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  • Research
  • Open Access

Two positive solutions for quasilinear elliptic equations with singularity and critical exponents

Boundary Value Problems20182018:97

https://doi.org/10.1186/s13661-018-1018-7

  • Received: 25 November 2017
  • Accepted: 16 June 2018
  • Published:

Abstract

In this paper, we consider the quasilinear elliptic equation with singularity and critical exponents
$$ \textstyle\begin{cases} -\Delta_{p}u-\mu \frac{ \vert u \vert ^{p-2}u}{ \vert x \vert ^{p}}=Q(x) \frac{ \vert u \vert ^{p^{*}(t)-2}u}{ \vert x \vert ^{t}}+\lambda u^{-s}, &\text{in }\Omega , \\ u>0, & \text{in }\Omega , \\ u=0, &\text{on }\partial \Omega , \end{cases} $$
where \(\Delta_{p}= \operatorname {div}(|\nabla u|^{p-2}\nabla u)\) is a p-Laplace operator with \(1< p< N\). \(p^{*}(t):=\frac{p(N-t)}{N-p}\) is a critical Sobolev–Hardy exponent. We deal with the existence of multiple solutions for the above problem by means of variational and perturbation methods.

Keywords

  • Quasilinear
  • Singularity
  • Critical
  • Sobolev–Hardy exponent

1 Introduction and preliminaries

The main goal of this paper is to consider the following singular boundary value problem:
$$ \textstyle\begin{cases} -\Delta_{p}u-\mu \frac{ \vert u \vert ^{p-2}u}{ \vert x \vert ^{p}}=Q(x) \frac{ \vert u \vert ^{p^{*}(t)-2}u}{ \vert x \vert ^{t}}+\lambda u^{-s}, &\text{in }\Omega , \\ u>0, & \text{in } \Omega , \\ u=0, & \text{on } \partial \Omega , \end{cases} $$
(1.1)
where Ω is a bounded domain in \(\mathbb{R}^{N}\), \(\Delta_{p}= \operatorname {div}( \vert \nabla u \vert ^{p-2}\nabla u)\) is a p-Laplace operator with \(1< p< N\). \(\lambda >0\), \(0< s<1\), \(0\leq t< p\), and \(0\leq \mu <\bar{ \mu }:=(\frac{N-p}{p})^{p}\). \(p^{*}(t):=\frac{p(N-t)}{N-p}\) is a critical Sobolev–Hardy exponent, \(Q(x)\in C(\overline{\Omega })\) and \(Q(x)\) is positive on Ω̅.
In recent years, the elliptic boundary value problems with critical exponents and singular potentials have been extensively studied [2, 6, 7, 1023, 25, 26, 28, 3034]. In [19], Han considered the following quasilinear elliptic problem with Hardy term and critical exponent:
$$ \textstyle\begin{cases} -\Delta_{p}u-\mu \frac{ \vert u \vert ^{p-2}u}{ \vert x \vert ^{p}}=Q(x) \vert u \vert ^{p^{*} -2}u+ \lambda \vert u \vert ^{p-2}u, & \text{in }\Omega , \\ u=0, & \text{on } \partial \Omega , \end{cases} $$
(1.2)
where \(1< p< N\). The existence of multiple positive solutions for (1.2) was established. Furthermore, Hsu [21] studied the following quasilinear equation:
$$ \textstyle\begin{cases} -\Delta_{p}u-\mu \frac{ \vert u \vert ^{p-2}u}{ \vert x \vert ^{p}}=Q(x) \vert u \vert ^{p^{*} -2}u+ \lambda f(x) \vert u \vert ^{q-2}u, & \text{in }\Omega , \\ u=0, & \text{on } \partial \Omega , \end{cases} $$
(1.3)
where \(1< q< p< N\). We should point out that the authors of [19, 21] both investigated the effect of \(Q(x)\). If \(p=2\), \(\mu =0\), and \(t=0\), Liao et al. [27] proved the existence of two solutions for problem (1.1) by the constrained minimizer and perturbation methods.

Compared with [2, 4, 8, 12, 19, 21, 22, 29], problem (1.1) contains the singular term \(\lambda u^{-s}\). Thus, the functional corresponding to (1.1) is not differentiable on \(W_{0}^{1,p}(\Omega )\). We will remove the singularity by the perturbation method. Our idea comes from [24, 27].

Definition 1.1

A function \(u\in W_{0}^{1,p}(\Omega )\) is a weak solution of problem (1.1) if, for every \(\varphi \in W_{0}^{1,p} ( \Omega )\), there holds
$$ \int_{\Omega } \biggl( \vert \nabla u \vert ^{p-2} \nabla u \nabla \varphi -\mu \frac{ \vert u \vert ^{p-2}u \varphi }{ \vert x \vert ^{p}} \biggr) \,dx = \int_{\Omega } \biggl( \frac{Q(x)(u^{+})^{p^{*}(t)-1}\varphi }{ \vert x \vert ^{t}} + \lambda \bigl(u^{+} \bigr)^{-s} \varphi \biggr) \,dx. $$
The energy functional corresponding to (1.1) is defined by
$$ I_{\lambda ,\mu }(u)=\frac{1}{p} \int_{\Omega } \biggl( \vert \nabla u \vert ^{p}- \mu \frac{ \vert u \vert ^{p}}{ \vert x \vert ^{p}} \biggr) \,dx-\frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{\lambda }{1-s} \int_{\Omega } \bigl(u^{+} \bigr)^{1-s}\,dx. $$
Throughout this paper, Q satisfies
\((Q_{1})\)
\(Q(0)=Q_{M}=\max_{x\in \overline{\Omega }}Q(x)\) and there exists \(\beta \geq p(b(\mu )-\frac{N-p}{p})\) such that
$$ Q(x)-Q(0)=o \bigl( \vert x \vert ^{\beta } \bigr), \quad \text{as } x \rightarrow 0, $$
where \(b(\mu )\) is given in Sect. 1.
In this paper, we use the following notations:
  1. (i)

    \(\Vert u \Vert ^{p}= \int_{\Omega } ( \vert \nabla u \vert ^{p}-\mu \frac{ \vert u \vert ^{p}}{ \vert x \vert ^{p}} ) \,dx\) is the norm in \(W_{0}^{1,p}(\Omega )\), and the norm in \(L^{p}(\Omega )\) is denoted by \(\vert \cdot \vert _{p}\);

     
  2. (ii)

    \(C,C_{1},C_{2},C_{3},\ldots \) denote various positive constants;

     
  3. (iii)

    \(u^{+}_{n} (x)=\max \{u_{n},0\}\), \(u^{-}_{n} (x)=\max \{0,-u _{n}\}\);

     
  4. (iv)
    We define
    $$ \partial B_{r}= \bigl\{ u \in W_{0}^{1,p}(\Omega ): \Vert u \Vert =r \bigr\} , \quad\quad B_{r}= \bigl\{ u \in W_{0}^{1,p}(\Omega ): \Vert u \Vert \leq r \bigr\} . $$
    Let S be the best Sobolev–Hardy constant
    $$ S:=\inf_{u\in W^{1,p}_{0}(\Omega )\backslash \{0\}}\frac{ \int_{\Omega }( \vert \nabla u \vert ^{p}-\mu \frac{ \vert u \vert ^{p}}{ \vert x \vert ^{p}})\,dx}{ ( \int_{\Omega }\frac{ \vert u \vert ^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx ) ^{\frac{p}{p^{*}(t)}}}. $$
    (1.4)
     

Our main result is the following theorem.

Theorem 1.1

Suppose that \((Q_{1})\) is satisfied. Then there exists \(\Lambda >0\) such that, for every \(\lambda \in (0,\Lambda )\), problem (1.1) has at least two positive solutions.

The following well-known Brézis–Lieb lemma and maximum principle will play fundamental roles in the proof of our main result.

Proposition 1.1

([3])

Suppose that \({u_{n}}\) is a bounded sequence in \(L^{p}(\Omega )\) (\(1\leq p<\infty \)), and \(u_{n}(x)\rightarrow u(x)\) a.e. \(x \in \Omega \), where \(\Omega \subset \mathbb{R}^{N}\) is an open set. Then
$$ \lim_{n\rightarrow \infty } \biggl( \int_{\Omega } \vert u_{n} \vert ^{p}\,dx- \int_{\Omega } \vert u_{n}-u \vert ^{p}\,dx \biggr) = \int_{\Omega } \vert u \vert ^{p}\,dx. $$

Proposition 1.2

([23])

