Ground State Solution of Critical p-Biharmonic Equation Involving Hardy Potential

Abstract

In this paper, we consider the following critical p-biharmonic equation involving Hardy potential

$$\begin{aligned} \begin{aligned} \Delta _p^2u-\Delta _qu-\mu \frac{|u|^{p-2}u}{|x|^{2p}}=|u|^{p_{*}-2}u,\quad {{x\in \mathbb {R}^{N}}}, \end{aligned} \end{aligned}$$

where $2\leqslant p<\frac{N}{2},0<\mu <\mu _{N,p}=\left( \frac{(p-1)N(N-2p)}{p^2}\right) ^p,\Delta _p^2u=\Delta (|\Delta u|^{p-2}\Delta u), q=p^{{*}}=\frac{Np}{N-p},$ and $p_{{*}}=\frac{Np}{N-2p}.$ The existence of ground state solution to above equation is established by using the Nehari manifold and some analysis techniques. Our result extends the existing results in the literature.

1 Introduction

In this paper, we study the following critical p-biharmonic equation involving Hardy potential:

$$\begin{aligned} \begin{aligned} \Delta _p^2u-\Delta _qu-\mu \frac{|u|^{p-2}u}{|x|^{2p}}=|u|^{p_{*}-2}u,\ \ x\in \mathbb {R}^{N}, \end{aligned} \end{aligned}$$

(1.1)

where $2\leqslant p<\frac{N}{2},0<\mu <\mu _{N,p}=\left( \frac{(p-1)N(N-2p)}{p^2}\right) ^p,$ $\Delta _p^2u=\Delta (|\Delta u|^{p-2}\Delta u)$ is p-biharmonic operator and $\Delta _qu=div(|\nabla u|^{q-2}\nabla u)$ is q-Laplace operator, $q=p^{{*}}=\frac{Np}{N-p},$ and $p_{{*}}=\frac{Np}{N-2p}$ denotes the critical Sobolev exponent.

In recent years, the nonlinear elliptic equations with singularities become an interesting topic. It arises from physical modeling, such as non-Newtonian fluid, viscous fluids, elastic mechanic, boundary layer, see, for instance, [1]. Recently, the existence and multiplicity of ground state solutions, positive solutions and sign-changing solutions of p-biharmonic equations with singular potential have been studied extensively. For more related works, we refer to [2,3,4,5,6,7,8,9,10,11,12] and the references therein. In particular, Dhifli–Alsaedi [3] studied the following p-biharmonic equation:

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} \Delta _p^2u-\mu \frac{|u|^{p-2}u}{|x|^{2p}}-\Delta _pu=g(x)u^{m-1}+\lambda f(x)u^{q-1}, &{}x\in \mathbb {R}^N,\\ u(x)>0, &{}x\in \mathbb {R}^N, \end{array}\right. } \end{aligned} \end{aligned}$$

(1.2)

where $0<m<1<p<q<p_{*},N>2p$, $\lambda >0$. Under some appropriate conditions on functions f and g, the authors showed that Eq. (1.2) has at least two positive solutions by using the fibering maps and Nehari manifold.

Yang–Zhang–Liu [13] dealt with the following p-biharmonic equation:

$$\begin{aligned} \begin{aligned} \Delta _p^2u-\mu \frac{|u|^{p-2}u}{|x|^{2p}}=a(x)|u|^{r-2}u, \ \ x\in \mathbb {R}^N, \end{aligned} \end{aligned}$$

(1.3)

where $1<p<\frac{N}{2},p<r<p_{*}.$ By applying the method of invariant sets of descending flow, the existence of sign-changing solutions of Eq. (1.3) was obtained. By using Nehari manifold, Su–Shi [14] investigated the existence of ground state solutions for the following equation:

$$\begin{aligned} \begin{aligned} \Delta ^2u-\mu \frac{u}{|x|^{4}}-\Delta _pu=|u|^{2_{*}-2}u, \ \ x\in \mathbb {R}^N, \end{aligned} \end{aligned}$$

(1.4)

where $N\geqslant 5$, $p=2^{*}=\frac{2N}{N-2}$, $2_{*}=\frac{2N}{N-4}$. Furthermore, Su–Liu–Feng [10] established the existence of ground state solutions for the thin film epitaxy equation via the generalized versions of Lions-type theorem.

Inspired by the above-mentioned works, it is natural to ask a question whether Eq. (1.1) admits ground state solution? As far as we know, there is no result about the ground state solution for p-biharmonic equations with Hardy potential in current literature. Therefore, in the present paper, we shall give a positive answer to the above question. Our main result is the following theorem.

Theorem 1.1

Assume that $2\leqslant p<\frac{N}{2}$, $q=p^{{*}}=\frac{Np}{N-p}$ and $\mu \in (0,\mu _{N,p})$ hold. Then, Eq. (1.1) has at least a ground state solution.

