Sign-Changing Solutions to the Schrödinger Equations Coupled with a Neutral Scalar Field

Abstract

Here we consider the problem of the Schrödinger equation coupled with a neutral scalar field. Through the methods of descending flow invariant sets via the Ljusternik–Schnirelman theory, we rigorously demonstrate the existence of infinitely many sign-changing solutions.

1. Introduction and Main Result

In this paper, we investigate the existence of sign-changing solutions to the Schrödinger equation coupled with a neutral scalar field in ${\mathbb{R}}^2$ as follows

$$\left\{\begin{array}{c}\left(i{\partial }_t+{\displaystyle \frac{1}{2m}}\mathrm{\Delta }\right)\psi -{\displaystyle \frac{q}{4m^2}}{|\psi |}^2\psi -\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)N\psi =0,\hfill \\ \left({\partial }_{tt}-\mathrm{\Delta }+{\kappa }^2{q}^2\right)N+q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){|\psi |}^2=0,\hfill \end{array}\right.$$

(1)

where $\psi :{\mathbb{R}}^{1,2}\to \mathbb{C}$ corresponds to a wave function of the underlying particle and $N:{\mathbb{R}}^{1,2}\to \mathbb{R}$ represents a neutral scalar field. The constants $m,\kappa ,q>0$ stand for the mass of the particle, the Chern–Simons coupling constant and the Maxwell coupling constant.

Moreover, problem (1) originates from a non-relativistic Abelian Maxwell–Chern–Simons model proposed in [1], where the neutral scalar field N is naturally embedded within this framework. We begin by systematically deriving the problem (1) through the Lagrangian formulation. Consider the $(2+1)$-dimensional Minkowski spacetime ${\mathbb{R}}^{1,2}$ with metric signature $(1,-1,-1)$. Following the framework established in [1], the Lagrangian density characterizing the non-relativistic Maxwell–Chern–Simons–Higgs model in this spacetime is defined as

$$\begin{array}{cc}\hfill {\mathcal{L}}^R\left(\varphi ,N,A_0,A_1,A_2\right)& =-{\displaystyle \frac{1}{4q}}{F}_{\mu \nu }{F}^{\mu \nu }+{\displaystyle \frac{\kappa }{4}}{\epsilon }^{\mu \nu \alpha }{F}_{\mu \nu }{A}_{\alpha }\hfill \\ \hfill +& \overline{D_{\mu }\varphi }{D}^{\mu }\varphi +{\displaystyle \frac{1}{2q}}{\partial }_{\mu }N{\partial }^{\mu }N-{\displaystyle \frac{1}{c^2}}{|\varphi |}^2{\left(N+{\displaystyle \frac{1}{\kappa c}}{v}^2\right)}^2-{\displaystyle \frac{q}{2c^2}}{\left({|\varphi |}^2+\kappa cN\right)}^2,\hfill \end{array}$$

where Greek letters denote space-time indices $0,1,2$ and Latin letters stand for spatial indices $1,2,\varphi :\mathbb{R}\times {\mathbb{R}}^2\to \mathbb{C}$ is a charged scalar field, $N:\mathbb{R}\times {\mathbb{R}}^2\to \mathbb{R}$ is a neutral scalar field, $A_{\mu }:\mathbb{R}\times {\mathbb{R}}^2\to \mathbb{R}$ are the components of the gauge potential, $F_{\mu \nu }={\partial }_{\nu }{A}_{\mu }-{\partial }_{\mu }{A}_{\nu },q>0$ is the gauge coupling constant, $\kappa >0$ is the Chern–Simons coupling constant, c is the velocity of the light, v is a real constant, ${\epsilon }^{\mu \nu \alpha }$ is the Levi-Civita tensor and $D_{\mu }={\partial }_{\mu }+{\displaystyle \frac{i}{c}}{A}_{\mu }$. The Lagrangian density ${\mathcal{L}}^R$ can be decomposed into three fundamental components. The first term corresponds to the standard Maxwell Lagrangian density and the second term represents the Chern–Simons contribution. The third component arises from the matter field $\varphi$, which couples to the gauge field through the covariant derivative $D_{\mu }$. This matter sector also includes a neutral scalar field N whose mass is degenerate with the gauge field mass, a characteristic feature of this model. For more details in the physical aspects, we refer the readers to [2,3,4,5,6,7,8,9,10].

We now devote to finding standing waves of problem (1), i.e., solutions of the form

$$\begin{array}{c}\hfill \psi (x,t)=e^{i\omega t}u\left(x\right),N(x,t)=N\left(x\right),\omega \in \mathbb{R},\end{array}$$

then problem (1) becomes

$$\left\{\begin{array}{c}-{\displaystyle \frac{1}{2m}}\mathrm{\Delta }u+\omega u+{\displaystyle \frac{q}{4m^2}}{|u|}^{2}u+\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)Nu=0,\hfill \\ -\mathrm{\Delta }N+{\kappa }^2{q}^{2}N+q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)u^2=0.\hfill \end{array}\right.$$

(2)

The energy functional $\mathfrak{I}:H^1\left({\mathbb{R}}^2\right)\times H^1\left({\mathbb{R}}^2\right)\to \mathbb{R}$ for problem (2) can be defined as

$$\begin{array}{cc}\hfill \mathfrak{I}(u,N)=& {\displaystyle \frac{1}{4m}}{\int }_{{\mathbb{R}}^2}{|\nabla u|}^{2}dx+{\displaystyle \frac{1}{2\omega }}{\int }_{{\mathbb{R}}^2}{u}^{2}dx+{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}N{u}^{2}dx+{\displaystyle \frac{q}{16m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx\hfill \\ \hfill & +{\displaystyle \frac{1}{4q}}{\int }_{{\mathbb{R}}^2}{|\nabla N|}^{2}dx+{\displaystyle \frac{{\kappa }^{2}q}{4}}{\int }_{{\mathbb{R}}^2}{\int }_{{\mathbb{R}}^2}{|N|}^{2}dx.\hfill \end{array}$$

(3)

Evidently, the functional $\mathfrak{I}$ is well defined and has two variables u and N. To avoid the difficulties caused by two variables, we reduce the study of (3) to a functional of the only variable u by using the following lemma.

Lemma 1.

([1]). For any fixed $u\in H^1\left({\mathbb{R}}^2\right)$, there exists a unique $N_u\in H^1\left({\mathbb{R}}^2\right)$, which solves the following equation

$$\begin{array}{c}\hfill \left(-\mathrm{\Delta }+{\kappa }^2{q}^2\right)N+q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){|u|}^2=0.\end{array}$$

(4)

Moreover, the map $\mathcal{N}:u\in H^1\left({\mathbb{R}}^2\right)\mapsto N\left(u\right):=N_u$ is continuously differentiable and

$$N_u=-q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){|u|}^2{\int }_{{\mathbb{R}}^2}{G}^{\kappa q}(x-y)u^2\left(y\right)dy\le 0\mathrm{in}{\mathbb{R}}^2,$$

where the integral kernel $G^{\mu }\left(x\right)$ is the Yukawa potential defined by

$$\begin{array}{c}\hfill G^{\mu }\left(x\right)={\int }_0^{\infty }{\displaystyle \frac{1}{4\pi t}}{e}^{-\left({\displaystyle \frac{{|x|}^2}{4t}}+{\mu }^{2}t\right)}dt.\end{array}$$

(5)

Also, we have

$$\begin{array}{c}\hfill {\left({\int }_{{\mathbb{R}}^2}{N}_u^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{{\kappa }^{2}q}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)·{\left({\int }_{{\mathbb{R}}^2}{u}^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}\end{array}$$

(6)

and then

$$\begin{array}{c}\hfill \left|{\int }_{{\mathbb{R}}^2}{N}_u{u}^{2}dx\right|=-{\int }_{{\mathbb{R}}^2}{N}_u{u}^{2}dx\le {\displaystyle \frac{1}{{\kappa }^{2}q}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}{u}^{4}dx.\end{array}$$

(7)

Then, for any fixed $u\in H^1\left({\mathbb{R}}^2\right)$, $N_u$ solves (4). Multiplying (4) by $N_u$ and integrating by parts, it obtains

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}{|\nabla N_u|}^{2}dx+{\kappa }^2{q}^2{\int }_{{\mathbb{R}}^2}{N}_u^{2}dx=-q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}{N}_u{u}^{2}dx.\end{array}$$

(8)

Let

$$I\left(u\right):=\mathfrak{I}(u,N_u).$$

Substituting (8) into (3) yields

$$\begin{array}{c}\hfill I\left(u\right)={\displaystyle \frac{1}{4m}}{\int }_{{\mathbb{R}}^2}{|\nabla u|}^{2}dx+{\displaystyle \frac{\omega }{2}}{\int }_{{\mathbb{R}}^2}{u}^{2}dx+{\displaystyle \frac{q}{16m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx-{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx.\end{array}$$

For any $\phi \in H^1\left({\mathbb{R}}^2\right)$, its derivative functional is

$$\begin{array}{c}\hfill 〈I^{\prime }\left(u\right),\phi 〉={\displaystyle \frac{1}{2m}}{\int }_{{\mathbb{R}}^2}\nabla u\nabla \phi dx+\omega {\int }_{{\mathbb{R}}^2}u\phi dx+{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{|u|}^{2}u\phi dx-\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u\phi dx.\end{array}$$

J. Han et al. in [1] confined the work space of problem (2) to the radially symmetric Sobolev space $H_r^1\left({\mathbb{R}}^2\right)$. They got the existence of standing wave solutions through an application of the Mountain Pass Theorem in their variational framework. By using the Nehari manifold technique with Moser iteration methods, Kang et al. in [11] established the existence of a regular, radially symmetric ground state solution for problem (2) in the non-radial symmetric space $H^1\left({\mathbb{R}}^2\right)$. Cao et al. in [12] replaced $\omega$ in problem (2) with $V\left(x\right)$, where $V\left(x\right)$ satisfies the following conditions:

(V₁)

$V\in \mathcal{C}({\mathbb{R}}^2,\mathbb{R})$ and ${inf}_{x\in {\mathbb{R}}^2}V\left(x\right)>0$.

(V₂)

There are constants $a>0,n>1,\theta >0$ and $V_{\infty }>0$, such that

$$V\left(x\right)=V(|x|)=V_{\infty }+{\displaystyle \frac{a}{r^n}}+O\left({\displaystyle \frac{1}{r^{n+\theta }}}\right)\mathrm{as}r=|x|\to +\infty .$$

By constructing solutions with multiple peaks, ref. [12] obtained that the number of non-radial solutions of problem (2) goes to infinity as the Maxwell coupling constant tends to infinity.

