Positive Solutions for Elliptic Problems Involving Hardy–Sobolev–Maz’ya Terms

Abstract

In the present paper, we study the semilinear elliptic problem $\displaystyle -\Delta u -\mu \frac{u}{|y|^{2}}=\frac{|u|^{2^{*}(s)-2}u}{|y|^{s}}+ f(x,u)$ in bounded domain. Replacing the Ambrosetti–Rabinowitz condition by general superquadratic assumptions and the nonquadratic assumption, we establish the existence results of positive solutions.

1 Introduction and Main Results

In this paper, we deal with the following semilinear elliptic problem with Dirichlet boundary value conditions

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u -\mu \frac{u}{|y|^{2}}=\frac{|u|^{2^{*}(s)-2}u}{|y|^{s}}+f(x,u), &{}\quad \mathrm {in} \quad \Omega , \\ u>0, &{} \quad \mathrm {in} \quad \Omega , \\ u=0, &{}\quad \mathrm {on} \quad \partial \Omega , \end{array}\right. } \end{aligned}$$

(1.1)

where $f\in C(\overline{\Omega }\times \mathbb {R}, \mathbb {R})$, $\Omega$ is a smooth bounded domain in $\mathbb {R}^{N}=\mathbb {R}^{k}\times \mathbb {R}^{N-k}$ with $2\le k<N$, a point $x\in \mathbb {R}^{N}$ is denoted as $x=(y,z)\in \mathbb {R}^{k}\times \mathbb {R}^{N-k}$ and $(0,z^{0})\in \Omega$, $0\le \mu <\bar{\mu }=\frac{(k-2)^{2}}{4}$ for $k>2$, and $\mu =0$ for $k=2$. The so-called Hardy–Sobolev critical exponent is denoted as $2^{*}(s)=\frac{2(N-s)}{N-2}$, where $0\le s<2$. Clearly, $2^{*}=2^{*}(0)=\frac{2N}{N-2}$ is the Sobolev critical exponent. F(x, t) is the primitive function of f(x, t) defined as $\displaystyle F(x,t)=\int _{0}^{t}f(x,s)\hbox {d}s$. $H^{1}_{0}(\Omega )$ is the Sobolev space with its equivalent norm

$$\begin{aligned} \Vert u\Vert =\left( \int _{\Omega }\left( |\nabla u|^{2}-\mu \frac{u^{2}}{|y|^{2}} \right) \hbox {d}x\right) ^{\frac{1}{2}} \end{aligned}$$

due to the Hardy inequality

$$\begin{aligned} C_{k}\int _{\mathbb {R}^{N}}\frac{u^{2}}{|y|^{2}}\hbox {d}x\le \int _{\mathbb {R}^{N}}|\nabla u|^{2}\hbox {d}x,~~~\forall u\in D^{1,2}(\mathbb {R}^{N}), \end{aligned}$$

where $C_{k}=(\frac{k-2}{2})^{2}$ is the best constant and is not attained. Let $S_{\mu }$ be the best Hardy–Sobolev constant defined as

$$\begin{aligned} S_{\mu }=\inf _{u\in D^{1,2}(\mathbb {R}^{N}\backslash (0,z^{0})), u\ne 0}\frac{\int _{\mathbb {R}^{N}}\left( |\nabla u|^{2} -\mu \frac{u^{2}}{|y|^{2}} \right) \hbox {d}x}{\left( \int _{\mathbb {R}^{N}}\frac{|u|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x \right) ^{\frac{2}{2^{*}(s)}}}. \end{aligned}$$

(1.2)

When $k=N$, (1.1) becomes

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u -\mu \frac{u}{|x|^{2}}=\frac{|u|^{2^{*}(s)-2}u}{|x|^{s}}+f(x,u), &{}\quad \mathrm {in} \quad \Omega , \\ u>0, &{}\quad \mathrm {in} \quad \Omega , \\ u=0, &{} \quad \mathrm {on} \quad \partial \Omega , \end{array}\right. } \end{aligned}$$

(1.3)

after the work of Brezis and Nirenberg [3], there are many papers concerning the Dirichlet problem with critical exponents (see [1, 6,7,8,9, 11, 17, 19, 26]). When $\mu =0$ and $s=0$, problem (1.3) becomes the well-known Brezis–Nirenberg problem and is studied extensively; for example, Nguyen and Lu [23] established the existence of nontrivial nonnegative solutions in dimension two involving exponential nonlinearities which had subcritical or critical exponential growth and did not satisfy the (AR) condition. When $\mu \ne 0$, the problem has its singularity at 0 and attracts much attention. For instance, Ding and Tang [12] studied the existence of positive solutions for $N\ge 3$ and $0\le s<2$. Kang and Peng [18] showed the existence of positive solutions replacing f(x, u) by $\lambda |u|^{q-2}u$ with $q>2$ for $0\le s<2$.

When $2\le k<N$, the singularity of the problem is more complicated. Very recently, it attracts more attention. Bhakta and Sandeep [2] studied the regularity, Palais–Smale characterization and existence of solutions in some special bounded domain and proved nonexistence of nontrivial solution with $f(x,u)=0$. Ganguly and Sandeep [14] researched the existence and nonexistence of sign-changing solutions for the Brezis–Nirenberg type problem in the hyperbolic space, which is closely related to Hardy–Sobolev–Maz’ya equations. Yang [28] showed the existence of positive solutions for $N\ge 3$ with Neumann boundary condition and f(x, u) satisfying some conditions. Wang and Wang [27] showed that the existence of infinitely many solutions replacing f(x, u) by $\lambda u$ for $N>6+s$. More details about Hardy–Sobolev–Maz’ya equations and elliptic equations in $\mathbb {H}^{n}$ (n-dimensional Hyperbolic space) can be seen in [4, 5, 13, 22] and their references.

In order to get a nontrivial solution, the Mountain Pass Lemma [25] is generally exploited, when the equation involves superlinearity. To use this lemma, the authors assume that f(x, t) satisfies the well-known Ambrosetti–Rabinowitz (AR) condition, that is, for some $\rho >2$, $M>0$, for a.e. $x\in \overline{\Omega }$ and all $|t|\ge M$, there holds

$$\begin{aligned} 0<\rho F(x,t)\le f(x,t)t. \end{aligned}$$

It is known that the (AR) condition plays an important role in ensuring that any Cerami (Ce) sequence of the functional is bounded. But this condition is very restrictive, and there are many functions which do not satisfy the (AR) condition, for example

$$\begin{aligned} f(x,t)=2t\ln (1+|t|). \end{aligned}$$

The main purpose of this present paper is to establish the existence of positive solutions for problem (1.1) with $2\le k<N$, $0\le \mu <\bar{\mu }$ under the case $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$ and f(x, t) satisfying different conditions which are weaker than the (AR) condition but play the same role as the (AR) condition. Here are the main results of this paper:

Theorem 1

Suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$, $2\le k<N$, $0\le \mu <\bar{\mu }$, and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$. $f\in C(\overline{\Omega }\times \mathbb {R}^{+}, \mathbb {R})$ satisfies

$(f_{1})$ :

$f(x,t)\ge 0$ for $t\ge 0$ and $f(x,t)=0$ for $t\le 0$. $\displaystyle \limsup _{t\rightarrow 0^{+}}\frac{f(x,t)}{t}<\lambda$ uniformly for $x\in \overline{\Omega }$, where $0<\lambda<$$\lambda _{1}$ and $\lambda _{1}$ is the first eigenvalue of $-\Delta -\mu |y|^{-2}$,

$(f_{2})$ :

$\displaystyle \lim _{t\rightarrow +\infty }\frac{f(x,t)}{t^{2^{*}(s)-1}}=0$ uniformly for $x\in \overline{\Omega }$,

$(f_{3})$ :

there exist a positive constant $\sigma$, a nonempty open subset $\omega$ with $(0,z^{0})\in \omega \subset \Omega$, and a nonempty open interval $I\subset (0,+\infty )$, so that $f(x,t)\ge \sigma >0$ for almost everywhere $x\in \omega$ and for all $t\in I$.

Then, problem (1.1) admits at least one positive solution.

Remark 1

Firstly, when $k=N$, problem (1.1) has been researched in [20], in which f(x, t) did not satisfy the (AR) condition. Secondly, there are many examples satisfying the assumptions of Theorem 1. For instance, we may take $f(x,t)=\lambda t$ with $0<\lambda <\lambda _{1}$, or $f(x,t)=\lambda t^{q}$ with $\lambda >0$ and $1<q<2^{*}(s)-1$.

Theorem 2

Suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$, $2\le k<N$, $0\le \mu <\bar{\mu }$ and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$. $f\in C(\overline{\Omega }\times \mathbb {R}^{+}, \mathbb {R}^+)$ satisfies $(f_{3})$ and

$(f_{4})$ :

$\displaystyle \lim _{t\rightarrow 0^{+}}\frac{f(x,t)}{t}=0$ uniformly for $x\in \overline{\Omega }$,

$(f_{5})$ :

$\displaystyle \lim _{t\rightarrow +\infty }\frac{f(x,t)}{t^{2^{*}-1}}=0$ uniformly for $x\in \overline{\Omega }$,

$(f_{6})$ :

$\displaystyle \lim _{t\rightarrow +\infty }\frac{f(x,t)}{t}=+\infty$ uniformly for $x\in \overline{\Omega }$,

$(f_{7})$ :

$|f(x,t)|^{\tau }\le a_{1}\widetilde{F}(x,t)|t|^{\tau }$ for some $a_{1}>0$, $\tau >1$ and $(x,t)\in \overline{\Omega }\times \mathbb {R}^{+}$ with t large enough, where $\widetilde{F}(x,t)=\frac{1}{2}f(x,t)t-F(x,t)$.

Then, problem (1.1) possesses at least a positive solution.

Remark 2

Firstly, $(f_{6})$ and $(f_{7})$ can lead to $\widetilde{F}(x,t)=\frac{1}{2}f(x,t)t-F(x,t)\rightarrow +\,\infty$ uniformly in $x\in \overline{\Omega }$ as $t\rightarrow +\,\infty$. Secondly, there are also many functions satisfying the conditions of Theorem 2. For example, one may take $f(x,t)=\lambda t^{q}$ with $\lambda >0$ and $1<q<2^{*}-1$. Thirdly, when $k=N$, $\mu \ne 0$, and $s\ne 0$, Ding and Tang [12] obtained the existence of positive solutions with f(x, u) satisfying a global (AR) condition. Here, we obtain the similar results as those in [12] when $2\le k<N$. Thus, our results complete the existence of positive solutions for elliptic problem with Hardy–Sobolev critical exponents.

Theorem 3

Suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$, $2\le k<N$, $0\le \mu <\bar{\mu }$, and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$. $f\in C(\overline{\Omega }\times \mathbb {R}^{+}, \mathbb {R})$ satisfies $(f_{1})$, $(f_{3})$, $(f_{5})$, $(f_{6})$ and

$(f_{8})$ :

there exist two constants $\theta \ge 1$, $\theta _{0}>0$ such that $\theta H(x,t)\ge H(x,st)-\theta _{0}$ for all $x\in \overline{\Omega }$, $t\ge 0$ and $s\in [0,1]$, where $H(x,t)=f(x,t)t-2F(x,t)$ and $F(x,t)=\int _{0}^{t}f(x,s)\hbox {d}s$.

Then, problem (1.1) admits at least one positive solution.

Remark 3

A condition similar to $(f_{8})$ was introduced by Jeanjean [16]. We can easily verify that when $\theta =1$, $(f_{8})$ means that $\frac{f(x,t)}{t}$ is nondecreasing with respect to $t\ge 0$, which leads to the (AR) condition. Thus, $(f_{8})$ gives a more general monotonicity when $\theta >1$. Moreover, one can find some examples that satisfy $(f_{8})$ but $\frac{f(x,t)}{t}$ is not monotone. For example, let

$$\begin{aligned} F(x,t)=t^{2}\ln (1+t^{2})+t\sin t, \end{aligned}$$

it follows that

$$\begin{aligned} f(x,t)=2 t\ln (1+t^{2})+\frac{2t^{3}}{1+t^{2}}+\sin t+t\cos t, \end{aligned}$$

then

$$\begin{aligned} H(x,t)=2(t^{2}-1)+\frac{2}{1+t^{2}}+t^{2}\cos t-t\sin t. \end{aligned}$$

Let $\theta =1000$, we can prove by some simple computation that f(x, t) satisfies $(f_{8})$ but $\frac{f(x,t)}{t}$ is not monotone any more.

