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國立高雄大學應用數學系

碩士論文

Multiplicity of positive solutions for singular elliptic

equations

奇異橢圓方程正解的多解性

研究生:林家宇撰

指導教授:吳宗芳教授

中華民國一百零八年十一月

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奇異橢圓方程正解的多解性

指導教授:吳宗芳 教授 國立高雄大學應用數學系 學生:林家孙 國立高雄大學應用數學系 摘要 在這篇文章中,我們探討涉及凹凸非線性的不定半線性橢圓方程正解的存在性與多 解性: 其 中 ( for ) 位 能 且 與參數 。我們假設函數 f,g 和 V 滿足適當的條件。 關鍵字:半線性橢圓問題、凹凸非線性、變分法、深井勢

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Multiplicity of positive solutions for singular elliptic

equations

Advisor: Professor Tsung-fang Wu Institute of Department of Applied Mathematics

National University of Kaohsiung

Student: Chia-yu Lin

Institute of Department of Applied Mathematics National University of Kaohsiung

ABSTRACT

In this paper, we study the existence, multiplicity of positive solutions for the following indefinite semilinear elliptic equations involving concave-convex nonlinearities:

{−∆𝑢 + 𝑉𝜆,𝜇(𝑥)𝑢 = 𝑓(𝑥)|𝑢|

𝑞−2𝑢 + 𝑔(𝑥)|𝑢|𝑝−2𝑢 𝑖𝑛 ℝ𝑁,

𝑢 ≥ 0, 𝑖𝑛 ℝ𝑁,

where 1 < q < 2 < p < 2∗ (2= 2𝑁

𝑁−2 for 𝑁 ≥ 3) the potential 𝑉𝜆,𝜇(x) = λV

+(𝑥) − 𝜇𝑉−(𝑥) with 𝑉± = max *±𝑉, 0+ and the parameter λ > 0. We assume that the functions f, g and V satisfy suitable conditions with the potential V and the weight function g without the assumptions of infinite limits.

Keywords: Semilinear elliptic problems, Concave-convex nonlinearities, Vari-

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Contents

1 Introduction 1 2 Preliminaries 3 2.1 Inequalities . . . 3 2.1.1 Hölder inequality . . . 3 2.1.2 Hardy inequality . . . 4 2.1.3 Sobolev inequality . . . 10

2.2 Variational methods and Theorems . . . 13

2.3 Implicit Function Theorem . . . 16

2.4 Mean Value Theorem . . . 17

2.5 Brezis-Lieb Lemma . . . 20

2.6 Variational Settings . . . 21

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1

Introduction

In this paper, we consider the existence, multiplicity and concentration of positive solu-tions of the following concave-convex elliptic equasolu-tions:

u + V ; (x) u = f (x)jujq 2u + g (x)jujp 2u in RN;

u 0; in RN; (E )

whereN 3; 1 < q < 2 < p < 2 2 = N2N2 ;the potential V ; (x) = V+(x) V (x)

with V = maxf V; 0g and the parameter > 0: We assume that the functions f; g and V satisfy the following conditions:

(F 1) f 2 Lq

RN (q = p

p q) with kf+kLq > 0; where f+(x) = maxff (x) ; 0g ;

(G1) g 2 L1 RN with kg+k1 > 0; where g+(x) = maxfg (x) ; 0g ; (V 1) V+

is a nonnegative continuous function on RN and 0 < V (x) 1

jxj2 for all

x2 RN

n f0g ;

(V 2) there exists b > 0 such that the set fV+ < b

g = x 2 RN

j V+(x) < b is

non-empty and has …nite measure;

(V 3) = int x2 RN j V+(x) = 0 is nonempty and has a smooth boundary with = x2 RN

j V+(x) = 0 :

This type of hypothesis was …rst introduced by Bartsch and Wang [17] in a study on a nonlinear Schrödinger equation. The potential V with V satisfying (V 1) (V 3); is called the steep well potential.

In recent years, Equation (E ) has been widely studied under variant assumptions on V and g: Most of the literature has focused on the equation for V being a positive constant and g being a positive weight function. Existence and multiplicity results have been obtained in many papers; see, e.g., [29], [31], [32], [37], [40], [42] and the references therein. In all the above results, it must be assumed that 0 < g1 = limjxj!+1g(x):Note

that this main assumption is crucial to using the concentration compactness method. In addition, the case of a positive potential V has been studied in Chabrowski and Bezzera do Ó [21] and Goncalves and Miyagaki [28], who have investigated the following equation:

u + V (x) u = h (x)jujq 2u +bg(x) jujp 2u in RN;

u 0 in RN; Ebh

where 1 < q < 2 < p < 2 and bg 2 C RN . They obtained some existence and

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In [28], the following conditions were assumed: b V 1 V (x) a0 > 0; x2 RN; b V 2 V (x)! 1 as jxj ! 1; (H1) h 0 and h 2 L22 q RN \ L1 RN :

Then it was proved that Equation Ebh has at least two positive solutions under the

assumption that khk

L 2

2 q is su¢ ciently small.

In [21], the following conditions were assumed: b

V 3 V (x) is positive, locally Hölder continuous and bounded in RN; (H2) h is a positive constant.

Next, it was demonstrated that Equation Ebh has at least one positive solution under

the assumption that h is su¢ ciently small.

Motivated by the above works, in the present paper we consider Equation (E ) with a more general potential V and weight function g without the assumptions of in…nite limits. Moreover, the norm kfkLq is not su¢ eiently small. The existence, multiplicity

and concentration of positive solutions of Equation (E ) are established via a variational method. Our main results are as follows:

Theorem 1.1 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 3) as well as the following conditions:

(V 4) 0 < jfV+ < b gj1 2p < A := S((N 2) 2 4 )1=2 (N 2)(p q)1=2 2 q kg+k 1 2 q p 2 kf+kLq p 2 1=2(p q) ; where S is a best Sobolev constant for the imbedding of D1;2

RN into L2

RN :

(G2) there exists a nonempty open set g such that g > 0 on g:

Then for each 0 < < (N42)2 there exist > 0 such that for every > ; Equation (E ) has two positive solutions u+; and u ; :

This paper is organized as follows. We …rst outline the preliminaries and the vari-ational notations and we also sort out some proofs in Section 2. In addition we prove Theorem 1.1 in Sections 3.

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2

Preliminaries

In this section,we introduce some inequalities and some de…nition of nouns.We also give some simple proofs of inequalities that will be frequently used in later chapters.

2.1

Inequalities

2.1.1 Hölder inequality

Theorem 2.1 (Hölder inequality) If 1 p 1 and 1

p + 1

q = 1; then

kfgk1 kfkpkgkq;

the equality hold i¤ almost all x 2 X jfjp kfkpp = jgj q kgkqq : Proof. If p = 1; q = 1; then Z Ejfgj esssup E jfj Z Ejgj :

In case kfkp =kgkq = 1; by Young’s inequality implies that

Z Ejfgj Z E jfjp p + jgjq q = kfkpp p + kgkqq q = 1 p + 1 q = 1 =kfkpkgkq: For general case, we assume that f1 = kfkf

p and g1 = g

kgkq;then kf1kp =kg1kq = 1:

Therefore, by the above case,we have Z E jf1g1j 1; thus Z E jf1g1j = Z E f kfkp g kgkq 1 =) Z E jfgj kfkpkgkq:

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2.1.2 Hardy inequality

Theorem 2.2 (Hardy inequality for N = 1,[1])

The classical integral inequality announced by G. H. Hardy in 1920 is given by Z 1 0 Fp(x) dx p p 1 pZ 1 0 fp(x) dx; (1)

where 0 < F (x) = 1xR0xf (t) dt < 1; p > 1; x > 0; f is a nonnegative measurable function on (0; 1) and Landau showed in [10] that the constant p 1p

p

is the best and that equality is only possible if f = 0.

Proof. Let p > 1; f is positive and continuous with compact support in (0; +1) and F is positive and di¤erentiable on [0; 1).

Setting u = Fp and dv = dx implies du = pFp 1F0dxand v = x:

Consider (0; A) with 0 < A < 1 and F (x) = x1

Rx

0 f (t) dt implies xF (x) =

Rx

0 f (t) dt,then di¤erentiate both sides of equation gives xF0(x) + F (x) = f (x). Then

use the integration by parts,we have Z A 0 Fpdx = AFp(A) p Z A 0 Fp 1(x) [f (x) F (x)]dx = AFp(A) p Z A 0 Fp 1(x) f (x) dx + p Z A 0 Fpdx: Thus, Z A 0 Fpdx = AF p(A) 1 p + p p 1 Z A 0 Fp 1(x) f (x) dx; (2) by Hölder inequality, Z A 0 Fp 1(x) f (x) dx Z A 0 F(p 1)q(x) dx 1 q Z A 0 fp(x) dx 1 p (3) where 1q +1p = 1:

Putting (3) into (2) ; then Z A 0 Fpdx AF p(A) 1 p + p p 1 Z A 0 F(p 1)q(x) dx 1 q Z A 0 fp(x) dx 1 p (4) and we have AFp(A)! 0; as A ! 1: Thus (4) implies to

Z 1 0 Fpdx p p 1 Z 1 0 Fp(x) dx (1 1p) Z 1 0 fp(x) dx 1 p :

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Since R01Fp(x) dx (1 1 p)> 0; then Z 1 0 Fpdx 1 p p p 1 Z 1 0 fp(x) dx 1 p ; Thus Z 1 0 Fpdx p p 1 pZ 1 0 fp(x) dx:

Theorem 2.3 Let 1 < p < 1 and set F (x) = R0xf (t) dt. Then for all f such thtat x f (x)2 Lp(0; 1), where < (1=p0) = 1 1=p, we have Z 1 0 jF (x)j p xp( 1)dx Cp; Z 1 0 jf (x)j p xp dx (5)

for some positive constant Cp;, which independent of f ,and given by

Cp; =

1 p0

p

(6) and equality can only be attained by f = 0:

Proof. We may assume ,without loss of generality, that f is real-valued and nonneg-ative,since the theorem will if we prove it for jfj : For < 1=p0,we have, by Hölder

inequality, F (x) x p01 = x p01 Z x 0 f (t) dt = x p01 Z x 0 f (t) t t dt x p01 Z x 0 fptpdt 1 p Z x 0 t p0dt 1 p0 = x p01 Z x 0 fptpdt 1 p 1 p0+ 1t p0+1 jx0 1 p0 = (1 p0)p01 Z x 0 fpt pdt 1 p ;

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which tends to 0 as x ! 0. Using integration by parts, it follows that with 0 < X < 1; Z X 0 Fp(x) xp( 1)dx = Fp x p p01 p 1 p0 X 0 p p 1 p0 Z X 0 Fp 1(x) f (x) xp( 1)dx 1 1 p0 Z X 0 Fp 1(x) f (x) xp( 1)dx = 1 1 p0 Z X 0 Fp0p (x) f (x) x p p0( 1)+ dx = 1 1 p0 Z X 0 Fp(x) xp( 1) 1 p0 (fp(x) xp )1pdx

and by the Hölder inequality, Z X 0 Fp(x) xp( 1)dx 1 1 p0 Z X 0 Fp(x) xp( 1)dx 1 p0 Z X 0 fp(x) xp dx 1 p =) Z X 0 Fp(x) xp( 1)dx 1 p 1 1 p0 Z X 0 fp(x) xp dx 1 p =) Z X 0 Fp(x) xp( 1)dx 1 1 p0 p Z X 0 fp(x) xp dx:

The inequality (5) with Cp; = p10 p

, following on allowing X ! 1.

