Open Access

Existence and multiplicity of positive solutions for p-Laplacian elliptic equations

Boundary Value Problems20162016:125

https://doi.org/10.1186/s13661-016-0632-5

Received: 27 April 2016

Accepted: 27 June 2016

Published: 8 July 2016

Abstract

We study a p-Laplacian elliptic equation with Hardy term and Hardy-Sobolev critical exponent, where the nonlinearity is \((p-1)\)-sublinear near zero and \((p^{\ast}(s)-1)\)-sublinear near infinity (\(p^{\ast}(s)=\frac{p(N-s)}{N-p}\) is the Hardy-Sobolev critical exponent). By using variational methods and some analysis techniques, we obtain the existence and multiplicity of positive solutions for the p-Laplacian elliptic equation. To the best of our knowledge, no result has been published concerning the existence and multiplicity of positive solutions for the p-Laplacian elliptic equation.

Keywords

p-Laplacian elliptic equationsHardy termHardy-Sobolev critical exponentvariational methods

MSC

35A1535J91

1 Introduction and main results

In this paper, we will study the existence and multiplicity of positive solutions for the following p-Laplacian elliptic equation:
$$ \textstyle\begin{cases} -\triangle_{p}u-\mu\frac{\vert u\vert ^{p-2}u}{\vert x\vert ^{p}}=\frac{\vert u\vert ^{p^{\ast }(s)-2}}{\vert x\vert ^{s}}u+\lambda f(x,u), & x\in\Omega\setminus\{0\},\\ u=0,& x\in\partial\Omega. \end{cases} $$
(1.1)
Here, \(\Omega\subset\mathbb{R}^{N}\) (\(N\geq3\)) is an open bounded domain with smooth boundary Ω and \(0\in\Omega\), \(p\in(1,N)\), \(s\in[0,p)\), \(\lambda,\mu\in\mathbb{R}^{+}\), \(\triangle_{p}u:= \operatorname {div}(\vert \nabla u\vert ^{p-2}\nabla u)\) is the p-Laplacian differential operator, \(p^{\ast}(s)=\frac{p(N-s)}{N-p}\) is the Hardy-Sobolev critical exponent, \(p^{\ast}=p^{\ast}(0)=\frac{Np}{N-p}\) is the Sobolev critical exponent, and we have the function \(f:\Omega\times\mathbb{R}\rightarrow \mathbb{R}\).
Let
$$\Vert u\Vert := \biggl( \int_{\Omega} \biggl(\vert \nabla u\vert ^{p}-\mu \frac{\vert u\vert ^{p}}{\vert x\vert ^{p}} \biggr)\,dx \biggr)^{\frac{1}{p}}, \quad u\in W^{1,p}_{0}( \Omega), $$
which is well defined on the Sobolev space \(W^{1,p}_{0}(\Omega)\) by the Hardy inequality [1]. From [2], we know \(\Vert u\Vert \) is comparable with the standard Sobolev norm of \(W^{1,p}_{0}(\Omega)\), but it is not a norm since the triangle inequality or subadditivity may fail. The following best Hardy-Sobolev constant will be useful in this paper:
$$ A_{\mu,s}(\Omega):=\inf_{u\in W^{1,p}_{0}(\Omega)\setminus\{0\}}\frac{\Vert u\Vert ^{p}}{ (\int _{\Omega} \frac{\vert u\vert ^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx )^{\frac{p}{p^{\ast}(s)}}}. $$
(1.2)

In recent decades, there were many authors [1, 317] who have studied the existence or multiplicity of solutions for elliptic equations with the operator \(-\triangle-\frac{\mu}{ \vert x\vert ^{2}}\) (\(0\leq\mu< (\frac{N-2}{2} )^{2}\)). But most of the authors only considered the case \(s=0\).

Next we only state some most related results of (1.1). Han [18] obtained the existence of multiplicity of positive solutions for the following equation:
$$ \textstyle\begin{cases} -\triangle_{p}u-\mu\frac{\vert u\vert ^{p-2}u}{\vert x\vert ^{p}}=Q(x)\vert u\vert ^{p^{\ast }-2}u+\lambda \vert u\vert ^{p-2}u, &x\in\Omega,\\ u=0, & x\in\partial\Omega, \end{cases} $$
(1.3)
where \(Q(x)\ge0\) is a bounded function on Ω̅. The authors [19] only studied (1.3) in the special cases where \(Q(x)\equiv1\) and \(\mu=0\). The authors [2] studied the following equation:
$$ \textstyle\begin{cases} -\triangle_{p}u-\mu\frac{u^{p-1}}{\vert x\vert ^{p}}=\frac{\vert u\vert ^{p^{\ast }(s)-1}}{\vert x\vert ^{s}}+\vert u\vert ^{p^{\ast}-1},& x\in\mathbb{R}^{N},\\ u\in D^{p}_{1}(\mathbb{R}^{N}), \end{cases} $$
(1.4)
where \(D^{p}_{1}(\mathbb{R}^{N})\) is defined as the completion of \(C_{c}^{\infty}(\mathbb{R}^{N})\), and they obtained a positive solution \(u\in D^{p}_{1}(\mathbb{R}^{N})\cap C^{1}(\mathbb{R}^{N}\setminus\{0\})\) for any \(0< s< p\) and \(\mu\in(-\infty,\mu_{1})\), where \(\mu_{1}:= (\frac {N-p}{p} )^{p}\). Later, the authors [20] obtained a nontrivial solution of a more general case than (1.4) by the ideas in [2]. Kang [21] obtained one positive solution for the following equation:
$$ \textstyle\begin{cases} -\triangle_{p}u-\mu\frac{\vert u\vert ^{p-2}u}{\vert x\vert ^{p}}=\frac{\vert u\vert ^{p^{\ast }(s)-2}}{\vert x\vert ^{s}}u+\lambda\frac{\vert u\vert ^{q-2}u}{\vert x\vert ^{t}}, &x\in\Omega \setminus\{0\},\\ u=0, & x\in\partial\Omega, \end{cases} $$
(1.5)
where \(0\leq t< p\), \(p\leq q< p^{\ast}(t)\).

Inspired by the above results, we shall study the existence and multiplicity of positive solutions for (1.1) with the nonlinearity f being \((p-1)\)-sublinear at zero and \((p^{\ast}(s)-1)\)-sublinear at infinity (see the following \((A_{1})\)), which is different from the above results. Due to the lack of compactness of the embeddings in \(W^{1,p}_{0}(\Omega)\hookrightarrow L^{p^{\ast}}(\Omega)\), \(W^{1,p}_{0}(\Omega)\hookrightarrow L^{p}(\Omega, \vert x\vert ^{-p}\,dx)\), and \(W^{1,p}_{0}(\Omega)\hookrightarrow L^{p^{\ast}(s)}(\Omega, \vert x\vert ^{-s}\,dx)\), we cannot use the standard variational argument directly. The corresponding energy functional fails to satisfy the classical Palais-Smale ((PS)) condition in \(W^{1,p}_{0}(\Omega)\). But we can establish a local (PS) condition in a suitable range, so the existence result can be obtained by constructing a minimax level within this range and the mountain pass lemma in [3, 22].

Let \(\Vert \cdot \Vert _{p}\) be the norm in \(L^{p}(\Omega)\) and \(F(x,t):=\int^{t}_{0}f(x,s)\,ds\), \(x\in\Omega\), \(t\in\mathbb{R}\). Let \(a(\mu)\) and \(b(\mu)\) be zeros of the function
$$f(t)=(p-1)t^{p}-(N-p)t^{p-1}+\mu, \quad t\geq0, 0\leq\mu< \mu_{1}:= \biggl(\frac{N-p}{p} \biggr)^{p} $$
satisfying \(0\leq a(\mu)<\frac{N-p}{p}<b(\mu)\); see [23]. To state our results, we make the following assumptions:
\((\textbf{A}_{\textbf{1}})\)
\(f\in C(\overline{\Omega} \times\mathbb{R}^{+},\mathbb{R})\), \(f(x,0)\equiv0\), and
$$\lim_{t\rightarrow 0^{+}}\frac{f(x,t)}{t^{p-1}}=+\infty, \quad\quad \lim _{t\rightarrow \infty}\frac{f(x,t)}{t^{p^{\ast}(s)-1}}=0 \quad\text{uniformly for } x\in\overline{ \Omega}. $$
\((\textbf{A}_{\textbf{2}})\)

\(f:\Omega\times\mathbb {R}^{+}\rightarrow \mathbb{R}\) is nondecreasing with respect to the second variable.