Theorem 4

Suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$, $2\le k<N$, $0\le \mu <\bar{\mu }$, and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$. $f\in C(\overline{\Omega }\times \mathbb {R}^{+}, \mathbb {R})$ satisfies $(f_{1})$, $(f_{3})$ and

$(f_{9})$ :

there exists $q\in [2, 2^{*})$ such that $\displaystyle \lim _{t\rightarrow +\infty }\frac{f(x,t)}{t^{q-1}}=0$ uniformly for $x\in \overline{\Omega }$,

$(f_{10})$ :

there exist constants $D>0$, $L>0$, and $\delta >\frac{N(q-2)}{2}$, such that

$$\begin{aligned} \frac{f(x,t)t-2F(x,t)}{|t|^{\delta }}\ge D, \end{aligned}$$

for $t\ge L$ and a.e. $x\in \overline{\Omega }$.

Then, problem (1.1) admits at least one positive solution.

Remark 4

A nonquadratic condition similar to $(f_{10})$ was introduced in [10]. Although $(f_{10})$ is weaker than the (AR) condition, it can guarantee the boundedness of the (Ce) sequence. There are also many functions that satisfy $(f_{10})$ but do not satisfy the (AR) condition. For example, $f(x,t)=2t\ln (1+t^{2})+\frac{2t^{3}}{1+t^{2}},~t\in \mathbb {R}$.

2 Proof of Theorems

To verify our main results, we make use of the following notations.

• The dual space of a Banach space E will be denoted by $E'$.

• $L^{p}(\Omega , |y|^{-s}\hbox {d}x)$ denotes the weighted Sobolev space.

• $\rightarrow$(resp. $\rightharpoonup$) denotes the strong (resp. weak) convergence.

• C, $C_{i}$ (i=0, 1, 2 ...) will denote various positive constants, and their values can vary from line to line.

In order to study the positive solutions of problem (1.1), we first consider the existence of nontrivial solutions to the problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\Delta u -\mu \frac{u}{|y|^{2}}=\frac{(u^{+})^{2^{*}(s)-1}}{|y|^{s}}+f(x,u^{+}), &{} \quad \mathrm {in} \quad \Omega , \\ u>0, &{} \quad \mathrm {in} \quad \Omega , \\ u=0, &{} \quad \mathrm {on} \quad \partial \Omega , \end{array}\right. } \end{aligned}$$

(2.1)

where $u^{+}=\max \{u,0\}$. The energy functional corresponding to problem (2.1) is given by

$$\begin{aligned} I(u)=\frac{1}{2}\int _{\Omega }\left( |\nabla u|^{2}-\mu \frac{u^{2}}{|y|^{2}}\right) \hbox {d}x-\frac{1}{2^{*}(s)}\int _{\Omega }\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x -\int _{\Omega }F(x,u^{+})\hbox {d}x, \end{aligned}$$

(2.2)

for $u\in H_{0}^{1}(\Omega )$. Clearly, I is well defined and is $C^{1}$ smooth thanks to the Hardy–Sobolev–Maz’ya inequality [21]

$$\begin{aligned} \left( \int _{\mathbb {R}^{N}}\frac{|u|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\right) ^{\frac{2}{2^{*}(s)}}\le S_{\mu }^{-1}\int _{\mathbb {R}^{N}}\left( |\nabla u|^{2}-\mu \frac{u^{2}}{|y|^{2}}\right) \hbox {d}x, \end{aligned}$$

(2.3)

where $S_{\mu }=S(\mu ,N,k,s)$ is the best constant defined in (1.2). By the existence of the one-to-one correspondence between the critical points of I and the weak solutions of problem (2.1), we know that if u is a weak solution of problem (2.1), there holds

$$\begin{aligned} \langle I'(u),v\rangle= & {} \int _{\Omega }\left( (\nabla u,\nabla v)-\mu \frac{uv}{|y|^{2}}\right) \hbox {d}x -\int _{\Omega }\frac{(u^{+})^{2^{*}(s)-1}v}{|y|^{s}}\hbox {d}x\\&-\int _{\Omega }f(x,u^{+})v\hbox {d}x=0, \end{aligned}$$

for any $v\in H^{1}_{0}(\Omega )$.

Before proving our main results, we need the following lemmas. First, it is necessary to give the estimates below. From [2, 4, 5, 22], when $s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$, the best constant $S_{\mu }$ given in (1.2) can be achieved by the following form of the extremal function

$$\begin{aligned} U(y,z)=c(\mu ,k,N)\frac{|y|^{\frac{\sqrt{(k-2)^{2}-4\mu }-(k-2)}{2}}}{((1+|y|)^{2}+|z|^{2})^{\frac{1}{2^{*}(s)-2}}}, \end{aligned}$$

where $c(\mu ,k,N)$ is a constant. In order to guarantee $s\ge 0$, we suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$ in this article. For $(0,z^{0})\in \Omega$, we can choose $\rho$, $R>0$, satisfying $B_{\rho }(0,z^{0})\subset \Omega \subset B_{R}(0,z^{0})$. Let $\varphi \in C_{0}^{\infty }(\Omega )$ be a cutoff function such that $0\le \varphi (x)\le 1$ and

$$\begin{aligned} \varphi (x)={\left\{ \begin{array}{ll} 1,&{} \quad x\in B_{\frac{\rho }{2}}(0,z^{0}),\\ 0,&{} \quad x\not \in B_{\rho }(0,z^{0}). \end{array}\right. } \end{aligned}$$

Denote $T=\frac{\sqrt{(k-2)^{2}-4\mu }-(k-2)}{2}$. Set $u_{\varepsilon }^{*}=\varepsilon ^{\frac{2-N}{2}}U(\frac{y}{\varepsilon },\frac{z-z^{0}}{\varepsilon })$ for $\varepsilon >0$. Then,

$$\begin{aligned} u_{\varepsilon }^{*}(x)=c(\mu ,k,N)\varepsilon ^{\frac{1}{2^{*}(s)-2}}\frac{|y|^{T}}{((\varepsilon +|y|)^{2}+|z-z^{0}|^{2})^{\frac{1}{2^{*}(s)-2}}} \end{aligned}$$

is also an extremal function of $S_{\mu }$ and solves the equation

$$\begin{aligned} -\Delta u -\mu \frac{u}{|y|^{2}}=\frac{|u|^{2^{*}(s)-2}u}{|y|^{s}},\quad \mathrm {in} \quad \mathbb {R}^{N}. \end{aligned}$$

For $\varepsilon >0$, we define $u_{\varepsilon }=\varphi (x)u_{\varepsilon }^{*}(x)$. We have the following estimates for $u_{\varepsilon }$.

Lemma 2.1

Suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$, $2\le k<N$, $0\le \mu <\bar{\mu }$, and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$, then the following estimates hold

$$\begin{aligned}&\int _{\Omega }|\nabla u_{\varepsilon }|^{2}\hbox {d}x = \int _{\mathbb {R}^{N}}|\nabla u_{\varepsilon }^{*}|^{2}\hbox {d}x+O\left( \varepsilon ^{\frac{2}{2^{*}(s)-2}} \right) , \end{aligned}$$

(2.4)

$$\begin{aligned}&\int _{\Omega }\frac{u_{\varepsilon }^{2}}{|y|^{2}}\hbox {d}x =\int _{\mathbb {R}^{N}}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x+O\left( \varepsilon ^{\frac{2}{2^{*}(s)-2}} \right) , \end{aligned}$$

(2.5)

$$\begin{aligned}&\int _{\Omega }\frac{|u_{\varepsilon }|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x = \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x +O \left( \varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}\right) . \end{aligned}$$

(2.6)

Proof

First, we estimate (2.5). There holds

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \int _{\Omega }\frac{u_{\varepsilon }^{2}}{|y|^{2}}\hbox {d}x&{}=&{}\displaystyle \int _{\Omega }\frac{(\varphi u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x =\int _{\Omega }\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x-\int _{\Omega }(1-\varphi ^{2})\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x \\ &{}=&{}\displaystyle \int _{\mathbb {R}^{N}}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x-\int _{\mathbb {R}^{N}{\setminus }\Omega }\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x -\int _{\Omega }(1-\varphi ^{2})\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x \\ &{}=&{}\displaystyle \int _{\mathbb {R}^{N}}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x-\int _{\mathbb {R}^{N}{\setminus }\Omega }\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x -\int _{\Omega {\setminus } B_{\frac{\rho }{2}}(0,z^{0})}(1-\varphi ^{2})\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x. \end{array}\right. \end{aligned}$$

Then, we have

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \int _{\mathbb {R}^{N}{\setminus } B_{\rho }(0,z^{0})}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x &{}=&{}\displaystyle \varepsilon ^\frac{2}{2^{*}(s)-2}\int _{\mathbb {R}^{N}{\setminus } B_{\rho }(0,z^{0})}\frac{|y|^{2T-2}}{((\varepsilon +|y|)^{2}+|z-z^{0}|^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}x\\ &{}=&{}\displaystyle \varepsilon ^\frac{2}{2^{*}(s)-2}\int _{\mathbb {R}^{N}{\setminus } B_{\rho }(0)}\frac{|y|^{2T-2}}{((\varepsilon +|y|)^{2}+|z|^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}x \\ &{}\le &{}\displaystyle \varepsilon ^\frac{2}{2^{*}(s)-2}\bigg (\int ^{+\infty }_{\frac{\rho }{2}}\int ^{+\infty }_{\frac{\rho }{2}}\frac{r^{2T+k-3}t^{N-k-1}}{((\varepsilon +r)^{2}+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}r\hbox {d}t\\ &{}&{} \quad +\displaystyle \int ^{\frac{\rho }{2}}_{0}\int ^{+\infty }_{\frac{\rho }{2}}\frac{r^{2T+k-3}t^{N-k-1}}{((\varepsilon +r)^{2}+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}r\hbox {d}t\\ \end{array}\right. \end{aligned}$$

$$\begin{aligned} \left. ~~~~~~~~~~~~~~~~~~~~~~~~\begin{array}{rcl}&{}&{}+\displaystyle \int ^{+\infty }_{\frac{\rho }{2}}\int ^{\frac{\rho }{2}}_{0}\frac{r^{2T+k-3}t^{N-k-1}}{((\varepsilon +r)^{2}+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}r\hbox {d}t\bigg )\\ &{}=&{}\displaystyle \varepsilon ^{2T+N-2-\frac{2}{2^{*}(s)-2}}\bigg (\int ^{+\infty }_{\frac{\rho }{2\varepsilon }}r^{2T+k-3}\hbox {d}r \int ^{+\infty }_{\frac{\rho }{2\varepsilon }}\frac{t^{N-k-1}}{((1+r)^{2}+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}t\\ &{}&{}+\displaystyle \int ^{\frac{\rho }{2\varepsilon }}_{0}r^{2T+k-3}\hbox {d}r \int ^{+\infty }_{\frac{\rho }{2\varepsilon }}\frac{t^{N-k-1}}{((1+r)^{2}+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}t\\ &{}&{}+\displaystyle \int ^{+\infty }_{\frac{\rho }{2\varepsilon }}r^{2T+k-3}\hbox {d}r \int ^{\frac{\rho }{2\varepsilon }}_{0}\frac{t^{N-k-1}}{((1+r)^{2}+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}t\bigg )\\ &{}\le &{}\displaystyle \varepsilon ^{2T+N-2-\frac{2}{2^{*}(s)-2}}\bigg (\int ^{+\infty }_{\frac{\rho }{2\varepsilon }}\frac{r^{2T+k-3}}{(1+r)^{\frac{4}{2^{*}(s)-2}-N+k}}\hbox {d}r \int ^{+\infty }_{0}\frac{t^{N-k-1}}{(1+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}t\\ &{}&{}+\displaystyle \int ^{\frac{\rho }{2\varepsilon }}_{0}\frac{r^{2T+k-3}}{(1+r)^{\frac{4}{2^{*}(s)-2}-N+k}}\hbox {d}r \int ^{+\infty }_{0}\frac{t^{N-k-1}}{(1+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}t\\ &{}&{}+\displaystyle \int ^{+\infty }_{\frac{\rho }{2\varepsilon }}\frac{r^{2T+k-3}}{(1+r)^{\frac{4}{2^{*}(s)-2}-N+k}}\hbox {d}r \int ^{+\infty }_{0}\frac{t^{N-k-1}}{(1+t^{2})^{\frac{2}{2^{*}(s)-2}}}\hbox {d}t\bigg )\\ &{}\le &{}\displaystyle C\varepsilon ^{2T+N-2-\frac{2}{2^{*}(s)-2}}\bigg (\int ^{+\infty }_{\frac{\rho }{2\varepsilon }}\frac{r^{2T+k-3}}{(1+r)^{\frac{4}{2^{*}(s)-2}-N+k}}\hbox {d}r\\ &{}&{}+\displaystyle \int ^{\frac{\rho }{2\varepsilon }}_{0}\frac{r^{2T+k-3}}{(1+r)^{\frac{4}{2^{*}(s)-2}-N+k}}\hbox {d}r+\int ^{+\infty }_{\frac{\rho }{2\varepsilon }} \frac{r^{2T+k-3}}{(1+r)^{\frac{4}{2^{*}(s)-2}-N+k}}\hbox {d}r\bigg )\\ &{}\le &{}\displaystyle C\varepsilon ^{\frac{2}{2^{*}(s)-2}}. \end{array}\right. \end{aligned}$$

