Existence of Nontrivial Solutions for Boundary Value Problems of Fourth-Order Differential Equations

Abstract

This article investigates the solvability problem of fourth-order differential equations with two-point boundary conditions; specifically, conclusions regarding sign-changing solutions are obtained. The methods used in this article are fixed-point theorems on lattices. Firstly, under some sublinear conditions, the existence of three nontrivial solutions is demonstrated, including a sign-changing solution, a negative solution and a positive solution. Secondly, under some unilaterally asymptotically linear and superlinear conditions, the existence of at least one sign-changing solution is proved. Finally, this article provides several specific examples to illustrate the obtained conclusions.

1. Introduction

In this paper, we discuss fourth-order differential equations with two-point boundary conditions:

$$\left\{\begin{array}{c}{x}^{\left(4\right)}\left(t\right)+{\alpha }_1{x}^{″}\left(t\right)-{\alpha }_{2}x\left(t\right)=g\left(t,x\left(t\right)\right),t\in [0,1],\hfill \\ x\left(0\right)=x\left(1\right)=0,\hfill \\ x^{″}\left(0\right)=x^{″}\left(1\right)=0,\hfill \end{array}\right.$$

(1)

where $g:[0,1]\times (-\infty ,+\infty )\to (-\infty ,+\infty )$ is a continuous function, and ${\alpha }_1,{\alpha }_2\in R$ satisfy ${\alpha }_1<2{\pi }^2, {\alpha }_2\ge -{\alpha }_1^2/4, {\alpha }_2/{\pi }^4+{\alpha }_1/{\pi }^2<1.$

Fourth-order differential equations have important applications in some fields, such as physics, engineering technology, and complex dynamic systems. They can be used to simulate and predict various natural phenomena, such as beam vibration, plate bending, etc. They can also be used to solve some engineering problems, such as structural analysis, control theory, etc. By establishing a suitable fourth-order differential equation model, it is possible to effectively analyze and design engineering systems and improve the stability of the systems. Based on the important applications of fourth-order differential equations, many scholars have studied the existence of solutions to fourth-order differential equations, providing theoretical support for some practical problems. There are many methods for studying the existence of solutions to differential equations, such as fixed-point theorems, upper and lower solution methods, iterative methods, fixed-point index theory, Mawhin superposition theory and so on. For applications of these methods, readers can refer to [1,2,3,4,5]. Currently, there are many studies on fourth-order differential equations in the literature, such as [6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22].

In [6], Ma and Wang discussed the following fourth-order differential equation:

$$\left\{\begin{array}{c}{\phi }^{\left(4\right)}\left(x\right)=h(x,u\left(x\right)),x\in [0,1],\hfill \\ \phi \left(0\right)=\phi \left(1\right)=0,\hfill \\ {\phi }^{″}\left(0\right)={\phi }^{″}\left(1\right)=0.\hfill \end{array}\right.$$

(2)

Problem (2) is the deformation model of an elastic beam, where $u\left(x\right)$ represents the weight of the elastic beam under the action of gravity, and the boundary condition describes that both endpoints of the beam are simply supported. The existence of at least one positive solution to Problem (2) is obtained by fixed-point theorems.

In [7], Yao investigated the following fourth-order boundary value problem:

$$\left\{\begin{array}{c}{\phi }^{\left(4\right)}\left(y\right)=f\left(y,\phi \left(y\right),{\phi }^{″}\left(y\right)\right),\hfill \\ \phi \left(0\right)=\phi \left(1\right)=0,\hfill \\ {\phi }^{″}\left(0\right)={\phi }^{″}\left(1\right)=0.\hfill \end{array}\right.$$

(3)

Multiple nontrivial solutions of Problem (3) have been given by the method of the Guo–Krasnosel’skii fixed-point theorem.

In [8], Cui and Kang studied the following fourth-order differential equation with a two-point boundary condition:

$$\left\{\begin{array}{c}{h}^{\left(4\right)}\left(t\right)+\alpha h^{″}\left(t\right)=f\left(h\left(t\right)\right),\hfill \\ h\left(0\right)=h\left(1\right)=0,\hfill \\ h^{″}\left(0\right)=h^{″}\left(1\right)=0.\hfill \end{array}\right.$$

(4)

From the bounded cone and positive operators, under some conditions, the authors proved that Problem (4) has at least a sign-changing solution.

In [9], Lu and Sun investigated fourth-order differential equations with a parameter as follows:

$$\left\{\begin{array}{c}{\psi }^{\left(4\right)}\left(s\right)=\lambda g\left(s,\psi \left(s\right)\right),\hfill \\ \psi \left(0\right)={\psi }^{\prime }\left(0\right)=0,\hfill \\ \psi \left(1\right)={\psi }^{\prime }\left(1\right)=0.\hfill \end{array}\right.$$

(5)

The authors used topological degree theory and fixed-point theory to obtain the existence results of positive solutions for Problem (5) when f can take negative values.

In [10,11], J.R.L Webb used a new method to give new existence theories and show the nonexistence of positive solutions for some nonlinear differential equations. These new theorems were proved by the relevant properties of Green’s function and the fixed-point theorem. These theorems have been to fourth-order boundary value problems. In [12,13], the authors considered the existence of positive solutions to some fourth-order differential equations with two-point boundary conditions, but the methods used were different. Lv et al. used the Guo–Krasnosel’skii fixed-point theorem in [12]. In [13], Younis et al. used fixed points for graphical Bc-Kannan contractions. In [14], Perov’s fixed-point theorem is used to study the uniqueness of nontrivial solutions for some fully fourth-order differential equations with two-point boundary conditions. In [15,16,17], in the case where the corresponding Green’s function can change sign, the authors used fixed-point theorems or fixed-point index theory to study the existence of positive solutions for some fourth-order differential equations with three-point boundary conditions.

In [18], combined with the calculation of the topological degree, Li et al. used the fixed-point index to obtain conclusions on the existence of at least three nontrivial solutions for some nonlinear operator equations, of which one is a positive solution, one is a negative solution, and the other is a sign-changing solution. Then, the conclusions were applied to Problem (1). In [19], Li also studied Problem (1). Under certain conditions, Li used the fixed-point index to conclude that there is at least one positive solution to Problem (1).

There are currently some studies on using fixed-point theorems to study sign-changing solutions for boundary value problems of differential equations, most of which are related to second-order or third-order differential equations, such as references [23,24,25,26]. In the literature on the study of solutions to fourth-order differential equations, the majority focuses on the existence of positive solutions, with only a few studies on sign-changing solutions [8,18].

Inspired by references [8,18,19], we continue to investigate the existence of nontrivial solutions to Problem (1), particularly the existence of sign-changing solutions. Eigenvalues are important indicators of linear operators, so in this article, we propose assumptions related to the eigenvalues of the corresponding linear operators. Firstly, under sublinear conditions related to the eigenvalues of the corresponding linear operators, we prove the existence of multiple nontrivial solutions, including a sign-changing solution, a negative solution and a positive solution. Secondly, we also provide the condition that the nonlinear term unilaterally satisfies asymptotic linearity, which is also related to the eigenvalues of the corresponding linear operators. At the same time, we also provide the condition that the nonlinear term is monotonically increasing. Under those conditions, we prove that Problem (1) has at least a sign-changing solution. Thirdly, under the superlinear condition, which is related to the eigenvalues of the corresponding linear operators, we also prove that Problem (1) has at least a sign-changing solution.

