1 Introduction and main results

In this paper, we are concerned with the multiplicity of positive solutions for the nonlocal problem

$$ \textstyle\begin{cases} - (a-b \int _{\Omega } \vert \nabla u \vert ^{2}\,dx ) \Delta u= \lambda \vert u \vert ^{q-2}u, & x\in \Omega , \\ u=0, & x\in \partial \Omega , \end{cases} $$
(1.1)

where Ω is a smooth bounded domain in \(\mathbb{R}^{N}\) \((N\ge 3)\), \(a,b>0\), \(1< q<2\) and \(\lambda >0\) is a parameter.

In the past two decades, the following Kirchhoff type problem on a bounded domain

$$ \textstyle\begin{cases} - (a+b \int _{\Omega } \vert \nabla u \vert ^{2}\,dx ) \Delta u=f(x,u), & x\in \Omega , \\ u=0, & x\in \partial \Omega , \end{cases} $$
(1.2)

has attracted great attention of many researchers. The Kirchhoff type problem is often viewed as nonlocal due to the presence of the term \(b\int _{\Omega } |\nabla u|^{2}\,dx\) which implies that such a problem is no longer a pointwise identity. By using the variational method, there are many interesting results of positive solutions to (1.2), see e.g. [1, 4, 5, 9, 11] and the references therein.

If we replace \(b\int _{\Omega } |\nabla u|^{2}\,dx\) with \(-b\int _{\Omega } |\nabla u|^{2}\,dx\), then (1.2) turns out to be the following new nonlocal one:

$$ \textstyle\begin{cases} - (a-b \int _{\Omega } \vert \nabla u \vert ^{2}\,dx ) \Delta u=f(x,u), & x\in \Omega , \\ u=0, & x\in \partial \Omega . \end{cases} $$
(1.3)

This kind of problem involving negative nonlocal term not only presents some interesting difficulties different from Kirchhoff type problem but also has its own physical and mechanical motivation, see [8, 14]. Yin and Liu [16] considered problem (1.3) when \(f(x,u)=|u|^{p-2}u\) with \(2< p<2^{*}\) and showed the existence of two nontrivial solutions. Based on [16], Wang and Yang [15] further obtained the existence of infinitely many sign-changing solutions. In [17], the authors extended the results of [16] to a general case of nonlinear terms. For \(f(x,u)=\lambda |u|^{-\gamma }\) with \(0<\gamma <1\), [6] got the multiplicity of positive solutions to (1.3). In [10], we proved that problem (1.3) possesses at least one positive solution when \(N=3\), \(f(x,u)=\lambda f(x)|u|^{p-2}u\) with \(3< p<4\) and \(f(x)\in L^{\frac{{6}}{{6-p}}}(\Omega )\) may change sign. In particular, Duan et al. [3] and Lei et al. [7] proved that there exists \(\lambda _{*}>0\) such that, for each \(\lambda \in (0,\lambda _{*})\), problem (1.1) has two positive solutions by using the minimization argument and the mountain pass theorem.

From the works described before, it is important and interesting to ask whether the multiplicity of positive solutions to problem (1.1) can be established by other methods? In the present paper, we shall give an affirmative answer. The main technique applied here is a separation argument for the Nehari-type set of problem (1.1), which has been firstly introduced by Tarantello [13] and later refined by Sun and Li [12].

Let \(H:= H_{0}^{1}(\Omega )\) and \(L^{s}(\Omega )\) be the standard Sobolev spaces endowed with the standard norms \(\|\cdot \|\) and \(|\cdot |_{p}\), respectively. Denote by → and ⇀ the strong and weak convergence, respectively. We use \(o_{n}(1)\) to denote a quantity such that \(o_{n}(1)\to 0\) as \(n\to \infty \). C and \(C_{i}\) denote various positive constants which may vary from line to line. We say that \(I\in C^{1}(H,\mathbb{R})\) satisfies the Palais–Smale condition at level \(c\in \mathbb{R}\) (\((\text{PS})_{c}\) in short) if any sequence \(\{u_{n}\}\subset H\) such that \(I(u_{n})\to c\) and \(I'(u_{n})\to 0\) in \(H^{-1}\) as \(n\to \infty \) has a convergent subsequence. S denotes the best constant in the Sobolev embedding \(H\hookrightarrow L^{2^{*}}(\Omega )\), that is,

$$ S=\inf_{u\in H\setminus \{0\} } \frac{ {\int \vert \nabla u \vert ^{2}\,dx}{} }{{ (\int \vert u \vert ^{2^{*}}\,dx )^{2/2^{*}} } }>0. $$

Associated with problem (1.1), we define the energy functional

$$ I_{b}(u)= \frac{a}{2} \Vert u \Vert ^{2} - \frac{b}{4} \Vert u \Vert ^{4}- \frac{{\lambda }}{q}{} \int _{\Omega } \vert u \vert ^{q}\,dx. $$

Then \(I_{b} \in C^{1}(H,\mathbb{R})\). Recall that a function \(u\in H\) is called a weak solution to (1.1) if, for any \(\phi \in H\), there holds

$$ \bigl(a-b \Vert u \Vert ^{2} \bigr){} \int _{\Omega }\nabla u {} \nabla {\phi }\,dx - \lambda \int _{\Omega } \vert u \vert ^{q-2}u \phi \,dx=0. $$

