# Huygens’ principle for the Dirac equation in spacetime of non-constant
curvature

Karen Yagdjian

###### Abstract

In this article we give sufficient and necessary conditions for the validity
of the Huygens’ principle for the Dirac equation in the non-constant curvature
spacetime of the Friedmann-Lemaître-Robertson-Walker models of cosmology. The
Huygens’ principle discussed for the equation of a field with mass $m=0$ as
well as a massive spin-$\frac{1}{2}$ field undergoing a red shift of its
wavelength as the universe expands.

Keywords  Dirac equation $\cdot$ Einstein & de Sitter model $\cdot$ FLRW
models $\cdot$ Huygens’ principle $\cdot$

School of Mathematical and Statistical Sciences,
University of Texas RGV, 1201 W. University Drive,

Edinburg, TX 78539, USA

e-mail<EMAIL_ADDRESS>
##  1 Introduction

In this article we give sufficient and necessary conditions for the validity
of the Huygens’ principle for the Dirac equation in the non-constant curvature
spacetime of the Friedmann-Lemaître-Robertson-Walker models of cosmology. We
use the definition of the Huygens’ principle due to Hadamard [23] as the
absence of tails. Thus, the field equations satisfy the Huygens’ principle if
and only if the solution has no tail, that is, the solution depends on the
source distributions on the past null cone of the field only and not on the
sources inside the cone.

The Dirac equation and its quantization in curved spacetime are of great
interest due to the role of spin-$\frac{1}{2}$ particles in astrophysics and
cosmology. Recent observational confirmation of the expansion of the universe
and the quantum field theory demand a detailed investigation of the solutions
of the Dirac equation in curved spacetime (see, e.g., [4, 19] and bibliography
therein). The standard models of Cosmology provide such backgrounds, which
form a family of curved backgrounds of FLRW models. For the de Sitter
spacetime in [25] a fundamental solution of the Dirac operator and an explicit
formula for the solution of the Cauchy problem are obtained. In [26] an
examination of these explicit formulas gave an answer to such an interesting
question in the physics of fundamental particles as a validity of the Huygens’
principle in the de Sitter spacetime. It was proved in [24] that the Klein-
Gordon equation in the de Sitter universe obeys the Huygens’ principle only
for the particle with the mass $\sqrt{2}$ (in the system of units with
$c=\hbar=H=1$, where $c$ is the speed of light, $\hbar$ is the Planck’s
constant, and $H$ is the Hubble constant), while for the zero mass fields the
incomplete Huygens’ principle defined in [24] holds. In the case of a
spatially curved spacetime the last was revealed in [16]. In fact, the
fundamental solutions to the operators can provide information such as the
Huygens’ principle, that is impossible to obtain by numerical solutions of
differential equations.

For the Dirac equation in a curved four-dimensional spacetime, the Huygens’
principle is generally violated by its solutions, due to the mass term in the
equation and the curvature of spacetime [9, 23]. The presence or absence of
tails for waves has been established for some spacetime metrics, including
constant curvature metrics [23]. In fact, the study of the Huygens’ principle
has important applications to quantum field theory and cosmology, especially
in the inflationary theories of the early universe. The fact that the support
of the commutator or the anticommutator-distribution, respectively, lies on
the null-cone if and only if the Huygens’ principle holds for the
corresponding equation [15, 18] shows significance of the Huygens’ principle
for quantum field theory. The Huygens’ principle has been studied in the
context of cosmology for classical fields [7] and for gravitational waves in a
curved background (see, e.g., [14]).

For the Dirac operator in the cosmological context, the Huygens’ principle is
discussed in [2, 5, 12]. The violations of the Huygens’ principle [11] may
have a consequence for relativistic quantum communication. For the fields
obeying the Huygens’ principle communication through massless fields is
confined to the light cone. The violation of the Huygens’ principle makes
possible a leakage of information towards the inside of the light cone [11].
According to [5] the violation of the Huygens’ principle has unexpected
consequences in the propagation of information from the early Universe to the
current era [8]. In [5] are studied conformal and minimal couplings of a test
massless scalar field in a cosmological background. The conclusion made in [5]
is that the signals received today generically contains overlapped information
about the past from both timelike and light connected events.

In the theory of partial differential equations the Huygens’ principle has
been used in the estimates of the Bahouri-Gérard concentration compactness
method and Strichartz estimates (see, e.g., [13, 20]) as well as in the study
of global solvability of nonlinear hyperbolic equations (see, e.g., [1]).

Our consideration is based on the constructed in [27] the fundamental solution
of the Dirac operator and the explicit formulas for the solution of the Cauchy
problem for the Dirac equation in the FLRW spacetime with both accelerating
and decelerating expansion or contraction. The spatially flat FLRW models
considered in [27] had the metric tensor that, in Cartesian coordinates, is
written as follows

$(g_{\mu\nu})=\left(\begin{array}[]{ccccc}1&0&0&0\\\ 0&-a^{2}(t)&0&0\\\
0&0&-a^{2}(t)&0\\\ 0&0&0&-a^{2}(t)\\\
\end{array}\right),\quad\mu,\nu=0,1,2,3,$ (1)

where the scale factor $a(t)=a_{0}t^{\ell}$, $\ell\in{\mathbb{R}}$, $t>0$, and
$x\in{\mathbb{R}}^{3}$, $x_{0}=t$. If $\ell<0$ the spacetime is contracting.
In the case of $\ell>1$ the expansion is accelerating (with horizon), while
for $0<\ell<1$ the expansion is decelerating. In the case of the Milne
spacetime [6, 10, 21] $\ell=1$. The FLRW spacetimes with the scale factors
$a(t)=a_{0}t^{2/3}$ and $a(t)=a_{0}t^{1/2}$ are modeling the matter dominated
universe and the radiation dominated universe, respectively (see, e.g., [17]).

It was proved by Wünsch [23] that if the massive ($m\not=0$) Dirac equation
obeys the Huygens’ principle, then spacetime has a constant curvature.
Accordingly, it was admitted in [27] that the mass term of the field can be
changing in time and vanishing at future infinity. The problem of the time
variation of the spin-$\frac{1}{2}$ massive particle in the early universe has
been studied in physical literature (see [22] and the references therein). In
the present paper, the decay assumption on the mass term has been made in the
context of the expanding universe. More exactly, the model is determined by
the Dirac operator

${\mathscr{D}}(t,\partial_{t},\partial_{x}):=\displaystyle
i{\gamma}^{0}\partial_{t}+i\frac{1}{a(t)}{\gamma}^{1}\partial_{x_{1}}+i\frac{1}{a(t)}{\gamma}^{2}\partial_{x_{2}}+i\frac{1}{a(t)}{\gamma}^{3}\partial_{x_{3}}+i\frac{3\dot{a}(t)}{2a(t)}{\gamma}^{0}-mt^{-1}{\mathbb{I}}_{4}\,,$
(2)

where $m\in{\mathbb{C}}$ and where the contravariant gamma matrices are

$\displaystyle\gamma^{0}=\left(\begin{array}[]{ccccc}{\mathbb{I}}_{2}&{\mathbb{O}}_{2}\\\
{\mathbb{O}}_{2}&-{\mathbb{I}}_{2}\\\
\end{array}\right),\quad\gamma^{k}=\left(\begin{array}[]{ccccc}{\mathbb{O}}_{2}&\sigma^{k}\\\
-\sigma^{k}&{\mathbb{O}}_{2}\\\ \end{array}\right),\quad k=1,2,3\,.$

Here $\sigma^{k}$ are the Pauli matrices

$\displaystyle\sigma^{1}=\left(\begin{array}[]{ccccc}0&1\\\ 1&0\\\
\end{array}\right),\quad\sigma^{2}=\left(\begin{array}[]{ccccc}0&-i\\\ i&0\\\
\end{array}\right),\quad\sigma^{3}=\left(\begin{array}[]{ccccc}1&0\\\ 0&-1\\\
\end{array}\right)\,,$

and ${\mathbb{I}}_{n}$, ${\mathbb{O}}_{n}$ denote the $n\times n$ identity and
zero matrices, respectively.

This model includes the equation of a neutrino with $m=0$ as well as a massive
spin-$\frac{1}{2}$ field undergoing a red shifting of its wavelength as the
universe expands. Thus, in the present paper we consider the Dirac equation in
the spacetime with the metric tensor (1), that is,

${\mathscr{D}}(t,\partial_{t},\partial_{x})\Psi=F\,.$

Recall that a retarded fundamental solution for the Dirac operator (2) is a
four-dimensional matrix
${\mathcal{E}}^{ret}={\mathcal{E}}^{ret}\left(x,t;x_{0},t_{0};m\right)$ with
the operator-valued entries that solves the equation

${\mathscr{D}}(t,\partial_{t},\partial_{x}){\mathcal{E}}\left(x,t;x_{0},t_{0};m\right)=\delta(x-x_{0})\delta(t-t_{0}){\mathbb{I}}_{4},\qquad(x,t),(x_{0},t_{0})\in{\mathbb{R}}^{3}\times{\mathbb{R}}_{+},$
(5)

and with the support in the chronological future (causal future)
$D_{+}(x_{0},t_{0})$ of the point
$(x_{0},t_{0})\in{\mathbb{R}}^{3}\times{\mathbb{R}}_{+}$. The advanced
fundamental solution (propagator)
${\mathcal{E}}^{adv}={\mathcal{E}}^{adv}(x,t;x_{0},t_{0};$ $m)$ solves the
equation (5) and has the support in the chronological past (causal past)
$D_{-}(x_{0},t_{0})$. The forward and backward light cones are defined as the
boundaries of

$D_{\pm}\left(x_{0},t_{0}\right):=\left\\{(x,t)\in{\mathbb{R}}^{3}\times{\mathbb{R}}_{+}\,;\,\left|x-x_{0}\right|\leq\pm\left(\phi(t)-\phi(t_{0})\right)\right\\}\,,$

where $\phi(t):=\frac{1}{1-\ell}t^{1-\ell}$ if $\ell\not=1$. In fact, any
intersection of $D_{-}(x_{0},t_{0})$ with the hyperplane $0<t=const<t_{0}$
determines the so-called dependence domain for the point $(x_{0},t_{0})$,
while the intersection of $D_{+}(x_{0},t_{0})$ with the hyperplane
$t=const>t_{0}>0$ is the so-called domain of influence of the point
$(x_{0},t_{0})$.

In [27] was defined the right co-factor

$\displaystyle{\mathscr{D}}^{co}(t,\partial_{t},\partial_{x})$
$\displaystyle:=$ $\displaystyle
it^{-\frac{\ell}{2}}\gamma^{0}(t^{im}\gamma^{U}+t^{-im}\gamma^{L})\frac{\partial}{\partial
t}+it^{-\frac{3\ell}{2}}\sum_{k=1}^{3}\gamma^{k}(t^{im}\gamma^{U}+t^{-im}\gamma^{L})\frac{\partial}{\partial
x_{k}}$ (6)

of the Dirac operator ${\mathscr{D}}(t,\partial_{t},\partial_{x})$ of (2) such
that the composition
${\mathscr{D}}(t,\partial_{t},\partial_{x}){\mathscr{D}}^{co}(t,\partial_{t},\partial_{x})$
was a diagonal matrix of operators. Here, in order to distinguish upper and
lower 2-spinors, the upper-left corner and lower-right corner matrices,

$\displaystyle\gamma^{U}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}{\mathbb{I}}_{2}&{\mathbb{O}}_{2}\\\
{\mathbb{O}}_{2}&{\mathbb{O}}_{2}\\\
\end{array}\right)=\frac{1}{2}({\mathbb{I}}_{4}+\gamma^{0}),\qquad\gamma^{L}=\left(\begin{array}[]{cc}{\mathbb{O}}_{2}&{\mathbb{O}}_{2}\\\
{\mathbb{O}}_{2}&{\mathbb{I}}_{2}\\\
\end{array}\right)=\frac{1}{2}({\mathbb{I}}_{4}-\gamma^{0})\,,$ (11)

respectively, will be used.

Let $\Delta$ be the Laplace operator in ${\mathbb{R}}^{3}$. Denote by
${\mathcal{E}}^{w}(x,t)$ the distribution that is the fundamental solution to
the Cauchy problem for the wave equation in the Minkowski spacetime

${\mathcal{E}}^{w}_{tt}-\Delta{\mathcal{E}}^{w}=0\,,\quad{\mathcal{E}}^{w}(x,0)=\delta(x)\,,\quad{\mathcal{E}}^{w}_{t}(x,0)=0\,.$

According to Theorem 1.2 [27], for every positive $\varepsilon>0$ and
$t>\varepsilon$ the fundamental solution
${\mathcal{E}}_{+}(x,t;x_{0};m;\varepsilon)$ to the Cauchy problem, that is, a
distribution satisfying

$\displaystyle\cases{{\mathscr{D}}(t,\partial_{t},\partial_{x}){\mathcal{E}}_{+}(x,t;x_{0};m;\varepsilon)={\mathbb{O}}_{4},\cr{\mathcal{E}}_{+}(x,\varepsilon;x_{0};m;\varepsilon)=\delta(x-x_{0}){\mathbb{I}}_{4}\,,}$

is given as follows

$\displaystyle{\mathcal{E}}_{+}(x,t;x_{0};m;\varepsilon)$ $\displaystyle=$
$\displaystyle-i\varepsilon^{1+\frac{\ell}{2}-im}(1-\ell)^{-1}{\mathscr{D}}^{co}(x,t,\partial_{t},\partial_{x})\gamma^{0}$
$\displaystyle\times\int_{0}^{\phi(t)-\phi(\varepsilon)}\left(\begin{array}[]{ccccc}K_{1}\left(r,t;m;\varepsilon\right){\mathbb{I}}_{2}&{\mathbb{O}}_{2}\\\
{\mathbb{O}}_{2}&K_{1}\left(r,t;-m;\varepsilon\right){\mathbb{I}}_{2}\\\
\end{array}\right){\mathcal{E}}^{w}(x-x_{0},r)\,dr,$

where

$\displaystyle K_{1}\left(r,t;m;\varepsilon\right)$ $\displaystyle:=$
$\displaystyle
2^{2i\frac{m}{1-\ell}}\phi(\varepsilon)^{2i\frac{m}{1-\ell}-1}\left(\left(\phi(t)+\phi(\varepsilon)\right)^{2}-r^{2}\right)^{-i\frac{m}{1-\ell}}F\left(i\frac{m}{1-\ell},i\frac{m}{1-\ell};1;\frac{\left(\phi(t)-\phi(\varepsilon)\right)^{2}-r^{2}}{\left(\phi(t)+\phi(\varepsilon)\right)^{2}-r^{2}}\right)\,.$

Henceforth, $F\left(\alpha,\beta;\gamma;z\right)$ is the hypergeometric
function (see, e.g., [3]). In order to write the solution to the Cauchy
problem in an explicit form, we use the integral operator

$\displaystyle{\cal K}_{1}(x,t,D_{x};m;\varepsilon)[\varphi](x,t)$
$\displaystyle:=$
$\displaystyle-i\varepsilon^{1+\frac{\ell}{2}-im}(1-\ell)^{-1}\int_{0}^{\phi(t)-\phi(\varepsilon)}K_{1}\left(r,t;m;\varepsilon\right)\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr\,,\quad\varphi\in
C_{0}^{\infty}({\mathbb{R}}^{n}).$

Theorem 1.3 [27] in the case of source free equation provides the
representation of the solution to the Cauchy problem

$\displaystyle\cases{{\mathscr{D}}(t,\partial_{t},\partial_{x})\Psi(x,t)=0\,,\quad
t>\varepsilon>0\,,\cr\Psi(x,\varepsilon)=\Phi(x)\,,}$ (15)

with $m\in{\mathbb{C}}$, as follows

$\Psi(x,t)={\mathscr{D}}^{co}(t,\partial_{t},\partial_{x})\gamma^{0}\left(\begin{array}[]{ccccc}{\cal
K}_{1}(x,t,D_{x};m;\varepsilon){\mathbb{I}}_{2}&{\mathbb{O}}_{2}\\\
{\mathbb{O}}_{2}&{\cal K}_{1}(x,t,D_{x};-m;\varepsilon){\mathbb{I}}_{2}\\\
\end{array}\right)[\Phi](x,t),\quad t>\varepsilon>0\,.$ (16)

We say that the equation (15) obeys the Huygens’ principle if the solution
$\Psi$ vanishes at all points which cannot be reached from the support of
initial data $\Phi$ by a null geodesic. The main result of this paper is the
following theorem.

