Energy quantization for a singular super-Liouville boundary value problem

Abstract

In this paper, we develop the blow-up analysis and establish the energy quantization for solutions to super-Liouville type equations on Riemann surfaces with conical singularities at the boundary. In other problems in geometric analysis, the blow-up analysis usually strongly utilizes conformal invariance, which yields a Noether current from which strong estimates can be derived. Here, however, the conical singularities destroy conformal invariance. Therefore, we develop another, more general, method that uses the vanishing of the Pohozaev constant for such solutions to deduce the removability of boundary singularities.

1 Introduction

Many problems with a noncompact symmetry group, like the conformal group, are limit cases where the Palais–Smale condition no longer applies, and therefore, solutions may blow up at isolated singularities, see for instance [31]. Therefore, a blow-up analysis is needed, and this has become one of the fundamental tools in the geometric calculus of variations. This usually depends on the fact that the invariance yields an associated Noether current whose algebraic structure can be turned into estimates. In the case of conformal invariance this Noether current is a holomorphic quadratic differential. For harmonic map type problems, finiteness of the energy functional in question implies that that differential is in $L^1$. This then can be used to obtain fundamental estimates. For other problems, however, like (super-) Liouville equations, finiteness of the energy functional is not sufficient to get the $L^1$ bound of that differential and hence this is an extra assumption leading to the removability of local singularities (Prop 2.6, [23]).

But for (super-) Liouville equations on surfaces with conical singularities, we do not even have conformal invariance, because the scaling behavior at the singularities is different from that at regular points, see [27]. It turns out, however, that for an important class of two-dimensional geometric variational problems, there is a condition that is weaker than conformal invariance, the vanishing of a so-called Pohozaev constant (i.e. the Pohozaev identity), that is not only sufficient but also necessary for the blow-up analysis. This Pohozaev constant on one hand measures the extent to which the Pohozaev identity fails and on the other hand provides a characterization of the singular behavior of a solution at an isolated singularity. Such kind of quantity appears also in other two dimensional geometric variational problems and can be applied to study the qualitative asymptotic behaviour of solutions defined on degenerating surfaces [20, 40, 41]. This vanishing condition is already known to play a crucial role in geometric analysis (see e.g. [34]), but for super-Liouville equations, as mentioned, this identity by itself suffices for the blow-up analysis.

In this paper, we shall apply this strategy to the blow-up analysis of the (super-)Liouville boundary problem on surfaces with conical singularities. To this purpose, let M be a compact Riemann surface with nonempty boundary $\partial M$ and with a spin structure. We also denote this compact Riemann surface as $(M,\mathcal {A}, g)$, where g is its Riemannian metric with the conical singularities of divisor

$$\begin{aligned} \mathcal {A}= \sum ^m_{j=1}\alpha _j q_j \end{aligned}$$

(for definition of $\mathcal {A}$, see Sect. 2). Associated to the metric g, one can define the gradient $\nabla$ and the Laplace operator $\Delta$ in the usual way.

We then have our main object of study, the super-Liouville functional that couples a real-valued function u and a spinor $\psi$ on M

$$\begin{aligned} E_B\left( u,\psi \right) =\int _{M}\left\{ \frac{1}{2} \left| \nabla u\right| ^2+K_gu+\left\langle (\not {D}+e^u)\psi ,\psi \right\rangle _g -e^{2u}\right\} dv+\int _{\partial M}\{h_gu-ce^u \}d\sigma , \end{aligned}$$

(1)

where $K_g$ is the Gaussian curvature in M, and $h_g$ is the geodesic curvature of $\partial M$ and c is a given positive constant. The Dirac operator $\not {D}$ is defined by $\not {D}\psi :=\sum _{\alpha =1}^2e_{_\alpha }\cdot \nabla _{e_\alpha }\psi ,$ where $\left\{ e_1,e_2\right\}$ is an orthonormal basis on TM, $\nabla$ is the Levi–Civita connection on M with respect to g and $\cdot$ denotes Clifford multiplication in the spinor bundle $\Sigma M$ of M. Finally, $\langle \cdot ,\cdot \rangle _g$ is the natural Hermitian metric on $\Sigma M$ induced by g. We also write $|\cdot |_g^2$ as $\langle \cdot , \cdot \rangle _g$. For the geometric background, see [28] or [19].

The Euler-Lagrange system for $E_B(u,\psi )$ with Neumann / chirality boundary conditions is

(2)

Here $B^{\pm }$ are the chirality operators (see Sect. 2 for the definition).

When $\psi =0$ and (M, g) is a closed smooth Riemann surface, we obtain the classical Liouville functional

$$\begin{aligned} E\left( u \right) =\int _{M}\left\{ \frac{1}{2} \left| \nabla u\right| ^2+K_gu -e^{2u}\right\} dv. \end{aligned}$$

The Euler-Lagrange equation for E(u) is the Liouville equation

$$\begin{aligned} -\Delta _g u = \displaystyle 2e^{2u} -K_g. \end{aligned}$$

Liouville [32] studied this equation in the plane, that is, for $K_g=0$. The Liouville equation comes up in many problems of complex analysis and differential geometry of Riemann surfaces, for instance the prescribing curvature problem. The interplay between the geometric and analytic aspects makes the Liouville equation mathematically very interesting.

When $\psi \ne 0$ and (M, g) again is a closed smooth Riemann surface, we obtain the super-Liouville functional

$$\begin{aligned} E\left( u,\psi \right) =\int _{M}\left\{ \frac{1}{2} \left| \nabla u\right| ^2+K_gu+\left\langle (\not {D}+e^u)\psi ,\psi \right\rangle _g -e^{2u}\right\} dv. \end{aligned}$$

The Euler-Lagrange system for $E(u,\psi )$ is

The supersymmetric version of the Liouville functional and equation have been studied extensively in the physics literature, see for instance [1, 15, 33]. As all supersymmetric functionals that arise in elementary particle physics, it needs anticommuting variables.

Motivated by the super-Liouville functional, a mathematical version of this functional that works with commuting variables only, but otherwise preserves the structure and the invariances of it, was introduced in [21]. That model couples the bosonic scalar field to a fermionic spinor field. In particular, the super-Liouville functional is conformally invariant, and it possesses a very interesting mathematical structure.

The analysis of classical Liouville type equations was developed in [2, 4, 29, 30] etc, and the corresponding analysis for super-Liouville equations in [21, 23, 26]. In particular, the complete blow-up theory for sequences of solutions was established, including the energy identity for the spinor part, the blow-up value at blow-up points and the profile for a sequence of solutions at the blow-up points. For results by physicists about super-Liouville equations, we refer to [1, 15, 33] etc.

When $(M,\mathcal {A}, g)$ is a closed Riemann surface (without boundary) with conical singularities of divisor $\mathcal {A}$ and with a spin structure, we obtain that

$$\begin{aligned} E\left( u,\psi \right) =\int _{M}\left\{ \frac{1}{2} \left| \nabla u\right| ^2+K_gu+\left\langle (\not {D}+e^u)\psi ,\psi \right\rangle _g -e^{2u}\right\} dv_g. \end{aligned}$$

The Euler-Lagrange system for $E(u,\psi )$ is

(3)

This system is closely related to the classical Liouville equation, or the prescribing curvature equation on M with conical singularites (see [12, 36]). In addition, [3, 4, 6,7,8, 35] studied the blow-up theory of the following Liouville type equations with singular data:

$$\begin{aligned} -\Delta _g u=\lambda \frac{Ke^u}{\int _M K e^{u}dg}-4\pi \left( \sum _{j=1}^{m}\alpha _j\delta _{q_j}- f\right) , \end{aligned}$$

where (M, g) is a smooth surface and the singular data appear in the equation, which is the asymptotic behavior associated to the m interior punctures. For system (3), [27] provides an analytic foundation and the blow-up theory.

For Liouville boundary problems on (M, g) with or without conical singularites, there are also lots of results on the blow-up analysis, see [9, 16, 22, 43, 44]. For super-Liouville boundary problems on a smooth Riemann surface M, the corresponding results can be found in [24, 25].

In this paper, we aim to provide an analytic foundation and to establish the blow-up analysis for the system (2). Our main result is the following energy quantization property for solutions to (2):

Theorem 1.1

Let $(u_n,\psi _n)$ be a sequence of solutions of (2) with energy conditions:

$$\begin{aligned} \int _{M}e^{2u_n}dg<C,~~~~~~\int _{M}|\psi _n|_g^4dg<C. \end{aligned}$$

Define

$$\begin{aligned} \Sigma _1=\left\{ x\in M,\text { there is a sequence }y_n\rightarrow x\text { such that }u_n(y_n)\rightarrow +\infty \right\} . \end{aligned}$$

If $\Sigma _1\ne \emptyset$, then the possible values of

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\{ \int _{M}2e^{2u_n}-e^{u_n}|\psi _n|^2_{g}dv_g+\int _{\partial M}ce^{u_n}d\sigma _g\right\} \end{aligned}$$

are

$$\begin{aligned} 4\pi m_1+2\pi m_2+\sum _{j\in J_1} 4\pi (1+\alpha _j)+\sum _{j\in J_1} 2\pi (1+\alpha _j), \end{aligned}$$

where $m_1, m_2$ are nonnegative integers, $J_1\subset \{1,\dots , l\}$ and $J_2 \subset \{l+1, \dots , m\}$. Here $\{1,\dots , l\}$ and $\{l+1, \dots , m\}$ are the indexing sets of the conical singularities of M. $J_1$ and $J_2$ could be empty.

From the energy quantization property, one can deduce the concentration properties of conformal volume and the compactness of solutions. It turns out that understanding of this property is the key step to study existence from a variational point of view by a refined Moser–Trudinger inequality, see e.g. [10, 11, 18].

If we assume that the points $q_1, q_2, \ldots , q_l$ are in $M^o$ for $1\le l<m$ and the points $q_{l+1}, q_{l+2}, \ldots , q_m$ are on $\partial M$ for the surface $(M,\mathcal {A},g)$ with the divisor $\mathcal {A}=\Sigma _{j=1}^{m}\alpha _jq_j$, $\alpha _j>0$, we have the following Gauss–Bonnet formula

$$\begin{aligned} \frac{1}{2\pi }\int _{M}K_gdv_g+\frac{1}{2\pi }\int _{\partial M}h_gd\sigma _g = \mathcal {X}(M) + |\mathcal {A}|, \end{aligned}$$

where $\mathcal {X}(M)=2-2g_M$ is the topological Euler characteristic of M itself, $g_M$ is the genus of M and

$$\begin{aligned} |\mathcal {A}|= \sum _{j=1}^{l}\alpha _j+\sum _{j=l+1}^{m}\frac{\alpha _j}{2} \end{aligned}$$

is the degree of $\mathcal {A}$, see [36]. From (2) we obtain that

$$\begin{aligned}&\int _{M}2e^{2u_n}-e^{u_n}|\psi _n|^2_gdv_g+\int _{\partial M}ce^{u_n}d\sum _{g}=\int _{M}K_gdv_g+\int _{\partial M}h_gd\sigma _g\\&\quad =2\pi (\mathcal {X}(M) + |\mathcal {A}|). \end{aligned}$$

Then we can use Theorem 1.1 to get the following:

Theorem 1.2

Let $(M,\mathcal {A},g)$ be as above, and $(u_n,\psi _n)$ be as in Theorem 1.1. Then the value in Theorem 1.1 satisfies

$$\begin{aligned} 4\pi m_1+2\pi m_2+\sum _{j\in J_1} 4\pi (1+\alpha _j)+\sum _{j\in J_1} 2\pi (1+\alpha _j)=4\pi (1-g_M)+ 2\pi |\mathcal {A}|. \end{aligned}$$

In particular,

• (i) If

  $$\begin{aligned} 4\pi (1-g_M)+ 2\pi |\mathcal {A}|= 2\pi , \end{aligned}$$

  then the blow-up set $\Sigma _1$ contains at most one point.

• (ii) If

  $$\begin{aligned} 4\pi (1-g_M)+ 2\pi |\mathcal {A}| < 2\pi , \end{aligned}$$

  then the blow-up set $\Sigma _1=\emptyset$.

To show Theorem 1.1, a key step is to compute the blow-up value

$$\begin{aligned} m(p)=\lim _{R\rightarrow 0}\lim _{n\rightarrow \infty }\left\{ \int _{B^M_R(p)}(2e^{2u_n}-e^{u_n}|\psi _n|_{g}^2-K_{g})dv_{g}+\int _{\partial M\cap B^M_R(p)}(ce^{u_n}-h_{g})d\sigma _{g}\right\} , \end{aligned}$$

at the blow-up point $p\in \Sigma _1$ for a blow-up sequence $(u_n,\psi _n)$. Here $B^M_R(p)$ is a geodesic ball of (M, g) at p. For this purpose, we need to study the following local super-Liouville boundary value problem (see Sect. 3):

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u(x) &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2u(x)}-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2, \quad &{}\text {in } D^+_{r}, \\ \not {D}\Psi &{}=&{} -V(x)|x|^{\alpha }e^{u(x)}\Psi , \quad &{}\text {in } D^+_{r}, \\ \frac{\partial u}{\partial n}&{} = &{} cV(x)|x|^{\alpha }e^{u(x)},\quad &{} \text { on } L_r, \\ B^{\pm }\Psi &{}=&{} 0,\quad &{} \text { on } L_r. \end{array} \right. \end{aligned}$$

(4)

Here $\alpha \ge 0$, V(x) is in $C^{1}_{loc}(D^+_r\cup L_r)$ and satisfies $0< a\le V(x)\le b$. $L_r$ and $S_r^+$ here and in the sequel are portions of $\partial D^+_r$, which are defined in section 3. Then we have the following Brezis–Merle type concentration compactness theorem:

Theorem 1.3

Let $(u_n,\Psi _n)$ be a sequence of regular solutions to (4) satisfying

$$\begin{aligned} \int _{D^+_r}|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4dx+\int _{L_r}|x|^{\alpha }e^{u_n}ds<C. \end{aligned}$$

Define

$$\begin{aligned} \Sigma _1= & {} \left\{ x\in D^+_r\cup L_r,\text { there is a sequence }y_n\rightarrow x\text { such that }u_n(y_n)\rightarrow +\infty \right\} ,\\ \Sigma _2= & {} \left\{ x\in D^+_r\cup L_r,\text { there is a sequence }y_n\rightarrow x\text { such that }\left| \Psi _n(y_n)\right| \rightarrow +\infty \right\} . \end{aligned}$$

Then, we have $\Sigma _2\subset \Sigma _1$. Moreover, $(u_n,\Psi _n)$ admits a subsequence, still denoted by $( u_n,\Psi _n),$ that satisfies

• a) $|\Psi _n|$ is bounded in $L_{loc}^{\infty } ( (D^+_r\cup L_r)\backslash \Sigma _2)$ .

• b) For $u_n$, one of the following alternatives holds:

  • i) $u_n$ is bounded in $L^{\infty }_{loc} (D^+_r\cup L_r).$

  • ii) $u_n$ $\rightarrow -\infty$ uniformly on compact subsets of $D^+_r\cup L_r$.

  • iii) $\Sigma _1$ is finite, nonempty and either

    $$\begin{aligned} u_n\text { is bounded in }L_{loc}^\infty ((D^+_r\cup L_r)\backslash \Sigma _1) \end{aligned}$$

    (5)

    or

    $$\begin{aligned} u_n\rightarrow -\infty \text { uniformly on compact subsets of } (D^+_r\cup L_r)\backslash \Sigma _1. \end{aligned}$$

    (6)

To show the quantization property of the blow-up value, we need to rule out (5) in the above theorem. When the spinor field is vanishing, namely, in the case of Liouville-type problems, a technique based on Pohozaev identity was introduced in [7] to prove the vanishing of the mass in the neck region for blow-up sequences of solutions with interior conical singularities. In the case of super-Liouville-type problems, we need to overcome some new difficulties caused by the spinor part. For the case of interior conical singularities, this was achieved in [27]. In the present paper, we shall handle the boundary conical singularities case. To this end, the decay estimates of the spinor part $\Psi _n$, the Pohozaev identity of the local coupled system (4) and the energy identity of $\Psi _n$, which means there is no energy contribution on the neck domain, play the essential roles. The corresponding theorem is the following:

Theorem 1.4

Let $(u_n,\Psi _n)$ be a sequence of regular solutions to (4) satisfying

$$\begin{aligned} \int _{D^+_r}|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4dx+\int _{L_r}|x|^{\alpha }e^{u_n}ds<C. \end{aligned}$$

Denote by $\Sigma _1=\{x_1, x_2,\ldots , x_l\}$ the blow-up set of $u_n$. Then there are finitely many solutions $(u^{i,k},\Psi ^{i,k})$ that satisfy

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u^{i,k} &{}=&{} \displaystyle 2|x|^\alpha e^{2u^{i,k}}-|x|^\alpha e^{u^{i,k}}\left\langle \Psi ^{i,k} ,\Psi ^{i,k} \right\rangle -1 ,\qquad &{}\text {in } S^2, \\ \not {D}\Psi ^{i,k} &{}=&{}\displaystyle -|x|^\alpha e^{u^{i,k}}\Psi ^{i,k}, \quad &{}\text {in } S^2,\\ \end{array} \right. \end{aligned}$$

(7)

for $i=1,2,\ldots , I$, and $k=1,2,\ldots , K_i$, and $\alpha \ge 0$, or there are finitely many solutions $(u^{j,l},\Psi ^{j,l})$ that satisfy

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u^{j,l} &{}=&{} \displaystyle 2|x|^\alpha e^{2u^{j,l}}-|x|^\alpha e^{u^{j,l}}\left\langle \Psi ^{j,l} ,\Psi ^{j,l} \right\rangle -1 ,\quad &{}\text { in } S^2_{c'}, \\ \not {D}\Psi ^{j,l} &{}=&{}\displaystyle -|x|^\alpha e^{u^{j,l}}\Psi ^{j,l}, \quad &{}\text { in } S^2_{c'},\\ \displaystyle \frac{\partial u^{j,l}}{\partial n}&{} = &{} c|x|^\alpha e^{u^{j,l}}-c', \quad &{}\text { on } \partial S^2_{c'},\\ B^\pm \Psi ^{j,l} &{}=&{} 0, \quad &{}\text { on } \partial S^2_{c'}, \end{array} \right. \end{aligned}$$

(8)

for $j=1,2,\ldots , J$, and $l=1,2,\cdots , L_j$, and $\alpha \ge 0$. Here $S^2_{c'}$ is a portion of the sphere cut out by a 2-plane with constant geodesic curvature $c'$. After selection of a subsequence, $\Psi _n$ converges in $C_{loc}^{\infty }$ to $\Psi$ on $(B_r^+\cup L_r)\backslash \Sigma _1$ and we have the energy identity:

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{D^+_r}|\Psi _n|^4dv=\int _{D^+_r}|\Psi |^4dv+\sum _{i=1}^{I} \sum _{k=1}^{K_i}\int _{S^2}|\Psi ^{i,k}|^4dv+\sum _{j=1}^{J} \sum _{l=1}^{L_j}\int _{S^2_{c'}}|\Psi ^{j,l}|^4dv. \end{aligned}$$

(9)

A crucial step in proving the above theorem is to show the removability of isolated singularities at the boundary, which is equivalent to the vanishing of the Pohozaev constant (see Theorem 4.5). Once the energy identity for the spinor part (9) is established, we can then obtain

Theorem 1.5

Let $(u_n, \Psi _n)$ be solutions as in Theorem 1.3. Assume that $(u_n, \Psi _n)$ blows up and the blow-up set $\Sigma _1\ne \emptyset$. Then

$$\begin{aligned} u_n\rightarrow -\infty \quad \text { uniformly on compact subsets of } (D^+_r\cup L_r)\setminus \Sigma _1. \end{aligned}$$

Furthermore,

$$\begin{aligned}&\int _{D^+_r(0)}[2V(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2]\phi dx+ \int _{L_r}cV(x)|x|^{\alpha }e^{u_n}\\&\quad \rightarrow \sum _{x_i\in \Sigma _1}m(x_i)\phi (x_i) \end{aligned}$$

for every $\phi \in C^\infty _o(D^+_r\cup L_r)$ and $m(x_i)>0$.

In the end, with the help of the Pohozaev identity (see Proposition 4.1) and the Green function at some singular points, we have the following:

Theorem 1.6

Let $(u_n, \Psi _n)$ be solutions as in Theorem 1.3. Assume that $(u_n, \Psi _n)$ blows up and the blow-up set $\Sigma _1\ne \emptyset$. Let $p\in \Sigma _1$ and assume that p is the only blow-up point in $\overline{D_{\delta _0}^+(p)}$ for some $\delta _0>0$. If there exists a positive constant C such that

$$\begin{aligned} \max _{S_{\delta _0}^+(p)}u_n-\min _{S_{\delta _0}^+(p)}u_n\le C, \end{aligned}$$

then the blow-up value $m(p)=4\pi$ when $p\notin L_{\delta _0}(p)$, $m(p)=2\pi$ when $p\in L_{\delta _0}(p)\setminus \{0\}$, and $m(p)=2\pi (1+\alpha )$ when $p=0$.

2 Preliminaries

In this section, we will first recall the definition of surfaces with conical singularities, following [36]. Then we shall recall the chirality boundary condition for the Dirac operator $\not {D}$, see e.g. [17]. In particular, we will see that under the chirality boundary conditions $B^{\pm }$, the Dirac operator $\not {D}$ is self-adjoint.

A conformal metric g on a Riemannian surface M (possibly with boundary) has a conical singularity of order $\alpha$ (a real number with $\alpha >-1$) at a point $p\in M\cup \partial M$ if in some neighborhood of p

$$\begin{aligned} g=e^{2u}|z-z(p)|^{2\alpha }|dz|^2 \end{aligned}$$

where z is a coordinate of M defined in this neighborhood and u is smooth away from p and continuous at p. The point p is then said to be a conical singularity of angle $\theta =2\pi (\alpha +1)$ if $p\notin \partial M$ and a corner of angle $\theta =\pi (\alpha +1)$ if $p\in \partial M$. For example, a (somewhat idealized) American football has two singularities of equal angle, while a teardrop has only one singularity. Both these examples correspond to the case $-1<\alpha <0$; in case $\alpha >0$, the angle is larger than $2\pi$, leading to a different geometric picture. Such singularities also appear in orbifolds and branched coverings. They can also describe the ends of complete Riemann surfaces with finite total curvature. If (M, g) has conical singularities of order $\alpha _1, \alpha _2, \ldots , \alpha _m$ at $q_1, q_2, \ldots , q_m$, then g is said to represent the divisor $\mathcal {A}= \Sigma ^m_{j=1}\alpha _j q_j$. Importantly, the presence of such conical singularities destroys conformal invariance, because the conical points are different from the regular ones.

The chirality boundary condition for the Dirac operator $\not {D}$ is a natural boundary condition for spinor part $\psi$. Let M be a compact Riemann surface with $\partial M\ne \emptyset$ and with a fixed spin structure, admitting a chirality operator G, which is an endomorphism of the spinor bundle $\Sigma M$ satisfying:

$$\begin{aligned} G^2=I,\qquad \langle G\psi ,G\varphi \rangle =\langle \psi , \varphi \rangle , \end{aligned}$$

and

$$\begin{aligned} y \nabla _{X}(G\psi )=G\nabla _{X}\psi ,\qquad X\cdot G\psi =-G(X\cdot \psi ), \end{aligned}$$

for any $X\in \Gamma (TM), \psi , \varphi \in \Gamma (\Sigma M)$. Here I denotes the identity endomorphism of $\Sigma M$ and $\Gamma (\cdot )$ denotes the space of sections of a given bundle.

We usually take $G=\gamma (\omega _2)$, which denotes the Clifford multiplication by the complex volume form $\omega _2=ie_1e_2$, where $e_1,e_2$ is a local orthonormal frame on M.

Denote by

$$\begin{aligned} S:=\Sigma M|_{\partial M} \end{aligned}$$

the restricted spinor bundle with induced Hermitian product.

Let $\overrightarrow{n}$ be the outward unit normal vector field on $\partial M$. One can verify that $\overrightarrow{n}G:\Gamma (S)\rightarrow \Gamma (S)$ is a self-adjoint endomorphism satisfying

$$\begin{aligned} (\overrightarrow{n}G)^2=I,\qquad \langle \overrightarrow{n} G\psi ,\varphi \rangle =\langle \psi , \overrightarrow{n} G\varphi \rangle , \end{aligned}$$

Hence, we can decompose $S=V^{+}\bigoplus V^{-}$, where $V^{\pm }$ is the eigensubbundle corresponding to the eigenvalue $\pm 1$. One verifies that the orthogonal projection onto the eigensubbundle $V^{\pm }$:

$$\begin{aligned}&\displaystyle B^{\pm }:L^2(S)\rightarrow L^2(V^{\pm })\\&\displaystyle \psi \mapsto \frac{1}{2}(I\pm \overrightarrow{n}G)\psi , \end{aligned}$$

defines a local elliptic boundary condition for the Dirac operator $\not {D}$ , see e.g. [17]. We say that a spinor $\psi \in L^2(\Gamma (\Sigma M))$ satisfies the chirality boundary conditions $B^{\pm }$ if

$$\begin{aligned} B^\pm \psi |_{\partial M}=0. \end{aligned}$$

It is well known (see e.g. [17]) that if $\psi ,\phi \in L^2(\Gamma (\Sigma M))$ satisfy the chirality boundary conditions $B^\pm$ then

$$\begin{aligned} \langle \overrightarrow{n}\cdot \psi ,\varphi \rangle =0, \quad \text { on } \partial M. \end{aligned}$$

In particular,

$$\begin{aligned} \int _{\partial M}\langle \overrightarrow{n}\cdot \psi ,\varphi \rangle =0. \end{aligned}$$

(10)

It follows that the Dirac operator $\not {D}$ is self-adjoint under the chirality boundary conditions $B^{\pm }$.

It may be helpful if we recall that on a surface the (usual) Dirac operator $\not {D}$ can be seen as the (doubled) Cauchy-Riemann operator. Consider $\mathbb {R}^2$ with the Euclidean metric $ds^2+dt^2$. Let $e_1=\frac{\partial }{\partial s}$ and $e_2=\frac{\partial }{\partial t}$ be the standard orthonormal frame. A spinor field is simply a map $\Psi :\mathbb {R}^2\rightarrow \Delta _2=\mathbf {C}^2$, and the actions of $e_1$ and $e_2$ on spinor fields can be identified by multiplication with matrices

$$\begin{aligned} e_1=\left( \begin{matrix}0&{} i\\ i&{}0 \end{matrix}\right) , \quad e_2=\left( \begin{matrix}0&{} 1\\ -1&{}0 \end{matrix}\right) . \end{aligned}$$

If $\Psi :=\left( \begin{matrix} \displaystyle f \\ \displaystyle g\end{matrix}\right) :\mathbb {R}^2\rightarrow \mathbf {C}^2$ is a spinor field, then the Dirac operator is

$$\begin{aligned} \not {D}\Psi =\displaystyle \left( \begin{matrix}0&{} i\\ i&{}0 \end{matrix}\right) \left( \begin{matrix} \displaystyle \frac{\partial f}{\partial s}\\ \displaystyle \frac{\partial g}{\partial s} \end{matrix}\right) + \left( \begin{matrix}0&{} 1\\ -1&{}0 \end{matrix}\right) \left( \begin{matrix} \displaystyle \frac{\partial f}{\partial t} \\ \displaystyle \frac{\partial g}{\partial t} \end{matrix}\right) = 2i\left( \begin{matrix} \displaystyle \frac{\partial g}{\partial z} \\ \displaystyle \frac{\partial f}{\partial \bar{z}}\end{matrix}\right) , \end{aligned}$$

where

$$\begin{aligned} \frac{\partial }{\partial z}=\frac{1}{2} \left( \frac{\partial }{\partial s} - i\frac{\partial }{\partial t}\right) , \quad \frac{\partial }{\partial \bar{z}}=\frac{1}{2} \left( \frac{\partial }{\partial s} + i\frac{\partial }{\partial t}\right) . \end{aligned}$$

Therefore, the elliptic estimates developed for (anti-) holomorphic functions can be used to study the Dirac equation.

