The Sign-Changing Solution for Fractional (p,q)-Laplacian Problems Involving Supercritical Exponent

Abstract

In this article, we consider the following fractional $(p,q)$-Laplacian problem ${(-\Delta )}_p^{s_1}u+{(-\Delta )}_q^{s_2}u+{V\left(x\right)\left(\right|u|}^{p-2}u+{\left|u\right|}^{q-2}{u)=f\left(u\right)+\lambda |u|}^{r-2}u,$ where $x\in {\mathbb{R}}^N$, ${(-\Delta )}_p^{s_1}$ is the fractional p-Laplacian operator (${(-\Delta )}_q^{s_2}$ is similar), $0<s_1<s_2<1<p<q<\frac{N}{s_2}$, $q_{s_2}^{*}=\frac{Nq}{N-s_{2}q}$, $r\ge q_{s_2}^{*}$, f is a $C^1$ real function and V is a coercive function. By using variational methods, we prove that the above problem admits a sign-changing solution if $\lambda >0$ is small.

1. Introduction

For the fractional $(p,q)$-Laplacian problem

$${(-\Delta )}_p^{s_1}u+{(-\Delta )}_q^{s_2}u+{V\left(x\right)\left(\right|u|}^{p-2}u+{\left|u\right|}^{q-2}{u)=f\left(u\right)+\lambda |u|}^{r-2}u,$$

(1)

we will prove it admits a sign-changing solution, where $x\in {\mathbb{R}}^N$, $0<s_1<s_2<1<p<q<\frac{N}{s_2}$, $q_{s_2}^{*}=\frac{Nq}{N-s_{2}q}$, $r\ge q_{s_2}^{*}$, $V:{\mathbb{R}}^N\to \mathbb{R}$ is a positive continuous function and $f:\mathbb{R}\to \mathbb{R}$ is a continuous real function.

1.1. Physical Background

The definition of the fractional p-Laplacian can be seen in [1]. For the physical background, we refer the reader to [2,3,4,5,6,7]. These papers tell us that the fractional p-Laplacian can describe financial markets, optimization, phase transformation, semi-permeable film, anomalous diffusion and minimal surface problems. Problem (1) models two different materials and it is called the double-phase equation (see e.g., [8]).

1.2. Related Works and Our Main Results

Recently, many authors have been concerned with the fractional $(p,q)$-Laplacian equations. For the critical and supercritical cases, the existence of multiple solutions is obtained in [8]. In the meantime, for the problem

$${(-\Delta )}_p^{s}u+{(-\Delta )}_q^{s}u+{V\left(x\right)\left(\right|u|}^{p-2}u+{\left|u\right|}^{q-2}u)=K\left(x\right)f\left(u\right),x\in {\mathbb{R}}^N,$$

(2)

where $s\in (0,1)$, $1<p<q<\frac{N}{s}$, $V:{\mathbb{R}}^N\to \mathbb{R}$, Isernian [9] showed that it admits a positive ground state solution. In 2022, the existence of the least energy sign-changing solution of problem (2) was given by Cheng et al. in [10]. We also quote the papers [11,12,13] for the p-Laplacian or fractional p-Laplacian in a bounded domain. For other results, please see [1,8,14,15,16,17,18,19,20] and the references therein.

The main goal of the present paper is to investigate the problem in (1). We assume that

$\left(V\right)$

$V\in C({\mathbb{R}}^N,\mathbb{R})$, $0<V_0=\underset{x\in {\mathbb{R}}^N}{inf}V\left(x\right)$ and $\underset{\left|x\right|\to +\infty }{lim}V\left(x\right)=+\infty$, where $V_0>0$ is a constant.

$\left(f_1\right)$

$f\in C^1(\mathbb{R},\mathbb{R})$ and $\underset{\left|t\right|\to 0}{lim}\frac{f\left(t\right)}{{\left|t\right|}^{p-1}}=0$.

$\left(f_2\right)$

there is $\sigma \in (p,q_{s_2}^{*})$, $C>0$ such that $\left|f\left(t\right)\right|\le C\left(\right|t{|}^{p-1}+\left|t{|}^{\sigma -1}\right)$.

