A modified singular Trudinger–Moser inequality

Abstract

Let $\varOmega \subset \mathbb{R}^{2}$ be a smooth bounded domain, $W_{0}^{1,2}(\varOmega )$ be the standard Sobolev space. Assuming certain conditions on a function $g:\mathbb{R}\rightarrow \mathbb{R}$, we derive a modified singular Trudinger–Moser inequality, which was originally established by Adimurthi and Sandeep (Nonlinear Differ. Equ. Appl. 13:585–603, 2007), namely,

$$ \sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{2}\leq 1} \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e ^{4\pi (1-\gamma ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx, $$

(1)

where $0<\gamma <1$. Following Yang and Zhu (J. Funct. Anal. 272:3347–3374, 2017), we prove that the extremal functions for the supremum in (1) exist. The proof is based on a blow-up analysis.

1 Introduction

Let Ω be a smooth bounded domain in $\mathbb{R}^{2}$, and $W_{0}^{1,2}(\varOmega )$ be the completion of $C^{\infty }_{0}(\varOmega )$ under the norm $\|u\|_{W_{0}^{1,2}(\varOmega )}=(\int _{\varOmega }| \nabla u|^{2}\,dx)^{{1}/{2}}$. For $1\leq p<2$, the standard Sobolev embedding theorem states that $W_{0}^{1,p}(\varOmega )\hookrightarrow L ^{q}(\varOmega )$ for all $1< q\leq {2p}/{(2-p)}$; while if $p>2$, we have $W_{0}^{1,p}(\varOmega )\hookrightarrow C^{0}(\overline{\varOmega })$. As a borderline of the Sobolev embeddings, the classical Trudinger–Moser inequality [21,22,23, 26, 33] says

$$ \sup_{u\in W_{0}^{1,2}(\varOmega ),\|\nabla u\|_{2}\leq 1} \int _{\varOmega }e ^{\alpha u^{2}}\,dx < +\infty ,\quad \forall \alpha \leq 4\pi . $$

(2)

Moreover, these integrals are still finite for any $\alpha >4\pi$, but the supremum is infinity. Here and in the sequel, for any real number $q\geq 1$, $\|\cdot \|_{q}$ denotes the $L^{q}(\varOmega )$-norm with respect to the Lebesgue measure.

A function $u_{0}$ is called an extremal function for the Trudinger–Moser inequality (2) if $u_{0}$ belongs to $W_{0}^{1,2}(\varOmega )$, $\|\nabla u_{0}\|_{2}\leq 1$ and

$$ \int _{\varOmega }e^{\alpha u_{0}^{2}}\,dx= \sup_{u\in W_{0}^{1,2}(\varOmega ),\|\nabla u\|_{2}\leq 1} \int _{\varOmega }e ^{\alpha u^{2}}\,dx. $$

An interesting question on Trudinger–Moser inequalities is whether or not extremal functions exist. The existence of extremal functions for (2) was obtained by Carleson–Chang [5] when Ω is a unit ball, and by Struwe [24] when Ω is close to the ball in the sense of measure. Then Flucher [12] extended this result when Ω is a general bounded smooth domain in $\mathbb{R}^{2}$. Later, Lin [16] generalized the existence result when Ω is an arbitrary dimensional domain. For recent developments, we refer the reader to Yang [28].

Using a rearrangement argument and a change of variables, Adimurthi–Sandeep [2] generalized the Trudinger–Moser inequality (1) to a singular version as follows:

$$\begin{aligned} \sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{2}\leq 1} \int _{\varOmega }\frac{e ^{4\pi (1-\gamma ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx< \infty . \end{aligned}$$

(3)

This inequality is also sharp in the sense that all integrals are still finite when $\alpha >1-\gamma$, but the supremum is infinity. Clearly, if $\gamma =0$, (3) reduces to (1). Following the lines of Flucher [12], in Csato and Roy [9], they adopt the concentration–compactness alternative by Lions [17] and deduced that the existence of extremals for such singular functionals. Later, (3) was extend to the entire $\mathbb{R}^{N}$ by Adimurthi and Yang [4]. Meanwhile, Souza and do Ó modified the singular to another version in $\mathbb{R}^{N}$ in [10]. When Ω is the unit ball $\mathbb{B}$, (3) was improved by Yuan and Zhu [32]. Similarly, an analog is also be proved by Yuan and Huang by using the method of symmetrization in [31]. Such singular Trudinger–Moser inequalities play an important role in the study of partial differential equations and conformal geometry; see [2, 4, 10, 14, 27] and [6] for details.

Recently, using a method of energy estimates in [19], Mancini–Martinazzi [20] reproved Carleson–Chang’s result. For applications of this method, we refer the reader to Yang [29]. Using the same idea, they proved that the supremum

$$ \sup_{u\in W_{0}^{1,2}(\mathbb{B}),\|\nabla u\|_{2}\leq 1} \int _{\mathbb{B}}\bigl(1+g(u)\bigr)e^{4\pi u^{2}}\,dx $$

(4)

can be achieved for certain smooth function $g:\mathbb{R}\rightarrow \mathbb{R}$, where $\mathbb{B}$ is a unit ball. On the other hand, in Yang and Zhu [30], one studied the following singular form:

$$ \sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{1,\alpha }\leq 1} \int _{\varOmega }\frac{e^{\beta u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx, $$

(5)

and they verified there exists some function $u_{0}$ to achieve this supremum for any $\beta <4\pi (1-\gamma )$, where

$$ \Vert u \Vert _{1,\alpha }= \biggl( \int _{\varOmega } \vert \nabla u \vert ^{2}\,dx-\alpha \int _{\varOmega }u^{2}\,dx \biggr)^{1/2}, $$

and α satisfies

$$ \alpha < \inf_{u\in W_{0}^{1,2}(\varOmega ),u\not \equiv 0} \frac{ \Vert \nabla u \Vert _{2}^{2}}{ \Vert u \Vert _{2}^{2}}. $$

Motivated by the above results, in this paper, we make a combination of (4) and (5) under the case $\alpha =0$ to discuss a new version of the singular Trudinger–Moser inequality. We are aim to prove two main results: One is to explain the new supremum is finite; the other is to discuss the existence of extremals for such functionals. In our proof, unlike the previous energy estimate procedure in [19, 20, 29], we mainly employ the method of blow-up analysis as in [11, 14, 15, 18] to prove the supremum in the following (9) can be achieved. Based on Mancini–Martinazzi [20] (see pages 3 and 4), we assume the function g in (9) satisfies

$$ \begin{aligned} &g\in C^{1}(\mathbb{R}), \qquad \inf_{\mathbb{R}}g>-1, \qquad g(-t)=g(t), \\ &\lim_{|t|\rightarrow \infty }t^{2}g(t)=0, \qquad g^{\prime }(t)>0\quad (\forall t>0). \end{aligned} $$

(6)

In the proof, we derive

$$ -\Delta u_{\varepsilon } =\frac{1}{\lambda _{\varepsilon }}\biggl(1+g(u_{ \varepsilon })+ \frac{g'(u_{\varepsilon })}{8\pi (1-\gamma -\varepsilon )u_{\varepsilon }}\biggr) u_{\varepsilon } e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}=\frac{1}{\lambda _{\varepsilon }} \bigl(1+h(u _{\varepsilon }) \bigr) u_{\varepsilon } e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}} $$

for some $\lambda _{\varepsilon }\in \mathbb{R}$, where we set

$$ h(t):=g(t)+\frac{g'(t)}{8\pi (1-\gamma -\varepsilon )t},\quad t\in \mathbb{R} \setminus \{0\}. $$

(7)

We further assume

$$ \inf_{(0,+\infty )}h(t)>-1, \qquad \sup _{(0,+\infty )}h(t)< +\infty , \quad \text{and} \quad \lim _{t\rightarrow \infty }t^{2} h(t)=0. $$

(8)

Comparing the conditions on the function g in Mancini–Martinazzi [20], one can see some differences. In this note, we assume $g^{\prime }(t)>0$ ($\forall t>0$), which is used in the Lemma 4. Moreover, the assumptions in (6) and (8) implies that $\lim_{|t|\rightarrow \infty } g(t)=0$ in [20]. Our main conclusion can be stated as the following two theorems, respectively.

Theorem 1

Let Ω be a smooth bounded domain in $\mathbb{R}^{2}$ and $W_{0}^{1,2}(\varOmega )$ be the usual Sobolev space. Let $0<\gamma <1$ be fixed. Suppose $g\in C^{1}(\mathbb{R})$ satisfies the hypotheses in (6) and (8). Then the supremum

$$ \varLambda _{4\pi (1-\gamma )}:= \sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{2}\leq 1} \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e ^{4\pi (1-\gamma ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx< \infty . $$

(9)

Theorem 2

Let Ω be a smooth bounded domain in $\mathbb{R}^{2}$ and $W_{0}^{1,2}(\varOmega )$ be the usual Sobolev space. Let $0<\gamma <1$ be fixed. Suppose $g\in C^{1}(\mathbb{R})$ satisfies the hypotheses in (6) and (8). Then, for any $\beta \leq 4\pi (1-\gamma )$, the supremum

$$ \sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{2}\leq 1} \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e ^{\beta u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx $$

can be attained by some function $u_{0}\in W_{0}^{1,2}(\varOmega )\cap C _{\mathrm{loc}}^{1}(\overline{\varOmega }\setminus \{0\})\cap C^{0}(\overline{ \varOmega })$.

In order to prove the critical singular Trudinger–Moser inequality, we firstly discuss the existence of extremal functions for a subcritical one, which is based on a direct method variation. We derive a different Euler–Lagrange equation on which the analysis is performed. The essential problem is the presence of the function g. To meet the necessary of our proof, we assume g satisfies certain conditions. Then following Yang and Zhu [30], we define maximizing sequences of functions by using a more delicate scaling. The existence of singular term $|x|^{-2\gamma }$ with $0<\gamma <1$ causes exact asymptotic behavior of certain maximizing sequence near the blow-up point. Unlike in [28], we employ two different classification theorems of Chen and Li [7, 8] to get the desired bubble. And our method in dealing with the bubble is also different from Yang–Zhu [30] because of the function g. We refer to Adimurthi and Druet [1], Carleson–Chang [5], Li [15], Struwe [24], Adimurthi and Struwe [3], Iula and Mancini [13], Yang [28], Lu and Yang [18], respectively.

2 Proof of Theorem 1

We divide the proof into several steps as follows.

2.1 Existence of maximizers for $\varLambda _{4\pi (1-\gamma -\varepsilon )}$ and the Euler–Lagrange equation

In this subsection, we shall prove that maximizers for the subcritical singular Trudinger–Moser functionals exist.

