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Solutions for a class of quasilinear Schrödinger equations with critical

Sobolev exponents

Zhouxin Li and Yimin Zhang

Citation: J. Math. Phys. 58, 021501 (2017); doi: 10.1063/1.4975009

View online: http://dx.doi.org/10.1063/1.4975009

View Table of Contents: http://aip.scitation.org/toc/jmp/58/2

Published by the American Institute of Physics

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JOURNAL OF MATHEMATICAL PHYSICS 58, 021501 (2017)

Solutions for a class of quasilinear Schrödinger equations

with critical Sobolev exponents

Zhouxin Li1,a) and Yimin Zhang2,a)

1School of Mathematics and Statistics, Central South University, Changsha 410083,

People’s Republic of China

2Department of Mathematics, School of Science, Wuhan University of Technology,

Wuhan 430070, People’s Republic of China

(Received 15 October 2015; accepted 11 January 2017; published online 2 February 2017)

In this paper, we study a class of quasilinear Schrödinger equations involving a critical

exponent which arises in plasma physics. By using the change of variable and varia-

tional approaches combining the concentration-compactness principle, the existence

of a positive solution which has a local maximum point and decays exponentially is

obtained. Published by AIP Publishing. [http://dx.doi.org/10.1063/1.4975009]

I. INTRODUCTION

In this paper, we are interested in the existence of a positive solution for the following quasilin-

ear Schrödinger equation:

−ε2∆u+V(x)u−kαε2(∆(|u|2α))|u|2α−2u=|u|q−2u+|u|2∗(2α)−2u,u>0,x∈RN,(1.1)

where V(x)is a given potential, ε > 0 is a real parameter, kis a real constant, q≥2, α > 1/2 is

a constant, 2∗(2α)=2∗×2α, and 2∗=2N

N−2is the critical Sobolev exponent. Solutions of problem

(1.1) are related to the standing wave solutions of certain quasilinear Schrödinger equations which

arise in several physical phenomena such as the theory of superﬂuid ﬁlm in plasma physics, see

Refs. 7and 12 and the references therein for more backgrounds. For this reason, problem (1.1) is

still called a quasilinear Schrödinger equation.

For α=1, the existence of nontrivial solutions for the quasilinear schrödinger equation was

extensively studied in the past decades. We can see Refs. 4,5, and 13 for the subcritical case

and Refs. 14,16,19, and 20 for the critical case. In Ref. 13, by using a change of variable, they

transform the equation to a semilinear one, then the existence of solutions was obtained via varia-

tional methods under diﬀerent types of potentials V(x). This method is signiﬁcant and was widely

used in the studies of this kind of problems. For general α, there are few results for this case, as far

as we know, just Refs. 1,2,12, and 15. In Ref. 12, for α > 1

2, 2 <q+1<2∗(2α), Liu and Wang

considered Equation (1.1) without a critical term, that is,

−∆u+V(x)u−kα(∆(|u|2α))|u|2α−2u=λ|u|q−2u,u>0,x∈RN,(1.2)

where λ > 0 is a parameter. They obtained the existence of a solution for problem (1.2) by using

the method of Lagrange multiplier. In Ref. 15, the ﬁbering method was employed to obtain the

existence of at least one or sometimes two standing wave solutions for the following quasilinear

Schrödinger equation:

−∆u+V(x)u−kα(∆(|u|2α))|u|2α−2u=µf(x)|u|q−2u,u>0,x∈RN,

where V(x)=λg(x)and g(x)ia a positive function and µand λare positive numbers. Employing

the change of variable method just as in Ref. 13, the authors of Ref. 1obtained the existence of at

least one positive solution for problem (1.2) by using variational approaches. Moreover, in Ref. 2,

a)Electronic addresses: lzx@math.pku.edu.cn and zhangym802@126.com

0022-2488/2017/58(2)/021501/15/$30.00 58, 021501-1 Published by AIP Publishing.

021501-2 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

for V(x)=µ,α > 1

2, they obtained the unique existence of a positive radial solution of problem

(1.2) under some suitable conditions and λ=1. All these results are obtained when the nonlinear

term satisﬁes subcritical growth. If the right side of Equation (1.2) has a critical term, as far as we

know, there are no results for general α.16

In this paper, our aim is to study the existence of positive solutions of (1.1) with general

α > 1/2 and at 2∗(2α)growth. Problem of (1.1) at 2∗(2α)growth has two diﬃculties. First, the

embedding H1(RN)↩→L2∗(2α)(RN)is not compact, so it is hard to prove the Palais-Smale (PS in

short) condition. Second, even if we can obtain the compactness result of PS sequence, it only holds

at some level of a positive upper bound and it is diﬃcult for us to prove that the functional has such

a minimax level.

We assume that V(x)is locally Hölder continuous and

(V)∃V∞>V0>0 such that minx∈RNV(x)=V0and lim|x|→∞ V(x)=V∞.

Under assumption (V), we deﬁne a space

XBu∈H1(RN):RN

V(x)u2<∞

with the norm ∥u∥2

X=RN|∇u|2+RNV(x)u2.

