\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 100, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/100\hfil Fundamental solutions]
{Fundamental solutions of two multidimensional
elliptic equations}

\author[I. Garipov, R. Mavlyaviev \hfil EJDE-2018/100\hfilneg]
{Ilnur Garipov, Rinat Mavlyaviev}

\address{Ilnur Garipov \newline
Kazan (Volga region) Federal University,
 Kazan, Russia}
\email{ilnur\_garipov@mail.ru}

\address{Rinat Mavlyaviev \newline
Kazan (Volga region) Federal University,
Kazan, Russia}
\email{mavly72@mail.ru}

\dedicatory{Communicated by Vicentiu D. Radulescu}

\thanks{Submitted  December 4, 2017. Published April 28, 2018.}
\subjclass[2010]{35A08}
\keywords{Fundamental solution; hypergeometric Gauss function;
\hfill\break\indent  confluent Horn function; transmutation operator}

\begin{abstract}
 We construct fundamental solutions for two-multidimensional elliptic
 equations. The solutions are written in explicit form  via
 hypergeometric Gauss functions for $\lambda=0$ and via confluent
 Horn functions for $\lambda\neq 0$. It is proved
 that the fundamental solutions found possess power-type
 singularity  $\rho^{2-n}$ as $\rho\to 0$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks


\section{Introduction}

The practical value of mixed-type equations was first highlighted by
Chaplygin  \cite{Chaplygin} in 1902.
Investigations of boundary-value problems for mixed-type equations was begun by
Tricomi in his works  \cite{Tricomi1,Tricomi2}. He stated and
solved the first boundary problem for the equation
$$
y\frac{\partial^2u}{\partial x^2}+ \frac{\partial^2u}{\partial y^2}=0.
$$

Holmgren \cite{Holmgren} simplified the solution of Tricomi problems for the equation
$$
y^m\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0.
$$

In his doctoral thesis Gellerstedt \cite{Gellerstedt3} solved the Tricomi
problem for the equation
$$
y^m\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}-cu
=F(x,y),
$$
and, in \cite{Gellerstedt1,Gellerstedt2} he generalized Tricomi's
results.

A systematic study of mixed-type equations attracted authors' attention
since the middle of 40s of the past century after indication by
Frankl of the possibility of their application in transonic and supersonic gas
dynamics and hydrodynamics
\cite{Frankl2,Frankl1,Frankl0}.

Essential contribution to the development of the theory of mixed-type
equations was made by the mathematicians
Germain, Bader \cite{Germain1}, Bitsadze \cite{Bizadze},
Babenko \cite{Babenko},
Volkodavov \cite{Volkodavov},
Keldysh \cite{Keldish}, Vekua \cite{Vekua} and others.

During recent years, the main interests focus shifted towards practical
applications of mathematical models in various fields of sciences
\cite{Ghergu,Frigon}. The monograph \cite{Ghergu} contains many
examples of applications of mathematical models in biology, chemistry, and
population genetics. In paper \cite{Frigon}, the results on g-differential
equations are applied to some mathematical models.

In [27] the parametric Stokes phenomena studied for Gauss hypergeometric
differential equation from the viewpoint of the alien calculus.
In the present article, for the degenerate elliptic equation
\begin{equation}\label{GM_00}
L_\lambda (u) \equiv x_n^m\sum_{i=1}^{n-1}
\Big(\frac{\partial^2u}{\partial x_i^2}+\lambda^2 u\Big)+
\frac{\partial^2u}{\partial x_n^2}=0,\quad (m>0,\; n>2),
\end{equation}
and for the elliptic equation
\begin{equation}\label{GM_01}
T_\lambda (u) \equiv e^{x_n}\sum_{i=1}^{n-1}
\Big(\frac{\partial^2u}{\partial x_i^2}+\lambda^2 u\Big)+
\frac{\partial^2u}{\partial x_n^2}=0,\quad (n>2)
\end{equation}
fundamental solutions are constructed. The fundamental solutions are
written in explicit form via hypergeometric Gauss functions for
$\lambda=0$ and confluent Horn functions for $\lambda\neq 0$.
It is proved that the fundamental solutions obtained possess power-type
singularity (further we will use power singularity term)
 $\rho^{2-n}$ as $\rho\to 0$.

