A Solution of a Boundary Value Problem with Mixed Conditions for a Four-Dimensional Degenerate Elliptic Equation

Abstract

The solvability issues of counterpart Holmgren’s boundary value problem with mixed conditions for a degenerate four-dimensional second-order Gellerstedt equation $H\left(u\right)\equiv y^m{z}^k{t}^l{u}_{xx}+x^n{z}^k{t}^l{u}_{yy}+x^n{y}^m{t}^l{u}_{zz}+x^n{y}^m{z}^k{u}_{tt}=0,$ $m,n,k,l\equiv const>0,$ are studied in the finite domain $R_4^{+}$, where the values of normal derivatives are set on the piecewise smooth part of the boundary and the values of the desired function are set on the remaining part of the boundary. The main results of the work are the proof of the uniqueness of the considered problem solution by using an energy integral’s method and the construction of the solution of counterpart Holmgren’s boundary value problem in explicit form by means of Green’s function method, containing the hypergeometric Lauricella’s function $F_A^{\left(4\right)}$. Using the corresponding fundamental solution for the considered generalized Gellerstedt equation of elliptic type, we construct Green’s function. In addition, formulas of differentiation, some adjacent relations, decomposition formulas, and various properties of Lauricella’s hypergeometric functions were used to establish the main results for the aforementioned problem.

1. Introduction

Differential equations with degenerations, the solutions of which are expressed using special functions, are used in various applied problems. The first fundamental results for differential equations, in the study of which the hypergeometric Gauss function participated, belong to F. Tricomi [1]. Later, Tricomi’s problems found their development in the applied works of F.I. Frankl [2]. Further in [3], Dirichlet and N boundary value problems for a two-dimensional degenerate elliptic equation with one line of degeneracy were studied in detail; the existence of these problems’ solutions was proved by constructing the single-layer potentials. It is known that hypergeometric functions are used to solve such problems of mathematical physics as heat conduction problems, superstring theory, gas dynamics, celestial mechanics, the theory of electromagnetic oscillations, and aerodynamics [4,5,6,7,8,9,10,11]. Appel’s hypergeometric functions of two variables arise in solving physical problems of quantum mechanics of atomic systems [12].

Boundary value problems for a degenerate elliptic equation of two variables were studied by S. Gellerstedt using potential theory [13]. The articles [14,15,16] are devoted to finding fundamental solutions and constructing the double-layer potential for the generalized biaxially symmetric Helmholtz equation. Fundamental solutions for a three-dimensional elliptic equation with singular coefficients were constructed in [17].

In [18], we found sixteen fundamental solutions for the four-dimensional generalized Gellerstedt equation

$$H\left(u\right)=y^m{z}^k{t}^l{u}_{xx}+x^n{z}^k{t}^l{u}_{yy}+x^n{y}^m{t}^l{u}_{zz}+x^n{y}^m{z}^k{u}_{tt}=0, m,n,k,l\equiv const>0$$

(1)

and applying appropriate fundamental solutions the solvability of Dirichlet and Neumann boundary value problems in an infinite area $R_{+}^4=\left\{\left(x,y,z,t\right):x>0,y>0,z>0,t>0\right\}$ was studied in [19,20]; the uniqueness and existence of the solution were proved.

In the course of the realization of the work purpose, the method used in [21,22] was applied, where solutions of boundary value problems for three-dimensional equations with singular coefficients were successfully constructed. In this paper, the correct solvability of the analogue of the Holmgren’s boundary value problem for Equation (1) is studied in the finite domain. The Green’s function of the problem under study is constructed using the corresponding fundamental solution of the Equation (1).

2. Preliminary Information

In this section, we recall some results and formulas related to Lauricella’s hypergeometric functions which will be used later in this work.

A multidimensional hypergeometric function $F_A^{\left(n\right)}$ in the case of four variables has the form [23] (p. 114 (1))

$$F_A^{(4)}(a;b_1,b_2,b_3,b_4;c_1,c_2,c_3,c_4;x,y,z,t)={\displaystyle \sum _{m,n,p,q}^{\infty }\frac{{\left(a\right)}_{m+n+p+q}{\left(b_1\right)}_m{\left(b_2\right)}_n{\left(b_3\right)}_p{\left(b_4\right)}_q}{{\left(c_1\right)}_m{\left(c_2\right)}_n{\left(c_3\right)}_p{\left(c_4\right)}_{q}m!n!p!q!}}{x}^m{y}^n{z}^p{t}^q,$$

(2)

moreover $F_A^{(4)}$ uniformly converges at the condition $\left|x\right|+\left|y\right|+\left|z\right|+\left|t\right|<1$ [23] (p. 115).

Late we will use the following formula [24] for differentiating hypergeometric functions of four variables:

$$\begin{array}{l}\frac{{\partial }^{l_1+l_2+l_3+l_4}}{\partial x^{l_1}\partial y^{l_2}\partial z^{l_3}\partial t^{l_4}}{F}_A^{\left(4\right)}\left(a;b_1,b_2,b_3,b_4;c_1,c_2,c_3,c_4;x,y,z,t\right)=\frac{{\left(a\right)}_{l_1+l_2+l_3+l_4}{\left(b_1\right)}_{l_1}{\left(b_2\right)}_{l_2}{\left(b_3\right)}_{l_3}{\left(b_4\right)}_{l_4}}{{\left(c_1\right)}_{l_1}{\left(c_2\right)}_{l_2}{\left(c_3\right)}_{l_3}{\left(c_4\right)}_{l_4}}\\ \times F_A^{\left(4\right)}\left(a+l_1+l_2+l_3+l_4;b_1+l_1,b_2+l_2,b_3+l_3,b_4+l_4;c_1+l_1,c_2+l_2,c_3+l_3,c_4+l_4;x,y,z,t\right),\\ l_1,l_2,l_3,l_4\in N_0=\left\{0,1,2,\dots \right\}.\end{array}$$

(3)

The validity of adjacent relations for multidimensional Lauricella’s hypergeometric functions was proved in [18] (p. 635):

$$\begin{array}{l}\frac{b_1}{c_1}{x}_1{F}_A^{\left(n\right)}\left(a+1;b_1+1,b_2,\dots ,b_n;c_1+1,c_2,\dots ,c_n;x_1,\dots ,x_n\right)+\\ +\frac{b_2}{c_2}{x}_2{F}_A^{\left(n\right)}\left(a+1;b_1,b_2+1,\dots ,b_n;c_1,c_2+1,\dots ,c_n;x_1,\dots ,x_n\right)+\\ +\dots +\frac{b_n}{c_n}{x}_n{F}_A^{\left(n\right)}\left(a+1;b_1,b_2,\dots ,b_n+1;c_1,c_2,\dots ,c_n+1;x_1,\dots ,x_n\right)=\\ =F_A^{\left(n\right)}\left(a+1;b_1,\dots ,b_n;c_1,\dots ,c_n;x_1,\dots ,x_n\right)-F_A^{\left(n\right)}\left(a;b_1,\dots ,b_n;c_1,\dots ,c_n;x_1,\dots ,x_n\right).\end{array}$$

(4)

We will also use the decomposition formula of many arguments [25]:

$$\begin{array}{l}{F}_A^{\left(r\right)}\left[a,b_1,\dots ,b_r;c_1,\dots ,c_r;x_1,\dots ,x_r\right]={\displaystyle \sum _{m_2,\dots ,m_r=0}^{\infty }}\frac{{\left(a\right)}_{m_2+\dots +m_r}{\left(b_1\right)}_{m_2+\dots +m_r}{\left(b_2\right)}_{m_2}\cdots {\left(b_r\right)}_{m_r}}{m_2!\cdots m_r!{\left(c_1\right)}_{m_2+\dots +m_r}{\left(c_2\right)}_{m_2}\cdots {\left(c_r\right)}_{m_r}}\\ \times x_1^{m_2+\cdots +m_r}{x}_2^{m_2}\cdots x_r^{m_r}{}_2{F}_1\left(a+m_2+\dots +m_r,b_1+m_2+\cdots +m_r;c_1+m_2+\cdots +m_r;x_1\right)\\ \times F_A^{\left(r-1\right)}\left[a+m_2+\cdots +m_r,b_2+m_2,\dots ,b_r+m_r;c_2+m_2,\dots ,c_r+m_r;x_2,\dots ,x_r\right],\\ \left(r\in \mathbb{N}\backslash \left\{1\right\};\mathbb{N}:=\left\{1,2,3,\dots \right\}\right).\end{array}$$

(5)

The Boltz auto transformation formula will be used [24] (p. 113)

$$F\left(a,b;c;x\right)={\left(1-x\right)}^{-b}F\left(c-a,b;c;\frac{x}{x-1}\right),$$

(6)

which is valid for the Gauss hypergeometric function

$$F\left(a,b;c;x\right)={\displaystyle \sum _{n=0}^{\infty }}\frac{{\left(a\right)}_n{\left(b\right)}_n}{{\left(c\right)}_{n}n!}{x}^n, c\ne 0, -1, -2, \dots ,$$

where

$${\left(a\right)}_n=a\left(a+1\right)\left(a+2\right)\dots \left(a+n-1\right) =\frac{\Gamma \left(a+n\right)}{\Gamma \left(a\right)} \left(n=0,1,2,3,\dots \right)$$