Our main result can be stated.

Theorem 1.

Supposed that $m,\kappa ,q$ and ω are given positive parameters. Then there exist infinitely many sign-changing solutions $(u,N)\in {\cap }_{k=0}^{\infty }\left(H_r^k\left({\mathbb{R}}^2\right)\times H_r^k\left({\mathbb{R}}^2\right)\right)$ for problem (2).

Remark 1.

To the best of our knowledge, there is no literature considering the existence of sign-changing solutions for Schrödinger equation coupled with a neutral scalar field. Due to the presence of non local terms N, the following standard splitting approach for energy functional I doesn’t usually apply in this situation:

$$\begin{array}{c}\hfill I\left(u\right)=I\left(u^{+}\right)+I\left(u^{-}\right),〈I^{\prime }\left(u\right),u^{\pm }〉=〈I^{\prime }\left(u^{\pm }\right),u^{\pm }〉.\end{array}$$

The reason is that the nonlocal term typically take the following form

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\ne {\int }_{{\mathbb{R}}^2}|N_{u^{+}}|{\left(u^{+}\right)}^{2}dx+{\int }_{{\mathbb{R}}^2}|N_{u^{-}}|{\left(u^{-}\right)}^{2}dx.\end{array}$$

We overcome this problem by adapting a technique from [13], which is called the descending flow invariant sets. For more details, we refer the readers to [14,15,16,17,18]. Moreover, we consider this problem in the radial space $H_r^1\left({\mathbb{R}}^2\right)$ to restore compactness. Thus, the infinitely many sign-changing solutions in Theorem 1 are also radially symmetric.

The paper is organized as follows. In Section 2, we provided some concepts and give some useful Lemmas which are crucial to prove our results. We construct invariant subsets of descending flow in Section 3. Section 4 is devoted to the proof of Theorem 1.

In this paper, we utilize the following notations.

• The work space is

  $$H_r^1\left({\mathbb{R}}^2\right):=\left\{u\in L^2\left({\mathbb{R}}^2\right):|\nabla u|\in L^2\left({\mathbb{R}}^2\right),u\mathrm{is}\mathrm{radial}\right\}$$

  endowed with the norm

  $$\langle u,v\rangle ={\displaystyle \frac{1}{2m}}{\int }_{{\mathbb{R}}^2}\nabla u\nabla vdx+\omega {\int }_{{\mathbb{R}}^2}uvdx\mathrm{and}\parallel u\parallel :={\langle u,u\rangle }^{{\displaystyle \frac{1}{2}}}.$$
• $L^p\left({\mathbb{R}}^2\right)$ is the Lebesgue space endowed with the norm ${\parallel u\parallel }_p:={\left({\int }_{{\mathbb{R}}^2}{|u|}^{p}dx\right)}^{{\displaystyle \frac{1}{p}}}.$
• The embedding $H_r^1\left({\mathbb{R}}^2\right)\hookrightarrow L^p\left({\mathbb{R}}^2\right)$ is compact for $p\in (2,+\infty )$.
• For $c\in \mathbb{R}$,

  $$J^c:=\left\{u\in H_r^1\left({\mathbb{R}}^2\right):I\left(u\right)\le c\right\}\mathrm{and}{K}_c:=\left\{u\in H_r^1\left({\mathbb{R}}^2\right):I\left(u\right)=c,I^{\prime }\left(u\right)=0\right\}.$$
• $B_R:=\left\{x\in {\mathbb{R}}^2:|x|\le R\right\}$ and $B_R^c:={\mathbb{R}}^2\setminus B_R$.
• C denotes a positive constant that may be different in different places.

2. Preliminaries

To facilitate the reader’s understanding of the Schrödinger equations coupled with a neutral scalar field, we first provide additional details on the properties of the non-local term $N_u$ in the following lemma.

Lemma 2.

([11]). For every $u\in H^1\left({\mathbb{R}}^2\right)$, the following statements hold.

(i)

$N_{tu}=t^2{N}_u$ for any $t\in \mathbb{R}$ and the map $u\to N_u$ is ${\mathcal{C}}^1(H^1\left({\mathbb{R}}^2\right),\mathbb{R})$;

(ii)

$\parallel N_u{\parallel }_{L^{\infty }}\hspace{0.17em}\le C{\parallel u\parallel }_{L^{\infty }}$ and $\parallel \nabla N_u{\parallel }_2\hspace{0.17em}\le {C\parallel u\parallel }_{L^{\infty }}{\parallel \nabla u\parallel }_2$;

(iii)

$\parallel \nabla G^{\kappa q}{\parallel }_1\hspace{0.17em}\le {\displaystyle \frac{\sqrt{\pi }}{2\kappa q}}\mathrm{\Gamma }\left(2^{-1}\right)$ and $\parallel \nabla N_u{\parallel }_2\hspace{0.17em}\le C{\parallel u\parallel }^2$ for any $p\in [1,+\infty )$;

(iv)

If u is radially symmetric, then $N_u$ is also radially symmetric.

For the convenience of the readers, we summarize the properties of a genus. We refer the readers to [19] for the proof of the next proposition.

Property 1.

Let A and B be closed symmetric subsets of X which do not contain the origin. Then (i)–(v) below hold.

(i)

If there is an odd continuous mapping from A to B, then $\gamma \left(A\right)\le \gamma \left(B\right)$;

(ii)

If there is an odd homeomorphism from A onto B, then $\gamma \left(A\right)=\gamma \left(B\right)$;

(iii)

If $\gamma \left(B\right)<\infty$, then $\gamma \left(\overline{A\setminus B}\right)\ge \gamma \left(A\right)-\gamma \left(B\right)$;

(iv)

If A is compact, then $\gamma \left(A\right)<\infty$ and $\gamma \left(N_{\delta }\left(A\right)\right)=\gamma \left(A\right)$ for $\delta >0$ small enough;

(v)

The n-dimensional sphere $S_n$ has genus equal to $n+1$.

Next we give a technical lemma to make our proof smoothly in Section 3.

Lemma 3.

If $u_n\to u_0$ in $H_r^1\left({\mathbb{R}}^2\right)$, we can get that $N_{u_n}\to N_{u_0}$ in $H_r^1\left({\mathbb{R}}^2\right)$ and then $N_{u_n}\to N_{u_0}$ in $L^p\left({\mathbb{R}}^2\right)$ for $2<p<+\infty$.

Proof. 

From $u_n\to u_0$, we know that

$$\begin{array}{c}\hfill \left\{u_n\right\}\mathrm{is}\mathrm{bounded}\mathrm{in}{H}_r^1\left({\mathbb{R}}^2\right)\end{array}$$

(9)

and

$$u_n\to u_0\mathrm{in}{L}^p\left({\mathbb{R}}^2\right)\mathrm{for}2<p<+\infty ,u_n\to u_0\mathrm{in}{L}_{loc}^p\left({\mathbb{R}}^2\right)\mathrm{for}1<p<+\infty .$$

It follows from (6)–(9) that $\left\{N_{u_n}\right\}$ is bounded in $H_r^1\left({\mathbb{R}}^2\right)$. Then there exists $N_0\in H_r^1\left({\mathbb{R}}^2\right)$ such that

• $N_{u_n}\rightharpoonup N_0$ in $H_r^1\left({\mathbb{R}}^2\right)$,
• $N_{u_n}\to N_0$ in $L^p\left({\mathbb{R}}^2\right)$ for $2<p<+\infty$.

Next we want to claim $N_0=N_{u_0}$. Let $\phi \in {\mathcal{C}}_0^{\infty }\left({\mathbb{R}}^2\right)$ be a test function. By $N_{u_n}\rightharpoonup N_0$ in $H_r^1\left({\mathbb{R}}^2\right)$ and $u_n\to u_0\mathrm{in}{L}_{loc}^p\left({\mathbb{R}}^2\right)\mathrm{for}1<p<+\infty$, as $n\to \infty$, one sees

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}\nabla N_{u_n}\nabla \phi dx+{\kappa }^2{q}^2{\int }_{{\mathbb{R}}^2}{N}_{u_n}\phi dx\to {\int }_{{\mathbb{R}}^2}\nabla N_0\nabla \phi dx+{\kappa }^2{q}^2{\int }_{{\mathbb{R}}^2}{N}_0\phi dx\end{array}$$

(10)

and

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}|u_n{|}^2\phi dx\to {\int }_{{\mathbb{R}}^2}{|u_0|}^2\phi dx.\end{array}$$

(11)

From (10) and (11), it obtains

$${\int }_{{\mathbb{R}}^2}\nabla N_0\nabla \phi dx+{\kappa }^2{q}^2{\int }_{{\mathbb{R}}^2}{N}_0\phi dx=-q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}{|u_0|}^2\phi dx,$$

which deduces that

$$(-\mathrm{\Delta }+{\kappa }^2{q}^2)N_0+q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){|u_0|}^2=0.$$

Combining with

$$(-\mathrm{\Delta }+{\kappa }^2{q}^2)N_{u_0}+q\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){|u_0|}^2=0,$$

by using the Lax–Milgram Lemma, we can get that $N_0=N_{u_0}$. From $u_n\to u_0$ and $N_{u_n}\to N_0$ in $L^p\left({\mathbb{R}}^2\right)$ for $2<p<+\infty$, it has

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^2}{N}_{u_n}{u}_n^{2}dx-{\int }_{{\mathbb{R}}^2}{N}_{u_0}{u}_0^{2}dx\hfill \\ \hfill =& {\int }_{{\mathbb{R}}^2}\left(N_{u_n}-N_{u_0}\right)u_n^{2}dx+{\int }_{{\mathbb{R}}^2}{N}_{u_0}\left(u_n^2-u_0^2\right)dx\hfill \\ \hfill \le & {\left({\int }_{{\mathbb{R}}^2}{\left(N_{u_n}-N_{u_0}\right)}^{3}dx\right)}^{{\displaystyle \frac{1}{3}}}{\left({\int }_{{\mathbb{R}}^2}{u}_n^{3}dx\right)}^{{\displaystyle \frac{2}{3}}}\hfill \\ \hfill & +{\left({\int }_{{\mathbb{R}}^2}{N}_{u_0}^{3}dx\right)}^{{\displaystyle \frac{1}{3}}}{\left({\int }_{{\mathbb{R}}^2}{\left(u_n+u_0\right)}^{3}dx\right)}^{{\displaystyle \frac{1}{3}}}{\left({\int }_{{\mathbb{R}}^2}{\left(u_n-u_0\right)}^{3}dx\right)}^{{\displaystyle \frac{1}{3}}}\to 0.\hfill \end{array}$$

(12)

Together with (8), we can get that $N_{u_n}\to N_{u_0}$ in $H_r^1\left({\mathbb{R}}^2\right)$. □

The following lemma plays a crucial role in the proof of Theorem 1.

