# On a nonhomogeneous Kirchhoff type elliptic system with the singular
Trudinger-Moser growth

Shengbing Deng111 E-mail address<EMAIL_ADDRESS>(S. Deng),
<EMAIL_ADDRESS>(X. Tian) and Xingliang Tian
School of Mathematics and Statistics, Southwest University, Chongqing, 400715,
P.R. China

###### Abstract

The aim of this paper is to study the multiplicity of solutions for the
following Kirchhoff type elliptic systems

$\displaystyle\left\\{\begin{array}[]{ll}-m\left(\sum^{k}_{j=1}\|u_{j}\|^{2}\right)\Delta
u_{i}=\frac{f_{i}(x,u_{1},\ldots,u_{k})}{|x|^{\beta}}+\varepsilon
h_{i}(x),&\mbox{in}\ \ \Omega,\ \ i=1,\ldots,k,\\\\[5.69054pt]
u_{1}=u_{2}=\cdots=u_{k}=0,&\mbox{on}\ \ \partial\Omega,\end{array}\right.$

where $\Omega$ is a bounded domain in $\mathbb{R}^{2}$ containing the origin
with smooth boundary, $\beta\in[0,2)$, $m$ is a Kirchhoff type function,
$\|u_{j}\|^{2}=\int_{\Omega}|\nabla u_{j}|^{2}dx$, $f_{i}$ behaves like
$e^{\beta s^{2}}$ when $|s|\rightarrow\infty$ for some $\beta>0$, and there is
$C^{1}$ function $F:\Omega\times\mathbb{R}^{k}\to\mathbb{R}$ such that
$\left(\frac{\partial F}{\partial u_{1}},\ldots,\frac{\partial F}{\partial
u_{k}}\right)=\left(f_{1},\ldots,f_{k}\right)$,
$h_{i}\in\left(\big{(}H^{1}_{0}(\Omega)\big{)}^{*},\|\cdot\|_{*}\right)$. We
establish sufficient conditions for the multiplicity of solutions of the above
system by using variational methods with a suitable singular Trudinger-Moser
inequality when $\varepsilon>0$ is small.

_Keywords:_ Kirchhoff type elliptic systems; multiple solutions; singular
Trundinger-Moser inequality.

_2020 Mathematics Subject Classification:_ 35J50, 35J57.

##  1 Introduction

In last decades, a great attention has been focused on the study of problems
involving exponential growth nonlinearities, which is related to the famous
Trudinger-Moser inequality. Let $\Omega$ be a bounded domain in
$\mathbb{R}^{2}$, and denote with $H_{0}^{1}(\Omega)$ the standard first order
Sobolev space given by

$H_{0}^{1}(\Omega)=cl\Big{\\{}u\in C^{\infty}_{0}(\Omega)\ :\
\int_{\Omega}|\nabla u|^{2}{\rm d}x<\infty\Big{\\}},\quad\ \
\|u\|=\left(\int_{\Omega}|\nabla u|^{2}{\rm d}x\right)^{\frac{1}{2}}.$

This space is a limiting case for the Sobolev embedding theorem, which yields
$H_{0}^{1}(\Omega)\hookrightarrow L^{p}(\Omega)$ for all $1\leq p<\infty$, but
one knows by easy examples that $H_{0}^{1}(\Omega)\not\subseteq
L^{\infty}(\Omega)$ such as, $u(x)=\log(1-\log|x|)$ in $B_{1}(0)$. Hence, one
is led to look for a function $g(s):\mathbb{R}\to\mathbb{R}^{+}$ with maximal
growth such that

$\sup\limits_{u\in H_{0}^{1}(\Omega),\|u\|\leq 1}\int_{\Omega}g(u){\rm
d}x<\infty.$

It was shown by Trudinger [29] and Moser [26] that the maximal growth is of
exponential type. More precisely, named the Trudinger-Moser inequality that

$\exp(\alpha u^{2})\in L^{1}(\Omega),\quad\forall\ u\in H_{0}^{1}(\Omega),\ \
\forall\ \alpha>0,$

and

$\displaystyle\sup\limits_{u\in H_{0}^{1}(\Omega),\|u\|\leq
1}\int_{\Omega}\exp(\alpha u^{2}){\rm d}x<\infty,\quad\mbox{if}\ \alpha\leq
4\pi,$

where $4\pi$ is the sharp constant in the sense that the supremum in the left
is $\infty$ if $\alpha>4\pi$.

In order to treat the system problems, here we give some definitions. For all
$1\leq p<\infty$, we define $L^{p}(\Omega,\mathbb{R}^{k})$ as

$L^{p}(\Omega,\mathbb{R}^{k}):=\underbrace{L^{p}(\Omega)\times\cdots\times
L^{p}(\Omega)}_{k},$

where $L^{p}(\Omega)$ is the standard $L^{p}$-space, and since

$\frac{1}{k}\left(\sum^{k}_{i=1}|u_{i}|^{p}\right)\leq\left(\sum^{k}_{i=1}|u_{i}|^{2}\right)^{\frac{p}{2}}\leq
k^{p}\left(\sum^{k}_{i=1}|u_{i}|^{p}\right),$ (1.1)

we can know that $L^{p}(\Omega,\mathbb{R}^{k})$ is well defined and for $U\in
L^{p}(\Omega,\mathbb{R}^{k})$, we define
$\|U\|_{p}=\left(\int_{\Omega}|U|^{p}dx\right)^{1/p}$ where
$|U|=(\sum^{k}_{i=1}|u_{i}|^{2})^{1/2}$. Moreover we denote

$H^{1}_{0}(\Omega,\mathbb{R}^{k}):=\underbrace{H_{0}^{1}(\Omega)\times\cdots\times
H_{0}^{1}(\Omega)}_{k},$

be the Sobolev space modeled in $L^{2}(\Omega,\mathbb{R}^{k})$ with the scalar
product

$\ \langle U,V\rangle=\sum^{k}_{i=1}\int_{\Omega}\nabla u_{i}\nabla v_{i}dx,$

where $U,V\in L^{2}(\Omega,\mathbb{R}^{k})$, to which corresponds the norm
$\|U\|=\langle U,U\rangle^{1/2}=(\sum^{k}_{i=1}\|u_{i}\|^{2})^{1/2}$, then
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$ is well defined and also is a Hilbert
space. For all $1\leq p<\infty$, by the inequality (1.1) and the Sobolev
embedding theorem, we can know that the embedding
$H^{1}_{0}(\Omega,\mathbb{R}^{k})\hookrightarrow L^{p}(\Omega,\mathbb{R}^{k})$
is compact and $H^{1}_{0}(\Omega,\mathbb{R}^{k})\nsubseteq
L^{\infty}(\Omega,\mathbb{R}^{k})$, where
$L^{\infty}(\Omega,\mathbb{R}^{k}):=L^{\infty}(\Omega)\times\cdots\times
L^{\infty}(\Omega)$. In Section 2, we will establish the Trudinger-Moser type
inequality for $H^{1}_{0}(\Omega,\mathbb{R}^{k})$.

Now, we begin to state our problem. Let $\Omega$ be a bounded domain in
$\mathbb{R}^{2}$ containing the origin with smooth boundary, we study the
multiplicity of solutions for the following Kirchhoff type systems

$\displaystyle\left\\{\begin{array}[]{ll}-m\left(\sum^{k}_{j=1}\|u_{j}\|^{2}\right)\Delta
u_{i}=\frac{f_{i}(x,u_{1},\ldots,u_{k})}{|x|^{\beta}}+\varepsilon
h_{i}(x),&\mbox{in}\ \ \Omega,\ \ i=1,\ldots,k,\\\\[5.69054pt]
u_{1}=u_{2}=\cdots=u_{k}=0,&\mbox{on}\ \ \partial\Omega,\end{array}\right.$
(1.4)

where $\beta\in[0,2)$, $m$ is a continuous Kirchhoff type function,
$h_{i}\in\big{(}\big{(}H^{1}_{0}(\Omega)\big{)}^{*},\|\cdot\|_{*}\big{)}\backslash\\{0\\}$
for some $i\in\\{1,\ldots,k\\}$, $\varepsilon$ is a small positive parameter,
and $f_{i}$ has the maximal growth which allows treating (1.4) variationally
in the Sobolev space $H^{1}_{0}(\Omega,\mathbb{R}^{k})$. We shall consider the
variational situation in which

$(f_{1}(x,U),\ldots,f_{k}(x,U))=\nabla F(x,U)$

for some function $F:\Omega\times\mathbb{R}^{k}\rightarrow\mathbb{R}$ of class
$C^{1}$, where $\nabla F$ stands for the gradient of $F$ in the variables
$U=(u_{1},\ldots,u_{k})\in\mathbb{R}^{k}$. We then rewrite (1.4) in the matrix
form as

$-m\left(\|U\|^{2}\right)\Delta U=\frac{\nabla
F(x,U)}{|x|^{\beta}}+\varepsilon H(x),$ (1.5)

where $\Delta U=(\Delta u_{1},\ldots,\Delta u_{k})$, $\frac{\nabla
F(x,U)}{|x|^{\beta}}=\Big{(}\frac{f_{1}(x,U)}{|x|^{\beta}},\ldots,\frac{f_{k}(x,U)}{|x|^{\beta}}\Big{)}$
and $H(x)=\big{(}h_{1}(x),\ldots,h_{k}(x)\big{)}$.

System (1.4) is called nonlocal because of the term
$m\left(\sum^{k}_{j=1}\|u_{j}\|^{2}\right)$ which implies that the equation in
(1.4) is no longer a pointwise identity. As we will see later the presence of
the term $m\left(\sum^{k}_{j=1}\|u_{j}\|^{2}\right)$ provokes some
mathematical difficulties which makes the study of such a class of problems
particularly interesting. Moreover, System (1.4) with $k=1$ has a physical
appeal which is generalization of a model introduced in 1883 by Kirchhoff
[25]. There are many results about the existence and multiplicity of solutions
for Kirchhoff problems by many mathematicians, we refer to [7, 10, 11, 12, 20,
21, 22, 23, 27] and the references therein. When $k=1$, $\beta=0$ and
$\varepsilon=0$, system (1.4) become the following Kirchhoff type problem

$\displaystyle\left\\{\begin{array}[]{ll}-m\Big{(}\int_{\Omega}|\nabla
u|^{2}dx\Big{)}\Delta u=f(x,u),&\mbox{in}\ \ \Omega,\\\\[5.69054pt]
u=0,&\mbox{on}\ \ \partial\Omega,\end{array}\right.$ (1.8)

where the Kirchhoff function $m:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$
satisfies

$(\overline{M}_{1})$ there exists $m_{0}>0$ such that $m(t)\geq m_{0}$ for all
$t\geq 0$ and $M(t+s)\geq M(t)+M(s)$ for all $s,t\geq 0$, where
$M(t)=\int^{t}_{0}m(\tau)d\tau$ is the primitive of $m$.

$(\overline{M}_{2})$ there exist constants $a_{1},a_{2}>0$ and $t_{0}>0$ such
that for some $\sigma\in\mathbb{R}$, $m(t)\leq a_{1}+a_{2}t^{\sigma},\forall
t\geq t_{0}$.

$(\overline{M}_{3})$ $\frac{m(t)}{t}$ is nonincreasing for $t>0$.

Moreover, the nonlinearity $f:\Omega\times\mathbb{R}\to\mathbb{R}$ is
continuous and satisfies

$(\overline{F}_{1})$ there exist constants $s_{0},K_{0}>0$ such that
$F(x,s)\leq K_{0}f(x,s),\
\forall(x,s)\in\Omega\times\left[s_{0},+\infty\right)$.

$(\overline{F}_{2})$ for each $x\in\Omega,\frac{f(x,s)}{s^{3}}$ is increasing
for $s>0$.

$(\overline{F}_{3})$ there exists
$\beta_{0}>\frac{2}{\alpha_{0}d^{2}}m\left(4\pi/\alpha_{0}\right)$ such that
$\lim_{s\rightarrow+\infty}\frac{sf(x,s)}{\exp\left(\alpha_{0}s^{2}\right)}\geq\beta_{0}$
uniformly in $x\in\Omega$.
Under these assumptions, by using minimax techniques with the Trudinger-Moser
inequality, Figueiredo and Severo [20] obtained the existence of ground state
solution of (1.8). We note that hypothesis $(\overline{F}_{2})$ is necessary
to obtain precise information about the minimax level of the energy functional
associated to problem (1.8), they show the existence of the least energy
solution. Recently, Naimen and Tarsi [27] studied the existence and
multiplicity of solutions for problem (1.8) with $m(t)=1+\alpha t$ under some
weaker assumptions than those in [20].

On the other hand, we mention that the existence of solutions for elliptic
equations involving critical exponential nonlinearities and a small
nonhomogeneous term was considered by many authors, see [5, 6, 17, 24, 30] and
the references therein. In the whole Euclidean space $\mathbb{R}^{N}$, for
$N$-Laplacian problems in [17], for $N$-Laplacian problem with the nonlinear
term involving critical Hardy exponential growth and the nonhomogeneous term
in [5, 30]. What’s more, Lam and Lu [24] established the existence and
multiplicity of nontrivial solutions for the nonuniformly elliptic equations
of $N$-Laplacian type. Moreover, Manasses de Souza [16] has studied the
existence of solutions for a singular class of elliptic systems involving
critical exponential growth in a bounded domain of $\mathbb{R}^{2}$. To the
best of our knowledge, there are no results for (1.4) with Kirchhoff function
and exponential growth nonlinearity.

The main purpose of the present paper is to consider the multiplicity of
solutions of system (1.4) and overcome the lack of compactness due to the
presence of exponential growth terms as well as the degenerate nature of the
Kirchhoff coefficient.

Let us introduce the precise assumptions under which our problem is studied.
For this, we define $M(t)=\int^{t}_{0}m(\tau)d\tau$, the primitive of $m$ so
that $M(0)=0$. The hypotheses on Kirchhoff function
$m:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ are the following:

($M_{1}$) there exists $m_{0}>0$ such that $m(t)\geq m_{0}$ for all $t\geq 0$;

($M_{2}$) $m(t)$ is nondecreasing for $t\geq 0$;

($M_{3}$) there exists $\theta>1$ such that $\theta M(t)-m(t)t$ is
nondecreasing for $t\geq 0$.

######  Remark 1.1.

By $(M_{1})$, we can get that $M(t)$ is increasing for $t\geq 0$.

######  Remark 1.2.

From $(M_{3})$, we have that

$\theta\mathcal{M}(t)-M(t)t\geq 0,\ \ \forall t\geq 0.$ (1.9)

######  Remark 1.3.

A typical example of a function $m$ satisfying the conditions
$(M_{1})-(M_{3})$ is given by $m(t)=m_{0}+at^{\theta-1}$ with
$\theta>1,m_{0}>0$ and $a\geq 0$. Another example is $m(t)=1+\ln(1+t)$.

