National University of Kaohsiung Repository System:Item 310360000Q/10841

國立高雄大學應用數學系碩士班

碩士論文

半線性橢圓方程中加權函數的影響

Effect of Weight Function for a Semilinear Elliptic Equation

研究生：戴偉恭撰

指導教授：吳宗芳

致謝詞

首先要感謝我的指導教授吳宗芳教授，在高雄大學的這兩年期間，都是跟著 吳老師在學習；老師採用的是美式風格，不太過問我們準備的如何，而每當我們 有問題時，再去找老師討論，會有適時的引導，讓我們知道該怎麼去解決，而往 往得到的也是更多相關的書籍或文章來閱讀。也因為老師這樣的方式，我們的知 識不會只侷限於某一個方向，而是自己組織而成，屬於自己的知識。 與我同門的陳奕廷同學，本身是一個對數學十分有熱忱的人，他總是不斷的 去充實各種相關的數學知識，而他對數學界的人物更是瞭如指掌，有一些大人物 來學校演講時，他都會告訴我演講者的相關資料。我在學習的過程中，由於相關 的知識以及基礎不太扎實，往往遇到許多簡單的問題，是得繁雜的去查閱基礎科 目的書，而陳奕廷卻能立刻告訴我該去找哪些方向；我在面對的許多問題，也因 為有他，而可以與他探究，藉著他豐富的知識以及相關的想法，來解決相關方面 的問題。 我在高大這兩年的同學們，感謝吳信宏同學、陳婉伶同學、曾詩婷同學、林 常宇同學、蔡宗穎同學以及童若雯同學，我們一起相互扶持，讀書增進知識，也 偶爾透過一些娛樂來放鬆，讓在高大的這兩年過的很快樂。 而我也要感謝我的口委們，吳宗芳教授、陳晴玉教授以及陳冠如教授，因為 有你們，讓我的論文能更精進，知道如何改進。 最後我要感謝我的家人們，在這兩年內的支持與鼓勵，我才能順利完成碩士 的學業。

Effect of Weight Function for a

Semilinear Elliptic Equation

by

Wei-kung Tai

Advisor

Tsung-fang Wu

Department of Applied Mathematics,

National University of Kaohsiung

Kaohsiung, Taiwan 811, R.O.C.

July 2009

i

Contents

1. Introduction 1

2. Notations and Preliminaries 2

3. Estimate of energy 9

ii

半線性橢圓方程中加權函數的影響

指導教授：吳宗芳 教授 國立高雄大學應用數學系 學生：戴偉恭 國立高雄大學應用數學系碩士班 摘要 本論文中，我們觀察半線性橢圓方程中具有變號的加權函數裡，結合了凹凸 非線性項的解個數。透過 Nehari 流形，我們證明出在這有界的值域下，方程

( )

Eλ,μ _{至少會有兩個解。} 關鍵字：半線性、Nehari 流形.

iii

Effect of Weight Function for a Semilienear Elliptic

Equation

Advisor: Dr. Tsung-fang Wu Institute of Applied mathematics National University of Kaohsiung

Student: Wei-kung Tai Institute of Applied mathematics National University of Kaohsiung

ABSTRACT

In this thesis, we study the combined effect of concave and convex nonlinearities on the number of solutions for a semilinear elliptic equation with sign-changing weight functions. With the help of the Nehari manifold, we prove that there are at least two solutions for the equation

( )

Eλ,μ in bounded domains.

1

Introduction

1_{In this thesis, we consider the multiplicity results of solutions of the following}

sub-superlinear elliptic equation: 8 < : ¡¢ = ()jj¡2 + ()jj¡2 in ­ _{¸ 0  6´ 0} in ­  = 0 on ­ ()

where ­ is a smooth bounded domain in R_{ 1    2    2}¤ ₍₂¤ ₌ 2

 ¡2 if  ¸ 3 2¤ _{=1 if  = 1 2)  = (}

+ ¡)   =

¡

₊ _¡¢ with _§ _{§¸ 0}

and the weight functions   are satisfying the following conditions: () _() = ₊+_()

¡ ¡¡() with §() = maxf§ ()  0g 6´ 0 and

 ()_{2 }¡_­¢_{, }

= _¡ ,  2 ( 2¤);

() () = ++() ¡ ¡¡() with §() = maxf§ ()  0g 6´ 0 and

 ()_{2 }¡_­¢_{, }

 = _¡ ,  2 ( 2¤)

