1 Introduction

In the past decades, the following Kirchhoff-type equations

$$\begin{aligned} -\left( a+b\displaystyle \int \limits _{{\mathbb {R}}^3}|\nabla u|^2\mathrm{{d}}x\right) \Delta u + V(x)u =f(x,u),\ \ x\in {\mathbb {R}}^3, \end{aligned}$$
(1.1)

has been investigated by many authors, where \(V:{\mathbb {R}}^3\rightarrow {\mathbb {R}}\), \(f\in \mathcal {C}({\mathbb {R}}^3 \times {\mathbb {R}}, {\mathbb {R}})\) and \(a,b>0\) are constants. If \(V(x)\equiv 0\) and replace \({\mathbb {R}}^3\) by a bounded domain \(\Omega \subset {\mathbb {R}}^3\) in (1.1), we then obtain the following Kirchhoff Dirichlet problem

$$\begin{aligned} \left\{ \begin{array}{ll} - \left( a + b\displaystyle \int \limits _\Omega |\nabla u|^2\mathrm{{d}}x\right) \Delta u = f(x,u), \ \ &{}x\in \Omega , \\ u = 0, &{}x\in \partial \Omega . \end{array} \right. \end{aligned}$$
(1.2)

Equation (1.2) is related to the stationary analogue of the following equation

$$\begin{aligned} \rho \frac{\partial ^2u}{\partial ^2t}-\left( \frac{P_0}{h} +\frac{E}{2L}\int \limits ^L_0\Big |\frac{\partial u}{\partial x}\Big |^2\mathrm{{d}}x\right) \frac{\partial ^2u}{\partial ^2x}=0, \end{aligned}$$

which is proposed by Kirchhoff in [19] as an extension of the classical D’Alembert’s wave equations for free vibration of elastic strings. After the pioneer work of Lions [20], where a functional analysis approach was proposed to the equation

$$\begin{aligned} \left\{ \begin{array}{ll} u_{tt} - \left( a + b \displaystyle \int \limits _\Omega |\nabla u|^2\mathrm{{d}}x\right) \Delta u = f(x,u), \ \ {} &{}x\in \Omega , \\ u = 0, \ \ {} &{}x\in \partial \Omega , \end{array} \right. \end{aligned}$$
(1.3)

equation (1.3) began to call attention of several researchers, see [2, 5, 8] and the references therein.

Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. In (1.2), u denotes the displacement, f(xu) is the external force, b is the initial tension and a is related to the intrinsic properties of the string. We point out that such nonlocal problems also appear in other fields as biological systems, where u describes a process which depends on the average of itself, for example, population density. For more mathematical and physical background of (1.2), we refer the reader to the papers [1, 2, 5, 15, 16, 19, 21] and the references therein.

Mathematically, Eq. (1.1) is a nonlocal problem as the appearance of the nonlocal term \(\int \limits _{{\mathbb {R}}^3}|\nabla u|^2\mathrm{{d}}x \Delta u\), which implies that (1.1) is not a pointwise identity. This causes some mathematical difficulties which make the study of (1.1) particularly interesting. A lot of interesting results on the existence of positive solutions, multiple solutions, semiclassical state solutions and sign-changing solutions for (1.1) are obtained in last decade, see for examples, [6, 9, 11, 12, 15,16,17,18, 21, 22, 24, 26,27,28] and the references therein.

In particular, Chen, Kuo and Wu [6] studied the following nonlinear Kirchhoff-type equation with indefinite nonlinearity

$$\begin{aligned} \displaystyle \left\{ \begin{array}{lll} -\displaystyle \left( a + b \displaystyle \int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x\right) \Delta u =\lambda f(x)|u|^{q-2}u+g(x)|u|^{p-2}u, &{}x\in \Omega , \\ u=0, &{}\text {on}\ \partial \Omega , \end{array} \right. \end{aligned}$$
(1.4)

where \(a,b>0\), \(\Omega \) is a smooth bounded domain in \({\mathbb {R}}^N\) with \(1<q<2<p<2^*\) (\(2^*=\frac{2N}{N-2}\) if \(N\ge 3\), \(2^*=+\infty \) if \(N= 1,2\)), \(\lambda >0\) is a parameter, the weight functions \(f, g\in \mathcal {C}(\Omega )\) satisfy \(f^+(x):=\max \{f(x),0\}\not \equiv 0\) and \(g^+(x):=\max \{g(x),0\}\not \equiv 0\). By using Nehari manifold and fibering map, the authors proved the existence of multiple positive solutions for Eq. (1.4). We point out that Kirchoff-type equations with potential well and indefinite nonlinearities were also investigated in [26, 30].

Recently, Figueiredo et al [13] investigated ground states of elliptic problems over cones. As an application, the authors [13] proved the following Kirchhoff-type equation

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -M\left( \int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x\right) \Delta u =b(x)|u|^{r-2}u, \ x\in \Omega ,\\ u \in H^{1}_0(\Omega ), \end{array} \right. \end{aligned}$$
(1.5)

has a positive ground state solution provided \(b^+(x):=\max \{b(x),0\}\not \equiv 0\) and \(r\in (4,6)\), where \(\Omega \subset {\mathbb {R}}^3\) is a bounded domain with smooth boundary, \(M:[0,+\infty )\rightarrow [0,+\infty )\) is a monotone increasing \(\mathcal {C}^1\) function such that \(M(0):=m_0>0\) and \(t\mapsto \frac{M(t)}{t}\) is increasing on \((0,+\infty )\).

Based on the above results, a natural question is whether Eq. (1.5) has sign-changing solutions with b(x) is a sign-changing function. The present paper is devoted to this aspect and partially answers this question. More precisely, we devoted to study the existence of sign-changing solutions for the following Kirchhoff-type equation

$$\begin{aligned} \left\{ \begin{array}{ll} -\left( a+b\displaystyle \int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x\right) \Delta u =\left( h^+(x)+\lambda h^-(x)\right) |u|^{p-2}u, \ \ \ {} &{}x\in \Omega , \\ u = 0, &{}x\in \partial \Omega , \end{array} \right. \end{aligned}$$
(1.6)

where a, \(b>0\), \(\Omega \subset {\mathbb {R}}^3\) is a bounded domain with smooth boundary, the potential \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function, \(\lambda >0\) is a parameter, and

$$\begin{aligned} h^+(x)=\max \big \{h(x),0\big \},\ \ \ h^-(x)=\min \big \{h(x),0\big \}. \end{aligned}$$

Throughout this paper, we denote \(H^1_0(\Omega )\) the usual Sobolev space equipped with the inner product and norm

$$\begin{aligned} (u,v)=\displaystyle \int \limits _{\Omega }\nabla u \nabla v\mathrm{{d}}x,\ \ \Vert u\Vert =(u,u)^{1/2}. \end{aligned}$$

Define the energy functional \(I_{b,\lambda }: H^1_0(\Omega )\rightarrow {\mathbb {R}}\) by

$$\begin{aligned} I_{b,\lambda }(u):=\frac{a}{2}\int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x+\frac{b}{4}\left( \,\,\int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x\right) ^2-\frac{1}{p}\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^{p}\mathrm{{d}}x. \end{aligned}$$
(1.7)

Obviously, the functional \(I_{b,\lambda }\) is well-defined and belongs to \(\mathcal {C}^1(H^1_0(\Omega ),{\mathbb {R}})\). Moreover, for any \(u, \varphi \in H^1_0(\Omega )\), we have

$$\begin{aligned} \langle I'_\lambda (u),\varphi \rangle =a\int \limits _{\Omega }\nabla u \nabla \varphi \mathrm{{d}}x +b\int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x\int \limits _{\Omega }\nabla u \nabla \varphi \mathrm{{d}}x -\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^{p-2}u\varphi \mathrm{{d}}x. \end{aligned}$$
(1.8)

In the case \(h(x)\equiv 1\), by constrained minimization method, Figueiredo and Nascimento [12] and Shuai [25] proved the existence of least energy sign-changing solution for Eq. (1.6). The authors first proved the following set

$$\begin{aligned} \mathcal {M}_{b,\lambda }=\Big \{u\in H^1_0(\Omega ),\ u^\pm \ne 0\ \ \text {and}\ \ \langle I_{b,\lambda }'(u),u^+\rangle =\langle I_{b,\lambda }'(u),u^-\rangle =0\Big \} \end{aligned}$$
(1.9)

is nonempty, which is a crucial step. Then, the authors sought a minimizer of the energy functional \(I_{b,\lambda }\) restricted on \(\mathcal {M}_{b,\lambda }\) and proved the minimizer is a sign-changing solution of (1.6) by quantitative deformation lemma. In the first step, the authors proved that, for each \(u\in H^1_0(\Omega )\) with \(u^\pm \ne 0\), there exists a unique pair \((s,t)\in {\mathbb {R}}_+\times {\mathbb {R}}_+\) such that \(su^+ +tu^- \in \mathcal {M}_{b,\lambda }\), see Lemma 2.3 in [12] and Lemma 2.1 in [25]. However, if h(x) is a sign-changing continuous function, this fact does not hold for all \(u\in H^1_0(\Omega )\) with \(u^\pm \ne 0\), but rather in some part of it. A direct observation is that, a necessary condition for \(u\in \mathcal {M}_{b,\lambda }\) is \(u^+, u^-\in \mathcal {A}\), where

$$\begin{aligned} \mathcal {A}:=\Big \{u\in H^1_0(\Omega )\setminus \{0\}:\ \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^{p}\mathrm{{d}}x>0\Big \}. \end{aligned}$$
(1.10)

Thus, the method that used in [12, 25] cannot be applied to Eq. (1.6), we need some crucial modifications.

Our first main result can be stated as follows.

Theorem 1.1

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function, \(p\in (4,6)\) and \(\lambda >0\), then Eq. (1.6) possesses one least energy sign-changing solution \(u_{b,\lambda }\), which has precisely two nodal domains. Moreover, \(I_{b,\lambda }(u_{b,\lambda })>2c_{b,\lambda }\), where

$$\begin{aligned} c_{b,\lambda }:=\inf \limits _{u\in \mathcal {N}_{b,\lambda }}I_{b,\lambda }(u) \end{aligned}$$
(1.11)

and

$$\begin{aligned} \mathcal {N}_{b,\lambda }:=\Big \{u\in H^1_0(\Omega ) \setminus \{0\} : \langle I_{b,\lambda }'(u),u\rangle =0\Big \}. \end{aligned}$$
(1.12)

Theorem 1.1 implies that, the energy of any sign-changing solution of Eq. (1.6) is larger than two times the least energy, this property is called energy doubling by Weth in [29]. It is obvious that the least energy of the sign-changing solution \(u_{b,\lambda }\) obtained in Theorem 1.1 depends on b and \(\lambda \). We next focus on the convergence property of \(u_{b,\lambda }\) as \(b\rightarrow 0^+\) or \(\lambda \rightarrow +\infty \). Our main results in this direction can be stated as follows.

Theorem 1.2

If the assumptions of Theorem 1.1 hold, for any sequence \(\{b_n\}\) with \(b_n \rightarrow 0^+\) as \(n\rightarrow \infty \), there exists a subsequence, still denoted by \(\{b_n\}\), such that \(u_{b_n,\lambda }\rightarrow u_{0,\lambda }\) strongly in \(H^1_0(\Omega )\) as \(n\rightarrow \infty \), where \(u_{b_n,\lambda }\) denote the least energy sign-changing solution of Eq. (1.6) with \(b=b_n\) obtained by Theorem 1.1, and \(u_{0,\lambda }\) is a least energy sign-changing solution of the following equation

$$\begin{aligned} \left\{ \begin{array}{ll} -a\Delta u =\left( h^+(x)+\lambda h^-(x)\right) |u|^{p-2}u, \ {} &{} x\in \Omega , \\ u=0, \ {} &{} \text {on}\ \partial \Omega , \end{array} \right. \end{aligned}$$
(1.13)

which changes sign only once.

The proof of Theorem 1.2 includes three steps, we first prove \(\{u_{b_n,\lambda }\}\) is bounded in \(H^1_0(\Omega )\), then we prove \(u_{b_n,\lambda }\rightarrow u_{0,\lambda }\) strongly in \(H^1_0(\Omega )\), and we finally prove that \(u_{0,\lambda }\) is just a least energy sign-changing solution of (1.13).

Theorem 1.3

If the assumptions of Theorem 1.1 hold, for any sequence \(\{\lambda _n\}\) with \(\lambda _n \rightarrow +\infty \) as \(n\rightarrow \infty \), there exists a subsequence, still denoted by \(\{\lambda _n\}\), such that \(u_{b,\lambda _n}\rightarrow \bar{u}\) strongly in \(H^1_0(\Omega )\) as \(n\rightarrow \infty \), where \(u_{b,\lambda _n}\) denote the least energy sign-changing solution of Eq. (1.6) with \(\lambda =\lambda _n\) obtained by Theorem 1.1, and \(\bar{u}\) is a least energy sign-changing solution of following equation

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\left( a+b\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x\right) \Delta u =h^+(x)|u|^{p-2}u, \ &{} x\in \Omega \setminus \Omega ^- ,\\ u=0,\ {} &{}x\in \Omega ^-,\\ u=0,\ {} &{}x\in \partial \Omega . \end{array} \right. \end{aligned}$$
(1.14)

which changes sign only once, here \(\Omega ^-:=\{x\in \Omega \ |\ h(x)<0\}\).

Next, we study the existence of multi-bump sign-changing solutions for Eq. (1.6). We now assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function satisfying

(\(h_1\))    \(\Omega ^+:=\{x\in \Omega \ |\ h(x)>0\}=\Omega {\setminus } \overline{\Omega ^-}\);

(\(h_2\))    the set \(\Omega ^+\) is the union of k (\(k\ge 2\)) open connected and disjoint Lipschitz components, that is

$$\begin{aligned} \displaystyle \Omega ^+=\cup _{i=1}^{k}\Omega _{i}\ \ \ \text {and}\ \ \ dist(\Omega _i,\Omega _j)>0 \ \ \text {for} \ \ i\ne j;\ \ i,j=1,2,\ldots ,k. \end{aligned}$$
(1.15)

Theorem 1.4

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \((h_1)\)\((h_2)\) hold. If \(p\in (4,6)\), then, for any non-empty subset \(\Gamma \subset \{1,2,\ldots ,k\}\) with

$$\begin{aligned} \Gamma =\Gamma _1\cup \Gamma _2\cup \Gamma _3 \ \ \text {and}\ \ \Gamma _i\cap \Gamma _j=\varnothing \ \ \text {for}\ \ i\ne j,\ i,j=1,2,3, \end{aligned}$$
(1.16)

there exists a constant \(\Lambda _\Gamma >0\) such that for \(\lambda \ge \Lambda _\Gamma \), Eq. (1.6) has a sign-changing multi-bump solution \(u_{b,\lambda }\), which possesses the following property: for any sequence \(\{\lambda _n\}\) with \(\lambda _n \rightarrow +\infty \) as \(n\rightarrow \infty \), there exists a subsequence, still denoted by \(\{\lambda _n\}\), such that \(u_{b,\lambda _n}\rightarrow u\) strongly in \(H^1_0(\Omega )\) as \(n\rightarrow \infty \), where u solves the following equation

$$\begin{aligned} \left\{ \begin{array}{ll} -\left( a+b\displaystyle \int \limits _{\Omega _\Gamma }|\nabla u|^2\mathrm{{d}}x\right) \Delta u =h^+(x)|u|^{p-2}u, \ {} &{} x\in \Omega _\Gamma =\cup _{i\in \Gamma }\Omega _i,\\ u=0, \ {} &{} x\in \Omega \setminus \Omega _\Gamma ,\\ u=0, \ {} &{} x\in \partial \Omega . \end{array} \right. \end{aligned}$$
(1.17)

Moreover, \(u|_{\Omega _i}\) is positive for \(i\in \Gamma _1\), \(u|_{\Omega _i}\) is negative for \(i\in \Gamma _2\), and \(u|_{\Omega _i}\) changes sign exactly once for \(i\in \Gamma _3\).

