Existence of Multiple Solutions to a Transmission Problem

Abstract

The transmission problem has emerged as an important component in various application fields, including industry, agriculture, and commerce. Simultaneously, it is also one of the most common challenges encountered in daily activities such as eating, drinking, living, and transportation. We have observed that numerous theories have been proposed regarding nonlocal transmission problems. However, how to find critical points of nonconvex functionals remains one of the major difficulties when one employs variational methods to establish the existence of solutions for nonlocal problems. This paper is devoted to the study of a class of weighted nonlocal transport equations. In order to establish the existence of solutions, we utilize algebraic analysis and combinatorial approximation techniques to overcome the nonconvexity of the corresponding Euler–Lagrange functional, and then prove the existence of a positive solution with the help of the strong maximum principle. Finally, by combining algebraic analysis with existing results, we obtain the existence of infinite solutions.

1. Background and Main Results

Transmission problems exist in every aspect of industrial production and daily life. In theoretical analysis, the researchers often model the problem an appropriate nonlocal model equations, and study the existence of solutions to the corresponding equations.

In 1876, Kirchhoff [1] proposed the nonlocal model

$$\begin{array}{c}\hfill \varrho {\displaystyle \frac{{\partial }^2\xi }{\partial t^2}}-\varphi \left({\int }_{\mathrm{\Omega }}|\nabla \xi (t,x){|}^2\mathrm{d}x\right){\displaystyle \frac{{\partial }^2\xi }{\partial x^2}}=f(t,x,\xi )\end{array}$$

(1)

with homogeneous term $f(t,x,\xi )=0$ and

$$\varphi \left({\int }_{\mathrm{\Omega }}|\nabla \xi (t,x){|}^2\mathrm{d}x\right)=\left({\displaystyle \frac{p_0}{h}}+{\displaystyle \frac{E}{2L}}{\int }_0^L|{\displaystyle \frac{\partial \xi }{\partial x}}{|}^2\mathrm{d}x\right),$$

where the constants $p_0$, h, E and L are all positive, and $\xi =\xi (x,t)$ represents the vibration of a particle. On the one hand, Equation (1) can be used to describe extremely thin string vibration and thin rod bending or torsion without external influences, but its solution is not easily obtained in practice. Although mechanical mechanics theory has made great progress, it remains a great challenge to obtain the classical solution of Equation (1) in its application. On the other hand, due to Kirchhoff’s other brilliant achievements in physics and electromagnetism, the research on Kirchhoff’s nonlocal model has been downplayed in mechanical mechanics for more than a hundred years. Until around 1990, researchers started to rediscover Kirchhoff’s work. Arosio et al. [2,3], as members of the Spagnolo team, called Equation (1) the Kirchhoff-type equation. In recent years, extensive applied research has been conducted on nonlocal problems, with representative works such as references [4,5,6]. In the study of universal gravitation and electromagnetism, two particles exhibit repulsion when they are relatively close, but they produce attraction when they are relatively far. From the perspective of entropy increase and decrease, the stiffness coefficient is regarded as a positive modulus when attraction occurs, or rather a negative modulus when repulsion occurs. That is to say, when ${\displaystyle \frac{p_0}{h}}$ and ${\displaystyle \frac{E}{L}}$ in Equation (1) have opposite signs, it can be regarded as a Kirchhoff-type problem with a negative modulus [7,8]. Furthermore, whether the boundary ∂Ω of the defined domain Ω or the endpoints of the interval are fixed or not, its steady-state equation can be expressed as

$${\displaystyle -\left(a-b{\int }_{\mathrm{\Omega }}{|\nabla u|}^2\mathrm{d}x\right)\Delta u=f(x,u),x\in \mathrm{\Omega },}$$

(2)

where Ω $\subset {\mathbf{R}}^N(N\ge 1)$, and $a,b>0$. In recent years, research on problem (2) has received increasing attention. Indeed, Yagdjian in [9] concluded that problem (1) is stably solvable with the conditions

$${\displaystyle \varphi \left({\int }_{\mathrm{\Omega }}|\nabla \xi (t,x){|}^2\mathrm{d}x\right)=\left(-1+{\int }_{\mathrm{\Omega }}|\nabla \xi (t,x){|}^2\mathrm{d}x\right).}$$

(3)

If the solution has the form $\xi (t,x)=u\left(x\right)e^{-\varpi t}$, problem (3) can be transformed into problem (2), which motivates us to study the existence and properties of solutions to the problem (2) under different conditions.

A strongly singular case is given in [8], and the necessary and sufficient conditions for the existence of positive solutions are obtained. Concretely, there exists a positive constant ${\lambda }_{\ast }$ such that the following problem:

$$\left\{\begin{array}{cc}{\displaystyle -\left(a-b{\int }_{\mathrm{\Omega }}{|\nabla u|}^2\mathrm{d}x\right)\Delta u=\lambda \frac{g\left(x\right)}{u^q},}\hfill & x\in \mathrm{\Omega },\hfill \\ u=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}\right.$$

has two positive solutions when $\lambda \in (0,{\lambda }_{\ast })$. Moreover, reference [8] summarizes the current research status for the problem (2) with different nonlinear terms $f(x,u)$. On the whole space, reference [10] discusses the existence of one, two, and three solutions for the problem

$$-\left(a-b{\int }_{{\mathbb{R}}^3}{|\nabla u|}^{2}dx\right)\Delta u={\displaystyle \frac{u^3}{\left|x\right|}}+\lambda f\left(x\right),\mathrm{in}{\mathbb{R}}^3\setminus \left\{0\right\}.$$

(4)

The main conclusion is that problem (4) admits a solution for any constant $\lambda >0$, and the problem (4) has at least three solutions for $\lambda >0$ that are small enough.

For the case of variable exponent, researchers such as Chu et al. [11,12,13,14] and Vetro et al. [15,16] have done a lot of valuable work. Liu et al. [17,18] extended the problem (2) to the Heisenberg group. In [19], the existence of multiple solutions to the problem (2) has been extended to fractional-order problems. There are many results for singularity and critical exponents, sign-change potential and coupled systems; see, for instance, [20,21,22]. For more details on problem (2), we refer readers to the introductions in [8,19], where we can learn about the research status over the past decade and understand the existence and multiplicity of solutions to the problem (2) under the different conditions of $f(x,u)$. Nevertheless, research on Kirchhoff-type, elliptic, and other types of variational problems still cannot be ignored, and we recommend references [23,24,25,26,27,28,29] to interested readers.

Inspired by the above literature, this article considers the following nonlocal transmission problem:

$$\left\{\begin{array}{cc}{\displaystyle -\left(a-b{\int }_{\mathrm{\Omega }}{|\nabla u|}^2\mathrm{d}x\right)\Delta u=\lambda g\left(x\right)u^3,}\hfill & x\in \mathrm{\Omega },\hfill \\ u=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}\right.$$

(5)

where $a,b,\lambda$ are arbitrary positive numbers and Ω $\subset {\mathbf{R}}^N(N=1,2,3,4)$ is a domain with smooth boundary. Especially when $N=1$, the smooth boundary means that the one-way derivative of the function exists at the endpoints of the interval, and $g\in C\left(\overline{\mathrm{\Omega }}\right)$ is a weight function. Our proposed framework offers a rigorous, novel methodology for investigating the existence of critical points in non-convex functionals, addressing longstanding challenges in variational analysis. The main conclusions are as follows:

Theorem 1.  

