Normalized solutions of linearly coupled Choquard system with potentials

Meng Li∗

###### Abstract

In this paper, we consider the existence of solutions for the linearly coupled
Choquard system with potentials

$\displaystyle\left\\{\begin{aligned} &-\Delta
u+\lambda_{1}u+V_{1}(x)u=\mu_{1}(I_{\alpha}\star|u|^{p})|u|^{p-2}u+\beta(x)v,\\\
&-\Delta
v+\lambda_{2}v+V_{2}(x)v=\mu_{2}(I_{\alpha}\star|v|^{q})|v|^{q-2}u+\beta(x)u,\end{aligned}\right.\quad
x\in\mathbb{R}^{N},$

under the constraint

$\displaystyle\int_{\mathbb{R}^{N}}u^{2}dx=\xi^{2},~{}\int_{\mathbb{R}^{N}}v^{2}dx=\eta^{2},$

where
$I_{\alpha}=\frac{1}{|x|^{N-\alpha}},~{}\alpha\in(0,N),~{}1+\frac{\alpha}{N}<p,~{}q<\frac{N+\alpha}{N-2},~{}\mu_{1}>0,~{}\mu_{2}>0$
and $\beta(x)$ is a fixed function.
Keywords: Linearly coupled Choquard system, normalized solutions, ground state
solutions

000 ∗ Corresponding author.
AMS Subject Classification: 58F11,35Q30.
E-mails<EMAIL_ADDRESS>
##  1 Introduction

In this paper, we consider the existence of solutions for the linearly coupled
Choquard system with potentials

$\displaystyle\left\\{\begin{aligned} &-\Delta
u+\lambda_{1}u+V_{1}(x)u=\mu_{1}(I_{\alpha}\star|u|^{p})|u|^{p-2}u+\beta(x)v,\\\
&-\Delta
v+\lambda_{2}v+V_{2}(x)v=\mu_{2}(I_{\alpha}\star|v|^{q})|v|^{q-2}u+\beta(x)u,\end{aligned}\right.\quad
x\in\mathbb{R}^{N},$ (1.1)

under the constraint

$\displaystyle\int_{\mathbb{R}^{N}}u^{2}dx=\xi^{2},~{}\int_{\mathbb{R}^{N}}v^{2}dx=\eta^{2},$
(1.2)

where
$I_{\alpha}=\frac{1}{|x|^{N-\alpha}},~{}\alpha\in(0,N),~{}1+\frac{\alpha}{N}<p,~{}q<\frac{N+\alpha}{N-2},~{}\mu_{1}>0,~{}\mu_{2}>0$
and $\beta(x)$ is a fixed function.

If $\beta(x)=0$ and $V_{1}(x)=V_{2}(x)=0$, then the system (1.1) can be
reduced to the scalar Choquard equation

$\displaystyle-\Delta u+\lambda u=\mu(I_{\alpha}\star|u|^{p})|u|^{p-2}u.$
(1.3)

Physical motivations arise from the case $N=3$ and $\alpha=2$. In 1954, Pekar
[29] described a polaron at rest in the quantum theory. In 1976, to model an
electron trapped in its own hole, Choquard [24] considered (1.3) as an
approximation to Hartree-Fock theory of one-component plasma. In particular
cases, Penrose [30] investigated the selfgravitational collapse of a quantum
mechanical wave function. The system of weakly coupled equations has been
widely considered in recent years and it has applications especially in
nonlinear optics [26, 27]. Furthermore, nonlocal nonlinearities have attracted
considerable interest as a means of eliminating collapse and stabilizing
multidimensional solitary waves. It appears naturally in optical systems [25]
and is known to influence the propagation of electromagnetic waves in plasmas
[8].

In (1.1), if $\lambda_{1}$ and $\lambda_{2}$ are given, some literature call
it as fixed frequency problem. There are some related articles. In [40], the
authors studied the existence and nonexistence results for a class of linearly
coupled Choquard system (1.1) with $V_{1}(x)=V_{2}(x)=0$ for critical case. In
[39], the authors were concerned with the following linearly coupled Choquard-
type system

$\displaystyle\left\\{\begin{aligned} &-\Delta
u+\lambda_{1}u=\mu_{1}(I_{\alpha}\star F(u))F^{\prime}(u)+\beta v,\\\ &-\Delta
v+\lambda_{2}v=\mu_{2}(I_{\alpha}\star F(v))F^{\prime}(v)+\beta
u,\end{aligned}\right.\quad x\in\mathbb{R}^{N}.$

Under some suitable conditions on $F^{\prime}$, the existence of vector
solutions was proved by using variational methods. Moreover, they proved the
asymptotic behavior of the solutions as $\beta\to 0^{+}$.

Condition (1.2) is called as the normalization condition, which imposes a
normalization on the $L^{2}$-masses of $u$ and $v$. The solutions to the
Schrödiger system (1.1) under the constraint (1.2) are usually referred as
normalized solutions. In order to obtain the solution to the fractional
Schrödiger system (1.1) satisfying the normalization condition (1.2), one need
to consider the critical point of the functional $E_{V}(u,v)$ on
$S_{\xi}\times S_{\eta}$(see (1.4) and (1.5)). And then, $\lambda_{1}$ and
$\lambda_{2}$ appear as Lagrange multipliers with respect to the mass
constraint, which cannot be determined a priori, but are part of the unknown.
Some literature called this problem as fixed mass problem.

Recently, the normalized solutions of nonlinear Schrödinger equations and
systems has attracted many researchers, see more details [1, 2, 3, 4, 5, 6,
11, 13, 15, 16, 17, 21, 32, 33, 37, 43]. In particular, for the normalized
solutions of Choquard equations, there are many papers such as [18, 19, 20,
41, 42]. For the normalized solutions of nonlinearly coupled Choquard system,
there are many papers, for example [34, 35].

Up to our knowledge, there is no paper about the normalized solutions of
linearly coupled Choquard system. In our paper, we first consider the
normalized solutions of linearly coupled Choquard system (1.1) without
potentials, and then consider the normalized solutions of linearly coupled
Choquard system (1.1) with different potentials.

The corresponding functional of (1.1) is

$\displaystyle\begin{split}E_{V}(u,v)=&\frac{1}{2}\int_{\mathbb{R}^{N}}(|\nabla
u|^{2}+|\nabla
v|^{2})dx+\frac{1}{2}\int_{\mathbb{R}^{N}}(V_{1}(x)|u|^{2}+V_{2}(x)|v|^{2})dx\\\
&-\frac{\mu_{1}}{2p}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{N-\alpha}}dydx-\frac{\mu_{2}}{2q}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|u(x)|^{q}|u(y)|^{q}}{|x-y|^{N-\alpha}}dydx\\\
&-\int_{\mathbb{R}^{N}}\beta(x)uvdx.\end{split}$ (1.4)

If $V_{1}(x)=V_{2}(x)=0$, then we denote that $E_{V}(u,v)=E(u,v)$. Denote the
$L^{2}$-sphere

$\displaystyle S_{\xi}=\\{u\in
H^{1}(\mathbb{R}^{N}):\|u\|_{2}^{2}=\xi^{2}\\}\text{~{}and~{}}S_{\eta}=\\{u\in
H^{1}(\mathbb{R}^{N}):\|u\|_{2}^{2}=\eta^{2}\\}.$ (1.5)

First, we consider the linearly coupled Choquard system (1.1) without
potentials, i.e., $V_{1}(x)=V_{2}(x)=0$.

If $1+\frac{\alpha}{N}<p,q<1+\frac{\alpha+2}{N}$, then the Choquard terms are
$L^{2}$-subcritical and $E(u,v)$ is bounded from below on $S_{\xi}\times
S_{\eta}$. Let $\beta=\beta(x)$ be a fixed constant, we focus on the problem

$\displaystyle e(\xi,\eta):=\inf_{S_{\xi}\times S_{\eta}}E(u,v)$ (1.6)

######  Theorem 1.1.

Suppose that $\beta>0$, $V_{1}(x)=V_{2}(x)=0$ and
$1+\frac{\alpha}{N}<p,q<1+\frac{\alpha+2}{N}$. Then there exists a solution
$(u,v,\lambda_{1},\lambda_{2})$ of (1.1), where $u$ and $v$ are positive and
radial.

If $1+\frac{\alpha+2}{N}<p,~{}q<\frac{N+\alpha}{N-2}$, then the Choquard terms
are $L^{2}$-supercritical and $E(u,v)$ is unbounded from below on
$S_{\xi}\times S_{\eta}$. Therefore, (1.6) doesn’t work. We adopt mountain
pass structure to obtain the solutions of (1.1). Denote that
$\delta_{p}=\frac{N(p-1)-\alpha}{2p}$ and

$\displaystyle h(x)=\frac{1}{2}x-\frac{C}{2p}x^{p\delta_{p}}.$ (1.7)

######  Theorem 1.2.

Assume that $1+\frac{\alpha+2}{N}<p=q<\frac{N+\alpha}{N-2}$ and
$V_{1}(x)=V_{2}(x)=0$. Let $\beta(x)$ satisfy

  * (i)

$\beta(x)=\beta(|x|)>0$,

  * (ii)

$\beta(x),~{}x\cdot\nabla\beta(x)$ are bounded, moreover,
$\|\beta(x)\|_{\infty}<\frac{\max\\{h(x)\\}}{2\xi\eta}$,

  * (iii)

$2\beta(x)+\frac{x\cdot\nabla\beta(x)}{\delta_{p}}\geq 0$.

Then there exists a solution $(u,v,\lambda_{1},\lambda_{2})$ of (1.1), where
$u$ and $v$ are positive and radial for some $\lambda_{1},\lambda_{2}>0$.

######  Remark 1.1.

In order to simplicity some notations, we only consider $p=q$. But for $p\neq
q$, we just do some technical generalization.

Next, we consider the linearly coupled Choquard system (1.1) with different
potentials.

For the $L^{2}$-subcritical case, i.e,
$1+\frac{\alpha}{N}<p,~{}q<1+\frac{\alpha+2}{N}$, we give two different
potentials:

  * •

(V1):$V(x)\in C(\mathbb{R}^{N},\mathbb{R})$,
$\lim\limits_{|x|\rightarrow\infty}V(x)=0$ and $V(x)<0$ for all
$x\in\mathbb{R}^{N}$;

  * •

(V2):$V(x)\in C(\mathbb{R}^{N},\mathbb{R})$ and
$\lim\limits_{|x|\rightarrow\infty}V(x)=\infty$.

######  Theorem 1.3.

Assume that $V_{1}(x)$ and $V_{2}(x)$ satisfy (V1),
$\max\\{1+\frac{\alpha}{N},2\\}<p,q<1+\frac{\alpha+2}{N}$ and $\beta>0$. Then
there exists a ground state solution for $(u,v,\lambda_{1},\lambda_{2})\in
H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})\times\mathbb{R}\times\mathbb{R}$ of the system (1.1),
where $u$ and $v$ are positive.

We define

$\displaystyle\tilde{H}_{i}:=\\{u\in
H^{1}(\mathbb{R}^{N}):\int_{\mathbb{R}^{N}}V_{i}(x)|u|^{2}dx<\infty\\}$

for $i=1,2$.

######  Theorem 1.4.

Assume that $V_{1}(x)$ and $V_{2}(x)$ satisfy (V2),
$\max\\{1+\frac{\alpha}{N},2\\}<p,q<1+\frac{\alpha+2}{N}$ and $\beta>0$. Then
there exists a ground state solution
$(u,v,\lambda_{1},\lambda_{2})\in\tilde{H}_{1}\times\tilde{H}_{2}\times\mathbb{R}\times\mathbb{R}$
for the system (1.1), where $u$ and $v$ are positive.

######  Theorem 1.5.

Assume that $V_{1}(x)$ satisfies (V1) and $V_{2}(x)$ satisfies (V2),
$\max\\{1+\frac{\alpha}{N},2\\}<p,q<1+\frac{\alpha+2}{N}$ and $\beta>0$. Then
the system (1.1) has a solution $(u,v,\lambda_{1},\lambda_{2})\in
H^{1}(\mathbb{R}^{N})\times\tilde{H}_{2}\times\mathbb{R}\times\mathbb{R}$,
where $u$ and $v$ are positive.

######  Remark 1.2.

In the above theorem, if $V_{1}(x)$ satisfies (V2) and $V_{2}(x)$ satisfies
(V1), then the theorem still holds.

In this paper, the essential difficulty is to obtain the $L^{2}$-strong
convergence of critical sequence due to the constraint (1.2).

For $1+\frac{\alpha}{N}<p,q<1+\frac{\alpha+2}{N}$ and $V_{1}(x)=V_{2}(x)=0$,
the key step is proved the subadditivity of $e(\xi,\eta)$(see Lemma 3.2),
combining with Brézis-Lieb lemma, it can lead to the $L^{2}$-strong
convergence of minimizing sequence of $e(\xi,\eta)$.

For $1+\frac{\alpha+2}{N}<p=q<\frac{N+\alpha}{N-2}$ and $V_{1}(x)=V_{2}(x)=0$,
applying the properties of the functional $E(u,v)$, we construct the mountain
pass structure and obtain the bounded Palais-Smale sequence satisfying Pohozev
identity in limit case. In addition, the sign of Lagrange multipliers plays an
important role in $L^{2}$-strong convergence of Palais-Smale sequence. Hence
the difficulty can be turned into showing the sign of Lagrange multipliers. By
$E^{\prime}(u,v)\to 0$ and Pohozev identity in limit case, we can verify the
sign of one of Lagrange multipliers. But for the other, by the fact the
linearly coupled system (1.1) has no semi-trivial solutions, we can obtain the
sign of the other Lagrange multiplier. Therefore, the $L^{2}$-strong
convergence of Palais-Smale sequence is established.

For $1+\frac{\alpha}{N}<p,q<1+\frac{\alpha+2}{N}$ and $V_{1}(x)\neq
0,V_{2}(x)\neq 0$, if $V_{1}(x)$ and $V_{2}(x)$ are trapping potentials, then
the $L^{2}$-strong convergence of Palais-Smale sequence can be given by the
compact embedding lemma, which can be seen in [31]. If $V_{1}(x)$ or
$V_{2}(x)$ is not trapping potential, then the key step is to establish the
split lemma for $e_{V}(\xi,\eta)$, see Lemma 5.3 and Lemma 5.6, combining with
Brézis-Lieb lemma, it can lead to the $L^{2}$-strong convergence of minimizing
sequence of $e_{V}(\xi,\eta)$.

In our paper, there is a unsolved problem, which is the existence of
normalized solutions for the linearly coupled with potentials for
$L^{2}$-supercritical case. Indeed, by comparing the energy, the proper
linking geometry can be constructed and the bounded Palais-Smale sequence
$\\{(u_{n},v_{n})\\}$ can be obtained. However, we can not verity the strong
convergence of the Palais-Smale sequence $\\{(u_{n},v_{n})\\}$. Similar to
Lemma 3.1 in [7], there exists a $(u,v)$ such that
$(u_{n},v_{n})\rightharpoonup(u,v)$ weakly in $H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})$. Moreover, there exist an integer $k\geq 0$, $k$
sequences of points $y^{j}_{n}(1\leq j\leq k)$ with
$|y^{j}_{n}|\rightarrow\infty$ as $n\rightarrow\infty$ and nontrivial
solutions $(w_{1j},w_{2j})(1\leq j\leq k)$ for the system (1.1) with
$V_{1}(x)=V_{2}(x)=0$ such that

$\displaystyle\|u_{n}-u-\sum_{i=1}^{k}w_{1i}(\cdot-
y_{n}^{i})\|_{H^{1}(\mathbb{R}^{N})}\rightarrow
0,~{}\|v_{n}-v-\sum_{i=1}^{k}w_{2i}(\cdot-
y_{n}^{i})\|_{H^{1}(\mathbb{R}^{N})}\rightarrow 0.$

By the upper bound of the energy, we can exclude the case $k\geq 2$. Since the
uniqueness of the solutions for the system (1.1) with $V_{1}(x)=V_{2}(x)=0$ is
not clear, we can not exclude the case $k=1$ by comparing the energy.