For the simplicity of notation, we let m=2α, ¯q=q/m, and kα=1 in this paper. Set

g(t)=|t|q−2t+|t|2∗m−2tand G(t)=t

0

g(s)ds.

We formulate problem (1.1) in the variational structure in the space Xas follows:

I(u)=ε2

2RN

(1+m|u|2(m−1))|∇u|2+1

2RN

V(x)u2−RN

G(u).

Note that Iis lower semicontinuous on X; we deﬁne that u∈Xis a weak solution for (1.1) if

u∈X∩L∞(RN)and it is a critical point of I.

First, for an arbitrary ε > 0, we have

Theorem 1.1. Assume that q ∈(2m,2∗m)and that condition (V) holds. Moreover, assume that

one of the following conditions holds

(i) 1<m<2and ¯q>4

N−2+2

m;

(ii) m ≥2and ¯q>2∗−1.

Then for ε > 0small enough, problem (1.1) has a positive weak solution uε∈X∩L∞(RN)with

lim

ε→0∥uε∥X=0and uε(x)≤Cexp(−β

ε|x−xε|),

where C >0, β > 0are constants, x ε∈RNis a local maximum point of uε.

Next, we consider the case ε=1. We have the following result:

Theorem 1.2. Assume that all conditions in Theorem 1.1 hold and that ε=1, then problem

(1.1) has a positive weak solution u1∈X∩L∞(RN).

This paper is organized as follows. In Section II, we ﬁrst use a change of variable to reformulate

the problem, then we modify the functional in order to regain the PS condition. In Section III, we

prove that the functional satisﬁes the PS condition, which is a crucial job of this paper. Finally, in

Section IV, we prove the main theorems, which involves the construction of a mountain pass level at

a certain height.

021501-3 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

II. PRELIMINARIES

Since Iis lower semicontinuous on X, we follow the idea in Refs. 5and 13 and make the

change of variables v=f−1(u), where fis deﬁned by

f(0)=0,

f′(v)=(1+m|f(v)|2(m−1))−1/2,on [0,+∞),

f(v)=−f(−v),on (−∞,0].

The above function f(t)and its derivative satisfy the following properties (see Refs. 1,2, and 13):

Lemma 2.1. For m >1, we have

(1) f is uniquely deﬁned, C2and invertible;

(2) |f′(t)| ≤1for all t ∈R;

(3) |f(t)| ≤|t|for all t ∈R;

(4) f (t)/t→1as t →0;

(5) |f(t)| ≤m1/2m|t|1/mfor all t ∈R;

(6) 1

mf(t)≤t f ′(t)≤f(t)for all t >0;

(7) f (t)/m

√t→m1/2mas t →+∞.

According to Ref. 6(see Corollary 2.1 and Proposition 2.2 in it, note that the embedding in

Corollary 2.1 of Ref. 6is also compact.), we have

Lemma 2.2. The map: v→ f(v)from X into Lr(RN)is continuous for 1≤r≤2∗m and is

compact for 1≤r<2∗m.

Using this change of variable, we rewrite the functional I(u)to

J(v)=I(f(v)) =ε2

2RN

|∇v|2+1

2RN

V(x)f2(v)−RN

G(f(v)).

The critical point of Jis the weak solution of the equation

−ε2∆v+V(x)f(v)f′(v)=g(f(v)) f′(v),x∈RN.(2.1)

Now we deﬁne a suitable modiﬁcation of the functional Jin order to regain the Palais-Smale

condition. In this time, we make use of the method in Ref. 17.

Let lbe a positive constant such that

l=sup{s>0 : g(t)

t≤V0

kfor every 0 ≤t≤s}(2.2)

for some k> θ/(θ−2)with θ∈(2m,q]. We deﬁne the functions,

γ(s)=

g(s),s>0,

0,s≤0¯γ(s)=

γ(s),0≤s≤l,

V0

ks,s>l

and

p(x,s)=χR(x)γ(s)+(1−χR(x)) ¯γ(s),

P(x,s)=s

0

p(x,t)dt,

where χRdenotes the characteristic function of the set BR(the ball centered at 0 and with radius R

in RN) and R>0 is suﬃciently large such that

min

BR

V(x)<min

∂BR

V(x).

By deﬁnition, the function p(x,s)is measurable in x, of class Cin sand satisﬁes

(p1) 0 < θP(x,s)≤p(x,s)sfor every x∈BRand s∈R+.

021501-4 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

(p2) 0 ≤2P(x,s)≤p(x,s)s≤1

kV(x)s2for every x∈Bc

RBRN\BRand s∈R+.

Now we study the existence of solutions for the deformed equation

−ε2∆v+V(x)f(v)f′(v)=p(x,f(v)) f′(v),x∈RN.(2.3)

The corresponding functional of (2.3) is given by

¯

J(v)=ε2

2RN

|∇v|2+1

2RN

V(x)f2(v)−RN

P(x,f(v)).

