Multiple existence of sign changing solutions for coupled nonlinear Schrödinger equations

Nonlinear Analysis 73 (2010) 2580–2593

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Multiple existence of sign changing solutions for coupled nonlinear Schrödinger equations Norimichi Hirano Department of Mathematics, Graduate School of Environment and Information Sciences, Yokohama National University, Tokiwadai, Hodogayaku, Yokohama, Japan

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Article history: Received 22 November 2009 Accepted 10 June 2010

abstract In this paper, we consider the multiple existence of sign changing solutions of coupled nonlinear Schrödinger equations

 MSC: primary 35B40 35B45 secondary 35J40

−∆u + µ1 u = λ1 u3 + βv 2 u in R3 −∆v + µ2 v = λ2 v 3 + β u2 v in R3

where µ1 , µ2 , λ1 , λ2 are positive and β < 0.

(P)

© 2010 Elsevier Ltd. All rights reserved.

Keywords: Coupled Schrödinger equations Sign changing solutions

1. Introduction In the present paper, we consider the multiple existence of nonradial sign changing solutions of coupled Schrödinger equations



−∆u + µ1 u = λ1 u3 + β uv 2 −∆v + µ2 v = λ2 v 3 + β u2 v

in R3 in R3 ,

(P)

where µ1 , µ2 , λ1 , λ2 are positive numbers and β < 0 is a coupling constant. In recent years, a lot of work has been devoted to the study of coupled nonlinear Schrödinger equations. Coupled nonlinear Schrödinger equations (P) models physical phenomena such as the binary mixture of Bose–Einstein condensates, propagation of pulses in nonlinear optical fiber (cf. [1–5]). In the case that the coupling constant β is positive, problem (P) was studied by Ambrosetti and Colorado [6], Maia et al. [7] and Lin and Wei [8]. They proved the existence of a ground state solution, i.e., a positive solution whose energy level is the minimal among the possible nontrivial solutions. The ground state solution is known to be radial (cf. [9]). On the other hand, in case that the coupling constant β is negative, it is known that there is no ground state solution of (P). In this case, the existence of radial positive solutions was established in [10]. The existence of positive solutions was also established in [11] for the case that µ1 = µ2 . Especially, they showed that if β < −1, there exists a nonradial positive solution of (P). For the case that µ1 6= µ2 , the existence of multiple positive solutions of problem (P) was considered in Lin and Wei [12] and the author [13]. They showed the existence of positive solutions which have multiple spikes, and showed that the spikes of the solutions form polygons in two dimensional subspaces of R3 or polyhedrons in R3 . On the other hand, little seems to be known about the existence of sign changing solutions for (P). In the present paper, we show the multiple existence of nonradial sign changing solutions of (P) for β < 0, µ1 6= µ2 and λ1 6= λ2 . To state our main results, we need a few notations. For each θ ∈ [0, 2π ) and x = (x1 , x2 , x3 ) ∈ R3 , we define a

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N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

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mapping σ (θ ) : R3 −→ R3 by

σ (θ )x = ((cos θ )x1 − (sin θ )x2 , (sin θ )x1 + (cos θ )x2 , x3 ).
We denote by li the half lines defined by l1 = {(x, 0, 0) : x > 0} and li = σ i−1 π2 l1 for 2 ≤ i ≤ 4. We also denote bye li the


half lines defined by e l1 = {(x, x, 0) : x > 0} ande li = σ i−1 π2 (e l1 ) for 2 ≤ i ≤ 4. We put H = H 1 (R3 ) and H = H × H. We set µ0 = 1 and denote by k·kµi the norm of H defined by kuk2µi =

(|∇ u|2 + µi |u|2 )dx for u ∈ H and i ∈ {0, 1, 2}. The Hilbert space H is equipped with the norm defined by kUk = kU kµ1 + kV k2µ2 for U = (U , V ) ∈ H. We recall that for each i ∈ {0, 1, 2}, the problem  −∆u + µi u = u3 in R3 (Pi ) u(x) −→ 0 as |x| −→ ∞ 2

R

2

R3

has a positive radial solution, denoted by Ui,0 (cf. [14,15]). The function Ui,0 is the unique positive smooth solution of (Pi ) up to translation. Moreover we know that Ui,0 satisfies Ii (Ui,0 ) = ci = min {Ii (v) : v ∈ Si }, where Ii is the functional associated with problem (Pi ) defined by Ii (v) =

1

kvk2µi −

1

2 4 and Si is the set defined by

|v|44

for v ∈ H

Si = v ∈ H \ {0} : kvk2µi = |v|44



(1.1)

for i ∈ {0, 1, 2}.



From the definition of Si , we find Ii (v) =

kvk2µi 4

=

|v|44 4

≥ ci for i ∈ {0, 1, 2} and v ∈ Si .

(1.2)

For each x ∈ R3 and i ∈ {0, 1, 2}, we put Ui,x (·) = Ui,0 (· − x). In the following, we consider, for simplicity of notations and the arguments, the problem



−∆ u + µ 1 u = λ 2 u 3 + β u v 2 −∆ v + µ 2 v = v 3 + β u 2 v

in R3 in R3

(1.3)

instead of problem (P). That is we consider the case λ1 = λ2 and λ2 = 1. The general case can be obtained by a slight modification of the arguments below. We can now state our main results.

√


−1 √

π Theorem 1. Let k ∈ N and suppose that µ1 > 1 − sin 4k problem (1.3) possesses a solution Uk = (U , V ) ∈ H of the form

Uk = ( U , V ) =

λ

−1

4k 4k X X (−1)i−1 U1,wi + u, U2,zi + v i=1

µ2 . Then there exists λk > 0 such that for each λ ∈ (0, λk ),

! (1.4)

i =1



π π where wi , zi ∈ li such that wi = σ i−1 2k w1 , zi = σ i−1 2k z1 for 1 ≤ i ≤ 4k, z1 = aw1 for some a > 1, and u, v ∈ H with kuk2µ1 < λ−2 c1 and kvk2µ2 < c2 .

Remark 1. By the invariance of problem (1.3) under translation and rotation, each function U obtained by translation and/or rotation of the function Uk is also a solution of (1.3). For each solution Uk = (U , V ), U and V have 4k spikes and the spikes form a regular 4k polygon in a two dimensional subspace of R3 . The function U changes the sign. It is not clear if we can expect a solution U = (U , V ) such that both of U and V change signs. The approach employed here does not work to find such a solution. It is natural to ask if there is a sign changing solution whose spikes form a polyhedron in R3 . We have Theorem 2. There exists λ0 > 0 such that for each λ ∈ (0, λ0 ), the problem (1.3) has a solution U0 = (U , V ) ∈ H such U0 has the form

U0 =

λ

−1

8 8 X X (−1)n(i) U1,wi + u, U2,zi + v i=1

! (1.5)

i=1

where {zi }8i=1 forms a cube in R3 centered at 0, zi = awi for each 1 ≤ i ≤ 4k with a > 1, n(i) ∈ {0, 1} such that n(i) 6= n(j) if wi and wj are adjacent vertices of the cube, and u, v ∈ H with kuk2µ1 < λ−2 c1 and kvk2µ2 < c2 . To avoid unnecessary complexity, we give a proof of Theorem 1 for the case k = 1 in Sections 2 and 3. The proofs of Theorem 1 for k 6= 2 and the proof of Theorem 2 are slight modifications of that for the case k = 1 of Theorem 1. The sketch of proofs for these cases are given in Section 4.