Note that the resulting inequality is strict, unless there are constant A; B 6= 0, such that AFp(x) xp( 1) = Bfp(x) xp . But this mean that f (x) = F0(x) is a power of x and R1

0 f

p(x) xp dx

is divergent. Consequently, (5) is strict inequality for f 6= 0:

To prove that the constant Cp; is the best in (5), we choose f (x) = x 1=p+ (0;a)(x),where

+ > 0; a > 0and (0;a)is the characteristic function of (0; a). Then x f (x) 2 Lp(0;

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and Z 1 0 fp(x) xpdx = Z a 0 x 1p+ p xpdx = Z a 0 x 1+ p+ pdx = 1 p + px p+ p a 0 = a p+ p p ( + ); and F (x) = 8 < : 1 +p01 x +p01 if x a; 1 +p01 a +p01 if x > a: and Z 1 0 Fp(x) xp( 1)dx = Z 0 1 +p10x +p01 !p xp( 1)dx + Z 1 1 + p10a +p01 !p xp( 1)dx = 1 +p10 !p 1 p ( + )x p( + ) 0 + 1 + p10 !p p 1 p 1 p p + 1x p p+1 j1 = 1 +p10 !p 1 p ( + ) p( + ) 1 p p + 1 p +p = p( + ) p +p10 p " 1 ( + ) 1 1 + 1 p # = p( + ) p +p10 p " 1 ( + )+ 1 1 p0 # :

Thus,we obtain that R1 0 F p(x) xp( 1)dx R1 0 f p(x) xpdx = 1 +p10 p " 1 + 1 + p0 # : which tends to p10 p

as ! .It follows that the constant p10 p

in (5) is the best.

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Theorem 2.4 (Hardy inequality for N 3,[11])

The standard Hardy inequality involving the distance to the origin asserts that when N 3, 1 < p < N and u 2 W1;p RN Z RN jujp jxjpdx CN;p Z RNjruj p dx where constant CN;p= Npp p is the best.

Theorem 2.5 Let 1 p <1; n > 1; and 1 + n

p 6= 0: Let f be di¤erentiable a.e. in

Rn;

and such that jxj 1+np f (x) ! 0 as jxj ! 0+ if 1 + n

p < 0; and as jxj ! 1 if 1 + np > 0: Then Z Rn jxj 1 jf (x)j pdx 1 + n p pZ Rn jxj 1 j(x r) f (x)j pdx (7) where rf = @x@f1; ; @f @xn : The constant 1 + n p p is the best.

Proof. Let 0 < ; N < 1, and choose polar coordinates x = rw; r = jxj ; w 2 Sn 1. Then use integration by parts,we obtain

Z N r 1jf (rw)j prn 1dr = jf (rw)j p p ( 1) + n N Z N rp( 1)+n p ( 1) + n @ @r jf (rw)j p dr Thus, Z N r 1jf (rw)j prn 1dr jf (rw)j p p ( 1) + n N = Z N rp( 1)+n p ( 1) + n @ @rjf (rw)j p dr Z N rp( 1)+n p ( 1) + n @ @r jf (rw)j p dr 1 p ( 1) + n Z N rp( 1)+n @ @rjf (rw)j p dr:

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we next let ! 0 and N ! 1, then use the hypothesis and the Hölder inequality to obtain Z 1 0 r 1jf (rw)j prn 1dr 1 p ( 1) + n Z 1 0 rp( 1)+n @ @r jf (rw)j p dr = 1 p ( 1) + n Z 1 0 rp( 1)+npjf (rw)jp 1 @ @rjf (rw)j dr = 1 1 + np Z 1 0 rp( 1)+njf (rw)jp 1 @ @r jf (rw)j dr 1 1 + np Z 1 0 rp( 1)+njf (rw)jp 1 @ @rf (rw) dr = 1 1 + np Z 1 0 r( 1)(p 1)jf (rw)jp 1rnp01 r @ @rf (rw) r n 1 p dr 1 1 + n p Z 1 0 r( 1)pjf (rw)jprn 1dr 1 p0 Z 1 0 rp @ @rf (rw) p rn 1dr 1 p : Therefore, since r @ @rf = (x r) f, implies that Z Rnjxj ( 1)p jf (x)jpdx 1 + n p 1 Z Rnjxj ( 1)p jf (x)jpdx 1 p0 Z Rnjxj ( 1)p j(x r) f (x)jpdx 1 p

This completes the prove of (7) :

To prove that the constant in (7) is the best, we consider the radial functions f (x) = f (r) ; r =jxj ; which satisfy the hypothesis of the Theorem 2.5.Then (7) becomes Z Sn 1 Z 1 0 r 1jf (r)j prn 1drdw 1 + n p pZ Sn 1 Z 1 0 r 1j(r r) f (r)j prn 1drdw =) Z Sn 1 Z 1 0 r 1f (r) pdrdw 1 p0 pZ Sn 1 Z 1 0 jr f0(r)jpdrdw where = + n 1 p .

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Suppose p10 = 1 + n

p < 0, and let f (r) =

Rr

0 (t) dt; where for some a > 0;

(t) = t 1p+ (0;a)(t) ; + > 0; so that f (r) = 8 < : 1 +1 p0 r +p01 if r a; 1 +1 p0 a +p01 if r > a:

Then f satis…es the hypothesis and as in proof of Theorem 2.3, lim ! R1 0 f p(r) rp( 1)dr R1 0 f 0(r)p x pdr ! = 1 + 1 p0 p = 1 1 + np p :

The constant is therefore the best. The case 1 + np > 0 is treated similarly.

Note that we choose = 0 in Theorem 2.5 yield the following corollary and we use the notation jrf (x)j = Pni=1

@f @xi

2 12

: Corollary 2.6 The inequality

Z Rn jf (x)jp jxjp dx p p n pZ Rnjrf (x)j p dx holds for all f 2 C1

0 (Rnn f0g) if n < p < 1 and for all f 2 C01 if 1 p < n:

2.1.3 Sobolev inequality

Theorem 2.7 (Sobolev inequality)

Let be a bounded domain in Rn, and 1 p <

1.there exists S (n; p; ) (the constant that will frequently be used in later chapters) such that for u 2 W01;p( )

Skuk L np n p( ) krukLp( ) (8) where p < n and S = 12n1=p p 1 n p 1 p 1 8 < : 1 + n2 (n) n p 1 + n + n p 9 = ; 1 n

The inequality sign holds in (8) if u has the form: u (x) =ha + bjxjpp1

i1 np

; where jxj = (x21+ ::: + x2n)

1

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Proof. Because that C1

0( ) is dense in W 1;p 0 ( ).

We prove …rst for u 2 C1

0( ) and will later justify why the proof actually extends to

the larger space. CaseI: p = 1

Fix an index i 2 f1; 2; :::; ng and observe u (x) =

Z xi 1

Diu (x1; x2; :::; t; :::xn) dt;

simplify denote Diu (x1; x2; :::; t; :::xn) = Diu then implies

ju (x)j = Z xi 1 Diudt Z xi 1jD iuj dt Z 1 1jD iuj dt:

Take the n 1 th root of the result to yield altogether ju (x)jnn1 Z 1 1jD iuj dt n n 1 Yn i=1 Z 1 1jD iuj dxi 1 n 1 :

Thus, we integrate over the x1-axis and subsequently apply the Hölder inequality

Z 1 1 ju (x)jnn1 dx 1 Z 1 1 n Y i=1 Z 1 1 jDiuj dxi 1 n 1 dx1 = Z 1 1jD 1uj dx1 1 n 1 "Z 1 1 n Y i=2 Z 1 1jD iuj dxi 1 n 1 dx1 # Z 1 1 jD1uj dx1 1 n 1 Yn i=2 Z 1 1 Z 1 1 jDiuj dxi n 1 n 1 dx1 ! 1 n 1 = Z 1 1jD 1uj dx1 1 n 1 Z 1 1 Z 1 1jD 2uj dx2dx1 1 n 1 n Y i=3 Z 1 1 Z 1 1jD iuj dxi dx1 1 n 1

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then, we integrate over the x2-axis and using Hölder inequality again Z 1 1 Z 1 1ju (x)j n n 1 dx 1dx2 Z 1 1 Z 1 1 jD2uj dx1dx2 1 n 1 (Z 1 1 Z 1 1jD 1uj dx1 1 n 1 Yn i=3 Z 1 1 Z 1 1jD iuj dxi dx1 1 n 1 dx2 ) Z 1 1 Z 1 1 jD2uj dx1dx2 1 n 1 Z 1 1 Z 1 1 jD1uj dx1 n 1 n 1 dx2 ! 1 n 1 n Y i=3 Z 1 1 Z 1 1 Z 1 1jD iuj dxidx1dx2 ! 1 n 1

then,we integrate over the xi-axis for i 2 f1; 2; :::; ng,

Z 1 1 Z 1 1 ju (x)jnn1 dx 1 dxn n Y i=1 Z 1 1 Z 1 1 jDiuj dx1 dxn 1 n 1 : In other words if we restrict to ;we obtain that

kukL n n 1( ) n n 1 n Y i=1 Z jDiuj dx 1 n 1 ; or still kukL n n 1( ) n Y i=1 Z jDiuj dx 1 n 1 n n X i=1 Z jDiuj dx 1 n n X i=1 Z jDuj dx = Z jDuj dx = krukL1( ): CaseII: 1 < p < 1

Let r > 1 be a constant to be speci…ed. We use previous CaseI kjujrkL n n 1 Z jD jujrj dx r Z jujr 1jDuj dx: Let q be such that 1

q + 1

p = 1 and using the Hölder inequality

Z jujr nn1 dx n 1 n r Z juj(r 1)qdx 1 q Z jDujpdx 1 p : We have q = p 1p Choose r = n 1n p p in order to have r (n 1) q = n 1n r.

