1 Introduction and Main Results

In this paper we study the nonlinear Schrödinger–Poisson system

$$\begin{aligned} {\left\{ \begin{array}{ll}\displaystyle i\partial _t\Psi -\Delta \Psi +\gamma \phi (x)\Psi =af(|\Psi |^2)\Psi ,&{} x\in {\mathbb {R}}^3, \\ -\Delta \phi =|\Psi |^2, &{}x\in {\mathbb {R}}^3, \end{array}\right. } \end{aligned}$$
(1.1)

where \(\Psi : {\mathbb {R}}\times {\mathbb {R}}^3\rightarrow {\mathbb {C}}\) is the time-dependent wave function, \(\gamma , a\in {\mathbb {R}}\) are parameters, the nonlinear term f simulates the interaction between many particles or external nonlinear perturbations. The nonlinear Schrödinger–Poisson system (1.1) attracted much attention in the last decade, starting from the fundamental contribution [13]. System (1.1) has many physical motivations, it derived from the approximation of the Hartree-Fock equation that describes a quantum mechanical of many particles, and is highly beneficial in the quantum description of the ground states of nonrelativistic atoms and molecules [34, 35, 39], and also arises in semiconductor theory [18].

When we are concerned with the standing wave solutions \(\Psi (t, x) =e^{-i\lambda t}u(x)\), \(\lambda \in {\mathbb {R}}\), then \(u: {\mathbb {R}}^3\rightarrow {\mathbb {R}}\) must verify

$$\begin{aligned} {\left\{ \begin{array}{ll}\displaystyle -\Delta u+\lambda u+\gamma \phi u=af(u), &{}x \in {\mathbb {R}}^{3},\\ -\Delta \phi =u^2, &{}x \in {\mathbb {R}}^{3}. \end{array}\right. } \end{aligned}$$
(1.2)

At this time, there are two possible choices to deal with (1.2). One can fix \(\lambda \in {\mathbb {R}}\) and to look for solutions as critical points of the associated energy functional

$$\begin{aligned} J_{\lambda }(u)=\frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\lambda }{2}\int _{{\mathbb {R}}^3}|u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}\frac{|u(x)|^2|u(y)|^2}{|x-y|}dxdy -a\int _{{\mathbb {R}}^3}F(u)dx. \end{aligned}$$

where \(F(u)=\int ^u_0f(s)ds\) is the primitive integral of f. Alternatively, one can search for solutions of Eq. (1.2) with prescribed \(L^2\)-norm. At this point, the parameter \(\lambda \in {\mathbb {R}}\) cannot longer be fixed but instead appears as a Lagrange multiplier. Analogous to the first case, the solutions of (1.2) with \(\Vert u\Vert _2^2 =m > 0\) can be obtained as critical points of the energy functional

$$\begin{aligned} J(u)=\frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4} \int _{{\mathbb {R}}^3}\int _{{\mathbb {R}}^3}\frac{|u(x)|^2|u(y)|^2}{|x-y|}dxdy-a\int _{{\mathbb {R}}^3}F(u)dx. \end{aligned}$$

under the constraint \(L^2\)-sphere \(S_m:=\{u\in H^1({\mathbb {R}}^3): \Vert u\Vert _2^2=m^2\}.\) It is easy to check that J is a well-defined and \(C^1\) functional on \(S_m.\) This approach is relevant from the physical point of view, in particular, since the \(L^2\)-norm is a preserved quantity of the evolution and since the variational characterization of such solutions is often a strong help to analyze their orbital stability, see for example, [6, 9,10,11, 29,30,31] and references therein.

As far as we know, the first work for normalized solution to Eq. (1.2) in the case \(\gamma = 1,\) and \(f(u) = |u|^{p-2}u\) is due to Sánchez and Soler [41]. They showed that there exists a normalized solution of (1.2) provided that m is sufficiently small and \(p = \frac{8}{3}\). Since then, there are some further studies for problem (1.2) in mass subcritical case. In this case, the corresponding functional is bounded from below on \(S_m,\) then a global minimizer can be obtained for some m. See for instance, [10, 11, 31, 32]. When the nonlinearity f in (1.2) is mass supercritical, the constrained functional \(J|_{S_m}\) is no longer bounded from below and coercive. In this case, using a mountain-pass argument on \(S_m,\) Bellazzini, Jeanjean and Luo [12] proved the existence and the instability of standing waves for \(m > 0\) sufficiently small. Bartsch and de Valeriola [6], Luo [38] studied the multiplicity of normalized solutions of (1.2).

At the same time, normalized solutions for Schrödinger–Poisson–Slater equation with general nonlinearity in case \(\gamma =-1,\) has also attracted much more attention. Xie, Chen and Shi [49] showed the existence and multiplicity results of solutions when f satisfies \(\lim _{t\rightarrow 0} f (t)/ t = 0\) and \(\lim _{|t|\rightarrow \infty }F(t)/|t|^{10/3} =\infty \) under some mild conditions on f. Recently, Chen Tang and Yuan [17] investigated the existence of normalized solution by some new analytical techniques in case that f satisfies \(\lim _{t\rightarrow 0} F (t)/ t^2 = 0\) and \(\lim _{|t|\rightarrow \infty }F(t)/|t|^{10/3} =\infty \).

Very recently, Wang and Qian [45] obtained the existence of normalized ground states and infinitely many radial solutions for (1.2) with Sobolev subcritical term f, by constructing a particular bounded Palais-Smale sequence when \(\gamma < 0, a > 0.\) Meanwhile, they obtained the nonexistence result in the case \(\gamma< 0, a < 0\) and the existence result when\(\gamma > 0, a < 0\) via variational methods. In [29], Jeanjean and Trung Le specialized in the existence of normalized solutions for problem (1.2) with \(L^2\)-supercritical growth:

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u +\gamma (|x|^{-1}*|u|^2) u=\lambda u+a|u|^{p-2}u, &{}~~ \text{ in }~{\mathbb {R}}^3,\\ \displaystyle \int _{{\mathbb {R}}^N}|u|^2dx=c^2, &{} \end{array}\right. } \end{aligned}$$
(1.3)

where \( u \in H^1({\mathbb {R}}^3), \gamma \in {\mathbb {R}}, a\in {\mathbb {R}}\) and \(p\in (\frac{10}{3},6]\). The authors dealt with the following cases:

  1. (a)

    If \(\gamma <0\) and \(a>0\), both in the Sobolev subcritical case \(p\in (\frac{10}{3},6)\) and in the Sobolev critical case \(p=6,\) they showed that there exists a \(c_1>0\) such that, for any \(c\in (0, c_1),\) (1.3) admits two solutions \(u^+_c\) and \(u^-_c\), which can be characterized as a local minimum and a mountain pass critical point of the associated energy functional, respectively.

  2. (b)

    In the case \(\gamma <0\) and \(a<0,\) they proved that, for any \(p\in (\frac{10}{3}, 6]\) and any \(c>0,\) (1.3) has a solution which is a global minimizer.

  3. (c)

    Finally, in the case \(\gamma>0, a>0\) and \(p=6\), they showed that (1.3) does not exist positive solutions.

When \(\gamma =1\), \(p\in (\frac{10}{3},6)\), Bellazzini, Jeanjean and Luo [12] studied the existence of normalized solutions of (1.3) by a mountain-pass argument as \(c > 0\) is sufficiently small and nonexistence as \(c> 0\) is not small. In [31], Jeanjean and Luo considered the existence of minimizers with \(L^2\)-norm for (1.3) when \(p\in [3, \frac{10}{3}],\) and they showed a threshold value of \(c> 0\) separating existence and nonexistence of minimizers. For more results on normalized solutions of Schrödinger–Poisson systems, we refer to [1, 19, 27, 29, 31, 33, 37, 38, 50,51,52,53] and references therein.

After the above literature review, we find that, only the article [29] has considered the existence of normalized solutions of (1.3) in the case \(p\in (\frac{10}{3},6]\), and \(\gamma <0\); and the no-existence of normalized solution of (1.3) with \(p=6, \gamma >0\) and \(a>0.\) Therefore, a natural and important question arising is how to obtain normalized solutions to system (1.3) in the case \(\gamma >0\), and in the presence of Sobolev critical exponent and mixed nonlinearities: \(a|u|^{p-2}u+|u|^4u\)? here \(a|u|^{p-2}u\) is a subcritical perturbation term with \(p\in (2,6)\) and \(a>0\) a parameter. We notice that, this kind of critical nonlinearities has been used by Soave [42], Wei and Wu [47] to search for the normalized solutions for the Schrödinger equation

$$\begin{aligned} -\Delta u =\lambda u +a |u|^{p-2}u+|u|^{2^*-2}u~~~\text{ in }~~{\mathbb {R}}^N, \end{aligned}$$

with the prescribed \(L^2\)-norm \(\int _{{\mathbb {R}}^3} |u|^2dx=c^2\). But for the Schrödinger–Poisson system in presence of the Sobolev critical term \(|u|^4u\), coupled with a subcritical perturbation term \(a|u|^{p-2}u\), the existence of normalized solutions has not been studied in the existing literature, as far as we know. For more studies of existence of normalized solutions of the Schrödinger equation, see for example [28, 30, 42, 43, 54] and references therein.

Motivated by the works mentioned above, in this paper we focuss on studying the Schrödinger–Poisson system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u +\gamma \phi u= \lambda u+\mu |u|^{q-2}u+|u|^4u,&{}~~ \text{ in }~{\mathbb {R}}^3, \\ -\Delta \phi =u^2,&{}~~ \text{ in }~{\mathbb {R}}^3, \end{array}\right. } \end{aligned}$$
(1.4)

having prescribed \(L^2\)-norm

$$\begin{aligned} \int _{{\mathbb {R}}^3} |u|^2dx=a^2, \end{aligned}$$
(1.5)

where \(\lambda \in {\mathbb {R}}\) is an undetermined parameter, \(a>0\) and \(\mu ,\gamma >0\) are parameters, \(\mu |u|^{q-2}u\) is a subcritical perturbation term with \(q\in (2,6)\). For this purpose, applying the reduction argument introduced in [40], system (1.4) is equivalent to the following single equation

$$\begin{aligned} -\Delta u +\gamma \phi _u u= \lambda u+\mu |u|^{q-2}u+|u|^4u,~~~x\in {\mathbb {R}}^3, \end{aligned}$$
(1.6)

where \(\phi _u(x)=\frac{1}{4\pi }\int _{{\mathbb {R}}^3}\frac{|u(y)|^2}{|x-y|}dy.\) We shall look for solutions to (1.4)–(1.5), as a critical points of the action functional

$$\begin{aligned} I_{\mu }(u)=\frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4} \int _{{\mathbb {R}}^3}\phi _u|u|^2dx-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|u|^qdx -\frac{1}{6}\int _{{\mathbb {R}}^3}|u|^6dx, \end{aligned}$$

under the \(L^2\)-norm constrained manifold

$$\begin{aligned} S(a):=\left\{ u\in H({\mathbb {R}}^3):\Psi (u)=\frac{1}{2}a^2\right\} , \end{aligned}$$

where \(\Psi (u)=\frac{1}{2}\int _{{\mathbb {R}}^3}u^2dx\). Physically, such type of solutions are the so-called normalized solutions to (1.4)–(1.5). In order to state our main results, we introduce some of the constants from the following Gagliardo-Nirenberg-Sobolev (GNS) inequality. That is, there exists a best constant C(p) depending on p such that for any \(u\in H^1({\mathbb {R}}^3),\)

$$\begin{aligned} \Vert u\Vert _p^p\le C(p)\Vert u\Vert ^{(1-\gamma _p)p}_2 \Vert \nabla u\Vert _2^{\gamma _pp}, \end{aligned}$$
(1.7)

where \(\gamma _p= \frac{3(p-2)}{2p}.\) The constant C(p) can be achieved by function \(Q_p\), see [43]. For the problem obtained from (1.4)–(1.5) by removing the critical exponent term and the nonlocal term, we obtain one of its normalized solutions by rescaling \(Q_p\). From \(\gamma _pp = 2\), we get \(p = \frac{10}{3}\) which is called the \(L^2\)-critical exponent for problem (1.4)–(1.5). Before presenting the existence result, we give the definition of ground states. If \(u^*\) is a solution to (1.4)–(1.5) having minimal energy among all the solutions which belongs to S(a) : 

$$\begin{aligned} (I_{\mu }|_{S(a)})' ( u^* ) = 0~~\text{ and }~~ I_{\mu }( u^*) = \inf \{(I_{\mu }|_{S(a)})' ( u ) = 0,~~ \text{ and }~~ u \in S(a)\},\end{aligned}$$

we say that \(u^*\) is a ground state of (1.4)–(1.5).

The following are the main results of this paper. In the \(L^2\)-subcritical case: \(2<q<\frac{10}{3}\), we have the following existence result of the normalized ground state solutions.

Theorem 1.1

Let \(2<q<\frac{10}{3}\), \(\gamma >0\), and assume that \(0<a<\min \{\alpha _1,\alpha _2\}\), where

$$\begin{aligned} \alpha _1:=\left\{ \frac{2q}{C(q)\mu (6-q\delta _q)}{\left( \frac{(2-q\delta _q)S^3}{6-q\delta _q}\right) } ^{\frac{2-q\delta _q}{4}}\right\} ^{\frac{1}{q(1-\delta _q)}}, \end{aligned}$$

and

$$\begin{aligned} \alpha _2:=\left\{ \frac{4}{C(q)\mu \delta _q(6-q\delta _q)}{\left( \frac{q\delta _qS^{\frac{3}{2}}}{2-q\delta _q}\right) } ^{\frac{2-q\delta _q}{2}}\right\} ^{\frac{1}{q(1-\delta _q)}}, \end{aligned}$$

where S is defined in (2.1). Then there exists \(\tilde{\mu }>0\) such that \(\mu >\tilde{\mu }\), problem (1.4)–(1.5) has a couple of solutions \((u_a,\lambda _a)\in S(a)\times {\mathbb {R}}\). Moreover,

$$\begin{aligned} I_\mu (u_a)=\inf _{u \in \mathcal {P}(a)}I_\mu (u)=\inf _{u \in \mathcal {P}(a)^+}I_\mu (u)=\inf _{u \in D_k}I_\mu (u), \end{aligned}$$

for some suitable small constant \(k>0\), where \(\mathcal {P}(a)\) is the Pohozaev manifold defined in Lemma 2.5, the set \(\mathcal {P}(a)^+\) is defined in (3.1), and

$$\begin{aligned} D_k=\{u \in S(a):\Vert \nabla u\Vert _{L^2({\mathbb {R}}^3)}<k\}. \end{aligned}$$

In the \(L^2\)-critical case: \(q=\frac{10}{3}\), we have the following conclusion.

Theorem 1.2

Let \(q=\frac{10}{3}\), \(\mu >0\), and assume that \(0<a<\min \{\alpha _3,\alpha _4\}\), where

$$\begin{aligned} \alpha _3:=\left( \frac{q}{2\mu C(q)}\right) ^{\frac{1}{q(1-\delta _q)}}, \end{aligned}$$

and

$$\begin{aligned} \alpha _4:=\left( \frac{k^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) ^{\frac{1}{3}}, \end{aligned}$$

where \({\tilde{C}}\) is defined as (2.3), and k is defined as

$$\begin{aligned} k=\min \left\{ \left( \frac{q(4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}})^{\frac{q(1-\delta _q)}{3}}}{32\mu C(q)}\right) ^{\frac{6}{q(1-\delta _q)}}, ~~\left( \frac{3 }{64}S^3\right) ^{\frac{1}{2}}\right\} . \end{aligned}$$

Then there exist \(\tilde{\gamma _1},\tilde{\gamma _2}>0\) such that \(0<\gamma <\min \{\tilde{\gamma _1},\tilde{\gamma _2}\}\), problem (1.4)–(1.5) has a couple of solutions \((u_a,\lambda _a)\in S(a)\times {\mathbb {R}}\). Moreover,

$$\begin{aligned} I_\mu (u_a)=\inf _{u \in \mathcal {P}(a)}I_\mu (u)=\inf _{u \in \mathcal {P}(a)^-}I_\mu (u), \end{aligned}$$

where \(\mathcal {P}(a)^-\) is defined in (3.2).

In the \(L^2\)-supercritical case: \(\frac{10}{3}<q<6\), we have the following existence result.

Theorem 1.3

Let \(\frac{10}{3}<q<6\), \(\mu >0\), and assume that \(0<a<\alpha _5\), where

$$\begin{aligned} \alpha _5:=\left( \frac{{k^*}^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) ^{\frac{1}{3}}, \end{aligned}$$

where \(k^*\) is defined in (5.1). Then there exist \(\tilde{\gamma _1},\tilde{\gamma _2}>0\) such that \(0<\gamma <\min \{\tilde{\gamma _1},\tilde{\gamma _2}\}\), problem (1.4)–(1.5) has a couple of solutions \((u_a,\lambda _a)\in S(a)\times {\mathbb {R}}\). Moreover,

$$\begin{aligned} I_\mu (u_a)=\inf _{u \in \mathcal {P}(a)}I_\mu (u)=\inf _{u \in \mathcal {P}(a)^-}I_\mu (u). \end{aligned}$$

Remark 1.1

In [29] Jeanjean and Le only studied the no-existence of normalized solutions of problem (1.4)–(1.5) with \(\gamma >0\) and \(\mu =0\). The existence of normalized solutions for (1.4)–(1.5) in the case \(\gamma>0,\mu >0\) and \(q\in (2,6)\) has not been studied in the existing literature. Theorems 1.11.3 provide a complete description of the existence of normative solutions in the \(L^2\)-subcritical, \(L^2\)-critical and \(L^2\)-supercritical perturbation \(\mu |u|^{q-2}u\), respectively.

Remark 1.2

In Theorem 1.1, we assume that the parameter \(\mu >0\) is large enough, so as to ensure that the Lagrange multiplier sequence \(\lambda _n\rightarrow \lambda <0\) as \(n\rightarrow \infty \), which plays a crucial role in our proof of the \(H^1\)-convergence of (PS)-sequence \(\{u_n\}\subset S(a)\). In Theorem 1.2 and Theorem 1.3, it is necessary for the parameter \(\gamma >0\) to be appropriately small so that the Mountain Pass level is strictly less than \(\frac{1}{3} S^{\frac{3}{2}}\). This characteristic is completely different from the critical Schrödinger–Poisson system without \(L^2\)-mass constrained, see for example [25, 26, 46, 55].

In order to prove Theorems 1.11.3, we apply the constrained variational methods. Note that the Sobolev critical terms \(|u|^4u\) is \(L^2\)-supercritical, the functional \(I_{\mu }\) is always unbounded from below on S(a),  and this causes difficulty to treat the existence of normalized solutions on the \(L^2\)- constraint. One of the main difficulties is to prove the convergence of constrained Palais-Smale sequences: Indeed, the Sobolev critical term \(|u|^4u\) and nonlocal convolution term \(\gamma \phi _u u\), make it more complex to estimate the critical value of mountain pass, and has to consider how the interaction between the nonlocal term and the mixed nonlinearities. In particularly, the energy balance between these competing terms needs to be controlled through moderate adjustments of parameter \(\gamma >0.\) Another obstacle is that sequences of approximated Lagrange multipliers have to be controlled, since \(\lambda \) is not prescribed; and moreover, weak limits of Palais-Smale sequences could leave the constraint, since the embeddings \(H^1({\mathbb {R}}^3)\hookrightarrow L^2({\mathbb {R}}^3)\) and also \(H^1_{rad}({\mathbb {R}}^3)\hookrightarrow L^2({\mathbb {R}}^3)\) are not compact.