Let $a,b,\lambda$ be arbitrary positive numbers; if $N=1,2,3,4$ and $g\left(x\right)$ is non-zero and non-negative, then problem (5) has at least one positive solution and one negative solution.

Theorem 2.  

Let $a,b,\lambda$ be arbitrary positive numbers; if $N=1,2,3$ and $g\left(x\right)$ is non-zero and non-negative, then problem (5) has infinite solutions ${\left\{u_n\right\}}_{n=1}^{\infty }$. Moreover, it holds that

$${\int }_{\mathrm{\Omega }}|\nabla u_{n+1}{|}^2\mathrm{d}x\ge {\int }_{\mathrm{\Omega }}{|\nabla u_n|}^2\mathrm{d}x$$

and

$${\displaystyle \underset{n\to \infty }{\lim }}{\int }_{\mathrm{\Omega }}{|\nabla u_n|}^2\mathrm{d}x={\displaystyle \frac{a}{b}}.$$

2. Preliminaries

This paper will present the proof of the main conclusions under the framework of Hilbert space $H_0^1\left(\mathrm{\Omega }\right)$. To this end, let X be a linear normed space, and $X^{\ast }$ denote its dual space, $\mathrm{\Phi }\left(u\right)=⟨\mathrm{\Phi },u⟩$ with $\mathrm{\Phi }\in X^{\ast }$ and $u\in X^{\ast }$. $L^s\left(\mathrm{\Omega }\right)$ represents the Lebesgue integrable space, and ${\parallel ·\parallel }_s$ denotes its norm. The basic knowledge required is as follows:

Definition 1  

([30]). A sequence ${\left\{u_i\right\}}_{i=1}^{\infty }\subset X$ is said to be strongly convergent if there is an $u_0\in X$ such that ${\displaystyle \underset{i\to \infty }{\lim }{\parallel u_i-u_0\parallel }_X=0}$, which is denoted as ${\displaystyle \underset{i\to \infty }{\lim }{u}_i=u_0}$ or simply $u_i\to u_0$. $u_0$ is called the strong limit of ${\left\{u_i\right\}}_{i=1}^{\infty }$, and we say that ${\left\{u_i\right\}}_{i=1}^{\infty }$ converges strongly to $u_0$.

A sequence ${\left\{u_i\right\}}_{i=1}^{\infty }\subset X$ is said to be weakly convergent if there is an $u_0\in X$ such that for every $\mathrm{\Phi }\in X^{\ast }$, it holds ${\displaystyle \underset{i\to \infty }{\lim }\mathrm{\Phi }\left(u_i\right)=\mathrm{\Phi }\left(u_0\right)}$, which is denoted as $u_i\rightharpoonup u_0$. The element $u_0$ is called the weak limit of ${\left\{u_i\right\}}_{i=1}^{\infty }$, and we say that ${\left\{u_1\right\}}_{i=1}^{\infty }$ converges weakly to $u_0$.

Proposition 1  

([30]). For any $\mathrm{\Phi }\in X^{\ast }$, we have

$$\mathrm{\Phi }(\alpha u+\beta \phi )=⟨\mathrm{\Phi },\alpha u+\beta \phi ⟩=\alpha ⟨\mathrm{\Phi },u⟩+\beta ⟨\mathrm{\Phi },\phi ⟩,\forall u,\phi \in X;\forall \alpha ,\beta \in \mathbf{R}.$$

Lemma 1 

([31,32]). Let X be a Banach space; then, X is a reflexive space if and only if the bounded sequence in X has a weakly convergent subsequence.

Proposition 2  

([30,32,33]). Let X be a Banach space, ${\left\{u_i\right\}}_{i=0}^{\infty }\subset X$. If ${\displaystyle u_k\rightharpoonup u_0}$, then there exist ${\lambda }_{\mu j}\in [0,1]$, $j=1,\dots ,N_k$, such that ${\displaystyle {\displaystyle \sum _{j=1}^{N_k}}{\lambda }_{kj}=1,k=1,2,\dots }$, and ${\displaystyle {\displaystyle \sum _{j=1}^{N_k}}{\lambda }_{kj}{u}_{k+j}\to u_0}$.

Definition 2  

([31,32]). Let the functional I be defined on X; then, I is said to be differentiable on $u\in X$ if there exists $\mathrm{\Phi }\in X^{\ast }$ such that

$$I\left(\phi \right)=I\left(u\right)+⟨\mathrm{\Phi },\phi -u⟩{+o(\parallel \phi -u\parallel }_X),\phi ,u\in X,$$

which is written as $I^{\prime }\left(u\right)=\mathrm{\Phi }$. The functional I is said to be differentiable on X if for every $u\in X$, there exists $I^{\prime }\left(u\right)\in X^{\ast }$ and $I^{\prime }:X\to X,u\mapsto I^{\prime }\left(u\right)$ continuous. This is written as $I\in C^1(X,\mathbf{R})$ and u is called the critical point of I if and only if $I^{\prime }\left(u\right)=0$.

Definition 3  

([30,31,32]). The space $H_0^1\left(\mathrm{\Omega }\right)$ is a Hilbert space, which is the completion of a linear normed space $C_0^1\left(\mathrm{\Omega }\right)$ under the norm

$$\parallel u\parallel ={\left({\int }_{\mathrm{\Omega }}{|\nabla u|}^2\mathrm{d}x\right)}^{{\displaystyle \frac{1}{2}}}={\left({\int }_{\mathrm{\Omega }}{\displaystyle \sum _{i=1}^N}|{\displaystyle \frac{\partial u}{\partial x_i}}{|}^2\mathrm{d}x\right)}^{{\displaystyle \frac{1}{2}}}.$$

Proposition 3  

([30,31,32]). Define the operation $(·,·)$ by

$$(u,\phi )={\int }_{\mathrm{\Omega }}\nabla u·\nabla \phi \mathrm{d}x,\forall u,\phi \in H_0^1\left(\mathrm{\Omega }\right),$$

then, $H_0^1\left(\mathrm{\Omega }\right)$ forms an inner product space, and $(·,·)$ is an inner product.