If \(b=0\), Eq. (1.6) does not depend on the nonlocal term \(\int \limits _{\Omega }|\nabla u|^2\mathrm{{d}}x\Delta u\) any more. In this case, Eq. (1.6) becomes to the following semilinear elliptic equation

$$\begin{aligned} \left\{ \begin{array}{ll} - a \Delta u =\left( h^+(x)+\lambda h^-(x)\right) |u|^{p-2}u,\ &{}x\in \Omega ,\\ u=0, \ {} &{} x\in \partial \Omega , \end{array} \right. \end{aligned}$$
(1.18)

Under the conditions (\(h_1\))–(\(h_2\)), separate the components of \(\Omega ^+\) arbitrarily into three families, i.e.,

$$\begin{aligned} \Omega ^+=\left( \cup _{i=1}^I \widetilde{\omega }_i\right) \cup \left( \cup _{j=1}^J \widehat{\omega }_j\right) \cup \left( \cup _{i=k}^K \overline{\omega }_k\right) , \end{aligned}$$

by using constrained minimization method, Gir\(\tilde{a}\)o and Gomes [14] proved the existence of multi-bump nodal solution \(u_\lambda \) for Eq. (1.18) if \(\lambda >0\) large enough. Moreover, for any sequence \(\{\lambda _n\}\) with \(\lambda _n \rightarrow +\infty \) as \(n\rightarrow \infty \), there exists a subsequence, still denoted by \(\{\lambda _n\}\), such that \(u_{\lambda _n} \rightarrow u\) strongly in \(H^1_0(\Omega )\) as \(n\rightarrow \infty \), where u solves the following equation

$$\begin{aligned} \left\{ \begin{array}{ll} - a \Delta u =h^+(x)|u|^{p-2}u, \ {} &{}x\in \Omega ^+,\\ u=0, \ {} &{} x\in \Omega \setminus \Omega ^+, \end{array} \right. \end{aligned}$$
(1.19)

here \(u|_{\widetilde{\omega }_i}\) changes sign exactly once for \(i=1,2,\ldots ,I\), \(u|_{\widehat{\omega }_j}\) is positive for \(j=1,2,\ldots ,J\), \(u|_{\overline{\omega }_k}\equiv 0\) for \(k=1,2,\ldots ,K\). We refer the reader to [4] for multiple positive solutions for Eq. (1.18).

However, we cannot apply the same method that used in [14] to Eq. (1.6), because Kirchhoff-type equation depends on the global information of its solution. Different from the method used in [14], we first construct a special minimax value of the energy functional; Then, by careful analysis of the deformation flow to the energy functional, we prove the existence of multi-bump sign-changing solutions for Eq. (1.6); Finally, we show that the multi-bump sign-changing solutions are localized near the components of \(\Omega ^+\) and converge to the solutions (1.17) with prescribed sign properties. We remark that our method also can be used to study the existence of multi-bump sign-changing solutions for Eq. (1.18).

The paper is organized as follows. In Sect. 2, we give some primarily results. In Sect. 3, we prove Theorems 1.11.3. In Sect. 4 and 5, we devote to proving Theorem 1.4.

2 Some preliminary results

In this section, we give some preliminary results.

Lemma 2.1

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \(p\in (4,6)\). If \(u\in \mathcal {A}\), then there exists a unique \(t>0\) such that \(tu\in \mathcal {N}_{b,\lambda }\), where \(\mathcal {A}\) is defined by (1.10), \(\mathcal {N}_{b,\lambda }\) is defined by (1.12).

Proof

For \(u\in \mathcal {A}\), we define

$$\begin{aligned} V_u(t)&=\langle I'_{b,\lambda }(tu),tu\rangle =at^2\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x+bt^4\left( \,\,\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x\right) ^2\\&\quad -t^p\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^p\mathrm{{d}}x. \end{aligned}$$

Since \(u\in \mathcal {A}\), then

$$\begin{aligned} \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^p\mathrm{{d}}x>0. \end{aligned}$$

Therefore, \(V_u(t)>0\) for \(t>0\) small enough and \(V_u(t)<0\) for \(t<0\) large enough, since \(p\in (4,6)\). Thus, there exists \(t_0>0\) such that \(I'_{b,\lambda }(t_0u)=0\), that is \(t_0u\in \mathcal {N}_{b,\lambda }\).

Assume \(t_1\), \(t_2>0\) such that \(t_1u,~t_2u\in \mathcal {N}_{b,\lambda }\), that is

$$\begin{aligned} at_1^2\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x+bt_1^4\left( \,\,\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x \right) ^2 -t_1^p\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^p\mathrm{{d}}x=0 \end{aligned}$$
(2.1)

and

$$\begin{aligned} at_2^2\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x+bt_2^4\left( \,\,\int \limits _\Omega |\nabla u|^2\mathrm{{d}}x\right) ^2 -t_2^p\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^p\mathrm{{d}}x=0. \end{aligned}$$
(2.2)

It follows from (2.1) and (2.2) that

$$\begin{aligned} a\left( \frac{1}{t_1^2}-\frac{1}{t_2^2}\right) \int \limits _\Omega |\nabla u|^2\mathrm{{d}}x=(t_1^{p-4}-t_2^{p-4})\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^p\mathrm{{d}}x, \end{aligned}$$

which implies \(t_1=t_2\). \(\square \)

Lemma 2.2

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \(p\in (4,6)\), if \(u\in H^1_0(\Omega )\) with \(u^{\pm }\in \mathcal {A}\), then there is a unique pair \((s_u,t_u)\) of positive numbers such that \(s_uu^+ +t_uu^-\in \mathcal {M}_{b,\lambda }\).

Proof

We prove the lemma by two steps.

Step 1:   Define \(\overrightarrow{F}(s,t):=\left( f_1(s,t),f_2(s,t)\right) \), where

$$\begin{aligned} \left\{ \begin{array}{ll} f_1(s,t)=as^2\Vert u^+\Vert ^2+bs^4\Vert u^+\Vert ^4+bs^2t^2\Vert u^+\Vert ^2\Vert u^-\Vert ^2 -s^p\displaystyle \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x,\\ f_2(s,t)=at^2\Vert u^-\Vert ^2+bt^4\Vert u^-\Vert ^4+bs^2t^2\Vert u^+\Vert ^2\Vert u^-\Vert ^2 -t^p\displaystyle \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x. \end{array} \right. \end{aligned}$$
(2.3)

Since \(u^{\pm }\in \mathcal {A}\), then

$$\begin{aligned} \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x>0\ \ \text {and}\ \ \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x>0. \end{aligned}$$

We deduce that there exist \(0<r<R\) such that

$$\begin{aligned} \left\{ \begin{array}{ll} f_1(r,t)>0 \ \ \text {and}\ \ f_1(R,t)<0 \ \ \text {for all}\ \ t\in [r,R],\\ f_2(s,r)>0 \ \ \text {and}\ \ f_2(s,R)<0 \ \ \text {for all}\ \ s\in [r,R], \end{array} \right. \end{aligned}$$
(2.4)

since \(p\in (4,6)\). Then, by using Miranda lemma [23], we conclude that there exists \((s_u,t_u)\in {\mathbb {R}}_+\times {\mathbb {R}}_+\) such that

$$\begin{aligned} f_1(s_u,t_u)=0 \ \ \text {and}\ \ f_2(s_u,t_u)=0, \end{aligned}$$

which implies that \(s_uu^+ +t_uu^-\in \mathcal {M}_{b,\lambda }\).

Step 2:   We prove \((s_u,t_u)\) is unique.

Case 1: \(u\in \mathcal {M}_{b,\lambda }\). Suppose \((\bar{s},\bar{t})\ne (1,1)\) be another pair of positive numbers such that \(\bar{s}u^+ +\bar{t}u^-\in \mathcal {M}_{b,\lambda }\), then

$$\begin{aligned} \left\{ \begin{array}{ll} a\bar{s}^2\Vert u^+\Vert ^2+b\bar{s}^4\Vert u^+\Vert ^4+b\bar{s}^2\bar{t}^2\Vert u^+\Vert ^2\Vert u^-\Vert ^2 =\bar{s}^p\displaystyle \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x,\\ a\bar{t}^2\Vert u^-\Vert ^2+b\bar{t}^4\Vert u^-\Vert ^4+b\bar{s}^2\bar{t}^2\Vert u^+\Vert ^2\Vert u^-\Vert ^2 =\bar{t}^p\displaystyle \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x. \end{array} \right. \end{aligned}$$

Without loss of generality, we assume \(\bar{s}\ge \bar{t}>0\), then

$$\begin{aligned} \left\{ \begin{array}{ll} a\displaystyle \frac{1}{\bar{s}^2}\Vert u^+\Vert ^2+b\Vert u^+\Vert ^4+b\Vert u^+\Vert ^2\Vert u^-\Vert ^2 \ge \bar{s}^{p-4}\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x,\\ a\displaystyle \frac{1}{\bar{t}^2}\Vert u^-\Vert ^2+b\Vert u^-\Vert ^4+b\Vert u^+\Vert ^2\Vert u^-\Vert ^2 \le \bar{t}^{p-4}\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x. \end{array} \right. \end{aligned}$$

Since \(u\in \mathcal {M}_{b,\lambda }\), we have

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle a\Vert u^+\Vert ^2+b\Vert u^+\Vert ^4+b\Vert u^+\Vert ^2\Vert u^-\Vert ^2 =\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x,\\ \displaystyle a\Vert u^-\Vert ^2+b\Vert u^-\Vert ^4+b\Vert u^+\Vert ^2\Vert u^-\Vert ^2 =\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x. \end{array} \right. \end{aligned}$$

Thus, we conclude that

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle a\left( \displaystyle \frac{1}{\bar{s}^2}-1\right) \Vert u^+\Vert ^2 \ge \left( \bar{s}^{p-4}-1\right) \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x,\\ \displaystyle a\left( \displaystyle \frac{1}{\bar{t}^2}-1\right) \Vert u^-\Vert ^2 \le \left( \bar{t}^{p-4}-1\right) \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x, \end{array} \right. \end{aligned}$$

which implies \(1\ge \bar{s}\ge \bar{t}\ge 1\). Thus, \((\bar{s},\bar{t})=(1,1)\).

Case 2: \(u\not \in \mathcal {M}_{b,\lambda }\) but \(u^{\pm }\in \mathcal {A}\), then by Step 1, we know that there exists \((s_u, t_u)\in {\mathbb {R}}_+\times {\mathbb {R}}_+\) such that \(s_u u^+ +t_u u^-\in \mathcal {M}_{b,\lambda }\). Assume that \((s_u', t_u')\in {\mathbb {R}}_+\times {\mathbb {R}}_+\) also satisfying \(s_u' u^+ +t_u' u^-\in \mathcal {M}_{b,\lambda }\). Hence we have

$$\begin{aligned} \frac{s'_u}{s_u}s_uu^+ + \frac{t'_u}{t_u}t_uu^- \in \mathcal {M}_{b,\lambda }. \end{aligned}$$
(2.5)

Since \(s_u u^+ +t_u u^-\in \mathcal {M}_{b,\lambda }\), by the arguments of case 1, we deduce that

$$\begin{aligned} \frac{s'_u}{s_u}=\frac{t'_u}{t_u}=1. \end{aligned}$$

Thus, \(s'_u=s_u\) and \(t'_u=t_u\). \(\square \)

Lemma 2.3

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \(p\in (4,6)\), suppose that \(u^{\pm }\in \mathcal {A}\) such that

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle a\Vert u^+\Vert ^2+b\Vert u^+\Vert ^4+b\Vert u^+\Vert ^2\Vert u^-\Vert ^2 \le \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x,\\ \displaystyle a\Vert u^-\Vert ^2+b\Vert u^-\Vert ^4+b\Vert u^+\Vert ^2\Vert u^-\Vert ^2 \le \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x. \end{array} \right. \end{aligned}$$

Then the unique pair \((s_u,t_u)\) of positive numbers obtained in Lemma 2.2 satisfies \(0<s_u,t_u\le 1\).

Proof

Suppose that \(s_u\ge t_u>0\), since \(s_u u^+ +t_u u^-\in \mathcal {M}_b\), then we have

$$\begin{aligned}&a s_u^2\Vert u^+\Vert ^2 +bs_u^4\left( \,\,\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\right) ^2 +bs_u^4\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x \int \limits _{\Omega }|\nabla u^-|^2\mathrm{{d}}x\nonumber \\&\quad \ge a s_u^2\Vert u^+\Vert ^2 +bs_u^4\left( \,\,\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\right) ^2 +bs_u^2t_u^2\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\int \limits _{\Omega }|\nabla u^-|^2\mathrm{{d}}x \nonumber \\&\quad =s_u^p \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x. \end{aligned}$$
(2.6)

On the other hand,

$$\begin{aligned} a\Vert u^+\Vert ^2 +b\left( \,\,\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\right) ^2 +b\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\int \limits _{\Omega }|\nabla u^-|^2\mathrm{{d}}x \le \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x. \end{aligned}$$
(2.7)

Combine (2.6) and (2.7), we then get

$$\begin{aligned} \left( \frac{1}{s_u^2}-1\right) a\Vert u^+\Vert ^2\ge (s_u^{p-4}-1)\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x. \end{aligned}$$

Therefore, we must have \(s_u\le 1\). Then the proof is completed. \(\square \)

Lemma 2.4

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \(p\in (4,6)\). If \(u^{\pm }\in \mathcal {A}\), then the vector \((s_u,t_u)\) which obtained in Lemma 2.2 is the unique maximum point of the function \(\phi :({\mathbb {R}}_+\times {\mathbb {R}}_+)\rightarrow {\mathbb {R}}\) defined by \(\phi (s,t):=I_{b,\lambda }(su^+ + tu^-)\).

Proof

From the proof of Lemma 2.2, \((s_u,t_u)\) is the unique critical point of \(\phi \) in \({\mathbb {R}}_+\times {\mathbb {R}}_+\). Since \(p\in (4,6)\), we deduce that \(\phi (s,t)\rightarrow -\infty \) uniformly as \(|(s,t)|\rightarrow +\infty \), so it is sufficient to check that the maximum point is not achieved on the boundary of \({\mathbb {R}}_+\times {\mathbb {R}}_+\).