To overcome these difficulties, we shall employ Jeanjean’s theory [28] by showing that the mountain pass geometry of \(I_{\mu }|_{S(a)}\) allows to construct a Palais-Smale sequence of functions satisfying the Pohozaev identity, to obtain the boundedness, which is the first step to show strong \(H^1\)-convergence. To restore the loss of compactness caused by the critical growth, we shall utilize the concentration-compactness principle, mountain pass theorem and energy analysis to get the existence of normalized ground states of (1.4)–(1.5), by showing that, suitably combining some of the main ideas from [15, 42], compactness can be derived in the present setting.

This paper is organized as follows: In Sect. 2 we summarize some preliminary results which will often be used in the rest the paper. In Sect. 3, we investigate the existence of normalized ground state solutions for system (1.4)–(1.5) under the \(L^2\)-subcritical perturbation case: \(q\in (2, \frac{10}{3})\) and complete the proof Theorem 1.1. In Sect. 4, we address the presence of the normalized ground state solutions for system (1.4)–(1.5) in \(L^2\)-critical perturbation case: \(q=\frac{10}{3}\) and prove Theorem 1.2, by employing manifold and mountain road theorems. In Sect. 5, we tackle the existence of the normalized ground state solutions for problem (1.4)–(1.5) under \(L^2\)-supercritical perturbation case: \(q\in (\frac{10}{3}, 6)\) and prove Theorem 1.3.

Notations. Throughout this paper, we denote \(B_r(z)\) the open ball of radius r with center at z in \({\mathbb {R}}^3\), and \(\Vert u\Vert _p\) is the usual norm of the space \(L^p({\mathbb {R}}^3)\) for \(p\ge 1\). Moreover, we denote by \(C,C_i > 0, i=1,2,\cdots ,\) different positive constants whose values may vary from line to line and are not essential to the problem.

2 Preliminary Stuff

In this section, we will give the functional space setting and introduce some notations and useful preliminary results, which are important to proving our Theorems. Let \(H^1({\mathbb {R}}^3)\) be the completion of \(C_0^\infty ({\mathbb {R}}^3)\) with respect to the norm

$$\begin{aligned} \Vert u\Vert _H=\left( \int _{{\mathbb {R}}^3}|\nabla u|^2+|u|^2dx\right) ^{\frac{1}{2}}. \end{aligned}$$

And the homogeneous Sobolev space \(D^{1,2}({\mathbb {R}}^3)\) is defined by

$$\begin{aligned}D^{1,2}({\mathbb {R}}^3)=\left\{ u\in L^6({\mathbb {R}}^3):\int _{{\mathbb {R}}^3}|\nabla u|^2dx<+\infty \right\} ,\end{aligned}$$

endowed with the norm

$$\begin{aligned}\Vert u\Vert ^2:=\Vert u\Vert ^2_{D^{1,2}({\mathbb {R}}^3)}=\Vert \nabla u\Vert _2^2=\int _{{\mathbb {R}}^3}|\nabla u|^2dx.\end{aligned}$$

The work space \(H^1_{rad}({\mathbb {R}}^3)\) is defined by

$$\begin{aligned} H^1_{rad}({\mathbb {R}}^3):=\left\{ u\in H^1({\mathbb {R}}^3): ~u~ \text{ is } \text{ radially } \text{ symmetric } \text{ and } \text{ decreasing }\right\} . \end{aligned}$$

Let \({\mathbb {H}} = H\times {\mathbb {R}}\) with usual scalar product

$$\begin{aligned} \langle \cdot ,\cdot \rangle _{{\mathbb {H}}}=\langle \cdot ,\cdot \rangle _{H}+\langle \cdot ,\cdot \rangle _{{\mathbb {R}}}, \end{aligned}$$

and the corresponding norm

$$\begin{aligned}\Vert (\cdot ,\cdot )\Vert ^2_{{\mathbb {H}}}=\Vert \cdot ,\cdot \Vert _{H}^2+|\cdot ,\cdot |_{{\mathbb {R}}}^2.\end{aligned}$$

We denote the best Sobolev constant S by

$$\begin{aligned} S=\inf _{u\in D^{1,2}({\mathbb {R}}^3)\backslash \{0\}}\frac{\Vert \nabla u\Vert _2^2}{(\int _{{\mathbb {R}}^3}|u|^6dx)^{\frac{1}{3}}}. \end{aligned}$$
(2.1)

It is well know that S is achieved by

$$\begin{aligned} U_{\varepsilon }(x)=\frac{C^*\varepsilon ^{\frac{1}{2}}}{(\varepsilon ^2+|x|^2)^{\frac{1}{2}}}, \end{aligned}$$
(2.2)

for any \(\varepsilon >0\) and \(C^*\) being normalized constant such that (see [15]):

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla U_{\varepsilon }|^2dx=\int _{{\mathbb {R}}^3}|U_{\varepsilon }|^6dx=S^{\frac{3}{2}}. \end{aligned}$$

In the following, we recall some useful inequalities, which play an important part in the proof of our main results.

Proposition 2.1

(Hardy–Littlewood–Sobolev inequality [34]) Let \(l,r>1\) and \(0<\mu <N\) be such that \(\frac{1}{r}+\frac{1}{l}+\frac{\mu }{N}=2, f\in L^r({\mathbb {R}}^N)\) and \(h\in L^l({\mathbb {R}}^N)\). Then there exists a constant \(C(N,\mu ,r,l)>0\) such that

$$\begin{aligned} \left| \int _{{\mathbb {R}}^{N}}\int _{{\mathbb {R}}^{N}}f(x)h(y)|x-y|^{-\mu }dxdy\right| \le C(N,\mu ,r,l)\Vert f\Vert _r\Vert h\Vert _l. \end{aligned}$$

From Proposition 2.1, with \(l= r = \frac{6}{5}\), we have that:

$$\begin{aligned} \int _{{\mathbb {R}}^3}\phi _uu^2dx\le \int _{{\mathbb {R}}^3}\left( \frac{1}{|x|}*u^2\right) u^2dx\le {\tilde{C}}\Vert u\Vert _{\frac{12}{5}}^4. \end{aligned}$$
(2.3)

Next, we introduce the following Gagliardo-Nirenberg inequality.

Lemma 2.2

([43]) Let \(p\in (2, 6).\) Then there exists a constant \(C(p)>0\) such that

$$\begin{aligned} \Vert u\Vert _p^p\le {C}(p)\Vert \nabla u\Vert _2^{p\delta _{p}}\Vert u\Vert _2^{p(1-\delta _{p})},~~~\forall u\in H^1({\mathbb {R}}^3), \end{aligned}$$
(2.4)

where \(\delta _{p}=\frac{3(p-2)}{2p}.\)

Lemma 2.3

(Lemma 5.1 [23]) If \(u_n\rightharpoonup u\) in \(H^1_{rad}({\mathbb {R}}^3)\), then

$$\begin{aligned} \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx\rightarrow \int _{{\mathbb {R}}^3}\phi _{u}u^2dx, \end{aligned}$$
(2.5)

and

$$\begin{aligned} \int _{{\mathbb {R}}^3}\phi _{u_n}u_n\varphi dx\rightarrow \int _{{\mathbb {R}}^3}\phi _{u}u\varphi dx,~~\forall \varphi \in H^1_{rad}({\mathbb {R}}^3). \end{aligned}$$
(2.6)

In the sequel, we define a useful fiber map (e.g. [42]) preserving the \(L^2\)-norm

$$\begin{aligned} (\iota \star u)(x):=e^{\frac{3\iota }{2}}u(e^{\iota }x),~~~~ x\in {\mathbb {R}}^3,~~\iota \in {\mathbb {R}}. \end{aligned}$$
(2.7)

By simple calculation, we can infer that

$$\begin{aligned} \Vert (\iota \star u)\Vert ^2_2=\Vert u\Vert ^2_2, \end{aligned}$$
(2.8)
$$\begin{aligned} \Vert (\iota \star u)\Vert ^q_q=e^{q\delta _q\iota }\Vert u\Vert ^q_q, \end{aligned}$$
(2.9)

and

$$\begin{aligned} \Vert \nabla (\iota \star u)(x)\Vert ^2_2=e^{2\iota }\Vert \nabla (\iota \star u)(x)\Vert ^2_2. \end{aligned}$$
(2.10)

Next, we define a auxiliary functional \(E:{\mathbb {H}}\rightarrow {\mathbb {R}}\) by

$$\begin{aligned} \begin{aligned} E(u,\iota ):&=I_{\mu } ((\iota \star u))\\&=\frac{1}{2}e^{2\iota }\Vert \nabla u\Vert ^2_2+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx-\frac{\mu }{q}e^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -\frac{1}{6}e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned} \end{aligned}$$
(2.11)

Besides, we have the fact that

$$\begin{aligned}q\delta _{q}\left\{ \begin{array}{ll}<2,&{}\text{ as }~~2<q<\bar{q}; \\ =2,&{}\text{ as }~~q=\bar{q};\\ >2,&{}\text{ as }~~ \bar{q}<q<6, \end{array}\right. \end{aligned}$$

where \(\bar{q}:=\frac{10}{3}\) is the \(L^2\)-critical exponent.

The Pohozaev manifold plays an important role in the proof of our main results, so we introduce it below [22].

Proposition 2.4

Let \(u\in H^1({\mathbb {R}}^3)\cap L^\infty ({\mathbb {R}}^3)\) be a weak solution of (1.4), then u satisfies the equality

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{5\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\frac{3\lambda }{2}\int _{{\mathbb {R}}^3}|u|^2dx+\frac{3\mu }{q}\int _{{\mathbb {R}}^3}|u|^qdx +\frac{1}{2}\int _{{\mathbb {R}}^3}|u|^6dx.\nonumber \\ \end{aligned}$$
(2.12)

Lemma 2.5

Let \(u\in H^1({\mathbb {R}}^3)\) be a weak solution of (1.4)–(1.5), then we can construct the following Pohozaev manifold

$$\begin{aligned} \mathcal {P}(a)=\{u\in S(a): P_\mu (u)=0\}, \end{aligned}$$

where

$$\begin{aligned} P_\mu (u)=\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\mu \delta _{q}\int _{{\mathbb {R}}^3}|u|^qdx -\int _{{\mathbb {R}}^3}|u|^6dx.\qquad \end{aligned}$$
(2.13)

Proof

Since u is the weak solution of (1.4)–(1.5), we have that

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{5\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\frac{3\lambda }{2}\int _{{\mathbb {R}}^3}|u|^2dx+\frac{3\mu }{q}\int _{{\mathbb {R}}^3}|u|^qdx +\frac{1}{2}\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$

Moreover, since u is the weak solution of system (1.4)–(1.5), we have

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u|^2dx+\gamma \int _{{\mathbb {R}}^3}\phi _uu^2dx=\lambda \int _{{\mathbb {R}}^3}|u|^2dx+\mu \int _{{\mathbb {R}}^3}|u|^qdx + \int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$

Combining with (2.13) and the above equality, we obtain that

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\mu \delta _{q}\int _{{\mathbb {R}}^3}|u|^qdx +\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$

The proof is completed. \(\square \)

We define \(\varphi _u(\iota ):=E(u,\iota )\) for any \(u \in S(a)\) and \(\iota \in {\mathbb {R}}\), then

$$\begin{aligned} \begin{aligned} (\varphi _u)'(\iota )&=e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx-\mu \delta _qe^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx\\&=\int _{{\mathbb {R}}^3}|\nabla (\iota \star u)|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{(\iota \star u)}|\iota \star u|^2dx-\mu \delta _q\int _{{\mathbb {R}}^3}|\iota \star u|^qdx -\int _{{\mathbb {R}}^3}|\iota \star u|^6dx\\&=P_\mu ((\iota \star u)). \end{aligned} \end{aligned}$$
(2.14)

Moreover, by direct calculation, we have

$$\begin{aligned}{} & {} (\varphi _u)''(\iota ) =2e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx\nonumber \\{} & {} -\mu q{\delta _q}^2e^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -6e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$
(2.15)

Therefore, we have the following lemma:

Lemma 2.6

For any \(u \in S(a)\), \(\iota \in {\mathbb {R}}\) is a critical point of \(\varphi _u(\iota )\) if and only if \((\iota \star u) \in \mathcal {P}(a)\). Particularly, \( u \in \mathcal {P}(a) \) if and only if 0 is a critical point for \(\varphi _u(\iota )\).

Finally, we state the following well-known embedding result.

Lemma 2.7

([44]) Let \(N\ge 2\). The embedding \(H^1_{rad}({\mathbb {R}}^N)\hookrightarrow L^p({\mathbb {R}}^N)\) is compact for any \(2< p < 2^*\).

Remark 2.8

([11]) The map \((u,\iota )\in {\mathbb {H}}\rightarrow (\iota \star u) \in H \) is continuous.

3 \(L^2\)-Subcritical Perturbation Case

In this section, we shall address the \(L^2\)-subcritical perturbation case: \(2<q<\frac{10}{3}\) and provide the proof of Theorem 1.1. First, we think about a decomposition of \(\mathcal {P}(a)\) as in [42, 43]. By Lemma  2.6, we define the following sets:

$$\begin{aligned} \begin{aligned} \mathcal {P}(a)^+&:=\left\{ u \in \mathcal {P}(a):2\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx>\mu q{\delta _q}^2\int _{{\mathbb {R}}^3}|u|^qdx +6\int _{{\mathbb {R}}^3}|u|^6dx\right\} \\&=\{u \in \mathcal {P}(a):(\varphi _u)''(0)>0\}, \end{aligned} \qquad \end{aligned}$$
(3.1)
$$\begin{aligned} \begin{aligned} \mathcal {P}(a)^0&:=\left\{ u \in \mathcal {P}(a):2\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\mu q{\delta _q}^2\int _{{\mathbb {R}}^3}|u|^qdx +6\int _{{\mathbb {R}}^3}|u|^6dx\right\} \\&=\{u \in \mathcal {P}(a):(\varphi _u)''(0)=0\}, \end{aligned} \qquad \end{aligned}$$
(3.2)
$$\begin{aligned} \begin{aligned} \mathcal {P}(a)^-&:=\left\{ u \in \mathcal {P}(a):2\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx<\mu q{\delta _q}^2\int _{{\mathbb {R}}^3}|u|^qdx +6\int _{{\mathbb {R}}^3}|u|^6dx\right\} \\&=\{u \in \mathcal {P}(a):(\varphi _u)''(0)<0\}. \end{aligned}\nonumber \\ \end{aligned}$$
(3.3)

We can easily get that

$$\begin{aligned} \mathcal {P}(a)=\mathcal {P}(a)^+\cup \mathcal {P}(a)^0\cup \mathcal {P}(a)^-. \end{aligned}$$

Next, we will give some lemmas, which are useful for the proof of Theorem 1.1.

Lemma 3.1

Let \(2<q<\frac{10}{3}\), \(\mu ,\gamma >0\), and \(0<a<\alpha _1\), where

$$\begin{aligned} \alpha _1:=\left\{ \frac{2q}{C(q)\mu (6-q\delta _q)}{\left( \frac{(2-q\delta _q)S^3}{6-q\delta _q}\right) } ^{\frac{2-q\delta _q}{4}}\right\} ^{\frac{1}{q(1-\delta _q)}}. \end{aligned}$$

Then \(\mathcal {P}(a)^0=\varnothing \) and \(\mathcal {P}(a)\) is a smooth manifold of codimension 2 in \(H({\mathbb {R}}^3)\).

Proof

Suppose by contradiction that \(\mathcal {P}(a)^0\ne \varnothing \). Taking \(u \in \mathcal {P}(a)^0\), one has

$$\begin{aligned} 2\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\mu q{\delta _q}^2\int _{{\mathbb {R}}^3}|u|^qdx+6\int _{{\mathbb {R}}^3}|u|^6dx, \end{aligned}$$

and

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\mu \delta _q\int _{{\mathbb {R}}^3}|u|^qdx+\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$

Since \(\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx\ge 0\), so combining the above equalities with the GNS inequality (2.4) and (2.1), we can infer to

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u|^2dx \le \frac{6-q\delta _q}{2-q\delta _q}\int _{{\mathbb {R}}^3}|u|^6dx \le \frac{6-q\delta _q}{(2-q\delta _q)S^3}\left( \int _{{\mathbb {R}}^3}|\nabla u|^2dx\right) ^3, \end{aligned}$$

and

$$\begin{aligned}{} & {} \int _{{\mathbb {R}}^3}|\nabla u|^2dx \le \frac{\mu \delta _q(6-q\delta _q)}{4}\int _{{\mathbb {R}}^3}|u|^qdx\\{} & {} \le \frac{\mu \delta _q(6-q\delta _q)}{4}C(q)a^{q(1-\delta _q)}\left( \int _{{\mathbb {R}}^3}|\nabla u|^2dx\right) ^{\frac{q\delta _q}{2}}. \end{aligned}$$

By simple calculation and the fact \(q\delta _q<2\), we have

$$\begin{aligned} \begin{aligned} a^{q(1-\delta _q)}&\ge \frac{4}{\mu \delta _qC(q)(6-q\delta _q)}\left( \int _{{\mathbb {R}}^3}|\nabla u|^2dx\right) ^{\frac{2-q\delta _q}{2}}\\&\ge \frac{4}{\mu \delta _qC(q)(6-q\delta _q)} \left( \frac{(2-q\delta _q)S^3}{6-q\delta _q}\right) ^{\frac{2-q\delta _q}{4}}\\&\ge \frac{2q}{\mu C(q)(6-q\delta _q)} \left( \frac{(2-q\delta _q)S^3}{6-q\delta _q}\right) ^{\frac{2-q\delta _q}{4}}\\&:=\alpha _1^{q(1-\delta _q)},\\ \end{aligned} \end{aligned}$$

which contradicts to \(a<\alpha _1\).

Then, we verify that \(\mathcal {P}(a)\) is a smooth manifold of codimension 2 in \(H({\mathbb {R}}^3)\). Let

$$\begin{aligned} \mathcal {P}(a)=\{u \in H:P_\mu (u)=0,G(u)=0\}, \end{aligned}$$

for \(G(u)=\Vert u\Vert ^2_2-a^2\), with \(P_\mu \) and G of class \(C^1\) in H. Hence, we need to show that the differential \((dG(u),dP_\mu (u)):H\rightarrow {\mathbb {R}}^2\) is surjective, for every \(u \in \mathcal {P}(a)\). For this purpose, we will prove that for every \(u \in \mathcal {P}(a)\), there exists \(\varphi \in T_uS\), where

$$\begin{aligned} T_u S:=\{v\in E:~(u,v)_H = 0\}, \end{aligned}$$

which is the tangent space of \(\mathcal {S}\) at a point \(u \in \mathcal {S}\). Then, one has \(dP_\mu (u)[\varphi ]\ne 0\). Once the existence of \(\varphi \) is established, the system

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle dG(u)[\alpha \varphi +\beta u]=x\\ dP_\mu (u)[\alpha \varphi +\beta u]=y \end{array}\right. } \end{aligned}$$

that is

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \beta a^2=x\\ \alpha dP_\mu (u)[\varphi ]+\beta dP_\mu (u)[u]=y \end{array}\right. } \end{aligned}$$

is solvable with respect to \(\alpha \), \(\beta \) for every \((x,y)\in {\mathbb {R}}^2\), so the surjective is proved. Next, suppose by contrary that for \(u \in \mathcal {P}(a)\) such that a tangent vector \(\varphi \) does not exist, that is, \(dP_\mu (u)[\varphi ]=0\) for every \(\varphi \in T_uS\). Then u is a constrained critical point for the functional \(P_\mu (u)\) on S(a). Thus, by the Lagrange multipliers rule, there exists \(\nu \in {\mathbb {R}}\) such that

$$\begin{aligned} -\triangle u+\frac{\gamma }{2}\phi _uu=\nu u+\frac{\mu q\delta _q}{2}|u|^{q-2}u+3|u|^4u,~~~\text{ in }~~{\mathbb {R}}^3. \end{aligned}$$

Then we can conclude the following Pohozaev type identity:

$$\begin{aligned} 2\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx=\mu q{\delta _q}^2\int _{{\mathbb {R}}^3}|u|^qdx+6\int _{{\mathbb {R}}^3}|u|^6dx, \end{aligned}$$

which is contradiction to the fact that \(u \in \mathcal {P}(a)\). \(\square \)

In virtue of the GNS inequality (2.4) and (2.1), for every \(u \in H({\mathbb {R}}^3)\bigcap S(a)\), we have

$$\begin{aligned} \begin{aligned} I_\mu (u)&=\frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4} \int _{{\mathbb {R}}^3}\phi _u u^2dx-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|u|^qdx - \frac{1}{6}\int _{{\mathbb {R}}^3}|u|^{6}dx\\&\ge \frac{1}{2}\Vert \nabla u\Vert ^2_2-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2-\frac{1}{6S^3}\Vert \nabla u\Vert ^6_2\\&:=g(\Vert \nabla u\Vert _2), \end{aligned} \end{aligned}$$
(3.4)

where

$$\begin{aligned} g(t)= \frac{1}{2}t^2-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}t^{q\delta _q}-\frac{1}{6S^3}t^6. \end{aligned}$$

By the fact \(q\delta _q<2\), we can derive that \(g(0^+)=0^-\) and \(g(+\infty )=-\infty \).