Then from above, one has

$$\begin{aligned} \int _{\mathbb {R}^{N}{\setminus } B_{\rho }(0,z^{0})}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x= O \left( \varepsilon ^{\frac{2}{2^{*}(s)-2}}\right) . \end{aligned}$$

(2.7)

By the same method, we get

$$\begin{aligned} \int _{\mathbb {R}^{N}{\setminus } B_{R}(0,z^{0})}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x= O \left( \varepsilon ^{\frac{2}{2^{*}(s)-2}}\right) . \end{aligned}$$

Since

$$\begin{aligned} \int _{\mathbb {R}^{N}{\setminus } B_{R}(0,z^{0})}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x\le \int _{\mathbb {R}^{N}{\setminus }\Omega }\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x \le \int _{\mathbb {R}^{N}{\setminus } B_{\rho }(0,z^{0})}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x, \end{aligned}$$

we obtain

$$\begin{aligned} \int _{\Omega }\frac{u_{\varepsilon }^{2}}{|y|^{2}}\hbox {d}x =\int _{\mathbb {R}^{N}}\frac{(u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x+O\left( \varepsilon ^{\frac{2}{2^{*}(s)-2}} \right) . \end{aligned}$$

Next, we estimate (2.6). In fact, there holds

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \int _{\Omega }\frac{|u_{\varepsilon }|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x &{}=&{} \displaystyle \int _{\Omega }\frac{|\varphi u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\\ &{}=&{} \displaystyle \int _{\Omega }\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }(1-\varphi ^{2^{*}(s)})\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x \\ &{}=&{} \displaystyle \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\mathbb {R}^{N}{\setminus }\Omega }\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x -\int _{\Omega }(1-\varphi ^{2^{*}(s)})\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x \\ &{}=&{} \displaystyle \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\mathbb {R}^{N}{\setminus }\Omega }\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x -\int _{\Omega {\setminus } B_{\frac{\rho }{2}}(0,z^{0})}(1-\varphi ^{2^{*}(s)})\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x. \end{array}\right. \end{aligned}$$

Since

$$\begin{aligned} \int _{\mathbb {R}^{N}{\setminus } B_{R}(0,z^{0})}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\le \int _{\mathbb {R}^{N}{\setminus }\Omega }\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x \le \int _{\mathbb {R}^{N}{\setminus } B_{\rho }(0,z^{0})}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x, \end{aligned}$$

using the method similar to (2.7), one gets

$$\begin{aligned} \int _{\mathbb {R}^{N}{\setminus } B_{R}(0,z^{0})}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x= O \left( \varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}\right) . \end{aligned}$$

Thus, we deduce

$$\begin{aligned} \int _{\Omega }\frac{|u_{\varepsilon }|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x=\int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x +O \left( \varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}\right) . \end{aligned}$$

Now, we estimate (2.4). Observe that

$$\begin{aligned} \int _{\Omega }|\nabla u_{\varepsilon }|^{2}\hbox {d}x=\int _{\Omega }|\nabla (\varphi u_{\varepsilon }^{*})|^{2}\hbox {d}x =\int _{\Omega }|u_{\varepsilon }^{*}|^{2}|\nabla \varphi |^{2}\hbox {d}x +\int _{\Omega }\left( \nabla u_{\varepsilon }^{*},\nabla (\varphi ^{2}u_{\varepsilon }^{*})\right) \hbox {d}x, \end{aligned}$$

and $\displaystyle -\Delta u_{\varepsilon }^{*} -\mu \frac{u_{\varepsilon }^{*}}{|y|^{2}}=\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)-2}u_{\varepsilon }^{*}}{|y|^{s}}$, one has

$$\begin{aligned} \int _{\Omega }\left( \nabla u_{\varepsilon }^{*},\nabla (\varphi ^{2}u_{\varepsilon }^{*})\right) \hbox {d}x=\mu \int _{\Omega }\frac{(\varphi u_{\varepsilon }^{*})^{2}}{|y|^{2}}\hbox {d}x+\int _{\Omega }\varphi ^{2}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x. \end{aligned}$$

When $x\in B_{\frac{\rho }{2}}(0,z^{0})$, one has $\nabla \varphi =0$, then

$$\begin{aligned} \int _{\Omega }|u_{\varepsilon }^{*}|^{2}|\nabla \varphi |^{2}\hbox {d}x=O\left( \varepsilon ^{\frac{2}{2^{*}(s)-2}} \right) . \end{aligned}$$

Therefore, one has

$$\begin{aligned} \left. \begin{array}{rcl} &{}&{}\displaystyle \int _{\Omega }|\nabla u_{\varepsilon }|^{2}\hbox {d}x\\ &{}&{}\quad =\displaystyle \int _{\Omega }|u_{\varepsilon }^{*}|^{2}|\nabla \varphi |^{2}\hbox {d}x +\int _{\Omega }\varphi ^{2}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\mu \int _{\Omega }\varphi ^{2}\frac{|u_{\varepsilon }^{*}|^{2}}{|y|^{2}}\hbox {d}x\\ &{}&{}\quad \le \displaystyle C\varepsilon ^{\frac{2}{2^{*}(s)-2}}+\int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x +C\varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}+\mu \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2}}{|y|^{2}}\hbox {d}x+C\varepsilon ^{\frac{2}{2^{*}(s)-2}}\\ &{}&{}\quad =\displaystyle \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\mu \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2}}{|y|^{2}}\hbox {d}x +C\varepsilon ^{\frac{2}{2^{*}(s)-2}}+C\varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}\\ &{}&{}\quad =\displaystyle \int _{\mathbb {R}^{N}}|\nabla u_{\varepsilon }^{*}|^{2}\hbox {d}x+C_{2}\varepsilon ^{\frac{2}{2^{*}(s)-2}} \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \left. \begin{array}{rcl} &{}&{}\displaystyle \int _{\Omega }|\nabla u_{\varepsilon }|^{2}\hbox {d}x\\ &{}&{}\quad =\displaystyle \int _{\Omega }|u_{\varepsilon }^{*}|^{2}|\nabla \varphi |^{2}\hbox {d}x +\int _{\Omega }\varphi ^{2}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\mu \int _{\Omega }\varphi ^{2}\frac{|u_{\varepsilon }^{*}|^{2}}{|y|^{2}}\hbox {d}x\\ &{}&{}\quad \ge \displaystyle C\varepsilon ^{\frac{2}{2^{*}(s)-2}}+\int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x +C\varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}+\mu \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2}}{|y|^{2}}\hbox {d}x+C\varepsilon ^{\frac{2}{2^{*}(s)-2}}\\ &{}&{}\quad =\displaystyle \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\mu \int _{\mathbb {R}^{N}}\frac{|u_{\varepsilon }^{*}|^{2}}{|y|^{2}}\hbox {d}x +C\varepsilon ^{\frac{2}{2^{*}(s)-2}}+C\varepsilon ^{\frac{2^{*}(s)}{2^{*}(s)-2}}\\ &{}&{}\quad =\displaystyle \int _{\mathbb {R}^{N}}|\nabla u_{\varepsilon }^{*}|^{2}\hbox {d}x+C_{1}\varepsilon ^{\frac{2}{2^{*}(s)-2}}. \end{array}\right. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \int _{\mathbb {R}^{N}}|\nabla u_{\varepsilon }^{*}|^{2}\hbox {d}x+C_{1}\varepsilon ^{\frac{2}{2^{*}(s)-2}}\le \int _{\Omega }|\nabla u_{\varepsilon }|^{2}\hbox {d}x\le \int _{\mathbb {R}^{N}}|\nabla u_{\varepsilon }^{*}|^{2}\hbox {d}x+C_{2}\varepsilon ^{\frac{2}{2^{*}(s)-2}}. \end{aligned}$$

The proof is completed. $\square$

For convenience, it is necessary to get the following estimates. Set

$$\begin{aligned} v_{\varepsilon }=u_{\varepsilon }/\left( \int _{\Omega }\frac{u_{\varepsilon }^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\right) ^{\frac{1}{2^{*}(s)}}. \end{aligned}$$

Clearly,

$$\begin{aligned} \left. \int _{\Omega }\frac{v_{\varepsilon }^{2^{*}(s)}}{|y|^{s}}\hbox {d}x=1. \right. \end{aligned}$$

Then, the following results can be obtained by the methods used in [15],

$$\begin{aligned} \displaystyle S_{\mu }+C_{7}\varepsilon ^{\frac{2}{2^{*}(s)-2}}\le \Vert v_{\varepsilon }\Vert ^{2}\le S_{\mu }+C_{8}\varepsilon ^{\frac{2}{2^{*}(s)-2}}, \end{aligned}$$

(2.8)

and

$$\begin{aligned} \int _{\Omega }v_{\varepsilon }^{q}\hbox {d}x={\left\{ \begin{array}{ll} O(\varepsilon ^{\frac{q}{2^{*}(s)-2}}),~~~~~~~~~~~~~~1<q<\frac{2N}{2N+\sqrt{(k-2)^{2}-4\mu }-(k-2)},\\ O(\varepsilon ^{\frac{q}{2^{*}(s)-2}}|\ln \varepsilon |),~~~~~~~q=\frac{2N}{2N+\sqrt{(k-2)^{2}-4\mu }-(k-2)},\\ O(\varepsilon ^{Tq+N-\frac{q}{2^{*}(s)-2}}),~~~~\frac{2N}{2N+\sqrt{(k-2)^{2}-4\mu }-(k-2)}<q<2^{*}. \end{array}\right. } \end{aligned}$$

(2.9)

We will use the function $v_{\varepsilon }$ as a test function to estimate I(u) below.

Lemma 2.2

Suppose $N\ge 2k-2-2\sqrt{(k-2)^{2}-4\mu }$, $0\le \mu <\bar{\mu }$, and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$. Assume that $(f_{1})$, $(f_{3})$, and $(f_{5})$ hold. There exists $u'\in H^{1}_{0}(\Omega )$ with $u'\not \equiv 0$, such that

$$\begin{aligned} \sup _{t\ge 0}I(tu')<\frac{2-s}{2(N-s)}S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}. \end{aligned}$$

Proof

Considering the functions

$$\begin{aligned}&g(t)=I(tv_{\varepsilon })=\frac{1}{2}t^{2}\Vert v_{\varepsilon }\Vert ^{2}-\frac{t^{2^{*}(s)}}{2^{*}(s)} -\int _{\Omega }F(x,tv_{\varepsilon })\hbox {d}x,\\&\tilde{g}(t)=\frac{1}{2}t^{2}\Vert v_{\varepsilon }\Vert ^{2}-\frac{t^{2^{*}(s)}}{2^{*}(s)}. \end{aligned}$$

Note that $g(0)=0$, $g(t)>0$ for $t>0$ small enough, and $\displaystyle \lim _{t\rightarrow +\infty }g(t)=-\infty$. It follows that $\displaystyle \sup _{t\ge 0}g(t)$ can be achieved by some $t_{\varepsilon }>0$.