\section{Finding fundamental solutions of equation \eqref{GM_00}}

For $\lambda=0$, the equation \eqref{GM_00} takes the form
\begin{equation}\label{GM1_1}
L_0 (u) \equiv x_n^m\sum_{i=1}^{n-1}\frac{\partial^2u}{\partial x_i^2}+
\frac{\partial^2u}{\partial x_n^2}=0,\quad (m>0, n>2).
\end{equation}
Following the works \cite{Hasanov,Hasanov1,Hasanov2,Mavlyaviev},
we search for a solution for equation \eqref{GM1_1} in the form
\begin{equation}\label{GM1_2}
u(x_1,x_2, \dots, x_n)=P\,\omega(\xi),
\end{equation}
where
\begin{gather}\label{GM1_3}
P=\rho^{-(\mu+n-2)}, \quad \mu=\frac{m}{m+2},\quad
\xi=\frac{\rho^2-\rho_1^2}{\rho^2},\\
\rho^2=\sum_{i=1}^{n-1}\Big(x_i-x_i^{(0)}\Big)^2
+\frac{4}{(m+2)^2}\Big(x_n^{\frac{m+2}{2}}
 -\Big(x_n^{(0)}\Big)^{\frac{m+2}{2}} \Big)^2, \nonumber\\
\rho_1^2=\sum_{i=1}^{n-1}\Big(x_i-x_i^{(0)}\Big)^2
 +\frac{4}{(m+2)^2}\Big(x_n^{\frac{m+2}{2}}+\Big(x_n^{(0)}\Big)^{\frac{m+2}{2}}
 \Big)^2, \nonumber
\end{gather}
while $\omega(\xi)$ is a function yet unknown.

By substituting the function \eqref{GM1_2} into \eqref{GM1_1}, we obtain
the equation
\begin{equation}\label{GM1_4}
\xi(1-\xi)\omega_{\xi\xi}+
\Big(\mu-\Big(\frac{\mu+n-2}{2}+\frac{\mu}{2}+1\Big)\xi\Big)\omega_{\xi}
-\frac{\mu+n-2}{2}\frac{\mu}{2}\omega = 0.
\end{equation}
Comparing \eqref{GM1_4} with Gauss equation
\begin{equation}\label{GM1_4_0}
\xi(1-\xi)\omega_{\xi\xi}+
\bigl(\delta-(\alpha+\beta+1)\xi\bigr)\omega_{\xi}-\alpha\beta\omega = 0,
\end{equation}
which in a neighborhood of the point $\xi=0$ has two linearly independent solutions
$$
\omega_1=F(\alpha,\beta;\delta;\xi), \quad
\omega_2=\xi^{1-\delta}F(\alpha-\delta+1,\beta-\delta+1;2-\delta;\xi),
$$
where
$$
F(\alpha,\beta;\delta;\xi)=
\sum_{k=0}^{\infty}
\frac{(\alpha)_{k}(\beta)_{k}}{(\delta)_{k}}
\frac{\xi^{k}}{k!},
$$
being hypergeometric Gauss function  \cite{Erdelyi}, and taking into account
\eqref{GM1_3}, we conclude that for  equation \eqref{GM1_4} the following
functions are solutions
\begin{gather*}
\omega_1=
F\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;
\frac{\rho^2-\rho_1^2}{\rho^2}\Big), \\
\omega_2
=\Big(\frac{\rho^2-\rho_1^2}{\rho^2}\Big)^{1-\mu}
F\Big(\frac{\mu+n-2}{2}+1-\mu,\frac{\mu}{2}+1-\mu;2-\mu;
\frac{\rho^2-\rho_1^2}{\rho^2}\Big).
\end{gather*}
Consequently, solutions of  \eqref{GM1_1} are given by the
functions
\begin{gather}\label{GM1_5}
q_{01}(M,M_0) = C_1
\rho^{-(\mu+n-2)}F\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;
\frac{\rho^2-\rho_1^2}{\rho^2}\Big), \\
\label{GM1_6}
q_{02}(M,M_0) = C_2
\rho^{-(\mu+n-2)}\Big(\frac{\rho^2-\rho_1^2}{\rho^2}\Big)^{1-\mu}
F\Big(\frac{n-\mu}{2},1-\frac{\mu}{2};2-\mu;\frac{\rho^2-\rho_1^2}{\rho^2}\Big),
\end{gather}
where $C_1$ and $C_2$ are some constants.

Using transformation formula in \cite{Erdelyi},
\begin{equation}\label{GM1_5_0}
F(\alpha,\beta;\delta;\xi)=(1-\xi)^{-\alpha}
F\Big(\alpha,\delta-\beta;\delta;\frac{\xi}{\xi-1}\Big)
\end{equation}
we write the solutions \eqref{GM1_5}, \eqref{GM1_6} in the
form
\begin{gather}\label{GM1_5_1}
q_{01}(M,M_0) = C_1
\rho_1^{-(\mu+n-2)}F\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;
\frac{\rho_1^2-\rho^2}{\rho_1^2}\Big), \\
\label{GM1_6_1}
\begin{aligned}
&q_{02}(M,M_0) \\
&= C_{21} \rho_1^{-(\mu+n-2)}\Big(\frac{\rho_1^2-\rho^2}{\rho_1^2}\Big)^{1-\mu}
F\Big(\frac{n-\mu}{2},1-\frac{\mu}{2};2-\mu;\frac{\rho_1^2-\rho^2}{\rho_1^2}\Big).
\end{aligned}
\end{gather}