The paper is organized as follows. In the Section 2, we give some
preliminaries. In the Section 3, we prove Theorem 1.1. In the Section 4, we
prove Theorem 1.2. In the Section 5, we prove Theorem 1.3, Theorem 1.4 and
Theorem 1.5.

##  2 Preliminaries

In this section, we give some preliminaries which will be used in later.

For simplicity, denote that

$B(u,p)=\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u|^{p})|u|^{p}dx=\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|u(x)|^{p}|u(y)|^{p}}{|x-y|^{N-\alpha}}dydx.$

First, we list some inequalities such as Gagliardo-Nirenberg inequality and
Hardy-Littewood-Sobolev inequality [23].

######  Lemma 2.1.

  * (1)

Let $p\in[2,2^{*})$. If $N\geq 3$ and $p\geq 2$ if $N=1,2$. Then,

$\displaystyle\|f\|_{p}\leq C_{N,p}\|\nabla
f\|_{2}^{\gamma_{p}}\|f\|_{2}^{1-\gamma_{p}}$ (2.1)

with $\gamma_{p}=N(\frac{1}{2}-\frac{1}{p})$.

  * (2)

Let $t,r>1$ and $\alpha\in(0,N)$ with
$\frac{1}{t}+\frac{1}{r}=1+\frac{\alpha}{N}$, $f\in L^{t}(\mathbb{R}^{N})$ and
$h\in L^{r}(\mathbb{R}^{N})$. Then there exists a sharp constant
$C(N,\alpha,t,r)>0$ such that

$\displaystyle\bigg{|}\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|f(x)||h(y)|}{|x-y|^{N-\alpha}}dxdy\bigg{|}\leq
C(N,\alpha,t,r)\|f\|_{t}\|h\|_{r}.$ (2.2)

Moreover, if $t=r=\frac{2N}{N+\alpha}$, then

$\displaystyle
C(N,\alpha):=C(N,\alpha,t,r)=\pi^{\frac{N-\alpha}{2}}\frac{\Gamma(\frac{\alpha}{2})}{\Gamma(\frac{N+\alpha}{2})}\bigg{\\{}\frac{\Gamma(\frac{N}{2})}{\Gamma(N)}\bigg{\\}}^{-\frac{\alpha}{N}}.$

######  Lemma 2.2 (Weak Young inequality [23]).

Let $N\in\mathbb{N},~{}\alpha\in(0,N),~{}p,r>1$ and
$\frac{1}{p}=\frac{\alpha}{N}+\frac{1}{r}$. If $v\in L^{p}(\mathbb{R}^{N})$,
then $I_{\alpha}\star v\in L^{r}(\mathbb{R}^{N})$ and

$\bigg{(}\int_{\mathbb{R}^{N}}|I_{\alpha}\star
v|^{r}\bigg{)}^{\frac{1}{r}}\leq
C(N,\alpha,p)\bigg{(}\int_{\mathbb{R}^{N}}|v|^{p}\bigg{)}^{\frac{1}{p}}.$

In particular, we can set $p=\frac{N}{\alpha}$ and $r=+\infty$.

In (2.2), take $t=r=\frac{2N}{N+\alpha}$ and $f=h=|u|^{p}$, together with
(2.1), we have

$\displaystyle B(u,p)\leq C(N,\alpha)\|u\|_{\frac{2Np}{N+\alpha}}^{2p}\leq
C(N,s,\alpha,p)\|\nabla u\|_{2}^{2p\delta_{p}}\|u\|_{2}^{2p(1-\delta_{p})},$
(2.3)

where $\delta_{p}=\frac{N(p-1)-\alpha}{2p}$. Furthermore, it is easy to
observe that $B(u,p)$ is well defined for
$p\in(1+\frac{\alpha}{N},\frac{N+\alpha}{N-2})$.

######  Lemma 2.3.

[22] Assume that $N>2s,~{}\alpha\in(0,N)$ and
$r\in[1+\frac{\alpha}{N},\frac{N+\alpha}{N-2s}]$. Suppose that
$\\{u_{n}\\}_{n=1}^{\infty}\subset H^{s}(\mathbb{R}^{N})$ satisfy
$u_{n}\rightharpoonup u$ weakly in $H^{s}(\mathbb{R}^{N})$ as
$n\rightarrow\infty$. Then

$\displaystyle\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u_{n}|^{p})|u_{n}|^{p-2}u_{n}\varphi\rightarrow\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u|^{p})|u|^{p-2}u\varphi\text{~{}weakly
in }H^{-s}(\mathbb{R}^{N})\text{~{}as~{}}n\rightarrow\infty,$

for any $\varphi\in H^{s}(\mathbb{R}^{N})$.

Next we give Brézis-Lieb lemma for the nonlocal term of the functional, which
can be seen in [28].

######  Lemma 2.4.

Let $N\in\mathbb{N},~{}\alpha\in(0,N),~{}p\in[1,\frac{2N}{N+\alpha})$ and
$\\{u_{n}\\}$ be a bounded sequence in $L^{\frac{2Np}{N+\alpha}}$. If
$u_{n}\to u$ almost everywhere on $\mathbb{R}^{N}$ as $n\to\infty$, then

$\lim_{n\to\infty}\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u_{n}|^{p})|u_{n}|^{p}-\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u_{n}-u|^{p})|u_{n}-u|^{p}=\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u|^{p})|u|^{p}.$

######  Lemma 2.5.

[23] Let $f,g$ and $h$ be three Lebesgue measurable non-negative functions on
$\mathbb{R}^{N}$. Then, with

$\Psi(f,g,h)=\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}f(x)g(x-y)h(y)dxdy,$

we have

$\Psi(f,g,h)\leq\Psi(f^{*},g^{*},h^{*}),$

where $f^{*}$ is the Schwartz rearrangement of $f$.

Next we consider the single equation

$\displaystyle\left\\{\begin{aligned} &-\Delta u+\lambda
u=\mu(I_{\alpha}\star|u|^{p})|u|^{p-2}u~{}\quad x\in\mathbb{R}^{N},\\\
&\int_{\mathbb{R}^{N}}u^{2}dx=c^{2}.\end{aligned}\right.$ (2.4)

The corresponding functional with (2.4) is

$\displaystyle F_{\mu}(u)=\frac{1}{2}\|\nabla
u\|_{2}^{2}-\frac{\mu}{2p}B(u,p).$

Denote that

$\displaystyle\begin{split}&m(c,\mu):=\inf_{u\in S_{c}}F_{\mu}(u)\text{ for
}1+\frac{\alpha}{N}<p<1+\frac{\alpha+2}{N},\\\
&n(c,\mu):=\inf_{\mathcal{N}_{c}}F_{\mu}(u)\text{ for
}1+\frac{\alpha+2}{N}<p<\frac{N+\alpha}{N-2},\end{split}$ (2.5)

where

$\displaystyle G(u)=\|\nabla
u\|_{2}^{2}-\mu\delta_{p}B(u,p)\text{~{}and~{}}\mathcal{N}_{c}=\\{u\in
S_{c}:G(u)=0\\}.$

We collect some properties about the scalar equation which can be seen in [10,
19, 20].

######  Lemma 2.6.

  * (i)

Assume that $1+\frac{\alpha}{N}<p<1+\frac{\alpha+2}{N}$. Then

    1. 1.

$m(c,\mu)<0$ for all $c>0$.

    2. 2.

$m(c,\mu)<m(c-\alpha,\mu)+m(\alpha,\mu)$ for any $\alpha\in(0,c)$.

  * (ii)

Assume that $1+\frac{\alpha+2}{N}<p<\frac{N+\alpha}{N-2}$. Then $n(c,\mu)$ is
strictly decreasing with respect to $c$.

######  Lemma 2.7.

[36] Suppose that $N=3,4,5$, if $u$ is a positive solution of

$\displaystyle-\Delta
u+u=\int_{\mathbb{R}^{N}}\frac{|u(y)|^{2}}{|x-y|^{N-2}}dyu,\quad
u\in\mathbb{R}^{N},$

then $u$ is unique up to translations.

##  3 $L^{2}$-subcritical case

In this section, we prove the existence of solutions for $L^{2}$-subcritical
case without potential. In order to show Theorem 1.1, we need some lemmata.

######  Lemma 3.1.

$E(u,v)$ is coercive and bounded from below on $S_{\xi}\times S_{\eta}$.

###### Proof.

By (2.3), it holds that

$\displaystyle B(u,p)\leq C(N,\alpha,p)\|\nabla
u\|_{2}^{2p\delta_{p}}\xi^{2p(1-\delta_{p})},$ $\displaystyle B(v,q)\leq
C(N,\alpha,q)\|\nabla v\|_{2}^{2q\delta_{q}}\eta^{2q(1-\delta_{q})},$

where $\delta_{p}=\frac{N(p-1)-\alpha}{2p}$. Thus,

$\displaystyle E(u,v)\geq$ $\displaystyle\frac{1}{2}(\|\nabla
u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})-\frac{\mu_{1}C(N,\alpha,p)\xi^{2p(1-\delta_{p})}}{2p}\|\nabla
u\|_{2}^{2p\delta_{p}}$
$\displaystyle-\frac{\mu_{2}C(N,\alpha,q)\eta^{2q(1-\delta_{q})}}{2q}\|\nabla
v\|_{2}^{2q\delta_{q}}-\beta\xi\eta.$

Since $1+\frac{\alpha}{N}<p,q<1+\frac{\alpha+2}{N}$, $2p\delta_{p}<2$ and
$2q\delta_{q}<2$. Hence we complete the proof. ∎

From Lemma 3.1, there exists a minimizing sequence $\\{(u_{n},v_{n})\\}$ for
$E|_{S_{\xi}\times S_{\eta}}$. Due to $\beta>0$ and
$\|\nabla|u|\|_{2}^{2}\leq\|\nabla u\|_{2}^{2}$, we can assume that
$u_{n},v_{n}\geq 0$.

In the following, we give the subadditivity of $e(\xi,\eta)$.

######  Lemma 3.2.

There holds that $e(\xi,\eta)\leq
e(\xi_{1},\eta_{1})+e(\xi-\xi_{1},\eta-\eta_{1})$.

###### Proof.

Denote that
$\gamma=(\xi,\eta),~{}\gamma_{1}=(\xi_{1},\eta_{1})\in[0,\xi]\times[0,\eta]$
and $S(\gamma):=S_{\xi}\times S_{\eta}$, one need to show that

$\displaystyle e(\gamma)\leq e(\gamma_{1})+e(\gamma-\gamma_{1}).$

For any $\varepsilon>0$, we may find
$\varphi_{\varepsilon},\psi_{\varepsilon}\in
C_{c}^{\infty}(\mathbb{R}^{N})\times C_{c}^{\infty}(\mathbb{R}^{N})$ such that

$\displaystyle supp\varphi_{\varepsilon}\cap
supp\psi_{\varepsilon}=\emptyset,~{}\varphi_{\varepsilon}\in
S(\gamma_{1}),~{}E(\varphi_{\varepsilon})\leq e(\gamma_{1})+\varepsilon,$
$\displaystyle\psi_{\varepsilon}\in
S(\gamma-\gamma_{1}),~{}E(\psi_{\varepsilon})\leq
e(\gamma-\gamma_{1})+\varepsilon.$

Set $u_{\varepsilon}=\varphi_{\varepsilon}+\psi_{\varepsilon}$. Since
$supp\varphi_{\varepsilon}\cap supp\psi_{\varepsilon}=\emptyset$, we can
observe that $u_{\varepsilon}\in S(\gamma)$. Moreover,

$\displaystyle e(\gamma)\leq E(u_{\varepsilon})\leq
E(\varphi_{\varepsilon})+E(\psi_{\varepsilon})\leq
e(\gamma_{1})+e(\gamma-\gamma_{1})+2\varepsilon.$

Since $\varepsilon$ is arbitrary, we can establish the lemma. ∎

######  Remark 3.1.

If $e(\gamma)$ is achieved, then
$e(\gamma)<e(\gamma_{1})+e(\gamma-\gamma_{1})$.

Let $u^{*}$ be the Schwartz rearrangement of $u$. Then $\|\nabla
u^{*}\|_{2}^{2}\leq\|\nabla u\|_{2}^{2}$. Denote that $S_{c}^{r}:=S_{c}\cap
H_{r}^{1}(\mathbb{R}^{N})$.

######  Lemma 3.3.

There holds that

$\displaystyle\inf_{S_{\xi}^{r}\times S_{\eta}^{r}}E(u,v)=\inf_{S_{\xi}\times
S_{\eta}}E(u,v).$

###### Proof.

On the one hand, since $S_{\xi}^{r}\times S_{\eta}^{r}\subset S_{\xi}\times
S_{\eta}$, it can be derived that

$\displaystyle\inf_{S_{\xi}^{r}\times
S_{\eta}^{r}}E(u,v)\geq\inf_{S_{\xi}\times S_{\eta}}E(u,v).$

On the other hand, for any $(u,v)\in S_{\xi}\times S_{\eta}$, let
$(u^{*},v^{*})$ be the Schwartz rearrangement of $(u,v)$. By Lemma 2.5, it
holds that $B(u^{*},p)\geq B(u,p)$. Therefore,

$\displaystyle E(u^{*},v^{*})=$ $\displaystyle\frac{1}{2}(\|\nabla
u^{*}\|_{2}^{2}+\|\nabla
v^{*}\|_{2}^{2})-\frac{\mu_{1}}{2p}B(u^{*},p)-\frac{\mu_{2}}{2q}B(v^{*},q)-\beta\int_{\mathbb{R}^{N}}u^{*}v^{*}dx$
$\displaystyle\leq$ $\displaystyle\frac{1}{2}(\|\nabla u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})-\frac{\mu_{1}}{2p}B(u,p)-\frac{\mu_{2}}{2q}B(v,q)-\beta\int_{\mathbb{R}^{N}}uvdx,$

hence the lemma holds. ∎

######  Remark 3.2.

From the above lemma, we can suppose that $(u_{n},v_{n})\in
H_{r}^{1}(\mathbb{R}^{N})\times H_{r}^{1}(\mathbb{R}^{N})$. From Lemma 3.1, it
can be deduced that $(u_{n},v_{n})$ is bounded in
$H_{r}^{1}(\mathbb{R}^{N})\times H_{r}^{1}(\mathbb{R}^{N})$. Therefore, there
exists $(u_{0},v_{0})\in H_{r}^{1}(\mathbb{R}^{N})\times
H_{r}^{1}(\mathbb{R}^{N})$ such that
$(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ weakly in
$H_{r}^{1}(\mathbb{R}^{N})\times H_{r}^{1}(\mathbb{R}^{N})$ and strongly in
$L^{t}(\mathbb{R}^{N})\times L^{t}(\mathbb{R}^{N})$ for $t\in(2,2_{s}^{*})$.
By (2.3), it follows that $B(u_{n},p)\rightarrow B(u_{0},p)$ and
$B(v_{n},q)\rightarrow B(v_{0},q)$.