Define the Nehari type set of (1.1)

$$ \Lambda _{b}= \bigl\{ u \in H: \bigl\langle I'_{b}(u),u {} \bigr\rangle = 0 \bigr\} = \biggl\{ u \in H: a \Vert u \Vert ^{2}-b \Vert u \Vert ^{4}={} \lambda \int _{\Omega } \vert u \vert ^{q}\,dx \biggr\} , $$

and then decompose \(\Lambda _{b}\) into three subsets:

$$\begin{aligned}& \Lambda _{b}^{0}= \bigl\{ u\in \Lambda _{b}: a(2-q) \Vert u \Vert ^{2}- b(4-q) \Vert u \Vert ^{4}=0 \bigr\} , \\& \Lambda _{b}^{+}= \bigl\{ u\in \Lambda _{b}: a(2-q) \Vert u \Vert ^{2}- b(4-q) \Vert u \Vert ^{4}>0 \bigr\} , \\& \Lambda _{b}^{-}= \bigl\{ u\in \Lambda _{b}: a(2-q) \Vert u \Vert ^{2}- b(4-q) \Vert u \Vert ^{4}< 0 \bigr\} . \end{aligned}$$

It is important to notice that there exists a norm gap in \(\Lambda _{b}\):

$$ \Vert \tilde{u} \Vert ^{2}>\frac{{a(2-q)}}{{b(4-q)}}> \Vert u \Vert ^{2} \quad \text{for all} \ u\in \Lambda _{b}^{+}, \tilde{u}\in \Lambda _{b}^{-}. {}$$
(1.4)

Set

$$ T_{b}=\frac{{2(2-q)}}{{(4-q)^{2}}}\frac{{a^{\frac{{4-q}}{2}}}}{{b^{\frac{{2-q}{} }{2}}}} \frac{S^{\frac{q}{2}}}{{ \vert \Omega \vert ^{\frac{{{2^{*}}-q}}{{2^{*}}}}}}. $$

Our main results are as follows.

Theorem 1.1

Assume that \(\lambda \in (0,T_{b})\), then problem (1.1) has at least two positive solutions \(u_{*} \in \Lambda _{b}^{+}\), \(\tilde{u}_{*} \in \Lambda _{b}^{-}\) with \(\|u_{*}\|<\|\tilde{u}_{*}\|\).

Moreover, as a by-product of our arguments, we regard b as a parameter and obtain the blow-up behavior of the solution \(\tilde{u}_{b} \in \Lambda _{b}^{-}\) and the asymptotic behavior of the other one \(u_{b} \in \Lambda _{b}^{+}\) of problem (1.1) as \(b\searrow 0\). Namely, we have the following theorem.

Theorem 1.2

Assume that \(\{b_{n}\}\) is a sequence satisfying \(b_{n}\searrow 0\) as \(n\to \infty \). Then there exists a subsequence, still denoted by \(\{b_{n}\}\), such that

(i) \(\|\tilde{u}_{b_{n}}\|\to \infty \) as \(n\to \infty \).

(ii) \({u}_{b_{n}}\to {u}_{0}\) in H as \(n\to \infty \), where \({u}_{0}\) is a positive solution of the problem

$$ \textstyle\begin{cases} -a\Delta u=\lambda \vert u \vert ^{q-2}u, & x\in \Omega , \\ u=0, & x\in \partial \Omega . \end{cases} $$
(1.5)

Remark 1.3

Compared with [3, 7], we adapt a new method to show the existence and multiplicity of positive solutions to problem (1.1). In particular, we obtain the blow-up and the asymptotic behavior of these solutions. As far as we know, such phenomena about the solutions to (1.1) are first observed, which reveals some relationship between the nonlocal problem (1.1) and the classical semilinear problem (1.5).

The paper is organized as follows. In Sect. 2, we present some preliminaries. Sections 3 and 4 are devoted to the proofs of Theorems 1.1 and 1.2, respectively.

2 Preliminaries

Lemma 2.1

Let \(\lambda \in (0,T_{b})\). Then and \(\Lambda _{b}^{0}=\{0\}\).

Proof

For any \(u\in H\), \(u\neq 0\), we define

$$ h(t)=at^{2-q} \Vert u \Vert ^{2}-bt^{4-q} \Vert u \Vert ^{4},\quad \forall \ t>0. $$

It is easy to see that \(g(t)\) attains its maximum value at \(t_{\max }=[\frac{{a(2-q)}}{{b(4-q)\|u\|^{2}}}]^{1/2}\) with

$$ h(t_{\max })=\frac{ {2(2-q)}}{{(4-q)^{2}}}\frac{{a^{\frac{{4-q}}{{2}}}}{} }{{b^{\frac{{2-q}}{{2}}}}} \Vert u \Vert ^{q} . $$

We note that, by Hölder’s inequality, for \(\lambda \in (0,T_{b})\), there holds

$$ \lambda \int _{\Omega } \vert u \vert ^{q}\,dx \le \lambda \vert \Omega \vert ^{\frac{{{2^{*}}-q}}{{2^{*}}}}S^{-q/2} \Vert u \Vert ^{q} < h(t_{\max }). $$

It follows that there are two and only two positive constants \(t^{+}=t^{+}(u)\) and \(t^{-}=t^{-}(u)\) such that

$$ h\bigl(t^{+}\bigr)=\lambda \int _{\Omega } \vert u \vert ^{q}\,dx={ {h \bigl(t^{-}\bigr)}} \quad \text{and} \quad h' \bigl(t^{+}\bigr)< 0< h'\bigl(t^{-}\bigr). $$

Equivalently, we obtain \(t^{+}u\in {{\Lambda ^{-}_{b}}}\) and \(t^{-}u\in {{\Lambda ^{+}_{b}}}\).