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Hence, Z juj(nn 1pp) n n 1 dx n 1 n r Z juj(nn 1pp 1)( p p 1) dx p 1 p Z jDujpdx 1 p ; thus Z jujnnppdx n 1 n p 1 p rkDukLp( ) = n 1 n p p kDukLp( ):

Note: The best constant of Sobolev inequality S were proformed by G.Talenti in [3].

2.2

Variational methods and Theorems

De…nition 2.8 (Lower/Upper Semicontinuous ,[7])

Let X be a metric space.A function f : X ! R is said to be lower semicontinuous at a point x 2 X if f (x) lim inf

n!1 f (xn)whenever xn! x as n ! 1: f is said to be lower

semicontinuous on X if it is lower semicontinuous at each point of X:

A function f : X ! R is said to be upper semicontinuous at a point x 2 X if f (x) lim sup

n!1

f (xn)whenever xn ! x as n ! 1: f is said to be upper semicontinuous

on X if it is upper semicontinuous at each point of X: Example 2.9 Let f : R ! R be de…ned as

f (x) = 1; if x 0; 1; if x > 0:

Then f is lower semicontinuous at x = 0 but not upper semicontinuous at x = 0: Theorem 2.10 (Strong form of Ekeland’s Variational Principle,[6][7][26] )

Let (X; d) be a complete metric space and E : X ! R [ f+1g be a bounded below and lower semicontinuous and 6 0. Let > 0 and any u 2 X be given such that

E (u) inf

X E +2; (9)

Then given > 0 there exists v 2 X such that

E (v) E (u) (10)

d (u; v) (11)

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Proof. For notational simpli…cation let us put d (x; y) = (1= )d(x; y). Let us de…ne a partial order in X by

u v () E (u) E (v) d (u; v) : It is straightforward that:

(i) (re‡exivity) u u;

(ii) (antisymmetry)u v and v u imply u = v; (iii) (transitivity) u v and v w imply u w; all these three properties for all u; v; w in X.

Now we de…ne a sequence fSng of subsets of X as follows.Start with u1 = u and

de…ne S1 =fu 2 X : u u1g; u2 2 S1 s.t. E (u2) inf S1 E + 22 and inductively Sn =fu 2 X : u ung; un+12 Sn s.t. E (un+1) inf Sn E + 2n+1:

Clearly S1 S2 S3 :::Each Sn is closed: let xj 2 Sn with xj ! x 2 X.We have

E(xj) E(un) d (xj; un). Taking limits using the lower semicontinuity of E and the

continuity of d we conclude that x 2 Sn. Now we prove that the diameters of these sets

go to zero: diamSn ! 0. Indeed, take an arbitrary point x 2 Sn.On one hand, x un

implies

E(x) E(un) d (x; un) : (13)

On the other hand, we observe that x belongs also to Sn 1. So it is one of the points

which entered in the competition when we picked un. So

E(un) E(x) +

2n: (14)

From (13) and (14) we get

d (x; un) 2 n 8x 2 Sn;

which gives diamSn 2 n+1. Now we claim that the unique point in the intersection of

the Sn’s satis…es conditions (10) (12). Let then 1

T

n=1

Sn = fvg. Since v 2 S1, (10) is

clear. Now let w 6= v. We cannot have w v, because otherwise w would belong to the intersection of the Sn’s. So w 6 v, which means that

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thus proving (12) :Finally to prove (11) we write d (u; un) n 1 X j=1 d (uj; uj+1) n 1 X j=1 2 j and take limits as n ! 1:

Corollary 2.11 If V is a Banach space and E 2 C1(V ) is bounded below, there exists

a minimizing sequence fvmg for E in V such that

E (vm)! inf

V E; DE (vm)! 0 in V as m ! 1:

Proof. Choose a sequence f"mg of numbers "m > 0; "m ! 0 as m ! 1:For m 2 N

choose um 2 V such that

E (um) inf V E + "

2 m:

For " = "2

m; = "m; u = um determine an element vm = v according to Theorem

2.10,satisfying

E (vm) E (vm+ w) + "mkwkV

for all w 2 V: Hence

kDE (vm)kV = lim sup 06=kwkV!0

E (vm) E (vm+ w)

kwkV

"m ! 0;

as claimed.

Theorem 2.12 (Lagrange Multipliers) Suppose that f : Rn

! R and g : Rn

! R are C1 functions. Suppose that for

some c0 2 R there is an x0 2 Rn such that g(x0) = c0: Let S = g 1(c0). Suppose that

fjS has a local maximum or local minimum at some point x1; and that Dg (x1) 6= 0:

(Dg (x1) =rg (x1) :) then there is a scalar such that

rf (x1) = rg (x1) :

De…nition 2.13 (Gâteaux derivative)

Let f : V (x) ! Y be a map. Consider the subset S, a convex subset (not necessarily open) of a Banach space X, let Y be a Banach space and V (x) a relative neighborhood of x 2 S. We say that the map is Gâteaux di¤erentiable at x in the admissible direction h2 Sx, if there exist a map T 2 L(X; Y ) such that:

f (x + v) f (x) = Tv+ o( ); ! 0

and v = khkh , and in some neighborhood oh zero. The map T is called the Gâteaux derivativeof f at x in the (admissible) direction h, if T does not depend on the admissi-ble direction we de…ne the Gâteaux di¤erential dGf (x; v) = f0(x)v. Where f0(x)v = Tv.

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De…nition 2.14 (Fréchet derivative)

Let U (x) be a neighborhood of x in X:Consider the map f : U (x) ! Y: f is Fréchet di¤erentiable at x if there is T 2 L (X; Y ) such that

f (x + h) f (x) = T h + o (khk) as h ! 0:

Lemma 2.15 Let fung be a (PS)c–sequence. Then there exist a u 2 E and a

subse-quence funkg such that

unk * u weakly in E; unk ! u strongly in L p loc R N ; for 1 p < N + 2 N 2: and unk ! u a.e..

Note that the Lemma 2.15 indicates the relationship between Palais-Smale sequences and critical points of J. We refer to [44] for a detailed proof.

2.3

Implicit Function Theorem

Theorem 2.16 (Implicit Function Theorem) Now suppose that X; Y , and W are three Banach spaces, k 1, A X Y is an open set, (x0; y0) is a point in A, and

f : A! W is a Ck map such f (x0; y0) = 0. Assume that D2f (x0; y0) D(f (x0; )(y0) :

Y ! W is a bounded invertible linear transformation. Then there is an open neighbor-hood U0 of x0 in X such that for all connected open neighborhoods U of x0 contained in

U0, there is a unique continuous function u : U ! Y such that u(x0) = yo; (x; u(x))2 A

and f (x; u(x)) = 0 for all x 2 U. Moreover u is necessarily Ck and

Du(x) = D2f (x; u(x)) 1D1f (x; u(x)) : for all x 2 U (15)

Proof. By replacing f by (x; y) ! D2f (x0; y0) 1f (x; y) if necessary, we may assume

with out loss of generality that W = Y and D2f (x0; y0) = IY. De…ne F : A ! X Y

by F (x; y) (x; f (x; y))for all (x; y) 2 A. Notice that DF (x; y) = I D1f (x; y)

0 D2f (x; y)

which is invertible i¤ D2f (x; y) is invertible and if D2f (x; y) is invertible then

DF (x; y) 1 = I D1f (x; y) D2f (x; y)

1

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Since D2f (x0; y0) = I is invertible, the implicit function theorem guarantees that there

exists a neighborhood U0 of x0 and V0 of y0 such that U0 V0 A, F (U0 V0) is

open in X Y, F j(U0 V0) has a C

k–inverse which we call F 1. Let

2(x; y) y for

all (x; y) 2 X Y and de…ne Ck–function u0 on U0 by u0(x) 2 F 1(x; 0). Since

F 1(x; 0) = (~x; u0(x)) i¤ (x; 0) = F (~x; u0(x)) = (~x; f (~x; u0(x))), it follows that x = ~x

and f (x; u0(x)) = 0. Thus (x; u0(x)) = F 1(x; 0)2 U0 V0 A and f (x; u0(x)) = 0 for

all x 2 U0. Moreover, u0 is Ck being the composition of the Ck–functions, x ! (x; 0),

F 1, and

2. So if U U0 is a connected set containing x0, we may de…ne u u0jU to

show the existence of the functions u as described in the statement of the theorem. The only statement left to prove is the uniqueness of such a function u.

Suppose that u1 : U ! Y is another continuous function such that u1(x0) = y0, and

(x; u1(x))2 A and f(x; u1(x)) = 0for all x 2 U. Let

O fx 2 Uju(x) = u1(x)g = fx 2 Uju0(x) = u1(x)g :

Clearly O is a (relatively) closed subset of U which is not empty since x0 2 O. Because

U is connected, if we show that O is also an open set we will have shown that O = U or equivalently that u1 = u0 on U: So suppose that x 2 O, i.e. u0(x) = u1(x). For ~x near

x2 U, 0 = 0 0 = f (~x; u0(~x)) f (~x; u1(~x)) = R(~x)(u1(~x) u0(~x)) (16) where R(~x) Z 1 0 D2f ((~x; u0(~x) + t(u1(~x) u0(~x)))dt: (17)

From (17) and the continuity of u0 and u1, limx!x~ R(~x) = D2f (x; u0(x)) which is

invert-ible. Thus R(~x) is invertible for all ~xsu¢ ciently close to x. Using (16), this last remark implies that u1(~x) = u0(~x)for all ~xsu¢ ciently close to x. Since x 2 O was arbitrary, we

have shown that O is open.

