# Existence of Two Solutions for a Critical Elliptic Problem with Nonlocal Term inℝ4

Authors: Khadidja Sabri; Atika Matallah; Safia Benmansour; Mohammed El Mokhtar Ould El Mokhtar
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this paper, we prove the existence of two positive solutions for a critical elliptic problem with nonlocal term and Sobolev exponent in dimension four.


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## Body

## 1. Introduction

In this work, we are mainly concerned by the existence and the multiplicity of solutions for the following critical elliptic nonlocal problem:(1)Pa,b−a∫Ω∇u2dx+bΔu=u3+finΩ,u=0on∂Ω,where Ω is a smooth bounded domain of ℝ4, a and b are positive constants, and f belongs to H−1H−1isthedualofH01Ω satisfying suitable condition specified afterward.In our setting, the Laplacian operator is associated to Kirchhoff terma∫Ω∇u2dx+b, which contains an integral over the entire domain Ω, this implies that the equation in Pa,b is no longer a pointwise identity and so the problem turns to be nonlocal. This fact brings some mathematical difficulties in the search of the solution, and the solvability of this kind of problems has been under various authors’ attention; so, some classical investigations can be seen in the works [1, 2] and the references therein.Such nonlinear Kirchhoff’s equations can be used for describing the dynamic for an axially moving string and was first formulated by Kirchhoff himself [3] in 1883, he take into account the changes in length of the strings produced by transverse vibrations, and his model can be seen as a generalization of the classical D’Alembert wave equation for free vibrations of elastic strings.Problems which involve nonlocal operator have been widely studied due to their numerous and relevant applications in various fields of sciences. In particular, Kirchhoff type problems proved to be valuable tools for modeling several physical and biological phenomena, and many works have been made to ensure the existence of solutions for such problems; we quote in particular the article of Lions [4]. Since this famous paper, very fruitful development has given rise to many works on this advantageous axis, and in most of them, the used approach relies on topological methods. However, just few improvements were held concerning the multiplicity of solutions. In [5], Maia obtained a multiplicity of solutions for a class of px-Choquard equations with a nonlocal and nondegenerate Kirchhoff term by using truncation arguments and Krasnoselskii’s genus. In [6], Vetro studied the existence of two different notions of solutions by using Galerkin approximation method, jointly with the theory of pseudomonotone operators.With this regard, variational approach was solicited instead of topological methods to solve this kind of problems and also to prove the existence of multiple solutions; we refer interested readers to the works [7–10].We begin by giving an overview about the previous research related to the problemPa,b which can be written in the more general form,
(2)P~a,b−a∫Ω∇u2dx+bΔu=φλ,x,uinΩ,u=0on∂Ω,whereΩ is a smooth bounded domain of ℝN,N≥3.Without nonlocal terma=0, much interest has grown on problems involving critical exponents, and there are many publications dealing with the existence of solutions, starting from the celebrated paper by Brézis and Nirenberg [11] when φλ,x,u=u2∗−2u+λu and 2∗=2N/N−2 are the critical Sobolev exponent. For convenience of the reader, we give a brief summary of these results: they established existence results in dimension N=3 when Ω is a ball namely, and they ensure the existence of a positive constant λ0 such that the 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 sufficiently small, i.e., λ≤λ0 and no positive solution for λ>λ0 and Ω a star-shaped domain.Whenφλ,x,u=u2∗−2u+λuq,1<q<2∗, Ambrosetti et al. [12] established a multiplicity result in a bounded domain of ℝN,N≥3 indeed, they ensured the existence of a positive constant λ0 such that the problem P~0,1 admits two positive solutions for λ∈0,λ∗, a positive solution for λ=λ∗ and no positive solution for λ>λ∗.For the nonhomogeneous case, namely, whenφλ,x,u=u2∗−2u+fx, Tarantello [13] proved the existence of at least two solutions when f satisfies
(3)f1∫Ωfudx<4N−2N−2N+2N+2/4uN+2/4,for allu∈H01Ω,∫Ωu2∗dx=1.We emphases that the extension of the previous results to the nonlocal case, namely, for elliptic problems driven by Kirchhoff type operator are not obvious in high dimensionsN≥4. Therefore, no improvement was hold concerning the multiplicity of solutions in this case.For the casea>0 and φλ,x,u=μu2∗−2u+λu, Naimen in [14] treated the problem P~a,b for N=3 and obtained homologous results than the ones obtained by Brézis and Niremberg [11] in the nonlocal case under a suitable condition on a.In dimension four, Naimen in [15] used variational methods to explore problem P~a,b and showed that P~a,b admits a positive solution when a>0,b≥0.In the same order of ideas and still in the nonlocal casea>0, Lei et al. in [16] and Liao et al. in [17] extended the findings of [12] to a more general setting, namely, with the Kirchhoff operator; they established a multiplicity result in dimensions three and four, respectively.Benmansour and Bouchekif [8] generalized the results obtained by Tarantello [13] to the nonlocal case in dimension three. Indeed, they have shown the existence of two solutions under a sufficient condition on f by introducing the Nehari manifold.A natural question is to know whether the multiplicity result persists in the case of dimension four.In the current paper, our main purpose inspired by [8] is to see that the result obtained in [8] can be extended to dimension four. We emphases that our results are new and complement the above works.In order to studyPa,b, we shall work with the space H=H01Ω endowed with the norm u=∫Ω∇u2dx1/2, we use also the following notation: u−≔uH−1,up≔∫Ωupdx1/p for 1≤p<∞, C and Ci denote generic positive constants whose exact values are not important, BH0,r≔u∈H:u<r is the ball of center 0 and radius r, on1 denotes any quantity which tends to zero as n tends to infinity, and S is the best Sobolev constant defined by
(4)S≔infu∈H\0u2u42.To state the main results, we define(5)γa,b,f=b3/2infu∈Hu4=1233/2u6u44−au41/2−∫Ωfb3/2udx,where b>0, a<S−2 a small enough positive number and f belongs to H−1\0.The main results are concluded as the following theorems, which are news for the case whena≠0.Theorem 1.
Assume thatγa,b,f>0. Then, the problem Pa,b admits a local minimal solution u0 with Iu0<0. Furthermore u0≥0 for f≥0.Theorem 2.
Assume thatγa,b,f>0. Then, the problem Pa,b admits another solution u1 with Iu1>0. Furthermore, u1≥0 for f≥0.Notice thatγa,b,f≥0 if
(6)∫Ωfudx<2b33/2u6u4−au41/2forallu∈H.Moreover, the assumptionγa,b,f>0 certainly holds if f satisfies certain conditions, for example,
(7)f2∫Ωfu<2b33/2u3,for allu∈H with u4=1. Indeed, we have γ0,b,f is achieved and strictly positive if f satisfies f2 (see Lemma 2.2 in [13]). In order as
(8)2b33/2u6u441/2≤2b33/2u6u44−au41/2forallu∈H\0,then, γa,b,f≥γ0,b,f>0.This paper is structured as follows: in Section2, we give some basic results useful for what follows. Section 3 is devoted to the proofs of our main results.