Next, we prove that \(\Lambda _{b}^{0}=\{0\}\). Arguing by contradiction, we assume that there exists \(w\in \Lambda _{b}^{0}\) satisfying \(w\neq 0\). Then we have \(a(2-q)\|w\|^{2}-b(4-q)\|w\|^{4}=0\). This yields \(b\|w\|^{2}=\frac{{a(2-q)}}{{4-q}}\). For \(\lambda \in (0,T_{b})\), it follows from \(w\in \Lambda _{b}\) and Hölder’s inequality that

$$\begin{aligned} 0&=a \Vert w \Vert ^{2}-b \Vert w \Vert ^{4}- \lambda \int _{\Omega } \vert w \vert ^{q}\,dx \\ &\ge \frac{{2a}}{{4-q}} \Vert w \Vert ^{2}-\lambda \vert \Omega \vert ^{\frac{{{2^{*}}-q}{} }{{2^{*}}}}S^{-q/2} \Vert w \Vert ^{q} \\ &= \Vert w \Vert ^{q} \biggl[ \frac{{2(2-q)}}{{(4-q)^{2}}} \frac{{a^{\frac{{4-q}}{{2}}}}{} }{{b^{\frac{{2-q}}{{2}}}}} \biggl(\frac{{4-q}}{{2-q}} \biggr)^{\frac{{q}{} }{{2}}} - \lambda \vert \Omega \vert ^{\frac{{{2^{*}}-q}}{{2^{*}}}}S^{\frac{-q{} }{2}} \biggr]>0, \end{aligned}$$

which makes no sense. This ends the proof. □

Lemma 2.2

Given \(u\in \Lambda _{b}^{\pm }\), there exist \(\rho _{u}>0\) and a differential function \(g_{\rho _{u}}:B_{\rho _{u}}(0)\to \mathbb{R}^{+}\) defined for \(w\in H\), \(w\in B_{\rho _{u}}(0)\) satisfying

$$ g_{\rho _{u}}(0)=1,\quad g_{\rho _{u}}(w) (u-w)\in \Lambda _{b}^{\pm }, $$

and

$$ {}\bigl\langle g'(0),\phi {}\bigr\rangle = \frac{ {(2a-4 b \Vert u \Vert ^{2}){} \int _{\Omega } \nabla u\nabla \phi \,dx -q\lambda \int _{\Omega } \vert u \vert ^{q-2}u\phi \,dx} }{{a(2-q) \Vert u \Vert ^{2}-b(4-q) \Vert u \Vert ^{4}}}. $$

Proof

We only give the proof for the case \(u\in \Lambda _{b}^{-}\). In a similar way, one can prove the other case \(u\in \Lambda _{b}^{+}\). Fix \(u\in \Lambda _{b}^{-}\) and define \(F:\mathbb{R}^{+}\times H \to \mathbb{R}\) by

$$ F(t,w)=at \Vert u-w \Vert ^{2}-b t^{3} \Vert u-w \Vert ^{4}-\lambda t^{q-1}{} \int _{\Omega } \vert u-w \vert ^{q} \,dx. $$

By \(u\in \Lambda _{b}^{-}\subset \Lambda _{b}\), we easily see that \(F(1,0)=0\) and

$$ F_{t}(1,0)=a(2-q) \Vert u \Vert ^{2}-b(4-q) \Vert u \Vert ^{4}< 0. $$

Then, we are able to use the implicit function theorem for F at the point \((1,0)\) and get \(\overline{\rho }>0\) and a differential functional \(g=g(w)>0\) defined for \(w\in H\), \(\|w\|<\overline{\rho }\) such that

$$ g(0)=1, \qquad g(w) (u-w)\in {\Lambda _{b}}, \quad \forall w\in H, \Vert w \Vert < \overline{\rho }. $$

Thanks to the continuity of g, we can take \(\rho >0\) possibly smaller (\(\rho <\overline{\rho }\)) such that, for any \(w\in H\), \(\|w\|<{\rho }\), there holds

$$ g(w) (u-w) \in \Lambda _{b}^{-}. $$

Moreover, for any \(\phi \in H\), \(r>0\), it follows from

$$\begin{aligned} &F(1,0+r\phi )-F(1,0) \\ &\quad =a \Vert u-r\phi \Vert ^{2}-b \Vert u-r\phi \Vert ^{4}-\lambda \int _{ \Omega } \vert u-r\phi \vert ^{q} \,dx-a \Vert u \Vert ^{2}+b \Vert u \Vert ^{4}+\lambda \int _{\Omega } \vert u \vert ^{p} \,dx \\ &\quad =-a{} \int _{\Omega } \bigl(2r\nabla u\nabla \phi -r^{2} \vert \nabla \phi \vert ^{2} \bigr) \,dx-\lambda \int _{\Omega } { \bigl( \vert u-r\phi \vert ^{q}- \vert u \vert ^{q} \bigr) \,dx} \\ &\qquad {}+b \biggl[2{} \int _{\Omega } \vert \nabla u \vert ^{2} \,dx{} \int _{\Omega } \bigl(2r\nabla u\nabla \phi -r^{2} \vert \nabla \phi \vert ^{2} \bigr) \,dx\\ &\qquad {}- \biggl({} \int _{\Omega } \bigl(2r\nabla u\nabla \phi -r^{2} \vert \nabla \phi \vert ^{2} \bigr) \,dx \biggr)^{2} \biggr] \end{aligned}$$

that

$$\begin{aligned} & {}\langle F_{w},\phi {}\rangle |_{t=1, w=0} \\ &\quad =\lim_{r\to 0} \frac{{F(1,0+r\phi )-F(1,0)}}{{r}} \\ &\quad =-\bigl(2a-4b \Vert u \Vert ^{2}\bigr){} \int _{\Omega } \nabla u\nabla \phi \,dx +q\lambda \int _{\Omega } { \vert u \vert ^{p-2}u\phi } \,dx. \end{aligned}$$

Consequently, we derive

$$ {}\bigl\langle g'(0),\phi {}\bigr\rangle = - \frac{{{}\langle F_{w}, \phi {}\rangle }}{{F_{t}}}|_{t=1 ,w=0} = \frac{{{(2a-4b \Vert u \Vert ^{2}){} \int _{\Omega } \nabla u\nabla \phi \,dx}-q{} \lambda \int _{\Omega } \vert u \vert ^{q-2}u\phi \,dx}}{{a(2-q) \Vert u \Vert ^{2}-b(4-q) \Vert u\Vert ^{4}}}. $$

This completes the proof. □

Lemma 2.3

If \(\lambda \in (0,T_{b})\), then we have

(i) the functional \(I_{b}\) is coercive and bounded from below on \(\Lambda _{b}\);

(ii) \(\inf_{\Lambda _{b}^{+}\cup \Lambda _{b}^{0}}I_{b}=\inf_{ \Lambda _{b}^{+}}I_{b} \in (-\infty ,0)\).

