The existence or nonexistence of solutions for some equations involving weighted critical exponents on the unit ball

Abstract

In this paper, we extend the results of Brézis and Willem (J Funct Anal 255:2286–2298, 2008) to the equation with single or double weighted critical exponents, including Hardy–Sobolev, Sobolev and Hénon–Sobolev exponents. More precisely, we establish the existence or nonexistence of equation with different coefficient which has an important impact.

1 Introduction

In this paper we consider the following equation

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\Delta u+h\left( \frac{|x|}{\lambda }\right) \frac{u}{\lambda ^2}=|x|^{\alpha _1} |u|^{2^*(\alpha _1)-2}u+\mu |x|^{\alpha _2} |u|^{2^*(\alpha _2)-2}u\ &{}\mathrm{in}\ B,\\ u \in H^1_{0,r}(B), &{} \end{array} \right. \end{aligned}$$

(1.1)

where $N \geqslant 3, \alpha _1>\alpha _2>-2, 2^*(\alpha _i)=\frac{2(N+\alpha _i)}{N-2}, i=1,2$ $\mu \in \mathbb {R}$, $B:=\{x\in \mathbb {R}^N: \ |x|<1\}$ is the unit ball in $\mathbb {R}^N$, $H_{0,r}^1(B)$ is the completion of $C^\infty _{0,r}(B)$ with the norm

$$\begin{aligned} \Vert u\Vert =\bigg (\int _{B}|\nabla u|^2 dx\bigg )^\frac{1}{2}, \end{aligned}$$

where $C^\infty _{0,r}(B)$ is the set of radial functions in $C^\infty _{0}(B)$. Let

$$\begin{aligned} L^p(B; |x|^\alpha )=\left\{ u: B\rightarrow \mathbb {R}: \ \ u \ \text{ is } \text{ measurable }, \ \int _{B} |x|^\alpha |u|^pdx<\infty \right\} \end{aligned}$$

be the weighted Lebesgue space with the norm

$$\begin{aligned} \Vert u\Vert _{L^p(B;|x|^\alpha )}:=\left( \int _B|x|^\alpha |u|^pdx\right) ^{\frac{1}{p}}, 1\le p<\infty . \end{aligned}$$

It holds that

$$\begin{aligned} H^1_{0,r}(B)\hookrightarrow L^p(B; |x|^\alpha ) \end{aligned}$$

(1.2)

with $\alpha \geqslant -2$ is continuous for all $1 \leqslant p\leqslant 2^*(\alpha ):=\frac{2(N+\alpha )}{N-2}$ and it is compact for all $1 \leqslant p< 2^*(\alpha )$, see [20, 21]. The compact embedding of (1.2) for $\alpha >0$ was first proved in [17]. In [23, 24] we have confirmed that $2^*(\alpha )$ is exactly the upper critical exponent of the embedding (1.2) by proving that there is no embedding from $H^1_{0,r}(B)$ into $L^p(B; |x|^\alpha )$ for any $p>2^*(\alpha )$ and (1.2) is not compact as $p=2^*(\alpha )$. It is known that $2^*(\alpha )$ is Hardy (resp.,Hardy–Sobolev, Sobolev) critical exponent as $\alpha =-2$(resp., $-2<\alpha <0$, $\alpha =0$), see [11, 23]. In [23, 24], we named $2^*(\alpha )$ as Hénon–Sobolev critical exponent for $\alpha >0$ due to Hénon [14] first raised a semilinear elliptic model involving $|x|^\alpha$ with $\alpha >0$. Therefore there are two critical terms in (1.1).

For $\alpha _1=0$ and $\mu =0$, (1.1) reduces as

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\Delta u+h\left( \frac{|x|}{\lambda }\right) \frac{u}{\lambda ^2}=|u|^{2^*(0)-2}u \ &{}\mathrm{in} \ B,\\ u=0 \ &{}\mathrm{on}\ \partial B. \end{array} \right. \end{aligned}$$

(1.3)

For the case of $h\left( \frac{|x|}{\lambda }\right) \frac{1}{\lambda ^2}:=a(x) \leqslant 0$, (1.3) has been treated extensively since the great work [5] of Brézis and Nirenberg. In [4], Brézis raised seven open problems and the fourth one read as

Q4. Assume $a(x) \geqslant 0$ on B. Find conditions on a(x) (hopefully a necessary and sufficient condition!) which guarantee that (1.3) has a solution.

In [18], Passaseo gave a partial answer to Q4. Under some conditions on a(x), Passaseo proved the existence of positive solutions (1.3). In [6, 7], Brézis and Peletier studied (1.3) with $N=3$ and

$$\begin{aligned} h(|x|)=\frac{K}{(1+|x|^2)^2}, \ \ K>0. \end{aligned}$$

(1.4)

They proved the existence and non-existence of solutions based on different region of value $\lambda$. In [8], Brézis and Willem studied Q4 for the case of $N \geqslant 3$ with more general assumptions on h. In the present paper we will extend the results of Brézis and Willem to the equation (1.1) with one or two weighted critical exponents. For other related works we refer to [3] with unbounded domain $\mathbb {R}^N$, to [1] with ball or annular domain, to [16] with p-Laplacian and to [24] with multiple weighted critical exponents.

In Sect. 2 we consider the non-existence of solutions of (1.1) applying the ODE theory. In Sect. 3 we are interested in the existence results of (1.1) with single weighted critical exponent ($\mu =0$).

2 Nonexistence

In this section, we are interested in (1.1) with multiple Hénon–Sobolev critical exponents as $\alpha _1>\alpha _2 \geqslant 0$. We will prove the nonexistence of solutions of (1.1) with different value $\lambda$, the methods depend on the ODE theory.

We assume that

$$\begin{aligned} (\mathrm {h}_1) \ \ h\in L_\mathrm{loc}^\infty \ \ \mathrm{and} \ r^2h(r) \ \mathrm{is nondecreasing on} \ [0,1]. \end{aligned}$$

It follows from $(\mathrm {h}_1)$ that $\displaystyle \lim _{r\rightarrow 1^-} r^2h(r)=h(1^-)$ exists. The function (1.4) satisfies $(\mathrm {h}_1)$.

Theorem 2.1

Assume that h satisfies $(\mathrm {h}_1)$. Then (1.1) has only the trivial solution in each of the following cases:

$(\mathrm {i})$$\lambda \geqslant 1$ if $\mu <0$;

$(\mathrm {ii})$ there exists $\lambda ^*= \lambda ^*(h) \in (0,1)$ and $\lambda >\lambda ^*$ if $\mu \geqslant 0$.

Next we consider the following equation

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta u+h(|x|)u=|x|^{\alpha _1} |u|^{2^*(\alpha _1)-2}u+\mu |x|^{\alpha _2} |u|^{2^*(\alpha _2)-2}u\ \ &{}\mathrm{in}\ B,\\ u\in H_{0,r}^1(B). \end{array} \right. \end{aligned}$$

(2.1)

Assume

$$\begin{aligned} (\mathrm {h}_2) \ \ \ h\in L^\infty (0, 1) \ \mathrm{and} \ \ r^2h(r) \ \mathrm{is nondecreasing on}\ (0,\delta ) \ \mathrm{for some} \ \delta \in (0,1). \end{aligned}$$

We remark that the function (1.4) also satisfies $(\mathrm {h}_2)$.

Theorem 2.2

For $\mu \geqslant 0$ and $\delta \in (0, 1)$, there exists $K_1=K_1(\delta , \alpha _1, \alpha _2, N)>0$ such that, if h satisfies $(\mathrm {h}_2)$ and $\Vert h\Vert _\infty \leqslant K_1$, then (2.1) has only trivial solution.

For $N=3$, a sharper conclusion will be obtained.

Theorem 2.3

Assume $N=3,$ $\mu \geqslant 0$ and $h \in L^\infty (0,1)$. There exists $K_1=K_1(\alpha _1, \alpha _2, N)>0$ such that if $\Vert h\Vert _\infty \leqslant K_1$ then (2.1) has only trivial solution.

Remark 2.4

For the case $\alpha _1=\mu =0$ in Theorem 2.3, the conclusion has been proved by Brézis and Willem in [8]. In addition, when $\alpha _1=\mu =0, N=3, h=-\lambda$ and $0<\lambda <\frac{\pi ^2}{4}$, Brézis and Nirenberg first prove the solution $u=0$ of (2.1) in [5]. In [24], we extend the results of Brézis and Nirenberg to the case of $\alpha _1>0$.

Now we begin to prove Theorems 2.1–2.3. We follow some arguments in [8] with modifications.

Under $(\mathrm {h}_1)$, for $\alpha _1>\alpha _2 \geqslant 0$ and $\lambda >0$, by Brézis-Kato theorem and Sobolev embedding theorem we have a fact that any a solution u of (1.1) must satisfy $u\in C^1(\bar{B})$, furthermore, $u\in L^\infty (B)$.