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N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

2. Preliminaries We denote by Br (x) the open ball in R3 centered at x ∈ R3 with radius r > 0. For each p ≥ 1, we denote by |·|p the norm R of the space Lp (R3 ). For each u, v ∈ H = H 1 (R3 ), we put hu, vi = R3 uv dx. We denote by hh·, ·iiµi the inner product of H defined by hhu, viiµi =

(∇ u · ∇v + µi uv)dx for u, v ∈ H and i ∈ {0, 1, 2}. For i ∈ {0, 1, 2}, A ⊂ H and u ∈ H , ku − Ak2µi  stands for the distance of u from A, i.e., ku − Ak2µi = inf ku − vk2µi : v ∈ A . Let Ω be an open subset of R3 and u ∈ H 1 (R3 ). We identify the restriction u|Ω of u on Ω with the function u ∈ L6 (RN ) defined by  u(x) for x ∈ Ω u|Ω (x) = 0 for x ∈ Ω c . R

R3

For the norm of H 1 (Ω ), we use the same notation k·kµi as that for the norm of H 1 (R3 ). That is

kuk2µi =

Z

(|∇ u|2 + µi |u|2 )dx for i ∈ {1, 2} and u ∈ H 1 (Ω ). F

Each function u ∈ H 1 (Ω ) is identified with the function in L6 (R3 ), denoted again by u, which is an extension of u with u(x) = 0 on Ω c . For each u ∈ H, we set u+ (x) = max {u(x), 0} for x ∈ R3 . For each a ∈ R, and a functional F : H → R, we denote by F a the level set defined by F a = {v ∈ H : F a (v) ≤ a}. Let i ∈ {0, 1, 2}. Then it is easy to see that for each u ∈ H \ {0}, there exists a unique positive number t such that tu ∈ Si (cf. [16]). We define a mapping Ni : H \ {0} → Si √ √ by Ni (u) = tu. It follows from the definitions of Ui,0 that Ui,0 (x) = µi U0,0 ( µi x) on R3 . In the following, we assume that

µ1 , µ2 satisfy the assumption of Theorem 1 for k = 1. That is we assume that

 √ µ1 > 1 −

−1 √ µ2 . Then one can see 2

√1

that c1 =

√ √ µ1 c0 > c2 = µ2 c0 .

(2.1)

It is also known that there exists c > 0 such that U0,0 (x) · |x| exp(|x|) −→ c ,

as |x| −→ ∞

(2.2)

(cf. [15]). From (2.2), we can choose R0 > 0 such that for each i ∈ {1, 2}, √

√

2−1 c |x|−1 e− µi |x| ≤ Ui,0 (x) ≤ 2c |x|−1 e− µi |x|





for all |x| ≥ R0 ,

(2.3)

and

Z BR (0)

  ∇ Ui,0 2 + µi Ui,0 2 ≥ 2ci .

(2.4)

0

We define a functional Φ : H −→ R associated with problem (1.3) by

Φ (U) =

1 2

1

β

4

2

(kU k2µ1 + kV k2µ2 ) − (λ2 |U |44 + |V+ |44 ) −

= Φ1 (U) + Φ2 (U) for U = (U , V ) ∈ H,

Z R3

U 2 V+2

where

Φ1 (U) =

1 2

kU k2µ1 −

λ2 4

|U |44 −

1

Z

4

R3

β U 2V 2

and

Φ2 (U) =

1 2

kV k2µ2 −

1 4

|V+ |44 −

1 4

Z R3

β U 2 V+2 .

Then we have

   ∇u Φ (U) −∆U + µ1 U − λ2 U 3 − β UV 2 ∇ Φ (U) = = ∇v Φ (U) −∆V + µ2 V − V+3 − β U 2 V+ 

for U = (U , V ) ∈ H. Then each
critical point  U = (U , V ) ∈ H of Φ is asolution of (1.3) with V ≥ 0. In the following, we simply write σ instead of σ π2 . Let G1 = σ− , σ−2 , σ−3 , γ1 , γ2 , G2 = σ+ , σ+2 , σ+3 , γ1 , γ2 be groups of linear mappings on H, where

σ+ (u)(x) = u(σ x), σ− (u)(x) = −u(σ x), γ1 (u)(x) = u(−x1 , x2 , x3 ) and γ2 (u)(x) = u(x1 , x2 , −x3 )

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

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for u ∈ H and x = (x1 , x2 , x3 ) ∈ R3 . Let H1 = {u ∈ H : g (u) = u for g ∈ G1 } ,

H2 = {u ∈ H : g (u) = u for g ∈ G2 }

and

H0 = H1 × H2 . One can see by invariance of the functional Φ under the group action G1 × G2 on H, each critical point of Φ in H0 is a solution of (1.3). We put



M = (U , V ) ∈ H0 : U 6≡ 0, V 6≡ 0, kU k2µ1 = λ2 |U |44 + β

Z R3

U 2 V 2 , kV k2µ2 = |V+ |44 + β



Z R3

U 2 V+2 .

Then if U = (U , V ) ∈ H1 × H2 is a critical point of Φ such that U 6≡ 0, V 6≡ 0 and V ≥ 0, then U is contained in M . Moreover we have that U is sign changing. We note that from the definition,

Φ1 (U) =

1 4

kU k2µ1 ,

Φ2 (U) =

1

for U = (U , V ) ∈ M .

kV k2µ2

4

(2.5)

It is also easy to verify that

h∇ Φ (U), (U , V )i = h∇u Φ (U), U i + h∇v Φ (U), V i = 0,

for each U = (U , V ) ∈ M .

(2.6)

For each r ≥ 0, we put Er ,0 = Br (0), Er ,1 = x = (x1 , x2 , x3 ) ∈ R3 : x1 ≥ 0, x2 ≥ 0, |x| ≥ r ,





and Er ,i = σ i−1 (Er ,1 ) for 2 ≤ i ≤ 4. We simply write E1 , . . . , E4 instead of E0,1 , . . . , E0,4 , respectively. We put Er ,1 (θ ) = σ (θ )(Er ,1 ) for r > 0 and θ ∈ [0, 2π ). For simplicity, we put Fi = E0,i − π4 for 1 ≤ i ≤ 4. Let θ ∈ [0, 2π ), u ∈ H 1 (Ei (θ )) and v ∈ H 1 (R3 ). Since u is extended to R3 by putting u = 0 on Ei (θ )c , we can define the distance ku − vkµ2 of u from v by




ku − vk2µ1 =

Z Ei (θ)

(|∇(u − v)|2 + µk |u − v|2 )dx +

Lemma 1. There exist δ0 > 0 and ε0 ∈ 0,

c2 6




Z Ei (θ )c

(|∇v|2 + µk |v|2 )dx.

satisfying the following conditions:

2 (a) For each i ∈ {1, 2} and U ∈ H such that infz∈R3 U − Ui,z µ < ε0 , there exists a unique element x of R3 such that i







