Ground states for planar axially Schrödinger–Newton system with an exponential critical growth

Abstract

In this paper, we study the following planar Schrödinger–Newton system:

$$ \left \{ \textstyle\begin{array}{l} -\Delta u+ V(x)u+\lambda\phi u= f(x,u)\quad \textrm{in } \mathbb{R}^{2},\\ \Delta\phi=u^{2}\quad\textrm{in } \mathbb{R}^{2}, \end{array}\displaystyle \right . $$

where V, f are axially symmetric about x, V is positive, and f is super-linear at zero and exponential critical at infinity. Using a weaker condition

$$ \biggl[\frac{f(x,u)}{u^{3}}-\frac{f(x,tu)}{(tu)^{3}} \biggr]\operatorname {sign}(1-t)+ \theta V(x)\frac{ \vert 1-t^{2} \vert }{(tu)^{2}}\geq0,\quad \forall x\in \mathbb{R}^{2}, t>0, u\neq0 $$

with $\theta\in[0,1)$ instead of the Nehari type monotonic condition on $\frac{f(x,u)}{|u|^{3}}$, we obtain a ground state solution of the above problem via variational methods.

1 Introduction and main results

In the present paper, we are concerned with the wave solutions of the Schrödinger–Newton system

$$ \left \{ \textstyle\begin{array}{l} -i\psi_{t} -\Delta\psi+W(x)\psi+\lambda\phi u=g(x,\psi)\quad \textrm{in } \mathbb{R}^{d},\\ \Delta\phi= \vert \psi \vert ^{2} \quad\textrm{in } \mathbb{R}^{d}, \end{array}\displaystyle \right . $$

(1.1)

where $\psi:\mathbb{R}^{d}\times\mathbb{R}\rightarrow\mathbb{C}$ is the wave function, $W(x)$ is a real external potential, $\lambda>0$ is a parameter. Problems of the type (1.1) arise in many problems from physics. We refer the readers to [15], therein (1.1) appears in a quantum mechanical context in the case $d\leq3$.

A standing wave solution of (1.1) is a solution of the form $\psi(x,t)=e^{-iEt}u(x)$ and its existence reduces (1.1) to the system

$$ \left \{ \textstyle\begin{array}{l} -\Delta u+V(x)u+\lambda\phi u=f(x,u)\quad \textrm{in } \mathbb{R}^{d},\\ \Delta\phi=u^{2}\quad \textrm{in } \mathbb{R}^{d}, \end{array}\displaystyle \right . $$

(1.2)

where $V(x)=W(x)-E$, $g(x,e^{-iEt}u)=f(x,u)e^{-iEt}$. For the case $d=3$, it is called the Schrödinger–Poisson system and it has been well studied. For the existence, multiplicity, and concentration, we refer the readers to [2, 3, 9, 10, 13, 20] and the references therein. For Kirchhoff type equations involving subcritical and critical growth in three dimensions, please see [19] and the references therein. We also quote the paper [12] for Hardy–Schrödinger–Kirchhoff systems.

However, much less is known about the case $d=2$. For $\Delta\phi =u^{2}$, in $\mathbb{R}^{2}$, one has

$$ \phi(x)=\frac{1}{2\pi} \int_{\mathbb{R}^{2}}\ln\bigl( \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2}\,dy. $$

(1.3)

Substituting it into (1.2), we obtain the integro-differential equation

$$ -\Delta u+V(x)u+\frac{\lambda}{2\pi}\bigl(\ln\bigl( \vert \cdot \vert \bigr)\ast u^{2}\bigr)u=f(x,u) \quad\textrm{in } \mathbb{R}^{2}. $$

(1.4)

For simplicity, throughout this paper, let $\lambda=2\pi$. The approach for $d=3$ cannot be easily adapted to $d=2$ since

$$ \frac{1}{4} \int_{\mathbb{R}^{2}} \int_{\mathbb{R}^{2}}\ln \bigl( \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy\,dx, $$

(1.5)

which is the functional associated with the third term in (1.4), is sign-changing, and is neither bounded from above nor from below on $H^{1}(\mathbb{R}^{2})$. This difficulty has been overcome recently in [7] or [16]. For

$$ -\Delta u+\bigl(\ln\bigl( \vert \cdot \vert \bigr)\ast u^{2}\bigr)u=\mu u \quad\textrm{in } \mathbb{R}^{2}, $$

(1.6)

by introducing the following subspace of $H^{1}(\mathbb{R}^{2})$

$$ X:=\biggl\{ u\in H^{1}\bigl(\mathbb{R}^{2}\bigr) : \int_{\mathbb{R}^{2}}\ln \bigl(1+ \vert x \vert \bigr)u^{2} \,dx< \infty\biggr\} $$

endowed with the norm

$$ \Vert u \Vert ^{2}= \int_{\mathbb{R}^{2}} \bigl( \vert \nabla u \vert ^{2}+u^{2}+ \ln \bigl(1+ \vert x \vert \bigr)u^{2} \bigr)\,dx, $$

Stubbe considered the $L^{2}$-constraint minimization problem and proved that (1.6) admits a ground state.

