Multiplicity of Solutions for Kirchhoff-Type Problem with Two-Superlinear Potentials

Abstract

In this paper we consider a class of Kirchhoff-type problem with 2-superlinear potentials. The existence of one positive solution and one negative solution will be established by using iterative technique and the Mountain Pass theorem, and a sign changing solution will be obtained by combining iterative technique and the Nehari method.

1 Introduction

In this paper we consider the following Kirchhoff-type problem

$$\begin{aligned} \left\{ \begin{array}{ll} -(a+b\int _{\Omega }|\nabla u|^{2}\mathrm{d}x)\Delta u=f(x,u) , &{}\qquad \text{ in } \Omega ,\\ u=0, &{}\qquad \text{ on } \partial \Omega , \end{array} \right. \end{aligned}$$

(1.1)

where $\Omega \subset {\mathbb {R}}^N$, $N\ge 3$, is a bounded smooth domain. The problem (1.1) is related to the stationary analogue of the equation

$$\begin{aligned} u_{tt}-\left( a+b\int _{\Omega }|\nabla u|^{2}\mathrm{d}x\right) \Delta u=f(x,u) \end{aligned}$$

(1.2)

proposed by Kirchhoff [8] as an existence of the classical D’Alembert’s wave equations for free vibration of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. Equation (1.2) received much attention only after Lions [12] introduced an abstract framework to the problem. In recent years, the Kirchhoff-type problem on a bounded domain $\Omega \subset {\mathbb {R}}^{N}$ or on ${\mathbb {R}}^{N}$ has been studied by many authors, see [1,2,3, 5,6,7, 9,10,11, 14, 15, 17, 19, 21,22,28] and references therein. To obtain the existence of solution by applying the Mountain Pass theorem or Morse theory, the authors have to impose a 4-superlinear or 4-asymptotically linear growth condition on the nonlinearity. For example, Perera and Zhang [17] considered the case where $f(x,\cdot )$ is asymptotically linear near zero and asymptotically 4-linear at infinity; they obtained a nontrivial solution of the problem by using the Yang index and critical group. For the cases when $f(x,\cdot )$ is 4-sublinear, 4-superlinear and asymptotically 4-linear at infinity, Zhang and Perera [26] obtained the existence of multiple and sign changing solutions by using variational methods and invariant sets of descent flow. He and Zou [5, 6] obtained infinitely many solutions by using the local minimax methods and the fountain theorems under the 4-superlinear condition. Sun and Liu [20] obtained nontrivial solutions via Morse theory when the nonlinearity is superlinear near zero but asymptotically 4-linear at infinity, and the nonlinearity is asymptotically linear near zero but 4-superlinear at infinity. Recently, Li et al. [9] discussed the existence of positive solutions to the 2-superlinear Kirchhoff problem in ${\mathbb {R}}^{N}$, $N\ge 3$; they obtained at least one positive radial solution by using truncation technique, Pohozaev-type identity and variational methods. Later, Zhang et al. [28] considered the 2-superlinear Kirchhoff problem in a smooth bounded convex domain; they established the existence of one positive solution using iterative technique, Pohozaev-type identity and variational methods.

Motivated by Li et al. [9] and Zhang et al. [28], in this paper we consider the 2-superlinear Kirchhoff problem in a bounded smooth domain but not necessarily convex. By using the iterative technique proposed in Figuereido et al. [4] and the Mountain Pass theorem, one positive solution and one negative solution for (1.1) will be obtained. Moreover, we will get a sign changing solution by combining the iterative technique and the Nehari method.

We make the following assumptions:

• (H0) $f\in C^{1}(\overline{\Omega }\times {\mathbb {R}},{\mathbb {R}})$ and there exist $a_{1}>0$, $p\in (1,\frac{N+2}{N-2})$ such that

  $$\begin{aligned} f'(x,t)\le a_{1}(1+|t|^{p-1}), \forall x\in \overline{\Omega }, t\in {\mathbb {R}}, \end{aligned}$$

  where $f'=\frac{\partial f}{\partial t}$;

• (H1) $\lim _{t\rightarrow 0}\frac{f(x,t)}{t}=0$ uniformly for $x\in \overline{\Omega }$;

• (H2) There exist constant $\theta >2$ and $t_{0}>0$ such that

  $$\begin{aligned} 0<\theta F(x,t)<tf(x,t), \forall x\in \overline{\Omega }, ~~~~|t|\ge t_{0} \end{aligned}$$

  where $F(x,t)=\int _{0}^{t}f(x,s)\mathrm{d}s$;

• (H3) $f'(x,t)>\frac{f(x,t)}{t}$ for all $t\ne 0, x\in \overline{\Omega }$.

Our main result is the following theorem.

Theorem 1.1

Assume that (H0)–(H3) hold. Then, there exists a constant $b_{0}>0$ such that for any $b\in [0,b_{0}]$, the problem (1.1) has at least three nontrivial solutions; among them, one is positive, one is negative, and one is sign changing.

The paper is organized as follows. In Sect. 2 we will prove the existence of positive and negative solutions for (1.1) by using the Mountain Pass theorem and the iterative technique. In Sect. 3, the existence of sign changing solution for (1.1) will be proved via the Nehari method and the iterative technique, and we will prove Theorem 1.1.

2 Positive and Negative Solutions

The aim of this section is to prove the positive and negative solutions for (1.1). We consider only the existence of positive solutions for (1.1), and the existence of negative solutions can be done by a similar argument.

Let $E=H_{0}^{1}(\Omega )$ be the usual Sobolev space equipped with the following inner product and norm,

$$\begin{aligned} \langle u,v\rangle =\int _{\Omega }\nabla u\cdot \nabla v\mathrm{d}x, \Vert u\Vert =\left( \int _{\Omega }|\nabla u|^{2}\mathrm{d}x\right) ^{\frac{1}{2}}. \end{aligned}$$

Let

$$\begin{aligned} f_{+}(x,t)= {\left\{ \begin{array}{ll} f(x,t),&{} \text {if }t\ge 0,\\ 0, &{}\text {if }t<0, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} F_{+}(x,t)=\int _{0}^{t}f_{+}(x,s)\mathrm{d}s, \end{aligned}$$

then we can conclude easily that $f_{+}(x,t)$ also satisfies (H0), (H1), (H2) for $t\ge t_{0}$ and (H3) for $t>0$.