(7)

is the Pochhammer symbol. Here $\Gamma \left(\cdot \right)$ is Euler gamma function, which has the property:

$${\left(\lambda \right)}_{m+n}={\left(\lambda \right)}_m{\left(\lambda +m\right)}_n.$$

(8)

For Lauricella’s hypergeometric function [23] (p. 114 (4))

$$\begin{array}{l}{F}_D^{\left(n\right)}\left(a;b_1,b_2,b_3,\mathrm{...},b_n;c;x_1,x_2,\dots ,x_n\right)={\displaystyle \sum _{m_1,\dots ,m_n=0}^{\infty }\frac{{\left(a\right)}_{m_1+\dots +m_n}{\left(b_1\right)}_{m_1}\dots {\left(b_n\right)}_{m_n}}{{\left(c\right)}_{m_1+\dots +m_n}{m}_1!\dots m_n!}{x}_1^{m_1}\dots x_n^{m_n}},\\ \left(\left|x_1\right|<1, \left|x_2\right|<1, \dots , \left|x_n\right|<1\right),\end{array}$$

following formula [24] (p. 117) holds, when n = 1:

$$\begin{array}{l}{F}_D^{\left(n\right)}\left(a;b_1,b_2,b_3,\dots ,b_n;c;1,1,\dots ,1\right)=\frac{\Gamma \left(c\right)\Gamma \left(c-a-b_1-b_2-\dots -b_n\right)}{\Gamma \left(c-a\right)\Gamma \left(c-b_1-b_2-\dots -b_n\right)}, n=1,2,\dots \\ \mathrm{Re}\left(c-a-b_1-b_2-\dots -b_n\right)>0, c\ne 0,-1,-2,\dots .\end{array}$$

(9)

3. An Analogue of the Holmgren’s Boundary Value Problem

3.1. Problem Statement and Main Results

Let D be a finite simply connected domain in $R_4^{+}$ bounded by hyperplanes

$$\begin{gathered} S_1=\left\{\left(0,y,z,t\right): x=0 , 0<y<b, 0<z<c, 0<t<e \right\}, \\ {S}_2=\left\{\left(x,0,z,t\right): 0<x<a, y=0, 0<z<c, 0<t<e\right\}, \\ {S}_3=\left\{\left(x,y,0,t\right): 0<x<a, 0<y<b, z=0, 0<t<e\right\}, \\ {S}_4=\left\{\left(x,y,z,0\right): 0<x<a, 0<y<b, 0<z<c, t=0\right\} \end{gathered}$$

and smooth surface $S_5$. Here $a, b, c, e$ are positive numbers, $A\left(a,0,0,0\right),$ $B\left(0,b,0,0\right),$ $C\left(0,0,c,0\right),$ $E\left(0,0,0,e\right)$ and $O\left(0,0,0,0\right)$.

Problem N.

Find a regular solution $u\left(x,y,z,t\right)$ of the Equation (1) from the class $C\left(\overline{D}\right)\cap C^1\left(D\cup S_1\cup S_2\cup S_3\cup S_4\right)\cap C^2\left(D\right)$ satisfying the conditions

$${\left.\frac{\partial }{\partial x}u\left(x,y,z,t\right)\right|}_{x=0}={\nu }_1\left(y,z,t\right), \left(y,z,t\right)\in S_1,$$

(10)

$${\left.\frac{\partial }{\partial y}u\left(x,y,z,t\right)\right|}_{y=0}={\nu }_2\left(x,z,t\right), \left(x,z,t\right)\in S_2,$$

(11)

$${\left.\frac{\partial }{\partial z}u\left(x,y,z,t\right)\right|}_{z=0}={\nu }_3\left(x,y,t\right), \left(x,y,t\right)\in S_3,$$

(12)

$${\left.\frac{\partial }{\partial t}u\left(x,y,z,t\right)\right|}_{t=0}={\nu }_4\left(x,y,z\right), \left(x,y,z\right)\in S_4,$$

(13)

$$u\left(x,y,z,t\right)=\phi \left(x,y,z,t\right), \left(x,y,z,t\right)\in \overline{S_5},$$

(14)

where ${\nu }_1\left(y,z,t\right)$, ${\nu }_2\left(x,z,t\right)$, ${\nu }_3\left(x,y,t\right)$, ${\nu }_4\left(x,y,z\right)$, $\phi \left(x,y,z,t\right)$ are given continuous functions, moreover the functions ${\nu }_i \left(i=\overline{1, 4}\right)$ at the origin of coordinates can tend to infinity of the integrable order.

Theorem 1.

Let ${\nu }_1\left(y,z,t\right)\in C\left(S_1\right),$  ${\nu }_2\left(x,z,t\right)\in C\left(S_2\right),$  ${\nu }_3\left(x,y,t\right)\in C\left(S_3\right),$   ${\nu }_4\left(x,y,z\right)\in C\left(S_4\right),$  $\phi \left(x,y,z,t\right)\in C^2\left(S_5\right)$, then the solution of the problem $N$ has unique solution and presented in the following form:

$$\begin{array}{ll}u\left(x_0,y_0,z_0,t_0\right)=& -{\displaystyle \underset{S_1}{\iiint }}{y}^m{z}^k{t}^l{\nu }_1\left(y,z,t\right)G_1\left(0,y,z,t;x_0,y_0,z_0,t_0\right)dydzdt\\ & -{\displaystyle \underset{S_2}{\iiint }}{x}^n{z}^k{t}^l{\nu }_2\left(x,z,t\right)G_1\left(x,0,z,t;x_0,y_0,z_0,t_0\right)dxdzdt\\ & -{\displaystyle \underset{S_3}{\iiint }}{x}^n{y}^m{t}^l{\nu }_3\left(x,y,t\right)G_1\left(x,y,0,t;x_0,y_0,z_0,t_0\right)dxdydt\\ & -{\displaystyle \underset{S_4}{\iiint }}{x}^n{y}^m{z}^k{\nu }_4\left(x,y,z\right)G_1\left(x,y,z,0;x_0,y_0,z_0,t_0\right)dxdydz\\ & -{\displaystyle \underset{S_5}{\iiint }}{x}^n{y}^m{z}^k{t}^l\phi \left(\sigma \right)A_S\left[G_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right]d\sigma .\end{array}.$$

(15)

3.2. Proof of Results

3.2.1. Uniqueness of Solution

Let $u\left(x,y,z,t\right)$ be the solution of a homogeneous problem N, i.e., satisfying the conditions (10)–(14) when

$${\nu }_1\left(y,z,t\right)={\nu }_2\left(x,z,t\right)={\nu }_3\left(x,y,t\right)={\nu }_4\left(x,y,z\right)=\phi \left(x,y,z,t\right)=0.$$

(16)

By $D_{\epsilon }$, we denote the domain strictly lying inside the domain $D$, $D_{\epsilon }\subset D$, where the distance from the border $\partial D_{\epsilon }$ to the border $\partial D$ is equal to $\epsilon$. Here $\partial D=S_1\cup S_2\cup S_3\cup S_4\cup S_5$.

By applying easily verifiable equalities to the solution $u\left(x,y,z,t\right)$ of the Equation (1)

$$\begin{array}{l}{y}^m{z}^k{t}^{l}u{u}_{xx}=\frac{\partial }{\partial x}\left(y^m{z}^k{t}^{l}u{u}_x\right)-y^m{z}^k{t}^l{u}_x^2, x^n{z}^k{t}^{l}u{u}_{yy}=\frac{\partial }{\partial y}\left(x^n{z}^k{t}^{l}u{u}_y\right)-x^n{z}^k{t}^l{u}_y^2,\\ x^n{y}^m{t}^{l}u{u}_{zz}=\frac{\partial }{\partial z}\left(x^n{y}^m{t}^{l}u{u}_z\right)-x^n{y}^m{t}^l{u}_z^2, x^n{y}^m{z}^{k}u{u}_{tt}=\frac{\partial }{\partial t}\left(x^n{y}^m{z}^{k}u{u}_t\right)-x^n{y}^m{z}^k{u}_t^2,\end{array}$$

it is not difficult to ascertain, that

$$\begin{array}{l}0={\displaystyle \int {\displaystyle \underset{D_{\epsilon }}{\iiint }uH\left(u\right)dxdydzdt}}={\displaystyle \int {\displaystyle \underset{D_{\epsilon }}{\iiint }\left[\frac{\partial }{\partial x}\left(y^m{z}^k{t}^{l}u{u}_x\right)\right.+}}\frac{\partial }{\partial y}\left(x^n{z}^k{t}^{l}u{u}_y\right)+\frac{\partial }{\partial z}\left(x^n{y}^m{t}^{l}u{u}_z\right)\\ \left.+\frac{\partial }{\partial t}\left(x^n{y}^m{z}^{k}u{u}_t\right)dxdydzdt\right]-{\displaystyle \int {\displaystyle \underset{D_{\epsilon }}{\iiint }\left(y^m{z}^k{t}^l{u}_x^2+x^n{z}^k{t}^l{u}_y^2+x^n{y}^m{t}^l{u}_z^2+x^n{y}^m{z}^k{u}_t^2\right)dxdydzdt}}.\end{array}$$