$\left(f_3\right)$

there exists $\theta \in (q,q_{s_2}^{*})$ such that $0<\theta F\left(t\right)\le f\left(t\right)t$ for all $\left|t\right|>0$, where $F\left(t\right)={\int }_0^{t}f\left(s\right)ds$.

$\left(f_4\right)$

the map $t\mapsto \frac{f\left(t\right)}{{\left|t\right|}^{q-1}}$ is strictly increasing for all $\left|t\right|>0$.

 Theorem 1. 

Let $0<s_1<s_2<1<p<q<\frac{N}{s_2}$, $q_{s_2}^{*}=\frac{Nq}{N-s_{2}q}$, $r\ge q_{s_2}^{*}$, $\left(V\right)$ and $\left(f_1\right)-\left(f_4\right)$ hold. Then there is ${\lambda }_0>0$ such that Problem (1) possesses one sign-changing solution $w_0$ when $\lambda \in (0,{\lambda }_0)$.

 Remark 1. 

$r\ge q_{s_2}^{*}$ is called critical or supercritical. We do not confirm that $w_0$ is a least energy sign-changing solution.

1.3. Our Motivations and Novelties

Like [21], a natural question for us is

• For the $(p,q)$-Laplacian problem (1), does there exist a sign-changing solution?
• Our motivation in this paper is to give this question an affirmative answer. It is different from [22,23] since (10), (11) and (15) are new and crucial.

1.4. Methods

We summarize our methods here. We adopt the idea from [22] or [24] to cut off the functional (see (4)). Then we shall prove (6) admits a minimizer $w_0$. Furthermore, we need to prove that the minimizer $w_0$ is a critical point of $I_{\lambda ,K}$ (see (5)). Finally, we borrow the idea from [22] to make a $L^{\infty }$-estimation such that $\parallel w_0{\parallel }_{L^{\infty }\left({\mathbb{R}}^N\right)}\le K$, which implies that we do not make any truncations.

1.5. Organization

This paper is organized as follows. Section 2 provides some preliminaries. Section 3 is divided into two parts, which will prove Theorem 1. The last Section is the conclusions and our future direction. Throughout this paper, we use the standard notations.

• C or $C_i$ ($i=1,2,...$) denote some positive constants (possibly different from line to line) and $C(·)$ denotes some positive constant only dependent on ·.
• ${\parallel ·\parallel }_{L^l}$ ($1<l<\infty$) is the standard norm in the usual Lebesgue space $L^l\left({\mathbb{R}}^N\right)$.
• For a function $u\left(x\right)$, $u^{+}\left(x\right):=max\{u\left(x\right),0\}$, $u^{-}\left(x\right):=min\{u\left(x\right),0\}$. Clearly, $u=u^{+}+u^{-}$.
• ${\mathbb{R}}^{+}:=[0,+\infty )$.

2. Preliminary Results

From now on, we always assume that $\left(V\right)$ and $\left(f_1\right)-\left(f_4\right)$ hold unless a special statement is made. We continue to use the notations and work space $W^{s,p}\left({\mathbb{R}}^N\right)$ as in [8]. Since the potential V is coercive, we introduce the subspace

$$E=\left\{u\in W^{s_1,p}\left({\mathbb{R}}^N\right)\bigcap W^{s_2,q}\left({\mathbb{R}}^N\right):{\int }_{{\mathbb{R}}^N}V\left(x\right){\left(\right|u|}^p+{\left|u\right|}^q)dx<\infty \right\}$$

equipped with the following norm $\parallel u\parallel ={\parallel u\parallel }_1+{\parallel u\parallel }_2$, where

$${\parallel u\parallel }_1={\left({\left[u\right]}_{s_1,p}^p+{\int }_{{\mathbb{R}}^N}V\left(x\right){\left|u\right|}^{p}dx\right)}^{\frac{1}{p}},{\parallel u\parallel }_2={\left({\left[u\right]}_{s_2,q}^q+{\int }_{{\mathbb{R}}^N}V\left(x\right){\left|u\right|}^{q}dx\right)}^{\frac{1}{q}}.$$