Set $u(r):=u(|x|)$ with $r=|x|$. Then (1.1) can be reset as

$$\begin{aligned} \left\{ \begin{array}{llll} \displaystyle -u''-\frac{N-1}{r}u'{+}h\left( \frac{r}{\lambda }\right) \frac{u}{\lambda ^2}{=}r^{\alpha _1} |u|^{2^*(\alpha _1)-2}u{+}\mu r^{\alpha _2} |u|^{2^*(\alpha _2)-2}u, \ r{\in } (0, 1),\\ u'(0)=u(1)=0. \end{array} \right. \qquad \end{aligned}$$

(2.2)

Applying the classical Emden transformation

$$\begin{aligned} u(r)=e^{\frac{N-2}{2}t}w(t), \ \ \ t=-\ln r, \end{aligned}$$

(2.3)

then (2.2) can be reduced as

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -w''+\frac{(N-2)^2}{4}w+H_\lambda (t) w=|w|^{2^*(\alpha _1)-2}w+\mu |w|^{2^*(\alpha _2)-2}w, \ \ t>0,\\ w(0)=0 \end{array} \right. \end{aligned}$$

(2.4)

with

$$\begin{aligned}&|w(t)| \leqslant e^{\frac{2-N}{2}t}\Vert u\Vert _{L^\infty (B)}, \ \ |w'(t)|\nonumber \\&\leqslant e^{\frac{2-N}{2}t} \left( \frac{N-2}{2}\Vert u\Vert _{L^\infty (B)}+e^{-t}\Vert u'\Vert _{L^\infty (B)}\right) , \end{aligned}$$

(2.5)

where

$$\begin{aligned}H_\lambda (t)=\frac{e^{-2t}}{\lambda ^2}h\left( \frac{e^{-t}}{\lambda }\right) .\end{aligned}$$

By (2.5), we see that

$$\begin{aligned} \lim _{t\rightarrow \infty }w(t)=0, \ \ \ \ \ \lim _{t\rightarrow \infty }w'(t)=0. \end{aligned}$$

(2.6)

Let

$$\begin{aligned} t_\lambda =-\ln \lambda . \end{aligned}$$

(2.7)

Lemma 2.5

Let h satisfy $(\mathrm {h}_1)$ and let $w:[0,\infty )\rightarrow \mathbb {R}$ be a solution of (2.4). Then

$$\begin{aligned} w'(0)^2 \leqslant -2\int _0^{t_\lambda }H_\lambda (t)w(t)w'(t)dt+h(1^-)w(t_\lambda )^2, \ \ 0<\lambda <1. \end{aligned}$$

(2.8)

Proof

Multiplying (2.4) by $w'$ and integrating on $(0,\infty )$, using (2.6), we get

$$\begin{aligned} \frac{1}{2}(w'(0))^2+\int _0^\infty H_\lambda (t) w(t)w'(t)dt=0. \end{aligned}$$

(2.9)

Now we decompose the integral interval of second term of (2.9) as

$$\begin{aligned} \int _0^\infty H_\lambda (t) ww'dt=\int _0^{t_\lambda } H_\lambda (t) w w'dt+\int _{t_\lambda }^\infty H_\lambda (t) ww'dt. \end{aligned}$$

(2.10)

Integration by parts, we obtain

$$\begin{aligned} \int _{t_\lambda }^\infty H_\lambda (t) w w'dt =-\frac{1}{2} H_\lambda (t_\lambda ^+) w(t_\lambda )^2-\frac{1}{2}\int _{t_\lambda }^\infty w(t)^2d(H_\lambda (t)). \end{aligned}$$

(2.11)

It follows from $(\mathrm {h}_1)$ and (2.7) that $H_\lambda (t)$ is non-increasing on $(t_\lambda ,\infty )$ and $H(t_\lambda ^+)=h(1^-)$. Thus

$$\begin{aligned} \int _{t_\lambda }^\infty H_\lambda w w'dt \geqslant -\frac{1}{2} h(1^-) w(t_\lambda )^2. \end{aligned}$$

(2.12)

Combining with (2.9), (2.10) and (2.12), we obtain the desired conclusion that

$$\begin{aligned} w'(0)^2 \leqslant -2\int _0^{t_\lambda } H_\lambda (t) w(t)w'(t)dt+h(1^-)w(t_\lambda )^2. \end{aligned}$$

The proof is complete. $\square$

Lemma 2.6

([8, Lemma 2.2]) Assume $A \geqslant 0$, $B>0$, $L>0$ and $w\in C^1([0,L])$ satisfies $w(0)=0$,

$$\begin{aligned} w'(t)^2\leqslant A^2+2B^2\int _0^t|w w'|d s \ \ \ \mathrm{for} \ \ 0 \leqslant t\leqslant L. \end{aligned}$$

Then

$$\begin{aligned} |w(t)| \leqslant \frac{A}{B}\left( e^{Bt}-1\right) \ \ |w'(t)|\leqslant A e^{Bt}, \ \ \ for \ 0 \leqslant t\leqslant L. \end{aligned}$$

Lemma 2.7

Assume $(\mathrm {h}_1)$ and $\mu \geqslant 0$. Then for $\frac{1}{2}<\lambda <1$ and $0\leqslant t \leqslant t_\lambda$, any a solution $w:[0,\infty )\rightarrow \mathbb {R}$ of (2.4) satisfies

$$\begin{aligned}&|w(t)| \leqslant \frac{1}{c_0} |w'(0)| \left( e^{c_0 t}-1\right) , \end{aligned}$$

(2.13)

$$\begin{aligned}&|w'(t)| \leqslant |w'(0)|e^{c_0t}, \end{aligned}$$

(2.14)

where

$$\begin{aligned} c_0=\sup _{1 \leqslant r <2}\left( \frac{(N-2)^2}{4}+r^2|h(r)|\right) ^{\frac{1}{2}}. \end{aligned}$$

Proof

By (2.4) we obtain that

$$\begin{aligned} \frac{w'(t)^2}{2}= & {} \frac{w'(0)^2}{2}+\int _0^tw'w''ds\\= & {} \frac{w'(0)^2}{2}+\int _0^t \left( \frac{(N-2)^2}{4}ww'+h\left( \frac{e^{-s}}{\lambda }\right) \frac{e^{-2s}w w'}{\lambda ^2}\right) d s\\&-\int _0^t|w|^{2^*(\alpha _1)-2}w w' d s-\mu \int _0^t|w|^{2^*(\alpha _2)-2}ww'ds\\\leqslant & {} \frac{w'(0)^2}{2}+\int _0^t\left[ \frac{(N-2)^2}{4} +\frac{e^{-2s}}{\lambda ^2}\left| h\left( \frac{e^{-s}}{\lambda }\right) \right| \right] |w(s)w'(s)|d s. \end{aligned}$$

For $\frac{1}{2}<\lambda <1$ and $0\leqslant t \leqslant t_\lambda$, we have

$$\begin{aligned} \frac{e^{-s}}{\lambda }\in \left[ \frac{e^{-t_\lambda }}{\lambda },\frac{e^0}{\lambda }\right] \subset [1,2). \end{aligned}$$

It follows that

$$\begin{aligned} w'(t)^2 \leqslant w'(0)^2+2 c_0^2 \int _0^t|w w'|dx, \end{aligned}$$

where

$$\begin{aligned} c_0= \sup _{1\leqslant r<2} \left( \frac{(N-2)^2}{4}+r^2|h(r)|\right) ^{\frac{1}{2}}. \end{aligned}$$

Applying Lemma 2.6 with $A=|w'(0)|$ and $B=c_0$, we obtain (2.13) and (2.14). $\square$

Proof of Theorem 2.1

For $\lambda \geqslant 1, \mu <0$. Multiplying (1.1) by $\sum _{i=1}^N x_i\frac{\partial u}{\partial x_i}$ and integrating on B, we obtain

$$\begin{aligned} \left. \begin{array}{llll} 0=&{}\displaystyle \frac{N-2}{2}\int _B|\nabla u|^2dx-\frac{N-2}{2}\int _B \left( |x|^{\alpha _1} |u|^{2^*(\alpha _1)}dx+\mu |x|^{\alpha _2} |u|^{2^*(\alpha _2)}\right) dx\\ &{} \displaystyle +\frac{1}{2}\int _{\partial B}\left| \frac{\partial u}{\partial n}\right| ^2d\sigma +\frac{N}{2}\int _B h\left( \frac{|x|}{\lambda }\right) \frac{1}{\lambda ^2} u^2 dx\\ &{}\displaystyle +\frac{1}{2}\int _B \frac{1}{\lambda ^2} \left[ \displaystyle \sum _{i=1}^N x_i \frac{ \partial }{\partial x_i}h\left( \frac{|x|}{\lambda }\right) \right] u^2 dx. \end{array} \right. \end{aligned}$$

(2.15)

Since u satisfies

$$\begin{aligned} \int _B |\nabla u|^2+\int _B h\left( \frac{|x|}{\lambda }\right) \frac{u^2}{\lambda ^2}dx=\int _B |x|^{\alpha _1} |u|^{2^*(\alpha _1)}dx+\mu \int _B |x|^{\alpha _2} |u|^{2^*(\alpha _2)}dx, \nonumber \\ \end{aligned}$$

(2.16)

it follows from (2.15) and (2.16) that

$$\begin{aligned} -\int _B \left( h\left( \frac{|x|}{\lambda }\right) +\frac{1}{2}\sum _{i=1}^N x_i \frac{\partial }{\partial x_i}h\left( \frac{|x|}{\lambda }\right) \right) \frac{u^2}{\lambda ^2}dx=\frac{1}{2}\int _{\partial B}\left| \frac{\partial u}{\partial n}\right| ^2d\sigma . \end{aligned}$$

Since

$$\begin{aligned} h\left( \frac{|x|}{\lambda }\right) +\frac{1}{2}\sum _{i=1}^N x_i \frac{\partial }{\partial x_i}h\left( \frac{|x|}{\lambda }\right) = \frac{1}{s} (s^2 h(s))', \ \ \ s=\frac{|x|}{\lambda }, \end{aligned}$$

it follows from $(\mathrm {h}_1)$ that $u = 0$.