First, we claim that $t_{\varepsilon }$ is bounded. By

$$\begin{aligned} 0=g'(t_{\varepsilon })=t_{\varepsilon }\left( \Vert v_{\varepsilon }\Vert ^{2}-t_{\varepsilon }^{2^{*}(s)-2}-\frac{1}{t_{\varepsilon }}\int _{\Omega }f(x,tv_{\varepsilon })v_{\varepsilon }\hbox {d}x \right) , \end{aligned}$$

we have $\displaystyle \Vert v_{\varepsilon }\Vert ^{2}=t_{\varepsilon }^{2^{*}(s)-2}+\frac{1}{t_{\varepsilon }}\int _{\Omega }f(x,tv_{\varepsilon })v_{\varepsilon }\hbox {d}x\ge t_{\varepsilon }^{2^{*}(s)-2}$; therefore, one gets

$$\begin{aligned} t_{\varepsilon }\le \Vert v_{\varepsilon }\Vert ^{\frac{2}{2^{*}(s)-2}}\triangleq t_{\varepsilon }^{0}. \end{aligned}$$

By (2.8), we get

$$\begin{aligned} t_{\varepsilon }\le C_{12}. \end{aligned}$$

(2.10)

Now, we prove that $t_{\varepsilon }$ is bounded below under $(f_{1})$ and $(f_{5})$. Obviously, one has $|f(x,t)t|\le \varepsilon |t|^{2^{*}}+Ct^{2}$, then $\displaystyle \Vert v_{\varepsilon }\Vert ^{2}\le t_{\varepsilon }^{2^{*}(s)-2}+\varepsilon \int _{\Omega }|t_{\varepsilon }|^{2^{*}-2}|v_{\varepsilon }|^{2^{*}}\hbox {d}x +C\int _{\Omega }|v_{\varepsilon }|^{2}\hbox {d}x.$ Due to $H^{1}_{0}(\Omega )\hookrightarrow L^{2^{*}}(\Omega )$ and (2.8), we can obtain $\displaystyle \int _{\Omega }|v_{\varepsilon }|^{2^{*}}\hbox {d}x\le C\Vert v_{\varepsilon }\Vert ^{2^{*}}\le C_{13}(2S_{\mu })^{\frac{2^{*}}{2}}$ for $\varepsilon >0$ small enough. From (2.8), one has $\displaystyle \int _{\Omega }|v_{\varepsilon }|^{2}\hbox {d}x\rightarrow 0, ~~\mathrm {as}~\varepsilon \rightarrow 0$. From above, combining with (2.9), we deduce that

$$\begin{aligned} t_{\varepsilon }^{2^{*}(s)-2}\ge C. \end{aligned}$$

(2.11)

From (2.10) and (2.11), we obtain that $t_{\varepsilon }$ is bounded for $\varepsilon >0$ small enough.

Secondly, we compute $\displaystyle \sup _{t\ge 0}g(t)$. Now we claim that

$$\begin{aligned} \Vert v_{\varepsilon }\Vert ^ {\frac{2\cdot 2^{*}(s)}{2^{*}(s)-2}}\le S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}+C_{14}\varepsilon ^{\frac{2}{2^{*}(s)-2}}. \end{aligned}$$

(2.12)

In order to prove, we first verify the following inequality

$$\begin{aligned} (a+b)^{r}\le a^{r}+r(a+1)^{r-1}b,~~a>0,~r\ge 1,~0\le b\le 1. \end{aligned}$$

Indeed, set $\phi (x)=(a+x)^{r}-a^{r}-r(a+1)^{r-1}x$, for $0\le x\le 1$. Obviously, $\phi '(x)\le 0$ for all $0\le x\le 1$, so $\phi (b)\le \phi (0)=0$; then, the inequality above holds. Then, let $a=S_{\mu }$, $b=C\varepsilon ^{\frac{2}{2^{*}(s)-2}}$, $r=\frac{2^{*}(s)}{2^{*}(s)-2}$, and combining with (2.8), we get (2.12).

It is easy to get that $\tilde{g}(t)$ attains its maximum at $t_{\varepsilon }^{0}$ and is increasing in the interval $[0,t_{\varepsilon }^{0}]$, and combining with (2.12) we conclude that

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle I(t_{\varepsilon }v_{\varepsilon })=g(t_{\varepsilon })&{}\le &{}\displaystyle \tilde{g}(t_{\varepsilon }^{0})-\int _{\Omega }F(x,t_{\varepsilon }v_{\varepsilon })\hbox {d}x\\ &{}=&{}\displaystyle \frac{2-s}{2(N-s)}\Vert v_{\varepsilon }\Vert ^{\frac{2\cdot 2^{*}(s)}{2^{*}(s)-2}}-\int _{\Omega }F(x,t_{\varepsilon }v_{\varepsilon })\hbox {d}x\\ &{}\le &{}\displaystyle \frac{2-s}{2(N-s)}S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}+C_{14}\varepsilon ^{\frac{2}{2^{*}(s)-2}} -\int _{\Omega }F(x,t_{\varepsilon }v_{\varepsilon })\hbox {d}x. \end{array}\right. \end{aligned}$$

Therefore, in order to verify that $\displaystyle \sup _{t\ge 0}I(tu')<\frac{2-s}{2(N-s)}S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}$, it is sufficient to show that

$$\begin{aligned} \displaystyle C_{14}\varepsilon ^{\frac{2}{2^{*}(s)-2}}-\int _{\Omega }F(x,t_{\varepsilon }v_{\varepsilon })\hbox {d}x<0, \end{aligned}$$

for $\varepsilon >0$ small enough. To this purpose, we prove

$$\begin{aligned} \displaystyle \lim _{\varepsilon \rightarrow 0^{+}}\varepsilon ^{-\frac{2}{2^{*}(s)-2}}\int _{\Omega }F(x,t_{\varepsilon }v_{\varepsilon })\hbox {d}x=+\infty . \end{aligned}$$

(2.13)

In fact, if there exists m(t) such that $f(x,t)\ge m(t)>0$, combining with the definition of $v_{\varepsilon }$, (2.6) and the boundedness of $t_{\varepsilon }$, we only need to verify

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0^{+}}\varepsilon ^{-\frac{2}{2^{*}(s)-2}}\int _{|x|<R}M\left( \frac{C_{\varepsilon }\varepsilon ^{\frac{1}{2^{*}(s)-2}}|y|^{T}}{\big ((\varepsilon +|y|)^{2} +|z|^{2}\big )^{\frac{1}{2^{*}(s)-2}}}\right) \hbox {d}x=+\infty , \end{aligned}$$

(2.14)

where $B_R(0,z^0)\subset \Omega$, without loss of generality, we assume $R>2$ and $\displaystyle M(t)=\int _{0}^{t}m(s)\hbox {d}s$ is the primitive function of m(t) and

$$\begin{aligned} C_{\varepsilon }=t_{\varepsilon }\left( \int _{\Omega }\frac{u_{\varepsilon }^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\right) ^{-\frac{1}{2^{*}(s)}} \end{aligned}$$

By (2.6), (2.10), and (2.11), we obtain that $C_{\varepsilon }$ is bounded. Through computation, one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle &{}&{}\varepsilon ^{-\frac{2}{2^{*}(s)-2}}\displaystyle \int _{|x|<R}M\left( \frac{C_{\varepsilon }\varepsilon ^{\frac{1}{2^{*}(s)-2}}|y|^{T}}{\big ((\varepsilon +|y|)^{2} +|z|^{2}\big )^{\frac{1}{2^{*}(s)-2}}}\right) \hbox {d}x\\ &{}&{}\quad \ge \displaystyle \varepsilon ^{-\frac{2}{2^{*}(s)-2}}\int ^{\frac{R}{2}}_{0}\int ^{\frac{R}{2}}_{0}M\left( \frac{C_{\varepsilon }\varepsilon ^{\frac{1}{2^{*}(s)-2}}r^{T}}{\big ((\varepsilon +r)^{2} +s^{2}\big )^{\frac{1}{2^{*}(s)-2}}}\right) r^{k-1}s^{N-k-1}\hbox {d}r\hbox {d}s\\ &{}&{}\quad =\displaystyle \varepsilon ^{N-\frac{2}{2^{*}(s)-2}}\int ^{\frac{R}{2\varepsilon }}_{0}\int ^{\frac{R}{2\varepsilon }}_{0}M\left( \frac{C_{\varepsilon }\varepsilon ^{T-\frac{1}{2^{*}(s)-2}}r^{T}}{\big ((1+r)^{2} +s^{2}\big )^{\frac{1}{2^{*}(s)-2}}}\right) r^{k-1}s^{N-k-1}\hbox {d}r\hbox {d}s\\ &{}&{}\quad =\displaystyle \varepsilon ^{N-\frac{2}{2^{*}(s)-2}}\int ^{\frac{R}{2\varepsilon }}_{0}r^{k-1}(1+r)^{N-k}\hbox {d}r\\ &{}&{}\qquad \displaystyle \int ^{\frac{R}{2\varepsilon (1+r)}}_{0}M\left( \frac{C_{\varepsilon }\varepsilon ^{T-\frac{1}{2^{*}(s)-2}}r^{T}}{\big (1+r)^{\frac{2}{2^{*}(s)-2}} (1+\rho ^{2})^{\frac{1}{2^{*}(s)-2}}}\right) \rho ^{N-k-1}\hbox {d}\rho . \end{array}\right. \end{aligned}$$

Thus, we can deduce that for $R'=\frac{R}{2}>1$, (2.14) is equivalent to

$$\begin{aligned}&\frac{\varepsilon ^{N}}{\varepsilon ^{\frac{2}{2^{*}(s)-2}}}\int ^{\frac{R'}{\varepsilon }}_{0}r^{k-1}(1+r)^{N-k}\hbox {d}r\nonumber \\&\quad \int ^{\frac{R'}{\varepsilon (1+r)}}_{0}M\left( \frac{C_{\varepsilon }\varepsilon ^{T-\frac{1}{2^{*}(s)-2}}r^{T}}{\big (1+r)^{\frac{2}{2^{*}(s)-2}} (1+\rho ^{2})^{\frac{1}{2^{*}(s)-2}}}\right) \rho ^{N-k-1}\hbox {d}\rho \rightarrow \displaystyle +\infty , \end{aligned}$$

(2.15)

as $\varepsilon \rightarrow 0^{+}$. Then, if we prove that

$$\begin{aligned}&\frac{\varepsilon ^{N}}{\varepsilon ^{\frac{2}{2^{*}(s)-2}}}\int ^{\frac{1}{\varepsilon }}_{0}r^{k-1}(1+r)^{N-k}\hbox {d}r\nonumber \\&\quad \int ^{\frac{1}{\varepsilon (1+r)}}_{0}M\left( \frac{C_{\varepsilon }\varepsilon ^{T-\frac{1}{2^{*}(s)-2}}r^{T}}{\big (1+r)^{\frac{2}{2^{*}(s)-2}} (1+\rho ^{2})^{\frac{1}{2^{*}(s)-2}}}\right) \rho ^{N-k-1}\hbox {d}\rho \rightarrow \displaystyle +\infty . \end{aligned}$$

(2.16)

as $\varepsilon \rightarrow 0^{+}$, then it is easy to check that (2.15) is established.

