# Existence of Solutions for ap-Kirchhoff-Type Problem with Critical Exponent

Authors: Hayat Benchira; M. El Mokhtar Ould El Mokhtar; Atika Matallah
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this study, we prove the existence of a positive solution for ap-Kirchhoff-type problem with Sobolev exponent.


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## Body

## 1. Introduction and Main Results

In this study, we are concerned with the followingp-Kirchhoff-type problem:(1)−auθ−1p+bdiv∇up−2∇u=up∗−2u+λup−2u,inΩ,u=0,on∂Ω,Pλwhere Ω⊂ℝN is a bounded domain, 1<p<N, a, b>0, θ>1, . is the usual norm in W01,pΩ given by up=∫Ω∇updx, λ is a parameter, and p∗=pN/N−p is the critical Sobolev exponent corresponding to the noncompact embedding of W01,pΩ into Lp∗Ω.Since equationPλ contains an integral over Ω, it is no longer a pointwise identity; therefore, it is often called a nonlocal problem. It is called also nondegenerate if b>0 and a≥0, while it is named degenerate if b=0 and a>0.Such nonlocal elliptic problem such asPλ is related to the original Kirchhoff’s equation in [1] which was first introduced by Kirchhoff as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model takes into account the changes in length of the strings produced by transverse vibrations.Much interest has grown on problems involving critical exponents, starting from the celebrated paper by Brézis and Nirenberg [2]. They considered problem Pλ with a=0, b=1, and p=2. From their results, it came out that the space dimension N was going to play a crucial role. They established existence results in dimension N=3 when Ω is a ball, namely, they ensure the existence of a positive constant λ0 such that problem P˜0,1 admits a positive solution for λ∈λ0/4,λ1, where λ1 is the first eigenvalue of the operator −Δ. In higher dimensions, N≥4, they proved the existence of a positive solution for λ∈0,λ1 and no positive solution for λ>λ1 or λ≤0 and Ω is a starshaped domain.Moreover, by using the concentration compactness principle [3], the results of [2] were extended to the quasi-linear cases by Guedda and Veron [4]. Precisely, they proved that if N≥p2, then the quasi-linear Brezis–Nirenberg problem,(2)div∇up−2∇u=up∗−2u+λup−2u,inΩ,u=0,on∂Ω,has a positive solution if λ∈0,λ1, where λ1 is the first eigenvalue of(3)−div∇up−2∇u=λup−2u,inΩ,with Dirichlet boundary condition and no positive solution for λ≥λ1 or and Ω star shaped.In the last few years, great attention has been paid to the study of Kirchhoff problems involving critical nonlinearities. This problems create many difficulties in applying variational methods, we refer the readers to [5–13] and the references therein. More precisely, Naimen in [10] generalized the results of [2] for 3- and 4-dimensional Kirchhoff-type equations. For larger dimensional case, Figueiredo [7] considers the case N≥3 if λ>0 is sufficiently large.The main results in the present paper can be considered as the extension of the work of [4] for a p-Kirchhoff problem with large range of N. The competing effect of the nonlocal term with the critical nonlinearity and the lack of compactness of the embedding of W01,pΩ into Lp∗Ω, prevents us from using the variational methods in a standard way. So, we need more delicate estimates.To the best of our knowledge, many of the results are new forp>1 and even in the case θ=2. Our results and setting are more general and delicate, it is difficult to obtain the solution in the degenerate case when θ<p∗/p.Our technique is based on variational methods and concentration compactness argument [3], and we need to estimate the energy levels.In this paper, we define the best Sobolev constant for the imbeddingW01,pΩ ↪Lp∗Ω as(4)S≔infu∈W01,pΩ0up∫Ωup∗dxp/p∗.Then, we obtain the following existence result.Theorem 1.
Letb>0, N≥p2, and λ∈0,bλ1. If θ=N/N−p and 0<a<S−θ or θ<N/N−p and a>0, then Pλ has a positive solution.Remark 1.
