Infinitely many solutions for p-harmonic equation with singular term

Abstract

In this paper, we study the following p-harmonic problem involving the Hardy term:

$$\mathrm{\Delta }\left({|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u\right)-\lambda \frac{{|u|}^{p-2}u}{{|x|}^{2p}}=f(x,u),\text{in }\mathrm{\Omega },u=\frac{\partial u}{\partial n}=0\text{on }\partial \mathrm{\Omega },$$

where Ω is an open bounded domain containing the origin in ${\mathbb{R}}^N$, $1<p<\frac{N}{2}$ and $0\le \lambda <{[N(p-1)(N-2p)/p^2]}^p$. By using the variational method, we prove that the above problem has infinitely many solutions with positive energy levels.

AMS Subject Classification:35J60, 35J65.

1 Introduction

The main purpose of this paper is to show the existence of infinitely many solutions for the following p-harmonic equation:

$$\{\begin{array}{cc}\mathrm{\Delta }({|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u)-\lambda \frac{{|u|}^{p-2}u}{{|x|}^{2p}}=f(x,u),\hfill & x\in \mathrm{\Omega },\hfill \\ u=\frac{\partial u}{\partial n}=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}$$

(1.1)

where Ω is an open bounded domain containing the origin in ${\mathbb{R}}^N$, the boundary ∂ Ω is smooth. $1<p<\frac{N}{2}$, $0\le \lambda <\overline{\lambda }={[N(p-1)(N-2p)/p^2]}^p$. $\frac{\partial }{\partial n}$ is the outer normal derivative. Nonlinearity $f(x,u)$ satisfies the following conditions:

(f₁) $f(x,u)$ is continuous on $\mathrm{\Omega }\times \mathbb{R}$ and limits subcritical growing at infinity; that is,

$$\underset{u\to \mathrm{\infty }}{lim}\frac{f(x,u)}{{|u|}^{p^{\ast }-1}}=0,\text{uniformly for }x\in \mathrm{\Omega },$$

(1.2)

where $p^{\ast }=Np/(N-2p)$ is the critical exponent of Sobolev’s embedding $W_0^{2,p}(\mathrm{\Omega })\hookrightarrow L^s(\mathrm{\Omega })$.

(f₂) For any $x\in \mathrm{\Omega }$, $f(x,u)$ satisfies

$$\underset{u\to 0}{lim}\frac{f(x,u)u}{{|u|}^p}=0\text{and}\underset{u\to \mathrm{\infty }}{lim}\frac{f(x,u)u}{{|u|}^p}=+\mathrm{\infty }.$$

(1.3)

(f₃) Denote $G(x,t)=f(x,t)t-p{\int }_0^{t}f(x,s)ds$. For all $x\in \mathrm{\Omega }$, there exists a constant $M\ge 0$ such that

$$G(x,t_1)\le G(x,t_2)\text{for any }M\le |t_1|\le |t_2|.$$

(1.4)

(f₄) $f(x,u)$ is odd with respect to u.

By the Hardy-Rellich inequality (see [1, 2]), we know that

$${\int }_{\mathrm{\Omega }}\frac{{|u|}^p}{{|x|}^{2p}}dx\le \frac{1}{\overline{\lambda }}{\int }_{\mathrm{\Omega }}{|\mathrm{\Delta }u|}^{p}dx.$$

(1.5)

Obviously, for any $\lambda \in [0,\overline{\lambda })$,

$$(1-\frac{\lambda }{\overline{\lambda }}){\int }_{\mathrm{\Omega }}{|\mathrm{\Delta }u|}^{p}dx\le {\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}})dx\le {\int }_{\mathrm{\Omega }}{|\mathrm{\Delta }u|}^{p}dx.$$

In $W_0^{2,p}(\mathrm{\Omega })$, for$\lambda \in [0,\overline{\lambda })$, we define

$$\parallel u\parallel ={\parallel u\parallel }_{W_0^{2,p}(\mathrm{\Omega })}:={\left({\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}})dx\right)}^{\frac{1}{p}},$$

this norm is equivalent to ${({\int }_{\mathrm{\Omega }}{|\mathrm{\Delta }u|}^{p}dx)}^{1/p}$.

A weak solution of the problem (1.1) is a critical point of the energy functional

$$I(u)=\frac{1}{p}{\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}})dx-{\int }_{\mathrm{\Omega }}F(x,u)dx,u\in W_0^{2,p}(\mathrm{\Omega }),$$

(1.6)

where $F(x,u)={\int }_0^{u}f(x,s)ds$. It is easy to check that $I(u)$ is a continuous even functional.

Biharmonic equations can describe the static form change of a beam or the sport of a rigid body. For example, this type of equation furnishes a model for studying traveling wave in suspension bridges (see [3]). By using variational arguments, many authors investigated nonlinear biharmonic equations under Dirichlet boundary conditions or Navier boundary conditions and got interesting results (see [4–10]).