Soon afterwards, in [8], Cingolani and Weth processed successfully the two dimensional Schrödinger–Newton equations with nonlinear term $|u|^{p-2}u$, $p\geq4$. Du and Weth [11] provided some results about $p>2$ and $p\geq3$. The key tool is Pohozaev type identity (see [11, Lemma 2.4]). Chen, Shi, and Tang [4] used the same idea to obtain a ground state but they could deal with the general nonlinearity $f(u)$. Simultaneously, Chen and Tang [5] investigated the existence of an axially symmetric Nehari type ground state and nontrivial solution for

$$ -\Delta u+V(x)u+\bigl(\ln\bigl( \vert \cdot \vert \bigr)\ast u^{2}\bigr)u=f(x,u) \quad\textrm{in } \mathbb{R}^{2}, $$

(1.7)

where V, f is axially symmetric about x. Please see [6, 17] for further results about two dimensional Schrödinger–Newton equations with the axially symmetric assumptions. Recently, when $V(x)=1$, Alves and Figueiredo [1] proved that (1.4) admits a positive ground state, where f is a continuous function with the exponential critical growth.

In this paper, motivated by the papers [1] and [5], we shall study the existence of ground state solutions of planar problem (1.1) with an exponential critical growth. In order to state our main result, we assume that

($V_{1}$):

$V\in C(\mathbb{R}^{2},\mathbb{R})$, $\inf_{x\in \mathbb{R}^{2}} V(x)>0$, $V(x):=V(x_{1},x_{2})=V(|x_{1}|,|x_{2}|)$ for all $x\in\mathbb{R}^{2}$.

($V_{2}$):

There exists a sequence $\{t_{n}\}\subset(0,\infty )$ such that $t_{n}\rightarrow\infty$ and

$$\sup_{x\in\mathbb{R}^{2},n\in\mathbb{N}}\frac {V(t^{-1}_{n}x)}{V(x)}< \infty. $$

($f_{1}$):

$f\in C^{1}(\mathbb{R}^{2}\times\mathbb{R},\mathbb {R})$, $f(x,u):=f((x_{1},x_{2}),u)=f((|x_{1}|,|x_{2}|),u)$.

($f_{2}$):

$f(x,u)=o(|u|)$ as $u\rightarrow0$, uniformly in $x\in\mathbb{R}^{2}$.

($f_{3}$):

There exists $\alpha_{0}>0$ such that

$$ \lim_{|u|\rightarrow\infty}\frac{f(x,u)}{\exp(\alpha u^{2})}=0 \quad\textrm {for } \alpha> \alpha_{0},\qquad \lim_{|u|\rightarrow\infty}\frac{f(x,u)}{\exp(\alpha u^{2})}=+\infty \quad\textrm{for } \alpha< \alpha_{0}. $$

($f_{4}$):

There exists $\theta\in[0,1)$ such that

$$\biggl[\frac{f(x,\tau)}{\tau^{3}}-\frac{f(x,t\tau)}{(t\tau)^{3}} \biggr]\operatorname{sign}(1-t)+ \theta V(x)\frac{ \vert 1-t^{2} \vert }{(t\tau)^{2}}\geq0,\quad \forall x \in\mathbb{R}^{2}, t>0, \tau\neq0; $$

($f_{5}$):

$\inf_{x\in\mathbb{R}^{2},u\neq0}\frac {F(x.u)}{u^{2}}>-\infty$, where $F(u)=\int^{u}_{0}f(t)\,dt$.

Remark 1.1

A simple example of satisfying the hypotheses of ($V_{1}$)–($V_{2}$) is the function $V(x)=1+|x_{2}|[1+\sin(\pi|x_{1}|)]$ with $t_{n}=n$. Here we also give an example which satisfies ($f_{1}$)–($f_{5}$):

$$ f(x,u)=\bigl({K(x) \vert u \vert ^{3}u-V(x) \vert u \vert ^{\frac{3}{2}}u+V(x) \vert u \vert u}\bigr)\exp\biggl(\frac{\frac {1}{2}-\frac{\theta}{2}}{m}\pi u^{2}\biggr), $$

where $K\in(\mathbb{R}^{2},\mathbb{R})$ is axially symmetric and $\inf_{x\in\mathbb{R}^{2}} K(x)>0$, V satisfies ($V_{1}$) and ($V_{2}$). But it does not satisfy the Nehari type monotonic condition

$$ \frac{f(x,u)}{ \vert u \vert ^{3}} \textrm{ is a strictly increasing function of } u\in\mathbb{R} \setminus\{0\}. $$

Now we state our main result as follows.

Theorem 1

For$d=2$, suppose that ($V_{1}$), ($V_{2}$) and ($f_{1}$)–($f_{5}$) are satisfied. Then, for any$\alpha\in(0,\frac{\pi(1-\theta)}{m})$, wheremis the least energy (it will be defined in (2.22)), θis from ($f_{3}$), (1.7) possesses a ground state solution.

Remark 1.2

The condition $\alpha\in(0,\frac{\pi(1-\theta)}{m})$ is used to prove the minimizing sequence of m is bounded, and please see Lemma 3.3. Up to now, we have not been able to remove it.

The paper is organized as follows. Section 2 is to establish the variational setting and to give some preliminaries. Section 3 is to prove the existence of ground states. Throughout the paper, we always assume that ($V_{1}$), ($V_{2}$) and ($f_{1}$)–($f_{5}$) hold and make use of the following notations:

• C, $C_{i}$ ($i=0,1,2,\ldots$) for positive constants (possibly different from line to line).