######  Remark 1.4.

Here, we compare assumptions $(\overline{M}_{1})-(\overline{M}_{3})$ in [20]
as shown before with our present assumptions $(M_{1})-(M_{3})$. From
$(\overline{M}_{3})$, we can obtain $(M_{3})$ with $\theta=2$. Indeed, for any
$0<t_{1}<t_{2}$,

$\begin{split}2M(t_{1})-m(t_{1})t_{1}&=2M(t_{2})-2\int^{t_{2}}_{t_{1}}m(s)ds-\frac{m(t_{1})t^{2}_{1}}{t_{1}}\\\
&\leq
2M(t_{2})-\frac{m(t_{2})(t^{2}_{2}-t^{2}_{1})}{t_{2}}-\frac{m(t_{2})t^{2}_{1}}{t_{2}}\\\
&=2M(t_{2})-m(t_{2})t_{2},\end{split}$

thus $2M(t)-m(t)t$ is nondecreasing for $t\geq 0$. From $(M_{1})-(M_{2})$, we
can obtain $(\overline{M}_{1})$. Indeed, by $m(t)$ is nondecreasing for $t\geq
0$, we have $\int^{t+s}_{t}m(\tau)d\tau\geq\int^{s}_{0}m(\tau)d\tau$ for all
$s,t\geq 0$, then it holds that
$\int^{t}_{0}m(\tau)d\tau+\int^{t+s}_{t}m(\tau)d\tau\geq\int^{t}_{0}m(\tau)d\tau+\int^{s}_{0}m(\tau)d\tau$,
i.e. $M(t+s)\geq M(t)+M(s)$. Then from (1.9), we can get $M(t)\geq
M(1)t^{\theta}$ for $t\leq 1$, and $M(t)\leq M(1)t^{\theta}$ for $t\geq 1$,
thus $M(t)\leq C_{1}t^{\theta}+C_{2}$ for some $C_{1},C_{2}>0$.

Motivated by pioneer works of Adimurthi [1], de Figueiredo et al. [15] and
J.M. do Ó [18], we treat the so-called subcritical case and also the critical
case. They say that a function $f:(\Omega,\mathbb{R})\to\mathbb{R}$ has
subcritical growth on $\Omega\subset\mathbb{R}^{2}$ if

$\lim_{|u|\rightarrow\infty}\frac{|f(x,u)|}{\exp\left(\alpha
u^{2}\right)}=0,\text{ uniformly on }\Omega,\ \forall\alpha>0,$

and $f$ has critical growth on $\Omega$ if there exists $\alpha_{0}>0$ such
that

$\lim_{|u|\rightarrow\infty}\frac{|f(x,u)|}{\exp\left(\alpha
u^{2}\right)}=0,\text{ uniformly on }\Omega,\ \forall\alpha>\alpha_{0},$

and

$\lim_{|u|\rightarrow\infty}\frac{|f(x,u)|}{\exp\left(\alpha
u^{2}\right)}=\infty,\text{ uniformly on }\Omega,\ \forall\alpha<\alpha_{0}.$

Throughout this paper, we assume the following hypotheses on the function
$f_{i}:\Omega\times\mathbb{R}^{k}\rightarrow\mathbb{R}$ and $F$:

($F_{0}$) $f_{i}$ is continuous and $f_{i}(x,0,\ldots,0)=0$,
$F(x,0,\ldots,0)=0$ uniformly on $x\in\Omega$.

($F_{1}$) $\lim\sup_{|U|\rightarrow
0}\frac{2F(x,U)}{|U|^{2}}<\lambda_{1}m_{0}$ uniformly on $\Omega$, where
$\lambda_{1}=\inf_{U\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})\setminus\\{0\\}}\frac{\|U\|^{2}}{\int_{\Omega}|U|^{2}/|x|^{\beta}dx}>0;$

($F_{2}$) there exist constants $S_{0},M_{0}>0$ such that $0<F(x,U)\leq
M_{0}|\nabla F(x,U)|$, for all $|U|\geq S_{0}$ uniformly on $\Omega$;

($F_{3}$) there exists $\mu>2\theta$ such that $0<\mu F(x,U)\leq U\cdot\nabla
F(x,U)$, for all
$(x,U)\in\Omega\times\mathbb{R}^{k}\setminus\\{\mathbf{0}\\}$;

We say that $U\in H^{1}_{0}(\Omega,\mathbb{R}^{k})$ is a weak solution of
problem (1.4) it holds

$\ m(\|U\|^{2})\int_{\Omega}\nabla U\cdot\nabla\Phi
dx=\int_{\Omega}\frac{\Phi\cdot\nabla
F(x,U)}{|x|^{\beta}}dx+\varepsilon\int_{\Omega}\Phi\cdot Hdx,\ \ \forall\
\Phi\in H^{1}_{0}(\Omega,\mathbb{R}^{k})\\\ .$

Since $f_{i}(x,0,\ldots,0)=0$, $U\equiv\mathbf{0}$ is the trivial solution of
problem (1.4). Thus, our aim is to obtain nontrivial solutions. Now, the main
results of this work can state as follows.

######  Theorem 1.1.

Assume $f_{i}$ has subcritical growth at $\infty$, that is,

$\lim_{|U|\to\infty}\frac{|f_{i}(x,U)|}{e^{\alpha|U|^{2}}}=0,\ \
\text{uniformly on }x\in\Omega,\ \ \forall\alpha>0.$ (1.10)

Moreover, assume $(M_{1})$, $(M_{3})$ and $(F_{1})-(F_{3})$, then there exists
$\varepsilon_{sc}>0$ such that for each $0<\varepsilon<\varepsilon_{sc}$,
problem (1.4) has at least two nontrivial weak solutions. One of them with
positive energy, while the other one with negative energy.

######  Theorem 1.2.

Assume $f_{i}$ has critical growth at $\infty$, that is, if there exists
$\alpha_{0}>0$ such that

$\lim_{|u_{i}|\to\infty}\frac{|f_{i}(x,U)|}{e^{\alpha|U|^{2}}}=\begin{cases}0,\
\ &\forall\alpha>\alpha_{0},\\\\[3.0pt] +\infty,\ \
&\forall\alpha<\alpha_{0},\end{cases}$ (1.11)

uniformly on $x\in\Omega$ and $u_{j}$ where
$j\in\\{1,\ldots,k\\}\backslash\\{i\\}$. Moreover, suppose $(M_{1})-(M_{3})$,
$(F_{1})-(F_{3})$ hold and

$(F_{4})$ if for some $i\in\\{1,\ldots,k\\}$, there exists $\eta_{0}$ such
that

$\liminf_{|u_{i}|\rightarrow\infty}\frac{u_{i}f_{i}(x,0,\ldots,0,u_{i},0,\ldots,0)}{e^{\alpha_{0}|u_{i}|^{2}}}\geq\eta_{0}>\frac{(2-\beta)^{2}m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)}{\alpha_{0}d^{2-\beta}e},$

uniformly on $\Omega$, where $d$ is the radius of the largest open ball
contained in $\Omega$ centered at the origin. Then there exists
$\varepsilon_{c}>0$ such that for each $0<\varepsilon<\varepsilon_{c}$,
problem (1.4) has at least two nontrivial weak solutions. One of them with
positive energy, while the other one with negative energy.

######  Remark 1.5.

If $\beta=\varepsilon=0,\ k=1$, for $(F_{4})$, in [20], the author replaced
$e$ with 2, therefore, in order to get this improvement on the growth of the
nonlinearity $f_{i}$ at $\infty$, it is crucial in our argument to use a new
sequence in [13].

######  Remark 1.6.

When $m\equiv 1$, $k=1$, $\beta=\varepsilon=0$, problems with critical growth
involving the Laplace operator in bounded domains of $\mathbb{R}^{2}$ have
been investigated in [3, 4, 6, 15], quasilinear elliptic problems with
critical growth for $N$-Laplacian in bounded domains of $\mathbb{R}^{N}$ have
been studied in [1, 18]. Moreover, for the problems with critical growth in
bounded domains in $\mathbb{R}^{2}$ and $f$ satisfied (see examples in [1, 15,
18]) the asymptotic hypothesis

$\displaystyle\liminf_{|u|\rightarrow\infty}\frac{uf(x,u)}{e^{\alpha_{0}u^{2}}}\geq\eta_{0}^{\prime}>\frac{2}{\alpha_{0}d^{2}},$
(1.12)

and for Kirchhoff problem, in [20]

$\displaystyle\liminf_{|u|\rightarrow\infty}\frac{uf(x,u)}{e^{\alpha_{0}u^{2}}}\geq\eta_{0}^{\prime\prime}>\frac{2m\big{(}\frac{4\pi}{\alpha_{0}}\big{)}}{\alpha_{0}d^{2}}.$
(1.13)

What’s more, when $m\equiv 1$, de Souza studied this problem in [16] and he
assumed the hypothesis

$\displaystyle\lim\inf_{|U|\rightarrow\infty}\frac{u_{i}f_{i}(x,U)}{e^{2^{k-1}\alpha_{0}|U|^{2}}}\geq\eta_{0}^{\prime\prime\prime}>\frac{(2-\beta)^{2}}{2^{k-1}\alpha_{0}d^{2-\beta}e},$
(1.14)

for some $i\in\\{1,2,\ldots,k\\}$. Motivated by [2] and [28], where they
proved a version of Trudinger-Moser inequality with singular weight and
studied the existence of positive weak solutions for the following semilinear
and homogeneous elliptic problem

$\displaystyle\left\\{\begin{array}[]{ll}-\Delta
u=\frac{f(x,u)}{|x|^{\beta}},&\mbox{in}\ \ \Omega,\\\\[5.69054pt]
u=0,&\mbox{on}\ \ \partial\Omega.\end{array}\right.$

In the present paper, we improve and complement some of the results cited
above for singular and nonhomogeneous case and extend the results to systems.
Moreover, thanks to de Figueiredo, do Ó and Ruf of [13] have constructed a
proper sequence which makes the hypotheses (1.12) and (1.13) can be improved
to $(F_{4})$ in Theorem 1.2. And using the improvement of the Young’s
inequality which will be introduced in Lemma 2.2, (1.14) can be improved
better in $(F_{4})$.

######  Remark 1.7.

On the basis of assumption $(F_{0})$, if we further assume that for any
$u_{j}\leq 0$ where $j\in\\{1,\ldots,k\\}$, $f_{i}(x,u_{1},\ldots,u_{k})\equiv
0$ for all $i\in\\{1,\ldots,k\\}$, uniformly in $x\in\Omega$, and $h_{i}\geq
0$ for all $i\in\\{1,\ldots,k\\}$, and $(F_{4})$ changes to

$\liminf_{u_{1},\ldots,u_{k}\rightarrow+\infty}\frac{U\cdot\nabla
F(x,U)}{e^{\alpha_{0}|U|^{2}}}\geq\eta_{0}>\frac{(2-\beta)^{2}m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)}{\alpha_{0}d^{2-\beta}e},$

where $U=(u_{1},\ldots,u_{k})$, then by using Maximum principle, we can proof
the solutions obtained in Theorem 1.2 are entire positive, i.e. each of the
component is positive. A typical example is
$F(x,U)=|U|^{\mu}\exp(\alpha_{0}|U|^{2})\prod^{k}_{i=1}{\rm sign}u_{i}$, where
${\rm sign}t=0$ if $t\leq 0$ and ${\rm sign}t=1$ if $t>0$.

This paper is organized as follows: Section 2 contains some technical results.
In Section 3, we present the variational setting in which our problem will be
treated. Section 4 is devoted to show some properties of the Palais-Samle
sequences. Finally, we split Section 5 into two subsections for the
subcritical and critical cases, and we complete the proofs of our main
results. Hereafter, $C,C_{0},C_{1},C_{2}...$ will denote positive (possibly
different) constants.

##  2 Some preliminary results

Now, we introduce some famous inequalities as follows, and inspired by those
inequalities, we conclude some similar forms of inequalities. In this paper,
we shall use the following version of the Trudinger-Moser inequality with a
singular weight due to Adimurthi-Sandeep [2]:

######  Lemma 2.1.

Let $\Omega$ be a bounded domain in $\mathbb{R}^{2}$ containing the origin and
$u\in H^{1}_{0}(\Omega)$. Then for every $\alpha>0$, and $\beta\in[0,2)$,

$\displaystyle\int_{\Omega}\frac{e^{\alpha|u|^{2}}}{|x|^{\beta}}<\infty.$

Moreover, there exists constant $C(\Omega)$ depending only on $\Omega$ such
that

$\displaystyle\sup_{||\nabla u||_{2}\leq
1}\int_{\Omega}\frac{e^{\alpha|u|^{2}}}{|x|^{\beta}}\leq C(\Omega),$

if and only if $\frac{\alpha}{4\pi}+\frac{\beta}{2}\leq 1$.