Last, we will prove that (2.16) holds under $(f_{3})$. By $(f_{3})$, one gets

$$\begin{aligned} f(x,t)\ge \sigma \chi _{I}(t)\triangleq m(t), \end{aligned}$$

for almost everywhere $x\in \omega$ and for all $t\ge 0$, where $\chi _{I}$ is the characteristic function of $I(I\subset (0,+\infty ))$. Thus, for some constants $\eta >0$ and $B>0$, it follows that

$$\begin{aligned} M(t)\ge \eta >0, \end{aligned}$$

for all $t\ge B$. Then, we obtain

$$\begin{aligned} M\left( \frac{C_{\varepsilon }\varepsilon ^{T-\frac{1}{2^{*}(s)-2}}r^{T}}{\big (1+r)^{\frac{2}{2^{*}(s)-2}}(1+\rho ^{2}) ^{\frac{1}{2^{*}(s)-2}}}\right) \ge \eta , \end{aligned}$$

for all $\rho$ satisfying $\displaystyle \rho \le C'_{\varepsilon }\frac{r^\frac{T(2^{*}(s)-2)}{2}}{1+r}\varepsilon ^{\frac{T(2^{*}(s)-2)-1}{2}}$, where $0<r<\varepsilon ^{-1}$ and $C'_{\varepsilon }$ is bounded and related to B and $C_{\varepsilon }$. Then, it leads to

$$\begin{aligned} \left. \begin{array}{rcl} &{}&{}\displaystyle \varepsilon ^{N-\frac{2}{2^{*}(s)-2}}\int ^{\frac{1}{\varepsilon }}_{0}r^{k-1}(1+r)^{N-k}\hbox {d}r\int ^{\frac{1}{\varepsilon (1+r)}}_{0}M\left( \frac{C_{\varepsilon }\varepsilon ^{T-\frac{1}{2^{*}(s)-2}}r^{T}}{\big (1+r)^{\frac{2}{2^{*}(s)-2}} (1+\rho ^{2})^{\frac{1}{2^{*}(s)-2}}}\right) \rho ^{N-k-1}\hbox {d}\rho \\ &{}&{}\ge \displaystyle \eta \varepsilon ^{N-\frac{2}{2^{*}(s)-2}}\int ^{\frac{1}{\varepsilon }}_{0}r^{k-1}(1+r)^{N-k}\hbox {d}r\int ^{C'_{\varepsilon }\frac{r^\frac{T(2^{*}(s)-2)}{2}}{1+r}\varepsilon ^{\frac{T(2^{*}(s)-2)-1}{2}}}_{0} \rho ^{N-k-1}\hbox {d}\rho \\ &{}&{}\ge C\displaystyle \eta \varepsilon ^{N-\frac{2}{2^{*}(s)-2}+\frac{T(2^{*}(s)-2)-1}{2}(N-k)}\int ^{\frac{1}{\varepsilon }}_{0}r^{k-1+\frac{T(2^{*}(s)-2)(N-k)}{2}}\hbox {d}r\\ &{}&{}=\displaystyle C_{18}\varepsilon ^{\frac{N-k}{2}-\frac{2}{2^{*}(s)-2}}, \end{array}\right. \end{aligned}$$

for $\displaystyle N\ge 2k-2-\sqrt{(k-2)^{2}-4\mu }$ and $\displaystyle s=2-\frac{N-2}{N-k+\sqrt{(k-2)^{2}-4\mu }}$; then, one has $\frac{N-k}{2}-\frac{2}{2^{*}(s)-2}<0$. Therefore, (2.16) holds. Then, we complete the proof of Lemma 2.2. $\square$

Proof of Theorem 1

From the continuity of embeddings

$$\begin{aligned} \displaystyle H_{0}^{1}(\Omega )\hookrightarrow L^{q}(\Omega )(1\le q\le 2^{*})~\mathrm {and}~ H_{0}^{1}(\Omega )\hookrightarrow L^{2^{*}(s)}(\Omega ,|y|^{-s}\hbox {d}x), \end{aligned}$$

there exist $C_{19}$, $C_{20}>0$ such that

$$\begin{aligned} \displaystyle \int _{\Omega }|u|^{q}\hbox {d}x\le C_{19}\Vert u\Vert ^{q},~~ \int _{\Omega }\frac{|u|^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\le C_{20}\Vert u\Vert ^{2^{*}(s)}. \end{aligned}$$

(2.17)

It follows from $(f_{1})$ and $(f_{2})$ that

$$\begin{aligned} |F(x,t)|\le \frac{1}{2}\lambda |t|^{2}+C_{21}|t|^{2^{*}(s)}, \end{aligned}$$

(2.18)

for all $t\in \mathbb {R}^{+}$ and $x\in \overline{\Omega }$. By (2.17) and (2.18), one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle I(u)&{}=&{}\displaystyle \frac{1}{2}\Vert u\Vert ^{2}-\frac{1}{2^{*}(s)}\int _{\Omega }\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }F(x,u^{+})\hbox {d}x\\ &{}\ge &{}\displaystyle \frac{1}{2}\Vert u\Vert ^{2}-\frac{1}{2^{*}(s)}\int _{\Omega }\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\frac{\lambda }{2}\int _{\Omega }|u|^{2}\hbox {d}x -C_{22}\int _{\Omega }|u|^{2^{*}(s)}\hbox {d}x\\ &{}\ge &{}\displaystyle \frac{1}{2}\Vert u\Vert ^{2}-\frac{C_{23}}{2^{*}(s)}\Vert u\Vert ^{2^{*}(s)}-\frac{\lambda }{2\lambda _{1}}\Vert u\Vert ^{2}-C_{24}\Vert u\Vert ^{2^{*}(s)}, \end{array}\right. \end{aligned}$$

for $0<\lambda <\lambda _{1}$; therefore, there exists $\alpha >0$ such that $I(u)\ge \alpha >0$ for all $\Vert u\Vert =r$, where $r>0$ small enough. For any $u\in H^{1}_{0}(\Omega )$ with $u^{+}\not \equiv 0$, together with the nonnegativity of F(x, t), one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle I(tu)&{}=&{}\displaystyle \frac{1}{2}t^{2}\Vert u\Vert ^{2}-\frac{1}{2^{*}(s)}t^{2^{*}(s)}\int _{\Omega }\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }F(x,tu^{+})\hbox {d}x\\ &{}\le &{}\displaystyle \frac{1}{2}t^{2}\Vert u\Vert ^{2}-\frac{1}{2^{*}(s)}t^{2^{*}(s)}\int _{\Omega }\frac{(u^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x, \end{array}\right. \end{aligned}$$

then $\displaystyle \lim _{t\rightarrow +\infty }I(tu)\rightarrow -\infty$. Thus, we can find $t'>0$ such that $I(t'u)<0$ when $\Vert t'u\Vert >r$. According to the Mountain Pass Lemma (see [25]), there exists a sequence $\{u_{n}\}\subset H^{1}_{0}(\Omega )$, such that as $n\rightarrow \infty$,

$$\begin{aligned} \left. \begin{array}{rcl} I(u_{n})\rightarrow c>\alpha ~~\mathrm {and}~~~(1+\Vert u_{n}\Vert )\Vert I'(u_{n})\Vert \rightarrow 0~\mathrm {in}~\big (H^{1}_{0}(\Omega )\big )', \end{array}\right. \end{aligned}$$

where

$$\begin{aligned}&c=\inf _{\gamma \in \Gamma }\max _{t\in [0,1]}I(\gamma (t)),\\&\Gamma =\{\gamma \in C([0,1], H^{1}_{0}(\Omega ))| \gamma (0)=0, \gamma (1)=t'u\}. \end{aligned}$$

It is easy to obtain that $(f_{2})$ leads to $(f_{5})$, then Lemma 2.2 holds if we replace $(f_{5})$ by $(f_{2})$. By the definition of c and Lemma 2.2, we obtain

$$\begin{aligned} 0<\alpha<c=\inf _{\gamma \in \Gamma }\max _{t\in [0,1]}I(\gamma (t))\le \max _{t\in [0,1]}I(tt'u)\le \sup _{t\ge 0}I(tu) <\frac{2-s}{2(N-s)}S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}. \end{aligned}$$

First, we claim that $\{u_{n}\}$ is bounded in $H_{0}^{1}(\Omega )$. Indeed, by $(f_{2})$, for any $\varepsilon >0$, there exists $M>0$, such that

$$\begin{aligned}&|F(x,t)|\le \varepsilon |t|^{2^{*}(s)},~x\in \Omega ,~t\ge M;~~|F(x,t)|\le C_{1}(\varepsilon ),~t\in (0,M];\\&|f(x,t)t|\le \varepsilon |t|^{2^{*}(s)},~x\in \Omega ,~t\ge M;~~|f(x,t)t|\le C_{2}(\varepsilon ),~t\in (0,M]. \end{aligned}$$

Thus, we have

$$\begin{aligned} |F(x,t)|\le C_{1}(\varepsilon )+\varepsilon |t|^{2^{*}(s)},~~|f(x,t)t|\le C_{2}(\varepsilon )+\varepsilon |t|^{2^{*}(s)}, \end{aligned}$$

(2.19)

for any $(x,t)\in \overline{\Omega }\times \mathbb {R}^{+}$. Then, for $\xi \in (2,2^{*}(s))$, one has

$$\begin{aligned} F(x,t)-\frac{1}{2}f(x,t)t\le F(x,t)-\frac{1}{\xi }f(x,t)t\le C_{3}(\varepsilon )+\varepsilon |t|^{2^{*}(s)}, \end{aligned}$$

(2.20)

for any $(x,t)\in \overline{\Omega }\times \mathbb {R}^{+}$. Set $l(x,t)\triangleq |y|^{-s}|t|^{2^{*}(s)-1}+f(x,t)$, we claim that l(x, t) satisfies the (AR) condition. By (2.20), one easily gets

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \xi L(x,t)-l(x,t)t&{}=&{}\displaystyle \left( \frac{\xi }{2^{*}(s)}-1\right) |y|^{-s}|t|^{2^{*}(s)}+\big (\xi F(x,t)-f(x,t)t\big )\\ &{}\le &{}\displaystyle \left( \frac{\xi }{2^{*}(s)}-1\right) |y|^{-s}|t|^{2^{*}(s)}+\xi C_{4}(\varepsilon )+\xi \varepsilon |t|^{2^{*}(s)}\\ &{}=&{}\displaystyle \bigg (\left( \frac{\xi }{2^{*}(s)}-1 \right) |y|^{-s}+\xi \varepsilon \bigg )|t|^{2^{*}(s)}+\xi C_{4}(\varepsilon ), \end{array}\right. \end{aligned}$$

so for $\varepsilon >0$ sufficiently small, there exists $M'>0$, such that

$$\begin{aligned} 0\le \xi L(x,t)\le l(x,t)t,~~x\in \overline{\Omega }{\setminus }\{(0,z^{0})\},~t\ge M', \end{aligned}$$

where $\displaystyle L(x,t)=\int ^{t}_{0}l(x,s)\hbox {d}s$. Moreover, by $(f_{2})$, we obtain

$$\begin{aligned} L(x,t)-\frac{1}{\xi }l(x,t)t\le \max _{{x\in \overline{\Omega }{\setminus }\{(0,z^{0})\}},0\le t\le M'}\bigg (F(x,t)-\frac{1}{\xi }f(x,t)t\bigg )\triangleq M_{0}. \end{aligned}$$

It follows from the inequalities above that

$$\begin{aligned} L(x,t)-\frac{1}{\xi }l(x,t)t\le M_{0}, \quad \mathrm {for}~\mathrm {all} \quad x\in \overline{\Omega }{\setminus }\{(0,z^{0})\},~t\ge 0. \end{aligned}$$

(2.21)

Then, one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle c+1+o(1)&{}\ge &{}\displaystyle I(u_{n})-\frac{1}{\xi }\langle I'(u_{n}),u_{n} \rangle \\ &{}=&{}\displaystyle \left( \frac{1}{2}-\frac{1}{\xi }\right) \Vert u_{n}\Vert ^{2}+\left( \frac{1}{\xi }-\frac{1}{2^{*}(s)}\right) \int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\\ &{}&{}-\displaystyle \int _{\Omega }\left( F(x,u_{n}^{+})-\frac{1}{\xi }f(x,u_{n}^{+})u_{n}^{+}\right) \hbox {d}x\\ &{}\ge &{}\displaystyle \left( \frac{1}{2}-\frac{1}{\xi }\right) \Vert u_{n}\Vert ^{2}-\int _{\Omega }\bigg (L(x,u_{n}^{+})-\frac{1}{\xi }l(x,u_{n}^{+})u_{n}^{+}\bigg )\hbox {d}x\\ &{}\ge &{}\displaystyle \left( \frac{1}{2}-\frac{1}{\xi }\right) \Vert u_{n}\Vert ^{2}-M_{0}|\Omega |. \end{array}\right. \end{aligned}$$

Thus, $\{u_{n}\}$ is bounded. Due to the continuity of embedding $\displaystyle H_{0}^{1}(\Omega )\hookrightarrow L^{2^{*}(s)}(\Omega )$, we have $\displaystyle \int _{\Omega }|u_{n}|^{2^{*}(s)}\hbox {d}x\le C<\infty$. Up to a subsequence, still denoted by $\{u_{n}\}$, there exists $u_{0}\in H^{1}_{0}(\Omega )$ satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} u_n\rightharpoonup u_{0}, &{}\quad \mathrm {weakly~in}~H_{0}^{1}(\Omega ),\\ u_n\rightarrow u_{0},&{} \quad \mathrm {strongly~in}~L^{p}(\Omega ),~1\le p<2^{*},\\ u_n(x)\rightarrow u_{0}(x), &{}\quad \mathrm {a.e.~in}~\Omega ,\\ u_{n}^{2^{*}(s)-1}\rightharpoonup u_{0}^{2^{*}(s)-1}, &{}\quad \mathrm {weakly~in}~\left( L^{2^{*}(s)}(\Omega ,|y|^{-s}\hbox {d}x)\right) ', \end{array}\right. } \end{aligned}$$

