\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 177, pp. 1--20.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/177\hfil Euler-Poisson-Darboux equation]
{General form of the Euler-Poisson-Darboux equation
and application of the transmutation method}

\author[E. L. Shishkina, S. M. Sitnik \hfil EJDE-2017/177\hfilneg]
{Elina L. Shishkina, Sergei M. Sitnik}

\address{Elina L. Shishkina \newline
Voronezh State University,
Faculty of Applied Mathematics, Informatics and Mechanics,
Universitetskaya square, 1,
Voronezh 394006, Russia} 
\email{ilina\_dico@mail.ru}


\address{Sergei M. Sitnik \newline
Belgorod State National Research University,
Belgorod, Russia. \newline
RUDN University, 6 Miklukho-Maklaya st,
Moscow, Russia}
\email{pochtasms@gmail.com}

\dedicatory{Communicated by Ludmila S. Pulkina}

\thanks{Submitted May 22, 2017. Published July 11, 2017.}
\subjclass[2010]{26A33, 44A15}
\keywords{Bessel operator; Euler-Poisson-Darboux equation;  Hankel transform;
\hfill\break\indent transmutation operators}

\begin{abstract}
 In this article, we find solution representations in the compact integral
 form  to the Cauchy problem for a general form of the Euler-Poisson-Darboux
 equation with Bessel operators via generalized translation and spherical
 mean operators for all values of the parameter $k$, including also not
 studying before exceptional odd negative values.  We use a Hankel transform
 method to prove results in a unified way. Under additional conditions we
 prove that a distributional solution is a classical one too.
 A transmutation property for connected generalized spherical mean is proved
 and importance of applying transmutation methods for differential equations
 with Bessel operators is emphasized. The paper also contains a short historical
 introduction on differential equations with Bessel operators and a rather
 detailed reference list of monographs and papers on mathematical theory and
 applications of this class of differential equations.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

The classical Euler-Poisson-Darboux (EPD)  equation is defined by
\begin{equation}\label{ClEPD}
\frac{{\partial }^2u}{\partial t^2}+\frac{k}{t}\frac{\partial u}{\partial t}
 =\sum^n_{i=1}\frac{{\partial }^2u}{\partial x^2_i},\quad
 u=u(x,t;k),\quad x\in\mathbb{R}^n,\; t>0,\; -\infty<k<\infty.
\end{equation}
The operator acting by variable $t$ is  the  Bessel operator
 and we will be denoted by (see, for example, \cite[p. 3]{Kipr})
$$
(B_k)_t=\frac{{\partial }^2}{\partial t^2}+\frac{k}{t}\frac{\partial }{\partial t}.
$$
When $n=1$  equation \eqref{ClEPD} appeared in Leonard Euler's work
 (see \cite[p. 227]{Euler}) and later was studied by Sim\'eon Denis Poisson
 \cite{Poisson}, by Gaston Darboux  \cite{Darboux} and by Bernhard Riemann
 \cite{Riman}.

For the Cauchy problem corresponding to \eqref{ClEPD}
we add initial conditions
\begin{equation}\label{USL0}
    u(x,0;k)=f(x),\quad \frac{\partial u(x,t;k)}{\partial t}\Big|_{t=0}=0.
\end{equation}

Interest in the multidimensional equation \eqref{ClEPD} has increased significantly
after Alexander Weinstein's papers \cite{Weinstein0}--\cite{Weinstein4}.
There the Cauchy problem  for  \eqref{ClEPD}
is considered with  $k\in\mathbb{R}$, the first initial condition being non-zero
and the second initial condition equal  zero.
A solution to the Cauchy problem \eqref{ClEPD}--\eqref{USL0}  in the classical
sense was obtained in \cite{Weinstein0}--\cite{Young}, and in the distributional
sense in \cite{Bresters1,Bresters2,CSh}.
Tersenov \cite{Tersenov} solved  the Cauchy problem  for  \eqref{ClEPD}
in the general form where the first and the second conditions are non-zero.
Different problems for the equation \eqref{ClEPD} with many applications
to gas dynamics, hydrodynamics, mechanics, elasticity and plasticity
and so on  were also studied in
\cite{Aks,Barabash,Blum}, \cite{Car1}--\cite{Car3}, \cite{Chap}--\cite{Dza2},
\cite{Fal}--\cite{Fran3}, \cite{Glushak2}--\cite{Sit7},
\cite{Ivan3}, \cite{Kuzmin}--\cite{Lev3},
\cite{LPSh2,Mizes,Pul1,Pul2}, \cite{Smirnov}--\cite{Vekua1}, \cite{ZhTr}.
Of course this list of references is incomplete.


In this article we consider the singular hyperbolic differential equation,
with respect to all variables,
 which is a generalization of multidimensional
Euler-Poisson-Darboux (EPD) equation \eqref{ClEPD}:
\begin{equation}\label{EPD0}
\frac{{\partial }^2u}{\partial t^2}+\frac{k}{t}\frac{\partial u}{\partial t}
=(\Delta_\gamma)_x\ u,\quad  u=u(x,t;k), \quad k\in\mathbb{R},\quad t>0,
\end{equation}
with the  singular elliptic operator defined by
\begin{equation}\label{LapBess1}
(\Delta_\gamma)_x=\sum_{i=1}^{n} (B_{\gamma_i})_{x_i}
=\sum_{i=1}^{n}\Big( \frac{\partial ^{2} }{\partial x_{i}^{2} }
 +\frac{\gamma _{i} }{x_{i} } \frac{\partial}{\partial x}\Big)
=\sum_{i=1}^{n}\frac{1}{x_i^{\gamma_i}}\frac{\partial}{\partial
x_i}x_i^{\gamma_i}\frac{\partial}{\partial x_i}
\end{equation}
under the natural restrictions
$$
\gamma_i>0, \quad  x=(x_1,\dots,x_n),\quad x_i>0,\quad i=1,2,\dots,n,
$$
together with  initial conditions \eqref{USL0}.

We call equation \eqref{EPD0}  the \emph{Euler-Poisson-Darboux equation in
 general form}.

Let us  emphasize that singular differential equations with the operator
\eqref{LapBess1} including equations \eqref{ClEPD} and \eqref{EPD0} were
thoroughly studied in many papers by  Kipriyanov's school, the results are
partially systemized in his monograph \cite{Kipr}. In accordance with
Kipriyanov's terminology the operator \eqref{LapBess1} is classified as
$B$-elliptic operator (sometimes also the term Laplace-Bessel operator is used),
and equations \eqref{ClEPD} and  \eqref{EPD0} are classified as $B$-hyperbolic
 equations.
In connection with results of this scientific school let us mention  papers
 of Ivanov \cite{Ivan1,Ivan2,Ivan3} in which important
problems for EPD equation were solved, such as generalizations to homogeneous
symmetric Riemann spaces, energy equipartition property, equations with
a product of EPD-type multipliers. Also note papers
\cite{Barabash,Shi5,LPSh2} on application of spherical mean and generalized
translation operators, generalized mean value theorems. Differential
equations of EPD type are applied in the study of fractional powers of EPD,
generalized EPD operators and connected generalized Riesz--type potentials,
 cf. \cite{Sh1}--\cite{Sh4}.

Another important approach to differential equations with Bessel operators
is based on application of the transmutation theory. This method is
essential in the study of singular problems with use of special classes
of transmutations such as Sonine, Poisson, Buschman--Erd\'elyi ones and
different forms of fractional integrodifferential operators,
cf. \cite{Car1}--\cite{Castilio1}, \cite{Sit8}--\cite{Sit7},
\cite{Kra}--\cite{Kra3}, \cite{Sit5}--\cite{Sit3}.
Abstract differential equations with Bessel operators were studied, and
in fact were mostly initiated, in the famous monograph \cite{CSh}.
See also recent papers \cite{Glushak1}--\cite{Glushak3}.

Considering the Cauchy problem \eqref{EPD0}--\eqref{USL0} in more details,
David Fox in \cite{Fox} (cf. also  \cite[p. 243]{CSh} and \cite{Stellmacher})
proved solution uniqueness for $k\geq 0$ and find a solution representation
in the explicit form for all $k$ except odd negative values.
The explicit solution was found via Lauricella functions in fact as $n$-times
series, which is not convenient  for applications and numerical solving.
In all the above references the case $k\neq -1,-3,-5,\dots$ was expelled and
not studied.
So in \cite{Barabash}, \cite{Shi5}--\cite{LPSh2} different approaches
from those used in \cite{Fox} to the solution of this Cauchy problem were considered.

In this paper we find solution representations to the above Cauchy problem
in the compact integral form via generalized translation and spherical
 mean operators for all values of the parameter $k$, including also not
studying before exceptional odd negative values.
 We use a Hankel transform method to prove results in a unified way.
Under additional conditions we prove that a distributional solution
is a classical one too.