## 2. Some Preliminary Results

We consider the energy functional associated to problemPa,b defined for u∈H and given by
(9)Iu=a4u4+b2u2−14u44−∫Ωfudx,forallu∈H.Observe thatI∈C1H,ℝ, whose derivative at the point u∈H is given by
(10)I′u,v=au2+b∫Ω∇u∇vdx−∫Ωu3vdx−∫Ωfvdx=0,forallv∈H.Obviously, ifu∈H is a critical point of the functional I; then, u is a weak solution of problem Pa,b.In general,I is not bounded from below on H, to overcome this and achieve a multiplicity result, the key argument is to use an appropriate manifold called in mathematical literature the Nehari manifold, it is a suitable manifold who has a pertinent property to prove the distinction of two solutions. Indeed, a minimizer in this set may give rise to solution of the corresponding equation. This so called Nehari manifold is defined by
(11)N≪u∈H:I′u,u=0.Lemma 3.
Assume thatb>0,a≥0, and f∈H−1\0. Then, the functional I is coercive and bounded from below on N.Proof.
Foru∈N, we have
(12)au4−u44=∫Ωfudx−bu2.
Therefore(13)Iu=a4u4+b2u2−14u44−∫Ωfudx=b4u2−34∫Ωfudx≥b4u2−34f_u.
Thus,I is coercive and bounded from below on N.
Lethut=Itu for t∈ℝ∗ and u∈H. These maps are known as fibering maps and were first introduced by Drábek and Pohozaev [18]. The set N is closely linked to the behavior of hut, for more details, see for example [19] or [20].
It is natural to splitN into three subsets:
(14)N+≔u∈N:hu′′1>0,N0≔u∈N:hu′′1=0,(15)N−≔u∈N:hu′′1<0,
wherehu′′t=3t2au4+bu2−3t2u44. These subsets correspond to local minima, points of inflexion, and local maxima of I, respectively.Definition 4.
A sequenceun⊂H is said to be a Palais Smale sequence at level c (PSc sequence in short) for I if
(16)Iun=c+on1andI′un=on1inH−1.
I verifies Palais Smale condition at level c (PSc condition in short) if any PSc sequence has a convergent subsequence in H.Next, foru≠0,b>0, and a, a small enough positive number set
(17)Φut=bu2t−u44−au4t3,then hu′t=Φut−∫Ωfudx. Easy computations show that Φu is concave and achieves its maximum at the point tmaxu where
(18)tmaxu=bu23u44−3au41/2.That is,(19)maxt≥0Φut=Φutmaxu=2b33/2u6u44−au41/2,(20)hu′tmaxu=2b33/2u6u44−au41/2−∫Ωfudx.Now, foru∈H\0 set Ψu=hu′tmaxu, that is
(21)γa,b,f=infu4=1Ψu.Fixt1>0, then, for t≥t1, we have Ψtu=tΨu,(22)t1γa,b,f≤tγa,b,f=tinfu4=1Ψu=inftu4=tΨtu=infv4=tΨv,so
(23)t1γa,b,f≤infv4≥t1Ψv.This is crucial for the following.Lemma 5.
Assume thatγa,b,f>0, then, N0=0.Proof.
Arguing by contradiction we assume that there existsu∈N0\0, i.e., u≠0 verifies
(24)au4+bu2−u44−∫Ωfudx=0,(25)3au4+bu2−3u44=0.
From (24) and (25), we derive that
(26)23bu2−∫Ωfudx=0,
As0≤a<S−2, we get from (24), (26), and the definition of S(27)3a−S−2u4+bu2≤0,that is
(28)u23a−S−2u2+b≤0.
Thus, asu≠0 and a<S−2, we derive that
(29)u2≥b3S−2−a>0,
therefore, from (25), we obtain u4≥t1, with
(30)t1=ab29S−2−a2+b29S−2−a1/4.
Then, from (23), (25), and (26), we get
(31)0<t1γa,b,f≤infu4≥t1Ψu≤Ψu=2b33/2u6u44−au41/2−∫Ωfudx=2b33/2u6b/3u21/2−∫Ωfudx=2b3u2−∫Ωfudx=0,which yields to a contradiction.Lemma 6.
Assume thatγa,b,f>0, then, for all u∈H\0, there exists unique positive value tu+=t+u such that
(32)tu+>tumax,tu+u∈N−andItu+u=maxt≥tmaxuItu.
Moreover, if∫Ωfudx>0, then, there exists unique positive value tu−=t−u such that
(33)0<tu−<tmaxu,tu−u∈N+andItu−u=inf0≤t≤tmaxuItu.Proof.
We havehut=Itu,hu′t=Φut−∫Ωfudx, and Φu is concave and achieves its maximum at the point tmaxu. If γa,b,f>0; then, there exists a unique tu+>0, such that hu′tu+=Φutu+−∫Ωfudx and hu′′tu+<0, which implies that tu+u∈N− and Itu+u≥Itu for all t≥tmaxu. Moreover, if ∫Ωfudx>0, then, there exists a unique tu−>0, such that hu′tu−=Φutu−−∫Ωfudx and hu′′tu+>0, which implies that tu−u∈N+ and Itu+u≤Itu for all t≤tmaxu.
Set(34)E1=u∈H:u=0oru<t+uu,(35)E2=u∈H\0:u>t+uu.
