# Existence Results for a Class of the Quasilinear Elliptic Equations with the Logarithmic Nonlinearity

Authors: Zhoujin Cui; Zisen Mao; Wen Zong; Xiaorong Zhang; Zuodong Yang
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, the nonlinear quasilinear elliptic problem with the logarithmic nonlinearity−div∇up−2∇u=axφpulogu+hxψpu in Ω⊂Rn was studied. By means of a double perturbation argument and Nehari manifold, the authors obtain the existence results.


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## Body

## 1. Introduction

In this paper, we consider the existence of solution to the following problem(1)−div∇up−2∇u=axφpulogu+hxψpu,inΩ,where Ω⊂Rn, φpz=zp−2z,ψpz=zp−1z,p>2, and n≥1. We always suppose that ax is a sign-changing function; hx≥0 is a ∈C1 function.Equations of the above form are mathematical models occurring in studies of thep-Laplace equation, generalized reaction-diffusion theory [1], non-Newtonian fluid theory [2, 3], non-Newtonian filtration theory [4, 5], and the turbulent flow of a gas in porous medium [6]. In the non-Newtonian fluid theory, the pair p is a characteristic quantity of the medium. Media with p>2 are called dilatant fluids, and those with p<2 are called pseudoplastics. If p=2, they are Newtonian fluids. When p≠2, the problem becomes more complicated since certain nice properties inherent to the case p=2 seem to be lose or at least difficult to verify. The main differences between p=2 and p≠2 can be founded in [7, 8].In recent years, logarithmic nonlinearity is widely used in pseudo-parabolic equations which describe the mathematical and physical phenomena. Equations of the type1 have been studied by many authors when p=2 (see, for example, [9–12] and the reference therein). To do so, the authors always use the nice properties of Δ, such that, maximum principle and comparison principle and so on. Meanwhile, existence and structure of solutions for such equations with p>1 in bounded domains have also attracted much interest (see [13, 14]).In the following discussion, we consider two different situations. Firstly, we consider the existence of positive solution for problem1 with Neumann boundary conditions. In this case, suppose that Ω=BR=BR0⊂Rn,ax>0,hx≥0 are also radial functions, ax=ax, hx=hx in BR. Our strategy in the study of problem 1 is to adopt a double perturbation argument. First, following [15, 16] (see also [17]), for each 0<ε<1, we consider a family of approximate problems
(2)−div∇up−2∇u=axφpulogu2+εu+εu+ε+hxψpuinBR,u>0,inBR,∂vu=0,on∂BR.Then, it is natural to look for a family of solutions of2 and then to pass the limit as ε→0 to obtain a solution to 1.For each0<r<R, define ArR≔BR\B¯r. Consider the second family of problems
(3)−div∇up−2∇u=axφpulogu2+εu+εu+ε+hxψpu,inArR,u>0,inArR,u=θ,on∂Br,∂vu=0,on∂BR.Here,θ>0 is an appropriate constant. When r→0+, we get a solution to 2. The role of problem 3 is that we cannot use Poincare inequality to solve 2 directly by variational methods.Secondly, we consider the multiple solutions for problem1 with Dirichlet boundary conditions. In this case, we consider ax is a sign-changing function, hx=0. The method is based on Nehari manifold and logarithmic Sobolev inequality.By modification of the methods given in [18–22], we obtain the following results.Theorem 1.
Letax>0,hx≥0 be any radial C1 function. Then, problem 1 has a positive radial solution u∈C1B¯R\0∩CB¯R.Remark 2.
Theorem1 is valid even if we change the logarithm by a more general singular function. In fact, suppose g:0,1→R is a smooth function such that
(4)lims→0+gs=−∞,lims→0+gssm=1,for some m∈0,1. Then, we can perturb g by a family gε of smooth functions decreasing in ε, such that gε0=0 and gεs→gs pointwise in s∈0,∞ as ε→0. This perturbation can be done in such a way that gε0≥0 for some εn>0, and then, all the results in Section 2 hold with little modification.Theorem 3.
Lethx=0,ax∈CΩ¯ and changes sign in Ω¯, satisfying
(5)maxΩ¯ax≤1μ,where μ=nLp/peexpmp2Ωn/ne,Ωn is the volume of Ω in Rn. Then, 1 possesses at least two nontrivial solutions.The paper is organized as follows. In Section2, we construct a sub- and a supersolution for 3 and finish the proof of Theorem 1. In Section 3, we prove Theorem 3 by the method of Nehari manifold and logarithmic Sobolev inequality.