For $\lambda <1$ and $\mu \geqslant 0$. We may assume $\frac{1}{2}<\lambda <1$. The inequality (2.8) from Lemma 2.5 implies that

$$\begin{aligned} w'(0)^2 \leqslant 2 \int _0^{t_\lambda }|H_\lambda (t)||w(t)||w'(t)|dt+h(1^-)w(t_\lambda )^2. \end{aligned}$$

(2.17)

Inserting (2.13) and (2.14) from Lemma 2.7 into (2.17), we have

$$\begin{aligned} w'(0)^2 \leqslant w'(0)^2 K(\lambda ), \end{aligned}$$

(2.18)

where

$$\begin{aligned} K(\lambda )=2c_0\int _0^{t_\lambda }\left| \frac{e^{c_0t}-1}{c_0}\right| \left| e^{c_0t}\right| dt +h(1^-)\frac{1}{c_0^2}(e^{c_0t_\lambda }-1)^2. \end{aligned}$$

It is obvious that $t_\lambda \downarrow 0$ and $K(\lambda )\rightarrow 0$ as $\lambda \uparrow 1$. It follows that there exists $\lambda ^*\in (\frac{1}{2},1)$ such that $K(\lambda )<1$ as $\lambda ^*<\lambda <1$. Using this fact, the inequality (2.18) leads to $w'(0)=0$ for $\lambda ^*<\lambda <1$. By the uniqueness of the Cauchy problem, we complete the proof. $\square$

Proof of Theorem 2.2

We will argue in the same way as proving Theorem 2.1. Let $u \in H_{0,r}^1(B)$ be a solution of (2.1). Using the same transformation (2.3), (2.1) becomes

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -w''(t)+\frac{(N-2)^2}{4}w(t) + h\left( e^{-t}\right) e^{-2t}w(t) \\ \ \ \ \ \ \ \ \ \ \ =|w(t)|^{2^*(\alpha _1)-2}w(t)+\mu |w(t)|^{2^*(\alpha _1)-2}w(t), \ \ t>0,\\ w(0)=0,\end{array} \right. \end{aligned}$$

(2.19)

Then w satisfies (2.5). Set $H(t)=e^{-2t}h(e^{-t})$ and $T_\delta =- \ln \delta$. Multiplying equation (2.19) by $w'$ and integrating on $(0,\infty )$, we deduce that

$$\begin{aligned} \frac{1}{2}(w'(0))^2+\int _0^\infty H(t) w(t)w'(t)dt=0. \end{aligned}$$

(2.20)

Rewriting

$$\begin{aligned} \int _0^\infty H(t) w(t)w'(t)dt=\int _0^{T_\delta } H(t) w(t)w'(t)dt+\int _{T_\delta }^\infty H(t) w(t)w'(t)dt.\qquad \end{aligned}$$

(2.21)

Integrating by parts, we obtain

$$\begin{aligned} \int _{T_\delta }^\infty H(t) w(t)w'(t)dt =-\frac{1}{2} H(T_\delta ^+) w(T_\delta )^2-\frac{1}{2}\int _{T_\delta }^\infty w(t)^2d(H(t)). \end{aligned}$$

By $(\mathrm {h}_2)$, we see that H(t) is non-increasing on $(T_\delta ,\infty )$ and $H(T_\delta ^+)=\delta ^2h(\delta ^-)$. Hence

$$\begin{aligned} \int _{T_\delta }^\infty H w w'dt \geqslant -\frac{1}{2} \delta ^2 h(\delta ^-) w(T_\delta )^2. \end{aligned}$$

(2.22)

It follows from (2.20), (2.21), (2.22) that

$$\begin{aligned} (w'(0))^2 \leqslant -2\int _0^{T_\delta } H(t) w(t)w'(t)dt+h(\delta ^-)\delta ^2w(T_\delta )^2. \end{aligned}$$

(2.23)

The estimates (2.13) and (2.14) are still valid on $(0,T_\delta )$ with

$$\begin{aligned} c_0:=\left( \frac{(N-2)^2}{4}+\Vert h\Vert _\infty \right) ^{\frac{1}{2}}. \end{aligned}$$

(2.24)

Combining with (2.23) and (2.13), (2.14), we deduce that

$$\begin{aligned} w'(0)^2 \leqslant w'(0)^2 \Vert h\Vert _\infty \left[ \frac{2}{c_0} \int _0^{T_\delta }(e^{c_0t}-1)e^{c_0t}dt +\frac{\delta ^2}{c_0^2}(e^{c_0T_\delta }-1)^2\right] . \end{aligned}$$

(2.25)

Hence there exists $K_1>0$ such that $w'(0)=0$ as $\Vert h\Vert _\infty \leqslant K_1$. $\square$

Proof of Theorem 2.3

Similar with the proofs of Theorem 2.1 and Theorem 2.2, we have

$$\begin{aligned} w'(0)^2 \leqslant 2\int _0^\infty |H(t)||w(t)||w'(t)|dt. \end{aligned}$$

Notice that in (2.24) $c_0=\left( \frac{1}{4}+\Vert h\Vert _\infty \right) ^{1/2}$ for $N=3$ so that $c_0\in (1/2,1)$ when $\Vert h\Vert _\infty$ is small. Therefore

$$\begin{aligned} |H(t)|\leqslant \Vert h\Vert _\infty e^{-2t}, \ \ |w(t)|\leqslant \frac{|w'(0)|}{c_0}e^{c_0t}, \ \ |w'(t)| \leqslant |w'(0)|e^{c_0t} \end{aligned}$$

and then

$$\begin{aligned} w'(0)^2 \leqslant \frac{2}{c_0}|w'(0)|^2 \Vert h\Vert _\infty \int _0^\infty e^{2(c_0-1)t}dt. \end{aligned}$$

Using the fact that $c_0<1$, the conclusion of theorem is proved. $\square$

3 Existence for the case of $\mu =0$

In this section we prove the existence of nontrivial solutions for the case $\mu =0$ in (1.1). We reformulate (1.1) as follows by setting $\alpha :=\alpha _1$,

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\Delta u+h\left( \frac{|x|}{\lambda }\right) \frac{u}{\lambda ^2}=|x|^{\alpha } |u|^{2^*(\alpha )-2}u \ &{}\mathrm{in} \ B,\\ u \in H_{0,r}^1(B), \end{array} \right. \end{aligned}$$

(3.1)

where $N \geqslant 3$, $\alpha >-2$, $2^*(\alpha )=\frac{2(N+\alpha )}{N-2}$ is the Hardy–Sobolev or Sobolev or Hénon–Sobolev critical exponent. The potential h satisfies

$(\mathrm {h}_3)$$h:[0,\infty ) \rightarrow [0,\infty )$ is such that $h\ne 0$ on a set of positive measure and

$$\begin{aligned}&h \in L_\mathrm{loc}^{N/2}([0,\infty ),s^{N-1}),\end{aligned}$$

(3.2)

$$\begin{aligned}&\quad \lim _{\lambda \rightarrow 0}\lambda ^{N-2}\int _0^{\frac{1}{\lambda }}h(s)s^{N-1}ds=0. \end{aligned}$$

(3.3)

We remark that $h\in L^{N/2}([0,\infty ),s^{N-1})$ satisfies (3.2), and (3.3) if $\lim _{s\rightarrow \infty }s^2h(s)=0$. Hence the function

$$\begin{aligned} h(s)=\frac{1}{1+s^3} \end{aligned}$$

satisfies (3.2) and (3.3). We will prove the following theorem.

Theorem 3.1

Assume that h satisfies $(\mathrm {h}_3)$ and $\alpha >-2$. Then there exists $\lambda _0>0$ such that (3.1) has a nonnegative solution for $0<\lambda <\lambda _0$.

Remark 3.2

Since $h\geqslant 0$, it is not possible to prove the existence of solutions of (3.1) by the global minimization as in [5, 24], the main difficulty is to estimate the energy level of quotient is less than some number, which guarantee the holds of local $\mathrm {(PS)}_c$. For $\alpha =0$ and $h\in L^{N/2}([0,\infty ),s^{N-1}ds)$, the existence of positive solutions was obtained by Passaseo in [18] using the constrained minimization and in [8], Brézis and Willem obtain the nontrivial solution under $(\mathrm {h}_3)$. Theorem 3.1 extends the result in [8] to the case $\alpha >-2$.

Remark 3.3

In Theorem 3.1 a positive solution can be obtained via strong maximum principle if (3.2) is replaced by

$$\begin{aligned} h\in L_\mathrm{loc}^\infty ([0,\infty ),s^{N-1}). \end{aligned}$$

The approach for proving Theorem 3.1 is from [8] and [16]. Define the manifolds

$$\begin{aligned}&\Gamma (B):=\left\{ u\in H_{0,r}^1(B): \ \int _B |x|^\alpha |u|^{2^*(\alpha )}dx=1\right\} ,\\&\Gamma (\mathbb {R}^N):=\left\{ u\in D_r^{1,2}(\mathbb {R}^N): \ \int _{\mathbb {R}^N} |x|^\alpha |u|^{2^*(\alpha )}dx=1\right\} , \end{aligned}$$

and the functionals for $\lambda >0$,

$$\begin{aligned}&\varphi _\lambda (u)=\int _{\mathbb {R}^N}|\nabla u|^2dx + h\left( \frac{|x|}{\lambda }\right) \frac{u^2}{\lambda ^2}dx, \ \ \ \ u \in D_r^{1,2}(\mathbb {R}^N), \\&\psi _\lambda (u)=\int _{\mathbb {R}^N}a_\lambda (|x|)|x|^\alpha |u|^{2^*(\alpha )}dx, \ \ \ a_\lambda (|x|)=\frac{|x|}{\lambda +|x|}, \ \ u \in L^{2^*(\alpha )}(\mathbb {R}^N; |x|^\alpha ), \end{aligned}$$

where

$$\begin{aligned}&D_r^{1,2}(\mathbb {R}^N) = \left\{ u \in D^{1,2}(\mathbb {R}^N): \ u \ \hbox {is radial} \right\} , \ \ \ \Vert u\Vert _{D_r^{1,2}}:=\Vert \nabla u\Vert _{L^2(\mathbb {R}^N)}, \\&L^{2^*(\alpha )}(\mathbb {R}^N; |x|^\alpha )=\left\{ u: \mathbb {R}^N \rightarrow \mathbb {R}: \ \hbox { is measurable}, \ \int _{\mathbb {R}^N} |x|^\alpha |u|^{2^*(\alpha )}dx<\infty \right\} . \end{aligned}$$

Under $(\mathrm {h}_3)$, the functional $\varphi _\lambda$ is well defined but not necessarily finite on $D_r^{1,2}(\mathbb {R}^N)$. We will prove

$$\begin{aligned} \Sigma _\lambda =\inf \left\{ \varphi _\lambda (u): \ u\in \Gamma (B), \ \psi _\lambda (u) \geqslant \frac{1}{2}\right\} \end{aligned}$$

is a critical value of $\varphi |_{\Gamma (B)}$. We shall estimate the values

$$\begin{aligned}&\Upsilon _\lambda =\inf \left\{ \varphi _\lambda (u): \ u\in \Gamma (B), \ \psi _\lambda (u)=\frac{1}{2}\right\} , \\&\Upsilon =\inf \left\{ \varphi _1(u): \ u \in \Gamma (\mathbb {R}^N), \ \psi _1(u)=\frac{1}{2}\right\} . \end{aligned}$$

Consider the weighted critical equation

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta u=|x|^\alpha u^{2^*(\alpha )-1}, \ u>0\ \text {in} \ \mathbb {R}^N, \\ u\in D^{1,2}_r(\mathbb {R}^{N}) \end{array} \right. \end{aligned}$$

(3.4)

with $\alpha >-2$. By [12, 13, 15], we have the following key result.

