# Ground-State Solutions for a Class ofN-Laplacian Equation with Critical Growth

Authors: Guoqing Zhang; Jing Sun
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We investigate the existence of ground-state solutions for a class ofN-Laplacian equation with critical growth in ℝN. Our proof is based on a suitable Trudinger-Moser inequality, Pohozaev-Pucci-Serrin identity manifold, and mountain pass lemma.


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## Body

## 1. Introduction

Consider the followingN-Laplacian equation:
(1.1)-ΔNu+|u|N-2u=f(u),inℝN,u>0,inRN,u∈W1,N(RN),
where N≥2. ΔNu=div(|∇u|N-2∇u) is the N-Laplacian, the nonlinear term f(u) has critical growth.The interest in these problems lies in that fact that the order of the Laplacian is the same as the dimensionN of the underlying space. The classical case of this problem that N=2, and the the problem (1.1) reduces to
(1.2)-Δu+u=f(u),inR2,
has been treated by Atkinson and Peletier [1] and Berestycki and Lions [2]. They obtained the existence of ground-state solution which the nonlinear term f(u) is subcritical growth.  Alves et al. [3] extend their results to the critical growth. As N≠2, do Ó and Medeiros [4] consider the following N-Laplacian equation problem:
(1.3)-ΔNu=g(u),inRN,
where g:ℝ↦ℝ has a subcritical growth and obtain a mountain pass characterization of the ground-state solution for the problem (1.3). In the present paper, we will improve and complement some of the results cited above.Assume the functionf:ℝ↦ℝ is continuous and satisfies the following conditions:     (g1)
lims→0+(f(s)/s|s|N-2)=0;(g2)
There exist constantsα0,b1,b2>0 such that |f(s)|≤b1|s|N-1+b2[exp(α0|s|N/(N-1))-SN-2(α0,s)], where SN-2(α0,s)=∑k=0N-2‍(α0k/k!)|s|Nk/(N-1);(g3)
There existλ>0 and q>N such that f(s)≥λsq-1, for every s≥0.Remark 1.1.
Condition(g2) implies that f has a critical growth with critical exponent α0.
Consider the energy functionalI:W1,N(ℝN)↦ℝ(1.4)I(u)=1N∫RN‍(|∇u|N+|u|N)dx-∫RN‍F(u)dx,
where F(s)=∫0s‍f(t)dt. By a ground-state solution, we mean a solution such ω∈W1,N(ℝN) such that I(ω)≤I(u) for every nontrivial solution u of the problem (1.1). Let Cq>0 denote the best constant of Sobolev embeddings:
(1.5)W1,N(RN)↪Lq(RN),
for q∈(N,+∞), that is,
(1.6)Cq[∫RN‍|u|qdx]N/q≤∫RN‍(|∇u|N+|u|N)dx,
for all u∈W1,N(ℝN).Now we state our main theorem in this paper.Theorem 1.2.
Iff satisfies (g1),(g2), and (g3), with
(1.7)λ>(q-Nq)(q-N)/NCqq/N,
then the problem (1.1) possesses a nontrivial ground-state solution.In this paper, we complement some results [4] from subcritical case to the critical case. Furthermore, the ground-state solution to the problem (1.1) is obtained without assuming that the function
(1.8)s↦f(s)|s|N-1
is increasing for s>0 (see [5]), and the so-called Ambrosetti-Rabinowitz condition: there exists θ>N, such that for all x∈ℝN,
(1.9)0<θF(x,u)≤uf(x,u).The paper is organized as follows. Section2 contains some technical results which allows us to give a variational approach for our results. In Section 3, we prove our main results.


