應用巴萊斯麥爾理論解決非線性雙偶合薛丁格方程

國立高雄大學 應用數學系

碩士論文

A Palais-Smale Approach the Two Coupled Nonlinear

Schrodinger Equations

應用巴萊斯麥爾理論解決非線性雙偶合薛丁格方程

研究生：王曉筠 撰

指導教授：吳宗芳

中華民國 九十七 年 七 月

致謝辭

踏出高雄大學的校門，燦爛的陽光灑落，兩年前從台北來求學的黃毛丫頭至現今的 我，想起高大辛苦指導我的師長們，心中滿懷感激。 這篇論文得以完成，首先非常感謝我的指導老師-吳宗芳老師，由於老師的指導與 期勉，無論是研究方向及觀念的導正與指引，老師無一不煞費苦心殷殷指導，要不是老 師幾番地直指問題的核心，以及不時地討論並指點方向，我可能無法完成這篇論文，且 忽略許多重要的細節。當老師的學生真的非常幸福，不管課務公務多麼繁忙，總是給予 學生最多的時間以及幫助。碩一剛開始，老是跟不上老師的腳步，對於隨之丟下來的問 題，總是沒辦法完全的透析理解，有時會搞不懂為什麼要從這個問題開始解決，直到碩 二才察覺到，原來老師這樣做都是有原因，看似沒有相關的拼圖，在老師的循序指點之 下，從混沌無頭緒慢慢地出現了雛形，而自己在不知不覺中，已經緩緩地打開研究的大 門；在老師的課堂上，學習到嚴謹與幽默揉為一體的風格，使我在待人接物上有更深一 層領悟。感謝老師這一路來的細心指引和叮嚀，讓我受益無窮，在此向老師獻上由衷的 敬意與無限的感激。 感謝我的口試老師們：感謝陳晴玉老師，無比認真地審閱我的論文，提出許多缺漏 之處，以及該修正的瑕疵，讓這篇論文更臻豐富完善。感謝陳冠如老師，在百忙之中用 心閱讀我的論文後，給予許多寶貴的意見，幫助我將論文粗糙的面貌，打理得有模有樣， 使得整篇論文有著更完善的架構。 感謝郭岳承老師，在老師的課堂之下，除了學習到線性代數的精髓，還學習到了豁 達的人生態度。感謝黃錦輝老師，雖然沒有機會和老師學習統計方面的專業，但是幾堂 課後的羽球指導，以及幾次充滿哲學的晤談，讓我學習到課堂沒教的事情。感謝王偉仲 老師，雖然我很沒悟性地退出，但是在老師諒解以及影響下，對於沒有程式語言基礎的 我，開啟了重要的那扇窗。 感謝施信宏老師，為我的實變數函數論紮下根基，開拓了我的視野。感謝劉晉良老 師，當初在迷惘不知該走那條路時，給予我鼓勵與指點。感謝鄭斯恩老師，總是和善地 幫忙學生的大小事。感謝所有指導過我的老師，由於您們專業知識的教導，讓我獲得許 多學術知識。 感謝從高中到現在的好友芙民和怡如，在無數的白天與夜晚裡，聽我訴說著一個又 一個突發奇想出來的話題，扮演我生活的最佳支援。 感謝陪我東奔西跑的戰友巍元，總是不計任何距離，可以陪伴我做任何事情，人生 有如此的活力泉源彌足幸福。 感謝陪我走過數年的大學好友們：佩婷、作君、雅婷、明采、巧鈴、俊賢、宣鋒、 詩亭、蕙君、御宸、如意，不管生活上還是課業上，總是願意提供大大小小的幫助，有 你們的陪伴以及協助，讓我感到份外溫暖。

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A Palais-Smale Approach the Two

Coupled Nonlinear Schrodinger

Equations

by

Hsiao-Yun Wang

Advisor

Tsung-Fang Wu

Department of Applied Mathematics,

National University of Kaohsiung

Kaohsiung, Taiwan 811, R.O.C.

July 2008

Contents

1 Introduction

1

2 Preliminary and Nehari Manifolds

3

3 Existence of Least Energy Solutions

15

3.1 0<β<

1/ 4 1/ 2 1

λ

λ

+

···

15

3.2 β<0···

17

應用巴萊斯麥爾理論解決非線性雙偶合薛丁格方程

指導教授：吳宗芳 教授 國立高雄大學應用數學系 學生：王曉筠 國立高雄大學應用數學系 摘要 本篇論文主要內容為，探討半線性雙偶合橢圓方程式最低能量解的存在性問題: 當參數 λ≦1 與β<

β

₀。 關鍵字: 巴萊斯麥爾理論、Nehari 流形

( )

3 2 3 3 2 3 , 0 in , 0 in , , 0 u u u uv v v v u v E u v λ β

β

λ

β

⎧ Δ − + + = ⎪Δ − + + = ⎨ ⎪ _> ⎩ \ \

A Palais-Smale Approach the Two Coupled Nonlinear

Schrodinger Equations

Adviser: Professor Tsung-Fang Wu Institute of Department of Applied Mathematics

National University of Kaohsiung

Student: Hsiao-Yun Wang

Institute of Department of Applied Mathematics National University of Kaohsiung

ABSTRACT

In this thesis, we consider the existence of ground state solutions of the following two coupled semilinear elliptic equations:

where the parameters λ≦1 and β< .

