1 Introduction

The theory of the existence, regularity, and the rigidity for the proper harmonic maps between balls in hyperbolic metrics has been studied by Li and Tam [11,12,13] and Li and Wang [14, 15], Wang [28], etc. The complex case in Bergman metrics was studied by Donanelly [3] and Li and Ni [18] and Li and Simon [19], etc.. The main purpose of the paper is to study the asymptotic expansions for proper harmonic maps between balls in Bergman metrics in \({\mathbb {C}}^n\) and \({\mathbb {C}}^m\), respectively.

Let \(B_{n}\) be the unit ball in \({\mathbb {C}}^{n}\). The Bergman kernel on \( B_{n}\) is given by \(K(z,w)=(1-\langle z,w\rangle )^{-n-1}\). The Bergman metric is defined by

$$\begin{aligned} g=\sum _{i,j=1}^{n}g_{i,{\overline{j}}}dz^{i}\otimes d{\overline{z}}^{j},\quad g_{i{\overline{j}}}={\frac{\partial ^{2}}{\partial z_{i}\partial {\overline{z}} _{j}}}{\frac{\log K(z,z)}{n+1}}. \end{aligned}$$
(1.1)

The Laplace–Beltrami operator \(\Delta _g\) in the Bergman metric in \(B_{n}\) is defined by

$$\begin{aligned} \Delta _{B_n}=-\Delta _g=\sum _{i,j=1}^{n}g^{i{\overline{j}}}{\frac{\partial ^{2}}{\partial z_{i}\partial {\overline{z}}_{j}}}=(1-|z|^{2})\sum _{i,j=1}^{n}(\delta _{i,j}-z_{i}{\overline{z}}_{j}){\frac{\partial ^{2}}{\partial z_{i}\partial {\overline{z}}_{j}}}. \end{aligned}$$
(1.2)

Let h be the Bergman metric on the unit ball \(B_{m}\) in \({\mathbb {C}}^{m}\) with the Christoffel symbols \(\Gamma _{t\gamma }^{s}\).

A \(C^{2}\) map \(u:B_{n}\rightarrow B_{m}\) is harmonic in Bergman metrics if the tension field

$$\begin{aligned} \tau ^{s}\left[ u\right] =\Delta _{B_n}u^{s}+\sum _{\beta ,\gamma =1}^{m}\sum _{i,j=1}^{n}\Gamma _{\beta \,\gamma }^{s}g^{i{\bar{j}}}\partial _{i}u^{\beta }\partial _{{\overline{j}}}u^{\gamma }=0,\quad \text { for }1\le s\le m. \end{aligned}$$
(1.3)

If \(u:B_{n}\rightarrow B_{m}\) is a proper harmonic map and if \(\phi _{0}(z)=u(z)|_{ \partial B_n} \in C^1(\partial B_{n})\), then

$$\begin{aligned} |\phi _{0}|^{2}=1\quad \hbox {and}\quad \langle {\tilde{Y}}_{j}\phi _{0},\phi _{0}\rangle =0,\quad \hbox { on }\partial B_{n},\quad \hbox {for all }1\le j\le n, \end{aligned}$$
(1.4)

where

$$\begin{aligned} {\tilde{Y}}_{j}={\frac{\partial }{\partial z_{j}}}-{\frac{{\overline{z}}_{j}}{ |z|^{2}}}R\quad \hbox {and}\quad R=\sum _{k=1}^{n}z_{k}{\frac{\partial }{ \partial z_{k}}}. \end{aligned}$$
(1.5)

Let

$$\begin{aligned} |{\overline{\partial }}_{b}u|^{2}=\sum _{j=1}^{n}\left| z\right| ^{2} \big |\overline{{\tilde{Y}}_{j}}u\big |^{2},\quad |\partial _{b}u|^{2}=\sum _{j=1}^{n}\left| z\right| ^{2}\big |{\tilde{Y}}_{j}u\big | ^{2}\quad \end{aligned}$$
(1.6)

and

$$\begin{aligned} E_{b}[u]=|\partial _{b}u|^{2}+|{\overline{\partial }} _{b}u|^{2}. \end{aligned}$$
(1.7)

Let \(\phi _{0}:\partial B_{n}\rightarrow \partial B_{m}\). The Dirichlet boundary value problem for a proper harmonic map in Bergman metric is given by

$$\begin{aligned} \tau [u]=0,\quad \hbox {in }B_{n}\quad \hbox {and}\quad u=\phi _{0},\quad \hbox { on }\partial B_{n}. \end{aligned}$$
(1.8)

Assuming that \(\phi _0 \in C^\infty (\partial B_n)\) and \(E_b[\phi _0]\ne 0\) on \( \partial B_n\), the existence and regularity were studied by Li and Tam [11,12,13], Li and Wang [14, 15], Wang [28] for real case. The complex case was studied by Donnelly [3] (also see Li and Ni [18]). Their result can be stated as follows:

Theorem 1.1

Let \(\phi _0 :\partial B_{n}\rightarrow \partial B_{m}\) be a \( C^{\infty }\) map with \(E_{b}[\phi _0 ]\ne 0\) and satisfying (1.4). Then the Dirchlet boundary value problem (1.8) has a unique solution \(u\in C^{n,\alpha }({\overline{B}}_{n})\) for any \(0<\alpha <1\).

The question about the structure of u associated with its regularity is always interesting, the problem is the so-called the asymptotic expansion of u.

For the linear case: The asymptotic expansion for the solution of

$$\begin{aligned} \Delta _{B_n}u=0,\quad \hbox {in }B_{n}\quad \hbox {and}\quad u=\phi \quad \hbox {on }\partial B_{n} \end{aligned}$$
(1.9)

was studied by Graham [5]. He proved the following theorem.

Theorem 1.2

If \(\phi \in C^{\infty }(\partial B_{n})\), then the unique solution u of (1.9) can be written by

$$\begin{aligned} u(z)=A(z)+B(z)(1-|z|^{2})^{n}\log (1-|z|^{2}), \end{aligned}$$

where \(A,B\in C^{\infty }({\overline{B}}_{n})\).

For the fully nonlinear equation, the Fefferman equation with \(\rho >0\) is given by

$$\begin{aligned} J[\rho ]=(-1)^{n}\det \left[ \begin{matrix} \rho &{} \rho _{{\overline{j}}}\\ \rho _{i} &{} \rho _{i{\overline{j}}} \end{matrix} \right] =1\hbox { in }\Omega \quad \hbox { and}\quad \rho =0\hbox { on }\partial \Omega , \end{aligned}$$

where \(\Omega \) is a smoothly bounded pseudoconvex domain in \({\mathbb {C}} ^{n} \). The uniqueness and approximating solution with \(-\log (\rho )\) being plurisubharmonic were given by C. Fefferman [4] and the existence and uniqueness and regularity were established by Cheng and Yau [2]. When \( \Omega \) is strongly pseudoconvex, the maximum regularity is \(C^{n+1,\alpha }\) up to boundary for any \(0<\alpha <1\). The asymptotic expansion for the unique solution u was given by Lee and Melrose [9].

The harmonic map equation is a system of semi-linear equations. The question was asked by Li and Tam [12]: Does one have the asymptotic expansion for the solution of (1.8) similar to the one in [5] and [9]? The first purpose of this paper is to study this question. In order to state our result, we assume that \(\phi _0\in C^\infty (\partial B_n)\) and let

$$\begin{aligned} h[\phi _0]= \left[ \begin{array}{cc} \overline{{\tilde{Y}}} \phi _0 &{} - {\tilde{Y}}\phi _0 \\ - \overline{{\tilde{Y}}\phi _0} &{} {\tilde{Y}} {\overline{\phi }}_0\\ \end{array}\right] \end{aligned}$$
(1.10)

be a \((2m)\times (2n)\) matrix-valued function on \(\partial B_n\). For any \(f \in C(\partial B_n, {\mathbb {C}}^{2m})\), we let \(P_0[f](z)\) be the orthogonal projection of f(z) onto the range of \(h[\phi _0](z) h[\phi _0]^*(z)\). It is easy to see that if \(h[\phi _0] h[\phi _0]^* \) has constant rank, then \(P_0\) maps \( C^\infty (\partial B_n, {\mathbb {C}}^{2m})\) into \( C^\infty (\partial B_n, {\mathbb {C}}^{2m})\). Now we are ready to state our theorem.

Theorem 1.3

Let \(\phi _{0}:\partial B_{n}\rightarrow \partial B_{m}\) be a \( C^{\infty }\) map satisfying (1.4). If \(|\partial _{b}\phi _{0}|| {\overline{\partial }}_{b}\phi _{0}|>0\) on \(\partial B_{n}\) and \(P_0 (C^\infty (\partial B_n, {\textbf{C}}^{2m}))\subset C^\infty (\partial B_n, {\textbf{C}}^{2m})\), then the solution u of (1.8) has the following asymptotic expansion:

$$\begin{aligned} u(z)=\phi (z)+\sum _{\ell =1}^{\infty }\psi _{\ell }\Big ((1-|z|)^{(n+1)}\log (1-|z|)\Big )^{\ell }, \end{aligned}$$

where \(\phi \) and \(\psi _{\ell }\in C^{\infty }({\overline{B}}_{n})\). Here, the infinite sum is in the sense of asymptotic expansion.

Note that if \(|\partial _b \phi _0 | |{\overline{\partial }}_b \phi _0|=0\) on \(\partial B_n\), the following theorem was proved by Li and Ni [18] on \(\partial B_n\) and by Li and Son [20] on more general torsion-free pseudo-Hermitian closed manifolds.

Theorem 1.4

If \(\phi \in C^2(\partial B_n)\) and if \(|\partial _b \phi _0| |{\overline{\partial }}_b \phi _0 |=0\) on \(\partial B_n\), then \(\phi _0\) is either CR or anti-CR on \(\partial B_n\).

The second purpose of the paper is looking for a sufficient and necessary condition on any boundary map \(\phi _0:\partial B_n\rightarrow \partial B_m\) so that it becomes either CR or anti-CR on \(\partial B_n\). The results are stated in Theorem 2.2. Since the computation of \(\tau ^s[u]\) is very complicated, in order to prove the asymptotic expansion formula, we divide our computations into several parts which are presented in Sections 3–5. The proof of Theorem 1.3 is given in Section 6. In Section 7, we will prove a technical proposition (Proposition 6.3). In Section 8, we provide a theorem, with assumption that \(h[\phi _0] h[\phi _0]^*\) has rank \(2m-2\), the harmonic map u has better regularity.

2 Preliminary

On the unit ball \(B_{n}\subset {\mathbb {C}}^{n}\), we define the radial first order differential operators:

$$\begin{aligned} R=\sum _{k=1}^{n}z_{k}{\frac{\partial }{\partial z_{k}}},\quad |z|^{2}R_{0}=R, \end{aligned}$$
(2.1)

the tangential first order differential operators:

$$\begin{aligned} X_{j}={\frac{\partial }{\partial z_{j}}}-{\overline{z}}_{j}R,\quad {\tilde{Y}} _{j}={\frac{\partial }{\partial z_{j}}}-{\overline{z}}_{j}R_{0},\quad Y_{j}=|z| {\tilde{Y}}_{j}\quad \hbox {and}\quad T=R-{\overline{R}}. \end{aligned}$$
(2.2)

For any \(C^{1}\) map \(\phi :{\overline{B}}_{n}\setminus \{0\}\rightarrow {\overline{B}}_{m}\setminus \{0\}\), we define

$$\begin{aligned} |Y\phi |^{2}=|\partial _{b}\phi |^{2},\quad |{\overline{Y}}\phi |^{2}=| {\overline{\partial }}_{b}\phi |^{2}\quad \hbox {and}\quad E_{b}[\phi ]=|Y\phi |^{2}+|{\overline{Y}}\phi |^{2}. \end{aligned}$$
(2.3)

Lemma 2.1

Let \(\phi :\partial B_{n}\rightarrow \partial B_{m}\) be a \( C^{2}\) map with

$$\begin{aligned} \langle Y_{j}\phi ,\phi \rangle =0,\quad \hbox {on }\partial B_{n},\quad 1\le j\le n. \end{aligned}$$
(2.4)

Then

$$\begin{aligned} (n-1)\langle T\phi ,\phi \rangle =|Y\phi |^{2}-|{\overline{Y}}\phi |^{2}=| {\overline{\partial }}_{b}{\overline{\phi }}|^{2}-|{\overline{\partial }}_{b}\phi |^{2} \end{aligned}$$
(2.5)

and

$$\begin{aligned} \Big \langle \sum _{j=1}^{n}\left( Y_{j}{\overline{Y}}_{j}+{\overline{Y}}_{j}Y_{j}\right) \phi ,\phi \Big \rangle =-E_{b}\left[ \phi \right] . \end{aligned}$$
(2.6)

Proof

Notice that

$$\begin{aligned} Y_{j}=|z|{\frac{\partial }{\partial z_{j}}}-{\frac{{\overline{z}}_{j}}{|z|}} R,\quad \sum _{j=1}^{n}z_{j}Y_{j}=0\quad \hbox {and}\quad {\tilde{Y}}_{j}\left( \left| z\right| \right) =\overline{{\tilde{Y}}}_{j}\left( \left| z\right| \right) =0, \end{aligned}$$
(2.7)

one has

$$\begin{aligned}&\sum _{j=1}^{n}\left( {\overline{Y}}_{j}Y_{j}-Y_{j}{\overline{Y}} _{j}\right) \\&\quad =\sum _{j=1}^{n}\Big ( |z|{\overline{Y}}_{j}{\frac{\partial }{\partial z_{j}}}- {\frac{{\overline{z}}_{j}}{|z|}}{\overline{Y}}_{j}R-{\frac{{\overline{Y}}_{j} {\overline{z}}_{j}}{|z|}}R-|z|Y_{j}{\frac{\partial }{\partial {\overline{z}}_{j}} }+{\frac{z_{j}}{|z|}}Y_{j}{\overline{R}}+{\frac{Y_{j}z_{j}}{|z|}}{\overline{R}} \Big ) \\&\quad =\sum _{j=1}^{n}\Big ( -z_{j}{\overline{R}}{\frac{\partial }{\partial z_{j}}}+ {\overline{z}}_{j}R{\frac{\partial }{\partial {\overline{z}}_{j}}}-(1-{\frac{ |z_{j}|^{2}}{|z|^{2}}})(R-{\overline{R}})\Big ) \\&\quad =-(n-1)(R-{\overline{R}}). \end{aligned}$$

This implies (2.5) and (2.6) hold.

Since \(|\phi |^{2}=1\) on \(\partial B_{n}\) and \(\langle Y_{j}\phi ,\phi \rangle =0\), one has \(\left\langle {\overline{Y}}_{j}\phi ,\phi \right\rangle =0\). Therefore,

$$\begin{aligned} 0=\left\langle Y_{j}\phi ,Y_{j}\phi \right\rangle +\left\langle {\overline{Y}} _{j}Y_{j}\phi ,\phi \right\rangle \ \text {and}\quad 0=\left\langle \overline{ Y}_{j}\phi ,{\overline{Y}}_{j}\phi \right\rangle +\left\langle Y_{j}\overline{Y }_{j}\phi ,\phi \right\rangle . \end{aligned}$$

These imply

$$\begin{aligned} |Y_{j}\phi |^{2}=-\left\langle {\overline{Y}}_{j}Y_{j}\phi ,\phi \right\rangle \text { and }|{\overline{Y}}_{j}\phi |^{2}=-\left\langle Y_{j}{\overline{Y}} _{j}\phi ,\phi \right\rangle . \end{aligned}$$
(2.8)

By (2.5) and (2.8) , one obtains that

$$\begin{aligned} (n-1)\langle T\phi ,\phi \rangle =|Y\phi |^{2}-|{\overline{Y}}\phi |^{2}= \Big \langle \sum _{j=1}^{n}(Y_{j}{\overline{Y}}_{j}-{\overline{Y}}_{j}Y_{j})\phi ,\phi \Big \rangle \end{aligned}$$

and

$$\begin{aligned} \sum _{j=1}^{n}\left\langle \left( Y_{j}{\overline{Y}}_{j}+{\overline{Y}} _{j}Y_{j}\right) \phi ,\phi \right\rangle =-E_{b}\left[ \phi \right] . \end{aligned}$$

Therefore, the proof of the lemma is complete.

\(\square \)

Theorem 2.2

Let \(\phi :\partial B_{n}\rightarrow \partial B_{m}\) be a \( C^{2}\) map satisfying (2.4). Then \(\phi \) is CR or anti-CR if and only if \(\phi \) satisfies

$$\begin{aligned} \int _{\partial B_{n}}E_{b}[\phi ]d\sigma \le (n-1)\int _{\partial B_{n}}|\langle T\phi ,\phi \rangle |d\sigma . \end{aligned}$$
(2.9)

Proof

By Lemma 2.1, one has

$$\begin{aligned} (n-1)\langle T\phi ,\phi \rangle =|Y\phi |^{2}-|{\overline{Y}}\phi |^{2} \end{aligned}$$
(2.10)

and

$$\begin{aligned} E_{b}[\phi ]=|Y\phi |^{2}+|{\overline{Y}}\phi |^{2}. \end{aligned}$$
(2.11)

Therefore,

$$\begin{aligned} \int _{\partial B_{n}}E_{b}[\phi ]d\sigma \le (n-1)\int _{\partial B_{n}}|\langle T\phi ,\phi \rangle |d\sigma \end{aligned}$$

implies

$$\begin{aligned} \int _{\partial B_{n}}|Y\phi |^{2}+|{\overline{Y}}\phi |^{2}-\Big ||Y\phi |^{2}-| {\overline{Y}}\phi |^{2}\Big |d\sigma \le 0. \end{aligned}$$

But the integrand is a nonnegative continuous function on \(\partial B_{n}\), one can easily see that

$$\begin{aligned} |Y\phi |^{2}+|{\overline{Y}}\phi |^{2}-\Big ||Y\phi |^{2}-|{\overline{Y}}\phi |^{2}\Big |=0\quad \hbox { on }\partial B_{n}. \end{aligned}$$

This implies that for any \(z\in \partial B_{n}\), one has either \(|Y\phi (z)|=0\) or \(|{\overline{Y}}\phi (z)|=0\). Therefore,

$$\begin{aligned} {\overline{\partial }}_{b}\phi (z)=0\quad \hbox {or }\quad {\overline{\partial }} _{b}{\overline{\phi }}(z)=0,\quad z\in \partial B_{n}. \end{aligned}$$

Applying Theorem 3.1 in [18] or Lemma 4.2 in [20], one has \( {\overline{\partial }}_{b}\phi =0\) on \(\partial B_{n}\) or \({\overline{\partial }} _{b}{\overline{\phi }}=0\) on \(\partial B_{n}\), which means either \(\phi \) is CR or \({\overline{\phi }}\) is CR.

On the other hand, if \(\phi \) is CR or \({\overline{\phi }}\) is CR, then \( {\overline{\partial }}_{b}\phi =0\) or \({\overline{\partial }}_{b}{\overline{\phi }} =0\) on \(\partial B_{n}.\) By (2.10) and (2.11) , one can easily show that (2.9) holds and the proof of the theorem is complete. \(\square \)

We state and prove the following proposition for future application.

Proposition 2.3

Let \(\phi :\partial B_{n}\rightarrow \partial B_{m}\) be a \( C^{2}\) map satisfying (2.4). Then

$$\begin{aligned} -\big ((n+1)E_{b}[\phi ]\big )^{2}+\Big ((n-1)\sqrt{E_{b}[\phi ]^2+4n\langle T\phi ,\phi \rangle ^{2}}\Big )^{2}=-16n|\partial _{b}\phi |^{2}|\overline{ \partial }_{b}\phi |^{2}. \end{aligned}$$

Proof

Since

$$\begin{aligned}{} & {} -\big ((n+1)E_{b}[\phi ]\big )^{2}+\Big ((n-1)\sqrt{E_{b}[\phi ]^2+4n\langle T\phi ,\phi \rangle ^{2}}\Big )^{2} \\{} & {} \quad =-(n+1)^{2}E_{b}[\phi ]^{2}+(n-1)^{2}E_{b}[\phi ]^2+4n(n-1)^{2}\langle T\phi ,\phi \rangle ^{2} \\{} & {} \quad =4n\Big [ \Big (|\partial _{b}\phi |^{2}-|{\overline{\partial }}_{b}\phi |^{2} \Big )^{2}-E_{b}[\phi ]^{2} \Big ] \\{} & {} \quad =-16n|\partial _{b}\phi |^{2}|{\overline{\partial }}_{b}\phi |^{2}, \end{aligned}$$

the proof is complete. \(\square \)

3 Reformulation of the harmonic map equations

In this section, we introduce more notations and prove more lemmas which will be used later.

Definition 3.1

A differential operator \({\mathcal {D}}\) on \(C^{2}({\overline{B}}_{n} \setminus \{0\})\) is called a boundary operator if for every \(\phi \in C^{2}({\overline{B}} _{n} \setminus \{0\})\) with \(\phi (z)=\phi (\rho z)\), one has \({\mathcal {D}}\phi (z)=({\mathcal {D}} \phi )(\rho z) \).

