# Decay estimates for one Aharonov-Bohm solenoid in a uniform magnetic field
I: Schrödinger equation

Haoran Wang  Department of Mathematics, Beijing Institute of Technology,
Beijing 100081<EMAIL_ADDRESS>,  Fang Zhang  Department of
Mathematics, Beijing Institute of Technology, Beijing 100081;
<EMAIL_ADDRESS>and  Junyong Zhang  Department of Mathematics, Beijing
Institute of Technology, Beijing 100081<EMAIL_ADDRESS>
###### Abstract.

This is the first of a series of papers in which we investigate the decay
estimates for dispersive equations with Aharonov-Bohm solenoids in a uniform
magnetic field. In this first starting paper, we prove the local-in-time
dispersive estimates and Strichartz estimates for Schrödinger equation with
one Aharonov-Bohm solenoid in a uniform magnetic field. The key ingredient is
the construction of Schrödinger propagator, we provide two methods to
construct the propagator. The first one is combined the strategies of [17] and
[19, 18], and the second one is based on the Schulman-Sunada formula in sprit
of [26, 27] in which the heat kernel has been studied. In future papers, we
will continue investigating this quantum model for wave with one or multiple
Aharonov-Bohm solenoids in a uniform magnetic field.

##  1\. Introduction

Let us consider the electromagnetic Hamiltonian

$H_{A,V}=-(\nabla+iA(x))^{2}+V(x),$

where the electric scalar potential $V:\mathbb{R}^{n}\to\mathbb{R}$ and the
magnetic vector potential

$A(x)=(A^{1}(x),\ldots,A^{n}(x)):\,\mathbb{R}^{n}\to\mathbb{R}^{n}$

satisfies the Coulomb gauge condition

$\mathrm{div}\,A=0.$ (1.1)

In three dimensions, the magnetic vector potential $A$ produces a magnetic
field $B$, which is given by

$B=\mathrm{curl}(A)=\nabla\times A.$

In general dimensions $n\geq 2$, $B$ should be viewed as the matrix-valued
field $B:\mathbb{R}^{n}\to\mathcal{M}_{n\times n}(\mathbb{R})$ given by

$B:=DA-DA^{t},\quad B_{ij}=\frac{\partial A^{i}}{\partial
x_{j}}-\frac{\partial A^{j}}{\partial x_{i}}.$ (1.2)

The Schrödinger operators with electromagnetic potentials have been
extensively studied from the aspects of spectral and scattering theory, we
refer to Avron-Herbst-Simon [3, 4, 5] and Reed-Simon [23], in which many
important physical potentials (e.g. the constant magnetic field and the
Coulomb electric potential) are discussed. The purpose of our program here is
to study how the electric or magnetic potentials affect the short-time or
long-time behavior of the solutions for dispersive equations (e.g. the
Schrödinger, wave and Klein-Gordon).

Indeed, the program on the study of the decay estimates and Strichartz
estimates of the dispersive equations has a long history due to their
significance in analysis and PDEs fields. We refer to [7, 8, 9, 10, 11, 12,
13, 24] and the references therein for the the classical and important
Schrödinger and wave equations with electromagnetic potentials in mathematical
and physical fields. However, since the different potentials has different
effects, it is hard to provide a uniform argument to treat all of potentials,
and the picture of the program is too far to completed, especially for some
critical physic potentials. For example, the dispersive equations with the
Aharonov-Bohm potential, as a diffraction physic model and scaling critical
purely magnetic potential, has attracted more and more people to study from
mathematic viewpoint. In [16, 17], Fanelli, Felli, Fontelos, and Primo studied
the validity of the time-decay estimate for the Schrödinger equation
associated with the operator included the Aharonov-Bohm potential. However,
the argument in [16, 17] breaks down for wave equation due to the lack of
pseudoconformal invariance (which was used for Schrödinger equation). Very
recently, Fanelli, Zheng and the last author [18] established the Strichartz
estimate for wave equation by constructing the odd sin propagator. To solve
open problems, raised in the survey [15], about the the dispersive estimate
for other equations (e.g. Klein-Gordon), Gao, Yin, Zheng and the last author
[19] constructed the spectral measure and further proved the time decay and
Strichartz estimates of Klein-Gordon equation. The potential models in [16,
17, 18, 19] are scaling invariant and with no unbounded (at infinity)
perturbations, which is a special cases of (1.3) below with $B_{0}\equiv 0$.

In this paper, as a sequence of recent paper [19, 18], we study the decay and
Strichartz estimates for the Schrödinger equation in a mixed magnetic fields
on the plane pierced by one infinitesimally thin Aharonov-Bohm solenoid and
subjected to a perpendicular uniform magnetic field of constant magnitude
$B_{0}$. More precisely, we focus on the typical 2D purely magnetic model
Hamiltonians

$H_{\alpha,B_{0}}=-(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))^{2},$ (1.3)

where $A_{B}(x)$ is the Aharonov-Bohm potential in [2]

$A_{B}(x)=\alpha\Big{(}-\frac{x_{2}}{|x|^{2}},\frac{x_{1}}{|x|^{2}}\Big{)},\quad
x=(x_{1},x_{2})\in\mathbb{R}^{2}\setminus\\{0\\},$ (1.4)

with $\alpha\in\mathbb{R}$ representing the circulation of $A_{B}$ around the
solenoid and $A_{\mathrm{hmf}}(x)$ is the potential

$A_{\mathrm{hmf}}(x)=\frac{B_{0}}{2}(-x_{2},x_{1}),\quad B_{0}>0.$ (1.5)

From (1.2), the total magnetic filed is given by $B_{0}+\alpha\delta(x)$ where
$\delta$ is the Dirac delta function. From mathematics viewpoint, it is worth
emphasizing three key features that about the perturbations of the potentials
here. The first one is that the Aharonov-Bohm potential producing the singular
magnetic field has the same scaling of $\nabla$ which is homogenous degree of
$-1$, so that the perturbation of Aharonov-Bohm potential (1.4) is non-
trivial; the second one is that the magnetic poetential (1.5) is homogenous
degree of $1$ so that the Schrödinger operator (1.3) is not scaling invariant
so is the Schrödinger equation. The potential $A_{\mathrm{hmf}}(x)$ is
unbounded and the magnetic filed $B(x)=B_{0}$ from (1.2) is a uniform magnetic
field caused a trapped well. On one hand, due to the perturbations of magnetic
fields (1.5) are involved, the operator $H_{\alpha,B_{0}}$ has pure point
spectrum, the dispersive behavior of Schrödinger equation associated with
$H_{\alpha,B_{0}}$ becomes quite different from the model in [18, 19]. On the
other hand, because of the Aharonov-Bohm effect, a feature of the Mehler
kernel which is related to the Schrödinger kernel associated with pure uniform
magnetic field breaks down. Indeed, if $\alpha=0$, the Schrödinger kernel can
be written as

$e^{itH_{0,B_{0}}}(x,y)=\frac{B_{0}}{4\pi\sin(B_{0}t)}\exp\Big{\\{}\frac{B_{0}}{4i}\big{(}\cot(B_{0}t)|x-y|^{2}-2x\wedge
y\big{)}\Big{\\}},$

furthermore, one can write

$\begin{split}e^{itH_{0,B_{0}}}(x,y)=\frac{B_{0}}{4\pi\sin(B_{0}t)}&\exp\Big{\\{}\frac{B_{0}}{4i}\cot(B_{0}t)\big{(}|x|^{2}+|y|^{2}\big{)}\Big{\\}}\\\
&\times\exp\Big{\\{}i\frac{B_{0}y\cdot
R(B_{0}t)x}{2\sin(B_{0}t)}\Big{\\}},\end{split}$

where $R(\theta)$ is the usual $2\times 2$ rotation matrix given by

$R(\theta)=\left(\begin{array}[]{cc}\cos\theta&-\sin\theta\\\
\sin\theta&\cos\theta\end{array}\right).$

Therefore, see [20, Theorem 2], one can prove the Strichartz estimates

$\|e^{itH_{0,B_{0}}}f\|_{L^{q}_{t}((0,\frac{\pi}{B_{0}});L^{r}_{x}(\mathbb{R}^{2})}=(4\pi)^{1-\frac{4}{q}}\|e^{it\Delta}f\|_{L^{q}_{t}(\mathbb{R};L^{r}_{x}(\mathbb{R}^{2})}\leq
C\|f\|_{L^{2}(\mathbb{R}^{2})},$

where $(q,r)\in\Lambda_{S}$ defined in (1.9) below.

More precisely, we aim to study the dispersive behavior of the magnetic
Schrödinger equation

$\begin{cases}i\partial_{t}u(t,x)-H_{\alpha,B_{0}}u(t,x)=0,\\\
u(0,x)=u_{0}(x),\end{cases}$ (1.6)

where $H_{\alpha,B_{0}}$ is given in (1.3) with the perturbation of potentials
$A_{B}(x)$ and $A_{\mathrm{hmf}}(x)$.

Now we state our main results.

######  Theorem 1.1.

Let $H_{\alpha,B_{0}}$ be in (1.3) with the potentials being (1.4)-(1.5) and
let $u(t,x)$ be the solution of (1.6). Then there exists a constant $C>0$ such
that the dispersive estimate

$\|u(t,x)\|_{L^{\infty}(\mathbb{R}^{2})}\leq
C|\sin(tB_{0})|^{-1}\|u_{0}\|_{L^{1}(\mathbb{R}^{2})},\quad
t\neq\frac{k\pi}{B_{0}},\,k\in\mathbb{Z},$ (1.7)

and the Strichartz estimate holds for

$\|u(t,x)\|_{L^{q}([0,T];L^{p}(\mathbb{R}^{2}))}\leq
C\|u_{0}\|_{L^{2}(\mathbb{R}^{2})},$ (1.8)

where $T\in(0,\frac{\pi}{2B_{0}})$ and $(q,p)\in\Lambda_{S}$ with

$\Lambda_{S}:=\Bigg{\\{}(q,p)\in[2,+\infty]\times[2,+\infty):\frac{2}{q}=2\Big{(}\frac{1}{2}-\frac{1}{p}\Big{)}\Bigg{\\}}.$
(1.9)

######  Remark 1.2.

The decay estimate (1.7) is the periodic with period $\pi/B_{0}$. The endpoint
of the time interval $T$ in the Strichartz estimates (1.8) depends on the
coefficient $B_{0}$ of unbounded potential. Actually, if the unbounded
potentials disappear, i.e. $B_{0}=0$, one can take $T$ to be $+\infty$ safely,
that is, the global-in-time Strichartz estimates, which is corresponding to
the Laplacian with the Aharonov-Bohm potential (see Theorem 1.3 of [17]).
However, in the present paper, as mentioned above, we only obtain local-in-
time Strichartz estimates associated with the operator (1.3) due to the
unbounded potentials caused trapped well.