For v∈X, since

|∇(| f(v)|m)|2=m2|f(v)|2(m−1)

1+m|f(v)|2(m−1)|∇v|2≤m|∇v|2,(2.4)

we infer that |f(v)|m∈X. By Sobolev inequality, we have

∥f(v)∥2∗m=∥| f(v)|m∥1/m

2∗≤C∥∇(| f(v)|m)∥1/m

2≤C∥v∥1/m

X.(2.5)

It results that f(v)∈L2∗m(RN). Using interpolation inequality, we obtain that f(v)∈Lq(RN). Thus

¯

Jis well deﬁned on X. Let (vn)⊂X, v ∈Xwith vn→vin X. Then from Lemma 2.2, we infer that

V(x)f2(vn)→V(x)f2(v)in L1(RN)and that f(vn)→f(v)in Lq(RN). Thus ¯

Jis continuous on X.¯

J

is Gateaux-diﬀerentiable in Xand the G-derivative is

⟨¯

J′(v), ϕ⟩=ε2RN∇v∇ϕ+RN

V(x)f(v)f′(v)ϕ−RN

p(x,f(v)) f′(v)ϕ, ∀ϕ∈X.

Then if v∈X∩L∞(RN)is a critical point of ¯

J, and v(x)≤aBf−1(l),∀x∈Bc

R, we have u=

f(v)∈X∩L∞(RN)(note that we have |u|≤|v|and |∇u|≤|∇v|by the properties of f) is a solution

of (1.1).

III. COMPACTNESS OF PS SEQUENCE

In this section, we show that the functional ¯

Jsatisﬁes the PS condition which is a crucial job

and its proof is composed of four steps. Let Sdenote the best Sobolev constant, we have

Lemma 3.1. Assume that condition (V) holds and q ∈(2m,2∗m). Then ¯

J satisﬁes the PS condi-

tion at level cε<1

N m εNSN/2.

Proof. Let (vn)∈Ebe a PS sequence of ¯

Jat level cε, that is, (vn)satisﬁes

¯

J(vn)=ε2

2RN

|∇vn|2+1

2RN

V(x)f2(vn)−RN

P(x,f(vn)) =cε+o(1)(3.1)

and

⟨¯

J′(vn), ϕ⟩=ε2RN∇vn∇ϕ+RN

V(x)f(vn)f′(vn)ϕ

−RN

p(x,f(vn)) f′(vn)ϕ=o(1)∥ϕ∥X,∀ϕ∈X.(3.2)

We divide the proof into four steps.

Step 1: The sequence RN(|∇vn|2+V(x)f2(vn)) is bounded. Multiplying (3.1) by θ(θis given

in Section II) and using (p1)-(p2), we get

θε2

2RN

|∇vn|2+θ

2RN

V(x)f2(vn)

≤BR

p(x,f(vn)) f(vn)+θ

2kBc

R

V(x)f2(vn)+θcε+o(1).

021501-5 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

On the other hand, taking ϕ=f(vn)/f′(vn)in (3.2), we get

RN

ε2(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vn|2+RN

V(x)f2(vn)

=RN

p(x,f(vn)) f(vn)+o(∥vn∥X)≥BR

p(x,f(vn)) f(vn)+o(1)∥vn∥X.

Combining the above two inequalities, we get

(θ

2−m)ε2RN

|∇vn|2+(θ

2−θ

2k−1)RN

V(x)f2(vn)

≤θε2

2RN

|∇vn|2−RN

ε2(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vn|2

+θ

2RN

V(x)f2(vn)−θ

2kBc

R

V(x)f2(vn)−RN

V(x)f2(vn)

≤θcε+o(1)+o(1)∥vn∥X.(3.3)

Since θ > 2mand k>θ

θ−2, we get the conclusion from (3.3).

Step 2: For every δ > 0, there exists R1≥R>0 such that

lim sup

n→∞ Bc

2R1

(|∇vn|2+V(x)f2(vn)) < δ. (3.4)

We consider a cutoﬀfunction ψR1=0 on BR1,ψR1=1 on Bc

2R1, and |∇ψR1|≤C/R1on RNfor some

constant C>0. On the one hand, taking ϕ=f(vn)/f′(vn), we compute ⟨¯

J′(vn), ϕψR1⟩and get

o(1)∥vn∥X=RN

ε2(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vn|2ψR1

+RN

ε2ϕ∇vn∇ψR1+RN

V(x)f2(vn)ψR1

−RN

p(x,f(vn)) f(vn)ψR1

≥RN

ε2|∇vn|2ψR1+RN

ε2ϕ∇vn∇ψR1

+(1−1

k)RN

V(x)f2(vn)ψR1.(3.5)

On the other hand, by Hölder inequality,

RN

ϕ∇vn∇ψR1≤C

R1

∥∇vn∥L2(RN)∥ϕ∥L2(RN).(3.6)

Note that ∥∇vn∥L2(RN)is bounded, and

∥ϕ∥2

L2(RN)=RN

f2(vn)(1+m|f(vn)|2(m−1))

=RN

f2(vn)+mRN

|f(vn)|2m,(3.7)

by (2.5), ∥ϕ∥L2(RN)is also bounded. Therefore, it follows from (3.5)-(3.7) that

lim sup

n→∞ Bc

2R1

(|∇vn|2+V(x)f2(vn)) ≤C

R1

for R1suﬃciently large, which yields (3.4).