Proof

(i) For \(u\in \Lambda _{b}\), by Hölder’s inequality, we have

$$\begin{aligned} I_{b}(u)&=I_{b}(u)- \frac{1}{4}{} \bigl\langle I'_{b}(u),u {} \bigr\rangle \\ &= \frac{a}{4} \Vert u \Vert ^{2} - \lambda { \biggl( \frac{1{} }{q} - \frac{1}{4} \biggr) } {} \int _{\Omega } \vert u \vert ^{q} \,dx \\ &\ge \frac{a}{4} \Vert u \Vert ^{2}- \lambda { \biggl( \frac{1}{q} - \frac{1}{4} \biggr) } \vert \Omega \vert ^{\frac{{2^{*}-q}{} }{2^{*}}}S^{-q/2} \Vert u \Vert ^{q}, \end{aligned}$$

and the conclusion (i) follows.

(ii) For \(u\in \Lambda _{b}^{+} \), there holds

$$\begin{aligned} I_{b}(u)&=I_{b}(u)- \frac{1}{q}{} \bigl\langle I'_{b}(u),u {} \bigr\rangle \\ &=a { \biggl(\frac{1}{2} - \frac{1}{q} \biggr) } \Vert u \Vert ^{2} - b { \biggl(\frac{1}{4} - \frac{1}{q} \biggr) } \Vert u \Vert ^{4} \\ &< \frac{{-a(2-q) \Vert u \Vert ^{2}+b(4-q) \Vert u \Vert ^{4}}}{{4q}}< 0. \end{aligned}$$

This together with Lemma 2.1 gives that \(\inf_{\Lambda _{b}^{+}\cup \Lambda _{b}^{0}}I_{b}= \inf_{\Lambda _{b}^{+}}I_{b}<0\). Moreover, from (i) we infer that \(\inf_{\Lambda _{b}^{+}\cup \Lambda _{b}^{0}}I_{b} \neq -\infty \). Therefore, \(\inf_{\Lambda _{b}^{+}\cup \Lambda _{b}^{0}}I \in (-\infty ,0)\). □

Lemma 2.4

For all \(\lambda >0\), \(I_{b}\) satisfies the \((PS)_{c}\) condition at any level \(c<\frac{{a^{2}}}{{4b}}\).

Proof

The proof is similar to that of [16, Lemma 2]. We omit the details. □

3 Proof of Theorem 1.1

Lemma 3.1

Assume that \(\lambda \in (0,T_{b})\), then problem (1.1) has a positive solution \(u_{b}\) with \(u_{b} \in \Lambda _{b}^{+}\).

Proof

It is easily verified that the sets \(\Lambda _{b}^{+}\cup \Lambda _{b}^{0}\) and \(\Lambda _{b}^{-}\) are closed. Applying the Ekeland variational principle, we can derive a minimizing sequence \(\{u_{n}\}\subset \Lambda _{b}^{+}\cup \Lambda _{b}^{0}\) satisfying that

$$ \lim_{n\to \infty }I_{b}(u_{n})= \inf_{\Lambda _{b}^{+} \cup \Lambda _{b}^{0}}I_{b}< 0 {}$$
(3.1)

and

$$ I_{b}(z)\geq I_{b}(u_{n})- \frac{1}{n} \Vert z-u_{n} \Vert \quad \text{for all} \ z\in \Lambda _{b}^{+}\cup \Lambda _{b}^{0}. {}$$
(3.2)

Noting that \(I_{b}(|u|)=I_{b}(u)\), we may suppose that \(u_{n} \ge 0\) in Ω. By Lemma 2.3, \(\{u_{n}\}\) is bounded in H, and so we can assume

$$\begin{aligned}& u_{n}\rightharpoonup u_{b} \quad \text{in } H, \\& u_{n}\rightarrow u_{b} \quad \text{in } L^{s}(\Omega ), 2\leq s< 2^{*}, \\& u_{n}\rightarrow u_{b} \quad \text{a.e. in } \Omega . \end{aligned}$$

In what follows we prove that \(u_{b}\) is a positive solution to (1.1). The proof will be divided into four steps.

textbfStep 1: \(u_{b}\neq 0\).

By contradiction, we suppose that \(u_{b}=0\). Since \(u_{n}\in \Lambda _{b}^{+}\cup \Lambda _{b}^{0}\), we see that, for n large,

$$ a \Vert u_{n} \Vert ^{2}> \frac{ {4-q}}{{2-q} } b \Vert u_{n} \Vert ^{4}. $$

As a consequence, we derive

$$ I_{b}(u_{n})=\frac{{1}}{{2}}a \Vert u_{n} \Vert ^{2}-\frac{{1}}{{4}}b \Vert u_{n} \Vert ^{4}+o_{n}(1)\ge \biggl( \frac{ {4-q}}{{2(2-q)} } - \frac{1{} }{4} \biggr)b \Vert u_{n} \Vert ^{4}+o_{n}(1) >0, $$

which is a contradiction to (3.1). Thus, \(u_{b}\neq 0\).

textbfStep 2: There exists a constant \(C_{1}>0\) such that

$$ 2a \Vert u_{n} \Vert ^{2}-\lambda (4-q){} \int _{\Omega } \vert u_{n} \vert ^{q} \,dx< -C_{1}. {}$$
(3.3)