## 2. Proof of Theorem1

### 2.1. Sub- and Supersolution for3

Lemma 4.
Suppose thatθ>1. Then, the function u≡1 is a subsolution for 3 which does not depend on 0<ε≤1 and θ.Proof.
We just need to see that, sinceax>0,hx≥0 in BR, the following inequality holds independently of 0<ε≤1 and θ>1:
(6)axlog1+ε1+ε1+ε+hx≥log1=0,We proceed to find a supersolution for3. Denote by Xr, the following subspace of H1ArR:
(7)Xr≔u∈H1ArRu=0on∂Br.Forυ∈Xr, we define the expression:
(8)vr≔∫ArR∇v2dx1/2.Remark.
The expression⋅r defines a norm on Xr, and Xr,⋅r is a reflexive Banach space. Furthermore, by ([23], (7.44)), the Poincare inequality holds on Xr, that is, there exists η>0 such that
(9)∫ArRvpdx≤η∫ArR∇vpdx.Next, we work with the radial formulation forEε,r in the specific case that ε=1,
(10)−sn−1opu′′=sn−1asopulogu2+u+1u+1+sn−1hsψpu,inr<s<R,u>0,ur=θ,inr<s<R,u′R=0,where ϕps=sp−2s. Notice that
(11)logu2+u+1u+1≥0foru≥0.For simplicity, denote(12)fs,z=asopzlogz2+z+1z+1+hsψpz.Then, ifυ solves
(13)−sn−1opυ′′=sn−1fs,υ+θ,inr<s<R,υ>0,υr=θ,inr<s<R,υ′R=0,we will have that υ+θ is a solution of Eq. (10). In order to prove existence of such υ, we find a minimum of the functional in the sequel. Let S⊂Xr denote the set of symmetric functions with respect to the origin. We define Φ:S→R by
(14)Φv=1p∫rRsn−1v′pds+∫rRsn−1Fs,vssn−1ds,where Fs,υs=∫0tfs,z+θ+dz and z+≔maxz,0.Lemma 5.
The functionalΦ is C1, weakly lower semicontinuous and coercive so that there exist υ∈Xr such that
(15)Φv=minu∈XrΦuandΦ′v≡0.The proof is standard by (9). Also, since υ is a weak solution of (13), we have
(16)vs=∫rsop−1t1−n∫tRzn−1fz,vz+θdzdt,in which
(17)op−1u=u1/p−1,ifu≥0,−−u1/p−1,ifu<0.Then, we define(18)ur≔υ+0.Lemma 6.
Suppose thatθ>1. Then, the function u¯≡ur is a supersolution for 3 which does not depend on 0<ε≤1.Lemma 7.
There exists a constantM>0 such that ur∞≤M and the constant M does not depend on r∈0,R. Moreover, for each ρ∈0,R, there exist a constant Cρ and rρ∈0,R such that we have the following estimates:
(19)urc0ρ,R,urc1ρ,R,opur′c1ρ,R≤Cρ.Proof or Lemmas6 and 7 can be found in [18], we omit them here.


### 2.2. Existence of Solution for3

In this section, we use the sub- and supersolution from Section2.1 (u and ur, respectively) to obtain a solution for the problem 3. Define the function
(20)gεs,u≔sn−1asopulogu2+εu+εu+ε+sn−1hsψpu+bu,s∈r,R,u≥0,where we choose b in such a way that the function u→gεs,u is increasing in u for all s∈r,R. Now, starting with u0=u, we define a sequence un such that each un satisfies
(21)−sn−1opun+1′′+bun+1=gεs,un,inr<s<R,un+1>0,un+1r=θ,inr<s<R,un+1′R=0.Let us now recall Lemma 2.1 in [24],Lemma 8 (weak comparison principle).
LetΩ be a bounded domain in RNN≥2 with smooth boundary ∂Ω and θ:0,+∞→0,+∞ is continuous and nondecreasing, let u1,u2∈W1,pΩ satisfy
(22)∫Ω∇u1p−2∇u1∇υdx+∫Ωθu1υdx≤∫Ω∇u2p−2∇u2∇υdx+∫Ωθu2υdx.For all nonnegativeυ∈W01,mΩ. Then, the inequality
(23)u1≤u2,on∂Ω,implies that
(24)u1≤u2,inΩ.Lemma 9.
The sequenceun is nondecreasing and satisfies u0s≤uns≤un+1s≤urs for all s∈r,R and all n∈N.Proof.
We just need to see thatu0≤u1≤ur and the general case follows by induction in an analogous way. We have
(25)−sn−1opu0′′+bu0≤−sn−1opu1′′+bu1,inr<s<R,u0−u1r≤0,u0−u1′R=0.So, we can apply Lemma8 and obtain that u0≤u1 in r,R. On the other hand,
(26)−sn−1opu1′′+bu1≤−sn−1opur′′+bur,inr<s<R,u1−urr≤0,u1−ur′R=0.Again, Lemma8 implies u1≤ur in r,R.By Lemma9, we define the pointwise limit
(27)urεs≔limn→∞uns,s∈r,R,and we see that
(28)1≤urε≤urs,s∈r,R.The functionurε is in fact a solution of 3.Lemma 10.
The functionurε is a solution of 3, and it belongs to C1r,R.Proof.
For eachn∈N, we have
(29)uns=θ+∫rsop−1t1−n∫tRzn−1azopunlogun2+εun+εun+ε+hzυpundzdt.Since we have(30)1≤un≤M,foralln∈N,we obtain, as in Lemma 7, that ϕpu′pC1ρ,R is bounded. Then, for a subsequence that we still denote by un, we have the convergence
(31)un→urεinC1ρ,R.