Assume that \(\Omega \subset \mathbb{R} ^{N}\) is a bounded domain with smooth boundary, \(0\in \Omega \), \(u\in C^{1} (\Omega \backslash \{0\})\), \(u\geq 0\), \(u\not \equiv 0\), and
$$ -\Delta_{p} u\geq 0 \quad \textit{in } \Omega . $$
Then \(u>0\) in Ω.
By [22, 23], we assume that \(1< p< N\), \(0\leq t< p\), and \(0\leq \mu <\overline{ \mu }\). Then the limiting problem
$$ \textstyle\begin{cases} -\Delta_{p}u-\mu \frac{u^{p-1}}{ \vert x \vert ^{p}}= \frac{u^{p^{*}(t)-1}}{ \vert x \vert ^{t}}, \quad \text{in } \mathbb{R}^{N}\backslash \{0\}, \\ u>0,\quad \text{in } \mathbb{R}^{N}\backslash \{0\},\quad\quad u\in D^{1,p}(\mathbb{R}^{N}) \end{cases} $$
has positive radial ground states
$$V_{\epsilon }(x)=\epsilon^{\frac{p-N}{p}}U_{p,\mu }\biggl( \frac{x}{\epsilon }\biggr)=\epsilon^{\frac{p-N}{p}}U_{p,\mu }\biggl( \frac{ \vert x \vert }{\epsilon }\biggr) \quad \forall \epsilon >0 $$
that satisfy
$$\int_{\Omega } \biggl( \bigl\vert \nabla V_{\epsilon }(x) \bigr\vert ^{p}-\mu \frac{ \vert V_{ \epsilon }(x) \vert ^{p}}{ \vert x \vert ^{p}} \biggr) \,dx = \int_{\Omega } \biggl( \frac{ \vert V_{\epsilon }(x) \vert ^{p^{*}(t)}}{ \vert x \vert ^{t}} \biggr) \,dx=S ^{\frac{N-t}{p-t}}, $$
where the function \(U_{p,\mu }(x)=U_{p,\mu }( \vert x \vert )\) is the unique radial solution of the above limiting problem with
$$U_{p,\mu }(1)= \biggl( \frac{(N-t)(\overline{\mu }-\mu )}{N-p} \biggr) ^{\frac{1}{p^{*}(t)-p}}, $$
and
$$\begin{aligned}& \lim_{r\rightarrow 0^{+}} r^{a(\mu )}U_{p,\mu }(r)=c_{1} >0, \quad\quad \lim_{r\rightarrow 0^{+}} r^{a(\mu )+1} \bigl\vert U'_{p,\mu }(r) \bigr\vert =c _{1} a(\mu )\geq 0, \\& \lim_{r\rightarrow +\infty } r^{b(\mu )}U_{p,\mu }(r)=c_{2} >0, \quad\quad \lim_{r\rightarrow +\infty } r^{b(\mu )+1} \bigl\vert U'_{p,\mu }(r) \bigr\vert =c _{2} b(\mu )\geq 0, \\& c_{3} \leq U_{p,\mu }(r) \bigl( r^{\frac{a(\mu )}{\nu }}+r^{\frac{b( \mu )}{\nu }} \bigr) ^{\nu }\leq c_{4}, \quad \quad \nu :=\frac{N-p}{p}, \end{aligned}$$
where \(c_{i}\) (\(i=1, 2, 3, 4\)) are positive constants depending on N, μ, and p, and \(a(\mu )\) and \(b(\mu )\) are the zeros of the function
$$ h(t)=(p-1)t^{p} -(N-p)t^{p-1}+\mu ,\quad t\geq 0, $$
satisfying \(0\leq a(\mu )<\nu <b(\mu )\leq \frac{N-p}{p-1}\).
Take \(\rho >0\) small enough such that \(B(0,\rho )\subset \Omega \), and define the function
$$ u_{\epsilon }(x)=\eta (x)V_{\epsilon }(x)=\epsilon^{\frac{p-N}{p}} \eta (x)U_{p,\mu } \biggl(\frac{ \vert x \vert }{\epsilon } \biggr), $$
where \(\eta \in C_{0}^{\infty }(\Omega )\) is a cutoff function
$$ \eta (x)= \textstyle\begin{cases} 1, & \vert x \vert \leq \frac{\rho }{2}, \\ 0, & \vert x \vert >\rho . \end{cases} $$
The following estimates hold when \(\epsilon \longrightarrow 0\):
$$\begin{aligned}& \Vert u_{\epsilon } \Vert ^{p} =S^{\frac{N-t}{p-t}}+O \bigl( \epsilon^{b(\mu )p+p-N} \bigr), \\& \int_{\Omega }\frac{ \vert u_{\epsilon } \vert ^{p^{*} (t)}}{ \vert x \vert ^{t}}\,dx=S^{ \frac{N-t}{p-t}}+O \bigl( \epsilon^{b(\mu )p^{*} (t)-N+t} \bigr). \end{aligned}$$

2 Existence of the first solution of problem (1.1)

In this section, we will get the first solution which is a local minimizer in \(W_{0}^{1,p} (\Omega )\) for (1.1).

Lemma 2.1

There exist \(\lambda_{0}>0\), \(R, \rho >0\) such that, for every \(\lambda \in (0,\lambda_{0})\), we have
$$ I_{\lambda ,\mu }(u)| _{u\in \partial B_{R}}\geq \rho ,\quad\quad \inf _{u\in B_{R}}I_{\lambda ,\mu }(u)< 0. $$

Proof

We can deduce from Hölder’s inequality that
$$ \begin{aligned} I_{\lambda , \mu }(u)&\geq \frac{1}{p} \Vert u \Vert ^{p} -\frac{1}{p^{*} (t)}Q _{M} S^{-\frac{p^{*} (t)}{p}} \Vert u \Vert ^{p^{*} (t)}-\frac{\lambda }{1-s}C _{0} \Vert u \Vert ^{1-s} \\ &= \Vert u \Vert ^{1-s} \biggl( \frac{1}{p} \Vert u \Vert ^{-1+s+p}-\frac{1}{p^{*} (t)}Q_{M} S^{-\frac{p^{*} (t)}{p}} \Vert u \Vert ^{-1+s+p^{*} (t)}-\frac{\lambda }{1-s}C _{0} \biggr) , \end{aligned} $$
where \(C_{0}\) is a positive constant. Put \(f(x)=\frac{1}{p}x^{-1+s+p}-\frac{1}{p ^{*} (t)}Q_{M} S^{-\frac{p^{*} (t)}{p}}x^{-1+s+p^{*} (t)}\), we find that there is a constant \(R= [ \frac{p^{*} (t)S^{\frac{p^{*} (t)}{p}}(-1+s+p)}{pQ _{M} (-1+s+p^{*} (t))} ] ^{\frac{1}{p^{*} (t)-p}}>0\) such that \(f(R)=\max_{x>0}f(x)>0\). Letting \(\lambda_{0} =\frac{(1-s)f(R)}{C _{0}}\), we have that there is a constant \(\rho >0\) such that \(I_{\lambda ,\mu }(u)| _{u\in \partial B_{R}}\geq \rho \) for every \(\lambda \in (0, \lambda_{0})\).
For given R, choosing \(u\in B_{R}\) with \(u^{+}\neq 0\), we have
$$ \begin{aligned} \lim_{r\rightarrow 0}\frac{I_{\lambda ,\mu }(ru)}{r^{1-s}}&= \lim _{r\rightarrow 0}\frac{\frac{1}{p}r^{p} \Vert u \Vert ^{p}-\frac{\lambda r ^{1-s}}{1-s} \int_{\Omega }(u^{+})^{1-s} \,dx -\frac{r^{p^{*}(t)}}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx}{r^{1-s}} \\ &=-\frac{\lambda }{1-s} \int_{\Omega } \bigl(u^{+} \bigr)^{1-s} \,dx< 0, \end{aligned} $$
since \(p^{*} (t)>p>1>s>0\) for \(0\leq t< p\). For all \(u^{+}\neq 0\) such that \(I_{\lambda ,\mu }(ru)<0\) as \(r\rightarrow 0\), that is, \(\Vert u \Vert \) sufficiently small, we have
$$ \Gamma =\inf_{u\in B_{R}}I_{\lambda ,\mu }(u)< 0. $$
(2.1)
The proof of Lemma 2.1 is completed. □

Theorem 2.2

Problem (1.1) has a positive solution \(u_{1}\in W^{1.p}_{0}(\Omega )\) with \(I_{\lambda ,\mu }(u_{1})<0\) for \(\lambda \in (0,\lambda_{0})\), where \(\lambda_{0}\) is defined in Lemma 2.1.

