Quasilinear elliptic equations with critical growth via perturbation method

J. Differential Equations 254 (2013) 102–124

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Journal of Differential Equations www.elsevier.com/locate/jde

Quasilinear elliptic equations with critical growth via perturbation method Xiang-Qing Liu a , Jia-Quan Liu b , Zhi-Qiang Wang c,d,∗ a

Department of Mathematics, Yunnan Normal University, Kunming 650092, PR China LMAM, School of Mathematical Science, Peking University, Beijing 100871, PR China c Chern Institute of Mathematics, Nankai University, Tianjin 300071, PR China d Department of Mathematics and Statistics, Utah State University, Logan, UT 84322, USA b

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 13 February 2012 Revised 4 September 2012 Available online 25 September 2012

We consider a class of quasilinear Schrödinger equations which include the Modified Nonlinear Schrödinger Equations. A new perturbation approach is used to treat the critical exponent case giving new existence results. © 2012 Elsevier Inc. All rights reserved.

MSC: 35B05 35B45 Keywords: Quasilinear elliptic equations Critical exponent Perturbation methods

1. Introduction In this paper we consider existence of solutions of quasilinear elliptic equations of the form N 





D j a i j (u ) D i u −

i , j =1

where D i = ∂∂x and D s ai j (s) = i Equation (MNLS)

*

N 1 

2

D s ai j (u ) D i u D j u − V (x)u + f (u ) = 0,

in R N ,

(1.1)

i , j =1

d a ( s ). ds i j

This contains the so-called Modified Nonlinear Schrödinger

Corresponding author at: Department of Mathematics and Statistics, Utah State University, Logan, UT 84322, USA. E-mail address: [email protected] (Z.-Q. Wang).

0022-0396/$ – see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2012.09.006

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

  u − V (x)u + u  u 2 + f (u ) = 0,

103

in R N

(1.2)

which corresponds to the special case ai j (s) = (1 + 2s2 )δi j . This type of quasilinear equations appear in many models from mathematical physics (e.g., [5–7,16,19] and references therein). In mathematical work this class of quasilinear equations have been studied in recent years both for evolutionary problems (e.g., [12,9,11,17,18,20,30]) and stationary problems (e.g., [1–3,8,10,13,14,21,23–29,31,34,35] and references therein). The current paper is concerned with the critical exponent problem. We point out that due to the quasilinear term presence there is a different critical exponent than the classical Sobolev exponent. As observed in [24, Proposition 5.12] it turns out that if f (u ) = |u |q−2 u in Eq. (1.2), q = 2(2∗ ) = N4N −2 behaves as a critical exponent for the equations. Non-existence results can be formulated when q  2(2∗ ) by usinga version of the Pohozaev identity [32], for example, (1.2) has no positive solution in H 1 (R N ) with RN u 2 |∇ u |2 dx < ∞ provided that x · ∇ V (x)  0 and q  2(2∗ ). For equations like (1.2) a change of variables was first used in [10,23] to bring the equation to one of semilinear type. Then various techniques for semilinear problems can be adapted in this situation. This approach has been explored by several authors to treat critical exponent problems such as do Ó, Miyagaki, Soares [13] and Silva, Vieira [34]. However this approach does not work for the more general quasilinear problems (1.1). A Nehari manifold approach was taken recently by the authors of the current paper to treat quasilinear equations (1.1). By using subcritical approximations an existence theory was given in [27]. The approach demands certain monotonicity conditions for the structure of the equations. In the current paper we propose a new approach for the critical exponent problems, namely a perturbation method. This idea has been initially used successfully by the authors for subcritical problems in [26]. We shall apply the idea to critical exponent problems and give new results improving existing ones as well as our own in [27]. We first make the following assumptions.

( V ) For some γ ∈ (0, 1), V ∈ C γ (R N ), 0 < V (x)  V ∞ := lim|x|→∞ V (x). (a1 ) ai j ∈ C 1,γ (R), ai j (s) = a ji (s), i , j = 1, 2, . . . , N . N (a2 ) There exist constants c 0 , b > 0 and 4 < p < N4N −2 such that for s ∈ R, ξ = (ξi ) ∈ R





c 0 1 + s2 |ξ |2 

N 





ai j (s)ξi ξ j  c 0−1 1 + s2 |ξ |2

i , j =1

and

(−2 + b)

N 

ai j (s)ξi ξ j 

i , j =1

N 

D s a i j ( s ) s ξi ξ j  ( p − 2 − b )

i , j =1

N 

ai j (s)ξi ξ j .

i , j =1

(a3 ) There exists α ∈ [0, 2) such that as s → ∞, uniformly in ξ ∈ S N −1 , the unit sphere in R N , N 





ai j (s)ξi ξ j = 2s2 + O |s|α ,

i , j =1

N 1 

2



i , j =1

For nonlinear term we assume 4N

( f ) f (s) = |s| N −2 −2 s + |s| p −2 s. The nonlinearity is of critical growth as

4N N −2



D s ai j (s)sξi ξ j = 2s2 + O |s|α .

serves as the critical exponent for Eq. (1.1).

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X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

Here is the main result. Theorem 1.1. Under the assumptions ( V ), (a1 ), (a2 ), (a3 ) and ( f ) Eq. (1.1) has a ground state positive solution 2( N +2) u ∈ H 1 (R N ) ∩ L ∞ (R N ), provided p > p N + α , where p N = 4 if N  6 and p N = N −2 , if N = 3, 4, 5. A function u ∈ H 1 (R N ) with C 0∞ (R N ) it holds

  N RN



a i j (u ) D i u D j φ +

i , j =1

RN

u 2 |∇ u |2 dx < ∞ is called a weak solution of (1.1), if for all φ ∈

N 1 

2





D s ai j (u ) D i u D j u φ dx +

i , j =1

 V (x)u φ dx =

RN

f (u )φ dx.

RN

When we consider the problem (1.1) by using the classical critical point theory we encounter the difficulties caused by the lack of an appropriate working space. It seems there is no natural spaces in which the variational functional possesses both smoothness and compactness properties. For smoothness one would need to work in a space smaller than H 1 (R N ) to control the term involving the quasilinear term in (1.1), but it seems impossible to obtain bounds for (PS) sequences in this setting. On the other hand, as the natural critical exponent is N4N −2 this causes additional difficulties to use the classical critical point theory. There have been several ideas used in recent years to overcome the difficulties such as minimizations [25,31], Nehari method [24], change of variables [10,23]. While the methods used in [10,23] are most effective it works only for the special form of the quasilinear term. In order to overcome the difficulties, we shall use perturbation method which has been successfully used in subcritical problems recently [26]. In order to deal with the potential well case in our main result we need first to consider a problem which is regarded as one problem at infinity, namely when V (x) is a constant. The methods used for constant potentials works well also for the general periodic potentials, that is, when V (x) is periodic in each of its variables. Results of this nature are of independent interest and will be stated in Section 2. Using the results from Section 2 we will prove the main theorem in Section 3. We finish the paper with Appendix A which contains discussions concerning the condition p > p N + α , which indicates the necessity as far as the method is concerned. 2. Problems with constant potential V ∞ consider the case V = V ∞ a constant potential. That is to find a function u ∈ H 1 (R N ) with  We 2 u |∇ u |2 dx < ∞ satisfying for all φ ∈ C 0∞ (R N ) RN

  N RN

a i j (u ) D i u D j φ +

i , j =1

N 1 

2





D s ai j (u ) D i u D j u φ dx +

i , j =1

 V ∞ u φ dx =

RN

f (u )φ dx.