The fact that the number of positive solutions of equation () is af-fected by the concave and convex nonlinearities has been the focus of a great deal of research in recent years. If the weight functions  ´  ´ 1 Ambrosetti-Brezis-Cerami [1] have investigated equation (11)  They found

that there exists 0  0 such that equation (11) admits at least two

pos-itive solutions for  2 (0 0)  has a positive solution for  = 0 and no

positive solution for   0 Wu [8] proved that equation () has at least two positive solutions under the assumptions the weight functions  change sign in ­  ´ 1 and  is su¢ciently small. For more general results, de Figueiredo-Grossez-Ubilla [4] proved the following:

Theorem 1 Assume that the conditions () and () hold, and in addition

(1) there exists a nonempty open subset ­1 ½ ­ such that, on ­1  ()¸ 1

for some ₁  0 and  () is bounded from below;

(2) there exists a nonempty open subset ­2 ½ ­ such that, on ­2  ()¸ 2

for some ₂  0 and  () is bounded from below.

Then there exists   0 such that if  2 ¡0 ¢ then equation () has at 1_{Key words: Semilinear elliptic equations, Nehari manifold, Concave-convex}

nonlinear-ities

least two solutions. 8 < : ¡¢ =  () jj¡2 +  ()_jj¡2 in ­ _{¸ 0  6´ 0} in ­  = 0 on ­ ()

The main purpose of this a new method to improve Theorem 1. In partic-ular, we do this without assuming the conditions (1) and (2)  Our main result is the following.

Theorem 2 Assume that the conditions ()  () and Lemma 6 hold. Then equation () has at least two solutions.

Among the other interesting problems which are similar to equation (_) for  = 1, Brown-Zhang [2] have investigated the following equation:

½

¡¢ =  ()  +  () jj¡1 in ­

_{2 }1 0(­) 

(1) where ­ is a bounded domain in R

and   : ­ ! R are smooth functions which change sign in ­ They found existence and non-existence results for positive solutions of equation (1) as  changes.

This thesis is organized as follows. In section 2, we give some notations and preliminaries. In section 3, we prove that equation () has at least two solutions for  su¢ciently small.

2

Notations and Preliminaries

Associated with equation ()  we consider the energy functional , for each  2 1 0 (­)  _() = 1 2 Z ­ jrj2_¡ 1  Z ­ _jj_¡1  Z ­ _jj

It is well-known that the solutions of equation () are the critical points of the energy functional  (see Rabinowitz [6]).

As the energy functional is not bounded below on 01(­) it is useful

to consider the functional on the Nehari manifold M =

©

 _{2 0}1(­)_{n f0g j} ­_0 ()  ® = 0ª

Thus,  2 M if and only if kk21 ¡ Z ­ jj¡ Z ­ jj = 0

Note that M contains every nonzero solution of problem () 

We denote by  the best Sobolev constant for the imbedding of 01(­)

into (­) , 1    2¤is which is given

= inf 201(­) kk21 ¡R jj¢2  0 In particular, µZ ­jj   ¶1  · ¡ 1 2  kk_1 for all  2 ₀1(­) 

And we denote that

 = inf

12¤fg 

Moreover, we have the following results.

Lemma 3 The energy functional is coercive and bounded below on M Proof. _{For  2 M}we have kk2_1 =

R

­jj



 +R_­jjThen by the Sobolev and Hölder inequalities,

() = _{¡ 2} 2 kk 2 1¡ µ _{¡ }  ¶ Z ­ jj   ¸ _2¡ 2kk21¡ + µ _{¡ }  ¶ Z ­ +_jj ¸ _2¡ 2kk21¡ + µ _{¡ }  ¶ ° °+°°  ¡2 kk 1 , (2) for all +  0 +  0 ¡ ¸ 0 ¡¸ 0

Thus,  is coercive and bounded below on M De…ne _() =­_0 ()  ®=_kk2_1 ¡ Z ­ jj¡ Z ­ jj 3

Then for  2 M ­ 0_()  ® = 2_kk2_1 ¡  Z ­ jj¡  Z ­ jj = (2_{¡ ) kk}2_1 ¡ ( ¡ ) Z ­ jj   (3) = (2_{¡ ) kk}2_1 ¡ ( ¡ ) Z ­ jj (4)

Now we split M into three parts: M+_ = ©_{2 M} j ­ 0_()  ® 0ª M0_ = ©_{2 M} j ­ 0_()  ®= 0ª M¡_ = ©_{2 M} j ­ 0_()  ® 0ª