Step 3: There exists v∈Xsuch that

lim

n→ ∞ RN

p(x,f(vn)) f(vn)=RN

p(x,f(v)) f(v).(3.8)

021501-6 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

First, by step 1, there exists v∈Xsuch that up to a subsequence, vn→vweakly in Xand vn→v

a.e. in RN. Since we may replace vnby |vn|, we assume vn≥0 and v≥0. By (3.4), for any δ > 0,

there exists R1>0 suﬃciently large such that

lim sup

n→∞ Bc

2R1

(|∇vn|2+V(x)f2(vn)) ≤kδ.

Therefore, by (p2) we have

lim sup

n→∞ Bc

2R1

p(x,f(vn)) f(vn)≤lim sup

n→∞ Bc

2R1

V(x)

kf2(vn)≤δ, (3.9)

and by Fatou lemma,

Bc

2R1

p(x,f(v)) f(v)≤δ. (3.10)

Second, we prove that

B2R1

p(x,f(vn)) f(vn)→B2R1

p(x,f(v)) f(v).(3.11)

Then from this, (3.9)-(3.10), and the arbitrariness of δ, we get (3.8). In fact, since (vn)is bounded

in X, we have (f(vn)) is also bounded. Thus there exists a w∈Xsuch that f(vn)⇀ w in X,

f(vn)→win Lr(BR1)for 1 ≤r<2∗, and f(vn)→wa.e. in BR1. According to (2.4), (| f(vn)|m)

is also bounded in X. By a normal argument, we have |f(vn)|m⇀|w|min X,|f(vn)|m→|w|min

Lr(BR1)for 1 ≤r<2∗, and |f(vn)|m→|w|ma.e. in BR1. Applying Lions’ concentration compact-

ness principle11 to (| f(vn)|m)on ¯

BR1, we obtain that there exist two nonnegative measures µ, ν, a

countable index set K, positive constants {µk},{νk},k∈K, and a collection of points {xk},k∈K

in ¯

BR1, such that

(i) ν=|w|2∗m+

k∈K

νkδxk;

(ii) µ=|∇(|w|m)|2+

k∈K

µkδxk;

(iii) µk≥Sν2/2∗

k,

where δxkis the Dirac measure at xkand Sis the best Sobolev constant. We claim that νk=0 for

all k∈K. In fact, let xkbe a singular point of measures µand ν, as in Ref. 9, we deﬁne a function

φ∈C∞

0(RN)by

φ(x)=

1,Bρ(xk),

0,RN\B2ρ(xk),

φ≥0,|∇φ|≤1

ρ,B2ρ(xk)\Bρ(xk),

where Bρ(xk)is a ball centered at xkand with radius ρ > 0. We take ϕ=φf(vn)/f′(vn)as test

functions in ⟨¯

J′(vn), ϕ⟩and get

RN

ε2(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vn|2·φ

+RN

ε2∇vn∇φ·f(vn)/f′(vn)+RN

V(x)f2(vn)φ

−RN

p(x,f(vn)) f(vn)φ=o(1)∥vnφ∥X.(3.12)

Then Lions’ concentration compactness principle implies that

BR1

|∇|f(vn)|m|2φ→BR1

φdµ, BR1

|f(vn)|2∗mφ→BR1

φdν. (3.13)

021501-7 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

Since xkis singular point of ν, by the continuity of f, we have

f(vn(x))|(B2ρ\{xk}) → ∞

as ρ→0. Thus

1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1)=m−o(ρ)

on B2ρfor ρsuﬃciently small. Then by (2.4) we get from (3.12) that

BR1

ε2φdµ−BR1

φdν

=lim

n→∞ BR1

ε2|∇|f(vn)|m|2φ−BR1

|f(vn)|2∗mφ

≤lim

n→∞ BR1

mε2|∇vn|2φ−BR1

|f(vn)|2∗mφ

≤lim

n→∞ BR1

ε2(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vn|2φ

+o(ρ)BR1

ε2|∇vn|2φ−BR1

|f(vn)|2∗mφ

≤lim

n→∞ −BR1

ε2∇vn∇φ·f(vn)/f′(vn)+λBR1

|f(vn)|qφ

+o(ρ)BR1

ε2|∇vn|2φ+o(1)∥vnφ∥X.(3.14)

We prove that the last inequality in (3.14) tends to zero as ρ→0. By Hölder inequality, we have

lim

n→ ∞ BR1∇vn∇φ·f(vn)/f′(vn)

≤lim sup

n→∞ (BR1

|∇vn|2)1/2

·(BR1

|[ f(vn)/f′(vn)] · ∇φ|2)1/2

.(3.15)

Since |f(vn)/f′(vn)|2=f2(vn)+m|f(vn)|2m, using Hölder inequality we have

lim

n→ ∞ BR1

|[ f(vn)/f′(vn)] · ∇φ|2

≤Cρ∥w∥2

L2∗(B2ρ(xj)) +∥w∥2

L2∗m(B2ρ(xj))→0

as ρ→0. Thus we obtain that the right hand side of (3.15) tends to 0. On the other hand, since

q∈(2m,2∗m), by Lemma 2.2, we can prove that g(x,h(vn))h(vn)φ→g(x, w )wφ in L1(BR1)and

BR1g(x, w)w φ →0 as ρ→0. All these facts imply that the last inequality in (3.14) tends to zero as