For any given $w\in E$, let us consider the following problem

$$\begin{aligned} \left\{ \begin{array}{ll} -(a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x)\Delta u=f_{+}(x,u) , &{}\qquad \text{ in } \Omega ,\\ u=0, &{}\qquad \text{ on } \partial \Omega . \end{array} \right. \end{aligned}$$

(2.1)

Define $I_{w+}(u)$ by

$$\begin{aligned} I_{w+}(u)=\frac{1}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }F_{+}(x,u)\mathrm{d}x. \end{aligned}$$

By (H0), $I_{w+}\in C^{2}(E,{\mathbb {R}})$ is weakly lower semi-continuous, and it is clearly that the weak solution of the problem (2.1) corresponds to the critical point of the functional $I_{w+}$, see [18].

First, we prove that $I_{w+}(u)$ satisfies the (PS) condition.

Lemma 2.1

Assume that (H0), (H2) hold, then $I_{w+}(u)$ satisfies the (PS) condition.

Proof

Let $\{u_{n}\}$ be a sequence such that $|I_{w+}(u_{n})|\le M$ for some constant $M>0$ and $I'_{w+}(u_{n})\rightarrow 0$ as $n\rightarrow \infty$. By a standard argument (see [18]), it suffices to prove that $\{u_{n}\}$ is bounded. From (H0) and (H2), there exists $C_{0}>0$ such that

$$\begin{aligned} M+o(\Vert u_{n}\Vert )\ge & {} I_{w+}(u_{n})- \frac{1}{\theta }(I'_{w+}(u_{n}),u_{n}) \nonumber \\\ge & {} \left( \frac{1}{2}-\frac{1}{\theta }\right) a\int _{\Omega }|\nabla u_{n}|^{2}\mathrm{d}x +\int _{\Omega }\left( \frac{1}{\theta }f_{+}(x,u_{n})u_{n}-F_{+}(x,u_{n})\right) \mathrm{d}x \nonumber \\\ge & {} \left( \frac{1}{2}-\frac{1}{\theta }\right) a\Vert u_{n}\Vert ^{2}+\int _{u_{n}> t_{0}}\left( \frac{1}{\theta }f_{+}(x,u_{n})u_{n}-F_{+}(x,u_{n})\right) \mathrm{d}x-C_{0}\nonumber \\\ge & {} \left( \frac{1}{2}-\frac{1}{\theta }\right) a\Vert u_{n}\Vert ^{2}-C_{0}. \end{aligned}$$

(2.2)

This implies that $\{u_{n}\}$ is bounded in E. $\square$

Next, we prove that $I_{w+}$ has the geometry of the Mountain Pass theorem.

Lemma 2.2

Assume that (H0), (H1) hold, then there exist $\rho >0$ and $\alpha >0$, which are independent of w, such that for any $u\in E$ and $\Vert u\Vert =\rho$,

$$\begin{aligned} I_{w+}(u)\ge \alpha . \end{aligned}$$

Proof

By (H0) and (H1), for any $\varepsilon >0$, there exists $C_{\varepsilon }>0$ such that

$$\begin{aligned} |F_{+}(x,u(x))|\le \frac{\varepsilon }{2}|u|^{2}+C_{\varepsilon }|u|^{p+1}. \end{aligned}$$

Choose $\varepsilon$ small enough, by Sobolev inequality we have that

$$\begin{aligned} I_{w+}(u)\ge & {} \frac{a}{2}\int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }F_{+}(x,u)\mathrm{d}x \nonumber \\\ge & {} \frac{a}{2}\int _{\Omega }|\nabla u|^{2}\mathrm{d}x -\frac{\varepsilon }{2}\int _{\Omega }|u|^{2}\mathrm{d}x-C_{\varepsilon }\int _{\Omega }|u|^{p+1}\mathrm{d}x \nonumber \\\ge & {} \frac{1}{2}\left( a-\frac{\varepsilon }{\lambda _{1}}\right) \Vert u\Vert ^{2}-C_{\varepsilon }^{'}\Vert u\Vert ^{p+1}\nonumber \\\ge & {} \left( \frac{a}{4}-C_{1}^{'}\Vert u\Vert ^{p-1}\right) \Vert u\Vert ^{2} \end{aligned}$$

(2.3)

where $C_{1}^{'}$ is a constant independent of w and $\lambda _{1}$ is the first eigenvalue of $-\Delta$. Since $p>1$, then we can choose $\rho >0$ such that $C_{1}^{'}\rho ^{p-1}\le \frac{a}{8}$. Let $\alpha =\frac{a}{8}\rho ^{2}$, then by (2.3), for all $u\in \partial B_{\rho }(0)$, we have that $I_{w+}(u)\ge \alpha$. $\square$

Lemma 2.3

Assume that (H0), (H2) hold, then for any given positive function $v_{0}\in E$ with $\Vert v_{0}\Vert =1$, there exists $T>0$ such that for all $s\ge T$,

$$\begin{aligned} I_{w+}(sv_{0})\le 0. \end{aligned}$$

Proof

By (H2), there exist positive constants $C_{1}$, $C_{2}$ such that

$$\begin{aligned} F_{+}(x,t)\ge C_{1}|t|^{\theta }-C_{2}. \end{aligned}$$

Hence, for $s\ge 0$ we have

$$\begin{aligned} I_{w+}(sv_{0})= & {} \frac{1}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \Vert sv_{0}\Vert ^{2}-\int _{\Omega }F_{+}(x,sv_{0})\mathrm{d}x \\\le & {} \frac{1}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) s^{2}-C_{1}s^{\theta }\int _{\Omega }|v_{0}|^{\theta }\mathrm{d}x+ C_{2}|\Omega |, \end{aligned}$$

where $|\Omega |$ is the Lebesgue measure of $\Omega$, combining with $\theta >2$, we conclude that there exists $T>0$ such that $I_{w+}(sv_{0})<0$ for $s\ge T$. $\square$

Theorem 2.1

Assume (H0)–(H2) hold, then there exists a constant $b_{1}>0$ such that for $b\in [0, b_{1}]$, (1.1) has at least one positive solution and one negative solution.

Proof

The proof will be divided into three steps.

Step 1 For any given $w\in E$, (2.1) has a positive solution $u_{w}$ with $\Vert u_{w}\Vert \ge c_{1}$ for some constant $c_{1}>0$ independent of w and b.

