Existence of τ-antisymmetric solutions for a system in RN

Accepted Manuscript Existence of τ -antisymmetric solutions for a system in RN

Janete de Gamboa, Elisandra Gloss, Jiazheng Zhou

PII: DOI: Reference:

S0022-247X(18)30262-2 https://doi.org/10.1016/j.jmaa.2018.03.050 YJMAA 22126

To appear in:

Journal of Mathematical Analysis and Applications

Received date:

6 October 2017

Please cite this article in press as: J. de Gamboa et al., Existence of τ -antisymmetric solutions for a system in RN , J. Math. Anal. Appl. (2018), https://doi.org/10.1016/j.jmaa.2018.03.050

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Existence of τ -antisymmetric solutions for a system in RN Janete de Gamboa∗ Universidade de Bras´ılia, Departamento de Matem´ atica 70910-900, Bras´ılia-DF, Brazil

Elisandra Gloss† ‡ Universidade Federal da Para´ıba, Departamento de Matem´ atica 58051-900, Jo˜ ao Pessoa-PB, Brazil.

Jiazheng Zhou§¶ Universidade de Bras´ılia, Departamento de Matem´ atica 70910-900, Bras´ılia-DF, Brazil

26 de mar¸co de 2018 Abstract In this paper we prove the existence of a nontrivial τ -antisymmetric solution for the following system  −Δu + u = |u|2p−2 u + β(x)|v|p |u|p−2 u, in RN −Δv + ω 2 v = |v|2p−2 v + β(x)|u|p |v|p−2 v, in RN where N ≥ 3, 2 < 2p < 2∗ , ω > 0 is a positive constant and τ : RN → RN is a nontrivial orthogonal involution. Here β is a continuous function, invariant under τ , satisfying additional conditions, some of them related to ω. The basic tool employed here is the Concentration Compactness Principle. 2012 Mathematics Subject Classifications: 35J47, 35J50, 35Q55. Keywords: Elliptic system; Sign changing solution; Concentration Compactness Principle.

1

Introduction In this paper we study the existence of a ⎧ −Δu + u = ⎪ ⎪ ⎨ −Δv + ω 2 v = ⎪ u(τ x) = −u(x) and ⎪ ⎩ u(x) → 0 and

nontrivial solution for the following system |u|2p−2 u + β(x)|v|p |u|p−2 u, in RN |v|2p−2 v + β(x)|u|p |v|p−2 v, in RN v(τ x) = −v(x) v(x) → 0 when |x| → ∞,

(P)

where N ≥ 3, 2 < 2p < 2∗ , ω > 0 is a positive constant and τ : RN → RN is a nontrivial orthogonal involution, that is, a linear orthogonal transformation on RN such that τ = Id and τ 2 = Id, Id being the identity on RN . We also suppose that β : RN → R is a continuous function satisfying: (β1) lim|x|→∞ β(x) = β∞ > 0; (β2) β(x) ≥ β∞ for all x ∈ RN , with β ≡ β∞ ; (β3) β(τ x) = β(x) for all x ∈ RN ; ∗ E-mail

Address: [email protected] partially supported by Capes/Brazil ‡ E-mail Address: [email protected] § Corresponding author ¶ E-mail Address: [email protected] † Research

1

(β4) β∞ ≥ C(ω) for

 C(ω) =

where

(1/2)l(ω) − 1, (1/2)l(1/ω) − 1,

if ω ≥ 1; if ω ≤ 1,

   p  N 1 1 1 N l(ω) = 1 + 1− + 2 1− 1− ω 2p−N (p−1) ; 2 p ω 2 p

(β5) There exist C > 0 and 0 < γ < α such that β(x) − β∞ ≥ Ce−γ|x| for all x ∈ RN , where α is related to the exponential decay of the solution of the limiting problem (P1) (see (35)). The existence of solution for a differential equation of the form −Δu + V (x)u = f (u)

in

RN

(1)

has been studied by many authors. When V is a constant we cite, among many others, the famous papers [2] and [3] where Berestycki and Lions presented, what they called, “almost optimal” conditions on the nonlinearity to guarantee existence of solutions for this problem and they also studied multiplicity results. For a coercive potential V , in [18] Rabinowitz proved the existence of positive solutions for equation (1), with additional hypotheses on V and f . For sign changing solutions we cite [12], where the authors proved the existence of nodal solutions when V (∞) = 0. In [5, 6] it is proved that there exists a nodal solution that changes sign exactly once, for a bounded potential V that is positive and tends to a constant at infinity. With similar assumptions on V , in [16] the authors study the existence of nodal solutions for a equation like (1) with a nonlinearity of the form K(x)u3 /(1 + u2 ). The basic tool employed in these articles is the Concentration-Compactness Principle. Using a different approach, in [10, 11] the authors studied multiplicity of sign changing solutions in exterior domains. Still on sign changing solutions, we can mention [13], where the authors showed a multiplicity result for changing sign solutions of the problem −2 Δg u + u = |u|p−2 u

in

Hg1 M

with

u(τ x) = −u(x)

∀ x ∈ M,

(2)

for  small enough, where (M, g) is a Riemmanian manifold symmetric and τ : RN → RN is a linear orthogonal transformation on RN such that τ = Id. In [17] the authors considered the problem (2) in the particular case τ = −Id. For systems, we cite [4] where Brezis and Lieb proved existence of a nontrivial solution at the least energy level among all nontrivial solutions. For β constant, they considered a class of problems more general than  −Δu + u = |u|2p−2 u + β(x)|v|p |u|p−2 u, in RN (3) −Δv + ω 2 v = |v|2p−2 v + β(x)|u|p |v|p−2 v, in RN which includes our limiting problem (P1) (see also [9]). In [15], Maia, Montefusco and Pellacci proved the existence of a positive ground state solution of (3), for β = β∞ , when hypothesis (β4) is satisfied. They also studied the problem (3) for a more general nonlinearity and showed the existence of a positive least energy solution, with additional assumptions on β(x). About the existence of sign-changing solution, in [7] the authors worked with the the system ⎧ ⎨ −Δu + λ1 u = μ1 u3 + βuv 2 , in Ω −Δv + λ2 v = μ2 v 3 + βvu2 , in Ω (4) ⎩ u = v = 0, on ∂Ω where Ω ⊂ RN (N = 2, 3) is a smooth bounded domain. In this paper they studied the existence of signchanging solutions, both components change sign, and semi-nodal solutions, one component changes sign and the other one is positive, when β > 0 is small. Still we can mention [8], where was studied a elliptic system with critical exponent in a smooth bounded domain of RN . The authors showed the existence of a sign-changing solution, where one component changes sign and has exactly two nodal domains, while the other one is positive. In our work we study the system (3) and we prove the existence of a nontrivial τ −antisymmetric solution, that is, (u, v) satisfying (u(τ x), v(τ x)) = (−u(x), −v(x)) Our main result in this article is the following. 2

for all

x ∈ RN .