Proposition 4  

([30,31,32]). The space $H_0^1\left(\mathrm{\Omega }\right)$ is a reflexive Banach space. Moreover, $H_0^1\left(\mathrm{\Omega }\right)\subset H_0^1{\left(\mathrm{\Omega }\right)}^{\ast }$ by the Riesz Representation Theorem, thereby

$$⟨u,\phi ⟩=(u,\phi )={\int }_{\mathrm{\Omega }}\nabla u·\nabla \phi \mathrm{d}x,\forall u,\phi \in H_0^1\left(\mathrm{\Omega }\right).$$

Proposition 5  

([32,34]). Denoted by $2^{\ast }$ the critical exponent, $2^{\ast }=\infty$ with $N=1,2$ and $2^{\ast }={\displaystyle \frac{2N}{N-2}}$ when $N\ge 3$. Then, for ${\displaystyle \frac{2^{\ast }}{2^{\ast }-1}}<p\le q\le 2^{\ast }$ and $w\in L^{{\displaystyle \frac{qp}{q-p}}}\left(\mathrm{\Omega }\right)$, there is ${\displaystyle \frac{p}{p-1}}\in (1,2^{\ast })$ and

$$\begin{array}{cc}\hfill \left|{\int }_{\mathrm{\Omega }}vw\phi \mathrm{d}x\right|& \le {\left({\int }_{\mathrm{\Omega }}{\left|\phi \right|}^{{\displaystyle \frac{p}{p-1}}}\mathrm{d}x\right)}^{{\displaystyle \frac{p-1}{p}}}{\left({\int }_{\mathrm{\Omega }}{\left|v\right|}^p{\left|w\right|}^p\mathrm{d}x\right)}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill & \le {\left({\int }_{\mathrm{\Omega }}{\left|\phi \right|}^{{\displaystyle \frac{p}{p-1}}}\mathrm{d}x\right)}^{{\displaystyle \frac{p-1}{p}}}{\left[{\left({\int }_{\mathrm{\Omega }}{\left|v\right|}^{p·{\displaystyle \frac{q}{q-p}}}\mathrm{d}x\right)}^{{\displaystyle \frac{q-p}{q}}}{\left({\int }_{\mathrm{\Omega }}{\left|w\right|}^{p·{\displaystyle \frac{q}{p}}}\mathrm{d}x\right)}^{{\displaystyle \frac{p}{q}}}\right]}^{{\displaystyle \frac{1}{p}}}\hfill \\ \hfill & \le {\left({\int }_{\mathrm{\Omega }}{\left|\phi \right|}^{{\displaystyle \frac{p}{p-1}}}\mathrm{d}x\right)}^{{\displaystyle \frac{p-1}{p}}}{\left({\int }_{\mathrm{\Omega }}{\left|v\right|}^{{\displaystyle \frac{qp}{q-p}}}\mathrm{d}x\right)}^{{\displaystyle \frac{q-p}{qp}}}{\left({\int }_{\mathrm{\Omega }}{\left|w\right|}^q\mathrm{d}x\right)}^{{\displaystyle \frac{1}{q}}}\hfill \\ \hfill & =\parallel \phi {\parallel }_{{\displaystyle \frac{p}{p-1}}}\parallel v{\parallel }_{{\displaystyle \frac{qp}{q-p}}}\parallel w{\parallel }_q<+\infty ,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right).\hfill \end{array}$$

Moreover, the embedding $H_0^1\left(\mathrm{\Omega }\right)\hookrightarrow L^s\left(\mathrm{\Omega }\right)$ is compact for all $s\in [1,2^{\ast })$, and $H_0^1\left(\mathrm{\Omega }\right)\hookrightarrow L^{2^{\ast }}\left(\mathrm{\Omega }\right)$ is continuous but not compact; thus, for $1\le {\displaystyle \frac{qp}{q-p}}\le 2^{\ast }$, there is

$$\left|{\int }_{\mathrm{\Omega }}vw\phi \mathrm{d}x\right|\le C\parallel \phi \parallel \parallel v\parallel \parallel w\parallel ,\forall v,w,\phi \in H_0^1\left(\mathrm{\Omega }\right),$$

where $C=C(N,\mathrm{\Omega },p,q)>0$ is a constant depending only on known data.

Definition 4.  

A function $u\in H_0^1\left(\mathrm{\Omega }\right)$ is called a solution of the Equation (5) if

$$\left(a-b{\int }_{\mathrm{\Omega }}{|\nabla u|}^2\mathrm{d}x\right){\int }_{\Omega }\nabla u\nabla \phi \mathrm{d}x=\lambda {\int }_{\mathrm{\Omega }}g\left(x\right)u^3\phi \mathrm{d}x,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right).$$

3. Proof of Main Results

Now, we will transform the solution of problem (5) into a critical point of the energy functional under the variational framework, and obtain the existence of critical points using combined approximation.

3.1. Variational Framework

In this part, we give the variational framework by using the method in ([32], Section 8.5, pp. 508–510). Define the functional $I:H_0^1\left(\mathrm{\Omega }\right)\to \mathbf{R}$, $u\mapsto I\left(u\right)$ as follows:

$$I\left(u\right)={\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x,$$

then

$$\begin{array}{cc}\hfill I\left(\phi \right)=& {\displaystyle \frac{b}{4}}{\parallel \phi \parallel }^4-{\displaystyle \frac{a}{2}}{\parallel \phi \parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{\phi }^4\mathrm{d}x\hfill \\ \hfill =& {\displaystyle \frac{b}{4}}{\parallel u+(\phi -u)\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u+(\phi -u)\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{\left[u+(\phi -u)\right]}^4\mathrm{d}x\hfill \\ \hfill =& {\displaystyle \frac{b}{4}}{\left({\parallel u\parallel }^2+2⟨u,\phi -u⟩+{\parallel \phi -u\parallel }^2\right)}^2-{\displaystyle \frac{a}{2}}\left({\parallel u\parallel }^2+2⟨u,\phi -u⟩+{\parallel \phi -u\parallel }^2\right)\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{\left[u+(\phi -u)\right]}^4\mathrm{d}x\hfill \\ \hfill =& {\displaystyle \frac{b}{4}}{(\parallel u\parallel }^4+{4\parallel u\parallel }^2⟨u,\phi -u⟩+4{⟨u,\phi -u⟩}^2+4{\parallel \phi -u\parallel }^2⟨u,\phi -u⟩\hfill \\ & +{2\parallel u\parallel }^2{\parallel \phi -u\parallel }^2+{\parallel \phi -u\parallel }^4)-{\displaystyle \frac{a}{2}}\left({\parallel u\parallel }^2+2⟨u,\phi -u⟩+{\parallel \phi -u\parallel }^2\right)\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g\left[u^4+4u{(\phi -u)}^3+6u^2{(\phi -u)}^2+4u^3(\phi -u)+{(\phi -u)}^4\right]\mathrm{d}x\hfill \\ \hfill =& \left({\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x\right)-\left[\left(a-{b\parallel u\parallel }^2\right)⟨u,\phi -u⟩-\lambda {\int }_{\mathrm{\Omega }}g{u}^3(\phi -u)\mathrm{d}x\right]\hfill \\ \hfill & +\left(b{⟨u,\phi -u⟩}^2{+b\parallel \phi -u\parallel }^2⟨u,\phi -u⟩+{\displaystyle \frac{b}{2}}{\parallel u\parallel }^2{\parallel \phi -u\parallel }^2+{\displaystyle \frac{b}{4}}{\parallel \phi -u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel \phi -u\parallel }^2\right)\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g\left[4u{(\phi -u)}^3+6u^2{(\phi -u)}^2+{(\phi -u)}^4\right]\mathrm{d}x\hfill \\ \hfill =& I\left(u\right)-\left[\left(a-{b\parallel u\parallel }^2\right)⟨u,\phi -u⟩-{\int }_{\mathrm{\Omega }}\nabla \left[{(-\Delta )}^{-1}\left(\lambda g{u}^3\right)\right]·\nabla (\phi -u)\mathrm{d}x\right]+{\rm Y}(\phi ,u)\hfill \\ \hfill =& I\left(u\right)+⟨-\left(a-{b\parallel u\parallel }^2\right)u-{(-\Delta )}^{-1}\left(\lambda g{u}^3\right),\phi -u⟩+{\rm Y}(\phi ,u),\hfill \\ \hfill =& I\left(u\right)+⟨{\mathrm{\Phi }}_u,\phi -u⟩+{\rm Y}(\phi ,u),\hfill \end{array}$$