For ${[·]}_{s_1,p}$ and ${[·]}_{s_2,q}$, see [8]. Formally, the corresponding energy functional of (1) is

$$I\left(u\right)=\frac{1}{p}{\parallel u\parallel }_1^p+\frac{1}{q}{\parallel u\parallel }_2^q-{\int }_{{\mathbb{R}}^N}F\left(u\right)dx-\frac{\lambda }{r}{\int }_{{\mathbb{R}}^N}{\left|u\right|}^{r}dx,u\in E.$$

(3)

It is well-known that the functional I is not well-defined on E. In order to overcome this difficulty, similar to [22],

$$\begin{array}{c}\hfill h_{\lambda ,K}\left(t\right):=\left\{\begin{array}{cc}f\left(t\right)+{\lambda \left|t\right|}^{r-2}t,\hfill & \mathrm{if}\left|t\right|\le K,\hfill \\ f\left(t\right)+\lambda K^{r-\sigma }{\left|t\right|}^{\sigma -2}t,\hfill & \mathrm{if}\left|t\right|>K,\hfill \end{array}\right.\end{array}$$

(4)

where $K>0$, $\sigma \in (\theta ,q_{s_2}^{*})$.

Thus, it holds that

 (h₁)

$\underset{\left|t\right|\to 0}{lim}\frac{h_{\lambda ,K}\left(t\right)}{{\left|t\right|}^{p-1}}=0$, and for any $\epsilon >0$, there exists a positive constant $C\left(\epsilon \right)>0$ such that

$$|h_{\lambda ,K}{\left(t\right)|\le \epsilon |t|}^{p-1}+(1+\lambda K^{r-\sigma })C\left(\epsilon \right){\left|t\right|}^{\sigma -1},\forall t\in \mathbb{R}.$$

 (h₂)

$0<\theta H_{\lambda ,K}\left(t\right)\le h_{\lambda ,K}\left(t\right)t$$\forall \left|t\right|>0$, and $h_{\lambda ,K}\left(t\right)t-q{H}_{\lambda ,K}\left(t\right)$ is increasing with respect to t, for all $t>0$ where $H_{\lambda ,K}\left(t\right)={\int }_0^t{h}_{\lambda ,K}\left(s\right)ds$.

 (h₃)

the map $t\mapsto \frac{h_{\lambda ,K}\left(t\right)}{{\left|t\right|}^{q-1}}$ is strictly increasing for all $\left|t\right|>0$.

For the auxiliary functional

$$I_{\lambda ,K}\left(u\right)=\frac{1}{p}{\parallel u\parallel }_1^p+\frac{1}{q}{\parallel u\parallel }_2^q-{\int }_{{\mathbb{R}}^N}{H}_{\lambda ,K}\left(u\right)dx,$$

(5)

$\left(h_1\right)$ implies that $I_{\lambda ,K}\in C^1(E,\mathbb{R})$. We want to prove that

$$m_{\lambda ,K}:=\underset{u\in {\mathcal{M}}_{\lambda ,K}}{inf}{I}_{\lambda ,K}\left(u\right)$$

(6)

admits a minimizer, where the corresponding Nehari manifold ${\mathcal{M}}_{\lambda ,K}$ is given by

$${\mathcal{M}}_{\lambda ,K}=\{u\in E:u^{\pm }\ne 0,⟨I_{\lambda ,K}^{\prime }\left(u\right),u^{\pm }⟩=0\}.$$

(7)

3. Proof of the Main Results

3.1. Some Lemmas

To begin with, we give several lemmas that will be used in the sequel.

Lemma 1. 

