.In what follows, we treat only the case 0<q<p-1. The other case follows with similar arguments. From Lemma 5(2), we have Gs≤c1|s|r+1+c2|s|q+1, where p-1<r<p. Since c∈Lk(Ω), for some k>N/p and vn converges strongly to v in Ls(Ω) with 1≤s<p∗, then, we obtain(40)∫ΩcxGMvn→0asn→+∞,due to v≡0 a.e. in Ω. By Hölder’s inequality, we get(41)∫ΩhxFMvn≤Casn→+∞.Hence, by choosing M>0 large enough, we deduce that I(zn)→+∞, as n→+∞.
(b) Showing that I z n is bounded: To prove that Izn is bounded, we distinguish two cases: tn≤2/un and tn>2/un.
The Case  tn≤2/un. Here, we only handle the proof for q∈0,p-1. The other case follows as in the proof of Proposition 6(i). By the definition of (zn) and I∈C1(W01,p(Ω),R), we have I′tnun,tnun=0, which means that(42)tnpunp=∫Ωcxgtnuntnun+∫Ωhxftnuntnun.By the definition of I in (6), we have(43)pItnun=tnpunp-p∫ΩcxGtnun-p∫ΩhxFtnun=∫ΩcxHtnun+∫ΩhxKtnun,where the function H is defined in NQ and K(s)≔f(s)s-pF(s). Moreover, from Lemma 5(2), we have (44)∫ΩcxHtnun≤∫Ωcxgtnuntnun+p∫ΩcxGtnun≤c1∫Ωcxtnunr+1+c2∫Ωcxtnunq+1.By choosing r and q as in the proof of Proposition 6(i), we get(45)∫ΩcxHtnun≤C1cktnunr+1+C2cktnunq+1.By inequality (19) and Sobolev’s embedding, we get(46)∫ΩhxKtnun≤∫Ωhxftnuntnun+p∫Ωhx(Ftnuntnun≤c1hk+c2hktnun+c3hktnunp.Then, by (43), (45), and (46), we obtain(47)Itnun≤C,for all n≥0, where C is independent of n. Thus, Izn is bounded, which contradicts the fact that Izn is unbounded (see(a)).
The Case t n > 2 / u n. Here, we are proceeding the technique inspired by [21]. To this end, we need the following technical lemma.
Lemma 9 . Let Φ : R → R the nonnegative function defined as (48) Φ s = e - ϵ / s 2 , i f   s ≠ 0 , 0 , i f   s = 0 , with ϵ > 0 . Then, we have (i)
l i m s → 0 Φ ( s ) = l i m s → 0 Φ ′ s = 0 . (ii)
for any positive function  z in  Ω and  p>1,(49)limϵ→0∫Ω∫stzxτp+11-Φϵτununpdτdx=0,uniformlyinn∈N.
Proof. Obviously we have (i). To prove (ii), we follow the same approach given in [21] for the case p=2 and z(x)=1, which can be immediately generalized for any positive function z and p>1. □
Now, we resume the proof of Lemma8. From Lemma 7, we have H(s)≥σ, for s large enough and some σ>0 (which will be chosen later). Moreover, if 0<q<p-1, then from Lemma 4(2), we have for s sufficiently small,(50)Hs≥-C1sq+1.Then, by the continuity of H, we have for all s>-1/μ(51)Hs≥σΦϵs-C2sq+1.Letting 0<s<t, then we have(52)Ituntpunp-Isunspunp=-∫ΩcxGtuntpunp-Gsunspunp-∫ΩhxFtuntpunp-Fsunspunp=-∫Ωcx∫stddτGτunτpunpdτdx︸A+∫Ω-hxFtuntpunp-Fsunspunp︸B.Let us handle the two terms A and B, respectively. (53)A=-∫Ω∫stcxτpungτun-pτp-1Gτunτ2punpdτdx=-∫Ω∫stcxunpHτunτp+1dτdx.By using (51), we get(54)A≤∫Ω∫stcxunpC2unq+1τp-q-σΦϵτunτp+1dτdx≤∫ΩcxunpC2p-q-1unq+1sp-q-1-σ∫stΦϵτunτp+1dτdx.For the term B, we have(55)B≤C∫Ωhx1+tunptpunp+∫Ωhx1+sunpspunp≤C∫Ωhx1tnun+ununp+∫Ωhx1snun+ununp.By setting s≔1/un, we obtain(56)Ituntpun⩽Ivn+∫ΩcxC2p-q-1vnq+1-σ∫stΦϵτunτp+1unpdτdx+C∫Ωhx12+vnp+∫Ω+hx1+vnp⩽Ivn+C∫Ωcxvnq+1+∫Ωhx+2∫Ωhxvnp-σ∫Ωcxp1-1tnpunp+σ∫Ωcxp1-1tnpunp-σ∫Ω∫stcxΦϵτunτp+1unpdτdx⩽Ivn+C∫Ωcxvnq+1+∫Ωhx+2∫Ωhxvnp-σ∫Ωcxp1-1tnpunp-σ∫Ω∫stcxτp+11-ΦϵτununpdτdxBy the technical Lemma 9, we have(57)limϵ→0⁡∫Ω∫stcxτp+11-Φϵτununpdτdx=0,uniformlyinn∈N.Then, (58)Ituntpunp⩽1p-∫ΩcxGvn-∫ΩhxFvn+C∫Ωcxvnq+1+∫Ωhx+2∫Ωhxvnp-σ∫Ωcxp1-12p.We choose σ such that(59)σ>2p1+pChk2p-1∫Ωcxwhich gives(60)1p+Chk-σ∫Ωcxp1-12pdx<0.Since vn converges to 0 almost everywhere in Ω, weakly in W01,p(Ω), and strongly in Ls(Ω) for some 1≤s<p∗, then, we have(61)Itnun<0,inΩfornlargeenough.Hence, Izn is bounded. Therefore, this contradicts the fact that Izn is unbounded (see(a)).
Now, Ifq≥p-1, then from Lemma 4(1) and the continuity of H(s), we have for all s>-1/μ,(62)Hs≥σΦϵs-C1sp-1.Following the computations as in (52), we find exactly the same terms A and B. The term B is handled as in (55), whereas A is handled as follows: (63)A≤∫Ω∫stcxunpC1unp-1τ2-σΦϵτunτp+1,≤∫Ω∫stcxunpC1unp-1s-σΦϵτunτp+1.Moreover, since p-1k′<pk′<Np/N-p, then by using Sobolev embedding, the rest of the proof is similar to the case q∈(0,p-1). Hence, we have also the contradiction with the fact that I is unbounded (see(a)).To finish the proof of the Palais-Smale condition forI, we only need to show the following lemma.Lemma 10.
Any Palais-Smale sequence at the levelc~ of W01,p(Ω) has a strongly convergent subsequence.Proof.
Letun be a Palais-Smale sequence at the level c~ and then I′un→0 in W-1,p′(Ω), which means that (64)-Δpun-cxgun-hxfun→0inW-1,p′Ω.By Lemma 8, un is bounded in W01,p(Ω). Hence, un converges weakly to u in W01,p(Ω) and strongly in Ls(Ω) for some 1≤s<p∗. Therefore,(65)-Δpun→cxgu+hxfuinW-1,p′Ω.We know that the operator -Δp:W01,p(Ω)↦W-1,p′(Ω) is a homeomorphism ( [25]). Hence, from (65) we get (66)un→-Δp-1cxgu+hxfuinW01,pΩ.Therefore, by the uniqueness of the limit we have(67)un→u,inW01,pΩ.