Ifu is a solution of Pλ, we obtain(5)−div∇up−2∇u=gx,u,where gx,u=Mufx,u with Mu=auθ−1p+b−1 and fx,u=up∗−2u+λup−2u. By the extended Pohozaev identity in [4], we can get a nonexistence solutions for λ≤0 and Ω is starshaped.


## 2. Preliminary Results

In this study, we use the following notation:⟶ (resp ⇀) denotes strong (resp. weak) convergence, on1 denotes on1⟶0 as n⟶+∞, λ1 is the first eigenvalue of(6)−div∇up−2∇u,with Dirichlet boundary condition, and BRx0 is the ball centered at x0 and of radius R, u−=max−u,0.Recall that the infimumS is attained in ℝN by the functions of the form(7)vεx≔NεN−pp−1p−1N−p/p2ε+xp/p−1p−N/p,ε>0.Moreover,vε satisfies(8)∫ℝN∇vεpdx=∫ℝNvεp∗dx=Sp∗/p∗−p.LetR be a positive constant and set φ∈C0∞Ω such that 0≤φx≤1 for x≤R and φx≡1 for x≤R/2 and BR0⊂Ω. Set v˜εx=φxvεx, by taking zε=v˜ε∫Ωv˜εp∗dx−1/p∗ so that ∫Ωzεp∗dx=1.We have the well-known estimates asε⟶0:(9)zεp=S+OεN−p/p∫Ωzεpdx≥Cεp−1,ifN>p2,CεN−plnε,ifN=p2,CεN−p/p,ifN<p2,(see [14]).The energy functionalIλ:W01,pΩ⟶ℝ, corresponding to the problem Pλ, is given by(10)Iλu=aθpuθp+bpup−1p∗∫Ωup∗dx−λp∫Ωupdx,∀u∈W01,pΩ.Notice thatIλ is well defined in W01,pΩ and belongs to C1W01,pΩ,ℝ. We say that u∈W01,pΩ\0 is a weak solution of Yλ if, for any v∈W01,pΩ, there holds(11)Iλ′u,v=auθ−1p+b∫Ω∇up−2∇u∇vdx−∫Ωup∗−2uvdx−λ∫Ωup−2uvdx=0.Hence, a critical point of functionalIλ is a weak solution of problem Pλ.Definition 1.
Letc∈ℝ; a sequence un⊂W01,pΩ is called a PSc sequence (Palais–Smale sequence at level c) if(12)Iλun⟶candIλ′un⟶0asn⟶+∞.
Letc∈ℝ. We say that Iλ satisfies the Palais–Smale condition at level c if any PSc sequence contains a convergent subsequence in W01,pΩ.Lemma 1.
Leta,b>0, θ>1, σ≥0, and y˜=a/σSθ1/σ−1, for σ>1. For y≥0, we consider the function Ψ:ℝ+⟶ℝ∗, given by(13)Ψy=S−1yσ−aSθ−1y−b.Then,(1)
Ifσ=1, 0≤a<S−θ, and b>0, then the equation Ψy=0 has a unique positive solution,(14)y1=bSθ−1S−θ−a,andΨy≥0, for all y≥y1.(2)
Ifσ>1, then the equation Ψy=0 has a unique positive solution y2>y˜ and Ψy≥0, for all y≥y2.Proof.
(1)
Forσ=1, 0<a<S−θ, and b>0, we have(15)Ψy=Sθ−1S−θ−ay−b,that is, the equationΨy=0 has a unique positive solution,(16)y1=bS−θ−aSθ−1,andΨy≥0, for all y≥y1.(2)
Forσ>1, we have Ψ′y=σS−1yσ−1−aSθ−1 and(17)Ψ′′y=σσ−1S−1yσ−2>0,∀y>0.
Then,Ψ′y˜=0, Ψ′y<0, for y<y˜, and Ψ′y>0, for y>y˜. Hence, Ψ is concave function and(18)miny≥0Ψy=Ψy˜=−σ−1S−1aσSθσ/σ−1<0.