Let

$$\begin{aligned} r(z)=|z|-1,\quad {\mathcal {R}}=R+{\overline{R}},\quad {{\mathcal {L}}} _{0}=2\sum _{j=1}^{n}(Y_{j}{\overline{Y}}_{j}+{\overline{Y}}_{j}Y_{j}) \end{aligned}$$
(3.1)

and

$$\begin{aligned} L={\frac{\Delta _{B_n}}{1-|z|^{2}}}=\sum _{j=1}^{n}{\frac{\partial ^{2}}{ \partial z_{j}\partial {\overline{z}}_{j}}}-R{\overline{R}}. \end{aligned}$$
(3.2)

Proposition 3.2

With the above notations, one has

$$\begin{aligned} 4|z|^{2}L={{\mathcal {L}}}_{0}+2(n-1){\mathcal {R}}-(2r+r^{2}){\mathcal {R}} ^{2}+(2r+r^{2})T^{2}. \end{aligned}$$
(3.3)

Proof

One can verify that \(R,{\overline{R}},R_{0}\),and \({\overline{R}}_{0}\) satisfy the Leibnitz’s rule. Moreover, by (2.7)

$$\begin{aligned} L&=\sum _{j=1}^{n}\Big ({\frac{\partial }{\partial {\overline{z}}_{j}}}\big ({ \frac{\partial }{\partial z_{j}}}-{\overline{z}}_{j}R_{0}\big )+{\frac{\partial }{\partial {\overline{z}}_{j}}}\big ({\overline{z}}_{j}R_{0}\big )\Big )-R\overline{ R} \\&=\sum _{j=1}^{n}{\frac{\partial }{\partial {\overline{z}}_{j}}}{\tilde{Y}} _{j}+nR_{0}+{\overline{R}}({\frac{1}{|z|^{2}}})R+{\frac{{\overline{R}}R}{|z|^{2}} }-R{\overline{R}} \\&=\sum _{j=1}^{n}{\frac{\partial }{\partial {\overline{z}}_{j}}}{\tilde{Y}} _{j}+nR_{0}-R_{0}+{\frac{1-|z|^{2}}{|z|^{2}}}R{\overline{R}} \\&=\sum _{j=1}^{n}\overline{{\tilde{Y}}_{j}}{\tilde{Y}}_{j}+{\frac{1-|z|^{2}}{ |z|^{2}}}{\overline{R}}R+(n-1)R_{0}. \end{aligned}$$

Since

$$\begin{aligned} 1-|z|^{2}=-(1+|z|)r(z)=-(2+r(z))r(z)=-2r(z)-r(z)^{2} \end{aligned}$$
(3.4)

and

$$\begin{aligned} 2(R{\overline{R}}+{\overline{R}}R)=(R+{\overline{R}})^{2}-(R-{\overline{R}})^{2}={ {\mathcal {R}}}^{2}-T^{2}, \end{aligned}$$

one has

$$\begin{aligned} 4|z|^{2}L={{\mathcal {L}}}_{0}+2(n-1){{\mathcal {R}}}-(2r(z)+r(z)^{2}){{\mathcal {R}}} ^{2}+(2r(z)+r(z)^{2})T^{2}. \end{aligned}$$

Therefore, the proof is complete. \(\square \)

Lemma 3.3

\(Y_{j},{{\mathcal {L}}}_{0}\), R, \({\overline{R}}\), \({{\mathcal {R}}} ^{2}\), and \(T^{2}\) are boundary operators.

Proof

Let \(\phi (z)=\phi (sz)\). Then

$$\begin{aligned} R\phi (z)=R(\phi (sz))=\sum _{j=1}^{n}z_{j}{\frac{\partial }{\partial z_{j}}} \phi (sz)=\sum _{j=1}^{n}sz_{j}{\frac{\partial \phi }{\partial z_{j}}} (sz)=(R\phi )(sz). \end{aligned}$$

Similarly, one has \({\overline{R}}\phi (z)=({\overline{R}}\phi )(sz)\), \(R {\overline{R}}\phi (z)=(R{\overline{R}}\phi )(sz)\), T and \(T^{2}\) are boundary operators.

Since

$$\begin{aligned} Y_{j}(\phi (sz))=|z|{\tilde{Y}}_{j}(\phi (sz))=s|z|({\tilde{Y}}_{j}\phi )(sz)=(Y_{j}\phi )(sz), \end{aligned}$$

one has

$$\begin{aligned} {\overline{Y}}_{j}Y_{j}(\phi (sz))=({\overline{Y}}_{j}Y_{j}\phi )(sz),\quad { {\mathcal {L}}}_{0}(\phi (sz))=({{\mathcal {L}}}_{0}\phi )(sz). \end{aligned}$$

This proves that \({{\mathcal {L}}}_{0}\) is a boundary operator. Therefore, the lemma is proved. \(\square \)

Proposition 3.4

Let \(u\in C^{2} (B_{n},B_{m}) .\) Then u is harmonic if and only if

$$\begin{aligned} (1-|u|^{2})4\left| z\right| ^{2}Lu+P\left[ u\right] =0, \end{aligned}$$
(3.5)

where

$$\begin{aligned} P\left[ u\right]= & {} P_{1}\left[ u\right] +P_{2}\left[ u\right] , \end{aligned}$$
(3.6)
$$\begin{aligned} P_{1}\left[ u\right]= & {} 4\sum _{j=1}^{n}\left( \left\langle Y_{j}u,u\right\rangle {\overline{Y}}_{j}u+\left\langle {\overline{Y}} _{j}u,u\right\rangle Y_{j}u\right) \end{aligned}$$
(3.7)

and

$$\begin{aligned} P_{2}\left[ u\right]&=2( 1-\left| z\right| ^{2}) \left( \left\langle {\mathcal {R}}u,u\right\rangle {\mathcal {R}}u-\left\langle Tu,u\right\rangle Tu\right) . \end{aligned}$$
(3.8)

Proof

Let h be the Bergman metric on \(B_{m}\). Then the Christoffel symbols of \( (B_{m},h)\) are:

$$\begin{aligned} \Gamma _{t\gamma }^{s}[u]=(1-|u|^{2})^{-1}({\overline{u}}^{\gamma }\delta _{ts}+{\overline{u}}^{t}\delta _{\gamma s}). \end{aligned}$$
(3.9)

By (1.3), it is easy to see that u is harmonic in Bergman metrics if and only if

$$\begin{aligned} (1-|u(z)|^{2})Lu^{s}+\sum _{j=1}^{n}\sum _{t=1}^{m}\left( u_{{\overline{j}}}^{s} {\overline{u}}^{t}(X_{j}u^{t})+u_{j}^{s}{\overline{u}}^{t}({\overline{X}} _{j}u^{t})\right) =0,\ \ 1\le s\le m. \nonumber \\ \end{aligned}$$
(3.10)

Let

$$\begin{aligned} P^{s}\left[ u\right] =4|z|^{2}\sum _{j=1}^{n}\sum _{t=1}^{m}\left( u_{ {\overline{j}}}^{s}{\overline{u}}^{t}(X_{j}u^{t})+u_{j}^{s}{\overline{u}}^{t}( {\overline{X}}_{j}u^{t})\right) , \ \ 1\le s\le m. \end{aligned}$$
(3.11)

Then u is harmonic if and only if (3.5) holds.

Note that

$$\begin{aligned}&\sum _{j=1}^{n}\sum _{t=1}^{m}|z|^{2}u_{{\overline{j}}}^{s}{\overline{u}} ^{t}(X_{j}u^{t}) \\&\quad =\sum _{j=1}^{n}\sum _{t=1}^{m}|z|^{2}u_{{\overline{j}}}^{s}{\overline{u}}^{t}( {\tilde{Y}}_{j}u^{t}+{\overline{z}}_{j}(1-|z|^{2})R_{0}u^{t}) \\&\quad =\sum _{j=1}^{n}\sum _{t=1}^{m}|z|^{2}[\overline{{\tilde{Y}}}_{j}u^{s}+z_{j} {\overline{R}}_{0}u^{s}]{\overline{u}}^{t}[{\tilde{Y}}_{j}u^{t}+{\overline{z}} _{j}(1-|z|^{2})R_{0}u^{t}] \\&\quad =\sum _{j=1}^{n}\sum _{t=1}^{m}{\overline{u}}^{t}{\overline{Y}} _{j}u^{s}\,Y_{j}u^{t}+\sum _{t=1}^{m}(1-|z|^{2}){\overline{u}}^{t}{\overline{R}} u^s\,Ru^{t}, \end{aligned}$$

one has

$$\begin{aligned} P^{s}[u]&=4\sum _{j=1}^{n}\sum _{t=1}^{m}\left( |z|^{2}u_{{\overline{j}}}^{s} {\overline{u}}^{t}(X_{j}u^{t})+|z|^{2}u_{j}^{s}{\overline{u}}^{t}({\overline{X}} _{j}u^{t})\right) \\&=4\sum _{j=1}^{n}\sum _{t=1}^{m}\left( {\overline{Y}}_{j}u^{s}\,{\overline{u}} ^{t}Y_{j}u^{t}+Y_{j}u^{s}\,{\overline{u}}^{t}{\overline{Y}}_{j}u^{t}\right) \\&\quad +4(1-|z|^{2})\sum _{t=1}^{m}[{\overline{R}}u^{s}\,{\overline{u}} ^{t}Ru^{t}+Ru^{s}\,{\overline{u}}^{t}{\overline{R}}u^{t}]. \end{aligned}$$

Since

$$\begin{aligned}&4(\left\langle Ru,u\right\rangle {\bar{R}}u+\left\langle {\bar{R}} u,u\right\rangle Ru) \\&\quad =\left\langle {\mathcal {R}}u+Tu,u\right\rangle \left( {\mathcal {R}}u-Tu\right) +\left\langle {\mathcal {R}}u-Tu,u\right\rangle \left( {\mathcal {R}}u+Tu\right) \\&\quad =2\left\langle {\mathcal {R}}u,u\right\rangle {\mathcal {R}}u-2\left\langle Tu,u\right\rangle Tu, \end{aligned}$$

one has

$$\begin{aligned} P\left[ u\right] =P_{1}\left[ u\right] +P_{2}\left[ u\right] . \end{aligned}$$

Therefore, the proof of the proposition is complete. \(\square \)

4 Computation of \(( 1-\left| u\right| ^{2}) 4|z|^2Lu \)

4.1 Computation of the Linear Part: \(4|z|^2 L u\)

Let \(\phi _{k}\) be radial and smooth, and let

$$\begin{aligned} \phi (z)=\sum _{k=0}^{\infty }\phi _{k}r(z)^{k}. \end{aligned}$$
(4.1)

Define

$$\begin{aligned} D_{0}[\phi ]={{\mathcal {L}}}_{0}\phi _{0}+2(n-1)\phi _{1}\quad \ \end{aligned}$$
(4.2)

and for \(k\ge 1\) with \(\phi _{-1}=0\),

$$\begin{aligned} \quad D_{k}[\phi ]&=2\left( k+1\right) \left( n-k-1\right) \phi _{k+1}+[{{\mathcal {L}}}_{0}-k(5k-1-2n)]\phi _{k} \nonumber \\&\quad +[2T^{2}-(k-1)(4k-5)]\phi _{k-1}+[T^{2}-(k-2)^{2}]\phi _{k-2}. \end{aligned}$$
(4.3)

Lemma 4.1

Let \(\phi \in C^{\infty }( {\overline{B}}_{n},{\overline{B}} _{m}) \) with the asymptotic expansion (4.1) near \(\partial B_{n}.\) Then

$$\begin{aligned} 4|z|^{2}L\phi (z)=\sum _{k=0}^{\infty }D_{k}\left[ \phi \right] r^{k}\quad \text { near }\partial B_{n}. \end{aligned}$$
(4.4)

Proof

Notice that

$$\begin{aligned} {{\mathcal {R}}}r(z)=|z|=r(z)+1,\quad {{\mathcal {R}}}\log (-r)={\frac{1+r}{r}} \end{aligned}$$
(4.5)

and

$$\begin{aligned} {\mathcal {R}}^{2}r^{k}&=k{\mathcal {R}}(r^{k-1}+r^{k}) \nonumber \\&=k(1+r)((k-1)r^{k-2}+kr^{k-1}) \nonumber \\&=k(k-1)r^{k-2}+k(2k-1)r^{k-1}+k^{2}r^{k}, \end{aligned}$$
(4.6)

we have

$$\begin{aligned}&(2r+r^{2}){\mathcal {R}}^{2}r^{k} \nonumber \\&\quad =2k(k-1)r^{k-1}+2k(2k-1)r^{k}+2k^{2}r^{k+1} \nonumber \\&\qquad +k(k-1)r^{k}+k(2k-1)r^{k+1}+k^{2}r^{k+2} \nonumber \\&\quad =2k(k-1)r^{k-1}+k(5k-3)r^{k}+k(4k-1)r^{k+1}+k^{2}r^{k+2}. \end{aligned}$$
(4.7)

Since \({\mathcal {L}}_{0}\) and T are boundary operators and \({{\mathcal {R}}} \phi _{k}=0\), by (4.5) , ( 4.6) and (4.7) ,

$$\begin{aligned}&4|z|^{2}L(\phi _{k}r^{k}) \\&\quad ={{\mathcal {L}}}_{0}\phi _{k}r^{k}+T^{2}\phi _{k}(2r^{k+1}+r^{k+2})+2(n-1)\phi _{k}kr^{k-1}(1+r) \\&\qquad -2k(k-1)\phi _{k}r^{k-1}-k(5k-3)\phi _{k}r^{k}-k(4k-1)\phi _{k}r^{k+1}-k^{2}\phi _{k}r^{k+2} \\&\quad =[{{\mathcal {L}}}_{0}\phi _{k}-k(5k-1-2n)\phi _{k}]r^{k}+2k\phi _{k}(n-k)r^{k-1} \\&\qquad +[2T^{2}-k(4k-1)]\phi _{k}r^{k+1}+[T^{2}-k^{2}]\phi _{k}r^{k+2}. \end{aligned}$$

Then

$$\begin{aligned}&4|z|^{2}L\phi (z) \\&\quad =\sum _{k=0}^{\infty }[{{\mathcal {L}}}_{0}-k(5k-1-2n)]\phi _{k}r^{k}+\sum _{k=0}^{\infty }2(k+1)\phi _{k+1}(n-k-1)r^{k} \\&\qquad +\sum _{k=1}^{\infty }[2T^{2}-(k-1)(4k-5)]\phi _{k-1}r^{k}+\sum _{k=2}^{\infty }[T^{2}-(k-2)^{2}]\phi _{k-2}r^{k} \\&\quad ={{\mathcal {L}}}_{0}\phi _{0}+2(n-1)\phi _{1} \\&\qquad +\left( 4\left( n-2\right) \phi _{2}+\left[ {\mathcal {L}}_{0}-\left( 4-2n\right) \right] \phi _{1}+T^{2}\phi _{0}\right) r \\&\qquad +\sum _{k=2}^{\infty }\left[ 2\left( k+1\right) \left( n-k-1\right) \phi _{k+1}+\left( {{\mathcal {L}}}_{0}-k(5k-1-2n)\right) \phi _{k}\right] r^{k} \\&\qquad +\sum _{k=2}^{\infty }\left[ \left( 2T^{2}-(k-1)(4k-5)\right) \phi _{k-1}+(T^{2}-(k-2)^{2})\phi _{k-2}\right] r^{k} \\&\quad =\sum _{k=0}^{\infty }D_{k}[\phi ]r^{k}. \end{aligned}$$

Therefore, the proof of the lemma is complete. \(\square \)

We will use the following notations:

$$\begin{aligned} \psi _{\ell }\left( z\right) =\sum _{k=0}^{\infty }\psi _{\ell ,k}r^{k}\quad \hbox { with }\ \psi _{\ell ,k}=0\ \hbox { when }k<\kappa _{\ell }:=(n+1)\ell \end{aligned}$$
(4.8)

and

$$\begin{aligned} u=\phi +\sum _{\ell =1}^{\infty }\psi _{\ell }\left( z\right) \left( \log \left( -r\right) \right) ^{\ell }. \end{aligned}$$
(4.9)

Let

$$\begin{aligned} {\mathcal {D}}_{k}=D_{k}[\phi ]+2\left( n-2\left( k+1\right) \right) \psi _{1,k+1}-4\psi _{2,k+1}+{\tilde{\xi }}_{1,k}[\psi _{1},\psi _{2}] \end{aligned}$$
(4.10)

and

$$\begin{aligned} {\mathcal {D}}_{\ell ,k}\left[ u\right]= & {} D_{k}[\psi _{\ell }]+2(\ell +1)\Big ( \big (n-2\left( k+1\right) \big )\psi _{\ell +1,k+1} \nonumber \\{} & {} -(\ell +2)\psi _{\ell +2,k+1})+{\tilde{\xi }}_{\ell +1,k}\Big ), \end{aligned}$$
(4.11)

where \(D_{k}\) is defined by (4.2) and (4.3),

$$\begin{aligned} \xi _{\ell ,k}[\psi _{\ell }]&=(2n+1)\psi _{\ell ,k}+\psi _{\ell ,k-1}, \end{aligned}$$
(4.12)
$$\begin{aligned} \eta _{\ell ,k}\left[ \psi _{\ell }\right]&=\ell \Big (5\psi _{\ell ,k}+4\psi _{\ell ,k-1}+\psi _{\ell ,k-2}\Big ), \end{aligned}$$
(4.13)
$$\begin{aligned} \zeta _{\ell ,k}&=10k\psi _{\ell ,k}+8\left( k-1\right) \psi _{\ell ,k-1}+2\left( k-2\right) \psi _{\ell ,k-2} \end{aligned}$$
(4.14)

and

$$\begin{aligned} {\tilde{\xi }}_{\ell ,k}[\psi _{\ell },\psi _{\ell +1}]=\frac{1}{2}\ell \Big ( \xi _{\ell ,k}[\psi _{\ell }]-\eta _{\ell +1,k}[\psi _{\ell +1}]-\zeta _{\ell ,k}\Big ). \end{aligned}$$
(4.15)

Notice that

$$\begin{aligned} {{\mathcal {R}}}(\log (-r))^{\ell }=\ell {\frac{1+r}{r}}(\log (-r))^{\ell -1} \end{aligned}$$

and

$$\begin{aligned} {{\mathcal {R}}}^{2}(\log (-r))^{\ell }=\ell (\ell -1){\frac{(1+r)^{2}}{r^{2}}} (\log (-r))^{\ell -2}-{\frac{1+r}{r^{2}}}\ell (\log (-r))^{\ell -1}. \end{aligned}$$

Let \(v_{\ell }=\psi _{\ell }(\log (-r))^{\ell }.\) Then

$$\begin{aligned} 4|z|^{2}Lv_{\ell }&=(4|z|^{2}L\psi _{\ell })(\log (-r))^{\ell }+2(n-1)\psi _{\ell }{\mathcal {R}}(\log (-r))^{\ell } \\&\quad -2(2r+r^{2}){\mathcal {R}}\psi _{\ell }{\mathcal {R}}(\log (-r))^{\ell }-(2r+r^{2})\psi _{\ell }{\mathcal {R}}^{2}(\log (-r))^{\ell } \\&=(4|z|^{2}L\psi _{\ell })(\log (-r))^{\ell }+2(n-1)\psi _{\ell }\ell { \frac{1+r}{r}}(\log (-r))^{\ell -1} \\&\quad -2\left( 2+r\right) \left( 1+r\right) \ell {\mathcal {R}}\psi _{\ell }\left( \log \left( -r\right) ^{\ell -1}\right) \\&\quad -(2+r){\frac{(1+r)^{2}}{r}}\ell (\ell -1)\psi _{\ell }(\log (-r))^{\ell -2}+(2+r)\ell \psi _{\ell }({\frac{1+r}{r}})(\log (-r))^{\ell -1} \\&=(4|z|^{2}L\psi _{\ell })(\log (-r))^{\ell }+\Big (2n{\frac{1+r}{r}}\psi _{\ell }+(1+r)\psi _{\ell }\Big )\ell (\log (-r))^{\ell -1} \\&\quad -(2+r){\frac{(1+r)^{2}}{r}}\ell (\ell -1)\psi _{\ell }(\log (-r))^{\ell -2}\\&\quad -2\left( 2+r\right) \left( 1+r\right) \ell {\mathcal {R}}\psi _{\ell }\left( \log \left( -r\right) ^{\ell -1}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} 4|z|^{2}L\sum _{\ell =1}^{\infty }v_{\ell }&=\sum _{\ell =1}^{\infty }(4|z|^{2}L\psi _{\ell })(\log (-r))^{\ell } \\&\quad +\sum _{\ell =1}^{\infty }\Big ({\frac{2n}{r}}\psi _{\ell +1}+(2n+1+r)\psi _{\ell +1}\Big )(\ell +1)(\log (-r))^{\ell } \\&\quad -\sum _{\ell =1}^{\infty }(2+r){\frac{(1+r)^{2}}{r}}(\ell +2)(\ell +1)\psi _{\ell +2}(\log (-r))^{\ell } \\&\quad -\sum _{\ell =1}^{\infty }2\left( 2+r\right) \left( 1+r\right) {\mathcal {R}} \psi _{\ell +1}(\ell +1)(\log (-r))^{\ell } \\&\quad +2n{\frac{1+r}{r}}\psi _{1}+(1+r)\psi _{1}-(2+r){\frac{(1+r)^{2}}{r}}2\psi _{2}-2\left( 2+r\right) \left( 1+r\right) {\mathcal {R}}_{{}}\psi _{1}. \end{aligned}$$

Notice that \(\psi _{\ell ,k}=0\) when \(k<\kappa _{\ell }=\ell (n+1)\) and

$$\begin{aligned}&\frac{2n}{r}\psi _{\ell +1}+(2n+1+r)\psi _{\ell +1} \\&\quad =2n\sum _{k=0}^{\infty }\psi _{\ell +1,k+1}r^{k}+(2n+1+r)\sum _{k=0}^{\infty }\psi _{\ell +1,k}r^{k} \\&\quad =\sum _{k=0}^{\infty }(2n\psi _{\ell +1,k+1}+(2n+1)\psi _{\ell ,k})r^{k}+\sum _{k=1}^{\infty }\psi _{\ell +1,k-1}r^{k} \\&\quad =2n\psi _{\ell +1,1}+\sum _{k=1}^{\infty }(2n\psi _{\ell +1,k+1}+(2n+1)\psi _{\ell +1,k}+\psi _{\ell +1,k-1})r^{k} \\&\quad =\sum _{k=1}^{\infty }(2n\psi _{\ell +1,k+1}+\xi _{\ell +1,k})r^{k}, \end{aligned}$$

where \(\xi _{\ell +1,k}\) is defined in (4.12) . Moreover,

$$\begin{aligned}{} & {} 2\left( 2+r\right) \left( 1+r\right) {\mathcal {R}}\psi _{\ell +1} \\{} & {} \quad =2\left( 2+r\right) \left( 1+r\right) \sum _{k=0}^{\infty }\psi _{\ell +1,k} {\mathcal {R}}_{{}}r^{k} \\{} & {} \quad =\sum _{k=2}^{\infty }\left( 4\left( k+1\right) \psi _{\ell +1,k+1}+10k\psi _{\ell +1,k}+8\left( k-1\right) \psi _{\ell +1,k-1}+2\left( k-2\right) \psi _{\ell +1,k-2}\right) r^{k} \\{} & {} \quad =\sum _{k=0}^{\infty }\left( 4\left( k+1\right) \psi _{\ell +1,k+1}+\zeta _{\ell +1,k}\right) r^k, \end{aligned}$$

where \(\zeta _{\ell +1,k}\) is defined in (4.14).

$$\begin{aligned}&(2+r)\frac{(1+r)^{2}}{r}(\ell +2)\psi _{\ell +2} \\&\quad =(\ell +2)({\frac{2}{r}}+5+4r+r^{2})2\psi _{\ell +2}(z) \\&\quad =\sum _{k=0}^{\infty }(2(\ell +2)\psi _{\ell +2,k+1}+(\ell +2)(5\psi _{\ell +2,k}+4\psi _{\ell +2,k-1}+2\psi _{\ell +2,k-2}))r^{k} \\&\quad =\sum _{k=0}^{\infty }(2(\ell +2)\psi _{\ell +2,k+1}+\eta _{\ell +2,k})r^{k}, \end{aligned}$$

where \(\eta _{\ell +2,k}\) is defined in (4.13). Therefore, with \(\ell \ge 0\),

$$\begin{aligned}&2n\frac{1+r}{r}\psi _{\ell +1}+\left( 1+r\right) \psi _{\ell +1}-\left( \ell +2\right) \left( 2+r\right) \frac{( 1+r) ^{2}}{r}2\psi _{\ell +2} \nonumber \\&\quad =\sum _{k=0}^{\infty }( 2n\psi _{\ell +1,k+1}-2( \ell +2) \psi _{\ell +2,k+1}+\xi _{\ell +1,k}\left[ \psi _{\ell +1}\right] -\eta _{\ell +2,k}\left[ \psi _{\ell +2}\right] ) r^{k} \end{aligned}$$
(4.16)

and

$$\begin{aligned}{} & {} 4|z|^{2}L\sum _{\ell =1}^{\infty }v_{\ell } =\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }D_{k}[\psi _{\ell }]r^{k}(\log (-r))^{\ell } \\{} & {} \qquad +\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }(\ell +1)[2( n-2k-2) \psi _{\ell +1,k+1}+\xi _{\ell +1,k}-2(\ell +2)\psi _{\ell +2,k+1}\\{} & {} \qquad -\eta _{\ell +2,k}-\zeta _{\ell +1,k}] r^{k}(\log (-r))^{\ell } \\{} & {} \qquad +\sum _{k=1}^{\infty }(2\left( n-2k-2\right) \psi _{1,k+1}-4\psi _{2,k+1}+\xi _{1,k}[\psi _{1}]-\eta _{2,k}[\psi _{2}]-\zeta _{1,k})r^{k}. \end{aligned}$$

Therefore,

$$\begin{aligned}&4|z|^{2}L\Big (\phi (z)+\sum _{\ell =1}^{\infty }\psi _{\ell }(\log (-r))^{\ell }\Big ) \\&\quad =\sum _{k=0}^{\infty }\Big (D_{k}[\phi ]+2\left( n-2\left( k+1\right) \right) \psi _{1,k+1}-4\psi _{2,k+1}+{\tilde{\xi }}_{1,k}\Big )r^{k} \\&\qquad +\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }\Big (D_{k}[\psi _{\ell }]\\&\qquad +2(\ell +1)\left( \big (n-2\left( k+1\right) \big )\psi _{\ell +1,k+1}-(\ell +2)\psi _{\ell +2,k+1}+{\tilde{\xi }}_{\ell +1,k}\right) \Big )r^{k}(\log (-r))^{\ell }, \end{aligned}$$

where \({\tilde{\xi }}_{\ell +1,k}\) is defined in (4.15) .