Proposition 3

For any $0<\varepsilon <1-\beta$, there exists some $u_{\varepsilon }\in W_{0}^{1,2}(\varOmega )\cap C_{\mathrm{loc}}^{1}(\overline{\varOmega }\setminus \{0\})\cap C^{0}(\overline{\varOmega })$ satisfying $\|\nabla u\|_{2}=1$ and

$$ \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }} \,dx= \varLambda _{4\pi (1-\gamma -\varepsilon )}:= \sup_{\substack{u\in W_{0}^{1,2}(\varOmega ), \\ \Vert \nabla u \Vert _{2}\leq 1}} \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx. $$

(10)

Proof

This is based on a direct method of variation. For any $0<\beta <1$, let $0<\varepsilon <1-\gamma$ be fixed. We take a sequence of functions $u_{j}\in W_{0}^{1,2}(\varOmega )$ satisfying $\|\nabla u_{j}\|_{2}\leq 1$ and, as $j\rightarrow \infty$,

$$ \lim_{j\rightarrow \infty } \int _{\varOmega }\bigl(1+g(u_{j})\bigr) \frac{e^{4 \pi (1-\gamma -\varepsilon ) u_{j}^{2}}}{ \vert x \vert ^{2\gamma }}\,dx = \varLambda _{4\pi (1-\gamma -\varepsilon )}. $$

(11)

Since $u_{j}$ is bounded in $W_{0}^{1,2}(\varOmega )$, there exists some $u_{\varepsilon }\in W_{0}^{1,2}(\varOmega )$ such that up to a subsequence, assuming

$$ \begin{aligned}[b] &u_{j}\rightharpoonup u_{\varepsilon } \quad \text{weakly in } W_{0}^{1,2}(\varOmega ), \\ &u_{j}\rightarrow u_{\varepsilon }\quad \text{strongly in } L^{p}( \varOmega ), \forall p\geq 1, \\ &u_{j}\rightarrow u_{\varepsilon }\quad \text{a.e. in } \varOmega . \end{aligned} $$

Since

$$ 0\leq \int _{\varOmega } \vert \nabla u_{\varepsilon } \vert ^{2}\,dx \leq \limsup_{j\rightarrow \infty } \biggl( \int _{\varOmega } \vert \nabla u_{\varepsilon } \vert ^{2}\,dx \biggr)^{\frac{1}{2}} \biggl( \int _{\varOmega } \vert \nabla u_{j} \vert ^{2}\,dx \biggr)^{\frac{1}{2}}, $$

we have $0\leq \|\nabla u_{\varepsilon }\|_{2}\leq 1$. Note that

$$ \begin{aligned}[b] \int _{\varOmega } \bigl\vert \nabla (u_{\varepsilon }-u_{j}) \bigr\vert ^{2}\,dx &= \int _{\varOmega } \vert \nabla u_{\varepsilon } \vert ^{2}\,dx- \int _{\varOmega } \vert \nabla u_{j} \vert ^{2}+o _{j}(1) \\ &\leq 1- \int _{\varOmega } \vert \nabla u_{\varepsilon } \vert ^{2}+o_{j}(1). \end{aligned} $$

(12)

Following Hölder’s inequality, for any $1< p\leq \frac{1}{\gamma }$, $\delta >0$, $w>1$ and $w'=w/(w-1)$, we have

$$ \begin{aligned}[b] \int _{\varOmega }\bigl(1+g(u_{j})\bigr)^{p} \frac{1}{ \vert x \vert ^{2\gamma p}}e^{4\pi (1- \gamma -\varepsilon )p u_{j}^{2}}\,dx &\leq C \biggl( \int _{\varOmega }\frac{1}{ \vert x \vert ^{2 \gamma p}}e^{4\pi (1-\gamma -\varepsilon )p(1+\delta )w(u_{j}-u_{ \varepsilon })^{2}}\,dx \biggr)^{\frac{1}{w}} \\ &\quad {}\times \biggl( \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma p}}e^{4\pi (1- \gamma -\varepsilon )p(1+\frac{1}{4\delta })w'u_{\varepsilon }^{2} }\,dx \biggr)^{\frac{1}{w'}}. \end{aligned} $$

(13)

When p, $1+\delta$ and s are sufficiently close to 1, we have

$$ (1-\gamma -\varepsilon )p(1+\delta )w+\gamma w p< 1. $$

(14)

Combining (12), (13) and (14), we have by the singular Trudinger–Moser inequality (3)

$$ \bigl(1+g(u_{\varepsilon })\bigr) \vert x \vert ^{-2\gamma }e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}} \quad \text{is bounded in } L ^{p}(\varOmega ), $$

for some $p>1$. Note that

$$ \begin{aligned}[b] & \biggl\vert \bigl(1+g(u_{j}) \bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{j}^{2}}}{ \vert x \vert ^{-2 \gamma }} -\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{-2\gamma }} \biggr\vert \\ &\quad \leq C \vert x \vert ^{-2\gamma }\bigl(e^{4\pi (1-\gamma -\varepsilon ) u_{j}^{2}}+e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\bigr) \bigl\vert u_{j}^{2}-u _{\varepsilon }^{2} \bigr\vert , \\ &\qquad {}+ \vert x \vert ^{-2\gamma }\max \bigl\{ g'(u_{j}),g'(u_{\varepsilon }) \bigr\} \vert u_{j}-u_{ \varepsilon } \vert e^{4\pi (1-\gamma -\varepsilon ) u_{j}^{2}}. \end{aligned} $$

(15)

Since $u_{j}\rightarrow u_{\varepsilon }$ strongly in $L^{p}(\varOmega )$ for any $p>1$, in view of (6) and (8), we can conclude from (15) that

$$ \int _{\varOmega }\bigl(1+g(u_{j})\bigr) \vert x \vert ^{-2\gamma }e^{4\pi (1-\gamma -\varepsilon ) u_{j}^{2}}\,dx\rightarrow \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr) \vert x \vert ^{-2 \gamma }e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\,dx, $$

as $j\rightarrow \infty$. This together with (11) immediately leads to (10). Obviously $u_{\varepsilon }\not \equiv 0$. If $\|\nabla u_{\varepsilon }\|_{2}<1$, set $\widetilde{u}_{\varepsilon }=\frac{u_{\varepsilon }}{\|\nabla u_{\varepsilon }\|_{2}}$, then we obtain $\| \nabla \widetilde{u}_{\varepsilon }\|_{2}=1$. Since $0\leq u_{\varepsilon }<\widetilde{u}_{\varepsilon }$ and $u_{\varepsilon }\not \equiv 0$, it follows from (6) that

$$ \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }} \,dx< \int _{\varOmega }\bigl(1+g( \widetilde{u_{\varepsilon }})\bigr) \frac{e^{4\pi (1-\gamma -\varepsilon ) \widetilde{u}_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx\leq \varLambda _{4\pi (1-\gamma -\varepsilon )}, $$

which contradicts (10). Consequently, $\|\nabla u_{\varepsilon }\|_{2}=1$ holds. Furthermore, one can also check that $|u_{\varepsilon }|$ attains the supremum $\varLambda _{4\pi (1-\gamma -\varepsilon )}$. Thus, $u_{\varepsilon }$ can be chosen so that $u_{\varepsilon } \geq 0$. It is not difficult to see that $u_{\varepsilon }$ satisfies the following Euler–Lagrange equation:

$$ \textstyle\begin{cases} -\Delta u_{\varepsilon } =\lambda _{\varepsilon }^{-1} \vert x \vert ^{-2\gamma }(1+h(u _{\varepsilon }))u_{\varepsilon } e^{4\pi (1-\gamma -\varepsilon ) u _{\varepsilon }^{2}} &\text{in }\varOmega \subset \mathbb{R}^{2}, \\ u_{\varepsilon }\geq 0, \qquad \Vert \nabla u_{\varepsilon } \Vert _{2}=1& \text{in }\varOmega \subset \mathbb{R}^{2}, \\ \lambda _{\varepsilon }=\int _{\varOmega } \vert x \vert ^{-2\gamma }(1+h(u_{\varepsilon })) u_{\varepsilon }^{2}e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\,dx, \end{cases} $$

(16)

where $h(x)$ is defined as in (7). □

2.1.1 The case when $u_{\varepsilon }$ is uniformly bounded in Ω

The proof of Theorem 2 will be ended if we can find some $u_{0}\in W _{0}^{1,2}(\varOmega )\cap C_{\mathrm{loc}}^{1}(\overline{\varOmega }\setminus \{0\}) \cap C^{0}(\overline{\varOmega })$ satisfying $\| \nabla u_{0}\|_{2}=1$ and

$$ \int _{\varOmega }\bigl(1+g(u_{0})\bigr)\frac{e^{4\pi (1-\gamma ) u_{0}^{2}}}{ \vert x \vert ^{2 \gamma }} \,dx=\sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{2}\leq 1} \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e^{4\pi (1-\gamma ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx. $$

(17)

Since $u_{\varepsilon }$ is bounded in $W_{0}^{1,2}(\varOmega )$, we assume without loss of generality

$$ \begin{aligned} &u_{\varepsilon }\rightharpoonup u_{0} \quad \text{weakly in } W_{0}^{1,2}(\varOmega ), \\ &u_{\varepsilon }\rightarrow u_{0} \quad \text{strongly in } L ^{p}(\varOmega ), \forall p\geq 1, \\ &u_{\varepsilon }\rightarrow u_{0} \quad \text{a.e. in } \varOmega . \end{aligned} $$

(18)

Let $c_{\varepsilon }=u_{\varepsilon }(x_{\varepsilon })=\max_{\varOmega }u_{\varepsilon }$. If $c_{\varepsilon }$ is bounded, for any $u\in W_{0}^{1,2}(\varOmega )$ with $u\geq 0$, $\| \nabla u_{0}\|_{2}=1$, together with Lebesgue dominated convergence theorem gives

$$ \begin{aligned}[b] \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e^{4\pi (1-\gamma ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx &=\lim _{\varepsilon \rightarrow 0} \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u^{2}}}{ \vert x \vert ^{2\gamma }} \,dx \\ &\leq \lim_{\varepsilon \rightarrow 0} \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2 \gamma }} \,dx \\ &= \int _{\varOmega }\bigl(1+g(u_{0})\bigr)\frac{e^{4\pi (1-\gamma ) u_{0}^{2}}}{ \vert x \vert ^{2 \gamma }} \,dx. \end{aligned} $$

(19)

By the arbitrariness of $u\in W^{1,2}_{0}(\varOmega )$, we conclude that $u_{0}$ is the desired maximizer when $u_{\varepsilon }$ is uniformly bounded in Ω. Applying elliptic estimates to its Euler–Lagrange equation, one can deduce that $u_{0}\in W_{0}^{1,2}(\varOmega )\cap C _{\mathrm{loc}}^{1}(\overline{\varOmega }\setminus \{0\})\cap C^{0}(\overline{ \varOmega })$. And then (17) follows immediately.

2.2 Blowing up analysis

In this subsection, as in [1, 17], we will use the blow-up analysis to understand the asymptotic behavior of the maximizers $u_{\varepsilon }$. Assume $c_{\varepsilon }=u_{\varepsilon }(x_{ \varepsilon })\rightarrow \infty$ and we distinguish two cases to proceed.

Case 1. If $u_{0}\not \equiv 0$, the supremum in (9) can be attained by $u_{0}$ without difficulty. And the proof will just be divided into several simple steps.

Step 1. A similar estimate as in (13), one can easily check that $\frac{(1+g(u_{\varepsilon }))}{|x|^{2\gamma }}e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}$ is bounded in $L^{p}(\varOmega )$ ($p>1$).

Step 2. By the mean value theorem and the Hölder inequality, we have

$$ \lim_{\varepsilon \rightarrow 0} \int _{\varOmega } \vert x \vert ^{-2\gamma }e^{4 \pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\,dx= \int _{\varOmega } \vert x \vert ^{-2 \gamma }e^{4\pi (1-\gamma ) u_{0}^{2}} \,dx. $$

Step 3. Based on the above steps, one can easily check that

$$ \begin{aligned}[b] & \int _{\varOmega } \bigl\vert \bigl(1+g(u_{\varepsilon })\bigr) \vert x \vert ^{-2\gamma }e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon }^{2}}-\bigl(1+g(u_{0})\bigr) \vert x \vert ^{-2\gamma }e^{4\pi (1-\gamma ) u_{0}^{2}} \bigr\vert \,dx \\ &\quad \leq \bigl\vert g(u_{0})+1 \bigr\vert \int _{\varOmega } \bigl( \vert x \vert ^{-2\gamma }e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon }^{2}}- \vert x \vert ^{-2\gamma }e^{4\pi (1- \gamma ) u_{0}^{2}} \bigr)\,dx \\ &\qquad {}+ \int _{\varOmega } \vert x \vert ^{-2\gamma } e^{4\pi (1-\gamma -\varepsilon ) u _{\varepsilon }^{2}} \bigl\vert g(u_{\varepsilon })-g(u_{0}) \bigr\vert \,dx \\ &\quad =o_{\varepsilon }(1). \end{aligned} $$

Thus, we arrive at the conclusion that

$$ \lim_{\varepsilon \rightarrow 0} \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr) \vert x \vert ^{-2 \gamma }e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\,dx= \int _{\varOmega }\bigl(1+g(u_{0})\bigr) \vert x \vert ^{-2\gamma }e^{4\pi (1-\gamma ) u_{0} ^{2}}\,dx. $$

This together with (17) gives the desired result.