U − Ui,x 2 = inf U − Ui,z 2 . µ µ i

(2.7)

i

z∈R3

2 (b) For each i ∈ {1, 2}, θ ∈ [0, 2π ) and U ∈ H 1 (Ei (θ )) such that infz∈R3 U − Ui,z µ < ε0 , there exists a unique element x of i R3 such that





U − Ui,x 2 = inf U − Ui,z 2 . µ µ i

i

z∈R3

(2.8)

c +δ0 ,

2 ε0 , infz∈R3 U − Ui,z µ < 24 . i  Proof. (a) Let i ∈ {1, 2}. Then since the set Θi = Ui,z : z ∈ R3 is a smooth manifold in H and the principle curvature at

2 each point of Θi is the same, we have that there exists ε0 > 0 such that for each U ∈ H such that infz∈R3 U − Ui,z µ < 4ε0 , i there exists a unique element x ∈ R3 satisfying (2.7). (b) Let i ∈ {1, 2}, θ ∈ [0, 2π ) and U ∈ H 1 (Ei (θ )) such that

2 infz∈R3 U − Ui,z µ < ε0 . Then since H = H 1 (R3 ) is a closed linear space, we have that there exists a unique element (c) For each i ∈ {1, 2} and U ∈ Si ∩ Ii i

i

U of H such that



U − U 2 = inf kU − V k2 . µi µ i

V ∈H

2

2 Since U − U µ < ε0 , we have that infz∈R3 U − Ui,z µ < 4ε0 . Then by (a), we have that there exists a unique x ∈ R3 such i

2i

2

2

2

2 that U − Ui,x µ = infz∈R3 U − Ui,z µ . Then noting that U − Ui,x µ = U − U µ + U − Ui,x µ , we find that (2.8) i i i i i holds. (c) It is well known that for δ0 sufficiently small, the assertion holds. 

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N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

By Lemma 1, we can define mappings Pi , QE and QF as follows: for each i ∈ {1, 2},





2

Pi U = x for U ∈ W ∈ H : inf W − Ui,z µ < ε0 , i

z∈R3

2

2 where x is the unique element of R3 such that U − Ui,x µ = infz∈R3 U − Ui,z µ . i

i

Similarly, we define a mapping QE by

 QE U = x for U ∈



2 V ∈ H 1 (E1 ) : inf V − U2,z µ < ε0 ,



2

z∈R3

2

2 where x is the unique element of R3 such that U − U2,x µ = infz∈R3 U − U2,z µ . Mapping QF is defined by the same 2

2

way as above with E1 replaced by F1 . The following lemma is a consequence of the Sobolev embedding theorem.

Lemma 2. There exists m0 > 0 such that for each r ∈ (1, ∞) and V ∈ H \ {0} such that

Z Br (0)

R

Br (0)

(|∇ V |2 + µ2 |V |2 ) ≤

|V |44 > m0 .

R

Br (0)

|V |44 , (2.9)

Proof. By Sobolev’s embedding theorem, we have that there exists mQ > 0 such that for each r ∈ (1, ∞) and V ∈ H 1 (Br (0)) \ {0},

Z

4

1/2

|V | dx

Z ≤ mQ

Br (0)

Let r > 0 and V ∈ H such that

Z

4

Br (0)

R

Br (0)

1/2

|V | dx

(|∇ V |2 + µ2 |V |2 ) ≤

Z ≤ mQ

Br (0)

(|∇ V |2 + µ2 |V |2 )dx. Br (0)

|V |44 . Then

(|∇ V | + µ2 |V | )dx ≤ mQ 2

Br (0)

R

2

2 Therefore we find that the assertion holds with m0 = m− Q .

Z Br (0)

|V |44 dx.



3. Proof of Theorem 1 We may assume that the positive number ε0 , which is given by Lemma 1, is so small that there exists m1 > 0 such that

Z R3

4 X (−1)i−1 U1,σ i−1 z1 + u

!2

i =1

4 X

!2 U2,σ i−1 z1 + v

dx > m1 ,

(3.1)

i =1

for z1 ∈ {(x1 , x2 , 0) : x1 , x2 ∈ R} with |z1 | ≥ R0 , and u, v ∈ H with kukµ1 , kvkµ2 ≤ 4ε0 . 4c

We first observe the profile of function V ∈ S2 ∩ H2 ∩ I2 2 . 4c

Lemma 3. There exists R1 > R0 such that for each V ∈ S2 ∩ H2 ∩ I2 2 , one of the followings holds

Z

|V |4 dx ≥ BR (0) 1

kV k2µ2 ≥ 16c2 −

7 2

c2 ;

ε0 6

(3.2)

and

V =

4 X

U2,σ i−1 z1 + v,

(3.3)

i=1

where kvk2 ≤ 4ε0 , z1 = QE V ∈e l1 or z1 = QF V ∈ l1 . Proof. For r > 0, we put Ar = (Br +2 (0) \ Br (0)) ∪ x = (x1 , x2 , x3 ) ∈ R3 : min {|x1 | , |x2 |} ≤ 1, |x| ≥ r .





Then Acr consists of five disjoint sets Ar ,i , i = 0, 1, . . . , 4 such that Ar ,1 ⊂ E1 , Ar ,i = σ i−1 (Ar ,1 ) for 2 ≤ i ≤ 4 and Ar ,0 = Br (0). We also put Ar (θ ) = σ (θ )Ar

and

Ar ,i (θ ) = σ (θ )Ar ,i

for r > 0, θ ∈ [0, 2π ) and 1 ≤ i ≤ 4.

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

2585 4c

For given ε > 0, we can choose Rε ∈ (R0 + 2, ∞) so large that for each V ∈ S2 ∩ H2 ∩ I2 2 , there exists r ∈ (R0 + 2, Rε ) and θ ∈ [0, 2π ) satisfying that

Z Ar (θ)

(|∇ V |2 + µ2 |V |2 )dx < ε.

(3.4)

4c

Now let V ∈ S2 ∩ H2 ∩ I2 2 , and r ∈ (R0 + 2, ∞), θ ∈ [0, 2π ) satisfy (3.4). Let ϕ ∈ C ∞ (R, [0, 1]) such that

ϕ(x) = 0 for x ≤ 0,

ϕ(x) = 1 for x ≥ 1 and 0 ≤ ϕ 0 (x) ≤ 2 on R.

Next define a function ψi ∈ C (R3 , [0, 1]) by


ψi (x) = ϕ 1 − x − Ar ,i (θ ) for x ∈ R3 and 0 ≤ i ≤ 4, where x − Ar ,i (θ ) stands for the distance of x from the set Ar ,i (θ ). We put and e Vi (x) = ψi (x)V

Vi = V |Er ,i (θ)

for 0 ≤ i ≤ 4 and x ∈ R3 .