######  Remark 1.3.

Let

$A(x)=A_{B}(x)+A_{\mathrm{hmf}}(x),$

one can verify that $\mathrm{div}\,A=0$ satisfies (1.1). Then, one observe
that

$\begin{split}H_{\alpha,B_{0}}&=-(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))^{2}\\\
&=-\Delta+\big{(}\frac{B_{0}}{2}\big{)}^{2}|x|^{2}+\frac{\alpha^{2}}{|x|^{2}}+iB_{0}(-x_{2},x_{1})\cdot\nabla+i\frac{2\alpha}{|x|^{2}}(-x_{2},x_{1})\cdot\nabla+\alpha
B_{0}.\end{split}$

One will see that the operator is perturbed by the inverse-square potential
and harmonic oscillator. This phenomenon is natural because the unbounded
potential cause a trapped well, the energy cannot be dispersive for long time.
This is closely relate to the models with pure harmonic oscillators, i.e.
$H_{0,V}=-\Delta+|x|^{2}$, in which Koch and Tataru [22] proved the decay is
the periodic with period $\pi$

$\|e^{itH_{0,V}}\|_{L^{1}(\mathbb{R}^{n})\to L^{\infty}(\mathbb{R}^{n})}\leq
C|\sin t|^{-\frac{n}{2}}.$

The paper is organized as follows. In Section 2, in a preliminary step, we
recall the the self-adjoint extension of the operator $H_{\alpha,B_{0}}$, and
study the spectrum of $H^{F}_{\alpha,B_{0}}$ (which is the Friedrichs of
$H_{\alpha,B_{0}}$). In Section 3, we construct the Schrödinger propagator by
combing the strategies of [17] and [19, 18]. In Section 4, we construct the
Schrödinger propagator by using another method based on the Schulman-Sunada
formula. Finally, in Section 5, we prove the Theorem 1.1 by using the
representation of the Schrödinger propagator constructed in previous sections.

Acknowledgments: The authors thank L. Fanelli and P. Št’ovíček for helpful
discussions. This work is supported by National Natural Science Foundation of
China (12171031, 11901041, 11831004).

##  2\. preliminaries

In this section, we first recall the self-adjoint extension of the operator
$H_{\alpha,B_{0}}$, the Friedrichs extension, and then we study the spectrum
of $H^{F}_{\alpha,B_{0}}$ (which is the Friedrichs of $H_{\alpha,B_{0}}$).

###  2.1. Quadratic form and the self-adjoint extension

Define the space $\mathcal{H}_{\alpha,B_{0}}^{1}(\mathbb{R}^{2})$ as the
completion of
$\mathcal{C}_{c}^{\infty}(\mathbb{R}^{2}\setminus\\{0\\};\mathbb{C})$ with
respect to the norm

$\|f\|_{\mathcal{H}_{\alpha,B_{0}}^{1}(\mathbb{R}^{2})}=\Big{(}\int_{\mathbb{R}^{2}}|\nabla_{\alpha,B_{0}}f(x)|^{2}dx\Big{)}^{\frac{1}{2}}$

where

$\nabla_{\alpha,B_{0}}f(x)=\nabla f+i(A_{B}+A_{\mathrm{hmf}})f.$

The quadratic form $Q_{\alpha,B_{0}}$ associated with $H_{\alpha,B_{0}}$ is
defined by

$\begin{split}Q_{\alpha,B_{0}}:&\quad\quad\mathcal{H}_{\alpha,B_{0}}^{1}\to\mathbb{R}\\\
Q_{\alpha,B_{0}}(f)&=\int_{\mathbb{R}^{2}}|\nabla_{\alpha,B_{0}}f(x)|^{2}dx.\end{split}$

Then the quadratic form $Q_{\alpha,B_{0}}$ is positive definite, thus it
implies that the operator $H_{\alpha,B_{0}}$ is symmetric semi bounded from
below which admits a self-adjoint extension (Friedrichs extension)
$H^{F}_{\alpha,B_{0}}$ with the natural form domain

$\mathcal{D}=\Big{\\{}f\in\mathcal{H}_{\alpha,B_{0}}^{1}(\mathbb{R}^{2}):H^{F}_{\alpha,B_{0}}f\in
L^{2}(\mathbb{R}^{2})\Big{\\}}$

Even though the operator $H_{\alpha,B_{0}}$ has many other self-adjoint
extensions (see [14]) by the von Neumann extension theory, in this whole
paper, we use the simplest Friedrichs extension and briefly write
$H_{\alpha,B_{0}}$ as its Friedrichs extension $H^{F}_{A}$.

###  2.2. The spectrum of the operator $H_{\alpha,B_{0}}$

In this subsection, we modify the argument of [17] to obtain the eigenvalue
and eigenfunction of the Schrödinger operator

$H_{\alpha,B_{0}}=-(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))^{2},$

where the magnetic vector potentials are in (1.4) and (1.5). More precisely,
we will prove that

######  Proposition 2.1 (The spectrum for $H_{\alpha,B_{0}}$).

Let $H_{\alpha,B_{0}}$ be the self-adjoint Schrödinger operator in (1.3). Then
the eigenvalues of $H_{\alpha,B_{0}}$ are discrete and are given by

$\lambda_{k,m}=(2m+1+|k+\alpha|)B_{0}+(k+\alpha)B_{0},\quad
m,\,k\in\mathbb{Z},\,m\geq 0,$ (2.1)

and the (finite) multiplicity of $\lambda_{k,m}$ is

$\\#\Bigg{\\{}j\in\mathbb{Z}:\frac{\lambda_{k,m}-(j+\alpha)B_{0}}{2B_{0}}-\frac{|j+\alpha|+1}{2}\in\mathbb{N}\Bigg{\\}}.$

Furthermore, let $\theta=\frac{x}{|x|}$, the corresponding eigenfunction is
given by

$V_{k,m}(x)=|x|^{|k+\alpha|}e^{-\frac{B_{0}|x|^{2}}{4}}\,P_{k,m}\Bigg{(}\frac{B_{0}|x|^{2}}{2}\Bigg{)}e^{ik\theta}$
(2.2)

where $P_{k,m}$ is the polynomial of degree $m$ given by

$P_{k,m}(r)=\sum_{n=0}^{m}\frac{(-m)_{n}}{(1+|k+\alpha|)_{n}}\frac{r^{n}}{n!}.$

with $(a)_{n}$ ($a\in\mathbb{R}$) denoting the Pochhammer’s symbol

$\displaystyle(a)_{n}=\begin{cases}1,&n=0;\\\
a(a+1)\cdots(a+n-1),&n=1,2,\cdots\end{cases}$

######  Remark 2.2.

One can verify that the orthogonality holds

$\int_{\mathbb{R}^{2}}V_{k_{1},m_{1}}(x)V_{k_{2},m_{2}}(x)\,dx=0,\quad\text{if}\quad(k_{1},m_{1})\neq(k_{2},m_{2}).$

######  Remark 2.3.

Let $L^{\alpha}_{m}(t)$ be the generalized Laguerre polynomials

$L^{\alpha}_{m}(t)=\sum_{n=0}^{m}(-1)^{n}\Bigg{(}\begin{array}[]{c}m+\alpha\\\
m-n\\\ \end{array}\Bigg{)}\frac{t^{n}}{n!},$

and the well known orthogonality relation

$\int_{0}^{\infty}x^{\alpha}e^{-x}L^{\alpha}_{m}(x)L^{\alpha}_{n}(x)\,dx=\frac{\Gamma(n+\alpha+1)}{n!}\delta_{n,m},$

where $\delta_{n,m}$ is the Kronecker delta. Let
$\tilde{r}=\frac{B_{0}|x|^{2}}{2}$ and $\alpha_{k}=|k+\alpha|$, then

$P_{k,m}(\tilde{r})=\sum_{n=0}^{m}\frac{(-1)^{n}m(m-1)\cdots(m-(n-1))}{(\alpha_{k}+1)(\alpha_{k}+2)\cdots(\alpha_{k}+n)}\frac{\tilde{r}^{n}}{n!}=\Bigg{(}\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\Bigg{)}^{-1}L^{\alpha_{k}}_{m}(\tilde{r}).$ (2.3)

Therefore,

$\|V_{k,m}(x)\|^{2}_{L^{2}(\mathbb{R}^{2})}=\pi\Big{(}\frac{2}{B_{0}}\Big{)}^{\alpha_{k}+1}\Gamma(1+\alpha_{k})\Bigg{(}\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\Bigg{)}^{-1}.$ (2.4)

######  Remark 2.4.

Recall the Poisson kernel formula for Laguerre polynomials [1, (6.2.25)]: for
$a,b,c,\alpha>0$

$\begin{split}&\sum_{m=0}^{\infty}e^{-cm}\frac{m!}{\Gamma(m+\alpha+1)}L_{m}^{\alpha}(a)L_{m}^{\alpha}(b)\\\
&=\frac{e^{\frac{\alpha
c}{2}}}{(ab)^{\frac{\alpha}{2}}(1-e^{-c})}\exp\left(-\frac{(a+b)e^{-c}}{1-e^{-c}}\right)I_{\alpha}\left(\frac{2\sqrt{ab}e^{-\frac{c}{2}}}{1-e^{-c}}\right)\end{split}$
(2.5)

then this together with (2.3) gives

$\begin{split}&\sum_{m=0}^{\infty}e^{-cm}\frac{m!}{\Gamma(m+\alpha_{k}+1)}\Bigg{(}\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\Bigg{)}^{2}P_{k,m}(a)P_{k,m}(b)\\\
&=\frac{e^{\frac{\alpha_{k}c}{2}}}{(ab)^{\frac{\alpha_{k}}{2}}(1-e^{-c})}\exp\left(-\frac{(a+b)e^{-c}}{1-e^{-c}}\right)I_{\alpha_{k}}\left(\frac{2\sqrt{ab}e^{-\frac{c}{2}}}{1-e^{-c}}\right).\end{split}$
(2.6)

###### Proof.

Notice that the operator (1.3), in the polar coordinates $(r,\theta)$, has a
nice representation

$H_{\alpha,B_{0}}=-\partial_{r}^{2}-\frac{1}{r}\partial_{r}+\frac{1}{r^{2}}\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}.$

We want to solve the eigenfunction equation

$H_{\alpha,B_{0}}g(x)=\lambda g(x)$ (2.7)

in the domain 111Here we use the Friedrichs self-adjoint extension.  of
$H_{A,V}$. Define the projectors $P_{k}$ onto the eigenspaces of the angular
momentum as