###### Theorem 1.1

The solution of the Dirac equation (15) with $\ell\in{\mathbb{R}}$,
$\ell\not=1$, and the mass $m\in{\mathbb{C}}$ obeys the Huygens’ principle if
and only if the mass term takes the values $m=0,\pm i(\ell-1)$.

According to Theorem 1.1 for the spin-$\frac{1}{2}$ particle like neutrino,
the radiation and matter dominated universes are to some extent opaque unless
we accept that the mass of particle is either zero or decaying in time.
Theorem 1.1 guaranties that the Huygens’ principle is fulfilled and,
consequently, the tail does not contribute to the dissipation of the the
propagating field. In that sense for the spin-$\frac{1}{2}$ massless or
decaying in time imaginary massive fields with $m=\pm i(\ell-1)$, the matter
dominated universe is transparent.

A discussion of an imaginary mass parameter in the Dirac equation from the
physical point of view is given in [26].

The rest of this paper is organized as follows. In Section 2, we prove the
sufficiency part of Theorem 1.1 by the representation formulas obtained in
[27] for the solution of the Dirac equation in the FLRW spacetime. In Section
3 the proof of the necessity part of Theorem 1.1 is reduced to the
verification of the asymptotic (for large time) behavior of some integral. The
asymptotic analysis is carried out separately for the cases $\ell>1$ and
$\ell<1$ in Section 4 and Section 5, respectively.

##  2 Huygens’ principle. Sufficient Conditions

The fundamental solution ${\mathcal{E}}^{w}$ to the Cauchy problem for the
wave equation in the Minkowski spacetime can be written as
${\mathcal{E}}^{w}=\partial_{t}{\mathscr{V}}^{w}$, where

${\mathscr{V}}^{w}_{tt}-\Delta{\mathscr{V}}^{w}=0\,,\quad{\mathscr{V}}^{w}(x,0)=0\,,\quad{\mathscr{V}}^{w}_{t}(x,0)=\delta(x)\,.$

In order to write the solution to the Cauchy problem for (15) we use the
operator ${\cal K}_{1}(x,t,D_{x};m;\varepsilon)$ defined in (1). According to
Theorem 1.3 [27] in the case of source free equation, the solution to the
Cauchy problem (15) with $m\in{\mathbb{C}}$ is given by (16). For $m=0$ we
obtain $K_{1}\left(r,t;0;\varepsilon\right)=\phi(\varepsilon)^{-1}$ and,
consequently,

$\displaystyle{\cal K}_{1}(x,t,D_{x};0;\varepsilon)[\varphi](x,t)$
$\displaystyle:=$
$\displaystyle-i\varepsilon^{\frac{\ell}{2}}(1-\ell)^{-1}\int_{0}^{\phi(t)-\phi(\varepsilon)}\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr\,,\quad\varphi\in
C_{0}^{\infty}({\mathbb{R}}^{n}).$

Hence the solution

${\cal
K}_{1}(x,t,D_{x};0;\varepsilon)[\varphi](x,t)=-i\varepsilon^{\frac{\ell}{2}}(1-\ell)^{-1}\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,\phi(t)-\phi(\varepsilon))\varphi(y)\,dy\,,\quad\varphi\in
C_{0}^{\infty}({\mathbb{R}}^{n}),$

obeys the Huygens’ principle.

For the case of the values of mass $m=\pm i(1-\ell)$ we have

$\displaystyle K_{1}\left(r,t;i(1-\ell);\varepsilon\right)$ $\displaystyle=$
$\displaystyle
r^{2}\left(\frac{1}{2}\ell^{3}\varepsilon^{3\ell-3}-\frac{3}{2}\ell^{2}\varepsilon^{3\ell-3}-\frac{1}{2}\varepsilon^{3\ell-3}+\frac{3}{2}\ell\varepsilon^{3\ell-3}\right)$
$\displaystyle+\left(\frac{1}{2}\varepsilon^{3\ell-3}t^{2-2\ell}-\frac{1}{2}\ell\varepsilon^{3\ell-3}t^{2-2\ell}-\frac{1}{2}\ell\varepsilon^{\ell-1}+\frac{\varepsilon^{\ell-1}}{2}\right),$
$\displaystyle K_{1}\left(r,t;-i(1-\ell);\varepsilon\right)$ $\displaystyle=$
$\displaystyle-(\ell-1)t^{\ell-1}\,.$

In the next theorem we consider the family of more general kernels
$K_{1}\left(r,t;\pm m;\varepsilon\right)$, which leads to the Huygens’
principle.

###### Theorem 2.1

If the kernels $K_{1}\left(r,t;\pm m;\varepsilon\right)$ can be represented in
the form

$\displaystyle K_{1}\left(r,t;\pm m;\varepsilon\right)=r^{2}a_{\pm}\left(\pm
m;\varepsilon\right)+b_{\pm}\left(t;\pm m;\varepsilon\right)\,,$

then the Dirac equation (15) obeys the Huygens’ principle.

Proof. First, we consider the case of plus. According to (1) we have

$\displaystyle{\cal K}_{1}(x,t,D_{x};m;\varepsilon)[\varphi](x,t)$
$\displaystyle=$
$\displaystyle-i\varepsilon^{1+\frac{\ell}{2}-im}(1-\ell)^{-1}a_{+}\left(m;\varepsilon\right)\int_{0}^{\phi(t)-\phi(\varepsilon)}r^{2}\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr$
$\displaystyle-i\varepsilon^{1+\frac{\ell}{2}-im}(1-\ell)^{-1}b_{+}\left(t;m;\varepsilon\right)\int_{0}^{\phi(t)-\phi(\varepsilon)}\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr\,,\quad\varphi\in
C_{0}^{\infty}({\mathbb{R}}^{n}).$

For the last term of the previous relation we have

$\int_{0}^{\phi(t)-\phi(\varepsilon)}\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr=\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,\phi(t)-\phi(\varepsilon))\varphi(y)\,dy\,,\quad\varphi\in
C_{0}^{\infty}({\mathbb{R}}^{n}).$

This term obeys the Huygens’ principle since ${\mathscr{V}}^{w}$ does it. Next
we consider the first term of that relation (2):

$\displaystyle\int_{0}^{\phi(t)-\phi(\varepsilon)}r^{2}\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr$
$\displaystyle=$
$\displaystyle\left(r^{2}\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,r)\varphi(y)\,dy\right)_{r=\phi(t)-\phi(\varepsilon)}-2\int_{0}^{\phi(t)-\phi(\varepsilon)}r\left(\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,r)\varphi(y)\,{d}y\right)\,{d}r\,.$

In view of (16) and the structure of
${\mathscr{D}}^{co}(t,\partial_{t},\partial_{x})$, the following lemma
completes the proof of Theorem 2.1.

###### Lemma 2.2

The function

$\displaystyle\frac{\partial}{\partial
x_{j}}\int_{0}^{\phi(t)-\phi(\varepsilon)}r\left(\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,r)\varphi(y)\,dy\right)\,{d}r\,,\quad\varphi\in
C_{0}^{\infty}({\mathbb{R}}^{3}),\quad j=1,2,3,$

satisfies the Huygens’ principle.

Proof. Indeed, we apply the Kirchhoff’s formula and consider, for instance,
the case of $j=3$. Then up to an unimportant factor, the possible tail is:

$\displaystyle\frac{\partial}{\partial
x_{j}}\int_{0}^{\phi(t)-\phi(\varepsilon)}r\left(\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,r)\varphi(y)\,dy\right)\,{d}r$
$\displaystyle=$
$\displaystyle\int\\!\int\\!\int_{|y|\leqslant\phi(t)-\phi(\varepsilon)}\frac{\partial}{\partial
y_{3}}\varphi(x+y){d}y_{1}{~{}d}y_{2}{~{}d}y_{3}$ $\displaystyle=$
$\displaystyle\int_{y_{1}^{2}+y_{2}^{2}\leqslant(\phi(t)-\phi(\varepsilon))^{2}}\left\\{\varphi\left(x_{1}+y_{1},x_{2}+y_{2},x_{3}+\sqrt{(\phi(t)-\phi(\varepsilon))^{2}-y_{1}^{2}-y_{1}^{2}}\right)\right.$
$\displaystyle\left.-\varphi\left(x_{1}+y_{1},x_{2}+y_{2},x_{3}-\sqrt{(\phi(t)-\phi(\varepsilon))^{2}-y_{1}^{2}-y_{1}^{2}}\right)\right\\}{d}y_{1}{~{}d}y_{2}\,.$

For every $t>\varepsilon$ the points

$\left(x_{1}+y_{1},x_{2}+y_{2},x_{3}\pm\sqrt{(\phi(t)-\phi(\varepsilon))^{2}-y_{1}^{2}-y_{1}^{2}}\right)\in{\mathbb{R}}^{3},\quad\mbox{\rm
where}\quad y_{1}^{2}+y_{2}^{2}\leqslant(\phi(t)-\phi(\varepsilon))^{2}\,,$

belong to the sphere of the radius $(\phi(t)-\phi(\varepsilon))$ in
${\mathbb{R}}^{3}$, that is, the domain of integration does not intersect the
interior of the domain of dependence. Thus, the tail is empty and the theorem
is proved. $\square$

##  3 Huygens’ principle. The necessary conditions

The proof of the necessity part will be carried out in two steps. The first
step is the choice of the special initial spinor that is a radial spinor with
a support in small neighborhood of the origin. The second step is in the
establishing asymptotic behavior of the solution for large time at the spatial
origin. If the value of the solution at the spatial origin differs from zero
for the large time, then the Huygens’ principle is violated.

According to Theorem 1.3 [27], the solution to the Cauchy problem (15) with
$m\in{\mathbb{C}}$, is given by (16), where the right co-factor
${\mathscr{D}}^{co}(t,\partial_{t},\partial_{x})$ is given by (6), while
$\gamma^{U}$ and $\gamma^{L}$ are defined in (11). The proof of the necessity
part of Theorem 1.1 is based on the large time asymptotics of the tail of
solution. The initial data
$\Phi(x)=(\Phi_{0}(x),\Phi_{1}(x),\Phi_{2}(x),\Phi_{3}(x))^{T}$ will be chosen
radial having small support. Consider the solution of the Cauchy problem with
the radial function $\Phi(x)=\Phi(r)$,
supp$\,\Phi\subset\left\\{x\in{\mathbb{R}}^{n};|x|\leq\min\\{1/3,{\widetilde{\varepsilon}}/|\ell-1|\\}\right\\}$,
${\widetilde{\varepsilon}}\in(0,1)$:

$\displaystyle\Psi(x,t)$ $\displaystyle=$
$\displaystyle{\mathscr{D}}^{co}(t,\partial_{t},\partial_{x})\gamma^{0}\left(\begin{array}[]{cccc}{\cal
K}_{1}(x,t,D_{x};m;\varepsilon)[\Phi_{0}(x)]\\\ {\cal
K}_{1}(x,t,D_{x};m;\varepsilon)[\Phi_{1}(x)]\\\ {\cal
K}_{1}(x,t,D_{x};-m;\varepsilon)[\Phi_{2}(x)]\\\ {\cal
K}_{1}(x,t,D_{x};-m;\varepsilon)[\Phi_{3}(x)]\\\ \end{array}\right)(x,t),\quad
t>\varepsilon>0\,.$

If we choose the initial data

$\Phi(x)=(\Phi_{0}(x),0,0,0)^{T}\,,$

then the solution
$\Psi(x,t)=(\Psi_{0}(x,t),\Psi_{1}(x,t),\Psi_{2}(x,t),\Psi_{3}(x,t))^{T}$ is
given by

$\displaystyle\Psi(x,t)$ $\displaystyle=$
$\displaystyle\left(it^{-\frac{\ell}{2}}\gamma^{0}(t^{im}\gamma^{U}+t^{-im}\gamma^{L})\frac{\partial}{\partial
t}+it^{-\frac{3\ell}{2}}\sum_{k=1}^{3}\gamma^{k}(t^{im}\gamma^{U}+t^{-im}\gamma^{L})\frac{\partial}{\partial
x_{k}}\right)$ $\displaystyle\times\gamma^{0}\left(\begin{array}[]{cccc}{\cal
K}_{1}(x,t,D_{x};m;\varepsilon)[\Phi_{0}(x)]\\\ 0\\\ 0\\\ 0\\\
\end{array}\right)(x,t),\quad t>\varepsilon>0\,.$

The first component of $\Psi(x,t)$ is

$\displaystyle\Psi_{0}(x,t)$ $\displaystyle=$ $\displaystyle
it^{-\frac{\ell}{2}+im}\frac{\partial}{\partial t}\left({\cal
K}_{1}(x,t,D_{x};m;\varepsilon)[\Phi_{0}(x)]\right)(x,t)$ $\displaystyle=$
$\displaystyle
t^{-\frac{\ell}{2}+im}\varepsilon^{1+\frac{\ell}{2}-im}(1-\ell)^{-1}\frac{\partial}{\partial
t}\left(\int_{0}^{\phi(t)-\phi(\varepsilon)}K_{1}\left(r,t;m;\varepsilon\right)\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\Phi_{0}(y)\,dy\,dr\right),\quad
t>\varepsilon>0\,.$

If we choose the initial data

$\Phi(x)=(0,0,\Phi_{2}(x),0)^{T}\,,$

then the solution
$\Psi(x,t)=(\Psi_{0}(x,t),\Psi_{1}(x,t),\Psi_{2}(x,t),\Psi_{3}(x,t))^{T}$ is
given by