2 Proof of Theorem 1.1

The space $W_0^{2,p}(\mathbb {R}^N)$ is the completion of $C_0^{\infty }(\mathbb {R}^N)$, where the norm is $\Vert u\Vert ^p_0=\int _{\mathbb {R}^N}|\Delta {u}|^p\mathrm {d}x$. According to [15], $\mu _{N,p}$ is the best constant in the following Rellich inequality:

$$\begin{aligned} \begin{aligned} \mu _{N,p}\int _{\mathbb {R}^N}\frac{|u|^p}{|x|^{2p}}\mathrm {d}x \leqslant \int _{\mathbb {R}^N} |\Delta {u}|^p \mathrm {d}x, \ \ \forall u\in {{W_0^{2,p}(\mathbb {R}^N)}}. \end{aligned} \end{aligned}$$

(2.1)

For the above inequality, we refer to [16] for more details. We define

$$\begin{aligned} \begin{aligned} E=\left\{ u\in {W_0^{2,p}(\mathbb {R}^{N})}\cap W_0^{1,p}(\mathbb {R}^{N}) \Big | \int _{\mathbb {R}^N}\left( |\Delta {u}|^p-\mu \frac{|u|^p}{|x|^{2p}}\right) \mathrm {d}x<\infty \right\} . \end{aligned} \end{aligned}$$

For $\mu \in (0,\mu _{N,p}),$ E is equipped with the norm

$$\begin{aligned} \begin{aligned} \Vert u\Vert =\left[ \int _{\mathbb {R}^N}\left( |\Delta {u}|^p-\mu \frac{|u|^p}{|x|^{2p}}\right) \mathrm {d}x\right] ^{\frac{1}{p}}. \end{aligned} \end{aligned}$$

Furthermore, we denote the best Sobolev’s constant by

$$\begin{aligned} \begin{aligned} S_\mu :=\inf \limits _{u\in E\backslash \{0\}}\left\{ \frac{\Vert u\Vert ^{p}}{(\int _{\mathbb {R}^N}|u|^{p_{{*}}}\mathrm {d}x)^{\frac{p}{p_{{*}}}}}\right\} . \end{aligned} \end{aligned}$$

(2.2)

The energy functional $I(u):E\rightarrow \mathbb {R}$ associated with Eq. (1.1) can be given by

$$\begin{aligned} \begin{aligned} I(u)=\frac{1}{p}\Vert u\Vert ^p+\frac{1}{p^{{*}}}\int _{\mathbb {R}^N} |\nabla {u}|^{p^{{*}}}\mathrm {d}x-\frac{1}{p_{{*}}}\int _{\mathbb {R}^N}|u|^{p_{{*}}}\mathrm {d}x, \end{aligned} \end{aligned}$$

and the Nehari manifold of E is defined by

$$\begin{aligned} \begin{aligned} \aleph =\{u\in E\backslash \{0\}|\langle {I^\prime (u),u}\rangle =0\}. \end{aligned} \end{aligned}$$

Denote

$$\begin{aligned} \begin{aligned} c=\inf \limits _{\gamma \in {\Gamma }}\max \limits _{t\in [0,1]}I(\gamma (t)), ~~~~ \bar{c}=\inf \limits _{u\in \aleph }I(u), ~~~~ \bar{\bar{c}}=\inf \limits _{u\in E\setminus \{0\}}\sup \limits _{t\geqslant 0}I(tu), \end{aligned} \end{aligned}$$

where $\Gamma =\{\gamma \in {C([0,1],E)}|\gamma (0)=0,\;I(\gamma (1))<0\}$.

Lemma 2.1

Assume that the assumptions of Theorem 1.1 hold. Then, the following conclusions are true.

$\mathrm {(i)}$ For every $u\in E\setminus \{0\},$ there exists only one $t_u>0$ such that $t_uu\in {\aleph }$ and $I(t_uu)=\max \limits _{t>0}I(tu)$.

$\mathrm {(ii)}$ $c=\bar{c}=\bar{\bar{c}}$.

$\mathrm {(iii)}$ There exists a $(PS)_c$ sequence $\{u_n\}\subset \aleph$ of I with $c>0$.

Proof

(i) For any $u\in E\setminus \{0\}$ and $t\in (0,+\infty ),$ we set

$$\begin{aligned} \begin{aligned} g(t):=I(tu)=\frac{t^p}{p}\Vert u\Vert ^p+\frac{t^{p^{{*}}}}{p^{{*}}}\int _{\mathbb {R}^N}|\nabla {u}|^{p^{{*}}}\mathrm {d}x-\frac{t^{p_{{*}}}}{p_{{*}}}\int _{\mathbb {R}^N}|u|^{p_{{*}}}\mathrm {d}x. \end{aligned} \end{aligned}$$

Then, we have

$$\begin{aligned} \begin{aligned} g^{\prime }(t)= t^{p-1}\Vert u\Vert ^p + t^{p^{*}-1} \int _{\mathbb {R}^N} |\nabla {u}|^{p^{{*}}} \mathrm {d}x - t^{p_{{*}}-1} \int _{\mathbb {R}^N} |u|^{p_{{*}}} \mathrm {d}x. \end{aligned} \end{aligned}$$

By $p<p^{{*}}<p_{{*}},$ we know that $g^{\prime }(\cdot )>0$ for $t>0$ enough small, and $g^{\prime }(\cdot )<0$ for t enough large. Then, there exists $t_u>0$ such that $g^{\prime }(t_u)=0.$