In order to obtain the critical point of $E(u,v)$ on $S_{\xi}\times S_{\eta}$,
we need to show the strong convergence of minimizing sequence in
$L^{2}(\mathbb{R}^{N})\times L^{2}(\mathbb{R}^{N})$.

######  Lemma 3.4.

There holds that

$\displaystyle\|u_{0}\|_{2}^{2}=\xi^{2},~{}\|v_{0}\|_{2}^{2}=\eta^{2}.$

###### Proof.

From Remark 3.2, $(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ weakly in
$H_{r}^{1}(\mathbb{R}^{N})\times H_{r}^{1}(\mathbb{R}^{N})$. Consequently,

$\displaystyle\|u_{0}\|_{2}^{2}\leq\liminf_{n\rightarrow\infty}\|u_{n}\|_{2}^{2}=\xi^{2},~{}\|v_{0}\|_{2}^{2}\leq\liminf_{n\rightarrow\infty}\|v_{n}\|_{2}^{2}=\eta^{2}.$

Next we show that $"="$ holds by contradiction. Set
$\|u_{0}\|_{2}=\xi_{1}~{}\|v_{0}\|_{2}=\eta_{1}$. There exist three cases:

$\displaystyle\left\\{\begin{array}[]{ll}\textbf{Case
}1:&\hbox{$\xi_{1}<\xi,~{}\eta_{1}<\eta$;}\\\ \textbf{Case
}2:&\hbox{$\xi_{1}=\xi,~{}\eta_{1}<\eta$;}\\\ \textbf{Case
}3:&\hbox{$\xi_{1}<\xi,~{}\eta_{1}=\eta$.}\end{array}\right.$

Set $\hat{u}_{n}=u_{n}-u_{0}$ and $\hat{v}_{n}=v_{n}-v_{0}$, denote that
$\|\hat{u}_{n}\|_{2}^{2}:=\xi_{2}^{2}=\xi^{2}-\xi_{1}^{2},~{}\|\hat{v}_{n}\|_{2}^{2}:=\eta_{2}^{2}=\eta^{2}-\eta_{1}^{2}$.

For Case 1, we have $\xi_{2}>0$ and $\eta_{2}>0$. By Brézis-Lieb lemma in [9],
it holds

$\displaystyle\begin{split}e(\xi,\eta)+o_{n}(1)=E(u_{n},v_{n})&=E(u_{0},v_{0})+E(\hat{u}_{n},\hat{v}_{n})+o_{n}(1)\\\
&\geq e(\xi_{1},\eta_{1})+E(\hat{u}_{n},\hat{v}_{n}).\end{split}$ (3.1)

From Remark 3.2, $B(\hat{u}_{n},\hat{v}_{n})=0$. As a consequence,

$\displaystyle
E(\hat{u}_{n},\hat{v}_{n})=\frac{1}{2}(\|\nabla\hat{u}_{n}\|_{2}^{2}+\|\nabla\hat{v}_{n}\|_{2}^{2})-\beta\int_{\mathbb{R}^{N}}\hat{u}_{n}\hat{v}_{n}dx+o(1)\geq-\beta\xi_{2}\eta_{2}+o(1).$
(3.2)

Without loss of generality, we may assume that $\xi_{2}^{2}\geq\eta_{2}^{2}$.
For any $u\in S_{\xi_{2}}$, it yields that $(u,\frac{\eta_{2}}{\xi_{2}}u)\in
S_{\xi_{2}}\times S_{\eta_{2}}$. Therefore,

$\displaystyle E(u,\frac{\eta_{2}}{\xi_{2}}u)=$
$\displaystyle\frac{1}{2}\bigg{(}1+\frac{\eta_{2}^{2}}{\xi_{2}^{2}}\bigg{)}\|\nabla
u\|^{2}-\frac{\mu_{1}}{2p}B(u,p)-\frac{\mu_{2}}{2q}\frac{\eta_{2}^{2q}}{\xi_{2}^{2q}}B(u,q)-\beta\xi_{2}\eta_{2}$
$\displaystyle\leq$ $\displaystyle 2\bigg{(}\frac{1}{2}\|\nabla
u\|^{2}-\frac{\mu_{1}}{2}\frac{1}{2p}B(u,p)\bigg{)}-\beta\xi_{2}\eta_{2}$
$\displaystyle=$ $\displaystyle
2F_{\frac{\mu_{1}}{2}}(u)-\beta\xi_{2}\eta_{2}.$

Applying Lemma 2.6, let $u$ satisfy
$m(\xi_{2},\frac{\mu_{1}}{2})=F_{\frac{\mu_{1}}{2}}(u)<0$. Then

$\displaystyle e(\xi_{2},\eta_{2})\leq E(u,\frac{\eta_{2}}{\xi_{2}}u)\leq
2F_{\frac{\mu_{1}}{2}}(u)-\beta\xi_{2}\eta_{2}.$ (3.3)

By (3.1)-(3.3) and Lemma 3.2, we can derive that

$\displaystyle e(\xi,\eta)$ $\displaystyle\geq
e(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2})-2F_{\frac{\mu_{1}}{2}}(u)$
$\displaystyle>e(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2})$ $\displaystyle\geq
e(\xi,\eta),$

which is a contradiction. Hence Case 1 does not hold.

Next we show Case 2 does not hold by contradiction. For Case 2, we have
$\xi_{2}=0,~{}\eta_{2}>0$. Similar to Case 1, it can be derived that

$\displaystyle
e(\xi,\eta)=e(\xi,\eta_{1})+E(\hat{u}_{n},\hat{v}_{n})+o_{n}(1),$

by $B(u_{n},p)\rightarrow B(u_{0},p),B(v_{n},q)\rightarrow B(v_{0},q)$ and
$u_{n},v_{n}\geq 0$, we have

$\displaystyle
E(\hat{u}_{n},\hat{v}_{n})=\frac{1}{2}(\|\nabla\hat{u}_{n}\|_{2}^{2}+\|\nabla\hat{v}_{n}\|_{2}^{2})-\beta\int_{\mathbb{R}^{N}}\hat{u}_{n}\hat{u}_{n}dx+o(1)\geq
o(1).$

where
$0\leq\beta\int_{\mathbb{R}^{N}}\hat{u}_{n}\hat{v}_{n}dx\leq\beta\|\hat{u}_{n}\|_{2}\|\hat{v}_{n}\|_{2}=0$.
From $e(0,\eta_{2})=m(\eta_{2},\mu)=F_{\mu}(u)+o(1)<0$, we obtain that

$\displaystyle e(\xi,\eta)$ $\displaystyle\geq
e(\xi,\eta_{1})+E(\hat{u}_{n},\hat{v}_{n})\geq e(\xi,\eta_{1})+o(1)$
$\displaystyle=e(\xi,\eta_{1})+e(0,\eta_{2})-F_{\mu}(u)+o(1)>e(\xi,\eta),$

which is a contradiction. Hence Case 2 does not hold. Moreover, similar to the
proof of Case 2, Case 3 does not hold. Hence the lemma holds. ∎

###### Proof of Theorem 1.1.

From Lemma 3.4, $(u_{n},v_{n})\rightarrow(u_{0},v_{0})$ strongly in
$L^{2}(\mathbb{R}^{N})\times L^{2}(\mathbb{R}^{N})$. Applying $(u_{n},v_{n})$
is a minimizing sequence, there holds that

$\displaystyle e(\xi,\eta)\leq
E(u_{0},v_{0})\leq\liminf_{n\rightarrow\infty}E(u_{n},v_{n})=e(\xi,\eta),$

which implies that $(u_{0},v_{0})$ is a minimizer. Moreover, $(u_{0},v_{0})$
satisfies

$\displaystyle\left\\{\begin{aligned} &-\Delta
u_{0}+\lambda_{1}u_{0}=\mu_{1}(I_{\alpha}\star|u_{0}|^{p})|u_{0}|^{p-2}u+\beta
v_{0},\\\ &-\Delta
v_{0}+\lambda_{2}v_{0}=\mu_{2}(I_{\alpha}\star|v_{0}|^{q})|v_{0}|^{q-2}u+\beta
u_{0},\end{aligned}\right.$

with $\|u_{0}\|_{2}^{2}=\xi^{2}$ and $\|v_{0}\|_{2}^{2}=\eta^{2}$. Finally by
maximum principle and $u_{0},v_{0}\geq 0$, there holds that $u_{0},v_{0}>0$.
Hence the theorem is established. ∎

##  4 $L^{2}$-supercritical case

In this section, we mainly consider
$1+\frac{\alpha+2}{N}<p<\frac{\alpha+N}{N-2}$. We work in radial setting. For
any $(u,v)\in H^{1}_{r}(\mathbb{R}^{N})\times H^{1}_{r}(\mathbb{R}^{N})$,
define the map

$\displaystyle s\star(u,v)=(s\star u,s\star
v):=(e^{\frac{Ns}{2}}u(e^{sx}),e^{\frac{Ns}{2}}v(e^{sx})).$

It is obvious to observe that $\|s\star u\|_{2}^{2}=\|u\|_{2}^{2}$ and
$\|s\star v\|_{2}^{2}=\|v\|_{2}^{2}$. Denote that $S_{c}^{r}=S(c)\cap
H^{1}_{r}(\mathbb{R}^{N})$.

######  Lemma 4.1.

Let $(u,v)\in S^{r}_{\xi}\times S^{r}_{\eta}$. Then

$\displaystyle\lim_{s\rightarrow-\infty}\|\nabla(s\star
u)\|_{2}^{2}+\|\nabla(s\star
v)\|_{2}^{2}=0^{+},~{}\lim_{s\rightarrow\infty}\|\nabla(s\star
u)\|_{2}^{2}+\|\nabla(s\star v)\|_{2}^{2}=\infty,$

and

$\displaystyle\lim_{s\rightarrow-\infty}E(s\star(u,v))=-\lim_{s\rightarrow-\infty}\int_{\mathbb{R}^{N}}\beta(e^{-s}x)uvdx,~{}\lim_{s\rightarrow\infty}E(s\star(u,v))=-\infty.$

###### Proof.

By direct computation, the lemma can be established. Hence we omit the
details. ∎

######  Lemma 4.2.

Let $h(x)$ be defined as in (1.7). Then there holds

  * (i)

there exists a $x_{0}>0$ such that $h(0)=h(x_{0})=0$;

  * (ii)

there exists a $x_{1}>0$ such that $h(x_{1})=\max\limits_{x\geq 0}\\{h(x)\\}$.

###### Proof.

Direct computation shows that

$\displaystyle
h^{\prime}(x)=\frac{1}{2}-\frac{Cp\delta_{p}}{2p}x^{p\delta_{p}-1},$

due to $p\delta_{p}-1>0$, it is easily to obtain that $h^{\prime}(x)=0$ has a
unique solution, denote by $x_{1}$. Hence $h(x)$ is increasing in $(0,x_{1})$
and decreasing in $(x_{1},+\infty)$. Since
$h(0)=0,~{}h(x_{1})=\max\\{h(x)\\}>0$, there exists a $x_{0}>0$ such that
$h(0)=h(x_{0})=0$. Therefore, we can establish the lemma. ∎

By (2.3), it follows that

$\displaystyle\begin{split}\mu_{1}B(u,p)+\mu_{2}B(v,p)\leq&C(N,\alpha,p)(\mu_{1}\|\nabla
u\|_{2}^{2p\delta_{p}}\xi^{2p(1-\delta_{p})}+\mu_{2}\|\nabla
v\|_{2}^{2p\delta_{p}}\eta^{2p(1-\delta_{p})})\\\
=:&C_{\xi,\eta}(N,\alpha,p,\mu_{1},\mu_{2})(\|\nabla
u\|_{2}^{2p\delta_{p}}+\|\nabla v\|_{2}^{2p\delta_{p}})\\\
\leq&C_{\xi,\eta}(N,\alpha,p,\mu_{1},\mu_{2})(\|\nabla u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})^{p\delta_{p}}.\end{split}$ (4.1)

The next lemma enlightens the mountain pass structure of the problem.

######  Lemma 4.3.

There exist a constant $K_{1}>0$ sufficient small, a constant $K_{2}$
satisfying $K_{2}>K_{1}$ and
$\|\beta(x)\|_{\infty}<\frac{\max\\{h(x)\\}}{2\xi\eta}$. Define the sets

$\displaystyle\Omega=\\{(u,v)\in S^{r}_{\xi}\times S^{r}_{\eta},\|\nabla
u\|_{2}^{2}+\|\nabla v\|_{2}^{2}\leq K_{1}\\},$ $\displaystyle\Pi=\\{(u,v)\in
S^{r}_{\xi}\times S^{r}_{\eta},\|\nabla u\|_{2}^{2}+\|\nabla
v\|_{2}^{2}=K_{2}\\}.$

There holds

  * (i)

for any $(u,v)\in\Omega$, there exists $C_{1}>0$ such that $E(u,v)>-C_{1}$;

  * (ii)

$\sup\limits_{\Omega}E(u,v)<\inf\limits_{\Pi}E(u,v)$;

  * (iii)

$\inf\limits_{\Pi}E(u,v)>0$.

###### Proof.

(i) For any $(u,v)\in\Omega$, by (4.1), it follows that

$\displaystyle E(u,v)=$ $\displaystyle\frac{1}{2}(\|\nabla
u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})-\frac{1}{2p}(\mu_{1}B(u,p)+\mu_{2}B(v,p))-\int_{\mathbb{R}^{N}}\beta(x)uvdx$
$\displaystyle\geq$ $\displaystyle\frac{1}{2}(\|\nabla u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})-\frac{C_{\xi,\eta}(N,\alpha,p,\mu_{1},\mu_{2})}{2p}(\|\nabla
u\|_{2}^{2}+\|\nabla v\|_{2}^{2})^{p\delta_{p}}-\|\beta(x)\|_{\infty}\xi\eta.$

Taking $C=C_{\xi,\eta}(N,\alpha,p,\mu_{1},\mu_{2})$ and $x=\|\nabla
u\|_{2}^{2}+\|\nabla v\|_{2}^{2}\leq K_{1}<x_{0}$ in (1.7), by Lemma 4.2, we
get that

$E(u,v)\geq-\|\beta(x)\|_{\infty}\xi\eta:=-C_{1}.$

Therefore (i) holds.

(ii) If $(u_{1},v_{1})\in\Pi$ and $(u_{2},v_{2})\in\Omega$, then by (4.1),

$\displaystyle\begin{split}E(u_{1},v_{1})-E(u_{2},v_{2})\geq\frac{K_{2}}{2}-\frac{C_{\xi,\eta}(N,\alpha,p,\mu_{1},\mu_{2})K_{2}^{p\delta_{p}}}{2p}-\frac{K_{1}}{2}-2\|\beta(x)\|_{\infty}\xi\eta.\end{split}$
(4.2)

From Lemma 4.2, take $K_{2}=x_{1}$ and let $K_{1}$ be sufficiently small, if
$\|\beta(x)\|_{\infty}<\frac{h(K_{2})}{2\xi\eta}$, then

$E(u_{1},v_{1})-E(u_{2},v_{2})>0$

for any $(u_{1},v_{1})\in\Pi$ and $(u_{2},v_{2})\in\Omega$. Therefore (ii)
holds.