As a summary, we have proved the following proposition.

Proposition 4.2

Let \(\psi _{\ell ,k}=0\) when \(k<\kappa _{\ell }\) and let

$$\begin{aligned} u(z)=\sum _{k=0}^{\infty }\phi _{k}(z)r(z)^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }\psi _{\ell ,k}(z)r(z)^{k}(\log (-r(z)))^{\ell } \end{aligned}$$
(4.17)

in the sense of asymptotic expansion. Then

$$\begin{aligned} 4|z|^{2}Lu(z)=\sum _{k=0}^{\infty }{\mathcal {D}}_{k}r^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }{\mathcal {D}}_{\ell ,k}r^{k}(\log (-r))^{\ell }, \end{aligned}$$
(4.18)

where \({\mathcal {D}}_{k}\) and \(\mathcal {{\mathcal {D}}}_{\ell ,k}\) are defined in (4.10) and (4.11).

4.2 Computation of \((1-|u|^2) 4|z|^2 L u\)

Define

$$\begin{aligned} A_{p,q}\left[ \phi \right]&=\left\langle \phi _{p},\phi _{q}\right\rangle ,{\mathcal {A}}_{k}\left[ \phi \right] =\sum _{p+q=k}{\mathcal {A}}_{p,q}\left[ \phi \right] , \end{aligned}$$
(4.19)
$$\begin{aligned} {\mathcal {A}}_{\ell ,k}\left[ u\right]&=\sum _{\alpha +\beta =k}\left( \left\langle \phi _{\alpha },\psi _{\ell ,\beta }\right\rangle +\left\langle \psi _{\ell ,\alpha },\phi _{\beta }\right\rangle \right) +\sum _{\alpha +\beta =k}\sum _{s+t=\ell }\left\langle \psi _{s,\alpha },\psi _{t,\beta }\right\rangle , \end{aligned}$$
(4.20)
$$\begin{aligned} B_{k}\left[ u\right]&={\mathcal {D}}_{k}\left[ u\right] -\sum _{\alpha +\beta =k}{\mathcal {A}}_{\alpha }\left[ \phi \right] {\mathcal {D}}_{\beta }\left[ u \right] \end{aligned}$$
(4.21)

and

$$\begin{aligned} B_{\ell ,k}\left[ u\right]= & {} {\mathcal {D}}_{\ell ,k}\left[ u\right] -\sum _{\alpha +\beta =k} \Big ( {\mathcal {A}}_{\alpha }\left[ \phi \right] {\mathcal {D}}_{\ell ,\beta }\left[ u\right] + {\mathcal {A}}_{\ell ,\alpha } [u] {\mathcal {D}}_{\beta } [ u] \Big ) \nonumber \\{} & {} -\sum _{s+t=\ell }\sum _{\alpha +\beta =k} {\mathcal {A}}_{s,\alpha }\left[ u \right] {\mathcal {D}}_{t,\beta }\left[ u\right] . \end{aligned}$$
(4.22)

Proposition 4.3

Let u be the map defined by (4.17) with \( \left| \phi _{0}\right| =1\), then

$$\begin{aligned} 4(1-|u|^{2})|z|^{2}Lu=\sum _{k=0}^{\infty }B_{k}r^{k}+\sum _{\ell =1}^{\infty }\sum _{p=1}^{\infty }B_{\ell ,p}r^{p}(\log (-r))^{\ell }. \end{aligned}$$

Proof

Since

$$\begin{aligned} 1-|u(z)|^{2}&=1-\sum _{k=0}^{\infty }\sum _{\alpha +\beta =k}\left\langle \phi _{\alpha },\phi _{\beta }\right\rangle r^{k} \\&\quad -\sum _{\ell =0}^{\infty }\sum _{k=0}^{\infty }\sum _{\alpha +\beta =k}\left( \left\langle \phi _{\alpha },\psi _{\ell ,\beta }\right\rangle +\left\langle \psi _{\ell ,\alpha },\phi _{\beta }\right\rangle \right) r^{k}\left( \log \left( -r\right) \right) ^{\ell } \\&\quad -\sum _{\ell =0}^{\infty }\sum _{k=0}^{\infty }\sum _{\alpha +\beta =k}\sum _{s+t= \ell }\left\langle \psi _{s,\alpha },\psi _{t,\beta }\right\rangle r^{k}\left( \log \left( -r\right) \right) ^{\ell } \\&=-\sum _{k=1}^{\infty }{\mathcal {A}}_{k}\left[ \phi \right] r^{k}-\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }{\mathcal {A}}_{\ell ,k}r^{k}\left( \log \left( -r\right) \right) ^{\ell }, \end{aligned}$$

by Proposition 4.2,

$$\begin{aligned}&-4(1-|u|^{2})|z|^{2}Lu \nonumber \\&\quad =\Big ( \sum _{k=1}^{\infty }{\mathcal {A}}_{k}\left[ \phi \right] r^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }{\mathcal {A}}_{\ell ,k}\left[ u \right] r^{k}\left( \log \left( -r\right) \right) ^{\ell }\Big ) \cdot 4\left| z\right| ^{2}Lu \nonumber \\&\quad =\sum _{k=1}^{\infty }\sum _{\alpha +\beta =k}{\mathcal {A}}_{\alpha }\left[ \phi \right] {\mathcal {D}}_{\beta }\left[ u\right] r^{k} \nonumber \\&\qquad +\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }\sum _{\alpha +\beta =k}\left( {\mathcal {A}}_{\alpha }\left[ \phi \right] {\mathcal {D}}_{\ell ,\beta }\left[ u \right] +{\mathcal {A}}_{\ell ,\alpha }\left[ u\right] {\mathcal {D}}_{\beta }\left[ u\right] \right) r^{k}\left( \log \left( -r\right) \right) ^{\ell } \nonumber \\&\qquad +\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }\sum _{s+t=\ell }\sum _{\alpha +\beta =k}{\mathcal {A}}_{s,\alpha }\left[ u\right] {\mathcal {D}}_{t,\beta }\left[ u \right] r^{k}\left( \log \left( -r\right) \right) ^{\ell } \nonumber \\&\quad =-\sum _{k=1}^{\infty }B_{k}\left[ \phi \right] r^{k}-\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }B_{\ell ,k}r^{k}\left( \log \left( -r\right) \right) ^{\ell }. \end{aligned}$$
(4.23)

The proof of the proposition is complete. \(\square \)

It is easy to prove the following lemma:

Lemma 4.4

Let u be the map defined by (4.17). Let \({\mathcal {H}}\) and \({\mathcal {K}}_{{}}\) be operators on u such that

$$\begin{aligned} {\mathcal {H}}\left[ u\right] =\sum _{k=0}^{\infty }{\mathcal {H}}_{k}\left[ u \right] r^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }{\mathcal {H}}_{\ell ,k}\left[ u\right] r^{k}\left( \log \left( -r\right) \right) ^{\ell } \end{aligned}$$

and

$$\begin{aligned} {\mathcal {K}}\left[ u\right] =\sum _{k=0}^{\infty }{\mathcal {K}}_{k}\left[ u \right] r^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }{\mathcal {K}}_{\ell ,k}\left[ u\right] r^{k}\left( \log \left( -r\right) \right) ^{\ell }. \end{aligned}$$

Then

$$\begin{aligned} \left\langle \mathcal {{\mathcal {H}}}\left[ u\right] ,u\right\rangle {\mathcal {K}} \left[ u\right] =\sum _{k=0}^{\infty }W_{k}\left[ {\mathcal {H}}_{{}},{\mathcal {K}} _{{}},u\right] r^{k}+\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }W_{\ell ,k} \left[ {\mathcal {H}}_{{}},{\mathcal {K}}_{{}},u\right] r^{k}\left( \log \left( -r\right) \right) ^{\ell }, \end{aligned}$$

where

$$\begin{aligned} W_{k}\left( {\mathcal {H}}_{{}},{\mathcal {K}},u\right) =\sum _{p+q+j=k}\left\langle {\mathcal {H}}_{p}\left[ u\right] ,\phi _{q}\right\rangle {\mathcal {K}}_{j}\left[ u\right] \end{aligned}$$

and

$$\begin{aligned}&W_{\ell ,k}\left( {\mathcal {H}}_{{}},{\mathcal {K}}_{{}},u\right) \\&\quad =\sum _{p+q+j=k}\left( \left\langle {\mathcal {H}}_{p}\left[ u\right] ,\phi _{q}\right\rangle {\mathcal {K}}_{\ell ,j}\left[ u\right] +\left\langle {\mathcal {H}}_{\ell ,p}\left[ u\right] ,\phi _{q}\right\rangle {\mathcal {K}}_{j} \left[ u\right] +\left\langle {\mathcal {H}}_{p}\left[ u\right] ,\psi _{\ell ,q}\right\rangle {\mathcal {K}}_{j}\left[ u\right] \right) \\&\qquad +\sum _{p+q+j=k}\sum _{\alpha +\beta =\ell }\left( \left\langle {\mathcal {H}} _{\alpha ,p}\left[ u\right] ,\phi _{q}\right\rangle {\mathcal {K}}_{\beta ,j} \left[ u\right] +\left\langle {\mathcal {H}}_{\alpha ,p}\left[ u\right] ,\psi _{\beta ,q}\right\rangle {\mathcal {K}}_{j}\left[ u\right] \right. \\&\qquad \left. +\left\langle {\mathcal {H}}_{p}\left[ u\right] ,\psi _{\alpha ,q}\right\rangle {\mathcal {K}} _{\beta ,j}\left[ u\right] \right) \\&\qquad +\sum _{p+q+j=k}\sum _{\alpha +\beta +\gamma =\ell }\left\langle {\mathcal {H}} _{\alpha ,p}\left[ u\right] ,\psi _{\beta ,q}\right\rangle {\mathcal {K}} _{\gamma ,j}\left[ u\right] . \end{aligned}$$

5 Computation of P[u]

Since \({\mathcal {R}}\left( r^{k}\right) =kr^{k-1}(1+r)\),

$$\begin{aligned} {\mathcal {R}}\left( r^{k}(\log (-r)\right) ^{\ell })=kr^{k-1}(1+r)(\log (-r))^{\ell }+\ell r^{k-1}(1+r)(\log (-r))^{\ell -1} \nonumber \\ \end{aligned}$$
(5.1)

and \(\psi _{\ell ,k}=0\) when \(k<\kappa _{\ell },\) one has

$$\begin{aligned} {\mathcal {R}}u&=\sum _{k=0}^{\infty }\ k\phi _{k}(r^{k-1}+r^{k})+\sum _{k=1}^{\infty }\psi _{1,k}(r^{k-1}+r^{k}) \nonumber \\&\quad +\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }\psi _{\ell ,k}k(r^{k-1}+r^{k})(\log (-r))^{\ell }+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }\psi _{\ell +1,k}(\ell +1)(r^{k-1}+r^{k})(\log (-r))^{\ell } \nonumber \\&=\sum _{k=0}^{\infty }\Big (\left( k+1\right) \phi _{k+1}+k\phi _{k}+\psi _{1,k+1}+\psi _{1,k}\Big )r^{k} \nonumber \\&\quad +\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }\Big ( (k+1)\psi _{\ell ,k+1}+k\psi _{\ell ,k}+(\ell +1)(\psi _{\ell +1,k+1}+\psi _{\ell +1,k})\Big ) r^{k}\left( \log \left( -r\right) \right) ^{\ell } \nonumber \\&=\sum _{k=0}^{\infty }F_{k}\left[ u\right] r^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }F_{\ell ,k}\left[ u\right] r^{k}(\log (-r))^{\ell }, \end{aligned}$$
(5.2)

where

$$\begin{aligned} F_{k}\left[ u\right] =(k+1)\phi _{k+1}+k\phi _{k}+\psi _{1,k+1}+\psi _{1,k} \end{aligned}$$
(5.3)

and

$$\begin{aligned} F_{\ell ,k}\left[ u\right] =(k+1)\psi _{\ell ,k+1}+k\psi _{\ell ,k}+(\ell +1)(\psi _{\ell +1,k+1}+\psi _{\ell +1,k}). \end{aligned}$$
(5.4)

Since \(F_{\ell ,k}\left[ u\right] =0\) when \(k<\kappa _{\ell }-1\), applying Lemma 4.4, we have

$$\begin{aligned}&\left\langle {\mathcal {R}}_{{}}u,u\right\rangle {\mathcal {R}}_{{}}u \nonumber \\&\quad =\sum _{k=0}^{\infty }W_{k}\left[ {\mathcal {R}}_{{}},{\mathcal {R}}_{{}},u\right] r^{k}+\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }W_{\ell ,k}\left[ \mathcal { R}_{{}},{\mathcal {R}}_{{}},u\right] r^{k}\left( \log \left( -r\right) \right) ^{\ell }. \end{aligned}$$
(5.5)

Then

$$\begin{aligned}&2\left( 1-\left| z\right| ^{2}\right) \left\langle {\mathcal {R}} _{{}}u,u\right\rangle {\mathcal {R}}_{{}}u \nonumber \\&\quad =-2\left( 2r+r^{2}\right) \left\langle {\mathcal {R}}_{{}}u,u\right\rangle {\mathcal {R}}_{{}}u \nonumber \\&\quad =\sum _{k=1}^{\infty }E_{k}r^{k}+\sum _{\ell =0}^{\infty }\sum _{k=1}^{\infty }E_{\ell ,k}r^{k}\left( \log \left( -r\right) \right) ^{\ell }, \end{aligned}$$
(5.6)

where

$$\begin{aligned} E_{k}=-4W_{k-1}\left[ {\mathcal {R}}_{{}},{\mathcal {R}}_{{}},u\right] -2W_{k-2} \left[ {\mathcal {R}}_{{}},{\mathcal {R}}_{{}},u\right] \end{aligned}$$
(5.7)

and

$$\begin{aligned} E_{\ell ,k}=-4W_{\ell ,k-1}\left[ {\mathcal {R}}_{{}},{\mathcal {R}}_{{}},u\right] -2W_{\ell ,k-2}\left[ {\mathcal {R}}_{{}},{\mathcal {R}}_{{}},u\right] . \end{aligned}$$
(5.8)

Let

$$\begin{aligned} G_{k}&=-4W_{k-1}\left( T_{{}},T_{{}},u\right) -2W_{k-2}\left( T_{{}},T_{{}},u\right) , \end{aligned}$$
(5.9)
$$\begin{aligned} G_{\ell ,k}&=-4W_{\ell ,k-1}\left( T_{{}},T _{{}},u\right) -2W_{\ell ,k-2}\left( T_{{}},T _{{}},u\right) \end{aligned}$$
(5.10)

and let

$$\begin{aligned} C_{k}&=4\Big (\sum _{j}W_{k}\left( {\bar{Y}}_{j},Y_{j},u\right) +\sum _{j}W_{k}\left( Y_{j},{\bar{Y}}_{j},u\right) \Big ), \end{aligned}$$
(5.11)
$$\begin{aligned} C_{\ell ,k}&=4\Big (\sum _{j}W_{\ell ,k}\left( {\bar{Y}}_{j},Y_{j},u\right) +\sum _{j}W_{\ell ,k}\left( Y_{j},{\bar{Y}}_{j},u\right) \Big ). \end{aligned}$$
(5.12)

Then, by similar computations, one has that

$$\begin{aligned} 2(1-\left| z\right| ^{2})\left\langle Tu,u\right\rangle Tu=\sum _{k=1}^{\infty }G_{k}r^{k}+\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }G_{\ell ,k}r^{k}\Big (\log (-r)\Big )^{\ell } \end{aligned}$$
(5.13)

and

$$\begin{aligned} 4\sum _{j=1}^{n}( \langle Y_{j}u,u\rangle {\overline{Y}}_{j}u+\langle {\overline{Y}} _{j}u,u\rangle Y_{j}u )=\sum _{k=0}^{\infty }C_{k}r^{k}+\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }C_{\ell ,k}r^{k} ( \log ( -r ) ) ^{\ell }. \nonumber \\ \end{aligned}$$
(5.14)

Using equations (5.6), (5.13), (5.14), and (1.4), we have proved the following proposition.

Proposition 5.1

With notations above, one has

$$\begin{aligned} P\left[ u\right] =\sum _{k=1}^{\infty }\left( C_{k}+E_{k}-G_{k}\right) r^{k}+\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }\left( C_{\ell ,k}+E_{\ell ,k}-G_{\ell ,k}\right) r^{k}\left( \log \left( -r\right) \right) ^{\ell }. \end{aligned}$$

6 Proof of Theorem 1.3

Let

$$\begin{aligned} u\left( z\right) =\sum _{k=0}^{\infty }\phi _{k}r^{k}+\sum _{\ell =1}^{\infty }\sum _{k=1}^{\infty }\psi _{\ell ,k}r^{k}\left( \log \left( -r\right) \right) ^{\ell } \end{aligned}$$

which is given by (4.17), where \(\psi _{\ell ,k}=0\) for \( k<\kappa _{\ell }=\ell (n+1)\). It is well known from [3] and [18] that if \( u\in C^2({\overline{B}}_n)\) is proper harmonic with \(u|_{\partial B_n}=\phi _0\), then

$$\begin{aligned} |\phi _{0}(z)|^{2}=1\quad \hbox {and}\quad \langle Y_{j}\phi _{0},\phi _{0}\rangle =\langle {\overline{Y}}_{j}\phi _{0},\phi _{0}\rangle =0, j=1,\cdots ,n. \end{aligned}$$
(6.1)

Combining Proposition 4.3 and Proposition 5.1, one has proved the following theorem

Theorem 6.1

Under the assumption (6.1), one has

$$\begin{aligned}&4\left( 1-\left| z\right| ^{2}\right) \left| z\right| ^{2}Lu+P\left[ u\right] \\&\quad =\sum _{k=1}^{\infty }\left( B_{k}+C_{k}+E_{k}-G_{k}\right) r^{k}+\sum _{k=1}^{\infty }\sum _{\ell =1}^{\infty }\left( B_{\ell ,k}+C_{\ell ,k}+E_{\ell ,k}-G_{\ell ,k}\right) r^{k}\left( \log \left( -r\right) \right) ^{\ell }, \end{aligned}$$

where \(B_{k}\) and \(B_{\ell ,k}\) are given by (4.21) and (4.22); \(E_{k}\) and \(E_{\ell ,k}\) are given by (5.7) and (5.8); \( G_{k},G_{\ell ,k}\) are given by (5.9) and (5.10) and \(C_{k}\) and \(C_{\ell ,k}\) are given by (5.11), (5.12).

From the definition of \(B_{\ell ,k},E_{\ell ,k},G_{\ell ,k}\), and \(C_{\ell ,k} \), one has

$$\begin{aligned} B_{\ell ,k}=C_{\ell ,k}=E_{\ell ,k}=G_{\ell ,k}=0,\text { for }k<\kappa _{\ell }. \end{aligned}$$
(6.2)

Theorem 6.1 implies that if u is proper harmonic then (6.1) and

$$\begin{aligned} \left\{ \begin{array}{l} B_{k}+C_{k}+E_{k}-G_{k}=0,\quad k\ge 1 \\ B_{\ell ,k}+C_{\ell ,k}+E_{\ell ,k}-G_{\ell ,k}=0,\quad k\ge \kappa _{\ell } \end{array} \right. \end{aligned}$$
(6.3)

hold. We will solve \(\phi _k \) and \(\psi _{\ell , k}\) through the above system of equations.