If M be the upper-half Euclidean space $\mathbb {R}^{2}_{+}$, then the chirality operator is $G=ie_1e_2=\left( \begin{matrix}1&{} 0\\ 0&{}-1 \end{matrix}\right)$. Note that $\overrightarrow{n}=-e_2$, we get that

$$\begin{aligned} B^{\pm }=\frac{1}{2}(I\pm \overrightarrow{n}\cdot G)=\frac{1}{2}\left( \begin{matrix}1&{} \pm 1\\ \pm 1&{} 1 \end{matrix}\right) . \end{aligned}$$

By the standard chirality decomposition, we can write $\psi =\left( \begin{matrix}\psi _+\\ \psi _{-} \end{matrix}\right)$, and then the boundary condition becomes

$$\begin{aligned} \psi _{+}=\mp \psi _{-} \quad \text { on } \partial M. \end{aligned}$$

Without loss of generality, in the sequel, we shall only consider the chirality boundary condition $B=B^+$.

We have the following geometric property:

Proposition 2.1

The functional $E_B(u,\psi )$ is invariant under conformal diffeomorphisms $\varphi :M\rightarrow M$ preserving the divisor, that is, be $\varphi ^{*} \mathcal {A}= \mathcal {A}$. In other word, if we write that $\varphi ^*(g)=\lambda ^2 g$, where $\lambda >0$ is the conformal factor of the conformal map $\varphi$, and set

$$\begin{aligned} \displaystyle \widetilde{u}= & {} \displaystyle u\circ \varphi -ln\lambda , \nonumber \\ \displaystyle \widetilde{\psi }= & {} \displaystyle \lambda ^{-\frac{1}{2}}\psi \circ \varphi , \end{aligned}$$

(11)

then $E_B(\tilde{u},\tilde{\psi })=E_B(u,\psi )$. In particular, if $(u,\psi )$ is a solution of (2), so is $(\tilde{u},\tilde{\psi })$.

3 The local singular super-Liouville boundary problem

In this section, we shall first derive the local version of the super-Liouville boundary problem. Then we shall analyze the regularity of solutions under the small energy condition.

First we take a point $p\in M^o$, choose a small neighborhood $U(p) \subset M^o$, and define an isothermal coordinate system $x=(x_1,x_2)$ centered at p, such that p corresponds to $x=0$ and $g=e^{2\phi }|x|^{2\alpha }(dx_1^2+dx_2^2)$ in $D_{r}(0)=\{(x_1,x_2) \in \mathbb {R}^2 \mid x_1^2+x_2^2< r^2\}$, where $\phi$ is smooth away from p, continuous at p and $\phi (p)=0$. We can choose such a neighborhood small enough so that if p is a conical singular point of g, then $U(p)\cap {\mathcal {A}}=\{p\}$ and $\alpha >0$, while, if p is a smooth point of g, then $U(p)\cap {\mathcal {A}}=\emptyset$ and $\alpha =0$. Consequently, with respect to the isothermal coordinates, we can obtain the local version of the singular super-Liouville-type equations,

$$\begin{aligned} \left\{ \begin{array}{rcl} -\Delta u(x) &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2u(x)}-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2 \quad \\ \not {D}\Psi &{}=&{} -V(x)|x|^{\alpha }e^{u(x)}\Psi \end{array} \text { in } D_{r}(0), \right. \end{aligned}$$

(12)

which has no boundary condition since p is a interior point of M. Here $\Psi =|x|^{\frac{\alpha }{2}}e^{\frac{\phi (x)}{2}}\psi$, V(x) is a $C^{1,\beta }$ function and satisfies $0< a\le V(x)\le b$. The detailed arguments can be found in the section 3 of [27]. We also assume that $(u,\Psi )$ satisfy the energy condition:

$$\begin{aligned} \int _{D_r(0)}|x|^{2\alpha }e^{2u}+|\Psi |^4dx<+\infty . \end{aligned}$$

(13)

We put $D_r:=D_r(0)$. We say that $(u, \Psi )$ is a weak solution of (12) and (13), if $u\in W^{1,2}(D_r)$ and $\Psi \in W^{1,\frac{4}{3}}(\Gamma (\Sigma D_r))$ satisfy

$$\begin{aligned} \int _{D_r} \nabla u\nabla \phi dx= & {} \int _{D_r} (2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u |\Psi |^2)\phi dx, \\ \int _{D_r} \langle \Psi ,\not {D} \xi \rangle dx= & {} -\int _{D_r} V(x)|x|^{\alpha }e^u \langle \Psi ,\xi \rangle dx, \end{aligned}$$

for any $\phi \in C^\infty _0(D_r)$ and any spinor $\xi \in C^\infty \cap W_0^{1,\frac{4}{3}} (\Gamma (\Sigma D_r))$. A weak solution is a classical solution by the following:

Proposition 3.1

Let $(u,\Psi )$ be a weak solution of (12) and (13). Then $(u,\Psi )\in C^2(D_r)\times C^2(\Gamma (\Sigma D_r))$.

Note that when $\alpha =0$ this proposition is proved in [21] (see Proposition 4.1). When $\alpha >0$, this proposition is proved in [27] (see Proposition 3.1).

For $p\in \partial M$, we also can choose a small geodesic ball $U(p) \subset M$ and define an isothermal coordinate system $x=(x_1,x_2)$ centered at p, such that p corresponds to $x=0$ and $g=e^{2\phi }|x|^{2\alpha }(dx_1^2+dx_2^2)$ in $\overline{D}^+_{r}(0)=\{(s,t)\in \mathbb {R}^2 \mid s^2+t^2< r^2, t\ge 0 \}$, where $\phi$ is smooth away from p and continuous at p. We can choose such a geodesic ball small enough so that if p is a conical singular point of g, then $U(p)\cap {\mathcal {A}}=\{p\}$ and $\alpha >0$, while, if p is a smooth point of g, then $U(p)\cap {\mathcal {A}}=\emptyset$ and $\alpha =0$. Set $L_r=\partial D_r^+\cap \partial {\mathbb {R}}^2_+$, and $S^+_r=\partial D^+_r\cap \mathbb {R}^2_+$. Also in the sequel, we will set $L_r(x_0)=\partial D_r^+(x_0)\cap \partial {\mathbb {R}}^2_+$, and $S^+_r(x_0)=\partial D^+_r(x_0)\cap \mathbb {R}^2_+$. Consequently, with respect to the isothermal coordinates, $(u,\psi )$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u(x) &{}=&{} e^{2\phi (x)}|x|^{2\alpha }(2e^{2u(x)}-e^{u(x)}|\psi |^2(x) -K_g)\quad &{}\text {in } D^+_{r},\\ \not {D}\left( e^{\frac{\phi (x)}{2}}|x|^{\frac{\alpha }{2}}\psi \right) &{}=&{} -e^{\phi (x)}|x|^{\alpha }e^{u(x)}\left( e^{\frac{\phi (x)}{2}} |x|^{\frac{\alpha }{2}}\psi \right) \quad &{} \text {in } D^+_{r},\\ \displaystyle \frac{\partial u}{\partial n}&{} = &{} e^{\phi (x)}|x|^{\alpha }(ce^u-h_g),\quad &{} \text { on } L_r,\\ B\left( e^{\frac{\phi (x)}{2}}|x|^{\frac{\alpha }{2}}\psi \right) &{}=&{} 0,\quad &{} \text { on } L_r. \end{array} \right. \end{aligned}$$

(14)

Here $\Delta =\partial ^2_{x_1x_1}+\partial ^2_{x_2x_2}$ is the usual Laplacian, and the Dirac operator $\not {D}$ can be seen as doubled Cauchy–Riemann operator, B is the chirality boundary operator of spinors.

Note that the relation between the two Gaussian curvatures and between the two geodesic curvatures are respectively

$$\begin{aligned} \left\{ \begin{array}{rcl} -\Delta \phi &{}=&{} e^{2\phi }|x|^{2\alpha }K_g,\\ \frac{\partial \phi }{\partial n} &{}=&{} e^{\phi }|x|^{\alpha }h_g. \end{array} \right. \end{aligned}$$

By standard elliptic regularity we conclude that $\phi \in W^{2,p}_{loc}(D^+_r\cup L_r)$ for some $p>1$ if $\alpha \ge 0$ and if the curvature $K_g$ and $h_g$ of M is regular enough. Therefore, by Sobolev embedding, $\phi \in C^{1}_{loc}(D^+_r\cup L_r)$. If we denote $V(x)=e^{\phi }$, $W_1(x)=e^{2\phi }|x|^{2\alpha }K_g$ and $W_2(x)=e^{\phi }|x|^{\alpha }h_g$, then $0< a\le V(x)\le b$, $W_1(x)$ is in $L^p(D^+_r)$ and $W_2(x)$ is in $L^p(L_r)$ for all $p>1$ if the curvature $K_g$ and $h_g$ of M is regular enough. Therefore, the Eq. (14) can be rewritten as:

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u(x) &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2u(x)}-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2-W_1(x), \quad &{}\text {in } D^+_{r}, \\ \not {D}\Psi &{}=&{} -V(x)|x|^{\alpha }e^{u(x)}\Psi , \quad &{}\text {in } D^+_{r}, \\ \frac{\partial u}{\partial n}&{} = &{} cV(x)|x|^{\alpha }e^u-W_2,\quad &{} \text { on } L_r, \\ B(\Psi ) &{}=&{} 0,\quad &{} \text { on } L_r. \end{array} \right. \end{aligned}$$

Furthermore, let w(x) satisfy

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta w(x) &{}=&{} -W_1(x), \quad &{}\text {in } D^+_{r},\\ \frac{\partial u}{\partial n}&{} = &{} -W_2(x), \quad &{} \text { on } L_r, \\ w(x) &{}=&{} 0, \quad &{}\text { on } S^+_r. \end{array} \right. \end{aligned}$$

It is easy to see that w(x) is in $C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r)$. Setting $v(x)=u(x)-w(x)$, then $(v,\Psi )$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta v(x) &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2v(x)}-V(x)|x|^{\alpha }e^{v(x)}|\Psi |^2, \quad &{}\text {in } D^+_{r}, \\ \not {D}\Psi &{}=&{} -V(x)|x|^{\alpha }e^{v(x)}\Psi , \quad &{}\text {in } D^+_{r}, \\ \frac{\partial v}{\partial n}&{} = &{} cV(x)|x|^{\alpha }e^{v(x)},\quad &{} \text { on } L_r, \\ B(\Psi ) &{}=&{} 0,\quad &{} \text { on } L_r. \end{array} \right. \end{aligned}$$

Here $\alpha \ge 0$, V(x) is in $C^{1}_{loc}(D^+_r\cup L_r)$ and satisfies $0< a\le V(x)\le b$. Thus we get the local system (4) of the boundary problem (2).

As the interior case, we can also define $(u,\Psi )$ be a weak solution of (4) if $u\in W^{1,2}(D^+_r)$ and $\Psi \in W^{1,\frac{4}{3}}_B(\Gamma (\Sigma D^+_r))$ satisfy

$$\begin{aligned} \int _{D^+_r} \nabla u\nabla \phi dx= & {} \int _{D^+_r} (2V^2(x)|x|^{2\alpha }e^{2u(x)}-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2)\phi dx\\&+\int _{ L_r}(cV(x)|x|^{\alpha }e^{v(x)})\phi d\sigma \\ \int _{D^+_r} \langle \Psi ,\not {D}\xi \rangle dx= & {} -\int _{D^+_r} V(x)|x|^{\alpha }e^{v(x)}\langle \Psi ,\xi \rangle dx \end{aligned}$$

for any $\phi \in C_0^{\infty }({D^+_r\cup L_r})$ and any spinor $\xi \in C^\infty _0(\Gamma (\Sigma (D^+_r\cup L_r)))\cap W^{1,\frac{4}{3}}_B(\Gamma (\Sigma D^+_r))$. Here

$$\begin{aligned} W^{1,\frac{4}{3}}_B(\Gamma (\Sigma D^+_r))=\{\psi | \psi \in W^{1,\frac{4}{3}}(\Gamma (\Sigma D^+_r)), B\psi |_{L_r}=0 \}. \end{aligned}$$

For weak solutions of (4) we also have the following regularity result.

Proposition 3.2

Let $(u,\Psi )$ be a weak solution of (4) with the energy condition

$$\begin{aligned} \int _{D^+_r}|x|^{2\alpha }e^{2u}+|\Psi |^4 dv+\int _{L_r}|x|^{\alpha }e^u d\sigma <\infty . \end{aligned}$$

(15)

Then $u\in C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r)$ and $\Psi \in C^{2}(\Gamma (\Sigma D^+_r ))\cap C^{1}(\Gamma (\Sigma (D^+_r\cup L_r)))$.

Note that when $\alpha =0$ this proposition has been proved in [24]. When $\alpha >0$, to get the $L^1$ integral of $u^+$, we need a trick which was introduced in [7] and also was used in [27]. That is, by using the fact that for some $t>0$

$$\begin{aligned} \int _{D^+_r}\frac{1}{|x|^{2t\alpha }}dx\le C, \end{aligned}$$

we can choose $s=\frac{t}{t+1}\in (0,1)$ when $\alpha >0$ and $s=1$ when $\alpha =0$ such that

$$\begin{aligned} 2s\int _{D^+_r}u^+dx\le \int _{D^+_r}e^{2su}dx\le \left( \int _{D^+_r}|x|^{2\alpha }e^{2u}dx\right) ^s \left( \int _{D^+_r}|x|^{-2t\alpha }dx\right) ^{1-s}<\infty . \end{aligned}$$

Once we get the $L^1$ integral of $u^+$, we can get the conclusion of Proposition 3.2 by use the same argument in [24]. We omit the proof here.

We call $(u,\psi )$ a regular solution to (4) if $u\in C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r)$ and $\Psi \in C^{2}(\Gamma (\Sigma D^+_r))\cap C^{1}(\Gamma (\Sigma (D^+_r\cup L_r)))$.

Next we consider the convergence of a sequence of regular solutions to (4) under a smallness condition for the energy. We assume that $(u_n, \Psi _n)$ satisfy that

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u_n(x) &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2u_n(x)}-V(x)|x|^{\alpha }e^{u_n(x)}|\Psi _n|^2, \quad &{}\text {in } D^+_{r}, \\ \not {D}\Psi _n &{}=&{} -V(x)|x|^{\alpha }e^{u_n(x)}\Psi _n, \quad &{}\text {in } D^+_{r}, \\ \frac{\partial u_n}{\partial n}&{} = &{} cV(x)|x|^{\alpha }e^{u_n(x)},\quad &{} \text { on } L_r, \\ B(\Psi _n) &{}=&{} 0,\quad &{} \text { on } L_r, \end{array} \right. \end{aligned}$$

(16)

with the energy condition

$$\begin{aligned} \int _{D^+_r}|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4 dv+\int _{L_r}|x|^{\alpha }e^{u_n} d\sigma <C \end{aligned}$$

(17)

for some constant $C>0$. First, we study the small energy regularity, i.e. when the energy $\int _{D^+_r}|x|^{2\alpha }e^{2u_n}dx$ and $\int _{L_r}|x|^{\alpha }e^{u_n}dx$ are small enough, $u_n$ will be uniformly bounded from above. Our Lemma is:

Lemma 3.3

For $\varepsilon _1<\pi$, and $\varepsilon _2<\pi$. If a sequence of regular solutions $(u_n,\Psi _n)$ to (16) with

$$\begin{aligned} \int _{D^+_r}2V^2(x)|x|^{2\alpha }e^{2u_n}dx<\varepsilon _1, \quad |c|\int _{L_r}V(x)|x|^\alpha e^{u_n}d\sigma<\varepsilon _2, \quad \int _{D^+_r}|\Psi _n|^4dx<C \end{aligned}$$

for some fixed constant $C>0$, we have that $||u^+_n||_{L^\infty (\overline{D}^+_{\frac{r}{4}})}$ and $||\Psi _n||_{L^\infty (\overline{D}^+_{\frac{r}{8}})}$ are uniformly bounded.

Proof

As the same situation as in Proposition 3.2, we can no longer use the inequality $2\int {u^+_n}<\int {e^{2u_n}}$ to get the uniform bound of the $L^1$-integral of $u^+_n$ when $\alpha >0$. But notice that there exists a constant $t>0$ such that

$$\begin{aligned} \int _{D^+_r}\frac{1}{|x|^{2t\alpha }}dx\le C. \end{aligned}$$

Setting $s=\frac{t}{t+1}\in (0,1)$, then we obtain

$$\begin{aligned} 2s\int _{D^+_r}u^+_ndx\le \int _{D^+_r}e^{2su_n}dx\le \left( \int _{D^+_r}|x|^{2\alpha }e^{2u_n}dx\right) ^s \left( \int _{D^+_r}|x|^{-2t\alpha }dx\right) ^{1-s}<C. \end{aligned}$$

Then by a similar argument as in the proof of Lemma 3.5 in [24] we can prove this Lemma. $\square$

When the energy $\int _{D^+_r}2V^2(x)|x|^{2\alpha }e^{2u_n}+\int _{L_r}V(x)|x|^{\alpha }e^{u_n}ds$ is large, in general, blow-up phenomenon may occur, i.e., Theorem 1.3 holds.

Remark 3.4

Let $v_n=u_n+\alpha \log |x|$, then $(v_n, \Psi _n)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta v_n(x) &{}=&{} 2V^2(x)e^{2v_n(x)}-V(x)e^{v_n(x)}|\Psi _n|^2, \quad &{}\text {in } D^+_r, \\ \not {D}\Psi _n &{}=&{} -V(x)e^{v_n(x)}\Psi _n, \quad &{}\text { in } D^+_r,\\ \frac{\partial v_n}{\partial n}&{} = &{} cV(x)e^{v_n(x)}+ \pi \alpha \delta _{p=0},\quad &{} \text { on } L_r, \\ B\Psi _n &{}=&{} 0,\quad &{} \text { on } L_r, \end{array} \right. \end{aligned}$$

with the energy condition

$$\begin{aligned} \int _{D^+_r}e^{2v_n}+|\Psi _n|^4dx+\int _{L_r}e^{v_n}ds<C. \end{aligned}$$

Then, by using similar arguments as in [7], the two blow-up sets of $u_n$ and $v_n$ are the same. To show this conclusion, it is sufficient to show the point $x=0$ is a blow-up point for $u_n$ if and only if it is a blow-up point for $v_n$. In fact, if 0 is the only blow-up point for $v_n$ in a small neighbourhood $D^+_{\delta _0}\cup L_{\delta _0}$, that is, for any $\delta \in (0,\delta _0)$, $\exists C_\delta >0$, such that

$$\begin{aligned} \max _{\overline{D^+_{\delta _0}\setminus D^+_\delta }}v_n\le C_\delta ,\quad \text { and }\quad \max _{\overline{D^+_{\delta _0}}}v_n\rightarrow +\infty , \end{aligned}$$

(18)

then, it is easy to see that 0 is also the only blow-up point for $u_n$ in a small neighbourhood $D^+_{\delta _0}\cup L_{\delta _0}$, that is, for any $\delta \in (0,\delta _0)$, $\exists C_\delta >0$, such that

$$\begin{aligned} \max _{\overline{D^+_{\delta _0}\setminus D^+_\delta }}u_n\le C_\delta ,\quad \text { and }\quad \max _{\overline{D^+_{\delta _0}}}u_n\rightarrow +\infty . \end{aligned}$$

(19)

In converse, we assume that 0 is the only blow-up point for $u_n$ in a small neighbourhood $D^+_{\delta _0}\cup L_{\delta _0}$ such that (19) is holds. We argue by contradiction and suppose that there exists a uniform constant C, such that $v_n(x)\le C$ for any $x\in \overline{D}^+_{\delta _0}$. First, we can obtain that there exists a uniform constant C, such that $|\Psi _n|^2(x)\le C$ for any $x\in \overline{D}^+_{\frac{\delta _0}{2}}$. For this purpose, we extend $(v_n,\Psi _n)$ to the lower half disk $D^-_{r}$. Assume $\bar{x}$ is the reflection point of x about $\partial \mathbb {R}^2_+$, and define

$$\begin{aligned} v_n(\bar{x}) :=&v_n(x), \quad \bar{x} \in D^{-}_{r},\\ \Psi _n(\bar{x}) :=&ie_1\cdot \Psi _n(x), \quad \bar{x}\in D^{-}_{r},\\ A_n(x):=&\left\{ \begin{matrix} e^{v_n(x)},\quad x\in D^{+}_{r},\\ e^{v_n(\bar{x})}, \quad x\in D^-_{r}.\end{matrix} \right. \end{aligned}$$

Then $\Psi _n$ satisfies

$$\begin{aligned} \not {D}\Psi _n=-A_n(x)\Psi _n,\quad \text { in } D_r. \end{aligned}$$

Since $A_n(x)$ is uniformly bounded in $L^{\infty }(D_{\delta _0})$ and $\int _{D_{\delta _0}}|\Psi _n|^4dx<C$, we have $\Psi _n$ is uniformly bounded in $W^{1,\frac{4}{3}}(\Gamma (\Sigma D_{\frac{\delta _0}{2}}))$ and in particular $\Psi _n$ is uniformly bounded $C^\gamma (\Gamma ( \Sigma \overline{D}^+_{\frac{\delta _0}{2}}))$ for some $0<\gamma <1$. Further, since

$$\begin{aligned}&f_n(x):=2V_n^2(x)|x|^{2\alpha }e^{2u_n(x)}-V_n(x)|x|^{\alpha }e^{u_n(x)}|\Psi _n|^2\\&=2V_n^2(x)e^{2v_n(x)}-V_n(x)e^{v_n(x)}|\Psi _n|^2 \end{aligned}$$

and

$$\begin{aligned} g_n:=-V_n(x)|x|^{\alpha }e^{u_n(x)}\Psi _n =- V_n(x)e^{v_n(x)}\Psi _n \end{aligned}$$

are uniformly bounded in $\overline{D}^+_{\frac{\delta _0}{2}}$. Then by Harnack type inequality of Neumann boundary problem (see Lemma A.2 in [24]), it follows that $\inf _{\overline{D}^+_{\frac{\delta _0}{2}}}u_n\rightarrow +\infty$. Thus we get a contradiction since the blow-up set of $u_n$ is finite.

4 Removability of local sigularities

The Pohozaev identity is closely related to the removability of singularities. In this section, we shall first establish the Pohozaev identity for regular solutions to (4). Then for solutions defined on a domain with isolated singularity, we define a constant which is called the Pohozaev constant. The most important is that a necessary and sufficient condition for the removability of local singularities is the vanishing of Pohazaev constant.

Proposition 4.1

(Pohozaev identity) Let $(u,\Psi )$ be a regular solution of (4), that is $(u,\Psi )$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u(x) &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2u(x)}-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2, \quad &{}\text { in } D^+_{R}, \\ \not {D}\Psi &{}=&{} -V(x)|x|^{\alpha }e^{u(x)}\Psi , \quad &{}\text { in } D^+_{R}, \\ \frac{\partial u}{\partial n}&{} = &{} cV(x)|x|^{\alpha }e^{u(x)},\quad &{} \text { on } L_R, \\ B\Psi &{}=&{} 0,\quad &{} \text { on } L_R. \end{array} \right. \end{aligned}$$

Then we have the following Pohozaev identity

$$\begin{aligned}&R\int _{S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \nonumber \\&\quad = (1+\alpha )\left\{ \int _{D^+_R}\left( 2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u|\Psi |^2\right) dv+ \int _{L_R}cV(x)|x|^{\alpha }e^{u}ds \right\} \nonumber \\&\qquad -R\int _{S^+_R}V^2(x)|x|^{2\alpha }e^{2u}d\sigma +\int _{L_R}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^{u(s,0)}ds\nonumber \\&\qquad -\,cV(s,0)|s|^{\alpha }se^{u(s,0)}|^{s=R}_{s=-R} \nonumber \\&\qquad +\int _{D^+_R}x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u}dv -\int _{D^+_R}x\cdot \nabla V(x)|x|^{\alpha }e^u|\psi |^2dv \nonumber \\&\qquad + \frac{1}{4}\int _{S^+_R}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \right\rangle d\sigma +\frac{1}{4}\int _{S^+_R}\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma , \end{aligned}$$

(20)

where $\nu$ is the outward normal vector to $S^+_R$, and $\bar{x}$ is the reflection point of x about $\partial \mathbb {R}^2_+$.

Proof

The case of $\alpha =0$ and $V\equiv 1$ has already been treated in [25]. The calculation of the Pohozaev identity is standard. Since in the sequel we will need to calculate the Pohozaev identity for different equations, for reader’s convenience, we give the detailed proof for this general case.