 \section{Definitions and propositions}

We use the subset of the Euclidean space
$$
\mathbb{R}^n_+=\{x=(x_1,\ldots,x_n)\in\mathbb{R}^n:  x_1>0,\ldots, x_n>0\}.
$$
Let  $|x|=\big(\sum_{i=1}^n x_i^2\big)^{1/2}$ and $\Omega$ be
bounded  or unbounded open set in $\mathbb{R}^n$ symmetric with respect
to each hyperplane $x_i=0$, $i=1,\dots,n$,
$\Omega_+=\Omega\cap{\mathbb{R}}^n_+$ and
$\overline{\Omega}_+=\Omega\cap\overline{\mathbb{R}}^n_+$ where
$$
\overline{\mathbb{R}}^n_+=\{x=(x_1,\ldots,x_n)\in\mathbb{R}^n:
 x_1{\geq}0,\ldots, x_n{\geq}0\}.
$$
We consider the class $C^m(\Omega_+)$ consisting of $m$-times differentiable
on $\Omega_+$ functions and denote by $C^m(\overline{\Omega}_+)$ the subset
of functions from $C^m(\Omega_+)$ such that all derivatives  of these functions
with respect to $x_i$ for any $i=1,\dots,n$  are continuous up to $x_i=0$.
Function $f\in C^m(\overline{\Omega}_+)$ we will call \emph{even with respect to}
$x_i$, $i=1,\dots,n$ if $\frac{\partial^{2k+1}f}{\partial x_i^{2k+1}}\Big|_{x=0}=0$
for all nonnegative integer $k\leq \frac{m-1}{2}$ (see \cite[p. 21]{Kipr}).
The class $C^m_{ev}(\overline{\Omega}_+)$ consists of functions from
$C^m(\overline{\Omega}_+)$ even with respect to each variable $x_i$, $i=1,\dots,n$.
In the following we will denote $C^m_{ev}(\overline{\mathbb{R}}^n_+)$
by $C^m_{ev}$.
We set
$$
C^\infty_{ev}(\overline{\Omega}_+)=\cap C^m_{ev}(\overline{\Omega}_+)
$$
with intersection taken for all finite $m$ and
$C^\infty_{ev}(\overline{\mathbb{R}}_+)=C^\infty_{ev}$.
Let $\mathring{C}^\infty_{ev}(\overline{\Omega}_+)$
be the space of all functions  $f{\in}C^\infty_{ev}(\overline{\Omega}_+)$
with a compact support. Elements of $\mathring{C}^\infty_{ev}(\overline{\Omega}_+)$
we will call \emph{test functions} and use the notation
$\mathring{C}^\infty_{ev}(\overline{\Omega}_+)=\mathcal{D}_+(\overline{\Omega}_+)$.

As the space of basic functions we will use  the subspace of the space of
rapidly decreasing functions:
$$
 S_{ev}({\mathbb{R}}^{n}_+)=\big\{f\in C^\infty_{ev}:
\sup _{{x\in {\mathbb{R}}^{n}_+}}|x^{\alpha }D^{\beta }f(x)|<\infty ,
\forall \alpha ,\beta \in \mathbb {Z} _{+}^{n}\big\},
$$
where $\alpha=(\alpha_1,\dots,\alpha_n)$,
 $\beta=(\beta_1,\dots,\beta_n)$, $\alpha_1,\dots,\alpha_n,\beta_1,\dots,\beta_n$
are integer nonnegative numbers,
$x^\alpha= x_1^{\alpha_1} x_2^{\alpha_2} \ldots x_n^{\alpha_n}$,
${D}^\beta={D}^{\beta_1}_{x_1}\dots{D}^{\beta_n}_{x_n}$,
 ${D}_{x_j}=\frac{\partial}{\partial x_j}$.

We deal with multi-index $\gamma=(\gamma_1,{\ldots},\gamma_{n})$ consists
of positive fixed reals $\gamma_i>0$, $i=1,{\dots},n$,
$|\gamma|=\gamma_1{+}{\ldots}{+}\gamma_{n}$.
Let $L_p^{\gamma}(\Omega_+)$, $1{\leq}p{<}\infty$, be the space of  all
measurable in $\Omega_+$ functions even with respect to each variable $x_i$,
$i=1,\dots,n$ such that
$$
\int_{\Omega_+}|f(x)|^p x{}^\gamma dx<\infty,
$$
where
$$
 x{}^\gamma=\prod_{i=1}^n x_i^{\gamma_i}.
$$
For a real number $p\geq 1$, the $L_p{}^\gamma(\Omega_+)$-norm of $f$ is defined by
$$
\|f\|_{L_p{}^\gamma(\Omega_+)}=\Big(\int_{\Omega_+}|f(x)|^p x{}^\gamma dx\Big)^{1/p}.
$$

The weighted measure of $\Omega_+$ is denoted by
  ${\operatorname{meas}}_\gamma(\Omega)$ and is defined by
$$
{\operatorname{meas}}_\gamma(\Omega_+)=\int_{\Omega_+}x{}^\gamma dx.
$$
For every measurable function $f(x)$ defined on $\mathbb{R}^n_+$ we consider
$$
\mu_\gamma(f,t)={\operatorname{meas}}_\gamma\{x\in\mathbb{R}^n_+:
|f(x)|>t\}
=\int_{\{x:\,\,|f(x)|>t\}^+}x{}^\gamma dx
$$
where $\{x:|f(x)|>t\}^+=\{x{\in}\mathbb{R}^n_+:|f(x)|>t\}$.
 We will call the function   $\mu_\gamma=\mu_\gamma(f,t)$ a
{\it weighted distribution function} $|f(x)|$.

The space $L_\infty{}^\gamma(\Omega_+)$ is defined as a set of measurable
on  $\Omega_+$ and even with respect to each variable functions $f(x)$ such as
$$
\|f\|_{L_\infty{}^\gamma(\Omega_+)}=\operatorname{ess\,sup}_{x\in \Omega_+, \gamma}
 |f(x)|
=\inf_{a\in\Omega_+}\{\mu_\gamma(f,a)=0\}<\infty.
$$
For $1\leq p\leq\infty$  the $L_{p,{\rm loc}}{}^\gamma(\Omega_+)$ is the set
of functions $u(x)$ defined almost everywhere in $\Omega_+$ such that
$uf\in L_{p}{}^\gamma(\Omega_+)$ for any
$f\in\mathring{C}^\infty_{ev}(\overline{\Omega}_+)$.
Each function $u(x)\in L_{1,loc}{}^\gamma(\Omega_+)$ will be identified with
the functional $u\in \mathcal{D}_+'(\overline{\Omega}_+)$
acting according to the formula
\begin{equation}\label{RegDist}
(u,f)_\gamma=\int_{\mathbb{R}^n_+} u(x)\,f(x)\,x{}^\gamma\, dx,\quad
f\in \mathring{C} ^\infty_{ev}(\overline{\mathbb{R}}^n_+).
\end{equation}
Functionals $u\in \mathcal{D}_+'(\overline{\Omega}_+)$ acting by the formula
\eqref{RegDist} will be called  \emph{regular weighted functionals}.
All other functionals $u\in \mathcal{D}_+'(\overline{\Omega}_+)$ will be called
\emph{singular weighted functionals}.

We will use the regular weighted functional $(t^2-|x|^2)_{+,\gamma}^\lambda$
defined by
\begin{equation}\label{WF}
((t^2-|x|^2)_{+,\gamma}^\lambda,\varphi)_\gamma
=\int_{\{x{\in}\mathbb{R}^n_+:|x|<t\}}(t^2-|x|^2)^\lambda\varphi(x)x{}^\gamma
dx,\quad \varphi\in S_{ev},\quad \lambda{\in}\mathbb{C}.
\end{equation}

The symbol $j_\nu$ is  used for the normalized Bessel function:
$$
j_\nu(t)=\frac{2^\nu\Gamma(\nu+1)}{t^\nu}J_\nu(t),
$$
where  $J_{\nu}(t)$ is the Bessel function of the first kind of order
$\nu$ (see \cite{Watson}). The function $j_\nu(t)$ is even by $t$.