6.1 \(B_{k}+E_{k}-G_{k}+C_{k}=0\) when \(1\le k\le n\)

Since

$$\begin{aligned} B_{1}=-{\mathcal {A}}_{1}D_{0} \end{aligned}$$

and

$$\begin{aligned} B_{k}= & {} -\sum _{q=1}^{k}{\mathcal {A}}_{q}D_{k-q}[\phi ]=-{\mathcal {A}} _{k}D_{0}-\sum _{p=1}^{k-1}{{\mathcal {A}}}_{p}D_{k-p}[\phi ],\quad k\ge 2,\\ D_{k}[\phi ]= & {} 2e_{k+1}\phi _{k+1}+[{{\mathcal {L}}}_{0}-k(5k-1-2n)]\phi _{k} \\{} & {} +[2T^{2}-(k-1)(4k-5)]\phi _{k-1}+[T^{2}-(k-2)^{2}]\phi _{k-2},\quad k\ge 1, \end{aligned}$$

where \(e_{k}=k\left( n-k\right) .\)

For \(E_k\), one has

$$\begin{aligned} E_{1}=-4\langle \phi _{1},\phi _{0}\rangle \phi _{1} \end{aligned}$$

and

$$\begin{aligned} E_{k}&=-4\sum _{p+q+s=k-1}\langle F_{p},\phi _{q}\rangle F_{s}-2\sum _{p+q+s=k-2}\langle F_{p},\phi _{q}\rangle F_{s} \\&=-4k\langle \phi _{k},\phi _{0}\rangle \phi _{1}-4k\langle \phi _{1},\phi _{0}\rangle \phi _{k}+{\tilde{E}}_{k}[\phi _{0},\cdots ,\phi _{k-1}],\quad k>1. \end{aligned}$$

For \(G_{k}\), one has

$$\begin{aligned} G_{1}=-4\langle T\phi _{0},\phi _{0}\rangle T \phi _{0} \end{aligned}$$
(6.4)

and if \(k>1,\) then

$$\begin{aligned} G_{k}&=-2(\sum _{\alpha +\beta +\gamma =k-1}2\langle T\phi _{\alpha },\phi _{\beta }\rangle T\phi _{\gamma }+\sum _{\alpha +\beta +\gamma =k-2}\langle T\phi _{\alpha },\phi _{\beta }\rangle T\phi _{\gamma })\quad \nonumber \\&=G_{k}[\phi _{0},\cdots ,\phi _{k-1}]. \end{aligned}$$
(6.5)

For \(C_{k}\), one has

$$\begin{aligned} C_{k}&=\sum _{p=0}^{k-1}\sum _{q+\gamma =k-p}C[\phi _{p},\phi _{q},\phi _{\gamma }] \nonumber \\&=4\sum _{j=1}^{n}\Big (Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{k},\phi _{0}\rangle +{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{k},\phi _{0}\rangle +Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{k}\rangle +{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},\phi _{k}\rangle \Big ) \nonumber \\&\quad +4\sum _{p=1}^{k-1}\sum _{q+\gamma =k-p}C[\phi _{p},\phi _{q},\phi _{\gamma }] \nonumber \\&=-4\sum _{j=1}^{n}\Big (Y_{j}\phi _{0}\langle \phi _{k},Y_{j}\phi _{0}\rangle +{\overline{Y}}_{j}\phi _{0}\langle \phi _{k},{\overline{Y}}_{j}\phi _{0}\rangle -Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{k}\rangle -{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},\phi _{k}\rangle \Big ) \nonumber \\&\quad +4\sum _{j=1}^{n}(Y_{j}\phi _{0}{\overline{Y}}_{j}A_{k,0}+{\overline{Y}} _{j}\phi _{0}Y_{j}A_{k,0})+4\sum _{p=1}^{k-1}\sum _{q+\gamma =k-p}C[\phi _{p},\phi _{q},\phi _{\gamma }], \end{aligned}$$
(6.6)

where

$$\begin{aligned} C\left[ \phi _{p},\phi _{q},\phi _{\gamma }\right] =\sum _{j=1}^{n}\left( Y_{j}\phi _{p}\left\langle {\bar{Y}}_{j}\phi _{q},\phi _{\gamma }\right\rangle +{\bar{Y}}_{j}\phi _{p}\left\langle Y_{j}\phi _{q},\phi _{\gamma }\right\rangle \right) . \end{aligned}$$
(6.7)

6.1.1 Case 1. \(k=1\)

Since \(B_{1}+E_{1}-G_{1}+C_{1}=0,\) one has

$$\begin{aligned} 0&=-{\mathcal {A}}_{1}D_{0}-4\langle \phi _{1},\phi _{0}\rangle \phi _{1}+4\langle T\phi _{0},\phi _{0}\rangle T\phi _{0} \\&\quad +\sum _{j=1}^{n}\left( 4Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{1},\phi _{0}\rangle +4{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{1},\phi _{0}\rangle +4Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle +4{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},\phi _{1}\rangle \right) . \end{aligned}$$

Since \(\langle Y_{j}\phi _{0},\phi _{0}\rangle =\langle {\overline{Y}}_{j}\phi _{0},\phi _{0}\rangle =0\) and \(D_{0}={{\mathcal {L}}}_{0}\phi _{0}+2(n-1)\phi _{1}\), one has

$$\begin{aligned} 0&=-{\mathcal {A}}_{1}\langle D_{0},\phi _{0}\rangle -4\langle \phi _{1},\phi _{0}\rangle \langle \phi _{1},\phi _{0}\rangle +4\langle T\phi _{0},\phi _{0}\rangle \langle T\phi _{0},\phi _{0}\rangle \\&=-{\mathcal {A}}_{1}(\langle {{\mathcal {L}}}_{0}\phi _{0},\phi _{0}\rangle +2(n-1)\langle \phi _{1},\phi _{0}\rangle )-4A_{1,0}^{2}+4\langle T\phi _{0},\phi _{0}\rangle \langle T\phi _{0},\phi _{0}\rangle \\&=-{\mathcal {A}}_{1}(-2E_{b}[\phi _{0}]+2(n-1)A_{1,0})-4A_{1,0}^{2}+4\langle T\phi _{0},\phi _{0}\rangle ^{2}. \end{aligned}$$

Notice that \(E_{b}[\phi _{0}]>0,\ {\mathcal {A}}_{1}=\lim _{r\rightarrow 0^-}\frac{ 1-\left| u\right| ^{2}}{(-r)}>0\) and \(\langle T\phi _{0},\phi _{0}\rangle ^{2}\ge 0\), one can easily see that \(A_{1,0}\) is real and \({ {\mathcal {A}}}_{1}=2A_{1,0}\). Therefore,

$$\begin{aligned} 0=-2E_{b}[\phi _{0}]A_{1,0}+2nA_{1,0}^{2}-2\langle T\phi _{0},\phi _{0}\rangle ^{2} \end{aligned}$$

and thus

$$\begin{aligned} A_{1,0}={\frac{E_{b}[\phi _{0}]+\sqrt{E_{b}[\phi _{0}]^{2}+4n\langle T\phi _{0},\phi _{0}\rangle ^{2}}}{2n}},\quad n\ge 2. \end{aligned}$$
(6.8)

Therefore,

$$\begin{aligned} 0&=-{\mathcal {A}}_{1}({{\mathcal {L}}}_{0}\phi _{0}+2(n-1)\phi _{1})-2{\mathcal {A}} _{1}\phi _{1}+4\langle T\phi _{0},\phi _{0}\rangle T\phi _{0} \nonumber \\&\quad +\sum _{j=1}^{n}\left( 4Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{1},\phi _{0}\rangle +4{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{1},\phi _{0}\rangle +4Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle +4{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},\phi _{1}\rangle \right) \nonumber \\&=-2n{\mathcal {A}}_{1}\phi _{1}-{\mathcal {A}}_{1}{{\mathcal {L}}}_{0}\phi _{0}+4\langle T\phi _{0},\phi _{0}\rangle T\phi _{0}+\sum _{j=1}^{n}\left( 2Y_{j}\phi _{0}{\overline{Y}}_{j}{\mathcal {A}}_{1}+2{\overline{Y}}_{j}\phi _{0}Y_{j}{\mathcal {A}}_{1}\right) \nonumber \\&\quad -4\sum _{j=1}^{n}(Y_{j}\phi _{0}\langle \phi _{1},Y_{j}\phi _{0}\rangle + {\overline{Y}}_{j}\phi _{0}\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle )+4\sum _{j=1}^{n}(Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle +{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},\phi _{1}\rangle ). \end{aligned}$$
(6.9)

Let

$$\begin{aligned} h[\phi _{0}]=\left[ \begin{matrix} {\overline{Y}}\phi _{0} &{} -Y\phi _{0}\\ -{\overline{Y}}{\overline{\phi }}_{0} &{} Y {\overline{\phi }}_{0}\\ \end{matrix} \right] . \end{aligned}$$
(6.10)

Then

$$\begin{aligned} h[\phi _{0}]h[\phi _0 ]^{*}&=\left[ \begin{matrix} {\overline{Y}}\phi _{0} &{} -Y\phi _{0}\\ -{\overline{Y}}{\overline{\phi }}_{0} &{} Y {\overline{\phi }}_{0}\\ \end{matrix} \right] \left[ \begin{matrix} ({\overline{Y}}\phi _{0})^{*} &{} -({\overline{Y}}{\overline{\phi }}_{0})^{*} \\ -(Y\phi _{0})^{*} &{} (Y{\overline{\phi }}_{0})^{*}\\ \end{matrix} \right] \\&=\left[ \begin{matrix} {\overline{Y}}\phi _{0}({\overline{Y}}\phi _{0})^{*}+(Y\phi _{0})((Y\phi _{0})^{*} &{} -({\overline{Y}}\phi _{0})({\overline{Y}}{\overline{\phi }} _{0})^{*}-Y\phi _{0}(Y{\overline{\phi }}_{0})^{*}\\ -{\overline{Y}} {\overline{\phi }}_{0}({\overline{Y}}\phi _{0})^{*}-Y{\overline{\phi }} _{0}(Y\phi _{0})^{*} &{} ({\overline{Y}}{\overline{\phi }}_{0})({\overline{Y}} {\overline{\phi }}_{0})^{*}+(Y{\overline{\phi }}_{0})(Y{\overline{\phi }} _{0})^{*}\\ \end{matrix} \right] . \end{aligned}$$

Therefore, (6.9) implies that

$$\begin{aligned} \Big (n{\mathcal {A}}_{1}I_{2m}+2h[\phi _{0}]h[\phi _{0}]^{*}\Big )\left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix} \right] =\left[ \begin{matrix} \Phi _{0}\\ {\overline{\Phi }}_{0}\\ \end{matrix} \right] , \end{aligned}$$
(6.11)

where

$$\begin{aligned} \Phi _{0}[\phi _{0}]=-{\frac{{{\mathcal {A}}}_{1}}{2}}{{\mathcal {L}}}_{0}\phi _{0}+2\langle T\phi _{0},\phi _{0}\rangle T\phi _{0}+\sum _{j=1}^{n}\left( Y_{j}\phi _{0}\overline{ Y}_{j}{{\mathcal {A}}}_{1}+{\overline{Y}}_{j}\phi _{0}Y_{j}{{\mathcal {A}}}_{1}\right) . \end{aligned}$$

Notice that \(n{{\mathcal {A}}}_{1}I_{2m}+2h[\phi _{0}]h[\phi _{0}]^{*}\) is invertible, we have

$$\begin{aligned} \left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix} \right] =\Big (n{\mathcal {A}}_{1}I_{2m}+2h[\phi _{0}]h[\phi _{0}]^{*}\Big ) ^{-1}\left[ \begin{matrix} \Phi _{0}\\ {\overline{\Phi }}_{0}\\ \end{matrix} \right] . \end{aligned}$$
(6.12)

6.1.2 Case 2. \(1<k\le n\)

Notice that

$$\begin{aligned} D_{k}=2e_{k+1}\phi _{k+1}+{\tilde{D}}_{k}[\phi _{k},\phi _{k-1},\phi _{k-2}]. \end{aligned}$$

Then

$$\begin{aligned} B_{k}+E_{k}-G_{k}+C_{k}=0 \end{aligned}$$

implies that

$$\begin{aligned} 0&=-(A_{k,0}+A_{0,k})D_{0}-{\mathcal {A}}_{1}2e_{k}\phi _{k}-4kA_{k,0}\phi _{1}-2k{\mathcal {A}}_{1}\phi _{k} \\&\quad +4\sum _{j=1}^{n}\Big (-Y_{j}\phi _{0}\langle \phi _{k},Y_{j}\phi _{0}\rangle -{\overline{Y}} _{j} \phi _0 \langle \phi _{k},{\overline{Y}}_{j}\phi _{0}\rangle +Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{k}\rangle +{\overline{Y}} _{j} \phi _0 \langle Y_{j}\phi _{0},\phi _{k}\rangle \Big ) \\&\quad +4\sum _{j=1}^{n}(Y_{j}\phi _{0}{\overline{Y}}_{j}A_{k,0}+{\overline{Y}}_j \phi _{0}Y_{j}A_{k,0})+H_{k}[\phi _{0},\cdots ,\phi _{k-1}]. \end{aligned}$$

In particular,

$$\begin{aligned}&\langle H_{k}[\phi _{0},\cdots ,\phi _{k-1}],\phi _{0}\rangle \\&\quad =2{\textrm{Re}}\,A_{k,0}\,\langle D_{0},\phi _{0}\rangle +2e_{k}{\mathcal {A}}_{1}A_{k,0}+4kA_{k,0}{\mathcal {A}}_{1} \\&\quad =2{\textrm{Re}}\,A_{k,0}\,(-2E_{b}[\phi _{0}]+(n-1){\mathcal {A}} _{1})+2k(n+2-k){\mathcal {A}}_{1}A_{k,0}. \end{aligned}$$

This implies

$$\begin{aligned} {\textrm{Im}}\,A_{k,0}={\frac{{\textrm{Im}}\,\langle H_{k}[\phi _{0},\cdots ,\phi _{k-1}],\phi _{0}\rangle }{2k(n+2-k){{\mathcal {A}}}_{1}}} \end{aligned}$$

and

$$\begin{aligned} 2{\textrm{Re}}\,A_{k,0}={\frac{{\textrm{Re}}\,\langle H_{k}[\phi _{0},\cdots ,\phi _{k-1}],\phi _{0}\rangle }{(k(n+2-k)+n-1){{\mathcal {A}}}_{1}-2E_{b}[\phi _{0}]}}. \end{aligned}$$

Therefore, with \(k(n+2-k)+n-1=e_{k}+2k+n-1\),

$$\begin{aligned} A_{k,0}={\frac{1}{2}}{\frac{{\textrm{Re}}\,\langle H_{k}[\phi _{0},\cdots ,\phi _{k-1}],\phi _{0}\rangle }{(e_{k}+2k+n-1){\mathcal {A}}_{1}-2E_{b}[\phi _{0}]}}+i{\frac{{\textrm{Im}}\,\langle H_{k}[\phi _{0},\cdots ,\phi _{k-1}],\phi _{0}\rangle }{2k(n+2-k){\mathcal {A}}_{1}}}.\quad \qquad \end{aligned}$$
(6.13)

Therefore

$$\begin{aligned}&(e_{k}+k){\mathcal {A}}_{1}\phi _{k} \\&\quad =-2\sum _{j=1}^{n}\Big (Y_{j}\phi _{0}\langle \phi _{k},Y_{j}\phi _{0}\rangle +{\overline{Y}} _{j} \phi _0 \langle \phi _{k},{\overline{Y}}_{j}\phi _{0}\rangle -Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},\phi _{k}\rangle -{\overline{Y}} _{j} \phi _0 \langle Y_{j}\phi _{0},\phi _{k}\rangle \Big ) \\&\qquad +\Phi _{k-1}[\phi _{0},\cdots ,\phi _{k-1}]. \end{aligned}$$

This implies

$$\begin{aligned} \Big (k(n+1-k){\mathcal {A}}_{1}I_{2m}+2h[\phi _{0}]h[\phi _{0}]^{*}\Big ) \left[ \begin{matrix} \phi _{k}\\ {\overline{\phi }}_{k}\\ \end{matrix} \right] =\left[ \begin{matrix} \Phi _{k-1}\\ {\overline{\Phi }}_{k-1}\\ \end{matrix} \right] \end{aligned}$$
(6.14)

and

$$\begin{aligned} \left[ \begin{matrix} \phi _{k}\\ {\overline{\phi }}_{k}\\ \end{matrix} \right] =(k(n+1-k){\mathcal {A}}_{1}I_{2m}+2h[\phi _{0}]h[\phi _{0}]^{*})^{-1}\left[ \begin{matrix} \Phi _{k-1}\\ {\overline{\Phi }}_{k-1}\\ \end{matrix} \right] . \end{aligned}$$
(6.15)

6.2 Case: \(k\ge n+1\)

6.2.1 \(B_{\ell , k}+E_{\ell , k}-G_{\ell ,k}+C_{\ell , k}=0 \)

For \(\kappa _{\ell }\le k<\kappa _{\ell +1}\), notice that

$$\begin{aligned} B_{1,\kappa _{1}}=-D_{0}2{\textrm{Re}}\,\langle \psi _{1 ,\kappa _1},\phi _{0}\rangle -2{\mathcal {A}}_1 e_{\kappa _1}\psi _{1,\kappa _1} \end{aligned}$$
(6.16)

and

$$\begin{aligned} B_{\ell ,k}=-D_{0}2{\textrm{Re}}\,\langle \psi _{\ell ,k},\phi _{0}\rangle - {\mathcal {A}}_{1}{\mathcal {D}}_{\ell ,k-1}+{\tilde{B}}_{\ell ,k-1}, \end{aligned}$$
(6.17)

where

$$\begin{aligned} {\tilde{B}}_{\ell ,k-1}={\tilde{B}}_{\ell ,k-1}[\phi _{0},\cdots ,\phi _{k-1},\psi _{p,q}:p\le \ell ,q<k]. \end{aligned}$$
(6.18)

When \(k<\kappa _{\ell +1}\), one has

$$\begin{aligned} E_{\ell ,k}=-4k\langle \phi _{1},\phi _{0}\rangle \psi _{\ell ,k}-4k\langle \psi _{\ell ,k},\phi _{0}\rangle \phi _{1}+{\tilde{E}}_{\ell ,k-1}, \end{aligned}$$
(6.19)

where

$$\begin{aligned} {\tilde{E}}_{\ell ,k-1}={\tilde{E}}_{\ell ,k-1}[\phi _{0},\cdots ,\phi _{k-1},\psi _{p,q}:p\le \ell ,q<k]. \end{aligned}$$
(6.20)

Since

$$\begin{aligned} C_{\ell ,k}=4C[\phi _{0},\phi _{0},\psi _{\ell ,k}]+4C[\phi _{0},\psi _{\ell ,k},\phi _{0}]+{\tilde{C}}_{\ell ,k-1}, \end{aligned}$$
(6.21)

where

$$\begin{aligned} {\tilde{C}}_{\ell ,k-1}={\tilde{C}}_{\ell ,k-1}[\phi _{p},\psi _{s,q}:p<k,s\le \ell ,q<k]. \end{aligned}$$
(6.22)

In particular,

$$\begin{aligned} E_{1,\kappa _{1}}= & {} -4\kappa _{1}\langle \phi _{1},\phi _{0}\rangle \psi _{1,\kappa _{1}}-4\kappa _1\langle \psi _{1,\kappa _{1}},\phi _{0}\rangle \phi _{1}, \end{aligned}$$
(6.23)
$$\begin{aligned} G_{\ell ,k}= & {} {\tilde{G}}_{\ell ,k-1}[\phi _{0},\cdots ,\phi _{k-1},\psi _{\ell ,p}:p<k],\quad G_{1,\kappa _{1}}=0, \end{aligned}$$
(6.24)
$$\begin{aligned} C_{\ell ,k}= & {} C[\phi _{0},\psi _{\ell ,k},\phi _{0}]+C[\phi _{0},\phi _{0},\psi _{\ell ,k}]+{\tilde{C}}_{\ell ,k-1}[\cdots ] \end{aligned}$$
(6.25)

and

$$\begin{aligned} C_{1,\kappa _{1}}=C[\phi _{0},\psi _{\ell ,\kappa _1},\phi _{0}]+C[\phi _{0},\phi _{0},\psi _{\ell ,\kappa _1}]. \end{aligned}$$
(6.26)

Then

$$\begin{aligned} X_{\ell ,k}:&=B_{\ell ,k}+E_{\ell ,k}-G_{\ell ,k}+C_{\ell ,k} \nonumber \\&=-D_{0}2{\textrm{Re}}\,\langle \psi _{\ell ,k},\phi _{0}\rangle -\mathcal {A }_{1}{\mathcal {D}}_{\ell ,k-1}+{\tilde{B}}_{\ell ,k-1} \nonumber \\&\quad -2k{\mathcal {A}}_{1}\psi _{\ell ,k}-4k\langle \psi _{\ell ,k},\phi _{0}\rangle \phi _{1}+{\tilde{E}}_{\ell ,k-1}-G_{\ell ,k} \nonumber \\&\quad +4C[\phi _{0},\phi _{0},\psi _{\ell ,k}]+4C[\phi _{0},\psi _{\ell ,k},\phi _{0}]+{\tilde{C}}_{\ell ,k-1} \nonumber \\&=-D_{0}2{\textrm{Re}}\,\langle \psi _{\ell ,k},\phi _{0}\rangle -\mathcal {A }_{1}{\mathcal {D}}_{\ell ,k-1}-2k{\mathcal {A}}_{1}\psi _{\ell ,k}-4k\langle \psi _{\ell ,k},\phi _{0}\rangle \phi _{1} \nonumber \\&\quad +4C[\phi _{0},\phi _{0},\psi _{\ell ,k}]+4C[\phi _{0},\psi _{\ell ,k},\phi _{0}] \nonumber \\&\quad +{\tilde{B}}_{\ell ,k-1}+{\tilde{E}}_{\ell ,k-1}-G_{\ell ,k}+{\tilde{C}}_{\ell ,k-1}. \end{aligned}$$
(6.27)

6.2.2 \(B_{1, \kappa }+E_{1, \kappa }-G_{1, \kappa }+C_{1, \kappa }=0\)

Let \(\kappa =\kappa _{1}\). Then from the previous subsection, one has

$$\begin{aligned} X_{\kappa }:&=B_{1, \kappa }+E_{1, \kappa }-G_{1, \kappa }+C_{1, \kappa } \\&=-D_{0} 2{\textrm{Re}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle -2{\mathcal {A}} _{1} e_{\kappa } \psi _{1, \kappa } -2 \kappa {\mathcal {A}}_{1} \psi _{1, \kappa } -4\kappa \langle \psi _{1, \kappa }, \phi _{0}\rangle \phi _{1} \\&\quad +4C[\phi _{0}, \psi _{1, \kappa }, \phi _{0}]+4 C[\phi _{0}, \phi _{0}, \psi _{1, \kappa }] \\&=-D_{0} 2{\textrm{Re}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle -2{\mathcal {A}} _{1} \kappa (n+1-\kappa ) \psi _{1, \kappa } -4\kappa \langle \psi _{1, \kappa }, \phi _{0}\rangle \phi _{1} \\&\quad +4C[\phi _{0}, \psi _{1, \kappa }, \phi _{0}]+4 C[\phi _{0}, \phi _{0}, \psi _{1, \kappa }] \\&=-2 (({{\mathcal {L}}}_{0}\phi _0 +2(n-1)\phi _{1}) {\textrm{Re}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle +2\kappa \langle \psi _{1, \kappa }, \phi _{0}\rangle \phi _{1}) \\&\quad +4C[\phi _{0}, \psi _{1, \kappa }, \phi _{0}]+4 C[\phi _{0}, \phi _{0}, \psi _{1, \kappa }]. \end{aligned}$$

Notice that

$$\begin{aligned} \langle X_{\kappa }, \phi _{0}\rangle&=-2 ( (-2E_{b}[\phi _{0}] +(n-1)\mathcal {A }_{1}) {\textrm{Re}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle +\kappa \langle \psi _{1, \kappa }, \phi _{0}\rangle {\mathcal {A}} _{1}). \end{aligned}$$

Thus \(\langle X_{\kappa }, \phi _{0}\rangle =0\) if and only if

$$\begin{aligned} 0&= (-2E_{b}[\phi _{0}] +(n-1){\mathcal {A}}_{1}) {\textrm{Re}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle +\kappa \langle \psi _{1, \kappa }, \phi _{0}\rangle {\mathcal {A}}_{1} \\&= (-2E_{b}[\phi _{0}] +(\kappa + n-1){\mathcal {A}}_{1}) {\textrm{Re}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle +\kappa {\mathcal {A}}_{1} i {\textrm{Im}}\, \langle \psi _{1, \kappa }, \phi _{0}\rangle . \end{aligned}$$

Therefore,

$$\begin{aligned} \langle X_{\kappa }, \phi _{0}\rangle =0 \ \iff \ \langle \psi _{1, \kappa }, \phi _{0}\rangle =0. \end{aligned}$$
(6.28)

With the assumption \(\langle \psi _{1,\kappa },\phi _{0}\rangle =0\), one has

$$\begin{aligned} X_{\kappa }&=4C[\phi _{0},\psi _{1,\kappa },\phi _{0}]+4C[\phi _{0},\phi _{0},\psi _{1,\kappa }]. \end{aligned}$$

Then \(X_{\kappa }=0\) if and only if

$$\begin{aligned} h[\phi _{0}]h[\phi _{0}]^{*}\left[ \begin{matrix} \psi _{1,\kappa }\\ {\overline{\psi }}_{1,\kappa }\\ \end{matrix} \right] ={\textbf{0}}\in {\mathbb {C}}^{2m}. \end{aligned}$$
(6.29)

Proposition 6.2

If \(h[\phi _{0}]h[\phi _{0}]^{*}\) has rank \(2m-2\) and if \(\langle \psi _{1,\kappa },\phi _{0}\rangle =0\) and \(h[\phi _{0}]h[\phi _{0}]^{*}[\psi _{1,\kappa }^{t},{\overline{\psi }}_{1,\kappa }^{t}]^{t}=0\) then \(\psi _{1,\kappa }=0\).

