1 Introduction

This paper deals with the existence of normalized solutions to the following nonautonomous Kirchhoff-type equation:

$$\begin{aligned} \left\{ \begin{array}{ll} -\left( a+b\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) \Delta u-\lambda u=K(x)f(u), &{} x\in {\mathbb {R}}^3; \\ u\in H^1({\mathbb {R}}^3), \end{array} \right. \end{aligned}$$
(1.1)

where ab are positive real numbers, \(\lambda \) is unknown and will appear as a Lagrange multiplier, \(K\in {\mathcal {C}}({\mathbb {R}}^3,{\mathbb {R}}^+)\) and \(f\in {\mathcal {C}}({\mathbb {R}},{\mathbb {R}})\). This equation is related to the stationary analogue of the Kirchhoff equation

$$\begin{aligned} u_{tt}-\left( a+b\int _{{\mathbb {R}}^3}|\nabla u|^{2}{\mathrm {d}}x\right) \Delta u=g(x,t), \end{aligned}$$
(1.2)

The Kirchhoff equation has been introduced for the first time in 1883 by Kirchhoff [16] in dimension 1, without forcing term and with Dirichlet boundary conditions, in order to describe the transversal free vibrations of a clamped string in which the dependence of the tension on the deformation cannot be neglected. This is a quasi-linear partial differential equation, namely the nonlinear part of the equation contains as many derivatives as the linear differential operator. The Kirchhoff equation is an extension of the classical D’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. We refer to [1, 5, 6, 12] for the physical background on Kirchhoff’s model.

From the mathematical point of view, problem (1.1) is nonlocal since the appearance of the term \(\int _{{\mathbb {R}}^3}|\nabla u|^{2}{\mathrm {d}}x\) indicates that (1.1) is not a pointwise identity. This kind of problem has been paid much attention after the pioneering work of Lions [19], in which an abstract functional analysis framework was introduced.

If \(\lambda \in {\mathbb {R}}\) is a fixed parameter or even in the presence of an additional external and fixed potential V(x), the existence of solutions of problem (1.1) has been intensively studied during the last decade; see, for example, [2, 7, 13, 17, 18, 20,21,22, 24, 32] and the references therein. In this case, solutions can be obtained as critical points of the corresponding energy functional, but without any information on the \(L^2\)-norm of the solutions.

Nowadays, since physicists are interested in normalized solutions, mathematical researchers began to focus on solutions having a prescribed \(L^2\)-norm, that is, solutions which satisfy \(\Vert u\Vert _2^2=c>0\) for a priori given c. To the best of our knowledge, the study of solutions with prescribed norm was initiated by Jeanjean [14] in the framework of semilinear elliptic equations. We also refer to Bellazzini et al. [4] and Cingolani and Jeanjean [11] for normalized solutions of the Schrödinger–Poisson system. In the present paper, we are interested in the existence of solutions with \(L^2\)-prescribed norm and their qualitative properties in the framework of nonlocal Kirchhoff problems. Our analysis includes both the \(L^2\)-supercritical case and the \(L^2\)-subcritical growth.

Solutions of problem (1.1) with \(\Vert u\Vert _2^2=c>0\) can be obtained by looking for critical points of the following functional

$$\begin{aligned} I(u) = \frac{a}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x+\frac{b}{4} \left( \int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) ^2 -\int _{{\mathbb {R}}^3}K(x)F(u){\mathrm {d}}x \end{aligned}$$
(1.3)

on the manifold

$$\begin{aligned} {\mathcal {S}}_{c}=\left\{ u\in H^1({\mathbb {R}}^3):\Vert u\Vert _2^2=c\right\} , \end{aligned}$$
(1.4)

where \(F(u)=\int _0^uf(t){\mathrm {d}}t\). In this case, the parameter \(\lambda \in {\mathbb {R}}\) cannot be fixed but instead it appears as a Lagrange multiplier, and each critical point \(u_c\in {\mathcal {S}}_{c}\) of \(I|_{{\mathcal {S}}_{c}}\), corresponds to a Lagrange multiplier \(\lambda _c\in {\mathbb {R}}\) such that \((u_c, \lambda _c)\) solves (weakly) problem (1.1). In particular, if \(u_c\in {\mathcal {S}}_{c}\) is a solution of the constrained minimization problem

$$\begin{aligned} \sigma (c):=\inf _{u\in {\mathcal {S}}_{c}}I(u), \end{aligned}$$
(1.5)

then there exists \(\lambda _c\in {\mathbb {R}}\) such that \(I'(u_c)=\lambda _c u_c\), that is, \((u_c, \lambda _c)\) is a solution of problem (1.1).

To the best of our knowledge, solutions of problem (1.1) having a prescribed \(L^2\)-norm have been studied only if \(K(x)\equiv 1\); see, e.g., [28,29,30,31, 33]. Let us introduce and review the few known results in this respect. Ye [29] studied the existence and non-existence of normalized solutions to the special form of (1.1):

$$\begin{aligned} \left\{ \begin{array}{ll} -\left( a+b\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) \Delta u-\lambda u=|u|^{p-2}u, &{} x\in {\mathbb {R}}^3; \\ u\in H^1({\mathbb {R}}^3) \end{array} \right. \end{aligned}$$
(1.6)

for \(p\in (2,6)\), and showed that \(p=\frac{14}{3}\) is a \(L^2\)-critical exponent for problem (1.6), that is, for any given \(c>0\),

$$\begin{aligned} \left\{ \begin{array}{ll} \sigma (c)\in (-\infty ,0], &{} \text{ if }\ p\in (2,\frac{14}{3}); \\ \sigma (c) = -\infty , &{} \text{ if }\ p\in (\frac{14}{3},6). \end{array} \right. \end{aligned}$$
(1.7)

More precisely, for any \(p\in (2,\frac{14}{3})\), Ye [29] obtained the sharp existence of global constraint minimizers for (1.6) by solving the minimization problem (1.5). If \(p\in (\frac{10}{3},\frac{14}{3})\), Ye [29] found a local minimizer which is also a critical point of \(I|_{{\mathcal {S}}_{c}}\) by constructing a geometry of local minima for I. For the case \(p\in (\frac{14}{3},6)\), since the minimization problem (1.5) is not available due to (1.7), to look for a critical point of \(I|_{{\mathcal {S}}_{c}}\), Ye [29] took a minimum on a suitable submanifold

$$\begin{aligned} {\mathcal {M}}_c:=\left\{ u\in {\mathcal {S}}_c: J(u):=\frac{{\mathrm {d}}}{{\mathrm {d}}t}I(u^t)\Big |_{t=1}=0\right\} , \end{aligned}$$
(1.8)

and showed that \({\mathcal {M}}_c\) is a natural constraint of \(I|_{{\mathcal {S}}_{c}}\) by using the Lagrange multiplier method, where

$$\begin{aligned} u^t(x):=t^{3/2}u(tx), \ \ \ \ \forall \ t>0, u\in H^1({\mathbb {R}}^3), \end{aligned}$$
(1.9)

and \(u^t\in {\mathcal {S}}_{c}\) for all \(t>0\) if \(u\in {\mathcal {S}}_{c}\). Note that this method relies heavily on the power-type nonlinearity \(f(u)=|u|^{p-2}u\) with \(p\in (\frac{14}{3},6)\). Recently, in [28], Xie and Chen generalized this special case to general nonlinearities f satisfying \(\lim _{|t|\rightarrow \infty }\frac{F(t)}{|t|^{14/3}}=+\infty \). Using some ideas developed in [24, 29], Xie and Chen [28] proved the existence of normalized solutions for problem (1.1) with \(K(x)\equiv 1\) under additional growth and monotonicity conditions. For the \(L^2\)-critical case, Ye [30] proved that problem (1.6) with \(p=\frac{14}{3}\) has a solution \((u_c,\lambda _c)\) which satisfies \(\Vert u_c\Vert _2^2=c>c^*\) for some \(c^*>0\). We also recall that Ye [31] also analyzed the concentration behavior of solutions. If \(p\in (2,\frac{14}{3})\), by using a different method, Zeng and Zhang [33] proved the existence and uniqueness of normalized solutions for problem (1.6). Additionally, it was considered in [29,30,31, 33] the existence of normalized solutions for one-dimensional and two-dimensional autonomous Kirchhoff type equations with power-type nonlinearity.

Let us also emphasize that all of the strategies used in [28,29,30,31, 33] only work for autonomous problems, and fail to adapt directly to problem (1.1) with non-constant potential K(x). To the best of our knowledge, there are no results dealing with the non-autonomous abstract setting. The main purpose of this paper is to extend and complement the corresponding existence results in [29] to problem (1.1) in the presence of the variable potential K(x).

2 Main Results

Motivated by the above works, we first consider the \(L^2\)-supercritical case, and establish the existence of a critical point of I on \({\mathcal {S}}_c\) by considering the constrained minimization problem

$$\begin{aligned} m(c):=\inf _{u\in {\mathcal {M}}_c}I(u), \end{aligned}$$
(2.1)

where the definition of \({\mathcal {M}}_c\) is given by (1.8). To this end, we introduce the following assumptions:

  1. (F1)

    \(f\in {\mathcal {C}}({\mathbb {R}}, {\mathbb {R}})\), there exists \(\mu \in (\frac{14}{3},6)\) such that \(0\le f(t)t\le \mu F(t)\) for all \(t\in {\mathbb {R}}\), and \(\mathrm {meas}\{t\in {\mathbb {R}}: \mu F(t)-f(t)t=0\}=0\);

  2. (F2)

    there exists \(\theta \in (2,\frac{14}{3})\) such that \(\lim _{|t|\rightarrow 0}\frac{F(t)}{|t|^\theta }=0\) and \(\lim _{|t|\rightarrow \infty }\frac{F(t)}{|t|^{14/3}}=+\infty \);

  3. (F3)

    the mapping \(t\mapsto [f(t)t-\theta F(t)]/|t|^{11/3}t\) is nondecreasing on \((-\infty ,0)\) and \((0,+\infty )\);

  4. (K1)

    \(K\in {\mathcal {C}}({\mathbb {R}}^3, {\mathbb {R}}^+)\) and \(0<K_{\infty }:=\lim _{|y|\rightarrow \infty }K(y)\le K(x)\) for all \(x\in {\mathbb {R}}^3\);

  5. (K2)

    \(K\in {\mathcal {C}}^1({\mathbb {R}}^3, {\mathbb {R}}^+)\), \((6-\mu )K(x)+2\nabla K(x)\cdot x\ge 0\) for all \(x\in {\mathbb {R}}^3\), and

    \(3(\theta -2)K(tx)-2\nabla K(tx)\cdot (tx)\) is nonincreasing on \(t\in (0,\infty )\) for every \(x\in {\mathbb {R}}^3\).

Our first result establishes the following qualitative property.

Theorem 2.1

Assume that (K1), (K2) and (F1)–(F3) hold. Then for any \(c>0\), problem (1.1) has a couple of solution \(({\bar{u}}_c, \lambda _c)\in {\mathcal {S}}_{c}\times {\mathbb {R}}^-\) such that

$$\begin{aligned} {I}({\bar{u}}_c)=\inf _{u\in {\mathcal {M}}_c}{I}(u) =\inf _{u\in {\mathcal {S}}_{c}}\max _{t> 0}{I}(u^t)>0, \end{aligned}$$

where the definitions of \({\mathcal {M}}_c\) and \(u^t\) are given by (1.8) and (1.9).

Next, in the \(L^2\)-subcritical case, we find a global minimizer and a local minimizer of I which are critical points of \(I|_{{\mathcal {S}}_{c}}\) by solving the minimization problem (1.5) and constructing a geometry of local minima for I (see Lemma 4.6), respectively. To this end, in addition to (K1), we introduce the following assumptions:

  • (F4) \(f\in {\mathcal {C}}({\mathbb {R}}, {\mathbb {R}})\), \(f(t)=o(t)\) as \(t\rightarrow 0\) and there exist constants \({\mathcal {C}}>0\) and \(p\in (\frac{10}{3},\frac{14}{3})\) such that \(|f(t)|\le {\mathcal {C}}(1+|t|^{p-1})\);

  • (F5) there exists \(\mu _0\in (2, \frac{14}{3})\) such that \(f(t)t\ge \mu _0 F(t)> 0\) for all \(t\in {\mathbb {R}}\setminus \{0\}\);

  • (F6) there exists \(q_0\in (2,\frac{10}{3})\) such that \(\lim _{|t|\rightarrow 0}\frac{F(t)}{|t|^{q_0}}>0\);

  • (\(\hbox {F6}'\)) \(\lim _{|t|\rightarrow 0}\frac{F(t)}{|t|^{10/3}}=0\);

  • (K3) the mapping \(t\mapsto t^{(\mu _0-2)/2}K(tx)\) is nondecreasing on \( (0,\infty )\) for every \(x\in {\mathbb {R}}^3\).

Let

$$\begin{aligned} c_*:=\inf \left\{ c\in (0,+\infty ), \sigma (c)<0\right\} . \end{aligned}$$
(2.2)

We have the following statement.

Theorem 2.2

Assume that K and f satisfy (K1), (K3), (F4) and (F5).

  1. (i)

    If (F6) holds, then \(c_*=0\) and I admits a critical point \(u_c\) on \({\mathcal {S}}_c\), which is a negative global minimum of I when \(c>0\).

  2. (ii)

    If (\(\hbox {F6}'\)) holds, then \(c_*>0\), and there exists \(c_0\in (0, c_*)\) such that I admits a critical point \(u_c\) on \({\mathcal {S}}_{c}\) which is a local minimum of I when \(c\in (c_0,c_*)\), but \(u_c\) is a global minimum of I when \(c\in [c_*,+\infty )\). In particular,

    $$\begin{aligned} I(u_c) \left\{ \begin{array}{lll} >0, &{} {\mathrm{if}}\ c\in (c_0,c_*); \\ =0, &{} {\mathrm{if}}\ c=c_*; \\ <0, &{} {\mathrm{if}}\ c\in [c_*,\infty ). \end{array} \right. \end{aligned}$$
    (2.3)

Moreover, for the above critical point \(u_c\), there is a Lagrange multiplier \(\lambda _c\in {\mathbb {R}}\) such that \((u_c,\lambda _c)\) is a solution of problem (1.1).

Remark 2.3

Theorems 2.1 and 2.2 make a substantial improvement and extension to the main results in [29]. In particular, if \(K(x)\equiv 1\), the conclusion of Theorem 2.1 holds under hypotheses (F1)–(F3) with \(\theta =2\), and it reduces to the result of [28, Theorem 1.1].

Compared with the previous works, we have to overcome the essential difficulties that the variable potential K(x) gives rise when searching for normalized solutions of problem (1.1). These difficulties enforce the implementation of new ideas and techniques for the proof of Theorems 2.1 and 2.2. Let us point them out in more detail.

For the \(L^2\)-supercritical case, some useful remarks are stated in what follows.

  • When \(K(x)\equiv K_{\infty }\), Ye [29] showed that

    $$\begin{aligned} c\mapsto m(c)\ {\mathrm{is \ strictly\ decreasing\ on}}\ (0,+\infty ) \end{aligned}$$
    (2.4)

    by using the translation invariance of I and the homogeneity of f. Then Ye can exclude the vanishing and the dichotomy cases of the minimizing sequence \(\{u_n\}\) for \(m(c)=\inf _{u\in {\mathcal {M}}_c}{I}(u)\) in applying the concentration-compactness principle. However, the approach used in [29] is valid only for autonomous equations and it does not work any more for (1.1) with \(K\ne \) constant and more general f, see Remark 3.10 for more details. Unlike [29], by establishing some new inequalities, we prove that m(c) is nonincreasing and \(m(c)>m({\tilde{c}})\) for any \({\tilde{c}}>c\) provided m(c) is attained. To bypass the difficulty caused by the lack of compactness of Sobolev embedding \(H^1({\mathbb {R}}^3)\hookrightarrow L^s({\mathbb {R}}^3)\) for \(2\le s<6\), we compare the constrained minimum m(c) with the one of the “limit equation” (that is, problem (1.1) with \(K(x)=K_{\infty }\)), and by using some subtle analysis we prove that \(u_n\rightarrow {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\) (after a translation and extraction of a subsequence if necessary) as long as \(m(c)<m^{\infty }(c)\), and \({\bar{u}}\in {\mathcal {S}}(c)\) is a minimizer of m(c), where the definition of \(m^{\infty }(c)\) is given by (2.10).

  • To verify that \({\mathcal {M}}_c\) is a natural constraint on \({\mathcal {S}}_c\), we use a combination of the deformation lemma, some new inequalities and an intermediary theorem for continuous functions, other than the mountain pass theorem on \({\mathcal {S}}_c\) and the Lagrange multiplier method used in [28, 29], respectively.