Hence

$$\begin{array}{l}{\displaystyle \int {\displaystyle \underset{D_{\epsilon }}{\iiint }\left(y^m{z}^k{t}^l{u}_x^2+x^n{z}^k{t}^l{u}_y^2+x^n{y}^m{t}^l{u}_z^2+x^n{y}^m{z}^k{u}_t^2\right)dxdydzdt}=}\\ ={\displaystyle \int {\displaystyle \underset{D_{\epsilon }}{\iiint }\left[\frac{\partial }{\partial x}\left(y^m{z}^k{t}^{l}u{u}_x\right)+\frac{\partial }{\partial y}\left(x^n{z}^k{t}^{l}u{u}_y\right)+\right.}}\left.\frac{\partial }{\partial z}\left(x^n{y}^m{t}^{l}u{u}_z\right)+\frac{\partial }{\partial t}\left(x^n{y}^m{z}^{k}u{u}_t\right)\right] dxdydzdt.\end{array}$$

(17)

Applying the Gauss–Ostrogradsky formula to (17), then passing to the limit at $\epsilon \to 0$, we have

$$\begin{array}{l}{\displaystyle \int {\displaystyle \underset{D}{\iiint }\left(y^m{z}^k{t}^l{u}_x^2+x^n{z}^k{t}^l{u}_y^2+x^n{y}^m{t}^l{u}_z^2+x^n{y}^m{z}^k{u}_t^2\right)dxdydzdt}}=\\ ={\displaystyle \underset{\partial D}{\iiint }u\left[y^m{z}^k{t}^l{u}_x\cos \left(n,x\right)+x^n{z}^k{t}^l{u}_y\cos \left(n,y\right)+\right.}\left.x^n{y}^m{t}^l{u}_z\cos \left(n,z\right)+x^n{y}^m{z}^k{u}_t\cos \left(n,t\right)\right] dS,\end{array}$$

where $n$ is external normal to $\partial D$, $\cos \left(n,x\right)dS=dydzdt,$ $\cos \left(n,y\right)dS=dxdzdt,$ $\cos \left(n,z\right)dS=dxdydt,$ $\cos \left(n,t\right)dS=dxdydz.$

Taking into account the boundary conditions (10)–(14), we obtain

$$\begin{array}{l}{\displaystyle \int {\displaystyle \underset{D}{\iiint }}}\left(y^m{z}^k{t}^l{u}_x^2+x^n{z}^k{t}^l{u}_y^2+x^n{y}^m{t}^l{u}_z^2+x^n{y}^m{z}^k{u}_t^2\right)dxdydzdt=\\ ={\displaystyle \underset{S_1}{\iiint }}{y}^m{z}^k{t}^{l}u{\nu }_{1}dydzdt+{\displaystyle \underset{S_2}{\iiint }}{x}^n{z}^k{t}^{l}u{\nu }_{2}dxdzdt+{\displaystyle \underset{S_3}{\iiint }}{x}^n{y}^m{t}^{l}u{\nu }_{3}dxdydt+\\ +{\displaystyle \underset{S_4}{\iiint }}{x}^n{y}^m{z}^{k}u{\nu }_{4}dxdydz-{\displaystyle \underset{S_5}{\iiint }}{x}^n{y}^m{z}^k{t}^l\phi A_s\left[u\right]dS,\end{array}$$

(18)

where

$$A_S\left[\right]=y^m{z}^k{t}^l\cos (x,n)\frac{\partial }{\partial x}+x^n{z}^k{t}^l\cos (y,n)\frac{\partial }{\partial y}+x^n{y}^m{t}^l\cos (z,n)\frac{\partial }{\partial z}+x^n{y}^m{z}^k\cos (t,n)\frac{\partial }{\partial t}$$

(19)

With considering (16) from (18) we get

$${\displaystyle \int {\displaystyle \underset{D}{\iiint }}}\left(y^m{z}^k{t}^l{u}_x^2+x^n{z}^k{t}^l{u}_y^2+x^n{y}^m{t}^l{u}_z^2+x^n{y}^m{z}^k{u}_t^2\right)dxdydzdt=0.$$

(20)

The equality (20) implies that $u_x=u_y=u_z=u_t=0$, i.e., so $u\left(x,y,z,t\right)=const$. Since the problem is homogeneous, then $u\left(x,y,z,t\right)=0$ in the domain $\overline{D}$; thus, uniqueness of solution is proved. □

3.2.2. Existence of Solution

Green’s function of the boundary value problem N. Consider the case when the surface $S_5$ at $x>0, y>0, z>0, t>0$ coincides with a part of the surface

$$\frac{4}{{\left(n+2\right)}^2}{x}^{n+2}+\frac{4}{{\left(m+2\right)}^2}{y}^{m+2}+\frac{4}{{\left(k+2\right)}^2}{z}^{k+2}+\frac{4}{{\left(l+2\right)}^2}{t}^{l+2}={\rho }^2.$$

In this case, we show the existence of the problem solution by applying the Green’s function construction method.

Definition 1.

The function  $G_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)$ is called the Green’s function of the problem N for the equation $H\left(u\right)\equiv 0$ if it satisfies the conditions:

(1) 

This function is a regular solution of the Equation (1) inside the domain$D$except for the point$\left(x_0,y_0,z_0,t_0\right)$;

(2) 

It satisfies the boundary conditions

$$\begin{gathered} {\left.\frac{\partial }{\partial x}{G}_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{x=0}=0, {\left.\frac{\partial }{\partial y}{G}_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{y=0}=0, \\ {\left.\frac{\partial }{\partial z}{G}_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{z=0}=0,{\left.\frac{\partial }{\partial t}{G}_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{t=0}=0 \\ {\left.G_1\left(x,y,z;x_0,y_0,z_0\right)\right|}_{S_5}=0; \end{gathered}$$

(21)

(3)

The Green’s function is represented as

$$G_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)=g_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)+g_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right);$$

(22)

where

$$g_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)=\frac{k_1}{{\left(r^2\right)}^{\alpha +\beta +\gamma +\delta +1}}{F}_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)$$

is the fundamental solution of the Equation (1), while$g_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)$is a regular solution of the Equation (1) everywhere in the domain$D$,$F_A^{\left(4\right)}$is represented in the form (2)

$$\begin{array}{l}{k}_1=\frac{1}{4{\pi }^2}{\left(\frac{4}{n+2}\right)}^{2\alpha }{\left(\frac{4}{m+2}\right)}^{2\beta }{\left(\frac{4}{k+2}\right)}^{2\gamma }{\left(\frac{4}{l+2}\right)}^{2\delta }\times \\ \times \frac{\Gamma \left(\alpha +\beta +\gamma +\delta +1\right)\Gamma \left(\alpha \right)\Gamma \left(\beta \right)\Gamma \left(\gamma \right)\Gamma \left(\delta \right)}{\Gamma \left(2\alpha \right)\Gamma \left(2\beta \right)\Gamma \left(2\gamma \right)\Gamma \left(2\delta \right)} ,\end{array}$$

(23)

$$\alpha =\frac{n}{2\left(n+2\right)}, \beta =\frac{m}{2\left(m+2\right)}, \gamma =\frac{k}{2\left(k+2\right)}, \delta =\frac{l}{2\left(l+2\right)},$$

$$\xi =\frac{r^2-r_1^2}{r^2}, \eta =\frac{r^2-r_2^2}{r^2}, \zeta =\frac{r^2-r_3^2}{r^2}, \varsigma =\frac{r^2-r_4^2}{r^2},$$

$$\begin{gathered} \left.\begin{array}{c}{r}^2\\ r_1^2\\ r_2^2\\ r_3^2\\ r_4^2\end{array}\right\}={\left(\frac{2}{n+2}{x}^{\frac{n+2}{2}} \begin{array}{c}-\\ +\\ -\\ -\\ -\end{array} \frac{2}{n+2}{x}_0^{\frac{n+2}{2}}\right)}^2+{\left(\frac{2}{m+2}{y}^{\frac{m+2}{2}} \begin{array}{c}-\\ -\\ +\\ -\\ -\end{array} \frac{2}{m+2}{y}_0^{\frac{m+2}{2}}\right)}^2+ \\ +{\left(\frac{2}{k+2}{z}^{\frac{k+2}{2}} \begin{array}{c}-\\ -\\ -\\ +\\ -\end{array} \frac{2}{k+2}{z}_0^{\frac{k+2}{2}}\right)}^2+{\left(\frac{2}{l+2}{t}^{\frac{l+2}{2}} \begin{array}{c}-\\ -\\ -\\ -\\ +\end{array} \frac{2}{l+2}{t}_0^{\frac{l+2}{2}}\right)}^2. \end{gathered}$$

The regular part of the Green’s function, by virtue of conditions (21) and representation (22), must satisfy the following boundary conditions

$$\begin{gathered} {\left.g_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{S_5}=-{\left.g_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{S_5}, \\ {\left.\frac{\partial }{\partial x}{g}_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{x=0}=0, {\left.\frac{\partial }{\partial y}{g}_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{y=0}=0, \\ {\left. {\left.\frac{\partial }{\partial z}{g}_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{z=0}=0, \frac{\partial }{\partial t}{g}_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right|}_{t=0}=0. \end{gathered}$$