Let

$$\begin{aligned} c_{w}=\inf _{g\in \Gamma }\max _{u\in g([0,1])}I_{w+}(u), \end{aligned}$$

(2.4)

where

$$\begin{aligned} \Gamma =\{g\in C([0,1],E)\mid g(0)=0, g(1)=Tv_{0}\}. \end{aligned}$$

By Lemmas 2.1, 2.2, 2.3 and the well-known Mountain Pass theorem [18], $c_{w}$ is a critical value of $I_{w+}$, so there is a $u_{w}\in E$ such that $I_{w+}(u_{w})=c_{w}$ and $I_{w+}'(u_{w})=0$. Hence, $u_{w}$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -(a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x)\Delta u_{w}=f_{+}(x,u_{w}) , &{}\qquad \text{ in } \Omega ,\\ u_{w}=0, &{}\qquad \text{ on } \partial \Omega . \end{array} \right. \end{aligned}$$

(2.5)

Multiplying Eq. (2.5) by $u_{w}^{-}$ with $u_{w}^{-}=\min \{u_{w},0\}$ and integrating on $\Omega$, we get

$$\begin{aligned} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{w}^{-}|^{2}\mathrm{d}x=\int _{\Omega }f_{+}(x,u_{w})u_{w}^{-}=0, \end{aligned}$$

so $u_{w}^{-}\equiv 0$. This shows $u_{w}$ is a positive solution of (2.1).

On the other hand, by (H0) and (H1), given $\varepsilon >0$, there exists a positive constant $C_{\varepsilon }$, such that

$$\begin{aligned} |f_{+}(x,t)|\le \varepsilon |t|+C_{\varepsilon }|t|^{p}, \end{aligned}$$

then using Eq. (2.5), by Sobolev inequality we obtain

$$\begin{aligned} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x= & {} \int _{\Omega }f_{+}(x,u_{w})u_{w}\mathrm{d}x \\\le & {} \varepsilon \int _{\Omega }|u_{w}|^{2}\mathrm{d}x+C_{\varepsilon }\int _{\Omega }|u_{w}|^{p+1}\mathrm{d}x \\\le & {} \frac{\varepsilon }{\lambda _{1}}\int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x+C_{\varepsilon }'\left( \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x\right) ^{\frac{p+1}{2}}; \end{aligned}$$

thus,

$$\begin{aligned} \left( a-\frac{\varepsilon }{\lambda _{1}}\right) \Vert u_{w}\Vert ^{2}\le C_{\varepsilon }'\Vert u_{w}\Vert ^{p+1}, \end{aligned}$$

which implies that there exists a positive constant $c_{1}$, independent of w and b, such that

$$\begin{aligned} \Vert u_{w}\Vert \ge c_{1}. \end{aligned}$$

(2.6)

Step 2 We construct a bounded positive functions sequence $\{u_{n}\}$ in E such that $I_{u_{n-1}+}'(u_{n})=0$ for any $n\ge 2$.

We fix a constant $L>0$ throughout this paper. Let $b(R)=\frac{L}{R^{2}}$ for $R>0$, then for any $w\in E$ with $\Vert w\Vert \le R$, any positive function $v_{0}\in E$ with $\Vert v_{0}\Vert =1$ and $b\in [0,b(R)]$, by (2.4) and (H2), we have

$$\begin{aligned} c_{w}= & {} I_{w+}(u_{w}) \\\le & {} \max _{t\ge 0}I_{w+}(tv_{0}) \\\le & {} \max _{t\ge 0}\left\{ \frac{1}{2}(a+bR^{2})\Vert tv_{0}\Vert ^{2}-\frac{1}{2}\int _{\Omega }F_{+}(tv_{0})\mathrm{d}x\right\} \\\le & {} \max _{t\ge 0}\left\{ \frac{1}{2}(a+L)t^{2}-\frac{1}{2}\int _{\Omega }(C_{1}t^{\theta }|v_{0}|^{\theta }-C_{2})\mathrm{d}x\right\} \\\le & {} C_{1}(L), \end{aligned}$$

where $C_{1}(L)$ is a constant independent of b, R and w. On the other hand, since $\langle I_{w+}'(u_{w}),u_{w}\rangle =0$, by (H2),

$$\begin{aligned} c_{w}= & {} I_{w+}(u_{w})-\frac{1}{\theta }\langle I_{w+}'(u_{w}),u_{w}\rangle \\= & {} \left( \frac{1}{2}-\frac{1}{\theta }\right) \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x\\&+\int _{\Omega }\left( \frac{1}{\theta }f_{+}(x,u_{w}) u_{w}-F_{+}(x,u_{w})\right) \mathrm{d}x \\\ge & {} \left( \frac{1}{2}-\frac{1}{\theta }\right) \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x-C_{0}. \end{aligned}$$

Hence,

$$\begin{aligned} \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x\le \frac{c_{w}+C_{0}}{\left( \frac{1}{2}-\frac{1}{\theta }\right) \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) } \le \frac{2\theta (c_{w}+C_{0})}{(\theta -2)a}\le \frac{2\theta C_{1}'(L)}{(\theta -2)a}, \end{aligned}$$

where $C_{1}'(L)=C_{1}(L)+C_{0}$.

Set $R_{1}=\sqrt{\frac{2\theta C_{1}'(L)}{(\theta -2)a}}$ and $b_{1}=b(R_{1})$, then for any $w\in E$ with $\Vert w\Vert \le R_{1}$ and $b\in [0, b_{1}]$, $I_{w+}$ has a critical point $u_{w}$ with $u_{w}>0$ and $c_{1}\le \Vert u_{w}\Vert \le R_{1}$. Let $w=u_{1}$ for some $u_{1}\in E$ with $u_{1}>0$ and $\Vert u_{1}\Vert \le R_{1}$, then $I_{u_{1}+}$ has a critical point $u_{2}$ with $u_{2}>0$ and $c_{1}\le \Vert u_{2}\Vert \le R_{1}$. Again, let $w=u_{2}$, then $I_{u_{2}+}$ has a critical point $u_{3}$ with $u_{3}>0$ and $c_{1}\le \Vert u_{3}\Vert \le R_{1}$. By induction, we get a sequence $\{u_{n}\}$ with $I_{u_{n-1}+}'(u_{n})=0$, $u_{n}>0$ and $c_{1}\le \Vert u_{n}\Vert \le R_{1}$.