For the \(L^2\)-subcritical case, some remarks are as follows.

  • The key step to prove that \(\sigma (c)=\inf _{{\mathcal {S}}_c}I\) is achieved is to obtain the subadditivity inequality

    $$\begin{aligned} \sigma (c)<\sigma (\alpha )+\sigma (c-\alpha ), \ \ \ \ \forall \ 0<\alpha <c. \end{aligned}$$
    (2.5)

    To this end, Ye in [29] used the scaling \(t\mapsto u(t^{-2/3}x)\). But this kind of scaling is not suitable in our case, excepting the case when K(x) is a positive constant. Instead, we present another scaling \(t\mapsto t^{1/2}u(x/t)\), and we succeed to prove that (2.5) still holds under assumptions of Theorem 2.2.

  • Compared with [29], it is more complicated to find a local minimizer of I on \({\mathcal {S}}_c\) which is a critical point of \(I|_{{\mathcal {S}}_{c}}\) because K(x) is variable and f is non-homogeneous. For this purpose, we make some improvements of the method used in [29] and employ some subtle analysis in the proofs.

When \(K\in {\mathcal {C}}({\mathbb {R}}^3, {\mathbb {R}}^+)\) is bounded, f satisfies (F1) and (F2) (or (F4)), we deduce by a standard argument that \(I\in {\mathcal {C}}^1(H^1({\mathbb {R}}^3),{\mathbb {R}})\). Let us define the “limit equation” associated to problem (1.1) by

$$\begin{aligned} \left\{ \begin{array}{ll} -\left( a+b\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) \Delta u-\lambda u=K_{\infty }f(u), &{} x\in {\mathbb {R}}^3; \\ u\in H^1({\mathbb {R}}^3). \end{array} \right. \end{aligned}$$
(2.6)

Corresponding to (1.3), (1.5), (1.8) and (2.1), we define

$$\begin{aligned} I^{\infty }(u)= & {} \frac{a}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x+\frac{b}{4} \left( \int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) ^2 -\int _{{\mathbb {R}}^3}K_{\infty }F(u){\mathrm {d}}x, \end{aligned}$$
(2.7)
$$\begin{aligned} \sigma ^{\infty }(c)= & {} \inf _{u\in {\mathcal {S}}_{c}}I^{\infty }(u), \end{aligned}$$
(2.8)
$$\begin{aligned} {\mathcal {M}}_c^{\infty }= & {} \left\{ u\in {\mathcal {S}}_c: J^{\infty }(u): =\frac{{\mathrm {d}}}{{\mathrm {d}}t}I^{\infty }(u^t)\Big |_{t=1}=0\right\} \end{aligned}$$
(2.9)

and

$$\begin{aligned} m^{\infty }(c)=\inf _{u\in {\mathcal {M}}_c^{\infty }}{I}^{\infty }(u). \end{aligned}$$
(2.10)

Remark that all above conclusions on problem (1.1) in this paper are also true for the limit equation (2.6), since \(K(x)\equiv K_{\infty }\) satisfies (K1)–(K3).

Let \(a>0\) and \(b> 0\) be fixed. Throughout this paper we make use of the following notations:

\(\bullet \)\(H^1({\mathbb {R}}^3)\) denotes the usual Sobolev space equipped with the inner product and norm

$$\begin{aligned} (u, v)=\int _{{\mathbb {R}}^3}(\nabla u\cdot \nabla v+uv){\mathrm {d}}x, \ \ \Vert u\Vert =(u, u)^{1/2}, \ \ \forall \ u,v\in H^1({\mathbb {R}}^3); \end{aligned}$$

\(\bullet \)\(L^s({\mathbb {R}}^3)\ (1\le s< \infty )\) denotes the Lebesgue space with the norm \(\Vert u\Vert _s =\left( \int _{{\mathbb {R}}^3}|u|^s{\mathrm {d}}x\right) ^{1/s}\);

\(\bullet \) For any \(u\in H^1({\mathbb {R}}^3)\), \(u^t(x):=t^{3/2}u(tx)\) and \(u_t(x):=t^{1/2}u(x/t)\);

\(\bullet \) For any \(x\in {\mathbb {R}}^3\) and \(r>0\), \(B_r(x):=\{y\in {\mathbb {R}}^3: |y-x|<r \}\);

\(\bullet \)\(S=\inf _{u\in {\mathcal {D}}^{1,2}({\mathbb {R}}^3)\setminus \{0\}}\Vert \nabla u\Vert _2^2/\Vert u\Vert _6^2\);

\(\bullet \)\(C_1, C_2,\ldots \) denote positive constants possibly different in different places.

3 First Existence Result

In this section, we give the proof of Theorem 2.1.

Lemma 3.1

Assume that hypotheses (K1) and (K2) hold. Then

$$\begin{aligned}&h_0(x,t) := t^{\frac{3(\theta -2)}{2}}[K(t^{-1}x)-K(x)]\nonumber \\&\quad -\frac{2\left( 1-t^{\frac{3(\theta -2)}{2}}\right) }{3(\theta -2)} \nabla K(x)\cdot x \ge 0, \ \ \ \ \forall \ x\in {\mathbb {R}}^3,\ t>0, \end{aligned}$$
(3.1)
$$\begin{aligned}&t\mapsto K(tx) \ \ \text{ is } \text{ nonincreasing } \text{ on } \ (0, \infty )\ \text{ for } \text{ every }\ x\in {\mathbb {R}}^3, \end{aligned}$$
(3.2)

and

$$\begin{aligned} |\nabla K(x)\cdot x|\rightarrow 0 \ \ \text{ as } \ |x|\rightarrow \infty . \end{aligned}$$
(3.3)

Proof

For any \(x\in {\mathbb {R}}^3\), by (K2), we have

$$\begin{aligned} \begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}}t}h_0(x,t)&= \frac{t^{\frac{3(\theta -2)}{2}-1}}{2}\left[ 3(\theta -2)K(t^{-1}x)-2\nabla K(t^{-1}x)\cdot (t^{-1}x)\right. \\&\quad \left. -3(\theta -2)K(x)+2\nabla K(x)\cdot x\right] \\&\left\{ \begin{array}{ll} \ge 0, \ &{} t\ge 1,\\ \le 0, \ &{} 0<t<1, \end{array}\right. \end{aligned} \end{aligned}$$

which implies that \(h_0(x,t)\ge h_0(x,1)=0\) for all \(x\in {\mathbb {R}}^3\) and \(t> 0\), hence relation (3.1) holds. By (3.1) and the continuity of \(h_0(x,\cdot )\), we have

$$\begin{aligned} \lim _{t\rightarrow 0}h_0(x,t)=-\frac{2}{3(\theta -2)}\nabla K(x)\cdot x\ge 0, \ \ \ \ \forall \ x\in {\mathbb {R}}^3, \end{aligned}$$
(3.4)

which leads to (3.2). Since \(h_0(x,2)\ge 0\) for all \(x\in {\mathbb {R}}^3\), we have

$$\begin{aligned} 0\le -\nabla K(x)\cdot x\le \frac{2^{\frac{3\theta -8}{2}}3(\theta -2) [K(x/2)-K(x)]}{2^{\frac{3(\theta -2)}{2}}-1}, \ \ \ \ \forall \ x\in {\mathbb {R}}^3. \end{aligned}$$

Thus, relation (3.3) holds by letting \(|x|\rightarrow \infty \) in the above inequality. \(\square \)

Lemma 3.2

Assume that hypotheses (F1)–(F3) hold. Then

$$\begin{aligned} \begin{aligned} h_1(t,\tau )&:= \frac{2\left( 1-t^{7-3\theta /2}\right) }{14-3\theta }[f(\tau )\tau -\theta F(\tau )] -\frac{2}{3}F(\tau )+\frac{2}{3}t^{-3\theta /2}F(t^{3/2}\tau )\\&\ge 0, \ \ \ \ \forall \ t> 0, \ \tau \in {\mathbb {R}}\end{aligned} \end{aligned}$$
(3.5)

and

$$\begin{aligned} \frac{F(t)}{|t|^{11/3}t}\ \text{ is } \text{ nondecreasing } \text{ on } \text{ both }\ \ (-\infty ,0)\ \text{ and }\ (0, +\infty ). \end{aligned}$$
(3.6)

Proof

For any \(\tau \in {\mathbb {R}}\), by (F3), we have

$$\begin{aligned} \begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}}t}h_1(t,\tau )&= t^{6-3\theta /2}|\tau |^{14/3}\left[ \frac{f(t^{3/2}\tau )t^{3/2}\tau -\theta F(t^{3/2}\tau )}{|t^{3/2}\tau |^{14/3}} -\frac{f(\tau )\tau -\theta F(\tau )}{|\tau |^{14/3}}\right] \\&\left\{ \begin{array}{ll} \ge 0, \ &{} t\ge 1,\\ \le 0, \ &{} 0<t<1, \end{array}\right. \end{aligned} \end{aligned}$$

which implies that \(h_1(t,\tau )\ge h_1(1,\tau )=0\) for all \(t> 0\) and \(\tau \in {\mathbb {R}}\), that is, inequality (3.5) holds. By (F3) and (3.5), we obtain

$$\begin{aligned} h_1(0,\tau ):=\lim _{|t|\rightarrow 0}h(t,\tau )=\frac{2}{14-3\theta }\left[ f(\tau )\tau -\frac{14}{3}F(\tau )\right] \ge 0, \ \ \ \ \forall \ \tau \in {\mathbb {R}}. \end{aligned}$$
(3.7)

From (3.7), we derive

$$\begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}}t}\frac{F(t)}{|t|^{11/3}t} =\frac{3}{2|t|^{7}}\left[ f(t)t-\frac{14}{3}F(t)\right] \ge 0. \end{aligned}$$

This shows that property (3.6) holds. \(\square \)

Lemma 3.3

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then

$$\begin{aligned} \begin{aligned} h_2(x,t,\tau )&:= t^{-3}K(t^{-1}x)F(t^{3/2}\tau )-K(x)F(\tau ){+}\frac{3(1{-}t^4)}{8}K(x)[f(\tau )\tau {-}2F(\tau )]\\&\ \ \ \ -\frac{1-t^4}{4}\nabla K(x)\cdot x F(\tau )\\&\ge 0, \ \ \ \ \forall \ x\in {\mathbb {R}}^3,\ t> 0, \ \tau \in {\mathbb {R}}. \end{aligned} \end{aligned}$$
(3.8)

Proof

For any \(x\in {\mathbb {R}}^3\) and \(\tau \in {\mathbb {R}}\), by (K1), (K2), (F1), (F3), (3.2) and (3.6), we have

$$\begin{aligned} \frac{{\mathrm {d}}}{{\mathrm {d}}t}h_2(x,t,\tau )= & {} -\frac{1}{t^{4}}\nabla K(t^{-1}x)\cdot (t^{-1}x)F(t^{3/2}\tau )+\frac{3}{2t^{4}}K(t^{-1}x) \left[ f(t^{3/2}\tau )t^{3/2}\tau -2F(t^{3/2}\tau )\right] \nonumber \\&\ \ -\frac{3 t^{3}}{2}K(x)[f(\tau )\tau -2F(\tau )]+t^3\nabla K(x)\cdot x F(\tau )\nonumber \\= & {} \frac{3}{2t^{4}}K(t^{-1}x)\left[ f(t^{3/2}\tau )t^{3/2}\tau -\theta F(t^{3/2}\tau )\right] \nonumber \\&\ \ +\frac{1}{2t^{4}}\left[ 3(\theta -2)K(t^{-1}x)-2\nabla K(t^{-1}x)\cdot (t^{-1}x)\right] F(t^{3/2}\tau )\nonumber \\&\ \ -\frac{3t^{3}}{2}K(x)[f(\tau )\tau -\theta F(\tau )] -\frac{t^{3}}{2}\left[ 3(\theta -2)K(x)-2\nabla K(x)\cdot x\right] F(\tau )\nonumber \\= & {} \frac{t^{3}|\tau |^7}{2}\Bigg \{3K(t^{-1}x)\frac{f(t^{3/2}\tau )t^{3/2}\tau -\theta F(t^{3/2}\tau )}{|t^{3/2}\tau |^{14/3}} -3K(x)\frac{f(\tau )\tau -\theta F(\tau )}{|\tau |^{14/3}}\Bigg .\nonumber \\&\ \ +\left[ 3(\theta -2)K(t^{-1}x)-2\nabla K(t^{-1}x)\cdot (t^{-1}x)\right] \frac{F(t^{3/2}\tau )}{|t^{3/2}\tau |^{14/3}}\nonumber \\&\ \ -\Bigg .\left[ 3(\theta -2)K(x)-2\nabla K(x)\cdot x\right] \frac{F(\tau )}{|\tau |^{14/3}}\Bigg \}\nonumber \\&\left\{ \begin{array}{ll} \ge 0, \ &{} t\ge 1,\nonumber \\ \le 0, \ &{} 0<t<1. \end{array}\right. \end{aligned}$$

It follows that \(h_2(x, t,\tau )\ge h_2(x, 1,\tau )=0\) for all \(x\in {\mathbb {R}}^3\), \(t> 0\) and \(\tau \in {\mathbb {R}}\), hence relation (3.8) holds. \(\square \)

By the scaling (1.9), we have

$$\begin{aligned} I(u^t) = \frac{at^2}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x+\frac{bt^4}{4} \left( \int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) ^2 -t^{-3}\int _{{\mathbb {R}}^3}K(t^{-1}x)F(t^{3/2}u){\mathrm {d}}x \end{aligned}$$
(3.9)

and

$$\begin{aligned} I^{\infty }(u^t) = \frac{at^2}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x+\frac{bt^4}{4} \left( \int _{{\mathbb {R}}^3}|\nabla u|^2{\mathrm {d}}x\right) ^2 -t^{-3}\int _{{\mathbb {R}}^3}K_{\infty }F(t^{3/2}u){\mathrm {d}}x. \end{aligned}$$
(3.10)

Noting that \(J(u)=\frac{{\mathrm {d}}}{{\mathrm {d}}t}I(u^t)\Big |_{t=1}\) and \(J^{\infty }(u): =\frac{{\mathrm {d}}}{{\mathrm {d}}t}I^{\infty }(u^t)\Big |_{t=1}\), it follows from (3.9) and (3.10) that

$$\begin{aligned} \begin{aligned} J(u)&= a\Vert \nabla u\Vert _2^2+b\Vert \nabla u\Vert _2^4-\frac{3}{2}\int _{{{\mathbb {R}}}^3}K(x)\left[ f(u)u-2F(u)\right] {\mathrm {d}}x\\&\ \ \ \ +\int _{{{\mathbb {R}}}^3}\nabla K(x)\cdot x F(u){\mathrm {d}}x \end{aligned} \end{aligned}$$
(3.11)

and

$$\begin{aligned} J^{\infty }(u) = a\Vert \nabla u\Vert _2^2+b\Vert \nabla u\Vert _2^4 -\frac{3}{2}\int _{{{\mathbb {R}}}^3}K_{\infty }\left[ f(u)u-2F(u)\right] {\mathrm {d}}x. \end{aligned}$$
(3.12)

Inspired by [9, 10], we prove the following lemma.

Lemma 3.4

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then

$$\begin{aligned} I(u)\ge I\left( u^t\right) +\frac{1-t^{4}}{4}{J}(u)+\frac{a(1-t^2)^2}{4}\Vert \nabla u\Vert _2^2, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3), \ t > 0.\quad \quad \end{aligned}$$
(3.13)

Proof

By (1.3), (3.5), (3.6), (3.8), (3.9) and (3.11), we have

$$\begin{aligned} {I}(u)-{I}\left( u^t\right)= & {} \frac{a(1-t^2)}{2}\Vert \nabla u\Vert _2^2+\frac{b(1-t^4)}{4}\Vert \nabla u\Vert _2^4\nonumber \\&+\int _{{\mathbb {R}}^3}\left[ t^{-3}K(t^{-1}x)F\left( t^{3/2}u\right) -K(x)F(u)\right] {\mathrm {d}}x\nonumber \\= & {} \frac{1-t^{4}}{4}\left\{ a\Vert \nabla u\Vert _2^2+b\Vert \nabla u\Vert _2^4-\frac{3}{2}\int _{{{\mathbb {R}}}^3}K(x)\left[ f(u)u-2F(u)\right] {\mathrm {d}}x\right. \nonumber \\&\ \ \left. +\int _{{{\mathbb {R}}}^3}\nabla K(x)\cdot x F(u){\mathrm {d}}x\right\} +\frac{a(1-t^2)^2}{4}\Vert \nabla u\Vert _2^2\nonumber \\&\ \ +\int _{{{\mathbb {R}}}^3}\left\{ t^{-3}K(t^{-1}x)F(t^{3/2}u)-K(x)F(u)\right. \nonumber \\&\left. +\frac{3(1-t^4)}{8}K(x)[f(u)u-2F(u)]\right. \nonumber \\&\ \ \left. -\frac{1-t^4}{4}\nabla K(x)\cdot x F(u)\right\} {\mathrm {d}}x\nonumber \\\ge & {} \frac{1-t^{4}}{4}{J}(u)+\frac{a(1-t^2)^2}{4}\Vert \nabla u\Vert _2^2, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3), \ \ t > 0. \end{aligned}$$
(3.14)

This shows that (3.13) holds. \(\square \)

From Lemma 3.4, we have the following corollary. In what follows, the definitions of \({\mathcal {M}}_c\) and \(u^t\) are given by (1.8) and (1.9).