Moreover, we haveΨy˜<0 and limy⟶+∞Ψy=+∞; thus, from (18) and the concavity of Ψ, we can conclude that the equation Ψy=0 has a unique positive solution y2>y˜ and Ψy≥0, for all y≥y2.
Now, we will verify that the functionalIλ exhibits the Mountain Pass geometry.Lemma 2.
AssumeN>p>1, b>0, and λ∈0,bλ1. Suppose that θ=N/N−p and 0<a<S−θ or 1<θ<N/N−p and a>0. Then, there exist positive numbers δ1 and ρ1 such that, for all λ∈0,bλ1, we have(1)
Iλu≥δ1>0, with u=ρ1(2)
There existse∈W01,pΩ\0 such that e>ρ1 and Iλe<0Proof.
(1)
Letu∈W01,pΩ\0; by Sobolev and Young inequalities, it holds that(19)Iλu=aθpuθp+bpup−1p∗∫Ωup∗dx−λp∫Ωupdx≥−1p∗S−p∗/pup∗+aθpuθp+bpup−λpλ1up=−1p∗S−p∗/pup∗+aθpuθp+bλ1−λpλ1up.Letρ=u; since a>0, b>0, and λ<bλ1 and from (19), one has(20)Iλu≥−1p∗S−p∗/pρp∗+aθpρθp+bλ1−λpλ1ρp.Let us define(21)hρ=−1p∗S−p∗/pρp∗+aθpρθp+bλ1−λpλ1ρp,gρ=−S−p∗/pρp∗−p+aρθ−1p+bλ1−λλ1.Then,(22)h′ρ=−S−p∗/pρp∗−1+aρθp−1+bλ1−λλ1ρp−1=−S−p∗/pρp∗−p+aρθ−1p+bλ1−λλ1ρp−1.(i)
Let1<θ<N/N−p and a>0. We have g is strictly increasing on the interval:(23)0,aθ−1pp∗−pSp∗/p1/p∗−θp,and it is strictly decreasing on the interval(24)aθ−1pp∗−pSp∗/p1/p∗−θp,+∞,withg0=bλ1−λ/λ1, limx⟶+∞gx=−∞, and(25)gaθ−1pp∗−pSp∗/p1/p∗−θp=p∗−θpθ−1pS−p∗/paθ−1pp∗−pSp∗/pp∗−p/p∗−θp+bλ1−λλ1>0.Then, direct calculation shows that(26)maxρ≥0hρ=hρ1>0,hρ≥0for allρ≤ρ1withρ1>aθ−1pp∗−pSp∗/p1/p∗−θp.So, for allλ∈0,bλ1, we have(27)Iλu≥hρ1=δ1>0,withu=ρ1.(ii)
WhenN=θp/θ−1, one obtains p∗=θp. Let ρ=u; from (20), one hasIf0<a<S−θ, similar to (i), there exist ρ1, δ1>0 such that(29)ρ1=bλ1−λλ1S−θ−a1/θ−1p,δ1=hρ1=θ−1θpS−θ−a−p/θ−1pbλ1−λλ1θp/θ−1p.(2)
Foru∈W01,pΩ\0, t>0, we have(30)Iλtu≤−tp∗p∗∫Ωup∗dx+atθpθpuθp+btppup.(i)
Ifθ<N/N−p and t⟶+∞, then Iλtu⟶−∞. So, we can easily find e∈W01,pΩ\0 with e>ρ1, such that Iλe<0.(ii)
Ifθ=N/N−p and 0<a<S−θ, using (9) and taking ε1>0 small enough, then(31)Iλtzε≤atθpθpzεθp+btppzεp−tp∗p∗∫Ωzεp∗dx≤1θpazεθp−∫Ωzεp∗dxtθp+btppzεp≤1θpaSp∗/p∗−p+OεN−p/p−1θ−Sp∗/p∗−p+OεN/p−1tθp+btppSp∗/p∗−p+OεN−p/p−1≤1θpaSθp∗/p∗−p−Sp∗/p∗−ptθp+bpSp∗/p∗−ptp+OεN−p/p−1≤1θpa−S−θSθp∗/p∗−ptθp+bpSp∗/p∗−ptp+OεN−p/p−1,for all ε∈0,ε1. Then, it follows from the above inequality, Iλtzε⟶−∞ as t⟶+∞. Thus, choosing t0>0 sufficiently large and putting e:=t0zε, we have a function e∈W01,pΩ\0 satisfying e>ρ1, such that Iλe<0.Proof.