• $L^{s}(\mathbb{R}^{2}):=\{u:\mathbb{R}^{2}\rightarrow\mathbb {R}:\int_{\mathbb{R}^{2}}|u|^{s}\,dx<\infty\}$ and $\|\cdot\|_{s}$ denotes the usual $L^{s}$-norm in $L^{s}(\mathbb{R}^{2})$.

2 Variational setting and preliminaries

In this section, we begin our study by establishing the variational setting for (1.7). Let $H^{1}(\mathbb{R}^{2})$ be the usual fractional Sobolev space with the usual norm

$$\Vert u \Vert _{H^{1}}=\biggl( \int_{\mathbb{R}^{2}}\bigl( \vert \nabla u \vert ^{2}+u^{2} \bigr)\,dx\biggr)^{\frac{1}{2}} $$

and

$$H_{as}^{1}\bigl(\mathbb{R}^{2}\bigr):=\bigl\{ u\in H^{1}\bigl(\mathbb{R}^{2}\bigr): u(x):=u(x_{1},x_{2})=u \bigl( \vert x_{1} \vert , \vert x_{2} \vert \bigr), \forall x\in \mathbb{R}^{2}\bigr\} . $$

By ($V_{1}$) and ($f_{1}$), similar to [5], let E be defined as

$$ E:=\biggl\{ u\in H_{as}^{1}\bigl(\mathbb{R}^{2} \bigr): \int_{\mathbb {R}^{2}}V(x)u^{2}\,dx< \infty\biggr\} $$

endowed with the norm

$$ \Vert u \Vert _{E}=\biggl( \int_{\mathbb{R}^{2}}\bigl( \vert \nabla u \vert ^{2}+V(x)u^{2}+ \ln \bigl(1+ \vert x \vert \bigr)u^{2}\bigr)\,dx \biggr)^{\frac{1}{2}}. $$

Denote

$$\Vert u \Vert :=\biggl( \int_{\mathbb{R}^{2}}\bigl( \vert \nabla u \vert ^{2}+V(x)u^{2} \bigr)\,dx\biggr)^{\frac {1}{2}},\qquad \Vert u \Vert _{\ast}:=\biggl( \int_{\mathbb{R}^{2}}\ln\bigl(1+ \vert x \vert \bigr)u^{2} \,dx\biggr)^{\frac{1}{2}}. $$

According to [1, Lemma 2.1], we have the following.

Proposition 2.1

$E\hookrightarrow L^{t}(\mathbb{R}^{2})$is compact for all$t\in [2,\infty)$.

We formally formulate problem (1.7) in a variational way as follows:

$$ \begin{aligned}[b] I(u)={}&\frac{1}{2} \int_{\mathbb{R}^{2}}\bigl( \vert \nabla u \vert ^{2}+V(x)u^{2} \bigr)\,dx+\frac{1}{4} \int_{\mathbb{R}^{2}} \int_{\mathbb {R}^{2}}\ln\bigl( \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy\,dx \\ &- \int_{\mathbb{R}^{2}}F(x,u)\,dx,\quad u\in E. \end{aligned} $$

(2.1)

For simplicity of notations, denote

$$ I_{0}(u):= \int_{\mathbb{R}^{2}} \int_{\mathbb{R}^{2}}\ln \bigl( \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy\,dx. $$

Similar to [8], using $\ln(r)=\ln(1+r)-\ln(1+\frac{1}{r})$, $\forall r>0$, it holds that

$$\begin{aligned} I_{0}(u)&= \int_{\mathbb{R}^{2}} \int_{\mathbb{R}^{2}}\ln \bigl(1+ \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy\,dx \\ &\quad- \int_{\mathbb{R}^{2}} \int_{\mathbb{R}^{2}}\ln\biggl(1+\frac {1}{ \vert x-y \vert }\biggr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy\,dx \\ &:= I_{1}(u)-I_{2}(u). \end{aligned}$$

We give the following proposition which is used to estimate the nonlinearity.

Proposition 2.2

([1, Lemma 2.5])

For every$\alpha>0$and for all$u\in H^{1}(\mathbb{R}^{2})$, we have

$$ \exp\bigl(\alpha u^{2}\bigr)-1\in L^{1}\bigl( \mathbb{R}^{2}\bigr). $$

(2.2)

Moreover, if$\|\nabla u\|_{2}\leq1$, $\|u\|_{2}\leq M$, and$\alpha <4\pi$, then there exists$C>0$independent ofusuch that

$$ \int_{\mathbb{R}^{2}}\bigl[\exp\bigl(\alpha u^{2}\bigr)-1\bigr] \,dx\leq C. $$

(2.3)

Lemma 2.3

$I\in C^{1}(E,\mathbb{R})$.

Proof

Noting that $\ln(1+|x-y|)\leq\ln(1+|x|)-\ln(1+|y|)$, $\forall x,y\in \mathbb{R}^{2}$, we get

$$ \bigl\vert I_{1}(u) \bigr\vert \leq2 \Vert u \Vert ^{2}_{2} \Vert u \Vert ^{2}_{\ast}. $$

(2.4)

In view of $\ln(1+r)\leq r$, $\forall r>0$, jointly with the Hardy–Littlewood–Sobolev inequality [14], we obtain

$$ \bigl\vert I_{2}(u) \bigr\vert \leq C \Vert u \Vert ^{4}_{\frac{8}{3}}. $$

(2.5)

So $I_{0}$ is well defined in E.