Proof

By Lemma 2.1, we have
$$ \begin{aligned} & \frac{1}{p} \Vert u \Vert ^{p} - \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx\geq \rho , \quad \forall u \in \partial B_{R}, \\ &\frac{1}{p} \Vert u \Vert ^{p}-\frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx\geq 0, \quad \forall u \in B_{R}. \end{aligned} $$
(2.2)
From (2.1) we guarantee that there exists a minimizing sequence \(\{u_{n}\}\subset B_{R}\) such that \(\lim_{n\rightarrow \infty }I _{\lambda ,\mu }(u_{n})=\Gamma <0\). Obviously, the minimizing sequence is a closed convex set in \(B_{R}\). Going if necessary to a sequence still called \(\{u_{n}\}\), there exists \(u_{1}\in W_{0}^{1,p}(\Omega )\) such that
$$ \textstyle\begin{cases} u_{n}\rightharpoonup u_{1}, & \text{in } W_{0}^{1.p}(\Omega ), \\ u_{n}\longrightarrow u_{1}, & \text{in } L^{p'}(\Omega , \vert x \vert ^{-t}), \\ u_{n}(x)\longrightarrow u_{1}(x), & \text{a.e. in } \Omega , \end{cases}\displaystyle \quad 1\leq p'< p^{*}(t), $$
(2.3)
and
$$ \textstyle\begin{cases} \nabla u_{n}(x) \longrightarrow \nabla u_{1} (x), & \text{a.e. in } \Omega , \\ \frac{ \vert u_{n} \vert ^{p-2}u_{n}}{ \vert x \vert ^{p-1}}\rightharpoonup \frac{ \vert u_{1} \vert ^{p-2}u _{1}}{ \vert x \vert ^{p-1}}, & \text{in } L^{\frac{p}{p-1}}(\Omega ), \\ \int_{\Omega }\frac{ \vert u_{n} \vert ^{p^{*} (t)-2}u_{n}}{ \vert x \vert ^{t}}v \,dx \longrightarrow \int_{\Omega }\frac{ \vert u_{1} \vert ^{p^{*} (t)-2}u_{1}}{ \vert x \vert ^{t}}v \,dx, &\forall v\in W_{0}^{1,p}(\Omega ). \end{cases} $$
For \(s\in (0,1)\), applying Hölder’s inequality, we obtain that
$$\begin{aligned} \int_{\Omega } \bigl(u^{+}_{n} \bigr)^{1-s}\,dx- \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{1-s}\,dx &\leq \int_{\Omega } \bigl\vert \bigl(u^{+}_{n} \bigr)^{1-s}- \bigl(u^{+}_{1} \bigr)^{1-s} \bigr\vert \,dx \\ &\leq \int_{\Omega } \bigl\vert u_{n}^{+}-u_{1}^{+} \bigr\vert ^{1-s}\,dx \\ &\leq \bigl\vert u^{+}_{n}-u^{+}_{1} \bigr\vert _{p}^{1-s} \vert \Omega \vert ^{\frac{1+s}{p}}, \end{aligned}$$
thus,
$$ \int_{\Omega } \bigl(u^{+}_{n} \bigr)^{1-s}\,dx= \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{1-s}\,dx+o(1). $$
(2.4)
Let \(\omega_{n}=u_{n}-u_{1}\), by the Brézis–Lieb lemma, one has
$$\begin{aligned}& \int_{\Omega } \vert \nabla u_{n} \vert ^{p}\,dx= \int_{\Omega } \vert \nabla \omega_{n} \vert ^{p}\,dx+ \int_{\Omega } \vert \nabla u_{1} \vert ^{p}\,dx+o(1), \end{aligned}$$
(2.5)
$$\begin{aligned}& \int_{\Omega }Q(x) \frac{(u^{+}_{n})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx= \int_{\Omega }Q(x)\frac{(\omega^{+}_{n})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx+ \int_{\Omega }Q(x) \frac{(u^{+}_{1})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx+o(1). \end{aligned}$$
(2.6)
Noting that \(\Vert u_{1} \Vert ^{p} = \vert \nabla u_{1} \vert _{p}^{p} -\mu \vert u_{1} /x \vert _{p} ^{p} \), we have that
$$ \lim_{n\rightarrow \infty } \bigl( \Vert u_{n} \Vert ^{p} - \Vert \omega_{n} \Vert ^{p} \bigr)= \Vert u_{1} \Vert ^{p}. $$
If \(u_{1}=0\), then \(\omega_{n}=u_{n}\), it follows that \(\omega_{n} \in B_{R}\). If \(u_{1}\neq 0\), from (2.2), we derive that
$$ \frac{1}{p} \Vert \omega_{n} \Vert ^{p}- \frac{1}{p*(t)} \int_{\Omega }Q(x)\frac{(\omega^{+}_{n})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx\geq 0. $$
(2.7)
By (2.3)–(2.7), we have
$$\begin{aligned} \Gamma &=I_{\lambda ,\mu }(u_{n})+o(1) \\ &=\frac{1}{p} \Vert u_{n} \Vert ^{p}- \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u_{n}^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{ \lambda }{1-s} \int_{\Omega } \bigl(u_{n}^{+} \bigr)^{1-s}\,dx+o(1) \\ &=I_{\lambda ,\mu }(u_{1})+\frac{1}{p} \Vert \omega_{n} \Vert ^{p}-\frac{1}{p ^{*}(t)} \int_{\Omega }Q(x)\frac{(\omega_{n}^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{ \lambda }{1-s} \int_{\Omega } \bigl(\omega_{n}^{+} \bigr)^{1-s}\,dx+o(1) \\ &\geq I_{\lambda ,\mu }(u_{1})+o(1). \end{aligned}$$
Consequently, \(\Gamma \geq I_{\lambda ,\mu }(u_{1})\) as \(n\rightarrow \infty \). Since \(B_{R}\) is convex and closed, so \(u_{1}\in B_{R}\). We get that \(I_{\lambda ,\mu }(u_{1})=\Gamma <0\) from (2.1) and \(u_{1}\not \equiv 0\). It means that \(u_{1}\) is a local minimizer of \(I_{\lambda ,\mu }\).
Now, we claim that \(u_{1}\) is a solution of (1.1) and \(u_{1}>0\). Letting \(r>0\) small enough, and for every \(\varphi \in W^{1.p}_{0}(\Omega )\), \(\varphi \geq 0\) such that \((u_{1}+r\varphi )\in B_{R}\), one has
$$\begin{aligned} 0&< I_{\lambda ,\mu }(u_{1}+r\varphi )-I_{\lambda ,\mu }(u_{1}) \\ &=\frac{1}{p} \Vert u_{1}+r\varphi \Vert ^{p}- \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{((u_{1}+r\varphi )^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{\lambda }{1-s} \int_{\Omega } \bigl((u_{1}+r\varphi )^{+} \bigr)^{1-s}\,dx \\ &\quad{} -\frac{1}{p} \Vert u_{1} \Vert ^{p}+ \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u_{1}^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx+\frac{ \lambda }{1-s} \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{1-s}\,dx \\ &\leq \frac{1}{p} \Vert u_{1}+r\varphi \Vert ^{p}-\frac{1}{p} \Vert u_{1} \Vert ^{p}. \end{aligned}$$
(2.8)
Next we prove that \(u_{1}\) is a solution of (1.1). According to (2.8), we have
$$ \begin{aligned} &\frac{\lambda }{1-s} \int_{\Omega } \bigl[ \bigl((u_{1}+r\varphi )^{+} \bigr)^{1-s}- \bigl(u^{+}_{1} \bigr)^{1-s} \bigr] \,dx \\ &\quad \leq \frac{1}{p} \bigl[ \Vert u_{1}+r\varphi \Vert ^{p}- \Vert u_{1} \Vert ^{p} \bigr] - \frac{1}{p ^{*}(t)} \int_{\Omega }Q(x)\frac{ [ ((u_{1}+r\varphi )^{+})^{p^{*}(t)}-(u _{1}^{+})^{p^{*}(t)} ] }{ \vert x \vert ^{t}}\,dx. \end{aligned} $$
Dividing by \(r>0\) and taking limit as \(r\rightarrow 0^{+}\), we have
$$ \begin{aligned}[b] &\frac{\lambda }{1-s}\liminf_{r\rightarrow 0^{+}} \int_{\Omega } \frac{((u_{1}+r\varphi )^{+})^{1-s}-(u^{+}_{1})^{1-s}}{t}\,dx \\ &\quad \leq \int_{\Omega } \biggl( \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \varphi - \mu \frac{ \vert u_{1} \vert ^{p-2}u_{1}\varphi }{ \vert x \vert ^{p}} \biggr) \,dx \\ &\quad \quad{} - \int_{\Omega }Q(x)\frac{(u^{+}_{1})^{p^{*}(t)-1}\varphi }{ \vert x \vert ^{t}} \,dx. \end{aligned} $$
(2.9)
However,
$$ \frac{\lambda }{1-s} \frac{((u_{1}+r\varphi )^{+})^{1-s}-(u^{+}_{1})^{1-s}}{t}=\lambda \int_{\Omega } \bigl((u_{1}+\xi r \varphi )^{+} \bigr)^{-s}\varphi \,dx, $$
where \(\xi \longrightarrow 0^{+}\) and \(\lim_{r\rightarrow 0^{+}}((u _{1}+\xi r \varphi )^{+})^{-s}\varphi =(u_{1}^{+})^{-s}\varphi\) (\(\xi \rightarrow 0^{+}\)) a.e. \(x \in \Omega \). Since \(((u_{1}+\xi r \varphi )^{+})^{-s}\varphi \geq 0\). By Fatou’s lemma, we obtain that
$$ \lambda \int_{\Omega } \bigl(u_{1}^{+} \bigr)^{-s}\varphi \,dx\leq \frac{\lambda }{1-s}\liminf _{r\rightarrow 0^{+}} \int_{\Omega } \frac{((u_{1}+r\varphi )^{+})^{1-s}-(u^{+}_{1})^{1-s}}{t}\,dx. $$
Hence, from (2.9), we obtain that
$$ \begin{aligned}[b] & \int_{\Omega } \biggl( \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \varphi - \mu \frac{ \vert u_{1} \vert ^{p-2}u_{1} \varphi }{ \vert x \vert ^{p}} \biggr) \,dx-\lambda \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{-s}\varphi \,dx \\ &\quad{} - \int_{\Omega }Q(x)\frac{(u^{+}_{1})^{p^{*}(t)-1}\varphi }{ \vert x \vert ^{t}} \,dx \geq 0\end{aligned} $$
(2.10)
for \(\varphi \geq 0\). Since \(I_{\lambda ,\mu }(u_{1})<0\), combining with Lemma 2.1, we can derive that \(u_{1} \notin \partial B_{R}\), thus \(\Vert u_{1} \Vert < R\). There exists \(\delta_{1}\in (0,1)\) such that \((1+\theta )u_{1}\in B_{R}\) (\(\vert \theta \vert \leq \delta_{1}\)). Let \(h(\theta )=I_{\lambda ,\mu }((1+\theta )u_{1})\). Apparently, \(h(\theta )\) attains its minimum at \(\theta =0\). Note that
$$ \begin{aligned} h'(\theta )&=\frac{d}{d\theta } \bigl(I_{\lambda ,\mu }(1+\theta )u_{1} \bigr) \\ &=(1+\theta )^{p-1} \Vert u_{1} \Vert ^{p}-(1+ \theta )^{p^{*}(t)-1} \int_{\Omega }Q(x)\frac{(u^{+}_{1})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx-\lambda (1+ \theta )^{-s} \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{1-s}\,dx. \end{aligned} $$
Furthermore,
$$ h^{\prime}(\theta )| _{\theta =0}= \Vert u_{1} \Vert ^{p}- \int_{\Omega }Q(x)\frac{(u^{+}_{1})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx-\lambda \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{1-s}\,dx=0. $$
(2.11)
Define \(\Psi \in W_{0}^{1,p}(\Omega )\) by
$$ \Psi = \bigl(u_{1}^{+}+\varepsilon \psi \bigr)^{+}, \quad \text{for every }\psi \in W _{0}^{1,p}( \Omega )\text{ and }\varepsilon >0, $$
where \((u_{1}^{+}+t\psi )^{+}=\max \{u_{1}^{+}+t\psi ,0\}\). We deduce from (2.10) and (2.11) that
$$\begin{aligned} 0&\leq \int_{\Omega } \biggl( \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \Psi - \mu \frac{ \vert u_{1} \vert ^{p-2}u_{1}\Psi }{ \vert x \vert ^{p}} \biggr) \,dx- \int_{\Omega }Q(x)\frac{(u_{1}^{+})^{p^{*}(t)-1}\Psi }{ \vert x \vert ^{t}} \,dx \\ &\quad{} - \lambda \int_{\Omega } \bigl(u^{+}_{1} \bigr)^{-s}\Psi \,dx \\ &= \int_{\{x\mid u_{1}^{+} +\varepsilon \psi >0\}} \biggl[ \vert \nabla u_{1} \vert ^{p-2} \nabla u_{1}\nabla \bigl(u_{1}^{+} +\varepsilon \psi \bigr)-\mu \frac{ \vert u_{1} \vert ^{p-2}u _{1}(u_{1}^{+} +\varepsilon \psi )}{ \vert x \vert ^{p}} \\ &\quad{} -Q(x)\frac{(u^{+}_{1})^{p^{*}(t)-1}(u_{1}^{+} +\varepsilon \psi )}{ \vert x \vert ^{t}} -\lambda \bigl(u_{1}^{+} \bigr)^{-s} \bigl(u_{1}^{+} +\varepsilon \psi \bigr) \biggr] \,dx \\ &= \biggl( \int_{\Omega }- \int_{\{x\mid u_{1}^{+} +\varepsilon \psi \leq 0\}} \biggr) \biggl[ \vert \nabla u _{1} \vert ^{p-2}\nabla u_{1}\nabla \bigl(u_{1}^{+} +\varepsilon \psi \bigr)-\mu \frac{ \vert u _{1} \vert ^{p-2}u_{1}(u_{1}^{+} +\varepsilon \psi )}{ \vert x \vert ^{p}}\,dx \\ &\quad{} -Q(x)\frac{(u_{1}^{+})^{p^{*}(t)-1}(u_{1}^{+} +\varepsilon \psi )}{ \vert x \vert ^{t}} -\lambda \bigl(u^{+}_{1} \bigr)^{-s} \bigl(u_{1}^{+} +\varepsilon \psi \bigr) \biggr] \,dx \\ &\leq \Vert u_{1} \Vert ^{p} - \int_{\Omega }Q(x)\frac{(u_{1}^{+})^{p^{*} (t)}}{ \vert x \vert ^{t}}\,dx-\lambda \int_{\Omega } \bigl(u_{1}^{+} \bigr)^{1-s}\,dx +\varepsilon \int_{\Omega } \biggl[ \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \psi \\ &\quad{} -\mu \frac{ \vert u_{1} \vert ^{p-2}u_{1}\psi }{ \vert x \vert ^{p}}-Q(x)\frac{(u_{1} ^{+})^{p^{*}(t)-1}\psi }{ \vert x \vert ^{t}} -\lambda \bigl(u^{+}_{1} \bigr)^{-s}\psi \biggr] \,dx \\ &\quad{} - \int_{\{x\mid u_{1}^{+} +\varepsilon \psi \leq 0\}} \biggl[ \vert \nabla u_{1} \vert ^{p-2} \nabla u_{1}\nabla \bigl(u_{1}^{+} +\varepsilon \psi \bigr)-\mu \frac{ \vert u_{1} \vert ^{p-2}u _{1}(u_{1}^{+} +\varepsilon \psi )}{ \vert x \vert ^{p}} \biggr] \,dx \\ &\quad{} + \int_{\{x\mid u_{1}^{+} +\varepsilon \psi \leq 0\}} \biggl[ Q(x)\frac{(u _{1}^{+})^{p^{*}(t)-1}(u_{1}^{+} +\varepsilon \psi )}{ \vert x \vert ^{t}} + \lambda \bigl(u^{+}_{1} \bigr)^{-s} \bigl(u_{1}^{+} +\varepsilon \psi \bigr) \biggr] \,dx \\ &\leq \varepsilon \int_{\Omega } \biggl[ \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \psi - \mu \frac{ \vert u_{1} \vert ^{p-2}u_{1} \psi }{ \vert x \vert ^{p}}-Q(x) \frac{(u_{1}^{+})^{p ^{*}(t)-1}\psi }{ \vert x \vert ^{t}} -\lambda \bigl(u^{+}_{1} \bigr)^{-s}\psi \biggr] \,dx \\ &\quad{} -\varepsilon \int_{\{x\mid u_{1}^{+} +\varepsilon \psi \leq 0\}} \biggl[ \vert \nabla u_{1} \vert ^{p-2} \nabla u_{1}\nabla \psi -\mu \frac{ \vert u_{1} \vert ^{p-1}u_{1}\psi }{ \vert x \vert ^{p}} \biggr] \,dx. \end{aligned}$$
(2.12)
Since the measure of \(\{x\mid u_{1}^{+} +\varepsilon \psi \leq 0\}\rightarrow 0\) as \(\varepsilon \rightarrow 0\), we have
$$ \lim_{\varepsilon \rightarrow 0} \int_{\{x\mid u_{1}^{+} +\varepsilon \psi \leq 0\}} \biggl[ \vert \nabla u_{1} \vert ^{p-2} \nabla u_{1}\nabla \psi -\mu \frac{ \vert u_{1} \vert ^{p-2}u_{1} \psi }{ \vert x \vert ^{p}} \biggr] \,dx=0. $$
Dividing by ε and letting \(\varepsilon \rightarrow 0^{+}\) in (2.12), we deduce that
$$ \int_{\Omega } \biggl[ \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \psi - \mu \frac{ \vert u_{1} \vert ^{p-2}u_{1} \psi }{ \vert x \vert ^{p}}-Q(x) \frac{(u_{1}^{+})^{p ^{*}(t)-1}}{ \vert x \vert ^{t}}\psi -\lambda \bigl(u_{1}^{+} \bigr)^{-s}\psi \biggr] \,dx \geq 0. $$
Since \(\psi \in W^{1.p}_{0}(\Omega )\) is arbitrary, replacing ψ with −ψ, we have
$$ \begin{aligned}[b]& \int_{\Omega } \biggl[ \vert \nabla u_{1} \vert ^{p-2}\nabla u_{1}\nabla \psi - \mu \frac{ \vert u_{1} \vert ^{p-2}u_{1} \psi }{ \vert x \vert ^{p}} \\ &\quad{} -Q(x) \frac{(u_{1}^{+})^{p ^{*}(t)-1}\psi }{ \vert x \vert ^{t}}-\lambda \bigl(u_{1}^{+} \bigr)^{-s}\psi \biggr] \,dx=0, \quad \forall \psi \in W^{1.p}_{0}(\Omega ),\end{aligned} $$
(2.13)
which implies that \(u_{1}\) is a weak solution of problem (1.1). Putting the test function \(\psi =u_{1} ^{-}\) in (2.13), we obtain that \(u_{1} \geq 0\). Noting that \(I_{\lambda ,\mu }(u_{1})=\Gamma <0\), then \(u_{1}\not \equiv 0 \). In terms of the maximum principle, we have that \(u_{1}>0\), a.e. \(x\in \Omega \).