Lemma 4.

The functional I satisfies the $\left(PS\right)$ condition.

Proof. 

Suppose that there exist a constant c and a sequence $\left\{u_n\right\}\subset H_r^1\left({\mathbb{R}}^2\right)$ such that

$$I\left(u_n\right)\to c\mathrm{and}{I}^{\prime }\left(u_n\right)\to 0.$$

Then

$$\begin{array}{c}\hfill c+o\left(1\right)\ge I\left(u_n\right)-{\displaystyle \frac{1}{4}}\langle I^{\prime }\left(u_n\right),u_n\rangle ={\displaystyle \frac{1}{4}}{\parallel u_n\parallel }^2,\end{array}$$

(13)

which deduces that

$$\begin{array}{c}\hfill \left\{u_n\right\}\mathrm{is}\mathrm{bounded}\mathrm{in}{H}_r^1\left({\mathbb{R}}^2\right).\end{array}$$

(14)

Hence, there exists $u\in H_r^1\left({\mathbb{R}}^2\right)$ such that, up to a subsequence, as $n\to \infty$,

• $u_n\rightharpoonup u$ in $H_r^1\left({\mathbb{R}}^2\right)$,
• $u_n\to u$ in $L^p\left({\mathbb{R}}^2\right)$ for $2<p<+\infty$,
• $u_n\left(x\right)\to u\left(x\right)$ a.e. in ${\mathbb{R}}^2$.

Obviously, it sees

$$\begin{array}{cc}\hfill o\left(1\right)=〈I^{\prime }\left(u_n\right)-I^{\prime }\left(u\right),u_n-u〉=& \parallel u_n{-u\parallel }^2-\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|u_n-|N_u|u\right)(u_n-u)dx\hfill \\ \hfill & +{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}\left(|u_n{|}^2{u}_n-{|u|}^{2}u\right)(u_n-u)dx.\hfill \end{array}$$

(15)

From (14), we know that $\left\{N_{u_n}\right\}$ is bounded in $H_r^1\left({\mathbb{R}}^2\right)$. Then, up to a subsequence, there exists $N_u\in H_r^1\left({\mathbb{R}}^2\right)$ such that $N_{u_n}\rightharpoonup N_u$ in $H_r^1\left({\mathbb{R}}^2\right)$. Together with $u_n\to u$ in $L^p\left({\mathbb{R}}^2\right)$, we have

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|u_n-|N_u|u\right)(u_n-u)dx\hfill \\ \hfill =& {\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|-|N_u|\right)u_n(u_n-u)dx+{\int }_{{\mathbb{R}}^2}|N_u|{(u_n-u)}^{2}dx\hfill \\ \hfill \le & {\left({\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|-|N_u|\right)dx\right)}^{{\displaystyle \frac{1}{2}}}\parallel u_n{\parallel }_4·\parallel u_n{-u\parallel }_4+\parallel N_u{\parallel }_2·{\parallel u_n-u\parallel }_4^2\to 0\hfill \end{array}$$

(16)

and

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^2}\left(|u_n{|}^2{u}_n-{|u|}^{2}u\right)(u_n-u)dx\hfill \\ \hfill \le & {\int }_{{\mathbb{R}}^2}\left(|u_n{|}^2-{|u|}^2\right)u_n(u_n-u)dx+{\int }_{{\mathbb{R}}^2}{|u|}^2{(u_n-u)}^{2}dx\hfill \\ \hfill \le & {\left({\int }_{{\mathbb{R}}^2}{\left(|u_n{|}^2-{|u|}^2\right)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\parallel u_n{\parallel }_4·\parallel u_n{-u\parallel }_4+{\parallel u\parallel }_4^2·{\parallel u_n-u\parallel }_4^2\to 0.\hfill \end{array}$$

(17)

Combining (15)–(17), we see that $u_n\to u$ in $H_r^1\left({\mathbb{R}}^2\right)$. □

3. Invariant Subsets of Descending Flow

In order to find sign-changing solutions for problem (2), we present an operator A as follows: for each $u\in H_r^1\left({\mathbb{R}}^2\right)$, we see that

$$\begin{array}{c}\hfill -{\displaystyle \frac{1}{2m}}\mathrm{\Delta }v+\omega v+{\displaystyle \frac{q}{4m^2}}{|u|}^{2}v=\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)|N_u|u\end{array}$$

(18)

has a unique weak solution $v:=A\left(u\right)\in H_r^1\left({\mathbb{R}}^2\right)$. Obviously, the fixed points of A is the same as the critical points of problem (2).

Lemma 5.

The operator A is well defined, continuous, compact and odd.

Proof. 

Firstly, we claim that A is well defined. For $u\in H_r^1\left({\mathbb{R}}^2\right)$, we define

$$\begin{array}{c}\hfill J_0\left(v\right)={\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}{|\nabla v|}^2+\omega v^2\right)dx+{\displaystyle \frac{q}{8m^2}}{\int }_{{\mathbb{R}}^2}{u}^2{v}^{2}dx-\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|uvdx.\end{array}$$

(19)

From (6), Hölder and Sobolev inequalities, we have

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}|N_u|uvdx\le & {\left({\int }_{{\mathbb{R}}^2}{|N_u|}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{u}^2{v}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & {\displaystyle \frac{1}{{\kappa }^{2}q}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\parallel u\parallel }_4^2·{\parallel u\parallel }_4·{\parallel v\parallel }_4\le C\parallel v\parallel .\hfill \end{array}$$

Then (19) becomes

$$\begin{array}{cc}\hfill J_0\left(v\right)\ge & {\displaystyle \frac{1}{2}}{\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}{|\nabla v|}^2+\omega v^2\right)dx-\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|uvdx\hfill \\ \hfill \ge & {\displaystyle \frac{1}{2}}{\parallel v\parallel }^2-C\parallel v\parallel ,\hfill \end{array}$$

which implies $J_0$ is coercive. Moreover, $J_0$ is strictly convex. Indeed,

$$\begin{array}{cc}\hfill 〈J_0^{\prime }\left(v\right)-J_0^{\prime }\left(w\right),v-w〉={\int }_{{\mathbb{R}}^2}& \left({\displaystyle \frac{1}{2m}}{|\nabla (v-w)|}^2+\omega {(v-w)}^2\right)dx\hfill \\ \hfill & +{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^2{(v-w)}^{2}dx\ge {\parallel v-w\parallel }^2.\hfill \end{array}$$

Then by using Theorems 1.5.6 and 1.5.8 in [20], $J_0$ has a unique minimizer $v=A\left(u\right)\in H_r^1\left({\mathbb{R}}^2\right)$, which is the unique solution of (18). Hence, the operator A is well defined.

From (18), Hölder and Sobolev inequalities, one sees

$$\begin{array}{cc}\hfill {\displaystyle \frac{1}{2m}}{\parallel \nabla v\parallel }_2^2+\omega {\parallel v\parallel }_2^2+{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^2{v}^{2}dx=& \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|uvdx\hfill \\ \hfill \le & C\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)\parallel N_u{\parallel }_2·{\parallel u\parallel }_4·\parallel v\parallel ,\hfill \end{array}$$

which deduces that

$$\begin{array}{c}\hfill \parallel v\parallel \le C\parallel N_u{\parallel }_2·{\parallel u\parallel }_4.\end{array}$$

Then $J_0$ is bounded below and maps bounded sets into bounded sets.

Secondly, we claim that A is continuous. Let $\left\{u_n\right\}\subset H_r^1\left({\mathbb{R}}^2\right)$ such that $u_n\to u$ in $H_r^1\left({\mathbb{R}}^2\right)$. Define $v:=A\left(u\right)$ and $v_n:=A\left(u_n\right)$. Next we show that $v_n\to v$ in $H_r^1\left({\mathbb{R}}^2\right)$. Note that

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}{|\nabla (v_n-v)|}^2+\omega {(v_n-v)}^2+{\displaystyle \frac{q}{4m^2}}\left(u_n^2{v}_n-u^{2}v\right)(v_n-v)\right)& dx\hfill \\ \hfill ={\int }_{{\mathbb{R}}^2}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)\left(|N_{u_n}|u_n-|N_u|u\right)(v_n\hspace{0.17em}-& v)dx\hfill \end{array}$$

(20)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}\left(u_n^2{v}_n-u^{2}v\right)\left(v_n-v\right)dx={\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}\left(u_n^2-u^2\right)v_n(v_n-v)+u^2{(v_n-v)}^{2}dx.\end{array}$$

(21)

Since $u_n\to u$ in $H_r^1\left({\mathbb{R}}^2\right)$, we get

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}\left(u_n^2-u^2\right)v_n(v_n-v)dx\le & {\left({\int }_{{\mathbb{R}}^2}{(u_n+u)}^2{(u_n-u)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{v}_n^2{(v_n-v)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & \parallel u_n{+u\parallel }_4·\parallel v_n{\parallel }_4·\parallel v_n{-v\parallel }_4·C\parallel u_n-u\parallel \to 0.\hfill \end{array}$$

(22)

Combining with (21), it obtains

$$\begin{array}{c}\hfill {\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}\left(u_n^2{v}_n-u^{2}v\right)\left(v_n-v\right)dx\ge 0,\end{array}$$

together with (20), which deduces

$$\begin{array}{cc}\hfill \parallel v_n{-v\parallel }^2\le & \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|u_n-|N_u|u\right)(v_n-v)dx\hfill \\ \hfill \le & \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\left({\int }_{{\mathbb{R}}^2}{\left(|N_{u_n}|u_n-|N_u|u\right)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{(v_n-v)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & C\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\left({\int }_{{\mathbb{R}}^2}{\left(|N_{u_n}|u_n-|N_u|u\right)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\parallel v_n-v\parallel .\hfill \end{array}$$

(23)

