Branches of solutions for an asymptotically linear elliptic problem on RN

doi:10.4236/ojapps.2012.24B043

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Branches of solutions for an asymptotically linear

elliptic problem onԹܰ

Youyan Wan

Department of Mathematics

Jianghan University

Wuhan, Hubei, China

youyanwan@yahoo.com.cn

Abstract—We consider the following nonlinear schrК

dinger

equation

െο࢛+ࣅࢂ(࢞)࢛=ࢌ(࢞,࢛)with࢛אࡴ૚(Թࡺ)and ࢛ء૙,(כ)

whereࣅ>0and ࢌ(࢞,࢙)is asymptotically linear withrespect to ࢙at

origin and infinity. The potential ࢂ(࢞)satisfies ࢂ(࢞)൒ࢂ૙>0for

all ࢞אԹࡺand ࢂ(࢞)=ࢂ

|࢞|՜+λ

࢒࢏࢓ (λ)א(૙,+λ).We provethat

problem (כ)has two connected sets of positive and negative

solutions inԹ×ࢃ૛,࢖(Թࡺ)for some࢖א[૛,+λ)ת(ࡺ

૛,+λ).

Keywords-Bifurcation, asymptotically linear, Fredholm opera-

tor of index zero.

I. INTRODUCTION

In this paper, we consider the following nonlinear

SchrÖdingerequation

ቊെοݑ+ߣܸ(ݔ)ݑ=݂(ݔ,ݑ)

ݑאܪ1(Թܰ)

ܰ൒3

 (1.1)

where ɉ>0and the functions Vand ݂satisfy the following

assumptions:

(ܸ1) ܸ(ݔ)אܥ(ܴܰ,ܴ) and there existsܸ0>0such that

ܸ(ݔ)൒ܸ0>0 for all ݔאԹܰ;

(ܸ2) ܸ(ݔ)=ܸ (λ )א(0 ,+λ)

|ݔ|՜+λ

݈݅݉ and ݁ܽݏ{ݔאԹܰ:ܸ(ݔ)<

ܸ(λ)}>0;

(ܨ1)݂ (ݔ,ݏ)אܥ( Թܰ×Թ,Թ)and (ݔ,ή)אܥ1(Թ,Թ);

(ܨ2)there exist two functions ݄,݃אܮλ(Թܰ)such that

݂(ݔ,ݏ)

ݏ=݄(ݔ)

ݏ՜0

݈݅݉ and ݂(ݔ,ݏ)

ݏ=݃(ݔ)

|ݏ|՜+λ

݈݅݉ uniformly inݔאԹܰ,

where ݄and ݃satisfy

(G) Setting Ȟ=inf༌{׬(|׏ݑ|2+ܸ(ݔ)ݑ2)݀ݔ:ݑא

Թܰ

ܪ1(Թܰ) ܽ݊݀ ׬ݑ2݀ݔ=1}

Թܰ,there existsߙא(Ȟ,ܸ(λ))such

that ݃(ݔ)=݃

(ݔ)=ߙ

ݔאԹܰ

݂݅݊

|ݔ|՜+λ

݈݅݉ ;

(H) |݄ |λ<ߙܸ0

ܸ(λ).

(ܨ3) ݄(ݔ)൑݂(ݔ,ݏ)

ݏ൑݃(ݔ)for all(ݔ,ݏ )אԹܰ×Թ\{0}.

The existence of solutions of problem (1.1) has

beeninvestigated extensively. For problem (1.1)with

potentialwell and various conditions on ݂(ݔ,ݑ)ء݂ (ݑ),

several authorshave obtained the existence of solutions for

large ߣbyvariational methods,for example, [1], [2], [3], [5].

And other authors have got the existence of solutions forߣis

not necessarily large by concentration compactnessargument

and mountain pass geometry, for instance, [7], [8]. Stuart and

Zhou [10] havestudied how the positive and negative solutions

of problem(1.1)depend on ߣby topological methods.

Inspired by the results we mentioned above, the main object

of thisarticle is to investigate the relation between the positive

andnegative solutions of problem (1.1)and the parameterߣ,

where the potential need not be well potential and݂(ݔ,ݑ)is

asymptotically linear with respect to ݑat origin andinfinity.

For this purpose, we use the following global branch

theorem established in [10].

Theorem 1.1:Let ܺand ܻbe real Banach spaces, ࣜ(ܺ,ܻ)

be thespace of bounded linear operators from ܺinto ܻwith its

usualnorm, ܲ:Թ×ܺ՜Թdenote theprojection ܲ(ߣ,ݑ)=ߣ,

and

Ȱ0(ܺ,ܻ)=

{ܮאࣜ(ܺ,ܻ):ܮ ݅ݏ ܽ ܨݎ݄݁݀݋݈݉ ݋݌݁ݎܽݐ݋ݎ ݋݂ ݅݊݀݁ݔ ݖ݁ݎ݋}.

Let ܮאܥ1(ܬ,ࣜ(ܺ,ܻ))where ܬis an open interval and

ܮ(ߣ)אȰ0(ܺ ,ܻ)for allߣאܬ. Suppose that there exists ߣ0אܬ

suchthat dim݇݁ݎܮ(ߣ 0)is odd and

ܮԢ(ߣ0)݇݁ݎܮ(ߣ0)۩ݎ݁݃ܮ(ߣ0)=ܻ (1.2)

Letܭאܥ(ܺ,ܻ)be such that ܭ:ܺ՜ܻis compact and

lim||ݑ||ܺ՜0||ܭ(ݑ)||ܻ

||ݑ||ܺ=0. (1.3)

Let ܼ෨=ܼ׫{(ߣ0,0)}where ܼ={(ߣ,ݑ)אܬ×ܺ:ݑ്

0 ܽ݊݀ ܮ(ߣ)ݑ+ܭ(ݑ)=0}be considered with the metric

inherited from Թ×ܺ ,and let ࣝdenote the connected

component ofܼ෨containing (ߣ0,0). Then ࣝpossesses at least

one of the followingproperties:

(i)ࣝis an unbounded subset of Թ×ܺ;

(ii)ࣝҧת[ܬ×{0}]്{(ߣ0,0)}, whereࣝҧis the closure of ࣝin

ܬ×ܺ;

(iii) eitherݏݑ݌ܲࣝ=ݏݑ݌ܬor ݂݅݊ܲࣝ=݂݅݊ܬ .

REMARK 1.1:For ܭאܥ(ܺ,ܻ),the condition (1.3) is

equivalent to theproperties ܭ(0)=0 and ܭ:ܺ՜ܻ is

FrÉchetdifferentiable at zero with ܭԢ(0)=0.