Note (see \cite{Nigmedzyanova2}) that one can pass straightforwardly  to
solutions \eqref{GM1_5_1}, \eqref{GM1_6_1}
if one seeks a solution in the form
$$
u(x_1,x_2, \dots, x_n)=P\,\omega(\sigma),
$$
where
$$
P=\rho_1^{-(\mu+n-2)}, \quad
\mu=\frac{m}{m+2},\quad  \sigma=\frac{\rho^2}{\rho_1^2}.
$$

Now let us consider the case $\lambda\neq0$. Following the papers
\cite{Hasanov,Hasanov1}, we seek a solution of
equation \eqref{GM_00} in the form
\begin{equation}\label{GM2_1}
u(x_1,x_2, \dots, x_n)=P\omega(\xi,\eta),
\end{equation}
where
\begin{equation}\label{GM2_2}
P=\rho^{-(\mu+n-2)}, \quad  \xi=\frac{\rho^2-\rho_1^2}{\rho^2}, \quad
\eta=\frac{\lambda^2 \rho^2}{4},
\end{equation}
while $\omega(\xi,\eta)$ is a function yet unknown.

By substituting the function \eqref{GM2_1} into equation
\eqref{GM_00}, we obtain
\begin{equation}\label{GM2_3}
\begin{gathered}
\begin{aligned}
&\xi(1-\xi)\omega_{\xi\xi}+\xi\eta\omega_{\xi\eta}  +
(\mu-(\frac{\mu+n-2}{2}+\frac{\mu}{2}+1)\xi)\omega_{\xi} \\
&+\frac{\mu}{2}\eta\omega_{\eta}-\frac{\mu+n-2}{2}\frac{\mu}{2}\omega = 0,
\end{aligned}\\
\eta\omega_{\eta\eta}-\xi\omega_{\xi\eta}
+(1-\frac{\mu+n-2}{2})\omega_{\eta}+\omega=0.
\end{gathered}
\end{equation}
Comparing \eqref{GM2_3} with the system
\begin{equation}\label{GM2_4}
\begin{gathered}
\xi(1-\xi)\omega_{\xi\xi}+\xi\eta\omega_{\xi\eta}+
\bigl(\delta-(\alpha+\beta+1)\xi\bigr)\omega_{\xi}+
\beta\eta\omega_{\eta}-\alpha\beta\omega = 0,\\
\eta\omega_{\eta\eta}-\xi\omega_{\xi\eta}+
(1-\alpha)\omega_{\eta}+\omega=0,
\end{gathered}
\end{equation}
which possesses in vicinity of the point $\xi=0$ the two linearly
independent solutions
\begin{equation}\label{GM2_4_0}
\omega_1=H_3(\alpha,\beta;\delta;\xi,\eta), \quad
\omega_2=\xi^{1-\delta}H_3(\alpha-\delta+1,\beta-\delta+1;2-\delta;\xi,\eta),
\end{equation}
where
\begin{equation}\label{GM2_5}
H_3(\alpha,\beta;\delta;\xi,\eta)
=\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(\beta)_{k}}{(\delta)_{k}}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}
\end{equation}
is confluent with the Horn-Kummer function in $H_3$ \cite{Conway},
we conclude that the following functions are solutions of system \eqref{GM2_3},
\begin{gather}\label{GM2_6}
\omega_1=H_3\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;\xi,\eta\Big)
=H_3\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;
\frac{\rho^2-\rho_1^2}{\rho^2},\frac{\lambda^2\rho^2}{4}\Big), \\
\label{GM2_7}
\begin{aligned}
\omega_2
&=\xi^{1-\mu}H_3\Big(\frac{\mu+n-2}{2}+1-\mu,\frac{\mu}{2}+1-\mu;
2-\mu;\xi,\eta\Big) \\
&= \Big(\frac{\rho^2-\rho_1^2}{\rho^2}\Big)^{1-\mu}
H_3\Big(\frac{n-\mu}{2},1-\frac{\mu}{2};2-\mu;\frac{\rho^2-\rho_1^2}{\rho^2},
\frac{\lambda^2\rho^2}{4}\Big).
\end{aligned}
\end{gather}

Therefore, the solutions of equation \eqref{GM_00} are given by the functions
\begin{gather}\label{GM2_8}
q_{\lambda 1}(M,M_0)
= C_3\rho^{-(\mu+n-2)}H_3\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;
\frac{\rho^2-\rho_1^2}{\rho^2},\frac{\lambda^2\rho^2}{4}\Big), \\
\label{GM2_9}
\begin{aligned}
q_{\lambda 2}(M,M_0)
&= C_{4}\rho^{-(\mu+n-2)}\Big(\frac{\rho^2-\rho_1^2}{\rho^2}\Big)^{1-\mu}\\
&\quad\times H_3\Big(\frac{n-\mu}{2},1-\frac{\mu}{2};2-\mu;
\frac{\rho^2-\rho_1^2}{\rho^2},\frac{\lambda^2\rho^2}{4}\Big),
\end{aligned}
\end{gather}
where $C_3$ and $C_{4}$ are some constants.