Case 2. If $u_{0}\equiv 0$, in view of Eq. (16), it is important to figure out whether $\lambda _{\varepsilon }$ has a positive lower bound or not. For this purpose, we have the following.

Lemma 4

Let $\lambda _{\varepsilon }$ be as in (16). Then we have $\liminf_{\varepsilon \rightarrow 0}\lambda _{\varepsilon }>0$.

Proof

By an inequality $e^{t^{2}}\leq 1+t^{2}e^{t^{2}}$ for $t\geq 0$, it follows from (6) and (7) that

$$ \begin{aligned}[b] \lambda _{\varepsilon } &\geq \frac{1}{4\pi (1-\gamma -\varepsilon )} \int _{\varOmega }\bigl(1+h(u_{\varepsilon })\bigr)\frac{(e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}-1)}{ \vert x \vert ^{2\gamma }} \,dx \\ &\geq \frac{1}{4\pi (1-\gamma -\varepsilon )}\biggl( \int _{\varOmega }\bigl(1+g(u _{\varepsilon })\bigr) \frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx- \int _{\varOmega }\frac{(1+g(u_{\varepsilon }))}{ \vert x \vert ^{2 \gamma }}\,dx \\ & \quad {}+ \int _{\varOmega } \frac{g'(u_{\varepsilon })}{8\pi (1-\gamma - \varepsilon ) \vert x \vert ^{2\gamma }u_{\varepsilon }} \bigl(e ^{4\pi (1-\gamma - \varepsilon )u_{\varepsilon }^{2}}-1\bigr) \,dx\biggr) \\ &\geq \frac{1}{4\pi (1-\gamma -\varepsilon )} \biggl(\int _{\varOmega }\bigl(1+g(u _{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }} \,dx- \int _{\varOmega }\frac{(1+g(u_{\varepsilon }))}{ \vert x \vert ^{2 \gamma }}\,dx \biggr). \end{aligned} $$

This together with (10) leads to

$$ \liminf_{\varepsilon \rightarrow 0}\lambda _{\varepsilon }\geq \frac{1}{4 \pi (1-\gamma )} \biggl(\varLambda _{4\pi (1-\gamma )}- \int _{\varOmega }\frac{(1+g(0))}{ \vert x \vert ^{2 \gamma }}\,dx \biggr)>0. $$

Or equivalently, we have

$$ \frac{1}{\lambda _{\varepsilon }}\leq C. $$

(20)

Therefore, $\frac{1}{\lambda _{\varepsilon }}$ is uniformly bounded in Ω. This ends the proof of the lemma. □

2.2.1 Energy concentration phenomenon

Using the same argument as the one in step 2 of [28], we get the following concentration phenomenon, which is crucial in our blow-up analysis.

Proposition 5

For the function sequence $\{u_{\varepsilon }\}$, we have $u_{\varepsilon }\rightharpoonup 0$ weakly in $W^{1,2}_{0}(\varOmega )$ and $u_{\varepsilon }\rightarrow 0$ strongly in $L^{q}(\varOmega )$ for any $q>1$. Moreover, $|\nabla u_{\varepsilon }|^{2} \,dx\rightharpoonup \delta _{0}$ weakly in a sense of measure, where $\delta _{0}$ is the usual Dirac measure centered at the point 0.

Proof

Since $\|\nabla u_{\varepsilon }\|_{2}=1$, we have the same assumptions as in (18). Observe that

$$ \int _{\varOmega } \bigl\vert \nabla (u_{\varepsilon }-u_{0}) \bigr\vert ^{2}\,dx=1- \int _{\varOmega } \vert \nabla u_{0} \vert ^{2}\,dx+o(1). $$

(21)

Suppose $u_{0}\not \equiv 0$. In view of (21) and an obvious analog of (13), it follows that

$$ \bigl(1+g(u_{\varepsilon })\bigr) \vert x \vert ^{-2\gamma }e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}} \quad \text{is bounded in } L^{q}( \varOmega ), $$

for some $q>1$. Then applying elliptic estimates to (18), one can deduce that $u_{\varepsilon }$ is bounded in $W_{0}^{2,q}(\varOmega )$. Together with Sobolev embedding results, we conclude $u_{\varepsilon }$ is bounded in $C^{0}(\overline{\varOmega })$, which contradicts $c_{\varepsilon }\rightarrow \infty$. Therefore $u_{0}\equiv 0$ and (21) becomes

$$ \int _{\varOmega } \vert \nabla u_{\varepsilon } \vert ^{2}\,dx=1+o_{\varepsilon }(1). $$

(22)

We next prove $|\nabla u_{\varepsilon }|^{2} \,dx\rightharpoonup \delta _{x_{0}}$. If the statements were false, suppose $|\nabla u_{ \varepsilon }|^{2} \,dx\rightharpoonup \eta$ in a sense of measure. In view of $\eta \neq \delta _{x_{0}}$, there exists $r_{0}>0$ such that

$$ \lim_{\varepsilon \rightarrow 0} \int _{B_{r_{0}}(x_{0})} \vert \nabla u_{ \varepsilon } \vert ^{2} \,dx\leq \frac{\eta +1}{2}< 1. $$

In view of (22) and $u_{0}\equiv 0$, we can choose a cut-off function $\phi \in C^{1}_{0}(B_{r_{0}}(x_{0}))$, which is equal to 1 on $B_{r_{0}/2}(x_{0})$, then it follows that

$$ \limsup_{\varepsilon \rightarrow 0} \int _{B_{r_{0}}(x_{0})} \bigl\vert \nabla ( \phi u_{\varepsilon }) \bigr\vert ^{2} \,dx< 1. $$

By the singular Trudinger–Moser inequality (3), one sees that $(1+g(\phi u_{\varepsilon }))\frac{e^{4\pi (1-\gamma -\varepsilon ) ( \phi u_{\varepsilon })^{2}}}{|x|^{2\gamma }}$ is bounded in $L^{r}(B_{r_{0}}(x_{0}))$ for some $r>1$. Applying elliptic estimates to (16), one gets $u_{\varepsilon }$ is uniformly bounded in Ω, which contradicts $c_{\varepsilon }\rightarrow \infty$ again. Therefore $|\nabla u_{\varepsilon }|^{2} \,dx\rightharpoonup \delta _{x_{0}}$. Moreover, we get $u_{\varepsilon }\rightarrow 0$ in $C^{1}_{\mathrm{loc}}(\overline{\varOmega }\setminus \{0, x_{0}\})\cap C^{0}_{\mathrm{loc}}(\overline{ \varOmega }\setminus \{ x_{0}\})$.

In fact, we have $x_{0}=0$. Set $r_{0}=|x_{0}|/2$. Note that $\lambda _{\varepsilon }^{-1}|x|^{-2\gamma }(1+h(u_{\varepsilon }))u _{\varepsilon } e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}$ is bounded in $L^{q_{1}}(B_{r_{0}}(0))$ for some $q_{1}>1$. When $|x|>r_{0}$, by the classical Trudinger–Moser inequality (2), we recognize $\lambda _{\varepsilon }^{-1}|x|^{-2\gamma }(1+h(u_{ \varepsilon }))u_{\varepsilon } e^{4\pi (1-\gamma -\varepsilon ) u _{\varepsilon }^{2}}$ is bounded in $L^{q_{2}}(\varOmega \setminus B_{r _{0}}(0))$ for some $q_{2}>1$. Choose $q= \min \{q_{1}, q_{2}\}>1$, and we conclude $\lambda _{\varepsilon }^{-1}|x|^{-2\gamma }(1+h(u_{\varepsilon }))u_{\varepsilon } e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon } ^{2}}$ is bounded in $L^{q}(\varOmega )$. Then the elliptic estimate on the Euler–Lagrange equation (16) implies that $c_{\varepsilon }$ is bounded, which also makes a contradiction. Thus, we complete the proof of the proposition. □

2.2.2 Asymptotic behavior of $u_{\varepsilon }$ near the concentration point

Let

$$ r_{\varepsilon }=\sqrt{\lambda _{\varepsilon }}c_{\varepsilon }^{-1}e ^{-2\pi (1-\gamma -\varepsilon )c_{\varepsilon }^{2}}. $$

(23)

For any $0<\delta <1-\gamma$, in view of (8), we have by using the Hölder inequality and the singular Trudinger–Moser inequality (3),

$$ \begin{aligned}[b] \lambda _{\varepsilon } &= \int _{\varOmega } \vert x \vert ^{-2\gamma } \bigl(1+h(u_{\varepsilon })\bigr) u_{\varepsilon }^{2}e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\,dx \\ &\leq e^{4\pi \delta c_{\varepsilon }^{2}} \int _{\varOmega } \vert x \vert ^{-2 \gamma } \bigl(1+h(u_{\varepsilon })\bigr) u_{\varepsilon }^{2}e ^{4\pi (1- \gamma -\varepsilon -\delta ) u_{\varepsilon }^{2}}\,dx \\ &\leq C e^{4\pi \delta c_{\varepsilon }^{2}} \end{aligned} $$

for some constant C depending only on δ. This leads to

$$ r_{\varepsilon }^{2} e^{4\pi \mu c_{\varepsilon }^{2}}\leq C c_{ \varepsilon }^{-2}e^{4\pi (\delta +\mu )}e ^{-4\pi (1-\gamma -\varepsilon ) c_{\varepsilon }^{2}} \rightarrow 0, \quad \text{for } \forall 0< \mu < 1-\gamma , $$

(24)

as $\varepsilon \rightarrow 0$. To characterize the blow-up behavior more exactly, we need to divide the process into two cases as in [30].

Case 1. $r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon }\leq C$.

Let $\varOmega _{\varepsilon }=\{x\in \mathbb{R}^{2}:x_{\varepsilon }+r _{\varepsilon }^{1/(1-\gamma )} x\in \varOmega \}$. Define two blow-up sequences of function on $\varOmega _{\varepsilon }$ as

$$ \zeta _{\varepsilon }(x)=c_{\varepsilon }^{-1}u_{\varepsilon } \bigl(x_{ \varepsilon }+r_{\varepsilon }^{1/(1-\gamma )} x\bigr), \qquad \vartheta _{\varepsilon }(x)=c_{\varepsilon }\bigl(u_{\varepsilon }\bigl(x_{ \varepsilon }+r_{\varepsilon }^{1/(1-\gamma )} x\bigr)-c_{\varepsilon }\bigr). $$

A direct computation shows

$$\begin{aligned}& -\Delta \zeta _{\varepsilon }(x)=c_{\varepsilon }^{-2} \bigl\vert x+r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon } \bigr\vert ^{-2 \gamma }\bigl(1+h(u_{\varepsilon })\bigr) \zeta _{\varepsilon }e^{4\pi (1-\gamma -\varepsilon ) (u_{\varepsilon }^{2}-c_{\varepsilon }^{2})} \quad \text{in } \varOmega _{\varepsilon }, \end{aligned}$$

(25)

$$\begin{aligned}& -\Delta \vartheta _{\varepsilon }(x)= \bigl\vert x+r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon } \bigr\vert ^{-2 \gamma } \bigl(1+h(u_{\varepsilon })\bigr) \zeta _{\varepsilon }e^{4\pi (1-\gamma -\varepsilon ) (1+ \zeta _{\varepsilon })\vartheta _{\varepsilon }}\quad \text{in } \varOmega _{\varepsilon }. \end{aligned}$$

(26)