Li and Squassina in [11] considered the superlinear p-harmonic equation with Navier boundary conditions

$$\mathrm{\Delta }\left({|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u\right)=f(x,u),\text{in }\mathrm{\Omega },u=\mathrm{\Delta }u=0\text{on }\partial \mathrm{\Omega },$$

(1.7)

where Ω is an open bounded domain in ${\mathbb{R}}^N$ with a smooth boundary ∂ Ω. $p>1$, $N\ge 2p+1$ and $f:\mathrm{\Omega }\times \mathbb{R}\to \mathbb{R}$ is a Carathéodory function such that for some positive constant C,

$$|f(x,u)|\le C(1+{|u|}^{q-1})\text{for all }q\in [1,p^{\ast }).$$

(1.8)

By means of the Morse theory, they proved the existence of two nontrivial solutions to (1.7). After that, Li and Tang [12] considered a more general problem than (1.7). They got three solutions by the three critical points theorem which was obtained in [13]. If $f(x,u)=\lambda g(x,u)$, Candito and Bisci [14] established a well-determined interval of values of the parameter λ for which the problem (1.7) admits at least two distinct weak solutions. The authors in [15] considered the p-harmonic equation with Dirichlet conditions and obtained the existence of a nontrivial solution. Under the condition of (1.8), [16] proved the existence and multiplicity of weak solutions for the nonuniformly nonlinear problem

$$\mathrm{\Delta }(a(x,\mathrm{\Delta }u))=f(x,u),\text{in }\mathrm{\Omega },u=\frac{\partial u}{\partial n}=0\text{on }\partial \mathrm{\Omega },$$

(1.9)

where $|a(x,t)|\le c_0(h_0(x)+h_1(x){|t|}^{p-1})$ with $h_0(x)\ge 0$, $h_1(x)\ge 1$. $a(x,t)$ and $f(x,t)$ are odd with respect to the second variable. Furthermore, $f(x,t)$ is subcritical and satisfies the Ambrosetti-Rabinowitz condition, that is, there exists a constant $\theta >p$ such that

$$0<\theta {\int }_0^{t}f(x,s)ds\le f(x,t)t\text{for any }x\in \mathrm{\Omega }.$$

(1.10)

The aim of this paper is to obtain infinitely many solutions for the problem (1.1) when $2p<N$. The main difficulty lies in the fact that the embeddings $W_0^{2,p}(\mathrm{\Omega })\hookrightarrow L^{p^{\ast }}(\mathrm{\Omega })$ and $W_0^{2,p}(\mathrm{\Omega })\hookrightarrow L^p(\mathrm{\Omega },{|x|}^{-2p}dx)$ are not compact. Furthermore, the assumption (f₁) is not the usual subcritical growth (1.8). The condition (f₃) is weaker than the A-R condition (1.10). We use the concentration compactness principle (see [17, 18]) to overcome those difficulties. The following theorem is our main result.

Theorem 1.1 Assume $f(x,u)$ satisfies (f₁)-(f₄), then the problem (1.1) possesses infinitely many weak solutions and the corresponding critical values are positive.

This paper proceeds as follows. In the next section, we prove the energy functional $I(u)$ satisfies the Palais-Smale condition. In Section 3, by using the symmetric mountain pass theorem (see [19]), we get the main result of this paper. Throughout the paper, denote

$$S_{p,\lambda }:=\underset{u\in W_0^{2,p}(\mathrm{\Omega })\setminus \{0\}}{inf}\frac{{\parallel u\parallel }^p}{{\parallel u\parallel }_{p^{\ast }}^p}.$$

We use ${\parallel u\parallel }_q={({\int }_{\mathrm{\Omega }}{|u|}^{q}dx)}^{1/q}$ to denote the norm of $L^q(\mathrm{\Omega })$, C, $C_i$ stand for universal constants. We omit dx and Ω in the integrals if there is no other indication.

2 Palais-Smale condition

To show the (PS) sequence $\{u_n\}$ of the variational functional $I(u)$ is compact in $W_0^{2,p}(\mathrm{\Omega })$, we first prove the boundedness of $\{u_n\}$ by the analytic argument which has been used in [20]. Then, using the concentration compactness principle, we get the compactness of $\{u_n\}$.

Lemma 2.1 The condition (f₁) implies that for any $\epsilon >0$, there exists a positive constant $C_{\epsilon }$ such that

$$F(x,u)\le \epsilon {|u|}^{p^{\ast }}+C_{\epsilon }|u|,(x,u)\in \mathrm{\Omega }\times \mathbb{R}.$$

(2.1)

(f₂) implies that $F(x,u)=o(u^p)$ as $u\to 0$. Furthermore, for any $|u|\ge M>0$, there exists a small positive constant θ

$$F(x,u)\ge C{|u|}^{p+\theta }\mathit{\text{and}}f(x,u)u>0.$$

(2.2)

Lemma 2.2 Under the conditions (f₁), (f₂) and (f₃), if the sequence $\{u_n\}\in W_0^{2,p}(\mathrm{\Omega })$ satisfies

$$I(u_n)\to c,I^{\prime }(u_n)\to 0\mathit{\text{as }}n\to +\mathrm{\infty },$$

(2.3)

then the sequence $\{u_n\}$ is bounded in $W_0^{2,p}(\mathrm{\Omega })$.