Theorem 3.4

Let $\alpha >-2$. It holds that

$$\begin{aligned} \int _{\mathbb {R}^{N}}|\nabla u|^{2} dx \geqslant S_\alpha \left( \int _{\mathbb {R}^{N}}|x|^{\alpha }|u|^{2^*(\alpha )} dx\right) ^{\frac{2}{2^*(\alpha )}}, \ \ u \in D^{1,2}_r(\mathbb {R}^{N}). \end{aligned}$$

(3.5)

The best constant $S_\alpha$ can be achieved uniquely (up to dilations) by

$$\begin{aligned} U_{\alpha }(x)=\frac{C(\alpha ,N)}{\left( 1+|x|^{2+\alpha }\right) ^{\frac{N-2}{2+\alpha }}}, \ \ \ C(\alpha ,N)=\big [(N+\alpha )(N-2)\big ]^{\frac{N-2}{2(2+\alpha )}} \end{aligned}$$

(3.6)

and $U_{\alpha }$ is the unique (up to dilations) solution of (3.4) and

$$\begin{aligned} S_\alpha =(N+\alpha )(N-2)\bigg (\frac{\omega _N}{2+\alpha } \frac{\Gamma ^2\big (\frac{N+\alpha }{2+\alpha }\big )}{\Gamma \big (\frac{2(N+\alpha )}{2+\alpha }\big )}\bigg ) ^{\frac{2+\alpha }{N+\alpha }}. \end{aligned}$$

We give some remarks. For $\alpha =0$, Theorem 3.4 was proved by Aubin[2], Talenti[22] and $S_0$ was the best Sobolev constant on $D^{1,2}_r(\mathbb {R}^{N})$(see [22]). For $-2<\alpha <0$, Theorem 3.4 was established by Ghoussoub and Yuan [11], Lieb[15], and $S_\alpha$ was named as the best Hardy–Sobolev constant on $D^{1,2}_r(\mathbb {R}^{N})$(see [10]). As $\alpha >0$, these results could be found in [12, 13, 15] and $S_\alpha$ was named in [24] as the best Hénon–Sobolev constant.

The corresponding energy functional of (3.1) is defined as

$$\begin{aligned} \Phi (u)=\frac{1}{2}\int _B|\nabla u|^2dx{+}h\left( \frac{|x|}{\lambda }\right) \frac{|u|^2}{\lambda ^2}dx -\frac{1}{2^*(\alpha )}\int _B|x|^\alpha |u|^{2^*(\alpha )}dx, \ \ u\in H_{0,r}^1(B). \end{aligned}$$

We define

$$\begin{aligned} \Psi (u)=\frac{1}{2}\int _{\mathbb {R}^N}|\nabla u|^2dx -\frac{1}{2^*(\alpha )}\int _{\mathbb {R}^N}|x|^\alpha |u|^{2^*(\alpha )}dx. \end{aligned}$$

Lemma 3.5

If $u_n\rightharpoonup u$ in $D_r^{1,2}(\mathbb {R}^N)$, then

$$\begin{aligned} |x|^{\alpha }|u_n|^{2^*(\alpha )-2}u_n-|x|^{\alpha }|u_n-u|^{2^*(\alpha )-2}(u_n-u) {\rightarrow } |x|^{\alpha }|u|^{2^*(\alpha )-2}u\ \mathrm{in}\ (D_r^{1,2}(\mathbb {R}^N))^*. \end{aligned}$$

Proof

The proof is similar with the argument in [9, Lemma 3.3]. Denote $w_n=u_n-u$. We have

$$\begin{aligned} \left| |x|^{\alpha }|u_n|^{2^*(\alpha )-2}u_n-|x|^{\alpha }|w_n|^{2^*(\alpha )-2}w_n\right| \leqslant (2^*(\alpha )-1)\left( |u_n|+|u|\right) ^{2^*(\alpha )-2}|x|^\alpha |u|. \end{aligned}$$

For $T>0$ and $\varphi \in C_{0,r}^\infty (\mathbb {R}^N)$, applying the Hölder inequality and (3.5),

$$\begin{aligned}&\left| \int _{|x|>T}\left( |x|^{\alpha }|u_n|^{2^*(\alpha )-2}u_n -|x|^{\alpha }|w_n|^{2^*(\alpha )-2}w_n\right) \varphi dx\right| \\\leqslant & {} C\left[ \left( \int _{|x|>T}|x|^\alpha |u_n|^{2^*(\alpha )}dx \right) ^{\frac{2^*(\alpha )-2}{2^*(\alpha )}}+\left( \int _{|x|>T}|x|^\alpha |u|^{2^*(\alpha )}dx \right) ^{\frac{2^*(\alpha )-2}{2^*(\alpha )}}\right] \\&\ \ \ \times \left( \int _{|x|>T}|x|^\alpha |u|^{2^*(\alpha )}dx \right) ^{\frac{1}{2^*(\alpha )}}\left( \int _{|x|>T}|x|^\alpha |\varphi |^{2^*(\alpha )}dx \right) ^{\frac{1}{2^*(\alpha )}}\\\leqslant & {} C\Vert \varphi \Vert _{D_r^{1,2}}\left( \int _{|x|>T}|x|^\alpha |u|^{2^*(\alpha )}dx \right) ^{\frac{1}{2^*(\alpha )}}. \end{aligned}$$

Similarly, we get that

$$\begin{aligned} \left| \int _{|x|>T}\left( |x|^{\alpha }|u|^{2^*(\alpha )-2}u\right) \varphi dx\right| \leqslant C\Vert \varphi \Vert _{D_r^{1,2}}\left( \int _{|x|>T}|x|^\alpha |u|^{2^*(\alpha )}dx \right) ^{\frac{1}{2^*(\alpha )}}. \end{aligned}$$

Therefore, for any $\epsilon >0$, there exists $T>0$ such that, for any $\varphi \in C_{0,r}^\infty (\mathbb {R}^N)$, it holds

$$\begin{aligned} \left| \int _{|x|>T}\left( |x|^{\alpha }|u_n|^{2^*(\alpha )-2}u_n- |x|^{\alpha }|w_n|^{2^*(\alpha )-2}w_n -|x|^{\alpha }|u|^{2^*(\alpha )-2}u\right) \varphi dx\right| \leqslant \epsilon \Vert \varphi \Vert _{D^{1,2}} \end{aligned}$$

Applying [26, Proposition 5.4.7]. We obtain on $\bar{B}_T$ with $B_T:=\{x\in \mathbb {R}^N,|x|<T\}$ that

$$\begin{aligned}&\int _{|x|\leqslant T}|x|^{\alpha }|w_n|^{2^*(\alpha )-2}w_n\varphi dx \rightarrow 0, \\&\int _{|x|\leqslant T}|x|^{\alpha }|u_n|^{2^*(\alpha )-2}u_n\varphi dx \rightarrow \int _{|x|\leqslant T}|x|^{\alpha }|u|^{2^*(\alpha )-2}u\varphi dx. \end{aligned}$$

Hence

$$\begin{aligned}&\int _{|x|\leqslant T}\left( |x|^{\alpha }|u_n|^{2^*(\alpha )-2}u_n\varphi dx -|x|^{\alpha }|w_n|^{2^*(\alpha )-2}w_n \right) \varphi dx\\&\quad \rightarrow \int _{|x|\leqslant T}|x|^{\alpha }|u|^{2^*(\alpha )-2}u\varphi dx. \end{aligned}$$

The proof is complete. $\square$

Theorem 3.6

Let $\{v_n\}\subset D_r^{1,2}(\mathbb {R}^N)$ be a $\mathrm{(PS)_c}$ sequence of $\Psi$, i.e.

$$\begin{aligned} \Psi (v_n)\rightarrow c, \ \ \ \ \Psi '(v_n)\rightarrow 0 \ \ \mathrm{in}\ (D_r^{1,2}(\mathbb {R}^N))^*. \end{aligned}$$

Then, passing subsequence if necessary, there exist a finite sequence $\{u^0,u^1,u^2,\ldots , u^k\} \subset D_r^{1,2}(\mathbb {R}^N)$ of solutions for

$$\begin{aligned} -\Delta u=|x|^\alpha |u|^{2^*(\alpha )-2}u \ \ \mathrm{on} \ \mathbb {R}^N \end{aligned}$$

and k sequences $\{\lambda _n^i\}\subset \mathbb {R}_+$ such that $\lambda _n^i \rightarrow 0$ or $\infty$ and

$$\begin{aligned}&\displaystyle \left\| v_n-u^0-\displaystyle \sum _{i=1}^k (\lambda _n^i)^{\frac{2-N}{2}}u^i\left( \frac{\cdot }{\lambda _n^i}\right) \right\| _{D^{1,2}_r}\rightarrow 0, \\&\Vert v_n\Vert _{D^{1,2}_r}^2\rightarrow \displaystyle \sum _{i=0}^k \Vert u^i\Vert _{D^{1,2}_r}^2,\\&\Psi (v_n)=\displaystyle \sum _{i=0}^k \Psi (u^i)+o(1). \end{aligned}$$