First, we multiply the first equation by $x\cdot \nabla u$ and integrate over $D^+_R$ to obtain

$$\begin{aligned} -\int _{D^+_R}\triangle u x\cdot \nabla udv=\int _{D^+_R}2V^2(x)|x|^{2\alpha }e^{2u}x\cdot \nabla udv-\int _{D^+_R}V(x)|x|^\alpha e^u|\Psi |^2 x\cdot \nabla udv. \end{aligned}$$

It follows from direct computations that

$$\begin{aligned}&\int _{D^+_R}\triangle u x\cdot \nabla udv\\&\quad = R\int _{S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma +\int _{L_R}\frac{\partial u}{\partial n}(x\cdot \nabla u)ds\\&\quad = R\int _{S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma +\int _{L_R}cV(x)|x|^{\alpha }e^u(x\cdot \nabla u)ds\\&\quad = R\int _{S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma -(\alpha +1)\int _{L_R}cV(x)|x|^{\alpha }e^{u}ds\\&\qquad -\int _{L_R}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^{u(s,0)}ds+cV(s,0)|s|^{\alpha }se^{u(s,0)}|^{s=R}_{s=-R},\\&\int _{D^+_R}2V^{2}(x)|x|^{2\alpha }e^{2u}x\cdot \nabla udv\\&\quad = R\int _{S^+_R}V^2(x)|x|^{2\alpha }e^{2u}d\sigma -(2+2\alpha )\int _{D^+_R}V^2(x)|x|^{2\alpha }e^{2u}dv\\&\qquad -\int _{D^+_R}x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u}dv, \end{aligned}$$

and

Therefore we have

$$\begin{aligned}&R\int _{S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \nonumber \\&\quad = (1+\alpha )\int _{D^+_R}2V^2(x)|x|^{2\alpha }e^{2u}dv-(2+\alpha )\int _{D^+_R}V(x)|x|^{\alpha }e^u|\Psi |^2dv\nonumber \\&\qquad + (\alpha +1)\int _{L_R}cV(x)|x|^{\alpha }e^{u}ds \nonumber \\&\qquad -R\int _{S^+_R}V^2(x)|x|^{2\alpha }e^{2u}d\sigma +R\int _{S^+_R}V(x)|x|^{\alpha }e^u|\Psi |^2d\sigma \nonumber \\&\qquad +\int _{L_R}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^{u(s,0)}ds-cV(s,0)|s|^{\alpha }se^{u(s,0)}|^{s=R}_{s=-R} \nonumber \\&\qquad +\int _{D^+_R}x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u}dv -\int _{D^+_R}|x|^{\alpha }e^ux\cdot \nabla (V(x)|\Psi |^2)dv \end{aligned}$$

(21)

On the other hand, for $x\in \mathbb {R}^2_+$, we denote $x=x_1e_1+x_2e_2$ under the local orthonormal basis $\{e_1,e_2\}$ on $\mathbb {R}^2_+$. Using the Clifford multiplication relation

$$\begin{aligned} e_i\cdot e_j+e_j\cdot e_i=-2\delta _{ij},\text { for }1\le i,j\le 2 \end{aligned}$$

and

$$\begin{aligned} \left\langle \psi ,\varphi \right\rangle =\left\langle e_i\cdot \psi ,e_i\cdot \varphi \right\rangle \end{aligned}$$

for any spinors $\psi ,\varphi \in \Gamma (\Sigma M)$. We know that

$$\begin{aligned} \langle \psi ,e_i\cdot \psi \rangle +\langle e_i\cdot \psi ,\psi \rangle =0 \end{aligned}$$

(22)

for any $i=1,2$. Using the chirality boundary condition of $\Psi$, we extend $(u,\Psi )$ to the lower half disk $D^-_{R}$. Assume $\bar{x}$ is the reflection point of x about $\partial \mathbb {R}^2_+$, and define

$$\begin{aligned} u(\bar{x})&:= u(x), \quad \bar{x} \in D^{-}_{R}, \end{aligned}$$

(23)

$$\begin{aligned} \Psi (\bar{x})&:= ie_1 \cdot \Psi (x), \quad \bar{x}\in D^{-}_{R}. \end{aligned}$$

(24)

Then it follows from the argument in Lemma 3.4 of [24] that we obtain

$$\begin{aligned} \not {D}\psi =-A(x)\psi \quad \text { in } D_R. \end{aligned}$$

Here

$$\begin{aligned} A(x)=\left\{ \begin{matrix} V(x)|x|^{\alpha }e^{u(x)},\quad x\in D^{+}_{R},\\ V(\bar{x})|\bar{x}|^{\alpha }e^{u(\bar{x})}, \quad x\in D^-_{R}.\end{matrix} \right. \end{aligned}$$

Using the Schrödinger–Lichnerowicz formula , we have

$$\begin{aligned} -\triangle \Psi =-dA(x)\cdot \psi +A^2(x)\Psi \quad \text { in } D_{R}. \end{aligned}$$

(25)

Then we multiply (25) by $x\cdot \Psi$ (where $\cdot$ denotes the Clifford multiplication) and integrate over $D_R$ to obtain

$$\begin{aligned} \int _{D_R}\langle \triangle \Psi ,x\cdot \Psi \rangle dv=\int _{B_R}\langle dA(x)\cdot \Psi ,x\cdot \Psi \rangle dv-\int _{D_R}A^2(x)\langle \Psi ,x\cdot \Psi \rangle dv, \end{aligned}$$

and

$$\begin{aligned} \int _{D_R}\langle x\cdot \Psi ,\triangle \Psi \rangle dv=\int _{D_R} \langle x\cdot \Psi ,dA(x)\cdot \Psi \rangle dv-\int _{D_R}A^2(x)\langle x\cdot \Psi , \Psi \rangle dv . \end{aligned}$$

On the other hand, by partial integration,

$$\begin{aligned}&\int _{D_R}\langle \triangle \Psi ,x\cdot \Psi \rangle dv \\&\quad =\int _{D_R}div\langle \nabla \Psi ,x\cdot \Psi \rangle dv-\int _{D_R}\sum ^2_{\alpha =1}\langle \nabla _{e_\alpha }\Psi ,e_\alpha \cdot \Psi \rangle dv-\int _{D_R}\langle \nabla \Psi ,x\cdot \nabla \Psi \rangle \\&\quad =\int _{\partial D_R}\left\langle \frac{\partial \Psi }{\partial \nu },x\cdot \Psi \right\rangle d\sigma +\int _{D_R}\langle \not {D}\Psi ,\Psi \rangle dv-\int _{D_R}\langle \nabla \Psi ,x\cdot \nabla \Psi \rangle \\&\quad = \int _{\partial D_R}\left\langle \frac{\partial \Psi }{\partial \nu },x\cdot \Psi \right\rangle d\sigma -\int _{D_R} A(x)|\Psi |^2 dv-\int _{D_R}\langle \nabla \Psi ,x\cdot \nabla \Psi \rangle ,\\&\quad = \int _{ \partial D^+_R\cap \mathbb {R}^2_+}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \right\rangle d\sigma -2\int _{D^+_R} V(x)|x|^{\alpha }e^u|\Psi |^2 dv\\&\qquad -\int _{D_R}\langle \nabla \Psi ,x\cdot \nabla \Psi \rangle , \end{aligned}$$

and similarly

$$\begin{aligned}&\int _{D_R}\langle x\cdot \Psi ,\triangle \Psi \rangle =\int _{\partial D^+_R\cap \mathbb {R}^2_+}\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma -2\int _{D^+_R}V(x)|x|^{\alpha } e^u|\Psi |^2 dv\\&\quad -\int _{D_R}\langle x\cdot \nabla \Psi ,\nabla \Psi \rangle . \end{aligned}$$

Furthermore we also have

$$\begin{aligned}&\int _{D_R}\langle dA(x)\cdot \Psi ,x\cdot \Psi \rangle dv +\int _{D_R}\langle x\cdot \Psi , dA(x)\cdot \Psi \rangle dv\\&\quad =\int _{D_R}\sum ^2_{\alpha ,\beta =1}\langle \nabla _{e_\alpha }A(x)e_\alpha \cdot \Psi ,e_\beta \cdot \Psi \rangle x_\beta dv\\&\qquad +\int _{D_R} \sum ^2_{\alpha ,\beta =1}\langle e_{\beta }\cdot \Psi ,\nabla _{e_\alpha }A(x)e_\alpha \cdot \Psi \rangle x_\beta dv\\&\quad = 2\int _{D_R}\sum ^2_{\alpha =1}\langle \nabla _{e_\alpha }A(x)e_\alpha \cdot \Psi ,e_\alpha \cdot \Psi \rangle x_\alpha dv\\&\quad = 2\int _{D_R}x\cdot \nabla (A(x))|\Psi |^2dv\\&\quad =-2\int _{D_R}A(x)x\cdot \nabla (|\Psi |^2)dv-4\int _{D_R}A(x)|\Psi |^2dv+2R\int _{\partial D_R}A(x)|\Psi |^2dv\\&\quad = -4\int _{D^+_R}V(x)|x|^{\alpha }e^u x\cdot \nabla (|\Psi |^2)dv-8\int _{D^+_R}V(x)|x|^{\alpha }e^u |\Psi |^2dv\\&\qquad +4R\int _{\partial D^+_R\cap \mathbb {R}^2_+} V(x)|x|^{\alpha }e^u|\Psi |^2dv. \end{aligned}$$

Therefore we obtain

$$\begin{aligned}&R\int _{\partial D^+_R\cap \mathbb {R}^2_+}V(x)|x|^{\alpha }e^u|\Psi |^2d\sigma -\int _{D^+_R}V(x)|x|^{\alpha }e^ux\cdot \nabla (|\Psi |^2)dv\nonumber \\&\quad =\frac{1}{4}\int _{\partial D^+_R\cap \mathbb {R}^2_+}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \right\rangle d\sigma +\frac{1}{4}\int _{\partial D^+_R\cap \mathbb {R}^2_+}\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma \nonumber \\&\qquad +\int _{D^+_R} V(x)|x|^{\alpha }e^u|\Psi |^2 dv. \end{aligned}$$

(26)

Putting (21) and (26) together, we obtain our Pohozaev type identity (20). $\square$

Pohozaev type identity is shown to be closely related to the removability of local singularities of solutions. For a solution of (12) and (13), we defined in [27] the following Pohozaev constant:

Definition 4.2

([27]). Let $(u,\Psi )\in C^2(D_r\backslash \{0\})\times C^2(\Gamma (\Sigma (D_r\backslash \{0\})))$ be a solution of (12) and (13). For $0<R<r$, we define the Pohozaev constant with respect to the Eqs. (12) with the constraint (13) as follows:

$$\begin{aligned} C(u,\Psi )&:= R\int _{\partial D_R(0)} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \\&\quad -(1+\alpha )\int _{D_R(0)}(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u|\Psi |^2)dx \\&\quad +R\int _{\partial D_R(0)}V^2(x)|x|^{2\alpha }e^{2u}d\sigma -\frac{1}{2}\int _{\partial D_R(0)}\left\langle \frac{\partial \Psi }{\partial \nu }, x\cdot \Psi \right\rangle +\left\langle x\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma \\&\quad -\int _{D_R(0)}(|x|^{2\alpha }e^{2u}x\cdot \nabla (V^2(x))-|x|^{\alpha }e^u|\Psi |^2x\cdot \nabla V(x))dx \end{aligned}$$

where $\nu$ is the outward normal vector of $\partial D_R(0)$.

It is clear that $C(u,\Psi )$ is independent of R for $0<R<r$. Thus, the vanishing of the Pohozaev constant $C(u,\Psi )$ is equivalent to the Pohozaev identity

$$\begin{aligned}&R\int _{\partial D_R(0)} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \nonumber \\&\quad =(1+\alpha )\int _{D_R(0)}(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u|\Psi |^2)dx \nonumber \\&\qquad -R\int _{\partial D_R(0)}V^2(x)|x|^{2\alpha }e^{2u}d\sigma +\frac{1}{2}\int _{\partial D_R(0)}\left( \left\langle \frac{\partial \Psi }{\partial \nu }, x\cdot \Psi \right\rangle +\left\langle x\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle \right) d\sigma \nonumber \\&\qquad +\int _{D_R(0)}(|x|^{2\alpha }e^{2u}x\cdot \nabla (V^2(x))-|x|^{\alpha }e^u|\Psi |^2x\cdot \nabla V(x))dx \end{aligned}$$

(27)

for a solution $(u,\Psi )\in C^2(D_r(0))\times C^2(\Gamma (\Sigma D_r(0)))$ of (12) and (13).

We also proved in [27] that a local singularity is removable if the Pohozaev identity (27) holds, that is, iff the Pohozaev constant vanishes.

Theorem 4.3

[27]. Let $(u,\Psi )\in C^2(D_r\setminus \{0\})\times C^2(\Gamma (\Sigma (D_r\setminus \{0\})))$ be a solution of (12) and (13). Then there is a constant $\gamma < 2\pi (1+\alpha )$ such that

$$\begin{aligned} u(x)=- \frac{\gamma }{2\pi }\mathrm{log} |x| + h, \quad \mathrm{near }\ 0, \end{aligned}$$

where h is bounded near 0. The Pohozaev constant $C(u, \Psi )$ and $\gamma$ satisfy:

$$\begin{aligned} C(u, \Psi )=\frac{\gamma ^2}{4\pi }. \end{aligned}$$

In particular, $(u,\Psi )\in C^2(D_r)\times C^2(\Gamma (\Sigma D_r))$, i.e. the local singularity of $(u,\Psi )$ is removable, iff $C(u,\Psi )=0$.

For the singular boundary problem (4), we can define the Pohozaev constant in a similar way:

Definition 4.4

Let $(u,\Psi ) \in C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r\backslash \{0\})\times C^{2}(\Gamma (\Sigma D^+_r))\cap C^{1}(\Gamma (\Sigma (D^+_r\cup L_r\backslash \{0\})))$ be a solution of (4) and (15). For $0<R<r$, we define the Pohozaev constant with respect to the Eqs. (4) with the constraint (15) as follows:

$$\begin{aligned}&C_B(u,\Psi ) := R\int _{\partial D^+_R\cap \mathbb {R}^2_+} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \\&\quad -(1+\alpha )\int _{D^+_R}(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u|\Psi |^2)dv\\&\qquad - (\alpha +1)\int _{\partial D^+_R\cap \partial \mathbb {R}^2_+}cV(x)|x|^{\alpha }e^{u}ds \\&\quad +R\int _{\partial D^+_R\cap \mathbb {R}^2_+}V^2(x)|x|^{2\alpha }e^{2u}d\sigma -\int _{\partial D^+_R\cap \partial \mathbb {R}^2_+}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^uds\\&\qquad +cV(s,0)|s|^{\alpha }se^u|^{s=R}_{s=-R} \\&\quad -\int _{D^+_R}x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u}dv +\int _{D^+_R}x\cdot \nabla V(x)|x|^{\alpha }e^u|\psi |^2dv \\&\quad - \frac{1}{4}\int _{\partial D^+_R\cap \mathbb {R}^2_+}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \right\rangle d\sigma -\frac{1}{4}\int _{\partial D^+_R\cap \mathbb {R}^2_+}\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

The removability theorem of a local singularity at the boundary is following:

Theorem 4.5

(Removability of a local boundary singularity). Let $(u,\Psi ) \in C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r\backslash \{0\})\times C^{2}(\Gamma (\Sigma D^+_r))\cap C^{1}(\Gamma (\Sigma (D^+_r\cup L_r\backslash \{0\})))$ be a solution of (4) and (15), then there is a constant $\gamma < \pi (1+\alpha )$ such that

$$\begin{aligned} u(x)=- \frac{\gamma }{2\pi }\mathrm{log} |x| + h, \quad \mathrm{near }\ 0, \end{aligned}$$

where h is bounded near 0. The Pohozaev constant $C(u, \Psi )$ and $\gamma$ satisfy:

$$\begin{aligned} C(u, \Psi )=\frac{\gamma ^2}{2\pi }. \end{aligned}$$

In particular, $(u,\Psi ) \in C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r)\times C^{2}(\Gamma (\Sigma D^+_r))\cap C^{1}(\Gamma (\Sigma (D^+_r\cup L_r)))$, i.e. the local singularity of $(u,\Psi )$ is removable, iff $C(u,\Psi )=0$.

To prove Theorem 4.5, we need to derive the decay of spinor part $\Psi$ near the singular point. For the case of $\alpha =0$ and $V(x)=1$, this is shown in [25]. By using similar arguments, we can also get the following lemma for the general case:

Lemma 4.6

There are $0<\varepsilon _1 < 2\pi$ and $0<\varepsilon _2<\pi$ such that $(v,\phi )$ satisfy

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta v &{}=&{} 2V^2(x)|x|^{2\alpha }e^{2v}-V(x)|x|^{\alpha }e^v\left\langle \phi ,\phi \right\rangle ,&{}\qquad \text { in } B_{r_0}^+, \\ \not {D}\phi &{}=&{}\displaystyle -V(x)|x|^{\alpha }e^v\phi , &{}\qquad \text { in } B_{r_0}^+,\\ \frac{\partial v}{\partial n} &{}=&{} cV(x)|x|^{\alpha }e^v, &{}\qquad \text { on } L_{r_0}\backslash \{0\},\\ B\phi &{}=&{} 0, &{}\qquad \text { on } L_{r_0}\backslash \{0\}, \end{array} \right. \end{aligned}$$

with energy conditions

$$\begin{aligned} \int _{B^+_{r_0}}|x|^{2\alpha }e^{2v}dx\le \varepsilon _1<2\pi ,~~~ \int _{B^+_{r_0}}|\phi |^4dx\le C, ~~~ |c|\int _{L_{r_0}}|x|^{\alpha }e^{v}ds\le \varepsilon _2<\pi . \end{aligned}$$

Then for any $x\in \overline{B}^+_{\frac{r_0}{2}}$ we have

$$\begin{aligned} |\phi (x)||x|^{\frac{1}{2}}+|\nabla \phi (x)||x|^{\frac{3}{2}}\le C\left( \int _{B^+_{2|x|}}|\phi |^4dx\right) ^{\frac{1}{4}}. \end{aligned}$$

(28)

Furthermore, if we assume that $e^{2v}=O(\frac{1}{|x|^{2(1+\alpha )-\varepsilon }})$, then, for any $x\in \overline{B}^+_{\frac{r_0}{2}}$, we have

$$\begin{aligned} |\phi (x)||x|^{\frac{1}{2}}+|\nabla \phi (x)||x|^{\frac{3}{2}}\le C|x|^{\frac{1}{4C}}\left( \int _{B^+_{r_0}}|\phi |^4dx\right) ^{\frac{1}{4}}, \end{aligned}$$

(29)

for some positive constant C. Here $\varepsilon$ is any sufficiently small positive number.

Proof of Theorem 4.5:

By the conformal invariance, we assume without loss of generality that $\int _{B^+_{r}}|x|^{2\alpha }e^{2v}dx\le \varepsilon _1$ and $|c|\int _{L_{r}}|x|^{\alpha }e^{v}ds\le \varepsilon _2$ where $\varepsilon _1$ and $\varepsilon _2$ are as in Lemma 4.6. By standard potential analysis, it follows that there is a constant $\gamma$ such that

$$\begin{aligned} \lim _{|x|\rightarrow 0}\frac{u}{-\log |x|}=\frac{\gamma }{\pi }. \end{aligned}$$

By $\int _{D^+_r}|x|^{2\alpha }e^{2u}dx<C$, we obtain that $\gamma \le \pi (1+\alpha )$. Furthermore, by using Lemma 4.6 and by a similar argument as in the proof of Proposition 5.4 of [23], we can improve this to the strict inequality $\gamma <\pi (1+\alpha )$. Next we set

$$\begin{aligned} v(x)= & {} -\frac{1}{\pi }\int _{B^+_{r}}\log |x-y|(2V^2(y)|y|^{2\alpha }e^{2u}-V(y)|y|^{\alpha }e^u|\Psi |^2)dy\\&-\frac{1}{\pi }\int _{L_{r}}\log |x-y|(cV(y)|y|^{\alpha }e^{u})d\sigma \end{aligned}$$

and set $w=u-v$. Notice that v satisfies that

$$\begin{aligned} \left\{ \begin{array}{rlll} -\Delta v &{}= &{} 2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u|\Psi |^2, &{}\quad \text { in } D^+_r,\\ \frac{\partial v}{\partial n}&{}=&{} cV(x)|x|^{\alpha }e^u, &{}\quad \text { on } L_r, \end{array} \right. \end{aligned}$$

and w satisfies that

$$\begin{aligned} \left\{ \begin{array}{rlll} -\Delta w &{}= &{} 0, &{}\quad \text { in } D^+_r,\\ \frac{\partial v}{\partial n}&{}=&{} 0, &{}\quad \text { on } L_r\backslash \{0\}. \end{array} \right. \end{aligned}$$

We can check that

$$\begin{aligned} \lim _{|x|\rightarrow 0}\frac{v(x)}{-\log |x|}=0. \end{aligned}$$

Since we can extend w to $B_r\backslash \{0\}$ evenly to get a harmonic function w in $B_r\backslash \{0\}$, then we obtain that

$$\begin{aligned} \lim _{|x|\rightarrow 0}\frac{w(x)}{-\log |x|}=\lim _{|x|\rightarrow 0}\frac{u-v}{-\log |x|}=\frac{\gamma }{\pi }. \end{aligned}$$

Duo to w is harmonic in $B_1\backslash \{0\}$ we have

$$\begin{aligned} w=-\frac{\gamma }{\pi }\log |x|+w_0 \end{aligned}$$

with a smooth harmonic function $w_0$ in $B_r$. Therefore we have

$$\begin{aligned} u=-\frac{\gamma }{\pi }\log |x|+v+w_0 \quad \text { near } 0. \end{aligned}$$

To compute the Pohozaev constant of $(u,\Psi )$ we need the decay of the gradient of u near the singular point. We denote that $f_1(x):=2V^2(x)|x|^{2\alpha }e^{2u(x)}$, $f_2(x):=-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2(x)$ and $f_3(x):= cV(x)|x|^{\alpha }e^u$. Since each $f_i$ is $L^1$ integrable, we can obtain $e^{|v(x)|}\in L^p(D^+_r)$ for any $p\ge 1$ and $e^{|v(x)|}\in L^p(L_r)$ for any $p\ge 1$. Since

$$\begin{aligned} f_1(x)= & {} |x|^{-\frac{2\gamma }{\pi }+2\alpha }\left( 2V^2(x)e^{2w_0(x)+2v(x)}\right) ,\\ f_2(x)= & {} -|x|^{-\frac{\gamma }{\pi }+\alpha -1}\left( V(x)e^{w_0(x)+v(x)}|x||\Psi |^2(x)\right) , \end{aligned}$$

and

$$\begin{aligned} f_3(x)=|x|^{-\frac{\gamma }{\pi }+\alpha }(cV(x)e^{w_0(x)+v(x)}), \end{aligned}$$

we set $s_1=\frac{2\gamma }{\pi }-2\alpha$ and $s_2=\frac{\gamma }{\pi }-\alpha +1$. Then $\max \{s_1, s_2\}=s_2<2$. Since $|\Psi |\le C|x|^{-\frac{1}{2}}$ near 0 and $w_0(x)$ is smooth in $B_r$, we have by Hölder’s inequality that $f_1\in L^{t}(D^+_r)$ for any $t\in (1,\frac{2}{s_1})$ if $s_1>0$, and $f_1\in L^{t}(D^+_r)$ for any $t>1$ if $s_1\le 0$. For $f_2$, we have $f_2\in L^t(D^+_r)$ for any $t\in (1,\frac{2}{s_2})$ if $s_2>0$, and $f_2\in L^{t}(D^+_r)$ for any $t>1$ if $s_2\le 0$. For $f_3$, we have $f_3\in L^t(L_r)$ for any $t\in (1,\frac{2}{s_1})$ if $s_1>0$, and $f_3\in L^{t}(L_r)$ for any $t>1$ if $s_1\le 0$. Putting all together and by standard elliptic theory, we have v(x) is in $L^{\infty }(\overline{D}^+_r)$. On the other hand, since v(x) is in $L^{\infty }(\overline{D}^+_r)$, it follows from Lemma 4.6 that there exists a small $\delta _0>0$ such that

$$\begin{aligned} |\Psi |\le C|x|^{\delta _0-\frac{1}{2}}, \quad \text { near } 0, \end{aligned}$$

and

$$\begin{aligned} |\nabla \Psi |\le C|x|^{\delta _0-\frac{3}{2}}, \quad \text { near } 0. \end{aligned}$$

Next we estimate $\nabla v(x)$. If $s_1<0$ and $s_2<0$, then v(x) is in $C^1(\overline{B}^+_r)$. If $s_1>0$ or $s_2>0$, $\nabla v(x)$ will have a decay when $|x|\rightarrow 0$. Without loss of generality, we assume that $0<s_1<2$ and $0<s_2<2$. For any $x\in D^+_r$ we hanve

$$\begin{aligned}&|\nabla v(x)| \le \frac{1}{\pi }\int _{D^+_r}\frac{1}{|x-y|}(|f_1(y)|+|f_2(y)|)dy+ \frac{1}{\pi }\int _{L_r}\frac{1}{|x-y|}|f_3(y)|dy\\&\quad = \frac{1}{\pi }\int _{\{|x-y|\ge \frac{|x|}{2}\}\cap {D^+_r}}\frac{1}{|x-y|}(|f_1(y)|+|f_2(y)|)dy\\&\qquad + \frac{1}{\pi }\int _{\{|x-y|\le \frac{|x|}{2}\}\cap {D^+_r}}\frac{1}{|x-y|}(|f_1(y)|+|f_2(y)|)dy\\&\qquad + \frac{1}{\pi }\int _{\{|x-y|\ge \frac{|x|}{2}\}\cap {L_r}}\frac{1}{|x-y|}|f_3(y)|dy+ \frac{1}{\pi }\int _{\{|x-y|\le \frac{|x|}{2}\}\cap {L_r}}\frac{1}{|x-y|}|f_3(y)|dy\\&\quad = I_1+I_2+I_3+I_4. \end{aligned}$$

Fix $t\in (1,\frac{2}{s_2})$ and choose $0<\tau _1<1$ such that $\frac{\tau _1 t}{t-1}<2$. Hence, we have $0<\tau _1 <2-s_2$. Then by Hölder’s inequality we obtain

$$\begin{aligned} I_1\le & {} \left( \int _{\{|x-y|\ge \frac{|x|}{2}\}\cap {D^+_r}}\frac{1}{|x-y|^{\frac{\tau _1t}{t-1}}}dy\right) ^{\frac{t-1}{t}}\\&\times \left( \int _{\{|x-y|\ge \frac{|x|}{2}\}\cap {D^+_r}}\frac{1}{|x-y|^{(1-\tau _1)t}}(|f_1|+|f_2|)^tdy\right) ^{\frac{1}{t}} \le \frac{C}{|x|^{1-\tau _1}}. \end{aligned}$$

For $I_2$, since $y\in \{y||x-y|\le \frac{|x|}{2}\}$ implies that $|y|\ge \frac{|x|}{2}$, we can get that

$$\begin{aligned} I_2 \le C \int _{\{|x-y|\le \frac{|x|}{2}\}\cap {D^+_r}}\frac{1}{|x-y||y|^{s_2}}dy \le C|x|^{1-s_2}. \end{aligned}$$

Similarly, for $I_3$, we fix $t\in (1,\frac{2}{s_1})$ and choose $\tau _{2}>0$ such that $\frac{\tau _{2} t}{t-1}<1$, and hence we have $0<\tau _{2} <1-\frac{s_1}{2}$. By Holder’s inequality we obtain,

$$\begin{aligned} | I_{3}|\le & {} \frac{1}{\pi }\left( \int _{\{|x-y|\ge \frac{|x|}{2}\}\cap L_{r}}\frac{1}{|x-y|^{\frac{\tau _{2} t}{t-1}}}dy\right) ^{\frac{t-1}{t}} \left( \int _{\{|x-y|\ge \frac{|x|}{2}\}\cap L_{r}}\frac{1}{|x-y|^{t(1-\tau _{2})}}|f_3(y)|^tdy\right) ^{\frac{1}{t}}\\\le & {} \frac{C}{|x|^{1-\tau _{2}}}. \end{aligned}$$

For $I_{4}$ we have

$$\begin{aligned} |I_{4}|\le C\int _{\{|x-y|\le \frac{|x|}{2}\}\cap L_{r}}\frac{1}{|x-y|}\frac{1}{|y|^{\frac{s_1}{2}}}dy \le \frac{C}{|x|^{\frac{s_1}{2}}}\int _{\{|x-y|\le \frac{|x|}{2}\}\cap L_{r}}\frac{1}{|x-y|}dy\le \frac{C}{|x|^{\tau _{3}}}, \end{aligned}$$

for some $\tau _{3}$ with $0<\tau _{3}<1$. In conclusion, for all $x\in B_{r}^+(0)$ we have

$$\begin{aligned} |\nabla v(x)|\le \frac{C}{|x|^{1-\tau _{1}}}+\frac{C}{|x|^{1-\tau _{2}}}+\frac{C}{|x|^{\tau _{3}}} \end{aligned}$$

(30)

for suitable constants $0<\tau _{1} <2-s_2$, $0<\tau _{2} <1-\frac{s_1}{2}$ and $0<\tau _{3}<1$.