Proof

Since \(\langle \psi _{1,\kappa },\phi _{0}\rangle =0\), one has that \([\psi _{1,\kappa }^{t},{\overline{\psi }}_{1,\kappa }^{t}]\) is orthogonal to \([\phi _{0}^{t},0^{t}]\) and \([0^{t},{\overline{\phi }} _{0}^{t}]\) and \(h[\phi _{0}]h[\phi _{0}]^{*}[\phi _{0}^{t},0^{t}]^{t}=0\) and \(h[\phi _{0}]h[\phi _{0}]^{*}[0^{t},{\overline{\phi }}_{0}^{t}]^{*}=0\). This means that both of them are eigenvectors of \(h[\phi _{0}]h[\phi _{0}]^{*}\) associated with the eigenvalue 0. This implies that if \(h[\phi _{0}]h[\phi _{0}]^{*}\) has rank \(2m-2\) then \(h[\phi _{0}]h[\phi _{0}]^{*}[\psi _{1,\kappa }^{t},{\overline{\psi }}_{1,\kappa }^{t}]^{t}=0\) if and only if \(\psi _{1,\kappa }=0\). \(\square \)

6.2.3 \(B_{k}+E_{k}-G_{k}+C_{k}=0\) with \(\kappa \le k< 2 \kappa \)

Notice that

$$\begin{aligned} B_{k}= & {} -(A_{k, 0}+A_{0,k}) D_{0}-{{\mathcal {A}}}_{1} ( 2 e_{k} \phi _{k} +2(n-2k)\psi _{1,k} )+{\tilde{B}}_{k-1}[\phi _{p}, \psi _{1, p}: p<k], \\ E_{k}= & {} -4 \Big (\langle \phi _{1}, \phi _{0}\rangle (k \phi _{k} +\psi _{1, k}) +\langle k\phi _{k} +\psi _{1, k} , \phi _{0}\rangle \phi _{1}\Big )+{\tilde{E}} _{k-1}[\phi _{p}, \psi _{p} :p<k] \end{aligned}$$

and

$$\begin{aligned} C_{k}=4 C[\phi _{0}, \phi _{k}, \phi _{0}]+4 C[\phi _{0}, \phi _{0}, \phi _{k}]+ {\tilde{C}}[\phi _{p}: p<k]. \end{aligned}$$

Therefore,

$$\begin{aligned} 0&=B_{k}+E_{k}-G_{k}+C_{k} \nonumber \\&=-(A_{k, 0}+A_{0,k}) D_{0}-{\mathcal {A}}_{1} ( 2 e_{k} \phi _{k} +2(n-2k)\psi _{1,k} ) \nonumber \\&\quad -4 (\langle \phi _{1}, \phi _{0}\rangle (k \phi _{k} +\psi _{1, k}) +\langle k\phi _{k} +\psi _{1, k} , \phi _{0}\rangle \phi _{1}) \nonumber \\&\quad +4 C[\phi _{0}, \phi _{k}, \phi _{0}]+4 C[\phi _{0}, \phi _{0}, \phi _{k}] +\Phi _{k-1}[\phi _{p}:p<k] \nonumber \\&=-(A_{k, 0}+A_{0,k}) D_{0} -2{\mathcal {A}}_{1} k(n+1-k) \phi _{k} -2 \mathcal { A}_{1} (n+1-2k)\psi _{1,k} \nonumber \\&\quad -4 k A_{k,0} \phi _{1} - 4\langle \psi _{1, k} , \phi _{0}\rangle \phi _{1} \nonumber \\&\quad +4 C[\phi _{0}, \phi _{k}, \phi _{0}]+4 C[\phi _{0}, \phi _{0}, \phi _{k}] +\Phi _{k-1}[\phi _{p}:p<k]. \end{aligned}$$
(6.30)

In particular, when \(k=\kappa =n+1\), since \(\langle \psi _{1,\kappa },\phi _{0}\rangle =0\), one has

$$\begin{aligned} 0&=-(A_{\kappa ,0}+A_{0,\kappa })D_{0}+2\kappa {\mathcal {A}}_{1}\psi _{1,\kappa }-4\kappa A_{\kappa ,0}\phi _{1} \\&\quad +4C[\phi _{0},\phi _{\kappa },\phi _{0}]+4C[\phi _{0},\phi _{0},\phi _{\kappa }]+\Phi _{k-1}[\phi _{p}:p<\kappa ]. \end{aligned}$$

This implies that

$$\begin{aligned} 0&=-(A_{\kappa ,0}+A_{0,\kappa })(-2E_{b}[\phi _{0}]+2n{\mathcal {A}} _{1})-2\kappa i{\textrm{Im}}\,({\mathcal {A}}_{1}A_{\kappa ,0}) \nonumber \\&\quad +\langle \Phi _{\kappa -1},\phi _{0}\rangle . \end{aligned}$$
(6.31)

This implies

$$\begin{aligned} A_{\kappa ,0}={\frac{1}{4}}{\frac{{\textrm{Re}}\,\langle \Phi _{\kappa -1},\phi _{0}\rangle }{n{\mathcal {A}}_{1}-E_{b}[\phi _{0}]}}+{\frac{i}{2\kappa {\mathcal {A}}_1}}{\textrm{Im}}\,\langle \Phi _{\kappa -1},\phi _{0}\rangle . \end{aligned}$$
(6.32)

Thus,

$$\begin{aligned} \kappa {\mathcal {A}}_{1}\psi _{1,\kappa }+2(C[\phi _{0},\phi _{\kappa },\phi _{0}]+C[\phi _{0},\phi _{0},\phi _{\kappa }])={\tilde{\Theta }} _{\kappa -1}, \end{aligned}$$
(6.33)

where

$$\begin{aligned} {\tilde{\Theta }} _{\kappa -1}:=(A_{\kappa ,0}+A_{0,\kappa }){\frac{D_{0}}{2}} -2kA_{\kappa ,0}\phi _{1}-{\frac{1}{2}}\Phi _{\kappa -1}. \end{aligned}$$
(6.34)

We will combine this with \(B_{1,\kappa }+E_{1,\kappa }-G_{1,\kappa }+C_{1,\kappa }=0\) to solve \(\psi _{1,\kappa }\) and \(\phi _{\kappa }\) in the later subsection.

6.2.4 The case: \(k=\kappa \)

We rewrite (6.33) as follows:

$$\begin{aligned} \kappa {\mathcal {A}}_{1} I_{2m} \left[ \begin{matrix} \psi _{1,\kappa } \\ {\overline{\psi }}_{1,\kappa }\\ \end{matrix} \right] -2 h[\phi _{0}] h[\phi _{0}]^{*} \left[ \begin{matrix} \phi _{\kappa } \\ {\overline{\phi }}_{\kappa }\\ \end{matrix} \right] =\left[ \begin{matrix} {\tilde{\Phi }}_{\kappa -1}\\ \overline{{\tilde{\Phi }}}_{\kappa -1} \\ \end{matrix} \right] , \end{aligned}$$
(6.35)

where \({\tilde{\Phi }}_{\kappa -1}\) depends on \(\phi _0,\cdots ,\phi _{\kappa -1}\).

Let U be a \((2m)\times (2m)\) unitary matrix such that

$$\begin{aligned} U^{*}h[\phi _{0}]h[\phi _{0}]^{*}U=D(\lambda _{1},\cdots ,\lambda _{s},0,\cdots ,0), \end{aligned}$$
(6.36)

where \(D(\lambda _{1},\cdots ,\lambda _{s},0,\cdots ,0)\) is a diagonal matrix with positive eigenvalues \(\lambda _{1},\cdots ,\lambda _{s}\). Since

$$\begin{aligned} h\left[ \phi _{0}\right] h\left[ \phi _{0}\right] ^{*}\left[ \begin{array}{c} \phi _{0} \\ 0 \end{array} \right] =h\left[ \phi _{0}\right] h\left[ \phi _{0}\right] ^{*}\left[ \begin{array}{c} 0 \\ \overline{\phi _{0}} \end{array} \right] =0, \end{aligned}$$

we can choose U such that

$$\begin{aligned} U=\left( U_{1},\cdots ,U_{2m-2},\left[ \begin{array}{c} \phi _{0} \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ \overline{\phi _{0}} \end{array} \right] \right) . \end{aligned}$$
(6.37)

By (6.35) and (6.36), one has

$$\begin{aligned} \kappa {\mathcal {A}}_{1}U^{*}\left[ \begin{matrix} \psi _{1,\kappa }\\ {\overline{\psi }}_{1,\kappa }\\ \end{matrix} \right] -2D(\lambda _{1},\cdots ,\lambda _{s},0,\cdots ,0)U^{*}\left[ \begin{matrix} \phi _{\kappa }\\ {\overline{\phi }}_{\kappa }\\ \end{matrix} \right] =U^{*}\left[ \begin{matrix} {\tilde{\Phi }}_{\kappa -1}\\ \overline{{\tilde{\Phi }}}_{\kappa -1}\\ \end{matrix} \right] . \end{aligned}$$
(6.38)

By (6.29), one has

$$\begin{aligned} D(\lambda _{1},\cdots ,\lambda _{s},0,\cdots ,0)U^{*}\left[ \begin{matrix} \psi _{1,\kappa }\\ {\overline{\psi }}_{1,\kappa }\\ \end{matrix} \right] =\left[ \begin{matrix} 0\\ 0\\ \end{matrix} \right] . \end{aligned}$$
(6.39)

Let

$$\begin{aligned} {\textbf{x}}=U^{*}\left[ \begin{matrix} \psi _{1,\kappa }\\ {\overline{\psi }}_{1,\kappa }\\ \end{matrix} \right] ,\quad {\textbf{y}}=U^{*}\left[ \begin{matrix} \phi _{\kappa }\\ {\overline{\phi }}_{\kappa }\\ \end{matrix} \right] \quad \hbox {and }\ {\textbf{z}}=U^{*}\left[ \begin{matrix} {\tilde{\Phi }}_{\kappa -1}\\ \overline{{\tilde{\Phi }}}_{\kappa -1}\\ \end{matrix} \right] . \end{aligned}$$

Therefore, by (6.29) ,

$$\begin{aligned} x_{j}=0,\quad 1\le j\le s; \end{aligned}$$

and by (6.28) ,

$$\begin{aligned} x_{2m-1}=x_{2m}=0. \end{aligned}$$

By (6.33) and (6.34) ,

$$\begin{aligned} z_{2m-1}=\langle {\tilde{\Phi }}_{\kappa -1},\phi _{0}\rangle =0,\quad z_{2m}=\langle \overline{{\tilde{\Phi }}_{\kappa -1}},{\bar{\phi }}_{0}\rangle =0. \end{aligned}$$

By (6.38) ,

$$\begin{aligned} x_{j}={\frac{1}{k{{\mathcal {A}}}_{1}}}z_{j},\quad s<j\le 2m;\quad y_{j}=-{ \frac{z_{j}}{2\lambda _{j}}},\quad 1\le j\le s. \end{aligned}$$

Then

$$\begin{aligned} {\textbf{x}}={\frac{1}{\kappa {{\mathcal {A}}}_{1}}}\left[ \begin{matrix} 0 &{} 0\\ 0 &{} I_{2m-s}\\ \end{matrix} \right] {\textbf{z}}\quad \hbox {and}\quad \left[ \begin{array}{c} \psi _{1,\kappa } \\ {\overline{\psi }}_{1,\kappa } \end{array} \right] ={\frac{1}{\kappa {{\mathcal {A}}}_{1}}}U\left[ \begin{matrix} 0 &{} 0\\ 0 &{} I_{2m-s}\\ \end{matrix} \right] U^{*}\left[ \begin{array}{c} {\tilde{\Phi }}_{\kappa -1} \\ \overline{{\tilde{\Phi }}_{\kappa -1}} \end{array} \right] . \end{aligned}$$

In general, one can solve \(\psi _{1, \kappa }\) as follows. Let V(z) be the range of \(h[\phi _0] h[\phi _0]^*\) on \({\mathbb {C}}^{2m}\) and let \({\tilde{H}}(z)\) be the projection of \(\left[ \begin{matrix} {\tilde{\Phi }}_{\kappa -1} \\ \overline{{\tilde{\Phi }}}_{\kappa -1}\\ \end{matrix}\right] \) onto V(z). By (6.35) and \(h[\phi _0] h[\phi _0]^* \left[ \begin{matrix} \psi _{1, \kappa } \\ \overline{\psi _{1, \kappa }}\\ \end{matrix}\right] =0\), one can see that

$$\begin{aligned} \left[ \begin{matrix} \psi _{1, \kappa } \\ \overline{\psi _{1, \kappa }}\\ \end{matrix}\right] ={1\over \kappa {{{\mathcal {A}}}}_1} \Big ( \left[ \begin{matrix} {\tilde{\Phi }}_{\kappa -1} \\ \overline{{\tilde{\Phi }}}_{\kappa -1}\\ \end{matrix}\right] -{\tilde{H}}\Big ). \end{aligned}$$
(6.40)

We know from (6.32), \(A_{\kappa , 0}\) can be solved smoothly. Only part of entries of \(\phi _{\kappa }\) are uniquely determined by the equations (6.3). Since

$$\begin{aligned} -2D(\lambda _{1},\cdots ,\lambda _{s},0,\cdots ,0){{\textbf{y}}}={{\textbf{z}}}-\kappa {{\mathcal {A}}} _{1}{{\textbf{x}}}, \end{aligned}$$

one has

$$\begin{aligned} -2D(\lambda _{1},\cdots ,\lambda _{s},{{\mathcal {A}}}_{1},\cdots ,{{\mathcal {A}}} _{1}){{\textbf{y}}}={{\textbf{z}}}-\kappa {{\mathcal {A}}}_{1}{{\textbf{x}}}-2D(0,\cdots ,0,{{\mathcal {A}}}_{1},\cdots ,{ {\mathcal {A}}}_{1}){{\textbf{y}}}^{\prime } \end{aligned}$$

where

$$\begin{aligned} {{\textbf{y}}}^{\prime }=(0,\cdots ,0,y_{s+1},\cdots ,y_{2m})^t. \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} -2\Big (h[\phi _{0}]h[\phi _{0}]^{*}+UD(0,\cdots ,0,{\mathcal {A}}_{1},\cdots ,{\mathcal {A}}_{1})U^{*}\Big )\left[ \begin{matrix}\phi _{\kappa }\\ {\overline{\phi }} _{\kappa }\\ \end{matrix}\right] \\{} & {} \quad =\kappa {\mathcal {A}}_{1}I_{2m}\left[ \begin{matrix} \psi _{1,\kappa }\\ {\overline{\psi }}_{1,\kappa }\\ \end{matrix} \right] -\left[ \begin{matrix} {\tilde{\Phi }}_{\kappa -1}\\ \overline{{\tilde{\Phi }}}_{\kappa -1}\\ \end{matrix} \right] -2UD(0,\cdots ,0,{\mathcal {A}}_{1},\cdots ,{\mathcal {A}}_{1})U^{*}y^{\prime }. \end{aligned}$$

Note. We point out here that \(y_{s+1},\cdots ,y_{2m}\) above are with no restrictions. Therefore, \(\phi _{\kappa }\) can not be solved in general from \(B_{\kappa }+C_{\kappa }-G_{\kappa }+E_{\kappa }=0\).

Proposition 6.3

With the notation above, one has

$$\begin{aligned} {\textrm{Im}}\, \langle \Psi _{1, n+1}, \phi _0\rangle =0. \end{aligned}$$

Since the proof of Proposition 6.3 is complicated and very long, we move the proof to Section 7.

6.2.5 \(\kappa _{1}<k<\kappa _{2}-1\)

We start with the following proposition.

Proposition 6.4

$$\begin{aligned} \langle D_0, \phi _0\rangle =-16 |\partial _b \phi _0| |{\overline{\partial }}_b \phi _0| \Big [(n-1) \sqrt{E_b[\phi _0]^2+4n \langle T\phi _0,\phi _0\rangle ^2} +(n+1) E_b[\phi _0]\Big ]^{-1}. \end{aligned}$$

Proof

By (6.8) and Proposition 2.3, one has

$$\begin{aligned} \langle D_0, \phi _0\rangle= & {} -2E_b[\phi _0]+(n-1)2 A_{1,0} \\= & {} -2E_b[\phi _0]+(n-1) {\frac{E_b[\phi _0]+\sqrt{E_b[\phi _0]^2+4n \langle T\phi _0, \phi _0\rangle ^2}}{n}} \\= & {} -{\frac{n+1}{n}} E_b[\phi _0]+{\frac{n-1}{n}} \sqrt{E_b[\phi _0]^2+4n \langle T\phi _0, \phi _0\rangle ^2} \\= & {} {\frac{(n-1)^2 (\sqrt{E_b[\phi _0]^2+4n \langle T\phi _0, \phi _0\rangle ^2} )^2-(n+1)^2 E_b[\phi _0]^2}{n \Big ((n-1) \sqrt{E_b[\phi _0]^2+4n \langle T\phi _0, \phi _0\rangle ^2} +(n+1) E_b[\phi _0]\Big )}} \\= & {} {\frac{ -16 |\partial _b \phi _0| |{\overline{\partial }}_b \phi _0|}{ (n-1) \sqrt{E_b[\phi _0]^2+4n \langle T\phi _0, \phi _0\rangle ^2} +(n+1) E_b[\phi _0]}} \\\ne & {} 0,\quad \hbox { on } \partial B_n. \end{aligned}$$

The proof of the proposition is complete. \(\square \)

Notice that

$$\begin{aligned} {{\mathcal {D}}}_{\ell , k}= 2 e_{k+1}\psi _{\ell , k+1}+2(\ell +1) (n-2k-2) \psi _{\ell +1, k+1} -2(\ell +1)(\ell +2) \psi _{\ell +2, k+1} +{\tilde{D}}_{\ell , k-1} \end{aligned}$$

and

$$\begin{aligned} {\tilde{D}}_{\ell , k-1}:={\tilde{D}}_{\ell , k-1}[\phi _{p}, \psi _{s, q}: p<k, s\le \ell , q<k]. \end{aligned}$$

We have

$$\begin{aligned} 0&=X_{\ell , k} \nonumber \\&=-D_{0} 2{\textrm{Re}}\, \langle \psi _{\ell , k}, \phi _{0}\rangle -\mathcal {A }_{1} {\mathcal {D}}_{\ell , k-1} -2k {\mathcal {A}}_{1} \psi _{\ell , k} -4 k \langle \psi _{\ell , k}, \phi _{0}\rangle \phi _{1} \nonumber \\&\quad +4 C[\phi _{0}, \phi _{0}, \psi _{\ell , k}]+4C[\phi _{0}, \psi _{\ell , k}, \phi _{0}] \nonumber \\&\quad +{\tilde{B}}_{\ell , k-1} +{\tilde{E}}_{\ell , k-1} -G_{\ell , k} +{\tilde{C}} _{\ell , k-1} \nonumber \\&=- D_0 2{\textrm{Re}}\, \langle \psi _{\ell , k}, \phi _{0}\rangle -{\mathcal {A}} _{1} 2 e_{k} \psi _{\ell , k} -2k {\mathcal {A}}_{1} \psi _{\ell , k} -4 k \langle \psi _{\ell , k}, \phi _{0}\rangle \phi _{1} \nonumber \\&\quad +4 C[\phi _{0}, \phi _{0}, \psi _{\ell , k} ]+4C[\phi _{0}, \psi _{\ell , k}, \phi _{0}] \nonumber \\&\quad +{\tilde{B}}_{\ell , k-1} +{\tilde{E}}_{\ell , k-1} -G_{\ell , k} +{\tilde{C}} _{\ell , k-1} -{\mathcal {A}}_{1} {\tilde{D}}_{\ell , k-1} \nonumber \\&=-D_0 2{\textrm{Re}}\, \langle \psi _{\ell , k}, \phi _{0}\rangle -2 k (n+1-k) {\mathcal {A}}_{1} \psi _{\ell , k} -4 k \langle \psi _{\ell , k}, \phi _{0}\rangle \phi _{1} \nonumber \\&\quad +4 C[\phi _{0}, \phi _{0}, \psi _{\ell , k} ]+4C[\phi _{0}, \psi _{\ell , k}, \phi _{0}] +\Psi _{\ell , k-1}, \end{aligned}$$
(6.41)

where

$$\begin{aligned} \Psi _{\ell , k-1}:= {\tilde{B}}_{\ell , k-1} +{\tilde{E}}_{\ell , k-1} -G_{\ell , k} +{\tilde{C}}_{\ell , k-1} -{\mathcal {A}}_{1} {\tilde{D}}_{\ell , k-1}. \end{aligned}$$
(6.42)