For any $u\in E$ with $u^{\pm }\ne 0$, there is a unique pair $(s_u,t_u)$ of positive numbers such that $s_u{u}^{+}+t_u{u}^{-}\in {\mathcal{M}}_{\lambda ,K}$. Moreover,

$$I_{\lambda ,K}(s_u{u}^{+}+t_u{u}^{-})=\underset{(s,t)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}}{max}{I}_{\lambda ,K}(s{u}^{+}+t{u}^{-}).$$

Proof 

The proof is standard (see e.g., [21]). For $u\in E$ with $u^{\pm }\ne 0$, we can deduce from $\left(h_1\right)$ and $\left(h_2\right)$ that there exist $C_1(\lambda ,K),C_2(\lambda ,K)>0$ such that

$$H_{\lambda ,K}\left(u\right)\ge C_1{(\lambda ,K)\left|u\right|}^{\theta }-C_2(\lambda ,K){\left|u\right|}^p.$$

(8)

Let $s,t\ge 0$, ${\varphi }_{\lambda ,K}(s,t):=I_{\lambda ,K}(s{u}^{+}+t{u}^{-})$. One has

$$\begin{array}{cc}\hfill & {\varphi }_{\lambda ,K}(s,t)\hfill \\ \hfill \le & \frac{1}{p}\parallel s{u}^{+}+t{u}^{-}{\parallel }_1^p+\frac{1}{q}{\parallel s{u}^{+}+t{u}^{-}\parallel }_2^q\hfill \\ \hfill & +C_2(\lambda ,K)s^p\parallel u^{+}{\parallel }_{L^p}+C_2(\lambda ,K)t^p{\parallel u^{-}\parallel }_{L^p}\hfill \\ \hfill & -C_1(\lambda ,K)s^{\theta }\parallel u^{+}{\parallel }_{L^{\theta }}-C_1(\lambda ,K)t^{\theta }{\parallel u^{-}\parallel }_{L^{\theta }}.\hfill \end{array}$$

(9)

Denote ${\left({\mathbb{R}}^N\right)}^{+}:=\{x\in {\mathbb{R}}^N:u\left(x\right)\ge 0\}$ and ${\left({\mathbb{R}}^N\right)}^{-}:=\{x\in {\mathbb{R}}^N:u\left(x\right)<0\}$. It holds that

$$\begin{array}{cc}\hfill & \parallel s{u}^{+}+t{u}^{-}{\parallel }_1^p\hfill \\ \hfill =& s^p{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}dy{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}dx\frac{{\left|u^{+}\left(x\right)-u^{+}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}\hfill \\ \hfill & +{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}dy{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}dx\frac{{\left|s{u}^{+}\left(x\right)-t{u}^{-}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}\hfill \\ \hfill & +{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}dy{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}dx\frac{{\left|t{u}^{-}\left(x\right)-s{u}^{+}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}\hfill \\ \hfill & +t^p{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}dy{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}dx\frac{{\left|u^{-}\left(x\right)-u^{-}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}\hfill \\ \hfill & +s^p{\int }_{{\mathbb{R}}^N}V\left(x\right){\left|u^{+}\right|}^{p}dx+t^p{\int }_{{\mathbb{R}}^N}V\left(x\right){\left|u^{-}\right|}^{p}dx.\hfill \end{array}$$

(10)

Obviously,

$$\begin{array}{cc}\hfill & {\int }_{{\left({\mathbb{R}}^N\right)}^{-}}dy{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}dx\frac{{\left|s{u}^{+}\left(x\right)-t{u}^{-}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}\hfill \\ \hfill \le & {max\left\{\right|s|}^p{,\left|t\right|}^p\}{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}dy{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}dx\frac{{\left|u^{+}\left(x\right)-u^{-}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}.\hfill \end{array}$$

(11)

The result also holds for the third integral in (10). We can estimate the term $\parallel s{u}^{+}+t{u}^{-}{\parallel }_2^p$ similarly. Thus, we obtain

$${\varphi }_{\lambda ,K}(s,t)\to -\infty \mathrm{as}\left|(s,t)\right|\to +\infty .$$

(12)

Note that ${\varphi }_{\lambda ,K}(0,0)=0$. The continuity of ${\varphi }_{\lambda ,K}$ shows that it admits a global maximum point $(s_u,t_u)\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}$.

Next, similar to ([21], Lemma 2.3), we can prove that the maximum point cannot be achieved on the boundary of $\left(0,+\infty \right)\times \left(0,+\infty \right)$.

