\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2017 (2017), No. 199, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2017 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2017/199\hfil Schr\"odinger equations with magnetic fields] {Schr\"odinger equations with magnetic fields and Hardy-Sobolev critical exponents} \author[Z. Guo, M. Melgaard, W. Zou \hfil EJDE-2017/199\hfilneg] {Zhenyu Guo, Michael Melgaard, Wenming Zou} \address{Zhenyu Guo \newline School of Sciences, Liaoning Shihua University, Fushun 113001, China. \newline Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China} \email{guozy@163.com} \address{Michael Melgaard (corresponding author)\newline Department of Mathematics, University of Sussex, Brighton BN1 9QH, Great Britain} \email{m.melgaard@sussex.ac.uk, phone +44 (0) 1273 67 8933} \address{Wenming Zou \newline Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China} \email{wzou@math.tsinghua.edu.cn} \dedicatory{Communicated by Jerry Bona} \thanks{Submitted May 1, 2016. Published August 11, 2017.} \subjclass[2010]{35Q55, 35A15, 47J30, 81V10} \keywords{Ground state; magnetic field; concentration-compactness; \hfill\break\indent Hardy-Sobolev critical exponent} \begin{abstract} This article is motivated by problems in astrophysics. We consider nonlinear Schr\"odinger equations and related systems with magnetic fields and Hardy-Sobolev critical exponents. Under proper conditions, existence of ground state solutions to these equations and systems are established. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} Astrophysics pose a rich class of nonlinear problems, in particular, \begin{equation}\label{151001} (-i\nabla+A)^2u=\frac{|u|^{2^\ast(s)-2}u}{|x|^s},\quad u\in D^{1,2}_A(\mathbb{R}^N), \end{equation} with the Hardy-Sobolev term models the dynamics of galaxies; we refer to \cite{BadialeTarantello.2002, BattFaltenbacherHorst.1986} and the references therein. In the present paper we consider the semilinear stationary Schr\"odinger equation \eqref{151001} with a magnetic field and a Hardy-Sobolev critical exponents, but also \begin{equation}\label{151002} \begin{gathered} (-i\nabla+A)^2u-\lambda u=\displaystyle\frac{|u|^{2^\ast(s)-2}u}{|x|^s}, \quad u\in H^1_A(\Omega),\\ u=0,\quad\text{ on }\partial\Omega, \end{gathered} \end{equation} and related systems thereof, viz. \begin{equation}\label{151003} \begin{gathered} (-i\nabla+A)^2u=\mu_1\frac{|u|^{2^\ast(s)-2}u}{|x|^s} +\frac{\alpha\gamma} {2^\ast(s)}\frac{|u|^{\alpha-2}u|v|^\beta}{|x|^s},\\ (-i\nabla+B)^2v=\mu_2\frac{|v|^{2^\ast(s)-2}v}{|x|^s} +\frac{\beta\gamma} {2^\ast(s)}\frac{|u|^{\alpha}|v|^{\beta-2}v}{|x|^s},\\ u\in D^{1,2}_A(\mathbb{R}^N),\quad v\in D^{1,2}_B(\mathbb{R}^N), \end{gathered} \end{equation} and \begin{equation}\label{151004} \begin{gathered} (-i\nabla+A)^2u-\lambda_1u=\mu_1\frac{|u|^{2^\ast(s)-2}u}{|x|^s} +\frac{\alpha\gamma} {2^\ast(s)}\frac{|u|^{\alpha-2}u|v|^\beta}{|x|^s},\\ (-i\nabla+B)^2v-\lambda_2v=\mu_2\frac{|v|^{2^\ast(s)-2}v}{|x|^s} +\frac{\beta\gamma} {2^\ast(s)}\frac{|u|^{\alpha}|v|^{\beta-2}v}{|x|^s},\\ u\in H^{1}_A(\Omega),\ \ v\in H^{1}_B(\Omega),\quad u=v=0,\quad\text{on } \partial\Omega, \end{gathered} \end{equation} where $u,v:\mathbb{R}^N\to\mathbb{C},N\ge3$, $A=(A_1,\dots,A_N), B=(B_1,\dots,B_N):\mathbb{R}^N\to\mathbb{R}^N$ are magnetic vector potentials, $0\le s<2,\lambda,\lambda_1,\lambda_2,\mu_1,\mu_2,\gamma>0$, $\alpha,\beta>1$ with $\alpha+\beta=2^\ast(s):=\frac{2(N-s)}{N-2}$, and $\Omega$ is a smooth bounded domain containing the origin as an interior point. Set $-\Delta_A:=(-i\nabla+A)^2, \nabla_A:=\nabla+iA$, and \begin{gather*} D_A^{1,2}(\mathbb{R}^N):=\big\{u\in L^{2^\ast}(\mathbb{R}^N): |\nabla_Au|\in L^2(\mathbb{R}^N)\big\},\\ H_A^{1}(\Omega):=\big\{u\in L^{2}(\Omega):|\nabla_Au|\in L^2(\Omega)\big\}. \end{gather*} Then $-\Delta_Au=-\Delta u-iu\operatorname{div}A-2iA\cdot\nabla u+|A|^2u$, $D_A^{1,2}(\mathbb{R}^N)$ and $H_A^{1}(\Omega)$ are Hilbert spaces obtained by the closures of $C_c^\infty(\mathbb{R}^N,\mathbb{C})$ and $C_c^\infty(\Omega,\mathbb{C})$ with respect to scaler products \[ \operatorname{Re}\Big(\int_{\mathbb{R}^N}\nabla_Au\cdot \overline{\nabla_Av}\Big) \quad\text{and}\quad \operatorname{Re}\Big(\int_{\Omega}\nabla_Au\cdot\overline{\nabla_Av}\Big) \] respectively, where the bar denotes complex conjugation. Here and in the following, $\int\cdot$ means $\int\cdot\ \text{d}x$. We regard the range of function as $\mathbb{C}$, except the places where we emphasize that the range is $\mathbb{R}$. $L^p\big(\Omega,\frac{\text{d}x}{|x|^s}\big)$ denotes the space of $L^p$-integrable functions with respect to the measure $\frac{\text{d}x}{|x|^s}$, endowed with norm \[ |u|_{p,s}: =\Big(\int_{\Omega}\frac{|u|^{p}}{|x|^s}\Big)^{1/p}. \] For $\Omega=\mathbb{R}^N$, denote the $L^p$ norm by \[ |u|_{p,s,\mathbb{R}^N} : =\Big(\int_{\mathbb{R}^N}\frac{|u|^{p}}{|x|^s}\Big)^{1/p}. \] Write $|u|_{p}:=|u|_{p,0}$ and $|u|_{p,\mathbb{R}^N}:=|u|_{p,0,\mathbb{R}^N}$ for simplicity. Define \begin{gather*} \mu_s^A(\mathbb{R}^N):=\inf_{u\in D_A^{1,2}(\mathbb{R}^N)\setminus\{0\}} \frac{|\nabla_Au|^2_{2,\mathbb{R}^N}}{|u|^2_{2^\ast(s),s,\mathbb{R}^N}},\\ \mu_s^{A,\lambda}(\Omega):=\inf_{u\in H_A^{1}(\Omega)\setminus\{0\}} \frac{|\nabla_Au|^2_2-\lambda|u|_2^2}{|u|^2_{2^\ast(s),s}}. \end{gather*} The first existence results for this kind of problems with a magnetic potential (i.e., $A\in L^2_{\rm loc}$) were established in the seminal work \cite{EstebanLions.1989}. Leaving aside periodic and singular magnetic fields, a number of papers dealt with nonlinear Schr\"odinger equations with regular fields, for example, \cite{CaoTang.2006, EnstedtMelgaard.2012, Kurata.2000, Pankov.2003, Shirai08, Tang.2008}, including \cite{AlvesFigueiredo.2014, ArioliSzulkin.2003, ChabrowskiSzulkin.2005, Han.2006, LiangZhang.2011, Wang.2008} for the critical Sobolev exponent and \cite{ClappSzulkin.2013} for the critical Hardy exponent. As far as we know, there are no results for problems of this type with Hardy-Sobolev critical exponents, in particular for the system case. The Hardy-Sobolev term has the same homogeneity as the Laplacian but it does not belong to the Kato class and, therefore, the resulting functional lacks compactness. The present paper is mainly motivated by \cite{ArioliSzulkin.2003}; we apply existence results of ground state solutions