The proof is complete.Lemma 3.
AssumeN>p>1, a,b>0, λ∈0,bλ1, and θ≤N/N−p. Let c∈ℝ+ and un⊂W01,pΩ be a PSc sequence for Iλ; then, there exists a subsequence of un which we still denote by un and u∈W01,pΩ such that(32)un⇀uinW01,pΩ,with Iλ′u=0.Proof.
We have(33)Iλun⟶c,Iλ′un⟶0,that is,(34)c+on1=Iλun,on1v=Iλ′un,v,for any v∈W01,pΩ.
Then, asn⟶∞, it follows that(35)c+on1−1p∗on1un=Iλun−1p∗Iλ′un,un=ap∗−θpθpp∗unθp+bp∗−ppp∗unp−λp∗−ppp∗∫Ωunpdx,≥ap∗−θpθpp∗unθp+bp∗−ppp∗unp−λλ1p∗−ppp∗unp≥ap∗−θpθpp∗unθp+b−λλ1p∗−ppp∗unp.
Asλ<bλ1, we obtain that un is bounded in W01,pΩ. Up to a subsequence if necessary, there exists a function u∈W01,pΩ such that(36)un⇀uinW01,pΩ,un⇀uinLp∗Ω,un⟶uinLrΩ,for allr<p∗un⟶ua.e onΩ.
Then,(37)Iλ′un,v=0,for allv∈C0∞Ω,and thus, Iλ′u=0. This completes the proof of Lemma 4.
Now, we prove an important lemma which ensures the local compactness of the Palais–Smale sequence forIλ.
Fori=1,2¯, yi is defined in Lemma 2, and we define(38)Ci≔a1θp−1p∗Syi1/θ−1θ+b1p−1p∗Syi1/θ−1,a>0,b>0,(39)C∗≔C1,ifθ=NN−pand0<a<S−θ,C2,ifθ<NN−panda>0.Lemma 4.
Letb>0 and λ<bλ1, and suppose θ=N/N−p and 0≤a<S−θ or θ<N/N−p and a>0. If un⊂W01,pΩ is a PSc sequence for Iλ with c<C∗, then un contains a subsequence converging strongly in W01,pΩ.Proof.
By the proof of Lemma3, we have un is a bounded sequence in W01,pΩ. Hence, by the concentration compactness principle due to Lions [3], there exists a subsequence, still denoted by un, such that(40)∇unp⇀dη≥∇up+∑j∈Jηjδ˜j,unp∗⇀dγ˜=up∗+∑j∈Jγ˜jδ˜j,where J is an at most countable index set, δ˜j is the Dirac mass at xj and xj∈Ω¯, and ηjj∈J∪0andγ˜jj∈J∪0 are sets of nonnegative real numbers. Moreover,(41)ηj≥Sγ˜jp/p∗for allj∈J.
Forε>0, let ϕε,jx be a smooth cut-off function centered at xj such that 0≤ϕε,jx≤1, and(42)ϕε,jx=1,inBxj,ε,0,inΩ\Bxj,2ε,∇ϕε,jx≤2ε.
Sinceϕε,jun is bounded in W01,pΩ and Iλ′un⟶0 as n⟶∞, it holds by Hölder’s inequality:(43)0=limε⟶0limn⟶∞Iλ′un,ϕε,jun=limε⟶0limn⟶∞aunθ−1p+b∫Ω∇unp−2∇un∇ϕε,jundx−∫Ωunp∗unϕε,jundx−λ∫Ωunp−1ϕε,jundx≥b+aηjθ−1ηj−γ˜j.