In the following lemma, we prove thatN− is closed and disconnects H in exactly two connected components E1 and E2.Lemma 7.
Assume thatγa,b,f>0. Then
(i)
N− is closed(ii)
H\N−=E1∪E2(iii)
N+⊂E1Proof.
Letun⊂N− and u~=limn⟶∞un, then, u~∈N. Assume by contradiction that u~∉N−, then
(36)3aun4+bun2−3un44<0,(37)3au~4+bu~2−3u~44=0.
So,u~∈N0, this implies that u~=0.
From (36) and the definition of S, we get un2≥bS2/3, so u~2≥bS2/3, which yields to a contradiction.
Letu∈N− and v=u/u, then, t+u=1, and there exists unique t+v such that t+vv∈N−. As
(38)t+vv=t+uu1uu∈N−,then
(39)t+uu1u=t+u=1.
Thus ifu∈H\0 and,
(40)t+uu1u≠1,then, u∉N− and
(41)H\N−=E1∪E2.
Letu∈N+ then
(42)t−u=t−uu1u=1.
Sincet+u>t−u, it follows that
(43)t+u=t+uu1u>1.
So,u<t+u/u, and we conclude that N+⊂E1.By Lemma6, we know that N and N− are not empty, so we can define θ0≤θ1 with
(44)θ0≔infu∈NIuandθ1≔infu∈N−Iu.Lemma 8.
Assume thatγa,b,f>0, then, there exists t∗>0 such that
(45)−916bf_2≤θ0≤−b4t∗2f_2.Proof.
Letu∈N, then
(46)Iu=b4u2−34∫Ωfudx≥b4u2−34f_u≥−916bf_2.
Thusθ0≥−9/16bf_2.
Setu∗∈H the unique solution of the equation −Δu=f, it follows
(47)∫Ωfu∗dx=u∗2=f_2.
By Lemma6, there exists a unique positive value t∗=tu∗− such that t∗u∗∈N+. So
(48)It∗u∗=−3a4t∗4u∗4−b2t∗2u∗2+34t∗4u∗44≤b4t∗2u∗2−b2t∗2u∗2=−b4t∗2f_2,consequently
(49)−916bf_2≤θ0≤−b4t∗2f_2.The following lemma is needed for prove the existence of Palais Smale sequences.Lemma 9.
Assume thatγa,b,f>0. Then, for any u∈N\0, there exist ε>0 and a differentiable function ζ:BH0,ε⟶ℝ+\0 such that
(50)ζ0=1,ζvu−v∈N,∀v∈BH0,ε,(51)ζ′0,v=4au2+2b∫Ω∇u∇vdx−4∫Ωu3vdx−∫Ωfvdx3au4+bu2−3u44.Proof.
Letu∈N\0 and define:ℝ×H⟶ℝ as follows
(52)ζ,v=aζ3u−v4+bζu−v2−ζ3u−v44−∫Ωfu−vdx.
Clearly,1,0=0. Moreover, from Lemma 5, we derive that
(53)∂∂ζ1,0=3au4+bu2−3u44≠0.
Thus, we get our result by a straightforward application of the implicit function theorem to the function at the point1,0.Lemma 10.
Letθ∈θ0,θ1. There exist a Palais Smale sequences un⊂N such that
(54)Iun⟶θ,I′un⟶0.Proof.
Assumeθ=θ0, by Lemma 3, I is bounded from below in N, then by applying the Ekeland Variational Principle, we can obtain a minimizing sequence un⊂N satisfying
(55)θ0≤Iun<θ0+1n,(56)Iw≥Iun−1nw−un,for all w∈N. Thus, Iun⟶θ0.
By using Lemma8, we get for n large enough
(57)−b4t∗2f_2≥θ0+1n>Iun=b4un2−34∫Ωfundx,this implies that
(58)∫Ωfundx≥b3un2+b3t∗2f_2,then
(59)∫Ωfundx≥b3t∗2f_2and∫Ωfundx≥b3un2,and by Holder inequality, we get
(60)f_un≥b3t∗2f_2>0andf_un≥b3un2.
Consequently,un≠0 and
(61)b3t∗2f_≤un≤3bf_.
Now, we show thatI′un tend to 0 as n goes to +∞. Arguing by contradiction and fix n with I′un≠0.
Then, by Lemma9, there exist ε>0 and a function ζn:BH0,ε⟶ℝ such that wn=ζnvnun−vn∈N with vn=δI′un/I′un and 0<δ<ε. By (56) and the Taylor expansion of I, we have
(62)−1nwn−un≤Iwn−Iun≤I′un,wn−un+∘wn−un=ζnvn−1I′un,un−δζnvnI′un,I′unI′un−+∘wn−un.
Then(63)ζnvnI′un−≤ζnvn−1δI′un,un+wn−unnδ+∘wn−unδ.
We have(64)limδ⟶0ζnvn=1,limδ⟶0ζnvn−1δ=limδ⟶0ζnvn−ζn0δ≤ζn′0−,(65)limδ⟶0wn−unnδ=limδ⟶01nδζnvn−1un−ζnvnvn≤limδ⟶01nζnvn−1δun+ζnvnvnδ≤1nζn′0−un+1.
This, together with (61) implies
(66)I′un−≤C3nζn′0−+1,for a suitable constant C3>0. Now, we must show that ζn′0− is uniformly bounded in n: indeed, since un is a bounded sequence, we have from Lemma 8(67)ζn′0≤C43aun4+bun2−3un44,for a suitable constant C4>0. Assume by contradiction that for a subsequence still called un, we have
(68)bun2−3un44−aun4=on1.
Then, asa is a small enough positive number, we get
(69)b3un23/2=un44−aun43/2+on1.
So from (61), we derive that
(70)un≥b3t∗2f_.
Also, asun∈N, we get from (68)
(71)∫Ωfundx=2un44−aun4+on1,then
(72)0<t∗2f_b33/2γa,b,f≤b3un21/2Ψun≤un44−aun41/2Ψun≤2b33/2un3−2un44−aun43/2=on1,which is absurd. At this point, we conclude that I′un⟶0 in H−1.
Forθ=θ1, adopting exactly the same way as in the case where θ=θ0.
In the following, we will prove our results.