Proof

It is easy to see that $\{v_n\}$ is bounded in $D_r^{1,2}(\mathbb {R}^N)$. Passing if necessary to a subsequence, we assume that $v_n\rightharpoonup u^0$ in $D_r^{1,2}(\mathbb {R}^N)$ and $v_n(x)\rightarrow u^0(x)$ a.e. on $\mathbb {R}^N$. By Lemma 3.5, we have that $\Psi '(u^0)=0$. Set $v_n^1:=v_n-u^0$. Then $\left\{ v_n^1\right\}$ satisfies

$$\begin{aligned} \left. \begin{array}{ll} &{}\mathrm {i}) \ \ \ \displaystyle \lim _{n\rightarrow \infty }\left( \Vert v_n\Vert _{D^{1,2}_r}^2-\Vert v_n^1\Vert _{D^{1,2}_r}^2\right) =\Vert u^0\Vert _{D^{1,2}_r}^2, \\ &{}\mathrm {ii}) \ \ \Psi (v_n^1) \rightarrow c-\Psi (u^0), \\ &{}\mathrm {iii}) \ \Psi '(v_n^1) \rightarrow 0 \ \ \mathrm{in}\ \left( D_r^{1,2}(\mathbb {R}^N)\right) ^*. \end{array} \right. \end{aligned}$$

(3.7)

If $v_n^1 \rightarrow 0$ in $L^{2^*(\alpha )}(\mathbb {R}^N,|x|^\alpha )$, then it follows from $\Psi '(v_n^1)\rightarrow 0$ in $(D_r^{1,2}(\mathbb {R}^N))^*$ that $v_n^1\rightarrow 0$ in $D_r^{1,2}(\mathbb {R}^N)$ and the proof is complete. Assume that there exists $0<\delta <\left( \frac{S_\alpha }{2}\right) ^{\frac{N+\alpha }{2+\alpha }}$ such that for all n large,

$$\begin{aligned} \int _{\mathbb {R}^N}|x|^\alpha \left| v_n^1\right| ^{2^*(\alpha )}dx>\delta . \end{aligned}$$

Defining the Levy concentration function

$$\begin{aligned} Q_n(r):=\int _{B_r(0)}|x|^\alpha \left| v_n^1\right| ^{2^*(\alpha )}dx. \end{aligned}$$

It follows from $Q_n(0)=0$ and $Q_n(\infty )>\delta$ that there exists a sequence $\left\{ \lambda _n^1\right\} \subset (0, \infty )$ such that

$$\begin{aligned} \delta =\int _{B_{\lambda _n^1}(0)}|x|^\alpha |v_n^1|^{2^*(\alpha )}dx. \end{aligned}$$

We denote $u_n^1(x):=(\lambda _n^1)^{\frac{N-2}{2}}v_n^1(\lambda _n^1 x)$ and assume that $u_n^1$ converges weakly to $u^1$ in $D^{1,2}_r(\mathbb {R}^N)$ and converges $u^1$ a.e. on $\mathbb {R}^N$. We claim that $u^1\not =0$. Otherwise, suppose that $u^1=0$. We note that

$$\begin{aligned} \delta =\int _{B_{\lambda _n^1}(0)}|x|^\alpha |v_n^1|^{2^*(\alpha )}dx=\int _B |x|^\alpha |u_n^1|^{2^*(\alpha )}dx. \end{aligned}$$

(3.8)

By the Riesz-Fréchet representation theorem, there exists $f_n\in D_r^{1,2}(\mathbb {R}^N)$ such that

$$\begin{aligned} \left\langle \Psi '(v_n^1), w\right\rangle =\int _{\mathbb {R}^N}\nabla f_n \nabla w dx, \ \ \forall \ w \in D^{1,2}_r(\mathbb {R}^N). \end{aligned}$$

Then $g_n(x):=(\lambda _n^1)^\frac{N-2}{2}f_n(\lambda _n^1x)$ satisfies

$$\begin{aligned}&\langle \Psi '(u_n^1),w\rangle =\int _{\mathbb {R}^N}\nabla g_n\nabla w dx, \ \ \forall \ w\in D^{1,2}_r(\mathbb {R}^N), \end{aligned}$$

(3.9)

$$\begin{aligned}&\int _{\mathbb {R}^N}|\nabla g_n|^2dx=\int _{\mathbb {R}^N}|\nabla f_n|^2dx=o(1). \end{aligned}$$

(3.10)

Taking $\nu \in C_{0,r}^\infty (\mathbb {R}^N)$ such that $\mathrm{supp}\ \nu \in B$ and the measure of $\mathrm{supp}\ \nu$ is small enough. By Hölder inequality and (3.5), we get

$$\begin{aligned}&\int _{\mathrm{supp}\ \nu } |\nu |^2|x|^\alpha \left| u_n^1\right| ^{2^*(\alpha )}dx\\&\quad \leqslant S_\alpha ^{-1}\int _{\mathrm{supp}\ \nu }\left| \nabla (\nu u_n^1)\right| ^2dx\left( \int _{\mathrm{supp}\ \nu }|x|^\alpha \left| u_n^1\right| ^{2^*(\alpha )}dx\right) ^{\frac{2^*(\alpha )-2}{2^*(\alpha )}}. \end{aligned}$$

Hence, combining with $u_n^1\rightarrow 0$ in $L^2(B)$, (3.9) and (3.10), we have

$$\begin{aligned} \int _{\mathrm{supp}\ \nu }\left| \nabla (\nu u_n^1)\right| ^2dx= & {} \int _{\mathrm{supp}\ \nu } |\nu |^2\left| \nabla u_n^1\right| ^2dx+o(1)\\= & {} \int _{\mathrm{supp}\ \nu } \nabla u_n^1\nabla \left( \left| \nu \right| ^2u_n^1\right) dx+o(1)\\= & {} \int _{\mathrm{supp}\ \nu }\nabla g_n\nabla \left( |\nu |^2u_n^1\right) dx\\&+\int _{\mathrm{supp}\ \nu } |x|^\alpha \left| u_n^1\right| ^{2^*(\alpha )}|\nu |^2dx+o(1)\\\leqslant & {} S_\alpha ^{-1}\delta ^{\frac{2+\alpha }{N+\alpha }}\int _{\mathrm{supp}\ \nu }\left| \nabla \left( \nu u_n^1\right) \right| ^2dx+o(1)\\\leqslant & {} \frac{1}{2}\int _{\mathrm{supp}\ \nu }\left| \nabla \left( \nu u_n^1\right) \right| ^2dx+o(1). \end{aligned}$$

Thus we get $\nabla u_n^1\rightarrow 0$ in $L^2(B_r)$ with $0<r<1$ and by (3.5) we obtain $u_n^1\rightarrow 0$ in $L^{2^*(\alpha )}(B_r;|x|^\alpha )$. Using the radial lemma(see [19]), it is easy to see that $u_n^1 \rightarrow 0$ in $L^{2^*(\alpha )}(B_{r,1};|x|^\alpha )$, where $B_{r,1}:=\{x\in \mathbb {R}^N:0<r<|x|<1\}$. Furthermore $u_n^1\rightarrow 0$ in $L^{2^*(\alpha )}(B,|x|^\alpha )$, this contradicts to (3.8). Therefore $u^1\not =0$.

We claim that $\lambda _n^1 \rightarrow 0$ or $\infty$. Assume that $\lambda _n^1\rightarrow \kappa _\infty$ with $0<\kappa _\infty <\infty$. Since $u^1\not =0$, then there exists a ball $B_R$ such that $u^1\not =0$ in $B_R$. On one hand, by locally compact embedding, we deduce that

$$\begin{aligned} \int _{B_R}|u_n^1|^2dx\rightarrow \int _{B_R}|u^1|^2dx>0. \end{aligned}$$

(3.11)

On the other hand, using the facts that $0<\kappa _\infty <\infty$ and $v_n^1\rightharpoonup 0$ in $D^{1,2}(\mathbb {R}^N)$, we have

$$\begin{aligned} \int _{B_R}|u_n^1|^2dx=(\lambda _n^1)^{-2}\int _{B_{R\lambda _n^1}}|v_n^1|^2dx\rightarrow 0, \end{aligned}$$

a contradiction with (3.11). Thus $\lambda _n^1\rightarrow 0$ or $\infty$. It follows from (3.7) that $\Psi '(u^1)=0$. Combining with Lemma 3.5, the sequence

$$\begin{aligned} v_n^2(x)=v_n^1(x)-(\lambda _n^1)^{\frac{2-N}{2}}u^1\left( \frac{x}{\lambda _n^1}\right) \end{aligned}$$

satisfies

$$\begin{aligned}&\mathrm {i}) \ \ \ \left\| v_n^2\right\| _{D^{1,2}_r}^2=\Vert v_n\Vert _{D^{1,2}_r}^2-\Vert u^0\Vert _{D^{1,2}_r}^2 -\Vert u^1\Vert _{D^{1,2}_r}^2+o(1), \\&\mathrm {ii}) \ \ \Psi (v_n^2)=c-\Psi (u^0)-\Psi (u^1)+o(1),\\&\mathrm {iii}) \ \Psi '(v_n^2) \rightarrow 0 \ \mathrm{in} \ \left( D_r^{1,2}(\mathbb {R}^N)\right) ^*. \end{aligned}$$

For any a nontrivial critical point u of $\Psi$, using (3.5), we have

$$\begin{aligned} S_\alpha \left( \int _{\mathbb {R}^N}|x|^\alpha |u|^{2^*(\alpha )}dx\right) ^{\frac{2}{2^*(\alpha )}} \leqslant \int _{\mathbb {R}^N}|\nabla u|^2dx=\int _{\mathbb {R}^N}|x|^\alpha |u|^{2^*(\alpha )}dx. \end{aligned}$$

Thus

$$\begin{aligned} \Psi (u)\geqslant \frac{2+\alpha }{2(N+\alpha )}S_\alpha ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

(3.12)

Iterating the above procedure, we can construct the sequence $\{u^i\}, \{\lambda _n^i\}, \{u_n^i\}$. But the inequality (3.12) implies that only a finite number of iterations is allowed. $\square$

Lemma 3.7

Under the assumption $(\mathrm {h}_3)$, for any $\lambda >0$, we have $S_\alpha <\Upsilon \leqslant \Upsilon _\lambda$.