By (ܨ1)and (ܨ2)we may define a function ݇having the

following properties

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݇(ݔ,ݏ)=݃(ݔ)െ݂(ݔ,ݏ)

ݏ (1.4)

with ݇(ݔ,ݏ)=݃(ݔ)െ݄(ݔ )

ݏ՜0

݈݅݉ ,݇(ݔ,ݏ)=0

|ݏ|՜+λ

݈݅݉ uniformly

inݔאԹܰand 0൑݇(ݔ,ݏ)൑݃(ݔ)െ݄(ݔ).From the above

notation (1.4),problem(1.1)is equivalent to

െοݑെ݃(ݔ)ݑ+ߣܸ(ݔ)ݑ+݇(ݔ,ݑ)ݑ=0,ݑאܪ1(Թܰ)(1.5)To

prove the asymptotic bifurcation result, first we study the

following formal asymptotic linearization of (1.5):

ቊെοݑെ݃(ݔ)ݑ+ߣܸ(ݔ)ݑ=0

ݑאܪ1(Թܰ)

ߣ>0.  (1.6).

A number ߣ>0is said to be an eigenvalue of (1.6)if there

exists ݑאܪ1(Թܰ)\{0}such that

׬[׏ݑ ׏ݒെ݃ (ݔ)ݑݒ+

Թܰߣܸݑݒ ]݀ݔ=0 ݂݋ݎ ݈݈ܽ ݒאܪ1(Թܰ).

For the discussion of equation (1.5), we take advantage ofthe

additional regularity of solutions that follows from

ourassumptions(see Proposition 2.1 in [10]).

Proposition 1.1: (1) Assume that the

conditions (ܨ1) (ܨ2) (ܸ1) (ܸ2)hold and ݑאܪ1(Թܰ)satisfies

(1.5),then ݑאܹ2,݌(Թܰ)for all ݌א[2,+λ) andhence

ݑאܥ1(Թܰ)with ݑ(ݔ)=0

|ݔ|՜+λ

݈݅݉ and ׏ݑ (ݔ)=0

|ݔ|՜+λ

݈݅݉ .

(2) If ܸsatisfies (ܸ1) (ܸ2)and ݒאܪ1(Թܰ)is an

eigenfunction of (1.6), then ݒאܹ2,݌(Թܰ)for all ݌א[2,+λ).

Our first result concerning the linearized equation (1.6) is the

following :

Theorem 1.2: Assume that ܸand ݃satisfies (ܸ1) (ܸ2)and

(G), then

(i) there exists an unique eigenvalue ߣ=Ȧ(ߙ)of (1.6)

having a positive eigenfunction. FurthermoreȦ(ߙ)>1, and it

is simple in the sense that ker༌(ܣȦ(ߙ))=ݏ݌ܽ݊ {ݑ Ȧ(ߙ)}whereܣߣ

denotes the SchrÖdinger operator ܣߣݑ=െȟݑെ݃(ݔ)ݑ+

ߣܸݑ and ݑȦ(ߙ)>0onԹܰ. All other eigenvalues of (1.6) are

less thanȦ(ߙ )and their eigenfunctions change sign.

(ii)Ȧ(ߙ)is the unique value of ߣin theinterval [ߙ

ܸ(λ),+λ)

for which 0is theinfimum of the spectrum of the

SchrÖdinger operatorܣߣ.

Now we can state our main result concerning the

nonlinear problem (1.1).

Theorem 1.3: Let the conditions (ܨ1) (ܨ2) (ܨ3)(ܸ1) (ܸ2)

hold and fix

Pא[2,+λ)ת(N

2,+λ). Thenthere exist two

connected subsets σ+and σെof Թ×ܹ2,݌(Թܰ),

whoseelements (ߣ,ݑ) are, respectively, positive and

negativesolutions ofproblem (1.1), such that ݂݅݊{ߣ:(ߣ,ݑ)א

σ±}=

ܸ(λ)andݑ݌{ߣ:(ߣ,ݑ)אσ±}=Ȧ(ߙ), where Ȧ(ߙ)is given

by Theorem 1.2. Furthermore, σ±is bounded away from

the line of trivial solutions of Թ×{0}and if {(ߣ݊,ݑ݊)}ؿ

σ±with ߣ݊݊

՜ߣ> ߙ

ܸ(λ),then |ݑ݊(ݔ)|݊

՜λ

ݔאԹܰ

݉ܽݔ if and only

if ߣ=Ȧ(ߙ).

II.EIGENVALUE PROBLEM

In this section, we prove Theorem 1.2. It follows from

Proposition 1.1that anyeigenfunctionݑof equation (1.6)

belongs toܥ(Թܰ)תܪ2(Թܰ), and this leads us to introduce a

SchrÖdinger operator having ݑas an eigenfunction.Define

ܣߣ:ܦ(ܣߣ)=ܪ2(Թܰ)ؿܮ2(Թܰ)՜ܮ2(Թܰ)

byܣߣ=െοݑെ݃(ݔ)ݑ+ߣܸݑ

Then ܣߣis a self-adjoint operator in ܮ2(Թܰ)with spectrum

ߪ(ܣߣ)and essential spectrumߪ݁(ܣߣ)=[ߣܸ (λ)െ

ߙ,+λ).Furthermore, settingσ(ߣ)=݂݅݊ߪ (ܣߣ),we have

σ(ߣ)=inf༌{ܽߣ(ݑ):ݑאܪ1(Թܰ)ܽ݊݀ ׬ݑ2݀ݔ=1}

Թܰ>

െλ,whereܽߣ(ݑ)=׬[|׏ݑ|2െ݃(ݔ)ݑ2+ߣܸݑ2]݀ݔ

Թܰ.

Lemma 2.1: Suppose that ܸsatisfies(ܸ1) (ܸ2)andȞ<ߙ,

thenσ(1)<0. Moreover, there exists ߣ1>1such thatσ(ߣ)<

0for all ߣא(െλ,ߣ1] .

proof :SinceȞ<ߙ, there exists ݑאܪ1(Թܰ)\{0}such

that׬(|׏ݑ| 2+ܸ(ݔ)ݑ2)݀ݔ<ߙ׬ݑ2݀ݔ

Թܰ

Թܰ൑

׬݃(ݔ)ݑ2݀ݔ

Թܰ.This means that ܽ1(ݑ)<0and σ(1)<0.

Hence there existsݑ1אܪ1(Թܰ) ݓ݅ݐ݄ ׬ݑ12݀ݔ=1

Թܰsuch that

ܽ1(ݑ1)<0. By thedefinition ofܽߣ, we have

ܽߣ(ݑ1)െܽ1(ݑ1)=(ߣെ1)׬ܸ(ݔ)ݑ12݀ݔ

Թܰfor all ߣאԹ. (2.1)

By (ܸ1) (ܸ2), we have there exists ܥ>0such that ܸ(ݔ)൑

ܥfor all ݔאԹܰ. From (2.1), we haveܽߣ(ݑ1)൑ܽ1(ݑ1)+

ܥ(ߣെ1).Therefore choosingߣ1=1+െܽ1(ݑ1)

2ܥ >1,we

getܽߣ1(ݑ1)൑ܽ1(ݑ1)

2<0and σ(ߣ1)<0. Since σ(ߣ)is

increasing with respectto ߣאԹ, we have σ(ߣ)<0for all

ߣא(െλ,ߣ1].