Applying the transformation formula in \cite{Kapilevich},
\begin{equation}\label{GM2_8_0}
H_3(\alpha,\beta;\delta;\xi,\eta)
=(1-\xi)^{-\alpha}H_3\Big(\alpha,\delta-\beta;\delta;\frac{\xi}{\xi-1};
\eta(1-\xi)\Big),
\end{equation}
one can write solutions \eqref{GM2_8} and \eqref{GM2_9}
in the form
\begin{gather}\label{GM2_8_1}
q_{\lambda 1}(M,M_0)
= C_3\rho_1^{-(\mu+n-2)}H_3\Big(\frac{\mu+n-2}{2},\frac{\mu}{2};\mu;
\frac{\rho_1^2-\rho^2}{\rho_1^2},\frac{\lambda^2\rho_1^2}{4}\Big), \\
\label{GM2_9_1}
\begin{aligned}
&q_{\lambda 2}(M,M_0)\\
& = C_{41}\rho_1^{-(\mu+n-2)}
 \Big(\frac{\rho_1^2-\rho^2}{\rho_1^2}\Big)^{1-\mu}
 H_3\Big(\frac{n-\mu}{2},1-\frac{\mu}{2};2-\mu;
\frac{\rho_1^2-\rho^2}{\rho_1^2},\frac{\lambda^2\rho_1^2}{4}\Big).
\end{aligned}
\end{gather}

Note that one can arrive straightforwardly at the solutions
\eqref{GM2_8_1}, \eqref{GM2_9_1} if one seeks solution in the form
$$
u(x_1,x_2, \dots, x_n)=P\,\omega(\sigma,\tau),
$$	
where
$$
P=\rho_1^{-(\mu+n-2)}, \quad \mu=\frac{m}{m+2},\quad
 \sigma=\frac{\rho^2}{\rho_1^2}\quad \tau=\frac{\lambda^2 \rho^2}{4}.
$$

Let us consider some properties of solutions.

\begin{lemma} \label{lem1}
 Solutions \eqref{GM1_5}, \eqref{GM1_6} and
\eqref{GM2_8}, \eqref{GM2_9}
satisfy the following conditions
\begin{gather*}
\frac{\partial q_{01}(M,M_0)}{\partial x_n}\Big|_{x_n=0}=0,\quad
q_{02}(M,M_0)\Big|_{x_n=0}=0, \\
\frac{\partial q_{\lambda1}(M,M_0)}{\partial x_n}\Big|_{x_n=0}=0,\quad
q_{\lambda2}(M,M_0)\big|_{x_n=0}=0.
\end{gather*}
\end{lemma}

The proof os the above lemma is to  a straightforward calculation.

\begin{lemma} \label{lem2}
Solutions \eqref{GM1_5}, \eqref{GM1_6} and \eqref{GM2_8}, \eqref{GM2_9}
possess power singularity $\rho^{2-n}$ as $\rho\to 0$.
\end{lemma}

\begin{proof}
It can be carried out on the basis of analytic
continuation formulae for the hypergeometric Gauss
function \cite{Andrews} and confluent Horn-Kumner function \cite{Srivastava},
\begin{gather*}
\begin{aligned}
F(\alpha,\beta;\delta;\xi)
&= \frac{\Gamma(\delta)\Gamma(\beta-\alpha)}{\Gamma(\beta)
\Gamma(\delta-\alpha)}(-\xi)^{-\alpha}
F\Big(\alpha,1+\alpha-\delta;1+\alpha-\beta;\frac{1}{\xi}\Big) \\
&\quad +\frac{\Gamma(\delta)\Gamma(\alpha-\beta)}
{\Gamma(\alpha) \Gamma(\delta-\beta)} (-\xi)^{-\beta}
 F\Big(\beta,1+\beta-\delta;1+\beta-\alpha;\frac{1}{\xi}\Big),
\end{aligned} \\
\begin{aligned}
&H_3(\alpha,\beta;\delta;\xi,\eta)\\
&=\frac{\Gamma(\delta)\Gamma(\beta-\alpha)}{\Gamma(\beta)\Gamma(\delta-\alpha)}
(-\xi)^{-\alpha}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(1+\alpha-\beta)_{k-l}}{(1+\alpha-\delta)_{k-l}}
\frac{(1/\xi)^{k}}{k!}\frac{(-\xi\eta)^{l}}{l!} \\
&\quad +\frac{\Gamma(\delta)\Gamma(\alpha-\beta)}{\Gamma(\alpha)
 \Gamma(\delta-\beta)}(-\xi)^{-\beta}\Xi_2
 \Big(\beta,1+\beta-\delta;1+\beta-\alpha;\frac{1}{\xi};-\eta\Big),
\end{aligned}
\end{gather*}
where
$$
\Xi_2(\alpha,\beta;\delta;\xi,\eta)=
\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k}(\beta)_{k}}{(\delta)_{k+l}}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}
$$
is a Humbert function \cite{Srivastava}.
\end{proof}

