# Decay estimates for one Aharonov-Bohm solenoid in a uniform magnetic field
II: wave 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 second 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 our first starting paper [36], we have studied the
Strichartz estimates for Schrödinger equation with one Aharonov-Bohm solenoid
in a uniform magnetic field. The wave equation in this setting becomes more
delicate since a difficulty is raised from the square root of the eigenvalue
of the Schrödinger operator $H_{\alpha,B_{0}}$ so that we cannot directly
construct the half-wave propagator. An independent interesting result
concerning the Gaussian upper bounds of the heat kernel is proved by using two
different methods. The first one is based on establishing Davies-Gaffney
inequality in this setting and the second one is straightforward to construct
the heat kernel (which efficiently captures the magnetic effects) based on the
Schulman-Sunada formula. As byproducts, we prove optimal bounds for the heat
kernel and show the Bernstein inequality and the square function inequality
for Schrödinger operator with one Aharonov-Bohm solenoid in a uniform magnetic
field.

Key Words: Strichartz estimates, Davies-Gaffney inequality, wave equation,
Aharonov-Bohm solenoids, uniform magnetic field
AMS Classification: 42B37, 35Q40.

##  1\. Introduction

In this paper, as a sequence of recent papers [19, 21, 36], we study the decay
and Strichartz estimates for the wave equation 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 study
the wave equation

$\begin{cases}\partial_{tt}u(t,x)+H_{\alpha,B_{0}}u(t,x)=0,\\\
u(0,x)=u_{0}(x),\quad\partial_{t}u(0,x)=u_{1}(x),\end{cases}$ (1.1)

where the magnetic Schrödinger operator

$H_{\alpha,B_{0}}=-(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))^{2},$ (1.2)

is the same as the one considered in [36]. Here, $A_{B}(x)$ is the Aharonov-
Bohm potential (initially introduced in [3])

$A_{B}(x)=\alpha\Big{(}-\frac{x_{2}}{|x|^{2}},\frac{x_{1}}{|x|^{2}}\Big{)},\quad
x=(x_{1},x_{2})\in\mathbb{R}^{2}\setminus\\{0\\},$ (1.3)

where $\alpha\in\mathbb{R}$ represents the circulation of $A_{B}$ around the
solenoid; $A_{\mathrm{hmf}}(x)$ is given by

$A_{\mathrm{hmf}}(x)=\frac{B_{0}}{2}(-x_{2},x_{1}),\quad B_{0}>0,$ (1.4)

which generates the background uniform magnetic field.

We stress that the model is on $\mathbb{R}^{2}$ and the magnetic field $B$ is
given by

$B(x):=DA-DA^{t},\quad B_{ij}=\frac{\partial A^{i}}{\partial
x_{j}}-\frac{\partial A^{j}}{\partial x_{i}},\quad i,j=1,2.$ (1.5)

Hence, the generated magnetic field $B(x)=B_{0}+\alpha\delta(x)$ is actually a
superposition of the uniform field and the Aharonov-Bohm field, where $\delta$
is the usual Dirac delta. As mentioned in [36], the Aharonov-Bohm potential
that produces the singular magnetic field has the same homogeneity as $\nabla$
(homogenous of degree $-1$) so that the perturbation from the Aharonov-Bohm
potential (1.3) is critical; the potential $A_{\mathrm{hmf}}(x)$ is unbounded
at infinity and the uniform magnetic filed $B(x)=B_{0}$ from (1.5) generates a
trapped well. Moreover, due to the presence of the potential (1.4), the
spectrum of the operator $H_{\alpha,B_{0}}$ consists of pure point, and thus
the dispersive behavior of wave equation associated with $H_{\alpha,B_{0}}$
will be distinguished from the models in [19, 21].

The Hamiltonian $H_{\alpha,B_{0}}$ can be defined as a self-adjoint operator
on $L^{2}$, via Friedrichs’ Extension Theorem (see e.g. [22, Thm. VI.2.1] and
[28, X.3]), with a natural form domain, which in 2D turns out to be equivalent
to

$\mathcal{D}(H_{\alpha,B_{0}})\simeq\mathcal{H}^{1}_{\alpha,B_{0}}:=\left\\{f\in
L^{2}(\mathbb{R}^{2};\mathbb{C}):\int_{\mathbb{R}^{2}}\left|\big{(}\nabla+i(A_{B}+A_{\mathrm{hmf}}\big{)}f\right|^{2}\,dx<+\infty\right\\}.$

We refer to [36, Section 2] for the Friedrichs’ extension via quadratic forms
and to [15] for more about the self-adjoint extension theory. In what follows
and throughout, the operator $H_{\alpha,B_{0}}$ should be regarded as a self-
adjoint operator generated by the procedure of the Friedrichs’ extension.
Therefore, the half-wave propagator $e^{it\sqrt{H_{\alpha,B_{0}}}}$ can be
treated as one-parameter groups of operators on $L^{2}(\mathbb{R}^{2})$. This
allows to study a large class of dispersive estimates, such as time-decay
(perhaps local in time), Strichartz and local smoothing for dispersive
evolutions as (1.1). The validity of such properties has been central object
of deep investigation of dispersive equations in the last decades, due to
their relevance in the description of linear and nonlinear dynamics. To better
frame our results, let us briefly sketch the state of the art about these
problems.

Due to the significance of dispersive and Strichartz estimates in harmonic
analysis and partial differential equations, there are too much literature to
cite all of them here. But we would like to refer to [6, 7, 8, 10, 11, 13, 14,
31] and the references therein for various dispersive equations with
electromagnetic potentials in mathematics and physics. The dispersive
equations with the Aharonov-Bohm potential, as a diffraction physical model,
have attracted more and more researchers to study from the mathematical
perspective. In [17, 18], the authors studied the validity of the time decay
estimates for the Schrödinger equation with the Aharonov-Bohm potential.
However, due to the lack of pseudo-conformal invariance (which plays a
critical role in the Schrödinger case), the arguments of [17, 18] break down
for the wave equation. Very recently, Fanelli, Zheng and the last author [19]
established Strichartz estimate for the wave equation by constructing the odd
sine propagator. To solve open problems, raised in the survey [16] on the
dispersive estimates for other equations (e.g. Klein-Gordon, Dirac, etc.),
Gao, Yin, Zheng and the last author [21] constructed the spectral measure and
then applied to prove the time decay and Strichartz estimates for the Klein-
Gordon equation. The potential models in [17, 18, 19, 21] are all scaling-
invariant and without unbounded (at infinity) perturbations, which is a
special case of our model (1.2) (with $B_{0}\equiv 0$). In this paper, as
[36], we proceed to consider the wave equation in the magnetic fields mixed
with the Aharonov-Bohm and the uniform ones.

Before stating our main results, let us introduce some preliminary notations.
We define the magnetic Besov spaces as follows. Let $\varphi\in
C_{c}^{\infty}(\mathbb{R}\setminus\\{0\\})$ satisfy $0\leq\varphi\leq
1,\text{supp}\,\varphi\subset[1/2,1]$, and

$\sum_{j\in\mathbb{Z}}\varphi(2^{-j}\lambda)=1,\quad\varphi_{j}(\lambda):=\varphi(2^{-j}\lambda),\,j\in\mathbb{Z},\quad\phi_{0}(\lambda):=\sum_{j\leq
0}\varphi(2^{-j}\lambda).$ (1.6)

######  Definition 1.1 (Magnetic Besov spaces associated with
$H_{\alpha,B_{0}}$).

For $s\in\mathbb{R}$ and $1\leq p,r<\infty$, the homogeneous Besov norm
$\|\cdot\|_{\dot{\mathcal{B}}^{s}_{p,r}(\mathbb{R}^{2})}$ is defined by

$\|f\|_{\dot{\mathcal{B}}^{s}_{p,r}(\mathbb{R}^{2})}=\Big{(}\sum_{j\in\mathbb{Z}}2^{jsr}\|\varphi_{j}(\sqrt{H_{\alpha,B_{0}}})f\|_{L^{p}(\mathbb{R}^{2})}^{r}\Big{)}^{1/r}.$
(1.7)

In particular, for $p=r=2$, we have the Sobolev norm

$\begin{split}\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}:=\|f\|_{\dot{\mathcal{B}}^{s}_{2,2}(\mathbb{R}^{2})}.\end{split}$
(1.8)

######  Remark 1.2.

Alternatively, the Sobolev space can be defined by

$\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2}):=H_{\alpha,B_{0}}^{-\frac{s}{2}}L^{2}(\mathbb{R}^{2}),$

with the norm

$\begin{split}\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}&:=\big{\|}H_{\alpha,B_{0}}^{\frac{s}{2}}f\big{\|}_{L^{2}(\mathbb{R}^{2})}.\end{split}$
(1.9)

By the spectral theory of operators on $L^{2}$, the norms in (1.8) and (1.9)
are equivalent; see Proposition 2.5 below.

######  Definition 1.3.

A pair $(q,p)\in[2,\infty]\times[2,\infty)$ is said to be admissible, if
$(q,p)$ satisfies

$\frac{2}{q}\leq\frac{1}{2}-\frac{1}{p}.$ (1.10)

For $s\in\mathbb{R}$, we denote $(q,p)\in\Lambda^{W}_{s}$ if $(q,p)$ is
admissible and satisfies

$\frac{1}{q}+\frac{2}{p}=1-s.$ (1.11)

Now we state our main theorem.

######  Theorem 1.4.

Let $H_{\alpha,B_{0}}$ be as in (1.2) and $t\in I:=[0,T]$ with any finite $T$.
Then there exists a constant $C_{T}$ depending on $T$ such that

$\|e^{it\sqrt{H_{\alpha,B_{0}}}}f\|_{L^{\infty}(\mathbb{R}^{2})}\leq
C_{T}|t|^{-1/2}\|f\|_{\dot{\mathcal{B}}^{3/2}_{1,1}(\mathbb{R}^{2})},\quad
t\in I,\quad t\neq 0.$ (1.12)

Let $u(t,x)$ be the solution of (1.1) with initial data
$(u_{0},u_{1})\in\dot{\mathcal{H}}_{\alpha,B_{0}}^{s}(\mathbb{R}^{2})\times\dot{\mathcal{H}}_{\alpha,B_{0}}^{s-1}(\mathbb{R}^{2})$,
then the Strichartz estimates

$\|u(t,x)\|_{L^{q}(I;L^{p}(\mathbb{R}^{2}))}\leq
C_{T}\left(\|u_{0}\|_{\dot{\mathcal{H}}_{\alpha,B_{0}}^{s}(\mathbb{R}^{2})}+\|u_{1}\|_{\dot{\mathcal{H}}_{\alpha,B_{0}}^{s-1}(\mathbb{R}^{2})}\right)$
(1.13)

hold for $(q,p)\in\Lambda^{W}_{s}$ and $0\leq s<1$.

######  Remark 1.5.

The local-in-time decay estimate (1.12) is quite different from the
Schrödinger counterpart (see [36, Theorem 1.1])

$\begin{split}&\big{\|}e^{itH_{\alpha,B_{0}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\leq
C|\sin(tB_{0})|^{-1}\big{\|}f\big{\|}_{L^{1}\mathbb{R}^{2})},\quad
t\neq\frac{k\pi}{B_{0}},\,k\in\mathbb{Z},\end{split}$

which is similar to the harmonic oscillators (see Koch and Tataru [24]). The
period $\pi/B_{0}$ is essentially the Larmor period. However, for the wave
equation, provided that the data $f=\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f$
is localized at frequency scale $2^{j}$ with $j\in\mathbb{Z}$, we can prove
(see (5.11) below)

$\begin{split}\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})&e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim
2^{2j}\big{(}1+2^{j}t\big{)}^{-N}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})},\quad\text{for}\quad
2^{j}t\lesssim 1\end{split}$

and (see (5.9) below)

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim
2^{2j}\big{(}1+2^{j}t\big{)}^{-\frac{1}{2}}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})},\quad\text{for}\quad
2^{j}t\lesssim 1,\quad 2^{-j}|t|\leq\frac{\pi}{8B_{0}}.\end{split}$

The decay estimates for waves depend on the frequency. The Strichartz
estimates (1.13) is still local-in-time but the endpoint of the time interval
$T$ can beyond $\frac{\pi}{2B_{0}}$ which is the upper bound of $T$ for
Schrödinger’s Strichartz estimates. Due to the unbounded potentials caused a
trapped well, the Strichartz estimate is impossible to be global-in-time (for
example, the Strichartz estimates for dispersive equations on sphere or
torus), but still captures integration regularity behavior near $t=0$.

Now let us figure out some points in our proof.

  * •

As mentioned above, for the Schrödinger equation considered in [36], the
explicit eigenvalues and eigenfunctions of the operator $H_{\alpha,B_{0}}$ are
the key ingredients. In particular, the eigenvalues are given by

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

see (2.1) below. One feature of $\lambda_{k,m}$ is that $k$ and $m$ can be
separated in the series convergent argument. However, for the half-wave
propagator, this feature breaks down for the square root of $\lambda_{k,m}$.
Therefore, we cannot directly construct the wave propagator by following the
argument of [19, 36].

  * •

Due to the uniform magnetic field caused a trapped well, the spectral measure
will involve a factor $\sin(tB_{0})$ (which is a short-time decay but not
long-time). This lead to the failure of the spectral measure argument in [21].

  * •

To go around constructing the spectral measure, we turn to prove the Bernstein
inequality to deal with the low frequency. For the high frequency, we use the
classical subordination formula

$e^{-y\sqrt{H_{\alpha,B_{0}}}}=\frac{y}{2\sqrt{\pi}}\int_{0}^{\infty}e^{-sH_{\alpha,B_{0}}}e^{-\frac{y^{2}}{4s}}s^{-\frac{3}{2}}ds,\quad
y>0,$

which provides a connecting bridge between the Schrödinger propagator and the
half-wave propagator. This idea is originated from [27] and [12]. The
dispersive estimates proved in [36] are then used to address the high
frequency of the waves.

  * •

The Littlewood-Paley theory (including Bernstein inequality and the square
function inequality) associated with the Schrödinger operator
$H_{\alpha,B_{0}}$ are proved by establishing the Gaussian upper bounds for
the heat kernel.