We use the multidimensional  Hankel (Fourier-Bessel)  transform.
The multidimensional Hankel transform  of a function $f(x)$ is given by
 (see \cite{Bateman}):
$$
F_B[f](\xi)=(F_B)_x[f(x)](\xi)=\widehat{f}(\xi)
=\int_{\mathbb{R}^n_+}f(x)\,\mathbf{j}_\gamma(x;\xi)x{}^\gamma dx,
$$
where
$$
\mathbf{j}_\gamma(x;\xi)
=\prod_{i=1}^n j_{\frac{\gamma_i-1}{2}}(x_i\xi_i),\quad
\gamma_1>0,\dots,\gamma_n>0.
$$
For $f\in S_{ev}$ inverse multidimensional Hankel transform is defined by
$$
F^{-1}_B[\widehat{f}(\xi)](x)=f(x)=\frac{2^{n-|\gamma|}}{\prod_{j=1}^n\,
\Gamma^2\left(\frac{\gamma_j{+}1}{2}\right)}\int_{\mathbb{R}^n_+}
\mathbf{j}_\gamma(x,\xi)\widehat{f}(\xi)\xi{}^\gamma\,d\xi.
$$

We will deal with the singular Bessel differential operator $B_{\nu}$
(see, for example, \cite[p. 5]{Kipr}):
$$
(B_{\nu})_{t}=\frac{\partial^2}{\partial t^2}
+\frac{\nu}{t}\frac{\partial}{\partial t}
=\frac{1}{t^{\nu}}\frac{\partial}{\partial t}t^{\nu}
 \frac{\partial}{\partial t},\quad t>0.
$$
and the elliptical singular operator or the Laplace-Bessel operator
 $\Delta_\gamma$:
\begin{equation}\label{LapBess}
\Delta_\gamma
=(\Delta_\gamma)_x=\sum_{i=1}^{n} (B_{\gamma_i})_{x_i}
=\sum_{i=1}^{n}\Big( \frac{\partial ^{2} }{\partial x_{i}^{2} }
 +\frac{\gamma _{i} }{x_{i} } \frac{\partial}{\partial x}\Big)
=\sum_{i=1}^{n}\frac{1}{x_i^{\gamma_i}}\frac{\partial}{\partial
x_i}x_i^{\gamma_i}\frac{\partial}{\partial x_i}.
\end{equation}
The operator \eqref{LapBess} belongs to the class of B-elliptic operators
by  Kipriyanovs' classification  (see \cite{Kipr}).

The \emph{B--polyharmonic of order} $p$ function $f=f(x)$ is the function
$f{\in}C^{2p}_{ev}(\overline{\mathbb{R}}\,_n^+)$ such that
\begin{equation}\label{PolG1}
\Delta^p_\gamma f=0,
\end{equation}
where $\Delta_\gamma$ is operator \eqref{LapBess}.
The operator \eqref{PolG1} was considered in \cite{Kipr}.
The B-polyharmonic of order $1$ function we will call B-harmonic.

Using \cite[formulas 9.1.27]{AbramowitzStegunSpF} we obtain
\begin{equation}\label{BessBess}
 (B_{\nu})_t {j}_{\frac{\nu-1}{2}}(\tau t)=-\tau^2{j}_{\frac{\nu-1}{2}}(\tau t).
\end{equation}
We will use the generalized convolution operator defined  by
$$
(f*g)_\gamma=\int_{\mathbb{R}^n_+}f(y)({}^\gamma T^yg)(x)y{}^\gamma dy,
$$
where ${}^\gamma T^y$ is multidimensional generalized translation
$$
{}^\gamma T^y=^{\gamma_1} T_{x_1}^{y_1}\dots^{\gamma_n}T_{x_n}^{y_n},
$$
each one-dimensional operator $^{\gamma_i} T_{x_i}^{y_i}$, $i=1,\dots,n$
acts according to (see \cite{Levitan})
\begin{align*}
&^{\gamma_i} T_{x_i}^{y_i}f(x) \\
&=\frac{\Gamma(\frac{\gamma_i+1}{2})}
{\Gamma(\frac{\gamma_i}{2})\Gamma(\frac{1}{2})}
\int_0^\pi f(x_1,\dots,x_{i-1},\sqrt{x_i^2+y_i^2-2x_iy_i\cos\alpha_i},
x_{i+1},\dots,x_n) \\
&\quad\times \sin^{\gamma_i-1}\alpha_i\,d\alpha_i\,.
\end{align*}
Based on the multidimensional generalized translation ${}^\gamma T^y$
the weighted spherical mean $M^\gamma_t[f(x)]$ of a suitable function
is defined by the formula
\begin{equation}\label{05}
M^\gamma_t[f(x)]=\frac{1}{|S_1^+(n)|_\gamma}\int_{S^+_1(n)}
{}^\gamma T_x^{t\theta}f(x) \theta{}^\gamma dS,
\end{equation}
where
\[
\theta{}^\gamma=\prod_{i=1}^{n}\theta_i^{\gamma_i},\quad
S^+_1(n)=\{\theta:|\theta|=1,\theta{\in}\mathbb{R}^n_+\}, \quad
|S^+_1(n)|_\gamma=\frac{\prod_{i=1}^n{\Gamma\left(\frac{\gamma_i{+}1}{2}\right)}}
{2^{n-1}\Gamma\big(\frac{n{+}|\gamma|}{2}\big)}.
\]
It is easy to see that
\begin{equation}\label{MeanCond}
  M^\gamma_0[f(x)]=f(x),\quad
\frac{\partial}{\partial t}M^\gamma_t[f(x)]\Big|_{t=0}=0.
\end{equation}

\begin{lemma} \label{lem2.1}
Let $u\in S_{ev}$ then
\begin{equation}\label{SaprBessel}
F_B[\Delta_\gamma f](\xi)=-|\xi|^2F_B[f](\xi).
\end{equation}
\end{lemma}

\begin{proof}
We have
\begin{align*}
F_B[\Delta_\gamma f](\xi)
&=\int_{\mathbb{R}^n_+}[\Delta_\gamma f(x)]\,\mathbf{j}_\gamma(x;\xi)x{}^\gamma dx\\
&=\sum_{i=1}^{n}\int_{\mathbb{R}^n_+}
\Big[\frac{1}{x_i^{\gamma_i}}\frac{\partial}{\partial
x_i}x_i^{\gamma_i}\frac{\partial}{\partial x_i}f(x)\Big]
 \mathbf{j}_\gamma(x;\xi)x{}^\gamma dx.
\end{align*}
Integrating by parts by variable $x_i$ and using formula \eqref{BessBess}, we obtain
\begin{align*}
F_B[\Delta_\gamma f](\xi)
&=\sum_{i=1}^{n}\int_{\mathbb{R}^n_+} f(x)
\Big[\frac{1}{x_i^{\gamma_i}}\frac{\partial}{\partial
x_i}x_i^{\gamma_i}\frac{\partial}{\partial x_i}\mathbf{j}_\gamma(x;\xi)\Big]
 x{}^\gamma dx \\
&=\sum_{i=1}^{n} (-\xi_i^2)\int_{\mathbb{R}^n_+} f(x)\mathbf{j}_\gamma(x;\xi)
 x{}^\gamma dx  \\
&=-|\xi|^2\int_{\mathbb{R}^n_+} f(x)\,\mathbf{j}_\gamma(x;\xi)x{}^\gamma dx
 =-|\xi|^2F_B[f](\xi).
\end{align*}
\end{proof}

\begin{lemma} \label{lem2.2}
We have the  formula
\begin{equation}\label{FBOF}
\frac{(F_B)_x(t^2-|x|^2)_{+,\gamma}^{\frac{k-n-|\gamma|-1}{2}}}
{\Gamma\big(\frac{k-n-|\gamma|+1}{2}\big)}
=\frac{t^{{k-1}}\prod^n_{i=1}\Gamma\left(\frac{\gamma_i+1}{2}\right)}
{ 2^n\Gamma\left(\frac{k+1}{2}\right)}j_{\frac{k-1}{2}}(t|x|),
\end{equation}
where $(t^2-|x|^2)_{+,\gamma}^{\frac{k-n-|\gamma|-1}{2}}$ is defined by \eqref{WF}.
\end{lemma}

Formula \eqref{FBOF} is obtained similarly as \cite[(5) p. 291]{Gelfand}.


\begin{lemma} \label{lem2.3}
Let $u =u(x,t;k)$ denote the solution to \eqref{EPD0}. Then the solution
satisfies the next two important recursion formulas:
\begin{gather} \label{Rec1}
u(x,t;k)=t^{1-k}u(x,t;2-k), \\
\label{Rec2} u_t(x,t;k)=tu(x,t;2+k).
\end{gather}
\end{lemma}

This is  particular cases of  Weinstein's formulas which state that
for any equation of the form $u_{tt}+\frac{k}{t}u_t=X(u)$, in which $X$
is an operator does not depend on $t$ the relations
\eqref{Rec1} and \eqref{Rec2} hold (see \cite{Fox}).