Lemma \ref{lem2} implies the following result.


\begin{theorem} \label{thm1}
The functions \eqref{GM1_5} and \eqref{GM1_6}
are fundamental solutions of  equation \eqref{GM_00} for
$\lambda=0$, while  functions \eqref{GM2_8} and \eqref{GM2_9}
are for $\lambda\neq0$.
\end{theorem}


\section{Fundamental solutions of \eqref{GM_01}}

For $\lambda=0$,  equation \eqref{GM_01} takes the form
\begin{equation}\label{GM3_1}
T_0 (u) \equiv e^{x_n}\sum_{i=1}^{n-1}\frac{\partial^2u}{\partial x_i^2}+
\frac{\partial^2u}{\partial x_n^2}=0,\quad (m>0, n>2).
\end{equation}
We seek a solution of equation \eqref{GM3_1} in the form
\begin{equation}\label{GM3_2}
u(x_1,x_2, \dots, x_n)=P\omega(\sigma),
\end{equation}
where
\begin{gather}\label{GM3_3}
P=\rho_1^{-(n-1)}, \quad  \sigma=\frac{\rho^2}{\rho_1^2},\\
\rho^2=\sum_{i=1}^{n-1}\Big(x_i-x_i^{(0)}\Big)^2
 +4\Big(e^{\frac{x_n}{2}}-e^{\frac{x_n^{(0)}}{2}}\Big)^2,\nonumber\\
\rho_1^2=\sum_{i=1}^{n-1}\Big(x_i-x_i^{(0)}\Big)^2
+4\Big(e^{\frac{x_n}{2}}+e^{\frac{x_n^{(0)}}{2}}\Big)^2, \nonumber
\end{gather}
while $\omega(\sigma)$ is a function yet unknown.

Substituting  function \eqref{GM3_2} into \eqref{GM3_1}, we obtain
the equation
\begin{equation}\label{GM3_4}
\sigma(1-\sigma)\omega_{\sigma\sigma}+
(\frac{n}{2}-\Big(\frac{n-1}{2}+\frac{1}{2}+1\Big)\sigma\Big)\omega_{\sigma}
-\frac{n-1}{2}\frac{1}{2}\omega = 0.
\end{equation}

By  the changeof variable  $\chi=1-\sigma$ this equation can be represented
in the form
\begin{equation}\label{GM3_4_0}
\chi(1-\chi)\omega_{\chi\chi}+
\Big(1-\Big(\frac{n-1}{2}+\frac{1}{2}+1\Big)\chi\Big)
\omega_{\chi}-\frac{n-1}{2}\frac{1}{2}\omega = 0.
\end{equation}
Equation \eqref{GM3_4_0} has two independent solutions
\cite{Andrews},
\begin{gather*}
\omega_1= F\Big(\frac{n-1}{2},\frac{1}{2};1;\chi\Big), \\
\begin{aligned}
\omega_2&= F\Big(\frac{n-1}{2},\frac{1}{2};1;\chi\Big)\ln \chi \\
&\quad +\sum_{k=0}^{\infty}\frac{\Big(\frac{n-1}{2}\Big)_{k}
 \Big(\frac{1}{2}\Big)_{k}}{(k!)^2}\Big(\psi\Big(\frac{n-1}{2}+k\Big)
 +\psi\Big(\frac{1}{2}+k\Big)-2\psi(1+k)\Big) \chi^{k},
\end{aligned}
\end{gather*}
where
$$
\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}
$$
is logarithmic derivative of Euler gamma-function.

Consequently, solutions of equation \eqref{GM3_1} are given by the functions
\begin{gather}\label{GM3_5_1}
q_{01}(M,M_0) = C_1\rho_1^{-(n-1)}
F\Big(\frac{n-1}{2},\frac{1}{2};1;\frac{\rho_1^2-\rho^2}{\rho_1^2}\Big), \\
\label{GM3_6_1}
\begin{aligned}
q_{02}(M,M_0)
&= C_2\rho_1^{-(n-1)}\Bigl(F\Big(\frac{n-1}{2},\frac{1}{2};1;
 \frac{\rho_1^2-\rho^2}{\rho_1^2}\Big)\ln \frac{\rho_1^2-\rho^2}{\rho_1^2} \\
&\quad +\sum_{k=0}^{\infty}\frac{\Big(\frac{n-1}{2}\Big)_{k}
 \Big(\frac{1}{2}\Big)_{k}}{(k!)^2}
 \Big(\psi\Big(\frac{n-1}{2}+k\Big)
 +\psi\Big(\frac{1}{2}+k\Big) \\
&\quad -2\psi(1+k)\Big)
\Big(\frac{\rho_1^2-\rho^2}{\rho_1^2} \Big)^{k}\Bigr).
\end{aligned}
\end{gather}

Now consider the case $\lambda\neq0$. We seek a solution of equation
\eqref{GM_01} in the form
\begin{equation}\label{GM4_1}
u(x_1,x_2, \dots, x_n)=P\,\omega(\sigma,\tau),
\end{equation}
where
\begin{equation}\label{GM4_2}
P=\rho_1^{-(\mu+n-2)}, \quad \sigma=\frac{\rho^2}{\rho_1^2}, \quad
\tau=\frac{\lambda^2 \rho_1^2}{4},
\end{equation}
while $\omega(\sigma,\tau)$ is a function yet unknown.