In the following, we show the properties of the function g and give some technical lemmas.

Lemma 3.2

Let \(2<q<\frac{10}{3}\), \(\mu ,\gamma >0\), and \(0<a<\alpha _1\). Then the function g has a local strict minimum at negative level and a global strict maximum at positive level, and there exist two positive constants \(R_1\),\(R_2\) both depending on a, with \(R_1<R_2\), such that \(g(R_1)=g(R_2)=0\) and \(g(t)>0\) for \(t \in (R_1,R_2)\).

Proof

Note that

$$\begin{aligned} \begin{aligned} g(t)&= \frac{1}{2}t^2-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}t^{q\delta _q}-\frac{1}{6S^3}t^6\\&=t^{q\delta _q} \left( \frac{1}{2}t^{2-q\delta _q}-\frac{1}{6S^3}t^{6-q\delta _q} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\right) \\&=t^{q\delta _q}m(t), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} m(t)=\frac{1}{2}t^{2-q\delta _q}-\frac{1}{6S^3}t^{6-q\delta _q} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}. \end{aligned}$$

It is easy to see that \(g(t)>0\) if and only if \(m(t)>0\) for all \(t>0\). So, by direct calculation, we have

$$\begin{aligned} m'(t)=\frac{2-q\delta _q}{2}t^{1-q\delta _q}-\frac{6-q\delta _q}{6S^3}t^{5-q\delta _q}. \end{aligned}$$

Let \(m'(t)=0\), it follows that

$$\begin{aligned} t_1=\left( \frac{3S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{1}{4}}, \end{aligned}$$

and we know that m is strictly increasing on \((0,t_1)\) and decreasing on \((t_1,\infty )\). Moreover, the maximum value of m on \((0,+\infty )\) is

$$\begin{aligned} \begin{aligned} m(t_1)&=\frac{1}{2}\left( \frac{3S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{2-q\delta _q}{4}} -\frac{1}{6S^3}\left( \frac{3S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{6-q\delta _q}{4}} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\\&=\frac{2}{6-q\delta _q}\left( \frac{3S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{2-q\delta _q}{4}} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\\&>\frac{2}{6-q\delta _q}\left( \frac{S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{2-q\delta _q}{4}} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\\&=\frac{\mu }{q}C(q)\alpha _1^{q(1-\delta _q)}-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}. \end{aligned} \end{aligned}$$

By virtue of \(a<\alpha _1\), we deduce that there exist two constants \(R_1\) and \(R_2\) such that

$$\begin{aligned} g(t) {\left\{ \begin{array}{ll} <0,&{}\text{ if }~~t\in (0,R_1) ~\text{ or }~(R_2,\infty );\\ =0,&{}\text{ if }~~t=R_1 ~\text{ or }~R_2;\\ >0,&{}\text{ if }~~t\in (R_1,R_2). \end{array}\right. } \end{aligned}$$

Based on above analysis and the fact \(g(0^+)=0^-\), we infer that g(t) has a global maximum at positive level in \((R_1,R_2)\) and a local minimum at negative level in \((0,R_1)\). It is easy to see that \(R_1<t_1<R_2.\) Besides, by a simple calculation, we have

$$\begin{aligned} \begin{aligned} g'(t)=&q\delta _qt^{q\delta _q-1}\left( \frac{1}{2}t^{2-q\delta _q}-\frac{1}{6S^3}t^{6-q\delta _q} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\right) \\&+t^{q\delta _q}\left( \frac{2-q\delta _q}{2}t^{1-q\delta _q}-\frac{6-q\delta _q}{6S^3}t^{5-q\delta _q}\right) \\ =&t^{q\delta _q-1} \left( t^{2-q\delta _q}-\frac{1}{S^3}t^{6-q\delta _q} -\mu \delta _qC(q)a^{q(1-\delta _q)}\right) \\:=&t^{q\delta _q-1}h(t), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} h(t)=t^{2-q\delta _q}-\frac{1}{S^3}t^{6-q\delta _q} -\mu \delta _qC(q)a^{q(1-\delta _q)}. \end{aligned}$$

It is easy to see that \(g'(t)=0\) if and only if \(h(t)=0\) for \(t>0\). So, by direct calculation, we get

$$\begin{aligned} h'(t)=(2-q\delta _q)t^{1-q\delta _q}-\frac{6-q\delta _q}{S^3}t^{5-q\delta _q}. \end{aligned}$$

From \(h'(t)=0\), there exists a unique solution \(t_2>0\) with the expression:

$$\begin{aligned} t_2=\left( \frac{S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{1}{4}}, \end{aligned}$$

and we know that h is a strictly increasing on \((0,t_2)\) and decreasing on \((t_2,+\infty )\). Hence, h has at most two zeros on \((0,+\infty )\), which are necessarily the previously found local minimum and the global maximum of g. \(\square \)

Lemma 3.3

Let \(2<q<\frac{10}{3}\), \(\mu ,\gamma >0\), and \(0<a<\alpha _1\). Then for every \(u \in S(a)\), \(\varphi _u(\iota )\) has two critical points \(s_u<t_u \in {\mathbb {R}}\) and two zeros \(c_u<d_u\) with \(s_u<c_u<t_u<d_u\). Besides,

  1. (i)

    \(s_u\star u \in \mathcal {P}(a)^+\), \(t_u\star u \in \mathcal {P}(a)^-\), and if \(\iota \star u \in \mathcal {P}(a)\), then either \(\iota =s_u\) or \(\iota =t_u\);

  2. (ii)

    \(\Vert \nabla u\Vert _2\le R_1\) for every \(\iota <c_u\) and

    $$\begin{aligned} I_\mu (s_u\star u)=\min \{I_\mu (\iota \star u):\iota \in {\mathbb {R}}~\text{ and }~\Vert \nabla u\Vert _2\le R_1\}<0; \end{aligned}$$
    (3.5)
  3. (iii)

    we have

    $$\begin{aligned} I_\mu (t_u\star u)=\max \{I_\mu (\iota \star u):\iota \in {\mathbb {R}}\}>0, \end{aligned}$$
    (3.6)

    and \(\varphi _u(\iota )\) is strictly decreasing and concave on \((t_u,+\infty )\);

  4. (iv)

    the maps \(u \in \mathcal {P}(a)\mapsto s_u\times {\mathbb {R}}\) and \(u \in \mathcal {P}(a)\mapsto t_u\times {\mathbb {R}}\) are of class \(C^1\).

Proof

We claim that \(\varphi _u(\iota )\) has two critical points. In view of (3.4), one has

$$\begin{aligned} \begin{aligned} \varphi _u(\iota )&=I_{\mu } ((\iota \star u))\\&\ge \frac{1}{2}\Vert \nabla (\iota \star u)\Vert ^2_2-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\Vert \nabla (\iota \star u)\Vert ^{q\delta _q}_2 -\frac{1}{6S^3}\Vert \nabla (\iota \star u)\Vert ^6_2\\&=g(\Vert \nabla (\iota \star u)\Vert _2)=g(e^{\iota }\Vert \nabla u\Vert _2), \end{aligned} \end{aligned}$$

we can obtain \(\varphi _u(\iota )>0\) on \((\xi (R_1),\xi (R_2))\) from Lemma 3.2, where

$$\begin{aligned} \xi (R)=\log R-\log \Vert \nabla u\Vert _2. \end{aligned}$$

Since \(\varphi _u(\iota )\) is a \(C^2\) function, and by the fact that \(\varphi _u(-\infty )=0^-\), \(\varphi _u(+\infty )=-\infty \), it follows that \(\varphi _u(\iota )\) has at least critical points \(s_u\), \(t_u\) with \(s_u<t_u\). Moreover, we know that \(s_u\) is a local minimum point on \((-\infty ,\xi (R_1))\) at negative level and \(t_u\) is a global maximum point at positive level. Hence, we derive to

$$\begin{aligned} I_\mu (t_u\star u)=\max \{I_\mu (\iota \star u):\iota \in {\mathbb {R}}\}>0, \end{aligned}$$
$$\begin{aligned} \Vert \nabla (s_u\star u)\Vert _2={e^{s_u}}\Vert \nabla u\Vert _2\le {e^{\xi (R_1)}}\Vert \nabla u\Vert _2=R_1, \end{aligned}$$
(3.7)

and

$$\begin{aligned} I_\mu (s_u\star u)=\min \{I_\mu (\iota \star u):\iota \in {\mathbb {R}}~\text{ and }~\Vert \nabla u\Vert _2\le R_1\}<0. \end{aligned}$$

Arguing as in the proof of Lemma 3.2, we can deduce that \(\varphi _u(\iota )\) has no other critical points. In view of \((\varphi _u)''(s_u)\ge 0\), \((\varphi _u)''(t_u)\le 0\) and the fact that \(\mathcal {P}(a)^0=\varnothing \), we have \(s_u\star u \in \mathcal {P}(a)^+\) and \(t_u\star u \in \mathcal {P}(a)^-\).

Next, we claim that \(\varphi _u(\iota )\) has two zeros \(c_u<d_u\). Since \(\varphi _u(s_u)<0\), \(\varphi _u(t_u)>0\) and \(\varphi _u(+\infty )=-\infty \), it is easy to get that \(\varphi _u(\iota )\) has two zeros \(c_u<d_u\) with \(s_u<c_u<t_u<d_u\). Furthermore, \(\varphi _u(\iota )\) has no other zeros. Indeed, if \(\varphi _u(\iota )\) has other zeros, then it will have other critical point, which leads to a contradiction.

Recalling that

$$\begin{aligned}{} & {} (\varphi _u)''(\iota ) =2e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx\\{} & {} -\mu q{\delta _q}^2e^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -6e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx, \end{aligned}$$

we have \((\varphi _u)''(-\infty )=0^-\). Since \((\varphi _u)''(s_u)>0\) and \((\varphi _u)''(t_u)<0\), we get \((\varphi _u)''(\iota )\) has two zeros, which means that \(\varphi _u(\iota )\) has two inflection points. Arguing as before, \((\varphi _u)''(\iota )\) has exactly two inflection points. Hence, \(\varphi _u(\iota )\) is is strictly decreasing and concave on \((t_u,+\infty )\). The items (i)–(iii) are proved.

Finally, we will prove that maps \(u \in \mathcal {P}(a)\mapsto s_u\times {\mathbb {R}}\) and \(u \in \mathcal {P}(a)\mapsto t_u\times {\mathbb {R}}\) are of class \(C^1\). Applying the implicit function theorem, let \(\Phi (\iota ,u):=(\varphi _u)'(\iota )>0\), since \(\Phi (s_u,u)=0\) and \(\partial _\iota \Phi (s_u,u)>0\), we know that \(u \in \mathcal {P}(a)\mapsto s_u\times {\mathbb {R}}\) is of class \(C^1\). Similarly, we have \(u \in \mathcal {P}(a)\mapsto t_u\times {\mathbb {R}}\) is of class \(C^1\). \(\square \)

Thus, we can easily deduce the following conclusion.

Corollary 3.4

\(\sup _{u\in \mathcal {P}(a)^+}I_\mu (u)\le 0\le \inf _{u\in \mathcal {P}(a)^-}I_\mu (u)~~~\text{ and }~~~\mathcal {P}(a)^+\subset D_{R_1},\) where

$$\begin{aligned}D_{R_1}:=\{u \in S(a):\Vert \nabla u\Vert _2<R_1\}, ~~~\text{ for }~~~ R_1>0.\end{aligned}$$

Lemma 3.5

There holds that \(-\infty<m_\mu (a)=\inf _{u\in \mathcal {P}(a)}I_\mu (u)=\inf _{u\in \mathcal {P}(a)^+}I_\mu (u)<0,\) and

$$\begin{aligned} m_\mu (a)<\inf _{u\in \overline{D_{R_1}}\setminus D_{R_1-\rho }}I_\mu (u), \end{aligned}$$

for \(\rho >0\) small enough, where

$$\begin{aligned} m_\mu (a):=\inf _{u\in D_{R_1}}I_\mu (u). \end{aligned}$$

Proof

For \(u \in D_{R_1}\), in view of (3.4), we have

$$\begin{aligned} I_\mu (u)\ge g(\Vert \nabla u\Vert _2)\ge \min _{t \in [0,R_1]}g(t)>-\infty . \end{aligned}$$

Besides, for any \(u \in S(a)\), we get \(\Vert \nabla u\Vert _2<R_1\) and \(I_\mu (s_u\star u)<0\). Hence, we can infer to

$$\begin{aligned} m_\mu (a)<0. \end{aligned}$$

On one hand, since \(\mathcal {P}(a)^+\subset D_{R_1}\), we get that \(m_\mu (a)\le \inf _{\mathcal {P}(a)^+}I_\mu \). On the other hand, if \(u \in D_{R_1}\), then \(s_u\star u\in \mathcal {P}(a)^+\subset D_{R_1}\), and

$$\begin{aligned} I_\mu (s_u\star u)=\min \{I_\mu (\iota \star u):\iota \in {\mathbb {R}}~\text{ and }~\Vert \nabla u\Vert _2\le R_1\}\le I_\mu (u), \end{aligned}$$

which implies that \(\inf _{u\in \mathcal {P}(a)^+}I_\mu (u)\le m_\mu (a)\). Combining with the fact \(0\le \inf _{u\in \mathcal {P}(a)^-}I_\mu (u)\), we obtain

$$\begin{aligned} \inf _{u \in \mathcal {P}(a)}I_\mu (u)=\inf _{u \in \mathcal {P}(a)^+}I_\mu (u). \end{aligned}$$

Finally, due to the continuity of g and \(g(R_1)=0\), there exists \(\rho >0\) such that

$$\begin{aligned} g(t)\ge \frac{m_\mu (a)}{2},~~~t \in [R_1-\rho ,R_1]. \end{aligned}$$

Therefore, by (3.4), we have

$$\begin{aligned} I_\mu (u)\ge g(\Vert \nabla u\Vert _2)\ge \frac{m_\mu (a)}{2}\ge m_\mu (a), \end{aligned}$$

for any \(u \in {\overline{D}_{R_1}\setminus D_{R_1-\rho }}\). The proof is completed. \(\square \)

Proof of Theorem 1.1

First, we take a minimizing sequence \(\{v_n\}\subset H\cap S(a)\) for \(I_\mu |_{D_{R_1}}\) and assume that \(\{v_n\}\subset H_r\) are radially decreasing for every n. Otherwise, we can let \(v_n:=|v_n|^*\), which is the Schwarz rearrangement of \(|v_n|\). In view of Lemmas 3.3 and 3.5, we know that there exists a sequence \(\{s_{v_n}\}\) such that \(s_{v_n}\star v_n \in \mathcal {P}(a)^+\) and \(I_\mu (s_{v_n}\star v_n)\le I_\mu (v_n)\) for every n. Furthermore, we have \(s_{v_n}\star v_n\notin {\overline{D}_{R_1}{\setminus } D_{R_1-\rho }}\). Based on above analysis, we get a new minimizing sequence \(\{\overline{v}_n:=s_{v_n}\star v_n\}\) for \(I_\mu |_{D_{R_1}}\), satisfying

$$\begin{aligned} \overline{v}_n\in H_r\cap \mathcal {P}(a)^+~~~\text{ and }~~~\Vert \nabla \overline{v}_n\Vert _2\le R_1-\rho . \end{aligned}$$

By Ekeland’s variational principle, there exists a new minimizing sequence \(\{u_n\}\), with \(\Vert u_n-\overline{v}_n\Vert \rightarrow 0\) as \(n\rightarrow \infty \), which is also a PS sequence for \(I_\mu \) on S(a). Since \(\{u_n\}\subset D_{R_1}\), we see that \(\{u_n\}\) is bounded in H. So from \(\Vert u_n-\overline{v}_n\Vert \rightarrow 0\) and the boundedness of \(\{u_n\}\), we can obtain

$$\begin{aligned}P_\mu (u_n)=P_\mu (\overline{v}_n)+o_n(1)\rightarrow 0~~~~~~~~~\text{ as }~~n\rightarrow \infty .\end{aligned}$$

In fact,

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u_n|^2dx= & {} \int _{{\mathbb {R}}^3}|\nabla \overline{v}_n|^2dx+\int _{{\mathbb {R}}^3}|\nabla (u_n-\overline{v}_n)|^2dx+\int _{{\mathbb {R}}^3}\nabla u_n\nabla \overline{v}_ndx\\= & {} \int _{{\mathbb {R}}^3}|\nabla \overline{v}_n|^2dx+o_n(1),\\ \frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx= & {} \frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{\overline{v}_n}\overline{v}_n^2dx +\gamma \int _{{\mathbb {R}}^3}\phi _{(\overline{v}_n+\theta ^1_n(u_n-\overline{v}_n))}|\overline{v}_n+\theta ^1_n(u_n-\overline{v}_n)|(u_n-\overline{v}_n)dx\\= & {} \frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{\overline{v}_n}\overline{v}_n^2dx+o_n(1), \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^3}|u_n|^pdx&=\int _{{\mathbb {R}}^3}|\overline{v}_n|^pdx+\int _{{\mathbb {R}}^3}p|\overline{v}_n+\theta ^2_n(u_n-\overline{v}_n)|^{p-1}(u_n-\overline{v}_n)dx\\&=\int _{{\mathbb {R}}^3}|\overline{v}_n|^pdx+o_n(1), \end{aligned} \end{aligned}$$

for every \(p\in [2,6]\), where \(\theta ^1_n,~\theta ^2_n\in [0,1]\). Moreover, \(\{u_n\}\) satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle I_\mu (u_n)\rightarrow m_\mu (a)~~~\text{ as }~~n\rightarrow \infty \\ I'_\mu |_{S(a)}(u_n)\rightarrow 0~~~~\text{ as }~~n\rightarrow \infty \\ \end{array}\right. } \end{aligned}$$
(3.8)

Then, using the Lagrange multipliers rule, there exists a sequence \(\lambda _n\in {\mathbb {R}}\) such that

$$\begin{aligned} I'_\mu (u_n)-\lambda _n\Psi '(u_n)\rightarrow 0~~~\text{ in }~~H^{-1}. \end{aligned}$$
(3.9)

Since \(\{u_n\}\subset D_{R_1}\), we have \(\{u_n\}\) is bounded in H. So there exists \(u_a \in H\), such that, for some subsequence, \(u_n\rightharpoonup u_a\) in H. In the following, we will proceed with our argument in three steps.

