# Normalized solutions to the Chern-Simons-Schrödinger system: the
supercritical case

Liejun Shen  and  Marco Squassina  Liejun Shen,
Department of Mathematics, Zhejiang Normal University,
Jinhua, Zhejiang, 321004, People’s Republic of China<EMAIL_ADDRESS>Marco
Squassina,
Dipartimento di Matematica e Fisica
Università Cattolica del Sacro Cuore,
Via della Garzetta 48, 25133, Brescia, Italy<EMAIL_ADDRESS>
###### Abstract.

We are concerned with the existence of normalized solutions for a class of
generalized Chern-Simons-Schrödinger type problems with supercritical
exponential growth

$\left\\{\begin{array}[]{ll}\displaystyle-\Delta u+\lambda
u+A_{0}u+\sum\limits_{j=1}^{2}A_{j}^{2}u=f(u),\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\leavevmode\nobreak\
\partial_{1}A_{1}+\partial_{2}A_{2}=0,\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\\\
\displaystyle\int_{\mathbb{R}^{2}}|u|^{2}dx=a^{2},\end{array}\right.$

where $a\neq 0$, $\lambda\in\mathbb{R}$ is known as the Lagrange multiplier
and $f\in\mathcal{C}^{1}(\mathbb{R})$ denotes the nonlinearity that fulfills
the supercritical exponential growth in the Trudinger-Moser sense at infinity.
Under suitable assumptions, combining the constrained minimization approach
together with the homotopy stable family and elliptic regularity theory, we
obtain that the problem has at least a ground state solution.

###### Key words and phrases:

Normalized solutions, Chern-Simons-Schrödinger system, Trudinger-Moser
inequality, Constrained minimization approach, Ground state solution,
Variational method.

###### 2010 Mathematics Subject Classification:

35J20, 35J61, 35B06.

L. Shen was partially supported by NSFC (12201565). M. Squassina is member of
Gruppo Nazionale per l’Analisi Matematica, la Probabilita e le loro
Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM)

###### Contents

  1. 1 Introduction and main results
  2. 2 Preliminary results
  3. 3 The subcritical case
  4. 4 The critical case
  5. 5 The supercritical case
    1. 5.1 The case $\bar{\delta}=\delta$
    2. 5.2 The case $\bar{\delta}=2$

##  1\. Introduction and main results

In this article, we investigate the existence of solutions for the following
generalized Chern-Simons-Schrödinger (CSS in short) system/equation with
supercritical exponential growth

(1.1) $\left\\{\begin{array}[]{ll}\displaystyle-\Delta u+\lambda
u+A_{0}u+\sum\limits_{j=1}^{2}A_{j}^{2}u=f(u),\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\leavevmode\nobreak\
\partial_{1}A_{1}+\partial_{2}A_{2}=0,\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\end{array}\right.$

under the constraint

(1.2) $\int_{\mathbb{R}^{2}}|u|^{2}dx=a^{2},$

where $a\neq 0$, $\lambda\in\mathbb{R}$ is known as the Lagrange multiplier
and $f\in\mathcal{C}^{0}(\mathbb{R})$ denotes the nonlinearity that fulfills
the supercritical exponential growth in the Trudinger-Moser sense at infinity
which would be specified later.

In recent years, the following time-dependent CSS system in two spatial
dimension

(1.3)
$\begin{cases}\text{i}D_{0}\psi+(D_{1}D_{1}+D_{2}D_{2})\psi+g(|\psi|^{2})\psi=0,\\\
\partial_{0}A_{1}-\partial_{1}A_{0}=-\text{Im}(\overline{\psi}D_{2}\psi),\\\
\partial_{0}A_{2}-\partial_{2}A_{0}=\text{Im}(\overline{\psi}D_{1}\psi),\\\
\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|\psi|^{2},\\\ \end{cases}$

is usually exploited to describe the non-relativistic dynamics behavior of
massive number of particles in Chern-Simons gauge fields, where i stands for
the imaginary unit, $\partial_{0}=\frac{\partial}{\partial t}$,
$\partial_{1}=\frac{\partial}{\partial x_{1}}$,
$\partial_{2}=\frac{\partial}{\partial x_{2}}$ for
$(t,x_{1},x_{2})\in\mathbb{R}^{1+2}$, $\psi:\mathbb{R}^{1+2}\to\mathbb{C}$ is
the complex scalar field, $A_{j}:\mathbb{R}^{1+2}\to\mathbb{R}$ denotes the
gauge field, $D_{j}=\partial_{j}+\text{i}A_{j}$ is the covariant derivative
for $j=0,1,2$ and the function $g$ is the nonlinearity.

This model plays an important role in the study of the high-temperature
superconductor, Aharovnov-Bohm scattering, and quantum Hall effect, we refer
the reader to [25, 26, 27]. Moreover, there are some further physical
motivations for considering CSS system (1.3), see e.g. [18, 23, 40, 41].

For $j=0,1,2$, we always deal with $A_{j}(t,x)=A_{j}(x)$ for all
$(t,x_{1},x_{2})\in\mathbb{R}^{1+2}$. If the standing wave ansatz
$\psi(t,x)=e^{\text{i}\lambda t}u(x)$ with a given $\lambda\in\mathbb{R}$ for
$u:\mathbb{R}^{2}\to\mathbb{R}$, then (1.3) reduces to

(1.4) $\left\\{\begin{array}[]{ll}\displaystyle-\Delta u+\lambda
u+A_{0}u+\sum_{j=1}^{2}A_{j}^{2}u=f(u),\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\end{array}\right.$

where $f(u)=g(|u|^{2})u$. Let us suppose that the gauge field $A_{j}$
satisfies the Coulomb gauge condition $\sum_{j=0}^{2}\partial_{j}A_{j}=0$,
then (1.4) becomes the original CSS equation (1.1), namely

(1.5) $\left\\{\begin{array}[]{ll}\displaystyle-\Delta u+\lambda
u+A_{0}u+\sum_{j=1}^{2}A_{j}^{2}u=f(u),\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\leavevmode\nobreak\
\partial_{1}A_{1}+\partial_{2}A_{2}=0.\end{array}\right.$

Since $\partial_{1}A_{0}=A_{2}|u|^{2}$ and $\partial_{2}A_{0}=-A_{1}|u|^{2}$
in (1.5), there holds

$\Delta
A_{0}=\partial_{1}\big{(}A_{2}|u|^{2}\big{)}-\partial_{2}\big{(}A_{1}|u|^{2}\big{)},$

which leads to

(1.6)
$A_{0}[u](x)=\frac{x_{1}}{2\pi|x|^{2}}\ast\big{(}A_{2}|u|^{2}\big{)}-\frac{x_{2}}{2\pi|x|^{2}}\ast\big{(}A_{1}|u|^{2}\big{)}.$

Analogously, adopting
$\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2}$ and
$\partial_{1}A_{1}+\partial_{2}A_{2}=0$ in (1.5) to deduce that

$\Delta
A_{1}=\partial_{2}\bigg{(}\frac{|u|^{2}}{2}\bigg{)}\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ \Delta
A_{2}=-\partial_{1}\bigg{(}\frac{|u|^{2}}{2}\bigg{)}.$

From which, the components $A_{j}$ for $j=1,2$ in (1.5) can be represented as

(1.7)
$A_{1}[u](x)=\frac{x_{2}}{2\pi|x|^{2}}\ast\bigg{(}\frac{|u|^{2}}{2}\bigg{)}=-\frac{1}{4\pi}\int_{\mathbb{R}^{2}}\frac{(x_{2}-y_{2})u^{2}(y)}{|x-y|^{2}}dy,$
(1.8)
$A_{2}[u](x)=-\frac{x_{1}}{2\pi|x|^{2}}\ast\bigg{(}\frac{|u|^{2}}{2}\bigg{)}=\frac{1}{4\pi}\int_{\mathbb{R}^{2}}\frac{(x_{1}-y_{1})u^{2}(y)}{|x-y|^{2}}dy.$

When there is no misunderstanding, for simplicity we shall write shortly
$A_{j}$ in place of $A_{j}[u]$, for $j=0,1,2$. Further properties of $A_{j}$,
for $j=0,1,2$, will be introduced in Section 2 below.

Actually, if $u$ is radially symmetric in the standing wave ansatz
$\psi(t,x)=e^{\text{i}\lambda t}u(x)$, CSS system (1.3) becomes a single
equation. In [11], Byeon-Huh-Seok considered the standing waves of type

(1.9) $\begin{gathered}\psi(t,x)=u(|x|)e^{\text{i}\lambda
t},\leavevmode\nobreak\ A_{0}(t,x)=k(|x|),\hfill\\\
A_{1}(t,x)=\frac{x_{2}}{|x|^{2}}h(|x|),\leavevmode\nobreak\
A_{2}(t,x)=-\frac{x_{1}}{|x|^{2}}h(|x|),\hfill\\\ \end{gathered}$

where $k$ and $h$ are real value functions depending only on $|x|$. Note that
(1.9) satisfies the Coulomb gauge condition with $\varsigma=ct+n\pi$, where
$n$ is an integer and $c$ is a real constant. To look for solutions of CSS
system (1.3) of the type (1.9), it suffices to solve the following semilinear
elliptic equation

(1.10) $-\Delta u+\lambda
u+\bigg{(}\int_{|x|}^{\infty}\frac{h(s)}{s}u^{2}(s)ds+\frac{h^{2}(|x|)}{|x|^{2}}\bigg{)}u=f(u)\leavevmode\nobreak\
\text{in}\leavevmode\nobreak\ \mathbb{R}^{2},$

where $h(s)=\int_{0}^{s}\frac{r}{2}u^{2}(r)dr$.

Now, we have two kinds of CSS type equations (1.1) and (1.10) in hands and
there are two aspects to the studies for them. On the one hand, one can choose
the frequency $\lambda\in\mathbb{R}$ to be fixed. In this situation, the
existence, nonexistence and multiplicity of nontrivial solutions have been
considerably contemplated by a lot of mathematicians, see [7, 14, 33, 42, 36,
45, 47, 49, 46, 53, 17, 32] and the references therein for example.

On the other hand, one can find solutions for (1.1) or (1.10) with an unknown
frequency $\lambda\in\mathbb{R}$ which appears as a Lagrange multiplier. As we
all know, the mass of each solution to the Cauchy problem for system (1.3) is
conserved along time, namely
$\int_{\mathbb{R}^{2}}|\psi(t,\cdot)|^{2}dx=\int_{\mathbb{R}^{2}}|\psi(0,\cdot)|^{2}dx$
for any $t\in[0,T)$. In physics, there exists a crucial meaning concerning the
mass which is often adopted to represent the power supply in nonlinear optics
or the total number of atoms in Bose-Einstein condensation, see [19, 20, 44]
for example. As a consequence, taking physical point of views into account, it
is interesting to seek for solutions to (1.3) with prescribed mass.

This naturally leads to the study of solutions to (1.1)-(1.2) for $a>0$ given.
In this scenario, solutions to (1.1)-(1.2) are referred to as normalized
solutions. Indeed, solutions to (1.1)-(1.2) correspond to critical points of
the underlying energy functional $E$ restricted on $S(a)$, where

(1.11) $E(u)\triangleq\frac{1}{2}\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-\int_{\mathbb{R}^{2}}F(u)dx$

and

$S(a)\triangleq\bigg{\\{}u\in
H^{1}(\mathbb{R}^{2}):\int_{\mathbb{R}^{2}}|u|^{2}dx=a^{2}\bigg{\\}}.$

Moreover, from mathematical perspectives, the consideration of normalized
solutions turns out to be also meaningful since it benefits from understanding
dynamical properties of stationary solutions to equations like (1.3).

In recent years, the considerations for normalized solutions for (1.1) or
(1.10) have received more and more attentions. Speaking precisely, for
$f(u)=|u|^{p-2}u$, Byeon _et al._ [11] derived the existence of normalized
solution for each $a\neq 0$ if $p\in(2,3]$ and sufficiently small $|a|$ if
$p\in(3,4)$. Afterwards, the authors in [58, 35] generalized the results in
[11] to $p>4$. For the particular case $p=4$, Gou and Zhang [22] exhibited
some very interesting results. There exist some other related results on
normalized solution in [38, 43, 56, 57, 59] and the references therein.

It should be mentioned that the research interest could date back to the
Schrödinger equations with prescribed $L^{2}$-norm. In [28], with the help of
a minimax approach and compactness argument, Jenajean contemplated the
existence of solutions for the following Schrödinger problem

(1.12) $\left\\{\begin{array}[]{ll}\displaystyle-\Delta u+\lambda
u=g(u)&\text{in}\leavevmode\nobreak\ \mathbb{R}^{N},\\\
\displaystyle\int_{\mathbb{R}^{N}}|u|^{2}dx=a^{2}>0.\end{array}\right.$

Later on, there are some complements and generalizations in [30]. In [50], for
$g(t)=\mu|t|^{q-2}t+|t|^{p-2}t$ with $2<q\leq 2+\frac{4}{N}\leq p<2^{*}$,
Soave considered the existence of solutions for problem (1.12), where
$2^{*}=\frac{2N}{N-2}$ if $N\geq 3$ and $2^{*}=\infty$ if $N=2$. For this type
of combined nonlinearities, Soave [51] proved the existence of ground state
and excited solutions when $p=2^{*}$. As to more results for problem (1.12),
see e.g. [8, 29, 31, 37, 54] and the references therein.

Whereas, the planar case for critical problem (1.12) are definitely different
from $N\geq 3$. In reality, one can observe that $2^{*}=\infty$ if $N=2$ and
$H^{1}(\mathbb{R}^{2})\not\hookrightarrow L^{\infty}(\mathbb{R}^{2})$ which
makes the problems special and quite delicate. So, it is not easy to deal with
the nonlinearity involving a (super)critical exponential growth trivially. Due
to the Trudinger-Moser type inequality, we can say that a function $f$
possesses the _critical exponential growth_ at infinity if there exists a
constant $\alpha_{0}>0$ such that

(1.13) $\lim_{t\rightarrow+\infty}\frac{|f(t)|}{e^{\alpha
t^{2}}}=\begin{cases}0,&\forall\alpha>\alpha_{0},\\\
+\infty,&\forall\alpha<\alpha_{0}.\end{cases}$

The above definition was introduced by Adimurthi and Yadava in [1], see also
de Figueiredo, Miyagaki and Ruf [15] for example. As to the _subcritical
exponential growth_ at infinity for $f$, it says that

(1.14) $\lim_{t\rightarrow+\infty}\frac{|f(t)|}{e^{\alpha
t^{2}}}=0,\leavevmode\nobreak\ \forall\alpha>0.$

Because of the appearance of the critical exponential growth in (1.13), the
results for problem (1.12) with $N=2$ are fewer than those for $N\geq 3$
involving Sobolev critical growth. Very recently, Alves, Ji and Miyagaki [3]
firstly studied problem (1.12) in the case $N=2$ with $g$ satisfying (1.13).
Based on the ideas in [3], the authors investigated the existence of
normalized solutions for (1.1) and (1.10) in [56] and [59], respectively. In
this article, we continue to study the existence of normalized solutions for
(1.1) with supercritical exponential growth which is definitely differently
from (1.13).

In order to depict clearly the ideas how to handle the supercritical
exponential case, we are going to derive the existence of normalized solutions
for CSS equation (1.1) with (sub)critical exponential growth. Let us impose
the assumptions on $f$ as follows.

  * $(f_{1})$

$f\in\mathcal{C}^{1}(\mathbb{R})$ and there is a constant $\chi\geq 4$ such
that $f(s)=o(s^{\chi-1})$ as $s\to 0$;

  * $(f_{2})$

There is a constant $\theta>4$ such that $0<\theta F(s)\leq f(s)s$ for all
$s\neq 0$;

  * $(f_{3})$

The function $\bar{F}(s)=f(s)s-2F(s)$ satisfies

$\bar{F}(s)/|s|^{4}\leavevmode\nobreak\ \text{is strictly increasing
in}\leavevmode\nobreak\ (0,+\infty).$

Since we are interested in positive solutions for Eq. (1.1), without loss of
generality, we suppose that $f(s)\equiv 0$ for all $s\in(-\infty,0]$ until the
end of the present paper.

Now, we can state the first main result.

######  Theorem 1.1.

Suppose that $f$ satisfies (1.14) and $(f_{1})-(f_{3})$, then there is a small
$a^{*}>0$ such that problems (1.1)-(1.2) possess a couple weak solution
$(u_{0},\lambda_{0})\in H^{1}(\mathbb{R}^{2})\times\mathbb{R}^{+}$ for all
$a\in(0,a^{*}]$, where $\lambda_{0}>0$, $u_{0}(x)>0$ for all
$x\in\mathbb{R}^{2}$ and $E(u_{0})=m(a)$ with

(1.15) $m(a)\triangleq\inf_{u\in\mathcal{M}(a)}E(u),\leavevmode\nobreak\
\mathcal{M}(a)=\big{\\{}u\in S(a):J(u)=0\big{\\}}.$

Here the energy functional $J:S(a)\to\mathbb{R}$ is defined by

$J(u)=\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-\int_{\mathbb{R}^{2}}[f(u)u-2F(u)]dx.$

Given a $u\in S(a)$, it follows from $(f_{1})-(f_{2})$ that

$E(tu(t\cdot))=\frac{t^{2}}{2}\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-t^{-2}\int_{\mathbb{R}^{2}}F(tu)dx\to-\infty$

as $t\to+\infty$ and so we cannot find critical points for $E$ restricted on
$S(a)$ directly. According to the discussions in Section 2 below, we see that
$\mathcal{M}(a)$ is a natural constraint manifold.

Next, we deal with the critical exponential growth case. To the end, we
additionally suppose that

$(f_{4})$:

there exist some constants $s_{0}>0$, $M_{0}>0$ and $\vartheta\in(0,1]$ such
that

$0<s^{\vartheta}F(s)\leq M_{0}f(s)\leavevmode\nobreak\ \text{for
all}\leavevmode\nobreak\ s\geq s_{0};$

$(f_{5})$:

$\liminf\limits_{|s|\to+\infty}F(s)e^{-\alpha_{0}s^{2}}\geq\beta_{0}>0$, where
$\beta_{0}$ is an arbitrary but fixed constant.

Our second main result reads in the following.

######  Theorem 1.2.

Suppose that $f$ satisfies (1.13) and $(f_{1})-(f_{5})$, then there is a small
$a_{*}>0$ such that problems (1.1)-(1.2) admit a couple weak solution
$(u_{0},\lambda_{0})\in H^{1}(\mathbb{R}^{2})\times\mathbb{R}^{+}$ for all
$a\in(0,a_{*}]$, where $\lambda_{0}>0$, $u_{0}(x)>0$ for all
$x\in\mathbb{R}^{2}$ and $E(u_{0})=m(a)$.

######  Remark 1.3.

In light of the nonlinearity $f$ in (1.1) is more general than its
counterparts in [35, 38, 22], so we can never simply repeat their methods to
conclude Theorem 1.1. The critical exponential case has been studied in [56,
59], but there are the following three important contributions which can be
regarded as a partial motivation to contemplate the problems behind this
article.

(1) We do not require any compact condition which allows us to study a wider
class of CSS equations. Explaining more precisely, to overcome the loss of
compactness, authors in [59] look for solutions in the radially symmetric
subspace $H_{r}^{1}(\mathbb{R}^{2})$ which implies that the imbedding
$H^{1}_{r}(\mathbb{R}^{2})\hookrightarrow L^{p}(\mathbb{R}^{2})$ is compact
for all $2<p<+\infty$; conversely, the authors in [56] find solutions in the
work space $X=\\{u\in
H^{1}(\mathbb{R}^{2}):\int_{\mathbb{R}^{2}}|x|^{2}|u|^{2}dx<+\infty\\}$ which
reveals that the imbedding $X\hookrightarrow L^{p}(\mathbb{R}^{2})$ is compact
for all $2\leq p<+\infty$.

(2) Although with the above two types of compact imbedding results in hands,
authors in [56, 59] have to suppose that $a^{2}<1$ to recover the compactness.
With the help of some specific calculations, we would remove this unpleasant
restriction. In other words, we only need to suppose that the mass $a^{2}$ is
suitably small which is just caused by the Chern-Simons term
$\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx$. Actually, we could
show that Theorems 1.1 and 1.2 holds true for all $a\neq 0$ provided that it
is absent.

(3) When taking the energy estimate, the assumption

$(f_{5}^{\prime})$:

There are constants $p>2$ and a sufficiently large $\varsigma>0$ such that
$F(s)\geq\varsigma|s|^{p}$ for all $s\in\mathbb{R}$,

acts as a key role in [56, 59]. Nevertheless, as pointed out in [6], it covers
the essential feature of the critical exponential growth given in (1.13)
because $(f_{5}^{\prime})$ is a global condition which requires $f$ to be
$p$-superlinear growth for all $t\in\mathbb{R}$ and the parameter $\varsigma$
must be large. Therefore, we shall depend on the slightly modified Moser
sequence functions introduced in [6] to reach the aim. However, owing to the
presence of the Chern-Simons term
$\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx$, there are some
additional challenges that we are necessary to bring in some technical
calculations, see e.g. Lemma 4.1.

Finally, as the applications of Theorems 1.1 and 1.2, we are able to
investigate the existence of normalized solutions for CSS equation (1.1) with
supercritical exponential growth at infinity. Due to [4, 5], we suppose that
the nonlinearity $f$ has the form of the following particular type

(1.16) $f(t)=h(t)e^{\bar{\alpha}_{0}|t|^{\tau}},\leavevmode\nobreak\ \forall
t\in\mathbb{R},$

for some $\bar{\alpha}_{0}>0$ and $\tau\geq 2$. Hereafter, the
$\mathcal{C}^{1}$ function $h$ that vanishes in $(-\infty,0])$ satisfies

$(h_{1})$:

There is a constant $\chi\geq 4$ such that $h(t)=o(t^{\chi-1})$ as $t\to 0$;

$(h_{2})$:

There exists a $\theta>4$ such that $0<\theta H(t)\leq h(t)t$ for all $t\neq
0$, where $H(t)=\int_{0}^{t}h(s)\,ds$;

$(h_{3})$:

The function $\bar{H}(s)=h(s)s-2H(s)$ satisfies

$\bar{H}(s)/|s|^{4}\leavevmode\nobreak\ \text{is strictly increasing
in}\leavevmode\nobreak\ (0,+\infty).$

$(h_{4})$:

There exist $\delta\in[8\theta^{-1},2)$ and $\gamma,M>0$ such that
$0\leq|h(t)|\leq Me^{\gamma|t|^{\delta}}$ for all $t\in\mathbb{R}$.