Step 3 We prove that $u_{n}\rightarrow {\bar{u}}$ in E for some ${\bar{u}}\in E$ up to a subsequence and ${\bar{u}}$ is a positive solution of (1.1).

Since $\Vert u_{n}\Vert \le R_{1}$, then there exists ${\bar{u}}\in E$ such that $u_{n}\rightharpoonup {\bar{u}}$ in E and $u_{n}\rightarrow {\bar{u}}$ in $L^{p+1}(\Omega )$ up to a subsequence. By (H0), we have

$$\begin{aligned} I_{u_{n-1}+}'({\bar{u}})(u_{n}-{\bar{u}})= & {} \left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x \right) \int _{\Omega }\nabla {\bar{u}}\cdot \nabla (u_{n}-{\bar{u}})\mathrm{d}x\\&-\int _{\Omega }f_{+}(x,{\bar{u}})(u_{n}-{\bar{u}})\mathrm{d}x \\\rightarrow & {} 0. \end{aligned}$$

Thus,

$$\begin{aligned} 0= & {} \lim _{n\rightarrow \infty }[I_{u_{n-1}+}'(u_{n})(u_{n}-{\bar{u}})-I_{u_{n-1}+}'({\bar{u}})(u_{n}-{\bar{u}})] \\= & {} \lim _{n\rightarrow \infty } \left[ \left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla (u_{n}-{\bar{u}})|^{2}\mathrm{d}x\right. \\&\left. -\int _{\Omega } (f_{+}(x,u_{n})-f_{+}(x,{\bar{u}})) (u_{n}-{\bar{u}})\mathrm{d}x\right] \\= & {} \lim _{n\rightarrow \infty } \left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x\right) \Vert u_{n}-{\bar{u}}\Vert ^{2}, \end{aligned}$$

which means that $u_{n}\rightarrow {\bar{u}}$ in E as $n\rightarrow \infty$. Then, for any $\varphi \in E$, we have

$$\begin{aligned} 0= & {} \lim _{n\rightarrow \infty }I_{u_{n-1}+}'(u_{n})\varphi \\= & {} \lim _{n\rightarrow \infty } \left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x\right) \int _{\Omega }\nabla u_{n}\cdot \nabla \varphi \mathrm{d}x -\int _{\Omega }f_{+}(x,u_{n})\varphi \mathrm{d}x\\= & {} \left( a+b\int _{\Omega }|\nabla {\bar{u}}|^{2}\mathrm{d}x\right) \int _{\Omega }\nabla {\bar{u}}\cdot \nabla \varphi -\int _{\Omega }f_{+}(x,{\bar{u}})\varphi \mathrm{d}x\\= & {} I_{{\bar{u}}+}'({\bar{u}})\varphi . \end{aligned}$$

Hence, ${\bar{u}}$ is a critical point of $I_{{\bar{u}}+}$, and ${\bar{u}}$ satisfies

$$\begin{aligned} \left\{ \begin{array}{ll} -(a+b\int _{\Omega }|\nabla {\bar{u}}|^{2}\mathrm{d}x)\Delta {\bar{u}}=f_{+}(x,{\bar{u}}), &{}\qquad \text{ in } \Omega ,\\ {\bar{u}}=0, &{}\qquad \text{ on } \partial \Omega . \end{array} \right. \end{aligned}$$

(2.7)

Again by multiplying equation (2.7) by ${\bar{u}}^{-}$ with ${\bar{u}}^{-}=\min \{{\bar{u}},0\}$ and integrating on $\Omega$, we get

$$\begin{aligned} \left( a+b\int _{\Omega }|\nabla {\bar{u}}|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla {\bar{u}}^{-}|^{2}\mathrm{d}x=\int _{\Omega }f_{+}(x,{\bar{u}}){\bar{u}}^{-}\mathrm{d}x=0, \end{aligned}$$

which means that

$$\begin{aligned} {\bar{u}}^{-}\equiv 0. \end{aligned}$$

(2.8)

On the other hand, since $u_{n}\rightarrow {\bar{u}}$ in E and $\Vert u_{n}\Vert \ge c_{1}$, we have $\Vert {\bar{u}}\Vert \ge c_{1}$. Combined with (2.8) it proves that ${\bar{u}}$ is a positive solution of (2.7). Since $f_{+}(x,{\bar{u}})=f(x,{\bar{u}})$, ${\bar{u}}$ is also a positive solution of (1.1).

By a similar argument, we can prove that for $b\in [0,b_{1}]$, (1.1) also has at least one negative solution. $\square$

3 Sign Changing Solution

In this section, we first study the sign changing solution for (1.1) using the Nehari method proposed by Nehari[16] and then give the proof of Theorem 1.1.

For any $w\in E$, we consider the following problem

$$\begin{aligned} \left\{ \begin{array}{ll} -(a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x)\Delta u=f(x,u) , &{}\qquad \text{ in } \Omega ,\\ u=0, &{}\qquad \text{ on } \partial \Omega . \end{array} \right. \end{aligned}$$

(3.1)

The associated functional corresponding to (3.1) is $I_{w}:E\rightarrow {\mathbb {R}}$,

$$\begin{aligned} I_{w}(u)=\frac{1}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }F(x,u)\mathrm{d}x. \end{aligned}$$

By (H0), $I_{w}\in C^{2}(E,{\mathbb {R}})$ is weakly lower semi-continuous and the weak solution of the problem (3.1) corresponds to the critical point of the functional $I_{w}$, see [18].

Define

$$\begin{aligned} \begin{aligned} G_{w}(u)&=\langle I_{w}^{'}(u),u\rangle = \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }f(x,u)u\mathrm{d}x,\\ {\mathcal {N}}_{w}&=\{u\in H_{0}^{1}(\Omega ){\setminus }{\{0\}}\mid G_{w}(u)=0\},\\ \mathcal {S}_{w}&=\{u\in {\mathcal {N}}_{w} \mid u^{+}\in {\mathcal {N}}_{w}, u^{-}\in {\mathcal {N}}_{w}\}, \end{aligned} \end{aligned}$$

where $u^{+}=\max \{u,0\}$, $u^{-}=\min \{u,0\}$. The set ${\mathcal {N}}_{w}$ is called Nehari manifold. Obviously, any sign changing solutions of (3.1) must be on $\mathcal {S}_{w}$.