Using ($f_{1}$)–($f_{3}$), for each $\varepsilon>0$, we have

$$ \bigl\vert F(x,u) \bigr\vert \leq\varepsilon \vert u \vert ^{2}+C(\varepsilon) \vert u \vert ^{p}\bigl[\exp\bigl( \alpha \vert u \vert ^{2}\bigr)-1\bigr], $$

(2.6)

where $p>2$. Thus, using Hölder’s inequality with $s>1$, $\frac {1}{s}+\frac{1}{s'}=1$, we get

$$\begin{aligned} \int_{\mathbb{R}^{2}}F(x,u)\,dx&\leq\varepsilon \int_{\mathbb {R}^{2}} \vert u \vert ^{2}\,dx+C(\varepsilon) \int_{\mathbb{R}^{2}} \vert u \vert ^{p}\bigl[\exp \bigl( \alpha \vert u \vert ^{2}\bigr)-1\bigr]\,dx \\ &\leq\varepsilon \int_{\mathbb{R}^{2}} \vert u \vert ^{2}\,dx+C(\varepsilon) \biggl( \int _{\mathbb{R}^{2}} \vert u \vert ^{ps}\,dx \biggr)^{\frac{1}{s}} \biggl( \int_{\mathbb {R}^{2}}\bigl[\exp\bigl(s'\alpha \vert u \vert ^{2}\bigr)-1\bigr]\,dx \biggr)^{\frac{1}{s'}}. \end{aligned}$$

By Propositions 2.1 and 2.2, I is well defined in E. By [8, Lemma 2.2], $I_{0}\in C^{1}(E,\mathbb{R})$. It is easy to check that $\int_{\mathbb{R}^{2}}F(x,u)\,dx$ belongs to $C^{1}(E,\mathbb{R})$. Thus, $I\in C^{1}(E,\mathbb{R})$. □

Based on Lemma 2.3, we have

$$ \begin{aligned}[b] \bigl\langle I'(u),v\bigr\rangle ={}& \int_{\mathbb{R}^{2}}\bigl(\nabla u\nabla v+V(x)uv\bigr)\,dx+ \int_{\mathbb{R}^{2}} \int_{\mathbb{R}^{2}}\ln \bigl( \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2}u(x)v(x)\,dy\,dx \\ &- \int_{\mathbb{R}^{2}}f(x,u)v\,dx. \end{aligned} $$

(2.7)

Lemma 2.4

For every$u\in E$, we have

$$ I(u)\geq I(tu)+\frac{1-t^{4}}{4}\bigl\langle I'(u),u \bigr\rangle +\frac{(1-\theta )(1-t^{2})^{2}}{4} \Vert u \Vert ^{2},\quad \forall t\geq0. $$

(2.8)

Proof

Since the proof is similar to [5, Lemma 2.3], we omit it here. □

Now, we define the Nehari manifold

$$ \mathcal{N}:=\bigl\{ u\in E\setminus\{0\}:\bigl\langle I'(u),u\bigr\rangle =0 \bigr\} . $$

Since the Nehari type monotonic condition on $\frac{f(x,u)}{|u|^{3}}$ and super-cubic condition are not satisfied, we need to prove that $\mathcal{N}\neq\emptyset$. To the end, we introduce the following new set:

$$ \mathcal{E}:=\biggl\{ u\in E\setminus\{0\}: \int_{\mathbb {R}^{2}}V(x)u^{2}\,dx+I_{0}(u)< \int_{\mathbb{R}^{2}}f(x,u)u\,dx\biggr\} . $$

Lemma 2.5

$\mathcal{E}\neq\emptyset$.

Proof

Let $u\in E$ with $u\neq0$. $u_{t}(x):=u(tx)$. By ($V_{2}$), there exists $C_{1}>0$ such that

$$ V\bigl(t^{-1}_{n}x\bigr)\leq C_{1}V(x),\quad \forall x\in\mathbb{R}^{2}, n\in\mathbb{N}. $$

(2.9)

It follows that

$$\begin{aligned} & \int_{\mathbb {R}^{2}}V(x) (t_{n}u_{t_{n}})^{2} \,dx+I_{0}(t_{n}u_{t_{n}})- \int_{\mathbb {R}^{2}}f(x,t_{n}u_{t_{n}})t_{n}u_{t_{n}} \,dx \\ &\quad\leq C_{1} \Vert u \Vert ^{2}+I_{0}(u)-\ln t_{n} \Vert u \Vert ^{4}_{2}- \int_{\mathbb {R}^{2}}\frac{f(t^{-1}_{n}x,t_{n}u)t_{n}u}{t^{2}_{n}}\,dx. \end{aligned}$$