  * •

The heat kernel estimates for magnetic Schrödinger operators have its own
interest, we provide two methods to study the heat kernel. Unfortunately, due
to the fact $A(x)=A_{\mathrm{hmf}}(x)+A_{B}(x)\notin
L^{2}_{\mathrm{loc}}(\mathbb{R}^{2})$, Simon’s diamagnetic pointwise
inequality (see e.g. [29, Theorem B.13.2], [4]) cannot be directly used. Even
though we cannot recover all the magnetic effects to prove the optimal heat
kernel estimates, but we can prove

$\big{|}e^{-tH_{\alpha,B_{0}}}(x,y)\big{|}\lesssim\frac{1}{t}e^{-\frac{|x-y|^{2}}{Ct}},$

which is enough for proving the Bernstein inequality and the square function
inequality. We first prove the on-diagonal estimates and then extend to the
off-diagonal estimates by establishing the Davies-Gaffney inequality. The key
points are the argument of [9] and [20] applying to the magnetic operator
$H_{\alpha,B_{0}}$. To recover more magnetic effects, we use the Schulman-
Sunada formula from [34, 35] to construct the heat kernel and prove

$\Big{|}e^{-tH_{\alpha,B_{0}}}(x,y)\Big{|}\leq C\frac{B_{0}e^{-\alpha
tB_{0}}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}|x-y|^{2}}{4\tanh(tB_{0})}},$

which is better than the previous one. For more discussion on the heat kernel
estimates, we refer to the remarks in Section 3.

The paper is organized as follows. In Section 2, as a preliminary step, we
briefly recall the self-adjoint extension and the spectrum of the operator
$H_{\alpha,B_{0}}$, and prove the equivalence between Sobolev norm and a
special Besov norm. In Section 3, we construct the heat kernel and prove the
Gaussian upper bounds. In Section 4, we prove the Bernstein inequalities and
the square function inequality by using the heat kernel estimates. Finally, in
Section 5 and Section 6, we prove the dispersive estimate (1.12) and the
Strichartz estimate (1.13) in Theorem 1.4 respectively.

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

##  2\. preliminaries

In this section, we first repeat the preliminary section of [36] to recall two
known results about the Friedrichs self-adjoint extension of the operator
$H_{\alpha,B_{0}}$ and the spectrum of $H_{\alpha,B_{0}}$. Next, we use the
spectral argument to prove the equivalence between the Sobolev norm and a
special Besov norm.

###  2.1. Quadratic form and 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, which implies
that the operator $H_{\alpha,B_{0}}$ is symmetric semi bounded from below and
thus 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 [15]) 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 $H^{F}_{\alpha,B_{0}}$.

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

In this subsection, we exhibit the eigenvalues and eigenfunctions 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 (1.3) and (1.4).

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

Let $H_{\alpha,B_{0}}$ be the self-adjoint Schrödinger operator in (1.2). Then
the eigenvalues of $H_{\alpha,B_{0}}$ 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)

with (finite) multiplicity

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

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 $(a)_{n}$ ($a\in\mathbb{R}$) 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

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

then one has 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 [2, (6.2.25)]: for
$a,b,c,\alpha>0$

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

then this together with (2.3) gives

$\begin{split}&\sum_{m=0}^{\infty}e^{-cm}\frac{m!}{\Gamma(m+\alpha_{k}+1)}\Bigg{(}\begin{array}[]{c}m+\alpha_{k}\\\
m\\\ \end{array}\Bigg{)}^{2}P_{k,m}(a)P_{k,m}(b)\\\
&=\frac{e^{\frac{\alpha_{k}c}{2}}}{(ab)^{\frac{\alpha_{k}}{2}}(1-e^{-c})}\exp\left(-\frac{(a+b)e^{-c}}{1-e^{-c}}\right)I_{\alpha_{k}}\left(\frac{2\sqrt{ab}e^{-\frac{c}{2}}}{1-e^{-c}}\right).\end{split}$
(2.5)

We refer to [36] for the proof.

###  2.3. The Sobolev spaces

In this subsection, we will prove the equivalence of two norms.

######  Proposition 2.5 (Equivalent norms).

Let the Sobolev norm and Besov norm be defined in (1.9) and (1.7)
respectively. For $s\in\mathbb{R}$, then there exist positive constants $c,C$
such that

$c\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}\leq\|f\|_{\dot{\mathcal{B}}^{s}_{2,2}(\mathbb{R}^{2})}\leq
C\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})},$ (2.6)

and

$c\|f\|_{\mathcal{H}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}\leq\|f\|_{{\mathcal{B}}^{s}_{2,2}(\mathbb{R}^{2})}\leq
C\|f\|_{\mathcal{H}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}.$ (2.7)

###### Proof.

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 $H_{\alpha,B_{0}}$.

By the functional calculus, for any well-behaved functions $F$ (e.g. bounded
Borel measurable function) and $f\in L^{2}$, we can write

$F(H_{\alpha,B_{0}})f=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}F(\lambda_{k,m})c_{k,m}\tilde{V}_{k,m}(x).$

where

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

Then

$\|F(H_{\alpha,B_{0}})f\|_{L^{2}(\mathbb{R}^{2})}=\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}\big{|}F(\lambda_{k,m})c_{k,m}\big{|}^{2}\Big{)}^{1/2}.$ (2.8)

In particular, we have

$\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}=\|H^{\frac{s}{2}}_{\alpha,B_{0}}f\|_{L^{2}(\mathbb{R}^{2})}=\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}\big{|}\lambda^{\frac{s}{2}}_{k,m}c_{k,m}\big{|}^{2}\Big{)}^{1/2}.$

Let $\varphi\in C_{c}^{\infty}(\mathbb{R}\setminus\\{0\\})$ in (1.6). On the
one hand, by the definition and (2.8), we have

$\begin{split}\|f\|_{\dot{\mathcal{B}}^{s}_{2,2}(\mathbb{R}^{2})}&=\Big{(}\sum_{j\in\mathbb{Z}}2^{2js}\|\varphi_{j}(\sqrt{H_{\alpha,B_{0}}})f\|_{L^{2}(\mathbb{R}^{2})}^{2}\Big{)}^{1/2}\\\
&=\Big{(}\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}2^{2js}\big{|}\varphi\Big{(}\frac{\sqrt{\lambda_{k,m}}}{2^{j}}\Big{)}c_{k,m}\big{|}^{2}\Big{)}^{1/2}\\\
&\cong\Big{(}\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}\lambda_{k,m}^{s}\big{|}\varphi\Big{(}\frac{\sqrt{\lambda_{k,m}}}{2^{j}}\Big{)}c_{k,m}\big{|}^{2}\Big{)}^{1/2}\\\
&\lesssim\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}|\lambda_{k,m}^{\frac{s}{2}}c_{k,m}|^{2}\sum_{j\in\mathbb{Z}}\big{|}\varphi\Big{(}\frac{\sqrt{\lambda_{k,m}}}{2^{j}}\Big{)}\big{|}^{2}\Big{)}^{1/2}\\\
&\lesssim\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}|\lambda_{k,m}^{\frac{s}{2}}c_{k,m}|^{2}\Big{)}^{1/2}=\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}.\end{split}$

On the other hand, we have

$\begin{split}\|f\|_{\dot{\mathcal{H}}^{s}_{\alpha,B_{0}}(\mathbb{R}^{2})}&=\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}|\lambda_{k,m}^{\frac{s}{2}}c_{k,m}|^{2}\Big{)}^{1/2}\\\
&=\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}\Big{|}\sum_{j\in\mathbb{Z}}\varphi\Big{(}\frac{\sqrt{\lambda_{k,m}}}{2^{j}}\Big{)}\lambda_{k,m}^{\frac{s}{2}}c_{k,m}\Big{|}^{2}\Big{)}^{1/2}\\\
&\leq C\Big{(}\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}2^{2js}\big{|}\varphi\Big{(}\frac{\sqrt{\lambda_{k,m}}}{2^{j}}\Big{)}c_{k,m}\big{|}^{2}\Big{)}^{1/2}\\\
&\lesssim\Big{(}\sum_{j\in\mathbb{Z}}2^{2js}\|\varphi_{j}(\sqrt{H_{\alpha,B_{0}}})f\|_{L^{2}(\mathbb{R}^{2})}^{2}\Big{)}^{1/2}=\|f\|_{\dot{\mathcal{B}}^{s}_{2,2}(\mathbb{R}^{2})}.\end{split}$

In the above inequality, we have used the fact that, for a fixed $\lambda$,
there are only finite terms in the summation

$1=\sum_{j\in\mathbb{Z}}\varphi\big{(}\frac{\lambda}{2^{j}}\big{)}.$

Above all, we have proved (2.6). One can prove (2.7) similarly.

∎

##  3\. Heat kernel estimates

In this section, for our purpose of the Littlewood-Paley theory associated
with $H_{\alpha,B_{0}}$, we study the heat kernel estimates associated with
the magnetic Schrödinger operator $H_{\alpha,B_{0}}$. We provide two methods
to study the heat kernel. In the first method, we first combine the strategies
of [18, 21, 19] to construct the heat kernel by using the spectrum property in
Proposition 2.1. And then we use the representation of the heat kernel to
obtain the on-diagonal estimates. Finally we extend the on-diagonal bounds by
adding the Gaussian factor $\exp(-d^{2}(x,y)/Ct)$ to obtain the off-diagonal
Gaussian bounds. In the second one, we directly construct the heat kernel by
using the Schulman-Sunada formula in [34, 35] and then optimal the established
bounds.

###  3.1. Method I:

More precisely, we will first prove the following result.

######  Proposition 3.1.

Let $H_{\alpha,B_{0}}$ be the operator in (1.2) 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
heat equation

$\begin{cases}\big{(}\partial_{t}+H_{\alpha,B_{0}}\big{)}u(t,x)=0,\\\
u(0,x)=f(x).\end{cases}$ (3.1)

Then

$u(t,x)=e^{-tH_{\alpha,B_{0}}}f=\int_{\mathbb{R}^{2}}K_{H}(t;x,y)f(y)\,dy,\quad
t>0,$

where the kernel of the heat propagator $e^{-tH_{\alpha,B_{0}}}$ is given by

$K_{H}(t;x,y)=\Big{(}\frac{B_{0}e^{-\alpha tB_{0}}}{4\pi^{2}\sinh
tB_{0}}\Big{)}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})\cosh tB_{0}}{4\sinh
tB_{0}}}\sum_{k\in\mathbb{Z}}e^{ik(\theta_{1}-\theta_{2}+itB_{0})}I_{\alpha_{k}}\left(\frac{B_{0}r_{1}r_{2}}{2\sinh
tB_{0}}\right).$ (3.2)

Furthermore, there exists a constant $C$ such that

$|K_{H}(t;x,y)|\leq C\frac{B_{0}e^{(1-\alpha)B_{0}t}}{\sinh
tB_{0}}e^{-\frac{B_{0}(r_{1}-r_{2})^{2}}{4\tanh tB_{0}}}.$ (3.3)

######  Remark 3.2.

The argument is a bit different from the proof for the Schrödinger propagator.
In particular, at first glance, the factor $e^{-tB_{0}k}$ is a trouble in the
summation of the formula (3.2) when $k\in-\mathbb{N}$, but it converges due to
the factor $\sinh(tB_{0})$ in the modified Bessel function.

###### Proof.

We construct the representation formula (3.2) of the heat flow
$e^{-tH_{\alpha,B_{0}}}$ by combining the argument of [18] and [19, 21]. This
is close to the construction of Schrödinger flow in our previous paper [36],
however, we provide the details again for self-contained.

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

$c_{k,m}=\int_{\mathbb{R}^{2}}f(x)\overline{\tilde{V}_{k,m}(x)}\,dx.$ (3.4)

The solution $u(t,x)$ of (3.1) can be written as

$u(t,x)=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}u_{k,m}(t)\tilde{V}_{k,m}(x),$ (3.5)

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

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

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

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

Plugging (3.4) into the above expression yields

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

We write $f$ in a harmonic spherical expansion

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

where

$f_{k}(r_{2})=\frac{1}{2\pi}\int_{0}^{2\pi}f(r_{2},\theta_{2})e^{-ik\theta_{2}}\,d\theta_{2},\quad
r_{2}=|y|,$ (3.6)

we thus have

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

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

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

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

Using the formula (2.5) and (3.6), we obtain

$\displaystyle u(t,x)$
$\displaystyle=\Big{(}\frac{B_{0}}{4\pi^{2}}\Big{)}\int_{0}^{\infty}\int_{0}^{2\pi}\sum_{k\in\mathbb{Z}}e^{ik(\theta_{1}-\theta_{2})}\frac{B_{0}^{\alpha_{k}}e^{-t\beta_{k}}}{2^{\alpha_{k}}\Gamma(1+\alpha_{k})}(r_{1}r_{2})^{\alpha_{k}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4}}f(r_{2},\theta_{2})$
$\displaystyle\times\frac{2^{\alpha_{k}}e^{t\alpha_{k}B_{0}}}{(B_{0}r_{1}r_{2})^{\alpha_{k}}}\exp\left(-\frac{\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{2}e^{-2tB_{0}}}{1-e^{-2tB_{0}}}\right)I_{\alpha_{k}}\left(\frac{B_{0}r_{1}r_{2}e^{-tB_{0}}}{1-e^{-2tB_{0}}}\right)r_{2}\mathrm{d}r_{2}\mathrm{d}\theta_{2}$
$\displaystyle=\Big{(}\frac{B_{0}e^{-\alpha tB_{0}}}{4\pi^{2}\sinh
tB_{0}}\Big{)}\int_{0}^{\infty}\int_{0}^{2\pi}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})\cosh
tB_{0}}{4\sinh tB_{0}}}f(r_{2},\theta_{2})$
$\displaystyle\quad\times\sum_{k\in\mathbb{Z}}e^{ik(\theta_{1}-\theta_{2}+itB_{0})}I_{\alpha_{k}}\left(\frac{B_{0}r_{1}r_{2}}{2\sinh
tB_{0}}\right)r_{2}\mathrm{d}r_{2}\mathrm{d}\theta_{2}.$

Therefore, we obtain the heat kernel

$K_{H}(t;x,y)=\Big{(}\frac{B_{0}e^{-\alpha tB_{0}}}{4\pi^{2}\sinh
tB_{0}}\Big{)}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})\cosh tB_{0}}{4\sinh
tB_{0}}}\sum_{k\in\mathbb{Z}}e^{ik(\theta_{1}-\theta_{2}+itB_{0})}I_{\alpha_{k}}\left(\frac{B_{0}r_{1}r_{2}}{2\sinh
tB_{0}}\right),$

which gives (3.2).