### 2.3. Obtaining a Solution forEε

In this section, we pass the limit asr→0+ and then obtain a solution for 2.Lemma 11.
For a fixed0<ε≤1, the problem 2 has a solution uε which is obtained as the limit of urε as r→0+.Proof.
For simplicity, we omit the dependence onε>0 for urε. We know that
(32)urs=θ+∫rsop−1t1−n∫tRzn−1azopurlogur2+εur+εur+ε+hzυpurdzdt.Also, we have(33)1≤ur≤M,inr,R,1≤ur≤M,inr,R.As in Lemma7, we can prove, for each ρ∈0,R, there exist a constant Cρ>0 and rρ∈0,R such that we have the following estimates:
(34)urc0ρ,R,urc1ρ,R,opur′c1ρ,R≤Cρ.Then, from the compact imbeddingC1ρ,R→C0ρ,R, we see that there exist a sequence rn and uε defined on 0,R such that, if we define wn≔urn, then
(35)wn→uεinCloc10,R,wn→uεinC1ρ,R.


### 2.4. Concluding the Proof of Theorem1

Now, we would like to pass the limit in the familyuε obtained in Section 2.3 and get a solution to 1. In order to do that, we need some estimates like the ones in Lemma 7 independently of ε.First, we observe that the following estimate holds in0,R(36)1≤uε≤M.Notice that the familyuε0<ε≤1 satisfies ε>0 for urε. We know that
(37)uεs=uεR2+∫R/2sop−1t1−n∫tRzn−1azopuεloguε2+εuε+εuε+ε+hzυpuεdzdt,if s∈R/2,R,and
(38)uεs=uεR2−∫sR/2op−1t1−n∫tRzn−1azopuεloguε2+εuε+εuε+ε+hzυpuεdzdt,if s∈0,R/2.From Eqs. (36)–(38) we see, as in Lemma 7 that, for each for each ρ∈0,R, there exist a constant Cρ>0 and ερ∈0,R such that we have the following estimates:
(39)uεc0ρ,R,uεc1ρ,R,opuε′c1ρ,R≤Cpforallε∈0,ερ.Now, arguing as in Section2.4, we can find a function u which satisfies
(40)−sn−1opu′′=sn−1logu+sn−1hsuq,inr<s<R,u>0,u′R=0.inr<s<R,that is, u is a radial solution for the problem 1.We see thatu∈C10,R∩C0,R. Now, extend continuously u to the whole interval 0,R. Indeed, let ri be a sequence in 0,R/2 with ri→0 as i→∞. From Eq. (13) (after we have passed the limit in ε)
(41)uri=uR2−∫riR/2op−1t1−n∫tRzn−1azopulogu+hzυpudzdt.Then, ifrj>ri, we get
(42)urj−uri=∫rirjop−1t1−n∫tRzn−1azopulogu+hzopudzdt.From Eq. (36), we obtain that there exists a constant C>0 such that
(43)urj−uri≤Crj−ri,so uri is a Cauchy sequence in R. Let L be the limit of such sequence. By a similar argument, we conclude that if si is another sequence in 0,R/2 converging to 0, then we necessarily have usi→L. So, we have proved that
(44)limr→0ur=L,finishing the proof of Theorem 1.


## 2.1. Sub- and Supersolution for3

Lemma 4.
Suppose thatθ>1. Then, the function u≡1 is a subsolution for 3 which does not depend on 0<ε≤1 and θ.Proof.
We just need to see that, sinceax>0,hx≥0 in BR, the following inequality holds independently of 0<ε≤1 and θ>1:
(6)axlog1+ε1+ε1+ε+hx≥log1=0,We proceed to find a supersolution for3. Denote by Xr, the following subspace of H1ArR:
(7)Xr≔u∈H1ArRu=0on∂Br.Forυ∈Xr, we define the expression:
(8)vr≔∫ArR∇v2dx1/2.Remark.
The expression⋅r defines a norm on Xr, and Xr,⋅r is a reflexive Banach space. Furthermore, by ([23], (7.44)), the Poincare inequality holds on Xr, that is, there exists η>0 such that
(9)∫ArRvpdx≤η∫ArR∇vpdx.Next, we work with the radial formulation forEε,r in the specific case that ε=1,
(10)−sn−1opu′′=sn−1asopulogu2+u+1u+1+sn−1hsψpu,inr<s<R,u>0,ur=θ,inr<s<R,u′R=0,where ϕps=sp−2s. Notice that
(11)logu2+u+1u+1≥0foru≥0.For simplicity, denote(12)fs,z=asopzlogz2+z+1z+1+hsψpz.Then, ifυ solves
(13)−sn−1opυ′′=sn−1fs,υ+θ,inr<s<R,υ>0,υr=θ,inr<s<R,υ′R=0,we will have that υ+θ is a solution of Eq. (10). In order to prove existence of such υ, we find a minimum of the functional in the sequel. Let S⊂Xr denote the set of symmetric functions with respect to the origin. We define Φ:S→R by
(14)Φv=1p∫rRsn−1v′pds+∫rRsn−1Fs,vssn−1ds,where Fs,υs=∫0tfs,z+θ+dz and z+≔maxz,0.Lemma 5.
The functionalΦ is C1, weakly lower semicontinuous and coercive so that there exist υ∈Xr such that
(15)Φv=minu∈XrΦuandΦ′v≡0.The proof is standard by (9). Also, since υ is a weak solution of (13), we have
(16)vs=∫rsop−1t1−n∫tRzn−1fz,vz+θdzdt,in which
(17)op−1u=u1/p−1,ifu≥0,−−u1/p−1,ifu<0.Then, we define(18)ur≔υ+0.Lemma 6.
Suppose thatθ>1. Then, the function u¯≡ur is a supersolution for 3 which does not depend on 0<ε≤1.Lemma 7.
There exists a constantM>0 such that ur∞≤M and the constant M does not depend on r∈0,R. Moreover, for each ρ∈0,R, there exist a constant Cρ and rρ∈0,R such that we have the following estimates:
(19)urc0ρ,R,urc1ρ,R,opur′c1ρ,R≤Cρ.Proof or Lemmas6 and 7 can be found in [18], we omit them here.