(24)

Due to the conditions (24), the regular part of the Green’s function can be represented as

$$g_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)=-{\left(\frac{\rho }{R_0}\right)}^{2\left(\alpha +\beta +\gamma +\delta +1\right)}{g}_1\left(x,y,z,t;\overline{x_0},\overline{y_0},\overline{z_0},\overline{t_0}\right),$$

where

$$\begin{gathered} R_0^2=\frac{4}{{\left(n+2\right)}^2}{x}_0^{n+2}+\frac{4}{{\left(m+2\right)}^2}{y}_0^{m+2}+\frac{4}{{\left(k+2\right)}^2}{z}_0^{k+2}+\frac{4}{{\left(l+2\right)}^2}{t}_0^{l+2}, \\ {\overline{x_0}}^{\frac{n+2}{2}}=\frac{{\rho }^2}{R_0^2}{x}_0^{\frac{n+2}{2}}, {\overline{y_0}}^{\frac{m+2}{2}}=\frac{{\rho }^2}{R_0^2}{y}_0^{\frac{m+2}{2}}, {\overline{z_0}}^{\frac{k+2}{2}}=\frac{{\rho }^2}{R_0^2}{z}_0^{\frac{k+2}{2}}, {\overline{t_0}}^{\frac{l+2}{2}}=\frac{{\rho }^2}{R_0^2}{t}_0^{\frac{l+2}{2}}. \end{gathered}$$

By direct calculation, we can establish that the constructed function

$$G_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)=g_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)-{\left(\frac{\rho }{R_0}\right)}^{2\left(\alpha +\beta +\gamma +\delta +1\right)}{g}_1\left(x,y,z,t;\overline{x_0},\overline{y_0},\overline{z_0},\overline{t_0}\right)$$

satisfies all conditions of Definition 1. Thus, in the case when $a={\left(\frac{n+2}{2}\rho \right)}^{\frac{n+2}{2}}, b={\left(\frac{m+2}{2}\rho \right)}^{\frac{m+2}{2}}, c={\left(\frac{k+2}{2}\rho \right)}^{\frac{k+2}{2}}, e={\left(\frac{l+2}{2}\rho \right)}^{\frac{l+2}{2}}$, that is, when $S_5$ at $x>0, y>0, z>0, t>0$ coincides with 1/16 of the surface

$$\frac{4}{{\left(n+2\right)}^2}{x}^{n+2}+\frac{4}{{\left(m+2\right)}^2}{y}^{m+2}+\frac{4}{{\left(k+2\right)}^2}{z}^{k+2}+\frac{4}{{\left(l+2\right)}^2}{t}^{l+2}={\rho }^2$$

Proof of the solution existence.

Let the point $\left(x_0,y_0,z_0,t_0\right)\in D$. We cut out from the region $D$ a four-dimensional ball of small radius $\epsilon$ with the center at the point $\left(x_0,y_0,z_0,t_0\right)$, denote the remaining part of the region $D$ by $D^{\epsilon }$, in which the Green’s function $G_1\left(x,y,z;x_0,y_0,z_0\right)$ will be a regular solution of Equation (1), and denote the three-dimensional sphere of the cut ball by $C^{\epsilon }$. □

Consider the identity

$$\begin{array}{l}uH\left(w\right)-wH\left(u\right)=\frac{\partial }{\partial x}\left[y^m{z}^k{t}^l\left(u{w}_x-w{u}_x\right)\right]+\frac{\partial }{\partial y}\left[x^n{z}^k{t}^l\left(u{w}_y-w{u}_y\right)\right]+\\ +\frac{\partial }{\partial z}\left[x^n{y}^m{t}^l\left(u{w}_z-w{u}_z\right)\right]+\frac{\partial }{\partial t}\left[x^n{y}^m{z}^k\left(u{w}_t-w{u}_t\right)\right].\end{array}$$

Integrating the reduced identity over the domain $D$ and applying the Gauss–Ostrogradsky formula, we obtain the following formula:

$$\begin{array}{l}{\displaystyle \underset{D}{\iiint }{\displaystyle \int }}\left[uH\left(w\right)-wH\left(u\right)\right]dxdydzdt={\displaystyle \underset{\partial D}{\iiint }}\left[y^m{z}^k{t}^l\left(u{w}_x-w{u}_x\right)\cos \left(n,x\right)+\right.x^n{z}^k{t}^l\times \\ \times \left(u{w}_y-w{u}_y\right)\cos \left(n,y\right)+\left.x^n{y}^m{t}^l\left(u{w}_z-w{u}_z\right)\cos \left(n,z\right)+x^n{y}^m{z}^k\left(u{w}_t-w{u}_t\right)\cos \left(n,t\right)\right]dS.\end{array}$$

(25)

If $u\left(x,y,z,t\right)$ and w(x,y,z,t) are the solutions of Equation (1), then from Formula (25) we have

$${\displaystyle \underset{\partial D}{\iiint }\left(u{A}_s\left[w\right]-w{A}_s\left[u\right]\right)dS}=0,$$

(26)

where $A_s\left[u\right]$ and $A_s\left[w\right]$ determined by the Formula (19).

By applying the Formula (26), we get

$$\begin{array}{ll}{\displaystyle \underset{C^{\epsilon }}{\iiint }\left(u{A}_s\left[G_1\right]-G_1{A}_s\left[u\right]\right)dS}=& -{\displaystyle \underset{S_1}{\iiint }{y}^m{z}^k{t}^l{\nu }_1\left(y,z,t\right)G_1\left(0,y,z,t;x_0,y_0,z_0,t_0\right)dydzdt}-\\ & -{\displaystyle \underset{S_2}{\iiint }{x}^n{z}^k{t}^l{\nu }_2\left(x,z,t\right)G_1\left(x,0,z,t;x_0,y_0,z_0,t_0\right)dxdzdt}-\\ & -{\displaystyle \underset{S_3}{\iiint }{x}^n{y}^m{t}^l{\nu }_3\left(x,y,t\right)G_1\left(x,y,0,t;x_0,y_0,z_0,t_0\right)dxdydt}-\\ & -{\displaystyle \underset{S_4}{\iiint }{x}^n{y}^m{z}^k{\nu }_4\left(x,y,z\right)G_1\left(x,y,z,0;x_0,y_0,z_0,t_0\right)dxdydz}-\\ & -{\displaystyle \underset{S_5}{\iiint }{x}^n{y}^m{z}^k{t}^l\phi \left(\sigma \right)A_s\left[G_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right]d\sigma .}\end{array}$$

(27)

In (27), the following formulas for the functions $F_A^{\left(4\right)}$ contained in the representation of the Green’s function $G_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)$ (22) are valid:

$$\begin{gathered} F_A^{\left(4\right)}\left(a;b_1,b_2,b_3,b_4;c_1,c_2,c_3,c_4;0,y,z,t\right)=F_A^{\left(3\right)}\left(a;b_2,b_3,b_4;c_2,c_3,c_4;y,z,t\right), \\ {F}_A^{\left(4\right)}\left(a;b_1,b_2,b_3,b_4;c_1,c_2,c_3,c_4;x,0,z,t\right)=F_A^{\left(3\right)}\left(a;b_1,b_3,b_4;c_1,c_3,c_4;x,z,t\right), \\ {F}_A^{\left(4\right)}\left(a;b_1,b_2,b_3,b_4;c_1,c_2,c_3,c_4;x,y,0,t\right)=F_A^{\left(3\right)}\left(a;b_1,b_2,b_4;c_1,c_2,c_4;x,y,t\right), \\ {F}_A^{\left(4\right)}\left(a;b_1,b_2,b_3,b_4;c_1,c_2,c_3,c_4;x,y,z,0\right)=F_A^{\left(3\right)}\left(a;b_1,b_2,b_3;c_1,c_2,c_3;x,y,z\right). \end{gathered}$$

(28)