Proof We prove this lemma by contradiction. Without loss of generality, we assume that

$$\parallel u_n\parallel \to +\mathrm{\infty }\text{as }n\to +\mathrm{\infty }.$$

(2.4)

Set $w_n=u_n/\parallel u_n\parallel$, then $\parallel w_n\parallel =1$ for any $n\in \mathbb{N}$. As $n\to +\mathrm{\infty }$, there exists $w\in W_0^{2,p}(\mathrm{\Omega })$ such that

$$\begin{array}{c}{w}_n\rightharpoonup w\text{weakly in }{W}_0^{2,p}(\mathrm{\Omega }),\hfill \\ w_n\to w\text{strongly in }{L}^s(\mathrm{\Omega })\text{ for any }1\le s<p^{\ast },\hfill \\ w_n\to w\text{a.e. }x\in \mathrm{\Omega }.\hfill \end{array}$$

If $w\not\equiv 0$, we denote ${\mathrm{\Omega }}_0=\{x\in \mathrm{\Omega }:w=0\}$, set $\mathrm{\Omega }\setminus {\mathrm{\Omega }}_0$ is nonempty. (2.3) implies that

$${\parallel u_n\parallel }^p-{\int }_{\mathrm{\Omega }}f(x,u_n)u_{n}dx=o(1).$$

Therefore,

$$1-o(1)={\int }_{\mathrm{\Omega }}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}dx={\int }_{{\mathrm{\Omega }}_0}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}dx+{\int }_{\mathrm{\Omega }\setminus {\mathrm{\Omega }}_0}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}dx.$$

(2.5)

As $n\to +\mathrm{\infty }$, ${\int }_{\mathrm{\Omega }}{|u_n|}^{p}dx\le \parallel u_n\parallel \to 0$. So, $|u_n|\to +\mathrm{\infty }$ for $x\in \mathrm{\Omega }\setminus {\mathrm{\Omega }}_0$. It follows from (f₂) that

$$\underset{u_n\to \mathrm{\infty }}{lim}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}=\underset{u_n\to \mathrm{\infty }}{lim}\frac{f(x,u_n)u_n}{{|u_n|}^p}{|w_n|}^p=+\mathrm{\infty }.$$

Since $|\mathrm{\Omega }\setminus {\mathrm{\Omega }}_0|>0$, then

$${\int }_{\mathrm{\Omega }\setminus {\mathrm{\Omega }}_0}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}dx\to +\mathrm{\infty }\text{as }n\to +\mathrm{\infty }.$$

(2.6)

Using (2.1) and (2.2), we derive that

$$\begin{array}{rcl}{\int }_{{\mathrm{\Omega }}_0}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}dx& =& \frac{1}{{\parallel u_n\parallel }^p}({\int }_{{\mathrm{\Omega }}_0(|u_n|>M)}\cdots +{\int }_{{\mathrm{\Omega }}_0(|u_n|\le M)}\cdots )\\ \ge & -\frac{1}{{\parallel u_n\parallel }^p}{\int }_{{\mathrm{\Omega }}_0(|u_n|\le M)}|f(x,u_n)u_n|dx\\ \ge & -\frac{C}{{\parallel u_n\parallel }^p}(M+M^{p^{\ast }})|{\mathrm{\Omega }}_0|\\ \ge & -C_1\text{as }n\to +\mathrm{\infty }.\end{array}$$

(2.7)

From (2.6) and (2.7), we get

$${\int }_{\mathrm{\Omega }}\frac{f(x,u_n)u_n}{{\parallel u_n\parallel }^p}dx\to +\mathrm{\infty }\text{as }n\to +\mathrm{\infty }.$$

(2.8)

This contradicts (2.5).

If $w\equiv 0$, then $w_n\to 0$ strongly in $L^1(\mathrm{\Omega })$. Since $I(u_n)\to c$, we know that

$$c\leftarrow \frac{1}{p}{\parallel u_n\parallel }^p-\int F(x,u_n)dx\ge \frac{1}{p}{\parallel u_n\parallel }^p-\epsilon \int {|u_n|}^{p^{\ast }}dx-C_{\epsilon }\int |u_n|dx.$$

(2.9)

In (2.9), choose $\epsilon =1$, then

$$\frac{1}{p}{\parallel u_n\parallel }^p\le {\parallel u_n\parallel }_{p^{\ast }}^{p^{\ast }}+C_2{\parallel u_n\parallel }_1+c\le C_3{\parallel u_n\parallel }_{p^{\ast }}^{p^{\ast }}+c.$$

From (2.4), it follows that ${\parallel u_n\parallel }_{p^{\ast }}\to +\mathrm{\infty }$ as $n\to +\mathrm{\infty }$. Denote

$$A_n=M\frac{\parallel u_n\parallel }{{\parallel u_n\parallel }_{p^{\ast }}}\ge M{S}_{p,\lambda }^{1/p},$$

(2.10)

where M is any fixed positive constant. It is clear that

$$\frac{A_n}{\parallel u_n\parallel }=\frac{M}{{\parallel u_n\parallel }_{p^{\ast }}}\to 0(\text{as }n\to +\mathrm{\infty }).$$

Therefore, for n large enough, $A_n/\parallel u_n\parallel \in (0,1)$.

