The Singular Values of the Logarithmic Potential Transform on Bound States Spaces of Landau Hamiltonians on the Poincaré Disk

Abstract

In the present manuscript, we prove that the singular numbers of the Cauchy transform Lσfz=−1π∫Dfξξ−zlog1∣z−ξ∣1−ξξ¯σ−2dμξ (2) defined on the space L2,σD of complex-valued measurable functions, which are 1−ξξ¯σ−2dμξ-square integrable on D where σ>1 is a fixed parameter, are asymptotically ≈Ckm−4ν+1,ask→∞ where C is a constant.

1. Introduction

Let $\mathbb{D}$ be the complex unit disk endowed with its Lebesgue measure $\mu$ and let $\mathrm{\partial ID}$ be its boundary. Denote by $L^2\left(\mathbb{D},\mathit{d\mu }\right)$ the space of complex-valued measurable functions, which are $\mathit{d\mu }$ square integrable on $\mathbb{D}$. The logarithmic potential transform: $L^2\left(\mathbb{D}\right)\to L^2\left(\mathbb{D}\right)$ is defined by

$$\mathcal{L}\left[f\right]\left(z\right)=-\frac{1}{\pi }{\int }_{\mathbb{D}}\frac{f\left(\xi \right)}{\xi -z}\mathit{log}\left(\frac{1}{\mid z-\xi \mid }\right)\mathit{d\mu }\left(\xi \right).$$E1

This operator is very important as the transformed Cauchy and it often appears in Analysis [1].

The dimensional analysis [1, 2] and scaling arguments form an integral part of theoretical physics to solve some important problems without doing much calculation.

The logarithmic potential in physics forms an interesting one as it provides some unusual predictions about the system. Moreover, this potential can be used suitably to illustrate some of the important features of field theory such as dimensional regularization and renormalization. In most of our textbooks, this potential is not discussed in detail; although the calculations are quite simple to demonstrate some of its unique features. We have obtained the bound state energy of this logarithmic potential through uncertainty principle, phase space quantization, and the Hellmann-Feynman theorem.

In Ref. [3] the authors have been dealing with the restriction of $\mathcal{L}$ to the space $L_a^2\left(\mathbb{D}\right)$ of analytic $\mu$-square integrable on $\mathbb{D}$. They precisely have considered the projection operator $P_0$: $L^2\left(\mathbb{D}\right)\to L_a^2\left(\mathbb{D}\right)$ and they have proved that the singular values ${\lambda }_k$ of $\mathcal{L}{P}_0$, (which turn out to be eigenvalues of the operator $\sqrt{{\left({\mathcal{L}\mathcal{P}}_0\right)}^{\ast }\left(\mathcal{L}{P}_0\right)}$ behave like $k^{-1}$ as $k$ goes to $\infty$. They also concluded that $\mathcal{L}{P}_0$ belongs to the Schatten class $S_{1,\infty }$.

Now, consider the following weighted logarithmic potential transform

$${\mathcal{L}}_{\sigma }\left[f\right]\left(z\right)=-\frac{1}{\pi }{\int }_{\mathbb{D}}\frac{f\left(\xi \right)}{\xi -z}\mathit{log}\left(\frac{1}{\mid z-\xi \mid }\right){\left(1-\xi \overline{\xi }\right)}^{\sigma -2}\mathit{d\mu }\left(\xi \right),$$E2

defined on the space $L^{2,\sigma }\left(\mathbb{D}\right)$ of complex-valued measurable functions, which are ${\left(1-\xi \overline{\xi }\right)}^{\sigma -2}\mathit{d\mu }\left(\xi \right)$-square integrable on $\mathbb{D}$ where $\sigma >1$ is a fixed parameter. We observe that the subspace $L_a^{2,\sigma }\left(\mathbb{D}\right)$ of analytic functions on $\mathbb{D}$ and belonging to $L^{2,\sigma }\left(\mathbb{D}\right)$ coincides with the eigenspace

$${\mathcal{A}}_0^{\sigma }\left(\mathbb{D}\right)≔\left\{\psi \in L^{2,\sigma }\left(\mathbb{D}\right),{\mathrm{\Delta }}_{\sigma }\psi =0\right\},$$E3

of the second order differential operator

$${\mathrm{\Delta }}_{\sigma }≔-4\left(1-z\overline{z}\right)\left(\left(1-z\overline{z}\right)\frac{{\partial }^2}{\partial z\partial \overline{z}}-\sigma \overline{z}\frac{\partial }{\partial \overline{z}}\right),$$E4

known as the $\sigma$-weight Maass Laplacian and its discrete eigenvalues are given by

$${\epsilon }_m≔4m\left(\sigma -1-m\right),m=\mathrm{0,1,2},\dots ,⌊\left(\sigma -1\right)/2⌋,$$E5

with their corresponding eigenspaces

$${\mathcal{A}}_m^{\sigma }\left(\mathbb{D}\right)≔\left\{\psi \in L^{2,\sigma }\left(\mathbb{D}\right)\text{and}{\mathrm{\Delta }}_{\sigma }\psi ={\epsilon }_m^{\sigma }\psi \right\},$$E6

are here called generalized Bergman spaces since …

After noticing that, we here deal with analogous questions as in Ref. [3] in the context of the weighted Cauchy transform (2) and for its restriction to the space ${\mathcal{A}}_m^{\sigma }\left(\mathbb{D}\right)$. That is, we are concerned with the operator ${\mathcal{C}}_{\sigma }{P}_m^{\sigma }$ where $P_m^{\sigma }$ is the projection $L^{2,\sigma }\left(\mathbb{D}\right)\to {\mathcal{A}}_m^{\sigma }\left(\mathbb{D}\right)$. The results achieved are as follows:

Firstly, we find that the singular values of ${\mathcal{L}}_{\sigma }{P}_m^{\sigma }$. For $k\ne m$, it can be expressed as

$${\lambda }_k=\sqrt{J_1+J_2+J_3},$$

where

$$\begin{array}{c}{J}_1={\left(\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(2n+2k-2m+6-1\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(2n+2k-4m+4\nu +6\right)},\\ J_2={\left(\frac{{\alpha }_k^{\nu ,m}}{2\nu -m-1}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(4\nu -2m-1\right)\mathrm{\Gamma }\left(2n+2\right)}{\mathrm{\Gamma }\left(2n+4\nu -2m+1\right)},\end{array}$$

and

$$J_3=\frac{{\left(1+k-m\right)}_m{\alpha }_k^{\nu ,m}}{m!\left(k-m+1\right)\left(2\nu -m-1\right)}\left(\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(k-m+2\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(4\nu -k-3m\right)}\right).$$

For $k=m$ can be expressed as

$${\lambda }_k^2=\frac{{\alpha }_k^{\nu ,m}\left(2\left(2\nu -m\right)-1\right)}{8\left(\pi \left(2\nu -m+1\right)\right)}\sum _{n=0}^{\infty }\frac{B_n}{n+2\nu -m},$$E7

where

$$\begin{array}{c}{B}_n=\sum _{n=0}^{\infty }\frac{\mathrm{\Gamma }\left(-m+1\right)\mathrm{\Gamma }\left(2\nu -m\right)\mathrm{\Gamma }\left(2\left(\nu -m\right)+1\right)}{n!\mathrm{\Gamma }(2\left(\nu -m\right)\mathrm{\Gamma }\left(2\nu -m+2\right)},\\ {\alpha }_k^{\nu ,m}=\frac{\mathrm{\Gamma }\left(2\right)\mathrm{\Gamma }\left(2\left(m-\nu \right)+1\right)}{\mathrm{\Gamma }\left(m+1\right)\mathrm{\Gamma }\left(2+m-2\nu \right)}.\end{array}$$

Secondly, we show that these singular values behave like

$${\lambda }_k\sim C\sqrt{k^{m-4\nu +1}},\mathrm{as}k\to \infty ,$$

where $C$ is a constant.