Compared with references [8,18,19], this article has the following differences. On the one hand, the methods we use are the fixed-point theorems with lattices obtained by Sun and Liu [20,21,23,24], which are different from the methods in references [8,18,19]. And, the space used is also different from that in references [18,19]. On the other hand, under the sublinear condition, we improve the condition $\left(D_4\right)$ of Theorem 3 in reference [15] and propose conditions related to the eigenvalues of the corresponding linear operators. And, reference [18] did not discuss the existence of sign-changing solutions under asymptotic linear and superlinear conditions. Reference [19] did not investigate the existence of sign-changing solutions to Problem (1) but only studied the existence of positive solutions to Problem (1) when the nonlinear term is nonnegative. Problem (1) has improved the problem in reference [8] and is more general.

2. Preliminaries

For a detailed introduction to cones, we can refer to references [2,3,4,5]. For an introduction to the calculation of topological degrees on lattices and the knowledge of fixed-point theorems with a lattice, references [27,28,29] can be referred to. In the following, we will give some lemmas that will be used in the following sections.

Let P be a total cone of a Banach space E. For $\forall x,y\in E$, if $sup\{x,y\}$ and $inf\{x,y\}$ exist, then E is called a lattice.

For $u\in E$, let

$$u^{+}=sup\{u,\theta \},u^{-}=sup\{-u,\theta \},$$

where $u^{+}$ and $u^{-}$ are called the positive part and negative part of u, respectively.

For convenience, let $u_{+}=u^{+},u_{-}=-u^{-}$.

Definition 1 

([1,27,28,29]). Suppose that $Q:D\subset E\to E$ is a nonlinear operator. If ∃$q^{*}\in E$ such that

$$Qu=Q{u}_{+}+Q{u}_{-}+q^{*},\forall u\in D,$$

then Q is said to be quasi-additive on a lattice.

To study Problem (1), we need some lemmas that arise from reference [18,19].

The polynomial $y^2+{\alpha }_{1}y-{\alpha }_2$ has two roots:

$${\zeta }_1=(-{\alpha }_1+\sqrt{{{\alpha }_1}^2+4{\alpha }_2})/2,{\zeta }_2=(-{\alpha }_1-\sqrt{{{\alpha }_1}^2+4{\alpha }_2})/2.$$

Obviously, ${\zeta }_1\ge {\zeta }_2\ge -{\pi }^2$.

Let $K_i$ be the Green’s function of the linear boundary value problem

$$\left\{\begin{array}{c}-x^{″}\left(t\right)+{\zeta }_{i}x\left(t\right)=0,t\in [0,1],\hfill \\ x\left(0\right)=x\left(1\right)=0,\hfill \end{array}\right.$$

By [18,19], we have the following specific expression of $K_i(t,s)$. Let ${\omega }_i=\sqrt{|{\zeta }_i|}$.

If ${\zeta }_i>0$, then $K_i$ is given by

$$K_i\left(t,s\right)={\displaystyle \frac{1}{{\omega }_{i}sinh{\omega }_i}}\left\{\begin{array}{c}sinh{\omega }_{i}t·sinh{\omega }_i(1-s),0\le t\le s\le 1,\hfill \\ sinh{\omega }_{i}s·sinh{\omega }_i(1-t),0\le s\le t\le 1.\hfill \end{array}\right.$$

(6)

If ${\zeta }_i=0$, then $K_i$ is given by

$$K_i\left(t,s\right)=\left\{\begin{array}{c}t(1-s),0\le t\le s\le 1,\hfill \\ s(1-t),0\le s\le t\le 1.\hfill \end{array}\right.$$

(7)

If $-{\pi }^2<{\zeta }_i<0$, then $K_i$ is given by

$$K_i\left(t,s\right)={\displaystyle \frac{1}{{\omega }_{i}sin{\omega }_i}}\left\{\begin{array}{c}sin{\omega }_{i}t·sin{\omega }_i(1-s),0\le t\le s\le 1,\hfill \\ sin{\omega }_{i}s·sin{\omega }_i(1-t),0\le s\le t\le 1.\hfill \end{array}\right.$$

(8)

Lemma 1 

([18,19]). By the definition of $K_i(t,s)(i=1,2)$, we have

(i) 

$K_i(s,t):[0,1]\times [0,1]\to (-\infty ,+\infty )$ is continuous.

(ii) 

$K_i(t,s)>0,t,s\in (0,1).$

(iii) 

$K_i(t,s)=K_i(s,t),\forall t,s\in [0,1],i=1,2.$

(iv) 

$K_i(t,s)\le m_i{K}_i(t,t),\forall t,s\in [0,1]$, where $m_i=1$ if ${\zeta }_i\ge 0$, and $m_i={\displaystyle \frac{1}{sin\sqrt{|{\zeta }_i|}}}$ if $-{\pi }^2<{\zeta }_i<0$.

(v) 

$K_i(t,s)\ge d_i{K}_i(t,t)K_i(s,s),\forall t,s\in [0,1],$ where $d_i>0$ is a constant.

(vi) 

${\int }_0^1{K}_1\left(t,s\right)K_2\left(s,\tau \right)ds={\int }_0^1{K}_2\left(t,s\right)K_1\left(s,\tau \right)ds,t,\tau \in [0,1].$

Let $E=C[0,1]$. Take the maximum norm in E as $\parallel x\parallel$. Set $P=\{x\in E|x\left(s\right)\ge 0,s\in [0,1]\}$, and then $P\subset E$ is normal. Let $E_e=\{x\in E|\exists \lambda >0\mathrm{such}\mathrm{that}-\lambda e\le x\le \lambda e\},$ where $e\in P$. Define the norm ${\parallel x\parallel }_e=inf\{\lambda >0|-\lambda e\le x\le \lambda e\}$. From [24,25], we know that $P_e$ is a normal solid, and $\mathrm{int}{P}_e=\{x\in E|\exists \lambda >\mu >0\mathrm{such}\mathrm{that}\mu e\le x\le \lambda e\}.$

From [18,19], Problem (1) can be transformed into the following integral equation:

$$x\left(t\right)={\int }_0^{1}K\left(t,\tau \right)g(\tau ,x\left(\tau \right))d\tau ,t\in [0,1],$$

(9)

where

$$K\left(t,\tau \right)={\int }_0^1{K}_1\left(t,s\right)K_2\left(s,\tau \right)ds,t,\tau \in [0,1],$$

(10)

and $K_i(t,s)(i=1,2)$ is defined by (6) or (7) or (8).

By (9) and (10), the operators $T,K$ and G are defined as follows:

$$\left(Tx\right)\left(t\right)={\int }_0^{1}K\left(t,\tau \right)g(\tau ,x\left(\tau \right))d\tau ,t\in [0,1],$$

(11)

$$\left(Kx\right)\left(t\right)={\int }_0^{1}K\left(t,\tau \right)x\left(\tau \right)d\tau ,t\in [0,1],$$

(12)

$$\left(Gx\right)\left(t\right)=g\left(t,x\left(t\right)\right),\forall x\in E.$$

(13)

Obviously, $T=KG.$

Lemma 2. 