In view of ($f_{4}$), $t\geq0$, $\tau\neq0$, it holds that

$$ \begin{aligned}[b] &\frac{1-t^{4}}{4}\tau f(x,\tau)+F(x,t \tau)-F(x,\tau)+\frac{\theta V(x)}{4}\bigl(1-t^{2}\bigr)^{2} \tau^{2} \\ &\quad= \int^{1}_{t} \biggl[\frac{f(x,\tau)}{\tau^{3}}- \frac{f(x,s\tau)}{(s\tau )^{3}} \biggr]\operatorname{sign}(1-t)+\theta V(x)\frac{ \vert 1-t^{2} \vert }{(s\tau )^{2}} \tau^{3} \tau^{4}\,ds\geq0. \end{aligned} $$

(2.10)

Taking $t=0$, we obtain

$$ \frac{1}{4}\tau f(x,\tau)-F(x,\tau)+\frac{\theta V(x)}{4} \tau^{2}\geq 0,\quad \forall x\in\mathbb{R}^{2}, \tau\in\mathbb{R}. $$

(2.11)

By ($f_{5}$), one has

$$ F(x,\tau)\geq-C_{2}\tau^{2},\quad \forall x\in \mathbb{R}^{2}, \tau\in \mathbb{R}. $$

(2.12)

Thus, we get

$$ \begin{aligned}[b] \int_{\mathbb{R}^{2}}\frac {f(t^{-1}_{n}x,t_{n}u)t_{n}u}{t^{2}_{n}}\,dx\geq{}& \int_{\mathbb {R}^{2}} \biggl[\frac{4F(t^{-1}_{n}x,t_{n}u)}{t^{2}_{n}}-\theta V \bigl(t^{-1}_{n}x\bigr)u^{2} \biggr]\,dx \\ \geq&{-}4C_{2} \int_{\mathbb{R}^{2}}u^{2}\,dx-\theta C_{1} \int_{\mathbb {R}^{2}}V(x)u^{2}\,dx. \end{aligned} $$

(2.13)

So

$$\begin{aligned} & \int_{\mathbb{R}^{2}}V(x) (t_{n}u_{t_{n}})^{2} \,dx+ \int_{\mathbb {R}^{2}} \int_{\mathbb{R}^{2}}\ln \bigl( \vert x-y \vert \bigr) \bigl\vert t_{n}u_{t_{n}}(y) \bigr\vert ^{2} \bigl\vert t_{n}u_{t_{n}}(x) \bigr\vert ^{2}\,dx\,dy \\ &\quad- \int_{\mathbb{R}^{2}}f(x,t_{n}u_{t_{n}})t_{n}u_{t_{n}} \,dx\rightarrow -\infty, \end{aligned}$$

which implies that $\mathcal{E}\neq\emptyset$. □

The following lemma shows that $\mathcal{N}\neq\emptyset$.

Lemma 2.6

For any$u\in\mathcal{E}$, there exists unique$t>0$such that$tu\in \mathcal{N}$.

Proof

Given $u\in\mathcal{E}$, let $\gamma_{u}(t):=\langle I'(tu),tu\rangle$ for $t>0$. Then $tu\in\mathcal{N}$ if and only if $\gamma_{u}(t)=0$. Taking $\varepsilon>0$ sufficiently small, jointly with Sobolev embedding, we obtain

$$\begin{aligned} \gamma_{u}(t)\geq{}&t^{2} \Vert u \Vert ^{2}-t^{4}I_{2}(u)- \int_{\mathbb {R}^{2}}f(x,tu)tu\,dx \\ \geq{}&t^{2} \Vert u \Vert ^{2}-t^{4}C_{1} \Vert u \Vert ^{\frac{3}{2}}-t^{2}\varepsilon C_{2} \Vert u \Vert ^{2}-t^{p}C(\varepsilon) \int_{\mathbb{R}^{2}} \vert u \vert ^{p}\bigl[\exp \bigl( \alpha \vert tu \vert ^{2}\bigr)-1\bigr]\,dx \\ \geq{}& t^{2}(1-\varepsilon C_{2}) \Vert u \Vert ^{2}-t^{4}C_{1} \Vert u \Vert ^{\frac {3}{2}} \\ &-t^{p}C(\varepsilon) \biggl( \int_{\mathbb{R}^{2}} \vert u \vert ^{sp}\,dx \biggr)^{\frac{1}{s}} \biggl( \int_{\mathbb{R}^{2}}\biggl[\exp\biggl(\alpha s' \Vert tu \Vert ^{2}\biggl(\frac{u}{ \Vert u \Vert }\biggr)^{2}\biggr)-1 \biggr]\,dx \biggr)^{\frac{1}{s'}}. \end{aligned}$$

Choosing $t>0$ small such that $\alpha s'\|tu\|^{2}<4\pi$, it follows from Proposition 2.2 that there exists $\bar{t}>0$ small enough such that

$$ \gamma_{u}(t)>0 \quad\textrm{for all } 0< t< \bar{t}. $$

(2.14)

Now, by ($f_{4}$), one has

$$ f(x,t\tau)t\tau\geq f(x,\tau)\tau t^{4}-\theta V(x) \bigl(t^{2}-1\bigr) (t\tau )^{2},\quad \forall x\in \mathbb{R}^{2}, t\geq1, \tau\in\mathbb{R}, $$

(2.15)

which implies that

$$ \int_{\mathbb{R}^{2}}\bigl[\theta V(x) (tu)^{2}-f(x,tu)tu \bigr]\,dx\leq t^{4} \int_{\mathbb{R}^{2}}\bigl[\theta V(x)u^{2}-f(x,u)u\bigr]\,dx,\quad \forall t\geq1. $$