Then, we have the following results. Lemma 4 We have () If  _{2 M}+_ then R_­jj    0; () If _{2 M}0  then R ­jj    0 andR_­jj  0; () If  _{2 M}¡_ then R_­jj    0

Proof. The proof is immediate from (3) and (4) 

Lemma 5 Suppose that ₀ is a local minimizer for _ on M_ and that 0 2 M 0 Then 0 (0) = 0 in ¡1(­) 

Proof. If 0 is a local minimizer for  on M, then 0 is a solution of

the optimization problem,

minimize () subject to () = 0

Hence, by the theory of Lagrange multipliers, there exists  2 R such that 0 (0) =  0(0) Thus ­ _0 (0) 0 ® =  ­0_(0)  0 ®  (5) Since 0 2 M, ­ 0 (0) 0 ® = 0 _{and so kk}2_1 ¡ R ­jj   _¡ R ­jj   = 0 Hence, ­ 0_(0)  0 ® = (2_{¡ ) kk}2_1 ¡ ( ¡ ) Z ­ jj 4

Thus, if 0 2 M 0 ­

0_(0)  0

®

6= 0 and so by (5)  = 0 Hence the proof is complete.

Note that, if  2 M0

 by (3)  (4) and the Sobolev and Hölder inequali-ties, then kk21 = _{¡ } _{¡ 2} Z ­ jj· + µ _{¡ } _{¡ 2} ¶ ° °+°_°  ¡2 kk 1 and kk21 = _{¡ } 2_{¡ } Z ­ jj· + µ _{¡ } 2_{¡ } ¶ ° °+°_°  ¡2 kk 1 This implies µ ₊ µ _{¡ } 2_{¡ } ¶ ° °+°_° ¡  2 ¶ 1 2¡ · kk1 · µ + µ _{¡ } _{¡ 2} ¶ ° °+°_°  ¡  2 ¶ 1 2¡  for all  2 M0

 and + +  0Thus, if the submanifold M0 is nonempty, then the inequality

¡2₊ 2¡_{+ ¸} µ _{¡ 2} k+_k  ¶¡2µ ₂ ¡  k+_k  ¶2¡µ_ ¡   ¶¡

must hold and so, we have the following result. Lemma 6 If 0  ¡2₊ 2¡₊  ³ ¡2 k+_k  ´¡2³ 2¡ k+_k  ´2¡¡ ¡  ¢¡ , then the submanifold M0 =; for all ¡ ¸ 0 ¡ ¸ 0

For each  2 01(­) with

R ­jj    0 we write max= Ã (2_{¡ ) kk}2_1 (_{¡ )}R_­_jj ! 1 ¡2  0

Then the following lemma hold.

Lemma 7 _{For each  2 0}1(­) with R_­jj  0 we have

() if R_­jj· 0 then there is a unique ¡  max such that ¡2 M¡_

and

_¡¡_¢_{= sup}

¸0

_() ; 5

() if R_­jj  0 then there are unique 0  +  max  ¡ such that +_ 2 M+ ¡2 M¡ and  ¡ +¢= inf 0··max ()   ¡ ¡¢= sup ¸max ()  Proof. _{Fix  2 0}1(­) with R_­jj  0. Let

 () = 2¡_kk2_1¡ ¡

Z

­

jj for  ¸ 0

We have (0) = 0 and () ! ¡1 as  ! 1  () achieves its maximum at

max increasing for  2 [0 max) and decreasing for  2 (max1)  Moreover,

 (max) = Ã (2_{¡ ) kk}2_1 (_{¡ )}R_­jj   !2¡ ¡2 kk21 ¡ Ã (2_{¡ ) kk}2_1 (_{¡ )}R_­jj !¡ ¡2 Z ­ _jj = _kk_1 "µ 2_{¡ } _{¡ } ¶2¡ ¡2 ¡ µ 2_{¡ } _{¡ } ¶¡ ¡2# µ kk 1 R ­jj   ¶2¡ ¡2 ¸ kk1 µ _{¡ 2} _{¡ } ¶ µ 2_{¡ } _{¡ } ¶2¡ ¡2 Ã 1 ₊¡2 k+k  !2¡ ¡2  (6) ()R_­jj  _{· 0} There is a unique ¡ _{ }