Theorem 1. Assume (β1) − (β5). Then there exists a nontrivial solution (u, v) for (P). We use variational methods to prove the existence of solutions for problem (P). Our argument can be divided in several steps: In the first one, we present some preliminary results. We deal with the limiting problem (P1), for which, existence of a ground state solution is known from [15, Theorem 2.3]. In this result, the hypothesis (β4) is crucial to prove the positivity of this least energy solution. In Section 3 we prove some compactness results. After a Splitting Lemma, we show that the restriction to the Nehari manifold of the functional I, associated to problem (P), satisfies the Palais-Smale condition for levels c < min{mτ∞ , 2m∞ }, where m∞ is the least energy level for the problem (P1) and τ }. In the last step we prove our theorem. In order to do this, we use mτ∞ =: inf{I∞ (u, v) : (u, v) ∈ N∞ hypothesis (β5) to prove that a suitable level is below 2m∞ and than we show that it is a critical level for I.

2

Preliminary Results

We denote Hω := H 1 (RN ) with the norm v2ω = ∇v22 + ω 2 v22 and H = H 1 (RN ) × Hω with the norm (u, v)2H = u2 + v2ω , where  ·  is the usual norm in H 1 (RN ) and  · q is the usual norm in Lq (RN ). Moreover, we denote (·, ·)qq =  · qq +  · qq the norm of a pair in Lq (RN ) × Lq (RN ). Weak solutions of system (P) are critical points of the functional I : H → R defined by  1 1 1 I(u, v) = (u, v)2H − (u, v)2p − β(x)|u|p |v|p . (5) 2p 2 2p p RN The functional I is of class C 1 and we have   2p−2

 2p−2 I (u, v)(ϕ, ψ) = u, ϕ + v, ψw − |u| uϕ + |v| vψ − RN

RN



β(x) |v|p |u|p−2 uϕ + |u|p |v|p−2 vψ

for all (u, v), (ϕ, ψ) ∈ H. We denote by N the Nehari manifold for I, namely, N = {(u, v) ∈ H \ {(0, 0)} : I  (u, v)(u, v) = 0}, which is closed in H. In order to obtain τ -antisymmetric solutions for (P), we look for critical points of I in N τ = {(u, v) ∈ N : Tτ (u, v) = (u, v)}, where Tτ : H → H is given by Tτ (u(x), v(x)) = (−u(τ x), −v(τ x)). Denoting Hτ = {(u, v) ∈ H : Tτ (u, v) = (u, v)} we see that N τ = N ∩ Hτ is a closed manifold in H τ . The isometry Tτ : H → H induces a isometry T˜τ : H  → H  given by T˜τ (F ) = F ◦ Tτ . To prove the existence of solutions for (P), we need to consider the following problem  −Δu + u = |u|2p−2 u + β∞ |v|p |u|p−2 u, in RN (P1) −v + ω 2 v = |v|2p−2 v + β∞ |u|p |v p−2 v, in RN which is a limiting problem for (P) when |x| → ∞. The natural functional I∞ : H → R associated to (P1) is given by 1 1 β∞ uvpp . (6) I∞ (u, v) = (u, v)2H − (u, v)2p 2p − 2 2p p The functional I∞ is of class C 1 and   I∞ (u, v)(ϕ, ψ) = u, ϕ + v, ψw −

RN

|u|

2p−2

2p−2

uϕ + |v|

vψ − β∞



RN

|v|p |u|p−2 uϕ + |u|p |v|p−2 vψ

for all (u, v) and (ϕ, ψ) in H. The Nehari manifold is  N∞ = {(u, v) ∈ H \ {(0, 0)} : I∞ (u, v)(u, v) = 0}

3

with its τ invariant set given by τ = {(u, v) ∈ N∞ : Tτ (u, v) = (u, v)}. N∞

Defining m∞ := inf{I∞ (u, v) : (u, v) ∈ N∞ }, we have m∞ > 0. Indeed, since N∞ is closed and  (u, v)(u, v) = 0, (0, 0) ∈ / N∞ we get (u, v) ≥ c > 0 for all (u, v) ∈ N∞ . For (u, v) ∈ N∞ , given that I∞ we obtain p (u, v)2H = (u, v)2p 2p + 2β∞ uvp . So I∞ (u, v)

= =

 1 1  p (u, v)2H − (u, v)2p 2p + 2β∞ uvp 2 2p 1 1 p−1 (u, v)2H − (u, v)2H = (u, v)2H ≥ C > 0, 2 2p 2p

which implies m∞ ≥ C > 0.

(7)

This result will be used later. From [4] we know that there exists a nontrivial least energy solution for system (P1), however one of the coordinates functions could be zero. In [15, Theorem 2.3], the authors proved that under condition (β4) the problem (P1) has a ground state solution (u, v) ∈ H, that is I∞ (u, v) = m∞ , which is radial and satisfies u > 0 and v > 0.

3

Compactness results

The next lemma is a version of a classic result due to M. Struwe (see [20]) and it is necessary to study the behavior of (P S) sequences for I in N τ . This will be crucial to prove our existence result. Lemma 1. Let (un , vn ) be a sequence in N τ such that I(un , vn ) is bounded and

I|N τ (un , vn ) → 0.

Then (un , vn ) is a bounded (PS) sequence for I in H and one of the alternatives below occurs: u, v¯) = limn (un , vn ) is a nontrivial solution of (P). (a) Up to subsequences, (un , vn ) converges in H and (¯ u, v¯) of (P), two numbers (b) Replacing (un , vn ) by a subsequence if necessary, there exist a solution (¯ k1 , k2 ∈ N∪{0} with k1 +k2 ≥ 1, k1 +k2 sequences of points {ynj }n , in RN , k1 +k2 pairs of functions (uj , v j ), nontrivial solutions of (P1) such that Tτ (uj , v j ) = (uj , v j ) for k1 + 1 ≤ j ≤ k1 + k2 if k2 ≥ 1, satisfying (i) If k1 ≥ 1, for 1 ≤ j ≤ k1 we have τ ynj = ynj and |ynj | → ∞, as n → ∞; (ii) If k2 ≥ 1, for k1 + 1 ≤ j ≤ k1 + k2 we have τ ynj = ynj and |ynj | → ∞, as n → ∞; (iii) (un , vn )

=

(¯ u, v¯) +

k1 

(uj (· − ynj ), v j (· − ynj )) + Tτ (uj (· − ynj ), v j (· − ynj ))

j=1

+

k 1 +k2

(uj (· − ynj ), v j (· − ynj )) + o(1)

in

H,

j=k1 +1

where the first sum appears only if k1 ≥ 1 and the second one appears only if k2 ≥ 1; u, v¯) + 2 (iv) I(un , vn ) → I(¯

k1  j=1

I∞ (uj , v j ) +

k 1 +k2 j=k1 +1

I∞ (uj , v j ), where the first sum appears only if

k1 ≥ 1 and the second one appears only if k2 ≥ 1.

4

Proof. We divide this proof in some steps. Step 1: (un , vn ) is bounded in H. Indeed, by definition of N we have

I  (un , vn )(un , vn ) = 0,

(8)

which implies, un 2 + vn 2ω = (un , vn )2p 2p + 2

 RN

β(x)|un |p |vn |p

for all

n ∈ N.