where

$${\mathrm{\Phi }}_u=-\left(a-{b\parallel u\parallel }^2\right)u+{(-\Delta )}^{-1}\left(g{u}^3\right),$$

and ${\rm Y}(\phi ,u)$ is a nonlinear operator with $(\phi ,u)$ which is expressed as follows:

$$\begin{array}{cc}\hfill {\rm Y}(\phi ,u)=& (b{⟨u,\phi -u⟩}^2{+b\parallel \phi -u\parallel }^2⟨u,\phi -u⟩+{\displaystyle \frac{b}{2}}{\parallel u\parallel }^2{\parallel \phi -u\parallel }^2+{\displaystyle \frac{b}{4}}{\parallel \phi -u\parallel }^4\hfill \\ \hfill & -{\displaystyle \frac{a}{2}}{\parallel \phi -u\parallel }^2)+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g\left[4u{(\phi -u)}^3+6u^2{(\phi -u)}^2+{(\phi -u)}^4\right]\mathrm{d}x.\hfill \end{array}$$

Take the test function $v\in H_0^1\left(\mathrm{\Omega }\right)$ to get

$$\begin{array}{cc}\hfill ⟨{\mathrm{\Phi }}_u,v⟩=& ⟨-\left(a-{b\parallel u\parallel }^2\right)u-{(-\Delta )}^{-1}\left(\lambda g{u}^3\right),v⟩\hfill \\ \hfill =& -\left[\left(a-{b\parallel u\parallel }^2\right)⟨u,v⟩-{\int }_{\mathrm{\Omega }}\nabla \left[{(-\Delta )}^{-1}\left(\lambda g{u}^3\right)\right]·\nabla v\mathrm{d}x\right]\hfill \\ \hfill =& -\left[\left(a-{b\parallel u\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u\nabla v\mathrm{d}x-\lambda {\int }_{\mathrm{\Omega }}g{u}^{3}v\mathrm{d}x\right].\hfill \end{array}$$

(6)

Moreover, for any bounded functions $u_1,u_2\in H_0^1\left(\mathrm{\Omega }\right)$ with $\parallel u_1\parallel ,\parallel u_2\parallel \le M$, by using the results in Proposition 4, we have

$$\begin{array}{cc}\hfill \left|⟨{\mathrm{\Phi }}_{u_1}-{\mathrm{\Phi }}_{u_2},v⟩\right|=& |-\left[\left(a-b\parallel u_1{\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u_1\nabla v\mathrm{d}x-\lambda {\int }_{\mathrm{\Omega }}g{u}_1^{3}v\mathrm{d}x\right]\hfill \\ \hfill & +\left[\left(a-b\parallel u_2{\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u_2\nabla v\mathrm{d}x-\lambda {\int }_{\mathrm{\Omega }}g{u}_2^{3}v\mathrm{d}x\right]|\hfill \\ \hfill =& |a{\int }_{\mathrm{\Omega }}\nabla (u_2-u_1)\nabla v\mathrm{d}x+b\parallel u_1{\parallel }^2{\int }_{\mathrm{\Omega }}\nabla u_1\nabla v\mathrm{d}x-b{\parallel u_1\parallel }^2{\int }_{\mathrm{\Omega }}\nabla u_2\nabla v\mathrm{d}x\hfill \\ \hfill & +b\parallel u_1{\parallel }^2{\int }_{\mathrm{\Omega }}\nabla u_2\nabla v\mathrm{d}x-b{\parallel u_2\parallel }^2{\int }_{\mathrm{\Omega }}\nabla u_2\nabla v\mathrm{d}x+\lambda {\int }_{\mathrm{\Omega }}g(u_1^3-u_2^3)v\mathrm{d}x|\hfill \\ \hfill =& |a{\int }_{\mathrm{\Omega }}\nabla (u_2-u_1)\nabla v\mathrm{d}x+b{\parallel u_1\parallel }^2{\int }_{\mathrm{\Omega }}\nabla (u_1-u_2)\nabla v\mathrm{d}x\hfill \\ \hfill & +b(\parallel u_1{\parallel }^2-\parallel u_2{\parallel }^2){\int }_{\mathrm{\Omega }}\nabla u_2\nabla v\mathrm{d}x+\lambda {\int }_{\mathrm{\Omega }}g(u_1-u_2)(u_1^2+u_1{u}_2+u_2^2)v\mathrm{d}x|\hfill \\ \hfill \le & a\parallel v\parallel \parallel u_1-u_2\parallel +b\parallel v\parallel \parallel u_1{\parallel }^2\parallel u_1-u_2\parallel +b\parallel v\parallel \parallel u_2\parallel \parallel u_1+u_2\parallel \parallel u_1-u_2\parallel \hfill \\ \hfill & +{\displaystyle \frac{3\lambda }{2}}\underset{x\in \overline{\mathrm{\Omega }}}{\max }g\left(x\right){\int }_{\mathrm{\Omega }}\left|(u_1-u_2)(u_1^2+u_2^2)v\right|\mathrm{d}x\hfill \\ \hfill \le & \left[a+b\parallel u_1{\parallel }^2+\parallel u_1\parallel +\parallel u_2\parallel +b\parallel u_2\parallel (\parallel u_1\parallel +\parallel u_2\parallel )\right]\parallel v\parallel \parallel u_1-u_2\parallel \hfill \\ \hfill & +{\displaystyle \frac{3\lambda }{2}}\underset{x\in \overline{\mathrm{\Omega }}}{\max }g\left(x\right)\left[{\left({\int }_{\mathrm{\Omega }}{\left|u_1^2\right|}^{{\displaystyle \frac{qp}{q-p}}}\mathrm{d}x\right)}^{{\displaystyle \frac{q-p}{qp}}}+{\left({\int }_{\mathrm{\Omega }}{\left|u_2^2\right|}^{{\displaystyle \frac{qp}{q-p}}}\mathrm{d}x\right)}^{{\displaystyle \frac{q-p}{qp}}}\right]\hfill \\ \hfill & ·{\left({\int }_{\mathrm{\Omega }}{\left|v\right|}^{{\displaystyle \frac{p}{p-1}}}\mathrm{d}x\right)}^{{\displaystyle \frac{p-1}{p}}}{\left({\int }_{\mathrm{\Omega }}{|u_1-u_2|}^q\mathrm{d}x\right)}^{{\displaystyle \frac{1}{q}}},\hfill \end{array}$$