Keywords: Palais-Smale, Nehari Manifolds

( )

3 2 3 3 2 3 , 0 in , 0 in , , 0 u u u uv v v v u v E u v λ β

β

λ

β

⎧ Δ − + + = ⎪Δ − + + = ⎨ ⎪ _> ⎩ \ \ 0

β

1

Introduction

Bose–Einstein condensation (BEC) is a state of matter of bosons confined in an external potential and cooled to temperature very near to absolute zero (0 K or −273◦_{C). This} state of matter was first predicted by Satyendra Nath Bose in 1925. He predicted the occurrence of a phase transition in a gas of non-interacting atoms. The phase transition is associated with the condensation of atoms in the state of lowest energy and is the consequence of quantum statistical effects. In short, at very low temperature, a large fraction of atoms would suddenly go crashing down into the very lowest energy level and the atoms pilimg up in the bottom is what we call BEC.

The following nonlinear Schr¨odinger equation describes physical phenomena such as the condensation of the gas of atoms in the state of lowest temperatures

  

−∆u + u = u3 _{in R}3_,

u > 0.

It has been proved the existence of the least energy solution (ground state solution) i.e. a positive, radial, symmetric solution whose level energy is the minimal among the ones of every possible solution. As we will show some properties later.

The coupled nonlinear Schr¨odinger system is a model of a double condensate such the binary mixture of BEC with two hyperfine states, propagation of pulses in nonlinear optical fiber. Indeed, these phenomena are governed by the following systems:

                 −ε2_{∆u + λ} 1u = µ1u3+ βuv2 in Ω, −ε2_{∆v + λ} 2v = µ2v3+ βu2v in Ω, u, v > 0 in Ω, u = v = 0 on ∂Ω.

where Ω is a smooth and bounded domain of R3_.

The constants ε, λ, µ and β influence the behavior of the ground state solutions. µ1and

µ2 are self-coupling constants and the height of the ground state solutions is influenced by

them. λ1 and λ2 are also self-coupling constants which control the height and the breadth

of the ground state solutions. ε is a diffusion coefficient and the change of the breadth is according to ε. Finally, β is a coupling constant between the ground state solutions u and v; β > 0 means attractive interaction of ground state solutions u and v, on the other hand, β < 0 means repulsive interaction of ground state solutions u and v.

However, in this thesis, we consider the existence of ground state solutions of the following two coupled semilinear elliptic equations on unbounded domain R3 _{by using}

Palais-Smale conditions:          ∆u − u + u3_{+ βuv}2 _{= 0 in R}3_, ∆v − λv + v3_{+ βu}2_{v = 0 in R}3_, u > 0, v > 0, (Eλ,β)

where the parameters λ ≤ 1 and β < β0.

Under the assumption β 6= 0, our Problem (Eλ,β) can be regarded as a coupled problem of the following scalar semilinear elliptic equations:

  

−∆u + λu = u3 _{in R}3_,

u > 0, (Kλ)

where λ > 0. Clearly, the energy functional of equation (Kλ) is defined by

Iλ(u) = 1 2 µZ R3 |∇u|2 + λu2 ¶ −1 4 Z R3 u4 _{for u ∈ H}1¡_R3¢_.

We consider the Nehari minimization problem:

αλ = inf {Iλ(u) | u ∈ Mλ}

where Mλ = {u ∈ H1(R3) : hIλ0 (u) , ui = 0} . Moreover, by Wang-Wu [7, Lemma 7] every minimizing sequence {un} in Mλ for Iλ is a (PS)αλ–sequence in H

1_(R3_{) for I} λ that is Iλ(un) = αλ + o (1) and Iλ0 (un) = o (1) in H−1 ¡ R3¢_. 2

In Section 2, we extend the classical Nehari’s manifold approach to a system of two coupled semilinear elliptic equations in order to find the least energy solution of (Eλ,β) and use the fibering map to observe the behavior of solutions. We obtain some results in Lemma (3) and Lemma (4) to use later. And then by Ekeland variational principle and Implicit Function theorem, we know that there exists a sequence {(un, vn)} ∈ eNλ,β which satisfies the (P S)_θe_λ,β−sequence.

In Section 3, if the energy functional Jλ,β is bounded from below and satisfies the (P S)_θe_λ,β−condition (meaning any (P S)_θe_λ,β−sequence has a covergent subsequence), then

the value of Jλ,β at (u0, v0) is least energy. And then, we can establish the existence of

least energy solutions for Problem (Eλ,β) and it shows that for 0 < β < β0, ground state

solution exists and for β < 0, ground state solution does not exist.