obtained in \cite{ChenZou.2015, GhoussoubYuan.2000, GZ, ZZ} to extend \cite[Theorems 1.1 and 1.2]{ArioliSzulkin.2003} to the case of Hardy-Sobolev critical exponent and also systems; it is worth to emphasize that systems are not considered in \cite{ArioliSzulkin.2003}. First, we establish results for single equations. \begin{theorem}\label{t15101} If $A\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N)$, then $\mu_s^A(\mathbb{R}^N)$ is attained by a $u\in D_A^{1,2}(\mathbb{R}^N)\setminus\{0\}$ if and only if $\operatorname{curl}A\equiv0$, where $\operatorname{curl}A$ is the usual curl operator for $N=3$ and the $N\times N$ skew-symmetric matrix with entries $a_{jk}=\partial_jA_k-\partial_kA_j$ for $N\ge4$. \end{theorem} \begin{theorem}\label{t15102} Assume that \begin{itemize} \item[(A1)] $A\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N),\operatorname{curl}A\equiv0$ or \item[(A2)] $A\in L_{\rm loc}^2(\mathbb{R}^N,\mathbb{R}^N), A$ is continuous at $0$ \end{itemize} holds. Let $N\ge4$ and $\sigma(-\Delta_A-\lambda)\subset(0,+\infty)$, where $\sigma(\cdot)$ is the spectrum in $L^2(\mathbb{R}^N)$. Then $\mu_s^{A,\lambda}(\Omega)$ is attained by some $u\in H_A^1(\Omega)\setminus\{0\}$. \end{theorem} Second, we establish results for systems. For this purpose we define \begin{gather*} \bar\mu_s^{A,B}(\mathbb{R}^N):=\inf_{(u,v)\in D_{A,B}\setminus\{(0,0)\}} \frac{\|(u,v)\|_{D_{A,B}}^2} {\big(\int_{\mathbb{R}^N} \big(\mu_1\frac{|u|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}},\\ \bar\mu_s^{A,B}(\Omega):=\inf_{(u,v)\in H_{A,B}\setminus\{(0,0)\}} \frac{\|(u,v)\|_{H_{A,B}}^2} {\big(\int_{\Omega} \big(\mu_1\frac{|u|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}, \end{gather*} where $D_{A,B}:=D_A^{1,2}(\mathbb{R}^N)\times D_B^{1,2}(\mathbb{R}^N)$, endowed with norm $$ \|(u,v)\|_{D_{A,B}}^2:=|\nabla_Au|_{2,\mathbb{R}^N}^2 +|\nabla_Bv|_{2, \mathbb{R}^N}^2, $$ and $H_{A,B}:=H_A^{1}(\Omega)\times H_B^{1}(\Omega)$, endowed with norm $$ \|(u,v)\|_{H_{A,B}}^2:=|\nabla_Au|_{2}^2-\lambda_1|u|_2^2+|\nabla_Bv|_{2}^2 -\lambda_2 |v|_2^2. $$ Then we have the following result. \begin{theorem}\label{t15103} Assume that $A,B\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N)$ and \begin{itemize} \item[(A3)] $N\ge3,1<\alpha,\beta<2,\gamma>0$ holds. \end{itemize} Then $\bar\mu_s^{A,B}(\mathbb{R}^N)$ is attained by some $(u,v)\in D_{A,B}$ such that $u\not\equiv0,v\not\equiv0$ if and only if $\operatorname{curl}A\equiv0\equiv\operatorname{curl}B$. \end{theorem} \begin{theorem}\label{t15104} Assume that {\rm (A3)} is satisfied and \begin{itemize} \item[(A4)] $A,B\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N), \operatorname{curl}A\equiv0\equiv \operatorname{curl}B$, or \item[(A5)] $A,B\in L_{\rm loc}^2(\mathbb{R}^N,\mathbb{R}^N), A$ and $B$ are continuous at $0$ \end{itemize} holds. If $\sigma(-\Delta_A-\lambda_1), \sigma(-\Delta_B-\lambda_2) \subset(0,+\infty)$ and $N\ge4$, then $\bar\mu_s^{A,B}(\Omega)$ is attained by some $(u,v)\in H_{A,B}$ such that $u\not\equiv0$, $v\not\equiv0$. \end{theorem} The corresponding energy functionals $I:D_{A,B}\to\mathbb{R}$ and $E:H_{A,B}\to\mathbb{R}$ of \eqref{151003} and \eqref{151004} are \begin{align*} &I(u,v)\\ &=\frac{1}{2}\|(u,v)\|^2_{D_{A,B}} -\frac{1}{2^\ast(s)}\Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} + \gamma \int_{\mathbb{R}^N}\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big), \end{align*} and \[ E(u,v)=\frac{1}{2}\|(u,v)\|^2_{H_{A,B}} -\frac{1}{2^\ast(s)}\Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} + \gamma \int_{\Omega}\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big), \] respectively. Define \begin{align*} \mathcal{N}:=\Big\{&(u,v)\in D_{A,B}\setminus\{(0,0)\}: \|(u,v)\|^2_{D_{A,B}}\\ &=\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\gamma\int_{\mathbb{R}^N} \frac{|u|^\alpha|v|^\beta}{|x|^s}\Big\},\\ \mathcal{M}:=\Big\{&(u,v)\in H_{A,B}\setminus\{(0,0)\}: \|(u,v)\|^2_{H_{A,B}}\\ &=\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_{\Omega} \frac{|u|^\alpha|v|^\beta}{|x|^s}\Big\}, \end{align*} $M_0:=\inf_{(u,v)\in\mathcal{N}}I(u,v)$ and $M:=\inf_{(u,v)\in\mathcal{M}}E(u,v)$. By nontrivial solutions $(u,v)\in D_{A,B}$ of \eqref{151003}, we mean $u\neq0,v\neq0$. A solution of \eqref{151003} is called a ground state solution if $(u,v)\in\mathcal{N}$ and $I(u,v)=M_0$. A ground state solution is semi-trivial if it is of type $(u,0)$ or $(0,v)$. Similar definitions applies to \eqref{151004} and single equations \eqref{151001} and \eqref{151002}. For ground states, we obtain \begin{theorem}\label{t15105} If {\rm (A1)} holds, then \eqref{151001} has a nontrivial ground state solution with energy $M_1:=\frac{2-s}{2(N-s)}\big( \mu_s^A(\mathbb{R}^N)\big)^\frac{N-s}{2-s}$. \end{theorem} \begin{theorem}\label{t15106} Assume that {\rm (A1)} or {\rm (A2)} holds. If $N\ge4$ and $\sigma(-\Delta_A-\lambda)\subset(0,+\infty)$, then \eqref{151002} has a nontrivial ground state solution with energy given by $M_2:=\frac{2-s}{2(N-s)}\big( \mu_s^{A,\lambda}(\Omega)\big)^\frac{N-s}{2-s}$. \end{theorem} \begin{theorem}\label{t15107} If {\rm (A3)} and {\rm (A4)} hold, then \eqref{151003} has a nontrivial ground state solution with energy given by $M_0:=\frac{2-s}{2(N-s)}\big(\bar\mu_s^{A,B}(\mathbb{R}^N)\big)^\frac{N-s}{2-s}$. \end{theorem} \begin{theorem}\label{t15108} Assume that {\rm (A3)} and one of {\rm (A4)} and {\rm (A5)} hold. If $$ \sigma(-\Delta_A-\lambda_1),\sigma(-\Delta_B-\lambda_2)\subset(0,+\infty) $$ and $N\ge4$, then \eqref{151004} has a nontrivial ground state solution with energy $M:=\frac{2-s}{2(N-s)}\big(\bar\mu_s^{A,B}(\Omega)\big)^\frac{N-s}{2-s}$. \end{theorem} \begin{remark} \label{rmk1.1} \rm Although the symmetric and decaying information about ground state solutions of equations with magnetic fields is not known, the existence of ground state solutions is heavily dependent on that of equations without magnetic fields, under proper conditions, such as (A1)--(A5). \end{remark} Consider the nonlinear system \begin{equation}\label{151012} \begin{gathered} \mu_1k^{\frac{2^\ast(s)-2}{2}}+\frac{\alpha\gamma}{2^\ast(s)} k^{\frac{\alpha-2}{2}}l^{\beta/2}=1,\\ \mu_2l^{\frac{2^\ast(s)-2}{2}}+\frac{\beta\gamma}{2^\ast(s)} k^{\frac{\alpha}{2}}l^{\frac{\beta-2}{2}}=1,\\ k>0,\quad l>0. \end{gathered} \end{equation} \begin{theorem}\label{t15109} Assume