Motivated by [4, 5], for the function $f$ defined in (1.16) together with
$(h_{2})$, one could observe that it possesses the so-called _supercritical
exponential growth_ at infinity when $\tau>2$, for example

(1.17) $(\text{I})\leavevmode\nobreak\ \tau>2\leavevmode\nobreak\ \text{is
arbitrary if}\leavevmode\nobreak\ \bar{\alpha}_{0}>0\leavevmode\nobreak\
\text{is fixed;}\leavevmode\nobreak\ (\text{II})\leavevmode\nobreak\
\bar{\alpha}_{0}>0\leavevmode\nobreak\ \text{is arbitrary
if}\leavevmode\nobreak\ \tau\geq 2\leavevmode\nobreak\ \text{is fixed.}$

Let us exhibit the main result on this topic as follows.

######  Theorem 1.4.

Suppose that the nonlinearity $f$ defined in (1.16) requires
$(h_{1})-(h_{4})$, then there is a small $\bar{a}^{*}>0$ such that, for each
$\tau>2$, there exists a $\bar{\alpha}^{*}_{0}=\bar{\alpha}_{0}(\tau)>0$ such
that problems (1.1)-(1.2) possess a couple of weak solution
$(\bar{u}_{0},\bar{\lambda}_{0})$ for all fixed
$\bar{\alpha}_{0}\in(0,\bar{\alpha}_{0}^{*})$ and $a\in(0,\bar{a}^{*}]$.
Moreover, if in addition

$(h_{5})$:

there are some constants $\xi>0$ and $p>4$ such that $h(t)\geq\xi t^{p-1}$ for
all $t\in[0,+\infty)$,

then there is a small $\bar{a}_{*}>0$ such that, given a $\bar{\alpha}_{0}>0$,
there exist $\tau_{*}=\tau(\bar{\alpha}_{0})>2$ and ${\xi}_{0}>0$ such that
problems (1.1)-(1.2) admit a couple of weak solution
$(\bar{u}_{0},\bar{\lambda}_{0})$ for every fixed $\tau\in[2,\tau_{*})$,
${\xi}>{\xi}_{0}$ and $a\in(0,\bar{a}_{*}]$.

######  Remark 1.5.

Up to the best knowledge of us so far, except [48], there are not any
existence results for CSS equation with supercritical exponential critical
growth. As a consequence, it seems the first paper to consider the existence
of normalized solutions for supercritical CSS equation in $\mathbb{R}^{2}$. We
emphasize here that although the authors in [6] showed that the supercritical
Schrödinger equation with Stein-Weiss convolution parts has at least a
normalized solution, one cannot derive Theorem 1.4 trivially because of the
Chern-Simons term $\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx$.
What’s more, the absence of compact imbedding results in this paper cannot
also be ignored, see [6, Lemma 2.3] in detail.

Let us briefly sketch the proofs for our main results. To conclude the proofs
of Theorems 1.1 and 1.2, we firstly introduce some interesting results for the
Chern-Simons term $\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx$ in
the literature and the Trudinger-Moser inequality developed by Cao [12] and so
that the energy functional $E$, defined in (1.11), is of class
$\mathcal{C}^{1}$ over $S(a)$. Then, we shall investigate some properties of
$\mathcal{M}(a)$ including it is natural constrain manifold, namely any
minimizer of $m(a)$ in (1.15) is indeed a weak solution of Eq. (1.1) with a
suitable $\lambda\in\mathbb{R}$. In the meantime, with the help of the
homotopy stable family, we can prove that there is a sequence
$\\{u_{n}\\}\subset\mathcal{M}(a)$ which is a Palais-Smale sequence for $E$
restricted on $S(a)$ at the level $m(a)$. To get the proof of Theorem 1.1,
combing the Vanishing lemma in [55] and Brézis-Lieb lemma as well as the
monotonicity of $m(a)$, we have that $E(u_{0})=m(a)$ and
$\lim\limits_{n\to\infty}|\nabla u_{n}|_{2}^{2}=|\nabla u_{0}|_{2}^{2}$, where
$u_{0}$ is the weak limit of $\\{u_{n}\\}$ in $H^{1}(\mathbb{R}^{2})$ along a
subsequence. On the other hand, we derive the strong convergence of
$\\{u_{n}\\}$ in $H^{1}(\mathbb{R}^{2})$ by showing that the Lagrange
multiplier $\lambda_{0}$ associated with $u_{0}$ is positive. Compared with
the proof of Theorem 1.1, in order to prove Theorem 1.2, we make the following
three additional adjustments: (a) We find a upper bound by the Moser sequence
functions for $m(a)$ which is adopted to verify that the weak limit $u_{0}\neq
0$. (b) By taking advantage of $(f_{2})$ jointly with the monotonicity of
$m(a)$, we still prove that $E(u_{0})=m(a)$ and
$\lim\limits_{n\to\infty}|\nabla u_{n}|_{2}^{2}=|\nabla u_{0}|_{2}^{2}$. Let
us stress here that this idea is also suitable for the subcritical case in
Theorem 1.1 and it should be reviewed as one of the striking novelties the the
present article. (c) As an application of $\lim\limits_{n\to\infty}|\nabla
u_{n}|_{2}^{2}=|\nabla u_{0}|_{2}^{2}$, we shall derive that $\lambda_{0}>0$
which leads to that $\bar{a}\mapsto m(\bar{a})$ is strictly decreasing on a
right neighborhood of $\bar{a}$, where
$\bar{a}^{2}=|u_{0}|_{2}^{2}\in(0,a^{2}]$. According to it, we must have
$\bar{a}^{2}=a^{2}$ which reveals the strong convergence of $\\{u_{n}\\}$ in
$H^{1}(\mathbb{R}^{2})$ and so the proof is done.

Concerning the proof of Theorem 1.4, due to a variational method point of
view, we continue to search for critical points for the variational functional
$E:S(a)\to\mathbb{R}$. Unfortunately, we couldn’t even show that $E$ is well-
defined over $S(a)$ directly which is caused by the appearance the
nonlinearity $f$ involving the supercritical exponential growth (1.17) in the
Trudinger-Moser sense. Hence, the most imperative starting point is to deduce
that $E$ is of class $\mathcal{C}^{1}$. Given a fixed constant $R>0$, followed
by [4, 5], we shall consider an auxiliary equation which possesses a
(sub)critical exponential growth. Speaking it clearly, introducing a cutoff
function $f^{R,\bar{\delta}}$ which is given by

(1.18) $f^{R,\bar{\delta}}(t)=\left\\{\begin{array}[]{ll}0,&t\leq 0,\\\
h(t)e^{\bar{\alpha}_{0}t^{\tau}},&0\leq t\leq R,\\\
h(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}},&t\geq
R,\end{array}\right.$

and

$\bar{\delta}\triangleq\left\\{\begin{array}[]{ll}\delta,&\text{if the Case I
in \eqref{definition2} is considered},\\\ 2,&\text{if the Case II in
\eqref{definition2} is considered}.\end{array}\right.$

Here, the function $h$ appears in (1.16) and the constant $\delta\in(0,2)$ is
proposed in the condition $(h_{4})$.

In consideration of such a nonlinearity $f^{R,\bar{\delta}}$, let us study the
following auxiliary equation

(1.19) $\left\\{\begin{array}[]{ll}\displaystyle-\Delta u+\lambda
u+A_{0}u+\sum\limits_{j=1}^{2}A_{j}^{2}u=f^{R,\bar{\delta}}(u),\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\leavevmode\nobreak\
\partial_{1}A_{1}+\partial_{2}A_{2}=0,\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\end{array}\right.$

Using $(h_{4})$, it is ready to observe that $f^{R,\bar{\delta}}$ admits the
subcritical or critical exponential growth at infinity for every fixed $R>0$.
Consequently, with the help of Theorems 1.1 and 1.2, we are able to establish
the existence of normalized solutions for Eq. (1.19) by finding critical
points of the energy functional $E^{R,\bar{\delta}}$ which is defined by

(1.20) $E^{R,\bar{\delta}}(u)=\frac{1}{2}\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-\int_{\mathbb{R}^{2}}F^{R,\bar{\delta}}(u)dx$

over a suitable subset of $S(a)$, where $R>0$ is a fixed constant determined
later and $F^{R,{\bar{\delta}}}(t)=\int_{0}^{t}f^{R,{\bar{\delta}}}(s)ds$ for
all $t\in\mathbb{R}$. Furthermore, it could simply contemplate that if a
couple of weak solution $(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ of problems
(1.19) and (1.2) satisfying $|\bar{u}_{0}^{R}|_{\infty}\leq R$, then it is a
couple of weak solution of problems (1.1)-(1.2). Have this in mind, we invite
the reader to acquaint that one should construct such a couple of weak
solution $(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ to conclude the proof of
Theorem 1.4. As a consequence, it is necessary to get the following result.

######  Theorem 1.6.

Suppose that the nonlinearity $f$ defined in (1.16) requires
$(h_{1})-(h_{4})$, then for every fixed $R>0$, there is a small
$\bar{a}^{*}_{R}>0$ such that, for each $\tau>2$, there exists a
$\bar{\alpha}^{*}_{0}=\bar{\alpha}_{0}(\tau)>0$ such that problems (1.19) and
(1.2) admit a couple of weak solution
$(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ for every fixed
$\bar{\alpha}_{0}\in(0,\bar{\alpha}_{0}^{*})$ and $a\in(0,\bar{a}^{*}_{R}]$.
Moreover, if in addition

$(h_{5})$:

there are some constants $\xi>0$ and $p>4$ such that $h(t)\geq\xi t^{p-1}$ for
all $t\in[0,+\infty)$,

then there is a small $\bar{a}_{*}^{R}>0$ such that, given a
$\bar{\alpha}_{0}>0$, there exist $\tau_{*}=\tau(\bar{\alpha}_{0})>2$ and
${\xi}_{0}^{R}>0$ such that problems (1.19) and (1.2) have a couple of weak
solution $(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ for every fixed
$\tau\in[2,\tau_{*})$, ${\xi}>{\xi}_{0}^{R}$ and $a\in(0,\bar{a}_{*}^{R}]$.

Again the results in Theorem 1.4 and 1.6 are new for the normalized solutions
under the Chern-Simons-Schrödinger system setting. As described above, one of
the key ingredients in the proof of Theorem 1.4 is to derive the
$L^{\infty}$-estimate for $\bar{u}_{0}^{R}$. It is widely known that both the
elliptic regular theory and Nash-Moser iteration procedure are the effective
tools and we shall argue as [48] by using the former one to reach the aim.
Alternatively, we have to exhibit some nontrivial calculations to show that
$(A_{0}+A_{1}^{2}+A_{2}^{2})\in L^{\infty}(\mathbb{R}^{2})$ when
$\bar{u}_{0}^{R}\in H^{1}(\mathbb{R}^{2})$, which differs from the
counterparts in [48].

The outline of the present paper is organized as follows. In Section 2, we
introduce the variational settings and preliminaries. In Sections 3 and 4, we
will exhibit the proofs of Theorems 1.1 and 1.2, respectively. Section 5 is
mainly devoted to the supercritical exponential case for problems (1.1)-(1.2).

Notations. From now on in this paper, otherwise mentioned, we use the
following notations:

  * •

$C,C_{1},C_{2},\bar{C}_{1},\bar{C}_{2},\cdots$ denote any positive constant,
whose value is not relevant.

  * •

Let $(X,\|\cdot\|_{X})$ be a Banach space with dual space
$(X^{-1},\|\cdot\|_{X^{-1}})$.

  * •

$|\,\,\,|_{p}$ denotes the usual norm of the Lebesgue space
$L^{p}(\mathbb{R}^{2})$, for every $p\in[1,+\infty]$,
$\|\,\,\,\|_{H^{i}(\mathbb{R}^{2})}$ denotes the usual norm of the Sobolev
space $H^{i}(\mathbb{R}^{2})$ for $i=1,2$.

  * •

$o_{n}(1)$ denotes a real sequence with $o_{n}(1)\to 0$ as $n\to+\infty$ and
$\mathbb{R}^{+}\triangleq[0,+\infty)$.

  * •

“$\to$” and “$\rightharpoonup$” stand for the strong and weak convergence in
the related function spaces, respectively.

  * •

The tangent space $S(a)$ at $u\in H^{1}(\mathbb{R}^{2})$ is defined as

$\mathbb{T}_{u}=\left\\{v\in
H^{1}(\mathbb{R}^{2}):\int_{\mathbb{R}^{2}}uvdx=0\right\\}.$

  * •

The norm of the $C^{1}$ restriction functional $E|_{S(a)}^{\prime}$ at $u\in
H^{1}(\mathbb{R}^{2})$ is defined by

$\|E|_{S(a)}^{\prime}\|_{H^{-1}(\mathbb{R}^{2})}=\sup_{v\in\mathbb{T}_{u},\|v\|_{H^{1}(\mathbb{R}^{2})}=1}E^{\prime}(u)[v].$

##  2\. Preliminary results

In this section, we are going to exhibit some preliminary results adopted to
prove the main results. To begin with, we shall give some useful observations.
Owing to the second equation and the last two equations in (1.1), for all
$u\in H^{1}(\mathbb{R}^{2})$, one has

(2.1)
$\begin{gathered}\int_{\mathbb{R}^{2}}A_{0}|u|^{2}dx=2\int_{\mathbb{R}^{2}}A_{0}(\partial_{2}A_{1}-\partial_{1}A_{2})dx\hfill\\\
\ \ \ \
=2\int_{\mathbb{R}^{2}}(A_{2}\partial_{1}A_{0}-A_{1}\partial_{2}A_{0})dx=2\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}dx.\hfill\\\
\end{gathered}$

It follows from the well-known Hardy-Littlewood-Sobolev inequality [39,
Theorem 4.3], we can derive the following estimates to the gauge fields
$A_{j}$ for $j=0,1,2$.

######  Lemma 2.1.

(see [24, Propositions 4.2-4.3]) Assume $1<r<2$ and
$\frac{1}{r}-\frac{1}{\widehat{r}}=\frac{1}{2}$, then

$|A_{j}|_{\widehat{r}}\leq C_{r}|u|_{2r}^{2}\leavevmode\nobreak\
\text{for}\leavevmode\nobreak\ j=1,2,\leavevmode\nobreak\
|A_{0}|_{\widehat{r}}\leq C_{r}|u|_{2r}^{2}|u|_{4}^{2},$

where $C_{r}>0$ is a constant dependent of $r$.

With Lemma 2.1 in hand, one can easily conclude that

(2.2) $|A_{j}u|_{2}\leq|A_{j}|_{\widehat{r}}|u|_{\frac{r}{r-1}}\leq
C_{r}|u|_{2r}^{2}|u|_{\frac{r}{r-1}}\leq\bar{C}_{r}\|u\|_{H^{1}(\mathbb{R}^{2})}^{3},\leavevmode\nobreak\
\text{for}\leavevmode\nobreak\ j=1,2,$

because $2r>2$ and $r/(r-1)>2$, where $\bar{C}_{r}>0$ depends only on $r>1$.
We also need the following Brézis-Lieb type lemma for the Chern-Simons term.

######  Lemma 2.2.

(see [22, Lemma 2.4]) If $u_{n}\rightharpoonup u$ in $H^{1}(\mathbb{R}^{2})$
and $u_{n}\to u$ a.e. in $\mathbb{R}^{2}$, then one has $A_{j}[u_{n}]\to
A_{j}[u]$ a.e. for $j=1,2$,

(2.3)
$\left\\{\begin{array}[]{ll}\displaystyle\lim_{n\to\infty}\int_{\mathbb{R}^{2}}A_{0}[u_{n}]u_{n}\psi
dx=\int_{\mathbb{R}^{2}}A_{0}[u]u\psi dx,\leavevmode\nobreak\ \forall\psi\in
H^{1}(\mathbb{R}^{2}),\\\
\displaystyle\lim_{n\to\infty}\int_{\mathbb{R}^{2}}A_{j}^{2}[u_{n}]u_{n}\psi
dx=\int_{\mathbb{R}^{2}}A_{j}^{2}[u]u\psi dx,\leavevmode\nobreak\
\forall\psi\in H^{1}(\mathbb{R}^{2})\leavevmode\nobreak\
\text{\emph{with}}\leavevmode\nobreak\ j=1,2,\end{array}\right.$

and

(2.4)
$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}\big{[}A_{j}^{2}[u_{n}]|u_{n}|^{2}-A_{j}^{2}[u_{n}-u]|u_{n}-u|^{2}\big{]}dx=\int_{\mathbb{R}^{2}}A_{j}^{2}[u]|u|^{2}dx,\leavevmode\nobreak\
\text{for}\leavevmode\nobreak\ j=1,2.$

Caused by the presence of (sub)critical exponential growth at infinity
associated with the nonlinearity $f$, let us introduce the following type of
Trudinger-Moser inequality due to Cao [12].

######  Lemma 2.3.

For any $u\in H^{1}(\mathbb{R}^{2})$ and for all $\alpha>0$, there holds

(2.5) $\int_{\mathbb{R}^{2}}(e^{\alpha|u|^{2}}-1)dx<+\infty.$

Moreover, if $|\nabla u|_{2}\leq 1$, $|u|_{2}\leq\bar{M}$ for some
$\bar{M}\in(0,+\infty)$ and $\alpha<4\pi$, then there exists a constant
$C=C(\bar{M},\alpha)>0$ such that

(2.6) $\int_{\mathbb{R}^{2}}(e^{\alpha|u|^{2}}-1)dx\leq C.$

Combining (1.14) and $(f_{1})$, for all $\varepsilon>0$ and $\alpha>0$, there
is a constant $C_{\varepsilon}>0$ such that

(2.7)
$|f(s)|\leq\varepsilon|s|^{\chi-1}+C_{\varepsilon}|s|^{q-1}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R},$

where $q>2$ can be arbitrarily chosen later. Using $(f_{2})$, there holds

(2.8)
$|F(s)|\leq\varepsilon|s|^{\chi}+C_{\varepsilon}|s|^{q}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}.$

Similarly, if the nonlinearity $f$ has a critical exponential growth at
infinity with the critical exponent $\alpha_{0}$ appearing in (1.13). Then fix
$q>2$ as above, for all $\varepsilon>0$ and $\alpha>\alpha_{0}$, there is a
constant $C_{\varepsilon}>0$ such that

(2.9)
$|f(s)|\leq\varepsilon|s|^{\chi-1}+C_{\varepsilon}|s|^{q-1}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R},$

and

(2.10)
$|F(s)|\leq\varepsilon|s|^{\chi}+C_{\varepsilon}|s|^{q}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}.$

Now, recallng the imbedding $H^{1}(\mathbb{R}^{2})\hookrightarrow
L^{p}(\mathbb{R}^{2})$ is continuous for all $2\leq p<+\infty$ and adopting
(2.2), we can apply (2.8) and (2.10) in (2.5) to verify that the energy
functional $E$, defined in (1.11), is of $\mathcal{C}^{1}$ class on $S(a)$
whose derivative can be computed as

$E^{\prime}(u)[v]=\int_{\mathbb{R}^{2}}\big{[}\nabla u\nabla
v+(A_{1}^{2}+A_{2}^{2}+A_{0})uv\big{]}dx-\int_{\mathbb{R}^{2}}f(u)vdx$

for any $u,v\in S(a)$. In particular, with the help of (2.1), one has

$E^{\prime}(u)[u]=\int_{\mathbb{R}^{2}}\big{[}|\nabla
u|^{2}+3(A_{1}^{2}+A_{2}^{2})|u|^{2}\big{]}dx-\int_{\mathbb{R}^{2}}f(u)udx.$

Therefore if $(u,\lambda)\in S(a)\times\mathbb{R}$ is a critical point of $E$,
then the quintuplet $\big{(}u,A_{0}[u],A_{1}[u],A_{2}[u],\lambda\big{)}$ is a
(weak) solution of system (1.1).

Next, we shall introduce the so-called Pohoz̆aev identity for Eq. (1.1) which
is more general but easier than its counterpart in [11, Propostion 2.3].

######  Lemma 2.4.

Suppose that $f$ satisfies (1.14) and $(f_{1})$, or (1.13) and $(f_{1})$. Let
$u\in H^{1}(\mathbb{R}^{2})$ be a nontrivial weak solution of Eq. (1.1) with
some suitable $\lambda>0$, then it satisfies the so-called Pohoz̆aev identity
below

(2.11)
$\lambda\int_{\mathbb{R}^{2}}|u|^{2}dx+2\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}dx-2\int_{\mathbb{R}^{2}}F(u)dx=0.$

###### Proof.

We postpone its detailed proof in Lemma A.1 in the Appendix below. ∎

As a direct consequence of Lemma 2.4, we know that each nontrivial weak
solution of Eq. (1.1) is contained in the set $\mathcal{M}(a)$. Indeed, let
$u$ be a test function on Eq. (1.1), there holds

(2.12) $\int_{\mathbb{R}^{2}}\big{[}|\nabla
u|^{2}+3(A_{1}^{2}+A_{2}^{2})|u|^{2}\big{]}dx+\lambda\int_{\mathbb{R}^{2}}|u|^{2}dx-\int_{\mathbb{R}^{2}}f(u)udx=0.$

One immediately derives $J(u)=0$ if (2.12) minuses (5.23). In other words, the
set $\mathcal{M}(a)$ is a natural constraint to find normalized ground state
solutions for problems (1.1) and (1.2).

In the sequel, we mainly focus on establishing some properties for
$\mathcal{M}(a)$. Because the subcritical case for the nonlinearity $f$ is
much simpler than the critical one, let us only consider that $f$ possesses
the critical exponential growth at infinity. From now on until the end of this
section, without loss of generality, we would always suppose that $f$
satisfies (1.13) and $(f_{1})-(f_{3})$ for simplicity.

######  Lemma 2.5.

Let $a\neq 0$. For every $u\in S(a)$, there exists a unique $t_{u}>0$ such
that $t_{u}u(t_{u}\cdot)\in\mathcal{M}(a)$ and
$E(u_{t_{u}})=\max\limits_{t>0}E(u_{t})$, where $u_{t}=tu(t\cdot)$ for all
$t>0$. Moreover, the map $S(a)\to(0,+\infty)$ defined by $u\mapsto t_{u}$ is
continuous.