The remaining part sets out to prove the uniqueness. We divide it into two cases. Case 1: For $u\in {\mathcal{M}}_{\lambda ,K}$. If there exists a pair $\overline{s}>0,\overline{t}>0$ such that $\overline{s}{u}^{+}+\overline{t}{u}^{-}\in {\mathcal{M}}_{\lambda ,K}$. We shall discuss the case $0<\overline{t}\le \overline{s}$. Clearly,

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{h}_{\lambda ,K}\left(\overline{s}{u}^{+}\right)\overline{s}{u}^{+}dx\hfill \\ \hfill \le & {\overline{s}}^p{\parallel u^{+}\parallel }_1^p+{\overline{s}}^q{\parallel u^{+}\parallel }_2^q\hfill \end{array}$$

$$\begin{array}{cc}\hfill & +{\overline{s}}^p{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}\frac{|u^{+}\left(x\right)-u^{-}{\left(y\right)|}^{p-1}{u}^{+}\left(x\right)-{\left|u^{+}\left(x\right)\right|}^p}{{|x-y|}^{N+s_{1}p}}dxdy\hfill \\ \hfill & +{\overline{s}}^p{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}\frac{|u^{-}\left(x\right)-u^{+}{\left(y\right)|}^{p-1}{u}^{+}\left(y\right)-{\left|u^{+}\left(y\right)\right|}^p}{{|x-y|}^{N+s_{1}p}}dxdy\hfill \\ \hfill & +{\overline{s}}^q{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}\frac{|u^{+}\left(x\right)-u^{-}{\left(y\right)|}^{q-1}{u}^{+}\left(x\right)-{\left|u^{+}\left(x\right)\right|}^q}{{|x-y|}^{N+s_{2}q}}dxdy\hfill \\ \hfill & +{\overline{s}}^q{\int }_{{\left({\mathbb{R}}^N\right)}^{+}}{\int }_{{\left({\mathbb{R}}^N\right)}^{-}}\frac{|u^{-}\left(x\right)-u^{+}{\left(y\right)|}^{q-1}{u}^{+}\left(y\right)-{\left|u^{+}\left(y\right)\right|}^q}{{|x-y|}^{N+s_{2}q}}dxdy\hfill \\ \hfill & :={\overline{s}}^p{\parallel u^{+}\parallel }_1^p+{\overline{s}}^q{\parallel u^{+}\parallel }_2^q+{\overline{s}}^p{A}_1+{\overline{s}}^p{A}_2+{\overline{s}}^q{A}_3+{\overline{s}}^q{A}_4.\hfill \end{array}$$

$u\in {\mathcal{M}}_{\lambda ,K}$ gives

$${\int }_{{\mathbb{R}}^N}{h}_{\lambda ,K}\left(u^{+}\right)u^{+}dx={\parallel u^{+}\parallel }_1^p+{\parallel u^{+}\parallel }_2^q+A_1+A_2+A_3+A_4.$$

(13)

If $\overline{s}>1$, in view of $\left(h_3\right)$, we obtain

$$\begin{array}{cc}\hfill 0<& {\int }_{{\mathbb{R}}^N}\left[\frac{h_{\lambda ,K}\left(\overline{s}{u}^{+}\right)}{{\left(\overline{s}{u}^{+}\right)}^{q-1}}-\frac{h_{\lambda ,K}\left(u^{+}\right)}{{u^{+}}^{q-1}}\right]{\left(u^{+}\right)}^{q}dx\hfill \\ \hfill \le & ({\overline{s}}^{p-q}-1){\parallel u^{+}\parallel }_1^p+({\overline{s}}^{p-q}-1)A_1+({\overline{s}}^{p-q}-1)A_2\hfill \\ \hfill <& 0.\hfill \end{array}$$

(14)

This is a contradiction. Similarly, we can have $\overline{t}\ge 1$. Therefore, $\overline{s}=\overline{t}=1$. Case 2: For $u\notin {\mathcal{M}}_{\lambda ,K}$. Using the method in ([25], page 90), the desired conclusion is obtained. □

Lemma 2. 

There exists $\beta >0$ such that $\parallel u^{\pm }\parallel \ge \beta$ for all $u\in {\mathcal{M}}_{\lambda ,K}$.