Lemma 2.2:Let ܸsatisfies (ܸ1) (ܸ2).ForȞ<ߙ<ܸ(λ), if

we set ܵ׷={ߣ൒ ߙ

ܸ(λ):σ(ߣ)<0}andȦ(ߙ)=sup༌{ߣ :ߣאܵ} ,

thenȦ(ߙ)א(1,+λ).

Proof :From Lemma 2.1, we haveȦ(ߙ)>1. It is clear that ܵ

is an interval since σ(ߣ)is increasing in ߣ. Therefore,

ifȦ(ߙ)=+λ, we have ܵ=[ߙ

ܸ(λ),+λ)and for any integer

݊൒ ߙ

ܸ(λ), there existsݑ݊אܪ1(Թܰ)with׬ݑ݊

2݀ݔ=1

Թܰsuch

that

ܽ݊(ݑ݊)=׬[|׏ݑ݊|2െ݃(ݔ)ݑ݊

2+ܸ݊(ݔ)ݑ݊

2]݀ݔ

Թܰ<0. (2.2)

By condition (ܸ1)we see that (2.2) is impossible when݊൒ߙ

ܸ0.

Lemma 2.3:Assume that (ܸ1) (ܸ2)hold andȞ< ߙ<ܸ(λ).

Then ߣא[ߙ

ܸ(λ),+λ)and σ(ߣ)=0if and only if ߣ=Ȧ(ߙ)ˈ

whe r eȦ(ߙ)is given by Lemma 2.2.

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Proof :SinceȞ<ߙ<ܸ(λ)and σ(ߣ)is increasing in ߣ, it

follows from Lemma2.1 that σቀߙ

ܸ(λ)ቁ൑σ(1)<0. If

ߣאቂߙ

ܸ(λ),+λቁand σ(ߣ)=0,then ߣ> ߙ

ܸ(λ). Nowwe have

σ(ߣ)=݂݅݊ߪ(ܣߣ)=0and ݂݅݊ߪ݁(ܣߣ)=ߣܸ(λ)െߙ>

ܸ(λ)ܸ(λ)െߙ=0. Hence 0 is aneigenvalue of ܣߣand there

exists ݑߣאܥ(Թܰ)תܪ2(Թܰ)such that ݇݁ݎܣߣ=ݏ݌ܽ݊ {ݑ ߣ}

andݑߣ>0onԹܰ(see [9],Theorem 3.20] for example). We

mayassume that ׬ݑߣ2݀ݔ=1

Թܰsuch thatܽߣ(ݑߣ)=0. Then

from the definition of ܽߣwe have for any ߝ>0

ܽߣെߝ(ݑߣ)=െߝනܸݑߣ2݀ݔ

Թܰ൑െߝܸ0නݑ

ߣ2݀ݔ

Թܰ<0

and thismeans that ߣെߝאܵfor any ߝ>0. Thereforeߣ

=supS=Ȧ(ߙ ).

Conversely, ifߣ=Ȧ(ߙ), by Lemma 2.2 wehaveȦ(ߙ)>

1> ߙ

ܸ(λ). Hence it issufficient to proveȦ(ߙ)בܵ׫ܶ,whereܶ

׷={ߣ൒ ߙ

ܸ(λ): σ(ߣ)>0}. Indeed, if Ȧ(ߙ)אܵ, then

σ൫ Ȧ(ߙ)൯<0. By the proof of Lemma 2.1 we see that there

existsߣ2>Ȧ(ߙ)such that σ(ߣ)<0for all ߣא(െλ,ߣ2]. This

contradicts the definition of Ȧ(ߙ). On the other hand, ifȦ(ߙ)א

ܶ, then σ൫ Ȧ(ߙ)൯>0. By the definition of ܽߣand ܸ(ݔ)൑ܥ

for allݔאԹܰ, we see that for anyߝ>0and ݑאܪ1(Թܰ)

wit h ׬ݑ2݀ݔ=1

Թܰ,

ܽȦ(ߙ)െߝ(ݑ)=ܽȦ(ߙ)(ݑ)െߝනܸݑ2݀ݔ

Թܰ൒σ൫ Ȧ(ߙ)൯െߝܥ

Therefore we can chooseߝ=σ൫ Ȧ(ߙ)൯

2ܥ such

that ܽȦ(ߙ)െߝ(ݑߣ)൒σ൫ Ȧ(ߙ)൯

2>0for allݑאܪ1(Թܰ)with

׬ݑ2݀ݔ=1

Թܰ. This means that σ( Ȧ(ߙ)െߝ)>0and also

contradicts thedefinition ofȦ(ߙ).

Proof of Theorem 1.2(i) From Lemma2.2 and 2.3 we know

thatȦ(ߙ)>1and σ൫ Ȧ(ߙ)൯=݂݅݊ ߪ ൫ܣ Ȧ(ߙ)൯=0. Since

ߙ<ܸ(λ), we have ݂݅݊ ߪ݁൫ܣȦ(ߙ)൯=Ȧ(ߙ)ܸ(λ)െߙ>0.

Hence0 is aneigenvalue of ܣ Ȧ(ߙ)and there existsݑȦ(ߙ)א

ܥ(Թܰ)תܪ2(Թܰ)such that ݇݁ݎܣȦ(ߙ)=ݏ݌ܽ݊{ݑȦ(ߙ)}and

ݑȦ(ߙ)>0on Թܰ. Suppose now thatȦ1്Ȧ(ߙ)is another

eigenvalue of(1.6) with eigenfunctionݑ1אܪ1(Թܰ). Then 0is

an eigenvalue of ܣȦ1and σ(Ȧ1)=݂݅݊ߪ൫ܣȦ1൯൑0. It follows

thatȦ1൑ Ȧ(ߙ). Otherwise, if Ȧ1> Ȧ(ߙ)andσ(Ȧ1)൑0,we

divide two cases to deduce thecontradiction. One hand, if

Ȧ1> Ȧ(ߙ)and σ(Ȧ1)=0, it contradicts Lemma 2.3. On

theother hand, if Ȧ1> Ȧ(ߙ)and σ(Ȧ1)<0, by the proof of

Lemma 2.1 we see thatthere exists ߣ3>Ȧ1such that σ(ߣ)<

0for all ߣא(െλ,ߣ3]. This contradicts thedefinition ofȦ(ߙ).

ThereforeȦ(ߙ)isthe largest eigenvalue of (1.6). Furthermore,

integratingby parts, we have

(Ȧ(ߙ)െȦ1)׬ܸݑ1ݑȦ(ߙ)݀ݔ=0

Թܰ.

For Ȧ1<Ȧ(ߙ)and ܸ(ݔ)ݑȦ(ߙ)>0on Թܰ, it follows that ݑ1

changes sign.

(ii) This follows from Lemma 2.3.

III. THE TRUNCATED PROBLEM

Let Pא[2,+λ)ת(N

2,+λ)be fixed and we set

ܺ=ܹ2,݌(Թܰ)with ||ή||=||ή|| ܹ2,݌൫Թܰ൯, and

ܻ=ܮ݌(Թܰ)with|ή|݌=||ή||ܮ݌(Թܰ).