Step 1 We show that, up to subsequence, \(\lim _{n\rightarrow +\infty }\lambda _n=\lambda _a<0\). By (3.9) and the fact \(\{u_n\}\) is bounded in H, we get

$$\begin{aligned} I'_\mu (u_n)u_n-\lambda _n\Psi '(u_n)u_n=o_n(1). \end{aligned}$$
(3.10)

Then, we infer to

$$\begin{aligned} \lambda _n \Vert u_n\Vert ^2_2=\Vert \nabla u_n\Vert ^2_2+\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx-\mu \Vert u_n\Vert ^q_q-\Vert u_n\Vert ^6_6+o_n(1). \end{aligned}$$
(3.11)

Again by \(\{u_n\}\) is bounded in H, we see that \(\{\lambda _n\}\) is bounded. Thus, up to subsequence, there exists \(\lambda _a \in {\mathbb {R}}\) such that \(\lambda _n\rightarrow \lambda _a\in {\mathbb {R}}\). Next, we prove \(\lambda _a<0\). Before this, we show

$$\begin{aligned} \int _{{\mathbb {R}}^3}|u_n|^qdx\rightarrow \int _{{\mathbb {R}}^3}|u_a|^qdx\ne 0,~~\text{ i.e. }~~u_a\ne 0. \end{aligned}$$

Assume by contradiction that, \(\int _{{\mathbb {R}}^3}|u_n|^qdx\rightarrow 0\). In view of the proof of Lemma 3.2, we have \(\Vert \nabla u_n\Vert _2\le R_1<t_1\), and \(t_1=\left( \frac{3S^3(2-q\delta _q)}{6-q\delta _q}\right) ^{\frac{1}{4}}<S^{\frac{3}{4}}\). Then we deduce

$$\begin{aligned} \Vert \nabla u_n\Vert _2<S^{\frac{3}{4}}. \end{aligned}$$

From the definition of \(I_{\mu }\) and above inequality, we infer that

$$\begin{aligned}\begin{aligned} 0>m_\mu (a)=&\lim _{n\rightarrow \infty }I_{\mu }(u_n)\\ =&\lim _{n\rightarrow \infty }\bigg [\frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\frac{\gamma }{4} \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|u_n|^qdx-\frac{1}{6}\int _{{\mathbb {R}}^3}|u_n|^6dx\bigg ]\\ \ge&\lim _{n\rightarrow \infty }\bigg [\frac{1}{2}\Vert \nabla u_n\Vert ^2_2 -\frac{1}{6}S^{-3}\Vert \nabla u_n\Vert _2^{6}-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|u_n|^qdx\bigg ]\\ \ge&-\frac{\mu }{q} \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}|u_n|^qdx=0, \end{aligned} \end{aligned}$$

which is absurd.

We claim that there exists \(\tilde{\mu }>0\) independently on \(n \in {\mathbb {N}}\) such that, if \(\mu >\tilde{\mu }\), the lagrange multiplier \(\lambda _a<0\). In fact, since \(\{u_n\}\subset D_{R_1}\), by (2.3) and the GNS inequality (2.4), there exists \(T_1>0\) independently on \(n \in {\mathbb {N}}\) such that

$$\begin{aligned} \begin{aligned} T_1\le \int _{{\mathbb {R}}^3}|u_n|^qdx&\le {C}(q) \Vert \nabla u_n\Vert _2^{q\delta _{q}}\Vert u_n\Vert _2^{q(1-\delta _{q})}\\&\le {C}(q)R_1^{q\delta _{q}}a^{q(1-\delta _{q})}, \end{aligned} \end{aligned}$$
(3.12)

and

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx\le {\tilde{C}}\Vert u_n\Vert _{\frac{12}{5}}^4&\le {\tilde{C}}[C(12/5)]^{\frac{5}{3}}\Vert \nabla u_n\Vert _2\Vert u_n\Vert _2^3\\&\le {\tilde{C}}[C(12/5)]^{\frac{5}{3}}R_1a^3:=T_2, \end{aligned} \end{aligned}$$
(3.13)

where \(T_2=T_2(R_1,a)>0\). We define the constant

$$\begin{aligned} \tilde{\mu }:=\frac{3\gamma T_2}{4(1-\delta _q)T_1}. \end{aligned}$$
(3.14)

By (3.12)–(3.14) we have

$$\begin{aligned} \tilde{\mu }\ge \lim _{n\rightarrow +\infty }\left\{ \frac{3\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx}{4(1-\delta _q)\int _{{\mathbb {R}}^3}|u_n|^qdx}\right\} =\frac{3\gamma \int _{{\mathbb {R}}^3}\phi _{u_a}u_a^2dx}{4(1-\delta _q)\int _{{\mathbb {R}}^3}|u_a|^qdx}>0. \end{aligned}$$
(3.15)

By the fact \(P_\mu (u_n)\rightarrow 0\), (3.11), Lemma 2.7 and \(\delta _q<1\), if \(\mu >\tilde{\mu }\), then

$$\begin{aligned} \begin{aligned} \lambda _aa^2&=\lim _{n\rightarrow +\infty }\left( \int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx-\mu \int _{{\mathbb {R}}^3}|u_n|^qdx-\int _{{\mathbb {R}}^3}|u_n|^6dx\right) \\&=\lim _{n\rightarrow +\infty }\left( \frac{3}{4}\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx-\mu (1-\delta _q)\int _{{\mathbb {R}}^3}|u_n|^qdx\right) \\&=\frac{3}{4}\gamma \int _{{\mathbb {R}}^3}\phi _{u_a}u_a^2dx-\mu (1-\delta _q)\int _{{\mathbb {R}}^3}|u_a|^qdx\\&<\frac{3}{4}\gamma \int _{{\mathbb {R}}^3}\phi _{u_a}u_a^2dx-\tilde{\mu }(1-\delta _q)\int _{{\mathbb {R}}^3}|u_a|^qdx\le 0. \end{aligned} \end{aligned}$$

Thus, if \(\mu >\tilde{\mu }\), we have \(\lim _{n\rightarrow +\infty }\lambda _n=\lambda _a<0\).

Step 2 Since \(\lambda _a<0\), we define an equivalent norm of H as:

$$\begin{aligned} \Vert u\Vert ^2=\int _{{\mathbb {R}}^3}|\nabla u|^2dx-\lambda \int _{{\mathbb {R}}^3}|u|^2dx. \end{aligned}$$

In view of the fact \(u_n\rightharpoonup u_a\) in H and (3.9), then \(u_a\) satisfies

$$\begin{aligned}{} & {} \int _{{\mathbb {R}}^3}\nabla u_a \nabla v dx+\gamma \int _{{\mathbb {R}}^3}\phi _{u_a}u_a v dx-\lambda _a\int _{{\mathbb {R}}^3}u_a v dx \nonumber \\{} & {} \quad -\mu \int _{{\mathbb {R}}^3}|u_a|^{q-2}u_a v dx-\int _{{\mathbb {R}}^3}|u_a|^4u_a v dx=0, \end{aligned}$$
(3.16)

for \(\forall v \in H\). It follows from the Pohozaev identity that \(P_\mu (u_a)=0\). Let \(v_n=u_n-u_a\rightharpoonup 0\), by Brezis–Lieb Lemma [48], we conclude

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle \Vert \nabla v_n\Vert _2^2=\Vert \nabla u_n\Vert _2^2-\Vert \nabla u_a\Vert _2^2+o_n(1),\\ \Vert v_n\Vert ^6_6=\Vert u_n\Vert ^6_6-\Vert u_a\Vert ^6_6+o_n(1). \end{array}\right. } \end{aligned}$$
(3.17)

By the fact (2.5), Lemma 2.7 and \(P_\mu (v_n)=P_\mu (u_n)-P_\mu (u_a)\rightarrow 0\), we obtain

$$\begin{aligned} \Vert \nabla v_n\Vert _2^2=\Vert v_n\Vert ^6_6+o_n(1). \end{aligned}$$

Thus, for some subsequence, we suppose that

$$\begin{aligned} \Vert \nabla v_n\Vert _2^2=\Vert v_n\Vert ^6_6\rightarrow \tau . \end{aligned}$$

By using (2.1), we derive to

$$\begin{aligned} \tau ^{\frac{1}{3}}\le \frac{\tau }{S}, \end{aligned}$$

then, one has

$$\begin{aligned} \tau \ge S^{\frac{3}{2}}~~~\text{ or }~~~\tau =0. \end{aligned}$$

If \(\tau \ge S^{\frac{3}{2}}\), by (3.17), we have

$$\begin{aligned} \begin{aligned} m_\mu (a)&=\lim _{n\rightarrow +\infty }I_\mu (u_n)\\&=\lim _{n\rightarrow +\infty }\left( I_\mu (u_a)+\frac{1}{2}\Vert \nabla v_n\Vert _2^2-\frac{1}{6}\Vert v_n\Vert ^6_6\right) \\&=I_\mu (u_a)+\frac{1}{3}\tau \\&\ge I_\mu (u_a)+\frac{1}{3}S^{\frac{3}{2}}. \end{aligned} \end{aligned}$$
(3.18)

In what follows, we verify that \(\tau \ge S^{\frac{3}{2}}\), which will lead to a contradiction. In fact, by the GNS inequality (2.5) and \(P_\mu (u_a)=0\), we get

$$\begin{aligned} \begin{aligned} I_\mu (u_a)&=I_\mu (u_a)-\frac{1}{6}P_\mu (u_a)\\&=\frac{1}{3}\Vert \nabla u_a\Vert _2^2+\frac{5}{24}\gamma \int _{{\mathbb {R}}^3}\phi _{u_a}u_a^2dx-\mu \left( \frac{1}{q}-\frac{\delta _q}{6}\right) \Vert u_a\Vert ^q_q\\&\ge \frac{1}{3}\Vert \nabla u_a\Vert _2^2-\frac{\mu (6-q\delta _q)}{6q}\Vert u_a\Vert ^q_q\\&\ge \frac{1}{3}\Vert \nabla u_a\Vert _2^2-\frac{\mu (6-q\delta _q)}{6q}C(q)a^{q(1-\delta _q)}\Vert \nabla u_a\Vert _2^{q\delta _q}\\&:=f(\Vert \nabla u\Vert _2), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} f(t)=\frac{1}{3}t^2-\frac{\mu (6-q\delta _q)}{6q}C(q)a^{q(1-\delta _q)}t^{q\delta _q}. \end{aligned}$$

By \(f'(t)=0\), there exists a unique \(t_3>0\) such that

$$\begin{aligned} f'(t_3)=\frac{2}{3}t_3-\frac{\mu (6-q\delta _q)}{6q}C(q)a^{q(1-\delta _q)}q\delta _qt_3^{q\delta _q-1}=0, \end{aligned}$$

with

$$\begin{aligned} t_3=\left( \frac{3}{2}\frac{\mu (6-q\delta _q)}{6q}C(q)a^{q(1-\delta _q)}q\delta _q\right) ^{\frac{1}{2-q\delta _q}}. \end{aligned}$$

Then, we see that f(t) is strictly decreasing on \((0,t_3)\) and increasing on \((t_3,+\infty )\). Moreover, f(t) gets the minimum on \((0,+\infty )\), that is

$$\begin{aligned} f(t_3)=-\frac{2-q\delta _q}{2}\left( \frac{\mu (6-q\delta _q)}{6q}C(q)a^{q(1-\delta _q)}\right) ^{\frac{2}{2-q\delta _q}} \left( \frac{3q\delta _q}{2}\right) ^{\frac{q\delta _q}{2-q\delta _q}}. \end{aligned}$$

Define

$$\begin{aligned} \alpha _2:=\left\{ \frac{4}{C(q)\mu \delta _q(6-q\delta _q)}\left( \frac{q\delta _qS^{\frac{3}{2}}}{2-q\delta _q}\right) ^{\frac{2-q\delta _q}{2}}\right\} ^{\frac{1}{q(1-\delta _q)}.} \end{aligned}$$

Since \(a<\alpha _2\), we get

$$\begin{aligned} f(t)>-\frac{1}{3}S^{\frac{3}{2}}~~~\text{ on }~~(0,+\infty ). \end{aligned}$$
(3.19)

Combining (3.18) and (3.19), we infer that \(m_\mu (a)>0\), which is a contradiction.

Step 3 From the above analysis, we know that \(\tau =0\). In other words, we have

$$\begin{aligned} \Vert u_n\Vert ^6_6\rightarrow \Vert u_a\Vert ^6_6. \end{aligned}$$

Then, by (3.16), we have

$$\begin{aligned} I'_\mu (u_a)u_a-\lambda _a\Psi '(u_a)u_a=0. \end{aligned}$$
(3.20)

Combining (3.10) and (3.20), one has

$$\begin{aligned} \Vert u_n\Vert ^2\rightarrow \Vert u_a\Vert ^2. \end{aligned}$$

Since \(u_n\rightharpoonup u_a\) in H, we have \(u_n\rightarrow u_a\) in H. Moreover, by the fact that \(I_\mu (u_a)=\inf _{u \in \mathcal {P}(a)}I_\mu (u)\), we know that \(u_a\) is a ground state.

Finally, in view of Lemma 3.5, one has

$$\begin{aligned} I_\mu (u_a)=\inf _{u \in \mathcal {P}(a)}I_\mu (u)=\inf _{u\in D_{R_1}}I_\mu (u). \end{aligned}$$

The proof is completed. \(\square \)

4 \(L^2\)-Critical Perturbation Case

In this section, we shall address the \(L^2\)-critical perturbation case: \(q=\frac{10}{3}\) and provide the proof of Theorem 1.2. To begin with, we give some useful lemmas, and show that \(E(u,\iota )\) has the mountain pass geometry on \(S_r(a)\times {\mathbb {R}}\), where \(S_r(a)=H^1_{rad}({\mathbb {R}}^3)\cap S(a)\).

Lemma 4.1

Let \(q=\frac{10}{3}\), \(\mu ,\gamma >0\) and \(u\in S(a)\), then

  1. (i)

    \(\Vert \nabla (\iota \star u)\Vert _2\rightarrow 0^+\) and \(I_{\mu } ((\iota \star u))\rightarrow 0^+\) if \(\iota \rightarrow -\infty \);

  2. (ii)

    \(\Vert \nabla (\iota \star u)\Vert _2\rightarrow +\infty \) and \(I_{\mu } ((\iota \star u))\rightarrow -\infty \) if \(\iota \rightarrow +\infty \).

Proof

By (2.10), we have

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla (\iota \star u)|^2dx=e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla u|^2dx, \end{aligned}$$

Then, it is easy to obtain

$$\begin{aligned} \Vert \nabla (\iota \star u)\Vert _2\rightarrow 0^+~~~\text{ if }~~\iota \rightarrow -\infty , \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla (\iota \star u)\Vert _2\rightarrow +\infty ~~~\text{ if }~~\iota \rightarrow +\infty . \end{aligned}$$

From (2.11), we have

$$\begin{aligned} I_{\mu } ((\iota \star u)) =\frac{1}{2}e^{2\iota }\Vert u\Vert ^2+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx-\frac{\mu }{q}e^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -\frac{1}{6}e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$

From the fact \(q\delta _q=2\), it follows that

$$\begin{aligned} I_{\mu } ((\iota \star u))\rightarrow 0^+~~~\text{ if }~~\iota \rightarrow -\infty , \end{aligned}$$

and

$$\begin{aligned} I_{\mu } ((\iota \star u))\rightarrow -\infty ~~~\text{ if }~~\iota \rightarrow +\infty . \end{aligned}$$

The proof is completed. \(\square \)

Lemma 4.2

Let \(q=\frac{10}{3}\), \(\mu ,\gamma >0\), and assume that \(0<a<\min \{\alpha _3,\alpha _4\}\), where

$$\begin{aligned} \alpha _3:=\left( \frac{q}{2\mu C(q)}\right) ^{\frac{1}{q(1-\delta _q)}}, \end{aligned}$$

and

$$\begin{aligned} \alpha _4:=\left( \frac{k^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) ^{\frac{1}{3}}. \end{aligned}$$

There exist \(0<k_1<k_2<k\) such that

$$\begin{aligned} P_\mu (u),I_{\mu }(u)>0~~\text{ for } \text{ all }~~{u\in A_{k_1}}~~\text{ and }~~0< \sup _{u\in A_{k_1}} I_{\mu } (u) < \inf _{u\in B_{k_2}}I_{\mu }(u), \end{aligned}$$

where

$$\begin{aligned} A_{k}:= \{u\in S_r(a): \Vert \nabla u\Vert ^2_2\le k\}~~~~\text{ and }~~~~ B_{k}:= \{u\in S_r(a): \Vert \nabla u\Vert ^2_2=2k\}. \end{aligned}$$

Proof

Take \(k>0\), which will be determined later. Assume that \(u,v\in S_r(a)\) such that \(\Vert \nabla u\Vert ^2_2\le k\) and \(\Vert \nabla v\Vert ^2_2=2k\). By (2.1), the GNS inequality (2.4) and \(q\delta _q=2\), we derive to

$$\begin{aligned} \begin{aligned} P_\mu (u)&=\Vert \nabla u\Vert ^2_2+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\mu \delta _q\Vert u\Vert ^q_q-\Vert u\Vert ^6_6\\&\ge \Vert \nabla u\Vert ^2_2-C(q)\mu \delta _qa^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2-S^{-3}\Vert \nabla u\Vert ^6_2\\&=\left( 1-\frac{2C(q)\mu }{q}a^{q(1-\delta _q)}\right) \Vert \nabla u\Vert ^2_2-S^{-3}\Vert \nabla u\Vert ^6_2, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} I_\mu (u)&=\frac{1}{2}\Vert \nabla u\Vert ^2_2+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\frac{\mu }{q}\Vert u\Vert ^q_q-\frac{1}{6}\Vert u\Vert ^6_6\\&\ge \frac{1}{2}\Vert \nabla u\Vert ^2_2-C(q)\frac{\mu }{q}a^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2-\frac{1}{6}S^{-3}\Vert \nabla u\Vert ^6_2\\&=\left( \frac{1}{2}-\frac{C(q)\mu }{q}a^{q(1-\delta _q)}\right) \Vert \nabla u\Vert ^2_2-\frac{1}{6}S^{-3}\Vert \nabla u\Vert ^6_2. \end{aligned} \end{aligned}$$

If \(a<\alpha _3\), we can deduce that

$$\begin{aligned} P_\mu (u)>0~~\text{ and }~~I_{\mu }(u)>0, \end{aligned}$$

for \(k>0\) small enough. Next, if \(a<\alpha _4\), we have

$$\begin{aligned} I_\mu (v)-I_\mu (u) \ge&\frac{1}{2}\Vert \nabla v\Vert ^2_2-\frac{1}{2}\Vert \nabla u\Vert ^2_2-\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\frac{\mu }{q}\Vert v\Vert ^q_q-\frac{1}{6}\Vert v\Vert ^6_6\\ \ge&\frac{1}{2}\Vert \nabla v\Vert ^2_2-\frac{1}{2}\Vert \nabla u\Vert ^2_2-\frac{\gamma }{4}{\tilde{C}}[C(12/5)]^{\frac{5}{3}}a^3\Vert \nabla u\Vert _2\\&-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\Vert \nabla v\Vert ^{q\delta _q}_2-\frac{1}{6}S^{-3}\Vert \nabla v\Vert ^6_2\\ \ge&k-\frac{1}{2}k-\frac{\gamma }{4}{\tilde{C}}[C(12/5)]^{\frac{5}{3}}\left( \frac{k^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) k^{\frac{1}{2}}\\&-\frac{C(q)\mu }{q}\left( \frac{k^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) ^{\frac{q(1-\delta _q)}{3}}2k -\frac{1}{6}S^{-3}(2k)^3\\ =&\frac{1}{2}k-\frac{1}{16}k-\left( \frac{2C(q)\mu }{q\left( 4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}\right) ^{\frac{q(1-\delta _q)}{3}}} k^{\frac{q(1-\delta _q)}{6}}\right) k-\left( \frac{4}{3}S^{-3}k^2\right) k\\ \ge&\frac{5}{16}k>0. \end{aligned}$$

If we take

$$\begin{aligned} k=\min \left\{ \left( \frac{q(4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}})^{\frac{q(1-\delta _q)}{3}}}{32\mu C(q)}\right) ^{\frac{6}{q(1-\delta _q)}}, ~~\left( \frac{3 }{64}S^3\right) ^{\frac{1}{2}}\right\} , \end{aligned}$$
(4.1)

then, for \(0<k_1<k_2<k\) small enough and \(0<a<\min \{\alpha _3,\alpha _4\}\), we infer to

$$\begin{aligned} P_\mu (u),~I_{\mu }(u)~~\text{ for } \text{ all }~~u\in A_{k_1}~~\text{ and }~~0< \sup _{u\in A_{k_1}} I_{\mu } (u) < \inf _{u\in B_{k_2}}I_{\mu }(u). \end{aligned}$$

The proof is completed. \(\square \)

In the following, we study the characteristics of the mountain pass levels for \(E(u,\iota )\) and \(I_{\mu }(u)\). Here, we define a closed set \(I_{\mu }^d:=\{u\in S_r(a): I_{\mu }(u) \le d \}\).