ρ→0. Thus νk≥ε2µk. This means that either νk=0 or νk≥εNSN/2by virtue of Lions’ concen-

tration compactness principle. We claim that the latter is impossible. Indeed, if νk≥εNSN/2holds

for some k∈K, then

cε=lim

n→∞ ¯

J(vn)−1

2m⟨¯

J′(vn),f(vn)/f′(vn)⟩

≥lim

n→∞ (1

2m−1

2∗m)RN

|f(vn)|2∗m≥(1

2m−1

2∗m)RN

dν

≥(1

2m−1

2∗m)RN

|w|2∗m+(1

2m−1

2∗m)SN/2εN≥1

N m εNSN/2,

which is a contradiction. Thus νk=0 for all k∈K, and it implies that ∥f(vn)∥L2∗m(BR1)→

∥w∥L2∗m(BR1). By the uniform convexity of L2∗m(BR1), we have f(vn)→wstrongly in L2∗m(BR1).

021501-8 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

Finally, since p(x,f(vn)) f(vn)is sub-(2∗m)growth on B2R1\BR1, we conclude that (3.11) holds. This

proves (3.8).

Step 4: (vn)is compact in X. Since we have (3.8), the proof of the compactness is trivial. This

completes the proof of the lemma.

IV. PROOF OF MAIN RESULTS

Before we prove Theorem 1.1, we will show ﬁrst some properties about the change of variable

f.

Lemma 4.1. Let f 1(v)=|f(v)|m/v , v ,0and f1(0)=0, then f 1is continuous, odd, nonde-

creasing and

lim

v→0f1(v)=0and lim

|v|→+∞|f1(v)| =√m.(4.1)

Proof. In fact, by (6) of Lemma 2.1,

f′

1(v)=v−2(m|f(v)|m−2f(v)f′(v)v−|f(v)|m)≥0,

so f1is nondecreasing. By (4) of Lemma 2.1, f1(v)→0 as v→0. Finally, according to L’Hospital’s

principle,

lim

v→+∞f1(v)=lim

v→+∞

|f(v)|m

v=lim

v→+∞m|f(v)|m−2f(v)f′(v)=√m.

This shows that (4.1) holds.

Lemma 4.2. There exists d0>0such that

lim

v→+∞(√mv−fm(v)) ≥d0.

Proof. Assume that v > 0. Since by (6) of Lemma 2.1, f(v)≤m f ′(v)v, we have

√mv−fm(v)≥√mv−m f m−1(v)f′(v)v

=1+m f 2(m−1)(v)−√m f m−1(v)

1+m f 2(m−1)(v)

√mv

=√mv

1+m f 2(m−1)(v)+√m f m−1(v)1+m f 2(m−1)(v)

≥√mv

2(1+m f 2(m−1)(v)) ≥fm(v)

4m f 2(m−1)(v)

=1

4m f m−2(v)Bd(m, v ).(4.2)

In the last inequality, we have used the fact that √mv≥fm(v)and that m f 2(m−1)(v)>1 for v > 0

suﬃciently large.

If 1 <m<2, then d(m, v )→+∞as v→+∞. If m=2, then d(m, v )=1/8. If m>2, we

claim that √mv−fm(v)→0 as v→+∞is impossible. In fact, assume on the contrary, then using

L’Hospital’s principle, we get

0<lim

v→+∞

√mv−fm(v)

f2−m(v)=lim

v→+∞

√m−m f m−1(v)f′(v)

(2−m)f1−m(v)f′(v)

=lim

v→+∞

m

(2−m)f1−m(v)√m1+m f 2(m−1)(v)+m f m−1(v)

=1

2(2−m)<0.

021501-9 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

This is a contradiction. Thus for all m>1, there exists d0>0 such that there holds

lim

v→+∞(√mv−fm(v)) ≥d0.

This completes the proof.

Lemma 4.3. We have

(i) If 1<m<2, then

lim

v→+∞

√mv−fm(v)

f2−m(v)=1

2(2−m).

(ii) If m ≥2, then

lim

v→+∞

√mv−fm(v)

log f(v)≤

1

2,m=2,

0,m>2.

Proof. First, we prove part (i). According to (4.2) in Lemma 4.2, we have √mv−fm(v)→+∞

as v→+∞. Thus by L’Hospital’s principle, we get

lim

v→+∞

√mv−fm(v)

f2−m(v)=lim

v→+∞

√m−m f m−1(v)f′(v)

(2−m)f1−m(v)f′(v)=1

2(2−m).

Next, we prove part (ii). If there exists a constant C>0 such that √mv−fm(v)≤C, then

the conclusion holds. Otherwise, assume that √mv−fm(v)→+∞as v→+∞. Then again by

L’Hospital’s principle, we have

lim

v→+∞

√mv−fm(v)

log f(v)=lim

v→+∞

√m−m f m−1(v)f′(v)

f′(v)/f(v)=

1

2,m=2,

0,m>2.

This completes the proof.

To prove Theorem 1.1, it is crucial to prove that ¯

Jhas the mountain pass level cε<1

N m εNSN/2.