(2.16)

Therefore,

$$ \begin{aligned}[b] \gamma_{u}(t)&=t^{2} \Vert u \Vert ^{2}+t^{4}I_{0}(u)- \int_{\mathbb {R}^{2}}f(x,tu)tu\,dx \\ & \leq t^{2} \Vert u \Vert ^{2}+t^{4} \biggl[ \int_{\mathbb {R}^{2}}\bigl[V(x)u^{2}-f(x,u)u\bigr] \,dx+I_{0}(u) \biggr] \\ &\quad-\theta t^{2} \int_{\mathbb{R}^{2}} V(x)u^{2}\,dx,\quad \forall t\geq1. \end{aligned} $$

(2.17)

Thus, we have $\gamma_{u}(t)\rightarrow-\infty$, as $t\rightarrow\infty$. So there exists $t_{0}>0$ such that $\gamma_{u}(t_{0})=0$. Next, we shall prove that $t_{0}$ is unique. Suppose to the contrary that there are $t_{1}, t_{2}>0$ with $t_{1}\neq t_{2}$ such that $\gamma _{u}(t_{1})=\gamma_{u}(t_{2})=0$. For $t_{1}u\in E$, using Lemma 2.4, for all $t>0$, we have

$$ I(t_{1}u)\geq I(tt_{1}u)+\frac{(1-\theta)(1-t^{2})^{2}t^{2}_{1} \Vert u \Vert ^{2}}{4}. $$

(2.18)

Taking $t=\frac{t_{2}}{t_{1}}$, it yields that

$$ I(t_{1}u)\geq I(t_{2}u)+\frac{(1-\theta)(1-(\frac {t_{2}}{t_{1}})^{2})^{2}t^{2}_{1} \Vert u \Vert ^{2}}{4}. $$

(2.19)

Similarly, one has

$$ I(t_{2}u)\geq I(t_{1}u)+\frac{(1-\theta)(1-(\frac {t_{1}}{t_{2}})^{2})^{2}t^{2}_{2} \Vert u \Vert ^{2}}{4}. $$

(2.20)

We obtain $t_{1}=t_{2}$, so it is absurd. □

Since $u\in\mathcal{N}$, by Lemma 2.4, one has

$$ I(u)=I(u)-\frac{1}{4}\bigl\langle I'(u),u\bigr\rangle \geq\frac{1-\theta}{4} \Vert u \Vert ^{2}. $$

(2.21)

So we can define

$$ m:=\inf_{u\in\mathcal{N}}I(u). $$

(2.22)

Up to this stage, preparations have been made. We point out that we can define m without using the condition $\alpha\in(0,\frac{\pi(1-\theta )}{m})$. In the next section, taking full advantage of the condition $\alpha\in(0,\frac{\pi(1-\theta)}{m})$, we shall prove the existence of ground state solutions of (1.7).

3 Existence of ground states

In this section, with the additional condition $\alpha\in(0,\frac{\pi (1-\theta)}{m})$, we are devoted to showing that m is achieved and the minimizer is a ground state solution of equation (1.7).

Lemma 3.1

There exists$C>0$such that$\|u\|\geq C$for all$u\in\mathcal{N}$; furthermore, $m>0$.

Proof

Assume by contradiction that there is $\{u_{n}\} \subset\mathcal{N}$ such that $\|u_{n}\|\rightarrow0$. Obviously,

$$ \Vert u_{n} \Vert ^{2}+4\bigl\langle I'_{1}(u_{n}),u_{n}\bigr\rangle =4 \bigl\langle I'_{2}(u_{n}),u_{n}\bigr\rangle + \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx. $$

In view of $(f_{1})-(f_{3})$, combining Hölder’s inequality, it follows that

$$\begin{aligned} & \biggl\vert \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx \biggr\vert \\ &\quad\leq\varepsilon \int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{2} \,dx+C(\varepsilon) \int _{\mathbb{R}^{2}} \vert u_{n} \vert ^{p} \bigl[\exp\bigl(\alpha \vert u_{n} \vert ^{2}\bigr)-1 \bigr]\,dx \\ &\quad\leq\varepsilon \int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{2}\,dx\\&\qquad+ C(\varepsilon) \biggl( \int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{sp}\,dx \biggr)^{\frac{1}{s}} \biggl( \int_{\mathbb{R}^{2}}\biggl[\exp\biggl(\alpha s' \Vert u_{n} \Vert ^{2}\biggl(\frac{u_{n}}{ \Vert u_{n} \Vert } \biggr)^{2}\biggr)-1\biggr]\,dx \biggr)^{\frac{1}{s'}}. \end{aligned}$$

With Proposition 2.2 in hand, using the Sobolev embedding, it leads to

$$ \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n} \,dx=o_{n}(1). $$

By direct calculation, it holds that

$$ 4\bigl\langle I'_{2}(u_{n}),u_{n} \bigr\rangle \leq C \Vert u_{n} \Vert ^{4}_{\frac{8}{3}}=o_{n}(1). $$