To prove the uniqueness of $t_u$, let us assume that $0<\bar{t}<\bar{\bar{t}}$ satisfy $g^{\prime }(\bar{t})=g^{\prime }(\bar{\bar{t}})=0$. Then,

$$\begin{aligned} \begin{aligned} 0=\frac{g^{\prime }(\bar{t})}{\bar{t}^{p^{{*}}-1}}= \bar{t}^{p-p^{{*}}}\Vert u\Vert ^p+ \int _{\mathbb {R}^N}|\nabla {u}|^{p^{{*}}}\mathrm {d}x-\bar{t}^{p_{{{*}}}-p^{{*}}}\int _{\mathbb {R}^N}|u|^{p_{{*}}}\mathrm {d}x \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} 0=\frac{g^{\prime }(\bar{\bar{t}})}{\bar{\bar{t}}^{p^{*}-1}}=\bar{\bar{t}}^{p-p^{*}}\Vert u\Vert ^p+\int _{\mathbb {R}^N}|\nabla {u}|^{p^{*}}\mathrm {d}x-\bar{\bar{t}}^{p_{*}-p^{*}}\int _{\mathbb {R}^N}|u|^{p_{*}}\mathrm {d}x. \end{aligned} \end{aligned}$$

According to $0=\frac{g^{\prime }(\bar{t})}{\bar{t}^{p^{*}-1}}=\frac{g^{\prime }(\bar{\bar{t}})}{\bar{\bar{t}}^{p^{*}-1}}$, we obtain

$$\begin{aligned} \begin{aligned} (\bar{t}^{p-p^{*}}-\bar{\bar{t}}^{p-p^{*}})\Vert u\Vert ^{p}=(\bar{t}^{p_{*}-p^{*}}-\bar{\bar{t}}^{p_{*}-p^{*}})\int _{\mathbb {R}^N}|u|^{p_{*}}\mathrm {d}x. \end{aligned} \end{aligned}$$

On the one hand, since $p-p^{*}<0$ and $0<\bar{t}<\bar{\bar{t}},$ then $\bar{t}^{p-p^{*}}-\bar{\bar{t}}^{p-p^{*}}>0.$ On the other hand, by $p_{*}-p^{*}>0$ and $0<\bar{t}<\bar{\bar{t}}$, we have $\bar{t}^{p_{*}-p^{*}}-\bar{\bar{t}}^{p_{*}-p^{*}}<0$. This is a contradiction. Hence, for any $u\in E\setminus \{0\},$ there exists a unique $t_u>0$ such that $t_uu\in {\aleph }.$ So g(t) admits a unique critical point $t_u\in (0,+\infty )$ such that g(t) attains its maximum at $t_u.$

(ii) First, we prove $I(u)\geqslant {I(tu)}$ for $t\ge 0$. Let $u\in {\aleph }$. Then, we have

$$\begin{aligned} \begin{aligned} \langle {I^\prime (u),u}\rangle =0\Leftrightarrow \int _{\mathbb {R}^N}|\nabla {u}|^{p^*}dx=-\Vert u\Vert ^p+\int _{\mathbb {R}^N}|u|^{p_*}dx. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned} I(u)-I(tu)&= \frac{1}{p}(1-t^p)\Vert u\Vert ^p + \frac{1}{p^{*}}(1-t^{p^{*}}) \int _{\mathbb {R}^N} |\nabla {u}|^{p^{*}} \mathrm {d}x - \frac{1}{p_{*}}(1-t^{p_{*}}) \int _{\mathbb {R}^N} |u|^{p_{*}} \mathrm {d}x\\&= \left( \frac{1-t^p}{p} + \frac{t^{p^{*}}-1}{p^{*}}\right) \Vert u\Vert ^p + \left( \frac{1-t^{p^{*}}}{p^{*}} + \frac{t^{p_{*}}-1}{p_{*}}\right) \int _{\mathbb {R}^N} |u|^{p_{*}} \mathrm {d}x. \end{aligned} \end{aligned}$$

(2.3)

Since $2\leqslant {p}<p^{*}<p_{*},$ it is easy to see that

$$\begin{aligned} \begin{aligned} \frac{1-t^p}{p}+\frac{t^{p^{*}}-1}{p^{*}}\geqslant 0,~~ \mathrm{and}~~ \frac{1-t^{p^{*}}}{p^{*}}+\frac{t^{p_{*}}-1}{p_{*}}\geqslant 0. \end{aligned} \end{aligned}$$

Thus, we have $I(u)\geqslant {I(tu)}$ for $t\ge 0$. From (i), it is obvious that $c=\bar{c}$.

Next, we prove $\bar{c}=\bar{\bar{c}}$. By the definition of c, we can select a sequence $\{u_n\}\subset E$ such that

$$\begin{aligned} \begin{aligned} c\leqslant \max \limits _{t\geqslant 0}I(tu_n) \leqslant c+\frac{1}{n}, \ \ \forall n\in {\mathbb {N}^{*}}. \end{aligned} \end{aligned}$$

(2.4)

For any $u\in E\backslash \{0\}$ and $t>0$ large enough, we have $g(t)=I(tu)<0$ and then there exists $t_n=t(u_n)>0$ and $s_n>t_n$ such that

$$\begin{aligned} \begin{aligned} I(t_nu_n)= \max \limits _{t\geqslant 0}I(tu_n), \ \ I(s_nu_n)<0, \ \ \forall n\in {\mathbb {N}^{*}}. \end{aligned} \end{aligned}$$

(2.5)

Let $\gamma _n(\bar{t})=\bar{t}s_nu_n$, $\bar{t}\in [0,1]$, then $\gamma _n\in {\Gamma }.$ It follows from (2.4) and (2.5) that

$$\begin{aligned} \begin{aligned} \sup \limits _{\bar{t}\in [0,1]} I(\gamma _n(\bar{t}))= \max \limits _{t\geqslant 0}I(tu_n) <c+\frac{1}{n}, \ \ \forall n\in {\mathbb {N}^{*}}, \end{aligned} \end{aligned}$$

which indicates $\bar{\bar{c}}<c.$ Obviously, E can be separated into two parts by the manifold $\aleph$ as follows:

$$\begin{aligned} \begin{aligned} E^+=\{u\in E|\;\langle {I^{\prime }(u),u\rangle }>0\}\cup \{0\} \ \ \mathrm {and} \ \ E^-=\{u\in E|\;\langle {I^{\prime }(u),u\rangle }<0\}. \end{aligned} \end{aligned}$$