At this point we are ready to compute the Pohozaev constant $C(u, \Psi )$. We denote

$$\begin{aligned} \nabla u=-\frac{\gamma }{\pi }\frac{x}{|x|^2}+\nabla (w_0+v(x))=-\frac{\gamma }{\pi }\frac{x}{|x|^2}+\nabla \eta (x). \end{aligned}$$

By (30), we have

$$\begin{aligned}&r\int _{S^+_{r}} \left( \frac{1}{2}| \nabla u|^{2}-| \frac{\partial u}{\partial \nu }|^{2}\right) ds \\&\quad =r\int _{S^+_{r}}\frac{1}{2}\left[ \left( \frac{\gamma }{\pi }\right) ^2\frac{1}{|x|^2}-2\frac{\gamma }{\pi }\frac{x\cdot \nabla \eta }{|x|^2}+|\nabla \eta |^2\right] ds-r \int _{S_{r}^+} \left( -\frac{\gamma }{\pi }\frac{1}{|x|}+\frac{x\cdot \nabla \eta }{|x|}\right) ^2ds \\&\quad =r\int _{S^+_{r}}\left[ -\frac{1}{2}\left( \frac{\gamma }{\pi }\right) ^2\frac{1}{|x|^2}+\frac{\gamma }{\pi }\frac{x\cdot \nabla \eta }{|x|^2}+\frac{1}{2}|\nabla \eta |^2-\left( \frac{x\cdot \nabla \eta }{|x|}\right) ^2\right] d\sigma \\&\quad =-\frac{1}{2}\left( \frac{\gamma }{\pi }\right) ^2\pi + \frac{\gamma }{\pi }r\int _{S_{r}^+}\frac{x\cdot \nabla \eta }{|x|^2}+\frac{r}{2}\int _{S_{r}^+}|\nabla \eta |^2-r\int _{S_{r}^+}\left( \frac{x\cdot \nabla \eta }{|x|}\right) ^2\\&\quad = -\frac{\gamma ^2}{2\pi }+o_r(1), \end{aligned}$$

where $o_r(1)\rightarrow 0$ as $r\rightarrow 0$. We also have

$$\begin{aligned} (1+\alpha )\int _{D^+_r}2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^{u}|\Psi |^2 dx =o_r(1), \end{aligned}$$

and

$$\begin{aligned} r\int _{ S^+_r}V^2(x)|x|^{2\alpha }e^{2u}d\sigma =o_r(1), \end{aligned}$$

and

$$\begin{aligned} \int _{D^+_r}(|x|^{2\alpha }e^{2u}x\cdot \nabla (V^2(x))-|x|^{\alpha }e^{u}|\Psi |^2x\cdot \nabla V(x))dx=o_r(1), \end{aligned}$$

and

$$\begin{aligned} (\alpha +1)\int _{S^+_r}cV(x)|x|^{\alpha }e^{u}d\sigma -\int _{L_r}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^uds+cV(s,0)|s|^{\alpha }se^u|^{s=r}_{s=-r}=o_r(1), \end{aligned}$$

and

$$\begin{aligned} \int _{S^+_r}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \nabla \Psi \right\rangle d\sigma +\int _{S^+_r}\left\langle (x+\bar{x})\cdot \nabla \Psi ,\frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma =o_r(1). \end{aligned}$$

Putting all together and letting $r\rightarrow 0$, we get

$$\begin{aligned} C(u,\Psi )=\lim _{r\rightarrow 0}C(u,\Psi , r)=\frac{\gamma ^2}{2\pi }. \end{aligned}$$

Since $C(u,\Psi )=0$ for $(u,\Psi )$, therefore we get $\gamma =0$. This implies that the local singularity of $(u,\Psi )$ is removable. $\square$

5 Bubble energy

After a suitable rescaling at a boundary blow-up point, we will obtain a bubble, i.e. an entire solution on the upper half-plane $\mathbb {R}_+^2$ with finite energy. In this section, we will investigate such entire solutions. We will first show the asymptotic behavior of an entire solution and compute the bubble energy, and then show that an entire solution can be conformally extended to a spherical cap, i.e., the singularity at infinity is removable.

The considered equations are

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u &{}=&{} \displaystyle 2|x|^{2\alpha }e^{2u}-|x|^{\alpha }e^u\left\langle \psi ,\psi \right\rangle ,&{} \qquad \text { in } \mathbb {R}_+^2, \\ \not {D}\psi &{}=&{}\displaystyle -|x|^\alpha e^u\psi , &{}\qquad \text { in } \mathbb {R}_+^2,\\ \displaystyle \frac{\partial u}{\partial n} &{}=&{} c|x|^\alpha e^u, &{}\qquad \text { on } \partial \mathbb {R}_+^2,\\ B\psi &{}=&{} 0, &{}\qquad \text { on } \partial \mathbb {R}_+^2. \end{array} \right. \end{aligned}$$

(31)

The energy condition is

$$\begin{aligned} I(u,\psi )=\int _{\mathbb {R}_+^2}(|x|^{2\alpha }e^{2u}+|\psi |^4)dx+\int _{\partial \mathbb {R}_+^2}|x|^{\alpha }e^u ds<\infty . \end{aligned}$$

(32)

First, let us notice that if $(u,\psi )$ is a weak solution of (31) and (32) with $u\in H^{1,2}_{loc}(\mathbb {R}_+^2)$ and $\psi \in W^{1,\frac{4}{3}}_{loc}(\Gamma (\Sigma \mathbb {R}_+^2))$, by using similar arguments as in the proof of Proposition 3.2, we have $u^{+}\in L^{\infty }(\overline{\mathbb {R}}_+^2)$. Consequently, it follows that $u \in C^{2}_{loc}(\mathbb {R}^2_+)\cap C^{1}_{loc}(\overline{\mathbb {R}}^2_+)$ and $\psi \in C^{2}_{loc}(\Gamma (\Sigma \mathbb {R}^2_+))\cap C^{1}_{loc}(\Gamma (\Sigma \overline{\mathbb {R}}^2_+))$.

We call $(u,\psi )$ a regular solution of (31) and (32), if $u \in C^{2}_{loc}(\mathbb {R}^2_+)\cap C^{1}_{loc}(\overline{\mathbb {R}}^2_+)$ and $\psi \in C^{2}_{loc}(\Gamma (\Sigma \mathbb {R}^2_+))\cap C^{1}_{loc}(\Gamma (\Sigma \overline{\mathbb {R}}^2_+))$.

Next, we denote by $(v,\phi )$ the Kelvin transformation of $(u,\psi )$, i.e.

$$\begin{aligned} v(x)= & {} u\left( \frac{x}{|x|^2}\right) -2(1+\alpha )\ln |x|,\\ \phi (x)= & {} |x|^{-1} \psi \left( \frac{x}{|x|^2}\right) . \end{aligned}$$

Then $(v,\phi )$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta v &{}=&{} 2|x|^{2\alpha }e^{2v}-|x|^{\alpha }e^v\left\langle \phi ,\phi \right\rangle ,&{}\qquad \text { in } \mathbb {R}_+^2, \\ \not {D}\phi &{}=&{}\displaystyle -|x|^{\alpha }e^v\phi , &{}\qquad \text { in } \mathbb {R}_+^2,\\ \frac{\partial v}{\partial n} &{}=&{} c|x|^{\alpha }e^v, &{}\qquad \text { on } \partial \mathbb {R}_+^2\backslash \{0\},\\ B\phi &{}=&{} 0, &{}\qquad \text { on } \partial \mathbb {R}_+^2\backslash \{0\}. \end{array} \right. \end{aligned}$$

(33)

And, by change of variable, we can choose $r_0$ small enough such that $(v,\phi )$ satisfies

$$\begin{aligned} \int _{|x|\le r_0}|x|^{2\alpha }e^{2v}dx\le \varepsilon _1<2\pi ,\quad \int _{|x|\le r_0}|\phi |^4dx\le C,\quad |c|\int _{|s|\le r_0}|x|^{\alpha }e^{v}ds\le \varepsilon _2<\pi . \end{aligned}$$

(34)

Applying Lemma 4.6 to (33) and (34), and by the Kelvin transformation, we obtain the asymptotic estimate of the spinor $\psi (x)$

$$\begin{aligned} |\psi (x)|\le C|x|^{-\frac{1}{2}-\delta _0}\qquad \text {for}\ \ |x| \ \ \text {near}\ \ \infty , \end{aligned}$$

(35)

and

$$\begin{aligned} |\nabla \psi (x)|\le C|x|^{-\frac{3}{2}-\delta _0}\qquad \text {for}\ \ |x| \ \ \text {near}\ \ \infty , \end{aligned}$$

(36)

for some positive number $\delta _0$ provided that $e^{2v}=O(\frac{1}{|x|^{2(1+\alpha )-\varepsilon }})$, where $\varepsilon$ is any small positive number.

Denote

$$\begin{aligned} d =\int _{\mathbb {R}_+^2}2|x|^{2\alpha }e^{2u}-|x|^{\alpha }e^u|\psi |^2dx+\int _{\partial R_+^2}c|x|^{\alpha }e^uds, \end{aligned}$$

and

$$\begin{aligned} \xi _0=\int _{\mathbb {R}_+^2}e^u\psi dx. \end{aligned}$$

Next, we will show that $d=2(1+\alpha )\pi$ and $\xi _0$ is a well-defined constant spinor.

Proposition 5.1

Let $(u,\psi )$ be a regular solution of (31) and (32) and let c be a nonnegative constant. Then we have

$$\begin{aligned} u(x)= & {} -\frac{d}{\pi } \ln {|x|}+C+O(|x|^{-1}) \qquad \text {for}\quad |x| \quad \text {near}\quad \infty , \end{aligned}$$

(37)

$$\begin{aligned} \psi (x)= & {} -\frac{1}{2\pi }\frac{x}{|x|^2} (I+ie_1)\cdot \xi _0+o(|x|^{-1})\qquad \text {for}\quad |x| \quad \text {near}\quad \infty , \end{aligned}$$

(38)

where $\cdot$ is the Clifford multiplication, C is a positive universal constant, and I is the identity. In particular we have $d =2(1+\alpha )\pi$ and $\xi _0$ is well defined.

Proof

We shall apply standard potential analysis to prove this proposition. Similar arguments can be found in [13, 21, 22] and the references therein. The essential facts used in this case are the Pohozaev identity and the decay estimate for the spinor. For readers’ convenience, we sketch the proof here.

Step 1. $\lim _{|x|\rightarrow \infty }\frac{u(x)}{\ln |x|}=-\frac{d}{\pi }$ and $d>\pi (1+\alpha )$.

Let

$$\begin{aligned} w(x)= & {} \frac{1}{2\pi } \int _{\mathbb {R}^{2}_{+}}\left( \log |x-y|+\log |\overline{x}-y|-2\log |y|\right) \left( 2|y|^{2\alpha }e^{2u(y)}-|y|^{\alpha }e^{u(y)}|\psi (y)|^2\right) dy\\&+\frac{1}{2\pi } \int _{\partial \mathbb {R}^{2}_{+}}\left( \log |x-y|+\log |\overline{x}-y|-2\log |y|\right) c|y|^{\alpha }e^{u(y)}dy. \end{aligned}$$

where $\bar{x}$ is the reflection point of x about $\partial \mathbb {R}^2_+$. It is easy to check that w(x) satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} {\Delta } w &{}=&{} 2|x|^{2\alpha }e^{2u}-|x|^{\alpha }e^u|\psi |^2, &{}\qquad \text {in} \quad \mathbb {R}^{2}_{+},\\ \frac{\partial w}{\partial n}&{}=&{} -c|x|^{\alpha }e^u, &{}\qquad \text {on}\quad \partial \mathbb {R}^{2}_{+}. \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \lim _{|x|\rightarrow \infty }\frac{w(x)}{\ln |x|}=\frac{d}{\pi }. \end{aligned}$$

Consider $v(x)=u+w$. Then v(x) satisfies

$$\begin{aligned} \left\{ \begin{array}{rcl} {\Delta } v &{}=&{} 0, \qquad \text {in} \quad \mathbb {R}^{2}_{+},\\ \frac{\partial v}{\partial n}&{}=&{} 0, \qquad \text {on}\quad \partial \mathbb {R}^{2}_{+}. \end{array} \right. \end{aligned}$$

We extend v(x) to $\mathbb {R}^2$ by even reflection such that v(x) is harmonic in $\mathbb {R}^2$. From Lemma 5.1 we know $v(x)\le C(1+\ln (|x|+1))$ for some positive constant C. Thus v(x) is a constant. This completes the proof of Step 1. Since $\int _{\mathbb {R}_+^2}|x|^{2\alpha }e^{2u}dx<\infty$, we get that $d \ge \pi (1+\alpha )$. Furthermore, similarly as in the case of the usual Liouville or super-Liouville equation, we can show that $d>\pi (1+\alpha )$.

Step 2. The proof of (37) and $d =2\pi (1+\alpha )$.

Notice that we have shown $d>\pi (1+\alpha )$ in Step 2, we then can improve the estimates of $e^{2u}$ to

$$\begin{aligned} e^{2u}\le C|x|^{-2(1+\alpha )-\varepsilon } \quad \text { for } |x| \text { near } \infty . \end{aligned}$$

Therefore the asymptotic estimates (35) and (36) of the spinor $\psi (x)$ hold. By using the standard potential analysis we can obtain that

$$\begin{aligned} u(x)=-\frac{d}{\pi }\ln {|x|}+C+O(|x|^{-1}) \qquad \text {for}\quad |x| \quad \text {near}\quad \infty \end{aligned}$$

for some constant $C>0$. Thus we get the proof of (37).

Furthermore, we can show that $d=2\pi (1+\alpha )$. For sufficiently large $R>0$, the Pohozaev identity for the solution $(u,\psi )$ gives

$$\begin{aligned}&R\int _{ S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \nonumber \\&\quad = (1+\alpha )\int _{D^+_R}2|x|^{2\alpha }e^{2u}-|x|^{\alpha }e^u|\Psi |^2dv+ (\alpha +1)\int _{L_R}c|x|^{\alpha }e^{u}ds \nonumber \\&\qquad -R\int _{S^+_R}|x|^{2\alpha }e^{2u}d\sigma -c|s|^{\alpha }se^u|^{s=R}_{s=-R} \nonumber \\&\qquad + \frac{1}{4}\int _{S^+_R}\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \rangle d\sigma +\frac{1}{4}\int _{ S^+_R}\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

(39)

By the asymptotic estimates (35), (36) and (37) of $(u,\psi )$ we have

$$\begin{aligned} \lim _{R\rightarrow +\infty }R\int _{ S^+_R} \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma =\frac{d^2}{2\pi }, \end{aligned}$$

and

$$\begin{aligned} \lim _{R\rightarrow +\infty }R\int _{S^+_R}|x|^{2\alpha }e^{2u}d\sigma +c|s|^{\alpha }se^u|^{s=R}_{s=-R}=0, \end{aligned}$$

and

$$\begin{aligned} \lim _{R\rightarrow +\infty }\int _{S^+_R}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \right\rangle d\sigma +\int _{ S^+_R}\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma =0. \end{aligned}$$

Let $R\rightarrow +\infty$ in (39), we get that

$$\begin{aligned} \frac{d^2}{2\pi }=(1+\alpha )d. \end{aligned}$$

It follows that $d=2\pi (1+\alpha )$.

Step 3. The proof of (38).

Since $d=2\pi (1+\alpha )$ by Step 2, we can improve the estimate for $e^{2u}$ to

$$\begin{aligned} e^{2u}\le C|x|^{-4(1+\pi )} \qquad \text {for}\ \ |x| \ \ \text {near}\ \ \infty . \end{aligned}$$

(40)

This implies that the constant spinor $\xi _0$ is well defined. By using the chirality boundary condition of spinor, we extend $(u,\psi )$ to the lower half plane $\mathbb {R}^2_{+}$(see (23) and (24)) to get

$$\begin{aligned} \not {D}\psi =-A(x)\psi ,\quad \text { in } \mathbb {R}^2. \end{aligned}$$

Here A(x) is defined by

$$\begin{aligned} A(x)=\left\{ \begin{matrix} |x|^{\alpha }e^{u(x)},\quad x\in \mathbb {R}^{2}_{+},\\ |\bar{x}|^{\alpha }e^{u(\bar{x})}, \quad x\in \mathbb {R}^2_{-}.\end{matrix} \right. \end{aligned}$$

Define

$$\begin{aligned} \xi _1=\int _{\mathbb {R}^2}A(x)\psi dx. \end{aligned}$$

The constant spinor $\xi _1$ is also well defined. From the asymptotic estimates (35) and (40) and a similar argument in [27] we obtain

$$\begin{aligned} \psi (x)=-\frac{1}{2\pi }\frac{x}{|x|^2} \cdot \xi _1+o(|x|^{-1})\qquad \text {for} \ \ |x| \ \ \text {near}\ \ \infty . \end{aligned}$$

(41)

Since

$$\begin{aligned} \xi _1= & {} \int _{\mathbb {R}_+^2}A(x)\psi dx+\int _{\mathbb {R}_-^2}A(x)\psi dx\\= & {} \int _{\mathbb {R}_+^2}|x|^\alpha e^u\psi dx+\int _{\mathbb {R}_-^2}|\bar{x}|^{\alpha }e^{u(\bar{x})}ie_1\cdot \psi (\bar{x}) dx\\= & {} \int _{\mathbb {R}_+^2}|x|^{\alpha }e^u\psi dx+\int _{\mathbb {R}_+^2}|y|^{\alpha }e^{u(y)}ie_1\cdot \psi (y) dy\\= & {} (I+ie_1)\cdot \int _{\mathbb {R}_+^2}|x|^{\alpha }e^u\psi dx\\= & {} (I+ie_1)\cdot \xi _0. \end{aligned}$$

Hence we obtain from (41)

$$\begin{aligned} \psi (x)=-\frac{1}{2\pi }\frac{x}{|x|^2}(I+ie_1) \cdot \xi _0+o(|x|^{-1})\qquad \text {for}\quad |x| \quad \text {near}\quad \infty . \end{aligned}$$

Thus we finish the proof of Step 3 and we complete the proof of the Proposition. $\square$

Proposition 5.1 indicates that the singularity at infinity of regular solutions for (31) and (32) can be removed as in many other conformally invariant problems.

Theorem 5.2

Let $(u,\psi )$ be a regular solution of (31) and (32). Then $(u,\psi )$ extends conformally to a regular solution on a spherical cap $\mathbb {S}^2_{c^\prime }$, where $c^\prime$ is the geodesic curvature of $\partial \mathbb {S}^2_{c^\prime }$.

Proof

Let $(v,\phi )$ be the Kelvin transformation of $(u,\psi )$ as before. Then $(v,\phi )$ satisfies the system (33). To prove the theorem, by conformal invariance, it is sufficient to show that $(v,\phi )$ is regular on $\overline{\mathbb {R}}_+^2$. Applying Proposition 5.1, we get

$$\begin{aligned} v(x)=\left( \frac{d}{\pi }-2(1+\alpha )\right) \ln {|x|}+O(1) \qquad \text {for}\quad |x| \quad \text {near}\quad 0. \end{aligned}$$

(42)

Since $\alpha =2\pi (1+\alpha )$, it follows that v is bounded near the singularity 0. Recall that $\phi$ is also bounded near 0, we can apply elliptic theory to obtain that $(v,\phi )$ is regular on $\overline{\mathbb {R}}_+^2$. $\square$

6 Energy identity for spinors

The energy identity for spinor part of solutions to the super-Liouville equations on closed Riemann surfaces was derived in [23, 27]. In this section, we shall prove an analogue for the singular super-Liouville boundary problem, i.e. Theorem 1.4. For harmonic maps in dimension two and J-holomorphic curves as well as for solutions of certain nonlinear Dirac type equations, similar results are derived in [14, 38, 39, 42] and the references therein.

To prove Theorem 1.4, we shall derive the local estimate for the spinor part on an upper half annulus. Since we can extend $(u,\Psi )$ to the lower half disk $D_r^-$ by the chirality boundary condition of $\Psi$, the proof of this local estimate can be established by using the result of Lemma 3.1 of [23]. Here we just state the Lemma and omit the proof.

Lemma 6.1

Let $(u,\Psi )$ satisfies (4) and

$$\begin{aligned} \int _{D^+_r}|x|^{2\alpha }e^{2u}+|\Psi |^4dx+\int _{L_r}|x|^{\alpha }e^{u}ds<C. \end{aligned}$$

For $0<r_1<2r_1<\frac{r_2}{2}<r_2<r$, consider the annulus $A_{r_1,r_2}=\{x\in \mathbb {R}^2|r_1\le |x|\le r_2\}$ and the upper half annulus $A^+_{r_1,r_2}=A_{r_1,r_2}\cap \mathbb {R}_+^2$. Then we have

$$\begin{aligned}&\left( \int _{A^+_{2r_1,\frac{r_2}{2}}}|D\Psi |^{\frac{4}{3}}\right) ^{\frac{3}{4}}+\left( \int _{A^+_{2r_1,\frac{r_2}{2}}}|\Psi |^4\right) ^{\frac{1}{4}}\nonumber \\&\quad \le C_0 \left( \int _{A^+_{r_1,r_2}}|x|^{2\alpha }e^{2u}\right) ^{\frac{1}{2}}\left( \int _{A^+_{r_1,r_2}}|\Psi |^4\right) ^{\frac{1}{4}}+C\left( \int _{A^+_{r_1,2r_1}}|\Psi |^4\right) ^{\frac{1}{4}}\nonumber \\&\qquad +C\left( \int _{A^+_{\frac{r_2}{2}, r_2}}|\Psi |^4\right) ^{\frac{1}{4}} \end{aligned}$$

(43)

for a positive constant $C_0$ and some universal positive constant C.

Proof of Theorem 1.4

We will follow closely the argument for the energy identity of harmonic maps, see [14], or for super-Liouville equations, see [7, 23, 25, 27] . Since the blow-up set $\Sigma _1$ is finite, we can find small disk $D^+_{\delta _i}(x_i)$, which is centered at each blow-up point $x_i$, such that $D^+_{\delta _i}(x_i)\cap D^+_{\delta _j}(x_j)=\emptyset$ for $i\ne j, i,j=1,2,\ldots , P$, and on $(D^+_r\cup \L _r)\backslash \bigcup _{i=1}^{P}(D^+_{\delta _i}(x_i)\cup L_{\delta _i}(x_i))$, $\Psi _n$ converges strongly to $\Psi$ in $L^4$. So, we need to prove that there are $(u^{i,k},\xi ^{i,k})$, which are solutions of (7), $i=1,2,\ldots , I; k=1,2,\ldots , K_i$, such that

$$\begin{aligned} \lim _{\delta _i\rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{\delta _i}(x_i)}|\Psi _n|^4dv=\sum _{k=1}^{L_i}\int _{S^2}|\xi ^{i,k}|^4dv, \text { for } i=1,2,\ldots , I; \end{aligned}$$

(44)

or, we need to prove that there are $(u^{j,l},\xi ^{j,l})$, which are solutions of (8), $j=1,2,\ldots , J; l=1,2,\ldots , L_j$, such that

$$\begin{aligned} \lim _{\delta _j\rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{\delta _j}(x_i)}|\Psi _n|^4dv=\sum _{l=1}^{L_j}\int _{S^2_{c'}}|\xi ^{j,l}|^4dv, \text { for } j=1,2,\ldots , J; \end{aligned}$$

(45)

When $p\in (D^+_r)^o$, from [27], we know that (44) holds. So, without loss of generality, we assume that $p\in L_r$ and there is only one bubble at each blow-up point p. Furthermore, we may assume that $p=0$. The case of $p\ne 0$ can be handled in an analogous way and in fact this case is simpler, as $|x|^{\alpha }$ is a smooth function near p. Then what we need to prove is that there exists a bubble $(u,\xi )$ as (7), such that

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{\delta }}|\Psi _n|^4dv=\int _{S^2}|\xi |^4dv, \end{aligned}$$

(46)

or there exists a bubble $(u, \xi )$ as (8) such that such that

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{\delta }}|\Psi _n|^4dv=\int _{S^2_{c'}}|\xi |^4dv. \end{aligned}$$

(47)

Next we rescale functions $(u_n,\Psi _n)$ at the blow-up point $p=0$ and then try to get the bubble of $(u_n,\Psi _n)$. To this purpose, we let $x_n\in \overline{D}^+_{\delta }$ such that $u_n(x_n)=\max _{\overline{D}^+_{\delta }}u_n(x)$. Write $x_n=(s_n,t_n)$. It is clear that $x_n\rightarrow p$ and $u_n(x_n)\rightarrow +\infty$. Define $\lambda _n=e^{-\frac{u_n(x_n)}{\alpha +1}}$. We know $\lambda _n$, $|x_n|$ and $t_n$ converge to 0 as $n\rightarrow 0$, but their rates of converging to 0 may be different. Next we will distinguish three cases.

Case I. $\frac{|x_n|}{\lambda _n}=O(1)$ as $n\rightarrow +\infty$.