Then,γ˜j≥bηj+aηjθ. Therefore, by (41), we deduce that(44)γ˜j=0orS−1γ˜jp∗−p/p∗−aSθ−1γ˜jp/p∗θ−1−b≥0.
Assume by contradiction that there existsj0∈J such that γ˜j0≠0. Set y=γ˜j0p/p∗θ−1 and σ=p∗−p/θ−1p; then, by (44), we obtain(45)S−1yσ−aSθ−1y−b≥0.
It is clear thatσ≥1, thanks to θ≤N/N−p. So, from (45) and the definition of Ψ in Lemma 1, we obtain(46)Ψy=S−1yσ−aSθ−1y−b≥0.
We will discuss it in two cases:Case 1:θ=N/N−p and 0<a<S−θ.According to Lemma1, we have Ψy1=0 and Ψy≥0 if y≥y1 with(47)y1=bS−θ−aSθ−1,which implies that(48)Sγ˜j0p/p∗≥Sy11/θ−1≕K1.Case 2:θ<N/N−p and a>0. In this case, from Lemma 2, we get Ψy2=0 and Ψy≥0 if y≥y2 with(49)y2>aθ−1pp∗−pSθθ−1p/p∗−θp,which implies that(50)Sγ˜j0p/p∗≥Sy21/θ−1≕K2.
Hence, using (41), we deduce(51)ηj0≥Sγ˜j0p/p∗≥K1,ifθ=NN−pand0<a<S−θ,K2,ifθ<NN−panda>0.
By Young inequality, we have(52)c=limn⟶∞Iλun−1θpIλ′un,un=limn⟶∞θ−1θpbunp+p∗−θpθpp∗∫Ωunp∗−λθ−1θp∫Ωunpdx≥θ−1θpbup+ηj0+p∗−θpθpp∗∫Ωup∗+γj0−λθ−1θp∫Ωupdx≥θ−1θpb−λλ1up+p∗−θpθpp∗∫Ωup∗+θ−1θpbηj0+p∗−θpθpp∗γ˜j0.
We observe thatθ−1/θpb−λ/λ1>0, p∗−θp≥0, and thus, for i∈1,2, we obtain(53)c≥θ−1θpbηj0+p∗−θpθpp∗γ˜j0≥1p−1θpbKi+p∗−θpθpp∗Kip∗/pS−p∗/p≥1p−1θpbKi+p∗−θpθpp∗Kip∗/pS−p∗/p++p∗−θpθpp∗aKiθ−p∗−θpθpp∗aKiθ+1p∗bKi−1p∗bKi≥p∗−θpθpp∗aKiθ+1p−1p∗bKi−p∗−θpθpp∗bKi++p∗−θpθpp∗Kip∗/pS−p∗/p−p∗−θpθpp∗aKiθ≥p∗−θpθpp∗aKiθ+1p−1p∗bKi++p∗−θpθpp∗KiKip∗−p/pS−p∗/p−aKiθ−1−b≥p∗−θpθpp∗aSyi1/θ−1θ+1p−1p∗bSyi1/θ−1++p∗−θpθpp∗Syi1/θ−1S−1yip∗−p/pθ−1−aSθ−1yi−b≥C∗+p∗−θpθpp∗Syi1/θ−1Ψyi≥C∗,since Ψyi=0 for i∈1,2 and C∗ is defined in (39). It is a contradiction with c<C∗. Then, J is empty, which implies that(54)∫Ωunp∗dx⟶∫Ωup∗dx.
Now, setl=limun as n⟶+∞; then, we have(55)Iλ′un,un=aunθ−1p+bunp−∫Ωunp∗dx−λ∫Ωunpdx=on1,(56)Iλ′un,v=aunθ−1p+b∫Ω∇unp−2∇un∇vdx+2,(57)−∫Ωunp∗unvdx−λ∫Ωunp−2unvdx=on1,for any v∈W1,pΩ.
Letn⟶+∞; then, from (55) and (56), we deduce that(58)alθ−1p+blp−∫Ωup∗dx−λ∫Ωupdx=0,(59)alθ−1p+b∫Ω∇up−2∇u∇vdx−∫Ωup∗−2uvdx−λ∫Ωup−2uvdx=0.