(Q)

RN

In form, Eq. (1.1) has a variational structure given by the functional

J (u ) =

1

  N

2 RN

where F (s) = Let

s 0

f (t ) dt =

μ ∈ (0, 1],

4N N +2

ai j (u ) D i u D j u dx +

i , j =1

4N N −2 |s| N −2 4N

1



 V ∞ u 2 dx −

2 RN

F (u ) dx,

(2.1)

RN

+ 1p |s| p .

< s < min{4, N }. We consider the perturbed problem by looking for u ∈ X =

W 1,s (R N ) ∩ H 1 (R N ) such that for all φ ∈ X

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124





μ

 |∇ u |s−2 ∇ u ∇φ + |u |s−2 u φ dx +

RN

  N



2

D s ai j (u ) D i u D j u φ dx +

i , j =1

RN

=



N 

1

+

ai j (u ) D i u D j φ dx

i , j =1

RN



105

V ∞ u φ dx

RN

f (u )φ dx.

( Q )μ

RN

The corresponding functional is defined on X

J μ (u ) =

μ



s

  | Du |s + |u |s dx + J (u )

RN

=

μ



s

  1 | Du |s + |u |s dx +

  N

2

RN

RN

ai j (u ) D i u D j u dx +

i , j =1

1



 V ∞ u 2 dx −

2 RN

F (u ) dx.

RN

 (u ) = 0 if and only if u is a solution of problem ( Q ) . J μ is a C 1 functional on X and J μ μ  (u ) → 0 and J (u ) → c as n → ∞. Lemma 2.1. Let {un } ⊂ X be a (PS)c sequence of J μ with c > 0, i.e., J μ n μ n  (u ) = 0 and Then there exist a sequence { yn } ⊂ R N and u ∈ X , u = 0, such that u˜ n (·) = un (· + yn ) u, J μ J μ (u )  limn→∞ J μ (un ).  (u ) → 0, J (u ) = c + o (1), we have Proof. We first prove that {un } is bounded. By J μ n μ n

1  J (un ), un p μ

c + o(1)un   J μ (un ) −

 =

1 s

1

−





μ

p

 | Dun |s + |un |s dx

RN

+

  N  1 

+

2

i , j =1

RN

1 2

 C μ

−

1

−

1 p

a i j (u n ) −



 V ∞ un2 dx +

p

p

RN



 | Dun |s + |un |s dx +

RN

1



1 2p

−

D s ai j (un )un D i un D j un dx N −2



4N

|un | N −2 dx

4N RN







1 + un2 | Dun |2 dx +

RN

 un2 dx +

RN

 4N |un | N −2 dx .

RN

This proves the bound of un . Now there are two cases that may occur. (a) Vanishing. sup y ∈RN

 B 1 ( y)

un2 dx → 0 as n → ∞.

(b) Non-vanishing. There exist { yn } ⊂ R N ,

ν > 0 such that lim infn→∞

 B 1 ( yn )

un2 dx  ν .

106

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124



If Vanishing occurs by Lions lemma [22] we have particular,



f (un )un dx =

RN





RN

4N N −2

(|un |

RN

|un |q dx → 0 as n → ∞ for q ∈ (2,

sN ). N −s

In

+ |un | ) dx → 0 as n → ∞. Then p



 | Dun |s + |un |s dx

μ RN

+ 

  N

RN

N 1 

a i j (u n ) D i u n D j u n +

2

i , j =1





D s ai j (un ) D i un D j un un dx +

i , j =1

V ∞ un2 dx

RN





 f (un )un dx + J μ (un ), un → 0.

=

(2.2)

RN

By (a2 ) we have un → 0 in X and J μ (un ) → 0, which is a contradiction. Suppose next that Non-vanishing occurs. Since the functional J μ is translation invariant, without loss of generality, we may assume yn = 0, n = 1, 2, . . . . Up to a subsequence un u in X , un → u in q 2 2 Lloc (R N ), for 2  q  N4N −2 . In particular, B 1 (0) u dx = limn→∞ B 1 (0) un dx  ν > 0. Using the equation for un



μ



 |∇ un |s−2 ∇ un ∇φ + |un |s−2 un φ dx

RN

+  =

  N

RN



N 1 

ai j (un ) D i un D j φ dx +

i , j =1

2

RN

 D s ai j (un ) D i un D j un φ dx +

i , j =1

V ∞ un φ dx

RN

f (un )φ dx + o(1)φ,

(2.3)

RN

and the convergence of un we get that u satisfies Eq. ( Q )μ with all φ ∈ C 0∞ (R N ). By approximation, ( Q )μ holds for all φ ∈ X . Finally by the lower semi-continuity we have

J μ (u ) = J μ (u ) −

 =

1 s

1

−

1  J (u ), u p μ



p



 | Du |s + |u |s dx

μ RN

+

  N  1 

+

2

i , j =1

RN

1 2

−



 lim inf n→∞

1

a i j (u ) −

p

 2

V ∞ u dx + RN

s

1



p 1

−

−

1 p



μ RN

1 2p

1 p

−

D s ai j (u )u D i u D j u dx N −2

 RN



 | Dun |s + |un |s dx

4N

|u | N −2 dx

4N

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

+

  N  1 

+

2

i , j =1

RN

1 2

−

1

−

1



p

a i j (u n ) −



 V ∞ un2 dx +

p

1 p

RN





1

D s ai j (un )un D i un D j un dx

2p

N −2

−



4N

n→∞

 4N |un | N −2 dx

RN



= lim inf J μ (un ) −

107

1  J (un ), un = lim J μ (un ). n→∞ p μ

2

Define the mountain pass value [33]

d(μ) = inf sup J μ



γ ∈Γ t ∈[0,1]



γ (t ) ,

where

      Γ = γ ∈ C [0, 1], X  γ (0) = 0, J μ γ (1) < 0 . Lemma 2.2. There exists m > 0 independent of μ such that d(μ)  m. Proof. For

ρ > 0, let        2 2 2 2 Σρ = u ∈ X  1 + u | Du | + u dx = ρ .