Then, we give two useful algebraic inequalities that will be used
systematically in the rest of the paper as the following:

######  Lemma 2.2.

(Improvement of the Young’s inequality) Let $a_{1},\ldots,a_{k}>0$,
$p_{1},\ldots,p_{k}>1$ with
$\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{k}}=1$, then

$\displaystyle a_{1}a_{2}\cdots
a_{k}\leq\frac{a^{p_{1}}_{1}}{p_{1}}+\frac{a^{p_{2}}_{2}}{p_{2}}+\cdots+\frac{a^{p_{k}}_{k}}{p_{k}}.$
(2.1)

###### Proof.

We will use the mathematical induction to proof this. When $k=2$, by Young
inequality, we can know this conclusion is correct. Suppose that when $k=s-1$
the conclusion is correct, we are going to show that when $k=s$ the conclusion
is still correct. Let

$\displaystyle\frac{1}{q}=\sum^{s-1}_{i=1}\frac{1}{p_{i}},\ \
\frac{1}{q}+\frac{1}{p_{s}}=1,\ \ \mbox{then}\ \
\sum^{s-1}_{i=1}\frac{1}{p_{i}/q}=1.$

Thus,

$\displaystyle\prod^{s}_{i=1}a_{i}=\Big{(}\prod^{s-1}_{i=1}a_{i}\Big{)}a_{s}\leq\frac{1}{q}\Big{(}\prod^{s-1}_{i=1}a_{i}\Big{)}^{q}+\frac{1}{p_{s}}a^{p_{s}}_{s},$

by the mathematical induction, we can get that

$\displaystyle\frac{1}{q}\Big{(}\prod^{s-1}_{i=1}a_{i}\Big{)}^{q}=\frac{1}{q}\Big{(}\prod^{s-1}_{i=1}a^{q}_{i}\Big{)}\leq\frac{1}{q}\sum^{s-1}\Big{[}\frac{1}{p_{i}/q}\big{(}a^{q}_{i})^{p_{i}/q}\Big{]}=\sum^{s-1}\frac{1}{p_{i}}a^{p_{i}}_{i}.$

Therefore

$\displaystyle\prod^{s}_{i=1}a_{i}\leq\sum^{s-1}_{i=1}\frac{1}{p_{i}}a^{p_{i}}_{i}+\frac{1}{p_{s}}a^{p_{s}}_{s}=\sum^{s}_{i=1}\frac{1}{p_{i}}a^{p_{i}}_{i}.$

This lemma is proved. If we take $p_{1}=p_{2}=\cdots=p_{k}=k$, we can get that

$\displaystyle a_{1}a_{2}\cdots
a_{k}\leq\frac{1}{k}\sum^{k}_{i=1}a^{k}_{i}\leq\sum^{k}_{i=1}a^{k}_{i}.$ (2.2)

∎

######  Lemma 2.3.

Suppose $a_{1},a_{2},\ldots,a_{k}\geq 0$ with $a_{1}+a_{2}+\cdots+a_{k}<1$,
then there exist $p_{1},\ldots,p_{k}>1$ satisfying
$\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{k}}=1$, such that

$\displaystyle p_{i}a_{i}<1,\quad\mbox{for all}\ \ i=1,2,\ldots,k.$ (2.3)

Moreover, if $a_{1},a_{2},\ldots,a_{k}\geq 0$ satisfying
$a_{1}+a_{2}+\cdots+a_{k}=1$, then we can take $p_{i}=\frac{1}{a_{i}}$ such
that $\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{k}}=1$ and

$\displaystyle p_{i}a_{i}=1,\quad\mbox{for all}\ \ i=1,2,\ldots,k.$ (2.4)

###### Proof.

For the case $a_{1},a_{2},\ldots,a_{k}\geq 0$ with
$a_{1}+a_{2}+\cdots+a_{k}<1$. We also make use of the mathematical induction
as previous. When $k=2$, $a_{1},a_{2}\geq 0$ with $a_{1}+a_{2}<1$. If
$a_{2}=0$ (or $a_{1}=0$), then taking
$p_{1}=\left(\frac{1}{2}+\frac{1}{2a_{1}}\right)>1$ (or
$p_{2}=\left(\frac{1}{2}+\frac{1}{2a_{2}}\right)>1$), we can obtain
$p_{1}a_{1}<1,\ p_{2}a_{2}<1$, where $p_{2}=\frac{p_{1}}{p_{1}-1}$. If
$a_{1},a_{2}>0$, then taking
$p_{1}=\left(\frac{1}{2(1-a_{1})}+\frac{1}{2a_{2}}\right)>1$, we can obtain
$p_{1}a_{1}<1,\ p_{2}a_{2}<1$, where $p_{2}=\frac{p_{1}}{p_{1}-1}$. Suppose
that when $k=s-1$ the conclusion is correct, if $k=s$ the conclusion is still
correct, then the lemma follows. Let

$\displaystyle(a_{1}+a_{2}+\cdots+a_{s-1})+a_{s}<1,$

then there exist $q_{1},q_{2}>1$ satisfying
$\frac{1}{q_{1}}+\frac{1}{q_{2}}=1$ such that

$\displaystyle q_{1}(a_{1}+a_{2}+\cdots+a_{s-1})<1,\ \ q_{2}a_{s}<1.$

Then by the assumption, there exist $q_{3},q_{4},\ldots,q_{s+1}>1$ satisfying
$\frac{1}{q_{3}}+\frac{1}{q_{4}}+\cdots+\frac{1}{q_{s+1}}=1$ such that

$\displaystyle q_{3}(q_{1}a_{1})<1,\ q_{4}(q_{1}a_{2})<1,\ldots,\
q_{s+1}(q_{1}a_{s-1})<1,$

i.e.

$\displaystyle(q_{3}q_{1})a_{1}<1,\ (q_{4}q_{1})a_{2}<1,\ldots,\
(q_{s+1}q_{1})a_{s-1}<1.$

Taking $p_{1}=q_{1}q_{3},\ p_{2}=q_{1}q_{4},\ldots,\ p_{s-1}=q_{1}q_{s+1},\
p_{s}=q_{2}$, then it holds that

$\displaystyle\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{s}}=\frac{1}{q_{1}}\left(\frac{1}{q_{3}}+\frac{1}{q_{4}}+\cdots+\frac{1}{q_{s+1}}\right)+\frac{1}{q_{2}}=\frac{1}{q_{1}}+\frac{1}{q_{2}}=1,$

and

$\displaystyle p_{1}a_{1}<1,\ p_{2}a_{2}<1,\ldots,\ p_{s}a_{s}<1.$

For the case $a_{1},a_{2},\ldots,a_{k}\geq 0$ satisfying
$a_{1}+a_{2}+\cdots+a_{k}=1$. If $a_{i}>0$ for all $i\in\\{1,\ldots,k\\}$,
then we can take $p_{i}=\frac{1}{a_{i}}$ such that
$\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{k}}=1$ and (2.3) holds. If
$a_{i}=0$ for some $i\in\\{1,\ldots,k\\}$ which same as the case $(k-1)$. The
proof is complete. ∎

By the above inequalities, we begin to establish the singular Trudinger-Moser
type inequalities in $H^{1}_{0}(\Omega,\mathbb{R}^{k})$:

######  Lemma 2.4.

(Improvement of the Trudinger-Moser inequality) Let $\Omega$ be a bounded
domain in $\mathbb{R}^{2}$ containing the origin and
$U=(u_{1},\ldots,u_{k})\in H^{1}_{0}(\Omega,\mathbb{R}^{k})$. Then for every
$\alpha>0$, and $\beta\in[0,2)$,

$\displaystyle\int_{\Omega}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}<\infty.$
(2.5)

Moreover, it holds that

$\displaystyle\sup_{U\in H^{1}_{0}(\Omega,\mathbb{R}^{k}),\ \|U\|\leq
1}\int_{\Omega}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}\leq C(\Omega).$ (2.6)

if and only if $\frac{\alpha}{4\pi}+\frac{\beta}{2}\leq 1$, where $C(\Omega)$
is given in Lemma 2.1.

###### Proof.

Because $U=(u_{1},\ldots,u_{k})$, we can get
$|U|^{2}=\sum^{k}_{i=1}|u_{i}|^{2}$. Thus, by using (2.2) and Lemma 2.1, we
have

$\displaystyle\int_{\Omega}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}=\int_{\Omega}\frac{e^{\alpha|u_{1}|^{2}}\cdots
e^{\alpha|u_{k}|^{2}}}{|x|^{\beta}}\leq\sum^{k}_{i=1}\int_{\Omega}\frac{e^{k\alpha|u_{i}|^{2}}}{|x|^{\beta}}<\infty.$

Then we begin to proof (2.6). For each $U\in H^{1}_{0}(\Omega,\mathbb{R}^{k})$
satisfying $\|U\|\leq 1$, then $\|U\|^{2}=\sum^{k}_{i=1}\|u_{i}\|^{2}\leq 1$,
Lemma 2.3 shows that there exist $p_{1},\ldots,p_{k}>1$ satisfying
$\frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots+\frac{1}{p_{k}}=1$ such that
$p_{i}\|u_{i}\|^{2}\leq 1$ holds, for all $i=1,2,\ldots,k$. If
$\frac{\alpha}{4\pi}+\frac{\beta}{2}\leq 1$, then it also holds that
$\frac{p_{i}\|u_{i}\|^{2}\alpha}{4\pi}+\frac{\beta}{2}\leq 1$, for all
$i=1,2,\ldots,k$. Then by using Lemma 2.2, i.e. the improvement of the Young’s
inequality, and from Lemma 2.1 we have

$\displaystyle\begin{split}\int_{\Omega}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}=\int_{\Omega}\frac{e^{\alpha|u_{1}|^{2}}\cdots
e^{\alpha|u_{k}|^{2}}}{|x|^{\beta}}\leq\sum^{k}_{i=1}\int_{\Omega}\frac{e^{p_{i}\alpha|u_{i}|^{2}}}{p_{i}|x|^{\beta}}=\sum^{k}_{i=1}\int_{\Omega}\frac{e^{p_{i}\alpha\|u_{i}\|^{2}(\frac{u_{i}}{\|u_{i}\|})^{2}}}{p_{i}|x|^{\beta}}\leq\sum^{k}_{i=1}\frac{C(\Omega)}{p_{i}}=C(\Omega),\end{split}$

then (2.6) follows. If $\frac{\alpha}{4\pi}+\frac{\beta}{2}>1$, we take
$U=(u,0,\ldots,0)$, then Lemma 2.1 shows that the supremum for the integral in
(2.6) is infinite. Thus the proof is complete. ∎

######  Lemma 2.5.

Let $\\{U_{n}\\}$ be a sequence of functions in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$ with $\|U_{n}\|=1$ such that
$U_{n}\rightharpoonup U\neq 0$ weakly in $H^{1}_{0}(\Omega,\mathbb{R}^{k})$.
Then for any $0<p<\frac{2\pi(2-\beta)}{{(1-\|U\|^{2})}}$ and $\beta\in[0,2)$,
we have

$\sup_{n}\int_{\Omega}\frac{e^{p|U_{n}|^{2}}}{|x|^{\beta}}<\infty.$

###### Proof.

Since $U_{n}\rightharpoonup U\neq 0$ and $\|\nabla U_{n}\|_{2}=1$, we conclude
that

$\displaystyle\|U_{n}-U\|^{2}=1-2\langle U_{n},U\rangle+\|U\|^{2}\rightarrow
1-\|U\|^{2}<\frac{2\pi(2-\beta)}{p}$

Thus, for large $n$ we have

$\displaystyle\frac{p\|U_{n}-U\|^{2}}{4\pi}+\frac{\beta}{2}<1.$

Now we can choose $q>1$ close to 1 and $\epsilon>0$ such that

$\displaystyle\frac{qp(1+\epsilon^{2})\|U_{n}-U\|^{2}}{4\pi}+\frac{q\beta}{2}\leq
1.$

Lemmas 2.4 shows that

$\displaystyle\int_{\Omega}\frac{e^{qp(1+\epsilon^{2})|U_{n}-U|^{2}}}{|x|^{q\beta}}\leq
C(\Omega).$

Moreover, since

$\displaystyle p|U_{n}|^{2}\leq
p(1+\epsilon^{2})|U_{n}-U|^{2}+p(1+1/\epsilon^{2})|U|^{2},$

which can be proved by Young inequality, then it follows that

$\displaystyle e^{p|U_{n}|^{2}}\leq
e^{p(1+\epsilon^{2})|U_{n}-U|^{2}}e^{p(1+1/\epsilon^{2})|U|^{2}}.$

Consequently, by Hölder inequality,

$\displaystyle\begin{split}\int_{\Omega}\frac{e^{p|U_{n}|^{2}}}{|x|^{\beta}}&\leq\left(\int_{\Omega}\frac{e^{qp(1+\epsilon^{2})|U_{n}-U|^{2}}}{|x|^{q\beta}}\right)^{1/q}\left(\int_{\Omega}e^{rp(1+1/\epsilon^{2})|U|^{2}}\right)^{1/r}\leq
C\left(\int_{\Omega}e^{rp(1+1/\epsilon^{2})|U|^{2}}\right),\end{split}$

for large $n$, where $r=\frac{q}{q-1}$. By Lemma 2.4, we know the second term
in the last inequality is bounded, and this lemma is proved. ∎

######  Remark 2.1.

Lemma 2.5 is actually an expression that Concentration-compactness principle
for a singular Trudinger-Moser inequality which is better than ([16], Lemma
2.3). However, it is still not clear whether
$\frac{2\pi(2-\beta)}{1-\|U\|^{2}}$ is sharp or not. This is still an open
question.

######  Lemma 2.6.

If $V\in H^{1}_{0}(\Omega,\mathbb{R}^{k}),\ \alpha>0,\ q>0,\ \beta\in[0,2)$
and $\|V\|\leq N$ with $\frac{\alpha N^{2}}{4\pi}+\frac{\beta}{2}<1$, then
there exists $C=C(\alpha,N,q)>0$ such that

$\displaystyle\int_{\Omega}|V|^{q}\frac{e^{\alpha|V|^{2}}}{|x|^{\beta}}\leq
C\|V\|^{q}.$ (2.7)

###### Proof.

We consider $r>1$ close to 1 such that $\frac{r\alpha
N^{2}}{4\pi}+\frac{r\beta}{2}\leq 1$ and $sq\geq 1$, where $s=\frac{r}{r-1}$.
By using Hölder inequality and Lemma 2.4, we have

$\displaystyle\begin{split}\int_{\Omega}|V|^{q}\frac{e^{\alpha|V|^{2}}}{|x|^{\beta}}&\leq\left(\int_{\Omega}\frac{e^{r\alpha|V|^{2}}}{|x|^{r\beta}}\right)^{1/r}\|V\|^{q}_{qs}\leq
C(\Omega)\|V\|^{q}_{qs}.\end{split}$

Finally, using the continuous embedding
$H^{1}_{0}(\Omega,\mathbb{R}^{k})\hookrightarrow
L^{sq}(\Omega,\mathbb{R}^{k})$, we conclude that

$\displaystyle\int_{\Omega}|V|^{q}\frac{e^{\beta|V|^{2}}}{|x|^{\beta}}\leq
C\|V\|^{q}.$

∎

##  3 The variational framework

We now consider the functional $I$ given by

$I_{\varepsilon}(U)=\frac{1}{2}M(\|U\|^{2})-\int_{\Omega}\frac{F(x,U)}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U\cdot
H(x)dx.$

######  Lemma 3.1.

Under our assumptions we have that $I$ is well defined and $C^{1}$ on
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$. Moreover,

$\langle I_{\varepsilon}^{\prime}(U),\Phi\rangle_{*}=m(\|U\|^{2})\langle
U,\Phi\rangle-\int_{\Omega}\frac{\Phi\cdot\nabla{F(x,U)}}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}\Phi\cdot
Hdx,$

where $\Phi\in H^{1}_{0}(\Omega,\mathbb{R}^{k})$, here
$\langle\cdot,\cdot\rangle_{*}$ simply denotes the dual pairing between
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$ and its dual space
$\left(H^{1}_{0}(\Omega,\mathbb{R}^{k})\right)^{*}$.