Now we need to verify the inequality (3.3). To this end, it suffices to show

$|K_{H}(t;x,y)|\leq\frac{B_{0}e^{-\alpha B_{0}t}}{4\pi^{2}\sinh
tB_{0}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh
tB_{0}}}\sum_{k\in\mathbb{Z}}e^{-kB_{0}t}I_{\alpha_{k}}\left(\frac{B_{0}r_{1}r_{2}}{2\sinh
tB_{0}}\right).$ (3.7)

Let $z=\frac{B_{0}r_{1}r_{2}}{2\sinh tB_{0}}>0$ and notice the monotonicity of
the modified Bessel function $I_{\mu}(z)$ with respect to the order, in other
words, for fixed $z>0$,

$I_{\mu}(z)\leq I_{\nu}(z),\quad\mu\geq\nu.$

Recall $\alpha_{k}=|k+\alpha|$ and $\alpha\in(0,1)$, thus we show

$\displaystyle\sum_{k\in\mathbb{Z}}e^{-kB_{0}t}I_{\alpha_{k}}\left(z\right)=$
$\displaystyle\sum_{k\geq 0}e^{-kB_{0}t}I_{k+\alpha}(z)+\sum_{k\geq
1}e^{kB_{0}t}I_{k-\alpha}(z)$ $\displaystyle\leq$ $\displaystyle\sum_{k\geq
0}e^{-kB_{0}t}I_{k}(z)+e^{B_{0}t}\sum_{k\geq 0}e^{kB_{0}t}I_{k}(z)$
$\displaystyle\leq$ $\displaystyle e^{B_{0}t}\Big{(}\sum_{k\geq
0}e^{-kB_{0}t}I_{k}(z)+\sum_{k\geq 0}e^{kB_{0}t}I_{k}(z)\Big{)}$
$\displaystyle\leq$ $\displaystyle
e^{B_{0}t}\Big{(}\sum_{k\in\mathbb{Z}}e^{kB_{0}t}I_{|k|}(z)+I_{0}(z)\Big{)}$
$\displaystyle\leq$ $\displaystyle Ce^{B_{0}t}(e^{z}+e^{z\cosh(B_{0}t)})$
$\displaystyle\leq$ $\displaystyle Ce^{B_{0}t}e^{z\cosh(B_{0}t)}$
$\displaystyle\leq$ $\displaystyle
Ce^{B_{0}t}e^{\frac{B_{0}r_{1}r_{2}}{2|\tanh tB_{0}|}},$

where we use the formula [5, Eq. (9.6.19)]

$\sum_{k\in\mathbb{Z}}e^{kt}I_{|k|}(z)=e^{z\cosh(t)},\quad I_{0}(z)\leq
Ce^{z}.$

Combining with (3.7), we have verified (3.3). ∎

We next extend our result of the ”on-diagonal” kernel estimate

$|K_{H}(t;x,x)|\leq C\frac{B_{0}e^{(1-\alpha)B_{0}t}}{|\sinh tB_{0}|}$

to the ”off-diagonal”. Let $p_{t}(x,y)$ denote the heat kernel corresponding
to a second-order differential elliptic or sub-elliptic operator, then the
usual theory says that one can automatically improve on-diagonal bounds

$p_{t}(x,x)\leq\frac{C}{V(x,\sqrt{t})}$

to the typical Gaussian heat kernel upper bound

$p_{t}(x,y)\leq\frac{C}{V(x,\sqrt{t})}\exp\Big{(}-\frac{d^{2}(x,y)}{Ct}\Big{)}$

for all $t>0$ and $x,y$ ranging in the space where the operator acts, for an
appropriate function $V$.

For our specific operator $H_{\alpha,B_{0}}$, we prove that

######  Proposition 3.3.

Let $K_{H}(t;x,y)$ be in Proposition 3.1, then there exists a constant $C$
such that

$|K_{H}(z;x,y)|\leq
C(\operatorname*{Re}z)^{-1}\exp\Big{(}-\operatorname*{Re}\frac{d^{2}(x,y)}{Cz}\Big{)},$

for all $z\in\mathbb{C}_{+}$ and $x,y\in\mathbb{R}^{2}$. In particular,
$z=t>0$, then

$|K_{H}(t;x,y)|\leq Ct^{-1}\exp\Big{(}-\frac{d^{2}(x,y)}{Ct}\Big{)},\quad
t>0.$ (3.8)

######  Remark 3.4.

One usual way to prove the Gaussian bounds for the magnetic Schrödinger
operator is to apply the important diamagnetic inequality

$\Big{|}\Big{(}e^{t(\nabla+iA(x))^{2}}f\Big{)}(x)\Big{|}\leq\Big{(}e^{t\Delta}|f|\Big{)}(x),$
(3.9)

which relates estimates on the magnetic Schrödinger operator semigroup to
estimates on the free heat semigroup. The obvious disadvantage of using (3.9)
is that all the effects of the magnetic field are completely eliminated. To
our best knowledge, (3.9) is available for $A(x)\in L^{2}_{\mathrm{loc}}$, see
[29]. Unfortunately, our magnetic potential
$A(x)=A_{B}(x)+A_{\mathrm{hmf}}(x)\notin
L^{2}_{\mathrm{loc}}(\mathbb{R}^{2})$.

######  Remark 3.5.

To recover some magnetic effects, it would be tempting to prove

$\Big{|}\Big{(}e^{-tH_{\alpha,B_{0}}}f\Big{)}(x)\Big{|}\leq\Big{(}e^{-tH_{0,B_{0}}}|f|\Big{)}(x),$
(3.10)

or

$\Big{|}\Big{(}e^{-tH_{\alpha,B_{0}}}f\Big{)}(x)\Big{|}\leq\Big{(}e^{-tH_{\alpha,0}}|f|\Big{)}(x).$
(3.11)

If (3.10) was available,

$\big{|}e^{-tH_{\alpha,B_{0}}}(x,y)\big{|}\lesssim\big{|}e^{-tH_{0,B_{0}}}(x,y)\big{|}\lesssim\frac{1}{\sinh(B_{0}t)}e^{-\frac{B_{0}|x-y|^{2}}{4\tanh(B_{0}t)}},$
(3.12)

where we use the Mehler heat kernel of $e^{-tH_{0,B_{0}}}$ (e.g. [30, P168])

$e^{-tH_{0,B_{0}}}(x,y)=\frac{B_{0}}{4\pi\sinh(B_{0}t)}e^{-\frac{B_{0}|x-y|^{2}}{4\tanh(B_{0}t)}-\frac{iB_{0}}{2}(x_{1}y_{2}-x_{2}y_{1})}.$
(3.13)

If (3.11) was available, we obtain

$\big{|}e^{-tH_{\alpha,B_{0}}}(x,y)\big{|}\lesssim\big{|}e^{-tH_{\alpha,0}}(x,y)\big{|}\lesssim\frac{1}{t}e^{-\frac{|x-y|^{2}}{4t}}.$
(3.14)

We refer to [19, Proposition 3.2] for the last Gaussian upper bounds for
$e^{-tH_{\alpha,0}}(x,y)$. Since for $t\geq 0$, one has

$\sinh(t)\geq t,\quad t/\tanh(t)\geq 1,$

hence (3.14) is weaker than (3.12). Unfortunately, as pointed out in [26], the
semigroup generated by the magnetic Schrödinger operator is not Markovian, in
fact not even positivity preserving which is important in the theory of
comparison of heat semigroups. So the truth of (3.10) and (3.11) is not known,
we refer to [26].

######  Remark 3.6.

The Ganssian decay on the right side of (3.8) is the one of the heat kernel,
which is considerably weaker than the decay of the Mehler kernel (3.12).
Similarly as [26], we ask that how to prove

$|K_{H}(t;x,y)|\lesssim\frac{B_{0}e^{-\alpha B_{0}t}}{|\sinh
tB_{0}|}e^{-\frac{|x-y|^{2}}{C\tanh(B_{0}t)}},\quad t>0.$

The truth of this estimate would reveal a robust dependence of the magnetic
heat kernel on the magnetic field. In our case, we give a positive answer to
this problem by proving (3.30) in the subsequent subsection.

###### Proof.

We prove (3.8) by using [9, Theorem 4.2]. The Theorem claims that if
$(M,d,\mu,L)$ satisfies the Davies-Gaffney estimates, that is,

$|\langle
e^{-tL}f,g\rangle|\leq\|f\|_{L^{2}(U_{1})}\|g\|_{L^{2}(U_{2})}e^{-\frac{d^{2}(U_{1},U_{2})}{4t}}$

for all $t>0$, $U_{i}\subset M$ with $i=1,2$ and $f\in L^{2}(U_{1},d\mu)$,
$g\in L^{2}(U_{2},d\mu)$ and $d(U_{1},U_{2})=\inf\\{\rho=|x-y|:x\in U_{1},y\in
U_{2}\\}$. If, for some $K$ and $D>0$,

$e^{-tL}(x,x)\leq Kt^{-\frac{D}{2}},\qquad\forall t>0,\quad x\in M,$

then

$|e^{-zL}(x,y)|\leq
K(\operatorname*{Re}z)^{-\frac{D}{2}}\Big{(}1+\operatorname*{Re}\frac{d^{2}(x,y)}{4z}\Big{)}^{\frac{D}{2}}\exp\Big{(}-\operatorname*{Re}\frac{d^{2}(x,y)}{4z}\Big{)}$

for all $z\in\mathbb{C}_{+}$ and $x,y\in M$.

For our model $M=\mathbb{R}^{2}$ and $L=H_{\alpha,B_{0}}$, we need to verify
the on-diagonal estimates

$e^{-tH_{\alpha,B_{0}}}(x,x)\leq Kt^{-1},\qquad\forall t>0,\quad x\in M,$
(3.15)

and the Davies-Gaffney estimates

$|\langle
e^{-tH_{\alpha,B_{0}}}f,g\rangle|\leq\|f\|_{L^{2}(U_{1})}\|g\|_{L^{2}(U_{2})}e^{-\frac{d^{2}(U_{1},U_{2})}{4t}}.$
(3.16)

If this has been done, for $z=t>0$ and $D=2$ and $\epsilon>0$, then

$\begin{split}|e^{-tH_{\alpha,B_{0}}}(x,y)|&\leq
Ct^{-1}\Big{(}1+\frac{|x-y|^{2}}{4t}\Big{)}\exp\Big{(}-\frac{|x-y|^{2}}{4t}\Big{)}\\\
&\leq Ct^{-1}\exp\Big{(}-\frac{|x-y|^{2}}{(4+\epsilon)t}\Big{)}.\end{split}$

Therefore, it suffices to verify (3.15) and (3.16). Since $x=y$ and
$\frac{B_{0}e^{-\alpha B_{0}t}}{4\pi^{2}|\sinh tB_{0}|}\leq Ct^{-1}$, the
estimate (3.15) is a direct consequence of (3.3). However, the inequality
(3.16) is more complicated, this is a consequence of (3.17) below.

∎

######  Proposition 3.7 (Davies-Gaffney inequality).

Let $A$ and $B$ be two disjoint measurable sets in $\mathbb{R}^{2}$ and
suppose that $f\in L^{2}(A)$ and $g\in L^{2}(B)$ such that
$\mathrm{supp}(f)\subset A$ and $\mathrm{supp}(g)\subset B$. Then

$|\langle
e^{-tH_{\alpha,B_{0}}}f,g\rangle|\leq\|f\|_{L^{2}(A)}\|g\|_{L^{2}(B)}e^{-\frac{d^{2}(A,B)}{4t}}$
(3.17)

where $d(A,B)=\inf\\{\rho=|x-y|:x\in A,y\in B\\}$.

###### Proof.

Let $\rho=d(A,B)$ and define

$A_{\rho}=\\{x\in\mathbb{R}^{2}:d(x,A)<\rho\\},\quad
A^{c}_{\rho}=\mathbb{R}^{2}\setminus A_{\rho}$

where $d(x,A)=\inf\\{|x-y|:y\in A\\}$. Then $B\subset A^{c}_{\rho}$,
furthermore, by Cauchy-Schwartz inequality, we have

$\begin{split}|\langle
e^{-tH_{\alpha,B_{0}}}f,g\rangle|&\leq\Big{(}\int_{B}|e^{-tH_{\alpha,B_{0}}}f|^{2}dx\Big{)}^{1/2}\|g\|_{L^{2}(B)}\\\
&\leq\Big{(}\int_{A^{c}_{\rho}}|e^{-tH_{\alpha,B_{0}}}f|^{2}dx\Big{)}^{1/2}\|g\|_{L^{2}(B)}.\end{split}$

Therefore, (3.17) follows if we could prove

$\begin{split}\int_{A^{c}_{\rho}}|e^{-tH_{\alpha,B_{0}}}f|^{2}dx\leq\|f\|^{2}_{L^{2}(A)}e^{-\frac{d^{2}(A,B)}{2t}}.\end{split}$
(3.18)

To this end, for any fixed $s>t$ and $x\in\mathbb{R}^{2}$ and $\tau\in[0,s)$,
we define the function

$\xi(\tau,x):=\frac{d^{2}(x,A^{c}_{\rho})}{2(\tau-s)},$

and set

$J(\tau):=\int_{\mathbb{R}^{2}}\big{|}e^{-\tau
H_{\alpha,B_{0}}}f\big{|}^{2}e^{\xi(\tau,x)}\,dx.$ (3.19)

######  Lemma 3.8.

For the function defined in (3.19), we have that

$J(t)\leq J(0).$ (3.20)

We assume (3.20) to prove (3.18) by postponing the proof for a moment. Since
$x\in A^{c}_{\rho}$, one has $\xi(\tau,x)=0$, thus

$\begin{split}\int_{A^{c}_{\rho}}|e^{-tH_{\alpha,B_{0}}}f|^{2}dx\leq\int_{\mathbb{R}^{2}}\big{|}e^{-tH_{\alpha,B_{0}}}f\big{|}^{2}e^{\xi(t,x)}\,dx=J(t).\end{split}$
(3.21)

For $t=0$, since

$e^{\xi(0,x)}\leq\begin{cases}1,&\qquad x\in A^{c}_{\rho};\\\
e^{-\frac{d^{2}(x,A^{c}_{\rho})}{2s}},&\qquad x\in A,\end{cases}$

we see

$J(0)=\int_{\mathbb{R}^{2}}f(x)^{2}e^{\xi(0,x)}\,dx\leq\int_{A^{c}_{\rho}}|f(x)|^{2}\,dx+\exp\Big{(}-\frac{\rho^{2}}{2s}\Big{)}\int_{A}|f(x)|^{2}\,dx.$
(3.22)

By using (3.21), (3.20) and (3.22) and taking $s\to t+$, we obtain

$\begin{split}&\int_{A^{c}_{\rho}}|e^{-tH_{\alpha,B_{0}}}f|^{2}dx\leq J(t)\leq
J(0)\\\ &\leq
C\left(\int_{A^{c}_{\rho}}|f(x)|^{2}\,dx+\exp\Big{(}-\frac{\rho^{2}}{2s}\Big{)}\int_{A}|f(x)|^{2}\,dx\right)\\\
&\leq\int_{A^{c}_{\rho}}|f(x)|^{2}\,dx+\exp\Big{(}-\frac{\rho^{2}}{2t}\Big{)}\|f\|^{2}_{L^{2}(A)}.\end{split}$

Since $f\in L^{2}(A)$ and $\mathrm{supp}(f)\subset A$, then
$\int_{A^{c}_{\rho}}|f(x)|^{2}\,dx=0$ which implies (3.18).