The operator $K:E\to E_e$ and $K:P\setminus \left\{\theta \right\}\to \mathrm{int}{P}_e$, where $e\left(t\right)=K_1(t,t)$.

From (12), and combined with Lemma 1, for $x\in E$, we have

$$\left(Kx\right)\left(t\right)={\int }_0^{1}K(t,\tau )x\left(\tau \right)d\tau \le {\int }_0^{1}K(t,\tau )\left|x\left(\tau \right)\right|d\tau \le m_1{m}_2{\int }_0^1{K}_2(s,s)\left|x\left(s\right)\right|ds·K_1(t,t)=\lambda e\left(t\right),$$

(14)

where ${\displaystyle \lambda =m_1{m}_2{\int }_0^1{K}_2(s,s)\left|x\left(s\right)\right|ds,e\left(t\right)=K_1(t,t),}$ and

$$\left(Kx\right)\left(t\right)\ge {\int }_0^{1}K(t,\tau )(-|x\left(\tau \right)\left|\right)d\tau \ge -m_1{m}_2{\int }_0^1{K}_2(s,s)\left|x\left(s\right)\right|ds·K_1(t,t)=\lambda e\left(t\right)=-\lambda e\left(t\right).$$

(15)

From (14) and (15), we obtain $K:E\to E_e$.

From Lemma 1, for $x\in P\setminus \left\{\theta \right\}$, we have

$$\begin{array}{c}{\displaystyle \left(Kx\right)\left(t\right)={\int }_0^1{\int }_0^1{K}_1(t,\tau )K_2(s,\tau )x\left(s\right)d\tau ds}\hfill \\ {\displaystyle \ge {\int }_0^1{\int }_0^1{d}_1{K}_1(t,t)K_1(\tau ,\tau )·d_2{K}_2(s,s)K_2(\tau ,\tau )x\left(s\right)d\tau ds}\hfill \\ {\displaystyle =d_1{d}_2{K}_1(t,t){\int }_0^1{\int }_0^1{K}_1(\tau ,\tau )·K_2(s,s)K_2(\tau ,\tau )x\left(s\right)d\tau ds}\hfill \\ {\displaystyle =c_0{d}_1{d}_2{K}_1(t,t){\int }_0^1{K}_2(s,s)x\left(s\right)ds=\mu e\left(t\right),}\hfill \end{array}$$

(16)

where ${\displaystyle c_0={\int }_0^1{K}_1(\tau ,\tau )K_2(\tau ,\tau )d\tau }$, ${\displaystyle \mu =c_0{d}_1{d}_2{\int }_0^1{K}_2(s,s)x\left(s\right)ds,e\left(t\right)=K_1(t,t).}$

By (15) and (16), we have $K:P\setminus \left\{\theta \right\}\to \mathrm{int}{P}_e$.

Lemma 3 

([18]). The linear operator K has the eigenvalues

$${\displaystyle \frac{1}{{\eta }_1}},{\displaystyle \frac{1}{{\eta }_2}},\cdots ,{\displaystyle \frac{1}{{\eta }_n}},\cdots$$

and the algebraic multiplicity of ${\displaystyle \frac{1}{{\eta }_n}}$ is 1, where

$${\eta }_n={\left(n\pi \right)}^4-{\alpha }_1{\left(n\pi \right)}^2-{\alpha }_2,n=1,2,\cdots ,$$

and $0<{\eta }_1<{\eta }_2<\cdots .$

Lemma 4. 

$T:E_e\to E_e$ is quasi-additive on a lattice.

Proof. 

From the proof of Lemma 2, $T:E_e\to E_e$. For $\phi \in E_e$, we have

$${\phi }_{+}\left(s\right)=\left\{\begin{array}{c}\phi \left(s\right),\mathrm{if}\phi \left(s\right)\ge 0,\hfill \\ 0,\mathrm{if}\phi \left(s\right)<0.\hfill \end{array}\right.$$

$${\phi }_{-}\left(s\right)=\left\{\begin{array}{c}\phi \left(s\right),\mathrm{if}\phi \left(s\right)\le 0,\hfill \\ 0,\mathrm{if}\phi \left(s\right)>0.\hfill \end{array}\right.$$

Since $g(s,0)=0$, $g(s,\phi \left(s\right))=g(s,{\phi }_{+}\left(s\right)+{\phi }_{-}\left(s\right))=g(s,{\phi }_{+}\left(s\right))+g(s,{\phi }_{-}\left(s\right))$. From $T=KG$, we have $T\phi \left(s\right)=Kg(s,{\phi }_{+}\left(s\right))+Kg(s,{\phi }_{-}\left(s\right))=T{\phi }_{+}\left(s\right)+T\left({\phi }_{-}\right)\left(s\right)$. That is, $T\phi =T{\phi }_{+}+T{\phi }_{-}$.

First, we provide the condition that will be used later.

$\left({\mathrm{H}}_0\right)g(s,0)=0$ uniformly for $s\in [0,1]$; ${\displaystyle \underset{v\to 0}{lim}\frac{g(s,v)}{v}=\gamma }$ uniformly for $s\in [0,1]$. □

Lemma 5. 

Suppose that $\left({\mathrm{H}}_0\right)$ holds; then, the Fréchet derivative operator $T_{\theta }^{\prime }=\gamma K$.

Proof. 

From $\left({\mathrm{H}}_0\right)$, $\exists ϵ>0$ and $\delta >0$ such that

$$|{\displaystyle \frac{g(t,v)}{v}}-\gamma |<ϵ,\forall t\in [0,1],|v|<\delta .$$

For $x\in E_e$, we obtain

$$\begin{array}{c}\left|T\right(x\left(t\right))-T\theta -\gamma Kx(t\left)\right|\hfill \\ {\displaystyle ={\int }_0^{1}K(t,s)[g(s,x\left(s\right))-\gamma x\left(s\right)]ds}\hfill \\ {\displaystyle \le {\int }_0^{1}K(t,s)|g(s,x\left(s\right))-\gamma x\left(s\right)|ds}\hfill \\ \le {Mϵ\parallel x\parallel }_e·e\left(t\right),\hfill \end{array}$$

where $M=\underset{t\in [0,1]}{max}{\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )dsd\tau >0.$ □

Therefore,

$${\parallel Tx-T\theta -\gamma Kx\parallel }_e\le Mϵ{\parallel x\parallel }_e.$$

Obviously,

$$\underset{{\parallel x\parallel }_e\to 0}{lim}{\displaystyle \frac{{\parallel Tx-T\theta -\gamma Kx\parallel }_e}{{\parallel x\parallel }_e}}=0.$$

That is, $T_{\theta }^{\prime }=\gamma K.$

In the following lemmas, we assume that E is a Banach space, the cone $P\subset E$ is a normal solid, and $Q:E\to E$ is completely continuous and also a quasi-additive operator on a lattice.

3. Existence of Solutions Under Sublinear Conditions

Lemma 6 

([1,27]). Assume that

(i) 

∃ a linear operator $N_1$ with the spectral radius $r\left(N_1\right)<1$ and $q^{*}\in P,q_1\in P$ such that

$$-q^{*}\le Qu\le N_{1}u+q_1,\forall u\in P;$$

where $N_1$ is positive and bounded.