where ${\displaystyle \frac{2^{\ast }}{2^{\ast }-1}}<p\le q\le 2^{\ast }$ and ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{qp}{q-p}}\le {\displaystyle \frac{2^{\ast }}{2}}$. In particular, for $q=4$ and $p={\displaystyle \frac{4}{3}}$, we have ${\displaystyle \frac{qp}{q-p}}=2$ and ${\displaystyle \frac{p}{p-1}}=4$ when $1\le N\le 4$. By using the results in Proposition 4, there exists a positive constant $C=C(N,\mathrm{\Omega },{\displaystyle \frac{4}{3}},4)>0$, such that

$$\begin{array}{cc}\hfill \left|⟨{\mathrm{\Phi }}_{u_1}-{\mathrm{\Phi }}_{u_2},v⟩\right|\le & \left[a+b\parallel u_1{\parallel }^2+\parallel u_1\parallel +\parallel u_2\parallel +b\parallel u_2\parallel (\parallel u_1\parallel +\parallel u_2\parallel )\right]\parallel v\parallel \parallel u_1-u_2\parallel \hfill \\ \hfill & +{\displaystyle \frac{3\lambda }{2}}\underset{x\in \overline{\mathrm{\Omega }}}{\max }g\left(x\right)\left[{\left({\int }_{\mathrm{\Omega }}{u}_1^4\mathrm{d}x\right)}^{{\displaystyle \frac{1}{4}}}+{\left({\int }_{\mathrm{\Omega }}{u}_2^4\mathrm{d}x\right)}^{{\displaystyle \frac{1}{4}}}\right]·\hfill \\ \hfill & ·{\left({\int }_{\mathrm{\Omega }}{\left|v\right|}^4\mathrm{d}x\right)}^{{\displaystyle \frac{1}{4}}}{\left({\int }_{\mathrm{\Omega }}{|u_1-u_2|}^4\mathrm{d}x\right)}^{{\displaystyle \frac{1}{4}}}\hfill \\ \hfill \le & \left[a+2M+3b{M}^2+3\lambda MC\right]\parallel v\parallel \parallel u_1-u_2\parallel .\hfill \end{array}$$

Therefore, we have

$${\parallel {\mathrm{\Phi }}_{u_1}-{\mathrm{\Phi }}_{u_2}\parallel }_{H_0^1{\left(\mathrm{\Omega }\right)}^{\ast }}=\underset{v\in H_0^1\left(\mathrm{\Omega }\right),\parallel v\parallel \ne 0}{\sup }{\displaystyle \frac{\left|⟨{\mathrm{\Phi }}_{u_1}-{\mathrm{\Phi }}_{u_2},v⟩\right|}{\parallel v\parallel }}\le \left[a+2M+3b{M}^2+3\lambda MC\right]\parallel u_1-u_2\parallel$$

with $\parallel u_1\parallel ,\parallel u_2\parallel \le M$. This means that $\mathrm{\Phi }\left(u\right)$ is a continuous mapping. Applying Proposition 4 again, there exist three positive constants $C_1,C_2,C_3$, depending only on known data, such that

$$\begin{array}{cc}\hfill \left|{\rm Y}\right(\phi ,u\left)\right|=& |b{⟨u,\phi -u⟩}^2{+b\parallel \phi -u\parallel }^2⟨u,\phi -u⟩+{\displaystyle \frac{b}{2}}{\parallel u\parallel }^2{\parallel \phi -u\parallel }^2+{\displaystyle \frac{b}{4}}{\parallel \phi -u\parallel }^4\hfill \\ \hfill & -{\displaystyle \frac{a}{2}}{\parallel \phi -u\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g\left[4u{(\phi -u)}^3+6u^2{(\phi -u)}^2+{(\phi -u)}^4\right]\mathrm{d}x|\hfill \\ \hfill \le & {b\parallel u\parallel }^2{\parallel \phi -u\parallel }^2{+b\parallel u\parallel \parallel \phi -u\parallel }^3+{\displaystyle \frac{b}{2}}{\parallel u\parallel }^2{\parallel \phi -u\parallel }^2+{\displaystyle \frac{a}{2}}{\parallel \phi -u\parallel }^2\hfill \\ \hfill & +{\displaystyle \frac{\lambda }{4}}\underset{x\in \overline{\mathrm{\Omega }}}{\max }g\left(x\right)\left(C_1{\parallel u\parallel \parallel \phi -u\parallel }^3+C_2{\parallel u\parallel }^2{\parallel \phi -u\parallel }^2+C_3{\parallel \phi -u\parallel }^4\right)\hfill \\ \hfill =& o(\parallel \phi -u\parallel ).\hfill \end{array}$$

Hence, there exists $\mathrm{\Phi }={\mathrm{\Phi }}_u\in H_0^1{\left(\mathrm{\Omega }\right)}^{\ast }$,

$${\mathrm{\Phi }}_u=-\left(a-{b\parallel u\parallel }^2\right)u+{(-\Delta )}^{-1}\left(g{u}^3\right),$$

such that

$$I\left(\phi \right)=I\left(u\right)+⟨\mathrm{\Phi },\phi -u⟩+o(\parallel \phi -u\parallel ),\phi ,u\in H_0^1\left(\mathrm{\Omega }\right),$$

which implies that I is a differentiable functional and ${\mathrm{\Phi }}_u$ is the Fréchet derivative at u.

Consequently, it can be inferred from (6) that

$$\begin{array}{c}\hfill ⟨I^{\prime }\left(u\right),\phi ⟩=-\left[\left(a-{b\parallel u\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u\nabla \phi \mathrm{d}x-\lambda {\int }_{\mathrm{\Omega }}g{u}^3\phi \mathrm{d}x\right]=0,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right)\end{array}$$

if there exists $u\in H_0^1\left(\mathrm{\Omega }\right)$ such that $I^{\prime }\left(u\right)=0$. That is to say,

$$\begin{array}{c}\hfill \left(a-b{\int }_{\mathrm{\Omega }}{|\nabla u|}^2\mathrm{d}x\right){\int }_{\mathrm{\Omega }}\nabla u\nabla \phi \mathrm{d}x=\lambda {\int }_{\mathrm{\Omega }}g{u}^3\phi \mathrm{d}x,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right).\end{array}$$

In other words, the critical point of I is the solution to problem (5).

3.2. Proof of Theorem 1

Proof. 