The proof of Theorem 2.2 is completed. □

3 Existence of a solution of the perturbation problem

In order to find another solution, we consider the following problem:
$$ \textstyle\begin{cases} -\Delta_{p}u-\mu \frac{ \vert u \vert ^{p-2}u}{ \vert x \vert ^{p}}=Q(x) \frac{(u^{+})^{p^{*}(t)-1}}{ \vert x \vert ^{t}}+\lambda (u^{+}+\gamma )^{-s}, & \text{in }\Omega , \\ u=0, & \text{on }\partial \Omega , \end{cases} $$
(3.1)
where \(\gamma >0\) is small. The solution of (3.1) is equivalent to the critical point of the following \(C^{1}\)-functional on \(W^{1,p}_{0}( \Omega )\):
$$ I_{\gamma }(u)=\frac{1}{p} \Vert u \Vert ^{p}- \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u ^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{\lambda }{1-s} \int_{\Omega } \bigl[ \bigl(u ^{+}+\gamma \bigr)^{1-s}-\gamma^{1-s} \bigr]\,dx. $$
For every \(\varphi \in W_{0}^{1,p}(\Omega )\), the definition of weak solution \(u\in W^{1,p}_{0}(\Omega )\) gives that
$$ \int_{\Omega } \biggl( \vert \nabla u \vert ^{p-2}\nabla u\nabla \varphi -\mu \frac{ \vert u \vert ^{p-2}u \varphi }{ \vert x \vert ^{p}} \biggr) -\lambda \int_{\Omega } \bigl(u^{+}+\gamma \bigr)^{-s} \varphi - \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)-1}\varphi }{ \vert x \vert ^{t}}=0. $$
(3.2)

Lemma 3.1

For \(R, \rho >0\), suppose that \(\lambda <\lambda _{0}\), then \(I_{\gamma }\) satisfies the following properties:
  1. (i)

    \(I_{\gamma }(u)\geq \rho >0\) for \(u\in \partial B_{R}\);

     
  2. (ii)

    There exists \(u_{2}\in W^{1,p}_{0}(\Omega )\) such that \(\Vert u_{2} \Vert >R\) and \(I_{\gamma }(u_{2})<\rho \),

     
where R, ρ, and \(\lambda_{0}\) are given in Lemma 2.1.

Proof

(i) By the subadditivity of \(t^{1-s}\), we have
$$ \bigl(u^{+}+\gamma \bigr)^{1-s}- \gamma^{1-s}\leq \bigl(u^{+} \bigr)^{1-s}, \quad \forall u\in W^{1,p}_{0}(\Omega ), $$
(3.3)
which leads to
$$ I_{\gamma }(u)\geq I_{\lambda ,\mu }(u), \quad \forall u\in W ^{1,p}_{0}(\Omega ). $$
Hence, if \(\lambda <\lambda_{0}\) for \(\rho , \lambda_{0}>0\), we can obtain the conclusion from Lemma 2.1.
(ii) \(\forall u^{+} \in W^{1.p}_{0}(\Omega )\), \(u^{+}\neq 0\) and \(r>0\), which yields
$$\begin{aligned} I_{\gamma }(ru)&=\frac{r^{p}}{p} \Vert u \Vert ^{p}-r^{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{\lambda }{1-s} \int_{\Omega } \bigl[ \bigl(ru^{+}+\gamma \bigr)^{1-s}-\gamma^{1-s} \bigr]\,dx \\ &\leq \frac{r^{p}}{p} \Vert u \Vert ^{p}-r^{p^{*}(t)} \int_{\Omega }Q(x)\frac{(u^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx \\ &\rightarrow -\infty \quad (r\rightarrow +\infty ). \end{aligned}$$
Therefore, there exists \(u_{2}\) such that \(\Vert u_{2} \Vert >R\) and \(I_{\gamma }(u_{2})<\rho \).