Lemma 3.1

Assume that (H0)–(H3) hold, then for each $u\in E{\setminus }{\{0\}}$ there exists unique $t=t(u)>0$ such that $t(u)u\in {\mathcal {N}}_{w}$ .

Proof

Similar as Lemma 2.2, there exist $\alpha >0$ and $\delta >0$ such that $I_{w}(u)>0$ for all $u\in B_{\delta }(0){\setminus }\{0\}$ and $I_{w}(u)\ge \alpha$ for all $u\in \partial B_{\delta }(0).$

Next we prove that for any $u\in E{\setminus }{\{0\}}$, $I_{w}(tu)\rightarrow -\infty$, as $t\rightarrow \infty$. By (H2),

$$\begin{aligned} I_{w}(tu)= & {} \frac{t^{2}}{2} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }F(x,tu)\mathrm{d}x\\\le & {} \frac{t^{2}}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-C_{1}t^{\theta }\int _{\Omega }|u|^{\theta }\mathrm{d}x+C_{2}|\Omega |. \end{aligned}$$

Since $\theta >2$, we have $I_{w}(tu)\rightarrow -\infty$, as $t\rightarrow \infty$.

For each fixed $u\in E{\setminus }{\{0\}}$, let $g_{w}(t)=I_{w}(tu)$ for $t>0$, then from the above argument, $g_{w}(t)$ has at least one maximum point with maximum value greater than $\alpha$. We will prove that $g_{w}(t)$ has a unique critical point for $t>0$. Noticed that

$$\begin{aligned} g_{w}^{'}(t)= & {} (I_{w}^{'}(tu),u) =\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }t|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }f(x,tu)u\mathrm{d}x,\\ g_{w}^{''}(t)= & {} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }f^{'}(x,tu)u^{2}\mathrm{d}x, \end{aligned}$$

by (H3), for every critical point $\overline{t}$ of $g_{w}(t)$, we have

$$\begin{aligned} g_{w}^{''}(\overline{t})= & {} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x-\int _{\Omega }f^{'}(x,\overline{t}u)u^{2}\mathrm{d}x \\= & {} \int _{\Omega }\frac{f(x,\overline{t}u)u}{\overline{t}}\mathrm{d}x-\int _{\Omega }f^{'}(x,\overline{t}u)u^{2}\mathrm{d}x \\= & {} \frac{1}{\overline{t}^{2}}\int _{\Omega }f(x,\overline{t}u)\overline{t} u-f^{'}(x,\overline{t}u)(\overline{t}u)^{2}\mathrm{d}x<0. \end{aligned}$$

This means that every critical point of $g_{w}(t)$ must be a strict local maximum; hence, $g_{w}(t)$ has a unique critical point, which is denoted by t(u).

Finally, by

$$\begin{aligned} (I_{w}^{'}(t(u)u),t(u)u)=t(u)(I_{w}^{'}(t(u)u),u)=t(u)g_{w}^{'}(t(u))=0, \end{aligned}$$

we obtain that $t(u)u\in {\mathcal {N}}_{w}$. $\square$

Lemma 3.2

There exists a constant $c_{2}>0$ independent of w such that $\Vert u\Vert \ge c_{2}$ for all $u\in {\mathcal {N}}_{w}$.

Proof

It follows from $u\in {\mathcal {N}}_{w}$ that

$$\begin{aligned} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x=\int _{\Omega }f(x,u)u\mathrm{d}x. \end{aligned}$$

By (H0) and (H1), for given $\varepsilon >0$, there exists $C_{\varepsilon }>0$ such that

$$\begin{aligned} |f(x,u)|\le \varepsilon |u|+C_{\varepsilon }|u|^{p}, \end{aligned}$$

then we have

$$\begin{aligned} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u|^{2}\mathrm{d}x\le \varepsilon \int _{\Omega }|u|^{2}\mathrm{d}x +C_{\varepsilon }\int _{\Omega }|u|^{p+1}\mathrm{d}x. \end{aligned}$$

Using Sobolev inequality, we obtain

$$\begin{aligned} \left( a-\frac{\varepsilon }{\lambda _{1}}\right) \Vert u\Vert ^{2}\le C_{\varepsilon }^{'}\Vert u\Vert ^{p+1}, \end{aligned}$$

which implies that there exists a constant $c_{2}>0$ independent of w such that $\Vert u\Vert \ge c_{2}$ for all $u\in {\mathcal {N}}_{w}$. $\square$

Define $m_{1}=\inf _{\mathcal {S}_{w}}I_{w}$, then it is clear that

$$\begin{aligned} m_{1}\ge \inf _{\partial B_{\delta }(0)}I_{w}\ge \alpha >0. \end{aligned}$$

Lemma 3.3

$m_{1}$ is achieved at some $u_{w}\in \mathcal {S}_{w}$.

Proof

Let $\{u_{n}\}$ be a minimizing sequence on $\mathcal {S}_{w}$ such that $I_{w}(u_{n})\rightarrow m_{1}$, then

$$\begin{aligned}&\frac{1}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{n}|^{2}\mathrm{d}x-\int _{\Omega }F(x,u_{n})\mathrm{d}x=m_{1}+o(1)\le C_{3},\nonumber \\ \end{aligned}$$

(3.2)

$$\begin{aligned}&\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{n}|^{2}\mathrm{d}x-\int _{\Omega }f(x,u_{n})u_{n}\mathrm{d}x=0, \end{aligned}$$

(3.3)

where $C_{3}>0$ is a constant. From (3.2), (3.3) and (H2) we have that

$$\begin{aligned}&\frac{\theta -2}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{n}|^{2}\mathrm{d}x\le C_{3}\\&\quad +\int _{\Omega }(\theta F(x,u_{n})-f(x,u_{n})u_{n})\mathrm{d}x\le C_{4}; \end{aligned}$$

thus, $\{u_{n}\}$ is bounded in E. Then, up to a subsequence, $u_{n}\rightharpoonup u$ and $u_{n}^{\pm }\rightharpoonup u^{\pm }$ in E.