(ii) 

∃ a linear operator $N_2$ with the spectral radius $r\left(N_2\right)<1$ and $q_2\in P$ such that

$$Qu\ge N_{2}u-q_2,\forall u\in (-P);$$

where $N_2$ is positive and bounded.

(iii) 

$Q\theta =\theta$, the Fréchet derivative operator $Q_{\theta }^{\prime }$ exists, 1 is not an eigenvalue of $Q_{\theta }^{\prime }$, and the sum of the algebraic multiplicities for all eigenvalues of $Q_{\theta }^{\prime }$ lying in $(1,+\infty )$ is a non-zero even number.

(iv) 

$Q(P\setminus \{\theta \left\}\right)\subset \mathrm{int}P,Q(-P\setminus \{\theta \left\}\right)\subset \mathrm{int}(-P).$

Then, Q has at least three nontrivial fixed points: a sign-changing fixed point, a negative fixed point and a positive fixed point.

(H₁) ∃ a constant $\delta >0$ s.t.

$$\underset{\left|v\right|\to +\infty }{lim\; sup}{\displaystyle \frac{g(s,v)}{v}}\le {\eta }_1-\delta \mathrm{uniformly}\mathrm{for}s\in [0,1].$$

(H₂) ${\eta }_{2n_0}<\gamma <{\eta }_{2n_0+1},$ where γ is defined by $\left(H_0\right),$ and $n_0>0$ is a natural number.

(H₃) $\forall s\in [0,1]$, $g(s,v)v>0,v\ne 0$.

Theorem 1. 

Suppose that $\left({\mathrm{H}}_0\right)$ and $\left({\mathrm{H}}_1\right)-\left({\mathrm{H}}_3\right)$ hold; then, Problem (1) has at least three nontrivial solutions: a sign-changing solution, a positive solution and a negative solution.

Proof. 

We will show that (i)–(iv) from Lemma 6 hold. From Lemma 4, T is a quasi-additive operator. Obviously, $T:E_e\to E_e$ is completely continuous.

By $\left({\mathrm{H}}_1\right)$, $\exists R^{*}>0$ such that

$${\displaystyle \frac{g(t,v)}{v}}\le {\eta }_1-{\displaystyle \frac{\delta }{3}},\forall s\in [0,1],\left|v\right|\ge R^{*}.$$

(17)

By (17), we have

$$g(s,v)\le \left({\eta }_1-{\displaystyle \frac{\delta }{3}}\right)v+M_0,s\in [0,1],v\ge 0,$$

(18)

$$g(s,v)\ge \left({\eta }_1-{\displaystyle \frac{\delta }{3}}\right)v-M_0,s\in [0,1],v\le 0,$$

(19)

where $M_0=\underset{0\le s\le 1,\left|v\right|\le R^{*}}{max}\left|g(s,v)\right|>0$.

Let $\tilde{K}=\left({\eta }_1-{\displaystyle \frac{\delta }{3}}\right)K$. Obviously, $\tilde{K}$ is a bounded and positive linear operator. By Lemma 3, we have

$$r\left(K\right)=\underset{\lambda \in \left\{{\displaystyle \frac{1}{{\eta }_i}},i=1,2,\cdots \right\}}{sup}\left|\lambda \right|={\displaystyle \frac{1}{{\eta }_1}}.$$

Therefore,

$$r\left(\tilde{K}\right)=\left({\eta }_1-{\displaystyle \frac{\delta }{3}}\right)r\left(K\right)=1-{\displaystyle \frac{\delta }{3{\eta }_1}}<1.$$

(20)

We note ${\displaystyle {\phi }_0\left(t\right)=M_0{\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )dsd\tau }$. Similar to the proof of Lemma 2, we obviously know that ${\phi }_0\left(t\right)\in P_e$.

From (18), $\forall x\left(t\right)\in P_e$, we have

$$\begin{array}{c}{\displaystyle \left(Tx\right)\left(t\right)={\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )g\left(\tau ,x\left(\tau \right)\right)dsd\tau }\hfill \\ {\displaystyle \le {\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )\left[\left({\eta }_1-\frac{\delta }{3}\right)x\left(\tau \right)+M_0\right]dsd\tau }\hfill \\ {\displaystyle =\left({\eta }_1-\frac{\delta }{3}\right){\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )x\left(\tau \right)dsd\tau +M_0{\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )dsd\tau }\hfill \\ =\left({\eta }_1-{\displaystyle \frac{\delta }{3}}\right)Kx\left(t\right)+{\phi }_0\left(t\right)\hfill \\ =\tilde{K}x\left(t\right)+{\phi }_0\left(t\right).\hfill \end{array}$$

(21)

From (19), $\forall x\left(t\right)\in (-P_e)$, we have

$$\begin{array}{c}{\displaystyle \left(Tx\right)\left(t\right)={\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )g\left(\tau ,x\left(\tau \right)\right)dsd\tau }\hfill \\ {\displaystyle \ge {\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )\left[\left({\eta }_1-\frac{\delta }{3}\right)x\left(\tau \right)-M_0\right]dsd\tau }\hfill \\ {\displaystyle =\left({\eta }_1-\frac{\delta }{3}\right){\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )x\left(\tau \right)dsd\tau -M_0{\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )dsd\tau }\hfill \\ =\left({\eta }_1-{\displaystyle \frac{\delta }{3}}\right)Kx\left(t\right)-{\phi }_0\left(t\right)\hfill \\ =\tilde{K}x\left(t\right)-{\phi }_0\left(t\right)\hfill \end{array}$$

(22)

□

Thus, by (20)–(22), Lemma 6(i), (ii) are proved.

From Lemma 5 and (${\mathrm{H}}_0$), we have $T\theta =\theta$, $T_{\theta }^{\prime }=\gamma K$, so ${\displaystyle \frac{\gamma }{{\eta }_i}}$ is an eigenvalue of $T_{\theta }^{\prime }$. From (${\mathrm{H}}_2$), 1 is not an eigenvalue of $T_{\theta }^{\prime }$, and the sum of the algebraic multiplicities for all eigenvalues of $T_{\theta }^{\prime }$ lying in $(1,+\infty )$ is a non-zero even number. Hence, Lemma 6(iii) is satisfied.

By $\left({\mathrm{H}}_3\right)$, when $v>0$, $g(t,v)>0$ holds; when $v<0$, $g(t,v)<0$ holds. Combined with Lemma 2, we can easily obtain $T:P_e\setminus \left\{\theta \right\}\to \mathrm{int}{P}_e$ and $T:(-P_e)\setminus \left\{\theta \right\}\to \mathrm{int}(-P_e)$. Lemma 6(iv) is proved.

Hence, according to Lemma 6, Problem (1) has at least three nontrivial solutions: a sign-changing solution, a positive solution and a negative solution.

4. Existence of Solutions Under Unilaterally Asymptotically Linear Conditions

Lemma 7 

([28]). Assume that

(i) 

∃$q_1,q_2\in P$ and a linear operator L such that

$$Qu\le q_1,\forall u\in (-P);$$

$$Qu\le Lu+q_2,\forall u\in P;$$

where $L:E\to E$ is positive and bounded with $r\left(L\right)<1$.

(ii) 

$Q\theta =\theta$, the Fréchet derivative $Q_{\theta }^{\prime }$ of Q at θ exists, and 1 is not an eigenvalue of $Q_{\theta }^{\prime }$.