This completes the proof of Lemma 3.1. □

Lemma 3.2

Assume that \(0<\gamma <1\). Then \(I_{\gamma }\) satisfies the \((PS)_{c}\) condition with \(c<\frac{(p-t)}{p(N-t)}\frac{S ^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}}_{M}}-D\lambda^{ \frac{p}{p+s-1}}\), where
$$ D=\frac{p+s-1}{p} \biggl\{ \biggl( \frac{1}{1-s}+\frac{N-p}{p(N-t)} \biggr) C _{2} \biggl[ \frac{p}{(N-t)(1-s)} \biggr] ^{\frac{s-1}{p}} \biggr\} ^{ \frac{p}{p+s-1}}. $$

Proof

Choose \(\{\tau_{n}\}\subset W^{1,p}_{0}(\Omega )\) satisfying
$$ I_{\gamma }(\tau_{n})\rightarrow c ,\quad \text{and}\quad I_{\gamma } ^{\prime}(\tau_{n})\rightarrow 0 \quad (n \rightarrow \infty ). $$
(3.4)
We assert that \(\{\tau_{n}\}\) is bounded in \(W_{0}^{1,p}(\Omega )\). Otherwise, we assume that \(\lim_{n\rightarrow \infty } \Vert \tau_{n} \Vert \rightarrow \infty \). By (3.4), we have
$$\begin{aligned} c&=I_{\gamma }(\tau_{n})-\frac{1}{p^{*}(t)} \bigl\langle I'_{\gamma }(\tau _{n}),\tau_{n} \bigr\rangle +o(1) \\ &= \frac{1}{p} \Vert \tau_{n} \Vert ^{p}- \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(\tau_{n}^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{ \lambda }{1-s} \int_{\Omega } \bigl[ \bigl(\tau_{n}^{+} + \gamma \bigr)^{1-s}-\gamma^{1-s} \bigr]\,dx \\ &\quad{} -\frac{1}{p^{*}(t)} \Vert \tau_{n} \Vert ^{p}+ \frac{1}{p^{*}(t)} \int_{\Omega }Q(x)\frac{(\tau_{n}^{+})^{p^{*}(t)-1}\tau_{n}}{ \vert x \vert ^{t}}\,dx +\frac{\lambda }{p^{*}(t)} \int_{\Omega } \bigl(\tau_{n}^{+}+\gamma \bigr)^{-s}\tau_{n}\,dx+o(1) \\ &= \biggl(\frac{1}{p}-\frac{1}{p^{*}(t)} \biggr) \Vert \tau_{n} \Vert ^{p}- \frac{\lambda }{1-s} \int_{\Omega } \bigl[ \bigl(\tau_{n}^{+}+ \gamma \bigr)^{1-s}-\gamma^{-s} \bigr]\,dx \\ &\quad{} +\frac{ \lambda }{p^{*}(t)} \int_{\Omega } \bigl(\tau_{n}^{+}+\gamma \bigr)^{-s}\tau_{n}\,dx+o(1) \\ &\geq \frac{p-t}{p(N-t)} \Vert \tau_{n} \Vert ^{p}- \lambda \biggl(\frac{1}{1-s}+\frac{1}{p ^{*}(t)} \biggr) \int_{\Omega } \vert \tau_{n} \vert ^{1-s} \,dx +o(1) \\ &\geq \frac{p-t}{p(N-t)} \Vert \tau_{n} \Vert ^{p}- \lambda \biggl(\frac{1}{1-s}+\frac{1}{p ^{*}(t)} \biggr)C_{1} \Vert \tau_{n} \Vert ^{1-s}+o(1). \end{aligned}$$
The last inequality is absurd thanks to \(0<1-s<1\). That is, \(\{\tau_{n}\}\) is bounded in \(W_{0}^{1,p}(\Omega )\). Hence, up to a sequence, there exists a subsequence, still called \(\{\tau_{n}\}\). We assume that there exists \(\{\tau_{1}\}\in W_{0}^{1,p}(\Omega )\) such that
$$ \textstyle\begin{cases} \tau_{n}\rightharpoonup \tau_{1}, & \text{in } W_{0}^{1,p}(\Omega ) , \\ \tau_{n}\longrightarrow \tau_{1}, & \text{in } L^{p}(\Omega , \vert x \vert ^{-t}), \\ \tau_{n}(x)\longrightarrow \tau_{1}(x), & \text{a.e. in } \Omega , \\ \vert \tau_{n}(x) \vert \leq h(x), & \text{a.e. in } \Omega \text{ for all } n \text{ with } h(x)\in L^{1} (\Omega ). \end{cases}\displaystyle \quad1\leq p< p^{*}(t) , $$
Since
$$ \bigl\vert (\tau_{n} -\tau_{1}) \bigl( \tau_{n} ^{+} +\gamma \bigr)^{-s} \bigr\vert \leq \gamma^{-s} \bigl(h+ \vert \tau_{1} \vert \bigr), $$
it follows from the dominated convergence theorem that
$$ \lim_{n\rightarrow \infty } \int_{\Omega }(\tau_{n} -\tau_{1}) \bigl( \tau_{n} ^{+} +\gamma \bigr)^{-s}\,dx =0. $$
Furthermore, by \(\vert \tau_{1} \vert (\tau_{n}^{+} +\gamma )^{-s}\leq \vert \tau_{1} \vert \gamma^{-s}\), and applying the dominated convergence theorem again, we have
$$ \lim_{n\rightarrow \infty } \int_{\Omega } \bigl(\tau_{n}^{+} +\gamma \bigr)^{-s}\tau_{1} \,dx= \int_{\Omega } \bigl(\tau_{1}^{+} +\gamma \bigr)^{-s}\tau_{1} \,dx. $$
Thus, we deduce that
$$ \lim_{n\rightarrow \infty } \int_{\Omega } \bigl(\tau_{n}^{+} +\gamma \bigr)^{-s}\tau_{n} \,dx= \int_{\Omega } \bigl(\tau_{1}^{+} +\gamma \bigr)^{-s}\tau_{1} \,dx. $$
Now we prove that \(\tau_{n}\rightarrow \tau_{1}\) strongly in \(W_{0}^{1,p}(\Omega )\). Set \(\omega_{n}=\tau_{n}-\tau_{1}\). Since \(I^{\prime}_{\lambda ,\mu }(\tau_{n})\rightarrow 0\) in \((W_{0}^{1,p}( \Omega ))^{*}\), we have
$$ \Vert \tau_{n} \Vert ^{p}- \int_{\Omega }Q(x)\frac{(\tau^{+}_{n})^{p^{*}(t)-1}\tau_{n}}{ \vert x \vert ^{t}}\,dx- \lambda \int_{\Omega } \bigl(\tau^{+}_{n}+\gamma \bigr)^{-s}\tau_{n}\,dx=o(1). $$
According to the Brézis–Lieb lemma, together with (3.4), we have
$$ \begin{aligned}& \Vert \omega_{n} \Vert ^{p}+ \Vert \tau_{1} \Vert ^{p}- \int_{\Omega }Q(x) \frac{(\omega^{+}_{n})^{p^{*}(t)-1}\omega_{n}}{ \vert x \vert ^{t}}\,dx - \int_{\Omega }Q(x)\frac{(\tau^{+}_{1})^{p^{*}(t)-1}\tau_{1}}{ \vert x \vert ^{t}}\,dx \\ &\quad{} -\lambda \int_{\Omega } \bigl(\tau^{+}_{1}+\gamma \bigr)^{-s}\tau_{1}\,dx=o(1),\end{aligned} $$
and
$$ \lim_{n\rightarrow \infty } \bigl\langle I^{\prime}_{\gamma }( \tau_{n}),\tau_{1} \bigr\rangle = \Vert \tau_{1} \Vert ^{p}- \int_{\Omega }Q(x)\frac{(\tau^{+}_{1})^{p^{*}(t)-1}\tau_{1}}{ \vert x \vert ^{t}}\,dx -\lambda \int_{\Omega } \bigl(\tau^{+}_{1}+\gamma \bigr)^{-s}\tau_{1}\,dx=0. $$
Thus
$$\begin{aligned}& \lim_{n\rightarrow \infty } \Vert \omega_{n} \Vert ^{p}=\lim_{n\rightarrow \infty } \int_{\Omega }Q(x)\frac{(\omega^{+}_{n})^{p^{*}(t)-1}\omega _{n}}{ \vert x \vert ^{t}}\,dx=l, \\& \int_{\Omega }\frac{ \vert \omega_{n} \vert ^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx \geq \int_{\Omega }\frac{Q(x)}{Q_{M}} \frac{ \vert \omega_{n} \vert ^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx \geq \int_{\Omega }\frac{Q(x)}{Q_{M}}\frac{(\omega^{+}_{n})^{p^{*}(t)-1} \omega_{n}}{ \vert x \vert ^{t}}\,dx. \end{aligned}$$
Sobolev’s inequality implies that
$$ \Vert \omega_{n} \Vert ^{p}\geq S \biggl( \int_{\Omega }\frac{ \vert \omega_{n} \vert ^{p^{*} (t)}}{ \vert x \vert ^{t}}\,dx \biggr) ^{\frac{p}{p^{*} (t)}}. $$
Consequently, \(l\geq S(\frac{l}{Q_{M}})^{\frac{p}{p^{*}(t)}}\). We guarantee that \(l=0\). Otherwise, we suppose that
$$ l\geq \frac{S^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}}_{M}}. $$
It follows that
$$\begin{aligned} c&= I_{\gamma }(\tau_{n})-\frac{1}{p^{*}(t)} \bigl\langle I^{\prime}_{\gamma }( \tau_{n}),\tau_{n} \bigr\rangle +o(1) \\ &=\frac{(p-t)}{p(N-t)} \Vert \tau_{n} \Vert ^{p}- \frac{\lambda }{1-s} \int_{\Omega } \bigl[ \bigl(\tau_{n}^{+}+ \gamma \bigr)^{1-s}-\gamma^{-s} \bigr]\,dx +\frac{ \lambda }{p^{*}(t)} \int_{\Omega } \bigl(\tau_{n}^{+}+\gamma \bigr)^{-s}\tau_{n}\,dx+o(1) \\ &\geq \frac{(p-t)}{p(N-t)}\frac{S^{\frac{N-t}{p-t}}}{Q^{ \frac{N-p}{p-t}}_{M}}+\frac{p-t}{p(N-t)} \Vert \tau_{1} \Vert ^{p} -\lambda \biggl( \frac{1}{1-s}+ \frac{1}{p^{*} (t)} \biggr) \int_{\Omega } \vert \tau_{n} \vert ^{1-s} \,dx+o(1) \\ &\geq \frac{(p-t)}{p(N-t)}\frac{S^{\frac{N-t}{p-t}}}{Q^{ \frac{N-p}{p-t}}_{M}}+\frac{p-t}{p(N-t)} \Vert \tau_{1} \Vert ^{p} -\lambda \biggl( \frac{1}{1-s}+ \frac{1}{p^{*} (t)} \biggr) C_{2} \Vert \tau_{1} \Vert ^{1-s}+o(1) \\ &\geq \frac{(p-t)}{p(N-t)}\frac{S^{\frac{N-t}{p-t}}}{Q^{ \frac{N-p}{p-t}}_{M}}-D\lambda^{\frac{p}{p+s-1}}, \end{aligned}$$
which contradicts the condition of Lemma 3.2. Hence \(l=0\). Therefore \(\tau_{n}\rightarrow \tau_{1}\).