2.4

Mean Value Theorem

Theorem 2.17 (Chain Rule)

Let x be …xed and set y = f (x) : Assume we are given f : U (x) X ! Y and g : U (y) Y ! Z with f (U (x)) U (y) ;where X; Y and Z are Banach spaces, and U(x) ; U (y) are neighborhoods of x; y; respectively. Suppose that there are f0(x) and g0(f (x)) as Fréchet derivatives. Then H = g f is Fréchet di¤erentiable at x; and

(g f )0(x) = g0(f (x)) f0(x) : Proof. By hypothesis,

g (y + k) = g (y) + g0(y) k +kkk r1(k)

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We choose

k = f (x + h) f (x) = f0(x) h +khk r2(h)

where r1(h)! 0 as h ! 0:

This implies that

g (f (x + h)) = g (f (x) + k) = g (f (x)) + g0(f (x)) k +kkk r1(k) = g (f (x)) + g0(f (x)) [f0(x) h +khk r2(h)] +kkk r1(k) = g (f (x)) + g0(f (x)) f0(x) h +khk r (h) where r (h) = g0(f (x)) r 2(h) + kkkkhkr1(k) : We must prove r (h) = o (1) : In fact. kg0(f (x)) r2(h)k kg0(f (x))k kr2(h)k ! 0 as h ! 0;and kkk khkkr1(k)k = kf0(x) h +khk r 2(h)k khk kr1(k)k kf0(x) hk + kr 2(h)k khk khk kr1(k)k kf0(x)k kr1(k)k + kr2(h)k kr1(k)k ! 0; as h ! 0:

We obtain that r (h) = o (1) and consequently, (g f )0(x) = g0(f (x)) f0(x) :

Theorem 2.18 Let [a; b] be a bounded interval in R; and E a Banach space. Let f : [a; b] ! E and g : f : [a; b] ! R be continuous on [a; b] and di¤erentiable on (a; b) : Suppose that kf0(t)k g0(t) for a < t < b: Then

kf (a) f (b)k g (a) g (b) : Proof. It su¢ ces to prove

kf (d) f (c)k g (d) g (c) for a < c < d < b:

Suppose the contrary. There exists a0; b0 2 (a; b) ; a0 < b0; with

kf (a0) f (b0)k [g (a0) g (b0)] = M > 0: Set m = a0+b0 2 : Then, kf (b0) f (m)k [g (b0) g (m)] M 2 or kf (m) f (a0)k [g (m) g (a0)] M 2 :

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Denote by [a1; b1] one of the intervals on which the inequality holds. Repeating the

procedure we obtain a sequence of intervals [an; bn] such that

a0 ::: an bn ::: b0;

jbn anj = (b0 a0) =2n;

kf (bn) f (an)k [g (bn) g (an)]

M 2n:

It follows that an and bn converge to the same point w 2 (a; b) and that

M 2n kf (bn) f (w) + f (w) f (an)k [g (bn) g (w) + g (w) g (an)] kf (bn) f (w)k + kf (w) f (an)k [g (bn) g (w)] [g (w) g (an)] kf0(w) (bn w) + o (bn w)k + kf0(w) (w an) + o (w an)k [g0(w) (bn w) + o (bn w)] [g0(w) (w an) + o (w an)] kf0(w)k (jbn wj + jw anj) g0(w) (jbn wj + jw anj) +o (bn w) + o (w an) [kf0(w)k g0(w)] (bn an) + o (bn an) :

Divide both sides by bn an= (b0 a0) =2n; and let n ! 1: We obtain

M

b0 a0 kf

0(w)k g0(w) ;

so, kf0(w)k > g0(w) ;a contradiction.

Corollary 2.19 Let [a; b] be a bounded interval in R; and E a Banach space. Let f : [a; b]! E be continuous on [a; b] and di¤erentiable on (a; b) : Suppose there is a constant k such that kf0(t)k k for all t 2 (a; b) :Then

kf (b) f (a)k k (b a) : Proof. Take g (t) = kt in Theorem 2.18.

Theorem 2.20 (Mean Value Theorem) Let E; F be two Banach spaces , U E an open set,[a; b] U a segment, and let f : U ! F be di¤erential. Then,

kf (b) f (a)k sup

0 t 1kf

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Proof. Let h (t) = f ((1 t) a + tb) ; t2 [0; 1] ; k = sup0 t 1kf0((1 t) a + tb)k kb ak :

By Chain rule (Theorem 2.17) we have h0(t) = f ((1 t) a + tb) (b a) ;and kh0(t)k

k;8t 2 [0; 1] : By Corollary 2.19 , we obtain kf (b) f (a)k = kh (1) h (0)k k (1 0) = sup 0 t 1kf 0((1 t) a + tb)k kb ak :

2.5

Brezis-Lieb Lemma

Lemma 2.21 (Brezis-Lieb Lemma) Suppose,for some p, 0 < p < 1,fn! f a.e. in

and kfnkp c <1 for each n. Then lim

n!1 kfnk p p kfn fkpp exists, and lim n!1 kfnk p p kfn fk p p =kfkp

Proof. For > 0,there is c > 0 such that for 0 < p < 1 we have jja + bjp jajpj jajp + c jbjp:

In fact, we may assume that ab 6= 0 and let ' (t) = tp for t > 0,then

'0(t) = ptp 1: The Mean Value Theorem implies

ja + bjp jajp = ' (ja + bj) ' (jaj) = p p 1(ja + bj jaj) where lies between jaj and ja + bj : Thus, for p0 = p

p 1;

jja + bjp jajpj = p p 1(ja + bj jaj) p p 1jbj = p1=p0 1=p0 p 1 p1=p 1=p0jbj p (p 1)p0 p0 pjbjp p 1 = p p0 p+pjbj p p 1 : However, jaj + ja + bj 2jaj + jbj ;

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or

p (2

jaj + jbj)p cp[2pjajp +jbjp] ;

so

jja + bjp jajpj 0jajp+ c0jbjp:

Let fn = f + gn for each n.Then gn! 0 as n ! 1;and kgnk p

p c <1 for each n.Set

w;n(x) = (jjfn(x)j p jgn(x)j p jf (x)jpj jgn(x)j p )+ where a+ = max (a; 0) :We have w;n(x)! 0 a.e. in as n ! 1;and

jjfnjp jgnjp jfjpj jjfnjp jgnjpj + jfjp jgnj p + c jfjp+jfjp: That is, w;n c jfjp+jfjp 2 L1:

By Lebesque Dominate Convergence Theorem, Z w;n(x) dx! 0 as n ! 1: However, jjfnjp jgnjp jfjpj w;n+ jgnjp or Z jjfnj p jgnj p jfjpj Z w;n+ Z jgnj p ; and we have lim sup n!1 Z jjfnjp jgnjp jfjpj c for each > 0 or lim n!1 kfnk p p kfn fk p p =kfkp:

2.6

Variational Settings

In this section, we give the variational setting for Equation (E ) following [25], and we establish the compactness conditions. Let

X = u2 D1;2 RN j Z

RN

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be equipped with the inner product and norm hu; vi = Z RNrurv + V +uvdx; kuk = hu; ui1=2: For > 0; we also need the following inner product and norm

hu; vi = Z

RNrurv + V

+uvdx;

kuk = hu; ui1=2:

It is clear that kuk kuk for 1:Furthermore, by the Hardy-Sobolev inequality and 0 < V 1 jxj2; for every 0 < < (N 2)2 4 ; kuk2 = Z RNjruj 2 + V+u2dx Z RNjruj 2 + V+u2 V u2dx kuk2 Z RN u2 jxj2dx kuk2 4 (N 2)2 Z RNjruj 2 dx kuk2 4 (N 2)2kuk 2 = (N 2) 2 4 (N 2)2 kuk 2 : Hence,we have kuk2 Z RNjruj 2 + V ; u2dx (N 2)2 4 (N 2)2 kuk 2 for all 0: (18) Set X = (X; kuk ) : It follows from conditions (V 1) and (V 2) that we have

Z RN jruj 2 + u2 dx = Z RNjruj 2 dx + Z fV+<bg u2dx + Z fV+ bg u2dx and using the Hölder and Sobolev inequalities with p = 22 2; q = 22 and 1p + 1q = 1:

Z fV+<bg u2dx Z RN u2 dx 2 2 Z fV+<bg 1dx 2 2 2 = Z RN u2 dx 2 2 V+< b 2 2 2 S 2 V+ < b 2 2 2 Z RN jruj 2 dx:

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and since V+ b then V+ b 1, so Z fV+ bg u2dx 1 b Z RN V+u2dx: Hence, Z RN jruj 2 + u2 dx = Z RNjruj 2 dx + Z fV+<bg u2dx + Z fV+ bg u2dx 1 + S 2 V+< b 2 2 2 Z RNjruj 2 dx + 1 b Z RN V+u2dx max 1 + S 2 V+< b 2 2 2 ;1 b Z RN jruj 2 + V+u2dx; this implies that the imbedding X ,! H1

RN

is continuous, where the set fV+ b

g := x2 RN

j V+(x) b : Moreover, using the conditions (V 1) and (V 2) ; and the Hölder

inequality (with ~p = 22 2p; ~q = 2p 22 where 1p~+1q~ = 1) and Sobolev inequality, we have for any p 2 [2; 2 ), Z RNjuj p dx = Z RNjuj np 2 +n+pjuj np 2 ndx Z RNjuj 2 dx 2 p 2 2 Z RNjuj 2 dx p 2 2 2 Z fV+ bg u2dx + Z fV+<bg u2dx 2 p 2 2 S 2 Z RNjruj 2 dx 2 2 ! p 2 2 2 1 b Z RN V+u2dx + V+< b 2 2 2 S 2 Z RNjruj 2 dx 2 p 2 2 S 2 kuk2 p 2 2 2 max ( S2 b; V + < b 22 2 )!2 p 2 2 S 2 kuk2 p 2 2 2 max ( S2 b; V + < b 2 2 2 )!2 p 2 2 S pkukp = V+< b 2 p 2 S p kukp for 0; (19) where 0 := S2 b V +< b 22 2 : (20)

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We use the variational methods to …nd positive solutions of equation (E ) : Associated with the equation (E ) ; we consider the energy functional J ; : X ! RN

J ; (u) = 1 2 Z RNjruj 2 + V ; u2dx 1 q Z RN fjujqdx 1 p Z RN gjujpdx = 1 2kuk 2 2 Z RN V u2dx 1 q Z RN fjujqdx 1 p Z RN gjujpdx:

Because the energy functional J ; is not bounded below on X; it is useful to consider

the functional on the Nehari manifold M ; =

n

u2 Xn f0g j DJ0; (u) ; uE= 0o: and de…ne the notation of ; as following

; = inf u2M ;

J ; (u)

Thus, u 2 M ; if and only if

kuk2 Z RN V u2dx Z RN fjujqdx Z RN gjujpdx = 0: Note that M ; contains every non-zero solution of equation (E ) :

Next, we de…ne the Palais–Smale (simply by (PS)) sequences, (PS)–values, and (PS)– conditions in X for J ; as follows.

