Ground State Solutions of Nehari-Pohozaev Type for Schrödinger–Poisson–Slater Equation with Zero Mass and Critical Growth

Abstract

In this article, we study the Schrödinger–Poisson–Slater type equation with the critical growth and zero mass:

$$\begin{aligned} {\left\{ \begin{array}{ll} {-\Delta } u+\phi u=\mu |u|^{p-2}u+u^5, \ \ \ \ x\in {\mathbb {R}}^{3},\\ {-\Delta } \phi =u^2, \ \ \ \ x\in {\mathbb {R}}^{3}, \end{array}\right. } \end{aligned}$$

where $3<p<6$ and $\mu >0$. By combining a new perturbation method and the mountain pass theorem, Liu et al. [J. Diff. Eq., 266 (2019), 5912–5941] prove that the above equation has at least one positive ground state solution for $p \in (4, 6)$ and $\mu >0$ or $p \in (3, 4]$ if $\mu$ is sufficiently large. By using a much simpler method than the ones used in the above mentioned paper, together with subtle estimates and analyses, we obtain better results on the existence for a ground state solution of Nehari-Pohozaev type.

1 Introduction

In this paper, we are concerned with the existence of ground state solutions for the Schrödinger–Poisson–Slater problem with critical growth and zero mass

$$\begin{aligned} {\left\{ \begin{array}{ll} {-\Delta } u+\phi u=\mu |u|^{p-2}u+u^5, \ \ \ \ x\in {\mathbb {R}}^{3},\\ {-\Delta } \phi =u^2, \ \ \ \ x\in {\mathbb {R}}^{3}, \end{array}\right. } \end{aligned}$$

(1.1)

where $\mu >0$ and $3<p<6$.

The interest on this system stems from the Schrödinger-Poisson-Slater problem

$$\begin{aligned} {\left\{ \begin{array}{ll} {-\Delta } u+\omega u+ \phi u=\mu |u|^{p-2}u, \ \ \ \ x\in {\mathbb {R}}^{3},\\ {-\Delta } \phi =u^2, \ \ \ \ x\in {\mathbb {R}}^{3}, \end{array}\right. } \end{aligned}$$

(1.2)

where $\omega >0$, which is the Slater approximation of the exchange term in the Hartree-Fock model, see [22]. The local term $|u|^{p-2}u$ was introduced by Slater, with $p=\frac{8}{3}$ and $\mu$ is the so-called Slater constant (up to renormalization), see [24]. Of course, other exponents have been employed in various approximations. In recent years, problem (1.2) has been the object of intensive research, a lot of attention has been focused on the study of the existence of solutions, sign-changing solutions, ground states, radial and semiclassical states, see [2,3,4,5,6,7,8,9, 11,12,13, 17, 24, 26, 28,29,30, 32, 34,35,37] and the references therein. From a mathematical point of view, this model presents an interesting competition between local and nonlocal nonlinearities.This interaction yields to some non expected situations, as has been shown in the literature.

For problem (1.2), the parameter $\omega$ corresponds to the phase of the standing wave for the time-dependent equation. In the case $\omega =0$, i.e. the Schrödinger-Poisson-Slater problem with zero mass

$$\begin{aligned} {\left\{ \begin{array}{ll} {-\Delta } u+\phi u=\mu |u|^{p-2}u, \ \ \ \ x\in {\mathbb {R}}^{3},\\ {-\Delta } \phi =u^2, \ \ \ \ x\in {\mathbb {R}}^{3}, \end{array}\right. } \end{aligned}$$

(1.3)

one could only search the static solutions (not periodic ones). The static case has been motivated and studied in [14, 27] when $p < 3$ and $p\ge 3$, respectively. The absence of a phase term $\omega u$ makes the usual Sobolev space $H^1(\mathbb {R}^3)$ not to be a good framework for the problem (1.3). In [27], the following working space and the norm are introduced:

$$\begin{aligned} E:=\left\{ u\in \mathcal {D}^{1,2}(\mathbb {R}^3): \int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u^2(x)u^2(y)}{|x-y|}\textrm{d}x\textrm{d}y < \infty \right\} \end{aligned}$$

(1.4)

and

$$\begin{aligned} \Vert u\Vert _E:=\left[ \int _{\mathbb {R}^3}|\nabla u|^2\textrm{d}x +\left( \int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u^2(x)u^2(y)}{{4\pi }|x-y|}\textrm{d}x\textrm{d}y\right) ^{\frac{1}{2}}\right] ^{\frac{1}{2}}. \end{aligned}$$

The double integral expression is the so-called Coulomb energy of the wave. In that paper, Ruiz proved that $(E,\Vert \cdot \Vert _E)$ is a uniformly convex Banach space, and $E\hookrightarrow L^s(\mathbb {R}^3)$ for all $s\in [3, 6]$. Moreover, the author gave also the equivalence characterizations of the convergences in the space E.

Based on the above information, Ianni and Ruiz [14] proved that (1.3) has a positive solution with minimal energy among all nontrivial solutions provided $3<p<6$. In the arguments, they used a technique that dates back to Struwe and is usually named “monotonicity trick" (see [15, 16]), well-known arguments of concentration-compactness of Lions ([33]) and “Pohozaev identity". Lei and Lei [20] used variational methods obtained existence of ground state solution of the Nehari–Pohozaev type. By the new variational approach, there is a series of analytical results on the Schrödinger-Poisson systems in the literature (see [16, 23] and the references therein).

Further, Liu, Zhang and Huang [24] studied the existence of ground state solutions for (1.1) by combining a new perturbation method and the mountain pass theorem, the authors obtained the existence of positive ground state solutions. To be specific, they proved that (1.1) has at least one positive ground state solution for $p \in (4, 6)$ and $\mu >0$ or $p \in (3, 4]$ if $\mu$ is sufficiently large. Via a truncation technique and Krasnoselskii genus theory, Yang and Liu [34] obtained infinitely many solutions for (1.1) provided $\mu \in (0,\mu ^*)$ with some $\mu ^*>0$. Zheng, Lei and Liao [35] discussed the existence of positive ground-state solutions and the multiplicity of positive solutions for a more general Schrödinger-Poisson-Slater-type equation with critical growth. Recently, Lei, Lei and Suo [21] obtained a ground state solution for (1.1) with the Coulomb-Sobolev critical growth by employing compactness arguments.

In this paper, inspired by [14, 24, 27, 30], we obtain ground state solutions of (1.1) under weaker assumptions on $\mu$ by using a much simpler method than the ones used in [24]. In particular, we introduce some new test functions, which, together with subtle estimates and analyses, to obtain a good energy estimate of the mountain pass level such that the compactness of (PS) sequences at the energy level still holds, see Lemmas 3.7 and 3.8.

Since $E\hookrightarrow L^s(\mathbb {R}^3)$ for all $s\in [3, 6]$, so, we have that the associated energy functional to (1.1)

$$\begin{aligned} \Phi (u)= & {} \frac{1}{2}\int _{\mathbb {R}^3}|\nabla u|^2\textrm{d}x+\frac{1}{4}\int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u^2(x)u^2(y)}{4\pi |x-y|}\textrm{d}x\textrm{d}y\nonumber \\{} & {} \quad -\int _{\mathbb {R}^3}\left( \frac{\mu }{p}|u|^p+\frac{1}{6}|u|^6\right) \textrm{d}x \end{aligned}$$

(1.5)

is well-defined and $\mathcal {C}^1$. Our main result is the following:

Theorem 1.1

Assume that one of the following conditions holds:

• (i) $p\in (4,6)$ and $\mu >0$;

• (ii) $p=4$ and $\mu >\frac{7\sqrt{3}}{\pi }$;

• (iii) $p\in (3,4)$ and $\mu >\frac{3[6\pi ^2(p-3)]^{2(p-3)/3}p^4}{16(2p-3)^{(2p-3)/3}\mathcal {S}^{(5p-12)/6}} {\left[ \frac{839803\mathcal {S}^\frac{3}{2}}{468750\root 3 \of {2}\pi ^3}\sqrt{\frac{2}{5\root 3 \of {2}\pi }}\right] }^{\frac{6-p}{9}}$.