∎

Now it remains to prove (3.20) in Lemma 3.8.

###### The proof of Lemma 3.8.

Indeed, we need to prove that the function $J(\tau)$ defined in (3.19) is non-
increasing in $\tau\in[0,s)$. We closely follow the argument of the integrated
maximum principle [20, Theorem 12.1]. Furthermore, for all
$\tau,\tau_{0}\in[0,s)$, if $\tau>\tau_{0}$, then

$J(\tau)\leq J(\tau_{0}).$ (3.23)

which shows (3.20) by taking $\tau=t$ and $\tau_{0}=0$. Without loss of
generality, we assume $f\geq 0$ in (3.19). Indeed, if $f$ has a change sign,
we set $g=|e^{-\tau_{0}H_{\alpha,B_{0}}}f|\geq 0$, then

$|e^{-\tau
H_{\alpha,B_{0}}}f|=|e^{-(\tau-\tau_{0})H_{\alpha,B_{0}}}e^{-\tau_{0}H_{\alpha,B_{0}}}f|\leq
e^{-(\tau-\tau_{0})H_{\alpha,B_{0}}}g.$

Assume that (3.23) holds for $g\geq 0$, then

$\begin{split}J(\tau)&=\int_{\mathbb{R}^{2}}\big{(}e^{-\tau
H_{\alpha,B_{0}}}f\big{)}^{2}e^{\xi(\tau,x)}\,dx\\\
&\leq\int_{\mathbb{R}^{2}}\big{(}e^{-(\tau-\tau_{0})H_{\alpha,B_{0}}}g\big{)}^{2}e^{\xi(\tau,x)}\,dx\\\
&\leq\int_{\mathbb{R}^{2}}g^{2}(x)e^{\xi(\tau_{0},x)}dx\\\
&=\int_{\mathbb{R}^{2}}\big{(}e^{-\tau_{0}H_{\alpha,B_{0}}}f\big{)}^{2}e^{\xi(\tau_{0},x)}dx\\\
&=J(\tau_{0}).\end{split}$

From now on, we assume $f\geq 0$. By using [20, Theorem 5.23](which claims
$e^{-\tau H^{\Omega_{i}}_{\alpha,B_{0}}}f\to e^{-\tau H_{\alpha,B_{0}}}f$ in
$L^{2}$ as $i\to+\infty$ where
$\Omega_{0}\subset\Omega_{1}\subset\cdots\subset\Omega_{i}\subset\cdots\to\mathbb{R}^{2}$
as $i\to+\infty$ ), it suffices to show that, for any relatively compact open
set $\Omega\subset\mathbb{R}^{2}$, the function

$J_{\Omega}(\tau):=\int_{\Omega}\big{|}e^{-\tau
H^{\Omega}_{\alpha,B_{0}}}f\big{|}^{2}e^{\xi(\tau,x)}\,dx$

is non-increasing in $\tau\in[0,s)$, where $H^{\Omega}_{\alpha,B_{0}}$ is the
Dirichelet Laplace operator in $\Omega$

$H^{\Omega}_{\alpha,B_{0}}=H_{\alpha,B_{0}}\big{|}_{W^{2}_{0}(\Omega)\cap
D(H_{\alpha,B_{0}})}.$

To this end, we need to prove the derivative on $J_{\Omega}(\tau)$ w.r.t.
$\tau$ is non-positive.

By using [20, Theorem 4.9], the function
$u(t,\cdot)=e^{-tH^{\Omega}_{\alpha,B_{0}}}f$ is strongly differentiable in
$L^{2}(\Omega)$ and its strong derivative $\frac{du}{dt}$ in $L^{2}(\Omega)$
is given by

$\frac{du}{dt}=-H^{\Omega}_{\alpha,B_{0}}u.$

Then we have

$\begin{split}\frac{dJ_{\Omega}(\tau)}{d\tau}&=\operatorname*{Re}\frac{d}{d\tau}\langle
u,ue^{\xi(\tau,x)}\rangle\\\
&=\operatorname*{Re}\langle\frac{du}{d\tau},ue^{\xi(\tau,x)}\rangle+\operatorname*{Re}\langle
u,\frac{d(ue^{\xi(\tau,x)})}{d\tau}\rangle\\\
&=2\operatorname*{Re}\langle\frac{du}{d\tau},ue^{\xi(\tau,x)}\rangle+\langle|u|^{2},\frac{d(e^{\xi(\tau,x)})}{d\tau}\rangle\\\
&=2\operatorname*{Re}\langle-H^{\Omega}_{\alpha,B_{0}}u,ue^{\xi(\tau,x)}\rangle+\langle|u|^{2},\frac{d(e^{\xi(\tau,x)})}{d\tau}\rangle.\end{split}$
(3.24)

Since $e^{\xi(\tau,\cdot)}\in\mathrm{Lip}_{\mathrm{loc}}(\mathbb{R}^{2})$, one
has $e^{\xi(\tau,\cdot)}\in\mathrm{Lip}(\Omega)$. The solution $u(t,\cdot)\in
W^{1}_{0}(\Omega)$, hence $e^{\xi(\tau,\cdot)}u(t,\cdot)\in
W^{1}_{0}(\Omega)$. On the one hand, recall the operator

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

by using the Green formula, we obtain

$\displaystyle
2\operatorname*{Re}\langle-H^{\Omega}_{\alpha,B_{0}}u,ue^{\xi(\tau,x)}\rangle=2\langle(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))^{2}u,ue^{\xi(\tau,x)}\rangle$
(3.25)
$\displaystyle=-2\int_{\Omega}|(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))u|^{2}e^{\xi(\tau,x)}\,dx-2\operatorname*{Re}\int_{\Omega}\nabla
u\cdot\nabla\xi e^{\xi(\tau,x)}\bar{u}\,dx.$

On the other hand, we observe that

$\frac{d(e^{\xi(\tau,x)})}{dt}=e^{\xi(\tau,x)}\frac{\partial\xi}{\partial\tau},$

then

$\begin{split}\langle|u|^{2},\frac{d(e^{\xi(\tau,x)})}{dt}\rangle&=\int_{\Omega}|u|^{2}e^{\xi(\tau,x)}\frac{\partial\xi(\tau,x)}{\partial\tau}\,dx\\\
&\leq-\frac{1}{2}\int_{\Omega}|u|^{2}e^{\xi(\tau,x)}|\nabla\xi|^{2}\,dx.\end{split}$
(3.26)

This is because that we can verify that the function $\xi(\tau,x)$ satisfies

$\frac{\partial\xi}{\partial\tau}+\frac{1}{2}|\nabla\xi|^{2}=-\frac{d^{2}(x,A^{c}_{\rho})}{2(\tau-s)^{2}}+\frac{1}{2}\Bigg{(}\frac{d(x,A^{c}_{\rho})|\nabla
d(x,A^{c}_{\rho})|}{2(\tau-s)}\Bigg{)}^{2}\leq-\frac{3}{4}\frac{d^{2}(x,A^{c}_{\rho})}{2(\tau-s)^{2}}\leq
0,$

since $\|\nabla f\|_{L^{\infty}}\leq\|f\|_{\mathrm{Lip}}$ and the function
$x\mapsto d(x,E)$ is Lipschitz function with Lipschitz norm $1$, see [20,
Lemma 11.2 and Theorem 11.3]. Therefore, by collecting (3.24), (3.25) and
(3.26), we finally show

$\begin{split}\frac{dJ_{\Omega}(\tau)}{d\tau}&\leq-2\int_{\Omega}|(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))u|^{2}e^{\xi(\tau,x)}\,dx\\\
&\quad-2\int_{\Omega}\Big{(}\operatorname*{Re}\big{(}\nabla
u\cdot\nabla\xi\bar{u}\big{)}+\frac{1}{4}|u|^{2}|\nabla\xi|^{2}\Big{)}e^{\xi(\tau,x)}\,dx.\end{split}$

On the one hand, we notice that

$\operatorname*{Re}(\nabla u\cdot\nabla\xi\bar{u})=\frac{1}{2}\big{(}(\nabla
u\cdot\nabla\xi\bar{u})+(\nabla\bar{u}\cdot\nabla\xi
u)\big{)}=\frac{1}{2}\big{(}\nabla|u|^{2}\cdot\nabla\xi\big{)}=(\nabla|u|)\cdot\nabla\xi|u|.$

Therefore, we have

$\begin{split}\frac{dJ_{\Omega}(\tau)}{d\tau}&\leq-2\int_{\Omega}\big{(}|(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))u|^{2}-|\nabla|u||^{2}\big{)}e^{\xi(\tau,x)}\,dx\\\
&\quad-2\int_{\Omega}\Big{(}|\nabla|u||^{2}+(\nabla|u|)\cdot\nabla\xi|u|+\frac{1}{4}|u|^{2}|\nabla\xi|^{2}\Big{)}e^{\xi(\tau,x)}\,dx\\\
&=-2\int_{\Omega}\big{(}|(\nabla+i(A_{B}(x)+A_{\mathrm{hmf}}(x)))u|^{2}-|\nabla|u||^{2}\big{)}e^{\xi(\tau,x)}\,dx\\\
&\quad-2\int_{\Omega}\big{|}\nabla|u|+\frac{1}{2}|u|\nabla\xi\big{|}^{2}e^{\xi(\tau,x)}\,dx.\end{split}$

On the other hand, the diamagnetic inequality shows that

$\begin{split}|\nabla|u|(x)|=\Big{|}\operatorname*{Re}\big{(}\frac{\bar{u}(x)}{|u(x)|}\nabla
u(x)\big{)}\Big{|}&=\Big{|}\operatorname*{Re}\Big{(}\big{(}\nabla
u(x)+i(A_{B}+A_{\mathrm{hmf}})u(x)\big{)}\frac{\bar{u}(x)}{|u(x)|}\Big{)}\Big{|}\\\
&\leq\Big{|}\big{(}\nabla+i(A_{B}+A_{\mathrm{hmf}})\big{)}u(x)\Big{|}.\end{split}$

Therefore, we finally prove that

$\begin{split}\frac{dJ_{\Omega}(\tau)}{d\tau}&\leq 0\end{split}$

Consequently, (3.23) follows, which implies Lemma 3.8.

∎

###  3.2. Method II

In this subsection, we will use the Schulman-Sunada formula (which is the
second method we used in [36] to construct the Schrödinger propagator) to
reconstruct the heat propagator. For more about the Schulman-Sunada formula,
we refer the reads to [34, 35]. The representation and estimate of the heat
kernel capture more magnetic effects.

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 [34, 35]) enables us to compute
the heat propagator $e^{-tH_{\alpha,B_{0}}}$ on $M$ by using the heat
propagator $e^{-t\tilde{H}_{\alpha,B_{0}}}$ (see the operator
$\tilde{H}_{\alpha,B_{0}}$ in (3.31) below) on $\tilde{M}$. More precisely,
see [35, (1)], we have

$e^{-tH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})=\sum_{j\in\mathbb{Z}}e^{-t\tilde{H}_{\alpha,B_{0}}}(r_{1},\theta_{1}+2j\pi;r_{2},\theta_{2}),$
(3.27)

which is similar to the construction of wave propagator on $\mathbb{T}^{n}$,
see [32, (3.5.12)]. In the following subsections, we will use it to construct
the heat kernel.

######  Proposition 3.9.

Let $K_{H}(t;x,y)$ be the heat kernel in (3.2) of Proposition 3.1 . Then

$\begin{split}&e^{-tH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})\\\
&=\frac{B_{0}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(tB_{0})}}\\\
&\qquad\times\bigg{(}e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(tB_{0}+i(\theta_{1}-\theta_{2}))}e^{-\alpha
tB_{0}}e^{-i\alpha(\theta_{1}-\theta_{2})}\varphi(\theta_{1},\theta_{2})\\\
&\quad-\frac{\sin(\alpha\pi)}{\pi}\int_{-\infty}^{\infty}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(s)}\frac{e^{s\alpha}}{e^{i(\theta_{1}-\theta_{2})}e^{s+tB_{0}}+1}\,ds\bigg{)},\end{split}$
(3.28)

where

$\begin{split}\varphi(\theta_{1},\theta_{2})=\begin{cases}1,\qquad|\theta_{1}-\theta_{2}|\leq\pi\\\
e^{-2\pi\alpha i},\quad-2\pi<\theta_{1}-\theta_{2}\leq-\pi\\\ e^{2\pi\alpha
i},\quad\pi<\theta_{1}-\theta_{2}\leq 2\pi.\end{cases}\end{split}$ (3.29)

Furthermore, we obtain the estimate

$\begin{split}\Big{|}e^{-tH_{\alpha,B_{0}}}(x,y)\Big{|}\leq
C\frac{B_{0}e^{-\alpha
tB_{0}}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}|x-y|^{2}}{4\tanh(tB_{0})}}.\end{split}$
(3.30)

###### Proof.

Recall (3.2) and $\alpha_{k}=|k+\alpha|$, we have

$K_{H}(t;x,y)=\frac{B_{0}}{4\pi^{2}\sinh
tB_{0}}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})\cosh tB_{0}}{4\sinh
tB_{0}}}\sum_{k\in\mathbb{Z}}e^{ik(\theta_{1}-\theta_{2})}e^{-(k+\alpha)tB_{0}}I_{\alpha_{k}}\left(\frac{B_{0}r_{1}r_{2}}{2\sinh
tB_{0}}\right).$

The main obstacle is to take the summation in $k$. If $\alpha=0$, by using the
formula [5, Eq. 9. 6. 19]

$\sum_{k\in\mathbb{Z}}e^{kz}I_{|k|}(x)=e^{x\cosh(z)},$

we will see

$K_{H}(t;x,y)=\frac{B_{0}}{4\pi\sinh(B_{0}t)}\exp\left(-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(B_{0}t)}+\frac{B_{0}r_{1}r_{2}}{2\sinh(B_{0}t)}\cosh(B_{0}t+i(\theta_{1}-\theta_{2}))\right),\quad\text{if}\quad\alpha=0,$

which is exactly the same as the result (3.13) obtained from the Mehler
formula. Heuristically, if we can replace the above summation in $k$ by the
integration in $k$, then one can use the translation invariant of integration
to obtain further result. To this end, as did in [36, Section 4], we consider
the operator

$\tilde{H}_{\alpha,B_{0}}=-\partial_{r}^{2}-\frac{1}{r}\partial_{r}+\frac{1}{r^{2}}\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2},$
(3.31)

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

$\Big{(}-i\partial_{\theta}+\alpha+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\varphi(\theta)=\Big{(}\tilde{k}+\frac{B_{0}r^{2}}{2}\Big{)}^{2}\varphi(\theta).$
(3.32)

It worths to point out that $\tilde{k}\in\mathbb{R}$ is a real number while
$k\in\mathbb{Z}$. More importantly, we informally move the $\alpha$ in right
hand side of (3.32) to the $e^{i(\tilde{k}-\alpha)\theta}$, which will
simplify the eigenfunctions. Hence, similarly as [36] for the Schrödinger
kernel, we obtain the heat kernel of $e^{-t\tilde{H}_{\alpha,B_{0}}}$

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

where $x=(r_{1},\theta_{1})\in\tilde{M}$ and
$y=(r_{2},\theta_{2})\in\tilde{M}$. A key identity is from [34, (2.11)], we
state it in the following lemma.