Substituting  function \eqref{GM4_1} into equation \eqref{GM_01}, we get
\begin{equation}\label{GM4_3}
\begin{gathered}
\begin{aligned}
&\sigma(1-\sigma)\omega_{\sigma\sigma}-(1-\sigma)\tau\omega_{\sigma\tau}  +
(\frac{n}{2}-(\frac{n-1}{2} \\
&+\frac{1}{2}+1)\sigma)\omega_{\sigma}+\frac{1}{2}\tau\omega_{\tau}
 -\frac{n-1}{2}\frac{1}{2}\omega = 0,
\end{aligned}\\
\tau\omega_{\tau\tau}+(1-\sigma)\omega_{\sigma\tau}
+(1-\frac{n-1}{2})\omega_{\tau}+\omega=0.
\end{gathered}
\end{equation}
The change of variable $\chi=1-\sigma$ allows us to write the present system
in the form
\begin{equation}\label{GM4_3_1}
\begin{gathered}
\chi(1-\chi)\omega_{\chi\chi}+\chi\tau\omega_{\chi\tau}
+(1-(\frac{n-1}{2}+\frac{1}{2}+1)\chi)\omega_{\chi}
+\frac{1}{2}\tau\omega_{\tau}-\frac{n-1}{2}\frac{1}{2}\omega = 0,\\
\tau\omega_{\tau\tau}-\chi\omega_{\chi\tau}+(1-\frac{n-1}{2})
\omega_{\tau}+\omega=0.
\end{gathered}
\end{equation}