By substituting (28) in (22), we have

$$\begin{array}{l}{G}_1\left(0,y,z,t;x_0,y_0,z_0,t_0\right)=k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\beta ,\gamma ,\delta ;2\beta ,2\gamma ,2\delta ;{\eta }_{01}^{\left(x\right)},{\zeta }_{01}^{\left(x\right)},{\varsigma }_{01}^{\left(x\right)}\right)}{{\left(r_x^2\right)}^{\alpha +\beta +\gamma +\delta +1}}\\ -k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\beta ,\gamma ,\delta ;2\beta ,2\gamma ,2\delta ;{\eta }_{02}^{\left(x\right)},{\zeta }_{02}^{\left(x\right)},{\varsigma }_{02}^{\left(x\right)}\right)}{{\left(r_{sx}^2\right)}^{\alpha +\beta +\gamma +\delta +1}},\\ G_1\left(x,0,z,t;x_0,y_0,z_0,t_0\right)=k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\gamma ,\delta ;2\alpha ,2\gamma ,2\delta ;{\xi }_{01}^{\left(y\right)},{\zeta }_{01}^{\left(y\right)},{\varsigma }_{01}^{\left(y\right)}\right)}{{\left(r_y^2\right)}^{\alpha +\beta +\gamma +\delta +1}}\\ -k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\gamma ,\delta ;2\alpha ,2\gamma ,2\delta ;{\xi }_{02}^{\left(y\right)},{\zeta }_{02}^{\left(y\right)},{\varsigma }_{02}^{\left(y\right)}\right)}{{\left(r_{sy}^2\right)}^{\alpha +\beta +\gamma +\delta +1}},\\ G_1\left(x,y,0,t;x_0,y_0,z_0,t_0\right)=k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\beta ,\delta ;2\alpha ,2\beta ,2\delta ;{\xi }_{01}^{\left(z\right)},{\eta }_{01}^{\left(z\right)},{\varsigma }_{01}^{\left(z\right)}\right)}{{\left(r_z^2\right)}^{\alpha +\beta +\gamma +\delta +1}}\\ -k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\beta ,\delta ;2\alpha ,2\beta ,2\delta ;{\xi }_{02}^{\left(z\right)},{\eta }_{02}^{\left(z\right)},{\varsigma }_{02}^{\left(z\right)}\right)}{{\left(r_{sz}^2\right)}^{\alpha +\beta +\gamma +\delta +1}},\\ G_1\left(x,y,z,0;x_0,y_0,z_0,t_0\right)=k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\beta ,\gamma ;2\alpha ,2\beta ,2\gamma ;{\xi }_{01}^{\left(t\right)},{\eta }_{01}^{\left(t\right)},{\zeta }_{01}^{\left(t\right)}\right)}{{\left(r_t^2\right)}^{\alpha +\beta +\gamma +\delta +1}}\\ -k_1\frac{F_A^{\left(3\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\beta ,\gamma ;2\alpha ,2\beta ,2\gamma ;{\xi }_{02}^{\left(t\right)},{\eta }_{02}^{\left(t\right)},{\zeta }_{02}^{\left(t\right)}\right)}{{\left(r_{st}^2\right)}^{\alpha +\beta +\gamma +\delta +1}},\end{array}$$

(29)

where

$$\begin{gathered} r_x^2={\left.r^2\right|}_{x=0}, r_y^2={\left.r^2\right|}_{y=0}, r_z^2={\left.r^2\right|}_{z=0}, r_t^2={\left.r^2\right|}_{t=0}, \\ {\xi }_{0i}^{\left(y\right)}={\xi }_i\left|{}_{y=0}\right. , {\xi }_{0i}^{\left(z\right)}={\xi }_i\left|{}_{z=0}\right. , {\xi }_{0i}^{\left(t\right)}={\xi }_i\left|{}_{t=0}\right. ,{\eta }_{0i}^{\left(x\right)}={\eta }_i\left|{}_{x=0}\right. , {\eta }_{0i}^{\left(z\right)}={\eta }_i\left|{}_{z=0}\right. , {\eta }_{0i}^{\left(t\right)}={\eta }_i\left|{}_{t=0}\right. , \\ {\zeta }_{0i}^{\left(x\right)}={\zeta }_i\left|{}_{x=0}\right. , {\zeta }_{0i}^{\left(y\right)}={\zeta }_i\left|{}_{y=0}\right. , {\zeta }_{0i}^{\left(t\right)}={\zeta }_i\left|{}_{t=0}\right. , {\varsigma }_{0i}^{\left(x\right)}={\varsigma }_i\left|{}_{x=0}\right. , {\varsigma }_{0i}^{\left(y\right)}={\varsigma }_i\left|{}_{y=0}\right. , {\varsigma }_{0i}^{\left(z\right)}={\varsigma }_i\left|{}_{z=0}\right. , \\ {\xi }_i={\sigma }_i{\left(\frac{2}{n+2}\right)}^2{\left(x{x}_0\right)}^{\frac{n+2}{2}},{\eta }_i={\sigma }_i{\left(\frac{2}{m+2}\right)}^2{\left(y{y}_0\right)}^{\frac{m+2}{2}},{\zeta }_i={\sigma }_i{\left(\frac{2}{k+2}\right)}^2{\left(z{z}_0\right)}^{\frac{k+2}{2}}, \\ {\varsigma }_i={\sigma }_i{\left(\frac{2}{l+2}\right)}^2{\left(t{t}_0\right)}^{\frac{l+2}{2}}, \left(i=1,2\right),{\sigma }_1=-\frac{4}{r^2},{\sigma }_2=-\frac{4}{r_s^2}\frac{{\rho }^2}{R_0^2}, \\ {r}_s^2={\left(\rho -\frac{4}{{\left(n+2\right)}^2}\frac{{\left(x{x}_0\right)}^{\frac{n+2}{2}}}{\rho }\right)}^2+{\left(\rho -\frac{4}{{\left(m+2\right)}^2}\frac{{\left(y{y}_0\right)}^{\frac{m+2}{2}}}{\rho }\right)}^2+{\left(\rho -\frac{4}{{\left(k+2\right)}^2}\frac{{\left(z{z}_0\right)}^{\frac{k+2}{2}}}{\rho }\right)}^2+ \\ +{\left(\rho -\frac{4}{{\left(l+2\right)}^2}\frac{{\left(t{t}_0\right)}^{\frac{l+2}{2}}}{\rho }\right)}^2+\frac{\frac{4}{{\left(m+2\right)}^2}{y}_0^{m+2}+\frac{4}{{\left(k+2\right)}^2}{z}_0^{k+2}+\frac{4}{{\left(l+2\right)}^2}{t}_0^{l+2}}{{\rho }^2}\frac{4}{{\left(n+2\right)}^2}{x}^{n+2}+ \\ +\frac{\frac{4}{{\left(n+2\right)}^2}{x}_0^{n+2}+\frac{4}{{\left(k+2\right)}^2}{z}_0^{k+2}+\frac{4}{{\left(l+2\right)}^2}{t}_0^{l+2}}{{\rho }^2}\frac{4}{{\left(m+2\right)}^2}{y}^{m+2}+ \\ +\frac{\frac{4}{{\left(n+2\right)}^2}{x}_0^{n+2}+\frac{4}{{\left(m+2\right)}^2}{y}_0^{m+2}+\frac{4}{{\left(l+2\right)}^2}{t}_0^{l+2}}{{\rho }^2}\frac{4}{{\left(k+2\right)}^2}{z}^{k+2}+ \\ +\frac{\frac{4}{{\left(n+2\right)}^2}{x}_0^{n+2}+\frac{4}{{\left(m+2\right)}^2}{y}_0^{m+2}+\frac{4}{{\left(k+2\right)}^2}{z}_0^{k+2}}{{\rho }^2}\frac{4}{{\left(l+2\right)}^2}{t}^{l+2}-3{\rho }^2, \\ {r}_{sx}^2={\left.r_s^2\right|}_{x=0}, r_{sy}^2={\left.r_s^2\right|}_{y=0}, r_{sz}^2={\left.r_s^2\right|}_{z=0}, r_{st}^2={\left.r_s^2\right|}_{t=0}. \end{gathered}$$

The left part of the Equation (27) we divide into two integrals

$${\displaystyle \underset{C^{\epsilon }}{\iiint }\left(u{A}_s\left[G_1\right]-G_1{A}_s\left[u\right]\right)dS}={\displaystyle \underset{C^{\epsilon }}{\iiint }u{A}_s\left[G_1\right]dS}-{\displaystyle \underset{C^{\epsilon }}{\iiint }{G}_1{A}_s\left[u\right]}dS,$$

(30)

we calculate the first of these integrals

$$I={\displaystyle \underset{C^{\epsilon }}{\iiint }u{A}_s\left[G_1\right]dS}.$$

(31)

By virtue of (22), the integral (31) is converted as the following:

$$I=I_1+I_2={\displaystyle \underset{C^{\epsilon }}{\iiint }u{A}_s\left[g_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right]dS}+{\displaystyle \underset{C^{\epsilon }}{\iiint }u{A}_s\left[g_1^{\ast }\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right]dS}.$$

(32)

According to (19)

$$\begin{array}{l}{A}_s\left[g_1\right]=y^m{z}^k{t}^l\cos (x,n)\frac{\partial g_1}{\partial x}+x^n{z}^k{t}^l\cos (y,n)\frac{\partial g_1}{\partial y}+\\ +x^n{y}^m{t}^l\cos (z,n)\frac{\partial g_1}{\partial z}+x^n{y}^m{z}^k\cos (t,n)\frac{\partial g_1}{\partial t}.\end{array}$$

(33)