\((\textbf{A}_{\textbf{3}})\)

\(2\leq p< N\), \(N<\min\{pb(\mu ),p(1+p)\}\) and \(0\leq s\leq N-\frac{(N-p)(1+p)}{p}\).

\((\textbf{A}_{\textbf{3}}')\)

\(2\leq p< N\), \(pb(\mu)\leq N< p+\frac{p^{2}b(\mu)}{1+p}\) and \(N-pb(\mu)< s\leq N-\frac{(N-p)(1+p)}{p}\).

Remark 1.1

In \((A_{3})\) and \((A_{3}')\), we can easily check that \(N< p(1+p)\) implies \(N-\frac{(N-p)(1+p)}{p}>0\), \(N< p+\frac{p^{2}b(\mu)}{1+p}\) implies \(N-pb(\mu)< N-\frac{(N-p)(1+p)}{p}\). Besides, \(N-\frac{(N-p)(1+p)}{p}< p\) holds.

Now our results read as follows.

Theorem 1.1

If \(N\geq3\), \(0\leq s< p\), \(0\leq\mu<\mu_{1}\), \(1< p< N\) and \((A_{1})\) hold, then there exists \(\lambda^{\ast}>0\) such that (1.1) has at least one nontrivial positive solution \(u_{\lambda}\) for any \(\lambda\in(0,\lambda^{\ast})\).

Theorem 1.2

If \(N\geq3\), \(0\leq s< p\), \(0\leq\mu<\mu_{1}\), \((A_{1})\), \((A_{2})\) and (\((A_{3})\) or \((A_{3}')\)) hold, then there exists \(\lambda^{\ast}>0\) such that (1.1) has at least two nontrivial positive solutions for every \(\lambda\in(0,\lambda^{\ast})\).

Remark 1.2

We should mention that the above p-Laplacian problems studied in [2, 1821] are all not \((p-1)\)-sublinear at zero. Besides, our nonlinearity f is more general. To the best of our knowledge, our Theorems 1.1 and 1.2 are new.

Let \(D^{1,p}(\mathbb{R}^{N}):= \{u\in L^{p^{\ast}}(\mathbb{R}^{N}); \vert \nabla u\vert \in L^{p}(\mathbb{R}^{N}) \}\). A typical model of (1.1) is the following equation:
$$\textstyle\begin{cases} \textstyle\begin{array}{l@{\quad}l} -\triangle_{p}u-\mu\frac{u^{p-1}}{\vert x\vert ^{p}}=u^{p^{\ast }-1}, &\text{in } \mathbb{R}^{N}\setminus\{0\}, \\ u>0, & \text{in } \mathbb{R}^{N}\setminus\{0\}, \end{array}\displaystyle \\ u\in D^{1,p}(\mathbb{R}^{N}), \quad\quad \mu\in[0,\mu_{1}). \end{cases} $$
From [23], we see that this problem has radially symmetric ground states,
$$V_{\varepsilon}(x)=\varepsilon^{-\frac{N-p}{p}}U_{p,\mu} \biggl( \frac {x}{\varepsilon} \biggr)= \varepsilon^{-\frac{N-p}{p}}U_{p,\mu} \biggl( \frac{\vert x\vert }{\varepsilon} \biggr), \quad\forall \varepsilon>0, $$
and they satisfy
$$\int_{\mathbb{R}^{N}} \biggl(\bigl\vert \nabla V_{\varepsilon}(x)\bigr\vert ^{p}- \mu\frac{\vert V_{\varepsilon}(x)\vert ^{p}}{\vert x\vert ^{p}} \biggr)\,dx= \int_{\mathbb{R}^{N}} \bigl\vert V_{\varepsilon}(x)\bigr\vert ^{p^{\ast}}\,dx=A_{\mu,0}^{\frac{N}{p}}, $$
where \(U_{p,\mu}(x)=U_{p,\mu}(\vert x\vert )\) is the unique radial solution of this problem, satisfying
$$U_{p,\mu}(1)= \biggl(\frac{N(\mu_{1}-\mu)}{N-p} \biggr)^{\frac {1}{p^{\ast}-p}}. $$
Moreover,\(U_{p,\mu}\) has the following properties:
$$\begin{aligned}& \lim_{r\rightarrow0}r^{a(\mu)}U_{p,\mu}(r)=c_{1}>0,\quad\quad \lim_{r\rightarrow+\infty}r^{b(\mu)}U_{p,\mu}(r)=c_{2}>0, \\& \lim_{r\rightarrow0}r^{a(\mu)+1}U'_{p,\mu}(r)=c_{1}a( \mu)\geq0,\quad\quad \lim_{r\rightarrow+\infty}r^{b(\mu)+1}U'_{p,\mu }(r)=c_{2}b( \mu)>0, \end{aligned}$$
where \(c_{1} \) and \(c_{2}\) are positive constants depending on p and N; \(a(\mu)\) and \(b(\mu)\) are zeros of the function
$$f(t)=(p-1)t^{p}-(N-p)t^{p-1}+\mu, \quad t\geq0, 0\leq\mu< \mu_{1}, $$
satisfying \(0\leq a(\mu)<\frac{N-p}{p}<b(\mu)\); see [23]. The above results are useful in studying equation (1.1).

Remark 1.3

As \(\mu=0\) and \(s=0\), then \(b(\mu)=b(0)=\frac{N-p}{p-1}\). When \(p=2\) and \(0\leq\mu<\mu_{2}:= (\frac{N-2}{2} )^{2}\), it is well known that \(a(\mu)=\sqrt{\mu_{2}}-\sqrt{\mu_{2}-\mu}\) and \(b(\mu)=\sqrt{\mu_{2}}+\sqrt{\mu_{2}-\mu}\).

In Section 2, we will give the proof of Theorem 1.1. In Section 3, we first of all give some preliminary lemmas, and then we will complete the proof of Theorem 1.2.

2 Proof of Theorem 1.1

Let \(X:=W^{1,p}_{0}(\Omega)\) and \(u^{\pm}:=\max\{\pm u,0\}\). Note that the values of \(f(x,t)\) for \(t<0\) are irrelevant in Theorems 1.1-1.2, so we define
$$f(x,t)\equiv0, \quad x\in\Omega, t\leq0. $$
The functional corresponding of (1.1) is
$$\begin{aligned} I(u) =&\frac{1}{p} \int_{\Omega} \biggl(\vert \nabla u\vert ^{p}-\mu \frac{\vert u\vert ^{p}}{\vert x\vert ^{p}} \biggr)\, dx-\frac{1}{p^{\ast}(s)} \int_{\Omega} \frac{(u^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx \\ &{}-\lambda \int_{\Omega }F\bigl(x,u^{+}\bigr)\,dx, \quad u\in W^{1,p}_{0}(\Omega). \end{aligned}$$
By \((A_{1})\) and the Hardy inequalities (see [1]), we have \(I\in C^{1}(W^{1,p}_{0}(\Omega),\mathbb{R})\). Now it is well known that there is a one-to-one correspondence between the weak solutions of (1.1) and the critical points of I on \(W^{1,p}_{0}(\Omega)\). More precisely, we say \(u\in W^{1,p}_{0}(\Omega)\) is a weak solution of (1.1) if
$$\begin{aligned} \bigl\langle I'(u),v\bigr\rangle =& \int_{\Omega} \biggl(\vert \nabla u\vert ^{p-2}\nabla u \nabla v -\mu\frac{\vert u\vert ^{p-2}uv}{\vert x\vert ^{p}} \biggr)\,dx- \int_{\Omega}\frac{(u^{+})^{p^{\ast}(s)-1}}{\vert x\vert ^{s}}v\, dx-\lambda \int_{\Omega}f\bigl(x,u^{+}\bigr)v\,dx \\ =&0 \end{aligned}$$
for any \(v\in W^{1,p}_{0}(\Omega)\).