The paper is organized as follows: In Section 2, we review the definition of the weighted logarithmic potential transform, as well as some of its needed properties. Section 3 deals with some basic facts on the spectral theory of Mass Laplacians on the Poincaré disk. In Section 4, a precise description of the generalized Bergmann spaces is reviewed. Section 5 is devoted to the computation of the singular values of the weighted logarithmic potential transform. The asymptotic behavior of these singular values is established in Section 6.

2. The weighted logarithmic potential transform ${\mathcal{L}}_{\nu }$

2.1 The case $\nu =1$

Let $\mathbb{D}$ the complex unit disk endowed with its Lebesgue measure $\mu$ and let $\mathrm{\partial ID}$ its boundary denote by $L^2\left(\mathbb{D}\right)$ the space of complex-valued measurable functions on $\mathbb{D}$ with finite norm

$$\Vert f\Vert ={\int }_{\mathbb{D}}{\left|f\left(\xi \right)\right|}^2\mathit{d\mu }\left(\xi \right).$$E8

The Logarithmic Potential operator $\mathcal{L}:L^2\left(\mathbb{D}\right)\to L^2\left(\mathbb{D}\right)$ is defined by

$$\mathcal{L}\left[f\right]\left(z\right)={\int }_{\mathrm{I}\mathrm{D}}f\left(\xi \right)log\left(\frac{1}{\left|\xi -z\right|}\right)\mathit{d\mu }\left(\xi \right).$$E9

2.2 The case of $\nu \ge 1$

We fix a real parameter $\nu$ such that $2\nu >1$ and we consider the following weighted logarithmic potential transform

$${\mathcal{L}}_{\nu }\left[f\right]\left(z\right)={\int }_{\mathbb{D}}f\left(\xi \right)log\left(\frac{1}{\left|\xi -z\right|}\right){\left(1-\xi \overline{\xi }\right)}^{2\nu -2}\mathit{d\mu }\left(\xi \right),$$E10

defined on the space $L^{2,\nu }\left(\mathbb{D}\right)$ complex-valued measurable functions are ${\left(1-\xi \overline{\xi }\right)}^{2\nu -2}\mathit{d\mu }\left(\xi \right)$-square integrable on $\mathbb{D}$. As a convolution of $L^{2,\nu }$-functions with the compactly supported measure $\frac{{\left(1-\xi \right)}^{2\nu -2}}{\xi }{\mathbb{1}}_{\mathrm{D}}\mathit{d\mu }\left(\xi \right)$${\mathcal{L}}_{\nu }:L^{2,\nu }\left(\mathbb{D}\right)\to L^{2,\nu }\left(\mathbb{D}\right)$ is obviously bounded. Moreover, it is not hard to show that ${\mathcal{L}}_{\nu }$ is in fact compact [4]. This raises a question concerning the spectral picture of ${\mathcal{L}}_{\nu }$.

3. The Landau Hamiltonian $H_{\nu }$ on the Poincaré disk $\mathbb{D}$

Let $\mathbb{D}$ =$\left\{z\in \mathrm{ℂ},z\overline{z}<1\right\}$ be the complex unit disk with the Poincaré metric ${\mathit{ds}}^2=4{\left(1-z\overline{z}\right)}^{-2}\mathit{dzd}\overline{z}.$$\mathbb{D}$ is a complete Riemannian manifold with all sectional curvature equal $-1.$ It has an ideal boundary $\mathrm{\partial ID}$ identified with the circle $\left\{\omega \in \mathrm{ℂ},\omega \overline{\omega }=1\right\}.$ One refers to points $\omega \in \mathrm{\partial ID}$ as points at infinity. The geodesic distance between two points $z$ and $w$ is given by

$$coshd\left(z,w\right)=1+\frac{2\left(z-w\right)\left(\overline{z}-\overline{w}\right)}{\left(1-z\overline{z}\right)\left(1-w\overline{w}\right)}.$$E11

By Ref. [5] the Schrödinger operator on $\mathbb{D}$ with a constant magnetic field of strength proportional to $\nu >0$ can be written as$:$

$${\mathcal{L}}_{\nu }≔-{\left(1-{\left|z\right|}^2\right)}^2\frac{{\partial }^2}{\partial z\partial \overline{z}}-\mathit{\nu z}\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial z}+\nu \overline{z}\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial \overline{z}}+{\nu }^2{\left|z\right|}^2.$$E12

which is also called Maass Laplacian on the disk. A slight modification of ${\mathcal{L}}_{\nu }$ is given by the operator

$$H_{\nu }≔4{\mathcal{L}}_{\nu }-4{\nu }^2$$E13

acting in the Hilbert space

$$L^{2,0}\left(\mathbb{D}\right)≔\left\{\phi :\mathbb{D}\to \mathrm{ℂ},{\int }_{\mathbb{D}}{\left|\phi \left(z\right)\right|}^2{\left(1-{\left|z\right|}^2\right)}^{-2}\mathit{d\mu }\left(z\right)<+\infty \right\},$$E14

For our purpose, we shall consider the unitary equivalent realization ${\tilde{H}}_{\nu }$ of the operator $H_{\nu }$ in the Hilbert space

$$L^{2,\nu }\left(\mathbb{D}\right)≔\left\{\phi :\mathbb{D}\to \mathbb{C},{\int }_{\mathbb{D}}{\left|\phi \left(z\right)\right|}^2{\left(1-{\left|z\right|}^2\right)}^{2\nu -2}\mathit{d\mu }\left(z\right)<+\infty \right\},$$E15

which is defined by

$${\tilde{H}}_{\nu }≔Q_{\nu }^{-1}{H}_{\nu }{Q}_{\nu },$$E16

where $Q_{\nu }:L^{2,\nu }\left(\mathbb{D}\right)\to L^{2,0}\left(\mathbb{D}\right)$ is the unitary transformation defined by the map $\phi \mapsto Q_{\nu }\left[\phi \right]≔{\left(1-{\left|z\right|}^2\right)}^{-\nu }\phi .$ Different aspects of the spectral analysis of the operator ${\tilde{H}}_{\nu }$ have been studied by many authors. For instance, note that ${\tilde{H}}_{\nu }$ is an elliptic densely defined operator on the Hilbert space $L^{2,\nu }\left(\mathbb{D}\right)$ and admits a unique self-adjoint realization that we denote also by ${\tilde{H}}_{\nu }.$ The spectrum of ${\tilde{H}}_{\nu }$ in $L^{2,\nu }\left(\mathbb{D}\right)$ consists of two parts: $\left(i\right)$ a continuous part $\left[1,+\infty \right[$, which corresponds to scattering states, $\left(\mathit{ii}\right)$ a finite number of eigenvalues (hyperbolic Landau levels) of the form

$${\epsilon }_m^{\nu }≔4\left(\nu -m\right)\left(1-\nu +m\right),m=\mathrm{0,1,2},\cdots ,\left[\nu -\frac{1}{2}\right]$$E17

with infinite degeneracy, provided that $2\nu >1$. The eigenvalues in (17) correspond eigenfunctions, which are called bound states since the particle in such a state cannot leave the system without additional energy. A concrete description of these bound states spaces will be the goal of the next section.