\begin{lemma} \label{lem2.4}
 The weighted spherical mean $M^\gamma_t[f(x)]$ is the transmutation operator
(see the definition of transmutation operators in \cite{Car1,Car2,Car3}
or \cite{Sit2,Sit1}) intertwining $(\Delta_\gamma)_x$ and
$(B_{n+|\gamma|-1})_t$ for the
$f\in C^2_{ev}$:
\begin{equation}\label{OPPr}
(B_{n+|\gamma|-1})_tM^\gamma_t[f(x)]=M^\gamma_t[(\Delta_\gamma)_xf(x)].
\end{equation}
\end{lemma}

\begin{proof}
First of all we note that the  function
$f\in C^2_{ev}$  satisfies the equation
 \begin{equation}\label{EQ15}
\int_{B_t^+(n)}\,f(x)\,x{}^\gamma\, dx
=\int_0^t \lambda^{n+|\gamma|-1}\,d\lambda \int_{S_1^+(n)}
f(\lambda\theta )\theta{}^\gamma dS_\theta,
 \end{equation}
which can be easy obtained by  passing to spherical coordinates
$x=\lambda \theta$, $|\theta|=1$ in the left hand of \eqref{EQ15}.
From \eqref{EQ15} we obtain
\begin{equation}\label{EQ16}
\begin{aligned}
&|S^+_1(n)|_\gamma\int_0^t \lambda^{n+|\gamma|-1}M^\gamma_\lambda[f(x)] d\lambda\\
&= \int_0^t \lambda^{n+|\gamma|-1}d\lambda
 \int_{S^+_1(n)}(T^{\lambda y}f)(x)\,\,y{}^\gamma dS_y\\
&=\int_{B^+_r(n)}(T^{z}f)(x)z{}^\gamma dz.
\end{aligned}
\end{equation}
Let us apply the operator $\Delta_{\gamma}$ to both sides of  \eqref{EQ16}
 with respect to $x$, then we obtain
\begin{equation}\label{EQ17}
\begin{aligned}
&|S^+_1(n)|_\gamma \sum_{i=1}^n\int_0^t \lambda^{n+|\gamma|-1}B_{\gamma_i}
M^\gamma_\lambda[f(x)]d\lambda \\
&=\sum_{i=1}^n\int_{B^+_t(n)}(B_{\gamma_i})_{x_i}(T^{z}f)(x)z{}^\gamma dz \\
&=\sum_{i=1}^n\int_{B^+_t(n)}(B_{\gamma_i})_{z_i}T_x^{z}f(x)z{}^\gamma dz.
\end{aligned}
\end{equation}
We have the  Green formula
\begin{equation}\label{EQ18}
\int_{\overline{\Omega}^+} (v\Delta_\gamma w-w\Delta_\gamma v)x{}^\gamma dx
=\int_{\Gamma=\partial\overline{\Omega}^+}
\Big(v \frac{\partial w}{\partial \vec{\nu}}-
w \frac{\partial v}{\partial \vec{\nu}}\Big) x{}^\gamma\,d\Gamma_x\,,
\end{equation}
where $w,v\in C^2_{ev}(\overline{\Omega}^+)$, $\vec{\nu}$ is the outward
normal to the boundary $\Gamma=\partial\overline{\Omega}^+$ of the
$\overline{\Omega}^+$.
This formula was presented in \cite{Shi5}.

 By applying formula  \eqref{EQ18} to the right-hand side of  \eqref{EQ17}, we obtain
$$
\sum_{i=1}^n\int_{B^+_t(n)}(B_{\gamma_i})_{z_i}(T^{z}f)(x)z{}^\gamma dz
=\sum_{i=1}^n\int_{S^+_t(n)}\frac{\partial}{\partial z_i}(T^{z}f)(x)
\cos(\vec{\nu},\vec{e}_i)z{}^\gamma \,dS_z,
$$
where $\vec{e}_i$ is the direction of the axis $Oz_i$, $i=1,\dots,n$.

Now, by using the fact that the direction of the outward normal to the boundary
 of a ball with center the origin coincides with the direction of the position
vector of the point on the ball, we obtain the relation
\begin{align*}
\sum_{i=1}^n\int_{B^+_t(n)}(B_{\gamma_i})_{z_i}(T^{z}f)(x)z{}^\gamma dz
&= t^{n+|\gamma|-1}\int_{S^+_1(n)}\frac{\partial}{\partial t}(T^{t\theta}f)(x)
\theta{}^\gamma dS_\theta \\
&= |S_1^+(n)|_\gamma t^{n+|\gamma|-1}\frac{\partial}{\partial t}M^\gamma_t [f(x)].
\end{align*}
Returning to \eqref{EQ17}, we obtain
\begin{equation}\label{EQ19}
\sum_{i=1}^n\int_0^t \lambda^{n+|\gamma|-1}B_{\gamma_i}M^\gamma_\lambda[f(x)]d\lambda
=t^{n+|\gamma|-1}\frac{\partial}{\partial t}M^\gamma_t[f(x)].
\end{equation}
By differentiating relation \eqref{EQ19} with respect to $t$, we obtain
$$
\sum_{i=1}^n t^{n+|\gamma|-1}B_{\gamma_i}M^\gamma_t [f(x)]
=(n+|\gamma|-1)t^{n+|\gamma|-2}\frac{\partial}{\partial t}M^\gamma_t
[f(x)]+t^{n+|\gamma|-1}\frac{\partial^2}{\partial t^2}M^\gamma_t [f(x)]
$$
or
$$
\sum_{i=1}^n B_{\gamma_i}M^\gamma_t[ f(x)]
=\frac{n+|\gamma|-1}{t}\frac{\partial}{\partial t}M^\gamma_t[f(x)]
+\frac{\partial^2}{\partial t^2}M^\gamma_t[f(x)],
$$
and so
\begin{equation}\label{Eq20}
(\Delta_\gamma)_x M^\gamma_t[ f(x)]
=\frac{n+|\gamma|-1}{t}\frac{\partial}{\partial t}M^\gamma_t [f(x)]
+\frac{\partial^2}{\partial t^2}M^\gamma_t[ f(x)].
\end{equation}

Now let us consider $(\Delta_\gamma)_x M^\gamma_t[f(x)]$.
Using the commutativity of $B_{\gamma_i}$ and $T_{x_i}^{t\theta_i}$
(see \cite{Kipr}) we obtain
\begin{align*}
(\Delta_\gamma)_x M^\gamma_t[f(x)]
&= \frac{1}{|S_1^+(n)|_\gamma}\,(\Delta_\gamma)_x
\int_{S^+_1(n)}{}^\gamma T_x^{t\theta}f(x) \theta{}^\gamma dS_\theta \\
&=\frac{1}{|S_1^+(n)|_\gamma}\int_{S^+_1(n)}{}^\gamma T_x^{t\theta}
 [(\Delta_\gamma)_x f(x)]
\theta{}^\gamma dS_\theta= M^\gamma_t [(\Delta_\gamma)_x f(x)].
\end{align*}
which with \eqref{Eq20} gives \eqref{OPPr}.
\end{proof}

A similarly proof can be found in \cite{Shi5}.

\section{Transmutation method}

An important and powerful approach to differential equations with Bessel
 operators is based on application of the transmutation theory.
This method is essential in the study of singular problems with use of special
classes of transmutations such as Sonine, Poisson, Buschman-Erd\'elyi ones
and different forms of fractional integrodifferential operators,
cf. \cite{Car1}--\cite{Castilio1}, \cite{Sit8}--\cite{Sit7},
\cite{Kra}--\cite{Kra3}, \cite{Sit5}--\cite{Sit3}.

In this section we show how the transmutation method can be used to find
the solution of the Cauchy problem for the general Euler-Poisson-Darboux equation
\begin{gather} \label{GEQ142}
 (B_k)_tu=(\Delta _{\gamma })_x u, \quad u=u(x,t;k),\quad
 x\in\mathbb{R}^n_+,\; t>0,\; k\in\mathbb{R}, \\
\label{GEQ143} u(x,0;k)=f(x),\quad  u_{t} (x,0;k)=0.
\end{gather}

\begin{theorem} \label{thm3.1}
Let $f=f(x)$, $x\in\mathbb{R}^n_+$ be twice continuous differentiable function
even with respect of each variable. Then for the case
$k> n+|\gamma |-1$ the function
 \begin{equation}\label{Sol1}
u(x,t;k)=
\frac{2^{n}\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k{-}n{-}|\gamma|{+}1}{2}\big)
\prod_{i=1}^n{\Gamma\left(\frac{\gamma_i{+}1}{2}\right)}}
\int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)](1{-}|y|^2)^{\frac{k{-}n{-}
 |\gamma|{-}1}{2}}y{}^\gamma \,dy
\end{equation}
is the solution to  problem \eqref{GEQ142}--\eqref{GEQ143}.
The solution to  \eqref{GEQ142}--\eqref{GEQ143} for the case
$k=n{+}|\gamma |{-}1$ is the weighted spherical mean  $M^\gamma_t[f(x)]$.
\end{theorem}

\begin{proof}
Using Lemma \ref{lem2.4} we obtain  that the  weighted spherical mean
 of any twice continuously differentiable function
$f=f(x)$ even with respect to each of the independent variables
$x_{1} ,\ldots ,x_{n} $ on $\mathbb{R}_{+}^n$ satisfies the general
 Euler-Poisson-Darboux equation
$$
(B_{k} )_{t} M^\gamma_t[f(x)]=(\Delta _{\gamma } )_{x} M^\gamma_t[f(x)]{\kern 1pt} ,
\quad k=n+|\gamma |-1{\kern 1pt}
$$
and initial conditions (see \eqref{MeanCond})
$$
M^\gamma_0[f(x)]=f(x),\quad M^\gamma_t[f(x)]\Big|_{t=0}=0.
$$
It means the the weighted spherical mean $M^\gamma_t[f(x)]$ is the solution
to the problem \eqref{GEQ142}-\eqref{GEQ143} for the $k=n+|\gamma |-1$.