Then we have 4 X

4 X

I2 (Vi ) = I2

i =0

! = 4I2 (V1 ) and

Vi

i=0

4 X

I2 ( e V i ) = I2

i =0

4 X

! e Vi

= 4I2 ( e V1 ).

i=0

From the definition of Vi and (3.4), one can see that there exist constants C1 > 0 and C2 > 0 such that

|Vi |44 − |e Vi |44 ≤ C1 ε,

k Vi − e Vi k2µ2 < C2 ε and

kVi k2 − ke Vi k2µ2 < C1 ε µ2

(3.5)

for 0 ≤ i ≤ 4. Here we choose ε > 0 and k ∈ N such that 4c2 1 + k−2 1 C1 ε +


2 − C1 ε > max



 max

1 + k−2 1 4

C1 ε

,



7 2

c2 , 4c2 −

2

2

4c2 k − 1

ε0



24

,

(4c2 + C1 ε) <

(3.6)

ε0 , 3 × 24

(3.7)


2 (4c2 + C1 ε) < c2 + δ0

(3.8)

and kC1 ε ≤ min

nm

0

2

,

c2 o 8

.

(3.9)

Now we put R1 = Rε . We divide the proof into the following two cases. Case 1. Assume that kV0 k2µ2 ≤ |V0 |44 . If |V0 |44 < kC1 ε , then, denoting V0 = V |Br (0) , we have by (3.9) and Lemma 2 that

kV0 k2µ2 > |V0 |44 . Therefore we find that |V0 |44 ≥ kC1 ε holds. Then by (3.5), we have  

2 4 4 2

e

e e V0 µ ≤ V0 4 + 2C1 ε ≤ 1 + V0 4 . 2 k−1

2 4 Let t > 0 such that t e V0 ∈ S2 . That is e V0 = t 2 e V0 . Then we have µ2

2 4 t 2 = e V0 µ / e V0 4 ≤ 2





 1+

We have by (1.2) that I2 (t e V0 ) = 1 4

4 4 1 e V0 4 = 4 t e V0 4 ≥ 4t

1 4



2 k−1

4

.

4 te V0 4 ≥ c2 . Therefore we find c2

1 + k−2 1


2 .

It then follows from (3.6) that

4 |V0 |44 ≥ e V0 4 − C 1 ε ≥ That is (3.2) holds.

4c2 2

1 + k−1


2 − C1 ε ≥

7 2

c2 .

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N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

Case 2: Assume that kV0 k2µ2 > |V0 |44 . If |V1 |44 < kC1 ε , then again by (1.2) and (3.9) that

|V0 |44 4

= I2 ( V ) −

4 1X

4 i =1

|Vi |44 = I2 (V ) − |V1 |44 ≥ c2 − kC1 ε ≥

7c2 8

.

Then (3.2) holds. Now we assume that |V1 |44 ≥ kC1 ε . Since kV k2µ2 = |V |44 , we have by the assumption that kV1 k2µ2 < |V1 |44 and then



2

e

V1 µ ≤ 1 + 2



2

4 e V1 . 4

k−1

2

4

Let t > 0 such that e V1 µ = t 2 e V1 4 . By the same argument as above, we find 2

t2 ≤ 1 +

2

and I2 (t e V1 ) ≥ 4c2

k−1

for 1 ≤ i ≤ 4.

(3.10)

This implies that

4 te Vi 4 ≥ 4c2 for 1 ≤ i ≤ 4.

(3.11)

That is 4 × 4c2 ≥

4 X

 X  4  4  4   X X 4 1 4 4c2 e e | | ≥ Vi 4 − C 1 ε = t Vi 4 − C1 ε ≥ − C1 ε 4 4 Vi 44

i=1

i=1

i =1

t

t

i =1

and then t4 ≥

4c2

.

4c2 + C1 ε

(3.12)

Therefore we find from the inequalities above that

(1 − t ) ≤ 1 − t 4 ≤

C1 ε

if t ≤ 1

4c2

and t − 1 ≤ t 2 − 1 ≤

2 k−1

if t > 1.

(3.13)

On the other hand, we have that I2 ( t e V1 ) =

4 t e V1 4

4

1 + k−2 1 ≤ 4


2 4 e V1 4

≤

1 + k−2 1 4


2

1+
|V1 |44 + C1 ε ≤

2 2 k−1




4

(4c2 + C1 ε).

(3.14)

Then by (3.8), it follows that I2 (t e V1 ) ≤ c2 + δ0 . Then we have by (c) of Lemma 1, there exists z ∈ R3 and v ∈ H 1 (R3 ) such ε that kvk2µ2 ≤ 240 and t e V1 = U2,z + v . It then follows from (3.5), (3.7) and (3.13) that



2

V1 − U2,z 2 = V1 − te V1 + v µ µ2 2  

2

2

e ≤ 3 V1 − V1 µ + (1 − t )2 e V1 µ + kvk2µ2 2 2 !   2 C1 ε 2 ε0 ≤ 3 C1 ε + max , (4c2 + C1 ε) + 4c2 k − 1 24 ≤

ε0 6

.

(3.15)

On the other hand, by (3.6) we have 4 X i=1

kVi k2µ2 ≥

4 4 X 1 X 2

t e (ke Vi k2µ2 − C1 ε) ≥ 2 Vi µ − 4C1 ε

t

i =1

≥4

!

4c2 1 + k−2 1

≥ 16c2 −

ε0 6

,

− C1 ε

i=1

2

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

2587

2 ε ε and then kV0 k2µ2 ≤ 60 . Therefore noting that Ω \E (θ) U2,z µ < 60 by (3.15), we find r ,1 2 Z Z Z





V |E (θ) − U2,z 2 =

V1 − U2,z 2 +

V0 − U2,z 2 +

U2,z 2 1 µ µ µ µ R

2

2

Er ,1 (θ)

ε0

≤

+ 2 kV0 k2µ2 +

6

ε0

≤

+5·

6

ε0

2

Er ,0 ∩E1 (θ )

!

Z



U2,z 2

µ2

Er ,0

Z + Ω \E1 (θ )

2

Ω \E1 (θ )



U2,z 2

µ2

6

= ε0 .

Then by (b) of Lemma 1, there exists x ∈ R3 such that V |E1 (θ) = U2,x + v1 , where v1 ∈ H 1 (E1 (θ )) with kv1 k2µ2 < ε0 . Recalling that γ2 (V ) = V , we have that x = (x1 , x2 , 0) for some x1 , x2 ∈ R. Then we find that V =

4 X

U2,σ i−1 x + v,

i =1

where v = find by (2.4),

P4

i=1

σ i−1 v1 ∈ H. We will see that x 6= 0. If x = 0, noting that kV0 − vk2µ2 ≤ 2(kvk2µ2 + kV0 k2µ2 ) ≤ 9ε0 , we

9ε0 ≥ kV0 − vk2µ2 ≥

Z BR



4U2,0 2 = 16 µ 2

Z BR (0)

  ∇ U2,0 2 + µ2 U2,0 2 ≥ 32c2 .