###### Proof.

We have that $f_{i}$ is continuous and has subcritical (or critical) growth at
$\infty$, as defined in (1.10) (or (1.11)). Thus, given $\alpha>0$ (or
$\alpha>\alpha_{0}$), there exists $C>0$ such that $|f_{i}(x,U)|\leq
Ce^{\alpha|U|^{2}}$ for all $(x,U)\in\Omega\times\mathbb{R}^{k}$. Then,

$|\nabla F(x,U)|\leq\sum^{k}_{i=1}|f_{i}(x,U)|\leq C_{1}e^{\alpha|U|^{2}},\ \
\text{for all }(x,U)\in\Omega\times\mathbb{R}^{k}.\\\ $ (3.1)

By $(F_{1})$, given $\epsilon>0$ there exists $\delta>0$ such that

$|F(x,U)|\leq\frac{\lambda_{1}m_{0}-\epsilon}{2}|U|^{2}\ \ \text{always that
}|U|<\delta.\\\ $ (3.2)

Thus, using (3.1), (3.2) and $(F_{3})$, we have

$\displaystyle\int_{\Omega}\frac{|F(x,U)|}{|x|^{\beta}}dx\leq\frac{\lambda_{1}m_{0}-\epsilon}{2}\int_{\Omega}\frac{|U|^{2}}{|x|^{\beta}}dx+C_{1}\int_{\Omega}\frac{|U|e^{\alpha|U|^{2}}}{|x|^{\beta}}dx.$

Considering the continuous imbedding
$H^{1}_{0}(\Omega,\mathbb{R}^{k})\hookrightarrow L^{s}(\Omega,\mathbb{R}^{k})$
for $s\geq 1$ and using Lemma 2.6, it follows that
$\frac{F(x,U)}{|x|^{\beta}}\in L^{1}(\Omega)$ which implies that
$I_{\varepsilon}$ is well defined. And we can see that $I\in
C^{1}(H^{1}_{0}(\Omega,\mathbb{R}^{k}),\mathbb{R})$ with

$\langle I_{\varepsilon}^{\prime}(U),\Phi\rangle_{*}=m(||U||^{2})\langle
U,\Phi\rangle-\int_{\Omega}\frac{\Phi\cdot\nabla{F(x,U)}}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}\Phi\cdot
Hdx,$

for all $U,\Phi\in H^{1}_{0}(\Omega,\mathbb{R}^{k})$. ∎

From Lemma 3.1, we have that critical points of the functional
$I_{\varepsilon}$ are precisely weak solutions of problem (1.4). In the next
three lemmas we check that the functional $I_{\varepsilon}$ satisfies the
geometric conditions of the Mountain-pass theorem.

######  Lemma 3.2.

Suppose that $(M_{1})$ and $(F_{1}),\ (F_{3})$ hold and the function $f_{i}$
has subcritical (or critical) growth at $\infty$. Then for small
$\varepsilon$, there exist positive number $\rho_{\varepsilon}$ and
$\varsigma$ such that

$\displaystyle I_{\varepsilon}(U)\geq\varsigma,\ \ \forall U\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})\ \ \mbox{with}\ \ \|U\|=\rho_{\varepsilon}.$

Moreover, $\rho_{\varepsilon}$ can be chosen such that $\rho_{\varepsilon}\to
0$ as $\varepsilon\to 0$.

###### Proof.

By $(F_{1})$, given $\kappa>0$, there exists $\delta>0$ such that

$\displaystyle|\nabla F(x,U)|\leq(\lambda_{1}m_{0}-\kappa)|U|$

always that $|U|<\delta$. On the other hand, for $\alpha>0$ (subcritical case)
or $\alpha>\alpha_{0}$ (critical case), we have that there exists $C_{1}>0$
such that $|f_{i}(x,U)|\leq C_{1}|U|^{q-1}e^{\alpha|U|^{2}}$ for all
$|U|\geq\delta$ with $q>2$. Thus,

$\displaystyle|\nabla
F(x,U)|\leq\sum^{k}_{i=1}|f_{i}(x,U)|\leq(\lambda_{1}m_{0}-\kappa)|U|+C_{2}|U|^{q-1}e^{\alpha|U|^{2}},\
\ \forall(x,U)\in\Omega\times\mathbb{R}^{k}.$ (3.3)

Thus, by $(M_{1}),\ (F_{3})$ and (3.3),

$\displaystyle\begin{split}I_{\varepsilon}(U)=&\frac{1}{2}M(\|U\|^{2})-\int_{\Omega}\frac{F(x,U)}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U\cdot
H(x)dx\\\
\geq&\frac{1}{2}m_{0}\|U\|^{2}-\int_{\Omega}\frac{F(x,U)}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U\cdot
H(x)dx\\\
\geq&\frac{1}{2}m_{0}\|U\|^{2}-\frac{1}{\mu}\int_{\Omega}\frac{U\cdot\nabla
F(x,U)}{|x|^{\beta}}dx-\varepsilon\|U\|\|H\|_{*}\\\
\geq&\frac{1}{2}m_{0}\|U\|^{2}-\frac{\lambda_{1}m_{0}-\kappa}{\mu}\int_{\Omega}\frac{|U|^{2}}{|x|^{\beta}}dx-
C_{2}\int_{\Omega}|U|^{q}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}dx-\varepsilon\|U\|\|H\|_{*}\\\
\geq&\Big{(}\frac{m_{0}}{2}-\frac{m_{0}}{\mu}+\frac{\kappa}{\mu\lambda_{1}}\Big{)}\|U\|^{2}-C_{2}\int_{\Omega}|U|^{q}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}dx-\varepsilon\|U\|\|H\|_{*},\end{split}$

By Lemma 2.6, there exists $N>0$ such that $\alpha N^{2}/4\pi+\beta/2<1$ and
we take $\|U\|\leq N$, there exists $C>0$ such that
$\int_{\Omega}|U|^{q}\frac{e^{\alpha|U|^{2}}}{|x|^{\beta}}dx\leq C\|U\|^{q}$.
Therefore,

$\displaystyle
I_{\varepsilon}(U)\geq\Big{(}\frac{m_{0}}{2}-\frac{m_{0}}{\mu}+\frac{\kappa}{\mu\lambda_{1}}\Big{)}\|U\|^{2}-C_{3}\|U\|^{q}-\varepsilon\|U\|\|H\|_{*},$

Since $m_{0}>0,\ \mu>2\theta>2,\ q>2$, then for small $\varepsilon$, there
exists $\xi_{\varepsilon}>0$ such that
$\Big{(}\frac{m_{0}}{2}-\frac{m_{0}}{\mu}+\frac{\kappa}{\mu\lambda_{1}}\Big{)}\xi_{\varepsilon}^{2}-C_{3}\xi_{\varepsilon}^{q}-\varepsilon\xi_{\varepsilon}\|H\|_{*}>0$.
Consequently, taking $\rho_{\varepsilon}=\min\\{N,\xi_{\varepsilon}\\}>0$, we
can get that $I_{\varepsilon}(U)\geq\varsigma$ whenever
$\|U\|=\rho_{\varepsilon}$ where
$\varsigma:=\Big{(}\frac{m_{0}}{2}-\frac{m_{0}}{\mu}+\frac{\kappa}{\mu\lambda_{1}}\Big{)}\rho_{\varepsilon}^{2}-C_{3}\rho_{\varepsilon}^{q}-\varepsilon\rho_{\varepsilon}\|H\|_{*}>0$.
And it is worth noting that, $\rho_{\varepsilon}\rightarrow 0$ as
$\varepsilon\rightarrow 0$. ∎

######  Lemma 3.3.

Assume that $(M_{1}),\ (M_{3}),\ (F_{3})$ hold and the function $f_{i}$ has
subcritical (or critical) growth at $\infty$. Then there exists $E\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})$ with $\|E\|>\rho_{\varepsilon}$ such that

$\displaystyle I_{\varepsilon}(E)<\inf_{\|U\|=\rho_{\varepsilon}}I(U).$

###### Proof.

We shall make use of the polar coordinate representation

$\displaystyle U=(\nu,\phi)=(\nu,\phi_{1},\ldots,\phi_{k-1}),$

where $\nu\geq 1,\ -\pi\leq\phi_{1}\leq\pi,\
0\leq\phi_{2},\ldots,\phi_{k-1}\leq\pi$ and

$\displaystyle\begin{split}u_{1}&=\nu\sin(\phi_{1})\sin(\phi_{2})\cdots\sin(\phi_{k-1}),\\\
u_{2}&=\nu\cos(\phi_{1})\sin(\phi_{2})\cdots\sin(\phi_{k-1}),\\\
u_{3}&=\nu\cos(\phi_{2})\cdots\sin(\phi_{k-1}),\\\ \vdots\\\
u_{k}&=\nu\cos(\phi_{k-1}).\end{split}$

Substituting in $(F_{3})$, we get $0<\mu F(x,U)\leq\nu F_{\nu}(x,U)$ and hence

$\displaystyle F(x,U)\geq\left(\min_{|W|=1}F(x,W)\right)|U|^{\mu},\ \
\mbox{for all}\ \ x\in\Omega\ \ \mbox{and}\ \ |U|\geq 1.$

Hence, for all $U\in H^{1}_{0}(\Omega,\mathbb{R}^{k})\setminus\\{0\\}$ with
$|U|\geq 1$, we have that

$\displaystyle F(x,U)\geq C|U|^{\mu}.$

From (1.9), we have that $0<M(t)\leq M(1)t^{\theta}$ for all $t\geq 1$. Thus
we have

$\displaystyle I_{\varepsilon}(tU)\leq
C_{1}t^{2\theta}\|U\|^{2\theta}-C_{2}t^{\mu}\int_{K}\frac{|U|^{\mu}}{|x|^{\beta}}dx-t\varepsilon\int_{\Omega}U\cdot
Hdx,$

for $t$ large enough, where $C_{1},\ C_{2}>0,\ \mu>2\theta>2$, $K$ is the
compact subset of $\Omega$ which yields
$I_{\varepsilon}(tU)\rightarrow-\infty$ as $t\rightarrow+\infty$. Setting
$E=tU$ with $t$ large enough such that $I_{\varepsilon}(E)<0$ with
$\|E\|>\rho_{\varepsilon}$. Thus, the proof is finished. ∎

######  Lemma 3.4.

If $f_{i}$ has subcritical (or critical) growth at $\infty$, there exists
$\eta_{\varepsilon}>0$ and $V\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})\backslash\\{\mathbf{0}\\}$ such that
$I_{\varepsilon}(tV)<0$ for all $0<t<\eta_{\varepsilon}$. In particular,

$\displaystyle\inf_{\|U\|\leq\eta_{\varepsilon}}I_{\varepsilon}(U)<0.$

###### Proof.

Since
$h_{i}\in\big{(}\big{(}H^{1}_{0}(\Omega)\big{)}^{*},\|\cdot\|_{*}\big{)}\backslash\\{0\\}$
for some $i\in\\{1,\ldots,k\\}$, then by the Riesz representation theorem, the
problem

$\displaystyle-\Delta v_{i}=h_{i},\ \ x\in\Omega;\ \ v_{i}=0\ \ \mbox{on}\ \
\partial\Omega,$

has a unique nontrivial weak solution $v_{i}$ in $H^{1}_{0}(\Omega)$. Thus,

$\displaystyle\int_{\Omega}h_{i}v_{i}=\|v_{i}\|^{2}>0.$

Since $f_{i}(x,0,\ldots,0)=0$, by continuity, $(F_{3})$ and (1.9), $\theta>1$,
it follows that there exists $\eta_{\varepsilon}>0$ such that for all
$0<t<\eta_{\varepsilon}$,

$\displaystyle\begin{split}\frac{d}{dt}[I_{\varepsilon}((0,\ldots,tv_{i},\ldots,0))]=&m(t^{2}\|v_{i}\|^{2})t\|v_{i}\|^{2}-\int_{\Omega}\frac{v_{i}f_{i}(x,(0,\ldots,tv_{i},\ldots,0))}{|x|^{\beta}}-\varepsilon\int_{\Omega}h_{i}v_{i}\\\
\leq&C\|v_{i}\|^{2}t-\varepsilon\|v_{i}\|^{2}-\int_{\Omega}\frac{v_{i}f_{i}(x,(0,\ldots,tv_{i},\ldots,0))}{|x|^{\beta}}<0.\end{split}$

Using that $I_{\varepsilon}(\mathbf{0})=0$, it must hold that
$I_{\varepsilon}(tV)<0$ for all $0<t<\eta_{\varepsilon}$ where
$V=(0,\ldots,v_{i},\ldots,0)$. ∎

##  4 On Palais-Smale sequences

To prove that a Palais-Smale sequence converges to a weak solution of problem
(1.4), we need to establish the following lemma.