RN

If u ∈ Σρ , with some positive constants C , C 1 , we have for

J μ (u ) 

1

  N

2



C

ai j (u ) D i u D j u dx +

i , j =1

RN

1

ρ small



 2

V ∞ u dx −

2 RN



 (1 + u 2 )| Du |2 + u 2 dx − 

RN

F (u ) dx

RN





u 2 dx − C 

RN

4N

|u | N −2 dx RN

1

2N

 C ρ 2 − C 1 ρ N −2  C ρ 2 . 2

Let

γ ∈ Γ . We know there is a t 0 ∈ (0, 1) such that γ (t 0 ) ∈ Σρ . We have d(μ) = inf

sup J μ

γ ∈Γ t ∈[0,1]





1

γ (t )  inf J μ (u )  C ρ 2 := m. u ∈Σρ

2

2

Proposition 2.1. J μ has a positive critical point. Proof. By Lemma 2.2 and the Mountain Pass Lemma there exists a sequence {un } ⊂ X such that  (u ) → 0 and J (u ) → d(μ) > 0. By Lemma 2.1 there exists u ∈ X , u = 0, J  (u ) = 0 and Jμ n μ n μ + defined by replacing u with u + J μ (u )  limn→∞ J μ (un ) = d(μ). If we consider the functional J μ + . We obtain in F (u ), where u + = max{u , 0}, then the conclusions of Lemmas 2.1 and 2.2 hold for J μ + a nontrivial nonnegative critical point u of J μ , and also a critical point of J μ . By Lemma 2.6 below and by the maximum principle u is smooth and positive. 2

108

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

μn → 0 as n → ∞, J μ n (un ) = 0, J μn (un )  C , un u in H 1 (RN ) and

Proposition 2.2. Suppose that

un ∇ un u ∇ u in L (R ). Then u ∈ H 1 (R N ) ∩ L ∞ (R N ), J  (u ) = 0 (that is, u is a solution of ( Q )) and J (u )  lim infn→∞ J μn (un ). 2

N

Proof. We first show that u is a bounded function. The proof is almost the same as that of Proposition 3.1 of [26]. Recall the equation





 |∇ un |s−2 ∇ un ∇φ + |un |s−2 un φ dx

μn RN

+ 

  N

=

2

i , j =1

RN

  N

1

ai j (un ) D i un D j φ dx +

 D s ai j (un ) D i un D j un φ dx +

i , j =1

RN

V ∞ un φ dx

RN

f (un )φ dx

(2.4)

RN

for all φ ∈ X . Taking φ = un |un |s ) dx  C 1 . For a function R > 0, let ψ ∈ C 0∞ (R N ) such Choose 0 < T < K , r > 0 and





in the above formula and using J μn (un )  C we have μn RN (|∇ un |s + v and T > 0 define v T as v T = v if | v |  T , v T = ± T if ± v (x)  T . For that ψ  0, ψ(x) = 1 for |x|  R, ψ(x) = 0 for |x|  2R and |∇ψ|  R2 . take φ = unK |unT |2r ψ as the test function in (2.4). We have

 2r |∇ un |s unT  ψ dx + μn (2r + 1)

μn T |un | K



+ μn



|un | T

 2r |∇ un |s−2 ∇ un ∇ψ u K u T  dx + μn n

+

a i j (u n ) +

T |un | K i , j =1 N  

1 2



  N

RN

+  = RN

1 2



2r

D s ai j (un )un D i un D j un unT  ψ dx

(2r + 1)ai j (un ) +

|un | T i , j =1

+

 2r |un |s−2 un ψ unK unT  dx

RN N  







n

RN

+

 2r |∇ un |s unT  ψ dx



1 2





2r

D s ai j (un )un D i un D j un unT  ψ dx

2r

ai j (un ) D i un D j ψ unK unT  dx

i , j =1



N 

D s ai j (un )unK

 2r D i un D j un u T  ψ dx +



n

|un | K i , j =1

 2r 4N |un | N −2 −2 un unK unT  ψ dx +





2r

V ∞ un unK unT  ψ dx

RN

 2r |un | p −2 un unK unT  ψ dx.

(2.5)

RN

By the assumption (a3 ) for K large enough, D s ai j (un )unK > 0 for all |un |  K . By the lower semicontinuity, we obtain as n → ∞,

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

N  



1

a i j (u ) +

2

T |u | K i , j =1



2

|u | T i , j =1

  N 

1 2





2r

ai j (u ) D i u D j ψ u K u T  dx

i , j =1

RN

+

N 



2r

D s ai j (u )u K D i u D j u u T  ψ dx +

|u | K i , j =1

 2r 4N |u | N −2 −2 uu K u T  ψ dx +



2r

D s ai j (u )u D i u D j u u T  ψ dx

N    2r 1 (2r + 1)ai j (u ) + D s ai j (u )u D i u D j u u T  ψ dx

 +

+



109

RN





2r

V ∞ uu K u T  ψ dx

RN



 2r |u | p −2 uu K u T  ψ dx.

(2.6)

RN

Letting R → ∞, then K → ∞, we have N  



a i j (u ) +

|u | T i , j =1 N  

 +

1 2





(2r + 1)ai j (u ) +

|u | T i , j =1

+  



1 2

N 

|u |

1 V . 2 ∞

2



 T 2r u  dx +



2r



2r

D s ai j (u )u D i u D j u u T  dx



RN

Choose 0 <  <

1

D s ai j (u )u D i u D j u |u |2r dx +

|u | K i , j =1 4N N −2

2r

D s ai j (u )u D i u D j u u T  dx

V ∞ u 2 u T  dx

RN



 2r |u | p u T  dx.

(2.7)

RN

There exists C > 0 such that



 2r |u | p u T  dx  

RN





2r

u 2 u T  dx + C

RN



2r 4N  |u | N −2 u T  dx.

RN

By (a2 ), (2.6) and (2.7)



  2  T 2r

2

u |∇ u | u



 2

dx + (2r + 1)

|u | T

2

2r

u |∇ u | |u | dx  C

|u | T

2r 4N  |u | N −2 u T  dx.

RN

By the Sobolev imbedding theorem

 S RN





r 

u 2 u T 

2N N −2

NN−2 dx

  RN

  2  T r 2 ∇ u u   dx  Cr



2r 4N  |u | N −2 u T  dx.

RN

From here a Moser iteration argument gives the L ∞ estimate of u (see [26]).

110

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

Next we show that u solves Eq. ( Q ). Choose T > 0 such that u  L ∞ (RN ) < T and D s ai j (s)s  0 for |s|  T . Define w n such that w n = un

if un  − T , w n = − T if un  − T . Let ψ ∈ C 0∞ (R N ), ψ  0. Take φ = ψ e − M w n as the test function in (2.4), where M > 0 is a parameter. We have

 un − T

 RN

N  



Mai j (un ) −

un − T i , j =1

2





ai j (un ) D i un D j ψ e − M w n dx

i , j =1



1

D s ai j (un ) D i un D j un ψ e − M w n dx

2

D s ai j (un ) D i un D j un ψ e − M w n dx +

un − T 4N

|un | N −2 −2 un ψ e − M w n dx +

=

  N

RN

− 1

|∇ un |s−2 ∇ un ∇ψ e − M w n dx

RN

|un |s−2 un ψ e − M w n dx +

+ μn

+



|∇ un |s ψ e − M w n dx + μn

−μn M

RN



V ∞ un ψ e − M w n dx

RN



|un | p −2 un ψ e − M w n dx.

RN

Notice that





|∇ un |s ψ e − M w n dx  0,

−μn M un − T



μn

D s ai j (un ) D i un D j un ψ e − M w n dx  0,

un − T

|∇ un |s−2 ∇ un ∇ψ e − M w n dx + μn

RN



|un |s−2 un ψ e − M w n dx

RN

  C μn

 |∇ un |s + |un |s dx



s −1 s

1

 C μns → 0,

RN



 RN

V ∞ un ψ e − M w n dx →

RN



4N

|u | N −2 −2 u ψ e − Mu dx,

RN

|un | p −2 un ψ e − M w n dx →



|u | p −2 u ψ e − Mu dx.