In particular,

$$\begin{aligned} 0&=-\langle D_0, \phi _0\rangle 2{\textrm{Re}}\, \langle \psi _{\ell , k}, \phi _{0}\rangle -2 k (n+2-k) {\mathcal {A}}_{1} \langle \psi _{\ell , k}, \phi _{0}\rangle +\langle \Psi _{\ell , k-1}, \phi _{0}\rangle \\&=(2E_{b}[\phi _{0}] + [k(k-n-2)+1-n] {\mathcal {A}}_{1}) 2{\textrm{Re}}\, \langle \psi _{\ell , k}, \phi _{0}\rangle +{\textrm{Re}}\, \langle \Psi _{\ell , k-1}, \phi _0\rangle \\&\quad -2 k (n+2-k) {\mathcal {A}}_{1} i {\textrm{Im}}\, \langle \psi _{\ell , k}, \phi _{0}\rangle +i {\textrm{Im}}\,\langle \Psi _{\ell , k-1}, \phi _{0}\rangle . \end{aligned}$$

Case 1. \(k\ne n+2\). One has

$$\begin{aligned} {\textrm{Im}}\, \langle \psi _{\ell , k}, \phi _0\rangle ={\frac{{\textrm{Im}} \, \langle \Psi _{\ell , k-1}, \phi _0\rangle }{2n (n+2-k) {{\mathcal {A}}}_1}} \end{aligned}$$

and

$$\begin{aligned} {\textrm{Re}}\, \langle \psi _{\ell , k}, \phi _0\rangle ={\frac{{\textrm{Re}} \, \langle \Psi _{\ell , k-1}, \phi _0\rangle }{4 E_b[\phi _0]+2(k(k-n-2)+1-n) { {\mathcal {A}}}_1 }}. \end{aligned}$$

When \(k=n+2\) and \(\ell =1\), we have

$$\begin{aligned} 0&=-\langle D_0, \phi _0\rangle 2{\textrm{Re}}\, \langle \psi _{1, n+2}, \phi _{0}\rangle +\langle \Psi _{1, n+1}, \phi _{0}\rangle . \end{aligned}$$
(6.43)

By Proposition 6.3, one has

$$\begin{aligned} {\textrm{Re}}\, \langle \psi _{1, n+2}, \phi _0\rangle ={\frac{ \langle \Psi _{1, n+1}, \phi _0\rangle }{2\langle D_0, \phi _0\rangle }} \end{aligned}$$

and

$$\begin{aligned} \langle \Psi _{1, n+1}, \phi _0\rangle =\langle {\tilde{B}}_{1, n+1} , \phi _0\rangle +\langle {\tilde{E}}_{1, n+1} -G_{1, n+2}+ {\tilde{C}}_{1, n+1} - {\mathcal {A}}_{1} {\tilde{D}}_{1, n+1}, \phi _0\rangle .\nonumber \\ \end{aligned}$$
(6.44)

When \(\kappa _{1}<k<\kappa _{2}-1\), by (6.30), one has

$$\begin{aligned} 0&=B_{k}+E_{k}-G_{k}+C_{k} \nonumber \\&=-(A_{k,0}+A_{0,k})D_{0}-2{\mathcal {A}}_{1}k(n+1-k)\phi _{k}-2{\mathcal {A}} _{1}(n+1-2k)\psi _{1,k} \nonumber \\&\quad -4kA_{k,0}\phi _{1}-4\langle \psi _{1,k},\phi _{0}\rangle \phi _{1} \nonumber \\&\quad +4C[\phi _{0},\phi _{k},\phi _{0}]+4C[\phi _{0},\phi _{0},\phi _{k}]+\Phi _{k-1}[\phi _{p}:p<k]. \end{aligned}$$
(6.45)

In particular,

$$\begin{aligned} 0&=-2({\textrm{Re}}\,A_{k,0})\langle D_{0},\phi _{0}\rangle -2{\mathcal {A}} _{1}k(n+1-k)A_{k,0}-2{\mathcal {A}}_{1}(n+1-2k)\langle \psi _{1,k},\phi _{0}\rangle \\&\quad -2k{\mathcal {A}}_{1}A_{k,0}-2{\mathcal {A}}_{1}\langle \psi _{1,k},\phi _{0}\rangle +\langle \Phi _{k-1},\phi _{0}\rangle \\&=2({\textrm{Re}}\,A_{k,0})(-\langle D_{0},\phi _{0}\rangle -k(n+2-k) {\mathcal {A}}_{1})+{\textrm{Re}}\,\langle \Phi _{k-1},\phi _{0}\rangle \\&\quad -i2k(n+2-k){\textrm{Im}}\,A_{k,0}-2{\mathcal {A}}_{1}(n+2-2k)\langle \psi _{1,k},\phi _{0}\rangle +i{\textrm{Im}}\,\langle \Phi _{k-1},\phi _{0}\rangle . \end{aligned}$$

If \(k\ne n+2\), then

$$\begin{aligned} -\langle D_{0},\phi _{0}\rangle -k(n+2-k){{\mathcal {A}}}_{1}=2E_{b}[\phi _{0}]-(n-1){{\mathcal {A}}}_{1}+k(k-n-2){{\mathcal {A}}}_{1}\ne 0. \end{aligned}$$

Thus,

$$\begin{aligned} {\textrm{Re}}\,A_{k,0}={\frac{-{\textrm{Re}}\,\langle \Phi _{k-1},\phi _{0}\rangle +2(n+2-2k){\mathcal {A}}_{1}{\textrm{Re}}\,\langle \psi _{1,k},\phi _{0}\rangle }{ -2\langle D_{0},\phi _{0}\rangle +2k(k-n-2){{\mathcal {A}}}_{1}}} \end{aligned}$$

and

$$\begin{aligned} {\textrm{Im}}\,A_{k,0}={\frac{{\textrm{Im}}\,\langle \Phi _{k-1},\phi _{0}\rangle +2(2k-n-2){\mathcal {A}}_{1}{\textrm{Im}}\,\langle \psi _{1,k},\phi _{0}\rangle }{ 2k(n+2-k) {{{\mathcal {A}}}}_1 }.} \end{aligned}$$

If \(k=n+2\), by Proposition 6.4, one has \(\langle D_{0},\phi _{0}\rangle \ne 0\). Then

$$\begin{aligned} 2{\textrm{Re}}\,A_{n+2,0}={\frac{{\textrm{Re}}\,\langle \Phi _{n+1},\phi _{0}\rangle +2(n+2){{\mathcal {A}}}_{1}{\textrm{Re}}\,\langle \psi _{1,n+2},\phi _{0}\rangle }{\langle D_{0},\phi _{0}\rangle }} \end{aligned}$$

and

$$\begin{aligned} {\textrm{Im}}\,\langle \psi _{1,n+2},\phi _{0}\rangle =-{\frac{{\textrm{Im}} \,\langle \Phi _{n+1},\phi _{0}\rangle }{2(n+2){{\mathcal {A}}}_{1}}}. \end{aligned}$$

By (6.41), we have

$$\begin{aligned}{} & {} k(k-n-1)A_{1}\psi _{1,k}+2C[\phi _{0},\psi _{1,k},\phi _{0}]+2C[\phi _{0},\phi _{0},\psi _{1,k}] \nonumber \\{} & {} \quad ={\textrm{Re}}\,\langle \psi _{1,k},\phi _{0}\rangle +2k\langle \psi _{1,k},\phi _{0}\rangle \phi _{1}-{\frac{1}{2}}\Psi _{1,k-1}. \end{aligned}$$
(6.46)

From the above equation and the similar argument of subsection 6.1.2, we can solve \(\psi _{1,k}\). By (6.45),

$$\begin{aligned}{} & {} k(k-n-1)A_{1}\phi _{k}+2C[\phi _{0},\phi _{k},\phi _{0}]+2C[\phi _{0},\phi _{0},\phi _{k}] \\{} & {} \quad ={\textrm{Re}}\,A_{k,0}D_{0}-(2k-n-1){\mathcal {A}}_{1}\psi _{1,k}+2A_{k,0}\phi _{1}+2\langle \psi _{1,k},\phi _{0}\rangle \phi _{1}-\Phi _{k-1}. \end{aligned}$$

From the above equation and the similar argument of subsection 6.1.2, one can easily solve \(\phi _{k}\).

When \(k\ge \kappa _{2}\), one can solve all \(\phi _{k}\) and \(\psi _{\ell ,k}\) similarly without the complication for the case \(k=n+1\) and \(k=n+2\).

Remark

Let

$$\begin{aligned} \ell _{k}=\left\{ \begin{array}{ll} \ell , &{} \text {if }\kappa _{\ell }\le k<\kappa _{\ell +1},\\ \ell +2, &{} \text {if }k\ge \kappa _{\ell +1} \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} {\mathcal {G}}_{\ell ,k}=\{\phi _{j}:j\le k-1\}\cup \{\psi _{s,\alpha }:\ell \le s\le \ell _{k},\alpha \le k\}\cup \{\psi _{s,\alpha }:s\le \ell -1,\alpha \le k-\kappa _{1}\}. \end{aligned}$$

From the expression:

$$\begin{aligned}&B_{\ell ,k}\left[ u\right] =-\sum _{\alpha =1}^{k}{\mathcal {A}}_{\alpha }{\mathcal {D}}_{\ell ,k-\alpha }-\sum _{\alpha +\beta =k}^{{}}{\mathcal {A}} _{\ell ,\alpha }{\mathcal {D}}_{\beta }-\sum _{s=1}^{k-1}\sum _{\begin{array}{c} \alpha \ge k_{s}\\ k-\alpha \ge k_{\ell -s}-1 \end{array}}{\mathcal {A}}_{s,\alpha }{\mathcal {D}} _{\ell -s,k-\alpha }\\&\quad =-\sum _{\beta =\kappa _{\ell }-1}^{k-1}{\mathcal {A}}_{k-\beta }{\mathcal {D}} _{\ell ,\beta }-\sum _{\alpha =\kappa _{\ell }}^{k}{\mathcal {A}}_{\ell ,\alpha }{\mathcal {D}}_{k-\alpha }-\sum _{s=1}^{\ell -1}\sum _{\alpha =\kappa _{s} }^{k+1-\kappa _{\ell -s}}{\mathcal {A}}_{s,\alpha }{\mathcal {D}}_{\ell -s,k-\alpha }, \end{aligned}$$

one can see that \(B_{\ell ,k}\left[ u\right] \) depends on \(\phi _{j}\) and \(\psi _{s,\alpha }\in {\mathcal {G}}_{\ell ,k}.\) It is easy to see that \(G_{\ell ,k}\left[ u\right] ,\) \(E_{\ell ,k}\left[ u\right] \), \(C_{\ell ,k}\left[ u\right] \) and therefore, \(X_{\ell ,k}\) depend on \(\phi _{j}\) and \(\psi _{s,\alpha }\in {\mathcal {G}}_{\ell ,k}\).

Assume that \(\kappa _{\ell }\le k<\kappa _{\ell +1}\) and that we have solved \(\left\{ \phi _{j}:j\le k-1\right\} \ \)and \(\left\{ \psi _{s,\alpha } :s\le \ell ,\alpha \le k-1\right\} \) by equations

$$\begin{aligned} B_{\alpha }+E_{\alpha }-G_{\alpha }+C_{\alpha }=0,\quad X_{\alpha ,\beta }=0,\alpha \le \ell ,\beta <k. \end{aligned}$$

Then by \(X_{\alpha ,k}=0,\alpha =\ell ,\ell -1,\cdots ,1\), we can solve \(\psi _{\ell ,k},\psi _{\ell -1,k},\cdots , \psi _{1,k} \), respectively and by \(B_{k}+E_{k}-G_{k}+C_{k}=0\) we can solve \(\phi _{k}\). Therefore, if \(\phi _{1} ,\cdots ,\phi _{\kappa },\psi _{1,\kappa },\psi _{1,\kappa +1}, \phi _{\kappa +1}\) are given, then by Equation (6.3), we can compute all \(\left\{ \phi _{k}\right\} \) and \(\left\{ \psi _{\ell ,k}\right\} \) uniquely in the following order:

$$\begin{aligned} \psi _{1,\kappa +2}, \phi _{\kappa +2}, \cdots ,\psi _{1,\kappa _{2}-1}, \phi _{\kappa _{2}-1}, \psi _{2,\kappa _{2}}, \psi _{1,\kappa _{2}},\phi _{\kappa _{2}} ,\psi _{2,\kappa _{2}+1},\psi _{1,\kappa _{2}+1},\phi _{\kappa _{2}+1},\cdots . \end{aligned}$$

Now we come back to prove Theorem 1.3. We need to determine \(\phi _{n+1}\) and \(A_{n+2, 0}\) so that they agree with the same quantities determined from the coefficients of u.

As a summary of the above, we have constructed a map v:

$$\begin{aligned} v(z)=\phi (z)+\sum _{\ell =1}^{\infty }\psi _{\ell }(\log (1-|z|))^\ell \quad \hbox {with }\ \psi _{\ell }=\sum _{k=(n+1)\ell }^\infty \psi _{\ell , k} r^{k} \end{aligned}$$

such that \(\tau ^{s}[v]\) vanishes on \(\partial B_{n}\) in infinite orders. The construction shows that

$$\begin{aligned} h(z):=u(z)-\sum _{j=0}^{n }\phi _{j}(z)r(z)^{j}-\psi _{1,n+1}r(z)^{n+1}\log (-r)=O(r(z)^{n+1}). \end{aligned}$$

Let

$$\begin{aligned} g\left( z\right) =\frac{1}{8n}\int _{\left| z\right| }^{1}\left( 1-t^{2}\right) ^{n-1}t^{-2n+1}dt\quad \hbox {and}\quad G(z,w)=-g(\varphi _{z}(w)), \end{aligned}$$

where \(\varphi _{z}(w)\) is the Mobius transform for \(B_{n}\) with \(\varphi _{z}(0)=z\) and \(\varphi _{z}(z)=0\). Then \(G\left( z,w\right) \) is the Green’s function for \(\Delta _{B_n}=(1-|z|^{2})L\) on \(B_{n}\). It was proved by M. Stoll, Lemma 6.6 in [27] that the Dirichlet boundary value problem

$$\begin{aligned} \Delta _{B_n}h=f\quad \hbox {in }B_{n}\quad \hbox {and}\quad h=\phi \quad \hbox {on }\partial B_{n} \end{aligned}$$

has the unique solution

$$\begin{aligned} h(z)=\int _{\partial B_{n}}{\frac{(1-|z|^{2})^{n}}{|1-\langle z,w\rangle |^{2n}}}\phi (w)d\sigma (w)+\int _{B_{n}}G(z,w)f(w)dv(z). \end{aligned}$$

Let \(0<\alpha <1\) and let \(\Lambda ^{0,\alpha }(B_{n})\) be the space of all functions f in \(B_{n}\) with

$$\begin{aligned} (r(z)+r(z^{\prime }))^{\alpha }|f(z)-f(z^{\prime })|\le C(|r(z)-r(z^{\prime })|^{\alpha }+|z-z^{\prime }|^{\alpha }),\quad |r(z)|,|r(z^{\prime })|\ge 1/2 \end{aligned}$$

and \(\Lambda ^{k,\alpha }(B_{n})\) denote the space of all functions f in \(B_n\) with

$$\begin{aligned} (r {{{\mathcal {R}}}})^\alpha Y_j^{\beta }{\overline{Y}}_k^{\gamma }T^{\ell }f(z) \in \Lambda ^{0,\alpha }(B_{n}) \end{aligned}$$

for all \(\alpha +\beta +\gamma +2 \ell \le k\). A map \(u\in \Lambda ^{k,\alpha }(B_{n})\) iff every component of u belongs to \(\Lambda ^{k,\alpha }(B_{n})\).

Let \(r^{m}\Lambda ^{k,\alpha }(B_{n})\) denote the space of all functions (or maps ) f in \(B_{n}\) with

$$\begin{aligned} {\frac{f(z)}{r(z)^{m}}}\in \Lambda ^{k,\alpha }(B_{n}). \end{aligned}$$

The following theorem was proved by Lee and Melrose [9] (similar to Proposition 4.25 in [9] )

Proposition 6.5

With the notation above, if \((-\Delta _{B_n}+k_n) f\in r^m \Lambda ^{k, \alpha }(B_n)\) or \((-\Delta _{B_n} +k_n) f\in r^m \Lambda ^{k,0}(B_n)\)) then \(f\in r^m \Lambda ^{k+2, \alpha }(B_n)\) or (\(f \in r^m \Lambda ^{k+1,\gamma }(B_n)\) for any \(\gamma \) with \(0<\gamma \le 1/2\)). Here \(0<m< {\frac{1}{2}}(n+\sqrt{ n^2+2k_n})\) and \(k_n>0\).

Let A be any differential operator and let

$$\begin{aligned} W[A,u]=Au\langle {\overline{A}}u,u\rangle +{\overline{A}}u\langle Au,u\rangle ,\quad a[u]={\frac{1-|u|^{2}}{1-|z|^{2}}}. \end{aligned}$$

Let

$$\begin{aligned} V_{n}(z)=\sum _{j=0}^{n}\phi _{j}(z)r(z)^{j}+\psi _{1,n+1}r^{n+1}\log (1-|z|). \end{aligned}$$

Then

$$\begin{aligned} \tau [u]=0,\quad \hbox {and}\quad \tau [V_{n}]\in r(z)^{n+1}\Lambda ^{k,\alpha }(B_{n}) \end{aligned}$$

for any \(k\in {\mathbb {N}}\) and \(0<\alpha <1\). By (6.35), we have

$$\begin{aligned} h[\phi _{0}]h[\phi _{0}]^{*}\Big ({\frac{u-V_{n}}{r^{n+1}}}\Big )= O(r) ,\quad \hbox {near }\partial B_{n}. \end{aligned}$$

By (6.32), one has

$$\begin{aligned}{} & {} W[Y,V_{n}]-W[Y,u] \\{} & {} \quad =\sum _{j=1}^{n}\left( Y_{j}V_{n}\langle {\overline{Y}}_{j}V_{n},V_{n}\rangle +{\overline{Y}}_{j}V_{n}\langle Y_{j}V_{n},V_{n}\rangle -Y_{j}V_{n}\langle {\overline{Y}}_{j}u,u\rangle -{\overline{Y}}_{j}V_{n}\langle Y_{j}u,u\rangle \right) \\{} & {} \qquad +\sum _{j=1}^{n}\left( -Y_{j}(u-V_{n})\langle {\overline{Y}}_{j}u,u\rangle - {\overline{Y}}_{j}(u-V_{n})\langle Y_{j}u,u\rangle \right) \\{} & {} \quad =\sum _{j=1}^{n}\left( Y_{j}V_{n}\langle {\overline{Y}}_{j}V_{n},V_{n}\rangle +{\overline{Y}}_{j}V_{n}\langle Y_{j}V_{n},V_{n}\rangle -Y_{j}V_{n}\langle {\overline{Y}}_{j}V_{n},u\rangle -{\overline{Y}}_{j}V_{n}\langle Y_{j}V_{n},u\rangle \right) \\{} & {} \qquad +\sum _{j=1}^{n}\left( -Y_{j}V_{n}\langle {\overline{Y}}_{j}(u-V_{n}),u \rangle -{\overline{Y}}_{j}V_{n}\langle Y_{j}(u-V_{n}),u\rangle \right) \\{} & {} \qquad +\sum _{j=1}^{n}\left( -Y_{j}(u-V_{n})\langle {\overline{Y}} _{j}V_{n},u\rangle +{\overline{Y}}_{j}(u-V_{n})\langle Y_{j}V_{n},u\rangle \right) \\{} & {} \qquad +\sum _{j=1}^{n}\left( -Y_{j}(u-V_{n})\langle {\overline{Y}} _{j}(u-V_{n}),u\rangle +{\overline{Y}}_{j}(u-V_{n})\langle Y_{j}(u-V_{n}),u\rangle \right) \\{} & {} \quad =\sum _{j=1}^{n}\left( Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},V_{n}-u\rangle +{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},V_{n}-u\rangle \right) \\{} & {} \qquad +\sum _{j=1}^{n}\left( -Y_{j}\phi _{0}\langle {\overline{Y}} _{j}(u-V_{n}),\phi _{0}\rangle -{\overline{Y}}_{j}\phi _{0}\langle Y_{j}(u-V_{n}),\phi _{0}\rangle \right) \\{} & {} \qquad +O(r(z)^{n+1+\alpha }) \\{} & {} \quad =\sum _{j=1}^{n}\left( Y_{j}\phi _{0}\langle {\overline{Y}}_{j}\phi _{0},V_{n}-u\rangle +{\overline{Y}}_{j}\phi _{0}\langle Y_{j}\phi _{0},V_{n}-u\rangle \right) \\{} & {} \qquad +\sum _{j=1}^{n}\left( Y_{j}\phi _{0}\langle (u-V_{n}),{\overline{Y}}_{j}\phi _{0}\rangle +{\overline{Y}}_{j}\phi _{0}\langle (u-V_{n}),{\overline{Y}}_{j}\phi _{0}\rangle \right) \\{} & {} \qquad -\sum _{j=1}^{n}\left( Y_{j}\phi _{0}{\overline{Y}}_{j}\langle (u-V_{n}),\phi _{0}\rangle +{\overline{Y}}_{j}\phi _{0}Y_{j}\langle (u-V_{n}),\phi _{0}\rangle \right) +O(r(z)^{n+1+\alpha }) \\{} & {} \quad ={\tilde{H}}_{n+1}-\sum _{j=1}^{n}(Y_{j}\phi _{0}{\overline{Y}}_{j}A_{n+1,0}+ {\overline{Y}}_{j}\phi _{0}Y_{j}A_{n+1,0})+o(r(z)^{n+1}) \end{aligned}$$

with \({\tilde{H}}_{n+1}\in C^{\infty }({\overline{B}}_{n}\setminus \{0\})\). Since