We now investigate the convergence behavior of $\zeta _{\varepsilon }(x)$ and $\vartheta _{\varepsilon }(x)$. Assume $\lim_{\varepsilon \rightarrow 0} r_{\varepsilon }^{-1/(1-\gamma )} x _{\varepsilon }=-\bar{x}$. From (24), we have $r_{\varepsilon }\rightarrow 0$ obviously. Thus $\varOmega _{\varepsilon }\rightarrow \mathbb{R}^{2}$ as $\varepsilon \rightarrow 0$. In view of $| \zeta _{\varepsilon }(x)|\leq 1$ and $\Delta \zeta _{\varepsilon }(x) \rightarrow 0$ in $x \in \varOmega _{\varepsilon }\setminus \{\bar{x}\}$ as $\varepsilon \rightarrow 0$, we have by elliptic estimates that $\zeta _{\varepsilon }(x)\rightarrow \zeta (x)$ in $C^{1}_{\mathrm{loc}}( \mathbb{R}^{2}\setminus \{\bar{x}\})\cap C^{0}_{\mathrm{loc}}(\mathbb{R}^{2})$, where ζ is a bounded harmonic function in $\mathbb{R}^{2}$. Observe that $\zeta (x)\leq \limsup_{\varepsilon \rightarrow 0} \zeta _{\varepsilon }(x)\leq 1$ and $\zeta (0)=1$. It follows from the Liouville theorem that $\zeta \equiv 1$ on $\mathbb{R}^{2}$. Thus, we have

$$ \zeta _{\varepsilon }\rightarrow 1\quad \text{in } C^{1}_{\mathrm{loc}}\bigl( \mathbb{R}^{2}\setminus \{ \bar{x}\}\bigr)\cap C^{0}_{\mathrm{loc}}\bigl(\mathbb{R}^{2} \bigr) $$

(27)

as $\varepsilon \rightarrow 0$. Note also that

$$ \vartheta _{\varepsilon }(x)\leq \vartheta _{\varepsilon }(0)=0\quad \text{for all } x\in \varOmega _{\varepsilon }(x). $$

In view of (27), we conclude by applying elliptic estimates to (26) that

$$ \vartheta _{\varepsilon }\rightarrow \vartheta \quad \text{in } C^{1}_{\mathrm{loc}}\bigl(\mathbb{R}^{2}\setminus \{\bar{x} \}\bigr)\cap C^{0}_{\mathrm{loc}}\bigl( \mathbb{R}^{2}\bigr), $$

(28)

where ϑ is a distributional solution to

$$ -\Delta \vartheta = \vert x-\bar{x} \vert ^{-2\gamma }e^{8\pi (1-\gamma )\vartheta } \quad \text{in } \mathbb{R}^{2}\setminus \{\bar{x}\}. $$

Observe that

$$ \zeta _{\varepsilon }(x)=\frac{u_{\varepsilon }(x_{\varepsilon }+r_{ \varepsilon }^{1/(1-\gamma )} x)}{c_{\varepsilon }}\rightarrow 1\quad \text{in } C^{1}_{\mathrm{loc}}(\mathbb{B}_{R}\setminus \mathbb{B}_{1/R}), $$

(29)

as $\varepsilon \rightarrow 0$. Set $y=x_{\varepsilon }+r_{\varepsilon }^{1/(1-\gamma )} x$ with $|x-\bar{x}|\leq R$, and then we have

$$ \vert y \vert \leq r_{\varepsilon }^{1/(1-\gamma )} \vert x-\bar{x} \vert + \bigl\vert x_{\varepsilon }+r _{\varepsilon }^{1/(1-\gamma )}\bar{x} \bigr\vert \leq 2Rr_{\varepsilon }^{1/(1- \gamma )}. $$

Since $r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon }\leq C$, choose R big enough such that

$$ \bigl\vert x-r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon } \bigr\vert \leq R. $$

This together with (29) leads to

$$ \begin{aligned}[b] &\lim_{\varepsilon \rightarrow 0} \biggl\Vert \frac{u_{\varepsilon }(r_{\varepsilon }^{1/(1-\gamma )}x)}{c_{\varepsilon }} \biggr\Vert _{L^{\infty }(\mathbb{B}_{R} \setminus \mathbb{B}_{1/R}(\bar{x}))} \\ &\quad =\lim_{\varepsilon \rightarrow 0} \biggl\Vert \frac{u_{\varepsilon }(x_{ \varepsilon }+r_{\varepsilon }^{1/(1-\gamma )} (x-r_{\varepsilon } ^{-1/(1-\gamma )}x_{\varepsilon }))}{c_{\varepsilon }} \biggr\Vert _{L^{\infty }( \mathbb{B}_{R}\setminus \mathbb{B}_{1/R}(\bar{x}))} \\ &\quad =1. \end{aligned} $$

Combining with Fatou’s lemma, we obtain

$$ \begin{aligned}[b] & \int _{\mathbb{B}_{R}\setminus \mathbb{B}_{1/R}(\bar{x})} \vert x-\bar{x} \vert ^{-2 \gamma }e^{8\pi (1-\gamma )\vartheta } \,dx \\ &\quad \leq \limsup_{\varepsilon \rightarrow 0} \int _{\mathbb{B}_{R}\setminus \mathbb{B}_{1/R}(\bar{x})} \bigl\vert x+r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon } \bigr\vert ^{-2\gamma }e^{4\pi (1-\gamma - \varepsilon ) (1+\zeta _{\varepsilon })\vartheta _{\varepsilon }}\,dx \\ &\quad \leq \limsup_{\varepsilon \rightarrow 0}\frac{1}{\lambda _{\varepsilon }} \int _{\mathbb{B}_{2Rr_{\varepsilon }^{1/(1-\gamma )}}\setminus \mathbb{B}_{\frac{1}{2}Rr_{\varepsilon }^{-1/(1-\gamma )}}(0)}\bigl(1+h(u _{\varepsilon })\bigr) \frac{u_{\varepsilon }^{2}(y)}{ \vert y \vert ^{2\gamma }} e^{4 \pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}(y)}\,dy \\ &\quad \leq 1. \end{aligned} $$

(30)

Passing to the limit $R\rightarrow \infty$, we have

$$ \int _{\mathbb{R}^{2}} \vert x-\bar{x} \vert ^{-2\gamma }e^{8\pi (1-\gamma )\vartheta } \,dx\leq 1. $$

The uniqueness theorem obtained in [3] implies that

$$ \vartheta (x)=-\frac{1}{4\pi (1-\gamma )}\log \biggl(1+\frac{1}{1- \gamma } \vert x-\bar{x} \vert ^{2(1-\gamma )} \biggr). $$

(31)

Let $x=0$, and then

$$ \vartheta (0)=\lim_{\varepsilon \rightarrow 0}\vartheta _{\varepsilon }(0)=0. $$

Thus, it follows from (31) that $\bar{x}=0$. Namely,

$$ \vartheta (x)=-\frac{1}{4\pi (1-\gamma )}\log \biggl(1+\frac{1}{1- \gamma } \vert x \vert ^{2(1-\gamma )} \biggr). $$

(32)

Furthermore, we can get

$$ \int _{\mathbb{R}^{2}} \vert x \vert ^{-2\gamma }e^{8\pi (1-\gamma )\vartheta } \,dx=1. $$

(33)

Case 2. $r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon }\rightarrow +\infty$. Set

$$ \widetilde{\varOmega }_{\varepsilon }=\bigl\{ x\in \mathbb{R}^{2}:x_{\varepsilon }+r_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }x\in \varOmega \bigr\} . $$

Denote the blowing up functions on $\overline{\varOmega }_{\varepsilon }$

$$ \alpha _{\varepsilon }(x)=c_{\varepsilon }^{-1}u_{\varepsilon } \bigl(x_{ \varepsilon }+r_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }x\bigr), \qquad \beta _{\varepsilon }(x)=c_{\varepsilon } \bigl(u_{\varepsilon }\bigl(x_{\varepsilon }+r_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }x\bigr)-c_{\varepsilon }\bigr). $$

In view of (16), $\alpha _{\varepsilon }(x)$ is a distributional solution to the equation

$$ -\Delta \alpha _{\varepsilon }(x)=f_{\varepsilon }(u)\quad \text{in } \overline{\varOmega }_{\varepsilon }, $$

(34)

where

$$ f_{\varepsilon }=c_{\varepsilon }^{-2} \vert x_{\varepsilon } \vert ^{2 \gamma } \bigl\vert x _{\varepsilon }+r_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }x \bigr\vert ^{-2 \gamma } \bigl(1+h(u_{\varepsilon })\bigr)\alpha _{\varepsilon }e^{4\pi (1-\gamma -\varepsilon ) c_{\varepsilon }^{2}(\alpha _{\varepsilon }^{2}-1)}. $$

Since $r_{\varepsilon }^{-1/(1-\gamma )}x_{\varepsilon }\rightarrow + \infty$, we have $|x_{\varepsilon }|^{2 \gamma }|x_{\varepsilon }+r _{\varepsilon }|x_{\varepsilon }|^{\gamma }x|^{-2\gamma }=1+o_{\varepsilon }(1)$ clearly. Since $|\alpha _{\varepsilon }(x)|\leq 1$, we obtain $f_{\varepsilon }$ is bounded in $L^{p}$ ($p>1$) according to (8). Elliptic estimates and embedding theorem lead to $\alpha _{\varepsilon }\rightarrow \alpha$ in $C^{1}_{\mathrm{loc}}(\mathbb{R} ^{2})$, where α satisfies

$$ -\Delta \alpha (x)=0 \quad \text{in } \mathbb{R}^{2}. $$

Note that $\alpha \leq 1$ and $\alpha (0)=1$. Thus, together with the Liouville theorem, we obtain $\alpha \equiv 1$. Also we have

$$ -\Delta \beta _{\varepsilon }= \vert x_{\varepsilon } \vert ^{2 \gamma } \bigl\vert x_{\varepsilon }+r_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }x \bigr\vert ^{-2\gamma } \bigl(1+h(u _{\varepsilon })\bigr)\alpha _{\varepsilon }e^{4\pi (1-\gamma -\varepsilon ) \beta _{\varepsilon }(\alpha _{\varepsilon }+1)}\quad \text{in } \overline{\varOmega }_{\varepsilon }. $$

(35)

Applying elliptic estimates to (35), we conclude that $\beta _{\varepsilon }\rightarrow \beta$ in $C^{1}_{\mathrm{loc}}(\mathbb{R} ^{2})$, where β is a distributional solution to

$$ \textstyle\begin{cases} \beta (0)=0=\sup \beta , \\ \Delta \beta =-e^{8\pi (1-\gamma )\beta }\quad \text{in } \mathbb{R}^{2}. \end{cases} $$

(36)

For $0<\beta <1$, (36) follows from Chen and Li [6] that β satisfies

$$ \int _{\mathbb{R}^{2}}e^{8\pi (1-\gamma )\beta }\,dx\geq \frac{1}{1- \beta }>1. $$

Using a suitable change of variable $y=x_{\varepsilon }+r_{\varepsilon }|x_{\varepsilon }|^{\gamma }x$, for any $R>0$, we have

$$ \begin{aligned}[b] \int _{\mathbb{B}_{R}(\bar{x})}e^{8\pi (1-\gamma )\beta }\,dx &= \lim_{\varepsilon \rightarrow 0} \int _{\mathbb{B}_{R}(0)}\bigl(1+h(u_{ \varepsilon })\bigr)e^{4\pi (1-\gamma -\varepsilon ) (u_{\varepsilon }^{2}(x _{\varepsilon }+r_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }x)-c_{ \varepsilon }^{2})} \,dx \\ &\leq \lim_{\varepsilon \rightarrow 0}\frac{1}{\lambda _{\varepsilon }} \int _{\mathbb{B}_{Rr_{\varepsilon } \vert x_{\varepsilon } \vert ^{\gamma }}(x _{\varepsilon })} \bigl(1+h(u_{\varepsilon })\bigr) \frac{u_{\varepsilon }^{2}(y)}{ \vert y \vert ^{2 \gamma }}e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}(y)}\,dy \\ &\leq 1, \end{aligned} $$

(37)

which leads to a contradiction. Thus, it is impossible for Case 2 to happen.

2.2.3 Convergence away from the concentration point

To understand the convergence behavior away from the blow-up point $x_{0}=0$, we need to investigate how $c_{\varepsilon }u_{\varepsilon }$ converges. Similar to [1, 15], define $u_{\varepsilon , \tau }=\min \{\tau c_{\varepsilon }, u_{\varepsilon }\}$, then we have the following.