Then Problem (1.1) has a solution $\bar{u}\in E$ such that $\Phi (\bar{u})=\inf _{\mathcal {M}}\Phi >0$, where

$$\begin{aligned} \mathcal {M}:= \{u\in E\setminus \{0\}: J(u)=0\} \end{aligned}$$

(1.6)

and

$$\begin{aligned} J(u)=\frac{3}{2}\Vert \nabla u\Vert _2^2+\frac{3}{4}\int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u^2(x)u^2(y)}{4\pi |x-y|}\textrm{d}x\textrm{d}y -\frac{(2p-3)\mu }{p}\Vert u\Vert _p^p-\frac{3}{2}\Vert u\Vert _6^6.\nonumber \\ \end{aligned}$$

(1.7)

The set $\mathcal {M}$ was introduced by Ruiz [26], is usually named “Nehari-Pohozaev” manifold.

Throughout this paper, we let $u_t(x):=u(tx)$ for $t>0$, and denote the norm of $L^s(\mathbb {R}^3)$ by $\Vert u\Vert _s =\left( \int _{\mathbb {R}^3}|u|^s \textrm{d}x\right) ^{1/s}$ for $s\ge 2$, $B_r(x)=\{y\in \mathbb {R}^3: |y-x|<r \}$, and positive constants possibly different in different places, by $C_1, C_2,\cdots$.

2 Variational Framework and Preliminaries

In this section we establish some notations that will be used throughout the paper. Let E be defined by (1.4) and study some basic properties of it.

Set

$$\begin{aligned} N(u):=\int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u^2(x)u^2(y)}{4\pi |x-y|}\textrm{d}x\textrm{d}y \end{aligned}$$

and

$$\begin{aligned} \Vert u\Vert _{E}=\left[ \Vert \nabla u\Vert _2^2+\sqrt{N(u)}\right] ^{1/2}. \end{aligned}$$

Lemma 2.1

[27] $\Vert \cdot \Vert _E$ is a norm, and $(E, \Vert \cdot \Vert _E)$ is a uniformly convex Banach space. Moreover, $\mathcal {C}_0^{\infty }(\mathbb {R}^3)$ is dense in E.

Lemma 2.2

[31] Assume that $a,b>0$. Then there holds

$$\begin{aligned} a\Vert \nabla u\Vert _2^2+bN(u)\ge 2\sqrt{ab}\Vert u\Vert _3^3, \ \ \ \ \forall \ u\in E. \end{aligned}$$

(2.1)

Let $E_0$ denote the Banach space equipped with the norm defined by

$$\begin{aligned} \Vert u\Vert _0=\left( \Vert \nabla u\Vert _2^2+\Vert u\Vert _3^2\right) ^{1/2}. \end{aligned}$$

Then Lemma 2.2 shows that $E \hookrightarrow E_0$.

Let us define

$$\begin{aligned} \phi _{u}(x):=\frac{1}{|x|}*u^2=\int _{\mathbb {R}^3}\frac{u^2(y)}{4\pi |x-y|}\textrm{d}y, \ \ \ \ \forall \ x\in \mathbb {R}^3, \end{aligned}$$

(2.2)

then, $u\in E$ if and only if both $u, \phi _u\in \mathcal {D}^{1,2}(\mathbb {R}^3)$. In such a case, $-\triangle \phi =u^2$ in a weak sense, and

$$\begin{aligned} \int _{\mathbb {R}^3}\nabla \phi _{u}\cdot \nabla v\textrm{d}x=\int _{\mathbb {R}^3}u^2v\textrm{d}x, \ \ \ \ \forall \ v\in E, \end{aligned}$$

(2.3)

$$\begin{aligned} \int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u^2(x)u^2(y)}{4\pi |x-y|}\textrm{d}x\textrm{d}y=\int _{\mathbb {R}^3}\phi _{u}(x)u^2\textrm{d}x. \end{aligned}$$

(2.4)

Moreover, $\phi _{u}(x)>0$ when $u\ne 0$. By using Hardy-Littlewood-Sobolev inequality (see [18] or [19, page 98]), we have the following inequality:

$$\begin{aligned} \int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{|u(x)v(y)|}{|x-y|}\textrm{d}x\textrm{d}y \le \frac{8\root 3 \of {2}}{3\root 3 \of {\pi }}\Vert u\Vert _{6/5}\Vert v\Vert _{6/5}, \ \ \ \ u, v\in L^{6/5}(\mathbb {R}^3). \end{aligned}$$

(2.5)

Lemma 2.3

[27] Suppose that $\{u_n\}\subset E$. Then

• (i) $u_n\rightarrow \bar{u}$ in E if and only if $u_n\rightarrow \bar{u}$ and $\phi _{u_n}\rightarrow \phi _{\bar{u}}$ in $\mathcal {D}^{1,2}(\mathbb {R}^3)$;

• (ii) $u_n\rightharpoonup \bar{u}$ in E if and only if $u_n\rightharpoonup \bar{u}$ in $\mathcal {D}^{1,2}(\mathbb {R}^3)$ and $\sup N(u_n)<+\infty$. In such case, $\phi _{u_n}\rightharpoonup \phi _{\bar{u}}$ in $\mathcal {D}^{1,2}(\mathbb {R}^3)$.

As in [14, 27], we define

$$\begin{aligned} T: E^4\rightarrow \mathbb {R}, \ T(u,v,w,z):=\int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u(x)v(x)w(y)z(y)}{4\pi |x-y|}\textrm{d}x\textrm{d}y \end{aligned}$$

(2.6)

and

$$\begin{aligned} D: E^2\rightarrow \mathbb {R}, \ D(u,v):=\int _{\mathbb {R}^3}\int _{\mathbb {R}^3}\frac{u(x)v(y)}{4\pi |x-y|}\textrm{d}x\textrm{d}y. \end{aligned}$$

(2.7)

Lemma 2.4

[14] Suppose that $\{u_n\}, \{v_n\}, \{w_n\}\subset E$, $z\in E$. If $u_n\rightharpoonup \bar{u}, v_n\rightharpoonup \bar{v}, w_n\rightharpoonup \bar{w}$ in E, then

$$\begin{aligned} T(u_n,v_n,w_n,z)\rightarrow T(\bar{u},\bar{v},\bar{w},z). \end{aligned}$$

In view of Lemmas 2.1-2.4, (F1) implies that $\Phi$ defined by (1.5) is a well-defined of classes $\mathcal {C}^{1}$ functional in E, and that

$$\begin{aligned} \langle \Phi '(u), v \rangle= & {} \int _{\mathbb {R}^3}\nabla u\cdot \nabla v\textrm{d}x+\int _{\mathbb {R}^3}\phi _{u}(x)uv\textrm{d}x \nonumber \\{} & {} \quad -\int _{\mathbb {R}^3}\left( |u|^{p-2}+u^4\right) uv\textrm{d}x, \ \ \ \ u,v\in E. \end{aligned}$$

(2.8)

Therefore, the solutions of (1.5) are then the critical points of the reduced functional (1.5).