Proposition 4.3

Let \(q=\frac{10}{3}\), \(\mu ,\gamma >0\), and assume that \(0<a<\min \{\alpha _3,\alpha _4\}\). Take

$$\begin{aligned} \widetilde{\sigma }_{\mu }(a):=\inf _{\widetilde{\zeta }\in \widetilde{\Gamma _a}}\max _{t\in [0,1]}E(\widetilde{\zeta }(t)), \end{aligned}$$

where

$$\begin{aligned} \widetilde{\Gamma }_a=\{\widetilde{\zeta }\in C([0,1], S_r(a)\times {\mathbb {R}}):~\widetilde{\zeta }(0)\in (A_{k_1},0),~ \widetilde{\zeta }(1)\in (I_{\mu }^0,0)\}, \end{aligned}$$

and

$$\begin{aligned} \sigma _{\mu }(a):=\inf _{\zeta \in {\Gamma _a}}\max _{t\in [0,1]}I_\mu (\zeta (t)), \end{aligned}$$

where

$$\begin{aligned} {\Gamma }_a=\{{\zeta }\in C([0,1], S_r(a)):~{\zeta }(0)\in A_{k_1},~ \zeta (1)\in I_{\mu }^0\}. \end{aligned}$$

Then we have

$$\begin{aligned}\widetilde{\sigma }_{\mu }(a)={\sigma }_{\mu }(a).\end{aligned}$$

Proof

Since \(\Gamma _a\times \{0\}\subset \widetilde{\Gamma }_a,\) it is easy to know that \(\widetilde{\sigma }_{\mu }(a)\le {\sigma }_{\mu }(a)\). Then we only need to verify \(\widetilde{\sigma }_{\mu }(a)\ge {\sigma }_{\mu }(a)\). For \(\widetilde{\zeta }(t)=(\widetilde{\zeta }_1(t),\widetilde{\zeta }_2(t))\in \widetilde{\Gamma }_a\), one has,

$$\begin{aligned} \widetilde{\zeta }(0)=(\widetilde{\zeta }_1(0),\widetilde{\zeta }_2(0))\in (A_{k_1},0) ~~\text{ and }~~ \widetilde{\zeta }(1)=(\widetilde{\zeta }_1(1),\widetilde{\zeta }_2(1))\in (I_{\mu }^0,0). \end{aligned}$$

So, set \(\zeta (t)= (\widetilde{\zeta }_2(t)\star \widetilde{\zeta }_1(t))\), we have \(\zeta (t)\in \Gamma _a\), and so,

$$\begin{aligned} \max _{t\in [0,1]}E(\widetilde{\zeta }(t)) =\max _{t\in [0,1]}I_{\mu }(\widetilde{\zeta }_2(t)\star \widetilde{\zeta }_1(t))=\max _{t\in [0,1]}I_{\mu }(\zeta (t)), \end{aligned}$$

which implies that \(\widetilde{\sigma }_{\mu }(a)\ge {\sigma }_{\mu }(a)\). The proof is completed. \(\square \)

Next, we will verify the existence of the \((PS)_{\widetilde{\sigma }_{\mu }(a)}\) sequence for \(E(u, \iota )\) on \(S_r(a)\times {\mathbb {R}}\), which is demonstrated by a standard argument by using Ekeland’s variational principle and constructing pseudo-gradient flow (Proposition 2.2 [28]).

Proposition 4.4

Let \(\{\xi _n\}\subset \widetilde{\Gamma }_a\) be such that

$$\begin{aligned} \max _{t\in [0,1]}E(\xi _n(t))\le \widetilde{\sigma }_{\mu }(a)+\frac{1}{n}, \end{aligned}$$

then there exists a sequence \(\{(u_n, \iota _n)\}\subset S_r(a)\times {\mathbb {R}}\) satisfying

  1. (i)

    \(E(u_n,\iota _n)\in [\widetilde{\sigma }_{\mu }(a)-\frac{1}{n},\widetilde{\sigma }_{\mu }(a)+\frac{1}{n}];\)

  2. (ii)

    \(\min _{t\in [0,1]}\Vert (u_n, \iota _n)-\xi _n(t)\Vert _{{\mathbb {H}}}\le \frac{1}{\sqrt{n}}\);

  3. (iii)

    \(\Vert E'|_{S_r(a)\times {\mathbb {R}}}(u_n,\iota _n)\Vert \le \frac{2}{\sqrt{n}},\) i.e.,

    $$\begin{aligned} |\langle E'(u_n,\iota _n),z\rangle _{{\mathbb {H}}^{-1}\times {\mathbb {H}}}|\le \frac{2}{\sqrt{n}}\Vert z\Vert _{{\mathbb {H}}}, \end{aligned}$$

    for all

    $$\begin{aligned} z\in \widetilde{T}_{(u_n,\iota _n)}:=\{(z_1, z_2)\in {\mathbb {H}}: \langle u_n, z_1\rangle _{L^2} = 0\}. \end{aligned}$$

With the help of Proposition 4.4, we can obtain a \((PS)_{\sigma _{\mu }(a)}\) sequence for \(I_{\mu }(u)\) on \(S_r(a)\) in the following.

Proposition 4.5

Let \(q=\frac{10}{3}\), \(\mu ,\gamma >0\), and assume that \(0<a<\min \{\alpha _3,\alpha _4\}\). There exists a sequence \(\{w_n\}\subset S_r(a)\) such that

  1. (i)

    \(I_{\mu }(w_n)\rightarrow \sigma _{\mu }(a)\) as \(n\rightarrow \infty ;\)

  2. (ii)

    \(P_{\mu }(w_n)\rightarrow 0\) as \(n\rightarrow \infty ;\)

  3. (iii)

    \(I'_{\mu }|_{S_r(a)}(w_n)\rightarrow 0\) as \(n\rightarrow \infty \), i.e.,

    $$\begin{aligned}|\langle I_{\mu }'(w_n),z\rangle _{H^{-1}\times H}|\rightarrow 0, \end{aligned}$$

    uniformly for all \(h\in T_{w_n}\) and \(\Vert h\Vert \le 1\), where \(T_{w_n}:=\{h\in H:~\langle w_n,h\rangle _{L^2}=0\}\).

Proof

By Proposition 4.3, we have \(\widetilde{\sigma }_{\mu }(a)={\sigma }_{\mu }(a)\). Now, we take \(\{\xi _n = ((\xi _n)_1, 0)\}\in \widetilde{\Gamma }_a\) such that

$$\begin{aligned} \max _{t\in [0,1]}E(\xi _n(t))\le \widetilde{\sigma }_{\mu }(a)+\frac{1}{n}. \end{aligned}$$

From Proposition 4.4, we know that there exists a sequence \(\{(u_n,\iota _n)\}\subset S_r(a)\times {\mathbb {R}}\) such that as \(n\rightarrow \infty \), we have

$$\begin{aligned} E(u_n,\iota _n)\rightarrow {\sigma }_{\mu }(a), \end{aligned}$$
(4.2)
$$\begin{aligned} \iota _n\rightarrow 0, \end{aligned}$$
(4.3)
$$\begin{aligned} \partial _{\iota }E(u_n,\iota _n)\rightarrow 0. \end{aligned}$$
(4.4)

Let \( w_n =\iota _n\star u_n\), then \(I_{\mu }(w_n) = E(u_n, \iota _n)\), so item (i) follows.

Next, we show item (ii). Since

$$\begin{aligned} \begin{aligned} \partial _{\iota }E(u_n, \iota _n)&=e^{2\iota _n}\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\frac{\gamma }{4}e^{\iota _n}\int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx- \mu \delta _qe^{q\delta _q\iota _n}\int _{{\mathbb {R}}^3}|u_n|^qdx\\&\hspace{0.45cm}-e^{6\iota _n}\int _{{\mathbb {R}}^3}|u_n|^6dx\\&=\int _{{\mathbb {R}}^3}|\nabla (\iota _n\star u_n)|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{(\iota _n\star u_n)}(\iota _n\star u_n)^2dx-\mu \delta _q\int _{{\mathbb {R}}^3}|\iota _n\star u_n|^qdx\\&\hspace{0.45cm}-\int _{{\mathbb {R}}^3}|\iota _n\star u_n|^6dx\\&=P_{\mu }(w_n), \end{aligned} \end{aligned}$$

it follows that item (ii) holds.

To prove item (iii), we set \(h_n\in T_{w_n}\), then

$$\begin{aligned} \begin{aligned}&\langle I_{\mu }'(w_n),h_n\rangle _{H^{-1}\times H}\\&\hspace{0.45cm}=\int _{{\mathbb {R}}^3}\nabla w_n\nabla h_ndx+\gamma \int _{{\mathbb {R}}^3}\phi _{w_n}w_nh_ndx-\mu \int _{{\mathbb {R}}^3}|w_n|^{q-2}w_nh_ndx-\int _{{\mathbb {R}}^3}|w_n|^4w_nh_ndx\\&\hspace{0.45cm}=e^{-\frac{\iota _n}{2}}\int _{{\mathbb {R}}^3}\nabla u_n(x)\nabla h_n(e^{-\iota _n}x)dx +\gamma e^{-\frac{\iota _n}{2}}\int _{{\mathbb {R}}^3}\phi _{u_n(x)}u_n(x)h_n(e^{-\iota _n}x)dx\\&\hspace{1cm}-\mu e^{\frac{3(q-3)}{2}\iota _n}\int _{{\mathbb {R}}^3}|u_n(x)|^{q-2}u_n(x)h_n(e^{-\iota _n}x)dx -e^{\frac{9}{2}\iota _n}\int _{{\mathbb {R}}^3}|u_n(x)|^4u_n(x)h_n(e^{-\iota _n}x)dx\\&\hspace{0.45cm}=e^\iota _n\int _{{\mathbb {R}}^3}\nabla u_n(x)\nabla \left( e^{-\frac{3}{2}\iota _n} h_n(e^{-\iota _n}x)\right) dx +\gamma e^\iota _n \int _{{\mathbb {R}}^3}\phi _{u_n(x)}u_n(x)e^{-\frac{3}{2}\iota _n}h_n(e^{-\iota _n}x)dx\\&\hspace{1cm}-\mu e^{q\delta _q\iota _n}\int _{{\mathbb {R}}^3}|u_n(x)|^{q-2}u_n(x)e^{-\frac{3}{2}\iota _n}h_n(e^{-\iota _n}x)dx\\&\quad -e^{6\iota _n}\int _{{\mathbb {R}}^3}|u_n(x)|^4u_n(x)e^{-\frac{3}{2}\iota _n}h_n(e^{-\iota _n}x)dx.\\ \end{aligned} \end{aligned}$$

Let \(\widetilde{h}_n(x) = e^{-\frac{3}{2}\iota _n}h_n(e^{-\iota _n}x)\), we obtain

$$\begin{aligned} \langle I_{\mu }'(w_n),h_n\rangle _{H^{-1}\times H}=\langle E'(u_n,\iota _n),(\widetilde{h}_n,0)\rangle _{{\mathbb {H}}^{-1}\times {\mathbb {H}}}. \end{aligned}$$

Moreover, we get

$$\begin{aligned} \begin{aligned} \langle u_n,\widetilde{h}_n\rangle _{L^2}&=\int _{{\mathbb {R}}^3}u_n(x) e^{-\frac{3}{2}\iota _n}h_n(e^{-\iota _n}x)dx\\&=\int _{{\mathbb {R}}^3}e^{\frac{3}{2}\iota _n}u_n(e^{\iota _n}x) h_n(x)dx\\&=\int _{{\mathbb {R}}^3}w_n(x) h_n(x)dx=0. \end{aligned} \end{aligned}$$

Thus, we obtain \((\widetilde{h}_n, 0)\in \widetilde{T}_{(u_n,\iota _n)}\). On the other hand,

$$\begin{aligned} \begin{aligned} \Vert (\widetilde{h}_n, 0)\Vert _{{\mathbb {H}}}^2&= \Vert \widetilde{h}_n(x)\Vert _{H}^2\\&=\Vert {h}_n(x)\Vert _{2}^2+ e^{-2\iota _n}\Vert {h}_n(x)\Vert ^2\\&\le C\Vert h_n(x)\Vert ^2, \end{aligned} \end{aligned}$$

where the last inequality can be established by (4.3). So the item (iii) is proved. \(\square \)

Now, we construct the relationship between \(\sigma _\mu (a)\) and \(m_{\mu ,r}(a)\), where

$$\begin{aligned} m_{\mu ,r}(a)=\inf _{u\in \mathcal {P}_r(a)}I_\mu (u), \end{aligned}$$

and

$$\begin{aligned} \mathcal {P}_r(a)=\mathcal {P}(a)\cap S_r(a). \end{aligned}$$

Lemma 4.6

Let \(q=\frac{10}{3}\), \(\mu ,\gamma >0\), and assume that \(0<a<\min \{\alpha _3,\alpha _4\}\). Then we have

$$\begin{aligned} m_{\mu ,r}(a)=\inf _{u\in \mathcal {P}_r(a)^-}I_\mu (u)=\sigma _\mu (a)>0, \end{aligned}$$

where

$$\begin{aligned}\mathcal {P}_r(a)^-=\mathcal {P}(a)^-\cap S_r(a).\end{aligned}$$

Proof

In the following, we split the proof into four steps.

Step 1 We verify that for each \(u \in S_r(a)\), there exists a unique \(t_u \in {\mathbb {R}}\) such that \(t_u \star u \in \mathcal {P}_r(a)\), with \(t_u\) is the strict maximum point for the function \(\varphi _u(\iota )\) on \((0,+\infty )\) at positive level. Moreover, \(\mathcal {P}_r(a)=\mathcal {P}_r(a)^-\).

In fact, by Lemma 4.1, we have

$$\begin{aligned} \varphi _u(-\infty )=0^+~~~\text{ and }~~~\varphi _u(+\infty )=-\infty . \end{aligned}$$
(4.5)

Since \(\varphi _u(\iota )\) is a \(C^2\) function, we can deduce that \(\varphi _u(\iota )\) has at least one critical point \(t_u\), with \(t_u\) is a global maximum point at positive level. In view of Lemma 2.6, we have \(t_u \star u \in \mathcal {P}_r(a)\). Next, we prove that \(\varphi _u(\iota )\) has no other critical points. Indeed, recall \((\varphi _u)'(\iota )\) and \((\varphi _u)''(\iota )\) as follow:

$$\begin{aligned}(\varphi _u)'(\iota ) =e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx-\mu \delta _qe^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx, \end{aligned}$$

and

$$\begin{aligned}{} & {} (\varphi _u)''(\iota ) =2e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _uu^2dx\\{} & {} -\mu q{\delta _q}^2e^{q\delta _q\iota }\int _{{\mathbb {R}}^3}|u|^qdx -6e^{6\iota }\int _{{\mathbb {R}}^3}|u|^6dx. \end{aligned}$$

Assume by contradiction, there exists other critical point \(e_u \in {\mathbb {R}}\) with \(t_u<e_u\) and \(e_u\) is also a global maximum point of \(\varphi _u(\iota )\). Then, we see that there exists a critical point \(f_u\), such that \(t_u<f_u<e_u\) and \(f_u\) is a minimum point of \(\varphi _u(\iota )\). Consequently, we have

$$\begin{aligned}(\varphi _u)'(e_u)=0,~~(\varphi _u)'(f_u)=0,\\(\varphi _u)''(e_u)=-\frac{\gamma }{4}e^{e_u}\int _{{\mathbb {R}}^3}\phi _uu^2dx-4e^{6e_u}\int _{{\mathbb {R}}^3}|u|^6dx<0,\end{aligned}$$

and

$$\begin{aligned}(\varphi _u)''(f_u)=-\frac{\gamma }{4}e^{f_u}\int _{{\mathbb {R}}^3}\phi _uu^2dx-4e^{6f_u}\int _{{\mathbb {R}}^3}|u|^6dx<0,\end{aligned}$$

which is a contradiction.

Step 2 We show that \(I_\mu (u)\le 0\) implies \(P_\mu (u)<0\). In fact, since \(\varphi _u(0)=I_\mu (0\star u)=I_\mu (u)\le 0\), by the properties of the function \(\varphi _u(\iota )\) presented in Step 1 and by (4.5), we infer that \(t_u<0\). Besides, since

$$\begin{aligned} P_\mu (t_u\star u)=(\varphi _u)'(t_u)=0~~\text{ and }~~P_\mu (u)=P_\mu (0\star u)=(\varphi _u)'(0), \end{aligned}$$

we obtain that \(P_\mu (u)<0\).

Step 3 We claim that \(m_{\mu ,r}(a)=\sigma _\mu (a)\). Indeed, let \(u \in S_r(a)\), we take \(\iota ^-\ll 0\) and \(\iota ^+\gg 0\) such that \(\iota ^-\star u\in A_{k_1}\) and \(I_\mu (\iota ^+\star u)<0\), respectively. Then we can define a path

$$\begin{aligned} \zeta _u:t\in [0,1]\mapsto ((1-t)\iota ^-+t\iota ^+)\star u \in \Gamma _a. \end{aligned}$$
(4.6)

Hence, we get

$$\begin{aligned}\max _{t\in [0,1]}I_\mu (\zeta _u(t))\ge \sigma _\mu (a), \end{aligned}$$

and so, we have \(m_{\mu ,r}(a)\ge \sigma _\mu (a)\). Moreover, for any \(\widetilde{\zeta }(t)=(\widetilde{\zeta }_1(t),\widetilde{\zeta }_2(t))\in \widetilde{\Gamma }_a\), one has,

$$\begin{aligned} \widetilde{\zeta }(0)=(\widetilde{\zeta }_1(0),\widetilde{\zeta }_2(0))\in (A_{k_1},~0) ~~\text{ and }~~ \widetilde{\zeta }(1)=(\widetilde{\zeta }_1(1),\widetilde{\zeta }_2(1))\in (I_{\mu }^0,~0). \end{aligned}$$

Now, we define the function

$$\begin{aligned} \widetilde{P}_\mu (t)=P_\mu (\widetilde{\zeta }_2(t)\star \widetilde{\zeta }_1(t)). \end{aligned}$$

Since \((\widetilde{\zeta }_2(0)\star \widetilde{\zeta }_1(0))=\widetilde{\zeta }_1(0) \in A_{k_1}\) and \((\widetilde{\zeta }_2(1)\star \widetilde{\zeta }_1(1))=\widetilde{\zeta }_1(1) \in I_{\mu }^0\), in view of Lemma 4.2 and Step 2, we have

$$\begin{aligned} \widetilde{P}_\mu (0)=\widetilde{P}_\mu (\widetilde{\zeta }_1(0))>0, \end{aligned}$$

and

$$\begin{aligned} \widetilde{P}_\mu (1)=\widetilde{P}_\mu (\widetilde{\zeta }_1(1))<0. \end{aligned}$$

Since \(\widetilde{P}_\mu (t)\) is continuous and by Remark 2.8, we infer that there exists \(t^*\in (0,1)\) so as to \(\widetilde{P}_\mu (t^*)=0\), which implies that \((\widetilde{\zeta }_2(t^*)\star \widetilde{\zeta }_1(t^*))\in \mathcal {P}_r(a)\). Consequently, one has

$$\begin{aligned} \max _{t\in [0,1]}E(\widetilde{\zeta }(t))=\max _{t\in [0,1]}I_\mu (\widetilde{\zeta }_2(t)\star \widetilde{\zeta }_1(t))\ge \inf _{u \in \mathcal {P}_r(a)}I_\mu (u). \end{aligned}$$

Hence, we have \(\sigma _\mu (a)\ge m_{\mu ,r}(a)\). In conclusion, one has \(\sigma _\mu (a)= m_{\mu ,r}(a)\).