(iii) 

$Q_{(-P)}^{\prime }$ exists, 1 is not an eigenvalue of $Q_{(-P)}^{\prime }$ corresponding a positive eigenvector, and $r\left(Q_{(-P)}^{\prime }\right)>1$.

Then, Q has at least one nontrivial fixed point.

Lemma 8 

([28]). Assume that

(i) 

Q is strongly increasing on P and $-P$;

(ii) 

$Q_P^{\prime }$ and $Q_{(-P)}^{\prime }$ exist, with $r\left(Q_P^{\prime }\right)>1$ and $r\left(Q_{(-P)}^{\prime }\right)>1$; 1 is not an eigenvalue of $Q_P^{\prime }$ or $Q_{(-P)}^{\prime }$ corresponding a positive eigenvector;

(iii) 

$Q\theta =\theta$, the Fréchet derivative $Q_{\theta }^{\prime }$ of Q at θ is strongly positive, and $r\left(Q_{\theta }^{\prime }\right)<1$;

(iv) 

The Fréchet derivative $Q_{\infty }^{\prime }$ of Q at ∞ exists, and 1 is not an eigenvalue of $Q_{\infty }^{\prime }$; the sum of the algebraic multiplicities for all eigenvalues of $Q_{\infty }^{\prime }$ lying in $(1,+\infty )$ is an even number.

Then, Q has at least three nontrivial fixed points, including one sign-changing fixed point.

For convenience, we list some assumptions as follows:

${\displaystyle \left({\mathrm{H}}_4\right)\underset{v\to +\infty }{lim}\frac{g(s,v)}{v}={\rho }_1}$ uniformly for $s\in [0,1]$.

${\displaystyle \left({\mathrm{H}}_5\right)\underset{v\to -\infty }{lim}\frac{g(s,v)}{v}={\rho }_2}$ uniformly for $s\in [0,1]$.

Lemma 9. 

(i)  Assume that $\left({\mathrm{H}}_4\right)$ is satisfied, and then $T_{P_e}^{\prime }={\rho }_{1}K.$

(ii) 

Assume that $\left({\mathrm{H}}_5\right)$ is satisfied, and then $T_{\left(-P_e\right)}^{\prime }={\rho }_{2}K.$

(iii) 

Assume that $\left({\mathrm{H}}_4\right)$ and $\left({\mathrm{H}}_5\right)$ are satisfied and ${\rho }_1={\rho }_2=\rho$, and then $T_{\infty }^{\prime }=\rho K.$

Proof. 

In the beginning, we start by proving (i). By $\left({\mathrm{H}}_4\right)$, we find that $\forall ϵ>0$, $\exists H>0$ s.t.

$$|g(s,v)-{\rho }_{1}v|<ϵv,\forall s\in [0,1],v\ge H.$$

(23)

Let ${\Omega }_H=\{x\in P_e{|\parallel x\parallel }_e\le H\}$. So, we define

$$W=\underset{x\in {\Omega }_H}{sup}{\{\parallel Tx\parallel }_e{,\parallel Kx\parallel }_e\}.$$

(24)

For $x\in P_e$, we can define

$${\psi }_1\left(x\left(s\right)\right)=\left\{\begin{array}{c}x\left(s\right),x\left(s\right)<H,\hfill \\ H,x\left(s\right)\ge H.\hfill \end{array}\right.$$

$${\psi }_2\left(x\left(s\right)\right)=\left\{\begin{array}{c}H,x\left(s\right)<H,\hfill \\ x\left(s\right),x\left(s\right)\ge H.\hfill \end{array}\right.$$

Hence,

$$Tx\left(s\right)=T{\psi }_1\left(x\left(s\right)\right)+T{\psi }_2\left(x\left(s\right)\right)-TH,s\in [0,1],x\in P_e.$$

(25)

$$Kx\left(s\right)=K{\psi }_1\left(x\left(s\right)\right)+K{\psi }_2\left(x\left(s\right)\right)-TH,s\in [0,1],x\in P_e.$$

(26)

From (24)–(26), we obtain

$$\parallel Tx-{\rho }_1{Kx\parallel }_e\le 4S+{\parallel T{\psi }_2\left(x\right)-{\rho }_{1}K{\psi }_2\left(x\right)\parallel }_e.$$

(27)

By (23), we obtain

$$\begin{array}{c}|T{\psi }_2\left(x\left(t\right)\right)-{\rho }_{1}K{\psi }_2\left(x\left(t\right)\right)|\hfill \\ {\displaystyle =|{\int }_0^{1}K(t,s)[g(s,{\psi }_2\left(x\left(s\right)\right))-{\rho }_1{\psi }_2\left(x\left(s\right)\right)]ds|}\hfill \\ {\displaystyle \le {\int }_0^{1}K(t,s)|g(s,{\psi }_2\left(x\left(s\right)\right))-{\rho }_1{\psi }_2\left(x\left(s\right)\right)|ds}\hfill \\ \le {Mϵ\parallel x\parallel }_e·e\left(t\right),\hfill \end{array}$$

where $M=\underset{t\in [0,1]}{max}{\int }_0^1{\int }_0^1{K}_1(t,s)K_2(s,\tau )dsd\tau .$ Obviously, $M>0$.

So,

$$\parallel T{\psi }_2\left(x\right)-{\rho }_{1}K{\psi }_2{\left(x\right)\parallel }_e\le Mϵ{\parallel x\parallel }_e.$$

(28)

According to (27) and (28), it can be concluded that

$$\underset{{\parallel x\parallel }_e\to \infty ,x\in P_e}{lim}{\displaystyle \frac{\parallel Tx-{\rho }_1{Kx\parallel }_e}{{\parallel x\parallel }_e}}=0.$$

□

That is, $T_{P_e}^{\prime }={\rho }_{1}K$.

(i)

Similarly, we can easily obtain $T_{(-P_e)}^{\prime }={\rho }_{2}K.$

(ii)

By ${\rho }_1={\rho }_2=\rho$, we obtain $\forall ϵ>0$, $\exists H>0$ s.t.

$$|g(s,v)-{\rho }_{1}v|<ϵ|v|,\forall s\in [0,1],\left|v\right|\ge R.$$

Let ${\Omega }_H=\{x\in E_e{|\parallel x\parallel }_e\le H\}$. Then, we set

$$S=\underset{x\in {\Omega }_H}{sup}{\{\parallel Tx\parallel }_e{,\parallel Kx\parallel }_e\}.$$

For $x\in E_e$, we can define

$${\psi }_1\left(x\left(s\right)\right)=\left\{\begin{array}{c}x\left(s\right),\left|x\right(s\left)\right|\le H,\hfill \\ H,\left|x\right(s\left)\right|\ge H.\hfill \end{array}\right.$$

$${\psi }_2\left(x\left(s\right)\right)=\left\{\begin{array}{c}H,\left|x\right(s\left)\right|\le H,\hfill \\ x\left(s\right),\left|x\right(s\left)\right|\ge H.\hfill \end{array}\right.$$

Similar to (27) and (28), we easily have $T_{\infty }^{\prime }=\rho K.$

Theorem 2. 