By $p<p^{*}<p_{*}$, it easy to obtain that $I(u)\geqslant \frac{1}{p^{*}}\langle {I^\prime (u),u}\rangle$ for any $u\in E$. It follows that $I(u)\geqslant 0$ for all $u\in {E^+}$ and indicates that $E^+$ conclude a little ball which is around the origin. Thus, every $\gamma \in \Gamma$ has to cross $\aleph$ for $\gamma (0)\in E^+$ and $\gamma (1)\in E^-$. So $\bar{c}\leqslant \bar{\bar{c}}$. Combining with $\bar{\bar{c}}\leqslant c$ and $c=\bar{c}$, we have $c=\bar{c}=\bar{\bar{c}}$.

(iii) Set

$$\begin{aligned} \begin{aligned} \Phi (u)=\langle I'(u),u\rangle = \Vert u\Vert ^{p} + \int _{\mathbb {R}^{N}} |\nabla u|^{p^{*}} \mathrm {d}x - \int _{\mathbb {R}^{N}} |u|^{p_{{*}}} \mathrm {d}x, \end{aligned} \end{aligned}$$

then

$$\begin{aligned} \begin{aligned} \langle \Phi '(u),u\rangle = p\Vert u\Vert ^{p} + p^{*}\int _{\mathbb {R}^{N}} |\nabla u|^{p^{*}} \mathrm {d}x -p_{{*}} \int _{\mathbb {R}^{N}} |u|^{p_{{*}}} \mathrm {d}x. \end{aligned} \end{aligned}$$

According to (i), we have $\aleph \ne \emptyset$ and $\inf \limits _{u\in \aleph }I(u)=\bar{c}=c$. Applying Ekeland’s variational principle, there exists $\{u_n\}\subset \aleph$ and $\lambda _n\in \mathbb {R}$ such that $I(u_n)\rightarrow \overline{c}$ and $I'(u_n)-\lambda _n\Phi '(u_n)\rightarrow 0$, as $n\rightarrow \infty$. Then, we get

$$\begin{aligned} \begin{aligned} I(u_n)=I(u_n)-\frac{1}{p^*}\langle I'(u_n),u_n\rangle \geqslant \left( \frac{1}{p}-\frac{1}{p^{*}}\right) \Vert u\Vert ^{p}. \end{aligned} \end{aligned}$$

Therefore, $\{u_n\}$ is bounded in E. In view of

$$\begin{aligned} \begin{aligned} |\langle I'(u_n),u_n\rangle -\langle \lambda _n\Phi '(u_n),u_n\rangle |\le \Vert I'(u_n)-\lambda _n\Phi '(u_n)\Vert \Vert u_n\Vert , \end{aligned} \end{aligned}$$

then we have

$$\begin{aligned} \begin{aligned} |\langle I'(u_n),u_n\rangle -\langle \lambda _n\Phi '(u_n),u_n\rangle |\rightarrow 0, \end{aligned} \end{aligned}$$

as $n\rightarrow \infty .$ Since $|\langle I'(u_n),u_n\rangle |=0$ and $\langle \Phi '(u_n),u_n\rangle \ne 0$, we can easily get $\lambda _n\rightarrow 0.$ Applying Hölder’s and Sobolev’s inequalities, we know

$$\begin{aligned} \begin{aligned} \Vert \Phi '(u_n)\Vert&=\sup _{\Vert \varphi \Vert =1}|\langle \Phi '(u_n),\varphi \rangle |\\&=\sup _{\Vert \varphi \Vert =1}\left| p\int _{R^{N}}|\Delta u|^{p-2}\Delta u\Delta \varphi -p\mu \frac{|u|^{p-2}u\varphi }{|x|^{2p}}\hbox {d}x \right. \\&\quad \left. +p^*\int _{R^{N}}|\nabla u|^{p^*-2}\nabla u\nabla \varphi \hbox {d}x-p_{*}\int _{R^{N}}|u|^{p_{*}-2}u\varphi \hbox {d}x\right| \\&\le \sup _{\Vert \varphi \Vert =1} \left\{ \left| p\left( \int _{\mathbb {R}^{N}}|\Delta u|^{p}\mathrm {d}x\right) ^{\frac{p-1}{p}}\left( \int _{\mathbb {R}^{N}}|\Delta \varphi |^{p} \mathrm {d}x\right) ^{\frac{1}{p}}\right| \right. \\&\quad +{\left| {p\mu }\left( \int _{\mathbb {R}^{N}}\frac{|u|^{p}}{|x|^{2p}}\mathrm {d}x\right) ^{\frac{p-1}{p}}\left( \int _{\mathbb {R}^{N}}\frac{|\varphi |^{p}}{|x|^{2p}}\mathrm {d}x\right) ^{\frac{1}{p}}\right| }\\&\quad + \left| p^{{*}}\left( \int _{\mathbb {R}^{N}}|\nabla u|^{p^{*}}\mathrm {d}x\right) ^{\frac{p^{*}-1}{p^{*}}}\left( \int _{\mathbb {R}^{N}}|\nabla \varphi |^{p^{*}}\mathrm {d}x\right) ^{\frac{1}{p^{*}}}\right| \\&\quad \left. +\left| p_{{*}}\left( \int _{\mathbb {R}^{N}}|u|^{p_{{*}}}\mathrm {d}x\right) ^{\frac{p_{{*}}-1}{p_{{*}}}} \left( \int _{\mathbb {R}^{N}}|\varphi |^{p_{{*}}}\mathrm {d}x\right) ^{\frac{1}{p_{{*}}}}\right| \right\} \\&\le C. \end{aligned} \end{aligned}$$