$\displaystyle\Psi(x,t)$ $\displaystyle=$
$\displaystyle\left(it^{-\frac{\ell}{2}}\gamma^{0}(t^{im}\gamma^{U}+t^{-im}\gamma^{L})\frac{\partial}{\partial
t}+it^{-\frac{3\ell}{2}}\sum_{k=1}^{3}\gamma^{k}(t^{im}\gamma^{U}+t^{-im}\gamma^{L})\frac{\partial}{\partial
x_{k}}\right)$ $\displaystyle\times\gamma^{0}\left(\begin{array}[]{cccc}0\\\
0\\\ {\cal K}_{1}(x,t,D_{x};m;\varepsilon)[\Phi_{2}(x)]\\\ 0\\\
\end{array}\right)(x,t),\quad t>\varepsilon>0\,.$

The third component of $\Psi(x,t)$ is

$\displaystyle\Psi_{2}(x,t)$ $\displaystyle=$ $\displaystyle-
it^{-\frac{\ell}{2}-im}\frac{\partial}{\partial t}{\cal
K}_{1}(x,t,D_{x};-m;\varepsilon)[\Phi_{2}(x)](x,t)$ $\displaystyle=$
$\displaystyle-\varepsilon^{1+\frac{\ell}{2}+im}(1-\ell)^{-1}t^{-\frac{\ell}{2}-im}\frac{\partial}{\partial
t}\Bigg{(}\int_{0}^{\phi(t)-\phi(\varepsilon)}K_{1}\left(r,t;-m;\varepsilon\right)\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\Phi_{2}(y)\,dy\,dr\Bigg{)},\,\,t>\varepsilon>0\,.$

Denote either $\varphi(y):=\Phi_{0}(y)$ or $\varphi(y):=\Phi_{2}(y)$. We need
to find a large time asymptotics only for the integral

$\displaystyle\int_{0}^{\phi(t)-\phi(\varepsilon)}\left(\frac{\partial}{\partial
t}K_{1}\left(r,t;\pm
m;\varepsilon\right)\right)\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,r)\varphi(y)\,dy\,dr\,,$

since the term obtained by differentiation of the upper limit,

$\displaystyle t^{-\ell}K_{1}\left(\phi(t)-\phi(\varepsilon),t;\pm
m;\varepsilon\right)\int_{{\mathbb{R}}^{3}}{\mathcal{E}}^{w}(x-y,\phi(t)-\phi(\varepsilon))\varphi(y)\,dy\,,$

obeys the Huygens’ principle. Hence, we consider

$\displaystyle\int_{0}^{\phi(t)-\phi(\varepsilon)}\left(\frac{\partial}{\partial
t}K_{1}\left(r,t;\pm
m;\varepsilon\right)\right)\partial_{r}\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,r)\varphi(y)\,dy\,dr\,.$

Denote

$\widetilde{\Psi}(x,t):=\int_{0}^{\phi(t)-\phi(\varepsilon)}\left(\frac{\partial}{\partial
t}K_{1}\left(r,t;\pm
m;\varepsilon\right)\right)\partial_{r}{\mathscr{V}}_{\varphi}(x,r)\,dr\,,$

where the notation

${\mathscr{V}}_{\varphi}(x,r):=\int_{{\mathbb{R}}^{3}}{\mathscr{V}}^{w}(x-y,r)\varphi(y)\,dy\,$

has been used. Then

$\displaystyle\widetilde{\Psi}(x,t)$ $\displaystyle=$
$\displaystyle\left(\frac{\partial}{\partial
t}K_{1}\left(\phi(t)-\phi(\varepsilon),t;\pm
m;\varepsilon\right)\right){\mathscr{V}}_{\varphi}(x,\phi(t)-\phi(\varepsilon))\,dr$
$\displaystyle-\int_{0}^{\phi(t)-\phi(\varepsilon)}\left(\frac{\partial}{\partial
r}\frac{\partial}{\partial t}K_{1}\left(r,t;\pm
m;\varepsilon\right)\right){\mathscr{V}}_{\varphi}(x,r)\,dr\,.$

In particular, by the Kirchhoff’s formula we have
${\mathscr{V}}_{\varphi}(0,r)=r\varphi(r)$ and

$\displaystyle{\mathscr{V}}_{\varphi}(0,\phi(t)-\phi(\varepsilon))=(\phi(t)-\phi(\varepsilon))\varphi_{0}(\phi(t)-\phi(\varepsilon))=0$

for sufficiently large $t$, that is, if
$\phi(t)-\phi(\varepsilon)>{\widetilde{\varepsilon}}$. Consequently, for large
$t$ we have

$\displaystyle\widetilde{\Psi}(0,t)$ $\displaystyle=$
$\displaystyle-\int_{0}^{\phi(t)-\phi(\varepsilon)}\left(\frac{\partial}{\partial
r}\frac{\partial}{\partial t}K_{1}\left(r,t;\pm
m;\varepsilon\right)\right)r\varphi(r)\,dr$ (21) $\displaystyle=$
$\displaystyle\int_{0}^{\widetilde{\varepsilon}}\left(\frac{\partial}{\partial
t}K_{1}\left(r,t;\pm m;\varepsilon\right)\right)\frac{\partial}{\partial
r}(r\varphi(r))\,dr\,.$

The outline of the remaining part of the proof is as follows. In the next
sections we study the asymptotcs of the function $\frac{\partial}{\partial
t}K_{1}\left(r,t;\pm m;\varepsilon\right)$ as $t\to\infty$. The principal term
of this asymptotics for the cases of $m\not=0,\pm i(\ell-1)$ is the function
of $r$ that allows us to find the initial function $\varphi(r)$ such that the
scalar product (21) of $\frac{\partial}{\partial r}(r\varphi(r))$ with the
principal term differs from zero. Thus, $\widetilde{\Psi}(0,t)\not=0$ for
sufficiently large time, and the Huygens’ principle is violated for these
values of $m$.

It is easily seen that it suffices to study the asymptotic behavior of the
function

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-M}F\left(M,M;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$

for $M\in{\mathbb{C}}$, $M\not=0,\pm 1$, as $\tau\to 0$ if $\ell>1$ or as
$\tau\to\infty$ if $\ell<1$. For $M=0,\pm 1$ the Huygens’ principle holds.
Here $\tau=t^{1-\ell}$ while $A^{2}:=(\ell-1)^{2}r^{2}$ is sufficiently small,
say $A^{2}\leq 1/9$. It is important that

$0<\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}<1\quad\mbox{\rm
when}\quad\tau\in(0,1/2)\cup(2,\infty)\quad\mbox{\rm and }\quad
A^{2}\in[0,1/9],$

as well as

$\lim_{\tau\to
0}\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}=\lim_{\tau\to\infty}\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}=1\quad\mbox{\rm
uniformly on}\quad A^{2}\in[0,1/9]\,.$

Moreover, we will use the finite sum

$\displaystyle\sum_{k=N_{1}}^{N_{2}}\tau^{a_{k}}\left(h_{k}(A,M)+{\widetilde{h}}_{k}(A,M)(\ln(\tau))^{b_{k}}\right)$

of the terms of the asymptotic (as $t\to\infty$) series

$\displaystyle\sum_{k=-\infty}^{\infty}\tau^{a_{k}}\left(h_{k}(A,M)+{\widetilde{h}}_{k}(A,M)(\ln(\tau))^{b_{k}}\right)\,,$

where $a_{k}\in{\mathbb{C}}$, $k=0,\pm 1,\pm 2,\ldots\,$, and the real parts
$\Re a_{k}$, $k=0,\pm 1,\pm 2,\ldots$ are such that

$\ldots<\Re a_{k-1}<\Re a_{k}<\Re a_{k+1}<\ldots\,,\quad\lim_{k\to-\infty}\Re
a_{k}=-\infty\,,\quad\lim_{k\to\infty}\Re a_{k}=\infty\,.$

It is crucial that if the coefficient $h_{k}(A,M)$ or
${\widetilde{h}}_{k}(A,M)$ is independent of $A$, then

$\displaystyle\int_{0}^{\widetilde{\varepsilon}}h_{k}(A,M)\tau^{a_{k}}\frac{\partial}{\partial
r}(r\varphi(r))\,dr=0\quad\mbox{\rm
or}\quad\int_{0}^{\widetilde{\varepsilon}}{\widetilde{h}}_{k}(A,M)\tau^{a_{k}}\frac{\partial}{\partial
r}(r\varphi(r))\,dr=0$

for every function $\varphi\in C_{0}^{\infty}(0,\widetilde{\varepsilon})$. On
the other hand, it will be shown that all coefficients $h_{k}(A,M)$,
${\widetilde{h}}_{k}(A,M)$ are polynomials in $\sqrt{1-A^{2}}$ or
$\ln(1-A^{2})$ (if $\ell<1$) or rational in $1-A^{2}$ (if $\ell>1$) functions.
Thus, it is enough to prove the existence of the depending on $A$ coefficient
$h_{k}(A,M)$ or ${\widetilde{h}}_{k}(A,M)$. Then the existence of a function
$\varphi\in C_{0}^{\infty}(0,\widetilde{\varepsilon})$ such that

$\Psi(0,t)=\int_{0}^{\widetilde{\varepsilon}}\left(\frac{\partial}{\partial
t}K_{1}\left(r,t;\pm m;\varepsilon\right)\right)\frac{\partial}{\partial
r}(r\varphi(r))\,dr\not=0\quad\mbox{\rm for sufficiently large}\quad t\,,$

is evident. This completes the proof of the necessity part of Theorem 1.1.

##  4 Asymptotics of $\frac{\partial}{\partial t}K_{1}\left(r,t;\pm
m;\varepsilon\right)$ when $\ell>1$

In this section we show that if $\ell>1$, then the principal term of the
asymptotics of $\frac{\partial}{\partial t}K_{1}\left(r,t;\pm
m;\varepsilon\right)$ for the case of $M\not=0,\pm 1$ is a function of $r$.
Taking into account the scaling, henceforth we can suppose $\varepsilon=1$.
Denote

$\displaystyle M:=\frac{im}{\ell-1},\quad
A:=(\ell-1)r,\quad\tau:=t^{1-\ell}\to 0\quad\mbox{\rm as}\quad t\to\infty\,,$
$\displaystyle
z:=\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}=1-\frac{4\tau}{1-A^{2}}+\frac{8\tau^{2}}{\left(1-A^{2}\right)^{2}}-\frac{4\left(A^{2}+3\right)\tau^{3}}{\left(1-A^{2}\right)^{3}}+O\left(\tau^{4}\right)\quad\mbox{\rm
as}\quad\tau\to 0.$ (22)

For the function defined by (1) we obtain

$\displaystyle\frac{\partial}{\partial t}K_{1}\left(r,t;m;1\right)$
$\displaystyle=$ $\displaystyle
C(m,\ell)t^{-\ell}\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M-2}{\mathcal{F}}(A,M;\tau),\quad
C(m,\ell)\not=0\,,$ (23)

where

$\displaystyle{\mathcal{F}}(A,M;\tau)$ $\displaystyle:=$ $\displaystyle
2M\left(1-A^{2}-\tau^{2}\right)F\left(1-M,1-M;2;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)$
$\displaystyle-\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)\,.$

For the case of minus, $M:=-\frac{im}{\ell-1}$, we have similar representation
for the derivative $\frac{\partial}{\partial t}K_{1}\left(r,t;-m;1\right)$. We
note, that for the values of $M=0,\pm 1$, the function
$\frac{\partial}{\partial t}K_{1}\left(r,t;\pm m;1\right)$ is independent of
$r$:

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)=0\quad
if\quad M=0\,,$ (26)
$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)=4\tau\quad
if\quad M=1\,,$
$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)=-\frac{1}{4\tau^{2}}\quad
if\quad M=-1\,.$ (27)

This indicates the fact that for these values of $M$ the Dirac equation obeys
the Huygens’ principle.

###  4.1 The case of $\ell>1$ and $M\in{\mathbb{C}}$, $M\not=\frac{1}{2}k$,
$k=0,\pm 1,\pm 2,\ldots$

We note here that

$\displaystyle\left((\tau+1)^{2}-A^{2}\right)^{M}=\left(1-A^{2}\right)^{M}+2M\tau\left(1-A^{2}\right)^{M-1}-M\tau^{2}\left(1-A^{2}\right)^{M-2}\left(A^{2}-2M+1\right)+O\left(\tau^{3}\right)$
(28)

as $\tau\searrow 0$, that together with (23) allows us to ignore all terms of
the expansion of the function ${\mathcal{F}}(A,M;\tau)$ except the first one.

###### Proposition 4.1

Assume that $\ell>1$, $M\in{\mathbb{C}}$, $M\not=\frac{1}{2}k$, $k=0,\pm 1,\pm
2,\ldots$, and $A\in[0,1/2]$. The function ${\mathcal{F}}(A,M;\tau)$ (4) has
the following expansion at $\tau=0$:

$\displaystyle{\mathcal{F}}(A,M;\tau)$ $\displaystyle=$
$\displaystyle\tau\frac{2M\Gamma(2M)\left(A^{2}(1-2M)+1\right)}{(1-2M)\Gamma(M+1)^{2}}$
$\displaystyle+\tau^{2M}16^{M}\frac{\Gamma(1-2M)}{\Gamma(1-M)^{2}}\left(1-A^{2}\right)^{-2M}\left((A^{2}-1)+\tau
2(M-2)\right)+O\left(\tau^{2}\right)\left(\tau^{2M}+1\right)\,.$

Then the principal term in the series of the function
${\mathcal{F}}(A,M;\tau)$ is

$\displaystyle\tau^{2M}16^{M}2M\frac{\Gamma(-2M)}{\Gamma(1-M)^{2}}\left(1-A^{2}\right)^{1-2M}\quad
if\quad\Re(M)<\frac{1}{2}\,,$
$\displaystyle\tau\frac{2M\Gamma(2M)\left(A^{2}(1-2M)+1\right)}{(1-2M)\Gamma(M+1)^{2}}\quad
if\quad\Re(M)>\frac{1}{2}\,,$
$\displaystyle\tau\frac{(1+2iB)\left(2A^{2}B+i\right)\Gamma(1+2iB)}{2B\Gamma\left(\frac{3}{2}+iB\right)^{2}}-\tau^{1+2iB}\frac{2^{2+4iB}\Gamma(-2iB)}{\Gamma\left(\frac{1}{2}-iB\right)^{2}}\left({1-A^{2}}\right)^{-2iB}\,\,if\,\,M=\frac{1}{2}+iB,\,\,B\in{\mathbb{R}}\setminus\\{0\\}.$

Proof. In the function (4) we use the notation of $z$ from (22) and the
expansion

$\displaystyle
1-z=\frac{4\tau}{1-A^{2}}-\frac{8\tau^{2}}{(1-A^{2})^{2}}+O(\tau^{3})\,.$

The hypergeometric function is defined (see, e.g., [3, Sec.2.1]) by the series

$F\left(a,b;c;z\right)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}n!}z^{n}\,.$
(30)

If $z$ is fixed and $|z|<1$, then $F\left(a,b;c;z\right)$ is entire analytic
function of the parameters $a$, $b$, and $c$ in the complex plane
${\mathbb{C}}$. (See, e.g., [3, Sec.2.1.6].) If $\Re(c-a-b)>0$, then
$F\left(a,b;c;1\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$.