Corollary 3.5

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then

$$\begin{aligned} {I}(u) = \max _{t> 0}{I}\left( u^t\right) , \ \ \ \ \forall \ u\in {\mathcal {M}}_c. \end{aligned}$$
(3.15)

Lemma 3.6

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then for any \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\), there exists a unique \(t_u>0\) such that \(u^{t_u}\in {\mathcal {M}}_c\).

Proof

Let \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\) be fixed and define a function \(\zeta (t):=I\left( u^t\right) \) on \((0, \infty )\). Clearly, by (3.9) and (3.11), we have

$$\begin{aligned} \zeta '(t)=0&\Leftrightarrow \ \ at\Vert \nabla u\Vert _2^2+bt^3\Vert \nabla u\Vert _2^4 \nonumber \\&\quad -\frac{3}{2t^4}\int _{{{\mathbb {R}}}^3}K(t^{-1}x)\left[ f(t^{3/2}u)t^{3/2}u -2F(t^{3/2}u)\right] {\mathrm {d}}x\nonumber \\&\quad +\frac{1}{t^4}\int _{{{\mathbb {R}}}^3}\nabla K(t^{-1}x)\cdot (t^{-1}x) F(t^{3/2}u){\mathrm {d}}x=0\nonumber \\&\ \Leftrightarrow \ \ \frac{1}{t}J\left( u^t\right) =0 \ \ \Leftrightarrow \ \ u^t\in {\mathcal {M}}_c. \end{aligned}$$
(3.16)

Note that (3.2) and (3.6) lead to

$$\begin{aligned} K(t^{-1}x)F(t^{3/2}\tau )\le t^7 K(x)F(\tau ), \ \ \ \ \forall \ x\in {\mathbb {R}}^3,\ t\in (0,1), \ \tau \in {\mathbb {R}}. \end{aligned}$$
(3.17)

From (3.9) and (3.17), we deduce that

$$\begin{aligned} I(u^t) \ge \frac{at^2}{2}\Vert \nabla u\Vert _2^2+\frac{bt^4}{4}\Vert \nabla u\Vert _2^4 -t^{4}\int _{{\mathbb {R}}^3}K(x)F(u){\mathrm {d}}x, \ \ \ \ \forall \ t\in (0,1), \end{aligned}$$
(3.18)

which implies that \(\zeta (t)>0\) for \(t>0\) small. Moreover, by (K1), (K2), (F1), (F2) and (3.9), it is easy to verify that \(\lim _{t\rightarrow 0}\zeta (t)=0\) and \(\zeta (t)<0\) for t large. Therefore \(\max _{t\in (0, \infty )}\zeta (t)\) is achieved at \(t_u>0\) so that \(\zeta '(t_u)=0\) and \(u^{t_u}\in {\mathcal {M}}_c\).

Next, we claim that \(t_u\) is unique for any \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\). Otherwise, for any given \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\), there exist positive constants \(t_1\ne t_2\) such that \(u^{t_1}, u^{t_2} \in {\mathcal {M}}_c\), that is, \(J\left( u^{t_1}\right) =J\left( u^{t_2}\right) =0\). Then (3.13) implies

$$\begin{aligned} I\left( u^{t_1}\right)> & {} I\left( u^{t_2}\right) +\frac{t_1^{4}-t_2^{4}}{t_1^{4}}J\left( u^{t_1}\right) =I\left( u^{t_2}\right) \nonumber \\> & {} I\left( u^{t_1}\right) +\frac{t_2^{4}-t_1^{4}}{t_2^{4}}J\left( u^{t_2}\right) =I\left( u^{t_1}\right) . \end{aligned}$$
(3.19)

This contradiction shows that \(t_u> 0\) is unique for any \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\). \(\square \)

Combining Corollary 3.5 with Lemma 3.6, we obtain the following property.

Lemma 3.7

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then

$$\begin{aligned} \inf _{u\in {\mathcal {M}}_c}{I}(u) =m(c)=\inf _{u\in {\mathcal {S}}_{c}}\max _{t > 0}{I}\left( u^t\right) . \end{aligned}$$

Lemma 3.8

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then

  1. (i)

    there exists \(\rho _0>0\) such that \(\Vert \nabla u\Vert _2\ge \rho _0, \ \forall \ u\in {\mathcal {M}}_c\);

  2. (ii)

    \(m(c)=\inf _{u\in {\mathcal {M}}_c}{I}(u)>0\).

Proof

(i) By (F1), we deduce that

$$\begin{aligned} \frac{F(t)}{|t|^{\mu -1}t}\ {\mathrm{is \ nonincreasing\ on\ both}}\ \ (-\infty ,0)\ {\mathrm{and}}\ (0, +\infty ). \end{aligned}$$
(3.20)

From (3.6) and (3.20), we derive that for any \(s\in {\mathbb {R}}\),

$$\begin{aligned} \left\{ \begin{array}{ll} |t|^{\mu }F(s) \le F(st)\le |t|^{\frac{14}{3}}F(s),\ \ {\mathrm{if}}\ |t|\le 1;\\ |t|^{\frac{14}{3}}F(s) \le F(st)\le |t|^{\mu }F(s),\ \ {\mathrm{if}}\ |t|\ge 1, \end{array} \right. \end{aligned}$$
(3.21)

which implies that there is a constant \(C_0>0\) such that

$$\begin{aligned} 0\le F(t)\le C_0\left( |t|^{\frac{14}{3}}+|t|^{\mu }\right) , \ \ \ \ \forall \ t\in {\mathbb {R}}. \end{aligned}$$
(3.22)

By the Gagliardo-Nirenberg inequality, we have

$$\begin{aligned} \Vert u\Vert _s^s\le {\mathcal {C}}(s)\Vert \nabla u\Vert _2^{\frac{3(s-2)}{2}}\Vert u\Vert _2^{\frac{6-s}{2}}, \ \ \ \ \forall \ s\in (2,6). \end{aligned}$$
(3.23)

Since \({J}(u)=0, \ \forall u\in {\mathcal {M}}_c\), by (K1), (K2), (3.11), (3.22), (3.23) and the Sobolev inequality, we deduce that

$$\begin{aligned} a\Vert \nabla u\Vert _2^2\le & {} a\Vert \nabla u\Vert _2^2+b\Vert \nabla u\Vert _2^4\nonumber \\= & {} 2\int _{{{\mathbb {R}}}^3}K(x)\left[ f(u)u-2F(u)\right] {\mathrm {d}}x -2\int _{{{\mathbb {R}}}^3}\nabla K(x)\cdot x F(u){\mathrm {d}}x\nonumber \\\le & {} C_1\left( \Vert u\Vert _{\frac{14}{3}}^{\frac{14}{3}}+\Vert u\Vert _6^6\right) \nonumber \\\le & {} C_2\Vert \nabla u\Vert _2^{4}\Vert u\Vert _2^{2/3}+C_1S^{-3}\Vert \nabla u\Vert _2^6\nonumber \\= & {} C_2 c^{1/3}\Vert \nabla u\Vert _2^{4}+C_1S^{-3}\Vert \nabla u\Vert _2^6, \ \ \ \ \forall \ u\in {\mathcal {M}}_c, \end{aligned}$$
(3.24)

which concludes the proof of (i).

(ii) By (i) and (3.13) with \(t\rightarrow 0\), we have

$$\begin{aligned} I(u)=I(u)-\frac{1}{4}J(u)\ge \frac{a}{4}\Vert \nabla u\Vert _2^2 \ge \frac{a}{4}\rho _0^2, \ \ \ \ \forall \ u\in {\mathcal {M}}_c. \end{aligned}$$
(3.25)

Hence, \(m(c)=\inf _{u\in {\mathcal {M}}_c}I(u)>0\). \(\square \)

Lemma 3.9

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then m(c) is nonincreasing on \((0, \infty )\). In particular, if m(c) is achieved, then \(m(c)>m({\tilde{c}})\) for any \({\tilde{c}}>c\).

Proof

For any \(c_2>c_1>0\), there exists \(\{u_n\}\subset {\mathcal {M}}_{c_1}\) such that

$$\begin{aligned} I(u_n)<m(c_1)+\frac{1}{n}. \end{aligned}$$

Let \(\xi =\sqrt{c_2/c_1}\in (1,\infty )\) and \(v_n(x)=\xi ^{-1/2}u_n(\xi ^{-1} x)\). Then \(\Vert v_n\Vert _2^2=c_2\) and \(\Vert \nabla v_n\Vert _2=\Vert \nabla u_n\Vert _2\). By Lemma 3.6, there exists \(t_n>0\) such that \((v_n)^{t_n}\in {\mathcal {M}}_{c_2}\). Note that since \((6-\mu )K(x)+\nabla K(x)\cdot x\ge 0\) for all \(x\in {\mathbb {R}}^3\) then

$$\begin{aligned} t\mapsto t^{\frac{6-\mu }{2}}K(tx)\ {\mathrm{is\ nondecreasing\ on}}\ (0,+\infty )\ {\mathrm{for\ every}}\ x\in {\mathbb {R}}^3. \end{aligned}$$
(3.26)

Next, by (K2), (3.9), (3.13), (3.20) and (3.26), it follows that

$$\begin{aligned}&m(c_2)\nonumber \\&\quad \le I\left( (v_n)^{t_n}\right) \nonumber \\&\quad = \frac{at_n^2}{2}\Vert \nabla u_n\Vert _2^2+\frac{bt_n^4}{4}\Vert \nabla u_n\Vert _2^4 -t_n^{-3}\xi ^3\int _{{\mathbb {R}}^3}K(t_n^{-1}\xi x)F(t_n^{3/2}\xi ^{-1/2}u_n){\mathrm {d}}x\nonumber \\&\quad \le I\left( (u_n)^{t_n}\right) +t_n^{-3}\int _{{\mathbb {R}}^3}\left[ K(t_n^{-1}x)F(t_n^{3/2}u_n) -\xi ^{\frac{6-\mu }{2}}K(t_n^{-1}\xi x)\xi ^{\frac{\mu }{2}}F(t_n^{3/2}\xi ^{-1/2}u_n)\right] {\mathrm {d}}x\nonumber \\&\quad \le I(u_n)-\frac{a(1-t_n^2)^2}{4}\Vert \nabla u_n\Vert _2^2<m(c_1)+\frac{1}{n}, \end{aligned}$$

which shows that \(m(c_2)\le m(c_1)\) by letting \(n\rightarrow \infty \).

We now assume that m(c) is achieved, that is, there exists \(u\in {\mathcal {M}}_c\) such that \(I(u)=m(c)\) for any given \(c<{\tilde{c}}\). Let \({\tilde{\xi }}={\tilde{c}}/c\in (1,\infty )\) and \(v(x)={\tilde{\xi }}^{-1/2}u({\tilde{\xi }}^{-1} x)\). Then \(\Vert v\Vert _2^2={\tilde{c}}\) and \(\Vert \nabla v\Vert _2=\Vert \nabla u\Vert _2\). By Lemma 3.6, there exists \({\tilde{t}}>0\) such that \(v^{{\tilde{t}}}\in {\mathcal {M}}_{{\tilde{c}}}\). Then it follows from (K2), (3.9), (3.13), (3.20) and (3.26) that

$$\begin{aligned} m({\tilde{c}})\le & {} I\left( v^{{\tilde{t}}}\right) \nonumber \\= & {} \frac{a{\tilde{t}}^2}{2}\Vert \nabla u\Vert _2^2+\frac{b{\tilde{t}}^4}{4}\Vert \nabla u\Vert _2^4 -{\tilde{t}}^{-3}{\tilde{\xi }}^3\int _{{\mathbb {R}}^3}K({\tilde{t}}^{-1}{\tilde{\xi }} x)F({\tilde{t}}^{3/2}{\tilde{\xi }}^{-1/2}u){\mathrm {d}}x\nonumber \\\le & {} I\left( u^{{\tilde{t}}}\right) +{\tilde{t}}^{-3}\int _{{\mathbb {R}}^3}\left[ K({\tilde{t}}^{-1}x)F({\tilde{t}}^{3/2}u) -{\tilde{\xi }}^{\frac{6-\mu }{2}}K({\tilde{t}}^{-1}{\tilde{\xi }} x){\tilde{\xi }}^{\frac{\mu }{2}} F({\tilde{t}}^{3/2}{\tilde{\xi }}^{-1/2}u)\right] {\mathrm {d}}x\nonumber \\\le & {} I(u)-\frac{a(1-{\tilde{t}}^2)^2}{4}\Vert \nabla u\Vert _2^2<m(c). \end{aligned}$$

The proof is completed. \(\square \)

Remark 3.10

Lemmas 3.13.4 provide technical ingredients allowing generalization of the previous results to a new scenario when K(x) is variable and f(u) has a more general structure. But it is still difficult to prove that m(c) is strictly decreasing on \((0, \infty )\) in the same way as the proof of [29, Lemma 4.5], because it is still unknown that \(\{t_n\}\), involved in the proof of Lemma 3.9, is bounded from below for more general nonlinearity f.

Lemma 3.11

Assume that (K1), (K2) and (F1)–(F3) hold. Then \(m(c)\le m^{\infty }(c)\).

Proof

In view of Lemmas 3.6 and 3.8, we have \({\mathcal {M}}_c^{\infty }\ne \emptyset \) and \(m^{\infty }(c)>0\). Inspired by [8, 25], assume by contradiction that \(m(c)> m^{\infty }(c)\). Let \(\varepsilon :=m(c)-m^{\infty }(c)\). Then there exists \(u_{\varepsilon }^{\infty }\) such that

$$\begin{aligned} u_{\varepsilon }^{\infty }\in {\mathcal {M}}_c^{\infty } \ \ \ \ \text{ and } \ \ \ \ m^{\infty }(c)+\frac{\varepsilon }{2}>I^{\infty }(u_{\varepsilon }^{\infty }). \end{aligned}$$
(3.27)

In view of Lemma 3.6, there exists \(t_{\varepsilon }>0\) such that \((u_{\varepsilon }^{\infty })^{t_{\varepsilon }}\in {\mathcal {M}}_c\). Since \(K_{\infty } \le K(x)\) for all \({\mathbb {R}}^3\), it follows from (1.3), (2.7), (3.27) and Corollary 3.5 that

$$\begin{aligned} m^{\infty }(c)+\frac{\varepsilon }{2}>{I}^{\infty }(u_{\varepsilon }^{\infty }) \ge {I}^{\infty }\left( (u_{\varepsilon }^{\infty })^{t_{\varepsilon }}\right) \ge {I}\left( (u_{\varepsilon }^{\infty })^{t_{\varepsilon }}\right) \ge m(c). \end{aligned}$$

This contradiction shows that \(m(c)\le m^{\infty }(c)\). \(\square \)

Similarly to [25, Lemma 2.7], [24, Lemma 2.10] and [27], we have the following Brezis-Lieb type lemma.

Lemma 3.12

Assume that hypotheses (K1), (K2), (F1) and (F2) hold. If \(u_n\rightharpoonup {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\), then

$$\begin{aligned} {I}(u_n)={I}({\bar{u}})+{I}(u_n-{\bar{u}})+\frac{b}{2}\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla (u_n-{\bar{u}})\Vert _2^2+o(1) \end{aligned}$$
(3.28)

and

$$\begin{aligned} {J}(u_n)={J}({\bar{u}})+{J}(u_n-{\bar{u}})+b\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla (u_n-{\bar{u}})\Vert _2^2+o(1). \end{aligned}$$
(3.29)

Lemma 3.13

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. Then m(c) is achieved.