Lemma 6

For any $0<\tau <1$, we have

$$ \lim_{\varepsilon \rightarrow 0} \int_{\varOmega } \vert \nabla u_{\varepsilon ,\tau } \vert ^{2} \,dx= \tau . $$

Proof

Observe that $u_{\varepsilon }/c_{\varepsilon }=1+o_{ \varepsilon }(1)$ in $B_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })$. For any $0<\tau <1$, it follows from Eq. (16) and the divergence theorem that

$$ \begin{aligned}[b] \int _{\varOmega } \vert \nabla u_{\varepsilon ,\tau } \vert ^{2} \,dx &=\frac{1}{ \lambda _{\varepsilon }} \int _{\varOmega }\frac{u_{ \varepsilon ,\tau }u_{\varepsilon }}{ \vert x \vert ^{2\gamma }} \bigl(1+h(u_{\varepsilon })\bigr) e^{4\pi (1- \gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx \\ &\geq \frac{1}{\lambda _{\varepsilon }} \int _{B_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \frac{u _{\varepsilon ,\tau }u_{\varepsilon }}{ \vert x \vert ^{2\gamma }}\bigl(1+h(u_{\varepsilon }) \bigr) e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx+o_{\varepsilon }(1) \\ &=\tau \int _{B_{R}(0)}\frac{(1+h(u_{\varepsilon }))e^{4\pi (1-\gamma -\varepsilon )(u_{\varepsilon }^{2}(x_{\varepsilon }+r_{\varepsilon } ^{1/(1-\gamma )}y)-c_{\varepsilon }^{2})}}{ \vert y+r_{\varepsilon }^{-1/(1- \gamma )}x_{\varepsilon } \vert ^{2\gamma }}\,dy +o_{\varepsilon }(1). \end{aligned} $$

Hence

$$ \liminf_{\varepsilon \rightarrow 0} \int _{\varOmega } \vert \nabla u_{\varepsilon ,\tau } \vert ^{2} \,dx\geq \tau \int _{B_{R}(0)}e^{8\pi (1-\gamma )\vartheta }\,dy,\quad \forall R>0. $$

In view of (33), passing to the limit $R\rightarrow +\infty$, we obtain

$$ \liminf_{\varepsilon \rightarrow 0} \int _{\varOmega } \vert \nabla u_{\varepsilon ,\tau } \vert ^{2} \,dx\geq \tau . $$

(38)

Note that

$$ \bigl\vert \nabla (u_{\varepsilon }-\tau c_{\varepsilon })^{+} \bigr\vert ^{2}=\nabla (u _{\varepsilon }-\tau c_{\varepsilon })^{+} \cdot \nabla u_{\varepsilon } \quad \text{on } \varOmega $$

and

$$ (u_{\varepsilon }-\tau c_{\varepsilon })^{+}=\bigl(1+o_{\varepsilon }(1) \bigr) (1- \tau )c_{\varepsilon }\quad \text{in } B_{Rr_{\varepsilon } ^{1/(1-\gamma )}}(x_{0}). $$

Testing Eq. (16) by $(u_{\varepsilon }-\tau c_{\varepsilon })^{+}$, for any fixed $R>0$, simple computation shows that

$$ \begin{aligned}[b] \int _{\varOmega } \bigl\vert \nabla (u_{\varepsilon }-\tau c_{\varepsilon })^{+} \bigr\vert ^{2}\,dx &= \int _{\varOmega }(u_{\varepsilon }-\tau c_{\varepsilon })^{+} \frac{u_{ \varepsilon }}{\lambda _{\varepsilon } \vert x \vert ^{2\gamma }}\bigl(1+h(u_{\varepsilon })\bigr)e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx \\ &\geq \int _{B_{Rr_{\varepsilon }^{1/(1-\gamma )}(x_{\varepsilon })}}(u _{\varepsilon }-\tau c_{\varepsilon })^{+} \frac{u_{\varepsilon }(1+h(u _{\varepsilon }))}{\lambda _{\varepsilon } \vert x \vert ^{2\gamma }}e^{4\pi (1- \gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx \\ &=\bigl(1+o_{\varepsilon }(1)\bigr) (1-\tau ) \int _{B_{R(0)}}\zeta _{\varepsilon }\bigl(1+h(u_{\varepsilon }) \bigr)e^{4\pi (1-\gamma -\varepsilon ) \vartheta _{\varepsilon }^{2}}\,dx. \end{aligned} $$

By passing to the limit $\varepsilon \rightarrow 0$, we get

$$ \liminf_{\varepsilon \rightarrow 0} \int _{\varOmega } \bigl\vert \nabla (u_{\varepsilon }-\tau c_{\varepsilon })^{+} \bigr\vert ^{2}\,dx\geq (1-\tau ) \int _{B_{R(0)}}e^{8 \pi (1-\gamma )\vartheta }\,dx=1-\tau . $$

(39)

Since $|\nabla u_{\varepsilon ,\tau }|^{2}+|\nabla (u_{\varepsilon }- \tau c_{\varepsilon })^{+}|^{2}=|\nabla u_{\varepsilon }|^{2}$ almost everywhere, it follows that

$$ \int _{\varOmega } \bigl\vert \nabla (u_{\varepsilon }-\tau c_{\varepsilon })^{+} \bigr\vert ^{2}\,dx+ \int _{\varOmega } \vert \nabla u_{\varepsilon ,\tau } \vert ^{2}\,dx= \int _{\varOmega } \vert \nabla u_{\varepsilon } \vert ^{2}\,dx=1+o_{\varepsilon }(1). $$

(40)

Therefore, we end the proof of this lemma together with (38), (39) and (40). □

The following estimate is a byproduct of Lemma 6 and will be employed in the next section.

Lemma 7

We have

$$ \lim_{\varepsilon \rightarrow 0} \int _{\varOmega } \vert x \vert ^{-2\gamma }\bigl(1+g(u _{\varepsilon })\bigr)e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx =\bigl(1+g(0)\bigr) \int _{\varOmega } \vert x \vert ^{-2\gamma }\,dx +\lim _{\varepsilon \rightarrow 0}\frac{\lambda _{\varepsilon }}{c_{\varepsilon }^{2}}. $$

(41)

Proof

Let $0<\tau <1$ be fixed. By the definition of $u_{\varepsilon ,\tau }$, we can get

$$ \begin{aligned}[b] & \int _{u_{\varepsilon }\leq \tau c_{\varepsilon }}\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2 \gamma }} \,dx-\bigl(1+g(0)\bigr) \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\,dx \\ &\quad \leq \int _{\varOmega }\bigl(1+g(u_{\varepsilon ,\tau })\bigr)\frac{e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon ,\tau }^{2}}}{ \vert x \vert ^{2\gamma }} \,dx-\bigl(1+g(0)\bigr) \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\,dx \\ &\quad \leq \int _{\varOmega } \bigl\vert g(u_{\varepsilon ,\tau })-g(0) \bigr\vert \frac{e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon ,\tau }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx + \bigl\vert 1+g(0) \bigr\vert \int _{\varOmega }\frac{(e^{4\pi (1-\gamma -\varepsilon ) u_{ \varepsilon ,\tau }^{2}}-1)}{ \vert x \vert ^{2\gamma }}\,dx. \end{aligned} $$

(42)

Combining Lemma 6 and Proposition 5, we see that $u_{\varepsilon , \sigma }$ converges to 0 in $C^{1}_{\mathrm{loc}}(\overline{\varOmega }\setminus \{0\})$ as $\varepsilon \rightarrow 0$. Then from (3), one can deduce that

$$ \int _{\varOmega }\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon , \tau }^{2}}}{ \vert x \vert ^{2\gamma }} \bigl\vert g(u_{\varepsilon ,\tau })-g(0) \bigr\vert \,dx=o_{ \varepsilon }(1). $$

(43)

According to the Hölder inequality and the Lagrange theorem, we have

$$ \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\bigl(e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon ,\tau }^{2}}-1\bigr)\,dx =o_{\varepsilon }(1). $$

(44)

Inserting (43) and (44) into (42), one has

$$ \lim_{\varepsilon \rightarrow 0} \int _{u_{\varepsilon }\leq \tau c_{\varepsilon }}\bigl(1+g(u_{\varepsilon })\bigr)\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2 \gamma }} \,dx=\bigl(1+g(0)\bigr) \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\,dx. $$

(45)

Moreover, we calculate

$$ \begin{aligned}[b] & \int _{u_{\varepsilon }>\tau c_{\varepsilon }}\bigl(1+g(u_{\varepsilon })\bigr)\frac{e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }} \,dx \\ &\quad \leq \frac{1}{\tau ^{2}} \int _{u_{\varepsilon }>\tau c_{\varepsilon }}\frac{u _{\varepsilon }^{2}}{c_{\varepsilon }^{2}}\bigl(1+g(u_{\varepsilon })\bigr) \frac{e ^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx \\ &\quad \leq \frac{1}{\tau ^{2}}\frac{\lambda _{\varepsilon }^{2}}{c_{\varepsilon }^{2}}. \end{aligned} $$

(46)

Combining (45) and (46), we obtain

$$ \lim_{\varepsilon \rightarrow 0} \int _{\varOmega }\frac{(1+g(u_{\varepsilon })) e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}}{ \vert x \vert ^{2 \gamma }}\,dx\leq \bigl(1+g(0)\bigr) \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\,dx+\frac{1}{ \tau ^{2}}\liminf _{\varepsilon \rightarrow 0} \frac{\lambda _{\varepsilon }^{2}}{c_{\varepsilon }^{2}}. $$

It follows by letting $\tau \rightarrow 1$ that

$$ \lim_{\varepsilon \rightarrow 0} \int _{\varOmega }\frac{(1+g(u_{\varepsilon })) e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}}{ \vert x \vert ^{2 \gamma }}\,dx-\bigl(1+g(0)\bigr) \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\,dx\leq \liminf _{\varepsilon \rightarrow 0} \frac{\lambda _{\varepsilon }^{2}}{c _{\varepsilon }^{2}}. $$

(47)

On the other hand, in view of (16), we estimate

$$ \begin{aligned}[b] & \int _{\varOmega }\bigl(1+g(u_{\varepsilon })\bigr) \frac{e^{4\pi (1-\gamma - \varepsilon )u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx-\bigl(1+g(0)\bigr) \int _{\varOmega }\frac{1}{ \vert x \vert ^{2\gamma }}\,dx \\ &\quad \geq \int _{\varOmega } \frac{u_{\varepsilon }^{2}}{c_{\varepsilon }^{2}} \biggl(\bigl(1+g(u_{ \varepsilon }) \bigr) \frac{e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}-\bigl(1+g(0)\bigr)\frac{1}{ \vert x \vert ^{2\gamma }} \biggr)\,dx \\ &\quad =\frac{\lambda _{\varepsilon }}{c_{\varepsilon }^{2}}-\frac{1}{c_{ \varepsilon }^{2}} \int _{\varOmega }\frac{(1+g(0))u_{\varepsilon }^{2}}{ \vert x \vert ^{2 \gamma }}\,dx -\frac{1}{c_{\varepsilon }^{2}} \int _{\varOmega }\frac{u_{ \varepsilon }g'(u_{\varepsilon })}{8\pi (1-\gamma -\varepsilon ) \vert x \vert ^{2 \gamma }}e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx. \end{aligned} $$

Thus, by Proposition 5 and (6), (8), one can check that

$$ \limsup_{\varepsilon \rightarrow 0}\frac{\lambda _{\varepsilon }^{2}}{c _{\varepsilon }^{2}}\leq \lim _{\varepsilon \rightarrow 0} \int _{\varOmega } \vert x \vert ^{-2\gamma } \bigl(1+g(u_{\varepsilon })\bigr) e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon }^{2}}\,dx-\bigl(1+g(0)\bigr) \int _{\varOmega } \vert x \vert ^{-2\gamma }\,dx. $$

(48)

In view of (47) and (48), we complete the proof of Lemma 7. □

Corollary 8

If $\theta <2$, then $\frac{\lambda _{\varepsilon }}{c_{\varepsilon } ^{\theta }}\rightarrow \infty$ as $\varepsilon \rightarrow 0$.