###### Proof.

Let $u\in S(a)$ be fixed and define the function $\varsigma(t)=E(u_{t})$ for
any $t>0$. Since $u_{t}\in S(a)$, we simply have that

$\displaystyle\varsigma^{\prime}(t)=0$ $\displaystyle\Longleftrightarrow
t\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-t^{-3}\int_{\mathbb{R}^{2}}[f(tu)tu-2F(tu)]dx=0$
$\displaystyle\Longleftrightarrow\frac{1}{t}J(u_{t})=0\Longleftrightarrow
J(u_{t})=0\Longleftrightarrow u_{t}\in\mathcal{M}(a).$

We claim that $\varsigma(t)>0$ for some sufficiently small $t>0$ and
$\lim\limits_{t\to+\infty}\varsigma(t)=-\infty$. Firstly, let us recall the
celebrated Gagliardo-Nirenberg inequality

(2.13) $|u|_{p}^{p}\leq C_{p}|\nabla
u|_{2}^{p-2}|u|_{2}^{2},\leavevmode\nobreak\ \forall u\in
H^{1}(\mathbb{R}^{2})\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
2<p<+\infty.$

Let $t_{1}>0$ small enough satisfy $t_{1}^{2}|\nabla
u|_{2}^{2}<\frac{\pi}{\alpha}$ with $\alpha>\alpha_{0}$ appearing in (2.10),
then we can define $\bar{u}=\sqrt{\frac{\alpha}{\pi}}t_{1}u$ which shows that
$|\nabla\bar{u}|_{2}^{2}\leq 1$ and
$|\bar{u}|_{2}^{2}=\frac{\alpha}{\pi}t_{1}^{2}a^{2}<+\infty$, using (2.6) to
obtain

(2.14) $\int_{\mathbb{R}^{2}}(e^{2\alpha
t_{1}^{2}|u|^{2}}-1)dx=\int_{\mathbb{R}^{2}}(e^{2\pi|\bar{u}|^{2}}-1)dx\leq
C<+\infty$

which together with the Gagliardo-Nirenberg inequality (2.13) implies that

$\displaystyle\varsigma(t_{1})$
$\displaystyle\geq\frac{t^{2}_{1}}{2}\int_{\mathbb{R}^{2}}|\nabla
u|^{2}dx-\varepsilon t_{1}^{\chi-2}\int_{\mathbb{R}^{2}}|u|^{\chi}dx-
C_{\varepsilon}t_{1}^{q-2}\int_{\mathbb{R}^{2}}|u|^{q}(e^{\alpha
t_{1}^{2}|u|^{2}}-1)dx$
$\displaystyle\geq\frac{t_{1}^{2}}{2}\int_{\mathbb{R}^{2}}|\nabla
u|^{2}dx-\varepsilon a^{2}t_{1}^{\chi-2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}-\bar{C}_{\varepsilon}at_{1}^{q-2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u|^{2}dx\bigg{)}^{\frac{q-1}{2}},$

where we need to require $q>4$ in (2.10) in this situation. Therefore, we
choose $\varepsilon=\frac{1}{4a^{2}}$ if $\chi=4$ and $\varepsilon=1$ if
$\chi>4$ to see there is a sufficiently small $t_{2}\in(0,t_{1})$ such that
$\varsigma(t_{2})>0$. According to $(f_{2})$, it is very clear to show that
$\varsigma(t)\to-\infty$ as $t\to\infty$ since $\theta>4$. Thanks to the
claim, it permits us to find a such $t_{u}>0$ and its uniqueness follows
directly from $(f_{3})$.

Invoking from the above arguments, the map $u\mapsto t_{u}$ is well-defined.
Suppose that $\\{u_{n}\\}\subset S(a)$ is a sequence satisfying $u_{n}\to
u_{0}$ in $H^{1}(\mathbb{R}^{2})$, then we only need to conclude that
$t_{u_{n}}\to t_{u_{0}}$ in $(0,+\infty)$ along a subsequence, where
$t_{u_{n}}u_{n}(t_{u_{n}}\cdot),t_{u_{0}}u_{0}(t_{u_{0}}\cdot)\in\mathcal{M}(a)$.
Note that $t_{u_{0}}$ exists since $|u_{0}|_{2}^{2}=a^{2}>0$ which shows that
$u_{0}\neq 0$. On the one hand, we claim that $\\{t_{n}\\}$ is uniformly
bounded from above. Indeed, since $J(t_{u_{n}}u_{n}(t_{u_{n}}\cdot))=0$ and
$(f_{2})$, there holds

$\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}]dx=\int_{\mathbb{R}^{2}}\left[\frac{f(t_{n}u_{n})u_{n}-2F(t_{n}u_{n})}{(t_{n}u_{n})^{4}}\right]|u_{n}|^{4}dx\geq(\theta-2)\int_{\mathbb{R}^{2}}\frac{F(t_{n}u_{n})}{(t_{n}u_{n})^{4}}|u_{n}|^{4}dx.$

Using $(f_{2})$ again which implies that
$\lim\limits_{s\to+\infty}F(s)/s^{4}=+\infty$, then the right hand hand goes
to $+\infty$ as $n\to\infty$. However, the left hand can be controlled by a
universal constant independent of $n\in\mathbb{N}^{+}$ by (2.2). Thereby, the
claim concludes. On the other hand, we claim that $\\{t_{n}\\}$ is uniformly
bounded from below by a positive constant. Otherwise, we should suppose that
$t_{u_{n}}\to 0$ up to a subsequence. By exploiting a very similar
calculations in (2.14), we can certify that

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}]dx=0$

which is impossible and the claim is true. So, passing to a subsequence if
necessary, there is $t_{0}>0$ such that $t_{u_{n}}\to t_{0}$ as $n\to\infty$.
Owing to the Lemma A.2 in the Appendix below, we immediately conclude that
$J(u_{t_{0}})=0$ by $J(t_{u_{n}}u_{n}(t_{u_{n}}\cdot))=0$. This actually
indicates that $t_{0}u_{0}(t_{0}\cdot)\in\mathcal{M}(a)$. Recalling the above
fact that $t_{u_{0}}u_{0}(t_{u_{0}}\cdot)\in\mathcal{M}(a)$, we must have
$t_{0}=t_{u_{0}}$ determined by the uniqueness of the map $u_{0}\mapsto
t_{u_{0}}$ which means that $t_{u_{n}}\to t_{u_{0}}$ as $n\to\infty$. The
proof is completed. ∎

######  Lemma 2.6.

Let $a\neq 0$. Then, for all $u\in\mathcal{M}(a)$, there is a constant
$\varrho>0$ such that $|\nabla u|_{2}^{2}\geq\varrho$. In particular, it holds
that the minimization $m(a)>0$.

###### Proof.

Arguing it indirectly, we suppose that there is a sequence
$\\{u_{n}\\}\subset\mathcal{M}(a)$ such that $|\nabla u_{n}|_{2}^{2}\to 0$ as
$n\to\infty$. Clearly, without loss of generality, we can assume that
$\sup_{n\in\mathbb{N}^{+}}|\nabla u_{n}|_{2}^{2}\leq K$ with some
$K\in(0,\frac{2\pi}{\alpha})$, where $\alpha>\alpha_{0}$ comes from (2.9).
Denoting $\bar{u}_{n}=K^{-\frac{1}{2}}u_{n}$, then
$\sup_{n\in\mathbb{N}^{+}}|\nabla\bar{u}_{n}|_{2}^{2}\leq 1$ and
$|\bar{u}_{n}|^{2}_{2}=K^{-1}a^{2}<+\infty$. Adopting (2.6) with
$2K\alpha<4\pi$, we obtain

$\sup_{n\in\mathbb{N}^{+}}\int_{\mathbb{R}^{2}}(e^{2\alpha|u_{n}|^{2}}-1)dx=\sup_{n\in\mathbb{N}^{+}}\int_{\mathbb{R}^{2}}(e^{2K\alpha|\bar{u}_{n}|^{2}}-1)dx\leq
C<+\infty.$

Choosing $q>3$ in (2.9) and using the Gagliardo-Nirenberg inequality (2.13),
one has

$\displaystyle\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx$
$\displaystyle\leq\varepsilon\int_{\mathbb{R}^{2}}|u_{n}|^{\chi}dx+C_{\varepsilon}\int_{\mathbb{R}^{2}}|u_{n}|^{q}(e^{\alpha|u_{n}|^{2}}-1)dx$
$\displaystyle\leq\varepsilon a^{2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}+C_{\varepsilon}\bigg{(}\int_{\mathbb{R}^{2}}|u_{n}|^{2q}dx\bigg{)}^{\frac{1}{2}}\bigg{(}\int_{\mathbb{R}^{2}}(e^{2\alpha|u_{n}|^{2}}-1)dx\bigg{)}^{\frac{1}{2}}$
(2.15) $\displaystyle\leq\varepsilon a^{2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}+\bar{C}_{\varepsilon}a\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{q-1}{2}}.$

Since $\\{u_{n}\\}\subset\mathcal{M}(a)$ which gives that $J(u_{n})=0$, one
sees that

$\int_{\mathbb{R}^{2}}|\nabla u_{n}|^{2}dx\leq\varepsilon
a^{2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}+\bar{C}_{\varepsilon}a\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{q-1}{2}}$

which is absurd when $\chi>4$ and $q>3$. It still remains valid when $\chi=4$
by choosing $\varepsilon=\frac{1}{2a^{2}}>0$. So, we derive the first part of
this lemma. According to $(f_{2})$, for all $u\in\mathcal{M}(a)$,

$E(u)=E(u)-\frac{1}{\theta-2}J(u)\geq\frac{\theta-4}{2(\theta-2)}\int_{\mathbb{R}^{2}}|\nabla
u|^{2}dx$

showing the desired result. The proof is completed. ∎

######  Lemma 2.7.

Let $\\{u_{n}\\}\subset\mathcal{M}(a)$ be a minimizing sequence of $m(a)$,
then $\\{u_{n}\\}$ is uniformly bounded in $H^{1}(\mathbb{R}^{2})$. Moreover,
we further have that

(2.16) $\left\\{\int_{\mathbb{R}^{2}}F(u_{n})dx\right\\}\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\
\left\\{\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx\right\\}\leavevmode\nobreak\
\text{are uniformly bounded in}\leavevmode\nobreak\ n\in\mathbb{N}^{+}.$

###### Proof.

Since $J(u_{n})=0$ for all $\\{u_{n}\\}\subset\mathcal{M}(a)$, we conclude
from $(f_{2})$ that

$\displaystyle m(a)+o_{n}(1)$
$\displaystyle=E(u_{n})=E(u_{n})-\frac{1}{\theta-2}J(u_{n})$
$\displaystyle=\frac{\theta-4}{2(\theta-2)}\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}]dx+\frac{1}{\theta-2}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-\theta
F(u_{n})\right]dx$
$\displaystyle\geq\frac{\theta-4}{2(\theta-2)}\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}]dx$

and

$m(a)+o_{n}(1)=E(u_{n})-\frac{1}{2}J(u_{n})=\frac{1}{2}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-4F(u_{n})\right]dx\geq\frac{\theta-4}{2}\int_{\mathbb{R}^{2}}F(u_{n})dx.$

Since $F(s)\geq 0$ for all $s\in\mathbb{R}$, exploiting the definition of $J$,
we can finish the proof of this lemma. ∎

Next, we are going to show that any minimizer of the minimization problem
$m(a)=\inf_{u\in\mathcal{M}(a)}E(u)$ is a solution of Eq. (1.1) with a
suitable $\lambda\in\mathbb{R}$. In other words, we prove that each critical
point of $E|_{\mathcal{M}(a)}$ is indeed a critical point of $E|_{S(a)}$.
Taking this purpose into account, we need to introduce the following result.

######  Lemma 2.8.

(see [13, Corollary 4.1.2]) Let $X$ be a real Banach space and $U\subset X$ be
an open set. Suppose that $\tilde{h},g_{1},\cdots,g_{m}:U\to\mathbb{R}^{1}$
are $C^{1}$ functions and $x_{0}\in\mathcal{M}$ satisfying
$\tilde{h}(x_{0})=\inf\limits_{x\in\mathcal{M}}\tilde{h}(x)$ with

$\mathcal{M}=\\{x\in U:g_{i}(x)=0,\leavevmode\nobreak\ i=1,2,\cdots,m\\}.$

If $\\{g_{i}^{\prime}(x_{0})\\}_{i=1}^{m}$ is linearly independent, then there
exist $\lambda_{1},\cdots,\lambda_{m}\in\mathbb{R}$ such that

$\tilde{h}^{\prime}(x_{0})+\sum_{i=1}^{m}\lambda_{i}g_{i}^{\prime}(x_{0})=0.$

######  Lemma 2.9.

Let $a\neq 0$, then $\mathcal{M}(a)$ is a $\mathcal{C}^{1}$-manifold of
codimension 2 in $H^{1}(\mathbb{R}^{2})$. Moreover, it is a
$\mathcal{C}^{1}$-manifold of codimension 1 in $S(a)$.

###### Proof.

Given a $u\in\mathcal{M}(a)$, one sees that $J(u)=0$ and
$P(u)\triangleq\int_{\mathbb{R}^{2}}u^{2}dx-a^{2}=0$. Since
$f\in\mathcal{C}^{1}$, we see that $J$ and $P$ are of $C^{1}$ class. There
exist the following two possibilities: either (i) $J$ and $P$ are linearly
dependent; or (ii) $J$ and $P$ are linearly independent.

If (i) holds true, then for all fixed $u\in\mathcal{M}(a)$, it means that
there is a constant $\lambda^{*}\in\mathbb{R}$ satisfying

$2\int_{\mathbb{R}^{2}}[\nabla
u\nabla\psi+(A_{0}+A_{1}^{2}+A_{2}^{2})u\psi]dx+2\lambda^{*}\int_{\mathbb{R}^{2}}u\psi
dx=\int_{\mathbb{R}^{2}}[f^{\prime}(u)u-f(u)]\psi dx$

for all $\psi\in H^{1}(\mathbb{R}^{2})$, that is,

$\left\\{\begin{array}[]{ll}\displaystyle-\Delta
u+\lambda^{*}u+A_{0}u+\sum\limits_{j=1}^{2}A_{j}^{2}u=\frac{1}{2}[f^{\prime}(u)u-f(u)],\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\leavevmode\nobreak\
\partial_{1}A_{1}+\partial_{2}A_{2}=0,\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\end{array}\right.$

Using Lemma 2.4, we deduce that $u$ must satisfy the equality

$\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx=\frac{1}{2}\int_{\mathbb{R}^{2}}[f^{\prime}(u)u^{2}-3f(u)u+4F(u)]dx.$

Recalling that $J(u)=0$ for $u\in\mathcal{M}(a)$, using the above equality,
there holds

$\int_{\mathbb{R}^{2}}[f^{\prime}(u)u^{2}-5f(u)u+8F(u)]dx=0$

which is impossible by Lemma A.3 in the Appendix below. So, we have that (i)
cannot occur.

When (ii) holds true, then by Lemma 2.8, there are
$\lambda_{1},\lambda_{2}\in\mathbb{R}$ such that

$E^{\prime}(u)+\lambda_{1}J^{\prime}(u)+\lambda_{2}u=0\leavevmode\nobreak\
\text{in}\leavevmode\nobreak\ H^{-1}(\mathbb{R}^{2}).$

To derive the proof of this lemma, we shall show that $\lambda_{1}=0$. First
of all, we know that

$\begin{gathered}(1+2\lambda_{1})\int_{\mathbb{R}^{2}}[\nabla
u\nabla\psi+(A_{0}+A_{1}^{2}+A_{2}^{2})u\psi]dx+\lambda_{2}\int_{\mathbb{R}^{2}}u\psi
dx\hfill\\\ \ \ \ \ \ \ \ \
=\int_{\mathbb{R}^{2}}[\lambda_{1}f^{\prime}(u)u-(\lambda_{1}-1)f(u)]\psi
dx\hfill\\\ \end{gathered}$

for all $\psi\in H^{1}(\mathbb{R}^{2})$, that is,

$\left\\{\begin{array}[]{ll}\displaystyle-(1+2\lambda_{1})\Delta
u+\lambda_{2}u+(1+2\lambda_{1})A_{0}u+(1+2\lambda_{1})\sum\limits_{j=1}^{2}A_{j}^{2}u=\lambda_{1}f^{\prime}(u)u-(\lambda_{1}-1)f(u),\\\
\displaystyle\partial_{1}A_{2}-\partial_{2}A_{1}=-\frac{1}{2}|u|^{2},\leavevmode\nobreak\
\partial_{1}A_{1}+\partial_{2}A_{2}=0,\\\
\displaystyle\partial_{1}A_{0}=A_{2}|u|^{2},\leavevmode\nobreak\
\partial_{2}A_{0}=-A_{1}|u|^{2},\end{array}\right.$

It follows from Lemma 2.4 again that

$(1+2\lambda_{1})\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx=\int_{\mathbb{R}^{2}}[\lambda_{1}f^{\prime}(u)u^{2}-(3\lambda_{1}-1)f(u)u+2(2\lambda_{1}-1)F(u)]dx.$

Since $J(u)=0$ for $u\in\mathcal{M}(a)$ combined with the above equality, we
have

$\lambda_{1}\int_{\mathbb{R}^{2}}[f^{\prime}(u)u^{2}-5f(u)u+8F(u)]dx=0$

jointly with Lemma A.3 below again yields that $\lambda_{1}=0$. The proof is
completed. ∎

To look for minimizer of $m(a)$, we shall construct a Palais-Smale sequence
for $E$ restricted on $S(a)$ at the level $m(a)$ by using the minimax
principle based on the homotopy stable family of compact subsets of $S(a)$.
Due to [21], we have its definition as follows.

######  Definition 2.10.

Let $B$ be a closed subset of a set $Y\subset H^{1}(\mathbb{R}^{2})$. We say
that a class $\mathcal{G}$ of compact subsets of $Y$ is a _homotopy stable
family_ with the closed boundary $B$ provided that

  * •

every set in $\mathcal{G}$ is contained in $B$;

  * •

for any $A\in\mathcal{G}$ and each function $\eta\in\mathcal{C}([0,1]\times
Y,Y)$ satisfying $\eta(t,x)=x$ for all $(t,x)\in(\\{0\\}\times
Y)\cup([0,1]\times B)$, then $\eta(\\{1\\}\times A)\in\mathcal{G}$.

We borrow some ideas developed in [9, 52] to consider the functional
$\Psi:S(a)\to\mathbb{R}$ defined by

$\Psi=E\circ\zeta,$

where the map $\zeta:S(a)\to\mathcal{M}(a)$ is defined as $u\mapsto
t_{u}u(t_{u}\cdot)$ and $t_{u}>0$ is determined by Lemma 2.5. Since $u\mapsto
t_{u}$ is a continuous map, motivated by [52, Proposition 2.9], it allows us
to show that $\Psi$ is of $\mathcal{C}^{1}$ class over $S(a)$ and its
derivative is computed as

(2.17) $\Psi^{\prime}(u)[v]=E^{\prime}(\zeta(u))[v_{t_{u}}]$

for any $u\in S(a)$ and $v\in\mathbb{T}_{u}$, see Lemma A.4 in the Appendix in
detail.

######  Lemma 2.11.

Let $\mathcal{G}$ be a homotopy stable family of compact subsets of $S(a)$
with closed boundary $B$ and define

$m_{\mathcal{G}}\triangleq\inf_{A\in\mathcal{G}}\max_{u\in A}\Psi(u).$

Suppose that $B$ is contained in a connected component of $\mathcal{M}(a)$ and

$\max\\{\sup\Psi(B),0\\}<m_{\mathcal{G}}<+\infty,$

then there is a Palais-Smale sequence $\\{u_{n}\\}\subset\mathcal{M}(a)$ for
$E$ restricted on $S(a)$ at the level $m_{\mathcal{G}}$.

###### Proof.

The idea can date bake to [9] and we will exhibit the detailed proof for the
convinence of the reader. According to the definition of $m_{\mathcal{G}}$,
there is a minimizing sequence $\\{A_{n}\\}\subset\mathcal{G}$ such that

$\max_{u\in A_{n}}\Psi(u)=m_{\mathcal{G}}+o_{n}(1).$

Define a map $\eta:[0,1]\times S(a)\to S(a)$ by
$\eta(t,u)\triangleq(1-t+tt_{u})u((1-t+tt_{u})\cdot)$, where $t_{u}>0$ comes
from Lemma 2.5 and so $\eta\in\mathcal{C}([0,1]\times S(a))$. Moreover,
adopting Lemma 2.5 again, one can see that $\eta(t,u)=u$ for any
$(t,u)\in(\\{0\\}\times S(a))\cup([0,1]\times B)$ since
$B\subset\mathcal{M}(a)$ which yields that $t_{u}\equiv 1$. In light of
$\mathcal{G}$ is a homotopy stable family of compact subsets of $S(a)$ with
closed boundary $B$, then $D_{n}\triangleq\eta(\\{1\\}\times
A_{n})=\\{u_{t_{u}}:u\in A_{n}\\}\in\mathcal{G}$ by Definition 2.10. Using
Definition 2.10 again, one knows that $D_{n}\subset B\subset\mathcal{M}(a)$,
and for each $v\in D_{n}$, there is $u\in A_{n}$ such that $v=u_{t_{u}}$. In
view of Lemma 2.5, we have that $\Psi(v)=\Psi(u_{t_{u}})=E(\zeta(u))=\Psi(u)$
which shows that

$\max_{v\in D_{n}}\Psi(v)=\max_{u\in A_{n}}\Psi(u).$

Therefore, there is another minimizing sequence
$\\{D_{n}\\}\subset\mathcal{M}(a)$ such that

$\max_{v\in D_{n}}\Psi(v)=\max_{u\in A_{n}}\Psi(u)=m_{\mathcal{G}}+o_{n}(1).$

Thanks to the equivalent minimax principle [21, Theorem 3.2], it permits us to
derive a Palais-Smale sequence $\\{\bar{u}_{n}\\}\subset S(a)$ for $\Psi$
restricted on $S(a)$ at the level $m_{\mathcal{G}}$ and
$\text{dist}_{H^{1}(\mathbb{R}^{2})}(\bar{u}_{n},D_{n})=o_{n}(1)$.