Proof of Theorem 1.1

By the Sobolev and Hardy-Sobolev inequalities, we get
$$ \begin{aligned} &\Vert u\Vert ^{p}_{p}\leq C\Vert u\Vert ^{p},\quad\quad \int_{\Omega} \frac{\vert u\vert ^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx\leq C\Vert u\Vert ^{p^{\ast}(s)} \quad\text{and} \\ & \Vert u\Vert ^{p^{\ast}}_{p^{\ast}}\leq C \Vert u\Vert ^{p^{\ast}},\quad \forall u\in X, \end{aligned} $$
(2.1)
and it follows from (\(A_{1}\)) that
$$\begin{aligned}& \exists\delta>0 \quad\text{such that} \quad\bigl\vert F(x,t)\bigr\vert < \frac{t^{p^{\ast}(s)}}{p^{\ast}(s)\vert x^{s}\vert } \quad\text{for } t>\delta, \\& \exists M_{1}>0\quad \text{such that} \quad\bigl\vert F(x,t)\bigr\vert \leq M_{1}, \quad\forall t\in(0,\delta], \end{aligned}$$
uniformly for all \(x\in\overline{\Omega}\setminus\{0\}\). Thus, we get
$$ \bigl\vert F(x,t)\bigr\vert \leq M_{1}+\frac{t^{p^{\ast }(s)}}{p^{\ast}(s)\vert x\vert ^{s}},\quad \forall t\in\mathbb{R}, x\in\overline{\Omega}\setminus\{0\}. $$
(2.2)
By (2.1) and (2.2), we have
$$I(u)=\frac{1}{p}\Vert u\Vert ^{p}-\frac{1}{p^{\ast}(s)} \int_{\Omega} \frac{(u^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\, dx-\lambda \int_{\Omega}F\bigl(x,u^{+}\bigr)\,dx \geq \frac{1}{p}\Vert u\Vert ^{p}-C_{1}\Vert u\Vert ^{p^{\ast}(s)}-\lambda M_{1}\vert \Omega \vert $$
for all \(\lambda\in(0,1]\) and some \(C_{1}=\frac{C\mu}{p^{\ast}(s)}\), so there are \(\rho>0\) and \(\lambda^{\ast}\in(0,1]\) such that
$$I(u)>0 \quad\text{if } \Vert u\Vert =\rho\quad\text{and} \quad I(u)\geq-C_{2} \quad\text{if } \Vert u\Vert \leq\rho $$
for any \(0<\lambda<\lambda^{\ast}\), where \(C_{2}=C_{1}\rho^{p^{\ast}(s)}+\lambda^{\ast}M_{1}\vert \Omega \vert \). We choose \(u_{0}\in W^{1,p}_{0}(\Omega)\cap L^{\infty}(\Omega)\) such that \(u_{0}^{+}\neq0\). Let \(M_{2}:=\Vert u_{0}\Vert ^{p} /(\lambda \Vert u_{0}^{+}\Vert _{p}^{p})\). By \((A_{1})\), there is \(\delta_{1}\) such that
$$\bigl\vert F(x,t)\bigr\vert \geq\frac{2M_{2}}{p}\vert t\vert ^{p},\quad 0< t< \delta_{1}. $$
Hence, we get
$$\begin{aligned} I(ru_{0}) =&\frac{r^{p}}{p}\Vert u_{0}\Vert ^{p}-\frac{ r^{p^{\ast}(s)}}{p^{\ast}(s)} \int_{\Omega} \frac{(u_{0}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx-\lambda \int_{\Omega }F\bigl(x,ru^{+}_{0}\bigr)\,dx \\ \leq&\frac{r^{p}}{p}\Vert u_{0}\Vert ^{p}- \frac{2r^{p}}{p}\lambda M_{2}\bigl\Vert u^{+}_{0} \bigr\Vert _{p}^{p} =-\frac{r^{p}}{p}\Vert u_{0}\Vert ^{p}< 0 \end{aligned}$$
for any \(0<\lambda<\lambda^{\ast}\) and \(0< r<\min\{\rho,\delta_{1}/\Vert u^{+}_{0}\Vert _{\infty}\}\). So there is u small enough such that \({I(u)<0}\). We deduce that
$$\inf_{u\in\overline{B_{\rho}(0)}}I(u)< 0< \inf_{u\in\partial\overline {B_{\rho}(0)}}I(u). $$
By Ekeland’s variational principle in [24], there is a minimizing sequence \(\{u_{n}\}\subset\overline{B_{\rho}(0)}\) such that
$$I(u_{n})\leq\inf_{u\in\overline{B_{\rho}(0)}}I(u)+\frac{1}{n},\quad \quad I( \omega)\geq I(u_{n})-\frac{1}{n}\Vert \omega-u_{n} \Vert , \quad\omega\in\overline{B_{\rho}(0)}. $$
So, we have
$$\bigl\Vert I'(u_{n})\bigr\Vert \rightarrow0 \quad\text{and} \quad I(u_{n})\rightarrow c_{\lambda} \quad\text{as } n\rightarrow \infty, $$
where \(c_{\lambda}\) stands for the infimum of \(I(u)\) on \(\overline{B_{\rho}(0)}\). Note that \(\{u_{n}\}\) is bounded and \(\overline{B_{\rho}(0)}\) is a closed convex set, so there is \(u_{\lambda}\in\overline{B_{\rho}(0)}\subset W^{1,p}_{0}(\Omega)\). By [1], we have
$$\begin{aligned}& u_{n}\rightharpoonup u_{\lambda} \quad\text{weakly in } W^{1,p}_{0}(\Omega), \\& u_{n}\rightarrow u_{\lambda} \quad\text{strongly in } L^{\gamma}(\Omega), 1< \gamma< p^{\ast}, \\& u_{n}\rightarrow u_{\lambda} \quad\text{a.e. in } \Omega, \\& \nabla u_{n}\rightarrow\nabla u_{\lambda} \quad\text{a.e. in } \Omega, \\& \frac{u_{n}}{x}\rightharpoonup\frac{u_{\lambda}}{x} \quad\text{weakly in } L^{p}(\Omega), \\& \int_{\Omega} \frac{\vert u_{n}\vert ^{p^{\ast }(s)-2}u_{n}}{\vert x\vert ^{s}}v\,dx\rightarrow \int_{\Omega }\frac{\vert u_{\lambda} \vert ^{p^{\ast}(s)-2}u_{\lambda }}{\vert x\vert ^{s}}v\,dx,\quad \forall v\in W^{1,p}_{0}(\Omega). \end{aligned}$$
Thus, passing to the limit in \(\langle I'(u_{n}),v\rangle\), as \(n\rightarrow\infty\), we have
$$\int_{\Omega} \biggl(\vert \nabla u_{\lambda} \vert ^{p-2}\nabla u_{\lambda}\nabla v -\mu\frac{\vert u_{\lambda} \vert ^{p-2}u_{\lambda}v}{ \vert x\vert ^{p}} \biggr)\,dx - \int_{\Omega}\frac{(u^{+}_{\lambda})^{p^{\ast}(s)-1}v}{\vert x\vert ^{s}}\,dx- \lambda \int_{\Omega}f\bigl(x,u^{+}_{\lambda}\bigr)v\,dx=0 $$
for all \(v\in W^{1,p}_{0}(\Omega)\). That is, \(\langle I'(u_{\lambda}),v\rangle=0\). Therefore, \(u_{\lambda}\) is a critical point of I. Since \(\Vert u^{-}_{\lambda} \Vert ^{p}=-\langle I'(u_{\lambda}),u^{-}_{\lambda}\rangle=0\), \(u_{\lambda}=u^{+}_{\lambda}\geq0\). Moreover, by \((A_{1})\) and the boundedness of Ω, we have
$$\begin{aligned}& \exists M_{3}>0 \quad\text{such that} \quad\bigl\vert f(x,t)\bigr\vert < \frac{1}{\lambda}\frac {t^{p^{\ast}(s)-1}}{\vert x\vert ^{s}} \quad\text{for } t>M_{3}, \\& \exists\delta_{2}\in(0,M_{3}) \quad\text{such that} \quad\bigl\vert f(x,t)\bigr\vert >0 \quad\text{for } 0< t< \delta_{2}, \\& \exists M_{4}>0 \quad\text{such that} \quad\bigl\vert f(x,t)\bigr\vert \leq M_{4} \quad\text{for all } t\in[\delta_{2},M_{3}] \end{aligned}$$
for all \(x\in\overline{\Omega}\setminus\{0\}\). Therefore, we deduce that
$$ f(x,t)\geq-\frac{1}{\lambda}\frac{t^{p^{\ast }(s)-1}}{\vert x\vert ^{s}}-M_{4}t \delta_{2}^{-1}, \quad\forall t\in\mathbb{R}^{+}, x\in \overline{\Omega}\setminus\{0\}. $$
(2.3)
From (1.1) and (2.3), we have \(-\triangle_{p}u_{\lambda}+\lambda M_{4}\delta_{2}^{-1}u_{\lambda}\geq0\). By the strong maximum principle, we have \(u_{\lambda}>0\). So the proof of Theorem 1.1 is finished. □