De…nition 2.22 (i) For 2 R; a sequence fung is a (PS) –sequence in X for J ; if

J ; (un) = + o(1) and J 0

; (un) = o(1) strongly in X 1 as n ! 1:

(ii) 2 R is a (PS)–value in X for J ; if there exists a (PS) –sequence in X for J ; :

(iii) J ; satis…es the (PS) –condition in X if every (PS) –sequence in X for J ;

con-tains a convergent subsequence. We have the following results.

Lemma 2.23 The energy functional J ; is coercive and bounded below on M ; :

Fur-thermore, if u0 is a solution of Equation (E ) ; then

J ; (u0) 2 q 2pq (N 2)2 (N 2)2 4 (p 2) !q=(2 q)0 @(p q)jfV+ < bgj q(2 p) 2 p kf +kLq Sq 1 A 2=(2 q) :

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Proof. (i)If u 2 M ; ; then J ; (u) = 1 2 kuk 2 Z RN V u2dx 1 q Z RN fjujqdx 1 p Z RN gjujpdx = 1 2 kuk 2 Z RN V u2dx 1 q Z RN fjujqdx 1 p kuk 2 Z RN V u2dx Z RN fjujqdx = p 2 2p kuk 2 Z RN V u2dx p q pq Z RN fjujqdx p 2 2p kuk 2 p 2 2p kuk 2 ! +1 as kuk ! 1: Hence,J ; (u0)! +1 as kuk ! 1; that is J ; is coercive.

(ii)If u 2 M ; ;then, by the Hölder inequality (with p = qp; q = p qp where p1 +q1 =

1) , (18) and (19) ; J ; (u) = p 2 2p kuk 2 Z RN V u2dx p q pq Z RN fjujqdx = p 2 2p Z RNjruj 2 + V ; u2dx p q pq Z RN fjujqdx p 2 2p 1 4 (N 2)2 kuk 2 p q pq Z RN fjujqdx p 2 2p 1 4 (N 2)2 kuk 2 p q pq kfkLq Z RNjuj q pq dx q p p 2 2p 1 4 (N 2)2 kuk 2 p q pq kfkLq V + < b 22 pS p kukp q p p 2 2p 1 4 (N 2)2 kuk 2 p q pq V + < b q(2 p) 2 p S q kf+kLq kuk q (21) 2 q 2pq (N 2)2 (N 2)2 4 (p 2) !q=(2 q)0 @(p q)jfV+ < bgj q(2 p) 2 p kf +kLq Sq 1 A 2=(2 q) : (22)

Thus, J ; is coercive and bounded below on M ; : This completes the proof.

Remark 2.1 Proof that (21) (22) :

Proof. We let constants A; B and x = kuk ;where

A = (p 2)(N 2) 2 4 (N 2)2 and B = (p q) V +< b q(22 pp) S q kf+kLq ;

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then (21) become 2pAx2 B pqx q and (22) become 1 2p 1 pq A q 2 qB 2 2 q.

That is we just proof that A 2px 2 B pqx q 1 2p 1 pq A q 2 qB 2 2 q

Now,we assume that

f (x) = A 2px 2 B pqx q; then we have f0(x) = A px B px q 1; f00(x) = A p B p (q 1) x q 2.

So,the critical point of f (x) is x = 0 and x = AB

1 q 2 : Then (i) If x = 0 then f (x) = 0: (ii) If x = AB 1 q 2 then f00 A B 1 q 2 > 0; thus f (x) = A 2p A B 2 q 2 B pq A B q q 2 = 1 2p 1 pq A q 2 qB 2 2 q

This completes the proof.

The Nehari manifold M ; is closely linked to the behavior of the function of the form

hu : t! J ; (tu) for t > 0: Such maps are known as …bering maps and were introduced

by Drábek and Pohozaev in [24]. They are also discussed in Brown and Zhang [20] and Brown and Wu [18, 19]. If u 2 X; we have

hu(t) = t2 2 kuk 2 Z RN V u2dx t q q Z RN fjujqdx t p p Z RN gjujpdx; h0u(t) = t kuk2 Z RN V u2dx tq 1 Z RN fjujqdx tp 1 Z RN gjujpdx; h00u(t) = kuk2 Z RN V u2dx (q 1) tq 2 Z RN fjujqdx (p 1) tp 2 Z RN gjujpdx: It is easy to see that

th0u(t) =ktuk2 Z RN V (tu)2dx Z RN fjtujqdx Z RN gjtujpdx and so, for u 2 Xn f0g and t > 0; h0

u(t) = 0if and only if tu 2 M ; , i.e., positive critical

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In particular, h0

u(1) = 0 if and only if u 2 M ; : Thus, it is natural to split M ;

into three parts corresponding to local minima, local maxima and points of in‡ection. Accordingly, we de…ne

M+; = fu 2 M ; j h00u(1) > 0g ;

M0; = fu 2 M ; j h00u(1) = 0g ;

M ; = fu 2 M ; j h00u(1) < 0g :

We now derive some basic properties of M+; ; M0

; and M ; :

Lemma 2.24 Suppose that u0 is a local minimizer for J ; on M ; and that u0 2 M= 0; :

Then, J0; (u0) = 0 in X 1:

Proof. If u0 is a local minimizer of J ; on M ; then u0 is a solution of the optimization

problem

minimize J ; subject to =

D

J0; (u) ; uE= 0 Hence, by the Lagrange multipliers,there exists 2 R such that

J0; (u0) = 0 (u0) in X 1 This implies D J0; (u0) ; u0 E = h 0 (u0) ; u0i since u0 2 M ; ; that is ku0k 2 R RNV u 2 0dx R RNfju0j q dx RRN gju0j p dx = 0: Hence,we have h 0 (u0) ; u0i = 2 ku0k 2 Z RN V u20dx q Z RN fju0j q dx p Z RN gju0j p dx = 2 Z RN fju0j q dx + Z RN gju0j p dx q Z RN fju0j q dx p Z RN gju0j p dx = (2 q) Z RN fju0jqdx + (2 p) Z RN gju0jpdx6= 0:

Since h 0 (u0) ; u0i 6= 0; thus = 0:i.e.J 0

; (u0) = 0: This completes the proof.

Note: The proof is essentially the same as that in Brown and Zhang [20, Theorem 2.3] (or see Binding et al. [15]).

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For each u 2 M ; ;we have h00u(1) = Z RNjruj 2 + V ; u2dx (q 1) Z RN fjujqdx (p 1) Z RN gjujpdx = Z RNjruj 2 + V ; u2dx (q 1) Z RN fjujqdx (p 1) Z RNjruj 2 + V ; u2dx Z RN fjujqdx = (2 p) Z RNjruj 2 + V ; u2dx (q p) Z RN fjujqdx (23) = Z RNjruj 2 + V ; u2dx (q 1) Z RN jruj 2 + V ; u2dx Z RN gjujpdx (p 1) Z RN gjujpdx = (2 q) Z RNjruj 2 + V ; u2dx (p q) Z RN gjujpdx: (24)

Let 0 > 0 be as in (20) : Then, we have the following result.

Lemma 2.25 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 4) : Then, for each 0; we have M0; =;:

Proof. Suppose the contrary. Then, there exists 0 such that M0; 6= ;: Then, for

u2 M0; (i.e.h00u(1) = 0);by (23) we have h00u(1) = (2 p) Z RNjruj 2 + V ; u2dx (q p) Z RN fjujqdx = 0 =) (2 p) Z RNjruj 2 + V ; u2dx = (q p) Z RN fjujqdx (25) and by the Hölder inequality and (19) ; (18) and (25) ;

(N 2)2 4 (N 2)2 kuk 2 < Z RNjruj 2 + V ; u2dx = p q p 2 Z RN fjujqdx p q p 2 Z RN f+jujqdx (p q)kf+kLq jfV+< bgj q(2 p) 2 p Sq(p 2) kuk q and so kuk 0 @(p q) (N 2)2kf+kLq jfV+< bgj q(2 p) 2 p Sq(p 2) (N 2)2 4 1 A 1=(2 q) : (26)

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Similarly, using (24) and the Hölder and Sobolev inequalities, we have (N 2)2 4 (N 2)2 kuk 2 Z RNjruj 2 + V ; u2dx = p q 2 q Z RN gjujpdx p q 2 q g + 1 Z RN juj p dx (p q)kg+ k1jfV+< b gj22 p Sp(2 q) kuk p ; which implies kuk " Sp(2 q) (N 2)2 4 (p q) (N 2)2kg+k 1jfV+ < bgj 2 p 2 #1=(p 2) : (27)

Hence, combine (26) and (27) implies that

V+ < b 1 p 2 S (N 2) 2 4 1=2 (p q)1=2(N 2) " 2 q kg+k 1 2 q p 2 kf+kLq p 2#1=2(p q)

which is a contradiction with (V 4). This completes the proof. Lemma 2.26 (i) For any u2 M+; [ M0; ; we have RRNfjuj

q

dx > 0: (ii) For any u2 M ; ; we have RRNgjuj

p

dx > 0: Proof. (i) For any u 2 M+; [ M0

; ;since M0; = ; implies that u 2 M + ; that is h00u(1) > 0: By (23) ; we have h00u(1) = (2 p) Z RNjruj 2 + V ; u2dx (q p) Z RN fjujqdx > 0 =) (2 p) Z RNjruj 2 + V ; u2dx > (q p) Z RN fjujqdx then by (18) ; implies that

Z RN fjujqdx > p 2 p q Z RNjruj 2 + V ; u2dx p 2 p q (N 2)2 4 (N 2)2 ! kuk2

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Hence, RRNfjuj q

dx > 0:

(ii)For any u 2 M ; that is h00

u(1) < 0: By (24) ;we have h00u(1) = (2 q) Z RN jruj 2 + V ; u2dx (p q) Z RN gjujpdx < 0 =) (2 q) Z RN jruj 2 + V ; u2dx < (p q) Z RN gjujpdx then by (18) ; implies that

Z RN gjujpdx > 2 q p q Z RNjruj 2 + V ; u2dx p 2 p q (N 2)2 4 (N 2)2 ! kuk2 Hence, RRNgjuj p dx > 0:

To obtain a better understanding of the Nehari manifold and …bering maps, we consider the function mu : R+! R de…ned by

mu(t) = t2 q Z RN jruj 2 + V ; u2dx tp q Z RN gjujpdx for t > 0: Clearly tu 2 M ; if and only if mu(t) =