Taking advantage of $\bar{u}_{n}$, we shall prove that
$u_{n}\triangleq\zeta(\bar{u}_{n})\in\mathcal{M}(a)$ is a Palais-Smale
sequence for $\Psi$ restricted on $S(a)$ at the level $m_{\mathcal{G}}$. Since
$E(u_{n})=E(\zeta(\bar{u}_{n}))=\Psi(\bar{u}_{n})=m_{\mathcal{G}}+o_{n}(1)$
and $=m_{\mathcal{G}}<+\infty$, due to Lemma 2.7, one obtains that
$\\{u_{n}\\}$ is uniformly bounded in $n\in\mathbb{N}^{+}$. In particular,
$\\{|\nabla u_{n}|_{2}^{2}\\}$ is uniformly bounded from above by a universal
constant. Exploiting Lemma 2.6, one concludes that $\\{|\nabla
u_{n}|_{2}^{2}\\}$ is uniformly bounded from below by a universal constant.
Then, we denote $\bar{t}_{n}\triangleq t_{\bar{u}_{n}}>0$ such that
$t_{\bar{u}_{n}}\bar{u}_{n}(t_{\bar{u}_{n}}\cdot)\in\mathcal{M}(a)$ and so

$\bar{t}_{n}^{2}=\frac{|\nabla
u_{n}|_{2}^{2}}{|\nabla\bar{u}_{n}|_{2}^{2}}\in[C_{1},C_{2}],$

for some positive constants $C_{1},C_{2}$ independent of $n\in\mathbb{N}^{+}$,
where we have depended on Lemma 2.6 and
$\text{dist}_{H^{1}(\mathbb{R}^{2})}(\bar{u}_{n},D_{n})=o_{n}(1)$ with
$\\{D_{n}\\}\subset\mathcal{M}(a)$.

In view of the definition of $\mathbb{T}_{u}$, since

$\int_{\mathbb{R}^{2}}uvdx=\int_{\mathbb{R}^{2}}u_{t_{u}}v_{t_{u}}dx$

which implies that $v_{t_{u}}\in\mathbb{T}_{u_{t_{u}}}$ if, and only if,
$v\in\mathbb{T}_{u}$, then we know that the map
$\mathbb{T}_{u}\to\mathbb{T}_{u_{t_{u}}}$, defined by $v\mapsto v_{t_{u}}$, is
an isomorphism by Lemma 2.5 with the inverse $v\mapsto v_{t_{u}^{-1}}$. So,

$\displaystyle\|E|_{S(a)}^{\prime}(u_{n})\|_{H^{-1}(\mathbb{R}^{2})}$
$\displaystyle=\sup_{v\in\mathbb{T}_{u_{n}},\|v\|_{H^{1}(\mathbb{R}^{2})}=1}E^{\prime}(u_{n})[v]=\sup_{v\in\mathbb{T}_{u_{n}},\|v\|_{H^{1}(\mathbb{R}^{2})}=1}E^{\prime}(\zeta(\bar{u}_{n}))[(v_{\bar{t}_{n}^{-1}})_{\bar{t}_{n}}]$
$\displaystyle=\sup_{v\in\mathbb{T}_{u_{n}},\|v\|_{H^{1}(\mathbb{R}^{2})}=1}\Psi^{\prime}(\bar{u}_{n})[v_{\bar{t}_{n}^{-1}}]=\sup_{v_{\bar{t}_{n}^{-1}}\in\mathbb{T}_{\bar{u}_{n}},\|v_{\bar{t}_{n}^{-1}}\|_{H^{1}(\mathbb{R}^{2})}=1}\Psi^{\prime}(\bar{u}_{n})[v_{\bar{t}_{n}^{-1}}],$

where we have made full use of (2.17) in the third equality as well as
$v_{\bar{t}_{n}^{-1}}\in\mathbb{T}_{\bar{u}_{n}}$ if, and only if,
$v\in\mathbb{T}_{u_{n}}$ in the last equality. In addition, by
$\bar{t}_{n}^{2}\in[C_{1},C_{2}]$ and
$\|v_{\bar{t}_{n}^{-1}}\|_{H^{1}(\mathbb{R}^{2})}^{2}=t_{n}^{-2}|\nabla
v|_{2}^{2}+|v|_{2}^{2}<C<+\infty$, we derive that

$\|E|_{S(a)}^{\prime}(u_{n})\|_{H^{-1}(\mathbb{R}^{2})}\leq\bar{C}\sup_{w\in\mathbb{T}_{\bar{u}_{n}},\|w\|_{H^{1}(\mathbb{R}^{2})}=1}\Psi^{\prime}(\bar{u}_{n})[w].$

Recalling $\\{\bar{u}_{n}\\}\subset S(a)$ is a Palais-Smale sequence for
$\Psi$ restricted on $S(a)$ at the level $m_{\mathcal{G}}$, thereby we see
that $\\{u_{n}\\}\subset\mathcal{M}(a)$ is a Palais-Smale sequence for $E$
restricted on $S(a)$ at the level $m_{\mathcal{G}}$. ∎

As a corollary of Lemma 2.11, we are able to derive the result below.

######  Lemma 2.12.

Let $a\neq 0$. Then, there is a Palais-Smale sequence
$\\{u_{n}\\}\subset\mathcal{M}(a)$ for $E$ restricted on $S(a)$ at the level
$m(a)$.

###### Proof.

Let $\mathcal{G}$ be all singletons in $S(a)$ and $B=\emptyset$, then it is
simple to check that $\mathcal{G}$ is a homotopy stable family of compact
subsets of $S(a)$ without boundary. Moreover, we obtain

$m_{\mathcal{G}}=\inf_{A\in\mathcal{G}}\max_{u\in A}\Psi(u)=\inf_{u\in
S(a)}\max_{t>0}E(u_{t}).$

If we show that $m_{\mathcal{G}}=m(a)$, then we are done by Lemma 2.11. On the
one hand, for each $u\in S(a)$, there is unique $t_{u}>0$ such that
$t_{u}u(t_{u}\cdot)\in\mathcal{M}(a)$ and $\max_{t>0}E(u_{t})=E(u_{t_{u}})$ by
Lemma 2.5, then

$\inf_{u\in S(a)}\max_{t>0}E(u_{t})\geq\inf_{u\in\mathcal{M}(a)}E(u).$

On the other hand, we claim that, for all $t>0$, there holds

$\varpi(t,s)\triangleq\frac{1-t^{2}}{2}[f(s)s-2F(s)]+t^{-2}F(ts)-F(s)\geq
0,\leavevmode\nobreak\ \forall s\in\mathbb{R}.$

Indeed, it suffices to consider the case $s\geq 0$. Using some elementary
calculations, one sees that

$\displaystyle\frac{\partial}{\partial t}\varpi(t,s)$
$\displaystyle=t^{-3}[f(ts)ts-2F(ts)]-t[f(s)s-2F(s)]$
$\displaystyle=ts^{4}\left\\{\frac{f(ts)ts-2F(ts)}{(ts)^{4}}-\frac{f(s)s-2F(s)}{s^{4}}\right\\}$
$\displaystyle\left\\{\begin{array}[]{ll}\geq 0,&\text{if}\leavevmode\nobreak\
t\in[1,+\infty),\\\ <0,&\text{if}\leavevmode\nobreak\
t\in(0,1),\end{array}\right.$

where we have adopted $(f_{3})$ in the last inequality. Hence, we know that
$t\mapsto\varpi(t,s)$ is decreasing in $(0,1)$ and increasing in $(1,+\infty)$
for all $s\in\mathbb{R}$, respectively. It has that
$\varpi(t,s)\geq\min_{t\in(0,+\infty)}\varpi(t,s)=\varpi(1,s)=0$ for all
$s\in\mathbb{R}$ and the claim follows. Due to the claim, for all $u\in S(a)$
and $t>0$,

$E(u)-E(u_{t})-\frac{1-t^{2}}{2}J(u)=\int_{\mathbb{R}^{2}}\left[\frac{1-t^{2}}{2}[f(u)u-2F(u)]+t^{-2}F(tu)-F(u)\right]dx\geq
0.$

Given a $u\in\mathcal{M}(a)$, we immediately conclude that
$E(u)\geq\max_{t>0}E(u_{t})\geq\inf_{u\in S(a)}\max_{t>0}E(u_{t})$ and so

$\inf_{u\in\mathcal{M}(a)}E(u)\geq\inf_{u\in S(a)}\max_{t>0}E(u_{t}).$

As a consequence, we must have that $m_{\mathcal{G}}=m(a)$. The proof is
completed. ∎

######  Lemma 2.13.

Let $\\{u_{n}\\}\subset\mathcal{M}(a)$ be a Palais-Smale sequence for $E$
restricted on $S(a)$ at the level $m(a)$, then there is a uniformly bounded
sequence $\\{\lambda_{n}\\}\subset\mathbb{R}$ such that

(2.18) $E^{\prime}(u_{n})+\lambda_{n}u_{n}=o_{n}(1)\leavevmode\nobreak\
\emph{\text{in}}\leavevmode\nobreak\ H^{-1}(\mathbb{R}^{2}).$

###### Proof.

Since $\\{u_{n}\\}$ is uniformly bounded in $H^{1}(\mathbb{R}^{2})$ by Lemma
2.7 and $\|E|^{\prime}_{S(a)}(u_{n})\|_{H^{-1}(\mathbb{R}^{2})}\to 0$, we then
obtain that
$\|E^{\prime}(u_{n})-([E^{\prime}(u_{n})]u_{n})u_{n}\|_{H^{-1}(\mathbb{R}^{2})}\to
0$ and it means that

$\int_{\mathbb{R}^{2}}[\nabla
u_{n}\nabla\psi+(A_{0}+A_{1}^{2}+A_{2}^{2})u_{n}\psi]dx+\lambda_{n}\int_{\mathbb{R}^{2}}u_{n}\psi
dx-\int_{\mathbb{R}^{2}}f(u_{n})\psi dx\to 0$

for all $\psi\in H^{1}(\mathbb{R}^{2})$ and the above formula is the desired
result (5.10), where

$\displaystyle\lambda_{n}$
$\displaystyle\triangleq-\frac{1}{a^{2}}\left\\{\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+3(A_{1}^{2}+A_{2}^{2})u_{n}^{2}]dx-\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx\right\\}$
(2.19) $\displaystyle=\frac{2}{a^{2}}\left\\{\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx+3\int_{\mathbb{R}^{2}}F(u_{n})dx-\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx\right\\}.$

Here, we have exploited $J(u_{n})=0$ for all
$\\{u_{n}\\}\subset\mathcal{M}(a)$ in the second equality. According to Lemma
2.7, the sequence $\\{\lambda_{n}\\}\subset\mathbb{R}$ is uniformly bounded.
The proof is completed. ∎

With Lemmas 2.7 and 2.13 in hands, passing to a subsequence if necessary,
there exist a function $u_{0}\in H^{1}(\mathbb{R}^{2})$ and a constant
$\lambda_{0}\in\mathbb{R}$ such that

(2.20) $u_{n}\rightharpoonup u_{0}\leavevmode\nobreak\
\text{in}\leavevmode\nobreak\ H^{1}(\mathbb{R}^{2}),\leavevmode\nobreak\
u_{n}\to u_{0}\leavevmode\nobreak\ \text{in}\leavevmode\nobreak\
L^{p}_{\text{loc}}(\mathbb{R}^{2})\leavevmode\nobreak\ \text{for
all}\leavevmode\nobreak\ 2\leq p<+\infty\leavevmode\nobreak\
\text{and}\leavevmode\nobreak\ u_{n}\to u_{0}\leavevmode\nobreak\ \text{a.e.
in}\leavevmode\nobreak\ \mathbb{R}^{2}$

and

(2.21) $\lambda_{n}\to\lambda_{0}.$

To conclude this section, we introduce the following result.

######  Lemma 2.14.

Let $\\{u_{n}\\}\subset\mathcal{M}(a)$ be a Palais-Smale sequence for $E$
restricted on $S(a)$ at the level $m(a)$, up to s subsequence if necessary,
there holds

(2.22) $\lim_{n\to\infty}\int_{\mathbb{R}^{2}}f(u_{n})\psi
dx=\int_{\mathbb{R}^{2}}f(u_{0})\psi dx,\leavevmode\nobreak\ \forall\psi\in
C_{0}^{\infty}(\mathbb{R}^{2}),$

where $u_{0}\in H^{1}(\mathbb{R}^{2})$ comes from (2.20). In particular,
$u_{0}$ is a solution of (1.1) with $\lambda=\lambda_{0}$ in (2.21).

###### Proof.

Since $E(u_{n})=m(a)+o_{n}(1)$ and (5.10) jointly with (2.22), the proof is
standard and we omit it here, see e.g. [15]. ∎

##  3\. The subcritical case

In this section, we give the proof of Theorem 1.1. For simplicity, we shall
always suppose that the nonlinearity $f$ satisfies (1.14) and
$(f_{1})-(f_{3})$ throughout the present section.

First of all, since $f$ admits the subcritical exponential growth at infinity,
then we have the following Brézis-Lieb type lemma associated with $f$.

######  Lemma 3.1.

Let $\\{u_{n}\\}\in H^{1}(\mathbb{R}^{2})$ satisfies (2.20), going to a
subsequence if necessary, we have

(3.1)
$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}[F(u_{n})-F(u_{n}-u_{0})]dx=\int_{\mathbb{R}^{2}}F(u_{0})dx$
(3.2)
$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}[f(u_{n})u_{n}-f(u_{n}-u_{0})(u_{n}-u_{0})]dx=\int_{\mathbb{R}^{2}}f(u_{0})u_{0}dx$

###### Proof.

The proof is standard and we omit it here. ∎

As a direct consequence of Lemma 3.1, we immediately derive the results below
that play crucial roles in this section.

######  Lemma 3.2.

Let $\\{u_{n}\\}\in H^{1}(\mathbb{R}^{2})$ satisfies (2.20), going to a
subsequence if necessary, we have

(3.3) $\lim_{n\to\infty}[E(u_{n})-E(u_{n}-u_{0})]=E(u_{0})$

and

(3.4) $\lim_{n\to\infty}[J(u_{n})-J(u_{n}-u_{0})]=J(u_{0}).$

###### Proof.

Since $u_{n}\rightharpoonup u$ as $n\to\infty$, it infers from the basic
property of Hilbert space that

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}[|\nabla u_{n}|^{2}-|\nabla
u_{n}-\nabla u_{0}|^{2}]dx=\int_{\mathbb{R}^{2}}|\nabla u_{0}|^{2}dx$

which together with (2.3) and (3.1)-(3.2) yields the desired results. The
proof is completed. ∎

Before showing the detailed proof of Theorem 1.1, we need to certify that the
minimization $m(a)$ is non-increasing with respect to $a$.

######  Lemma 3.3.

For all $0<a_{1}<a_{2}$, there holds that $m(a_{1})\geq m(a_{2})$.

###### Proof.

According to the definition of $m(a_{1})$, for any $\epsilon>0$, there is a
$u\in\mathcal{M}(a)$ such that

$E(u)\leq m(a_{1})+\frac{\epsilon}{4}.$

Let us consider a cutoff function $\psi\in
C_{0}^{\infty}(\mathbb{R}^{2},[0,1])$ such that $\psi(x)=1$ if $|x|\leq 1$ and
$\psi(x)=0$ if $|x|\geq 2$ and define
$u_{\varepsilon}(x)\triangleq\psi(\varepsilon x)u(x)\in
H^{1}(\mathbb{R}^{2})\backslash\\{0\\}$, then $u_{\varepsilon}\to u$ in
$H^{1}(\mathbb{R}^{2})$ as $\varepsilon\to 0^{+}$. In view of Lemma 2.5, there
is a unique $t_{\varepsilon}\triangleq t_{u_{\varepsilon}}$ such that
$t_{u_{\varepsilon}}u_{\varepsilon}(t_{u_{\varepsilon}}\cdot)\in\mathcal{M}(a)$
and $t_{\varepsilon}\to t_{u}\triangleq t_{0}=1$ as $\varepsilon\to 0^{+}$.

As a consequence, we conclude that $E((u_{\varepsilon})_{t_{\varepsilon}})\to
E(u)$ as $\varepsilon\to 0^{+}$ with the help of Lemma A.2 in the Appendix
below and then one can fix a sufficiently small $\varepsilon_{0}>0$ to satisfy

$E((u_{\varepsilon_{0}})_{t_{\varepsilon_{0}}})\leq
E(u)+\frac{\epsilon}{4}\leq m(a_{1})+\frac{1}{2}\epsilon.$

Let $v\in C_{0}^{\infty}(\mathbb{R})$ satisfy ${\rm supp}{\hskip
1.42271pt}v\subset B_{1+\frac{4}{\varepsilon_{0}}}(0)\backslash
B_{\frac{4}{\varepsilon_{0}}}(0)$ and define

$v_{\varepsilon_{0}}=\frac{a_{2}^{2}-|u_{\varepsilon_{0}}|_{2}^{2}}{|v|_{2}^{2}}v.$

We choose
$w_{t}=u_{\varepsilon_{0}}+tv_{\varepsilon_{0}}(t\cdot)=u_{\varepsilon_{0}}+(v_{\varepsilon_{0}})_{t}$
for all $t\in(0,1)$, then it is simple to calculate that

$\text{dist}_{\mathbb{R}^{2}}({\rm supp}{\hskip
1.42271pt}u_{\varepsilon_{0}},{\rm supp}{\hskip
1.42271pt}(v_{\varepsilon_{0}})_{t})\geq\frac{2}{\varepsilon_{0}}\left(\frac{2}{t}-1\right)>\frac{2}{\varepsilon_{0}}>0$

and so $|w_{t}|_{2}^{2}=a_{2}^{2}$ which means that $w_{t}\in S(a_{2})$. It
simply invokes from Lemma 2.5 that there exists a $t_{w_{t}}>0$ such that
$(w_{t})_{t_{w_{t}}}\triangleq
t_{w_{t}}w_{t}(t_{w_{t}}\cdot)\in\mathcal{M}(a_{2})$. We claim that
$\\{t_{w_{t}}\\}$ is uniformly bounded from above in $t\in(0,1)$. Indeed,
$\|w_{t}\|^{2}_{H^{1}(\mathbb{R}^{2})}=\|u_{\varepsilon_{0}}\|^{2}_{H^{1}(\mathbb{R}^{2})}+\|(v_{\varepsilon_{0}})_{t}\|^{2}_{H^{1}(\mathbb{R}^{2})}\leq\|u_{\varepsilon_{0}}\|^{2}_{H^{1}(\mathbb{R}^{2})}+\|v_{\varepsilon_{0}}\|^{2}_{H^{1}(\mathbb{R}^{2})}<+\infty$,
we argue as the proof of Lemma 2.5 to reach the claim. Similarly,
$\|w_{t}\|^{2}_{H^{1}(\mathbb{R}^{2})}\geq\|u_{\varepsilon_{0}}\|^{2}_{H^{1}(\mathbb{R}^{2})}>0$,
we can derive that $\\{t_{w_{t}}\\}$ is uniformly bounded from below by a
positive constant in $t\in(0,1)$. All in all, there are constants
$T_{1},T_{2}>0$ independent of $t\in(0,1)$ such that $0<T_{1}\leq
t_{w_{t}}\leq T_{2}<+\infty$. So

$\text{dist}_{\mathbb{R}^{2}}\big{(}{\rm supp}{\hskip
1.42271pt}(u_{\varepsilon_{0}})_{t_{w_{t}}},{\rm supp}{\hskip
1.42271pt}((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}\big{)}\geq\frac{2}{t_{w_{t}}\varepsilon_{0}}\left(\frac{2}{t}-1\right)>\frac{2}{T_{2}\varepsilon_{0}}>0$

which indicates that

$\left\\{\begin{array}[]{ll}\displaystyle\int_{\mathbb{R}^{2}}|\nabla(w_{t})_{t_{w_{t}}}|^{2}dx=\int_{\mathbb{R}^{2}}|\nabla(u_{\varepsilon_{0}})_{t_{w_{t}}}|^{2}dx+\int_{\mathbb{R}^{2}}|\nabla((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}|^{2}dx,\\\
\displaystyle\int_{\mathbb{R}^{2}}F((w_{t})_{t_{w_{t}}})dx=\int_{\mathbb{R}^{2}}F((u_{\varepsilon_{0}})_{t_{w_{t}}})dx+\int_{\mathbb{R}^{2}}F(((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}})dx,\end{array}\right.$

and jointly with (1.7) and (1.8) yields that

$A_{j}[(w_{t})_{t_{w_{t}}}]=A_{j}[(u_{\varepsilon_{0}})_{t_{w_{t}}}]+A_{j}[((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}],\leavevmode\nobreak\
\text{for}\leavevmode\nobreak\ j=1,2.$

According to Lemma A.5 in the Appendix below, there holds

$\displaystyle\int_{\mathbb{R}^{2}}A_{j}^{2}[(w_{t})_{t_{w_{t}}}]|(w_{t})_{t_{w_{t}}}|^{2}dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{2}}(A_{j}^{2}[(u_{\varepsilon_{0}})_{t_{w_{t}}}]+A_{j}^{2}[((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}]+2A_{j}[(u_{\varepsilon_{0}})_{t_{w_{t}}}]A_{j}[((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}])(|(u_{\varepsilon_{0}})_{t_{w_{t}}}|^{2}+|((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}|^{2})dx$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}^{2}}(A_{j}^{2}[(u_{\varepsilon_{0}})_{t_{w_{t}}}]|(u_{\varepsilon_{0}})_{t_{w_{t}}}|^{2}+A_{j}^{2}[((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}]|((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}|^{2})dx+o_{t}(1),$

where $o_{t}(1)\to 0$ as $t\to 0^{+}$. Consequently, by
$(u_{\varepsilon_{0}})_{t_{\varepsilon_{0}}}\in\mathcal{M}(a)$, we obtain that

$E\big{(}(w_{t})_{t_{w_{t}}}\big{)}=E\big{(}(u_{\varepsilon_{0}})_{t_{w_{t}}}\big{)}+E\big{(}((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}\big{)}+o_{t}(1)\leq
E\big{(}(u_{\varepsilon_{0}})_{t_{\varepsilon_{0}}}\big{)}+E\big{(}((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}}\big{)}+o_{t}(1).$

On the other hand, one easily exploits $(f_{1})$ to verify that

$E(((v_{\varepsilon_{0}})_{t})_{t_{w_{t}}})=\frac{t^{2}t^{2}_{w_{t}}}{2}\int_{\mathbb{R}^{2}}[|\nabla
v_{\varepsilon_{0}}|^{2}+(A_{1}^{2}+A_{2}^{2})|v_{\varepsilon_{0}}|^{2}]dx-t^{-2}t^{-2}_{w_{t}}\int_{\mathbb{R}^{2}}F(tt_{w_{t}}v_{\varepsilon_{0}})dx=o_{t}(1).$

Finally, since $(w_{t})_{t_{w_{t}}}\in\mathcal{M}(a_{2})$, we could fix $t>0$
which is so close to $0^{+}$ such that

$\displaystyle m(a_{2})$ $\displaystyle\leq E((w_{t})_{t_{w_{t}}})\leq
E((u_{\varepsilon_{0}})_{t_{\varepsilon_{0}}})+\frac{\epsilon}{2}\leq
m(a_{1})+\epsilon$

finishing the proof since $\epsilon>0$ is arbitrary. ∎

Now, we are ready to exhibit the proof of Theorem 1.1.