## 2.2. Existence of Solution for3

In this section, we use the sub- and supersolution from Section2.1 (u and ur, respectively) to obtain a solution for the problem 3. Define the function
(20)gεs,u≔sn−1asopulogu2+εu+εu+ε+sn−1hsψpu+bu,s∈r,R,u≥0,where we choose b in such a way that the function u→gεs,u is increasing in u for all s∈r,R. Now, starting with u0=u, we define a sequence un such that each un satisfies
(21)−sn−1opun+1′′+bun+1=gεs,un,inr<s<R,un+1>0,un+1r=θ,inr<s<R,un+1′R=0.Let us now recall Lemma 2.1 in [24],Lemma 8 (weak comparison principle).
LetΩ be a bounded domain in RNN≥2 with smooth boundary ∂Ω and θ:0,+∞→0,+∞ is continuous and nondecreasing, let u1,u2∈W1,pΩ satisfy
(22)∫Ω∇u1p−2∇u1∇υdx+∫Ωθu1υdx≤∫Ω∇u2p−2∇u2∇υdx+∫Ωθu2υdx.For all nonnegativeυ∈W01,mΩ. Then, the inequality
(23)u1≤u2,on∂Ω,implies that
(24)u1≤u2,inΩ.Lemma 9.
The sequenceun is nondecreasing and satisfies u0s≤uns≤un+1s≤urs for all s∈r,R and all n∈N.Proof.
We just need to see thatu0≤u1≤ur and the general case follows by induction in an analogous way. We have
(25)−sn−1opu0′′+bu0≤−sn−1opu1′′+bu1,inr<s<R,u0−u1r≤0,u0−u1′R=0.So, we can apply Lemma8 and obtain that u0≤u1 in r,R. On the other hand,
(26)−sn−1opu1′′+bu1≤−sn−1opur′′+bur,inr<s<R,u1−urr≤0,u1−ur′R=0.Again, Lemma8 implies u1≤ur in r,R.By Lemma9, we define the pointwise limit
(27)urεs≔limn→∞uns,s∈r,R,and we see that
(28)1≤urε≤urs,s∈r,R.The functionurε is in fact a solution of 3.Lemma 10.
The functionurε is a solution of 3, and it belongs to C1r,R.Proof.
For eachn∈N, we have
(29)uns=θ+∫rsop−1t1−n∫tRzn−1azopunlogun2+εun+εun+ε+hzυpundzdt.Since we have(30)1≤un≤M,foralln∈N,we obtain, as in Lemma 7, that ϕpu′pC1ρ,R is bounded. Then, for a subsequence that we still denote by un, we have the convergence
(31)un→urεinC1ρ,R.


## 2.3. Obtaining a Solution forEε

In this section, we pass the limit asr→0+ and then obtain a solution for 2.Lemma 11.
For a fixed0<ε≤1, the problem 2 has a solution uε which is obtained as the limit of urε as r→0+.Proof.
For simplicity, we omit the dependence onε>0 for urε. We know that
(32)urs=θ+∫rsop−1t1−n∫tRzn−1azopurlogur2+εur+εur+ε+hzυpurdzdt.Also, we have(33)1≤ur≤M,inr,R,1≤ur≤M,inr,R.As in Lemma7, we can prove, for each ρ∈0,R, there exist a constant Cρ>0 and rρ∈0,R such that we have the following estimates:
(34)urc0ρ,R,urc1ρ,R,opur′c1ρ,R≤Cρ.Then, from the compact imbeddingC1ρ,R→C0ρ,R, we see that there exist a sequence rn and uε defined on 0,R such that, if we define wn≔urn, then
(35)wn→uεinCloc10,R,wn→uεinC1ρ,R.