3 Proof of Theorem 1.2

In this section, we will look for the second positive solution by a translated functional as in [3]. For fixed \(\lambda\in(0,\lambda^{\ast})\), we will look for the second solution of (1.1) of the form \(u=u_{\lambda}+v\), where \(u_{\lambda}\) is the first positive solution obtained in the previous section. The corresponding equation for v is
$$ \textstyle\begin{cases} -\bigtriangleup_{p} v-\mu\frac{\vert v\vert ^{p-2}v}{\vert x\vert ^{p}} \\ \quad =\frac {(u_{\lambda}+v)^{p^{\ast }(s)-1}}{\vert x\vert ^{s}}-\frac{u_{\lambda}^{p^{\ast }(s)-1}}{\vert x\vert ^{s}} +\lambda f(x,u_{\lambda}+v)-\lambda f(x,u_{\lambda}),& x\in\Omega\setminus\{0\},\\ v=0,& x\in\partial\Omega. \end{cases} $$
(3.1)
Let us define
$$\begin{aligned}& g(x,t)= \textstyle\begin{cases} \frac{(u_{\lambda}+t)^{p^{\ast}(s)-1}}{\vert x\vert ^{s}}-\frac {u_{\lambda }^{p^{\ast}(s)-1}}{\vert x\vert ^{s}} +\lambda f(x,u_{\lambda}+t)-\lambda f(x,u_{\lambda}), &t\geq0,\\ 0, & t< 0, \end{cases}\displaystyle \\& G(x,t)= \int_{0}^{t}g(x,s)\,ds, \end{aligned}$$
(3.2)
and
$$\begin{aligned} J(v) =&\frac{1}{p} \int_{\Omega} \biggl(\vert \nabla v\vert ^{p}-\mu \frac{\vert v\vert ^{p}}{\vert x\vert ^{p}} \biggr)\,dx- \int_{\Omega}G\bigl(x,v^{+}\bigr)\,dx \\ =&\frac{1}{p}\Vert v\Vert ^{p}-\frac{1}{p^{\ast}(s)} \int_{\Omega} \biggl(\frac {(u_{\lambda} +v^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}-\frac{u_{\lambda }^{p^{\ast}(s)}}{\vert x\vert ^{s}} -p^{\ast}(s)\frac{u_{\lambda}^{p^{\ast}(s)-1}v^{+}}{\vert x\vert ^{s}} \biggr)\,dx \\ &{}-\lambda \int_{\Omega} \bigl(F\bigl(x,u_{\lambda}+v^{+} \bigr)-F(x,u_{\lambda })-f(x,u_{\lambda})v^{+} \bigr)\,dx. \end{aligned}$$

Now, we have one-to-one correspondence between critical points of J in \(W_{0}^{1,p}(\Omega)\) and solutions of (3.1). That is, if \(v\in W_{0}^{1,p}(\Omega)\), \(v\not\equiv0\) is a critical point of J, then v is a solution of (3.1). Since \(\Vert v^{-}\Vert ^{p}=-\langle J'(v),v^{-}\rangle=0\), \(v=v^{+}\geq0\). Besides, by the maximum principle, \(v>0\) in Ω. Here, \(u=u_{\lambda}+v\) is a positive solution of (1.1) and \(u\neq u_{\lambda}\). If \(v=0\) is the only critical point of J in \(W_{0}^{1,p}(\Omega)\), we will get a contradiction. Then the existence of the second positive solution of (1.1) can be proved.

Lemma 3.1

\(v=0\) is a local minimum of J in \(W_{0}^{1,p}(\Omega)\).

Proof

For any \(v\in W_{0}^{1,p}(\Omega)\), we write \(v=v^{+}-v^{-}\). By J and direct computation, we have
$$ J(v)=\frac{1}{p}\bigl\Vert v^{-}\bigr\Vert ^{p}+I\bigl(u_{\lambda}+v^{+}\bigr)-I(u_{\lambda}). $$
(3.3)
Since \(u_{\lambda}\) is a local minimizer of I in \(W_{0}^{1,p}(\Omega)\), we have \(J(v)\geq\frac{1}{p}\Vert v^{-}\Vert ^{p}\) for \(\Vert v\Vert \leq\varepsilon\) with ε being small enough. □

Lemma 3.2

Suppose that \(1< p< N\), \((A_{1})\) and \((A_{2})\) hold, moreover, \(v=0\) is the only critical point of J. Let \(\{v_{n}\}\) be a \((PS)_{c}\) sequence with \(0< c<\frac{p-s}{p(N-s)}A^{\frac{N-s}{p-s}}_{\mu,s}\), then we have
$$v_{n}\rightarrow0 \quad\textit{in }W_{0}^{1,p}(\Omega) \textit{ as }n \rightarrow\infty. $$