######  Lemma 3.10.

For any $z\in\mathbb{C}$, one has

$\begin{split}\int_{\mathbb{R}}e^{z\tilde{k}}I_{|\tilde{k}|}(x)\,d\tilde{k}&=e^{x\cosh(z)}H(\pi-|\operatorname*{Im}z|)\\\
&+\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{z+s+i\pi}-\frac{1}{z+s-i\pi}\right)\,ds.\end{split}$
(3.33)

where

$H(x)=\begin{cases}1,\quad x>0,\\\ 0,\quad x\leq 0.\end{cases}$

is the Heaviside step function.

###### Proof.

111We would like to appreciate to Prof. P.
$\check{S}\check{t}$oví$\check{c}$ek for his helpful discussion.

To prove (3.33), we recall the representation of the modified Bessel function
of order $\nu$

$I_{\nu}(x)=\frac{1}{\pi}\int_{0}^{\pi}e^{x\cos
s}\cos(s\nu)\,ds-\frac{\sin(\nu\pi)}{\pi}\int_{0}^{\infty}e^{-x\cosh s-\nu
s}\,ds,\quad x>0.$ (3.34)

For fixed $x>0$, one has that

$I_{\nu}(x)\sim\frac{1}{\sqrt{2\pi\nu}}\Big{(}\frac{ex}{2\nu}\Big{)}^{\nu},\quad\nu\to+\infty$

decays very rapidly in the order $\nu$. Due to this fact, the LHS of (3.33) is
absolutely convergent, hence the dominated convergent theorem implies that the
LHS of (3.33) represents an entire function in $z$ (holomorphic everywhere on
$\mathbb{C}$). The RHS of (3.33) is also an entire function in $z$ as well but
it is less obvious. The RHS of (3.33) is surely holomorphic in $z$ everywhere
on $\mathbb{C}$ except the lines $\operatorname*{Im}z=\pm\pi$. On these lines,
there is no discontinuity for the RHS of (3.33). For example, we consider
$\operatorname*{Im}z=\pi$. In fact, if set

$\begin{split}F(z)=&e^{x\cosh(z)}H(\pi-|\operatorname*{Im}z|)\\\
&+\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{z+s+i\pi}-\frac{1}{z+s-i\pi}\right)\,ds,\end{split}$

then one can prove

$\lim_{\operatorname*{Im}z\to\pi-}F(z)=\lim_{\operatorname*{Im}z\to\pi+}F(z).$

Indeed, for $a\in\mathbb{R}$ and $\epsilon>0$, we need to prove

$\lim_{\epsilon\to 0}F(a+i\pi-i\epsilon)=\lim_{\epsilon\to
0}F(a+i\pi+i\epsilon),$

that is,

$\begin{split}&\lim_{\epsilon\to
0}\left(e^{x\cosh(a+i\pi-i\epsilon)}+\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{a+i\pi-i\epsilon+s+i\pi}-\frac{1}{a+i\pi-i\epsilon+s-i\pi}\right)\,ds\right)\\\
&=\lim_{\epsilon\to 0}\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{a+i\pi+i\epsilon+s+i\pi}-\frac{1}{a+i\pi+i\epsilon+s-i\pi}\right)\,ds.\end{split}$

By direct computation, we obtain

$\begin{split}&\lim_{\epsilon\to 0}\frac{1}{2\pi
i}\Big{[}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{a+i\epsilon+s+2i\pi}-\frac{1}{a-i\epsilon+s+2i\pi}\right)\,ds\\\
&\qquad+\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{a+s-i\epsilon}-\frac{1}{a+s+i\epsilon}\right)\,ds\Big{]}\\\
&=\lim_{\epsilon\to 0}\frac{1}{2\pi
i}\Big{[}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\frac{-2i\epsilon}{(a+s+2i\pi)^{2}+\epsilon^{2}}\,ds+\int_{-\infty}^{\infty}e^{-x\cosh(s)}\frac{2i\epsilon}{(a+s)^{2}+\epsilon^{2}}\,ds\Big{]}\\\
&=\lim_{\epsilon\to
0}\frac{1}{\pi}\Big{[}-\int_{-\infty}^{\infty}e^{-x\cosh(s)}\frac{\epsilon}{(a+s+2i\pi)^{2}+\epsilon^{2}}\,ds+\int_{-\infty}^{\infty}e^{-x\cosh(s)}\frac{\epsilon}{(a+s)^{2}+\epsilon^{2}}\,ds\Big{]}\\\
&=e^{-x\cosh(a)},\end{split}$

where we use the fact that the Poisson kernel is an approximation to the
identity, implying that, for any reasonable function $m(x)$

$m(x)=\lim_{\epsilon\to
0}\frac{1}{\pi}\int_{-\infty}^{\infty}m(y)\frac{\epsilon}{(x-y)^{2}+\epsilon^{2}}\,ds.$

Obviously, since $\cosh(a+i\pi)=-\cosh a$, we have

$\begin{split}&e^{x\cosh(a+i\pi)}=e^{-x\cosh(a)}\\\ =&\lim_{\epsilon\to
0}\frac{1}{2\pi
i}\Big{[}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{a+i\epsilon+s+2i\pi}-\frac{1}{a-i\epsilon+s+2i\pi}\right)\,ds\\\
&\qquad+\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{a+s-i\epsilon}-\frac{1}{a+s+i\epsilon}\right)\,ds\Big{]}.\end{split}$

Therefore the RHS of (3.33), $F(z)$, is an entire function in $z$ as well. As
a consequence, it suffices to verify the formula (3.33) for purely imaginary
value of $z$ only. Let $z=ib$ and recall (3.34), then

$\begin{split}\int_{\mathbb{R}}e^{ib\tilde{k}}I_{|\tilde{k}|}(x)\,d\tilde{k}&=\frac{1}{\pi}\int_{\mathbb{R}}e^{ib\tilde{k}}\int_{0}^{\pi}e^{x\cos
s}\cos(s|\tilde{k}|)\,ds\,d\tilde{k}\\\
&\qquad-\int_{\mathbb{R}}e^{ib\tilde{k}}\frac{\sin(|\tilde{k}|\pi)}{\pi}\int_{0}^{\infty}e^{-x\cosh
s-|\tilde{k}|s}\,ds\,d\tilde{k}.\end{split}$

The first term becomes that

$\begin{split}\frac{1}{\pi}\int_{\mathbb{R}}e^{ib\tilde{k}}\int_{0}^{\pi}e^{x\cos
s}\cos(s\tilde{k})\,ds\,&d\tilde{k}=\frac{1}{2\pi}\int_{\mathbb{R}}e^{ib\tilde{k}}\int_{\mathbb{R}}H(\pi-|s|)e^{x\cos
s}e^{-is\tilde{k}}\,ds\,d\tilde{k}\\\ &=e^{x\cos
b}H(\pi-|b|)=e^{x\cosh(z)}H(\pi-|\operatorname*{Im}z|).\end{split}$

The second term gives

$\begin{split}&-\int_{\mathbb{R}}e^{ib\tilde{k}}\frac{\sin(|\tilde{k}|\pi)}{\pi}\int_{0}^{\infty}e^{-x\cosh
s-|\tilde{k}|s}\,ds\,d\tilde{k}\\\ &=-\frac{1}{2\pi
i}\int_{0}^{\infty}e^{-x\cosh
s}\Big{(}\int_{0}^{\infty}\Big{(}e^{[i(b+\pi)-s]\tilde{k}}-e^{[i(b-\pi)-s]\tilde{k}}\Big{)}\,d\tilde{k}\\\
&\qquad+\int_{-\infty}^{0}\Big{(}e^{[i(b-\pi)+s]\tilde{k}}-e^{[i(b+\pi)+s]\tilde{k}}\Big{)}\,d\tilde{k}\Big{)}\,ds\\\
&=\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-x\cosh(s)}\left(\frac{1}{ib+s+i\pi}-\frac{1}{ib+s-i\pi}\right)\,ds.\end{split}$

Therefore, we have proved (3.33). ∎

Let $z=tB_{0}+i(\theta_{1}-\theta_{2})$, by using Lemma 3.10, we obtain

$\begin{split}&\tilde{K}_{H}(t;x,y)=\frac{B_{0}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(tB_{0})}}e^{-\alpha
tB_{0}}e^{-i\alpha(\theta_{1}-\theta_{2})}\\\
&\qquad\qquad\times\bigg{(}e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(z)}H\big{(}\pi-|\theta_{1}-\theta_{2}|\big{)}\\\
&\quad+\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(s)}\left(\frac{1}{z+s+i\pi}-\frac{1}{z+s-i\pi}\right)\,ds\bigg{)}.\end{split}$

By using (3.27) and letting $z_{j}=tB_{0}+i(\theta_{1}+2\pi j-\theta_{2})$, we
further show

$\begin{split}&e^{-tH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})=\sum_{j\in\mathbb{Z}}e^{-t\tilde{H}_{\alpha,B_{0}}}(r_{1},\theta_{1}+2j\pi;r_{2},\theta_{2}),\\\
&=\frac{B_{0}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(tB_{0})}}\sum_{j\in\mathbb{Z}}e^{-\alpha
z_{j}}\\\
&\times\bigg{(}e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(z_{j})}H\big{(}\pi-|\theta_{1}+2\pi
j-\theta_{2}|\big{)}\\\ &\quad+\frac{1}{2\pi
i}\int_{-\infty}^{\infty}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(s)}\left(\frac{1}{z_{j}+s+i\pi}-\frac{1}{z_{j}+s-i\pi}\right)\,ds\bigg{)}.\end{split}$
(3.35)

In the following, we consider the two summations. Since
$\theta_{1},\theta_{2}\in[0,2\pi)$, then
$\theta_{1}-\theta_{2}\in(-2\pi,2\pi)$, recall (3.29), hence we obtain

$\begin{split}&\sum_{j\in\mathbb{Z}}e^{-\alpha
z_{j}}e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(z_{j})}H\big{(}\pi-|\theta_{1}+2\pi
j-\theta_{2}|\big{)}\\\
&=e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(z)}e^{-\alpha
tB_{0}}\sum_{j\in\mathbb{Z}}e^{-i\alpha(\theta_{1}-\theta_{2}+2\pi
j)}H\big{(}\pi-|\theta_{1}+2\pi j-\theta_{2}|\big{)}\\\
&=e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(z)}e^{-\alpha
tB_{0}}e^{-i\alpha(\theta_{1}-\theta_{2})}\varphi(\theta_{1},\theta_{2}),\end{split}$

For the second summation term, 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

$\begin{split}&\sum_{j\in\mathbb{Z}}e^{-\alpha
z_{j}}\left(\frac{1}{z_{j}+s+i\pi}-\frac{1}{z_{j}+s-i\pi}\right)\\\
&=e^{-\alpha(tB_{0}+i(\theta_{1}-\theta_{2}))}\sum_{j\in\mathbb{Z}}\big{(}\frac{e^{-2\pi
i\alpha j}}{i(\theta_{1}-\theta_{2}+2j\pi+\pi)+s+tB_{0}}-\frac{e^{-2\pi
i\alpha j}}{i(\theta_{1}-\theta_{2}+2j\pi-\pi)+s+tB_{0}}\big{)}\\\
&=-ie^{-\alpha(tB_{0}+i(\theta_{1}-\theta_{2}))}\sum_{j\in\mathbb{Z}}\big{(}\frac{e^{-2\pi
i\alpha j}}{\sigma_{1}+2j\pi}-\frac{e^{-2\pi i\alpha
j}}{\sigma_{2}+2j\pi}\big{)}\\\
&=e^{-\alpha(tB_{0}+i(\theta_{1}-\theta_{2}))}\big{(}\frac{e^{i\alpha\sigma_{1}}}{e^{i\sigma_{1}}-1}-\frac{e^{i\alpha\sigma_{2}}}{e^{i\sigma_{2}}-1}\big{)}=\frac{e^{s\alpha}}{e^{i(\theta_{1}-\theta_{2}+\pi)}e^{s+tB_{0}}-1}\big{(}e^{i\alpha\pi}-e^{-i\alpha\pi}\big{)}\\\
&=-2i\sin(\alpha\pi)\frac{e^{s\alpha}}{e^{i(\theta_{1}-\theta_{2})}e^{s+tB_{0}}+1},\end{split}$

where $\sigma_{1}=(\theta_{1}-\theta_{2}+\pi)-i(s+tB_{0})$ and
$\sigma_{2}=(\theta_{1}-\theta_{2}-\pi)-i(s+tB_{0})$. Therefore, by using
(3.35), we show (3.28)