In view of the Gagliardo-Nirenberg inequality [1, 25] and Sobolev inequality [33], one has

$$\begin{aligned} \Vert u\Vert _s^s\le K_{GN}^s\Vert u\Vert _3^{6-s}\Vert \nabla u\Vert _2^{2s-6}, \ \ \ \ \forall \ u\in D^{1,3}(\mathbb {R}^3), \ \ s\in (3,6) \end{aligned}$$

(2.9)

and

$$\begin{aligned} \mathcal {S}\Vert u\Vert _6^2\le \Vert \nabla u\Vert _2^2, \ \ \ \ \forall \ u\in D^{1,3}(\mathbb {R}^3). \end{aligned}$$

(2.10)

where $K_{GN}>0$ is a constant and $\mathcal {S}$ is the best embedding constant.

We also state here, for convenience of the reader, an adaptation to the space E of a result due to P.-L. Lions, see [22, Lemma I.1]:

Lemma 2.5

If $u_n\rightharpoonup \bar{u}$ in $E_0$, and

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{y\in \mathbb {R}^N}\int _{B_1(y)}|u_n|^3\textrm{d}x=0, \end{aligned}$$

(2.11)

then

$$\begin{aligned} \Vert u_n\Vert _s\rightarrow 0, \ \ \ \ \forall \ s\in (3,6). \end{aligned}$$

(2.12)

3 Ground State Solutions

Set

$$\begin{aligned} g(t):=\frac{2(p-3)-(2p-3)t^3+3t^{2p-3}}{3p}, \ \ \ \ t>0. \end{aligned}$$

(3.1)

Then we have the following lemma by a simple computation.

Lemma 3.1

Assume that $p\in (3,6)$. Then $g(t)>g(1)=0$ for all $t\in (0,1)\cup (1,+\infty )$.

Lemma 3.2

Assume that $p\in (3,6)$ and $\mu >0$. Then

$$\begin{aligned}{} & {} \Phi (u)\ge \Phi (t^2u_t)+\frac{1-t^3}{3}J(u)\nonumber \\{} & {} \quad +\frac{(1-t^3)^2(2+t^3)}{6}\Vert u\Vert _6^6, \ \ \ \ \forall \ u\in E, \ \ t\ge 0. \end{aligned}$$

(3.2)

Proof

Note that

$$\begin{aligned} \Phi (t^2u_t)= & {} \frac{t^3}{2}\int _{\mathbb {R}^3}|\nabla u|^2\textrm{d}x+\frac{t^3}{4}\int _{\mathbb {R}^3}\phi _{u}(x)u^2\textrm{d}x \nonumber \\{} & {} \quad -\int _{\mathbb {R}^3}\left[ \frac{\mu t^{2p-3}}{p}|u|^p+\frac{t^9}{6}|u|^6\right] \textrm{d}x. \end{aligned}$$

(3.3)

Thus, by (1.5), (1.7), (3.1) and (3.3), one has

$$\begin{aligned} \Phi (u)-\Phi (t^2u_t)= & {} \frac{1-t^3}{2}\int _{\mathbb {R}^3}|\nabla u|^2\textrm{d}x+\frac{1-t^3}{4}\int _{\mathbb {R}^3}\phi _{u}(x)u^2\textrm{d}x\\{} & {} +{\int _{\mathbb {R}^3}\left[ \frac{\mu (t^{2p-3}-1)}{p}|u|^p+\frac{t^9-1)}{6}|u|^6 \right] \textrm{d}x}\\= & {} \frac{1-t^3}{3}J(u) +\int _{\mathbb {R}^3}\left[ \mu g(t)|u|^p+\frac{(1-t^3)^2(2+t^3)}{6}|u|^6\right] \textrm{d}x\\= & {} \frac{1-t^3}{3}J(u)+\mu g(t)\Vert u\Vert _p^p+\frac{(1-t^3)^2(2+t^3)}{6}\Vert u\Vert _6^6. \end{aligned}$$

This shows that (3.2) holds. $\square$

From Lemma 3.2, we have the following corollary immediately.

Corollary 3.3

Assume that $p\in (3,6)$ and $\mu >0$. Then for $u\in \mathcal {M}$,

$$\begin{aligned} \Phi (u) = \max _{t\ge 0}\Phi (t^2u_t). \end{aligned}$$

(3.4)

Lemma 3.4

Assume that $p\in (3,6)$ and $\mu >0$. Then for any $u\in E\setminus \{0\}$, there exists a unique $t(u)>0$ such that $t(u)^2u_{t(u)}\in \mathcal {M}$.

Proof

Let $u\in E\setminus \{0\}$ be fixed and define a function $\zeta (t):=\Phi (t^2u_t)$ on $[0, \infty )$. Clearly, by (3.3), we have

$$\begin{aligned} \zeta '(t){=}0\Leftrightarrow & {} \ \ {\frac{3t^2}{2}\Vert \nabla u\Vert _2^2{+}\frac{3t^2}{4}\int _{\mathbb {R}^3}\phi _{u}(x)u^2\textrm{d}x {-}\frac{(2p{-}3)\mu t^{2p{-}2}}{p}\Vert u\Vert _p^p {-}\frac{3t^8}{2}\Vert u\Vert _6^6{=}0} \\\Leftrightarrow & {} \ \ J(t^2u_t){=}0 \ \ \Leftrightarrow \ \ t^2u_t\in \mathcal {M}. \end{aligned}$$

It is easy to verify that $\zeta (0)=0$, $\zeta (t)>0$ for $t>0$ small and $\zeta (t)<0$ for t large. Therefore $\max _{t\in [0, \infty )}\zeta (t)$ is achieved at a $t_0=t(u)>0$ so that $\zeta '(t_0)=0$ and $t_0^2u_{t_0}\in \mathcal {M}$.

Next we claim that t(u) is unique for any $u\in E\setminus \{0\}$. In fact, for any given $u\in E\setminus \{0\}$, let $t_1, t_2>0$ such that $\zeta '(t_1)= \zeta '(t_2)=0$. Then $J(t_1^2u_{t_1})=J(t_2^2u_{t_2})=0$. Jointly with (3.2), we have

$$\begin{aligned} \Phi (t_1^2u_{t_1})\ge & {} \Phi (t_2^2u_{t_2})+\frac{t_1^3-t_2^3}{3t_1^3}J(t_1^2u_{t_1}) +\frac{(t_1^3-t_2^3)^2(2t_1^3+t_2^3)}{6t_1^9}\Vert u\Vert _6^6\nonumber \\= & {} \Phi (t_2^2u_{t_2})+\frac{(t_1^3-t_2^3)^2(2t_1^3+t_2^3)}{6t_1^9}\Vert u\Vert _6^6 \end{aligned}$$

(3.5)

and

$$\begin{aligned} \Phi (t_2^2u_{t_2})\ge & {} \Phi (t_1^2u_{t_1})+\frac{t_2^3-t_1^3}{3t_2^3}J(t_2^2u_{t_2}) +\frac{(t_2^3-t_1^3)^2(2t_2^3+t_1^3)}{6t_2^9}\Vert u\Vert _6^6\nonumber \\= & {} \Phi (t_1^2u_{t_1})+\frac{(t_2^3-t_1^3)^2(2t_2^3+t_1^3)}{6t_2^9}\Vert u\Vert _6^6. \end{aligned}$$

(3.6)

(3.5) and (3.6) imply $t_1=t_2$. Therefore, $t(u)> 0$ is unique for any $u\in E\setminus \{0\}$. $\square$

Both Corollary 3.3 and Lemma 3.4 imply the following lemma.