According to the previous description, in order to prove the existence of the solution to problem (5), it is only necessary to prove that functional I has the critical point. In fact,

$$\begin{array}{cc}\hfill \underset{\parallel u\parallel \to +\infty }{\lim }I\left(u\right)=& \underset{\parallel u\parallel \to +\infty }{\lim }\left({\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x\right)\hfill \\ \hfill \ge & \underset{\parallel u\parallel \to +\infty }{\lim }\left({\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2\right)\hfill \\ \hfill =& +\infty \hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I\left(u\right)=& {\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x\hfill \\ \hfill \ge & {\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2\hfill \\ \hfill \ge & -{\displaystyle \frac{a^2}{4b}},\hfill \end{array}$$

This indicates that the differentiable functional I is coercive and has a lower bound on $H_0^1\left(\mathrm{\Omega }\right)$. Since $I\left(u\right)=I(-u)=I\left(\right|u\left|\right)$, the minimization sequence of $I\left(u\right)$ can be a non-negative sequence. Thus, there exists a lower bound c and a non-negative minimization sequence ${\left\{u_n\right\}}_{n=1}^{\infty }\subset H_0^1\left(\mathrm{\Omega }\right)$, satisfying

$$\begin{array}{cc}\hfill \underset{u\in H_0^1\left(\mathrm{\Omega }\right)}{\inf }I\left(u\right)=& \underset{n\to \infty }{\lim }\left({\displaystyle \frac{b}{4}}\parallel u_n{\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u_n\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x\right):=c.\hfill \end{array}$$

It is easy to verify that the minimization sequence ${\left\{u_n\right\}}_{n=1}^{\infty }$ is bounded.

Now, we consider the positive sequence ${\left\{u_n\right\}}_{n=1}^{\infty }\subset H_0^1\left(\mathrm{\Omega }\right)$ with

$$\parallel u_n{\parallel }^2={\displaystyle \frac{a}{b}}·{\displaystyle \frac{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}}}{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}+\frac{\lambda }{b}}}}\in \left(0,{\displaystyle \frac{a}{b}}\right),$$

then

$$\begin{array}{cc}\hfill \underset{t\ge 0}{\min }I\left(t{u}_n\right)=& \underset{t\ge 0}{\min }\left({\displaystyle \frac{b}{4}}\parallel t{u}_n{\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel t{u}_n\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{\left(t{u}_n\right)}^4\mathrm{d}x\right)\hfill \\ \hfill =& \underset{t\ge 0}{\min }\left({\displaystyle \frac{b{A}_n^4+\lambda G_n}{4}}{t}^4-{\displaystyle \frac{a{A}_n^2}{2}}{t}^2\right)\hfill \\ \hfill =& -{\displaystyle \frac{a^2{A}_n^4}{4(b{A}_n^4+\lambda )}}=-{\displaystyle \frac{a^2}{4b}}·{\displaystyle \frac{\frac{A_n^4}{G_n}}{\frac{A_n^4}{G_n}+\frac{\lambda }{b}}}\hfill \\ \hfill =& -{\displaystyle \frac{a^2}{4b}}·{\displaystyle \frac{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}}}{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}+\frac{\lambda }{b}}}}\hfill \end{array}$$

if and only if $t=1$, where $A_n=\parallel u_n\parallel$ and $G_n={\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x$. So, ${\left\{u_n\right\}}_{n=1}^{\infty }$ is a minimization sequence. It is noted that

$$\begin{array}{c}\hfill I\left(u\right)={\displaystyle \frac{b}{4}}{\parallel u\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x\end{array}$$

is a locally convex functional when

$${\parallel u\parallel }^2\ge {\displaystyle \frac{a}{3b}}·{\displaystyle \frac{{\displaystyle \frac{{\parallel u\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x}}}{{\displaystyle \frac{{\parallel u\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}^4\mathrm{d}x}+\frac{\lambda }{b}}}},$$

we obtain that $I\left(u\right)$ is a convex functional in the closure formed by minimization sequence ${\left\{u_{k_j}\right\}}_{k=1}^{\infty }$ by

$$\parallel u_n{\parallel }^2={\displaystyle \frac{a}{b}}·{\displaystyle \frac{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}}}{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}+\frac{\lambda }{b}}}}>{\displaystyle \frac{a}{3b}}·{\displaystyle \frac{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}}}{{\displaystyle \frac{\parallel u_n{\parallel }^4}{{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x}+\frac{\lambda }{b}}}}.$$

With the help of the Lemma 1 and Proposition 3, the sequence ${\left\{u_n\right\}}_{n=1}^{\infty }$ has weakly convergent subsequences ${\left\{u_k\right\}}_{k=1}^{\infty }\subset {\left\{u_n\right\}}_{n=1}^{\infty }$; thus, there is a function $u_{\ast }\in H_0^1\left(\mathrm{\Omega }\right)$ such that $u_k\rightharpoonup u_{\ast }$. Concretely,

$$\begin{array}{c}\hfill \underset{k\to \infty }{\lim }\mathrm{\Phi }\left(u_k\right)=\underset{k\to \infty }{\lim }⟨\mathrm{\Phi },u_k⟩=\mathrm{\Phi }\left(u\right)=⟨\mathrm{\Phi },u⟩,\forall \mathrm{\Phi }\in H_0^1{\left(\mathrm{\Omega }\right)}^{\ast }.\end{array}$$

(7)

Since $u_k\rightharpoonup u_{\ast }$, Proposition 2 implies that there exists a positive integer $N_k$ and a sequence ${\left\{{\mu }_{kj}\right\}}_{j=1}^{N_k}\subset [0,1]$, such that

$${\displaystyle {\displaystyle \sum _{j=1}^{N_k}}{\mu }_{kj}=1,\forall k=1,2,\dots ,{\displaystyle \sum _{j=1}^{N_k}}{\mu }_{kj}{u}_{k+j}\to u_{\ast },u_{\ast }\left(x\right)\ge 0,\forall x\in \mathrm{\Omega }.}$$