For ݇defined in (1.4), it can be shown that (see [10],

Lemma B.1] )|݇(ݔ,ݑ)ݑ|݌

ห|ݑ|หե0asห|ݑ|ห՜+λ

uniformly in ݔאԹܰ. Hence in order to make use ofTheorem

1.1 we need to introduce the following truncatedproblem

െοݑെ݃(ݔ)ݑ+ߣܸ(ݔ)ݑ+߰݊(ݔ)݇(ݔ,ݑ)ݑ=0(3.1)

whereݑאܪ1(Թܰ)and

߰݊(ݔ)=൜1, ݂݅ |ݔ|൑݊,

0, ݂݅ |ݔ|>݊.

Define ܮ(ߣ):ܺ՜ܻby

ܮ(ߣ)ݑ=െοݑെ݃ (ݔ)ݑ+ߣܸ(ݔ)ݑ (3.2)

Using the inversion, ݑհݒ= ݑ

ห|ݑ|ห2, problem (3.1) is equivalent

to ܮ(ߣ)ݒ+ܭ݊(ݒ)=0 (3.3)

where

ܭ݊(ݒ)=൞߰݊݇൭ݔ,ݒ

ห|ݒ|ห2൱ݒ, ݂݋ݎ ݒאܺ\{0}

0, ݂݋ݎ ݒؠ0 

In the sequel we show that Theorem 1.1 is applicable tothe

inverted truncated problem (3.3). First, it follows

fromTheorem 4.3 of [6]thatܮ(ߣ)אȰ0(ܺ,ܻ)for all>ߙ

ܸ(λ),

where Ȱ0(ܺ,ܻ)is defined in Theorem 1.1 and

Y= ݇݁ݎܮ(ߣ)۩ݎ݃݁ܮ (ߣ)for allߣ> ߙ

ܸ(λ) (3.4)

From (3.4) and Theorem1.2 we can prove (1.2) holds with

ߣ0=Ȧ(ߙ)(see the proof of Theorem 4.2in [10]). By Remark

1.1 the following lemma(Lemma 3.2in [10]) can verify

condition (1.3) for ܭ݊defined in(3.3) .

Lemma 3.1: For all݊אܰ,ܭ݊אܥ(ܺ,ܻ)תܥ1(ܺ\{0},ܻ),

ܭ݊:ܺ՜ܻis compact and it is FrÉchetdifferentiable at 0

withܭ݊Ԣ(0)=0.

Now we can apply Theorem 1.1for ܮ(ߣ)defined by (3.2) on

the interval ܬ=(ߙ

ܸ(λ),+λ)and at the point ߣ0=Ȧ(ߙ). Set

ࣴ݊={(ߣ,ݒ)א(ߙ

ܸ(λ),+λ)×ܺ:ݒis a nontrivial solution of

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(3.3)}. Let ࣝ݊denote the connected component ofࣴ݊׫

{( Ȧ(ߙ),0)}containing ( Ȧ(ߙ),0). In order to get some initial

informationabout ࣝ݊from Theorem 1.1, we establish

someestimates about the solutions of (3.2) that will help us

toobtain more precise information about ࣝ݊.

Lemma 3.2: Let the conditions (ܸ1) (ܸ2)hold and ߙ=݈+

ߤSuppose thatȞ<ߙ<ܸ(λ)and 0൑ߤ< ݈ܸ0

ܸ(λ)െܸ0. Then, there

exists ܶ>0, such that for anyߚ> ߙ

ܸ(λ)there is ܰߚא

Գsatisfying ||V||൑ܶfor all (ߣ,ݒ)אࣴ݊with ߣ൒ߚand

݊൒ܰߚ.

Proof :Firstwe define the bound ܶ. Since0൑ߤ< ݈ܸ0

ܸ(λ)െܸ0and

ߙ=݈+ߤ, we candeduce that ߤ<ߙܸ0

ܸ(λ)and choose ߟ=

2ቀߙܸ0

ܸ(λ)െߤቁ>0.By (1.4) and (ܸ1)we know ݇(ݔ,0)=݈and

there exists ߜ1such that ݇(ݔ,ݏ)൒݈െߟ=1

2݈+1

2ቀߙെߙܸ0

ܸ(λ)ቁ൒

2݈>0for all |ݏ |൑ߜ1. By Sobolevembeding there is ܥ1

suchthat |ݑ |λ൑ܥ1ห|ݑ|ห.Setܶ=ܥ1

ߜ1.Suppose that ݒאܺand

ห|ݒ|ห൒ܶ. Then, for all ݔאԹܰ, we have|ݒ(ݔ)|

ห|ݒ|ห2൑|ݒ|λ

ห|ݒ|ห2൑ܥ1

ห|ݒ|ห൑

ܥ1

T=ߜ1. Consequently, ݇൬ݔ,|ݒ(ݔ)|

ห|ݒ|ห2൰൒݈െߟ. For anyߚ> ߙ

ܸ(λ),

we can choose ߜߚ=ܸ(λ)െߙ

ߚ>0. By (ܸ2), there existsܰߚא

Գ,such that ܸ(ݔ)=ܸ(λ)െߜߚfor all |ݔ |൒ܰߚ. Hence

ߣV(x) ൒ߚ൫ܸ(λ)െߜߚ൯=ߙ (3.5)

for all ߣ൒ߚand |ݔ|൒ܰߚ. On the otherhand, for ߣ൒ߚ,

݊൒ܰߚ,|ݔ|൑ܰߚ, and ห|ݒ|ห൒ܶ, we have

ܭ݊(ݒ)=߰݊݇൬ݔ,|ݒ|

ห|ݒ|ห2൰=݇൬ݔ,|ݒ |

ห|ݒ|ห2൰൒݈െߟ,

andെߙ+ߣܸ(ݔ)+ܭ݊(ݒ)൒െߙ+ ߙ

ܸ(λ)ܸ0+݈െߟ

=െߤെߟ+ߙ

ܸ(λ)ܸ0=1

2ቀߙܸ0

ܸ(λ)െߤቁ>0.

Combining the above inequality and (3.5), we get that

for(ߣ,ݒ)אࣴ݊with ߣ൒ߚand ݊൒ܰߚ,ifห|ݒ|ห൒ܶ, then

ݒ(ݔ)οݒ(ݔ)=ݒ2(ݔ) [െߙ+ߣܸ(ݔ)+߰݊݇൭ݔ,|ݒ|

ห|ݒ|ห2൱]൒0

for all ݔאԹܰ. The maximum principle now leads toa

contradiction.

Lemma 3.3: Fix ߚ> ߙ

ܸ(λ). Then, for any ߝא(0,ܸ(λ)െߙ

ߚ),

there exists ܥߝ>0such that|ݑ (ݔ)|൑|ݑ|λ݁െඥߦ(|ݔ|െܥߝ)for all

ݔאԹܰ,where ߦ=ߚ(ܸ(λ)െߝ)െߙ>0for all

(ߣ,ݑ )א

[ߚ ,+λ)with ݑܮ (ߣ)ݑ൑0onԹܰ.