R RNfjuj q dx. Moreover, m0u(t) = (2 q)t1 q Z RN jruj 2 + V ; u2dx (p q)tp q 1 Z RN gjujpdx (28) and so if tu 2 M ; ;then tq 1m0u(t) h00u(t) = (1 q) Z RNjruj 2 + V ; u2dx + (q 1) tq 2 Z RN fjujqdx + (p 1)tp 2 Z RN gjujpdx = (1 q) Z RNjruj 2 + V ; u2dx + (q 1) Z RNjruj 2 + V ; u2dx = 0 thus, tq 1m0 u(t) = h00u(t). Hence, tu 2 M + ; (or M ; ) if and only if m0u(t) > 0( or < 0):

Suppose u 2 Xn f0g with RRNgjuj p

dx > 0: Then, by (28), mu has a unique critical

point at t = tmax; (u) ;means that

m0u(t) = (2 q)t1 q Z RNjruj 2 + V ; u2dx (p q)tp q 1 Z RN gjujpdx = 0

(36)

implies that tmax; (u) = (2 q)R RNjruj 2 + V ; u2dx (p q)RRNgjuj p dx ! 1 p 2 > 0

and clearly muis strictly increasing on (0; tmax; (u))and strictly decreasing on (tmax; (u) ;1)

with limt!1mu(t) = 1. Moreover, if the functions f; g and V satisfy the conditions

(F 1) ; (G1) and (V 1) (V 3) ; and 0;then

mu(tmax; (u)) = tmax; (u)2 q Z RNjruj 2 + V ; u2dx tmax; (u)p q Z RN gjujpdx = (2 q) R RNjruj 2 + V ; u2dx (p q)RRNgjuj p dx !2 q p 2 Z RN jruj 2 + V ; u2dx (2 q)RRNjruj 2 + V ; u2dx (p q)R RNgjuj p dx ! Z RN gjujpdx = 2 q p q 2 q p 2 (2 q)R RNjruj 2 + V ; u2dx p q p 2 (p q)R RNgjuj p dx 2 q p 2 2 q p q p q p 2 (2 q)R RN jruj 2 + V ; u2dx p q p 2 (p q)RRNgjuj p dx 2 q p 2 = " 2 q p q 2 q p 2 2 q p q p q p 2# R RNjruj 2 + V ; u2dx p q p 2 R RNgjuj p dx 2 q p 2 = Z RNjruj 2 + V ; u2dx q=2 p 2 p q 2 q p q 2 q p 2 R RNjruj 2 + V ; u2dx p=2 R RNgjuj p dx !2 q p 2 > Z RN fjujqdx:

Thus, we have the following lemma.

Lemma 2.27 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 4) : Then, for each 0 and u 2 Xn f0g with

R RNgjuj p dx > 0; we have the following. (i) If RRNfjuj q

dx 0; then there is a unique t = t (u) > tmax; (u) such that t u 2

M ; and hu is increasing on (0; t ) and decreasing on (t ; 1). Moreover,

J ; t u = sup t 0

(37)

(ii) If R

RN fjuj q

dx > 0; then there are unique

0 < t+ = t+(u) < tmax; (u) < t = t (u)

such that t+u

2 M+; , t u 2 M ; , hu is decreasing on (0; t

+), increasing on (t+; t )

and decreasing on (t ; 1). Moreover, J ; t+u = inf

0 t tmax; (u)

J ; (tu) ; J ; t u = sup t t+

J ; (tu) : (30)

Proof. Fix u 2 Xn f0g with R

RNgjuj p

dx > 0: (i) SupposeRRN fjuj

q dx 0: Then, mu(t) = R RNfjuj q

dx has a unique solution t > tmax; (u).

such that mu(t ) = Z RN fjujqdx() t u 2 M ; () h0u t = 0 =) t q 1m0u t = h00u t < 0

and by h00t u(1) = (t )2h00u(t ) < 0: then we know that t u 2 M ; and h0u(t ) = 0.

Hence,hu has a unique critical point at t = t and h00u(t ) < 0. Thus (29) holds.

(ii) SupposeRRN fjuj q

dx > 0. Because mu(tmax; (u)) >

R RNfjuj q dx, the equation mu(t) = R RNfjuj q dx has ex-actly two solutions t+ < t

max; (u) < t

CaseI:

For t+2 (0; tmax; (u)) and mu(t) is strictly increasing on (0; tmax; (u)),that is

mu(t+) = Z RN fjujqdx() t+u2 M ; () h0u t+ = 0 =) t+ q 1m0u t+ = h00u t+ > 0 and by h00t+u(1) = (t+) 2

h00(t+) > 0: then we know that t+u

2 M+; :

CaseII:

For t 2 (tmax; (u) ;1) and mu(t)is strictly decreasing on (tmax; (u) ;1),that is

mu(t ) = Z RN fjujqdx() t u 2 M ; () h0u t = 0 =) t q 1m0u t = h00u t < 0 thus, t u 2 M ; and by h 00 t u(1) = (t ) 2

h00u(t ) < 0: then we know that t u 2 M ; : Hence, there has critical points at t = t+and t = t with h00u(t+) > 0and h00u(t ) < 0:

Thus, hu is decreasing on (0; t+) ; increasing on (t+; t ) and decreasing on (t+;1) :

(38)

We remark that it follows from Lemma 2.25 that M ; = M+; [ M ;

for all > 0. Furthermore, by Lemma 2.27, it follows that M+; and M ; are

non-empty and, by Lemma 2.23, we may de…ne

+

; = inf u2M+;

J ; (u) and ; = inf u2M ;

J ; (u) :

Moreover, we have the following result.

Theorem 2.28 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 4) : Then, for each 0; there exists bC0 > 0 such that +; < 0 < bC0 <

; :

In particular, +; = infu2M ; J ; (u) :

Proof. (i) Let u 2 M+; (i.e.h00

u(1) > 0) M ; : Then, by (23) ; h00u(1) = (2 p) Z RNjruj 2 + V ; u2dx (q p) Z RN fjujqdx > 0 =) (2 p) Z RNjruj 2 + V ; u2dx > (q p) Z RN fjujqdx =) p 2 p q Z RNjruj 2 + V ; u2dx < Z RN fjujqdx: (31) Thus, by (18) and (31) ; J ; (u) = p 2 2p Z RNjruj 2 + V ; u2dx p q pq Z RN fjujqdx < p 2 2p Z RNjruj 2 + V ; u2dx p q pq p 2 p q Z RNjruj 2 + V ; u2dx = (2 q) (p 2) 2pq Z RNjruj 2 + V ; u2dx < (2 q) (p 2) (N 2) 2 4 2pq (N 2)2 kuk 2 < 0 and so infu2M ; J ; (u) infu2M+

; J ; (u) < 0:

(ii) Let u 2 M ; : Then, by(21) ; J ; (u) p 2 2p 1 4 (N 2)2 kuk 2 p q pq V +< b q(22 pp) S q kf+kLq kuk q (32)

(39)

then by (24) u 2 M ; (i.e. h00u(1) < 0) ;we have h00u(1) = (2 q) Z RN jruj 2 + V ; u2dx (p q) Z RN gjujpdx < 0 =) 2 q p q Z RNjruj 2 + V ; u2dx < Z RN gjujpdx: (33)

and by(18) and (33) and the Sobolev inequality, (2 q) (N 2)2 4 (p q) (N 2)2 kuk 2 2 q p q Z RNjruj 2 + V u2dx < Z RN gjujpdx g+ 1 V + < b 22 pS p kukp; which implies kuk > " (N 2)2 4 (2 q) Sp (N 2)2(p q)kg+k 1jfV+< bgj 2 p 2 #1=(p 2) for all u 2 M ; : (34)

Then we substitute (34) into (32) ; we have J ; (u) kukq " (p 2) (N 2)2 4 2p (N 2)2 kuk 2 q p q pq V + < b q(2 p) 2 p S q kf+kLq # > " (2 q) (N 2)2 4 Sp (N 2)2(p q)kg+k 1jfV+ < bgj 2 p 2 #q=(p 2) 0 @p 2 2p " (N 2)2 4 (2 q) Sp (N 2)2(p q)kg+k 1jfV+ < bgj 2 p 2 #2 q p 2 p q pqSq V + < b q(2 p) 2 p kf+kLq 1 A : Thus, using the condition (V 4) and if 0;then

; > bC0 for some bC0 > 0: and we have inf u2M ; J ; (u) = inf u2M+; [M ; J ; (u) = inff inf u2M+; J ; (u) ; inf u2M ; J ; (u)g = inf u2M+; J ; (u) = +; : This completes the proof.

(40)

Lemma 2.29 For each u 2 M ; , there exists > 0 and a di¤erentiable function

: B (0; ") Xn f0g ! R such that (0) = 1; the function (v) (u v)2 M ; and

h 0(0) ; vi = 2 R RNrurv + V ; uvdx q R RNfjuj q 2 uvdx pRRNgjuj p 2 uvdx (2 q) RRNjruj 2 + V ; u2dx + (q p) R RNgjuj p dx ; (35) for all v 2 Xn f0g :

Proof. For u 2 M ; ;de…ned a function F : R Xn f0g ! R by

Fu( ; w) = J0; ( (u w)) ; (u w) = 2 Z RNjr (u w)j2+ V ; (u w) 2 dx q Z RN fju wjqdx p Z RN gju wjpdx Then Fu(1; 0) = J 0 ; (u) ; u = 0 and d d Fu(1; 0) = 2 Z RN jruj 2 + V ; u2dx q Z RN fjujqdx p Z RN gjujpdx = (2 q) Z RN jruj 2 + V ; u2dx + (q p) Z RN gjujpdx6= 0 and using L0Hôpital’s rule

lim t!0 R RNfju (w + tv)j q dx R RNfju wj q dx t = lim t!0q Z RN fju (w + tv)jq 2[u (w + tv)] ( v) dx = q Z RN fju wjq 2(u w) vdx; (36) Similarly, lim t!0 R RNgju (w + tv)j p dx R RNgju wj p dx t = p Z RN gju wjp 2(u w) vdx; (37)