(iii) Similar to (4.2), for any $(u_{1},v_{1})\in\Pi$, there holds

$\displaystyle E(u_{1},v_{1})\geq$
$\displaystyle\frac{K_{2}}{2}-\frac{C_{\xi,\eta}(N,\alpha,p,\mu_{1},\mu_{2})K_{2}^{p\delta_{p}}}{2p}-\|\beta(x)\|_{\infty}\xi\eta>0$

for $\|\beta(x)\|_{\infty}<\frac{h(K_{2})}{2\xi\eta}$. Hence (iii) holds. ∎

Define

$\displaystyle\Delta=\\{(u,v)\in S^{r}_{\xi}\times S^{r}_{\eta},\|\nabla
u\|_{2}^{2}+\|\nabla v\|_{2}^{2}\geq 2K_{2}~{}\text{and}~{}E(u,v)\leq 0\\}.$

Notice that $\Delta\neq\emptyset$ by Lemma 4.1. From Lemma 4.1 and Lemma 4.3,
if $(\bar{u},\bar{v})\in\Omega$ and $(\hat{u},\hat{v})\in\Delta$, then there
is a mountain path linking $(\bar{u},\bar{v})$ and $(\hat{u},\hat{v})$ and
passing through $\Pi$. Define

$\displaystyle\Gamma:=\\{\gamma:=(\gamma_{1}(t),\gamma_{2}(t))\in
C([0,1],S_{\xi}\times
S_{\eta}):\gamma(0)=(\bar{u},\bar{v}),\gamma(1)=(\hat{u},\hat{v})\\}.$

######  Lemma 4.4.

There exists a Palais-Smale sequence $(\bar{u}_{n},\bar{v}_{n})$ for $E$ on
$S^{r}_{\xi}\times S^{r}_{\eta}$ at the level

$\displaystyle
c:=\inf_{\gamma\in\Gamma}\max_{t\in[0,1]}E(\gamma(t))\geq\inf_{\Pi}E>0\geq\max\\{E(\bar{u},\bar{v}),E((\hat{u},\hat{v}))\\}$

satisfying additional condition

$\displaystyle\begin{split}&(\|\nabla\bar{u}_{n}\|_{2}^{2}+\nabla\bar{v}_{n}\|_{2}^{2})-\delta_{p}(\mu_{1}B(\bar{u}_{n},p)+\mu_{2}B(\bar{v}_{n},p))\\\
+&\int_{\mathbb{R}^{N}}(x\cdot\nabla\beta(x))\bar{u}_{n}(x)\bar{v}_{n}(x)dx=o_{n}(1)\end{split}$
(4.3)

with $o_{n}(1)\rightarrow 0$ as $n\rightarrow\infty$.

###### Proof.

We consider the auxiliary functional $\bar{E}$:

$\displaystyle\bar{E}:\mathbb{R}\times S^{r}_{\xi}\times
S^{r}_{\eta}\rightarrow\mathbb{R},$ $\displaystyle\bar{E}(s,u,v)=E(s\star
u,s\star v).$

Set

$\bar{\Gamma}=\\{\bar{\gamma}(t)=(s(t),\gamma_{1}(t),\gamma_{2}(t))\in
C([0,1],\mathbb{R}\times S^{r}_{\xi}\times
S^{r}_{\eta}):\bar{\gamma}(0)=(0,\bar{u},\bar{v}),\bar{\gamma}(1)=(0,\hat{u},\hat{v})\\},$

and define

$\displaystyle\bar{c}:=\inf_{\bar{\gamma}\in\bar{\Gamma}}\sup_{t\in[0,1]}\bar{E}(\bar{\gamma}(t)).$

We next show $c=\bar{c}$. On the one hand, since $\Gamma\subset\bar{\Gamma}$,
it is easy to find that $c\geq\bar{c}$. On the other hand, for any
$\bar{\gamma}(t)=(s(t),\gamma_{1}(t),\gamma_{2}(t))\in\bar{\Gamma}$, it holds
that

$\displaystyle(s(t)\star\gamma_{1}(t),s(t)\star\gamma_{2}(t))\in\Gamma~{}\text{and}~{}\bar{E}(\bar{\gamma}(t))=E(s(t)\star\gamma_{1}(t),s(t)\star\gamma_{2}(t)),$

hence $c\leq\bar{c}$. Therefore, $c=\bar{c}$.

Notice that

$\displaystyle\bar{E}(\bar{\gamma}(t))=E(s(t)\star\gamma_{1}(t),s(t)\star\gamma_{2}(t))=\bar{E}(0,s(t)\star\gamma_{1}(t),s(t)\star\gamma_{2}(t)).$

Since $\beta(x)\geq 0$, $\bar{E}(s,|u|,|v|)\leq\bar{E}(s,u,v)$. As a
consequence, we can choose the minimizing sequence
$\bar{\gamma}_{n}=(s_{n},\gamma_{1n},\gamma_{2n})$ for $\bar{c}$ satisfying

$\displaystyle\gamma_{1n},~{}\gamma_{2n}\geq 0\text{ a.e. in }\mathbb{R}^{N}.$

Using Theorem 3.2 in [12], there exists a Palais-Smale sequence
$(s_{n},u_{n},v_{n})\in\mathbb{R}\times S_{\xi}\times S_{\eta}$ for $\bar{E}$
at level $\bar{c}$ such that

  * (1)

$\lim\limits_{n\rightarrow\infty}\bar{E}(s_{n},u_{n},v_{n})=\bar{c}=c$,

  * (2)

$\lim\limits_{n\rightarrow\infty}|s_{n}|+dist((u_{n},v_{n}),(\gamma_{1n},\gamma_{2n}))=0$,

  * (3)

$\lim\limits_{n\rightarrow\infty}\|\nabla_{\mathbb{R}\times S_{\xi}\times
S_{\eta}}\bar{E}(s_{n},u_{n},v_{n})\|=0$.

From (1), we have $E(s_{n}\star u_{n},s_{n}\star v_{n})=c,~{}s_{n}\rightarrow
0$. From $\gamma_{1n},~{}\gamma_{2n}\geq 0$ and (2), it follows that
$u_{n},v_{n}\geq 0$. Therefore,
$\\{(\bar{u}_{n},\bar{v}_{n})\\}=:\\{(s_{n}\star u_{n},s_{n}\star v_{n})\\}$
is a Palais-Smale sequence for $E$. Similarly, for any
$(\varphi_{n},\psi_{n})\in H_{1}^{r}\times H_{1}^{r}$, setting
$(\tilde{\varphi}_{n},\tilde{\psi}_{n})=((-s_{n})\star\varphi_{n},(-s_{n})\star\psi_{n})$,
it is easily to find that $\int_{\mathbb{R}^{N}}\bar{u}_{n}\varphi_{n}=0$ and
$\int_{\mathbb{R}^{N}}\bar{v}_{n}\psi_{n}=0$ is equivalent to
$\int_{\mathbb{R}^{N}}u_{n}\tilde{\varphi}_{n}=0$ and
$\int_{\mathbb{R}^{N}}v_{n}\tilde{\psi}_{n}=0$. Therefore,

$\displaystyle
DE(\bar{u}_{n},\bar{v}_{n})[(\varphi_{n},\psi_{n})]=D\bar{E}(s_{n},u_{n},v_{n})[(0,\tilde{\varphi}_{n},\tilde{\psi}_{n})]+o(1)\|(\tilde{\varphi}_{n},\tilde{\psi}_{n})\|_{H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})},$

due to
$\|(\tilde{\varphi}_{n},\tilde{\psi}_{n})\|^{2}_{H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})}\leq
2\|(\varphi_{n},\psi_{n})\|^{2}_{H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})}$ for $n$ large, it can be deduced that
$\nabla_{S_{\xi}\times S_{\eta}}E(\bar{u}_{n},\bar{v}_{n})\rightarrow 0$. By
(3), it implies that

$\displaystyle D\bar{E}(s_{n},u_{n},v_{n})[(1,0,0)]\rightarrow
0,~{}~{}s_{n}\rightarrow 0\text{~{}as~{}}n\rightarrow\infty.$

We can compute that

$\displaystyle\partial_{s}(\int_{\mathbb{R}^{N}}\beta(x)e^{Ns}u(e^{s}x)v(e^{s}x)dx)$
$\displaystyle=\partial_{s}(\int_{\mathbb{R}^{N}}\beta(e^{-s}x)u(x)v(x)dx)$
$\displaystyle=-\int_{\mathbb{R}^{N}}(e^{-s}x\cdot\nabla\beta(e^{-s}x))u(x)v(x)dx,$

and

$\displaystyle\left\\{\begin{aligned}
&\partial_{s}\bigg{\\{}\frac{1}{2}(\|\nabla(s\star
u)\|_{2}^{2}+\|\nabla(s\star v)\|_{2}^{2})\bigg{\\}}=e^{2s}(\|\nabla
u\|_{2}^{2}+\|\nabla v\|_{2}^{2}),\\\ &\partial_{s}\frac{\mu_{1}B(s\star
u,p)+\mu_{2}B(s\star
v,p)}{2p}=\delta_{p}e^{2p\delta_{p}s}(\mu_{1}B(u,p)+\mu_{2}B(v,p)),\end{aligned}\right.$

by (3), it follows that

$\displaystyle(\|\nabla\bar{u}_{n}\|_{2}^{2}+\nabla\bar{v}_{n}\|_{2}^{2})-\delta_{p}(\mu_{1}B(\bar{u}_{n},p)+\mu_{2}B(\bar{v}_{n},p))$
$\displaystyle+$
$\displaystyle\int_{\mathbb{R}^{N}}(x\cdot\nabla\beta(x))\bar{u}_{n}(x)\bar{v}_{n}(x)dx\rightarrow
0~{}~{}\text{as~{}}n\rightarrow\infty.$

Thus we finish the proof. ∎

Next lemma gives the boundedness of Palais-Smale sequence
$\\{(\bar{u}_{n},\bar{v}_{n})\\}$.

######  Lemma 4.5.

Suppose that $\beta(x)$ and $x\cdot\nabla\beta(x)$ are bounded. Then
$\\{(\bar{u}_{n},\bar{v}_{n})\\}$ is bounded in
$H^{1}_{r}(\mathbb{R}^{N})\times H^{1}_{r}(\mathbb{R}^{N})$. Furthermore,
there exists a constant $\bar{C}>0$ such that

$\displaystyle\|\nabla\bar{u}_{n}\|_{2}^{2}+\|\nabla\bar{v}_{n}\|_{2}^{2}\geq\bar{C}.$

###### Proof.

By (4.3), it yields that

$\displaystyle 2p\delta_{p}c\leftarrow$ $\displaystyle
2p\delta_{p}E(\bar{u}_{n},\bar{v}_{n})-D\bar{E}(s_{n},u_{n},v_{n})[(1,0,0)]$
$\displaystyle=$
$\displaystyle(p\delta_{p}-1)(\|\nabla\bar{u}_{n}\|_{2}^{2}+\|\nabla\bar{v}_{n}\|_{2}^{2})-\int_{\mathbb{R}^{N}}(2p\delta_{p}\beta(x)+x\cdot\nabla\beta(x))\bar{u}_{n}(x)\bar{v}_{n}(x)dx$

as $n\rightarrow\infty$. Since $1+\frac{\alpha+2}{N}<p<\frac{\alpha+N}{N-2}$,
we have $p\delta_{p}-1>0$. By the boundedness of $\beta(x)$ and
$x\cdot\nabla\beta(x)$, there exist $C_{1}>0$ and $C_{2}>0$ such that

$\displaystyle
C_{2}\leq\|\nabla\bar{u}_{n}\|_{2}^{2}+\|\nabla\bar{v}_{n}\|_{2}^{2}\leq
C_{1}.$

Hence the desired results is obtained. ∎

Since $E^{\prime}|_{S_{\xi}\times
S_{\eta}}(\bar{u}_{n},\bar{v}_{n})\rightarrow 0$, there exist two sequence
$\\{\lambda_{1n}\\}\subset\mathbb{R}$ and
$\\{\lambda_{2n}\\}\subset\mathbb{R}$ such that

$\displaystyle\begin{split}&\int_{\mathbb{R}^{N}}[\nabla\bar{u}_{n}\nabla\varphi+\nabla\bar{v}_{n}\nabla\psi]dx-\mu_{1}\int_{\mathbb{R}^{N}}(I_{\alpha}\star(|\bar{u}_{n}|^{p}))|\bar{u}_{n}|^{p-2}\bar{u}_{n}\varphi
dx\\\
-&\mu_{2}\int_{\mathbb{R}^{N}}(I_{\alpha}\star(|\bar{v}_{n}|^{p}))|\bar{v}_{n}|^{p-2}\bar{v}_{n}\psi
dx-\int_{\mathbb{R}^{N}}\beta(x)(\bar{u}_{n}\psi+\bar{v}_{n}\varphi)dx\\\
+&\lambda_{1n}\int_{\mathbb{R}^{N}}\bar{u}_{n}\varphi
dx+\lambda_{2n}\int_{\mathbb{R}^{N}}\bar{v}_{n}\psi
dx=o_{n}(1)\|(\varphi,\psi)\|_{H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})}\end{split}$ (4.4)

for any $(\varphi,\psi)\in H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$
with $o_{n}(1)\rightarrow 0$ as $n\rightarrow\infty$.

######  Lemma 4.6.

Both $\\{\lambda_{1n}\\}$ and $\\{\lambda_{2n}\\}$ are bounded sequences.
Moreover, if $2\beta(x)+\frac{x\cdot\nabla\beta(x)}{\delta_{p}}\geq 0$, then
at least one of the sequences converges to a strict positive value.

###### Proof.

Taking $(\varphi,\psi)=(\bar{u}_{n},0)$ ($(\varphi,\psi)=(0,\bar{v}_{n})$,
respectively) in (4.4), we can obtain that

$\displaystyle-\lambda_{1n}\xi^{2}+o_{n}(1)=\|\nabla\bar{u}_{n}\|_{2}^{2}-\mu_{1}B(\bar{u}_{n},p)-\int_{\mathbb{R}^{N}}\beta(x)\bar{u}_{n}\bar{v}_{n}dx,$
(4.5)

and

$\displaystyle-\lambda_{2n}\eta^{2}+o_{n}(1)=\|\nabla\bar{v}_{n}\|_{2}^{2}-\mu_{2}B(\bar{v}_{n},p)-\int_{\mathbb{R}^{N}}\beta(x)\bar{u}_{n}\bar{v}_{n}dx.$
(4.6)

From the boundedness of $\beta(x)$ and $\\{(\bar{u}_{n},\bar{v}_{n})\\}$ in
$H^{1}_{r}(\mathbb{R}^{N})\times H^{1}_{r}(\mathbb{R}^{N})$, it is easy to
deduce that $\\{\lambda_{1n}\\}$ and $\\{\lambda_{2n}\\}$ are bounded.