Therefore, we have

$$\begin{aligned} \begin{aligned} \Vert I'(u_n)\Vert \le \Vert I'(u_n)-\lambda _n\Phi '(u_n)\Vert +|\lambda _n|\Vert \Phi '(u_n)\Vert =o(1). \end{aligned} \end{aligned}$$

It shows that $I'(u_n)\rightarrow 0$ and then $\{u_n\}\subset \aleph$ is the $(PS)_c$ sequence of I. Next, we prove that $c>0$. For any $u\in \aleph$, it follows that

$$\begin{aligned} \begin{aligned} I(u)&=I(u)-\frac{1}{p^{*}} \langle I'(u),u\rangle \\&=\frac{1}{p}\Vert u\Vert ^p+\frac{1}{p^*}\int _{\mathbb {R}^N} |\nabla {u}|^{p^*}\hbox {d}x-\frac{1}{p_*}\int _{\mathbb {R}^N}|u|^{p_*}\hbox {d}x\\&\quad -\frac{1}{p^*}\Vert u\Vert ^p-\frac{1}{p^*}\int _{\mathbb {R}^N} |\nabla {u}|^{p^*}\hbox {d}x+\frac{1}{p^*}\int _{\mathbb {R}^N}|u|^{p_*}\hbox {d}x\\&=\left( \frac{1}{p}-\frac{1}{p^{*}}\right) \Vert u\Vert ^p + \left( \frac{1}{p^{*}}-\frac{1}{p_{*}}\right) \int _{\mathbb {R}^N} |u|^{p_{*}}\mathrm {d}x\\&\geqslant \left( \frac{1}{p}-\frac{1}{p^{*}}\right) \Vert u\Vert ^p, \end{aligned} \end{aligned}$$

which implies our desired results. The proof is completed. $\square$

Lemma 2.2

Assume that the assumptions described in Theorem 1.1 hold. Let $\{u_n\}\subset \aleph$ be a $(PS)_c$ sequence of I with $c>0$. Then, there exists $C_{1}>0$ such that $\limsup \limits _{n\rightarrow \infty }\int _{\mathbb {R}^N}|u_n|^{p_{*}}\mathrm {d}x=C_{1}$.

Proof

It follows from the proof of Lemma 2.1-(iii) that $\{u_n\}$ is uniformly bounded in E. We divide the following proof into two steps.

Step 1. There is a constant $C>0$ independent of n such that $0\leqslant {Q_n}= \int _{\mathbb {R}^N}|u_n|^{p_{*}} \mathrm {d}x\leqslant C$, which means that $\{Q_n\}$ is a bounded sequence in $\mathbb {R}$. By Bolzano–Weierstrass theorem, we know that there is an accumulation point $Q_0$.

Let us define $H\subset [0,C]\subset \mathbb {R}$ be the set of all accumulation points of $\{Q_n\}$. By $Q_0\in {H}$, so $H\ne \emptyset$. It follows from the definition of the superior limit and H that $\limsup \limits _{n\rightarrow \infty }{Q_n}=\sup {H}$. Using $H\subset [0,C]$ and the supremum and infimum principle, we can get the existence of $\sup {H}$. Then, there is $C_1\in [0,C]$ such that $\limsup \limits _{n\rightarrow \infty }\int _{\mathbb {R}^N}|u_n|^{p_{*}}\mathrm {d}x=C_1$.

Step 2. We prove that $C_1>0.$ By contradiction, we assume that

$$\begin{aligned} \begin{aligned} \lim \limits _{n\rightarrow \infty } \int _{\mathbb {R}^N} |u_n|^{p_{*}} \mathrm {d}x=0. \end{aligned} \end{aligned}$$

(2.6)

From the Gagliardo–Nirenberg inequality, we have

$$\begin{aligned} \begin{aligned} \lim \limits _{n\rightarrow \infty } \int _{\mathbb {R}^N} |\nabla {u_n}|^{p^{*}} \mathrm {d}x \leqslant C\left( \lim \limits _{n\rightarrow \infty } \int _{\mathbb {R}^N} |\Delta {u_n}|^{p} \mathrm {d}x\right) ^{\frac{p^{*}}{2p}} \left( \lim \limits _{n\rightarrow \infty } \int _{\mathbb {R}^N} |{u_n}|^{p_{{*}}} \mathrm {d}x\right) ^{\frac{2p-p^{*}}{2p}}=0. \end{aligned} \end{aligned}$$

(2.7)

Combining (2.6), (2.7) and $\{u_{n}\}$ is a $(PS)_{c}$ sequence of I, we get

$$\begin{aligned} \begin{aligned} c+o(1)=\frac{1}{p}\Vert u_n\Vert ^p \end{aligned} \end{aligned}$$

and $\Vert u_n\Vert ^p=o(1)$, which indicates $c=0$. This contradicts $c>0$. Hence, we get $\limsup \limits _{n\rightarrow \infty }\int _{\mathbb {R}^N}|\nabla {u_n}|^{p_{*}}\mathrm {d}x=C_1>0$. The proof is completed. $\square$