$P_{k}f(r,\theta)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{ik(\theta-\theta^{\prime})}f(r,\theta^{\prime})d\theta^{\prime}:=f_{k}(r)e^{ik\theta},\quad
k\in\mathbb{Z}$

then it clear that the operator $H_{\alpha,B_{0}}$ commutes with the
projectors $P_{k}$. In the polar coordinates $(r,\theta)$, (2.7) implies that
for every $k\in\mathbb{Z}$,

$g^{\prime\prime}_{k}(r)+\frac{1}{r}g^{\prime}_{k}(r)-\frac{1}{r^{2}}\Big{(}k+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}g_{k}(r)=-\lambda
g_{k}(r),$ (2.8)

where

$g_{k}(r)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ik\theta}g(r,\theta)d\theta.$

Let

$\phi_{k}(s)=\Big{(}\frac{2s}{B_{0}}\Big{)}^{-\frac{|k+\alpha|}{2}}e^{\frac{s}{2}}g_{k}\Big{(}\sqrt{\frac{2s}{B_{0}}}\Big{)},$

then $\phi_{k}$ satisfies

$s\phi^{\prime\prime}_{k}(s)+(1+|k+\alpha|-s)\phi^{\prime}_{k}(s)-\frac{1}{2}\Big{(}1+|k+\alpha|+k+\alpha-\frac{\lambda}{B_{0}}\Big{)}\phi_{k}(s)=0.$
(2.9)

######  Lemma 2.5.

The Kummer Confluent Hypergeometric equation

$s\phi^{\prime\prime}(s)+(b-s)\phi^{\prime}(s)-a\phi(s)=0,\quad s>0,$

has two linearly independent solutions given by the Kummer function (also
called confluent hypergeometric function)

$M(a,b,s)=\sum_{n=0}^{\infty}\frac{(a)_{n}}{(b)_{n}}\frac{s^{n}}{n!},\quad
b\neq 0,-1,-2,\cdots$

where the Pochhammer’s symbol

$\displaystyle(a)_{n}=\begin{cases}1,&n=0;\\\
a(a+1)\cdots(a+n-1),&n=1,2,\cdots\end{cases}$

and the Tricomi function (also called confluent hypergeometric function of the
second kind)

$U(a,b,s)=\frac{\Gamma(1-b)}{\Gamma(a-b+1)}M(a,b,s)+\frac{\Gamma(b-1)}{\Gamma(a)}s^{1-b}M(a-b+1,2-b,s).$

By using Lemma 2.5, we have two linearly independent solutions of (2.9), hence
two linearly independent solutions of (2.8) are given by

$\displaystyle g_{k}^{1}(\lambda;r)$
$\displaystyle=r^{|\alpha+k|}M\Big{(}\beta(k,\lambda),\gamma(k),\frac{B_{0}r^{2}}{2}\Big{)}e^{-\frac{B_{0}r^{2}}{4}}$
$\displaystyle g_{k}^{2}(\lambda;r)$
$\displaystyle=r^{|\alpha+k|}U\Big{(}\beta(k,\lambda),\gamma(k),\frac{B_{0}r^{2}}{2}\Big{)}e^{-\frac{B_{0}r^{2}}{4}}$

with

$\displaystyle\beta(k,\lambda)$
$\displaystyle=\frac{1}{2}\Big{(}1+k+\alpha+|k+\alpha|-\frac{\lambda}{B_{0}}\Big{)},$
$\displaystyle\gamma(k)$ $\displaystyle=1+|k+\alpha|.$

Therefore, the general solution of (2.9) is given by

$\displaystyle g_{k}(r)=$ $\displaystyle
A_{k}g_{k}^{1}(\lambda;r)+B_{k}g_{k}^{2}(\lambda;r)$ (2.10)
$\displaystyle=r^{|\alpha+k|}e^{-\frac{B_{0}r^{2}}{4}}\Bigg{(}A_{k}M\Big{(}\beta(k,\lambda),\gamma(k),\frac{B_{0}r^{2}}{2}\Big{)}+B_{k}U\Big{(}\beta(k,\lambda),\gamma(k),\frac{B_{0}r^{2}}{2}\Big{)}\Bigg{)},$

where $A_{k},B_{k}$ are two constants which are dependent on $k$.

######  Lemma 2.6 ([6],Chap.13).

The following properties hold:

  * •

The two functions $M(a,b,z)$ and $U(a,b,z)$ are linearly dependent if and only
if $a\in-\mathbb{Z}_{+}$.

  * •

$M(a,b,z)$ is an entire function of $z$ and it is regular at $z=0$. However,
$U(a,b,z)$ is singular at the origin provided that $b>1$ and
$a\notin-\mathbb{Z}_{+}$ and it holds true that

$\lim_{z\rightarrow 0^{+}}z^{b-1}U(a,b,z)=\frac{\Gamma(b-1)}{\Gamma(a)}.$
(2.11)

If $b\in(1,2)$, the asymptotic behavior of $U(a,b,z)$ as $z\rightarrow 0^{+}$
is

$U(a,b,z)=\frac{\Gamma(1-b)}{\Gamma(a-b+1)}+\frac{\Gamma(b-1)}{\Gamma(a)}z^{1-b}+O(z^{2-b}).$

  * •

If $-a\notin\mathbb{N}$, the asymptotic behavior as $z\rightarrow+\infty$
holds true:

$M(a,b,z)=\frac{\Gamma(b)}{\Gamma(b-a)}(-z)^{-a}(1+O(z^{-1}))+\frac{\Gamma(b)}{\Gamma(a)}e^{z}z^{a-b}(1+O(z^{-1})),$
(2.12)

and

$U(a,b,z)=z^{-a}(1+O(z^{-1})).$

Now we use this asymptotic lemma to conclude more detail information about the
eigenvalues and eigenfunctions. Let

$m:=-\beta(k,\lambda)=-\frac{1}{2}\Big{(}1+k+\alpha+|k+\alpha|-\frac{\lambda}{B_{0}}\Big{)}.$

On one hand, from (2.12), if $m\notin\mathbb{N}$, the function

$M\big{(}-m,\gamma(k),\tilde{r}\big{)}\sim
e^{\tilde{r}}\tilde{r}^{-m-\gamma(k)},\quad\tilde{r}\rightarrow+\infty,$

is singular at $+\infty$; while if $m\in\mathbb{N}=\\{0,1,2,\cdots\\}$, then
$M(m,\gamma(k),\tilde{r})$ is in fact a polynomial of degree $m$ in
$\tilde{r}$, which we shall denote as $P_{k,m}$, i.e.,

$P_{k,m}(\tilde{r})=M(-m,1+|k+\alpha|,\tilde{r})=\sum_{n=0}^{m}\frac{(-m)_{n}}{(1+|k+\alpha|)_{n}}\frac{\tilde{r}^{n}}{n!}.$

On the other hand, note that $\gamma(k)=1+|k+\alpha|\geq 1$, from (2.11), it
follows that

$U\big{(}\beta(k,\lambda),\gamma(k),\tilde{r}\big{)}\sim\tilde{r}^{-|k+\alpha|},\quad\tilde{r}\rightarrow
0+,$

with the implicit constant depending only on $\lambda$ and $k$. Hence, by
letting $\tilde{r}=\frac{B_{0}r^{2}}{2}$ and using (2.10) and Lemma 2.6, we
have for fixed $k\in\mathbb{Z}$ that

$g_{k}(r)\sim
B_{k}r^{|\alpha+k|}e^{-\frac{B_{0}r^{2}}{4}}U\Big{(}\beta(k,\lambda),\gamma(k),\frac{B_{0}r^{2}}{2}\Big{)}\sim
B_{k}r^{-|\alpha+k|},\quad\text{as}\quad r\rightarrow 0+.$

and

$g_{k}(r)\sim
A_{k}r^{|\alpha+k|}e^{-\frac{B_{0}r^{2}}{4}}M\Big{(}\beta(k,\lambda),\gamma(k),\frac{B_{0}r^{2}}{2}\Big{)}\sim
A_{k}e^{\frac{B_{0}r^{2}}{4}}r^{-1+k+\alpha-\frac{\lambda}{B_{0}}},\quad\text{as}\quad
r\rightarrow\infty.$

We now conclude that one must have $B_{k}\equiv 0$. Indeed, otherwise, from
the fact that the eigenfunction $g\in D(H_{\alpha,B_{0}})$, thus we have that

$\int_{0}^{\infty}g_{k}^{2}(r)\frac{dr}{r}\leq\int_{\mathbb{R}^{2}}\frac{g^{2}(x)}{|x|^{2}}dx<\infty,$

which is obviously contradict with the fact that the integral
$\int_{0}^{\infty}r^{-2|k+\alpha|-1}dr$ is divergent for all $k\in\mathbb{Z}$
at $0$.

We next conclude that one must have $A_{k}\equiv 0$ if $m\notin\mathbb{N}$.
Actually, from the fact that the eigenfunction $g\in D(H_{\alpha,B_{0}})$
again, thus we have that

$\int_{0}^{\infty}g_{k}^{2}(r)\,rdr\leq\int_{\mathbb{R}^{2}}g^{2}(x)dx<\infty.$

However, we note that $B_{0}>0$ and

$\int_{0}^{\infty}g_{k}^{2}(r)\,rdr\geq\int_{1}^{\infty}e^{\frac{B_{0}r^{2}}{4}}\,rdr$

which is divergent. This is a contradiction.

Therefore, we must have

$\mathbb{N}\ni
m=-\frac{1}{2}\Big{(}1+k+\alpha+|k+\alpha|-\frac{\lambda}{B_{0}}\Big{)},$

and

$g_{k}(r)=r^{|k+\alpha|}e^{-\frac{B_{0}r^{2}}{4}}\,P_{k,m}\Big{(}\frac{B_{0}r^{2}}{2}\Big{)}.$

Therefore, we prove the function

$V_{k,m}(x)=|x|^{|k+\alpha|}e^{-\frac{B_{0}|x|^{2}}{4}}\,P_{k,m}\Big{(}\frac{B_{0}|x|^{2}}{2}\Big{)}e^{ik\theta},$

belongs to $D(H_{\alpha,B_{0}})$, thus providing an eigenfunction of the
operator $H_{\alpha,B_{0}}$. Thus from
$-m=\frac{1}{2}\Big{(}1+k+\alpha+|k+\alpha|-\frac{\lambda}{B_{0}}\Big{)}$, we
solve (2.7) to obtain the eigenvalues $\lambda$ of $H_{\alpha,B_{0}}$

$\lambda_{k,m}=(2m+1+|k+\alpha|)B_{0}+(k+\alpha)B_{0},\quad
k\in\mathbb{Z},\,m\in\mathbb{N}.$

∎

##  3\. The Schrödinger propagator

In this section, we construct the Schrödinger propagator by using the spectrum
property of Proposition 2.1. The combined strategies are from [17] and [19,
18].