This proof of Lemma 3.2 is finished. □

Lemma 3.3

For \(0< s<1\) and \(\lambda >0\) small enough, there exists \(u_{2}\in W_{0}^{1,p}(\Omega )\) such that
$$ \sup_{t\geq 0}I_{\lambda ,\mu }(t u_{2}) \leq \frac{(p-t)}{p(N-t)}\frac{S^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}} _{M}}-D \lambda^{\frac{p}{p-1+s}}, $$
(3.5)
where D is defined in Lemma 3.2.

Proof

For every \(r\geq 0\), we have
$$ I_{\gamma }(ru_{\epsilon })=\frac{r^{p}}{p} \Vert u_{\epsilon } \Vert ^{p}-\frac{r ^{p^{*}(t)}}{p^{*}(t)} \int_{\Omega } Q(x)\frac{(u_{\epsilon }^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx -\frac{ \lambda }{1-s} \int_{\Omega } \bigl[ \bigl(ru_{\epsilon }^{+}+\gamma \bigr)^{1-s}-\gamma^{1-s} \bigr]\,dx, $$
which implies that there exists a positive constant \(\epsilon_{0}\) such that
$$ \lim_{r\rightarrow 0}I_{\gamma }(ru_{\epsilon })=0,\quad \forall \epsilon \in (0,\epsilon_{0}), $$
and
$$ \lim_{r\rightarrow +\infty }I_{\gamma }(ru_{\epsilon })=-\infty , \quad \forall \epsilon \in (0,\epsilon_{0}), $$
where \(u_{\epsilon }\) is defined in Sect. 1. Let
$$\begin{aligned}& A_{\epsilon }(r)=\frac{r^{p}}{p} \Vert u_{\epsilon } \Vert ^{p}-\frac{r^{p^{*}(t)}}{p ^{*}(t)} \int_{\Omega }Q(x)\frac{(u_{\epsilon }^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx; \\& B_{\epsilon }(r)=-\frac{1}{1-s} \int_{\Omega } \bigl[ \bigl(ru_{\epsilon }^{+}+\gamma \bigr)^{1-s}-\gamma^{1-s} \bigr]\,dx, \end{aligned}$$
because of \(\lim_{r\rightarrow \infty }A_{\epsilon }(r)=-\infty \), \(A_{\epsilon }(0)=0\), and \(\lim_{r\rightarrow 0^{+}}A_{\epsilon }(r)>0\), so \(A_{\epsilon }(r)\) attains its maximum at some positive number. In fact, we let
$$ A^{\prime}_{\epsilon }(r)=r^{p-1} \Vert u_{\epsilon } \Vert ^{p}-r^{p^{*}(t)-1} \int_{\Omega }Q(x) \frac{(u^{+}_{\epsilon })^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx=0, $$
therefore
$$ r= \biggl( \frac{ \Vert u_{\epsilon } \Vert ^{p}}{ \int_{\Omega } Q(x)\frac{(u_{\epsilon }^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx} \biggr) ^{\frac{1}{p^{*}(t)-p}}:=T_{\epsilon }. $$
Noting that \(A'_{\epsilon }(r)>0\) for every \(0< r< T_{\epsilon }\) and \(A'_{\epsilon }(r)<0\) for every \(r>T_{\epsilon }\), our claim is proved. Thus, the properties of \(I_{\gamma }(ru_{\epsilon })\) at \(r=0\) and \(r=+\infty \) tell us that \(\sup_{r\geq 0}I_{\gamma }(ru_{ \epsilon })\) is attained for some \(r_{\epsilon }>0\).
From condition \((Q_{1})\), we have
$$ \biggl\vert \int_{\Omega }Q(x)\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx- \int_{\Omega }Q_{M}\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx \biggr\vert \leq \int_{\Omega } \bigl\vert Q(x)-Q(0) \bigr\vert \frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx =O \bigl(\epsilon^{\beta } \bigr). $$
It follows that
$$ \int_{\Omega }Q(x)\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx=Q(0)S ^{\frac{N-t}{p-t}}+O \bigl(\epsilon^{b(\mu )p^{*} (t)-N+t} \bigr)+O \bigl(\epsilon^{ \beta } \bigr). $$
(3.6)
By (3.6), we deduce that
$$\begin{aligned} A_{\epsilon }(T_{\epsilon })&=\frac{1}{p} \biggl[ \frac{ \Vert u_{\epsilon } \Vert ^{p}}{ \int_{\Omega }Q(x)\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx} \biggr] ^{\frac{p}{p^{*}(t)-p}} \Vert u_{\epsilon } \Vert ^{p} \\ &\quad{} -\frac{1}{p^{*}(t)} \biggl[ \frac{ \Vert u_{\epsilon } \Vert ^{p}}{ \int_{\Omega }Q(x)\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx} \biggr] ^{\frac{p^{*}(t)}{p^{*}(t)-p}} \int_{\Omega }Q(x)\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx \\ &=\frac{p-t}{p(N-t)} \biggl[ \frac{ \Vert u_{\epsilon } \Vert ^{p}}{ \int_{\Omega }Q(x)\frac{u_{\epsilon }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx} \biggr] ^{\frac{p}{p^{*}(t)-p}} \Vert u_{\epsilon } \Vert ^{p} \\ &\leq \frac{p-t}{p(N-t)}\frac{S^{\frac{N-t}{p-t}}}{(Q(0))^{ \frac{N-p}{p-t}}}+O \bigl(\epsilon^{b(\mu )p+p-N} \bigr)+O \bigl(\epsilon^{\beta } \bigr). \end{aligned}$$
(3.7)
Next, we will estimate \(B_{\epsilon }\). Here, we use the following inequality from [24, 27]:
$$ x^{1-s}-(x+y)^{1-s}\leq -(1-s)y^{\frac{1-s}{4}}x^{\frac{3(1-s)}{4}}, \quad 0< x< y. $$
(3.8)
Observe from (3.8) that
$$\begin{aligned} B_{\epsilon }(r_{\epsilon })&\leq \frac{1}{1-s} \int_{\{x\mid \vert x \vert \leq \epsilon^{\frac{1-s}{2p}}\}} \bigl[\gamma^{1-s}-(r_{ \epsilon }u_{\epsilon }+ \gamma )^{1-s} \bigr]\,dx \\ &\leq -C_{3} \int_{\{x\mid \vert x \vert \leq \epsilon^{\frac{1-s}{2p}}\}}(r_{\epsilon }u_{\epsilon })^{\frac{1-s}{4}} \,dx \\ &\leq -C_{3} \int_{\{x\mid \vert x \vert \leq \epsilon^{\frac{1-s}{2p}}\}\cap \{\eta (x)=1\}} \biggl[ r_{\epsilon }\epsilon^{-\frac{N-p}{p}}U_{p,\mu } \biggl(\frac{ \vert x \vert }{ \epsilon } \biggr) \biggr] ^{\frac{1-s}{4}}\,dx \\ &\leq -C_{4} \int_{0}^{\epsilon^{\frac{1-s-2p}{2p}}} \bigl[ \epsilon^{-\frac{N-p}{p}}U _{p,\mu }(y) \bigr] ^{\frac{1-s}{4}}y^{N-1}\epsilon^{N} \,dy \\ &\leq -C_{5}\epsilon^{-\frac{(N-p)(1-s)}{4p}+N} \int_{0}^{\epsilon^{\frac{1-s-2p}{2p}}}y^{-b(\mu )p+N-1}\,dy \\ &\leq -C_{5} \textstyle\begin{cases} \epsilon^{-\frac{(N-p)(1-s)}{4p}+N}, & b(\mu )>\frac{N}{p}, \\ \epsilon^{-\frac{(N-p)(1-s)}{4p}+N} \vert \ln \epsilon \vert , & b(\mu )=\frac{N}{p}, \\ \epsilon^{-\frac{(N-p)(1-s)}{4p}+N+\frac{(1-s-2p)(-b(\mu )p+N)}{2p}}, & b(\mu )< \frac{N}{p}. \end{cases}\displaystyle \end{aligned}$$
(3.9)
From (3.7) and (3.9), we find that there exists a positive constant \(\widetilde{\lambda }_{0}\) such that, for every \(\lambda \in (0, \widetilde{\lambda }_{0})\), one has
$$\begin{aligned} I_{\gamma }(r_{\epsilon }u_{\epsilon })&=A_{\epsilon }(r_{\epsilon })+ \lambda B_{\epsilon }(r_{\epsilon }) \\ &\leq \frac{p-t}{p(N-p)}\frac{S^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}} _{M}} +O \bigl(\epsilon^{b(\mu )p-N+p} \bigr)+O \bigl(\epsilon^{\beta } \bigr) \\ &\quad{} -C_{5} \textstyle\begin{cases} \epsilon^{-\frac{(N-p)(1-s)}{4p}+N}, & b(\mu )>\frac{N}{p}, \\ \epsilon^{-\frac{(N-p)(1-s)}{4p}+N} \vert \ln \epsilon \vert , & b(\mu )=\frac{N}{p}, \\ \epsilon^{-\frac{(N-p)(1-s)}{4p}+N+\frac{(1-s-2p)(-b(\mu )p+N)}{2p}}, & b(\mu )< \frac{N}{p}, \end{cases}\displaystyle \\ &< \frac{p-t}{p(N-p)}\frac{S^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}} _{M}}-D\lambda^{\frac{p}{p+s-1}}. \end{aligned}$$
This completes the proof of Lemma 3.3. □