To obtain the solution to \eqref{GEQ142}--\eqref{GEQ143}
for   $k>n+|\gamma |-1$ we will use the method of descent.
First, we will seek solution to the Cauchy problem \eqref{GEQ142}--\eqref{GEQ143}
for the case $k>n+|\gamma|$.

Let $\gamma'=(\gamma_1,\dots,\gamma_n,\gamma_{n+1}')$, $\gamma_{n+1}'>0$,
 $x'=(x_{1} ,\dots,x_{n+1})\in\mathbb{R}^{n+1}_+$
and
$$
(\Delta_{\gamma'})_{x'}=(B_{\gamma_1})_{x_{1} } +\dots+(B_{\gamma_n})_{x_{n} }
+(B_{\gamma_{n+1}'})_{x_{n+1} }.
$$
 Consider the equation of type \eqref{GEQ142},
$$
(B_k)_tu=(\Delta_{\gamma'})_{x'}u,\quad u=u(x',t;k),\quad x'\in\mathbb{R}^{n+1}_+,\;
t>0
$$
with the initial conditions
$$
u(x',0;k)=f_{1}(x') ,\quad  u_{t} (x',0;k)=0.
$$
When $k=n+|\gamma'|=n+|\gamma|+\gamma_{n+1}'$ the
 weighted spherical mean $M^\gamma_t[f_1(x')]$  is a solution to this
Cauchy problem:
\begin{gather} \label{GEQ144}
\begin{aligned}
&u(x',t;k) \\
&=\frac{1}{|S_{1}^{+} (n+1)|_{\gamma'} }
 \int_{S_{1}^{+} (n+1)} [^{\gamma_1}T^{ty_{1} } \dots {}^{\gamma_n}
T^{ty_{n} \gamma_{n+1}'}T^{ty_{n+1} } f_1(x)] (y')^{\gamma'} dS_{y'},
\end{aligned} \\
 y'=(y_1,\dots,y_n,y_{n+1}') \in\mathbb{R}^{n+1}_+, \nonumber \\
 |S_{1}^{+} (n+1)|_{\gamma'}
=\frac{\prod_{i=1}^n{\Gamma\big(\frac{\gamma_i{+}1}{2}\big)}
\Gamma\big(\frac{\gamma_{n+1}'{+}1}{2}\big)}
{2^{n}\Gamma\big(\frac{n{+}1{+}|\gamma|+\gamma_{n+1}'}{2}\big)}
=\frac{\prod_{i=1}^n{\Gamma\big(\frac{\gamma_i{+}1}{2}\big)}
\Gamma\big(\frac{k-n-|\gamma|{+}1}{2}\big)}{2^{n}\Gamma\left(\frac{k+1}{2}\right)}.
\nonumber
\end{gather}
Let us put $f_{1} (x_{1} ,\dots,x_{n} ,0)=f(x_{1} ,\dots,x_{n} )$, where
 $f$ is the function which appears in  initial conditions \eqref{GEQ143}.
In this way the $u$ defined by \eqref{GEQ144} becomes a function only
of $x_{1} ,\dots,x_{n} $ which satisfies equation \eqref{GEQ142} and  initial
conditions \eqref{GEQ143}.
We have
$$
u(x,t;k)=\frac{1}{|S_{1}^{+} (n+1)|_{\gamma'} } \int_{S_{1}^{+} (n+1)}
 [^{\gamma}T^{ty}f(x)](y')^{\gamma'} dS_{y'},\quad\gamma_{n+1}'=k-n-|\gamma|.
$$
Now we rewrite the integral over the part of the sphere $S_{1}^{+} (n+1)$
 as an integral over the part of ball
$B_{1}^{+} (n)=\{y{\in}\mathbb{R}^{n}_+:\sum_{i=1}^{n} y_i^2\leq 1\}$.
We write the surface integral over multiple integral:
\begin{align*}
\int_{S_{1}^{+} (n+1)} [^{\gamma}T^{ty}f(x)](y')^{\gamma'} dS_{y'}
&=\int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)](1-y_1^2-\dots
-y_{n}^2)^{\frac{\gamma_{n+1}'-1}{2}}y{}^\gamma dy \\
&=\int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)](1-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}
 y{}^\gamma dy,
\end{align*}
where $B_{1}^{+} (n)$ is a projection of the  $S_{1}^{+} (n+1)$ on the
equatorial plane $x_{n+1}=0$.
We have
\[ %\label{Sol1b}
u(x,t;k)=\frac{2^{n}\Gamma\left(\frac{k+1}{2}\right)}
{\prod_{i=1}^n{\Gamma\left(\frac{\gamma_i{+}1}{2}\right)}
\Gamma\big(\frac{k{-}n{-}|\gamma|{+}1}{2}\big)}
 \int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)](1{-}|y|^2)
^{\frac{k{-}n{-}|\gamma|{-}1}{2}}y{}^\gamma dy.
\]


Although \eqref{Sol1} was obtained as the solution to  \eqref{GEQ142}--\eqref{GEQ143}
for the case $k>n{+}|\gamma |$ the integral on its right-hand side converges
and for $k>n+|\gamma |-1$. We can verify by direct substitution
 \eqref{Sol1} in \eqref{GEQ142}--\eqref{GEQ143}  that \eqref{Sol1} satisfies
to the differential equation \eqref{GEQ142} and to the initial conditions
\eqref{GEQ143} for all values of $k$ which are greater than $n+|\gamma |-1$.
Let show it. Changing coordinates from $y$ to $y/t$ and using that
 $(B_{\gamma_i})_{x_i}^{\gamma_i}T_{x_i}^{y_i}
=(B_{\gamma_i})_{y_i}^{\gamma_i}T_{x_i}^{y_i}$ (see \cite{Levitan}) we obtain
\begin{equation}\label{RH}
\begin{aligned}
I&=(\Delta_\gamma)_x\int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)](1-|y|^2)
 ^{\frac{k-n-|\gamma|-1}{2}}y{}^\gamma dy \\
&=\sum_{i=1}^n (B_{\gamma_i})_{x_i}\int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)]
 (1-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}y{}^\gamma dy \\
&=t^{1-k}\sum_{i=1}^n \int_{B_{t}^{+} (n)}[(B_{\gamma_i})_{x_i}^{\gamma}T^{y}f(x)]
 (t^2-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}y{}^\gamma dy \\
&=t^{1-k}\sum_{i=1}^n \int_{B_{t}^{+} (n)}[(B_{\gamma_i})_{y_i}^{\gamma}
 T^{y}f(x)](t^2-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}y{}^\gamma dy,
\end{aligned}
\end{equation}
where $B_{t}^{+} (n)=\{y{\in}\mathbb{R}^{n}_+:\sum_{i=1}^{n} y_i^2\leq t\}$.