### 3.2. Second Critical Point

In this subsection, we use the geometrical structure ofI (see Proposition 6) and the standard lower semicontinuity argument; we show the existence of the second critical point. We state the result as follows.Theorem 11.
Assume thatck and hk are suitably small to ensure Proposition 6. Then, the functional I possesses a critical point v∈B0,ρ with Iv≤0.Proof.
SinceI0=0, then infv∈B(0,ρ)I(v)≤0. Moreover, if h≢0, then we obtain that infv∈B(0,ρ)I(v)<0. Indeed, we choose v∈C0∞(Ω) a positive function that satisfies cv>0 and hv>0. From the definition of I in (6), we have for t>0(68)Itv=tp1p∫Ω∇vp-∫ΩcxGtvtpvpvp-∫ΩhxFtvtpvpvp.If q≥p-1, then from Lemma 4(2), we have G(s)/sp→c<+∞ as s→0+. If 0<q<p-1, obviously, we have G(s)/sp→+∞ as s→0+. In addition, in both cases, we have Fs/sp→+∞ as s→0+. Hence, by using these limits, we get from (68) that I(tv)<0 for t>0 small enough.
Now, we setm≔infv∈B(0,ρ)I(v). Then, by Proposition 6(i), we have Iv≥β>0 for v=ρ. Moreover, there exists a sequence (vn)⊂B(0,ρ) such that Ivn converges to m. Since vn is bounded in W01,p(Ω), then there exists a subsequence denoted again vn such that vn converges to v weakly in W01,p(Ω) and strongly in Ls(Ω) for some 1≤s<p∗ respectively. Hence, we get(69)∫ΩhxFvn→∫ΩhxFv,∫ΩcxGvn→∫ΩcxGvasn→+∞. In addition, since vp≤liminfn→∞vnp, then I(v)≤m=infv∈B(0,ρ)I(v). Hence, we conclude that v is a local minimum of I in B(0,ρ).Remark 12.
By Section3.1, I has a critical point at the level c~; that is, there exists w in W01,p(Ω) such that Iw=c~ and I′w=0. Since Iw=c~>0≥Iv, where v∈B0,ρ is the second critical point given in the previous theorem, then w is different from v. Hence, we have two distinct solutions for the problem P.