Step 4 We claim that \(m_{\mu ,r}(a)>0\). Let \(u\in \mathcal {P}_r(a)\), we have \(P_\mu (u)=0\). By the GNS inequality (2.4) and (2.1), we infer to

$$\begin{aligned} \begin{aligned} \Vert \nabla u\Vert ^2_2&=-\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx+\mu \delta _{q}\Vert u\Vert ^q_q +\Vert u\Vert ^6_6\\&\le \mu \delta _{q}C(q)a^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2+S^{-3}\Vert \nabla u\Vert ^6_2. \end{aligned} \end{aligned}$$

And by \(q\delta _q=2\), one has

$$\begin{aligned} \left( 1-\mu \delta _{q}C(q)a^{q(1-\delta _q)}\right) \Vert \nabla u\Vert ^2_2\le S^{-3}\Vert \nabla u\Vert ^6_2. \end{aligned}$$

By \(a<\alpha _3\), there exists \(\rho >0\) such that

$$\begin{aligned} \inf _{u \in \mathcal {P}_r(a)}\Vert \nabla u\Vert ^2_2\ge \rho . \end{aligned}$$

So, for any \(u\in \mathcal {P}_r(a)\), it follows that

$$\begin{aligned} \begin{aligned} I_{\mu }(u)&=I_{\mu }(u)-\frac{1}{6}P_\mu (u)\\&=\frac{1}{3}\Vert \nabla u\Vert ^2_2+\frac{5\gamma }{24} \int _{{\mathbb {R}}^3}\phi _u u^2dx-\frac{2\mu }{3q}\int _{{\mathbb {R}}^3}|u|^qdx\\&\ge \frac{1}{3}\Vert \nabla u\Vert ^2_2-\frac{2\mu }{3q}C(q)a^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2\\&=\frac{1}{3}\left( 1-\frac{2\mu }{q}C(q)a^{q(1-\delta _q)}\right) \Vert \nabla u\Vert ^2_2\\&\ge \frac{1}{3}\left( 1-\frac{2\mu }{q}C(q)a^{q(1-\delta _q)}\right) \rho >0. \end{aligned} \end{aligned}$$

Consequently, we obtain \(\sigma _\mu (a)>0\), which completes the proof. \(\square \)

Next, we give an upper bounded estimate for the mountain pass level \(\sigma _\mu (a)\) in the following Lemma, which plays an important role in the proof of Theorem 1.2.

Lemma 4.7

Let \(q=\frac{10}{3}\), \(\mu >0\), and assume that \(0<a<\min \{\alpha _3,\alpha _4\}\). Then there exists \(\widetilde{\gamma _1}>0\), such that \( \sigma _{\mu }(a)<\frac{1}{3}S^{\frac{3}{2}}\) for \(\gamma \in (0,\widetilde{\gamma _1})\) small enough.

Proof

Recall (2.1) and (2.2), we have the best constant S is attained by

$$\begin{aligned} U_{\varepsilon }(x)=\frac{C^*\varepsilon ^{\frac{1}{2}}}{(\varepsilon ^2+|x|^2)^{\frac{1}{2}}} \end{aligned}$$

for any \(\varepsilon >0\) and \(C^*\) being normalized constant such that

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla U_{\varepsilon }|^2dx=\int _{{\mathbb {R}}^3}|U_{\varepsilon }|^6dx=S^{\frac{3}{2}}. \end{aligned}$$

We take

$$\begin{aligned} u_{\varepsilon }=\varphi U_{\varepsilon }, \end{aligned}$$

where \(\varphi (x)\in C_0^{\infty }(B_2(0))\) is a radial cutoff function such that \(0\le \varphi (x)\le 1\) and \(\varphi (x)\equiv 1\) on \(B_1(0).\) Let

$$\begin{aligned} v_{\varepsilon }=a\frac{u_{\varepsilon }}{\Vert u_{\varepsilon }\Vert _2}\in S(a)\cap H_{rad}^1({\mathbb {R}}^3). \end{aligned}$$

As showed in [15], we have

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u_{\varepsilon }|^2dx= S^{\frac{3}{2}}+O(\varepsilon ), \end{aligned}$$
(4.7)

and

$$\begin{aligned} \int _{{\mathbb {R}}^3}|u_{\varepsilon }|^6dx=S^{\frac{3}{2}}+O(\varepsilon ^{3}). \end{aligned}$$
(4.8)

From Lemma 7.1 [30], we have the following estimations:

$$\begin{aligned} \int _{{\mathbb {R}}^3}|u_{\varepsilon }|^qdx ={\left\{ \begin{array}{ll} O\left( \varepsilon ^{\frac{q}{2}}\right) ,&{}\text{ if }~~2<q<3;\\ O\left( \varepsilon ^{\frac{3}{2}}|\log \varepsilon |\right) ,&{}\text{ if }~~q=3;\\ O\left( \varepsilon ^{\frac{6-q}{2}}\right) ,&{}\text{ if }~~3<q<6. \end{array}\right. } \end{aligned}$$
(4.9)

and when \(q=2\), one has

$$\begin{aligned} \int _{{\mathbb {R}}^3}|u_{\varepsilon }|^2dx = C\varepsilon . \end{aligned}$$
(4.10)

Recall the function

$$\begin{aligned}{} & {} \varphi _{v_{\varepsilon }}(\iota ):=I_{\mu } ((\iota \star v_{\varepsilon })) =\frac{1}{2}e^{2\iota }\int _{{\mathbb {R}}^3}|\nabla v_{\varepsilon }|^2dx+\frac{\gamma }{4}e^{\iota }\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\nonumber \\{} & {} \quad -\frac{\mu }{q}e^{\frac{3(q-2)}{2}\iota }\int _{{\mathbb {R}}^3}|v_{\varepsilon }|^qdx -\frac{1}{6}e^{6\iota }\int _{{\mathbb {R}}^3}|v_{\varepsilon }|^6dx, \end{aligned}$$
(4.11)

Similar to the first step in proving Lemma 4.6, we can conclude that \(\varphi _{v_{\varepsilon }}\) can obtain its global positive maximum at some \(\iota _{\varepsilon }\). And so, by (2.14), we have

$$\begin{aligned} (\varphi )'_{v_{\varepsilon }}(\iota _{\varepsilon })=P_\mu (\iota _{\varepsilon }\star v_{\varepsilon })=0. \end{aligned}$$
(4.12)

By (4.12) and \(q\delta _q=2\), we deduce

$$\begin{aligned} \begin{aligned} e^{4\iota _{\varepsilon }}&=\frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^6} +\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\frac{\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx}{\Vert v_{\varepsilon }\Vert _6^6} -\mu \delta _{q}e^{(q\delta _{q}-2)\iota _{\varepsilon }}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}}\\&\ge \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}}-\mu \delta _{q} \frac{\Vert v_{\varepsilon }\Vert ^q_q}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}\Vert \nabla u_{\varepsilon }\Vert ^2_2}{a^{4}\Vert u_{\varepsilon }\Vert _{6}^{6}} -\mu \delta _{q}\frac{\Vert u_{\varepsilon }\Vert ^{6-q}_2\Vert u_{\varepsilon }\Vert ^q_q}{a^{6-q}\Vert u_{\varepsilon }\Vert _{6}^{6}}\\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}}{a^{4}\Vert u_{\varepsilon }\Vert _{6}^{6}} \left( \Vert \nabla u_{\varepsilon }\Vert ^2_2-\mu \delta _{q}\frac{\Vert u_{\varepsilon }\Vert ^{2-q}_2\Vert u_{\varepsilon }\Vert ^q_q}{a^{2-q}}\right) \\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}}{a^{4}\Vert u_{\varepsilon }\Vert _{6}^{6}} \left( \Vert \nabla u_{\varepsilon }\Vert ^2_2-\frac{\mu \delta _{q}}{a^{2-q}}\frac{\Vert u_{\varepsilon }\Vert ^q_q}{\Vert u_{\varepsilon }\Vert ^{q-2}_2}\right) . \end{aligned} \end{aligned}$$
(4.13)

In view of (4.7) and (4.8), we have that there exist positive constants \(C_1, C_2\) and \(C_3\) depending on s and q, such that

$$\begin{aligned} C_1\le (\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{6-q\delta _{q}}{4}}\le \frac{1}{C_1}, \end{aligned}$$
(4.14)
$$\begin{aligned} C_2\le (\Vert u_{\varepsilon }\Vert _{6}^{6})^{\frac{q\delta _{q}-2}{4}}\le \frac{1}{C_2}, \end{aligned}$$
(4.15)

and

$$\begin{aligned} \frac{\Vert u_{\varepsilon }\Vert ^{q}_q}{\Vert u_{\varepsilon }\Vert _2^{q(1-\delta _{q})}}=C_3\varepsilon ^{\frac{6-q}{4}}. \end{aligned}$$
(4.16)

Hence, by (4.13)–(4.16) and \(q\delta _q=2\), we obtain

$$\begin{aligned} e^{4\iota _{\varepsilon }} \ge C\frac{\Vert u_{\varepsilon }\Vert _2^{4}}{a^4}\left( C_1-\frac{\mu \delta _q}{a^{2-q}}C_3\varepsilon ^{\frac{2}{3}}\right) \ge C\frac{\Vert u_{\varepsilon }\Vert _2^{4}}{a^4}. \end{aligned}$$
(4.17)

In the following, we shall make an upper estimation of \(\max _{\iota \in {\mathbb {R}}}\varphi _{v_\varepsilon }(\iota )\). Firstly, we define the function \(\varphi ^0_{v_{\varepsilon }}(\iota )\) as follow:

$$\begin{aligned} \varphi ^0_{v_{\varepsilon }}(\iota ):= \frac{e^{2\iota }}{2}\int _{{\mathbb {R}}^3}|\nabla v_{\varepsilon }|^2dx-\frac{e^{6\iota }}{6} \int _{{\mathbb {R}}^3}|v_{\varepsilon }|^{6}dx. \end{aligned}$$
(4.18)

By simple calculation, we derive that the function \(\varphi ^0_{v_{\varepsilon }}(\iota )\) has a unique critical point \(\iota _\varepsilon ^0\), which is a strict maximum point given by

$$\begin{aligned} e^{\iota _\varepsilon ^0}=\left( \frac{\Vert \nabla v_\varepsilon \Vert ^2_2}{\Vert v_\varepsilon \Vert ^6_6}\right) ^{\frac{1}{4}}. \end{aligned}$$
(4.19)

Applying the fact that

$$\begin{aligned} \sup _{\theta \ge 0}\left( \frac{\theta ^{2}}{2}a-\frac{\theta ^{6}}{6} b\right) =\frac{1}{3}\left( \frac{a}{b^{\frac{1}{3}}}\right) ^{\frac{3}{2}}, \end{aligned}$$

for any fixed \(a,b>0\). In view of (4.7) and (4.8), we infer that

$$\begin{aligned} \begin{aligned} \varphi ^0_{{v}_{\varepsilon }}(\iota _{\varepsilon }^0)&=\frac{1}{3}\left( \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{(\Vert v_\varepsilon \Vert ^6_6)^{\frac{1}{3}}}\right) ^{\frac{3}{2}} =\frac{1}{3}\left( \frac{\Vert \nabla u_{\varepsilon }\Vert ^2_2}{(\Vert u_\varepsilon \Vert ^6_6)^{\frac{1}{3}}}\right) ^{\frac{3}{2}}\\&=\frac{1}{3}\left( \frac{S^{\frac{3}{2}}+O(\varepsilon )}{(S^{\frac{3}{2}}+O(\varepsilon ^2))^{\frac{1}{3}}}\right) ^{\frac{3}{2}}=\frac{1}{3}S^{\frac{3}{2}}+O(\varepsilon ). \end{aligned} \end{aligned}$$
(4.20)

Secondly, we make an estimation for \(\varphi _{v_\varepsilon }(\iota )\). By (2.3), (4.12) and Hölder inequality, we derive to

$$\begin{aligned} \begin{aligned} e^{4\iota _{\varepsilon }}&=\frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^6} +\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\frac{\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx}{\Vert v_{\varepsilon }\Vert _6^6} -\mu \delta _qe^{(q\delta _q-2)\iota _{\varepsilon }}\frac{\Vert v_{\varepsilon }\Vert ^q_q}{\Vert v_{\varepsilon }\Vert _{6}^6}\\&\le \frac{1}{\Vert v_{\varepsilon }\Vert _{6}^6}\left( \Vert \nabla v_{\varepsilon }\Vert ^2_2+\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\right) \\&\le \frac{1}{\Vert v_{\varepsilon }\Vert _{6}^6}\left( \Vert \nabla v_{\varepsilon }\Vert ^2_2+\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\widetilde{C}\Vert v_{\varepsilon }\Vert ^4_{\frac{12}{5}}\right) \\&\le \frac{1}{\Vert v_{\varepsilon }\Vert _{6}^6}\left( \Vert \nabla v_{\varepsilon }\Vert ^2_2+\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\widetilde{C}\Vert v_{\varepsilon }\Vert ^3_2\Vert v_{\varepsilon }\Vert _6\right) \\&=\frac{1}{a^4\Vert u_{\varepsilon }\Vert _{6}^6} \left( \Vert \nabla u_{\varepsilon }\Vert ^2_2\Vert u_{\varepsilon }\Vert ^4_2 +\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\widetilde{C}a^2\Vert u_{\varepsilon }\Vert ^5_2\Vert u_{\varepsilon }\Vert _6\right) . \end{aligned} \end{aligned}$$
(4.21)

By virtue of (4.7)–(4.8), (4.10) and (4.21), we can see that \(\iota _{\varepsilon }\) can not go to \(+\infty \), namely, there exists some \(\iota ^*\in {\mathbb {R}}\) such that

$$\begin{aligned} \iota _{\varepsilon }\le \iota ^*,~~~\text{ for } \text{ all }~~\varepsilon >0. \end{aligned}$$
(4.22)

Based on above analysis, by (2.3)–(2.4), (4.7)–(4.8), (4.17), (4.20), (4.22) and the fact \(q\delta _q=2\), we derive to

$$\begin{aligned} \begin{aligned} \sup _{\iota \in {\mathbb {R}}}\varphi _{v_{\varepsilon }}(\iota )&=\varphi _{v_{\varepsilon }}(\iota _{\varepsilon }) =\varphi ^{0}_{v_{\varepsilon }}(\iota _{\varepsilon }) +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx -\frac{\mu }{q}e^{q\delta _{q}\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^q_q\\&\le \sup _{\iota \in {\mathbb {R}}}\Psi ^{0}_{v_{\varepsilon }}(\iota ) +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx -\frac{\mu }{q}e^{2\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^q_q\\&\le \Psi ^{0}_{v_{\varepsilon }}(\iota _{v_{\varepsilon }^0}) +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\widetilde{C}\Vert v_{\varepsilon }\Vert ^4_{\frac{12}{5}} -\frac{C\mu }{q}\frac{\Vert u_{\varepsilon }\Vert ^2_2}{a^2}\Vert v_{\varepsilon }\Vert ^q_q\\&\le \frac{1}{3}S^{\frac{3}{2}}+O(\varepsilon ) +C\gamma \frac{a^4}{\Vert u_{\varepsilon }\Vert ^4_2}\Vert u_{\varepsilon }\Vert ^4_{\frac{12}{5}} -\frac{C\mu }{q}\frac{a^{q-2}}{\Vert u_{\varepsilon }\Vert ^{q-2}_2}\Vert u_{\varepsilon }\Vert ^q_q\\&\le \frac{1}{3}S^{\frac{3}{2}}+C^1\varepsilon +C^2\gamma \frac{\varepsilon ^{\frac{6}{5}\times \frac{5}{3}}}{\varepsilon ^{2}}-C^3\varepsilon ^{\frac{6-q}{4}}\\&=\frac{1}{3}S^{\frac{3}{2}}+C^1\varepsilon +C^2\gamma -C^3\varepsilon ^{\frac{2}{3}}<\frac{1}{3}S^{\frac{3}{2}}, \end{aligned} \end{aligned}$$
(4.23)

if we choose \(\gamma =\varepsilon ^\alpha \) for some constant \(\alpha \ge 1\), and use the fact \(0<\frac{6-q}{4}<1\).

Finally, by Lemma 4.1, we take \(\iota _1<0\) and \(\iota _2> 0\) such that \(\iota _1\star {v}_{\varepsilon } \in A_k\) and \(I_{\mu }(\iota _2\star {v}_{\varepsilon }) < 0\), respectively. Define a path

$$\begin{aligned} \eta _{{v}_{\varepsilon }}: t\in [0, 1] \mapsto ((1 - t)\iota _1 + t\iota _2)\star {v}_{\varepsilon } \in \Gamma _a. \end{aligned}$$

Consequently, by (4.23), we have that there exists some \(\widetilde{\gamma _1}>0\), such that

$$\begin{aligned} m_{\mu ,r}(a)=\sigma _{\mu }(a) \le \max _{t\in [0,1]}I_{\mu }(\eta _{{v}_{\varepsilon } }(t))\le \sup _{\iota \in {\mathbb {R}}}\varphi _{{v}_{\varepsilon }}(\iota )<\frac{1}{3}S^{\frac{3}{2}}, \end{aligned}$$

for \( \gamma \in (0,\widetilde{\gamma _1})\) small enough. \(\square \)

Now, based on the above preparation, we are ready to accomplish the proof of Theorem 1.2.