Using the formula for differentiating hypergeometric functions (3) we calculate the partial derivatives of the fundamental solution $g_1$ from (33):

$$\begin{array}{l}\frac{\partial g_1}{\partial x}=-\frac{\left(\alpha +\beta +\gamma +\delta +1\right)k_1{x}^{\frac{n}{2}}}{{\left(r^2\right)}^{\alpha +\beta +\gamma +\delta +2}}\frac{4}{n+2}\left(x^{\frac{n+2}{2}}-x_0^{\frac{n+2}{2}}\right)\times \\ \times F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +1;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)-2\frac{\left(\alpha +\beta +\gamma +\delta +1\right)k_1{x}^{\frac{n}{2}}}{{\left(r^2\right)}^{\alpha +\beta +\gamma +\delta +2}}\times \\ \left\{\frac{4}{n+2}\frac{{\left(\alpha \right)}_1}{{\left(2\alpha \right)}_1}{x}_0^{\frac{n+2}{2}}{F}_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha +1,\beta ,\gamma ,\delta ;2\alpha +1,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\right.+\\ \frac{2}{n+2}\left(x^{\frac{n+2}{2}}-x_0^{\frac{n+2}{2}}\right)\left[\frac{{\left(\alpha \right)}_1}{{\left(2\alpha \right)}_1}\right.\xi F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha +1,\beta ,\gamma ,\delta ;2\alpha +1,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\\ +\frac{{\left(\beta \right)}_1}{{\left(2\beta \right)}_1}\eta F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta +1,\gamma ,\delta ;2\alpha ,2\beta +1,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\\ +\frac{{\left(\gamma \right)}_1}{{\left(2\gamma \right)}_1}\zeta F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma +1,\delta ;2\alpha ,2\beta ,2\gamma +1,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\\ +\left.\left.\frac{{\left(\delta \right)}_1}{{\left(2\delta \right)}_1}\varsigma F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta +1;2\alpha ,2\beta ,2\gamma ,2\delta +1;\xi ,\eta ,\zeta ,\varsigma \right)\right]\right\}.\end{array}$$

(34)

In (34) using the formula of adjacent relations (4) we get the following

$$\begin{array}{l}\frac{\partial g_1}{\partial x}=-\frac{4}{n+2}{k}_1\left(\alpha +\beta +\gamma +\delta +1\right)x^{\frac{n}{2}}{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ \times \left\{x_0^{\frac{n+2}{2}}\right.F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha +1,\beta ,\gamma ,\delta ;2\alpha +1,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)+\\ +\left.\left(x^{\frac{n+2}{2}}-x_0^{\frac{n+2}{2}}\right)F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\right\}.\end{array}$$

The other derivatives are found similarly:

$$\begin{gathered} \begin{array}{l}\frac{\partial g_1}{\partial y}=-\frac{4}{m+2}{k}_1\left(\alpha +\beta +\gamma +\delta +1\right)y^{\frac{m}{2}}{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ \times \left\{y_0^{\frac{m+2}{2}}\right.F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta +1,\gamma ,\delta ;2\alpha ,2\beta +1,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)+\\ +\left.\left(y^{\frac{m+2}{2}}-y_0^{\frac{m+2}{2}}\right)F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\right\},\end{array} \\ \begin{array}{l}\frac{\partial g_1}{\partial z}=-\frac{4}{k+2}{k}_1\left(\alpha +\beta +\gamma +\delta +1\right)z^{\frac{k}{2}}{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ \times \left\{z_0^{\frac{k+2}{2}}\right.F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma +1,\delta ;2\alpha ,2\beta ,2\gamma +1,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)+\\ +\left.\left(z^{\frac{k+2}{2}}-z_0^{\frac{k+2}{2}}\right)F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\right\},\end{array} \\ \begin{array}{l}\frac{\partial g_1}{\partial t}=-\frac{4}{l+2}{k}_1\left(\alpha +\beta +\gamma +\delta +1\right)t^{\frac{l}{2}}{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ \times \left\{t_0^{\frac{l+2}{2}}\right.F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta +1;2\alpha ,2\beta ,2\gamma ,2\delta +1;\xi ,\eta ,\zeta ,\varsigma \right)+\\ +\left.\left(t^{\frac{l+2}{2}}-t_0^{\frac{l+2}{2}}\right)F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\right\}.\end{array} \end{gathered}$$

By substituting the partial derivatives into (33), we get

$$\begin{array}{l}{A}_s\left[g_1\right]=-\left(\alpha +\beta +\gamma +\delta +1\right)k_1{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -1}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)A_s\left[\ln r^2\right]-\\ -2k_1\left(\alpha +\beta +\gamma +\delta +1\right){\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\frac{2}{n+2}{x}_0^{\frac{n+2}{2}}{x}^{\frac{n}{2}}{y}^m{z}^k{t}^l\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha +1,\beta ,\gamma ,\delta ;2\alpha +1,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\cos \left(x,n\right)-\\ -2k_1\left(\alpha +\beta +\gamma +\delta +1\right){\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\frac{2}{m+2}{y}_0^{\frac{m+2}{2}}{x}^n{y}^{\frac{m}{2}}{z}^k{t}^l\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta +1,\gamma ,\delta ;2\alpha ,2\beta +1,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\cos \left(y,n\right)-\\ -2k_1\left(\alpha +\beta +\gamma +\delta +1\right){\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\frac{2}{k+2}{z}_0^{\frac{k+2}{2}}{x}^n{y}^m{z}^{\frac{k}{2}}{t}^l\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma +1,\delta ;2\alpha ,2\beta ,2\gamma +1,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)\cos \left(z,n\right)-\\ -2k_1\left(\alpha +\beta +\gamma +\delta +1\right){\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\frac{2}{l+2}{t}_0^{\frac{l+2}{2}}{x}^n{y}^m{z}^k{t}^{\frac{l}{2}}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta +1;2\alpha ,2\beta ,2\gamma ,2\delta +1;\xi ,\eta ,\zeta ,\varsigma \right)\cos \left(t,n\right).\end{array}$$

(35)

Substitute (35) into (32), then the first integral I₁ will be represented as a sum of integrals:

$$\begin{array}{l}{I}_1={\displaystyle \underset{C^{\epsilon }}{\iiint }u{A}_s\left[g_1\left(x,y,z,t;x_0,y_0,z_0,t_0\right)\right]dS}=I_{11}+I_{12}+I_{13}+I_{14}+I_{15}=\\ =-\left(\alpha +\beta +\gamma +\delta +1\right)k_1{\displaystyle \underset{C^{\epsilon }}{\iiint }}{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -1}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)A_s\left[\ln r^2\right]udS-\\ -2\left(\alpha +\beta +\gamma +\delta +1\right)\frac{2}{n+2}{x}_0^{\frac{n+2}{2}}{x}^{\frac{n}{2}}{k}_1{\displaystyle \underset{C^{\epsilon }}{\iiint }}{y}^m{z}^k{t}^l{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha +1,\beta ,\gamma ,\delta ;2\alpha +1,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)udydzdt-\\ -2\left(\alpha +\beta +\gamma +\delta +1\right)\frac{2}{m+2}{y}_0^{\frac{m+2}{2}}{y}^{\frac{m}{2}}{k}_1{\displaystyle \underset{C^{\epsilon }}{\iiint }}{x}^n{z}^k{t}^l{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta +1,\gamma ,\delta ;2\alpha ,2\beta +1,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)udxdzdt-\\ -2\left(\alpha +\beta +\gamma +\delta +1\right)\frac{2}{k+2}{z}_0^{\frac{k+2}{2}}{z}^{\frac{k}{2}}{k}_1{\displaystyle \underset{C^{\epsilon }}{\iiint }}{x}^n{y}^m{t}^l{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma +1,\delta ;2\alpha ,2\beta ,2\gamma +1,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)udxdydt-\\ -2\left(\alpha +\beta +\gamma +\delta +1\right)\frac{2}{l+2}{t}_0^{\frac{l+2}{2}}{t}^{\frac{l}{2}}{k}_1{\displaystyle \underset{C^{\epsilon }}{\iiint }}{x}^n{y}^m{z}^k{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -2}\times \\ F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta +1;2\alpha ,2\beta ,2\gamma ,2\delta +1;\xi ,\eta ,\zeta ,\varsigma \right)udxdydz.\end{array}$$

(36)

To calculate the integral (36), first we need to calculate the integral

$$\begin{array}{l}{I}_{11}=-\left(\alpha +\beta +\gamma +\delta +1\right)k_1{\displaystyle \underset{C^{\epsilon }}{\iiint }}{\left(r^2\right)}^{-\alpha -\beta -\gamma -\delta -1}\times \\ \times F_A^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;\xi ,\eta ,\zeta ,\varsigma \right)A_s\left[\ln r^2\right]udS.\end{array}$$

(37)

In (37), we move to the hyperspherical coordinates at $n=4$

$$\begin{gathered} x=x_0+\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3,y=y_0+\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3, \\ z=z_0+\epsilon \cos {\theta }_2\sin {\theta }_3,t=t_0+\epsilon \cos {\theta }_3,0\le {\theta }_1<2\pi ,0\le {\theta }_2\le \pi ,0\le {\theta }_3\le \pi . \end{gathered}$$

Then we have

$$\begin{array}{l}{I}_{11}=2\left(\alpha +\beta +\gamma +\delta +1\right)k_1{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}{\tilde{r}}^{-2\alpha -2\beta -2\gamma -2\delta -3}}}}{\epsilon }^3{\left(x_0+\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^n\times \\ \times {\left(y_0+\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^m{\left(z_0+\epsilon \cos {\theta }_2\sin {\theta }_3\right)}^k{\left(t_0+\epsilon \cos {\theta }_3\right)}^l\sin {\theta }_2{\sin }^2{\theta }_3\times \\ \times u\left[\left(x_0+\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right),\left(y_0+\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right),\right.\left.\left(z_0+\epsilon \cos {\theta }_2\sin {\theta }_3\right),\left(t_0+\epsilon \cos {\theta }_3\right)\right]\\ \times F_{As}^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;{\xi }_s,{\eta }_s,{\zeta }_s,{\varsigma }_s\right)d{\theta }_{1}d{\theta }_{2}d{\theta }_{3}d\epsilon ,\end{array}$$