For  integration over  $\overline{\Omega}^+$ the functions
 $w,v{\in}C^2_{ev}(\overline{\Omega}^+)$ we have the Green formula \eqref{EQ18}.
By applying  \eqref{EQ18} to the right-hand side of  \eqref{RH},
we obtain
$$
I=t^{1-k}\sum_{i=1}^n\int_{S^+_t(n)}
\big[\frac{\partial}{\partial y_i}^{\gamma}T^{y}f(x)\big]
(t^2-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}\cos(\vec{\nu},\vec{e}_i)\,y{}^\gamma \,dS,
$$
where $\vec{e}_i$ is the direction of the axis $Oy_i$, $i = 1,\dots,n$, and
thus $\cos(\vec{\nu},\vec{e}_i)=\frac{y_i}{t}$.
Now, by using the fact that the direction of the outward normal to the boundary
of a ball with center the origin coincides with the direction of the position
vector of the point on the ball, we obtain the relation
$$
I=\frac{1}{t^k}\frac{\partial}{\partial t}t^k\frac{\partial}{\partial t}
\int_{B^+_1(n)}\left[^{\gamma}T^{ty}f(x)\right](1-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}
\,y{}^\gamma \,dy.
$$
Given  \eqref{RH} and that $\frac{1}{t^k}\frac{\partial}{\partial t}
t^k\frac{\partial}{\partial t}=(B_k)_t$, we have
\begin{align*}
&(\Delta_\gamma)_x\int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}f(x)]
(1-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}y{}^\gamma dy \\
&=(B_k)_t\int_{B^+_1(n)}\left[^{\gamma}T^{ty}f(x)\right]
(1-|y|^2)^{\frac{k-n-|\gamma|-1}{2}}\,y{}^\gamma \,dy.
\end{align*}
It means that $u(x,t;k)$ defined by the formula \eqref{Sol1} indeed
satisfies to equation \eqref{GEQ142}  for $k>n+|\gamma |-1$.
Validity of the first and the second initial conditions follows from
\cite[(5.20) and (5.21)]{Levitan} respectively.
\end{proof}

\begin{theorem} \label{thm3.2}
 Let $f=f(x)$, $f\in C^{[\frac{n+|\gamma|-k}{2}]+2}_{ev}$.
Then for $k<n+|\gamma |-1$, $k\neq -1,-3,-5,\dots$ the function
\begin{equation}\label{34}
u(x,t;k)=t^{1-k}\Big(\frac{\partial}{t \partial t}\Big)^m(t^{k+2m-1}u(x,t;k+2m)),
\end{equation}
is the solution to  \eqref{GEQ142}--\eqref{GEQ143},
where $m$ is a minimum integer such that $m\geq \frac{n+|\gamma|-k-1}{2}$
and $u(x,t;k+2m)$ is the solution to the Cauchy problem
\begin{gather} \label{32}
   (B_{k+2m})_t u=(\Delta_\gamma)_x u, \\
\label{33}
u(x,0;k+2m)=\frac{f(x)}{(k+1)(k+3){\dots}(k+2m-1)},\quad u_t(x,0;k+2m)=0.
\end{gather}
\end{theorem}

\begin{proof}
 To proof that \eqref{34} is a solution of \eqref{GEQ142}--\eqref{GEQ143}
when $k<n+|\gamma|{-}1$, $k\neq -1$, $-3$, $-5$, \dots, we  use the
recursion formulas \eqref{Rec1} and \eqref{Rec2}. Let choose minimum integer
 $m$ such that $k+2m\geq n+|\gamma|-1$. Now we can write the solution of the
Cauchy problem
\begin{gather*}
   (B_{k+2m})_t u=(\Delta_\gamma)_x u, \\
u(x,0;k+2m)=g(x),\quad u_t(x,0;k+2m)=0,\quad g\in C^2_{ev}
\end{gather*}
by \eqref{Sol1}. We have
\begin{align*}
 &u(x,t;k+2m) \\
 &=\frac{2^{n}\Gamma\left(\frac{k+2m+1}{2}\right)}
 {\prod_{i=1}^n{\Gamma\left(\frac{\gamma_i{+}1}{2}\right)}\Gamma\big(\frac{k+2m-n-|\gamma|{+}1}{2}\big)}
 \int_{B_{1}^{+} (n)}[^{\gamma}T^{ty}g(x)](1-|y|^2)
 ^{\frac{k+2m-n-|\gamma|-1}{2}} y{}^\gamma dy,
\end{align*}
and, using \eqref{Rec1} we obtain
$$
t^{k+2m-1}u(x,t;k+2m)=u(x,t;2-k-2m).
$$
Applying \eqref{Rec2} to the last formula  $m$ times we obtain
$$
\Big(\frac{\partial}{t \partial t}\Big)^m(t^{k+2m-1}u(x,t;k+2m)=u(x,t;2-k).
$$
Applying again \eqref{Rec1} we can write
\begin{equation}\label{34b}
u(x,t;k)=t^{1-k}\Big(\frac{\partial}{t \partial t}\Big)^m(t^{k+2m-1}u(x,t;k+2m)),
\end{equation}
which gives the solution to \eqref{32}. Now we obtain the function $g$ such
that the \eqref{33} is true.
From \eqref{34} we have asymptotic relation
\[
u(x,t;k)=(k+1)(k+3)\dots(k+2m-1)u(x,t;k+2m)+Ctu(x,t;k+2m)+O(t^2),
\]
as $t\to 0$, where $C$ is a constant. Therefore, if
$$
g(x)=\frac{f(x)}{(k+1)(k+3){\dots}(k+2m-1)}
$$
then $u(x,t;k)$ defined by \eqref{34} satisfies to the initial conditions
\eqref{GEQ143}.

Let us recall that for $u(x,t;k+2m)$ to be a solution to \eqref{32}--\eqref{33}
it is sufficient that $f\in C^2_{ev}$.
 To be able to carry out the construction \eqref{34}, it is sufficient to
require that
$f\in C^{\left[\frac{n+|\gamma|-k}{2}\right]+2}_{ev}$.
\end{proof}

\begin{theorem} \label{thm3.3}
If $f$ is B-polyharmonic of order $\frac{1-k}{2}$ and even with respect to
each variable then one of the solutions to the Cauchy problem
\eqref{32}--\eqref{33}  for the  $k=-1,-3,-5,\dots$ is given by
 \begin{gather}\label{Ex1}
 u(x,t;k)=f(x),\quad k=-1, \\
\label{Ex2}
 u(x,t;k)=f(x)+\sum_{h=1}^{-\frac{k+1}{2}}\frac{\Delta^h_\gamma f}{(k+1)
\dots(k+2h-1)}\frac{t^{2h}}{2\cdot 4\cdots 2h},\quad k=-3,-5,\dots
\end{gather}
 \end{theorem}

\begin{proof}
Let us first take $k=-1$ and assume that
$\lim_{t\to 0}\frac{\partial^2u(x,t;-1)}{\partial t^2}$ exists. Let $t\to 0$ in
$$
(\Delta_\gamma)_x u^{-1}(x,t;k)=\frac{\partial^2 u(x,t;-1)}{\partial t^2}
-\frac{1}{t}\frac{\partial u(x,t;-1)}{\partial t},
$$
i.e.,
$$
(\Delta_\gamma)_x u(x,0;-1)=\lim_{t\to 0}\frac{\partial^2 u(x,t;-1)}{\partial t^2}{-}
\lim_{t\to 0}\frac{1}{t}\frac{\partial u(x,t;-1)}{\partial t}=0.
$$
We find that  $(\Delta_\gamma)_x u(x,0;-1)=0$ which shows that $f$ must be
B-harmonic. So the function $f$ satisfies \eqref{32}--\eqref{33}
 for the  $k={-}1$.

When $k=-3$ we have
$$
\lim_{t\to 0}\frac{\partial^2 u(x,t;-3)}{\partial t^2}
=\lim_{t\to 0}\frac{1}{t}\frac{\partial u(x,t;-3)}{\partial t}.
$$
From the general form of the  Euler-Poisson-Darboux equation for $k=-3$ we obtain
\begin{align*}
\lim_{t\to 0}(\Delta_\gamma)_x u(x,t;-3)
&=\lim_{t\to 0}\frac{\partial^2 u(x,t;-3)}{\partial t^2}
-3\lim_{t\to 0}\frac{1}{t}\frac{\partial u(x,t;-3)}{\partial t} \\
&=-2\lim_{t\to 0}\frac{1}{t}\frac{\partial u(x,t;-3)}{\partial t}.
\end{align*}
It is follows from \eqref{Rec2} that
$$
\frac{1}{t}\frac{\partial u(x,t;-3)}{\partial t}=u(x,t;-1)
$$
hence
 \begin{equation}\label{35}
   \lim_{t\to 0}(\Delta_\gamma)_x u(x,t;-3)=-2u(x,0;-1).
 \end{equation}
If  $\lim_{t\to 0}\frac{\partial^4 u(x,t;-3)}{\partial t^4}$ exists
 and all odd derivatives of $u(x,t;-3)$ tend to zero when $t\to 0$, then
 $\lim_{t\to 0}\frac{\partial^2 u(x,t;-1)}{\partial t^2}$ also exists.
Therefore, $\lim_{t\to 0}(\Delta_\gamma)_x u(x,t;-1)=0$  and by \eqref{35}
we have $\lim_{t\to 0}(\Delta_\gamma)_x^2 u(x,t;-3)=0$. This remark can be
easily generalized to include all  exceptional values.
So, in this case a solution to the Cauchy problem for the general form of
the Euler-Poisson-Darboux equation for the case  $k={-}3,{-}5,{\dots}$
is given by the formula
 $$
 u(x,t;k)=f(x)+\sum_{h=1}^{-\frac{k+1}{2}}\frac{\Delta^h_\gamma f}{(k+1)
\dots(k+2h-1)}\,\frac{t^{2h}}{2\cdot 4\cdots 2h},\quad k=-3,-5,\dots
 $$
 and as we proved earlier $ u(x,t;-1)=f(x)$.
\end{proof}


\section{Solution of the  singular Cauchy problem using the Hankel transform}

In this section we look for the solution
$u\in S_{ev}'(\mathbb{R}^{n}_{+})\times C^2(0,\infty)$ to the problem
\begin{gather} \label{GEQ1421}
(B_k)_tu=(\Delta _{\gamma })_x u, \quad u=u(x,t;k),\quad x\in\mathbb{R}^n_+,\; t>0, \\
\label{GEQ1431} u(x,0;k)=f(x),\quad u_{t} (x,0;k)=0.
\end{gather}
when $f(x)\in S_{ev}'(\mathbb{R}^{n}_{+})$,
$k\in\mathbb{R}\setminus\{-1,-3,-5,\dots\}$.