Proof

In view of Lemmas 3.6 and 3.8, we have \({\mathcal {M}}_c\ne \emptyset \) and \(m(c)>0\). Let \(\{u_n\}\subset {\mathcal {M}}_c\) be such that \({I}(u_n)\rightarrow m(c)\). Since \({J}(u_n)=0\), then it follows from (3.13) with \(t \rightarrow 0\) that

$$\begin{aligned} m(c)+o(1)= {I}(u_n)\ge \frac{a}{4}\Vert \nabla u_n\Vert _2^2. \end{aligned}$$
(3.30)

This relation together with \(\Vert u_n\Vert _2^2=c\), implies that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\). Passing to a subsequence, we have \(u_n\rightharpoonup {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\). Then \(u_n\rightarrow {\bar{u}}\) in \(L_{\mathrm {loc}}^s({\mathbb {R}}^3)\) for \(2\le s<6\) and \(u_n\rightarrow {\bar{u}}\) a.e. in \({\mathbb {R}}^3\). There are two possible cases: i) \({\bar{u}}=0\) and ii) \({\bar{u}}\ne 0\).

Case i) \({\bar{u}}=0\), namely \(u_n\rightharpoonup 0\) in \(H^1({\mathbb {R}}^3)\). Then \(u_n\rightarrow 0\) in \(L_{\mathrm {loc}}^s({\mathbb {R}}^3)\) for \(2\le s<6\) and \(u_n\rightarrow 0\) a.e. in \({\mathbb {R}}^3\). By (K1) and (3.3), it is easy to show that

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}[K_{\infty }-K(x)]F(u_n){\mathrm {d}}x= \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}\nabla K(x)\cdot x F(u_n){\mathrm {d}}x=0. \end{aligned}$$
(3.31)

From (1.3), (2.7), (3.11), (3.12) and (3.31), we deduce that

$$\begin{aligned} {I}^{\infty }(u_n)\rightarrow m(c), \ \ \ \ {J}^{\infty }(u_n)\rightarrow 0. \end{aligned}$$
(3.32)

From (3.12), (3.32), Lemmas 3.6 and  3.7 (i), we have

$$\begin{aligned} a\rho _0^2 \le a\Vert \nabla u_n\Vert _2^2+b\Vert \nabla u_n\Vert _2^4 =\frac{3}{2}\int _{{{\mathbb {R}}}^3}K_{\infty }\left[ f(u_n)u_n-2F(u_n)\right] {\hbox {dx}}. \end{aligned}$$
(3.33)

Using (3.22), (3.33) and Lions’ concentration-compactness principle [27, Lemma 1.21], we prove that there exist \(\delta >0\) and \(\{y_n\}\subset {\mathbb {R}}^3\) such that \(\int _{B_1(y_n)}|u_n|^2{\mathrm {d}}x> \delta \). Let \({\hat{u}}_n(x)=u_n(x+y_n)\). Then we have \(\Vert {\hat{u}}_n\Vert =\Vert u_n\Vert \) and

$$\begin{aligned} {J}^{\infty }({\hat{u}}_n)= o(1), \ \ \ \ {I}^{\infty }({\hat{u}}_n)\rightarrow m(c), \ \ \ \ \int _{B_1(0)}|{\hat{u}}_n|^2{\mathrm {d}}x> \delta . \end{aligned}$$
(3.34)

Therefore, there exists \({\hat{u}}\in H^1({\mathbb {R}}^3)\setminus \{0\}\) such that, up to a subsequence,

$$\begin{aligned} \left\{ \begin{array}{ll} {\hat{u}}_n\rightharpoonup {\hat{u}}, &{} \text{ in } \ H^1({\mathbb {R}}^3); \\ {\hat{u}}_n\rightarrow {\hat{u}}, &{} \text{ in } \ L_{\mathrm {loc}}^s({\mathbb {R}}^3), \ \forall \ s\in [1, 6);\\ {\hat{u}}_n\rightarrow {\hat{u}}, &{} \text{ a.e. } \text{ on } \ {\mathbb {R}}^3. \end{array} \right. \end{aligned}$$
(3.35)

Let \(w_n={\hat{u}}_n-{\hat{u}}\). Then (3.35) and Lemma 3.12 yield

$$\begin{aligned}&\Vert {\hat{u}}\Vert _2^2:={\hat{c}}\le c, \ \ \ \ \Vert w_n\Vert _2^2:={\hat{c}}_n\le c\ \text{ for } \text{ large }\ n\in {\mathbb {N}}, \end{aligned}$$
(3.36)
$$\begin{aligned}&{I}^{\infty }({\hat{u}}_n) = {I}^{\infty }({\hat{u}})+{I}^{\infty }(w_n) +\frac{b}{2}\Vert \nabla {\hat{u}}\Vert _2^2\Vert \nabla w_n\Vert _2^2+o(1) \end{aligned}$$
(3.37)

and

$$\begin{aligned} {J}^{\infty }({\hat{u}}_n) = {J}^{\infty }({\hat{u}})+{J}^{\infty }(w_n)+b\Vert \nabla {\hat{u}}\Vert _2^2\Vert \nabla w_n\Vert _2^2+o(1). \end{aligned}$$
(3.38)

Let

$$\begin{aligned} \Psi ^{\infty }(u):= & {} I^{\infty }(u)-\frac{1}{4}J^{\infty }(u)\nonumber \\= & {} \frac{a}{4}\Vert \nabla u\Vert _2^2 +\frac{1}{8}\int _{{\mathbb {R}}^3}K_{\infty }[3f(u)u-14F(u)]{\mathrm {d}}x, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3). \end{aligned}$$
(3.39)

By (3.7), we have \(\Psi ^{\infty }(u)> 0\) for all \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\). Moreover, it follows from (3.34), (3.37), (3.38) and (3.39) that

$$\begin{aligned} \Psi ^{\infty }(w_n)\le m(c)-\Psi ^{\infty }({\hat{u}})+o(1), \ \ \ \ {J}^{\infty }(w_n) \le -{J}^{\infty }({\hat{u}})+o(1). \end{aligned}$$
(3.40)

If there exists a subsequence \(\{w_{n_i}\}\) of \(\{w_n\}\) such that \(w_{n_i}=0\), then it follows from (3.36), Lemmas 3.9 and 3.11 that

$$\begin{aligned} m^{\infty }({\hat{c}}) \le {I}^{\infty }({\hat{u}})=m(c) \le m({\hat{c}})\le m^{\infty }({\hat{c}}), \ \ \ \ {J}^{\infty }({\hat{u}})=0, \end{aligned}$$

which, together with \(m^{\infty }(c) \le m^{\infty }({\hat{c}}) \le {I}^{\infty }({\hat{u}})=m(c)\le m^{\infty }(c)\), implies

$$\begin{aligned} {I}^{\infty }({\hat{u}})= m^{\infty }({\hat{c}})=m({\hat{c}})=m(c)= m^{\infty }(c), \ \ \ \ {J}^{\infty }({\hat{u}})=0. \end{aligned}$$
(3.41)

Next, we assume that \(w_n\ne 0\). We claim that \({J}^{\infty }({\hat{u}})\le 0\). Otherwise, if \({J}^{\infty }({\hat{u}})>0\), then (3.40) implies \({J}^{\infty }(w_n) < 0\) for large n. In view of Lemma 3.6, there exists \(t_n>0\) such that \((w_n)^{t_n}\in {\mathcal {M}}_{{\hat{c}}_n}^{\infty }\). Then it follows from (2.7), (3.12), (3.13), (3.39), (3.40), Lemmas 3.9 and 3.11 that

$$\begin{aligned} m(c)-\Psi ^{\infty }({\hat{u}})+o(1)\ge & {} \Psi ^{\infty }(w_n) = {I}^{\infty }(w_n)-\frac{1}{4}{J}^{\infty }(w_n)\nonumber \\\ge & {} {I}^{\infty }\left( {(w_n)}^{t_n}\right) -\frac{t_n^{4}}{4}{J}^{\infty }(w_n)\nonumber \\\ge & {} m^{\infty }({\hat{c}}_n)-\frac{t_n^{4}}{4}{J}^{\infty }(w_n)\nonumber \\\ge & {} m^{\infty }(c)+o(1)\ge m(c)+o(1), \end{aligned}$$

which is impossible due to \(\Psi ^{\infty }({\hat{u}})>0\). This shows that \({J}^{\infty }({\hat{u}})\le 0\). In view of Lemma 3.6, there exists \(t_{\infty }>0\) such that \({\hat{u}}^{t_{\infty }}\in {\mathcal {M}}_{{\hat{c}}}^{\infty }\). From (2.7), (3.12), (3.13), (3.34), (3.39), the weak semicontinuity of the norm, Fatou’s lemma, Lemmas 3.9 and 3.11, one has

$$\begin{aligned} m(c)= & {} \lim _{n\rightarrow \infty } \left[ {I}^{\infty }({\hat{u}}_n)-\frac{1}{4}{J}^{\infty }({\hat{u}}_n)\right] \nonumber \\= & {} \lim _{n\rightarrow \infty }\Psi ^{\infty }({\hat{u}}_n)\ge \Psi ^{\infty }({\hat{u}})\nonumber \\= & {} {I}^{\infty }({\hat{u}})-\frac{1}{4}{J}^{\infty }({\hat{u}})\ge {I}^{\infty }\left( {{\hat{u}}}^{t_{\infty }}\right) -\frac{t_{\infty }^{4}}{4}{J}^{\infty }({\hat{u}})\nonumber \\\ge & {} m^{\infty }({\hat{c}})-\frac{t_{\infty }^{4}}{4}{J}^{\infty }({\hat{u}}) \ge m({\hat{c}}) \ge m(c), \end{aligned}$$

which implies that (3.41) holds for \(w_n\ne 0\). From (3.36), (3.41), Lemmas 3.9 and 3.11, we deduce that

$$\begin{aligned} m^{\infty }(c)\le m^{\infty }({\hat{c}})={I}^{\infty }({\hat{u}})=m({\hat{c}})=m(c)\le m^{\infty }(c), \ \ \ \ {J}^{\infty }({\hat{u}})=0, \end{aligned}$$

which implies that (3.41) holds for \(w_n\ne 0\). Thus, \(m^{\infty }({\hat{c}})\) is achieved at \({\hat{u}}\). In view of Lemma 3.9, we deduce that \(\Vert {\hat{u}}\Vert _2^2={\hat{c}}=c\) due to \(m^{\infty }({\hat{c}})=m^{\infty }(c)\). By Lemma 3.6, there exists \({\hat{t}}>0\) such that \({\hat{u}}^{{\hat{t}}}\in {\mathcal {M}}_{c}\). Then it follows from (3.9), (3.13), (3.20), (3.26) and (3.41) that

$$\begin{aligned} m(c)\le & {} I\left( {\hat{u}}^{{\hat{t}}}\right) \le I^{\infty }\left( {\hat{u}}^{{\hat{t}}}\right) \nonumber \\\le & {} I^{\infty }({\hat{u}})-\frac{a(1-{\hat{t}}^2)^2}{4}\Vert \nabla {\hat{u}}\Vert _2^2 =m(c)-\frac{a(1-{\hat{t}}^2)^2}{4}\Vert \nabla {\hat{u}}\Vert _2^2, \end{aligned}$$

which implies that \({\hat{u}}\in {\mathcal {M}}_{c}\) and \(I({\hat{u}})=m(c)\). Hence, m(c) is achieved at \({\hat{u}}\in {\mathcal {M}}_c\).

Case ii) \({\bar{u}}\ne 0\). Let \(v_n=u_n-{\bar{u}}\). Then Lemma 3.12 yields

$$\begin{aligned}&\Vert {\bar{u}}\Vert _2^2:={\bar{c}}\le c, \ \ \ \ \Vert v_n\Vert _2^2:=c_n\le c\ \text{ for } \text{ large }\ n\in {\mathbb {N}}, \end{aligned}$$
(3.42)
$$\begin{aligned}&{I}(u_n)={I}({\bar{u}})+{I}(v_n)+\frac{b}{2}\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla v_n\Vert _2^2+o(1) \end{aligned}$$
(3.43)

and

$$\begin{aligned} {J}(u_n)={J}({\bar{u}})+{J}(v_n)+b\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla v_n\Vert _2^2+o(1). \end{aligned}$$
(3.44)

Let

$$\begin{aligned} \Psi (u):= & {} I(u)-\frac{1}{4}J(u)\nonumber \\= & {} \frac{a}{4}\Vert \nabla u\Vert _2^2 +\frac{1}{8}\int _{{\mathbb {R}}^3}K(x)[3f(u)u-14F(u)]{\mathrm {d}}x\nonumber \\&\ \ -\frac{1}{4}\int _{{\mathbb {R}}^3}\nabla K(x)\cdot xF(u){\mathrm {d}}x, \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3). \end{aligned}$$
(3.45)

By (3.4) and (3.7), one has \(\Psi (u)> 0\) for all \(u\in H^1({\mathbb {R}}^3)\setminus \{0\}\). Similarly to the proof of (3.40), we deduce that

$$\begin{aligned} \Psi (v_n)\le m(c)-\Psi ({\bar{u}})+o(1), \ \ \ \ {J}(v_n) \le -{J}({\bar{u}})+o(1). \end{aligned}$$
(3.46)

If there exists a subsequence \(\{v_{n_i}\}\) of \(\{v_n\}\) such that \(v_{n_i}=0\), then it follows from Lemma 3.9 that

$$\begin{aligned} m({\bar{c}}) \le {I}({\bar{u}})=m(c) \le m({\bar{c}}), \ \ \ \ {J}({\bar{u}})=0, \end{aligned}$$

which implies

$$\begin{aligned} {I}({\bar{u}})=m(c)=m({\bar{c}}), \ \ \ \ {J}({\bar{u}})=0. \end{aligned}$$
(3.47)

Next, we prove that (3.47) holds for \(v_n\ne 0\). To this end, we assume that \(v_n\ne 0\). We claim that \({J}({\bar{u}})\le 0\). Otherwise \({J}({\bar{u}})>0\), then (3.46) implies \({J}(v_n) < 0\) for large n. In view of Lemma 3.6, there exists \(t_n>0\) such that \((v_n)^{t_n}\in {\mathcal {M}}_{c_n}\). From (3.13), (3.45) and (3.46), we obtain

$$\begin{aligned} m(c)-\Psi ({\bar{u}})+o(1)\ge & {} \Psi (v_n) = {I}(v_n)-\frac{1}{4}{J}(v_n)\nonumber \\\ge & {} {I}\left( {(v_n)}^{t_n}\right) -\frac{t_n^{4}}{4}{J}(v_n) \ge m({\bar{c}}_n)\ge m(c)+o(1), \end{aligned}$$

which is impossible due to \(\Psi ({\bar{u}})>0\). This shows that \({J}({\bar{u}})\le 0\). In view of Lemma 3.6, there exists \({\tilde{t}}>0\) such that \({\bar{u}}^{{\tilde{t}}}\in {\mathcal {M}}_{{\bar{c}}}\). Then it follows from (3.13), (3.45), the weak semicontinuity of norm, Fatou’s lemma and Lemma 3.11 that

$$\begin{aligned} m(c)= & {} \lim _{n\rightarrow \infty } \left[ {I}(u_n)-\frac{1}{4}{J}(u_n)\right] =\lim _{n\rightarrow \infty } \Psi (u_n)\nonumber \\\ge & {} \Psi ({\bar{u}})={I}({\bar{u}})-\frac{1}{4}{J}({\bar{u}}) \nonumber \\\ge & {} {I}\left( {{\bar{u}}}^{{\tilde{t}}}\right) -\frac{{\tilde{t}}^{4}}{4}{J}({\bar{u}}) \ge m({\bar{c}})\ge m(c), \end{aligned}$$

which implies (3.47) holds for \(v_n\ne 0\). This shows that \(m({\bar{c}})\) is achieved at \({\bar{u}}\in {\mathcal {M}}_{{\bar{c}}}\). In view of Lemma 3.9, we have \(\Vert {\bar{u}}\Vert _2^2={\bar{c}}=c\) due to \(m(c)=m({\bar{c}})\). By Lemma 3.6, there exists \({\bar{t}}>0\) such that \({\bar{u}}^{{\bar{t}}}\in {\mathcal {M}}_{c}\). Then it follows from (3.9), (3.13), (3.20), (3.26) and (3.47) that

$$\begin{aligned} m(c)\le & {} I\left( {\bar{u}}^{{\bar{t}}}\right) \nonumber \\\le & {} I({\bar{u}})-\frac{a(1-{\bar{t}}^2)^2}{4}\Vert \nabla {\bar{u}}\Vert _2^2 =m(c)-\frac{a(1-{\bar{t}}^2)^2}{4}\Vert \nabla {\bar{u}}\Vert _2^2, \end{aligned}$$

which implies that \({\bar{u}}\in {\mathcal {M}}_{c}\) and \(I({\bar{u}})=m(c)\). Hence, m(c) is achieved at \({\bar{u}}\in {\mathcal {M}}_c\). \(\square \)

Inspired by [8, Lemma 2.14], we prove the following lemma.