######  _Proof of Theorem 1.1_.

According to Lemma 2.11, there is a Palais-Smale sequence for $E$ restricted
on $S(a)$ at the level $m(a)$ and we denoted it by $\\{u_{n}\\}$. Using Lemmas
2.7 and 2.13, we derive the two sequences appeared in (2.21) and (2.20). To
show the proof clearly, we split it into three steps.

Step 1. Without loss of generality, we could assume that the weak limit
$u_{0}\neq 0$.

Since $\\{u_{n}\\}\subset H^{1}(\mathbb{R}^{2})$ is uniformly bounded, for
some $\rho>0$, then we have the following two cases

$\text{Case
I}:\lim_{n\to\infty}\sup_{y\in\mathbb{R}^{2}}\int_{B_{\rho}(y)}|u_{n}|^{2}dx=0;\leavevmode\nobreak\
\text{Case
II}:\lim_{n\to\infty}\sup_{y\in\mathbb{R}^{2}}\int_{B_{\rho}(y)}|u_{n}|^{2}dx>0.$

If the Case I occurs, then $u_{n}\to 0$ in $L^{p}(\mathbb{R}^{2})$ for all
$2<p<+\infty$ by the Vanishing lemma. Exploiting some simple calculations, we
obtain

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}F(u_{n})dx=0\leavevmode\nobreak\
\text{and}\lim_{n\to\infty}\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx=0$

which indicate that

$m(a)=\lim_{n\to\infty}[E(u_{n})-\frac{1}{2}J(u_{n})]=\frac{1}{2}\lim_{n\to\infty}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-4F(u_{n})\right]dx=0.$

It is impossible by Lemma 2.6. So, the Case (II) must occur and there is a
sequence $\\{y_{n}\\}\subset\mathbb{R}^{2}$ such that
$\lim\limits_{n\to\infty}\int_{B_{\rho}(0)}|v_{n}|^{2}dx>0$, where
$v_{n}=u_{n}(\cdot-y_{n})$. Recalling both $E$ and $J$ are translation-
invariant, then $\\{v_{n}\\}\subset\mathcal{M}(a)$ is a Palais-Smale sequence
for $E$ restricted on $S(a)$ at the level $m(a)$. Repeating the previous
arguments, there is a $v\in H^{1}(\mathbb{R}^{2})$ such that
$v_{n}\rightharpoonup v$ in $H^{1}(\mathbb{R}^{2})$, $v_{n}\to v$ in
$L_{\text{loc}}^{p}(\mathbb{R}^{2})$ for all $2<p<+\infty$ and $v_{n}\to v$
a.e. in $\mathbb{R}^{2}$. Clearly, we deduce that $v\neq 0$. Regardless of the
abuse of notations, we adopt the sequence $\\{u_{n}\\}$ and $u_{0}\neq 0$. The
proof is done.

Step 2. $E(u_{0})=m(a)$ and $\lim\limits_{n\to\infty}|\nabla
u_{n}|_{2}^{2}=|\nabla u_{0}|_{2}^{2}$.

In view of lemma 2.14, $u_{0}$ is a solution of (1.1) with
$\lambda=\lambda_{0}$ in (2.21) and so $J(u_{0})=0$ by Lemma 2.4. Since
$\\{u_{n}\\}\subset\mathcal{M}(a)$ which is $J(u_{n})=0$ and then
$\lim\limits_{n\to\infty}J(u_{n}-u_{0})=0$ by (3.4). Thus, we have

$\displaystyle E(u_{n}-u_{0})$
$\displaystyle=E(u_{n}-u_{0})-\frac{1}{2}J(u_{n}-u_{0})+o_{n}(1)$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}^{2}}\left[f(u_{n}-u_{0})(u_{n}-u_{0})-4F(u_{n}-u_{0})\right]dx+o_{n}(1)$
(3.5)
$\displaystyle\geq\frac{\theta-4}{2}\int_{\mathbb{R}^{2}}F(u_{n}-u_{0})dx+o_{n}(1)\geq
o_{n}(1)$

It follows from the Fatou’s lemma that
$|u_{0}|_{2}^{2}\leq\lim\limits_{n\to\infty}|u_{n}|_{2}^{2}\leq a^{2}$. In
light of $u_{0}\neq 0$ by Step 1, we are derived from (3.3) and Lemma 3.3 that

(3.6)
$\lim_{n\to\infty}E(u_{n}-u_{0})+E(u_{0})=\lim_{n\to\infty}E(u_{n})=m(a)\leq
m(|u_{0}|_{2}^{2})\leq E(u_{0}).$

It follows from (3) and (3.6) that

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}F(u_{n}-u_{0})dx=0$

and so

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}f(u_{n}-u_{0})(u_{n}-u_{0})dx=0.$

Combining the above two formulas and
$\lim\limits_{n\to\infty}J(u_{n}-u_{0})=0$ again, we conclude that

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}[|\nabla u_{n}-\nabla
u_{0}|^{2}+(A_{1}^{2}[u_{n}-u_{0}]+A_{2}^{2}[u_{n}-u_{0}])|u_{n}-u_{0}|^{2}]dx=0.$

Furthermore, we obtain $\lim\limits_{n\to\infty}E(u_{n}-u_{0})=0$ jointly with
(3) gives that $E(u_{0})=m(a)$.

Step 3. There is a small $a^{*}>0$ such that $\lambda_{0}>0$ for all
$a\in(0,a^{*}]$. So, $|u_{0}|_{2}^{2}=a^{2}$.

It follows from (2) and (2.21) as well the calculations in Step 2 that

$\lambda_{0}=\frac{2}{a^{2}}\left\\{\int_{\mathbb{R}^{2}}|\nabla
u_{0}|^{2}dx+3\int_{\mathbb{R}^{2}}F(u_{0})dx-\int_{\mathbb{R}^{2}}f(u_{0})u_{0}dx\right\\}.$

Since $\\{u_{n}\\}$ is uniformly bounded in $H^{1}(\mathbb{R}^{2})$, saying it
$\sup_{n\in\mathbb{N}^{+}}\|u_{n}\|_{H^{1}(R^{2})}^{2}\leq\bar{K}$, then one
sees that $\|u_{0}\|_{H^{1}(R^{2})}^{2}\leq\bar{K}$ by Fatou’s lemma. Because
$\alpha>0$ is arbitrary in (2.7), we could choose it to satisfy
$2\bar{K}\alpha<4\pi$. Taking the very similar calculations in (2), for
$q=\frac{2+\chi}{2}\geq 3$, we obtain

$\int_{\mathbb{R}^{2}}f(u_{0})u_{0}dx\leq\varepsilon
a^{2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{0}|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}+\bar{C}_{\varepsilon}a\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{0}|^{2}dx\bigg{)}^{\frac{q-1}{2}}.$

Note that $\varepsilon=\frac{1}{2a^{2}}$ if $\chi=4$ and $\varepsilon=1$ if
$\chi>4$, then the constants $\bar{C}_{1/2a^{2}}$ and $\bar{C}_{1}$ are only
dependent of $f$, $\chi$ and $\alpha$. Let us define

$a^{*}\triangleq\left\\{\begin{array}[]{ll}\frac{1}{2\bar{C}_{1/2a^{2}}},&\text{if}\leavevmode\nobreak\
\chi=4,\\\
\min\left\\{\frac{1}{2\bar{C}_{1},\frac{1}{\sqrt{2}}}\right\\}\varrho^{\frac{4-\chi}{2}},&\text{if}\leavevmode\nobreak\
\chi>4,\end{array}\right.$

As a consequence, for all $a\in(0,a^{*}]$, we can infer that

$\lambda_{0}\geq\frac{6}{a^{2}}\int_{\mathbb{R}^{2}}F(u_{0})dx>0.$

Finally, combining (5.10), Lemma 2.14 and the calculations in Step 2, we have
that

$\displaystyle\lambda_{0}|u_{n}-u_{0}|_{2}^{2}$
$\displaystyle=\lambda_{0}|u_{n}|_{2}^{2}-\lambda_{0}|u_{0}|_{2}^{2}+o_{n}(1)=\lambda_{n}|u_{n}|_{2}^{2}-\lambda_{0}|u_{0}|_{2}^{2}+o_{n}(1)$
$\displaystyle=\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx-\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})u_{n}^{2}]dx-\int_{\mathbb{R}^{2}}f(u_{0})u_{0}dx$
$\displaystyle\ \ \ \ +\int_{\mathbb{R}^{2}}[|\nabla
u_{0}|^{2}+(A_{1}^{2}+A_{2}^{2})u_{0}^{2}]dx+o_{n}(1)$
$\displaystyle=o_{n}(1)$

showing that $|u_{0}|_{2}^{2}=\lim\limits_{n\to\infty}|u_{n}|_{2}^{2}=a^{2}$
since $\lambda_{0}\neq 0$. Thanks to Lemma 2.9, we finish the proof. ∎

##  4\. The critical case

In this section, we concentrate on the proof of Theorem 1.2. Since it is the
critical exponential case, it could be much complex than that of Theorem 1.1.
Nevertheless, we can still find a Palais-Smale sequence $\\{u_{n}\\}\subset
S(a)$ at the level $m(a)$. Thereby, we can exploit the results in Section 2
directly.

Alternatively, the nonlinearity $f$ has the critical exponential growth at
infinity as well as there is no compact result for
$H^{1}(\mathbb{R}^{2})\hookrightarrow L^{p}(\mathbb{R}^{2})$ with
$2<p<+\infty$, thus we have to pull the minimization value $m(a)$ defined in
(1.15) down to some threshold value which is crucial in restoring the
compactness. To the aim, motivated by [6], for the sufficiently integer $n\geq
2$, we require the constant $R_{n}\geq\frac{2+\log 2}{2(2-\log 2)}\rho$ to
satisfy

(4.1) $a^{2}=\frac{\rho^{2}}{16\log n}\left(2\log^{2}2+2\log
2+1-\frac{8}{n^{2}}\log
n-\frac{4}{n^{2}}\right)+\frac{(2R_{n}-\rho)(2R_{n}+3\rho)\log^{2}2}{48\log
n}.$

Here the constant $\rho>0$ is suitably large and meets that

(4.2)
$\rho>\frac{1}{\alpha_{0}\bar{\beta}_{0}}\sqrt{\frac{\pi(\theta-2)}{\theta}}.$

Moreover, it is simple to observe that

(4.3) $\lim_{n\to\infty}R_{n}=+\infty,\lim_{n\to\infty}\frac{R_{n}}{\log
n}=0\leavevmode\nobreak\ \text{and}\lim_{n\to\infty}\frac{R_{n}^{2}}{\log
n}=\frac{12a^{2}}{\log^{2}2}.$

By the above discussions, inspired by [2, 16, 34], we define the following
Moser type functions

(4.4)
$w_{n}(x)\triangleq\frac{1}{\sqrt{2\pi}}\left\\{\begin{array}[]{ll}\displaystyle\sqrt{\log
n},&0\leq|x|\leq\frac{\rho}{n},\\\
\displaystyle\frac{\log(\rho/|x|)}{\sqrt{\log
n}},&\frac{\rho}{n}\leq|x|\leq\frac{\rho}{2},\\\
\displaystyle\frac{2(R_{n}-|x|)\log 2}{(2R_{n}-\rho)\sqrt{\log
n}},&\frac{\rho}{2}\leq|x|\leq R_{n},\\\ \displaystyle 0,&|x|\geq
R_{n}.\end{array}\right.$

Clearly, $\\{w_{n}\\}\subset H^{1}(\mathbb{R}^{2})$ and one can calculate
standardly that

$\displaystyle\int_{\mathbb{R}^{2}}|w_{n}|^{2}dx$
$\displaystyle=\int_{0}^{\frac{\rho}{n}}r\log
ndr+\int_{\frac{\rho}{n}}^{\frac{\rho}{2}}\frac{\log^{2}(\rho/r)}{\log
n}rdr+\int_{\frac{\rho}{2}}^{R_{n}}\frac{4(R_{n}-r)^{2}\log^{2}2}{(2R_{n}-\rho)^{2}\log
n}rdr$ $\displaystyle=\frac{\rho^{2}}{16\log n}\left(2\log^{2}2+2\log
2+1-\frac{8}{n^{2}}\log
n-\frac{4}{n^{2}}\right)+\frac{(2R_{n}-\rho)(2R_{n}+3\rho)\log^{2}2}{48\log
n}$ $\displaystyle=a^{2}$

and

$\displaystyle\displaystyle\int_{\mathbb{R}^{2}}|\nabla w_{n}|^{2}dx$
$\displaystyle=\frac{1}{\log
n}\int_{\frac{\rho}{n}}^{\frac{\rho}{2}}\frac{1}{r}dr+\int_{\frac{\rho}{2}}^{R_{n}}\frac{(2\log
2)^{2}}{(2R_{n}-\rho)^{2}\log n}rdr=1-\frac{\log 2}{\log
n}+\frac{(2R_{n}+\rho)\log^{2}2}{2(2R_{n}-\rho)\log n}\leq 1.$

Due to the negative impact caused by the Chern-Simons term in the energy
functional $E$, defined by (1.11), we shall follow the ideas exploited in [49,
Lemma 3.10] to establish the following result.

######  Lemma 4.1.

Let $\\{w_{n}\\}\subset H^{1}(\mathbb{R}^{2})$ be defined as (4.4), then,
passing to a subsequence if necessary, it holds that

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}(A_{1}^{2}[w_{n}]+A_{2}^{2}[w_{n}])w_{n}^{2}dx=0.$

###### Proof.

Recalling the definition of $\\{w_{n}\\}$, we are able to see that

$\int_{|x|\leq\frac{\rho}{n}}|w_{n}|^{4}dx=\frac{\log^{2}n}{(2\pi)^{2}}\int_{|x|\leq\frac{\rho}{n}}dx=\frac{\rho^{2}\log^{2}n}{4\pi
n^{2}}=o_{n}(1),$

and

$\displaystyle\int_{\frac{\rho}{n}\leq|x|\leq\frac{\rho}{2}}|w_{n}|^{4}dx$
$\displaystyle=\frac{1}{(2\pi)^{2}\log^{2}n}\int_{\frac{\rho}{n}\leq|x|\leq\frac{\rho}{2}}\log^{4}(\rho/|x|)dx=\frac{1}{2\pi\log^{2}n}\int_{\frac{\rho}{n}}^{\frac{\rho}{2}}\log^{4}(\rho/r)rdr$
$\displaystyle=\frac{\rho^{2}}{8\pi\log^{2}n}(2r^{4}-4r^{3}+6r^{2}-6r+4)e^{2r}\bigg{|}_{\log\frac{1}{n}}^{\log\frac{1}{2}}=o_{n}(1).$

In view of (4.3), there holds

$\begin{gathered}\int_{\frac{\rho}{2}\leq|x|\leq
R_{n}}|w_{n}|^{4}dx=\frac{16\log^{4}2}{(2\pi)^{2}\log^{2}n}\int_{\frac{\rho}{2}\leq|x|\leq
R_{n}}\frac{(R_{n}-|x|)^{4}}{(2R_{n}-\rho)^{4}}dx=\frac{8\log^{4}2}{\pi\log^{2}n}\int_{\frac{\rho}{2}}^{R_{n}}\frac{(R_{n}-r)^{4}}{(2R_{n}-\rho)^{4}}rdr\hfill\\\
\ \ \ \
=\frac{4\log^{4}2}{15\pi\log^{2}n}\big{[}15(r-R_{n})^{4}r^{2}-20(r-R_{n})^{3}r^{3}+15(r-R_{n})^{2}r^{4}-6(r-R_{n})r^{5}+r^{6}\big{]}\bigg{|}_{\frac{\rho}{2}}^{R_{n}}\hfill\\\
\ \ \ \
=\frac{4\log^{4}2}{15\pi(2R_{n}-\rho)^{4}\log^{2}n}\bigg{[}R_{n}^{6}-15\bigg{(}\frac{\rho}{2}-R_{n}\bigg{)}^{4}\bigg{(}\frac{\rho}{2}\bigg{)}^{2}+20\bigg{(}\frac{\rho}{2}-R_{n}\bigg{)}^{3}\bigg{(}\frac{\rho}{2}\bigg{)}^{3}\hfill\\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
-15\bigg{(}\frac{\rho}{2}-R_{n}\bigg{)}^{2}\bigg{(}\frac{\rho}{2}\bigg{)}^{4}+6\bigg{(}\frac{\rho}{2}-R_{n}\bigg{)}\bigg{(}\frac{\rho}{2}\bigg{)}^{5}-\bigg{(}\frac{\rho}{2}\bigg{)}^{6}\bigg{]}\hfill\\\
\ \ \ \
=\frac{4\log^{4}2}{15\pi(2R_{n}-\rho)^{4}\log^{2}n}\big{[}R_{n}^{6}+o_{n}(1)\big{]}=o_{n}(1).\hfill\\\
\end{gathered}$

As a consequence of the above calculations, we derive $|w_{n}|_{4}^{4}\to$ as
$n\to\infty$. Since $|w_{n}|_{2}^{2}=a^{2}$, one will see that
$|w_{n}|_{p}^{p}\to 0$ for each $2<p\leq 4$ by the interpolation inequality.
Due to (2.2), we can finish the proof of this lemma. ∎

Then, with (4.1) in hands, it permits us to take the estimate for the
minimization value $m(a)$ as follows.

######  Lemma 4.2.

Assume that $f$ satisfies (1.13) and $(f_{1})-(f_{5})$, then there holds

(4.5) $m(a)<c^{*}\triangleq\frac{2\pi}{\alpha_{0}}.$

###### Proof.

Due to the definition of $m(a)$, it is obvious to conclude that
$m(a)\leq\inf\limits_{u\in S_{a}}\max\limits_{t>0}E(u_{t})$. Since $w_{n}\in
S(a)$ for all integer $n\geq 2$, then it suffices to show that
$\max\limits_{t>0}E((w_{n_{0}})_{t})<c^{*}$ for some integer $n_{0}\geq 2$.
Suppose it by a contradiction, we assume that
$\max\limits_{t>0}E((w_{n})_{t})\geq c^{*}$ for all integer $n\geq 2$. It
follows from some direct calculations in Lemma 2.5 that there is a $t_{n}>0$
such that

$J((w_{n})_{t_{n}})=0\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
E((w_{n})_{t_{n}})=\max\limits_{t>0}E((w_{n})_{t})\geq c^{*}.$

Combining $|\nabla w_{n}|_{2}^{2}\leq 1$ and $J((w_{n})_{t_{n}})=0$ jointly
with $(f_{2})$, one has that

(4.6)
$t_{n}^{4}+t_{n}^{4}\int_{\mathbb{R}^{2}}(A_{1}^{2}[w_{n}]+A_{2}^{2}[w_{n}])w_{n}^{2}dx\geq\frac{\theta-2}{\theta}\int_{\mathbb{R}^{2}}F\left(t_{n}w_{n}\right)dx.$

Since $F(s)\geq 0$ for all $s\in\mathbb{R}$, using $|\nabla w_{n}|_{2}^{2}\leq
1$ again and $E((w_{n})_{t_{n}})\geq c^{*}$, there holds

(4.7)
$t_{n}^{2}+t_{n}^{2}\int_{\mathbb{R}^{2}}(A_{1}^{2}[w_{n}]+A_{2}^{2}[w_{n}])w_{n}^{2}dx\geq
2c^{*},\leavevmode\nobreak\ \text{for all integer}\leavevmode\nobreak\ n\geq
2.$

Invoking from $\left(f_{5}\right)$, for all
$\varepsilon\in\left(0,\beta_{0}\right)$, there is a constant
$R_{\varepsilon}>0$ such that

$F(s)\geq\left(\beta_{0}-\varepsilon\right)e^{\alpha_{0}s^{2}},\forall s\geq
R_{\varepsilon},$

which together with (4.6) implies that

(4.8)
$\begin{gathered}t_{n}^{4}+t_{n}^{4}\int_{\mathbb{R}^{2}}(A_{1}^{2}[w_{n}]+A_{2}^{2}[w_{n}])w_{n}^{2}dx\geq\frac{\theta-2}{\theta}F\left(\frac{t_{n}\sqrt{\log
n}}{\sqrt{2\pi}}\right)\int_{B_{\frac{\rho}{n}}(0)}dx\hfill\\\ \ \ \ \
\geq\frac{\left(\theta-2\right)\left(\bar{\beta}_{0}-\varepsilon\right)}{\theta}\left(e^{\alpha_{0}t_{n}^{2}(2\pi)^{-1}\log
n}\right)\left|B_{\frac{\rho}{n}}(0)\right|=\frac{\pi(\theta-2)\left(\bar{\beta}_{0}-\varepsilon\right)\rho^{2}}{\theta}e^{\left[\alpha_{0}t_{n}^{2}(2\pi)^{-1}-2\right]\log
n}.\hfill\\\ \end{gathered}$

So, with the help of (2.2), we obtain

(4.9) $4\log
t_{n}+\log[\bar{C}_{r}(a^{2}+2)]\geq\log\bigg{[}\frac{\pi(\theta-2)\left(\bar{\beta}_{0}-\varepsilon\right)\rho^{2}}{\theta}\bigg{]}+[\alpha_{0}t_{n}^{2}(2\pi)^{-1}-2]\log
n.$

If $\\{t_{n}\\}$ is unbounded, up to a subsequence if necessary, we can assume
that $t_{n}\to+\infty$ and then

$\frac{4\log t_{n}+\log[\bar{C}_{r}(a^{2}+2)]}{t_{n}^{2}}\geq
t_{n}^{-2}\log\bigg{[}\frac{\pi(\theta-2)\left(\bar{\beta}_{0}-\varepsilon\right)\rho^{2}}{\theta}\bigg{]}+[\alpha_{0}(2\pi)^{-1}-2t_{n}^{-2}]\log
n$

which yields a contradiction if we tend $n\to\infty$. Thereby, passing to a
subsequence if necessary, there exists a positive constant $t_{0}^{2}\geq
2c^{*}$ determined by Lemma 4.1 and (4.7) such that

$\lim_{n\to\infty}t_{n}^{2}=t_{0}^{2}\geq 2c^{*}.$

Moreover, we derive $t_{0}^{2}=2c^{*}$. Otherwise, we arrive at a
contradiction by letting $n\to\infty$ in (4.9). Let us tend $n\to\infty$ and
then $\varepsilon\to 0^{+}$ in (4.8), using Lemma 4.1 again, there holds

$(2c^{*})^{2}=t_{0}^{4}\geq\frac{\pi(\theta-2)\bar{\beta}_{0}\rho^{2}}{\theta}$

which contradicts with (4.2). The proof is completed. ∎

To deal with the lack of compactness, we also need the following type of
compact result.