RN

Choose M large enough such that Mai j (s) −

n→∞



4N

|un | N −2 −2 un ψ e − M w n dx → 

V ∞ u ψ e − Mu dx,

RN

RN

lim inf



N  

un − T i , j =1

1 D a (s) 2 s ij

Mai j (un ) −

1 2

> 0. By lower semi-continuity we have



D s ai j (un ) D i un D j un ψ e − M w n dx

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

N  

 

1

Mai j (u ) −

2

u − T i , j =1

=

  N 

RN

1

Mai j (u ) −

2

i , j =1

111



D s ai j (u ) D i u D j u ψ e − Mu dx

D s ai j (u ) D i u D j u ψ e − Mu dx.

Altogether we have

  N RN



a i j (u ) D i u D j ψ e

− Mu



dx +

2

i , j =1

 

4N

RN

D s a i j (u ) D i u D j u ψ e

− Mu

 dx +

i , j =1

RN



|u | N −2 −2 u ψ e − Mu dx +

  N

1

V ∞ u ψ e − Mu dx

RN

|u | p −2 u ψ e − Mu dx.

RN

Let φ  0, φ ∈ C 0∞ (R N ). Choose ψn ∈ C 0∞ (R N ) such that ψn → φ e Mu in H 1 (R N ) and {ψn } is bounded in L ∞ (R N ). Take ψ = ψn in the above inequality. Letting n → ∞ we have for all φ ∈ C 0∞ (R N ), φ  0,

  N RN

ai j (u ) D i u D j φ dx +

i , j =1

 

4N

RN

2

 D s ai j (u ) D i u D j u φ dx +

i , j =1

RN



|u | N −2 −2 u φ dx +

  N

1

V ∞ u φ dx

RN

|u | p −2 u φ dx.

RN

The opposite inequality can be obtained in a similar way by setting φ = ψ e M v n where v n = un if un  T , v n = T if un  T . Thus we have proved that Eq. ( Q ) holds for all ψ ∈ C 0∞ (R N ), ψ  0. By approximation ( Q ) holds for all ψ ∈ H 1 (R N ) ∩ L ∞ (R N ). Finally we estimate J (u ),

J (u ) = J (u ) −

=

1  J (u ), u p

  N  1 2

i , j =1

RN

 +

1 2

−

1

−

1



p

a i j (u ) −



1 2p

 V ∞ u 2 dx +

p

1 p

D s ai j (u )u D i u D j u dx

−

RN

  lim inf n→∞

1 s

−

1

N −2



4N

|u | N −2 dx

4N RN





μn

p



 | Dun |s + |un |s dx

RN

+

  N  1

RN

i , j =1

2

−

1 p

a i j (u n ) −

1 2p

D s ai j (un )un D i un D j un dx

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X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

 +

1

−

2

1



 V ∞ un2 dx +

p

1

−

p

N −2

 4N |un | N −2 dx

4N

RN

RN





= lim inf J μn (un ) − n→∞

1  J μn (un ), un = lim J μn (un ). n→∞ p 

Let S be the best Sobolev constant, i.e., S = infu ∈C ∞ (RN )  0 (

Lemma 2.3. Suppose that



RN RN

|∇ u |2 dx 2

∗

|u |2 dx) 2∗

2

where 2∗ =

2N . N −2

N

1 μn → 0 as n → ∞, J μ n (un ) = 0, J μn (un ) → c ∈ (0, 2N S 2 ). Then there exist

{ yn } ⊂ R N , u = 0, u ∈ H 1 (R N ) ∩ L ∞ (R N ) such that up to a subsequence, u˜ n (·) = un (· + yn ) u in H 1 (R N ) and u˜ n ∇ u˜ n u ∇ u in L 2 (R N ), J  (u ) = 0 and J (u )  lim infn→∞ J μn (un ).  (u ) = 0, J (u ) → c we have Proof. Since J μ n μn n n

c + 1  J μn (un ) −

 =

1 s

1

−



1  J (un ), un p μn 



μn

p

 | Dun |s + |un |s dx

RN

+

  N  1

RN



+

2

i , j =1

1 2



− 

 C μn

1

−

1 p

a i j (u n ) −



 V ∞ un2 dx +

p

1 p

RN



s

s





| Dun | + |un | dx +

RN



1

D s ai j (un )un D i un D j un dx

2p

−

N −2



4N

|un | N −2 dx

4N RN



1 + un2



 2

|∇ un | dx +

RN

 un2 dx +

RN

|u n |

4N N −2

dx .

RN

Two cases may occur for the sequence {un }:



(a) Vanishing, i.e., sup y ∈RN B ( y ) un2 dx → 0, as n → ∞. 1  (b) Non-vanishing. There exist { yn } ⊂ R N and ν > 0 such that lim infn→∞ B ( y ) un2 dx  ν . 1 n 4N

Suppose that Vanishing occurs. Since {un } is bounded in L 2 (R N ) and L N −2 (R N ), we have un → 0 T in L q (R N ) for all 2 < q < N4N −2 . Given a function v, let v be the truncated function of v as before. We have

 |u n |

4N −2 N −2

 un unT

dx  T

RN

4N

|un | N −2 −1 dx → 0.

RN

Take φ = unT in the equation. We obtain for any T > 0

 |un | T







1 + un2 |∇ un |2 dx + |un | T

un2 dx → 0,

n → ∞.

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

113

Consequently as n → ∞



 un2 dx 

|un | T

RN



N  

a i j (u n ) +

|un | T i , j =1

1 2

un4 dx → 0, |un | T



1 T2

un2 |∇ un |2 dx → 0, |un | T







D s ai j (un )un D i un D j un dx  C



1 + T 2 |∇ un |2 dx → 0,

|un | T

N  



T2

|∇ un |2 dx +

|un | T

RN



1



|∇ un |2 dx  

un2 dx +

a i j (u n ) +

|un | T i , j =1





1 2

D s ai j (un )un D i un D j un dx



4un2 1 + o T (1) |∇ un |2 dx

= |un | T







un2 |∇ un |2 dx + o T (1) −

=4



4un2 1 + o T (1) |∇ un |2 dx

|un | T

RN



un2 |∇ un |2 dx + o T (1) + on (1),

=4 RN

where o T (1) → 0 as T → ∞ independent of n. Hence we have

  N 

a i j (u n ) +

RN

i , j =1

1 2



 un2 |∇ un |2 dx + o(1).

D s ai j (un )un D i un D j un dx = 4

(2.8)

RN

Take φ = un in the equation. We obtain



  |∇ un |s + |un |s dx + 4

μn RN



 un2 |∇ un |2 dx =

RN

4N

|un | N −2 dx + o(1).

RN

Hence

 |u n |

4N N −2



 dx  4

RN

un2 |∇ un |2 dx + o(1)

RN



S

|u n |

4N N −2

NN−2 dx

+ o(1).