(2.22)

as $n\rightarrow \infty$. By $(f_{2})$, for any $\varepsilon >0$, there exists $a(\varepsilon )>0$ such that

$$\begin{aligned} |F(x,t)|\le \frac{1}{2C_{24}}\varepsilon |t|^{2^{*}(s)}+a(\varepsilon )~~~\mathrm {for}~(x,t)\in \Omega \times \mathbb {R}^{+}. \end{aligned}$$

Set $\delta =\frac{\varepsilon }{2a(\varepsilon )}>0$. When $E\subset \Omega$, meas $E<\delta$, one gets

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \left| \int _{E}F(x,u_{n}^{+})\hbox {d}x\right| &{}\le &{}\displaystyle \int _{E}|F(x,u_{n}^{+})|\hbox {d}x\\ &{}\le &{}\displaystyle \int _{E}a(\varepsilon )\hbox {d}x+\frac{1}{2C_{24}}\varepsilon \int _{E}|u_{n}|^{2^{*}(s)}\hbox {d}x\\ &{}\le &{}\displaystyle a(\varepsilon ){ meas} E+\displaystyle \frac{1}{2C_{24}}\varepsilon C_{24}\\ &{}\le &{}\displaystyle \varepsilon . \end{array}\right. \end{aligned}$$

Hence, $\displaystyle \bigg \{\int _{\Omega }F(x,u_{n}^{+})\hbox {d}x, n\in N\bigg \}$ is equi-absolutely continuous. It follows from the Vitali Convergence Theorem that

$$\begin{aligned} \int _{\Omega }F(x,u_{n}^{+})\hbox {d}x\rightarrow \int _{\Omega }F(x,u_{0}^{+})\hbox {d}x, \end{aligned}$$

(2.23)

as $n\rightarrow \infty$. Applying the same method, one has

$$\begin{aligned} \int _{\Omega }f(x,u_{n}^{+})u_{n}\hbox {d}x\rightarrow \int _{\Omega }f(x,u_{0}^{+})u_{0}\hbox {d}x \end{aligned}$$

(2.24)

By (2.22) and (2.24), we have

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \lim _{n\rightarrow +\infty }\langle I'(u_{n}),v\rangle &{}=&{}\displaystyle \int _{\Omega }\left( (\nabla u_{0},\nabla v)-\mu \frac{u_{0}v}{|y|^{2}}\right) \hbox {d}x-\int _{\Omega }\frac{(u_{0}^{+})^{2^{*}(s)-1}v}{|y|^{s}}\hbox {d}x\\ &{}&{}-\displaystyle \int _{\Omega }f(x,u_{0}^{+})v\hbox {d}x=0, \end{array}\right. \end{aligned}$$

(2.25)

for all $v\in H^{1}_{0}(\Omega )$. Thus, $u_{0}$ is a critical point of I, that is, $u_{0}$ is a solution of problem (1.1). Now we verify that $u_{0}\not \equiv 0$. Let $v=u_{0}$ in (2.25), we get

$$\begin{aligned} \Vert u_{0}\Vert ^{2}-\int _{\Omega }\frac{(u_{0}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }f(x,u_{0}^{+})u_{0}\hbox {d}x=0. \end{aligned}$$

(2.26)

Set $w_{n}=u_{n}-u_{0}$, then we have

$$\begin{aligned} \int _{\Omega }|\nabla u_{n}|^{2}\hbox {d}x=\int _{\Omega }|\nabla u_{0}|^{2}\hbox {d}x+\int _{\Omega }|\nabla w_{n}|^{2}\hbox {d}x+o(1). \end{aligned}$$

(2.27)

From Br$\acute{\mathrm {e}}$zis–Lieb’s lemma (see [3]), it follows that

$$\begin{aligned}&\int _{\Omega }\frac{u_{n}^{2}}{|y|^{2}}\hbox {d}x=\int _{\Omega }\frac{u_{0}^{2}}{|y|^{2}}\hbox {d}x+\int _{\Omega }\frac{w_{n}^{2}}{|y|^{2}}\hbox {d}x+o(1). \end{aligned}$$

(2.28)

$$\begin{aligned}&\int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x=\int _{\Omega }\frac{(u_{0}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x +\int _{\Omega }\frac{(w_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+o(1). \end{aligned}$$

(2.29)

By (2.23) and (2.27)–(2.29), one has

$$\begin{aligned} I(u_{n})=I(u_{0})+\frac{1}{2}\Vert w_{n}\Vert ^{2}-\frac{1}{2^{*}(s)}\int _{\Omega }\frac{(w_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x=c+o(1). \end{aligned}$$

(2.30)

Since $\langle I'(u_{n}),u_{n}\rangle =o(1)$, combining with (2.23), (2.26), one has

$$\begin{aligned} \Vert w_{n}\Vert ^{2}-\int _{\Omega }\frac{(w_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x=o(1). \end{aligned}$$

We may assume that as $n\rightarrow \infty$,

$$\begin{aligned} \Vert w_{n}\Vert ^{2}\rightarrow b, ~~~~~~~ ~~~~~\int _{\Omega }\frac{(w_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\rightarrow b. \end{aligned}$$

Clearly, $b\ge 0$. We now suppose that $u_{0}\equiv 0$. If $b=0$, then from (2.30), $c=I(0)=0$, which contradicts with $c>0$. If $b\ne 0$, we have from the definition of $S_{\mu }$ that

$$\begin{aligned} \Vert w_{n}\Vert ^{2}=\int _{\Omega }\left( |\nabla w_{n}|^{2}-\mu \frac{w_{n}^{2}}{|y|^{2}}\right) \hbox {d}x\ge S_{\mu }\left( \int _{\Omega }\frac{(w_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\right) ^{\frac{2}{2^{*}(s)}}, \end{aligned}$$

and $b\ge S_{\mu }b^{\frac{2}{2^{*}(s)}}$, together with (2.30), we deduce

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle c+o(1)&{}=&{}\displaystyle I(u_{0})+\frac{1}{2}\Vert w_{n}\Vert ^{2}-\frac{1}{2^{*}(s)}\int _{\Omega }\frac{(w_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+o(1)\\ &{}=&{}\displaystyle I(u_{0})+o(1)+\left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) b\\ &{}\ge &{}\displaystyle \frac{2-s}{2(N-s)}S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}+o(1), \end{array}\right. \end{aligned}$$

which contradicts $\displaystyle c<\frac{2-s}{2(N-s)}S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}$. Therefore, $u_{0}\not \equiv 0$ and $u_{0}$ is a nontrivial solution of problem (2.1). Then, by $\langle I'(u_{0}),u_{0}^{-}\rangle =0$ where $\displaystyle u_{0}^{-}=\min \{u_{0},0\}$, one has $\Vert u_{0}^{-}\Vert =0$, which implies that $u_{0}\ge 0$. From (2.25), we get $\displaystyle \int _{\Omega }(\nabla u_{0},\nabla v)\hbox {d}x\ge 0$ for any $v\in H^{1}_{0}(\Omega )$, which means $\displaystyle -\triangle u_{0}\ge 0$ in $\Omega$. By the strong maximum principle, we know $u_{0}$ is a positive solution of problem (1.1). Therefore, Theorem 1 holds. $\square$

Proof of Theorem 2

Obviously, $(f_{4})$ and $(f_{5})$ can ensure that I has a mountain pass geometry and then there exists a $(Ce)_{c}$ sequence $\{u_{n}\}$, that is,

$$\begin{aligned} I(u_{n})\rightarrow c~~\mathrm {and}~~~(1+\Vert u_{n}\Vert )\Vert I'(u_{n})\Vert \rightarrow 0~~~\mathrm {as}~ n\rightarrow \infty . \end{aligned}$$

We claim that $\{u_{n}\}$ is bounded. In fact, there exists $n_{0}>0$, such that for $n\ge n_{0}$, one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle c+1 &{}\ge &{}\displaystyle I(u_{n})-\frac{1}{2}\langle I'(u_{n}),u_{n} \rangle \\ &{}=&{}\displaystyle \left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\displaystyle \int _{\Omega }\widetilde{F}(x,u_{n}^{+})\hbox {d}x\\ &{}\ge &{}\displaystyle \int _{\Omega }\widetilde{F}(x,u_{n}^{+})\hbox {d}x. \end{array}\right. \end{aligned}$$

(2.31)

Set

$$\begin{aligned} g(r):=\inf \bigg \{\widetilde{F}(x,u): x\in \overline{\Omega },~|u|\ge r\bigg \}. \end{aligned}$$

By Remark 2, one deduces $g(r)\rightarrow +\infty$ as $r\rightarrow +\infty$. For $0\le a<b$, let

$$\begin{aligned} \Omega _{n}(a,b):=\bigg \{x\in \overline{\Omega },~a\le u_{n}^{+}(x)<b\bigg \} \end{aligned}$$

and

$$\begin{aligned} C_{a}^{b}:=\inf \bigg \{\frac{\widetilde{F}(x,u)}{u^{2}}:x\in \overline{\Omega },~a\le |u(x)|<b\bigg \}; \end{aligned}$$

thus for all $x\in \Omega _{n}(a,b)$, one obtains

$$\begin{aligned} \widetilde{F}\big (x,u_{n}^{+}(x)\big )\ge C_{a}^{b}(u_{n}^{+}(x))^{2}. \end{aligned}$$

It follows from (2.31) that

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle c+1&{}\ge &{}\displaystyle \int _{\Omega _{n}(0,b)}\widetilde{F}(x,u_{n}^{+})\hbox {d}x +\int _{\Omega _{n}(b,+\infty )}\widetilde{F}(x,u_{n}^{+})\hbox {d}x\\ &{}\ge &{}\displaystyle C_{0}^{b}\int _{\Omega _{n}(a,b)}|u_{n}^{+}|^{2}\hbox {d}x+g(b)|\Omega _{n}(b,+\infty )|. \end{array}\right. \end{aligned}$$

(2.32)

Arguing directly, assume $\Vert u_{n}\Vert \rightarrow +\infty$. Set $\displaystyle v_{n}=\frac{u_{n}}{\Vert u_{n}\Vert }$, then $\Vert v_{n}\Vert =1$ and $\displaystyle \int _{\Omega }|v_{n}|^{s}\hbox {d}x\le M_{s}$ for all $s\in [2,2^{*}]$. Using (2.32),

$$\begin{aligned} |\Omega _{n}(b,+\infty )|\le \frac{c+1}{g(b)}\rightarrow 0, \end{aligned}$$

(2.33)

uniformly in $n\ge n_{0}$ as $b\rightarrow +\infty$. And for any fixed $0\le a<b$,

$$\begin{aligned} \int _{\Omega _{n}(a,b)}(v_{n}^{+})^{2}\hbox {d}x=\frac{1}{\Vert u_{n}\Vert ^{2}}\int _{\Omega _{n}(a,b)}(u_{n}^{+})^{2}\hbox {d}x\le \frac{C}{\Vert u_{n}\Vert ^{2}}\rightarrow 0, \end{aligned}$$

(2.34)

as $n\rightarrow \infty$. It follows from (2.33) and the Hölder inequality that for $r\in [2,2^{*}]$, $p\in (r,2^{*})$, and a suitable constant $C_{*}$,

$$\begin{aligned} \int _{\Omega _{n}(b,+\infty )}(v_{n}^{+})^{r}\hbox {d}x\le C_{*}|\Omega _{n}(b,\infty )|^{\frac{p-r}{p}}\rightarrow 0, \end{aligned}$$

(2.35)

uniformly in $n\ge n_{0}$ as $b\rightarrow +\infty$. Let $\varepsilon >0$, by $(f_{4})$, there exists $a_{\varepsilon }>0$ such that $\displaystyle |f(x,u)|\le \frac{\varepsilon }{M_{2}}|u|$ for all $|u|<a_{\varepsilon }$. Consequently,

$$\begin{aligned} \int _{\Omega _{n}(0,a_{\varepsilon })}\frac{f(x,u_{n}^{+})}{u_{n}^{+}}(v_{n}^{+})^{2}\hbox {d}x\le \int _{\Omega _{n}(0,a_{\varepsilon })}\frac{\varepsilon }{M_{2}}(v_{n}^{+})^{2}\hbox {d}x\le \varepsilon , \end{aligned}$$