Proof 

For $u\in {\mathcal{M}}_{\lambda ,K}$, we only prove the result for $u^{+}$. It is easy to check that

$$\begin{array}{cc}\hfill & {\int }_{{\mathbb{R}}^N}{\int }_{{\mathbb{R}}^N}\frac{{|u\left(x\right)-u\left(y\right)|}^{p-2}(u\left(x\right)-u\left(y\right))\left(u^{+}\left(x\right)-u^{+}\left(y\right)\right)}{{|x-y|}^{N+p{s}_1}}dxdy\hfill \\ \hfill \ge & {\int }_{{\mathbb{R}}^N}{\int }_{{\mathbb{R}}^N}\frac{{\left|u^{+}\left(x\right)-u^{+}\left(y\right)\right|}^p}{{|x-y|}^{N+p{s}_1}}dxdy.\hfill \end{array}$$

(15)

Combining with $\left(h_1\right)$, it is shown that

$$\parallel u^{+}{\parallel }_1^p+\parallel u^{+}{\parallel }_2^q\le \epsilon {\int }_{{\mathbb{R}}^N}|u^{+}{|}^{p}dx+(1+\lambda K^{r-\sigma })C\left(\epsilon \right){\int }_{{\mathbb{R}}^N}{\left|u^{+}\right|}^{\sigma }dx.$$

(16)

Choosing $\epsilon =\frac{V_0}{2}$, we obtain

$$\frac{1}{2}\parallel u^{+}{\parallel }_1^p+{\parallel u^{+}\parallel }_2^q\le C{\parallel u^{+}\parallel }^{\sigma },$$

(17)

which implies the desired conclusion. □

Lemma 3. 

$m_{\lambda ,K}>0$.

Proof 

For $u\in {\mathcal{M}}_{\lambda ,K}$, with $\left(h_2\right)$, (10), (15) in hand, we have

$$\begin{array}{cc}\hfill I_{\lambda ,K}\left(u\right)=& I_{\lambda ,K}\left(u\right)-\frac{1}{\theta }⟨I_{\lambda ,K}^{\prime }\left(u\right),u^{+}⟩-\frac{1}{\theta }⟨I_{\lambda ,K}^{\prime }\left(u\right),u^{-}⟩\hfill \\ \hfill \ge & \left(\frac{1}{p}-\frac{1}{\theta }\right)\parallel u^{+}{\parallel }_1^p+\left(\frac{1}{q}-\frac{1}{\theta }\right){\parallel u^{+}\parallel }_2^q\hfill \\ \hfill & +\left(\frac{1}{p}-\frac{1}{\theta }\right)\parallel u^{-}{\parallel }_1^p+\left(\frac{1}{q}-\frac{1}{\theta }\right){\parallel u^{-}\parallel }_2^q.\hfill \end{array}$$

(18)

Jointly with Lemma 2, we derive $m_{\lambda ,K}>0$. □

Lemma 4. 

$m_{\lambda ,K}$ is achieved.

Proof 

Let $\left\{u_n\right\}$ be a minimization sequence. Since $\theta >q>p$, similar to (18), we obtain

$$\begin{array}{cc}\hfill m_{\lambda ,K}+o_n\left(1\right)& \ge \left(\frac{1}{p}-\frac{1}{\theta }\right)\parallel u_n^{+}{\parallel }_1^p+\left(\frac{1}{q}-\frac{1}{\theta }\right){\parallel u_n^{+}\parallel }_2^q\hfill \\ \hfill & +\left(\frac{1}{p}-\frac{1}{\theta }\right)\parallel u_n^{-}{\parallel }_1^p+\left(\frac{1}{q}-\frac{1}{\theta }\right){\parallel u_n^{-}\parallel }_2^q.\hfill \end{array}$$

(19)

Thus, $\left\{u_n^{\pm }\right\}$ is bounded in E. Using ([26], Lemma 2.5), up to a subsequence, there exists a $u_0\in E$ satisfying $u_0^{\pm }\ne 0$ and

$$\underset{n\to \infty }{lim}{\int }_{{\mathbb{R}}^N}{h}_{\lambda ,K}\left(u_n^{\pm }\right)u_n^{\pm }dx={\int }_{{\mathbb{R}}^N}{h}_{\lambda ,K}\left(u_0^{\pm }\right)u_0^{\pm }ddx.$$