The notation $u\in S_{ev}'(\mathbb{R}^{n}_{+})
\times C^2(0,\infty)$ means that $u(x,t;k)$ belongs to
$S_{ev}'(\mathbb{R}^{n}_{+})$ by variable $x$ and belongs to $C^2(0,\infty)$
by variable $t$.

\begin{theorem} \label{thm4.1}
There exists the solution from the class
$u\in S_{ev}'(\mathbb{R}^{n}_{+})\times C^2(0,\infty)$
to the problem  \eqref{GEQ1421}--\eqref{GEQ1431} when $k\neq -1,-3,-5,\dots$
and it is defined by the formula
\begin{equation}\label{Sol0}
u(x,t;k)=\frac{ 2^n t^{1-k}\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-n-|\gamma|+1}{2}\big)\prod^n_{i=1}
\Gamma\left(\frac{\gamma_i+1}{2}\right)}((t^2-|x|^2)_{+,\gamma}
 ^{\frac{k-n-|\gamma|-1}{2}}*f(x))_\gamma.
\end{equation}
The solution  \eqref{Sol0} is unique for $k\geq0$, and not unique for $k<0$.
\end{theorem}

\begin{proof}
Applying multidimensional Hankel transform to \eqref{GEQ1421} with respect
 to the  variables $x_1,\dots,x_n$ only and using \eqref{SaprBessel} we obtain
\begin{gather}\label{22}
\Big(|\xi|^2+\frac{\partial^2}{\partial t^2}+\frac{k}{t}
\frac{\partial}{\partial t}\Big)\widehat{u}(\xi,t)=0, \\
\label{23}
\lim_{t\to 0}\widehat{u}(\xi,t;k)
=\widehat{f}(\xi),\quad \lim_{t\to 0}
\frac{\partial \widehat{u}(\xi,t;k)}{\partial t}=0,
\end{gather}
where $\xi=(\xi_l,\xi_2,{\dots},\xi_n){\in}\mathbb{R}^n_+$
corresponds to $x=(x_1,{\dots},x_n){\in}\mathbb{R}^n_+$,
$|\xi|^2=\xi^2_1{+}\xi_2^2+\dots +\xi_n^2$,
$$
\widehat{u}(\xi,t;k)=(F_B)_x[u(x,t;k)](\xi)
=\int_{\mathbb{R}^n_+}u(x,t;k)\,\mathbf{j}_\gamma(x;\xi)x{}^\gamma dx
$$
and $\widehat{f}(\xi)=F_B[f](\xi)$.

In \cite{Bresters1} the solution  $\widehat{G}^k(\xi,t)$ to the Cauchy problem
\begin{gather*}
\Big(|\xi|^2+\frac{\partial^2}{\partial t^2}
 +\frac{k}{t}\frac{\partial}{\partial t}\Big)\widehat{G}^k(\xi,t)=0, \\
\lim_{t\to 0}\widehat{G}^k(\xi,t)=1,\quad
\lim_{t\to 0}\frac{\partial \widehat{G}^k(\xi,t)}{\partial t}=0
\end{gather*}
was obtained and it has the form
\begin{equation}\label{221}
    \widehat{G}^k(\xi,t)=j_{\frac{k-1}{2}}(|\xi| t),
\end{equation}
for $ k\geq 0$,
\begin{equation}\label{222}
\widehat{G}^k(\xi,t)=j_{\frac{k-1}{2}}(|\xi| t)
+A t^{\frac{1-k}{2}}J_{\frac{1-k}{2}}(|\xi| t),
\end{equation}
for $ k<0$, $k\neq -1,-3,-5,\dots$,
\begin{equation}\label{223}
        \widehat{G}^k(\xi,t)=B t^{\frac{1-k}{2}}J_{\frac{1-k}{2}}
(|\xi| t)-\frac{\pi 2^{\frac{k-1}{2}}}{\Gamma\left(\frac{1-k}{2}\right)}
(|\xi| t)^{\frac{1-k}{2}} Y_{\frac{1-k}{2}}(|\xi| t),
\end{equation}
for $k=-1,-3,-5,\dots$.

In \eqref{221}--\eqref{223} $A$ and $B$ are arbitrary complex numbers and
$Y_\nu(z)$ is a Bessel functions of the  the second kind.
The solutions \eqref{222}, \eqref{223} depend on the constants $A$ and $B$
since they are is not unique (see \cite{Bresters1}).
When $\widehat{G}^k(\xi,t)$ is found, the solution of \eqref{22}--\eqref{23} is
$$
\widehat{u}(\xi,t;k)=\widehat{G}^k(\xi,t)\cdot \widehat{f}(\xi)
$$
and the solution to \eqref{GEQ1421}--\eqref{GEQ1431} is then given by
$$
u(x,t;k)=((F^{-1}_B)_\xi[\widehat{G}^k(\xi,t)]*f(x))_\gamma
=({G}^k(x,t)*f(x))_\gamma.
$$

We are looking for the solution when $k\neq -1,-3,-5,\dots$ and when $A=0$.
The  obtained solution will be unique for $k\geq 0$ and will be one of the
possible solutions for $k<0$, $k\neq -1,-3,-5,\dots$. So we are interested
in case when $\widehat{G}^k(\xi,t)=j_{\frac{k-1}{2}}(|\xi| t)$
Using \eqref{FBOF} we can find $(F^{-1}_B)_\xi[j_{\frac{k-1}{2}}(|\xi| t)](x)$:
\begin{align*}
{G}^k(x,t)
&=(F^{-1}_B)_\xi[j_{\frac{k-1}{2}}(|\xi|\cdot t)](x)\\
&=\frac{ 2^n t^{1-k}\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-n-|\gamma|+1}{2}\big)\prod^n_{i=1}
\Gamma\left(\frac{\gamma_i+1}{2}\right)}
(t^2-|x|^2)_{+,\gamma}^{\frac{k-n-|\gamma|-1}{2}}.
\end{align*}
Then the solution to  \eqref{GEQ1421}--\eqref{GEQ1431} has the form
\begin{equation}\label{Sol}
u(x,t;k)
=\frac{ 2^n t^{1-k}\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-n-|\gamma|+1}{2}\big)\prod^n_{i=1}
\Gamma\left(\frac{\gamma_i+1}{2}\right)}
\Big((t^2-|x|^2)_{+,\gamma}^{\frac{k-n-|\gamma|-1}{2}}*f(x)\Big)_\gamma,
\end{equation}
for $k\neq -1,-3,-5,\dots$.

Since $(t^2-|x|^2)_{+,\gamma}^{\lambda}$ has its support in the interior
of the part of the sphere $S_1^+(n)$ when $x_1\geq 0,\dots,x_n\geq0$,
we may conclude that the convolution exists for arbitrary $\varphi(x)\in S'_+$.
In \cite{Fox} it was shown that the solution to the singular Cauchy problem
\eqref{GEQ1421}--\eqref{GEQ1431} is unique when $k$ is  nonnegative and not
unique when  $k$ is negative.
\end{proof}

\begin{corollary} \label{coro4.1}
For $k>n+|\gamma|-1$ when $f\in C^2_{ev}$  the solution to
\eqref{GEQ1421}--\eqref{GEQ1431}
exists in the classical sense and is defined by
\begin{equation}\label{SolCon}
\begin{aligned}
& u(x,t;k) \\
&=\frac{ 2^n \Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-n-|\gamma|+1}{2}\big)\prod^n_{i=1}
\Gamma\left(\frac{\gamma_i+1}{2}\right)}
\int_{B^+_1(n)}(1-|y|^2)^{\frac{k-n-|\gamma|-1}{2}} {}^\gamma T^{t y}f(x)y^\gamma dy,
\end{aligned}
\end{equation}
which coincides with formula \eqref{Sol1}.
\end{corollary}

\begin{proof}
In the case when $k>n+|\gamma|-1$ and $f(x)$ is continuous and even with
respect to all variables the integral in \eqref{Sol} exists in the classical sense.
So, taking in \eqref{Sol}  usual function $(t^2-|x|^2)^{\lambda}$ instead
of the weighted generalized function  $(t^2-|x|^2)_{+,\gamma}^{\lambda}$,
passing to the integral over  the part of the ball
$B_t^+=\{x{\in}\mathbb{R}^n_+:|x|{<}t\}$ and changing the variables by formula
$x=ty$ we obtain \eqref{SolCon}.
\end{proof}