Now we claim that $u^{+}\not \equiv 0$ and $u^{-}\not \equiv 0$. In fact, if $u^{+}\equiv 0$, then $u_{n}^{+}\rightarrow 0$ in $L^{p+1}(\Omega )$, so by $u_{n}^{+}\in {\mathcal {N}}_{w}$ and (H0), we get

$$\begin{aligned} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{n}^{+}|^{2}\mathrm{d}x=\int _{\Omega }f(x,u_{n}^{+})u_{n}^{+}\mathrm{d}x\rightarrow 0, \end{aligned}$$

this is a contradiction with Lemma 3.2. Similarly, we can prove that $u^{-}\not \equiv 0$.

By Lemma 3.1, there exist $t,s>0$, such that $tu^{+}\in {\mathcal {N}}_{w}$ and $su^{-}\in {\mathcal {N}}_{w}$, and hence $tu^{+}+su^{-}\in \mathcal {S}_{w}$. Furthermore, since $I_{w}(u)$ is weakly lower semi-continuous, we get

$$\begin{aligned} m_{1}\le I_{w}(tu^{+}+su^{-})= & {} I_{w}(tu^{+})+I_{w}(su^{-}) \nonumber \\\le & {} \liminf _{n\rightarrow \infty }I_{w}(tu_{n}^{+})+\liminf _{n\rightarrow \infty }I_{w}(su^{-})\nonumber \\\le & {} \liminf _{n\rightarrow \infty }I_{w}(u_{n}^{+})+\liminf _{n\rightarrow \infty }I_{w}(u_{n}^{-})\nonumber \\\le & {} \liminf _{n\rightarrow \infty }(I_{w}(u_{n}^{+})+I_{w}(u_{n}^{-}))\nonumber \\= & {} \liminf _{n\rightarrow \infty }(I_{w}(u_{n}))=m_{1}.\nonumber \\ \end{aligned}$$

Let $u_{w}=tu^{+}+su^{-}$, then $I_{w}(u_{w})=m_{1}$. $\square$

In what follows we prove that the minimizer $u_{w}$ of $I_{w}$ on $\mathcal {S}_{w}$ is a critical point of $I_{w}$, here we use an argument similar as [13].

Lemma 3.4

If $I_{w}(u_{w})=m_{1}$ for some $u_{w}\in \mathcal {S}_{w}$, then $u_{w}$ is a critical point of $I_{w}$.

Proof

If $u_{w}$ is not a critical point of $I_{w}$, then there exists $\varphi \in C_{0}^{\infty }(\Omega )$ such that

$$\begin{aligned} \langle I_{w}^{'}(u_{w}),\varphi \rangle \le -1; \end{aligned}$$

thus, there exists $\varepsilon _{0}>0$ such that for $|t-1|\le \varepsilon _{0}$, $|s-1|\le \varepsilon _{0}$, and $|\sigma |\le \varepsilon _{0}$,

$$\begin{aligned} \langle I_{w}^{'}(tu_{w}^{+}+su_{w}^{-}+\sigma \varphi ),\varphi \rangle \le -\frac{1}{2}. \end{aligned}$$

Let $\eta =\eta (t,s)\ge 0, (t,s)\in T=[\frac{1}{2},\frac{3}{2}]\times [\frac{1}{2},\frac{3}{2}]$ be a cut-off function such that $\eta (t,s)=1$, if $|t-1|\le \frac{1}{2}\varepsilon _{0}$ and $|s-1|\le \frac{1}{2}\varepsilon _{0}$, $\eta (t,s)=0$, if $|t-1|\ge \varepsilon _{0}$ or $|s-1|\ge \varepsilon _{0}$. If $|t-1|\le \varepsilon _{0}$ and $|s-1|\le \varepsilon _{0}$, then

$$\begin{aligned}&I_{w}(tu_{w}^{+}+su_{w}^{-}+\varepsilon _{0}\eta (t,s)\varphi ) \nonumber \\&\quad =I_{w}(tu_{w}^{+}+su_{w}^{-})+\int _{0}^{1}\langle I_{w}^{'}(tu_{w}^{+}+su_{w}^{-}+\mu \varepsilon _{0}\eta (t,s)\varphi ),\varepsilon _{0}\eta (t,s)\varphi \rangle \mathrm{d}\mu \nonumber \\&\quad \le I_{w}(tu_{w}^{+}+su_{w}^{-})-\frac{1}{2}\varepsilon _{0}\eta (t,s). \end{aligned}$$

(3.4)

For $|t-1|\ge \varepsilon _{0}$ or $|s-1|\ge \varepsilon _{0}$, since $\eta (t,s)=0$, the above estimate is trivial. Since $u_{w}\in \mathcal {S}_{w}$, for $(t,s)\ne (1,1)$, we have $I_{w}(tu_{w}^{+}+su_{w}^{-})<I_{w}(u_{w})$. Then, by (3.4), for $(t,s)\ne (1,1)$,

$$\begin{aligned} I_{w}(tu_{w}^{+}+su_{w}^{-}+\varepsilon _{0}\eta (t,s)\varphi )\le I_{w}(tu_{w}^{+}+su_{w}^{-})<I_{w}(u_{w}), \end{aligned}$$

for $(t,s)=(1,1)$,

$$\begin{aligned} I_{w}(u_{w}^{+}+u_{w}^{-}+\varepsilon _{0}\eta (t,s)\varphi )\le I_{w}(u_{w})-\frac{1}{2}\varepsilon _{0}\eta (1,1) =I_{w}(u_{w})-\frac{1}{2}\varepsilon _{0}. \end{aligned}$$

Hence,

$$\begin{aligned} \sup _{(t,s)\in T}I_{w}(tu_{w}^{+}+su_{w}^{-}+\varepsilon _{0}\eta (t,s)\varphi )<I_{w}(u_{w})=m_{1}, \end{aligned}$$

which implies that for all $(t,s)\in T$, $tu_{w}^{+}+su_{w}^{-}+\varepsilon _{0}\eta (t,s)\varphi \notin \mathcal {S}_{w}.$ We will show that there must be $(t_{0},s_{0})\in T$ such that $t_{0}u_{w}^{+}+s_{0}u_{w}^{-}+\varepsilon _{0}\eta (t_{0},s_{0})\varphi \in \mathcal {S}_{w},$ then this gives a contradiction.