(38)

where a multi-argument decomposition Formula (5) was applied to the function $F_{As}^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;{\xi }_s,{\eta }_s,{\zeta }_s,{\varsigma }_s\right)$:

$$\begin{array}{l}{F}_{As}^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;{\xi }_s,{\eta }_s,{\zeta }_s,{\varsigma }_s\right)=\\ ={\displaystyle \sum _{k_2,k_3,k_4}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2\right)}_{k_2+k_3+k_4}{\left(\alpha \right)}_{k_2+k_3+k_4}{\left(\beta \right)}_{k_2}{\left(\gamma \right)}_{k_3}{\left(\delta \right)}_{k_4}}{{\left(2\alpha \right)}_{k_2+k_3+k_4}{\left(2\beta \right)}_{k_2}{\left(2\gamma \right)}_{k_3}{\left(2\delta \right)}_{k_4}{k}_2!k_3!k_4!}\\ \times {\displaystyle \sum _{i,j,k=0}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4\right)}_{i+j+k}{\left(\beta +k_2\right)}_{i+j}{\left(\gamma +k_3\right)}_{i+k}{\left(\delta +k_4\right)}_{j+k}}{{\left(2\beta +k_2\right)}_{i+j}{\left(2\gamma +k_3\right)}_{i+k}{\left(2\delta +k_4\right)}_{j+k}i!j!k!}\\ \times {\xi }_s^{k_2+k_3+k_4}{\eta }_s^{k_2+i+j}{\zeta }_s^{k_3+i+k}{\varsigma }_s^{k_4+j+k}\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4,\alpha +k_2+k_3+k_4;2\alpha +k_2+k_3+k_4;{\xi }_s\right)\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j,\beta +k_2+i+j;2\beta +k_2+i+j;{\eta }_s\right)\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j+k,\gamma +k_3+i+k;2\gamma +k_3+i+k;{\zeta }_s\right)\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j+k,\delta +k_4+j+k;2\delta +k_4+j+k;{\varsigma }_s\right).\end{array}$$

We use the Boltz auto transformation Formula (6) in the resulting decomposition

$$\begin{array}{l}{F}_{As}^{\left(4\right)}\left(\alpha +\beta +\gamma +\delta +2;\alpha ,\beta ,\gamma ,\delta ;2\alpha ,2\beta ,2\gamma ,2\delta ;{\xi }_s,{\eta }_s,{\zeta }_s,{\varsigma }_s\right)=\\ ={\displaystyle \sum _{k_2,k_3,k_4}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2\right)}_{k_2+k_3+k_4}{\left(\alpha \right)}_{k_2+k_3+k_4}{\left(\beta \right)}_{k_2}{\left(\gamma \right)}_{k_3}{\left(\delta \right)}_{k_4}}{{\left(2\alpha \right)}_{k_2+k_3+k_4}{\left(2\beta \right)}_{k_2}{\left(2\gamma \right)}_{k_3}{\left(2\delta \right)}_{k_4}{k}_2!k_3!k_4!}\\ \times {\displaystyle \sum _{i,j,k=0}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4\right)}_{i+j+k}{\left(\beta +k_2\right)}_{i+j}{\left(\gamma +k_3\right)}_{i+k}{\left(\delta +k_4\right)}_{j+k}}{{\left(2\beta +k_2\right)}_{i+j}{\left(2\gamma +k_3\right)}_{i+k}{\left(2\delta +k_4\right)}_{j+k}i!j!k!}\\ \times \frac{{\xi }_s^{k_2+k_3+k_4}}{{\left(1-{\xi }_s\right)}^{\alpha +k_2+k_3+k_4}}\frac{{\eta }_s^{k_2+i+j}}{{\left(1-{\eta }_s\right)}^{\beta +k_2+i+j}}\frac{{\zeta }_s^{k_3+i+k}}{{\left(1-{\zeta }_s\right)}^{\gamma +k_3+i+k}}\frac{{\varsigma }_s^{k_4+j+k}}{{\left(1-{\varsigma }_s\right)}^{\delta +k_4+j+k}}\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4,\alpha +k_2+k_3+k_4;2\alpha +k_2+k_3+k_4;\frac{{\xi }_s}{{\xi }_s-1}\right)\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j,\beta +k_2+i+j;2\beta +k_2+i+j;\frac{{\eta }_s}{{\eta }_s-1}\right)\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j+k,\gamma +k_3+i+k;2\gamma +k_3+i+k;\frac{{\zeta }_s}{{\zeta }_s-1}\right)\\ \times F\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j+k,\delta +k_4+j+k;2\delta +k_4+j+k;\frac{{\varsigma }_s}{{\varsigma }_s-1}\right),\\ {\xi }_s=-\frac{1}{{\tilde{r}}^2}{\left(\frac{4}{n+2}\right)}^2{\left(x_0{}^2+x_0\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{n+2}{2}}, {\eta }_s=-\frac{1}{{\tilde{r}}^2}{\left(\frac{4}{m+2}\right)}^2{\left(y_0{}^2+y_0\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{m+2}{2}}\\ {\zeta }_s=-\frac{1}{{\tilde{r}}^2}{\left(\frac{4}{k+2}\right)}^2{\left(z_0{}^2+z_0\epsilon \cos {\theta }_2\sin {\theta }_3\right)}^{\frac{k+2}{2}}, {\varsigma }_s=-\frac{1}{{\tilde{r}}^2}{\left(\frac{4}{l+2}\right)}^2{\left(t_0{}^2+t_0\epsilon \cos {\theta }_3\right)}^{\frac{l+2}{2}},\\ {\tilde{r}}^2=\frac{4}{{\left(n+2\right)}^2}{\left[{\left(x_0+\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{n+2}{2}}-x_0^{\frac{n+2}{2}}\right]}^2+\frac{4}{{\left(m+2\right)}^2}{\left[{\left(y_0+\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{m+2}{2}}-y_0^{\frac{m+2}{2}}\right]}^2\\ +\frac{4}{{\left(k+2\right)}^2}{\left[{\left(z_0+\epsilon \cos {\theta }_2\sin {\theta }_3\right)}^{\frac{k+2}{2}}-z_0^{\frac{k+2}{2}}\right]}^2+\frac{4}{{\left(l+2\right)}^2}{\left[{\left(t_0+\epsilon \cos {\theta }_3\right)}^{\frac{l+2}{2}}{-}_0^{\frac{l+2}{2}}\right]}^2.\end{array}$$

After the transformations, we get

$$\begin{array}{l}{F}_{As}^{\left(4\right)}={\tilde{r}}^{2\alpha +2\beta +2\gamma +2\delta }{\left[{\tilde{r}}^2+{\left(\frac{4}{n+2}\right)}^2{\left(x_0{}^2+x_0\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{n+2}{2}}\right]}^{-\alpha }\times \\ \times {\left[{\tilde{r}}^2+{\left(\frac{4}{m+2}\right)}^2{\left(y_0{}^2+y_0\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{m+2}{2}}\right]}^{-\beta }{\left[{\tilde{r}}^2+{\left(\frac{4}{k+2}\right)}^2{\left(z_0{}^2+z_0\epsilon \cos {\theta }_2\sin {\theta }_3\right)}^{\frac{k+2}{2}}\right]}^{-\gamma }\\ \times {\left[{\tilde{r}}^2+{\left(\frac{4}{l+2}\right)}^2{\left(t_0{}^2+t_0\epsilon \cos {\theta }_3\right)}^{\frac{l+2}{2}}\right]}^{-\delta }\mathsf{{\rm P}},\end{array}$$