that {\rm (A4)} and \begin{itemize} \item[(A6)] $N\ge4$, $1<\alpha,\beta<2$, and \[ \gamma\ge\frac{2(N-s)(2-s)}{(N-2)^2}\max\Big\{\frac{\mu_1}{\alpha} \Big(\frac {2-\beta} {2-\alpha}\Big)^{\frac{2-\beta}{2}}, \frac{\mu_2}{\beta}\Big(\frac{2-\alpha} {2-\beta}\Big) ^{\frac{2-\alpha}{2}}\Big\}. \] \end{itemize} If $A=B$, then \eqref{151003} has a nontrivial ground state solution $(\sqrt{k_0}U,\sqrt{l_0}U)$ with energy $M_0=\frac{2-s}{2(N-s)}(k_0+l_0)\big(\mu_s(\mathbb{R}^N)\big)^\frac{N-s}{2-s}$, where $U$ is a nontrivial ground state solution of \eqref{151001}, obtained in Theorem \ref{t15105}, \begin{equation}\label{151013} (k_0,l_0)\ \text{satisfies \eqref{151012} and } k_0=\min\{k:(k,l) \text{ is a solution of \eqref{151012}}\}. \end{equation} That is, $M_0$ is attained at $(\sqrt{k_0}U,\sqrt{l_0}U)$. \end{theorem} \begin{theorem}\label{t15110} Assume that {\rm (A6)} and either {\rm (A4)} or {\rm (A5)} holds. If $A=B$, $\lambda_1=\lambda_2=\lambda$, and $\sigma(-\Delta_A-\lambda)\subset(0,+\infty)$, then \eqref{151004} has a nontrivial ground state solution $(\sqrt{k_0}\omega,\sqrt{l_0}\omega)$ with energy $M=\frac{2-s}{2(N-s)}(k_0+l_0)\big( \mu_s^{A,\lambda} (\Omega)\big)^\frac{N-s}{2-s}$, where $(k_0,l_0)$ satisfies \eqref{151013} and $\omega$ is a nontrivial ground state solution of \eqref{151002}, obtained in Theorem \ref{t15106}. That is, $M$ is attained at $(\sqrt{k_0}\omega,\sqrt{l_0}\omega)$. \end{theorem} \begin{remark} \label{rmk1.2} \rm By \cite[Lemma 1.1]{Leinfelder.1983}, we see that the above theorems also hold when conditions (A1) and (A4) are replaced with (A1') and (A4') respectively: \begin{itemize} \item[(A1')] $A\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N)$, there exists $\varphi\in W_{\rm loc}^{1,N}(\mathbb{R}^N,\mathbb{R})$ such that $\nabla \varphi=A$, \item[(A4')] $A,B\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N)$, there exist $\varphi,\psi\in W_{\rm loc}^{1,N}(\mathbb{R}^N,\mathbb{R})$ such that $\nabla \varphi=A, \nabla\psi=B$. \end{itemize} For more details, we refer to \cite[Theorem 3.7]{EstebanLions.1989} and the proof of Theorem \ref{t15103} in this paper. \end{remark} The paper is organized as follows. In Section 2, we establish several auxiliary results for the proof of our main results; key ingredients are Lemma~\ref{l15101} and Lemma~\ref{l15105}, not found elsewhere. The latter is proven by using Ekeland's variational principle. In Section 3, we discuss the attainability of the infimum defined above by applying the method of concentration-compactness. The existence of ground state solution to the Schr\"odinger problems is studied in Section 4. Finally, in Section 5 we consider a magnetic field in three dimensions as an application of some of the above theorems. \section{Preliminaries} Define \[ \mu_s(\mathbb{R}^N):=\inf_{u\in D^{1,2}(\mathbb{R}^N)\setminus\{0\}} \frac{|\nabla u|_{2,\mathbb{R}^N}^2}{|u|^2_{2^\ast(s),s,\mathbb{R}^N}}, \] where $D^{1,2}(\mathbb{R}^N)=\{u\in L^{2^\ast(\mathbb{R}^N)} :|\nabla u|\in L^2(\mathbb{R}^N)\}$. Then, by \cite{GhoussoubYuan.2000}, $\mu_s(\mathbb{R}^N)$ is attained by functions of form \[%\label{151005} y_{\varepsilon}(x):=\big((N-s)(N-2)\big)^{\frac{N-2}{2(2-s)}} \varepsilon^\frac{N-2}{2} \big(\varepsilon^{2-s}+|x|^{2-s}\big)^{-\frac{N-2}{2-s}}, \] where $\varepsilon>0$. The function $y_\varepsilon$ is a positive solution of $-\Delta u=\frac{|u|^{2^\ast(s)-2}u}{|x|^s}$, and moreover, \[ |\nabla y_\varepsilon|_{2,\mathbb{R}^N}^2 =|y_\varepsilon|_{2^\ast(s),s,\mathbb{R}^N} ^{2^\ast(s)} =\big(\mu_s(\mathbb{R}^N)\big)^\frac{N-s}{2-s}. \] Define \[ \bar\mu_s(\mathbb{R}^N):=\inf_{(u,v)\in D\setminus\{(0,0)\}} \frac{\|(u,v)\|_{D}^2} {\big(\int_{\mathbb{R}^N} \big(\mu_1\frac{|u|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}, \] where $D:=D^{1,2}(\mathbb{R}^N)\times D^{1,2}(\mathbb{R}^N)$, endowed with norm $$ \|(u,v)\|_{D}^2:=|\nabla u|_{2,\mathbb{R}^N}^2 +|\nabla v|_{2, \mathbb{R}^N}^2. $$ Then, by Lemma \ref{l15105}, \cite{ChenZou.2015, GZ} ($s=0$) and \cite{ZZ} ($00$, let $\Omega_\varepsilon:=\Omega\cap B_\varepsilon$ and $\Omega_\varepsilon^c:=\Omega\setminus \Omega_\varepsilon$, where $B_\varepsilon$ is the ball centered at 0 with radius $\varepsilon$. Noting $N-\frac{2^\ast s}{2^\ast-p}>0$, we have \begin{equation}\label{151006} \begin{aligned} \int_{\Omega_\varepsilon}\frac{|u_n-u|^p}{|x|^s} &\le\Big( \int_{\Omega_\varepsilon}|u_n-u|^{2^\ast}\Big)^\frac{p}{2^\ast} \Big( \int_{\Omega_\varepsilon}|x|^{-\frac{2^\ast s}{2^\ast-p}} \Big)^\frac{2^\ast-p}{2^\ast}\\ &\le C\Big( \int_0^\varepsilon r^{-\frac{2^\ast s}{2^\ast-p}}r^{N-1} \text{d}r\Big)^\frac{2^\ast-p}{2^\ast}\\ &=O\big(\varepsilon^\frac{(N-2)(2^\ast(s)-p)}{2}\big). \end{aligned} \end{equation} On the other hand, for any $x\in\Omega_\varepsilon^c$, there exists a constant $C_\varepsilon>0$ such that $\frac{1}{|x|^s}\le C_\varepsilon$. It follows from Rellich-Kondrachov Compactness Theorem that $\int_{\Omega_\varepsilon^c}\frac{|u_n-u|^p}{|x|^s}=o(1)$. Combining this and \eqref{151006}, we get that $\lim_{n\to\infty}|u_n-u|_{p,s}^p=0$. \end{proof} \begin{remark} \label{rmk2.1} \rm If $\sigma(-\Delta_A-\lambda_1), \sigma(-\Delta_B-\lambda_2)\subset(0,+\infty)$, then by Lemma \ref{l15101}, it is standard to see that the quantities $$ \mu_s^{A}(\mathbb{R}^N),\mu_s^{A,\lambda_1}(\Omega), \mu_s^{B,\lambda_2}(\Omega), \bar\mu_s^{A,B}(\mathbb{R}^N),\mbox{ and } \bar\mu_s^{A,B}(\Omega) $$ are strictly positive. \end{remark} \begin{lemma}\label{l15102} If $A,B\in L^N_{\rm loc}(\mathbb{R}^N,\mathbb{R}^N)$, then $\mu_s^A(\mathbb{R}^N)=\mu_s(\mathbb{R}^N)$ and $\bar\mu_s^{A,B}(\mathbb{R}^N)=\bar\mu_s(\mathbb{R}^N)$. \end{lemma} \begin{proof} We only prove the latter equality. For any $(u,v)\in D_{A,B}\setminus\{(0,0)\}$, by the diamagnetic inequality, we have \begin{align*} &\frac{|\nabla_Au|_{2,\mathbb{R}^N}^2+|\nabla_Bv|_{2,\mathbb{R}^N}^2} {\big(\int_{\mathbb{R}^N}\big(\mu_1\frac{|u|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}\\ &\ge \frac{\big|\nabla|u|\big|_{2,\mathbb{R}^N}^2 +\big|\nabla|v|\big| _{2,\mathbb{R}^N}^2} {\big(\int_{\mathbb{R}^N}\big(\mu_1\frac{|u|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}\\ &\ge \bar\mu_s(\mathbb{R}^N), \end{align*} which implies that $\bar\mu^{A,B}_s(\mathbb{R}^N)\ge\bar\mu_s(\mathbb{R}^N)$. Define \begin{equation}\label{151026} \big(U_\varepsilon(x),V_\varepsilon(x)\big):=\Big(\varepsilon ^{-\frac{N-2}{2}} U\big(\frac{x}{\varepsilon}\big),\varepsilon^{-\frac{N-2}{2}} V\big(\frac{x}{\varepsilon}\big)\Big) \end{equation} and \[ \big(u_\varepsilon(x),v_\varepsilon(x)\big):=\big(\phi(x)U_\varepsilon(x) , \phi(x)V_\varepsilon(x)\big), \] where $(U,V)$ achieves $\bar\mu_s(\mathbb{R}^N)$ with \eqref{11}, and $\phi\in C_0^1(B_{2})$ is a cut-off function satisfying $\phi\equiv1$ on $B_1$. Then, a direct computation yields \begin{gather} \int_\Omega|\nabla u_\varepsilon|^2 \le\int_{\mathbb{R}^N}|\nabla U|^2+O(\varepsilon^{N-2}),\label{151021}\\ \int_\Omega|\nabla v_\varepsilon|^2 \le\int_{\mathbb{R}^N}|\nabla V|^2+O(\varepsilon^{N-2}),\label{151022}\\ \int_\Omega\frac{|u_\varepsilon|^{2^\ast(s)}}{|x|^s} \ge\int_{\mathbb{R}^N}\frac {|U|^{2^\ast(s)}}{|x|^s} +O(\varepsilon^{N-s}),\label{151023}\\ \int_\Omega\frac{|v_\varepsilon|^{2^\ast(s)}}{|x|^s} \ge\int_{\mathbb{R}^N}\frac{ |V|^{2^\ast(s)}}{|x|^s} +O(\varepsilon^{N-s}),\label{151024}\\ \int_\Omega\frac{|u_\varepsilon|^\alpha |v_\varepsilon|^\beta}{|x|^s} \ge\int_{\mathbb{R}^N}\frac{ |U|^\alpha |V|^\beta}{|x|^s}+O(\varepsilon^{N-s}). \label{151025} \end{gather} It follows from $\{u_\varepsilon\}$ that it is bounded in $L^{2^\ast}(\mathbb{R}^N)$ and $u_\varepsilon\to 0$ a.e. in $\mathbb{R}^N$ as $\varepsilon\to0$ that for any $\varphi\in L^\frac{2^\ast}{2^\ast-1}(\mathbb{R}^N)$, \[ \Big|\int_{\mathbb{R}^N}u_\varepsilon\varphi\Big| \le\Big(\int_{\mathbb{R}^N} u_\varepsilon^{2^\ast}\Big)^\frac{1}{2^\ast} \Big(\int_{\mathbb{R}^N} |\varphi|^\frac{2^\ast}{2^\ast-1} \Big)^\frac{2^\ast-1}{2^\ast}\to0, \] i.e., $u_\varepsilon\rightharpoonup0$ weakly in $L^{2^\ast}(\mathbb{R}^N)$. Hence, $u_\varepsilon^2\rightharpoonup0$ weakly in $L^\frac{2^\ast}{2}(\mathbb{R}^N)$. Since $|A|^2\in L_{\rm loc}^\frac{N}{2}(\mathbb{R}^N) = \big(L_{\rm loc}^\frac{2^\ast}{2}(\mathbb{R}^N)\big)'$, the dual space of $L_{\rm loc}^\frac{2^\ast}{2}(\mathbb{R}^N)$, we obtain \[ \int_{\mathbb{R}^N}|Au_\varepsilon|^2=\langle|A|^2,u_\varepsilon^2\rangle\to0, \] where the duality product is taken with respect to $L^\frac{N}{2}(\mathbb{R}^N)$ and $L^\frac{2^\ast}{2}(\mathbb{R}^N)$. Similarly, we have $\int_{\mathbb{R}^N}|Bv_\varepsilon|^2\to0$ as $\varepsilon\to0$. Let $\delta>0$. For $\varepsilon$ small enough, noting that $u_\varepsilon$ and $v_\varepsilon$ are real-valued, by \eqref{151021}--\eqref{151025}, we have \begin{align*} \bar\mu_s^{A,B}(\mathbb{R}^N)&\leq \frac{|\nabla_Au_\varepsilon|_{2,\mathbb{R}^N}^2 +|\nabla_Bv_\varepsilon|_{2,\mathbb{R}^N}^2} {\big(\int_{\mathbb{R}^N}\big(\mu_1\frac{|u_\varepsilon|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v_\varepsilon|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u_\varepsilon|^\alpha|v_\varepsilon|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}\\ &= \frac{\int_{\mathbb{R}^N}\big(|\nabla u_\varepsilon|^2 +|Au_\varepsilon|^2+|\nabla v_\varepsilon|^2 +|Bv_\varepsilon|^2\big)}{\big(\int_{\mathbb{R}^N} \big(\mu_1\frac{|u_\varepsilon|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v_\varepsilon|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u_\varepsilon|^\alpha|v_\varepsilon|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}\\ &\le \frac{\int_{\mathbb{R}^N}\big(|\nabla U|^2+|\nabla V|^2 +|Au_\varepsilon|^2+|Bv_\varepsilon|^2\big)+O(\varepsilon^{N-2})} {\big(\int_{\mathbb{R}^N} \big(\mu_1\frac{|U|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|V|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|U|^\alpha|V|^\beta}{|x|^s}\big)+O(\varepsilon^{N-s}) \big)^\frac{2}{2^\ast(s)}}\\ &\le \bar\mu_s(\mathbb{R}^N)+\delta, \end{align*} which implies that $\bar\mu_s^{A,B}(\mathbb{R}^N)\le\bar\mu_s(\mathbb{R}^N)$. Therefore, $\bar\mu_s^{A,B}(\mathbb{R}^N)=\bar\mu_s(\mathbb{R}^N)$. \end{proof} \begin{lemma}\label{l15105} The following conclusions hold. \begin{itemize} \item[(i)] $\mu_s^A(\mathbb{R}^N)$ is attained if and only if \eqref{151001} has a nontrivial ground state solution; \item[(ii)] $\mu_s^{A,\lambda}(\Omega)$ is attained if and only if \eqref{151002} has a nontrivial ground state solution; \item[(iii)] $\bar\mu_s^{A,B}(\mathbb{R}^N)$ is attained by $(u,v)\in D_{A,B}$ with $u\not\equiv0, v\not\equiv0$ if and only if \eqref{151003} has a nontrivial ground state solution; \item[(iv)] $\bar\mu_s^{A,B}(\Omega)$ is attained by $(u,v)\in H_{A,B}$ with $u\not\equiv0, v\not\equiv0$ if and only if \eqref{151004} has a nontrivial ground state solution. \end{itemize} \end{lemma} \begin{proof} We only prove (iv). Setting \[ F(u,v):= \frac{\|(u,v)\|_{H_{A,B}}^2} {\big(\int_{\Omega} \big(\mu_1\frac{|u|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|v|^{2^\ast(s)}}{|x|^s} +\gamma\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) \big)^\frac{2}{2^\ast(s)}}, \] then $\bar\mu_s^{A,B}(\Omega)=\inf_{(u,v)\in H_{A,B}\setminus\{(0,0)\}}F(u,v)$ and $F(tu,tv)=F(u,v)$ for any $t\in\mathbb{R}$. Obviously, for any $(u,v)\in H_{A,B}\setminus\{(0,0)\}$, there exists an unique \[ t_{u,v}=\Big(\frac{\|(u,v)\|^2_{H_{A,B}}}{\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_{\Omega} \frac{|u|^\alpha|v|^\beta}{|x|^s}}\Big)^\frac{1}{2^\ast(s)-2} \] such that $(t_{u,v}u,t_{u,v}v)\in\mathcal{M}$. Therefore, \begin{align*} \bar\mu_s^{A,B}(\Omega) &=\inf_{(u,v)\in H_{A,B}\setminus\{(0,0)\}}F(t_{u,v}u, t_{u,v}v)\\ &=\inf_{(u,v)\in\mathcal{M}}F(u,v) =\inf_{(u,v)\in\mathcal{M}}\|(u,v)\| ^\frac{2^\ast(s)-2}{2^\ast(s)}_{H_{A,B}}. \end{align*} Noting that $M=\frac{2-s}{2(N-s)}\inf_{(u,v)\in\mathcal{M}}\|(u,v)\|^2_{H_{A,B}}$, we see that $\bar\mu_s^{A,B}(\Omega)$ is attained if and only if $M$ is attained. Assume that \eqref{151004} has a nontrivial ground state solution, i.e., $M$ is attained by a nontrivial element in $H_{A,B}$. Then, $\bar\mu_s^{A,B}(\Omega)$ is attained by some $(u,v)\in H_{A,B}$ with $u\not\equiv0$ and $v\not\equiv0$. On the other hand, assume that $\bar\mu_s^{A,B}(\Omega)$ is achieved by a nontrivial element in $H_{A,B}$. Then, there exists $(u,v)\in H_{A,B}$ with $u\not\equiv0$ and $v\not\equiv0$ such that $M=\inf_{\mathcal{M}}E=E(u,v)$. It remains to show that $(u,v)$ is a solution of \eqref{151004}. It is easy to see that $E|_{\mathcal{M}}\in C^1(\mathcal{M},\mathbb{R})$ is bounded below. By Ekeland's variational principle (e.g. \cite{Willem.1996}), for $\varepsilon,\delta>0$, there exists $(u',v')\in\mathcal{M}$ such that \begin{equation}\label{151018} E(u',v')\le E(u,v)+2\varepsilon,\ \|E'(u',v')\|_{H'_{A,B}} <\frac{8\varepsilon}{\delta},\ \|(u',v')-(u,v)\|_{H_{A,B}}\le2\delta. \end{equation} Choosing $\varepsilon_n=\frac{1}{n}$ and $\delta_n=\frac{1}{\sqrt{n}}$ in \eqref{151018}, there exists $\{(u_n,v_n)\}$ such that $(u_n,v_n)\to(u,v)$ in $H_{A,B}$, $E'(u_n,v_n)\to0$ in $H_{A,B}'$, and $E(u_n,v_n)\to E(u,v)$, as $n\to\infty$. Hence, $E'(u,v)=0$ in $H_{A,B}'$, that is $(u,v)$ is a solution of \eqref{151004}. \end{proof} \section{Attainability of the infimum} Since the proofs of Theorems \ref{t15101} and \ref{t15102} are similar to that of \cite[Theorems 1.1 and 1.2]{ArioliSzulkin.2003} and easier than that of Theorems \ref{t15103} and \ref{t15104} in the present paper, we only prove Theorems \ref{t15103} and \ref{t15104} here. Note that systems are not treated in \cite{ArioliSzulkin.2003} and the concentration-compactness arguments therein, going back to Willem \cite{Willem.1996}, has to be combined with new arguments in order to treat these systems. \begin{proof}[Proof of Theorem \ref{t15103}] (Necessary condition) Let $(u,v)$ be a minimizer of $\bar\mu_s^{A,B}(\mathbb{R}^N)$ normalized by $\mu_1|u|_{2^\ast(s),s,\mathbb{R}^N}+\mu_2|v|_{2^\ast(s),s, \mathbb{R}^N} +\gamma\int_{\mathbb{R}^N}\frac{|u|^\alpha|v|^\beta}{|x|^s}=1$. By the diamagnetic inequality and Lemma \ref{l15102}, we have \begin{align*} \bar\mu_s^{A,B}(\mathbb{R}^N) &=\int_{\mathbb{R}^N}\big(|\nabla_Au|^2+|\nabla_Bv|^2 \big)\\ &\ge \int_{\mathbb{R}^N}\big(\big|\nabla|u|\big|^2 +\big|\nabla|v|\big|^2\big)\\ &\ge\bar\mu_s(\mathbb{R}^N) =\bar\mu_s^{A,B}(\mathbb{R}^N), \end{align*} which means the above inequality must be equality and \begin{gather*} |\nabla_Au|=\big|\nabla|u|\big| =\Big|\operatorname{Re} \Big(\nabla u\frac{\overline{u}}{|u|}\Big)\Big| =\Big|\operatorname{Re}\Big((\nabla u+iAu)\frac{\overline{u}}{|u|}\Big)\Big|,\\ |\nabla_Bv|=\big|\nabla|v|\big| =\Big|\operatorname{Re}\Big(\nabla v\frac{\overline{v}}{|v|}\Big)\Big| =\Big|\operatorname{Re}\Big((\nabla v+iBv)\frac{\overline{v}}{|v|}\Big)\Big|. \end{gather*} Then, we deduce that $\operatorname{Im}\big(\nabla u\frac{\overline{u}}{|u|}\big)=0$ and $\operatorname{Im}\big(\nabla v\frac{\overline{v}}{|v|}\big)=0$, which are equivalent to $A=-\operatorname{Im}\big(\frac{\nabla u}{u}\big)$ and $B=-\operatorname{Im}\big(\frac{\nabla v}{v}\big)$. Since $\operatorname{curl}\big(\frac{\nabla u}{u}\big)=0$ and $\operatorname{curl}\big(\frac{\nabla v}{v}\big)=0$, we infer that $\operatorname{curl}A=0$ and $\operatorname{curl}B=0$. (Sufficient condition) Assume that $\operatorname{curl}A=0$ and $\operatorname{curl}B=0$. By \cite[Lemma 1.1]{Leinfelder.1983}, there exist $\varphi,\psi\in W_{\rm loc}^{1,N}(\mathbb{R}^N,\mathbb{R})$ such that $\nabla \varphi=A, \nabla\psi=B$. Let $$ \big(u_\varepsilon(x),v_\varepsilon(x)\big)=\big(U_\varepsilon(x) e^{-i\varphi(x)},V_\varepsilon(x)e^{-i\psi(x)}\big), $$ where $\varepsilon>0$ and $(U_\varepsilon,V_\varepsilon)$ is defined in \eqref{151026}. It follows from Lemma \ref{l15102} that $(u_\varepsilon,v_\varepsilon)$ is a minimizer for $\bar\mu_s^{A,B}(\mathbb{R}^N)$. \end{proof} \begin{lemma}\label{l15106} If {\rm (A1)} or {\rm (A2)} holds, $N\ge4$, and $\sigma(-\Delta_A-\lambda_1), \sigma(-\Delta_B-\lambda_2)\subset(0,+\infty)$, then $\bar\mu_s^{A,B}(\Omega)<\min\Big\{\mu_1^{-\frac{2}{2^\ast(s)}} \mu_s^{A,\lambda_1}(\Omega),\ \mu_2^{-\frac{2}{2^\ast(s)}}\mu_s^{B,\lambda_2} (\Omega)\Big\}$. \end{lemma} \begin{proof} By Theorem \ref{t15102}, we assume that $u_{\mu_1}$ achieves $\mu_s^{A,\lambda_1}(\Omega)$ with $|u_{\mu_1}|_{2^\ast(s),s}=\big(\frac{\mu_s^{A,\lambda_1} (\Omega)}{\mu_1}\big) ^\frac{1}{2^\ast(s)-2}$. Define $t(\epsilon):=t_{u_{\mu_1},\epsilon u_{\mu_1}}$, i.e., $$ t(\epsilon)=\Big(\frac{\|u_{\mu_1}\|^2_{H_A}+\epsilon^2 \|u_{\mu_1}\|^2_{H_B}} {\big(\mu_1+\mu_2|\epsilon|^{2^\ast(s)}+\gamma |\epsilon|^\beta\big) |u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s}}\Big) ^{\frac{1}{2^\ast(s)-2}}, $$ where $\|u\|^2_{H_A}:=|\nabla_Au|_2^2-\lambda_1|u|^2_2$ and $\|u\|^2_{H_B}:=|\nabla_Bu|_2^2-\lambda_2|u|^2_2$. It is easy to see that $\big(t(\epsilon)u_{\mu_1},t(\epsilon)\epsilon u_{\mu_1}\big)\in\mathcal{M}$. Noting that $\|u_{\mu_1}\|^2_{H_A}=\mu_1|u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s}$ and $t(0)=1$, we deduce that $$ \lim_{\epsilon\to0}\frac{t'(\epsilon)}{|\epsilon|^{\beta-2}\epsilon} =-\frac{\gamma\beta} {\big(2^\ast(s)-2\big)\mu_1}, $$ that is, $$ t'(\epsilon)=-\frac{\gamma\beta|\epsilon|^{\beta-2}\epsilon} {\big(2^\ast(s)-2\big)\mu_1}\big(1+o(1)\big), \quad\text{as}\ \epsilon\to0. $$ Then $$ t(\epsilon)=1-\frac{\gamma|\epsilon|^{\beta}} {\big(2^\ast(s)-2\big)\mu_1}\big(1+o(1)\big),\quad\text{as } \epsilon\to0, $$ and hence, $$ t(\epsilon)^{2^\ast(s)}=1-\frac{2^\ast(s)\gamma|\epsilon|^{\beta}} {\big(2^\ast(s)-2\big)\mu_1}\big(1+o(1)\big), \quad\text{as } \epsilon\to0. $$ Thus, we have \begin{align*} &t(\epsilon)^{2^\ast(s)} \big(\mu_1+\mu_2|\epsilon|^{2^\ast(s)} +\gamma |\epsilon|^\beta\big)|u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s}\\ &=\Big(1-\frac{2^\ast(s)\gamma|\epsilon|^{\beta}} {\big(2^\ast(s)-2\big)\mu_1} \big(1+o(1)\big)\Big) \big(\mu_1+\mu_2|\epsilon|^{2^\ast(s)} +\gamma |\epsilon|^\beta\big)|u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s}\\ &=\mu_1|u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s}- \frac{\gamma|\epsilon|^\beta}{2^\ast(s)}|u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s} +o(|\epsilon|^\beta)\\ &<\mu_1|u_{\mu_1}|^{2^\ast(s)}_{2^\ast(s),s}\quad \text{for $ |\epsilon|$ small enough}. \end{align*} Therefore, \begin{align*} \bar\mu_s^{A,B}(\Omega) &=\inf_{(u,v)\in\mathcal{M}}\Big(\mu_1|u|^{2^\ast(s)} _{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)} _{2^\ast(s),s} +\gamma\int_\Omega\frac {|u|^\alpha|v|^\beta}{|x|^s} \Big)^\frac{2^\ast(s)-2}{2^\ast(s)}\\ &\le \Big(t(\epsilon)^{2^\ast(s)} \big(\mu_1+\mu_2|\epsilon|^{2^\ast(s)}+\gamma |\epsilon|^\beta\big)|u_{\mu_1} |^{2^\ast(s)}_{2^\ast(s),s}\Big) ^\frac{2^\ast(s)-2}{2^\ast(s)}\\ &<\mu_1^\frac{2^\ast(s)-2}{2^\ast(s)}|u_{\mu_1}|^{2^\ast(s)-2}_{2^\ast(s),s}\\ &=\mu_1^{-\frac{2}{2^\ast(s)}}\mu_s^{A,\lambda_1}(\Omega). \end{align*} Similarly, $\bar\mu_s^{A,B}(\Omega)<\mu_2^{-\frac{2}{2^\ast(s)}} \mu_s^{B,\lambda_2}(\Omega)$. \end{proof} \begin{proof}[Proof