\section{Case $x$ is one-dimensional}

In this section we concentrate on the case when $x$ is one-dimensional.
Then problems and constructed above solutions are simplified.
For these problems we consider below  some illustrative examples with
explicit solution representations and some visual graphs using the Wolfram Alpha.
In this case we have the Cauchy problem
\begin{gather}\label{OneD1}
\frac{\partial^2 u}{\partial x^2}+\frac{\gamma}{x}\frac{\partial u}{\partial x}
=\frac{\partial^2 u}{\partial t^2}+\frac{k}{t}\frac{\partial u}{\partial t}, \\
\label{OneD2}
    u(x,0;k)=f(x),\quad \frac{\partial u(x,t;k)}{\partial t}\Big|_{t=0}=0,
\quad f(x)\in C_{ev}^2(\overline{\mathbb{R}}^{1}_{+}).
\end{gather}
When $k>\gamma>0$ the solution to \eqref{OneD1}--\eqref{OneD2} is given by
 (see \eqref{Sol1})
\begin{equation}\label{SolEx}
u(x,t;k)=\frac{ 2\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-\gamma}{2}\big)\Gamma\left(\frac{\gamma+1}{2}\right)}
\int_0^1(1-y^2)^{\frac{k-\gamma-2}{2}}{}^\gamma T^{ty}f(x)y{}^\gamma
dy.
\end{equation}
When $k<\gamma$ the solution to \eqref{OneD1}--\eqref{OneD2} is found
by the formulas \eqref{34}, \eqref{Ex1} or \eqref{Ex2}.


\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\linewidth]{fig1} % RisEx01.jpg
\end{center}
\caption{$u(x,t;k)={j}_{-\frac{1}{6}}(x){j}_{\frac{3}{4}}(t)$.}
\label{Ris1}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\linewidth]{fig2} % RisEx02.jpg
\end{center}
\caption{$u(x,t;k)={j}_{-\frac{1}{6}}(x){j}_{\frac{3}{4}}(t)$.} \label{Ris2}
\end{figure}

\begin{example} \label{examp1} \rm
We are looking for the solution to
\begin{gather*}
\frac{\partial^2 u}{\partial x^2}+\frac{\gamma}{x}\frac{\partial u}{\partial x}
=\frac{\partial^2 u}{\partial t^2}+\frac{k}{t}\frac{\partial u}{\partial t},
\quad k>\gamma>0, \\
 u(x,0;k)={j}_{\frac{\gamma-1}{2}}(x),\quad
\frac{\partial u(x,t;k)}{\partial t}\Big|_{t=0}=0,\quad
f(x)\in C_{ev}^2(\overline{\mathbb{R}}^{1}_{+}).
\end{gather*}
By \eqref{SolEx} we obtain
$$
u(x,t;k)=\frac{ 2\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-\gamma}{2}\big)\Gamma\left(\frac{\gamma+1}{2}\right)}
\int_0^1(1-y^2)^{\frac{k-\gamma-2}{2}}{}^\gamma
T^{ty}{j}_{\frac{\gamma-1}{2}}(x)y{}^\gamma dy.
$$
The next formula is valid
$$
T^{ty}{j}_{\frac{\gamma-1}{2}}(x)
={j}_{\frac{\gamma-1}{2}}(x){j}_{\frac{\gamma-1}{2}}(ty)
$$
and so
$$
u(x,t;k)={j}_{\frac{\gamma-1}{2}}(x)\,t^{\frac{1-\gamma}{2}}
\frac{ 2^{\frac{\gamma+1}{2}}\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-\gamma}{2}\big)}\int_0^1(1-y^2)^{\frac{k-\gamma-2}{2}}
J_{\frac{\gamma-1}{2}}(ty)\,y^{\frac{\gamma+1}{2}} dy.
$$
Using \cite[formula 2.12.4.6]{IR2},
\begin{equation}\label{Prud}
\int_0^a x^{\nu+1}(a^2-x^2)^{\beta-1}J_\nu(cx)dx=\frac{2^{\beta-1}a^{\beta+\nu}}{c^{\beta}}
\Gamma(\beta)J_{\beta+\nu}(ac),
\end{equation}
for $a>0$, $\operatorname{Re}\beta>0$, $\operatorname{Re}\nu>-1$,
we obtain
\begin{equation}\label{Resh}
u(x,t;k)={j}_{\frac{\gamma-1}{2}}(x){j}_{\frac{k-1}{2}}(t).
\end{equation}

The plot of \eqref{Resh} when $k=\frac{5}{2}$ and $\gamma=\frac{2}{3}$
is presented on Figure \ref{Ris1}, and obtained through the Wolfram-Alpha.
We can continue the solution to negative values of $x$ and $t$ as an even function.
The plot of such continuation is presented on Figure \ref{Ris2}.
\end{example}

If we denote
\begin{equation}\label{OOOS}
\begin{gathered}
    ^{k,\gamma}T^t_xf(x)
=C(\gamma,k)\int_0^1(1-y^2)^{\frac{k-\gamma-2}{2}}T^{ty}f(x)y{}^\gamma dy, \\
C(\gamma,k)=\frac{ 2\Gamma\left(\frac{k+1}{2}\right)}
{\Gamma\big(\frac{k-\gamma}{2}\big)\Gamma\left(\frac{\gamma+1}{2}\right)}.
\end{gathered}
\end{equation}
We can consider the operator \eqref{OOOS} as a generalized translation
operator (see \cite{Lev3}). For this operator the next property holds
$$
^{k,\gamma}T^t_x{j}_{\frac{\gamma-1}{2}}(x)
={j}_{\frac{\gamma-1}{2}}(x){j}_{\frac{k-1}{2}}(t)\,.
$$

\begin{example} \label{examp2} \rm
 The solution to
\begin{gather*}
\frac{\partial^2 u}{\partial x^2}
+\frac{\gamma}{x}\frac{\partial u}{\partial x}
=\frac{\partial^2 u}{\partial t^2}+\frac{k}{t}\frac{\partial u}{\partial t},\quad
1-\gamma\leq k<\gamma,\; k\neq -1,-3,-5,\dots, \quad \gamma>\frac{1}{2}, \\
  u(x,0;k)={j}_{\frac{\gamma-1}{2}}(x),\quad
\frac{\partial u(x,t;k)}{\partial t}\Big|_{t=0}=0.
\end{gather*}
is given by \eqref{OneD1} where $m=1$,
$$
u(x,t;k)=\frac{1}{t^{k}}\,\frac{\partial}{ \partial t}(t^{k+1}u(x,t;k+2)),
$$
and $u(x,t;k+2)$ is the solution to the Cauchy problem
\begin{gather*}
   (B_{k+2})_t u=(\Delta_\gamma)_x u, \\
u(x,0;k+2)=\frac{{j}_{\frac{\gamma-1}{2}}(x)}{k+1},\quad u_t(x,0;k+2)=0.
\end{gather*}
Using the previous example we obtain
\begin{gather*}
u(x,t;k+2)=\frac{1}{k+1}{j}_{\frac{\gamma-1}{2}}(x){j}_{\frac{k+1}{2}}(t),\\
u(x,t;k)=\,_0F_1\Big(\frac{\gamma +1}{2};-\frac{x^2}{4}\Big)
\, _0F_1\Big(;\frac{k+1}{2};-\frac{t^2}{4}\Big).
\end{gather*}

The plot of \eqref{Resh} when $k=\frac{1}{3}$ and $\gamma=\frac{3}{2}$
is presented on Figure \ref{Ris3} obtained through the Wolfram-Alpha.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.7\linewidth]{fig3} % RisEx03.jpg
\end{center}
\caption{$u(x,t;k)=\, _0F_1\left(;\frac{2}{3};-\frac{t^2}{4}\right)
\, _0F_1\left(;\frac{5}{4};-\frac{x^2}{4}\right)$.} \label{Ris3}
\end{figure}

We can continue the solution to negative values of $x$ and $t$ as an even function.
 The plot of such continuation is presented on Figure \ref{Ris4}.
\end{example}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.8\linewidth]{fig4} % RisEx04.jpg
\end{center}
\caption{$u(x,t;k)=\, _0F_1\left(;\frac{2}{3};-\frac{t^2}{4}\right)
\, _0F_1\left(;\frac{5}{4};-\frac{x^2}{4}\right)$.}\label{Ris4}
\end{figure}

\subsection*{Acknowledgements}
 Sergei M. Sitnik was supported by the Ministry of Education and Science
 of the Russian Federation (Agreement 02.A03.21.0008).



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