Thus, one has

$$ \bigl\langle I'_{1}(u_{n}),u_{n} \bigr\rangle =o_{n}(1). $$

Therefore, we obtain

$$ \begin{aligned}[b] \Vert u_{n} \Vert ^{2}\leq{}& 4\bigl\langle I'_{1}(u_{n}),u_{n} \bigr\rangle + \int_{\mathbb {R}^{2}}f(x,u_{n})u_{n}\,dx \\ \leq{}& o_{n}(1)+\varepsilon \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{2} \,dx+C(\varepsilon) \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{p} \bigl[\exp\bigl(\alpha \vert u_{n} \vert ^{2}\bigr)-1 \bigr]\,dx. \end{aligned} $$

(3.1)

That is,

$$ \begin{aligned}[b] &(1-\varepsilon C) \Vert u_{n} \Vert ^{2} \\ &\quad \leq o_{n}(1)+ C(\varepsilon) \biggl( \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{sp}\,dx \biggr)^{\frac{1}{s}} \biggl( \int_{\mathbb{R}^{2}}\biggl[\exp\biggl(\alpha s' \Vert u_{n} \Vert ^{2}\biggl(\frac{u_{n}}{ \Vert u_{n} \Vert } \biggr)^{2}\biggr)-1\biggr]\,dx \biggr)^{\frac{1}{s'}}. \end{aligned} $$

(3.2)

Noting that $\|u_{n}\|\rightarrow0$, using Proposition 2.2 again, we get

$$ (1-\varepsilon C) \Vert u_{n} \Vert ^{2} \leq C(\varepsilon) \Vert u_{n} \Vert ^{p}, $$

(3.3)

which is ridiculous. Combining with (2.21), we have $m>0$. □

Next, we give the following lemma which shall be used later.

Lemma 3.2

For every$u\in E$, it holds that$I_{1}(u)\geq\frac{1}{16}\|u\| ^{2}_{2}\|u\|^{2}_{\ast}$.

Proof

The proof is similar to [5, Lemma 2.2]. Let

$$\varLambda_{1}:=\bigl\{ (x_{1},x_{2})\in \mathbb{R}^{2},x_{1}>0,x_{2}\geq0\bigr\} ,\qquad \varLambda_{3}:=\bigl\{ (x_{1},x_{2})\in \mathbb{R}^{2},x_{1}< 0,x_{2}\leq0\bigr\} . $$

For any $(x,y)\in\varLambda_{1}\times\varLambda_{3}$, it holds that

$$ \vert x-y \vert =\sqrt{ \vert x \vert ^{2}+ \vert y \vert ^{2}-2x\cdot y}\geq\sqrt{ \vert x \vert ^{2}+ \vert y \vert ^{2}}\geq \vert x \vert . $$

Thus,

$$\begin{aligned} I_{1}(u)&= \int_{\mathbb{R}^{2}} \int_{\mathbb{R}^{2}}\ln \bigl(1+ \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy\,dx \\ & \geq \int_{\varLambda_{3}} \int_{\varLambda_{1}}\ln \bigl(1+ \vert x-y \vert \bigr) \bigl\vert u(y) \bigr\vert ^{2} \bigl\vert u(x) \bigr\vert ^{2}\,dy \,dx \\ &\geq \int_{\varLambda_{3}} \bigl\vert u(y) \bigr\vert ^{2}\,dy \int_{\varLambda_{1}}\ln \bigl(1+ \vert x \vert \bigr) \bigl\vert u(x) \bigr\vert ^{2}\,dx \\ &=\frac{1}{16} \Vert u \Vert ^{2}_{2} \Vert u \Vert ^{2}_{\ast}. \end{aligned}$$

 □

Let $\{u_{n}\}\subset\mathcal{N}$ be a minimizing sequence of m. On the additional condition $\alpha\in(0,\frac{\pi(1-\theta)}{m})$, we want to prove that $\{u_{n}\}$ is bounded in E.

Lemma 3.3

If$\alpha\in(0,\frac{\pi(1-\theta)}{m})$, we have$\{u_{n}\}$is bounded inE.

Proof

Similar to (2.21), $\{\|u_{n}\|\}$ is bounded. Similar to (2.5), $\{I_{2}(u_{n})\}$ is bounded. Next, we want to estimate the $\{I_{1}(u_{n})\}$. Note that

$$ \biggl\vert \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx \biggr\vert \leq\varepsilon \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{2} \,dx+C(\varepsilon) \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{p} \bigl[\exp\bigl(\alpha \vert u_{n} \vert ^{2}\bigr)-1 \bigr]\,dx. $$

(3.4)

For the second term on the right, using Hölder’s inequality with $s'>1$ and $s'\approx1$, it holds that

$$\begin{aligned} & \int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{p} \bigl[\exp\bigl(\alpha \vert u_{n} \vert ^{2}\bigr)-1 \bigr]\,dx \\ &\quad\leq \biggl( \int_{\mathbb{R}^{2}} \vert u_{n} \vert ^{sp}\,dx \biggr)^{\frac{1}{s}} \biggl( \int_{\mathbb{R}^{2}}\biggl[\exp\biggl(\alpha s' \Vert u_{n} \Vert ^{2}\biggl(\frac{u_{n}}{ \Vert u_{n} \Vert } \biggr)^{2}\biggr)-1\biggr]\,dx \biggr)^{\frac{1}{s'}}. \end{aligned}$$