Proof

Let \(\{v_{n}\}\) be a sequence in \(W_{0}^{1,p}(\Omega)\) such that
$$ J(v_{n})\rightarrow c< \frac{p-s}{p(N-s)}A^{\frac{N-s}{p-s}}_{\mu,s} \quad\text{and} \quad J'(v_{n})\rightarrow0 \quad\text{in } \bigl(W_{0}^{1,p}(\Omega) \bigr)^{\ast}. $$
(3.4)
By (3.3) and (3.4), we have
$$\begin{aligned}& J(v_{n})=\frac{1}{p}\bigl\Vert v_{n}^{-} \bigr\Vert ^{p}+I\bigl(u_{\lambda }+v_{n}^{+} \bigr)-I(u_{\lambda})=c+o(1), \\& \bigl\langle J'(v_{n}),u_{\lambda}+v_{n}^{+} \bigr\rangle = \int_{\Omega}\bigl\vert \nabla v_{n}^{-} \bigr\vert ^{p-2}\nabla v_{n}^{-}\nabla u_{\lambda}\,dx+ \bigl\langle I'\bigl(u_{\lambda}+v_{n}^{+} \bigr),u_{\lambda}+v_{n}^{+}\bigr\rangle =o(1)\bigl\Vert u_{\lambda }+v_{n}^{+}\bigr\Vert , \end{aligned}$$
(3.5)
which yields
$$\begin{aligned}& J(v_{n})-\frac{1}{p}\bigl\langle J'(v_{n}),u_{\lambda}+v_{n}^{+} \bigr\rangle \\& \quad=\frac{1}{p} \biggl(\bigl\Vert v_{n}^{-} \bigr\Vert ^{p}- \int_{\Omega}\bigl\vert \nabla v_{n}^{-} \bigr\vert ^{p-2}\nabla v_{n}^{-}\nabla u_{\lambda}\,dx-\bigl\langle I'\bigl(u_{\lambda}+v_{n}^{+} \bigr),u_{\lambda}+v_{n}^{+}\bigr\rangle \biggr)+I \bigl(u_{\lambda}+v_{n}^{+}\bigr)-I(u_{\lambda}) \\& \quad\leq c+1+o(1)\bigl\Vert u_{\lambda}+v_{n}^{+} \bigr\Vert . \end{aligned}$$
Therefore, we have
$$ \begin{aligned}[b] &\frac{1}{p} \biggl(\bigl\Vert v_{n}^{-}\bigr\Vert ^{p}- \int_{\Omega}\bigl\vert \nabla v_{n}^{-} \bigr\vert ^{p-2}\nabla v_{n}^{-}\nabla u_{\lambda}\,dx \biggr)+ \biggl(\frac{1}{p}- \frac{1}{p^{\ast}(s)} \biggr) \int_{\Omega}\frac{(u_{\lambda }+v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx \\ & \quad\quad{} +\lambda \int_{\Omega} \biggl[\frac{1}{p}f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr) \bigl(u_{\lambda }+v_{n}^{+}\bigr)-F \bigl(x,u_{\lambda}+v_{n}^{+}\bigr) \biggr]\,dx \\ &\quad\leq I(u_{\lambda})+c+1+o(1)\bigl\Vert u_{\lambda}+v_{n}^{+} \bigr\Vert . \end{aligned} $$
(3.6)
By \((A_{1})\) and the boundedness of Ω, for any \(\varepsilon>0\), there is \(M_{5}=M_{5}(\varepsilon)> 0\) such that
$$\begin{aligned}& \bigl\vert f(x,t)t\bigr\vert \leq\varepsilon\frac{\vert t\vert ^{p^{\ast}(s)}}{\vert x\vert ^{s}} \quad\text{for } x \in\Omega\setminus\{0\} \text{ and } \vert t\vert >M_{5}, \\& \bigl\vert f(x,t)t\bigr\vert \leq C_{3}(\varepsilon) \quad\text{for } x\in \Omega\text{ and } \vert t\vert \in[0,M_{5}]; \\& \bigl\vert F(x,t)\bigr\vert \leq\frac{\varepsilon}{p}\frac{\vert t\vert ^{p^{\ast}(s)}}{\vert x\vert ^{s}} \quad\text{for } x\in\Omega\setminus\{0\} \text{ and } \vert t\vert >M_{5}, \\& \bigl\vert F(x,t)\bigr\vert \leq C_{4}(\varepsilon) \quad\text{for } x\in \Omega\text{ and } \vert t\vert \in[0,M_{5}], \end{aligned}$$
where \(C_{3}(\varepsilon), C_{4}(\varepsilon)>0\). Thus, we have
$$\begin{aligned}& \bigl\vert f(x,t)t\bigr\vert \leq C_{3}(\varepsilon)+\varepsilon \frac{\vert t\vert ^{p^{\ast }(s)}}{\vert x\vert ^{s}}, \quad (x,t) \in\bigl(\Omega\setminus\{0\} \bigr)\times\mathbb{R}, \end{aligned}$$
(3.7)
$$\begin{aligned}& \bigl\vert F(x,t)\bigr\vert \leq C_{4}(\varepsilon)+ \frac{\varepsilon}{p}\frac{\vert t\vert ^{p^{\ast }(s)}}{\vert x\vert ^{s}}, \quad (x,t) \in\bigl(\Omega\setminus\{0\} \bigr) \times\mathbb{R}. \end{aligned}$$
(3.8)
Let \(C(\varepsilon)=\frac{1}{p}C_{3}(\varepsilon)+C_{4}(\varepsilon)\), by (3.7) and (3.8), we have
$$ F(x,t)-\frac{1}{p}f(x,t)t\leq C(\varepsilon)+\frac{2\varepsilon}{p} \frac {\vert t\vert ^{p^{\ast}(s)}}{\vert x\vert ^{s}},\quad (x,t) \in \bigl(\Omega\setminus\{0\}\bigr)\times\mathbb{R}. $$
(3.9)
By (3.6) and (3.9), we have
$$\begin{aligned}& \biggl(\frac{p-s}{p(N-s)}-\frac{2\lambda\varepsilon}{p} \biggr) \int_{\Omega}\frac{(u_{\lambda}+v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx \\& \quad \leq\lambda C(\varepsilon) \vert \Omega \vert -\frac{1}{p}\bigl\Vert v_{n}^{-} \bigr\Vert ^{p}+C_{5}\bigl\Vert v_{n}^{-} \bigr\Vert ^{p-1}+C_{6}+o(1)\bigl\Vert u_{\lambda}+v_{n}^{+}\bigr\Vert , \end{aligned}$$
where \(C_{5}=\frac{1}{p}\Vert u_{\lambda} \Vert \) and \(C_{6}=I(u_{\lambda})+c+1\). Let \(\varepsilon=\frac{p-s}{4(N-s)\lambda}\), then we have
$$\int_{\Omega}\frac{(u_{\lambda}+v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx\leq C_{7}\bigl\Vert v_{n}^{-}\bigr\Vert ^{p-1}+C_{8} +o(1)\bigl\Vert u_{\lambda}+v_{n}^{+}\bigr\Vert , $$
where \(C_{7}=\frac{2p(N-s)}{p-s}C_{5}\) and \(C_{8}=\frac{2p(N-s)}{p-s}(\lambda C(\varepsilon)\vert \Omega \vert +C_{6})\). Together with (3.3), (3.5), and (3.8), we have
$$\begin{aligned}& \frac{1-\varepsilon}{p}\bigl\Vert v_{n}^{-}\bigr\Vert ^{p}+\frac{1}{p} \bigl[(1-\varepsilon)\bigl\Vert v_{n}^{+}\bigr\Vert ^{p} -\overline{C_{\varepsilon}} \Vert u_{\lambda} \Vert ^{p}-(1-\varepsilon)\bigl\Vert v_{n}^{+}\bigr\Vert ^{p-1} \bigr] \\& \quad\leq\frac{1}{p}\bigl\Vert v_{n}^{-}\bigr\Vert ^{p}+\frac{1}{p} \bigl[(1-\varepsilon)\bigl\Vert v_{n}^{+}\bigr\Vert ^{p} -\overline{C_{\varepsilon}} \Vert u_{\lambda} \Vert ^{p} \bigr] \\& \quad\leq\frac{1}{p}\bigl\Vert v_{n}^{-}\bigr\Vert ^{p} +\frac{1}{p}\bigl\vert \bigl(\bigl\Vert v_{n}^{+}\bigr\Vert -\Vert u_{\lambda} \Vert \bigr)\bigr\vert ^{p} \\& \quad\leq\frac{1}{p}\bigl\Vert v_{n}^{-}\bigr\Vert ^{p}+\frac{1}{p}\bigl\Vert u_{\lambda}+v_{n}^{+} \bigr\Vert ^{p} \\& \quad=\frac{1}{p^{\ast}(s)} \int_{\Omega}\frac{(u_{\lambda }+v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx+ \lambda \int_{\Omega}F\bigl(x,u_{\lambda}+v_{n}^{+} \bigr)\,dx+J(v_{n})+I(u_{\lambda })+o(1) \\& \quad\leq C_{9}\bigl\Vert v_{n}^{-}\bigr\Vert ^{p-1}+C_{10}+o(1)\bigl\Vert u_{\lambda}+v_{n}^{+} \bigr\Vert , \end{aligned}$$
where the second inequality is due to the elementary inequality