For every $n\in \mathbb{N}$, we define a sequence of $t_n$ as follows:

$$I(t_n{u}_n)=\underset{t\in [0,1]}{max}I(t{u}_n).$$

We claim that

$$I(t_n{u}_n)\to +\mathrm{\infty }(\text{as }n\to +\mathrm{\infty }).$$

(2.11)

In fact, by (2.1)-(2.10) and the definition of $t_n$, we have

$$\begin{array}{rcl}I(t_n{u}_n)& \ge & I\left(\frac{A_n}{\parallel u_n\parallel }{u}_n\right)\ge \frac{1}{p}{A}_n^p-\epsilon \int {\left|\frac{A_n}{\parallel u_n\parallel }{u}_n\right|}^{p^{\ast }}dx-C_{\epsilon }{A}_n\int |w_n|dx\\ =& \frac{1}{p}{A}_n^p-\epsilon M^{p^{\ast }}-C_{\epsilon }{A}_n\int |w_n|dx.\end{array}$$

If $A_n\to +\mathrm{\infty }$ as $n\to +\mathrm{\infty }$, then (2.11) follows from the above estimate. If $A_n$ is bounded for any n, that is, there exists a constant $K>0$ such that $A_n\le K$, then

$$\begin{array}{rcl}I(t_n{u}_n)& \ge & \frac{1}{p}{A}_n^p-\epsilon M^{p^{\ast }}-C_{\epsilon }{A}_n\int |w_n|dx\ge \frac{1}{p}{S}_{p,\lambda }{M}^p-\epsilon M^{p^{\ast }}-C_{\epsilon }K\int |w_n|dx\\ \to & \frac{1}{p}{S}_{p,\lambda }{M}^p-\epsilon M^{p^{\ast }}\text{as }n\to +\mathrm{\infty },\end{array}$$

where we use the fact that ${\parallel w_n\parallel }_1\to 0$ as $n\to +\mathrm{\infty }$. The positive constant ε is arbitrary, thus

$$\underset{n\to +\mathrm{\infty }}{lim}I(t_n{u}_n)\ge \frac{1}{p}{S}_{p,\lambda }{M}^p\text{for any positive constant }M.$$

(2.11) follows from the above inequality.

From (2.3), we know that

$$pc+o(1)=pI(u_n)-〈I^{\prime }(u_n),u_n〉=\int (f(x,u_n)u_n-pF(x,u_n))dx.$$

(2.12)

By the definition of $t_n$, we obtain $\frac{d}{dt}I(t{u}_n){|}_{t=t_n}=0$ and $t_n\in [0,1]$, then $|t_n{u}_n|\le |u_n|$. The condition (f₃) and (2.11) derive that

$$\begin{array}{rcl}\int (f(x,u_n)u_n-pF(x,u_n))dx& \ge & \int (f(x,t_n{u}_n)t_n{u}_n-pF(x,t_n{u}_n))dx\\ =& pI(t_n{u}_n)-〈I^{\prime }(t_n{u}_n),t_n{u}_n〉\\ =& pI(t_n{u}_n)-t_n\frac{d}{dt}I(t{u}_n){|}_{t=t_n}\\ \to & +\mathrm{\infty }.\end{array}$$

Which is a contradiction to (2.12). Therefore, the sequence $\{u_n\}$ is bounded in $W_0^{2,p}(\mathrm{\Omega })$. □

Under the conditions of Lemma 2.2, we know $\parallel u_n\parallel \le C$. Therefore, there exists a subsequence, still denoted by $\{u_n\}$, and some $u\in W_0^{2,p}(\mathrm{\Omega })$ such that

$$\begin{array}{c}{u}_n\rightharpoonup u\text{weakly in }{W}_0^{2,p}(\mathrm{\Omega }),\hfill \\ u_n\rightharpoonup u\text{weakly in }{L}^p(\mathrm{\Omega },{|x|}^{-2p}dx),\hfill \\ u_n\to u\text{strongly in }{L}^s(\mathrm{\Omega })\text{ for any }1\le s<p^{\ast },\hfill \\ u_n\to u\text{a.e. }x\in \mathrm{\Omega }.\hfill \end{array}$$

Obviously, u is a weak solution of (1.1). Now we prove that $u\not\equiv 0$. According to the concentration compactness principle (see [17, 18]), there exists a subsequence, still denoted by $\{u_n\}$, at most countable set J, a set of different points ${\{x_j\}}_{j\in J}\subset \overline{\mathrm{\Omega }}\setminus \{0\}$ and two positive number sequences ${\{{\mu }_j\}}_{j\in J\cup \{0\}}$, ${\{{\nu }_j\}}_{j\in J\cup \{0\}}$ such that