In this case, we define the rescaling functions

$$\begin{aligned} \left\{ \begin{array}{rcl} \widetilde{u}_n(x)&{}=&{}u_n(\lambda _nx)+(1+\alpha )\ln {\lambda _n}\\ \widetilde{\Psi }_n(x)&{}=&{}\lambda ^{\frac{1}{2}}_n\Psi _n(\lambda _nx) \end{array} \right. \end{aligned}$$

for any $x \in \overline{D}^+_{\frac{\delta }{2\lambda _n} }$. Then $(\widetilde{u}_n(x),\widetilde{\Psi }_n(x))$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle \widetilde{u}_n(x)&{}=&{} 2V^2(\lambda _n x)|x|^{2\alpha }e^{2\widetilde{u}_n(x)}-V(\lambda _n x)|x|^\alpha e^{\widetilde{u}_n(x)}|\widetilde{\Psi }_n(x)|^2, &{} \text { in } D^+_{\frac{\delta }{2\lambda _n} },\\ \not {D}\widetilde{\Psi }_n(x)&{}=&{}-V(\lambda _n x)|x|^{\alpha }e^{\widetilde{u}_n(x)}\widetilde{\Psi }_n(x), &{} \text { in } D^+_{\frac{\delta }{2\lambda _n} },\\ \frac{\partial \widetilde{u}_n(x)}{\partial n}&{}=&{}cV(\lambda _n x)|x|^\alpha e^{\widetilde{u}_n(x)}, &{} \text { on } L_{\frac{\delta }{2\lambda _n} },\\ B\widetilde{\Psi }_n(x)&{}=&{} 0, &{} \text { on } L_{\frac{\delta }{2\lambda _n} }, \end{array} \right. \end{aligned}$$

with the energy condition

$$\begin{aligned} \int _{D^+_{\frac{\delta }{2\lambda _n} }}|x|^{2\alpha }e^{2\widetilde{u}_n(x)}+|\widetilde{\Psi }_n(x)|^4dv+\int _{L_{\frac{\delta }{2\lambda _n} }}|x|^\alpha e^{\widetilde{u}_n(x)}d\sigma <C. \end{aligned}$$

We know that

$$\begin{aligned} \max _{\bar{D}^+_{\frac{\delta }{2\lambda _n}}}\widetilde{u}_n(x)= \widetilde{u}_n(\frac{x_n}{\lambda _n})=u_n(x_n)+(\alpha _n+1)\ln \lambda _n=0. \end{aligned}$$

Notice that the maximum point of $\widetilde{u}_n(x)$, i.e. $\frac{x_n}{\lambda _n}$, is bounded, namely $|\frac{x_n}{t_n}|\le C$. So by taking a subsequence, we can assume that $\frac{x_n}{t_n}\rightarrow x_0\in \bar{\mathbb {R}}^2_+$ with $|x_0|\le C$. Therefore it follows from Theorem 1.3 that, by passing to a subsequence, $(\widetilde{u}_n,\widetilde{\Psi }_n)$ converges in $C^{2}_{loc}(\mathbb {R}^2_+)\cap C^{1}_{loc}(\bar{\mathbb {R}}^2_+)\times C^{2}_{loc}(\Gamma (\Sigma {\mathbb {R}}^2_+))\cap C^{1}_{loc}(\Gamma (\Sigma {\bar{\mathbb {R}}}^2_+))$ to some $(\widetilde{u},\widetilde{\Psi })$ satisfying

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta \widetilde{u} &{}=&{} 2V^2(0)|x|^{2\alpha }e^{2\widetilde{u}}-V(0)|x|^{\alpha }e^{\widetilde{u}}| \widetilde{\Psi }|^2, &{}\quad \text { in } \mathbb {R}^2_+,\\ \not {D}\widetilde{\Psi } &{}=&{} -V(0)|x|^{\alpha }e^{\widetilde{u}}\widetilde{\Psi }, &{}\quad \text { in } \mathbb {R}^2_+,\\ \frac{\partial \widetilde{u}}{\partial n}&{}=&{}cV(0)|x|^\alpha e^{\widetilde{u}}, &{} \quad \text { on } \partial \mathbb {R}^2_+,\\ B\widetilde{\Psi }&{}=&{} 0, &{} \quad \text { on } \partial \mathbb {R}^2_+ \end{array} \right. \end{aligned}$$

(48)

with the energy condition $\int _{\mathbb {R}^2_+}(|x|^{2\alpha }e^{2\widetilde{u}}+|\widetilde{\Psi }|^4)dx +\int _{\partial \mathbb {R}^2_+} |x|^\alpha e^{\widetilde{u}}d\sigma < \infty$. By Proposition 5.1, there holds

$$\begin{aligned} \int _{\mathbb {R}^2_+}(2V^2(0)|x|^{2\alpha }e^{2\widetilde{u}}-V(0)|x|^{\alpha }e^{\widetilde{u}}|\widetilde{\Psi }|^2)dx+\int _{\partial \mathbb {R}^2_+} cV(0)|x|^\alpha e^{\widetilde{u}}d\sigma =2\pi (1+\alpha ). \end{aligned}$$

By the removability of a global singularity (Theorem 5.2), we get a bubbling solution on $S^2_{c'}$.

Case II. $\frac{|x_n|}{\lambda _n}\rightarrow +\infty$ as $n\rightarrow +\infty$.

In this case, we must have

$$\begin{aligned} \overline{u}_n(y_n):=u_n(x_n)+(\alpha +1)\ln |x_n|=(\alpha +1)\ln |x_n|-(\alpha +1)\ln \lambda _n\rightarrow +\infty . \end{aligned}$$

(49)

Therefore we can rescale twice to get the bubble. First, we define the rescaling functions

$$\begin{aligned} \left\{ \begin{array}{rcl} \overline{u}_n(x)&{}=&{}u_n(|x_n|x)+(\alpha +1)\ln |x_n|\\ \overline{\Psi }_n(x)&{}=&{}|x_n|^{\frac{1}{2}}\Psi _n(|x_n|x) \end{array} \right. \end{aligned}$$

for any $x\in \overline{D}^+_{\frac{\delta }{2|x_n|}}$. Then $(\overline{u}_n(x),\overline{\Psi }_n(x))$ satisfies that

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle \overline{u}_n(x)&{}=&{} 2V^2(|x_n|x)|x|^{2\alpha }e^{2\overline{u}_n(x)}-V(|x_n|x)|x|^\alpha e^{\overline{u}_n(x)}|\overline{\Psi }_n(x)|^2, &{} \text { in } D^+_{\frac{\delta }{2|x_n|}},\\ \not {D}\overline{\Psi }_n(x)&{}=&{}-V(|x_n|x)|x|^\alpha e^{\overline{u}_n(x)}\overline{\Psi }_n(x), &{} \text { in } D^+_{\frac{\delta }{2|x_n|}},\\ \frac{\partial \overline{u}_n(x)}{\partial \nu }&{}=&{}cV(|x_n|x)|x|^\alpha e^{\overline{u}_n(x)}, &{} \text { on } L_{\frac{\delta }{2|x_n|}},\\ B\overline{\Psi }_n(x)&{}=&{} 0, &{} \text { on } L_{\frac{\delta }{2|x_n|}}. \end{array} \right. \end{aligned}$$

Set that $y_n=\frac{x_n}{|x_n|}$. We assume that $y_0=\lim _{n\rightarrow \infty }\frac{x_n}{|x_n|}$. By (49), we know $y_0$ is a blow-up point of $(\overline{u}_n, \overline{\Psi }_n)$. We can set $\delta _n=e^{-\overline{u}_n(y_n)}$, and $\rho _n=\frac{e^{-u_n(x_n)}}{|x_n|^\alpha }=\lambda _n(\frac{\lambda _n}{|x_n|})^\alpha$. It is clear that $\delta _n\rightarrow 0$, $\rho _n\rightarrow 0$ and $\frac{|x_n|}{\rho _n}\rightarrow +\infty$ as $n\rightarrow \infty$. We define the rescaling functions

$$\begin{aligned} \left\{ \begin{array}{rcl} \widetilde{u}_n(x)&{}=&{}\overline{u}_n(\delta _nx+y_n)+\ln \delta _n=u_n(x_n+\rho _nx)-u_n(x_n)\\ \widetilde{\Psi }_n(x)&{}=&{}=\delta _n^{\frac{1}{2}}\overline{\Psi }_n(\delta _nx+y_n)=\rho _n^{\frac{1}{2}}\Psi _n(x_n+\rho _n x) \end{array} \right. \end{aligned}$$

for any x such that $y_n+\delta _n x \in \overline{D}^+_{R}(y_n)$ with any $R>1$. By a direct computation, we have

$$\begin{aligned} \Omega _n=\{x\in \mathbb {R}^2 | y_n+\delta _n x \in \overline{D}^+_{R}(y_n)\}=\{x\in \mathbb {R}^2 | x_n+\rho _n x \in \overline{D}^+_{R|x_n|}(x_n)\}. \end{aligned}$$

We set $L_n=\partial \Omega _n \cap \{x\in \mathbb {R}^2 | t= -\frac{t_n}{\rho _n} \}$. Then $(\widetilde{u}_n(x),\widetilde{\Psi }_n(x))$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle \widetilde{u}_n(x)&{}=&{} 2V^2(x_n+\rho _nx)|\frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x|^{2\alpha }e^{2\widetilde{u}_n(x)}\\ &{} &{} -V(x_n+\rho _nx)|\frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x|^\alpha e^{\widetilde{u}_n(x)}|\widetilde{\Psi }_n(x)|^2, &{} \text { in } \Omega _n ,\\ \not {D}\widetilde{\Psi }_n(x)&{}=&{}-V(x_n+\rho _nx)|\frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x|^\alpha e^{\widetilde{u}_n(x)}\widetilde{\Psi }_n(x), &{} \text { in } \Omega _n ,\\ \frac{\partial \widetilde{u}_n(x)}{\partial n}&{}=&{}cV(x_n+\rho _nx)|\frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x|^\alpha e^{\widetilde{u}_n(x)}, &{} \text { on } L_n ,\\ B\widetilde{\Psi }_n(x)&{}=&{} 0, &{} \text { on } L_n, \end{array} \right. \end{aligned}$$

with the energy condition

$$\begin{aligned} \int _{\Omega _n}\left| \frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x\right| ^{2\alpha }e^{2\widetilde{u}_n(x)}+\left| \widetilde{\Psi }_n(x)|^4dv+\int _{L_n}|\frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x\right| ^\alpha e^{\widetilde{u}_n(x)}d\sigma <C. \end{aligned}$$

It is clear that

$$\begin{aligned} \widetilde{u}_n(x)\le \max _{\Omega _n}\widetilde{u}_n(x)= \widetilde{u}_n(0)=0 . \end{aligned}$$

Now we proceed by distinguishing two subcases.

Case II.1 $\frac{t_n}{\rho _n}\rightarrow +\infty$ as $n\rightarrow \infty$.

Notice that $|\frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x|\rightarrow 1$ as $n\rightarrow \infty$ in $C^0_{loc}(\mathbb {R}^2)$. It follows from Theorem 1.3 that, by passing to a subsequence, $(\widetilde{u}_n,\widetilde{\Psi }_n)$ converges in $C^{2}_{loc}(\mathbb {R}^2)\times C^{2}_{loc}(\Gamma (\Sigma {\mathbb {R}}^2))$ to some $(\widetilde{u},\widetilde{\Psi })$ satisfying

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta \widetilde{u} &{}=&{} 2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}| \widetilde{\Psi }|^2, &{}\quad \text { in } \mathbb {R}^2,\\ \not {D}\widetilde{\Psi } &{}=&{} -V(0)e^{\widetilde{u}}\widetilde{\Psi }, &{}\quad \text { in } \mathbb {R}^2,\\ \end{array} \right. \end{aligned}$$

(50)

with the energy condition $\int _{\mathbb {R}^2}e^{2\widetilde{u}}+|\widetilde{\Psi }|^4dx <\infty$. By Proposition 6.4 in [21], there holds

$$\begin{aligned} \int _{\mathbb {R}^2}(2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}|\widetilde{\Psi }|^2)dx=4\pi . \end{aligned}$$

By the removability of a global singularity (Theorem 6.5 in [21]), we get a bubbling solution on $S^2$.

Case II.2 $\frac{t_n}{\rho _n}\rightarrow \Lambda$ as $n\rightarrow \infty$.

Similar in the Case II.1, we have from Theorem 1.3 that, by passing to a subsequence, $(\widetilde{u}_n,\widetilde{\Psi }_n)$ converges in $C^{2}_{loc}(\mathbb {R}^2_{-\Lambda })\cap C^{1}_{loc}(\bar{\mathbb {R}}^2_{-\Lambda })\times C^{2}_{loc}(\Gamma (\Sigma {\mathbb {R}}^2_{-\Lambda }))\cap C^{1}_{loc}(\Sigma \bar{\mathbb {R}}^2_{-\Lambda })$ to some $(\widetilde{u},\widetilde{\Psi })$ satisfying

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta \widetilde{u} &{}=&{} 2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}| \widetilde{\Psi }|^2, &{}\quad \text { in } \mathbb {R}^2_{-\Lambda },\\ \not {D}\widetilde{\Psi } &{}=&{} -V(0)e^{\widetilde{u}}\widetilde{\Psi }, &{}\quad \text { in } \mathbb {R}^2_{-\Lambda },\\ \frac{\partial \widetilde{u}}{\partial n}&{}=&{}cV(0) e^{\widetilde{u}}, &{}\quad \text { on } \partial \mathbb {R}^2_{-\Lambda } ,\\ B\widetilde{\Psi } &{}=&{} 0, &{} \quad \text { on } \partial \mathbb {R}^2_{-\Lambda }, \end{array} \right. \end{aligned}$$

(51)

with the energy condition $\int _{\mathbb {R}^2_{-\Lambda }}e^{2\widetilde{u}}+|\widetilde{\Psi }|^4dx+\int _{\partial \mathbb {R}^2_{-\Lambda }}e^{\widetilde{u}}d\sigma <\infty$. By Proposition 6.4 in [21], there holds

$$\begin{aligned} \int _{\mathbb {R}^2_{-\Lambda }}(2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}|\widetilde{\Psi }|^2)dx+\int _{\partial \mathbb {R}^2_{-\Lambda }}cV(0)e^{\widetilde{u}}d\sigma =2\pi . \end{aligned}$$

By the removability of a global singularity (Theorem 6.5 in [25]), we get a bubbling solution on $S^2_{c'}$.

It is well know, in order to prove (46) or (47), we need to prove that there is no any energy of $\Psi _n$ in the neck domain, i.e.

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{R\rightarrow +\infty }\lim _{n\rightarrow \infty }\int _{A^+_{\delta ,R,n}}|\Psi _n|^4dv=0, \end{aligned}$$

(52)

where $A^+_{\delta ,R,n}$ is the neck domain which is defined latter. To this purpose, we shall proceed separately for Case I, Case II.1 and Case II.2.

For Case I, we define the neck domain is

$$\begin{aligned} A^+_{\delta , R, n}=\{x\in \mathbb {R}^2_+|\lambda _n R\le |x|\le \delta \}. \end{aligned}$$

We have two claims.

Claim 1 For any $\varepsilon >0$, there is an $N>1$ such that for any $n\ge N$, we have

$$\begin{aligned} \int _{D^+_r\setminus D^+_{e^{-1}r}}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4)+\int _{\partial (D^+_r\setminus D^+_{e^{-1}r})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}<\varepsilon ; \quad \forall r\in [e\lambda _n R, \delta ]. \end{aligned}$$

To prove this claim, we note two facts. The first fact is: for any $T>0$, there exists some N(T) such that for any $n\ge N(T)$, we have

$$\begin{aligned} \int _{D^+_\delta \setminus D^+_{\delta e^{-T}}}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4)+\int _{\partial ( D^+_\delta \setminus D^+_{\delta e^{-T}})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}<\varepsilon . \end{aligned}$$

(53)

Actually, since $(u_n,\Psi _n)$ has no blow-up point in $\overline{D}^+_{\delta }\backslash \{p\}$, then $|\Psi _n|$ is uniformly bounded in $\overline{D^+_\delta \backslash D^+_{\delta e^{-T}}}$ , and $u_n$ will either be uniformly bounded in $\overline{D^+_\delta \backslash D^+_{\delta e^{-T}}}$ or uniformly tend to $-\infty$ in $\overline{D^+_\delta \backslash D^+_{\delta e^{-T}}}$. So if $u_n$ uniformly tends to $-\infty$ in $\overline{D_\delta \backslash D_{\delta e^{-T}}}$, it is clear that, for any given $T>0$, we have an N(T) big enough such that when $n\ge N(T)$

$$\begin{aligned} \int _{D^+_\delta \backslash D^+_{\delta e^{-T}}}(|x|^{2\alpha }e^{2u_n}+\int _{\partial (D^+_\delta \backslash D^+_{\delta e^{-T}})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}<\frac{\varepsilon }{2}. \end{aligned}$$

Moreover, since $\Psi _n$ converges to $\Psi$ in $L_{loc}^4((D^+_r\cap L_r)\setminus \Sigma _1)$ and hence

$$\begin{aligned} \int _{D^+_\delta \backslash D^+_{\delta e^{-T}}}|\Psi _n|^{4} \rightarrow \int _{D^+_\delta \backslash D^+_{\delta e^{-T}}}|\Psi |^{4}. \end{aligned}$$

For any small $\varepsilon >0$, we may choose $\delta >0$ small enough such that $\int _{D^+_\delta }|\Psi |^{4}<\frac{\varepsilon }{4}$, then for any given $T>0$, we have an N(T) big enough such that when $n\ge N(T)$

$$\begin{aligned} \int _{D^+_\delta \backslash D^+_{\delta e^{-T}}}|\Psi _n|^{4}<\frac{\varepsilon }{2}. \end{aligned}$$

Consequently, we get (53).

If $(u_n, \Psi _n)$ is uniformly bounded in $\overline{D^+_\delta \backslash D^+_{\delta e^{-T}}}$, then we know $(u_n, \Psi _n)$ converges to a weak solution $(u,\Psi )$ strongly on compact sets of $D^+_{\delta }\setminus \{p\}$. Therefore, we can also choose $\delta >0$ small enough such that, for any given $T>0$, there exists an N(T) big enough, when $n\ge N(T)$, (53) holds.

The second fact is: For any small $\varepsilon >0$, and $T>0$, we may choose an N(T) such that when $n\ge N(T)$

$$\begin{aligned}&\int _{D^+_{\lambda _nRe^T}\setminus D^+_{\lambda _nR}}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4)+\int _{\partial (D^+_{\lambda _nRe^T}\setminus D^+_{\lambda _nR})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}\\&\quad = \int _{D^+_{Re^T}\setminus D^+_{R}}(|x|^{2\alpha }e^{2\widetilde{u}_n}+|\widetilde{\Psi }_n|^4)+\int _{\partial (D^+_{Re^T}\setminus D^+_{R})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{\widetilde{u}_n} < \varepsilon , \end{aligned}$$

if R is big enough.

Now we can prove the claim. We argue by contradiction by using the above two facts. If there exists $\varepsilon _0>0$ and a sequence ${r_n}$, $r_n\in [e\lambda _nR,\delta ]$, such that

$$\begin{aligned} \int _{D^+_{r_n}\setminus D^+_{e^{-1}r_n}}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^{4})+\int _{\partial (D^+_{r_n}\setminus D^+_{e^{-1}r_n})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}\ge \varepsilon _0. \end{aligned}$$

Then, by the above two facts, we know that $\frac{\delta }{r_n}\rightarrow +\infty$ and $\frac{\lambda _nR}{r_n}\rightarrow 0$, in particular, $r_n\rightarrow 0$ as $n\rightarrow +\infty$. Rescaling again, we set

$$\begin{aligned} \left\{ \begin{array}{rcl} v_n(x)&{}=&{} u_n(r_nx)+(1+\alpha )\ln r_n, \\ \varphi _n(x)&{}=&{} r_n^{\frac{1}{2}}\Psi (r_nx)\\ \end{array} \right. \end{aligned}$$

for any $x\in D^+_{\frac{\delta }{r_n}}\setminus D^+_{\frac{\lambda _n R}{r_n}}$.

It is clear that

$$\begin{aligned} \int _{D^+_1\setminus D^+_{e^{-1}}}(|x|^{2\alpha }e^{2v_n}+|\varphi _n|^4)+\int _{\partial (D^+_1\setminus D^+_{e^{-1}})\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{v_n}\ge \varepsilon _0. \end{aligned}$$

(54)

And $(v_n,\varphi _n)$ satisfies for any $R>0$

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle v_n(x) &{}=&{} 2V^2(r_nx)|x|^{2\alpha }e^{2v_n(x)}-V(r_nx)|x|^{\alpha }e^{v_n(x)} |\varphi _n(x)|^2, &{}\quad \text { in } \left( D^+_{\frac{\delta }{r_n}}\setminus D^+_{\frac{\lambda _nR}{r_n}}\right) , \\ \not {D}\varphi _n(x) &{}=&{} -V(r_nx)|x|^{\alpha }e^{v_n(x)}\varphi _n(x), &{} \quad \text { in } \left( D^+_{\frac{\delta }{r_n}}\setminus D^+_{\frac{\lambda _nR}{r_n}}\right) ,\\ \frac{\partial v_n(x)}{\partial n}&{}=&{} cV(r_nx)|x|^{\alpha }e^{v_n(x)}, &{}\quad \text { on } \partial \left( D^+_{\frac{\delta }{r_n}}\setminus D^+_{\frac{\lambda _nR}{r_n}}\right) \cap \partial \mathbb {R}^2_+,\\ B\varphi _n(x)&{}=&{}0, &{}\quad \text { on } \partial \left( D^+_{\frac{\delta }{r_n}}\setminus D^+_{\frac{\lambda _nR}{r_n}}\right) \cap \partial \mathbb {R}^2_+. \end{array} \right. \end{aligned}$$

According to Theorem 1.3, there are three possible cases:

• 1. There exists some $q\in Q_n=(D^+_{\frac{\delta }{r_n}}\setminus D^+_{\frac{\lambda _nR}{r_n}})$ and energy concentration occurs near the point q, namely along some subsequence we have

  $$\begin{aligned} \lim _{n\rightarrow \infty } \int _{D_r(q)\cap Q_n }(|x|^{2\alpha }e^{2v_n}+|\varphi _n|^{4})+\int _{ D_r(q)\cap \partial Q_n\cap \{t=0\}}|x|^{\alpha }e^{v_n}\ge \varepsilon _0>0 \end{aligned}$$

  for any small $r>0$. In such a case, we still obtain the second “bubble” by the rescaling argument. Thus we get a contradiction.

• 2. For any $R>0$, there is no blow-up point in $D^+_R\setminus D^+_{\frac{1}{R}}$ and $v_n\rightarrow -\infty$ uniformly in $\overline{D^+_R\setminus D^+_{\frac{1}{R}}}$. Then, it is clear that $\varphi _n$ converges to a spinor $\varphi$ in $L_{loc}^4(\overline{\mathbb {R}_+^2}\setminus \{0\})$ which satisfies

  $$\begin{aligned} \left\{ \begin{array}{rcll} \not {D}\varphi &{}=&{} 0, &{} \text { in } \mathbb {R}^2_+,\\ B \varphi &{}=&{} 0, &{}\text { on } \partial \mathbb {R}^2_+\setminus \{0\}. \end{array} \right. \end{aligned}$$

  We translate $\varphi$ to be a harmonic spinor on $\mathbb {R}_+^2 \setminus \{0\}$ satisfying the corresponding chirality boundary condition and then extend it as in (24) to a harmonic spinor $\overline{\varphi }$ on $\mathbb {R}^2\setminus \{0\}$ with bounded energy, i.e., $||\overline{\varphi }||_{L^4(\mathbb {R}^2)} < \infty$. As discussed in [23], $\overline{\varphi }$ conformally extends to a harmonic spinor on $S^2$. By the well known fact that there is no nontrivial harmonic spinor on $S^2$, we have that $\overline{\varphi } \equiv 0$ and hence $\varphi _n$ converges to 0 in $L_{loc}^4(\mathbb {R}^2_+\setminus \{0\})$. This will contradict (54)

• 3. For any $R>0$, there is no blow-up point in $(D^+_R\setminus D^+_{\frac{1}{R}})$ and $(v_n, \varphi _n)$ is uniformly bounded in $(D^+_R\setminus D^+_{\frac{1}{R}})$. In such a case $(v_n, \varphi _n)$ will converge to $(v,\varphi )$ strongly on $(D^+_R\setminus D^+_{\frac{1}{R}})$ and $(v,\varphi )$ satisfying

  $$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle v &{}=&{} 2V^2(0)|x|^{2\alpha }e^{2v}-V(0)|x|^{\alpha }e^{v}|\varphi |^2, &{} \quad \text { in }\mathbb {R}^2_+,\\ \not {D}\varphi &{}=&{} -V(0)|x|^{\alpha }e^{v}\varphi , &{} \quad \text { in }\mathbb {R}^2_+,\\ \frac{\partial v}{\partial n}&{}=&{} cV(0)|x|^{\alpha }e^{v}, &{}\quad \text { on } \partial \mathbb {R}^2_+\setminus \{0\},\\ B\varphi &{}=&{} 0, &{}\quad \text { on } \partial \mathbb {R}^2_+\setminus \{0\} \end{array} \right. \end{aligned}$$

  with finite energy. It is clear that $(v,\varphi )$ is regular.