2

Preliminary and Nehari Manifolds

In this section, we have given some results which will be used for later sections. Problem (Eλ,β) is posed in the framework of the Sobolev space H = H1(R3) × H1(R3) with the standard norm k(u, v)k_H = ¡kuk2+ kvk2_λ¢1/2 = µZ R3 |∇u|2+ Z R3 u2+ Z R3 |∇v|2+ λ Z R3 v2 ¶_1/2 .

Moreover, a pair of functions (u, v) ∈ H is said to be a weak solution of problem (Eλ,β) if Z R3 ∇u∇ϕ1+ Z RN uϕ1+ Z R3 ∇v∇ϕ2+ λ Z R3 vϕ2− Z R3 u3ϕ1− Z R3 v3ϕ2 −β Z R3 uv2_ϕ 1 − β Z R3 u2_vϕ 2 = 0 for all (ϕ1, ϕ2) ∈ H. (1)

Thus, the corresponding energy functional of problem (Eλ,β) is defined by

Jλ,β(u, v) = 1 2k(u, v)k 2 H − 1 4 µZ R3 u4₊ Z R3 v4 ¶ − β 2 Z R3 u2_v2

for (u, v) ∈ H. As the energy functional Jλ,β is not bounded below on H, it is useful to consider the functional on the Nehari manifold (see [6]). Because we will find the positive solution of Problem (Eλ,β). According to the change of (1) and the parameters, we will consider the Nehari manifold as follows (or see Lin-Wei [5]):

e Nλ,β =   (u, v) ∈ T : kuk2 =R_R3u4+ β R R3u2v2 kvk2_λ =R_R3v4+ β R R3u2v2    where T = {(u, v) ∈ H : u 6= 0 and v 6= 0} . Thus, if (u, v) ∈ eNλ,β then

k(u, v)k2_H − Z R3 u4₋ Z R3 v4 _{− 2β} Z R3 u2_v2 _{= 0. (2)} 4

It is well known that equation (K1) in R3 has a unique (up to translation), radial

solution (see [1, 2]), which we denote w0. Moreover, wλ(x) := λ1/2w0

³√

λx

´

is the unique (up to translation) solution of Equation (Kλ) in R3and αλ = λ1/2α1. Since wλis a solution of Equation (Kλ) in R3, it holds Z R3 |∇wλ|2 + λw2λ = Z R3 w4_λ = 4αλ. (3)

Moreover, by the Pohozaev identity (see [1, 3]) it follows that 1 2 Z R3 |∇wλ|2+ 3λ 2 Z R3 w2 λ = 3 4 Z R3 w4 λ. (4)

Collecting the preceding indentities, we have

λ Z R3 w2 λ = 1 4 Z R3 w4 λ (5) and _Z R3 |∇wλ|2 = 3 4 Z R3 w4 λ. (6) Thus kw0k2_λ = Z R3 |∇w0|2+ λ Z R3 w2 0 = µ 3 4+ 1 4λ ¶ Z R3 w4 0. Let A = 3

4 + 14λ, clearly A ≤ 1 for all λ ≤ 1. We consider the fibering map h :

R+_{× R}+ _{→ R defined by} h (s, t) = Jλ,β ³√ sw0, √ tw0 ´ = s 2kw0k 2₊ t 2kw0k 2 λ− s2 4 Z R3 w4 0− t2 4 Z R3 w4 0 − stβ 2 Z R3 w4 0 = s 2 Z R3 w4 0 + t 4A Z R3 w4 0 − s2 4 Z R3 w4 0− t2 4 Z R3 w4 0 − stβ 2 Z R3 w4 0 = R R3w₀4 2 · s + tA − s 2 2 − t2 2 − stβ ¸ . Moreover, ∇h (s, t) = R R3w4₀ 2   1 − s − tβ A − t − sβ   . 5

Similar to the argument of Lemma 2.1 (Claim 2) in Lin-Wei [5], h (s, t) has a unique maximum (or critical) point at (s, t) = (smax, tmax) where

  smax tmax   = 1 1 − β2   1 − βA A − β   ₍₇₎

such that h (smax, tmax) ≥ h (s, t) for all s, t ≥ 0, and

¡√ smaxw0, √ tmaxw0 ¢ ∈ eNλ,β. (8)

Now we consider the Nehari minimization problem: e θλ,β = inf n Jλ,β(u, v) | (u, v) ∈ eNλ,β o .

Then we have the following results.