max such that  (¡) =

R ­jj   and 0_(¡_{)  0} Now, (2_{¡ )}°°¡°°2 1 ¡ ( ¡ ) Z ­  ¯ ¯¡_¯_¯_ = ¡¡¢1+ · (2_{¡ )}¡¡¢1¡_kk2 1 ¡ ( ¡ ) ¡ ¡¢¡¡1 Z ­ _jj ¸ = ¡¡¢1+0¡¡¢ 0 6

and ­ _0 ¡¡¢ ¡® = ¡¡¢2_kk2_1 ¡ ¡ ¡¢ Z ­ jj  _¡¡¡¢ Z ­ jj   = ¡¡¢ · ¡¡¢_¡ Z ­ jj ¸ = 0 Thus, ¡_{2 M}¡

 Since for   max we have

(2_{¡ ) kk}2_1 ¡ ( ¡ ) Z ­ jj  0 2 2()  0 and  () = kk 2 1¡ ¡1 Z ­ jj¡ ¡1 Z ­ jj = 0for  = ¡ Thus, (¡) = sup_¸0() Moreover,

_¡¡¢ _{¸ }()_¸  2 2 kk 2 1 ¡   Z ­ _jj _{for all  ¸ 0}

Similar to the argument in the function  (), we obtain

()¸ _{¡ 2} 2 µ kk1 R ­jj   ¶ 2 ¡2  0 ()R_­_jj  0 By (6) and  (0) = 0  Z ­ _jj_{· +}°°+°°  ¡2 kk 1  _kk_1 µ _{¡ 2} _{¡ } ¶ µ 2_{¡ } _{¡ } ¶2¡ ¡2 Ã ₁ ₊¡2 k+k  !2¡ ¡2 ·  (max) 

there are unique + _{and }¡ _{such that 0  }+_{ }

max ¡ ¡+¢= Z ­ jj   = ¡¡¢ 7

and

0¡+¢ 0  0¡¡¢

We have +_{2 M}+_ ¡_{2 M}¡_ and (¡)¸ ()¸ (+) for each  2 [+_{ }¡_{] and }

(+)· () for each  2 [0 +]  Thus,

 ¡ +¢= inf 0··max ()   ¡ ¡¢= sup ¸max ()  This completes the proof.

For each  2 1 0(­) with R ­jj    0 we write max= Ã (_{¡ )}R_­jj (_{¡ 2) kk}2_1 ! 1 2¡  0 (7)

Then we have the following lemma.

Lemma 8 _{For each  2 0}1(­) with R_­jj 

  0 we have

() if R_­jj · 0 then there is a unique 0  +  max such that

+_ 2 M+ and  ¡ +¢= inf ¸0() ;

() if R_­jj  0 then there are unique 0  +  max  ¡ such that

+_ 2 M+ ¡2 M¡and  ¡ +¢ = inf 0··max () ;  ¡ ¡¢ = sup ¸max ()  Proof. _{Fix  2 }1 0 (­) with R ­jj    0. Let  () = 2¡_kk2_1 ¡ ¡ Z ­ jj   for   0 (8) Clearly, () ! ¡1 as  ! 0+ and () ! 0 as  ! 1. Since 0() = (2_{¡ ) }1¡_kk2_{ ¡ ( ¡ ) }¡¡1 Z ­ jj we have 0() = 0 at  = max  0

()  0 _{for  2 [0 }max) and  0

()  0 for  2 ¡max1

¢

 Then  () achieves its maximum at max increasing for

 ₂ ¡0 max

¢

and decreasing for  2 ¡max1

¢

 Similar to the argument in Lemma 7, we can obtain the results of Lemma 8.

3

Estimate of energy

By Lemma 6, we write M_= M+_{[ M}¡_ and de…ne

 = inf 2 () ; + = inf 2M+ () ; ¡= inf 2M¡ ()  First, we consider the following semilinear elliptic equation:

½

¡¢ =  () jj¡2 in £ 0_{·  2 }1

0 (£) 

( e) where £ is a domain in R_{ 1    2}¤ _{and  : £ ! R is a continuous}

positive function. Associated with equation ³e_´ we consider the energy functional () = 1 2 Z £jrj 2 _¡ 1  Z £  ()_jj

and the minimization problem

_(£) = inf_f() j  2 N(£)g 

where N(£) = f 2 01(£)n f0g j h0()  i = 0g  Then we have the following well-known result.