0

0

Since ε0 < c2 /6, this is a contradiction and thus x 6= 0. By the symmetry of the space H2 , it follows that x ∈ li or x ∈e li holds for some 1 ≤ i ≤ 4. This completes the proof.  We next choose R2 ∈ (R1 , ∞) such that

Z BR

1

4 4 X U2,σ i−1 x + v < c2 (0) i=1

(3.16)

for x ∈ {(x1 , x2 , 0) : x1 , x2 ∈ R} with |x| ≥ R2 , and v ∈ H with kvk2µ2 < 4ε0 . We next observe the profile of the function U ∈ S1 ∩ H1 . Lemma 4. (1) For each U ∈ S1 ∩ H1 , I1 (U ) ≥ 4c1 . 4c +δ (2) There exists δ1 ∈ (0, δ0 ) such that each U ∈ S1 ∩ H1 ∩ I1 1 1 has the form U =

4 4 X X (−1)i−1 U1,σ i−1 w + u or U = − (−1)i−1 U1,σ i−1 w + u, i=1

i =1

where w = P1 (U |F1 ) ∈ l1 with |w| ≥ R2 and kukµ1 ≤ 4ε0 . 2

4c1 +δ0

Proof. (1) Let U ∈ S1 ∩ H1 ∩ I1 Ui = U |Fi for 1 ≤ i ≤ 4. Then U =

4 X

Ui

and

. By the definition of H1 , we have that U = 0 on (x1 , x2 , x3 ) ∈ R3 : x1 = ±x2 . Put

c +δ0 /4

Ui ∈ S1 ∩ I11





for 1 ≤ i ≤ 4.

i =1 c +δ /4

4c +δ

Then since I1 (Ui ) ≥ c1 for 1 ≤ i ≤ 4, the assertion follows. (2) Let U ∈ S1 ∩ H1 ∩ I1 2 with δ ∈ (0, δ0 ). Then Ui ∈ S1 ∩ I11 0 for each 1 ≤ i ≤ 4. Then by (c) of Lemma 1, we have that each Ui has the form Ui = U1,σ i−1 w + ui , where w = P1 U1 and kui k2µ1 < ε0 . Since U ∈ H1 , we find by the invariance of H1 under the group action G1 that w ∈ l1 . By a standard argument, 4c1 +δ1

one can see that |w| → ∞, as δ → 0. Then we can choose δ1 ∈ (0, δ0 ) so small that for each U ∈ S1 ∩ H1 ∩ I1 This completes the proof.  Here we define

B1 (U ) = w1 for U ∈ S1 ∩ H1 ∩ I14c1 +δ1 , where w1 = P1 (U |F1 ). Then B1 (U ) ∈ l1 = {(x1 , 0, 0) : x1 > 1}.

, |w| ≥ R2 .

2588

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593 4c

On the other hand, for each V ∈ S2 ∩ H2 ∩ I2 2 , we define

B2 ( V ) =

Z

 0

|V |4 dx ≥

if BR (0) 1



z1

7 2

c2

otherwise

where z1 is the vector given in (3.3) of Lemma 3, i.e., z1 = QE V ∈ e l1 or z1 = QF V ∈ l1 . One can see by Lemma 3 that B2 is 4c well defined on S2 ∩ H2 ∩ I2 2 . Here we choose λ0 ∈ (0, 1) such that 2 λ− 0 δ1 > 4c2 ,

2 2 2 16λ− 0 c1 > α2 (16c2 ) + 1

and

2 |β| m1 λ− 0

8

> c2 ,

(3.17)

where α > 0 such that |u|24 ≤ α kuk2µ2 for u ∈ H. Then we have that for each λ ∈ (0, λ0 ), u ∈ S1 if and only if λ−1 u ∈

S1,λ = u ∈ H \ {0} : kuk2µ1 = λ2 |u|44 , and then





( inf

kuk2µ1 2

−

λ2 |u|44 4

) = λ−2 inf {I1 (u) : u ∈ S1 } = λ−2 c1 .

: u ∈ S1,λ

In the following, we fix λ ∈ (0, λ0 ) and put

Λ = {U = (U , V ) ∈ M : B1 (N1 (U )), B2 (N2 (V )) ∈ l1 with |B2 (N2 (V ))| ≥ |B1 (N1 (U ))|} ∩ Φ 4(λ

−2 c +c ) 1 2

.

In the following lemma, we will see the shape of function (U , V ) ∈ Λ. Lemma 5. Let U = (U , V ) ∈ Λ. Then there exist s, t ∈ (0, 1) such that U has the form

N 1 U = sλ U =

4 4 X X (−1)i−1 U1,σ i−1 w1 + u and N2 V = tV = U2,σ i−1 z1 + u, i=1

(3.18)

i =1

where w1 , z1 ∈ l1 , |z1 | ≥ |w1 | ≥ R2 , kuk2µ1 < 4ε0 and kvk2µ2 < 4ε0 . Proof. Let U = (U , V ) ∈ Λ. Then noting that

kU k2µ1 < kU k2µ1 − β

Z R3

U 2 V 2 dx = λ2 |U |44 ,

(3.19)

we have that N1 U = sλU ∈ S1 ∩ H1 for some s ∈ (0, 1). Then by (1) of Lemma 4, 4c1 ≤ I1 (sλU ) =

ksλU k2µ1 4

<

λ2 kU k2µ1 4

= λ2 Φ1 (U) ≤ 4λ2 (λ−2 c1 + c2 ).

(3.20)

Then by (3.17), we find I1 (N1 U ) ≤ 4c1 + δ1 . Therefore by (2) Lemma 4, we obtain that N1 U has the form in (3.18). On the other hand, we have

Φ2 (V ) =

kV k2µ2 4

≤ Φ ( U) −

kU k2µ2 4

≤ 4c2 .

(3.21)

By the same argument as above, we have that tV ∈ S2 for some s ∈ (0, 1). Then by the inequality above, N2 V = sV ∈ S2 4c ∩ I2 2 . Since |B2 (N2 (V ))| ≥ |B1 (N1 (U ))| ≥ R2 and B2 (N2 (V )) ∈ l1 , we have by the definition of B2 that the assertion (3.3) of Lemma 3 holds with z1 ∈ l1 . That is N2 (V ) has the form in (3.18). This completes the proof.  Lemma 6. mΛ = inf {Φ (U) : U ∈ Λ} < 4(λ−2 c1 + c2 ). Proof. Put U = λ−1

P4 P4 i+1 U1,σ i−1 w1 , V = i=1 (−1) i=1 U2,σ i−1 z1 and U = (U , V ), where w1 , z1 ∈ l1 and |z1 | ≥ |w1 | ≥ R2 .  −1 √ √ 1 Since µ1 > 1 − √ µ2 , we can choose a > 1 such that 2   √ √ µ1 1 √ > µ2 and c = 2 1 − > 2. 

a

a

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

2589


In the following, we assume that z1 = aw1 . Then since min σ i−1 w1 − σ j−1 z1 : 1 ≤ i, j ≤ 4 = 1 − 1a |z1 |, we have by (2.3) that there exists C0 > 0 such that for |z1 | sufficiently large, !2 !2 Z 4 4 X X √

2 2 i−1 U ,V = (−1) U1,σ i−1 w1 U2,σ i−1 z1 ≤ C0 e−c µ2 |z1 | . R3

i=1

i=1

Since λ2 |U |44 → 4λ−2 c1 and |V |44 → 4c2 as |z1 | → ∞, we find that for |z1 | sufficiently large, the matrix



2 4  λ |U |4 A(U) =   Z β U 2V 2

β

Z



U 2V 2 R3

  

|V |44

R3

(3.22)

is invertible, i.e. there exists (s, t ) ∈ R+ × R+ such that

 2 s t2

= A(U)−1



kU k2µ1 kV k2µ2



.