######  Lemma 4.1.

Assume that $(M_{1}),\ (M_{3}),\ (F_{3})$ hold and $f_{i}$ has subcritical (or
critical) growth at $\infty$. Let $\\{U_{n}\\}\subset
H^{1}_{0}(\Omega,\mathbb{R}^{k})$ be the Palais-Smale sequence for functional
$I_{\varepsilon}$ at finite level. Then there exist $C>0$ such that

$\displaystyle\|U_{n}\|\leq C,\ \ \int_{\Omega}\frac{\left|U_{n}\cdot\nabla
F(x,U_{n})\right|}{|x|^{\beta}}dx\leq C\ \ \mbox{and}\ \
\int_{\Omega}\frac{F(x,U_{n})}{|x|^{\beta}}dx\leq C.$

###### Proof.

Let $\\{U_{n}\\}\subset H^{1}_{0}(\Omega,\mathbb{R}^{k})$ be a sequence such
that $I_{\varepsilon}(U_{n})\rightarrow c$ and
$I_{\varepsilon}^{\prime}(U_{n})\rightarrow 0$, where $|c|<\infty$, then we
can take this as follows:

$I_{\varepsilon}(U_{n})=\frac{1}{2}M(\|U_{n}\|^{2})-\int_{\Omega}\frac{F(x,U_{n})}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U_{n}\cdot
Hdx=c+\delta_{n},$ (4.1)

where $\delta_{n}\rightarrow 0$ as $n\rightarrow\infty$, and

$\langle
I_{\varepsilon}^{\prime}(U_{n}),U_{n}\rangle_{*}=m(\|U_{n}\|^{2})\|U_{n}\|^{2}-\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U_{n}\cdot
Hdx=o(\|U_{n}\|).$ (4.2)

Then for $n$ large enough, by $(M_{1}),\ (F_{3})$ and (1.9), it holds that

$\displaystyle\begin{split}C+\|U_{n}\|\geq&I_{\varepsilon}(U_{n})-\frac{1}{\mu}\langle
I^{\prime}_{\varepsilon}(U_{n}),U_{n}\rangle_{*}\\\
=&\frac{1}{2}M(\|U_{n}\|^{2})-\frac{1}{\mu}m(\|U_{n}\|^{2})\|U_{n}\|^{2}+\frac{1}{\mu}\int_{\Omega}\frac{\left(U_{n}\cdot\nabla
F(x,U_{n})-\mu F(x,U_{n})\right)}{|x|^{\beta}}dx\\\
&-\frac{\mu-1}{\mu}\varepsilon\int_{\Omega}U_{n}\cdot Hdx\\\
\geq&\frac{1}{2\theta}\Big{[}\theta
M(\|U_{n}\|^{2})-m(\|U_{n}\|^{2})\|U_{n}\|^{2}\Big{]}+\Big{(}\frac{1}{2\theta}-\frac{1}{\mu}\Big{)}m(\|U_{n}\|^{2})\|U_{n}\|^{2}\\\
&-\frac{\mu-1}{\mu}\varepsilon\|U_{n}\|\|H\|_{*}\\\
\geq&\Big{(}\frac{1}{2\theta}-\frac{1}{\mu}\Big{)}m(\|U_{n}\|^{2})\|U_{n}\|^{2}-\frac{\mu-1}{\mu}\varepsilon\|U_{n}\|\|H\|_{*}\\\
\geq&\frac{\mu-2\theta}{2\theta\mu}m_{0}\|U_{n}\|^{2}-\frac{\mu-1}{\mu}\varepsilon\|U_{n}\|\|H\|_{*},\end{split}$

for some $C>0$. Since $\mu>2\theta>2,\ m_{0}>0,\ \varepsilon>0$, we obtain
$\|U_{n}\|$ is bounded. From (4.1) and (4.2), it can be concluded directly
that there exist $C>0$ such that $\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})}{|x|^{\beta}}dx\leq C$ and
$\int_{\Omega}\frac{F(x,U_{n})}{|x|^{\beta}}dx\leq C$. Condition $(F_{3})$
implies $U_{n}\cdot\nabla F(x,U_{n})\geq 0$ for all $x\in\Omega$, thus we have
$\int_{\Omega}\frac{\left|U_{n}\cdot\nabla
F(x,U_{n})\right|}{|x|^{\beta}}dx\leq C$. ∎

In order to show that the limit of a sequence in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$ is a weak solution of problem (1.4) we will
use the following convergence result due to Figueiredo-do Ó-Ref [14] and the
dominated result due to do Ó-Medeiros-Severo [19].

######  Lemma 4.2.

[14] Let $\Omega\in\mathbb{R}^{2}$ be a bounded domain and
$f:\Omega\times\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function,
$\beta\in[0,2)$. Then for any sequence $\\{u_{n}\\}$ in $L^{1}(\Omega)$ such
that

$\displaystyle u_{n}\rightarrow u\ \ in\ \ L^{1}(\Omega),\ \
\frac{f(x,u_{n})}{|x|^{\beta}}\in L^{1}(\Omega)\ \ \mbox{and}\ \
\int_{\Omega}\frac{|f(x,u_{n})u_{n}|}{|x|^{\beta}}dx\leq C,$

up to a sequence we have

$\displaystyle\frac{f(x,u_{n})}{|x|^{\beta}}\rightarrow\frac{f(x,u)}{|x|^{\beta}}\
\ \mbox{in}\ \ L^{1}(\Omega).$

######  Lemma 4.3.

[19] Let $\\{u_{n}\\}$ be a sequence of functions in $H^{1}_{0}(\Omega)$
strongly convergent. Then there exists a subsequence $\\{u_{n_{k}}\\}$ of
$\\{u_{n}\\}$ and $g\in H^{1}_{0}(\Omega)$ such that $|u_{n_{k}}(x)|\leq g(x)$
almost everywhere in $\Omega$.

######  Lemma 4.4.

Assume $(M_{1}),\ (M_{3}),\ (F_{2}),\ (F_{3})$ hold and $f_{i}$ has
subcritical (or critical) growth at $\infty$. Let $\\{U_{n}\\}\subset
H^{1}_{0}(\Omega,\mathbb{R}^{k})$ be the Palais-Smale sequence for functional
$I_{\varepsilon}$ at finite level, then there exists $U\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})$ such that

$\frac{f_{i}(x,U_{n})}{|x|^{\beta}}\rightarrow\frac{f_{i}(x,U)}{|x|^{\beta}}\
\ \mbox{in}\ \ L^{1}(\Omega),\ \ \mbox{for all}\ \ i=1,\ldots,k.$ (4.3)

and

$U_{n}\cdot H\rightarrow U\cdot H\ \ \mbox{in}\ \ L^{1}(\Omega).$ (4.4)

Moreover,

$\frac{F(x,U_{n})}{|x|^{\beta}}\rightarrow\frac{F(x,U)}{|x|^{\beta}}\ \
\mbox{in}\ \ L^{1}(\Omega).$ (4.5)

###### Proof.

According to Lemma 4.1, we know that $\\{U_{n}\\}$ is bounded in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$, then up to a subsequence, for some $U\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})$ such that $U_{n}\rightharpoonup U$ weakly in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$, $U_{n}\rightarrow U$ in
$L^{p}(\Omega,\mathbb{R}^{k})$ for all $p\geq 1$ and $U_{n}(x)\rightarrow
U(x)$ almost everywhere in $\Omega$. Consequently, by Lemmas 4.1 and 4.2, we
have

$\frac{f_{i}(x,U_{n})}{|x|^{\beta}}\rightarrow\frac{f_{i}(x,U)}{|x|^{\beta}}\
\ \mbox{in}\ \ L^{1}(\Omega),\ \ \mbox{for all}\ \ i=1,\ldots,k.$

Since

$\displaystyle\int_{\Omega}|U_{n}\cdot H-U\cdot
H|dx\leq\int_{\Omega}|H||U_{n}-U|dx\leq\|H\|_{2}\|U_{n}-U\|_{2}\rightarrow 0,$

we can conclude that

$U_{n}\cdot H\rightarrow U\cdot H\ \ \mbox{in}\ \ L^{1}(\Omega).$

Then by Lemma 4.3, there exists $g_{i}\in L^{1}(\Omega)$ such that
$\frac{|f_{i}(x,U_{n})|}{|x|^{\beta}}\leq g_{i}$ almost everywhere in
$\Omega$. From $(F_{2})$ we can conclude that

$\displaystyle|F(x,U_{n})|\leq\sup_{\Omega\times[-S_{0},S_{0}]}|F(x,U_{n}(x))|+M_{0}|\nabla
F(x,U_{n})|\ \ \mbox{a.e. in}\ \ \Omega.$

Thus, by the generalized Lebesgue dominated convergence theorem, we get

$\frac{F(x,U_{n})}{|x|^{\beta}}\rightarrow\frac{F(x,U)}{|x|^{\beta}}\ \
\mbox{in}\ \ L^{1}(\Omega).$

∎

##  5 Proof of the main results

In order to obtain a weak solution with positive energy, according to Lemmas
3.2 and 3.3, let

$c_{M,\varepsilon}=\inf_{\gamma\in\Upsilon}\max_{t\in[0,1]}I_{\varepsilon}(\gamma(t))>0,$
(5.1)

be the minimax level of $I_{\varepsilon}$, where $\Upsilon=\\{\gamma\in
C\big{(}[0,1],H^{1}_{0}(\Omega,\mathbb{R}^{k})\big{)}:\gamma(0)=\mathbf{0},I_{\varepsilon}(\gamma(1))<0\\}$.
Therefore, using the Mountain-pass theorem, there exists a sequence
$\\{U_{n}\\}\subset H^{1}_{0}(\Omega,\mathbb{R}^{k})$ satisfying

$I_{\varepsilon}(U_{n})\rightarrow c_{M,\varepsilon}\ \ \mbox{and}\ \
I^{\prime}_{\varepsilon}(U_{n})\rightarrow 0.$ (5.2)

And in order to obtain another weak solution with negative energy, by Lemmas
3.2 and 3.4, we take $\eta_{\varepsilon}\leq\rho_{\varepsilon}$ and so we have
that

$-\infty<c_{0,\varepsilon}:=\inf_{\|V\|\leq\rho_{\varepsilon}}I_{\varepsilon}(V)<0,$
(5.3)

where $\rho_{\varepsilon}$ is given as in Lemma 3.2. Since
$\overline{B}_{\rho_{\varepsilon}}$ is a complete metric space with the metric
given by the norm of $H^{1}_{0}(\Omega,\mathbb{R}^{k})$, convex and the
functional $I_{\varepsilon}$ is of class $C^{1}$ and bounded below on
$\overline{B}_{\rho_{\varepsilon}}$, by the Ekeland variational principle,
there exists a sequence $\\{V_{n}\\}$ in $\overline{B}_{\rho_{\varepsilon}}$
such that

$I_{\varepsilon}(V_{n})\rightarrow c_{0,\varepsilon}\ \ \mbox{and}\ \
I^{\prime}_{\varepsilon}(V_{n})\rightarrow 0.$ (5.4)

###  5.1 Subcritical case: Proof of Theorem 1.1

In this subsection, we assume that $f_{i}$ has subcritical growth at $\infty$
satisfying (1.10) and proof Theorem 1.1.

######  Lemma 5.1.

The functional $I_{\varepsilon}$ satisfies the Palais-Smale condition at any
finite level $c$.

###### Proof.

Let $\\{U_{n}\\}\subset H^{1}_{0}(\Omega,\mathbb{R}^{k})$ be a sequence such
that $I_{\varepsilon}(U_{n})\rightarrow c$ and
$I_{\varepsilon}^{\prime}(U_{n})\rightarrow 0$. Lemma 4.1 shows that
$\\{U_{n}\\}$ is bounded in $H^{1}_{0}(\Omega,\mathbb{R}^{k})$, then we can
get a subsequence still labeled by $\\{U_{n}\\}$, for some $U\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})$ such that

$\displaystyle U_{n}\rightharpoonup U\ \ \mbox{in}\ \
H^{1}_{0}(\Omega,\mathbb{R}^{k});\quad U_{n}\rightarrow U\ \ \mbox{in}\ \
L^{q}(\Omega,\mathbb{R}^{k})\ \mbox{for all}\ \ q\geq 1.$

Since

$\begin{split}\langle
I_{\varepsilon}^{\prime}(U_{n}),U_{n}-U\rangle_{*}=&m(\|U_{n}\|^{2})\langle
U_{n},U_{n}-U\rangle-\int_{\Omega}\frac{(U_{n}-U)\cdot\nabla
F(x,U_{n})}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}(U_{n}-U_{0})\cdot
Hdx.\end{split}$ (5.5)

From $I_{\varepsilon}^{\prime}(U_{n})\rightarrow 0$ in
$\big{(}H^{1}_{0}(\Omega,\mathbb{R}^{k})\big{)}_{*}$, we have $\langle
I^{\prime}_{\varepsilon}(U_{n}),U_{n}-U\rangle_{*}\rightarrow 0$. Meanwhile,
Lemma 4.1 shows that $\|U_{n}\|$ is bounded, i.e. $\|U_{n}\|^{2}\leq C_{0}$
for some $C_{0}>0$, then by subscritical condition, the Hölder’s inequality
and Lemma 2.2, it follows that

$\displaystyle\begin{split}\left|\int_{\Omega}\frac{(U_{n}-U)\cdot\nabla
F(x,U_{n})}{|x|^{\beta}}dx\right|&\leq\int_{\Omega}\frac{|U_{n}-U||\nabla
F(x,U_{n})|}{|x|^{\beta}}dx\\\ &\leq
C_{1}\int_{\Omega}|U_{n}-U|\frac{e^{\alpha|U_{n}|^{2}}}{|x|^{\beta}}dx\\\
&\leq
C_{2}\|U_{n}-U\|_{\frac{r}{r-1}}\left(\int_{\Omega}\frac{e^{r\alpha|U_{n}|^{2}}}{|x|^{r\beta}}dx\right)^{\frac{1}{r}}\\\
&\leq
C_{2}\|U_{n}-U\|_{\frac{r}{r-1}}\left(\sum^{k}_{i=1}\int_{\Omega}\frac{e^{kr\alpha\|U_{n}\|^{2}\left(\frac{u^{i}_{n}}{\|U_{n}\|}\right)^{2}}}{|x|^{r\beta}}dx\right)^{\frac{1}{r}}\end{split}$

for some $C_{1},C_{2}>0$, where $\alpha=\frac{4\pi(1-r\beta/2)}{krC_{0}}$ and
$r>1$ sufficiently close to 1 such that $r\beta<2$. Since
$\frac{kr\alpha\|U_{n}\|^{2}}{4\pi}+\frac{r\beta}{2}\leq 1$, then by Lemma 2.1
and $H^{1}_{0}(\Omega,\mathbb{R}^{k})\hookrightarrow
L^{s}(\Omega,\mathbb{R}^{k})$ is compact for $s\geq 1$, the fourth term in the
last inequality converges to zero. Thus from (4.4), in (5.5), it must be that

$\displaystyle m(\|U_{n}\|^{2})\langle U_{n},U_{n}-U\rangle\rightarrow 0\ \
\mbox{as}\ \ n\rightarrow+\infty.$

Because $(M_{1})$: $m(t)\geq m_{0}>0$ for $t\geq 0$ and $U_{n}\rightharpoonup
U$ in $H^{1}_{0}(\Omega,\mathbb{R}^{k})$, it must be that

$\displaystyle\langle U_{n},U_{n}-U\rangle\rightarrow 0,$

which means $\|U_{n}\|^{2}\rightarrow\|U\|^{2}$. By Radon’s Theorem,
$U_{n}\rightarrow U$ strongly in $H^{1}_{0}(\Omega,\mathbb{R}^{k})$. This
proof is complete. ∎

Proof of Theorem 1.1. By (5.2), (5.4) and Lemma 5.1, there exists
$\varepsilon_{sc}>0$ such that for each $0<\varepsilon<\varepsilon_{sc}$,
using Minimax principle, there exist critical points $U_{M,\varepsilon}$ for
$I_{\varepsilon}$ at level $c_{M,\varepsilon}$ and $V_{0,\varepsilon}$ for
$I_{\varepsilon}$ at level $c_{0,\varepsilon}$. We claim that
$U_{M,\varepsilon}\neq\mathbf{0}$. In fact, suppose by contradiction that
$U_{M,\varepsilon}\equiv\mathbf{0}$. We can know that
$0<c_{M,\varepsilon}=\lim_{n\rightarrow\infty}I_{\varepsilon}(U_{n})=I_{\varepsilon}(U_{M,\varepsilon})=I(\mathbf{0})=0$,
what is absurd. Similarly, we have $V_{0,\varepsilon}\neq\mathbf{0}$. In the
end, we claim $U_{M,\varepsilon}\neq V_{0,\varepsilon}$. Suppose by
contradiction that $U_{M,\varepsilon}\equiv V_{0,\varepsilon}$, then
$0>c_{0,\varepsilon}=\lim_{n\rightarrow\infty}I_{\varepsilon}(V_{n})=I(V_{0,\varepsilon})=I(U_{M,\varepsilon})=\lim_{n\rightarrow\infty}I_{\varepsilon}(U_{n})=c_{M,\varepsilon}>0$,
what is absurd. Thus, the proof of Theorem 1.1 is complete. ∎

###  5.2 Critical case: Proof of Theorem 1.2

In this subsection, we assume that $f_{i}$ has critical growth at $\infty$
satisfying (1.11) and give the proof Theorem 1.2.