Proof of Theorem  1.2

Take a PS sequence \(\{u_n\}\) as in Proposition 4.5, we have

$$\begin{aligned} I'_{\mu }|_{S_r(a)}(u_n)\rightarrow 0~~\text{ as }~~n\rightarrow \infty , \end{aligned}$$

Using the Lagrange multipliers rule, we have that there exists a sequence \(\{\lambda _n\}\in {\mathbb {R}}\) such that

$$\begin{aligned} I_{\mu }'(u_n)-\lambda _n\Psi '(u_n)\rightarrow 0~~\text{ in }~~H^{-1}. \end{aligned}$$
(4.24)

Again by Proposition 4.5, we get

$$\begin{aligned} I_{\mu }(u_n)\rightarrow \sigma _\mu (a)~~\text{ as }~~n\rightarrow \infty . \end{aligned}$$

which means \(I_{\mu }(u_n)\) is bounded. So we deduce to

$$\begin{aligned} \frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\frac{\gamma }{4} \int _{{\mathbb {R}}^3}\phi _{u_n} {u_n}^2dx-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|u_n|^qdx -\frac{1}{6}\int _{{\mathbb {R}}^3}|u_n|^{6}dx\le C.\nonumber \\ \end{aligned}$$
(4.25)

From \(P_\mu (u_n)\rightarrow 0\) and \(q\delta _q=2\), we infer that

$$\begin{aligned} |I_{\mu }(u_n)+P_\mu (u_n)|\le C, \end{aligned}$$

that is,

$$\begin{aligned} \frac{3}{2}\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\frac{\gamma }{2}\int _{{\mathbb {R}}^3}\phi _{u_n}u^2_n dx -\frac{3\mu }{q}\int _{{\mathbb {R}}^3}|u_n|^qdx-\frac{7}{6}\int _{{\mathbb {R}}^3}|u_n|^6dx\ge -C.\nonumber \\ \end{aligned}$$
(4.26)

Combining (4.25)–(4.26), one has

$$\begin{aligned} \frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{u_n}u^2_n dx+\frac{2}{3}\int _{{\mathbb {R}}^3}|u_n|^6dx\le 4C. \end{aligned}$$
(4.27)

Thus, we know that \(\int _{{\mathbb {R}}^3}\phi _{u_n}u^2_n dx\) and \(\int _{{\mathbb {R}}^3}|u_n|^6dx\) are bounded. Then, by (4.25) and the GNS inequality (2.4), we infer to

$$\begin{aligned} \begin{aligned} C&\ge \frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla {u_n}|^2dx-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|{u_n}|^qdx\\&\ge \frac{1}{2}\Vert \nabla u_n\Vert ^2_2-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\Vert \nabla u_n\Vert ^{q\delta _q}_2\\&=\left( \frac{1}{2}-\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\right) \Vert \nabla u_n\Vert ^2_2. \end{aligned} \end{aligned}$$

Since \(a<\alpha _3\), it is easy to get \(\Vert \nabla u_n\Vert _2\le R^*\) for some \(R^*>0\) independently on \(n \in {\mathbb {N}}\). Consequently, we obtain that \(\{u_n\}\) is bounded in \(H^1({\mathbb {R}}^3)\), and so, up to subsequence, there exists \(u_a\) such that

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle u_n\rightharpoonup u_a~~\text{ in }~~H^1({\mathbb {R}}^3),\\ u_n\rightarrow u_a~~\text{ in }~~L^p({\mathbb {R}}^3),~\forall p\in (2,6). \end{array}\right. } \end{aligned}$$

From (4.24), we have

$$\begin{aligned}{} & {} \lambda _n\Vert u_n\Vert _2^2=\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx\nonumber \\{} & {} -\mu \int _{{\mathbb {R}}^3}|u_n|^qdx-\int _{{\mathbb {R}}^3}|u_n|^6dx+o_n(1), \end{aligned}$$
(4.28)

that is

$$\begin{aligned} \lambda _n=\frac{1}{a^2}\left[ \int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx-\mu \int _{{\mathbb {R}}^3}|u_n|^qdx-\int _{{\mathbb {R}}^3}|u_n|^6dx\right] +o_n(1). \end{aligned}$$

From the boundedness of \(\{u_n\}\) in \(H^1({\mathbb {R}}^3)\), we have that \(\{\lambda _n\}\) is bounded. Now, we verify

$$\begin{aligned} \int _{{\mathbb {R}}^3}|u_n|^qdx\rightarrow \int _{{\mathbb {R}}^3}|u_a|^qdx\ne 0,~~\text{ i.e. }~~u_a\ne 0. \end{aligned}$$

Assume by contradiction that, \(\int _{{\mathbb {R}}^3}|u_n|^qdx\rightarrow 0\). By (2.3) and the interpolation inequality, we obtain \(\int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx\rightarrow 0\). Combining with

$$\begin{aligned} P_\mu (u_n)=\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx-\mu \delta _q\int _{{\mathbb {R}}^3}|u_n|^qdx-\int _{{\mathbb {R}}^3}|u_n|^6dx=o_n(1), \end{aligned}$$

we get

$$\begin{aligned} \lambda _n=\frac{1}{a^2}\left[ \frac{3\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx+\mu (\delta _q-1)\int _{{\mathbb {R}}^3}|u_n|^qdx\right] +o_n(1). \end{aligned}$$

So, \(\lambda _n\rightarrow 0\) as \(n\rightarrow \infty \). Then (4.28) becomes

$$\begin{aligned} \int _{{\mathbb {R}}^3}|\nabla u_n|^2dx-\int _{{\mathbb {R}}^3}|u_n|^6dx=o_n(1). \end{aligned}$$

Set

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx=\lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}|u_n|^6dx=l, \end{aligned}$$

then,

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }I_\mu (u_n)&=\lim _{n\rightarrow \infty }\left[ \frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx+\frac{\gamma }{4} \int _{{\mathbb {R}}^3}\phi _u u^2dx-\frac{\mu }{q}\int _{{\mathbb {R}}^3}|u|^qdx -\frac{1}{6}\int _{{\mathbb {R}}^3}|u|^{6}dx\right] \\&=\lim _{n\rightarrow \infty }\left[ \frac{1}{2}\int _{{\mathbb {R}}^3}|\nabla u|^2dx-\frac{1}{6}\int _{{\mathbb {R}}^3}|u|^{6}dx\right] \\&=\frac{1}{2}l-\frac{1}{6}l\\&=\frac{1}{3}l. \end{aligned} \end{aligned}$$
(4.29)

Since \(I_{\mu }(u_n)\rightarrow \sigma _\mu (a)~~\text{ as }~~n\rightarrow \infty \), and \(\sigma _\mu (a)<\frac{1}{3}S^{\frac{3}{2}}\) from Lemma 4.7 and (4.29), we obtain

$$\begin{aligned} l<S^{\frac{3}{2}}. \end{aligned}$$

On the other hand, by virtue of the Sobolev inequality (2.1), we have

$$\begin{aligned} S\le \frac{\lim _{n\rightarrow \infty }\Vert \nabla u_n\Vert _2^2}{(\lim _{n\rightarrow \infty }\int _{{\mathbb {R}}^3}|u_n|^6dx)^{\frac{1}{3}}}=\frac{l}{l^{\frac{1}{3}}}=l^{\frac{2}{3}}, \end{aligned}$$

which leads to a contradiction. Hence, \(\int _{{\mathbb {R}}^3}|u_n|^qdx\rightarrow \int _{{\mathbb {R}}^3}|u_a|^qdx\ne 0\). Then, by the boundedness of \(\{\lambda _n\}\), up to subsequence, there exists \(\lambda _a\) such that \(\lambda _n\rightarrow \lambda _a\). Consequently, by \(\Vert \nabla u_n\Vert _2\le R^*\), (2.3), the GNS inequality (2.4) and \(q\delta _q=2\), we have

$$\begin{aligned} \begin{aligned} T_3\le \int _{{\mathbb {R}}^3}|u_n|^qdx&\le {C}(q) \Vert \nabla u_n\Vert _2^2\Vert u_n\Vert _2^{q(1-\delta _{q})}\\&\le C(q){R^*}^2a^{q(1-\delta _{q})}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx\le \widetilde{C}\Vert u_n\Vert _{\frac{12}{5}}^4&\le \widetilde{C}[C(12/5)]^{\frac{5}{3}}\Vert \nabla u_n\Vert _2\Vert u_n\Vert _2^3\\&\le \widetilde{C}[C(12/5)]^{\frac{5}{3}}{R^*}a^3\\&:=T_4, \end{aligned} \end{aligned}$$

where \( T_3>0\) and \(T_4= T_4(R^*,a)\). We define the positive constant

$$\begin{aligned} \widetilde{\gamma _2}:=\frac{4\mu (1-\delta _q)T_3}{3T_4}. \end{aligned}$$

So, we get

$$\begin{aligned} \widetilde{\gamma _2}\le \lim _{n\rightarrow \infty }\frac{4\mu (1-\delta _q)\int _{{\mathbb {R}}^3}|u_n|^qdx}{3\int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx} =\frac{4\mu (1-\delta _q)\int _{{\mathbb {R}}^3}|u_a|^qdx}{3\int _{{\mathbb {R}}^3}\phi _{u_a}u_a^2dx}. \end{aligned}$$

In view of (4.24) and \(\{u_n\}\) is bounded in \(H({\mathbb {R}}^3)\), we have

$$\begin{aligned} \lambda _n\Vert u_n\Vert _2^2=\Vert \nabla u_n\Vert _2^2+\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx-\mu \Vert u_n\Vert _q^q-\Vert u_n\Vert _6^6+o_n(1). \end{aligned}$$

Combining with \(P_\mu (u_n)\rightarrow 0\) and Lemma 2.7, if \(\gamma \in (0,\widetilde{\gamma _2})\), one has

$$\begin{aligned} \lambda _aa^2= & {} \lim _{n\rightarrow \infty }\left( \Vert \nabla u_n\Vert _2^2+\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx-\mu \Vert u_n\Vert _q^q-\Vert u_n\Vert _6^6\right) \\= & {} \lim _{n\rightarrow \infty }\left( \frac{3}{4}\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}{u_n}^2dx+\mu (\delta _q-1)\Vert u_n\Vert _q^q\right) \\= & {} \frac{3}{4}\gamma \int _{{\mathbb {R}}^3}\phi _{u_a}{u_a}^2dx+\mu (\delta _q-1)\Vert u_a\Vert _q^q\\{} & {} <\frac{3}{4}\widetilde{\gamma _2}\int _{{\mathbb {R}}^3}\phi _{u_a}{u_a}^2dx+\mu (\delta _q-1)\Vert u_a\Vert _q^q\le 0, \end{aligned}$$

which proves that \(\lim _{n\rightarrow \infty }\lambda _n=\lambda _a<0\). Let \(v_n=u_n-u_a\rightharpoonup 0\), by (3.17), (2.5), Lemma 2.7 and \(P_\mu (v_n)=P_\mu (u_n)-P_\mu (u_a)\rightarrow 0\), we infer to

$$\begin{aligned} \Vert \nabla v_n\Vert _2^2=\Vert v_n\Vert _6^6+o_n(1). \end{aligned}$$

Up to subsequence, we assume that

$$\begin{aligned} \Vert \nabla v_n\Vert _2^2=\Vert v_n\Vert _6^6\rightarrow \tau . \end{aligned}$$

So, by (2.1), we have

$$\begin{aligned} \tau ^{\frac{1}{3}}{S}\le {\tau }, \end{aligned}$$

that is,

$$\begin{aligned} \tau \ge S^{\frac{3}{2}}~~~\text{ or }~~~\tau =0. \end{aligned}$$

If \(\tau \ge S^{\frac{3}{2}}\), in view of (3.17), we derive as

$$\begin{aligned} \begin{aligned} \sigma _\mu (a)&=\lim _{n\rightarrow \infty }I_\mu (u_n)\\&=\lim _{n\rightarrow \infty }\left( I_\mu (u_a)+\frac{1}{2}\Vert \nabla v_n\Vert _2^2-\frac{1}{6}\Vert v_n\Vert _6^6\right) \\&=I_\mu (u_a)+\frac{1}{3}\tau \\&\ge I_\mu (u_a)+\frac{1}{3}S^{\frac{3}{2}}. \end{aligned} \end{aligned}$$

Besides,

$$\begin{aligned} I_\mu (u_a)=I_\mu (u_a)+\frac{1}{2}P_\mu (u_a)=\frac{\gamma }{8}\int _{{\mathbb {R}}^3}\phi _{u_a}{u_a}^2dx+\frac{1}{3}\Vert u_a\Vert _6^6>0, \end{aligned}$$

which is contradicted to Lemma 4.7. Thus, we have \(\tau =0\). By a similar argument as in the end of the proof of Theorem 1.1, we infer to

$$\begin{aligned} u_n\rightarrow u_a~~\text{ in }~~H^1({\mathbb {R}}^3). \end{aligned}$$

Next, we claim that \(m_\mu (a)=m_{\mu ,r}(a)\). Since \(\mathcal {P}_r(a)\subset \mathcal {P}(a)\), it is easy to see that \(m_\mu (a)\le m_{\mu ,r}(a)\). Then, we only need to verify \(m_\mu (a)\ge m_{\mu ,r}(a)\). Suppose by contradiction, there exists \(w \in \mathcal {P}(a)\backslash S_r(a)\) such that

$$\begin{aligned} I_\mu (w)<\inf _{\mathcal {P}_r(a)}I_\mu (u). \end{aligned}$$
(4.30)

Then, let \(v:=|w|^*\), by virtue of the schwarz rearrangement, it follows that

$$\begin{aligned} I_\mu (v)\le I_\mu (w)~~~\text{ and }~~~P_\mu (v)\le P_\mu (w)=0. \end{aligned}$$

If \(P_\mu (v)=0\), we know \( v \in \mathcal {P}(a)\), \(v:=|w|^*\in \mathcal {P}_r(a)\) and

$$\begin{aligned} I_\mu (v)\ge \inf _{\mathcal {P}_r(a)}I_\mu (u)>I_\mu (w)\ge I_\mu (v), \end{aligned}$$

which is a contradiction. If \(P_\mu (v)<0\), we see that \((\varphi _v)'(0)=P_\mu (v)<0\), by the claim of Step 1 of the proof of Lemma 4.6, we have that \(t_v<0\). Since \(t_v\star v \in \mathcal {P}_r(a)\), by (4.29), we deduce to

$$\begin{aligned} \begin{aligned} I_\mu (w)&<I_\mu (t_v\star v)=I_\mu (t_v\star v)-\frac{1}{2}P_\mu (t_v\star v)\\&=\frac{\gamma }{8}e^{t_v}\int _{{\mathbb {R}}^3}\phi _{v}{v}^2dx+\frac{1}{3}e^{6t_v}\int _{{\mathbb {R}}^3}|v|^6dx\\&=\frac{\gamma }{8}e^{t_v}\int _{{\mathbb {R}}^3}\phi _{w}{w}^2dx+\frac{1}{3}e^{6t_v}\int _{{\mathbb {R}}^3}|w|^6dx\\&\le e^{t_v}\left( I_\mu (w)-\frac{1}{2}P_\mu (w)\right) \\&=e^{t_v}I_\mu (w)<I_\mu (w), \end{aligned} \end{aligned}$$

which leads to a contradiction. Again by the the claim of step 1 of the proof of Lemma 4.6, we have \(\mathcal {P}(a)=\mathcal {P}(a)^-\). Consequently, we get

$$\begin{aligned} I_\mu (u_a)=\sigma _\mu (a)=m_{\mu ,r}(a)=m_\mu (a)=\inf _{\mathcal {P}(a)}I_\mu (u)=\inf _{\mathcal {P}(a)^-}I_\mu (u), \end{aligned}$$

and \(u_a\) is a ground state. \(\square \)

5 \(L^2\)-Supcritical Perturbation Case

In this section, we consider the \(L^2\)-supercritical case: \(\frac{10}{3}<q<6\) and prove Theorem 1.3. For the sake of convenience, we still utilize the notations and definitions in Section 4.

In Lemma 4.1, the conclusion remains valid when \(\frac{10}{3}<q<6\). In the following, we show that \(E(u,\iota )\) has the mountain pass geometry on \(S_r(a)\times {\mathbb {R}}\).

Lemma 5.1

Let \(\frac{10}{3}<q<6\), \(\mu ,\gamma >0\), and assume that \(0<a<\alpha _5\), where

$$\begin{aligned} \alpha _5:=\left( \frac{{k^*}^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) ^{\frac{1}{3}}. \end{aligned}$$

There exist \(0<k^*_1<k^*_2<k^*\) such that

$$\begin{aligned} P_\mu (u),I_{\mu }(u)~~\text{ for } \text{ all }~~{u\in A_{k^*_1}}~~\text{ and }~~0< \sup _{u\in A_{k^*_1}} I_{\mu } (u) < \inf _{u\in B_{k^*_2}}I_{\mu }(u), \end{aligned}$$

where

$$\begin{aligned} A_{k^*}:= \{u\in S_r(a): \Vert \nabla u\Vert ^2_2\le k^*\}~~~~\text{ and }~~ B_{k^*}:= \{u\in S_r(a): \Vert \nabla u\Vert ^2_2=2k^*\} \end{aligned}$$

Proof

Take \(k^*>0\), which will be determined later. Suppose that \(u,v\in S_r(a)\) such that \(\Vert \nabla u\Vert ^2_2\le k^*\) and \(\Vert \nabla v\Vert ^2_2=2k^*\). The proof here is similar to Lemma 4.2, which we briefly outline.

$$\begin{aligned} \begin{aligned} P_\mu (u)&=\Vert \nabla u\Vert ^2_2+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\mu \delta _q\Vert u\Vert ^q_q-\Vert u\Vert ^6_6\\&\ge \Vert \nabla u\Vert ^2_2-C(q)\mu \delta _qa^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2-S^{-3}\Vert \nabla u\Vert ^6_2, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} I_\mu (u)&=\frac{1}{2}\Vert \nabla u\Vert ^2_2+\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\frac{\mu }{q}\Vert u\Vert ^q_q-\frac{1}{6}\Vert u\Vert ^6_6\\&\ge \frac{1}{2}\Vert \nabla u\Vert ^2_2-C(q)\frac{\mu }{q}a^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2-\frac{1}{6}S^{-3}\Vert \nabla u\Vert ^6_2. \end{aligned} \end{aligned}$$

Moreover,

$$\begin{aligned} I_\mu (v)-I_\mu (u)\ge & {} \frac{1}{2}\Vert \nabla v\Vert ^2_2-\frac{1}{2}\Vert \nabla u\Vert ^2_2-\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _uu^2dx-\frac{\mu }{q}\Vert v\Vert ^q_q-\frac{1}{6}\Vert v\Vert ^6_6\\\ge & {} \frac{1}{2}\Vert \nabla v\Vert ^2_2-\frac{1}{2}\Vert \nabla u\Vert ^2_2-\frac{\gamma }{4}{\tilde{C}}[C(12/5)]^{\frac{5}{3}}a^3\Vert \nabla u\Vert _2\\{} & {} -\frac{\mu }{q}C(q)a^{q(1-\delta _q)}\Vert \nabla u\Vert ^{q\delta _q}_2-\frac{1}{6}S^{-3}\Vert \nabla u\Vert ^6_2\\\ge & {} k^*-\frac{1}{2}k^*-\frac{\gamma }{4}{\tilde{C}}[C(12/5)]^{\frac{5}{3}}\left( \frac{{k^*}^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) {k^*}^{\frac{1}{2}}\\{} & {} -\frac{C(q)\mu }{q} \left( \frac{{k^*}^{\frac{1}{2}}}{4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}}\right) ^{\frac{q(1-\delta _q)}{3}}(2{k^*})^{\frac{q\delta _q}{2}} -\frac{1}{6}S^{-3}(2{k^*})^3\\= & {} \frac{7}{16}{k^*}-\left( \frac{2^{\frac{q\delta _q}{2}}C(q)\mu }{q\left( 4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}}\right) ^{\frac{q(1-\delta _q)}{3}}} {k^*}^{\frac{2q-6}{3}}\right) {k^*}-\left( \frac{4}{3}S^{-3}{k^*}^2\right) {k^*}\\\ge & {} \frac{5}{16}{k^*}>0, \end{aligned}$$

for \(a<\alpha _5\), and we take

$$\begin{aligned} {k^*}=\min \left\{ \left( \frac{q(4\gamma {\tilde{C}}[C(12/5)]^{\frac{5}{3}})^{\frac{q(1-\delta _q)}{3}}}{16\mu C(q)2^{\frac{q\delta _q}{2}}}\right) ^{\frac{3}{2q-6}}, ~~\left( \frac{3 }{64}S^3\right) ^{\frac{1}{2}}\right\} , \end{aligned}$$
(5.1)

then, for \(0<k^*_1<k^*_2<k^*\) small enough and \(0<a<\alpha _5\), we have

$$\begin{aligned} P_\mu (u),~I_{\mu }(u)~~\text{ for } \text{ all }~~u\in A_{k^*_1}~~\text{ and }~~0< \sup _{u\in A_{k^*_1}} I_{\mu } (u) < \inf _{u\in B_{k^*_2}}I_{\mu }(u). \end{aligned}$$

The proof is completed. \(\square \)

Lemma 5.2

Let \(\frac{10}{3}<q<6\), \(\mu ,\gamma >0\), and assume that \(0<a<\alpha _5\). Then we have

  1. (i)

    There exists a sequence \(\{w_n\}\in S_r(a)\) such that

    $$\begin{aligned} I_{\mu }(w_n)\rightarrow \sigma _{\mu }(a)~~~ \text{ as }~~~n\rightarrow \infty , \end{aligned}$$
    (5.2)
    $$\begin{aligned} P_{\mu }(w_n)\rightarrow 0~~~ \text{ as }~~~n\rightarrow \infty , \end{aligned}$$
    (5.3)
    $$\begin{aligned} I'_{\mu }|_{S_r(a)}(w_n)\rightarrow 0~~~ \text{ as }~~~n\rightarrow \infty . \end{aligned}$$
    (5.4)
  2. (ii)

    \(\sigma _\mu (a)=m_{\mu ,r}(a)>0\), where \(\sigma _\mu (a)\) and \(m_{\mu ,r}(a)\) is defined in Section 4.