Note that

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}{\left(|N_{u_n}|u_n-|N_u|u\right)}^{2}dx=& {\int }_{{\mathbb{R}}^2}({\left(|N_{u_n}|u_n-|N_u|u_n\right)}^2+{\left(|N_u|u_n-|N_u|u\right)}^2\hfill \\ \hfill & +2\left(|N_{u_n}|u_n-|N_u|u_n\right)\left(|N_u|u_n-|N_u|u\right))dx.\hfill \end{array}$$

(24)

By Lemma 3, Hölder and Sobolev inequalities, one obtains

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}{\left(|N_{u_n}|u_n-|N_u|u_n\right)}^{2}dx\le & {\left({\int }_{{\mathbb{R}}^2}{\left(|N_{u_n}|-|N_u|\right)}^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{u}_n^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & C{\parallel |N_{u_n}|-|N_u|\parallel }^2·{\parallel u_n\parallel }_4^2\to 0,\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}{\left(|N_u|u_n-|N_u|u\right)}^{2}dx\le & {\left({\int }_{{\mathbb{R}}^2}{|N_u|}^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}·{\parallel u_n-u\parallel }_4^2\hfill \\ \hfill \le & {\left({\int }_{{\mathbb{R}}^2}{|N_u|}^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}·C{\parallel u_n-u\parallel }^2\to 0\hfill \end{array}$$

(26)

and

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}& \left(|N_{u_n}|u_n-|N_u|u_n\right)\left(|N_u|u_n-|N_u|u\right)dx\hfill \\ \hfill \le & {\left({\int }_{{\mathbb{R}}^2}{\left(|N_u|-|N_u|\right)}^2{u}_n^2\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{|N_u|}^2{\left(u_n-u\right)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & {\parallel |N_{u_n}|-|N_u|\parallel }_4·\parallel u_n{\parallel }_4·\parallel N_u{\parallel }_4·C\parallel u_n-u\parallel \to 0.\hfill \end{array}$$

(27)

From (24)–(27), we know

$${\int }_{{\mathbb{R}}^2}{\left(|N_u|u_n-|N_u|u\right)}^{2}dx\to 0,$$

together with (23), which implies $\parallel v_n-v\parallel \to 0$. That is, $v_n\to v$ in $H_r^1\left({\mathbb{R}}^2\right)$.

Thirdly, we claim that A is compact. Assume that $\left\{u_n\right\}$ is a bounded sequence. Let $v_n=A\left(u_n\right)$ and then $\left\{v_n\right\}$ is a bounded sequence. Therefore, there exist $u,v\in H_r^1\left({\mathbb{R}}^2\right)$ such that $u_n\rightharpoonup u$ in $H_r^1\left({\mathbb{R}}^2\right)$ and $v_n\rightharpoonup v$ in $H_r^1\left({\mathbb{R}}^2\right)$. Then $u_n\to u\mathrm{in}{L}^p\left({\mathbb{R}}^2\right)\mathrm{and}{u}_n\to u\mathrm{in}{L}^q\left({\mathbb{R}}^2\right)\mathrm{for}2<p<\infty .$ Therefore,

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla v_n\nabla \phi +\omega v_n\phi \right)dx\to {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla v\nabla \phi +\omega v\phi \right)dx.\end{array}$$

(28)

Moreover,

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}{u}_n^2{v}_n\phi dx-{\int }_{{\mathbb{R}}^2}uv\phi dx=& {\int }_{{\mathbb{R}}^2}(u_n^2-u^2)v_n\phi dx+{\int }_{{\mathbb{R}}^2}{u}^2(v_n-v)\phi dx\hfill \\ \hfill \le & {\left({\int }_{{\mathbb{R}}^2}{(u_n^2-u^2)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{u}^2{\phi }^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\left({\int }_{{\mathbb{R}}^2}{u}^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{(v_n-v)}^2{\phi }^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & \parallel u_n{+u\parallel }_4\parallel u_n{-u\parallel }_4\parallel v_n{\parallel }_4{\parallel \phi \parallel }_4+{\parallel u\parallel }_4^2\parallel v_n{-v\parallel }_4{\parallel \phi \parallel }_4\to 0\hfill \end{array}$$

(29)

and

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}|N_{u_n}|u_n\phi dx-{\int }_{{\mathbb{R}}^2}|N_u|u\phi dx=& {\int }_{{\mathbb{R}}^2}|N_{u_n}|(u_n-u)\phi dx+{\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|-|N_u|\right)u\phi dx\hfill \\ \hfill \le & \parallel N_{u_n}{\parallel }_4^2\parallel u_n{-u\parallel }_4{\parallel \phi \parallel }_4+\parallel |N_{u_n}|-|N_u{|\parallel }_4^2·{\parallel u\parallel }_4·{\parallel \phi \parallel }_4\to 0.\hfill \end{array}$$

(30)

For any $\phi \in H_r^1\left({\mathbb{R}}^2\right)$, it gets

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla v_n\nabla \phi +\omega v_n\phi +{\displaystyle \frac{q}{4m}}{u}_n^2{v}_n\phi \right)dx=\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u_n\phi dx.\end{array}$$

(31)

From (28)–(31), we obtains

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla v\nabla \phi +\omega v\phi +{\displaystyle \frac{q}{4m^2}}{u}^{2}v\phi \right)dx=\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u\phi dx,\end{array}$$

which deduces $v=A\left(u\right)$. Thus,

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}{\left|\nabla (v_n-v)\right|}^2+\omega {(v_n-v)}^2+{\displaystyle \frac{q}{4m^2}}(u_n^2{v}_n-u^{2}v)(v_n-v)\right)& dx\hfill \\ \hfill =\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}\left(|N_{u_n}|u_n-|N_u|u\right)& (v_n-v)dx.\hfill \end{array}$$

Similar to the proof above, we see that $\parallel v_n-v\parallel \to 0$, that is, $A\left(u_n\right)\to A\left(u\right)$ in $H_r^1\left({\mathbb{R}}^2\right)$ as $n\to \infty$.

At last, by using Lemma 1, we can easily get that $A(-u)=-A\left(u\right)$. That is, A is odd. □

The following lemma is crucial for the construction of the descending flow.

Lemma 6.

(i) 

$\langle I^{\prime }\left(u\right),u-A\left(u\right)\rangle \ge {\parallel u-A\left(u\right)\parallel }^2$ for all $u\in H_r^1\left({\mathbb{R}}^2\right)$;

(ii) 

$\parallel I^{\prime }\left(u\right)\parallel \hspace{0.17em}\le \hspace{0.17em}\parallel u-A\left(u\right)\parallel ·\left(1+{C\parallel u\parallel }^2\right)$ for all $u\in H_r^1\left({\mathbb{R}}^2\right)$.

Proof. 

(i) From (18) and $v=A\left(u\right)$, one sees

$$\begin{array}{c}\hfill -{\displaystyle \frac{1}{2m}}\mathrm{\Delta }A\left(u\right)+\omega A\left(u\right)+{\displaystyle \frac{q}{4m^2}}{|u|}^{2}A\left(u\right)=\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)|N_u|u.\end{array}$$

(32)

Multiplying (32) by u and integrating by parts, it gets

$$\begin{array}{c}\hfill {\int }_{{\mathbb{R}}^2}{\displaystyle \frac{1}{2m}}\nabla A\left(u\right)\nabla u+\omega A\left(u\right)u+{\displaystyle \frac{q}{4m^2}}{|u|}^{2}A\left(u\right)udx={\int }_{{\mathbb{R}}^2}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)|N_u|u^{2}dx.\end{array}$$

(33)

Then

$$\begin{array}{cc}\hfill 〈I^{\prime }\left(u\right),u〉=& {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}{|\nabla u|}^2+\omega u^2+{\displaystyle \frac{q}{4m^2}}{u}^4\right)dx-\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\hfill \\ \hfill =& {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla u\nabla (u-A\left(u\right))+\omega u(u-A\left(u\right))+{\displaystyle \frac{q}{4m^2}}{|u|}^{2}u(u-A\left(u\right))\right)dx.\hfill \end{array}$$

Similarly, we obtain

$$\begin{array}{c}\hfill 〈I^{\prime }\left(u\right),A\left(u\right)〉={\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla A\left(u\right)\nabla (u-A\left(u\right))+\omega A\left(u\right)(u-A\left(u\right))+{\displaystyle \frac{q}{4m^2}}{|u|}^{2}A\left(u\right)(u-A\left(u\right))\right)dx.\end{array}$$

Thus,

$$\begin{array}{cc}\hfill 〈I^{\prime }\left(u\right),u-A\left(u\right)〉=& {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}{|\nabla (u-A\left(u\right))|}^2+\omega {(u-A\left(u\right))}^2+{\displaystyle \frac{q}{4m^2}}{|u|}^2{(u-A\left(u\right))}^2\right)dx\hfill \\ \hfill \ge & {\parallel u-A\left(u\right)\parallel }^2.\hfill \end{array}$$

(34)

($ii$) For any $\phi \in H_r^1\left({\mathbb{R}}^2\right)$, we also have

$$\begin{array}{cc}\hfill 〈I^{\prime }\left(u\right),\phi 〉=& {\int }_{{\mathbb{R}}^2}\left({\displaystyle \frac{1}{2m}}\nabla (u-A\left(u\right))\nabla \phi +\omega (u-A\left(u\right))\phi +{\displaystyle \frac{q}{4m^2}}{|u|}^2(u-A\left(u\right))\phi \right)dx\hfill \\ \hfill =& 〈u-A\left(u\right),\phi 〉+{\int }_{{\mathbb{R}}^2}{\displaystyle \frac{q}{4m^2}}{|u|}^2(u-A\left(u\right))\phi dx\hfill \end{array}$$

(35)

and

$$\begin{array}{cc}\hfill {\int }_{{\mathbb{R}}^2}{|u|}^2(u-A\left(u\right))\phi dx\le & {\left({\int }_{{\mathbb{R}}^2}{u}^{4}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{(u-A\left(u\right))}^{4}dx\right)}^{{\displaystyle \frac{1}{4}}}{\left({\int }_{{\mathbb{R}}^2}{\phi }^{4}dx\right)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill \le & {C\parallel u\parallel }^2·\parallel u-A\left(u\right)\parallel ·\parallel \phi \parallel .\hfill \end{array}$$

(36)

Combining (35) and (36), one gets

$$\parallel I^{\prime }\left(u\right)\parallel \hspace{0.17em}\le \hspace{0.17em}\parallel u-A\left(u\right)\parallel \left(1+{C\parallel u\parallel }^2\right)\mathrm{for}\mathrm{all}u\in H_r^1\left({\mathbb{R}}^2\right).$$

□

Lemma 7.