(3.23)

That is (s, t ) satisfies

 kU k2µ1 = s2 λ2 |U |44 + β t 2 U 2 , V 2

and

kV k2µ2 = t 2 |V |44 + β s2

Z R3

U 2V 2,

(3.24)

and then (sU , tV ) ∈ M . From the boundedness of kU k2µ1 and kV k2µ2 , one can see that there exists M0 > 0 such that s2 , t 2 ≤ M0 for |z1 | sufficiently large. We will see that

Φ (sU , tV ) < 4(λ−2 c1 + c2 ) for |z1 | sufficiently large. Since s2 ≤ M0 , we can see √



 kV k2µ2 = t 2 |V |44 + β s2 U 2 , V 2 ≥ t 2 |V |44 − |β| M0 U 2 , V 2 ≥ t 2 |V |44 − |β| C0 M0 e−c µ2 |z1 | .

That is

kV k2µ2 + |β| C0 M0 e−c

2

t ≤

√

µ2 |z1 |

|V |44

.

Recalling that U2,z is a solution of (P2 ) for each z ∈ R3 , we have 2

kV kµ2

2 4

X

U2,σ i−1 z1 =

i=1

µ2 X

2 = 4 U2,0 µ + hhU2,σ i−1 z1 , U2,σ j−1 z1 iiµ 2

i6=j

2 = 4 U2,0 µ + 2

XD

2

E

U23,σ i−1 z , U2,σ j−1 z1 . 1

i6=j

Then noting that

|V |44 =

X XD U2,0 4 + 4 U3 4

i6=j

2,σ i−1 z1

E E X D , U2,σ j−1 z1 + 12 U22,σ i−1 z , U2,σ j−1 z1 U2,σ k−1 z1 , 1

i6=j,i6=k

we find 2

ktV kµ2 ≤

kV k4µ2 + |β| C0 M0 e−c

16c2 +

≤

|V |44 PD i6=j

U23,σ i−1 z 1

√

µ2 |z1 |

, U2,σ j−1 z1

! E

16c2 +

16c2 + 4

PD i6=j

PD i6=j

U23,σ i−1 z 1

, U2,σ j−1 z1 + |β| C0 M0 e

U23,σ i−1 z , U2,σ j−1 z1 1

E

E

√ −c µ2 |z1 |

! .

2590

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

√  Since min σ i−1 z1 − σ j−1 z1 : i 6= j = 2z1 , we have by (2.3) that there exists C1 > 0 such that for |z1 | sufficiently large, E √ √ XD U23,σ i−1 z , U2,σ j−1 z1 ≥ C1 e− 2 µ2 |z1 | . (3.25) 1

i6=j

Since c >

√

2, it follows that there exists C2 > 0 such that for |z1 | sufficiently large,

ktV k2µ2 < 16c2 − C2 e−

√ √ 2 µ2 |z1 |

.

On the other hand, we have 2

kU kµ 1

2 4

E D  X

2

−1 X

i−1 = λ (−1)i+j U13,σ i−1 w , U1,σ j−1 w1 (−1) U1,σ i−1 w1 = λ−2 4 U1,0 µ + 1 1

i=1 µ1

and

λ |U | = λ 2

4 4

! D E D E X X 2 3 j +k i+j U1,σ i−1 w , U1,σ j−1 w1 U1,σ k−1 w1 . (−1) 4c1 + 4 (−1) U1,σ i−1 w , U1,σ j−1 w1 + 12

−2

1

i6=j

1

i6=j,i6=k

Noting that |z1 | = a |w1 |, again by (2.3), we have that there exists C3 > 0 such that

√ X D E 2√ i +j 3 (−1) U1,σ i−1 w1 , U1,σ j−1 w1 ≤ C3 e− a µ1 |z1 | i6=j

(3.26)

 √ D E X 2+2 √ µ1 |z1 | j +k 2 (−1) U1,σ i−1 w , U1,σ j−1 w1 U1,σ k−1 w1 ≤ C3 e a . 12 1 i6=j,i6=k

(3.27)

and

Then noting that 2

s ≤

kU k2µ1 + |β| C0 M0 e−c

√

µ2 |z1 |

|U |44

,

we have that there exists C4 > 0 such that for |z1 | sufficiently large, 2

ksU kµ1 ≤

kU k2µ1 kU k2µ1 + |β| C0 M0 e−c

√

µ2 |z1 |




|U |44 

√

≤ λ−2 16c1 + C4 e−

2√ µ1 |z1 | a

+ e− c

√

µ2 |z 1 |



.

Therefore we find

Φ (sU , tV ) ≤ 4(λ−2 c1 + c2 ) − √

Since

µ1

a

>

1 4

C2 e−

√ √

2 µ2 |z1 |

+

λ −2 4

 C4

√

e−

2√ µ1 |z1 | a

+ e− c

√ µ2 |z1 |



.

√ √ µ2 and c > 2, Φ (sU , tV ) < 4(λ−2 c1 + c2 ) holds for |z1 | sufficiently large. This completes the proof.



Lemma 7. For U ∈ Λ, |B2 (N2 (V ))| > |B1 (N1 (U ))|. Proof. Let U = (U , V ) ∈ Λ and suppose that |B1 (N1 (U ))| = |B2 (N2 (V ))|. Since (3.19) holds, we have that there exists t ∈ (0, 1) such that kU k2µ1 = t 2 λ2 |U |44 . That is t λU ∈ S1 . Since t λU ∈ S1 ∩ H1 , we have by (1) of Lemma 4 that

kλtU k2µ1 ≥ 16c1 . On the other hand, we have from the definition of t, Lemma 5 and (3.1) that t2 =

kU k2µ1 λ2 | U |

4 4

=

2

kU kµ1

kU k2µ1 kU k2µ1 R ≤ . 2 kU kµ1 + |β| λ−2 m1 − β R3 U 2 V 2

Since kU k2µ1 ≤ 4(λ−2 c1 + c2 ) ≤ 4(λ−2 + 1)c1 and λ−2 ≥ 1, we have t2 ≤

kU k2µ1 2

kU kµ 1 +

|β| λ−2 m

1

= 1

1+

|β|λ−2 m1 kU k2µ 1

1

≤ 1+

|β|λ−2 m1 4(λ−2 c1 +c2 )

1

≤ 1+

|β|m1 8c1

.

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

2591

Then by (3.17), we obtain

Φ (U) ≥ Φ1 (U) =

kU k2µ1 4

≥

ktU k2µ1

  |β| m1 ≥ 4λ−2 c1 1 + > 4(λ−2 c1 + c2 ).

4t 2

8c1

This contradicts the assumption. Therefore we obtain the assertion.



Lemma 8. Let {Un } ⊂ Λ be a sequence such that limn−→∞ Φ (Un ) = mΛ . Then lim h∇ Φ (Un ), Ψ i = 0

n−→∞

for all Ψ = (ξ , η) ∈ (C0∞ (R3 ) ∩ H1 ) × (C0∞ (R3 ) ∩ H2 ).