Assume that conditions $\left({\mathrm{H}}_0\right),\left({\mathrm{H}}_4\right)$ and $\left({\mathrm{H}}_5\right)$ hold. And, we also assume that

(i) 

$0<{\rho }_1<{\eta }_1$;

(ii) 

${\rho }_2>{\eta }_1,{\rho }_1\ne {\eta }_n,n=1,2,\cdots ;$

(iii) 

$\gamma \ne {\eta }_n,n=1,2,\cdots$

Then, Problem (1) has at least a nontrivial solution.

Proof. 

We will show that Lemma 7 holds. From Lemma 4, T is quasi-additive, and $T:E_e\to E_e$ is also completely continuous.

Since ${\rho }_2>0$, by $\left({\mathrm{H}}_5\right)$, we find that ∃ a constant $C_1>0$ s.t.

$$g(s,v)\le C_1,\forall s\in [0,1],v\le 0.$$

(29)

From the definition of T and K, combined with (29), for any $x\in (-P_e)$, we obtain

$$Tx\left(s\right)\le K{C}_1,\forall s\in [0,1],x\in (-P_e).$$

(30)

Obviously, we know that $K{C}_1\in P_e.$

Since $0<{\rho }_1<{\eta }_1$, then $\exists m>0$ s.t.

$$0<{\rho }_1+m<{\eta }_1.$$

(31)

From $\left({\mathrm{H}}_4\right)$ and (31), $\exists C_2>0$ s.t.

$$g(s,v)\le ({\rho }_1+m)v+C_2,\forall s\in [0,1],v\ge 0.$$

(32)

By (32), for $x\in P_e$, we have

$$Tx\left(t\right)\le ({\rho }_1+m)Kx\left(t\right)+K{C}_2,\forall t\in [0,1],x\in P_e.$$

(33)

□

Let $B=({\rho }_1+m)K$. Then, $r\left(B\right)=({\rho }_1+m)r\left(K\right)<{\eta }_{1}r\left(K\right)=1$.

According to (30) and (33), Lemma 7(i) holds.

From Lemma 5 and $\left({\mathrm{H}}_1\right)$, we have $T\theta =\theta$, $T_{\theta }^{\prime }=\gamma K$. Since $\gamma \ne {\eta }_n,n=1,2,\cdots$, and $r\left(T_{\theta }^{\prime }\right)=\gamma r\left(K\right)$, 1 is not an eigenvalue of $T_{\theta }^{\prime }$. Hence, Lemma 7(ii) holds.

By Lemma 9, we know that $T_{(-P_e)}^{\prime }={\rho }_{2}K.$ Then, $r\left(T_{(-P_e)}^{\prime }\right)={\rho }_{2}r\left(K\right)$. Since ${\rho }_2>{\eta }_1$, 1 is not an eigenvalue of $T_{(-P_e)}^{\prime }$, and apparently, $r\left(T_{\left(-P_e\right)}^{\prime }\right)>1$. So, Lemma 7(iii) holds.

Therefore, from Lemma 7, we can infer that Problem (1) has at least a nontrivial solution.

Theorem 3. 

Assume that conditions $\left({\mathrm{H}}_0\right)$, $\left({\mathrm{H}}_4\right)$ and $\left({\mathrm{H}}_5\right)$ hold, and ${\rho }_1={\rho }_2=\rho$. And, the following assumptions are also given:

(i) 

$g(s,v)$ is strictly increasing about v;

(ii) 

$0<\gamma <{\eta }_1$;

(iii) 

$\exists k>0$ such that

$${\eta }_k<\rho <{\eta }_{k+1},$$

where K is an even number.

Then, Problem (1) has at least three nontrivial solutions, including a sign-changing solution.

Proof. 

We will show that (i)–(iv) from Lemma 8 hold.

By Lemma 4, T is quasi-additive. It is obvious that $T:E_e\to E_e$ is completely continuous.

(i)

By Lemma 2, we have $K(P_e\setminus \left\{\theta \right\})\subset \mathrm{int}{P}_e$. Combined with condition (i), it is easy to find that the operator T is strongly increasing on $P_e$ or $(-P_e)$.

(ii)

By Lemma 9, we know that $T_{P_e}^{\prime }=T_{(-P_e)}^{\prime }=\rho K$. Then, ${\displaystyle r\left(T_{P_e}^{\prime }\right)=r\left(T_{(-P_e)}^{\prime }\right)=\rho r\left(K\right)=\frac{\rho }{{\eta }_1}>1}$. Since ${\displaystyle \frac{\rho }{{\eta }_1}},{\displaystyle \frac{\rho }{{\eta }_2}},{\displaystyle \frac{\rho }{{\eta }_3}},\cdots$ are eigenvalues of $T_{P_e}^{\prime }$ or $T_{(-P_e)}^{\prime }$, by condition (ii), we know that 1 is not an eigenvalue of $T_{P_e}^{\prime }$ or $T_{(-P_e)}^{\prime }$.

(iii)

By Lemma 5, we have $T_{\theta }^{\prime }=\gamma K$. Then, $r\left(T_{\theta }^{\prime }\right)=\gamma r\left(K\right)$. By condition (ii), $r\left(T_{\theta }^{\prime }\right)<1$. Since $K(P_e\setminus \left\{\theta \right\})\subset \mathrm{int}{P}_e$, $T_{\theta }^{\prime }$ is strongly positive.

(iv)

According to Lemma 9, we have $T_{\infty }^{\prime }=\rho K$. Hence, $r\left(T_{\infty }^{\prime }\right)=\rho r\left(K\right).$ In addition, by condition (iii), 1 is not an eigenvalue of $T_{\infty }^{\prime }$, and the sum of the algebraic multiplicities for all eigenvalues of $T_{\infty }^{\prime }$ lying in $(1,+\infty )$ is an even number.

□

By the above proof and Lemma 8, we find that Problem (1) has at least three nontrivial solutions, including a sign-changing solution.

5. Existence of Solutions Under Superlinear Conditions

Suppose that a linear operator $L:P\to P$ is positive and continuous with $r\left(L\right)>0$. Let $L^{*}$ be a conjugate operator of L, $P^{*}$ be a conjugate cone of P. By the Krein–Rutmann theorem, $\exists \phi \in P\setminus \left\{\theta \right\},h^{*}\in P^{*}\setminus \left\{\theta \right\}$ s.t.

$$L\phi =r\left(L\right)\phi ,L^{*}{h}^{*}=r\left(L\right)h^{*}.$$

(34)

Let $\xi >0$. We define

$$P\left(h^{*},\xi \right)=\left\{x\in P\mid h^{*}\left(x\right)\ge \xi \parallel x\parallel \right\},$$

and then $P\left(h^{*},\delta \right)$ is a cone and $P\left(h^{*},\xi \right)\subset E$.

Definition 2 

([23,24]). Suppose that a linear operator L is positive. If $\exists \phi \in P\setminus \left\{\theta \right\},h^{*}\in P^{*}\setminus \left\{\theta \right\}$ and $\xi >0$ such that (34) is satisfied, and L maps P to $P\left(h^{*},\xi \right)$, then we say that L satisfies the $\mathrm{H}$ condition.

From [1,29], the following lemmas can be easily obtained.

Lemma 10. 