Lemma 3.14

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. If \({\bar{u}}\in {\mathcal {M}}_c\) and \({I}({\bar{u}})=m(c)\), then \({\bar{u}}\) is a critical point of \({I}\big |_{{\mathcal {S}}_{c}}\).

Proof

Assume that \(I|_{{\mathcal {S}}_c}'({\bar{u}})\ne 0\). Then there exist \(\delta >0\) and \(\varrho >0\) such that

$$\begin{aligned} u\in {\mathcal {S}}_c, \ \ \Vert u-{\bar{u}}\Vert \le 3\delta \Rightarrow \Vert I|_{{\mathcal {S}}_c}'(u)\Vert \ge \varrho . \end{aligned}$$
(3.48)

Similarly to [25, (2.59)], we prove that

$$\begin{aligned} \lim _{t\rightarrow 1}\left\| {\bar{u}}^t-{\bar{u}}\right\| =0. \end{aligned}$$
(3.49)

Thus, there exists \(\delta _1\in (0, 1/4)\) such that

$$\begin{aligned} |t-1|<\delta _1\Rightarrow \left\| {\bar{u}}^t-{\bar{u}}\right\| < \delta . \end{aligned}$$
(3.50)

In view of (3.13), we have

$$\begin{aligned} I\left( {\bar{u}}^t\right) \le I({\bar{u}})-\frac{a(1-t^2)^2}{4}\Vert \nabla {\bar{u}}\Vert _2^2 = m(c)-\frac{a(1-t^2)^2}{4}\Vert \nabla {\bar{u}}\Vert _2^2, \ \ \ \ \forall \ t> 0.\nonumber \\ \end{aligned}$$
(3.51)

From (F3), (3.2), (3.11), (3.16) and (3.17), we deduce that there exist \(T_1\in (0,1)\) and \(T_2\in (1, \infty )\) such that

$$\begin{aligned} J\left( {\bar{u}}^{T_1}\right) >0, \ \ \ \ J\left( {\bar{u}}^{T_2}\right) <0. \end{aligned}$$
(3.52)

Let \(\varepsilon :=\min \{a(1-T_1^2)^2\Vert \nabla {\bar{u}}\Vert _2^2/12, a(1-T_2^2)^2\Vert \nabla {\bar{u}}\Vert _2^2/12, 1, \varrho \delta /8\}\) and \({\mathcal {S}}:=B({\bar{u}}, \delta )\cap {\mathcal {S}}_c\). By [27, Lemma 2.3] or [23], there exists a deformation \(\eta \in {\mathcal {C}}([0, 1]\times {\mathcal {S}}_c, {\mathcal {S}}_c)\) such that

  1. (i)

    \(\eta (1, u)=u\) if \(I(u)<m(c)-2\varepsilon \) or \(I(u)>m(c)+2\varepsilon \);

  2. (ii)

    \(\eta \left( 1, I^{m(c)+\varepsilon }\cap {\mathcal {S}} \right) \subset I^{m(c)-\varepsilon }\);

  3. (iii)

    \(I(\eta (1, u))\le I(u), \ \forall \ u\in {\mathcal {S}}_c\);

  4. (iv)

    \(\eta (1, u)\) is a homeomorphism of \({\mathcal {S}}_c\).

By Corollary 3.5, \(I\left( {\bar{u}}^t\right) \le I({\bar{u}})=m(c)\) for \(t> 0\). Thus, by (3.50) and ii) we obtain

$$\begin{aligned} I\left( \eta \left( 1, {\bar{u}}^t\right) \right) \le m(c)-\varepsilon , \ \ \ \ \forall \ t> 0, \ \ |t-1|< \delta _1. \end{aligned}$$
(3.53)

On the other hand, by iii) and (3.51), one has

$$\begin{aligned} I\left( \eta \left( 1, {\bar{u}}^t\right) \right)\le & {} I\left( {\bar{u}}^t\right) \nonumber \\\le & {} m(c)-\frac{a(1-t^2)^2}{4}\Vert \nabla {\bar{u}}\Vert _2^2 \nonumber \\\le & {} m(c)-\delta _2\Vert \nabla {\bar{u}}\Vert _2^2, \ \ \ \ \forall \ t> 0, \ \ |t-1|\ge \delta _1, \end{aligned}$$
(3.54)

where

$$\begin{aligned} \delta _2:=\frac{a}{4}\min \left\{ (1-T_1^2)^2, (1-T_1^2)^2\right\} >0. \end{aligned}$$

Combining (3.53) with (3.54), we have

$$\begin{aligned} \max _{t\in [T_1, T_2]}I\left( \eta \left( 1, {\bar{u}}^t\right) \right) <m(c). \end{aligned}$$
(3.55)

Define \(\Psi _0(t):=J\left( \eta \left( 1, {\bar{u}}^t\right) \right) \) for \(t> 0\). It follows from (3.51) and i) that \(\eta (1, {\bar{u}}^t)={\bar{u}}^t\) for \(t=T_1\) and \(t=T_2\), which, together with (3.52), implies

$$\begin{aligned} \Psi _0(T_1)=J\left( {\bar{u}}^{T_1}\right) >0, \ \ \ \ \Psi _0(T_2)=J\left( {\bar{u}}^{T_1}\right) <0. \end{aligned}$$

Since \(\Psi _0(t)\) is continuous on \((0, \infty )\), then \(\eta \left( 1, {\bar{u}}^t\right) \cap {\mathcal {M}}_c\ne \emptyset \) for some \(t_0\in [T_1, T_2]\), contradicting the definition of m(c). \(\square \)

Lemma 3.15

Assume that hypotheses (K1), (K2) and (F1)–(F3) hold. If \({\bar{u}}\in {\mathcal {S}}_{c}\) is a critical point of \({I}\big |_{{\mathcal {S}}_{c}}\), then \(J({\bar{u}})=0\), and there exists \(\lambda _c<0\) such that \(I'({\bar{u}})-\lambda _c u=0\).

Proof

Since \(({I}|_{{\mathcal {S}}_{c}})'({\bar{u}})=0\), there exists \(\lambda _c\in {\mathbb {R}}\) such that \(I'({\bar{u}})-\lambda _c {\bar{u}}=0\), and so

$$\begin{aligned} \langle I'({\bar{u}})-\lambda _c {\bar{u}}, u\rangle =a\Vert \nabla {\bar{u}}\Vert _2^2+b\Vert \nabla {\bar{u}}\Vert _2^4-\int _{{\mathbb {R}}^3}K(x)f({\bar{u}}){\bar{u}}{\mathrm {d}}x -\lambda _c \Vert {\bar{u}}\Vert _2^2=0.\nonumber \\ \end{aligned}$$
(3.56)

Moreover, \({\bar{u}}\) satisfies the following Pohozaev identity:

$$\begin{aligned} {\mathcal {P}}({\bar{u}}):=\frac{a}{2}\Vert \nabla {\bar{u}}\Vert _2^2+\frac{b}{2}\Vert \nabla {\bar{u}}\Vert _2^4 -\int _{{\mathbb {R}}^3}[3K(x)+\nabla K(x)\cdot x]F({\bar{u}}){\mathrm {d}}x -\frac{3\lambda _c}{2} \Vert {\bar{u}}\Vert _2^2=0. \end{aligned}$$
(3.57)

Then relations (3.11), (3.56) and (3.57) yield

$$\begin{aligned} J({\bar{u}})=\frac{3}{2}\langle I'({\bar{u}})-\lambda _c {\bar{u}}, {\bar{u}}\rangle -{\mathcal {P}}({\bar{u}})=0. \end{aligned}$$

Noting that \(\Vert {\bar{u}}\Vert _2^2=c\), it follows from (K2), (F1), (3.56) and (3.57) that

$$\begin{aligned} \begin{aligned} 2\lambda c&=\ \int _{{\mathbb {R}}^3}\left\{ K(x)f({\bar{u}}){\bar{u}}-[6K(x)+2\nabla K(x)\cdot x]F({\bar{u}})\right\} {\mathrm {d}}x\\&=\ \int _{{\mathbb {R}}^3}\left\{ K(x)[f({\bar{u}}){\bar{u}}-\mu F({\bar{u}})] -[(6-\mu )K(x)+2\nabla K(x)\cdot x]F({\bar{u}})\right\} {\mathrm {d}}x<0, \end{aligned} \end{aligned}$$

and so \(\lambda _c<0\). This completes the proof. \(\square \)

Proof of Theorem 2.1

In view of Lemmas 3.73.8 and 3.133.15, for any \(c>0\) there exists \({\bar{u}}_c\in {\mathcal {M}}_c\) such that

$$\begin{aligned} I({\bar{u}}_c)=m(c)=\inf _{u\in {\mathcal {S}}_{c}}\max _{t> 0}I(({\bar{u}}_c)^t)>0, \ \ \ \ I'({\bar{u}}_c)=0, \end{aligned}$$

and there exists Lagrange multiplier \(\lambda _c\in {\mathbb {R}}^-\) such that \(({\bar{u}}_c,\lambda _c)\) is a solution of problem (1.1). \(\square \)

4 Second Existence Result and Qualitative Properties of Solutions

In this section, we give the proof of Theorem 2.2.

4.1 Global Minimizers on the Constraint \({\mathcal {S}}_c\)

Lemma 4.1

Assume that hypotheses (K1) and (F4) hold. Then

  1. (i)

    for any \(c>0\), \(\sigma (c)=\inf _{u\in {\mathcal {S}}_{c}}I(u)\) is well defined and \(\sigma (c)\le 0\);

  2. (ii)

    for any \(c>0\), \(\sigma (c)<0\) if (F6) holds;

  3. (iii)

    there exists \({\mathcal {C}}_0>0\) such that \(\sigma (c)<0\) for any \(c>{\mathcal {C}}_0\) if (F5) holds.

Proof

(i) Using (F4), for any \(\varepsilon >0\), there exists \(C_{\varepsilon }>0\) such that

$$\begin{aligned} |f(t)t|+|F(t)|\le \varepsilon |t|^2+C_{\varepsilon }|t|^{p}, \ \ \ \ \forall \ t\in {\mathbb {R}}. \end{aligned}$$
(4.1)

By (1.3), (3.23) and (4.1), we have

$$\begin{aligned} \begin{aligned} I(u)&\ \ge \ \frac{a}{2}\Vert \nabla u\Vert _2^2+\frac{b}{4}\Vert \nabla u\Vert _2^4 -\sup _{{\mathbb {R}}^3}K\left( \varepsilon \Vert u\Vert _{2}^{2}+C_{\varepsilon }\Vert u\Vert _{p}^{p}\right) \\&\ \ge \ \frac{a}{2}\Vert \nabla u\Vert _2^2+\frac{b}{4}\Vert \nabla u\Vert _2^4 -\sup _{{\mathbb {R}}^3}K\left( \varepsilon \Vert u\Vert _2^{2}+C_{\varepsilon }{\mathcal {C}}(p)\Vert \nabla u\Vert _2^{\frac{3(p-2)}{2}}\Vert u\Vert _2^{\frac{6-p}{2}}\right) ,\\&\ \ \ \ \ \ \ \ \forall \ u\in {\mathcal {S}}_{c},\ c>0. \end{aligned} \end{aligned}$$
(4.2)

This relation together with \(0<3(p-2)/2<4\), shows that I is bounded from below on \({\mathcal {S}}_{c}\) for any \(c>0\), that is, \(\sigma (c)\) is well defined. Noting that \(u^t\in {\mathcal {S}}_{c}\) for all \(u\in {\mathcal {S}}_{c}\), from (3.9) and (4.1), we deduce that \(I(u^t)\rightarrow 0\) as \(t\rightarrow 0\), and so \(\sigma (c)\le 0\) for any \(c>0\).

(ii) By (F4) and (F6), there exist \(\delta _0,\varrho _0>0\) such that

$$\begin{aligned} |F(t)|\ge \delta _0|t|^{q_0}, \ \ \ \ \forall \ |t|\le \varrho _0. \end{aligned}$$
(4.3)

For any \(c>0\), we choose a function \(u_0\in {\mathcal {C}}_0^{\infty }({\mathbb {R}}^3,[-\varrho _0,\varrho _0])\) satisfying \(\Vert u_0\Vert _2^2=c\). Then it follows from (3.9) and (4.3) that

$$\begin{aligned} I(u_0^t)\le \frac{at^2}{2}\Vert \nabla u_0\Vert _2^2+\frac{bt^4}{4} \Vert \nabla u_0\Vert _2^4 -K_{\infty }\delta _0t^{3(q_0-2)/2}\Vert u_0\Vert _{q_0}^{q_0}, \ \ \ \ \forall \ 0<t\le 1. \end{aligned}$$
(4.4)

Since \(0<3(q_0-2)/2<2\), (4.4) implies that \(I({u_0}^t)<0\) for small \(t\in (0,1)\). Jointly with the fact that \(\Vert {u_0}^t\Vert _2=\Vert {u_0}^t\Vert _2\), we have \(\sigma (c)\le \inf _{t\in (0,1]}I({u_0}^t)<0\).

(iii) We set the following scaling

$$\begin{aligned} u_t(x):=t^{1/2}u(x/t), \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3),\ t>0. \end{aligned}$$
(4.5)

Note that (F5) implies

$$\begin{aligned} \frac{F(t)}{|t|^{\mu _0-1}t}\ \ {\mathrm{is\ nondecreasing\ on}}\ (-\infty ,0)\ {\mathrm{and}}\ (0,\infty ). \end{aligned}$$
(4.6)

For any \(u\in {\mathcal {S}}_1\), by (1.3), (4.5) and (4.6), we have

$$\begin{aligned} \begin{aligned} I(u_t)&\ = \ \frac{at^2}{2}\Vert \nabla u\Vert _2^2+\frac{bt^4}{4}\Vert \nabla u\Vert _2^4 -t^3\int _{{{\mathbb {R}}}^3}K(tx)F\left( t^{1/2}u\right) {\mathrm {d}}x\\&\ \le \ \frac{at^2}{2}\Vert \nabla u\Vert _2^2+\frac{bt^4}{4}\Vert \nabla u\Vert _2^4 -K_{\infty }t^{3+\mu _0/2}\int _{{{\mathbb {R}}}^3}F(u){\mathrm {d}}x, \ \ \ \ \forall \ t>1, \end{aligned} \end{aligned}$$
(4.7)

which, together with \(3+\mu _0/2>4\), implies that \(I(u_t)\rightarrow -\infty \) as \(t\rightarrow +\infty \). Since \(\Vert u_t\Vert _2^2=t^4\Vert u\Vert _2^2=t^4\) for \(u\in {\mathcal {S}}_1\) and \(t>0\), there exists \({\mathcal {C}}_0>0\) such that \(\sigma (c)<0\) for any \(c>{\mathcal {C}}_0\). \(\square \)

Noting that Lemma 4.1 implies

$$\begin{aligned} \left\{ c\in (0,+\infty ), \sigma (c)<0\right\} \ne \emptyset , \end{aligned}$$
(4.8)

we have

$$\begin{aligned} c_*=\inf \left\{ c\in (0,+\infty ), \sigma (c)<0\right\} \end{aligned}$$

is well-defined.

Lemma 4.2

Assume that hypotheses (K1) and (F4) hold. Then the following properties hold:

  1. (i)

    if (F6) holds, then \(c_*=0\);

  2. (ii)

    if (F5) and (\(\hbox {F6}'\)) hold, then \(c_*\in (0,+\infty )\); moreover, \(\sigma (c)=0\) for any \(c\in (0,c_*]\) and \(\sigma (c)<0\) for any \(c>c_*\).

Proof

(i) Obviously, (i) follows directly from Lemma 4.1 (ii).