Proof. Fix δ > 0 such that mΛ + δ < 4(λ−2 c1 + c2 ) and put Λδ =

U ∈ Λ : Φ (U) + δ < 4(λ−2 c1 + c2 ) . Let





U = (U , V ) ∈ Λδ . Since kV k2µ2 = |V |44 + β U 2 , V 2 , we have by (3.21) that





 |β| U 2 , V 2 ≤ |V |44 ≤ α22 kV k4µ2 ≤ α22 (16c2 )2 , 4c2

where α2 is the Sobolev constant. On the other hand, noting that tV ∈ S2 ∩ H2 ∩ I2 Lemma 3 that

for some t ∈ (0, 1), we find by (3.3) of

|V |44 ≥ kV k2µ2 ≥ ktV k2µ2 ≥ 15c2 . We also have by (3.20) that λ2 |U |44 ≥ 16λ−2 c1 . Then we find by (3.17) and the inequalities above,

|A(U)| = λ2 |U |44 |V |44 − (βhU 2 , V 2 i)2
≥ |V |44 λ2 |U |44 − |V |44 ≥ 15c2 (16λ−2 c1 − α22 (16c2 )2 ) ≥ 15c2 , where |A(U)| is the determinant of matrix A(U) defined in (3.22). Let Ψ = (ξ , η) ∈ (C0∞ (R3 ) ∩ H1 ) × (C0∞ (R3 ) ∩ H2 ). Then we have that |A(U + ε Ψ )| ≥ c2 for ε > 0 sufficiently small. Therefore there exist M0 > 0 and sε , tε ∈ R+ such that sε , tε ∈ (0, M0 ) and

 2 sε

tε2

= A(U + ε Ψ )

kU + εξ k2µ1 kV + εηk2µ2

−1

! (3.28)

for ε > 0 sufficiently small. That is (sε (U + εξ ), tε (V + εη)) ∈ M . From (3.28), one can see that there exists C > 0 such that

|sε − 1| ≤ C ε and

|tε − 1| ≤ C ε for ε > 0 sufficiently small.

We will see that (sε (U + εξ ), tε (V + εη)) ∈ Λ. Since Φ (U , V ) + δ < 4(λ

Φ (sε (U + εξ ), tε (V + εη)) =

1 4

(3.29) −2

c1 + c2 ), we have that

ksε (U + εξ )k2µ1 + ktε (V + εη)k2µ2 < 4(λ−2 c1 + c2 )

(3.30)

for ε > 0 sufficiently small. As in the proof of Lemma 5, we find that there exists (e s,e t ) ∈ (0, 1) × (0, 1) such that λe ssε (U + εξ ) ∈ S1 and e ttε (V + εη) ∈ S2 . Then we have 1 4

ksε (U + εξ )k2µ1 ≥

1 4

ke ssε (U + εξ )k2µ1 ≥ 4λ−2 c1 .

This implies I2 (N2 (V + εη)) =

2

e ttε (V + εη) µ

2

4

≤

ktε (V + εη)k2µ2 4

≤ 4c2 .

On the other hand, noting that U ∈ Λ, we have that N2 V has the form

N2 (V ) =

4 X

U2,σ i−1 z1 + v,

i=1

where kvk2µ2 < 4ε0 and z1 = B2 (N2 V ) ∈ l1 with |z1 | ≥ R2 . Then by (3.16), we have

Z BR (0) 1

|N2 (V )|4 dx < c2 .

(3.31)

2592

N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

Then by (3.29), we have

Z

|N2 (V + εη)|44 ≤

BR (0)

Z

1

BR (0)

|tε (V + εη)|44 < 2c2 for ε sufficiently small.

1

Therefore by (3.31) and the inequality above, we find that (3.3) of Lemma 3 holds, i.e.,

N2 (V + εη) =

4 X

U2,σ i−1 zε + vε

i=1

where zε = B2 (N2 (V + εη)) ∈ l1 and kvε k2µ2 < 4ε0 . Since tε → 1 as ε → 0, we find that B2 (N2 (V + εη)) → B2 (N2 (V )) and B1 (N1 (U + εξ )) → B1 (N1 (U )). Recall that |B2 (N2 (V ))| > |B1 (N1 (U ))| by Lemma 7. Then we find

|B2 (N2 (V + εη))| > |B1 (N1 (U + εξ ))|

for ε sufficiently small.

(3.32)

Thus we find by (3.30) and (3.32) that there exists ε > 0 such that for each ε ∈ (0, ε), (sε (U + εξ ), tε (V + εη)) ∈ Λ. One can see from the argument above that ε can be chosen independent of the choice of U ∈ Λδ . Now let {Un } = {(Un , Vn )} ⊂ M be a sequence such that limn→∞ Φ (Un ) = mΛ . Let {εn } ⊂ R+ such that limn→∞ εn = 0 and

Φ (Un ) ≤ mΛ + εn2 for n ≥ 1. Then

Φ (Un ) − Φ (sεn (Un + εn ξ ), tεn (Vn + εn η)) ≤ εn2 for n ≥ 1. Then noting that (3.29) holds, we have by subtracting subsequences that there exists (a, b) ∈ R × R such that a = limn→∞ (sεn − 1)/εn and b = limn→∞ (tεn − 1)/εn . Then it follows that lim h∇u Φ (Un ), aUn + ξ i + h∇v Φ (Un ), bVn + ηi ≤ 0.

n−→∞

Then since (2.6) holds, we have lim h∇u Φ (Un ), ξ i + h∇v Φ (Un ), ηi ≤ 0.

n−→∞

Since Ψ = (ξ , η) ∈ (C0∞ (R3 ) ∩ H1 ) × (C0∞ (R3 ) ∩ H2 ) is arbitrary, the assertion follows.



Proof of Theorem 1 for k = 1. Let {Un } = {(Un , Vn )} ⊂ Λ be a sequence such that limn→∞ Φ (Un ) = mΛ . By subtracting subsequences, we may assume that Un = (Un , Vn ) → U = (U , V ) ∈ H0 weakly in H and strongly in L4loc (R3 ) × L4loc (R3 ). Then by Lemma 8, we have that

h∇ Φ (U), θi = lim h∇ Φ (Un ), θi = 0 for all θ ∈ (C0∞ (R3 ) ∩ H1 ) × (C0∞ (R3 ) ∩ H2 ). n−→∞

That is U is a solution of (1.3). To prove the assertion, it is sufficient to show that U 6≡ 0 and V 6≡ 0. By Lemma 5, we have that each Un = (Un , Vn ) satisfies that there exists (sn , tn ) ∈ (0, 1) × (0, 1) such that N1 Un = sn Un , N2 Vn = tn Vn and that

N1 (Un ) = λ

−1

4 X (−1)i−1 U1,σ i−1 wn + un

! and N2 (Vn ) =

i=1

4 X

U2,σ i−1 zn + vn ,

i =1

where wn , zn ∈ l1 and |zn | ≥ |wn | ≥ R2 and kun k2µ1 , kvn k2µ2 < 4ε0 . kU k2

Then by (3.20), we have that 4n ≥ 4λ−2 c1 for all n ≥ 1. Now suppose that limn→∞ |zn | = ∞. Then we can choose (θn , rn , εn , kn ) ∈ (R+ )3 × N such that limn→∞ rn = limn→∞ kn = ∞, limn→∞ εn = 0, and (3.4)–(3.9) hold with Vi , ε and k replaced by Vn |Ern ,i (θn ) , εn and kn , respectively. Then we have that (3.10) and (3.11) hold with t = tn , Vi = Vn |Ern ,i (θn ) and

e Vi = ψ(x)Vn . Therefore we find that 2 4 X Vn |Er ,i (θn ) lim inf Φ2 (Un ) ≥ lim inf ≥ 4c2 . n−→∞

n−→∞

i=1

4

Since (3.20) holds with U replaced by Un , we obtain that limn→∞ Φ (Un ) = limn→∞ Φ1 (Un ) + Φ2 (Un ) = 4(λ−2 c1 + c2 ). This contradicts the assumption. Therefore we have that there exists z ∈ R3 such that zn −→ z in R3 . Since |wn | ≤ |zn | for each n ≥ 1, we may assume that wn −→ w ∈ R3 . It then follows that U = (U , V ) has the form U =λ

−1

! 4 X i−1 (−1) U1,σ i−1 w + u i=1

and

V =

4 X

U2,σ i−1 z + v,

i=1

with kuk2µ1 , kvk22 < 4ε0 . That is U 6≡ 0 and V 6≡ 0. This completes the proof.