For $a<b$ and $\alpha >0$, if $u\in H_r^1\left({\mathbb{R}}^2\right)$, $I\left(u\right)\in [a,b]$ and $\parallel I^{\prime }\left(u\right)\parallel \hspace{0.17em}\ge \alpha$, then there exists $\beta >0$ such that $\parallel u-A\left(u\right)\parallel \hspace{0.17em}\ge \beta$.

Proof. 

Let $\mu >4$. For $u\in H_r^1\left({\mathbb{R}}^2\right)$, we have

$$\begin{array}{cc}\hfill I\left(u\right)-{\displaystyle \frac{1}{\mu }}〈u,u-A\left(u\right)〉=& \left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{\mu }}\right){\parallel u\parallel }^2+{\displaystyle \frac{q}{16m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx\hfill \\ \hfill & -{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx+{\displaystyle \frac{1}{\mu }}\langle u,A\left(u\right)\rangle .\hfill \end{array}$$

(37)

From (33), one sees

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{\mu }}\langle u,A\left(u\right)\rangle ={\displaystyle \frac{1}{\mu }}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx-{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{|u|}^{2}A\left(u\right)udx.\end{array}$$

Combining with (37), it has

$$\begin{array}{cc}\hfill I\left(u\right)-{\displaystyle \frac{1}{\mu }}\langle u,u-A\left(u\right)\rangle =& \left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{\mu }}\right){\parallel u\parallel }^2+\left({\displaystyle \frac{1}{\mu }}-{\displaystyle \frac{1}{4}}\right)\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\hfill \\ \hfill & +{\displaystyle \frac{q}{16m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx-{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{|u|}^{2}A\left(u\right)udx\hfill \\ \hfill =& \left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{\mu }}\right){\parallel u\parallel }^2+\left({\displaystyle \frac{1}{\mu }}-{\displaystyle \frac{1}{4}}\right)\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\hfill \\ \hfill & +\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{\mu }}\right){\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx-{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{|u|}^2\left(A\left(u\right)-u\right)udx\hfill \\ \hfill \ge & \left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{\mu }}\right){\parallel u\parallel }^2-{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{|u|}^2\left(A\left(u\right)-u\right)udx\hfill \\ \hfill & +\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{\mu }}\right){\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx.\hfill \end{array}$$

(38)

By (38), Hölder and Young inequalities, we know

$$\begin{array}{cc}\hfill & \left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{\mu }}\right){\parallel u\parallel }^2+\left({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{\mu }}\right){\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx\hfill \\ \hfill \le & I\left(u\right)-{\displaystyle \frac{1}{\mu }}〈u,u-A\left(u\right)〉+{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{|u|}^2(u-A\left(u\right))udx\hfill \\ \hfill \le & |I\left(u\right)|+{\displaystyle \frac{1}{\mu }}\parallel u\parallel ·\parallel u-A\left(u\right)\parallel +{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\left({\int }_{{\mathbb{R}}^2}{u}^{4}dx\right)}^{{\displaystyle \frac{3}{4}}}{\left({\int }_{{\mathbb{R}}^2}{(u-A\left(u\right))}^{4}dx\right)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill \le & |I\left(u\right)|+{\displaystyle \frac{1}{\mu }}\parallel u\parallel ·\parallel u-A\left(u\right)\parallel +{\displaystyle \frac{3}{4}}·{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx+{\displaystyle \frac{1}{4}}·{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{(u-A\left(u\right))}^{4}dx.\hfill \end{array}$$

Then,

$$\begin{array}{c}\hfill \left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{\mu }}\right){\parallel u\parallel }^2\le |I\left(u\right)|+{\displaystyle \frac{1}{\mu }}\parallel u\parallel ·\parallel u-A\left(u\right)\parallel +{\displaystyle \frac{1}{4}}·{\displaystyle \frac{1}{\mu }}·{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{(u-A\left(u\right))}^{4}dx.\end{array}$$

It follows from Sobolev inequality that

$$\begin{array}{c}\hfill {\parallel u\parallel }^2\le C\left({|I\left(u\right)|+\parallel u\parallel ·\parallel u-A\left(u\right)\parallel +\parallel u-A\left(u\right)\parallel }^4\right).\end{array}$$

(39)

Assume that there exists $\left\{u_n\right\}\in H_r^1\left({\mathbb{R}}^2\right)$ satisfying $I\left(u_n\right)\in [a,b]$ and $\parallel I^{\prime }\left(u_n\right)\parallel \hspace{0.17em}\ge \alpha$ such that $\parallel u_n-A\left(u_n\right)\parallel \to 0$. Combining with (39), we get that $\{\parallel u_n\parallel \}$ is bounded. Then by Lemma 6$\left(ii\right)$, one has $I^{\prime }\left(u_n\right)\to 0$, which is a contradiction with $\parallel I^{\prime }\left(u_n\right)\parallel \hspace{0.17em}\ge \alpha$. □

Next we define the non-negative and non-positive cones as follows

$$P^{+}:=\left\{u\in H_r^1\left({\mathbb{R}}^3\right):u\ge 0\right\}\mathrm{and}{P}^{-}:=\left\{u\in H_r^1\left({\mathbb{R}}^3\right):u\le 0\right\}.$$

Then for $\epsilon >0$,

$$P_{\epsilon }^{+}:=\left\{u\in H_r^1\left({\mathbb{R}}^3\right):\mathrm{dist}(u,P^{+})<\epsilon \right\}\mathrm{and}{P}_{\epsilon }^{-}:=\left\{u\in H_r^1\left({\mathbb{R}}^3\right):\mathrm{dist}(u,P^{-})<\epsilon \right\},$$

where $\mathrm{dist}(u,P^{\pm })={inf}_{v\in P^{\pm }}\parallel u-v\parallel$; obviously, $P_{\epsilon }^{-}=-P_{\epsilon }^{+}$.

Lemma 8.

There exists ${\epsilon }_0>0$ such that

$$\begin{array}{c}\hfill A\left(P_{\epsilon }^{\pm }\right)\subset P_{\epsilon }^{\pm }\mathrm{for}\mathrm{all}0<\epsilon <{\epsilon }_0.\end{array}$$

Moreover, every nontrivial solution $u\in P_{\epsilon }^{+}$ and $u\in P_{\epsilon }^{-}$ of problem (2) are positive and negative.

Proof. 

For every $u\in H_r^1\left({\mathbb{R}}^2\right)$, denote

$$u=u^{+}+u^{-},\mathrm{where}{u}^{+}:=max\{u,0\}\mathrm{and}{u}^{-}:=min\{u,0\}.$$

It follows from Sobolev embedding theorem that there exists a constant $C_q>0$ such that

$$\begin{array}{c}\hfill \parallel u^{\pm }{\parallel }_q=\underset{w\in P^{\mp }}{inf}{\parallel u-w\parallel }_q\le C_q\underset{w\in P^{\mp }}{inf}\parallel u-w\parallel =C_q\mathrm{dist}(u,P^{\mp }).\end{array}$$

(40)

We claim that $A\left(P_{\epsilon }^{-}\right)\subset P_{\epsilon }^{-}$ and the other case can be gained by the similar way. Notice that

$$\begin{array}{c}\hfill \mathrm{dist}(v,P^{-})\le \parallel v^{+}\parallel .\end{array}$$

From (6), Hölder and Sobolev inequalities, we know that

$$\begin{array}{cc}\hfill \mathrm{dist}(v,P^{-})·\parallel v^{+}\parallel \le & \parallel v^{+}{\parallel }^2=\langle v,v^{+}\rangle \hfill \\ \hfill =& \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u{v}^{+}dx-{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^{2}v{v}^{+}dx\hfill \\ \hfill =& \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u{v}^{+}dx-{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}^2{\left(v^{+}\right)}^{2}dx\hfill \\ \hfill \le & \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u{v}^{+}dx\hfill \\ \hfill \le & \left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\left({\int }_{{\mathbb{R}}^2}{|N_u|}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_{{\mathbb{R}}^2}{u}^2{\left(v^{+}\right)}^{2}dx\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill \le & \left(1+{\displaystyle \frac{\kappa q}{2m}}\right)\parallel N_u{\parallel }_2·{\parallel u\parallel }_4·{\parallel v^{+}\parallel }_4\hfill \\ \hfill \le & \left(1+{\displaystyle \frac{2m}{\kappa q}}\right)·{\displaystyle \frac{1}{{\kappa }^{2}q}}\left(1+{\displaystyle \frac{2m}{\kappa q}}\right){\parallel u\parallel }_4^2·{\parallel u\parallel }_4·{\parallel v^{+}\parallel }_4\hfill \\ \hfill \le & {\displaystyle \frac{1}{{\kappa }^{2}q}}{\left(1+{\displaystyle \frac{2m}{\kappa q}}\right)}^2{\parallel u\parallel }_4^3·C\parallel v^{+}\parallel ,\hfill \end{array}$$

which implies

$$\begin{array}{c}\hfill \mathrm{dist}(v,P^{-})\le {\displaystyle \frac{C}{{\kappa }^{2}q}}{\left(1+{\displaystyle \frac{2m}{\kappa q}}\right)}^2{\parallel u\parallel }_4^3\le {\displaystyle \frac{C{C}_q^3}{{\kappa }^{2}q}}{\left(1+{\displaystyle \frac{2m}{\kappa q}}\right)}^2{\left(\mathrm{dist}(u,P^{-})\right)}^3.\end{array}$$

Let ${\epsilon }_0\in \left(0,\sqrt{{\displaystyle \frac{{\kappa }^{2}q}{2C{C}_q^3}}}{\left(1+{\displaystyle \frac{2m}{\kappa q}}\right)}^{-1}\right)$. For any $u\in P_{\epsilon }^{+}$ with $\epsilon \in (0,{\epsilon }_0)$, one sees

$$\mathrm{dist}(A\left(u\right),P^{-})\le {\displaystyle \frac{1}{2}}\mathrm{dist}(u,P^{-}),$$

which implies $A\left(P_{\epsilon }^{-}\right)\subset P_{\epsilon }^{-}$. If there exists $u\in P_{\epsilon }^{-}$ such that $A\left(u\right)=u$, we know that $\mathrm{dist}(u,P^{-})=0$, that is $u\in P^{-}$. By using the maximum principle, $u<0$ in ${\mathbb{R}}^3$ if $u\not\equiv 0$. □

The continuity of the operator A alone is insufficient to derive a descending flow formulation for the functional I. Next, we construct a locally Lipschitz continuous operator B on $E:=H_r^1\left({\mathbb{R}}^2\right)\setminus K$, where K is the set of fixed points of the operator A. The operator B preserves the essential properties of A.