Proof :Since ܸ(ݔ)=ܸ(λ)

|ݔ|՜+λ

݈݅݉ , for any ߝא(0,ܸ(λ)െ

ߚ), there exists ܥߝ>0suchthat ܸ(ݔ)൒ܸ(λ)െߝfor all

|ݔ |൒ܥߝ. Set ݍ(ݔ)=|ݑ|λ݁െඥߦ(|ݔ|െܥߝ)െݑ(ݔ) and ȳߝ=

{ݔאԹܰ: |ݔ|>ܥߝ ܽ݊݀ ݍ(ݔ)<0}.For all ݔאȳߝ, we have

ݑ(ݔ)>0and0൒ܮ(ߣ)ݑ=െοݑെߙݑ+ߣܸ(ݔ)ݑ

൒െοݑെߙݑ+ߚ(ܸ(λ)െߝ)ݑ=െοݑ+ߦݑ,

since ൒ߚ. By direct calculations, we have forݔאȳߝ

οݍ(ݔ)=|ݑ|λ݁ඥߦܥߝ൬ߦെܰെ1

|ݔ|ඥߦ൰݁െඥߦ|ݔ|െοݑ

൑|ݑ|λ݁ඥߦܥߝߦ݁െඥߦ|ݔ|െߦݑ =ߦݍ(ݔ)<0.

Since ݍ(ݔ)՜0as |ݔ|՜+λand ݍ(ݔ)൒0for |ݔ|=ܥߝ, we

have ݍ(ݔ)൒0for ݔאȳߝ. If ȳߝ്׎, the weakmaximum

principle (Theorem 8.1 in[4]) implies that ݍ(ݔ)൒0in ȳߝ, a

contradiction. Thus we see that|ݑ (ݔ)|൑|ݑ|λ݁െඥߦ(|ݔ|െܥߝ)for

all|ݔ|൒ܥߝ.Replacing ݑby െݑ , we get the above inequality for

|ݑ (ݔ)|.Hence, we complete the proof.

Lemma 3.4:For each ݊אԳ, there exists an open

neighborhood ܷof(Ȧ(ߙ),0)אԹ×ܺ, such that ݑ2>0on Թܰ

for all (ߣ,ݒ)אܷתࣴ݊.

Proof :By contradiction, suppose that there exists asequence

{(ߣ݅,ݑ݅)}ؿࣴ݊such thatߣ݅݅

՜Ȧ(ߙ)and ||ݑ||݅݅

՜0andeach

continuous functionݑ݅has at least one zero in Թܰ. We prove

this leadsto a contradiction.

Setting ݖ݅=ݑ݅

||ݑ||݅, by the definition ofࣴ݊, we

have ܮ(ߣ݅)ݖ݅+ܭ݊(ݑ ݅)

||ݑ||݅=0on Թܰ.Since ܸ(ݔ)אܮλ(Թܰ), by

Lemma 3.1, wefind that

ܮ൫Ȧ(ߙ)൯ݖ݅=(Ȧ(ߙ)െߣ݅)ܸ (ݔ)ݖ݅െܭ݊(ݑ݅)

||ݑ||݅݅

՜0in ܻ.

On the other hand, by passing to a subsequence, we may

supposethat ݖ݅֊ݖweakly in ܺ. SinceȦ(ߙ)>1>ߙ

ܸ(λ), we

have ܮ൫Ȧ(ߙ)൯אȰ0(ܺ,ܻ). By Lemma 3.5 of [10], we know

that ݖ݅݅

՜ݖ strongly inܺ. This means

that ห|ݖ|ห=1and ܮ൫Ȧ(ߙ)൯ݖ݅=0. By Lemma3.3, wehave

ݖאܪ1(Թܰ). According to Theorem 1.2(i), we may as well

suppose thatZ>0on Թܰ. Sinceߣ݅݅

՜Ȧ(ߙ)and Ȧ(ߙ)>1,

choosingߝ=Ȧ(ߙ)െ1

2>0, there exists ݅0אԳsuch that ߣ݅൒1+

ߝfor all ݅൒݅0. By (ܸ2)and ߙ<ܸ(λ), thereexists ܴ൒݊such

that ܸ(ݔ)൒ߙfor all|ݔ |൒ܴ. Hence, for all ݅൒݅0and |ݔ|൒ܴ,

we haveെߙ+ߣܸ݅(ݔ)൒െߙ+(1+ߝ)ߙ>0.Sinceܴ൒݊, we

have for |ݔ|>ܴ,߰݊(ݔ)=0and

0=ܮ(ߣ݅)ݖ݅=െοݖ݅+൫െߙ+ߣܸ݅(ݔ)൯ݖ݅ (3.6)

190 Cop

Recalling that ܺؿܥ(Թܰ), we have ߜ= ݖ(ݔ)>0

|ݔ|൑ܴ

݂݅݊ .

Since ݖ݅݅

՜ݖ strongly in ܺ, we have that

|ݖ݅(ݔ)െݖ(ݔ)|݅

՜0

|ݔ|൑ܴ

MAX༌and there exists ݅1൒݅0such that

ݖ݅(ݔ)൒1

2ߜfor all |ݔ|൑ܴand݅൒݅1. Since ݖ݅(ݔ)൒1

2ߜfor

|ݔ|=ܴand ݖ݅(ݔ)=0

|ݔ|՜+λ

݈݅݉ , we get ݖ݅(ݔ)൒0on ߲ܤܴܿ.

Now we can apply the strong maximumprinciple (Theorem

8.19 in [4]) to equation (3.6) on theregion ܤܴܿand deduce

thatݖ݅>0in ܤܴܿfor all݅൒݅1. This contradicts that each ݑ݅

has at least onezero in Թܰ.

Now we can give more information about ࣝ݊.

Theorem 3.1: Let ࣝ݊denote the connected component

ofࣴ݊׫{(Ȧ(ߙ),0)}containing the point(Ȧ(ߙ),0). Then:

(i) ݑ2>0on Թܰand ߣ൑Ȧ(ߙ)for all (ߣ,ݑ)א

ࣝ݊\{(Ȧ(ߙ),0)};

(ii) for any ߚ> ߙ

ܸ(λ), there exist ܶ>0and ܰߚאԳsuch that,

for all ݊൒ܰߚ,

݂݅݊ܲ ࣝ݊׷= ݂݅݊{ ߣ: (ߣ,ݑ )אࣝ݊}<ߚand||ݑ||൑ܶ

for all (ߣ,ݒ )אࣝ݊with ߣ൒ߚ.