$$\begin{array}{c}{|\mathrm{\Delta }{u}_n|}^p\rightharpoonup d\mu \ge {|\mathrm{\Delta }u|}^p+\sum _{j\in J}{\mu }_j{\delta }_{x_j}+{\mu }_0{\delta }_0\text{weakly in the sense of measure},\hfill \\ {|u_n|}^{p^{\ast }}\rightharpoonup d\nu ={|u|}^{p^{\ast }}+\sum _{j\in J}{\nu }_j{\delta }_{x_j}+{\nu }_0{\delta }_0\text{weakly in the sense of measure},\hfill \\ \frac{{|u_n|}^p}{{|x|}^{2p}}\rightharpoonup d\gamma =\frac{{|u|}^p}{{|x|}^{2p}}+{\gamma }_0{\delta }_0\text{weakly in the sense of measure},\hfill \\ S_{p,\lambda }{\nu }_0^{\frac{p}{p^{\ast }}}\le {\mu }_0-\lambda {\gamma }_0,\text{and}{S}_{p,0}{\nu }_j^{\frac{p}{p^{\ast }}}\le {\mu }_j\text{for }j\in J,\hfill \end{array}$$

where ${\delta }_{x_j}$ is the unit Dirac measure at $x_j$.

Lemma 2.3 Assume $f(x,u)$ satisfies (f₁), the sequence $\{u_n\}$ is bounded in $W_0^{2,p}(\mathrm{\Omega })$ and satisfies (2.3). Then the set $J=\{x_1,x_2,\dots ,x_j\}\subset \overline{\mathrm{\Omega }}$ is finite, ${\mu }_0-\lambda {\gamma }_0=0$.

Proof We first prove that ${\mu }_j=0$ for any $x_j\in J$.

Let $\epsilon >0$ be small enough such that $0\notin B_{\epsilon }(x_j)$ and $B_{\epsilon }(x_i)\cap B_{\epsilon }(x_j)=\mathrm{\varnothing }$ for $i\ne j$. ${\varphi }_j(x)\in [0,1]$ is a cutting-off function in ${\mathcal{C}}_0^{\mathrm{\infty }}(\mathrm{\Omega })$. ${\varphi }_j(x)\equiv 1$ for $|x-x_j|<\epsilon /2$, ${\varphi }_j(x)\equiv 0$ for $|x-x_j|\ge \epsilon$, and $|\mathrm{\Delta }{\varphi }_j|\le 4/{\epsilon }^2$. Since

$$\begin{array}{rcl}〈I^{\prime }(u_n),u_n{\varphi }_j〉& =& \int {|\mathrm{\Delta }{u}_n|}^p{\varphi }_{j}dx+2\int {|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n\mathrm{\nabla }{u}_n\mathrm{\nabla }{\varphi }_{j}dx\\ +\int u_n{|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n\mathrm{\Delta }{\varphi }_{j}dx\\ -\int \lambda \frac{{|u_n|}^p}{{|x|}^{2p}}{\varphi }_{j}dx-\int f(x,u_n)u_n{\varphi }_{j}dx,\end{array}$$

(2.13)

we deduce that

(2.14)

(2.15)

By Hölder’s inequality,

$$\begin{array}{c}\underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}|\int {|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n\mathrm{\nabla }{u}_n\mathrm{\nabla }{\varphi }_{j}dx|\hfill \\ \le \underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}{\parallel \mathrm{\Delta }{u}_n\parallel }_{L^p(B_{\epsilon }(x_j))}^{p-1}{\parallel \mathrm{\nabla }{u}_n\parallel }_{L^{Np/(N-p)}(B_{\epsilon }(x_j))}{\parallel \mathrm{\nabla }{\varphi }_j\parallel }_{L^N(B_{\epsilon }(x_j))}\hfill \\ \le C\underset{\epsilon \to 0}{lim}{\parallel \mathrm{\Delta }{u}_n\parallel }_{L^p(B_{\epsilon }(x_j))}^p=0,\hfill \\ \underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}|\int {|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n{u}_n\mathrm{\Delta }{\varphi }_{j}dx|\hfill \\ \le \underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}{\parallel \mathrm{\Delta }{u}_n\parallel }_{L^p(B_{\epsilon }(x_j))}^{p-1}{\parallel u_n\parallel }_{L^{p^{\ast }}(B_{\epsilon }(x_j))}{\parallel \mathrm{\Delta }{\varphi }_j\parallel }_{L^{N/2}(B_{\epsilon }(x_j))}\hfill \\ \le C\underset{\epsilon \to 0}{lim}{\parallel u_n\parallel }_{L^{p^{\ast }}(B_{\epsilon }(x_j))}=0.\hfill \end{array}$$

Using (f₁), (2.14) and (2.15), we have

$$\underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}\int |f(x,u_n)u_n{\varphi }_j|dx\le \underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}(\epsilon \int {|u_n|}^{p^{\ast }}{\varphi }_{j}dx+C_{\epsilon }\int |u_n|{\varphi }_{j}dx)=0.$$

By the above estimates, (2.13) becomes

$$0=\underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}〈I^{\prime }(u_n),u_n{\varphi }_j〉\ge {\mu }_j.$$

(2.16)

Therefore, for any $j\in J$, ${\mu }_j=0$, which implies that J is finite.