Proof

In contrast, we have $\lambda _{\varepsilon }/c_{\varepsilon }^{2}\rightarrow 0$ as $\varepsilon \rightarrow 0$. For any $\nu \in W_{0}^{1,2}(\varOmega )$ with $\|\nabla \nu \|_{2}\leq 1$, clearly, it is impossible for (41) to hold since $\nu \not \equiv 0$. □

Lemma 9

For any function $\phi \in C_{0}^{1}(\varOmega )$, we have

$$ \lim_{\varepsilon \rightarrow 0} \int _{\varOmega }\bigl(1+h(u_{\varepsilon })\bigr) \lambda _{\varepsilon }^{-1}c_{\varepsilon }u_{\varepsilon } \vert x \vert ^{-2 \gamma } e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}\phi \,dx= \phi (0). $$

(49)

Proof

To see this, let $\phi \in C_{0}^{1}(\varOmega )$ be fixed. Write for simplicity

$$ \omega _{\varepsilon }=\bigl(1+h(u_{\varepsilon })\bigr)\lambda _{\varepsilon } ^{-1}c_{\varepsilon }u_{\varepsilon } \vert x \vert ^{-2\gamma } e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon }^{2}}. $$

Clearly

$$ \begin{aligned}[b] \int _{\varOmega }\omega _{\varepsilon }\phi \,dx &= \int _{\{u_{\varepsilon }< \tau c_{\varepsilon }\}}\omega _{\varepsilon }\phi \,dx+ \int _{\{u_{\varepsilon }\geq \tau c_{\varepsilon }\}\setminus B_{R _{r_{\varepsilon }}^{1/(1-\gamma )}}(x_{\varepsilon })} \omega _{\varepsilon }\phi \,dx \\ &\quad {}+ \int _{\{u_{\varepsilon }\geq \tau c_{\varepsilon }\}\cap B_{R_{r_{ \varepsilon }}^{1/(1-\gamma )}}(x_{\varepsilon })}\omega _{\varepsilon }\phi \,dx. \end{aligned} $$

(50)

Given $0<\tau <1$, we estimate the three integrals on the right hand of (50), respectively. Note that $u_{\varepsilon }\rightarrow 0$ in $L^{q}$ ($\forall q>1$). This together with Lemma 6 and Corollary 8 gives

$$ \begin{aligned}[b] \int _{\{u_{\varepsilon }< \tau c_{\varepsilon }\}}\omega _{\varepsilon }\phi \,dx &\leq \lambda _{\varepsilon }^{-1}c_{\varepsilon } \Bigl(\sup_{ \varOmega } \bigl\vert \phi \bigl(1+h(u_{\varepsilon })\bigr) \bigr\vert \Bigr) \int _{\{u_{\varepsilon }< \tau c_{\varepsilon }\}} u_{\varepsilon } \vert x \vert ^{-2 \gamma } e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon ,\tau }^{2}}\,dx \\ &\leq C\lambda _{\varepsilon }^{-1}c_{\varepsilon } \int _{\{u_{\varepsilon }< \tau c_{\varepsilon }\}} u_{\varepsilon } \vert x \vert ^{-2 \gamma }e^{4\pi (1-\gamma -\varepsilon )u_{\varepsilon ,\tau }^{2}} \,dx \\ &=o_{\varepsilon }(1). \end{aligned} $$

(51)

Now we consider in $B_{R_{r_{\varepsilon }}^{1/(1-\gamma )}}(x_{ \varepsilon })\subset \{x\in \varOmega \mid u_{\varepsilon }\geq \tau c _{\varepsilon }\}$ for sufficiently small $\varepsilon >0$. One can deduce from (33) that

$$ \begin{aligned}[b] \int _{\{u_{\varepsilon }\geq \tau c_{\varepsilon }\}\cap B_{R_{r_{ \varepsilon }}^{1/(1-\gamma )}}(x_{\varepsilon })}\omega _{\varepsilon }\phi \,dx &=\phi (0) \bigl(1+o_{\varepsilon }(1)\bigr) \int _{B_{R\backslash 1/R}(0)} \vert x \vert ^{-2 \gamma }e^{8\pi \vartheta } \,dx \\ &=\phi (0) \bigl(1+o_{\varepsilon }(1)+o_{R}(1)\bigr). \end{aligned} $$

(52)

On the other hand, we calculate

$$ \begin{aligned}[b] \int _{\{u_{\varepsilon }\geq \tau c_{\varepsilon }\}\backslash B_{R _{r_{\varepsilon }}^{1/(1-\gamma )}}(x_{\varepsilon })} \omega _{\varepsilon }\phi \,dx &\leq \frac{C}{\tau } \biggl(1- \int _{B_{R_{r_{\varepsilon }}^{1/(1-\gamma )}}(x_{\varepsilon })} \frac{u _{\varepsilon }^{2}}{ \lambda _{\varepsilon }}\frac{e^{4\pi (1-\gamma -\varepsilon ) u_{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx \biggr) \\ &=\frac{C}{\tau } \biggl(1- \int _{B_{R}(0)}\frac{e^{8\pi (1-\gamma ) \vartheta }}{ \vert x \vert ^{2\gamma }}\,dx \biggr). \end{aligned} $$

Hence, we derive by (33) that

$$ \lim_{R\rightarrow \infty }\lim_{\varepsilon \rightarrow 0} \int _{\{u_{\varepsilon }\geq \tau c_{\varepsilon }\}\backslash B_{R _{r_{\varepsilon }}^{1/(1-\gamma )}}(x_{\varepsilon })} \omega _{\varepsilon }\phi \,dx =0. $$

(53)

Inserting (51)–(53) to (50), we conclude (49) finally. □

In particular, we propose, by letting $\phi =1$,

$$ \omega _{\varepsilon }(x)= \bigl(1+h(u_{\varepsilon })\bigr) \lambda _{\varepsilon }^{-1}c_{\varepsilon }u_{\varepsilon } \vert x \vert ^{-2\gamma } e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon }^{2}} \quad \text{is bounded in } L^{1}(\varOmega ), $$

(54)

which will be used in the following proof.

We now prove that $c_{\varepsilon }u_{\varepsilon }$ converges to a Green function in distributional sense when $\varepsilon \rightarrow 0$, where $\delta _{0}$ stands for the Dirac measure centered at 0. More precisely, we have

Lemma 10

$c_{\varepsilon }u_{\varepsilon }\rightarrow G$ in $C_{\mathrm{loc}}^{1}(\overline{ \varOmega }\setminus \{0\})$ and weakly in $W_{0}^{1,q}(\varOmega )$ for all $1< q<2$, where $G\in C^{1}(\overline{\varOmega }\setminus \{0\})$ is a distributional solution satisfying the following:

$$ \textstyle\begin{cases} -\Delta G=\delta _{0} & \textit{in } \varOmega , \\ G=0 & \textit{on } \partial \varOmega . \end{cases} $$

(55)

Moreover, G takes the form

$$ G(x)=-\frac{1}{2\pi }\log \vert x \vert +A_{0}+ \xi (x), $$

(56)

where $\xi (x)\in C^{1}(\overline{\varOmega })$ and $A_{0}$ is a constant depending on 0.

Proof

By Eq. (16), $c_{\varepsilon }u_{\varepsilon }$ is a distributional solution to

$$ -\Delta (c_{\varepsilon }u_{\varepsilon })=\omega _{\varepsilon } \quad \text{in } \varOmega . $$

(57)

It follows from (54) that $\omega _{\varepsilon }$ is bounded in $L^{1}(\varOmega )$. Using the argument in Struwe ([25], Theorem 2.2), one concludes that $c_{\varepsilon }u_{\varepsilon }$ is bounded in $W_{0}^{1,q}(\varOmega )$ for all $1< q<2$. Hence, we can assume, for any $1< q<2$, $r>1$, that

$$ \begin{aligned}[b] &c_{\varepsilon }u_{\varepsilon }\rightharpoonup G \quad \text{weakly in } W_{0}^{1,q}(\varOmega ), \\ &c_{\varepsilon }u_{\varepsilon }\rightarrow G \quad \text{strongly in } L^{r}(\varOmega ). \end{aligned} $$

Testing (57) by $\phi \in C^{1}_{0}(\varOmega )$, we deduce

$$ \int _{\varOmega }\nabla (c_{\varepsilon }u_{\varepsilon })\nabla \phi \,dx= \int _{\varOmega }\phi \lambda _{\varepsilon }^{-1}c_{\varepsilon }u_{ \varepsilon } \bigl(1+h(u_{\varepsilon })\bigr) \vert x \vert ^{-2\gamma } e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon }^{2}}. $$

Let $\varepsilon \rightarrow 0$ and it yields by (55)

$$ \int _{\varOmega }\nabla G\nabla \phi \,dx =\phi (0), $$

which implies that $-\Delta G=\delta _{0}$ in a distributional sense. Since $\Delta (G+\frac{1}{2\pi }\log |x|)\in L^{p}(\varOmega )$ for any $p>2$, (56) follows from the elliptic solution immediately. Applying elliptic estimates to Eq. (57), we arrive at the conclusion

$$ c_{\varepsilon }u_{\varepsilon }\rightarrow G\quad \text{in } C _{\mathrm{loc}}^{1}\bigl(\overline{\varOmega }\setminus \{0\}\bigr). $$

(58)

Thus, the two assertions holds. □

2.3 Upper bound calculates by means of capacity estimate

In this subsection, we aim to derive an upper bound of the integrals $\int _{\varOmega }(1+g(u_{\varepsilon }))|x|^{-2\gamma } e^{4\pi (1- \gamma -\varepsilon ) u_{\varepsilon }^{2}}\,dx$. Analogous to the one obtained in [15], we mainly use the capacity estimate. Now choose a proper δ to ensure that $B_{2\delta }\subset \varOmega$, and then construct a new function space

$$ \mathscr{M}_{\varepsilon }(\rho _{\varepsilon },\sigma _{\varepsilon })= \bigl\{ u|u\in W^{1,2}\bigl(\mathbb{B}_{\delta }(x_{\varepsilon }) \setminus \mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })\bigr): u|_{ \partial \mathbb{B}_{\delta }(x_{\varepsilon })}= \rho _{\varepsilon }, u|_{\partial \mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })}=\sigma _{\varepsilon }\bigr\} $$

where

$$ \rho _{\varepsilon }=\sup_{\partial \mathbb{B}_{\delta }(x_{\varepsilon })}u_{\varepsilon },\qquad \sigma _{\varepsilon }= \inf_{\partial \mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{ \varepsilon })} u_{\varepsilon }. $$

Define

$$ \varLambda _{\varepsilon }= \inf_{u\in \mathscr{M}_{\varepsilon }(\rho _{\varepsilon }, \sigma _{\varepsilon })} \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla u \vert ^{2}\,dx. $$

Clearly, the infimum $\varLambda _{\varepsilon }$ can be attained by the sequence $u_{k}\in \mathscr{M}$ as $k\rightarrow \infty$. By the proof of the Poincaré inequality, we infer that $u_{k}$ is bounded in $W_{0}^{1,2}(\varOmega )$. Without loss of generality, there exists some function $t\in W^{1,2}(\varOmega )$ such that up to a subsequence. As $k\rightarrow \infty$, we have $u_{k}\rightharpoonup t$ weakly in $W^{1,2}(\varOmega )$, $u_{k}\rightarrow t$ in $L^{p}_{\mathrm{loc}}(\varOmega )$ for any $p>0$ and $u_{k}\rightarrow t$ a.e. in Ω. Besides, for $t\in \mathscr{M}_{\varepsilon }(\rho _{\varepsilon },\sigma _{\varepsilon })$, we have

$$ \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla t \vert ^{2}\,dx \leq \lim_{k\rightarrow \infty } \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla u_{k} \vert ^{2}\,dx= \varLambda _{\varepsilon } $$

and

$$ \varLambda _{\varepsilon }\leq \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla t \vert ^{2}\,dx. $$

Through the method of variation, we see that there exists some harmonic function $t(x)$ to reach the $\varLambda _{\varepsilon }$ which satisfies the following:

$$ \textstyle\begin{cases} \Delta t =0 \quad \text{in } \mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon }), \\ t|_{\partial \mathbb{B}_{\delta }(x_{\varepsilon })}=\rho _{\varepsilon }, \\ t|_{\partial \mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })}=\sigma _{\varepsilon }. \end{cases} $$