4. The bound states spaces ${\mathcal{A}}_{\nu ,m}^2\left(\mathbb{D}\right)$

Here, we consider the eigenspace

$${\mathcal{A}}_{\nu ,m}^2\left(\mathbb{D}\right)≔\left\{\mathrm{\Phi }:\mathbb{D}\to \mathbb{C},\mathrm{\Phi }\in L^{2,\nu }\left(\mathbb{D}\right)\text{and}{\tilde{H}}_{\nu }\mathrm{\Phi }={\epsilon }_m^{\nu }\mathrm{\Phi }\right\}.$$E18

See Refs. [6, 7], for the following proposition.

Proposition 4.1.Let$2\nu >1$and$m=\mathrm{0,1,2},\cdots ,\left[\nu -\frac{1}{2}\right].$Then, we have.

$\left(i\right)$an orthogonal basis of ${\mathcal{A}}_{\nu ,m}^2\left(\mathbb{D}\right)$is given by the functions

$$\begin{array}{c}{\varphi }_k^{\nu ,m}\left(z\right)≔{\left|z\right|}^{\mid m-k\mid }{\left(1-{\left|z\right|}^2\right)}^{-m}{e}^{-i\left(m-k\right)argz}\\ {\times }_2{F}_1\left(-m+\frac{m-k+\mid m-k\mid }{2},2\nu -m+\frac{\mid m-k\mid -m+k}{2},1+|m-k|;{\left|z\right|}^2\right)\end{array}$$E19

$k=\mathrm{0,1,2},\cdots ,$in terms of a terminating${}_2{F}_1$Gauss hypergeometric function.

$\left(\mathit{ii}\right)$the norm square of${\varphi }_k^{\nu ,m}$in$L^{2,\nu }\left(\mathbb{D}\right)$is given by

$${‖{\varphi }_k^{\nu ,m}‖}^2=\frac{\pi {\left(\mathrm{\Gamma }\left(1+\left|m-k\right|\right)\right)}^2}{\left(2\left(\nu -m\right)-1\right)}\frac{\mathrm{\Gamma }\left(m-\frac{\mid m-k\mid +m-k}{2}+1\right)\mathrm{\Gamma }\left(2\nu -m-\frac{\mid m-k\mid +m-k}{2}\right)}{\mathrm{\Gamma }\left(m+\frac{\mid m-k\mid -m+k}{2}+1\right)\mathrm{\Gamma }\left(2\nu -m+\frac{\mid m-k\mid -m+k}{2}\right)}.$$E20

Corollary 4.1.The functions$\left\{{\mathrm{\Phi }}_k^{\nu ,m}\right\},k=\mathrm{0,1,2},\dots ,$given by

$${\mathrm{\Phi }}_k^{\nu ,m}\left(z\right)≔{\left(-1\right)}^k{\left(\frac{2\left(\nu -m\right)-1}{\pi }\right)}^{\frac{1}{2}}{\left(\frac{k!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\right)}{m!\mathrm{\Gamma }\left(2\left(\nu -m\right)+k\right)}\right)}^{\frac{1}{2}}$$E21

$$\times {\left(1-{\left|z\right|}^2\right)}^{-m}{\overline{z}}^{m-k}{P}_k^{\left(m-k,2\left(\nu -m\right)-1\right)}\left(1-2z\overline{z}\right),$$E22

in terms of Jacobi polynomials constitute an orthonormal basis of ${\mathcal{A}}_m^{2,\nu }\left(\mathbb{D}\right)$.

Proof. Write the connection between the ${}_2{F}_1$-sum and the Jacobi polynomial

$$P_k^{\alpha ,\beta }\left(u\right)=\frac{{\left(1+\alpha \right)}_k}{k!}.{}_2{F}_1\left(-k,1+\alpha +\beta +k,1+\alpha ;\frac{1-u}{2}\right),$$

then the functions

$${\varphi }_k^{\nu ,m}\left(z\right)=\frac{{\left(-1\right)}^{min\left(m,k\right)}}{{\left(1-{\left|z\right|}^2\right)}^m}{\left|z\right|}^{\left|m-k\right|}{e}^{-i\left(m-k\right)argz}{P}_{min\left(m,k\right)}^{\left(\left|m-k\right|,2\left(\nu -m\right)-1\right)}\left(1-2z\overline{z}\right),$$E23

constitute an orthonormal basis of ${\mathcal{A}}_{\nu ,m}^2.$ The norm square of ${\varphi }_k^{\nu ,m}$ in $L^{2,\nu }\left(\mathbb{D}\right)$ is given by

$${‖{\varphi }_k^{\nu ,m}‖}^2=\frac{\pi }{\left(2\left(\nu -m\right)-1\right)}\frac{\left(m\vee k\right)!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\wedge k\right)}{\left(m\wedge k\right)!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\vee k\right)}.$$E24

Here, $m\wedge k≔min\left(m,k\right)$ and $m\vee k≔max\left(m,k\right).$ Thus, the set of functions

$${\mathrm{\Phi }}_k^{\nu ,m}≔\frac{{\varphi }_k^{\nu ,m}}{‖{\varphi }_k^{\nu ,m}‖},k=\mathrm{0,1,2},\dots$$E25

is an orthonormal basis of ${\mathcal{A}}_{\nu ,m}^2\left(\mathbb{D}\right)$ and can be rewritten as.

$${\mathrm{\Phi }}_k^{\nu ,m}\left(z\right)={\left(-1\right)}^k{\left(\frac{2\left(\nu -m\right)-1}{\pi }\right)}^{\frac{1}{2}}{\left(\frac{k!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\right)}{m!\mathrm{\Gamma }\left(2\left(\nu -m\right)+k\right)}\right)}^{\frac{1}{2}}$$E26

$$\times {\left(1-{\left|z\right|}^2\right)}^{-m}{\overline{z}}^{m-k}{P}_k^{\left(m-k,2\left(\nu -m\right)-1\right)}\left(1-2z\overline{z}\right)$$E27

by making appeal to the identity $\left(S,p.63\right):$

$$\frac{\mathrm{\Gamma }\left(m+1\right)}{\mathrm{\Gamma }\left(m-s+1\right)}{P}_m^{\left(-s,\alpha \right)}\left(u\right)=\frac{\mathrm{\Gamma }\left(m+\alpha +1\right)}{\mathrm{\Gamma }\left(m-s+\alpha +1\right)}{\left(\frac{u-1}{2}\right)}^s{P}_{m-s}^{\left(s,\alpha \right)}\left(u\right),1\le s\le m$$E28

for $s=m-k,$$t=1-2{\left|z\right|}^2$_, and $\alpha =2\left(\nu -m\right)-1.\dots \square$.