## 3. Proofs of the Main Results

The proof of our main results is divided in two parts.


### 3.1. Existence of a Solution inN+

In this subsection, we prove thatI has a solution in N+.Proposition 11.
Assume thatγa,b,f>0. Then, the minimization problem
(73)θ0=infu∈NIu,attaints its infimum at a point u0∈N+. Moreover, u0 is a local minimizer for I in H.Proof.
By using Lemma10, there exists a bounded minimizing sequence un⊂N+ such that Iun⟶θ0 and I′un⟶0 in H−1. So, we deduce that un is bounded in H.
Passing to a subsequence if necessary, we haveun⇀u0 weakly in H, then, I′u0,w=0, for all w∈H. In addition, from (60), we get ∫Ωfu0dx>0. So, u0 is a weak solution for Pa,b and u0∈N.
Thus(74)θ0≤Iu0=b4u02−34∫Ωfu0dx≤lim¯n⟶∞Iun=θ0,then Iu0=θ0=infu∈NIu. It follows that un converges strongly to u0 in H, then, u0∈N+ and necessarily
(75)t−u0=1<tmaxu0.
To conclude thatu0 is a local minimum of I, let us recall that we have from Lemma 6(76)Isu≥Iu0forevery0<s<tmaxu0.
Chooseε>0 sufficiently small to have
(77)1<tmaxu0−wforw<ε,and tw satisfying twu0−w∈N for every w<ε. Since tw⟶1 as w⟶0, we can always assume that
(78)tw<tmaxu0−wforeverywsuchthatw<ε,so twu0−w∈N+ and for 0<s<tmaxu0−w, we have
(79)Isu0−w≥Itwu0−w≥Iu0,from (75), we can take s=1 and conclude that Iu0−w≥Iu0, for all w∈H such that w<ε. Thus, u0 is a local minimum of I.
Iff≥0, we have ∫Ωfu0dx≤∫Ωfu0dx and clearly Iu0≤Iu0, and from (75) necessarily t−u0≥1. Therefore, as Iu0=infu∈N+Iu, we get
(80)It−u0u0≤Iu0≤Iu0≤It−u0u0,so, we can always take u0≥0.