(9)

Then we get, I(un , vn )

= = =



 1 1 2p 2 2 p p (un  + vn ω ) − (un , vn )2p + 2 β(x)|un | |vn | 2 2p RN   1 1  − un 2 + vn 2ω 2 2p p−1 (un , vn )2H for all n ∈ N 2p

and the boundedness of I(un , vn ) implies that (un , vn ) is bounded in H. Step 2:

I  (un , vn ) → 0,

Defining F : H → R by F (u, v) = I  (u, v)(u, v) = u2 + v2w −

in



RN

H .

(10)

|u|2p + |v|2p − 2

 RN

β(x)|u|p |v|p

we have F ∈ C 1 (H, R) with F  (u, v)(ϕ, ψ)

=

 2p−2

|u| 2 u, ϕ + 2 v, ψw − 2p uϕ + |v|2p−2 vψ RN  p p−2

−2p β(x) |v| |u| uϕ + |u|p |v|p−2 vψ RN

for all (u, v), (ϕ, ψ) ∈ H. Note that N = F −1 (0)\{0}. Moreover, for (u, v) ∈ N , by (9) we have F  (u, v) = 0 because 

 2p  2 p p β(x)|u| |v| = 2(1 − p)(u, v)2H < 0. F (u, v)(u, v) = 2(u, v)H − 2p (u, v)2p + 2 RN

It follows that N is a C 1 manifold of H. If Hτ := {(u, v) ∈ H : Tτ (u, v) = (u, v)} then Hτ is a closed subspace of H and then, N τ is a closed submanifold of N in Hτ . So T(u,v) N τ = {(ϕ, ψ) ∈ Hτ : F  (u, v)(ϕ, ψ) = 0} is the tangent space to N τ at (u, v) and it follows from [21, Theorem 8.5] the existence of a sequence (λn ) in R satisfying (11) I  (un , vn ) − λn F  (un , vn )Hτ → 0. By Step 1 we know that (un , vn ) is bounded in H and we get I  (un , vn )(un , vn ) − λn F  (un , vn )(un , vn ) → 0.

(12)

Given that (un , vn ) ∈ N , we obtain I  (un , vn )(un , vn ) = 0

5

for all

n ∈ N.

(13)

Moreover, 

F (un , vn )(un , vn )

=

2(un , vn )2H

 − 2p

(un , vn )2p 2p



p

+2 RN

β(x)|un | |vn |

p



= (2 − 2p)(un , vn )2H .

Since 0 ∈ / N , we have constants such that 0 < c ≤ (un , vn )H ≤ C and so 0 < c˜ ≤ |F  (un , vn )(un , vn )| ≤ C˜

n ∈ N.

for all

(14)

By (12), (13) and (14) we conclude that λn → 0 as n → ∞. On the other hand, since (un , vn ) is bounded we get (F  (un , vn )) bounded in Hτ , which implies λn F  (un , vn )Hτ → 0

as

n → ∞.

(15)

Thus, using (11) and (15) we see that (un , vn ) is a (PS) sequence for I in Hτ . But, since Tτ is a isometry, by (β3) we get I  (un , vn ) = T˜τ ◦ I  ◦ Tτ (un , vn ) = T˜τ ◦ I  (un , vn ) = I  (un , vn ) ◦ Tτ . Then I  (un , vn )(ϕ, ψ) = 0 for all (ϕ, ψ) ∈ Hτ⊥ which implies that (un , vn ) is a (PS) sequence for I in H and so (10) is valid. u, v¯) in H where (¯ u, v¯) is a critical point for I, which is a solution for (P). Step 3: (un , vn ) (¯ u, v¯) ∈ H such that Using that (un , vn ) is bounded in H, up to a subsequence, there exists (¯ un u ¯ in H 1 (RN ) un → u ¯ in Lqloc (RN )

and

and ¯(x) un (x) → u

and

and

vn v¯ in Hw ,

vn → v¯ in Lqloc (RN ),

for

1 ≤ q < 2∗

vn (x) → v¯(x) for almost every x ∈ RN .

These convergences ensure that I  (un , vn )(ϕ, ψ) → I  (¯ u, v¯)(ϕ, ψ)

as

n→∞

u, v¯)(ϕ, ψ) = 0. By density of Cc∞ in H 1 (RN ) we get for any ϕ, ψ ∈ Cc∞ , and so (10) implies that I  (¯ I  (¯ u, v¯) = 0.

(16)

Moreover, since (un , vn ) ∈ N τ we have u(x), v¯(x)) Tτ (¯

= =

(−¯ u(τ x), −¯ v (τ x)) = lim (−un (τ x), −vn (τ x)) n→∞

lim Tτ (un (x), vn (x)) = lim (un (x), vn (x)) = (¯ u(x), v¯(x))

n→∞

n→∞

for almost every x ∈ RN , and this finishes Step 3. u, v¯) in H we obtain (¯ u, v¯) ∈ N τ and At this point, if we have the strong convergence (un , vn ) → (¯ then, we conclude the proof of the lemma, because (a) is valid. On the contrary, if (un , vn ) → (¯ u, v¯) in H, we need a few more steps. We define u1n (x) = un (x) − u ¯(x)

vn1 (x) = vn (x) − v¯(x)

and

for

x ∈ RN .

(17)

¯ and v¯ are τ -antisymmetric, we get u1n and vn1 also τ -antisymmetric. Since un , vn , u Step 4: I∞ (u1n , vn1 ) = I(un , vn ) − I(¯ u, v¯) + o(1). ¯ in H 1 (RN ) and vn v in Hw , we obtain Indeed, since un u u1n 2 = un 2 − ¯ u2 + o(1)

vn1 2w = vn 2w − ¯ v 2w + o(1).

and

6

(18)

Once we have (un , vn ) bounded in H and H 1 (RN ) → Lr (RN ) for 2 ≤ r ≤ 2∗ it follows that {un } and {vn } are bounded in L2p (RN ). So, the Brezis-Lieb Lemma implies that    |u1n |2p = |un |2p − |¯ u|2p + o(1) and RN RN RN    1 2p 2p |vn | = |vn | − |¯ v |2p + o(1). (19) RN

RN

It remains for us to verify that   |u1n |p |vn1 |p = β∞ RN

RN

RN

β(x)|un |p |vn |p −

 RN

β(x)|¯ u|p |¯ v |p + o(1).

(20)

Here we use an argument similar to that one used in [21, Lemma 8.1] (see also [5]). By Lemma 4 on Appendix, for F (t) = |t|r , r > 1, we know that there exists C = C(r) > 0 satisfying |F (t − s) − F (t) − F (s)| ≤ C (|F  (t)s| + |F  (s)t|)

for all

t, s ∈ R.

(21)

So, for all n ∈ N we get ||un |p |vn |p − |un − u ¯|p |vn − v¯|p − |¯ u|p |¯ v |p |

≤

|−|un |p (|vn − v¯|p − |vn |p − |¯ v |p )| + |−|vn − v¯|p (|un − u ¯|p − |un |p − |¯ u|p )| +|un |p |¯ v |p + |vn − v¯|p |¯ u|p + |¯ u|p |¯ v |p

≤

Cp |un |p (|vn |p−1 |¯ v | + |¯ v |p−1 |vn |) +Cp |vn − v¯|p (|un |p−1 |¯ u| + |¯ u|p−1 |un |) +|un |p |¯ v |p + |vn − v¯|p |¯ u|p + |¯ u|p |¯ v |p .