Theorem 3.4

For \(0<\gamma <1\), there is \(\lambda^{*}>0\) such that \(\lambda \in (0,\lambda^{*})\), problem (3.1) admits a positive solution \(\tau_{\gamma }\in W_{0}^{1,p}(\Omega )\) satisfying \(I_{\gamma }(\tau_{\gamma })>\rho \), where ρ is given in Lemma 2.1.

Proof

Let \(\lambda^{*}=\min \{\lambda_{0}, \widetilde{\lambda }_{0}\}\), then Lemmas 3.13.3 hold for \(0<\lambda <\lambda^{*}\). Based on Lemma 3.1, we know that \(I_{\gamma }\) satisfies the geometry of the mountain pass lemma [1]. Therefore, there is a sequence \(\{\tau_{n}\}\subset W_{0}^{1,p}(\Omega )\) such that
$$ I_{\gamma }(\tau_{n})\rightarrow c_{\gamma }>\rho >0, \quad\quad I_{\gamma }^{\prime}( \tau_{n})\rightarrow 0, $$
(3.10)
where
$$\begin{aligned}& c_{\gamma }=\inf_{\phi \in \Phi }\max_{ r\in [0,1]}I_{\gamma } \bigl(\phi (r) \bigr), \\& \Phi = \bigl\{ \phi \in C \bigl([0,1], W_{0}^{1,p}(\Omega ) \bigr):\phi (0)=0, \phi (1)=u_{2} \bigr\} . \end{aligned}$$
So, according to Lemmas 3.1 and 3.3, one has
$$ \begin{aligned}[b] 0&< \rho < c_{\gamma }\leq \max _{r\in [0,1]}I_{\gamma }(ru_{2}) \leq \sup _{r\geq 0}I_{\gamma }(ru_{2}) \\ &< \frac{p-t}{p(N-p)}\frac{S^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}} _{M}}-D\lambda^{\frac{p}{p+s-1}}. \end{aligned} $$
(3.11)
From Lemma 3.2, note that \(\{\tau_{n}\}\) has a convergent subsequence, still denoted by \(\{\tau_{n}\}\) (\(\{\tau_{n}\}\subset W_{0}^{1,p}( \Omega )\)). Assume that \(\lim_{n\rightarrow \infty }\tau_{n}= \tau_{\gamma }\) in \(W_{0}^{1,p}(\Omega )\). Hence, combining (3.10) and (3.11), we have
$$ I_{\gamma }(\tau_{\gamma })=\lim_{n\rightarrow \infty }I_{\gamma }( \tau_{n})=c_{\gamma }>\rho >0, $$
which implies that \(\tau_{\gamma }\not \equiv 0\). By the continuity of \(I_{\gamma }^{\prime}\), we know that \(\tau_{\gamma }\) is a solution of (3.1). Furthermore, \(\tau_{\gamma }\geq 0\). Hence, applying the strong maximum principle, we obtain that \(\tau_{\gamma }\) is a positive solution of (3.1). □

4 Existence of the second solution of problem (1.1)

Theorem 4.1

For \(\lambda \in (0, \lambda^{*})\), problem (1.1) possesses a positive solution \(\tau_{1}\) satisfying \(I_{\lambda ,\mu }(\tau_{1})>0\), where \(\lambda^{*}\) is given in Theorem 3.4.