Proof : The first step is to show that if (ߣ,ݑ)א

ࣝ݊\{(Ȧ(ߙ),0)}, we haveߣ൑Ȧ(ߙ). Since (ߣ,ݑ )solves

equation (3.3), we have

െοݑെߙݑ+ߣܸ(ݔ)ݑ+߰݊(ݔ)݇൬ݔ,ݑ

ห|ݑ|ห2൰ݑ=0 (3.7)

By Lemma 3.3, we know ݑאܪ1(Թܰ). Hence

byTheorem1.2 (ii) and (3.7) ,

0=݂݅݊ቐන[|׏ݒ |2െߙݒ2+Ȧ(ߙ)ܸݒ2]݀ݔ:ݒ

Թܰאܪ1(Թܰ)ቑ

൑න[|׏ݑ|

2െߙݑ2+Ȧ(ߙ)ܸݑ2+߰݊(ݔ)݇൭ݔ, |ݑ|

ห|ݑ|ห2൱ݑ2]݀ݔ

Թܰ

= (Ȧ(ߙ)െߣ)׬ܸ(ݔ)ݑ2݀ݔ

Թܰ.

Since ׬ܸ(ݔ)ݑ2݀ݔ൒ܸ0

Թܰ׬ݑ2݀ݔ>0

Թܰ, we see that ߣ൑

Ȧ(ߙ).

The next step is to show that if (ߣ,ݑ)אࣝ݊\{(Ȧ(ߙ),0)},

thenݑ2>0onԹܰ. Set

࣫={(ߣ,ݑ)אࣝ݊: ݑ2>0 onԹܰ}ת{(Ȧ(ߙ),0)}.

We prove that ࣫is both anopen and closed subset of ࣝ݊, then

by theconnectedness ofࣝ݊we have࣫=ࣝ݊.

First we prove that ࣫is open in ࣝ݊. Given(ߣ,ݑ)א࣫, we

show that there exists an openneighborhood ܷof (ߣ,ݑ)in

Թܰ×ܺsuchthat ܷתࣝ݊ؿ࣫. For (ߣ,ݑ)=(Ȧ(ߙ),0)this is

established in Lemma 3.4 . For(ߣ,ݑ)אࣝ݊with ݑ2>0 onԹܰ,

wehave that ݑdoes not change sign sinceܺؿܥ(Թܰ).We

suppose that ݑ>0on Թܰ, the case ݑ<0beingsimilar. By

(ܸ2)and ߣ> ߙ

ܸ(λ), there exist ݎ>0andܴ൒݊such that for all

ߟwith|ߣെߟ |൑ݎ,െߙ+ߟܸ (ݔ)>0for all |ݔ|൒ܴ.Let

ߜ= ݑ(ݔ)

|ݔ|൑ܴ

݂݅݊ . Then, ߜ>0andthere exists an open

neighborhood ܷof (ߣ,ݑ)inԹܰ×ܺsuch that |ߣെߟ|൑ݎ

and ݒ(ݔ)൒ߜ2

|ݔ|൑ܴ

݂݅݊ for all (ߟ,ݒ)אܷ. Sinceܴ൒݊, for

|ݔ|>ܴ,

we have ߰݊(ݔ)=0andfor (ߟ,ݒ)אܷתࣴ݊,ܮ(ߟ)ݒ=െοݒ+

൫െߙ+ߟܸ (ݔ)൯ݒ=0.As the proof of Lemma 3.4, the

maximum principle impliesthat ݒ(ݔ)>0for |ݔ|>ܴand then

ݒ>0onԹܰ. Henceܷתࣝ݊ؿ࣫and ࣫is open.

Now we show that ࣫is closed in ࣝ݊. Supposethat (ߣ ,ݑ)א

ࣝ݊and there exists a sequence{(ߣ݅,ݑ݅)}ؿ࣫such thatߣ݅݅

՜ߣ

and ||ݑ݅െݑ||݅

՜0. Ifݑ=0, we must have ߣ=Ȧ(ߙ)since

ࣝ݊ת[Թ×{0}]={(Ȧ(ߙ),0)}and so (ߣ ,ݑ )א࣫. If ݑ്0,

passing to a subsequence, we may as wellsuppose that ݑ݅>0

on Թܰfor all݅אԳ. Itfollows that ݑ൒0on Թܰandെοݑ+

ܿ+ݑ=ܿെݑ൒0 on Թܰ,where ܿ(ݔ)=െߙ+ߣܸ(ݔ)+

+߰݊݇ቀݔ,|ݑ|

||ݑ||2ቁ. Bythe strong maximum principle we have

ݑ>0on Թܰ. Hence(ߣ,ݑ)א࣫and ࣫is closed inࣝ݊.

Now we know that࣫=ࣝ݊. We claim that thismeans that

case (ii) in Theorem 1.1 cannot occur. Indeed, ifࣝ݊has the

property (ii), there exist ߣאܬ\{Ȧ(ߙ)}and a

sequence{(ߣ݅,ݑ݅)}ؿࣝ݊such thatߣ݅݅

՜ߣand ||ݑ݅||݅

՜0.

Settingݖ݅=ݑ݅

||ݑ݅||and arguing as in the proof of Lemma 3.4, we

may assume thatݖ݅݅

՜ݖ strongly in ܺand ܮ(ߣ)ݖ=0with

ห|ݖ|ห=1. By Lemma 3.3, we know ݖאܪ1(Թܰ). It follows

fromTheorem1.2 (i) that ߣ<Ȧ(ߙ)and ݖchanges sign on Թܰ.

On the other hand, since ࣫=ࣝ݊, the sequence ݖ݅can be

chosen sothat ݖ݅>0on Թܰ. Then we have ݖ൒0on Թܰand

this contradicts the earlier conclusion.

For any ߚ>ߙ

ܸ(λ), by Lemma 3.2 there exists ܰߚאԳsuch

thatห|ݒ|ห൑ܶfor all (ߣ ,ݒ )אࣝ݊with ߣ൒ߚand ݊൒ܰߚ. Thus,

if ݊൒ܰߚand݂݅݊ܲࣝ݊൒ߚ, we deduce that ࣝ݊is boundedin

Թ×ܺ. Since we have shown that ࣝ݊has not the property (ii) in

Theorem 1.1, we must haveࣝ݊satisfying ݂݅݊ܲࣝ݊=݂݅݊ܬ=

ܸ(λ). This contradicts our earlier assumption݂݅݊ܲࣝ݊൒ߚ>

ܸ(λ)and we complete the proof.

IV. PROOF OF THEOREM 1.3

In this section, by using the globalbifurcation result for the

inverted truncated problem (3.3), we first prove the bifurcation

result for the following invertedproblem

Cop

ܮ(ߣ)ݒ+ܭ(ݒ)=0, (4.1)

Whereܭ(ݒ)=൝݇൬ݔ,|ݒ|

ห|ݒ|ห2൰ݒ, ݂݋ݎ ݒאܺ\{0}

0, ݂݋ݎ ݒؠ0. 

Set ࣴ={(ߣ,ݒ)א(ߙ

ܸ(λ),+λ)×ܺ: ݒis a nontrivial

solution of (4.1)}andࣴ෨={(ߣ,ݒ)אࣴ:ݒ2>0 ݋݊ Թܰ}.