N. Hirano / Nonlinear Analysis 73 (2010) 2580–2593

2593

4. On the proofs of Theorem 1 for k ≥ 2 and Theorem 2 The proof of Theorem 1 for k ≥ 2 is a slight modification of that for k = 1. We just note the role of the condition

  π −1 √ √ µ1 > 1 + sin µ2

(4.1)

2k


π

(−1)i+1 U1,σ i−1 w1 and V = i=1 U1,σ i−1 z1 , where σk = σ 4k , w1 , z1 ∈ l1 and k k |z1 | ≥ |w1 | > 0. We put (s, t ) ∈ (0, 1) × (0, 1) such that Then U = (sU , tV ) ∈ M. As in the case of k = 1, we need to show −2 that Φ (sU , tV ) < 4k(λ c1 + c2 ) for λ sufficiently small. By (4.1), we can choose a > 1 such that   √ π  µ1 1 √ . > µ2 and c = 2 1 − > 2 sin a a 4k n o
j −1 π |z1 |, we have, instead of (3.25), that Since min σki−1 z1 − σk z1 : i 6= j = 2 sin 4k   X π √ U 3 i−1 , U2,σ j−1 z ≥ C1 e−2 sin( 4k ) µ2 |z1 | for |z1 | sufficiently large. 2,σ z for the case k ≥ 2. Let k ≥ 2, U = λ

i6=j

k

1

k

P4k

P4k −1

i=1

1

Similarly we have, instead of (3.26) and (3.27), that there exists C3 > 0 such that

  X 2 sin( π ) √ 4k i+j 3 µ1 |z1 | a U i−1 , U1,σ j−1 w ≤ C3 e− (−1) 1,σk w1 1 i6=j k and

2(sin( π ))+sin sin   4k X a j+m 2 (−1) U i−1 , U2,σ j−1 w U2,σ m−1 w ≤ C3 e 12 2,σk w1 1 1 k i6=j,i6=m k



π

4(k−1)

!

√

µ1 |z 1 |

,

for |z1 | sufficiently large. Then using inequalities above instead of (3.25)–(3.27), we can prove that Φ (sU , tV ) < 4k(λ−2 c1 + c2 ) for |z1 | sufficiently large. The rest of the argument is the same argument as that for k = 1. The proof of Theorem 2 is also a slight modification of the proof of Theorem 1. We define a mapping ρ± : H → H by

ρ+ (u)(x) = u(x1 , −x3 , x2 ), ρ− (u)(x) = −u(x1 , −x3 , x2 ) for u ∈ H and x ∈ R3 .   2 Set G∗1 = σ− , σ−2 , σ−3 , ρ− , ρ− , ρ−3 , γ1 , γ2 and G∗2 = σ+ , σ+2 , σ+3 , ρ+ , ρ+2 , ρ+3 , γ1 , γ2 . Instead of H1 , H2 and H0 , we will work on the set

 H1∗ = u ∈ H : g (u) = u for g ∈ G∗1 ,

 H2∗ = u ∈ H : g (u) = u for g ∈ G∗2

and

H∗0 = H1∗ × H2∗ . Then by the same argument in Sections 2 and 3, we can find a solution U0 ∈ H∗0 of the form (1.5). We cannot find a solution of the form (1.5) such that {wi } and {zi } form polyhedrons different from cube. In fact, it is necessary that (−1)n(i) 6= (−1)n(j) if wi and wj are adjacent vertices of a polyhedron. This requirement is satisfied only by cube. Acknowledgement The author expresses his hearty thanks to the referee for reading this article very carefully and giving valuable comments. References [1] M. Mitchell, D.N. Christodoulides, T.H. Coskun, M. Segev, Theory of incoherent self-focusing in biased photorefractive media, Phys. Rev. Lett. 78 (1997) 646–649. [2] C.R. Menyuk, Nonlinear pulse propagation in birefringent optical fibers, IEEE J. Quantum Electron. 23 (1987) 174–176. [3] C.R. Menyuk, Pulse propagation in an elliptically birefringent Kerr medium, IEEE J. Quantum Electron. 25 (1989) 2674–2682. [4] A. Pomponio, Coupled nonlinear Schrödinger systems with potentials, J. Differential Equations 227 (2006) 258–281. [5] A.B. Shabat, V.E. Zakharov, Exact theory of two-dimensional self-focusing and onedimensional self modulation of waves in nonlinear median, Sov. Phys. JETP 34 (1972) 62–69. [6] A. Ambrosetti, E. Colorado, Bounded and ground states of coupled nonlinear Schrödinger equations, C. R. Math. Acad. Sci. Paris 342 (2006) 453–458. [7] L.A. Maia, E. Montefusco, B. Pellacci, Positive solutions for a weakly coupled nonlinear Schrödinger system, preprint. [8] T.C. Lin, J. Wei, Ground state of n coupled nonlinear Schrödinger equations in r n , Comm. Math. Phys. 255 (2005) 629–653. [9] W.C. Troy, Symmetry properties in systems of semilinear elliptic equations, J. Differential Equations 42 (1981) 400–413. [10] Z.Q. Wang, T. Bartsch, J. Wei, Bound states for a coupled Schrödinger system, J. Fixed Point Theory Appl. 2 (2007) 353–367. [11] J. Wei, T. Weth, Existence of nonraidal symmetric bound states for a system of coupled Schrödinger equations, Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl. 18 (2007) 279–293. [12] T.C. Lin, J. Wei, Solitary and self-similar solutions of two-component system of nonlinear Schrödinger equations, Physica D 2 (2006) 99–115. [13] N. Hirano, Multiple existence of nonradial positive solutions for a coupled nonlinear Schrödinger system, NoDEA Nonlinear Differential Equations Appl. 16 (2) (2009) 159–188. [14] B. Gidas, W.M. Ni, L. Nirenberg, Symmetry of positive solutions fo nonlinear elliptic equation in r n , Adv. Math. Suppl. Stud. 7 (1981) 369–402. [15] M.K. Kwong, Uniqueness of positive solutions of −δ u − u + up = 0 in r n , Arch. Ration. Mech. Anal. 105 (1989) 243–266. [16] X. Zhu, A perturbation result on positive entire solutions of a semilinear elliptic equation, J. Differential Equations 92 (1991) 163–178.