There is a formula ([3, Sec.2.10]) that ties together the values of
$F\left(a,b;c;z\right)$ at the points $z=0$ and $z=1$:

$\displaystyle F\left(a,b;c;z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}F\left(a,b;a+b-c+1;1-z\right)$
$\displaystyle+(1-z)^{c-a-b}\frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}F\left(c-a,c-b;c-a-b+1;1-z\right)\,,$

where $|\arg(1-z)|<\pi$, $\quad|1-z|<1$, and $c-a-b\not=\pm 1,\pm 2,\ldots\,$.
Hence,

$\displaystyle F\left(-M,-M;1;z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(1+2M)}{\Gamma(1+M)}F\left(-M,-M;-2M;1-z\right)+(1-z)^{1+2M}\frac{\Gamma(-2M-1)}{\Gamma(-M)^{2}}$
$\displaystyle\hskip 85.35826pt\times
F\left(1+M,1+M;2+2M;1-z\right),\,\,1+2M\not=\pm 1,\pm 2,\ldots\,,$
$\displaystyle F\left(1-M,1-M;2;z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(2M)}{\Gamma(M)^{2}}F\left(1-M,1-M;-2M+1;1-z\right)$
$\displaystyle+(1-z)^{2M}\frac{\Gamma(-2M)}{\Gamma(1-M)^{2}}F\left(1+M,1+M;2M+1;1-z\right),\quad
2M\not=\pm 1,\pm 2,\ldots\,.$

It follows

$\displaystyle
F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2M+1)}{\Gamma(M+1)^{2}}\left(1-\frac{2M}{1-A^{2}}\tau\right)$
$\displaystyle-\left(1-A^{2}\right)^{-2M-1}\tau^{2M+1}4^{2M+1}\frac{\Gamma(-2M)}{(2M+1)\Gamma(-M)^{2}}\left(1-\frac{2M}{1-A^{2}}\tau\right)+\tau^{2M}O\left(\tau^{3}\right)+O\left(\tau^{2}\right)\,.$

Further,

$\displaystyle
F\left(1-M,1-M;2;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle-\tau\frac{4(M-1)^{2}\Gamma(2M)}{\left(A^{2}-1\right)(1-2M)\Gamma(M+1)^{2}}+\frac{\Gamma(2M)}{\Gamma(M+1)^{2}}+O\left(\tau^{2}\right)$
$\displaystyle+\left(\frac{1-A^{2}}{\tau}\right)^{-2M}\left(\frac{4^{2M+1}\left(-M^{2}+M+1\right)\Gamma(1-2M)}{(2M+1)2\left(A^{2}-1\right)M\Gamma(1-M)^{2}}\tau-\frac{4^{2M}\Gamma(1-2M)}{2M\Gamma(1-M)^{2}}+O\left(\tau^{2}\right)\right)\,.$

Consequently, (4.1) holds. Proposition is proved. $\square$

###### Lemma 4.1

Assume that $\ell>1$, $M\in{\mathbb{C}}$, $M\not=\frac{1}{2}k$, $k=0,\pm 1,\pm
2,\ldots$, and $A\in[0,1/2]$. Then the principal term in the asymptotic
expansion of the function $\frac{1}{C(m,\ell)}\frac{\partial}{\partial
t}K_{1}\left(r,t;m;1\right)$ is

$\displaystyle
t^{2M(1-\ell)-\ell}16^{M}2M\frac{\Gamma(-2M)}{\Gamma(1-M)^{2}}\left(1-A^{2}\right)^{-1-M}\quad
if\quad\Re(M)<\frac{1}{2}\,,$ $\displaystyle
t^{1-2\ell}\frac{2M\Gamma(2M)}{(1-2M)\Gamma(M+1)^{2}}\left(A^{2}(1-2M)+1\right)\left(1-A^{2}\right)^{M-2}\quad
if\quad\Re(M)>\frac{1}{2}\,,$ $\displaystyle
t^{1-2\ell}\frac{(1+2iB)\Gamma(1+2iB)}{2B\Gamma\left(\frac{3}{2}+iB\right)^{2}}\left(2A^{2}B+i\right)\left(1-A^{2}\right)^{M-2}-t^{(1+2iB)(1-\ell)-\ell}\frac{2^{2+4iB}\Gamma(-2iB)}{\Gamma\left(\frac{1}{2}-iB\right)^{2}}\left({1-A^{2}}\right)^{M-2-2iB}$
$\displaystyle\hskip
227.62204ptif\,\,M=\frac{1}{2}+iB,\,\,B\in{\mathbb{R}}\setminus\\{0\\}.$

Proof. It follows from Proposition 4.1 and (28). $\square$

###  4.2 The case of $\ell>1$ and $M=n$, $n=1,2,\ldots$

In this case according to the definition (1) we have

$\frac{\partial}{\partial
t}K_{1}\left(r,t;m;\varepsilon\right)=C(n,\ell,\varepsilon)(1-\ell)t^{-\ell}\frac{\partial}{\partial\tau}\left((\tau+1)^{2}-A^{2}\right)^{n}F\left(-n,-n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\,,\quad
C(n,\ell,\varepsilon)\not=0.$

###### Lemma 4.2

For $n=1,2,\ldots$ the asymptotic expansion of the function
$\frac{\partial}{\partial t}K_{1}\left(r,t;m;\varepsilon\right)$ contains the
term

$-C(n,\ell,\varepsilon)(1-\ell)t^{1-2\ell}\frac{4n^{2}\Gamma(2n)}{(2n-1)\Gamma(n+1)^{2}}\left(1-A^{2}\right)^{n-2}\left(A^{2}(2n-1)-1\right)\,.$

Proof. For $n=1,2,\ldots$ the hypergeometric function $F(-n,-n;1;z)$ is the
following polynomial

$F(-n,-n;1;z)=1+\sum_{j=1}^{n}\left(\frac{\Gamma(n+1)}{\Gamma(j+1)\Gamma(n+1-j)}\right)^{2}z^{j}\,.$
(33)

We use (22) and obtain

$\displaystyle\frac{\partial}{\partial t}K_{1}\left(r,t;m;\varepsilon\right)$
$\displaystyle=$ $\displaystyle
C(n,\ell,\varepsilon)(1-\ell)t^{-\ell}\frac{\partial}{\partial\tau}\Bigg{[}\left((\tau+1)^{2}-A^{2}\right)^{n}$
$\displaystyle\times\left(1+\sum_{j=1}^{n}\left(1+\frac{4\tau}{A^{2}-1}+\frac{8\tau^{2}}{\left(A^{2}-1\right)^{2}}+\frac{4\left(A^{2}+3\right)\tau^{3}}{\left(A^{2}-1\right)^{3}}+O(\tau^{4})\right)^{j}\left(\frac{\Gamma(n+1)}{\Gamma(j+1)\Gamma(n+1-j)}\right)^{2}\right)\Bigg{]}$
$\displaystyle=$
$\displaystyle-C(n,\ell,\varepsilon)(1-\ell)t^{-\ell}\left\\{\tau\frac{4n^{2}\left(1-A^{2}\right)^{n-2}\left(A^{2}(2n-1)-1\right)\Gamma(2n)}{(2n-1)\Gamma(n+1)^{2}}+O\left(\tau^{2}\right)\right\\}$

for $n=1,2,\ldots$. The lemma is proved. $\square$

###  4.3 The case of $\ell>1$ and $M=-n$, $n=1,2,\ldots$

If $M=-n$, $n=1,2,\ldots$, then according to the definition (1) we have

$K_{1}\left(r,t;m;\varepsilon\right):=C(n,\ell,\varepsilon)\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\,,\quad
C(n,\ell,\varepsilon)\not=0\,,$

and

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-2n(\tau+1)\left((\tau+1)^{2}-A^{2}\right)^{-n-1}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle+4n^{2}\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{-n-2}F\left(n+1,n+1;2;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\,.$

###### Lemma 4.3

The principal term of the expansion of

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$

as $\tau\to 0$ is

$\displaystyle\cases{\displaystyle-\frac{1}{4}\tau^{-2},\quad\mbox{\rm
if}\quad
n=1\,,\cr\displaystyle-4^{1-2n}n^{2}\left(1-A^{2}\right)^{n-1}\frac{\Gamma(2n)}{[\Gamma(n+1)]^{2}}\tau^{-2n},\quad\mbox{\rm
if}\quad n=2,3,4\ldots\,.}$

Proof. For $n=1$ we use (30) and $F(1,1;1;z)=\frac{1}{1-z}$. Hence

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-1}F\left(1,1;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)=-\frac{1}{4\tau^{2}}\,.$

For $n=2$,

$F\left(2,2;1;z\right)=\sum_{n=0}^{\infty}\frac{\Gamma(2+n)\Gamma(2+n)}{\Gamma(1+n)}\frac{z^{n}}{n!}=\sum_{n=0}^{\infty}(1+n)^{2}z^{n}=\frac{z+1}{(1-z)^{3}}$

and, consequently,

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-2}F\left(2,2;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)=-\frac{3\left(1-A^{2}\right)}{32\tau^{4}}-\frac{1}{32\tau^{2}}\,.$
(35)

Consider the case of $n=3,4,\ldots\,$. There is a typo in the first statement
of Lemma 6.3 [26]. The correct version is

$\displaystyle\lim_{x\searrow 0}x^{2n-1}F(n,n;1;1-x)$ $\displaystyle=$
$\displaystyle F(1-n,1-n;1;1)=\frac{\Gamma(2n-1)}{[\Gamma(n)]^{2}},\quad
n=2,3,\ldots\,,$ $\displaystyle\lim_{x\searrow 0}x^{2n-2}F(n,n;2;1-x)$
$\displaystyle=$ $\displaystyle
F(2-n,2-n;2;1)=\frac{\Gamma(2n-2)}{[\Gamma(n)]^{2}},\quad n=2,3,\ldots\,.$

For the proof we appeal to [3, (14) Sec. 2.1] and to elementary relations,
which can be verified, for instance, by (30) and the multiplication of the
series in the left-hand sides of the relations

$\displaystyle(1-z)^{2n-1}F(n,n;1;z)=F(1-n,1-n;1;z)\quad\mbox{\rm for
all}\quad 0<z<1,\quad n=1,2,3,\ldots\,,$ (36)
$\displaystyle(1-z)^{2n-2}F(n,n;2;z)=F(2-n,2-n;2;z)\quad\mbox{\rm for
all}\quad 0<z<1,\quad n=1,2,3,\ldots$

or by [3, (14) Sec. 2.10]. Then with $a=b=n$, $c=1$, it holds $a+b-c=2n-1\geq
3$ and we have

$\displaystyle F(n,n,1;z)$ $\displaystyle=$
$\displaystyle(1-z)^{-2n+1}\frac{\Gamma(2n-1)}{[\Gamma(n)]^{2}}\sum_{k=0}^{2n-2}\frac{[(-n+1)_{k}]^{2}}{(-2n+2)_{k}k!}(1-z)^{k}\,,$

while with $a=b=n$, $c=2$, it holds $a+b-c=2n-2\geq 3$ and we have

$\displaystyle F(n,n,2;z)$ $\displaystyle=$
$\displaystyle(1-z)^{-2n+2}\frac{\Gamma(2n-2)}{[\Gamma(n)]^{2}}\sum_{k=0}^{2n-2}\frac{[(-n+2)_{k}]^{2}}{(-2n+3)_{k}k!}(1-z)^{k}\,,$

where $-\pi<\arg(1-z)<\pi$. According to (22) we can write

$1-z=1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}},\quad\lim_{\tau\to
0}(1-z)=0\,.$

Thus, from (4.3) we derive

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-2n(\tau+1)\left((\tau+1)^{2}-A^{2}\right)^{-n-1}\left(\frac{\Gamma(2n-1)}{[\Gamma(n)]^{2}}+o(1)\right)\left(1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)^{1-2n}$
$\displaystyle+4n^{2}\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{-n-2}\left(\frac{\Gamma(2n)}{[\Gamma(n+1)]^{2}}+o(1)\right)\left(1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)^{-2n}\,.$

Furthermore, there are easily verified, for instance, by induction, the
following expansions

$\displaystyle\left(1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)^{-2n}$
$\displaystyle=$
$\displaystyle\left(\frac{\tau}{1-A^{2}}\right)^{-2n}\left(4^{-2n}-\frac{4^{1-2n}n\tau}{A^{2}-1}+O\left(\tau^{2}\right)\right),\quad
n>1,$
$\displaystyle\left(1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)^{1-2n}$
$\displaystyle=$
$\displaystyle\left(\frac{\tau}{1-A^{2}}\right)^{1-2n}\left(4^{1-2n}+\frac{2^{3-4n}(1-2n)\tau}{A^{2}-1}+O\left(\tau^{2}\right)\right),\quad
n>2,$ $\displaystyle\left((\tau+1)^{2}-A^{2}\right)^{-n-1}$ $\displaystyle=$
$\displaystyle\left(1-A^{2}\right)^{-n-1}-2\tau\left((n+1)\left(1-A^{2}\right)^{-n-2}\right)+O\left(\tau^{2}\right),$
$\displaystyle\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{-n-2}$
$\displaystyle=$
$\displaystyle-\left(1-A^{2}\right)^{-n-1}+2(n+2)\tau\left(1-A^{2}\right)^{-n-2}+O\left(\tau^{2}\right).$

Hence,

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-2n(\tau+1)\Big{[}\left(1-A^{2}\right)^{-n-1}-2\tau(n+1)\left(1-A^{2}\right)^{-n-2}+O\left(\tau^{2}\right)\Big{]}$
$\displaystyle\times\left(\frac{\Gamma(2n-1)}{[\Gamma(n)]^{2}}+o(1)\right)\left(\frac{\tau}{1-A^{2}}\right)^{1-2n}\left(4^{1-2n}+\frac{2^{3-4n}(1-2n)\tau}{A^{2}-1}+O\left(\tau^{2}\right)\right)$
$\displaystyle+4n^{2}\Big{[}-\left(1-A^{2}\right)^{-n-1}+2(n+2)\tau\left(1-A^{2}\right)^{-n-2}+O\left(\tau^{2}\right)\Big{]}$
$\displaystyle\times\left(\frac{\Gamma(2n)}{[\Gamma(n+1)]^{2}}+o(1)\right)\left(\frac{\tau}{1-A^{2}}\right)^{-2n}\left(4^{-2n}-\frac{4^{1-2n}n\tau}{A^{2}-1}+O\left(\tau^{2}\right)\right)\,.$

The principal term of the last asymptotics for $n=3,4,\ldots$ is

$\displaystyle-4^{1-2n}n^{2}\left(1-A^{2}\right)^{n-1}\frac{\Gamma(2n)}{[\Gamma(n+1)]^{2}}\tau^{-2n}\,.$

This completes the proof of the lemma. $\square$

###  4.4 The case of $\ell>1$ and $M=\frac{1}{2}+n$, $n=0,1,2,\ldots$

The following lemma is decisive for the asymptotics of the function
$\frac{\partial}{\partial t}K_{1}\left(r,t;m;\varepsilon\right)$.