### 3.3. Boundedness of Solutions

Now, to finish the proof of our main result, it remains to show the boundedness of the solutions. Therefore, we show the following result.Proposition 13.
Any solutionu of the problem P′ belongs to L∞(Ω).Proof.
Ifu≤1, it is over. Otherwise, we begin by writing the problem P′ as follows:(70)-Δpu=ax1+up-1,where(71)ax=cxgu+hxfu1+up-1.Then, by Theorem 2.4 in [26], we can deduce the boundedness of u if we show that a belongs to Lp/N(1-ϵ)(Ω), for some ϵ∈]0,1[. Indeed, from (19) and Lemma 5, we obtain(72)ax≤Ccxur-p+1+1+hx.Let m>1 and m′ its conjugate. By using Hölder’s inequality in (72), we obtain(73)∫Ωaxp/N1-ϵ≤Ccxp/N1-ϵmur-p+1p/N1-ϵm′+hp/N1-ϵm+1. By choosing 0<ϵ<1-N-pr-p+1/N2-p/kN, we have(74)pN1-ϵm≤k,r-p+1pN1-ϵm′<NpN-p.Hence, the terms c(x)p/N(1-ϵ)m, h(x)p/N1-ϵm, and u(r-p+1)p/N(1-ϵ)m′ are finite (recall that c,h∈LkΩ for some k>N/p).