If the concentration is at the origin, let ${\varphi }_0(x)\in [0,1]$ be a cutting-off function in ${\mathcal{C}}_0^{\mathrm{\infty }}(\mathrm{\Omega })$. ${\varphi }_0(x)\equiv 1$ for $|x|<\epsilon /2$, ${\varphi }_0(x)\equiv 0$ for $|x|\ge \epsilon$, and $|\mathrm{\Delta }{\varphi }_0|\le 4/{\epsilon }^2$. Choose $\epsilon >0$ small enough such that $x_j\notin B_{\epsilon }(0)$ for all $j\in J$. Then

(2.17)

(2.18)

Similarly, we can get

(2.19)

(2.20)

From equalities (2.17)-(2.20), we get

$$0=\underset{\epsilon \to 0}{lim}\underset{n\to +\mathrm{\infty }}{lim}〈I^{\prime }(u_n),u_n{\varphi }_0〉\ge {\mu }_0-\lambda {\gamma }_0.$$

(2.21)

On the other hand, ${\mu }_0-\lambda {\gamma }_0\ge S_{p,\lambda }{\nu }_0^{\frac{p}{p^{\ast }}}\ge 0$, thus ${\mu }_0-\lambda {\gamma }_0=0$. The proof is complete. □

Lemma 2.4 Assume $f(x,u)$ satisfies (f₁), the sequence $\{u_n\}$ satisfying (2.3) is bounded in $W_0^{2,p}(\mathrm{\Omega })$. Then there exist some $u\in W_0^{2,p}(\mathrm{\Omega })$ and a subsequence, still denoted by $\{u_n\}$, such that

$$\begin{array}{c}{u}_n\to u\text{strongly in }{L}^{p^{\ast }}({\mathrm{\Omega }}_{ϵ}),\hfill \\ \mathrm{\nabla }{u}_n\to \mathrm{\nabla }u\text{strongly in }{L}^{\frac{Np}{N-p}}({\mathrm{\Omega }}_{ϵ}),\hfill \end{array}$$

where ${\mathrm{\Omega }}_{ϵ}=\mathrm{\Omega }\setminus {\bigcup }_{i=0}^j{\overline{B}}_{ϵ}(x_i)$, $x_i\in J=\{x_1,x_2,\dots ,x_j\}\cup \{0\}$.

Proof Lemma 2.3 implies that J is finite. Choose $\phi \in C_0^{\mathrm{\infty }}(\mathrm{\Omega })$ with $\phi (x_i)=0$ for any $x_i\in J\cup \{0\}$. Then

$$\begin{array}{rcl}\int {|\phi u_n|}^{p^{\ast }}dx& =& \int {|\phi u|}^{p^{\ast }}dx+\sum _{i=1}^j{\nu }_i{\phi }^{p^{\ast }}(x_i)+{\nu }_0{\phi }^{p^{\ast }}(0)\\ =& \int {|\phi u|}^{p^{\ast }}dx.\end{array}$$

(2.22)

Since $\phi u_n\to \phi u$ a.e. in Ω, we get that

$$u_n\to u\text{strongly in }{L}^{p^{\ast }}({\mathrm{\Omega }}_{ϵ}).$$

Similarly,

$$\mathrm{\nabla }{u}_n\to \mathrm{\nabla }u\text{strongly in }{L}^{\frac{Np}{N-p}}({\mathrm{\Omega }}_{ϵ}).$$

 □

By using the Lebesgue decomposition theorem, we have

$$d\mu -\lambda d\gamma ={|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}}+d\sigma ,$$

(2.23)

where $({|\mathrm{\Delta }u|}^p-\lambda {|u|}^p{|x|}^{-2p})\perp d\sigma$. Then $d\sigma \ge {\sum }_{j\in J}{\mu }_j{\delta }_{x_j}+({\mu }_0-\lambda {\gamma }_0){\delta }_0$.

Lemma 2.5 Assume $f(x,u)$ satisfies (f₁)-(f₃), the sequence $\{u_n\}\in W_0^{2,p}(\mathrm{\Omega })$ satisfies (2.3). Then there exist $u\in W_0^{2,p}(\mathrm{\Omega })$ and a subsequence, still denoted by $\{u_n\}$, such that $u_n$ converges to u strongly in $W_0^{2,p}(\mathrm{\Omega })$.

Proof We first prove that

$$(\sigma -\sum _{j\in J}{\mu }_j{\delta }_{x_j}-({\mu }_0-\lambda {\gamma }_0){\delta }_0)(\overline{\mathrm{\Omega }})=0.$$

(2.24)

Claim for any closed set $F\subset \overline{\mathrm{\Omega }}\setminus \overline{J}$, $\overline{J}=J\cup \{0\}$, $\sigma (F)=0$. In fact, denote $r=dist(F,\overline{J})>0$. Then, by using a finite covering theorem, there exist finite open balls $B_{r/4}(z_i)$, with $z_i\in F$, $i=1,\dots ,m$, such that $F\subset {\bigcup }_{i=1}^m{B}_{r/4}(z_i)\subset \overline{\mathrm{\Omega }}\setminus \overline{J}$. Let $\varrho (t)$ be a smooth cutting-off function. $0\le \varrho (t)\le 1$ for any $t\in [0,\mathrm{\infty })$. $\varrho (x)\equiv 1$ for $0\le t\le 1/2$, and $\varrho \equiv 0$ for $t\ge 1$. Denote ${\eta }_{\epsilon ,y}=\varrho (|x-y|/\epsilon )$, then