(ii) We first prove that \(c^*>0\). From (F4), (F6\('\)) and (3.23), we obtain

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^3}F(u){\mathrm {d}}x&\le C_3\left( \Vert u\Vert _{\frac{10}{3}}^{\frac{10}{3}}+\Vert u\Vert _p^{p}\right) \\&\le C_4\left( \Vert \nabla u\Vert _2^{2}\Vert u\Vert _2^{\frac{4}{3}} +\Vert \nabla u\Vert _2^{\frac{3(p-2)}{2}}\Vert u\Vert _2^{\frac{6-p}{2}}\right) , \ \ \ \ \forall \ u\in H^1({\mathbb {R}}^3). \end{aligned} \end{aligned}$$
(4.9)

It is easy to check that there exists a constant \(\epsilon >0\) small enough such that

$$\begin{aligned} \epsilon ^{\frac{2}{3}}\le \frac{a}{4C_4}\ \ \text{ and }\ \ \epsilon ^{\frac{6-p}{14-3p}}\le C_4^{\frac{4}{3p-14}}\frac{4a}{14-3p}\left( \frac{4b}{3p-10}\right) ^{\frac{3p-10}{14-3p}}. \end{aligned}$$
(4.10)

For any \(u\in {\mathcal {S}}_{\epsilon }\), namely \(\Vert u\Vert _2^2=\epsilon \), it follows from (1.3), (4.9) and (4.10) that

$$\begin{aligned} \begin{aligned} I(u)&\ \ge \ \frac{a}{2}\Vert \nabla u\Vert _2^2+\frac{b}{4}\Vert \nabla u\Vert _2^4 -C_4\left( \Vert u\Vert _2^{\frac{4}{3}}\Vert \nabla u\Vert _2^{2} +\Vert u\Vert _2^{\frac{6-p}{2}}\Vert \nabla u\Vert _2^{\frac{3(p-2)}{2}}\right) \\&\ \ge \ \frac{1}{4}\Vert \nabla u\Vert _2^2\left( a+b\Vert \nabla u\Vert _2^2-C_4\epsilon ^{\frac{6-p}{4}}\Vert \nabla u\Vert _2^{\frac{3p-10}{2}}\right) , \end{aligned} \end{aligned}$$
(4.11)

By Young’s inequality and (4.10), we have

$$\begin{aligned} \begin{aligned} C_4\epsilon ^{\frac{6-p}{4}}\Vert \nabla u\Vert _2^{\frac{3p-10}{2}}&\ = \ \left( \frac{4b}{3p-10}\right) ^{\frac{3p-10}{4}}\Vert \nabla u\Vert _2^{\frac{3p-10}{2}}\cdot \left( \frac{3p-10}{4b}\right) ^{\frac{3p-10}{4}}C_4\epsilon ^{\frac{6-p}{4}}\\&\ \le \ b\Vert \nabla u\Vert _2^2+C_4^{\frac{4}{14-3p}}\frac{14-3p}{4}\left( \frac{3p-10}{4b}\right) ^{\frac{3p-10}{14-3p}} \epsilon ^{\frac{6-p}{14-3p}}\\&\ \le \ b\Vert \nabla u\Vert _2^2+a, \end{aligned} \end{aligned}$$
(4.12)

which, together with (4.11), leads to \(I(u)\ge 0\) for \(u\in {\mathcal {S}}_{\epsilon }\). On the other hand, by Lemma 4.1 (i), one has \(\sigma (c)\le 0\) for \(c>0\). This shows that \(c_*=\inf \left\{ c\in (0,+\infty ), \sigma (c)<0\right\} >0\). Moreover, from the definition of \(c_*\) and Lemma 4.1 (i), we deduce that \(\sigma (c)=0\) for any \(c\in (0,c_*]\) and \(\sigma (c)<0\) for any \(c>c_*\). \(\square \)

Lemma 4.3

Assume that (K1), (K3), (F4), (F5) and either of (F6) and (\(\mathrm{{F6}}'\)) hold. Then

  1. (i)

    for any \(c>0\), \(\sigma (c)\) is continuous;

  2. (ii)

    for any \(c>c_*\),

    $$\begin{aligned} \sigma (c)<\sigma (\alpha )+\sigma (c-\alpha ), \ \ \ \ \forall \ 0<\alpha <c. \end{aligned}$$
    (4.13)

Proof

(i) For any \(c>0\), let \(c_n>0\) and \(c_n\rightarrow c\). For every \(n\in {\mathbb {N}}\), let \(u_n\in {\mathcal {S}}_{c_n}\) such that \(I(u_n)<\sigma (c_n)+\frac{1}{n}\le \frac{1}{n}\). Then (4.2) implies that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\), moreover, we have

$$\begin{aligned} \sigma (c) \le I\left( \sqrt{\frac{c}{c_n}}u_n\right) = I(u_n)+o(1)\le \sigma (c_n)+o(1). \end{aligned}$$
(4.14)

On the other hand, given a minimization sequence \(\{v_n\}\subset {\mathcal {S}}_{c}\) for I, we have

$$\begin{aligned} \sigma (c_n)\le I\left( \sqrt{\frac{c_n}{c}}v_n\right) \le I(v_n)+o(1)=\sigma (c)+o(1), \end{aligned}$$

which, jointly to (4.14), gives \(\lim _{n\rightarrow \infty }\sigma (c_n)=\sigma (c)\).

(ii) Note that (1.3) and (4.6) lead to

$$\begin{aligned} \begin{aligned} I(u_t)&\ = \ \frac{at^2}{2}\Vert \nabla u\Vert _2^2+\frac{bt^4}{4}\Vert \nabla u\Vert _2^4 -t^3\int _{{{\mathbb {R}}}^3}K(tx)F\left( t^{1/2}u\right) {\mathrm {d}}x\\&\ \le \ \frac{at^2}{2}\Vert \nabla u\Vert _2^2+\frac{bt^4}{4}\Vert \nabla u\Vert _2^4 -t^{4}\int _{{{\mathbb {R}}}^3}t^{(\mu _0-2)/2}K(tx)F(u){\mathrm {d}}x\\&\ \le \ t^4I(u)+\frac{at^2(1-t^2)}{2}\Vert \nabla u\Vert _2^2 \ < \ t^4I(u), \ \ \ \ \forall \ u \in {\mathcal {S}}_{c},\ c>0,\ t>1, \end{aligned} \end{aligned}$$
(4.15)

where the definition of \(u_t\) is given by (4.5).

Let \(\{u_n\}\subset {\mathcal {S}}_{c}\) be such that \(I(u_n)\rightarrow \sigma (c)\) for any \(c>0\). Since \(\Vert (u_n)_t\Vert _2^2=t^4\Vert u_n\Vert _2^2=t^4c\) for all \(t>0\), it follows from (4.15) that

$$\begin{aligned} \begin{aligned} \sigma (t^4c)&\ \le \ I((u_n)_t) \le t^4I(u_n)-\frac{at^2(t^2-1)}{2}\Vert \nabla u_n\Vert _2^2\\&\ \le \ t^4\sigma (c)+o(1), \ \ \ \ \forall \ t>1, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \sigma (tc)\le t\sigma (c),\ \ \ \ t>1,\ c>0. \end{aligned}$$
(4.16)

Let \(\{u_n\}\subset {\mathcal {S}}_{c}\) be such that \(I(u_n)\rightarrow \sigma (c)\) for any \(c>c^*\). We claim that exists a constant \(\rho _0>0\) such that

$$\begin{aligned} \liminf _{n\rightarrow \infty }\Vert \nabla u_n\Vert _2>\rho _0. \end{aligned}$$
(4.17)

Otherwise, if (4.17) is not true, then up to a subsequence, \(\Vert \nabla u_n\Vert _2\rightarrow 0\), and so (4.2) yields \(0>\sigma (c)=\lim _{n\rightarrow \infty }I(u_n)=0\). This contradiction shows that (4.17) holds. Since \(\Vert (u_n)_t\Vert _2^2=t^4\Vert u_n\Vert _2^2=t^4c\) for all \(t>0\), it follows from (4.15) and (4.17) that

$$\begin{aligned} \begin{aligned} \sigma (t^4c)&\ \le \ I((u_n)_t) \le t^4I(u_n)-\frac{at^2(t^2-1)}{2}\Vert \nabla u_n\Vert _2^2\\&\ \le \ t^4\sigma (c)-\frac{at^2(t^2-1)}{2}\rho _0^2+o(1), \ \ \ \ \forall \ t>1, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \sigma (tc)< t\sigma (c),\ \ \ \ t>1,\ c>c^*. \end{aligned}$$
(4.18)

If \(\alpha >c^*\) and \(c-\alpha >c^*\), we conclude from (4.18) that for any \(c>c^*\),

$$\begin{aligned} \sigma (c)=\frac{\alpha }{c}\sigma (c)+\frac{c-\alpha }{c}\sigma (c)<\sigma (\alpha )+\sigma (c-\alpha ), \ \ \ \ \forall \ 0<\alpha <c. \end{aligned}$$

If \(\alpha \le c^*\) or \(c-\alpha \le c^*\), then Lemma 4.2 (ii) implies that \(\sigma (c-\alpha )=0\) or \(\sigma (\alpha )=0\), and we deduce easily from (4.18) that (4.13) holds for any \(c>c^*\). This completes the proof. \(\square \)

Lemma 4.4

Assume that hypotheses (K1), (K3), (F4) and (F5) hold. Then the following properties hold:

  1. (i)

    \(\sigma (c)\le \sigma ^{\infty }(c)\) for any \(c>0\);

  2. (ii)

    if (F6) holds, then \(\sigma (c)\) has a minimizer for any \(c>0\);

  3. (iii)

    if (\(\mathrm{{F6}}'\)) holds, then \(\sigma (c)\) has a minimizer for any \(c\ge c_*\) and \(\sigma (c_*)=0\).

Proof

(i) Let \(\{u_n\}\subset {\mathcal {S}}_{c}\) be such that \(I^{\infty }(u_n)\rightarrow \sigma ^{\infty }(c)\) for any \(c>0\). Since \(K_{\infty }\le K(x)\) for all \(x\in {\mathbb {R}}^3\), it follows from (1.3) and (2.7) that

$$\begin{aligned} \sigma (c)\le I(u_n)\le I^{\infty }(u_n)=\sigma ^{\infty }(c)+o(1), \end{aligned}$$

which implies that \(\sigma (c)\le \sigma ^{\infty }(c)\) for any \(c>0\).

(ii) In view of Lemma 4.1 (ii), we have \(\sigma (c)<0\) for any \(c>0\). Let \(\{u_n\}\subset {\mathcal {S}}_{c}\) be such that \(I(u_n)\rightarrow \sigma (c)\) for any \(c>0\). Then (4.2) implies that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\). Thus, we can assume that for some \({\bar{u}}\in H^1({\mathbb {R}}^3)\) and up to a subsequence, \(u_n\rightharpoonup {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\). Here, we distinguish two cases: a) \({\bar{u}}\ne 0\) and b) \({\bar{u}}= 0\).

Case a): \({\bar{u}}\ne 0\). Then \(u_n\rightharpoonup {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\), \(u_n\rightarrow {\bar{u}}\) in \(L_{\mathrm {loc}}^s({\mathbb {R}}^3)\) for \(2\le s<6\) and \(u_n\rightarrow {\bar{u}}\) a.e. in \({\mathbb {R}}^3\). By Lemmas 3.12 and 4.3, we have

$$\begin{aligned} \begin{aligned} \sigma (c)&\ =\ \lim _{n\rightarrow \infty }I(u_n)= I({\bar{u}})+\lim _{n\rightarrow \infty }\left[ I(u_n-{\bar{u}}) +\frac{b}{2}\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla (u_n-{\bar{u}})\Vert _2^2\right] \\&\ \ge \ \sigma (\Vert {\bar{u}}\Vert _2^2)+\lim _{n\rightarrow \infty }\sigma (\Vert u_n-{\bar{u}}\Vert _2^2)\\&\ =\ \sigma (\Vert {\bar{u}}\Vert _2^2) +\sigma (c-\Vert {\bar{u}}\Vert _2^2). \end{aligned} \end{aligned}$$
(4.19)

If \(\Vert {\bar{u}}\Vert _2^2<c\), then relation (4.19) and Lemma 4.3 (ii) imply \(\sigma (c)\ge \sigma (\Vert {\bar{u}}\Vert _2^2)+\sigma (c-\Vert {\bar{u}}\Vert _2^2)>\sigma (c)\), which is impossible. This shows \(\Vert {\bar{u}}\Vert _2^2=c=\Vert u_n\Vert _2^2\). Moreover, it follows from (4.19) that \(u_n\rightarrow {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\), and so \(\sigma (c)=\lim _{n\rightarrow \infty }I(u_n)= I({\bar{u}})\). Hence, \({\bar{u}}\) is a minimizer of \(\sigma (c)\) for any \(c>0\) if \({\bar{u}}\ne 0\).

Case b): \({\bar{u}}=0\), that is, \(u_n\rightharpoonup 0\) in \(H^1({\mathbb {R}}^3)\). Then \(u_n\rightarrow 0\) in \(L_{\mathrm {loc}}^s({\mathbb {R}}^3)\) for \(2\le s<6\) and \(u_n\rightarrow 0\) a.e. in \({\mathbb {R}}^3\). By (K1), it is easy to check that

$$\begin{aligned} \int _{{\mathbb {R}}^3}[K(x)-K_{\infty }]F(u_n){\mathrm {d}}x=o(1). \end{aligned}$$
(4.20)

Then (1.3), (2.7) and (4.20) imply

$$\begin{aligned} I^{\infty }(u_n)\rightarrow \sigma (c). \end{aligned}$$
(4.21)

We claim that

$$\begin{aligned} \delta := \limsup _{n\rightarrow \infty }\sup _{y\in {\mathbb {R}}^3}\int _{B_1(y)}|u_n|^2{\mathrm {d}}x>0. \end{aligned}$$
(4.22)

In fact, if \(\delta =0\), then by Lions’ concentration-compactness principle [27, Lemma 1.21], one has \(u_n\rightarrow 0\) in \(L^s({\mathbb {R}}^3)\) for \(2<s<6\), and so (4.1) implies that \(\int _{{\mathbb {R}}^3}F(u_n){\mathrm {d}}x\rightarrow 0\). Then by (1.3), we have

$$\begin{aligned} 0>\sigma (c)=\lim _{n\rightarrow \infty }I(u_n) =\lim _{n\rightarrow \infty }\left( \frac{a}{2}\Vert \nabla u_n\Vert _2^2+\frac{b}{4}\Vert \nabla u_n\Vert _2^4\right) \ge 0, \end{aligned}$$

which is impossible. It follows that \(\delta >0\) and there exists \(\{y_n\}\subset {\mathbb {R}}^3\) such that

$$\begin{aligned} \int _{B_{1+\sqrt{3}} (y_n)}|u_n|^2{\mathrm {d}}x\ge \frac{\delta }{2}. \end{aligned}$$
(4.23)

Let \({\hat{u}}_n(x)=u_n(x+y_n)\). Then (4.21) leads to

$$\begin{aligned} {\hat{u}}_n\in {\mathcal {S}}_{c}, \ \ \ \ I^{\infty }({\hat{u}}_n)\rightarrow \sigma (c). \end{aligned}$$
(4.24)

In view of (4.23), we may assume that there exists \({\hat{u}}\in H^1({\mathbb {R}}^3)\setminus \{0\}\) such that, passing to a subsequence,

$$\begin{aligned} \left\{ \begin{array}{ll} {\hat{u}}_n\rightharpoonup {\hat{u}}, &{} \text{ in } \ H^1({\mathbb {R}}^3); \\ {\hat{u}}_n\rightarrow {\hat{u}}, &{} \text{ in } \ L_{\mathrm {loc}}^s({\mathbb {R}}^3), \ \forall \ s\in [1, 6);\\ {\hat{u}}_n\rightarrow {\hat{u}}, &{} \text{ a.e. } \text{ on } \ {\mathbb {R}}^3. \end{array} \right. \end{aligned}$$
(4.25)

Then it follows from (4.24), (4.25), (i), Lemmas 3.12 and 4.3 that

$$\begin{aligned} \begin{aligned} \sigma (c)&\ =\ \lim _{n\rightarrow \infty }I^{\infty }({\hat{u}}_n)\\&\ =\ I^{\infty }({\hat{u}})+\lim _{n\rightarrow \infty }\left[ I^{\infty }({\hat{u}}_n-{\hat{u}}) +\frac{b}{2}\Vert \nabla {\hat{u}}\Vert _2^2\Vert \nabla ({\hat{u}}_n-{\hat{u}})\Vert _2^2\right] \\&\ \ge \ \sigma ^{\infty }(\Vert {\hat{u}}\Vert _2^2)+\lim _{n\rightarrow \infty }\sigma ^{\infty }(\Vert {\hat{u}}_n-{\hat{u}}\Vert _2^2)\\&\ =\ \sigma ^{\infty }(\Vert {\hat{u}}\Vert _2^2) +\sigma ^{\infty }(c-\Vert {\hat{u}}\Vert _2^2) \ge \sigma (\Vert {\hat{u}}\Vert _2^2) +\sigma (c-\Vert {\hat{u}}\Vert _2^2), \end{aligned} \end{aligned}$$
(4.26)

If \(\Vert {\hat{u}}\Vert _2^2<c\), then (4.26) and Lemma 4.3 (ii) imply \(\sigma (c)\ge \sigma (\Vert {\hat{u}}\Vert _2^2)+\sigma (c-\Vert {\hat{u}}\Vert _2^2)>\sigma (c)\) which is impossible. Hence, \(\Vert {\hat{u}}\Vert _2^2=c\), moreover, it follows from (4.26) that

$$\begin{aligned} {\hat{u}}_n\rightarrow {\hat{u}}\ \text{ in }\ H^1({\mathbb {R}}^3), \ \ \sigma (c)\le I({\hat{u}})\le I^{\infty }({\hat{u}})=\lim _{n\rightarrow \infty }I^{\infty }({\hat{u}}_n)=\sigma (c). \end{aligned}$$
(4.27)

This shows that \({\hat{u}}\) is a minimizer of \(\sigma (c)\) for any \(c>0\).