Then, since β ∈ L∞ (RN ), for R > 0 we obtain   

  p p 1 p 1 p p p  β(x) |un | |vn | − |un | |vn | − |¯ u| |¯ v|     RN  ≤ C ||un |p |vn |p − |un − u ¯|p |vn − v¯|p − |¯ u|p |¯ v |p | B(0,R) 

+C |un |p (|vn |p−1 |¯ v | + |¯ v |p−1 |vn | + |¯ v |p ) N R \B(0,R) 

+C |vn − v¯|p (|un |p−1 |¯ u| + |¯ u|p−1 |un | + |¯ u|p ) + |¯ u|p |¯ v |p RN \B(0,R)

for all n ∈ N. Since we have {un } and {vn } bounded in L2p (RN ) and u ¯, v¯ ∈ L2p (RN ), given ε > 0 we can choose R > 0 sufficiently large such that the sum of the integrals given above, on RN \B(0, R), is less than ε/2. After that, the Rellich-Kondrachov theorem gives us the compactness of the imbedding H 1 (B(0, R)) → L2p (B(0, R)) which implies  ε C for large n. ||un |p |vn |p − |un − u ¯|p |vn − v¯|p − |¯ u|p |¯ v |p | < 2 B(0,R) So we obtain

 RN

β(x)|u1n |p |vn1 |p



p

= RN

p

β(x)|un | |vn | −

 RN

β(x)|¯ u|p |¯ v |p + o(1).

(22)

On the other hand, using condition (β1), given ε > 0 we can choose R > 0 satisfying |β(x) − β∞ | < ε for |x| > R. Since β ∈ L∞ (RN ), we get       1 p 1 p 1 p 1 p  (β(x) − β∞ )|un | |vn |  ≤ C |un | |vn | + |β(x) − β∞ ||u1n |p |vn1 |p  RN

≤

B(0,R) RN \B(0,R) p p Cu1n L2p (B(0,R)) vn1 L2p (B(0,R))

7

+ εu1n p2p vn1 p2p .

Using again the boundedness of (u1n ) and (vn1 ) in L2p (RN ) and the strong convergence u1n , vn1 → 0 in L2p (B(0, R)) we obtain   β(x)|u1n |p |vn1 |p = β∞ |u1n |p |vn1 |p + o(1). (23) RN

RN

From (22) and (23) it follows (20) and we conclude Step 4.  I∞ (u1n , vn1 ) → 0 in H  as n → ∞.

Step 5:

At first, as in (23) we observe that  [I∞ (u1n , vn1 ) − I  (u1n , vn1 )](ϕ, ψ) = −

 RN



(β∞ − β(x)) |vn1 |p |u1n |p−2 u1n ϕ + |u1n |p |vn1 |p−2 vn1 ψ → 0

as n → ∞, uniformly in (ϕ, ψ) in H with (ϕ, ψ)H ≤ 1. By Steps 2 and 3, to prove Step 5 we need to verify u, v¯) + o(1) in H  . (24) I  (u1n , vn1 ) = I  (un , vn ) − I  (¯ We have I  (u1n , vn1 )(ϕ, ψ)

=

I  (un , vn )(ϕ, ψ) − I  (¯ u, v¯)(ϕ, ψ) 

+ |un |2p−2 un ϕ − |u1n |2p−2 u1n ϕ − |¯ u|2p−2 u ¯ϕ N R

+ |vn |2p−2 vn ψ − |vn1 |2p−2 vn1 ψ − |¯ v |2p−2 v¯ψ N R

+ β(x) |vn |p |un |p−2 un ϕ − |vn1 |p |u1n |p−2 u1n ϕ − |¯ v |p |¯ u|p−2 u ¯ϕ N R

+ β(x) |un |p |vn |p−2 vn ψ − |u1n |p |vn1 |p−2 vn1 ψ − |¯ u|p |¯ v |p−2 v¯ψ . RN

By [21, Lemma 8.1] we get    |un − u ¯|2p−2 (un − u ¯)ϕ = |un |2p−2 un ϕ − |¯ u|2p−2 u ¯ϕ + o(1) N N N R R R    |vn |2p−2 vn ψ − |¯ v |2p−2 v¯ψ + o(1) |vn − v¯|2p−2 (vn − v¯)ψ = RN

RN

RN

and using similar arguments, like in Step 4, we see that    β(x)|vn − v¯|p |un − u ¯|p−2 (un − u ¯)ϕ = β(x)|vn |p |un |p−2 un ϕ − RN

and  RN

RN

β(x)|un − u ¯|p |vn − v¯|p−2 (vn − v¯)ψ =

and

 RN

β(x)|un |p |vn |p−2 vn ψ −

uniformly in (ϕ, ψ)H ≤ 1. Thus we get (24) and conclude Step 5. Step 6: There exist a sequence {yn1 } in RN and (u1 , v 1 ) ∈ H satisfying • |yn1 | → ∞; • (u1n (· + yn1 ), vn1 (· + yn1 )) (u1 , v 1 ) in H; • (u1 , v 1 ) is a nontrivial critical point for I∞ . 

We define δu := lim sup sup

n→∞ y∈RN

8

B(y,1)

|u1n (x)|2 dx

RN

 RN

β(x)|¯ v |p |¯ u|p−2 u ¯ϕ + o(1)

β(x)|¯ u|p |¯ v |p−2 v¯ψ + o(1)



and δv := lim sup sup

n→∞ y∈RN

B(y,1)

|vn1 (x)|2 dx.

We note that if δu = δv = 0, from the concentration compactness lemma due to P. L. Lions (see [14]), we get u1n , vn1 → 0 in Lq (RN ) for 2 < q < 2∗ .   Moreover, since I∞ (u1n , vn1 ) → 0 and (u1n , vn1 ) is bounded in H we see that I∞ (u1n , vn1 )(u1n , vn1 ) → 0 as n → ∞. So    1 2p   |un | + |vn1 |2p dx + 2β∞ (u1n , vn1 )(u1n , vn1 ) + |u1n |p |vn1 |p dx → 0 (u1n , vn1 )2H = I∞ RN

RN

which means that (un , vn ) → (¯ u, v¯) in H and contradicts our hypothesis. Thus we have δu + δv > 0 and without loss of generality we can suppose that δu > 0. Then there exists a sequence {yn } ⊂ RN satisfying  δu for all n ∈ N. (25) |u1n (x)|2 dx > 2 B(yn ,1) Now we can distinguish two cases: Case 1: The sequence {yn − τ yn } is bounded. In this case we consider Γ = {x ∈ RN : τ x = x} and write RN = Γ⊕Γ⊥ . Then we define yn1 = PΓ (yn ) the orthogonal projection to Γ, that is yn1 = (yn +τ yn )/2. Case 2: The sequence {yn − τ yn } is not bounded. Here we can assume that |yn − τ yn | → ∞ and take yn1 = yn . In any case it follows from (25) that there exists r > 0 such that  δu for all n ∈ N. |u1n (x)|2 dx > 2 1 ,r) B(yn

(26)