### 3.2. Existence of a Solution inN−

The following part is devoted to prove the existence of a second solutionu1 such that Iu1=θ1=infv∈N−Iv. First, we determine the good level for covering the Palais Smale condition.We have the following important result.Lemma 12.
Assume thatγa,b,f>0. Then, I satisfies the PSc condition for c<ca,b∗ with
(81)ca,b∗=θ0+b24S−2−a.Proof.
Letun be a PSc sequence with c<ca,b∗, then
(82)c+on1=Iun−14I′un,un=b4un2−34∫Ωfundx≥b4un2−34f_un.
Hence,un is a bounded sequence in H. Thus for a subsequence still denoted un and we can find u1∈H such that un⇀u1 in H,∫Ωfundx⟶∫Ωfu1dx and un⟶u1 a.e in Ω. Therefore, u1≠0,u1∈N, and Iu1≥θ1.
Letwn=un−u1. From Brézis-Lieb Lemma [21], one has
(83)un2=wn2+u12+on1,un4=wn4+2wn2u12+u14+on1,(84)un44=wn44+u144+on1,this implies that
(85)on1=I′un,un=awn4+bwn2−wn44−∫Ωfwndx+2awn2u12.
Assume thatwn⟶l with l>0, then, by (85) and the Sobolev inequality, we obtain
(86)S−2−al4≥bl2+2al2u12.this implies that
(87)S−2−al4−bl2≥0.
Hence(88)l2≥bS−2−a.
On the other hand, we have(89)c+on1=Iun−14I′un,un=b4un2−34∫Ωfundx=b4wn2+b4u12−34∫Ωfu1dx=b4wn2+Iu1−14I′u1,u1=b4wn2+Iu1,consequently, we obtain
(90)c≥θ0+b24S−2−a,which is a contradiction. Therefore, l=0 and un⟶u1 strongly in H.
Now, it is natural to show thatθ1<ca,b∗. As it is well know, seeHL, the best Sobolev constant S defined above is attained in ℝ4 by
(91)Uεx=εε2+x2−1,ε>0andx∈ℝ4.
Forx0∈Ω let Vεx=Uεx−x0ϕ∈C0∞Ω such that ϕx=1 for x∈Bx0r,ϕx=0 for x∈ℝ4\Bx02r, and 0≤ϕ≤1,∇ϕ≤C. Now, we shall give some useful estimates of the extremal functions Uε. Let Vεx=ϕxε+x−x02−2 and uε=Vεxuε4. The following estimates are obtained in [22] as ε tends to 0(92)uε42=1,uε2=S+Oε,C1ε4−q≤∫Ωuεqdx≤C2ε4−qforq>2.
LetΩ′⊂Ω a set of positive measure such that u0>0 on Ω′ (if not replace u0 and f by −u0 and −f, respectively).Lemma 13.
Letb>0,0≤a<S−2 a small enough positive number. Assume that f∈H−1\0 satisfies γa,b,f>0; then, for every t>0 and a.e. x0∈Ω′⊂Ω, there exists ε0 such that Iu0+tuε<ca,b∗ for every 0<ε<ε0.Proof.
Forε small enough, let us consider the functional g defined by
(93)gt=bt22uε2−t44uε44.
We have by (92)
(94)supt≥0gt=b24S−2+Oε.
On the other hand, sinceu0∈N+ is a solution of problem Pa,b, we have u0≤C, Iu0=θ0 and I′u0,tuε=0.
Then we obtain by (92)
(95)Iu0+tuε=a4u0+tuε4+b2u0+tuε2−14u0+tuε44−∫Ωfu0+tuεdx=a4u04+b2u02−14u044−∫Ωfu0dx+at44uε4+bt22uε2−t44uε44+au02+b∫Ω∇u0∇tuεdx−∫Ωu03tuεdx−t∫Ωfuεdx+t2a∫Ω∇u0∇uε2dx+a2u02uε2−32∫Ωu02uε2dx+t3auε2∫Ω∇u0∇uεdx−∫Ωuε3u0dx≤Iu0+I′u0,tuε+at44uε4+bt22uε2−t44uε44+t2aC1+t3aC2−C3ε≤θ0+ht,where
(96)ht=at44uε4+bt22uε2−t44uε44+at2C1+t3aC2−C3ε.
Sincea<S−2, we have ht⟶−∞ as t goes to ∞ and ht⟶0 as t goes to 0. This implies that there exist 0<T1<T2 such that
(97)supt>0Iu0+tuε≤supT1<t<T2Iu0+tuε≤θ0+bt22uε2−t44uε44+aT244uε4+aT22C1+T13aC2−C3ε≤θ0+b24S−2+Oε+aT22C1+T13C2−T13C3ε.
Then, fora=εσ with σ>1, we conclude
(98)Iu0+tuε<θ0+b24S−2=c0,b∗≤ca,b∗.
Therefore,(99)supt>0Iu0+tuε<ca,b∗,for a small enough positive number.Proposition 14.
Letb>0,0≤a<S−2 a small enough positive number. Assume that f∈H−1\0 satisfies γa,b,f>0, then, I has a local minimizer u1 on N− such that Iu1=θ1. Moreover u1 is a local minimizer for I on H.Proof.
By Lemma6, for every u∈H such that u=1, there exists unique t+u>0 such that t+uu∈N− and It+u≥Itu, for all t≥tmaxu. Then, for suitable constant T3>0, we deduce that
(100)0<t+u<T3,forallusuchthatu=1.
Therefore fort0>0 carefully chosen, the estimate (97) holds for ε small enough.
Thus, we derive that(101)u0+t0uε2>C12≥t+u0+t0uεu0+t0uε2,forallε>0smallenough.
Then, from Lemma7, we conclude that u0+t0uε∈E2.
Set(102)Γ=ξ:0,1⟶Hcontinuous,ξ0=u0,ξ1=u0+t0uε.
It is obvious thatξ:0,1⟶H given by ξt=u0+tt0uε belongs to Γ. We conclude from Lemma 13 that
(103)c=infξ∈Γmaxt∈0,1Iξt<c0,b∗.
As the range of anyξ∈Γ intersects N−, one has
(104)c≥θ1=infN−I,
From Lemma10, we can obtain a minimizing sequence un⊂N− such that
(105)Iun⟶θ1andI′un⟶0.
We also deduce thatθ1<c0,b∗.
Consequently, we obtain a subsequence still denotedun, and we can find u1∈H such that
(106)un⟶u1stronglyinH.
This implies thatu1 is a critical point for I,u1∈N− and Iu1=θ1.
Finally, forf≥0, let t+u1>0 satisfying t+u1u1∈N−. From Lemma 6, we have
(107)It+u1u1≥Iu1=maxt≥tmaxItu1≥It+u1u1≥It+u1u1.
So we conclude thatu1≥0.