Remark 1 By Theorem 1 in Hirano-Shioji [8] , for each β ∈ (0, β1) , there exists a ground

state solution (u0, v0) ∈ eNλ,β of (Eλ,β) such that

e

θλ,β < α1+ αλ. (9)

Proof. We find that s, t ∈ (0, 1) . Therefore, we have e θλ,β < Jλ,β(su, tv) = 1 4 ¡ s2_kuk2_{+ t}2_kvk2 λ ¢ < 1 4 ¡ kuk2+ kvk2_λ¢= α1+ αλ. Proposition 2 We have e θλ,β ≤ 1 + A2_{− 2βA} 1 − β2 α1, for all β < A. (10) Proof. By (7) , (8) e θλ,β ≤ Jλ,β ¡√ smaxw0, √ tmaxw0 ¢ = R R3w₀4 2 · smax+ tmaxA − s2 max 2 − t2 max 2 − smaxtmaxβ ¸ 6

and by (3) so that α1 = 1₄ R R3w4₀ then e θλ,β ≤ 2α1 · smax+ tmaxA − s2 max 2 − t2 max 2 − smaxtmaxβ ¸ .

Since 0 < β < A and A ≤ 1 we have

smax+ tmaxA − s2 max 2 − t2 max 2 − smaxtmaxβ = 1 − βA 1 − β2 + A2 _{− βA} 1 − β2 − (1 − βA)2 2 (1 − β2₎2 − (A − β)2 2 (1 − β2₎2 − β (1 − βA) (1 − β) (1 − β2₎2 = 1 + A2− 2βA 1 − β2 − 1 − βA 2 (1 − β2₎− A (A − β) 2 (1 − β2₎ = 1 + A2− 2βA 2 (1 − β2₎ and so e θλ,β ≤ 1 + A2_{− 2βA} 1 − β2 α1.

This completed the proof.

Moreover, we have the following basic properties on eNλ,β.

Lemma 3 We have

(i) Jµ,β is coercive and bounded below on eNλ,β; (ii) if 0 < β < λ1/4

1+λ1/2, then there exist c, d0 > 0 such that

min ½Z R3 u4_, Z R3 v4 ¾ ≥ c and _Z R3 u4 Z R3 v4 − β2 µZ R3 u2v2 ¶₂ > d0

for all (u, v) ∈ eNλ,β with Jλ,β(u, v) < α1+ αλ; (iii) if β < 0, then _Z R3 u4 _≥ Z R3 |∇u|2+ u2 _{≥ 4α} 1 7

and Z R3 v4 _≥ Z R3 |∇v|2+ λv2 _{≥ 4α} λ

for all (u, v) ∈ eNλ,β.

In particular, if R_R3u2v2 > 0, then

R

R3u4 > 4α1 and

R

R3v4 > 4α.

Proof. (i) If (u, v) ∈ eNλ,β, then by (2)

Jλ,β(u, v) = 1

4k(u, v)k

2

H (11)

Thus, Jλ,β is coercive and bounded below on eNλ,β. (ii) Let 0 < β < λ1/4

1+λ1/2. For (u, v) ∈ eNλ,β, then

S₄2 µZ R3 u4 ¶_1/2 ≤ kuk2 ≤ Z R3 u4+ β µZ R3 u4 ¶_1/2µZ R3 v4 ¶_1/2 and λ14S2 4 µZ R3 v4 ¶_1/2 ≤ kvk2_λ ≤ Z R3 v4_{+ β} µZ R3 u4 ¶_1/2µZ R3 v4 ¶_1/2 . This implies µZ R3 u4 ¶_1/2 + β µZ R3 v4 ¶_1/2 ≥ S2 4 (12) and β µZ R3 u4 ¶_1/2 + µZ R3 v4 ¶_1/2 ≥ λ14S2 4. (13) Since Jλ,β(u, v) = 1 4k(u, v)k 2 H < α1+ αλ = 1 4S 4 4 ¡ 1 + λ1/2¢_,

by the Sobolev inequality µZ R3 u4 ¶_1/2 + λ14 µZ R3 v4 ¶_1/2 <¡1 + λ1/2¢_S2 4. (14)

Because the set which constitutes of the preceding indentities needs to be a nonempty set. We let X =¡R_RNu4

¢_1/2

and Y =¡R_RNv4

¢_1/2

for simplification. So we have

Pλ,β = n (X, Y ) ∈ R × R : X + βY ≥ S2 4, βX + Y ≥ λ 1 4S2 4 and X + λ14Y ≤ ¡ 1 + λ1/2¢_S2 4 o . Figure 1: Pλ,β Since 0 < β < λ1/4 1+λ1/2, it is clear that 1 < 1 + λ1/2 _{< λ}1/4_β−1 ₍₁₅₎ and λ1/4 _{< λ}−1/4_{+ λ}1/4_{< β}−1_{. (16)}

So we can conclude that Pλ,β 6= ∅ if (15) and (16) hold. And then there exists c > 0 such that

min ½Z R3 u4_, Z R3 v4 ¾ ≥ c. 9

Moreover, by the H˝older inequality, there exists d0 > 0 such that Z R3 u4 Z R3 v4 − β2 µZ R3 u2v2 ¶₂ > Z R3 u4 Z R3 v4 _{− β}2 "µZ R3 u4 ¶1 2 µZ R3 v4 ¶1 2 #2 = Z R3 u4 Z R3 v4 _{− β}2 Z R3 u4 Z R3 v4 > ¡1 − β2¢ Z R3 u4 Z R3 v4 > d0.