Lemma 2.3

Assume that the assumptions described in Theorem 1.1 hold. Let $\{u_n\}\subset E$ be a $(PS)_{c}$ sequence of I at $c>0$, and $u_n\rightharpoonup 0$ weakly in E. Then, there exists $\varepsilon >0$ satisfying that

$$\begin{aligned} \begin{aligned} either~~ \lim _{n\rightarrow \infty } \int _{B_{1}(0)} |u_{n}|^{p_{*}} \mathrm {d}x=0 ~~or~~ \limsup _{n\rightarrow \infty } \int _{B_{1}(0)} |u_{n}|^{p_{*}} \mathrm {d}x\geqslant \varepsilon , \end{aligned} \end{aligned}$$

where $B_{1}(0)$ denotes a sphere with a center at 0 and radius of 1.

Proof

Let $\{u_n\}$ be a $(PS)_c$ sequence of I at $c>0$. For any $\varphi \in E$, we have

$$\begin{aligned} \begin{aligned} \langle I'(u_{n}),\varphi \rangle&= \int _{\mathbb {R}^{N}} |\Delta u_n|^{p-2} \Delta u_n\Delta \varphi \mathrm {d}x -\mu \int _{\mathbb {R}^{N}} \frac{|u_n|^{p-2}u_n\varphi }{|x|^{2p}} \mathrm {d}x \\ {}&\quad + \int _{\mathbb {R}^{N}} |\nabla u_n|^{p^{*}-2} \nabla u_n\nabla \varphi \mathrm {d}x - \int _{\mathbb {R}^{N}} |u_n|^{p_{{*}}-2}u_n\varphi \mathrm {d}x. \end{aligned} \end{aligned}$$

(2.8)

Let $\psi \in C_0^{\infty }(\mathbb {R}^{N})$ be a cutoff function satisfying $supp(\psi )=\overline{B_2(0)}$ and $\psi =1$ in $B_1(0)$. The embedding

$$\begin{aligned} \begin{aligned} E\hookrightarrow L^{r}{(\overline{B_2(0)}}) \end{aligned} \end{aligned}$$

is compact for all $r\in [2,p_{{*}})$.

Step 1. According to Rellich’s compactness theorem, Sobolev’s inequality and Hölder’s inequality, we obtain

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^{N}} |\Delta u_n|^{p-2}\Delta u_n \Delta (\psi ^{p}u_n) \mathrm {d}x= \int _{\mathbb {R}^{N}} |\Delta (\psi u_n)|^{p} \mathrm {d}x+o(1) \end{aligned} \end{aligned}$$

(2.9)

and

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^{N}} |\nabla u_n|^{p^{*}-2} \nabla u_n\nabla (\psi ^{p}u_n) \mathrm {d}x = \int _{\mathbb {R}^{N}} \psi ^{p}|\nabla u_n|^{p^{*}} \mathrm {d}x +o(1). \end{aligned} \end{aligned}$$

(2.10)

According to Hölder’s inequality, one gets

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^{N}}|u_n|^{p_{{*}}}\psi ^{p}\hbox {d}x&=\int _{\mathbb {R}^{N}}(\psi u_n)^{p}|u_n|^{p_{*}-p}\hbox {d}x \leqslant \int _{B_2(0)}|\psi u_n|^{p}|u_n|^{p_{*}-p}\hbox {d}x\\&\leqslant \left( \int _{B_2(0)} |\psi u_n|^{p_{{*}}} \mathrm {d}x\right) ^{\frac{p}{p_{*}}} \left( \int _{B_2(0)} |u_n|^{p_{*}}\mathrm {d}x \right) ^{\frac{p_{{*}}-p}{p_{*}}}\\&\le \Vert \psi u_{n}\Vert _{L^{p_{*}}(\mathbb {R}^{N})}^{p}\left( \int _{B_2(0)}|u_n|^{p_{*}}\hbox {d}x\right) ^{\frac{p_{*}-p}{p_{*}}}\\&\leqslant \frac{1}{S_\mu } \Vert \psi u_n\Vert ^{p} \Vert u_n\Vert _{L^{p_{{*}}}(B_2(0))} ^{p_{{*}}-p}. \end{aligned} \end{aligned}$$

(2.11)

We choose $\varphi =\psi ^{p}u_n$ in (2.8), and there holds

$$\begin{aligned} \begin{aligned} \langle I'(u_n),\psi ^{p}u_n\rangle&= \int _{\mathbb {R}^{N}} |\Delta u_n|^{p-2} \Delta u_n\Delta (\psi ^{p}u_n) \mathrm {d}x -\mu \int _{\mathbb {R}^{N}} \frac{|u_n|^{p-2}u_n(\psi ^{p}u_n)}{|x|^{2p}}\mathrm {d}x\\&\quad + \int _{\mathbb {R}^{N}} |\nabla u_n|^{p^{*}-2} \nabla u_n\nabla (\psi ^{p}u_n) \mathrm {d}x - \int _{\mathbb {R}^{N}} |u_n|^{p_{*}-2}u_n(\psi ^{p}u_n) \mathrm {d}x. \end{aligned} \end{aligned}$$