By (4.3), (4.5) and (4.6), it yields that

$\displaystyle\lambda_{1n}\xi^{2}+\lambda_{2n}\eta^{2}+o_{n}(1)$
$\displaystyle=$
$\displaystyle-(\|\nabla\bar{u}_{n}\|_{2}^{2}+\|\nabla\bar{v}_{n}\|_{2}^{2})+(\mu_{1}B(\bar{u}_{n},p)+\mu_{2}B(\bar{v}_{n},p))+\int_{\mathbb{R}^{N}}2\beta(x)\bar{u}_{n}\bar{v}_{n}dx$
$\displaystyle=$
$\displaystyle\bigg{(}\frac{1}{\delta_{p}}-1\bigg{)}(\|\nabla\bar{u}_{n}\|_{2}^{2}+\|\nabla\bar{v}_{n}\|_{2}^{2})+\int_{\mathbb{R}^{N}}\bigg{(}2\beta(x)+\frac{x\cdot\nabla\beta(x)}{\delta_{p}}\bigg{)}\bar{u}_{n}\bar{v}_{n}dx.$

Since $\delta_{p}<1$ and
$2\beta(x)+\frac{x\cdot\nabla\beta(x)}{\delta_{p}}\geq 0$, by Lemma 4.5, it
can be inferred that

$\displaystyle\lambda_{1n}\xi^{2}+\lambda_{2n}\eta^{2}+o_{n}(1)\geq\bigg{(}\frac{1}{\delta_{p}}-1\bigg{)}C_{2}>0$

for $n$ large. Hence the lemma holds. ∎

From the above lemma, there exists a subsequence of $\\{\lambda_{1n}\\}$
(resp. $\\{\lambda_{2n}\\}$) and $\lambda_{1}\in\mathbb{R}$ (resp.
$\lambda_{2}\in\mathbb{R}$) such that $\lambda_{1n}\rightarrow\lambda_{1}$
(resp. $\lambda_{2n}\rightarrow\lambda_{2}$) as $n\rightarrow\infty$. The
signs of $\lambda_{1}$ and $\lambda_{2}$ play an important role in the strong
convergence of $\bar{u}_{n},~{}\bar{v}_{n}$ in $H^{1}(\mathbb{R}^{N})$.

######  Lemma 4.7.

If $\lambda_{1}>0~{}(\text{resp.~{}}\lambda_{2}>0)$, then
$\bar{u}_{n}\rightarrow\bar{u}~{}(\text{resp.~{}}\bar{v}_{n}\rightarrow\bar{v})$
strongly in $H^{1}(\mathbb{R}^{N})$.

###### Proof.

By Lemma 4.5, $\\{(\bar{u}_{n},\bar{v}_{n})\\}$ is bounded in
$H^{1}_{r}(\mathbb{R}^{N})\times H^{1}_{r}(\mathbb{R}^{N})$. By (2.3), there
exists a $(\bar{u},\bar{v})\in H^{1}_{r}(\mathbb{R}^{N})\times
H^{1}_{r}(\mathbb{R}^{N})$ such that

$\displaystyle\left\\{\begin{aligned}
&(\bar{u}_{n},\bar{v}_{n})\rightharpoonup(\bar{u},\bar{v})\text{~{}weakly in
}H^{1}_{r}(\mathbb{R}^{N})\times H^{1}_{r}(\mathbb{R}^{N}),\\\
&B(\bar{u}_{n},p)\rightarrow B(\bar{u},p),~{}B(\bar{v}_{n},p)\rightarrow
B(\bar{v},p),\\\
&(\bar{u}_{n},\bar{v}_{n})\rightarrow(\bar{u},\bar{v})\text{~{}a.e. in
}\mathbb{R}^{N}.\end{aligned}\right.$ (4.7)

From (4.4), we have

$\displaystyle\langle
E^{\prime}(\bar{u}_{n},\bar{v}_{n})-\lambda_{1n}(\bar{u}_{n},0),(\bar{u}_{n},0)\rangle\rightarrow\langle
E^{\prime}(\bar{u},\bar{v})-\lambda_{1}(\bar{u},0),(\bar{u},0)\rangle~{}~{}~{}\text{as
}n\rightarrow\infty.$

Combining with (4.7), it holds that

$\displaystyle\|\nabla\bar{u}_{n}\|_{2}^{2}+\lambda_{1n}\|\bar{u}_{n}\|_{2}^{2}-\int_{\mathbb{R}^{N}}\beta(x)\bar{u}_{n}\bar{v}_{n}\rightarrow\|\nabla\bar{u}\|_{2}^{2}+\lambda_{1}\|\bar{u}\|_{2}^{2}-\int_{\mathbb{R}^{N}}\beta(x)\bar{u}\bar{v}.$

From $\beta(x)$ is sufficiently small and $\lambda_{1}>0$, we can obtain that

$\displaystyle\|\nabla\bar{u}_{n}\|_{2}^{2}+\lambda_{1n}\|\bar{u}_{n}\|_{2}^{2}\rightarrow\|\nabla\bar{u}\|_{2}^{2}+\lambda_{1}\|\bar{u}\|_{2}^{2}.$
(4.8)

From the low semi-continuous of $L^{2}$-norm and Fatou’s Lemma, it follows
that

$\displaystyle\lim\inf_{n\rightarrow\infty}\|\nabla\bar{u}_{n}\|_{2}^{2}\geq\|\nabla\bar{u}\|_{2}^{2},~{}\lim\inf_{n\rightarrow\infty}\|\bar{u}_{n}\|_{2}^{2}\geq\|\bar{u}\|_{2}^{2},$

combining with (4.8), and due to $\lambda_{1}>0$, hence

$\displaystyle\lim_{n\rightarrow\infty}\|\nabla\bar{u}_{n}\|_{2}^{2}=\|\nabla\bar{u}\|_{2}^{2}~{}\text{and}~{}\lim_{n\rightarrow\infty}\|\bar{u}_{n}\|_{2}^{2}=\|\bar{u}\|_{2}^{2},$

For $\lambda_{2}>0$, the proof is similar. Thus we complete the proof. ∎

###### Proof of Theorem 1.2.

From Lemma 4.6, we may assume that $\lambda_{1}>0$. By Lemma 4.7,
$\bar{u}_{n}\rightarrow\bar{u}$ strongly in $H^{r}_{1}(\mathbb{R}^{N})$. If
$\lambda_{2}>0$, then $\bar{v}_{n}\rightarrow\bar{v}$ strongly in
$H^{r}_{1}(\mathbb{R}^{N})$. Hence we remain to show $\lambda_{2}>0$. We argue
it by contradiction and assume that $\lambda_{2}\leq 0$. By the weak limit and
(4.4), $\bar{v}$ satisfies

$\displaystyle-\Delta\bar{v}=-\lambda_{2}\bar{v}+\mu_{2}(I_{\alpha}\star
v)|\bar{v}|^{q-2}\bar{v}+\beta(x)\bar{u}\bar{v}\geq 0,$

due to $\bar{u},~{}\bar{v}\geq 0$ and $\beta(x)\geq 0$. By Liouville-type
Lemma in [14], we get that $\bar{v}\equiv 0$. However, $(\bar{u},0)$ can not
be the solution of (1.1), hence there exists a contradiction. Therefore,
$\lambda_{2}>0$. By Lemma 4.7, $\bar{v}_{n}\rightarrow\bar{v}$ strongly in
$H^{1}_{r}(\mathbb{R}^{N})$. Moreover, by maximum principle, $u,~{}v>0$. Thus
the theorem is established. ∎

##  5 $V(x)$ for $\max\\{1+\frac{\alpha}{N},2\\}<p,q<1+\frac{\alpha+2}{N}$

In this section, we consider the $L^{2}$-subcritical case for (1.1) with
different potentials.

###  5.1 Proof of Theorem 1.4

In this subsection, we consider the case $V_{1}(x)$ and $V_{2}(x)$ satisfy
(V2).

######  Lemma 5.1.

[31] Let $V_{1}(x)$ and $V_{2}(x)$ satisfy (V2). Then
$\tilde{H}_{1}\times\tilde{H}_{2}\hookrightarrow L^{p}\times L^{q}$ for
$p,q\in[2,2^{*})$ is compact.

Similar to Lemma 3.1, we obtain that

$\displaystyle E_{V}(u,v)\geq$ $\displaystyle\frac{1}{2}(\|\nabla
u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})-\frac{\mu_{1}C(N,\alpha,p)\xi^{2p(1-\delta_{p})}}{2p}\|\nabla
u\|_{2}^{2p\delta_{p}}$ $\displaystyle-$
$\displaystyle\frac{\mu_{2}C(N,\alpha,q)\eta^{2q(1-\delta_{q})}}{2q}\|\nabla
v\|_{2}^{2q\delta_{q}}-\beta\xi\eta+\frac{1}{2}\int_{\mathbb{R}^{N}}(V_{1}(x)|u|^{2}+V_{2}(x)|v|^{2})dx.$

Since $1+\frac{\alpha}{N}<p,q<1+\frac{\alpha+2}{N}$ and
$(u,v)\in\tilde{H_{1}}\times\tilde{H_{2}}$, $E_{V}(u,v)$ is also bounded from
below. Furthermore, there exists a bounded minimizing sequence
$\\{(u_{n},v_{n})\\}\in\tilde{H_{1}}\times\tilde{H_{2}}$.

###### Proof of theorem 1.4.

Notice that $\\{(u_{n},v_{n})\\}\in\tilde{H_{1}}\times\tilde{H_{2}}$ is
bounded. Hence there exists a
$(u_{0},v_{0})\in\tilde{H_{1}}\times\tilde{H_{2}}$ such that
$(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ weakly in
$\tilde{H_{1}}\times\tilde{H_{2}}$. By Lemma 5.1 and (2.3), we can obtain that
$B(u_{n},p)\to B(u_{0},p)$ and $B(v_{n},q)\to B(v_{0},q)$. Moreover,
$(u_{n},v_{n})\to(u_{0},v_{0})$ strongly in $\tilde{H_{1}}\times\tilde{H_{2}}$
due to

$\displaystyle e_{V}(\xi,\eta)\leq
E_{V}(u_{0},v_{0})\leq\liminf_{n\rightarrow\infty}E_{V}(u_{n},v_{n})=e_{V}(\xi,\eta).$

Thus we establish the theorem.

∎

###  5.2 Proof of Theorem 1.3

In this subsection, we focus on the case that $V_{1}(x)$ and $V_{2}(x)$
satisfy (V1).

######  Lemma 5.2.

Assume that $V_{1}(x)$ and $V_{2}(x)$ satisfy (V1). Then there holds

  * (i)

$e_{V}(\xi,\eta)\leq e(\xi,\eta)\leq 0$; if $\xi+\eta>0$, then
$e_{V}(\xi,\eta)<e(\xi,\eta)<0$.

  * (ii)

every minimizing sequence for $e_{V}(\xi,\eta)$ is bounded in
$H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$,

  * (iii)

$e_{V}(\xi,\eta)$ is continuous about $(\xi,\eta)$. Furthermore,
$e_{V}(\xi,\eta)\leq e_{V}(\xi-\xi_{1},\eta-\eta_{1})+e(\xi_{1},\eta_{1})$ for
$\xi_{1}>0$ and $\eta_{1}>0$,

  * (iv)

$e_{V}(\xi,\eta)$ is nondecreasing with respect to $(\xi,\eta)$.

###### Proof.

(i) holds immediately from $V_{1}(x),~{}V_{2}(x)<0$. And the proof of (ii) is
similar to the proof of Lemma 3.1. If (iii) holds, combining with (i), then
(iv) holds. Hence we only need to show that (iii) holds.

Assume that $|\xi_{n}-\xi|+|\eta_{n}-\eta|=o_{n}(1)$, for any $\varepsilon>0$,
there exists a sequence $\\{(u_{n},v_{n})\\}\in S_{\xi_{n}}\times
S_{\eta_{n}}$ such that

$\displaystyle E_{V}(u_{n},v_{n})\leq e_{V}(\xi_{n},\eta_{n})+o_{n}(1).$ (5.1)

Let $\hat{u}=\frac{\xi}{\xi_{n}}u_{n}$ and
$\hat{v}=\frac{\eta}{\eta_{n}}v_{n}$. Then $(\hat{u},\hat{v})\in S_{\xi}\times
S_{\eta}$ and

$\displaystyle E_{V}(\hat{u},\hat{v})=$
$\displaystyle\frac{1}{2}\bigg{(}\frac{\xi^{2}}{\xi_{n}^{2}}\|\nabla
u_{n}\|^{2}_{2}+\frac{\eta^{2}}{\eta_{n}^{2}}\|\nabla
v_{n}\|^{2}_{2}\bigg{)}-\frac{\xi}{\xi_{n}}\frac{\eta}{\eta_{n}}\int_{\mathbb{R}^{N}}\beta
u_{n}v_{n}dx$
$\displaystyle+\frac{1}{2}\int_{\mathbb{R}^{N}}\bigg{(}\frac{\xi^{2}}{\xi_{n}^{2}}V_{1}(x)|u_{n}|^{2}+\frac{\eta^{2}}{\eta_{n}^{2}}V_{2}(x)|v_{n}|^{2}\bigg{)}dx$
$\displaystyle-\frac{\xi^{2p}}{\xi_{n}^{2p}}\frac{\mu_{1}}{2p}B(u_{n},p)-\frac{\eta^{2q}}{\eta_{n}^{2q}}\frac{\mu_{2}}{2q}B(v_{n},q),$

by (5.1) and $|\xi_{n}-\xi|+|\eta_{n}-\eta|=o_{n}(1)$, it follows that

$\displaystyle e_{V}(\xi,\eta)\leq
E_{V}(\hat{u},\hat{v})=E_{V}(u_{n},v_{n})+o_{n}(1)\leq
e_{V}(\xi_{n},\eta_{n})+o_{n}(1),$

consequently, $e_{V}(\xi,\eta)\leq e_{V}(\xi_{n},\eta_{n})+o_{n}(1)$. Vice
versa. Hence $e_{V}(\xi,\eta)$ is continuous about $(\xi,\eta)$.

Next we show the subadditivity of $e_{V}(\xi,\eta)$, which is similar to Lemma
3.2.