From \eqref{GM2_6} and \eqref{GM2_7} it is evident that, for $\mu=1$,
the solutions of the system coincide. Moreover, the solutions
\eqref{GM2_4_0} of system \eqref{GM2_4}
coincide not only for $\delta=1$, but also for any natural
$\delta.$ Indeed, let $\delta=p$ be a natural number. Then the functions
\begin{align*}
\Phi_1
&=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\delta)}
H_3(\alpha,\beta;\delta;\xi,\eta) \\
&=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\delta)}
\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(\beta)_{k}}{(\delta)_{k}}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&=\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+k-l)
 \Gamma(\beta+k)}{\Gamma(\delta+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}
\end{align*}
and
\begin{align*}
\Phi_2
&=\frac{\Gamma(\alpha+1-\delta)\Gamma(\beta+1-\delta)}
 {\Gamma(2-\delta)}
\xi^{1-\delta} H_3(\alpha-\delta+1,\beta-\delta+1;2-\delta;\xi,\eta) \\
&=\frac{\Gamma(\alpha+1-\delta)\Gamma(\beta+1-\delta)}
 {\Gamma(2-\delta)}
\xi^{1-\delta} \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha+1-\delta)_{k-l}(\beta+1-\delta)_{k}}{(2-\delta)_{k}}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&= \xi^{1-\delta}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-\delta+k-l)
 \Gamma(\beta+1-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&= \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-\delta+k-l)
 \Gamma(\beta+1-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\xi^{1-\delta+k}}{k!}\frac{\eta^{l}}{l!}
\end{align*}
are equal to each other. To find the second solution in this case,
suppose that $\alpha$ and $\beta$ are nonnegative integers.
In a way analogous to that in \cite{Andrews}, we consider
the limit
\begin{align*}
&\lim_{\delta\to  p}\frac{\Phi_1-\Phi_2}{\delta-p} \\
&=\frac{\partial}{\partial \delta}(\Phi_1-\Phi_2)\Big|_{\delta=p} \\
&=-\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+k-l)\Gamma(\beta+k)
 \Gamma'(p+k)}{\Gamma(p+k)\Gamma(p+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad + \xi^{1-p}\ln\xi \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-p+k-l)
 \Gamma(\beta+1-p+k)}{\Gamma(2-p+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad + \xi^{1-p}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-p+k-l)\Gamma(\beta+1-p+k)}
 {\Gamma(2-p+k)} \\
&\quad\times
\Big(\frac{\Gamma'(\alpha+1-p+k-l)}{\Gamma(\alpha+1-p+k-l)}+
\frac{\Gamma'(\beta+1-p+k)}{\Gamma(\beta+1-p+k)}
\Big)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad -\lim_{\delta\to  p} \xi^{1-\delta}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-\delta+k-l)\Gamma(\beta+1-\delta+k)}
{\Gamma(2-\delta+k)}
\frac{\Gamma'(2-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}.
\end{align*}
The first of the series can be written in the form
\begin{align*}
&-\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+k-l)\Gamma(\beta+k)
 \Gamma'(p+k)}{\Gamma(p+k)\Gamma(p+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
& =-\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(p)}
 \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(\beta)_{k}}{(p)_{k}}\psi(p+k)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}.
\end{align*}
The second series is equal to
$$
\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(p)}\ln \xi\,
H_3(\alpha,\beta;p;\xi,\eta).
$$
We rewrite the third series as
\begin{align*}
&\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+k-l)\Gamma(\beta+k)}
{\Gamma(p+k)}
\Big(
\frac{\Gamma'(\alpha+k-l)}{\Gamma(\alpha+k-l)}+
\frac{\Gamma'(\beta+k)}{\Gamma(\beta+k)}
\Big)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
& =\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(p)}
 \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(\beta)_{k}}{(p)_{k}}
\Big(
\psi(\alpha+k-l)+
\psi(\beta+k)
\Big)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}
\end{align*}
because $\frac{1}{\Gamma(2-p+k)}=0$  for $k=0, 1, \dots, p-2$.
From the formula
$$
\frac{\Gamma'(1-z)}{(\Gamma(1-z))^2}
=\frac{\Gamma'(z)}{\Gamma(z)\Gamma(1-z)}
 +\cos\pi z\Gamma(z),
$$
we obtain
$$
\lim_{\delta\to  p} \frac{\Gamma'(2-\delta+k)}
{(\Gamma(2-\delta+k))^2}=
(-1)^{p-k-1}\Gamma(p-k-1).
$$
It follows that the fourth addend can be rewritten in the form
\begin{align*}
&-\lim_{\delta\to  p} \xi^{1-\delta}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-\delta+k-l)\Gamma(\beta+1-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\Gamma'(2-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad -\lim_{\delta\to  p} \xi^{1-\delta}\sum_{k=0}^{p-2}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-\delta+k-l)\Gamma(\beta+1-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\Gamma'(2-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad -\lim_{\delta\to  p} \xi^{1-\delta}\sum_{k=p-1}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-\delta+k-l)\Gamma(\beta+1-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\Gamma'(2-\delta+k)}{\Gamma(2-\delta+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&=- \xi^{1-p}\sum_{k=0}^{p-1}\sum_{l=0}^{\infty}
\Gamma(\alpha+1-p+k-l)\Gamma(\beta+1-p+k)
(-1)^{p-k-1}\Gamma(p-k-1)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad -\xi^{1-p}\sum_{k=p-1}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-p+k-l)\Gamma(\beta+1-p+k)}{\Gamma(2-p+k)}
\frac{\Gamma'(2-p+k)}{\Gamma(2-p+k)}
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&=- \sum_{k=0}^{\infty}\sum_{l=0}^{\infty}(-1)^{p-k-1}
\Gamma(\alpha+1-p+k-l)\Gamma(\beta+1-p+k)
\Gamma(p-k-1)
\frac{\xi^{1-p+k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad -\sum_{k=p-1}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+1-p+k-l)\Gamma(\beta+1-p+k)}{\Gamma(2-p+k)}
\frac{\Gamma'(2-p+k)}{\Gamma(2-p+k)}
\frac{\xi^{1-p+k}}{k!}\frac{\eta^{l}}{l!} \\
&= \sum_{k=1}^{p-1}\sum_{l=0}^{\infty}(-1)^{k-1}(k-1)!
\Gamma(\alpha-k-l)\Gamma(\beta-k)
\frac{\xi^{-k}}{(p-k-1)!}\frac{\eta^{l}}{l!} \\
&\quad -\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{\Gamma(\alpha+k-l)\Gamma(\beta+k)}{\Gamma(p+k)}
\psi(1+k)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&= \sum_{k=1}^{p-1}\sum_{l=0}^{\infty}(-1)^{k-1}(k-1)!
\Gamma(\alpha-k-l)\Gamma(\beta-k)
\frac{\xi^{-k}}{(p-k-1)!}\frac{\eta^{l}}{l!} \\
&\quad -\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(p)}
\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(\beta)_{k}}{(p)_{k}}
\psi(1+k)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!}.
\end{align*}
Consequently, when $\alpha$ and $\beta$ are not negative integers,
the second solution has the form
\begin{align*}
 \omega_2
&=H_3(\alpha,\beta;p;\xi,\eta)\ln\xi
+\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\alpha)_{k-l}(\beta)_{k}}{(p)_{k}}
\Big(\psi(\alpha+k-l) \\
&\quad +\psi(\beta+k)-\psi(p+k)-\psi(1+k)\Big)
\frac{\xi^{k}}{k!}\frac{\eta^{l}}{l!} \\
&\quad +\frac{\Gamma(p)}{\Gamma(\alpha)\Gamma(\beta)}\sum_{k=1}^{p-1}
\sum_{l=0}^{\infty}(-1)^{k-1}(k-1)!
\Gamma(\alpha-k-l)\Gamma(\beta-k)
\frac{\xi^{-k}}{(p-k-1)!}\frac{\eta^{l}}{l!}.
\end{align*}
Thus, a solution of system \eqref{GM4_3} is given by the functions
\begin{equation}\label{GM4_6}
\omega_1=H_3(\frac{n-1}{2},\frac{1}{2};1;
\frac{\rho_1^2-\rho2^2}{\rho^2},\frac{\lambda^2r_1^2}{4}),
\end{equation}
and
\begin{equation}\label{GM4_7}
\begin{aligned}
\omega_2
&= H_3\Big(\frac{n-1}{2},\frac{1}{2};1;\frac{\rho_1^2-\rho^2}{\rho_1^2},
\frac{\lambda^2r_1^2}{4}\Big)\ln(\Big(frac{\rho_1^2-\rho^2}{\rho_1^2}\Big) \\
&\quad +\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\frac{n-1}{2})_{k-l}(\frac{1}{2})_{k}}{(1)_{k}}
\Big(\psi(\frac{n-1}{2}+k-l)+\psi(\frac{1}{2}+k)-2\psi(1+k)\Big)\\
&\quad \times \frac{(\frac{\rho_1^2-\rho^2}{\rho_1^2})^{k}}{k!}
\frac{(\frac{\lambda^2r_1^2}{4})^{l}}{l!}.
\end{aligned}
\end{equation}