We define a new sequence {(wn1 , zn1 )} in H by (wn1 , zn1 ) := (u1n (· + yn1 ), vn1 (· + yn1 )). Since (wn1 , zn1 ) is bounded, up to subsequences, we have (wn1 , zn1 ) (u1 , v 1 ) in H,

wn1 → u1 in L2 (B(0, r))

So, by using (26) we get   |u1 (x)|2 dx = lim B(0,r)

n→∞

B(0,r)

and

(wn1 (x), zn1 (x)) → (u1 (x), v 1 (x)) a.e. in RN .

|wn1 (x)|2 dx = lim



n→∞

1 ,r) B(yn

|u1n (x)|2 dx ≥

δu >0 2

and u1 = 0. Once (u1n , vn1 ) (0, 0) in H, the sequence {yn1 } is not bounded and we can assume that  (u1 , v 1 ) = 0 and finish this step. |yn1 | → ∞. Like in Step 3 we see that I∞ Step 7: There exists a sequence {(u2n , vn2 )} in H, τ -antisymmetric, such that

or Moreover

I∞ (u2n , vn2 ) = I∞ (u1n , vn1 ) − I∞ (u1 , v 1 ) + o(1),

if Case 1 occurs;

(27)

I∞ (u2n , vn2 ) = I∞ (u1n , vn1 ) − 2I∞ (u1 , v 1 ) + o(1),

if Case 2 occurs.

(28)

or (ϕ, ψ) = (u2n , vn2 ).

(29)

 I∞ (u2n , vn2 )(ϕ, ψ) = o(1)

Indeed, if Case 1 occurs we have

τ yn1

=

for all ϕ, ψ ∈ Cc∞ (RN ) yn1

and so u1 and v1 are τ -antisymmetric. Defining

(u2n (x), vn2 (x)) = (u1n (x) − u1 (x − yn1 ), vn1 (x) − v 1 (x − yn1 ))

for x ∈ RN

we get {(u2n , vn2 )} τ -antisymmetric and as in Steps 4 and 5 we see that (27) and (29) hold. 9

If Case 2 occurs, as in [16] for the scalar case, we consider σn (x) = u1 (x − yn1 ) − u1 (τ x − yn1 ) and

γn (x) = v 1 (x − yn1 ) − v 1 (τ x − yn1 )

where yn1 = yn . After that, defining   (u2n , vn2 ) = u1n − σn , vn1 − γn we have {(u2n , vn2 )} τ -antisymmetric for all n ∈ N. At first we will verify that (28) holds. We have (u2n , vn2 )2H = (u1n , vn1 )2H − 2 (u1n , vn1 ), (σn , γn )H + (σn , γn )2H . As in [5, Lemma 2.3], without the function ξ which simplifies our calculations, we can see that

(u1n , vn1 ), (σn , γn )H = 2(u1 , v 1 )2H + o(1) So

and

(σn , γn )2H = 2(u1 , v 1 )2H + o(1).

(u2n , vn2 )2H = (u1n , vn1 )2H − 2(u1 , v 1 )2H + o(1).

Now we estimate the L2p norm. Using again (21), we see that    1 2p   2 2p−1  |un | − |u2n |2p − |σn |2p  ≤ C |un | |σn | + |σn |2p−1 |u2n | RN

RN

for any n ∈ N. Once u1n is τ -antisymmetric we get   1

|un (x + yn ) − u1 (x)|2p−1 + |u1 (τ x + τ yn1 − yn1 )|2p−1 |u1 |. |u2n |2p−1 |σn | ≤ C RN

RN

Then, for any R > 0 we obtain  1 1 2p−1 1 |u2n |2p−1 |σn | ≤ C u1n (· + yn ) − u1 2p−1 L2p (B(0,R)) u 2p + C u L2p (B(τ y 1 −y 1 ,R)) u 2p n

RN

+C

u1n (·

+ yn ) −

u1 2p−1 u1 L2p (RN \B(0,R)) 2p

+C

n

u1 2p−1 u1 L2p (RN \B(0,R)) . 2p

Because (u1n ) is bounded in L2p (RN ), u1 ∈ L2p (RN ), u1n (·+yn ) → u1 in L2p (B(0, R)) and |τ yn1 −yn1 | → ∞, we see that  lim |u2n |2p−1 |σn | = 0. (30) n→∞

Similarly, we show that

RN

 lim

n→∞

RN

|σn |2p−1 |u2n | = 0.

Using the same arguments we prove this for vn2 , which allows us to write 2p 1 1 2p (u2n , vn2 )2p 2p = (un , vn )2p − (σn , γn )2p + o(1).

It is easily seen that and so

1 1 2p (σn , γn )2p 2p = 2(u , v )2p + o(1) 1 1 2p 1 1 2p (u2n , vn2 )2p 2p = (un , vn )2p − 2(u , v )2p + o(1).

For the last term of I∞ (u2n , vn2 ), as we made in Step 4 we use (21) to get   p p |σn | |γn | = 2 |u1 |p |v 1 |p + o(1). RN

RN

10

(31)

After that we estimate   1 1p  |un vn | − |u2n vn2 |p − |σn γn |p 

 ≤

RN

RN



  |vn1 |p |u1n |p − |u2n |p − |σn |p 

+ R +  ≤

N

  |σn |p |vn1 |p − |vn2 |p − |γn |p 

RN

RN



  |vn2 |p |σn |p + |u2n |p |vn1 |p − |vn2 |p 



|vn1 |p |u2n |p−1 |σn | + |u2n ||σn |p−1

+ R

N

+



|σn |p |vn2 |p−1 |γn | + |vn2 ||γn |p−1 

RN



 |vn2 |p |σn |p + |u2n |p |vn1 |p−1 + |vn2 |p−1 |γn | .

With similar arguments as used to prove (30) we obtain that each integral on the right side of the last inequality goes to 0 as n → ∞. Then      2 2 p 1 1 p p 1 1 p |un vn | = |un vn | − |σn γn | + o(1) = |un vn | − 2 |u1 v 1 |p + o(1) RN

RN

RN

RN

RN

which concludes the proof of (28). Moreover, it shows us that  (u2n , vn2 )(u2n , vn2 ) I∞

= =

=

2 2 p (u2n , vn2 )2 − (u2n , vn2 )2p 2p − 2β∞ un · vn p 1 1 p (u1n , vn1 )2 − (u1n , vn1 )2p 2p − 2β∞ un · vn p + o(1)   1 1 p − 2β u · v  −2 (u1 , v 1 )2 − (u1 , v 1 )2p ∞ p 2p   I∞ (u1n , vn1 )(u1n , vn1 ) − 2I∞ (u1 , v 1 )(u1 , v 1 ) + o(1).