More precisely, we will prove the following result.

######  Proposition 3.1.

Let $H_{\alpha,B_{0}}$ be the operator in (1.3) and suppose
$x=r_{1}(\cos\theta_{1},\sin\theta_{1})$ and
$y=r_{2}(\cos\theta_{2},\sin\theta_{2})$. Let $u(t,x)$ be the solution of the
Schrödinger equation

$\begin{cases}\big{(}i\partial_{t}-H_{\alpha,B_{0}}\big{)}u(t,x)=0,\\\
u(0,x)=f(x).\end{cases}$

Then

$u(t,x)=e^{-itH_{\alpha,B_{0}}}f=\int_{\mathbb{R}^{2}}K_{S}(x,y)f(y)\,dy,$

where $t\neq\frac{k\pi}{B_{0}},\,k\in\mathbb{Z}$. Let
$\theta=\theta_{1}-\theta_{2}-tB_{0}\in\mathbb{R},$ there exists an integer
$j_{0}$ satisfying $\theta+2j_{0}\pi\in[-\pi,\pi]$. Define

$\chi(\theta,j_{0})=\begin{cases}\begin{array}[]{ll}1,&\hbox{if\,
$|\theta+2j_{0}\pi|<\pi$ ;}\\\ e^{-i2\pi\alpha}+1,&\hbox{if\,
$\theta+2j_{0}\pi=-\pi;$}\\\ e^{i2\pi\alpha}+1,&\hbox{if
\,$\theta+2j_{0}\pi=\pi$.}\end{array}\end{cases}$

Then the kernel of Schrödinger propagator $e^{-itH_{\alpha,B_{0}}}$ has the
representation

$\begin{split}K_{S}(x,y)&=\frac{B_{0}e^{-itB\alpha}}{8\pi^{2}i\sin(tB_{0})}e^{\frac{iB_{0}(r_{1}^{2}+r_{2}^{2})}{4\tan(tB_{0})}}\\\
\times&\Big{[}e^{\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cos(\theta_{1}-\theta_{2}-tB_{0})}e^{-i\alpha(\theta_{1}-\theta_{2}-tB_{0}+2j_{0}\pi)}\chi(\theta,j_{0})\\\
&-\frac{\sin(\pi\alpha)}{\pi}\int_{\mathbb{R}}e^{-\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cosh
s}\frac{e^{-\alpha
s}}{1+e^{-s+i(\theta_{1}-\theta_{2}-tB_{0})}}\,ds\Big{]}.\end{split}$ (3.1)

###### Proof.

We construct the representation formula for the kernel of the Schrödinger flow
$e^{-itH_{\alpha,B_{0}}}$ by combining the argument of [17] and [18, 19].

Our starting point is the Proposition 2.1. Let $\tilde{V}_{k,m}$ be the
$L^{2}$-normalization of $V_{k,m}$ in (2.2), then the eigenfunctions
$\Big{\\{}\tilde{V}_{k,m}\Big{\\}}_{k\in\mathbb{Z},m\in\mathbb{N}}$ form an
orthonormal basis of $L^{2}(\mathbb{R}^{2})$ corresponding to the
eigenfunctions of $H_{\alpha,B_{0}}$.

We expand the initial data $f(x)\in L^{2}$ as

$f(x)=\sum_{k\in\mathbb{Z},\atop m\in\mathbb{N}}c_{k,m}\tilde{V}_{k,m}(x)$

where

$c_{k,m}=\int_{\mathbb{R}^{2}}f(x)\overline{\tilde{V}_{k,m}(x)}\,dx.$ (3.2)

The solution $u(t,x)$ of (1.6) can be written as

$u(t,x)=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}u_{k,m}(t)\tilde{V}_{k,m}(x),$ (3.3)

where $u_{k,m}(t)$ satisfies the ODE

$\begin{cases}&iu^{\prime}_{k,m}(t)=\lambda_{k,m}u_{k,m}(t),\\\
&u_{k,m}(0)=c_{k,m},\quad k\in\mathbb{Z},\,m\in\mathbb{N}.\end{cases}$

Thus we obtain $u_{k,m}(t)=c_{k,m}e^{-it\lambda_{k,m}}$. Therefore the
solution (3.3) becomes

$u(t,x)=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}c_{k,m}e^{-it\lambda_{k,m}}\tilde{V}_{k,m}(x).$

Plugging (3.2) into the above expression yields

$u(t,x)=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}e^{-it\lambda_{k,m}}\left(\int_{\mathbb{R}^{2}}f(y)\overline{\tilde{V}_{k,m}(y)}dy\right)\tilde{V}_{k,m}(x).$

We write $f$ in a harmonic spherical expansion

$f(y)=\sum_{k\in\mathbb{Z}}f_{k}(r_{2})e^{ik\theta_{2}},$

where

$f_{k}(r_{2})=\frac{1}{2\pi}\int_{0}^{2\pi}f(r_{2},\theta_{2})e^{-ik\theta_{2}}\,d\theta_{2},\quad
r_{2}=|y|,$ (3.4)

we thus have

$\displaystyle u(t,x)$ $\displaystyle=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}e^{-it\lambda_{k,m}}\frac{V_{k,m}(x)}{\|V_{k,m}\|^{2}_{L^{2}}}\Bigg{(}\int_{0}^{\infty}f_{k}(r_{2})e^{-\frac{B_{0}r_{2}^{2}}{4}}\,P_{k,m}\Big{(}\frac{B_{0}r_{2}^{2}}{2}\Big{)}r_{2}^{1+\alpha_{k}}\mathrm{d}r_{2}\Bigg{)}$
$\displaystyle=\Big{(}\frac{B_{0}}{2\pi}\Big{)}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}\frac{B_{0}^{\alpha_{k}}e^{-it\beta_{k}}}{2^{\alpha_{k}}\Gamma(1+\alpha_{k})}\Bigg{[}\sum_{m=0}^{\infty}\Bigg{(}\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\Bigg{)}e^{-2itmB_{0}}$
$\displaystyle\times\Bigg{(}\int_{0}^{\infty}f_{k}(r_{2})(r_{1}r_{2})^{\alpha_{k}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}P_{k,m}\left(\frac{B_{0}r_{2}^{2}}{2}\right)P_{k,m}\left(\frac{B_{0}r_{1}^{2}}{2}\right)r_{2}\mathrm{d}r_{2}\Bigg{)}\Bigg{]},$

where $\alpha_{k}=|k+\alpha|$ and we use (2.4), (2.1), (2.2) and

$\begin{split}\lambda_{k,m}&=(2m+1+|k+\alpha|)B_{0}+(k+\alpha)B_{0}\\\
&:=2mB_{0}+\beta_{k}\end{split}$

with $\beta_{k}=(1+|k+\alpha|)B_{0}+(k+\alpha)B_{0}>0$.

Notice that $P_{k,m}$ can be expressed as (see e.g. [1, (6.2.15)])

$P_{k,m}\left(\frac{r^{2}}{2}\right)=\frac{\Gamma(1+\alpha_{k})}{\Gamma(1+\alpha_{k}+m)}e^{\frac{r^{2}}{2}}r^{-\alpha_{k}}2^{\frac{\alpha_{k}}{2}}\int_{0}^{\infty}e^{-s}s^{m+\frac{\alpha_{k}}{2}}J_{\alpha_{k}}(\sqrt{2s}r)ds,$

in terms of Bessel functions $J_{\alpha_{k}}$ of order $\alpha_{k}$.

Hence, we have

$\displaystyle u(t,x)=$
$\displaystyle\frac{B_{0}}{2\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}\frac{B_{0}^{\alpha_{k}}e^{-it\beta_{k}}}{2^{\alpha_{k}}\Gamma(1+\alpha_{k})}\Bigg{[}\sum_{m\in\mathbb{N}}\left(\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\right)e^{-2itmB_{0}}$
$\displaystyle\times\Big{(}\frac{\Gamma(1+\alpha_{k})}{\Gamma(1+\alpha_{k}+m)}\Big{)}^{2}\Bigg{(}\int_{0}^{\infty}f_{k}(r_{2})(r_{1}r_{2})^{\alpha_{k}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}e^{\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{2}}\Big{(}\frac{2}{B_{0}r_{2}^{2}}\Big{)}^{\frac{\alpha_{k}}{2}}$
$\displaystyle\times\Big{(}\frac{2}{B_{0}r_{1}^{2}}\Big{)}^{\frac{\alpha_{k}}{2}}\Big{(}\int_{0}^{\infty}\int_{0}^{\infty}e^{-s_{1}-s_{2}}(s_{1}s_{2})^{m+\frac{\alpha_{k}}{2}}J_{\alpha_{k}}(\sqrt{2B_{0}s_{1}}r_{1})J_{\alpha_{k}}(\sqrt{2B_{0}s_{2}}r_{2})ds_{1}ds_{2}\Big{)}r_{2}dr_{2}\Bigg{)}\Bigg{]}$
$\displaystyle=$
$\displaystyle\frac{B_{0}}{2\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}e^{-it\beta_{k}}\Gamma(1+\alpha_{k})\Bigg{[}\sum_{m\in\mathbb{N}}\left(\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\right)\frac{e^{-2itmB_{0}}}{(\Gamma(1+\alpha_{k}+m))^{2}}$
$\displaystyle\times\Bigg{(}\int_{0}^{\infty}f_{k}(r_{2})e^{\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}\Big{(}\int_{0}^{\infty}\int_{0}^{\infty}e^{-s_{1}-s_{2}}(s_{1}s_{2})^{m+\frac{\alpha_{k}}{2}}$
$\displaystyle\times
J_{\alpha_{k}}(\sqrt{2B_{0}s_{1}}r_{1})J_{\alpha_{k}}(\sqrt{2B_{0}s_{2}}r_{2})ds_{1}ds_{2}\Big{)}r_{2}dr_{2}\Bigg{)}\Bigg{]},\quad\big{(}\text{variable
changes:}\,s_{1}\to s_{1}^{2},\,s_{2}\to s_{2}^{2}\big{)}$ $\displaystyle=$
$\displaystyle\frac{2B_{0}}{\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}e^{-it\beta_{k}}\Bigg{[}\int_{0}^{\infty}f_{k}(r_{2})e^{\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}e^{i\alpha_{k}(tB_{0}+\frac{\pi}{2})}$
$\displaystyle\times\Bigg{(}\int_{0}^{\infty}\int_{0}^{\infty}\frac{s_{1}s_{2}}{e^{s_{1}^{2}+s_{2}^{2}}}\Bigg{(}\sum_{m\in\mathbb{N}}\frac{(-1)^{m}e^{-i(tB_{0}+\frac{\pi}{2})(2m+\alpha_{k})}}{\Gamma(1+m)\Gamma(1+\alpha_{k}+m)}(s_{1}s_{2})^{2m+\alpha_{k}}\Bigg{)}$
$\displaystyle\times
J_{\alpha_{k}}\big{(}\sqrt{2B_{0}}s_{1}r_{1}\big{)}J_{\alpha_{k}}\big{(}\sqrt{2B_{0}}s_{2}r_{2}\big{)}ds_{1}ds_{2}\Bigg{)}r_{2}dr_{2}\Bigg{]}.$