######  Lemma 4.3.

Let $f$ meets (1.13) and $(f_{1})-(f_{4})$. If the sequence
$\\{u_{n}\\}\subset H^{1}(\mathbb{R}^{2})$ satisfying $u_{n}\to 0$ in
$L^{p}(\mathbb{R}^{2})$ for all $2<p<+\infty$, $u_{n}\to 0$ a.e. in
$\mathbb{R}^{2}$ and

(4.10) $\sup_{n\in\mathbb{N}}\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx\leq C,$

then we have that $F(u_{n})\to 0$ in $L^{1}(\mathbb{R}^{2})$ as $n\to\infty$
along a subsequence.

###### Proof.

The proof is standard and its detail is omitted, we refer the reader to [15]
for example. ∎

Note that the map $a\mapsto m(a)$ is non-increasing by Lemma 3.3, whereas
taking into account that the nonlinearity $f$ admits the critical exponential
growth as the infinity, we are going to prove the following result which is
obviously stronger.

######  Lemma 4.4.

Let $\tilde{a}\neq 0$ and $\tilde{u}\in\mathcal{M}(\tilde{a})$ be a minimizer
of $m(\tilde{a})$. If there is a $\tilde{\lambda}>0$ such that $\tilde{u}$ is
a solution of Eq. (1.1) with $\lambda=\tilde{\lambda}$, then $m(\tilde{a})$ is
strictly decreasing in the right neighborhood of $\tilde{a}$.

###### Proof.

We follow the method exploited in [22, Proposition 5.1] to exhibit the proof.
For each $t,\gamma>0$, define $\tilde{u}_{t,\gamma}\triangleq\gamma
t\tilde{u}(t\cdot)$. With the help of the two functionals
$E,J:S(a)\to\mathbb{R}$ before, we define the following two new functionals
$\Phi_{E}(t,\gamma)\triangleq E(\tilde{u}_{t,\gamma})$ and
$\Phi_{J}(t,\gamma)\triangleq J(\tilde{u}_{t,\gamma})$ which are given by

$\Phi_{E}(t,\gamma)=\frac{t^{2}}{2}\int_{\mathbb{R}^{2}}[\gamma^{2}|\nabla\tilde{u}|^{2}+\gamma^{6}(A_{1}^{2}[\tilde{u}]+A_{2}^{2}[\tilde{u}])|\tilde{u}|^{2}]dx-t^{-2}\int_{\mathbb{R}^{2}}F(\gamma
t\tilde{u})dx$

and

$\Phi_{J}(t,\gamma)=t^{2}\int_{\mathbb{R}^{2}}[\gamma^{2}|\nabla\tilde{u}|^{2}+\gamma^{6}(A_{1}^{2}[\tilde{u}]+A_{2}^{2}[\tilde{u}])|\tilde{u}|^{2}]dx-t^{-2}\int_{\mathbb{R}^{2}}[f(\gamma
t\tilde{u})\gamma t\tilde{u}-2F(\gamma t\tilde{u})]dx.$

Since $\tilde{u}\in\mathcal{M}(\tilde{a})$ which shows that $J(\tilde{u})=0$,
then we have that $\frac{\partial\Phi_{E}}{\partial t}(1,1)=J(\tilde{u})=0$
and

$\displaystyle\frac{\partial\Phi_{E}}{\partial\gamma}(1,1)$
$\displaystyle=\int_{\mathbb{R}^{2}}|\nabla\tilde{u}|^{2}dx+3\int_{\mathbb{R}^{2}}(A_{1}^{2}[\tilde{u}]+A_{2}^{2}[\tilde{u}])|\tilde{u}|^{2}dx-\int_{\mathbb{R}^{2}}f(\tilde{u})\tilde{u}dx$
$\displaystyle=-\tilde{\lambda}\int_{\mathbb{R}^{2}}|\tilde{u}|^{2}dx=-\tilde{\lambda}\tilde{a}^{2}<0,$

where we choose $\tilde{u}$ as a text function of Eq. (1.1) with $u=\tilde{u}$
and $\lambda=\tilde{\lambda}$ as well as (2.1).

Moreover, combining $J(\tilde{u})=0$ and Lemma A.3 in the Appendix below, we
can compute that

$\displaystyle\frac{\partial^{2}\Phi_{E}}{\partial t^{2}}(1,1)$
$\displaystyle=\int_{\mathbb{R}^{2}}[|\nabla\tilde{u}|^{2}dx+(A_{1}^{2}[\tilde{u}]+A_{2}^{2}[\tilde{u}])|\tilde{u}|^{2}]dx-\int_{\mathbb{R}^{2}}[f^{\prime}(\tilde{u})\tilde{u}^{2}-4f(\tilde{u})\tilde{u}+6F(\tilde{u})]dx$
$\displaystyle=-\int_{\mathbb{R}^{2}}[f^{\prime}(\tilde{u})\tilde{u}^{2}-5f(\tilde{u})\tilde{u}+8F(\tilde{u})]dx<0.$

Thanks to the above three facts, for all $|\iota_{t}|>0$ small enough and
$\iota_{\gamma}>0$, there holds

(4.11)
$E(\tilde{u}_{1+\iota_{t},1+\iota_{\gamma}})=\Phi_{E}(1+\iota_{t},1+\iota_{\gamma})<\Phi_{E}(1,1)=E(\tilde{u}).$

According to $J(\tilde{u})=0$ again, one sees that $\Phi_{J}(1,1)=0$. Again
adopting Lemma A.3 in the Appendix below and $J(\tilde{u})=0$, there holds

$\displaystyle\frac{\partial\Phi_{J}}{\partial t}(1,1)$
$\displaystyle=2\int_{\mathbb{R}^{2}}[|\nabla\tilde{u}|^{2}+(A_{1}^{2}[\tilde{u}]+A_{2}^{2}[\tilde{u}])|\tilde{u}|^{2}]dx-\int_{\mathbb{R}^{2}}[f^{\prime}(\tilde{u})\tilde{u}^{2}-3f(\tilde{u})\tilde{u}+4F(\tilde{u})]dx$
$\displaystyle=-\int_{\mathbb{R}^{2}}[f^{\prime}(\tilde{u})\tilde{u}^{2}-5f(\tilde{u})\tilde{u}+8F(\tilde{u})]dx<0.$

It therefore concludes from the implicit function theorem that there exist a
constant $\varepsilon>0$ and a continuous function
$g:[1-\varepsilon,1+\varepsilon]\to\mathbb{R}$ satisfying $g(1)=1$ such that
$\Phi_{J}(g(\gamma),\gamma)=0$ for every
$\gamma\in[1-\varepsilon,1+\varepsilon]$. Particularly, there holds
$\tilde{u}_{g(1+\varepsilon),1+\varepsilon}\in\mathcal{M}((1+\varepsilon)^{2}\tilde{a})$.
Consequently, we infer from (4.11) that

$m((1+\varepsilon)^{2}\tilde{a})\leq\inf_{u\in\mathcal{M}((1+\varepsilon)^{2}\tilde{a})}E(u)\leq
E(\tilde{u}_{g(1+\varepsilon),1+\varepsilon})<E(\tilde{u})=m(\tilde{a})$

showing the desired result. The proof is completed. ∎

At this point, we can exhibit the detailed proof of Theorem 1.2.

######  _Proof of Theorem 1.2_.

Arguing as before, we can derive that there exists a Palais-Smale sequence
$\\{u_{n}\\}\subset\mathcal{M}(a)$ for $E$ restricted on $S(a)$ at the level
$m(a)$. In fact, all of the conclusions in Section 2 remain true in this
situation and we shall exploit them directly when there is no
misunderstanding. Following the proof of Theorem 1.1, we continue to divide
the proof into three steps.

Step I. The weak limit $u_{0}\neq 0$.

In view of the Step 1 in the proof of Theorem 1.1, the proof will be done by
ruling out the Case I. Suppose it by contradiction that, if the Case I holds
true, then $u_{n}\to 0$ in $L^{p}(\mathbb{R}^{2})$ for all $2<p<+\infty$.
Since we obtain (4.10) by Lemma 2.7, then $F(u_{n})\to 0$ in
$L^{1}(\mathbb{R}^{2})$ as $n\to\infty$ by Lemma 4.3. Hence, taking
$E(u_{n})\to m(a)$, (2.2) and Lemma 4.2 into account, we have that

$\limsup_{n\to\infty}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx=2\limsup_{n\to\infty}E(u_{n})=2m(a)<\frac{4\pi}{\alpha_{0}}.$

Thereby, we shall choose $\alpha>\alpha_{0}$ sufficiently close to
$\alpha_{0}$ and $\nu$ sufficiently close to 1 in such a way that
$\frac{1}{\nu}+\frac{1}{\nu^{\prime}}=1$ and

$|\nabla u_{n}|_{2}^{2}<\frac{4\pi(1-\epsilon)}{\nu\alpha}\leavevmode\nobreak\
\text{for some suitable}\leavevmode\nobreak\ \epsilon\in(0,1).$

Define $\bar{u}_{n}=\sqrt{\frac{\nu\alpha}{4\pi(1-\epsilon)}}u_{n}$, then
$|\nabla\bar{u}_{n}|_{2}^{2}<1$ and
$|\bar{u}_{n}|_{2}^{2}=\frac{\nu\alpha}{4\pi(1-\epsilon)}a^{2}<+\infty$. With
these observations, we adopt (2.6) to obtain

$\sup_{n\in\mathbb{N}^{+}}\int_{\mathbb{R}^{2}}(e^{\nu\alpha|u_{n}|^{2}}-1)dx=\sup_{n\in\mathbb{N}^{+}}\int_{\mathbb{R}^{2}}(e^{4\pi(1-\epsilon)|\bar{u}_{n}|^{2}}-1)dx\leq
C<+\infty.$

which together with (2.9) with $q>2$ gives that

$\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx\leq
C\int_{\mathbb{R}^{2}}|u_{n}|^{\chi}dx+C\bigg{(}\int_{\mathbb{R}^{2}}|u_{n}|^{q\nu^{\prime}}dx\bigg{)}^{\frac{1}{\nu^{\prime}}}.$

Recalling $J(u_{n})=0$, we take advantage of the above fact and Lemma 4.3 to
arrive at $|\nabla u_{n}|_{2}^{2}\to 0$. However, it contradicts with Lemma
2.6 and so the Case I cannot occur.

Step II. $E(u_{0})=m(a)$ and $\lim\limits_{n\to\infty}|\nabla
u_{n}|_{2}^{2}=|\nabla u_{0}|_{2}^{2}$.

Since $u_{0}\neq 0$, one easily sees that
$0<|u_{0}|_{2}^{2}\triangleq\tilde{a}^{2}\leq\liminf\limits_{n\to\infty}|u_{n}|^{2}_{2}=a^{2}$
by Fatou’s lemma. Recalling Lemma 3.3 still holds true in this case,
$J(u_{0})=0$ by Lemmas 2.4 and 2.14, it then concludes from $(f_{2})$ and
Fatou’s lemma again that

$\displaystyle m(a)$ $\displaystyle\leq m(\tilde{a})\leq
E(u_{0})=E(u_{0})-\frac{1}{2}J(u_{0})=\frac{1}{2}\int_{\mathbb{R}^{2}}\left[f(u_{0})u-4F(u_{0})\right]dx$
$\displaystyle\leq\frac{1}{2}\liminf_{n\to\infty}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-4F(u_{n})\right]dx=\liminf_{n\to\infty}[E(u_{n})-\frac{1}{2}J(u_{n})]$
$\displaystyle=\liminf_{n\to\infty}E(u_{n})=m(a)$

showing that $E(u_{0})=m(a)=m(\tilde{a})$ and

(4.12)
$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-4F(u_{n})\right]dx=\int_{\mathbb{R}^{2}}\left[f(u_{0})u_{0}-4F(u_{0})\right]dx.$

With $E(u_{0})=m(a)$ in hands, combining Fatou’s lemma and $J(u_{0})=0$, we
also obtain that

$\displaystyle m(a)$
$\displaystyle=\lim_{n\to\infty}E(u_{n})=\limsup_{n\to\infty}[E(u_{n})-\frac{1}{\theta-2}J(u_{n})]$
$\displaystyle=\limsup_{n\to\infty}\left\\{\frac{\theta-4}{2(\theta-2)}\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}]dx+\frac{1}{\theta-2}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-\theta
F(u_{n})\right]dx\right\\}$
$\displaystyle\geq\liminf_{n\to\infty}\left\\{\frac{\theta-4}{2(\theta-2)}\int_{\mathbb{R}^{2}}[|\nabla
u_{n}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}]dx+\frac{1}{\theta-2}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-\theta
F(u_{n})\right]dx\right\\}$
$\displaystyle\geq\frac{\theta-4}{2(\theta-2)}\int_{\mathbb{R}^{2}}[|\nabla
u_{0}|^{2}+(A_{1}^{2}+A_{2}^{2})|u_{0}|^{2}]dx+\frac{1}{\theta-2}\int_{\mathbb{R}^{2}}\left[f(u_{0})u_{0}-\theta
F(u_{0})\right]dx$
$\displaystyle=E(u_{0})-\frac{1}{\theta-2}J(u_{0})=E(u_{0})=m(a).$

As a direct consequence of the above equality, we derive

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx=\int_{\mathbb{R}^{2}}|\nabla u_{0}|^{2}dx,\leavevmode\nobreak\
\lim_{n\to\infty}\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}dx=\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u_{0}|^{2}dx,$

and

(4.13) $\lim_{n\to\infty}\int_{\mathbb{R}^{2}}\left[f(u_{n})u_{n}-\theta
F(u_{n})\right]dx=\int_{\mathbb{R}^{2}}\left[f(u_{0})u_{0}-\theta
F(u_{0})\right]dx.$

Hence, the Step II concludes.

Step III. There is a small $a_{*}>0$ such that $\lambda_{0}>0$ for all
$a\in(0,a_{*}]$. So, $|u_{0}|_{2}^{2}=a^{2}$.

Since $|\nabla u_{n}-\nabla u_{0}|_{2}^{2}=o_{n}(1)$, we could continue to
choose $\alpha>\alpha_{0}$ and $\nu,\nu^{\prime}>1$ adopted in Step II to
satisfy

$|\nabla u_{n}-\nabla
u_{0}|_{2}^{2}<\frac{4\pi}{\nu\alpha(1+\bar{\epsilon})^{4}}\leavevmode\nobreak\
\text{for some suitable}\leavevmode\nobreak\ \bar{\epsilon}\in(0,1).$

Define
$\tilde{u}_{n}=\sqrt{\frac{\nu\alpha(1+\bar{\epsilon})^{4}}{4\pi}}(u_{n}-u_{0})$,
then $|\nabla\tilde{u}_{n}|_{2}^{2}<1$ and
$|\tilde{u}_{n}|_{2}^{2}\leq\frac{\nu\alpha(1+\bar{\epsilon})^{4}}{4\pi}4a^{2}<+\infty$.
Besides, for the above fixed $\bar{\epsilon}\in(0,1)$, we need the following
two types of Young’s inequality

$|a+b|^{2}\leq(1+\bar{\epsilon})|a|^{2}+(1+\bar{\epsilon}^{-1})|b|^{2},\leavevmode\nobreak\
\forall a,b\in\mathbb{R}$

and

$e^{a+b}-1\leq\frac{1}{1+\bar{\epsilon}}\big{[}e^{(1+\bar{\epsilon})a}-1\big{]}+\frac{\bar{\epsilon}}{1+\bar{\epsilon}}\big{[}e^{(1+\bar{\epsilon}^{-1})b}-1\big{]},\leavevmode\nobreak\
\forall a,b\in\mathbb{R}.$

By means of the above facts together with (2.6), we derive

$\displaystyle\int_{\mathbb{R}^{2}}(e^{\nu\alpha u_{n}^{2}}-1)dx$
$\displaystyle\leq\frac{1}{1+\bar{\epsilon}}\int_{\mathbb{R}^{2}}(e^{\frac{4\pi}{(1+\bar{\epsilon})^{2}}|\tilde{u}_{n}|^{2}}-1)dx+\frac{\bar{\epsilon}}{1+\bar{\epsilon}}\int_{\mathbb{R}^{2}}(e^{\nu\alpha(1+\bar{\epsilon}^{-1})^{2}u_{0}^{2}}-1)dx$
$\displaystyle\leq\frac{C_{1}}{1+\bar{\epsilon}}+\frac{C_{2}\bar{\epsilon}}{1+\bar{\epsilon}}\leq
C_{3}<+\infty,\leavevmode\nobreak\ \forall n\in\mathbb{N}^{+}.$

Letting $q=\frac{2+\chi}{2}\geq 3$ in (2.9), we then make full use of the
above fact and (2.13) to deduce that

$\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx\leq\varepsilon
a^{2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}+\tilde{C}_{\varepsilon}a\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx\bigg{)}^{\frac{q-1}{2}}.$

It follows from (4.12) and (4.13) that

$\lim_{n\to\infty}\int_{\mathbb{R}^{2}}f(u_{n})u_{n}dx=\int_{\mathbb{R}^{2}}f(u_{0})u_{0}dx.$

From which, adopting $|\nabla u_{n}|_{2}^{2}\to|\nabla u_{0}|_{2}^{2}$ as
$n\to\infty$, there holds

$\int_{\mathbb{R}^{2}}f(u_{0})u_{0}dx\leq\varepsilon
a^{2}\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{0}|^{2}dx\bigg{)}^{\frac{\chi-2}{2}}+\tilde{C}_{\varepsilon}a\bigg{(}\int_{\mathbb{R}^{2}}|\nabla
u_{0}|^{2}dx\bigg{)}^{\frac{q-1}{2}}.$

Repeating the calculations of Step 2 in the proof of Theorem 1.1, we could
determine a suitable $a_{*}>0$ such that $\lambda_{0}>0$ for all
$a\in(0,a_{*}]$. At this stage, we reach a conclusion that
$u_{0}\in\mathcal{M}(\tilde{a})$ is a minimizer of $m(\tilde{a})$ and it is a
solution of Eq. (1.1) with $\lambda=\lambda_{0}>0$. Owing to Lemma 4.4,
$m(\tilde{a})$ is strictly decreasing in the right neighborhood of
$\tilde{a}$. So, we must have $\tilde{a}^{2}=a^{2}$. Otherwise, there is a
sufficiently small $\varepsilon>0$ such that $(1+\varepsilon)^{2}\tilde{a}<a$
and $(1+\varepsilon)^{2}\tilde{a}$ locates in a right neighborhood of
$\tilde{a}$. Therefore, there holds
$m(\tilde{a})>m((1+\varepsilon)^{2}\tilde{a})\geq m(a)$ which is impossible
because of the Step II. So $u_{0}\in\mathcal{M}(a)$ is a minimizer of $m(a)$
and $(u_{0},\lambda_{0})$ is a couple of weak solutions of problems
(1.1)-(1.2) when $a\in(0,a_{*}]$. The proof is completed. ∎

##  5\. The supercritical case

In this section, we shall turn to the supercritical case for problems
(1.1)-(1.2) and it is the main topic in the present article. Note that if
there is no misunderstanding, it is always supposed that the $\mathcal{C}^{1}$
function $h$ that vanishes in $(-\infty,0]$ satisfies $(h_{1})-(h_{3})$ for
simplicity.

To conclude Theorem 1.6, as explained before, we begin with verifying the
necessary assumptions for Theorems 1.1 and 1.2.

Now, we shall deduce that the nonlinearities $f^{R,{\delta}}$ and $f^{R,2}$
given in (1.18) satisfy some growth conditions which are the counterparts of
(2.7), (2.8), (2.9) and (2.10), respectively. On the one hand, for every fixed
$R>0$, $q>2$, $\alpha>0$ and $\varepsilon>0$, there is a positive constant
$M_{\varepsilon}^{R}$ which is dependent of $R>0$ and $\varepsilon>0$ such
that

(5.1)
$|f^{R,\delta}(s)|\leq\varepsilon|s|^{\chi-1}+M_{\varepsilon}^{R}|s|^{q-1}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}$

which together with $(h_{2})$ implies that

(5.2)
$|F^{R,\delta}(s)|\leq\varepsilon|s|^{\chi}+M_{\varepsilon}^{R}|s|^{q}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}.$

Actually, exploiting $(h_{1})$ and $(h_{3})$ with $\delta\in(0,2)$, we have
that

$\lim_{s\to 0}\frac{f^{R,\delta}(s)}{s^{\chi-1}}=\lim_{s\to
0}\frac{h(s)}{s^{\chi-1}}=0\leavevmode\nobreak\ \text{uniformly
in}\leavevmode\nobreak\ R>0$

and

$0\leq\lim_{|s|\to+\infty}\frac{|f^{R,\delta}(s)|}{e^{\alpha|s|^{2}}-1}\leq\lim_{|s|\to+\infty}\frac{Me^{(\gamma+\bar{\alpha}_{0}R^{\tau-\delta})|s|^{\delta}}}{e^{\alpha|s|^{2}}-1}=0$

implying the desired result (5.1). Similarly, we fix $R>0$, $q>2$ and
$\varepsilon>0$ to find a $\bar{M}>M$ which is independent of $R$ such that

(5.3)
$|f^{R,2}(s)|\leq\varepsilon|s|^{\chi-1}+\bar{M}|s|^{q-1}(e^{(\gamma+\bar{\alpha}_{0}R^{\delta-2})|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R},$

and

(5.4)
$|F^{R,2}(s)|\leq\varepsilon|s|^{\chi}+\bar{M}|s|^{q}(e^{(\gamma+\bar{\alpha}_{0}R^{\delta-2})|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}.$

Moreover, we have to show the counterparts of $(f_{2})$ and $(f_{3})$ for the
nonlinearity $f^{R,\bar{\delta}}$ which are exhibited as follows.