Theorem 9 () If 1    2 there exists a positive solution  2 N(£) for ³

e

 ´

such that () = (£)  0;

() If 2    2¤_{ there exists a positive solution }

 2 N(£) for ³ e  ´ such that () = (£)  0

By the conditions () and ()  the sets f 2 ­ j  ()  0g and

f 2 ­ j  ()  0g are open in RThus we may choose two domains £+

and £_+ in R such that £_+ ½ f 2 ­ j  ()  0g and £_+ ½ f 2 ­ j  ()  0g 

Furthermore, we have the following result. Theorem 10 We have

() there exists   0 such that  · + 

(¡2)2  (£+)  0 for all +  0 +  0 ¡¸ 0 ¡¸ 0; () _¡(¡2)(2¡)_2 µ (2¡)2 +(¡)k+k ¶ 2 ¡2 · ¡ for all +  0 +  0 ¡ ¸ 0 _{¡ ¸ 0} 9

Proof. () Let  2 N(£+) be a positive solution of ³ e  ´ such that () = (£+)  0Then kk 2 1 = Z £_+ +()_jj   = Z ­ ()jj    0 for all +  0 ¡ ¸ 0

Set  = +()  0 as de…ned by Lemma 7. Hence 2 M+ and

() = _{¡ 2} 2 kk 2 1¡ _{¡ }  Z ­ jj  ¡ 2 2 kk 2 1¡ _{¡ 2}  kk 2 1 = _¡(¡ 2) (2 ¡ )  2  2 kk 2 1 = (¡ 2)  2   (£+)  0

This yields that

 · +_ 

(_{¡ 2) }2



 (£+)  0 for all + 0 +  0 ¡¸ 0 ¡¸ 0

(9) () _{Let  2 M}¡_By (3)  (4)and the Hölder and Sobolev inequalities,

_{¡ } _{¡ 2} Z ­ jj  kk21  _{¡ } 2_{¡ } Z ­ jj · + _{¡ } 2_{¡ } ° °+°°  ¡2 kk 1 This implies kk1  µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶ 1 ¡2 for all  2 M¡ (10) 10

By (2) in the proof of Lemma 3, ()¸ _{¡ 2} 2 kk 2 1 ¡ + _{¡ }  ° °+°°   ¡2 kk 1 ¸ kk1 · _{¡ 2} 2 kk 2¡ 1 ¡ + _{¡ }  °° +°°  ¡2 ¸  µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶  ¡2 2 4¡ 2 2 µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶2¡ ¡2 ¡ + _{¡ }  ° °+°°  ¡  2 3 5  µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶  ¡2 2 4¡ 2 2 µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶2¡ ¡2 ¡ _¡ 2 µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶2¡ ¡2 3 5 = _¡(¡ 2) (2 ¡ ) 2 µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶ 2 ¡2  Thus, ¡_{ ¸ ¡}(¡ 2) (2 ¡ ) 2 µ (2_{¡ ) }2 ₊(_{¡ ) k}+_k  ¶ 2 ¡2 for all + 0 +  0 ¡¸ 0 ¡ ¸ 0

This completes the proof.

4

Proof of Theorem 2

First, we will use the idea of Tarantello [7] to obtain the following results.

Lemma 11 _{For each  2 }(­), there exist   0 and a di¤erentiable

function  :  (0; ) ½ 1

0 (­) ! R+ such that  (0) = 1 the function

 () (_{¡ ) 2 }(­) and h0(0)  _{i =} 2 R ­rr ¡  R ­jj ¡2_ ¡ R­jj ¡2_{)} (2_{¡ )}R_­jrj2_{¡ ( ¡ )}R_­jj  (11) for all  2 1 0(­)

Proof. _{For  2 }(­), de…ne a function  : R £ 01(­)! R by

( ) = ­ _0 ((_{¡ )) ( ¡ )}® = 2 Z ­jr( ¡ )j 2 ¡  Z ­ j ¡ j  _{¡ } Z ­ j ¡ j  Then (1 0) = ­ 0 ()  ® = 0and  (1 0) = 2 Z ­ jrj2_{¡ } Z ­ jj¡  Z ­ jj = (2_{¡ )} Z ­jrj 2 _{¡ ( ¡ )} Z ­ jj  _{6= 0}