Therefore, the following conclusion can be drawn from the properties of convex functionals:

$$\begin{array}{cc}\hfill c\le I\left(u_{\ast }\right)=& {\displaystyle \frac{b}{4}}\parallel u_{\ast }{\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u_{\ast }\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}_{\ast }^4\mathrm{d}x\hfill \\ \hfill =& {\displaystyle \frac{b}{4}}\parallel \underset{k\to \infty }{\lim }{\displaystyle \sum _{j=1}^{N_k}}{\mu }_{kj}{u}_{k+j}{\parallel }^4-{\displaystyle \frac{a}{2}}\parallel \underset{k\to \infty }{\lim }{\displaystyle \sum _{j=1}^{N_k}}{\mu }_{kj}{u}_{k+j}{\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g|\underset{k\to \infty }{\lim }{\displaystyle \sum _{j=1}^{N_k}}{\mu }_{kj}{u}_{k+j}{|}^4\mathrm{d}x\hfill \\ \hfill \le & {\displaystyle \frac{b}{4}}\underset{k\to \infty }{\lim }{\displaystyle \sum _{j=1}^{N_k}}\parallel u_{k+j}{\parallel }^4-{\displaystyle \frac{a}{2}}\underset{k\to \infty }{\lim }{\displaystyle \sum _{j=1}^{N_k}}\parallel u_{k+j}{\parallel }^2+{\displaystyle \frac{\lambda }{4}}\underset{k\to \infty }{\lim }{\int }_{\mathrm{\Omega }}g|u_{k+j}{|}^4\mathrm{d}x\hfill \\ \hfill =& \underset{k\to \infty }{\lim }\left({\displaystyle \frac{b}{4}}\parallel u_{k+j}{\parallel }^4-{\displaystyle \frac{a}{2}}\parallel u_{k+j}{\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g|u_{k+j}{|}^4\mathrm{d}x\right)\hfill \\ \hfill =& \underset{n\to \infty }{\lim }\left({\displaystyle \frac{b}{4}}\parallel u_n{\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u_n\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}_n^4\mathrm{d}x\right)\hfill \\ \hfill =& c.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill \underset{u\in H_0^1\left(\mathrm{\Omega }\right)}{\inf }I\left(u\right)=c=I\left(u_{\ast }\right)={\displaystyle \frac{b}{4}}\parallel u_{\ast }{\parallel }^4-{\displaystyle \frac{a}{2}}{\parallel u_{\ast }\parallel }^2+{\displaystyle \frac{\lambda }{4}}{\int }_{\mathrm{\Omega }}g{u}_{\ast }^4\mathrm{d}x.\end{array}$$

Because $I\left(u_{\ast }\right)$ is a minimum value, we just replace $\phi$ with $(u+\phi )$ and $(u-\phi )$, respectively; then, there will be a conclusion that

$$\begin{array}{cc}\hfill I(u+\phi )-I\left(u_{\ast }\right)=& -\left[\left(a-b\parallel u_{\ast }{\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u_{\ast }\nabla \phi \mathrm{d}x-\lambda {\int }_{\mathrm{\Omega }}g{u}_{\ast }^3\phi \mathrm{d}x\right]+o(\parallel \phi \parallel )\hfill \\ \hfill \ge & 0,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right),\hfill \\ \hfill I(u-\phi )-I\left(u_{\ast }\right)=& \left[\left(a-b\parallel u_{\ast }{\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u_{\ast }\nabla \phi \mathrm{d}x-\lambda {\int }_{\mathrm{\Omega }}g{u}_{\ast }^3\phi \mathrm{d}x\right]+o(\parallel \phi \parallel )\hfill \\ \hfill \ge & 0,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right).\hfill \end{array}$$

Comparing the above two equations, it can be seen that

$$\begin{array}{c}\hfill \left(a-b\parallel u_{\ast }{\parallel }^2\right){\int }_{\mathrm{\Omega }}\nabla u_{\ast }\nabla \phi \mathrm{d}x=\lambda {\int }_{\mathrm{\Omega }}g{u}_{\ast }^3\phi \mathrm{d}x,\forall \phi \in H_0^1\left(\mathrm{\Omega }\right).\end{array}$$

From Definition 4, we obtain that $u_{\ast }$ is a nonzero and non-negative solution of problem (5).

Taking the test function $\phi =u_{\ast }$ to get that

$$\begin{array}{c}\hfill \left(a-b\parallel u_{\ast }{\parallel }^2\right){\parallel u_{\ast }\parallel }^2=\lambda {\int }_{\mathrm{\Omega }}g{u}_{\ast }^4\mathrm{d}x>0,\end{array}$$

we have $(a-b\parallel u_{\ast }{\parallel }^2)>0$. Hence

$$\left\{\begin{array}{cc}{\displaystyle -\Delta u_{\ast }=\frac{\lambda g\left(x\right)u_{\ast }^3}{a-b{\int }_{\mathrm{\Omega }}{|\nabla u_{\ast }|}^2\mathrm{d}x}\gneqq 0,}\hfill & x\in \mathrm{\Omega },\hfill \\ u_{\ast }=0,\hfill & x\in \partial \mathrm{\Omega }.\hfill \end{array}\right.$$

Because the identity $u_{\ast }(\partial \mathrm{\Omega })=0$, the function $u_{\ast }\left(x\right)$ cannot reach non-negative maximum and non-positive minimum values inside Ω by the strong maximum principle. Thus, $u_{\ast }$ is positive. Since $-u_{\ast }$ also satisfies the equation, problem (5) has at least one positive solution and one negative solution. □

3.3. Proof of Theorem 2

The proof of Theorem 2 is directly tied to the existence of infinite solutions to the following problem:

$$\left\{\begin{array}{cc}{\displaystyle \mathcal{L}\left(v\right)\equiv -{\displaystyle \sum _{i,j=1}^N}{\left(a_{ij}\left(x\right)v_{x_i}\right)}_{x_j}+c\left(x\right)v=f(x,v),}\hfill & x\in \mathrm{\Omega },\hfill \\ u=0,\hfill & x\in \partial \mathrm{\Omega }\hfill \end{array}\right.$$

(8)

where $\mathcal{L}\left(v\right)$ denotes a uniformly elliptic operator, the function $a_{ij}\left(x\right)\in C^1\left(\mathrm{\Omega }\right)$ with its first-order derivatives satisfying Hölder continuity on Ω, and the function $c\left(x\right)$ is non-negative and satisfies the Hölder continuity condition on Ω.

Assumption 1 

([35], Section 3). Let $F(x,z)={\int }_0^{z}f(x,\tau )d\tau$ and assume that

$\left(f_1\right)$ $f(x,0)=0$ and f is local Hölder continuous on $\overline{\mathrm{\Omega }}\times \mathbf{R}$;

$\left(f_2\right)$ There exist three positive constants $a_1,a_2,a_3$, such that $\left|f(x,z)\right|\le a_1+a_2{\left|z\right|}^s$ with $N>2$, $\left|f(x,z)\right|\le a_3{e}^{a\left(z\right)}$ with $N=2$, where $s\in (1,{\displaystyle \frac{N+2}{N-2}})$, $a\left(z\right)$ satisfies ${\displaystyle \underset{z\to \infty }{\lim }}{\displaystyle \frac{a\left(z\right)}{z^2}}=0$;

$\left(f_3\right)$ $f(x,z)=o\left(\right|z\left|\right)$ uniformly with respect to $x\in \overline{\mathrm{\Omega }}$ when $z\to 0$;

$\left(f_4\right)$ ${\displaystyle \frac{f(x,z)}{z}}\to \infty$ uniformly with respect to $x\in \overline{\mathrm{\Omega }}$ when $z\to +\infty$ and $z\to -\infty$;

$\left(f_5\right)$ There exists a positive constant $a_4$ such that $F(x,z)\le \theta f(x,z)z$ with $\left|x\right|\ge a_4$, where $\theta \in [0,{\displaystyle \frac{1}{2}})$;

$\left(f_6\right)$ f is odd with respect to z on $\overline{\mathrm{\Omega }}$; thus, $f(x,z)=-f(x,-z),\forall x\in \overline{\mathrm{\Omega }}$.