Corollary 4.2.The$L^2-$eigenspace${\mathcal{A}}_{\nu ,0}^2\left(\mathbb{D}\right)$, corresponding to$m=0$in$\left(3.1\right)$and associated with the bottom energy${\epsilon }_0^{\nu }=0$in$\left(2.6\right),$reduces further to the weighted Bergman space consisting of holomorphic functions$\varphi$:$\mathbb{D}\to \mathbb{C}$such that

$$\underset{\mathbb{D}}{\int }{\left|\varphi \left(z\right)\right|}^2{\left(1-{\left|z\right|}^2\right)}^{2\nu -2}\mathit{d\mu }\left(z\right)<+\infty .$$E29

5. Computation of the singular values ${\lambda }_k$

Elements of this basis are given in terms of Jacobi polynomials as

$${\varphi }_k^{\nu ,m}\left(z\right)=\frac{{\left(-1\right)}^{min\left(m,k\right)}}{{\left(1-{\left|z\right|}^2\right)}^m}{\left|z\right|}^{\left|m-k\right|}{e}^{-i\left(m-k\right)argz}{P}_{min\left(m,k\right)}^{\left(\left|m-k\right|,2\left(\nu -m\right)-1\right)}\left(1-2z\overline{z}\right).$$E30

The norm square of ${\varphi }_k^{\nu ,m}$ in $L^{2,\nu }\left(\mathbb{D}\right)$ is given by

$${\rho }_k^{\nu ,m}=\frac{\pi }{\left(2\left(\nu -m\right)-1\right)}\frac{\left(m\vee k\right)!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\wedge k\right)}{\left(m\wedge k\right)!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\vee k\right)}.$$E31

Here, $m\wedge k≔min\left(m,k\right)$ and $m\vee k≔max\left(m,k\right).$ Let us introduce the notation. The set of functions

$${\gamma }_k^{\nu ,m}≔\frac{{\left(-1\right)}^{m\wedge k}}{\sqrt{{\rho }_k^{\nu ,m}}},k=\mathrm{0,1,2},\dots$$E32

So that we consider the elements

$${\mathrm{\Phi }}_k^{\nu ,m}\left(z\right)≔{\gamma }_k^{\nu ,m}\frac{1}{{\left(1-z\overline{z}\right)}^m}{\left|z\right|}^{\left|m-k\right|}{e}^{-i\left(m-k\right)argz}{P}_{min\left(m,k\right)}^{\left(\left|m-k\right|,2\left(\nu -m\right)-1\right)}\left(1-2z\overline{z}\right).$$E33

5.1 The action ${\mathcal{L}}_{\nu }$

Lemma 5.1.We set$z=\rho e^{\mathit{it}}$, and$I=-{\int }_0^{2\pi }{e}^{i\left(k-m\right)\theta }\mathit{log}\left(|z-{\mathit{re}}^{\mathit{i\theta }}|\right)\frac{\mathit{d\theta }}{2\pi }$, we have

$$\left\{\begin{array}{ll}I=-log\left(\rho \wedge r\right)& k=m,\\ I=\frac{e^{i\left(k-m\right)t}}{2\mid m-k\mid }\left({\left(\frac{r}{\rho }\right)}^{m-k}\wedge {\left(\frac{r}{\rho }\right)}^{m-k}\right)& k\ne m,\end{array}\right.$$E34

Proof. By ref. [3], it remains to prove that this lemma for $k<m$.

We have

$${\int }_0^{2\pi }{e}^{i\left(k-m\right)\theta }\mathit{log}\left(\left|\rho e^{\mathit{it}}-{\mathit{re}}^{\mathit{i\theta }}\right|\right)\mathit{d\theta }=-{\int }_0^{2\pi }{e}^{i\left(m-k\right)\left(-\theta \right)}\mathit{log}\left(\left|{\mathit{re}}^{i\left(-t\right)}-\rho e^{i\left(-\theta \right)}\right|\right)d\left(-\theta \right)$$E35

The function $\theta \to e^{i\left(m-k\right)\left(-\theta \right)}\mathit{log}\left(\left|{\mathit{re}}^{i\left(-t\right)}-\rho e^{i\left(-\theta \right)}\right|\right)$ is a periodic mapping with the period equal $2\pi$, then

$$\begin{array}{c}{\int }_0^{2\pi }{e}^{i\left(k-m\right)\theta }\mathit{log}\left(\left|\rho e^{\mathit{it}}-{\mathit{re}}^{\mathit{i\theta }}\right|\right)\mathit{d\theta }\\ =-{\int }_0^{2\pi }{e}^{i\left(m-k\right)\left(-\theta \right)}\mathit{log}\left(\left|{\mathit{re}}^{i\left(-t\right)}-\rho e^{i\left(-\theta \right)}\right|\right)d\left(-\theta \right)\\ =\frac{e^{i\left(k-m\right)t}}{2\left(m-k\right)}\times \left({\left(\frac{r}{\rho }\right)}^{m-k}\wedge {\left(\frac{r}{\rho }\right)}^{m-k}\right).\end{array}$$

□

Lemma 5.2.For all$\lambda \in \partial D$.${\mathcal{L}}_{\nu }$commutes with the rotations$R_{\lambda }$, where

$$\left(R_{\lambda }f\right)\left(z\right)=f\left(\mathit{\lambda z}\right).$$

Proof. We observe that

$$R_{\lambda }{\varphi }_k^{\nu ,m}\left(z\right)={\lambda }^{k-m}{\varphi }_k^{\nu ,m}\left(z\right),\forall k\ne m.$$

□

Corollary 5.1.${\left\{{\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\right\}}_{k=0}^{\infty }$are orthonormal in$L^{2,\nu }\left(\mathbb{D}\right)$.

Proof. As $R_{\lambda }$ is an isometry of $L^{2,\nu }\left(\mathbb{D}\right)$,

$$\begin{array}{c}\left({\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right),{\mathcal{L}}_{\nu }\left({\varphi }_j^{\nu ,m}\right)\right)\\ =\left(R_{\lambda }{\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right),R_{\lambda }{\mathcal{L}}_{\nu }\left({\varphi }_j^{\nu ,m}\right)\right)\\ =\left({\mathcal{L}}_{\nu }{R}_{\lambda }\left({\varphi }_k^{\nu ,m}\right),{\mathcal{L}}_{\nu }{R}_{\lambda }\left({\varphi }_j^{\nu ,m}\right)\right)\\ =\overline{{\lambda }^{j-k}}\left({\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right),{\mathcal{L}}_{\nu }\left({\varphi }_j^{\nu ,m}\right)\right),\mathit{\text{if}}m>k,\end{array}$$

or

$$={\lambda }^{k-j}\left({\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right),{\mathcal{L}}_{\nu }\left({\varphi }_j^{\nu ,m}\right)\right),\mathit{\text{if}}m<k$$

For all $\lambda \in \partial D$, since $\lambda \ne 0$, we have

$$\left({\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right),{\mathcal{L}}_{\nu }\left({\varphi }_j^{\nu ,m}\right)\right)=0\mathit{\text{if}}j\ne k.$$

□

Lemma 5.3.If we denote${}^1{\varphi }_k^{\nu ,m}\left(z\right),\text{if}k>m$and${}^2{\varphi }_k^{\nu ,m}\left(z\right),\text{if}k<m$, we have

$${\mathcal{L}}_{\nu }\left({}^1{\varphi }_k^{\nu ,m}\right)\left(z\right)=\frac{\mathrm{\Gamma }\left(k+1\right)\mathrm{\Gamma }\left(2\nu -m\right)}{\mathrm{\Gamma }\left(m+1\right)\mathrm{\Gamma }\left(2\nu -k\right)}{\mathcal{L}}_{\nu }\left({}^2{\varphi }_k^{\nu ,m}\right)\left(z\right).$$

