Solutions for a class of quasilinear Schrödinger equations with critical Sobolev exponents

Abstract

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 variational approaches combining the concentration-compactness principle, the existence of a positive solution which has a local maximum point and decays exponentially is obtained.

Full text

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
XBu∈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
2RN
(1+m|u|2(m−1))|∇u|2+1
2RN
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
2RN
|∇v|2+1
2RN
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
2RN
|∇v|2+1
2RN
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⟩=ε2RN∇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
2RN
|∇vn|2+1
2RN
V(x)f2(vn)−RN
P(x,f(vn)) =cε+o(1)(3.1)
and
⟨¯
J′(vn), ϕ⟩=ε2RN∇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
2RN
|∇vn|2+θ
2RN
V(x)f2(vn)
≤BR
p(x,f(vn)) f(vn)+θ
2kBc
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)ε2RN
|∇vn|2+(θ
2−θ
2k−1)RN
V(x)f2(vn)
≤θε2
2RN
|∇vn|2−RN
ε2(1+m(m−1)| f(vn)|2(m−1)
1+m|f(vn)|2(m−1))|∇vn|2
+θ
2RN
V(x)f2(vn)−θ
2kBc
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)+mRN
|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)+λBR1
|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)√m1+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−2Bω
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∗mBω(√mtωvω)2∗−(fm(tωvω))2∗
≤1
mBω
(√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∗mBω(√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−2Bω
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∗mBω(√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−2Bω
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−2B2ω\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−2B2ω\Bω
H(tωvω)
≥ − 1
2ωN−2B2ω\Bω
V(x)f2(tωvω)−1
2∗mωN−2B2ω\Bω
|√mtωvω|2∗
≥ − C5
ωN−2B2ω\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
ε2RN\BR
|∇(vε−a)+|2+ε2RN\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
2RN
|∇v|2+1
2RN
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=ε2RN(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)
=ε2RN(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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