of Theorem \ref{t15104}] Since the proof under the assumption (A4) is similar to that of Theorem \ref{t15103}, we only prove it under the assumption (A5). Setting $$ \theta(x):=-\sum_{j=1}^NA_j(0)x_j,\ \ \vartheta(x):=-\sum_{j=1}^NB_j(0)x_j, $$ we have \begin{gather*} \nabla\theta(x)=\big(-A_1(0),\dots,-A_N(0)\big)=-A(0),\\ \nabla\vartheta(x)=\big(-B_1(0),\dots,-B_N(0)\big)=-B(0), \end{gather*} which imply that $(\nabla\theta+A)(0)=0$ and $(\nabla\vartheta+B)(0)=0$. Then, continuity ensures that there exists $\delta>0$ satisfying \begin{equation}\label{151007} \big|(\nabla\theta+A)(x)\big|^2\le\frac{\lambda_3}{2},\quad \big|(\nabla\vartheta+B)(x)\big|^2\le\frac{\lambda_3}{2},\quad\ \forall|x|<\delta, \end{equation} where $\lambda_3=\min\{\lambda_1,\lambda_2\}$. There exists $\rho>0$ such that $B_\rho\subset\Omega$. Let $2r:=\min\{\delta,\rho\}$ and $$ \big(u_\varepsilon(x),v_\varepsilon(x)\big) =\big(\phi(x)U_\varepsilon(x) e^{i\theta(x)}, \phi(x)V_\varepsilon(x)e^{i\vartheta(x)}\Big), $$ where $\phi\in C_0^1(B_{2r})$ is a cut-off function such that $\phi(x)=1$ in $B_r$ and $(U_\varepsilon,V_\varepsilon)$ is defined by \eqref{151026}. By \eqref{151021}, \eqref{151022} and \eqref{151007}, we deduce that \begin{align*} &\int_\Omega \big(|\nabla_Au_\varepsilon|^2-\lambda_1|u_\varepsilon|^2 +|\nabla_Bv_\varepsilon|^2-\lambda_2|v_\varepsilon|^2\big) \\ &= \int_\Omega\big( |\nabla(\phi U_\varepsilon)|^2 +\phi^2U_\varepsilon^2|\nabla\theta+A|^2-\lambda_1 \phi^2U_\varepsilon^2\big)\\ &\quad +\int_\Omega\big( |\nabla(\phi V_\varepsilon)|^2 +\phi^2V_\varepsilon^2|\nabla\vartheta+B|^2-\lambda_2 \phi^2V_\varepsilon^2\big)\\ &\le \int_{\mathbb{R}^N}\big(|\nabla U|^2+|\nabla V|^2\big) +O(\varepsilon^{N-2})+\frac{\lambda_3}{2}\int_{B_{2r}}\phi^2U_\varepsilon^2 \\ &\quad -\lambda_1\int_{B_{2r}}\phi^2U_\varepsilon^2 +\frac{\lambda_3}{2}\int_{B_{2r}}\phi^2V_\varepsilon^2 -\lambda_2\int_{B_{2r}}\phi^2V_\varepsilon^2\\ &\le \int_{\mathbb{R}^N}\big(|\nabla U|^2+|\nabla V|^2\big) +O(\varepsilon^{N-2})-\frac{\lambda_3}{2}\int_{B_{r}}(U_\varepsilon^2 +V_\varepsilon^2). \end{align*} Since \begin{equation}\label{151016} \begin{aligned} \int_{B_r}|U_\varepsilon|^2 &\ge\int_{|x|\le r}\varepsilon^{2-N}\big|U\big(\frac {x}{\varepsilon} \big)\big|^2\text{d}x\\ &=\varepsilon^2\int_{\mathbb{R}^N}|U(y)|^2\text{d}y -\varepsilon^2\int_{|y| \ge\frac{r}{\varepsilon}}|U(y)|^2\text{d}y\\ &\ge C\varepsilon^2-C\varepsilon^2 \int_{|y|\ge\frac{r}{\varepsilon}}|y|^{4-2N}\text{d}y\\ &=C\varepsilon^2+O(\varepsilon^{N-2}), \end{aligned} \end{equation} and \[ \int_{B_r}|V_\varepsilon|^2 \ge C\varepsilon^2+O(\varepsilon^{N-2}), \] by \eqref{151023}--\eqref{151025}, we have \begin{equation}\label{151010} \begin{aligned} \bar\mu_s^{A,B}(\Omega) &\le \frac{\int_\Omega\big(|\nabla_Au_\varepsilon|^2 -\lambda_1|u_\varepsilon|^2 +|\nabla_Bv_\varepsilon|^2 -\lambda_2|v_\varepsilon|^2\big)} {\big(\mu_1|u_\varepsilon|^{2^\ast(s)} _{2^\ast(s),s} +\mu_2|v_\varepsilon|^{2^\ast(s)} _{2^\ast(s),s} +\gamma\int_\Omega\frac{|u_\varepsilon|^\alpha|v_\varepsilon|^\beta} {|x|^s}\big)^\frac{2}{2^\ast(s)}}\\ &\le \frac{\int_{\mathbb{R}^N}\big(|\nabla U|^2+|\nabla V|^2\big)-C\varepsilon^2 +O(\varepsilon^{N-2})} {\big(\int_{\mathbb{R}^N} \big(\mu_1\frac{|U|^{2^\ast(s)}}{|x|^s} +\mu_2\frac{|V|^{2^\ast(s)}}{|x|^s}+\gamma\frac{|U|^\alpha|V|^\beta}{|x|^s}\big) +O(\varepsilon^{N-s})\big)^\frac{N-2}{N-s}}\\ &<\bar\mu_s(\mathbb{R}^N). \end{aligned} \end{equation} Let $\{(u_n,v_n)\}$ be a minimizing sequence for $\bar\mu_s^{A,B}(\Omega)$ normalized as $$ \mu_1|u_n|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v_n|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u_n|^\alpha|v_n|^\beta}{|x|^s}=1; $$ that is, \begin{equation}\label{151009} |\nabla_Au_n|^2_2-\lambda_1|u_n|^2_2+|\nabla_Bv_n|^2_2 -\lambda_2|v_n|^2_2 =\bar\mu_s^{A,B}(\Omega)+o(1). \end{equation} Noting that $\{u_n\}$ is bounded in $H_A^1(\Omega)$ and $\{v_n\}$ is bounded in $H_B^1(\Omega)$, by Lemma \ref{l15101}, we may extract two subsequences-still denoted by $\{u_n\}$ and $\{v_n\}$-such that \begin{gather*} u_n\rightharpoonup u\quad \text{ weakly in } H_A^1(\Omega),\\ v_n\rightharpoonup v\quad \text{ weakly in } H_B^1(\Omega)\\ u_n\to u,\quad v_n\to v\quad \text{ strongly in } L^2(\Omega),\\ u_n\to u,\quad v_n\to v\quad \text{ a.e. on } \Omega, \end{gather*} with \[ \mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\le1. \] Setting $w_n:=u_n-u$ and $z_n:=v_n-v$, then $w_n\rightharpoonup0$ weakly in $H_A^1(\Omega)$, $z_n\rightharpoonup0$ weakly in $H_B^1(\Omega)$ and $w_n\to0, z_n\to0$ a.e. on $\Omega$. It follows from diamagnetic inequality and \eqref{151009} that \begin{gather*} |\nabla_Au_n|_2^2+|\nabla_Bv_n|_2^2\ge\big|\nabla|u_n|\big|_2^2 + \big|\nabla|v_n|\big|_2^2\ge\bar\mu_s(\mathbb{R}^N), \\ \bar\mu_s^{A,B}(\Omega)+\lambda_1|u_n|^2_2+\lambda_2|v_n|^2_2+o(1) \ge\bar\mu_s(\mathbb{R}^N). \end{gather*} By \eqref{151010}, we see that $\lambda_1|u|^2_2+\lambda_2|v|^2_2\ge\bar\mu_s(\mathbb{R}^N) -\bar\mu_s^{A,B}(\Omega)>0$, which means that $(u,v)\not\equiv(0,0)$. Since $w_n\rightharpoonup0$ weakly in $H_A^1(\Omega)$ and $z_n\rightharpoonup0$ weakly in $H_B^1(\Omega)$, we have \begin{gather*} \begin{aligned} |\nabla_Au_n|_2^2 &= \int_\Omega|\nabla_Aw_n|^2+\int_\Omega|\nabla_Au|^2 +2\operatorname{Re} \Big(\int_\Omega\nabla_Aw_n\cdot\overline{\nabla_Au}\Big)\\ &= |\nabla_Aw_n|_2^2+|\nabla_Au|_2^2+o(1), \end{aligned}\\ |\nabla_Bv_n|_2^2= |\nabla_Bz_n|_2^2+|\nabla_Bv|_2^2+o(1). \end{gather*} Then, \eqref{151009} yields \begin{equation}\label{151027} \bar\mu_s^{A,B}(\Omega)=|\nabla_Aw_n|_2^2+|\nabla_Au|_2^2-\lambda_1|u|_2^2 +|\nabla_Bz_n|_2^2+|\nabla_Bv|_2^2-\lambda_2|v|_2^2+o(1). \end{equation} The Brezis-Lieb Lemma guarantees that \begin{align*} 1&= \mu_1|u+w_n|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v+z_n|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u+w_n|^\alpha|v+z_n|^\beta}{|x|^s}\\ &= \mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\\ &\quad +\mu_1|w_n|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|z_n|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|w_n|^\alpha|z_n|^\beta}{|x|^s}+o(1). \end{align*} Noting \begin{gather*} \mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\le 1, \\ \mu_1|w_n|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|z_n|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|w_n|^\alpha|z_n|^\beta}{|x|^s}\le1, \end{gather*} we