######  Lemma 5.1.

Let $f$ be defined in (1.16) with the $\mathcal{C}^{1}$ function $h$
satisfying $(h_{2})$. Then, for all fixed $R>0$ one has

$0<F^{R,\bar{\delta}}(t)\leq\theta f^{R,\bar{\delta}}(t)t,\leavevmode\nobreak\
\forall t\in\mathbb{R}\backslash\\{0\\}.$

###### Proof.

The reader can refer to [4, 5], whereas we should exhibit the detailed proof
for the completeness. For all $t\in[0,R]$, we can see that
$f^{R,\bar{\delta}}(t)=f(t)$ and so $F^{R,\bar{\delta}}(t)=F(t)$, then the
lemma is done for $t\in[0,R]$. Indeed, for all $t\in[0,R]$, one exploits
$(h_{2})$ to deduce that

$F^{R,\bar{\delta}}(t)=F(t)=\int_{0}^{t}h(s)e^{\bar{\alpha}_{0}s^{\tau}}ds\leq
e^{\bar{\alpha}_{0}t^{\tau}}\int_{0}^{t}h(s)ds=e^{\bar{\alpha}_{0}t^{\tau}}H(t)\leq\theta
e^{\bar{\alpha}_{0}t^{\tau}}h(t)t=\theta f^{R,\bar{\delta}}(t)t.$

Given a $t\in[R,+\infty)$, then

$\displaystyle F^{R,\bar{\delta}}(t)$
$\displaystyle=\int_{0}^{t}f^{R,\bar{\delta}}(s)ds=\int_{0}^{R}f^{R,\bar{\delta}}(s)ds+\int_{R}^{t}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds$
$\displaystyle=F^{R,\bar{\delta}}(R)+\int_{R}^{t}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds$
$\displaystyle=F^{R,\bar{\delta}}(R)+\int_{0}^{t}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds-\int_{0}^{R}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds.$

We note that

$F^{R,\bar{\delta}}(R)=\int_{0}^{R}h(s)e^{\bar{\alpha}_{0}s^{\tau}}ds\leq\int_{0}^{R}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds$

leading to

$F^{R,\bar{\delta}}(t)\leq\int_{0}^{t}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds\leq
e^{\alpha_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}\int_{0}^{t}h(s)ds=e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}H(t).$

Thereby, it follows from $(h_{2})$ that

$\theta F^{R,\bar{\delta}}(t)\leq\theta
e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}H(t)\leq
e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}h(t)t=f^{R,\bar{\delta}}(t)t,\leavevmode\nobreak\
\forall t\geq R.$

The proof is completed. ∎

######  Lemma 5.2.

Let $f$ be defined in (1.16) with the $\mathcal{C}^{1}$ function $h$
satisfying $(h_{2})-(h_{4})$. Then, for all fixed $R>0$ one has

$\bar{F}^{R,\bar{\delta}}(t)/t^{4}\leavevmode\nobreak\ \text{is strictly
increasing in}\leavevmode\nobreak\ (0,+\infty),$

where
$\bar{F}^{R,\bar{\delta}}(t)=f^{R,\bar{\delta}}(t)t-2F^{R,\bar{\delta}}(t)$
for all $t\in\mathbb{R}$.

###### Proof.

First of all, one simply observes that $f^{R,\bar{\delta}}(t)=f(t)$ for all
$t\in[0,R]$, then $(f^{R,\bar{\delta}})^{\prime}(t)=f^{\prime}(t)$ and
$F^{R,\bar{\delta}}(t)=F(t)$ for all $t\in[0,R]$. Then, for any fixed
$t\in(0,R]$,

$\displaystyle F(t)$
$\displaystyle=\int_{0}^{t}h(s)e^{\bar{\alpha}_{0}s^{\tau}}ds=H(t)e^{\bar{\alpha}_{0}t^{\tau}}-\int_{0}^{t}H(s)de^{\bar{\alpha}_{0}s^{\tau}}=H(t)e^{\bar{\alpha}_{0}t^{\tau}}-\bar{\alpha}_{0}\tau\int_{0}^{t}H(s)s^{\tau-1}e^{\bar{\alpha}_{0}s^{\tau}}ds$
$\displaystyle\geq H(t)e^{\bar{\alpha}_{0}t^{\tau}}-\bar{\alpha}_{0}\tau
H(t)e^{\bar{\alpha}_{0}t^{\tau}}\int_{0}^{t}s^{\tau-1}ds=H(t)e^{\bar{\alpha}_{0}t^{\tau}}-\bar{\alpha}_{0}H(t)t^{\tau}e^{\bar{\alpha}_{0}t^{\tau}}.$

Hence, for all $t\in(0,R]$, using $(h_{2})$, we deduce that

$\begin{gathered}(f^{R,\bar{\delta}})^{\prime}(t)t^{2}-5f^{R,\bar{\delta}}(t)t+8F^{R,\bar{\delta}}(t)=f^{\prime}(t)t^{2}-5f(t)t+8F(t)\hfill\\\
\ \ \ \
\geq\big{[}h^{\prime}(t)t^{2}-5h(t)t+8H(t)\big{]}e^{\bar{\alpha}_{0}t^{\tau}}+\bar{\alpha}_{0}t^{\tau}e^{\bar{\alpha}_{0}t^{\tau}}[\tau
h(t)t-8H(t)]\hfill\\\ \ \ \ \
>\bar{\alpha}_{0}t^{\tau}e^{\bar{\alpha}_{0}t^{\tau}}(\tau\theta-8)H(t)>0,\hfill\\\
\end{gathered}$

where we have applied a very similar result in Lemma A.3 in the Appendix below
to $(h_{3})$ in the last second inequality. So, $\forall t\in(0,R]$, it has
that

(5.5)
$\frac{d}{dt}\left(\frac{\bar{F}^{R,\bar{\delta}}(t)}{t^{4}}\right)=\frac{(f^{R,\bar{\delta}})^{\prime}(t)t^{2}-5f^{R,\bar{\delta}}(t)t+8F^{R,\bar{\delta}}(t)}{t^{5}}>0,$

showing the conclusions for all $t\in(0,R]$.

We next consider the case for all $t\in[R,+\infty)$. Firstly, one can
calculate that

$\displaystyle F^{R,\bar{\delta}}(R)$
$\displaystyle=\int_{0}^{R}h(s)e^{\bar{\alpha}_{0}s^{\tau}}ds=\int_{0}^{R}e^{\bar{\alpha}_{0}s^{\tau}}dH(s)=H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\bar{\alpha}_{0}\tau\int_{0}^{R}H(s)s^{\tau-1}e^{\bar{\alpha}_{0}s^{\tau}}ds$
$\displaystyle\geq
H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\bar{\alpha}_{0}H(R)e^{\bar{\alpha}_{0}R^{\tau}}\int_{0}^{R}\tau
s^{\tau-1}ds=H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\bar{\alpha}_{0}R^{\tau}H(R)e^{\bar{\alpha}_{0}R^{\tau}}.$

Similarly, one has that

$\displaystyle\int_{R}^{t}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds$
$\displaystyle=\int_{R}^{t}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}dH(s)=H(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}-H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\int_{R}^{t}H(s)de^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}$
$\displaystyle=H(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}-H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\bar{\alpha}_{0}R^{\tau-\bar{\delta}}\bar{\delta}\int_{R}^{t}H(s)s^{\bar{\delta}-1}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds$
$\displaystyle\geq
H(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}-H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\bar{\alpha}_{0}R^{\tau-\bar{\delta}}H(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}\int_{R}^{t}\bar{\delta}s^{\bar{\delta}-1}ds$
$\displaystyle\geq
H(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}-H(R)e^{\bar{\alpha}_{0}R^{\tau}}-\bar{\alpha}_{0}R^{\tau-\bar{\delta}}H(t)t^{\bar{\delta}}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}+\bar{\alpha}_{0}R^{\tau}H(R)e^{\bar{\alpha}_{0}R^{\tau}}.$

From which, we obtain

$\displaystyle F^{R,\bar{\delta}}(t)$
$\displaystyle=F^{R,\bar{\delta}}(R)+\int_{R}^{t}h(s)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}s^{\bar{\delta}}}ds\geq
H(t)e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}-\bar{\alpha}_{0}R^{\tau-\bar{\delta}}H(t)t^{\bar{\delta}}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}$

Recalling $\delta\in[8\theta^{-1},2)$ in $(h_{4})$ which implies that
$\bar{\delta}-8\theta\geq 0$, adopting $(h_{2})-(h_{3})$, so there holds

$\begin{gathered}(f^{R,\bar{\delta}})^{\prime}(t)t^{2}-5f^{R,\bar{\delta}}(t)t+8F^{R,\bar{\delta}}(t)\hfill\\\
\ \ \ \
\geq\big{[}h^{\prime}(t)t^{2}-5h(t)t+8H(t)\big{]}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}+\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}[\bar{\delta}h(t)t-8H(t)]\hfill\\\
\ \ \ \
>\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}e^{\bar{\alpha}_{0}R^{\tau-\bar{\delta}}t^{\bar{\delta}}}(\bar{\delta}-8\theta)H(t)\geq
0,\hfill\\\ \end{gathered}$

yielding that (5.5) remains true for all $t\in[R,+\infty)$. The proof is
completed. ∎

With (5.1), (5.2), (5.3), and (5.4) as well as Lemmas 5.1-5.2 in hands, we
could proceed as the proofs in Section 2 to get that, for all fixed $R>0$,
there are two sequences $\\{u_{n}\\}\subset S_{a}$ and
$\\{\lambda_{n}^{R,\bar{\delta}}\\}\subset\mathbb{R}$ such that

(5.6)
$E^{R,\bar{\delta}}(u_{n})=m^{R,\bar{\delta}}(a)+o_{n}(1)\leavevmode\nobreak\
\text{as}\leavevmode\nobreak\ n\to\infty,$ (5.7) $-\Delta
u_{n}+\lambda_{n}^{R,\bar{\delta}}u_{n}+A_{0}u_{n}+\sum_{j=1}^{2}A_{j}^{2}[u_{n}]u_{n}-f^{R,\bar{\delta}}(u_{n})=o_{n}(1)\leavevmode\nobreak\
\text{as}\leavevmode\nobreak\ n\to\infty,$ (5.8)
$J^{R,\bar{\delta}}(u_{n})=o_{n}(1)\leavevmode\nobreak\
\text{as}\leavevmode\nobreak\ n\to\infty,$

where $E^{R,\bar{\delta}}:S(a)\to\mathbb{R}$ is given by (1.20) and

(5.9)
$m^{R,\bar{\delta}}(a)\triangleq\inf_{u\in\mathcal{M}^{R,\bar{\delta}}(a)}E^{R,\bar{\delta}}(u)>0,\leavevmode\nobreak\
\mathcal{M}^{R,\bar{\delta}}(a)=\big{\\{}u\in
S(a):J^{R,\bar{\delta}}(u)=0\big{\\}}.$

Here the auxiliary energy functional $J^{R,\bar{\delta}}:S(a)\to\mathbb{R}$ is
defined by

$J^{R,\bar{\delta}}(u)=\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-\int_{\mathbb{R}^{2}}[f^{R,\bar{\delta}}(u)u-2F^{R,\bar{\delta}}(u)]dx.$

Combining (5.6), (5.7) and (5.8) as well as Lemma 5.1, we can repeat the proof
of Lemmas 2.7 and 2.13 to see that the sequence $\\{u_{n}\\}\subset S(a)$ is
uniformly bounded in $H^{1}(\mathbb{R}^{2})$ and

(5.10)
$\lambda_{n}^{R,\bar{\delta}}=\frac{2}{a^{2}}\left\\{\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx+3\int_{\mathbb{R}^{2}}F^{R,\bar{\delta}}(u_{n})dx-\int_{\mathbb{R}^{2}}f^{R,\bar{\delta}}(u_{n})u_{n}dx\right\\}.$

Moreover, we can derive that (2.20) and (2.21) remain true in some suitable
forms in this situation.

In what follows, we shall handle the case $\bar{\delta}=\delta$ and
$\bar{\delta}=2$ in two subsections.

###  5.1. The case $\bar{\delta}=\delta$

In this subsection, we know that the nonlinearity
$f^{R,\bar{\delta}}=f^{R,\delta}$. According to the above discussions, we can
give the proof of the first part (subcritical exponential case) of Theorem
1.6.

######  _Proof of Theorem 1.6_.

The proof is totally similar to that of Theorem 1.1, so we omit it here. ∎

Before presenting the proof of Theorem 1.4, we have to study the
$L^{\infty}$-estimate for the nontrivial solution
$(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ established in Theorem 1.6. We first
show that any sequence satisfying (5.6), (5.7) and (5.8) is uniformly bounded
in $n\in\mathbb{N}^{+}$ and $R>0$ for some suitable $\bar{\alpha}_{0}>0$.

######  Lemma 5.3.

There is a constant $\bar{\alpha}_{0}^{*}=\frac{1}{R^{\tau-\delta}}>0$ such
that for all $\bar{\alpha}_{0}\in(0,\bar{\alpha}_{0}^{*})$ and $\tau>2$, the
sequence $\\{u_{n}\\}\subset H^{1}(\mathbb{R}^{2})$ satisfying (5.7), (5.8)
and (5.8) is uniformly bounded in $n\in\mathbb{N}^{+}$ and $R>0$, that is,
there is a constant $\Sigma>0$ independent of $n\in\mathbb{N}^{+}$ and $R>0$
such that

(5.11) $\sup_{n\in\mathbb{N}}|\nabla
u_{n}|_{2}^{2}\leq\Sigma<+\infty\leavevmode\nobreak\
\text{\emph{and}}\leavevmode\nobreak\
0<\lambda_{n}^{R,\delta}\leq\Sigma<+\infty.$

###### Proof.

We claim that there is a constant $\bar{m}(a)>0$ which is independent of $R>0$
such that

(5.12) $0<m^{R,{\delta}}(a)\leq\bar{m}(a)<+\infty,\leavevmode\nobreak\ \forall
R>0.$

Indeed, in view of the definition of $f^{R,\delta}$, one concludes that
$F^{R,\delta}(t)\geq H(t)$ for all $t\in\mathbb{R}$, and so,
$E^{R,\delta}(u)\leq I(u)$ for all $u\in H^{1}(\mathbb{R}^{2})$, where the
energy functional $I:H^{1}(\mathbb{R}^{2})\to\mathbb{R}$ is defined by

(5.13) $I(u)=\frac{1}{2}\int_{\mathbb{R}^{2}}[|\nabla
u|^{2}+(A_{1}^{2}+A_{2}^{2})|u|^{2}]dx-\int_{\mathbb{R}^{2}}H(u)dx.$

Here, it is enough to choose the constant $\bar{m}(a)$ to be a minimization of
$I|_{\mathcal{M}(a)}$.

Secondly, we improve (5.1) and (5.2) in the sense that the positive constant
$M_{\varepsilon}^{R}$ is independent of $R>0$ under some suitable
$\bar{\alpha}_{0}>0$. Indeed, for each
$\bar{\alpha}_{0}\in(0,\bar{\alpha}_{0}^{*})$ with
$\bar{\alpha}_{0}^{*}=\frac{1}{R^{\tau-\delta}}>0$, by $(h_{4})$ with
$\delta\in(0,2)$, one has that

$|f^{R,\delta}(s)|\leq
Me^{\gamma|s|^{\delta}}e^{\bar{\alpha}_{0}^{*}R^{\tau-\delta}|s|^{\delta}}\leq
Me^{(\gamma+1)|s|^{\delta}},\quad\forall s\in\mathbb{R}.$

Thus, given $\alpha>0$, there is $M_{1}=M_{1}(\alpha)>0$ independent of $R>0$,
such that

$|f^{R,\delta}(s)|\leq M_{1}e^{\alpha|s|^{2}},\leavevmode\nobreak\ \forall
s\in\mathbb{R}.$

Therefore, there is $M_{\varepsilon}>0$ independent of $R>0$ such that

(5.14)
$|f^{R,\delta}(s)|\leq\varepsilon|s|^{\chi-1}+M_{\varepsilon}|s|^{q-1}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}$

and

(5.15)
$|F^{R,\delta}(s)|\leq\varepsilon|s|^{\chi}+M_{\varepsilon}|s|^{q}(e^{\alpha|s|^{2}}-1),\leavevmode\nobreak\
\forall s\in\mathbb{R}.$

Finally, by virtue of (5.12) and (5.14)-(5.15), we could repeat the
calculations exploited in Lemma 2.7 to derive the first part of (5.11). From
it and (5.12) combined with (5.6), we obtain the second part of (5.11). Hence,
the proof of this lemma is completed. ∎

Thanks to the growth conditions (5.14) and (5.15) with respect to the
nonlinearity $f^{R,\delta}$, we recall the choice of $\bar{a}^{*}_{R}>0$ which
is very similar to its counterpart in the Step 3 in the proof of Theorem 1.1,
it would conclude that $\bar{a}^{*}_{R}$ is independent of $R>0$. Furthermore,
we have the following result.

######  Lemma 5.4.

Let $(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ established by Theorem 1.6 for
all fixed $R>0$, then there is a constant
$\bar{\alpha}_{0}^{*}=\frac{1}{R^{\tau-\delta}}>0$ such that for all
$\bar{\alpha}_{0}\in(0,\bar{\alpha}_{0}^{*})$ and $\tau>2$, we have

$\sup_{R>0}|A_{0}[\bar{u}_{0}^{R}]+A_{1}^{2}[\bar{u}_{0}^{R}]+A_{2}^{2}[\bar{u}_{0}^{R}]|_{\infty}\leq
C_{A},$

where $C_{A}\in(0,+\infty)$ is a constant independent of $R>0$.

###### Proof.

According to the above observations, we derive that the constant
$a\in(0,\bar{a}^{*}_{R}]$ is independent of $R>0$. On the one hand, using
(1.7) and (1.8) for $j=1,2$, there holds

$\displaystyle|A_{j}[\bar{u}_{0}^{R}]|_{\infty}$
$\displaystyle\leq\frac{1}{4\pi}\int_{\mathbb{R}^{2}}\frac{|\bar{u}_{0}^{R}|^{2}}{|x-y|}dy=\frac{1}{4\pi}\int_{|x-y|<1}\frac{|\bar{u}_{0}^{R}|^{2}}{|x-y|}dy+\frac{1}{4\pi}\int_{|x-y|\geq
1}\frac{|\bar{u}_{0}^{R}|^{2}}{|x-y|}dy$
$\displaystyle\leq\frac{1}{\sqrt[3]{4\pi}}\left(\int_{\mathbb{R}^{2}}|\bar{u}_{0}^{R}|^{6}dy\right)^{\frac{1}{3}}+\frac{1}{4\pi}\int_{\mathbb{R}^{2}}|\bar{u}_{0}^{R}|^{2}dy\leq\frac{1}{\sqrt[3]{4\pi}}\Sigma^{\frac{2}{3}}a^{\frac{2}{3}}+\frac{1}{4\pi}a^{2},$

where we have used (2.13) and (5.11). We exploit (1.6) to see that

$|A_{0}[\bar{u}_{0}^{R}]|_{\infty}\leq\frac{|A_{j}[\bar{u}_{0}^{R}]|_{\infty}}{\pi}\int_{\mathbb{R}^{2}}\frac{|\bar{u}_{0}^{R}|^{2}}{|x-y|}dy.$

Combining the above two formulas, we can finish the proof of this lemma. ∎

At this position, we are ready to give the proof of Theorem 1.6.

######  _Proof of Theorem 1.4_.

Firstly, with (5.11) and (5.14) in hands, we can easily prove that
$|f^{R,\delta}(\bar{u}_{0}^{R})|_{2}$ is uniformly bounded in $R>0$.