Lemma 3.5

Assume that $p\in (3,6)$ and $\mu >0$. Then

$$\begin{aligned} \inf _{u\in \mathcal {M}}\Phi (u):=m_0=\inf _{u\in E\setminus \{0\}}\max _{t\ge 0}\Phi (t^2u_t). \end{aligned}$$

Lemma 3.6

Assume that $p\in (3,6)$ and $\mu >0$. Then

• (i) there exists $\rho _0>0$ such that $\Vert \nabla u\Vert _2^2\ge \rho _0, \ \forall \ u\in \mathcal {M}$;

• (ii) $m_0=\inf _{u\in \mathcal {M}}\Phi (u)>0$.

Proof

Since $J(u)=0, \ \forall u\in \mathcal {M}$, by (1.7), (2.1), (2.9), (2.10) and the Young inequality, it has

$$\begin{aligned} \frac{3}{4}\Vert \nabla u\Vert _2^2+\frac{3}{2}\Vert u\Vert _3^3\le & {} \frac{3}{2}\int _{\mathbb {R}^3}|\nabla u|^2\textrm{d}x+\frac{3}{4}\int _{\mathbb {R}^3}\phi _{u}(x)u^2\textrm{d}x\nonumber \\= & {} \frac{(2p-3)\mu }{p}\Vert u\Vert _p^p+\frac{3}{2}\Vert u\Vert _6^6\nonumber \\\le & {} \frac{3}{2}\Vert u\Vert _3^3+C_1\Vert u\Vert _6^6\end{aligned}$$

(3.7)

$$\begin{aligned}\le & {} \frac{3}{2}\Vert u\Vert _3^3+\frac{C_1}{\mathcal {S}^3}\Vert \nabla u\Vert _2^6, \end{aligned}$$

(3.8)

where $C_1$ is a positive constant. This implies

$$\begin{aligned} \Vert \nabla u\Vert _2^2\ge \rho _0:=\frac{\sqrt{3}\mathcal {S}^{\frac{3}{2}}}{2\sqrt{C_1}}, \ \ \ \ \forall \ u\in \mathcal {M}. \end{aligned}$$

(3.9)

From (1.5), (1.7), (3.7) and (3.9), we have

$$\begin{aligned} \Phi (u)= & {} \Phi (u)-\frac{1}{3}J(u)\nonumber \\= & {} \frac{2(p-3)\mu }{3p}\Vert u\Vert _p^p+\frac{1}{3}\Vert u\Vert _6^6\nonumber \\\ge & {} \frac{3}{4C_1}\Vert \nabla u\Vert _2^2\nonumber \\\ge & {} \frac{3\sqrt{3}\mathcal {S}^{\frac{3}{2}}}{8C_1\sqrt{C_1}}, \ \ \ \ \forall \ u\in \mathcal {M}. \end{aligned}$$

This shows that $m_0=\inf _{u\in \mathcal {M}}\Phi (u)>0$. $\square$

Now as in [10], we define functions $U_n(x):=\Theta _n(|x|)$, where

$$\begin{aligned} \Theta _n(r)=\root 4 \of {3} {\left\{ \begin{array}{ll} \sqrt{\frac{n}{1+n^2r^2}}, \ \ {} &{} 0\le r< 1;\\ \sqrt{\frac{n}{1+n^2}}(2-r), \ \ {} &{} 1\le r< 2;\\ 0, \ \ {} &{} r\ge 2. \end{array}\right. } \end{aligned}$$

(3.10)

Computing directly, we have

$$\begin{aligned} \Vert \nabla U_n\Vert _2^2= & {} \int _{\mathbb {R}^3}|\nabla U_n|^2\textrm{d}x =4\pi \int _{0}^{+\infty }r^2|\Theta _n'(r)|^2\textrm{d}r\nonumber \\= & {} 4\sqrt{3}\pi \left[ \int _{0}^{1}\frac{n^5r^4}{\left( 1+n^2r^2\right) ^3}\textrm{d}r +\frac{n}{1+n^2}\int _{1}^{2}r^2\textrm{d}r\right] \nonumber \\= & {} 4\sqrt{3}\pi \left[ \int _{0}^{n}\frac{s^4}{\left( 1+s^2\right) ^3}\textrm{d}s +\frac{7n}{3(1+n^2)}\right] \nonumber \\= & {} \mathcal {S}^{\frac{3}{2}}+4\sqrt{3}\pi \left[ -\int _{n}^{+\infty }\frac{s^4}{\left( 1+s^2\right) ^3}\textrm{d}s+\frac{7n}{3(1+n^2)}\right] , \end{aligned}$$

(3.11)

$$\begin{aligned} \Vert U_n\Vert _{6}^{6}= & {} \int _{\mathbb {R}^3}|U_n|^{6}\textrm{d}x =4\pi \int _{0}^{+\infty }r^2|\Theta _n(r)|^{6}\textrm{d}r\nonumber \\= & {} 12\sqrt{3}\pi \left[ \int _{0}^{1}\frac{n^3r^2}{\left( 1+n^2r^2\right) ^3}\textrm{d}r +\left( \frac{n}{1+n^2}\right) ^3 \int _{1}^{2}r^2(2-r)^{6}\textrm{d}r\right] \nonumber \\= & {} 12\sqrt{3}\pi \left[ \int _{0}^{n}\frac{s^2}{\left( 1+s^2\right) ^3}\textrm{d}s +\left( \frac{n}{1+n^2}\right) ^3\int _{0}^{1}s^{6}(2-s)^2\textrm{d}s\right] \nonumber \\= & {} \mathcal {S}^{\frac{3}{2}}+12\sqrt{3}\pi \left[ -\int _{n}^{+\infty }\frac{s^2}{\left( 1+s^2\right) ^3}\textrm{d}s+\frac{23}{126}\left( \frac{n}{1+n^2}\right) ^3\right] , \end{aligned}$$

(3.12)

$$\begin{aligned} \Vert U_n\Vert _q^q= & {} \int _{\mathbb {R}^3}|U_n|^q\textrm{d}x=4\pi \int _{0}^{+\infty }r^{2}|\Theta _n(r)|^q\textrm{d}r\nonumber \\= & {} 4(\root 4 \of {3})^q\pi \left[ \int _{0}^{1}\frac{n^{q/2}r^{2}}{\left( 1+n^2r^2\right) ^{q/2}}\textrm{d}r +\left( \frac{n}{1+n^2}\right) ^{q/2}\int _{1}^{2}r^{2}(2-r)^q\textrm{d}r\right] \nonumber \\= & {} 4(\root 4 \of {3})^q\pi \left[ \frac{1}{n^{(6-q)/2}}\int _{0}^{n}\frac{s^{2}}{(1+s^2)^{q/2}}\textrm{d}s +\left( \frac{n}{1+n^2}\right) ^{q/2}\int _{0}^{1}s^{q}(2-s)^2\textrm{d}s\right] \nonumber \\= & {} 4(\root 4 \of {3})^q\pi \left[ \frac{1}{n^{(6-q)/2}}\int _{0}^{n}\frac{s^{2}\textrm{d}s}{(1+s^2)^{q/2}} +\frac{q^2+7q+14}{(q+1)(q+2)(q+3)}\left( \frac{n}{1+n^2}\right) ^{\frac{q}{2}}\right] \ \ \ \ \ \ \end{aligned}$$

(3.13)

and

$$\begin{aligned} \Vert U_n\Vert _{12/5}^{12/5}= & {} 4(\root 4 \of {3})^{12/5}\pi \left[ \frac{1}{n^{9/5}}\int _{0}^{n}\frac{s^{2}}{(1+s^2)^{6/5}}\textrm{d}s +\frac{2285}{5049}\left( \frac{n}{1+n^2}\right) ^{6/5}\right] .\nonumber \\ \end{aligned}$$

(3.14)

Both (2.5), (3.11) and (3.14) imply that $U_n\in E$ for all $n\in \mathbb {N}$.