Proof. Just use

$$\frac{\mathrm{\Gamma }\left(m+1\right)}{\mathrm{\Gamma }\left(m-s+1\right)}{P}_m^{\left(-s,\alpha \right)}\left(u\right)=\frac{\mathrm{\Gamma }\left(m+\alpha +1\right)}{\mathrm{\Gamma }\left(m-s+\alpha +1\right)}{\left(\frac{u-1}{2}\right)}^s{P}_{m-s}^{\left(s,\alpha \right)}\left(u\right),1\le s\le m.$$E36

□

Proposition 5.1.The action of the operator$\mathcal{L}$on a basis element${\varphi }_k^{\nu ,m}$is of the form:

If $k=m$, We put $z=\rho e^{\mathit{i\theta }}$ then

$${\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\left(z\right)=\frac{{\alpha }_k^{\nu ,m}}{2\left(2\nu -m+1\right)}\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_3{F}_2\left(\begin{array}{c}-m+1,2\nu -m,2\nu -m+1\\ 2\left(\nu -m\right),2\nu -m+2\end{array}|1-{\rho }^2\right).$$E37

If$k\ne m$then

$${\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\left(z\right)=\frac{{\mathit{\pi \gamma }}_k^{\nu ,m}{e}^{i\left(k-m\right)t}}{2\left(k-m\right)}\left(I_3+I_4\right),$$

where

$$I_3=\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}{\rho }^{k-m+2}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{cc}-m+1,2\left(\nu -m\right)+k& \\ 2+k-m& \end{array}|{\rho }^2\right),$$

and

$$I_4=\frac{{\alpha }_k^{\nu ,m}}{2\nu -m-1}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{cc}-m+1,2\nu -m-1& \\ 2\left(\nu -m\right),& \end{array}|{\rho }^2\right).$$

Proof. For $k=m$, we have

$$\begin{array}{rcl}{\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\left(z\right)& =& \frac{{\left(-1\right)}^m}{\pi }\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }}{\int }_{\mathbb{D}}{\left(1-{\left|\xi \right|}^2\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2{\left|\xi \right|}^2\right)log\left(\left|z-\xi \right|\right)\mathit{d\mu }\left(\xi \right)\\ & =& {\left(-1\right)}^m\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }}{\int }_0^1{\left(1-r^2\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)log\left(\rho \wedge r\right){\mathit{dr}}^2\\ & =& \frac{{\left(-1\right)}^m}{2}\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }}{\int }_0^1{\left(1-t\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2t\right)log\left({\rho }^2\vee t\right)\mathit{dt}\\ & =& \frac{{\left(-1\right)}^m}{2}\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }}\left[I_1+I_2\right].\end{array}$$

where

$$I_1={\int }_0^{{\rho }^2}{\left(1-t\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2t\right)log\left({\rho }^2\vee t\right)\mathit{dt},$$

and

$$I_2={\int }_{{\rho }^2}^1{\left(1-t\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2t\right)log\left(t\right)\mathit{dt}.$$

Calculus of $I_1$.

$$I_1=log\left({\rho }^2\right){\int }_{{\rho }^2}^1{\left(1-t\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2t\right)\mathit{dt}.$$

We use the formula

$$P_k^{\left(\alpha ,\beta \right)}\left(u\right)={\frac{{\left(1+\alpha \right)}_k}{k!}}_2{F}_1\left(\begin{array}{cc}-k,1+\alpha +\beta +k& \\ 1+\alpha & \end{array}|\frac{1-u}{2}\right).$$

We have

$$I_1=log\left({\rho }^2\right){\int }_0^{{\rho }^2}{{\left(1-t\right)}^{2\nu -m-2}}_2{F}_1\left(\begin{array}{cc}-m,2\nu -m& \\ 1& \end{array}|t\right)\mathit{dt}.$$

By Ref. [8], we have

$$\int x^{c-1}{{\left(1-x\right)}^{b-c-1}}_2{F}_1\left(\begin{array}{cc}a,b& \\ c& \end{array}|x\right)\mathit{dx}=\frac{1}{c}{x}^c{{\left(1-x\right)}^{b-c}}_2{F}_1\left(\begin{array}{cc}a+1,b& \\ c+1& \end{array}|x\right),$$

implies that

$$I_1=log\left({\rho }^2\right){\rho }^2{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{cc}-m+1,2\nu -m& \\ 2& \end{array}|{\rho }^2\right).$$

Calculus of $I_2$.

$$I_2={\int }_{{\rho }^2}^1{\left(1-t\right)}^{2\nu -m-2}{P}_m^{\left(0,2\left(\nu -m\right)-1\right)}\left(1-2t\right)log\left(t\right)\mathit{dt}.$$

Use the previous formula in Ref. [8] and the integration by part gives

$$\begin{array}{rcl}{I}_2& =& {\left[t{{\left(1-t\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|t\right)log\left(t\right)\right]}_{{\rho }^2}^1-{\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|t\right)\mathit{dt}\\ & =& -{\rho }^{2}log\left({\rho }^2\right){{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|{\rho }^2\right)-{\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|t\right)\mathit{dt}.\end{array}$$

Calculus of

$${\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m}}_2{F}_1\left(\begin{array}{cc}-m+1,2\nu -m& \\ 2& \end{array}|t\right)\mathit{dt}.$$

Use the following formula, which has place in [9]

$${}_2{F}_1\left(\begin{array}{c}a,b\\ c\end{array}|t\right)=\begin{array}{l}{\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }\left(c-a-b\right)}{\mathrm{\Gamma }\left(c-a\right)\mathrm{\Gamma }\left(c-b\right)}}_2{F}_1\left(\begin{array}{c}a,b\\ a+b-c+1\end{array}|1-t\right)\\ +\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }\left(a+b-c\right)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{{\left(1-t\right)}^{c-a-b}}_2{F}_1\left(\begin{array}{c}a,b\\ a+b-c+1\end{array}|1-t\right).\end{array}$$

We put $a=1-m$, $b=2\nu -m$, $c=2$ and use the formula Boher-Mollerup, for $z\in {\mathrm{IR}}_{+}^{\ast }$,

$$\mathrm{\Gamma }\left(z\right)=\frac{e^{-\mathit{\gamma z}}}{z}\prod _{n=1}^{\infty }{\left(1+\frac{z}{n}\right)}^{-1}{e}^{-\frac{z}{n}},$$

which implies $\frac{1}{\mathrm{\Gamma }\left(1-m\right)}=0$, then

$${}_2{F}_1\left(\begin{array}{cc}-m+1,2\nu -m& \\ 2& \end{array}|t\right)={\frac{2\mathrm{\Gamma }\left(2\left(m-\nu \right)+1\right)}{m!\mathrm{\Gamma }\left(2+m-2\nu \right)}}_2{F}_1\left(\begin{array}{cc}-m+1,2\nu -m& \\ 2\left(\nu -m\right)& \end{array}|1-t\right),$$

implies that

$${\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|t\right)\mathit{dt}=\frac{2\mathrm{\Gamma }\left(2\left(m-\nu \right)+1\right)}{m!\mathrm{\Gamma }\left(2+m-2\nu \right)}{\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\left(\nu -m\right)\end{array}|1-t\right)\mathit{dt}.$$