###### Lemma 4.4

For $A\in[0,1/2]$ and all $n=1,2,3,\ldots$

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{\frac{1}{2}+n}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-\tau(2n+1)^{2}\left(1-A^{2}\right)^{n-\frac{3}{2}}\left(2A^{2}n-1\right)\frac{\Gamma(2n)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}+O\left(\tau^{2}\right)\,.$

Proof. Indeed,

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{\frac{1}{2}+n}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle(2n+1)(\tau+1)\left((\tau+1)^{2}-A^{2}\right)^{n-\frac{1}{2}}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle+(2n+1)^{2}\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{n-\frac{3}{2}}F\left(\frac{1}{2}-n,\frac{1}{2}-n;2;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\,.$

According to Lemma 7.1 [26], for $z\searrow 0$ the following formulas hold

$\displaystyle F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;1-z\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2n+2)}{[\Gamma(\frac{3}{2}+n)]^{2}}-z\left(-n-\frac{1}{2}\right)^{2}\frac{\Gamma(2n+1)}{[\Gamma(\frac{3}{2}+n)]^{2}}+O(z^{2}),\hskip
11.38092ptn=0,1,2,3,\ldots\,,$ $\displaystyle
F\left(-n+\frac{1}{2},-n+\frac{1}{2};2;1-z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(2n+1)}{[\Gamma(\frac{3}{2}+n)]^{2}}-z\left(-n+\frac{1}{2}\right)^{2}\frac{\Gamma(2n)}{[\Gamma(\frac{3}{2}+n)]^{2}}+O(z^{2}),\hskip
11.38092ptn=1,2,3,\ldots\,.$

Then

$\displaystyle
F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;1-\frac{4\tau}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2n+2)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}-\tau\frac{\left(n+\frac{1}{2}\right)^{2}4\Gamma(2n+1)}{\left((\tau+1)^{2}-A^{2}\right)\Gamma\left(n+\frac{3}{2}\right)^{2}}+O\left(\tau^{2}\right),$
$\displaystyle
F\left(\frac{1}{2}-n,\frac{1}{2}-n;2;1-\frac{4\tau}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2n+1)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}-\tau\frac{\left(\frac{1}{2}-n\right)^{2}4\Gamma(2n)}{\left((\tau+1)^{2}-A^{2}\right)\Gamma\left(n+\frac{3}{2}\right)^{2}}+O\left(\tau^{2}\right)\,.$

Hence,

$\displaystyle(\tau+1)\left((\tau+1)^{2}-A^{2}\right)^{n-\frac{1}{2}}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle+(2n+1)\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{n-\frac{3}{2}}F\left(\frac{1}{2}-n,\frac{1}{2}-n;2;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle(\tau+1)\left((\tau+1)^{2}-A^{2}\right)^{n-\frac{1}{2}}\Bigg{[}\frac{\Gamma(2n+2)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}-\tau\frac{\left(n+\frac{1}{2}\right)^{2}4\Gamma(2n+1)}{\left((\tau+1)^{2}-A^{2}\right)\Gamma\left(n+\frac{3}{2}\right)^{2}}+O\left(\tau^{2}\right)\Bigg{]}$
$\displaystyle+(2n+1)\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{n-\frac{3}{2}}\Bigg{[}\frac{\Gamma(2n+1)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}-\tau\frac{\left(\frac{1}{2}-n\right)^{2}4\Gamma(2n)}{\left((\tau+1)^{2}-A^{2}\right)\Gamma\left(n+\frac{3}{2}\right)^{2}}+O\left(\tau^{2}\right)\Bigg{]}$
$\displaystyle=$
$\displaystyle-\tau(2n+1)\left(1-A^{2}\right)^{n-\frac{3}{2}}\left(2A^{2}n-1\right)\frac{\Gamma(2n)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}+O\left(\tau^{2}\right)\,.$

Lemma is proved. $\square$

###  4.5 The case of $\ell>1$ and $M=-\frac{1}{2}-n$, $n=0,1,2,\ldots$

For $n=1,2,\ldots$, from (23) we obtain the following relation

$\frac{\partial}{\partial
t}K_{1}\left(r,t;m;1\right)=C(n,\ell,\varepsilon)t^{-\ell}\left(\left(1+\tau\right)^{2}-A^{2}\right)^{-\frac{5}{2}-n}{\mathcal{F}}\left(A,-\frac{1}{2}-n;\tau\right),\quad
C(n,\ell,\varepsilon)\not=0\,.$

###### Lemma 4.5

There exist numbers $C_{0}(\ell,\varepsilon)\not=0$ and
$C_{1}(n,\ell,\varepsilon)\not=0$ such that if $n=0$, then

$\frac{\partial}{\partial
t}K_{1}\left(r,t;m;1\right)=C_{0}(\ell,\varepsilon)(1-A^{2})^{-\frac{1}{2}}t^{-1}+t^{-\ell}\ln(t)O\left(1\right),$

while

$\frac{\partial}{\partial
t}K_{1}\left(r,t;m;1\right)=C_{1}(n,\ell,\varepsilon)\left(1-A^{2}\right)^{-\frac{1}{2}-n}t^{-\ell}(1+o(1))\quad\mbox{
as}\quad t\to\infty\quad\mbox{ if}\quad n=1,2,\ldots\,.$

Proof. First, we use (28) :

$\displaystyle\left(\left(1+\tau\right)^{2}-A^{2}\right)^{-n-\frac{5}{2}}$
$\displaystyle=$
$\displaystyle\left(1-A^{2}\right)^{-n-\frac{5}{2}}-(2n+5)\tau\left(1-A^{2}\right)^{-n-\frac{7}{2}}+\frac{1}{2}(2n+5)\tau^{2}\left(1-A^{2}\right)^{-n-\frac{9}{2}}\left(A^{2}+2n+6\right)+O\left(\tau^{3}\right)\,.$

If we plug $M=-\frac{1}{2}-n$, $n=0,1,2,\ldots$ in the definition (4) of
${\mathcal{F}}(A,M;\tau)$, then

$\displaystyle{\mathcal{F}}\left(A,-\frac{1}{2}-n;\tau\right)$
$\displaystyle=$
$\displaystyle-(2n+1)\left(1-A^{2}-\tau^{2}\right)F\left(n+\frac{3}{2},n+\frac{3}{2};2;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)$
$\displaystyle-\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)F\left(n+\frac{1}{2},n+\frac{1}{2};1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right),\quad
n=0,1,2,\ldots$

Denote

$z:=\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\in(0,1),\quad
1-z=\frac{4\tau}{(\tau+1)^{2}-A^{2}}\to 0\quad\mbox{\rm as}\quad\tau\to 0.$

For $n=1,2,3,\ldots$, the following asymptotics

$\displaystyle\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)F\left(n+\frac{1}{2},n+\frac{1}{2};1;\frac{\left(1-\tau\right)^{2}-A^{2}}{\left(1+\tau\right)^{2}-A^{2}}\right)$
$\displaystyle+(2n+1)\left(1-\tau^{2}-A^{2}\right)F\left(n+\frac{3}{2},n+\frac{3}{2};2;\frac{\left(1-\tau\right)^{2}-A^{2}}{\left(1+\tau\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle(2n+1)\left(1-\tau^{2}-A^{2}\right)(1-z)^{-(2n+1)}\left\\{\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}+(1-z)O(1)\right\\}$

is proved in [26, subsection 7.1, p.30]. In particular,

$\displaystyle\lim_{\tau\to
0}(1-z)^{(2n+1)}\Bigg{[}\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)F\left(n+\frac{1}{2},n+\frac{1}{2};1;z\right)$
$\displaystyle+(2n+1)\left(1-\tau^{2}-A^{2}\right)F\left(n+\frac{3}{2},n+\frac{3}{2};2;z\right)\Bigg{]}$
$\displaystyle=$
$\displaystyle(2n+1)\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}\left(1-A^{2}\right),\quad
n=1,2,3,\ldots\,.$

Together with

$\displaystyle\left(1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)^{-2n-1}$
$\displaystyle=$
$\displaystyle\left(1-A^{2}\right)^{2n+1}\tau^{-2n-1}4^{-2n-1}+\tau^{-2n}O\left(1\right)$

the relation (4.5) proves the second statement of Lemma 4.5. For the case of
$n=0$ we apply (23), (4), and the result of Section 8 [26]:

$\displaystyle\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(1-\tau\right)^{2}-A^{2}}{\left(1+\tau\right)^{2}-A^{2}}\right)+\left(1-\tau^{2}-A^{2}\right)F\left(\frac{3}{2},\frac{3}{2};2;\frac{\left(1-\tau\right)^{2}-A^{2}}{\left(1+\tau\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)\left(\frac{1}{\pi}\left[2\psi(1)-2\psi\left(\frac{1}{2}\right)-\ln(z)\right]+z\ln(z)O(1)\right)$
$\displaystyle+\left(1-\tau^{2}-A^{2}\right)\left(\frac{4z^{-1}}{\pi}-O(1)\ln(z)\right)$
$\displaystyle=$
$\displaystyle\left(1-A^{2}\right)^{2}\frac{1}{\pi}\tau^{-1}-O(1)\ln(t)\,.$

The lemma is proved. $\square$

###  4.6 The case of $\ell>1$ and $M=\frac{1}{2}$

###### Lemma 4.6

For $M=\frac{1}{2}$,

$\displaystyle\frac{\partial}{\partial\tau}\left(((\tau+1)^{2}-A^{2})^{\frac{1}{2}}F\left(-\frac{1}{2},-\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=-4\frac{1}{\pi}\left(1-A^{2}\right)^{-3/2}\tau\left(A^{2}-\ln\left(4-4A^{2}\right)+\ln(\tau)+1\right)+O\left(\tau^{2}\right)\,,$

and the principal term of the expansion is
$-\frac{4}{\pi}\left(1-A^{2}\right)^{-3/2}\tau\ln(\tau)+O\left(\tau\right)\,.$

Proof. We apply the results of subsection 7.3 [26] together with (23) and
$M-2-3/2=-4$. $\square$

###  4.7 The case of $\ell>1$ and $M=-\frac{1}{2}$

###### Lemma 4.7

The following asymptotic expansion holds:

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)=-\frac{1}{\pi\sqrt{1-A^{2}}\tau}+O\left(\tau\right)\quad\mbox{as}\quad\tau\to
0\,.$

Proof. We have

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-\left((\tau+1)(\tau+1)^{2}-A^{2}\right)^{-\frac{3}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle+\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{-\frac{5}{2}}F\left(\frac{3}{2},\frac{3}{2};2;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\,.$

For $F\left(\frac{1}{2},\frac{1}{2};1;z\right)$ we apply [3, (12) Sec. 2.10]
with $a=b=\frac{1}{2}$, $c=1$, $m=0$, and obtain

$\displaystyle F\left(\frac{1}{2},\frac{1}{2};1;z\right)$ $\displaystyle=$
$\displaystyle\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{[(\frac{1}{2})_{n}]^{2}}{[n!]^{2}}\left[h_{n}^{\prime\prime}-\ln(1-z)\right](1-z)^{n}$
$\displaystyle=$
$\displaystyle\frac{4\ln\left(2\right)-\ln\left(1-z\right)}{\pi}+\frac{2-4\ln(2)+\ln(1-z)}{4\pi}(z-1)$
$\displaystyle+\frac{3(12\ln(2)-7-3\ln(1-z))}{64\pi}(z-1)^{2}+O\left((z-1)^{3}\right)\,.$

If $c-a-b=-m$, then we apply [3, (14) Sec. 2.10]:

$\displaystyle F(a,b,a+b-m;z)\frac{1}{\Gamma(a+b-m)}$ $\displaystyle=$
$\displaystyle\frac{\Gamma(m)(1-z)^{-m}}{\Gamma(a)\Gamma(b)}\sum_{n=0}^{m-1}\frac{(a-m)_{n}(b-m)_{n}}{(1-m)_{n}n!}(1-z)^{n}$
$\displaystyle+\frac{(-1)^{m}}{\Gamma(a-m)\Gamma(b-m)}\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(n+m)_{n}n!}\left[\bar{h}_{n}-\ln(1-z)\right](1-z)^{n}\,,$

where $-\pi<\arg(1-z)<\pi,\quad a,b,\neq 0,-1,-2,\dots$, (45), and
$\sum_{n=0}^{m-1}$ is set zero if $m=0$.

