Segregated vector solutions for nonlinear Schrödinger systems with electromagnetic potentials

In this paper, we study the following nonlinear Schrodinger system in \begin{document}$\mathbb{R}^3$\end{document} \begin{document}$\left\{ \begin{array}{*{35}{l}} {{(\frac{\nabla }{i}-A(y))}^{2}}u+{{\lambda }_{1}}(|y|)u=|u{{|}^{2}}u+\beta |v{{|}^{2}}u,x'>where \begin{document}$A(y)=A(|y|)∈ C^1(\mathbb{R}^3,\mathbb{R})$\end{document} is bounded, \begin{document}$λ_1(|y|),λ_2(|y|)$\end{document} are continuous positive radial potentials, and \begin{document}$β∈ \mathbb{R}$\end{document} is a coupling constant. We proved that if \begin{document}$A(y),λ_1(y),λ_2(y)$\end{document} satisfy some suitable conditions, the above problem has infinitely many non-radial segregated solutions.


(1.3)
Here ψ(y, t) takes on complex values, is the Planck constant, i is the imaginary unit, G : R 3 → R denotes an electric potential, A = (A 1 , A 2 , A 3 ) : R 3 → R 3 is a 1786 JING YANG vector potential with magnetic field, B = curlA, 2 ≤ p ≤ 5, µ, ν > 0, 2 ≤ q ≤ 5 and p + q ≤ 6. Moreover, the Schrödinger operator is defined by This type of (1.2) arises in different physical theories, e.g. the description of Bose-Einstein condensates, plasma physics and nonlinear optics (see [1,30] and the references therein). When we are always focused on finding standing wave solutions for (1.2), namely, waves of the form ψ(y, t) = e − iEt u(y) for some function u : R N → C, one is led to solve the following complex equation in R 3 i ∇ − A(y) 2 u + (G(y) − E)u = f (y, |u|)u.
For simplicity, letting V (y) = G(y) − E, then the above problem can be rewritten as which received a lot of interest in recent years. For the existence of ground state solutions, multi-bump solutions, infinitely many solutions or asymptotic behavior of the solutions to (1.4), one can refer to [6, 10, 15-18, 20, 29] and the references therein.
Before we close this introduction, let us outline the main idea in the proof of Theorem (1.1).
For any function K(y) > 0, the sobolev space H 1 K (R 3 , C) is endowed with the standard norm which is induced by the inner product Denote H to be the product space It is well known that the following problem has a unique solution, denoted by U , which is non-degenerate and satisfies for some c > 0 (see [19]), U (r)re r → c, U (r)re r → −1, as r = |y| → ∞.
Let U c : R 3 → C be a least-energy solution of the following equation Then by energy comparison (see [18]), one has U c (y) = e iσ U (y) for some choice of σ ∈ [0, 2π]. Moreover, form [11,12], we know that U c is nondegenerate.
In this paper, we will use the least energy solutions of problem where A 0 = (A 0,1 , A 0,2 , A 0,3 ) is a constant vector, to build up the approximate solutions for (1.1). It is easy to find that u is a solutions of (1.7) if and only if e −iA0·y u(y) is a solution of (1.6). From the non-degeneracy of U c , we can infer that e iA0·y U c is non-degenerate. Define and where r ∈ [r 0 k ln k, r 1 k ln k] for some constant r 1 > r 0 > 0. Furthermore, we let H λ1 = u : u ∈ H 1 λ1 (R 3 , C), u is even in y j , j = 2, 3, u(r cos θ, r sin θ, y 3 ) = u(r cos(θ + 2iπ k ), r sin(θ + 2iπ k ), y 3 ) . and we can define H λ2 similarly. Set To prove Theorem 1.1, it suffices to verify the following result: Under the assumption of Theorem (1.1), there is an integer k 0 > 0, such that for any integer k ≥ k 0 , (1.1) has a solution (u k , v k ) of the form The rest of the paper is organized as follows. In Section 2, we will carry out a reduction procedure. We prove our main result in Section 3. Finally, in Appendix, some basic estimates and an energy expansion for the functional corresponding to problem (1.1) will be established.
2. The finite-dimensional reduction. In this section, we carry out a finitedimensional reduction. Write where µ > 0 is a small and M is a large constant depending on a, b, m, A 1 , A 2 (seen in the Appendix). Let (2.1) Now we can expand It is easy to check that is a bounded bi-linear functional in E. Thus, there exists a bounded linear operator L from E to E such that for any (u, v), (ϕ, ψ) ∈ E, Using the above discussion, we can get the following results.
Lemma 2.1. There exists a constant C > 0, independent of k, such that for any Now we consider the invertibility of L.
Proof. Arguing by contradiction, we suppose that there are k → ∞, and Next, for simplicity, we will use r replace r k . By symmetry, we see from (2.4) In particular, Obviously, these estimates (2.4),(2.5) and (2.6) are also true in Ω 1 .
Applying the same argument on Ω 1 , we can prove that as . As a result, Note that by Lemma A.2, we get Thus, it follows from (2.3) that where we used the fact that for some constant C > 0 independent of k.
Therefore, from the above estimates, we have proved our result.

3.
Proof of the main result. In this section, we come to prove the main theorem.
Proof of Theorem 1.4. Let (ϕ r,σ , ψ r,σ ) be the map obtained in Proposition 2.4. Define F (r) = I(Z r + ϕ r,σ , Z * r + ψ r,σ ), ∀ r ∈ D k . Applying the same argument used in [9,26], we can easily check for k sufficiently large, if r is a critical point of F (r), then (Z r + ϕ r,σ , Z * r + ψ r,σ ) is a critical point of I(u, v).
It follows from Lemmas 2.3, 2.5 and A.3 that For any β < β * , where β * is defined in Lemma 2.2, we can take k 0 > 0 such that Assume that (3.1) is achieved by some r k ∈ D k . We will prove that r k is an interior point of D k . When o(1)β + O β 2 ln k > 0, we let Then But we can easily check g(t) has a maximum point t k satisfying As a result, the function | ln k| m k m for some constant c 1 > 0 depending only on a, b, A 1 , m.
When o(1)β + O β 2 ln k ≤ 0, we definē Then we can still verify that the maximum ofḡ 1 (r) is Next we show that the maximum can not be on the boundary of D k . Assume that r k = m 2π − µ k ln k and o(1)β + O β 2 ln k > 0. Then which yields a contradiction to (3.4).
Assume that r k = M k ln k. Then if M is large enough. This is also a contradiction to (3.3).
So we have proved that r k is an interior point of D k for large k and then r k is a critical point of F (r). (A.1) Next, we estimate each term in (A.1). Firstly, Using Lemma A.2, one has k i,j=1 Moreover, similar to the above two equalities, it is easy to check Secondly, noticing that where η j = iσ + iA(x j )(y − x j ), we find But applying the same argument as before, we calculate cos(A(y 1 )(y j − y 1 )) Ω1 U 3 y 1 U y j + kO k r e −3r π k .