where

$$\begin{array}{l}\mathsf{{\rm P}}={\displaystyle \sum _{k_2,k_3,k_4}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2\right)}_{k_2+k_3+k_4}{\left(\alpha \right)}_{k_2+k_3+k_4}{\left(\beta \right)}_{k_2}{\left(\gamma \right)}_{k_3}{\left(\delta \right)}_{k_4}}{{\left(2\alpha \right)}_{k_2+k_3+k_4}{\left(2\beta \right)}_{k_2}{\left(2\gamma \right)}_{k_3}{\left(2\delta \right)}_{k_4}{k}_2!k_3!k_4!}\times \\ \times {\displaystyle \sum _{i,j,k=0}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4\right)}_{i+j+k}{\left(\beta +k_2\right)}_{i+j}{\left(\gamma +k_3\right)}_{i+k}{\left(\delta +k_4\right)}_{j+k}}{{\left(2\beta +k_2\right)}_{i+j}{\left(2\gamma +k_3\right)}_{i+k}{\left(2\delta +k_4\right)}_{j+k}i!j!k!}\\ \times {\left[1-\frac{{\tilde{r}}^2}{{\tilde{r}}^2+{\left(\frac{4}{n+2}\right)}^2{\left(x_0{}^2+x_0\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{n+2}{2}}}\right]}^{k_2+k_3+k_4}{\left[1-\frac{{\tilde{r}}^2}{{\tilde{r}}^2+{\left(\frac{4}{m+2}\right)}^2{\left(y_0{}^2+y_0\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{m+2}{2}}}\right]}^{k_2+i+j}\\ \times {\left[1-\frac{{\tilde{r}}^2}{{\tilde{r}}^2+{\left(\frac{4}{k+2}\right)}^2{\left(z_0{}^2+z_0\epsilon \cos {\theta }_2\sin {\theta }_3\right)}^{\frac{k+2}{2}}}\right]}^{k_3+i+k}{\left[1-\frac{{\tilde{r}}^2}{{\tilde{r}}^2+{\left(\frac{4}{l+2}\right)}^2{\left(t_0{}^2+t_0\epsilon \cos {\theta }_3\right)}^{\frac{l+2}{2}}}\right]}^{k_4+j+k}\\ \times F\left(\alpha -\beta -\gamma -\delta -2,\alpha +k_2+k_3+k_4;2\alpha +k_2+k_3+k_4;1-\frac{{\tilde{r}}^2}{{\left(\frac{4}{n+2}\right)}^2{\left(x_0{}^2+x_0\epsilon \sin {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{n+2}{2}}}\right)\\ \times F\left(-\alpha +\beta -\gamma -\delta -2-k_3-k_4,\beta +k_2+i+j;2\beta +k_2+i+j;1-\frac{{\tilde{r}}^2}{{\left(\frac{4}{m+2}\right)}^2{\left(y_0{}^2+y_0\epsilon \cos {\theta }_1\sin {\theta }_2\sin {\theta }_3\right)}^{\frac{m+2}{2}}}\right)\\ \times F\left(-\alpha -\beta +\gamma -\delta -2-k_2-k_4-j,\gamma +k_3+i+k;2\gamma +k_3+i+k;1-\frac{{\tilde{r}}^2}{{\left(\frac{4}{k+2}\right)}^2{\left(z_0{}^2+z_0\epsilon \cos {\theta }_2\sin {\theta }_3\right)}^{\frac{k+2}{2}}}\right)\\ \times F\left(-\alpha -\beta -\gamma +\delta -2-k_2-k_3-i,\delta +k_4+j+k;2\delta +k_4+j+k;1-\frac{{\tilde{r}}^2}{{\left(\frac{4}{l+2}\right)}^2{\left(t_0{}^2+t_0\epsilon \cos {\theta }_3\right)}^{\frac{l+2}{2}}}\right).\end{array}$$

(39)

In (39), we pass to the limit at $\epsilon \to 0$

$$\begin{array}{l}\underset{\epsilon \to 0}{\lim }\mathsf{{\rm P}}={\displaystyle \sum _{k_2,k_3,k_4}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2\right)}_{k_2+k_3+k_4}{\left(\alpha \right)}_{k_2+k_3+k_4}{\left(\beta \right)}_{k_2}{\left(\gamma \right)}_{k_3}{\left(\delta \right)}_{k_4}}{{\left(2\alpha \right)}_{k_2+k_3+k_4}{\left(2\beta \right)}_{k_2}{\left(2\gamma \right)}_{k_3}{\left(2\delta \right)}_{k_4}{k}_2!k_3!k_4!}\\ \times {\displaystyle \sum _{i,j,k=0}^{\infty }}\frac{{\left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4\right)}_{i+j+k}{\left(\beta +k_2\right)}_{i+j}{\left(\gamma +k_3\right)}_{i+k}{\left(\delta +k_4\right)}_{j+k}}{{\left(2\beta +k_2\right)}_{i+j}{\left(2\gamma +k_3\right)}_{i+k}{\left(2\delta +k_4\right)}_{j+k}i!j!k!}\\ \times F\left(\alpha -\beta -\gamma -\delta -2,\alpha +k_2+k_3+k_4;2\alpha +k_2+k_3+k_4;1\right)\\ \times F\left(-\alpha +\beta -\gamma -\delta -2-k_3-k_4,\beta +k_2+i+j;2\beta +k_2+i+j;1\right)\\ \times F\left(-\alpha -\beta +\gamma -\delta -2-k_2-k_4-j,\gamma +k_3+i+k;2\gamma +k_3+i+k;1\right)\\ \times F\left(-\alpha -\beta -\gamma +\delta -2-k_2-k_3-i,\delta +k_4+j+k;2\delta +k_4+j+k;1\right).\end{array}$$

(40)

In (40), we calculate the values of the hypergeometric functions by the formula (9), then we have

$$\begin{array}{l}F\left(\alpha -\beta -\gamma -\delta -2,\alpha +k_2+k_3+k_4;2\alpha +k_2+k_3+k_4;1\right)=\\ =\frac{\Gamma \left(2\alpha +k_2+k_3+k_4\right)\Gamma \left(\beta +\gamma +\delta +2\right)}{\Gamma \left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4\right)\Gamma \left(\alpha \right)},\\ F\left(-\alpha +\beta -\gamma -\delta -2-k_3-k_4,\beta +k_2+i+j;2\beta +k_2+i+j;1\right)=\\ =\frac{\Gamma \left(2\beta +k_2+i+j\right)\Gamma \left(\alpha +\gamma +\delta +2+k_3+k_4\right)}{\Gamma \left(\alpha +\beta +\gamma +\delta +2+k_3+k_2+k_4+i+j\right)\Gamma \left(\beta \right)},\\ F\left(-\alpha -\beta +\gamma -\delta -2-k_2-k_4-j,\gamma +k_3+i+k;2\gamma +k_3+i+k;1\right)\\ =\frac{\Gamma \left(2\gamma +k_3+i+k\right)\Gamma \left(\alpha +\beta +\delta +2+k_2+k_4+j\right)}{\Gamma \left(\alpha +\beta +\delta +\gamma +2+k_2+k_4+k_3+i+j+k\right)\Gamma \left(\gamma \right)},\\ F\left(-\alpha -\beta -\gamma +\delta -2-k_2-k_3-i,\delta +k_4+j+k;2\delta +k_4+j+k;1\right)=\\ =\frac{\Gamma \left(2\delta +k_4+j+k\right)\Gamma \left(\alpha +\beta +\gamma +2+k_2+k_3+i\right)}{\Gamma \left(\alpha +\beta +\gamma +\delta +2+k_2+k_3+k_4+i+j+k\right)\Gamma \left(\delta \right)}.\end{array}$$

(41)

We substitute (41) in (40), calculating the double sum based on gamma function properties (7) and (8), we get

$$\underset{\epsilon \to 0}{\lim }\mathsf{{\rm P}}=\frac{\Gamma \left(2\alpha \right)\Gamma \left(2\beta \right)\Gamma \left(2\gamma \right)\Gamma \left(2\delta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)\Gamma \left(\gamma \right)\Gamma \left(\delta \right)\Gamma \left(\alpha +\beta +\gamma +\delta +2\right)}.$$

(42)

Taking into account the previous calculations and transformations, from (38) we get the following:

$$\begin{array}{l}\underset{\epsilon \to 0}{\lim }{I}_{11}=4{\pi }^2{k}_1\frac{\Gamma \left(2\alpha \right)\Gamma \left(2\beta \right)\Gamma \left(2\gamma \right)\Gamma \left(2\delta \right)}{\Gamma \left(\alpha +\beta +\gamma +\delta +1\right)\Gamma \left(\alpha \right)\Gamma \left(\beta \right)\Gamma \left(\gamma \right)\Gamma \left(\delta \right)}\times \\ \times {\left(\frac{4}{n+2}\right)}^{-2\alpha }{\left(\frac{4}{m+2}\right)}^{-2\beta }{\left(\frac{4}{k+2}\right)}^{-2\gamma }{\left(\frac{4}{l+2}\right)}^{-2\delta }u\left(x_0,y_0,z_0,t_0\right).\end{array}$$

(43)

Taking into account the value of the coefficient $k_1$ from (23), we have

$$\underset{\epsilon \to 0}{\lim }{I}_{11}=u\left(x_0,y_0,z_0,t_0\right).$$

(44)

By performing similar calculations, we can make sure that

$$\underset{\epsilon \to 0}{\lim }{I}_{12}=\underset{\epsilon \to 0}{\lim }{I}_{13}=\underset{\epsilon \to 0}{\lim }{I}_{14}=\underset{\epsilon \to 0}{\lim }{I}_{15}=\underset{\epsilon \to 0}{\lim }{I}_2=0.$$

(45)

Let us consider the second integral of (30) ${\displaystyle \underset{C^{\epsilon }}{\iiint }{G}_1{A}_s\left[u\right]}dS$. Using the algorithm used in the calculation (31), it is not difficult to prove that

$$\underset{\epsilon \to 0}{\lim }{\displaystyle \underset{C^{\epsilon }}{\iiint }{G}_1{A}_s\left[u\right]}dS=0.$$

(46)

Thus, from (27) we get the solution of the problem N (15). Theorem 1 is proved.

4. Conclusions

In this paper, we have constructed a Green’s function of the Holmgren’s problem counterpart for a degenerate second order elliptic Gellerstedt equation, with the help of which an explicit solution of the problem is obtained and a uniqueness of the problem solution is proved. By applying the methods used in this paper and selecting the corresponding fundamental solutions, it is also possible to study the solvability of a number of different boundary value problems for Equation (1).