(2.9)

RN



4N

4N

N

This implies that either RN |un | N −2 dx → 0 or RN |un | N −2 dx  S 2 + o(1). The former implies   μn RN (|∇ un |s + |un |s ) dx + 4 RN un2 |∇ un |2 dx → 0 and J μn (un ) → 0. The latter gives

J μn (un ) =

μn



s RN



 |∇ un |s + |un |s dx +



RN

un2 |∇ un |2 dx −

N −2



4N

|un | N −2 dx + o(1)

4N RN

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X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

 = μn

1 s

−

1





 1 |∇ un |s + |un |s dx +

4

1 2N

4N

|un | N −2 dx + o(1)

2N

RN



 RN

N

S 2 + o(1).

In both cases we arrive at a contradiction. Thus Vanishing cannot occur. Now assume Non-vanishing occurs. Since the functionals J μ and J are translation invariant,  without loss of generality, we assume that yn = 0, n = 1, 2, . . . , and B (0) un2 dx  ν > 0. Up to 1

a subsequence, un u in H 1 (R N ), un ∇ un u ∇ u in L 2 (R N ). By Proposition 2.2, J  (u ) = 0 and J (u )  limn→∞ J μn (un ). Finally B (0) u 2 dx = limn→∞ B (0) un2 dx  ν , u = 0. 2 1

1

Lemma 2.4. Assume that (a3 ) holds and p > p N + α , where p N = 4 if N  6 and p N = Then there exist μ0 , δ > 0 such that d(μ) 

1 2N

S

N 2

2( N +2) N −2

if N = 3, 4, 5.

− δ for μ ∈ (0, μ0 ).

Proof. This follows essentially the original idea of Brezis and Nirenberg [4]. We look for a positive N

1 function such that supt 0 J (tu )  2N S 2 − δ for some δ > 0. One of the possible candidates for u is u  = φ w  , where φ is a smooth cut-off function such that φ(x) = 1 if |x|  1, φ(x) = 0 if |x|  2 and |∇φ|  2, and

( N ( N − 2) )

w  (x) =

( + |x|2 )

N −2 8

N −2 4

,

 > 0.

In [27] it was proved that for p > p N + α there exists C 0 > 0 such that as

1

sup J (tu  ) 

N

1

S 2 − C 0  2 − 8 p ( N −2 ) .

2N

t 0

Now we fix such an

N

 is small enough (2.10)

 > 0. It is easy to show that there exist 0 < T 1 < T 2 such that

sup J μ (tu  ) = t 0

sup

sup J (tu  ) =

J μ (tu  ),

t ∈[ T 1 , T 2 ]

t 0

sup

t ∈[ T 1 , T 2 ]

J (tu  ).

Hence

d(μ)  sup J μ (tu  ) t 0

=



sup

 C μ +

for

μ small. We take δ =

C0 2

N

s

t ∈[ T 1 , T 2 ]

 C μ +



μ

1 2N 1

N

 t

s

|∇ u  | + |u  | dx + J (tu  )



s

s



RN

sup

t ∈[ T 1 , T 2 ]

1

N

2N

S2 −

J (tu  ) N

1

S 2 − C 0  2 − 8 p ( N −2 )

C0 2

N

1

 2 − 8 p ( N −2 )

 2 − 8 p(N −2) . 2

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

Remark 2.1. In Appendix A we present another choice of the curve 1 2N

N 2

115

γ : [0, 1] → C 0∞ (RN ) which satisfies

S − δ for some δ > 0. The construction of such that γ (0) = 0, J (γ (1)) < 0 and supt 0 J (γ (t ))  a curve is related to the methods of change of variables, see [10,13,23,34]. We prove that such an inequality is not possible for p  p N + α . Theorem 2.1. Assume that (a3 ) holds and p > p N + α . Then Eq. ( Q ) has a positive solution u such that J (u )  lim infn→∞ d(μn ) 

1 2N

S

N 2

− δ and u ∈ H 1 (R N ) ∩ L ∞ (R N ).

+ (which is obtained by replacing u by u in F of J ) and Proof. If we consider the functional J μ + μ + define the mountain pass value of J μ

d+ (μ) = inf

γ ∈Γ +

Γ

+

+ sup J μ

t ∈[0,1]





γ (t ) ,

    + = γ ∈ C [0, 1], X  γ (0) = 0, J μ γ (1) < 0 , 



then the conclusion of Lemma 2.4 holds for d+ (μ), that is d+ (μ) 

1 2N

S

N 2

(2.11)

− δ for μ ∈ (0, μ0 ). By N

1 S 2 − δ for μ ∈ (0, μ0 ). Proposition 2.1, J μ has a positive critical point u μ with J μ (u μ )  d+ (μ) < 2N By Proposition 2.2 and Lemma 2.4, there exists a sequence μn such that μn → 0 and un = u μn u

in H 1 (R N ), un ∇ un u ∇ u in L 2 (R N ), u = 0, J  (u ) = 0 and J (u )  lim infn→∞ J μn (un )  u ∈ H 1 (R N ) ∩ L ∞ (R N ). 2

1 2N

S

N 2

− δ,

Now we consider solutions with least energy. Define



    d∗ (μ) = inf J μ (u )  J μ (u ) = 0, u  0, u = 0 , 











d = inf J (u )  u ∈ H 1 R N ∩ L ∞ R N , J  (u ) = 0, u  0, u = 0 .

(2.12)

Lemma 2.5. (1) 0 < d  lim infμ→0 d∗ (μ). (2) d is achieved. Proof. (1) First it is easy to see that by condition (a2 ) there is d > 0. Next suppose that μn → 0, lim infn→∞ d∗ (μn ) = lim infμ→0 d∗ (μ). We have a positive critical point un of J μn with J μn (un )  d∗ (μn ) + n1 . By Proposition 2.2 and Lemma 2.4, up to a subsequence and a translation, there exists u ∈ H 1 (R N )∩ L ∞ (R N ) such that un u = 0 in H 1 (R N ), un ∇ un u ∇ u in L 2 (R N ), J  (u ) = 0 and J (u )  lim infn→∞ J μn (un ). Hence u is a nonnegative critical point of J , and d  J (u )  lim infn→∞ J μn (un ) = lim infμ→0 d∗ (μ). (2) Let {un } ⊂ H 1 (R N ) ∩ L ∞ (R N ), J  (un ) = 0, un  0 and limn→∞ J (un ) = d. We can show that, up to a subsequence and a translation, there exists u = 0, u ∈ H 1 (R N ) ∩ L ∞ (R N ) such that un u in H 1 (R N ), un ∇ un u ∇ u in L 2 (R N ), J  (u ) = 0 and J (u )  limn→∞ J (un ). Hence J (u ) = d. The proof is similar to that of Proposition 2.2 and Lemma 2.3. We omit it here. 2 Finally we discuss the regularity of the weak solutions. Lemma 2.6. Weak solutions of (Q) are classical ones in the sense that they are C 2,α for some nonnegative solutions u = 0 are strictly positive.