For any $\varepsilon>0$, we can seek
$(\varphi_{1\varepsilon},\psi_{1\varepsilon}),(\varphi_{1\varepsilon},\psi_{1\varepsilon})\in
C_{c}(\mathbb{R}^{N})\times C_{c}(\mathbb{R}^{N})$ such that

$\displaystyle\begin{split}&(\varphi_{1\varepsilon},\psi_{1\varepsilon})\in
S_{\xi-\xi_{1}}\times
S_{\eta-\eta_{1}},~{}E_{V}(\varphi_{1\varepsilon},\psi_{1\varepsilon})\leq
e_{V}(\xi-\xi_{1},\eta-\eta_{1})+\varepsilon,\\\
&(\varphi_{2\varepsilon},\psi_{2\varepsilon})\in S_{\xi_{1}}\times
S_{\eta_{1}},~{}E(\varphi_{2\varepsilon},\psi_{2\varepsilon})\leq
e(\xi_{1},\eta_{1})+\varepsilon,\end{split}$ (5.2)

Set
$(u_{\varepsilon,n}(x),v_{\varepsilon,n}(x)):=(\varphi_{1\varepsilon}(x)+\varphi_{2\varepsilon}(x-n\textbf{e}_{1}),\psi_{1\varepsilon}(x)+\psi_{2\varepsilon}(x-n\textbf{e}_{1}))$,
where $\textbf{e}_{1}=(1,0,...,0)$. Since
$(\varphi_{1\varepsilon},\psi_{1\varepsilon})$ and
$(\varphi_{2\varepsilon},\psi_{2\varepsilon})$ have compact support, there
holds that

$(u_{\varepsilon,n}(x),v_{\varepsilon,n}(x))\in S_{\xi}\times S_{\eta}$

for large $n$. Therefore,

$\displaystyle e_{V}(\xi,\eta)\leq
E_{V}(u_{\varepsilon,n}(x),v_{\varepsilon,n}(x))=E_{V}(\varphi_{1\varepsilon}(x),\psi_{1\varepsilon}(x))+E_{V}(\varphi_{2\varepsilon}(x-n\textbf{e}_{1}),\psi_{2\varepsilon}(x-n\textbf{e}_{1})).$

Due to
$\lim\limits_{|x|\rightarrow\infty}V_{1}(x)=\lim\limits_{|x|\rightarrow\infty}V_{2}(x)=0$,
it means that
$E_{V}(\varphi_{2\varepsilon}(x-n\textbf{e}_{1}),\psi_{2\varepsilon}(x-n\textbf{e}_{1}))\rightarrow
E(\varphi_{2\varepsilon},\psi_{2\varepsilon})$ as $n\rightarrow\infty$.
Together with (5.2), there holds that

$\displaystyle e_{V}(\xi,\eta)\leq$ $\displaystyle
E_{V}(\varphi_{1\varepsilon}(x),\psi_{1\varepsilon}(x))+E(\varphi_{2\varepsilon},\psi_{2\varepsilon})\leq
e_{V}(\xi-\xi_{1},\eta-\eta_{1})+e(\xi_{1},\eta_{1})+2\varepsilon$

for any $\varepsilon>0$. Due to the arbitrary of $\varepsilon>0$, hence (iii)
holds. ∎

######  Lemma 5.3.

Let $\\{(u_{n},v_{n})\\}\subset S_{\xi}\times S_{\eta}$ be a minimizing
sequence for $e_{V}(\xi,\eta)$ and there exists $(u_{0},v_{0})\in
H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$ such that
$(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ in $H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})$. Set $\xi_{1}=\|u_{0}\|_{2},~{}\eta_{1}=\|v_{0}\|_{2}$.
If $\xi_{1}^{2}+\eta_{1}^{2}<\xi^{2}+\eta^{2}$, then there exist
$\\{y_{n}\\}\subset\mathbb{R}^{N}$ and $(\varphi_{0},\psi_{0})\in
H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})\setminus{(0,0)}$ such that

$\displaystyle\left\\{\begin{aligned} &|y_{n}|\rightarrow\infty,\\\
&(u_{n}(\cdot+y_{n}),v_{n}(\cdot+y_{n}))\rightharpoonup(\varphi_{0},\psi_{0})\text{~{}weakly
in~{}}H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N}),\end{aligned}\right.\text{~{}~{}as~{}}n\rightarrow\infty,$
(5.3)

and

$\displaystyle\lim_{n\rightarrow\infty}\|u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n})\|_{2}^{2}=0,~{}\lim_{n\rightarrow\infty}\|v_{n}-v_{0}-\psi_{0}(\cdot-
y_{n})\|_{2}^{2}=0,$ (5.4)

where
$\|\varphi_{0}\|_{2}^{2}=\xi_{2}^{2}=\xi^{2}-\xi_{1}^{2},~{}\|\psi_{0}\|_{2}^{2}=\eta_{2}^{2}=\eta^{2}-\eta_{1}^{2}$.
Furthermore,

$\displaystyle e_{V}(\xi,\eta)=e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2}),$
(5.5)

and

$\displaystyle
E_{V}(u_{0},v_{0})=e_{V}(\xi_{1},\eta_{1}),~{}E(\varphi_{0},\psi_{0})=e(\xi_{2},\eta_{2}).$
(5.6)

###### Proof.

We divide the proof into three steps.

Step 1: We find $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and
$(\varphi_{0},\psi_{0})$ such that (5.3) holds.

We claim that

$\displaystyle\delta_{0}:=\liminf_{n\rightarrow\infty}\sup_{y\in\mathbb{R}^{N}}\int_{y+B(0,1)}[|u_{n}-u_{0}|^{2}+|v_{n}-v_{0}|^{2}]dx>0,$

where $B(0,1):=\\{x\in\mathbb{R}^{N}:|x|\leq 1\\}$. The claim can be shown by
contradiction. Suppose that $\delta_{0}=0$. Then
$(u_{n},v_{n})\to(u_{0},v_{0})$ strongly in $L^{s}(\mathbb{R}^{N})\times
L^{t}(\mathbb{R}^{N})$ for $s,t\in(2,2^{*})$ (see [38]). By (2.3), we have
$B(u_{n},p)\to B(u_{0},p)$ and $B(v_{n},q)\to B(v_{0},q)$. Since
$(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ weakly in
$H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$, we can verify that

$\int_{\mathbb{R}^{N}}(V_{1}(x)(u_{n}-u_{0})^{2}+V_{2}(x)(v_{n}-v_{0})^{2})dx\rightarrow
0.$

Thus,

$\displaystyle e_{V}(\xi,\eta)$ $\displaystyle=E_{V}(u_{n},v_{n})+o(1)$
$\displaystyle\geq E_{V}(u_{0},v_{0})+E_{V}(u_{n}-u_{0},v_{n}-v_{0})+o(1)$
$\displaystyle=E_{V}(u_{0},v_{0})+\frac{1}{2}(\|\nabla(u_{n}-u_{0})\|_{2}^{2}+\|\nabla(v_{n}-v_{0})\|_{2}^{2})$
$\displaystyle~{}~{}-\int_{\mathbb{R}^{N}}\beta(u_{n}-u_{0})(v_{n}-v_{0})dx+o(1)$
$\displaystyle\geq
E_{V}(u_{0},v_{0})-\beta\|(u_{n}-u_{0})\|_{2}\|(v_{n}-v_{0})\|_{2}$
$\displaystyle=E_{V}(u_{0},v_{0})-\beta\xi_{2}\eta_{2},$

together with (3.3), we can obtain that

$\displaystyle e_{V}(\xi,\eta)>E_{V}(u_{0},v_{0})+e(\xi_{2},\eta_{2})+o(1)\geq
e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2}),$

which is a contradiction with Lemma 5.2 (iii). Hence $\delta_{0}>0$.

From $\delta_{0}>0$ and $(u_{n},v_{n})\rightarrow(u_{0},v_{0})$ in
$L_{loc}^{2}(\mathbb{R}^{N})\times L_{loc}^{2}(\mathbb{R}^{N})$, we can find a
sequence $\\{y_{n}\\}\subset\mathbb{R}^{N}$ such that

$\displaystyle\int_{y_{n}+B(0,1)}(|u_{n}-u_{0}|^{2}+|v_{n}-v_{0}|^{2})dx\rightarrow
c_{0}>0~{}\text{
and}~{}~{}|y_{n}|\rightarrow\infty,\text{~{}as~{}}n\to\infty.$

Denote that
$(u_{n}(\cdot+y_{n}),v_{n}(\cdot+y_{n}))\rightharpoonup(\varphi_{0},\psi_{0})$
weakly in $H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$. Due to
$c_{0}>0$, $(\varphi_{0},\psi_{0})\neq(0,0)$. Therefore, $\\{y_{n}\\}$ and
$(\varphi_{0},\psi_{0})$ satisfy (5.3).

Since $y_{n}\rightarrow\infty$, we have

$\displaystyle\|u_{n}-u_{0}-\varphi_{0}(\cdot-y_{n})\|_{2}^{2}$
$\displaystyle=$
$\displaystyle\|u_{n}\|_{2}^{2}+\|u_{0}\|_{2}^{2}+\|\varphi_{0}\|_{2}^{2}-2\langle
u_{n},u_{0}\rangle_{L^{2}}-2\langle
u_{n}(\cdot+y_{n}),\varphi_{0}\rangle_{L^{2}}+o(1)$ $\displaystyle=$
$\displaystyle\|u_{n}\|_{2}^{2}-\|u_{0}\|_{2}^{2}-\|\varphi_{0}\|_{2}^{2}+o(1),$

and

$\displaystyle\|v_{n}-v_{0}-\psi_{0}(\cdot-y_{n})\|_{2}^{2}$ $\displaystyle=$
$\displaystyle\|v_{n}\|_{2}^{2}+\|v_{0}\|_{2}^{2}+\|\psi_{0}\|_{2}^{2}-2\langle
v_{n},v_{0}\rangle_{L^{2}}-2\langle
v_{n}(\cdot+y_{n}),\psi_{0}\rangle_{L^{2}}+o(1)$ $\displaystyle=$
$\displaystyle\|v_{n}\|_{2}^{2}-\|v_{0}\|_{2}^{2}-\|\psi_{0}\|_{2}^{2}+o(1),$

hence it yields that

$\displaystyle\xi_{2}^{2}:=\|\varphi_{0}\|_{2}^{2}\leq\liminf_{n\rightarrow\infty}(\|u_{n}\|^{2}_{2}-\|u_{0}\|^{2}_{2})=\xi^{2}-\xi_{1}^{2},$
$\displaystyle\eta_{2}^{2}:=\|\psi_{0}\|_{2}^{2}\leq\liminf_{n\rightarrow\infty}(\|u_{n}\|^{2}_{2}-\|u_{0}\|^{2}_{2})=\eta^{2}-\eta_{1}^{2}.$

From $c_{0}>0$, it follows that $\xi_{2}^{2}+\eta_{2}^{2}>0$.

Step 2: We show $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and
$(\varphi_{0},\psi_{0})$ such that (5.4) holds.

Let
$\delta_{1}^{2}=\lim\limits_{n\rightarrow\infty}\|u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n})\|_{2}^{2}$ and
$\delta_{2}^{2}=\lim\limits_{n\rightarrow\infty}\|v_{n}-v_{0}-\psi_{0}(\cdot-
y_{n})\|_{2}^{2}$. Then $\delta_{1}^{2}=\xi^{2}-\xi_{1}^{2}-\xi_{2}^{2}$ and
$\delta_{2}^{2}=\eta^{2}-\eta_{1}^{2}-\eta_{2}^{2}$. We next show that
$\delta_{1}=0$ and $\delta_{2}=0$ by contradiction. Suppose that
$\delta_{1}^{2}+\delta_{2}^{2}>0$. Then

$\displaystyle\quad\|\nabla u_{n}\|_{2}^{2}-\|\nabla
u_{0}\|_{2}^{2}-\|\nabla\varphi_{0}\|_{2}^{2}-\|\nabla(u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n}))\|_{2}^{2}$ $\displaystyle=-2\|\nabla
u_{0}\|_{2}^{2}-2\|\nabla\varphi_{0}\|_{2}^{2}+2\langle\nabla u_{n},\nabla
u_{0}\rangle+2\langle\nabla u_{n}(\cdot+y_{n}),\nabla\varphi_{0}\rangle$
$\displaystyle=o(1).$

Similarly,

$\displaystyle\quad\|\nabla v_{n}\|_{2}^{2}-\|\nabla
v_{0}\|_{2}^{2}-\|\nabla\psi_{0}\|_{2}^{2}-\|\nabla(v_{n}-v_{0}-\psi_{0}(\cdot-
y_{n}))\|_{2}^{2}$ $\displaystyle=-2\|\nabla
v_{0}\|_{2}^{2}-2\|\nabla\psi_{0}\|_{2}^{2}+2\langle\nabla v_{n},\nabla
v_{0}\rangle+2\langle\nabla v_{n}(\cdot+y_{n}),\nabla\psi_{0}\rangle$
$\displaystyle=o(1).$

Furthermore,

$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)(|u_{n}|^{2}-|u_{0}|_{2}^{2}-|\varphi_{0}|^{2}-|(u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n}))|^{2})dx=o(1),$
$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{N}}V_{2}(x)(|v_{n}|^{2}-|v_{0}|_{2}^{2}-|\psi_{0}|^{2}-|(v_{n}-v_{0}-\psi_{0}(\cdot-
y_{n}))|^{2})dx=o(1).$

In addition, by Brézis-Lieb Lemma, there holds that

$\displaystyle B(u_{n},p)=$ $\displaystyle B(u_{0},p)+B(u_{n}-u_{0},p)+o(1)$
$\displaystyle=$ $\displaystyle
B(u_{0},p)+B(\varphi_{0},p)+B(u_{n}-u_{0}-\varphi_{0}(\cdot-y_{n}),p)+o(1),$

and

$\displaystyle B(v_{n},q)$ $\displaystyle=B(v_{0},q)+B(v_{n}-v_{0},q)+o(1)$
$\displaystyle=$ $\displaystyle
B(v_{0},q)+B(\psi_{0},q)+B(v_{n}-v_{0}-\psi_{0}(\cdot-y_{n}),q)+o(1).$

As a consequence,

$\displaystyle E_{V}(u_{n},v_{n})=$ $\displaystyle
E_{V}(u_{0},v_{0})+E_{V}(\varphi_{0},\psi_{0})$ $\displaystyle-
E_{V}(u_{n}-u_{0}-\varphi_{0}(\cdot-y_{n}),v_{n}-v_{0}-\psi_{0}(\cdot-
y_{n}))+o(1).$

Since $(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ weakly in
$H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$,
$|y_{n}|\rightarrow\infty$ and

$\lim_{|x|\rightarrow\infty}V_{1}(x)=\lim_{|x|\rightarrow\infty}V_{2}(x)=0,$

we have

$\displaystyle\int_{\mathbb{R}^{N}}V_{1}(x)|u_{n}-u_{0}-\varphi_{0}(x-y_{n})|^{2}+V_{2}(x)|v_{n}-v_{0}-\psi_{0}(x-y_{n})|^{2}dx\rightarrow
0.$ (5.7)

Notice that

$\displaystyle E_{V}(u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n}),v_{n}-v_{0}-\psi_{0}(\cdot-y_{n}))$ $\displaystyle=$ $\displaystyle
E(u_{n}-u_{0}-\varphi_{0}(\cdot-y_{n}),v_{n}-v_{0}-\psi_{0}(\cdot-y_{n}))$
$\displaystyle+\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)|u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n})|^{2}+V_{2}(x)|v_{n}-v_{0}-\psi_{0}(\cdot-y_{n})|^{2}dx,$

together with (5.7), it yields that

$\displaystyle\liminf_{n\rightarrow\infty}E_{V}(u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n}),v_{n}-v_{0}-\psi_{0}(\cdot-y_{n}))$ $\displaystyle=$
$\displaystyle\liminf_{n\rightarrow\infty}E(u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n}),v_{n}-v_{0}-\psi_{0}(\cdot-y_{n}))$ $\displaystyle\geq$ $\displaystyle
e(\delta_{1},\delta_{2}),$

Similarly, we have

$\displaystyle\liminf_{n\rightarrow\infty}E_{V}(\varphi_{0}(\cdot-
y_{n}),\psi_{0}(\cdot-y_{n}))\geq e(\xi_{2},\eta_{2}).$

Consequently,

$\displaystyle e_{V}(\xi,\eta)\geq
e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2})+e(\delta_{1},\delta_{2}).$

By Remark 3.1, we have
$e(\xi_{2}+\delta_{1},\eta_{2}+\delta_{2})<e(\xi_{2},\eta_{2})+e(\delta_{1},\delta_{2})$
for $\delta_{1}^{2}+\delta_{2}^{2}>0$. By Lemma 5.2 (iii), it yields that

$\displaystyle e_{V}(\xi,\eta)$ $\displaystyle\geq
e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2})+e(\delta_{1},\delta_{2})$
$\displaystyle>e_{V}(\xi_{1},\eta_{1})+e(\xi_{2}+\delta_{1},\eta_{2}+\delta_{2})$
$\displaystyle\geq
e_{V}(\xi_{1}+\xi_{2}+\delta_{1},\eta_{1}+\eta_{2}+\delta_{2})$
$\displaystyle=e_{V}(\xi,\eta),$

which is a contradiction. Hence $\delta_{1}^{2}+\delta_{2}^{2}=0$.