For the function $F\left(\frac{3}{2},\frac{3}{2};2;z\right)$, we set
$a=b=\frac{3}{2}$, $c=2$, $m=1$, and obtain

$\displaystyle F\left(\frac{3}{2},\frac{3}{2};2;z\right)$ $\displaystyle=$
$\displaystyle\frac{4}{\pi}(1-z)^{-1}-\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{[(\frac{1}{2})_{n}]^{2}}{(n+1)_{n}n!}\left[\bar{h}_{n}-\ln(1-z)\right](1-z)^{n}$
$\displaystyle=$
$\displaystyle-\frac{4}{\pi(z-1)}+\frac{3-2\ln(4)+\ln(1-z)}{\pi}+\frac{3(-17+12\ln(4)-6\ln(1-z))}{16\pi}(z-1)$
$\displaystyle-\frac{15(-14+10\ln(4)-5\ln(1-z))}{64\pi}(z-1)^{2}+O\left((z-1)^{3}\right)\,.$

Taking into account (22) we write

$\displaystyle
F\left(\frac{1}{2},\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{\log\left(4\left(1-A^{2}\right)\right)-\log(\tau)}{\pi}+\frac{\left(-\log\left(4\left(1-A^{2}\right)\right)+\log(\tau)+2\right)}{\pi\left(A^{2}-1\right)}\tau+O\left(\tau^{2}\right)\,,$
$\displaystyle
F\left(\frac{3}{2},\frac{3}{2};2;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{1-A^{2}}{\pi\tau}+\frac{5-\log\left(4\left(1-A^{2}\right)\right)+\log(\tau)}{\pi}\,,$
$\displaystyle-\frac{\left(4A^{2}+18\log\left(1-A^{2}\right)-18\log(\tau)-47+36\log(2)\right)}{4\pi\left(1-A^{2}\right)}\tau+O\left(\tau^{2}\right)\,.$

On the other hand,

$\displaystyle-(\tau+1)\left((\tau+1)^{2}-A^{2}\right)^{-\frac{3}{2}}$
$\displaystyle=$
$\displaystyle-\frac{1}{\left(1-A^{2}\right)^{3/2}}+\frac{\left(A^{2}+2\right)}{\left(1-A^{2}\right)^{5/2}}\tau+O\left(\tau^{2}\right)\,,$
$\displaystyle\left(A^{2}+\tau^{2}-1\right)\left((\tau+1)^{2}-A^{2}\right)^{-\frac{5}{2}}$
$\displaystyle=$
$\displaystyle-\frac{1}{\left(1-A^{2}\right)^{3/2}}+\frac{5}{\left(1-A^{2}\right)^{5/2}}\tau+O\left(\tau^{2}\right)\,.$

Then (4.7) imply

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-\frac{1}{\pi\sqrt{1-A^{2}}\tau}+O\left(\tau\right)\,.$

The lemma is proved. $\square$

##  5 Asymptotics of $\frac{\partial}{\partial t}K_{1}\left(r,t;\pm
m;\varepsilon\right)$ when $\ell<1$

We appeal to the function $K_{1}\left(r,t;m;\varepsilon\right)$ of (1). Taking
into account the scaling, henceforth we can suppose $\varepsilon=1$. Assume
$\ell<1$ and denote

$M:=\frac{im}{\ell-1},\quad
A:=(\ell-1)r,\quad\tau:=t^{1-\ell}\to\infty\quad\mbox{\rm as}\quad
t\to\infty,\quad
z=\frac{\left(t^{1-\ell}-1\right)^{2}-A^{2}}{\left(t^{1-\ell}+1\right)^{2}-A^{2}}=\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\,.$

Then, according to (1), we write

$K_{1}\left(r,t;m;1\right):=C(m,\ell)\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right),\quad
C(m,\ell)\not=0\,,$

and with the constant $C_{1}(m,\ell)=(1-\ell)C(m,\ell)$ obtain

$\frac{\partial}{\partial
t}K_{1}\left(r,t;m;1\right)=C_{1}(m,\ell)t^{-\ell}\frac{\partial}{\partial\tau}\left(\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)\right)\,.$

Further, we consider

$\frac{\partial}{\partial\tau}\left(\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)\right)=\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M-2}{\mathcal{F}}(A,M;\tau)\,,$

where

$\displaystyle{\mathcal{F}}(A,M;\tau)$ $\displaystyle:=$ $\displaystyle
2M\left(1-A^{2}-\tau^{2}\right)F\left(1-M,1-M;2;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)$
$\displaystyle-\left(1+\tau\right)\left(\left(1+\tau\right)^{2}-A^{2}\right)F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)\,,$

similar to (23) and (4), respectively. We note that any term of the
asymptotics that is independent of $r$ obeys the Huygens’ principle and can be
neglected. (Compare with (26),(26), and (27).)

###  5.1 The case of $\ell<1$ and $M\in{\mathbb{C}}$, $2M\not=0,\pm 1,\pm
2,\ldots$

###### Proposition 5.1

Assume that $\ell<1$ and $M\in{\mathbb{C}}$, $2M\not=0,\pm 1,\pm 2,\ldots$,
and $A\in[0,1/3]$. Then the following asymptotic expansion

$\displaystyle\frac{\partial}{\partial\tau}\left(\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M}F\left(-M,-M;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle\tau^{2M}\Bigg{[}-\frac{2M\Gamma(1-2M)\Gamma(2M)}{\Gamma(-2M)[\Gamma(M+1)]^{2}}\tau^{-1}+\frac{4(M-1)\left(A^{2}(2M-1)-1\right)\Gamma(2M)}{(1-2M)[\Gamma(M)]^{2}}\tau^{-3}+O\left(\frac{1}{\tau^{5}}\right)\Bigg{]}$
$\displaystyle-\frac{4^{2M+1}\Gamma(1-2M)}{2(2M+1)M[\Gamma(-M)]^{2}}\tau^{-2}-\frac{3\left(4^{2M+1}(M+1)\left(A^{2}(2M+3)+1\right)\Gamma(1-2M)\right)}{2(2M+3)(2M+1)M[\Gamma(-M)]^{2}}\tau^{-4}+O\left(\frac{1}{\tau^{5}}\right)$

holds as $\tau\to\infty$.

Proof. It is easily seen that

$1-z=1-\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}=\frac{4}{\tau}-\frac{8}{\tau^{2}}+\frac{4\left(A^{2}+3\right)}{\tau^{3}}-\frac{16\left(A^{2}+1\right)}{\tau^{4}}+\frac{4\left(A^{4}+10A^{2}+5\right)}{\tau^{5}}+O\left(\frac{1}{\tau^{6}}\right)$
(42)

as $\tau\to\infty$. We apply the previous expansion and the relations (4.1),
(4.1) to obtain

$\displaystyle
F\left(-M,-M;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{2M\Gamma(2M)}{\Gamma(M+1)^{2}}-\frac{4M^{2}\Gamma(2M)}{\Gamma(M+1)^{2}}\tau^{-1}-\frac{8M^{2}\Gamma(2M)}{(1-2M)\Gamma(M)^{2}}\tau^{-2}$
$\displaystyle+\frac{4M^{2}\left(A^{2}(6M-3)+2M(2M(M+1)+1)-1\right)\Gamma(2M)}{3(1-2M)\Gamma(M+1)^{2}}\tau^{-3}$
$\displaystyle-\frac{16M^{4}\left(A^{2}(6M-9)+M\left(M^{2}+M-1\right)-4\right)\Gamma(2M)}{3(2M-3)(1-2M)\Gamma(M+1)^{2}}\tau^{-4}+O\left(\frac{1}{\tau^{5}}\right)$
$\displaystyle+\tau^{-2M}\Bigg{[}\frac{4^{2M+1}\Gamma(1-2M)}{2M(2M+1)\Gamma(-M)^{2}}\tau^{-1}-\frac{2^{4M+2}\Gamma(1-2M)}{(2M+1)\Gamma(-M)^{2}}\tau^{-2}$
$\displaystyle+\frac{4^{2M+1}\left(A^{2}(4M(M+2)+3)+4M(M+1)^{2}+1\right)\Gamma(1-2M)}{2(2M+3)M(2M+1)\Gamma(-M)^{2}}\tau^{-3}$
$\displaystyle-\frac{16^{M+1}(M+1)\left(A^{2}(6M+9)+2M(M+2)+3\right)\Gamma(1-2M)}{6(2M+3)(2M+1)\Gamma(-M)^{2}}\tau^{-4}$
$\displaystyle+\frac{4^{2M+1}(M+1)\Gamma(1-2M)}{6(2M+3)(2M+5)M(2M+1)\Gamma(-M)^{2}}$
$\displaystyle\times\Big{(}3A^{4}(2M+1)(2M+3)(2M+5)+6A^{2}(2M+3)(2M+5)(2M(M+1)+1)$
$\displaystyle+2M(2M(M+3)+3)(2M(M+3)+7)+9\Big{)}\tau^{-5}+O\left(\frac{1}{\tau^{6}}\right)\Bigg{]}$

and

$\displaystyle
F\left(1-M,1-M;2;\frac{(1-\tau)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2M)}{\Gamma(M+1)^{2}}+\frac{4(M-1)^{2}\Gamma(2M)}{\left((1-2M)\Gamma(M+1)^{2}\right)}\tau^{-1}-\frac{4(M-1)((M-2)M+2)\Gamma(2M)}{\left((1-2M)\Gamma(M+1)^{2}\right)}\tau^{-2}$
$\displaystyle+\frac{4(M-1)\left(3A^{2}(M-1)(2M-3)+M(2M(2(M-4)M+17)-45)+27\right)\Gamma(2M)}{3(2M-3)(1-2M)\Gamma(M+1)^{2}}\tau^{-3}$
$\displaystyle-\frac{8(M-1)\Gamma(2M)}{3\left((2M-3)(1-2M)\Gamma(M+1)^{2}\right)}$
$\displaystyle\hskip
71.13188pt\times\left(3A^{2}(2M-3)((M-2)M+2)+M(M(M((M-4)M+11)-23)+30)-18\right)\tau^{-4}$
$\displaystyle+O\left(\frac{1}{\tau^{5}}\right)+\tau^{-2M}\Bigg{[}-\frac{16^{M}\Gamma(1-2M)}{2M\Gamma(1-M)^{2}}+\frac{4^{2M+1}((M-1)M-1)\Gamma(1-2M)}{2M(2M+1)\Gamma(1-M)^{2}}\tau^{-1}$
$\displaystyle-\frac{2^{4M}\left(M\left(A^{2}(2M+1)+2(M-2)M+1\right)+4\right)\Gamma(1-2M)}{M(2M+1)\Gamma(1-M)^{2}}\tau^{-2}$
$\displaystyle+\frac{4^{2M+1}\Gamma(1-2M)}{6(2M+3)M(2M+1)\Gamma(1-M)^{2}}$
$\displaystyle\hskip
71.13188pt\times\left(3A^{2}(2M+1)(2M+3)((M-1)M-1)+(M-1)M\left(4M^{3}-4M+15\right)-27\right)\tau^{-3}$
$\displaystyle-\frac{16^{M}\Gamma(1-2M)}{6(2M+3)M(2M+1)\Gamma(1-M)^{2}}$
$\displaystyle\times\Big{(}3A^{4}M(2M+1)^{2}(2M+3)+6A^{2}(2M+3)(M(4M((M-1)M-1)+11)+8)$
$\displaystyle+M(2M(2M(2M((M-1)M+1)+5)-47)+57)+144\Big{)}\tau^{-4}+O\left(\frac{1}{\tau^{5}}\right)\Bigg{]}$

as $\tau\to\infty$. On the other hand,

$\displaystyle\left(\left(1+\tau\right)^{2}-A^{2}\right)^{M-2}$
$\displaystyle=$
$\displaystyle\tau^{2M-4}\Bigg{[}1+\frac{2(M-2)}{\tau}-\frac{(M-2)\left(A^{2}-2M+5\right)}{\tau^{2}}$
$\displaystyle-\frac{2\left((M-3)(M-2)\left(3A^{2}-2M+5\right)\right)}{3\tau^{3}}$
$\displaystyle+\frac{(M-3)(M-2)\left(3A^{4}-12\left(A^{2}+2\right)M+42A^{2}+4M^{2}+35\right)}{6\tau^{4}}+O\left(\frac{1}{\tau^{5}}\right)\Bigg{]}$

as $\tau\to\infty$. Then we put the last expressions into (4) and obtain
(5.1). The proposition is proved. $\square$

###### Corollary 5.1

If $\Re(M)>-\frac{1}{2}$, then in the expansion (5.1) there is the term
containing $A$:

$\tau^{2M-3}\frac{4(M-1)\left(A^{2}(2M-1)-1\right)\Gamma(2M)}{(1-2M)[\Gamma(M)]^{2}}\,.$

If $\Re(M)<-\frac{1}{2}$, then in the expansion (5.1) there is the term
containing $A$:

$-\frac{3\left(4^{2M+1}(M+1)\left(A^{2}(2M+3)+1\right)\Gamma(1-2M)\right)}{2(2M+3)(2M+1)M[\Gamma(-M)]^{2}}\tau^{-4}\,.$

If $M=-\frac{1}{2}+iB,\,\,B\in{\mathbb{R}}\setminus\\{0\\}$, then in the
expansion (5.1) there are the terms containing $A$:

$-\frac{1}{4\tau^{4}}\Bigg{[}\frac{8(2B+3i)(2A^{2}(B+i)+i)}{[\Gamma(-\frac{1}{2}+iB)]^{2}\Gamma(3-2iB)}\tau^{2iB}+\frac{3i4^{2iB}(2B-i)(2A^{2}(B-i)-i)}{(B-i)[\Gamma\left(\frac{1}{2}-iB\right)]^{2}\Gamma(1+2iB)}\Bigg{]}\,.$

###  5.2 The case of $\ell<1$ and $M=n$, $n=2,3,\ldots$

###### Lemma 5.2

The function

$\frac{\partial}{\partial\tau}\left(\left(\left(1+\tau\right)^{2}-A^{2}\right)^{n}F\left(-n,-n;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)\right)$
(43)

for $n=2,3,\ldots$ is a polynomial in $\tau$ and $A^{2}$, which varies in
$A^{2}$ and contains at least one term $\tau^{a}A^{2b}$ with $a,b\not=0$.

Proof. To verify the first statement we just use (33). For $M=n$, where
$n=2,3,4,\ldots$ we put
$z=\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}$ in
(33), and obtain for the function of (43) the polynomial

$\displaystyle\frac{\partial}{\partial\tau}\left(\left(\left(1+\tau\right)^{2}-A^{2}\right)^{n}\Bigg{[}1+\sum_{j=1}^{n}\left(\frac{\Gamma(n+1)}{\Gamma(j+1)\Gamma(-j+n+1)}\right)^{2}\left(\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)^{j}\Bigg{]}\right)$
$\displaystyle=$ $\displaystyle
2n(\tau+1)\left(\left(1+\tau\right)^{2}-A^{2}\right)^{n-1}$
$\displaystyle+2\sum_{j=1}^{n}\left(\frac{n!}{j!(n-j)!}\right)^{2}\Bigg{[}j(\tau-1)\left(\left(\tau-1\right)^{2}-A^{2}\right)^{j-1}\left(\left(\tau+1\right)^{2}-A^{2}\right)^{n-j}$
$\displaystyle+(n-j)(\tau+1)\left(\left(\tau-1\right)^{2}-A^{2}\right)^{j}\left(\left(\tau+1\right)^{2}-A^{2}\right)^{n-j-1}\Bigg{]}\,.$

Thus, the result is a polynomial in $\tau$ and $A^{2}$. The coefficient of
$A^{2n-1}$ vanishes since

$n+\sum_{j=1}^{n}(n-2j)\left(\frac{\Gamma(n+1)}{\Gamma(j+1)\Gamma(-j+n+1)}\right)^{2}=0\,.$

At $A=\tau-1$ the polynomial equals to
$[(2n+2)\tau+(2n-2)]\left(4\tau\right)^{n-1}$, while at $A=\tau+1$ it takes
value $(2n\tau-2n)\left(4\tau\right)^{n-1}$. Thus, there is at least one
nonvanishing term of the plynomial containing $\tau^{a}A^{2b}$ with
$a,b\not=0$. We can use such term with the greatest $a$, in fact $a=2n-3$. The
lemma is proved. $\square$

###  5.3 The case of $\ell<1$ and $M=-n$, $n=1,2,3,\ldots$

In the case of $M=-n$, with $n=1,2,3,\ldots$, we appeal to (4.3) to prove the
next lemma.