(iii) In view of Lemma 4.1 (iii), we have \(\sigma (c)<0\) for any \(c>c_*\). Arguing as in the proof of (ii), we deduce that \(\sigma (c)\) is attained for any \(c>c_*\). Let \(c_n=c_*+\frac{1}{n}\). Note that Lemma 4.2 (ii) leads to \(\sigma (c_n)<0\) for every \(n\in {\mathbb {N}}\). Arguing as in the proof of (ii), there exists \(\{u_n\}\subset {\mathcal {S}}_{c_n}\) such that

$$\begin{aligned} I(u_n)= \sigma (c_n)<0 \ \ {\mathrm{for\ every}}\ n\in {\mathbb {N}}. \end{aligned}$$
(4.28)

By the definition of \(c_*\) and Lemma 4.3 (i), we have \(I(u_n)=\sigma (c_n)\rightarrow \sigma (c_*)=0\). Then (4.2) implies that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\). We then may assume that there exists \({\bar{u}}\in H^1({\mathbb {R}}^3)\) such that, up to a subsequence, \(u_n\rightharpoonup {\bar{u}}\) in \(H^1({\mathbb {R}}^3)\). To prove that \(\sigma (c_*)\) is attained, we need to distinguish two cases: a) \({\bar{u}}\ne 0\) and b) \({\bar{u}}=0\).

Case a): \({\bar{u}}\ne 0\). By Lemmas 3.124.2 and 4.3, we have

$$\begin{aligned} \begin{aligned} 0&\ =\ \sigma (c^*)=\lim _{n\rightarrow \infty }I(u_n)= I({\bar{u}})+\lim _{n\rightarrow \infty }\left[ I(u_n-{\bar{u}}) +\frac{b}{2}\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla (u_n-{\bar{u}})\Vert _2^2\right] \\&\ \ge \ \sigma (\Vert {\bar{u}}\Vert _2^2)+\lim _{n\rightarrow \infty }\sigma (\Vert u_n-{\bar{u}}\Vert _2^2)\\&\ =\ \sigma (\Vert {\bar{u}}\Vert _2^2) +\sigma (c^*-\Vert {\bar{u}}\Vert _2^2)=0, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert \nabla (u_n-{\bar{u}})\Vert _2=0, \end{aligned}$$
(4.29)

and so a standard argument shows that \(I({\bar{u}})=\lim _{n\rightarrow \infty }I(u_n)=\sigma (c^*)=0\). To prove that \({\bar{u}}\) is a minimizer of \(\sigma (c^*)\), it suffices to show that \(\Vert {\bar{u}}\Vert _2^2=c^*\). By contradiction, let us assume that \(\Vert {\bar{u}}\Vert _2^2<c^*\). Let

$$\begin{aligned} t^*=\left( \frac{c^*}{\Vert {\bar{u}}\Vert _2^2}\right) ^{\frac{1}{4}}. \end{aligned}$$

Then \(t^*>1\), and \(\Vert {\bar{u}}_{t^*}\Vert _2^2=(t^*)^4\Vert {\bar{u}}\Vert _2^2=c^*\) by the scaling (4.5). Thus it follows from (4.15) that

$$\begin{aligned} 0 =\sigma (c^*)\le I({\bar{u}}_{t^*}) \le (t^*)^4I({\bar{u}})+\frac{a(t^*)^2[1-(t^*)^2]}{2}\Vert \nabla {\bar{u}}\Vert _2^2 < (t^*)^4I({\bar{u}})=0. \end{aligned}$$
(4.30)

This contradiction shows that \(\Vert {\bar{u}}\Vert _2^2=c^*\). Hence, \({\bar{u}}\in {\mathcal {S}}_{c^*}\) and \(I({\bar{u}})=\sigma (c^*)=0\).

Case b): \({\bar{u}}=0\). By (2.7) and (4.20), we deduce that \(I^{\infty }(u_n)\rightarrow 0\). We claim that (4.22) holds. In fact, if \(\delta =0\), by Lions’ concentration-compactness principle [27, Lemma 1.21], we obtain that \(u_n\rightarrow 0\) in \(L^s({\mathbb {R}}^3)\) for \(2<s<6\). By (F4) and (F6\('\)), for any \(\varepsilon >0\), there exists \(C_{\varepsilon }>0\) such that

$$\begin{aligned} |F(t)|\le \varepsilon |t|^{\frac{10}{3}}+C_{\varepsilon }|t|^{p}, \ \ \ \ \forall \ t\in {\mathbb {R}}. \end{aligned}$$
(4.31)

Then (4.31) implies \(\int _{{\mathbb {R}}^3}F(u_n){\mathrm {d}}x\rightarrow 0\). Jointly with \(I(u_n)\rightarrow 0\), we have \(\Vert \nabla u_n\Vert _2\rightarrow 0\). From (1.3), (3.23) and (4.31), we deduce

$$\begin{aligned} I(u_n) \ge \frac{1}{4}\Vert \nabla u_n\Vert _2^2\left( 2a+b\Vert \nabla u_n\Vert _2^2 -4\varepsilon \Vert u_n\Vert _2^{\frac{4}{3}}-4C_{\varepsilon }\Vert u_n\Vert _2^{\frac{6-p}{2}}\Vert \nabla u_n\Vert _2^{\frac{3p-10}{2}}\right) , \ \ \ \ \forall \ n\in {\mathbb {N}}, \end{aligned}$$

which, together with \(3p-10>0\) and arbitrariness of \(\varepsilon \), implies that \(I(u_n)\ge 0\) for large \(n\in {\mathbb {N}}\). This contradicts (4.28), and thus (4.22) holds. Let \({\hat{u}}_n(x)=u_n(x+y_n)\). From (4.22), there exists \({\hat{u}}\in H^1({\mathbb {R}}^3)\setminus \{0\}\) such that, passing to a subsequence, \({\hat{u}}_n\rightharpoonup {\hat{u}}\) in \(H^1({\mathbb {R}}^3)\). Similarly to the proof of (4.26), we have

$$\begin{aligned} \begin{aligned} 0&\ =\ \sigma (c^*)=\lim _{n\rightarrow \infty }I^{\infty }({\hat{u}}_n)\\&\ =\ I^{\infty }({\hat{u}})+\lim _{n\rightarrow \infty }\left[ I^{\infty }({\hat{u}}_n-{\hat{u}}) +\frac{b}{2}\Vert \nabla {\hat{u}}\Vert _2^2\Vert \nabla ({\hat{u}}_n-{\hat{u}})\Vert _2^2\right] \\&\ \ge \ \sigma ^{\infty }(\Vert {\hat{u}}\Vert _2^2)+\lim _{n\rightarrow \infty }\sigma ^{\infty }(\Vert {\hat{u}}_n-{\hat{u}}\Vert _2^2)\\&\ =\ \sigma ^{\infty }(\Vert {\hat{u}}\Vert _2^2) +\sigma ^{\infty }(c^*-\Vert {\hat{u}}\Vert _2^2) \ge \sigma (\Vert {\hat{u}}\Vert _2^2) +\sigma (c^*-\Vert {\hat{u}}\Vert _2^2)=0, \end{aligned} \end{aligned}$$
(4.32)

which implies

$$\begin{aligned} \lim _{n\rightarrow \infty }\Vert \nabla ({\hat{u}}_n-{\hat{u}})\Vert _2=0. \end{aligned}$$

Proceeding as in the proof of Case a), we deduce that \({\hat{u}}\in {\mathcal {S}}_{c^*}\) and \(I({\hat{u}})=\sigma (c^*)=0\).

\(\square \)

4.2 Local Minimizers on the Constraint \({\mathcal {S}}_c\)

In this subsection, we shall look for a local minimizer of I on the constraint \({\mathcal {S}}_c\), which is a critical point of \(I|_{{\mathcal {S}}_c}\).

For \(k>0\), set

$$\begin{aligned} {\mathcal {S}}_c(k):=\left\{ u\in {\mathcal {S}}_{c}: \Vert \nabla u\Vert _2^2=k\right\} . \end{aligned}$$

For any \(0<c<c_*\) and \(u\in {\mathcal {S}}_{c}\), it follows from (1.3), (3.23) and (4.31) that

$$\begin{aligned} I(u) \ge \frac{1}{4}\Vert \nabla u\Vert _2^2\left( 2a+b\Vert \nabla u\Vert _2^2 -4\varepsilon c_*^{\frac{2}{3}}-4C_{\varepsilon }c_*^{\frac{6-p}{4}}\Vert \nabla u\Vert _2^{\frac{3p-10}{2}}\right) . \end{aligned}$$
(4.33)

Since \(3p-10>0\) and \(\varepsilon \) is arbitrary, there exists \(k_0>0\) independent of c such that

$$\begin{aligned} I^{\infty }(u)\ge I(u) \ge \frac{ak}{4}>0, \ \ \ \ \forall \ u\in {\mathcal {S}}_c(k),\ 0<k\le k_0. \end{aligned}$$
(4.34)

In view of Lemma 4.1 (i), we have

$$\begin{aligned} {\bar{\sigma }}(c):=\inf _{u\in {\mathcal {S}}_{c}\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k)}I(u) \ge \inf _{u\in {\mathcal {S}}_{c}}I(u)=\sigma (c)>-\infty , \ \ \ \ \forall \ c>0. \end{aligned}$$
(4.35)

Lemma 4.5

Assume that hypotheses (K1), (K3), (F4), (F5) and (\(\mathrm{{F6}}'\)) hold. Then \({\bar{\sigma }}(c)\) is continuous on \((0,+\infty )\) and for any \(c>0\),

$$\begin{aligned} {\bar{\sigma }}(c)<{\bar{\sigma }}(\alpha )+{\bar{\sigma }}(c-\alpha ), \ \ \ \ \forall \ 0<\alpha <c. \end{aligned}$$
(4.36)

Proof

Similarly to the proof of Lemma 4.3, we deduce that \({\bar{\sigma }}(c)\) is continuous for any \(c>0\). Let \(\{u_n\}\subset {\mathcal {S}}_{c}\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_{c}(k)\) be such that \(I(u_n)\rightarrow {\bar{\sigma }}(c)\) for any \(c>0\). Then (4.2) implies that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\). Noting that \(\Vert \nabla u_n\Vert _2^2>k_0\) and \(\Vert (u_n)_t\Vert _2^2=t^4\Vert u_n\Vert _2^2=t^4c\) for all \(n\in {\mathbb {N}}\) and \(t>0\), it follows from (4.15) that

$$\begin{aligned} \begin{aligned} {\bar{\sigma }}(t^4c)&\ \le \ I((u_n)_t) \le t^4I(u_n)-\frac{at^2(t^2-1)}{2}\Vert \nabla u_n\Vert _2^2\\&\ \le \ t^4{\bar{\sigma }}(c)-\frac{at^2(t^2-1)}{2}k_0^2+o(1), \ \ \ \ \forall \ t>1,\ c>0, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} {\bar{\sigma }}(tc)< t{\bar{\sigma }}(c),\ \ \ \ t>1,\ c>0. \end{aligned}$$
(4.37)

Then it follows from (4.37) that

$$\begin{aligned} {\bar{\sigma }}(c)=\frac{\alpha }{c}{\bar{\sigma }}(c)+\frac{c-\alpha }{c}{\bar{\sigma }}(c)<{\bar{\sigma }}(\alpha )+{\bar{\sigma }}(c-\alpha ), \ \ \ \ \forall \ 0<\alpha <c, \end{aligned}$$

which completes the proof. \(\square \)

Lemma 4.6

(Geometry of local minima) Assume that hypotheses (K1), (K3), (F4), (F5) and (\(\hbox {F6}'\)) hold. Then there exists \(0<c_0<c_*\) such that

$$\begin{aligned} 0<{\bar{\sigma }}(c)< \frac{a}{4}k_0\le \inf _{u\in {\mathcal {S}}_c(k_0)}I(u), \ \ \ \ \forall \ c\in (c_0, c_*). \end{aligned}$$
(4.38)

Proof

By the definition of \(c_*\) (see (2.2)), Lemma 4.1 (i) and (4.35), we obtain

$$\begin{aligned} {\bar{\sigma }}(c) \ge \sigma (c)=0, \ \ \ \ \forall \ 0<c<c_*. \end{aligned}$$
(4.39)

We first prove that \({\bar{\sigma }}(c)>0\) for all \(0<c<c_*\). If not, then \({\bar{\sigma }}({\bar{c}})=0\) for some \({\bar{c}}\in (0,c_*)\), and so there exists \(\{u_n\}\subset {\mathcal {S}}_{{\bar{c}}}\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_{{\bar{c}}}(k)\) such that \(I(u_n)\rightarrow 0\) and \(\Vert \nabla u_n\Vert _2^2> k_0\). Since \(\Vert (u_n)_t\Vert _2^2=t^4\Vert u_n\Vert _2^2=t^4{\bar{c}}\) by the scaling (4.5), it follows from (4.15) that

$$\begin{aligned} \begin{aligned} {\bar{\sigma }}(t^4{\bar{c}})&\le \lim _{n\rightarrow \infty }I((u_n)_t) \le \lim _{n\rightarrow \infty }\left[ t^4I(u_n)-\frac{at^2(t^2-1)}{2}\Vert \nabla u_n\Vert _2^2\right] \\&\le -\frac{ak_0t^2(t^2-1)}{2}<0, \ \ \ \ \forall \ 1<t<\left( \frac{c_*}{{\bar{c}}}\right) ^{1/4}, \end{aligned} \end{aligned}$$

which contradicts (4.39). Hence, \({\bar{\sigma }}(c)>0\) for all \(0<c<c_*\).

Next, inspired by [29, (3.9)], we prove that there exists \(c_0>0\) such that the last inequality of (4.38) holds. In view of Lemma 4.4 (iii), there exists \(u_*\in {\mathcal {S}}_{c_*}\) such that \(I(u_*)=\sigma (c_*)=0\). From the continuity of \(I(tu_*)\) on \(t\in (0,\infty )\), we deduce that there exists a constant \(t_0\in (0,1)\) sufficiently close to 1 such that

$$\begin{aligned} I(tu_*)<\frac{a}{4}k_0,\ \ \ \ \ \forall \ t_0\le t\le 1, \end{aligned}$$
(4.40)

which, together with (4.34), implies

$$\begin{aligned} tu_*\in {\mathcal {S}}_{tc_*}\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_{tc_*}(k), \ \ \ \ \ \forall \ t_0\le t\le 1. \end{aligned}$$
(4.41)

Letting \(c_0=t_0c_*\), it follows from (4.34), (4.40) and (4.41) that for any \(c\in (c_0,c_*)\), there exists \(v_*\in {\mathcal {S}}_{c}\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_{c}(k)\) such that

$$\begin{aligned} {\bar{\sigma }}(c)\le I(v_*)< \frac{a}{4}k_0\le \inf _{u\in {\mathcal {S}}_{c}(k_0)}I(u). \end{aligned}$$

The proof is completed. \(\square \)

Define

$$\begin{aligned} {\bar{\sigma }}^{\infty }(c):=\inf _{u\in {\mathcal {S}}_{c}\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k)}I^{\infty }(u) \ge \inf _{u\in {\mathcal {S}}_{c}}I^{\infty }(u)=\sigma ^{\infty }(c)>-\infty , \ \ \ \ \forall \ c>0. \end{aligned}$$
(4.42)

Lemma 4.7

Assume that hypotheses (K1), (K3), (F4), (F5) and (\(\mathrm{{F6}}'\)) hold. Then

  1. (i)

    \({\bar{\sigma }}(c)\le {\bar{\sigma }}^{\infty }(c)\) for any \(c>0\);

  2. (ii)

    for any \(c\in (c_0, c_*)\), problem (1.1) admits a solution \(({\bar{u}}_c,{\bar{\lambda }}_c)\in \left( {\mathcal {S}}_c\setminus \bigcup _{0<k< k_0}{\mathcal {S}}_c(k)\right) \times {\mathbb {R}}\) such that \(I({\bar{u}}_c)={\bar{\sigma }}(c)>0\).