Then, we would show that there is a constant $C_{0}>0$ which is independent of
$R>0$ such that $|\bar{u}_{0}^{R}|_{\infty}\leq C_{0}$. To obtain it, since
$\bar{u}_{0}^{R}$ is a nontrivial solution of Eq. (1.1) with a suitable
$\bar{\lambda}_{0}^{R}$, namely there holds

$-\Delta\bar{u}_{0}^{R}+\bar{\lambda}_{0}^{R}\bar{u}_{0}^{R}+(A_{0}[\bar{u}_{0}^{R}]+A_{1}^{2}[\bar{u}_{0}^{R}]+A_{2}^{2}[\bar{u}_{0}^{R}])\bar{u}_{0}^{R}=f^{R,\delta}(\bar{u}_{0}^{R})\leavevmode\nobreak\
\mbox{in}\leavevmode\nobreak\ \mathbb{R}^{2}.$

Since $h$ is nonnegative, without loss of generality, we can assume that
$\bar{u}_{0}^{R}\geq 0$ for all $x\in\mathbb{R}^{2}$, and so,

$-\Delta\bar{u}_{0}^{R}+\bar{u}_{0}^{R}\leq(1+\bar{\lambda}_{0}^{R})\bar{u}_{0}^{R}+|A_{0}[\bar{u}_{0}^{R}]+A_{1}^{2}[\bar{u}_{0}^{R}]+A_{2}^{2}[\bar{u}_{0}^{R}]|\bar{u}_{0}^{R}+f^{R,\delta}(\bar{u}_{0}^{R})\leavevmode\nobreak\
\mbox{in}\leavevmode\nobreak\ \mathbb{R}^{2}.$

According to (5.11), we have that $\bar{\lambda}_{a}^{R}\leq\bar{C}<+\infty$
for some constant $\bar{C}>0$ which is independent of $R>0$. As a consequence
of Lemma 5.4, the Lax-Milgram theorem gives the existence of a function
$\bar{w}_{0}^{R}\in H^{2}(\mathbb{R}^{2})$ such that

$-\Delta\bar{w}_{0}^{R}+\bar{w}_{0}^{R}=(1+\bar{\lambda}_{0}^{R})\bar{u}_{0}^{R}+|A_{0}[\bar{u}_{0}^{R}]+A_{1}^{2}[\bar{u}_{0}^{R}]+A_{2}^{2}[\bar{u}_{0}^{R}]|\bar{u}_{0}^{R}+f^{R,\delta}(\bar{u}_{0}^{R})\leavevmode\nobreak\
\mbox{in}\leavevmode\nobreak\ \mathbb{R}^{2}.$

Next, we fix the test function

$v_{r}(x)=\phi(x/r)(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})^{+}(x)\in
H^{1}(\mathbb{R}^{2}),$

where $\phi\in C_{0}^{\infty}(\mathbb{R}^{2})$ satisfies

$0\leq\phi(x)\leq 1,\leavevmode\nobreak\ \forall
x\in\mathbb{R}^{2};\leavevmode\nobreak\ \phi(x)=1,\leavevmode\nobreak\ \forall
x\in B_{1}(0);\leavevmode\nobreak\ \mbox{and}\leavevmode\nobreak\
\phi(x)=0\leavevmode\nobreak\ \forall x\in B^{c}_{2}(0).$

Using the function test $v_{r}$ on
$-\Delta(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})+(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})\leq
0$ in $\mathbb{R}^{2}$, we get the inequality below

$\int_{\mathbb{R}^{2}}\nabla\big{(}\bar{u}_{0}^{R}-\bar{w}_{0}^{R})\nabla
v_{r}+(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})v_{r}\big{]}dx\leq 0,$

Since

$v_{r}\to(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})^{+}\leavevmode\nobreak\
\mbox{as}\leavevmode\nobreak\ r\to+\infty\leavevmode\nobreak\
\mbox{in}\leavevmode\nobreak\ H^{1}(\mathbb{R}^{2}),$

by the Lebesgue’s Dominated Convergence theorem, we arrive at

$\int_{\mathbb{R}^{2}}|\nabla(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})^{+}|^{2}+|(\bar{u}_{0}^{R}-\bar{w}_{0}^{R})^{+}|^{2}\,dx\leq
0,$

implying that

$0\leq\bar{u}_{0}^{R}\leq\bar{w}_{0}^{R},\leavevmode\nobreak\ \forall
x\in\mathbb{R}^{2}.$

By using the continuous Sobolev embedding
$H^{2}(\mathbb{R}^{2})\hookrightarrow L^{\infty}(\mathbb{R}^{2})$, there is a
$C_{4}>0$ independent of $R>0$ such that

$|\bar{w}_{0}^{R}|_{\infty}\leq
C_{4}\|\bar{w}_{0}^{R}\|_{H^{2}(\mathbb{R}^{2})},\leavevmode\nobreak\ \forall
R>0$

which together with the last fact gives that

$|\bar{u}_{0}^{R}|_{\infty}\leq
C_{4}\|\bar{w}_{0}^{R}\|_{H^{2}(\mathbb{R}^{2})},\quad\forall R>0.$

On the other hand, by Brézis [10, Theorem 9.25], there is a $C_{5}>0$
independent of $R>0$ such that

$\|\bar{w}_{0}^{R}\|_{H^{2}(\mathbb{R}^{2})}\leq
C_{5}\big{|}(1+\bar{\lambda}_{0}^{R})\bar{u}_{0}^{R}+|A_{0}[\bar{u}_{0}^{R}]+A_{1}^{2}[\bar{u}_{0}^{R}]+A_{2}^{2}[\bar{u}_{0}^{R}]|\bar{u}_{0}^{R}+f^{R,\delta}(\bar{u}_{0}^{R})\big{|}_{2},\leavevmode\nobreak\
\forall R>0,$

from where it follows that

$\|\bar{w}_{0}^{R}\|_{H^{2}(\mathbb{R}^{2})}\leq C_{6},\leavevmode\nobreak\
\forall R>0$

for some $C_{6}>0$ independent of $R>0$. Have this in mind, we must have

$|\bar{u}_{0}^{R}|_{\infty}\leq C_{0},\quad\forall R>0,$

for some $C_{0}>0$ independent of $R>0$, showing the desired result.

Finally, we shall conclude that $\bar{u}_{0}^{R}\in H^{1}(\mathbb{R}^{2})$ is
a nontrivial solution of the original Eq. (1.1) with
$\lambda=\bar{\lambda}_{0}^{R}$ by choosing $R=C_{0}>0$ in (1.18) and hence
$\bar{\alpha}_{0}^{*}=C_{0}^{\delta-\tau}>0$. In other words, the couple
$(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ is a weak solution of problems
(1.1)-(1.2). The proof is completed. ∎

###  5.2. The case $\bar{\delta}=2$

In this subsection, we have the nonlinearity $f^{R,\bar{\delta}}=f^{R,2}$.
Firstly, we derive the following estimate for the mountain-pass value
$m^{R,2}(a)$ defined in (5.9).

######  Lemma 5.5.

Suppose $(h_{5})$ additionally, then there is a sufficiently large
$\xi_{0}^{R}>0$ such that for all $\xi>\xi_{0}^{R}$, there holds

(5.16) $m^{R,2}(a)<\frac{2\pi}{\gamma+\bar{\alpha}_{0}R^{\delta-2}}.$

###### Proof.

Given $a>0$, define $\psi(x)=a\pi^{-\frac{1}{2}}e^{-\frac{1}{2}|x|^{2}}$ for
all $x\in\mathbb{R}^{2}$ and so $|\nabla\psi|_{2}^{2}=|\psi|_{2}^{2}=a^{2}$.
By (2.2),

$\int_{\mathbb{R}^{2}}(A_{1}^{2}[\psi]+A_{2}^{2}[\psi])|\psi|^{2}dx\leq
16\bar{C}_{r}^{2}a^{6}.$

For all $t>0$, since $f^{R,2}(t)\geq h(t)\geq\xi t^{p-1}$ for all $t>0$, we
have that

$\displaystyle E^{R,2}(\psi_{t})$
$\displaystyle\leq\max_{t>0}\bigg{(}\frac{t^{2}}{2}\int_{\mathbb{R}^{2}}[|\nabla\psi|^{2}+(A_{1}^{2}+A_{2}^{2})|\psi|^{2}]dx-\frac{\xi
t^{p-2}}{p}\int_{\mathbb{R}^{2}}|\psi|^{p}dx\bigg{)}$
$\displaystyle\leq\frac{p-4}{2(p-2)}\left[\frac{p}{(p-2)\xi}\right]^{\frac{2}{p-4}}(1+16\bar{C}_{r}^{2}a^{3})^{\frac{2(p-2)}{p-4}}\pi^{\frac{p-2}{p-4}}a^{-\frac{2p}{p-4}},$

where we have used $|\psi|_{p}^{p}\geq a^{p}\pi^{-\frac{p-2}{2}}$. By Lemma
2.5, one sees that $m^{R,2}(a)\leq\max\limits_{t>0}E^{R,2}(\psi_{t})$. Let us
choose the constant $\xi_{0}(R)$ ro satisfy

$\xi_{0}^{R}=\frac{p(1+16\bar{C}_{r}^{2}a^{3})^{p-2}}{(p-2)a^{p}}\pi^{\frac{p-2}{2}}\left[\frac{(p-4)(\gamma+\bar{\alpha}_{0}R^{\delta-2})}{4\pi(p-2)}\right]^{\frac{p-4}{2}}$

which indicates the desired result. The proof is completed. ∎

Now, we can exhibit the proof of the second part (critical exponential case)
of Theorem 1.6.

######  _Proof of Theorem 1.6_.

(Completed). Due to Lemma 5.5, the proof is totally similar to that of Theorem
1.2 and so we just show the counterpart of Step I in this situation.

If the case I occurs, then we can derive

$\limsup_{n\to\infty}\int_{\mathbb{R}^{2}}|\nabla
u_{n}|^{2}dx=2\limsup_{n\to\infty}E^{R,2}(u_{n})=2m^{R,2}(a)<\frac{4\pi}{\gamma+\bar{\alpha}_{0}R^{\delta-2}}.$

Choosing $\nu>1$ sufficiently close to 1 in such a way that
$\frac{1}{\nu}+\frac{1}{\nu^{\prime}}=1$ and

$|\nabla
u_{n}|_{2}^{2}<\frac{4\pi(1-\epsilon)}{\nu(\gamma+\bar{\alpha}_{0}R^{\delta-2})}\leavevmode\nobreak\
\text{for some suitable}\leavevmode\nobreak\ \epsilon\in(0,1).$

Define
$\bar{u}_{n}=\sqrt{\frac{\nu(\gamma+\bar{\alpha}_{0}R^{\delta-2})}{4\pi(1-\epsilon)}}u_{n}$,
then $|\nabla\bar{u}_{n}|_{2}^{2}<1$ and
$|\bar{u}_{n}|_{2}^{2}=\frac{\nu(\gamma+\bar{\alpha}_{0}R^{\delta-2})}{4\pi(1-\epsilon)}a^{2}<+\infty$.
Using the above facts, we depend on (2.6) to see that

$\int_{\mathbb{R}^{2}}(e^{\nu(\gamma+\bar{\alpha}_{0}R^{\delta-2})u_{n}^{2}}-1)dx=\int_{\mathbb{R}^{2}}(e^{4\pi(1-\epsilon)\bar{u}_{n}^{2}}-1)dx\leq
C<+\infty,$

The remaining part is trivial and we omit it here. The proof is completed. ∎

Then, arguing as before, we begin considering the $L^{\infty}$-estimate for
the nontrivial solution $(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ established
in Theorem 1.6. To the end, we prove the following result.

######  Lemma 5.6.

Let $(h_{5})$ be satisfied. There is a constant $\tau^{*}=2+\frac{1}{R}>0$,
then for all for every $\alpha>0$, $\tau\in[2,\tau^{*})$ and $R>e$, each
sequence $\\{u_{n}\\}\subset H^{1}(\mathbb{R}^{2})$ satisfying (5.6), (5.7)
and (5.8) is uniformly bounded in $n\in\mathbb{N}$ and $R>e$, that is, there
is a constant $\Pi>0$ independent of $n\in\mathbb{N}$ and $R>0$ such that

(5.17) $\sup_{n\in\mathbb{N}}|\nabla
u_{n}|_{2}^{2}\leq\Pi<+\infty\leavevmode\nobreak\
\text{\emph{and}}\leavevmode\nobreak\
0<\lambda_{n}^{R,\delta}\leq\Pi<+\infty,$

provided $\xi_{0}>0$ in $(h_{5})$ is sufficiently large, where the constant
$\Pi>0$ independent of $R>e$ satisfies

(5.18) $\Pi<\frac{\pi}{2\alpha e^{\frac{1}{e}}}.$

###### Proof.

We claim that there is a constant $\tilde{m}(a)>0$ which is independent of
$R>0$ such that

(5.19) $0<m^{R,{2}}(a)\leq\tilde{m}(a)<+\infty,\leavevmode\nobreak\ \forall
R>0.$

Indeed, in view of the definition of $f^{R,2}$, one concludes that
$F^{R,2}(t)\geq H(t)$ for all $t\in\mathbb{R}$, and so, $E^{R,2}(u)\leq I(u)$
for all $u\in H^{1}(\mathbb{R}^{2})$, where the energy functional
$I:H^{1}(\mathbb{R}^{2})\to\mathbb{R}$ is defined by (5.13). Here, it is
enough to choose the constant $\tilde{m}(a)$ to be a minimization of
$I|_{\mathcal{M}(a)}$. Exploiting the very similar calculations in Lemma 5.5,
we can find such a $\Pi>0$ in (5.18) to satisfy

(5.20) $\tilde{m}(a)<\frac{\theta-4}{2(\theta-3)}\Pi.$

Secondly, we shall improve (5.3) and (5.4). Obviously,
$\lim\limits_{R\to+\infty}R^{\frac{1}{R}}=1$ and the function
$R^{\frac{1}{R}}$ is strictly decreasing in $R\in(e,+\infty)$, then
$0<R^{\frac{1}{R}}\leq e^{\frac{1}{e}}$ for all $R\in(e,+\infty)$. Consider
$\tau^{*}=2+\frac{1}{R}>0$ for the fixed $R>e$, $\varepsilon>0$ and $q\geq 2$,
there is a constant $M_{\varepsilon}>0$ independent of $R>e$ such that

(5.21)
$|f^{R,2}(s)|\leq\varepsilon|s|^{\chi-1}+M_{\varepsilon}|s|^{q-1}(e^{2\alpha
e^{\frac{1}{e}}|s|^{2}}-1),\leavevmode\nobreak\ \forall s\in\mathbb{R}.$

Actually, using $(h_{1})$, by some elementary calculations,

$0\leq\frac{|f^{R,2}(s)|}{|s|^{\chi-s}}\leq\frac{|h(s)|e^{\alpha
R^{\tau-2}|s|^{2}}}{|s|^{\chi-1}}\leq\frac{|h(s)|e^{\alpha
R^{\tau^{*}-2}|s|^{2}}}{|s|^{\chi-1}}\leq\frac{|h(s)|e^{\alpha
R^{\frac{1}{R}}|s|^{2}}}{|s|^{\chi-1}}\leq\frac{|h(s)|e^{\alpha
e^{\frac{1}{e}}|s|^{2}}}{|s|^{\chi-1}}\to 0$

uniformly in $s\to 0$. On the other hand, there is a $s_{0}>0$ independent of
$R>e$ such that

$\displaystyle|f^{R,2}(s)|$ $\displaystyle\leq
Me^{\gamma|s|^{\delta}}e^{\alpha e^{\frac{1}{e}}|s|^{2}}\leq\bar{M}e^{2\alpha
e^{\frac{1}{e}}|s|^{2}}\leq\tilde{M}|t|^{q}(e^{2\alpha
e^{\frac{1}{e}}t^{2}}-1),\leavevmode\nobreak\ \forall|s|\geq s_{0},$

where the constants $\tilde{M}>\bar{M}>M$ are independent of $R>e$ and we have
used $(h_{3})$.

$0\leq f^{R,\delta}(t)\leq Me^{\gamma
t^{\delta}}e^{\alpha_{0}^{*}R^{\tau-\delta}t^{\delta}}\leq
Me^{(\gamma+1)t^{\delta}}\leq M_{1}e^{\alpha t^{2}},\leavevmode\nobreak\
\forall t\in\mathbb{R},$

where $M_{1}>0$ is a constant independent of $R>0$. Similarly, there holds

(5.22)
$|F^{R,\delta}(s)|\leq\varepsilon|s|^{\chi}+M_{\varepsilon}|s|^{q}(e^{2\alpha
e^{\frac{1}{e}}|s|^{2}}-1),\leavevmode\nobreak\ \forall s\in\mathbb{R}.$

Finally, we show (5.17). In fact, we make full use (5.6) and (5.8) jointly
with Lemma 5.1 to get

$\displaystyle\int_{\mathbb{R}^{2}}|\nabla u_{n}|^{2}dx$
$\displaystyle=2E^{R,2}(u_{n})-\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u_{n}|^{2}dx+\int_{\mathbb{R}^{2}}F(u_{n})dx+o_{n}(1)$
$\displaystyle\leq
2E^{R,2}(u_{n})+\frac{2}{\theta-4}\big{[}E^{R,2}(u_{n})-\frac{1}{2}J^{R,2}(u_{n})\big{]}+o_{n}(1)$
$\displaystyle=\frac{2(\theta-3)}{\theta-4}E^{R,2}(u_{n})+o_{n}(1)=\frac{2(\theta-3)}{\theta-4}m^{R,2}(a)+o_{n}(1)$

which together with (5.19) and (5.20) indicates the first part of (5.17). The
remaining part is trivial, we omit it here. The proof is completed. ∎

######  _Proof of Theorem 1.4_.

(Completed). We claim that $|f^{R,2}(\bar{u}_{0}^{R})|_{2}$ is uniformly
bounded in $R>e$. Let us contemplate
$\bar{w}_{0}^{R}=\Pi^{-\frac{1}{2}}\bar{u}_{0}^{R}$, then
$|\nabla\bar{w}_{0}^{R}|^{2}_{2}\leq 1$ and
$|\bar{w}_{0}^{R}|^{2}_{2}=\Pi^{-1}a^{2}<+\infty$. According to the definition
of $\tau^{*}$, we deduce that (5.21) still remains true. Since $q\geq 2$ and
(5.18) gives that

$8\alpha e^{\frac{1}{e}}\Pi<4\pi,$

we then apply (2.6) to conclude that

$\int_{\mathbb{R}^{2}}|\bar{u}_{0}^{R}|^{2(q-1)}(e^{4\alpha
e^{\frac{1}{e}}|\bar{u}_{a}^{R}|^{2}}-1)dx\leq\bigg{(}\int_{\mathbb{R}^{2}}|\bar{u}_{0}^{R}|^{4(q-1)}dx\bigg{)}^{\frac{1}{2}}\bigg{(}\int_{\mathbb{R}^{2}}(e^{8\alpha
e^{\frac{1}{e}}\Pi|\bar{w}_{0}^{R}|^{2}}-1)dx\bigg{)}^{\frac{1}{2}}\leq C$

for some $C>0$ independent of $R>e$. Besides, $2(\chi-1)\geq 6$, it simply
calculates that

$\int_{\mathbb{R}^{2}}|\bar{u}_{0}^{R}|^{2(\chi-1)}dx\leq C$

for some $C>0$ independent of $R>e$. Exploiting (5.21), we could derive the
claim. Repeating the remaining parts in proof of Theorem 1.4 in Subsection
5.1, we must have

$|\bar{u}_{0}^{R}|_{\infty}\leq\bar{C}_{0},\quad\forall R>e.$

for some $\overline{C}_{0}>0$ independent of $R>e$. Finally, we obtain that
the couple $(\bar{u}_{0}^{R},\bar{\lambda}_{0}^{R})$ is a weak solutions to
problems (1.1)-(1.2) by choosing $R=\tilde{C}_{0}=\max\\{\bar{C}_{0},e\\}>0$
in (1.18) and then $\tau^{*}=2+\frac{1}{\tilde{C}_{0}}>0$. The proof is
completed. ∎

## Appendix

Lemma A.1. Suppose that $f$ satisfies (1.14) and $(f_{1})$, or (1.13) and
$(f_{1})$. Let $u\in H^{1}(\mathbb{R}^{2})$ be a nontrivial weak solution of
Eq. (1.1) with some suitable $\lambda>0$, then it satisfies the so-called
Pohoz̆aev identity below

$\lambda\int_{\mathbb{R}^{2}}|u|^{2}dx+2\int_{\mathbb{R}^{2}}(A_{1}^{2}+A_{2}^{2})|u|^{2}dx-2\int_{\mathbb{R}^{2}}F(u)dx=0.$

###### Proof.

Since $u\neq 0$, taking (1.6) and (1.7)-(1.8) into account, define

$b(x)=\frac{f(u(x))}{u(x)}-(A_{0}[u(x)]+A_{1}^{2}[u(x)]+A_{2}^{2}[u(x)])-\lambda,$

by means of (2.5) and using a similar arguments in Lemma 5.4, we can conclude
that $b(x)\in L_{\text{loc}}^{1}(\mathbb{R}^{2})$ and $u$ satisfies the
following elliptic equation

$-\Delta u=b(x)u.$

In view of the classic Brézis-Kato theorem, one would conclude that $u\in
L_{\text{loc}}^{s}(\mathbb{R}^{2})$ for all $1\leq s<+\infty$. Thus $u\in
W^{2.s}_{\text{loc}}(\mathbb{R}^{2})$ for all $1\leq s<+\infty$ by the
Caldéron-Zygmund inequality.

To derive the Pohoz̆aev type identity for Eq. (1.1), we use a truncation
argument due to Kavian (see e.g. [55, Appendix B]). Let $\psi\in
C^{\infty}([0,+\infty),[0,1])$ such that $\psi(r)=1$ for $r\in[0,1]$ and
$\psi(r)=0$ for $r\in[2,+\infty)$. Define
$\psi_{n}(x)\triangleq\psi(|x|^{2}/n^{2})$ on $\mathbb{R}^{2}$ for
$n\in\mathbb{N}$. Then there exists $C_{1}>0$ such that

$0\leq\psi_{n}\leq C_{1}\leavevmode\nobreak\ \text{and}\leavevmode\nobreak\
|x||\nabla\psi_{n}(x)|\leq C_{1}.$

Multiplying Eq. (1.1) by $\psi_{n}(x,\nabla u)$, we have for every
$n\in\mathbb{N}$,

(5.23) $0=\big{\\{}-\Delta u+\lambda
u+\left(A_{0}[u]+A_{1}^{2}[u]+A_{2}^{2}[u]\right)u-f(u)\big{\\}}\psi_{n}(x,\nabla
u).$

For every $n\in\mathbb{N}$, by virtue of the properties of the divergence, it
is clear to compute that

$\left\\{\begin{array}[]{ll}-\psi_{n}(x,\nabla u)\Delta
u=-\text{div}\bigg{\\{}\bigg{[}\nabla u(x,\nabla u)-x\frac{|\nabla
u|^{2}}{2}\bigg{]}\psi_{n}\bigg{\\}}-\frac{|\nabla
u|^{2}}{2}(x,\nabla\psi_{n})+(x,\nabla u)(\nabla\psi_{n},\nabla u),\\\
\psi_{n}(x,\nabla
u)u=\frac{1}{2}\text{div}[xu^{2}\psi_{n}]-u^{2}\psi_{n}-\frac{1}{2}u^{2}(x,\nabla\psi_{n}),\\\
-\psi_{n}(x,\nabla
u)f(u)=-\text{div}(x\psi_{n}F(u))+2\psi_{n}F(u)+F(u)(x,\nabla\psi_{n}),\end{array}\right.$

and by setting $\phi_{u}=A_{0}[u]+A_{1}^{2}[u]+A_{2}^{2}[u]$ that

$\psi_{n}(x,\nabla
u)\phi_{u}u=\frac{1}{2}\text{div}\big{[}\phi_{u}xu^{2}\psi_{n}\big{]}-\phi_{u}u^{2}\psi_{n}-\frac{1}{2}\phi_{u}u^{2}(x,\nabla\psi_{n})-\frac{1}{2}u^{2}\psi_{n}(x,\nabla\phi_{u}).$

It follows from the divergence theorem and (5.23) that

$\begin{gathered}\int_{\partial B_{2n}(0)}\bigg{\\{}\frac{|(x,\nabla
u)|^{2}}{2n}-n|\nabla u|^{2}-n\lambda
u^{2}-n\phi_{u}u^{2}+2nF(u)\bigg{\\}}\psi_{n}d\sigma\hfill\\\ \ \ \ \
=-\int_{B_{2n}(0)}\bigg{\\{}\lambda
u^{2}+\phi_{u}u^{2}+\frac{1}{2}(x,\nabla\phi_{u})u^{2}-2F(u)\bigg{\\}}\psi_{n}dx\hfill\\\