###### Lemma 5.3

For $n=2,3,4,\ldots$, in the expansion of the function

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
(44)

as $\tau\to\infty$ there exists the term

$\displaystyle-\frac{2^{1-2n}\left(1-A^{2}\right)^{n-1}\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}\Gamma(n)}\frac{1}{\tau^{2n}}\,.$

Proof. The case of $M=-1,-2\,$ is given by Lemma 4.3 and (35). For the case of
$n=3,4,\ldots$ we appeal to [3, (14) section 2.1] and to elementary relation
(36). Hence, by definition (30) we have

$F(n,n;1;z)=(1-z)^{1-2n}F(1-n,1-n;1;z)=(1-z)^{1-2n}\sum_{k=0}^{n-1}\frac{[(1-n)_{k}]^{2}}{[k!]^{2}}z^{k}\,.$

Then

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n}F\left(n,n;1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$
$\displaystyle=$ $\displaystyle 4^{1-2n}\tau^{-2n}P(A;n,\tau)\,,$

where

$\displaystyle P(A;n,\tau)$ $\displaystyle:=$
$\displaystyle\tau\sum_{k=0}^{n-1}\frac{\left[(1-n)_{k}\right]^{2}}{[k!]^{2}}\left(A^{2}((n-1)(\tau+1)-2k)-\left(\tau^{2}-1\right)(2k+(n-1)(\tau-1))\right)$
$\displaystyle+4^{1-2n}(1-2n)\sum_{k=0}^{n-1}\frac{\left[(1-n)_{k}\right]^{2}}{[k!]^{2}}\left(\left((\tau-1)^{2}-A^{2}\right)^{k}\left((\tau+1)^{2}-A^{2}\right)^{-k+n-1}\right)$

is polynomial in $\tau$ and $A^{2}$. We set $\tau=0$ and obtain

$P(A;n,0)=-\frac{2^{1-2n}\left(1-A^{2}\right)^{n-1}\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}\Gamma(n)}\,.$

Hence, the asymptotic expansion of the function (44) contains the term

$-\tau^{-2n}\frac{2^{1-2n}\left(1-A^{2}\right)^{n-1}\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}\Gamma(n)}\,.$

The lemma is proved. $\square$

Thus, in the asymptotic expansion of (44) we can choose the term with greatest
degree in $\tau$ among all containing variable $A^{2}$.

###  5.4 The case of $\ell<1$ and $M=\frac{1}{2}+n$, $n=1,2,\ldots$

###### Lemma 5.4

For $M=\frac{3}{2}$ ($n=1$) we have

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-\frac{3}{2}}F\left(\frac{3}{2},\frac{3}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)\\\
=-\frac{1}{4\pi\tau^{2}}+\frac{6A^{2}-6\ln(\tau)+5-6\ln(4)}{16\pi\tau^{4}}+O\left(\frac{1}{\tau^{5}}\right).$

For $n=2,3,4,\ldots\,$, in the asymptotic expansion of the function

$\displaystyle\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n-\frac{1}{2}}F\left(n+\frac{1}{2},n+\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)$

as $\tau\to\infty$ there exists the term with $\tau^{-4}$ depending on $A$,
namely, this term is

$\frac{3\Gamma(2n)\left(4^{-2n-1}(2n-1)\left(2A^{2}(n-1)-1\right)\right)}{[\Gamma(n+\frac{1}{2})]^{2}(n-1)}\frac{1}{\tau^{4}}\,.$

Proof. For $n=1,2,3,4,\ldots$, we have to discuss

$\frac{\partial}{\partial\tau}\left(\left((\tau+1)^{2}-A^{2}\right)^{-n-\frac{1}{2}}F\left(n+\frac{1}{2},n+\frac{1}{2};1;\frac{(\tau-1)^{2}-A^{2}}{(\tau+1)^{2}-A^{2}}\right)\right)\,.$

We make the change of variable $\lambda=\tau^{-1}$ and consider the asymptotic
series about $\lambda=0$:

$\displaystyle\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{-n-\frac{1}{2}}F\left(n+\frac{1}{2},n+\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle(2n+1)\lambda^{2n}\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)^{-n-\frac{5}{2}}{\mathcal{F}}(A,n;\lambda)\,,$

where has been used the notation

$\displaystyle{\mathcal{F}}(A,n;\lambda)$ $\displaystyle:=$
$\displaystyle(\lambda+1)\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)F\left(n+\frac{1}{2},n+\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)$
$\displaystyle-\lambda((2n+1))\left(\left(A^{2}-1\right)\lambda^{2}+1\right)F\left(n+\frac{3}{2},n+\frac{3}{2};2;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)\,.$

Denote

$z:=\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\to
1\quad\mbox{as}\quad\lambda\to 0\,,$

then we can use (42) as $\lambda=\tau^{-1}\to 0$. According to [3, (14) Sec.
2.10] for $n=1,2,\ldots$ with $a=b=n+\frac{1}{2}$, $c=1$, $c-a-b=-2n$, and
$m=2n$, we have :

$\displaystyle F\left(n+\frac{1}{2},n+\frac{1}{2};1;z\right)$ $\displaystyle=$
$\displaystyle(1-z)^{-2n}\frac{\Gamma(2n)}{[\Gamma(n+\frac{1}{2})]^{2}}\sum_{k=0}^{2n-1}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(1-2n)_{k}k!}(1-z)^{k}$
$\displaystyle+\frac{1}{[\Gamma(\frac{1}{2}-n)]^{2}}\sum_{k=0}^{\infty}\frac{[(n+\frac{1}{2})_{k}]^{2}}{(k+2n)_{k}k!}\left[\bar{h}_{k}-\ln(1-z)\right](1-z)^{k}$
$\displaystyle=$
$\displaystyle(1-z)^{-2n}\frac{\Gamma(2n)}{[\Gamma(n+\frac{1}{2})]^{2}}\sum_{k=0}^{2n-1}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(1-2n)_{k}k!}(1-z)^{k}+|\ln(1-z)|O(1)\,,$

where $-\pi<\arg(1-z)<\pi,\quad a,b,\neq 0,-1,-2,\dots$

$\bar{h}_{n}=\psi(1+n)+\psi(1+n+m)-\psi(a+n)-\psi(b+n)\,.$ (45)

The function $\psi(z)$ is the logarithmic derivative of the gamma function:
$\psi(z)=\frac{\Gamma^{\prime}(z)}{\Gamma(z)}$.

Similarly, since $a=b=n+\frac{3}{2}$, $c=2$, $c-a-b=-m$ , $m=2n+1$, we have

$\displaystyle F\left(n+\frac{3}{2},n+\frac{3}{2};2;z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(2n+1)(1-z)^{-2n-1}}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{2n}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}$
$\displaystyle+\frac{(-1)}{[\Gamma(\frac{1}{2}-n)]^{2}}\sum_{k=0}^{\infty}\frac{[(n+\frac{3}{2})_{k}]^{2}}{(k+2n+1)_{k}k!}\left[\bar{h}_{k}-\ln(1-z)\right](1-z)^{k}$
$\displaystyle=$
$\displaystyle(1-z)^{-2n-1}\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{2n}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+|\ln(1-z)|O(1)\,.$

Hence,

$\displaystyle{\mathcal{F}}(A,n;\lambda)$ $\displaystyle=$
$\displaystyle(1-z)^{-2n-1}\Bigg{[}(\lambda+1)\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)$
$\displaystyle\times\Bigg{\\{}\frac{\Gamma(2n)}{[\Gamma(n+\frac{1}{2})]^{2}}\sum_{k=0}^{2n-1}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(1-2n)_{k}k!}(1-z)^{k+1}+(1-z)^{2n+1}|\ln(1-z)|O(1)\Bigg{\\}}$
$\displaystyle-\lambda((2n+1))\left(\left(A^{2}-1\right)\lambda^{2}+1\right)\Bigg{\\{}\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{2n}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+(1-z)^{2n+1}|\ln(1-z)|O(1)\Bigg{\\}}\Bigg{]}$
$\displaystyle=$
$\displaystyle(1-z)^{-2n-1}\frac{\Gamma(2n)}{[\Gamma(n+\frac{1}{2})]^{2}}$
$\displaystyle\times\Bigg{[}(\lambda+1)\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)\Bigg{\\{}\sum_{k=0}^{2n-1}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(1-2n)_{k}k!}(1-z)^{k+1}+(1-z)^{2n+1}|\ln(1-z)|O(1)\Bigg{\\}}$
$\displaystyle-\lambda((2n+1))\left(\left(A^{2}-1\right)\lambda^{2}+1\right)\Bigg{\\{}\frac{2n}{[(n+\frac{1}{2})]^{2}}\sum_{k=0}^{2n}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+(1-z)^{2n+1}|\ln(1-z)|O(1)\Bigg{\\}}\Bigg{]}\,.$

Since $n\geq 2$, we can truncate the sums as follows:

$\displaystyle{\mathcal{F}}(A,n;\lambda)$ $\displaystyle=$
$\displaystyle(1-z)^{-2n-1}\frac{\Gamma(2n)}{[\Gamma(n+\frac{1}{2})]^{2}}$
$\displaystyle\times\Bigg{[}(\lambda+1)\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)\Bigg{\\{}\sum_{k=0}^{2}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(1-2n)_{k}k!}(1-z)^{k+1}+(1-z)^{4}|O(1)\Bigg{\\}}$
$\displaystyle-\lambda((2n+1))\left(\left(A^{2}-1\right)\lambda^{2}+1\right)\Bigg{\\{}\frac{2n}{[(n+\frac{1}{2})]^{2}}\sum_{k=0}^{3}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+(1-z)^{4}O(1)\Bigg{\\}}\Bigg{]}\,.$

It is evident that for $n\geq 2$ the following asymptotic expansion holds

$\displaystyle(2n+1)\lambda^{2n}\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)^{-n-\frac{5}{2}}\left(1-\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)^{-2n-1}$
$\displaystyle=$ $\displaystyle
4^{-2n-1}(2n+1)\frac{1}{\lambda}+4^{-2n-1}(2n-3)(2n+1)+2^{-4n-3}(2n-3)(2n+1)\left(-A^{2}+2n-4\right)\lambda+O\left(\lambda^{2}\right)\,.$

On the other hand

$\displaystyle(\lambda+1)\left(\lambda\left(-A^{2}\lambda+\lambda+2\right)+1\right)\Bigg{\\{}\sum_{k=0}^{2}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(1-2n)_{k}k!}(1-z)^{k+1}+(1-z)^{4}O(1)\Bigg{\\}}$
$\displaystyle-\lambda((2n+1))\left(\left(A^{2}-1\right)\lambda^{2}+1\right)\Bigg{\\{}\frac{2n}{[(n+\frac{1}{2})]^{2}}\sum_{k=0}^{3}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+(1-z)^{4}O(1)\Bigg{\\}}$
$\displaystyle=$
$\displaystyle\frac{\lambda(\lambda+1)}{(n-1)\left(\lambda\left(A^{2}(-\lambda)+\lambda+2\right)+1\right)^{2}}$
$\displaystyle\times\Bigg{[}\Big{(}\lambda^{2}\left(-8A^{2}(n-1)+8n^{3}-44n^{2}+78n-41\right)$
$\displaystyle+4\left(A^{2}-1\right)^{2}\lambda^{4}(n-1)+4\left(A^{2}-1\right)\lambda^{3}(n-1)(2n-5)-4\lambda(n-1)(2n-5)+4(n-1)\Big{)}$
$\displaystyle-\frac{\lambda
n(2n+1)\left(\left(A^{2}-1\right)\lambda^{2}+1\right)}{12\left(n+\frac{1}{2}\right)^{2}}\Bigg{\\{}\frac{6\lambda^{2}(2n-1)(3-2n)^{2}}{n\left(\lambda\left(A^{2}(-\lambda)+\lambda+2\right)+1\right)^{2}}$
$\displaystyle+\frac{12\lambda(1-2n)^{2}}{n\left(\left(A^{2}-1\right)\lambda^{2}-2\lambda-1\right)}+\frac{\lambda^{3}(2n-1)(3-2n)^{2}(5-2n)^{2}}{(n-1)n\left(\left(A^{2}-1\right)\lambda^{2}-2\lambda-1\right)^{3}}+24\Bigg{\\}}\Bigg{]}$

implies

$\displaystyle(2n+1)\lambda^{2n}\left(\lambda\left(A^{2}(-\lambda)+\lambda+2\right)+1\right)^{-n-\frac{5}{2}}{\mathcal{F}}(A,n;\lambda)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2n)}{[\Gamma(n+\frac{1}{2})]^{2}}\left(4^{-2n}-\frac{3\left(4^{-2n-1}(2n-1)\left(2A^{2}(n-1)-1\right)\right)}{n-1}\lambda^{2}+o\left(\lambda^{2}\right)\right)\,.$

This completes the proof of the second statement of the lemma. The proof of
the first statement of the lemma is straightforward. $\square$

###  5.5 The case of $\ell<1$ and $M=-\frac{1}{2}-n$, $n=1,2,\ldots$

Taking into account the scaling, we from now on can suppose $\varepsilon=1$.
Assume $\ell<1$ and denote

$K_{1}\left(r,t;m;1\right):=C(m,\ell)\left(\left(1+\tau\right)^{2}-A^{2}\right)^{\frac{1}{2}+n}F\left(-\frac{1}{2}-n,-\frac{1}{2}-n;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right),\quad
C(m,\ell)\not=0\,.$

Consider the function

$\frac{\partial}{\partial
t}K_{1}\left(r,t;m;1\right):=C(m,\ell)(1-\ell)t^{-\ell}\frac{\partial}{\partial\tau}\left(\left(1+\tau\right)^{2}-A^{2}\right)^{\frac{1}{2}+n}F\left(-\frac{1}{2}-n,-\frac{1}{2}-n;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right).$

###### Lemma 5.5

In the asymptotic expansion of the function

$\displaystyle\frac{\partial}{\partial\tau}\left(\left(1+\tau\right)^{2}-A^{2}\right)^{\frac{1}{2}+n}F\left(-\frac{1}{2}-n,-\frac{1}{2}-n;1;\frac{\left(\tau-1\right)^{2}-A^{2}}{\left(\tau+1\right)^{2}-A^{2}}\right)$

as $\tau\to\infty$ there is a term containing $A^{2}$:

$\tau^{2n-2}\frac{2^{2n+1}\left(2A^{2}n-1\right)\Gamma(n)}{\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}\,.$

Proof. We consider the series about $\lambda=0$ after the change of variable
$\lambda=\tau^{-1}$:

$\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{n+\frac{1}{2}}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)\,.$

We remind that $\varphi\in C_{0}^{\infty}(0,1)$ implies
$A^{2}=(1-\ell)^{2}r^{2}>constant>0$. Then,

$\displaystyle\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{n+\frac{1}{2}}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-(2n+1)\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)^{n-\frac{5}{2}}\lambda^{-2n-2}{\mathscr{F}}(\lambda;n,A)\,,$

where

$\displaystyle{\mathscr{F}}(\lambda;n,A)$ $\displaystyle:=$
$\displaystyle(\lambda+1)\left(\lambda\left(A^{2}(-\lambda)+\lambda+2\right)+1\right)^{2}F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;1-\frac{4\lambda}{(\lambda+1)^{2}-(A\lambda)^{2}}\right)$
$\displaystyle+\lambda(2n+1)\left(\left(A^{2}-1\right)\lambda^{2}+1\right)\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)F\left(\frac{1}{2}-n,\frac{1}{2}-n;2;1-\frac{4\lambda}{(\lambda+1)^{2}-(A\lambda)^{2}}\right)\,.$

We consider the function ${\mathscr{F}}(\lambda;n,A)$ around the point
$\lambda=0$. In order to substitute the following expansion

$\displaystyle
z:=1-\frac{4\lambda}{(\lambda+1)^{2}-(A\lambda)^{2}}=1-4\lambda+8\lambda^{2}-4\left(A^{2}+3\right)\lambda^{3}+O\left(\lambda^{4}\right)$
(47)

in the series for the function ${\mathscr{F}}(\lambda;n,A)$ we appeal to [3,
(12) Sec. 2.10]:

$\displaystyle F(a,b;a+b+m;z)\frac{1}{\Gamma(a+b+m)}$ $\displaystyle=$
$\displaystyle\frac{\Gamma(m)}{\Gamma(a+m)\Gamma(b+m)}\sum_{n=0}^{m-1}\frac{(a)_{n}(b)_{n}}{(1-m)_{n}n!}(1-z)^{n}$
$\displaystyle+\frac{(1-z)^{m}(-1)^{m}}{\Gamma(a)\Gamma(b)}\sum_{n=0}^{\infty}\frac{(a+m)_{n}(b+m)_{n}}{(n+m)!n!}\left[h_{n}^{\prime\prime}-\ln(1-z)\right](1-z)^{n}\,,$

where $-\pi<\arg(1-z)<\pi$, $a,b,\neq 0,-1,2,\ldots$,

$h_{n}^{\prime\prime}=\psi(n+1)+\psi(n+m+1)-\psi(a+n+m)-\psi(b+n+m),$

and the term $\sum_{n}^{m-1}$ in the expression for $F(a,b;a+b+m;z)$ is to be
interpreted as zero when $m=0$.