(2.36)

for all n. Combining $(f_{7})$, (2.31) with (2.35), we can take $b_{\varepsilon }$ so large that

$$\begin{aligned} \left. \begin{array}{rcl} &{}&{}\displaystyle \int _{\Omega _{n}(b_{\varepsilon },+\infty )}\frac{f(x,u_{n}^{+})}{u_{n}^{+}}(v_{n}^{+})^{2}\hbox {d}x\\ &{}&{}\quad \le \displaystyle \bigg (\int _{\Omega _{n}(b_{\varepsilon },+\infty )}\bigg |\frac{f(x,u_{n}^{+})}{u_{n}^{+}}\bigg |^{\tau }\hbox {d}x\bigg )^{\frac{1}{\tau }} \bigg (\int _{\Omega _{n}(b_{\varepsilon },+\infty )}(v_{n}^{+})^{\alpha }\hbox {d}x\bigg )^{\frac{2}{\alpha }}\\ &{}&{}\quad \le \displaystyle \bigg (\int _{\Omega _{n}(b_{\varepsilon },+\infty )}a_{1}\tilde{F}(x,u_{n}^{+})\hbox {d}x\bigg )^{\frac{1}{\tau }} \bigg (\int _{\Omega _{n}(b_{\varepsilon },+\infty )}(v_{n}^{+})^{\alpha }\hbox {d}x\bigg )^{\frac{2}{\alpha }}\\ &{}&{}\quad \le \displaystyle \varepsilon , \end{array}\right. \end{aligned}$$

(2.37)

for all $n\ge n_{0}$, where $\displaystyle \alpha =\frac{2\tau }{\tau -1}$. Note that there is $\gamma =\gamma (\varepsilon )>0$ independent of n such that $|f(x,u_{n}^{+})|\le \gamma u_{n}^{+}$ for $x\in \Omega _{n}(a_{\varepsilon },b_{\varepsilon })$. By (2.34), there exists $n_{1}>0$ such that

$$\begin{aligned} \displaystyle \int _{\Omega _{n}(a_{\varepsilon },b_{\varepsilon })}\frac{f(x,u_{n}^{+})}{u_{n}^{+}}(v_{n}^{+})^{2}\hbox {d}x\le \gamma \int _{\Omega _{n}(a_{\varepsilon },b_{\varepsilon })}(v_{n}^{+})^{2}\hbox {d}x\le \varepsilon , \end{aligned}$$

(2.38)

for all $n\ge n_{1}$. Now (2.36), (2.37), and (2.38) imply that

$$\begin{aligned} \int _{\Omega }\frac{f(x,u_{n}^{+})u_{n}^{+}}{\Vert u_{n}\Vert ^{2}}\hbox {d}x=\int _{\Omega }\frac{f(x,u_{n}^{+})}{u_{n}^{+}}(v_{n}^{+})^{2}\hbox {d}x<3\varepsilon , \end{aligned}$$

(2.39)

for $n\ge \max \{n_{0}, n_{1}\}$. Since $\Vert v_{n}\Vert =1$, passing to a subsequence, there exists $v\in H^{1}_{0}(\Omega )$, such that

$$\begin{aligned} {\left\{ \begin{array}{ll} v_n\rightharpoonup v, &{} \quad \mathrm {weakly~in}~H_{0}^{1}(\Omega ),\\ v_n\rightarrow v,&{} \quad \mathrm {strongly~in}~L^{p}(\Omega ),~1\le p<2^{*},\\ v_n(x)\rightarrow v(x),&{} \quad \mathrm {a.e.~in}~\Omega . \end{array}\right. } \end{aligned}$$

Set $\Omega '=\{x\in \overline{\Omega }:v^{+}(x)\ne 0\}$, if meas $\Omega '>$0, then $u_{n}^{+}(x)\rightarrow +\infty$ for a.e. $x\in \Omega '$. By $(f_{6})$, it is easy to get that for any $M>0$, there exists $C(M)>0$ such that

$$\begin{aligned} f(x,u_{n}^{+})u_{n}^{+}\ge M(u_{n}^{+})^{2}-C(M) \end{aligned}$$

for all $x\in \overline{\Omega }$ and n large enough. Hence,

$$\begin{aligned} \int _{\Omega }\frac{f(x,u_{n}^{+})u_{n}^{+}}{\Vert u_{n}\Vert ^{2}}\hbox {d}x\ge M\int _{\Omega }(v_{n}^{+})^{2}\hbox {d}x-\frac{C(M)}{\Vert u_{n}\Vert ^{2}}, \end{aligned}$$

then

$$\begin{aligned} 0=\lim _{n\rightarrow \infty }\int _{\Omega }\frac{f(x,u_{n}^{+})u_{n}^{+}}{\Vert u_{n}\Vert ^{2}}\hbox {d}x\ge M\int _{\Omega }(v^{+})^{2}\hbox {d}x>0, \end{aligned}$$

which is a contradiction. Hence, meas $\Omega '$=0. Therefore, $v^{+}(x)=0$ a.e. $x\in \overline{\Omega }$. By Remark 2, for any $m>0$, there exists $L_0>0$ such that

$$\begin{aligned} tf(x,t)-2F(x,t)\ge m>0, \end{aligned}$$

for $t>L_0$. It follows from $(f_{4})$ and $(f_{5})$ that $\displaystyle |F(x,t)|\le C_{25}(t^{2}+t^{2^{*}})$ for all $(x,t)\in \overline{\Omega }\times \mathbb {R}^{+}$. Hence, we have, for $x\in \overline{\Omega }$ and $|t|\le L_0$,

$$\begin{aligned} |tf(x,t)-2F(x,t)|\le C_{26}t^{2}, \end{aligned}$$

where $C_{26}=2C_{25}(1+L_0^{2^{*}-2})$. The two inequalities above show that

$$\begin{aligned} tf(x,t)-2F(x,t)\ge -C_{27}t^{2}, \end{aligned}$$

(2.40)

for all $(x,t)\in \overline{\Omega }\times \mathbb {R}^{+}$. From $(1+\Vert u_{n}\Vert )\Vert I'(u_{n})\Vert \rightarrow 0$, one has $\langle I'(u_{n}), u_{n}\rangle \rightarrow 0$, that is,

$$\begin{aligned} \Vert u_{n}\Vert ^{2}-\int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }f(x,u_{n}^{+})u_{n}^{+}\hbox {d}x=o(1), \end{aligned}$$

and combining with (2.31), one deduces

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle c+1+o(1) &{}\ge &{}\displaystyle I(u_{n})-\frac{1}{2}\langle I'(u_{n}),u_{n}\rangle \\ &{}=&{}\displaystyle \left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \Vert u_{n}\Vert ^{2}-\left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \int _{\Omega }f(x,u_{n}^{+})u_{n}^{+}\hbox {d}x\\ &{}&{}+\displaystyle \frac{1}{2}\int _{\Omega }\big (f(x,u_{n}^{+})u_{n}^{+}-2F(x,u_{n}^{+})\big )\hbox {d}x. \end{array}\right. \end{aligned}$$

Consequently, together with (2.39) and (2.40), we deduce

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \frac{1}{\Vert u_{n}\Vert ^{2}}\bigg (c+1+o(1)\bigg )&{}\ge &{}\displaystyle \left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) -\frac{1}{{\Vert u_{n}\Vert ^{2}}}\left( \frac{1}{2} -\frac{1}{2^{*}(s)}\right) \int _{\Omega }f(x,u_{n}^{+})u_{n}^{+}\hbox {d}x\\ &{}&{}+\displaystyle \frac{1}{2\Vert u_{n}\Vert ^{2}}\int _{\Omega }\big (f(x,u_{n}^{+})u_{n}^{+}-2F(x,u_{n}^{+})\big )\hbox {d}x\\ &{}\ge &{}\displaystyle \left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) -3\left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \varepsilon -C_{27}\int _{\Omega }(v_{n}^{+})^{2}\hbox {d}x, \end{array}\right. \end{aligned}$$

which implies $\displaystyle 0\ge \frac{1}{2}-\frac{1}{2^{*}(s)}$ as $n\rightarrow \infty$, a contradiction. Thus, $\{u_{n}\}$ is bounded.

It is obvious that $(f_{4})$ leads to $(f_{1})$ and then Lemma 2.2 is also true if we replace $(f_{1})$ by $(f_{4})$. Thanks to $(f_{4})$, $(f_{5})$, and Lemma 2.2, similar to the proof of Theorem 1, we obtain a positive solution of problem (1.1). $\square$

Proof of Theorem 3

Due to $(f_{1})$, $(f_{5})$, and Lemma 2.2, one obtains that the proof of Theorem 3 is similar to the proof of Theorem 2 and we only need to prove that the $(Ce)_{c}$ sequence is bounded. In fact, let $\{u_{n}\}\subset H_{0}^{1}(\Omega )$ be a $(Ce)_{c}$ sequence, that is,

$$\begin{aligned} \left. \begin{array}{rcl} I(u_{n})\rightarrow c ~~{ and}~~ (1+\Vert u_{n}\Vert )\Vert I'(u_{n})\Vert \rightarrow 0~~~{ as}~ n\rightarrow \infty . \end{array}\right. \end{aligned}$$

(2.41)

Assume that $\{u_{n}\}$ is unbounded, there is a subsequence of $\{u_{n}\}$ (still denoted by $\{u_{n}\}$) satisfying $\Vert u_{n}\Vert \rightarrow +\infty$. Set $\omega _{n}=\frac{u_{n}}{\Vert u_{n}\Vert }$, then $\Vert \omega _{n}\Vert =1$. Then, there exists $\omega \in H_{0}^{1}(\Omega )$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} \omega _n\rightharpoonup \omega ,&{} \quad \mathrm {weakly~in}~H_{0}^{1}(\Omega ),\\ \omega _n\rightarrow \omega ,&{} \quad \mathrm {strongly~in}~L^{p}(\Omega ),~1\le p<2^{*},\\ \omega _n(x)\rightarrow \omega (x),&{} \quad \mathrm {a.e.~in}~\Omega , \end{array}\right. } \end{aligned}$$

(2.42)

as $n\rightarrow \infty$. We claim that $\omega ^{+}=0$. It follows from (2.41) that

$$\begin{aligned} \Vert u_{n}\Vert ^{2}-\int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }f(x,u_{n}^{+})u_{n}^{+}\hbox {d}x=o(1), \end{aligned}$$

which implies

$$\begin{aligned} \int _{\Omega }f(x,u_{n}^{+})u_{n}^{+}\hbox {d}x=\Vert u_{n}\Vert ^{2}-\int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+o(1)\le \Vert u_{n}\Vert ^{2}+o(1). \end{aligned}$$

Then,

$$\begin{aligned} \int _{\Omega }\frac{f(x,u_{n}^{+})u_{n}^{+}}{\Vert u_{n}\Vert ^{2}}\hbox {d}x\le 1+o(1). \end{aligned}$$

(2.43)

For $\displaystyle x\in \Omega ^{+}:=\{x\in \overline{\Omega }: \omega ^{+}(x)>0\}$, $u_{n}^{+}(x)\rightarrow +\infty ~$as$~n\rightarrow \infty$. Combining with $(f_{6})$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{f(x,u_{n}^{+})}{u_{n}^{+}}(\omega ^{+}_{n})^{2}=+\infty ,~~\mathrm {a.e.}~x\in \Omega ^{+}. \end{aligned}$$

(2.44)

If meas $\Omega ^{+}>0$, using the Fatou’s lemma, (2.44) implies that as $n\rightarrow \infty$,

$$\begin{aligned} \int _{\omega >0}\frac{f(x,u_{n}^{+})}{u_{n}^{+}}(\omega _{n}^{+})^{2}\hbox {d}x\rightarrow +\infty . \end{aligned}$$