Next we need to remove the singularities of $(v,\varphi )$ and then obtain the second bubble of the system. Consequently we get a contradiction. To this purpose, let us use the Pohozaev identity of $(u_n, \Psi _n)$ in $D^+_\delta$, it follows for any $\rho$ with $r_n\rho <\delta$

$$\begin{aligned}&r_n\rho \int _{ S^+_{r_n\rho }} \left| \frac{\partial u_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u_n|^2d\sigma \\&\quad = (1+\alpha )\int _{D^+_{r_n\rho }}2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2dv\\&\qquad + (\alpha +1)\int _{L_{r_n\rho }}cV(x)|x|^{\alpha }e^{u_n}ds \\&\qquad -r_n\rho \int _{S^+_{r_n\rho }}V^2(x)|x|^{2\alpha }e^{2u_n}d\sigma \\&\qquad +\int _{ L_{r_n \rho }}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^{u_n}ds-cV((s,0))|s|^{\alpha }se^{u_n}|^{s=r_n\rho }_{s=-r_n\rho } \\&\qquad +\int _{D^+_{r_n\rho }}x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u_n}dv -\int _{D^+_{r_n\rho }}x\cdot \nabla V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2dv \\&\qquad + \frac{1}{4}\int _{S^+_{r_n\rho }}\left\langle \frac{\partial \Psi _n}{\partial \nu },(x+\bar{x})\cdot \Psi _n\right\rangle d\sigma +\frac{1}{4}\int _{ S^+_{r_n\rho }}\left\langle (x+\bar{x})\cdot \Psi _n, \frac{\partial \Psi _n}{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

Hence for rescaling functions $(v_n, \varphi _n)$ we have

$$\begin{aligned}&\rho \int _{ S^+_{\rho }} \left| \frac{\partial v_n}{\partial \nu }|^2-\frac{1}{2}|\nabla v_n\right| ^2d\sigma \\&\quad = (1+\alpha )\int _{D^+_{\rho }}2V^2(r_nx)|x|^{2\alpha }e^{2v_n}-V(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv\\&\qquad + (\alpha +1)\int _{L_{\rho }}cV(r_nx)|x|^{\alpha }e^{v_n}ds \\&\qquad -\rho \int _{S^+_{\rho }}V^2(r_nx)|x|^{2\alpha }e^{2v_n}d\sigma +\int _{ L_{\rho }}c\frac{\partial V((r_ns,0))}{\partial s}|s|^{\alpha }se^{v_n}ds\\&\qquad -cV((r_ns,0))|s|^{\alpha }se^{v_n}|^{s=\rho }_{s=-\rho } \\&\qquad +\int _{D^+_{\rho }}x\cdot (\nabla V^2)(r_nx)|x|^{2\alpha }e^{2v_n}dv -\int _{D^+_{\rho }}x\cdot (\nabla V)(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv \\&\qquad + \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \varphi _n}{\partial \nu },(x+\bar{x})\cdot \varphi _n\right\rangle d\sigma +\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \varphi _n, \frac{\partial \varphi _n}{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

This implies that the associated Pohozaev constant of $(v_n,\varphi _n)$ satisfies

$$\begin{aligned}&C(v_n, \varphi _n) = C(v_n, \varphi _n,\rho )\\&\quad = \rho \int _{ S^+_{\rho }} \left| \frac{\partial v_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla v_n|^2d\sigma \\&\qquad -(1+\alpha )\int _{D^+_{\rho }}2V^2(r_nx)|x|^{2\alpha }e^{2v_n}-V(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv\\&\qquad - (\alpha +1)\int _{L_{\rho }}cV(r_nx)|x|^{\alpha }e^{v_n}ds \\&\qquad +\rho \int _{S^+_{\rho }}V^2(r_nx)|x|^{2\alpha }e^{2v_n}d\sigma -\int _{ L_{\rho }}c\frac{\partial V((r_ns,0))}{\partial s}|s|^{\alpha }se^{v_n}ds\\&\qquad +cV((r_ns,0))|s|^{\alpha }se^{v_n}|^{s=\rho }_{s=-\rho } \\&\qquad -\int _{D^+_{\rho }}x\cdot (\nabla V^2)(r_nx)|x|^{2\alpha }e^{2v_n}dv +\int _{D^+_{\rho }}x\cdot (\nabla V)(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv \\&\qquad - \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \varphi _n}{\partial \nu },(x+\bar{x})\cdot \varphi _n\right\rangle d\sigma -\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \varphi _n, \frac{\partial \varphi _n}{\partial \nu }\right\rangle d\sigma \\&\quad = 0. \end{aligned}$$

Since, for any $\rho >0$, $\int _{D^+_\rho }|x|^{2\alpha }e^{2v_n}+|\varphi _n|^4dv+\int _{L_\rho }|x|^{\alpha }e^{v_n}ds<C$, it is easy to check that

$$\begin{aligned} \lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{\rho }}x\cdot (\nabla V^2)(r_nx)|x|^{2\alpha }e^{2v_n}dv +\int _{D^+_{\rho }}x\cdot (\nabla V)(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv=0, \end{aligned}$$

and

$$\begin{aligned} \lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty }\int _{ L_{\rho }}c\frac{\partial V((r_ns,0))}{\partial s}|s|^{\alpha }se^{v_n}ds=0. \end{aligned}$$

This implies that

$$\begin{aligned} 0= & {} \lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty } C(v_n, \varphi _n,\rho )\\= & {} \lim _{\rho \rightarrow 0}C(v,\varphi ,\rho )-(1+\alpha )\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{r}}2V^2(r_nx)|x|^{2\alpha }e^{2v_n}-V(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv\\&-(1+\alpha )\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{L_{r}}cV(r_nx)|x|^{\alpha }e^{v_n}ds\\= & {} C(v,\varphi )-(1+\alpha )\beta . \end{aligned}$$

Here

$$\begin{aligned} \beta= & {} \lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\left[ \int _{D^+_{r}}2V^2(r_nx)|x|^{2\alpha }e^{2v_n}-V(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv\right. \\&\qquad \qquad \qquad \left. +\int _{L_{r}}cV(r_nx)|x|^{\alpha }e^{v_n}ds\right] , \end{aligned}$$

and $C(v,\varphi )=C(v,\varphi ,\rho )$ is the Pohozaev constant of $(v,\varphi )$, i.e.

$$\begin{aligned} C(v, \varphi )= & {} \rho \int _{ S^+_{\rho }} \left| \frac{\partial v}{\partial \nu }|^2-\frac{1}{2}|\nabla v\right| ^2d\sigma \\&-(1+\alpha )\left[ \int _{D^+_{\rho }}2V^2(0)|x|^{2\alpha }e^{2v}-V(0)|x|^{\alpha }e^{v}|\varphi |^2dv+ \int _{L_{\rho }}cV(0)|x|^{\alpha }e^{v}ds \right] \\&+\rho \int _{S^+_{\rho }}V^2(0)|x|^{2\alpha }e^{2v}d\sigma +cV(0)|s|^{\alpha }se^{v}|^{s=\rho }_{s=-\rho } \\&- \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \varphi }{\partial \nu },(x+\bar{x})\cdot \varphi \right\rangle d\sigma -\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \varphi , \frac{\partial \varphi }{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

On the other hand, we use the fact that $(v_n,\varphi _n)$ converges to $(v,\varphi )$ in $C^2_{loc}(\mathbb {R}^2_+)\cap C^1_{loc}(\overline{\mathbb {R}}^2_+\backslash \{0\})\times C^2_{loc}(\Gamma (\Sigma \mathbb {R}^2_+))\cap C^1_{loc}(\Gamma (\Sigma \overline{\mathbb {R}}^2_+\backslash \{0\}))$ again to get

$$\begin{aligned}&\int _{D^+_{\rho }}2V^2(r_nx)|x|^{2\alpha }e^{2v_n}-V(r_nx)|x|^{\alpha }e^{v_n}|\varphi _n|^2dv+\int _{L_{\rho }}cV(r_nx)|x|^{\alpha }e^{v_n}ds\\&\quad \rightarrow \int _{D^+_{\rho }}2V^2(0)|x|^{2\alpha }e^{2v}-V(0)|x|^{\alpha }e^{v}|\varphi |^2dv+\int _{L_{\rho }}cV(0)|x|^{\alpha }e^{v}ds+\beta \end{aligned}$$

as $n\rightarrow \infty$. By using Green’s representation formula for $u_n$ in $D^+_\rho$ and then take $n\rightarrow \infty$, we have

$$\begin{aligned} v(x)=\frac{\beta }{\pi }\ln \frac{1}{|x|}+\phi (x)+\gamma (x), \end{aligned}$$

where

$$\begin{aligned} \phi (x)= & {} \frac{1}{\pi }\int _{D_\rho ^+}\ln \frac{1}{|x-y|}(2V^2(0)|y|^{2\alpha }e^{2v(y)}-V(0)|y|^{\alpha }e^{v(y)}|\varphi |^2(y))dy\\&+\frac{1}{\pi }\int _{L_r}\ln \frac{1}{|x-y|}(cV(0)|y|^{\alpha }e^{v(y)})dy, \end{aligned}$$

and

$$\begin{aligned} \gamma (x)=\frac{1}{\pi }\int _{S^+_\rho }\ln \frac{1}{|x-y|}\frac{\partial v}{\partial \nu }+\frac{(x-y)\cdot \nu }{|x-y|^2}v(y)dy. \end{aligned}$$

It is clear that $\gamma (x)$ is in $C^1(\overline{D^+_\rho })$ and $\phi$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle \phi &{}=&{} 2V^2(0)|x|^{2\alpha }e^{2v}-V(0)|x|^{\alpha }e^{v}|\varphi |^2, &{} \quad \text { in }D_\rho ^+,\\ \frac{\partial \phi }{\partial \nu }&{}=&{} cV(0)|x|^{\alpha }e^{v}, &{}\quad \text { on } L_\rho .\\ \end{array} \right. \end{aligned}$$

By similar arguments as the proof of Proposition 4.5, we can obtain that

$$\begin{aligned} C(v,\varphi )=\frac{\beta ^2}{2\pi }, \end{aligned}$$

This implies that

$$\begin{aligned} (1+\alpha )\beta =\frac{\beta ^2}{2\pi }. \end{aligned}$$

Noticing that $\int _{D^+_\rho }|x|^{2\alpha }e^{2v}dx<\infty$, we have $\beta \le (1+\alpha )\pi$. Therefore we obtain that $\beta =0$, i.e. $C(v,\varphi )=0$, and the singularity at 0 of $(v,\varphi )$ is removed by Proposition 4.5. Furthermore, the singularity at $\infty$ of $(v,\varphi )$ is also removed by Theorem 5.2. Thus we get another bubble on $S^2_{c'}$, and we get a contradiction to the assumption that $m=1$. Concequently we complete the proof of the claim 1.

Claim 2 We can separate $A^+_{\delta , R, n}$ into finitely many parts

$$\begin{aligned} A^+_{\delta , R, n}=\bigcup _{k=1}^{N_k}A^+_k \end{aligned}$$

such that on each part

$$\begin{aligned} \int _{A^+_k}|x|^{2\alpha }e^{2u_n}\le \frac{1}{4\Lambda ^2},\quad k=1,2,\ldots ,N_k. \end{aligned}$$

where $N_k\le N_0$ for $N_0$ is a uniform integer for all n large enough, $A^+_k=D^+_{r^{k-1}}\setminus D^+_{r^k}$, $r^0=\delta , r^{N_k}=\lambda _nR$, $r^k<r^{k-1}$ for $k=1,2,\ldots , N_k$, and $C_0$ is a constant as in Lemma 6.1.

The proof of this claim is very similar to those in [23, 25, 41] and the argument is now standard, so we omit it.

Now we apply Claim 1 and Claim 2 to prove (52). Let $\varepsilon >0$ be small, and let $\delta$ be small enough, and let R and n be big enough. We apply Lemma 6.1 to each part $A_k^+$ to obtain

$$\begin{aligned} \left( \int _{A^+_l}|\Psi _n|^4\right) ^{\frac{1}{4}}\le & {} C_0 \left( \int _{D^+_{er^{l-1}}\setminus D^+_{e^{-1}r^l}}|x|^{2\alpha }e^{2u_n}\right) ^{\frac{1}{2}}\left( \int _{D^+_{er^{l-1}}\setminus D^+_{e^{-1}r^l}}|\Psi _n|^4\right) ^{\frac{1}{4}}\\&+C\left( \int _{D^+_{er^{l-1}}\setminus D^+_{r^{l-1}}}|\Psi _n|^4\right) ^{\frac{1}{4}}+C\left( \int _{D^+_{r^{l}}\setminus D^+_{e^{-1}r^l}}|\Psi _n|^4\right) ^{\frac{1}{4}}\\\le & {} C_0 \left( \left( \int _{A^+_l}|x|^{2\alpha }e^{2u_n}\right) ^{\frac{1}{2}}+\varepsilon ^{\frac{1}{2}}+\varepsilon ^{\frac{1}{2}}\right) \left( \left( \int _{A^+_l}|\Psi _n|^4\right) ^{\frac{1}{4}}+\varepsilon ^{\frac{1}{4}}+\varepsilon ^{\frac{1}{4}}\right) +C\varepsilon ^{\frac{1}{4}}\\\le & {} C_0 \left( \int _{A^+_l}|x|^{2\alpha }e^{2u_n}\right) ^{\frac{1}{2}}\left( \int _{A^+_l}|\Psi _n|^4\right) ^{\frac{1}{4}}+C\left( \varepsilon ^{\frac{1}{4}}+\varepsilon ^{\frac{1}{2}}+\varepsilon ^{\frac{3}{4}}\right) \\\le & {} \frac{1}{2} \left( \int _{A^+_l}|\Psi _n|^4\right) ^{\frac{1}{4}}+C\left( \varepsilon ^{\frac{1}{4}}+\varepsilon ^{\frac{1}{2}}+\varepsilon ^{\frac{3}{4}}\right) . \end{aligned}$$

Therefore we have

$$\begin{aligned} \left( \int _{A^+_l}|\Psi _n|^4\right) ^{\frac{1}{4}}\le C\left( \varepsilon ^{\frac{1}{4}}+\varepsilon ^{\frac{1}{2}}+\varepsilon ^{\frac{3}{4}}\right) . \end{aligned}$$

Since $\varepsilon$ is small, we may assume $\varepsilon \le 1$. Then we get

$$\begin{aligned} \left( \int _{A^+_l}|\Psi _n|^4\right) ^{\frac{1}{4}}\le C\varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(55)

With similar arguments, and using (55), we have

$$\begin{aligned} \left( \int _{A^+_l}|\nabla \psi _n|^{\frac{4}{3}}\right) ^{\frac{3}{4}}\le C\varepsilon ^{\frac{1}{4}}. \end{aligned}$$

(56)

Summing up (55) and (56) on $A^+_l$ we get

$$\begin{aligned} \int _{A^+_{\delta , R,n}}|\Psi _n|^4+\int _{A^+_{\delta ,R,n}}|\nabla \psi _n|^{\frac{4}{3}}=\sum _{l=1}^{N_0}\int _{A^+_l}|\Psi _n|^4+|\nabla \psi _n|^{\frac{4}{3}}\le C\varepsilon ^{\frac{1}{3}}. \end{aligned}$$

(57)

Thus we have shown (52) in the first case.

For Case II, according the blow-up process, we define the neck domain is

$$\begin{aligned} A^+_{S, R, n}=\{x\in \mathbb {R}^2_+|\rho _n R\le |x-x_n|\le |x_n|S \}. \end{aligned}$$

Notice that

$$\begin{aligned}&\int _{D^+_{\delta }}|\Psi _n|^4dv=\int _{D^+_{\frac{\delta }{|x_n|}}}|\overline{\Psi }_n|^4dv\\&\quad = \int _{D^+_{\frac{\delta }{|x_n|}}\backslash D^+_{R_1}(y_n)}|\overline{\Psi }_n|^4dv+\int _{D^+_{R_1}(y_n)\backslash D^+_{\delta _n R_2}(y_n) }|\overline{\Psi }_n|^4dv+\int _{D^+_{\delta _n R_2}(y_n)}|\overline{\Psi }_n|^4dv\\&\quad = \int _{D^+_{\frac{\delta }{|x_n|}}\backslash D^+_{R_1}(y_n)}|\overline{\Psi }_n|^4dv+\int _{D^+_{|x_n|R_1}(x_n)\backslash D^+_{|x_n|\delta _n R_2}(x_n) }|\Psi _n|^4dv+\int _{D^+_{\delta _n R_2}(y_n)}|\overline{\Psi }_n|^4dv. \end{aligned}$$

Duo to the assumption that $(u_n,\Psi _n)$ has only one bubble at the blow-up point $p=0$, $(\overline{u}_n,\overline{\Psi }_n)$ also has only one bubble at its blow-up point $y_0$. Therefore, we have

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{R_1\rightarrow \infty }\lim _{n\rightarrow \infty }\int _{D^+_{\frac{\delta }{|x_n|}}\backslash D^+_{R_1}(y_n)}|\overline{\Psi }_n|^4dv=0. \end{aligned}$$

While $D^+_{\delta _n R_2}(y_n)$ is a bubble domain, we know to prove (52) it is sufficient to prove that

$$\begin{aligned} \lim _{S\rightarrow \infty }\lim _{R\rightarrow \infty }\lim _{n\rightarrow \infty }\int _{A^+_{S, R, n}}|{\Psi }_n|^4dv=0. \end{aligned}$$

(58)

To prove (58), by using the similar argument as the case 1, we have the following facts:

Fact II.1 For any small $\varepsilon >0$, and $T>0$, there exists some N(T) such that for any $n\ge N(T)$ we have

$$\begin{aligned} \int _{D^+_{|x_n|S}(x_n) \setminus D^+_{|x_n|Se^{-T}}(x_n)}\left( |x|^{2\alpha }e^{2u_n}+|\Psi _n|^4\right) +\int _{\partial ( D^+_{|x_n|S}(x_n) \setminus D^+_{|x_n|Se^{-T}}(x_n))\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}<\varepsilon , \end{aligned}$$

for sufficiently large S.

Fact II.2 For any small $\varepsilon >0$, and $T>0$, we may choose an N(T) such that when $n\ge N(T)$

$$\begin{aligned}&\int _{D^+_{\rho _nRe^T}(x_n)\setminus D^+_{\rho _nR}(x_n)}\left( |x|^{2\alpha }e^{2u_n}+|\Psi _n|^4\right) +\int _{\partial (D^+_{\rho _nRe^T}(x_n)\setminus D^+_{\rho _nR}(x_n))\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}\\&\quad = \int _{(D_{Re^T}\setminus D_{R})\cap \{t>-\frac{t_n}{\rho _n}\}}\left( \left| \frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x\right| ^{2\alpha }e^{2\widetilde{u}_n}+|\widetilde{\Psi }_n|^4\right) \\&\qquad +\int _{(D_{Re^T}\setminus D_{R})\cap \{t=-\frac{t_n}{\rho _n}\}}\left| \frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x\right| ^{\alpha }e^{\widetilde{u}_n}\\&\quad < \varepsilon , \end{aligned}$$

if R is large enough.

Buy using the above two facts, we need to prove the following claim:

Claim II.1 For any $\varepsilon >0$, there is an $N>1$ such that for any $n\ge N$, we have

$$\begin{aligned}&\int _{D^+_r(x_n)\setminus D^+_{e^{-1}r}(x_n)}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^4)\\&\quad +\int _{\partial (D^+_r(x_n)\setminus D^+_{e^{-1}r}(x_n))\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}<\varepsilon ; \quad \forall r\in [e\rho _n R, |x_n|S]. \end{aligned}$$

Proof of Claim II.1 We assume by a contradiction that there exists $\varepsilon _0>0$ and a sequence ${r_n}$, $r_n\in [e\rho _n R, |x_n|S]$, such that

$$\begin{aligned} \int _{D^+_{r_n}(x_n)\setminus D^+_{e^{-1}r_n}(x_n)}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^{4})+\int _{\partial (D^+_{r_n}(x_n)\setminus D^+_{e^{-1}r_n}(x_n))\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n}\ge \varepsilon _0. \end{aligned}$$

Then, by Facts II.1 and II.2, we know that $\frac{|x_n|S}{r_n}\rightarrow +\infty$ and $\frac{\rho _n R}{r_n}\rightarrow 0$, in particular, $r_n\rightarrow 0$ as $n\rightarrow +\infty$. We assume that $\Lambda =\lim _{n\rightarrow \infty }\frac{t_n}{r_n}$. Here $\Lambda$ is either a nonnegative real number or $+\infty$. Next we proceed by distinguishing two cases.

Case II.1 $\Lambda >0$.

In this case, we note that $D_{r_n\rho }(x_n)$ is in $\mathbb {R}^2_+$ when n is sufficient small and $0<\rho <\Lambda$. We define the rescaling functions again

$$\begin{aligned} \left\{ \begin{array}{rcl} v_n(x)&{}=&{} u_n(r_nx+x_n)+\ln (r_n|x_n|^{\alpha }), \\ \varphi _n(x)&{}=&{} r_n^{\frac{1}{2}}\Psi (r_nx+x_n)\\ \end{array} \right. \end{aligned}$$

for any $r_nx+x_n\in D^+_{|x_n|S}(x_n)\setminus D^+_{\rho _n R}(x_n)$. Then $(v_n(x), \varphi _n(x))$ satisfies that

$$\begin{aligned}&\int _{(D_1\setminus D_{e^{-1}})\cap \{t>-\frac{t_n}{r_n}\}}\left( \left| \frac{x_n}{|x_n|}+\frac{r_n}{|x_n|}x\right| ^{2\alpha }e^{2v_n}+|\varphi _n|^4\right) \nonumber \\&\quad +\int _{(D_1\setminus D_{e^{-1}})\cap \left\{ t=-\frac{t_n}{r_n}\right\} }\left| \frac{x_n}{|x_n|}+\frac{r_n}{|x_n|}x\right| ^{\alpha }e^{v_n}\ge \varepsilon _0. \end{aligned}$$

(59)

Note that $(v_n,\varphi _n)$ satisfies for any $R>0$ and $S>0$

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle v_n(x) &{}=&{} 2V^2(r_nx+x_n)|\frac{x_n}{|x_n|}+\frac{r_n}{|x_n|}x|^{2\alpha }e^{2v_n(x)}\\ &{} &{}-V(r_nx+x_n)|\frac{x_n}{|x_n|}+\frac{r_n}{|x_n|}x|^{\alpha }e^{v_n(x)} |\varphi _n(x)|^2, &{} \text { in } \left( D_{\frac{|x_n|S}{r_n}}\setminus D_{\frac{\rho _nR}{r_n}}\right) \cap \left\{ t>-\frac{t_n}{r_n}\right\} , \\ \not {D}\varphi _n(x) &{}=&{} -V(r_nx+x_n)|\frac{x_n}{|x_n|}+\frac{r_n}{|x_n|}x|^{\alpha }e^{v_n(x)}\varphi _n(x), &{} \text { in } \left( D_{\frac{|x_n|S}{r_n}}\setminus D_{\frac{\rho _nR}{r_n}}\right) \cap \left\{ t>-\frac{t_n}{r_n}\right\} ,\\ \frac{\partial v_n(x)}{\partial n}&{}=&{} cV(r_nx+x_n)|\frac{x_n}{|x_n|}+\frac{r_n}{|x_n|}x|^{\alpha }e^{v_n(x)}, &{} \text { on } \left( D_{\frac{|x_n|S}{r_n}}\setminus D_{\frac{\rho _nR}{r_n}}\right) \cap \left\{ t=-\frac{t_n}{r_n}\right\} ,\\ B\varphi _n(x)&{}=&{}0, &{} \text { on } \left( D_{\frac{|x_n|S}{r_n}}\setminus D_{\frac{\rho _nR}{r_n}}\right) \cap \left\{ t=-\frac{t_n}{r_n}\right\} . \end{array} \right. \end{aligned}$$

According to Theorem 1.3, there are three possible cases. Similar to the Case I, we can rule out the first and the second possible cases. If the third case happens, then there is no blow-up point in $(D_R\setminus D_{\frac{1}{R}})\cap \{t\ge -b\}$ for any $R>0$ and any $b<\Lambda$. Furthermore $(v_n, \varphi _n)$ will converge to $(v,\varphi )$ strongly on $(D_R\setminus D_{\frac{1}{R}})\cap \{t\ge -b\}$. If $\Lambda >0$, then $(v,\varphi )$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle v &{}=&{} 2V^2(0)e^{2v}-V(0)e^{v}|\varphi |^2, &{} \quad \text { in }\mathbb {R}^2_\Lambda \setminus \{0\},\\ \not {D}\varphi &{}=&{} -V(0)e^{v}\varphi , &{} \quad \text { in }\mathbb {R}^2_\Lambda \setminus \{0\},\\ \frac{\partial v}{\partial n}&{}=&{} cV(0)e^{v}, &{}\quad \text { on } \partial \mathbb {R}^2_\Lambda (\text { in the case of } \Lambda< +\infty ), \\ B\varphi &{}=&{} 0, &{}\quad \text { on } \partial \mathbb {R}^2_\Lambda (\text { in the case of } \Lambda < +\infty ) \end{array} \right. \end{aligned}$$

(60)

with finite energy.

Since $D_{r_n\rho }(x_n)$ contains completely in $\mathbb {R}^2_+$ when n is sufficient small and $0<\rho <\Lambda$, we know that the origin is actually an interior singular point of $(v,\varphi )$ to (60). Then this local singular can be removed by using the similar arguments in the case II of [27]. After removing the local singularity 0, we can remove the singularity at $\infty$ of $(v,\varphi )$ to (60) by Theorem 5.2. Thus we get another bubble on $S^2_{c'}$, and we get a contradiction to the assumption that $m=1$. Concequently we complete the proof of the claim II.1.

Case II.2 $\Lambda =0$.

In this case, noticing that $x_n=(s_n,t_n)$ and $\lim _{n\rightarrow \infty }\frac{|x_n|}{r_n}=+\infty$, we have $\lim _{n\rightarrow \infty }\frac{|s_n|}{t_n}=+\infty$ and $\lim _{n\rightarrow \infty }\frac{|s_n|}{r_n}=+\infty$. We set $x'_n=(s_n, 0)$. Then we define the rescaling functions in this case

$$\begin{aligned} \left\{ \begin{array}{rcl} v_n(x)&{}=&{} u_n(r_nx+x'_n)+\ln (r_n|s_n|^{\alpha }), \\ \varphi _n(x)&{}=&{} r_n^{\frac{1}{2}}\Psi (r_nx+x'_n)\\ \end{array} \right. \end{aligned}$$

for any $r_nx+x'_n\in D^+_{|x_n|S}(x'_n)\setminus D^+_{\rho _n R}(x'_n)$. Since that

$$\begin{aligned}&\int _{D^+_{\frac{3}{2} r_n}(x'_n)\setminus D^+_{\frac{1}{2}e^{-1}r_n}(x'_n)}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^{4})+\int _{\partial (D^+_{\frac{1}{2} r_n}(x'_n)\setminus D^+_{\frac{3}{2} e^{-1}r_n}(x'_n))\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n} \\&\quad \ge \int _{D^+_{r_n}(x_n)\setminus D^+_{e^{-1}r_n}(x_n)}(|x|^{2\alpha }e^{2u_n}+|\Psi _n|^{4})+\int _{\partial (D^+_{r_n}(x_n)\setminus D^+_{e^{-1}r_n}(x_n))\cap \partial \mathbb {R}^2_+}|x|^{\alpha }e^{u_n} \\&\quad \ge \varepsilon _0, \end{aligned}$$

we have that $(v_n(x), \varphi _n(x))$ satisfies that

$$\begin{aligned}&\int _{D^+_{\frac{3}{2}}\setminus D^+_{\frac{ e^{-1}}{2}}}\left( \left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}+|\varphi _n|^4\right) \nonumber \\&+\int _{\partial \left( D^+_{\frac{3}{2}}\setminus D^+_{\frac{e^{-1}}{2}}\right) \cap \{t=0\}}\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}\ge \varepsilon _0. \end{aligned}$$

(61)

Note that $(v_n,\varphi _n)$ satisfies for any $R>0$ and $S>0$

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle v_n(x) &{}=&{} 2V^2(r_nx+x'_n)|\frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x|^{2\alpha }e^{2v_n(x)}\\ &{} &{}-V(r_nx+x'_n)|\frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x|^{\alpha }e^{v_n(x)} |\varphi _n(x)|^2, &{} \text { in } \left( D^+_{\frac{|x_n|S}{r_n}}\setminus D^+_{\frac{\rho _nR}{r_n}}\right) , \\ \not {D}\varphi _n(x) &{}=&{} -V(r_nx+x'_n)|\frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x|^{\alpha }e^{v_n(x)}\varphi _n(x), &{} \text { in } \left( D^+_{\frac{|x_n|S}{r_n}}\setminus D^+_{\frac{\rho _nR}{r_n}}\right) ,\\ \frac{\partial v_n(x)}{\partial n}&{}=&{} cV(r_nx+x'_n)|\frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x|^{\alpha }e^{v_n(x)}, &{} \text { on } \partial \left( D^+_{\frac{|x_n|S}{r_n}}\setminus D^+_{\frac{\rho _nR}{r_n}}\right) \cap \partial \mathbb {R}^2_+,\\ B\varphi _n(x)&{}=&{}0, &{} \text { on } \partial \left( D^+_{\frac{|x_n|S}{r_n}}\setminus D^+_{\frac{\rho _nR}{r_n}}\right) \cap \partial \mathbb {R}^2_+. \end{array} \right. \end{aligned}$$

According to Theorem 1.3, there are three possible cases. From (61), we can rule out the first and the second possible cases by using the similar arguments of Case I. Next we assume that the third case happens, i.e. there is no blow-up point in $\overline{D^+_R\setminus D^+_{\frac{1}{R}}}$ for any $R>0$. Furthermore $(v_n, \varphi _n)$ will converge to $(v,\varphi )$ strongly on $\overline{D^+_R\setminus D^+_{\frac{1}{R}}}$, and $(v,\varphi )$ satisfies

$$\begin{aligned} \left\{ \begin{array}{rcll} -\triangle v &{}=&{} 2V^2(0)e^{2v}-V(0)e^{v}|\varphi |^2, &{} \quad \text { in }\mathbb {R}^2_+,\\ \not {D}\varphi &{}=&{} -V(0)e^{v}\varphi , &{} \quad \text { in }\mathbb {R}^2_+,\\ \frac{\partial v}{\partial n}&{}=&{} cV(0)e^{v}, &{}\quad \text { on } \partial \mathbb {R}^2_+\setminus \{0\}, \\ B\varphi &{}=&{} 0, &{}\quad \text { on } \partial \mathbb {R}^2_+\setminus \{0\} \end{array} \right. \end{aligned}$$

(62)

with finite energy.