Let us consider the following family of functions in Ref. 3:

v∗

ω(x)=[N(N−2)ω2](N−2)/4

[ω2+|x|2](N−2)/2,

which solves the equation −∆u=u2∗−1in RNand satisﬁes ∥∇v∗

ω∥2

L2=∥v∗

ω∥2∗

L2∗=SN/2. Let ωbe

such that 2ω < Rand let ηω(x)∈[0,1]be a positive smooth cutoﬀfunction with ηω(x)=1 in Bω,

ηω(x)=0 in BR\B2ω. Let vω=ηωv∗

ω. For all ω > 0, there exists tω>0 such that ¯

J(tωvω)<0 for

all t>tω. Deﬁne the class of paths

Γ={γ∈C([0,1],X):γ(0)=0, γ(1)=tωvω},

and the minimax level

cε=inf

γ∈Γmax

t∈[0,1]

¯

J(γ(t)).

Let tωbe such that

¯

J(tωvω)=max

t≥0

¯

J(tvω).

Note that the sequence (vω)is uniformly bounded in X, then if ¯

J(tωvω)→0 as tω→0, we are done;

on the other hand, if tω→+∞, then ¯

J(tωvω)→ −∞, which is impossible, so it remains to consider

the case where the sequence (tω)is upper and lower bounded by two positive constants. According

to Ref. 3, we have, as ω→0,

∥∇vω∥2

L2=SN/2+O(ωN−2),∥vω∥2∗

L2∗=SN/2+O(ωN).

021501-10 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

Let a∈(0,ε(N−2)/2

2√m),b∈(2ε(N−2)/2

√m,+∞)be such that tω∈[a,b],∀ω∈(0,ω0), where ω0>0 small

enough. By computing d

dt ¯

J(tvω)=0, we obtain tω=ε(N−2)/2

√m+o(1). Let

H(v)=−1

2V(x)f2(v)+λ

q|f(v)|q−1

2∗m|√mv|2∗+1

2∗m|f(v)|2∗m,

then by (4.1) and (4) of Lemma 2.1, for m>1, we have

lim

|v|→+∞H(v)/|v|2∗=0 and lim

v→0H(v)/v2=−1

2V(x).

Thus H(v)is sub-(2∗)growth.

The following proposition is important to the computation of a mountain pass level cε<

1

N m εnSN/2.

Proposition 4.4. Under the assumptions of Theorem 1.1, there exists a function τ=τ(ω)such

that limω→0τ(ω)= +∞and for ωsmall enough,

RN

H(tωvω)≥τ(ω)·ωN−2.

Proof. We divide the proof into three steps.

Step 1: We prove that

1

ωN−2Bω

H(tωvω)≥τ1(ω)(4.3)

with limω→0τ1(ω)= +∞.

By the deﬁnition of vω, for x∈Bω, there exist constants c2≥c1>0 such that for ωsmall

enough, we have

c1ω−(N−2)/2≤vω(x)≤c2ω−(N−2)/2

and

c1ω−(N−2)/2≤fm(vω(x)) ≤c2ω−(N−2)/2.(4.4)

On the one hand, by (7) of Lemma 2.1, (4.4) and the continuity of V(x)in ¯

Bω, there exists C1>0

such that

Bω

V(x)f2(tωvω)≤C1ωN−2

mN−2

2=C1ω(2∗

2−1

m)(N−2).(4.5)

Similarly, there exists C2>0 such that

Bω

fq(tωvω)≥C2ωN−q

mN−2

2=C2ω(2∗

2−¯q

2)(N−2),(4.6)

where ¯q=q/m. On the other hand, since the function ζ(t)=t2∗is convex, using (5) of Lemma 2.1

and Hölder inequality, we have

1

2∗mBω(√mtωvω)2∗−(fm(tωvω))2∗

≤1

mBω

(√mtωvω)2∗−1[√mtωvω−fm(tωvω)]

≤1

m(Bω

(√mtωvω)2∗)(2∗−1)/2∗(Bω

[√mtωvω−fm(tωvω)]2∗)1/2∗.(4.7)

Case 1: 1 <m<2. From (4.7) and (i) of Lemma 4.3, we obtain that there exists C3>0 such

that

1

2∗mBω(√mtωvω)2∗−(fm(tωvω))2∗≤C3ω[N−(2

m−1)N−2

22∗]1

2∗=C3ω(1−1

m)(N−2).(4.8)

021501-11 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

Combining (4.5), (4.6), and (4.8), we have

1

ωN−2Bω

H(tωvω)≥ −C1ω(2∗

2−1

m−1)(N−2)+C2ω(2∗

2−¯q

2−1)(N−2)−C3ω−1

m(N−2)Bτ1(ω).

It is obvious that 2∗

2−1

m−1>−1

m. Using the condition (i) in Theorem 1.1, we have 2∗

2−¯q

2−1<

−1

m. It results that τ1(ω)→+∞as ω→0.