Lemma 3.7

Assume that condition (i) or (ii) in Theorem 1.1 holds. Then there exists a positive integer $\hat{n}$ such that

$$\begin{aligned} m_0\le \sup _{t>0}\Phi \left( t^2(U_{\hat{n}})_t\right) <\frac{1}{3}\mathcal {S}^{\frac{3}{2}}. \end{aligned}$$

(3.15)

Proof

By (2.5), (3.3), (3.11), (3.12), (3.13) and (3.14), we have

$$\begin{aligned}{} & {} \Phi \left( (t^2(U_n)_t\right) \nonumber \\{} & {} \quad = \frac{t^3}{2}\Vert \nabla U_n\Vert _2^2+\frac{t^3}{4}N(U_n) -\frac{\mu t^{2p-3}}{p}\Vert U_n\Vert _p^p-\frac{t^{9}}{6}\Vert U_n\Vert _6^6\nonumber \\{} & {} \quad< \frac{t^3}{2}\left[ \mathcal {S}^{\frac{3}{2}}+\frac{28\sqrt{3}\pi n}{3(1+n^2)}\right] \nonumber \\{} & {} \qquad +4\root 3 \of {4\pi }t^3\left[ \frac{1}{n^{9/5}}\int _{0}^{n}\frac{s^{2}}{(1+s^2)^{6/5}}\textrm{d}s +\frac{2285}{5049}\left( \frac{n}{1+n^2}\right) ^{\frac{6}{5}}\right] ^{\frac{5}{3}}\nonumber \\{} & {} \qquad -\frac{4(\root 4 \of {3})^p\pi \mu t^{2p-3}}{pn^{(6-p)/2}}\int _{0}^{n}\frac{s^{2}}{(1+s^2)^{p/2}}\textrm{d}s \nonumber \\{} & {} \qquad -\frac{t^{9}}{6}\left[ \mathcal {S}^{\frac{3}{2}}+12\sqrt{3}\pi \left( -\frac{1}{3n^3}+\frac{23}{126}\left( \frac{n}{1+n^2}\right) ^3\right) \right] \nonumber \\{} & {} \quad < \mathcal {S}^{\frac{3}{2}}\left( \frac{t^3}{2}-\frac{t^{9}}{6}\right) +\frac{\sqrt{3}\pi }{n^3}t^9+\frac{29\sqrt{3}\pi }{6n}t^3\nonumber \\{} & {} \qquad -\frac{4(\root 4 \of {3})^p\pi \mu t^{2p-3}}{pn^{(6-p)/2}}\int _{0}^{n}\frac{s^{2}}{(1+s^2)^{p/2}}\textrm{d}s, \ \ \ \ \ \forall \ n\ge 100. \end{aligned}$$

(3.16)

Under condition (i) or (ii) of Theorem 1.1, there are three cases to distinguish.

Csae 1. $t\in [2,+\infty )$, $p\in (3,6)$ and $\mu >0$. It follows from (3.16) that

$$\begin{aligned} \Phi \left( (t^2(U_n)_t\right)< & {} \mathcal {S}^{\frac{3}{2}}\left( \frac{t^3}{2}-\frac{t^{9}}{6}\right) +O\left( \frac{1}{n^3}\right) t^9+O\left( \frac{1}{n}\right) t^3-O\left( \frac{1}{n^{(6-p)/2}}\right) t^{2p-3}\nonumber \\< & {} 0, \ \ \ \ n\rightarrow \infty . \end{aligned}$$

(3.17)

Csae 2. $t\in (0,2)$, $p\in (4,6)$ and $\mu >0$. It follows from (3.16) that

$$\begin{aligned} \Phi \left( (t^2(U_n)_t\right)< & {} \mathcal {S}^{\frac{3}{2}}\left( \frac{t^3}{2}-\frac{t^{9}}{6}\right) +O\left( \frac{1}{n^3}\right) +O\left( \frac{1}{n}\right) -\mu \left[ O\left( \frac{1}{n^{(6-p)/2}}\right) \right] \nonumber \\\le & {} \frac{1}{3}\mathcal {S}^{\frac{3}{2}}-\mu \left[ O\left( \frac{1}{n^{(6-p)/2}}\right) \right] , \ \ \ \ n\rightarrow \infty . \end{aligned}$$

(3.18)

Csae 3. $t\in (0,2)$, $p=4$ and $\mu >\frac{7\sqrt{3}}{\pi }$. It follows from (3.16) that

$$\begin{aligned} \Phi \left( (t^2(U_n)_t\right)< & {} \mathcal {S}^{\frac{3}{2}}\left( \frac{t^3}{2}-\frac{t^{9}}{6}\right) +\frac{5\sqrt{3}\pi t^3}{n}-\frac{3\pi \mu t^5}{n} \int _{0}^{n}\frac{s^{2}}{(1+s^2)^{2}}\textrm{d}s\nonumber \\= & {} \mathcal {S}^{\frac{3}{2}}\left( \frac{t^3}{2}-\frac{t^{9}}{6}\right) +\frac{5\sqrt{3}\pi }{n}t^3-\frac{3\pi ^2\mu }{4n} t^5+O\left( \frac{1}{n^3}\right) \nonumber \\\le & {} \frac{1}{3}\mathcal {S}^{\frac{3}{2}}-O\left( \frac{1}{n}\right) , \ \ \ \ n\rightarrow \infty . \end{aligned}$$

(3.19)

Case 1-Case 3 imply that there exists a positive integer $\hat{n}>100$ such that (3.15) holds. $\square$

Set

$$\begin{aligned} \kappa ^3:={\frac{839803\mathcal {S}^\frac{3}{2}}{468750\root 3 \of {2}\pi ^3}\sqrt{\frac{2}{5\root 3 \of {2}\pi }}} \end{aligned}$$

(3.20)

and

$$\begin{aligned} w=\kappa e^{-|x|}. \end{aligned}$$

(3.21)

Then $w\in H^1(\mathbb {R}^3)$, and

$$\begin{aligned} \Vert \nabla w\Vert _2^2=\int _{\mathbb {R}^3}|\nabla w|^2\textrm{d}x=4\pi ^2\kappa ^2\int _{0}^{+\infty }r^2e^{-2r}\textrm{d}r =\pi ^2\kappa ^2, \end{aligned}$$

(3.22)

$$\begin{aligned} \Vert w\Vert _s^s=\int _{\mathbb {R}^3}|w|^s\textrm{d}x=4\pi ^2\kappa ^s\int _{0}^{+\infty }r^2e^{-sr}\textrm{d}r =\frac{8\pi ^2\kappa ^s}{s^3}, \ \ \ \ \forall \ s\in [2,6] \end{aligned}$$