(iii) For (u, v) ∈ eNλ,β and β < 0, then we have Z R3 |∇u|2+ u2 = Z R3 u4+ β Z R3 u2v2 ≤ Z R3 u4 and _Z R3 |∇v|2+ λv2 = Z R3 v4 + β Z R3 u2v2 ≤ Z R3 v4.

By the routine computations, there exist s0, t0 > 0 such that s0u0 ∈ M1, t0v0 ∈ Mλ. Moreover, s0 = ÃR R3|∇u0| 2_{+ u}2 0 R R3u4₀ !1 2 ≤ 1 and t0 = ÃR R3|∇v0| 2_{+ λv}2 0 R R3v4₀ !1 2 ≤ 1. Thus, α1 ≤ I1(s0u0) = µ 1 2− 1 4 ¶ s2₀ µZ R3 |∇u0|2+ u20 ¶ ≤ 1 4 µZ R3 |∇u|2+ u2 ¶ and αλ ≤ Iλ(t0v0) = µ 1 2 − 1 4 ¶ t2 0 µZ R3 |∇v0|2+ λv02 ¶ ≤ 1 4 µZ R3 |∇v|2+ λv2 ¶ . 10

This implies Z R3 |∇u|2 + u2 _{≥ 4α} 1 and Z R3 |∇v|2+ λv2 _{≥ 4α} λ. In particular, if R_R3u2v2 > 0, then R R3u4 > 4α1 and R R3v4 > 4α.

Throughout this section, we denote that the Sλ are the best Sobolev constants for the imbedding of H1_(R3_{) into L}4_(R3_{) is given by}

Sλ = inf u∈H1_(R3_)\{0} R R3|∇u| 2_{+ λu2} ¡R R3u4 ¢_1/2 > 0. In particular, Sλ ¡R R3u4 ¢_1/2 ≤R_R3|∇u|

2_{+ λu2 for all u ∈ H1(R3) \ {0} . Furthermore,}

we have the following results.

Lemma 4 For each 0 < β < λ1/4

1+λ1/2 there exists a positive number σ such that

Z

R3

u2_v2 _{≥ σ (17)}

for all (u, v) ∈ eNλ,β with Jλ,β(u, v) < b for some b < α1+ αλ. Proof. Since kuk2 = Z R3 u4+ β Z R3 u2v2 and kvk2_λ = Z R3 v4 + β Z R3 u2v2.

There exist s, t ∈ R+ _{such that}

1 < s2 ₌ kuk 2 R R3u4 = 1 +β R R3u2v2 R R3u4 < 1 + β µ R R3v4 R R3u4 ¶1/2 and 1 < t2 ₌ kvk 2 λ R R3v4 = 1 + β R R3u2v2 R R3v4 < 1 + β µR R3u4 R R3v4 ¶1/2 . 11

Then su ∈ M1 and tv ∈ Mλ. Thus b > Jλ,β(u, v) = 1 4 £ kuk2+ kvk2_λ¤ = 1 4 ·µ s2− β R R3u2v2 R R3u4 ¶ kuk2+ µ t2− β R R3u2v2 R R3v4 ¶ kvk2_λ ¸ = 1 4 ( £ ksuk2+ ktvk2_λ¤− β Z R3 u2v2 " kuk2 R R3u4 + kvk 2 λ R R3v4 #) ≥ (α1+ αλ) − βR_R3u2v2 4 Ã kuk2 R R3u4 + kvk 2 λ R R3v4 ! ≥ (α1+ αλ) − βR_R3u2v2 4c ¡ kuk2+ kvk2_λ¢ ≥ (α1+ αλ) − βR_R3u2v2 c (α1+ αλ) and so β (α1 + αλ) c Z R3 u2_v2 _{≥ (α} 1+ αλ) − b. Moveover, (α1+ αλ) − b > 0. Thus Z R3 u2_v2 _≥ c (α1+ αλ − b) β (α1+ αλ) .

This completed the proof.

Theorem 5 If 0 < β < λ1/4

1+λ1/2 and Problem (Eλ,β) has a nontrivial solution (u0, v0) ∈

e

Nλ,β, then u0 > 0 and v0 > 0.

Proof. By Lemma 3 (iii) .

Moreover, we have the following results.

Lemma 6 If 0 < β < λ1/4

1+λ1/2, then for each (u, v) ∈ eNλ,β, there exist ² > 0 and a

differentiable function (s, t) : B (0; ²) ⊂ H → R+_×R+ _{such that (s (0, 0) , t (0, 0)) = (1, 1),}

the function (s (eu, ev) (u − eu) , t (eu, ev) (v − ev)) ∈ eNλ,β for all (eu, ev) ∈ B (0; ²) . Furthermore,   hs0(0, 0) , (φ, ψ)i ht0_{(0, 0) , (φ, ψ)i}   =   f1(u, v, φ, ψ) R R3v4− βf2(u, v, φ, ψ) R R3u2v2 −βf1(u, v, φ, ψ) R R3u2v2+ f2(u, v, φ, ψ) R R3u4   2R_R3u4 R R3v4− 2β2 ¡R R3u2v2 ¢₂ (18) 12

where f1(u, v, φ, ψ) = 2 Z R3 ∇u∇φ + uφ − 4 Z R3 u3_{φ − 2β} Z R3 uv2_{φ + u}2_vψ and f2(u, v, φ, ψ) = 2 Z R3 ∇v∇ψ + vψ − 4 Z R3 v3_{ψ − 2β} Z R3 uv2_{φ + u}2_vψ for all (φ, ψ) ∈ H.