$$\vert a-b\vert ^{t}\geq(1-\varepsilon)a^{t}- \overline{C_{\varepsilon}}b^{t}, \quad t\geq1, a, b>0. $$
Here, \(C_{9}= (\frac{1}{p^{\ast}(s)}+\frac{\lambda\varepsilon }{p} )C_{7}\) and \(C_{10}=\lambda C_{4}(\varepsilon)\vert \Omega \vert + (\frac{1}{p^{\ast}(s)} +\frac{\lambda\varepsilon}{p} )C_{8}+I(u_{\lambda})+c+o(1)\). Since \(\Vert v_{n}^{-}\Vert ^{p}+\Vert v_{n}^{+}\Vert ^{p}=\Vert v_{n}\Vert ^{p}\), we get
$$\Vert v_{n}\Vert ^{p}-C_{11}\bigl\Vert v_{n}^{+}\bigr\Vert ^{p-1}-C'_{11} \bigl\Vert v_{n}^{-}\bigr\Vert ^{p-1}\leq C_{12}+o(1)\Vert u_{\lambda} \Vert , $$
where \(C_{11}=1+o(1)\), \(C'_{11}=\frac{C_{9}p}{1-\varepsilon}\), \(C_{12}=\frac{\overline{C_{\varepsilon}} \Vert u_{\lambda} \Vert ^{p}+pC_{10}}{1-\varepsilon}\). So we get
$$\Vert v_{n}\Vert ^{p}-C_{13}\Vert v_{n}\Vert ^{p-1}\leq C_{12}+o(1)\Vert u_{\lambda} \Vert , $$
where \(C_{13}=C_{11}+C'_{11}\). It shows that \(\{v_{n}\}\) is bounded in \(W_{0}^{1,p}(\Omega)\), going if necessary to a subsequence, we have
$$ \begin{aligned} &v_{n}\rightharpoonup v_{0} \quad\text{weakly in } W^{1,p}_{0}(\Omega), \\ &v_{n}\rightarrow v_{0} \quad\text{strongly in } L^{\gamma}(\Omega), 1< \gamma< p^{\ast}, \\ &v_{n}\rightarrow v_{0} \quad\text{a.e. in } \Omega, \end{aligned} $$
(3.10)
as \(n\rightarrow\infty\).
Since \(v_{n}\) is bounded in \(W^{1,p}_{0}(\Omega)\), it follows from the Sobolev embedding theorem that there is \(M'>0\) such that \(\Vert u_{\lambda}+v_{n}^{+}\Vert ^{p^{\ast}(s)}_{p^{\ast }(s)}\leq M'\). Let measE denote the measure of E. By \((A_{1})\), for any \(\varepsilon>0\), there is \(C_{14}(\varepsilon)>0\) such that
$$\bigl\vert f(x,t)t\bigr\vert \leq C_{14}(\varepsilon)+ \frac{\varepsilon}{2M'}\vert t\vert ^{p^{\ast }(s)}, \quad (x,t) \in\overline{\Omega} \times\mathbb{R}. $$
Let \(\delta=\frac{\varepsilon}{2C_{14}(\varepsilon)}>0\), if \(E\subset\Omega\), \(\operatorname {meas}E<\delta\), we have
$$\begin{aligned} \biggl\vert \int_{E}f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr) \bigl(u_{\lambda}+v_{n}^{+}\bigr)\,dx\biggr\vert \leq& \int_{E}\bigl\vert f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr) \bigl(u_{\lambda }+v_{n}^{+}\bigr)\bigr\vert \,dx \\ \leq& \int_{E}C_{14}(\varepsilon)\,dx+\frac{\varepsilon}{2M'} \int_{E}\bigl\vert u_{\lambda}+v_{n}^{+} \bigr\vert ^{p^{\ast}(s)}\,dx \\ \leq& C_{14}(\varepsilon)\operatorname {meas}E+\frac{\varepsilon}{2}< \varepsilon. \end{aligned}$$
By the Vitali theorem, we have
$$\int_{\Omega}f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr) \bigl(u_{\lambda}+v_{n}^{+}\bigr)\,dx \rightarrow \int_{\Omega}f\bigl(x,u_{\lambda}+v_{0}^{+} \bigr) \bigl(u_{\lambda }+v_{0}^{+}\bigr)\,dx \quad\text{as }n \rightarrow\infty. $$
Hence,
$$ \begin{aligned}[b] \int_{\Omega}f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr) (u_{\lambda}+v_{n})\,dx&= \int_{\Omega}f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr) \bigl(u_{\lambda}+v_{n}^{+}\bigr)\,dx- \int_{\Omega}f(x,u_{\lambda}) \bigl(v_{n}^{-} \bigr)\,dx \\ &\rightarrow \int_{\Omega}f\bigl(x,u_{\lambda}+v_{0}^{+} \bigr) (u_{\lambda }+v_{0})\,dx \quad\text{as } n\rightarrow\infty. \end{aligned} $$
(3.11)
By the same method, we get
$$ \begin{aligned} &\int_{\Omega}f\bigl(x,u_{\lambda}+v_{n}^{+} \bigr)\omega\,dx\rightarrow \int_{\Omega}f\bigl(x,u_{\lambda}+v_{0}^{+} \bigr)\omega\,dx, \\ &\int_{\Omega}F\bigl(x,u_{\lambda}+v_{n}^{+} \bigr)\,dx\rightarrow \int_{\Omega }F\bigl(x,u_{\lambda}+v_{0}^{+} \bigr)\,dx \end{aligned} $$
(3.12)
as \(n\rightarrow\infty\) for \(\omega\in W^{1,p}_{0}(\Omega)\). Similar to the proof of Theorem 1.1, we have
$$0=\lim_{n\rightarrow\infty}\bigl\langle J'(v_{n}), \omega\bigr\rangle =\bigl\langle J'(v_{0}),\omega\bigr\rangle $$
for \(\omega\in W^{1,p}_{0}(\Omega)\), which implies that \(J'(v_{0})=0\). Therefore, \(v_{0}\) is a critical point of J in \(W^{1,p}_{0}(\Omega)\). By the assumption that \(v=0\) is the only critical point of J, we have \(v_{0}\)=0. Now, we want to prove \(v_{0}\rightarrow0\) strongly in \(W^{1,p}_{0}(\Omega)\). By (3.10), (3.12), and the Brezis-Leib Lemma (see [25]), we have
$$J(v_{n})=\frac{1}{p}\bigl\Vert v_{n}^{-} \bigr\Vert ^{p}+I\bigl(u_{\lambda}+v_{n}^{+} \bigr) -I(u_{\lambda})=\frac{1}{p}\Vert v_{n}\Vert ^{p}-\frac{1}{p^{\ast}(s)} \int_{\Omega}\frac{(v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\, dx+o(1). $$
Therefore,
$$\bigl\langle J'(v_{n}),v_{n}\bigr\rangle = \Vert v_{n}\Vert ^{p}- \int_{\Omega}\frac{(v_{n}^{+})^{p^{\ast }(s)}}{\vert x\vert ^{s}}\,dx+o(1)\rightarrow0. $$
In fact, \(\Vert v_{n}\Vert ^{p}\rightarrow0\) as \(n\rightarrow\infty\). If not, then there is a subsequence (still denoted by \(v_{n}\)) such that
$$\lim_{n\rightarrow\infty} \Vert v_{n}\Vert ^{p}=k,\quad\quad \lim_{n\rightarrow\infty} \int_{\Omega} \frac{(v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx=k,\quad k>0. $$
By (1.2), we get
$$\Vert v_{n}\Vert ^{p}\geq A _{\mu,s} \biggl( \int_{\Omega} \frac{(v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx \biggr)^{\frac{p}{p^{\ast }(s)}},\quad \text{for all } n\in\mathbb{N}. $$
Then \(k\geq A_{\mu,s}k^{\frac{p}{p^{\ast}(s)}}\), i.e., \(k\geq A_{\mu,s}^{\frac{N-s}{p-s}}\). Thus, we have
$$\begin{aligned} c =&o(1)+J(v_{n})=\frac{1}{p}\Vert v_{n}\Vert ^{p}-\frac{1}{p^{\ast}(s)} \int_{\Omega}\frac{(v_{n}^{+})^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx+o(1) \\ =&\frac{p-s}{p(N-s)}k+o(1) \\ \geq&\frac{p-s}{p(N-s)}A _{\mu,s}^{\frac{N-s}{p-s}}. \end{aligned}$$
It is a contradiction. So \(v_{n}\rightarrow0\) strongly in \(W_{0}^{1,p}(\Omega)\) as \(n\rightarrow \infty\). □

