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

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.

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, 10–23, 25, 26, 28, 30–34]. 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:

• (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 $\vert \cdot \vert _{p}$;

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

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

• (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:

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

• (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 3.1–3.3 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.