By the change $1-t=s$, we get

$${\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|t\right)\mathit{dt}=\frac{2\mathrm{\Gamma }\left(2\left(m-\nu \right)+1\right)}{m!\mathrm{\Gamma }\left(2+m-2\nu \right)}{\int }_0^{1-{\rho }^2}{t^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\left(\nu -m\right)\end{array}|t\right)\mathit{dt}.$$

In [8], p. 44,

$$\int {x^{\alpha -1}}_2{F}_1\left(\begin{array}{cc}a,b& \\ c& \end{array}|-t\right)\mathit{dx}={\frac{x^{\alpha }}{\alpha }}_3{F}_2\left(\begin{array}{cc}a,b,\alpha & \\ c,\alpha +1& \end{array}|-t\right)+\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(a-\alpha \right)\mathrm{\Gamma }\left(b-\alpha \right)\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }\left(c-\alpha \right)}$$

Since $a=1-m$, $b=2\nu -m$, $c=2\left(\nu -m\right)$, and $\alpha =2\nu -m+1$ we have

$$\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(a-\alpha \right)\mathrm{\Gamma }\left(b-\alpha \right)\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }\left(c-\alpha \right)}=0$$

and by the change $t=-s$

$$\begin{array}{l}{\int }_0^{1-{\rho }^2}{t^{2\nu -m+1}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\left(\nu -m\right)\end{array}|t\right)\mathit{dt}\\ ={\left(-1\right)}^m{\int }_0^{{\rho }^2}{t^{2\nu -m}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\left(\nu -m\right)\end{array}|-t\right)\mathit{dt}\\ ={\left(-1\right)}^m{\frac{{\left({\rho }^2-1\right)}^{2\nu -m+1}}{2\nu -m+1}}_3{F}_2\left(\begin{array}{c}-m+1,2\nu -m,2\nu -m+1\\ 2\left(\nu -m\right),2\nu -m+2\end{array}|1-{\rho }^2\right).\end{array}$$

we set ${\alpha }_k^{\nu ,m}=\frac{2\mathrm{\Gamma }\left(2\left(m-\nu \right)+1\right)}{m!\mathrm{\Gamma }\left(2+m-2\nu \right)}$. We get

$$I_2=\begin{array}{l}-{\rho }^{2}log\left({\rho }^2\right){{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\end{array}|{\rho }^2\right)\\ +{\left(-1\right)}^m{\alpha }_k^{\nu ,m}{\frac{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}{2\nu -m+1}}_3{F}_2\left(\begin{array}{c}-m+1,2\nu -m,2\nu -m+1\\ 2\left(\nu -m\right),2\nu -m+2\end{array}|1-{\rho }^2\right).\end{array}$$

Finally

$${\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\left(z\right)=\frac{{\alpha }_k^{\nu ,m}}{2\left(2\nu -m+1\right)}\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_3{F}_2\left(\begin{array}{c}-m+1,2\nu -m,2\nu -m+1\\ 2\left(\nu -m\right),2\nu -m+2\end{array}|1-{\rho }^2\right).$$

Now if $k>m$, set $z=\rho e^{\mathit{it}}$.

$$\begin{array}{c}{\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\left(z\right)={\gamma }_k^{\nu ,m}{\int }_{\mathbb{D}}{\left(1-{\left|\xi \right|}^2\right)}^{2\nu -m-2}{\xi }^{k-m}log\left(\frac{1}{\mid z-\xi \mid }\right)P_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2{\left|\xi \right|}^2\right)\mathit{d\mu }\left(\xi \right)\\ ={\gamma }_k^{\nu ,m}{\int }_0^1{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m+1}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right){\int }_0^{2\pi }{e}^{i\left(k-m\right)\theta }log\left(\frac{1}{\mid z-r^{\mathit{i\theta }}\mid }\right)\mathit{\text{dθdr}}\\ =\frac{{\mathit{\pi \gamma }}_k^{\nu ,m}{e}^{i\left(k-m\right)t}}{2\left(k-m\right)}{\int }_0^1{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)\left({\left(\frac{r}{\rho }\right)}^{k-m}\wedge {\left(\frac{\rho }{r}\right)}^{k-m}\right){\mathit{dr}}^2\\ =\frac{{\mathit{\pi \gamma }}_k^{\nu ,m}{e}^{i\left(k-m\right)t}}{2\left(k-m\right)}\left({\int }_0^{\rho }{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)\left({\left(\frac{r}{\rho }\right)}^{k-m}\wedge {\left(\frac{\rho }{r}\right)}^{k-m}\right){\mathit{dr}}^2\right.\\ +{\int }_{\rho }^1{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)\left({\left(\frac{r}{\rho }\right)}^{k-m}\wedge {\left(\frac{\rho }{r}\right)}^{k-m}\right){\mathit{dr}}^2.\end{array}$$

We set

$$I_3={\int }_0^{\rho }{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)\left({\left(\frac{r}{\rho }\right)}^{k-m}\wedge {\left(\frac{\rho }{r}\right)}^{k-m}\right){\mathit{dr}}^2.$$

and

$$I_4={\int }_{\rho }^1{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)\left({\left(\frac{r}{\rho }\right)}^{k-m}\wedge {\left(\frac{\rho }{r}\right)}^{k-m}\right){\mathit{dr}}^2.$$

Calculus of $I_3$.

$$I_3=\frac{{\rho }^{m-k}{\left(1+k-m\right)}_m}{m!}{\int }_0^{{\rho }^2}{t}^{k-m}{{\left(1-t\right)}^{2\nu -m-2}}_2{F}_1\left(\begin{array}{cc}-m,2\left(\nu -m\right)+k& \\ 1+k-m& \end{array}|t\right)\mathit{dt}.$$

By the formula

$$\int x^{c-1}{{\left(1-x\right)}^{b-c-1}}_2{F}_1\left(\begin{array}{cc}a,b& \\ c& \end{array}|x\right)\mathit{dx}=\frac{1}{c}{x}^c{{\left(1-x\right)}^{b-c}}_2{F}_1\left(\begin{array}{cc}a+1,b& \\ c+1& \end{array}|x\right),$$

we have

$$I_3=\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}{\rho }^{k-m+2}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{cc}-m+1,2\left(\nu -m\right)+k& \\ 2+k-m& \end{array}|{\rho }^2\right)$$

Calculus of $I_4$.

$$\begin{array}{c}{I}_4={\int }_{\rho }^1{\left(1-r^2\right)}^{2\nu -m-2}{r}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2r^2\right)\left({\left(\frac{r}{\rho }\right)}^{k-m}\wedge {\left(\frac{\rho }{r}\right)}^{k-m}\right){\mathit{dr}}^2\\ =\frac{{\rho }^{k-m}{\left(1+k-m\right)}_m}{2m!}{\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m-2}}_2{F}_1\left(\begin{array}{cc}-m,2\left(\nu -m\right)+k& \\ 1+k-m& \end{array}|t\right.\mathit{dt}.\end{array}$$