## 2.4. Concluding the Proof of Theorem1

Now, we would like to pass the limit in the familyuε obtained in Section 2.3 and get a solution to 1. In order to do that, we need some estimates like the ones in Lemma 7 independently of ε.First, we observe that the following estimate holds in0,R(36)1≤uε≤M.Notice that the familyuε0<ε≤1 satisfies ε>0 for urε. We know that
(37)uεs=uεR2+∫R/2sop−1t1−n∫tRzn−1azopuεloguε2+εuε+εuε+ε+hzυpuεdzdt,if s∈R/2,R,and
(38)uεs=uεR2−∫sR/2op−1t1−n∫tRzn−1azopuεloguε2+εuε+εuε+ε+hzυpuεdzdt,if s∈0,R/2.From Eqs. (36)–(38) we see, as in Lemma 7 that, for each for each ρ∈0,R, there exist a constant Cρ>0 and ερ∈0,R such that we have the following estimates:
(39)uεc0ρ,R,uεc1ρ,R,opuε′c1ρ,R≤Cpforallε∈0,ερ.Now, arguing as in Section2.4, we can find a function u which satisfies
(40)−sn−1opu′′=sn−1logu+sn−1hsuq,inr<s<R,u>0,u′R=0.inr<s<R,that is, u is a radial solution for the problem 1.We see thatu∈C10,R∩C0,R. Now, extend continuously u to the whole interval 0,R. Indeed, let ri be a sequence in 0,R/2 with ri→0 as i→∞. From Eq. (13) (after we have passed the limit in ε)
(41)uri=uR2−∫riR/2op−1t1−n∫tRzn−1azopulogu+hzυpudzdt.Then, ifrj>ri, we get
(42)urj−uri=∫rirjop−1t1−n∫tRzn−1azopulogu+hzopudzdt.From Eq. (36), we obtain that there exists a constant C>0 such that
(43)urj−uri≤Crj−ri,so uri is a Cauchy sequence in R. Let L be the limit of such sequence. By a similar argument, we conclude that if si is another sequence in 0,R/2 converging to 0, then we necessarily have usi→L. So, we have proved that
(44)limr→0ur=L,finishing the proof of Theorem 1.


## 3. Proof of Theorem3

### 3.1. Preliminaries

In this section, we consider the multiple solutions for problem1 with Dirichlet boundary conditions. In this case, we consider ax is a sign-changing function, hx=0. Moreover, it is necessary to note that the presence of the logarithmic nonlinearity leads to some difficulties in deploying the potential well method. In order to handle this situation, we need the following logarithmic Sobolev inequality which was introduced by [25].Proposition 12.
Letp>1, μ>0, and u∈W1,pΩ\0. Then, we have
(45)p∫RnuxploguxuLpRndx+nplogpμenLp∫Rnuxpdx≤μ∫Rn∇uxpdx,where
(46)Lp=pnp−1ep−1π−p/2Γn/2+1Γnp−1/p+1p/n.Remark.
Ifu∈W01,pΩ then, by defining ux=0 for x∈Rn\Ω, we derive
(47)p∫Ωuxploguxupdx+nplogpμenLp∫Ωuxpdx≤μ∫Ω∇uxpdx,
for any real numberμ>0.We start by considering the energy functionalJ by
(48)Ju=1p∇upp−1p∫Ωaxuplogudx+1p2∫Ωaxupdx,in which up=uLpΩ.Lemma 13.
Foru∈H01Ω and ∫Ωaxupdx=0, let M=maxΩ¯ax, then it holds
(49)Ju≥1p−Mμp∇upp,in which μ=nLp/peexpmp2Ωn/ne.Proof.
Using the fact∫Ωaxupdx=0, we have
(50)Ju=1p∇upp−1p∫Ωaxuploguxupdx.
Letu¯x=ux/up, then
(51)∫Ωaxuploguxupdx=∫Ω1axuplogu¯dx+∫Ω2axuplogu¯dx,where
(52)Ω1=x∈Ω,u¯x<1,andΩ2=x∈Ω,u¯x≥1.
By direct calculations, we know(53)∫Ω1axuplogu¯dx≤MΩn2eupp.
Also, by logarithmic Sobolev inequality (47) and (51), we have
(54)∫Ω2axuplogu¯dx≤M∫Ωuplogu¯dx+Ωn2eupp≤Mμp∫Ω∇uxpdx−np2logpμenLp−MΩn2eupp.
Then, combining (50), (51), (53), and (54), we have
(55)Ju≥1p−Mμp∇upp+np3logpμenLp−ΩnpeMupp.
Takingμ=nLp/peexpmp2Ωn/ne in (55), then
(56)np3logpμenLp−Ωnpe=0,we know (49).Lemma 14.
[19] Let um be a sequence in W01,pΩ. If um⇀u0 and um↛u0 in W01,pΩ, then
(57)Ju0<lim¯m→∞Jum.Ifum→u0 in W01,pΩ, then
(58)Ju0=limm→∞Jum.