$\begin{split}&e^{-tH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})\\\
&=\frac{B_{0}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(tB_{0})}}\\\
&\qquad\times\bigg{(}e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(tB_{0}+i(\theta_{1}-\theta_{2}))}e^{-\alpha
tB_{0}}e^{-i\alpha(\theta_{1}-\theta_{2})}\varphi(\theta_{1},\theta_{2})\\\
&\quad-\frac{\sin(\alpha\pi)}{\pi}\int_{-\infty}^{\infty}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(s)}\frac{e^{s\alpha}}{e^{i(\theta_{1}-\theta_{2})}e^{s+tB_{0}}+1}\,ds\bigg{)}.\end{split}$
(3.36)

To prove (3.30), we first note that

$\begin{split}\cosh(tB_{0}+i\theta)&=\cos\theta\cosh(tB_{0})+i\sin\theta\sinh(tB_{0}),\\\
|x-y|^{2}&=r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2}),\end{split}$

the first term is controlled by

$\begin{split}&\frac{B_{0}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(tB_{0})}}e^{\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(tB_{0})\cos(\theta_{1}-\theta_{2}))}e^{-\alpha
tB_{0}}\\\ &\leq C\frac{B_{0}e^{-\alpha
tB_{0}}}{4\pi\sinh(tB_{0})}e^{-\frac{B_{0}|x-y|^{2}}{4\tanh(tB_{0})}}.\end{split}$
(3.37)

For the second term, we aim to prove

$\displaystyle\Big{|}\int_{\mathbb{R}}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh
s}\frac{e^{\alpha s}}{1+e^{s+i(\theta_{1}-\theta_{2})+tB_{0}}}\,ds\Big{|}\leq
C,$ (3.38)

where $C$ is a constant independent of $t$, $r_{1},r_{2}$ and
$\theta_{1},\theta_{2}$. To this end, let $\theta=\theta_{1}-\theta_{2}$, we
write

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

then we just need to verify that

$\displaystyle\int_{0}^{\infty}\Big{|}\frac{e^{-\alpha
s}}{1+e^{-s+i\theta}}+\frac{e^{\alpha s}}{1+e^{s+i\theta}}\Big{|}\,ds\lesssim
1,$ (3.39)

and

$\displaystyle\Big{|}\int_{0}^{\infty}\Big{(}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(s-tB_{0})}-e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(-s-tB_{0})}\Big{)}\frac{e^{\alpha
s}}{1+e^{s+i\theta}}\,ds\Big{|}\lesssim 1,$ (3.40)

where the implicit constant is independent of $\theta$. We first prove (3.39).
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 $\cosh x-1\sim\frac{x^{2}}{2},\sinh x\sim x$, 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

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

We next prove (3.40). For convenience, we denote

$f(s)=e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}.\cosh(s-tB_{0})}$

Then, by noting that $|f(\pm s)|\leq 1,$ we obtain

$\displaystyle\Big{|}\int_{1}^{\infty}\big{(}f(s)-f(-s)\big{)}\frac{e^{\alpha
s}}{1+e^{s+i\theta}}\,ds\Big{|}$
$\displaystyle\leq\int_{1}^{\infty}\Big{|}\frac{e^{\alpha
s}}{1+e^{s+i\theta}}\Big{|}ds$
$\displaystyle=\int_{1}^{\infty}\Big{|}\frac{e^{(\alpha-1)s}}{e^{-s}+e^{i\theta}}\Big{|}ds$
$\displaystyle\leq\frac{1}{1-e^{-1}}\int_{1}^{\infty}e^{(\alpha-1)s}ds$
$\displaystyle\lesssim 1,$

hence, for (3.40), it suffices to prove

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

Since

$\displaystyle f^{\prime}(\pm s)$
$\displaystyle=\mp\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\sinh(\pm
s-tB_{0})f(\pm s)$
$\displaystyle=\mp\frac{B_{0}r_{1}r_{2}}{2}\Big{(}\frac{\pm\sinh
s\cosh(tB_{0})}{\sinh(tB_{0})}-\cosh
s\Big{)}e^{-\frac{B_{0}r_{1}r_{2}}{2\sinh(tB_{0})}\cosh(\pm s-tB_{0})},$

then for $0<s<1,$ there exists a constant $C$ such that $|f^{\prime}(s)|\leq
C$. Hence, by differential mean value theorem, $|f(s)-f(-s)|\leq Cs$ for
$0<s<1$, thus we have

$\displaystyle\int_{0}^{1}|f(s)-f(-s)|\Big{|}\frac{e^{\alpha
s}}{1+e^{s+i\theta}}\Big{|}\,ds\leq C\int_{0}^{1}\frac{se^{\alpha
s}}{e^{s}-1}ds\lesssim 1.$

Therefore, we prove (3.38). By collecting (3.36), (3.37) and (3.38), we
finally obtain

$\begin{split}&\Big{|}e^{-tH_{\alpha,B_{0}}}(r_{1},\theta_{1};r_{2},\theta_{2})\Big{|}\\\
&\leq C\frac{B_{0}e^{-\alpha
tB_{0}}}{4\pi\sinh(tB_{0})}\left(e^{-\frac{B_{0}|x-y|^{2}}{4\tanh(tB_{0})}}+e^{-\frac{B_{0}(r_{1}^{2}+r_{2}^{2})}{4\tanh(tB_{0})}}\right).\end{split}$

which implies (3.30).

∎

##  4\. Bernstein inequalities and square function inequalities

In this section, we prove the Bernstein inequalities and the square function
inequality associated with the Schrödinger operator $H_{\alpha,B_{0}}$ by
using the heat kernel estimates showed in the previous section.

######  Proposition 4.1 (Bernstein inequalities).

Let $\varphi(\lambda)$ be a $C^{\infty}_{c}$ bump function on $\mathbb{R}$
with support in $[\frac{1}{2},2]$, then it holds for any $f\in
L^{q}(\mathbb{R}^{2})$ and $j\in\mathbb{Z}$

$\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{p}(\mathbb{R}^{2})}\lesssim
2^{2j\big{(}\frac{1}{q}-\frac{1}{p}\big{)}}\|f\|_{L^{q}(\mathbb{R}^{2})},\quad
1\leq q\leq p\leq\infty.$ (4.1)

###### Proof.

Let $\psi(x)=\varphi(\sqrt{x})$ and $\psi_{e}(x):=\psi(x)e^{2x}$. Then
$\psi_{e}$ is a $C^{\infty}_{c}$-function on $\mathbb{R}$ with support in
$[\frac{1}{4},4]$ and then its Fourier transform $\hat{\psi}_{e}$ belongs to
Schwartz class. We write

$\displaystyle\varphi(\sqrt{x})$ $\displaystyle=\psi(x)=e^{-2x}\psi_{e}(x)$
$\displaystyle=e^{-2x}\int_{\mathbb{R}}e^{ix\cdot\xi}\hat{\psi}_{e}(\xi)\,d\xi$
$\displaystyle=e^{-x}\int_{\mathbb{R}}e^{-x(1-i\xi)}\hat{\psi}_{e}(\xi)\,d\xi.$

Therefore, by the functional calculus, we obtain

$\displaystyle\varphi(\sqrt{H_{\alpha,B_{0}}})=\psi(H_{\alpha,B_{0}})=e^{-H_{\alpha,B_{0}}}\int_{\mathbb{R}}e^{-(1-i\xi)H_{\alpha,B_{0}}}\hat{\psi}_{e}(\xi)\,d\xi,$

furthermore,

$\displaystyle\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})=\psi(2^{-2j}H_{\alpha,B_{0}})=e^{-2^{-2j}H_{\alpha,B_{0}}}\int_{\mathbb{R}}e^{-(1-i\xi)2^{-2j}H_{\alpha,B_{0}}}\hat{\psi}_{e}(\xi)\,d\xi.$

By using (3.8) in Proposition 3.3 with $t=2^{-2j}$, we have

$\displaystyle\Big{|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})(x,y)\Big{|}$
$\displaystyle\lesssim
2^{4j}\int_{\mathbb{R}^{2}}e^{-\frac{2^{2j}|x-z|^{2}}{C}}e^{-\frac{2^{2j}|y-z|^{2}}{C}}\,dz\int_{\mathbb{R}}\hat{\psi}_{e}(\xi)\,d\xi$
$\displaystyle\lesssim
2^{2j}\int_{\mathbb{R}^{2}}e^{-\frac{|2^{j}x-z|^{2}}{C}}e^{-\frac{|2^{j}y-z|^{2}}{C}}\,dz$
$\displaystyle\lesssim 2^{2j}(1+2^{j}|x-y|)^{-N},\quad\forall N\geq 1.$

where we use the fact that
$|\alpha-z|^{2}+|\beta-z|^{2}\geq\frac{1}{2}|\alpha-\beta|^{2}$ with
$\alpha,\beta\in\mathbb{R}^{2}$ and

$\displaystyle\int_{\mathbb{R}^{2}}e^{-\frac{|\alpha-z|^{2}}{C}}e^{-\frac{|\beta-z|^{2}}{C}}\,dz$
$\displaystyle\lesssim
e^{-\frac{|\alpha-\beta|^{2}}{4C}}\int_{\mathbb{R}^{2}}e^{-\frac{|\alpha-z|^{2}}{2C}}e^{-\frac{|\beta-z|^{2}}{2C}}\,dz$
$\displaystyle\lesssim
e^{-\frac{|\alpha-\beta|^{2}}{4C}}\leq(1+|\alpha-\beta|)^{-N},\forall N\geq
1.$

By Young’s inequality, it follows (4.1). ∎

######  Proposition 4.2 (The square function inequality).

Let $\\{\varphi_{j}\\}_{j\in\mathbb{Z}}$ be a Littlewood-Paley sequence given
by(1.6). Then for $1<p<\infty$, there exist constants $c_{p}$ and $C_{p}$
depending on $p$ such that

$c_{p}\|f\|_{L^{p}(\mathbb{R}^{2})}\leq\Big{\|}\Big{(}\sum_{j\in\mathbb{Z}}|\varphi_{j}(\sqrt{H_{{\alpha},B_{0}}})f|^{2}\Big{)}^{\frac{1}{2}}\Big{\|}_{L^{p}(\mathbb{R}^{2})}\leq
C_{p}\|f\|_{L^{p}(\mathbb{R}^{2})}.$ (4.2)

###### Proof.

By using (3.8) in Proposition 3.3, Proposition 4.2 follows from the Rademacher
functions argument in [33]. We also refer the reader to [1] for result that
the square function inequality (4.2) can be derived from the heat kernel with
Gaussian upper bounds. ∎

##  5\. The decay estimates

In this section, we mainly prove the decay estimate (1.12). The first key
ingredient is the following Proposition about the subordination formula from
[27, 12].

######  Proposition 5.1.

If $\varphi(\lambda)\in C_{c}^{\infty}(\mathbb{R})$ is supported in
$[\frac{1}{2},2]$, then, for all $j\in\mathbb{Z},t,x>0$ with $2^{j}t\geq 1$,
we can write

$\begin{split}&\varphi(2^{-j}\sqrt{x})e^{it\sqrt{x}}\\\
&=\rho\big{(}\frac{tx}{2^{j}},2^{j}t\big{)}+\varphi(2^{-j}\sqrt{x})\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}tsx}\,ds,\end{split}$
(5.1)

where $\rho(s,\tau)\in\mathcal{S}(\mathbb{R}\times\mathbb{R})$ is a Schwartz
function and and $\chi\in C^{\infty}(\mathbb{R}\times\mathbb{R})$ with
$\text{supp}\,\chi(\cdot,\tau)\subseteq[\frac{1}{16},4]$ such that

$\sup_{\tau\in\mathbb{R}}\big{|}\partial_{s}^{\alpha}\partial_{\tau}^{\beta}\chi(s,\tau)\big{|}\lesssim_{\alpha,\beta}(1+|s|)^{-\alpha},\quad\forall\alpha,\beta\geq
0.$ (5.2)

If this has been done, then by the spectral theory for the non-negative self-
adjoint operator $H_{\alpha,B_{0}}$, we can have the representation of the
micolocalized half-wave propagator

$\begin{split}&\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}\\\
&=\rho\big{(}\frac{tH_{\alpha,B_{0}}}{2^{j}},2^{j}t\big{)}+\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}tsH_{\alpha,B_{0}}}\,ds.\end{split}$
(5.3)

###### Proof.

This result is original from [27] and the authors of [12] provided an
independent proof of the key formula. For the self-contained and the
convenience of the reader, we follow the idea of [12] to provide the details
of the proof.