(59)

Obviously, the solution of (59) can be expressed as

$$ t(x)=a\log \vert x-x_{0} \vert +b. $$

One can check that

$$ \textstyle\begin{cases} a=\frac{\sigma _{\varepsilon }-\rho _{\varepsilon }}{\log \delta - \log Rr_{\varepsilon }^{1/(1-\gamma )}}, \\ b=\frac{\sigma _{\varepsilon }\log Rr_{\varepsilon }^{1/(1-\gamma )} -\rho _{\varepsilon }\log \delta }{\log Rr_{\varepsilon }^{1/(1-\gamma )}-\log \delta }. \end{cases} $$

(60)

Thus, $t(x)$ can be expressed as

$$ t(x)=\frac{\sigma _{\varepsilon }(\log \delta -\log \vert x-x_{\varepsilon } \vert )-\rho _{\varepsilon }(\log Rr_{\varepsilon }^{1/(1-\gamma )}-\log \vert x-x _{\varepsilon } \vert )}{\log \delta -\log Rr_{\varepsilon }^{1/(1-\gamma )}}. $$

With a direct computation, it is easy to check that

$$ \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla t \vert ^{2}\,dx= \frac{2 \pi (\sigma _{\varepsilon }-\rho _{\varepsilon })^{2}}{\log \delta - \log Rr_{\varepsilon }^{1/(1-\gamma )}}. $$

(61)

According to (23), we have

$$ \log \delta -\log Rr_{\varepsilon }^{1/(1-\gamma )}=\log \delta - \log R +\frac{2\pi (1-\gamma -\varepsilon )c_{\varepsilon }^{2}}{1- \gamma }- \frac{1}{2(1-\gamma )}\log \frac{\lambda _{\varepsilon }}{c _{\varepsilon }^{2}}. $$

(62)

Furthermore, Lemma 10 and (31) show that

$$ \sigma _{\varepsilon }=c_{\varepsilon }+\frac{1}{c_{\varepsilon }} \biggl(-\frac{1}{4\pi (1-\gamma )}\log \biggl(1+\frac{\pi }{1-\gamma }R^{2(1- \gamma )} \biggr)+o(1) \biggr) $$

(63)

and

$$ \rho _{\varepsilon }=\frac{1}{c_{\varepsilon }} \biggl(- \frac{1}{2 \pi }\log \delta +A_{0}+o(1) \biggr), $$

(64)

where $o(1)\rightarrow 0$ by letting $\varepsilon \rightarrow 0$ and $\delta \rightarrow 0$ in succession. Set $u_{\varepsilon }^{*}= \max \{\rho _{\varepsilon },\min \{u_{\varepsilon },\sigma _{\varepsilon }\}\}$. From $u_{\varepsilon }^{*}\in \mathscr{M}_{\varepsilon }( \rho _{\varepsilon },\sigma _{\varepsilon })$, one can easily check that

$$ \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla t \vert ^{2}\,dx= \varLambda _{\varepsilon }\leq \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \bigl\vert \nabla u_{\varepsilon }^{*} \bigr\vert ^{2}\,dx. $$

(65)

Observe that $|\nabla u_{\varepsilon }^{*}|\leq |\nabla u_{\varepsilon }|$ a.e. in $\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })$ if ε is sufficiently small. Thus, it follows

$$ \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \bigl\vert \nabla u_{\varepsilon }^{*} \bigr\vert ^{2}\,dx\leq \int _{\mathbb{B}_{\delta }(x_{\varepsilon })\setminus \mathbb{B}_{Rr _{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla u_{\varepsilon } \vert ^{2}\,dx. $$

(66)

In view of (61), (65) and (66), it can be inferred that

$$ \begin{aligned}[b] 2\pi (\sigma _{\varepsilon }-\rho _{\varepsilon })^{2} &\leq \biggl(1- \int _{\varOmega \setminus \mathbb{B}_{\delta }(x_{\varepsilon })} \vert \nabla u_{\varepsilon } \vert ^{2}\,dx- \int _{\mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla u_{\varepsilon } \vert ^{2}\,dx \biggr) \\ &\quad {}\times \bigl(\log \delta -\log Rr_{\varepsilon }^{1/(1- \gamma )} \bigr). \end{aligned} $$

(67)

Since $c_{\varepsilon }u_{\varepsilon }\rightarrow G$ in $C^{1}_{\mathrm{loc}}(\overline{ \varOmega }\backslash \{0\})$, we obtain the conclusion through integrating by parts:

$$ \begin{aligned}[b] \int _{\varOmega \setminus \mathbb{B}_{\delta }(x_{\varepsilon })} \vert \nabla u_{\varepsilon } \vert ^{2}\,dx &=\frac{1}{c_{\varepsilon }^{2}} \int _{\varOmega \setminus \mathbb{B}_{\delta }(x_{\varepsilon })} \vert \nabla G_{\varepsilon } \vert ^{2}\,dx \\ &=-\frac{1}{c_{\varepsilon }^{2}} \biggl( \int _{\varOmega \setminus \mathbb{B}_{\delta }(x_{\varepsilon })}G\Delta G \,dx+ \int _{\partial \mathbb{B}_{\delta }(x_{\varepsilon })}G \frac{\partial G}{\partial \nu }\,ds \biggr) \\ &=-\frac{1}{c_{\varepsilon }^{2}} \biggl(\frac{1}{2\pi }\log \delta -A _{0}+o_{\varepsilon }(1)+ o_{\delta }(1) \biggr). \end{aligned} $$

(68)

Observe that $\vartheta _{\varepsilon }\rightarrow \vartheta$ in $C^{1}_{\mathrm{loc}}(\mathbb{R}^{2}\setminus \{0\})$, and

$$ u_{\varepsilon }=\frac{\vartheta _{\varepsilon }(x)}{c_{\varepsilon }}+c _{\varepsilon }\quad \text{in } \mathbb{B}_{Rr_{\varepsilon } ^{1/(1-\gamma )}}(x_{\varepsilon }). $$

(69)

A direct computation shows that

$$ \begin{aligned}[b] \int _{\mathbb{B}_{R}(0)} \vert \nabla \vartheta \vert ^{2}\,dx &= \int _{0}^{R}\frac{2 \pi }{4(1-\gamma )^{2}(1+\frac{\pi }{1-\gamma } \vert r \vert ^{2(1-\gamma )})^{2}}r ^{-4\gamma } \,dr \\ &=\frac{1}{4\pi (1-\gamma )}\log \frac{\pi }{1-\gamma }+\frac{1}{2 \pi }\log R- \frac{1}{4\pi (1-\gamma )}+O\biggl(\frac{1}{R^{2(1-\gamma )}}\biggr). \end{aligned} $$

(70)

Then it follows from (69) and (70) that

$$ \begin{aligned}[b] \int _{\mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \vert \nabla u_{\varepsilon } \vert ^{2}\,dx &=\frac{1}{c_{\varepsilon }^{2}} \int _{\mathbb{B}_{Rr_{\varepsilon }^{1/(1-\gamma )}}(x_{\varepsilon })} \bigl\vert \nabla \vartheta _{\varepsilon }(x) \bigr\vert ^{2}\,dx \\ &=\frac{1}{c_{\varepsilon }^{2}} \biggl( \int _{\mathbb{B}_{R}(0)} \bigl\vert \nabla \vartheta (y) \bigr\vert ^{2} \,dy+o_{\varepsilon }(1) \biggr) \\ &=\frac{1}{4\pi c_{\varepsilon }^{2}(1-\gamma )}\log \frac{\pi }{1- \gamma }+\frac{1}{2\pi c_{\varepsilon }^{2}}\log R- \frac{1}{4\pi c _{\varepsilon }^{2}(1-\gamma )}+\frac{o(1)}{c_{\varepsilon }^{2}}. \end{aligned} $$

This together with (62)–(64) and (68), we obtain

$$ -2\pi A_{0}-\frac{\log (1+\frac{\pi }{1-\gamma }R^{2(1-\gamma )} )}{1- \gamma } \leq -2\log R+\frac{(1-\log \frac{\lambda _{\varepsilon }}{c _{\varepsilon }^{2}}-\log \frac{\pi }{1-\gamma })}{2(1-\gamma )}+o(1). $$

Hence,

$$ \limsup_{\varepsilon \rightarrow 0}\frac{\lambda _{\varepsilon }}{c _{\varepsilon }^{2}} \leq \frac{\pi }{1-\gamma }e^{4\pi (1-\gamma )A _{0}+1}. $$

In view of Lemma 7, we arrive at the conclusion

$$ \varLambda _{4\pi (1-\gamma )} \leq \bigl(1+g(0)\bigr) \int _{\varOmega } \vert x \vert ^{-2\gamma }\,dx + \frac{\pi }{1-\gamma }e^{4\pi (1-\gamma )A_{0}+1}. $$

(71)

2.4 Completion of the proof of Theorem 1

As a consequence, if $c_{\varepsilon }\rightarrow \infty$, it follows from (71) that $\varLambda _{4\pi (1-\gamma )}$ is bounded. Otherwise, we can find the extremal function $u_{0}$ which satisfies (17). Therefore, necessarily

$$ \sup_{u\in W_{0}^{1,2}(\varOmega ), \Vert \nabla u \Vert _{2}\leq 1} \int _{\varOmega }\bigl(1+g(u)\bigr)\frac{e ^{4\pi (1-\gamma ) u^{2}}}{ \vert x \vert ^{2\gamma }}\,dx< \infty . $$

3 Proof of Theorem 2

3.1 Test function computation

Similar to [30], we construct a blow-up sequence $\phi _{\varepsilon }\in W_{0}^{1,2}(\varOmega )$ with $\|\nabla \phi _{\varepsilon }\|_{2}=1$. For sufficiently small $\varepsilon >0$, there exists

$$ \int _{\varOmega } \vert x \vert ^{-2\gamma }\bigl(1+g(\phi _{\varepsilon })\bigr)e^{4\pi (1- \gamma )\phi _{\varepsilon }^{2}}\,dx >\bigl(1+g(0)\bigr) \int _{\varOmega } \vert x \vert ^{-2 \gamma }\,dx + \frac{\pi }{1-\gamma }e^{4\pi (1-\gamma )A_{0}+1}. $$

(72)

Then we will find (72) is a contradiction to (71), so that $c_{\varepsilon }$ has to be bounded, which means the blow-up cannot take place. Furthermore, Theorem 2 follows immediately from what we have proved according to the elliptic estimates. For this purpose we set

$$\begin{aligned} \phi _{\varepsilon }(x)=\textstyle\begin{cases} c+\frac{1}{c}(-\frac{1}{4\pi (1-\gamma )}\log (1+\frac{\pi }{1-\gamma }\frac{ \vert x \vert ^{2(1-\gamma )}}{\varepsilon ^{2(1-\gamma )}})+b), & \text{for } x\in \overline{\mathbb{B}}_{ R\varepsilon }, \\ \frac{G-\xi \eta }{c}, & \text{for } x\in \mathbb{B}_{ 2R\varepsilon }\setminus \overline{\mathbb{B}}_{ R \varepsilon }, \\ \frac{G}{c}, & \text{for } x\in \varOmega \setminus \mathbb{B}_{ 2R\varepsilon }, \end{cases}\displaystyle \end{aligned}$$

(73)

where $\eta \in C_{0}^{1}(\mathbb{B}_{2R\varepsilon })$ is a cut-off function satisfying $\eta =1$ on $\mathbb{B}_{R\varepsilon }$, and $|\nabla \eta |\leq \frac{2}{R\varepsilon }$. And G is given as in (56). b and c are constants which depend only on ε, to be determined later. To ensure $\phi _{\varepsilon }\in W_{0}^{1,2}(\varOmega )$, we let

$$ c+\frac{1}{c}\biggl(-\frac{1}{4\pi (1-\gamma )}\log \biggl(1+\frac{\pi }{1-\gamma } \frac{ \vert x \vert ^{2(1-\gamma )}}{\varepsilon ^{2(1-\gamma )}}\biggr)+b\biggr)= \frac{1}{c}\biggl(-\frac{1}{2\pi } \log R\varepsilon +A_{0}\biggr), $$

which leads to

$$ 2\pi c^{2}=-\log \varepsilon -2\pi b+2\pi A_{0}+ \frac{1}{2(1-\gamma )} \log \frac{\pi }{1-\gamma }+O\biggl( \frac{1}{R^{2(1- \gamma )}}\biggr). $$