(3.23)

and

$$\begin{aligned} \Vert w\Vert _{12/5}^4=\left( \int _{\mathbb {R}^3}|w|^{12/5}\textrm{d}x\right) ^{\frac{5}{3}} =\left[ 8\pi ^2\kappa ^{12/5}\left( \frac{5}{12}\right) ^3\right] ^{\frac{5}{3}} =\left( \frac{5}{6}\right) ^{5}\pi ^3\root 3 \of {\pi }\kappa ^4. \nonumber \\ \end{aligned}$$

(3.24)

Lemma 3.8

Assume that condition (iii) in Theorem 1.1 holds. Then

$$\begin{aligned} m_0\le \sup _{t>0}\Phi \left( t^2w_t\right) <\frac{1}{3}\mathcal {S}^{\frac{3}{2}}. \end{aligned}$$

(3.25)

Proof

Both (2.5) and (3.24) imply

$$\begin{aligned} N(w)=\int _{\mathbb {R}^3}\phi _{w}(x)w^2\textrm{d}x\le {\frac{2\root 3 \of {2}}{3\root 3 \of {\pi }}\Vert w\Vert _{12/5}^4} ={\frac{2\root 3 \of {2}}{3}\left( \frac{5}{6}\right) ^{5}\pi ^3\kappa ^4}. \end{aligned}$$

(3.26)

From (2.5), (3.20), (3.22), (3.23), (3.26) and condition (iii) in Theorem 1.1, we have

$$\begin{aligned} \Phi (t^2w_t)= & {} \frac{t^3}{2}\Vert \nabla w\Vert _2^2+\frac{t^3}{4}N(w) -\frac{\mu t^{2p-3}}{p}\Vert w\Vert _p^p-\frac{t^{9}}{6}\Vert w\Vert _6^6\nonumber \\\le & {} \frac{\pi ^2\kappa ^2t^3}{2}+{\frac{\root 3 \of {2}t^3}{6}}\left( \frac{5}{6}\right) ^{5}\pi ^3\kappa ^4 -\frac{8\pi ^2\kappa ^p\mu t^{2p-3}}{p^4}-\frac{8\pi ^2\kappa ^6t^{9}}{6^4}\nonumber \\= & {} \left[ \frac{\pi ^2\kappa ^2t^3}{2}-\frac{8\pi ^2\kappa ^p\mu t^{2p-3}}{p^4}\right] +\left[ {\frac{\root 3 \of {2}t^3}{6}}\left( \frac{5}{6}\right) ^{5}\pi ^3\kappa ^4-\frac{8\pi ^2\kappa ^6t^{9}}{6^4}\right] \nonumber \\\le & {} \frac{(p-3)\pi ^2}{2p-3}\kappa ^{(p-6)/2(p-3)}\left[ \frac{3p^4}{16(2p-3)\mu }\right] ^{\frac{3}{2(p-3)}}\nonumber \\{} & {} \quad +{\frac{78125\root 3 \of {2}\pi ^3\kappa ^3}{839803}\sqrt{\frac{5\root 3 \of {2}\pi }{2}}}\nonumber \\= & {} \frac{(p-3)\pi ^2}{2p-3}\kappa ^{(p-6)/2(p-3)}\left[ \frac{3p^4}{16(2p-3)\mu }\right] ^{\frac{3}{2(p-3)}} +\frac{1}{6}\mathcal {S}^{\frac{3}{2}}\nonumber \\< & {} \frac{1}{3}\mathcal {S}^{\frac{3}{2}}. \end{aligned}$$

(3.27)

This shows that (3.25) holds. $\square$

Lemma 3.9

Assume that $p\in (3,6)$ and $\mu >0$. If $u_n\rightharpoonup \bar{u}$ in E, then

$$\begin{aligned} \Phi (u_n)=\Phi (\bar{u})+\Phi (u_n-\bar{u})+o(1), \end{aligned}$$

(3.28)

$$\begin{aligned} \langle \Phi '(u_n), u_n\rangle =\langle \Phi '(\bar{u}), \bar{u}\rangle +\langle \Phi '(u_n-\bar{u}), u_n-\bar{u}\rangle +o(1) \end{aligned}$$

(3.29)

and

$$\begin{aligned} J(u_n)=J(\bar{u})+J(u_n-\bar{u})+o(1). \end{aligned}$$

(3.30)

Proof

Set

$$\begin{aligned}{} & {} I_1(u):= \int _{{\mathbb {R}}^3}|\nabla u|^2\textrm{d}x, \ \ \ \ I_2(u):= \int _{{\mathbb {R}}^3}\phi _{u}(x)u^2\textrm{d}x, \nonumber \\{} & {} I_3(u):= \int _{\mathbb {R}^3}\left( \frac{\mu }{p}|u|^p+\frac{1}{6}|u|^6\right) \textrm{d}x. \end{aligned}$$

(3.31)

Let $v_n=u_n-\bar{u}$. Then $u_n\rightharpoonup \bar{u}$ and $v_n\rightharpoonup 0$ in E. From (2.4), (2.6), (2.7), (3.31) and Lemma 2.4, we have

$$\begin{aligned} I_2(u_n)= & {} D((\bar{u}+v_n)^2,(\bar{u}+v_n)^2)\nonumber \\= & {} D(\bar{u}^2,\bar{u}^2)+D(v_n^2,v_n^2)+4D(\bar{u}^2,\bar{u}v_n)+4D(v_n^2,\bar{u}v_n)\nonumber \\{} & {} \ \ +4D(\bar{u}v_n,\bar{u}v_n)+2D(\bar{u}^2,v_n^2)\nonumber \\= & {} D(\bar{u}^2,\bar{u}^2)+D(v_n^2,v_n^2)+o(1)\nonumber \\= & {} I_2(\bar{u})+I_2(v_n)+o(1) \end{aligned}$$

(3.32)

and

$$\begin{aligned} \langle I_2'(u_n), u_n\rangle= & {} 4D((\bar{u}+v_n)^2,(\bar{u}+v_n)^2)\nonumber \\= & {} 4D(\bar{u}^2,\bar{u}^2)+4D(v_n^2,v_n^2)+o(1)\nonumber \\= & {} \langle I_2'(\bar{u}, \bar{u}\rangle +\langle I_2'(v_n), v_n\rangle +o(1). \end{aligned}$$

(3.33)

By (3.32), (3.33) and the Brezis-Lieb lemma, one can easily prove that

$$\begin{aligned} \Phi (u_n)=\Phi (\bar{u})+\Phi (v_n)+o(1) \end{aligned}$$

(3.34)

and

$$\begin{aligned} \langle \Phi '(u_n), u_n\rangle= & {} \langle \Phi '(v_n), v_n\rangle +\langle \Phi '(\bar{u}), \bar{u}\rangle +o(1). \end{aligned}$$

Note that

$$\begin{aligned} J(u)=2\langle \Phi '(u), u\rangle -3\Phi (u)+\Vert \nabla u\Vert _2^2-\frac{1}{2}I_2(u), \end{aligned}$$

(3.35)

then from (3.28), (3.29) and (3.35), we can prove that (3.30) holds. $\square$

Lemma 3.10

Assume that the conditions in Theorem 1.1 hold. Then $m_0$ is achieved.