Proof

Let \(\{\tau_{\gamma }\}\) be a family of positive solutions of (1.1), we will show that \(\{\tau_{\gamma }\}\) has a uniform lower bound. Indeed, we denote
$$\begin{aligned}& d(r)=r^{p^{*}(t)-1}+\frac{\lambda }{(r+p-1)^{s}}; \\& \text{case~(i)}\quad 0< r< 1, \quad d(r)\geq \frac{\lambda }{(1+p-1)^{s}}=\frac{ \lambda }{p^{s}}; \\& \text{case~(ii)} \quad r\geq 1, \quad d(r)\geq 1. \end{aligned}$$
Therefore, for every \(\gamma \in (0,1)\), \(r\geq 0\), we get
$$ r^{p^{*}(t)-1}+\frac{\lambda }{(r+\gamma )^{s}}\geq r^{p^{*}(t)-1}+\frac{ \lambda }{(r+p-1)^{s}} \geq \min \biggl\{ 1,\frac{\lambda }{p^{s}} \biggr\} . $$
Recall that e is a weak solution of the following problem:
$$ \textstyle\begin{cases} -\Delta_{p} u-\mu \frac{ \vert u \vert ^{p-2}u}{ \vert x \vert ^{p}}=1, & \text{in } \Omega , \\ u=0, &\text{on } \partial \Omega , \end{cases} $$
so \(e(x)>0\) in Ω. According to the comparison principle, we have
$$ \tau_{\gamma }\geq \min \{1,Q_{m}\}\min \biggl\{ 1, \frac{\lambda }{p^{s}} \biggr\} e>0, $$
(4.1)
where \(Q_{m} =\min_{x\in Q}Q(x)>0\). Since \(\{\tau_{\gamma }\}\) are solutions of problem (3.1), one has
$$ \Vert \tau_{\gamma } \Vert ^{p}- \int_{\Omega }Q(x)\frac{\tau^{p^{*}(t)}_{\gamma }}{ \vert x \vert ^{t}}\,dx-\lambda \int_{\Omega }(\tau_{\gamma }+\gamma )^{-s} \tau_{\gamma }\,dx=0. $$
(4.2)
Combining with (3.3), (4.2), and Theorem 3.4, we have
$$\begin{aligned}& \frac{p-t}{p(N-p)} \frac{S^{\frac{N-t}{p-t}}}{Q^{\frac{N-p}{p-t}}_{M}}-D \lambda^{\frac{p}{p+s-1}} \\& \quad >I_{\gamma }(\tau_{\gamma })-\frac{1}{p^{*}(t)} \bigl\langle I^{\prime}_{\gamma }(\tau_{\gamma }),\tau_{\gamma } \bigr\rangle \\& \quad =\frac{p-t}{p(N-t)} \Vert \tau_{\gamma } \Vert ^{p}+ \frac{\lambda }{p^{*}(t)} \int_{\Omega }(\tau_{\gamma }+\gamma )^{-s} \tau_{\gamma }\,dx-\frac{\lambda }{1-s} \int_{\Omega } \bigl[(\tau_{\gamma }+\gamma )^{1-s} - \gamma^{1-s} \bigr]\,dx \\& \quad \geq \frac{p-t}{p(N-t)} \Vert \tau_{\gamma } \Vert ^{p}-\frac{\lambda }{1-s} \int_{\Omega } \bigl[(\tau_{\gamma }+\gamma )^{1-s}- \gamma ^{1-s} \bigr]\,dx \\& \quad =\frac{p-t}{p(N-t)} \Vert \tau_{\gamma } \Vert ^{p}- \frac{\lambda C_{6}}{1-s} \Vert \tau_{\gamma } \Vert ^{1-s}, \end{aligned}$$
since \(s\in (0,1)\), so \(\{\tau_{\gamma }\}\) is bounded in \(W_{0}^{1,p}( \Omega )\). Going if necessary to a subsequence, also called \(\{\tau_{\gamma }\}\), there exists \(\tau_{1}\in W_{0}^{1,p}(\Omega )\) such that
$$ \textstyle\begin{cases} \tau_{\gamma }\rightharpoonup \tau_{1}, & \text{in } W_{0}^{1.p}(\Omega ), \\ \tau_{\gamma }\longrightarrow \tau_{1}, & \text{in } L^{p'}(\Omega , \vert x \vert ^{-t}), \\ \tau_{\gamma }(x)\longrightarrow \tau_{1}(x), & \text{a.e. in } \Omega . \end{cases}\displaystyle \quad1\leq p'< p^{*}(t), $$
(4.3)
Now, we show that \(\tau_{\gamma }\rightarrow \tau_{1}\) in \(W_{0}^{1.p}( \Omega )\) as \(\gamma \rightarrow 0\). Set \(w_{\gamma }=\tau_{\gamma }- \tau_{1}\), then \(\Vert w_{\gamma } \Vert \rightarrow 0\) as \(\gamma \rightarrow 0\); otherwise, there exists a subsequence (still denoted by \(w_{\gamma }\)) such that \(\lim_{\gamma \rightarrow 0} \Vert w_{\gamma } \Vert =l>0\). Since \(0\leq \frac{ \tau_{\gamma }}{(\tau_{\gamma }+\gamma )^{s}}\leq \tau_{\gamma }^{1-s}\), applying Hölder’s inequality and (4.3), we have
$$\begin{aligned} \int_{\Omega }\tau_{\gamma }(\tau_{\gamma }+\gamma )^{-s}\,dx&\leq \int_{\Omega }\tau_{\gamma }^{1-s}\,dx \leq \int_{\Omega } \vert w_{\gamma } \vert ^{1-s} \,dx+ \int_{\Omega } \vert \tau_{1} \vert ^{1-s} \,dx \\ &= \vert w_{\gamma } \vert _{p}^{1-s} \vert \Omega \vert ^{\frac{1+s}{p}}+ \int_{\Omega } \vert \tau_{1} \vert ^{1-s} \,dx \\ &\leq \int_{\Omega } \vert \tau_{1} \vert ^{1-s} \,dx+o(1). \end{aligned}$$
Similarly,
$$ \int_{\Omega } \vert \tau_{1} \vert ^{1-s} \,dx\leq \int_{\Omega }\tau_{\gamma }( \tau_{\gamma }+\gamma )^{-s}\,dx+o(1). $$
Therefore
$$ \lim_{\gamma \rightarrow 0} \int_{\Omega }\tau_{\gamma }(\tau_{\gamma }+\gamma )^{-s}\,dx= \int_{\Omega }\tau_{1}^{1-s}\,dx. $$
It follows from \(\langle I_{\gamma }^{\prime}(\tau_{\gamma }),\tau_{\gamma }\rangle =0\) and the Brézis–Lieb lemma that
$$ \Vert w_{\gamma } \Vert ^{p}+ \Vert \tau_{1} \Vert ^{p}- \int_{\Omega }Q(x)\frac{w_{\gamma }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx - \int_{\Omega }Q(x)\frac{\tau_{1}^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx-\lambda \int_{\Omega }\tau_{1}^{1-s}\,dx =o(1). $$
(4.4)
Note that \(\tau_{\gamma }\rightharpoonup \tau_{1}\) as \(\gamma \rightarrow 0^{+}\). Choose the test function \(\varphi =\phi \in W_{0}^{1,p}( \Omega )\cap C_{0}(\Omega )\) in (3.2). Letting \(\gamma \rightarrow 0^{+}\) and using (4.1), we deduce that \(\tau_{1} \geq \min \{1,Q_{m}\}\min \{1,\frac{\lambda }{p^{s}}\}e>0\), and
$$ \int_{\Omega } \biggl( \vert \nabla \tau_{1} \vert ^{p-2}\nabla \tau_{1}\nabla \phi -\mu \frac{ \vert \tau_{1} \vert ^{p-2}\tau_{1} \phi }{ \vert x \vert ^{p}} \biggr) \,dx= \int_{\Omega }Q(x)\frac{\tau_{1}^{p^{*}(t)-1}}{ \vert x \vert ^{t}} \phi \,dx+\lambda \int_{\Omega }\tau_{1}^{-s}\phi \,dx. $$
(4.5)
We show that (4.5) holds for every \(\phi \in W_{0}^{1,p}(\Omega )\). In fact, since \(W_{0}^{1,p}(\Omega )\cap C_{0}(\Omega )\) is dense in \(W_{0}^{1,p}(\Omega )\), then for every \(\phi \in W_{0}^{1,p}(\Omega )\), there exists a sequence \(\{\phi_{n}\}\subset W_{0}^{1,p}(\Omega ) \cap C_{0}(\Omega )\) such that \(\lim_{n\rightarrow \infty }\phi_{n}= \phi \). For \(m, n \in \mathbb{N^{+}}\) large enough, replacing ϕ with \(\phi_{n}-\phi_{m}\) in (4.5) yields
$$ \begin{aligned}[b] & \int_{\Omega } \biggl( \vert \nabla \tau_{1} \vert ^{p-2} \nabla \tau_{1}\nabla (\phi_{n}- \phi_{m})-\mu \frac{ \vert \tau_{1} \vert ^{p-2}\tau_{1} \vert \phi_{n}-\phi_{m} \vert }{ \vert x \vert ^{p}}\, \biggr)\,dx \\ &\quad = \int_{\Omega }Q(x)\frac{\tau_{1}^{p^{*}(t)}}{ \vert x \vert ^{t}} \vert \phi_{n}- \phi _{m} \vert \,dx+\lambda \int_{\Omega }\tau^{-s}_{1} \vert \phi_{n}-\phi_{m} \vert \,dx. \end{aligned} $$
(4.6)
On the one hand, using \(\phi_{n}\rightarrow \phi \) and (4.6), we have that \(\{\frac{\phi_{n}}{\tau_{1}}\} \) is a Cauchy sequence in \(L^{p}(\Omega )\), hence there exists \(\nu \in L^{p}(\Omega )\) such that \(\lim_{n\rightarrow \infty }\frac{\phi_{n}}{\tau^{s}_{0}}=\nu \), which implies that \(\lim_{n\rightarrow \infty } \frac{\phi_{n}}{\tau^{s}_{0}}=\nu \) in measure. By Riesz’s theorem, without loss of generality, choose a subsequence of \(\{\frac{\phi_{n}}{ \tau^{s}_{0}}\}\), still denoted by \(\{\frac{\phi_{n}}{\tau^{s}_{0}}\}\), such that
$$ \lim_{n\rightarrow \infty }\frac{\phi_{n}}{\tau^{s}_{0}}=\nu (x), \quad \text{a.e. } x\in \Omega . $$
(4.7)
On the other hand, from (4.7), we have that \(\nu =\frac{\phi }{\tau ^{s}_{0}}\), which leads to
$$ \lim_{n\rightarrow \infty } \int_{\Omega }\frac{\phi_{n}(x)}{\tau^{s}_{0}}\,dx= \int_{\Omega }\frac{\phi (x)}{\tau^{s}_{0}}\,dx. $$
Therefore, we deduce that (4.5) holds for \(\phi \in W_{0}^{1,p}( \Omega )\). Setting \(\phi =\tau_{1}\) in (4.5), we have
$$ \Vert \tau_{1} \Vert ^{p}- \int_{\Omega }Q(x)\frac{\tau_{1}^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx-\lambda \int_{\Omega }\tau_{1}^{1-s}\,dx=0. $$
(4.8)
Together with (4.4), we obtain that
$$ \Vert w_{\gamma } \Vert ^{p}- \int_{\Omega }Q(x)\frac{w_{\gamma }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx=o(1). $$
(4.9)
Hence
$$ \lim_{\gamma \rightarrow 0^{+}} \Vert w_{\gamma } \Vert ^{p}= \lim_{\gamma \rightarrow 0^{+}} \int_{\Omega }Q(x)\frac{w_{\gamma }^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx=l>0. $$
Since
$$ \int_{\Omega }\frac{ \vert w_{\gamma } \vert ^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx\geq \int_{\Omega }\frac{Q(x)}{Q_{M}} \frac{ \vert w_{\gamma } \vert ^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx \geq \int_{\Omega }\frac{Q(x)}{Q_{M}} \frac{(w_{\gamma }^{+})^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx. $$
Then \(l\geq \frac{S^{\frac{N-t}{p-t}}}{Q_{M}^{\frac{N-p}{p-t}}}\). By (4.8), we have
$$\begin{aligned} I_{\lambda ,\mu }(\tau_{1})&=\frac{1}{p} \Vert \tau_{1} \Vert ^{p}-\frac{1}{p ^{*} (t)} \int_{\Omega }Q(x)\frac{\tau_{1}^{p^{*}(t)}}{ \vert x \vert ^{t}}\,dx-\frac{\lambda }{1-s} \int_{\Omega }\tau_{1}^{1-s}\,dx \\ &=\frac{p-t}{p(N-t)} \Vert \tau_{1} \Vert ^{p}-\lambda \biggl(\frac{1}{1-s}-\frac{1}{p ^{*}(t)} \biggr) \int_{\Omega }\tau_{1}^{1-s}\,dx \\ &\geq \frac{p-t}{p(N-t)} \Vert \tau_{1} \Vert ^{p}- \lambda \biggl(\frac{1}{1-s}+\frac{1}{p ^{*} (t)} \biggr)C_{2} \Vert \tau_{1} \Vert ^{1-s} \\ &>-D\lambda^{\frac{p}{p+s-1}}. \end{aligned}$$
(4.10)
At the same time, it follows from (4.4) and (4.9) that
$$\begin{aligned} I_{\lambda ,\mu }(\tau_{1})&=I_{\gamma }(\tau_{\gamma })- \frac{p-t}{p(N-t)} \Vert w_{\gamma } \Vert ^{p}+o(1) \\ &< \frac{p-t}{p(N-t)} \biggl( \frac{S^{\frac{N-t}{p-t}}}{Q_{M}^{ \frac{N-p}{p-t}}}-l \biggr) -D \lambda^{\frac{p}{p-1+s}} \\ &\leq -D \lambda^{\frac{p}{p-1+s}}, \end{aligned}$$
which contradicts (4.10). Therefore, we deduce that
$$ I_{\lambda ,\mu }(\tau_{1})=\lim_{\gamma \rightarrow 0}I_{\gamma }( \tau_{\gamma })>\rho >0. $$
Consequently, problem (1.1) has two different solutions \(u_{1}\) and \(\tau_{1}\). Furthermore, \(\tau_{1} \not \equiv 0\), together with the maximum principle, we conclude that \(\tau_{1} >0\) a.e. \(x\in \Omega \). That is, \(\tau_{1}\) is a positive solution of problem (1.1).

The proof of Theorem 4.1 is completed. □

Remark 4.1

In order to apply the Brézis–Lieb lemma, we need to establish the convergence results for the sequences with gradient terms [5, 9]. Furthermore, the strong maximum principle for a p-Laplace operator is also used.

Declarations

Acknowledgements

We would like to thank the referees for their valuable comments and suggestions to improve our paper.

Availability of data and materials

Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

Funding

This project is supported by the Natural Science Foundation of Shanxi Province (201601D011003), NSFC (11401583) and the Fundamental Research Funds for the Central Universities (16CX02051A).

Authors’ contributions

All authors contributed equally to this work. All authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

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)
Department of Mathematics, School of Science, North University of China, Taiyuan, China
(2)
School of Data Sciences, Zhejiang University of Finance and Economics, Hangzhou, China
(3)
China Academy of Financial Research, Zhejiang University of Finance and Economics, Hangzhou, China

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