As the previous

$$\begin{array}{l}{\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m-2}}_2{F}_1\left(\begin{array}{c}-m,2\left(\nu -m\right)+k\\ 1+k-m\end{array}|t\right.\mathit{dt}\\ ={\alpha }_k^{\nu ,m}{\int }_{{\rho }^2}^1{{\left(1-t\right)}^{2\nu -m-2}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\left(\nu -m\right)\end{array}|1-t\right)\mathit{dt}\end{array}$$

$$\begin{array}{c}={\left(-1\right)}^m{\alpha }_k^{\nu ,m}{\int }_{{\rho }^2-1}^0{t^{2\nu -m-2}}_2{F}_1\left(\begin{array}{c}-m+1,2\nu -m\\ 2\left(\nu -m\right)\end{array}|t\right.\mathit{dt}\end{array}$$

$$\begin{array}{c}=\frac{{\alpha }_k^{\nu ,m}}{2\nu -m-1}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_3{F}_2\left(\begin{array}{c}-m+1,2\nu -m,2\nu -m-1\\ 2\left(\nu -m\right),2\nu -m\end{array}|1-{\rho }^2\right.\end{array}$$

also

$${}_3{F}_2\left(\begin{array}{cc}-m+1,2\nu -m,2\nu -m-1& \\ 2\left(\nu -m\right),2\nu -m& \end{array}|1-{\rho }^2\right)={}_2{F}_1\left(\begin{array}{cc}-m+1,2\nu -m-1& \\ 2\left(\nu -m\right)& \end{array}|1-{\rho }^2\right)$$

Now if $k<m$. We have

$${\varphi }_k^{\nu ,m}\left(z\right)={\left(-1\right)}^k\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }\frac{k!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\right)}{m!\mathrm{\Gamma }\left(2\left(\nu -m\right)+k\right)}}{\left(1-{\left|z\right|}^2\right)}^{-m}{\overline{z}}^{m-k}{P}_k^{\left(m-k,2\left(\nu -m\right)-1\right)}\left(1-2{\left|z\right|}^2\right).$$

By the formula

$$\frac{\mathrm{\Gamma }\left(m+1\right)}{\mathrm{\Gamma }\left(m-s+1\right)}{P}_m^{\left(-s,\alpha \right)}\left(u\right)=\frac{\mathrm{\Gamma }\left(m+\alpha +1\right)}{\mathrm{\Gamma }\left(m-s+\alpha +1\right)}{\left(\frac{u-1}{2}\right)}^s{P}_{m-s}^{\left(s,\alpha \right)}\left(u\right),1\le s\le m,$$E38

and put $s=m-k$ and $\alpha =2\left(\nu -m\right)-1$, we have

$$P_k^{\left(m-k,2\left(\nu -m\right)-1\right)}\left(1-2{\left|z\right|}^2\right)=\frac{m!\mathrm{\Gamma }\left(k+\alpha +1\right)}{k!\mathrm{\Gamma }\left(m+\alpha +1\right)}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2{\left|z\right|}^2\right),$$

substituting in the expression of ${\varphi }_k^{\nu ,m}\left(z\right)$, we get

$${\varphi }_k^{\nu ,m}\left(z\right)={\left(-1\right)}^m\sqrt{\frac{2\left(\nu -m\right)-1}{\pi }\frac{m!\mathrm{\Gamma }\left(2\left(\nu -m\right)+k\right)}{k!\mathrm{\Gamma }\left(2\left(\nu -m\right)+m\right)}}{\left(1-{\left|z\right|}^2\right)}^{-m}{z}^{k-m}{P}_m^{\left(k-m,2\left(\nu -m\right)-1\right)}\left(1-2{\left|z\right|}^2\right),$$

it is the same formula for $k>m$, which proves the same formula of ${\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\left(z\right)$ if $k>m$.

Remark 5.1.By the previous formula in [9], we have

$${}_2{F}_1\left(\begin{array}{cc}-m+1,2\left(\nu -m\right)+k& \\ 2\left(\nu -m\right)& \end{array}|{\rho }^2\right)={\frac{k!\mathrm{\Gamma }\left(2+k-m\right)}{\mathrm{\Gamma }\left(1-2\left(\nu -m\right)\right)}}_2{F}_1\left(\begin{array}{cc}-m+1,2\left(\nu -m\right)+k& \\ 2\left(\nu -m\right)& \end{array}|1-{\rho }^2\right).$$

5.2 The spectrum of ${\mathcal{L}}_{\nu }$

Proposition 5.2.If$k\ne m$, then

$${\lambda }_k=\sqrt{J_1+J_2+J_3}.$$

where

$$\begin{array}{c}{J}_1={\left(\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(2n+2k-2m+6-1\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(2n+2k-4m+4\nu +6\right)},\\ J_2={\left(\frac{{\alpha }_k^{\nu ,m}}{2\nu -m-1}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(4\nu -2m-1\right)\mathrm{\Gamma }\left(2n+2\right)}{\mathrm{\Gamma }\left(2n+4\nu -2m+1\right)}.\end{array}$$

and

$$J_3=\frac{{\left(1+k-m\right)}_m{\alpha }_k^{\nu ,m}}{m!\left(k-m+1\right)\left(2\nu -m-1\right)}\left(\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(k-m+2\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(4\nu -k-3m\right)}\right),$$

If$k=m$then

$${\lambda }_k^2=\frac{{\alpha }_k^{\nu ,m}\left(2\left(2\nu -m\right)-1\right)}{8\left(\pi \left(2\nu -m+1\right)\right)}\sum _{n=0}^{\infty }\frac{B_n}{n+2\nu -m}.$$E39

where

$$B_n=\sum _{n=0}^{\infty }\frac{\mathrm{\Gamma }\left(-m+1\right)\mathrm{\Gamma }\left(2\nu -m\right)\mathrm{\Gamma }\left(2\left(\nu -m\right)+1\right)}{n!\mathrm{\Gamma }(2\left(\nu -m\right)\mathrm{\Gamma }\left(2\nu -m+2\right)}.$$

Proof. If $k\ne m$. We have

$$\left({\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)\right)\left(z\right)=\frac{{\mathit{\pi \gamma }}_k^{\nu ,m}\left(I_3+I_4\right)}{2\left(k-m\right)}{e}^{i\left(k-m\right)t}.$$

We set ${H}=\left(L^2\left(\mathbb{D}\right),{\left(1-{\left|\xi \right|}^2\right)}^{2\nu -2}\mathit{d\mu }\left(\xi \right)\right)$, $I_3=I_3\left(\rho \right)$, and $I_4=I_4\left(\rho \right)$ we have

$$\begin{array}{c}{\lambda }_k^2={⟨{\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right),{\mathcal{L}}_{\nu }\left({\varphi }_k^{\nu ,m}\right)⟩}_{{H}}\\ =\frac{{\pi }^2{\gamma }_k^{\nu ,m}}{\left(k-m\right)}{\int }_0^1{\left(I_3\left(\rho \right)+I_4\left(\rho \right)\right)}^2\mathit{\rho d\rho }.\end{array}$$

Calculus of ${\int }_0^1{\left(I_3\left(\rho \right)\right)}^2\mathit{\rho d\rho }$.

$$I_3\left(\rho \right)=\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}{\rho }^{k-m+2}{{\left(1-{\rho }^2\right)}^{2\nu -m-1}}_2{F}_1\left(\begin{array}{cc}-m+1,2\left(\nu -m\right)+k& \\ 2+k-m& \end{array}|{\rho }^2\right).$$