(41)

then using Gâteaux derivative, (36) and (37) d dwFu( ; w) = lim t!0 Fu( ; w + tv) Fu( ; w) t = lim t!0 1 t 2 Z RNjr (u (w + tv))j2+ V ; (u (w + tv))2dx q Z RN fju (w + tv)jqdx p Z RN gju (w + tv)jpdx 2 Z RNjr (u w)j2+ V ; (u w) 2 dx + q Z RN fju wjqdx + p Z RN gju wjpdx = 2 2 Z RNjr (u w)j jrvj + V ; (u w) vdx q q Z RN fju wjq 2(u w) vdx p p Z RN gju wjp 2(u w) vdx: Therefore, we have d dwFu(1; 0) = 2 Z RN jruj jrvj + V ; uvdx q Z RN fjujq 2uvdx p Z RN gjujp 2uvdx: According to the implicit function theorem , there exists > 0 and a di¤erentiable function : B (0; ") Xn f0g ! R such that (0) = 1;

h 0(0) ; vi = 2 R RNrurv + V ; uvdx q R RNfjuj q 2 uvdx pRRNgjuj p 2 uvdx (2 q) R RNjruj 2 + V ; u2dx + (q p) R RNgjuj p dx and Fu( (v) ; v) = 0 for all v 2 B (0; ") ; which is equivalent to J0; ( (v) (u v)) ; (v) (u v) = 0 for all v 2 B (0; ") ; that is, (v) (u v)2 M ; :

Lemma 2.30 For each u 2 M ; ;there exist > 0 and a di¤erential function : B (0; ") Xn f0g ! R such that (0) = 1; the function (v) (u v)2 M ; and

D 0 (0) ; vE= 2 R RNrurv + V ; uvdx q R RNfjuj q 2 uvdx pR RNgjuj p 2 uvdx (2 q) R RNjruj 2 + V ; u2dx + (q p) R RNgjuj p dx ; (38) for all v 2 Xn f0g :

(42)

Proof. Similar to the argument in Lemma 2.29, there exist > 0 and a di¤erential function : B (0; ") Xn f0g ! R such that (0) = 1; the function (v) (u v)2 M ; for all v 2 Xn f0g. Since u 2 M ; means that h00u(1) < 0;that is

h00u(1) =kuk2 Z RN V u2dx (q 1) Z RN fjujqdx (p 1) Z RN gjujpdx < 0: Thus, by the continuity of the function , then we have

h00 (v)(u v)(1) = (v) (u v) 2 Z RN V (v) (u v) 2dx (q 1) Z RN f (v) (u v) qdx (p 1) Z RN g (v) (u v) pdx < 0

if su¢ ciently small,this implies that (v) (u v)2 M ; : Proposition 2.31 For 0; then

(i) there exists a minizing sequence fung M ; such that

J ; (un) = +; + o (1) ;

J0; (un) = o (1) in X 1;

(ii) there exists a minizing sequence fung M ; such that

J ; (un) = ; + o (1) ;

J0; (un) = o (1) in X 1:

Proof. (i) By Lemma 2.23 and Eklend variational principle.There exists a minizing sequence fung M ; such that

J ; (un) < +; + 1 n (39) and J ; (un) < J ; (w) + 1 n kw unk for each w 2 M ; (40) By taking n large, from Theorem 2.28 , we have

J ; (un) = p 2 2p Z RNjru nj2+ V ; u2ndx p q pq Z RN fjunjqdx < +; + 1 n < ; 2 < 0: (41)

(43)

This implies V+ < b q(2 p) 2 p S q kf+kLq kuk q Z RN fjunj q dx > pq p q ; 2 > 0: (42) Consequently, un6= 0 and by(42) we have

kuk > ppqq 2; V+ < b q(2 p) 2 p Sq kf+kLq1 1 q ; (43)

and putting together (41) (42) and (18), we obtain

p 2 2p (N 2)2 4 (N 2)2 kuk 2 ! < p 2 2p Z RNjru nj 2 + V ; u2ndx < p q pq Z RN fjunj q dx < p q pq V + < b q(2 p) 2 p S q kf+kLq kuk q ; implies that kuk < " p 2 2p (N 2)2 4 (N 2)2 ! p q pq V + < b q(2 p) 2 p S q kf+kLq # 1 2 q : (44) Now we will show that

J0; (un)

X 1 ! 0 as n ! 1:

Applying Lemma 2.29 with unto obtain the function n: B (0; ")! R+for some n > 0;

such that n(w) (un w)2 M ; :Choose 0 < < n:Let u 2 Xn f0g with u 6 0 and let

w = kuku: We set = n(w ) (un w ) : Since 2 M ; , we deduce from (40) that

J ; J ; (un)

1

n un ;

and by mean value theorem , we have D J0; (un) ; un E + o un 1 n un (45) and D J0; (un) ; un E = DJ0; (un) ; n(w ) (un w ) un E = DJ0; (un) ; n(w ) (un w ) un w + w E = DJ0; (un) ; w E +DJ0; (un) ; n(w ) (un w ) (un w ) E = D J0; (un) ; w E + ( n(w ) 1) D J0; (un) ; (un w ) E

(44)

then (45) become to D J0; (un) ; w E + ( n(w ) 1)DJ0; (un) ; (un w ) E 1 n un + o un (46) From n(w ) (un w )2 M ; we have J0; ( n(w ) (un w )) ; n(w ) (un w ) = 0 implies that D J0; ; (un w ) E = 0 substitute into (46) ; it follows that

J0; (un) ; u kuk + ( n(w ) 1) D J0; (un) J 0 ; ; (un w ) E 1 n un + o un : Thus, J0; (un) ; u kuk un n + o un +( n(w ) 1) DJ0; (un) J 0 ; ; (un w ) E ; (47) since un = k n(w ) (un w ) unk = k n(w ) (un) n(w ) w unk = kun[ n(w ) 1] n(w ) w k j n(w ) 1j kunk + j n(w )j ; and lim !0 j n(w ) 1j = lim !0 j n(w ) n(0)j lim !0 j n(0 + ) n(0)j = 0 n(0) :

If we let ! 0 in (47) for a …xed n, then by (44) we can …nd a constant C > 0; which is independent of ; such that

J0; (un) ; u kuk C n 1 + 0 n(0) :

(45)

We are done once we show that 0n(0) is uniformly bounded in n. By(35) ; (44) and Hölder inequality, we have

h 0(0) ; vi bkvk (2 q) R RNjruj 2 + V ; u2dx + (q p) R RNgjuj p dx for some b > 0: We only need to show that

(2 q) Z RNjruj 2 + V ; u2dx + (q p) Z RN gjujpdx > c (48) for some c > 0 and n large.We argue by way of contradiction.Assume that there exists a subsequence fung such that

(2 q) Z RNjru nj2+ V ; u2ndx + (q p) Z RN gjunjpdx = o (1) : (49)

Combing (49) with (43) ;we can …nd a suitable constant d > 0 such that Z

RN

gjunjpdx d for n su¢ ciently large. (50)

In addition, (49) ; and the fact that un2 M ; also give

Z RN fjunjqdx = Z RNjru nj2+ V ; u2ndx Z RN gjunjpdx = p 2 2 q Z RN gjunjpdx + o (1) ; (51) and by (18) we have that

(N 2)2 4 (N 2)2 kuk 2 Z RNjru nj2+ V ; u2ndx = Z RN gjunj p dx + Z RN fjunj q dx = 2 q p 2 Z RN fjunj q dx + Z RN fjunj q dx + o (1) = p q p 2 Z RN fjunjqdx + o (1) V+ < b q(2 p) 2 p S q kf+kLq kuk q + o (1) Thus, kuk " (N 2)2 (N 2)2 4 ! V+ < b q(2 p) 2 p S q kf+kLq # 1 2 q : (52)

(46)

Let I ; : M ; ! R and K (p; q) = 2 qp q p 1 p 2 p 2 2 q , then we have I ; (un) = K (p; q) R RNjrunj2+ V ; u2ndx p 1 R RNgjunj p dx ! 1 p 2 Z RN fjunjqdx = 2 q p q p 1 p 2 p 2 2 q 0 B @ p q 2 q p 1 R RNgjunjpdx p 1 R RNgjunj p dx 1 C A 1 p 2 p 2 2 q Z RN gjunjpdx + o (1) = o (1) : (53)

However,by (50)and (52). We get

I ; (un) = K (p; q) R RNjrunj 2 + V ; u2ndx p 1 R RNgjunjpdx ! 1 p 2 Z RN fjunj q dx K (p; q) 0 B @ (N 2)2 4 (N 2)2 p 1 kunk2(p 1) S pkg+k 1jfV+ < bgj 2 p 2 ku nkp 1 C A 1 p 2 V+< b q(2 p) 2 p S q kf+kLq kunkq = K (p; q) (N 2) 2 4 (N 2)2 !p 1 p 2 S p p 2 g+ 1 p 2 1 V + < b 2 (p(2 2)p) kunk V+< b q(2 p) 2 p S q kf+kLq kunkq = kunkq 8 < :K (p; q) (N 2)2 4 (N 2)2 !p 1 p 2 S p p 2 g+ 1 p 2 1 V +< b 2 (p(2 2)p) kunk1 q V+ < b q(2 p) 2 p S q kf+kLq kunk q 8 > < > :K (p; q) 2 4 (N 2) 2 4 (N 2)2 !p 1 Sp g+ 1 1 V +< b ( 2 p) 2 3 5 1 p 2 " (N 2)2 (N 2)2 4 ! V+< b q(2 p) 2 p S q kf+kLq #1 q 2 q V+ < b q(2 p) 2 p S q kf+kLq 6= o (1) :

(47)

J0; (un) ; u kuk C n 1 + 0 n(0) :

We have that it contradicts to (53). Consequently,this show that fung is a (PS) +

-sequence for J ; .

(ii) Similarly, we can use Lemma 2.30 to prove (ii). We will omit deltailed proof here.

3

Proof of Theorem 1.1

Theorem 3.1 Suppose that 2 < p < 2 and the functions f; g and V satisfy the con-ditions (F 1) ; (G1) and (V 1) (V 5) : Then for each 0; the functional J ; has a

minimizer u+; 2 M+; M ; and it satis…es

(i) J ; u+ = + = infu2M ; J ; (u) < 0;

(ii) u+ is a positive solution of equation (E ; ) :

Proof. By Theorem 2.28 and the Ekeland variational principle [26] (or see Wu [40, Proposition 1]), there exists fung M+; such that it is a (PS) +–sequence for J ; .