Lemma 9.

There exist a locally Lipschitz continuous operator $B:E\to H_r^1\left({\mathbb{R}}^2\right)$ such that

(i)

$B\left(P_{\epsilon }^{+}\right)\subset \mathrm{int}\left(P_{\epsilon }^{+}\right)$ and $B\left(P_{\epsilon }^{-}\right)\subset \mathrm{int}\left(P_{\epsilon }^{-}\right)$;

(ii)

${\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel \le \parallel u-B\left(u\right)\parallel \le 2\parallel u-A\left(u\right)\parallel$;

(iii)

$〈I^{\prime }\left(u\right),u-B\left(u\right)〉\ge {\displaystyle \frac{1}{2}}{\parallel u-A\left(u\right)\parallel }^2$;

(iv)

B is odd.

Proof. 

The approach adopted here resembles the techniques employed in Lemma 4.1 of [21] and Lemma 2.1 of [22]. Let ${\mathrm{\Delta }}_1,{\mathrm{\Delta }}_2$ with

$$\begin{array}{c}\hfill {\mathrm{\Delta }}_1\left(u\right)={\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel ,{\mathrm{\Delta }}_2\left(u\right)={\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel {\left(1+{C\parallel u\parallel }^2\right)}^{-1},\end{array}$$

(41)

where C is as in $\left(ii\right)$ of Lemma 6. For each $u\in E$, we choose $r\left(u\right)\in (0,1)$ such that for any $v,w\in N\left(u\right):=\{z\in H_r^1\left({\mathbb{R}}^3\right):\parallel z-u\parallel \hspace{0.17em}\le r\left(u\right)\}$,

$$\begin{array}{c}\hfill \parallel A\left(v\right)-A\left(w\right)\parallel \le min\{{\mathrm{\Delta }}_1\left(v\right),{\mathrm{\Delta }}_2\left(v\right),{\mathrm{\Delta }}_1\left(w\right),{\mathrm{\Delta }}_2\left(w\right)\}.\end{array}$$

(42)

Define $\mathcal{F}$ as a locally finite open refinement of $\left\{N\right(u),u\in E\}$, and let

$$\mathcal{W}:=\left\{F\in \mathcal{F}:F\cap \overline{P_{\epsilon }^{+}}\ne \varnothing ,F\cap \overline{P_{\epsilon }^{-}}\ne \varnothing ,F\cap \overline{P_{\epsilon }^{+}}\cap \overline{P_{\epsilon }^{-}}=\mathrm{\varnothing }\right\}$$

and

$$\mathcal{U}:=\bigcup _{F\in \mathcal{F}\setminus \mathcal{W}}F\cup \bigcup _{F\in \mathcal{W}}\left\{F\setminus \overline{P_{\epsilon }^{+}},F\setminus \overline{P_{\epsilon }^{-}}\right\}.$$

Hence, $\mathcal{U}$ admits a locally finite open refinement of $\left\{N\right(u),u\in E\}$.

Next we will construct the operator B. For any $U\in \mathcal{U}$, let ${\rho }_U\left(u\right)=dist(u,E\setminus U)$ and $\{{\pi }_U:U\in \mathcal{U}\}$ be the standard partition of unity subordinated to $\mathcal{U}$ with

$${\pi }_U\left(u\right):={\left(\sum _{F\in \mathcal{U}}{\rho }_F\left(u\right)\right)}^{-1}{\rho }_U\left(u\right).$$

If $U\cap \overline{P_{\epsilon }^{+}}\ne \varnothing$ and $U\cap \overline{P_{\epsilon }^{-}}\ne \varnothing$, then $U\cap \overline{P_{\epsilon }^{+}}\cap \overline{P_{\epsilon }^{-}}\ne \varnothing$. Then, we can choose ${\overline{\rho }}_U\in U$ such that ${\overline{\rho }}_U\in U\cap \overline{P_{\epsilon }^{+}}\cap \overline{P_{\epsilon }^{-}}$ if $U\cap \overline{P_{\epsilon }^{\pm }}\ne \varnothing$. The operator B can be defined by

$$B\left(u\right)=\sum _{U\in \mathcal{U}}{\pi }_U\left(u\right)A\left({\overline{\rho }}_U\right).$$

Evidently, B is locally Lipschitz continuous and $\left(i\right)$ holds.

Notice that

$$\begin{array}{c}\hfill \parallel B\left(u\right)-A\left(u\right)\parallel \le \sum _{U\in \mathcal{U}}{\pi }_U\left(u\right)∥A\left({\overline{\rho }}_U\right)-A\left(u\right)∥.\end{array}$$

(43)

It follows from (41)–(43) that

$$\begin{array}{c}\hfill \parallel B\left(u\right)-A\left(u\right)\parallel \le {\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel \end{array}$$

(44)

and

$$\begin{array}{c}\hfill \parallel B\left(u\right)-A\left(u\right)\parallel <{\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel {\left(1+{C\parallel u\parallel }^2\right)}^{-1}\end{array}$$

(45)

for $u\in E$. By (44), one has

$$\parallel u-B\left(u\right)\parallel \ge \parallel u-A\left(u\right)\parallel -\parallel A\left(u\right)-B\left(u\right)\parallel \ge {\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel ,$$

namely,

$$\parallel u-A\left(u\right)\parallel \le 2\parallel u-B\left(u\right)\parallel .$$

Also,

$$\begin{array}{c}\hfill \parallel u-B\left(u\right)\parallel \ge \parallel u-A\left(u\right)\parallel -\parallel A\left(u\right)-B\left(u\right)\parallel \ge {\displaystyle \frac{1}{2}}\parallel u-A\left(u\right)\parallel \end{array}$$

(46)

and then

$$\begin{array}{c}\hfill \parallel u-A\left(u\right)\parallel \ge {\displaystyle \frac{1}{2}}\parallel u-B\left(u\right)\parallel .\end{array}$$

(47)

Combining (46) and (47), $\left(ii\right)$ holds.

In order to obtain $\left(iii\right)$, it follows from $\left(ii\right)$ of Lemma 6, (34) and (44) that

$$\begin{array}{cc}\hfill 〈I^{\prime }\left(u\right),u-B\left(u\right)〉=& 〈I^{\prime }\left(u\right),u-A\left(u\right)〉+〈I^{\prime }\left(u\right),A\left(u\right)-B\left(u\right)〉\hfill \\ \hfill \ge & 〈I^{\prime }\left(u\right),u-A\left(u\right)〉-\parallel I^{\prime }\left(u\right)\parallel ·\parallel A\left(u\right)-B\left(u\right)\parallel \hfill \\ \hfill \ge & {\displaystyle \frac{1}{2}}{\parallel u-A\left(u\right)\parallel }^2.\hfill \end{array}$$

(48)

Furthermore, it is easy to see that A is odd and I is even when f is odd. Since $\overline{P_{\epsilon }^{+}}=-\overline{P_{\epsilon }^{-}}$, replacing B by $\tilde{B}={\displaystyle \frac{1}{2}}\left(B\left(u\right)-B(-u)\right)$ yields $\left(iv\right)$. □

Consider the following initial value problem

$$\left\{\begin{array}{c}{\displaystyle \frac{d}{dt}}\sigma (t,u)=-V\left(\sigma \right),\hfill \\ \sigma (0,t)=u\in H_r^1\left({\mathbb{R}}^2\right),\hfill \end{array}\right.$$

(49)

where $V=id-B$. By utilizing the local Lipschitz property of B, the theory governing the existence and uniqueness of solutions to ODE ensures that Equation (49) possesses a unique solution, represented as $\sigma (t,u)$, which is defined on its maximal existence interval $[0,T(u\left)\right)$. It follows from $\left(iii\right)$ of Lemma 9 that $I\left(\sigma \right(t,u\left)\right)$ is strictly decreasing in $[0,T(u\left)\right)$. Let $W:=\overline{P_{\epsilon }^{+}}\cup \overline{P_{\epsilon }^{-}}$. It follows from Lemma 4 that I satisfies the $\left(PS\right)$ condition. If $K_c\setminus W=\varnothing$, then there exists $\delta >0$ such that

$$N_{\delta }\left(K_c\right)\subset W,$$

where

$$N_{\delta }\left(K_c\right):=\left\{u\in H_r^1\left({\mathbb{R}}^2\right):\mathrm{dist}(u,K_c)<\delta \right\}.$$

In order to construct sign-changing solutions based on the invariant sets and minimax method, we need the following deformation lemma, which is a variant of Lemma 2.4 in [23].

Lemma 10.

If $K_c\setminus W=\varnothing$, then there exists ${\epsilon }_1>0$ and a continuous map $\sigma \in \mathcal{C}([0,1]\times H_r^1\left({\mathbb{R}}^2\right),H_r^1\left({\mathbb{R}}^2\right))$ such that, for $0<\epsilon <{\epsilon }_1$,

(i)

$\sigma (0,u)=u$ for $u\in H_r^1\left({\mathbb{R}}^2\right)$;

(ii)

$\sigma (t,u)=u$ for $t\in [0,1],u\notin I^{-1}[c-{\epsilon }_1,c+{\epsilon }_1]$;

(iii)

$\sigma \left(1,I^{c+\epsilon }\cup W\setminus N_{\delta }\left(K_c\right)\right)\subset I^{c-\epsilon }\cup W$;

(iv)

$\sigma (t,\overline{P_{\epsilon }^{+}})\subset \overline{P_{\epsilon }^{+}}$ and $\sigma (t,\overline{P_{\epsilon }^{-}})\subset \overline{P_{\epsilon }^{-}}$ for $t\in [0,1]$;

(v)

$\sigma (t,-u)=-\sigma (t,u)$ for all $(t,u)\in [0,1]\times H_r^1\left({\mathbb{R}}^2\right)$.

Proof. 