## 3.1. First Critical Point: Palais-Smale Condition

In this subsection, we prove thatI satisfies the Palais-Smale condition at the level c~. Precisely, we show that any Palais-Smale sequence at the level c~ for I is bounded in W01,p(Ω), and, then, it has a strongly convergent subsequence.They key point to prove the boundedness of the Palais-Smale sequence at the levelc~ in W01,p(Ω) is to show that g verifies the nonquadraticity condition at infinity NQ. Indeed, we have the following lemma.Lemma 7.
The functiong defined in (3) verifies the nonquadraticity condition at infinity NQ.Proof.
To proveNQ, we show that H is increasing and unbounded for s sufficiently large. We recall that H(s)=g(s)s-pG(s). Then, by simple calculations, we get(30)H′s=Cμs1+μsp-2ln⁡1+μsq-11-pln⁡1+μsμs+q,where C=(p-1)q-p+1/μq. Thus, H is increasing for s large enough. Moreover, H is unbounded. Indeed, by contradiction, if H is bounded, then there exists a positive constant M such that(31)Hs≤M,forslargeenough.In addition, from the definition of H and using integration by parts on G, we get(32)Hs=-C1μln⁡1+μsq1+μsp-1+qC∫0s1+μtp-1ln⁡1+μtq-1dt.By choosing δ∈p-1,p, we obtain(33)Hssδ=-1μln⁡1+μsq1+μsp-1sδ+qC∫0s1+μtp-1ln⁡1+μtq-1sδ≤Msδ.When s→+∞, we obtain H(s)/sδ→+∞ and M/sδ→0. Hence, we have a contradiction. As a conclusion, the function g verifies NQ.Lemma 8.
Letun be a Palais-Smale sequence at the level c~ for I in W01,p(Ω). Then, un is bounded in W01,p(Ω).Proof.
Letun be a Palais-Smale sequence at the level c~ for I in W01,p(Ω). We prove by contradiction that un is bounded in W01,p(Ω). We assume that un is unbounded in W01,p(Ω), that is, un→+∞.
For all integern≥0, we define(34)Izn≔max0≤t≤1⁡Itun,wherezn=tnun,tn∈0,1.We are going to prove that I(zn)→+∞ and also Izn is bounded, which is the desired contradiction.
(a) Showing that I ( z n ) → + ∞: We set vn≔un/un; then (vn) is bounded in W01,p(Ω). Hence, there exists a subsequence denoted again vn such that vn converges weakly and strongly to v in W01,p(Ω) and in Ls(Ω) for some 1≤s<p∗, respectively. Moreover, vn also converges to v almost everywhere in Ω. Recall that p∗≔Np/N-p is Sobolev conjugate.
Now, we claim by contradiction that v ≡ 0 a.e. in Ω.
Sinceun is Palais-Smale type sequence, then we have(35)Iun→c~,I′un∗→0.Hence,(36)∫Ω∇unp-2∇un∇φ-∫Ωcxgunφ-∫Ωhxfunφ=ϵn,for all φ∈W01,p(Ω) and for some ϵn→0 as n→+∞. We divide both sides of (36) by unp-1, to obtain(37)∫Ωcxgununp-1φ=ϵnunp-1+∫Ω∇vnp-2∇vn∇φ+∫Ωhxfununp-1φ.On the one hand, since vn converges weakly to v in W01,p(Ω) and by the inequality (19), then for n large enough the second and the third terms of the right-hand side of (37) are bounded.
On the other hand, ifv≢0 in Ω, then cv≢0 in Ω. Now, we choose φ∈W01,p(Ω) such that cvφ>0 in Ωφ and cvφ≡0 in Ω∖Ωφ, with |Ωφ|>0. Since vnun=un in Ω, then by using Lemma 4(3), we obtain(38)liminfcxgununp-1φ=liminfcxvnp-1gvnunvnunp-1φ=+∞inΩφ.Hence, by using Fatou’s lemma in (37) we obtain the unbounded term in the left-hand side of (37). Hence, the claim (i.ev≡0 a.e. in Ω.)
Sinceun→+∞, then there exists M>0 such that un>M, for n large enough. Moreover, we have(39)Izn≥IMunun=IMvn=Mpp-∫ΩcxGMvn-∫ΩhxFMvn.In what follows, we treat only the case 0<q<p-1. The other case follows with similar arguments. From Lemma 5(2), we have Gs≤c1|s|r+1+c2|s|q+1, where p-1<r<p. Since c∈Lk(Ω), for some k>N/p and vn converges strongly to v in Ls(Ω) with 1≤s<p∗, then, we obtain(40)∫ΩcxGMvn→0asn→+∞,due to v≡0 a.e. in Ω. By Hölder’s inequality, we get(41)∫ΩhxFMvn≤Casn→+∞.Hence, by choosing M>0 large enough, we deduce that I(zn)→+∞, as n→+∞.
(b) Showing that I z n is bounded: To prove that Izn is bounded, we distinguish two cases: tn≤2/un and tn>2/un.