## 3.1. Existence of a Solution inN+

In this subsection, we prove thatI has a solution in N+.Proposition 11.
Assume thatγa,b,f>0. Then, the minimization problem
(73)θ0=infu∈NIu,attaints its infimum at a point u0∈N+. Moreover, u0 is a local minimizer for I in H.Proof.
By using Lemma10, there exists a bounded minimizing sequence un⊂N+ such that Iun⟶θ0 and I′un⟶0 in H−1. So, we deduce that un is bounded in H.
Passing to a subsequence if necessary, we haveun⇀u0 weakly in H, then, I′u0,w=0, for all w∈H. In addition, from (60), we get ∫Ωfu0dx>0. So, u0 is a weak solution for Pa,b and u0∈N.
Thus(74)θ0≤Iu0=b4u02−34∫Ωfu0dx≤lim¯n⟶∞Iun=θ0,then Iu0=θ0=infu∈NIu. It follows that un converges strongly to u0 in H, then, u0∈N+ and necessarily
(75)t−u0=1<tmaxu0.
To conclude thatu0 is a local minimum of I, let us recall that we have from Lemma 6(76)Isu≥Iu0forevery0<s<tmaxu0.
Chooseε>0 sufficiently small to have
(77)1<tmaxu0−wforw<ε,and tw satisfying twu0−w∈N for every w<ε. Since tw⟶1 as w⟶0, we can always assume that
(78)tw<tmaxu0−wforeverywsuchthatw<ε,so twu0−w∈N+ and for 0<s<tmaxu0−w, we have
(79)Isu0−w≥Itwu0−w≥Iu0,from (75), we can take s=1 and conclude that Iu0−w≥Iu0, for all w∈H such that w<ε. Thus, u0 is a local minimum of I.
Iff≥0, we have ∫Ωfu0dx≤∫Ωfu0dx and clearly Iu0≤Iu0, and from (75) necessarily t−u0≥1. Therefore, as Iu0=infu∈N+Iu, we get
(80)It−u0u0≤Iu0≤Iu0≤It−u0u0,so, we can always take u0≥0.