The proof of this lemma is similar to that of Propositions 4.34.5 and Lemmas 4.24.6 utilizing \(q\delta _q>2\), and thus it is omitted here.

Now, we make an upper bounded estimation for the mountain pass level \(\sigma _\mu (a)\) in the following.

Lemma 5.3

Let \(\frac{10}{3}<q<6\), \(\mu ,\gamma >0\), and assume that \(0<a<\alpha _5\). Then we have \( \sigma _{\mu }(a)<\frac{1}{3}S^{\frac{3}{2}}\) for \(\gamma \in (0,\widetilde{\gamma _1})\) small enough, where \(\widetilde{\gamma _1}\) is defined in Lemma 4.7.

Proof

As in the proof of Lemma 4.7, we conclude that \(\varphi _{v_\varepsilon }(\iota )\) achieves its global positive maximum at some \(\iota _\varepsilon \), and the critical point \(\iota _\varepsilon \) is unique. In view of \((\varphi _{v_{\varepsilon }})'(\iota _{\varepsilon })=P_{\mu }(\iota _{\varepsilon }\star v_{\varepsilon })=0\), one has

$$\begin{aligned} \begin{aligned} e^{6\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^6_6&=e^{2\iota _{\varepsilon }}\Vert \nabla v_{\varepsilon }\Vert ^2_2 +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx -\mu \delta _qe^{q\delta _q}\Vert v_{\varepsilon }\Vert ^q_q\\&\le e^{2\iota _{\varepsilon }}\Vert \nabla v_{\varepsilon }\Vert ^2_2 +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\\&=e^{2\iota _{\varepsilon }}\left( \Vert \nabla v_{\varepsilon }\Vert ^2_2 +\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\right) \\&\le e^{2\iota _{\varepsilon }} 2\max \left\{ \Vert \nabla v_{\varepsilon }\Vert ^2_2, \frac{\gamma }{4}e^{-\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\right\} . \end{aligned} \end{aligned}$$
(5.5)

Then, we consider the following possible cases.

Case 1 If \(\Vert \nabla v_{\varepsilon }\Vert ^2_2> \frac{\gamma }{4}e^{-\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\), we have

$$\begin{aligned} e^{6\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^6_6\le e^{2\iota _{\varepsilon }}2\Vert \nabla v_{\varepsilon }\Vert ^2_2, \end{aligned}$$

that is,

$$\begin{aligned} e^{4\iota _{\varepsilon }}\le \frac{2\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert ^6_6}. \end{aligned}$$
(5.6)

By \((\varphi _{v_{\varepsilon }})'(\iota _{\varepsilon })=0\), we infer that

$$\begin{aligned} \begin{aligned} e^{4\iota _{\varepsilon }}&=\frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^6} +\frac{\gamma }{4}\frac{e^{-\iota _{\varepsilon }} \int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx}{\Vert v_{\varepsilon }\Vert _6^6} -\mu \delta _{q}e^{(q\delta _{q}-2)\iota _{\varepsilon }}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}}\\&\ge \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}}-\mu \delta _{q} \left( \frac{2\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}}\right) ^{\frac{q\delta _{q}-2}{4}}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}\Vert \nabla u_{\varepsilon }\Vert ^2_2}{a^{4}\Vert u_{\varepsilon }\Vert _{6}^{6}}-\mu \delta _{q} \left( \frac{2\Vert u_{\varepsilon }\Vert ^{4}_2\Vert \nabla u_{\varepsilon }\Vert ^2_2}{a^{4}\Vert u_{\varepsilon }\Vert ^{6}_{6}}\right) ^{\frac{q\delta _{q}-2}{4}} \frac{\Vert u_{\varepsilon }\Vert _{2}^{6-q}\Vert u_{\varepsilon }\Vert _{q}^{q}}{a^{6-q}\Vert u_{\varepsilon }\Vert _{6}^{6}}\\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}}{a^{4} \Vert u_{\varepsilon }\Vert _{6}^{6}}\left[ (\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{6-q\delta _{q}}{4}}- \frac{\mu \delta _{q}2^{\frac{q\delta _{q}-2}{4}}a^{q(1-\delta _{q})}\Vert u_{\varepsilon }\Vert ^q_q}{\Vert u_{\varepsilon }\Vert _2^{q(1-\delta _{q})} (\Vert u_{\varepsilon }\Vert _{6}^{6})^{\frac{q\delta _{q}-2}{4}}}\right] ,\\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}}{a^{4} \Vert u_{\varepsilon }\Vert _{6}^{6}}\left[ (\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{6-q\delta _{q}}{4}}- \frac{\mu \delta _{q}2^{\frac{q\delta _{q}-2}{4}}a^{q(1-\delta _{q})}}{(\Vert u_{\varepsilon }\Vert _{6}^{6})^{\frac{q\delta _{q}-2}{4}}} \frac{\Vert u_{\varepsilon }\Vert ^q_q}{\Vert u_{\varepsilon }\Vert _2^{q(1-\delta _{q})}}\right] . \end{aligned} \end{aligned}$$
(5.7)

By virtue of (4.14)–(4.16) and (5.7), we get

$$\begin{aligned} e^{4\iota _{\varepsilon }}\ge C\frac{\Vert u_{\varepsilon }\Vert ^{4}_{2}}{a^{4}}\left[ C_1 -\mu \delta _{q}a^{q(1-\delta _{q})}2^{\frac{q\delta _{q}-2}{4}}\frac{C_3}{C_2}\varepsilon ^{ \frac{6-q}{4}}\right] \ge C\frac{\Vert u_{\varepsilon }\Vert ^{4}_{2}}{a^{4}}. \end{aligned}$$
(5.8)

Case 2 If \(\Vert \nabla v_{\varepsilon }\Vert ^2_2\le \frac{\gamma }{4}e^{-\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx\), we have

$$\begin{aligned} e^{6\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^6_6\le 2 e^{2\iota _{\varepsilon }}\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx, \end{aligned}$$

that is,

$$\begin{aligned} e^{5\iota _{\varepsilon }}\le \frac{\gamma }{2}\frac{\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx}{\Vert v_{\varepsilon }\Vert ^6_6}. \end{aligned}$$
(5.9)

Again by \((\varphi _{v_{\varepsilon }})'(\iota _{\varepsilon })=0\), (2.3) and Hölder inequality, we infer to

$$\begin{aligned} e^{4\iota _{\varepsilon }}&=\frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}} +\frac{\gamma }{4}e^{-\iota _{\varepsilon }}\frac{\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx}{\Vert v_{\varepsilon }\Vert _{6}^{6}} -\mu \delta _{q}e^{(q\delta _{q}-2)\iota _{\varepsilon }}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}}\nonumber \\&\ge \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}}-\mu \delta _{q}\left( \frac{\gamma }{2}\frac{\int _{{\mathbb {R}}^3} \phi _{v_{\varepsilon }}v_{\varepsilon }^2dx}{\Vert v_{\varepsilon }\Vert _{6}^{6}}\right) ^{\frac{q\delta _{q}-2}{5}}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \nonumber \\&\ge \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}} -\mu \delta _{q}\left( \frac{\gamma }{2}\frac{ \widetilde{C}\Vert v_{\varepsilon }\Vert _{\frac{12}{5}}^4}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \right) ^{\frac{q\delta _{q}-2}{5}}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \nonumber \\&\ge \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}} -\mu \delta _{q}\left( \frac{\gamma }{2}\frac{\widetilde{C} \Vert v_{\varepsilon }\Vert ^{3}_2\Vert v_{\varepsilon }\Vert _{6}}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \right) ^{\frac{q\delta _{q}-2}{5}}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}}\nonumber \\&\ge \frac{\Vert \nabla v_{\varepsilon }\Vert ^2_2}{\Vert v_{\varepsilon }\Vert _{6}^{6}} -\mu \delta _{q}\left( \frac{\gamma \widetilde{C}}{2}\right) ^{\frac{q\delta _{q}-2}{5}}a^{\frac{3(q\delta _{q}-2)}{5}} \left( \frac{1}{\Vert v_{\varepsilon }\Vert _6^5}\right) ^{\frac{q\delta _{q}-2}{5}}\frac{\Vert v_{\varepsilon }\Vert _{q}^{q}}{\Vert v_{\varepsilon }\Vert _{6}^{6}} \nonumber \\&\ge \frac{\Vert u_{\varepsilon }\Vert _{2}^{4}\Vert \nabla u_{\varepsilon }\Vert ^2_2}{a^4\Vert u_{\varepsilon }\Vert _{6}^{6}} -\mu \delta _{q}\left( \frac{\gamma \widetilde{C}}{2}\right) ^{\frac{q\delta _{q}-2}{5}}a^{\frac{5(q-6)-2(q\delta _q-2)}{5}} \frac{\Vert u_{\varepsilon }\Vert _{q}^{q}\Vert u_{\varepsilon }\Vert _{2}^{4-q(1-\delta _q)}}{\Vert u_{\varepsilon }\Vert _{6}^{q\delta _q+4}} \nonumber \\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}}{a^{4}\Vert u_{\varepsilon }\Vert _{6}^{6}}\nonumber \\&\bigg [(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{6-q\delta _{q}}{4}} -\mu \delta _{q}\left( \frac{\gamma \widetilde{C}}{2}\right) ^{\frac{q\delta _{q}-2}{5}}a^{\frac{5q-2q\delta _q-26}{5}} \frac{\Vert u_{\varepsilon }\Vert _{q}^{q}\Vert u_{\varepsilon }\Vert _{2}^{q(\delta _q-1)}}{\Vert u_{\varepsilon }\Vert _{6}^{q\delta _q-2}(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}}\bigg ]\nonumber \\&=\frac{\Vert u_{\varepsilon }\Vert _2^{4}(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}}{a^{4}\Vert u_{\varepsilon }\Vert _{6}^{6}}\nonumber \\&\bigg [(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{6-q\delta _{q}}{4}} -\mu \delta _{q}\left( \frac{\gamma \widetilde{C}}{2}\right) ^{\frac{q\delta _{q}-2}{5}}\frac{a^{\frac{5q-2q\delta _q-26}{5}}}{\Vert u_{\varepsilon }\Vert _{6}^{q\delta _q-2}(\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}} \frac{\Vert u_{\varepsilon }\Vert _{q}^{q}}{\Vert u_{\varepsilon }\Vert _{2}^{q(1-\delta _q)}}\bigg ], \end{aligned}$$
(5.10)

By (4.7)–(4.8), we deduce that there exists constant \(C_4>0\) such that

$$\begin{aligned} C_4\le (\Vert \nabla u_{\varepsilon }\Vert ^2_2)^{\frac{q\delta _{q}-2}{4}}\Vert u_{\varepsilon }\Vert _{6}^{q\delta _{q}-2}\le \frac{1}{C_4}. \end{aligned}$$
(5.11)

So, in view of (4.14)–(4.16) and (5.11), we get

$$\begin{aligned} \begin{aligned} e^{4\iota _{\varepsilon }}&\ge C\frac{\Vert u_{\varepsilon }\Vert ^{4}_{2}}{a^{4}}\left[ C_1 -\mu \delta _{q}\left( \frac{\gamma \widetilde{C}}{2}\right) ^{\frac{q\delta _{q}-2}{5}}a^{\frac{5q-2q\delta _q-26}{5}} \frac{C_3}{C_4}\varepsilon ^{\frac{6-q}{4}}\right] \\&\ge \frac{C\Vert u_{\varepsilon }\Vert ^{4}_{2}}{a^{4}}. \end{aligned} \end{aligned}$$
(5.12)

Based on above analysis, we will make an upper estimation for \(\varphi _{v_\varepsilon }(\iota )\). Firstly, as in Lemma 4.7, we can define the function \(\varphi _{v_\varepsilon }^0(\iota )\) and make an estimation for \(\varphi _{v_\varepsilon }^0(\iota )\), that is

$$\begin{aligned} \varphi _{v_\varepsilon }^0(\iota ^0_{\varepsilon })=\frac{1}{3}S^{\frac{3}{2}}+O(\varepsilon ). \end{aligned}$$
(5.13)

where \(\iota ^0_{\varepsilon }\) is a unique strict maximum point of \(\varphi _{v_\varepsilon }^0(\iota )\). Secondly, we make an estimation for \(\varphi _{v_\varepsilon }(\iota )\). By virtue of (4.22), we know that there exists some \(\iota ^*\in {\mathbb {R}}\) such that

$$\begin{aligned} \iota _{\varepsilon }\le \iota ^*,~~~\text{ for } \text{ all }~~\varepsilon >0. \end{aligned}$$

So, by (2.3)–(2.4), (5.11)–(5.13), (4.9)–(4.10), (4.16) and above inequality, we obtain

$$\begin{aligned} \sup _{\iota \in {\mathbb {R}}}\varphi _{v_{\varepsilon }}(\iota )= & {} \varphi _{v_{\varepsilon }}(\iota _{\varepsilon })\nonumber \\= & {} \varphi ^{0}_{v_{\varepsilon }}(\iota _{\varepsilon }) +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx- \frac{\mu }{q}e^{q\delta _{q}\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^q_q\nonumber \\\le & {} \sup _{\iota \in {\mathbb {R}}}\Psi ^{0}_{v_{\varepsilon }}(\theta ) +\frac{\gamma }{4}e^{\iota _{\varepsilon }}\int _{{\mathbb {R}}^3}\phi _{v_{\varepsilon }}v_{\varepsilon }^2dx- \frac{\mu }{q}e^{q\delta _{q}\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^q_q\nonumber \\\le & {} \Psi ^{0}_{v_{\varepsilon }}(\iota _{\varepsilon ,0})+\frac{\gamma }{4}\widetilde{C}\Vert v_{\varepsilon }\Vert _{\frac{12}{5}}^4 -\frac{\mu }{q}e^{q\delta _{q}\iota _{\varepsilon }}\Vert v_{\varepsilon }\Vert ^q_q\nonumber \\\le & {} \frac{1}{3}S^{\frac{3}{2}}+O(\varepsilon )+C\gamma \frac{ a^4}{\Vert u_{\varepsilon }\Vert _2^4}\Vert u_{\varepsilon }\Vert _{\frac{12}{5}}^4-\frac{C\mu a^{q(1-\delta _{q})}}{q}\frac{\Vert u_{\varepsilon }\Vert ^q_q}{\Vert u_{\varepsilon }\Vert _2^{q(1-\delta _{q})}}\nonumber \\\le & {} \frac{1}{3}S^{\frac{3}{2}}+C^1\varepsilon +C^2\gamma \frac{\Vert u_{\varepsilon }\Vert _{\frac{12}{5}}^4}{\Vert u_{\varepsilon }\Vert _2^4} -C^3\frac{\Vert u_{\varepsilon }\Vert ^q_q}{\Vert u_{\varepsilon }\Vert _2^{q(1-\delta _{q})}}\nonumber \\= & {} \frac{1}{3}S^{\frac{3}{2}}+C^1\varepsilon +C^2\gamma \frac{\varepsilon ^{\frac{6}{5}\times \frac{5}{3}}}{\varepsilon ^{2}} -C^3\varepsilon ^{ \frac{6-q}{4}} \nonumber \\= & {} \frac{1}{3}S^{\frac{3}{2}}+C^1\varepsilon +C^2\gamma -C^3\varepsilon ^{ \frac{6-q}{4}} \nonumber \\{} & {} \quad <\frac{1}{3}S^{\frac{3}{2}}, \end{aligned}$$
(5.14)

if we choose \(\gamma =\varepsilon ^\alpha \) for some constant \(\alpha \ge 1\), and using the fact \(0<\frac{6-q}{4}<1\).

Since \( {v}_{\varepsilon } \in S_r(a) \), from Lemma 5.2 we take \(\iota _3<0\) and \(\iota _4> 0\) such that \(\iota _3\star {v}_{\varepsilon } \in A_k\) and \(I_{\mu }(\iota _4\star {v}_{\varepsilon }) < 0\), respectively. We define a path

$$\begin{aligned} \eta ^*_{{v}_{\varepsilon } }: t\in [0, 1] \mapsto ((1 - t)\iota _3 + t\iota _4)\star {v}_{\varepsilon } \in \Gamma _a. \end{aligned}$$

Consequently, by (5.14), we obtain that there exists some \(\widetilde{\gamma }_1>0\), such that

$$\begin{aligned} \sigma _{\mu }(a) \le \max _{t\in [0,1]}I_{\mu }(\eta ^*_{{v}_{\varepsilon } }(t))\le \sup _{\iota \in {\mathbb {R}}}\varphi _{{v}_{\varepsilon }}(\iota )<\frac{1}{3}S^{\frac{3}{2}}, \end{aligned}$$

for \( \gamma \in (0,\widetilde{\gamma }_1)\) small enough, which completes the proof. \(\square \)

Now, we are ready to prove Theorem 1.3.

Proof of Theorem  1.3

By virtue of (5.4), we have that there exists a sequence \(\{\lambda _n\}\in {\mathbb {R}}\) such that

$$\begin{aligned} I_{\mu }'(u_n)-\lambda _n\Psi '(u_n)\rightarrow 0~~\text{ in }~~H^{-1}. \end{aligned}$$
(5.15)

Then we claim that \(\{u_n\}\) is bounded in H. Indeed, by (5.2)and (5.3), we have

$$\begin{aligned} |2I_{\mu }(u_n)+P_\mu (u_n)|\le C, \end{aligned}$$
(5.16)

that is,

$$\begin{aligned}{} & {} 2\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx+\frac{3}{4}\gamma \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx -\frac{\mu (3q-2)}{2q}\int _{{\mathbb {R}}^3}|u_n|^qdx\nonumber \\{} & {} \quad -\frac{4}{3}\int _{{\mathbb {R}}^3}|u_n|^6dx\ge -C. \end{aligned}$$
(5.17)

By (5.17) and the bounded of \(I_{\mu }(u_n)\), we infer to

$$\begin{aligned} -C\le -\frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx -\frac{\mu (3q-10)}{2q}\int _{{\mathbb {R}}^3}|u_n|^qdx-\frac{2}{3}\int _{{\mathbb {R}}^3}|u_n|^6dx+4C, \end{aligned}$$

it follows that

$$\begin{aligned} \frac{\gamma }{4}\int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx +\frac{\mu (3q-10)}{2q}\int _{{\mathbb {R}}^3}|u_n|^qdx+\frac{2}{3}\int _{{\mathbb {R}}^3}|u_n|^6dx\le 5C, \end{aligned}$$

which implies that

$$\begin{aligned} \int _{{\mathbb {R}}^3}\phi _{u_n}u_n^2dx,~~~\int _{{\mathbb {R}}^3}|u_n|^qdx~~\text{ and }~~\int _{{\mathbb {R}}^3}|u_n|^6dx \end{aligned}$$

are all bounded. Thus, we deduce that \(\int _{{\mathbb {R}}^3}|\nabla u_n|^2dx\) is also bounded. For convenience, we still take \(\Vert \nabla u_n\Vert _2\le R^*\). We can proceed exactly as in the proof of Theorem 1.2 utilizing Lemmas 5.25.3, so complete the proof of Theorem 1.3. \(\square \)