Next we will remove two singular points at 0 and at $\infty$, and consequently we get the second bubble of the considered system. Thus we get a contradiction. To this purpose, let us computate the Pohozaev constant of $(v, \varphi )$. Let start with the Pohozaev identity of $(u_n, \Phi _n)$. We multiply all terms in (4) by $(x-x'_n)\cdot \nabla u_n$ and integrate over $D^+_{r_n\rho }(x'_n)$. It follows for any sufficient small $\rho$ that

$$\begin{aligned}&r_n\rho \int _{S^+_{r_n\rho }(x'_n)} \left| \frac{\partial u_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u_n|^2d\sigma \\&\quad = \int _{D^+_{r_n\rho }(x'_n)}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dv+ \int _{L_{r_n\rho }(x'_n)}cV(x)|x|^{\alpha }e^{u_n}ds \\&\qquad -r_n\rho \int _{S^+_{r_n\rho }(x'_n)}V^2(x)|x|^{2\alpha }e^{2u_n}d\sigma +\int _{L_{r_n\rho }(x'_n)}c\frac{\partial (V(s,0)|s|^{\alpha })}{\partial s}(s-s_n)e^{u_n}ds\\&\qquad -cV(s,0)|s|^{\alpha }(s-s_n)e^{u_n}|^{s=x'_n+r_n\rho }_{s=x'_n-r_n\rho } \\&\qquad +\int _{D^+_{r_n\rho }(x'_n)}(x-x'_n)\cdot \nabla (V^2(x)|x|^{2\alpha })e^{2u_n}dv \\&\qquad -\int _{D^+_{r_n\rho }(x'_n)}(x-x'_n)\cdot \nabla (V(x)|x|^{\alpha })e^{u_n}|\Psi _n|^2dv \\&\qquad + \frac{1}{4}\int _{S^+_{r_n\rho }(x'_n)}\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x}-2x'_n)\cdot \Psi \right\rangle d\sigma +\frac{1}{4}\int _{S^+_{r_n\rho }(x'_n)}\left\langle (x+\bar{x}-2x'_n)\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma \end{aligned}$$

Hence for rescaling functions $(v_n, \varphi _n)$ we have

$$\begin{aligned}&\rho \int _{ S^+_{\rho }} \left| \frac{\partial v_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla v_n|^2d\sigma \nonumber \\&\quad = \int _{D^+_{\rho }}2V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}\nonumber \\&\qquad -V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}|\varphi _n|^2dv\nonumber \\&\qquad +\int _{L_{\rho }}cV(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}ds \nonumber \\&\qquad -\rho \int _{S^+_{\rho }}V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}d\sigma \nonumber \\&\qquad +\int _{ L_{\rho }}c\frac{\partial ( V((r_ns+s_n,0))\left| \frac{r_n}{|s_n|}s+\frac{s_n}{|s_n|}\right| ^{\alpha })}{\partial s}se^{v_n}ds\nonumber \\&\qquad -cV((r_ns+s_n,0))\left| \frac{r_n}{|s_n|}s+\frac{s_n}{|s_n|}\right| ^{\alpha }se^{v_n}|^{s=\rho }_{s=-\rho } \nonumber \\&\qquad +\int _{D^+_{\rho }}x\cdot \nabla \left( V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }\right) e^{2v_n}dv\nonumber \\&\qquad -\int _{D^+_{\rho }}x\cdot \nabla (V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha })e^{v_n}|\varphi _n|^2dv \nonumber \\&\qquad + \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \varphi _n}{\partial \nu },(x+\bar{x})\cdot \varphi _n\right\rangle d\sigma +\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \varphi _n, \frac{\partial \varphi _n}{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

(63)

Since the associated Pohozaev constant of $(v_n,\varphi _n)$ is

$$\begin{aligned}&C(v_n, \varphi _n) = C(v_n, \varphi _n,\rho )\\&\quad = \rho \int _{ S^+_{\rho }} \left| \frac{\partial v_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla v_n|^2d\sigma \\&\qquad -\int _{D^+_{\rho }}2V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}\\&\qquad -V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}|\varphi _n|^2dv\\&\qquad -\int _{L_{\rho }}cV(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}ds \\&\qquad +\rho \int _{S^+_{\rho }}V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}d\sigma \\&\qquad -\int _{ L_{\rho }}c\frac{\partial \left( V((r_ns+s_n,0))\left| \frac{r_n}{|s_n|}s+\frac{s_n}{|s_n|}\right| ^{\alpha }\right) }{\partial s}se^{v_n}ds\\&\qquad +cV((r_ns+s_n,0))\left| \frac{r_n}{|s_n|}s+\frac{s_n}{|s_n|}\right| ^{\alpha }se^{v_n}|^{s=\rho }_{s=-\rho } \\&\qquad -\int _{D^+_{\rho }}x\cdot \nabla \left( V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }\right) e^{2v_n}dv\\&\qquad +\int _{D^+_{\rho }}x\cdot \nabla \left( V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }\right) e^{v_n}|\varphi _n|^2dv \\&\qquad - \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \varphi _n}{\partial \nu },(x+\bar{x})\cdot \varphi _n\right\rangle d\sigma -\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \varphi _n, \frac{\partial \varphi _n}{\partial \nu }\right\rangle d\sigma , \end{aligned}$$

we have from (63) that

$$\begin{aligned} C(v_n, \varphi _n) = C(v_n, \varphi _n,\rho )=0. \end{aligned}$$

Since $|\frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x|^{2\alpha }$ is a smooth function in $\overline{D^+_\rho }$, by the energy condition,

$$\begin{aligned} \int _{D^+_\rho }\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}+|\varphi _n|^4dv+\int _{L_\rho }\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}ds<C, \end{aligned}$$

we can easily to check that

$$\begin{aligned}&\lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{\rho }}x\cdot \nabla \left( V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }\right) e^{2v_n}dv\\&\qquad +\int _{D^+_{\rho }}x\cdot \nabla \left( V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }\right) e^{v_n}|\varphi _n|^2dv =0, \end{aligned}$$

and

$$\begin{aligned} \lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty }\int _{ L_{\rho }}c\frac{\partial \left( V((r_ns+s_n,0))\left| \frac{r_n}{|s_n|}s+\frac{s_n}{|s_n|}\right| ^{\alpha }\right) }{\partial s}se^{v_n}ds=0. \end{aligned}$$

This implies that

$$\begin{aligned} 0= & {} \lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty } C(v_n, \varphi _n,\rho ) = \lim _{\rho \rightarrow 0}C(v,\varphi ,\rho )\\&-\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_{r}}2V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}\\&-V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}|\varphi _n|^2dv\\&-\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{L_{r}}cV(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}ds\\= & {} C(v,\varphi )-\beta . \end{aligned}$$

Here

$$\begin{aligned} \beta= & {} \lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\left[ \int _{D^+_{r}}2V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}\right. \\&\qquad \qquad \left. -V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}|\varphi _n|^2dv\right] \\&+\lim _{r\rightarrow 0}\lim _{n\rightarrow \infty }\int _{L_{r}}cV(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}ds. \end{aligned}$$

and $C(v,\varphi )=C(v,\varphi ,\rho )$ is the Pohozaev constant of $(v,\varphi )$, i.e.

$$\begin{aligned} C(v, \varphi )= & {} \rho \int _{ S^+_{\rho }} \left| \frac{\partial v}{\partial \nu }|^2-\frac{1}{2}|\nabla v\right| ^2d\sigma \\&-\int _{D^+_{\rho }}2V^2(0)e^{2v}-V(0)e^{v}|\varphi |^2dv+ \int _{L_{\rho }}cV(0)e^{v}ds \\&+\rho \int _{S^+_{\rho }}V^2(0)e^{2v}d\sigma +cV((0,0))se^{v}|^{s=\rho }_{s=-\rho } \\&- \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \varphi }{\partial \nu },(x+\bar{x})\cdot \varphi \right\rangle d\sigma -\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \varphi , \frac{\partial \varphi }{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

On the other hand, we use the fact that $(v_n,\varphi _n)$ converges to $(v,\varphi )$ in $C^2_{loc}(\mathbb {R}^2_+)\cap C^1_{loc}(\overline{\mathbb {R}}^2_+\backslash \{0\})\times C^2_{loc}(\Gamma (\Sigma \mathbb {R}^2_+))\cap C^1_{loc}(\Gamma (\Sigma \overline{\mathbb {R}}^2_+\backslash \{0\}))$ again to get

$$\begin{aligned}&\int _{D^+_{r}}2V^2(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{2\alpha }e^{2v_n}-V(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}|\varphi _n|^2dv\\&\qquad +\int _{L_{r}}cV(r_nx+x'_n)\left| \frac{x'_n}{|s_n|}+\frac{r_n}{|s_n|}x\right| ^{\alpha }e^{v_n}ds\\&\quad \rightarrow \int _{D^+_{\rho }}2V^2(0)e^{2v}-V(0)e^{v}|\varphi |^2dv+\int _{L_{\rho }}cV(0)e^{v}ds+\beta \end{aligned}$$

as $n\rightarrow \infty$. By using Green’s representation formula for $u_n$ in $D^+_\rho$ and then take $n\rightarrow \infty$, we have

$$\begin{aligned} v(x)=\frac{\beta }{\pi }\ln \frac{1}{|x|}+\phi (x)+\gamma (x), \end{aligned}$$

with $\phi$ being a bounded term and $\gamma (x)$ being a regular term. Consequently, we can obtain that

$$\begin{aligned} C(v,\varphi )=\frac{\beta ^2}{2\pi }. \end{aligned}$$

This implies that

$$\begin{aligned} \beta =\frac{\beta ^2}{2\pi }. \end{aligned}$$

Noticing that $\int _{D^+_\rho }e^{2v}dx<\infty$, we have $\beta \le \pi$. Therefore we obtain that $\beta =0$, i.e. $C(v,\varphi )=0$, and the singularity at 0 of $(v,\varphi )$ is removed by Propostion 4.5. Furthermore, the singularity at $\infty$ of $(v,\varphi )$ is also removed by Theorem 5.2. Thus we get another bubble on $S^2_{c'}$, and we get a contradiction to the assumption that $m=1$. Consequently we complete the proof of Claim II.1.

Next , similarly to Case I. we can prove the following:

Claim II.2 We can separate $A^+_{S, R, n}(x_n)$ into finitely many parts

$$\begin{aligned} A^+_{S, R, n}=\bigcup _{k=1}^{N_k}A^+_k \end{aligned}$$

such that on each part

$$\begin{aligned} \int _{A^+_k}|x|^{2\alpha }e^{2u_n}\le \frac{1}{4\Lambda ^2},\quad k=1,2,\ldots ,N_k. \end{aligned}$$

where $N_k\le N_0$ for $N_0$ is a uniform integer for all n large enough, $A^+_k=D^+_{r^{k-1}}(x_n)\setminus D^+_{r^k}(x_n)$, $r^0=\delta , r^{N_k}=\lambda _nR$, $r^k<r^{k-1}$ for $k=1,2,\ldots , N_k$, and $C_0$ is a constant as in Lemma 6.1.

Then, by using Claim II.1 and Claim II.2 we can complete the proof of the result. $\square$

7 Blow-up behavior

In this section, we will show that $u_n\rightarrow -\infty$ uniformly on compact subset of $(D^+_r\cup L_r)\setminus \Sigma _1$ in means of the energy identity for spinors. Thus we rule out the possibility that $u_n$ is uniformly bounded in $L^{\infty }_{loc}((D^+_r\cup L_r)\setminus \Sigma _1)$ in Theorem 1.3. The following is the proof of Theorem 1.5.

Proof of Theorem 1.5:

We prove the results by contradiction. Assume that the conclusion of the theorem is false. Then by Theorem 1.3, $u_n$ is uniformly bounded in $L_{loc}^{\infty }((D^+_r\cup L_r)\setminus \Sigma _1)$. Thus we know that $(u_n, \Psi _n)$ converges in $C^2$ on any compact subset of $(D^+_r\cup L_r)\setminus \Sigma _1$ to $(u,\Psi )$, which satisfies that

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta u(x) &{}=&{} 2u^2(x)|x|^{2\alpha }e^{2u(x)}-V(x)|x|^{\alpha }e^{u(x)}|\Psi |^2, \quad &{}\text {in } D^+_{r}\setminus \Sigma _1, \\ \not {D}\Psi &{}=&{} -V(x)|x|^{\alpha }e^{u(x)}\Psi , \quad &{}\text {in } D^+_{r}\setminus \Sigma _1, \\ \frac{\partial u}{\partial n}&{} = &{} cV(x)|x|^{\alpha }e^{u(x)},\quad &{} \text { on } L_r\setminus \Sigma _1, \\ B(\Psi ) &{}=&{} 0,\quad &{} \text { on } L_r\setminus \Sigma _1. \end{array} \right. \end{aligned}$$

(64)

with bounded energy

$$\begin{aligned} \int _{D^+_r}(|x|^{2\alpha }e^{2u}+|\Psi |^4)dx +\int _{L_r}|x|^\alpha e^u ds< +\infty . \end{aligned}$$

Since the blow-up set $\Sigma _1$ is not empty, we can take a point $p \in \Sigma _1$. Choose a small $\delta _0>0$ such that p is the only point of $\Sigma _1$ in $\overline{D}_{2\delta _0} (p)\cap (D^+_r\cup L_r)=\{p\}$. If p is the interior point of $D^+_r$, then we can choose $\delta _0$ sufficiently small such that $D_{2\delta _0}(p)\subset (D^+_r\cup L_r)$. Hence by Theorem 1.3 in [27] we can get a contradiction.

Next we assume that p is on $L_r$. Without loss of generality, we assume that $p=0$. The case of $p\ne 0$ can be dealed with in an analogous way.

We shall first show that the limit $(u,\Psi )$ is regular at the isolated singularity $p=0$, i.e. $u\in C^{2}(D^+_r)\cap C^{1}(D^+_r\cup L_r)$ and $\Psi \in C^{2}(\Gamma (\Sigma D^+_r ))\cap C^{1}(\Gamma (\Sigma (D^+_r\cup L_r)))$ for some small $r>0$. To this end, we shall using Theorem 4.5 for removability of a local singularity to remove the singularity. We know that the Phohozaev constant, denote $C_B(u, \Psi )$, of $(u, \Psi )$ at $p=0$ is

$$\begin{aligned} C_B(u,\Psi )&:= C_B(u,\Psi ,\rho ) = \rho \int _{S^+_\rho } \left| \frac{\partial u}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u|^2d\sigma \\&\quad -(1+\alpha )\int _{D^+_\rho }(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^u|\Psi |^2)dv\\&\quad - (\alpha +1)\int _{L_\rho }cV(x)|x|^{\alpha }e^{u}ds \\&\quad +\rho \int _{S^+_\rho }V^2(x)|x|^{2\alpha }e^{2u}d\sigma -\int _{L_\rho }c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^uds\\&\quad +cV(s,0)|s|^{\alpha }se^u|^{s=\rho }_{s=-\rho } \\&\quad -\int _{D^+_\rho }x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u}dv +\int _{D^+_\rho }x\cdot \nabla V(x)|x|^{\alpha }e^u|\psi |^2dv \\&\quad - \frac{1}{4}\int _{S^+_\rho }\left\langle \frac{\partial \Psi }{\partial \nu },(x+\bar{x})\cdot \Psi \right\rangle d\sigma -\frac{1}{4}\int _{S^+_\rho }\left\langle (x+\bar{x})\cdot \Psi , \frac{\partial \Psi }{\partial \nu }\right\rangle d\sigma \end{aligned}$$

for any $0<\rho <\delta _0$. On the other hand, since $(u_n,\Psi _n)$ are the regular solution, the Pohozaev constant $C_B(u_n, \Psi _n)=C_B(u_n, \Psi _n,\rho )$ satisfies

$$\begin{aligned} 0 = C(u_n, \Psi _n)= & {} C(u_n, \Psi _n,\rho )\\= & {} \rho \int _{ S^+_{\rho }} \left| \frac{\partial u_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla u_n|^2d\sigma \\&- (1+\alpha )\int _{D^+_{\rho }}2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2dv\\&- (\alpha +1)\int _{L_{\rho }}cV(x)|x|^{\alpha }e^{u_n}ds \\&+\rho \int _{S^+_{\rho }}V^2(x)|x|^{2\alpha }e^{2u_n}d\sigma -\int _{ L_{\rho }}c\frac{\partial V(s,0)}{\partial s}|s|^{\alpha }se^{u_n}ds\\&+cV((s,0))|s|^{\alpha }se^{u_n}|^{s=\rho }_{s=-\rho } \\&-\int _{D^+_{\rho }}x\cdot \nabla (V^2(x))|x|^{2\alpha }e^{2u_n}dv +\int _{D^+_{\rho }}x\cdot \nabla V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2dv \\&- \frac{1}{4}\int _{S^+_{\rho }}\left\langle \frac{\partial \Psi _n}{\partial \nu },(x+\bar{x})\cdot \Psi _n\right\rangle d\sigma -\frac{1}{4}\int _{ S^+_{\rho }}\left\langle (x+\bar{x})\cdot \Psi _n, \frac{\partial \Psi _n}{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

Let $n\rightarrow \infty$ and $\rho \rightarrow 0$, by using that $(u_n,\Psi _n)$ converges to $(u,\Psi )$ regularly on any compact subset of $\overline{D}^+_{2\delta _0} \setminus \{0\}$ and that the energy condition (17), to get

$$\begin{aligned} 0= & {} \lim _{\rho \rightarrow 0}\lim _{n\rightarrow \infty }C(u_n,\Psi _n, \rho )=\lim _{\rho \rightarrow 0}C(u,\Psi , \rho ) \\&-(1+\alpha )\lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\left\{ \int _{D^+_\delta }(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx\right. \\&\qquad \qquad \qquad \left. +\int _{L_{\delta }}cV(x)|x|^{\alpha }e^{u_n}ds\right\} \\= & {} C(u,\Psi )-(1+\alpha )\beta , \end{aligned}$$

where

$$\begin{aligned} \beta =\lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\left\{ \int _{D^+_\delta }(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx+\int _{L_{\delta }}cV(x)|x|^{\alpha }e^{u_n}ds\right\} . \end{aligned}$$

Moreover, we can also assume that

$$\begin{aligned}&(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2 )dx+ cV(x)|x|^{\alpha }e^{u_n}ds\\&\quad \rightarrow \nu =(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^{u}|\Psi |^2)dx+ cV(x)|x|^{\alpha }e^{u}ds+\beta \delta _{p=0} \end{aligned}$$

in the sense of distributions in $D^+_\delta \cup L_\delta$ for any small $\delta >0$. Then, applying similar arguments as in the proof of the local singularity removability in Claim I.1, Theorem 1.4, we can show that $C_B(u,\Psi )=0$, $\beta = 0$ and hence $(u, \Psi )$ is a regular solution of (4) on $D^+_{2\delta _0}$ with bounded energy

$$\begin{aligned} \int _{D^+_{2\delta _0}}(|x|^{2\alpha } e^{2u}+|\Psi |^4)dx+ \int _{L_{2\delta }}|x|^{\alpha }e^{u}ds< +\infty . \end{aligned}$$

Hence, we can choose some small $\delta _1 \in (0, \delta _0)$ such that for any $\delta \in (0, \delta _1)$,

$$\begin{aligned} \int _{D^+_{\delta }}(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^{u}|\Psi |^2)dx +\int _{L_\delta } cV(x)|x|^{\alpha }e^{u}ds< \mathrm{min} \left\{ \frac{1+\alpha }{10}, \frac{1}{10}\right\} . \end{aligned}$$

(65)

Next, as in the proof of Theorem 1.4, we rescale $(u_n,\Psi _n)$ near $p=0$. We let $x_n\in \overline{D}^+_{\delta _1}$ such that $u_n(x_n)=\max _{\overline{D}^+_{\delta _1}}u_n(x)$. Write $x_n=(s_n,t_n)$. Then $x_n\rightarrow p=0$ and $u_n(x_n)\rightarrow +\infty$. Define $\lambda _n=e^{-\frac{u_n(x_n)}{\alpha +1}}$. It is clear that $\lambda _n$, $|x_n|$ and $t_n$ converge to 0 as $n\rightarrow 0$. we will proceed by distinguishing the following three cases:

Case I. $\frac{|x_n|}{\lambda _n}=O(1)$ as $n\rightarrow +\infty$.

In this case, the rescaling functions are

$$\begin{aligned} \left\{ \begin{array}{rcl} \widetilde{u}_n(x)&{}=&{}u_n(\lambda _nx)+(1+\alpha )\ln {\lambda _n}\\ \widetilde{\Psi }_n(x)&{}=&{}\lambda ^{\frac{1}{2}}_n\Psi _n(\lambda _nx) \end{array} \right. \end{aligned}$$

for any $x \in \overline{D}^+_{\frac{\delta _1}{2\lambda _n}}$. Moreover, by passing to a subsequence, $(\widetilde{u}_n,\widetilde{\Psi }_n)$ converges in $C^{2}_{loc}(\mathbb {R}^2_+)\cap C^{1}_{loc}(\bar{\mathbb {R}}^2_+)\times C^{2}_{loc}(\Gamma (\Sigma {\mathbb {R}}^2_+))\cap C^{1}_{loc}(\Gamma (\Sigma {\bar{\mathbb {R}}}^2_+))$ to some $(\widetilde{u},\widetilde{\Psi })$ satisfying

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta \widetilde{u} &{}=&{} 2V^2(0)|x|^{2\alpha }e^{2\widetilde{u}}-V(0)|x|^{\alpha }e^{\widetilde{u}}| \widetilde{\Psi }|^2, &{}\quad \text { in } \mathbb {R}^2_+,\\ \not {D}\widetilde{\Psi } &{}=&{} -V(0)|x|^{\alpha }e^{\widetilde{u}}\widetilde{\Psi }, &{}\quad \text { in } \mathbb {R}^2_+,\\ \frac{\partial \widetilde{u}}{\partial n}&{}=&{}cV(0)|x|^\alpha e^{\widetilde{u}}, &{} \quad \text { on } \partial \mathbb {R}^2_+,\\ B\widetilde{\Psi }&{}=&{} 0, &{} \quad \text { on } \partial \mathbb {R}^2_+ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \int _{\mathbb {R}^2_+}(2V^2(0)|x|^{2\alpha }e^{2\widetilde{u}}-V(0)|x|^{\alpha }e^{\widetilde{u}}|\widetilde{\Psi }|^2)dx+\int _{\partial \mathbb {R}^2_+} cV(0)|x|^\alpha e^{\widetilde{u}}d\sigma =2\pi (1+\alpha ). \end{aligned}$$

Then for $\delta \in (0, \delta _1)$ small enough, $R>0$ large enough and n large enough, we have

$$\begin{aligned}&\int _{D^+_{\delta }}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx+\int _{L_{\delta }}cV(x)|x|^{\alpha }e^{u_n}ds\nonumber \\&\quad =\int _{D_{\lambda _n R}}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx+\int _{L_{\lambda _n R}}cV(x)|x|^{\alpha }e^{u_n}ds\nonumber \\&\qquad +\int _{D^+_\delta \backslash D^+_{\lambda _n R}}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx+\int _{L_{\delta }\setminus L_{\lambda _n R}}cV(x)|x|^{\alpha }e^{u_n}ds\nonumber \\&\quad \ge \int _{D^+_{R}}(2V^2(\lambda _nx)|x|^{2\alpha }e^{2\widetilde{u}_n}-V(\lambda _nx)|x|^{\alpha }e^{\widetilde{u}_n}|\widetilde{\Psi }_n|^2)+\int _{L_{R}}cV(\lambda _n x)|x|^{\alpha }e^{\widetilde{u}_n}ds\nonumber \\&\qquad -\int _{D^+_\delta \backslash D^+_{t_nR}}V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2 \nonumber \\&\quad \ge 2\pi (1+\alpha ) - \frac{1+\alpha }{10}. \end{aligned}$$

(66)

Here in the last step, we use the fact from Theorem 1.4 that the neck energy of the spinor field $\Psi _n$ is converging to zero. We remark that in the above estimate, if there are multiple bubbles then we need to decompose $D^+_\delta \backslash D^+_{\lambda _nR}$ further into bubble domains and neck domains and then apply the no neck energy result in Theorem 1.4 to each of these neck domains.

On the other hand, we fix some $\delta \in (0, \delta _1)$ small such that (66) holds and then let $n\rightarrow \infty$ to conclude that

$$\begin{aligned} 2\pi (1+\alpha ) - \frac{1+\alpha }{10}\le & {} \int _{D^+_{\delta }}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx\\&+\int _{L_{\delta }}cV(x)|x|^{\alpha }e^{u_n}ds\\= & {} -\int _{D^+_{\delta }}\Delta u_n= -\int _{\partial B_{\delta }}\frac{\partial u_n}{\partial n}\\\rightarrow & {} -\int _{\partial D^+_{\delta }}\frac{\partial u}{\partial n} = -\int _{D^+_{\delta }}\Delta u\\= & {} \int _{D^+_{\delta }}(2V^2(x)|x|^{2\alpha }e^{2u}-V(x)|x|^{\alpha }e^{u}|\Psi |^2)dx\\&+\int _{L_{\delta }}cV(x)|x|^{\alpha }e^{u}ds < \frac{1+\alpha }{10} \end{aligned}$$

Here in the last step, we have used (65). Thus we get a contradiction and finish the proof of the Theorem in this case.

Case II. $\frac{|x_n|}{\lambda _n}\rightarrow +\infty$ as $n\rightarrow +\infty$.

In this case, as in the arguments in Theorem 1.4, we can rescale twice to get the bubble. First, we define the rescaling functions

$$\begin{aligned} \left\{ \begin{array}{rcl} \bar{u}_n(x)&{}=&{}u_n(|x_n|x)+(\alpha +1)\ln {|x_n|}\\ \bar{\Psi }_n(x)&{}=&{}|x_n|^{\frac{1}{2}}\Psi _n(|x_n|x) \end{array} \right. \end{aligned}$$

for any $x\in \overline{D^+}_{\frac{\delta _1}{2|x_n|}}$. Set $y_n:=\frac{x_n}{|x_n|}$. Due to $\bar{u}_n(y_n)\rightarrow +\infty$, we set that $\delta _n=e^{-\bar{u}_n(y_n)}$ and define the rescaling function

$$\begin{aligned} \left\{ \begin{array}{rcl} \widetilde{u}_n(x)&{}=&{}\bar{u}_n(\delta _nx+y_n)+\ln {\delta _n}\\ \widetilde{\Psi }_n(x)&{}=&{}\delta _n^{\frac{1}{2}}\bar{\Psi }_n(\delta _nx+y_n) \end{array} \right. \end{aligned}$$

for any $\delta _nx+y_n\in \overline{D^+}_{\frac{\delta _1}{2|x_n|}}$. Denote that $\rho _n=\frac{e^{-u_n(x_n)}}{|x_n|^\alpha }=\lambda _n(\frac{\lambda _n}{|x_n|})^\alpha$ and $x_n=(s_n, t_n)$.