Since $I\left(u\right)$ satisfies $\left(PS\right)$ condition, there exists ${\epsilon }_0>0$ such that

$$\parallel I^{\prime }\left(u\right)\parallel \ge \alpha \mathrm{for}u\in I^{-1}\left([c-{\epsilon }_0,c+{\epsilon }_0]\right)\setminus N_{{\displaystyle \frac{\delta }{2}}}\left(K_c\right).$$

From Lemmas 7 and 9, there exists $\beta >0$ such that

$$\begin{array}{c}\hfill \parallel u-B\left(u\right)\parallel \hspace{0.17em}\ge \beta \mathrm{for}u\in I^{-1}\left([c-{\epsilon }_0,c+{\epsilon }_0]\right)\setminus N_{{\displaystyle \frac{\delta }{2}}}\left(K_c\right).\end{array}$$

(50)

Define

$$X_1:=\{u\in H_r^1\left({\mathbb{R}}^2\right):u\notin I^{-1}\left([c-{\epsilon }_0,c+{\epsilon }_0]\right)\mathrm{or}u\in N_{{\displaystyle \frac{\delta }{4}}}\left(K_c\right)\},$$

$$X_2:=\{u\in H_r^1\left({\mathbb{R}}^2\right):u\in I^{-1}\left([c-{\epsilon }_0,c+{\epsilon }_0]\right)\setminus N_{{\displaystyle \frac{\delta }{2}}}\left(K_c\right)\},$$

$$\psi \left(u\right):={\displaystyle \frac{\mathrm{dist}(u,X_2)}{\mathrm{dist}(u,X_1)+\mathrm{dist}(u,X_2)}}.$$

Consider

$$\left\{\begin{array}{c}{\displaystyle \frac{d}{dt}}\xi (t,u)=-\psi \left(\xi (t,u)\right)V\left(\xi (t,u)\right),\hfill \\ \xi (0,u)=u\in H_r^1\left({\mathbb{R}}^2\right).\hfill \end{array}\right.$$

(51)

Obviously, problem (51) has a unique solution $\xi (·,u)\in \mathcal{C}({\mathbb{R}}^{+},H_r^1\left({\mathbb{R}}^2\right))$. Let $\sigma (t,u)=\xi \left({\displaystyle \frac{\epsilon }{2{\beta }^2}}t,u\right)$. Evidently, $\left(i\right)$, $\left(ii\right)$ and $\left(v\right)$ hold.

For $\left(iii\right)$, let $u\in I^{c+\epsilon }\setminus W$. We can easily get that $I\left(\sigma \right(t,u\left)\right)$ is decreasing for $t\ge 0$ from problem (51). If there exists $t_0\in \left[0,{\displaystyle \frac{2\epsilon }{{\beta }^2}}\right]$ such that $I\left(\xi (t_0,u)\right)<c-\epsilon$, then

$$I\left(\sigma (1,u)\right))=I\left(\xi \left({\displaystyle \frac{2\epsilon }{{\beta }^2}},u\right)\right)\le I\left(\xi (t_0,u)\right)<c-\epsilon .$$

Then $\left(iii\right)$ hold. Otherwise, $I\left(\xi \right(t,u\left)\right)\ge c-\epsilon$ for every $t\in \left[0,{\displaystyle \frac{2\epsilon }{{\beta }^2}}\right]$. That is,

$$\begin{array}{c}\hfill \xi (t,u)\in I^{-1}\left([c-{\epsilon }_0,c+{\epsilon }_0]\right)\mathrm{for}\mathrm{every}t\in \left[0,{\displaystyle \frac{2\epsilon }{{\beta }^2}}\right].\end{array}$$

(52)

Next we claim for every $t\in \left[0,{\displaystyle \frac{2\epsilon }{{\beta }^2}}\right]$,

$$\begin{array}{c}\hfill \xi (t,u)\notin N_{{\displaystyle \frac{\delta }{2}}}\left(K_c\right).\end{array}$$

(53)

Indeed, if there exists $t_1\in \left[0,{\displaystyle \frac{2\epsilon }{{\beta }^2}}\right]$ such that $\xi (t,u)\in N_{{\displaystyle \frac{\delta }{2}}}\left(K_c\right)$, combining with $u\notin N_{\delta }\left(K_c\right)$, we have

$${\displaystyle \frac{4\epsilon }{{\beta }^2}}\le \parallel \xi (t,u)-u\parallel \le {\int }_0^{t_1}\parallel {\xi }^{\prime }(s,u)\parallel ds\le t_1\le {\displaystyle \frac{2\epsilon }{{\beta }^2}}$$

a contradiction. From (52), (53) and the definition of $\psi \left(u\right)$, one sees

$$\begin{array}{c}\hfill \psi \left(\xi (t,u)\right)\equiv 1\mathrm{for}\mathrm{any}t\in \left[0,{\displaystyle \frac{2\epsilon }{{\beta }^2}}\right].\end{array}$$

(54)

By (51), (54) and Lemma 6, it obtains

$$\begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}}I\left(\xi (t,u)\right)=& 〈I^{\prime }\left(\xi (t,u)\right),{\xi }^{\prime }(t,u))〉\hfill \\ \hfill =& -〈I^{\prime }\left(\xi (t,u)\right),\psi \left(\xi (t,u)\right)V\left(\xi (t,u)\right)〉\hfill \\ \hfill =& -〈I^{\prime }\left(\xi (t,u)\right),V\left(\xi (t,u)\right)〉\hfill \\ \hfill =& -〈I^{\prime }\left(\xi (t,u)\right),\xi (t,u)-B\left(\xi (t,u)\right)〉\hfill \\ \hfill \le & -∥\xi (t,u))-B(\xi (t,u))∥\hfill \\ \hfill \le & -{\beta }^2.\hfill \end{array}$$

(55)

Then we can get

$$\begin{array}{cc}\hfill I\left(\sigma (1,u)\right)=I\left(\xi \left({\displaystyle \frac{2\epsilon }{{\beta }^2}},u\right)\right)=& I\left(u\right)+{\int }_0^{{\displaystyle \frac{2\epsilon }{{\beta }^2}}}{\displaystyle \frac{d}{dt}}I\left(\xi (t,u)\right)dt\hfill \\ \hfill \le & c+\epsilon -{\displaystyle \frac{2\epsilon }{{\beta }^2}}·{\beta }^2=c-\epsilon ,\hfill \end{array}$$

a contradiction.

At last, $\left(iv\right)$ can be obtained by using $\left(i\right)$ of Lemma 9; see [24] for a detailed proof. □

4. Proof of Theorem 1

Proof of Theorem 1.

Since $H_r^1\left({\mathbb{R}}^2\right)\hookrightarrow L^2\left({\mathbb{R}}^2\right)$ and $L^2\left({\mathbb{R}}^2\right)$ is a separable Hilbert space, $H_r^1\left({\mathbb{R}}^2\right)$ has countable orthogonal basis $\left\{e_i\right\}$. For any $n\in \mathbb{N}$, let $X_n:=\{e_1,e_2,\dots ,e_n\}$. In view of Lemma 3.5 in [1], we know that there exists $R_n=R\left(X_n\right)>0$ such that

$$\underset{B_{R_n}^c\cap X_n}{sup}I<0,$$

where $B_{R_n}^c:=H_r^1\left({\mathbb{R}}^2\right)\setminus B_{R_n}$ and $B_{R_n}:=\{u\in H_r^1\left({\mathbb{R}}^2\right):\parallel u\parallel \le R_n\}$. Define

$$G_n:=\left\{h\in \mathcal{C}(B_{R_n}\cap E_n,H_r^1\left({\mathbb{R}}^2\right):h\mathrm{is}\mathrm{odd}\mathrm{and}h=\mathrm{id}\mathrm{on}\partial B_{R_n}\cap E_n\right\}.$$

Obviously, $G_n\ne \varnothing$ for the reason that $\mathrm{id}\in G_n$. Set

$${\mathrm{\Gamma }}_k:=\left\{h(B_{R_n}\cap E_n\setminus Y:h\in G_n,n\ge k,Y=-Y\mathrm{is}\mathrm{open}\mathrm{and}\gamma \left(Y\right)\le n-k\right\}.$$

From Proposition 9.18 in [25], ${\mathrm{\Gamma }}_k$ has the following properties:

(a)

${\mathrm{\Gamma }}_k\ne \varnothing$;

(b)

${\mathrm{\Gamma }}_{k+1}\subset {\mathrm{\Gamma }}_k$ for $k\ge 2$;

(c)

If $\psi \in \mathcal{C}(\mathbb{R},\mathbb{R})$ is odd and $\psi =id$ on $\partial B_{R_n}\cap E_n$, then $\psi \left(A\right)\in {\mathrm{\Gamma }}_k$ if $A\in {\mathrm{\Gamma }}_k$ for $k\ge 2$;

(d)

If $A\in {\mathrm{\Gamma }}_k$ and $Z=-Z$ is open and $\gamma \left(Z\right)\le s<k$ and $k-s\ge 2$, then $A\setminus Z\in {\mathrm{\Gamma }}_{k-s}$.