Using the results of Step 5 and Step 6 we obtain  (u2n , vn2 )(u2n , vn2 ) = o(1), I∞

which is part of (29). Now we fix ϕ, ψ ∈ Cc∞ (RN ) and R > 0 such that supp(ϕ) ∪ supp(ψ) ⊂ B(0, R). Since u1n , vn1 , u2n and vn2 0 in H 1 (RN ), we get u1n , vn1 , u2n and vn2 → 0 in L2p (B(0, R)) and so 

2 2p−2 2  |un | (u2n , vn2 )(ϕ, ψ) = (u2n , vn2 ), (ϕ, ψ)H − un ϕ + |vn2 |2p−2 vn2 ψ I∞ B(0,R)  2 p 2 p−2 2

−β∞ |vn | |un | un ϕ + |u2n |p |vn2 |p−2 vn2 ψ = o(1), B(0,R)

and we finish Step 7. Step 8:Conclusion. Suppose that (u2n , vn2 ) → 0 in H. If Case 1 occurs in Step 6, using Step 4 and equation (27) we have proved that u, v¯) + I∞ (u1 , v 1 ) + I∞ (u2n , vn2 ) + o(1) = I(¯ u, v¯) + I∞ (u1 , v 1 ) + o(1) I(un , vn ) = I(¯ with ¯ − u1 (· − yn1 ), vn − v¯ − v 1 (· − yn1 )) o(1) = (u2n , vn2 ) = (u1n − u1 (· − yn1 ), vn1 − v 1 (· − yn1 )) = (un − u which implies

u, v¯) + (u1 (· − yn1 ), v 1 (· − yn1 )) + o(1). (un , vn ) = (¯

Then the lemma will be proved with k1 = 0 and k2 = 1. If Case 2 occurs in Step 6, using Step 4 and equation (28) we have u, v¯) + 2I∞ (u1 , v 1 ) + I∞ (u2n , vn2 ) + o(1) = I(¯ u, v¯) + 2I∞ (u1 , v 1 ) + o(1) I(un , vn ) = I(¯ 11

with ¯ − σn , vn − v¯ − γn ) = (un , vn ) − (¯ u, v¯) − (σn , γn ). o(1) = (u2n , vn2 ) = (u1n − σn , vn1 − γn ) = (un − u Since

(σn , γn ) = (u1 (· − yn1 ) − u1 (τ · −yn1 ), v 1 (· − yn1 ) − v 1 (τ · −yn1 ))

in H

we get     u, v¯) + u1 (· − yn1 ), v 1 (· − yn1 ) − u1 (τ · −yn1 ), v 1 (τ · −yn1 ) + o(1) (un , vn ) = (¯

in H

and the lemma will be proved with k1 = 1 and k2 = 0. If (u2n , vn2 )  0 in H, we repeat the process made before. Thus we show the existence of sequences j+1  j+1 j+1 in RN such that |ynj | → +∞, of sequences {(uj+1 n , vn )}n in H satisfying limn I∞ (un , vn ) = 0, j j and nontrivial solutions for (P1) (u , v ) satisfying {ynj }n

j+1 u, v¯) − I∞ (uj+1 n , vn ) = I(un , vn ) − I(¯

j 

δi I∞ (ui , v i ) + o(1)

i=1

where δi = 1 if {yni }n satisfies Case 1 in Step 6 and δi = 2 if {yni } satisfies Case 2 in Step 6. Since I(un , vn ) is bounded and, by (7), I∞ (ui , v i ) ≥ m∞ > 0 for all i = 1, · · · , j we get j+1 I∞ (uj+1 n , vn )

≤C−

j 

δi m∞ + o(1) ≤ C − jm∞ + o(1) for all j.

i=1

Thus the iteration must terminate at some finite index. We will use Lemma 1 in order to find conditions under which the functional I satisfies the Palais-Smale condiction. Lemma 2. The functional I|N τ satisfies (P S)c for all c < min{mτ∞ , 2m∞ }. Proof. Let {(un , vn )} ⊂ N τ be a (P S)c sequence for I|N τ with c < min{mτ∞ , 2m∞ }, that is I(un , vn ) → c and I|N τ (un , vn ) → 0. It follows from Lemma 1 that {(un , vn )} is a bounded (P S)c sequence for I in H. Thus, up to a subsequence, we can assume that (un , vn ) (u0 , v0 ) in H and so (u0 , v0 ) is a critical point for I. In particular,  2p  2 0 = I (u0 , v0 )(u0 , v0 ) = (u0 , v0 )H − (u0 , v0 )2p − 2 β(x)|u0 |p |v0 |p dx, RN

which implies (u0 , v0 )2H = (u0 , v0 )2p 2p + 2 Then I(u0 , v0 ) =

1 1 1 (u0 , v0 )2H − (u0 , v0 )2p 2p − 2 2p p

 RN

 RN

β(x)|u0 |p |v0 |p dx.

β(x)|u0 |p |v0 |p dx =

p−1 (u0 , v0 )2H ≥ 0. 2p

(32)

If there is no subsequence such that (un , vn ) → (u0 , v0 ) in H, then, by Lemma 1 we obtain k1 , k2 ≥ 0, with k1 + k2 ≥ 1, k1 solutions (uj , v j ), j = 1, · · · , k1 and k2 τ −antisymetric solutions (uj , v j ), for j = k1 + 1, ..., k1 + k2 of problem (P1), satisfying c

= = ≥

lim I(un , vn )

n→∞

I(u0 , v0 ) + 2 2k1 m∞ +

k1 

j=1 k2 mτ∞

I∞ (uj , v j ) + ≥

k 1 +k2

I∞ (uj , v j )

j=k1 +1 τ min{m∞ , 2m∞ },

which contradicts our hypothesis. Therefore, up to subsequences, (un , vn ) → (u0 , v0 ) in H and we are done.

12

4

Proof of the main result

The next result shows that N τ is not a empty set. One can find the proof in [15, Lemma 3.1] but we include it here for completeness. Lemma 3. For each pair (u, v) ∈ Hτ \{(0, 0)} there exists a unique number t¯ = t¯(u,v) > 0 such that (t¯u, t¯v) ∈ N τ . Moreover, I(t¯u, t¯v) is the maximum value for the function t → I(tu, tv) for t ≥ 0. The same result holds for I∞ and N∞ . Proof. Consider (u, v) ∈ Hτ \{(0, 0)} and for t ≥ 0 define g(t) := I(tu, tv) =

t2p t2p t2 (u, v)2H − (u, v)2p 2p − 2 2p p



β(x)|u|p |v|p dx.

RN

We observe that g  (t) = I  (tu, tv)(u, v) and so (tu, tv) ∈ N τ if and only if g  (t) = 0 and t > 0. Once  2p  2 2p−1 2p−1 g (t) = t(u, v)H − t (u, v)2p − 2t β(x)|u|p |v|p dx, RN

for t¯ > 0 we get g  (t¯) = 0 when   (u, v)2H = t¯2(p−1) (u, v)2p + 2 2p For this t¯ we have (t¯u, t¯v) ∈ N τ and so we see that    ¯ 2 2p−1 ¯ ¯ ¯ + 2 (u, v)2p tg (t) = t(u, v)H − (2p − 1)t 2p

RN

β(x)|u|p |v|p dx .

p

RN

p



β(x)|u| |v| dx

(33)

= 2(1 − p)t¯(u, v)2H < 0.