According to the implicit function theorem, there exist   0 and a dif-ferentiable function  :  (0; ) ½ 1 0 ¡ R¢ ! R such that  (0) = 1, h0(0)  _{i =} 2 R ­rr ¡  R ­jj ¡2_ ¡ R­jj ¡2_{)} (2_{¡ )}R_­jrj2_{¡ ( ¡ )}R_­jj   and (() ) = 0 for all  2 (0; ) which is equivalent to ­ _0 (()(_{¡ )) ()( ¡ )}®= 0_{for all  2 (0; )} that is  () ( ¡ ) 2 (­)  12

Lemma 12 _{For each  2 }¡ (­), there exist   0 and a di¤erentiable

function ¡ :  (0; ) _{½ }1

0(­) ! R+ such that ¡(0) = 1 the function

¡() (_{¡ ) 2 }¡ (­) and ­ (¡)0(0)  ®= 2 R ­rr ¡  R ­jj ¡2_ ¡ R­jj ¡2_{)} (2_{¡ )}R_­jrj2_{¡ ( ¡ )}R_­jj  (12) for all  2 1 0(­)

Proof. Similiar to the argument in Lemma 11, there exist   0 and a di¤erentiable function  :  (0; ) ½ 1

0

¡ R¢

! R such that  (0) = 1 and

¡() (_{¡ ) 2 }(­) for all  2 (0; ). Since ­ 0_()  ®= (2_{¡ ) kk}2_1 ¡ ( ¡ ) Z ­ jj    0

Thus, by the continuity of the functions 0_ and ¡, we have ­ 0_(¡()(_{¡ )) }¡()(_{¡ )}® = (2_{¡ )}°°¡() (_{¡ )}°°2 1 ¡ ( ¡ ) Z ­  ¯ ¯¡_{() (¡ )}¯_¯_{  0}

if  is su¢ciently small; this implies that ¡() (_{¡ ) 2 }¡ (­) 

Proposition 13 Let ¤ =³ ¡2 k+_k  ´¡2³ 2¡ k+_k  ´2¡¡ ¡  ¢¡

Proof. , then for ¡2₊ 2¡_{+ 2 (0 ¤) and ¡ ¸ 0 , ¡ ¸ 0}

() _{there exists a minimizing sequence fg ½ }(­)such that

() = (­) +  (1)

_0 () =  (1) in ¡1(­) ; 13

() _{there exists a minimizing sequence f}g ½ _¡ (­) such that

() = ¡(­) +  (1)

_0 (_) =  (1) in ¡1_{(­) }

() By Lemma 3, Lemma 6 and the Ekeland variational principle [3], there exists a minimizing sequence fg ½ (­) such that

()  (­) + 1  (13) and ()  () + 1 k ¡ k for each  2 (­)  (14)

By taking  large, from Theorem 10 (), we have

() = _{¡ 2} 2 kk 2 1 ¡ µ _{¡ }  ¶ Z ­ jj  (­) + 1   (_{¡ 2) }2   (£+)  0 (15) This implies ° °+°° ¡  2 k_k 1  Z ­ +_jj  1 + Z ­ jj ¡ (_{¡ 2)} +(¡ ) 2__(£+)  (16) Consequently  6= 0 and putting together (15), (16) and the Hölder inequality, we obtain kk1  · ¡_(¡ 2) +(¡ ) 2__(£+)°°+°°¡1    2 ¸1   (17) 14

and kk_1  · 2(_{¡ )} (_{¡ 2)}+ ° °+°_°  ¡2 ¸ 1 2¡  (18)

Now, we will show that ° °0 () ° ° ¡1 ! 0 as  ! 1

Applying Lemma 11 with  to obtain the functions  : (0; ) ! R+ for some   0, such that ()(¡ ) 2 (­). Choose 0    . Let  2 01(­)with  6´ 0 and let  = _kk

1. We set  = () (¡).

Since _{ 2 }(­), we deduce from (14) that

()¡ ()¸ ¡ 1  ° °_{¡ } ° ° 1

and by the mean value theorem, we have ­ _0 () (¡ ) ® + (°°_{¡ } ° ° 1)¸ ¡ 1  ° °_{ ¡ } ° ° 1 Thus, ­ _0 ()¡ ® + (_()¡ 1) ­ _0 () (¡ ) ® ¸ ¡_1 °°¡  ° ° 1 + ( ° °_{¡ } ° ° 1). (19)

From _() (¡ )2 (­)and (19), it follows that ¡ ¿ _0 ()  kk1 À + (_()¡ 1) ­ _0 ()¡ 0 () (¡ ) ® ¸ ¡_1°° ¡  ° ° 1 + ( ° °_{¡ } ° ° 1). 15

Thus, ¿ _0 ()  kk1 À · ° °_{¡ } ° ° 1  + (°°_{ ¡ } ° ° 1)  +(()¡ 1)  ­ _0 ()¡ 0 () (¡ ) ® , (20) since _° °_{¡ } ° ° 1 ·  j()j + j()¡ 1j kk_1 and lim !0 j()¡ 1j  · k 0 (0)k .