Since (see e.g. [1, (4.5.2)])

$\sum_{m=0}^{\infty}\frac{(-1)^{m}e^{-i(tB_{0}+\frac{\pi}{2})(2m+\alpha_{k})}}{\Gamma(1+\alpha_{k}+m)\Gamma(1+m)}(s_{1}s_{2})^{2m+\alpha_{k}}=J_{\alpha_{k}}(2s_{1}s_{2}e^{-i(tB_{0}+\frac{\pi}{2})}),$

then we have

$\displaystyle
u(t,x)=\frac{2B_{0}}{\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}e^{it(\alpha_{k}B_{0}-\beta_{k})+i\alpha_{k}\frac{\pi}{2}}\int_{0}^{\infty}e^{\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}f_{k}(r_{2})G_{k,t}(r_{1},r_{2})r_{2}dr_{2}$

with

$\begin{split}G_{k,t}(r_{1},r_{2})=\int_{0}^{\infty}\int_{0}^{\infty}&\frac{s_{1}s_{2}}{e^{s_{1}^{2}+s_{2}^{2}}}J_{\alpha_{k}}(2s_{1}s_{2}e^{-i(tB_{0}+\frac{\pi}{2})})\\\
&\times
J_{\alpha_{k}}(\sqrt{2B_{0}}r_{1}s_{1})J_{\alpha_{k}}(\sqrt{2B_{0}}r_{2}s_{2})ds_{1}ds_{2}.\end{split}$

Using formula (see [28, formula (1), P.395])

$\displaystyle\int_{0}^{\infty}$ $\displaystyle
e^{-p^{2}t^{2}}J_{\nu}(at)J_{\nu}(bt)tdt=\frac{1}{2p^{2}}e^{-\frac{a^{2}+b^{2}}{4p^{2}}}I_{\nu}\Big{(}\frac{ab}{2p^{2}}\Big{)},$
$\displaystyle\qquad\operatorname*{Re}\nu>-1,\quad|\arg p|<\frac{\pi}{4},\quad
I_{\nu}(r)=e^{-\frac{1}{2}\nu\pi i}J_{\nu}(re^{i\frac{\pi}{2}}),$ (3.5)

with
$t=s_{2},p=1,a=\sqrt{2B_{0}}r_{2},b=2s_{1}e^{-i(tB_{0}+\frac{\pi}{2})},\nu=\alpha_{k}$,
we get

$\displaystyle\int_{0}^{\infty}e^{-s_{2}^{2}}J_{\alpha_{k}}\big{(}\sqrt{2B_{0}}r_{2}s_{2}\big{)}$
$\displaystyle
J_{\alpha_{k}}\Big{(}2s_{1}s_{2}e^{-i(tB_{0}+\frac{\pi}{2})}\Big{)}s_{2}ds_{2}$
$\displaystyle=\frac{1}{2}e^{-\frac{B_{0}r_{2}^{2}+2s_{1}^{2}e^{-i(2tB_{0}+\pi)}}{2}}I_{\alpha_{k}}\big{(}\sqrt{2B_{0}}r_{2}s_{1}e^{-i(tB_{0}+\frac{\pi}{2})}\big{)},$

where $I_{\alpha_{k}}$ denotes the modified Bessel function of order
$\alpha_{k}$. Hence

$\displaystyle G_{k,t}(r_{1},r_{2})$
$\displaystyle=\frac{1}{2}\int_{0}^{\infty}e^{-s_{1}^{2}}J_{\alpha_{k}}(\sqrt{2B_{0}}r_{1}s_{1})e^{-\frac{B_{0}r_{2}^{2}+2s_{1}^{2}e^{-i(2tB_{0}+\pi)}}{2}}I_{\alpha_{k}}\big{(}\sqrt{2B_{0}}r_{2}s_{1}e^{-i(tB_{0}+\frac{\pi}{2})}\big{)}s_{1}ds_{1}$
$\displaystyle=\frac{1}{4B_{0}}\int_{0}^{\infty}e^{-\frac{s_{1}^{2}}{2B_{0}}}J_{\alpha_{k}}(r_{1}s_{1})e^{-\frac{B_{0}r_{2}^{2}+\frac{s_{1}^{2}}{B_{0}}e^{-i(2tB_{0}+\pi)}}{2}}I_{\alpha_{k}}(r_{2}s_{1}e^{-i(tB_{0}+\frac{\pi}{2})})s_{1}ds_{1}$
$\displaystyle=\frac{1}{4B_{0}}e^{-i\alpha_{k}\frac{\pi}{2}}e^{-\frac{B_{0}r_{2}^{2}}{2}}\int_{0}^{\infty}e^{-\frac{s_{1}^{2}}{2B_{0}}(1+e^{-i(2tB_{0}+\pi)})}J_{\alpha_{k}}(r_{1}s_{1})J_{\alpha_{k}}(r_{2}s_{1}e^{-itB_{0}})s_{1}ds_{1}$
$\displaystyle=\frac{1}{4B_{0}}e^{-i\alpha_{k}\frac{\pi}{2}}e^{-\frac{B_{0}r_{2}^{2}}{2}}\frac{B_{0}}{1+e^{-i(2tB_{0}+\pi)}}e^{-\frac{B_{0}r_{1}^{2}+B_{0}r_{2}^{2}e^{-2itB_{0}}}{2(1+e^{-i(2tB_{0}+\pi)})}}I_{\alpha_{k}}(\frac{B_{0}r_{1}r_{2}e^{-itB_{0}}}{1+e^{-i(2tB_{0}+\pi)}})$
$\displaystyle=\frac{1}{4(1+e^{-i(2tB_{0}+\pi)})}e^{-i\alpha_{k}\frac{\pi}{2}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{2(1+e^{-i(2tB_{0}+\pi)})}}I_{\alpha_{k}}\Big{(}\frac{B_{0}r_{1}r_{2}e^{-itB_{0}}}{1+e^{-i(2tB_{0}+\pi)}}\Big{)}$

where we have used the fact
$I_{\nu}(r)=e^{-i\frac{\pi}{2}\nu}J_{\nu}(re^{i\frac{\pi}{2}})$ in the third
equality and (3) with
$t=s_{1},p^{2}=\frac{1+e^{-i(2tB_{0}+\pi)}}{2B_{0}},a=r_{1},b=r_{2}e^{-itB_{0}},\nu=\alpha_{k}$
in the last equality. Finally, we use the simple fact (for $tB_{0}\neq
k\pi,k\in\mathbb{Z}$)

$\frac{1}{1+e^{-i(2tB_{0}+\pi)}}=\frac{e^{itB_{0}}}{2i\sin tB_{0}}$

to obtain

$\displaystyle u(t,x)=$
$\displaystyle\frac{2B_{0}}{\pi}\frac{e^{itB_{0}}}{8i\sin
tB_{0}}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}e^{it(\alpha_{k}B_{0}-\beta_{k})}$
$\displaystyle\times\Bigg{[}\int_{0}^{\infty}e^{\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}f_{k}(r_{2})e^{-\frac{B_{0}(r_{2}^{2}+r_{1}^{2})}{4}\cdot\frac{e^{itB_{0}}}{i\sin
tB_{0}}}I_{\alpha_{k}}\Bigg{(}\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\Bigg{)}r_{2}dr_{2}\Bigg{]}$
$\displaystyle=$ $\displaystyle\frac{B_{0}e^{-itB_{0}\alpha}}{4\pi
i\sin(tB_{0})}\sum_{k\in\mathbb{Z}}e^{ik\theta_{1}}e^{-itB_{0}k}\Bigg{(}\int_{0}^{\infty}f_{k}(r_{2})e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})\cos(tB_{0})}{4i\sin(tB_{0})}}I_{\alpha_{k}}\Bigg{(}\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\Bigg{)}r_{2}dr_{2}\Bigg{)},$

where we use $\beta_{k}=(k+\alpha)B_{0}+(1+\alpha_{k})B_{0}$ in the last line.
Recalling $f_{k}$ in (3.4), we finally have

$\begin{split}u(t,x)=&\frac{B_{0}e^{-itB_{0}\alpha}}{8\pi^{2}i\sin(tB_{0})}\int_{0}^{\infty}\int_{0}^{2\pi}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4i\tan(tB_{0})}}\\\
&\times\sum_{k\in\mathbb{Z}}\Bigg{(}e^{ik(\theta_{1}-\theta_{2}-tB_{0})}I_{\alpha_{k}}\bigg{(}\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\bigg{)}\Bigg{)}f(r_{2},\theta_{2})r_{2}dr_{2}d\theta_{2}.\end{split}$
(3.6)

Now we consider the summation in $k$. Let
$z=\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}$ and
$\theta=\theta_{1}-\theta_{2}-tB_{0}$ and recall the following integral
representation [28] of the modified Bessel function $I_{\nu}$

$I_{\nu}(z)=\frac{1}{\pi}\int_{0}^{\pi}e^{z\cos s}\cos(\nu
s)ds-\frac{\sin(\pi\nu)}{\pi}\int_{0}^{\infty}e^{-z\cosh s}e^{-s\nu}ds.$ (3.7)

Recall $\alpha_{k}=|\alpha+k|$ and $\alpha\in(0,1)$, we need to consider

$\frac{1}{\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta}\int_{0}^{\pi}e^{z\cos
s}\cos(\alpha_{k}s)ds,$ (3.8)

and

$\frac{1}{\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta}\sin(\pi\alpha_{k})\int_{0}^{\infty}e^{-z\cosh
s}e^{-s\alpha_{k}}ds.$ (3.9)

Similarly as in [18, Proposition 3.1], we use the Poisson summation formula

$\sum_{j\in\mathbb{Z}}\delta(x-2\pi
j)=\sum_{k\in\mathbb{Z}}\frac{1}{2\pi}e^{ikx},$

to obtain

$\begin{split}\frac{1}{\pi}\sum_{k\in\mathbb{Z}}e^{ik\theta}\cos(\alpha_{k}s)=\sum_{j\in\mathbb{Z}}\big{[}e^{is\alpha}\delta(\theta+s+2j\pi)+e^{-is\alpha}\delta(\theta-s+2j\pi)\big{]}.\end{split}$