Fix \(\bar{t}>0\), since

$$\begin{aligned} \phi (s,\bar{t})&=I_{b,\lambda }(su^+ +\bar{t}u^-)\\&=\frac{a s^2}{2}\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x+\frac{bs^4}{4}\left( \,\,\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\right) ^2-s^p \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^+|^p \mathrm{{d}}x\\&\quad +\frac{bs^2\bar{t}^2}{2}\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\int \limits _{\Omega }|\nabla u^-|^2\mathrm{{d}}x\\&\quad +\frac{a\bar{t}^2}{2}\int \limits _{\Omega }|\nabla u^-|^2\mathrm{{d}}x+\frac{b\bar{t}^4}{4}\left( \,\,\int \limits _{\Omega }|\nabla u^+|^2\mathrm{{d}}x\right) ^2-{\bar{t}}^p \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u^-|^p \mathrm{{d}}x \end{aligned}$$

is an increasing function with respect to s if \(s>0\) small enough, therefore the pair \((0,\bar{t})\) is not a maximum point of \(\phi \) in \({\mathbb {R}}_+\times {\mathbb {R}}_+\). \(\square \)

By Lemma 2.2, we now define

$$\begin{aligned} m_{b,\lambda }:=\inf \Big \{I_{b,\lambda }(u):u\in \mathcal {M}_{b,\lambda }\Big \}. \end{aligned}$$
(2.8)

Lemma 2.5

Assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \(p\in (4,6)\), then \(m_{b,\lambda }>0\) is achieved.

Proof

For every \(u\in \mathcal {M}_{b,\lambda }\), we have \(\langle I'_{b,\lambda }(u),u\rangle =0\). Then, by using Sobolev embedding theorem, one gets

$$\begin{aligned} a\Vert u\Vert ^2&\le a\int \limits _\Omega |\nabla u|^2 \mathrm{{d}}x +b\left( \,\,\int \limits _\Omega |\nabla u|^2 \mathrm{{d}}x\right) ^2=\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u|^{p}\mathrm{{d}}x\nonumber \\&\le \int \limits _{\Omega }h^+(x)|u|^{p}\mathrm{{d}}x\le \Vert h^+(x)\Vert _{L^\infty (\Omega )}\int \limits _{\Omega }|u|^{p}\mathrm{{d}}x\nonumber \\&\le C\Vert u\Vert ^p. \end{aligned}$$
(2.9)

Thus, there exists a constant \(\alpha >0\) such that \(\Vert u\Vert ^2\ge \alpha \). Therefore

$$\begin{aligned} I_{b,\lambda }(u)=I_{b,\lambda }(u)-\frac{1}{p}\langle I'_{b,\lambda }(u),u\rangle \ge \left( \frac{1}{2}-\frac{1}{p}\right) a\Vert u\Vert ^2\ge \left( \frac{1}{2}-\frac{1}{p}\right) a\alpha , \ \ \text {for each}\ \ u\in \mathcal {M}_{b,\lambda }, \end{aligned}$$

which implies \(m_{b,\lambda }\ge \left( \frac{1}{2}-\frac{1}{p}\right) a\alpha >0\).

Let \(\{u_n\}\subset \mathcal {M}_{b,\lambda }\) be a sequence such that \(I_{b,\lambda }(u_n)\rightarrow m_{b,\lambda }\). Then \(\{u_n\}\) is bounded in \(H^1_0(\Omega )\), up to a subsequence, still denote by \(\{u_n\}\), such that \(u_n^{\pm }\rightharpoonup u_{b,\lambda }^{\pm }\) weakly in \(H^1_0(\Omega )\). Since \(u_n\in \mathcal {M}_{b,\lambda }\), we have \(\langle I'_{b,\lambda }(u_n),u_n^\pm \rangle =0\), that is

$$\begin{aligned} a\int \limits _\Omega |\nabla u_n^\pm |^2 \mathrm{{d}}x +b\int \limits _\Omega |\nabla u_n|^2 \mathrm{{d}}x\int \limits _\Omega |\nabla u_n^\pm |^2 \mathrm{{d}}x=\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_n^\pm |^{p}\mathrm{{d}}x. \end{aligned}$$
(2.10)

Similar as (2.9) there exist a constant \(\mu >0\) such that \(\Vert u_n^\pm \Vert ^2\ge \mu \) for all \(n\in {\mathbb N}\). Since \(u_n\in \mathcal {M}_{b,\lambda }\), thus

$$\begin{aligned} \mu \le \Vert u^\pm _n\Vert ^2<\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_n^\pm |^{p}\mathrm{{d}}x \le \int \limits _{\Omega }h^+(x)|u_n^\pm |^{p}\mathrm{{d}}x. \end{aligned}$$

By the compactness of the embedding \(H^1_0(\Omega )\hookrightarrow L^{q}(\Omega )\) for \(2\le q <6\), we get

$$\begin{aligned} \int \limits _\Omega h^+(x)|u_{b,\lambda }^\pm |^p \mathrm{{d}}x\ge \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_{b,\lambda }^\pm |^{p}\mathrm{{d}}x \ge \mu . \end{aligned}$$
(2.11)

Hence, \(u^\pm _{b,\lambda }\in \mathcal {A}\). By the weak semicontinuity of norm, we have

$$\begin{aligned} a\Vert u_{b,\lambda }^\pm \Vert ^2+b\int \limits _\Omega |\nabla u_{b,\lambda }|^2\mathrm{{d}}x\int \limits _\Omega |\nabla u_{b,\lambda }^\pm |^2\mathrm{{d}}x\le \liminf \limits _{n\rightarrow \infty } \big \{a\Vert u_n^\pm \Vert ^2+b\int \limits _\Omega |\nabla u_n|^2\mathrm{{d}}x\int \limits _\Omega |\nabla u_n^\pm |^2\mathrm{{d}}x\big \}. \end{aligned}$$
(2.12)

It follows from (2.10) that

$$\begin{aligned} a\Vert u_{b,\lambda }^\pm \Vert ^2+b\int \limits _\Omega |\nabla u_{b,\lambda }|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_{b,\lambda }^\pm |^2\mathrm{{d}}x\le \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_{b,\lambda }^\pm |^{p}\mathrm{{d}}x. \end{aligned}$$
(2.13)

From (2.13) and Lemma 2.3, there exists \((\bar{s},\bar{t})\in (0,1]\times (0,1]\) such that

$$\begin{aligned} \overline{u}_{b,\lambda }:=\bar{s} u_{b,\lambda }^+ + \bar{t} u_{b,\lambda }^-\in \mathcal {M}_{b,\lambda }. \end{aligned}$$

Hence

$$\begin{aligned} m_{b,\lambda }&\le I_{b,\lambda }(\overline{u}_{b,\lambda })=I_{b,\lambda }(\overline{u}_{b,\lambda })-\frac{1}{p}\langle I'_{b,\lambda }(\overline{u}_{b,\lambda }),\overline{u}_{b,\lambda }\rangle \nonumber \\&=\left( \frac{1}{2}-\frac{1}{p}\right) a\int \limits _\Omega |\overline{u}_{b,\lambda }|^2\mathrm{{d}}x+\left( \frac{1}{4}-\frac{1}{p}\right) b\left( \,\,\int \limits _\Omega |\overline{u}_{b,\lambda }|^2\mathrm{{d}}x\right) ^2\nonumber \\&=\left( \frac{1}{2}-\frac{1}{p}\right) a\left[ \Vert \bar{s}u^+_{b,\lambda }\Vert ^2+\Vert \bar{t}u^-_{b,\lambda }\Vert ^2\right] +\left( \frac{1}{4}-\frac{1}{p}\right) b\left[ \Vert \bar{s}u^+_{b,\lambda }\Vert ^2+\Vert \bar{t}u^-_{b,\lambda }\Vert ^2\right] ^2\nonumber \\&\le \left( \frac{1}{2}-\frac{1}{p}\right) a\left[ \Vert u^+_{b,\lambda }\Vert ^2+\Vert u^-_{b,\lambda }\Vert ^2\right] +\left( \frac{1}{4}-\frac{1}{p}\right) b\left[ \Vert u^+_{b,\lambda }\Vert ^2+\Vert u^-_{b,\lambda }\Vert ^2\right] ^2\nonumber \\&\le \liminf \limits _{n\rightarrow \infty }\left[ I_{b,\lambda }(u_n)-\frac{1}{p}\langle I'_{b,\lambda }({u}_n),{u}_n\rangle \right] =m_{b,\lambda }, \end{aligned}$$
(2.14)

which implies that \(\bar{s}=\bar{t}=1\). Thus, \(\overline{u}_{b,\lambda }=u_{b,\lambda }\) and \(I_{b,\lambda }(u_{b,\lambda })=m_{b,\lambda }\). \(\square \)

3 Proof of Theorems 1.11.3.

The main aim of this section is to prove Theorems 1.11.3. We first prove that the minimizer \(u_{b,\lambda }\) to the minimization problem (2.8) is indeed a sign-changing solution of Eq. (1.6), that is, \(I'_{b,\lambda }(u_{b,\lambda })=0\).

Proof of Theorem 1.1

Using the quantitative deformation lemma, we prove that \(I'_{b,\lambda }(u_{b,\lambda })=0\).

It is clear that \(\langle I'_{b,\lambda }(u_{b,\lambda }),u^+_{b,\lambda }\rangle =0=\langle I'_{b,\lambda }(u_{b,\lambda }),u^-_{b,\lambda }\rangle \). If \((s,t)\in {\mathbb {R}}_+\times {\mathbb {R}}_+\) and \((s,t)\ne (1,1)\), it follows from Lemma 2.4 that

$$\begin{aligned} I_{b,\lambda }(su_{b,\lambda }^+ +t u_{b,\lambda }^-)<I_{b,\lambda }(u_{b,\lambda }^+ + u_{b,\lambda }^-)=m_{b,\lambda }. \end{aligned}$$
(3.1)

If \(I'_{b,\lambda }(u_{b,\lambda })\ne 0\), then there exist \(\delta >0\) and \(\rho >0\) such that

$$\begin{aligned} \Vert I'_{b,\lambda }(v)\Vert \ge \rho , \ \ \text {for all}\ \ \Vert v-u_{b,\lambda }\Vert \le 3\delta . \end{aligned}$$

Let \(D:=(\frac{1}{2},\frac{3}{2})\times (\frac{1}{2},\frac{3}{2})\) and \(g(s,t):=su_{b,\lambda }^+ +tu_{b,\lambda }^-\). It follows from Lemma 2.4 again that

$$\begin{aligned} \bar{m}_{b,\lambda }:=\max \limits _{\partial D}I_{b,\lambda }\circ g< m_{b,\lambda } \end{aligned}$$
(3.2)

For \(\varepsilon :=\min \{(m_{b,\lambda }-\bar{m}_{b,\lambda })/2, \rho \delta /8\}\) and \(S:=B(u_{b,\lambda },\delta )\), [see [31], Lemma 2.3] yields a deformation \(\eta \) such that

(a)    \(\eta (1,u)=u\) if \(u\not \in I_{b,\lambda }^{-1}([m_{b,\lambda }-2\varepsilon ,m_{b,\lambda }+2\varepsilon ])\cap S_{2\delta }\);

(b)    \(\eta (1,I_{b,\lambda }^{m_{b,\lambda }+\varepsilon }\cap S)\subset I_{b,\lambda }^{m_{b,\lambda }-\varepsilon }\);

(c)    \(I_{b,\lambda }\left( \eta (1,u)\right) \le I_{b,\lambda }(u)\) for all \(u\in H_0^1(\Omega )\).

It is clear that

$$\begin{aligned} \max \limits _{(s,t)\in \bar{D}}I_{b,\lambda }\left( \eta (1,g(s,t))\right) <m_{b,\lambda }. \end{aligned}$$
(3.3)

We now prove that \(\eta (1,g(D))\cap \mathcal {M}_{b,\lambda }\ne \varnothing \), contradicting to the definition of \(m_{b,\lambda }\). Let us define \(h(s,t):=\eta (1,g(s,t))\) and

$$\begin{aligned}&\Psi _0(s,t):=\left( I'_{b,\lambda }(su_{b,\lambda }^+ +tu_{b,\lambda }^-)u_{b,\lambda }^+,I'_{b,\lambda }(su_{b,\lambda }^+ +tu_{b,\lambda }^-)u_{b,\lambda }^-\right) ,\\&\Psi _1(s,t):=\left( \displaystyle \frac{1}{s}I'_{b,\lambda }\left( h(s,t)\right) h^+(s,t),\displaystyle \frac{1}{t}I'_{b,\lambda }\left( h(s,t)\right) h^-(s,t)\right) . \end{aligned}$$

Lemma 2.2 and the the degree theory now yields deg\((\Psi _0,D,0)=1\). It follows from (3.2) that \(g=h\) on \(\partial D\). Consequently, we obtain deg\((\Psi _1,D,0)=\)deg\((\Psi _0,D,0)=1\). Therefore, \(\Psi _1(s_0,t_0)=0\) for some \((s_0,t_0)\in D\), so that \(\eta (1,g(s_0,t_0))=h(s_0,t_0)\in \mathcal {M}_{b,\lambda }\), which is a contradiction. From this, \(u_{b,\lambda }\) is a critical point of \(I_{b,\lambda }\), and so, a sign-changing solution for equation (1.6).

Now, we show that \(u_{b,\lambda }\) has exactly two nodal domains. The proof on the number of nodal domains follows the arguments in Bartsch [3] and Castro et al. [7]. To this end, we assume by contradiction that

$$\begin{aligned} u_{b,\lambda }=u_1+u_2+u_3 \end{aligned}$$

with

$$\begin{aligned} u_i\ne 0,\ u_1\ge 0,\ u_2\le 0\ \ \text {and}\ \ \text {suppt}(u_i)\cap \text {suppt}(u_j)=\varnothing ,\ \ \text {for}\ i\ne j,\ \ i,j=1,2,3 \end{aligned}$$

and

$$\begin{aligned} \langle I'_{b,\lambda }(u_{b,\lambda }),u_i\rangle =0,\ \ \text {for}\ i=1,2,3. \end{aligned}$$
(3.4)

Setting \(v:=u_1+u_2\), we see that \(v^+ =u_1\) and \(v^- =u_2\), i.e. \(v^\pm \ne 0\). Then, we can conclude \(v^\pm \in \mathcal {A}\). By Lemma 2.2, there exists a unique pair \((s_v,t_v)\) of positive numbers such that

$$\begin{aligned} s_v v^+ + t_v v^-\in \mathcal {M}_{b,\lambda }, \end{aligned}$$

or equivalently,

$$\begin{aligned} s_v u_1+t_v u_2\in \mathcal {M}_{b,\lambda }. \end{aligned}$$

And so,

$$\begin{aligned} I_{b,\lambda }(s_v u_1+t_v u_2)\ge m_{b,\lambda }. \end{aligned}$$
(3.5)

Moreover, using the fact that \(\langle I_{b,\lambda }'(u_{b,\lambda }),u_i\rangle =0\) for \(i=1,2,3\), it follows that

$$\begin{aligned} \langle I'_{b,\lambda }(v),v^\pm \rangle < 0. \end{aligned}$$

From Lemma 2.3, we have that

$$\begin{aligned} (s_v,t_v)\in (0,1]\times (0,1]. \end{aligned}$$

On the other hand,

$$\begin{aligned} 0&=\frac{1}{4}\langle I'_{b,\lambda }(u_{b,\lambda }),u_3\rangle =\frac{a}{4}\int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x+\frac{b}{4}\left( \,\,\int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x\right) ^2+\frac{b}{4}\int \limits _\Omega |\nabla u_1|^2\mathrm{{d}}x\int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x\nonumber \\&\quad +\frac{b}{4}\int \limits _\Omega |\nabla u_2|^2\mathrm{{d}}x\int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x -\frac{1}{4}\int \limits _\Omega \left( h^+(x)+\lambda h^-(x)\right) |u_3|^{p}\mathrm{{d}}x\nonumber \\&<I_{b,\lambda }(u_3)+\frac{b}{4}\int \limits _\Omega |\nabla u_1|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x +\frac{b}{4}\int \limits _\Omega |\nabla u_2|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x. \end{aligned}$$
(3.6)