If we let  ! 0 in (20) for a …xed , then by (18) we can …nd a constant

  0, independent of , such that ¿ _0 ()  kk1 À · _(1 +_k0_(0)_k)

We are done once we show that k0(0)k is uniformly bounded in . By (11), (18) and the Hölder inequality, we have

h0(0) i · _kk1 ¯ ¯(2 ¡ )R­jrj 2 _{¡ ( ¡ )}R_­jj  ¯¯ for some   0

We only need to show that ¯ ¯ ¯ ¯(2 ¡ ) Z ­jrj 2 _{¡ ( ¡ )} Z ­ jj   ¯ ¯ ¯ ¯   (21) 16

for some   0 and  large enough. We argue by contradiction. Assume that there exists a subsequence fg such that

(2_{¡ )} Z ­ jrj2_{¡ ( ¡ )} Z ­ jj = (1) (22)

Combining (22) with (17), we can …nd a suitable constant   0 such that Z

­

jj 

_{¸  for  su¢ciently large.} (23)

In addition to (22), and the fact that 2 (­) also give Z ­ jj =kk2_1 ¡ Z ­ jj = _{¡ 2} 2_{¡ } Z ­ jj + (1) and kk1 · · +( _{¡ } _{¡ 2}) ° °+°° ¡  2 ¸ 1 2¡ + (1) (24) Let  :  ! R be given by () = ( ) Ã kk2(¡1)1 R ­jj  ! 1 ¡2 ¡ Z ­ jj where ( ) =³2¡ ¡ ´¡1 ¡2³_¡2 2¡ ´

. Then () = 0for all  2 0 .

This implies () = ( ) Ã kk2(¡1)_1 R ­jj  ! 1 ¡2 ¡ Z ­ jj = µ 2_{¡ } _{¡ } ¶¡1 ¡2 µ_¡ 2 2_{¡ } ¶0 B @ ³ ¡ 2¡ ´¡1¡R ­jj ¢¡1 R ­jj  1 C A 1 ¡2 ¡₂¡ 2 ¡  Z ­ jj + (1) = (1) (25) However, by (23), (24) and ¡2₊ 2¡_{+ 2 (0 ¤) } () ¸ ( ) Ã kk2(¡1)_1 R ­jj  ! 1 ¡2 ¡ + ° °+°_°   ¡2 k_k 1 ¸ kk  1 ( ( )2¡ _ 1¡ 2¡ + ·µ _{¡ } _{¡ 2} ¶ ° °+°°  ¡2 ¸1¡ 2¡ ¡ + ° °+°°  ¡2 ) 

this contradicts (25). We get ¿ _0 ()  kk1 À ·  

This completes the proof of ().

() Similarly, using Theorem 10() and Lemma 12, we can prove (). We will omit the details here.

Now, we establish the existence of a local minimum for  on M+(­). Theorem 14 Let ¤  0 as in Proposition 13 , then for ¡2₊ 2¡_{+ 2 (0 ¤)} and _{¡¸ 0 , ¡ ¸ 0 the functional }has a minimizer +0 in M

+ (­) and it satis…es ()  ¡ +₀¢= (­) = +_(­) ;

() +₀ is a nontrivial and nonnegative solution of equation ()  18

Proof. _{Let f}g ½ M(­) be a minimizing sequence for  on M(­) such that

() = (­) +  (1) and 0 () =  (1) in ¡1(­) 

Then by Lemma 3 and the compact imbedding theorem, there exists a sub-sequence fg and +0 2 01(­) such that

 +0 weakly in  1 0(­) and ! +0 strongly in   (­) for 1    2¤ (26) First, we claim that R_­

¯ ¯+

0

¯ ¯

_{6= 0 If not, by (26) we can conclude that}

Z ­ jj! Z ­ ¯¯+0¯¯   = 0_{as  ! 1} Thus, _Z ­ jrj2 = Z ­ jj +  (1)  and () = µ 1 2 ¡ 1  ¶ Z ­jr j 2  +  (1) 

this contradicts ()! (­)  0as  ! 1 Moreover,

 (1) =­_0 ()   ®

=­_0 (0)  