(74)

Now we calculate

$$ \begin{aligned}[b] \int _{\mathbb{B}_{R\varepsilon }} \vert \nabla \phi _{\varepsilon } \vert ^{2}\,dx &= \int _{\mathbb{B}_{R}}\frac{ \vert x \vert ^{2-4\gamma }}{4 c^{2}(1-\gamma )^{2}(1+\frac{ \pi }{1-\gamma } \vert x \vert ^{2-4\gamma })^{2}}\,dx \\ &= \int _{0}^{\frac{\pi }{1-\gamma }R^{2-2\gamma }}\frac{t \,dt}{4 \pi c^{2} (1-\gamma )(1+t)^{2}}\,dt \\ &=\frac{1}{4\pi c^{2}(1-\gamma ) } \biggl(\log \frac{\pi }{1-\gamma }-1+ \log R^{2-2\gamma }+O \biggl(\frac{1}{R^{2-2\gamma }}\biggr) \biggr). \end{aligned} $$

(75)

On the other hand

$$ \begin{aligned}[b] \int _{\varOmega \setminus \mathbb{B}_{R\varepsilon }} \vert \nabla \phi _{\varepsilon } \vert ^{2}\,dx &= \frac{1}{c^{2}}\biggl( \int _{\varOmega \setminus \mathbb{B}_{R\varepsilon }} \vert \nabla G \vert ^{2}\,dx+ \int _{\mathbb{B}_{2R\varepsilon }\setminus \mathbb{B}_{R\varepsilon }} \bigl\vert \nabla (\xi \eta ) \bigr\vert ^{2}\,dx \\ &\quad {}-2 \int _{\mathbb{B}_{2R\varepsilon }\setminus \mathbb{B}_{R\varepsilon }}\nabla G\nabla (\xi \eta )\,dx \biggr) \\ &= \frac{1}{c^{2}}\biggl(- \int _{\varOmega \setminus \mathbb{B}_{R\varepsilon }}G \Delta G \,dx- \int _{\partial \mathbb{B}_{R\varepsilon }}G \frac{\partial G}{\partial \nu }\,ds \\ &\quad {}+ \int _{\mathbb{B}_{2R\varepsilon }\setminus \mathbb{B}_{R\varepsilon }} \bigl\vert \nabla (\xi \eta ) \bigr\vert ^{2}\,dx -2 \int _{\mathbb{B}_{2R\varepsilon }\setminus \mathbb{B}_{R\varepsilon }} \nabla G\nabla (\xi \eta )\,dx \biggr). \end{aligned} $$

Observe that $\xi (x)=O(|x|)$ as $x\rightarrow 0$. Since η is a cut-off function, it yields $|\nabla (\xi \eta )|=O(1)$ as $\varepsilon \rightarrow 0$. Then we have

$$ \int _{\mathbb{B}_{2R\varepsilon }\setminus \mathbb{B}_{R\varepsilon }} \bigl\vert \nabla (\xi \eta ) \bigr\vert ^{2}\,dx=O\bigl(R^{2}\varepsilon ^{2}\bigr),\qquad \int _{\mathbb{B}_{2R\varepsilon }\setminus \mathbb{B}_{R\varepsilon }} \nabla G \nabla (\xi \eta )\,dx=O(R\varepsilon ), $$

which together with (56) leads to

$$ \int _{\varOmega \setminus \mathbb{B}_{R\varepsilon }} \vert \nabla \phi _{\varepsilon } \vert ^{2}\,dx =\frac{1}{c^{2}} \biggl( -\frac{1}{2\pi } \log (R \varepsilon )+A_{0}+O(R\varepsilon ) \biggr). $$

(76)

Combining (75) and (76), a delicate but straightforward calculation shows

$$ \int _{\varOmega } \vert \nabla \phi _{\varepsilon } \vert ^{2}\,dx=\frac{1}{c^{2}} \biggl( -\frac{\log \varepsilon }{2\pi }- \frac{1}{4\pi (1-\gamma ) }+\frac{1}{4 \pi (1-\gamma ) }\log \frac{\pi }{1-\gamma }+A_{0}+O \biggl(\frac{1}{R^{2-2 \gamma }}\biggr) \biggr). $$

Put $\|\nabla \phi _{\varepsilon }\|_{2}=1$. It yields

$$ c^{2}=A_{0}-\frac{1}{2\pi }\log \varepsilon +\frac{1}{4\pi (1-\gamma ) }\log \frac{\pi }{1-\gamma }-\frac{1}{4\pi (1-\gamma ) }+O \biggl(\frac{1}{R ^{2-2\gamma }}\biggr). $$

(77)

Together with (74) and (77), we are led to

$$ b=\frac{1}{4\pi (1-\gamma ) }+O\biggl(\frac{1}{R^{2-2\gamma }}\biggr). $$

(78)

For all $x\in \mathbb{B}_{R\varepsilon }$, it follows from (77) and (78) that

$$ \begin{aligned}[b] 4\pi (1-\gamma )\phi _{\varepsilon }^{2} &\geq 4\pi (1-\gamma ) c^{2}+8 \pi (1- \gamma ) b-2\log \biggl(1+\frac{\pi \vert x \vert ^{2(1-\gamma )}}{(1- \gamma )\varepsilon ^{2(1-\gamma )}} \biggr) \\ &=1+4\pi (1-\gamma )A_{0}+\log \frac{\pi }{1-\gamma }- 2(1-\gamma ) \log \varepsilon \\ &\quad {}-2\log \biggl(1+\frac{\pi \vert x \vert ^{2(1-\gamma )}}{(1-\gamma ) \varepsilon ^{2(1-\gamma )}} \biggr)+O\biggl(\frac{1}{R^{2-2\gamma }}\biggr). \end{aligned} $$

(79)

Note that $\|\frac{\phi _{\varepsilon }(x)}{c}\|_{L^{\infty }(B_{R \varepsilon })}\rightarrow 1$ by passing to the limit $\varepsilon \rightarrow 0$. When $r\leq R\varepsilon$, there exists

$$ \biggl\vert \frac{\phi _{\varepsilon }(x)}{c} \biggr\vert = \biggl\vert 1+ \frac{-\log (1+\pi \frac{r^{2}}{ \varepsilon ^{2}})+b}{c^{2}} \biggr\vert \rightarrow 1. $$

as $\varepsilon \rightarrow 0$. Since $\phi _{\varepsilon }(x)\sim c$ in $\mathbb{B}_{R\varepsilon }$ and $g(c)=o(\frac{1}{c^{2}})$, we conclude $g(\phi _{\varepsilon }(\xi _{\varepsilon }))= o(\frac{1}{c^{2}})$ as $\varepsilon \rightarrow 0$, where $\xi _{\varepsilon }\in \mathbb{B} _{R\varepsilon }$. Combining with the mean value theorem, it follows from (79) that

$$\begin{aligned} \begin{aligned}[b] \int _{\mathbb{B}_{R\varepsilon }}\bigl(1+g(\phi _{\varepsilon })\bigr) \frac{e ^{4\pi (1-\gamma ) \phi _{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx &=\bigl(1+g\bigl( \phi _{\varepsilon }(\xi _{\varepsilon })\bigr)\bigr) \int _{\mathbb{B}_{R\varepsilon }}\frac{e^{4\pi (1-\gamma ) \phi _{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx \\ &\geq \biggl(1+o\biggl(\frac{1}{c^{2}}\biggr)\biggr)\frac{\pi }{1-\gamma }e^{1+4\pi (1- \gamma )A_{0}+ O(\frac{1}{R^{2-2\gamma }})} \\ &\quad {}\times \int _{0}^{R}\frac{2\pi r^{1-2\gamma }}{(1+\frac{\pi }{1- \gamma }r^{2-2\gamma })^{2}}\,dr \\ &=\frac{\pi }{(1-\gamma )}e^{1+4\pi (1-\gamma )A_{0}} +O\biggl(\frac{1}{R ^{2-2\gamma }}\biggr)+o \biggl(\frac{1}{c^{2}}\biggr). \end{aligned} \end{aligned}$$

(80)

Furthermore, $\frac{G}{c_{\varepsilon }}\geq 0$ a.e. in $\varOmega \setminus \mathbb{B}_{R\varepsilon }$, by using the inequality $e^{t}\geq t+1$, $\forall t\geq 0$, we estimate

$$ \begin{aligned}[b] &\int _{\varOmega \setminus \mathbb{B}_{R\varepsilon }}\bigl(1+g(\phi _{\varepsilon })\bigr) \frac{e^{4\pi (1-\gamma ) \phi _{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx\\ &\quad \geq \int _{\varOmega \setminus \mathbb{B}_{2R\varepsilon }}\bigl(1+g( \phi _{\varepsilon })\bigr) \frac{1+4\pi (1-\gamma )\phi _{\varepsilon }^{2}}{ \vert x \vert ^{2 \gamma }}\,dx \\ &\quad \geq \int _{\varOmega }\bigl(1+g(0)\bigr) \vert x \vert ^{-2\gamma } \,dx+O \bigl((R\varepsilon ) ^{2-2\gamma }\log ^{2}(R\varepsilon ) \bigr) \\ &\qquad {}+\frac{4\pi (1-\gamma )}{c^{2}} \int _{\varOmega } \bigl(1+g(0)\bigr) \vert x \vert ^{-2\gamma }G^{2}\,dx +O \bigl((R\varepsilon ) ^{2-2\gamma } \bigr). \end{aligned} $$

(81)

Observe that

$$ O \bigl((R\varepsilon ) ^{2-2\gamma } \bigr)=O \bigl((R\varepsilon ) ^{2-2\gamma }\log ^{2}(R\varepsilon ) \bigr)=O\biggl( \frac{1}{R^{2-2\gamma }}\biggr). $$

This together with (80) and (81) yields

$$\begin{aligned} \begin{aligned}[b] &\int _{\varOmega }\bigl(1+g(\phi _{\varepsilon })\bigr) \frac{e^{4\pi (1-\gamma ) \phi _{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx \\ &\quad \geq \bigl(1+g(0)\bigr) \int _{ \varOmega } \vert x \vert ^{-2\gamma }\,dx + \frac{\pi }{(1-\gamma )}e^{4\pi (1-\gamma )A_{0}+1} \\ &\qquad {}+\frac{4\pi (1-\gamma )}{c^{2}} \int _{\varOmega } \frac{(1+g(0))G^{2}}{ \vert x \vert ^{2 \gamma }}\,dx+O\biggl(\frac{1}{R^{2-2\gamma }} \biggr)+o\biggl(\frac{1}{c^{2}}\biggr). \end{aligned} \end{aligned}$$

(82)

Recalling (77) and the choice $R=-\log \varepsilon ^{1/(1- \gamma )}$, one can deduce that $\frac{1}{R^{2-2\gamma }}=o(\frac{1}{c ^{2}})$. Therefore, we conclude from (82) that

$$ \int _{\varOmega }\bigl(1+g(\phi _{\varepsilon })\bigr) \frac{e^{4\pi (1-\gamma ) \phi _{\varepsilon }^{2}}}{ \vert x \vert ^{2\gamma }}\,dx>\bigl(1+g(0)\bigr) \int _{\varOmega } \vert x \vert ^{-2 \gamma }\,dx + \frac{\pi }{1-\gamma }e^{4\pi (1-\gamma )A_{0}+1}. $$

for sufficiently small $\varepsilon >0$.

3.2 Completion of the proof of Theorem 2

Comparing (71) with (72), we arrive at the final conclusion that $c_{\varepsilon }$ must be bounded. Then applying elliptic estimates to (16), we can get the desired extremal function. This ends the proof of Theorem 2.