α > 0. And

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X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

Proof. First our solutions are L ∞ functions by the proof of Proposition 2.2. By Corollary 4.23 on page 95 of [15] the solutions are of C β for some β > 0. By Theorem 4.24 of [15] they are of C 1,β . Finally using Schauder estimates we have C 2,α regularity for some α > 0. Now by the strong maximum principle, if u  0 and u = 0 we have u is strictly positive. 2 Remark 2.2. (1) We remark that the Hölder continuity of V and D s ai j is only used here to get the C 2,α property of the solutions. (2) Our method allows to treat periodic potentials, i.e., if we replace V ∞ by V (x) which is assumed to be periodic in each of the variables x1 , . . . , x N . The same result still holds. We omit the details. 3. Problems with potential well V (x) We look for u ∈ H 1 (R N ) with

  N RN

a i j (u ) D i u D j φ +

i , j =1

 RN

u 2 |∇ u |2 dx < ∞ such that for all φ ∈ C 0∞ (R N )



N 1 

2





D s ai j (u ) D i u D j u φ dx +

i , j =1

V (x)u φ dx =

RN

f (u )φ dx.

(P)

RN

The corresponding functional is given by

1

I (u ) =

  N

2 RN

Let X=W

μ ∈ (0, 1], 1, s

ai j (u ) D i u D j u dx +

i , j =1

1



 V (x)u 2 dx −

2 RN

F (u ) dx.

(3.1)

RN

4N < N +2 1 N

s < min{4, N }. We consider the perturbed problem ( P )μ by looking for u ∈ (R ) ∩ H (R ) such that for all φ ∈ X N



μ



 |∇ u |s−2 ∇ u ∇φ + |u |s−2 u φ dx +

RN

  N

 +  =

N 1 

2

RN

ai j (u ) D i u D j φ dx

i , j =1

RN

 D s ai j (u ) D i u D j u φ dx +

i , j =1

V (x)u φ dx

RN

f (u )φ dx.

( P )μ

RN

The corresponding functional is defined on X

I μ (u ) =

μ





 | Du |s + |u |s dx + J (u )

s RN

=

μ



s



 1 | Du |s + |u |s dx +

  N

2

RN

RN

i , j =1

 ai j (u ) D i u D j u dx + RN

 V (x)u 2 dx −

F (u ) dx.

RN

 (u ) = 0 if and only if u is a solution of problem ( P ) . Then I μ is a C 1 functional on X and J μ μ The following Lemmas 3.1 and 3.2 are counterparts of Lemmas 2.1 and 2.2.

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

117

 (u ) → 0 and I (u ) → c = 0 as n → ∞. Then Lemma 3.1. Let {un } ⊂ X be a (PS) sequence of I μ , i.e., I μ n μ n N  (u ) = there exist a sequence { yn } ⊂ R and u ∈ X , u = 0, such that either { yn } = 0 for all n, un u in X , I μ  (u ) = 0 and 0 and I μ (u )  limn→∞ I μ (un ), or { yn } is unbounded, and u˜ n (·) = un (· + yn ) u in X , J μ J μ (u )  limn→∞ I μ (un ). + which is defined by replacing u by u + in Define the mountain pass value for the functional I μ F (u ) of the definition of I μ

+ c + (μ) = inf sup I μ γ ∈Γ



t ∈[0,1]



γ (t ) ,

    + Γ = γ ∈ C [0, 1], X  γ (0) = 0, I μ γ (1) < 0 . 



(3.2)

Lemma 3.2. There exists m > 0 independent of μ such that c + (μ)  m. Proposition 3.1. If c + (μ) < d+ (μ), then I μ has a positive critical point u with I μ (u )  c + (μ). Proof. By Lemma 3.2 and the Mountain Pass Lemma, there exists a sequence {un } ⊂ X such that + ) (u ) → 0 and I + (u ) → c + (μ). The conclusions of Lemma 3.1 hold for the functional I + too. (Iμ n μ n μ That is, there exist a sequence { yn } ⊂ R N and u = 0, u ∈ X such that either yn = 0 for all n, un u + ) (u ) = 0 and I + (u )  lim + in X , ( I μ n→∞ I μ (un ); or { yn } is unbounded and u n (· + yn ) u in X , and μ + + + J μ (u )  limn→∞ I μ (un ) = c μ . In the latter case, u is positive and hence a positive critical point of J μ . + ) (u ) = 0. Again u is We have d+ (μ)  J μ (u )  c + (μ) which is a contradiction. So un u in X , ( I μ positive and hence a positive critical point of I μ , with I μ (u )  limn→∞ I μ (un ). 2 We will estimate c + (μ). Let u be a positive critical point of J such that u ∈ H 1 (R N ) ∩ L ∞ (R N ), J (u ) = d. Using the elliptic equation satisfied by u, with the regularity estimates we may derive the following Pohozaev identity [32]:

  N

N −2 2

  ai j (u ) D i u D j u dx = N

i , j =1

RN

F (u ) −

V∞ 2

u 2 dx.

RN

Lemma 3.3. lim supμ→0 c + (μ) < d. Proof. Let u be a strictly positive critical point of J with J (u ) = d. Define ut (x) = u ( xt ), x ∈ R N , t > 0      and u 0 = 0. We have RN |∇ ut |s dx = t N −s RN |∇ u |s dx, RN |∇ ut |2 dx = t N −2 RN |∇ u |2 dx, RN |ut |s dx =    t N RN |u |s dx, RN |ut |2 dx = t N RN |u |2 dx. Hence ut → 0 in X as t → 0,

J (u t ) =

1

  N

2 RN

i , j =1

1

= t N −2

RN



1 2

−

1



 V ∞ ut2 dx −

2 RN

ai j (u ) D i u D j u dx − t N

i , j =1

N −2

RN

F (u ) −

1 2

V ∞ u 2 dx

RN

  N

2N

F (ut ) dx

RN

 

  N

2



ai j (ut ) D i ut D j ut dx +

i , j =1

ai j (u ) D i u D j u dx = J (u ) = d.

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X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

There exist 0 < T 1 < T 2 such that supt 0 J (ut ) = sup T 1 t  T 2 J (ut ), supt 0 I (ut ) = sup T 1 t  T 2 I (ut ), supt 0 I μ (ut ) = sup T 1 t  T 2 I μ (ut ). We have

c + (μ)  sup I μ (ut ) = t 0

= O (μ) +  O (μ) +

sup

T 1 t  T 2

I μ (u t )

sup

I (u t )

sup

J (u t ) −

T 1 t  T 2

T 1 t  T 2

= O (μ) + d −



1 2

inf

T 1 t  T 2

1 2

 inf



T 1 t  T 2



V ∞ − V (x) ut2 dx

RN





V ∞ − V (x) ut2 dx.

RN

Since u > 0, ut is strictly positive, and V (x) = V ∞ , δ := have lim supμ→0 c + (μ)  d − δ . 2

1 2

inf T 1 t  T 2

 RN

( V ∞ − V (x))ut2 dx > 0. We

Proposition 3.2. There exists μ0 ∈ (0, 1] such that I μ has a positive critical point u with I μ (u )  c + (μ) for μ ∈ (0, μ0 ). Proof. By Lemma 2.5, d  lim infμ→0 d∗ (μ). By Lemma 3.3, lim supμ→0 c + (μ) < d hence c + (μ) < d∗ (μ) for μ small and I μ has a positive critical point by Proposition 3.1. 2 Remark 3.1. We can prove c + (μ) < d∗ (μ) for positive critical point for μ ∈ (0, 1].