Proof. Define a function F : H × R2 _{→ R}2 _{given by}

F(u,v)(w1, w2, s, t) =   F1(w1, w2, s, t) F2(w1, w2, s, t)   where F1(w1, w2, s, t) = s ku − w1k2_H1 − s3 Z R3 (u − w1)4− βst2 Z R3 (u − w1)2(v − w2)2 and F2(w1, w2, s, t) = t kv − w2k2_H1 − t3 Z R3 (v − w2)4− βs2t Z R3 (u − w1)2(v − w2)2.

Since (u, v) ∈ eNλ,β, by Lemma 3 (ii) we have F (0, 0, 1, 1) = (0, 0) and ¯ ¯ ¯ ¯ ¯ ¯ ∂ ∂sF1(0, 0, 1, 1) ∂t∂F1(0, 0, 1, 1) ∂ ∂sF2(0, 0, 1, 1) ∂t∂F2(0, 0, 1, 1) ¯ ¯ ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ ¯ ¯ −2R_R3u4 −2β R R3u2v2 −2βR_R3u2v2 −2 R R3v4 ¯ ¯ ¯ ¯ ¯ ¯ ≥ 4d0

where d0 > 0 as in Lemma 3 (ii). According to the Implicit Function Theorem, there

exists a differentiable function

(s, t) : B (0; ²) ⊂ H → R+× R+

such that (s (0) , t (0)) = (1, 1) and

F(u,v)(w1, w2, s (eu, ev) , t (eu, ev)) = (0, 0) for all (eu, ev) ∈ B (0; ²) .

This is equivalent to (s (eu, ev) (u − eu) , t (eu, ev) (v − ev)) ∈ eNλ,β.

Proposition 7 If 0 < β < λ1/4

1+λ1/2, then there exists {(un, vn)} ⊂ eNλ,β such that

Jλ,β(un, vn) = eθλ,β + o (1) ,

J0

λ,β(un, vn) = o (1) in H∗ as n → ∞.

Proof. By the Ekeland variational principle [4], there exists a minimizing sequence

{(un, vn)} ⊂ eNλ,β such that

Jλ,β(un, vn) < eθλ,β+ 1 n (19) and Jλ,β(un, vn) < Jλ,β(w1, w2) + 1 nk(w1, w2) − (un, vn)kH (20) for all (w1, w2) ∈ eNλ,β. By application of Lemma 6 for (un, vn) , we obtain the function

(sn, tn) : B (0; ²n) → R+× R+ for some ²n > 0

such that (s (eu, ev) (un− eu) , t (eu, ev) (vn− ev)) ∈ eNλ,β. Choose 0 < ρ < ²n. Let (u, v) ∈ H with u 6≡ 0, v 6≡ 0 and let (φρ, ψρ) = _k(u,v)k(ρu,ρv)

H1. We set

yρ = s (φρ, ψρ) (un− φρ) and zρ= t (φρ, ψρ) (vn− ψρ) . Since (yρ, zρ) ∈ eNλ,β, we deduce from (20) that

Jλ,β(yρ, zρ) − Jλ,β(un, vn) ≥ − 1

nk(yρ, zρ) − (un, vn)kH and by the Mean Value Theorem, we have

­ J0 λ,β(un, vn) , (yρ, zρ) − (un, vn) ® + o¡k(yρ, zρ) − (un, vn)k_H ¢ ≥ −1 n k(yρ, zρ) − (un, vn)kH. Thus, −­J0 λ,β(un, vn) , (φρ, ψρ) ® +­J0 λ,β(un, vn) , ([s (φρ, ψρ) − 1] (un− φρ) , [t (φρ, ψρ) − 1] (vn− ψρ)) ® ≥ −1 nk(yρ, zρ) − (un, vn)kH + o ¡ k(yρ, zρ) − (un, vn)k_H ¢ . (21) 14

By the fact that (yρ, zρ) = (s (φρ, ψρ) (un− φρ) , t (φρ, ψρ) (vn− ψρ)) ∈ eNλ,β and (21), we have −ρ ¿ J0 λ,β(un, vn) , (u, v) k(u, v)k_H À +­J0 λ,β(un, vn) − Jλ,β0 (yρ, zρ) , ([s (φρ, ψρ) − 1] (un− φρ) , [t (φρ, ψρ) − 1] (vn− ψρ)) ® ≥ −1 nk(yρ, zρ) − (un, vn)kH + o ¡ k(yρ, zρ) − (un, vn)k_H ¢ . Thus, ¿ J_λ,β0 (un, vn) , (u, v) k(u, v)k_H À ≤ k(yρ, zρ) − (un, vn)kH nρ + o¡k(yρ, zρ) − (un, vn)k_H ¢ ρ +­J_λ,β0 (un, vn) − Jλ,β0 (yρ, zρ) , ([s (φρ, ψρ) − 1] (un− φρ) , [t (φρ, ψρ) − 1] (vn− ψρ)) ® .