Firstly, we give a conclusion that functional $I_{\varepsilon}$ satisfies the
Palais-Smale condition if Palais-Smale sequence less than appropriate level:

######  Lemma 5.2.

If $\\{V_{n}\\}$ is a Palais-Smale sequence for $I_{\varepsilon}$ at any
finite level with

$\displaystyle\liminf_{n\rightarrow\infty}\|V_{n}\|^{2}<\frac{2\pi(2-\beta)}{\alpha_{0}},$
(5.6)

then $\\{V_{n}\\}$ possesses a strongly subsequence.

###### Proof.

Let $\\{V_{n}\\}\subset H^{1}_{0}(\Omega,\mathbb{R}^{k})$ such that
$I_{\varepsilon}(V_{n})\rightarrow c$ and
$I^{\prime}_{\varepsilon}(V_{n})\rightarrow 0$ in
$\big{(}H^{1}_{0}(\Omega,\mathbb{R}^{k})\big{)}^{*}$. By Lemma 4.1,
$\|V_{n}\|\leq C$ for some $C>0$, thus, up to a sequence, for some $V\in
H^{1}_{0}(\Omega,\mathbb{R}^{k})$

$\displaystyle\begin{split}&V_{n}\rightharpoonup V\ \ \mbox{in}\ \
H^{1}_{0}(\Omega,\mathbb{R}^{k}).\end{split}$

Taking $V_{n}=V+W_{n}$, it follows that $W_{n}\rightharpoonup 0$ in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$ and by the Brézis-Lieb Lemma (see [9]), we
get

$\displaystyle\begin{split}\|V_{n}\|^{2}=\|V\|^{2}+\|W_{n}\|^{2}+o_{n}(1).\end{split}$
(5.7)

By $V_{n}\rightharpoonup V$ which means $\langle
V_{n},V\rangle\rightarrow\langle V,V\rangle=\|V\|^{2}$. Therefore, (5.7) can
be replaced by

$\displaystyle\begin{split}\|V_{n}\|^{2}=\langle
V_{n},V\rangle+\|W_{n}\|^{2}+o_{n}(1).\end{split}$ (5.8)

By $I_{\varepsilon}^{\prime}(V_{n})\rightarrow 0$ in
$\big{(}H^{1}_{0}(\Omega,\mathbb{R}^{k})\big{)}^{*}$ and (4.4), (5.8), we can
get

$\displaystyle\begin{split}\langle
I_{\varepsilon}^{\prime}(V_{n}),V_{n}-V\rangle_{*}&=m(\|V_{n}\|^{2})\langle
V_{n},V_{n}-V\rangle-\int_{\Omega}\frac{(V_{n}-V)\cdot\nabla
F(x,V_{n})}{|x|^{\beta}}+o_{n}(1)\\\
&=m(\|V_{n}\|^{2})\|W_{n}\|^{2}-\int_{\Omega}\frac{W_{n}\cdot\nabla
F(x,V_{n})}{|x|^{\beta}}+o_{n}(1),\end{split}$

that is,

$\displaystyle\begin{split}m(\|V_{n}\|^{2})\|W_{n}\|^{2}=\int_{\Omega}\frac{W_{n}\cdot\nabla
F(x,V_{n})}{|x|^{\beta}}+o_{n}(1).\end{split}$ (5.9)

From (5.6), there exists $\zeta>0$ such that
$\alpha_{0}\|V_{n}\|^{2}<\zeta<2\pi(2-\beta)$ for $n$ sufficiently large and
also, there exist $\alpha>\alpha_{0}$ close to $\alpha_{0}$ and $q>q$ close to
1 such that $q\alpha\|V_{n}\|^{2}<\zeta<2\pi(2-q\beta)$ for $n$ sufficiently
large. Then by (3.1), we have

$\displaystyle\begin{split}\left|\int_{\Omega}\frac{W_{n}\cdot\nabla
F(x,V_{n})}{|x|^{\beta}}\right|\leq
C_{1}\int_{\Omega}|W_{n}|\frac{e^{\alpha|V_{n}|^{2}}}{|x|^{\beta}},\end{split}$

and by the Hölder’s inequality and Lemma 2.4, we can get that

$\displaystyle\begin{split}\int_{\Omega}|W_{n}|\frac{e^{\alpha|V_{n}|^{2}}}{|x|^{\beta}}\leq
C_{1}\|W_{n}\|_{s}\bigg{(}\int_{\Omega}\frac{e^{q\alpha|V_{n}|^{2}}}{|x|^{q\beta}}\bigg{)}^{1/r}\leq
C_{2}\|W_{n}\|_{s},\end{split}$

where $s=\frac{q}{q-1}$. By the compact embedding
$H^{1}_{0}(\Omega,\mathbb{R}^{k})\hookrightarrow L^{s}(\Omega,\mathbb{R}^{k})$
for $s\geq 1$, we conclude that

$\displaystyle\begin{split}\int_{\Omega}\frac{W_{n}\cdot\nabla
F(x,V_{n})}{|x|^{\beta}}\rightarrow 0.\end{split}$

Thus, this together with (5.9) and $(M_{1})$, we get that
$\|W_{n}\|\rightarrow 0$ and the result follows. ∎

Then, in order to get a more precise information about the minimax level
$c_{M,\varepsilon}$, let us consider the following sequence which was
introduced in [13]: for $n\in\mathbb{N}$ set $\delta_{n}=\frac{2\log n}{n}$,
and let

$\displaystyle
y_{n}(t)=\frac{1}{\sqrt{2\pi}}\left\\{\begin{array}[]{ll}\frac{t}{n^{1/2}}(1-\delta_{n})^{1/2},&{\rm
if}\ \ 0\leq t\leq n,\\\\[5.69054pt]
\frac{1}{\big{[}n(1-\delta_{n})\big{]}^{1/2}}\log\frac{A_{n}+1}{A_{n}+e^{-(t-n)}}+\big{[}n(1-\delta_{n})\big{]}^{1/2},&{\rm
if}\ \ \ t\geq n,\\\\[5.69054pt] \end{array}\right.$

where $A_{n}$ is defined as $A_{n}=\frac{1}{en^{2}}+O(\frac{1}{n^{4}})$. The
sequence of function $\\{y_{n}\\}$ satisfies the following properties:

$\displaystyle\left\\{\begin{array}[]{ll}\\{y_{n}\\}\subset
C\big{(}[0,+\infty)\big{)},\ \mbox{piecewise differentiable, with}\
y_{n}(0)=0\ \mbox{and}\ y^{\prime}_{n}(t)\geq 0;\\\\[5.69054pt]
\int^{+\infty}_{0}|y^{\prime}_{n}(t)|^{2}=1;\\\\[5.69054pt]
\lim_{n\rightarrow\infty}\int^{+\infty}_{0}e^{y^{2}_{n}(t)-t}dt=1+e.\end{array}\right.$

Now, let $y_{n}(t)=2\sqrt{\pi}\widehat{G}_{n}(e^{-t/2})$ with $|x|=e^{-t/2}$,
define a function $\widehat{G}_{n}(x)=\widehat{G}_{n}(|x|)$ on
$\overline{B_{1}(0)}$, which is nonnegative and radially symmetric. Moreover,
we have

$\displaystyle\int_{B_{1}(0)}|\nabla\widehat{G}_{n}(x)|^{2}dx=\int^{+\infty}_{0}|y^{\prime}_{n}(t)|^{2}=1.$

Therefore $\|\widehat{G}_{n}\|=1$. Let $\tau=\frac{2-\beta}{2}$, then
$\widehat{G}_{n}$ defines another function nonnegative and radially symmetric
$\tilde{G}_{n}$ as follows:

$\displaystyle\widehat{G}_{n}(\varrho)=\tau^{1/2}\tilde{G}_{n}(\varrho^{1/\tau})\
\ \mbox{for}\ \ \varrho\in[0,1].$

Note that

$\displaystyle\int^{1}_{0}|\widehat{G}^{\prime}_{n}(\varrho)|^{2}\varrho
d\varrho=\int^{1}_{0}|\tilde{G}^{\prime}_{n}(\varrho)|^{2}\varrho d\varrho.$

Therefore, $\|\widehat{G}_{n}\|=\|\tilde{G}_{n}\|$. The open ball $B_{d}(0)$
is contained in $\Omega$, where $d$ was given in $(F_{4})$. Considering

$\displaystyle\mathscr{G}_{n,d}(x):=\big{(}G_{n,d}(x),0,\ldots,0\big{)},\ \ \
\mbox{where}\ \ \ G_{n,d}(x):=\tilde{G}_{n}\left(\frac{x}{d}\right),$ (5.12)

then $\mathscr{G}_{n,d}(x)$ belongs to $H^{1}_{0}(\Omega,\mathbb{R}^{k})$ with
$\|\mathscr{G}_{n,d}\|=1$, and the support of $\mathscr{G}_{n,d}$ contained in
$B_{d}(0)$.

######  Remark 5.1.

If condition $(F_{4})$ holds, we define $\mathscr{G}^{\prime}_{n,d}(x)$ that
$i$-th component is set to $G_{n,d}(x)$, and the remaining components are set
to $0$, i.e.,
$\mathscr{G}^{\prime}_{n,d}(x)=(0,\ldots,0,G_{n,d}(x),0,\ldots,0)$, then given
$\delta>0$ there exists $s_{\delta}>0$ such that

$\displaystyle\begin{split}\mathscr{G}^{\prime}_{n,d}\cdot\nabla
F(x,\mathscr{G}^{\prime}_{n,d})=G_{n,d}f_{i}(x,\mathscr{G}^{\prime}_{n,d})\geq(\eta_{0}-\delta)\exp\left(\alpha_{0}|\mathscr{G}^{\prime}_{n,d}|^{2}\right)=(\eta_{0}-\delta)\exp\left(\alpha_{0}|G_{n,d}|^{2}\right),\end{split}$

$\forall x\in\Omega,\ \ |\mathscr{G}^{\prime}_{n,d}|=|G_{n,d}|\geq
s_{\delta}$. This is the same as type (5.16) below, therefore without lose of
generality, we can assume that $i=1$ in $(F_{4})$.

######  Lemma 5.3.

For any $0<\epsilon<1$, we have that for $x\in B_{\frac{d}{\varpi(n)}}(0)$
with $\varpi(n)=\exp\left\\{\frac{n^{(1+\epsilon)/2}}{2}\right\\}$,

$\displaystyle|\mathscr{G}_{n,d}(x)|\geq\frac{1}{2\sqrt{\pi}}n^{\frac{\epsilon}{2}}\left(1-\frac{2\log
n}{n}\right)^{\frac{1}{2}},$ (5.13)

where $\mathscr{G}_{n,d}$ is given in (5.12).

###### Proof.

For $x\in B_{\frac{d}{\varpi(n)}}(0)$,

$\displaystyle\left|\mathscr{G}_{n,d}(x)\right|=\left|G_{n,d}(x)\right|=\left|\tilde{G}_{n}\left(\frac{x}{d}\right)\right|=\left|\tilde{G}_{n}(y)\right|,$

where $y=\frac{x}{d}\in B_{\frac{1}{\varpi(n)}}(0)$. Moreover,

$\displaystyle\left|\tilde{G}_{n}(y)\right|=\left|\tilde{G}_{n}(|y|)\right|=\frac{1}{2\sqrt{\pi}}y_{n}(-2\log(|y|))=\frac{1}{2\sqrt{\pi}}y_{n}(t),$

where $t=-2\log(|y|)\in(n^{\frac{1+\epsilon}{2}},+\infty)$. Noticing that,
$y_{n}(t)\geq\big{[}n(1-\delta_{n})\big{]}^{\frac{1}{2}}=n^{\frac{1}{2}}\left(1-\frac{2\log
n}{n}\right)^{\frac{1}{2}}$ if $t\geq n$. Moreover, in
$(n^{\frac{1+\epsilon}{2}},n)$,

$\displaystyle\begin{split}y_{n}(t)\geq\frac{n^{\frac{1+\epsilon}{2}}}{n^{\frac{1}{2}}}(1-\delta_{n})^{\frac{1}{2}}=n^{\frac{\epsilon}{2}}\left(1-\frac{2\log
n}{n}\right)^{\frac{1}{2}}.\end{split}$

The proof is complete. ∎

######  Lemma 5.4.

If conditions $(M_{1}),\ (M_{3})$ and $(F_{3}),\ (F_{4})$ hold, then

$\displaystyle\max_{t\geq
0}\left[\frac{1}{2}M(t^{2})-\int_{\Omega}\frac{F(x,t\mathscr{G}_{n,d})}{|x|^{\beta}}\right]<\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right).$

###### Proof.

Suppose by contradiction, that for all $n\in\mathbb{N}$, we have

$\displaystyle\max_{t\geq
0}\left[\frac{1}{2}M(t^{2})-\int_{\Omega}\frac{F(x,t\mathscr{G}_{n,d})}{|x|^{\beta}}\right]\geq\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right).$

By Lemmas 3.2 and 3.3, for each $n$ there exists $t_{n}>0$ such that

$\displaystyle\frac{1}{2}M(t^{2}_{n})-\int_{\Omega}\frac{F(x,t_{n}\mathscr{G}_{n,d})}{|x|^{\beta}}=\max_{t\geq
0}\left[\frac{1}{2}M(t^{2})-\int_{\Omega}\frac{F(x,t\mathscr{G}_{n,d})}{|x|^{\beta}}\right]$

From this, and using $(F_{3})$, one has $M(t^{2}_{n})\geq
M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)$. By $(M_{1})$, which implies
that $M:[0,+\infty)\rightarrow[0,+\infty)$ is an increasing bijection and so

$\displaystyle t^{2}_{n}\geq\frac{2\pi(2-\beta)}{\alpha_{0}}.$ (5.14)