The starting point of the proof is the subordination formula

$e^{-y\sqrt{x}}=\frac{y}{2\sqrt{\pi}}\int_{0}^{\infty}e^{-sx-\frac{y^{2}}{4s}}s^{-\frac{3}{2}}ds,\quad
x,y>0.$ (5.4)

Indeed, let $|\xi|=\sqrt{x}$, we have

$\begin{split}e^{-y\sqrt{x}}&=e^{-y|\xi|}=\int_{\mathbb{R}}e^{-2\pi
it\cdot\xi}\int_{\mathbb{R}}e^{2\pi it\cdot\eta}e^{-y|\eta|}\,d\eta\,dt\\\
&=\int_{\mathbb{R}}e^{-2\pi it\cdot\xi}\Big{(}\int_{-\infty}^{0}e^{2\pi
it\cdot\eta}e^{y\eta}\,d\eta+\int_{0}^{\infty}e^{2\pi
it\cdot\eta}e^{-y\eta}\,d\eta\Big{)}\,dt\\\ &=2\int_{\mathbb{R}}e^{-2\pi
it\cdot\xi}\frac{y}{y^{2}+(2\pi t)^{2}}\,dt=2\int_{\mathbb{R}}e^{-2\pi
iyt\cdot\xi}\frac{1}{1+(2\pi t)^{2}}\,dt\\\ &=2\int_{\mathbb{R}}e^{-2\pi
iyt\cdot\xi}\int_{0}^{\infty}e^{-r(1+(2\pi t)^{2})}\,dr\,dt\\\
&=2\int_{0}^{\infty}e^{-r}\int_{\mathbb{R}}e^{-2\pi iyt\cdot\xi}e^{-r(2\pi
t)^{2}}\,dt\,dr\\\
&=2\int_{0}^{\infty}\frac{e^{-r}e^{-\frac{\xi^{2}y^{2}}{4r}}}{\sqrt{4\pi
r}}\,dr=2\int_{0}^{\infty}\frac{e^{-r}e^{-\frac{xy^{2}}{4r}}}{\sqrt{4\pi
r}}\,dr,\quad s=\frac{y^{2}}{4r},\xi=\sqrt{x},\\\
&=\frac{y}{2\sqrt{\pi}}\int_{0}^{\infty}e^{-\frac{y^{2}}{4s}}e^{-xs}s^{-\frac{3}{2}}\,ds\\\
\end{split}$

where we use

$\alpha^{-\frac{n}{2}}e^{-\pi|\xi|^{2}/\alpha}=\int_{\mathbb{R}^{n}}e^{-2\pi
ix\xi}e^{-\pi\alpha|x|^{2}}dx.$

To obtain $e^{it\sqrt{x}}$, we extend (5.4) by setting $y=\epsilon-it$ with
$\epsilon>0$

$\begin{split}e^{it\sqrt{x}}&=\lim_{\epsilon\to 0^{+}}e^{-(\epsilon-
it)\sqrt{x}}\\\ &=\lim_{\epsilon\to 0}\frac{\epsilon-
it}{2\sqrt{\pi}}\int_{0}^{\infty}e^{-xs}e^{\frac{(it-\epsilon)^{2}}{4s}}s^{-\frac{3}{2}}ds,\quad
s=r(\epsilon-it)\\\ &=\lim_{\epsilon\to 0}\frac{\sqrt{\epsilon-
it}}{2\sqrt{\pi}}\int_{0}^{\infty}e^{rx(it-\epsilon)}e^{\frac{it-\epsilon}{4r}}r^{-\frac{3}{2}}dr\\\
&=\lim_{\epsilon\to 0}\frac{\sqrt{\epsilon-
it}}{2\sqrt{\pi}}\int_{0}^{\infty}e^{itrx}e^{-\epsilon
rx}e^{\frac{it}{4r}}e^{-\frac{\epsilon}{4r}}r^{-\frac{3}{2}}dr\\\
&=\lim_{\epsilon\to 0}\frac{\sqrt{\epsilon-
it}}{2\sqrt{\pi}}I_{\epsilon,\epsilon x}(tx,t),\end{split}$ (5.5)

where

$I_{\epsilon,\delta}(a,t):=\int_{0}^{\infty}e^{ira}e^{-\delta
r}e^{\frac{it}{4r}}e^{-\frac{\epsilon}{4r}}r^{-\frac{3}{2}}dr.$

By the dominate convergence theorem, we have that

$\begin{split}e^{it\sqrt{x}}&=\lim_{\epsilon\to 0}\frac{\sqrt{\epsilon-
it}}{2\sqrt{\pi}}I_{\epsilon,\epsilon
x}(tx,t)=\sqrt{\frac{t}{4\pi}}e^{-\frac{\pi}{4}i}\lim_{\epsilon\to
0}I_{\epsilon,\epsilon x}(tx,t).\end{split}$

Thus it suffices to consider the oscillation integral

$\begin{split}\lim_{\epsilon\to 0}I_{\epsilon,\epsilon
x}(a,t)=I_{0,0}(a,t)=\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}r^{-\frac{3}{2}}dr.\end{split}$
(5.6)

######  Lemma 5.2.

Let

$\begin{split}I(a,t)=\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}r^{-\frac{3}{2}}dr.\end{split}$

Then we can write

$I(a,t)=\tilde{\rho}(a,t)+\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}\tilde{\chi}(r)\,dr,$
(5.7)

where $\tilde{\chi}(r)\in C_{0}^{\infty}(r)$ and
$\mathrm{supp}\,\tilde{\chi}\subset[\frac{1}{16},4]$ and $\tilde{\rho}(a,t)$
satisfies

$\big{|}\partial_{a}^{\alpha}\partial_{t}^{\beta}\tilde{\rho}(a,t)\big{|}\leq
C_{N,\alpha,\beta}(a+t)^{-N},\quad\frac{1}{4}\leq\frac{a}{t}\leq 4,\,t\geq
1,\forall N\geq 1.$ (5.8)

We now assume this lemma to prove (5.1). By (5.5) and (5.6) and noticing

$I(a,t)=2^{\frac{j}{2}}I(2^{-j}a,2^{j}t),$

we have that

$\varphi(2^{-j}\sqrt{x})e^{it\sqrt{x}}=\sqrt{\frac{t}{4\pi}}e^{-\frac{\pi}{4}i}\varphi(2^{-j}\sqrt{x})2^{\frac{j}{2}}I\big{(}\frac{tx}{2^{j}},2^{j}t\big{)}.$

By the support of $\varphi$, one has $2^{2j-2}\leq x\leq 2^{2j+2}$ , hence
$\frac{1}{4}\leq\frac{tx}{2^{j}}/2^{j}t=x/2^{2j}\leq 4$. Note the condition
$2^{j}t\geq 1$. Therefore, by using this lemma, we prove

$\begin{split}&\varphi(2^{-j}\sqrt{x})e^{it\sqrt{x}}\\\
=&\frac{1}{\sqrt{4\pi}}e^{-\frac{\pi}{4}i}\big{(}2^{j}t\big{)}^{\frac{1}{2}}\varphi(2^{-j}\sqrt{x})\Big{(}\tilde{\rho}\big{(}\frac{tx}{2^{j}},2^{j}t\big{)}+\int_{0}^{\infty}\tilde{\chi}(s)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}txs}\,ds\Big{)}.\end{split}$

We need consider this expression when $2^{j}t\geq 1$. To this end, let
$\phi\in C^{\infty}([0,+\infty)$ satisfies $\phi(t)=1$ if $t\geq 1$ and
$\phi(t)=0$ if $0\leq t\leq\frac{1}{2}$, then set

$\rho\big{(}\frac{tx}{2^{j}},2^{j}t\big{)}=e^{-\frac{\pi}{4}i}\big{(}2^{j}t\big{)}^{\frac{1}{2}}\varphi(2^{-j}\sqrt{x})\tilde{\rho}\big{(}\frac{tx}{2^{j}},2^{j}t\big{)}\phi(2^{j}t).$

This together with (5.8) shows

$\big{|}\partial_{a}^{\alpha}\partial_{t}^{\beta}{\rho}(a,t)\big{|}\leq
C_{N,\alpha,\beta}(1+(a+t))^{-N},\quad\forall N\geq 0.$

which implies $\rho(a,t)\in\mathcal{S}(\mathbb{R}_{+}\times\mathbb{R}_{+})$.
Set

$\chi\big{(}s,2^{j}t\big{)}=e^{-\frac{\pi}{4}i}\tilde{\chi}\big{(}s\big{)}\phi(2^{j}t),$

then $\chi$ satisfies (5.2). then we finally write

$\begin{split}&\varphi(2^{-j}\sqrt{x})e^{it\sqrt{x}}\\\
=&\rho\big{(}\frac{tx}{2^{j}},2^{j}t\big{)}+\big{(}2^{j}t\big{)}^{\frac{1}{2}}\varphi(2^{-j}\sqrt{x})\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}txs}\,ds,\end{split}$

which proves (5.1) as desired. ∎

###### The proof of Lemma 5.2.

To prove (5.7), we divide the integral into three pieces. Let $\beta(r)$ be a
function in $C^{\infty}(\mathbb{R})$ compact supported in $[\frac{1}{2},2]$
such that

$1=\sum_{j\in\mathbb{Z}}\beta_{j}(r),\quad\beta_{j}(r)=\beta(2^{-j}r).$

Corresponding to $\beta_{j}$, we decompose

$I(a,t)=\sum_{j\in\mathbb{Z}}I_{j}(a,t)=I_{l}(a,t)+I_{m}(a,t)+I_{h}(a,t)$

where

$\begin{split}I_{l}(a,t)=\sum_{j\leq-5}I_{j}(a,t),\quad&I_{m}(a,t)=\sum_{-4\leq
j\leq 1}I_{j}(a,t),\quad I_{h}(a,t)=\sum_{j\geq 2}I_{j}(a,t),\\\
&I_{j}(a,t)=\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}\beta_{j}(r)r^{-\frac{3}{2}}dr.\end{split}$

Define the phase function $\phi_{a,t}(r)=ra+\frac{t}{4r}$, then

$\phi^{\prime}_{a,t}(r)=a-\frac{t}{4r^{2}}.$

Define

$\tilde{\rho}(a,t)=I_{l}(a,t)+I_{h}(a,t),$

we aim to prove (5.8). We first consider $I_{h}(a,t)$.

$\begin{split}\big{|}\partial_{a}^{\alpha}\partial_{t}^{\beta}I_{h}(a,t)\big{|}&=\sum_{j\geq
2}\Big{|}\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}\beta_{j}(r)r^{-\frac{3}{2}+\alpha}\big{(}\frac{1}{4r}\big{)}^{\beta}dr\Big{|}\\\
&\leq C\sum_{j\geq
2}\int_{0}^{\infty}\Big{|}\frac{d}{dr}\Big{[}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\frac{d}{dr}\Big{)}^{N-1}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\beta_{j}(r)r^{-\frac{3}{2}+\alpha-\beta}\Big{)}\Big{]}\Big{|}\,dr.\end{split}$

Due to that $\mathrm{supp}\,\beta_{j}\subset[2^{j-1},2^{j+1}]$ with $j\geq 2$
implies $r\in[2,\infty)$ and the assumption $\frac{a}{t}\geq\frac{1}{4}$, we
see that

$\big{|}\phi^{\prime}_{a,t}(r)\big{|}=\big{|}a-\frac{t}{4r^{2}}\big{|}\geq\frac{a+t}{16}.$

For choosing $N>\alpha$, we notice the fact that

$\Big{|}\Big{(}\frac{d}{dr}\Big{)}^{K}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\Big{)}\Big{|}\leq
C_{K}(a+t)^{-1}r^{-K},\quad t,r\in[1,+\infty)$

to obtain

$\Big{|}\frac{d}{dr}\Big{[}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\frac{d}{dr}\Big{)}^{N-1}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\beta_{j}(r)r^{-\frac{3}{2}+\alpha-\beta}\Big{)}\Big{]}\Big{|}\leq
C_{N}(a+t)^{-N}2^{-\frac{3}{2}j}.$

Therefore, we prove

$\big{|}\partial_{a}^{\alpha}\partial_{t}^{\beta}I_{h}(a,t)\big{|}\leq
C_{N}(a+t)^{-N}\sum_{j\geq 2}2^{-\frac{1}{2}j}\,\leq C_{N}(a+t)^{-N},\quad
N\geq\alpha.$

Next we consider $I_{l}(a,t)$.

$\big{|}\partial_{a}^{\alpha}\partial_{t}^{\beta}I_{l}(a,t)\big{|}=\sum_{j\leq-5}\Big{|}\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}\beta_{j}(r)r^{-\frac{3}{2}+\alpha}\big{(}\frac{1}{4r}\big{)}^{\beta}dr\Big{|}.$
$\begin{split}&\sum_{j\leq-5}\Big{|}\int_{0}^{\infty}e^{\frac{it}{4r}}e^{ira}{\beta}_{j}(r)r^{-\frac{3}{2}+\alpha-\beta}dr\Big{|}\\\
&\leq
C_{N}\sum_{j\leq-5}\int_{0}^{\infty}\Big{|}\frac{d}{dr}\Big{[}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\frac{d}{dr}\Big{)}^{N-1}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}{\beta}_{j}(r)r^{-\frac{3}{2}+\alpha-\beta}\Big{)}\Big{]}\Big{|}\,dr.\end{split}$

Due to that $\mathrm{supp}\,\beta_{j}\subset[2^{j-1},2^{j+1}]$ with $j\leq-5$
implies $r\in(0,\frac{1}{16}]$ and the assumption $\frac{a}{t}\leq 4$, we see
that

$\big{|}\phi^{\prime}_{a,t}(r)\big{|}=\big{|}a-\frac{t}{4r^{2}}\big{|}=\frac{|4r^{2}a-t|}{4r^{2}}\geq\frac{a+t}{32r^{2}}.$

For choosing $N>\alpha$, we notice the fact that

$\Big{|}\Big{(}\frac{d}{dr}\Big{)}^{K}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\Big{)}\Big{|}\leq
C_{K}(a+t)^{-1}r^{2-K},\quad t\in[1,+\infty),r\in(0,\frac{1}{16}]$

to obtain

$\begin{split}&\Big{|}\frac{d}{dr}\Big{[}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}\frac{d}{dr}\Big{)}^{N-1}\Big{(}\frac{1}{\phi^{\prime}_{a,t}(r)}{\beta}_{j}(r)r^{-\frac{3}{2}+\alpha-\beta}\Big{)}\Big{]}\Big{|}\\\
&\leq
C_{N}(a+t)^{-N}2^{j\big{(}-\frac{3}{2}+\alpha-\beta+N\big{)}}.\end{split}$

Therefore, for large enough $N$ such that $-\frac{1}{2}+\alpha-\beta+N>0$, we
prove

$\begin{split}\big{|}\partial_{a}^{\alpha}\partial_{t}^{\beta}I_{h}(a,t)\big{|}&\leq
C_{N}(a+t)^{-N}\sum_{j\leq-5}2^{j\big{(}-\frac{1}{2}+\alpha-\beta+N\big{)}}\\\
&\leq C_{N}(a+t)^{-N},\end{split}$

where we use the assumption that $\frac{a}{t}\leq 4$ and $t\geq 1$. In sum, we
prove (5.8). Let

$\tilde{\chi}(r)=\sum_{j=-4}^{1}\beta_{j}(r)r^{-\frac{3}{2}},$

then $\tilde{\chi}(r)\in C_{0}^{\infty}(r)$ and
$\mathrm{supp}\,\tilde{\chi}\subset[\frac{1}{16},4]$. Hence we have

$I_{m}(a,t)=\int_{0}^{\infty}e^{ira}e^{\frac{it}{4r}}\tilde{\chi}(r)dr.$

Therefore, we complete the proof of Lemma 5.2. ∎

###  5.1. Decay estimates for the microlocalized half-wave propagator

In this subsection, we mainly prove the following results

######  Proposition 5.3.

Let $2^{-j}|t|\leq\frac{\pi}{8B_{0}}$ and $\varphi$ be in (1.6), then

$\begin{split}\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})&e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim
2^{2j}\big{(}1+2^{j}|t|\big{)}^{-\frac{1}{2}}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})}.\end{split}$
(5.9)

In particular, for $0<t<T$ with any finite $T$, there exists a constant
$C_{T}$ depending on $T$ such that

$\begin{split}\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})&e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\leq
C_{T}2^{2j}\big{(}1+2^{j}|t|\big{)}^{-\frac{1}{2}}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})}.\end{split}$
(5.10)

######  Remark 5.4.

The finite $T$ can be choosen beyond $\frac{\pi}{B_{0}}$. If we could prove
(5.10), then (1.12) follows

$\begin{split}&\big{\|}e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\leq\sum_{j\in\mathbb{Z}}\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\leq
C_{T}|t|^{-\frac{1}{2}}\sum_{j\in\mathbb{Z}}2^{\frac{3}{2}j}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})}\leq
C_{T}|t|^{-\frac{1}{2}}\|f\|_{\dot{\mathcal{B}}^{3/2}_{1,1}(\mathbb{R}^{2})}.\end{split}$

We estimate the microlocalized half-wave propagator

$\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}$

by considering two cases that: $|t|2^{j}\geq 1$ and $|t|2^{j}\lesssim 1$. In
the following argument, we can choose $\tilde{\varphi}\in
C_{c}^{\infty}((0,+\infty))$ such that $\tilde{\varphi}(\lambda)=1$ if
$\lambda\in\mathrm{supp}\,\varphi$ and $\tilde{\varphi}\varphi=\varphi$. Since
$\tilde{\varphi}$ has the same property of $\varphi$, without confusion, we
drop off the tilde above $\varphi$ for brief. Without loss of generality, in
the following argument, we assume $t>0$.