Proof

(i) Similarly to the proof of Lemma 4.4 (i), we get \({\bar{\sigma }}(c)\le {\bar{\sigma }}^{\infty }(c)\) for any \(c>0\).

(ii) By Ekeland’s variational principle, there exists a sequence \(\{u_n\}\subset {\mathcal {S}}_c\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k)\) such that

$$\begin{aligned} {\left\{ \begin{array}{ll} {\bar{\sigma }}(c)\le I(u_n)\le {\bar{\sigma }}(c)+\frac{1}{n},\\ I(v)\ge I(u_n)-\frac{1}{n}\Vert u_n-v\Vert , \ \ \ \ \forall \ v\in {\mathcal {S}}_c\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k). \end{array}\right. } \end{aligned}$$
(4.43)

Then (4.2) implies that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\). Thus, there exists \({\bar{u}}_c\in H^1({\mathbb {R}}^3)\) such that, up to a subsequence, \(u_n\rightharpoonup {\bar{u}}_c\) in \(H^1({\mathbb {R}}^3)\), \(u_n\rightarrow {\bar{u}}_c\) in \(L_{\mathrm {loc}}^s({\mathbb {R}}^3)\) for \(2\le s<6\) and \(u_n\rightarrow {\bar{u}}_c\) a.e. in \({\mathbb {R}}^3\). For brevity, we denote \({\bar{u}}_c\) by \({\bar{u}}\). We complete our proof in two steps as follows.

Step 1. We prove that if \({\bar{u}}\ne 0\), then \(I({\bar{u}})={\bar{\sigma }}(c)\) and there exists \({\bar{\lambda }}\in {\mathbb {R}}\) such that \(I'({\bar{u}})-{\bar{\lambda }}u=0\) for any \(c\in (c_0, c_*)\).

Similarly to the proof of [29, (3.17)], we have

$$\begin{aligned} |\langle I'(u_n)-\lambda _nu_n,\varphi \rangle |\le \frac{C}{n}, \ \ \ \ \forall \ \varphi \in {\mathcal {C}}_0^{\infty }({\mathbb {R}}^3), \end{aligned}$$
(4.44)

where \(\lambda _n=-\frac{\langle I'(u_n),u_n\rangle }{c}\) and \(C>0\) is a constant independent of n. Since \(\{\lambda _n\}\) is bounded, we may assume that up to a subsequence, \(\lambda _n\rightarrow {\bar{\lambda }}\) for some \(\lambda \in {\mathbb {R}}\).

Now, we claim that

$$\begin{aligned} {\mathrm{up\ to\ a\ subsequence,}}\ \Vert \nabla (u_n-{\bar{u}})\Vert _2\rightarrow 0\ {\mathrm{if}}\ {\bar{u}}\ne 0. \end{aligned}$$
(4.45)

Assume by contradiction that \(\liminf _{n\rightarrow \infty }\Vert \nabla (u_n-{\bar{u}})\Vert _2^2>0\). Then it follows from the scaling (4.5) that there exists \(T_0>1\) such that

$$\begin{aligned} \Vert \nabla {\bar{u}}_{t}\Vert _2^2=t^2\Vert \nabla {\bar{u}}\Vert _2^2>k_0, \ \ \ \ \Vert \nabla (u_n-{\bar{u}})_{t}\Vert _2^2=t^2\Vert \nabla (u_n-{\bar{u}})\Vert _2^2>k_0, \ \ \ \ \forall \ t>T_0. \end{aligned}$$
(4.46)

By (4.15), (4.43) and Lemma 3.12, we have

$$\begin{aligned} \begin{aligned} {\bar{\sigma }}(c)+o(1)&\ =\ I(u_n)= I({\bar{u}})+I(u_n-{\bar{u}}) +\frac{b}{2}\Vert \nabla {\bar{u}}\Vert _2^2\Vert \nabla (u_n-{\bar{u}})\Vert _2^2\\&\ \ge \ \frac{1}{t^4}\left[ I({\bar{u}}_t)+I((u_n-{\bar{u}})_t)\right] +\frac{at^2(t^2-1)}{2t^4}\left[ \Vert \nabla {\bar{u}}\Vert _2^2+\Vert \nabla (u_n-{\bar{u}})\Vert _2^2\right] \\&\ \ge \ \frac{1}{t^4}\left[ {\bar{\sigma }}(t^4\Vert {\bar{u}}\Vert _2^2)+{\bar{\sigma }}(t^4c-t^4\Vert {\bar{u}}\Vert _2^2)\right] +\frac{at^2(t^2-1)}{2t^4}\Vert \nabla u_n\Vert _2^2+o(1)\\&\ \ge \ \frac{1}{t^4}{\bar{\sigma }}(t^4c)+\frac{at^2(t^2-1)}{2t^4}k_0+o(1), \ \ \ \ \forall \ t\ge T_0. \end{aligned} \end{aligned}$$
(4.47)

From (4.2), we deduce

$$\begin{aligned} \liminf _{t\rightarrow +\infty }\frac{{\bar{\sigma }}(t^4c)}{t^4}\ge 0. \end{aligned}$$
(4.48)

By (4.48), there exists \(T_1\ge T_0\) such that

$$\begin{aligned} \frac{1}{T_1^4}{\bar{\sigma }}(T_1^4c)\ge -\frac{ak_0}{16}, \ \ \ \ \frac{aT_1^2(T_1^2-1)}{2t^4}k_0\ge \frac{3ak_0}{8}. \end{aligned}$$
(4.49)

Then (4.38), (4.47) and (4.49) lead to

$$\begin{aligned} \frac{ak_0}{4}> {\bar{\sigma }}(c)\ge \frac{1}{T_1^4}{\bar{\sigma }}(T_1^4c)+\frac{aT_1^2(T_1^2-1)}{2t^4}k_0 \ge \frac{5ak_0}{16}. \end{aligned}$$
(4.50)

This contradiction shows that (4.45) holds. Thus, from (F4), (F5), (4.43), (4.44) and (4.45), we derive

$$\begin{aligned} {\bar{\sigma }}(c)=\lim _{n\rightarrow \infty }I(u_n)=I({\bar{u}}) \end{aligned}$$
(4.51)

and

$$\begin{aligned} \begin{aligned} 0&\ =\ \lim _{n\rightarrow \infty }\langle I'(u_n)-\lambda _nu_n,\varphi \rangle \\&\ =\ \lim _{n\rightarrow \infty }\left[ a\int _{{\mathbb {R}}^3}\nabla u_n\cdot \nabla \varphi {\mathrm {d}}x +b\Vert \nabla u_n\Vert _2^2\int _{{\mathbb {R}}^3}\nabla u_n\cdot \nabla \varphi {\mathrm {d}}x\right. \\&\ \ \ \ \ \ -\left. \int _{{\mathbb {R}}^3}K(x)f(u_n)\varphi {\mathrm {d}}x -\lambda _n\int _{{\mathbb {R}}^3} u_n\varphi {\mathrm {d}}x\right] \\&\ =\ a\int _{{\mathbb {R}}^3}\nabla {\bar{u}}\cdot \nabla \varphi {\mathrm {d}}x {+}b\Vert \nabla {\bar{u}}\Vert _2^2\int _{{\mathbb {R}}^3}\nabla {\bar{u}}\cdot \nabla \varphi {\mathrm {d}}x {-}\int _{{\mathbb {R}}^3}K(x)f({\bar{u}})\varphi {\mathrm {d}}x {-}{\bar{\lambda }}\int _{{\mathbb {R}}^3}{\bar{u}}\varphi {\mathrm {d}}x\\&\ =\ \langle I'({\bar{u}})-{\bar{\lambda }}{\bar{u}},\varphi \rangle , \ \ \ \ \forall \ \varphi \in {\mathcal {C}}_0^{\infty }({\mathbb {R}}^3). \end{aligned} \end{aligned}$$
(4.52)

By Fatou’s lemma, we have \(\Vert {\bar{u}}\Vert _2^2\le \liminf _{n\rightarrow \infty }\Vert u_n\Vert _2^2=c\). In view of (4.51) and (4.52), to finish the proof of Step 1, it suffices to show that \(\Vert {\bar{u}}\Vert _2^2=c\).

Suppose by contradiction that \(\Vert {\bar{u}}\Vert _2^2<c\). As in [29, (3.14)], for any \(\varepsilon >0\), there exists \(\delta =\delta (\varepsilon )>0\) such that if \(u\in \bigcup _{k_0-\delta \le k\le k_0+\delta }{\mathcal {S}}_c(k)\), then

$$\begin{aligned} I(u)\ge \inf _{u\in {\mathcal {S}}_c(k_0)}I(u)-\varepsilon , \end{aligned}$$

and so

$$\begin{aligned} u_n\in {\mathcal {S}}_c\setminus \bigcup _{0\le k\le k_0+\delta }{\mathcal {S}}_c(k), \ \mathrm{for}\ n\ \mathrm{large\ enough}, \end{aligned}$$
(4.53)

that is, \(\{u_n\}\) stays away from the boundary. Jointly with (4.45), we have \(\Vert \nabla {\bar{u}}\Vert _2^2>k_0\). Let \({\bar{t}}:=(c/\Vert {\bar{u}}\Vert _2^2)^{1/4}>1\). By the scaling (4.5), one has \(\Vert {\bar{u}}_{{\bar{t}}}\Vert _2^2={\bar{t}}^4\Vert {\bar{u}}\Vert _2^2\) and \(\Vert \nabla {\bar{u}}_{{\bar{t}}}\Vert _2^2={\bar{t}}^2\Vert \nabla {\bar{u}}\Vert _2^2>k_0\). Then it follows from (4.15) and (4.51) that

$$\begin{aligned} \begin{aligned} {\bar{t}}^4{\bar{\sigma }}(c)&\ = \ {\bar{t}}^4I({\bar{u}})\ge I({\bar{u}}_{{\bar{t}}})+\frac{a{\bar{t}}^2({\bar{t}}^2-1)}{2}\Vert \nabla {\bar{u}}\Vert _2^2\\&\ \ge \ {\bar{\sigma }}(c)+\frac{a{\bar{t}}^2({\bar{t}}^2-1)}{2}k_0, \end{aligned} \end{aligned}$$
(4.54)

which implies

$$\begin{aligned} {\bar{\sigma }}(c) \ge \frac{a\sqrt{c}k_0}{2(\Vert {\bar{u}}\Vert +\sqrt{c})} >\frac{ak_0}{4}, \ \ \ \ \forall \ c\in (c_0, c_*). \end{aligned}$$

This contradicts (4.38). Hence, we have \(\Vert {\bar{u}}\Vert _2^2=c\). Moreover, (4.52) implies that \(({\bar{u}},{\bar{\lambda }}) \in \left( {\mathcal {S}}_c\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k)\right) \times {\mathbb {R}}\) solves problem (1.1). The proof of Step 1 is completed.

Step 2. We prove that \({\bar{u}}\ne 0\).

Assume by contradiction that \({\bar{u}}=0\), that is, \(u_n\rightharpoonup 0\) in \(H^1({\mathbb {R}}^3)\). Then \(u_n\rightarrow 0\) in \(L_{\mathrm {loc}}^s({\mathbb {R}}^3)\) for \(2\le s<6\) and \(u_n\rightarrow 0\) a.e. in \({\mathbb {R}}^3\). Then (1.3), (2.7), (4.20) and (4.43) imply

$$\begin{aligned} I^{\infty }(u_n)\rightarrow {\bar{\sigma }}(c). \end{aligned}$$
(4.55)

We first prove that \(\{u_n\}\) is not vanishing, that is, relation (4.22) holds. Otherwise, if \(\{u_n\}\) is vanishing, by Lions’ concentration-compactness principle [27, Lemma 1.21], one has \(u_n\rightarrow 0\) in \(L^s({\mathbb {R}}^3)\) for \(2<s<6\), and so (4.31) implies that \(\int _{{\mathbb {R}}^3}F(u_n){\mathrm {d}}x\rightarrow 0\). Since \(\{u_n\}\subset {\mathcal {S}}_c\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k)\), by (1.3), (4.38) and (4.43), we have

$$\begin{aligned} \begin{aligned} \frac{ak_0}{2}+\frac{bk_0^2}{4}&\ \le \ \lim _{n\rightarrow \infty }\left( \frac{a}{2}\Vert \nabla u_n\Vert _2^2+\frac{b}{4}\Vert \nabla u_n\Vert _2^4\right) \\&\ =\ \lim _{n\rightarrow \infty }I(u_n)={\bar{\sigma }}(c)<\frac{ak_0}{4}, \end{aligned} \end{aligned}$$

which is impossible. Hence, we have (4.22) holds, and so there exists \(\{y_n\}\subset {\mathbb {R}}^3\) such that (4.23) holds. Let \({\tilde{u}}_n(x)=u_n(x+y_n)\). Then (4.55) leads to

$$\begin{aligned} {\tilde{u}}_n\in {\mathcal {S}}_c\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k), \ \ \ \ I^{\infty }({\tilde{u}}_n)\rightarrow {\bar{\sigma }}(c). \end{aligned}$$
(4.56)

We may assume that there exists \({\tilde{u}}\in H^1({\mathbb {R}}^3)\setminus \{0\}\) such that, passing to a subsequence,

$$\begin{aligned} \left\{ \begin{array}{ll} {\tilde{u}}_n\rightharpoonup {\tilde{u}}, &{} \text{ in } \ H^1({\mathbb {R}}^3); \\ {\tilde{u}}_n\rightarrow {\tilde{u}}, &{} \text{ in } \ L_{\mathrm {loc}}^s({\mathbb {R}}^3), \ \forall \ s\in [1, 6);\\ {\tilde{u}}_n\rightarrow {\tilde{u}}, &{} \text{ a.e. } \text{ on } \ {\mathbb {R}}^3. \end{array} \right. \end{aligned}$$
(4.57)

Similarly to the proof of [29, (3.17)], we have

$$\begin{aligned} |\langle (I^{\infty })'({\tilde{u}}_n)-\lambda _n{\tilde{u}}_n,\varphi \rangle |\le \frac{C}{n}, \ \ \ \ \forall \ \varphi \in {\mathcal {C}}_0^{\infty }({\mathbb {R}}^3), \end{aligned}$$
(4.58)

where \(\lambda _n\) and \(C>0\) are the same as that of (4.44). Arguing as in (4.45), (4.51) and (4.52), we prove that

$$\begin{aligned} {\mathrm{up\ to\ a\ subsequence}}\ \Vert \nabla ({\tilde{u}}_n-{\tilde{u}})\Vert _2\rightarrow 0 \end{aligned}$$
(4.59)

and

$$\begin{aligned} {\bar{\sigma }}(c)=\lim _{n\rightarrow \infty }I^{\infty }({\tilde{u}}_n)=I^{\infty }({\tilde{u}}). \end{aligned}$$
(4.60)

Moreover, as in the proof of Step 1, we have \(\Vert {\tilde{u}}\Vert _2^2=c\) and \(\Vert \nabla {\tilde{u}}\Vert _2^2>k_0\). Hence, \({\tilde{u}} \in {\mathcal {S}}_c\setminus \bigcup _{0<k\le k_0}{\mathcal {S}}_c(k)\). Since \(K(x)\ge (\not \equiv ) K_{\infty }\) and \(\Vert {\tilde{u}}\Vert _2^2=c>0\), it follows that there exist \(r>0\) and \({\bar{x}}_1, {\bar{x}}_2\in {\mathbb {R}}^3\) such that

$$\begin{aligned} K(x)-K_{\infty }>0, \ \ \ \ \forall \ |x-{\bar{x}}_1|<r, \ \ \ \ |{\tilde{u}}(x)|>0, \ \ \ \ \forall \ |x-{\bar{x}}_2|<r. \end{aligned}$$

Let \({\hat{u}}(x)={\tilde{u}}(x-{\bar{x}}_1+{\bar{x}}_2)\). Then \(\Vert {\hat{u}}\Vert _2^2=c\), \(\Vert \nabla {\hat{u}}\Vert _2^2>k_0\) and \(I^{\infty }({\hat{u}})=I^{\infty }({\tilde{u}})\), which together with (1.3), (2.7), implies that

$$\begin{aligned} {\bar{\sigma }}(c)=I^{\infty }({\hat{u}})>I({\hat{u}})\ge {\bar{\sigma }}(c). \end{aligned}$$
(4.61)

This contradiction shows that \({\bar{u}}\ne 0\). \(\square \)

Proof of Theorem  2.2

Note that if \(u_c\) is a critical point of \({I}|_{{\mathcal {S}}_{c}}\), then there exists \(\lambda _c\in {\mathbb {R}}\) such that \(I'(u_c)-\lambda _c u_c=0\). Combining Lemma 4.24.4 and 4.7, we conclude the proof of Theorem 2.2. \(\square \)