Next we derive that a solution of equation \eqref{GM_01} is provided
by the functions
\begin{equation}\label{GM4_8}
q_{\lambda 1}(M,M_0) = C_1\rho_1^{-(n-1)}H_3(\frac{n-1}{2},\frac{1}{2};1;
\frac{\rho_1^2-\rho^2}{\rho_1^2},\frac{\lambda^2r_1^2}{4}),
\end{equation}
and
\begin{equation}\label{GM4_9}
\begin{aligned}
q_{\lambda 2}(M,M_0)
&= C_2\rho_1^{-(n-1)}\Bigl( H_3\Big(\frac{n-1}{2},\frac{1}{2};1;
 \frac{\rho_1^2-\rho^2}
{\rho_1^2},\frac{\lambda^2r_1^2}{4}\Big)\ln(\frac{\rho_1^2-\rho^2}{\rho_1^2}) \\
&\quad +\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}
\frac{(\frac{n-1}{2})_{k-l}(\frac{1}{2})_{k}}{(1)_{k}}
\Big(\psi(\frac{n-1}{2}+k-l)+\psi(\frac{1}{2}+k)\\
&\quad -2\psi(1+k)\Big)
\frac{(\frac{\rho_1^2-\rho^2}{\rho_1^2})^{k}}{k!}
\frac{(\frac{\lambda^2r_1^2}{4})^{l}}{l!}\Bigl).
\end{aligned}
\end{equation}
Let us consider some properties of the solutions.

\begin{lemma} \label{lem3}
The solutions \eqref{GM3_5_1} and
\eqref{GM4_8} satisfy
$$
\frac{\partial q_{01}(M,M_0)}{\partial x_n}\Big|_{x_n=0}=0, \quad
\frac{\partial q_{\lambda1}(M,M_0)}{\partial x_n}\Big|_{x_n=0}=0.
$$
\end{lemma}

The proof of the above lemma is a straightforward calculation.

\begin{lemma} \label{lem4}
The solutions \eqref{GM3_5_1}
and \eqref{GM4_8} possess a singularity $\rho^{2-n}$ as $\rho\to 0$.
\end{lemma}

The proof of the above lemma is  carried out analogously to that in Lemma 2.
We omit it.
Lemma \ref{lem4} implies the following result.


\begin{theorem} \label{thm2}
The functions \eqref{GM3_5_1} and \eqref{GM4_8}
are fundamental solutions of equations \eqref{GM_01} with
$\lambda=0$ and with $\lambda\neq0$, respectively.
\end{theorem}

\subsection*{Conclusions}

Nigmedzyanova \cite{Negmedzianova1} obtained a fundamental
solution of the equation \eqref{GM_00} by using generalized shift
operator technique \cite{Levitan}. To this end, by the change of variables
$$
\xi_{i}=x_{i}, \quad j=\overline{1,n-1}, \quad
\xi_{n}=\frac{2}{m+2}x_{n}^{\frac{m+2}{2}}
$$
she reduced equation \eqref{GM_00} to the form
\begin{equation}\label{GM5}
\sum_{i=1}^{n-1}\frac{\partial^2u}{\partial \xi_i^2}
+\frac{\partial^2u}{\partial \xi_n^2}+
\frac{m}{m+2}\frac{1}{\xi_n}\frac{\partial u}{\partial \xi_n}
+\lambda^2 u=0.
\end{equation}
Clearly, for no value of $m$, equation \eqref{GM_00} is reducible to
equation \eqref{GM5}. Therefore, in the present article, along with
equation \eqref{GM_00} equation \eqref{GM_01} was considered.

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\end{document}