Let $Q:=H_r^1\left({\mathbb{R}}^2\right)\setminus W$, together with Lemma 8, all sign-changing solutions are contained in Q. First, we claim $A\cap Q\ne \varnothing$ for any $A\in {\mathrm{\Gamma }}_k$ with $k\ge 2$. Now we consider the attracting domain in $H_r^1\left({\mathbb{R}}^2\right)$ as follows

$$\begin{array}{c}\hfill \mathcal{D}:=\left\{u\in H_r^1\left({\mathbb{R}}^3\right):\sigma (t,u)\to 0\mathrm{as}t\to \infty \right\}.\end{array}$$

As 0 is a local minimum of I, together with the continuous dependence of ODE on initial data, one can get that $\mathcal{D}$ is open. Also, $\partial \mathcal{D}$ is an invariant set and $\overline{P_{\epsilon }^{+}}\cap \overline{P_{\epsilon }^{-}}\subset \mathcal{D}$. By (7), one has

$$\begin{array}{cc}\hfill I\left(u\right)=& {\displaystyle \frac{1}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{q}{16m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx-{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\hfill \\ \hfill \ge & -{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\ge -{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\parallel u\parallel }_4^4.\hfill \end{array}$$

(56)

From (40), it gets

$$\begin{array}{c}\hfill \parallel u^{\pm }{\parallel }_4\le C·\mathrm{dist}(u,P_{\epsilon }^{\mp })\le C{\epsilon }_0\mathrm{for}\mathrm{any}u\in P_{\epsilon }^{+}\cap P_{\epsilon }^{-}.\end{array}$$

(57)

Combining (56) and (57), one obtains

$$\begin{array}{c}\hfill \underset{P_{\epsilon }^{+}\cap P_{\epsilon }^{-}}{inf}I\left(u\right)>-\infty .\end{array}$$

(58)

From Lemma 8 and (58), there exists $\sigma \in \mathcal{C}([0,1]\times H_r^1\left({\mathbb{R}}^2\right),H_r^1\left({\mathbb{R}}^2\right))$ such that

$$\begin{array}{c}\hfill \sigma (t,u)\to 0\mathrm{as}t\to T\left(u\right)\mathrm{for}u\in \overline{P_{\epsilon }^{+}}\cap \overline{P_{\epsilon }^{-}},\end{array}$$

which implies $P_{\epsilon }^{+}\cap P_{\epsilon }^{-}\cap \partial \mathcal{D}\ne \varnothing$ and then

$$\begin{array}{c}\hfill \gamma \left(\left(P_{\epsilon }^{+}\cap P_{\epsilon }^{-}\right)\cap \partial \mathcal{D}\right)\le 1.\end{array}$$

(59)

Combining with

$$\begin{array}{cc}\hfill I\left(u\right)=& {\displaystyle \frac{1}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{q}{16m^2}}{\int }_{{\mathbb{R}}^2}{u}^{4}dx-{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_u|u^{2}dx\hfill \\ \hfill \ge & {\displaystyle \frac{1}{2}}{\parallel u\parallel }^2-{\displaystyle \frac{1}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\parallel u\parallel }_4^4\ge {\displaystyle \frac{1}{2}}{\parallel u\parallel }^2-{\displaystyle \frac{C}{4}}\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\parallel u\parallel }^4,\hfill \end{array}$$

(60)

we can also get

$$\begin{array}{c}\hfill I\left(u\right)>0\mathrm{for}\mathrm{any}u\in \left(\overline{P_{\epsilon }^{+}}\cap \overline{P_{\epsilon }^{-}}\right)\setminus \left\{0\right\}.\end{array}$$

Let $A\in {\mathrm{\Gamma }}_k$ for $k\ge 2$, that is $A=h(B_{R_n}\cap E_n\setminus Y)$ where $\gamma \left(Y\right)\le m-k$. Define

$$\mathcal{O}:=\left\{u\in B_{R_n}\cap E_n:h\in \mathcal{D}\right\}.$$

Evidently, $\mathcal{O}$ is a bounded open symmetric set satisfying $0\in \mathcal{O}$ and $\mathcal{O}\subset B_{R_n}\cap E_n$. Then, $\gamma (\partial \mathcal{O})\subset \partial \mathcal{D}$ by the definition of h. Hence,

$$\begin{array}{c}\hfill h(\partial \mathcal{O}\setminus Y)\subset A\cap \partial \mathcal{D}.\end{array}$$

(61)

By using Proposition 1, we get

$$\begin{array}{c}\hfill \gamma (\partial \mathcal{O}\setminus Y)\ge \gamma \left(h\right(\partial \mathcal{O}\setminus Y\left)\right)\ge \gamma (\partial \mathcal{O}\setminus Y)\ge \gamma (\partial \mathcal{O})-\gamma \left(Y\right)\ge k.\end{array}$$

(62)

From (59), (61) and (62), we get

$$\begin{array}{c}\hfill \gamma (A\cap Q\cap \partial \mathcal{D})\ge \gamma (A\cap \partial \mathcal{D})-\gamma \left(W\cap \partial \mathcal{D}\right)\ge k-1\ge 1\mathrm{for}k\ge 2.\end{array}$$

Thus, $A\cap Q\cap \partial \mathcal{D}\ne \varnothing$, that is, $A\cap Q\ne \varnothing$. Therefore, a minimax value $c_k$ for $k=2,3,\dots$ can be defined as follows

$$c_k:=\underset{A\in {\mathrm{\Gamma }}_k}{inf}\underset{A\cap Q}{sup}I\left(u\right).$$

It follows from (60) that there exist $\rho ,\alpha >0$ such that

$$\underset{\partial B_{\rho }}{inf}I\ge \alpha .$$

Meanwhile, $\partial B_{\rho }\subset \mathcal{D}$, together with $A\cap Q\cap \partial \mathcal{D}\ne \varnothing$, gets

$$\underset{A\cap Q}{sup}I\ge \underset{\partial \mathcal{D}}{inf}I\ge \underset{\partial B_{\rho }}{inf}I\ge \alpha >0,\mathrm{for}\mathrm{any}A\in {\mathrm{\Gamma }}_k.$$

Therefore, $c_k\ge \alpha >0$. Furthermore, $c_{k+1}\ge c_k$ by using $\left(b\right)$.

Next, we will show

$$\begin{array}{c}\hfill K_{c_k}\cap Q\ne \varnothing \mathrm{for}\mathrm{all}k\ge 2,\end{array}$$

which implies that there exists a sign-changing critical point $u_k$ such that $I\left(u_k\right)=c_k$. We argue by contradiction and suppose $K_{c_k}\cap Q=\varnothing$. It follows from Lemma 10 that there exist $\epsilon >0$ and $\sigma \in \mathcal{C}\left([0,1]\times H_r^1\left({\mathbb{R}}^2\right),H_r^1\left({\mathbb{R}}^2\right)\right)$ such that

$$\begin{array}{c}\hfill \sigma (1,·)\mathrm{is}\mathrm{odd},\sigma (1,u)=u\mathrm{for}{I}^{c_k-2\epsilon },\sigma (1,I^{c_k+\epsilon })\subset I^{c_k-\epsilon }.\end{array}$$

(63)

By the definition of $c_k$, there exists $A\in {\mathrm{\Gamma }}_k$ such that

$$c_k\le \underset{A\cap Q}{sup}I\le c_k+\epsilon .$$

Combining with (63), one obtains

$$\begin{array}{c}\hfill c_k\le \underset{\sigma (1,A)\cap Q}{sup}I\le c_k-\epsilon ,\end{array}$$

which is a contradiction.

At last, we claim that $c_k\to \infty$ as $k\to \infty$, which deduces that $I\left(u\right)$ has infinitely many sign-changing solutions. We argue by contradiction and suppose $c_k\to c<\infty$ as $k\to \infty$. It follows from Lemma 4 that $K_c$ is nonempty and compact. Denote ${\overline{K}}_c=K_c\cap Q$, then ${\overline{K}}_c$ is nonempty. In fact, let $\left\{u_k\right\}$ be sign-changing solutions for problem (2) with $I\left(u_k\right)=c_k$. Then by (6), one sees

$$\parallel u_k{\parallel }^2+{\displaystyle \frac{q}{4m^2}}{\int }_{{\mathbb{R}}^2}{u}_k^{4}dx=\left(1+{\displaystyle \frac{\kappa q}{2m}}\right){\int }_{{\mathbb{R}}^2}|N_{u_k}|u_k^{2}dx\le {\displaystyle \frac{1}{{\kappa }^{2}q}}{\left(1+{\displaystyle \frac{\kappa q}{2m}}\right)}^2{\parallel u_k\parallel }_4^4,$$

together with Sobolev embedding theorem, there exists a constant $C>0$ such that $\parallel u_k\parallel \ge C$. By Lemma 4, there exists $u\in K_c$ such that $u_k\to u$. Clearly, u is sign-changing and then ${\overline{K}}_c\ne \varnothing$.

Suppose $\gamma \left({\overline{K}}_c\right)=m$. By using $0\notin {\overline{K}}_c$ and Proposition 1, there exists an open neighborhood N in $H_r^1\left({\mathbb{R}}^2\right)$ with ${\overline{K}}_c\subset N$ such that $\gamma \left(N\right)=m$. From Lemma 10, there are $ϵ>0$ and $\sigma \in \mathcal{C}([0,1]\times H_r^1\left({\mathbb{R}}^2\right),H_r^1\left({\mathbb{R}}^2\right)$ such that $\sigma (1,·)$ is odd, $\sigma (1,u)=u$ for $u\in I^{c-2\epsilon }$ and

$$\begin{array}{c}\hfill \sigma \left(1,I^{c+\epsilon }\cup W\setminus N\right)\subset I^{c-\epsilon }\cup W.\end{array}$$

(64)

Since $c_k\to c$ as $k\to \infty$, there exists k sufficiently large such that

$$\begin{array}{c}\hfill c_k\ge c-{\displaystyle \frac{1}{2}}\epsilon .\end{array}$$

(65)

Choose $A\subset {\mathrm{\Gamma }}_{k+m}$ such that

$$\underset{A\cap Q}{sup}I<c+\epsilon .$$

Then $A=h(B_R\cup X_n\setminus Y)\subset I^{c+\epsilon }\cup W$, where $h\in G_n,n\ge k+m,\gamma \left(Y\right)\le n-(m+k)$. It follows from (64) that

$$\begin{array}{c}\hfill \sigma (1,A\setminus N)\subset I^{c-\epsilon }\cup W.\end{array}$$

(66)

Denote $Y_1:=Y\cup h^{-1}\left(N\right)$ and then $Y_1$ is symmetric and open. Moreover,

$$\gamma \left(Y_1\right)\le \gamma \left(Y\right)+\gamma \left(h^{-1}\left(N\right)\right)\le n-(k+m)+m=n-k,$$

together with $\left(c\right)$ and $\left(d\right)$, it obtains $\overline{A}:=\sigma (1,h(B_R\cap X_n\setminus Y_1))\in {\mathrm{\Gamma }}_k$. Hence, from (66), we have

$$\begin{array}{c}\hfill c_k\le \underset{\overline{A}\cup Q}{sup}\le \underset{\sigma (1,\overline{A\setminus N})\cap Q}{sup}I\le \underset{I^{c-\epsilon }\cap Q}{sup}I\le c-\epsilon ,\end{array}$$

which is a contradiction with (65) and we complete the proof. □

5. Discussion

This paper aims to find infinitely many sign-changing solutions for the Schrödinger equation coupled with an neutral scalar field. The main method in this paper is descending flow invariant sets with the Ljusternik–Schnirelman theory.

Two promising directions for future research are as follows:

(1)

Existence of least energy sign-changing solutions for the Schrödinger equation coupled with an neutral scalar field;

(2)

Existence of normalized solutions for the Schrödinger equation coupled with an neutral scalar field.