Taking the test functionv=u in (59), we obtain(60)alθ−1p+bup−∫Ωup∗dx−λ∫Ωupdx=0.
Therefore, equalities (58) and (59) imply that u=l. Consequently un converges strongly in W1,pΩ, which is the desired result.


### 2.1. Proof of the Main Result

By Lemma4, Iλ satisfies the Palais–Smale condition at level c for any c<C∗. So, the existence of the positive solution follows immediately from the following estimates.Lemma 5.
LetN≥p2, b>0, and λ<bλ1. Suppose that θ=N/N−p and 0<a<S−θ or θ<N/N−p and a>0. Then,(61)supt≥0Iλtzε<C∗.Proof.
Employing estimate (9), we define the following functions:(62)gt=Iλtzε=aθptθpzεθp+bptpzεp−1p∗tp∗∫Ωzεp∗dx−λptp∫Ωzεpdx=aθptθpzεθp+bptpzεp−1p∗S−p∗/pzεp∗tp∗−1p∗∫Ωzεp∗dx−S−p∗/pzεp∗tp∗−λptp∫Ωzεpdx,ht=−1p∗S−p∗/pzεp∗tp∗+aθptθpzεθp+bptpzεp.
Note thatlimt⟶∞gt=−∞ and gt>0 when t is close to 0 so that supt≥0gt is attained for some Tε>0. Furthermore, from g′Tε=0, it follows that(63)−Tεp∗−p∫Ωzεp∗dx+aTεθ−1pzεθp+bzεp−λ∫Ωzεpdx=0,and therefore,(64)Tεp∗−p∫Ωzεp∗dx=aTεθ−1pzεθp+bzεp−λ∫Ωzεpdx≥b−λλ1zεp.
Chooseε small enough so that, by (9), we have Tε≥t0, for some t0>0.
Besides, it holds(65)Tεp∗−θp∫Ωzεp∗dx=azεθp+bTεθ−1pzεp−λTεθ−1p∫Ωzεpdx≤azεθp+bTεθ−1pzεp.
Forθ<p∗/p, a>0, and b>0, we have by (9)(66)Tεp∗−θp∫Ωzεp∗dx≤azεθp+bzεpt0θ−1p.
Then, forε small enough, the above estimates yield Tε<t1 for some t1>0 (independently of ε).
Forθ=p∗/p, 0<a<S−θ, and b>0 and for ε small enough, we have, by (63),(67)Tεθ−1p=bzεp−λ∫Ωzεpdx∫Ωzεp∗dx−azεθp,which implies that Tε is bounded above, for all ε>0, that is, there exists a positive real number t2>0 (independently of ε).
Now, we estimategTε. It follows from h′t=0 that(68)−S−p∗/pzεp∗tp∗−1+atθp−1zεθp+btp−1zεp=0,that is,(69)−S−p∗/pzεp∗−ptp∗−p−atθ−1pzεθ−1p−b=0.
Sety=tθ−1pS1−θzεθ−1p, σ=p∗−p/θ−1p≥1, and(70)y∗=y1,ifp∗=θpand0≤a<S−θ,y2,ifp∗>θpanda≥0.
Then, by (69) and the definition of Ψ, we obtain(71)−S−1yσ−aSθ−1y−b=−Ψy=0,which implies from the proof of Lemma 1 that Ψy∗=0. Therefore, h′t∗=0, where t∗=S1/pzε−1y∗1/θ−1p. As Ψy is concave, then h′t is convex, so(72)maxt≥0ht=ht∗=−1p∗S−p∗/pzεp∗t∗p∗+aθpzεθpt∗θp+bpzεpt∗p.
Byh′t∗=0, we have(73)S−p∗/pzεp∗t∗p∗=azεθpt∗θp+bzεpt∗p.