On the other hand,

$\displaystyle\frac{d}{dt}\left[\frac{1}{2}M(t^{2})-\int_{\Omega}\frac{F(x,t\mathscr{G}_{n,d})}{|x|^{\beta}}\right]\bigg{|}_{t=t_{n}}=0,$

from which we obtain

$\displaystyle m(t^{2}_{n})t^{2}_{n}$
$\displaystyle=\int_{\Omega}\frac{t_{n}\mathscr{G}_{n,d}\cdot\nabla
F(x,t_{n}\mathscr{G}_{n,d})}{|x|^{\beta}}dx=\int_{\Omega}\frac{t_{n}G_{n,d}f_{1}(x,t_{n}\mathscr{G}_{n,d})}{|x|^{\beta}}dx.$
(5.15)

By Remark 5.1 and $(F_{4})$, given $\delta>0$ there exists $s_{\delta}>0$ such
that

$\displaystyle
u_{1}f_{1}(x,u_{1},0,\ldots,0)\geq(\eta_{0}-\delta)e^{\alpha_{0}|u_{1}|^{2}},\
\ \ \forall\ x\in\Omega,\ \ |u_{1}|\geq s_{\delta}.$ (5.16)

Lemma 5.3 shows that for any $0<\epsilon<1$, $t_{n}|\mathscr{G}_{n,d}|\geq
s_{\delta}$ in $B_{\frac{d}{\varpi(n)}}(0)\subset\Omega$ for $n$ sufficiently
large, where $\varpi(n)=\exp\left\\{\frac{n^{(1+\epsilon)/2}}{2}\right\\}.$
Thus, by (5.15) and (5.16), we have

$\displaystyle\begin{split}m(t^{2}_{n})t^{2}_{n}\geq(\eta_{0}-\delta)\int_{B_{\frac{d}{\varpi(n)}}(0)}\frac{e^{\alpha_{0}|t_{n}G_{n,d}|^{2}}}{|x|^{\beta}}dx&=(\eta_{0}-\delta)\Big{(}\frac{d}{\varpi(n)}\Big{)}^{2-\beta}\int_{B_{1}(0)}\frac{e^{\beta_{0}|t_{n}\tilde{G}_{n}|^{2}}}{|x|^{\beta}}dx\\\
&=2\pi(\eta_{0}-\delta)\Big{(}\frac{d}{\varpi(n)}\Big{)}^{2-\beta}\int^{1}_{0}e^{\alpha_{0}|t_{n}\tilde{G}_{n}(\sigma)|^{2}}\sigma^{1-\beta}d\sigma.\end{split}$

By performing the change of variable $\sigma=\tau^{\frac{2}{2-\beta}}$, we get

$\displaystyle
m(t^{2}_{n})t^{2}_{n}\geq\frac{4\pi}{2-\beta}(\eta_{0}-\delta)\Big{(}\frac{d}{\varpi(n)}\Big{)}^{2-\beta}\int^{1}_{0}e^{\frac{2\alpha_{0}|t_{n}\tilde{G}_{n}(\tau)|^{2}}{2-\beta}}\tau
d\tau.$

Meanwhile, setting $\tau=e^{-t/2}$, we obtain

$\displaystyle\begin{split}m(t^{2}_{n})t^{2}_{n}\geq\frac{2\pi}{2-\beta}(\eta_{0}-\delta)\Big{(}\frac{d}{\varpi(n)}\Big{)}^{2-\beta}\int^{+\infty}_{0}e^{\frac{\alpha_{0}|t_{n}y_{n}(t)|^{2}}{2\pi(2-\beta)}}e^{-t}dt.\end{split}$

Consequently,

$\displaystyle\begin{split}m(t^{2}_{n})t^{2}_{n}&\geq\frac{2\pi}{2-\beta}(\eta_{0}-\delta)\Big{(}\frac{d}{\varpi(n)}\Big{)}^{2-\beta}\int^{+\infty}_{n}e^{\frac{\alpha_{0}t_{n}^{2}(n-2\log
n)}{2\pi(2-\beta)}}e^{-t}dt\\\
&=\frac{2\pi}{2-\beta}(\eta_{0}-\delta)d^{2-\beta}\exp\left\\{\frac{\alpha_{0}t_{n}^{2}(n-2\log
n)}{2\pi(2-\beta)}-(2-\beta)\log\varpi(n)-n\right\\}\\\
&=\frac{2\pi}{2-\beta}(\eta_{0}-\delta)d^{2-\beta}\exp\left\\{\frac{\alpha_{0}t_{n}^{2}(n-2\log
n)}{2\pi(2-\beta)}-\frac{(2-\beta)n^{\frac{1+\epsilon}{2}}}{2}-n\right\\}\\\
&=\frac{2\pi}{2-\beta}(\eta_{0}-\delta)d^{2-\beta}\exp\left\\{\left[\frac{\alpha_{0}t_{n}^{2}}{2\pi(2-\beta)}-1\right]n-\frac{(2-\beta)n^{\frac{1+\epsilon}{2}}}{2}-\frac{\alpha_{0}t_{n}^{2}}{\pi(2-\beta)}\log
n\right\\}.\end{split}$ (5.17)

From this

$\displaystyle\begin{split}1\geq\frac{2\pi}{2-\beta}(\eta_{0}-\delta)d^{2-\beta}\exp\left\\{t^{2}_{n}n\left[\frac{\alpha_{0}\big{(}1-\frac{2\log
n}{n}\big{)}}{2\pi(2-\beta)}-\frac{(2-\beta)n^{\frac{1+\epsilon}{2}}+2n}{2t^{2}_{n}n}-\frac{\log{\big{(}m(t^{2}_{n})t^{2}_{n}\big{)}}}{t^{2}_{n}n}\right]\right\\},\end{split}$
(5.18)

thus, $\\{t_{n}\\}$ is bounded. Otherwise, noting that, from (1.9), making use
of the property of $M$ and $m$, we would have that

$\displaystyle\begin{split}t^{2}_{n}n\left[\frac{\alpha_{0}\big{(}1-\frac{2\log
n}{n}\big{)}}{2\pi(2-\beta)}-\frac{(2-\beta)n^{\frac{1+\epsilon}{2}}+2n}{2t^{2}_{n}n}-\frac{\log{\big{(}m(t^{2}_{n})t^{2}_{n}\big{)}}}{t^{2}_{n}n}\right]\rightarrow+\infty,\end{split}$

which is a contradiction with (5.18). Therefore $\\{t_{n}\\}$ has a
subsequence convergent, from (5.14), for some
$t_{0}^{2}\geq\frac{2\pi(2-\beta)}{\alpha_{0}}$, $t_{n}\rightarrow t_{0}$.
Moreover, using (5.17), we must have
$\frac{\alpha_{0}t^{2}_{0}}{2\pi(2-\beta)}-1\leq 0$ and therefore,

$\displaystyle t^{2}_{n}\rightarrow\frac{2\pi(2-\beta)}{\alpha_{0}}.$ (5.19)

At this point, following arguments as in [15], we are going to estimate (5.15)
more exactly. For this, in view of (5.16), for $0<\delta<\eta_{0}$ and
$n\in\mathbb{N}$ we set

$D_{n,\delta}:=\\{x\in B_{d}(0):t_{n}G_{n,d}\geq s_{\delta}\\}\ \ \mbox{and}\
\ E_{n,\delta}:=B_{d}(0)\backslash D_{n,\delta}.$

Thus, by splitting the integral (5.6) on $D_{n,\delta}\ \mbox{and}\
E_{n,\delta}$, and using (5.16), it follows that

$\displaystyle\begin{split}M(t^{2}_{n})t^{2}_{n}\geq&(\eta_{0}-\delta)\int_{B_{d}(0)}\frac{e^{\alpha_{0}t^{2}_{n}G^{2}_{n,d}}}{|x|^{\beta}}dx-(\eta_{0}-\delta)\int_{E_{n,\delta}}\frac{e^{\alpha_{0}t^{2}_{n}G^{2}_{n,d}}}{|x|^{\beta}}dx\\\
&+\int_{E_{n,\delta}}\frac{t_{n}G_{n,d}f_{1}(x,t_{n}\mathscr{G}_{n,d})}{|x|^{\beta}}dx.\end{split}$
(5.20)

Since $G_{n,d}(x)\rightarrow 0$ for almost everywhere $x\in B_{d}(0)$, we have
that the characteristic functions $\chi_{E_{n,\delta}}$ satisfy

$\chi_{E_{n,\delta}}\rightarrow 1\ \ \mbox{a.e. in}\ \ B_{d}(0)\ \ \mbox{as}\
\ n\rightarrow+\infty.$

Moreover, $t_{n}G_{n,d}<s_{\delta}$ in $E_{n,\delta}$. Thus, invoking the
Lebesgue dominated convergence theorem, we obtain

$\int_{E_{n,\delta}}\frac{e^{\alpha_{0}t^{2}_{n}G^{2}_{n,d}}}{|x|^{\beta}}dx\rightarrow\pi
d^{2}\ \ \mbox{and}\ \
\int_{E_{n,\delta}}\frac{t_{n}G_{n,d}f_{1}(x,t_{n}\mathscr{G}_{n,d})}{|x|^{\beta}}dx\rightarrow
0,\ \ \mbox{as}\ \ n\rightarrow+\infty.$

Noting that

$\begin{split}\int_{B_{d}(0)}\frac{e^{\alpha_{0}t^{2}_{n}G^{2}_{n,d}}}{|x|^{\beta}}dx=d^{2-\beta}\int_{B_{1}(0)}\frac{e^{\alpha_{0}t^{2}_{n}\tilde{G}^{2}_{n}}}{|x|^{\beta}}dx=2\pi
d^{2-\beta}\int^{1}_{0}{e^{\alpha_{0}t^{2}_{n}\tilde{G}^{2}_{n}(\sigma)}}\sigma^{2-\beta}d\sigma,\end{split}$

By performing the change of variable $\sigma=\tau^{\frac{2}{2-\beta}}$, we get

$\displaystyle\int_{B_{d}(0)}\frac{e^{\alpha_{0}t^{2}_{n}G^{2}_{n,d}}}{|x|^{\beta}}dx=\frac{4\pi}{2-\beta}d^{2-\beta}\int^{1}_{0}e^{\frac{2\alpha_{0}|t_{n}\tilde{G}_{n}(\tau)|^{2}}{2-\beta}}\tau
d\tau.$

Meanwhile, setting $\tau=e^{-t/2}$ and using (5.14), we obtain

$\displaystyle\begin{split}\int_{B_{d}(0)}\frac{e^{\alpha_{0}t^{2}_{n}G^{2}_{n,d}}}{|x|^{\beta}}dx=&\frac{2\pi}{2-\beta}d^{2-\beta}\int^{+\infty}_{0}e^{\frac{\alpha_{0}|t_{n}y_{n}(t)|^{2}}{2\pi(2-\beta)}}e^{-t}dt\\\
\geq&\frac{2\pi}{2-\beta}d^{2-\beta}\int^{+\infty}_{0}e^{y^{2}_{n}(t)-t}dt.\end{split}$

Passing to limit in (5.20), we obtain that

$\displaystyle
m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\frac{2\pi(2-\beta)}{\alpha_{0}}\geq(\eta_{0}-\delta)\left[\frac{2\pi}{2-\beta}d^{2-\beta}(1+e)-\frac{2\pi}{2-\beta}d^{2-\beta}\right]=(\eta_{0}-\delta)\frac{2\pi
e}{2-\beta}d^{2-\beta},$ (5.21)

and doing $\delta\rightarrow 0^{+}$, we get
$\eta_{0}\leq\frac{(2-\beta)^{2}m\big{(}\frac{2\pi(2-\beta)}{\alpha_{0}}\big{)}}{\alpha_{0}d^{2-\beta}e}$,
which contradicts $(F_{4})$. Thus, this lemma is proved. ∎

Now, we establish an estimate for the minimax level.

######  Lemma 5.5.

If conditions $(M_{1}),\ (M_{3})$ and $(F_{3})-(F_{4})$ hold, then for small
$\varepsilon$, it holds that

$\displaystyle
c_{M,\varepsilon}<\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right),$

where $c_{M,\varepsilon}$ is given as in (5.1).

###### Proof.

Since $\|\mathscr{G}_{n,d}\|=1$, as in the proof of Lemma 3.3, we have that
$I(t\mathscr{G}_{n,d})\rightarrow-\infty$ as $t\rightarrow+\infty$.
Consequently, $c_{M,\varepsilon}\leq\max_{t\geq
0}I_{\varepsilon}(t\mathscr{G}_{n,d}),\ \forall\ n\in\mathbb{N}$. Thus, from
Lemma 5.4, taking $\varepsilon$ sufficiently small, we can get what we
desired. ∎

######  Lemma 5.6.

If $f_{i}$ has critical growth at $\infty$, $(M_{1})-(M_{3})$ and
$(F_{1})-(F_{4})$ satisfy, then for small $\varepsilon$, problem (1.4) has one
nontrivial mountain-pass type solution $U_{M,\varepsilon}$ at level
$c_{M,\varepsilon}$, where $c_{M,\varepsilon}$ is given as in (5.1).

###### Proof.

From (5.2) and Lemma 4.1, there exists a bounded Palais-Smale consequence
$\\{U_{n}\\}$ for $I_{\varepsilon}$ at level $c_{M,\varepsilon}$. And, up to a
subsequence, for some $U\in H^{1}_{0}(\Omega,\mathbb{R}^{k})$, one has

$\displaystyle U_{n}\rightharpoonup U\ \ \mbox{in}\ \
H^{1}_{0}(\Omega,\mathbb{R}^{k});\qquad U_{n}\rightarrow U\ \ \mbox{in}\ \
L^{s}(\Omega,\mathbb{R}^{k})\ \mbox{for all}\ \ s\geq 1.$ (5.22)

By Lemma 4.1, $\|U_{n}\|\leq C$ for some $C>0$, and
$\|U\|\leq\lim\inf_{n\rightarrow\infty}\|U_{n}\|\leq C$. And by $(F_{3})$, we
have that for small $\varepsilon$, it holds

$\displaystyle\begin{split}&\frac{1}{2\theta}\int_{\Omega}\bigg{[}\frac{U\cdot\nabla
F(x,U)-2\theta F(x,U)}{|x|^{\beta}}-\varepsilon(2\theta-1)U\cdot
H\bigg{]}dx\\\ &\geq\frac{1}{2\theta}\int_{\Omega}\frac{U\cdot\nabla
F(x,U)-2\theta
F(x,U)}{|x|^{\beta}}dx-\frac{\varepsilon(2\theta-1)\|U\|\|H\|_{*}}{2\theta}\geq
0.\end{split}$ (5.23)

Next, we will make some claims as follows.

Claim 1. $U\neq\mathbf{0}$.