Moreover, Lemma 2.23, fung is bounded in X : Therefore, there exist a subsequence

fung and u+ in X such that

un * u+ weakly in X ;

un ! u+ strongly in Lrloc R

N ; for 2 r < 2 :

Moreover, J0; u+ = 0: First, we claim that u+ 6 0: Suppose the contrary, then by (18) ; the condition (F 1) ; the Egorov theorem and the Hölder inequality, we have

Z

RN

fjunjqdx! 0 as n ! 1;

which implies that Z RNjru nj 2 + V ; u2ndx = Z RN gjunjpdx + o (1) and J ; (un) = 1 2 Z RNjru nj2+ V ; u2ndx 1 q Z RN fjunjqdx 1 p Z RN gjunjpdx (p 2) (N 2)2 4 2p (N 2)2 kunk 2 + o (1) ;

(48)

this contradicts limn!1J ; (un) = ; < 0: Thus, by Lemma 2.26 (i) ;

R

RN f u

+ q > 0:

In particular, u+ is a nontrivial solution of Equation (E ) : Now we prove that un! u+

strongly in X : Suppose the contrary, then by (18) ; Z RN ru + 2+ V+ u+ 2dx < lim inf Z RNjru nj2+ V+u2ndx; and Z RN fjunjqdx! Z RN f u+ qdx; so inf un2M ; J ; (un) J ; u+ = 1 2 1 p Z RN ru + 2+ V ; u+ 2dx 1 q 1 p Z RN f u+ qdx < lim inf n!1 1 2 1 p Z RNjru nj 2 + V ; (un) 2 dx 1 q 1 p n!1lim Z RN fjunj q dx = lim inf n!1 1 2 1 p Z RNjru nj 2 + V ; (un) 2 dx 1 q 1 p Z RN fjunj q dx < lim n!1J ; (un) = +:

This contradicts infu2M ; J ; (u) =

+. Consequently, u

n ! u+ strongly in X and

J ; u+ = +:

Moreover,we have u+ 2 M+; . In fact,if u+ 2 M ; , by Lemma 2.27, there are unique t+0 and t0 such that t

+ 0u + 2 M+; and t0u + 2 M ; ,we have t + 0 < t0 = 1. Since d dthu+ t + 0 = d dtJ ; t + 0u+ = 0 and d2 dt2hu+ t + 0 = d2 dt2J ; t + 0u+ > 0;

there exists t and t+0 < t < t0 such that J ; t+0u + < J ; tu+ . By Lemma 2.27, J ; t+0u + < J ; tu+ J ; t0u + = J ; u+ ;

which contradicts to local minimizer u+. Because J ; u+ = J ; u+ and u+ 2

M+; ; by Lemma 2.24 and Theorem 2.28, we may assume that u+ is a positive solution of Equation (E ) and J ; u+ = += infu2M ; J ; (u) < 0:This completes the proof.

Proposition 3.2 Suppose that 2 < p < 2 and the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 2) : Then for each D > 0 there exists 0 = (D) > 0

(49)

Proof. Let fung be a (P S) -sequence with < D: By Lemma 2.23, there exists C

such that kunk C : Thus, there exist a subsequence fung and u0 in X such that

un * u0 weakly in X ;

un ! u0 strongly in Lrloc R

N for 2 r < 2 :

Moreover, J ; (u0) = 0: Thus, by the condition (F 1), Hölder inequality and Egorov’s

theorem, then

8 > 0; 9 > 0, suppose that there exists a set E is a subset of RN

and jEj < 1, (a) Since u0 is …nit a.e.in E and uqn ! u

q

0 by Egorov’s theorem there exists a close subset

F E such that jEnF j < and uqn ! uq0 on F , so we have jjunj q ju0j q j < (b) If kfkq <1; where q = p p q and q > 0, thenR Ecjfj q < q ; (c)* f 2 Lq ) jfj 2 L1 then there exist > 0 such that R

EnF f q < q ; 8 jEnF j < Z RN fjunjqdx Z RN fju0jqdx Z RNjf ju nj q fju0j q j dx = Z F fjjunjq ju0jqj dx + Z EnF fjjunjq ju0jqj dx + Z Ec fjjunjq ju0jqj dx Z F f dx +kjunjq ju0jqkp Z EnF jfjq 1 q +kjunjq ju0jqkp Z Ecjfj q 1 q jF jp1 kfk q + 2C + 2C jEjp1 kfk q + 4M : Hence, we have Z RN fjunjqdx! Z RN fju0jqdx (54)

Now, we prove that un ! u0 strongly in X : Let vn= un u0:It follows from (V 2) that

Z RN v2ndx = Z fV+ bg vn2dx + Z fV+<bg vn2dx 1 bkvnk 2 + o (1) : Then, by the Hölder and Sobolev inequalities, we have

Z RN jv njpdx Z RNjv nj2dx 2 p 2 2 Z RNjv nj2 dx p 2 2 2 1 bkvnk 2 2 p 2 2 S 2 Z RNjrv nj2dx 2 2 !p 2 2 2 + o (1) 1 b 2 p 2 2 S 2 (p 2) 2 2 kvnkp + o (1) : (55)

(50)

and by (54) ; we let hn= fjun u0jq (a) hn ! 0; (b) let n:= 2q 1f (ju0j q junj q )! := 2pf ju0j q ; we have n! ; (c) hn n :

Since (a) to (c) and by Generalized Lebesgue Dominated Convergence Theorem,we have Z

jhn 0j ! 0 as n ! 1;

that implies Z

RN

fjvnjqdx! 0:

By the conditions (F 1) ; (G1) and Brezis-Lieb Lemma, we have lim n!1 Z junjp+ Z jun u0jp = Z ju0jp; implies J ; (u0) = lim n!1[J ; (un) J ; (vn)] : Hence, J ; (vn) = J ; (un) J ; (u0) + o (1) and J ; (vn) = o (1) :

Consequently, by this together with (54) and Lemma 2.23, we obtain 1 2 1 p Z RN gjvnjp = 1 2 1 p Z RN gjvnjpdx 1 q 1 2 Z RN fjvnjqdx + o (1) = J ; (vn) 1 2 D J0; (vn) ; vn E + o (1) = J ; (u0) + o (1) D + K0+ o (1) ; where K0 = 2 q 2pq (N 2)2 (N 2)2 4 (p 2) !q=(2 q)0 @(p q)jfV+ < bgj q(2 p) 2 p kf +kLq Sq 1 A 2=(2 q) and so Z RN gjvnj p 2p p 2(D + K0) + o (1) : (56)

(51)

Because J0; (vn) ; vn = o (1) ; it follows from (18) ; (55) and (56) that o (1) = Z RN jrv nj2+ V vn2 dx Z RN gjvnjpdx (N 2)2 4 (N 2)2 kvnk 2 Z RN gjvnjpdx (p 2)=p Z RN gjvnjpdx 2=p (N 2)2 4 (N 2)2 kvnk 2 kgk2=p1 Z RN gjvnj p dx (p 2)=p Z RNjv nj p dx 2=p kvnk2 2 6 4(N 2) 2 4 (N 2)2 kgk 2 p 1 2p (D + K0) p 2 (p 2)=p0 @ 1 b 2 p 2 2 S 2 (p 2) 2 2 1 A 2 p3 7 5 + o (1) : Thus, there exists = (D0) > 0 such that vn ! 0 strongly in X for > : This

completes the proof.

Note that if the functions f; g and V satisfy the conditions as in Theorem 1.1, then by Lemma 2.27, we may choose 2 C1

0 ( g) ;which the function

h (t) = J ; (t ) = t2 2 Z g jr j2dx t q q Z g fj jqdx t p p Z g gj jpdx

have t0 > 0 and D0 > 0 are independent of such that t0 2 M ; for all 0 and

sup

t 0

h (t) = h t0 = D0 > 0;

this implies D0 for 0: Moreover, we have the following result.

Theorem 3.3 Suppose that the functions f; g and V satisfy the conditions as in Theo-rem 1.1. Then for each > 0; the functional J ; has a minimizer u ; in M ; and it

satis…es

(i) J ; u ; = ; ;

(ii) u ; is a positive solution of equation (E ; ) :

Proof. By the Ekeland variational principle [26] (or see Wu [40, Proposition 1]), there exists fung M ; such that it is a (PS) –sequence for J ; . Then, by Proposition 3.2,

there exists a subsequence fung and u 2 M ; is a non-zero solution of equation (E ) ;

such that un ! u strongly in X and J ; u = : Because J ; u = J ; u

and u 2 M ; ; by Lemma 2.24 we may assume that u is a positive solution of equation (E ) :

(52)

We can now complete the proof of Theorem 1.1: by Theorems 2.28, 3.1 and 3.3, Equation (E ) has two positive solutions, u+; ; u ; ; such that u+; 2 M+; and u ; 2 M ; with

J ; u+; = +

; < 0 < bC0 < J ; u ; = ; :

This completes the proof of Theorem 1.1.

References

[1] Iddrisu Mohammed Muniru , Okpoti Christopher Adjei, and Gbolagade Kazeem Alagbe, SOME PROOFS OF THE CLASSICAL INTEGRAL HARDY INEQUAL-ITY, Korean J. Math. 22 (2014), No. 3

[2] G. H. Hardy, Note on a theorem of Hilbert, Math. Z. 6 (1920), 314-317.

[3] G. Talenti, Best constant in Sobolev inequality, Ann. Mat. Pura Appl. 110 (1976), 353-372

[4] R.L.Wheeden and A.Zygmund: Measure and Integral - An introduction to Real Analysis, Marcel Dekker, 1977.

[5] M.Willem,Minimax Theorems,Birkhäuser,Boston,1996

[6] M.Struwe,Variational methods (Applications to nonlinear PDE and Hamiltonian systems), Springer-Verlag, 1990

[7] Saleh Almezel, Qamrul Hasan Ansari, Mohamed Amine Khamsi,Topics in Fixed Point Theory,Springer Science & Business Media, 2013

[8] David G. Costa,An Invitation to Variational Methods In Di¤erential Equa-tions,Springer Science & Business Media,2010

[9] Hwai-Chiuan Wang,Nonlinear Analysis,National Tsinf Hua University,2003

[10] Landau, E.: A note on a theorem concerning series of positive terms. J. Lond. Math. Soc. 1,38–39 (1926)

[11] A. A. Balinsky, W. D. Evans, and R. T. Lewis. The analysis and geometry of Hardy’s inequality.Universitext. Springer, Cham, 2015.

[12] A. Ambrosetti, G. J. Azorero, I. Peral, Multiplicity results for some nonlinear elliptic equations, J. Funct. Anal. 137 (1996), 219–242.

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