## 2. The Variational Framework

For1≤p≤∞, Lp(ℝN) denotes the Lebesgue spaces with the norm ∥u∥Lp(ℝN)=(∫ℝN‍|u|pdx)1/p, W1,p(ℝN) denotes the Sobolev spaces with the norm ∥u∥W1,p(ℝN)=(∫ℝN‍(|∇u|p+|u|p)dx)1/p. As p=N, we have the following version of Trudinger-Moser inequality.Lemma 2.1 (see [6]).
IfN≥2, α>0 and u∈W1,N(ℝN), then
(2.1)∫RN‍[exp⁡(α|u|N/(N-1))-SN-2(α,u)]dx<∞.
Moreover, if ∥∇u∥LN(ℝN)N≤1,∥u∥LN(ℝN)≤M<∞ and α<αN, then there exists a constant C, which depends only on N, M, and α, such that
(2.2)∫RN‍[exp⁡(α|u|N/(N-1))-SN-2(α,u)]dx≤C(N,M,α),
where αN=NωN-11/(N-1) and ω1/(N-1) is the measure of the unit sphere in ℝN.In the sequel, since we seek positive solutions, and assume thatf(s)=0 for s≤0. Consider the following minimization problem:
(2.3)min⁡{1N∫RN‍|∇u|Ndx:∫RN‍G(u)dx=0},
where g(s)=f(s)-s|s|N-2, G(s)=∫0s‍g(t)dt=F(s)-(1/N)sN. Since the problem (1.1) is an autonomous problem, under the Schwarz symmetric process, we can minimize the problem (2.3) on the space Wrad1,N(ℝN), the subspace of W1,N(ℝN) formed by radially symmetric functions. Indeed, let u* be the Schwarz symmetrization of u, we have
(2.4)∫RN‍G(u*)dx=∫RN‍G(u)dx,∫RN‍|∇u*|Ndx≤∫RN‍|∇u|Ndx.
Hence, we can minimize the problem (2.3) on the space Wrad1,N(ℝN) (see [7]). Now, we defined the following notations
(2.5)m=inf{I(u):uisnontrivialsolutionoftheproblem(1.1)}A=inf⁡{1N∫RN|∇u|Ndx:∫RN‍G(u)dx=0‍}b=infγ∈Γmaxt∈[0,1]I(γ(t)),
where Γ={γ∈C([0,1],Wrad1,N(ℝN)):γ(0)=0,I(γ(1))<0}.We recall that Pohozaev-Pucci-Serrin identity shows that any solutionsu of the problem (1.1) should satisfies the Pohozaev-Pucci-Serrin identity:
(2.6)(N-p)∫RN‍|∇u|pdx=Np∫RN‍G(u)dx.Then, asp=N, we have ∫ℝN‍G(u)dx=0.Hence, we have the Pohozaev identity manifold:(2.7)P={u∈W1,N(RN)∖{0}:(N-p)∫RN‍|∇u|pdx=Np∫RN‍G(u)dx}={u∈W1,N(RN)∖{0}:∫RN‍G(u)dx=0}.So, we have(2.8)A=inf⁡u∈P1N∫RN‍|∇u|Ndx,m=inf⁡u∈τI(u),
where τ={u∈W1,N(ℝN)∖{0}:I′(u)=0}.In what follows, we will show thatA is attained, and afterwards we prove that
(2.9)m=A=b,
thereby proving that the problem (1.1) has a ground-state solution.