Lemma 3.3

[21]

If \(1< p< N\), \(0 \leq s< p\) and \(0\leq\mu<\mu_{1}\), then the limiting problem
$$ \textstyle\begin{cases} -\triangle_{p}u-\mu\frac{u^{p-1}}{\vert x\vert ^{p}}=\frac {u^{p^{\ast }(s)-1}}{\vert x\vert ^{s}}, &\textit{in } \mathbb{R}^{N}\setminus\{ 0\},\\ u>0, &\textit{in } \mathbb{R}^{N}\setminus\{0\},\\ u\in D^{1,p}(\mathbb{R}^{N}), \end{cases} $$
(P)
has radially symmetric ground states,
$$\widetilde{V}_{\varepsilon}(x) :=\varepsilon^{-\frac{N-p}{p}} \widetilde{U}_{p,\mu} \biggl(\frac {x}{\varepsilon} \biggr)= \varepsilon^{-\frac{N-p}{p}}\widetilde{U}_{p,\mu} \biggl(\frac {\vert x\vert }{\varepsilon} \biggr), \quad \forall\varepsilon>0, $$
and it satisfies
$$\int_{\mathbb{R}^{N}} \biggl(\bigl\vert \nabla\widetilde{V}_{\varepsilon}(x) \bigr\vert ^{p}- \mu\frac{\vert \widetilde{V}_{\varepsilon }(x)\vert ^{p}}{\vert x\vert ^{p}} \biggr)\,dx= \int_{\mathbb{R}^{N}} \frac{\vert \widetilde{V}_{\varepsilon }(x)\vert ^{p^{\ast}(s)}}{\vert x\vert ^{s}}\,dx=A_{\mu ,s}^{\frac{N-s}{p-s}}, $$
where \(\widetilde{U}_{p,\mu}(x)=\widetilde{U}_{p,\mu}(\vert x\vert )\) is the unique radial solution of (P), satisfying
$$\widetilde{U}_{p,\mu}(1)= \biggl(\frac{(N-s)(\mu_{1}-\mu)}{N-p} \biggr)^{\frac{1}{p^{\ast}(s)-p}}. $$
Moreover, \(\widetilde{U}_{p,\mu}\) has the following properties:
$$\begin{aligned}& \lim_{r\rightarrow0}r^{a(\mu)}\widetilde{U}_{p,\mu}(r)=c_{1}>0,\quad\quad \lim_{r\rightarrow+\infty}r^{b(\mu)}\widetilde{U}_{p,\mu}(r)=c_{2}>0, \\& \lim_{r\rightarrow0}r^{a(\mu)+1}\widetilde{U}'_{p,\mu}(r)=c_{1}a( \mu)\geq0, \quad\quad \lim_{r\rightarrow+\infty}r^{b(\mu)+1}\widetilde{U}'_{p,\mu }(r)=c_{2}b( \mu)>0, \end{aligned}$$
where \(c_{1}\) and \(c_{2}\) are positive constants depending on p and N; \(a(\mu)\) and \(b(\mu)\) are zeros of the function
$$f(t)=(p-1)t^{p}-(N-p)t^{p-1}+\mu, \quad t\geq0, 0\leq\mu< \mu_{1}, $$
satisfying \(0\leq a(\mu)<\frac{N-p}{p}<b(\mu)<\frac{N-p}{p-1}\).
Since \(u_{\lambda}>0\) is a solution of (1.1), similar to the proof of Theorem 1.1 in [26], there are constants \(R>0\) and \(r_{0}>0\) such that \(B_{2R}(0)\subset\Omega\) and
$$ 0< r_{0}\leq u_{\lambda}(x), \quad \forall x\in B_{2R}(0) \setminus\{0\}. $$
(3.13)
Let \(\varphi\in C_{0}^{\infty}(\Omega)\) such that \(0\leq\varphi(x)\leq1\) and
$$\varphi(x):= \textstyle\begin{cases} 1, & \vert x\vert \leq R,\\ 0, & \vert x\vert \geq 2R, \end{cases} $$
where \(B_{2R}(0)\subset\Omega\). Set \(v_{\varepsilon}(x)=\varphi(x)\widetilde {V}_{\varepsilon}(x)\), \(\varepsilon>0\), where \(\widetilde{V}_{\varepsilon}(x)\) is defined in Lemma 3.3. Then we can get the following results by the method used in [27]:
$$\begin{aligned}& \Vert v_{\varepsilon} \Vert ^{p}= A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl(\varepsilon^{b(\mu)p+p-N} \bigr), \end{aligned}$$
(3.14)
$$\begin{aligned}& \int_{\Omega}\frac{\vert v_{\varepsilon} \vert ^{p^{\ast }(s)}}{\vert x\vert ^{s}}\,dx= A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl(\varepsilon^{b(\mu)p^{\ast }(s)+s-N} \bigr), \end{aligned}$$
(3.15)
$$\begin{aligned}& \int_{\Omega}\frac{\vert v_{\varepsilon} \vert ^{r}}{\vert x\vert ^{s}}\,dx=O \bigl(\varepsilon ^{p-s} \bigr), \quad \frac{N-s}{b(\mu)}< r< p^{\ast}(s). \end{aligned}$$
(3.16)

Lemma 3.4

For \(\gamma\geq2\), \(1\leq t\leq\gamma-1\), \(\forall a,b>0\), there exists a positive constant C such that
$$(a+b)^{\gamma}\geq a^{\gamma}+b^{\gamma}+Ca^{\gamma-t}b^{t}. $$

Proof

To prove this lemma, we only need to prove
$$(1+x)^{\gamma}\geq1+x^{\gamma}+Cx^{t},\quad 0< x< \infty. $$
Let \(\gamma=k+\theta\), \(t=m+\eta\), where \(k\geq2\), \(1\leq m\leq k-1\) are integral numbers and \(0\leq\eta\leq\theta<1\) are real numbers. Clearly,
$$\begin{aligned} (1+x)^{\gamma} =&(1+x)^{k+\theta}=(1+x)^{k}(1+x)^{\theta} \geq\bigl(1+x^{k}+Cx^{m}\bigr) (1+x)^{\theta} \\ \geq&1+x^{k+\theta}+Cx^{m}(1+x)^{\theta} \\ \geq&1+x^{k+\theta}+Cx^{m}x^{\eta}=1+x^{\gamma}+Cx^{t}. \end{aligned}$$
 □

Lemma 3.5

If \(N\geq3\), \(0\leq s< p\), \(0\leq\mu<\mu_{1}\), \(1< p< N\), \((A_{1})\), \((A_{2})\), \((A_{3})\) (or \((A_{3}')\)), and \(f(x,0)\equiv0\) hold, then there is \(v_{\ast}\in W^{1,p}_{0}(\Omega)\), \(v_{\ast}\not\equiv0\), such that
$$\sup_{t\geq0} J(tv_{\ast})< \frac{p-s}{p(N-s)}A_{\mu,s}^{\frac{N-s}{p-s}}. $$