Thus, t¯ is a local maximum point for g, which is global in [0, ∞) because there is no other critical point for g in (0, ∞) and, since g(0) = 0 and g(t) > 0 for small t > 0, t = 0 is a local minimum for g. The results that we prove in the propositions below are crucial to show that mτ is a critical value of I . Proposition 1. Suppose (β1) − (β4). Then mτ ≤ mτ∞ . τ Proof. Taking (u, v) ∈ N∞ , by the Lemma 3 we have

I∞ (u, v) = max I∞ (tu, tv). t≥0

Moreover, there exists an unique t¯ = t¯u,v > 0 so that t¯(u, v) ∈ N τ , and this t¯ satisfies I(t¯u, t¯v) = max I(tu, tv). t≥0

Now

mτ ≤

I(t¯u, t¯v) = 12 ||(t¯u, t¯v)||2H −

1 ¯ ¯ 2p 2p ||(tu, tv)||2p

≤

1 ¯ ¯ 2 2 ||(tu, tv)||H

=

I∞ (t¯u, t¯v) ≤ I∞ (u, v).

−

1 ¯ ¯ 2p 2p ||(tu, tv)||2p

−

1 p

 RN

−

1 p

 RN

β(x)|t¯u|p |t¯v|p dx

β∞ |t¯u|p |t¯v|p dx

Thus, mτ ≤ mτ∞ as we desired. Proposition 2. Suppose (β1) − (β5). Then mτ := inf{I(u, v) : (u, v) ∈ N τ } < 2m∞ .

(34)

Proof. Consider (¯ u, v¯) ∈ N∞ a ground state solution solution for (P1) (See [15]). We know that u ¯, v¯ > 0 are radial functions such that I∞ (¯ u, v¯) = m∞ . Moreover u ¯ and v¯ have the exponential decay (see [9]) u ¯(x) + v¯(x) + |∇¯ u(x)| + |∇¯ v (x)| ≤ Ce−α|x| 13

for all

x ∈ RN

(35)

for some constants C, α > 0. Now, for y ∈ RN we define   (wy (x), zy (x)) = u ¯(x − y) − u ¯(x − τ y), v¯(x − y) − v¯(x − τ y) . It is easily seen that (wy , zy ) ∈ Hτ \(0, 0) for y = τ y. Denoting by ty the number given by Lemma 3 for (wy , zy ), we have (ty wy , ty zy ) ∈ N τ . Our job here is to verify that I(ty wy , ty zy ) < 2m∞ for some y, which implies that mτ < 2m∞ . At first, we take y ∈ Γ⊥ , where Γ = ker(Id − τ ) as in Lemma 1, which gives us y = −τ y.

(36)

We will prove that ty is bounded for large |y|. More precisely, that 0 < t1 < ty < t2 for constants t1 and t2 . We use some steps. Step 1: We get (wy , zy )2p 2p

=

2(¯ u, v¯)2p 2p + oy (1),

where |oy (1)| ≤ Ce−α|y| . Using the result given by [1, Lemma 2.4] for the function t → |t|2p we get     |wy (x)|2p − |¯ u(x − y)|2p−1 u u(x − y)|2p − |¯ u(x − τ y)|2p  ≤ 4p |¯ ¯(x − τ y) + u ¯(x − y)|¯ u(x − τ y)|2p−1 ¯ is radial, we obtain for x ∈ RN . Because u            2p 2p 2p 2p 2p     |w | dx − 2 |¯ u | dx = |w | dx − |¯ u (x − y)| dx − |¯ u (x − τ y)| dx y  N y    R RN RN RN RN 

≤ 4p |¯ u(x − y)|2p−1 u ¯(x − τ y) + |¯ u(x − τ y)|2p−1 u ¯(x − y) dx N R = 8p |¯ u(x)|2p−1 |¯ u(x + y − τ y)|dx. RN

Using the exponential decay of u ¯ and (36) we see that    2p−1 −(2p−1)α|x| −α|x+y−τ y| −α|y−τ y| |¯ u(x)| |¯ u(x + y − τ y)|dx ≤ C e e dx ≤ Ce RN

≤

RN −α|y−τ y|

Ce

e−(2p−2)α|x| dx RN

≤ Ce−α|y| .

This gives us the desired result for wy and using the same arguments we prove for zy . Step 2: We have

(wy , zy )2H = 2(¯ u, v¯)2H + oy (1)

where |oy (1)| ≤ Ce−α|y| . In order to prove Step 2 is enough to show that   u ¯(x − y)¯ u(x − τ y)dx + v¯(x − y)¯ v (x − τ y)dx = oy (1), and N RN  R  ∇¯ u(x − y)∇¯ u(x − τ y)dx + ∇¯ v (x − y)∇¯ v (x − τ y)dx = oy (1). RN

We will lead with the first integral. We have   u ¯(x)¯ u(x + y − τ y)dx ≤ C¯ u1/2 ∞ RN

RN

RN

e−α|x| e−(α/2)|x+y−τ y| dx ≤ Ce−α|y−τ y|/2 = Ce−α|y| .

Using the exponential decay of v¯, |∇¯ u| and |∇¯ v | we see that this inequality is also valid for these functions and so Step 2 is proved.

14

Step 3: We have   p p β(x)|wy | |zy | dx = 2 RN



p p

RN

β(x + y)¯ u v¯ + oy (1)

and RN

[β∞ − β(x + y)]¯ up v¯p dx ≤ −Ce−γ|y|

where |oy (1)| ≤ Ce−α|y| and γ is given in (β5). Indeed, 

p

RN

p

β(x)|wy | |zy | dx

 = 

RN

= RN

β(x)|¯ u(x − y) − u ¯(x − τ y)|p |¯ v (x − y) − v¯(x − τ y)|p dx  β(x + y)¯ up v¯p dx + β(x + τ y)¯ up v¯p dx + oy (1) RN

where oy (1) satisfies

|oy (1)| ≤ Ce−α|y| .

Since u ¯ and v¯ are symmetric and (β3) holds, we get   p p β(x + y)¯ u v¯ dx = RN

β(x + τ y)¯ up v¯p dx.

RN

The last inequality in Step 3 follows from (β5) because   e−γ|x| u ¯p v¯p dx ≤ C 0< RN

RN

e−α|x| dx < ∞.

Step 4: There exist t1 , t2 ∈ R such that 0 < t1 < ty < t2

for |y| large enough.

Indeed, by (33) we have (wy , zy )2H



 p p β(x)|w | |z | dx y y RN    2(p−1) 2p (wy , zy )2p + 2β∞ RN |wy |p |zy |p dx . ≤ ty 2(p−1)

= ty

(wy , zy )2p 2p + 2



Then, from the previous steps it follows that ≥ t2(p−1) y

2(¯ u, v¯)2H + oy (1) .  2(¯ u, v¯)2p ¯p v¯p dx + oy (1) 2p + 4β∞ RN u

Once |oy (1)| ≤ min{1, (¯ u, v¯)2p H } for large |y|, we obtain 0<

(¯ u, v¯)2H ≤ t2(p−1) .  y pv p dx + 1 2(¯ u, v¯)2p + 4β u ¯ ¯ ∞ N 2p R

(37)

On the other hand, since I  (ty wy , ty zy )(ty wy , ty zy ) = 0, using the condition (β2) we get     2p 2p 2p |w + 2β | + |z | t |wy |p |zy |p dx t2y (wy , zy )2H ≥ t2p y y ∞ y y RN