On the other hand, by Lemma 3 (ii) and (18) we can find a constant C > 0, independent of ρ, such that lim ρ→0 |s (φρ, ψρ) − 1| ρ ≤ ks 0_{(0, 0)k ≤ C,} lim ρ→0 |t (φρ, ψρ) − 1| ρ ≤ kt 0_{(0, 0)k ≤ C} and k(yρ, zρ) − (un, vn)k_H ≤ C (ρ + max {|s (φρ, ψρ) − 1| , |t (φρ, ψρ) − 1|}) .

If we let ρ → 0 in (21) for a fixed n and use the fact that (yρ, zρ) → (un, vn) in H, we get ¿ J0 λ,β(un, vn) , (u, v) k(u, v)k_H À ≤ C n.

This shows that Jλ,β(un, vn) = eθλ,β + o (1) and Jλ,β0 (un, vn) = o (1) in H∗ as n → ∞.

3

Existence of Least Energy Solutions

In this section, we establish the existence of least energy solutions for Problem (Eλ,β).

3.1

0 < β <

_1+λλ1/41/2

Theorem 8 Problem (Eλ,β) has a least energy solution (u0, v0) such that u0 > 0 and

v0 > 0 in R3.

Proof. By Proposition 7, there exists a sequence {(un, vn)} ⊂ eNλ,β such that

Jλ,β(un, vn) → eθλ,β and Jλ,β0 (un, vn) = o (1) in H∗. (22) Then by Lemma 3 (i) there exists a subsequence {(un, vn)} and (u0, v0) ∈ H such that

(un, vn) * (u0, v0) weakly in H, (23)

(un, vn) → (u0, v0) strongly in Lloc4 (R3) × L4loc(R3), (24)

(un, vn) → (u0, v0) a.e. in R3. (25)

From (23) and (25) , we can find a weak solution of Problem (Eλ,β) . Since (0, 0) also is a solution of Problem (Eλ,β) , we still need to show (u0, v0) 6= (0, 0) . By Lemma 4 there

exist yn∈ R,and d0, R ∈ R+ such that

Z B(yn,R) u2 nvn2 ≥ d0 for all n ∈ N. (26) Now, we define (un, vn) = (un(x + yn) , vn(x + yn)) . (27) From the translation invariance of the functional, we get that {(un, vn)} also satisfies (un, vn) ∈ eNλ,β,

Jλ,β(un, vn) → eθλ,β and Jλ,β0 (un, vn) = o (1) in H∗.

So that, by the same argument for {(un, vn)} , there exist a subsequence {(un, vn)} and (u0, v0) ∈ H such that

(un, vn) → (u0, v0) weakly in H.

Moreover, (26) implies that Z

B(0,R)

u2

nv2n ≥ d0 > 0 for all n ∈ N. (28)

Indeed, from (27) is follows Z B(0,R) u2 nv2n = Z B(0,R) u2 n(x + yn) vn2(x + yn) = Z B(yn,R) u2_nv2_n,

and (28) follows easily from (26) . Since

(un, vn) → (u0, v0) strongly in L4loc(R3) × L4loc(R3), we pass the limit into (28) and we obtain

Z B(0,R)

u2

0v20 ≥ d0,

which implies that (u0, v0) ∈ eNλ,β. Now, we will show that, in fact Jλ,β(u0, v0) = eθλ,β. By the Fatou lemma

e θλ,β ¡ R3¢ = lim n→∞Jλ,β(un, vn) = lim n→∞ µ 1 2− 1 4 ¶ µZ R3 u4 n+ Z R3 v4 n+ 2β Z R3 u2 nv2n ¶ ≥ µ 1 2− 1 4 ¶ µZ R3 u4 0+ Z R3 v4 0+ 2β Z R3 u2 0v20 ¶ = Jλ,β(u0, v0) .

Thus, (u0, v0) is a least energy solution of Problem (Eλ,β) in R3. By the maximum principle

u0 > 0 and v0 > 0 in R3.

3.2

β < 0

Theorem 9 Problem (Eλ,β) in R3 does not admit any least energy solution and eθλ,β =

α1+ αλ.