The Case  tn≤2/un. Here, we only handle the proof for q∈0,p-1. The other case follows as in the proof of Proposition 6(i). By the definition of (zn) and I∈C1(W01,p(Ω),R), we have I′tnun,tnun=0, which means that(42)tnpunp=∫Ωcxgtnuntnun+∫Ωhxftnuntnun.By the definition of I in (6), we have(43)pItnun=tnpunp-p∫ΩcxGtnun-p∫ΩhxFtnun=∫ΩcxHtnun+∫ΩhxKtnun,where the function H is defined in NQ and K(s)≔f(s)s-pF(s). Moreover, from Lemma 5(2), we have (44)∫ΩcxHtnun≤∫Ωcxgtnuntnun+p∫ΩcxGtnun≤c1∫Ωcxtnunr+1+c2∫Ωcxtnunq+1.By choosing r and q as in the proof of Proposition 6(i), we get(45)∫ΩcxHtnun≤C1cktnunr+1+C2cktnunq+1.By inequality (19) and Sobolev’s embedding, we get(46)∫ΩhxKtnun≤∫Ωhxftnuntnun+p∫Ωhx(Ftnuntnun≤c1hk+c2hktnun+c3hktnunp.Then, by (43), (45), and (46), we obtain(47)Itnun≤C,for all n≥0, where C is independent of n. Thus, Izn is bounded, which contradicts the fact that Izn is unbounded (see(a)).
The Case t n > 2 / u n. Here, we are proceeding the technique inspired by [21]. To this end, we need the following technical lemma.
Lemma 9 . Let Φ : R → R the nonnegative function defined as (48) Φ s = e - ϵ / s 2 , i f   s ≠ 0 , 0 , i f   s = 0 , with ϵ > 0 . Then, we have (i)
l i m s → 0 Φ ( s ) = l i m s → 0 Φ ′ s = 0 . (ii)
for any positive function  z in  Ω and  p>1,(49)limϵ→0∫Ω∫stzxτp+11-Φϵτununpdτdx=0,uniformlyinn∈N.
Proof. Obviously we have (i). To prove (ii), we follow the same approach given in [21] for the case p=2 and z(x)=1, which can be immediately generalized for any positive function z and p>1. □
Now, we resume the proof of Lemma8. From Lemma 7, we have H(s)≥σ, for s large enough and some σ>0 (which will be chosen later). Moreover, if 0<q<p-1, then from Lemma 4(2), we have for s sufficiently small,(50)Hs≥-C1sq+1.Then, by the continuity of H, we have for all s>-1/μ(51)Hs≥σΦϵs-C2sq+1.Letting 0<s<t, then we have(52)Ituntpunp-Isunspunp=-∫ΩcxGtuntpunp-Gsunspunp-∫ΩhxFtuntpunp-Fsunspunp=-∫Ωcx∫stddτGτunτpunpdτdx︸A+∫Ω-hxFtuntpunp-Fsunspunp︸B.Let us handle the two terms A and B, respectively. (53)A=-∫Ω∫stcxτpungτun-pτp-1Gτunτ2punpdτdx=-∫Ω∫stcxunpHτunτp+1dτdx.By using (51), we get(54)A≤∫Ω∫stcxunpC2unq+1τp-q-σΦϵτunτp+1dτdx≤∫ΩcxunpC2p-q-1unq+1sp-q-1-σ∫stΦϵτunτp+1dτdx.For the term B, we have(55)B≤C∫Ωhx1+tunptpunp+∫Ωhx1+sunpspunp≤C∫Ωhx1tnun+ununp+∫Ωhx1snun+ununp.By setting s≔1/un, we obtain(56)Ituntpun⩽Ivn+∫ΩcxC2p-q-1vnq+1-σ∫stΦϵτunτp+1unpdτdx+C∫Ωhx12+vnp+∫Ω+hx1+vnp⩽Ivn+C∫Ωcxvnq+1+∫Ωhx+2∫Ωhxvnp-σ∫Ωcxp1-1tnpunp+σ∫Ωcxp1-1tnpunp-σ∫Ω∫stcxΦϵτunτp+1unpdτdx⩽Ivn+C∫Ωcxvnq+1+∫Ωhx+2∫Ωhxvnp-σ∫Ωcxp1-1tnpunp-σ∫Ω∫stcxτp+11-ΦϵτununpdτdxBy the technical Lemma 9, we have(57)limϵ→0⁡∫Ω∫stcxτp+11-Φϵτununpdτdx=0,uniformlyinn∈N.Then, (58)Ituntpunp⩽1p-∫ΩcxGvn-∫ΩhxFvn+C∫Ωcxvnq+1+∫Ωhx+2∫Ωhxvnp-σ∫Ωcxp1-12p.We choose σ such that(59)σ>2p1+pChk2p-1∫Ωcxwhich gives(60)1p+Chk-σ∫Ωcxp1-12pdx<0.Since vn converges to 0 almost everywhere in Ω, weakly in W01,p(Ω), and strongly in Ls(Ω) for some 1≤s<p∗, then, we have(61)Itnun<0,inΩfornlargeenough.Hence, Izn is bounded. Therefore, this contradicts the fact that Izn is unbounded (see(a)).
Now, Ifq≥p-1, then from Lemma 4(1) and the continuity of H(s), we have for all s>-1/μ,(62)Hs≥σΦϵs-C1sp-1.Following the computations as in (52), we find exactly the same terms A and B. The term B is handled as in (55), whereas A is handled as follows: (63)A≤∫Ω∫stcxunpC1unp-1τ2-σΦϵτunτp+1,≤∫Ω∫stcxunpC1unp-1s-σΦϵτunτp+1.Moreover, since p-1k′<pk′<Np/N-p, then by using Sobolev embedding, the rest of the proof is similar to the case q∈(0,p-1). Hence, we have also the contradiction with the fact that I is unbounded (see(a)).To finish the proof of the Palais-Smale condition forI, we only need to show the following lemma.Lemma 10.
Any Palais-Smale sequence at the levelc~ of W01,p(Ω) has a strongly convergent subsequence.Proof.
Letun be a Palais-Smale sequence at the level c~ and then I′un→0 in W-1,p′(Ω), which means that (64)-Δpun-cxgun-hxfun→0inW-1,p′Ω.By Lemma 8, un is bounded in W01,p(Ω). Hence, un converges weakly to u in W01,p(Ω) and strongly in Ls(Ω) for some 1≤s<p∗. Therefore,(65)-Δpun→cxgu+hxfuinW-1,p′Ω.We know that the operator -Δp:W01,p(Ω)↦W-1,p′(Ω) is a homeomorphism ( [25]). Hence, from (65) we get (66)un→-Δp-1cxgu+hxfuinW01,pΩ.Therefore, by the uniqueness of the limit we have(67)un→u,inW01,pΩ.