For $0\le \varepsilon \le \varepsilon _{0}$, define $h_{\varepsilon }:T\rightarrow H_{0}^{1}(\Omega )$ by

$$\begin{aligned} h_{\varepsilon }(t,s)=tu_{w}^{+}+su_{w}^{-}+\varepsilon \eta (t,s)\varphi \end{aligned}$$

and $H_{\varepsilon }:T\rightarrow {\mathbb {R}}^{2}$ by

$$\begin{aligned} H_{\varepsilon }(t,s)=(G_{w}(h_{\varepsilon }(t,s)^{+}),G_{w}(h_{\varepsilon }(t,s)^{-})). \end{aligned}$$

For all $(t,s)\in \partial T$, $\eta (t,s)=0$, then $h_{\varepsilon }(t,s)=tu_{w}^{+}+su_{w}^{-}$; thus, for all $(t,s)\in \partial T$ and $0\le \varepsilon \le \varepsilon _{0}$,

$$\begin{aligned} H_{\varepsilon }(t,s)= & {} (G_{w}((tu_{w}^{+}+su_{w}^{-})^{+}),G_{w}((tu_{w}^{+}+su_{w}^{-})^{-})) \\= & {} (G_{w}(tu_{w}^{+}),G_{w}(su_{w}^{-})) \\\ne & {} (0,0). \end{aligned}$$

Hence, by the homotopy invariance of Brouwer degree, we have

$$\begin{aligned} \deg (H_{\varepsilon _{0}}(t,s),T,(0,0))=\deg (H_{0}(t,s),T,(0,0)). \end{aligned}$$

(3.5)

Next we shall prove that

$$\begin{aligned} \deg (H_{0}(t,s),T,(0,0))=1. \end{aligned}$$

Notice that $H_{0}(t,s)=(G_{w}(tu_{w}^{+}),G_{w}(su_{w}^{-})),$ and we denote

$$\begin{aligned} a(t)=G_{w}(tu_{w}^{+}), b(s)=G_{w}(su_{w}^{-}). \end{aligned}$$

By $u_{w}^{+}\in {\mathcal {N}}_{w}$ and (H3) we have

$$\begin{aligned} a^{'}(1)= & {} \langle G_{w}^{'}(u_{w}^{+}),u_{w}^{+}\rangle \nonumber \\= & {} \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }2|\nabla u_{w}^{+}|^{2}\mathrm{d}x-\int _{\Omega }\left[ f^{'}(x,u_{w}^{+})(u_{w}^{+})^{2}+f(x,u_{w}^{+})u_{w}^{+} \right] \mathrm{d}x\\= & {} \int _{\Omega }2f(x,u_{w}^{+})u_{w}^{+}\mathrm{d}x-\int _{\Omega } \left[ f^{'}(x,u_{w}^{+})(u_{w}^{+})^{2}+f(x,u_{w}^{+})u_{w}^{+}\right] \mathrm{d}x\\= & {} \int _{\Omega }\left[ f(x,u_{w}^{+})u_{w}^{+}-f^{'}(x,u_{w}^{+})(u_{w}^{+})^{2}\right] \mathrm{d}x<0. \end{aligned}$$

Similarly, $b^{'}(1)<0$. By Lemma 3.1, $(t,s)=(1,1)$ is the unique solution of $H_{0}(t,s)=(a(t),b(s))=(0,0)$. Then, by the definition of Brouwer degree, clearly we have

$$\begin{aligned} \deg (H_{0}(t,s),T,(0,0))=1. \end{aligned}$$

Thus, by (3.5) we have

$$\begin{aligned} \deg (H_{\varepsilon _{0}}(t,s),T,0)=1\ne 0. \end{aligned}$$

Therefore, there must exists $(t_{0},s_{0})\in T$ such that

$$\begin{aligned} H_{\varepsilon _{0}}(t_{0},s_{0})&= (G_{w}((t_{0}u_{w}^{+}+s_{0}u_{w}^{-}+\varepsilon _{0}\eta (t_{0},s_{0})\varphi )^{+}),\\&\quad G_{w}((t_{0}u_{w}^{+}+s_{0}u_{w}^{-}+\varepsilon _{0}\eta (t_{0},s_{0})\varphi )^{-}))\\&= (0,0), \end{aligned}$$

which means that $(t_{0}u_{w}^{+}+s_{0}u_{w}^{-}+\varepsilon _{0}\eta (t_{0},s_{0})\varphi \in \mathcal {S}_{w}.$$\square$

Now we can state and prove the existence of sign changing solution for (1.1).

Theorem 3.1

Assume (H0)–(H3) hold, then there exists a constant $b_{2}>0$ such that for any $b\in [0, b_{2}]$, (1.1) has at least one sign changing solution.

Proof

First we fix a function $v\in E$ with $v^{+}\ne 0$ and $v^{-}\ne 0$. Thanks to Lemma 3.3 and Lemma 3.4, we get a minimizer $u_{w}$ of $I_{w}$ on $\mathcal {S}_{w}$ and $I'_{w}(u_{w})=0$. For the fixed constant $L>0$ as in the proof of Theorem 2.1, recall that $b(R)=\frac{L}{R^{2}}$ for $R>0$. By Lemma 3.1, (H2), and notice that $u_{w}$ is a minimizer of $I_{w}$ on $\mathcal {S}_{w}$, for $w\in E$ with $\Vert w\Vert \le R$ and $b\in [0,b(R)]$, clearly we have

$$\begin{aligned} I_{w}(u_{w})\le & {} \sup _{t,s>0}I_{w}(tv^{+}+sv^{-})\\\le & {} \sup _{t>0}\left( \frac{t^{2}}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla v^{+}|^{2}\mathrm{d}x -C_{1}t^{\theta }\int _{\Omega }|v^{+}|^{\theta }+C_{2}|\Omega |\right) \\&+\sup _{s>0}\left( \frac{s^{2}}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla v^{-}|^{2}\mathrm{d}x -C_{1}s^{\theta }\int _{\Omega }|v^{-}|^{\theta }+C_{2}|\Omega |\right) \\\le & {} \sup _{t>0}\left( \frac{t^{2}}{2}(a+L)\int _{\Omega }|\nabla v^{+}|^{2}\mathrm{d}x -C_{1}t^{\theta }\int _{\Omega }|v^{+}|^{\theta }+C_{2}|\Omega |\right) \\&+\sup _{s>0}\left( \frac{s^{2}}{2}(a+L)\int _{\Omega }|\nabla v^{-}|^{2}\mathrm{d}x -C_{1}s^{\theta }\int _{\Omega }|v^{-}|^{\theta }+C_{2}|\Omega |\right) \\\le & {} C_{2}(L), \end{aligned}$$