Lemma 4.1:Let {(ߣ݊,ݑ݊)}ؿࣴ

෨be a sequencesuch that

ߣ݊݊

՜ߣ> ߙ

ܸ(λ)and||ݒ݊||݊

՜0. Then ߣ=Ȧ(ߙ).

Proof :First we show that

|ܭቆ |ݒ݊|

ห|ݒ݊|ห2ቇ|݌

ห|ݒ݊|ห݊

՜0 (4.2)

for all ݌א(1,+λ). Since(ߣ݊,ݒ݊)solves(4.1), we have

ݒ݊ܮ(ߣ݊)ݒ݊൑0on Թܰ. From ߣ݊݊

՜ߣ> ߙ

ܸ(λ), we may choose

ߚ= ߙ

ܸ(λ)+1

2ቀߣെ ߙ

ܸ(λ)ቁ> ߙ

ܸ(λ)such thatߣ݊൒ߚfor ݊large. It

follows from Lemma 3.3 that there exist ܮ>0and ߛ>0

which areindependent of ݊such that

|ݒ݊(ݔ)|൑ܮ|ݒ݊|λ݁െߛ|ݔ|for allݔאԹܰ. (4.3)

By the Sobolev embedding, there is ܥ>0which is

independent of ݊such that |ݒ݊|λ൑ܥห|ݒ݊|ห. Thus, we have

|ܭቆ |ݒ݊|

ห|ݒ݊|ห2ቇ|

ห|ݒ݊|ห൑ܮܥ݁െߛ|ݔ|for allݔאԹܰ. (4 .4)

Conseuqently, for every fixed ݔאԹܰ,

|ܭቆ |ݒ݊|

ห|ݒ݊|ห2ቇ|

ห|ݒ݊|ห݊

՜0 (4.5)

Combining (4.3) and (4.5), it follows from the

dominatedconvergence that (4.2) holds. Let ߱݊=ݒ݊

||ݒ݊||. Since

ݒ݊2>0on Թܰ, passing to a subsequence, we mayassume that

߱݊>0 on Թܰfor all ݊אԳand for some ߱אܺwith ߱൒0

on Թܰ,߱݊֊߱weakly in ܺ. By (4.2) wededuce that

ܮ(ߣ)߱݊=ܮ(ߣ݊)߱݊െ(ߣ݊െߣ)ܸ(ݔ)߱݊݊

՜0 ݅݊ ܻ

Since ߣ>ߙ

ܸ(λ), we have ܮ(ߣ)אȰ0(ܺ,ܻ). By Lemma 3.5 of

[1] we know that ߱݊݊

՜߱strongly in ܺ. So ܮ(ߣ)߱=0and

ห|߱|ห=1. By Lemma 3.3 we have ߱אܪ1(Թܰ). This means

that ߣis an eigenvalue of equation (1.6) and its

correspondingeigenfunction߱does not change sign. Thus it

follows from Theorem 1.2 (i) thatߣ=Ȧ(ߙ).

Theorem 4.1: Let ࣝdenote the connected component

ofࣴ෨׫{( Ȧ(ߙ),0)}containing{( Ȧ(ߙ),0)}. The following hold.

(i) ࣝis bounded with ݂݅݊ܲࣝ= ߙ

ܸ(λ)and ݏݑ݌ܲࣝ= Ȧ(ߙ).

(ii) If

{(ߣ݊,ݒ݊)}ؿࣝwith ߣ݊݊՜+λ

݈݅݉ =Ȧ(ߙ),then ||ݒ݊||

݊՜+λ

݈݅݉ =

Proof : (i) First we show that if (ߣ,ݒ)אࣴ෨, then ߣ<Ȧ(ߙ).

Sinceݒ്0solves equation (4.1), by Lemma3.3, we knowݒא

ܪ1(Թܰ)and then we deduce thatݑ= ݒ

ห|ݒ|ห2solves equation (1.5).

By Theorem 1.2 (ii) we have

0=݂݅݊ቐන[|׏߱ |2െߙ߱2+ߣܸ߱2]݀ݔ:߱

Թܰאܪ1(Թܰ) ܽ݊݀ න߱

2݀ݔ=1}

Թܰቑ

൑1

׬ݑ2݀ݔ

Թܰ[(Ȧ(ߙ)െߣ)නܸ

(ݔ)ݑ2݀ݔ

Թܰെන݇(ݔ,ݑ)ݑ2݀ݔ]

Թܰ

Thus(Ȧ(ߙ)െߣ)׬ܸ(ݔ)ݑ2݀ݔ

Թܰ൒׬݇(ݔ,ݑ)ݑ2݀ݔ

Թܰ. We claim

that ׬݇(ݔ,ݑ)ݑ2݀ݔ

Թܰ>0. Indeed, since݇(ݔ,ݏ)=݈>0

ݏ՜0

݈݅݉

uniformly in ݔאԹܰ,there exists ߜ>0such that ݇(ݔ,ݏ)൒݈2

for all |ݏ|൑ߜ and ݔאԹܰ. ByProposition 1.1 we have

ݑ(ݔ)

|ݔ|՜+λ

݈݅݉ =0and there exists ܴ>0such that |ݑ(ݔ)|൑

ߜfor|ݔ|൒ܴ. Since (ߣ,ݒ)אࣴ෨, we haveݒ2>0on Թܰand

ݑ2>0on Թܰ. Thereforewe deduce that

׬݇(ݔ,ݑ)ݑ2݀ݔ

Թܰ൒׬݇(ݔ,ݑ)ݑ2݀ݔ

|ݔ|൒ܴ ൒݈2׬ݑ2݀ݔ

|ݔ|൒ܴ .

It follows thatߣ<Ȧ(ߙ). Hence{ߣ: (ߣ,ݒ)אࣝሚ\{(Ȧ(ߙ),0)}ؿ

(ߙ

ܸ(λ),Ȧ(ߙ))and ݏݑ݌ܲࣝ= Ȧ(ߙ).

Secondly we show that ห|ݒ|ห൑ܶfor all (ߣ,ݒ)אࣴ෨, where

ܶ>0is defined in Lemma 3.2. Bycontradiction, if (ߣ,ݒ)א

ࣴand ห|ݒ|ห൒ܶ, similar to the proof of Lemma 3.2, we have

for allݔאԹܰ,ܭ൬ݒ(ݔ)

ห|ݒ|ห2൰൒݈െߟ, whereߟ=1

2ቀߙܸ0

ܸ(λ)െߤቁ>0.

Consequently,

ݒοݒ=[െߙ+ߣܸ(ݔ)+݇൭ݔ,ݒ(ݔ)

ห|ݒ|ห2൱]ݒ2(ݔ)

൒ቂെߤ+ߙܸ0

ܸ(λ)െߟቃݒ2(ݔ)=ߟݒ2(ݔ)൒0.

The maximum principle leads to a contradiction. Therefore we

seethat ࣴ෨is bounded in Թ×ܺand sois ࣝ.