®

+  (1) _{for all  2 0}1(­)  Thus, +_{0 2 M} is a nonzero solution of equation () and 

¡

+₀¢ _¸ (­)  We now prove that 

¡ +₀¢= (­)  Since  ¡ +₀¢ = 1 2°° + 0°° 2 1 ¡ 1  Z ­ ¯¯+0¯¯  _¡ 1  Z ­ ¯¯+0¯¯   = µ 1 2 ¡ 1  ¶ °_°+ 0 ° °2 1 + µ 1 ¡ 1  ¶ Z ­  ¯ ¯+ 0 ¯ ¯  · lim inf !1 µµ 1 2 ¡ 1  ¶ kk2_1 + µ 1  ¡ 1  ¶ Z ­ jj ¶ = lim inf !1 () = (­)  19

Thus,  ¡ +₀¢ = (­)  Moreover, we have +0 2 M + (­)  In fact, if +

0 2 M¡(­)  by Lemma 7, there are unique 

+

0 and ¡0 such that +0+0 2

M+_(­) and ¡₀+_{0 2 M}¡_(­)  we have +₀  ¡₀ = 1 Since

  ¡ + 0+0 ¢ = 0and  2 2 ¡ + 0+0 ¢  0

there exists +₀  ¹_{· }¡₀ such that  ¡ +₀+₀¢  ¡_¹ +₀¢ By Lemma 7,  ¡ +₀+₀¢  ¡_¹ +₀¢_{· } ¡ ¡₀+₀¢=  ¡ +₀¢

which is a contradiction. Since _¡+ 0 ¢ = _¡¯¯+ 0 ¯ ¯¢ and¯¯+ 0 ¯ ¯ 2 M+ (­)  by Lemma 5 we may assume that +₀ is a solution of equation ().

Next, we establish the existence of a local minimum for  on M¡(­) Theorem 15 Let ¤  0 as in Propostion 13, then for ¡2₊ 2¡_{+ 2 (0 ¤) and} _{¡ ¸ 0 , ¡ ¸ 0 the functional } has a minimizer ¡0 in M¡(­) and it

satis…es

()  ¡

¡₀¢= ¡_(­) ;

() ¡₀ is a nontrivial and nonnegative solution of equation () 

Proof. _{By Proposition 13 (), there exists a minimizing sequence f}g for

 on M¡ (­)such that

() = ¡_(­) +  (1) and 0 () =  (1) in ¡1(­) 

By Lemma 3 and the compact imbedding theorem, there exists a subsequence fg and ¡0 2 M¡(­)such that

 ¡0 weakly in  1 0(­) and ! ¡0 strongly in  _(­) for 1 ·   2¤ Since  (1) =­_0 ()   ® =­_0 (0)   ® +  (1) _{for all  2 0}1(­) and 0  ­0_()   ® = (2_{¡ ) k}k 2 1 ¡ ( ¡ ) Z ­ jj   ¸ (2 ¡ ) k0k2_1 ¡ ( ¡ ) Z ­ j0j 20

Thus, ¡_{0 2 M}¡_(­) is a nonzero solution of equation () We now prove that _{ ! }¡

0 strongly in 01(­)

Suppose otherwise, then°°¡₀°°

1  lim inf !1 kk1 and so ° °¡ 0 ° °2 1 ¡ Z ­  ¯ ¯¡ 0 ¯ ¯ _¡ Z ­  ¯ ¯¡ 0 ¯ ¯   lim inf !1 µ kk 2 1 ¡ Z ­ jj  _¡ Z ­ jj   ¶ = 0

this contradicts ¡_{0 2 M}¡_(­). Hence  ! ¡0 strongly in 01(­) This

implies ()!  ¡ ¡₀¢ = ¡_(­) _{as  ! 1} Since  ¡ ¡₀¢ = ¡¯¯¡0 ¯ ¯¢ and ¯¯¡ 0 ¯ ¯ 2 M¡ (­) by Lemma 5 we may assume that ¡₀ is a solution of equation ().

Now, we complete the proof of Theorem 2: By Theorems 14 and 15, for the equation () there exist two solutions +0 and ¡0 such that +0 2

M+

(­)  ¡0 2 M¡(­). Since M

+

(­)\ M¡(­) = ;, this implies that

+₀ and ¡₀ are di¤erent.

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