Proof

It is obvious that $S_\alpha \leqslant \Upsilon$. Assume that $S_\alpha =\Upsilon$. Then there exists a sequence $\{u_n\}\subset \Gamma (\mathbb {R}^N)$ satisfying

$$\begin{aligned}&\displaystyle \int _{\mathbb {R}^N}| \nabla u_n|^2dx+h(|x|)|u_n|^2dx\rightarrow S_\alpha \ \ \mathrm{as} \ \ n\rightarrow \infty ,\nonumber \\&\displaystyle \int _{\mathbb {R}^N}a_1 (|x|)|x|^\alpha |u_n|^{2^*(\alpha )}dx=\frac{1}{2}, \end{aligned}$$

(3.13)

$$\begin{aligned}&\displaystyle \int _{\mathbb {R}^N}|x|^\alpha |u_n|^{2^*(\alpha )}dx=1. \end{aligned}$$

(3.14)

By the definition of $S_\alpha$ in Theorem 3.4, the nonnegativity of h implies

$$\begin{aligned} \int _{\mathbb {R}^N}|\nabla u_n|^2dx \rightarrow S_\alpha , \ \ \ \int _{\mathbb {R}^N}|x|^\alpha |u_n|^{2^*(\alpha )}dx=1. \end{aligned}$$

Define

$$\begin{aligned} S(u_n):=\int _{\mathbb {R}^N}|\nabla u_n|^2dx. \end{aligned}$$

Applying the Ekeland principle(see [25, Theorem 8.5]), there exists Palas-Smale sequence for $S\big |_{\Gamma (\mathbb {R}^N)}$ at the level $S_\alpha$, i.e., there exist $\{\beta _n\}\subset \mathbb {R}_+$ and $\tilde{u}_n\subset D^{1,2}_r(\mathbb {R}^N)$ such that as $n\rightarrow \infty$

$$\begin{aligned}&\Vert \tilde{u}_n-u_n\Vert _{D_r^{1,2}} \rightarrow 0 , \end{aligned}$$

(3.15)

$$\begin{aligned}&S(\tilde{u}_n)\rightarrow S_\alpha , \ \ \ -\Delta \tilde{u}_n -\beta _n |x|^\alpha |\tilde{u}_n|^{2^*(\alpha )-2}\tilde{u}_n\rightarrow 0\ \mathrm{in}\ (D^{1,2}_r(\mathbb {R}^N))^*. \end{aligned}$$

(3.16)

It follows that

$$\begin{aligned}&\beta _n\rightarrow S_\alpha , \end{aligned}$$

(3.17)

$$\begin{aligned}&\Psi (v_n)\rightarrow \frac{2+\alpha }{2(N+\alpha )}S_\alpha ^{\frac{N+\alpha }{2+\alpha }},\ \ \ \Psi '(v_n) \rightarrow 0, \end{aligned}$$

(3.18)

where $v_n:=\beta _n^{\frac{1}{2^*(\alpha )-2}}\tilde{u}_n$. From (3.18), it is easy to see that $\{v_n\}$ is bounded in $D^{1,2}_r(\mathbb {R}^N)$, passing to a subsequence such that $v_n\rightharpoonup v^0$ in $D^{1,2}_r(\mathbb {R}^N)$. It follows from Theorem 3.6 that there exist k functions $v^1,v^2,\ldots , v^k \in D^{1,2}_r(\mathbb {R}^N)$ and k sequences $\{\lambda _n^i\}\subset \mathbb {R}_+$ satisfying

$$\begin{aligned} -\Delta v^i=|x|^\alpha |v^i|^{2^*(\alpha )-2}v^i \ \ \ \mathrm{in} \ \mathbb {R}^N \end{aligned}$$

(3.19)

for $i=0, 1, \ldots , k$, $\lambda _n^i \rightarrow 0\ \mathrm{or} \ \infty$ and

$$\begin{aligned}&\Psi (v_n)= \frac{2+\alpha }{2(N+\alpha )}\sum _{i=0}^k \int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx+o(1), \end{aligned}$$

(3.20)

$$\begin{aligned}&\Vert v_n\Vert _{D^{1,2}_r}^2\rightarrow \displaystyle \sum _{i=0}^k \Vert v^i\Vert _{D^{1,2}_r}^2, \end{aligned}$$

(3.21)

$$\begin{aligned}&\left\| v_n-v^0-\displaystyle \sum _{i=1}^k \left( \lambda _n^i\right) ^{\frac{2-N}{2}}v^i\left( \frac{\cdot }{\lambda _n^i}\right) \right\| _{D^{1,2}_r} \rightarrow 0. \end{aligned}$$

(3.22)

Multiplying the equation (3.19) by $(v^i)^+$ and $(v^i)^-$, combining with (3.5) and the uniqueness of solution of (3.4), for any $i=0,1, \ldots , k$, one of the following cases holds:

$$\begin{aligned}&\displaystyle \int _{\mathbb {R}^N}|\nabla v^i|^2dx=\int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx=0,\\&\displaystyle \int _{\mathbb {R}^N}|\nabla v^i|^2dx=\int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx =S_\alpha ^{\frac{N+\alpha }{2+\alpha }}, \\&\displaystyle \int _{\mathbb {R}^N}|\nabla v^i|^2dx =\int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx \geqslant 2S_\alpha ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

If $v^0\not =0$, then it follows from (3.18) and (3.20) that $k=0$. By (3.21), we get that $v_n\rightarrow v^0$ in $D^{1,2}_r(\mathbb {R}^N)$ and then $u_n\rightarrow u:=S_\alpha ^{\frac{1}{2-2^*(\alpha )}}v^0 \not =0$. By (3.15) and the first limit in (3.16), we get that $S_\alpha$ is arrived at u. The key Theorem 3.4 implies that u is positive. Combining with assumption $h\not =0$, we obtain a contradiction as

$$\begin{aligned} S_\alpha <\int _{\mathbb {R}^N}|\nabla u|^2+h(|x|)|u|^2dx = S_\alpha . \end{aligned}$$

If $v^0=0$, then it follows from (3.18) and (3.20) that $k=0,1$.

(i) The case of $k=0$. By (3.20), (3.21) and (3.15), we have $u_n \rightarrow 0$ in $D^{1,2}_r(\mathbb {R}^N)$ and this contradicts to the fact that

$$\begin{aligned}\int _{\mathbb {R}^N}|x|^\alpha |u_n|^{2^*(\alpha )}dx=1.\end{aligned}$$

(ii) The case of $k=1$. We distinguish the cases $\lambda _n^1\rightarrow 0$ and $\lambda _n^1\rightarrow \infty$. Define

$$\begin{aligned} w_n(x):=(\lambda _n^1)^{\frac{2-N}{2}}v^1\left( \frac{x}{\lambda _n^1}\right) . \end{aligned}$$

(ii-1) As $\lambda _n^1\rightarrow 0$. It follows from (3.13), (3.15) and (3.17) that

$$\begin{aligned} \int _{\mathbb {R}^N}a_1(|x|)|x|^\alpha |v_n|^{2^*(\alpha )}dx =S_\alpha ^{\frac{N+\alpha }{2+\alpha }}\left( \frac{1}{2}+o(1)\right) . \end{aligned}$$

However

$$\begin{aligned} \int _{\mathbb {R}^N}a_1(|x|)|x|^\alpha |v_n|^{2^*(\alpha )}dx= & {} \int _{\mathbb {R}^N}a_1(|x|)|x|^\alpha |w_n|^{2^*(\alpha )}dx+o(1)\ (\mathrm{by}\ (3.22))\\= & {} \int _{\mathbb {R}^N}a_1(\lambda _n^1|x|)|x|^\alpha |v^1(x)|^{2^*(\alpha )}dx+o(1)\\= & {} o(1), \end{aligned}$$

where using the fact that $\lim _{n\rightarrow \infty }a_1(\lambda _n^1|x|)=0$ a.e. on $\mathbb {R}^N$ and Lebesgue Theorem. Hence we get a contradiction.

(ii-2) As $\lambda _n^1\rightarrow \infty$. We have

$$\begin{aligned} S_\alpha ^{\frac{N+\alpha }{2+\alpha }}\left( \frac{1}{2}+o(1)\right)= & {} \int _{\mathbb {R}^N}a_1(|x|)|x|^\alpha |v_n|^{2^*(\alpha )}dx\ (\mathrm{by}\ (3.13),\ (3.15)\ \mathrm{and}\ (3.17))\\= & {} \int _{\mathbb {R}^N}a_1(|x|)|x|^\alpha |w_n|^{2^*(\alpha )}dx+o(1)\ (\mathrm{by}\ (3.22))\\= & {} \int _{\mathbb {R}^N}a_1(|x|)|x|^\alpha \left| (\lambda _n^1)^{\frac{2-N}{2}}v^1\left( \frac{x}{\lambda _n^1}\right) \right| ^{2^*(\alpha )}dx+o(1)\\= & {} \int _{\mathbb {R}^N}a_1(\lambda _n^1|x|)|x|^\alpha |v^1(x)|^{2^*(\alpha )}dx+o(1)\\= & {} \int _{\mathbb {R}^N}|x|^\alpha |v^1(x)|^{2^*(\alpha )}dx+o(1)\ (\mathrm{by Lebesgue Theorem}) \\= & {} \int _{\mathbb {R}^N}|x|^\alpha |w_n(x)|^{2^*(\alpha )}dx+o(1)\ (\mathrm{by} \ (3.13),\ (3.15)\ \mathrm{and}\ (3.17))\\= & {} \int _{\mathbb {R}^N}|x|^\alpha |v_n(x)|^{2^*(\alpha )}dx+o(1)\\= & {} S_\alpha ^{\frac{N+\alpha }{2+\alpha }}\left( 1+o(1)\right) \ (\mathrm{by } .(3.14), (3.15)) \end{aligned}$$

where using the fact that $\lim _{n\rightarrow \infty }a_\lambda (\lambda _n^1|x|)=1$ a.e. on $\mathbb {R}^N$. This leads to a contradiction. Therefore $S_\alpha <\Upsilon$.