(20)

Here, we only prove $u_0^{\pm }\ne 0$. If not, in consideration of (10), we deduce that

$$0=⟨I_{\lambda ,K}^{\prime }\left(u_n\right),u_n^{\pm }⟩\ge \frac{1}{p}\parallel u_n^{\pm }{\parallel }_1^p+\frac{1}{q}{\parallel u_n^{\pm }\parallel }_2^q+o_n\left(1\right).$$

(21)

This is in contradiction with Lemma 2. Based on Lemma 1, there exist $s_0,t_0>0$ such that

$$⟨I_{\lambda ,K}^{\prime }(s_0{u_0}^{+}+t_0{u_0}^{-}),s_0{u_0}^{+}⟩=⟨I_{\lambda ,K}^{\prime }(s_0{u_0}^{+}+t_0{u_0}^{-}),t_0{u_0}^{-}⟩=0.$$

(22)

Obviously, similar to (10), $u_n\in {\mathcal{M}}_{\lambda ,K}$ shows that $⟨I_{\lambda ,K}^{\prime }\left(u_0\right),u_0^{+}⟩\le 0$. Similarly, $⟨I_{\lambda ,K}^{\prime }\left(u_0\right),u_0^{-}⟩\le 0$. In the spirit of ([27], Lemma 3.2), we have $0<s_0,t_0\le 1$. Using (10) again, we find that

$$\parallel s_0{u_0}^{+}+t_0{u_0}^{-}{\parallel }_1^p\le {\parallel {u_0}^{+}+{u_0}^{-}\parallel }_1^p.$$

(23)

Taking into account $\left(h_2\right)$, we have

$$\begin{array}{cc}\hfill & m_{\lambda ,K}\hfill \\ \hfill \le & I_{\lambda ,K}\left(s_0{u_0}^{+}+t_0{u_0}^{-}\right)-\frac{1}{q}⟨I^{\prime }\left(s_0{u_0}^{+}+t_0{u_0}^{-}\right),\left(s_0{u_0}^{+}+t_0{u_0}^{-}\right)⟩\hfill \\ \hfill \le & \underset{n\to \infty }{lim\; inf}\left(I_{\lambda ,K}\left(u_n\right)-\frac{1}{q}⟨I_{\lambda ,K}^{\prime }\left(u_n\right),u_n⟩\right)\hfill \\ \hfill =& \underset{n\to \infty }{lim}{I}_{\lambda ,K}\left(u_n\right)=m_{\lambda ,K}.\hfill \end{array}$$

(24)

It indicates that $s_0{u_0}^{+}+t_0{u_0}^{-}$ is the minimizer. □

3.2. Proof of Theorem 1

We are devoted to proving Theorem 1 in this section. From Lemmas 1–4, we find that (6) possesses a minimizer $w_0:=s_0{u_0}^{+}+t_0{u_0}^{-}$. There are two methods to ensure that the minimizer $w_0$ is a critical point of $I_{\lambda ,K}$. One method can be seen in ([21], Section 3). The other method can be used as ([28], lemma 3.6). Using a standard Moser iteration (see e.g., [22] or [23]), we can draw the conclusion that ${\lambda }_0>0$ such that $\parallel w_0{\parallel }_{L^{\infty }\left({\mathbb{R}}^N\right)}\le K$ when $\lambda \in (0,{\lambda }_0)$. Thus, $w_0$ is a sign-changing solution of the initial problem (1), which means that in (4), we do not make any truncations.

4. Conclusions and Future Studies

With the above analysis made, the following conclusions can be drawn. Under the assumptions of Theorem 1, problem (1) admits a sign-changing solution. As mentioned in Remark 1, the sign-changing solution $w_0$ does not necessarily mean that it is a least energy sign-changing solution. Our future work will study the ground state or least energy sign-changing solution to (1). Maybe it is an open problem since it appears as the supercritical term ${\left|u\right|}^{r-2}u$.