Then, by using (3.4), we can calculate that

$$\begin{aligned} I_{b,\lambda }(s_v u_1+t_v u_2)&= \frac{as_v^2}{4}\Vert u_1\Vert ^2+\left( \frac{1}{4}-\frac{1}{p}\right) {s_v^p}\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_1|^{p}\mathrm{{d}}x +\frac{at_v^2}{4}\Vert u_2\Vert ^2\nonumber \\&\quad \quad +\left( \frac{1}{4}-\frac{1}{p}\right) {t_v^p}\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_2|^{p}\mathrm{{d}}x\nonumber \\&\le \frac{a}{4}\Vert u_1\Vert ^2+\left( \frac{1}{4}-\frac{1}{p}\right) \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_1|^{p}\mathrm{{d}}x +\frac{a}{4}\Vert u_2\Vert ^2\nonumber \\&\quad \quad +\left( \frac{1}{4}-\frac{1}{p}\right) \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |u_2|^{p}\mathrm{{d}}x\nonumber \\&=I_{b,\lambda }(u_1)+I_{b,\lambda }(u_2)+\frac{b}{2}\int \limits _\Omega |\nabla u_1|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_2|^2\mathrm{{d}}x\nonumber \\&\quad \quad +\frac{b}{4}\int \limits _\Omega |\nabla u_1|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x +\frac{b}{4}\int \limits _\Omega |\nabla u_2|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x. \end{aligned}$$
(3.7)

Then, from (3.5), (3.6) and (3.7), we have

$$\begin{aligned} m_{b,\lambda }&\le I_{b,\lambda }(s_v u_1+t_v u_2)< I_{b,\lambda }(u_1)+I_{b,\lambda }(u_2)+I_{b,\lambda }(u_3)+\frac{b}{2}\int \limits _\Omega |\nabla u_1|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_2|^2\mathrm{{d}}x\\&\quad +\frac{b}{2}\int \limits _\Omega |\nabla u_1|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x +\frac{b}{2}\int \limits _\Omega |\nabla u_2|^2\mathrm{{d}}x \int \limits _\Omega |\nabla u_3|^2\mathrm{{d}}x\\&= I_{b,\lambda }(u_{b,\lambda })=m_{b,\lambda }, \end{aligned}$$

which is a contradiction. This way, \(u_3=0\), and \(u_{b,\lambda }\) has exactly two nodal domains.

Recall that \(c_{b,\lambda }\) and \(\mathcal {N}_{b,\lambda }\) are defined by (1.11) and (1.12), respectively. Then, similar as the proof of Lemma 2.5, for each \(b>0\), we can deduce that there exists \(v_{b,\lambda }\in \mathcal {N}_{b,\lambda }\) such that \(I_{b,\lambda }(v_{b,\lambda })=c_{b,\lambda }>0\). By Corollary 2.9 in [15], the critical points of the functional \(I_{b,\lambda }\) on \(\mathcal {N}_{b,\lambda }\) are critical points of \(I_{b,\lambda }\) in \(H_0^1(\Omega )\), we conclude that \(I'_{b,\lambda }(v_{b,\lambda })=0\). Thus, \(v_{b,\lambda }\) is a ground state solution of (1.6).

On the other hand, suppose that \(u_{b,\lambda }=u_{b,\lambda }^+ +u^-_{b,\lambda }\) is a least energy sign-changing solution for Eq. (1.6). By Lemma 2.1, there is unique \(\bar{s}>0\), \(\bar{t}>0\) such that

$$\begin{aligned} \bar{s}u_{b,\lambda }^+\in \mathcal {N}_{b,\lambda }\ \ \text {and}\ \ \bar{t}u_{b,\lambda }^+\in \mathcal {N}_{b,\lambda }. \end{aligned}$$

Then, by Lemma 2.4, we get

$$\begin{aligned} 2c_{b,\lambda }\le I_{b,\lambda }(\bar{s}u_{b,\lambda }^+)+I_{b,\lambda }(\bar{t}u_{b,\lambda }^-) <I_{b,\lambda }(\bar{s}u_{b,\lambda }^+ +\bar{t}u_{b,\lambda }^-)\le I_{b,\lambda }(u_{b,\lambda }^+ +u_{b,\lambda }^-)=m_{b,\lambda }, \end{aligned}$$

that is \(m_{b,\lambda }>2c_{b,\lambda }\). This completes the proof. \(\square \)

Now, we are in a situation to prove Theorem 1.2. In the following, we regard \(b>0\) as a parameter in equation (1.6). We shall analyze the convergence property of \(u_{b,\lambda }\) as \(b\rightarrow 0^+\).

Proof of Theorem 1.2

For any \(b>0\) and \(\lambda >0\), denote \(u_{b,\lambda }\in H^1_0(\Omega )\) the least energy sign-changing solution of (1.6) obtained in Theorem 1.1, which changes sign only once.

Step 1. We claim that, for any sequence \(\{b_n\}\) with \(b_n \rightarrow 0^+\) as \(n\rightarrow \infty \), \(\{u_{b_n,\lambda }\}\) is bounded in \(H^1_0(\Omega )\).

Choose a nonzero function \(\varphi \in \mathcal {C}_0^\infty (\Omega )\) with \(\varphi ^\pm \in \mathcal {A}\). Since \(p\in (4,6)\), then, for any \(b\in [0,1]\), there exists a pair \((\tau _1,\tau _2)\) of positive numbers, which does not depend on b, such that

$$\begin{aligned} \left\{ \begin{array}{ll} a\tau _1^2\Vert \varphi ^+\Vert ^2 +b\tau _1^4\left( \,\,\displaystyle \int \limits _{\Omega }|\nabla \varphi ^+|^2\mathrm{{d}}x\right) ^2 + bB_{\varphi } \tau _1^2 \tau _2^2-\tau _1^p\displaystyle \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |\varphi ^+|^p\mathrm{{d}}x<0,\\ a\tau _2^2\Vert \varphi ^-\Vert ^2 +b\tau _2^4\left( \,\,\displaystyle \int \limits _{\Omega }|\nabla \varphi ^-|^2\mathrm{{d}}x\right) ^2 + bB_{\varphi } \tau _1^2 \tau _2^2-\tau _2^p\displaystyle \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |\varphi ^-|^p\mathrm{{d}}x<0, \end{array} \right. \end{aligned}$$

where \(B_{\varphi }=\int \limits _{\Omega }|\nabla \varphi ^+|^2\mathrm{{d}}x\int \limits _{\Omega }|\nabla \varphi ^-|^2\mathrm{{d}}x\). In view of Lemma 2.2 and Lemma 2.3, for any \(b\in [0,1]\), there exists a unique pair \(\left( s_\varphi (b),t_\varphi (b)\right) \in (0,1]\times (0,1]\) such that

$$\begin{aligned} \bar{\varphi }:=s_\varphi (b)\tau _1\varphi ^+ +t_\varphi (b)\tau _2\varphi ^-\in \mathcal {M}_{b,\lambda }. \end{aligned}$$
(3.8)

Thus, for any \(b\in [0,1]\), we have

$$\begin{aligned} I_{b,\lambda }(u_{b,\lambda })&\le I_{b,\lambda }(\bar{\varphi })=I_{b,\lambda } (\bar{\varphi })-\frac{1}{4}\langle I'_{b,\lambda }(\bar{\varphi }),\bar{\varphi }\rangle \nonumber \\&=\frac{a}{4}\Vert \bar{\varphi }\Vert ^2 +\left( \frac{1}{4}-\frac{1}{p}\right) \int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |\bar{\varphi }|^p\mathrm{{d}}x\nonumber \\&\le \frac{a}{4}\Vert \bar{\varphi }\Vert ^2+\left( \frac{1}{4}-\frac{1}{p}\right) \int \limits _{\Omega }h^+(x)|\bar{\varphi }|^p\mathrm{{d}}x\nonumber \\&\le \frac{a}{4}\Vert \tau _1\varphi ^+\Vert ^2+\frac{a}{4}\Vert \tau _2\varphi ^-\Vert ^2 +\left( \frac{1}{4}-\frac{1}{p}\right) \int \limits _{\Omega }h^+(x)\left( \tau _1^p|\varphi ^+|^p+\tau _2^p|\varphi ^-|^p\right) \mathrm{{d}}x\nonumber \\&:=C_0, \end{aligned}$$
(3.9)

where \(C_0\) does not depend on b. For n large enough, it follows that

$$\begin{aligned} C_0+1\ge I_{b_n,\lambda }(u_{b_n,\lambda })=I_{b_n,\lambda }(u_{b_n,\lambda }) -\frac{1}{4}\langle I'_{b_n,\lambda }(u_{b_n,\lambda }),u_{b_n,\lambda }\rangle \ge \frac{a}{4}\Vert u_{b_n,\lambda }\Vert ^2, \end{aligned}$$
(3.10)

which implies \(\{u_{b_n,\lambda }\}\) is bounded in \(H_0^1(\Omega )\).

Step 2. There exists a subsequence of \(\{b_n\}\), still denoted by \(\{b_n\}\), such that

$$\begin{aligned} u_{b_n,\lambda }\rightharpoonup u_{0,\lambda }\ \text {weakly in}\ H_0^1(\Omega ). \end{aligned}$$

Then, \(u_{0,\lambda }\) is a weak solution of (1.13). Since \(u_{b_n,\lambda }\) is the least energy sign-changing solution of (1.6) with \(b=b_n\), then by the compactness of the embedding \(H_0^1(\Omega )\hookrightarrow L^q(\Omega )\) for \(2\le q<6\), we deduce that \(u_{b_n,\lambda }\rightarrow u_{0,\lambda }\) strongly in \(H_0^1(\Omega )\) as \(n\rightarrow \infty \). In fact,

$$\begin{aligned} \Vert u_{b_n,\lambda }-u_{0,\lambda }\Vert ^2&=\langle I'_{b_n,\lambda }(u_{b_n,\lambda })-I'_{0,\lambda }(u_{0,\lambda }),u_{b_n,\lambda }-u_{0,\lambda }\rangle -b_n \int \limits _{\Omega }|\nabla u_{b_n,\lambda }|^2\mathrm{{d}}x \int \limits _{\Omega }\nabla u_{b_n,\lambda }\left( \nabla u_{b_n,\lambda }-\nabla u_{0,\lambda }\right) \mathrm{{d}}x\\&\quad +\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) \left[ |u_{b_n,\lambda }|^{p-2}u_{b_n,\lambda } -|u_{0,\lambda }|^{p-2}u_{0,\lambda }\right] \left( u_{b_n,\lambda }- u_{0,\lambda }\right) \mathrm{{d}}x, \end{aligned}$$

and the right hand of last equality tend to zero as \(n\rightarrow \infty \). Then, by the same arguments as (2.11), we conclude \(u_{0,\lambda }^\pm \ne 0\), hence \(u_{0,\lambda }\) is sign-changing solution of equation (1.13).

Step 3. Suppose that \(v_0\) is a least energy sign-changing solution of (1.13), the existence of \(v_0\) was proved by Vladimir in [32]. By Lemma 2.2, for each \(b_n>0\), there is a unique pair \(\left( s_{b_n},t_{b_n}\right) \) of positive numbers such that

$$\begin{aligned} s_{b_n}v_0^+ +t_{b_n}v_0^-\in \mathcal {M}_{b_n,\lambda }. \end{aligned}$$

Then, we have

$$\begin{aligned}&a(s_{b_n})^2\Vert v_0^+\Vert ^2+b_n(s_{b_n})^4\left( \,\,\int \limits _{\Omega }|\nabla v_0^+|^2\mathrm{{d}}x\right) ^2+ b_n(s_{b_n} t_{b_n})^2\int \limits _{\Omega }|\nabla v_0^+|^2\mathrm{{d}}x\int \limits _{\Omega }|\nabla v_0^-|^2\mathrm{{d}}x \nonumber \\&\quad =(s_{b_n})^p\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |v^+_0|^p\mathrm{{d}}x \end{aligned}$$
(3.11)

and

$$\begin{aligned}&a(t_{b_n})^2\Vert v_0^-\Vert ^2+b_n(t_{b_n})^4\left( \,\,\int \limits _{\Omega }|\nabla v_0^-|^2\mathrm{{d}}x\right) ^2+ b_n(s_{b_n} t_{b_n})^2\int \limits _{\Omega }|\nabla v_0^+|^2\mathrm{{d}}x\int \limits _{\Omega }|\nabla v_0^-|^2\mathrm{{d}}x \nonumber \\&\quad =(t_{b_n})^p\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |v^-_0|^p\mathrm{{d}}x. \end{aligned}$$
(3.12)

Recall that \(v_0^\pm \) satisfies

$$\begin{aligned} a\Vert v_0^+\Vert ^2=\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |v^+_0|^p\mathrm{{d}}x \end{aligned}$$

and

$$\begin{aligned} a\Vert v_0^-\Vert ^2=\int \limits _{\Omega }\left( h^+(x)+\lambda h^-(x)\right) |v^-_0|^p\mathrm{{d}}x. \end{aligned}$$

Up to a subsequence, one can easily deduce that

$$\begin{aligned} (s_{b_n},t_{b_n})\rightarrow (1,1), \ \ \text {as}\ n\rightarrow \infty . \end{aligned}$$
(3.13)

It follows from (3.13) and Lemma 2.4 that

$$\begin{aligned} I_{0,\lambda }(v_0)\le I_{0,\lambda }(u_{0,\lambda })&=\lim \limits _{n\rightarrow \infty }I_{b_n,\lambda }(u_{b_n,\lambda }) =m_{b_n,\lambda } \nonumber \\&\le \lim \limits _{n\rightarrow \infty }I_{b_n,\lambda }\left( s_{b_n}v_0^+ +t_{b_n}v_0^-\right) =I_{0,\lambda }(v_0^+ +v_0^-)=I_{0,\lambda }(v_0), \end{aligned}$$
(3.14)

which implies \(u_{0,\lambda }\) is a least energy sign-changing solution of Eq. (1.13). This completes the proof of Theorem 1.2. \(\square \)

Proof of Theorem 1.3

For arbitrary \(b>0\), let \(u_{b, \lambda _n}\in H^1_0(\Omega )\) is a least energy sign-changing solution for Eq. (1.6) with \(\lambda =\lambda _n\), which is obtained by Theorem 1.1. Obviously,

$$\begin{aligned} m_{b,0}\ge m_{b,\lambda },\ \ \text {for each}\ \ \lambda >0. \end{aligned}$$
(3.15)

Therefore

$$\begin{aligned} m_{b,0}\ge m_{b,\lambda _n}&=I_{b, \lambda _n}(u_{b,\lambda _n})\\&=I_{b, \lambda _n}(u_{b,\lambda _n})-\frac{1}{p}\langle I'_{b, \lambda _n}(u_{b,\lambda _n}),u_{b,\lambda _n}\rangle \\&=\left( \frac{1}{2}-\frac{1}{p}\right) a\int \limits _\Omega |\nabla u_{b,\lambda _n}|^2\mathrm{{d}}x+\left( \frac{1}{4}-\frac{1}{p}\right) b\left( \,\,\int \limits _\Omega |\nabla u_{b,\lambda _n}|^2\mathrm{{d}}x\right) ^2, \end{aligned}$$

which implies that \(\{u_{b,\lambda _n}\}\) is bounded in \(H^1_0(\Omega )\). Up to a subsequence, we may suppose there exists \(u_{b,0}\in H^1_0(\Omega )\) such that \(u_{b,\lambda _n}\rightharpoonup u_{b,0}\) weakly in \(H^1_0(\Omega )\).