Therefore we have

$\displaystyle\eqref{equ:term1}$
$\displaystyle=\sum_{j\in\mathbb{Z}}\int_{0}^{\pi}e^{z\cos
s}\Bigg{(}e^{is\alpha}\delta(\theta+s+2\pi
j)+e^{-is\alpha}\delta(\theta-s+2\pi j)\Bigg{)}ds$
$\displaystyle=e^{z\cos(\theta)}e^{-i\alpha\theta}\sum_{j\in\mathbb{Z}}\chi_{[-\pi,\pi]}(\theta+2\pi
j)e^{-i2\pi j\alpha}.$ (3.10)

Recall $\theta=\theta_{1}-\theta_{2}-tB_{0}\in\mathbb{R}$ and observe that

$\theta+2\pi j_{0}\in(-\pi,\pi)\implies\theta+2\pi(j_{0}\pm
m)\notin(-\pi,\pi),|m|\geq 1,$

the summation in $j$ is only one term (except for $\theta+2j_{0}\pi=\pm\pi$).
It’s easily to verify that the identity (3.10) are the same at
$\theta+2j_{0}\pi=\pm\pi.$ For all $\theta\in\mathbb{R},$ there exists
$j_{0}\in\mathbb{Z}$ such that $\theta+2j_{0}\pi\in[-\pi,\pi].$ Therefore, we
get

$\displaystyle\eqref{equ:term1}=e^{z\cos(\theta)}e^{-i\alpha\theta}e^{-i2\pi
j_{0}\alpha}\left\\{\begin{array}[]{ll}1,&\hbox{$|\theta+2j_{0}\pi|<\pi$ ;}\\\
e^{-i2\pi\alpha}+1,&\hbox{$\theta+2j_{0}\pi=-\pi;$}\\\
e^{i2\pi\alpha}+1,&\hbox{$\theta+2j_{0}\pi=\pi$.}\end{array}\right.$

Recall again that $\alpha_{k}=|\alpha+k|,k\in\mathbb{Z}$ and $\alpha\in(0,1)$,
we have

$\displaystyle\alpha_{k}=|\alpha+k|=\begin{cases}k+\alpha,&k\geq 0;\\\
-\alpha-k,&k\leq-1,\end{cases}$

and hence

$\displaystyle\sin(\pi\alpha_{k})=\sin(\pi|\alpha+k|)=\begin{cases}\cos
k\pi\sin(\pi\alpha)=e^{ik\pi}\sin(\pi\alpha),&k\geq 0;\\\ -\cos
k\pi\sin(\pi\alpha)=-e^{ik\pi}\sin(\pi\alpha),&k\leq-1.\end{cases}$

We can compute the summation as follows

$\displaystyle\sum_{k\in\mathbb{Z}}\sin(\pi|\alpha+k|)e^{-s|\alpha+k|}e^{ik\theta}$
$\displaystyle\quad=\sin(\pi\alpha)\sum_{k\geq
0}e^{ik\pi}e^{-s(k+\alpha)}e^{ik\theta}-\sin(\pi\alpha)\sum_{k\leq-1}e^{ik\pi}e^{s(k+\alpha)}e^{ik\theta}$
$\displaystyle\quad=\sin(\pi\alpha)\Big{(}e^{-\alpha s}\sum_{k\geq
0}e^{ik(is+\pi+\theta)}-e^{\alpha s}\sum_{k\geq
1}e^{ik(is-\pi-\theta)}\Big{)}$
$\displaystyle\quad=\sin(\pi\alpha)\Big{(}\frac{e^{-\alpha
s}}{1-e^{i(is+\pi+\theta)}}-\frac{e^{\alpha
s}e^{i(is-\pi-\theta)}}{1-e^{i(is-\pi-\theta)}}\Big{)}$
$\displaystyle\quad=\sin(\pi\alpha)\Big{(}\frac{e^{-\alpha
s}}{1+e^{-s+i\theta}}+\frac{e^{\alpha s}}{1+e^{s+i\theta)}}\Big{)}.$

Therefore, we see

$\displaystyle\eqref{equ:term2}$ $\displaystyle=$
$\displaystyle\frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}e^{-z\cosh
s}\Big{(}\frac{e^{-\alpha s}}{1+e^{-s+i\theta}}+\frac{e^{\alpha
s}}{1+e^{s+i\theta)}}\Big{)}\,ds$ $\displaystyle=$
$\displaystyle\frac{\sin(\alpha\pi)}{\pi}\int_{-\infty}^{\infty}e^{-z\cosh
s}\frac{e^{-\alpha s}}{1+e^{-s+i\theta}}\,ds.$

So we put the two terms together and recall
$\theta=\theta_{1}-\theta_{2}-tB_{0}$ to obtain

$\begin{split}K_{S}(x,y)&=\frac{B_{0}e^{-itB\alpha}}{8\pi^{2}i\sin(tB_{0})}e^{\frac{iB_{0}(r_{1}^{2}+r_{2}^{2})}{4\tan(tB_{0})}}\\\
\times&\Big{[}e^{\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cos(\theta_{1}-\theta_{2}-tB_{0})}e^{-i\alpha(\theta_{1}-\theta_{2}-tB_{0}+2j_{0}\pi)}\chi(\theta,j_{0})\\\
&-\frac{\sin(\pi\alpha)}{\pi}\int_{\mathbb{R}}e^{-\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cosh
s}\frac{e^{-\alpha
s}}{1+e^{-s+i(\theta_{1}-\theta_{2}-tB_{0})}}\,ds\Big{]},\end{split}$

which gives (3.1).

##  4\. The Construction of Schrödinger propagator based on the Schulman-
Sunada formula

In this section, we will use the Schulman-Sunada formula to reconstruct the
Schrödinger propagator. We refer the reads to [26, 27].

Let $M=\mathbb{R}^{2}\setminus\\{\vec{0}\\}=(0,+\infty)\times\mathbb{S}^{1}$
where $\mathbb{S}^{1}$ is the unit circle. The universal covering space of $M$
is $\tilde{M}=(0,+\infty)\times\mathbb{R}$, then $M=\tilde{M}/\Gamma$ where
the structure group $\Gamma=2\pi\mathbb{Z}$ acts in the second factor of the
Cartesian product. Then Schulman’s ansatz (see [26, 27]) enables us to compute
the Schrödinger propagator $e^{-itH_{\alpha,B_{0}}}$ on $M$ by using the
Schrödinger propagator $e^{-it\tilde{H}_{\alpha,B_{0}}}$ (see the operator
$\tilde{H}_{\alpha,B_{0}}$ in (4.2) below) on $\tilde{M}$. More precisely, see
[27, (1)], we have

$e^{-itH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})=\sum_{j\in\mathbb{Z}}e^{-it\tilde{H}_{\alpha,B_{0}}}(r_{1},\theta_{1}+2j\pi;r_{2},\theta_{2}).$
(4.1)

This is similar to the construction of wave propagator on $\mathbb{T}^{n}$,
see [25, (3.5.12)]. In the following subsections, we will construct the
Schrödinger propagator $e^{-it\tilde{H}_{\alpha,B_{0}}}$.

###  4.1. The eigenfunctions and eigenvalues

Before we construct the propagator $e^{-it\tilde{H}_{\alpha,B_{0}}}$, let us
give some remarks about the difference and advantages between
$\tilde{H}_{\alpha,B_{0}}$ and $H_{\alpha,B_{0}}$. First, we recall the proof
of Proposition 2.1. From (1.3), in the polar coordinates $(r,\theta)\in M$,
then

$H_{\alpha,B_{0}}=-\partial_{r}^{2}-\frac{1}{r}\partial_{r}+\frac{1}{r^{2}}\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2},$

which acts on $L^{2}(M,rdr\,d\theta)$. For $V_{k,m}(x)$ in (2.2) and
$\lambda_{k,m}$ in (2.1), we have show that

$H_{\alpha,B_{0}}V_{k,m}(x)=\lambda_{k,m}V_{k,m}(x).$

We remark here that we choose $e^{ik\theta}$ as an eigenfunction of the
operator $\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}$
on $L^{2}_{\theta}([0,2\pi))$ which satisfies that

$\begin{cases}\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\varphi(\theta)=\Big{(}k+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\varphi(\theta)\\\
\varphi(0)=\varphi(2\pi).\end{cases}$

Instead, we consider the operator

$\tilde{H}_{\alpha,B_{0}}=-\partial_{r}^{2}-\frac{1}{r}\partial_{r}+\frac{1}{r^{2}}\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2},$
(4.2)

which acts on $L^{2}(\tilde{M},rdr\,d\theta)$. We emphasize that the variable
$\theta\in\mathbb{R}$ while not compact manifold $\mathbb{S}^{1}$. Then we
choose $e^{i(\tilde{k}-\alpha)\theta}$ as an eigenfunction of the operator
$\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}$ on
$L^{2}_{\theta}(\mathbb{R})$ which satisfies that

$\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\varphi(\theta)=\Big{(}\tilde{k}+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\varphi(\theta).$
(4.3)

It worths to point out that $\tilde{k}\in\mathbb{R}$ is a real number while
$k\in\mathbb{Z}$. More important, we inform move the $\alpha$ in right hand
side of (4.3) to the $e^{i(\tilde{k}-\alpha)\theta}$ which will simplify the
eigenfunctions.