Since

$${}_2{F}_1\left(\begin{array}{cc}-m+1,2\left(\nu -m\right)+k& \\ 2+k-m& \end{array}|{\rho }^2\right)=\sum _{n=0}^{\infty }\frac{{\left(-m+1\right)}_n{\left(2\left(\nu -m\right)+k\right)}_n}{{\left(2+k-m\right)}_n}\frac{{\rho }^{2n}}{n!},$$

then

$${\left(I_3\left(\rho \right)\right)}^2={\left(\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}\right)}^2\sum _{n=0}^{\infty }{A}_n{\rho }^{2n}{\left(1-{\rho }^2\right)}^{4\nu -2m-2}.$$

where

$$A_n=\frac{1}{n!}\sum _{i=0}^n\frac{{\left(-m+1\right)}_i{\left(-m+1\right)}_{n-i}{\left(2\left(\nu -m\right)+k\right)}_i{\left(2\left(\nu -m\right)+k\right)}_{n-i}}{{\left(2\left(\nu -m\right)\right)}_i{\left(2\left(\nu -m\right)\right)}_{n-i}}.$$

Thus

$$J_1={\int }_0^1{\left(I_3\left(\rho \right)\right)}^2\mathit{\rho d\rho }={\left(\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}\right)}^2\sum _{n=0}^{\infty }{A}_n{\int }_0^1{\rho }^{2n+2k-2m+6-1}{\left(1-{\rho }^2\right)}^{4\nu -2m-1-1}\mathit{d\rho }.$$

Use the fact that

$${\int }_0^1{t}^{\alpha -1}{\left(1-t\right)}^{\beta -1}\mathit{dt}=\frac{\mathrm{\Gamma }\left(\alpha \right)\mathrm{\Gamma }\left(\beta \right)}{\mathrm{\Gamma }\left(\alpha +\beta \right)},$$

implies

$${\int }_0^1{\left(I_3\left(\rho \right)\right)}^2\mathit{\rho d\rho }={\left(\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(2n+2k-2m+6-1\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(2n+2k-4m+4\nu +6\right)}.$$E40

Calculus of ${\int }_0^1{\left(I_4\left(\rho \right)\right)}^2\mathit{\rho d\rho }$.

In the same,

$$J_2={\int }_0^1{\left(I_4\left(\rho \right)\right)}^2\mathit{\rho d\rho }={\left(\frac{{\alpha }_k^{\nu ,m}}{2\nu -m-1}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(4\nu -2m-1\right)\mathrm{\Gamma }\left(2n+2\right)}{\mathrm{\Gamma }\left(2n+4\nu -2m+1\right)}.$$E41

Calculus of $2{\int }_0^1\left(I_3\left(\rho \right)\right)\left(I_4\left(\rho \right)\right)\mathit{\rho d\rho }$.

$$J_3=2{\int }_0^1\left(I_3\left(\rho \right)\right)\left(I_4\left(\rho \right)\right)\mathit{\rho d\rho }=\frac{{\left(1+k-m\right)}_m{\alpha }_k^{\nu ,m}}{m!\left(k-m+1\right)\left(2\nu -m-1\right)}\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(k-m+2\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(4\nu -k-3m\right)}.$$E42

If $k=m$.

Since

$${\left({}_3{F}_2\left(\begin{array}{cc}-m+1,2\nu -m,2\left(\nu -m\right)+1& \\ 2\left(\nu -m\right),2\nu -m+2& \end{array}|1-{\rho }^2\right)\right)}^2=\sum _{n=0}^{\infty }\frac{\mathrm{\Gamma }\left(-m+1\right)\mathrm{\Gamma }\left(2\nu -m\right)\mathrm{\Gamma }\left(2\left(\nu -m\right)+1\right)}{n!\mathrm{\Gamma }(2\left(\nu -m\right)\mathrm{\Gamma }\left(2\nu -m+2\right)}{\left(1-{\rho }^2\right)}^n.$$E43

$${\lambda }_k^2=\frac{{\alpha }_k^{\nu ,m}\left(2\left(2\nu -m\right)-1\right)}{8\left(\pi \left(2\nu -m+1\right)\right)}\sum _{n=0}^{\infty }{B}_n{\int }_0^1{\left(1-{\rho }^2\right)}^{n+2\nu -m-1}\mathit{d\rho }=\frac{{\alpha }_k^{\nu ,m}\left(2\left(2\nu -m\right)-1\right)}{8\left(\pi \left(2\nu -m+1\right)\right)}\sum _{n=0}^{\infty }\frac{B_n}{n+2\nu -m},$$E44

where

$$B_n=\sum _{n=0}^{\infty }\frac{\mathrm{\Gamma }\left(-m+1\right)\mathrm{\Gamma }\left(2\nu -m\right)\mathrm{\Gamma }\left(2\left(\nu -m\right)+1\right)}{n!\mathrm{\Gamma }(2\left(\nu -m\right)\mathrm{\Gamma }\left(2\nu -m+2\right)}.$$

6. Asymptotic behavior of singular values ${\lambda }_k$ as $k\to \infty$

Proposition 6.1.

$${\lambda }_k\sim C\sqrt{k^{m-4\nu +1}},\mathit{as}k\to \infty ,$$

where $C$ is a constant.

Proof. If $k>m$, then

$${\lambda }_k=\sqrt{J_1+J_2+J_3},$$

where

$$\begin{array}{c}{J}_1={\left(\frac{{\left(1+k-m\right)}_m}{m!\left(k-m+1\right)}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(2n+2k-2m+6-1\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(2n+2k-4m+4\nu +6\right)},\\ J_2={\left(\frac{{\alpha }_k^{\nu ,m}}{2\nu -m-1}\right)}^2\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(4\nu -2m-1\right)\mathrm{\Gamma }\left(2n+2\right)}{\mathrm{\Gamma }\left(2n+4\nu -2m+1\right)}\end{array}$$

and

$$J_3=\frac{{\left(1+k-m\right)}_m{\alpha }_k^{\nu ,m}}{m!\left(k-m+1\right)\left(2\nu -m-1\right)}\left(\sum _{n=0}^{\infty }{A}_n\frac{\mathrm{\Gamma }\left(k-m+2\right)\mathrm{\Gamma }\left(4\nu -2m-1\right)}{\mathrm{\Gamma }\left(4\nu -k-3m\right)}\right).$$

The limit of ${\lambda }_k$ as $k\to \infty$.

We use the formula

$$\frac{\mathrm{\Gamma }\left(k+a\right)}{k=b}\sim k^{a-b}$$

we have

$$J_1\sim {\left(\frac{k^{-1-m}}{m!}\right)}^2\sum _{n=0}^{\infty }{A}_n\mathrm{\Gamma }\left(4\nu -2m-1\right){\left(2k\right)}^{2m-4\nu -1}\sim k^{-4\nu -1}{2}^{2m-4\nu -1}\mathrm{\Gamma }\left(4\nu -2m-1\right)\sum _{n=0}^{\infty }\frac{A_n}{m!}.$$E45

$$J_2={\mathcal{O}}_{k\sim \infty }\left(1\right)$$E46

In the same

$$J_3\sim k^{m-4\nu +1}\frac{{\alpha }_k^{\nu ,m}\mathrm{\Gamma }\left(4\nu -2m-1\right)}{m!\left(2\nu -m-1\right)}\sum _{n=0}^{\infty }{A}_n.$$E47

Therefore

$${\lambda }_k\sim C\sqrt{k^{m-4\nu +1}},$$

where $C$ is a constant.