## 3.2. Existence of a Solution inN−

The following part is devoted to prove the existence of a second solutionu1 such that Iu1=θ1=infv∈N−Iv. First, we determine the good level for covering the Palais Smale condition.We have the following important result.Lemma 12.
Assume thatγa,b,f>0. Then, I satisfies the PSc condition for c<ca,b∗ with
(81)ca,b∗=θ0+b24S−2−a.Proof.
Letun be a PSc sequence with c<ca,b∗, then
(82)c+on1=Iun−14I′un,un=b4un2−34∫Ωfundx≥b4un2−34f_un.
Hence,un is a bounded sequence in H. Thus for a subsequence still denoted un and we can find u1∈H such that un⇀u1 in H,∫Ωfundx⟶∫Ωfu1dx and un⟶u1 a.e in Ω. Therefore, u1≠0,u1∈N, and Iu1≥θ1.
Letwn=un−u1. From Brézis-Lieb Lemma [21], one has
(83)un2=wn2+u12+on1,un4=wn4+2wn2u12+u14+on1,(84)un44=wn44+u144+on1,this implies that
(85)on1=I′un,un=awn4+bwn2−wn44−∫Ωfwndx+2awn2u12.
Assume thatwn⟶l with l>0, then, by (85) and the Sobolev inequality, we obtain
(86)S−2−al4≥bl2+2al2u12.this implies that
(87)S−2−al4−bl2≥0.
Hence(88)l2≥bS−2−a.
On the other hand, we have(89)c+on1=Iun−14I′un,un=b4un2−34∫Ωfundx=b4wn2+b4u12−34∫Ωfu1dx=b4wn2+Iu1−14I′u1,u1=b4wn2+Iu1,consequently, we obtain
(90)c≥θ0+b24S−2−a,which is a contradiction. Therefore, l=0 and un⟶u1 strongly in H.
Now, it is natural to show thatθ1<ca,b∗. As it is well know, seeHL, the best Sobolev constant S defined above is attained in ℝ4 by
(91)Uεx=εε2+x2−1,ε>0andx∈ℝ4.
Forx0∈Ω let Vεx=Uεx−x0ϕ∈C0∞Ω such that ϕx=1 for x∈Bx0r,ϕx=0 for x∈ℝ4\Bx02r, and 0≤ϕ≤1,∇ϕ≤C. Now, we shall give some useful estimates of the extremal functions Uε. Let Vεx=ϕxε+x−x02−2 and uε=Vεxuε4. The following estimates are obtained in [22] as ε tends to 0(92)uε42=1,uε2=S+Oε,C1ε4−q≤∫Ωuεqdx≤C2ε4−qforq>2.
LetΩ′⊂Ω a set of positive measure such that u0>0 on Ω′ (if not replace u0 and f by −u0 and −f, respectively).Lemma 13.
Letb>0,0≤a<S−2 a small enough positive number. Assume that f∈H−1\0 satisfies γa,b,f>0; then, for every t>0 and a.e. x0∈Ω′⊂Ω, there exists ε0 such that Iu0+tuε<ca,b∗ for every 0<ε<ε0.Proof.
Forε small enough, let us consider the functional g defined by
(93)gt=bt22uε2−t44uε44.
We have by (92)
(94)supt≥0gt=b24S−2+Oε.
On the other hand, sinceu0∈N+ is a solution of problem Pa,b, we have u0≤C, Iu0=θ0 and I′u0,tuε=0.
Then we obtain by (92)
(95)Iu0+tuε=a4u0+tuε4+b2u0+tuε2−14u0+tuε44−∫Ωfu0+tuεdx=a4u04+b2u02−14u044−∫Ωfu0dx+at44uε4+bt22uε2−t44uε44+au02+b∫Ω∇u0∇tuεdx−∫Ωu03tuεdx−t∫Ωfuεdx+t2a∫Ω∇u0∇uε2dx+a2u02uε2−32∫Ωu02uε2dx+t3auε2∫Ω∇u0∇uεdx−∫Ωuε3u0dx≤Iu0+I′u0,tuε+at44uε4+bt22uε2−t44uε44+t2aC1+t3aC2−C3ε≤θ0+ht,where