Since \(\{u_{b,\lambda _n}\}\) is bounded in \(H^1_0(\Omega )\), it follows from (3.15) that

$$\begin{aligned} -\frac{\lambda _n}{p}\int \limits _{\Omega }h^{-}(x)|u_{b,\lambda _n}|^{p} d x&={I}_{b,\lambda _{n}}\left( u_{b,\lambda _n}\right) -\frac{a}{2} \int \limits _{\Omega }\left| \nabla u_{b,\lambda _n}\right| ^{2} d x-\frac{b}{4}\left( \,\,\int \limits _{\Omega }\left| \nabla u_{b,\lambda _n}\right| ^{2} d x\right) ^{2}\\&\ \ \ \ +\frac{1}{p}\int \limits _{\Omega } h^{+}(x)\left| u_{b,\lambda _n}\right| ^{p} d x\\&\le C. \end{aligned}$$

Therefore

$$\begin{aligned} -\frac{1}{p}\int \limits _{\Omega }h^{-}(x)|u_{b,0}|^{p} d x&= \liminf _{ n\rightarrow \infty }\left[ -\frac{1}{p}\int \limits _{\Omega }h^{-}(x)|u_{b,\lambda _n}|^{p} d x\right] \\&=\liminf _{ n\rightarrow \infty }\left[ \frac{1}{\lambda _n}\left( -\frac{\lambda _n}{p}\int \limits _{\Omega }h^{-}(x)|u_{b,\lambda _n}|^{p} d x\right) \right] =0, \end{aligned}$$

which implies \(u_{b,0}=0\) on \(\Omega ^-\).

On the other hand, since \(\langle {I}'_{b,\lambda _{n}}(u_{b,\lambda n})-I'_{b,0}(u_{b,0}), u_{b,\lambda _n}-u_{b,0}\rangle =0\), then

$$\begin{aligned} \begin{aligned}&a\int \limits _{\Omega }\left| \nabla u_{b,\lambda _n}-\nabla u_{b,0}\right| ^{2} d x+b \int \limits _{\Omega }\left| \nabla u_{b,\lambda _n}\right| ^{2} d x \int \limits _{\Omega }\left| \nabla u_{b,\lambda _n}-\nabla u_{b,0}\right| ^{2} d x \\&\quad =b\left( \,\,\int \limits _{\Omega }|\nabla u_{b,0}|^{2} d x-\int \limits _{\Omega }\left| \nabla u_{b,\lambda _n}\right| ^{2} d x\right) \int \limits _{\Omega } \nabla u_{b,0}\left( \nabla u_{b,\lambda _n}-\nabla u_{b,0}\right) d x \\&\qquad +\int \limits _{\Omega }\left( h^{+}(x)+\lambda h^{-}(x)\right) \left( \left| u_{b,\lambda _n}\right| ^{p-2} u_{b,\lambda _n}-|u_{b,0}|^{p-2} u_{b,0}\right) \left( u_{b,\lambda _n}-u_{b,0}\right) d x, \end{aligned} \end{aligned}$$
(3.16)

the right hand of (3.16) tend to zero as \(n\rightarrow \infty \) since \(u_{b,\lambda _{n}}\rightharpoonup u_{b,0}\) weakly in \(H_{0}^{1}(\Omega )\), which implies \(u_{b,n} \rightarrow u_{b,0}\) strongly in \(H_{0}^{1}\left( \Omega \right) \). Therefore

$$\begin{aligned} \langle I'_{b,0}\left( u_{b,0}\right) ,\varphi \rangle = \liminf \limits _{n \rightarrow \infty } \langle I'_{b,\lambda _{n}}\left( u_{b,\lambda n}\right) , \varphi \rangle =0,\ \ \text {for each}\ \ \varphi \in H^1_0(\Omega ), \end{aligned}$$

which implies \(u_{b,0}\) is a solution of Eq. (1.14). By a similar method that used in [25], one can prove the existence of least energy sign-changing solution for equation (1.14). Suppose \(v_{b,0}\) is a least energy sign-changing solution for Eq. (1.14), by Lemma  2.2, for each \(\lambda _{n}>0\), there exist a unique pair of positive numbers \(\left( s_{\lambda _{n}}, t_{\lambda _{n}}\right) \) such that

$$\begin{aligned} s_{\lambda _{n}} v_{b,0}^{+}+t_{\lambda _{n}} v_{b,0}^{-} \in \mathcal {M}_{b,\lambda _n}. \end{aligned}$$

That is

$$\begin{aligned} a(s_{\lambda _{n}})^{2}\Vert v_{b,0}^{+}\Vert ^{2}+b\left( s_{\lambda _{n}}\right) ^{4}&\left( \,\,\int \limits _{\Omega }|\nabla v_{b,0}^{+}|^{2} \mathrm{{d}}x\right) ^{2} +b\left( s_{\lambda _{n}} t_{\lambda _{n}}\right) ^{2} \int \limits _{\Omega }\left| \nabla v_{b,0}^{+}\right| ^{2} d x \int \limits _{\Omega }\left| \nabla v_{b,0}^{-}\right| ^{2} d x\nonumber \\&=s_{\lambda _{n}}^p\int \limits _{\Omega }\left( h^+(x)+\lambda _n h^-(x)\right) |v_{b,0}^{+}|^p d x, \end{aligned}$$
(3.17)

and

$$\begin{aligned} a(t_{\lambda _{n}})^{2}\Vert v_{b,0}^{-}\Vert ^{2}+b\left( t_{\lambda _{n}}\right) ^{4}&\left( \,\,\int \limits _{\Omega }|\nabla v_{b,0}^{-}|^{2} \mathrm{{d}}x\right) ^{2} +b\left( s_{\lambda _{n}} t_{\lambda _{n}}\right) ^{2} \int \limits _{\Omega }\left| \nabla v_{b,0}^{+}\right| ^{2} d x \int \limits _{\Omega }\left| \nabla v_{b,0}^{-}\right| ^{2} d x\nonumber \\&=t_{\lambda _{n}}^p\int \limits _{\Omega }\left( h^+(x)+\lambda _n h^-(x)\right) |v_{b,0}^{-}|^p d x, \end{aligned}$$
(3.18)

Recall that \(v_{b,0}^{\pm }\) satisfying

$$\begin{aligned} a\Vert v_{0,\lambda }^{+}\Vert ^{2}+b\Vert v_{0,\lambda }^{+}\Vert ^{4}=\int \limits _{\Omega }h^{+}(x) |v_{b,0}^{+}|^p d x\ \ \text {and}\ \ a\Vert v_{0,\lambda }^{-}\Vert ^{2}+b\Vert v_{0,\lambda }^{-}\Vert ^{4}=\int \limits _{\Omega }h^{+}(x) |v_{b,0}^{-}|^p d x. \end{aligned}$$
(3.19)

It follows from (3.17)–(3.19) that

$$\begin{aligned} \left( s_{\lambda _{n}}, t_{\lambda _{n}}\right) \rightarrow (1,1), \quad \text {as}\ n \rightarrow \infty . \end{aligned}$$
(3.20)

Therefore, by (3.20) and Lemma 2.4, we can deduce that

$$\begin{aligned} \begin{aligned} I_{b,0}\left( v_{b,0}\right)&\le I_{b,0}\left( u_{b,0}\right) =\lim _{n \rightarrow \infty } I_{b,\lambda _{n}}\left( u_{b,\lambda _{n}}\right) \\&\le \lim _{n \rightarrow \infty } I_{b,\lambda _{n}}\left( s_{\lambda _{n}} v_{b,0}^{+}+t_{\lambda _{n}} v_{b,0}^{-}\right) =I_{b,0}\left( v_{b,0}^{+}+v_{b,0}^{-}\right) =I_{b,0}\left( v_{b,0}\right) . \end{aligned} \end{aligned}$$
(3.21)

Therefore, we conclude that \(u_{b,0}\) is a least energy sign-changing solution for Eq. (1.14), which changes sign once. The proof is completed. \(\square \)

4 A special minimax value for the energy functional

In this section, we assume \(h:\overline{\Omega }\rightarrow {\mathbb {R}}\) is a sign-changing continuous function and \((h_1)\)\((h_2)\) hold.

We first state a result on the existence of solutions for Eq. (1.17).

Theorem 4.1

(Theorem1.2, [10]) Suppose that \(4<p<6\) and \((h_1)\)\((h_2)\) hold. Then, for any non-empty subset \(\Gamma \subset \{1,2,\ldots ,k\}\) satisfies (1.16), Eq. (1.17) has a nontrivial solution \(u\in H^1_0(\Omega )\) with \(u|_{\Omega _i}\) is positive for \(i\in \Gamma _1\), \(u|_{\Omega _i}\) is negative for \(i\in \Gamma _2\), \(u|_{\Omega _i}\) changes sign exactly once for \(i\in \Gamma _3\), and \(u\equiv 0\) on \(\Omega \setminus \Omega _\Gamma \). Furthermore, u is the least energy solution among all solutions with these sign properties, that is, u achieves the following extremum

$$\begin{aligned} m_{\Gamma }:=\inf \left\{ I_{\Gamma }(u)\bigg | \displaystyle \begin{array}{ll} \displaystyle \text { u is a solution of }(1.17) \text { with u }|_{\Omega _i} \text { is positive for i }\in \Gamma _1, u|_{\Omega _i} is \\ \text { negative for i }\in \Gamma _2 \text { and u }|_{\Omega _i} \text { changes sign exactly once for i}\in \Gamma _3. \end{array} \right\} \end{aligned}$$
(4.1)

The functional \(I_{\Gamma }:H^1_0(\Omega _\Gamma )\rightarrow {\mathbb {R}}\) is defined by

$$\begin{aligned} I_{\Gamma }(u):=\frac{1}{2}\int \limits _{\Omega _\Gamma }a|\nabla u|^2 \mathrm{{d}}x +\frac{b}{4}\left( \,\,\int \limits _{\Omega _\Gamma }|\nabla u|^2\mathrm{{d}}x\right) ^2-\int \limits _{\Omega _\Gamma }h^+(x)|u|^p\mathrm{{d}}x. \end{aligned}$$
(4.2)

Without loss of generality, we next only consider the case \(\Gamma _1=\{1\}\), \(\Gamma _2=\{2\}\), \(\Gamma _3=\{3\}\) for simplicity. In this case, \(\Gamma =\cup _{i=1}^3\Gamma _i=\{1,2,3\}\) and

$$\begin{aligned} \displaystyle \Omega _\Gamma =\cup _{i=1}^{3}\Omega _{i} \ \ \text {with}\ \ dist(\Omega _i,\Omega _j)>0 \ \ \text {for} \ \ i\ne j,\ \ i,j=1,2,3. \end{aligned}$$

We can choose open sets \(\Omega ^\rho _{i}:=\big \{x\in \Omega \,\ dist(x,\Omega _{i})<\rho \big \}\) for \(i=1,2,3\) with smooth boundary such that

$$\begin{aligned} \Omega _{i}\subset \subset \Omega ^\rho _{i} \ \ \text {and}\ \ \ dist(\Omega ^\rho _i,\Omega ^\rho _j)>0 \ \ \text {for} \ \ i\ne j,\ \ i,j=1,2,3. \end{aligned}$$

We denote \(\Omega ^\rho :=\cup ^3_{i=1}\Omega ^\rho _{i}\) and define

$$\begin{aligned} \widehat{I}_{b,\lambda }(u):=\frac{a}{2}\int \limits _{\Omega ^\rho }|\nabla u|^{2}\mathrm{{d}}x +\frac{b}{4}\left( \,\,\int \limits _{\Omega ^\rho }|\nabla u|^2\mathrm{{d}}x\right) ^2-\frac{1}{p}\int \limits _{\Omega ^\rho }\left( h^+(x)+\lambda h^-(x)\right) |u|^p\mathrm{{d}}x,\ \ u\in H_0^1(\Omega ^\rho ). \end{aligned}$$
(4.3)

Now, we consider the following constraint minimization problem

$$\begin{aligned} \widehat{m}_{\lambda }:=\inf _{u \in \widehat{\mathcal {M}}_{b,\lambda }}\widehat{I}_{b,\lambda }(u), \end{aligned}$$

where

$$\begin{aligned} \widehat{\mathcal {M}}_{b,\lambda }:=\Big \{u \in H_0^1(\Omega ^\rho ) \ \big |\ \langle \widehat{I}_{b,\lambda }'(u), u_{i}\rangle =0 \ \text {for}\ i=1,2,\ {}&u_1^+\ne 0, u_2^-\ne 0\ \\&\text {and}\ \langle \widehat{I}_{b,\lambda }'(u), u^\pm _{3}\rangle =0, u_{3}^\pm \ne 0\Big \}. \end{aligned}$$

Combining the approach applied in Sect. 2 in [10] and that used in the proof of Theorem 1.1, we deduce that there exists \(v_{\lambda }\in H_0^1(\Omega ^\rho )\) such that

$$\begin{aligned} \widehat{I}_{b,\lambda }(v_\lambda )=\widehat{m}_{\lambda }\ \ \text {and}\ \ \widehat{I}_{b,\lambda }'(v_\lambda )=0. \end{aligned}$$

Proposition 4.2

Suppose \(\lambda _n\rightarrow +\infty \) as \(n\rightarrow \infty \) and \(\{v_{\lambda _n}\}\subset H_0^1(\Omega ^\rho )\) satisfying

$$\begin{aligned} \widehat{I}_{b,\lambda _n}(v_{\lambda _n})=\widehat{m}_{\lambda _n}\ \ \text {and}\ \ \widehat{I}'_{b,\lambda _n}(v_{\lambda _n})=0. \end{aligned}$$

then, up to a subsequence, there exists \(v\in H_0^1(\Omega ^\rho )\) such that

(i)    \(v_{n} \rightarrow v\) strongly in \(H_0^1(\Omega ^\rho )\), where we write \(v_{\lambda _n}\) as \(v_n\) for simplicity;

(ii)    \(v=0\) in \(\Omega ^\rho {\setminus } \Omega _\Gamma \) and v is a solution to Eq. (1.17);

(iii)    \(\displaystyle \widehat{I}_{b,\lambda _n}(v_n)\rightarrow \widehat{I}_{b,0}(v)=\frac{a}{2}\int \limits _{\Omega _\Gamma }|\nabla v|^{2}\mathrm{{d}}x +\frac{b}{4}\left( \,\,\int \limits _{\Omega _\Gamma }|\nabla v|^2\mathrm{{d}}x\right) ^2 -\int \limits _{\Omega _\Gamma } h^+(x)|v|^p \mathrm{{d}}x\).

