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.