Now we modify the argument of Proposition 2.1to solve

$\tilde{H}_{\alpha,B_{0}}g(x)=\lambda g(x)$ (4.4)

to obtain the eigenfunctions of $\tilde{H}_{\alpha,B_{0}}$. Define the Fourier
transform $F_{\theta\to\tilde{k}}$ with respect to the variable $\theta$

$F_{\theta\to\tilde{k}}f(r,\tilde{k})=\frac{1}{2\pi}\int_{\mathbb{R}}e^{i\tilde{k}\theta}f(r,\theta)\,d\theta:=\hat{f}(r,\tilde{k}),\quad\tilde{k}\in\mathbb{R}.$
(4.5)

By taking the Fourier transform of (4.4), in contrast to (2.8) we obtain

$\hat{g}^{\prime\prime}(r,\tilde{k})+\frac{1}{r}\hat{g}^{\prime}(r,\tilde{k})-\frac{1}{r^{2}}\Big{(}\tilde{k}+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\hat{g}(r,\tilde{k})=-\lambda\hat{g}(r,\tilde{k}).$
(4.6)

Let

$\psi_{\tilde{k}}(s)=\Big{(}\frac{2s}{B_{0}}\Big{)}^{-\frac{|\tilde{k}|}{2}}e^{\frac{s}{2}}\hat{g}(\Big{(}\sqrt{\frac{2s}{B_{0}}},\tilde{k}\Big{)},$

then $\psi_{\tilde{k}}(s)$ satisfies

$s\psi^{\prime\prime}_{\tilde{k}}(s)+(1+|\tilde{k}|-s)\psi^{\prime}_{\tilde{k}}(s)-\frac{1}{2}\Big{(}1+|\tilde{k}|+\tilde{k}-\frac{\lambda}{B_{0}}\Big{)}\psi_{\tilde{k}}(s)=0.$
(4.7)

which is same to (2.9) by replacing $|k+\alpha|$ by $\tilde{k}$. By using
Lemma 2.5 again, we have two linearly independent solutions of (4.7), hence
two linearly independent solutions of (4.6) are given by

$\displaystyle\hat{g}^{1}(\lambda;r,\tilde{k})$
$\displaystyle=r^{|\tilde{k}|}M\Big{(}\tilde{\beta}(\tilde{k},\lambda),\tilde{\gamma}(\tilde{k}),\frac{B_{0}r^{2}}{2}\Big{)}e^{-\frac{B_{0}r^{2}}{4}}$
$\displaystyle\hat{g}^{2}(\lambda;r,\tilde{k})$
$\displaystyle=r^{|\tilde{k}|}U\Big{(}\tilde{\beta}(\tilde{k},\lambda),\tilde{\gamma}(\tilde{k}),\frac{B_{0}r^{2}}{2}\Big{)}e^{-\frac{B_{0}r^{2}}{4}}$

with

$\displaystyle\tilde{\beta}(\tilde{k},\lambda)$
$\displaystyle=\frac{1}{2}\Big{(}1+\tilde{k}+|\tilde{k}|-\frac{\lambda}{B_{0}}\Big{)},$
$\displaystyle\tilde{\gamma}(\tilde{k})$ $\displaystyle=1+|\tilde{k}|.$

Therefore, the general solution of (4.7) is given by

$\hat{g}(r,\tilde{k})=A_{\tilde{k}}\hat{g}^{1}(\lambda;r,\tilde{k})+B_{\tilde{k}}\hat{g}^{2}(\lambda;r,\tilde{k}).$

where $A_{\tilde{k}},B_{\tilde{k}}$ are two constants which depend on
$\tilde{k}\in\mathbb{R}$. Let

$m:=-\tilde{\beta}(\tilde{k},\lambda)=-\frac{1}{2}\Big{(}1+\tilde{k}+|\tilde{k}|-\frac{\lambda}{B_{0}}\Big{)}.$

Similar as above, we use this asymptotic Lemma 2.6 to conclude that
$m\in\mathbb{N}$ again, we omit the details.

Therefore, we must have

$\mathbb{N}\ni
m=-\frac{1}{2}\Big{(}1+\tilde{k}+|\tilde{k}|-\frac{\lambda}{B_{0}}\Big{)},$

and

$\hat{g}(r,\tilde{k})=r^{|\tilde{k}|}e^{-\frac{B_{0}r^{2}}{4}}\,P_{\tilde{k}-\alpha,m}\Big{(}\frac{B_{0}r^{2}}{2}\Big{)}.$

Therefore, we obtain a complete set of generalized eigenfunctions of
$\tilde{H}_{\alpha,B_{0}}$

$\Big{\\{}U_{m}(x,\tilde{k}):m\in\mathbb{N},\tilde{k}\in\mathbb{R}\Big{\\}}$

where

$U_{m}(x,\tilde{k})=|x|^{|\tilde{k}|}e^{-\frac{B_{0}|x|^{2}}{4}}\,P_{\tilde{k}-\alpha,m}\Bigg{(}\frac{B_{0}|x|^{2}}{2}\Bigg{)}e^{i(\tilde{k}-\alpha)\theta}$
(4.8)

which belongs to $L^{2}(\tilde{M},rdr\,d\theta)$. Thus from
$-m=\frac{1}{2}\Big{(}1+\tilde{k}+|\tilde{k}|-\frac{\lambda}{B_{0}}\Big{)}$,
we solve (4.4) to obtain the eigenvalues $\lambda$ of
$\tilde{H}_{\alpha,B_{0}}$

$\lambda_{\tilde{k},m}=(2m+1+|\tilde{k}|+\tilde{k})B_{0},\quad\tilde{k}\in\mathbb{R},\,m\in\mathbb{N}.$

We obtain analogue of (2.6) by using (2.5) and (2.3)

$\begin{split}&\sum_{m=0}^{\infty}e^{-cm}\frac{m!}{\Gamma(m+|\tilde{k}|+1)}\Bigg{(}\begin{array}[]{c}m+|\tilde{k}|\\\
m\\\
\end{array}\Bigg{)}^{2}P_{\tilde{k}-\alpha,m}(a)P_{\tilde{k}-\alpha,m}(b)\\\
&=\frac{e^{\frac{|\tilde{k}|c}{2}}}{(ab)^{\frac{|\tilde{k}|}{2}}(1-e^{-c})}\exp\left(-\frac{(a+b)e^{-c}}{1-e^{-c}}\right)I_{|\tilde{k}|}\left(\frac{2\sqrt{ab}e^{-\frac{c}{2}}}{1-e^{-c}}\right).\end{split}$

###  4.2. Construction of $e^{-it\tilde{H}_{\alpha,B_{0}}}$

Now we construct the propagator $e^{-it\tilde{H}_{\alpha,B_{0}}}$ by using the
above eigenfunctions. Let $\tilde{U}_{m}(x,\tilde{k})$ be the
$L^{2}$-normalization of $U_{m}(x,\tilde{k})$ in (4.8). We wirte initial data
$f(x)\in L^{2}$ as

$f(x)=\sum_{m\in\mathbb{N}}\int_{\mathbb{R}}c_{m}(\tilde{k})\tilde{U}_{m}(x,\tilde{k})\,d\tilde{k}$

where

$c_{m}(\tilde{k})=\int_{\mathbb{R}^{2}}f(x)\overline{\tilde{U}_{m}(x,\tilde{k})}\,dx.$

Using the Fourier transform (4.5), we solve

$\begin{cases}\big{(}i\partial_{t}-\tilde{H}_{\alpha,B_{0}}\big{)}u(t,x)=0,\quad(t,x)\in\mathbb{R}\times\tilde{M}\\\
u(0,x)=f(x).\end{cases}$

to obtain

$u(t,x)=\sum_{m\in\mathbb{N}}\int_{\mathbb{R}}e^{-it\lambda_{\tilde{k},m}}\left(\int_{\mathbb{R}^{2}}f(y)\overline{\tilde{U}_{m}(y,\tilde{k})}\,dy\right)\tilde{U}_{m}(x,\tilde{k})\,d\tilde{k}.$

By repeating the the proof of Proposition 3.1, we similarly show that

$\begin{split}u(t,x)=&\frac{B_{0}}{8\pi^{2}i\sin(tB_{0})}\int_{0}^{\infty}\int_{\mathbb{R}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4i\tan(tB_{0})}}\\\
&\times\int_{\mathbb{R}}\Bigg{(}e^{i(\tilde{k}-\alpha)(\theta_{1}-\theta_{2}-tB_{0})}I_{|\tilde{k}|}\bigg{(}\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\bigg{)}\Bigg{)}\,d\tilde{k}f(r_{2},\theta_{2})r_{2}dr_{2}d\theta_{2}.\end{split}$

In contrast to (3.6), the difference here is that we replace the summation in
$k\in\mathbb{Z}$ by integration on $\tilde{k}\in\mathbb{R}$. Hence the kernel
of $e^{-it\tilde{H}_{\alpha,B_{0}}}$ is

$\begin{split}\tilde{K}_{S}(x,y)=&\frac{B_{0}}{4\pi
i\sin(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4i\tan(tB_{0})}}\\\
&\times\int_{\mathbb{R}}\Bigg{(}e^{i(\tilde{k}-\alpha)(\theta_{1}-\theta_{2}-tB_{0})}I_{|\tilde{k}|}\bigg{(}\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\bigg{)}\Bigg{)}\,d\tilde{k},\end{split}$

where $x=(r_{1},\theta_{1})\in\tilde{M}$ and
$y=(r_{2},\theta_{2})\in\tilde{M}$.

Now, instead of summing in $k$ as before, we consider the integration in
$\tilde{k}$. Again by letting $z=\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}$ and
$\theta=\theta_{1}-\theta_{2}-tB_{0}$ and using (3.7), we compute that

$\begin{split}&\frac{1}{\pi}\int_{\mathbb{R}}e^{i\tilde{k}\theta}\int_{0}^{\pi}e^{z\cos
s}\cos(|\tilde{k}|s)dsd\tilde{k}\\\
&=\frac{1}{2\pi}e^{z\cos\theta}\Big{(}\chi_{[0,\pi]}(\theta)+\chi_{[0,\pi]}(-\theta)\Big{)}=e^{z\cos\theta}\chi_{[-\pi,\pi]}(\theta),\end{split}$

and

$\begin{split}&\frac{1}{\pi}\int_{\mathbb{R}}e^{i\tilde{k}\theta}\sin(\pi|\tilde{k}|)\int_{0}^{\infty}e^{-z\cosh
s}e^{-s|\tilde{k}|}ds\,d\tilde{k}\\\
&=\frac{1}{\pi}\int_{0}^{\infty}e^{-z\cosh
s}\Big{(}\int_{0}^{\infty}e^{i\tilde{k}\theta}\frac{e^{i\pi\tilde{k}}-e^{-i\pi\tilde{k}}}{2i}e^{-s\tilde{k}}d\tilde{k}\\\
&\qquad+\int_{-\infty}^{0}e^{i\tilde{k}\theta}\frac{e^{-i\pi\tilde{k}}-e^{i\pi\tilde{k}}}{2i}e^{s\tilde{k}}d\tilde{k}\Big{)}\,ds\\\
&=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-z\cosh
s}\big{(}\frac{1}{\theta+\pi+is}-\frac{1}{\theta-\pi+is}\big{)}ds\end{split}$

Therefore, we obtain

$\begin{split}&\tilde{K}_{S}(x,y)=\frac{B_{0}}{8\pi^{2}i\sin(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4i\tan(tB_{0})}}e^{-i\alpha\theta}\\\
&\times\Big{(}e^{z\cos\theta}\chi_{[-\pi,\pi]}(\theta)-\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-z\cosh
s}\big{(}\frac{1}{\theta+\pi+is}-\frac{1}{\theta-\pi+is}\big{)}ds\Big{)}.\end{split}$