Applying (2.9)–(2.11) and

$$\begin{aligned} \begin{aligned} \Vert \psi u_n\Vert =\left( \int _{\mathbb {R}^{N}} |\Delta (\psi u_n)|^{p} \mathrm {d}x -\mu \int _{\mathbb {R}^{N}} \frac{|\psi u_n|^{p}}{|x|^{2p}}\mathrm {d}x\right) ^{\frac{1}{p}}, \end{aligned} \end{aligned}$$

one has

$$\begin{aligned} \begin{aligned} \frac{1}{S_\mu } \Vert \psi u_n\Vert ^{p} \Vert u_n\Vert _{L^{p_{*}}(B_2(0))} ^{p_{{*}}-p}&\ge C\Vert \psi u_n\Vert ^{p} + \int _{\mathbb {R}^{N}} \psi ^{p}|\nabla u_n|^{p^{*}} \mathrm {d}x+o(1)\\&\ge C\Vert \psi u_n\Vert ^{p}+o(1). \end{aligned} \end{aligned}$$

(2.12)

Step 2. In this step, we split our following proof into two aspects: (I) $\limsup \limits _{n\rightarrow \infty }\Vert \psi u_n\Vert >0$ and (II) $\lim \limits _{n\rightarrow \infty }\Vert \psi u_n\Vert =0$.

Case (I). According to $\limsup \limits _{n\rightarrow \infty }\Vert \psi u_n\Vert >0$ and (2.12), we get

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{{*}}} \mathrm {d}x \geqslant \left( C_2S_\mu \right) ^{\frac{p_{*}}{p_{*}-p}} >0. \end{aligned} \end{aligned}$$

(2.13)

Similar to Step 1 of Lemma 2.2, there exists $0\leqslant C_3<\infty$ such that

$$\begin{aligned} \begin{aligned} C_3= \limsup \limits _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{{*}}} \mathrm {d}x. \end{aligned} \end{aligned}$$

In view of (2.13), we get $C_3>0$. Set $D_1:= \limsup \limits _{n\rightarrow \infty } \int _{B_2(0)\backslash \overline{B_1(0)}}|u_n|^{p_{*}} \mathrm {d}x$, we have

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{{*}}} \mathrm {d}x \leqslant D_1+ \limsup _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{{*}}} \mathrm {d}x. \end{aligned} \end{aligned}$$

(2.14)

According to the range of $D_{1}$, there are three subcases.

Case (I-1). If $D_1=0$, by (2.14), then

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{{*}}} \mathrm {d}x = \limsup _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{*}} \mathrm {d}x=C_3>0. \end{aligned} \end{aligned}$$

(2.15)

Case (I-2). If $D_1\in (0,C_3)$, then there exists $C_4=C_3-D_1>0$ satisfying

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{*}} \mathrm {d}x \geqslant C_4>0. \end{aligned} \end{aligned}$$

(2.16)

Case (I-3). If $D_1=C_3= \limsup \limits _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{{*}}} \mathrm {d}x$. Then, we have the following two subsubcases: (1) $\lim \limits _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{{*}}} \mathrm {d}x$ exists, and (2) $\lim \limits _{n\rightarrow \infty }\int _{B_1(0)} |u_n|^{p_{{*}}}\mathrm {d}x$ does not exist. If (1) happens, then (2.14) turns into

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{{*}}} \mathrm {d}x = D_1+ \limsup _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{{*}}} \mathrm {d}x. \end{aligned} \end{aligned}$$

Substituting $D_1=C_3= \limsup \limits _{n\rightarrow \infty } \int _{B_2(0)} |u_n|^{p_{*}}\mathrm {d}x$ into above equality, we can see that

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{{*}}} \mathrm {d}x = \lim _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{*}} \mathrm {d}x=0. \end{aligned} \end{aligned}$$

(2.17)

If (2) happens, it follows that $\limsup \limits _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{*}} \mathrm {d}x> \liminf \limits _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{*}}\mathrm {d}x \geqslant 0$, which indicates that there exists $C_5>0$ such that

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{*}}\mathrm {d}x =C_5>0. \end{aligned} \end{aligned}$$

(2.18)

Case (II). From $\lim \limits _{n\rightarrow \infty }\Vert \psi u_n\Vert =0$ and Sobolev’s inequality, we get

$$\begin{aligned} \begin{aligned} 0=\lim _{n\rightarrow \infty } \frac{\Vert \psi u_n\Vert ^{p}}{S_\mu } \geqslant \lim _{n\rightarrow \infty } \left( \int _{\mathbb {R}^{N}} |\psi u_n|^{p_{*}}\mathrm {d}x\right) ^{\frac{p}{p_{*}}} \geqslant \lim _{n\rightarrow \infty } \left( \int _{B_1(0)} |u_n|^{p_{*}}\mathrm {d}x \right) ^{\frac{p}{p_{*}}}, \end{aligned} \end{aligned}$$

which indicates

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty } \int _{B_1(0)} |u_n|^{p_{*}} \mathrm {d}x=0. \end{aligned} \end{aligned}$$

(2.19)

In conclusion, setting $\varepsilon =\min \{C_3,C_4,C_5\}$ and combining (2.15)–(2.19), we can deduce

$$\begin{aligned} \begin{aligned} \mathrm {either}~~ \lim _{n\rightarrow \infty } \int _{B_{1}(0)} |u_{n}|^{p_{*}} \mathrm {d}x=0 ~~\mathrm {or}~~ \limsup _{n\rightarrow \infty } \int _{B_{1}(0)} |u_{n}|^{p_{*}} \mathrm {d}x\geqslant \varepsilon , \end{aligned} \end{aligned}$$

The proof is completed. $\square$

In order to obtain our main result, we also give the following general version Brezis–Lieb lemma.