(2.25)

(2.26)

The sequence $\{{\eta }_{\epsilon ,y}{u}_n\}$ is still bounded in $W_0^{2,p}(\mathrm{\Omega })$. Define

$$\eta (x)=\sum _{i=1}^m{\eta }_{r/4,z_i}(x),$$

(2.27)

then $\eta (x)\in C_0^{\mathrm{\infty }}(\overline{\mathrm{\Omega }}\setminus \overline{J})$ and $0\le \eta (x)\le m$. Furthermore,

$$F\subset \bigcup _{i=1}^m{B}_{\frac{r}{4}}(z_i)\subset spt\eta (x)\subset {\mathrm{\Omega }}_{\frac{r}{4}},$$

(2.28)

where $spt\eta (x)$ denotes the support of $\eta (x)$. As $n\to +\mathrm{\infty }$, (2.3) implies

$$\begin{array}{rcl}0& \leftarrow & 〈I^{\prime }(u_n)-I^{\prime }(u),(u_n-u)\eta 〉\\ =& {\int }_{\mathrm{\Omega }}\eta ({|\mathrm{\Delta }{u}_n|}^p-{|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n\mathrm{\Delta }u-{|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u\mathrm{\Delta }{u}_n+{|\mathrm{\Delta }u|}^p)dx\\ -\lambda {\int }_{\mathrm{\Omega }}\frac{\eta }{{|x|}^{2p}}({|u_n|}^p-{|u_n|}^{p-2}{u}_{n}u-{|u|}^{p-2}u{u}_n+{|u|}^p)dx\\ +2{\int }_{\mathrm{\Omega }}\mathrm{\nabla }\eta \mathrm{\nabla }(u_n-u)({|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n-{|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u)dx\\ +{\int }_{\mathrm{\Omega }}\mathrm{\Delta }\eta (u_n-u)({|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n-{|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u)dx\\ -{\int }_{\mathrm{\Omega }}\eta (u_n-u)(f(x,u_n)-f(x,u))dx\\ =& I+\mathit{II}+\mathit{III}+\mathit{IV}+V.\end{array}$$

As $n\to +\mathrm{\infty }$,

$$\begin{array}{rcl}I+\mathit{II}& =& {\int }_{\mathrm{\Omega }}\eta ({|\mathrm{\Delta }{u}_n|}^p-{|\mathrm{\Delta }{u}_n|}^{p-2}\mathrm{\Delta }{u}_n\mathrm{\Delta }u-{|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u\mathrm{\Delta }{u}_n+{|\mathrm{\Delta }u|}^p)dx\\ -\lambda {\int }_{\mathrm{\Omega }}\frac{\eta }{{|x|}^{2p}}({|u_n|}^p-{|u_n|}^{p-2}{u}_{n}u-{|u|}^{p-2}u{u}_n+{|u|}^p)dx\\ =& {\int }_{\mathrm{\Omega }}\eta ({|\mathrm{\Delta }{u}_n|}^p-{|\mathrm{\Delta }u|}^p-\lambda \frac{{|u_n|}^p-{|u|}^p}{{|x|}^{2p}})dx+o(1)\\ =& {\int }_{\mathrm{\Omega }}\eta d\sigma +o(1),\end{array}$$

where we have used (2.23). Lemma 2.2 implies that $\parallel u_n\parallel \le M$. Hölder’s inequality, (2.25)-(2.28) and Lemma 2.4 deduce that

By (2.1), for any $\epsilon >0$, we have

$$\begin{array}{rcl}|V|& =& |{\int }_{\mathrm{\Omega }}(f(x,u_n)-f(x,u))(u_n-u)\eta dx|\\ \le & {\int }_{\mathrm{\Omega }}(|f(x,u_n)|+|f(x,u)|)|u_n-u|\eta dx\\ \le & {\int }_{\mathrm{\Omega }}(C_{\epsilon }+\epsilon ({|u_n|}^{p^{\ast }-1}+{|u|}^{p^{\ast }-1}))|u_n-u|\eta dx\\ \le & C_{\epsilon }{\int }_{\mathrm{\Omega }}|u_n-u|\eta dx+\epsilon ({\parallel u_n\parallel }_{p^{\ast }}^{p^{\ast }-1}+{\parallel u\parallel }_{p^{\ast }}^{p^{\ast }-1}){\left({\int }_{spt(\eta )}{|u_n-u|}^{p^{\ast }}dx\right)}^{\frac{1}{p^{\ast }}}\\ \le & C_{\epsilon }{\int }_{\mathrm{\Omega }}|u_n-u|dx+\epsilon C(M)\\ \to & 0\text{as }n\to +\mathrm{\infty },\epsilon \to 0.\end{array}$$

Therefore, as $n\to +\mathrm{\infty }$,

$$0\le \sigma (F)\le \sigma (spt(\eta ))\le {\int }_{\mathrm{\Omega }}\eta d\sigma \le |\mathit{III}|+|\mathit{IV}|+|V|=0.$$

(2.29)

Since the set F is arbitrary, (2.24) is obtained.