Hence with $m=2n+2,\quad n=1,2,\ldots$, we obtain

$\displaystyle F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;z\right)$
$\displaystyle=$
$\displaystyle\frac{\Gamma(2n+2)}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{2n+1}\frac{[(-n-\frac{1}{2})_{k}]^{2}}{(-2n-1)_{k}k!}(1-z)^{k}+(1-z)^{2n+2}|\ln(1-z)|O(1)\,,$

and with $m=2n+1,\quad n=1,2,\ldots$, we derive

$\displaystyle F\left(\frac{1}{2}-n,\frac{1}{2}-n;2;z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{2n}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+(1-z)^{2n+1}|\ln(1-z)O(1)\,.$

We keep the terms of order $(1-z)^{2}$ and of order $(1-z)$ in the finite sums

$\displaystyle F\left(-n-\frac{1}{2},-n-\frac{1}{2};1;z\right)$
$\displaystyle=$
$\displaystyle(2n+1)\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{2}\frac{[(-n-\frac{1}{2})_{k}]^{2}}{(-2n-1)_{k}k!}(1-z)^{k}+(1-z)^{3}|\ln(1-z)|O(1)\,,$

and

$\displaystyle F\left(\frac{1}{2}-n,\frac{1}{2}-n;2;z\right)$ $\displaystyle=$
$\displaystyle\frac{\Gamma(2n+1)}{[\Gamma(n+\frac{3}{2})]^{2}}\sum_{k=0}^{1}\frac{[(\frac{1}{2}-n)_{k}]^{2}}{(-2n)_{k}k!}(1-z)^{k}+(1-z)^{2}|\ln(1-z)|O(1)\,,$

respectively. On the other hand

$\displaystyle(\lambda+1)\left(\lambda\left(A^{2}(-\lambda)+\lambda+2\right)+1\right)^{2}=1+5\lambda+\left(10-2A^{2}\right)\lambda^{2}+\left(10-6A^{2}\right)\lambda^{3}+O\left(\lambda^{4}\right)\,,$
(48)
$\displaystyle\lambda(2n+1)\left(\left(A^{2}-1\right)\lambda^{2}+1\right)\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)=(2n+1)\lambda+(4n+2)\lambda^{2}+O\left(\lambda^{4}\right)\,,$
(49)

and

$\displaystyle\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)^{n-\frac{5}{2}}$
$\displaystyle=$ $\displaystyle
1+(2n-5)\lambda-\frac{1}{2}(2n-5)\left(A^{2}-2n+6\right)\lambda^{2}+O\left(\lambda^{3}\right)\,.$
(50)

We substitute in
$-(2n+1)\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)^{n-\frac{5}{2}}{\mathscr{F}}(\lambda;n,A)$
the relations (47), (48), (49), (50) and obtain

$-(2n+1)\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)^{n-\frac{5}{2}}{\mathscr{F}}(\lambda;n,A)=-\frac{(2n+1)\Gamma(2n+2)}{\Gamma\left(n+\frac{3}{2}\right)^{2}}+\frac{2^{2n+1}\left(2A^{2}n-1\right)\Gamma(n)}{\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}\lambda^{2}+O\left(\lambda^{3}\right)\,.$

The lemma is proved. $\square$

###  5.6 The case of $\ell<1$ and $M=\frac{1}{2}$

Similar to what was done in the previous subsection, we replace $\tau$ with
$\lambda=\tau^{-1}\to 0$ as $\tau\to\infty$.

###### Lemma 5.6

In the expansion for small $\lambda\searrow 0$ of the function

$\displaystyle\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)$

there is a term $-(4\pi)^{-1}\left(6A^{2}+3\right)\lambda^{2}\ln(\lambda)$,
which is depending on $A$.

Proof. Indeed,

$\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)\\\
=\left(\left(1-A^{2}\right)\lambda^{2}+2\lambda+1\right)^{-\frac{5}{2}}{\mathscr{F}}(A,\lambda)\,,$

where

$\displaystyle{\mathscr{F}}(A,\lambda)$ $\displaystyle:=$
$\displaystyle\Bigg{\\{}(-A^{2}\lambda^{3}-A^{2}\lambda^{2}+\lambda^{3}+3\lambda^{2}+3\lambda+1)F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)$
(51)
$\displaystyle-(A^{2}\lambda^{3}-\lambda^{3}+\lambda)F\left(\frac{3}{2},\frac{3}{2};2;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\Bigg{\\}}\,.$

Then it is easy to see the following asymptotic expansions

$\displaystyle\left(\left(1-A^{2}\right)\lambda^{2}+2\lambda+1\right)^{-\frac{5}{2}}$
$\displaystyle=$ $\displaystyle
1-5\lambda+\frac{5}{2}\left(A^{2}+6\right)\lambda^{2}-\frac{35}{2}\left(A^{2}+2\right)\lambda^{3}+O\left(\lambda^{4}\right)\,,$
$\displaystyle
1-\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}$
$\displaystyle=$ $\displaystyle
4\lambda-8\lambda^{2}+4\left(A^{2}+3\right)\lambda^{3}+O\left(\lambda^{4}\right)\,.$
(52)

Hence, according to (4.7) and (4.7), we have

$F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)\\\
=\frac{2\ln(2)-\ln\left(\lambda\right)}{\pi}+\frac{2\ln(2)-\ln\left(\lambda\right)}{\pi}\lambda+\frac{-4A^{2}-9+2\ln(2)-\ln\left(\lambda\right)}{4\pi}\lambda^{2}+O\left(\lambda^{3}\right)$

and

$\displaystyle
F\left(\frac{3}{2},\frac{3}{2};2;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{\pi\lambda}+\frac{5-2\ln(2)+\ln(\lambda)}{\pi}+\frac{-4A^{2}+47-18\ln(4)+18\ln(\lambda)}{4\pi}\lambda$
$\displaystyle+\frac{4A^{2}+84-39\ln(4)+39\ln(\lambda)}{4\pi}\lambda^{2}+O\left(\lambda^{3}\right)\,.$

It remains to substitute obtained expansions into (51) to derive

$\displaystyle{\mathscr{F}}(A,\lambda)$ $\displaystyle=$
$\displaystyle\frac{-1+\ln(4)-\ln(\lambda)}{\pi}+\frac{5(-1+\ln(4)-\ln(\lambda))}{\pi}\lambda$
$\displaystyle+\frac{-4A^{2}-4A^{2}\ln(4)-52+86\ln(2)-43\ln(\lambda)+4A^{2}\ln(\lambda)}{4\pi}\lambda^{2}+O\left(\lambda^{3}\right)\,.$

Thus,

$\displaystyle\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle\Bigg{\\{}1-5\lambda+\frac{5}{2}\left(A^{2}+6\right)\lambda^{2}-\frac{35}{2}\left(A^{2}+2\right)\lambda^{3}+O\left(\lambda^{4}\right)\Bigg{\\}}$
$\displaystyle\times\Bigg{\\{}\frac{-1+\ln(4)-\ln(\lambda)}{\pi}+\frac{5(-1+\ln(4)-\ln(\lambda))}{\pi}\lambda$
$\displaystyle+\frac{-4A^{2}-4A^{2}\ln(4)-52+86\ln(2)-43\ln(\lambda)+4A^{2}\ln(\lambda)}{4\pi}\lambda^{2}+O\left(\lambda^{3}\right)\Bigg{\\}}\,.$

Finally, we obtain

$\displaystyle\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{-\frac{1}{2}}F\left(\frac{1}{2},\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle\frac{-\ln(\lambda)-1+\ln(4)}{\pi}+\frac{-\left(6A^{2}+3\right)\ln(\lambda)-14A^{2}+12A^{2}\ln(2)-12+6\ln(2)}{4\pi}\lambda^{2}+O\left(\lambda^{3}\right)\,.$

The lemma is proved. $\square$

###  5.7 The case of $\ell<1$ and $M=-\frac{1}{2}$

We appeal to the arguments which have been used in subsection 5.5 and consider
around the point $\lambda=0$ the function
$\left((\lambda+1)^{2}-A^{2}\lambda^{2}\right)^{-\frac{5}{2}}{\mathscr{F}}(\lambda;0,A)$,
where ${\mathscr{F}}(\lambda;0,A)$ is defined in (5.5).

###### Lemma 5.7

In the expansion of the function (5.5) for small $\lambda\searrow 0$ there is
a term $(-2A^{2}-3+4\ln(2))/\pi$ depending on $A$.

Proof. Indeed, we have

$\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{1/2}F\left(-\frac{1}{2},-\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)=-\frac{1}{\lambda^{2}\left(-A^{2}\lambda^{2}+\lambda^{2}+2\lambda+1\right)^{3/2}}{\mathscr{F}}(\lambda,A)\,,$

where

$\displaystyle{\mathscr{F}}(\lambda,A)$ $\displaystyle:=$
$\displaystyle\Bigg{\\{}\left(-A^{2}\lambda^{3}-A^{2}\lambda^{2}+\lambda^{3}+3\lambda^{2}+3\lambda+1\right)F\left(-\frac{1}{2},-\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)$
$\displaystyle+\left(A^{2}\lambda^{3}-\lambda^{3}+\lambda\right)F\left(\frac{1}{2},\frac{1}{2};2;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\Bigg{\\}}\,.$

Furthermore, according to subsection 5.5, we can write

$\displaystyle F\left(-\frac{1}{2},-\frac{1}{2};1;z\right)$ $\displaystyle=$
$\displaystyle\frac{4}{\pi}+\frac{z-1}{\pi}+\frac{-2\ln(1-z)-5+8\ln(2)}{16\pi}(z-1)^{2}$
$\displaystyle-\frac{3(-2\ln(1-z)-5+8\ln(2))}{64\pi}(z-1)^{3}+O\left((z-1)^{4}\right)$

and

$\displaystyle F\left(\frac{1}{2},\frac{1}{2};2;z\right)$ $\displaystyle=$
$\displaystyle\frac{4}{\pi}+\frac{4\ln(2)-\ln(1-z)-3}{\pi}(z-1)-\frac{3(-6\ln(1-z)-17+24\ln(2))}{16\pi}(z-1)^{2}$
$\displaystyle+\frac{15(-5\ln(1-z)-14+20\ln(2))}{64\pi}(z-1)^{3}+O\left((z-1)^{4}\right)\,.$

We substitute the relation (52) in the expansions for
$F\left(-\frac{1}{2},-\frac{1}{2};1;z\right)$ and
$F\left(\frac{1}{2},\frac{1}{2};2;z\right)$ and derive

$\displaystyle
F\left(-\frac{1}{2},-\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{4}{\pi}-\frac{4}{\pi}\lambda+\frac{-2\ln(\lambda)+3+4\ln(2)}{\pi}\lambda^{2}+\frac{-4A^{2}+2\ln(\lambda)-3-4\ln(2)}{\pi}\lambda^{3}+O\left(\lambda^{4}\right)\,,$
$\displaystyle
F\left(\frac{1}{2},\frac{1}{2};2;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}}\right)$
$\displaystyle=$
$\displaystyle\frac{4}{\pi}-\frac{4(-\ln(\lambda)-3+2\ln(2))}{\pi}\lambda+\frac{10\ln(\lambda)+19-20\ln(2)}{\pi}\lambda^{2}$
$\displaystyle+\frac{4A^{2}\ln(\lambda)+16A^{2}-8A^{2}\ln(2)+15\ln(\lambda)+26-30\ln(2)}{\pi}\lambda^{3}+O\left(\lambda^{4}\right)\,.$

On the other hand,

$\displaystyle-\frac{1}{\lambda^{2}\left(-A^{2}\lambda^{2}+\lambda^{2}+2\lambda+1\right)^{3/2}}$
$\displaystyle=$
$\displaystyle-\frac{1}{\lambda^{2}}+\frac{3}{\lambda}-\frac{3}{2}\left(A^{2}+4\right)+\frac{5}{2}\left(3A^{2}+4\right)\lambda-\frac{15}{8}\left(A^{4}+12A^{2}+8\right)\lambda^{2}$
$\displaystyle+\frac{21}{8}\left(5\left(A^{2}+4\right)A^{2}+8\right)\lambda^{3}+O\left(\lambda^{4}\right)\,.$

Thus, we obtain

$\displaystyle\frac{\partial}{\partial\lambda}\left(\left(\left(\frac{1}{\lambda}+1\right)^{2}-A^{2}\right)^{1/2}F\left(-\frac{1}{2},-\frac{1}{2};1;\frac{\left(\frac{1}{\lambda}-1\right)^{2}-A^{2}}{\left(1+\frac{1}{\lambda}\right)^{2}-A^{2}}\right)\right)$
$\displaystyle=$
$\displaystyle-\frac{4}{\pi\lambda^{2}}+\frac{-2A^{2}-3+4\ln(2)-2\ln(\lambda)}{\pi}+O\left(\lambda^{2}\right)\,.$

The lemma is proved. $\square$

The Huygens’ principle’s is a local property, that is, it can be verified for
the small time. On the other hand, the proof of the necessity part presented
in this paper is based on the large time asymptotics. In fact, by taking into
account the scaling invariance of the operator (2), the proof can be easily
modified to the small time by the implementing a large auxiliary parameter. In
this paper, we avoid that modification, which creates an unnecessary
cumbersome, even though it could be a useful tool for the further
generalizations.

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