## 3. The Proof of Theorem1.2

In this section, we prove thatA is attained, the equality (2.9) is satisfied. Hence the proof of Theorem 1.2 is obtained.In the following, we consider the following minimax value:(3.1)c=inf⁡0≠v∈W1,N(RN)max⁡t≥0I(tv).Now, we show a sufficient condition, on a sequence{vn} to get a convergence like F(vn)→F(v) in L1(ℝN).Lemma 3.1.
Assume thatf satisfies (g1) and(g2), and let {vn} be a sequence in Wrad1,N(ℝN) such that ∥∇vn∥LN(ℝN)N≤1,∥vn∥LN(ℝN)≤M<∞, then we have
(3.2)∫RNF(vn)dx⟶∫RNF(v)dx,
where vn⇀v in W1,N(ℝN).Proof.
Without loss of generality, we assume that there existv∈Wrad1,N(ℝN) such that
(3.3)vn⇀v,inWrad1,N(ℝN),vn⇀v,a.e.inRN.
Let v* is the Schwarz symmetrization of v, then we have
(3.4)∫ℝNv[exp(α0|v|N/(N-1))-SN-2(α0,v)]dx=∫ℝNv*[exp(α0|v*|N/(N-1))-SN-2(α0,v*)]dx,∫RN|v|Ndx=∫RN|v*|Ndx.
From(g1), we obtain that for ϵ>0, there exists δ>0, such that
(3.5)f(s)≤ϵ|s|N-1,for|s|<δ,
so, we have
(3.6)F(s)≤ϵN|s|N,for|s|<δ.
From(g2), we obtain
(3.7)F(s)≤C1|s|N+C2|s|[exp⁡(α0|s|N/(N-1))-SN-2(α0,u)],for|s|≥δ.
There two estimates yield
(3.8)F(s)≤C3|s|N+C2|s|[exp⁡(α0|s|N/(N-1))-SN-2(α0,u)],fors>0.
On one hand, from Lemma2.1, we obtain that there exists a constant C, which depends only on N, M, and α such that
(3.9)∫RN‍exp⁡(α|vn|N/(N-1))≤C.
When ∥∇vn∥LN(ℝN)N≤1,∥vn∥LN(ℝN)≤M<∞ and α<αN. Hence, we have
(3.10)∫|x|≤rF(vn)≤C3∫RN|vn*|Ndx+C2∫|x|≤r|vn*|[exp⁡(α0|vn*|N/(N-1))-SN-2(α0,vn*)]dx≤C3MN+C2∫|x|≤r|vn*|exp⁡(α0|vn*|N/(N-1))dx≤C3MN+C3(∫|x|≤r|vn*|μ)1/μ(∫|x|≤rexp⁡(βα0|vn*|N/(N-1))dx)1/β≤C3MN+C4M(∫|x|≤rexp⁡(βα0|vn*|N/(N-1))dx)1/β≤C5,
where Ci(i=2,3,4,5) are positive constants, the continuous imbedding W1,N(ℝN)↪Lμ(ℝN), 1/μ+1/β=1 and βα0<αN.
Then, by Dominated convergence theorem, we obtain(3.11)∫|x|≤r‍F(vn)dx⟶∫|x|≤r‍F(v)dx.
On the other hand,(3.12)∫|x|>r‍F(vn)dx≤C3∫|x|>r‍|vn|Ndx+C2∫|x|>r‍|vn|[exp(α0|vn|N/(N-1))-SN-2(α0,vn)]dx∫|x|>r‍|vn|[exp⁡(α0|vn|N/(N-1))-SN-2(α0,vn)]dx=∑j=N-1∞‍α0jj!∫|x|>r‍|vn|⋅|vn|Nj/(N-1)dx=∑j=N-1∞‍α0jj!∫|x|>r‍|vn*||vn*|Nj/(N-1)dx,
where vn* is the Schwarz symmetrization of vn. Notice that the estimate
(3.13)∫|x|>r‍1|x|1+Nj/(N-1)dx=ωN-1∫r∞‍tN-1t1+Nj/(N-1)dt=(ωN-1Nj/(N-1)-N+1)rN-1-Nj/(N-1)≤ωN-1r,
for all j≥N-1, together with the Radial Lemma [4] leads to
(3.14)∑j=N-1∞‍α0jj!∫|x|>r‍|vn*||vn*|Nj/(N-1)dx≤M(NωN-1)1/N∑j=N-1∞‍α0jj!(NωN-1)j/(N-1)MNj/(N-1)∫|x|>r‍|x|-1-Nj/(N-1)dx≤C(N)r.
Thus, given δ>0, there exists r>0 such that
(3.15)∫|x|>r‍|vn|Ndx<δ,∫|x|>r‍[exp⁡(α0|vn|N/(N-1))-SN-2(α0,vn)]dx<δ.
Which implies that
(3.16)∫|x|>r‍F(vn)dx≤Cδ,∫|x|>r‍F(v)dx≤Cδ.
Using the estimate(3.17)|∫RN‍(F(vn)-F(v))dx|≤|∫|x|≤r‍(F(vn)-F(v))dx|+|∫|x|>r‍(F(vn)-F(v))dx|,
we get
(3.18)lim⁡n→∞|∫RN‍F(vn)-∫RN‍F(v)dx|≤Cδ,
Hence, we obtain that
(3.19)∫RN‍F(vn)dx⟶∫RN‍F(v)dx.Lemma 3.2.