To prove that, it suffices to verify

$$ 2a\liminf_{n\to \infty } \Vert u_{n} \Vert ^{2}< \lambda (4-q){} \int _{\Omega } \vert u_{b} \vert ^{q} \,dx . $$

By \(u_{n} \in \Lambda _{b}^{+}\cup \Lambda _{b}^{0}\),

$$ 2a\liminf_{n\to \infty } \Vert u_{n} \Vert ^{2}\le \lambda (4-q){} \int _{\Omega } \vert u_{b} \vert ^{q} \,dx. $$

Suppose to the contrary that

$$ 2a\liminf_{n\to \infty } \Vert u_{n} \Vert ^{2}= \lambda (4-q){} \int _{\Omega } \vert u_{b} \vert ^{q} \,dx. $$

Then we can assume \(\|u_{n}\|^{2}\to A>0\) as \(n\to \infty \), where A satisfies

$$ \lambda \int _{\Omega } \vert u_{b} \vert ^{q} \,dx= \frac{ {2aA}}{{4-q} }. $$

Combining this with \(\{u_{n}\} \subset \Lambda _{b}\), we have

$$ 0=aA-bA^{2}- \frac{ {2aA}}{{4-q} }. $$

It follows that

$$ A= \frac{ {a(2-q)}}{{b(4-q)} }, $$

which leads to a contradiction

$$\begin{aligned} 0&< { \frac{ {2(2-q)}}{{(4-q)^{2}}}\frac{{a^{\frac{{4-q}}{{2}}}}{} }{{b^{\frac{{2-q}}{{2}}}}} \Vert u_{n} \Vert ^{q}-\lambda \int _{\Omega } \vert u_{n} \vert ^{q} \,dx} \\ &\to { \frac{ {2(2-q)}}{{(4-q)^{2}}}\frac{{a^{\frac{{4-q}}{{2}}}}{} }{{b^{\frac{{2-q}}{{2}}}}} \biggl[\frac{{a(2-q)}}{{b(4-q)}} \biggr]^{q/2} - \frac{ {2a}}{{4-q} }\frac{ {a(2-q)}}{{b(4-q)} } } \\ &= { \frac{{2a^{2}(2-q)}}{{b(4-q)^{2}}} \biggl[ \biggl(\frac{{2-q}{} }{{4-q}} \biggr)^{q/2}-1 \biggr]< 0 } \end{aligned}$$

when \(\lambda \in (0,T_{b})\). Thus, (3.3) holds.

textbfStep 3: \(I'_{b}(u_{n}) \to 0\) in \(H^{-1}\).

Let \(0 <\rho <\rho _{n} \equiv \rho _{u_{n}}\), \(g_{n}\equiv g_{u_{n}}\), where \(\rho _{u_{n}}\) and \(g_{u_{n}}\) are given as in Lemma 2.2 with \(u=u_{n}\). Let \(w_{\rho }=\rho u\) with \(\|u\|=1\). Fix n and set \(z_{\rho }=g_{n}(w_{\rho })(u_{n}-w_{\rho })\). By \(z_{\rho }\in \Lambda _{b}^{+}\), we have from (3.2) that

$$ I_{b}(z_{\rho })-I_{b}(u_{n})\ge - \frac{{1}}{{n}} \Vert z_{ \rho }-u_{n} \Vert . $$

Then, by the mean value theorem,

$$ {}\bigl\langle I'_{b}(u_{n}),z_{\rho }-u_{n} {}\bigr\rangle +o\bigl( \Vert z_{\rho }-u_{n} \Vert \bigr)\ge - \frac{{1}}{{n}} \Vert z_{\rho }-u_{n} \Vert . $$

Hence, we derive

$$ {}\bigl\langle I'_{b}(u_{n}),-w_{\rho }+ \bigl( g_{n}(w_{\rho })-1 \bigr) (u_{n}-w_{ \rho }) {}\bigr\rangle \ge - \frac{{1}}{{n}} \Vert z_{\rho }-u_{n} \Vert +o\bigl( \Vert z_{\rho }-u_{n} \Vert \bigr), $$

and thus,

$$ -\rho {}\bigl\langle I'_{b}(u_{n}),u {} \bigr\rangle + \bigl( g_{n}(w_{\rho })-1 \bigr){}\bigl\langle I'_{b}(u_{n}),u_{n}-w_{\rho } {}\bigr\rangle \ge - \frac{{1}}{{n}} \Vert z_{\rho }-u_{n} \Vert +o\bigl( \Vert z_{\rho }-u_{n} \Vert \bigr), $$

from which it follows that

$$ {}\bigl\langle I'_{b}(u_{n}),u {}\bigr\rangle \le \frac{{1}}{{n}} \frac{{ \Vert z_{\rho }-u_{n} \Vert }}{{\rho }}+ \frac{{o( \Vert z_{ \rho }-u_{n} \Vert )}}{{\rho }} + \frac{{g_{n}(w_{\rho })-1}{} }{{\rho }}{}\bigl\langle I'_{b}(u_{n}),u_{n}-w_{\rho } {}\bigr\rangle . {}$$
(3.4)

By Step 2, Lemma 2.2, and the boundedness of \(\{u_{n}\}\), one sees that

$$ \Vert z_{\rho }-u_{n} \Vert = \bigl\Vert \bigl(g_{n}(w_{\rho })-1 \bigr) (u_{n}-w_{\rho })-w_{ \rho } \bigr\Vert \le \bigl\vert g_{n}(w_{\rho })-1 \bigr\vert C_{2}+\rho $$

and

$$ \lim_{\rho \to 0} \frac{{ \vert g_{n}(w_{\rho })-1 \vert }}{{ \rho }}={}\bigl\langle g_{n}'(0),u{}\bigr\rangle \le \bigl\Vert g'_{n}(0) \bigr\Vert \le C_{3}. $$

Therefore, for fixed n, we deduce by taking \(\rho \to 0\) in (3.4) that

$$ {}\bigl\langle I'_{b}(u_{n}),u{}\bigr\rangle \le \frac{{C}}{{n}}, $$

which provides that \(I'_{b}(u_{n})\rightarrow 0\) as \(n\to \infty \).

textbfStep 4: \(u_{b}\) is a positive solution of problem (1.1) and \(u_{b}\in \Lambda _{b}^{+}\).