μ ∈ (0, 1]. In fact d∗ (μ) is achieved and I μ has a

The following three results are counterparts of Proposition 2.2 and Lemma 2.3. Proposition 3.3. Suppose that

 (u ) = 0, I (u )  C , u u in H 1 (R N ) and μn → 0 as n → ∞, I μ n μn n n n

un ∇ un u ∇ u in L (R ). Then u ∈ H 1 (R N ) ∩ L ∞ (R N ), I  (u ) = 0 and I (u )  lim infn→∞ I μn (un ). 2

N

Proposition 3.4. Suppose that

N

 (u ) = 0, I (u )  C with C ∈ (0, 1 S 2 ), and for μn → 0 as n → ∞, I μ n μn n 2N n

an unbounded sequence { yn } ⊂ R N , u˜ n = un (· + yn ) u in H 1 (R N ) and u˜ n ∇ u˜ n u ∇ u in L 2 (R N ). Then u ∈ H 1 (R N ) ∩ L ∞ (R N ), J  (u ) = 0 and J (u )  lim infn→∞ I μn (un ). Lemma 3.4. Suppose that

N

1  (u ) = 0, lim μn → 0 as n → ∞, I μ n n→∞ I μn (un ) = c ∈ (0, 2N S 2 ). Then there n

exist { yn } ⊂ R N , u = 0, u ∈ H 1 (R N )∩ L ∞ (R N ) such that up to a subsequence, either yn = 0 for all n, un u in H 1 (R N ), un ∇ un u ∇ u in L 2 (R N ), I  (u ) = 0 and I (u )  lim infn→∞ I μn (un ); or { yn } is unbounded, u˜ n (·) = un (· + yn ) u in H 1 (R N ) and u˜ n ∇ u˜ n u ∇ u in L 2 (R N ), J  (u ) = 0 and J (u )  lim infn→∞ I μn (un ). Now we give the proof of Theorem 1.1. Proof of Theorem 1.1. Choose μn → 0. By Proposition 3.2 I μn has a positive critical point un such that I μn (un )  c + (μ). By Lemma 3.3, lim supn→∞ I μn (un )  lim supn→∞ c + (μ) < d. By the proof of N

1 Lemma 2.4, d < 2N S 2 . By Lemma 3.4, there exist a sequence { yn } ⊂ R N and u = 0, u ∈ H 1 (R N ) ∩ ∞ N L (R ) such that, up to a subsequence, u˜ n = un (· + yn ) u in H 1 (R N ) and u˜ n ∇ u˜ n u ∇ u in L 2 (R N ). Moreover, we have either { yn } = 0 for all n and I  (u ) = 0, I (u )  lim infn→∞ I μn (un ), or { yn } is unbounded in R N and J  (u ) = 0, J (u )  lim infn→∞ I μn (un ) < d, which is a contradiction. Hence { yn } = 0 for all n and I  (u ) = 0, u is a positive critical point of I . 2

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

119

Remark 3.2. In [27] we have studied the critical exponent problems (1.1) by using the Nehari method. We establish solutions for subcritical exponent problems first then pass to limit to get solutions for the critical case. A key condition for the Nehari method to work is the following monotonicity (H ) |s|2− p (ai j (s) + 12 sD s ai j (s))ξi ξ j is decreasing in s ∈ (0, ∞) and increasing in s ∈ (−∞, 0). Here p is the same one appearing in (a2 ) and ( f ). This is a global restriction on the terms involved. Besides this, in [27] we also need D s ai j (s)s  0. Acknowledgment The authors are grateful to the referee’s helpful comments. Appendix A Our main result has a condition on the subcritical power p depending on the dimension N, that 2( N +2) is, p N + α < p < N4N −2 where p N = 4 for N  6 and p N = N −2 for N = 3, 4, 5. We do not know whether this is necessary. We give some discussion here which may indicate the necessity of the condition as far as the method is concerned. We show that the mountain pass value is no less than the least threshold for compactness of (PS) sequences provided 4 < p  p N + α for N = 3, 4, 5. For simplicity we consider the Modified Nonlinear Schrödinger Equation 4N

u + u u 2 − V (x)u + |u | N −2 −2 u + |u | p −2 u = 0, In this case ai j (s) = (1 + 2s2 )δi j and C 0∞ (R N ) such that for  > 0 small

α = 0. In the proof of Lemma 2.4 we choose a function u  ∈

sup I (tu  )  sup J (tu  )  t 0

x ∈ RN .

t 0

1

N

2N

N

1

S 2 − C 0  2 − 8 p ( N −2 ) .

Of course, there are other choices. In this section we give such an example, which was used in [13]. The corresponding functional

I (u ) =

1



2





1 + 2u 2 |∇ u |2 dx +



1

V (x)u 2 dx −

2

RN

N −2



4N

RN

1

4N

|u | N −2 dx −

 |u | p dx.

p

RN

(A.1)

RN

Let u = f ( v ) where f is defined by

f  (t ) = 

1 1 + 2 f 2 (t )

,

t  0,

f (t ) = − f (−t ),

t  0.

(A.2)

After the change of variables, from I (u ) we obtain the following functional

L(v ) =

1

 |∇ v |2 dx +

2 RN

1

 V (x) f 2 ( v ) dx −

2 RN

N −2



4N



  4N  f ( v ) N −2 dx − 1

p

RN

   f ( v ) p dx. (A.3)

RN

The MNSE is reduced to a semilinear equation

  4N −2   p −2    v + f  ( v ) − V (x) f ( v ) +  f ( v ) N −2 f ( v ) +  f ( v ) f ( v ) = 0,

x ∈ RN .

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X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

This idea was initially used in [10,23] for existence of solutions of subcritical problems. If we can find a function v ∈ C 0∞ (R N ) such that supt 0 L (t v ) <

1 2N

N

S 2 , then by setting ut = f (t v ) we have

1

sup I (ut ) = sup L (t v ) < t 0

N

2N

t 0

(A.4)

S2.

One of the possible candidates is as follows. Let ψ = φ w  , where

w  (x) =

( N ( N − 2) ) ( + |x|2 )

N −2 4

N −2 2

 > 0,

,

(A.5)

and φ is a truncation function such that φ(x) = 1 if |x|  R  and φ(x) = 0 if |x|  2R  . R  is a 1

parameter satisfying R   − 2 → ∞ as  → 0 and R   R for some constant R > 0. For example, one can choose R  =  α , α ∈ ( 14 , 12 ), as in [13]. In the following we show that if p > p N then we obtain the desired estimates (see Lemma A.1). However for p  p N we have the following estimates which indicate the approach does not work to have the mountain pass value under the threshold of compactness. 1

Proposition A.1. Suppose that R   − 2 → ∞ as  → 0 and R   R for some R > 0. Then for 4 < p  p N = 2( N +2) (N = 3, 4, 5), as  → 0, N −2

sup L (t ψ )  t 0

1 2N

N

S 2 + C0

N −2 4

|ln  |.

First we need the following elementary estimates. Lemma A.1. The following properties hold. 1

1

(1) | f (t )|  2 4 |t | 2 for all t ∈ R. (2)

f (t ) 1

t2

1 4

→ 2 , as t → +∞.