have \begin{align*} 1&\leq \Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big)^\frac{2}{2^\ast(s)}\\ &\quad +\Big(\mu_1|w_n|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|z_n|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|w_n|^\alpha|z_n|^\beta}{|x|^s}\Big)^\frac{2}{2^\ast(s)} +o(1)\\ &\leq \Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big)^\frac{2}{2^\ast(s)}\\ &+\frac{1}{\bar\mu_s(\mathbb{R}^N)}\big(\big|\nabla|w_n|\big|_2^2 +\big|\nabla|z_n|\big|_2^2\big) +o(1)\\ &\leq \Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big)^\frac{2}{2^\ast(s)}\\ &\quad +\frac{1}{\bar\mu_s(\mathbb{R}^N)}\big(|\nabla_Aw_n|_2^2 + |\nabla_Bz_n|_2^2\big) +o(1). \end{align*} It follows from \eqref{151010}, \eqref{151027} and $\bar\mu_s^{A,B}(\Omega)>0$ that \begin{align*} &|\nabla_A u|_2^2-\lambda_1|u|_2^2+|\nabla_Bv|_2^2-\lambda_2|v|_2^2\\ &\leq \bar\mu_s^{A,B}(\Omega)\Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega \frac{|u|^\alpha|v|^\beta}{|x|^s}\Big)^\frac{2}{2^\ast(s)}\\ &\quad +\Big(\frac{\bar\mu_s^{A,B}(\Omega)}{\bar\mu_s(\mathbb{R}^N)}-1\Big) \big(|\nabla_Aw_n|_2^2+|\nabla_Bz_n|_2^2\big)+o(1)\\ &< \bar\mu_s^{A,B}(\Omega)\Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big)^\frac{2}{2^\ast(s)} +o(1), \end{align*} which, combining with $(u,v)\not\equiv(0,0)$, implies $$ \frac{|\nabla_Au|^2_{2}-\lambda_1|u|^2_2 +|\nabla_Bv|^2_{2}-\lambda_2|v|_2^2} {\big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s}+\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_{\Omega}\frac{|u|^\alpha|v|^\beta}{|x|^s}\big) ^\frac{2}{2^\ast(s)}} \le\bar\mu_s^{A,B}(\Omega). $$ Then, $\bar\mu_s^{A,B}(\Omega)$ is attained by $(u,v)$. It remains to show that $(u,v)$ can not be the type of $(u,0)$ or $(0,v)$. Suppose by contradiction that $\bar\mu_s^{A,B}(\Omega)$ is attained by $(u,0)$. Then \[ \bar\mu_s^{A,B}(\Omega) =\frac{|\nabla_Au|^2_{2}-\lambda_1|u|^2_2} {\mu_1^\frac{2}{2^\ast(s)} |u|^2_{2^\ast(s),s}}\ge\mu_1^{-\frac{2}{2^\ast(s)}} \mu_s^{A,\lambda_1}(\Omega), \] which contradicts to Lemma \ref{l15106}. Hence, $(u,v)$ can not be the type of $(u,0)$. Similarly, it can not be $(0,v)$, which completes the proof. \end{proof} \begin{remark} \label{rmk3.1} \rm Even if $\bar\mu_s^{A,B}(\Omega)\le0$, it is also attained. Indeed, by \eqref{151027} and $\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\le1$, we obtain \begin{align*} &|\nabla_A u|_2^2-\lambda_1|u|_2^2+|\nabla_Bv|_2^2-\lambda_2|v|_2^2\\ & \le\mu_s^{A,B}(\Omega)\\ &\leq \mu_s^{A,B}(\Omega)\Big(\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\gamma\int_\Omega\frac{|u|^\alpha|v|^\beta}{|x|^s}\Big). \end{align*} \end{remark} \section{Ground states for the equations} By Lemma \ref{l15105}, Theorems \ref{t15105}--\ref{t15108}, follow from Theorems \ref{t15101}--\ref{t15104} respectively. Considering \eqref{151003}, by Theorem \ref{t15105}, we assume that $u_{\mu_1}$ and $v_{\mu_2}$ are ground state solutions of $-\Delta_Au=\mu_1\frac{|u|^{2^\ast(s)-2}u}{|x|^s}$ and $-\Delta_Bv=\mu_1\frac{|v|^{2^\ast(s)-2}v}{|x|^s}$, respectively. It follows from Lemma \ref{l15102} that the ground state energies are \[ M_{\mu_1}:=\frac{2-s}{2(N-s)}\mu_1^{-\frac{N-2}{2-s}} \big( \mu_s(\mathbb{R}^N)\big)^\frac{N-s}{2-s},\quad M_{\mu_2}:=\frac{2-s}{2(N-s)}\mu_2^{-\frac{N-2}{2-s}} \big( \mu_s(\mathbb{R}^N)\big)^\frac{N-s}{2-s}. \] We claim that if $\gamma<0$, then \eqref{151003} has no nontrivial ground state solution, which is the reason that we only consider the case $\gamma>0$ in this paper. In fact, if $\gamma<0$, then \[ |\nabla_Au|^2_{2,\mathbb{R}^N}-\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} =\frac{\alpha\gamma}{2^{\ast}(s)}\int_{\mathbb{R}^N}\frac{|u|^\alpha|v|^\beta} {|x|^s}\le0, \] which implies \[ \mu_1|u|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} \ge|\nabla_Au|^2_{2,\mathbb{R}^N} \ge\mu_s^A(\mathbb{R}^N)|u|^{2}_{2^\ast(s), s,\mathbb{R}^N}. \] If $u\in D^{1,2}_A(\mathbb{R}^N)\setminus\{0\}$, then $|u|_{2^\ast(s),s,\mathbb{R}^N}\ge\big(\frac{\mu_s^A(\mathbb{R}^N)}{\mu_1}\big) ^\frac{1}{2^\ast(s)-2}$, which yields that $|\nabla_Au|^2_{2,\mathbb{R}^N}\ge\mu_s^A(\mathbb{R}^N) \big(\frac{\mu_s^A(\mathbb{R}^N)}{\mu_1}\big)^\frac{2}{2^\ast(s)-2}$. Therefore, $$ M_{\mu_1} =\frac{2-s}{2(N-s)}\mu_1^{-\frac{N-2}{2-s}} \big( \mu_s^A(\mathbb{R}^N)\big)^\frac{N-s}{2-s} \le\frac{2-s}{2(N-s)}|\nabla_Au|^2_{2,\mathbb{R}^N}. $$ Similarly, $M_{\mu_2}\le\frac{2-s}{2(N-s)}|\nabla_Bv|^2_{2,\mathbb{R}^N}$ for any $v\in D^{1,2}_B(\mathbb{R}^N)\setminus\{0\}$. Suppose that $(u,v)$ is a ground state solution of \eqref{151003}. Then \begin{align*} M_0&=I(u,v) =\frac{2-s}{2(N-s)}\big(|\nabla_Au|^2_{2,\mathbb{R}^N} +|\nabla_Bv|^2_{2,\mathbb{R}^N}\big)\\ &\ge\begin{cases}M_{\mu_2},&\text{if } u=0,v\neq0,\\ M_{\mu_1}+M_{\mu_2},&\text{if }u\neq0,v\neq0,\\ M_{\mu_1},&\text{if }u\neq0,v=0. \end{cases} \end{align*} It can be seen that $M_0\le\min\{M_{\mu_1},\ M_{\mu_2}\}$, which means that \eqref{151003} has no nontrivial ground state solution. Define \begin{align*} \mathcal{N}':=\Big\{&(u,v)\in D_{A,A}: u\not\equiv0,v\not\equiv0,\\ &|\nabla_Au|_{2,\mathbb{R}^N}^2=\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\frac{\alpha\gamma}{2^\ast(s)}\int_{\mathbb{R}^N} \frac{|u|^\alpha|v|^\beta}{|x|^s},\\ &|\nabla_Av|_{2,\mathbb{R}^N}^2=\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\frac{\beta\gamma}{2^\ast(s)}\int_{\mathbb{R}^N} \frac{|u|^\alpha|v|^\beta} {|x|^s}\Big\}, \end{align*} \begin{align*} \mathcal{M}':=\Big\{&(u,v)\in H_{A,A}: u\not\equiv0,v\not\equiv0,\\ &|\nabla_Au|_{2}^2-\lambda|u|^2_2=\mu_1|u|^{2^\ast(s)}_{2^\ast(s),s} +\frac{\alpha\gamma}{2^\ast(s)}\int_{\Omega} \frac{|u|^\alpha|v|^\beta}{|x|^s},\\ &|\nabla_Av|_{2}^2-\lambda|v|^2_2=\mu_2|v|^{2^\ast(s)}_{2^\ast(s),s} +\frac{\beta\gamma}{2^\ast(s)}\int_{\Omega} \frac{|u|^\alpha|v|^\beta}{|x|^s}\Big\}, \end{align*} $M_0':=\inf_{(u,v)\in\mathcal{N}'}I(u,v)$ and $M':=\inf_{(u,v)\in\mathcal{M}'}E(u,v)$. It can be seen from Theorems \ref{t15105} and \ref{t15106} that \begin{equation}\label{151014} |\nabla_Au|_{2,\mathbb{R}^N}^2\ge\Big(\frac{2(N-s)}{2-s}M_1\Big)^{\frac{2-s}{N-s}} |u|^{2}_{2^\ast(s),s,\mathbb{R}^N},\quad \forall u\in D_A^{1,2}(\mathbb{R}^N) \end{equation} and \[ |\nabla_Au|_{2}^2-\lambda|u|_2^2\ge\Big(\frac{2(N-s)}{2-s}M_2 \Big) ^{\frac{2-s}{N-s}} |u|^{2}_{2^\ast(s),s},\quad \forall u\in H_A^{1}(\Omega). \] Define functions: \begin{equation}\label{12} \begin{gathered} F_1(k,l):=\mu_1k^{\frac{2^\ast(s)-2}{2}}+\frac{\alpha\gamma}{2^\ast(s)} k^{\frac{\alpha-2}{2}}l^{\beta/2}-1,\quad k>0,l\ge0;\\ F_2(k,l):=\mu_2l^{\frac{2^\ast(s)-2}{2}}+\frac{\beta\gamma}{2^\ast(s)} k^{\frac{\alpha}{2}}l^{\frac{\beta-2}{2}}-1,\quad k\ge0,l>0. \end{gathered} \end{equation} Following the arguments as in \cite[Lemma 2.4]{ChenZou.2015} or \cite[Proposition 2.2]{GZ}, we have the following result. \begin{proposition}\label{p2} If {\rm (A6)} holds, then \begin{equation}\label{22} \begin{gathered} k+l\le k_0+l_0,\\ F_1(k,l)\ge0,\quad F_2(k,l)\ge0,\\ k,l\ge0,\quad (k,l)\neq(0,0) \end{gathered} \end{equation} has a unique solution $(k,l)=(k_0,l_0)$, where $(k_0,l_0)$ is defined by \eqref{151013}. \end{proposition} \begin{proof}[Proof of Theorem \ref{t15109}] Recalling \eqref{151012}, we see that $(\sqrt{k_0}U,\sqrt{l_0}U)\in\mathcal{N}'$, that $(\sqrt{k_0}U,\sqrt{l_0}U)$ is a nontrivial solution of \eqref{151003}, and that \begin{equation}\label{24} M_0'\le I(\sqrt{k_0}U,\sqrt{l_0}U)=\Big(\frac{1}{2} -\frac{1}{2^\ast(s)}\Big) (k_0+l_0)|\nabla_AU|_{2,\mathbb{R}^N}^2=(k_0+l_0)M_1. \end{equation} On the other hand, assume that $\{(u_n,v_n)\}\subset\mathcal{N}'$ is a minimizing sequence for $M_0'$, that is, $I(u_n,v_n)\to M_0'$ as $n\to\infty$. Define $$ c_n=|u_n|^2_{2^\ast(s),s,\mathbb{R}^N},\quad d_n=|v_n|^2_{2^\ast(s),s,\mathbb{R}^N}, $$ and by \eqref{151014}, we obtain \begin{equation}\label{25} \begin{split} \Big(\frac{2(N-s)M_1}{2-s}\Big)^{\frac{2-s}{N-s}}c_n &\le|\nabla_Au_n|^2 _{2,\mathbb{R}^N}\\& =\mu_1|u_n|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\frac{\alpha\gamma}{2^\ast(s)} \int_{\mathbb{R}^N} \frac{|u_n|^\alpha|v_n|^\beta}{|x|^s}\\ &\le \mu_1c_n^{\frac{2^\ast(s)}{2}} +\frac{\alpha\gamma}{2^\ast(s)} c_n^{\frac{\alpha}{2}}d_n^{\beta/2},\\ \Big(\frac{2(N-s)M_1}{2-s}\Big)^{\frac{2-s}{N-s}}d_n &\le|\nabla_Av_n|^2 _{2,\mathbb{R}^N}\\ &=\mu_2|v_n|^{2^\ast(s)}_{2^\ast(s),s,\mathbb{R}^N} +\frac{\beta\gamma}{2^\ast(s)} \int_{\mathbb{R}^N}\frac{|u_n|^\alpha |v_n|^\beta}{|x|^s}\\ &\le \mu_2d_n^{\frac{2^\ast(s)}{2}}+\frac{\beta\gamma}{2^\ast(s)} c_n^{\frac{\alpha}{2}}d_n^{\beta/2}. \end{split} \end{equation} Dividing both sides of the inequalities by $\big(\frac{2(N-s)M_1}{2-s}\big)^{\frac{2-s}{N-s}}c_n$ and $\big(\frac{2(N-s)M_1}{2-s}\big)^{\frac{2-s}{N-s}}d_n$, respectively, and setting $$ \tilde{c}_n=\frac{c_n}{\big(\frac{2(N-s)M_1}{2-s}\big)^{\frac{N-2}{N-s}}}, \quad \tilde{d}_n=\frac{d_n}{\big(\frac{2(N-s)M_1}{2-s}\big)^{\frac{N-2}{N-s}}}, $$ we have \begin{gather*} \mu_1\tilde{c}_n^{\frac{2^\ast(s)-2}{2}}+\frac{\alpha\gamma}{2^\ast(s)} \tilde{c}_n^{\frac{\alpha-2}{2}}\tilde{d}_n^{\beta/2}\ge1,\\ \mu_2\tilde{d}_n^{\frac{2^\ast(s)-2}{2}}+\frac{\beta\gamma}{2^\ast(s)} \tilde{c}_n^{\frac{\alpha}{2}}\tilde{d}_n^{\frac{\beta-2}{2}}\ge1, \end{gather*} i.e., $F_1(\tilde{c}_n,\tilde{d}_n)\ge0$ and $F_2(\tilde{c}_n,\tilde{d}_n)\ge0$. Then, Proposition \ref{p2} ensures that $\tilde{c}_n+\tilde{d}_n\ge k_0+l_0$, which means that \begin{equation}\label{26} c_n+d_n\ge(k_0+l_0)\Big(\frac{2(N-s)M_1}{2-s}\Big)^{\frac{N-2}{N-s}}. \end{equation} It follows from \eqref{24}, \eqref{25} and $I(u_n,v_n)=\frac{2-s}{2(N-s)}\|(u_n,v_n)\|_{D_{A,A}}^2$ that \begin{align*} \Big(\frac{2(N-s)M_1}{2-s}\Big)^{\frac{2-s}{N-s}}(c_n+d_n) &\leq \frac{2(N-s)}{2-s}I(u_n,v_n)\\ &= \frac{2(N-s)}{2-s}M_0'+o(1)\\ &\leq \frac{2(N-s)}{2-s}(k_0+l_0)M_1+o(1). \end{align*} Combining this with \eqref{26}, we obtain $$ c_n+d_n\to (k_0+l_0)\Big(\frac{2(N-s)M_1}{2-s}\Big)^{\frac{N-2}{N-s}},\quad \text{as } n\to\infty. $$ Therefore, \begin{align*} M_0'&=\lim_{n\to \infty}I(u_n,v_n)\\ &\ge\lim_{n\to \infty}\frac{2-s}{2(N-s)} \Big(\frac{2(N-s)B_1}{2-s}\Big)^{\frac{2-s}{N-s}}(c_n+d_n)=(k_0+l_0)M_1. \end{align*} By \eqref{24}, we have \begin{equation}\label{151028} M_0'=(k_0+l_0)M_1=I(\sqrt{k_0}U,\sqrt{l_0}U). \end{equation} Theorem \ref{t15107} ensures that $M_0$ is attained by a nontrivial ground state solution $(u,v)\in \mathcal{N}$ of \eqref{151003} with $B=A$. It is easy to see that $(u,v)\in\mathcal{N}'$, which implies that \[ M_0=I(u,v)\ge\inf_{(\tilde{u},\tilde{v})\in\mathcal{N}'}I(\tilde{u},\tilde{v}) =M_0'. \] Obviously, $M_0\le M_0'$ follows from $\mathcal{N}'\subset\mathcal{N}$. Therefore, $M_0=M_0'$, and combining this with \eqref{151028}, we see that $(\sqrt{k_0}U,\sqrt{l_0}U)$ is a ground state solution of \eqref{151003}. By \eqref{151028}, Theorem \ref{t15105} and Lemma \ref{l15102}, we have $$ M_0=\frac{2-s}{2(N-s)}(k_0+l_0)\big(\mu_s(\mathbb{R}^N)\big)^\frac{N-s}{2-s}. $$ \end{proof} The proof Theorem \ref{t15110} is similar to that of Theorem \ref{t15109}, it is omitted. \section{Application in three dimensions} In this section, we consider a constant magnetic field in dimension 3 as an application of Theorems \ref{t15101} and \ref{t15103}. \subsection*{Constant magnetic field} Let $\tilde{A}:\mathbb{R}^3\to\mathbb{R}^3$ by $\tilde{A}(x_1,x_2,x_3):=(-x_2,x_1,0)$, which is called constant magnetic potential as $\operatorname{curl}\tilde{A}=2\neq0$. Theorem \ref{t15101} guarantees that $\mu_s^{r\tilde{A}}(\mathbb{R}^3)$ is not achieved, and then \[ (-i\nabla+r\tilde{A})^2u=\frac{|u|^{2^\ast(s)-2}u}{|x|^s},\quad u\in D^{1,2}_{\tilde{A}}(\mathbb{R}^N) \] has no ground state solution, where $r$ is a nonzero real number. By Theorem \ref{t15103}, we obtain that under condition (A3), $\bar{\mu}_s^{r_1\tilde{A},r_2\tilde{A}}(\mathbb{R}^3)$ is not attained, and thus, \begin{gather*} (-i\nabla+r_1\tilde{A})^2u=\mu_1\frac{|u|^{2^\ast(s)-2}u}{|x|^s} +\frac{\alpha\gamma} {2^\ast(s)}\frac{|u|^{\alpha-2}u|v|^\beta}{|x|^s},\\ (-i\nabla+r_2\tilde{A})^2v=\mu_2\frac{|v|^{2^\ast(s)-2}v}{|x|^s} +\frac{\beta\gamma} {2^\ast(s)}\frac{|u|^{\alpha}|v|^{\beta-2}v}{|x|^s},\\ u,\ v\in D^{1,2}_{\tilde{A}}(\mathbb{R}^3) \end{gather*} has no ground state solution, where $r_1$ and $r_2$ are nonzero real numbers. \subsection*{Acknowledgments} This research was supported by the NSFC (Nos. 11371212, 11271386). The authors thank the anonymous referee for pointing out an error. Melgaard kindly acknowledges the hospitality of Tsinghua University during his visits in 2014 and 2015. \begin{thebibliography}{00} \bibitem{AlvesFigueiredo.2014} C.~O. Alves, G.~M. 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