It follows from Step 3, Lemmas 2.3 and 2.4 that, along a subsequence, \(u_{n}\to u_{b}\) in H with \(I_{b}(u_{b})<0\) and \(I'_{b}(u_{b})=0\). Hence, \(u_{b}\ge 0\) is a weak solution to problem (1.1) satisfying \(u_{b}\in \Lambda _{b}^{+}\). The standard elliptic regularity argument and the strong maximum principle imply that \(u_{b}\) is positive. Thus we complete the proof of Lemma 3.1. □

Lemma 3.2

Assume that \(\lambda \in (0,T_{b})\), then problem (1.1) has a positive solution \(\tilde{u}_{b}\) with \(\tilde{u}_{b} \in {{\Lambda ^{-}_{b}}}\).

Proof

Similar to the proof of Lemma 3.1, one can construct a bounded and nonnegative sequence \(\{\tilde{u}_{n}\}\subset \Lambda _{b}^{-}\) satisfying that

$$\begin{aligned} (\mbox{i})&\quad \lim_{n\to \infty }I_{b}(\tilde{u}_{n})= \inf_{ \Lambda _{b}^{-}}I_{b}, \\ (\text{ii})&\quad I_{b}(z)\geq I_{b}( \tilde{u}_{n})-\frac{1}{n} \Vert z-u_{n} \Vert , \quad \text{for all } z\in \Lambda _{b}^{-}, \\ (\text{iii})&\quad \tilde{u}_{n}\rightharpoonup \tilde{u}_{b} \quad \text{in } H, \\ (\text{iv})&\quad \tilde{u}_{n}\rightarrow \tilde{u}_{b} \quad \text{in } L^{s}( \Omega ),\quad 2\leq s< 2^{*}, \\ (\mbox{v})&\quad \tilde{u}_{n}\rightarrow \tilde{u}_{b} \quad \text{a.e. in } \Omega . \end{aligned}$$

Without loss of generality, we may suppose that \(0\in \Omega \). Take a cut-off function \(\varphi (x)\in C_{0}^{\infty }(\Omega )\) satisfying \(0 \leq \varphi \leq 1\) in Ω and \(\varphi (x) \equiv 1\) near zero. Define

$$ v_{\varepsilon }(x)= \varphi (x)\frac{ { (N(N-2) )^{(N-2)/4}{ \varepsilon }^{(N-2)/2}}}{{(\varepsilon ^{2} + \vert x \vert ^{2})^{1/2}} }. $$

It is known that (see [2])

$$ \Vert v_{\varepsilon } \Vert ^{2}=S^{N/2}+O\bigl( \varepsilon ^{N-2}\bigr). $$

Firstly, we prove the following upper bound for \(\inf \nolimits _{\Lambda _{b}^{-}}I_{b}\):

$$ \inf_{\Lambda _{b}^{-}}I_{b}^{-} \leq \sup_{t>0}I_{b}(u_{b}+tv_{ \varepsilon })< {\frac{{a^{2}}}{{4b}}}, {}$$
(3.5)

where \(u_{b}\) is the first positive solution obtained in the previous subsection. By \(u_{b}\in \Lambda _{b}^{+}\) and (1.4), we easily see that \(a-b\|u_{b}\|^{2}>0\). Since \(u_{b}\) is a positive solution of (1.1), we also have

$$ 0={}\bigl\langle I_{b}'(u_{b}),tv_{\varepsilon }{} \bigr\rangle =t \bigl( a-b \Vert u_{b} \Vert ^{2} \bigr){} \int _{\Omega } \nabla u_{b}\nabla {v_{ \varepsilon }}\,dx-t\lambda \int _{\Omega } u_{b}^{q-1}v_{ \varepsilon } \,dx, {}$$
(3.6)

from which it follows that

$$ {} \int _{\Omega } \nabla u_{b}\nabla {v_{\varepsilon }}\,dx= \frac{{\lambda \int _{\Omega } u_{b}^{q-1}v_{\varepsilon }\,dx}{} }{{a-b \Vert u_{b} \Vert ^{2}}}>0. {}$$
(3.7)

To proceed, set \(w_{\varepsilon }=u_{b}+Rv_{\varepsilon }\) with \(R>1\). By (3.7), we have

$$ \Vert w_{\varepsilon } \Vert ^{2}= \Vert u_{b} \Vert ^{2}+2R{} \int _{\Omega }{} \nabla u_{b}\nabla v_{\varepsilon }\,dx +R^{2} \Vert v_{\varepsilon } \Vert ^{2} \ge \Vert u_{b} \Vert ^{2}+R^{2}S^{3/2}+O( \varepsilon ). {}$$
(3.8)

Let \(h(t)\) be defined as in Lemma 2.1. As can be seen from the proof of Lemma 2.1, we have that \(h(t_{\varepsilon })=\lambda \int _{\Omega }|{\frac{w_{\varepsilon }{} }{{\|w_{\varepsilon }\|}}}|^{q}\,dx\) and \(h'(t_{\varepsilon })<0\), where \(t_{\varepsilon }=t^{+} ({\frac{w_{\varepsilon }}{{\|w_{\varepsilon } \|}}} )\). From the structure of h and \(\int _{\Omega }|{\frac{w_{\varepsilon }}{{\|w_{\varepsilon }\|}}}|^{q}\,dx>0\), it follows that \(t_{\varepsilon }\) is uniformly bounded by a suitable positive constant \(C_{1}\), \(\forall R\ge 1\) and \(\forall \varepsilon >0\).