Case 2:m≥2. Note that for any δ∈(0,m), limv→+∞log f(v)/fδ(v)=0,we have log f(v)≤

fδ(v)for v > 0 large enough. Thus for ω > 0 small enough, from (4.7) and (ii) of Lemma 4.3, we

get

1

2∗mBω(√mtωvω)2∗−(fm(tωvω))2∗≤C′

3ω[N−δ

mN−2

22∗]1

2∗=C′

3ω1

2(1−δ

m)(N−2).(4.9)

Combining (4.5), (4.6), and (4.9), we have

1

ωN−2Bω

H(tωvω)≥ −C1ω(2∗

2−1

m−1)(N−2)+C2ω(2∗

2−¯q

2−1)(N−2)−C′

3ω−1

2(1+δ

m)(N−2)Bτ1(ω).

Since m≥2, we have 2∗

2−1

m−1>−1

2(1+δ

m). From condition (ii) in Theorem 1.1, there exists a

δ=δ(N,¯q)>0 (depends on Nand ¯q) small enough such that 2∗

2−¯q

2−1<−1

2(1+δ

m). It results

that τ1(ω)→+∞as ω→0.

Cases 1 and 2 show that (4.3) holds.

Step 2: We prove that there exists C4>0 such that

1

ωN−2B2ω\Bω

H(tωvω)≥ −C4ω(2∗

2−1

m−1)(N−2)Bτ2(ω).(4.10)

Note that for x∈B2ω\Bω, we have

vω(x)≤v∗

ω(x)≤c2ω−(N−2)/2.(4.11)

Since ηωis a positive smooth cutoﬀfunction, without the loss of generality, we may assume that ηω

is such that

B2ω\Bω

|vω|2∗≤B2ω\Bω

V(x)f2(vω).

By (4.1) and (4.11), we have fm(vω(x)) ≤c2ω−(N−2)/2for x∈B2ω\Bω. Thus

1

ωN−2B2ω\Bω

H(tωvω)

≥ − 1

2ωN−2B2ω\Bω

V(x)f2(tωvω)−1

2∗mωN−2B2ω\Bω

|√mtωvω|2∗

≥ − C5

ωN−2B2ω\Bω

V(x)f2(tωvω)

≥ −C4ωN−2

mN−2

2−(N−2)=−C4ω(2∗

2−1

m−1)(N−2),

where C4>0, C5>0 are constants. This shows that (4.10) holds.

Step 3: To conclude, let τ(ω)=τ1(ω)+τ2(ω), we have τ(ω)→+∞as ω→0. This implies the

conclusion of the proposition.

Proof of Theorem 1.1. By (4) and (7) of Lemma 2.1, it is easy to verify that ¯

Jhas the moun-

tain pass geometry. Lemma 3.1 shows that ¯

Jsatisﬁes the PS condition. We prove that ¯

Jhas the

mountain pass level cε<1

N m εNSN/2. Let

F(t)=ε2

2∥∇(tvω)∥2

L2−1

2∗m∥√mtvω∥2∗

L2∗.

Then we have

F(t)≤F(t0)=1

N m εNSN/2+O(ωN−2),∀t≥0,(4.12)

021501-12 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

where t0=ε(N−2)/2

√m. By (4.12) and Proposition 4.4, we have

¯

J(tωvω)=F(tωvω)−RN

H(tωvω)

≤1

N m εNSN/2+O(ωN−2)−τ(ω)ωN−2

<1

N m εNSN/2.

This shows that ¯

J(v)has a nontrivial critical point vε∈X, which is a weak solution of (2.3).

We prove that vεis also a weak solution of (2.1). First, we can argue as the proof of Proposition

2.1 in Ref. 17 to obtain that

lim

ε→0max

x∈∂BR

vε(x)=0.

Thus there exists ε0>0 such that for all ε∈(0, ε0), we have vε(x)≤aBf−1(l),∀|x|=R, where l

is given in (2.2). Secondly, we prove that

vε(x)≤a,∀ε∈(0, ε0)and ∀x∈RN\BR.(4.13)

Taking

ϕ=

(vε−a)+,x∈RN\BR,

0,x∈BR

as a test function in ⟨¯

J′(vε), ϕ⟩=0, we get

ε2RN\BR

|∇(vε−a)+|2+ε2RN\BR(V(x)−p(x,f(vε))

f(vε))f(vε)f′(vε)(vε−a)+=0.(4.14)

By (p2), we have

V(x)−p(x,f(vε))

f(vε)>0,∀x∈RN\BR.

Therefore, all terms in (4.14) must be equal to zero. This implies vε≤ain RN\BR. This proves

(4.13). Thus vεis a solution of problem (2.1).

To complete the proof, we deduce as the proof for Theorem 4.1 in Ref. 10 to obtain that

vε|BR∈L∞(BR). Thus uε=f(vε)∈X∩L∞(RN)is a nontrivial weak solution of (1.1). Finally, by

Proposition 4.5 in the following, we have limε→0∥uε∥X=0 and uε(x)≤Ce−β

ε|x−xε|.This completes

the proof.

We prove the norm estimate and the exponential decay.

Proposition 4.5. Let vε∈X∩L∞(RN)be a solution of (2.1) and let uε=f(vε), then we have

lim

ε→0∥uε∥X=0and uε(x)≤Ce−β

ε|x−xε|,

where C >0,β > 0are constants.

Proof. First, let x0∈BRbe such that V(x0)=V0. Deﬁne J0:X→Rby

J0(v)=1

2RN

|∇v|2+1

2RN

V0f2(v)−RN

G(f(v)).