Case 1: $t2^{j}\lesssim 1$. We remark that we consider $t2^{j}\lesssim 1$
while not $t2^{j}\leq 1$, this will be used to extend the time interval. By
the spectral theorem, one has

$\|e^{it\sqrt{H_{\alpha,B_{0}}}}\|_{L^{2}(\mathbb{R}^{2})\to
L^{2}(\mathbb{R}^{2})}\leq C.$

Indeed, by the functional calculus, for $f\in L^{2}$, we can write

$\begin{split}e^{it\sqrt{H_{\alpha,B_{0}}}}f&=\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}e^{it\sqrt{\lambda_{k,m}}}c_{k,m}\tilde{V}_{k,m}(x).\end{split}$

where

$\begin{split}c_{k,m}=\int_{\mathbb{R}^{2}}f(y)\overline{\tilde{V}_{k,m}(y)}dy.\end{split}$

Then

$\begin{split}\|e^{it\sqrt{H_{\alpha,B_{0}}}}f\|_{L^{2}(\mathbb{R}^{2})}=\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}\big{|}e^{it\sqrt{\lambda_{k,m}}}c_{k,m}\big{|}^{2}\Big{)}^{1/2}=\Big{(}\sum_{k\in\mathbb{Z},\atop
m\in\mathbb{N}}\big{|}c_{k,m}\big{|}^{2}\Big{)}^{1/2}=\|f\|_{L^{2}(\mathbb{R}^{2})}.\end{split}$

Together with this, we use the Bernstein inequality (4.1) to prove

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim
2^{j}\|e^{it\sqrt{H_{\alpha,B_{0}}}}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{2}(\mathbb{R}^{2})}\\\
&\lesssim
2^{j}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{2}(\mathbb{R}^{2})}\lesssim
2^{2j}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})}.\end{split}$

In this case $0<t\lesssim 2^{-j}$, we have

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\quad\lesssim
2^{2j}(1+2^{j}t)^{-N}\|\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\|_{L^{1}(\mathbb{R}^{2})},\quad\forall
N\geq 0.\end{split}$ (5.11)

Case 2: $t2^{j}\geq 1$. In this case, we can use (5.3) to obtain the
micolocalized half-wave propagator

$\begin{split}&\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}\\\
&=\rho\big{(}\frac{tH_{\alpha,B_{0}}}{2^{j}},2^{j}t\big{)}+\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}tsH_{\alpha,B_{0}}}\,ds.\end{split}$

We first use the spectral theorems and the Bernstein inequality again to
estimate

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\rho\big{(}\frac{tH_{\alpha,B_{0}}}{2^{j}},2^{j}t\big{)}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}.\end{split}$

Indeed, since $\rho\in\mathcal{S}(\mathbb{R}\times\mathbb{R})$, then

$\big{|}\rho\big{(}\frac{t\lambda_{k,m}}{2^{j}},2^{j}t\big{)}\big{|}\leq
C(1+2^{j}t)^{-N},\quad\forall N\geq 0.$

Therefore, we use the Bernstein inequality and the spectral theorems to show

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\rho\big{(}\frac{tH_{\alpha,B_{0}}}{2^{j}},2^{j}t\big{)}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim
2^{j}\Big{\|}\rho\big{(}\frac{tH_{\alpha,B_{0}}}{2^{j}},2^{j}t\big{)}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\Big{\|}_{L^{2}(\mathbb{R}^{2})}\\\
&\lesssim
2^{j}(1+2^{j}t)^{-N}\Big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\Big{\|}_{L^{2}(\mathbb{R}^{2})}\\\
&\lesssim
2^{2j}(1+2^{j}t)^{-N}\Big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\Big{\|}_{L^{1}(\mathbb{R}^{2})}.\end{split}$

Next we use the dispersive estimates of Schrödinger propagator (see [36,
Theorem 1.1])

$\begin{split}&\big{\|}e^{itH_{\alpha,B_{0}}}f\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\leq
C|\sin(tB_{0})|^{-1}\big{\|}f\big{\|}_{L^{1}\mathbb{R}^{2})},\quad
t\neq\frac{k\pi}{B_{0}},\,k\in\mathbb{Z},\end{split}$

to estimate

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}tsH_{\alpha,B_{0}}}f\,ds\big{\|}_{L^{\infty}(\mathbb{R}^{2})}.\end{split}$

For $0<t<T_{0}<\frac{\pi}{2B_{0}}$, then $\sin(tB_{0})\sim tB_{0}$

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}tsH_{\alpha,B_{0}}}f\,ds\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)|\sin(2^{-j}tsB_{0})|^{-1}\,ds\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\big{\|}_{L^{1}(\mathbb{R}^{2})}.\end{split}$

Since $s\in[\frac{1}{16},4]$ (the compact support of $\chi$ in $s$) and
$B_{0}>0$, if $2^{-j}t\leq\frac{\pi}{8B_{0}}$, then

$\begin{split}&\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})\big{(}2^{j}t\big{)}^{\frac{1}{2}}\int_{0}^{\infty}\chi(s,2^{j}t)e^{\frac{i2^{j}t}{4s}}e^{i2^{-j}tsH_{\alpha,B_{0}}}f\,ds\big{\|}_{L^{\infty}(\mathbb{R}^{2})}\\\
&\lesssim\big{(}2^{j}t\big{)}^{\frac{1}{2}}(2^{-j}t)^{-1}\int_{0}^{\infty}\chi(s,2^{j}t)\,ds\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\big{\|}_{L^{1}(\mathbb{R}^{2})}\\\
&\lesssim
2^{2j}\big{(}2^{j}t\big{)}^{-\frac{1}{2}}\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\big{\|}_{L^{1}(\mathbb{R}^{2})}\\\
&\lesssim
2^{2j}\big{(}1+2^{j}t\big{)}^{-\frac{1}{2}}\big{\|}\varphi(2^{-j}\sqrt{H_{\alpha,B_{0}}})f\big{\|}_{L^{1}(\mathbb{R}^{2})}.\end{split}$
(5.12)

Collecting (5.11) and (5.12), it gives (5.9). To prove (5.10), we consider
$0<t<T$. For any $T>0$, there exists $j_{0}$ such that
$2^{-j_{0}}T\leq\frac{\pi}{8B_{0}}$ with $j_{0}\in\mathbb{Z}_{+}$. For $j\leq
j_{0}$, then $2^{j}t\lesssim 1$, then one has (5.10) from the first case.
While for $j\geq j_{0}$, if $2^{j}t\lesssim 1$, one still has (5.10) from the
first case. Otherwise, i.e. $2^{j}t\geq 1$, one has (5.10) from the second
case, since we always have $2^{-j}t\leq\frac{\pi}{8B_{0}}$ for $j\geq j_{0}$
and $0<t\leq T$.

##  6\. Strichartz estimate

In this section, we prove the Strichartz estimates (1.13) in Theorem 1.4 by
using (5.10). To this end, we need a variety of the abstract Keel-Tao’s
Strichartz estimates theorem ([23]).

######  Proposition 6.1.

Let $(X,\mathcal{M},\mu)$ be a $\sigma$-finite measured space and
$U:I=[0,T]\rightarrow B(L^{2}(X,\mathcal{M},\mu))$ be a weakly measurable map
satisfying, for some constants $C$ may depending on $T$, $\alpha\geq 0$,
$\sigma,h>0$,

$\begin{split}\|U(t)\|_{L^{2}\rightarrow L^{2}}&\leq C,\quad
t\in\mathbb{R},\\\ \|U(t)U(s)^{*}f\|_{L^{\infty}}&\leq
Ch^{-\alpha}(h+|t-s|)^{-\sigma}\|f\|_{L^{1}}.\end{split}$ (6.1)

Then for every pair $q,p\in[1,\infty]$ such that
$(q,p,\sigma)\neq(2,\infty,1)$ and

$\frac{1}{q}+\frac{\sigma}{p}\leq\frac{\sigma}{2},\quad q\geq 2,$

there exists a constant $\tilde{C}$ only depending on $C$, $\sigma$, $q$ and
$r$ such that

$\Big{(}\int_{I}\|U(t)u_{0}\|_{L^{r}}^{q}dt\Big{)}^{\frac{1}{q}}\leq\tilde{C}\Lambda(h)\|u_{0}\|_{L^{2}}$

where $\Lambda(h)=h^{-(\alpha+\sigma)(\frac{1}{2}-\frac{1}{p})+\frac{1}{q}}$.

###### Proof.

This is an analogue of the semiclassical Strichartz estimates for Schrödinger
in [25, 38]. We refer to [37] for the proof. ∎

Now we prove the Strichartz estimates (1.13). Recall $\varphi$ in (1.6) and
Littlewood-Paley frequency cutoff $\varphi_{k}(\sqrt{H_{\alpha,B_{0}}})$, for
each $k\in\mathbb{Z}$, we define

$u_{k}(t,\cdot)=\varphi_{k}(\sqrt{H_{\alpha,B_{0}}})u(t,\cdot).$

where $u(t,x)$ is the solution of (1.1). Then, for each $k\in\mathbb{Z}$,
$u_{k}(t,x)$ solves the Cauchy problem

$\partial_{t}^{2}u_{k}+H_{\alpha,B_{0}}u_{k}=0,\quad
u_{k}(0)=f_{k}(z),~{}\partial_{t}u_{k}(0)=g_{k}(z),$

where $f_{k}=\varphi_{k}(\sqrt{H_{\alpha,B_{0}}})u_{0}$ and
$g_{k}=\varphi_{k}(\sqrt{H_{\alpha,B_{0}}})u_{1}$. Since
$(q,p)\in\Lambda_{s}^{W}$ in definition 1.3, then $q,p\geq 2$. Thus, by using
the square-function estimates (4.2) and the Minkowski inequality, we obtain

$\|u(t,x)\|_{L^{q}(I;L^{p}(\mathbb{R}^{2}))}\lesssim\Big{(}\sum_{k\in\mathbb{Z}}\|u_{k}(t,x)\|^{2}_{L^{q}(I;L^{p}(\mathbb{R}^{2}))}\Big{)}^{\frac{1}{2}},$
(6.2)

where $I=[0,T]$. Denote the half-wave propagator
$U(t)=e^{it\sqrt{H_{\alpha,B_{0}}}}$, then we write

$\begin{split}u_{k}(t,z)=\frac{U(t)+U(-t)}{2}f_{k}+\frac{U(t)-U(-t)}{2i\sqrt{H_{\alpha,B_{0}}}}g_{k}.\end{split}$
(6.3)

By using (6.2) and (6.3), we complete the proof of (1.13) after taking
summation in $k\in\mathbb{Z}$ if we could prove

######  Proposition 6.2.

Let $f=\varphi_{k}(\sqrt{H_{\alpha,B_{0}}})f$ for $\varphi_{k}$ in (1.6) and
$k\in\mathbb{Z}$. Then

$\|U(t)f\|_{L^{q}(I;L^{p}(\mathbb{R}^{2}))}\leq
C_{T}2^{ks}\|f\|_{L^{2}(\mathbb{R}^{2})},$ (6.4)

where the admissible pair $(q,p)\in[2,+\infty]\times[2,+\infty)$ and $s$
satisfy (1.10) and (1.11).

###### Proof.

Since $f=\varphi_{k}(\sqrt{\mathrm{H}})f$, then

$U(t)f=\varphi_{k}(\sqrt{H_{\alpha,B_{0}}})e^{it\sqrt{H_{\alpha,B_{0}}}}f:=U_{k}f.$

By using the spectral theorem, we see

$\|U_{k}(t)f\|_{L^{2}(\mathbb{R}^{2})}\leq C\|f\|_{L^{2}(\mathbb{R}^{2})}.$

By using (5.10), we obtain

$\begin{split}\|U_{k}(t)U_{k}^{*}(s)f\|_{L^{\infty}(\mathbb{R}^{2})}&=\|U_{k}(t-s)f\|_{L^{\infty}(\mathbb{R}^{2})}\\\
&\leq
C_{T}2^{\frac{3}{2}k}\big{(}2^{-k}+|t-s|\big{)}^{-\frac{1}{2}}\|f\|_{L^{1}(\mathbb{R}^{2})},\end{split}$

Then the estimates (6.1) for $U_{k}(t)$ hold for $\alpha=3/2$, $\sigma=1/2$
and $h=2^{-k}$. Hence, Proposition 6.1 gives

$\|U(t)f\|_{L^{q}(I;L^{p}(\mathbb{R}^{2}))}=\|U_{k}(t)f\|_{L^{q}(I;L^{p}(\mathbb{R}^{2}))}\leq
C_{T}2^{k[2(\frac{1}{2}-\frac{1}{p})-\frac{1}{q}]}\|f\|_{L^{2}(\mathbb{R}^{2})}.$

which implies (6.4) since $s=2(\frac{1}{2}-\frac{1}{p})-\frac{1}{q}$. ∎

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