Finally, taking $u\in \Gamma (B)$ such that $\psi _\lambda (u)=\frac{1}{2}$ and set $v_\lambda (x):=\lambda ^{\frac{N-2}{2}}u(\lambda x)$ if $|x|\leqslant \frac{1}{\lambda }$ and $v_\lambda (x)=0$ for $|x|>\frac{1}{\lambda }$. Since

$$\begin{aligned}&\psi _1(v_\lambda )=\psi _\lambda (u)=\frac{1}{2}, \ \ \varphi _1(v_\lambda )=\varphi _\lambda (u), \\&\int _{\mathbb {R}^N}|x|^\alpha |v_\lambda |^{2^*(\alpha )}dx =\int _{\{|y|\leqslant 1\}}|y|^\alpha |u|^{2^*(\alpha )}dy =1, \end{aligned}$$

it follows from the definitions of $\Upsilon$ and $\Upsilon _\lambda$ that $\Upsilon \leqslant \Upsilon _\lambda$. $\square$

Theorem 3.8

Assume $(\mathrm {h}_3)$ and $\lambda >0.$ Let $\{v_n\}\subset H_{0,r}^1(B)$ be a $\mathrm{(PS)_c}$ sequence of $\Phi$, i.e.

$$\begin{aligned} \Phi (v_n)\rightarrow c, \ \ \ \ \Phi '(v_n)\rightarrow 0 \ \ \mathrm{in} \ (H_{0,r}^1(B))^*. \end{aligned}$$

Then, passing subsequence if necessary, there exist a solution $v^0\in H_{0,r}^1(B)$ of (3.1), a finite sequence $\{u^1,u^2,\ldots , u^k\} \subset D_r^{1,2}(\mathbb {R}^N)$ of solutions for

$$\begin{aligned} -\Delta u=|x|^\alpha |u|^{2^*(\alpha )-2}u\ \ \mathrm{on}\ \mathbb {R}^N \end{aligned}$$

and k sequences $\{\lambda _n^i\}\subset \mathbb {R}_+$ such that $\lambda _n^i \rightarrow 0$ and

$$\begin{aligned}&\left\| v_n-v^0-\displaystyle \sum _{i=1}^k (\lambda _n^i)^{\frac{2-N}{2}}u^i\left( \frac{\cdot }{\lambda _n^i}\right) \right\| _{D^{1,2}_r}\rightarrow 0,\\&\Vert v_n\Vert ^2\rightarrow \Vert v^0\Vert +\displaystyle \sum _{i=1}^k \Vert u^i\Vert _{D^{1,2}_r}^2,\\&\Phi (v_n)=\Phi (v^0)+\displaystyle \sum _{i=1}^k \Psi (u^i)+o(1). \end{aligned}$$

Proof

The proof is similar with Theorem 3.6, but there need to make modify and we give a sketch proof. The boundedness of $\{v_n\}$ in $H_{0,r}^1(B)$ is obvious and which implies there exists a subsequence such that $v_n\rightharpoonup v^0$ in $H_{0,r}^1(B)$ and $v_n(x)\rightarrow v^0(x)$ a.e. on B. Combining with $(\mathrm {h}_3)$ and Lemma 3.5, it is obvious that $\Phi '(v_0)=0$ and $v_n^1:=v_n-v^0$ satisfies

$$\begin{aligned} \left. \begin{array}{ll} &{}\mathrm {i}) \ \ \ \displaystyle \lim _{n\rightarrow \infty }\left( \Vert v_n\Vert ^2-\Vert v_n^1\Vert ^2\right) =\Vert v^0\Vert ^2,\\ &{}\mathrm {ii}) \ \ \Psi (v_n^1) \rightarrow c-\Phi (v^0),\\ &{} \mathrm {iii)} \ \Psi '(v_n^1) \rightarrow 0 \ \ \mathrm{in}\ (H_{0,r}^1(B))^*. \end{array} \right. \end{aligned}$$

(3.23)

If $v_n^1\rightarrow 0$ in $L^{2^*(\alpha )}(B,|x|^\alpha )$, then the proof is complete. Otherwise we assume that

$$\begin{aligned} \int _B|x|^\alpha |v_n^1|^{2^*(\alpha )}dx>\delta \end{aligned}$$

for some $0<\delta <\left( \frac{S_\alpha }{2}\right) ^{\frac{N+\alpha }{2+\alpha }}$. Defining the Levy concentration function

$$\begin{aligned} Q_n(r):=\int _{B_r(0)}|x|^\alpha |v_n^1|^{2^*(\alpha )}dx. \end{aligned}$$

Since $Q_n(0)=0$ and $Q_n(1)>\delta$, there exists a sequence $\{\lambda _n^1\}\subset (0,1)$ such that

$$\begin{aligned} \delta =\int _{B_{\lambda _n^1}(0)}|x|^\alpha |v_n^1|^{2^*(\alpha )}dx. \end{aligned}$$

We assume that $u_n^1(x):=(\lambda _n^1)^{\frac{N-2}{2}}v_n^1(\lambda _n^1x)$ converges weakly to $u^1$ in $D^{1,2}_r(\mathbb {R}^N)$ and a.e. on $\mathbb {R}^N$. Using the Riesz-Fréchet representation theorem, Hölder inequality, inequality (3.5) and the radial lemma(see [19]), we can prove that $u^1\not =0$. Set $\lambda _n^1\rightarrow \lambda _0^1$. If $\lambda _0^1>0$, since $v_n^1\rightharpoonup 0$ in $H_{0,r}^1(B)$, we get $u_n^1\rightharpoonup 0$ in $D_r^{1,2}(\mathbb {R}^N)$, this is impossible. If $\lambda _n^1\rightarrow 0$, from (3.23), then we have $\Psi '(u^1)=0$. The sequence

$$\begin{aligned} v_n^2(x)=v_n^1(x)-(\lambda _n^1)^{\frac{2-N}{2}}u^1\left( \frac{x}{\lambda _n^1}\right) \end{aligned}$$

satisfies

$$\begin{aligned}&\mathrm {i}) \ \ \ \Vert v_n^2\Vert _{D^{1,2}_r}^2=\Vert v_n\Vert ^2-\Vert v^0\Vert ^2-\Vert u^1\Vert _{D^{1,2}_r}^2+o(1)\\&\mathrm {ii}) \ \ \Psi (v_n^2)=c-\phi (v^0)-\Psi (u^1)+o(1),\\&\mathrm {iii}) \ \Psi '(v_n^2)\rightarrow 0\ \mathrm{in}\ (D_r^{1,2}(\mathbb {R}^N))^*. \end{aligned}$$

Similar with Theorem 3.6, there exists a finite number of sequence such that the conclusions of theorem hold. $\square$

Lemma 3.9

Assume $(\mathrm {h}_3)$. Then for any $\lambda >0$, we have $S_\alpha <\Sigma _\lambda$ and

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+} \Sigma _\lambda =S_\alpha . \end{aligned}$$

(3.24)

Proof

(1) We first prove $S_\alpha <\Sigma _\lambda$ using a similar argument as in Lemma 3.7. Assume that $S_\alpha =\Sigma _\lambda$, then there exists a sequence $\{u_n\}\subset \Gamma (B)$ satisfying

$$\begin{aligned}&\int _B|\nabla u_n|^2+h\left( \frac{|x|}{\lambda }\right) \frac{1}{\lambda ^2}|u_n|^2dx\rightarrow S_\alpha , \\&\int _B a_\lambda (|x|)|x|^\alpha |u_n|^{2^*(\alpha )}dx \geqslant \frac{1}{2},\\&\int _B|x|^\alpha |u_n|^{2^*(\alpha )}dx=1. \end{aligned}$$

Since h is nonnegative and $\lambda >0$, we get

$$\begin{aligned} \int _B|\nabla u_n|^2dx \rightarrow S_\alpha ,\ \ \ \int _B|x|^\alpha |u_n|^{2^*(\alpha )}dx=1. \end{aligned}$$

Let

$$\begin{aligned} \phi (u)=\frac{1}{2}\int _B|\nabla u|^2dx-\frac{1}{2^*(\alpha )}\int _B|x|^\alpha |u|^{2^*(\alpha )}dx \end{aligned}$$

and

$$\begin{aligned} \bar{S}(u_n):=\int _B|\nabla u_n|^2dx. \end{aligned}$$

Applying the Ekeland principle(see [25, Theorem 8.5]), there exists a (PS) sequence for $\bar{S}\big |_{\Gamma (B)}$ at the level $S_\alpha$, i.e. there exist $\{\beta _n\} \subset \mathbb {R}_+$ and $\tilde{u}_n\subset H_{0,r}^1(B)$ such that

$$\begin{aligned}&\Vert \tilde{u}_n-u_n\Vert \rightarrow 0\ \mathrm{as}\ n\rightarrow \infty , \\&\bar{S}(\tilde{u}_n)\rightarrow S_\alpha , \ \ \ -\Delta \tilde{u}_n -\beta _n |x|^\alpha |\tilde{u}_n|^{2^*(\alpha )-2}\tilde{u}_n \rightarrow 0 \ \mathrm{in} \ (H_{0,r}^1(B))^*. \end{aligned}$$

Set $v_n:=\beta _n^{\frac{1}{2^*(\alpha )-2}}\tilde{u}_n$, then

$$\begin{aligned}&\beta _n\rightarrow S_\alpha , \ \ \\&\phi (v_n)\rightarrow \frac{2+\alpha }{2(N+\alpha )}S_\alpha ^{\frac{N+\alpha }{2+\alpha }}, \ \ \phi '(v_n)\rightarrow 0. \end{aligned}$$