On the other hand, we can infer from (3.8) that there exists \(\varepsilon _{1}>0\) such that

$$ \Vert w_{\varepsilon } \Vert ^{2}\ge \Vert u_{b} \Vert ^{2}+\frac{1}{2}R^{2}S^{3/2}, \quad \forall \varepsilon \in (0,\varepsilon _{1}). $$

Thus, we can find \(R_{1}\ge 1\) such that \(\|w_{\varepsilon }\|>C_{1}\), \(\forall R\ge R_{1}\), and \(\forall \varepsilon \in (0,\varepsilon _{1})\).

Let

$$ E_{1}= \biggl\{ u: u=0 \text{ or } \Vert u \Vert < t^{+} \biggl(\frac{{u}}{{ \Vert u \Vert }} \biggr) \biggr\} \quad \text{and} \quad E_{2}= \biggl\{ u: \Vert u \Vert >t^{+} \biggl(\frac{{u}}{{ \Vert u \Vert }} \biggr) \biggr\} . $$

Note that \(H-\Lambda _{b}^{-}=E_{1}\cup E_{2}\) and \(\Lambda _{b}^{+} \subset E_{1}\). Since \(u_{b} \in \Lambda _{b}^{+}\), by the continuity of \(t^{+}(u)\), one sees that \(u_{b}+tR_{1}v_{\varepsilon }\) for \(t\in (0,1)\) must intersect \(\Lambda _{b}^{-}\), and consequently

$$ \inf_{\Lambda _{b}^{-}}I_{b}\leq \sup_{t>0}I_{b}(u_{b}+tv_{ \varepsilon }). $$

Hence (3.5) will follow if we show that

$$ \sup_{t>0}I_{b}(u_{b}+tv_{\varepsilon })< { \frac{{a^{2}}}{{4b}}}. $$

By the mean value theorem, we can get \(\delta (x)\in [0,1]\) satisfying

$$ \bigl(u_{b}(x)+tv_{\varepsilon }(x) \bigr)^{q}-u_{b}^{q}(x)=q \bigl(u_{b}(x)+ \delta (x)tv_{\varepsilon }(x) \bigr)^{q-1}tv_{\varepsilon }(x)\ge qtu_{b}^{q-1}(x)v_{ \varepsilon }(x) {}$$
(3.9)

for any \(x\in \Omega \). By (3.6), (3.7), and (3.9),

$$\begin{aligned} &I_{b}(u_{b}+tv_{\varepsilon }) \\ &\quad = { \frac{{a}}{{2}} \Vert u_{b} \Vert ^{2}+at{} \int _{\Omega }\nabla u_{b}\nabla {v_{\varepsilon }}\,dx+\frac{{a}}{{2}}t^{2} \Vert v_{\varepsilon } \Vert ^{2}-\frac{{b}}{{4}} \Vert u_{b} \Vert ^{4}-b t^{2} \biggl({} \int _{\Omega }\nabla u_{b}\nabla {v_{\varepsilon }}\,dx \biggr)^{2} } \\ & \qquad {} -\frac{{b}}{{4}}t^{4} \Vert v_{\varepsilon } \Vert ^{4} -b t \Vert u_{b} \Vert ^{2}{} \int _{\Omega }\nabla u_{b}\nabla {v_{ \varepsilon }}\,dx -\frac{{b}}{{2}}t^{2} \Vert u_{b} \Vert ^{2} \Vert v_{\varepsilon } \Vert ^{2} \\ &\qquad {} -b t^{3} \Vert v_{\varepsilon } \Vert ^{2} {} \int _{\Omega }{} \nabla u_{b}\nabla {v_{\varepsilon }}\,dx -\frac{{\lambda }}{{q}}{} \int _{\Omega } (u_{b}+tv_{\varepsilon })^{q} \,dx \\ &\quad \le I_{b}(u_{b})+ { \frac{{a}}{{2}}t^{2} \Vert v_{ \varepsilon } \Vert ^{2} -\frac{{b}}{{4}}t^{4} \Vert v_{\varepsilon } \Vert ^{4}} - { \frac{{b}}{{2}}t^{2} \Vert u_{b} \Vert ^{2} \Vert v_{\varepsilon } \Vert ^{2}} \\ &\qquad {} - {\frac{{\lambda }}{{q}}{} \int _{ \Omega } \bigl[ (u_{b}+tv_{\varepsilon })^{q}-u_{b}^{q}-qtu_{b}^{q-1}v_{ \varepsilon } \bigr]\,dx} \\ &\quad \le { \frac{{a}}{{2}}t^{2} \Vert v_{\varepsilon } \Vert ^{2} -\frac{{b}{} }{{4}}t^{4} \Vert v_{\varepsilon } \Vert ^{4}} - { \frac{{b}}{{2}}t^{2} \Vert u_{b} \Vert ^{2} \Vert v_{\varepsilon } \Vert ^{2}}, \end{aligned}$$

which implies that there exists \(t_{1}>0\) small enough such that

$$ \sup_{0< t< t_{1}}I_{b}(u_{b}+tv_{\varepsilon })< { \frac{{a^{2}}}{{4b}}}. $$

Thus, we only need to consider the case of \(t\ge t_{1}\). Since

$$\begin{aligned} \sup_{t\ge t_{1}}I_{b}(u_{b}+tv_{\varepsilon })& \le { \sup_{t>0} \biggl\{ \frac{{a}}{{2}}t^{2} \Vert v_{ \varepsilon } \Vert ^{2} -\frac{{b}}{{4}}t^{4} \Vert v_{\varepsilon } \Vert ^{4} \biggr\} - \frac{{b}}{{2}}t_{1}^{2} \Vert u_{b} \Vert ^{2} \Vert v_{\varepsilon } \Vert ^{2} } \\ &= \frac{ {a^{2}}}{{4b} }-\frac{{b}}{{2}}t_{1}^{2} \Vert u_{b} \Vert ^{2} \Vert v_{\varepsilon } \Vert ^{2}< \frac{ {a^{2}}}{{4b} }, \end{aligned}$$

we deduce that (3.5) holds.