RN

which implies that t2(p−1) y

≤

2(¯ u, v¯)2H + 1  2p (¯ u, v¯)2p + 2β∞ RN u ¯p v¯p dx

for large |y|. This prove the boundedness of {ty } for large |y|. It remains for us to show that I(wy , zy ) < 2m∞ 15

(38)

for |y| large enough. For simplicity of notation we let ty := t. Using the results we just proved, we have I(twy , tzy )

= = ≤

 t2p   t2p  t2  (¯ u, v¯)2H + oy (1) − (¯ u, v¯)2p + o (1) − β(x)|wy |p |zy |p dx y 2p 2 2p p RN   t2p u, t¯ v) + β∞ u ¯p v¯p dx − β(x + y)¯ up v¯p dx + oy (1) 2I∞ (t¯ p RN RN u, t¯ v ) − Ce−γ|y| + C  e−α|y| 2I∞ (t¯

for large |y|. Since (¯ u, v¯) ∈ N∞ , by Lemma 3 we get   I(twy , tzy ) ≤ 2I∞ (¯ u, v¯) + e−α|y| C  − Ce(α−γ)|y| < 2m∞ for |y| large enough, which concludes the proof of this proposition. This next result ensures the existence of a (PS) sequence for I in H at the level mτ . Proposition 3. There exists a sequence (un , vn ) ⊂ N τ satisfying I(un , vn ) → mτ

and

I  (un , vn ) → 0 in H  .

Proof. Since that N τ is closed in Hτ and (0, 0) ∈ / N τ then N τ is bounded from below, this is, there exist ρ > 0 such that (39) (u, v)H ≥ ρ > 0, for all (u, v) ∈ N and so, using (9) we get I(u, v) =

p−1 2 p−1 (u, v)2H ≥ ρ > 0. 2p 2p

Let (un , vn ) ⊂ N τ be a minimizing sequence, this is, (un , vn ) is such that I(un , vn ) → mτ . As in Step 2 of Lemma 1 we denote F (u, v) = I  (u, v)(u, v). We can apply the Ekeland variational principle in Hτ and obtain sequences (wn , zn ) ⊂ N τ and (λn ) ⊂ R such that I(wn , zn ) → mτ and I  (wn , zn ) − λn F  (wn , zn )Hτ = o(1).

(40)

This means that (wn , zn ) is a (PS) sequence for I|N τ and as in Step 2 of Lemma 1 we see that (wn , zn ) is a (PS) sequence for I in H. Hence, without loss of generality, we can assume that I(un , vn ) → mτ

and

I  (un , vn ) → 0

which concludes this proof. Now we are able to prove our main result.

Proof of Theorem 1: We observe that from Proposition 3 there exists (un , vn ) ⊂ N τ a (PS) sequence for I at level mτ . By Proposition 1 we have mτ ≤ mτ∞ . In order to conclude the proof of Theorem 1, we will consider two cases: Case 1: mτ < mτ∞ . In this case, by Proposition 2 we have mτ < inf{mτ∞ , 2m∞ } and by Lemma 2 we know that I|N τ satisfies the (P S)c condition for c < inf{mτ∞ , 2m∞ }. Then, up to a subsequence, (un , vn ) → (u, v) in H for some (u, v) ∈ H\{0}. Thus (u, v) ∈ N τ is a critical point for I in H and consequently is a nontrivial solution for problem (P). Case 2: mτ = mτ∞ . If the (P S)mτ sequence for I, (un , vn ) given by Proposition 3, possesses a convergent subsequence in H we are done. On the contrary, by Lemma 1 we have τ

u, v¯) + 2 m = lim I(un , vn ) = I(¯ n→∞

k1 

j

j

I∞ (u , v ) +

j=1

k 1 +k2 j=k1 +1

16

I∞ (uj , v j ) ≥ I(¯ u, v¯) + 2k1 m∞ + k2 mτ∞ .

By (32) we know that I(¯ u, v¯) ≥ 0 and then, by Proposition 2 we get k1 = 0, k2 = 1 and I(¯ u, v¯) = C||(¯ u, v¯)||2H = 0. So we can write I(un , vn ) → I∞ (u1 , v 1 ) τ and I∞ (u1 , v 1 ) = mτ = mτ∞ , where (u1 , v 1 ) ∈ N∞ . By the same argument of Proposition 1, due to 1 1 τ ¯ ¯ Lemma 3, there exists t > 0 so that t(u , v ) ∈ N and we have

mτ ≤ I(t¯u1 , t¯v 1 ) ≤ I∞ (t¯v 1 , t¯v 1 ) ≤ I∞ (u1 , v 1 ) = mτ . τ The uniquiness of t˜ > 0 such that t˜(u1 , v 1 ) ∈ N∞ , which is the unique number satisfying

I∞ (t˜u1 , t˜v 1 ) = max I∞ (tu1 , tv 1 ), t≥0

implies that t¯ = 1. So (u1 , v 1 ) ∈ N τ and I(u1 , v 1 ) = mτ . Therefore, (u1 , v 1 ) is a critical point of I|N τ . Obviously, (¯ un , v¯n ) = (u1 , v 1 ), ∀n ∈ N, is a (P S)mτ sequence for I|N τ . By Lemma 1, it is a (P S)mτ sequence for I in H. Then we have ||I  (u1 , v 1 )||H  = ||I  (¯ un , v¯n )||H  → 0, that is ||I  (u1 , v 1 )||H  = 0 and 1 1 1 1 (u , v ) is a critical point for I in H. Consequently (u , v ) is a nontrivial solution for problem (P).

5

Appendix

We present here a easy and necessary inequality to prove the splitting lemma. The arguments used on the proof are similar to that ones in [1, Lemma 2.4]. Lemma 4. Consider F : R → R given by F (t) = |t|r with r > 1. Then there exists a positive constant C = C(r) satisfying |F (t − s) − F (t) − F (s)| ≤ C (|F  (t)s| + |F  (s)t|)

for all

t, s ∈ R.

(41)

Proof. The inequality obviously holds for (s, t) = (0, t), (s, t) = (s, 0) or s = t with any C > 0. For 0 < |s| ≤ |t|, since g(ζ) = ζ r−1 is a increasing function if ζ ≥ 0, we see that  t−s    1   1           |F (t − s) − F (t)| =  F (σ)dσ  = −s F (t − σs)dσ  ≤ r|s| |t − σs|r−1 dσ t

≤

2r−1 r|s|



0

1

0

0

|t|r−1 dσ = 2r−1 |sF  (t)|.

On the other hand we have |F (s)| = |s|r ≤ |t||s|r−1 =

1 |tF  (s)| ≤ 2r−1 |tF  (s)|. r

So the inequality in (41) holds in the case 0 < |s| ≤ |t|. If we take 0 < |t| < |s|, since F is an even function, we get |F (t − s) − F (t) − F (s)| = |F (s − t) − F (s) − F (t)| and for the results above we obtain |F (s − t) − F (s) − F (t)| ≤ 2r−1 (|tF  (s)| + |sF  (t)|) . Thus (41) holds with C = 2r−1 .

Acknowledgments The authors would like to thank professor Ol´ımpio Hiroshi Miyagaki for his valuable suggestions and comments.

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