So, from (73), we deduce that(74)maxt≥0ht=−1p∗azεθpt∗θp+bzεpt∗p+aθpzεθpt∗θp+bpzεpt∗p=a1θp−1p∗t∗θpzεθp+b1p−1p∗t∗pzεp=a1θp−1p∗Sθy∗θ/θ−1+b1p−1p∗Sy∗1/θ−1=C∗.
Consequently, by (9),(75)supt≥0Iλtzε≤supt≥0ht+1p∗S−p∗/pzεp∗−∫Ωzεp∗dxt1p∗−λpt0p∫Ωzεpdx≤C∗+C1S−p∗/pSp∗/p+OεN−p/p−1+−C2CεN−p/plnε,ifN=p2,CεN−p/p,ifN<p2,≤C∗+OεN−p/p−CεN−p/plnε,ifN=p2,CεN−p/p,ifN<p2,<C∗.which is the desired result.
Now, we can proof the existence of a Mountain Pass-type solution.Proof of Theorem 1.
Note thatIλ0=0, so from Lemma 2, Iλ satisfies the geometry conditions of the Mountain Pass Theorem [15]. Then, there exists a Palais–Smale sequence un at level c, such that(76)Iλun⟶c,Iλ′un⟶0,asn⟶+∞,with(77)0<c=infγ∈Γmaxt∈0,1Iλγt≤supt≥0Iλte<C∗,where(78)Γ=γ∈C0,1,W01,pΩ,γ0=0,γ1=e.
Using Lemma3, we have that un has a subsequence, still denoted by un, such that un⇀u in W01,pΩ as n⟶+∞. Hence, from Lemma 4 and 5, we have un⟶u in W01,pΩ as n⟶+∞. Hence, Iλ′u=0 and Iλu=c>0. So, as c>0=Iλ0, we can conclude that u is a nonzero solution of Yλ with positive energy. Now, we show that u>0 because(79)0=Iλ′u,u−=auθ−1p+b∫Ω∇up−2∇u∇u−dx+−∫Ωup∗−2uu−dx−λ∫Ωup−2uu−dx≥auθ−1p+b∫Ω∇u−pdx+∫Ωup∗dx+λ∫Ωupdx≥bu−p,which implies that u−=0. By the strong maximum principle [16], one has u>0. This completes the Proof of Theorem 1.


## 2.1. Proof of the Main Result

By Lemma4, Iλ satisfies the Palais–Smale condition at level c for any c<C∗. So, the existence of the positive solution follows immediately from the following estimates.Lemma 5.
LetN≥p2, b>0, and λ<bλ1. Suppose that θ=N/N−p and 0<a<S−θ or θ<N/N−p and a>0. Then,(61)supt≥0Iλtzε<C∗.Proof.
Employing estimate (9), we define the following functions:(62)gt=Iλtzε=aθptθpzεθp+bptpzεp−1p∗tp∗∫Ωzεp∗dx−λptp∫Ωzεpdx=aθptθpzεθp+bptpzεp−1p∗S−p∗/pzεp∗tp∗−1p∗∫Ωzεp∗dx−S−p∗/pzεp∗tp∗−λptp∫Ωzεpdx,ht=−1p∗S−p∗/pzεp∗tp∗+aθptθpzεθp+bptpzεp.
Note thatlimt⟶∞gt=−∞ and gt>0 when t is close to 0 so that supt≥0gt is attained for some Tε>0. Furthermore, from g′Tε=0, it follows that(63)−Tεp∗−p∫Ωzεp∗dx+aTεθ−1pzεθp+bzεp−λ∫Ωzεpdx=0,and therefore,(64)Tεp∗−p∫Ωzεp∗dx=aTεθ−1pzεθp+bzεp−λ∫Ωzεpdx≥b−λλ1zεp.
Chooseε small enough so that, by (9), we have Tε≥t0, for some t0>0.
Besides, it holds(65)Tεp∗−θp∫Ωzεp∗dx=azεθp+bTεθ−1pzεp−λTεθ−1p∫Ωzεpdx≤azεθp+bTεθ−1pzεp.