where $C_{2}(L)>0$ is a constant independent of b, R and w, and the last inequality follows from $\theta >2$. Since $u_{w}$ is a critical point of $I_{w}$, by (H2) we have

$$\begin{aligned}&\frac{1}{2}\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x\\&\quad =I_{w}(u_{w})+\int _{\Omega }F(x,u_{w})\mathrm{d}x\\&\quad \le C_{2}(L)+\int _{\Omega }F(x,u_{w})\mathrm{d}x \\&\quad \le C_{2}(L)+\frac{1}{\theta }\int _{\Omega }f(x,u_{w})u_{w}\mathrm{d}x+C_{0} \\&\quad =C_{2}'(L)+\frac{1}{\theta }\left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) \int _{\Omega }|\nabla u_{w}|^{2}\mathrm{d}x, \end{aligned}$$

where $C_{2}'(L)=C_{2}(L)+C_{0}$, then

$$\begin{aligned} \Vert u_{w}\Vert ^{2}\le \frac{C_{2}'(L)}{\left( \frac{1}{2}-\frac{1}{\theta }\right) \left( a+b\int _{\Omega }|\nabla w|^{2}\mathrm{d}x\right) } \le \frac{2\theta C_{2}'(L)}{(\theta -2)a}. \end{aligned}$$

(3.6)

Set $R_{2}=\sqrt{\frac{2\theta C_{2}'(L)}{(\theta -2)a}}$ and $b_{2}=b(R_{2})$. For any $w\in E$ with $\Vert w\Vert \le R_{2}$ and $0\le b\le b_{2}$, by Lemma 3.4 and (3.6), $I_{w}$ has a critical point $u_{w}\in \mathcal {S}_{w}$ with $\Vert u_{w}\Vert \le R_{2}$. Let $w=u_{1}$ for some $u_{1}\in E$ with $\Vert u_{1}\Vert \le R_{2}$, then $I_{u_{1}}$ has a critical point $u_{2}$ with $u_{2}\in \mathcal {S}_{u_{1}}$ and $\Vert u_{2}\Vert \le R_{2}$. Again, let $w=u_{2}$, then $I_{u_{2}}$ has a critical point $u_{3}\in \mathcal {S}_{u_{2}}$ with $\Vert u_{3}\Vert \le R_{2}$. By induction, we get a sequence $\{u_{n}\}$ with $I_{u_{n-1}}'(u_{n})=0$, $u_{n}\in \mathcal {S}_{u_{n-1}}$ and $\Vert u_{n}\Vert \le R_{2}$.

Since $\Vert u_{n}\Vert \le R_{2}$, we have $u_{n}\rightharpoonup \tilde{u}$ in E and $u_{n}\rightarrow \tilde{u}$ in $L^{p+1}(\Omega )$ up to a subsequence. Then, by (H0), we have

$$\begin{aligned} I_{u_{n-1}}'(\tilde{u})(u_{n}-\tilde{u})= & {} \left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x\right) \int _{\Omega }\nabla \tilde{u}\cdot \nabla (u_{n}-\tilde{u})\mathrm{d}x\\&-\int _{\Omega }f(x,\tilde{u})(u_{n}-\tilde{u})\mathrm{d}x \\\rightarrow & {} 0. \end{aligned}$$

Thus,

$$\begin{aligned} 0= & {} \lim _{n\rightarrow \infty }[I_{u_{n-1}}'(u_{n})(u_{n}-\tilde{u})-I_{u_{n-1}}'(\tilde{u})(u_{n}-\tilde{u})] \\= & {} \lim _{n\rightarrow \infty } \left[ (a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x)\int _{\Omega }|\nabla (u_{n} -\tilde{u})|^{2}\mathrm{d}x\right. \\&\left. -\int _{\Omega }(f(x,u_{n})-f(x,\tilde{u})) (u_{n}-\tilde{u})\mathrm{d}x\right] \\= & {} \lim _{n\rightarrow \infty } \left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x\right) \Vert u_{n}-\tilde{u}\Vert ^{2}, \end{aligned}$$

which means that $u_{n}\rightarrow \tilde{u}$ in E as $n\rightarrow \infty$. Then, for any $\varphi \in E$, we have

$$\begin{aligned} 0= & {} \lim _{n\rightarrow \infty }I_{u_{n-1}}'(u_{n})\varphi \\= & {} \lim _{n\rightarrow \infty }\left( a+b\int _{\Omega }|\nabla u_{n-1}|^{2}\mathrm{d}x\right) \int _{\Omega }\nabla u_{n}\cdot \nabla \varphi -\int _{\Omega }(f(x,u_{n})\varphi \\= & {} \left( a+b\int _{\Omega }|\nabla \tilde{u}|^{2}\mathrm{d}x\right) \int _{\Omega }\nabla \tilde{u}\cdot \nabla \varphi -\int _{\Omega }(f(x,\tilde{u})\varphi \\= & {} I_{\tilde{u}}'(\tilde{u})\varphi , \end{aligned}$$

so $\tilde{u}$ is a critical point of $I_{\tilde{u}}$, and $\tilde{u}$ satisfies (1.1). Since $u_{n}\in \mathcal {S}_{u_{n-1}}$, we have $u_{n}^{+}\in {\mathcal {N}}_{u_{n-1}}$ and $u_{n}^{-}\in {\mathcal {N}}_{u_{n-1}}$. By Lemma 3.2, $\Vert u_{n}^{+}\Vert \ge c_{2}$ and $\Vert u_{n}^{-}\Vert \ge c_{2}$; hence, $\Vert \tilde{u}^{+}\Vert \ge c_{2}$ and $\Vert \tilde{u}^{-}\Vert \ge c_{2}$. Therefore, $\tilde{u}$ is a sign changing solution of (1.1). $\square$

Now we give the proof of Theorem 1.1.

Proof of Theorem 1.1

Let $b_{0}=\min \{b_{1}, b_{2}\}$, then by Theorems 2.1 and 3.1, for any $b\in [0, b_{0}]$, the problem (1.1) has at least three nontrivial solutions; among them, one is positive, one is negative, and one is sign changing. $\square$