192 Cop

The next step is to prove that ݂݅݊ܲࣝ= ߙ

ܸ(λ). By

contradiction, supposethatߩ׷=݂݅݊ܲࣝ> ߙ

ܸ(λ). For anyߩҧא

(ߙ

ܸ(λ),ߩ), we setܣ={(ߣ,ݒ)א࣫: ߣ൒ߩҧ },ܣ1={( Ȧ(ߙ),0)},

ܣ2={(ߣ,ݒ)א࣫: ߣ=ߩҧ}, where ࣫=ࣴ

෨׫{( Ȧ(ߙ),0)}. We

claim that ܣis a compact subset of Թ×ܺ.

Indeed, let{(ߣ݊,ݒ݊)}ؿܣbe an infinitely sequence. Passing toa

subsequence, we may suppose that ݒ݊>0on Թܰforall

݊andߣ݊݊

՜ߣא[ߩҧ,Ȧ(ߙ)],ݒ݊֊ݒweakly in ܺ,||ݒ݊||݊

՜߬൒

0.To prove this claim, it is sufficient to prove ݒ݊݊

՜ݒstrongly

in ܺwith(ߣ ,ݒ )אܣ .If ߬൒0, then ݒ݊݊

՜ݒstrongly in ܺ. By

Lemma 4.1, we must have (ߣ,ݒ)=( Ȧ(ߙ),0)אܣ. If ്߬0,

similar to the proof ofLemma 4.1, we can deduce that||ݒ݊െ

ݒ||݊

՜0and hence (ߣ ,ݒ )אࣴ. The strong maximum implies

that (ߣ,ݒ )אܣ . By a result of Whyburn [11] (see Lemma C.2

in [10]), we can prove that there exists a connected subset ܣ0of

ܣsuchthatܣ0תܣ1്׎ ܽ݊݀ ܣ0תܣ2്׎ .It follows that

݂݅݊ܲ ܣ 0=ߩҧ<ߩ. But sinceܣ0ؿࣝ, we have ݂݅݊ܲࣝ൑ߩҧ<ߩ,

a contradiction.

(ii) By contradiction, suppose that there exists a

sequence{(ߣ݊,ݒ݊)}ؿࣝ\{( Ȧ(ߙ),0)}such that ߣ݊݊

՜ Ȧ(ߙ)and

ห|ݒ݊|ห൒ߜ>0for all ݊אԳ. Bythe proof of part (i) we

haveห|ݒ݊|ห൑ܶ.Passing to asubsequence, we may assume that

ݒ݊֊ݒweaklyin ܺ. Similar to the proof of Lemma 4.1, we

candeduce that ||ݒ݊െݒ||݊

՜0and ( Ȧ(ߙ),ݒ)אࣴ. Then by the

strong maximum principle wehave( Ȧ(ߙ),ݒ)אࣴ

෨. But at

thebeginning of the proof of part (i) we proved that ߣ<Ȧ(ߙ)

for all (ߣ,ݒ )אࣴ෨. This is a contradiction.

We have established the global properties of a connected

subset ofࣴ෨׫{( Ȧ(ߙ),0)}containing {( Ȧ(ߙ),0)}. However, in

order to maintainconnectedness under inversion, we need to

find a connected subset ofࣴ෨having similar properties.

Set ࣴ+={(ߣ,ݒ)אࣴ:ݒ>0 ݋݊ Թܰ}and ࣴെ={(ߣ,ݒ)א

ࣴ:ݒ<0 ݋݊ Թܰ}.

Corollary 4.1: Let the function ݂be odd. Then there exist

two bounded connectedsubsets ࣝ0+and ࣝ0െof ࣴ+and ࣴെ,

respectively, satisfying the followingproperties:

(i) ݂݅݊ܲࣝ0±=ߙ

ܸ(λ)and( Ȧ(ߙ),0)אࣝ0±

ത

(ii) ݏݑ݌ܲ ࣝ0±= Ȧ(ߙ)and0<ห|ݒ|ห൑ܶfor all (ߣ ,ݒ)אࣝ0±,

where ܶ>0isgiven by Lemma 3.2.

Proof : The proof is the same as the one of Corollary 5.3 in

[10].

Proof of Theorem1.3

Let ݂ܴand ݂ܮbe the odd functions defined by

݂ܴ(ݏ)=൜݂(ݔ,ݏ), ݂݋ݎ ݏ൒0,

െ݂ (ݔ,െݏ), ݂݋ݎ ݏ<0

and݂ܮ(ݏ)=൜െ݂(ݔ,െݏ), ݂݋ݎ ݏ൒0,

݂(ݔ,ݏ), ݂݋ݎ ݏ<0.

By corollary 4.1there exist two bounded connected

subsetsࣝ0+and ࣝ0െof positive or negativesolutions for problem

(1.1) with ݂ܴor݂ܮ(ݏ)respectively. Setting

σ±={൬ߣ, ݒ

ห|ݒ|ห2൰: (ߣ,ݒ)אࣝ0±},

it follows that σ±areconnected sets of (ߙ

ܸ(λ),Ȧ(ߙ))×

ܹ2,݌(Թܰ)consisting of, respectively, positive and negative

solutions of (1.1) with ݂݅݊ܲσ±=ߙ

ܸ(λ),ݏݑ݌ܲσ±= Ȧ(ߙ)and

ห|ݑ|ห൒1

ܶfor all (ߣ,ݒ)אσ±.

Suppose that {(ߣ݊,ݑ݊)}ؿσ±with ߣ݊݊

՜ߣ> ߙ

ܸ(λ)and

|ݑ݊(ݔ)|

ݔאԹܰ

݉ܽݔ ݊

՜λ.Thenห|ݑ݊|ห݊

՜+λby the Sobolev

embedding. Hence(ߣ݊,ݒ݊)אࣴ

෨with ݒ݊=ݑ݊

ห|ݑ݊|ห2and ห|ݒ݊|ห݊

՜0.

ByLemma 4.1 we have ߣ=Ȧ(ߙ). On the otherhand ,

if{(ߣ݊,ݑ݊)}ؿσ±with ߣ݊݊

՜ Ȧ(ߙ), by setting ݒ݊=ݑ݊

ห|ݑ݊|ห2we

know that(ߣ݊,ݒ݊)אࣝ. Then by Theorem 4.1(ii) we have

ห|ݒ݊|ห݊

՜0. This means thatห|ݑ݊|ห݊

՜+λ. We claim that

|ݑ݊|

ݔאԹܰ

݉ܽݔ ݊

՜+λ. Otherwise, passing to a subsequence, there

is ܥ>0such that|ݑ݊|

ݔאԹܰ

݉ܽݔ ൑ܥfor all݊אԳ. Since(ߣ݊,ݑ݊)is

a solution of problem (1.1) we have

ܮ(ߣ݊)ݑ݊+݇(ݔ,ݑ݊)ݑ݊=0 (4. 6)

By Lemma 3.3, we know that ݑ݊isbounded inܻ. Therefore,

{(െο+1)ݑ݊}is bounded in ܻby (4.6). Sinceെο+1:ܺ՜ܻ

isan isomorphism, this implies that {ݑ݊}is bounded in ܺ, a

contradiction.

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