It is easy to see that $\{v_n\}$ is bounded in $H_{0,r}^1(B)$, passing to a subsequence, that $v_n\rightharpoonup v^0$ in $H_{0,r}^1(B)$. Using Theorem 3.8 with $h=0$, there exist k functions $v^1,v^2,\ldots , v^k \in D^{1,2}_r(\mathbb {R}^N)$ such that

$$\begin{aligned} -\Delta v^i=|x|^\alpha |v^i|^{2^*(\alpha )-2}v^i \ \ \mathrm{in}\ \mathbb {R}^N, \ \ \end{aligned}$$

for $i=0,1, \ldots , k$,

$$\begin{aligned} \phi (v_n)=\phi (v^0)+\displaystyle \sum _{i=1}^k \Psi (v^i)+o(1)=\frac{2+\alpha }{2(N+\alpha )}\sum _{i=0}^k \int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx.\qquad \end{aligned}$$

(3.25)

Multiplying the equation by $(v^i)^+$, $(v^i)^-$ and using (3.5), for any $i=0,1,\ldots , k$, one of the following cases holds:

$$\begin{aligned}&\int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx=0,\\&\int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx=S_\alpha ^{\frac{N+\alpha }{2+\alpha }},\\&\int _{\mathbb {R}^N}|x|^\alpha |v^i|^{2^*(\alpha )}dx \geqslant 2S_\alpha ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

Similar with the arguments of Lemma 3.7, the case of $v^0\not =0$ is impossible, so $v^0=0$ and $k=0,1$. When $k=0$, we have $u_n\rightarrow 0$ in $H_{0,r}^1(B)$ and this is impossible since $\int _B|x|^\alpha |u_n|^{2^*(\alpha )}dx=1$. If $k=1$ and $\lambda _n^1\rightarrow 0$. Then

$$\begin{aligned} S_\alpha ^{\frac{N+\alpha }{2+\alpha }}\left( \frac{1}{2}+o(1)\right)\leqslant & {} \int _B a_\lambda (|x|)|x|^\alpha |v_n|^{2^*(\alpha )}dx\\= & {} \int _B a_\lambda (|x|)|x|^\alpha |(\lambda _n^1)^{\frac{2-N}{2}} v^1(\frac{x}{\lambda _n^1})|^{2^*(\alpha )}dx+o(1)\\\leqslant & {} C\lambda _n^1+o(1) \rightarrow 0, \end{aligned}$$

where $C>0$ is a constant. This leads to a contradiction. Therefore $S_\alpha <\Sigma _\lambda$.

(2) Now we prove the limit (3.24). Let $\epsilon >0$ and $u\in \Gamma (B)\cap C_{0,r}^\infty (B)$ be such that

$$\begin{aligned} \int _B|\nabla u|^2dx<S_\alpha +\epsilon . \end{aligned}$$

By $(\mathrm {h}_3)$ we have

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+}\int _Bh\left( \frac{|x|}{\lambda }\right) \frac{u^2}{\lambda ^2}dx=0. \end{aligned}$$

Hence we obtain

$$\begin{aligned} \lim _{\lambda \rightarrow 0^+}\varphi _\lambda (u)=\int _B|\nabla u|^2dx<S_\alpha +\epsilon . \end{aligned}$$

Since $\lim _{\lambda \rightarrow 0^+}\psi _\lambda (u)=1>\frac{1}{2}$ there exists $\delta >0$ such that for $0<\lambda <\delta$,

$$\begin{aligned} S_\alpha<\Sigma _\lambda <S_\alpha +\epsilon . \end{aligned}$$

Therefore $\lim _{\lambda \rightarrow 0^+}\Sigma _\lambda =S_\alpha$. $\square$

Proof of Theorem 3.1

By Lemma 3.7 and Lemma 3.9, there exists $\delta >0$ such that

$$\begin{aligned} S_\alpha<\Sigma _\lambda<\min \left\{ \Upsilon , 2^{\frac{2+\alpha }{N+\alpha }}S_\alpha \right\} \leqslant \Upsilon _\lambda , \ \ \ \forall \ 0<\lambda <\delta . \end{aligned}$$

(3.26)

Since $\Sigma _\lambda <\Upsilon _\lambda$, by Ekeland variational principle(see [25, Theorem 8.5]), there exists a Palais-Smale sequence for $\varphi _\lambda |_{\Gamma (B)}$ at the level $\Sigma _\lambda$. Namely, there exists a sequence $\{u_n\} \subset \Gamma (B)$ and $\{\theta _n\}\subset \mathbb {R}$ such that

$$\begin{aligned} \varphi _\lambda (u_n)\rightarrow \Sigma _\lambda ,\ \ -\Delta u_n+h\left( \frac{|x|}{\lambda }\right) \frac{u_n}{\lambda ^2}-\theta _n|x|^\alpha |u_n|^{2^*(\alpha )-2}u_n \rightarrow 0 \ \ \mathrm{in} \ (H_{0,r}^1(B))^*. \end{aligned}$$

It follows from $u_n\in \Gamma (B)$ that $\varphi (u_n)-\theta _n\rightarrow 0$ and $\theta _n\rightarrow \Sigma _\lambda$. Now define

$$\begin{aligned}v_n=\theta _n^{\frac{1}{2^*(\alpha )-2}}u_n.\end{aligned}$$

Then

$$\begin{aligned}&\left. \begin{array}{ll} \displaystyle \Phi (v_n)&{}\displaystyle =\frac{1}{2}\int _B|\nabla v_n|^2+h\left( \frac{|x|}{\lambda }\right) \frac{|v_n|^2}{\lambda ^2}dx-\frac{1}{2^*(\alpha )}\int _B |x|^\alpha |v_n|^{2^*(\alpha )}dx \\ &{} \displaystyle \rightarrow \frac{2+\alpha }{2(N+\alpha )}\Sigma _\lambda ^{\frac{N+\alpha }{2+\alpha }}, \end{array} \right. \end{aligned}$$

(3.27)

$$\begin{aligned}&-\Delta v_n+h\left( \frac{|x|}{\lambda }\right) \frac{v_n}{\lambda ^2}-|x|^\alpha |v_n|^{2^*(\alpha )-2}v_n\rightarrow 0 \ \mathrm{in} \ (H_{0,r}^1(B))^*. \end{aligned}$$

(3.28)

The relation (3.26) implies that

$$\begin{aligned} \frac{2+\alpha }{2(N+\alpha )}S_\alpha ^{\frac{N+\alpha }{2+\alpha }}<\frac{2+\alpha }{2(N+\alpha )}\Sigma _\lambda ^{\frac{N+\alpha }{2+\alpha }} <\frac{2+\alpha }{N+\alpha }S_\alpha ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

(3.29)

According to Theorem 3.8, we get the following decomposition:

$$\begin{aligned}&\displaystyle \Phi (v_n)=\Phi (v)+\frac{2+\alpha }{2(N+\alpha )}\sum _{i=1}^k\int _{\mathbb {R}^N}|x|^\alpha |w_i|^{2^*(\alpha )}dx+o(1),\nonumber \\&\displaystyle \Vert v_n\Vert ^2 \rightarrow \Vert v\Vert ^2+\displaystyle \sum _{i=1}^k \Vert w_i\Vert _{D^{1,2}_r}^2, \end{aligned}$$

(3.30)

where $w_i\in D^{1,2}_r(\mathbb {R}^N)$ is the solutions of

$$\begin{aligned} -\Delta w=|x|^\alpha |w|^{2^*(\alpha )-2} w\ \ \mathrm{in}\ \mathbb {R}^N \end{aligned}$$

(3.31)

and $v\in H_{0,r}^1(B)$ satisfies

$$\begin{aligned} -\Delta v+h\left( \frac{|x|}{\lambda }\right) \frac{v}{\lambda ^2} =|x|^\alpha |v|^{2^*(\alpha )-2}v. \end{aligned}$$

(3.32)

Hence

$$\begin{aligned} \displaystyle \Phi (v)+\frac{2+\alpha }{2(N+\alpha )}\sum _{i=1}^k\int _{\mathbb {R}^N}|x|^\alpha |w_i|^{2^*(\alpha )}dx=\frac{2+\alpha }{2(N+\alpha )}\Sigma _\lambda ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

(3.33)

Multiplying the equation (3.31) by $w_i^+$, $w_i^-$ and using (3.5), for any $i=0,1,\ldots , k$, one of the following cases holds:

$$\begin{aligned}&\displaystyle w_i=0 \ \Rightarrow \ \int _{\mathbb {R}^N}|x|^\alpha |w_i|^{2^*(\alpha )}dx=0, \end{aligned}$$

(3.34)

$$\begin{aligned}&\displaystyle w_i \ \mathrm{has\ a\ constant \ sign \ and}\ \int _{\mathbb {R}^N}|x|^\alpha |w_i|^{2^*(\alpha )}dx=S_\alpha ^{\frac{N+\alpha }{2+\alpha }}, \end{aligned}$$

(3.35)

$$\begin{aligned}&\displaystyle w_i \ \mathrm{changes\ a\ sign \ and} \ \int _{\mathbb {R}^N}|x|^\alpha |w_i|^{2^*(\alpha )}dx\geqslant 2S_\alpha ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

(3.36)

Similarly we have

$$\begin{aligned}&\displaystyle v=0 \ \Rightarrow \ \Phi (v)=0, \end{aligned}$$

(3.37)

$$\begin{aligned}&\displaystyle v \ \mathrm{has \ a \ constant \ sign \ and} \ \Phi (v)\geqslant \frac{2+\alpha }{2(N+\alpha )}S_\alpha ^{\frac{N+\alpha }{2+\alpha }}, \end{aligned}$$

(3.38)

$$\begin{aligned}&\displaystyle v \ \mathrm{changes \ a \ sign \ and } \ \Phi (v)\geqslant \frac{2+\alpha }{N+\alpha }S_\alpha ^{\frac{N+\alpha }{2+\alpha }}. \end{aligned}$$

(3.39)

It follows from (3.29) and (3.33) that the only possible case is (3.34) together (3.38). Combining with (3.30), we know $v_n\rightarrow v$ in $H_{0,r}^1(B)$. By (3.27) and (3.38), v is a constant sign solution and $\Phi (v)=\frac{2+\alpha }{2(N+\alpha )} \Sigma _\lambda ^{\frac{N+\alpha }{2+\alpha }}$, moreover by structure of (3.32), we can obtain the nonnegative solution v. $\square$