From (2.43), one has

$$\begin{aligned} 1+o(1)\ge \int _{\Omega }\frac{f(x,u_{n}^{+})u_{n}^{+}}{\Vert u_{n}\Vert ^{2}}\hbox {d}x\ge \int _{\Omega ^{+}}\frac{f(x,u_{n}^{+})}{u_{n}^{+}} (\omega _{n}^{+})^{2}\hbox {d}x\rightarrow +\infty , \end{aligned}$$

which is a contradiction; then, one has meas $\Omega ^{+}=0$, that is, $\omega ^{+}=0$. Set a sequence $\{t_{n}\}$ of real numbers such that $I(t_{n}u_{n}^{+})=\displaystyle \max _{t\in [0,1]}I(tu_{n}^{+})$. Let $v_{n}=\displaystyle S_{\mu }^{\frac{2^{*}(s)}{2(2^{*}(s)-2)}}\omega _{n}$, due to the continuity of embedding $H_{0}^{1}(\Omega )\hookrightarrow L^{2^{*}}(\Omega )$, we have $\displaystyle \int _{\Omega }|v_{n}|^{2^{*}}\hbox {d}x\le C_{23}<\infty$. By $(f_{5})$ and the same method as the proof of Theorem 1, one has

$$\begin{aligned} \int _{\Omega }F(x,v_{n}^{+})\hbox {d}x\rightarrow \int _{\Omega }F(x,0)\hbox {d}x=0, \end{aligned}$$

as $n\rightarrow \infty$. Because $\Vert u_{n}\Vert \rightarrow +\infty$ as $n\rightarrow \infty$, one has $\displaystyle \frac{S_{\mu }^{\frac{2^{*}(s)}{2(2^{*}(s)-2)}}}{\Vert u_{n}\Vert }\in [0,1]$ for n large enough. By the definition of $t_{n}$ and $v_{n}=\displaystyle \frac{S_{\mu }^{\frac{2^{*}(s)}{2(2^{*}(s)-2)}}}{\Vert u_{n}\Vert }u_{n}$, one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle I(t_{n}u_{n})\ge I(v_{n})&{}=&{}\displaystyle \frac{1}{2}\Vert v_{n}\Vert ^{2}-\frac{1}{2^{*}(s)}\int _{\Omega }\frac{(v_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega }F(x,v_{n}^{+})\hbox {d}x\\ &{}\ge &{}\displaystyle \frac{1}{2}\Vert v_{n}\Vert ^{2}-\frac{1}{2^{*}(s)}S_{\mu }^{-\frac{2^{*}(s)}{2}}\Vert v_{n}\Vert ^{2^{*}(s)}-\int _{\Omega }F(x,v^{+}_{n})\hbox {d}x\\ &{}=&{}\displaystyle \left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) S_{\mu }^{\frac{2^{*}(s)}{2^{*}(s)-2}}-\int _{\Omega }F(x,v^{+}_{n})\hbox {d}x, \end{array}\right. \end{aligned}$$

which implies that $I(t_{n}u_{n})\rightarrow +\infty$ as $n\rightarrow \infty$. Noting that $I(0)=0$, $I(u_{n})\rightarrow c$, thus $0<t_{n}<1$ when n is large enough. It follows that

$$\begin{aligned}&\int _{\Omega }\left( |\nabla (t_{n}u_{n})|^{2}-\mu \frac{(t_{n}u_{n})^{2}}{|y|^{2}}\right) \hbox {d}x-\int _{\Omega }\frac{(t_{n}u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\int _{\Omega } f(x,t_{n}u_{n}^{+})t_{n}u_{n}^{+}\hbox {d}x\\&\quad =\langle I'(t_{n}u_{n}),t_{n}u_{n}\rangle =t_{n}\frac{d I(t_{n}u_{n})}{\hbox {d}t}\mid _{t=t_{n}}=0. \end{aligned}$$

By $(f_{8})$, for $0\le t_{n}\le 1$, we have $\theta H(x,u_{n}^{+})\ge H(x,t_{n}u_{n}^{+})-\theta _{0}$, then

$$\begin{aligned} \left. \begin{array}{rcl} &{}&{}\displaystyle \int _{\Omega }\left( \frac{1}{2}f(x,u_{n}^{+})u_{n}^{+}-F(x,u_{n}^{+})\right) \hbox {d}x\\ &{}&{}\quad =\displaystyle \frac{1}{2}\int _{\Omega }H(x,u_{n}^{+})\hbox {d}x\\ &{}&{}\quad \ge \displaystyle \frac{1}{2\theta }\int _{\Omega }\bigg (H(x,t_{n}u_{n}^{+})-\theta _{0}\bigg )\hbox {d}x\\ &{}&{}\quad =\displaystyle \frac{1}{\theta }\int _{\Omega }\left( \frac{1}{2}f(x,t_{n}u_{n}^{+})t_{n}u_{n}^{+}-F(x,t_{n}u_{n}^{+})\right) \hbox {d}x-\frac{\theta _{0}}{2\theta }|\Omega |\\ &{}&{}\quad =\displaystyle \frac{1}{\theta }\left( \frac{1}{2}\Vert t_{n}u_{n}\Vert ^{2}-\frac{1}{2}\int _{\Omega }\frac{(t_{n}u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x -\int _{\Omega }F(x,t_{n}u_{n}^{+})\hbox {d}x\right) -\displaystyle \frac{\theta _{0}}{2\theta }|\Omega |\\ &{}&{}\quad =\displaystyle \frac{1}{\theta }I(t_{n}u_{n})+\displaystyle \frac{1}{\theta }\left( \frac{1}{2^{*}(s)} -\frac{1}{2}\right) \int _{\Omega }\frac{(t_{n}u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x-\frac{\theta _{0}}{2\theta }|\Omega |, \end{array}\right. \end{aligned}$$

which implies that

$$\begin{aligned}&\displaystyle \int _{\Omega }\left( \frac{1}{2}f(x,u_{n}^{+})u_{n}^{+}-F(x,u_{n}^{+})\right) \hbox {d}x+\frac{1}{\theta }\left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \int _{\Omega } \frac{(t_{n}u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\nonumber \\&\quad \ge \displaystyle \frac{1}{\theta }I(t_{n}u_{n})-\frac{\theta _{0}}{2\theta }|\Omega |\nonumber \\&\quad \rightarrow \displaystyle +\infty , \end{aligned}$$

(2.45)

as $n\rightarrow \infty$. But by $\theta \ge 1$ and $0\le t_{n}\le 1$, we have $\displaystyle \frac{t_{n}^{2^{*}(s)}}{\theta }\le 1$; then, it follows from (2.41) that

$$\begin{aligned} \left. \begin{array}{rcl} c+1+o(1)&{}\ge &{}\displaystyle I(u_{n})-\frac{1}{2}\langle I'(u_{n}),u_{n}\rangle \\ &{}\ge &{}\displaystyle \left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\\ &{}&{}\quad +\int _{\Omega }\left( \frac{1}{2}f(x,u_{n}^{+})u_{n}^{+}-F(x,u_{n}^{+})\right) \hbox {d}x\\ &{}\ge &{}\displaystyle \frac{1}{\theta }\left( \frac{1}{2}-\frac{1}{2^{*}(s)}\right) \int _{\Omega } \frac{(t_{n}u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\\ &{}&{} \quad +\displaystyle \int _{\Omega }\left( \frac{1}{2}f(x,u_{n}^{+})u_{n}^{+}-F(x,u_{n}^{+})\right) \hbox {d}x, \end{array}\right. \end{aligned}$$

which contradicts (2.45). Therefore, $\{u_{n}\}$ is bounded. $\square$

Proof of Theorem 4

It is obvious that $(f_{9})$ leads to $(f_{5})$ and then Lemma 2.2 is also true if we replace $(f_{5})$ by $(f_{9})$. Due to $(f_{1})$, $(f_{9})$, and Lemma 2.2, one obtains that the proof of Theorem 4 is similar to the proof of Theorem 2, and we only need to prove that the $(Ce)_{c}$ sequence is bounded. In fact, let $\{u_{n}\}\subset H_{0}^{1}(\Omega )$ be a $(Ce)_{c}$ sequence, that is,

$$\begin{aligned} \left. \begin{array}{rcl} I(u_{n})\rightarrow c ~~{ and}~~ (1+\Vert u_{n}\Vert )\Vert I'(u_{n})\Vert \rightarrow 0~~~{ as}~ n\rightarrow \infty . \end{array}\right. \end{aligned}$$

(2.46)

By $(f_{10})$, there exist positive constants D, $C_{28}>0$, such that

$$\begin{aligned} f(x,t)t-2F(x,t)\ge D|t|^{\sigma }-C_{28}, \end{aligned}$$

(2.47)

for all $t\in \mathbb {R}^{+}$ and a.e. $x\in \overline{\Omega }$. Together with (2.46), one has

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle 2c+1+o(1)&{}\ge &{}\displaystyle 2I(u_{n})-\langle I'(u_{n}), u_{n}\rangle \\ &{}=&{}\displaystyle \left( 1-\frac{2}{2^{*}(s)}\right) \int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\int _{\Omega }\bigg (f(x,u_{n}^{+})u_{n}-2F(x,u_{n}^{+})\bigg )\hbox {d}x\\ &{}\ge &{}\displaystyle \left( 1-\frac{2}{2^{*}(s)}\right) \int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+D\int _{\Omega }|u_{n}|^{\sigma }\hbox {d}x-C_{28}|\Omega |.\\ \end{array}\right. \end{aligned}$$

From above, we easily obtain that there exist constants $C_{29}$, $C_{30}>0$ such that

$$\begin{aligned} \displaystyle \int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x\le C_{29},~~~\int _{\Omega }|u_{n}|^{\sigma }\hbox {d}x\le C_{30}. \end{aligned}$$

By $(f_{9})$, for any $\varepsilon >0$, there exists $a(\varepsilon )>0$ such that

$$\begin{aligned} \displaystyle |F(x,t)|\le \varepsilon t^{q}+a(\varepsilon )~~~\mathrm {for}~(x,t)\in \Omega \times \mathbb {R}^{+}, \end{aligned}$$

then it follows that

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \frac{1}{2}\Vert u_{n}\Vert ^{2}-I(u_{n})&{}=&{}\displaystyle \frac{1}{2^{*}(s)}\int _{\Omega }\frac{(u_{n}^{+})^{2^{*}(s)}}{|y|^{s}}\hbox {d}x+\int _{\Omega }F(x,u_{n}^{+})\hbox {d}x\\ &{}\le &{}\displaystyle \frac{C_{29}}{2^{*}(s)}+C_{31}\int _{\Omega }|u_{n}|^{q}\hbox {d}x+a(\varepsilon )|\Omega |.\\ \end{array}\right. \end{aligned}$$

By the Gagliardo–Nirenberg interpolation inequality, one has

$$\begin{aligned} \int _{\Omega }|u_{n}|^{q}\hbox {d}x\le \left( \int _{\Omega }|u_{n}|^{\sigma }\hbox {d}x\right) ^{\frac{qt}{\sigma }}\left( \int _{\Omega }|u_{n}|^{2^{*}}\hbox {d}x\right) ^{\frac{q(1-t)}{2^{*}}}, \end{aligned}$$

where $0<\sigma \le q<2^{*}$, $\displaystyle \frac{1}{q}=\frac{t}{\sigma }+\frac{1-t}{2^{*}}$, and $t\in (0,1]$. Then, we deduce from above inequality and Sobolev inequality

$$\begin{aligned} \left. \begin{array}{rcl} \displaystyle \frac{1}{2}\Vert u_{n}\Vert ^{2}&{}\le &{}\displaystyle \frac{C_{29}}{2^{*}(s)}+a(\varepsilon )|\Omega |+C_{31}\int _{\Omega }|u_{n}|^{q}\hbox {d}x+I(u_{n})\\ &{}\le &{}\displaystyle \frac{C_{29}}{2^{*}(s)}+C_{31}\left( \int _{\Omega }|u_{n}|^{\sigma }\hbox {d}x\right) ^{\frac{qt}{\sigma }}\left( \int _{\Omega }|u_{n}|^{2^{*}}\hbox {d}x\right) ^{\frac{q(1-t)}{2^{*}}}+a(\varepsilon )|\Omega |+c+1\\ &{}\le &{}\displaystyle \frac{C_{29}}{2^{*}(s)}+C_{32}\Vert u_{n}\Vert ^{q(1-t)}+a(\varepsilon )|\Omega |+c+1.\\ \end{array}\right. \end{aligned}$$

Since by definition of q, we have $\displaystyle q(1-t)=\frac{2^{*}(q-\sigma )}{2^{*}-\sigma }$ with $\displaystyle \sigma >\frac{N(q-2)}{2}$, it follows that $\displaystyle q(1-t)<2$. Thus, $\{u_{n}\}$ is bounded. $\square$