Proof

It is easy to prove that \(\{v_n\}\) is bounded in \(H_0^1(\Omega ^\rho )\), since \(\widehat{m}_{\lambda _n}\le m_\Gamma \). Then, up to a subsequence, there exists \(v \in H_0^1(\Omega ^\rho )\) such that

$$\begin{aligned} \left\{ \displaystyle \begin{array}{ll} v_n\rightharpoonup v \ &{}\text {weakly in}\ \ H_0^1(\Omega ^\rho ), \\ v_n\rightarrow v \ &{}\text {strongly in}\ \ L^{q}(\Omega ^\rho )\ \ \text {for}\ \ 2\le q <6,\\ v_{n}\rightarrow v\ &{}\text {for a.e.}\ \ x\in \Omega ^\rho . \end{array} \right. \end{aligned}$$
(4.4)

We first prove \(v=0\) in \(\Omega ^\rho \setminus \Omega _\Gamma \). Set \(\Omega ^\rho _-=\big \{x\in \Omega ^\rho : h(x)< 0 \big \}\), since \(\{v_{\lambda _n}\}\) is bounded in \(H_0^1(\Omega ^\rho )\), then

$$\begin{aligned} -\frac{1}{p}\int \limits _{\Omega ^\rho _-}\lambda _n h^-(x)|v_{n}|^p \mathrm{{d}}x&= \widehat{I}_{b,\lambda _n}(v_n) -\frac{a}{2}\int \limits _{\Omega ^\rho }|\nabla v_n|^{2}\mathrm{{d}}x-\frac{b}{4}\left( \,\,\int \limits _{\Omega ^\rho }|\nabla v_n|^2\mathrm{{d}}x\right) ^2\nonumber \\&\quad +\frac{1}{p}\int \limits _{\Omega ^\rho }h^+(x)|v_n|^p\mathrm{{d}}x \le C. \end{aligned}$$
(4.5)

Therefore

$$\begin{aligned} -\int \limits _{\Omega ^\rho _-}h^-(x)|v|^p \mathrm{{d}}x= \liminf \limits _{n\rightarrow \infty }\left[ -\frac{1}{\lambda _n}\int \limits _{\Omega ^\rho _-}\lambda _n h^-(x)|v_n|^p \mathrm{{d}}x\right] =0, \end{aligned}$$

which indicates that \(v=0\) on \(\Omega ^\rho _-\). Thus, we conclude \(v=0\) in \(\Omega ^\rho \setminus \Omega _\Gamma \).

By using the fact \(\langle \widehat{I}_{b,\lambda _n}'(v_n) -\widehat{I}_{b,0}'(v),v_n-v\rangle =0\) that

$$\begin{aligned}&a\int \limits _{\Omega ^\rho }|\nabla v_n-\nabla v|^{2}\mathrm{{d}}x +b\int \limits _{\Omega ^\rho }|\nabla v_n|^2\mathrm{{d}}x\int \limits _{\Omega ^\rho }|\nabla v_n-\nabla v|^{2}\mathrm{{d}}x\\&\quad =b\left( \,\,\int \limits _{\Omega ^\rho }|\nabla v|^2\mathrm{{d}}x-\int \limits _{\Omega ^\rho }|\nabla v_n|^2\mathrm{{d}}x\right) \int \limits _{\Omega ^\rho }\nabla v(\nabla v_n-\nabla v)\mathrm{{d}}x\\&\qquad +\int \limits _{\Omega ^\rho }h^+(x)\left( |v_n|^{p-2}v_n-|v|^{p-2}v\right) (v_n-v)\mathrm{{d}}x +\int \limits _{\Omega ^\rho }\lambda _n h^-(x)|v_n|^{p-2}v_n(v_n-v)\mathrm{{d}}x. \end{aligned}$$

Obviously, the right hand of the last equality tend to zero as \(n\rightarrow \infty \), since \(\{v_n\}\) is bounded in \(H_0^1(\Omega ^\rho )\) and \(v=0\) in \(\Omega ^\rho {\setminus } \Omega _\Gamma \). Thus, \(v_n \rightarrow v\) strongly in \(H^{1}_0(\Omega ^\rho )\), and hence v is a solution of (1.17).

Finally, it is easy to conclude that (iii) from (i)–(ii). \(\square \)

Moreover, we have the following asymptotic behavior for \(\widehat{m}_{\lambda }\) as \(\lambda \rightarrow +\infty \).

Lemma 4.3

There holds that

(i)    \(0< \widehat{m}_{\lambda } \le m_{\Gamma }\), for all \(\lambda \ge 0\);

(ii)    \(\widehat{m}_{\lambda } \rightarrow m_{\Gamma }\), as \(\lambda \rightarrow +\infty \).

Proof

The proof of point (i) is trivial, so we omit the detail.

Now, we are going to prove point (ii). Let \(\{\lambda _n\}\) be a sequence with \(\lambda _n \rightarrow +\infty \) as \(n\rightarrow +\infty \). For each \(\lambda _n\), there exists \(v_{\lambda _n} \in H^1_0(\Omega ^\rho )\) with

$$\begin{aligned} \widehat{I}_{b,\lambda _n}(v_{\lambda _n})=\widehat{m}_{b,\lambda _n}\ \ \text {and}\ \ \widehat{I}_{b,\lambda _n}'(v_{\lambda _n})=0. \end{aligned}$$
(4.6)

We suppose, up to a subsequence, \(\{\widehat{I}_{b,\lambda _n}(v_{\lambda _n})\}\) converges, since \(\widehat{m}_{b,\lambda _n} \le m_{\Gamma }\). By using similar arguments as in Proposition 4.2, we know that there exists \(v\in H^1_0(\Omega ^\rho )\) such that

$$\begin{aligned} v_{\lambda _n}\rightarrow v \ \ \text {strongly in}\ \ H^1_0(\Omega ^\rho ) \ \ \text {as}\ \ n\rightarrow +\infty , \end{aligned}$$

and \((v|_{\Omega _1})^+\), \((v|_{\Omega _2})^-\), \((v|_{\Omega _3})^\pm \ne 0\). Moreover,

$$\begin{aligned} \widehat{m}_{b,\lambda _n}=\widehat{I}_{b,\lambda _n}(v_{\lambda _n})\rightarrow \widehat{I}_{b,0}(v), \end{aligned}$$
(4.7)

and

$$\begin{aligned} 0=\widehat{I}_{b,\lambda _n}'(v_{\lambda _n})\rightarrow \widehat{I}_{b,0}'(v). \end{aligned}$$
(4.8)

By the definition of \(m_{\Gamma }\), we have that

$$\begin{aligned} \lim _{n_i\rightarrow +\infty }\widehat{m}_{b,\lambda _n}= \widehat{I}_{b,0}(v)\ge m_{\Gamma }. \end{aligned}$$
(4.9)

By conclusion (i) of this Lemma, we know that \(\widehat{m}_{b,\lambda _n}\rightarrow m_{\Gamma }\) as \(n\rightarrow \infty \). \(\square \)

Next, we denote the solution of (1.17) given in Theorem 4.1 by \(v\in H_0^1(\Omega )\), that is

$$\begin{aligned} v\in H^1_0(\Omega _\Gamma ),~~~{I}_{\Gamma }(v)=m_{\Gamma },~~~{I}'_{\Gamma }(v)=0, \end{aligned}$$
(4.10)

and \(v_1=v|_{\Omega _1}\) is positive, \(v_2=v|_{\Omega _2}\) is negative, \(v_3=v|_{\Omega _3}\) changes sign exactly once. Obviously, there exist constants \(\tau _2>\tau _1>0\) such that

$$\begin{aligned} \tau _1\le \Vert v_1\Vert ,\,\,\,\Vert v_2\Vert ,\,\,\,\Vert v^+_3\Vert ,\,\,\,\Vert v^-_4\Vert \le \tau _2. \end{aligned}$$
(4.11)

We now define \(\gamma _0:[\frac{1}{2},\frac{3}{2}]^4\rightarrow H_0^1(\Omega )\) by

$$\begin{aligned} \gamma _0(t_1,t_2,t_3,t_4):=t_1v_1+t_2v_2+t_3v_3^++t_4v_3^- \end{aligned}$$
(4.12)

and

$$\begin{aligned} m_\lambda :=\inf \limits _{\gamma \in \Sigma _\lambda } \max \limits _{\textbf{t} \in \left[ \frac{1}{2},\frac{3}{2}\right] ^4}I_{b,\lambda }(\gamma (\textbf{t})), \end{aligned}$$
(4.13)

where

$$\begin{aligned} \Sigma _\lambda :&=\Big \{\gamma \in \mathcal {C}\left( \left[ \frac{1}{2},\frac{3}{2}\right] ^4,H^1_0(\Omega )\right) :\Vert \gamma (\textbf{t})\Vert \le 6\tau _2+\tau _1, \ (\gamma |_{\Omega _1^\rho })^+,\ (\gamma |_{\Omega _2^\rho })^-,\ (\gamma |_{\Omega _3^\rho })^\pm \ne 0\nonumber \\&\quad \text {and}\ \ \gamma =\gamma _0\ \ \text {on}\ \ \partial \left[ \frac{1}{2},\frac{3}{2}\right] ^4 \Big \}. \end{aligned}$$
(4.14)

Obviously, \(\gamma _0\in \Sigma _\lambda \), so \(\Sigma _\lambda \ne \varnothing \). Thus \(m_\lambda \) is well-defined.

Lemma 4.4

For any \(\gamma \in \Sigma _\lambda \), there exists an 4-tuple \(\textbf{t}^*=(t_1^*,t_2^*,t_3^*,t_4^*)\in D=(\frac{1}{2},\frac{3}{2})^4\) such that

$$\begin{aligned} \langle \widehat{I}'_{b,\lambda }(\gamma (\textbf{t}^*)|_{\Omega ^\rho }),\gamma _1^+(\textbf{t}^*)\rangle =\langle \widehat{I}'_{b,\lambda }(\gamma (\textbf{t}^*)|_{\Omega ^\rho }),\gamma _2^-(\textbf{t}^*)\rangle =0 \ \ \text {and} \ \ \langle \widehat{I}'_{b,\lambda }(\gamma (\textbf{t}^*)|_{\Omega ^\rho }),\gamma _3^\pm (\textbf{t}^*)\rangle =0, \end{aligned}$$

where \(\gamma _{i}(\textbf{t})=\gamma (\textbf{t})|_{\Omega ^\rho _{i}}\) for \(i=1,2,3\).

Proof

For each \(\gamma \in \Sigma _\lambda \), let us define \(\Psi :[\frac{1}{2},\frac{3}{2}]^4\rightarrow {\mathbb {R}}^4\) given by

$$\begin{aligned} \Psi (\textbf{t})=\left( \widehat{I}'_{b,\lambda }\left( \gamma (\textbf{t})|_{\Omega ^\rho }\right) \gamma _1^+(\textbf{t}),\ \widehat{I}'_{b,\lambda }\left( \gamma (\textbf{t})|_{\Omega ^\rho }\right) \gamma ^-_2(\textbf{t}),\ \widehat{I}'_{b,\lambda }\left( \gamma (\textbf{t})|_{\Omega ^\rho }\right) \gamma _3^+(\textbf{t}),\ \widehat{I}'_{b,\lambda }\left( \gamma (\textbf{t})|_{\Omega ^\rho }\right) \gamma _3^-(\textbf{t})\right) . \end{aligned}$$

Denote

$$\begin{aligned} \Psi _0(\textbf{t})=\left( \widehat{I}'_{b,\lambda }\left( \gamma _0(\textbf{t})\right) t_1v_1,\ \widehat{I}'_ {b,\lambda }\left( \gamma _0(\textbf{t})\right) t_2v_2,\ \widehat{I}'_{b,\lambda }\left( \gamma _0(\textbf{t})\right) t_3v_3^+,\ \widehat{I}'_{b,\lambda }\left( \gamma _0(\textbf{t})\right) t_4v_3^-\right) . \end{aligned}$$

Obviously,

$$\begin{aligned} \Psi (\textbf{t})=\Psi _0(\textbf{t})\ne 0, \ \ \text {for each}\ \ \textbf{t}\in \partial \left( \frac{1}{2},\frac{3}{2}\right) ^4. \end{aligned}$$

Therefore, we can verify that

$$\begin{aligned} deg(\Psi ,D,0)=deg(\Psi _0,D,0)=1. \end{aligned}$$

This implies that there exists \(\textbf{t}^*\in (\frac{1}{2},\frac{3}{2})^4\) such that \(\Psi (\textbf{t}^*)=0\). \(\square \)

Lemma 4.5

There holds that

(i)    \(\widehat{m}_\lambda \le m_\lambda \le m_{\Gamma }\) for all \(\lambda \ge 1\);

(ii)    \(m_\lambda \rightarrow m_{\Gamma }\) as \(\lambda \rightarrow +\infty \);

(iii)    There exists \(\varepsilon _0>0\) such that \(I_{b,\lambda }(\gamma (\textbf{t}))<m_{\Gamma }-\varepsilon _0\) for all \(\lambda > 0\), \(\gamma \in \Sigma _\lambda \) and \(\textbf{t}=(t_1,t_2,t_3,t_4)\in \partial [\frac{1}{2},\frac{3}{2}]^4\).

Proof

(i)   Since \(\gamma _0\in \Sigma _\lambda \), we have

$$\begin{aligned} m_\lambda \le \max \limits _{\textbf{t}\in [\frac{1}{2},\frac{3}{2}]^4}I_{b,\lambda }(\gamma _0(\textbf{t})) = I_{b,\lambda }(\gamma _0(1,1,1,1))=m_{\Gamma }, \end{aligned}$$

where we have used Lemma 2.2 in [10]. Recall that

$$\begin{aligned} \widehat{m}_{\lambda }:=\inf _{u \in \widehat{\mathcal {M}}_{b,\lambda }}\widehat{I}_{b,\lambda }(u). \end{aligned}$$

For each \(\gamma \in \Sigma _\lambda \), fix \(\textbf{t}^*\in (\frac{1}{2},\frac{3}{2})^4\) given by Lemma 4.4, then

$$\begin{aligned} \widehat{m}_{\lambda } \le \widehat{I}_{b,\lambda }(\gamma (\textbf{t}^*)|_{\Omega ^\rho }). \end{aligned}$$

Therefore,

$$\begin{aligned} \max \limits _{\textbf{t}\in [\frac{1}{2},\frac{3}{2}]^4}I_{b,\lambda }(\gamma (\textbf{t})) \ge \widehat{I}_{b,\lambda }(\gamma (\textbf{t}^*)|_{\Omega ^\rho })\ge \widehat{m}_\lambda , \ \ \text {for each}\ \ \gamma \in \Sigma _\lambda . \end{aligned}$$

Thus,

$$\begin{aligned} m_\lambda \ge \widehat{m}_\lambda . \end{aligned}$$

(ii)   Since \(\widehat{m}_\lambda \rightarrow m_{\Gamma }\) by Lemma 4.3 (ii), we have

$$\begin{aligned} m_\lambda \rightarrow m_{\Gamma } \ \ \text {as}\ \ \lambda \rightarrow +\infty . \end{aligned}$$

(iii)   For \(\textbf{t}=(t_1,t_2,t_3,t_4)\in \partial [\frac{1}{2},\frac{3}{2}]^4\), it holds \(\gamma (\textbf{t})=\gamma _0(\textbf{t})\) and hence

$$\begin{aligned} I_{b,\lambda }(\gamma (\textbf{t}))=I_{b,\lambda }(\gamma _0(\textbf{t}))\ \ \text {for}\ \ \textbf{t}=(t_1,t_2,t_3,t_4)\in \partial \left[ \frac{1}{2},\frac{3}{2}\right] ^4. \end{aligned}$$

By Lemma 2.2 in [10], we know that (1, 1, 1, 1) is the unique maximum point of \(\varphi (\textbf{t})=I_{b,0}(\gamma _0(\textbf{t}))\), which gives that

$$\begin{aligned} I_{b,\lambda }(\gamma (\textbf{t}))<m-\varepsilon _0\ \ \text {for}\ \ \textbf{t}=(t_1,t_2,t_3,t_4)\in \partial \left[ \frac{1}{2},\frac{3}{2}\right] ^4. \end{aligned}$$

where \(\varepsilon _0>0\) is a small constant. \(\square \)

5 Proof of Theorem 1.4.

In this section, we prove Theorem 1.4. More precisely, we show that the existence of sign-changing multi-bump solutions to Eq. (1.6) for large \(\lambda \), which converges to solutions of (1.17) with prescribed sign properties as \(\lambda \rightarrow +\infty \).