###### Proof.

Suppose by contradiction that $U\equiv\mathbf{0}$. Then Lemma 4.4 show that
$\int_{\Omega}U_{n}\cdot Hdx\rightarrow 0$ and
$\int_{\Omega}\frac{F(x,U_{n})}{|x|^{\beta}}dx\rightarrow 0$, thus

$\displaystyle\frac{1}{2}M(\|U_{n}\|^{2})\rightarrow
c_{M,\varepsilon}<\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right),$

then condition $(M_{1})$ implies

$\displaystyle\liminf_{n\rightarrow\infty}\|U_{n}\|^{2}<\frac{2\pi(2-\beta)}{\alpha_{0}}.$

By Lemma 5.2, we have $\|U_{n}\|^{2}\rightarrow 0$ and therefore
$I_{\varepsilon}(U_{n})\rightarrow 0$, what is absurd for $c_{M,\varepsilon}$
and we must have $U\neq\mathbf{0}$. ∎

Claim 2. Let $A:=\lim_{n\rightarrow\infty}\|U_{n}\|^{2}$, then $U$ is a weak
solution of

$\displaystyle\left\\{\begin{array}[]{ll}-m(A)\Delta U=\frac{\nabla
F(x,U)}{|x|^{\beta}}+\varepsilon H,&\mbox{in}\ \ \Omega,\\\\[5.69054pt]
U=0,&\mbox{on}\ \ \partial\Omega.\end{array}\right.$

###### Proof.

We define
$C^{\infty}_{0}(\Omega,\mathbb{R}^{k}):=C^{\infty}_{0}(\Omega)\times\cdots\times
C^{\infty}_{0}(\Omega)$. By $I_{\varepsilon}^{\prime}(U_{n})\rightarrow 0$ and
Lemma 4.4, we see that

$\displaystyle\begin{split}m(A)\int_{\Omega}\nabla U\cdot\nabla\Phi
dx-\int_{\Omega}\frac{\Phi\cdot\nabla
F(x,U)}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}\Phi\cdot
Hdx=0,\quad\forall\Phi\in C^{\infty}_{0}(\Omega,\mathbb{R}^{k}).\end{split}$

Since $C^{\infty}_{0}(\Omega)$ is dense in $H^{1}_{0}(\Omega)$, then
$C^{\infty}_{0}(\Omega,\mathbb{R}^{k})$ is also dense in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$, and we conclude this claim. ∎

Claim 3.
$A:=\lim_{n\rightarrow\infty}\|U_{n}\|^{2}<\|U\|^{2}+\frac{2\pi(2-\beta)}{\alpha_{0}}$.

###### Proof.

Suppose by contradiction that
$A\geq\|U\|^{2}+\frac{2\pi(2-\beta)}{\alpha_{0}}\geq\frac{2\pi(2-\beta)}{\alpha_{0}}$.
Therefore, from (5.2) and Lemma 4.4, we obtain

$\displaystyle\begin{split}c_{M,\varepsilon}=&\lim_{n\rightarrow\infty}\left[I_{\varepsilon}(U_{n})-\frac{1}{2\theta}\langle
I^{\prime}_{\varepsilon}(U_{n}),U_{n}\rangle_{*}\right]\\\
=&\frac{1}{2\theta}\lim_{n\rightarrow\infty}\Big{[}\theta
M(\|U_{n}\|^{2})-m(\|U_{n}\|^{2})\|U_{n}\|^{2}\Big{]}\\\
&+\frac{1}{2\theta}\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\\\
&+\frac{1}{2\theta}\int_{\Omega}\bigg{[}\frac{U\cdot\nabla F(x,U)-2\theta
F(x,U)}{|x|^{\beta}}-\varepsilon(2\theta-1)U\cdot H\bigg{]}dx,\\\ \end{split}$

thus, by $(M_{3})$ and (5.23)

$\displaystyle\begin{split}c_{M,\varepsilon}\geq&\frac{1}{2\theta}\lim_{n\rightarrow\infty}\Big{[}\theta
M(\|U_{n}\|^{2})-m(\|U_{n}\|^{2})\|U_{n}\|^{2}\Big{]}\\\
&+\frac{1}{2\theta}\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\\\
=&\frac{1}{2\theta}\Big{[}\theta
M(A)-m(A)A\Big{]}+\frac{1}{2\theta}\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\\\
\geq&\frac{1}{2\theta}\bigg{[}\theta
M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)-m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\frac{2\pi(2-\beta)}{\alpha_{0}}\bigg{]}\\\
&+\frac{1}{2\theta}\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\\\
=&\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\\\
&-\frac{1}{2\theta}\bigg{[}m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\frac{2\pi(2-\beta)}{\alpha_{0}}-\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\bigg{]}.\\\ \end{split}$

Here, we assert

$\displaystyle 0\geq
m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\frac{2\pi(2-\beta)}{\alpha_{0}}-\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx.$

Indeed, $\langle I^{\prime}(U_{n}),U_{n}\rangle_{*}\rightarrow 0$, (4.4) and
Claim 2 indicate that

$\displaystyle\begin{split}0=&m(A)A-\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U\cdot Hdx,\\\
0=&m(A)\|U\|^{2}-\int_{\Omega}\frac{U\cdot\nabla
F(x,U)}{|x|^{\beta}}dx-\varepsilon\int_{\Omega}U\cdot Hdx.\end{split}$

Subtracting the second equality from the first one, by $(M_{2})$ which implies
$m(t)$ and $m(t)t$ are nondecreasing for $t\geq 0$, we get

$\displaystyle\begin{split}0&=m(A)(A-\|U\|^{2})-\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\\\ &\geq
m(A-\|U\|^{2})(A-\|U\|^{2})-\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx\\\ &\geq
m\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\frac{2\pi(2-\beta)}{\alpha_{0}}-\lim_{n\rightarrow\infty}\int_{\Omega}\frac{U_{n}\cdot\nabla
F(x,U_{n})-U\cdot\nabla F(x,U)}{|x|^{\beta}}dx.\end{split}$

This shows the assertion. Noting Lemma 5.5, we conclude

$\displaystyle\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)\leq
c_{M,\varepsilon}<\frac{1}{2}M\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right),$

which is absurd. Thus this claim is proved. ∎

Claim 4. $A:=\lim_{n\rightarrow\infty}\|U_{n}\|^{2}=\|U\|^{2}$.

###### Proof.

By using semicontinuity of norm, we have $\|U\|^{2}\leq A$. We are going to
show that the case $\|U\|^{2}<A$ can not occur. Indeed, if $\|U\|^{2}<A$,
defining $Z_{n}=\frac{U_{n}}{\|U_{n}\|}$ and $Z_{0}=\frac{U}{A^{1/2}}$, we
have $Z_{n}\rightharpoonup Z_{0}\ \mbox{in}\ H^{1}_{0}(\Omega,\mathbb{R}^{k})$
and $\|Z_{0}\|<1$. Thus, by Lemma 2.6

$\displaystyle\sup_{n}\int_{\Omega}\frac{e^{p|Z_{n}|^{2}}}{|x|^{\beta}}dx<\infty,\
\ \forall p<\frac{2\pi(2-\beta)}{1-\|Z_{0}\|^{2}}.$ (5.25)

Since $A=\frac{A-\|U\|^{2}}{1-\|Z_{0}\|^{2}}$, from Claim 3 which follows that
$A<\frac{2\pi(2-\beta)}{\alpha_{0}(1-\|Z_{0}\|^{2})}$. Thus, there exists
$\zeta>0$ such that
$\alpha_{0}\|U_{n}\|^{2}<\zeta<\frac{2\pi(2-\beta)}{1-\|Z_{0}\|^{2}}$ for $n$
sufficiently large. For $q>1$ close to 1 and $\alpha>\alpha_{0}$ close to
$\alpha_{0}$ we still have
$q\alpha\|U_{n}\|^{2}<\zeta<\frac{2\pi(2-q\beta)}{1-\|Z_{0}\|^{2}}$ with
$q\beta<2$, and provoking (5.25), for some $C>0$ and $n$ large enough, we
conclude that

$\displaystyle\int_{\Omega}\frac{e^{q\alpha|U_{n}|^{2}}}{|x|^{q\beta}}dx\leq\int_{\Omega}\frac{e^{\zeta|Z_{n}|^{2}}}{|x|^{q\beta}}dx\leq
C.$

Hence, using (3.1), (5.22) and the Hölder’s inequality, we get

$\displaystyle\begin{split}\left|\int_{\Omega}\frac{(U_{n}-U)\cdot\nabla
F(x,U_{n})}{|x|^{\beta}}dx\right|&\leq
C_{1}\int_{\Omega}|U_{n}-U|\frac{e^{\alpha|U_{n}|^{2}}}{|x|^{\beta}}dx\\\
&\leq
C_{1}\|U_{n}-U\|_{\frac{q}{q-1}}\left(\int_{\Omega}\frac{e^{q\alpha|U_{n}|^{2}}}{|x|^{q\beta}}dx\right)^{1/q}\\\
&\leq C_{2}\|U_{n}-U\|_{\frac{q}{q-1}}\rightarrow 0,\end{split}$

as $n\rightarrow\infty.$ Since $\langle
I^{\prime}_{\varepsilon}(U_{n}),U_{n}-U\rangle_{*}\rightarrow 0$, by (4.4), it
follows that $m(\|U_{n}\|^{2})\langle U_{n},U_{n}-U\rangle\rightarrow 0$. On
the other hand,

$\displaystyle\begin{split}m(\|U_{n}\|^{2})\langle
U_{n},U_{n}-U\rangle&=m(\|U_{n}\|^{2})\|U_{n}\|^{2}-m(\|U_{n}\|^{2})\int_{\Omega}\nabla
U_{n}\cdot\nabla Udx\\\ &\rightarrow m(A)A-m(A)\|U\|^{2},\end{split}$

which implies that $A=\|U\|^{2}$ what is absurd. Thus, this claim is proved. ∎

Finalizing the proof of Lemma 5.6: Since $H^{1}_{0}(\Omega,\mathbb{R}^{k})$ is
uniformly convex Banach space and (5.22), Claim 4, by Radon’s Theorem,
$U_{n}\rightarrow U$ in $H^{1}_{0}(\Omega,\mathbb{R}^{k})$. Hence, by (5.2):
$I_{\varepsilon}^{\prime}(U_{n})\rightarrow 0$ and Lemma 4.4, we have

$m(\|U\|^{2})\int_{\Omega}\nabla
U\cdot\nabla\Phi=\int_{\Omega}\frac{\Phi\cdot\nabla{F(x,U)}}{|x|^{\beta}}dx+\varepsilon\int_{\Omega}H\cdot\Phi
dx,\ \ \forall\Phi\in C^{\infty}_{0}(\Omega,\mathbb{R}^{k}).$

Since $C^{\infty}_{0}(\Omega,\mathbb{R}^{k})$ is dense in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$, we conclude that $U_{M,\varepsilon}:=U$ is
a Mountain-pass type solution for problem (1.4) with
$I_{\varepsilon}(U_{M,\varepsilon})=c_{M,\varepsilon}>0$ and according to
Claim 1, the proof is complete. ∎

Finally, let us to find out a minimum type solution $V_{0,\varepsilon}$ with
$I_{\varepsilon}(V_{0,\varepsilon})=c_{0,\varepsilon}<0$, where
$c_{0,\varepsilon}$ is given as in (5.3).

######  Lemma 5.7.

For small $\varepsilon$, problem (1.4) has a nontrivial minimum type solution
$V_{0,\varepsilon}$ with
$I_{\varepsilon}(V_{0,\varepsilon})=c_{0,\varepsilon}<0$.

###### Proof.

Let $\rho_{\varepsilon}$ be as in Lemma 3.2. Then we can choose $\varepsilon$
sufficiently small such that

$\displaystyle\rho_{\varepsilon}<\left(\frac{2\pi(2-\beta)}{\alpha_{0}}\right)^{1/2}.$

Since $\overline{B}_{\rho_{\varepsilon}}$ is a complete metric space with the
metric given by the norm of $H^{1}_{0}(\Omega,\mathbb{R}^{k})$, convex and the
functional $I_{\varepsilon}$ is of class $C^{1}$ and bounded below on
$\overline{B}_{\rho_{\varepsilon}}$, by the Ekeland variational principle,
there exists a sequence $\\{V_{n}\\}$ in $\overline{B}_{\rho_{\varepsilon}}$
such that

$I_{\varepsilon}(V_{n})\rightarrow
c_{0,\varepsilon}=\inf_{\|V\|\leq\rho_{\varepsilon}}I_{\varepsilon}(V)<0\ \
\mbox{and}\ \ I^{\prime}_{\varepsilon}(V_{n})\rightarrow 0.$

Observing that

$\displaystyle\|V_{n}\|^{2}\leq\rho_{\varepsilon}^{2}<\frac{2\pi(2-\beta)}{\alpha_{0}},$

by Lemma 5.2, there exists a strongly convergent subsequence and therefore,
for some $V_{0,\varepsilon}$, $V_{n}\rightarrow V_{0,\varepsilon}$ strongly in
$H^{1}_{0}(\Omega,\mathbb{R}^{k})$. Consequently, we have $V_{0,\varepsilon}$
is a minimum type solution of problem (1.4) with
$I_{\varepsilon}(V_{0,\varepsilon})=c_{0,\varepsilon}<0$. We claim
$V_{0,\varepsilon}\neq\mathbf{0}$. Indeed, suppose by a contradiction
$V_{0,\varepsilon}=\mathbf{0}$, then
$0>c_{0,\varepsilon}=I_{\varepsilon}(V_{0,\varepsilon})=I_{\varepsilon}(\mathbf{0})=0$,
what is absurd and this lemma is proved. ∎

Proof of Theorem 1.2. By Lemmas 5.6 and 5.7, there exists $\varepsilon_{c}>0$
such that for each $0<\varepsilon<\varepsilon_{c}$, there exist nontrivial
critical points $U_{M,\varepsilon}$ for $I_{\varepsilon}$ at level
$c_{M,\varepsilon}$ and $V_{0,\varepsilon}$ for $I_{\varepsilon}$ at level
$c_{0,\varepsilon}$. In the end, we claim $U_{M,\varepsilon}\neq
V_{0,\varepsilon}$. Suppose by contradiction that $U_{M,\varepsilon}\equiv
V_{0,\varepsilon}$, then
$0>c_{0,\varepsilon}=\lim_{n\rightarrow\infty}I_{\varepsilon}(V_{n})=I(V_{0,\varepsilon})=I(U_{M,\varepsilon})=\lim_{n\rightarrow\infty}I_{\varepsilon}(U_{n})=c_{M,\varepsilon}>0$,
what is absurd. Thus, the proof of Theorem 1.2 is complete. ∎

Acknowledgements. The authors have been supported by NSFC 11971392, Natural
Science Foundation of Chongqing, China cstc2019jcyjjqX0022 and Fundamental
Research Funds for the Central Universities XDJK2019TY001.

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