Step 3: We show $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and
$(\varphi_{0},\psi_{0})$ satisfy (5.5) and (5.6).

Since $\delta_{1}=\delta_{2}=0$, we have

$\displaystyle e_{V}(\xi,\eta)=\lim_{n\rightarrow\infty}E_{V}(u_{n},v_{n})$
$\displaystyle=\lim_{n\rightarrow\infty}\big{(}E_{V}(u_{0},v_{0})+E_{V}(\varphi_{0}(\cdot+y_{n}),\psi_{0}(\cdot+y_{n}))\big{)}$
$\displaystyle\geq E_{V}(u_{0},v_{0})+E(\varphi_{0},\psi_{0})$
$\displaystyle\geq e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2}),$

together with Lemma 5.2 (iii), there holds that

$e_{V}(\xi,\eta)=e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2}),$

Furthermore, $E_{V}(u_{0},v_{0})=e_{V}(\xi_{1},\eta_{1})$ and
$E(\varphi_{0},\psi_{0})=e(\xi_{2},\eta_{2})$. Thus we finish the proof. ∎

###### Proof of Theorem 1.3.

Let $\\{(u_{n},v_{n})\\}$ be the minimizing sequence for $e_{V}(\xi,\eta)$. We
divide the proof into two steps.

Step 1: $\\{(u_{n},v_{n})\\}$ is strongly convergent in
$L^{2}(\mathbb{R}^{N})\times L^{2}(\mathbb{R}^{N})$.

By Lemma 5.2 (ii), we can find that $\\{(u_{n},v_{n})\\}$ is bounded in
$H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$. Hence there exists a
$(u_{0},v_{0})\in H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$ such that

$\displaystyle(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})\text{ weakly in
}H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N}).$

Since $\\{(u_{n},v_{n})\\}$ is the minimizing sequence for $e_{V}(\xi,\eta)$,
we have $dE_{V}|_{S_{\xi}\times S_{\eta}}(u_{n},v_{n})\rightarrow 0$.
Moreover, there exist two sequences $\lambda_{1n}\subset\mathbb{R}$ and
$\lambda_{2n}\subset\mathbb{R}$ such that

$\displaystyle\begin{split}&\int_{\mathbb{R}^{N}}(\nabla
u_{n}\nabla\varphi+\nabla
v_{n}\nabla\psi)dx+\int_{\mathbb{R}^{N}}(V_{1}(x)u_{n}\varphi+V_{2}(x)v_{n}\psi)dx\\\
-&\int_{\mathbb{R}^{N}}(\mu_{1}(I_{\alpha}\star|u_{n}|^{p})|u_{n}|^{p-2}u_{n}\varphi+\mu_{2}(I_{\alpha}\star|v_{n}|^{q})|v_{n}|^{q-2}v_{n}\psi)dx\\\
-&\int_{\mathbb{R}^{N}}\beta(x)(u_{n}\psi+v_{n}\varphi)dx+\lambda_{1n}\int_{\mathbb{R}^{N}}u_{n}\varphi
dx+\lambda_{2n}\int_{\mathbb{R}^{N}}v_{n}\psi dx\\\
=&o_{n}(1)\|(\varphi,\psi)\|_{H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})}\end{split}$ (5.8)

for any $(\varphi,\psi)\in H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$
with $o_{n}(1)\rightarrow 0$ as $n\rightarrow\infty$. Taking
$(\varphi,\psi)=(u_{n},0)$ and $(\varphi,\psi)=(0,v_{n})$ in (5.8), we can
derive that

$\displaystyle\int_{\mathbb{R}^{N}}|\nabla
u_{n}|^{2}dx+\int_{\mathbb{R}^{N}}V_{1}(x)|u_{n}|^{2}dx-\mu_{1}\int_{\mathbb{R}^{N}}(I_{\alpha}\star|u_{n}|^{p})|u_{n}|^{p}dx-\int_{\mathbb{R}^{N}}\beta
u_{n}v_{n}dx=-\lambda_{1n}\xi^{2}.$

and

$\displaystyle\int_{\mathbb{R}^{N}}|\nabla
v_{n}|^{2}dx+\int_{\mathbb{R}^{N}}V_{2}(x)|v_{n}|^{2}dx-\mu_{2}\int_{\mathbb{R}^{N}}(I_{\alpha}\star|v_{n}|^{q})|v_{n}|^{q}dx-\int_{\mathbb{R}^{N}}\beta
u_{n}v_{n}dx=-\lambda_{2n}\eta^{2}.$

Since $\\{(u_{n},v_{n})\\}$ is bounded in $H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})$, by (2.3), $B(u_{n},p)$ and $B(v_{n},q)$ are bounded.
Therefore, $\lambda_{1n}$ and $\lambda_{2n}$ are bounded in $\mathbb{R}$.
Moreover, there exist $\lambda_{1}\in\mathbb{R}$ and
$\lambda_{2}\in\mathbb{R}$ such that $\lambda_{1n}\rightarrow\lambda_{1}$ and
$\lambda_{2n}\rightarrow\lambda_{2}$. Therefore, $(u_{0},v_{0})$ satisfies

$\displaystyle\left\\{\begin{aligned} &-\Delta
u_{0}+V_{1}(x)u_{0}+\lambda_{1}u_{0}=\mu_{1}(I_{\alpha}\star|u_{0}|^{p})|u_{0}|^{p-2}u_{0}+\beta
v_{0},\\\ &-\Delta
v_{0}+V_{2}(x)v_{0}+\lambda_{2}v_{0}=\mu_{2}(I_{\alpha}\star|v_{0}|^{q})|v_{0}|^{q-2}v_{0}+\beta
u_{0},\end{aligned}\right.$

with $\|u_{0}\|_{2}^{2}=\xi_{1}^{2}$ and $\|v_{0}\|_{2}^{2}=\eta_{1}^{2}$. If
$\xi_{1}=\xi$ and $\eta_{1}=\eta$, then the step is finished. Otherwise,
$\xi_{1}^{2}+\eta_{1}^{2}<\xi^{2}+\eta^{2}$, i.e., there are three cases:

$\displaystyle\left\\{\begin{aligned} &\textbf{Case 1:
}\xi_{1}<\xi,~{}\eta_{1}<\eta,\\\ &\textbf{Case 2:
}\xi_{1}=\xi,~{}\eta_{1}<\eta,\\\ &\textbf{Case 3:
}\xi_{1}<\xi,~{}\eta_{1}=\eta.\end{aligned}\right.$

By Lemma 5.3, $(\varphi_{0},\psi_{0})$ satisfies

$\displaystyle\left\\{\begin{aligned}
&-\Delta\varphi_{0}+\lambda_{1}\varphi_{0}=\mu_{1}(I_{\alpha}\star|\varphi_{0}|^{p})|\varphi_{0}|^{p-2}\varphi_{0}+\beta\psi_{0},\\\
&-\Delta\psi_{0}+\lambda_{2}\psi_{0}=\mu_{2}(I_{\alpha}\star|\psi_{0}|^{q})|\psi_{0}|^{q-2}\psi_{0}+\beta\varphi_{0},\end{aligned}\right.$
(5.9)

with $\|\varphi_{0}\|_{2}^{2}=\xi^{2}-\xi_{1}^{2}=\xi_{2}^{2}$ and
$\|\psi_{0}\|_{2}^{2}=\eta^{2}-\eta_{1}^{2}=\eta_{2}^{2}$.

For Case 2, it is easy to find $\|\varphi_{0}\|_{2}^{2}=\xi^{2}-\xi_{1}^{2}=0$
and $\|\psi_{0}\|_{2}^{2}=\eta^{2}-\eta_{1}^{2}>0$, which implies that
$\varphi_{0}=0$ and $\psi_{0}\neq 0$. As a consequence, $(0,\psi_{0})$ is a
solution of (5.9), which is impossible for $\beta>0$. Hence Case 2 doesn’t
hold. Similarly, Case 3 also doesn’t hold.

For Case 1, take
$(\sqrt{u_{0}^{2}+\varphi_{0}^{2}},\sqrt{v_{0}^{2}+\phi_{0}^{2}})\in
S_{\xi}\times S_{\eta}$ as the test function, direct computation shows that

$\displaystyle\left\\{\begin{aligned}
&|\nabla\sqrt{u_{0}^{2}+\varphi_{0}^{2}}|^{2}\leq|\nabla
u_{0}|^{2}+|\nabla\varphi_{0}|^{2},~{}|\nabla\sqrt{v_{0}^{2}+\psi_{0}^{2}}|^{2}\leq|\nabla
v_{0}|^{2}+|\nabla\psi_{0}|^{2},\\\
&|\sqrt{u_{0}^{2}+\varphi_{0}^{2}}|^{p}\geq|u_{0}|^{p}+|\varphi_{0}|^{p},~{}|\sqrt{v_{0}^{2}+\phi_{0}^{2}}|^{q}\geq|v_{0}|^{q}|\phi_{0}|^{q},\\\
&\beta\sqrt{u_{0}^{2}+\varphi_{0}^{2}}\sqrt{v_{0}^{2}+\phi_{0}^{2}}\geq\beta
u_{0}v_{0}+\beta\varphi_{0}\psi_{0}\text{~{}for~{}}\beta>0.\end{aligned}\right.$
(5.10)

From the second inequality in (5.10), we can derive that

$\displaystyle\begin{split}B(\sqrt{u_{0}^{2}+\varphi_{0}^{2}},p)&=\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|\sqrt{u_{0}^{2}+\varphi_{0}^{2}}(x)|^{p}|\sqrt{u_{0}^{2}+\varphi_{0}^{2}}(y)|^{p}}{|x-y|^{N-\alpha}}dydx\\\
&\geq\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{(|u_{0}(x)|^{p}+|\varphi_{0}(x)|^{p})(|u_{0}(y)|^{p}+|\varphi_{0}(y)|^{p})}{|x-y|^{N-\alpha}}dydx\\\
&\geq\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|u_{0}(x)|^{p}|u_{0}(y)|^{p}}{|x-y|^{N-\alpha}}dydx+\int_{\mathbb{R}^{N}}\int_{\mathbb{R}^{N}}\frac{|\varphi_{0}(x)|^{p}|\varphi_{0}(y)|^{p}}{|x-y|^{N-\alpha}}dydx\\\
&=B(u_{0},p)+B(\varphi_{0},p),\end{split}$ (5.11)

similarly, there holds that

$\displaystyle B(\sqrt{v_{0}^{2}+\psi_{0}^{2}},q)\geq
B(v_{0},q)+B(\psi_{0},q).$ (5.12)

By (5.10), (5.11) and (5.12), it yields that

$\displaystyle\begin{split}e_{V}(\xi,\eta)&\leq
E_{V}(\sqrt{u_{0}^{2}+\varphi_{0}^{2}},\sqrt{v_{0}^{2}+\psi_{0}^{2}})\\\ &\leq
E_{V}(u_{0},v_{0})+E_{V}(\varphi_{0},\psi_{0})\\\
&=E_{V}(u_{0},v_{0})+E(\varphi_{0},\psi_{0})+\frac{1}{2}\int_{\mathbb{R}^{N}}[V_{1}(x)|\varphi_{0}|^{2}+V_{2}(x)|\psi_{0}|^{2}]dx.\end{split}$
(5.13)

Notice that

$\displaystyle\int_{\mathbb{R}^{N}}(V_{1}(x)|\varphi_{0}|^{2}+V_{2}(x)|\psi_{0}|^{2})dx<0.$
(5.14)

From Lemma 5.3, (5.13) and (5.14), it can be derived that

$\displaystyle e_{V}(\xi,\eta)$ $\displaystyle\leq
E_{V}(u_{0},v_{0})+E(\varphi_{0},\psi_{0})+\frac{1}{2}\int_{\mathbb{R}^{N}}(V_{1}(x)|\varphi_{0}|^{2}+V_{2}(x)|\psi_{0}|^{2})dx$
$\displaystyle<E_{V}(u_{0},v_{0})+E(\varphi_{0},\psi_{0})$
$\displaystyle=e_{V}(\xi_{1},\eta_{1})+e(\xi_{2},\eta_{2})$
$\displaystyle=e_{V}(\xi,\eta),$

which is impossible. Hence Case 1 doesn’t hold. In conclusion, $\xi_{1}=\xi$
and $\eta_{1}=\eta$. Therefore, we complete the proof of Step 1.

Step 2: $\\{(u_{n},v_{n})\\}$ is strongly convergent in
$H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$.

From Step 1, $(u_{n},v_{n})\rightarrow(u_{0},v_{0})$ strongly in
$L^{2}(\mathbb{R}^{N})\times L^{2}(\mathbb{R}^{N})$. Hence
$B(u_{n},p)\rightarrow B(u_{0},p)$ and $B(v_{n},q)\rightarrow B(v_{0},q)$ by
(2.3).

Notice that

$e_{V}(\xi,\eta)\leq
E_{V}(u_{0},v_{0})\leq\liminf_{n}E_{V}(u_{n},v_{n})=e_{V}(\xi,\eta),$

which implies that

$\displaystyle\|\nabla u_{0}\|_{2}^{2}+\|\nabla
v_{0}\|_{2}^{2}=\lim_{n\rightarrow\infty}\bigg{\\{}\|\nabla
u_{n}\|_{2}^{2}+\|\nabla v_{n}\|_{2}^{2}\bigg{\\}}.$

Hence Step 2 holds. Therefore, we establish the theorem. ∎

###  5.3 Proof of Theorem 1.5

In this subsection, we consider the case that $V_{1}(x)$ satisfies (V1) and
$V_{2}(x)$ satisfies (V2). We work on the space
$H^{1}(\mathbb{R}^{N})\times\tilde{H}_{2}$.

######  Lemma 5.4.

Assume that $V_{2}(x)$ satisfies (V2). Then $\tilde{H}_{2}\hookrightarrow
L^{p}(\mathbb{R}^{N})(p\in[2,2^{*}))$ is compact.