Suppose that the linear operator L is positive and bounded. And, the following assumptions are also given:

(i) 

$\exists c_1>r^{-1}\left(L\right),q_1\in P$, such that

$$Qu\ge c_{1}Lu-q_1,\forall u\in P;$$

(ii) 

$\exists 0<c_2<r^{-1}\left(L\right),q_2\in P$, such that

$$Qu\ge c_{2}Lx-q_2,\forall u\in (-P);$$

(iii) 

$Q\theta =\theta$, Q is Fréchet differentiable at θ, and 1 is not an eigenvalue of $Q_{\theta }^{\prime }$.

Then, Q has at least one nontrivial fixed point.

Lemma 11. 

Suppose that Lemma 10 holds, and also assume that the sum of the algebraic multiplicities for all eigenvalues of $Q_{\theta }^{\prime }$ lying in $(1,+\infty )$ is an even number, and

$$Q(P\setminus \{\theta \left\}\right)\subset \mathrm{int}P,Q(-P\setminus \{\theta \left\}\right)\subset \mathrm{int}(-P).$$

Then, A has at least two nontrivial fixed points: a sign-changing fixed point and a negative fixed point.

For convenience, the needed conditions are given as follows.

$\left({\mathrm{H}}_6\right)$$\exists \delta >0$ such that

$$\underset{v\to +\infty }{lim}{\displaystyle \frac{g(s,v)}{v}}\ge {\eta }_1+\delta ,\mathrm{uniformly}\mathrm{on}s\in [0,1],$$

(35)

$$\underset{v\to -\infty }{lim}{\displaystyle \frac{g(s,v)}{v}}\le {\eta }_1-\delta ,\mathrm{uniformly}\mathrm{on}s\in [0,1].$$

(36)

Lemma 12. 

K satisfies the $\mathrm{H}$ condition.

Proof. 

Let

$$\left(K^{*}x\right)\left(t\right)={\int }_0^1{K}^{*}(t,\tau )x\left(\tau \right)d\tau ,$$

where $K^{*}(t,\tau )=K(\tau ,t)$, and $K(\tau ,t)$ is defined by (10). So, the operator $K^{*}$ is the conjugate operator of K. □

According to Lemma 3, $r\left(K\right)={\displaystyle \frac{1}{{\eta }_1}}>0$. Hence, from the Krein–Rutmann theorem, we know that $\exists \omega \left(t\right)\in P_e\setminus \left\{\theta \right\},f^{*}\left(t\right)\in P_e^{*}\setminus \left\{\theta \right\}$ s.t.

$$\left(K\omega \right)\left(t\right)=r\left(K\right)\omega \left(t\right),$$

(37)

$$K^{*}{f}^{*}=r\left(K\right)f^{*}.$$

(38)

In fact, we can take $\omega \left(t\right)=sin\pi t$, and we can take $f^{*}\in P_e\setminus \left\{\theta \right\}$, where

$$f^{*}\left(x\right)={\int }_0^1\omega \left(t\right)x\left(t\right)dt,x\in E_e.$$

For $x\in E_e$, we obtain

$$\begin{array}{c}{\displaystyle \left(K^{*}{f}^{*}\right)\left(x\right)=f^{*}\left(Kx\right)={\int }_0^1\omega \left(t\right)\left({\int }_0^{1}K(t,\tau )x\left(\tau \right)d\tau \right)dt}\hfill \\ {\displaystyle ={\int }_0^1\left({\int }_0^{1}K(t,\tau )\omega \left(t\right)dt\right)x\left(\tau \right)d\tau }\hfill \\ {\displaystyle ={\int }_0^1\left({\int }_0^{1}K(\tau ,t)\omega \left(t\right)dt\right)x\left(\tau \right)d\tau }\hfill \\ {\displaystyle ={\int }_0^{1}x\left(\tau \right)\left(K\omega \right)\left(\tau \right)d\tau ={\int }_0^{1}x\left(\tau \right)r\left(K\right)\omega \left(\tau \right)d\tau }\hfill \\ {\displaystyle =r\left(K\right){\int }_0^1\omega \left(\tau \right)x\left(\tau \right)d\tau =r\left(K\right)f^{*}\left(x\right).}\hfill \end{array}$$

That is, $K^{*}{f}^{*}=r\left(K\right)f^{*}$.

By (6)–(8), combined with (10), we can easily obtain

$$K(t,\tau )\ge h\left(t\right)K(\eta ,\tau ),\forall t,\eta ,\tau \in [0,1],$$

(39)

where $h\left(t\right)={\displaystyle \frac{sinh{\omega }_{1}t·sinh{\omega }_1(1-t)}{{\omega }_1{sinh}^2{\omega }_1}}$ when ${\zeta }_1>0$; $h\left(t\right)=t(1-t)$ when ${\zeta }_1=0$; and $h\left(t\right)=sin{\omega }_{1}t·sin{\omega }_1(1-t)$ when $-{\pi }^2<{\zeta }_1<0$.

By (39), we obtain

$$\begin{array}{c}{\displaystyle f^{*}\left(Kx\right)={\int }_0^1\omega \left(t\right)\left({\int }_0^{1}K(t,\tau )x\left(\tau \right)d\tau \right)dt}\hfill \\ {\displaystyle \ge {\int }_0^1\omega \left(t\right)h\left(t\right)dt·{\int }_0^{1}K(\eta ,\tau )x\left(\tau \right)d\tau }\hfill \\ {\displaystyle =\left(Kx\right)\left(\tau \right){\int }_0^1\omega \left(t\right)h\left(t\right)dt.}\hfill \end{array}$$

(40)

By (14), we easily obtain

$${\parallel Kx\parallel }_e\le m_1{m}_2{\int }_0^1{K}_2(s,s)x\left(s\right)ds.$$

(41)

By (16), we have

$$Kx\left(t\right)\ge c_0{d}_1{d}_2{\int }_0^1{K}_2(s,s)x\left(s\right)ds·K_1(t,t).$$

(42)

By (40)–(42), we easily know that $\exists C>0$ s.t.

$$f^{*}\left(Kx\right)\ge C{\parallel Kx\parallel }_e.$$

(43)

So, by (37), (38) and (43), we prove that K satisfies the $\mathrm{H}$ condition.

Theorem 4. 

Assume that conditions $\left({\mathrm{H}}_0\right)$ and $\left({\mathrm{H}}_6\right)$ hold, and $\gamma \notin \left\{{\eta }_1,{\eta }_2,\cdots ,{\eta }_n,\cdots \right\}$. Then, Problem (1) has at least one nontrivial solution.

Proof. 