Lemma 2.4

(Brezis–Lieb Lemma, [17]) Let $\Omega$ be an open subset of $\mathbb {R}^{N}$, and let $\{u_n\}\subset L^{p}(\Omega )$, $1\le p<\infty$. If

(1) $\{u_n\}$ is bounded in $L^{p}(\Omega )$;

(2) $u_n\rightarrow u_0$ almost everywhere in $\Omega$,

then

$$\begin{aligned} \begin{aligned} \lim _{n\rightarrow \infty }(\Vert u_n\Vert _{L^p}^{p}-\Vert u_n-u_0\Vert _{L^p}^{p})=\Vert u_0\Vert _{L^p}^{p}. \end{aligned} \end{aligned}$$

(2.20)

The proof of Theorem 1.1

In view of Lemma 2.2, we have

$$\begin{aligned} \begin{aligned} \limsup _{n\rightarrow \infty } \int _{\mathbb {R}^{N}} |u_n|^{p_{*}} \mathrm {d}x =C_1>0. \end{aligned} \end{aligned}$$

(2.21)

Set $\delta =\min \{C_1,\frac{\varepsilon }{2}\}$, where $\varepsilon >0$ is taken in Lemma 2.3. By (2.21), there exists a sequence $\{r_n\}\subset \mathbb {R}^{+}$ such that for any $\delta '\in (0,\delta )$, one has $\limsup \limits _{n\rightarrow \infty } \int _{B_{r_n}(0)} |u_n|^{p_{*}} \mathrm {d}x=\delta '$. Define $\overline{u_n}:=r_{n}^{\frac{N-2p}{p}}u_n(r_nx)$. Then, $\overline{u_n}\in E$ and

$$\begin{aligned} \begin{aligned} I(\overline{u_n}) \rightarrow c, \ \ I'(\overline{u_n}) \rightarrow 0 \ \, \mathrm {and} \ \, \int _{B_1(0)} |\overline{u_n}|^{p_{*}} \mathrm {d}x=\delta ', \end{aligned} \end{aligned}$$

(2.22)

as $n\rightarrow \infty$. By (2.22), it is easy to check $\{\overline{u_n}\}$ is bounded in E. Without loss of generality, we suppose there exists $\overline{u}\in E$ such that $u_n\rightharpoonup u$ in E.

We next prove $\overline{u}\not \equiv 0$. Argue by contradiction. Let $u\equiv 0$. It follows from Lemma 2.3 that we have

$$\begin{aligned} \begin{aligned} \mathrm {either}~~\lim \limits _{n\rightarrow \infty }\int _{B_{1}(0)} |u_{n}|^{p_{*}}\mathrm {d}x=0 ~~\mathrm {or}~~\limsup \limits _{n\rightarrow \infty }\int _ {B_{1}(0)}|u_{n}|^{p_{*}}\mathrm {d}x \geqslant \varepsilon , \end{aligned} \end{aligned}$$

which contradicts to (2.22) since $0<\delta '<\delta =\min \{C_1,\frac{\varepsilon }{2}\}$.

Now, we prove $\overline{u_n}\rightarrow \overline{u}$ strongly in E. It follows from $\lim \limits _{n\rightarrow \infty }\langle I'(\overline{u_n}),\varphi \rangle =o(1)$, $\overline{u_n}\rightharpoonup \overline{u}$ in E and Lemma 2.1 that

$$\begin{aligned} \begin{aligned} \langle I'(\overline{u}),\varphi \rangle =0 \ \ \mathrm {and} \ \ \overline{u}\in \aleph . \end{aligned} \end{aligned}$$

Set

$$\begin{aligned} \begin{aligned} F(u):= \left( \frac{1}{p}-\frac{1}{p_{*}}\right) \Vert u_n\Vert ^{2} + \left( \frac{1}{p^{*}}-\frac{1}{p_{*}}\right) \int _{R^{N}} |\nabla u_n|^{p^{*}} \mathrm {d}x. \end{aligned} \end{aligned}$$

According to Lemma 2.4, Fatou lemma and $\overline{c}=c$, we have

$$\begin{aligned} c+o(1)&= \lim _{n\rightarrow \infty } I(\overline{u_n}) - \lim _{n\rightarrow \infty } \frac{1}{p_{*}} \langle I'(\overline{u_n}),\overline{u_n}\rangle = \lim _{n\rightarrow \infty } F(\overline{u_n})\nonumber \\&\geqslant F(\overline{u}) = I(\overline{u}) \geqslant \overline{c}=c. \end{aligned}$$

(2.23)

Thus, the above inequalities will be equalities. Applying $\lim \limits _{n\rightarrow \infty }F(\overline{u_n})=F(\overline{u})$ and Lemma 2.4 again, we get $\lim \limits _{n\rightarrow \infty }F(\overline{u_n})-\lim \limits _{n\rightarrow \infty }F(\overline{u_n}-\overline{u})=F(\overline{u})+o(1).$ So $\lim \limits _{n\rightarrow \infty }F(\overline{u_n}-\overline{u})=0$, which implies $\overline{u_n}\rightarrow \overline{u}$ strongly in E. Applying (2.23) again, we have $I'(\overline{u})=c$. Thus, $\overline{u}$ is a ground state solution of Eq. (1.1). $\square$