Similar to Lemma 3.1, we obtain that

$\displaystyle E_{V}(u,v)\geq$ $\displaystyle\frac{1}{2}(\|\nabla
u\|_{2}^{2}+\|\nabla
v\|_{2}^{2})-\frac{\mu_{1}C(N,\alpha,p)\xi^{2p(1-\delta_{p})}}{2p}\|\nabla
u\|_{2}^{2p\delta_{p}}$
$\displaystyle-\frac{\mu_{2}C(N,\alpha,q)\eta^{2q(1-\delta_{q})}}{2q}\|\nabla
v\|_{2}^{2q\delta_{q}}-\beta\xi\eta+\frac{1}{2}\int_{\mathbb{R}^{N}}(V_{1}(x)|u|^{2}+V_{2}(x)|v|^{2})dx.$

Since $\max\\{1+\frac{\alpha}{N},2\\}<p,q<1+\frac{\alpha+2}{N}$ and $(u,v)\in
H^{1}(\mathbb{R}^{N})\times\tilde{H_{2}}$, we have $E_{V}(u,v)$ is also
bounded from below. Moreover, there exists a bounded minimizing sequence
$\\{(u_{n},v_{n})\\}\in H^{1}(\mathbb{R}^{N})\times\tilde{H_{2}}$ with
$\|u_{n}\|_{2}=\xi,~{}\|v_{n}\|_{2}=\eta$. Therefore, there exists a
$(u_{0},v_{0})\in H^{1}(\mathbb{R}^{N})\times\tilde{H_{2}}$ such that
$(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ weakly in
$H^{1}(\mathbb{R}^{N})\times\tilde{H_{2}}$. Furthermore, by Lemma 5.4,
$\|v_{0}\|_{2}^{2}=\eta$.

######  Lemma 5.5.

There holds that $e_{V}(\xi,\eta)\leq
e_{V}(\xi_{1},\eta)+m(\xi-\xi_{1},\mu_{1})$, where $m(\xi-\xi_{1},\mu_{1})$ is
defined in (2.5).

###### Proof.

The proof is similar to Lemma 5.2 (iii). We omit the details. ∎

Similar to Lemma 5.3, we give the following splitting lemma.

######  Lemma 5.6.

Let $\\{(u_{n},v_{n})\\}\subset S_{\xi}\times S_{\eta}$ be a minimizing
sequence for $e_{V}(\xi,\eta)$ such that
$(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})$ in $H^{1}(\mathbb{R}^{N})\times
H^{1}(\mathbb{R}^{N})$ and let $\xi_{1}=\|u_{0}\|_{2},~{}\eta=\|v_{0}\|_{2}$.
If $\xi_{1}<\xi$, then there exist $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and
$w_{0}\in H^{1}(\mathbb{R}^{N})\setminus{0}$ such that

$\displaystyle|y_{n}|\rightarrow\infty,~{}~{}u_{n}(\cdot+y_{n})\rightharpoonup
w_{0}\quad\text{in~{}}H^{1}(\mathbb{R}^{N}),$ (5.15)
$\displaystyle\lim_{n\rightarrow\infty}\|u_{n}-u_{0}-w_{0}(\cdot-
y_{n})\|_{2}^{2}=0,$ (5.16)

and $\xi^{2}=\xi_{1}^{2}+\xi_{2}^{2}$, where $\|w_{0}\|_{2}=\xi_{2}$.
Furthermore,

$\displaystyle e_{V}(\xi,\eta)=e_{V}(\xi_{1},\eta)+m(\xi_{2},\mu_{1}),$ (5.17)

and

$\displaystyle
E_{V}(u_{0},v_{0})=e_{V}(\xi_{1},\eta),~{}E(w_{0},0)=F_{\mu_{1}}(w_{0})=m(\xi_{2},\mu_{1}).$
(5.18)

###### Proof.

We divide the proof into three steps.

Step 1: We find $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and $w_{0}$ such that
(5.15) holds.

We claim that

$\displaystyle\delta_{0}:=\liminf_{n\rightarrow\infty}\sup_{y\in\mathbb{R}^{N}}\int_{y+B(0,1)}|u_{n}-u_{0}|^{2}dx>0,$

where $B(0,1):=\\{x\in\mathbb{R}^{N}:|x|\leq 1\\}$. We prove the claim by
contradiction. In fact, we may assume that $\delta_{0}=0$. Then by (2.3),
$B(u_{n},p)\to B(u_{0},p)$. Hence

$\displaystyle e_{V}(\xi_{1},\eta)\leq
E_{V}(u_{0},v_{0})\leq\liminf_{n}E_{V}(u_{n},v_{n})=e_{V}(\xi,\eta).$

By Lemma 2.6 and Lemma 5.5, there holds that

$\displaystyle e_{V}(\xi_{1},\eta)\leq e_{V}(\xi,\eta)\leq
e_{V}(\xi_{1},\eta)+m(\xi-\xi_{1},\mu_{1})<e_{V}(\xi_{1},\eta),$

it is impossible. Hence $\delta_{0}>0$.

Furthermore, by the fact $u_{n}\rightarrow u_{0}$ in
$L_{loc}^{2}(\mathbb{R}^{N})$, we can find $\\{y_{n}\\}\subset\mathbb{R}^{N}$
such that

$\displaystyle\int_{y_{n}+B(0,1)}|u_{n}-u_{0}|^{2}dx\rightarrow
c_{0}>0~{}\text{ and}~{}~{}|y_{n}|\rightarrow\infty.$

Denote that $u_{n}(\cdot+y_{n})\rightarrow w_{0}$ weakly in
$H^{1}(\mathbb{R}^{N})$. Due to $c_{0}>0$, we can deduce that $w_{0}\neq 0$.
Therefore, $\\{y_{n}\\}$ and $w_{0}$ satisfy (5.15).

Since $y_{n}\rightarrow\infty$, we have

$\displaystyle\|u_{n}-u_{0}-w_{0}(\cdot-y_{n})\|_{2}^{2}$ $\displaystyle=$
$\displaystyle\|u_{n}\|_{2}^{2}+\|u_{0}\|_{2}^{2}+\|w_{0}\|_{2}^{2}-2\langle
u_{n},u_{0}\rangle_{L^{2}}-2\langle
u_{n}(\cdot+y_{n}),w_{0}\rangle_{L^{2}}+o(1)$ $\displaystyle=$
$\displaystyle\|u_{n}\|_{2}^{2}-\|u_{0}\|_{2}^{2}-\|w_{0}\|_{2}^{2}+o(1),$

in other words,

$\displaystyle\xi_{2}^{2}:=\|w_{0}\|_{2}^{2}\leq\liminf_{n\rightarrow\infty}(\|u_{n}\|^{2}_{2}-\|u_{0}\|^{2}_{2})=\xi^{2}-\xi_{1}^{2},$

From $c_{0}>0$, it follows that $\xi_{2}>0$.

Step 2: We show $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and $w_{0}$ such that
(5.16) holds.

Let $\delta_{1}^{2}=\lim\limits_{n\rightarrow\infty}\|u_{n}-u_{0}-w_{0}(\cdot-
y_{n})\|_{2}^{2}$. Then $\delta_{1}^{2}=\xi^{2}-\xi_{1}^{2}-\xi_{2}^{2}$. Next
we show that $\delta_{1}=0$ by contradiction. Indeed, if $\delta_{1}>0$, then

$\displaystyle\|\nabla u_{n}\|_{2}^{2}-\|\nabla u_{0}\|_{2}^{2}-\|\nabla
w_{0}\|_{2}^{2}-\|\nabla(u_{n}-u_{0}-w_{0}(\cdot-y_{n}))\|_{2}^{2}$
$\displaystyle=$ $\displaystyle-2\|\nabla u_{0}\|_{2}^{2}-2\|\nabla
w_{0}\|_{2}^{2}+2\langle\nabla u_{n},\nabla u_{0}\rangle+2\langle\nabla
u_{n}(\cdot+y_{n}),\nabla w_{0}\rangle$ $\displaystyle=$ $\displaystyle o(1).$

Furthermore,

$\displaystyle\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)(|u_{n}|^{2}-|u_{0}|_{2}^{2}-|w_{0}|^{2}-|(u_{n}-u_{0}-w_{0}(\cdot-
y_{n}))|^{2})dx=o(1).$

In addition, by Brézis-Lieb Lemma, there holds that

$\displaystyle B(u_{n},p)$ $\displaystyle=B(u_{0},p)+B(u_{n}-u_{0},p)+o(1)$
$\displaystyle=B(u_{0},p)+B(\varphi_{0},p)+B(u_{n}-u_{0}-\varphi_{0}(\cdot-
y_{n}),p)+o(1),$

and

$\int_{\mathbb{R}^{N}}\beta
u_{n}v_{n}=\int_{\mathbb{R}^{N}}\beta(u_{n}-u_{0})(v_{n}-v_{0})+\int_{\mathbb{R}^{N}}\beta
u_{0}v_{0}=o(1)+\int_{\mathbb{R}^{N}}\beta u_{0}v_{0}.$

Therefore,

$\displaystyle E_{V}(u_{n},v_{n})\geq$ $\displaystyle
E_{V}(u_{0},v_{0})+F_{\mu_{1}}(w_{0}(\cdot-
y_{n}))+\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)|w_{0}(\cdot-y_{n})|^{2}dx$
$\displaystyle+F_{\mu_{1}}(u_{n}-u_{0}-w_{0}(\cdot-
y_{n}))+\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)|u_{n}-u_{0}-w_{0}(\cdot-
y_{n})|^{2}dx.$

Since
$|y_{n}|\rightarrow\infty,~{}\lim\limits_{|x|\rightarrow\infty}V_{1}(x)=0$ and
$V_{1}(x)<0$, we can conclude that

$\displaystyle\int_{\mathbb{R}^{N}}V_{1}(x)|w_{0}(x-y_{n})|^{2}dx\rightarrow
0\text{~{}and}~{}\int_{\mathbb{R}^{N}}V_{1}(x)|u_{n}-u_{0}-w_{0}(x-y_{n})|^{2}dx\rightarrow
0,$

as $n\rightarrow\infty$, which implies that

$\displaystyle E_{V}(u_{n},v_{n})\geq
E_{V}(u_{0},v_{0})+F_{\mu_{1}}(w_{0}(\cdot-
y_{n}))+F_{\mu_{1}}(u_{n}-u_{0}-w_{0}(\cdot-y_{n}).$

Hence

$\displaystyle e_{V}(\xi,\eta)\geq
e_{V}(\xi_{1},\eta)+m(\xi_{2},\mu_{1})+m(\delta_{1},\mu_{1}).$

By Lemma 2.6 and Lemma 5.5, we can see that

$\displaystyle e_{V}(\xi,\eta)$
$\displaystyle=e_{V}(\xi_{1},\eta)+m(\xi_{2},\mu_{1})+m(\delta_{1},\mu_{1})$
$\displaystyle>e_{V}(\xi_{1},\eta)+m(\xi_{2}+\delta_{1},\mu_{1})$
$\displaystyle\geq e_{V}(\xi_{1}+\xi_{2}+\delta_{1},\eta)$
$\displaystyle=e_{V}(\xi,\eta),$

which is a contradiction. Hence $\delta_{1}=0$.

Step 3: We show $\\{y_{n}\\}\subset\mathbb{R}^{N}$ and
$(\varphi_{0},\psi_{0})$ satisfy (5.17) and (5.18).

Since $\delta_{1}=0$, we have

$\displaystyle e_{V}(\xi,\eta)$
$\displaystyle=\lim_{n\rightarrow\infty}E_{V}(u_{n},v_{n})$
$\displaystyle\geq\liminf_{n\rightarrow\infty}\bigg{(}E_{V}(u_{0},v_{0})+F_{\mu_{1}}(w_{0}(\cdot-
y_{n}))+\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)|w_{0}(\cdot-
y_{n})|^{2}dx\bigg{)}$ $\displaystyle\geq
E_{V}(u_{0},v_{0})+F_{\mu_{1}}(w_{0})$ $\displaystyle\geq
e_{V}(\xi_{1},\eta)+m(\xi_{2},\mu_{1}),$

together with Lemma 5.5, we can obtain that
$e_{V}(\xi,\eta)=e_{V}(\xi_{1},\eta)+m(\xi_{2},\mu_{1})$ and
$E_{V}(u_{0},v_{0})=e_{V}(\xi_{1},\eta),~{}F_{\mu_{1}}(w_{0})=m(\xi_{2},\mu_{1})$.
Thus we complete the proof. ∎

In the following, we give the $L^{2}-$convergence of $\\{u_{n}\\}$.

######  Lemma 5.7.

$\|u_{0}\|_{2}=\xi$.

###### Proof.

Take $(\sqrt{u_{0}^{2}+w_{0}^{2}},v_{0})\in S_{\xi}\times S_{\eta}$ as the
test function, where $w_{0}$ is defined as in Lemma 5.6. From
$\beta\int_{\mathbb{R}^{N}}\sqrt{u_{0}^{2}+w_{0}^{2}}v_{0}dx\geq\beta\int_{\mathbb{R}^{N}}u_{0}v_{0}dx$
for $w_{0}\neq 0$ and (5.11), we can derive that

$\displaystyle\begin{split}e_{V}(\xi,\eta)&\leq
E_{V}(\sqrt{u_{0}^{2}+w_{0}^{2}},v_{0})\\\ &\leq
E_{V}(u_{0},v_{0})+\frac{1}{2}\|\nabla
w_{0}\|_{2}^{2}-\frac{\mu_{1}}{2p}B(w_{0},p)+\frac{1}{2}\int_{\mathbb{R}^{N}}V_{1}(x)w_{0}^{2}dx,\\\
\end{split}$ (5.19)

due to $V_{1}(x)<0$, it means that

$\displaystyle\int_{\mathbb{R}^{N}}V_{1}(x)w_{0}^{2}dx<0,$

together with (5.19), by Lemma 5.6, it follows that

$\displaystyle e_{V}(\xi,\eta)$
$\displaystyle<E_{V}(u_{0},v_{0})+\frac{1}{2}\|\nabla
w_{0}\|_{2}^{2}-\frac{\mu_{1}}{2p}B(w_{0},p)$
$\displaystyle=e_{V}(\xi_{1},\eta)+F_{\mu_{1}}(w_{0})$
$\displaystyle=e_{V}(\xi_{1},\eta)+m(\xi_{2},\mu_{1})$
$\displaystyle=e_{V}(\xi,\eta),$

which is a contradiction. Hence $\|u_{0}\|_{2}=\xi$. In conclusion,
$(u_{n},v_{n})\rightarrow(u_{0},v_{0})$ strongly in
$L^{2}(\mathbb{R}^{N})\times L^{2}(\mathbb{R}^{N})$. ∎

Similar to the proof of Theorem 1.1, we prove Theorem 1.5.

###### Proof of Theorem 1.5.

By the facts

$\displaystyle\left\\{\begin{aligned}
&(u_{n},v_{n})\rightharpoonup(u_{0},v_{0})\text{~{}weakly
in~{}}H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N}),\\\
&(u_{n},v_{n})\rightarrow(u_{0},v_{0})\text{~{}strongly
in~{}}L^{2}(\mathbb{R}^{N})\times L^{2}(\mathbb{R}^{N}),\end{aligned}\right.$

and (2.3), it yields that $B(u_{n},p)\rightarrow B(u_{0},p)$ and
$B(v_{n},q)\rightarrow B(v_{0},q)$. Furthermore, it can be deduced that
$(u_{n},v_{n})\rightarrow(u_{0},v_{0})$ strongly in
$H^{1}(\mathbb{R}^{N})\times H^{1}(\mathbb{R}^{N})$. Hence we finish the
proof. ∎

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