### 3.2. Multiple Solutions

Inspired by [19], we seek the weak solutions of E by Nehari manifold. First, a simple calculation shows that Ju∈C1W01,pΩ,R, and its derivative is given by
(59)J′u,v=∫Ω∇up−2∇u⋅∇vdx−∫Ωaxφpuvlogudx,for all u,v∈W01,pΩ.From (49), Ju is not bounded on W01,pΩ, but we can prove that Ju is bounded from below on Nehari manifold
(60)N=u∈W1,pΩ\0:J′u,u=0,where , denotes the usual duality.It is clear that all nontrivial critical points ofJ must lie on N, and as we will see below, local minimizers on N are usually critical points of J. Also, we can see that
(61)u∈N⇔∫Ω∇updx−∫Ωaxuplogudx=0.Letu∈W1,pRn\0 and consider the real function j:λ→Jλu for λ>0 defined byjλ≔Jλu=λp/p∇upp−λp/p∫Ωaxuplogudx−λp/plogλ∫Ωaxupdx+λp/p2∫Ωaxupdx.Such maps are known as fibering maps which were introduced by Drabek and Pohozaev [26].Then, by direct calculations, we have(62)j′λ=λp−1p−1∇upp−λp−1p−1∫Ωaxuplogλudx,(63)j′′λ=λp−2p−2∇upp−λp−2p−2∫Ωaxuplogλudx−λp−2p−1∫Ωaxupdx.Lemma 15.
Letu∈W1,pΩ\0 and λ>0. Then, λu∈N if and only if j′λ=0.Proof.
First, by direct calculations, we know(64)λu∈N⇔λp−1p−1∇upp−∫Ωaxuplogλudx=0⇔λj′λ=0.
Sinceλ>0, then λu∈N if and only if j′λ=0.Then, ifu∈N, we have j′1=0 and j″1=−1/p−1∫Ωaxupdx.Thus, we can divideN into three subsets N+, N−, and N0, where
(65)N+=u∈N:∫Ωaxupdx>0,N−=u∈N:∫Ωaxupdx>0,N0=u∈N:∫Ωaxupdx>0.Lemma 16.
Ifu0 is a local minimizer for J on N and u0∉N0. Then, J′u0=0.Proof.
Ifu0 is a local minimizer for J on N, by Lagrange multipliers, there exists κ∈R such that
(66)J′u0=κχ′u0,where χu=∇upp−∫Ωaxuplogudx.
Sinceu0∈N, then
(67)J′u0,u0=0,andκχ′u0,u0=0.
On the other hand, fromu0∉N0, we can see
(68)χ′u0,u0=ju0″1=−λp−2p−1∫Ωaxupdx≠0.
Then,κ=0 and J′u0=0.Proposition 17.
BothN+ and N− are nonempty.Proof.
From (62), j′λ has a unique turning point at
(69)λu=exp∇upp−∫Ωaxuplogudx∫Ωaxupdx.
Sinceax is sign-changing, then we can take u1 such that
(70)∫Ωaxu1pdx<0,andthenλu1u1∈N+.
Also, we can takeu2 such that
(71)∫Ωaxu2pdx<0,andthenλu2u2∈N−.
Then, bothN+ and N− are nonempty.Just like [19], by Lemmas 13–16, we can get the following results.Lemma 18.
[19] N+ is bounded; J is bounded below on N+.Lemma 19.
[19] Every minimizing sequence for J on N− is bounded, 0∉N−¯, infu∈N−Ju>0.Proposition 20.
J has a minimizer on N+.Proof.
Letum⊆N+ be a minimizing sequence, i.e., limm→∞Jum=infu∈N+Ju<0.
By Lemma18, N+ is bounded; we may assume that
(72)um⇀u0,inW01,pΩ,andsoum→u0inLpΩ.
Sinceum⊆N+, we can get
(73)∫Ωaxumpdx<0,∫Ωaxu0pdx=limm→∞∫Ωaxumpdx=limm→∞p2∫Ωaxumpdx<0,∇umpp−∫Ωaxumplogumdx=0.
Supposeum↛u0 in W01,pΩ, then
(74)∇u0pp−∫Ωaxu0plogu0dx<limm→∞∇umpp−∫Ωaxumplogumdx=0.
Then, there exists(75)λu0=exp∇u0pp−∫Ωaxu0plogu0dx∫Ωaxu0pdx>1,such that λu0u0∈N+, and then, J attains minimum at λu0u0.
Hence(76)Jλu0u0<Ju0≤limm→∞Jum=infu∈N+Ju,which is impossible. Hence, um→u0 in W01,pΩ, u0∈N+, and Ju0=infu∈N+Ju<0, this means that u0 is a minimizer for J on N+.Proposition 21.
There exists a minimizer ofJ on N−.Proof.
Letum⊆N− be a minimizing sequence. By Lemma 19, um is bounded; we may assume that
(77)um⇀u0,inW01,pΩ,andsoum→u0inLpΩ.
SinceJum=1/p2∫Ωaxumpdx, by Lemma 19, we can get
(78)∫Ωaxu0pdx=limm→∞∫Ωaxumpdx>0.
Supposeum↛u0 in W01,pΩ, then
(79)∇u0pp−∫Ωaxu0plogu0dx<limm→∞∇umpp−∫Ωaxumplogumdx=0.
Then, there exists(80)λu0=exp∇u0pp−∫Ωaxu0plogu0dx∫Ωaxu0pdx<1,such that λu0u0∈N−, λu0um⇀λu0u0, but λu0um↛λu0u0 in W01,pΩ.
Hence(81)Jλu0u0<limm→∞Jλu0um.
Since the mapλ→Jλum attains its maximum at t=1,
(82)limm→∞Jλu0um≤limm→∞Jum=infu∈N−Ju.
This meansJλu0u0<infu∈N−Ju is impossible.
Hence,um→u0 in W01,pΩ, and this means that u0 is a minimizer for J on N−.Proof of Theorem 3.
Propositions20 and 21 show that the energy functional J has two minimizers u1 on N+ and u2 on N−. Next, by Lemma 16, J has two critical points u1 and u2 on W01,pΩ, which means that the problem 1 has at least two nontrivial solutions under the condition hx=0.