The numbersA and c satisfy the inequality A≤c.Proof.
For eachv∈W1,N(ℝN)∖{0}, since we only consider the nontrivial solutions of the problem (1.1), we divide them into two cases to consider.
Case 1. Let v+=max{v,0}≠0, we define the function h:ℝ→ℝ by
(3.20)h(t)=∫RN‍G(tv)dx=∫RN‍[F(tv)-tNvNN]dx.
By (g1), we obtain that there exists δ>0,0<c0<1 such that |s|<δ, and
(3.21)|f(s)|<c0|s|N-1.
Hence
(3.22)h(t)≤∫RN‍∫0tv‍c0|s|N-1dx-1N∫RN‍tNvNdx=c0N∫ℝN‍|tv|Ndx-1N∫ℝN‍|tv|Ndx,
we obtain that h(t)<0 for t small enough. On the other hand, by (g2), we obtain that h(t)>0 for t large enough. In this way, there exists t0>0 such that h(t0)=0, That is, t0v∈𝒫. Hence
(3.23)A≤1N∫RN‍|∇(t0v)|Ndx=I(t0v)≤max⁡t≥0I(tv).
Case 2. Let v+=max{v,0}=0, since f(s)=0 for all s<0, we obtain
(3.24)max⁡t≥0I(tv)=+∞.
As a consequence,
(3.25)A≤c.
Combining Cases 1 and 2, we obtain that A≤c.Lemma 3.3.
The numberA defined by (2.8) is positive, that is, A>0.Proof .
Clearly,A≥0. Assume by contradiction that A=0 and let {un} be a minimizing sequence in W1,N(ℝℕ) to A, that is,
(3.26)1N∫RN‍|∇un|Ndx⟶A=0with∫RN‍G(un)dx=0.
For each λn>0, set vn(x)=un(x/λn) satisfying
(3.27)1N∫RN‍|∇vn|Ndx=1N∫RN‍|∇vn|Ndx.
Similarly, we have
(3.28)∫ℝN‍G(vn)dx=λnN∫ℝN‍G(un)dx=0,∫RN‍|vn|Ndx=λnN∫RN‍|un|Ndx.
We choose λnN=1/∫ℝN‍|un|Ndx, so ∫ℝN‍|vn|Ndx=1. Then we get
(3.29)1N∫ℝN‍|∇vn|Ndx→A=0,∫RN‍|vn|Ndx=1,∫RN‍G(vn)dx=0.
In what follows, we study in the spaceWrad1,N(ℝN). Firstly, we assume that there exists v∈Wrad1,N(ℝN) such that vn⇀v in Wrad1,N(ℝN).
On one hand, since(1/N)∫ℝN‍|∇vn|N→A=0, then ∃N0>0, for all 0<ϵ<1, when n>N0, we have ∫ℝN‍|∇vn|Ndx<ϵ<1 and we also know that ∫ℝN‍|vn|Ndx=1, so ∥vn∥LN(ℝN)≤M<∞. From Lemma 3.1, we have
(3.30)∫RN‍F(vn)dx⟶∫RN‍F(v)dx.
Note that
(3.31)∫RN‍G(vn)dx=∫RN‍(F(vn)-1N|vn|N)dx=0,
so we have
(3.32)∫RN‍F(vn)dx=1N∫RN‍|vn|Ndx=1N.
Hence, we have
(3.33)∫RN‍F(v)dx=1N.
It implies thatv≠0. On the other hand, since vn⇀v in Wrad1,N(ℝN), and the space Wrad1,N(ℝN) is a reflexible Banach space, we have
(3.34)lim⁡n→∞inf⁡1N∫RN‍|∇vn|Ndx≥1N∫RN‍|∇v|Ndx≥0.
Since
(3.35)lim⁡n→∞inf⁡1N∫RN‍|∇vn|Ndx=A=0,
we get
(3.36)1N∫RN‍|∇v|Ndx=0.
From which it follows that v=0, we have an absurd. Hence, we have
(3.37)A>0.Lemma 3.4.
Ifλ>(q-N/q)(q-N)/NCqq/N, then c<1/N.Proof.
From(g3), we have f(s)>λsq-1, for all s≥0. Now we choose ψ∈Wrad1,N(ℝN) such that
(3.38)ψ≥0,‖ψ‖qN=Cq-1,‖ψ‖W1,N(RN)=1.
Hence, we have
(3.39)c≤max⁡t≥0I(tψ)=max⁡t≥0{1N∫RN‍(|∇(tψ)|N+|tψ|N)dx-∫RN‍F(tψ)dx}=max⁡t≥0{tNN-∫RN‍∫0tψ‍f(s)dsdx}≤max⁡t≥0{tNN-λ∫RN‍∫0tψ‍sq-1dsdx}=max⁡t≥0{tNN-λtqq∫RN‍ψqdx}.
Let K(t)=tN/N-(λtq/q)∫ℝN‍ψqdx, then K(t) is continuous function, we have
(3.40)K′(t)=tN-1-λtq-1∫RN‍ψqdx=0.
By a simple calculation, when t0=(1/λ∫ℝN‍ψqdx)1/(q-N)>0, we have