Proof. Let w1 and w2 be ground state solution of equations (E1,β) and (Eλ,β) . We define

w1(x) = w1(x + Rne) and w2(x) = w2(x − Rne) where e = (1, 0, 0) and Rn → ∞ as

n → ∞. Consider the cut-off function ξ ∈ C∞ _{such that 0 ≤ ξ ≤ 1 and}

ξ (t) =    1 0 t ∈£0,1 2 ¤ t ∈ [1, ∞) . And let ξRn,i(x) = ξ µ |x+(−1)i+1Rne| Rn ¶

for i = 1, 2 and let

un(x) = ξRn,1(x) w1(x + Rne) , vn(x) = ξRn,2(x) w2(x − Rne) . Then {(un, vn)} ⊂ H Z R3 u2_nv_n2 = ξ_R2_n,1(x) w2₁(x) ξ_R2_n,2(x) w2₂(x) = 0, (29) kunk2 = Z R3 u4 n+ o (1) , (30) kvnk2λ = Z R3 v4 n+ o (1) , (31) and Jλ,β(un, vn) = I (w1) + Iλ(w2) + o (1) = α1+ αλ+ o (1) .

By the routine computations and (29) − (31), there exist s(i)n ⊂ R+ i = 1, 2 such that

s(i)n = 1 + o(1) and ° °s(1) n un ° °2 H = Z R3 ¡ s(1)_n un ¢4 + β Z R3 ¡ s(1)_n un ¢2¡ s(2)_n vn ¢2 ; ° °s(2) n vn ° °2 H = Z R3 ¡ s(2) n vn ¢4 + β Z R3 ¡ s(1) n un ¢2¡ s(2) n vn ¢2 . 18

This implies { ³ s(1)n un, s(2)n vn ´ } ⊂ eNλ,β and e θλ,β ≤ Jλ,β(s(1)n un, s(2)n vn) = α1+ αλ+ o(1). Therefore, e θλ,β ≤ α1+ αλ.

Next we claim that eθλ,β ≥ α1+ αλ. In fact, let (u, v) ⊂ eNλ,β, then

Jλ,β(u, v) = 1 4k(u, v)k 2 H. By Lemma 3 (iii)

Jλ,β(u, v) ≥ α1 + αλ, for all (u, v) ∈ eNλ,β. Thus,

e

θλ,β ≥ α1+ αλ. This proves that

e

θλ,β = α1+ αλ.

Finally we prove that Problem (Eλ,β) does not admit any least energy solution. Suppose otherwise, then there exists (u0, v0) ∈ eNλ,β is a least energy solution of Problem (Eλ,β) such that

Jλ,β(u0, v0) = eθλ,β = α1+ αλ.

By Maximum principle, u0 > 0 and v0 > 0. Then by Lemma 3 (iii) again

Jλ,β(u0, v0) = 1 4k(u0, v0)k 2 H > α1+ αλ which is a contradiction. This completes the proof.

Lemma 10 For each ε > 0 there exists δ0, d0 > 0 such that for (u, v) ∈ eNλ,β with Jλ,β(u, v) ≤ eθλ,β+ δ0 we have (i) R_R3u2v2 < ε; (ii) R_R3u4 R RNv4− β2 ¡R R3u2v2 ¢₂ > d0.

Proof. (i) Suppose otherwise, there exist c0 > 0 and a sequence {(un, vn)} ⊂ eNλ,β such that Jλ,β(un, vn) = eθλ,β+ o (1) and _Z R3 u2_nv_n2 > c0. (32) Then kunk2H1 < Z R3 u4_n and kvnk2H1 < Z R3 v_n4.

By (32) and Lemma 3 (iii) e θλ,β+ o (1) = Jλ,β(un, vn) = 1 4 ¡ kuk2+ kvk2_λ¢ = 1 4 µZ R3 u4 n+ Z R3 v4 n ¶ −β 2 Z R3 u2 nv2n > 1 4 µZ R3 u4 n+ Z R3 v4 n ¶ +|β| c0 2 ≥ α1+ αλ+ |β| c0 2 . This contradicts Theorem 9.

(ii) By Lemma 3 (iii) and part (i) .

Proposition 11 For each minimizing sequence {(un, vn)} in eNλ,β for Jλ,β we have (i) {un} is a (PS)α1–sequence in H 1 _{for I} 1; (ii) {vn} is a (PS)αλ–sequence in H 1 _{for I} λ. 20

Proof. By Lemma 10 (i) , Z R3 u2 nv2n→ 0 as n → ∞. Then kunk2 = Z R3 u4 n+ o (1) and kvnk2λ = Z R3 v4 n+ o (1) . This implies Jλ,β(un, vn) = 1 2k(un, vn)k 2 H − 1 4 µZ R3 u4 n+ Z R3 v4 n ¶ + o (1) = I1(un) + Iλ(vn) = α1+ αλ.

Then by Lemma 3 (iii) , we can conclude that I1(un) = α1+o (1) and Iλ(vn) = αλ+o (1) . Moreover, by Wang-Wu [7, Lemma 7] {un} is a (PS)α1(R3)–sequence in H

1 0(R3) for I1 and {vn} is a (PS)αλ(R3)–sequence in H 1 0(R3) for Iλ. 21

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