Define

$$\begin{aligned} \mathcal {S}:=\Big \{u\in \mathcal {M}_{\Gamma }\ \big |\ \ I_{\Gamma }(u)=m_{\Gamma }\Big \}, \end{aligned}$$

where

$$\begin{aligned}&\mathcal {M}_{\Gamma }=\Big \{u \in H_{0}^{1}(\Omega _\Gamma )~|~\langle I'_{\Gamma }(u), u|_{\Omega _{i}}\rangle =0, i=1,2,\ (u|_{\Omega _{1}})^{+} \ne 0, (u|_{\Omega _{2}})^{-} \ne 0,\\&\quad \text {and}~\langle I'_{\Gamma }(u), (u|_{\Omega _{3}})^\pm \rangle =0, (u|_{\Omega _{3}})^\pm \ne 0\Big \}. \end{aligned}$$

Obviously, \(\mathcal {S}\) contains all least energy solutions of (1.17) with \(u|_{\Omega _1}\) is positive, \(u|_{\Omega _2}\) is negative, \(u|_{\Omega _3}\) changes sign exactly once. Moreover, we have the following Lemma.

Lemma 5.1

\(\mathcal {S}\) is compact in \(H_0^1(\Omega _\Gamma )\).

Proof

Let \(\{u_n\}\subset \mathcal {S}\), then \(\{u_n\}\) is a bounded \((PS)_{m_{\Gamma }}\) sequence of \(I_{\Gamma }\). Since \(I_{\Gamma }\) satisfies (PS)-condition, up to a subsequence, we may suppose \(u_n\rightarrow u_\infty \) strongly in \(H_0^1(\Omega _\Gamma )\). It follows that \(u_\infty \in \mathcal {M}_{\Gamma }\) and \(I_{\Gamma }(u_\infty )=\lim \limits _{n\rightarrow \infty } I_{\Gamma }(u_n)=m_{\Gamma }\). Therefore, \(u_\infty \in \mathcal {S}\). \(\square \)

Lemma 5.2

Let \(d>0\) be a fixed number and let \(\{u_n\}\subset \mathcal {S}^d\) be a sequence. Then, up to a subsequence, \(u_n\rightharpoonup u_0\in \mathcal {S}^{2d}\) weakly in \(H^1_0(\Omega )\) as \(n\rightarrow \infty \), where

$$\begin{aligned} \mathcal {S}^d:=\Bigl \{u\in H^1_0(\Omega ): dist_\lambda (u,\mathcal {S})\le d\Bigr \} \end{aligned}$$

and dist denotes the distance in \(H^1_0(\Omega )\).

Proof

Since \(\mathcal {S}\) is compact in \(H_0^1(\Omega )\), then there exists a sequence \(\{\bar{u}_n\}\subset \mathcal {S}\) such that

$$\begin{aligned} dist \left( u_n,\mathcal {S}\right) =dist\left( u_n,\bar{u}_n\right) \le d. \end{aligned}$$

By Lemma 5.1, there exists \(\bar{u}\in \mathcal {S}\) such that, up to a subsequence, \(\bar{u}_n\rightarrow \bar{u}\) strongly in \(H_0^1(\Omega )\). Hence, \(dist\left( \bar{u}_n,\bar{u}\right) \le d\) for n large enough. Thus, \(\{u_n\}\) is bounded and, up to a subsequence, \(u_n\rightharpoonup u_0\) weakly in \(H^1_0(\Omega )\). Since \(B_{2d}(\bar{u})\) is weakly closed in \(H^1_0(\Omega )\), therefore, \(u_0\in B_{2d}(\bar{u})\subset \mathcal {S}^{2d}\). \(\square \)

Lemma 5.3

Let \(d\in (0,\tau _1)\), where \(\tau _1\) is given by (4.11). Suppose that there exist a sequence \(\lambda _n>0\) with \(\lambda _n\rightarrow +\infty \), and \(\{u_n\}\subset \mathcal {S}^d\) satisfying

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } I_{b,\lambda _n}(u_n)\le m_\Gamma ,\,\,\,\,\,\,\lim \limits _{n\rightarrow \infty }I'_{b,\lambda _n}(u_n)=0. \end{aligned}$$

Then, up to a subsequence, \(\{u_n\}\) converges strongly in \(H^1_0(\Omega )\) to an element \(u\in \mathcal {S}\).

Proof

Since \(\lim \limits _{n\rightarrow \infty } I_{b,\lambda _n}(u_n)\le m_\Gamma \) and \(\lim \limits _{n\rightarrow \infty }I'_{b,\lambda _n}(u_n)=0\), we deduce that \(\{\Vert u_n\Vert \}\) and \(\{I_{\lambda _n}(u_n)\}\) are bounded. Up to a subsequence, we may assume that

$$\begin{aligned} I_{b,\lambda _n}(u_n)\rightarrow c\in (-\infty ,m_\Gamma ]. \end{aligned}$$

By using Proposition 4.2, there exists \(u \in H^1_0(\Omega )\) such that

$$\begin{aligned} u_n \rightarrow u \ \ \text {strongly in} \ \ H^1_0(\Omega ), \ \ u=0\ \text {in}\ \Omega \setminus \Omega _\Gamma \ \ \text {and}\ \ I_{b,\lambda _n}(u_n)\rightarrow I_{\Gamma }(u). \end{aligned}$$
(5.1)

Moreover, u is a solution to the following equation

$$\begin{aligned} \left\{ \displaystyle \begin{array}{ll} -\displaystyle \left( a+b\int \limits _{\Omega _\Gamma }|\nabla u|^2\mathrm{{d}}x\right) \Delta u =h^+(x)|u|^{p-2}u, \ {} &{} x\in \Omega _\Gamma ,\\ u=0, \ {} &{} x\in \Omega \setminus \Omega _\Gamma ,\\ u=0, \ {} &{} x\in \partial \Omega . \end{array} \right. \end{aligned}$$
(5.2)

Since \(\{u_n\}\subset \mathcal {S}^d\) and \(d\in (0,\tau _1)\), we deduce that \((u|_{\Omega _1})^+\ne 0\), \((u|_{\Omega _2})^-\ne 0\) and \((u|_{\Omega _3})^\pm \ne 0\). Consequently, \(I_\Gamma (u)\ge m\). The conclusion \(I_{\Gamma }(u)=m\) follows from the fact that \(I_{b,\lambda _n}(u_n)\rightarrow I_{\Gamma }(u)\le m_{\Gamma }\), Thus, \(u \in \mathcal {S}\) is proved. \(\square \)

Lemma 5.4

Let \(\tau _1>0\) be as in Lemma 5.3. Then, for \(\delta \in (0,d)\), there exist constants \(0<\sigma <1\) and \(\Lambda _1>0\) such that \(\Vert I_{b,\lambda }'(u)\Vert _{H^{-1}}\ge \sigma \) for any \(u\in I_{b,\lambda }^{m_\lambda }\cap (\mathcal {S}^\delta {\setminus } \mathcal {S}^{\frac{\delta }{2}})\) and \(\lambda \ge \Lambda _1\).

Proof

We argue by contradiction. Suppose that there exist a number \(\delta _0\in (0,d)\), a positive sequence \(\{\lambda _j\}\) with \(\lambda _j\rightarrow 0\), and a sequence of function \(\{u_j\}\subset I_{b,\lambda _j}^{m_{\lambda _j}}\cap (\mathcal {S}^{\delta _0}{\setminus } \mathcal {S}^{\frac{\delta _0}{2}})\) such that

$$\begin{aligned} \lim \limits _{j\rightarrow +\infty }I_{b,\lambda _j}'(u_j)=0. \end{aligned}$$

Up to a subsequence, we obtain

$$\begin{aligned} \{u_j\}\subset \mathcal {S}^{\delta _0},\,\,\,\,\lim \limits _{j\rightarrow \infty }I_{b,\lambda _j}(u_j)\le m_\Gamma . \end{aligned}$$

Hence, we can apply Lemma 5.3 and assert that there exists \(u\in \mathcal {S}\) such that \(u_j\rightarrow u\) strongly in \(H^1_0(\Omega )\). As a consequence, \(dist\left( u_j,\mathcal {S}\right) \rightarrow 0\) as \(j\rightarrow +\infty \). This contradict the fact that \(u_j\not \in \mathcal {S}^{\frac{\delta _0}{2}}\). \(\square \)

From now on, we fix a small constant \(\delta \in (0,d)\) and corresponding constants \(0<\sigma <1\) and \(\Lambda _1>0\) such that our Lemma 5.4 hold. For convenient, we next denote \(Q:=[\frac{1}{2},\frac{3}{2}]^4\).

Lemma 5.5

There exist \(\Lambda _2\ge \Lambda _1\) and \(\alpha >0\) such that for any \(\lambda \ge \Lambda _2\),

$$\begin{aligned} I_{b,\lambda }(\gamma _0(t_1,t_2,t_3,t_4))\ge m_\lambda -\alpha \,\,\hbox {implies that}\,\,\,\, \gamma _0(t_1,t_2,t_3,t_4)\in \mathcal {S}^{\frac{\delta }{2}}. \end{aligned}$$
(5.3)

Proof

Assume by contradiction that there exist \(\lambda _n\rightarrow \infty \), \(\alpha _n\rightarrow 0\) and \((t^{(n)}_1,t^{(n)}_2,t^{(n)}_3,t^{(n)}_4)\in Q\) such that

$$\begin{aligned} I_{b,\lambda }(\gamma _0(t^{(n)}_1,t^{(n)}_2,t^{(n)}_3,t^{(n)}_4))\ge m_{\lambda _n}-\alpha _n \,\,\hbox {and}\,\,\gamma _0(t^{(n)}_1,t^{(n)}_2,t^{(n)}_3,t^{(n)}_4)\not \in \mathcal {S}^{\frac{\delta }{2}}. \end{aligned}$$
(5.4)

Passing to a subsequence, we may assume that \((t^{(n)}_1,t^{(n)}_2,t^{(n)}_3,t^{(n)}_4)\rightarrow (\bar{t}_1,\bar{t}_2,\bar{t}_3,\bar{t}_4)\in Q\). Then, Lemma 4.5 implies that

$$\begin{aligned} I_{\Gamma }(\gamma _0(\bar{t}_1,\bar{t}_2,\bar{t}_3,\bar{t}_4))\ge \lim \limits _{n\rightarrow \infty }\left( m_{\lambda _n}-\alpha _n\right) =m_{\Gamma }. \end{aligned}$$

From Lemma 2.2 in [10], we can deduce that \((\bar{t}_1,\bar{t}_2,\bar{t}_3,\bar{t}_4)=(1,1,1,1)\) and hence

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert \gamma _0(t^{(n)}_1,t^{(n)}_2,t^{(n)}_3,t^{(n)}_4)-\gamma _0(1,1,1,1)\Vert =0. \end{aligned}$$

However, \(\gamma _0(1,1,1,1)=v\in \mathcal {S}\), which contradicts to (5.4). \(\square \)

Next, we set

$$\begin{aligned} \alpha _0:=\min \Big \{\frac{\alpha }{2},\frac{\varepsilon _0}{2},\frac{1}{8}\delta \sigma ^2\Big \}, \end{aligned}$$
(5.5)

where \(\delta \), \(\sigma \) are given in Lemma 5.4, \(\alpha \) is from Lemma 5.5, \(\varepsilon _0\) is from Lemma 4.5 (iii). By Lemma 4.4, there exists \(\Lambda _3\ge \Lambda _2\) such that

$$\begin{aligned} |m_\lambda -m_{\Gamma }|<\alpha _0 \,\, \text {for all}\ \ \lambda \ge \Lambda _3. \end{aligned}$$
(5.6)

Proposition 5.6

For each \(\lambda \ge \Lambda _3\), there exists a critical point \(u_\lambda \) of \(I_{b,\lambda }\) with \(u_\lambda \in \mathcal {S}^\delta \cap I_{b,\lambda }^{m_{\Gamma }}\).

Proof

Fix \(\lambda \ge \Lambda _3\). Assume by contradiction that there exists \(0<\rho _\lambda <1\) such that \(\Vert I_{b,\lambda }'(u)\Vert \ge \rho _\lambda \) on \(\mathcal {S}^\delta \cap I_\lambda ^{m_{\Gamma }}\). Then there exists a pseudo-gradient vector field \(T_\lambda \) in \(H^1_0(\Omega )\) which is defined on a neighborhood \(Z_\lambda \) of \(\mathcal {S}^\delta \cap I_{b,\lambda }^{m_{\Gamma }}\) such that for any \(u\in Z_\lambda \) there holds

$$\begin{aligned} \Vert T_\lambda (u)\Vert&\le 2\min \{1,\Vert I_{b,\lambda }'(u)\Vert \},\\ \langle I_{b,\lambda }'(u),T_\lambda (u)\rangle&\ge \min \{1, \Vert I'_{b,\lambda }(u)\Vert \}\Vert I_{b,\lambda }'(u)\Vert . \end{aligned}$$

Let \(\psi _\lambda \) be a Lipschitz continuous function on \(H^1_0(\Omega )\) such that \(0\le \psi _\lambda \le 1\), \(\psi _\lambda \equiv 1\) on \(S^\delta \cap I_{b,\lambda }^{m_{\Gamma }}\) and \(\psi _\lambda \equiv 0\) on \(H^1_0(\Omega ){\setminus } Z_\lambda \). Let \(\xi _\lambda \) be a Lipschitz continuous function on \({\mathbb {R}}\) such that \(0\le \xi _\lambda \le 1\), \(\xi _\lambda (t)\equiv 1\) if \(|t-m_{\Gamma }|\le \frac{\alpha }{2}\) and \(\xi _\lambda (t)\equiv 0\) if \(|t-m_{\Gamma }|\ge \alpha \). Define

$$\begin{aligned} e_\lambda (u):=\left\{ \displaystyle \begin{array}{ll} -\psi _\lambda (u)\xi _\lambda (I_{b,\lambda }(u))T_\lambda (u),\, &{}\text {if}\, \, u\in Z_\lambda , \\ 0,\, &{}\text {if}\, \, u\in H^1_0(\Omega )\setminus Z_\lambda . \end{array} \right. \end{aligned}$$
(5.7)

Then there exists a global solution \(\eta _\lambda : H^1_0(\Omega ) \times [0,+\infty )\rightarrow H^1_0(\Omega )\) for the initial value problem

$$\begin{aligned} \left\{ \displaystyle \begin{array}{ll} \frac{\textrm{d}}{\textrm{d}\theta }\eta _\lambda (u,\theta )=e_\lambda (\eta _\lambda (u,\theta )), \\ \eta _\lambda (u,0)=u. \end{array} \right. \end{aligned}$$
(5.8)

It is easy to see that \(\eta _\lambda \) has the following properties:

(1)  \(\eta _\lambda (u,\theta )=u\) if \(\theta =0\) or \(u\in H^1_0(\Omega ){\setminus } Z_\lambda \) or \(|I_{b,\lambda }(u)-m_\Gamma |\ge \alpha \).