Proof

By (3.2), \((A_{2})\), and Lemma 3.4, we have
$$g(x,l)\geq\frac{l^{p^{\ast}(s)-1}}{\vert x\vert ^{s}}+C\frac{l^{p-1} u_{\lambda}^{p^{\ast}(s)-p}}{\vert x\vert ^{s}}. $$
By \((A_{3})\) or \((A_{3}')\), we have \(p\geq2\) and \(s\leq N-\frac{(N-p)(1+p)}{p}\), which imply \(p^{\ast}(s)-1\geq2\) and \(1\leq p-1\leq(p^{\ast}(s)-1)-1\). Therefore,
$$G(x,tv_{\varepsilon})\geq\frac{t^{p^{\ast}(s)}}{p^{\ast}(s)}\frac {v_{\varepsilon}^{p^{\ast}(s)}}{\vert x\vert ^{s}} + \frac{C t^{p}}{p}\frac{v_{\varepsilon}^{p}u_{\lambda}^{p^{\ast}(s)-p}}{\vert x\vert ^{s}}. $$
From \((A_{3})\) (or \((A_{3}')\)), we have \(s>N-Pb(\mu)\), which implies \(p>\frac{N-s}{b(\mu)}\), so (3.16) holds. So by (3.13)-(3.16), we have
$$\begin{aligned} J(tv_{\varepsilon}) =&\frac{t^{p}}{p}\Vert v_{\varepsilon} \Vert ^{p}- \int_{\Omega}G(x,tv_{\varepsilon})\,dx \\ \leq&\frac{t^{p}}{p}\Vert v_{\varepsilon} \Vert ^{p}- \frac{t^{p^{\ast }(s)}}{p^{\ast}(s)} \int_{\Omega}\frac{\vert v_{\varepsilon} \vert ^{p^{\ast }(s)}}{\vert x\vert ^{s}}\,dx-C_{15} t^{p} \int_{\Omega}\frac{v_{\varepsilon}^{p}}{\vert x\vert ^{s}}\,dx \\ =&\frac{t^{p}}{p} \bigl(A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl(\varepsilon ^{b(\mu)p+p-N} \bigr) \bigr) -\frac{t^{p^{\ast}(s)}}{p^{\ast}(s)} \bigl(A^{\frac{N-s}{p-s}}_{\mu ,s}+O \bigl(\varepsilon^{b(\mu)p^{\ast}(s)+s-N} \bigr) \bigr) -C_{15}t^{p}O \bigl(\varepsilon^{p-s} \bigr), \end{aligned}$$
where \(C_{15}=\frac{Cr_{0}^{p^{\ast}(s)-p}}{p}\). Let
$$Q(t):=\frac{t^{p}}{p} \bigl(A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl( \varepsilon^{b(\mu)p+p-N} \bigr) \bigr) -\frac{t^{p^{\ast}(s)}}{p^{\ast }(s)} \bigl(A^{\frac{N-s}{p-s}}_{\mu ,s}+O \bigl(\varepsilon^{b(\mu)p^{\ast}(s)+s-N} \bigr) \bigr) -C_{15}t^{p}O \bigl(\varepsilon^{p-s} \bigr). $$
Clearly, the following equation:
$$0=Q'(t)=t^{p-1} \bigl[A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl(\varepsilon^{b(\mu)p+p-N} \bigr) -O \bigl(\varepsilon^{p-s} \bigr) \bigr] -t^{p^{\ast}(s)-1} \bigl[A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl( \varepsilon^{b(\mu)p^{\ast}(s)+s-N} \bigr) \bigr] $$
has only a positive root
$$t_{\varepsilon}= \biggl(\frac{A^{\frac{N-s}{p-s}}_{\mu,s}+O (\varepsilon^{b(\mu)p+p-N} ) -O (\varepsilon^{p-s} )}{ A^{\frac{N-s}{p-s}}_{\mu,s}+O (\varepsilon^{b(\mu)p^{\ast }(s)+s-N} )} \biggr)^{\frac{1}{p^{\ast}(s)-p}}. $$
We have
$$\begin{aligned} Q(t_{\varepsilon}) =&\frac{t_{\varepsilon}^{p}}{p} \bigl(A^{\frac {N-s}{p-s}}_{\mu,s}+O \bigl(\varepsilon^{b(\mu)p+p-N} \bigr) -O \bigl(\varepsilon^{p-s} \bigr) \bigr)-\frac{t_{\varepsilon}^{p^{\ast }(s)}}{p^{\ast}(s)} \bigl(A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl( \varepsilon^{b(\mu)p^{\ast }(s)+s-N} \bigr) \bigr) \\ =& \biggl(\frac{1}{p}-\frac{1}{p^{\ast}(s)} \biggr) \bigl[A^{\frac {N-s}{p-s}}_{\mu,s} +O \bigl(\varepsilon^{b(\mu)p^{\ast}(s)+s-N} \bigr) \bigr] \biggl[\frac {A^{\frac{N-s}{p-s}}_{\mu,s}+O (\varepsilon^{b(\mu )p+p-N} ) -O (\varepsilon^{p-s} )}{ A^{\frac{N-s}{p-s}}_{\mu,s}+O (\varepsilon^{b(\mu)p^{\ast }(s)+s-N} )} \biggr] ^{\frac{p^{\ast}(s)}{p^{\ast}(s)-p}} \\ =&\frac{p-s}{p(N-s)} \bigl[A^{\frac{N-s}{p-s}}_{\mu,s} +O \bigl( \varepsilon^{b(\mu)p^{\ast}(s)+s-N} \bigr) \bigr]^{-\frac{N-p}{p-s}} \bigl[A^{\frac{N-s}{p-s}}_{\mu,s}+O \bigl(\varepsilon^{b(\mu)p+p-N} \bigr) -O \bigl(\varepsilon^{p-s} \bigr) \bigr]^{\frac{N-s}{p-s}} \\ =&\frac{p-s}{p(N-s)}A_{\mu,s}^{\frac{N-s}{p-s}}+O \bigl(\varepsilon ^{b(\mu)p^{\ast}(s)+s-N} \bigr) +O \bigl(\varepsilon^{b(\mu)p+p-N} \bigr)-O \bigl( \varepsilon^{p-s} \bigr). \end{aligned}$$
By \(s>N-pb(\mu)\) (see \((A_{3})\) or \((A_{3}')\)), we have
$$b(\mu)p+p-N>p-s. $$
Since \(b(\mu)>\frac{N-p}{p}\) implies \(b(\mu)p^{\ast}(s)+s-N>b(\mu)p+p-N\), we have
$$b(\mu)p^{\ast}(s)+s-N>p-s. $$
Since \(Q(0)=0\) and \(\lim_{t\rightarrow+\infty}Q(t)=-\infty\), we have
$$\sup_{t\geq0}Q(t)=Q(t_{\varepsilon})< \frac{p-s}{p(N-s)}A_{\mu,s}^{\frac {N-s}{p-s}} $$
for \(\varepsilon>0\) sufficiently small. So we get
$$\sup_{t\geq0}J(tv_{\varepsilon})\leq\sup_{t\geq0}Q(t)< \frac {p-s}{p(N-s)}A_{\mu,s}^{\frac{N-s}{p-s}}, $$
for \(\varepsilon>0\) sufficiently small. It completes the proof if we let \(v_{\ast}=v_{\varepsilon}\) with \(\varepsilon>0\) being sufficiently small. □

Proof of Theorem 1.2

If \(v=0\) is the only critical point of J in \(W^{1,p}_{0}(\Omega)\). By Lemma 3.1, we know there is \(\alpha>0\) such that \(J(v)>\alpha\), \(\forall v\in\partial B_{\rho}=\{v\in W^{1,p}_{0}(\Omega), \Vert v\Vert =\rho\}\), where \(\rho>0\) is small enough. Lemma 3.5 implies that there is \(v_{\ast}\in W^{1,p}_{0}(\Omega)\) and \(v_{\ast}\not\equiv0\) such that
$$\sup_{t\geq0} J(tv_{\ast})< \frac{p-s}{p(N-s)}A_{\mu,s}^{\frac{N-s}{p-s}}. $$
By (3.8), we get \(\lim_{t\rightarrow\infty}J(tv_{\ast})\rightarrow-\infty\). Hence, we can choose \(t_{0}>0\) such that \(\Vert t_{0}v_{\ast} \Vert >\rho\) and \(J(t_{0}v_{\ast})<0\). By the mountain pass lemma in [22], there is a sequence \(\{v_{n}\}\subset W^{1,p}_{0}(\Omega)\) satisfying
$$J(v_{n})\rightarrow c\geq\alpha\quad\text{and} \quad J'(v_{n}) \rightarrow0, $$
where
$$\begin{aligned}& c=\inf_{h\in\Gamma} \max_{t\in[0,1]} J\bigl(h(t)\bigr), \\& \Gamma=\bigl\{ h\in C\bigl([0,1], X\bigr)\mid h(0)=0, h(1)=t_{0}v_{\ast} \bigr\} . \end{aligned}$$
We have
$$0< \alpha\leq c=\inf_{h\in\Gamma} \max_{t\in[0,1]} J \bigl(h(t)\bigr)\leq\max_{t\in[0,1]} J(tt_{0}v_{\ast}) \leq\sup_{t\geq0} J(tv_{\ast})< \frac{p-s}{p(N-s)}A_{\mu,s}^{\frac{N-s}{p-s}}, $$
and this together with Lemma 3.2 implies that \(v_{n}\rightarrow0\) strongly in \(W^{1,p}_{0}(\Omega)\) as \(n\rightarrow\infty\). Hence, we have \(0=J(0)=\lim_{n\rightarrow\infty}J(v_{n})=c\geq\alpha>0\), a contradiction. So, Theorem 1.2 holds. □

Declarations

Acknowledgements

We would like to thank the referee for his/her valuable comments, which have led to an improvement of the presentation of this paper. Research supported by National Natural Science Foundation of China (No. 11401011).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Anyang Normal University
(2)
School of Mathematical Sciences, University of Jinan

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