## 3.1. Preliminaries

In this section, we consider the multiple solutions for problem1 with Dirichlet boundary conditions. In this case, we consider ax is a sign-changing function, hx=0. Moreover, it is necessary to note that the presence of the logarithmic nonlinearity leads to some difficulties in deploying the potential well method. In order to handle this situation, we need the following logarithmic Sobolev inequality which was introduced by [25].Proposition 12.
Letp>1, μ>0, and u∈W1,pΩ\0. Then, we have
(45)p∫RnuxploguxuLpRndx+nplogpμenLp∫Rnuxpdx≤μ∫Rn∇uxpdx,where
(46)Lp=pnp−1ep−1π−p/2Γn/2+1Γnp−1/p+1p/n.Remark.
Ifu∈W01,pΩ then, by defining ux=0 for x∈Rn\Ω, we derive
(47)p∫Ωuxploguxupdx+nplogpμenLp∫Ωuxpdx≤μ∫Ω∇uxpdx,
for any real numberμ>0.We start by considering the energy functionalJ by
(48)Ju=1p∇upp−1p∫Ωaxuplogudx+1p2∫Ωaxupdx,in which up=uLpΩ.Lemma 13.
Foru∈H01Ω and ∫Ωaxupdx=0, let M=maxΩ¯ax, then it holds
(49)Ju≥1p−Mμp∇upp,in which μ=nLp/peexpmp2Ωn/ne.Proof.
Using the fact∫Ωaxupdx=0, we have
(50)Ju=1p∇upp−1p∫Ωaxuploguxupdx.
Letu¯x=ux/up, then
(51)∫Ωaxuploguxupdx=∫Ω1axuplogu¯dx+∫Ω2axuplogu¯dx,where
(52)Ω1=x∈Ω,u¯x<1,andΩ2=x∈Ω,u¯x≥1.
By direct calculations, we know(53)∫Ω1axuplogu¯dx≤MΩn2eupp.
Also, by logarithmic Sobolev inequality (47) and (51), we have
(54)∫Ω2axuplogu¯dx≤M∫Ωuplogu¯dx+Ωn2eupp≤Mμp∫Ω∇uxpdx−np2logpμenLp−MΩn2eupp.
Then, combining (50), (51), (53), and (54), we have
(55)Ju≥1p−Mμp∇upp+np3logpμenLp−ΩnpeMupp.
Takingμ=nLp/peexpmp2Ωn/ne in (55), then
(56)np3logpμenLp−Ωnpe=0,we know (49).Lemma 14.
[19] Let um be a sequence in W01,pΩ. If um⇀u0 and um↛u0 in W01,pΩ, then
(57)Ju0<lim¯m→∞Jum.Ifum→u0 in W01,pΩ, then
(58)Ju0=limm→∞Jum.