Forθ<p∗/p, a>0, and b>0, we have by (9)(66)Tεp∗−θp∫Ωzεp∗dx≤azεθp+bzεpt0θ−1p.
Then, forε small enough, the above estimates yield Tε<t1 for some t1>0 (independently of ε).
Forθ=p∗/p, 0<a<S−θ, and b>0 and for ε small enough, we have, by (63),(67)Tεθ−1p=bzεp−λ∫Ωzεpdx∫Ωzεp∗dx−azεθp,which implies that Tε is bounded above, for all ε>0, that is, there exists a positive real number t2>0 (independently of ε).
Now, we estimategTε. It follows from h′t=0 that(68)−S−p∗/pzεp∗tp∗−1+atθp−1zεθp+btp−1zεp=0,that is,(69)−S−p∗/pzεp∗−ptp∗−p−atθ−1pzεθ−1p−b=0.
Sety=tθ−1pS1−θzεθ−1p, σ=p∗−p/θ−1p≥1, and(70)y∗=y1,ifp∗=θpand0≤a<S−θ,y2,ifp∗>θpanda≥0.
Then, by (69) and the definition of Ψ, we obtain(71)−S−1yσ−aSθ−1y−b=−Ψy=0,which implies from the proof of Lemma 1 that Ψy∗=0. Therefore, h′t∗=0, where t∗=S1/pzε−1y∗1/θ−1p. As Ψy is concave, then h′t is convex, so(72)maxt≥0ht=ht∗=−1p∗S−p∗/pzεp∗t∗p∗+aθpzεθpt∗θp+bpzεpt∗p.
Byh′t∗=0, we have(73)S−p∗/pzεp∗t∗p∗=azεθpt∗θp+bzεpt∗p.
So, from (73), we deduce that(74)maxt≥0ht=−1p∗azεθpt∗θp+bzεpt∗p+aθpzεθpt∗θp+bpzεpt∗p=a1θp−1p∗t∗θpzεθp+b1p−1p∗t∗pzεp=a1θp−1p∗Sθy∗θ/θ−1+b1p−1p∗Sy∗1/θ−1=C∗.
Consequently, by (9),(75)supt≥0Iλtzε≤supt≥0ht+1p∗S−p∗/pzεp∗−∫Ωzεp∗dxt1p∗−λpt0p∫Ωzεpdx≤C∗+C1S−p∗/pSp∗/p+OεN−p/p−1+−C2CεN−p/plnε,ifN=p2,CεN−p/p,ifN<p2,≤C∗+OεN−p/p−CεN−p/plnε,ifN=p2,CεN−p/p,ifN<p2,<C∗.which is the desired result.
Now, we can proof the existence of a Mountain Pass-type solution.Proof of Theorem 1.
Note thatIλ0=0, so from Lemma 2, Iλ satisfies the geometry conditions of the Mountain Pass Theorem [15]. Then, there exists a Palais–Smale sequence un at level c, such that(76)Iλun⟶c,Iλ′un⟶0,asn⟶+∞,with(77)0<c=infγ∈Γmaxt∈0,1Iλγt≤supt≥0Iλte<C∗,where(78)Γ=γ∈C0,1,W01,pΩ,γ0=0,γ1=e.
Using Lemma3, we have that un has a subsequence, still denoted by un, such that un⇀u in W01,pΩ as n⟶+∞. Hence, from Lemma 4 and 5, we have un⟶u in W01,pΩ as n⟶+∞. Hence, Iλ′u=0 and Iλu=c>0. So, as c>0=Iλ0, we can conclude that u is a nonzero solution of Yλ with positive energy. Now, we show that u>0 because(79)0=Iλ′u,u−=auθ−1p+b∫Ω∇up−2∇u∇u−dx+−∫Ωup∗−2uu−dx−λ∫Ωup−2uu−dx≥auθ−1p+b∫Ω∇u−pdx+∫Ωup∗dx+λ∫Ωupdx≥bu−p,which implies that u−=0. By the strong maximum principle [16], one has u>0. This completes the Proof of Theorem 1.



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*Source: 2818091-2022-03-09.xml*