Let

c0=inf

γ∈Γ0

sup

t∈[0,1]

J0(γ(t)),

Γ0={v∈C([0,1],X):γ(0)=0,J0(γ(1)) <0}.

021501-13 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

Similar to the proof for estimate (2.4) in Ref. 17 (or Lemma 3.1 in Ref. 18), we can show that

cε≤εNc0+o(εN)by using the change of coordinates y=(x−x0)/ε. Arguing as for (3.3), and by

virtue of this energy estimate, we obtain

∥vε∥X≤θcε

min{( θ

2−m)ε2,(θ

2−θ

2k−1)} ≤2θc0

θ−2mεN−2+o(εN−2)

for ε > 0 suﬃciently small. Let uε=f(vε), then uε,0. Note that |∇uε|≤|vε|and |uε|≤|vε|, we

get limε→0∥uε∥X=0.

Second, similar to the proof for Theorem 4.1 in Ref. 10, we conclude that vε∈L∞(RN)

and by Ref. 8, we have vε∈C1,α(BR). Now let xεdenote the maximum point of vεin BRand

let

σBsup{s>0 : g(t)<V0tfor every t∈[0,s]}.

Then vε(xε)≥f−1(σ)for ε > 0 small. In fact, assume that vε(xε)<f−1(σ)for some ε > 0 suﬃ-

ciently small. According to the deﬁnition of l(see (2.2)) and σ, we have vε(x)≤f−1(l)<f−1(σ)

(note that k>1 in (2.2)), ∀x∈RN\BR. Thus

V(x)−g(f(vε))

f(vε)>0,∀x∈RN.

Since vε=f−1(uε)is a critical point of Jε, we choose ϕ=f(vε)/f′(vε)as a test function in

⟨J′

ε(vε), ϕ⟩=0 and get

0=ε2RN(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vε|2

+RN

V(x)f2(vε)−RN

g(f(vε)) f(vε)

=ε2RN(1+m(m−1)| f(vn)|2(m−1)

1+m|f(vn)|2(m−1))|∇vε|2

+RN(V(x)−g(f(vε))

f(vε))f2(vε).

It turns out that all terms in the above equality must be equal to zero, which means that vε≡0, a

contradiction.

Now let wε(x)=vε(xε+εx), then wεsolves the equation

−∆wε+V(xε+εx)f(wε)f′(wε)=g(f(wε)) f′(wε),x∈RN.

Note that limt→0+f(t)f′(t)

t=1 by the properties of fand that wε(x)→0 as |x|→+∞; we have,

there exists R0>0 such that for all |x|≥R0,

f(wε(x)) f′(wε(x)) ≥3

4wε(x)(4.15)

and

g(f(wε(x))) f′(wε(x)) ≤V0

2wε(x).(4.16)

Let ϕ(x)=Me−β|x|with β2<V0

4and Me−βR0≥wε(x)for all |x|=R0. It is easy to verify that for

x,0,

∆ϕ≤β2ϕ. (4.17)

Now deﬁne ψε=ϕ−wε. Using (4.15)-(4.17), we have

−∆ψε+V0

4ψε≥0,in |x|≥R0,

ψε≥0,in |x|=R0,

lim|x|→∞ψε=0.

021501-14 Z. Li and Y. Zhang J. Math. Phys. 58, 021501 (2017)

By the maximum principle, we have ψε≥0 for all |x|≥R0. Thus, we obtain that for all |x|≥

R0,

wε(x)≤ϕ(x)≤Me−β|x|.

Using the change of variable, we have that for all |x|≥R0,

vε(x)=wε(ε−1(x−xε)) ≤Me−β

ε|x−xε|.

Then by the regularity of vεon BRand note that f(t)≤tfor all t≥0, we have

uε(x)≤Ce−β

ε|x−xε|

for some C>0. This completes the proof.

Proof of Theorem 1.2. We consider the following equation:

−∆u+V(x)u−kα(∆(|u|2α))|u|2α−2u=|u|q−2u+|u|2∗(2α)−2u,u>0,x∈RN.(4.18)

Let y=εxwith ε∈(0, ε0),ε0is given by Theorem 1.1, then we can transform (4.18) into

−ε2∆u+¯

V(y)u−kαε2(∆(|u|2α))|u|2α−2u=|u|q−2u+|u|2∗(2α)−2u,u>0, y ∈RN.(4.19)

Here ¯

V(y)=V(y

ε)still has the properties given in assumption (V). Thus according to Theorem 1.1,

(4.19) has a positive weak solution uε(y)in X∩L∞(RN), which implies that (4.18) has a positive

weak solution u1(x)=uε(εx).

ACKNOWLEDGMENTS

The authors would like to express their sincere gratitude to the anonymous referee for his/her

valuable suggestions and comments. The ﬁrst author was supported by the Natural Science

Foundation of China (No. 11201488) and the Hunan Provincial Natural Science Foundation of

China (No. 14JJ4002). The second author was supported by the Natural Science Foundation of

China (No. 11471330) and the Fundamental Research Funds for the Central Universities(WUT:

2017 IVA 075).

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