Taking into account $\alpha\in(0,\frac{\pi(1-\theta)}{m})$, jointly with

$$ \frac{1-\theta}{4} \Vert u_{n} \Vert ^{2} \leq I(u_{n})\rightarrow m, $$

(3.5)

for n large enough, we obtain $\alpha s'\|u_{n}\|^{2}<4\pi$. So, by Proposition 2.2, we get

$$ \biggl\vert \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx \biggr\vert \leq\varepsilon \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{2} \,dx+C(\varepsilon)C \biggl( \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{sp}\,dx \biggr)^{\frac{1}{s}}. $$

(3.6)

Since

$$ \Vert u_{n} \Vert ^{2}+I_{1}(u_{n})=I_{2}(u_{n})+ \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx, $$

(3.7)

which yields that $\{I_{1}(u_{n})\}$ is bounded. And it follows from Lemma 3.2 that $\{u_{n}\}$ is bounded in E. □

Next, we claim that there are $R,\eta>0$ such that

$$ \liminf_{n\rightarrow\infty} \int_{B_{R}(y_{n})} \vert u_{n} \vert ^{2}\,dx \geq\eta. $$

(3.8)

If it is false, using Lions’ lemma (see [18, Lemma 1.21]), we get $u_{n}\rightarrow0$ in $L^{t}(\mathbb{R}^{2})$ for all $t\in[2,\infty )$. Noting that

$$ \bigl\vert I_{1}(u_{n}) \bigr\vert \leq2 \Vert u_{u} \Vert ^{2}_{2} \Vert u_{n} \Vert ^{2}_{\ast}=o_{n}(1),\qquad \bigl\vert I_{2}(u_{n}) \bigr\vert \leq C \Vert u_{n} \Vert ^{4}_{\frac{8}{3}}=o_{n}(1), $$

(3.9)

similar to (3.5), it holds that

$$ \begin{aligned}[b] \Vert u_{n} \Vert ^{2}={}&o_{n}(1)+ \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx \\ \leq{}&o_{n}(1)+\varepsilon \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{2} \,dx+C(\varepsilon)C \biggl( \int_{\mathbb {R}^{2}} \vert u_{n} \vert ^{sp}\,dx \biggr)^{\frac{1}{s}} \\ ={}&o_{n}(1), \end{aligned} $$

(3.10)

which contradicts Lemma 3.1.

Lemma 3.4

mis achieved and the minimizer is a weak solution of (1.7).

Proof

Now, we can assume that $u_{n}\rightharpoonup u_{0}\neq0$ in E, $u_{n}\rightarrow u_{0}$ in $L^{t}(\mathbb{R}^{2})$ for all $t\in [2,\infty)$ and $u_{n}(x)\rightarrow u_{0}(x)$ a.e. in $\mathbb {R}^{2}$. By a standard argument, one can deduce that $I'(u_{0})=0$. Obviously, we have

$$\begin{aligned}& \int_{\mathbb{R}^{2}}F(x,u_{n})\,dx= \int_{\mathbb{R}^{2}}F(x,u_{0})\,dx+o_{n}(1), \end{aligned}$$

(3.11)

$$\begin{aligned}& \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx= \int_{\mathbb {R}^{2}}f(x,u_{0})u_{0} \,dx+o_{n}(1). \end{aligned}$$

(3.12)

Here, we only check 3.12 since (3.11) is similar. We have already known that

$$ \bigl\vert f(x,u_{n})u_{n} \bigr\vert \leq\varepsilon \vert u_{n} \vert ^{2}+C(\varepsilon ) \vert u_{n} \vert ^{p}\biggl[\exp\biggl(\alpha \Vert u_{n} \Vert ^{2}\biggl(\frac{u_{n}}{ \Vert u_{n} \Vert } \biggr)^{2}\biggr)-1\biggr]. $$

(3.13)

Noting that $\alpha\in(0,\frac{\pi(1-\theta)}{m})$ and (3.5), we obtain that $\alpha\|u_{n}\|^{2}<4\pi$ for n large enough. By Proposition 2.2, there exists $C>0$ independent of n such that

$$ \int_{\mathbb{R}^{2}}\biggl[\exp\biggl(\alpha \Vert u_{n} \Vert ^{2}\biggl(\frac{u_{n}}{ \Vert u_{n} \Vert }\biggr)^{2}\biggr)-1 \biggr]\,dx\leq C. $$

It follows from [18, Lemma A.1] and Lebesgue dominated convergence theorem that

$$ \lim_{n\rightarrow\infty} \int_{\mathbb{R}^{2}}f(x,u_{n})u_{n}\,dx= \int _{\mathbb{R}^{2}}f(x,u_{0})u_{0}\,dx. $$

(3.14)

Thus, we have

$$ \begin{aligned} m={}&\lim_{n\rightarrow\infty} \biggl[I(u_{n})- \frac{1}{4}\bigl\langle I'(u_{n}),u_{n} \bigr\rangle \biggr] \\ \geq{}&\frac{1}{4} \Vert u_{0} \Vert ^{2}+ \int_{\mathbb{R}^{2}} \biggl[\frac {1}{4}f(x,u_{0})-F(x,u_{0}) \biggr]\,dx \\ ={}&I(u_{0})-\frac{1}{4}\bigl\langle I'(u_{0}),u_{0} \bigr\rangle \\ \geq{}& m. \end{aligned} $$

 □