1 (3) t − √ ln t − √1 f 2 (t ) → c 0 > 0 as t → +∞. 4 2

(4) f

4N N −2

2

N

N +2

2N

N +2

(t ) = 2 N −2 t N −2 − ct N −2 ln t + O (t N −2 ), as t → +∞.

Proof. By direct integration we obtain

t=



√ 1 f (t ) 1 + 2 f 2 (t ) + √ ln 2 f (t ) + 2 2 2 1

Then it is easy to obtain the conclusions from this.





1 + 2 f 2 (t ) .

(A.6)

2

Lemma A.2. Choose R  = 1 in the definition of φ . Then for p > p N , supt 0 L (t ψ ) <

1 2N

N

S2.

Proof. First there exist 0 < T 1 < T 2 such that supt 0 L (t ψ ) = sup T 1 t  T 2 L (t ψ ). For t ∈ [ T 1 , T 2 ], by Lemma A.1

L (t ψ ) =

1 2

 t2 RN

|∇ψ |2 dx +

1

 V (x) f 2 (t ψ ) dx −

2 RN

N −2



4N

  4N  f (t ψ ) N −2 dx − 1



p

RN

RN

   f (t ψ ) p dx

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124



1

 t2

 |∇ψ |2 dx + C

2

RN

4N

RN



+ C1

(t ψ )

N +2 N −2



N −2

t ψ dx −

2N

2 N −2 (t ψ ) N −2 dx t ψ  T



p

ln(t ψ ) dx − C 2

t ψ  T

(t ψ ) 2 dx,

t ψ  T

 . We estimate the errors in the above formula. By

where C , C 1 , C 2 , T are constants independent of direct computations



 N −2  N |∇ψ |2 dx = S 2 + O  2 ,

RN

 ψ dx ∼ C  RN





N +2

t ψ  T

N −2 4

,

N +2

(t ψ ) N −2 ln(t ψ ) dx 

There exists

N

121

(t w  ) N −2 ln(t w  ) dx ∼ C 

N −2 4

|ln  |.

RN

√

0 such that for   0 and |x|   , ψ = w   C  − 



p

p

N

N −2 4

and t ψ (x)  T . Hence

1

( w  ) 2 dx ∼ C 0  2 − 8 p ( N −2) .

(ψ ) 2 dx  √ |x| 

t ψ  T

1

Choose δ > 0 small such that t ψ (x)  T for |x|  δ  4 . Then





2N

 N 2N N ( w  ) N −2 dx = S 2 + O  4 .

(ψ ) N −2 dx  t ψ  T

|x|δ  1/4

It follows from the above estimates that for t ∈ [ T 1 , T 2 ],

 L (t ψ ) 

 since

N −2 4

>

N 2

1 2

1 2N

t2 −

N −2 2N

N

S2 −

1 2

2

2N



2 N −2 t N −2 N

N

S 2 + C

N −2 4

N

1

|ln  | − C 0  2 − 8 p ( N −2)

1

C 0  2 − 8 p ( N −2 ) ,

− 18 p ( N − 2), that is p > p N =

2( N +2) . N −2

2

Finally we give the proof of Proposition A.1. Proof of Proposition A.1. By Lemma A.1, we have

L (t ψ ) =

1 2

 t2 RN

|∇ψ |2 dx +

1

 V (x) f 2 (t ψ ) dx −

2 RN

N −2



4N

  4N  f (t ψ ) N −2 dx − 1



p

RN

RN

   f (t ψ ) p dx

122

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124



1

NN−2  2N N 2N N −2 |ψ | N −2 dx − 2 N −2 (t ψ ) N −2 dx

 t2 S 2

4N

RN

+

N −2 4N

RN

 c (t ψ )

N +2 N −2

ln(t ψ ) dx −

t ψ  T



1

p

p

2 4 (t ψ ) 2 dx,

p RN

where S is the Sobolev imbedding constant and T > 0. Let

1

L 0 (t ψ ) =

2

 t2 S

NN−2  2N N 2N N −2 |ψ | N −2 dx − 2 N −2 (t ψ ) N −2 dx 4N

RN

RN

and N −2 4

1

t  = 2− 2 S



−2 − N2N 2N |ψ | N −2 dx .

RN 1

Then t  → t 0 = 2− 2 as

sup L (t ψ )  L (t  ψ )  t 0

 (ψ ) t  ψ  T

1 2N

 → 0 and L 0 (t  ψ ) =

N +2 N −2



1 2N

N

S 2 . Then



N

N +2

S 2 + C1

(ψ ) N −2 ln(t  ψ ) dx −

t  ψ  T N +2

ln(t  ψ ) dx 

p

p

2 4 (t  ψ ) 2 dx,

p

( w  ) N −2 ln(t  w  ) dx  C 0 

N −2 4

|ln  |,

√ |x| 



p

(ψ ) 2 dx 

( w  ) 2 dx |x|2R 

RN



1

( N ( N − 2) ) 8 p ( N −2)

=

1

( + |x|2 ) 4 p ( N −2) 

|x|2R  N

1

= C  2 − 8 p ( N −2 )

:= C 

N 2

− 18 p ( N −2)

−1 2

dy

(1 + | y |2 ) 4 p ( N −2)

I ( , N , p ).

We estimate the integral I ( , N , p ) case by case. Case 1: either N = 4, 5 or N = 3 and p > 6. In this case converges as  → 0. We have



dx

1

| y |2R  

p

N

1 p(N 2

− 2) > N, the integral I ( , N , p )

1

(ψ ) 2 dx ∼ C  2 − 8 p ( N −2) .

(A.8)

RN

It follows from the above estimates that

sup L (t ψ )  t 0

(A.7)

RN



p



1

1 2N

N

S 2 + C0

N −2 4

N

1

|ln  | − C  2 − 8 p ( N −2) 

1 2N

N

S2 +

1 2

C0

N −2 4

|ln  |.

X.-Q. Liu et al. / J. Differential Equations 254 (2013) 102–124

123

Case 2: N = 3 and p = 6. We have



dy

I ( , 3, 6) = | y |2R  

(1 + | y |2 )3/2

−1 2

 1 ∼ C ln R   − 2

and



 1 ψ3 dx ∼ C  3/4 ln R   − 2 ∼ C 1  3/4 |ln  |.

R3

It follows from our estimates that

sup L (t ψ )  t 0

1

1

N

2N

S 2 + C 0  1/4 |ln  | − C 1  3/4 |ln  | 

N

2N

S2 +

1 2

C 0  1/4 |ln  |.

Case 3: N = 3 and 4 < p < 6. We have



dy

I ( , 3, p ) = | y |2R  

−1 2

(1 + | y |2 ) p /4

 1 3 − 1 p 2 ∼ C R − 2

and



p /2

ψ

1

dx = C  p /8 ( R  )3− 2 p  C  p /8 .

R3

It follows from our estimates again that

sup L (t ψ )  t 0

1 2N

1

N

S 2 + C 0  1/4 |ln  | − C 1  p /8 

In all these cases we have that supt 0 L (t ψ ) 

1 2N

S

N 2

2N

N

S2 +

1 2

C 0  1/4 |ln  |.

1

+ 12 C 0  4 |ln  |. 2

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