Notation 1 

([35], Section 3). If condition $\left(f_6\right)$ is satisfied, and if $f(x,z)$ behaves like $z^s$ plus a lower-order term as $z\to \infty$, then condition $\left(f_5\right)$ also holds. This conclusion includes the case where $N=1$, in which condition $\left(f_2\right)$ can be omitted directly.

In order to obtain the main conclusion in Theorem 2, we need the following existing results:

Lemma 2 

([35], Theorems 3.13 and 3.14). Under Assumption 1 and Notation 1, problem (8) has infinite solutions ${\left\{v_n\right\}}_{n=1}^{\infty }$, $\parallel v_{n+1}\parallel \ge \parallel v_n\parallel$ and ${\displaystyle \underset{n\to \infty }{\lim }}\parallel v_n\parallel =+\infty$.

Proof of Theorem 2. 

To obtain the main conclusion in Theorem 2, we consider the following problem:

$$\left\{\begin{array}{cc}{\displaystyle -\Delta v=\lambda g\left(x\right)v^3,}\hfill & x\in \mathrm{\Omega },\hfill \\ v=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}\right.$$

(9)

Then, $\mathcal{L}=-\Delta$ is an uniformly elliptic operator with $\mathcal{L}\left(v\right)=-\Delta v$. It is easy to see that $f(x,v)=\lambda g\left(x\right)v^3$ satisfies both Assumption 1 and Notation 1. Therefore, problem (9) has infinite solutions ${\left\{v_n\right\}}_{n=1}^{\infty }$ by Lemma 2, $\parallel v_{n+1}\parallel \ge \parallel v_n\parallel$ and ${\displaystyle \underset{n\to \infty }{\lim }}\parallel v_n\parallel =+\infty$.

Based on algebraic analysis methods, for each solution $v_n$, if there exists a constant $t_n\in \mathbf{R}$, satisfying the equation

$$\left\{\begin{array}{cc}{\displaystyle -\left(a-b{\int }_{\mathrm{\Omega }}{|\nabla \left(t_n{v}_n\right)|}^2\mathrm{d}x\right)\Delta \left(t_n{v}_n\right)=\lambda g\left(x\right){\left(t_n{v}_n\right)}^3,}\hfill & x\in \mathrm{\Omega },\hfill \\ t_n{v}_n=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}\right.$$

(10)

then $u_n=t_n{v}_n$ is a solution to problem (5).

Since $-\Delta v_n=\lambda g\left(x\right)v_n^3$ and $v_n(\partial \mathrm{\Omega })=0$ contains

$$\left\{\begin{array}{cc}{\displaystyle -t_n^2\Delta \left(t_n{v}_n\right)=\lambda g\left(x\right){\left(t_n{v}_n\right)}^3,}\hfill & x\in \mathrm{\Omega },\hfill \\ t_n{v}_n=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}\right.$$

the existence of $t_n$ in Equation (10) is equivalent to the existence of roots for the following algebraic equation

$$\begin{array}{c}\hfill a-b\parallel t_n{v}_n{\parallel }^2=t_n^2=a-b{t}_n^2{\parallel v_n\parallel }^2.\end{array}$$

So, for every $v_n$, an obvious result is that there exists a unique

$$t_n=\pm \sqrt{{\displaystyle \frac{a}{1+b\parallel v_n{\parallel }^2}}}$$

satisfying the above equation. Let

$$u_n=\pm \sqrt{{\displaystyle \frac{a}{1+b\parallel v_n{\parallel }^2}}}·v_n,$$

we can easily verify the following identity

$$\left\{\begin{array}{cc}{\displaystyle -\left(a-b{\int }_{\mathrm{\Omega }}{|\nabla u_n|}^2\mathrm{d}x\right)\Delta u_n=\lambda g\left(x\right)u_n^3,}\hfill & x\in \mathrm{\Omega },\hfill \\ u_n=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}\right.$$

Therefore, we conclude that problem (5) admits infinite solutions.

In addition, the fact ${\displaystyle \underset{n\to \infty }{\lim }}\parallel v_n\parallel =+\infty$ implies that

$$\underset{n\to \infty }{\lim }\parallel u_n{\parallel }^2=\underset{n\to \infty }{\lim }\parallel \pm \sqrt{{\displaystyle \frac{a}{1+b\parallel v_n{\parallel }^2}}}{v}_n{\parallel }^2=\underset{n\to \infty }{\lim }{\displaystyle \frac{a}{1+b\parallel v_n{\parallel }^2}}{\parallel v_n\parallel }^2={\displaystyle \frac{a}{b}}.$$

Furthermore, if we let $h\left(t\right)={\displaystyle \frac{at}{1+bt}}$, we deduce that

$$h^{\prime }\left(t\right)={\displaystyle \frac{a(1+bt)-bat}{{(1+bt)}^2}}={\displaystyle \frac{a}{{(1+bt)}^2}}>0,\forall t\in [0,+\infty ).$$

This implies that the function $h\left(t\right)$ monotonically increases on $[0,+\infty )$. Therefore, from the above result and $\parallel v_{n+1}\parallel \ge \parallel v_n\parallel$, we obtain

$$\parallel u_{n+1}{\parallel }^2={\displaystyle \frac{a}{1+b\parallel v_{n+1}{\parallel }^2}}\parallel v_{n+1}{\parallel }^2\ge {\displaystyle \frac{a}{1+b\parallel v_n{\parallel }^2}}\parallel v_n{\parallel }^2={\parallel u_n\parallel }^2,$$

This concludes that

$${\int }_{\mathrm{\Omega }}{|\nabla u_{n+1}|}^2\mathrm{d}x\ge {\int }_{\mathrm{\Omega }}{|\nabla u_n|}^2\mathrm{d}x$$

and

$${\displaystyle \underset{n\to \infty }{\lim }}{\int }_{\mathrm{\Omega }}{|\nabla u_n|}^2\mathrm{d}x={\displaystyle \frac{a}{b}}$$

by Definition 3. □

4. Conclusions

In contrast to the findings presented in [36], our work adopts a novel combinatorial approximation approach to investigate problem (5). Specifically, our method eliminates the need for any additional assumptions on $g\left(x\right)$. We rigorously prove that for dimensions $N=1,2,3,4$, the problem admits at least one positive solution and one negative solution.

Furthermore, for the cases $N=1,2,3$, we establish an appropriate connection between local and nonlocal equations through general algebraic analysis techniques. Furthermore, we employ algebraic methods to rigorously verify the monotonicity property $\parallel u_{n+1}\parallel \ge \parallel u_n\parallel$ and the asymptotic behavior ${\displaystyle \underset{n\to \infty }{\lim }}{\parallel u_n\parallel }^2={\displaystyle \frac{a}{b}}$.