(3.41)max⁡t>0K(t)=K(t0)=1N(1λ∫RN‍ψqdx)N/(q-N)-(λq∫RN‍ψqdx)(1λ∫RN‍ψqdx)q/(q-N)=q-NNqλ-N/(q-N)Cq(q/N)⋅(N/(q-N))<q-NNq(q-Nq)((q-N)/N)⋅(-N/(q-N))Cq(q/N)(-N/(q-N))Cqq/(q-N)=1N.
Hence, we have
(3.42)c<1N.Lemma 3.5.
The numberA is attained, that is, there exists u∈Wrad1,N(ℝN) such that A=∫ℝN‍|∇u|Ndx and ∫ℝN‍G(u)dx=0.Proof.
Let{un} be a minimizing sequence in Wrad1,N(ℝN) for A, that is,
(3.43)1N∫RN‍|∇un|Ndx⟶A(n⟶∞),∫RN‍G(un)dx=0.
Arguing as in Lemma 3.3, we assume that ∫ℝN‍|un|Ndx=1. From (3.43), Lemmas 3.3 and 3.4, we obtain
(3.44)lim⁡n→∞∫RN‍|∇un|Ndx=NA≤Nc<1.
From Lemma 3.1,
(3.45)∫RN‍F(un)dx⟶∫RN‍F(u)dx
where un⇀uinW1,N(ℝN),asn→∞.
By (3.43) and (3.45), we have
(3.46)∫ℝNF(un)dx=1N∫ℝN‍|un|Ndx=1N,∫RNF(u)dx=1N.
It implies that
(3.47)u≠0,(3.48)1N∫ℝN‍|∇u|Ndx≤limn→∞inf1N∫ℝN‍|∇un|Ndx=A,(3.49)∫ℝN‍|u|Ndx≤limn→∞inf∫ℝN‍|un|Ndx=1.
From (3.48) and (3.49), we have
(3.50)∫RN‍G(u)dx=∫RN‍F(u)dx-1N∫RN‍|u|Ndx=1N-1N∫RN‍|u|Ndx≥0.
If ∫ℝN‍G(u)dx≠0, from (3.50), we have ∫ℝN‍G(u)dx>0. Consider the function h defined in Lemma 3.2 relative to the function:
(3.51)h(t)=∫RN‍G(tu)dx.
We concludes that h(t)<0 for t small enough. On the other hand, h(1)=∫ℝN‍G(u)dx>0. In this way, we obtain that there is t0∈(0,1) such that h(t0)=0, that is,
(3.52)∫RN‍G(t0u)dx=0.
Hence, from (3.48),
(3.53)0<1N∫RN‍|∇(t0u)|Ndx=1Nt0N∫RN‍|∇u|Ndx≤t0NA<A.
However, from (3.52), we have t0u∈𝒫. Hence, we obtain
(3.54)1N∫RN‍|∇(t0u)|Ndx≥A.
Which is contradictory with (3.53).
Thus, we obtain(3.55)∫RN‍G(u)dx=0.
It implies u∈𝒫 and
(3.56)1N∫RN‍|∇u|Ndx≥A.
From (3.48) and (3.56), we obtain that
(3.57)1N∫RN‍|∇u|Ndx=A,
with ∫ℝN‍G(u)dx=0,u≠0.
We obtain thatA is attained.Proof of Theorem1.2.
From Lemma3.5, there is u∈Wrad1,N(ℝN)∖{0} such that
(3.58)1N∫RN|∇u|Ndx=A,∫RN‍G(u)dx=0.
we will prove that m=b=A.
By Lagrange multipliers, there existsρ∈ℝ, such that
(3.59)∫RN|∇u|N-2∇u∇vdx=ρ∫RNg(u)vdx,
for every v∈W1,N(ℝN).
Define the rescaled functionuρ1/N=u(ρ-1/Nx), which is a nontrivial solution of (1.1) with
(3.60)∫ℝN‍|∇uρ1/N|Ndx=∫ℝN‍|∇u|Ndx,∫RN‍G(uρ1/N)dx=ρ∫RN‍G(u)dx=0.
Thus, we have
(3.61)m≤I(uρ1/N)=1N∫RN‍|∇uρ1/N|Ndx-∫RN‍G(uρ1/N)dx=1N∫RN‍|∇u|Ndx=A.
So, we have
(3.62)m≤A.
For eachγ∈Γ, one has γ([0,1])∩𝒫≠∅ from [4]. We obtain that there exists t0∈[0,1] such that γ(t0)∈𝒫, that is, γ(t0) satisfied that ∫ℝN‍G(γ(t0))dx=0 and then
(3.63)A≤1N∫RN‍|∇γ(t0)|Ndx-1N∫RN‍G(γ(t0))dx=I(γ(t0)).
Hence A≤I(γ(t0))≤maxt∈[0,1]I(γ(t)) for every γ∈Γ, we obtain that
(3.64)A≤b.
From (3.62) and (3.64), we obtain that m≤A≤b.
On the other hand, for every nontrivial solutionω∈W1,N(ℝN) of the problem (1.1), there exists a path γω∈Γ such that ω∈γω([0,1]) and maxt∈[0,1]I(γω(t))=I(ω). Consequently, b≤I(ω),b≤m.
In conclusion, we obtain(3.65)m=A=b.
Hence, the function uρ1/N is a ground-state solution of the problem (1.1).



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