(96)ht=at44uε4+bt22uε2−t44uε44+at2C1+t3aC2−C3ε.
Sincea<S−2, we have ht⟶−∞ as t goes to ∞ and ht⟶0 as t goes to 0. This implies that there exist 0<T1<T2 such that
(97)supt>0Iu0+tuε≤supT1<t<T2Iu0+tuε≤θ0+bt22uε2−t44uε44+aT244uε4+aT22C1+T13aC2−C3ε≤θ0+b24S−2+Oε+aT22C1+T13C2−T13C3ε.
Then, fora=εσ with σ>1, we conclude
(98)Iu0+tuε<θ0+b24S−2=c0,b∗≤ca,b∗.
Therefore,(99)supt>0Iu0+tuε<ca,b∗,for a small enough positive number.Proposition 14.
Letb>0,0≤a<S−2 a small enough positive number. Assume that f∈H−1\0 satisfies γa,b,f>0, then, I has a local minimizer u1 on N− such that Iu1=θ1. Moreover u1 is a local minimizer for I on H.Proof.
By Lemma6, for every u∈H such that u=1, there exists unique t+u>0 such that t+uu∈N− and It+u≥Itu, for all t≥tmaxu. Then, for suitable constant T3>0, we deduce that
(100)0<t+u<T3,forallusuchthatu=1.
Therefore fort0>0 carefully chosen, the estimate (97) holds for ε small enough.
Thus, we derive that(101)u0+t0uε2>C12≥t+u0+t0uεu0+t0uε2,forallε>0smallenough.
Then, from Lemma7, we conclude that u0+t0uε∈E2.
Set(102)Γ=ξ:0,1⟶Hcontinuous,ξ0=u0,ξ1=u0+t0uε.
It is obvious thatξ:0,1⟶H given by ξt=u0+tt0uε belongs to Γ. We conclude from Lemma 13 that
(103)c=infξ∈Γmaxt∈0,1Iξt<c0,b∗.
As the range of anyξ∈Γ intersects N−, one has
(104)c≥θ1=infN−I,
From Lemma10, we can obtain a minimizing sequence un⊂N− such that
(105)Iun⟶θ1andI′un⟶0.
We also deduce thatθ1<c0,b∗.
Consequently, we obtain a subsequence still denotedun, and we can find u1∈H such that
(106)un⟶u1stronglyinH.
This implies thatu1 is a critical point for I,u1∈N− and Iu1=θ1.
Finally, forf≥0, let t+u1>0 satisfying t+u1u1∈N−. From Lemma 6, we have
(107)It+u1u1≥Iu1=maxt≥tmaxItu1≥It+u1u1≥It+u1u1.
So we conclude thatu1≥0.


## 4. Conclusion

In our work, we have searched the critical points as the minimizers of the energy functional associated to the problem on the constraint defined by the Nehari manifoldN, which is a solution of our problem. Under some sufficient conditions, we split N in two disjoint subsets N+ and N−. Thus, we consider the minimization problems on N+ and N−, respectively. If γa,b,f>0, then, the problem Pa,b has a local minimal solution u0 with Iu0<0. Furthermore, u0≥0 for f≥0 and if γa,b,f>0. The problem Pa,b has another solution u1 with Iu1>0. Furthermore, u1≥0 for f≥0.



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*Source: 2187810-2022-06-20.xml*