Secondly, we claim that \(\tilde{u}_{b}\neq 0\). If, to the contrary, \(\tilde{u}_{b}=0\), from \(\{\tilde{u}_{n}\}\subset \Lambda _{b}^{-}\) it then follows that

$$ a \Vert \tilde{u}_{n} \Vert ^{2}-b \Vert \tilde{u}_{n} \Vert ^{4}+o_{n}(1)=0. $$

As a consequence, we obtain \(\|\tilde{u}_{n}\|^{2}\to \frac{a}{b}\) as \(n\to \infty \). Furthermore,

$$ \inf_{\Lambda _{b}^{-}}I_{b}=\lim_{n\to \infty }I_{b}( \tilde{u}_{n})=\lim_{n\to \infty } { \biggl[ \frac{a}{2} \Vert \tilde{u}_{n} \Vert ^{2}- \frac{b}{4} \Vert \tilde{u}_{n} \Vert ^{4}-\lambda \int _{\Omega } \vert \tilde{u}_{n} \vert ^{q}\,dx \biggr]=\frac{a^{2}{} }{{4b}} }, $$

which contradicts (3.5). Hence, the claim holds. This time we can proceed as in the proof of Lemma 3.1 and deduce that \(\tilde{u}_{b}\) is a positive solution of problem (1.1) with \(\tilde{u}_{b}\in \Lambda _{b}^{-}\). The proof is complete. □

Proof of Theorem 1.1

This is an immediate consequence of (1.4), Lemmas 3.1 and 3.2. □

4 Proof of Theorem 1.2

Proof of Theorem 1.2

For any sequence \(\{b_{n}\}\) with \(b_{n}\searrow 0\), we can use Theorem 1.1 to obtain sequences \(\{u_{b_{n}}\}\subset {{\Lambda _{b_{n}}^{+}}}\) and \(\{\tilde{u}_{b_{n}}\} \subset {{\Lambda _{b_{n}}^{-}}}\) corresponding to positive solutions to problem (1.1) with \(b=b_{n}\) when \(\lambda \in (0,T_{b_{n}})\).

By \(\tilde{u}_{b_{n}} \in {{\Lambda _{b_{n}}^{-}}}\) and (1.4), we see that

$$ \lim_{n\to \infty } \Vert \tilde{u}_{b_{n}} \Vert ^{2}\ge \lim_{n\to \infty }\frac{{a(2-q)}{} }{{b_{n}(4-q)}}=\infty , $$

and conclusion (i) follows.

Next, we prove conclusion (ii) of Theorem 1.2. Note that

$$ I_{b}(u_{b_{n}})=\inf_{\Lambda _{b_{n}}^{+}\cup \Lambda _{b_{n}}^{0}}I_{b_{n}}< 0 $$

for all \(n\in \mathbb{N}\). Then, by Hölder’s inequality, we have

$$\begin{aligned} 0&\ge I_{b_{n}}(u_{b_{n}}) - \frac{{1}}{{4}}{}\bigl\langle I'_{b_{n}}(u_{b_{n}}),u_{b_{n}} {}\bigr\rangle \\ &\ge \biggl( \frac{1}{2} - \frac{1}{4} \biggr) \Vert u_{b_{n}} \Vert ^{2}- \lambda \biggl( \frac{1}{q} - \frac{1}{4} \biggr) \vert \Omega \vert ^{\frac{{{2^{*}}-q}}{{2^{*}}}}S^{\frac{{-q}}{2}} \Vert u_{b_{n}} \Vert ^{q}. \end{aligned}$$

Since \(1< q<2\), it follows that \(\{u_{b_{n}}\}\) is bounded in H. As a consequence, there exists a subsequence of \(\{b_{n}\}\) (still denoted by \(\{b_{n}\}\)) such that \(u_{b_{n}}\rightharpoonup u_{0}\) in H as \(n\to \infty \). Furthermore, we have that, for all \(v\in H\),

$$\begin{aligned} 0&=\lim_{n\to \infty } \bigl\langle I'_{b_{n}}(u_{b_{n}}),v {} \bigr\rangle \\ &=\lim_{n\to \infty } \biggl[ \bigl(a-b_{n} \Vert u_{b_{n}} \Vert ^{2}\bigr){} \int _{\Omega }\nabla u_{b_{n}} \nabla v\,dx - {} \lambda \int _{\Omega } u_{b_{n}}^{q-1}v\,dx \biggr] \\ &=a{} \int _{\Omega }\nabla u_{0} \nabla v\,dx - {} \lambda \int _{\Omega } u_{0}^{q-1}v\,dx, \end{aligned}$$

which implies that \(u_{0}\) is a positive solution to problem (1.5). To complete the proof, we only need to show that \(u_{b_{n}}\to u_{0}\) in H. This follows easily from

$$\begin{aligned} &a \Vert u_{b_{n}}-u_{0} \Vert ^{2} \\ &\quad =\bigl\langle I_{b_{n}}'(u_{b_{n}})-I_{0}'(u_{0}),u_{b_{n}}-u_{0} {} \bigr\rangle + b_{n} \int _{\Omega } \vert \nabla u_{b_{n}} \vert ^{2}\,dx{} \int _{\Omega }\nabla u_{b_{n}}\nabla (u_{b_{n}}-u_{0})\,dx \\ &\qquad {} +\lambda \int _{\Omega } \bigl(u_{b_{n}}^{q-1}-u_{0}^{q-1} \bigr) (u_{b_{n}}-u_{0})\,dx \\ &\quad \to 0, \end{aligned}$$

as \(n\to \infty \). Theorem 1.2 is thus proved. □