Lemma 2.3 means that ${\mu }_j=0$ for any $j\in J$ and ${\mu }_0-\lambda {\gamma }_0=0$, so $\sigma (\overline{\mathrm{\Omega }})=0$, which implies that as $n\to +\mathrm{\infty }$,

$${\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }{u}_n|}^p-\lambda \frac{{|u_n|}^p}{{|x|}^{2p}})dx\to {\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}})dx.$$

(2.30)

Together with $u_n\to u$ a.e. in Ω, we complete the proof. □

3 The proof of the main result

In this section, by using the following symmetric mountain pass theorem (see [19]), we give the proof of Theorem 1.1.

Lemma 3.1 (Symmetric mountain pass theorem)

Assume functional I satisfies the following conditions:

• (1) $I\in {\mathcal{C}}^1(W_0^{2,p}(\mathrm{\Omega }),\mathbb{R})$ is even and satisfies the Palais-Smale condition.

• (2) There exists a finite dimensional subspace $X\in W_0^{2,p}(\mathrm{\Omega })$ such that $I{|}_{X\cap \partial B_r}\ge a>0$.

• (3) There exists a sequence of the finite dimensional subspace $\{X_j\}$, $dim(X_j)=j$ and $r_j>0$ such that

  $$I(u)\le 0\mathit{\text{for any }}u\in X_j\setminus B_{r_j},j=1,2,\dots .$$

Then I has infinitely many different critical points, and the corresponding energy values are positive.

Proof of Theorem 1.1 We know the energy functional $I(u)$ is continuous and even. Lemma 2.5 implies that $I(u)$ satisfies the PS condition in $W_0^{2,p}(\mathrm{\Omega })$. In any finite dimensional subspace $X\subset W_0^{2,p}(\mathrm{\Omega })$, all norms are equivalent. For any $u\in X$, since $F(x,u)=o(u^p)$ as $u\to 0$, there exists a constant $q>p$ such that

$$\begin{array}{rcl}I(u)& =& \frac{1}{p}{\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}})dx-{\int }_{\mathrm{\Omega }}F(x,u)dx\\ \ge & \frac{1}{p}{\parallel u\parallel }^p-C{\parallel u\parallel }^q>0\text{for }u\text{ small enough}.\end{array}$$

(3.1)

Thus, there exists $r>0$ such that $I{|}_{X\cap \partial B_r}\ge a>0$. On the other hand, by using (2.2), we have

$$\begin{array}{rcl}I(u)& =& \frac{1}{p}{\int }_{\mathrm{\Omega }}({|\mathrm{\Delta }u|}^p-\lambda \frac{{|u|}^p}{{|x|}^{2p}})dx-{\int }_{\mathrm{\Omega }}F(x,u)dx\\ \le & \frac{1}{p}{\parallel u\parallel }^p-C{\int }_{\{x\in \mathrm{\Omega }:|u|>M\}}{|u|}^{p+\theta }dx-C^{\prime }\\ \le & \frac{1}{p}{\parallel u\parallel }^p-C{\parallel u\parallel }^{p+\theta }-C^{\prime }\to -\mathrm{\infty }\text{as }u\to \mathrm{\infty }.\end{array}$$

There exists a sequence of the finite dimensional subspace $\{X_j\}$, $dim(X_j)=j$ and $r_j>0$ such that

$$I(u)\le 0\text{for any }u\in X_j\setminus B_{r_j},j=1,2,\dots .$$

Finally, all the assumptions of Lemma 3.1 are satisfied. Hence, the problem (1.1) possesses infinitely many weak solutions, and the corresponding critical values are positive. □

Remark 3.1 Under the same conditions of Theorem 1.1, we can also prove that the following p-harmonic type equation with Navier boundary conditions:

$$\{\begin{array}{cc}\mathrm{\Delta }({|\mathrm{\Delta }u|}^{p-2}\mathrm{\Delta }u)-\lambda \frac{{|u|}^{p-2}u}{{|x|}^{2p}}=f(x,u),\hfill & x\in \mathrm{\Omega },\hfill \\ u=\mathrm{\Delta }u=0,\hfill & x\in \partial \mathrm{\Omega },\hfill \end{array}$$

(3.2)

has infinitely many solutions $\{u_n\}\in W^{2,p}(\mathrm{\Omega })\cap W_0^{1,p}(\mathrm{\Omega })$, where Ω containing the origin is an open bounded domain in ${\mathbb{R}}^N$, ∂ Ω is smooth. $1<p<\frac{N}{2}$, $0\le \lambda <\overline{\lambda }={[N(p-1)(N-2p)/p^2]}^p$.

Remark 3.2 All the results obtained above obviously hold if we choose $f(x,u)={|u|}^{q-2}u+{|u|}^{r-2}u$ with $p<q<r<p^{\ast }$.