$$\begin{aligned}{} & {} 4(W[R,V_{n}]-W[R,u]) \\{} & {} \quad ={\mathcal {R}}V_{n}\langle {\mathcal {R}}V_{n},V_{n}\rangle -{\mathcal {R}} u\langle {\mathcal {R}}u,u\rangle \\{} & {} \quad ={\mathcal {R}}(V_{n}-u)\langle {\mathcal {R}}V_{n},V_{n}\rangle +{\mathcal {R}} u\langle {\mathcal {R}}(V_{n}-u),V_{n}\rangle +{\mathcal {R}}u\langle {\mathcal {R}} u,V_{n}-u\rangle \\{} & {} \quad ={\mathcal {R}}(V_{n}-u)\langle {\mathcal {R}}V_{n},V_{n}\rangle +{\mathcal {R}} V_{n}\langle {\mathcal {R}}(V_{n}-u),V_{n}\rangle \\{} & {} \qquad +{\mathcal {R}}V_{n}\langle {\mathcal {R}}V_{n},V_{n}-u\rangle +O(r(z)^{2n-1}) \\{} & {} \quad =\frac{1}{2}{\mathcal {A}}_{1}{\mathcal {R}}(V_{n}-u)+\phi _{1}\langle \mathcal {R }(V_{n}-u),\phi _{0}\rangle +\phi _{1}\langle \phi _{1},V_{n}-u\rangle +O(r(z)^{n+1})+O(r(z)^{2n-1}) \\{} & {} \quad =\frac{1}{2}{\mathcal {A}}_{1}{\mathcal {R}}(V_{n}-u)+\phi _{1}{\mathcal {R}} \langle V_{n}-u,\phi _{0}\rangle +\phi _{1}\langle \phi _{1},V_{n}-u\rangle +O(r(z)^{n+1})+O(r(z)^{2n-1}) \end{aligned}$$

and

$$\begin{aligned}{} & {} 4(W[T,V_{n}]-W[T,u]) \\{} & {} \quad =TV_{n}\langle TV_{n},V_{n}\rangle -Tu\langle Tu,u\rangle \\{} & {} \quad =T(V_{n}-u)\langle TV_{n},V_{n}\rangle +Tu\langle T(V_{n}-u),V_{n}\rangle +Tu\langle Tu,V_{n}-u\rangle \\{} & {} \quad =T(V_{n}-u)\langle TV_{n},V_{n}\rangle +TV_{n}\langle T(V_{n}-u),V_{n}\rangle \\{} & {} \qquad +TV_{n}\langle TV_{n},V_{n}-u\rangle +O(r(z)^{2n-1}) \\{} & {} \quad =\langle T\phi _{0},\phi _{0}\rangle T(V_{n}-u)+T\phi _{0}\langle T(V_{n}-u),\phi _{0}\rangle \\{} & {} \qquad +T\phi _{0}\langle T\phi _{0},V_{n}-u\rangle +O(r(z)^{n+1})+O(r(z)^{2n-1}) \\{} & {} \quad =\langle T\phi _{0},\phi _{0}\rangle T(V_{n}-u)+T\phi _{0}\langle T(V_{n}-u),\phi _{0}\rangle +T\phi _{0}\langle V_{n}-u,T\phi _{0}\rangle \\{} & {} \qquad +T\phi _{0}\langle T\phi _{0},V_{n}-u\rangle +O(r(z)^{n+1})+O(r(z)^{2n-1}), \end{aligned}$$

one has

$$\begin{aligned}{} & {} -4\left| z\right| ^{2}\Delta _{B_n}\left( u\left( z\right) -V_n\right) \\{} & {} \quad =4\frac{1}{a[u]}\sum _{j}\left( W\left[ Y_{j},u\right] +W\left[ {\bar{Y}} _{j},u\right] \right) -4\frac{1}{a[V_n]}\sum _{j}\left( W\left[ Y_{j},V_n\right] +W\left[ {\bar{Y}}_{j},V_n\right] \right) \\{} & {} \qquad +\frac{2\left( 1-\left| z\right| ^{2}\right) }{a[u]}\left( W\left[ {\mathcal {R}}_{{}},u\right] -W\left[ {\mathcal {T}}_{{}},u\right] \right) -\frac{ 2\left( 1-\left| z\right| ^{2}\right) }{a[V_n]}\left( W\left[ {\mathcal {R}}_{{}},V_n\right] -W\left[ {\mathcal {T}}_{{}},V_n\right] \right) , \\{} & {} a[u]-a[V_{n}]=O(r(z)^{n}),\quad W[Y_{j},u-V_n],\ W[{\overline{Y}}_{j},u-V_n]=O(r(z)^{n+1}) \end{aligned}$$

and

$$\begin{aligned} W[R,u-V_n],\ W[{\overline{R}},u-V_n]=O(r^n). \end{aligned}$$

Therefore, our computations show that

$$\begin{aligned} \Delta _{B_n}\left( u-V_{n}\right) =O\left( r^{n+1}\right) . \end{aligned}$$

Let

$$\begin{aligned} \rho (z)=1-|z|^{2}\quad \hbox {and}\quad U_{k}(z)={\frac{u(z)-V_{n}(z)}{\rho (z)^{k}}}. \end{aligned}$$

Since \(\Delta _{B_{n}}=\rho (z)\sum _{j=1}^{n}X_{j}\frac{\partial }{\partial {\overline{z}}_{j}}\), one has

$$\begin{aligned}{} & {} \Delta _{B_{n}}U_{k}(z) \\{} & {} \quad ={\frac{\Delta _{B_{n}}(u-V_n )}{\rho ^{k}}}+\rho \sum _{j=1}^{n}[X_{j}(u-V_{n}){\frac{\partial }{\partial {\overline{z}}_{j}}} \rho ^{-k}+{\frac{\partial (u-V_{n})}{\partial {\overline{z}}_{j}}} X_{j}\rho ^{-k}] +(u-V)\Delta _{g}\rho ^{-k} \\{} & {} \quad ={\frac{\Delta _{B_{n}}(u-V_{n})}{\rho (z)^{k}}}+k\rho (z)^{-k}[\rho (z)R(u-V_{n})+\rho {\overline{R}}(u-V_{n})]+(u-V_n)\Delta _{g}\rho ^{-k}. \end{aligned}$$

Now since

$$\begin{aligned} \Delta _{B_{n}}\rho ^{-k}= & {} \rho ({\frac{\partial ^{2}}{\partial z_{j}\partial {\overline{z}}_{j}}}-R{\overline{R}})\rho ^{-k} \\= & {} -k\rho (-{\frac{\partial }{\partial z_{j}}}(\rho ^{-k-1}z_{j})+R(\rho ^{-k-1}|z|^{2})) \\= & {} -k\rho (-n\rho ^{-k-1}-(k+1)\rho ^{-k-2}|z|^{2}+\rho ^{-k-1}|z|^{2}+(k+1)\rho ^{-k-2}|z|^{4}) \\= & {} -k\rho ^{-k}(-(n-1)-\rho -(k+1)|z|^{2}) \\= & {} -k\rho ^{-k}(-(n-1)-k-1-\rho +(k+1)\rho ) \\= & {} -k\rho ^{-k}(-(n+k)+k\rho ). \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} \Delta _{B_{n}}U_{k}(z) ={\frac{\Delta _{B_{n}}(u-V_{n})}{\rho (z)^{k}}}+k\rho (z)^{-k}[\rho (z)R(u-V_{n})+\rho {\overline{R}}(u-V_{n})] \\{} & {} \qquad +k(n+k)(u-V_{n})\rho ^{-k}-k^{2}(u-V_{n})\rho ^{-k+1} \\{} & {} \quad ={\frac{\Delta _{B_{n}}(u-V_{n})}{\rho (z)^{k}}}+k\rho (z)^{-k}[\rho (z)R(u-V_{n})+\rho {\overline{R}}(u-V_{n})] +k(n+k)U_{k}-k^{2}\rho U_{k}. \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} [-\Delta _{B_{n}}+k(n+k)]U_{k}(z) \\{} & {} \quad =-{\frac{\Delta _{B_{n}}(u-V_{n})}{\rho (z)^{k}}}-k\rho (z)^{-k}[\rho (z)R(u-V_{n})+\rho {\overline{R}}(u-V_{n})]+k^{2}\rho U_{k} \\{} & {} \quad =r(z)^{n+1-k}f(z)-k\rho (z)^{-k}[\rho (z)R(u-V_{n})+\rho {\overline{R}} (u-V_{n})]+k^{2}\rho U_{k} \\{} & {} \quad =r(z)^{n+1-k}H(z), \end{aligned}$$

where \(f(z), H(z)\in L^{\infty }(B_{n})\cap C^{\infty }(B_{n}\setminus \{0\})\). Here we take \(1\le k\le n\). By Proposition 6.5, one has \(U_{k}\in r^{n+1-k}\Lambda ^{1,\alpha }(B_{n})\) for any \(\alpha \le 1/2\). This implies that

$$\begin{aligned} (-\Delta _{B_n}+k(n+k))U_{k}\in r(z)^{n+1-k}\Lambda ^{1,\alpha }({\overline{B}} _{n}). \end{aligned}$$

Thus, \(U_{k}\in r(z)^{n+1-k}\Lambda ^{3,\alpha }({\overline{B}}_{n})\). By iteration, one has that

$$\begin{aligned} U_{k}\in r^{n+1-k}\Lambda ^{m,\alpha }(B_{n}) \end{aligned}$$

for any \(m\in {\mathbb {N}}\). Notice that \(U_{k}={\frac{u-V_{n}}{r^{k}}}\), one has

$$\begin{aligned} u-V_{n}\in r^{n+1}\Lambda ^{m,\alpha }(B_{n}) \end{aligned}$$

for any \(m\in {\mathbb {N}}\). Let

$$\begin{aligned} u_{n+1}(z)={\frac{u(z)-V_{n}(z)}{r(z)^{n+1}}}. \end{aligned}$$

Then \(u_{n+1}\in \bigcap _{k\in {\mathbb {N}}}\Lambda ^{k,\alpha }(B_{n})\). In particular, \(u_{n+1}|_{\partial B_{n}}\in C^{\infty }(\partial B_{n})\). We let \(\phi _{n+1}(z)=u_{n+1}(z/|z|)\) and let the new \({\tilde{V}}_{n+1}\) be defined as

$$\begin{aligned} {\tilde{V}}_{n+1}=\sum _{j=0}^{n+1}\phi _{j}r^{j}+\psi _{1,n+1}r^{n+1}\log r. \end{aligned}$$

Repeating the above argument for \(\left\langle u-{\tilde{V}}_{n+1},\phi _{0}\right\rangle ,\) we have

$$\begin{aligned} r^{-n-2} \langle u-{\tilde{V}}_{n+1}, \phi _0\rangle |_{\partial B_n} \in C^\infty (\partial B_n). \end{aligned}$$

Let

$$\begin{aligned} A_{n+2,0}=r^{-n-2} \langle u-{\tilde{V}}_{n+1},\phi _{0} \rangle \vert _{\partial B_{n}}. \end{aligned}$$

Then we can solve \(\psi _{1,n+2}\) and \(\phi _{n+2}\) and then all the other \(\psi _{j,k}\) and \(\phi _{k}.\) Then one can find v above such that \(u(z)-v(z)\) vanishes on \(\partial B_{n}\) in infinite orders.

Let \(0\le \chi \le 1\), \(\chi (r(z))\in C^{\infty }({\overline{B}}_{n})\) with \(\chi (r(z))=1\) if \(r(z)>-1/2\) and \(\chi (r(z))=0\) if \(r(z)<-3/4\). Then \( \Delta _{B_n}(u-\chi v)\) is smooth in \({\overline{B}}_{n}\) and vanishes on \( \partial B_{n}\) in infinite orders. Thus,

$$\begin{aligned} F(z)=u(z)-\chi v(z)=\int _{B_{n}}G(z,w)(\Delta _{B_n}u-\chi v)dv(w)\in C^{\infty }({\overline{B}}_{n}). \end{aligned}$$

Therefore,

$$\begin{aligned} u(z)=F(z)+\chi (r(z))v(z). \end{aligned}$$

This proves Theorem 1.3. \(\square \)

7 Proof of Proposition 6.3

In this section, we will prove Proposition 6.3.

Proof

Since

$$\begin{aligned} \Psi _{1,k-1}:={\tilde{B}}_{1,k-1}+{\tilde{E}}_{1,k-1}-G_{1,k}+{\tilde{C}}_{1,k-1}. \end{aligned}$$
(7.1)

We will compute \({\textrm{Im}}\,\langle \Psi _{1,n+1},\phi _{0}\rangle \) term by term.

By (4.23)

$$\begin{aligned} B_{1,n+2}=-{{\mathcal {A}}}_{1}{{\mathcal {D}}}_{1,n+1}-{{\mathcal {A}}}_{2}[\phi ]{ {\mathcal {D}}}_{1,n}-2D_{0}{\textrm{Re}}\,\langle \psi _{1,n+2},\phi _{0}\rangle , \end{aligned}$$

where

$$\begin{aligned} {\mathcal {D}}_{1,n+1}=D_{n+1}[\psi _{1}]=-4(n+2)\psi _{1,n+2}+[{{\mathcal {L}}} _{0}-(n+1)(3n+4)]\psi _{1,n+1} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {D}}_{1,n}=D_{n}[\psi _{1}]=-2(n+1)\psi _{1,n+1}. \end{aligned}$$

Therefore,

$$\begin{aligned} {\tilde{B}}_{1,n+1}=-{{\mathcal {A}}}_{1}[{{\mathcal {L}}}_{0}-(n+1)(3n+4)]\psi _{1,n+1}+2{{\mathcal {A}}}_{2}(n+1)\psi _{1,n+1}. \end{aligned}$$

Therefore, By (6.28),

$$\begin{aligned} \langle {\tilde{B}}_{1,n+1},\phi _{0}\rangle =-\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle . \end{aligned}$$
(7.2)

Since

$$\begin{aligned} {\tilde{E}}_{1,n+1}&=E_{1,n+2}+2\left( n+2\right) {\mathcal {A}}_{1}\psi _{1,n+2}+4\left( n+2\right) \left\langle \psi _{1,n+2},\phi _{0}\right\rangle \phi _{1} \\&=-4W_{1,n+1}\left( {\mathcal {R}},{\mathcal {R}},u\right) -2W_{1,n}\left( {\mathcal {R}},{\mathcal {R}},u\right) \\&\quad +2\left( n+2\right) {\mathcal {A}}_{1}\psi _{1,n+2}+4\left( n+2\right) \left\langle \psi _{1,n+2},\phi _{0}\right\rangle \phi _{1} \\&=-4\left\langle F_{0},\phi _{0}\right\rangle F_{1,n+1}-4(\langle F_{0},\phi _{1}\rangle +\langle F_{1},\phi _{0}\rangle )F_{1,n} \\&\quad -4(\left\langle F_{1,n+1},\phi _{0}\right\rangle +\langle F_{1,n},\phi _{1}\rangle )F_{0}-4(\langle F_{1,n},\phi _{0}\rangle F_{1}-4\langle F_{0},\psi _{1,n+1}\rangle F_{0} \\&\quad -2\langle F_{0},\phi _{0}\rangle F_{1,n}-2\langle F_{1,n},\phi _{0}\rangle F_{0} \\&\quad +2\left( n+2\right) {\mathcal {A}}_{1}\psi _{1,n+2}+4\left( n+2\right) \left\langle \psi _{1,n+2},\phi _{0}\right\rangle \phi _{1} \\&=-2(n+1){\mathcal {A}}_{1}\psi _{1,n+1}-4(A_{1,1}+2A_{2,0}+A_{1,0})(n+1)\psi _{1,n+1} \\&\quad -4\left( n+1\right) \left\langle \psi _{1,n+1},\phi _{0}\right\rangle \phi _{1}-4(n+1)\langle \psi _{1,n+1},\phi _{1}\rangle \phi _{1} \\&\quad -4\langle (n+1)\psi _{1,n+1},\phi _{0}\rangle F_{1}-4\langle \phi _{1},\psi _{1,n+1}\rangle \phi _{1} \\&\quad -{\mathcal {A}}_{1}(n+1)\psi _{1,n+1}-2(n+1)\langle \psi _{1,n+1},\phi _{0}\rangle \phi _{1} \end{aligned}$$

and (6.28), one has

$$\begin{aligned} \langle {\tilde{E}}_{1,n+1},\phi _{0}\rangle =-2(n+1){\mathcal {A}}_{1}\langle \psi _{1,n+1},\phi _{1}\rangle -2{\mathcal {A}}_{1}\langle \phi _{1},\psi _{1,n+1}\rangle . \end{aligned}$$
(7.3)

Since

$$\begin{aligned} G_{1,n+2}&=-4W_{1,n+1}\left( T,T,u\right) -2W_{1,n}\left( T,T,u\right) \\&=-4\left( \langle T\phi _{0},\phi _{0}\rangle T\psi _{1,n+1}+\langle T\psi _{1,n+1},\phi _{0}\rangle T\phi _{0}+\left\langle T\phi _{0},\psi _{1,n+1}\right\rangle T\phi _{0}\right) \end{aligned}$$

and

$$\begin{aligned} \langle T\psi _{1,n+1},\phi _{0}\rangle =\langle \psi _{1,n+1},T\phi _{0}\rangle , \end{aligned}$$

we have

$$\begin{aligned}{} & {} \langle G_{1,n+2},\phi _{0}\rangle \nonumber \\{} & {} \quad =-8\langle T\phi _{0},\phi _{0}\rangle \left\langle \psi _{1,n+1},T\phi _{0}\right\rangle -4\left\langle T\phi _{0},\phi _{0}\right\rangle \left\langle T\phi _{0},\psi _{1,n+1}\right\rangle \nonumber \\{} & {} \quad =-8{\text {Re}}\left\langle T\phi _{0},\phi _{0}\right\rangle \left\langle \psi _{1,n+1},T\phi _{0}\right\rangle -4\left\langle T\phi _{0},\phi _{0}\right\rangle \left\langle \psi _{1,n+1},T\phi _{0}\right\rangle . \end{aligned}$$
(7.4)

Therefore,

$$\begin{aligned}{} & {} {\textrm{Im}}\,(\langle {\tilde{B}}_{1,n+1},\phi _{0}\rangle +\langle {\tilde{E}} _{1,n+1},\phi _{0}\rangle -\langle G_{1,n+2},\phi _{0}\rangle ) \nonumber \\{} & {} \quad =-{\mathcal {A}}_{1}{\textrm{Im}}\,\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle -2n{\mathcal {A}}_{1}{\textrm{Im}}\,\langle \psi _{1,n+1},\phi _{1}\rangle +4\langle T\phi _{0},\phi _{0}\rangle {\textrm{Im}} \,\langle \psi _{1,n+1},T\phi _{0}\rangle .\nonumber \\ \end{aligned}$$
(7.5)

Notice that

$$\begin{aligned} 0=h[\phi _{0}]^{*}\left[ \begin{matrix} \psi _{1,n+1}\\ {\overline{\psi }}_{1,n+1}\\ \end{matrix} \right] = \left[ \begin{matrix} ({\overline{Y}} \phi _0)^* \psi _{1,n+1} -({\overline{Y}} {\overline{\phi }}_0)^* {\overline{\psi }}_{1, n+1}\\ -(Y\phi _0)^* \psi _{1, n+1}+(Y {\overline{\phi }} _0)^*{\overline{\psi }}_{1,n+1}\\ \end{matrix} \right] . \end{aligned}$$

This implies for \(j=1,\cdots ,n,\) that

$$\begin{aligned} (Y_{j}\phi _{0})^{*}\psi _{1,n+1}=(Y_{j}{\overline{\phi }}_{0})^{*} {\overline{\psi }}_{1,n+1} \end{aligned}$$
(7.6)

and for any real-valued function A,

$$\begin{aligned} \langle \psi _{1,n+1},(Y_{j}\phi _{0})\overline{Y_{j}}A\rangle= & {} \langle (Y_{j}\phi _{0})^{*}\psi _{1,n+1},{\overline{Y}}_{j}A\rangle \\= & {} \langle (Y_{j}{\overline{\phi }}_{0})^{*}{\overline{\psi }}_{1,n+1}, {\overline{Y}}_{j}A\rangle \\= & {} \langle {\overline{\psi }}_{1,n+1},(Y_{j}{\overline{\phi }}_{0}){\overline{Y}} _{j}A\rangle \\= & {} \overline{\langle \psi _{1,n+1},({\overline{Y}}_{j}\phi _{0})Y_{j}A\rangle } . \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} \langle \psi _{1,n+1},Y_{j}\phi _{0}{\overline{Y}}_{j}A+{\overline{Y}}_{j}\phi _{0}Y_{j}A\rangle \nonumber \\{} & {} \quad =\langle \psi _{1,n+1},(Y_{j}\phi _{0}){\overline{Y}}_{j}A\rangle +\langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}Y_{j}A\rangle \nonumber \\{} & {} \quad =\overline{\langle \psi _{1,n+1},({\overline{Y}}_{j}\phi _{0})Y_{j}A\rangle } +\langle \psi _{1,n+1},({\overline{Y}}_{j}\phi _{0})Y_{j}A\rangle . \end{aligned}$$
(7.7)

Notice

$$\begin{aligned} \Theta _{j}[A,B,C]:=\sum _{j=1}^{n}(\langle Y_{j}A,B\rangle {\overline{Y}} _{j}C+\langle {\overline{Y}}_{j}A,B\rangle Y_{j}C),\quad \langle \Theta _{j}[A,B,\phi _{0}],\phi _{0}\rangle =0 \end{aligned}$$

and

$$\begin{aligned} {\tilde{C}}_{\ell ,k-1}=C_{\ell ,k-1}\left[ u\right] -4\sum _{j}\Theta _{j} \left[ \phi _{0},\phi _{0},\psi _{\ell ,k}\right] -4\sum _{j}\Theta _{j}\left[ \phi _{0},\psi _{\ell ,k},\phi _{0}\right] . \end{aligned}$$

Then with \(\langle \psi _{1,n+1},\phi _{0}\rangle =0\),

$$\begin{aligned} \langle {\tilde{C}}_{1,n+1},\phi _{0}\rangle= & {} 4\sum _{j}\langle \Theta _{j}[\phi _{0},\phi _{1},\psi _{1,n+1}]+\Theta _{j}[\phi _{0},\psi _{1,n+1},\phi _{1}],\phi _{0}\rangle \\{} & {} +4\sum _{j}\langle \Theta _{j}[\phi _{1},\phi _{0},\psi _{1,n+1}]+\Theta _{j}[\psi _{1,n+1},\phi _{0},\phi _{1}],\phi _{0}\rangle \end{aligned}$$

and

$$\begin{aligned} \langle Y_{j}\psi _{1,n+1},\phi _{0}\rangle =-\langle \psi _{1,n+1}, {\overline{Y}}_{j}\phi _{0}\rangle . \end{aligned}$$