Proof

We prove this lemma by using the strategy used in [30]. Let $\{u_n\}\subset \mathcal {M}$ be such that $\Phi (u_n)\rightarrow m_0$. Since $J(u_n)=0$, then it follows from (1.5) and (1.7) that

$$\begin{aligned} m_0+o(1)= \frac{2(p-3)\mu }{3p}\Vert u_n\Vert _p^p+\frac{1}{3}\Vert u_n\Vert _6^6 \end{aligned}$$

(3.36)

and

$$\begin{aligned} m_0+o(1)= \frac{1}{3}\Vert \nabla u_n\Vert _2^2+\frac{1}{4}N(u_n)-\frac{2(6-p)\mu }{9p}\Vert u_n\Vert _p^p. \end{aligned}$$

(3.37)

By (1.7) and $J(u_n)=0$, we have

$$\begin{aligned} \frac{3}{2}\Vert \nabla u_n\Vert _2^2+\frac{3}{4}N(u_n)= & {} \frac{(2p-3)\mu }{p}\Vert u_n\Vert _p^p+\frac{3}{2}\Vert u_n\Vert _6^6. \end{aligned}$$

(3.38)

Hence, (3.36) (3.38) show that $\{u_n\}$ is bounded in E. From (3.38), one has

$$\begin{aligned} \Vert \nabla u_n\Vert _2^2\le & {} \frac{2(2p-3)\mu }{3p}\Vert u_n\Vert _p^p+\Vert u_n\Vert _6^6. \end{aligned}$$

(3.39)

We claim that there exist a $\delta >0$ and a sequence $y_n\in \mathbb {R}^3$ such that

$$\begin{aligned} \liminf _{n\rightarrow \infty }\int _{B_1(y_n)}|u_n|^3\textrm{d}x> \delta . \end{aligned}$$

(3.40)

Indeed, suppose that (3.40) does not hold. Then we have

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

(3.41)

By Lemma 2.5, we have

$$\begin{aligned} \Vert u_n\Vert _p^p\rightarrow 0. \end{aligned}$$

(3.42)

Up to a subsequence, we assume that

$$\begin{aligned} \Vert \nabla u_n\Vert _2^2\rightarrow l_1\ge 0, \ \ \Vert u_n\Vert _6^6\rightarrow l_2\ge 0. \end{aligned}$$

(3.43)

Then it from (2.10), (3.39), (3.42) and (3.43) follows that

$$\begin{aligned} l_1=\lim _{n\rightarrow \infty }\Vert \nabla u_n\Vert _2^2\le \lim _{n\rightarrow \infty }\Vert u_n\Vert _6^6 \le \mathcal {S}^{-3}\lim _{n\rightarrow \infty }\Vert \nabla u_n\Vert _2^6=\mathcal {S}^{-3}l_1^3. \end{aligned}$$

(3.44)

If $l_1 > 0$, then (3.44) implies that $l_1\ge \mathcal {S}^{\frac{3}{2}}$, which, together with (3.37) and (3.42), implies that $m_0\ge \frac{1}{3}\mathcal {S}^{\frac{3}{2}}$. This contradicts with (3.15) and (3.25). Therefore, (3.40) holds.

Let $\hat{u}_n(x)=u_n(x+y_n)$. Then we have $\Vert \hat{u}_n\Vert _E=\Vert u_n\Vert _E$ and

$$\begin{aligned} J(\hat{u}_n)= 0, \ \ \ \ \Phi (\hat{u}_n)\rightarrow m_0, \ \ \ \ \liminf _{n\rightarrow \infty }\int _{B_1(0)}|\hat{u}_n|^3\textrm{d}x> \delta . \end{aligned}$$

(3.45)

Therefore, there exists $\bar{u}\in E\setminus \{0\}$ such that, passing to a subsequence,

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

(3.46)

Let $w_n=\hat{u}_n-\bar{u}$. Then (3.46) and Lemma 3.9 yield

$$\begin{aligned} \Phi (\hat{u}_n)=\Phi (\bar{u})+\Phi (w_n)+o(1) \end{aligned}$$

(3.47)

and

$$\begin{aligned} J(\hat{u}_n)=J(\bar{u})+J(w_n)+o(1). \end{aligned}$$

(3.48)

From (1.5), (1.7), (3.45), (3.47) and (3.48), one has

$$\begin{aligned} \frac{2(p-3)\mu }{3p}\Vert w_n\Vert _p^p+\frac{1}{3}\Vert w_n\Vert _6^6=m_0-\frac{2(p-3)\mu }{3p}\Vert \bar{u}\Vert _p^p-\frac{1}{3}\Vert \bar{u}\Vert _6^6+o(1) \end{aligned}$$

(3.49)

and

$$\begin{aligned} J(w_n) = -J(\bar{u})+o(1). \end{aligned}$$

(3.50)

If there exists a subsequence $\{w_{n_i}\}$ of $\{w_n\}$ such that $w_{n_i}=0$, then going to this subsequence, we have

$$\begin{aligned} \Phi (\bar{u})=m_0, \ \ \ \ J(\bar{u})=0, \end{aligned}$$

(3.51)

which implies the conclusion of Lemma 3.10 holds. Next, we assume that $w_n\ne 0$. In view of Lemma 3.4, there exists $t_n>0$ such that $t_n^2(w_n)_{t_n}\in \mathcal {M}$. We claim that $J(\bar{u})\le 0$. Otherwise, if $J(\bar{u})>0$, then (3.50) implies $J(w_n) < 0$ for large n. From (1.5), (1.7), (3.2) and (3.49), we obtain

$$\begin{aligned} m_0-\frac{2(p-3)\mu }{3p}\Vert \bar{u}\Vert _p^p-\frac{1}{3}\Vert \bar{u}\Vert _6^6+o(1)= & {} \frac{2(p-3)\mu }{3p}\Vert w_n\Vert _p^p+\frac{1}{3}\Vert w_n\Vert _6^6\nonumber \\= & {} \Phi (w_n)-\frac{1}{3}J(w_n)\nonumber \\\ge & {} \Phi \left( t_n^2(w_n)_{t_n}\right) -\frac{t_n^3}{3}J(w_n)\nonumber \\\ge & {} m_0-\frac{t_n^3}{3}J(w_n)\nonumber \\\ge & {} m_0, \end{aligned}$$

which implies $J(\bar{u})\le 0$ due to $\frac{2(p-3)\mu }{3p}\Vert \bar{u}\Vert _p^p+\frac{1}{3}\Vert \bar{u}\Vert _6^6>0$. Since $\bar{u}\in E\setminus \{0\}$, in view of Lemma 3.4, there exists $\bar{t}>0$ such that $\bar{t}^2\bar{u}_{\bar{t}}\in \mathcal {M}$. From (1.5), (1.7), (3.2), (3.45) and Fatou’s lemma, one has

$$\begin{aligned} m_0= & {} \lim _{n\rightarrow \infty } \left[ \Phi (\hat{u}_n)-\frac{1}{3}J(\hat{u}_n)\right] \nonumber \\= & {} \lim _{n\rightarrow \infty } \left[ \frac{2(p-3)\mu }{3p}\Vert u_n\Vert _p^p+\frac{1}{3}\Vert u_n\Vert _6^6\right] \nonumber \\\ge & {} \frac{2(p-3)\mu }{3p}\Vert \bar{u}\Vert _p^p+\frac{1}{3}\Vert \bar{u}\Vert _6^6\nonumber \\= & {} \Phi (\bar{u})-\frac{1}{3}J(\bar{u})\nonumber \\\ge & {} \Phi \left( \bar{t}^2\bar{u}_{\bar{t}}\right) -\frac{\bar{t}^3}{3}J(\bar{u})\nonumber \\\ge & {} m_0-\frac{\bar{t}^3}{3}J(\bar{u})\ge m_0, \end{aligned}$$

which implies (3.51) holds also. $\square$

Lemma 3.11

Assume that the conditions in Theorem 1.1 hold. If $\bar{u}\in \mathcal {M}$ and $\Phi (\bar{u})=m_0$, then $\bar{u}$ is a critical point of $\Phi$.