Finally, by using (4.1), we have that

$\begin{split}&e^{-itH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})\\\
&=\frac{B_{0}}{8\pi^{2}i\sin(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4i\tan(tB_{0})}}\sum_{j\in\mathbb{Z}}e^{-i\alpha(\theta+2j\pi)}\Big{(}e^{z\cos(\theta+2j\pi)}\chi_{[-\pi,\pi]}(\theta+2j\pi)\\\
&\qquad-\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-z\cosh
s}\big{(}\frac{1}{(\theta+2j\pi)+\pi+is}-\frac{1}{(\theta+2j\pi)-\pi+is}\big{)}ds\Big{)}.\end{split}$

Due to the period property of $\cos$ function and for all
$\theta\in\mathbb{R},$ there exists $j_{0}\in\mathbb{Z}$ such that
$\theta+2j_{0}\pi\in[-\pi,\pi),$ the first term in the big bracket becomes

$\displaystyle
e^{z\cos\theta}e^{-i\alpha\theta}\sum_{j\in\mathbb{Z}}e^{-i\alpha
2j\pi}\chi_{[-\pi,\pi]}(\theta+2j\pi)$
$\displaystyle=e^{z\cos\theta}e^{-i\alpha\theta}e^{-i2\pi
j_{0}\alpha}\times\left\\{\begin{array}[]{ll}1,&\hbox{$|\theta+2j_{0}\pi|<\pi$;}\\\
e^{-i2\pi\alpha}+1,&\hbox{$\theta+2j_{0}\pi=-\pi$;}\\\
e^{i2\pi\alpha}+1,&\hbox{$\theta+2j_{0}\pi=\pi$.}\end{array}\right.$

which is the same to (3.10). Hence it is the same to the first term of (3.1).

For the second term in the big bracket, we use the formula

$\sum_{j\in\mathbb{Z}}\frac{e^{-2\pi i\alpha j}}{\sigma+2\pi
j}=\frac{ie^{i\alpha\sigma}}{e^{i\sigma}-1},\quad\alpha\in(0,1),\quad\sigma\in\mathbb{C}\setminus
2\pi\mathbb{Z},$

to obtain

$\begin{split}\sum_{j\in\mathbb{Z}}e^{-2\pi i\alpha
j}&\big{(}\frac{1}{(\theta+2j\pi)+\pi-is}-\frac{1}{(\theta+2j\pi)-\pi-
is}\big{)}\\\
&=2\sin(\pi\alpha)\frac{e^{\alpha(s+i\theta)}}{1+e^{s+i\theta}}.\end{split}$

Now we consider the second term

$\begin{split}&-e^{-i\theta\alpha}\frac{2\sin(\pi\alpha)}{2\pi}e^{i\theta\alpha}\int_{-\infty}^{\infty}e^{-z\cosh
s}\frac{e^{\alpha s}}{1+e^{s+i\theta}}\,ds\\\
=&-\frac{\sin(\pi\alpha)}{\pi}\int_{-\infty}^{\infty}e^{-z\cosh
s}\frac{e^{-\alpha s}}{1+e^{-s+i\theta}}\,ds\end{split}$

Recall $\theta=\theta_{1}-\theta_{2}-tB_{0}$ and
$z=\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}$, we obtain

$\begin{split}K_{S}(x,y)&=\frac{B_{0}e^{-itB\alpha}}{8\pi^{2}i\sin(tB_{0})}e^{\frac{iB_{0}(r_{1}^{2}+r_{2}^{2})}{4\tan(tB_{0})}}\\\
\times&\Big{[}e^{\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cos(\theta_{1}-\theta_{2}-tB_{0})}e^{-i\alpha(\theta_{1}-\theta_{2}-tB_{0}+2j_{0}\pi)}\chi(\theta,j_{0})\\\
&-\frac{\sin(\pi\alpha)}{\pi}\int_{\mathbb{R}}e^{-\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cosh
s}\frac{e^{-\alpha
s}}{1+e^{-s+i(\theta_{1}-\theta_{2}-tB_{0})}}\,ds\Big{]},\end{split}$

which is exact same to (3.1). ∎

##  5\. proof of Theorem 1.1

In this section, we prove the main Theorem 1.1 by using (3.1). We first prove
the dispersive estimate (1.7). To this end, it is enough to prove

$\displaystyle\Big{|}\int_{\mathbb{R}}e^{-\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cosh
s}\frac{e^{-\alpha
s}}{1+e^{-s+i(\theta_{1}-\theta_{2}-tB_{0})}}\,ds\Big{|}\leq C,$

where $C$ is a constant independent of $t$, $r_{1},r_{2}$ and
$\theta_{1},\theta_{2}$.

We notice that

$\displaystyle\int_{\mathbb{R}}e^{-\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cosh
s}\frac{e^{-\alpha s}}{1+e^{-s+i(\theta_{1}-\theta_{2}-tB_{0})}}\,ds$
$\displaystyle=\int_{0}^{\infty}e^{-\frac{B_{0}r_{1}r_{2}}{2i\sin(tB_{0})}\cosh
s}\Big{(}\frac{e^{-\alpha
s}}{1+e^{-s+i(\theta_{1}-\theta_{2}-tB_{0})}}+\frac{e^{\alpha
s}}{1+e^{s+i(\theta_{1}-\theta_{2}-tB_{0})}}\Big{)}\,ds$

then we just need to verify that, for $\theta=\theta_{1}-\theta_{2}-tB_{0},$

$\displaystyle\int_{0}^{\infty}\Big{|}\frac{e^{-\alpha
s}}{1+e^{-s+i\theta}}+\frac{e^{\alpha s}}{1+e^{s+i\theta}}\Big{|}\,ds\lesssim
1,$

where the implicit constant is independent of $\theta$. In fact,

$\displaystyle\frac{e^{-\alpha s}}{1+e^{-s+i\theta}}+\frac{e^{\alpha
s}}{1+e^{s+i\theta}}$ $\displaystyle=\frac{\cosh(\alpha
s)e^{-i\theta}+\cosh((1-\alpha)s)}{\cos\theta+\cosh s}$
$\displaystyle=\frac{\cosh(\alpha
s)\cos\theta+\cosh((1-\alpha)s)-i\sin\theta\cosh(\alpha
s)}{2(\cos^{2}(\frac{\theta}{2})+\sinh^{2}(\frac{s}{2}))}$
$\displaystyle=\frac{2\cos^{2}(\frac{\theta}{2})\cosh(\alpha
s)+(\cosh((1-\alpha)s)-\cosh(\alpha
s))-2i\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\cosh(\alpha
s)}{2(\cos^{2}(\frac{\theta}{2})+\sinh^{2}(\frac{s}{2}))}.$

Since that $\cosh x-1\sim\frac{x^{2}}{2},\sinh x\sim x$, as $x\to 0;$ $\cosh
x\sim e^{x},\sinh x\sim e^{x}$, as $x\to\infty,$ we have

$\displaystyle\int_{0}^{\infty}\Big{|}\frac{\cos^{2}(\frac{\theta}{2})\cosh(\alpha
s)}{\cos^{2}(\frac{\theta}{2})+\sinh^{2}(\frac{s}{2})}\Big{|}\,ds\lesssim\int_{0}^{1}\frac{2|\cos(\frac{\theta}{2})|}{s^{2}+(2|\cos(\frac{\theta}{2})|)^{2}}ds+\int_{1}^{\infty}\
e^{(\alpha-1)s}ds\lesssim 1.$

Similarly, we obtain

$\displaystyle\int_{0}^{\infty}\Big{|}\frac{\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\cosh(\alpha
s)}{\cos^{2}(\frac{\theta}{2})+\sinh^{2}(\frac{s}{2})}\Big{|}\,ds\lesssim 1.$

Finally, we verify that

$\displaystyle\int_{0}^{\infty}\Big{|}\frac{\cosh((1-\alpha)s)-\cosh(\alpha
s)}{\cos^{2}(\frac{\theta}{2})+\sinh^{2}(\frac{s}{2})}\Big{|}\,ds$
$\displaystyle\lesssim\int_{0}^{1}\frac{|\frac{(1-\alpha)^{2}}{2}-\frac{\alpha^{2}}{2}|s^{2}}{s^{2}}ds+\int_{1}^{\infty}\big{(}e^{-\alpha
s}+e^{(\alpha-1)s}\big{)}ds\lesssim 1.$

Therefore, we obtain the bound (1.7) as desired.

Next we prove (1.8). For $T\in(0,\frac{\pi}{2B_{0}})$ and $t\in(0,T)$, we have

$\frac{2}{\pi}\leq\frac{\sin(tB_{0})}{tB_{0}}\leq 1,$

then the dispersive estimate (1.7) gives

$\|e^{-itH_{\alpha,B_{0}}}\|_{L^{1}(\mathbb{R}^{2})\rightarrow
L^{\infty}(\mathbb{R}^{2})}\lesssim\frac{1}{t},\quad\forall t\in(0,T],$ (5.1)

when $T\in(0,\frac{\pi}{2B_{0}}]$. On the other hand, by the spectral theory,
we have the energy estimate

$\|e^{-itH_{\alpha,B_{0}}}\|_{L^{2}(\mathbb{R}^{2})\rightarrow
L^{2}(\mathbb{R}^{2})}\lesssim 1.$ (5.2)

To prove the Strichartz estimates (1.8), we recall the abstract mechanism of
Keel-Tao [21]

######  Theorem 5.1.

(Keel-Tao [21]) Let $(X,d\mu)$ be a measure space and $H$ a Hilbert space.
Suppose that for each time $t\in\mathbb{R}$, $U(t):H\rightarrow L^{2}(X)$
which satisfies the energy estimate

$\|U(t)\|_{H\rightarrow L^{2}}\leq C,\quad t\in\mathbb{R}$

and that for some $\sigma>0$ either

$\|U(t)U(s)^{*}f\|_{L^{\infty}(X)}\leq C|t-s|^{-\sigma}\|f\|_{L^{1}(X)},\quad
t\neq s.$

Then the estimates hold

$\|U(t)f\|_{L^{q}_{t}L^{r}(X)}\lesssim\|f\|_{L^{2}(X)}.$

where

$(q,r)\in\Lambda:=\Big{\\{}(q,r)\in[2,+\infty]\times[2,+\infty):\frac{2}{q}=n(\frac{1}{2}-\frac{1}{r})\Big{\\}}.$

By making use of this theorem to
$U(t)=e^{-itH_{\alpha,B_{0}}}\chi_{[0,T]}(t)$, since (5.1) and (5.2), we
obtain (1.8). Therefore we finish the proof of Theorem 1.1.

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