Since \({{\mathcal {A}}}_{1}=2\langle \phi _{1},\phi _{0}\rangle \), one has

$$\begin{aligned} 2\langle Y_{j}\phi _{1},\phi _{0}\rangle =Y_{j}{{\mathcal {A}}}_{1}-2\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle . \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} 4\langle \Theta _{j}[\phi _{0},\phi _{1},\psi _{1,n+1}],\phi _{0}\rangle \\{} & {} \quad =4\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}}_{j}\psi _{1,n+1},\phi _{0}\rangle +4\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle Y_{j}\psi _{1,n+1},\phi _{0}\rangle \\{} & {} \quad =-4\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle -4\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle ,\\{} & {} 4\langle \Theta _{j}[\phi _{1},\phi _{0},\psi _{1,n+1}],\phi _{0}\rangle \\{} & {} \quad =4\langle Y_{j}\phi _{1},\phi _{0}\rangle \langle {\overline{Y}}_{j}\psi _{1,n+1},\phi _{0}\rangle +4\langle {\overline{Y}}_{j}\phi _{1},\phi _{0}\rangle \langle Y_{j}\psi _{1,n+1},\phi _{0}\rangle \\{} & {} \quad =4\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle +4\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad -{2}Y_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle -{2} {\overline{Y}}_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle ,\\{} & {} 4\langle \Theta _{j}[\phi _{0},\psi _{1,n+1},\phi _{1}],\phi _{0}\rangle \\{} & {} \quad =4\langle Y_{j}\phi _{1},\phi _{0}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle +4\langle {\overline{Y}}_{j}\phi _{1},\phi _{0}\rangle \langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \quad =-4\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle {\overline{Y}} _{j}\phi _{0},\psi _{1,n+1}\rangle +4\langle {\overline{Y}}_{j}\phi _{1},\phi _{0}\rangle \langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \qquad +{2}Y_{j}{\mathcal {A}}_{1}\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle +{2}{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle \end{aligned}$$

and

$$\begin{aligned}{} & {} 4\langle \Theta _{j}[\psi _{1,n+1},\phi _{0},\phi _{1}],\phi _{0}\rangle \\{} & {} \quad =4\langle Y_{j}\phi _{1},\phi _{0}\rangle \langle {\overline{Y}}_{j}\psi _{1,n+1},\phi _{0}\rangle +4\langle {\overline{Y}}_{j}\phi _{1},\phi _{0}\rangle \langle Y_{j}\psi _{1,n+1},\phi _{0}\rangle \\{} & {} \quad =4\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle +4\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad -{2}Y_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle -{2} {\overline{Y}}_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle . \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} 4\left\langle \Theta _{j}\left[ \phi _{0},\phi _{1},\psi _{1,n+1} \right] +\Theta _{j}\left[ \phi _{1},\phi _{0},\psi _{1,n+1}\right] ,\phi _{0}\right\rangle \\{} & {} \qquad +4\left\langle \Theta _{j}\left[ \phi _{0},\psi _{1,n+1},\phi _{1}\right] +\Theta _{j}\left[ \psi _{1,n+1},\phi _{0},\phi _{1}\right] ,\phi _{0}\right\rangle \\{} & {} \quad =-4\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle -4\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad +4\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle +4\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad -4\langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle \langle \phi _{1},Y_{j}\phi _{0}\rangle -4\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad +4\langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \phi _{1},Y_{j}\phi _{0}\rangle +4\langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad -2{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},{\overline{Y}} _{j}\phi _{0}\rangle -2Y_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \\{} & {} \qquad +2{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle +2Y_{j}{\mathcal {A}}_{1}\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \qquad -2{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},{\overline{Y}} _{j}\phi _{0}\rangle -2Y_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \\{} & {} \quad =-8{\textrm{Re}}\,\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle -8{\textrm{Re}}\,\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad +8\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle +8\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad -6{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},{\overline{Y}} _{j}\phi _{0}\rangle -6Y_{j}{\mathcal {A}}_{1}\langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \\{} & {} \qquad +4{\textrm{Re}}\,{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle +4{\textrm{Re}}\,Y_{j}{\mathcal {A}}_{1}\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \quad =-8{\textrm{Re}}\,\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle -8{\textrm{Re}}\,\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \\{} & {} \qquad +4{\textrm{Re}}\,{\overline{Y}}_{j}{\mathcal {A}}_{1}\langle Y_{j}\phi _{0},\psi _{1,n+1}\rangle +4{\textrm{Re}}\,Y_{j}{\mathcal {A}}_{1}\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \qquad +8\Big \langle \psi _{1,n+1},\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle Y_{j}\phi _{0}+\langle Y_{j}\phi _{0},\phi _{1}\rangle \overline{ Y}_{j}\phi _{0}\Big \rangle \\{} & {} \qquad -6\Big \langle \psi _{1,n+1},Y_{j}{\mathcal {A}}_{1}{\overline{Y}}_{j}\phi _{0}+ {\overline{Y}}_{j}{\mathcal {A}}_{1}Y_{j}\phi _{0}\Big \rangle . \end{aligned}$$

This coupling with (7.7) implies

$$\begin{aligned} {\textrm{Im}}\,\langle {\tilde{C}}_{1,n+1},\phi _{0}\rangle =8\sum _{j=1}^{n} {\textrm{Im}}\,\Big \langle \psi _{1,n+1},\langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle Y_{j}\phi _{0}+\langle Y_{j}\phi _{0},\phi _{1}\rangle {\overline{Y}}_{j}\phi _{0}\Big \rangle . \end{aligned}$$
(7.8)

Therefore, by (7.5) and (7.8)

$$\begin{aligned}{} & {} \frac{1}{2}{\textrm{Im}}\,\langle \Psi _{1,n+1},\phi _{0}\rangle \\{} & {} \quad =-{\frac{{\mathcal {A}}_{1}}{2}}{\textrm{Im}}\,\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle -{\textrm{Im}}\,\langle \psi _{1,n+1},n{\mathcal {A}} _{1}\phi _{1}\rangle -2{\textrm{Im}}\,\Big \langle \psi _{1,n+1},\langle T\phi _{0},\phi _{0}\rangle T_{0}\phi _{0}\Big \rangle \\{} & {} \qquad +4\sum _{j=1}^{n}{\textrm{Im}}\Big (\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle +\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \Big ) \\{} & {} \quad =-{\frac{{\mathcal {A}}_{1}}{2}}{\textrm{Im}}\,\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle -{\textrm{Im}}\,\langle \psi _{1,n+1},n{\mathcal {A}} _{1}\phi _{1}\rangle -2{\textrm{Im}}\,\Big \langle \psi _{1,n+1},\langle T\phi _{0},\phi _{0}\rangle T_{0}\phi _{0}\Big \rangle \\{} & {} \qquad +4\sum _{j=1}^{n}{\textrm{Im}}\Big (\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle +\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \Big ). \end{aligned}$$

Now compute \({\textrm{Im}}\,\langle \psi _{1,n+1},n{{\mathcal {A}}}_{1}\phi _{1}\rangle \). Since

$$\begin{aligned}{} & {} \Big \langle h[\phi _{0}]h[\phi _{0}]^{*}\left[ \begin{matrix} \psi _{1,n+1}\\ -{\overline{\psi }}_{1,n+1}\\ \end{matrix}\right] ,\left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix}\right] \Big \rangle \\{} & {} \quad =\Big \langle -h[\phi _{0}]h[\phi _{0}]^{*}\left[ \begin{matrix} \psi _{1,n+1}\\ {\overline{\psi }}_{1,n+1}\\ \end{matrix} \right] +2h[\phi _{0}]h[\phi _{0}]^{*}\left[ \begin{matrix} \psi _{1,n+1}\\ 0\\ \end{matrix} \right] ,\left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix} \right] \Big \rangle \\{} & {} \quad =2\Big \langle h[\phi _{0}]h[\phi _{0}]^{*}\left[ \begin{matrix} \psi _{1,n+1}\\ 0\\ \end{matrix} \right] ,\left[ \begin{matrix} \phi _{1}\\ -{\overline{\phi }}_{1}\\ \end{matrix} \right] \Big \rangle \\{} & {} \quad =2\Big \langle \left[ \begin{matrix} [{\overline{Y}}\phi _{0}({\overline{Y}}\phi _{0})^{*}+(Y\phi _{0})(Y\phi _{0})^{*}]\psi _{1,n+1}\\ -[{\overline{Y}}{\overline{\phi }}_{0}( {\overline{Y}}\phi _{0})^{*}+(Y{\overline{\phi }}_{0})(Y\phi _{0})^{*}]\psi _{1,n+1}\\ \end{matrix} \right] ,\left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix} \right] \Big \rangle \\{} & {} \quad =2\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle +\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) \\{} & {} \qquad -2\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}{\overline{\phi }}_{0}, {\overline{\phi }}_{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle +\langle Y_{j}{\overline{\phi }}_{0},{\overline{\phi }}_{1}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) \\{} & {} \quad =2\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle +\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1},\rangle \right) \\{} & {} \qquad -2\sum _{j=1}^{n}\left( \langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle +\langle \phi _{1},\overline{Y }_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) \\{} & {} \quad =2\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle -\langle \phi _{1}, {\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) \\{} & {} \qquad +4i{\textrm{Im}}\,\sum _{j=1}^{n}\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle , \end{aligned}$$

by (7.6), one has

$$\begin{aligned} \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle =\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle . \end{aligned}$$
(7.9)

Therefore,

$$\begin{aligned}{} & {} 2i{\textrm{Im}}\,\langle \psi _{1,n+1},n{\mathcal {A}}_{1}\phi _{1}\rangle \\{} & {} \quad =\langle \psi _{1,n+1},n{\mathcal {A}}_{1}\phi _{1}\rangle +\langle - {\overline{\psi }}_{1,n+1},n{\mathcal {A}}_{1}{\overline{\phi }}_{1}\rangle \\{} & {} \quad =\Big \langle \left[ \begin{matrix} \psi _{1,n+1}\\ -{\overline{\psi }}_{1,n+1}\\ \end{matrix} \right] ,n{\mathcal {A}}_{1}\left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix} \right] \Big \rangle \\{} & {} \quad =\Big \langle \left[ \begin{matrix} \psi _{1,n+1}\\ -{\overline{\psi }}_{1,n+1}\\ \end{matrix} \right] ,\left[ \begin{matrix} \Phi _{0}\\ {\overline{\Phi }}_{0}\\ \end{matrix} \right] -2h[\phi _{0}]h[\phi _{0}]^{*}\left[ \begin{matrix} \phi _{1}\\ {\overline{\phi }}_{1}\\ \end{matrix} \right] \Big \rangle \\{} & {} \quad =2i{\textrm{Im}}\,\Big \langle \psi _{1,n+1},\Phi _{0}\Big \rangle \\{} & {} \qquad -4\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle +\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \right) \\{} & {} \qquad +4\sum _{j=1}^{n}\left( \langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle +\langle \phi _{1},\overline{Y }_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}{\overline{\phi }}_{0}\rangle \right) \\{} & {} \quad =2i{\textrm{Im}}\,\langle \psi _{1,n+1},\Phi _{0}\rangle -4\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},\phi _{1}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle -\langle \phi _{1}, {\overline{Y}}_{j}\phi _{0}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \right) \\{} & {} \qquad -8i{\textrm{Im}}\,\sum _{j=1}^{n}\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \quad =2i{\textrm{Im}}\,\langle \psi _{1,n+1},\Phi _{0}\rangle +8i{\textrm{Im}} \,\sum _{j=1}^{n}\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \qquad -8i{\textrm{Im}}\,\sum _{j=1}^{n}\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle . \end{aligned}$$

Notice that

$$\begin{aligned} 0= & {} {{\mathcal {L}}}_{0}\langle \psi _{1,n+1},\phi _{0}\rangle =\langle { {\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle +\langle \psi _{1,n+1},{ {\mathcal {L}}}_{0}\phi _{0}\rangle \\{} & {} +4\sum _{j=1}^{n}\left( \langle Y_{j}\psi _{1,n+1}Y_{j}\phi _{0}\rangle +4\langle {\overline{Y}}_{j}\psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{j=1}^{n}\left( \langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle +\langle \psi _{1,n+1},{\overline{Y}}_{j}Y_{j}\phi _{0}\rangle \right) =\sum _{j=1}^{n}\left( \langle Y_{j}{\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle +\langle {\overline{Y}}_{j}\phi _{0},{\overline{Y}}_{j}\psi _{1,n+1}\rangle \right) . \end{aligned}$$

This implies

$$\begin{aligned} \sum _{j=1}^{n}\left( \langle \psi _{1,n+1},{\overline{Y}}_{j}Y_{j}\phi _{0}\rangle -\langle Y_{j}{\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \right) =\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},{\overline{Y}} _{j}\psi _{1,n+1}\rangle -\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) . \end{aligned}$$

Thus,

$$\begin{aligned}{} & {} 2i\sum _{j=1}^{n}{\textrm{Im}}\,\langle \psi _{1,n+1},{\overline{Y}} _{j}Y_{j}\phi _{0}\rangle -(n-1)\langle T\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \quad =\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},{\overline{Y}}_{j}\psi _{1,n+1}\rangle -\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) \end{aligned}$$

and

$$\begin{aligned}{} & {} 2i\sum _{j=1}^{n}{\textrm{Im}}\,\langle \psi _{1,n+1},Y_{j}\overline{ Y}_{j}\phi _{0}\rangle \\{} & {} \quad =2i\sum _{j=1}^{n}{\textrm{Im}}\,\langle \psi _{1,n+1},{\overline{Y}} _{j}Y_{j}\phi _{0}\rangle +(n-1)2i{\textrm{Im}}\,\langle \psi _{1,n+1},T\phi _{0}\rangle \\{} & {} \quad =\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},{\overline{Y}} _{j}\psi _{1,n+1}\rangle -\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) \\{} & {} \qquad +(n-1)\langle T\phi _{0},\psi _{1,n+1}\rangle +2i(n-1){\textrm{Im}} \,\langle \psi _{1,n+1},T\phi _{0}\rangle \\{} & {} \quad =\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},{\overline{Y}} _{j}\psi _{1,n+1}\rangle -\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) +(n-1)\langle \psi _{1,n+1},T\phi _{0}\rangle . \end{aligned}$$

Therefore,

$$\begin{aligned}{} & {} i{\textrm{Im}}\,\langle \psi _{1,n+1},{{\mathcal {L}}}_{0}\phi _{0}\rangle =(n-1)2{\textrm{Re}}\,\langle T\phi _{0},\psi _{1,n+1}\rangle \\{} & {} \quad +2\sum _{j=1}^{n}\left( \langle {\overline{Y}}_{j}\phi _{0},{\overline{Y}} _{j}\psi _{1,n+1}\rangle -\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) . \end{aligned}$$

Thus,

$$\begin{aligned} {\textrm{Im}}\,\langle \psi _{1,n+1},{{\mathcal {L}}}_{0}\phi _{0}\rangle =2\sum _{j=1}^{n}\left( {\textrm{Im}}\,\langle {\overline{Y}}_{j}\phi _{0}, {\overline{Y}}_{j}\psi _{1,n+1}\rangle -{\textrm{Im}}\,\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \right) . \end{aligned}$$

Notice that

$$\begin{aligned} \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle =\langle {\overline{\psi }} _{1,n+1},Y{\overline{\phi }}_{0}\rangle =\overline{\langle \psi _{1,n+1},{\overline{Y}} _{j}\phi _{0}\rangle }=\langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle \end{aligned}$$

and

$$\begin{aligned} \langle \psi _{1,n+1}{{\mathcal {L}}}_{0}\phi _{0}\rangle +\langle {{\mathcal {L}}} _{0}\psi _{1,n+1},\phi _{0}\rangle =-4(\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle +\langle {\overline{Y}}_{j}\psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle ). \end{aligned}$$

Then

$$\begin{aligned}{} & {} {\textrm{Im}}\,(\langle \psi _{1,n+1},{{{\mathcal {L}}}}_{0}\phi _{0}\rangle -\langle {{{\mathcal {L}}}}_{0}\psi _{1,n+1},\phi _{0}\rangle ) \\{} & {} \quad =2{\textrm{Im}}\,\langle \psi _{1,n+1},{{\mathcal {L}}}_{0}\phi _{0}\rangle +4\sum _{j=1}^{n}{\textrm{Im}}\,(\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle +\langle {\overline{Y}}_{j}\psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle ) \\{} & {} \quad =4\sum _{j=1}^{n}{\textrm{Im}}\,\langle {\overline{Y}}_{j}\phi _{0},\overline{ Y}_{j}\psi _{1,n+1}\rangle -4\sum _{j=1}^{n}{\textrm{Im}}\,\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle \\{} & {} \qquad +4\sum _{j=1}^{n}{\textrm{Im}}\,(\langle Y_{j}\psi _{1,n+1},Y_{j}\phi _{0}\rangle +\langle {\overline{Y}}_{j}\psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle ) \\{} & {} \quad =0. \end{aligned}$$

Since

$$\begin{aligned} \Phi _{0}=-{\frac{{{\mathcal {A}}}_{1}}{2}}{{\mathcal {L}}}_{0}\phi _{0}+2\langle T\phi _{0},\phi _{0}\rangle T\phi _{0}+\sum _{j=1}^{n}\left( Y_{j}\phi _{0} {\overline{Y}}_{j}{{\mathcal {A}}}_{1}+{\overline{Y}}_{j}\phi _{0}Y_{j}{{\mathcal {A}}} _{1}\right) \end{aligned}$$

and (7.7), one has

$$\begin{aligned}{} & {} {{\mathcal {A}}_{1}\over 2} {\textrm{Im}}\,\langle {{{\mathcal {L}}}}_{0}\psi _{1,n+1},\phi _{0}\rangle +2{\textrm{Im}}\,\langle \psi _{1,n+1},\langle T\phi _{0},\phi _{0}\rangle T\phi _{0}\rangle -{\textrm{Im}}\,\langle \psi _{1,n+1}.\Phi _{0}\rangle \\{} & {} \quad ={\frac{{\mathcal {A}}_{1}}{2}}{\textrm{Im}}\,(\langle \psi _{1,n+1},{ {\mathcal {L}}}_{0}\phi _{0}\rangle -\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle )-\sum _{j=1}^{n}{\textrm{Im}}\,\langle \psi _{1,n+1},Y_{j}\phi _{0}{\overline{Y}}_{j}{{\mathcal {A}}}_{1}+{\overline{Y}} _{j}\phi _{0}Y_{j}{{\mathcal {A}}}_{1}\rangle \\{} & {} \quad =0. \end{aligned}$$

Therefore, by (7.9)

$$\begin{aligned}{} & {} {1\over 2}{\textrm{Im}}\,\langle \Psi _{1,n+1},\phi _{0}\rangle \\{} & {} \quad =-{\frac{{\mathcal {A}}_{1}}{2}}{\textrm{Im}}\,\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle -{\textrm{Im}}\,\langle \psi _{1,n+1},n{\mathcal {A}} _{1}\phi _{1}\rangle +2{\textrm{Im}}\,\Big \langle \psi _{1,n+1},\langle T\phi _{0},\phi _{0}\rangle T\phi _{0}\Big \rangle \\{} & {} \qquad +4\sum _{j=1}^{n}{\textrm{Im}}\Big (\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle +\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \Big ) \\{} & {} \quad =-{\frac{{\mathcal {A}}_{1}}{2}}{\textrm{Im}}\,\langle {{\mathcal {L}}}_{0}\psi _{1,n+1},\phi _{0}\rangle +2{\textrm{Im}}\,\Big \langle \psi _{1,n+1},\langle T\phi _{0},\phi _{0}\rangle T\phi _{0}\Big \rangle -{\textrm{Im}}\,\langle \psi _{1,n+1},\Phi _{0}\rangle \\{} & {} \qquad +4\sum _{j=1}^{n}{\textrm{Im}}\Big (\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle {\overline{Y}}_{j}\phi _{0},\psi _{1,n+1}\rangle +\langle \phi _{1},Y_{j}\phi _{0}\rangle \langle \psi _{1,n+1},{\overline{Y}}_{j}\phi _{0}\rangle \Big ) \\{} & {} \qquad -4{\textrm{Im}}\,\sum _{j=1}^{n}\langle \phi _{1},{\overline{Y}}_{j}\phi _{0}\rangle \langle \psi _{1,n+1},Y_{j}\phi _{0}\rangle +4{\textrm{Im}} \,\sum _{j=1}^{n}\langle Y_{j}\phi _{0},\phi _{1}\rangle \langle {\overline{Y}} _{j}\phi _{0},\psi _{1,n+1},\rangle \\{} & {} \quad =0. \end{aligned}$$

The proof of the proposition is complete. \(\square \)

8 When \(h[\phi _0] h[\phi _0]^*\) has rank \(2m-2\)

When \(h[\phi _0] h[\phi _0]^*\) has rank \(2m-2\), the harmonic map u has better regularity. Precisely, we can state it as the following theorem.

Theorem 8.1

Let \(\phi _{0}:\partial B_{n}\rightarrow \partial B_{m}\) be a \( C^{\infty }\) map satisfying (1.4) and \(h[\phi _{0}]h[\phi _{0}]^{*} \) having the rank \(2m-2\). Then

(a) If \(E_{b}[\phi _{0}]>0\) on \(\partial B_{n}\) then the solution u of (1.8) belongs to \(C^{n+1, \alpha }({\overline{B}}_n)\) for any \(0<\alpha <1\);

(ii) If \(|\partial _b u| |{\overline{\partial }}_b u|> 0\) on \(\partial B_n\), then the solution u of (1.8) has the following asymptotic expansion:

$$\begin{aligned} u(z)=\phi (z)+\psi _{1}(1-|z|)^{n+2}\log (1-|z|)+\sum _{\ell =2}^{\infty }\psi _{\ell }\Big ((1-|z|)^{(n+1)}\log (1-|z|)\Big )^{\ell }, \end{aligned}$$

where \(\phi \) and \(\psi _{\ell }\in C^{\infty }({\overline{B}}_{n})\).

Proof

Since \(h[\phi _{0}]h[\phi _{0}]^{*}\) has rank \(2m-2\), one has \( \psi _{1,n+1}=0\). Thus, if \(E_{b}[\phi _{0}]\ne 0\) on \(\partial B_{n}\), one has \(u\in C^{n+1,\alpha }\) for all \(0<\alpha <1\). If \(|\partial _{b}u|| {\overline{\partial }}_{b}u|>0\) on \(\partial B_{n}\), by Theorem 1.3, one has Part (ii) of the theorem holds. \(\square \)