Proof

We prove this lemma by using the method introduced in [9]. Assume that $\Phi '(\bar{u})\ne 0$. Then there exist $\delta >0$ and $\varrho >0$ such that

$$\begin{aligned} \Vert u-\bar{u}\Vert \le 3\delta \Rightarrow \Vert \Phi '(u)\Vert \ge \varrho . \end{aligned}$$

(3.52)

Let $\{t_n\}\subset \mathbb {R}$ such that $t_n\rightarrow 1$. Since $t_n^2\bar{u}_{t_n}\rightharpoonup \bar{u}$ in E, then it follows from (2.7) and Lemma 2.4 that

$$\begin{aligned} \left\| \nabla \left( t_n^2\bar{u}_{t_n}\right) -\nabla \bar{u}\right\| _2^2= & {} \int _{\mathbb {R}^3}\left| \nabla \left( t_n^2\bar{u}_{t_n}\right) -\nabla \bar{u}\right| ^2\textrm{d}x \nonumber \\= & {} (t_n^3+1)\int _{\mathbb {R}^3}|\nabla \bar{u}|^2\textrm{d}x -2\int _{\mathbb {R}^3}\nabla \left( t_n^2\bar{u}_{t_n}\right) \cdot \nabla \bar{u}\textrm{d}x\nonumber \\= & {} o(1) \end{aligned}$$

(3.53)

and

$$\begin{aligned}{} & {} N\left( t_n^2\bar{u}_{t_n}-\bar{u}\right) \nonumber \\{} & {} \quad = D\left( (t_n^2\bar{u}_{t_n}-\bar{u})^2,(t_n^2\bar{u}_{t_n}-\bar{u})^2\right) \nonumber \\{} & {} \quad = D\left( (t_n^2\bar{u}_{t_n})^2,(t_n^2\bar{u}_{t_n})^2\right) +D\left( \bar{u}^2,\bar{u}^2\right) {-}4D\left( (t_n^2\bar{u}_{t_n})^2,(t_n^2\bar{u}_{t_n})\bar{u}\right) \nonumber \\{} & {} \qquad -4D\left( \bar{u}^2,(t_n^2\bar{u}_{t_n})\bar{u}\right) \nonumber \\{} & {} \qquad +4D\left( (t_n^2\bar{u}_{t_n})\bar{u},(t_n^2\bar{u}_{t_n})\bar{u}\right) +2D\left( (t_n^2\bar{u}_{t_n})^2,\bar{u}^2\right) \nonumber \\{} & {} \quad = D\left( (t_n^2\bar{u}_{t_n})^2,(t_n^2\bar{u}_{t_n})^2\right) -D\left( \bar{u}^2,\bar{u}^2\right) +o(1)\nonumber \\{} & {} \quad = (t_n^3-1)D\left( \bar{u}^2,\bar{u}^2\right) +o(1)\nonumber \\{} & {} \quad = o(1). \end{aligned}$$

(3.54)

Combining (3.53) with (3.54), one has

$$\begin{aligned} \lim _{t\rightarrow 1}\left\| t^2\bar{u}_t-\bar{u}\right\| _E=0. \end{aligned}$$

(3.55)

Thus, there exists $\delta _1>0$ such that

$$\begin{aligned} |t-1|<\delta _1\Rightarrow \Vert t^2\bar{u}_t-\bar{u}\Vert _E< \delta . \end{aligned}$$

(3.56)

In view of Lemma 3.1, one has

$$\begin{aligned} \Phi (t^2\bar{u}_t)\le & {} \Phi (\bar{u})-\frac{(1-t^3)^2(2+t^3)}{6}\Vert \bar{u}\Vert _6^6\nonumber \\= & {} m_0-\frac{(1-t^3)^2(2+t^3)}{6}\Vert \bar{u}\Vert _6^6, \ \ \ \ \forall \ t> 0. \end{aligned}$$

(3.57)

It follows from (1.7) that there exist $T_1\in (0,1)$ and $T_2\in (1, \infty )$ such that

$$\begin{aligned} J\left( T_1^2\bar{u}_{T_1}\right) >0, \ \ \ \ J\left( T_2^2\bar{u}_{T_2}\right) <0. \end{aligned}$$

(3.58)

Set $\Theta :=\inf _{t\in (0,T_1]\cup [T_2,+\infty )}\frac{(1-t^3)^2(2+t^3)}{6}\Vert \bar{u}\Vert _6^6$. Let $S:=B(\bar{u}, \delta )$ and $\varepsilon :=\min \{\Theta /24, 1, \varrho \delta /8\}$. Then [33, Lemma 2.3] yields a deformation $\eta \in \mathcal {C}([0, 1]\times E, E)$ such that

• (i) $\eta (1, u)=u$ if $\Phi (u)<m_0-2\varepsilon$ or $\Phi (u)>m_0+2\varepsilon$;

• (ii) $\eta \left( 1, \Phi ^{m_0+\varepsilon }\cap B(\bar{u}, \delta )\right) \subset \Phi ^{m_0-\varepsilon }$;

• (iii) $\Phi (\eta (1, u))\le \Phi (u), \ \forall \ u\in E$;

• (iv) $\eta (1, u)$ is a homeomorphism of E.

By Corollary 2.3, $\Phi (t^2\bar{u}_t)\le \Phi (\bar{u})=m_0$ for $t> 0$, then it follows from (3.56) and ii) that

$$\begin{aligned} \Phi (\eta (1, t^2\bar{u}_t)){} & {} \le m_0-\varepsilon , \ \ \ \ \forall \ t> 0, \ \ |t-1|< \delta _1. \end{aligned}$$

(3.59)

On the other hand, by iii) and (3.57), one has

$$\begin{aligned} \Phi (\eta (1, t^2\bar{u}_t))\le & {} \Phi (t^2\bar{u}_t)\le m_0-\frac{(1-t^3)^2(2+t^3)}{6}\Vert \bar{u}\Vert _6^6,\nonumber \\{} & {} \forall \ t> 0, \ \ |t-1|\ge \delta _1. \ \ \ \ \end{aligned}$$

(3.60)

Combining (3.59) with (3.60), we have

$$\begin{aligned} \max _{t\in [T_1, T_2]}\Phi (\eta (1, t^2\bar{u}_t))<m_0. \end{aligned}$$

(3.61)

Define $\Psi _0(t):=J\left( \eta \left( 1, t^2\bar{u}_t\right) \right)$ for $t> 0$. It follows from (3.60) and i) that $\eta (1, \bar{u}_t)=\bar{u}_t$ for $t=T_1$ and $t=T_2$, which, together with (3.58), implies

$$\begin{aligned} \Psi _0(T_1)=J\left( T_1^2\bar{u}_{T_1}\right) >0, \ \ \ \ \Psi _0(T_2)=J\left( T_2^2\bar{u}_{T_2}\right) <0. \end{aligned}$$

Since $\Psi _0(t)$ is continuous on $(0, \infty )$, then we have that $\eta \left( 1, t^2\bar{u}_t\right) \cap \mathcal {M}\ne \emptyset$ for some $t_0\in [T_1, T_2]$, contradicting to the definition of $m_0$. $\square$

Theorem 1.1 is a direct corollary of Lemmas 3.6, 3.10 and 3.11.