## 3.2. Second Critical Point

In this subsection, we use the geometrical structure ofI (see Proposition 6) and the standard lower semicontinuity argument; we show the existence of the second critical point. We state the result as follows.Theorem 11.
Assume thatck and hk are suitably small to ensure Proposition 6. Then, the functional I possesses a critical point v∈B0,ρ with Iv≤0.Proof.
SinceI0=0, then infv∈B(0,ρ)I(v)≤0. Moreover, if h≢0, then we obtain that infv∈B(0,ρ)I(v)<0. Indeed, we choose v∈C0∞(Ω) a positive function that satisfies cv>0 and hv>0. From the definition of I in (6), we have for t>0(68)Itv=tp1p∫Ω∇vp-∫ΩcxGtvtpvpvp-∫ΩhxFtvtpvpvp.If q≥p-1, then from Lemma 4(2), we have G(s)/sp→c<+∞ as s→0+. If 0<q<p-1, obviously, we have G(s)/sp→+∞ as s→0+. In addition, in both cases, we have Fs/sp→+∞ as s→0+. Hence, by using these limits, we get from (68) that I(tv)<0 for t>0 small enough.
Now, we setm≔infv∈B(0,ρ)I(v). Then, by Proposition 6(i), we have Iv≥β>0 for v=ρ. Moreover, there exists a sequence (vn)⊂B(0,ρ) such that Ivn converges to m. Since vn is bounded in W01,p(Ω), then there exists a subsequence denoted again vn such that vn converges to v weakly in W01,p(Ω) and strongly in Ls(Ω) for some 1≤s<p∗ respectively. Hence, we get(69)∫ΩhxFvn→∫ΩhxFv,∫ΩcxGvn→∫ΩcxGvasn→+∞. In addition, since vp≤liminfn→∞vnp, then I(v)≤m=infv∈B(0,ρ)I(v). Hence, we conclude that v is a local minimum of I in B(0,ρ).Remark 12.
By Section3.1, I has a critical point at the level c~; that is, there exists w in W01,p(Ω) such that Iw=c~ and I′w=0. Since Iw=c~>0≥Iv, where v∈B0,ρ is the second critical point given in the previous theorem, then w is different from v. Hence, we have two distinct solutions for the problem P.


## 3.3. Boundedness of Solutions

Now, to finish the proof of our main result, it remains to show the boundedness of the solutions. Therefore, we show the following result.Proposition 13.
Any solutionu of the problem P′ belongs to L∞(Ω).Proof.
Ifu≤1, it is over. Otherwise, we begin by writing the problem P′ as follows:(70)-Δpu=ax1+up-1,where(71)ax=cxgu+hxfu1+up-1.Then, by Theorem 2.4 in [26], we can deduce the boundedness of u if we show that a belongs to Lp/N(1-ϵ)(Ω), for some ϵ∈]0,1[. Indeed, from (19) and Lemma 5, we obtain(72)ax≤Ccxur-p+1+1+hx.Let m>1 and m′ its conjugate. By using Hölder’s inequality in (72), we obtain(73)∫Ωaxp/N1-ϵ≤Ccxp/N1-ϵmur-p+1p/N1-ϵm′+hp/N1-ϵm+1. By choosing 0<ϵ<1-N-pr-p+1/N2-p/kN, we have(74)pN1-ϵm≤k,r-p+1pN1-ϵm′<NpN-p.Hence, the terms c(x)p/N(1-ϵ)m, h(x)p/N1-ϵm, and u(r-p+1)p/N(1-ϵ)m′ are finite (recall that c,h∈LkΩ for some k>N/p).



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*Source: 6824502-2019-02-05.xml*