We will show that Lemma 10 holds. According to Lemma 4, T is quasi-additive. Obviously, $T:E_e\to E_e$ is completely continuous. □

By (35) and (36), $\exists \tilde{C}>0$ s.t.

$$g(s,v)\ge \left({\eta }_1+{\displaystyle \frac{1}{3}}\delta \right)v,s\in [0,1],v\ge \tilde{C},$$

(44)

$$g(s,v)\ge \left({\eta }_1-{\displaystyle \frac{1}{3}}\delta \right)v,s\in [0,1],v\le -\tilde{C}.$$

(45)

So, by (44) and (45), combined with the continuity of g, we know that $\exists w>0$ such that

$$g(s,v)\ge \left({\eta }_1+{\displaystyle \frac{1}{3}}\delta \right)v-w,s\in [0,1],v\ge 0,$$

(46)

$$g(s,v)\ge \left({\eta }_1-{\displaystyle \frac{1}{3}}\delta \right)v-w,s\in [0,1],v\le 0.$$

(47)

For $x\in P_e$, by (46), we obtain

$$Tx\left(t\right)\ge ({\eta }_1+{\displaystyle \frac{1}{3}}\delta )Kx\left(t\right)-Kw,$$

obviously, $Kw\in P_e.$

For $x\in (-P_e)$, by (47), we obtain

$$Tx\left(t\right)\ge ({\eta }_1-{\displaystyle \frac{1}{3}}\delta )Kx\left(t\right)-Kw,$$

and obviously, $Kw\in P_e.$

By $\left({\mathrm{H}}_0\right)$, according to Lemma 5, $T\theta =\theta$ and $T_{\theta }^{\prime }=\gamma K$. Since $\gamma \notin \left\{{\eta }_1,{\eta }_2,\cdots ,{\eta }_n,\cdots \right\}$, 1 is not an eigenvalue of $T_{\theta }^{\prime }$.

Hence, by the above proof, Lemma 10 holds. So, Problem (1) has at least one nontrivial solution.

Theorem 5. 

Let conditions $\left({\mathrm{H}}_0\right),\left({\mathrm{H}}_2\right),\left({\mathrm{H}}_3\right)$ and $\left({\mathrm{H}}_6\right)$ be satisfied. Then, Problem (1) has at least two nontrivial solutions: a negative solution and a sign-changing solution.

Proof. 

We will show that Lemma 11 holds. From $\left({\mathrm{H}}_3\right)$, we have

$$g(s,v)>0,\forall v>0.$$

(48)

$$g(s,v)<0,\forall v<0.$$

(49)

□

By Lemma 2 and (48) and (49), we know that

$$T(P_e\setminus \left\{\theta \right\})\subset \mathrm{int}{P}_e,T(-P_e\setminus \left\{\theta \right\})\subset \mathrm{int}(-P_e).$$

According to Lemma 5, $T_{\theta }^{\prime }=\gamma K$. So, $r\left(T_{\theta }^{\prime }\right)=\gamma r\left(K\right)$. Since ${\eta }_{2n_0}<\gamma <{\eta }_{2n_0+1}$, we know that 1 is not an eigenvalue of $T_{\theta }^{\prime }$, and the sum of the algebraic multiplicities for all eigenvalues of $A_{\theta }^{\prime }$ lying in $(1,+\infty )$ is an even number $2n_0$.

Based on the above proof, combined with Theorem 4, we conclude that Lemma 11 holds. Therefore, according to Lemma 11, we know that Problem (1) has at least two nontrivial solutions: a negative solution and a sign-changing solution.

6. Applications

We consider fourth-order differential equations with two-point boundary conditions as follows:

$$\left\{\begin{array}{c}{x}^{\left(4\right)}\left(s\right)=g(s,x\left(s\right)),s\in [0,1],\hfill \\ x\left(0\right)=x\left(1\right)=0,\hfill \\ x^{″}\left(0\right)=x^{″}\left(1\right)=0.\hfill \end{array}\right.$$

(50)

By simple calculations, we obtain ${\eta }_1\approx 97.41,{\eta }_2\approx 1558.55,{\eta }_3\approx 7890.14$.

Example 1. 

Take

$$g(s,v)=\left\{\begin{array}{cc}25v+s\sqrt{v},\hfill & s\in [0,1],v\in (16,+\infty ),\hfill \\ {\displaystyle \frac{s-1199}{15}}(v-1)+1599+3s,\hfill & s\in [0,1],v\in (1,16],\hfill \\ 1598v+(1+3s)v^{{\displaystyle \frac{5}{3}}},\hfill & s\in [0,1],v\in [-1,1],\hfill \\ {\displaystyle \frac{s+1}{63}}(v+1)-1599-3s,\hfill & s\in [0,1],v\in [-64,-1),\hfill \\ 25v+s\sqrt[3]{v},\hfill & s\in [0,1],v\in (-\infty ,-64).\hfill \end{array}\right.$$

(51)

We can choose $\delta =12.41$. From (51), we can easily find that $\gamma =1598$. It is easy to prove that Theorem 1 holds. So, according to Theorem 1, we conclude that Problem (50) has at least three nontrivial solutions, including a sign-changing solution, a negative solution and a positive solution.

Example 2. 

Take

$$g(s,v)=\left\{\begin{array}{cc}1600v+s\sqrt{v},\hfill & s\in [0,1],v\in (4,+\infty ),\hfill \\ {\displaystyle \frac{s+6309}{3}}(v-1)+91+s,\hfill & s\in [0,1],v\in (1,4],\hfill \\ 90v+(1+3s)v^{{\displaystyle \frac{5}{3}}},\hfill & s\in [0,1],v\in [-1,1],\hfill \\ {\displaystyle \frac{s+12709}{7}}(v+1)-91-s,\hfill & s\in [0,1],v\in [-8,-1),\hfill \\ 1600v+s\sqrt[3]{v},\hfill & s\in [0,1],v\in (-\infty ,-8).\hfill \end{array}\right.$$

(52)

From (52), we easily obtain

$$g(s,0)=0,\underset{v\to 0}{lim}{\displaystyle \frac{g(s,v)}{v}}=90,\underset{\left|v\right|\to \infty }{lim}{\displaystyle \frac{g(s,v)}{v}}=1600.$$

We can prove that Theorem 3 holds. So, according to Theorem 3, Problem (50) has at least three nontrivial solutions, including a sign-changing solution.

Example 3. 

Take

$$g(s,v)=\left\{\begin{array}{cc}120v+s\sqrt{v},\hfill & s\in [0,1],v\in (4,+\infty ),\hfill \\ {\displaystyle \frac{s-1121}{3}}(v-1)+1601+s,\hfill & s\in [0,1],v\in (1,4],\hfill \\ 1600v+(1+s)v^{{\displaystyle \frac{5}{3}}},\hfill & s\in [0,1],v\in [-1,1],\hfill \\ {\displaystyle \frac{-879-3s}{7}}(v+1)-1601-s,\hfill & s\in [0,1],v\in [-8,-1),\hfill \\ 90v+(1-s)\sqrt[3]{v},\hfill & s\in [0,1],v\in (-\infty ,-8).\hfill \end{array}\right.$$

(53)

From (53), we have

$$g(s,0)=0,\underset{v\to 0}{lim}{\displaystyle \frac{g(s,v)}{v}}=1600,$$

$$\underset{v\to +\infty }{lim}{\displaystyle \frac{g(s,v)}{v}}=120,\underset{v\to -\infty }{lim}{\displaystyle \frac{g(s,v)}{v}}=90.$$

We can easily prove that Theorem 5 holds. So, according to Theorem 5, Problem (50) has at least two nontrivial solutions, including a sign-changing solution.

7. Conclusions

Most of the existing literature on fourth-order differential equations focuses on the existence of positive solutions, with only a few studies on the existence of sign-changing solutions. This article mainly studies the existence of sign-changing solutions, so the research in this article is also very meaningful. This article discusses the existence of sign-changing solutions under sublinear, unilaterally asymptotically linear and superlinear conditions. The problem we need to further investigate is the existence of positive solutions under unilaterally asymptotically linear and superlinear conditions.