Case II.1 $\frac{t_n}{\rho _n}\rightarrow +\infty$ as $n\rightarrow \infty$.

Then, by passing to a subsequence, $(\widetilde{u}_n,\widetilde{\Psi }_n)$ converges in $C^{2}_{loc}(\mathbb {R}^2)\times C^{2}_{loc}(\Gamma (\Sigma {\mathbb {R}}^2))$ to some $(\widetilde{u},\widetilde{\Psi })$ satisfying

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta \widetilde{u} &{}=&{} 2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}| \widetilde{\Psi }|^2, &{}\quad \text { in } \mathbb {R}^2,\\ \not {D}\widetilde{\Psi } &{}=&{} -V(0)e^{\widetilde{u}}\widetilde{\Psi }, &{}\quad \text { in } \mathbb {R}^2,\\ \end{array} \right. \end{aligned}$$

with the bubble energy

$$\begin{aligned} \int _{\mathbb {R}^2}(2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}|\widetilde{\Psi }|^2)dx=4\pi . \end{aligned}$$

Therefore, for $\delta \in (0, \delta _1)$ small enough, $S, R>0$ large enough and n large enough, the fact that the neck energy of the spinor field $\Psi _n$ is converging to zero, we have

$$\begin{aligned}&\int _{D^+_{\delta }}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx+\int _{L_{\delta }}cV(x)|x|^{\alpha }e^{u_n}ds\\&\quad =\int _{D^+_{\frac{\delta }{|x_n|}}}(2V^2(|x_n|x)|x|^{2\alpha }e^{2\bar{u}_n}-V(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}|\bar{\Psi }_n|^2)dx\\&\qquad +\int _{L_{\frac{\delta }{|x_n|}}}cV(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}ds \\&\quad =\int _{D^+_{\frac{\delta }{|x_n|}}\setminus D^+_S(y_n)}(2V^2(|x_n|x)|x|^{2\alpha }e^{2\bar{u}_n}-V(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}|\bar{\Psi }_n|^2)dx \\&\qquad +\int _{ D^+_S(y_n)\setminus D^+_{\frac{\rho _n}{|x_n|}R}(y_n)}(2V^2(|x_n|x)|x|^{2\alpha }e^{2\bar{u}_n}-V(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}|\bar{\Psi }_n|^2)dx \\&\qquad +\int _{D^+_{\frac{\rho _n}{|x_n|}R}(y_n)}(2V^2(|x_n|x)|x|^{2\alpha }e^{2\bar{u}_n}-V(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}|\bar{\Psi }_n|^2)dx\\&\qquad +\int _{L_{\frac{\rho _n R}{|x_n|}}(y_n)}cV(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}ds\\&\qquad +\int _{L_S(y_n)\setminus L_{\frac{\rho _n R}{|x_n|}}(y_n)}cV(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}ds+\int _{L_{\frac{\delta }{|x_n|}}\setminus L_{S}(y_n)}cV(|x_n|x)|x|^{\alpha }e^{\bar{u}_n}ds\\&\quad \ge \int _{D_R \cap \left\{ t>-\frac{t_n}{\rho _n}\right\} }\left( 2V^2(x_n+\rho _nx)\left| \frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x\right| ^{2\alpha }e^{2\widetilde{u}_n(x)}\right. \\&\qquad \qquad \qquad \qquad \qquad \left. -V(x_n+\rho _nx)\left| \frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x|^{\alpha } e^{\widetilde{u}_n(x)}\right| \widetilde{\Psi }_n|^2\right) dx \\&\qquad + \int _{D_R \cap \left\{ t=-\frac{t_n}{\rho _n}\right\} }(cV(x_n+\rho _nx)\left| \frac{x_n}{|x_n|}+\frac{\rho _n}{|x_n|}x\right| ^{\alpha }e^{\widetilde{u}_n(x)}\\&\qquad -\int _{D^+_{|x_n|S}(x_n)\backslash D^+_{\rho _n R}(x_n)}V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2 - \int _{D^+_{\frac{\delta }{|x_n|}}\setminus D^+_S(y_n)}V(t_nx)|x|^{\alpha }e^{\bar{u}_n}|\bar{\Psi }_n|^2 \\&\quad \ge 4\pi - \frac{1}{10}. \end{aligned}$$

Then, applying similar arguments as in Case I, we get a contradiction, and finish the proof of the Theorem in this case.

Case II.2 $\frac{t_n}{\rho _n}\rightarrow \Lambda$ as $n\rightarrow \infty$.

Then, by passing to a subsequence, $(\widetilde{u}_n,\widetilde{\Psi }_n)$ converges in $C^{2}_{loc}(\mathbb {R}^2_{-\Lambda })\cap C^{1}_{loc}(\bar{\mathbb {R}}^2_{-\Lambda })\times C^{2}_{loc}(\Gamma (\Sigma {\mathbb {R}}^2_{-\Lambda }))\cap C^{1}_{loc}(\Sigma \bar{\mathbb {R}}^2_{-\Lambda })$ to some $(\widetilde{u},\widetilde{\Psi })$ satisfying

$$\begin{aligned} \left\{ \begin{array}{rcll} -\Delta \widetilde{u} &{}=&{} 2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}| \widetilde{\Psi }|^2, &{}\quad \text { in } \mathbb {R}^2_{-\Lambda },\\ \not {D}\widetilde{\Psi } &{}=&{} -V(0)e^{\widetilde{u}}\widetilde{\Psi }, &{}\quad \text { in } \mathbb {R}^2_{-\Lambda },\\ \frac{\partial \widetilde{u}}{\partial n}&{}=&{}cV(0) e^{\widetilde{u}}, &{}\quad \text { on } \partial \mathbb {R}^2_{-\Lambda } ,\\ B\widetilde{\Psi } &{}=&{} 0, &{} \quad \text { on } \partial \mathbb {R}^2_{-\Lambda }, \end{array} \right. \end{aligned}$$

with the bubble energy

$$\begin{aligned} \int _{\mathbb {R}^2_{-\Lambda }}(2V^2(0)e^{2\widetilde{u}}-V(0)e^{\widetilde{u}}|\widetilde{\Psi }|^2)dx+\int _{\partial \mathbb {R}^2_{-\Lambda }}cV(0)e^{\widetilde{u}}d\sigma =2\pi . \end{aligned}$$

Then, applying similar arguments as in Case II.1, we can get a contradiction, and finish the proof of the Theorem. $\square$

8 Blow-up Value

By means of Theorem 1.5, we can further compute the blow-up value at the blow-up point p, which is defined as

$$\begin{aligned} m(p)= & {} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\left\{ \int _{D^+_\delta (p)}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx\right. \\&\qquad \qquad \qquad \left. +\int _{L_{\delta }(p)}cV(x)|x|^{\alpha }e^{u_n}ds\right\} . \end{aligned}$$

We know from Theorem 1.5 that $m(p)> 0$. Now we shall determine the precise value of m(p) under a boundary condition.

Proof of Theorem 1.6:

When $p\notin L_{\delta _0}(p)$, It is clear that we can choose $\delta _0$ sufficiently small such that $\overline{D_{\delta _0}^+(p)}=\overline{D_{\delta _0}(p)}$. Then we have $m(p)=4\pi$ according to the arguments in [27]. Next we assume that $p\in L_{\delta _0}(p)$. Without loss of generality, we assume $p=0$. The case of $p\ne 0$ can be handled analogously.

By using the boundary condition, it follows that

$$\begin{aligned} 0\le u_n-\min _{ S^+_{\delta _0}}u_n\le C \end{aligned}$$

on $S^+_{\delta _0}$. Let $w_n$ be the unique solution of the following problem

$$\begin{aligned} \left\{ \begin{array}{rlll} -\Delta w_n &{}=&{} 0, &{}\text {in } D^+_{\delta _0},\\ \frac{\partial w_n}{\partial n} &{}=&{} 0, &{}\text {on }L_{\delta _0},\\ w_n &{}=&{} u_n-\min _{S^+_{\delta _0}}u_n, &{}\text {on }S^+_{\delta _0}. \end{array} \right. \end{aligned}$$

It follows from the maximum principle and the Hopf Lemma that $w_n$ is uniformly bounded in $\overline{D^+_{\delta _0}}$, and consequently $w_n$ is $C^2(D^+_{\delta _0})\cap C^1(D^+_{\delta _0}\cup L_{\delta _0})$. Now we set $v_n=u_n-\min _{S^+_{\delta _0}}u_n-w_n$. Then $v_n$ satisfies that

$$\begin{aligned} \left\{ \begin{array}{lcl} -\Delta v_n=2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2, &{}\text {in } D^+_{\delta _0},\\ \frac{\partial v_n}{\partial n}=cV(x)|x|^{\alpha }e^{u_n}, &{}\text {on }L_{\delta _0},\\ v_n=0, &{}\text {on }S^+_{\delta _0}, \end{array} \right. \end{aligned}$$

with the energy condition

$$\begin{aligned} \int _{D^+_{\delta _0}}(2V^2(x)|x|^{2\alpha }e^{2u_n}-V(x)|x|^{\alpha }e^{u_n}|\Psi _n|^2)dx+\int _{L_{\delta _0}}cV(x)|x|^{\alpha }e^{u_n}ds\le C. \end{aligned}$$

(67)

By Green’s representation formula, we have

$$\begin{aligned} v_n(x)= & {} \frac{1}{\pi }\int _{D^+_{\delta _0}}\log \frac{1}{|x-y|}(2V^2(y)|y|^{2\alpha }e^{2u_n}-V(y)|y|^{\alpha }e^{u_n}|\Psi _n|^2)dy \nonumber \\&+\frac{1}{\pi }\int _{L_{\delta _0}}\log \frac{1}{|x-y|}cV(y)|y|^{\alpha }e^{u_n}dy+R_n(x) \end{aligned}$$

(68)

where $R_n(x) \in C^1 (D^+_{\delta _0}\cup L_{\delta _0})$ is a regular term. By using Theorem 1.5, we know

$$\begin{aligned} v_n(x)\rightarrow \frac{m(p)}{\pi }\ln \frac{1}{|x|}+R(x), \text { in } C^{1}_{loc}((D^+_{\delta _0}\cup L_{\delta _0})\setminus \{0\}) \end{aligned}$$

(69)

for $R(x)\in C^1 (D^+_{\delta _0}\cup L_{\delta _0})$. On the other hand, we observe that $(v_n,\Psi _n)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{lcll} -\Delta v_n &{}=&{} 2K_n^2(x)|x|^{2\alpha }e^{2v_n}-K_n(x)|x|^{\alpha }e^{v_n}|\Psi _n|^2, &{}\text {in } D^+_{\delta _0},\\ \not {D}{\Psi _n} &{}=&{}-K_n(x)e^{v_n}{\Psi _n}, &{}\text {in } D^+_{\delta _0},\\ \frac{\partial v_n}{\partial n}&{}=&{}cK_n(x)|x|^{\alpha }e^{u_n}, &{}\text {on }L_{\delta _0},\\ B(\Psi _n)&{}=&{}0, &{}\text {on }L_{\delta _0}, \end{array} \right. \end{aligned}$$

where $K_n=V(x)e^{\min _{S^+_{\delta _0}}u_n+w_n}$. Noticing the Pohozaev identity of $(v_n,\Psi _n)$ in $D^+_{\delta _0}$ for $0<\delta <\delta _0$ is

$$\begin{aligned}&\delta \int _{S^+_\delta } \left| \frac{\partial v_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla v_n|^2d\sigma \nonumber \\&\quad = (1+\alpha )\left\{ \int _{D^+_\delta }(2K_n^2(x)|x|^{2\alpha }e^{2v_n}-K_n(x)|x|^{\alpha }e^{v_n}|\Psi _n|^2)dv +\int _{L_{\delta }}cK_n(x)|x|^{\alpha }e^{v_n}ds \right\} \nonumber \\&\qquad -\delta \int _{S^+_{\delta }}K_n^2(x)|x|^{2\alpha }e^{2v_n}d\sigma \nonumber \\&\qquad +\int _{L_{\delta }}c\frac{\partial K_n(s,0)}{\partial s}|s|^{\alpha }se^{v_n(s,0)}ds-cK_n(s,0)|s|^{\alpha }se^{v_n(s,0)}|^{s=\delta }_{s=-\delta } \nonumber \\&\qquad +\int _{D^+_\delta }x\cdot \nabla (K_n^2(x))|x|^{2\alpha }e^{2v_n}dv -\int _{D^+_\delta }x\cdot \nabla K_n(x)|x|^{\alpha }e^{v_n}|\Psi _n|^2dv \nonumber \\&\qquad + \frac{1}{4}\int _{S^+_{\delta }}\left\langle \frac{\partial \Psi _n}{\partial \nu },(x+\bar{x})\cdot \Psi _n\right\rangle d\sigma +\frac{1}{4}\int _{S^+_\delta }\left\langle (x+\bar{x})\cdot \Psi _n, \frac{\partial \Psi _n}{\partial \nu }\right\rangle d\sigma . \end{aligned}$$

(70)

We will take $n\rightarrow \infty$ first and then $\delta \rightarrow 0$ in (70). By using (69) we get

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\delta \int _{S^+_\delta } \left| \frac{\partial v_n}{\partial \nu }\right| ^2-\frac{1}{2}|\nabla v_n|^2d\sigma =\lim _{\delta \rightarrow 0}\delta \int _{S^+_\delta }\frac{1}{2} \left| \frac{\partial \left( \frac{m(p)}{\pi }\ln \frac{1}{|x|}\right) }{\partial \nu }\right| ^2d\sigma =\frac{1}{2\pi }m^2(p). \end{aligned}$$

By using $u_n\rightarrow -\infty$ uniformly on $S^+_{\delta }$, we also have

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty } \delta \int _{S^+_\delta }K_n^2(x)|x|^{2\alpha }e^{2v_n}d\sigma =\lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty } \delta \int _{S^+_\delta }V^2(x)|x|^{2\alpha }e^{2u_n}d\sigma =0, \end{aligned}$$

and

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty } cK_n(s,0)|s|^{\alpha }se^{v_n(s,0)}|^{s=\delta }_{s=-\delta } =0. \end{aligned}$$

By using the energy condition (67), we have

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\int _{D^+_\delta }\left( |x|^{2\alpha }e^{2v_n}x\cdot \nabla (K_n^2(x)\right) -|x|^{\alpha }e^{v_n}|\Psi _n|^2x\cdot \nabla K_n(x))dx=0, \end{aligned}$$

and

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty }\int _{L_{\delta }}c\frac{\partial K_n(s,0)}{\partial s}|s|^{\alpha }se^{v_n(s,0)}ds=0. \end{aligned}$$

Since $u_n\rightarrow -\infty$ uniformly in any compact subset of $(D^+_{\delta _0}\cup L_{\delta _0} )\backslash \{0\}$, and $|\Psi _n|$ is uniformly bounded in any compact subset of $(D^+_{\delta _0}\cup L_{\delta _0}) \backslash \{0\}$, we know

$$\begin{aligned} \left\{ \begin{array}{rcll} \not {D}\Psi &{}=&{} 0, &{} \text { in } D^+_{\delta _0},\\ B \Psi &{}=&{} 0, &{}\text { on } L_{\delta _0}\setminus \{0\}. \end{array} \right. \end{aligned}$$

We extend $\Psi$ a harmonic spinor $\overline{\Psi }$ on $D_{\delta _0}\setminus \{0\}$ with bounded energy, i.e., $||\overline{\Psi }||_{L^4(D_{\delta _0})} < \infty$. Since the local singularity of a harmonic spinor with finite energy is removable, we have $\overline{\Psi }$ is smooth in $D_{\delta _0}$. It follows that $\Psi$ is smooth in $D^+_{\delta _0}\cup L_{\delta _0}$. Therefore we obtain that

$$\begin{aligned} \lim _{\delta \rightarrow 0}\lim _{n\rightarrow \infty } \left( \frac{1}{4}\int _{S^+_{\delta }}\left\langle \frac{\partial \Psi _n}{\partial \nu },(x+\bar{x})\cdot \Psi _n\right\rangle d\sigma +\frac{1}{4}\int _{S^+_\delta }\left\langle (x+\bar{x})\cdot \Psi _n, \frac{\partial \Psi _n}{\partial \nu }\right\rangle d\sigma \right) =0. \end{aligned}$$

Putting all together, we obtain that

$$\begin{aligned} \frac{1}{2\pi }m^2(p)=(1+\alpha )m(p). \end{aligned}$$

It follows that $m(p)=2\pi (1+\alpha )$. Thus we finish the proof of Theorem 1.6. $\square$

9 Energy quantization for the global super-Liouville boundary problem

In this section, we will show the quantization of energy for a sequence of blowing-up solutions to the global super-Liouville boundary problem on a singular Riemann surface. Let $(M, \mathcal {A}, g)$ be a compact Riemann surface with conical singularities represented by the divisor $\mathcal {A}=\Sigma _{j=1}^{m}\alpha _j q_j$, $\alpha _j>0$ and with a spin structure. We assume that $\partial M$ is not empty and (M, g) has conical singular points $q_1, q_2, \ldots , q_m$ such that $q_1, q_2, \ldots , q_l$ are in $M^o$ for $1\le l<m$ and $q_{l+1}, q_{l+2}, \ldots , q_m$ are on $\partial M$. Writing $g=e^{2\phi }g_0$, where $g_0$ is a smooth metric on M, we can deduce from the results for the local super-Liouville equations:

Proof of Theorem 1.1:

Since $g=e^{2\phi }g_0$ with $g_0$ being smooth, then by the well known properties of $\phi$ (see e.g. [36] or [5], p. 5639), we know that $(u_n,\psi _n)$ satisfies

with the energy conditions:

$$\begin{aligned} \int _{M}e^{2(u_n+\phi )}dg_0+|e^{\frac{\phi }{2} }\psi _n|_{g_0}^4dv_{g_0}+\int _{\partial M}e^{u_n+\phi }d\sigma _{g_0}<C. \end{aligned}$$

If we define the blow-up set of $u_n + \phi$ as

$$\begin{aligned} \Sigma '_1=\left\{ x\in M,\text { there is a sequence }y_n\rightarrow x\text { such that }(u_n+\phi )(y_n)\rightarrow +\infty \right\} , \end{aligned}$$

then by Remark 3.4 and Remark 3.3 in [27], we have $\Sigma _1=\Sigma '_1$. By the blow-up results of the local system, it follows that one of the following alternatives holds:

• (i) $u_n$ is bounded in $L^{\infty }(M).$

• (ii) $u_n$ $\rightarrow -\infty$ uniformly on M.

• (iii) $\Sigma _1$ is finite, nonempty and

  $$\begin{aligned} u_n\rightarrow -\infty \text { uniformly on compact subsets of }M\backslash \Sigma _1. \end{aligned}$$

  Furthermore,

  $$\begin{aligned} \int _M(2e^{2(u_n+\phi )}-e^{u_n+\phi }|e^{\frac{\phi }{2}}\psi _n|_{g_0}^2)\varphi dv_{g_0}+\int _{\partial M}ce^{u_n+\phi }\varphi d\sigma _{g_0}\rightarrow \sum _{p_i\in \Sigma _1}m(p_i)\varphi (p_i) \end{aligned}$$

  for any smooth function $\varphi$ on M.

Next let $p=\frac{q}{q-1}>2$. Notice that

$$\begin{aligned}&||\nabla (u_n+\phi )||_{L^q(M,g_0)}\\&\quad \le \sup \left\{ |\int _M\nabla (u_n+\phi )\nabla \varphi dv_{g_0}||\varphi \in W^{1,p}(M,g_0),\right. \\&\qquad \qquad \qquad \left. \int _{M}\varphi dv_{g_0}=0,||\varphi ||_{W^{1,p}(M,g_0)}=1\right\} . \end{aligned}$$

Due to $||\varphi ||_{L^{\infty }(M,g_0)\le C}$ for any $\varphi \in W^{1,p}(M,g_0)$ with $\int _{M}\varphi dv_{g_0}=0$ and $||\varphi ||_{W^{1,p}(M,g_0)}=1$ by the Sobolev embedding theorem, we get that

$$\begin{aligned}&\left| \int _{M}\nabla (u_n+\phi )\nabla \varphi dv_{g_0}\right| \\&\quad = \left| -\int _M\Delta _{g_0} (u_n+\phi )\varphi dv_{g_0}+\int _{\partial M}\frac{\partial (u_n+\phi )}{\partial n}\varphi d\sigma _{g_0} \right| \\&\quad \le \int _{M}(2e^{2(u_n+\phi )}+e^{u_n+\phi }|e^{\frac{\phi }{2}} \psi _n|_{g_0}^2 + |K_{g_0}|)|\varphi |dg_0 +\int _{\partial M}( ce^{u_n+\phi }+|h_{g_0}|)|\varphi |d\sigma _{g_0} \\&\qquad + \sum _{j=1}^l \left| \int _M 2\pi \alpha _j \delta _{q_j}\varphi dv_{g_0}\right| + \sum _{j=1+1}^m \left| \int _{\partial M} \pi \alpha _j \delta _{q_j} \varphi d\sigma _{g_0}\right| \\&\quad \le C. \end{aligned}$$

This means that $u_n+\phi -\frac{1}{|M|}\int _{M}(u_n+\phi )dv_{g_0}$ is uniformly bounded in $W^{1,q}(M,g_0)$.

We define the Green function G by

$$\begin{aligned} \left\{ \begin{array}{l} - \Delta _{g_0} G=\sum _{p\in M^o\cap \Sigma _1}m(p)\delta _p-K_{g_0}-\sum _{j=1}^l 2\pi \alpha _j \delta _{q_j},\\ \frac{\partial G}{\partial n}=\sum _{p\in \partial M\cap \Sigma _1}m(p)\delta _p-h_{g_0}+\sum _{j=l+1}^m \pi \alpha _j \delta _{q_j},\\ \int _{M}Gdv_{g_0}=0. \end{array}\right. \end{aligned}$$

It is clear that $G\in W^{1,q}(M,g_0)\cap C_{loc}^2(M\backslash \Sigma _1)$ with $\int _{M}Gdg_0=0$ for $1<q<2$.

Now we take $R>0$ small such that, at each blow-up point $p\in \Sigma _1$, the geodesic ball of M, $B^{M}_R(p)$, satisfies $B^{M}_R(p)\cap (\Sigma _1 \cup \{q_1,q_2, \ldots , q_m\})=\{p\}$. Then we also have

$$\begin{aligned} G(x)=\left\{ \begin{array}{lcl} - \frac{1}{2\pi }m(p)\log d(x,p)+g(x), \quad &{}\text {if}&{} p\in M^o\cap (\Sigma _1 \setminus \{q_1,q_2, \ldots , q_m\}), \\ - \left( \frac{1}{2\pi }m(p)- \alpha _j \right) \log d(x,p)+g(x), \quad &{}\text {if}&{} p=q_j\in M^o\cap \Sigma _1 \cap \{q_1,q_2, \ldots , q_l\}, \\ - \left( \frac{1}{\pi }m(p)\right) \log d(x,p)+g(x), \quad &{}\text {if}&{} p\in \partial M\cap (\Sigma _1 \setminus \{q_{l+1},q_{l+2}, \ldots , q_m\}), \\ - \left( \frac{1}{\pi }m(p)+a_j\right) \log d(x,p)+g(x), \quad &{}\text {if}&{} p=q_j\in \partial M\cap \Sigma _1 \cap \{q_{l+1},q_{l+2}, \ldots , q_m\}, \end{array} \right. \end{aligned}$$

for $x\in B^M_R(p)\backslash \{p\}$ with $g\in C^2(B^M_R(p))$, where d(x, p) denotes the Riemannian distance between x and p with respect to $g_0$ and

$$\begin{aligned} m(p)= & {} \lim _{R\rightarrow 0}\lim _{n\rightarrow \infty }\left\{ \int _{B^M_R(p)}(2e^{2(u_n+\phi )}-e^{u_n+\phi }|e^{\frac{\phi }{2}}\psi _n|_{g_0}^2-K_{g_0})dv_{g_0}\right. \\&\qquad \qquad \qquad \left. +\int _{\partial M\cap B^M_R(p)}(ce^{u_n+\varphi }-h_{g_0})d\sigma _{g_0}\right\} . \end{aligned}$$

On the other hand, since for any $\varphi \in C^{\infty }(M)$

$$\begin{aligned}&\int _{M}\nabla (u_n+\phi -G)\nabla \varphi dv_{g_0} \\&\quad = -\int _{M}\Delta _{g_0}(u_n+\phi -G)\varphi dv_{g_0}+\int _{\partial M}\frac{\partial (u_n+\phi -G)}{\partial n}\varphi d\sigma _{g_0} \\&\quad = \int _{M}( 2e^{2(u_n+\phi )}-e^{u_n+\phi }|e^{\frac{\phi }{2}}\psi _n|^2_{g_0}-\sum _{p\in M^o\cap \Sigma _1}m(p)\delta _p)\varphi dv_{g_0}\\&\qquad +\int _{\partial M}\left( ce^{u_n+\phi }-\sum _{p\in \partial M\cap \Sigma _1}m(p)\delta _p\right) \varphi d\sigma _{g_0}\\&\quad \rightarrow 0,\quad \text { as } n\rightarrow \infty , \end{aligned}$$

by using the fact that $u_n+\phi -\frac{1}{|M|}\int _{M}(u_n+\phi )dg_0$ is uniformly bounded in $W^{1,q}(M,g_0)$, we get

$$\begin{aligned} u_n+\phi -\frac{1}{|M|}\int _{M}(u_n+\phi )dg_0\rightarrow G \end{aligned}$$

strongly in $C_{loc}^2(M\backslash \Sigma _1)$ and weakly in $W^{1,q}(M,g_0)$. Consequently we have

$$\begin{aligned} \max _{M^o\cap \partial B^M_{R}(p)}u_n-\min _{M^o\cap \partial B^M_{R}(p)}u_n\le C. \end{aligned}$$

Therefore we get the blow-up value $m(p)=4\pi$ when $p\in M^o\cap (\Sigma _1 \setminus \{q_1,q_2, \ldots , q_m\})$, $m(p)=4\pi (1+\alpha _j)$ when $p=q_j\in M^o\cap \Sigma _1 \cap \{q_1,q_2, \ldots , q_l\}$, $m(p)=2\pi$ when $p\in \partial M\cap (\Sigma _1 \setminus \{q_{l+1},q_{l+2}, \ldots , q_m\})$, and $m(p)=2\pi (1+\alpha _j)$ when $p=q_j\in \partial M\cap \Sigma _1 \cap \{q_{l+1},q_{l+2}, \ldots , q_m\}$. By using that

$$\begin{aligned}&\int _{M}2e^{2u_n}-e^{u_n}|\psi _n|^2_gdv_g+\int _{\partial M}ce^{u_n}d\sigma _{g}=\int _{M}2e^{2(u_n+\phi )}-e^{u_n+\phi }|e^{\frac{\phi }{2}}\psi _n|^2_{g_0}dv_{g_0}\\&\quad +\int _{\partial M}ce^{u_n}d\sigma _{g_0}, \end{aligned}$$

we get the conclusion of the Theorem. $\square$