## 3.2. Multiple Solutions

Inspired by [19], we seek the weak solutions of E by Nehari manifold. First, a simple calculation shows that Ju∈C1W01,pΩ,R, and its derivative is given by
(59)J′u,v=∫Ω∇up−2∇u⋅∇vdx−∫Ωaxφpuvlogudx,for all u,v∈W01,pΩ.From (49), Ju is not bounded on W01,pΩ, but we can prove that Ju is bounded from below on Nehari manifold
(60)N=u∈W1,pΩ\0:J′u,u=0,where , denotes the usual duality.It is clear that all nontrivial critical points ofJ must lie on N, and as we will see below, local minimizers on N are usually critical points of J. Also, we can see that
(61)u∈N⇔∫Ω∇updx−∫Ωaxuplogudx=0.Letu∈W1,pRn\0 and consider the real function j:λ→Jλu for λ>0 defined byjλ≔Jλu=λp/p∇upp−λp/p∫Ωaxuplogudx−λp/plogλ∫Ωaxupdx+λp/p2∫Ωaxupdx.Such maps are known as fibering maps which were introduced by Drabek and Pohozaev [26].Then, by direct calculations, we have(62)j′λ=λp−1p−1∇upp−λp−1p−1∫Ωaxuplogλudx,(63)j′′λ=λp−2p−2∇upp−λp−2p−2∫Ωaxuplogλudx−λp−2p−1∫Ωaxupdx.Lemma 15.
Letu∈W1,pΩ\0 and λ>0. Then, λu∈N if and only if j′λ=0.Proof.
First, by direct calculations, we know(64)λu∈N⇔λp−1p−1∇upp−∫Ωaxuplogλudx=0⇔λj′λ=0.
Sinceλ>0, then λu∈N if and only if j′λ=0.Then, ifu∈N, we have j′1=0 and j″1=−1/p−1∫Ωaxupdx.Thus, we can divideN into three subsets N+, N−, and N0, where
(65)N+=u∈N:∫Ωaxupdx>0,N−=u∈N:∫Ωaxupdx>0,N0=u∈N:∫Ωaxupdx>0.Lemma 16.
Ifu0 is a local minimizer for J on N and u0∉N0. Then, J′u0=0.Proof.
Ifu0 is a local minimizer for J on N, by Lagrange multipliers, there exists κ∈R such that
(66)J′u0=κχ′u0,where χu=∇upp−∫Ωaxuplogudx.
Sinceu0∈N, then
(67)J′u0,u0=0,andκχ′u0,u0=0.
On the other hand, fromu0∉N0, we can see
(68)χ′u0,u0=ju0″1=−λp−2p−1∫Ωaxupdx≠0.
Then,κ=0 and J′u0=0.Proposition 17.
BothN+ and N− are nonempty.Proof.
From (62), j′λ has a unique turning point at
(69)λu=exp∇upp−∫Ωaxuplogudx∫Ωaxupdx.
Sinceax is sign-changing, then we can take u1 such that
(70)∫Ωaxu1pdx<0,andthenλu1u1∈N+.
Also, we can takeu2 such that
(71)∫Ωaxu2pdx<0,andthenλu2u2∈N−.
Then, bothN+ and N− are nonempty.Just like [19], by Lemmas 13–16, we can get the following results.Lemma 18.
[19] N+ is bounded; J is bounded below on N+.Lemma 19.
[19] Every minimizing sequence for J on N− is bounded, 0∉N−¯, infu∈N−Ju>0.Proposition 20.
J has a minimizer on N+.Proof.
Letum⊆N+ be a minimizing sequence, i.e., limm→∞Jum=infu∈N+Ju<0.
By Lemma18, N+ is bounded; we may assume that
(72)um⇀u0,inW01,pΩ,andsoum→u0inLpΩ.
Sinceum⊆N+, we can get
(73)∫Ωaxumpdx<0,∫Ωaxu0pdx=limm→∞∫Ωaxumpdx=limm→∞p2∫Ωaxumpdx<0,∇umpp−∫Ωaxumplogumdx=0.
Supposeum↛u0 in W01,pΩ, then
(74)∇u0pp−∫Ωaxu0plogu0dx<limm→∞∇umpp−∫Ωaxumplogumdx=0.
Then, there exists(75)λu0=exp∇u0pp−∫Ωaxu0plogu0dx∫Ωaxu0pdx>1,such that λu0u0∈N+, and then, J attains minimum at λu0u0.
Hence(76)Jλu0u0<Ju0≤limm→∞Jum=infu∈N+Ju,which is impossible. Hence, um→u0 in W01,pΩ, u0∈N+, and Ju0=infu∈N+Ju<0, this means that u0 is a minimizer for J on N+.Proposition 21.
There exists a minimizer ofJ on N−.Proof.
Letum⊆N− be a minimizing sequence. By Lemma 19, um is bounded; we may assume that
(77)um⇀u0,inW01,pΩ,andsoum→u0inLpΩ.
SinceJum=1/p2∫Ωaxumpdx, by Lemma 19, we can get
(78)∫Ωaxu0pdx=limm→∞∫Ωaxumpdx>0.
Supposeum↛u0 in W01,pΩ, then
(79)∇u0pp−∫Ωaxu0plogu0dx<limm→∞∇umpp−∫Ωaxumplogumdx=0.
Then, there exists(80)λu0=exp∇u0pp−∫Ωaxu0plogu0dx∫Ωaxu0pdx<1,such that λu0u0∈N−, λu0um⇀λu0u0, but λu0um↛λu0u0 in W01,pΩ.
Hence(81)Jλu0u0<limm→∞Jλu0um.
Since the mapλ→Jλum attains its maximum at t=1,
(82)limm→∞Jλu0um≤limm→∞Jum=infu∈N−Ju.
This meansJλu0u0<infu∈N−Ju is impossible.
Hence,um→u0 in W01,pΩ, and this means that u0 is a minimizer for J on N−.Proof of Theorem 3.
Propositions20 and 21 show that the energy functional J has two minimizers u1 on N+ and u2 on N−. Next, by Lemma 16, J has two critical points u1 and u2 on W01,pΩ, which means that the problem 1 has at least two nontrivial solutions under the condition hx=0.



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*Source: 6545918-2020-12-19.xml*