(2)  \(\Vert \frac{\textrm{d}}{\textrm{d}\theta }\eta _\lambda (u,\theta )\Vert \le 2\).

(3)  \(\frac{\textrm{d}}{\textrm{d}\theta }I_{b,\lambda }(\eta _\lambda (u,\theta ))=\langle I_{b,\lambda }'(\eta _\lambda (u,\theta )), e_\lambda (\eta _\lambda (u,\theta ))\rangle \le 0\). \(\square \)

Claim 1

For any \((t_1,t_2,t_3,t_4)\in Q\), there exists \(\overline{\theta }=\theta (t_1,t_2,t_3,t_4)\in [0,+\infty )\) such that \(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\overline{\theta }) \in I_{b,\lambda }^{m_{\Gamma }-\alpha _0}\), where \(\alpha _0\) is given by (5.5).

Assume by contradiction that there exists \((t_1,t_2,t_3,t_4)\in Q\) such that

$$\begin{aligned} I_{b,\lambda }(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))>m_{\Gamma }-\alpha _0 \end{aligned}$$

for any \(\theta \ge 0\). Note that \(\alpha _0<\alpha \), we see, from Lemma 5.5, that \(\gamma _0(t_1,t_2,t_3,t_4)\in \mathcal {S}^{\frac{\delta }{2}}\). Moreover, since \(I_{b,\lambda }(\gamma _0(t_1,t_2,t_3,t_4))\le m_{\Gamma }\), we have, from the property (3) of \(\eta _\lambda \), that

$$\begin{aligned} m_{\Gamma }-\alpha _0<I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))\le I_{b,\lambda }(\gamma _0(t_1,t_2,t_3,t_4))\le m_{\Gamma } \end{aligned}$$

for \(\theta \ge 0\). This implies that \(\xi _\lambda (I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta )))\equiv 1\). If \(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta )\in \mathcal {S}^\delta \) for all \(\theta \ge 0\), we can deduce that

$$\begin{aligned} \psi _\lambda \left( \eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta )\right) \equiv 1\ \ \text {and}\ \ \Vert I'_{b,\lambda }(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))\Vert \ge \rho _\lambda \end{aligned}$$

for all \(\theta >0\). It follows that

$$\begin{aligned} I_{b,\lambda } \bigl (\eta _\lambda \bigl (\gamma _0(t_1,t_2,t_3,t_4),\frac{\alpha }{\rho _\lambda ^2}\bigr )\bigr )\le m_{\Gamma }-\int \limits _{0}^{\frac{\alpha }{\rho _\lambda ^2}} \rho _\lambda ^2dt\le m_{\Gamma }-{\alpha }, \end{aligned}$$

which is a contradiction. Thus, there exists \(\theta _3>0\) such that \(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _3)\not \in \mathcal {S}^\delta \). Note that \(\gamma _0(t_1,t_2,t_3,t_4)\in \mathcal {S}^{\frac{\delta }{2}}\), there exist \(0<\theta _1<\theta _2\le \theta _3\) such that

$$\begin{aligned} \eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _1)\in \partial \mathcal {S}^{\frac{\delta }{2}},\ \ \eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _2)\in \partial \mathcal {S}^\delta \end{aligned}$$

and

$$\begin{aligned} \eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta )\in \mathcal {S}^\delta \setminus \mathcal {S}^{\frac{\delta }{2}} \ \ \text {for all}\ \ \theta \in (\theta _1,\theta _2). \end{aligned}$$

By Lemma 5.4, we have that

$$\begin{aligned} \Vert I_{b,\lambda } '(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))\Vert \ge \sigma \ \ \text {for all}\ \ \theta \in (\theta _1,\theta _2). \end{aligned}$$

By using property (2) of \(\eta _\lambda \) we have

$$\begin{aligned} \frac{\delta }{2}\le \Vert \eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _2)-\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _1)\Vert \le 2 |\theta _2-\theta _1|. \end{aligned}$$

This implies that

$$\begin{aligned}&I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _2))\nonumber \\&\quad \le I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),0)) +\int \limits _{0}^{\theta _2}\frac{\textrm{d}}{\textrm{d}\theta }I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))\mathrm{{d}}\theta \nonumber \\&\quad<I_{b,\lambda } (\gamma _0(t_1,t_2,t_3,t_4)) +\int \limits _{\theta _1}^{\theta _2}\frac{\textrm{d}}{\textrm{d}\theta }I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))\mathrm{{d}}\theta \nonumber \\&\quad \le m_{\Gamma }-\sigma ^2(\theta _2-\theta _1)\le m_{\Gamma }-\frac{1}{4}\delta \sigma ^2\nonumber \\&\quad <m_{\Gamma }-\alpha _0, \end{aligned}$$
(5.9)

which is a contradiction. Thus, we finish the proof of Claim 1.

Now, we can define

$$\begin{aligned} T(t_1,t_2,t_3,t_4):=\inf \Big \{\theta \ge 0: I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))\le m_{\Gamma }-\alpha _0\Big \} \end{aligned}$$

and

$$\begin{aligned} \widetilde{\gamma }(t_1,t_2,t_3,t_4):=\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),T(t_1,t_2,t_3,t_4)). \end{aligned}$$

Then \(\Phi _\lambda (\widetilde{\gamma }(t_1,t_2,t_3,t_4))\le m_{\Gamma }-\alpha _0\) for all \((t_1,t_2,t_3,t_4)\in Q\).

Claim 2

\(\widetilde{\gamma }(t_1,t_2,t_3,t_4)=\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),T(t_1,t_2,t_3,t_4)) \in \Sigma _\lambda \).

For any \((t_1,t_2,t_3,t_4)\in \partial Q\), by (5.5)–(5.6), we have

$$\begin{aligned} I_{b,\lambda }(\gamma _0(t_1,t_2,t_3,t_4))\le I_{\Gamma }(\gamma _0(t_1,t_2,t_3,t_4))<m_{\Gamma }-\varepsilon _0 \le m_\Gamma -\alpha _0, \end{aligned}$$

which implies that \(T(t_1,t_2,t_3,t_4)=0\) and thus \(\widetilde{\gamma }(t_1,t_2,t_3,t_4)=\gamma _0(t_1,t_2,t_3,t_4)\) for \((t_1,t_2,t_3,t_4)\in \partial Q\).

By the definition of \(\Sigma _\lambda \) in (4.14), it suffices to prove that \(\Vert \widetilde{\gamma }(t_1,t_2,t_3,t_4)\Vert \le 6\tau _2+\tau _1\) for all \((t_1,t_2,t_3,t_4)\in Q\) and \(T(t_1,t_2,t_3,t_4)\) is continuous with respect to \((t_1,t_2,t_3,t_4)\).

For any \((t_1,t_2,t_3,t_4)\in Q\), we have \(T(t_1,t_2,t_3,t_4)=0\) if \(I_{b,\lambda } (\gamma _0(t_1,t_2,t_3,t_4))\le m_{\Gamma }-\alpha _0\), and hence \(\widetilde{\gamma }(t_1,t_2,t_3,t_4)=\gamma _0(t_1,t_2,t_3,t_4)\). By (4.11), we deduce that \(\Vert \widetilde{\gamma }(t_1,t_2,t_3,t_4)\Vert \le 6\tau _2< 6\tau _2+\tau _1\).

On the other hand, if \(I_{b,\lambda } (\gamma _0(t_1,t_2,t_3,t_4))>m_{\Gamma }-\alpha _0\), we can deduce that

$$\begin{aligned} I_{b,\lambda }(\gamma _0(t_1,t_2,t_3,t_4))> m_\lambda -\alpha , \end{aligned}$$

thus \(\gamma _0(t_1,t_2,t_3,t_4)\in \mathcal {S}^{\frac{\delta }{2}}\) and

$$\begin{aligned} m_{\Gamma }-\alpha _0<I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta ))< m_{\Gamma }+\alpha _0,\,\,\,\text {for all}\ \,\theta \in [0,T(t_1,t_2,t_3,t_4)). \end{aligned}$$

This implies that

$$\begin{aligned} \xi _\lambda (I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta )))\equiv 1 \ \ \text {for all}\ \ \theta \in [0,T(t_1,t_2,t_3,t_4)). \end{aligned}$$

Now, we are going to prove that \(\widetilde{\gamma }(t_1,t_2,t_3,t_4)\in \mathcal {S}^\delta \). Otherwise, if \(\widetilde{\gamma }(t_1,t_2,t_3,t_4)\not \in \mathcal {S}^\delta \), similar to the proof of Claim 1, we can find two constants \(0<\theta _1<\theta _2<T(t_1,t_2,t_3,t_4)\) such that (5.9) hold. This implies that \(I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),\theta _2))<m_{\Gamma }-\alpha _0\) which contradicts to the definition of \(T(t_1,t_2,t_3,t_4)\). Therefore,

$$\begin{aligned} \widetilde{\gamma }(t_1,t_2,t_3,t_4)=\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),T(t_1,t_2,t_3,t_4))\in \mathcal {S}^\delta . \end{aligned}$$

Thus there exists \(u\in \mathcal {S}\) such that \(\Vert \widetilde{\gamma }(t_1,t_2,t_3,t_4)-u\Vert \le \delta \le \tau _1\). It follows from (4.11) that

$$\begin{aligned} \Vert \widetilde{\gamma }(t_1,t_2,t_3,t_4)\Vert \le \Vert u\Vert +\tau _1 \le 6\tau _2+\tau _1. \end{aligned}$$

To prove the continuity of \(T(t_1,t_2,t_3,t_4)\), we fix arbitrarily \((t_1,t_2,t_3,t_4)\in Q\). First, we assume that \(I_{b,\lambda }(\widetilde{\gamma }(t_1,t_2,t_3,t_4))<m_{\Gamma }-\alpha _0\). In this case, we deduce directly that \(T(t_1,t_2,t_3,t_4)=0\) by the definition of \(T(t_1,t_2,t_3,t_4)\), which gives that

$$\begin{aligned} I_{b,\lambda }(\gamma _0(t_1,t_2,t_3,t_4))<m-\alpha _0. \end{aligned}$$

By the continuity of \(\gamma _0\), there exists \(r>0\) such that for any \((s_1,s_2,s_3,s_4)\in B_r(t_1,t_2,t_3,t_4)\cap Q\), we have \(I_{b,\lambda }(\gamma _0(s_1,s_2,s_3,s_4))<m_{\Gamma }-\alpha _0\). Thus, \(T(s_1,s_2,s_3,s_4)=0\), and hence T is continuous at \((t_1,t_2,t_3,t_4)\).

Now, we assume that \(I_{b,\lambda } (\widetilde{\gamma }(t_1,t_2,t_3,t_4))=m_{\Gamma }-\alpha _0\). From the previous proof we see that \(\widetilde{\gamma }(t_1,t_2,t_3,t_4)=\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),T(t_1,t_2,t_3,t_4))\in \mathcal {S}^\delta \), and so

$$\begin{aligned} \Vert I_{b,\lambda } '(\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),T(t_1,t_2,t_3,t_4)))\Vert \ge \rho _\lambda >0. \end{aligned}$$

Thus for any \(\omega >0\), we have

$$\begin{aligned} I_{b,\lambda } (\eta _\lambda (\gamma _0(t_1,t_2,t_3,t_4),T(t_1,t_2,t_3,t_4)+\omega ))<m_{\Gamma }-\alpha _0. \end{aligned}$$

By the continuity of \(\eta _\lambda \), there exists \(r>0\) such that

$$\begin{aligned} I_{b,\lambda } (\eta _\lambda (\gamma _0(s_1,s_2,s_3,s_4),T(t_1,t_2,t_3,t_4)+\omega )))<m_\Gamma -\alpha _0, \end{aligned}$$

for any \((s_1,s_2,s_3,s_4)\in B_r(t_1,t_2,t_3,t_4)\cap Q\). Thus, \(T(s_1,s_2,s_3,s_4)\le T(t_1,t_2,t_3,t_4)+\omega \). It follows that

$$\begin{aligned} 0\le \limsup \limits _{(s_1,s_2,s_3,s_4)\rightarrow (t_1,t_2,t_3,t_4)}T(s_1,s_2,s_3,s_4)\le T(t_1,t_2,t_3,t_4). \end{aligned}$$
(5.10)

If \(T(t_1,t_2,t_3,t_4)=0\), we immediately implies that

$$\begin{aligned} \lim \limits _{(s_1,s_2,s_3,s_4)\rightarrow (t_1,t_2,t_3,t_4)}T(s_1,s_2,s_3,s_4)=T(t_1,t_2,t_3,t_4). \end{aligned}$$

If \(T(t_1,t_2,t_3,t_4)>0\), we can similarly deduce that

$$\begin{aligned} I_{b,\lambda } (\eta _\lambda (\gamma _0(s_1,s_2,s_3,s_4),T(t_1,t_2,t_3,t_4)-\omega ))>m_{\Gamma }-\alpha _0. \end{aligned}$$

for any \(0<\omega <T(t_1,t_2,t_3,t_4)\).

By the continuity of \(\eta _\lambda \) again, we see that

$$\begin{aligned} \liminf \limits _{(s_1,s_2,s_3,s_4)\rightarrow (t_1,t_2,t_3,t_4)}T(s_1,s_2,s_3,s_4)\ge T(t_1,t_2,t_3,t_4). \end{aligned}$$
(5.11)

It follows from (5.10)–(5.11) that T is continuous at \((t_1,t_2,t_3,t_4)\). This completes the proof of Claim 2.

Thus, we have proved that \(\widetilde{\gamma }(t_1,t_2,t_3,t_4)\in \Sigma _\lambda \) and

$$\begin{aligned} \max \limits _{(t_1,t_2,t_3,t_4)\in Q}I_\lambda (\widetilde{\gamma }(t_1,t_2,t_3,t_4))\le m_{\Gamma }-\alpha _0, \end{aligned}$$

which contradicts the definition of \(m_{\Gamma }\). This completes the proof. \(\square \)

Proof of Theorem 1.4

We still prove Theorem 1.4 with \(\Gamma _1=\{1\}\), \(\Gamma _2=\{2\}\) and \(\Gamma _3=\{3\}\). For the general \(\Gamma \) verifying (1.16), the proof is very similar and just needs a slight modification.

By Proposition 5.6, there exists a solution \(u_\lambda \) for Eq. (1.6) with \(u_\lambda \in \mathcal {S}^{\delta }\cap I_{b,\lambda }^{m_{\Gamma }}\) for all \(\lambda \ge \Lambda _3\). Therefore, for any sequence \(\{\lambda _n\}\) with \(\lambda _n \rightarrow +\infty \) as \(n\rightarrow \infty \), there exists a sequence \(\{u_n\}\subset H^1_0(\Omega )\) such that

$$\begin{aligned} I_{b,\lambda _n}(u_n)\le m_\Gamma ,\,\,\,\,\,\,I'_{b,\lambda _n}(u_n)=0. \end{aligned}$$

By using Lemma 5.3, we can deduce that \(u_{\lambda _n}\rightarrow u\in \mathcal {S}\) strongly in \(H^1_0(\Omega )\). Thus, we complete the proof of Theorem 1.4. \(\square \)