\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 129, pp. 1--14.\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-2018/129\hfil Boundary condition of the volume potential]
{Boundary condition of the volume potential for an elliptic-parabolic
equation with a scalar parameter}

\author[T. Sh. Kal'menov, G. D. Arepova, D. D. Arepova \hfil EJDE-2018/129\hfilneg]
{Tynysbek Sh. Kal'menov, Gaukhar D. Arepova, Dana D. Arepova}

\address{Tynysbek Sh. Kal'menov \newline
Institute of Mathematics and Mathematical Modelling,
125 Pushkin street,
050010 Almaty, Kazakhstan. \newline
Al-Farabi Kazakh National University,
71 Al-Farabi Avenue,
050040 Almaty, Kazakhstan}
\email{kalmenov.t@mail.ru}

\address{Gaukhar D. Arepova \newline
Institute of Mathematics and Mathematical Modelling,
125 Pushkin street,
050010 Almaty, Kazakhstan. \newline
Al-Farabi Kazakh National University,
71 Al-Farabi Avenue,
050040 Almaty, Kazakhstan}
\email{arepovag@mail.ru}

\address{Dana D. Arepova \newline
Institute of Mathematics and Mathematical Modelling,
125 Pushkin street,
050010 Almaty, Kazakhstan}
\email{danaarepova@gmail.com}

\dedicatory{Communicated by Ludmila S. Pulkina}

\thanks{Submitted January 7, 2018. Published June 23, 2018.}
\subjclass[2010]{35M12}
\keywords{Boundary conditions; descent method; fundamental solutions,
\hfill\break\indent Elliptic-parabolic equation; Newton's potential; 
 volume heat potential; surface heat potential}

\begin{abstract}
 Using the descent method for the fundamental solution of the heat equation
 with a scalar parameter, we find the fundamental solution of the
 multidimensional Helmholtz equation in an explicit form.
 We also find a boundary condition of the volume potential for an
 elliptic-parabolic equation with a scalar parameter. In turn, this
 condition allows us to construct and study a new correct nonlocal
 (initial) Bitsadze-Samarsky type problem for an elliptic-parabolic
 equation with a scalar parameter.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Most of the references in this paper are devoted to systematic 
study of the boundary conditions of the Newton's potential \cite{KS1},
the heat potential \cite{KT2} and the surface heat potential \cite{KA3,KA4,KA5}. 
In this paper, we present a boundary condition for an elliptic-parabolic 
equation with a scalar parameter.

Let $\varepsilon^+_{n+1}(x,t,\lambda)$ be the fundamental solution of the 
heat equation with a scalar parameter
\begin{equation}  \label{e1.1}
\frac{\partial u(x,t)}{\partial t}-\Delta_x u(x,t)+\lambda u(x,t)=f^+(x,t), \quad 
x\in R^n, t>0
\end{equation}
and $\varepsilon^-_{n+1}(x,t,\lambda)$ be the fundamental solution of the 
Helmholtz equation
\begin{equation} \label{e1.2}
-\frac{\partial^2 u(x,t)}{\partial t^2}-\Delta_x u(x,t)+\lambda u(x,t)=f^- (x,t), 
\quad x\in R^n, t<0
\end{equation}
where $\lambda$ is an arbitrary complex number and $Re \lambda\geq0$.

Let $\Omega \in R^n$ be a bounded domain with smooth boundary $\partial \Omega$ 
and $D^+=\Omega \times[0,T]$ be a cylindrical domain.
$D^-\subset R^{n+1}$ is the domain at $t<0$ with smooth surface $\sigma$ and 
when $t=0$ bounded with the domain $\Omega$. We will also use the notation 
$D=D^+\cup\Omega\cup D^-$.

We define an elliptic-parabolic potential as
\begin{equation} \label{e1.3}
\begin{aligned}
u(x,t)
&=(L_B^{-1}f)(x,t) \\
&=\begin{cases}
\int_{D^+}\varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)f^+(\xi,\eta)dD^+ \\
+\int_{\Omega} \varepsilon^+_{n+1}(x-\xi,t,\lambda)\tau(\xi)d\xi,
& \text{if }t>0,\\[4pt]
\int_{D^-}\varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda)f^-(\xi,\eta)dD^-,
& \text{if }t<0
\end{cases}
\end{aligned}
\end{equation}
where the unknown function $\tau (x)$ is determined from the condition 
of continuity of the potential $L_B^{-1}f(x,t)$ when $t=0$.

\section{Main results}

First we find the fundamental solutions of the heat equation with a scalar 
parameter and Helmholtz equation.

\begin{lemma} \label{lem2.1}
The fundamental solution of the heat equation \eqref{e1.1} is a function
\begin{equation} \label{e2.1}
\varepsilon^+_{n+1} (x,t,\lambda)
=\Theta(t)\frac{e^{-\frac{|x|^2}{4t}}}{(2\sqrt{\pi t})^n}e^{-\lambda t}.
\end{equation}
\end{lemma}

\begin{proof}
A direct calculation shows that
\begin{align*}
&\Big(\frac{\partial}{\partial t}-\Delta_x +\lambda\Big)
 \varepsilon^+_{n+1}(x,t,\lambda) \\
&= \Big[\big(\frac{\partial}{\partial t}-\Delta_x\big)
 \Theta(t)\frac{e^{-\frac{|x|^2}{4t}}}{(2\sqrt{\pi t})^n}\Big]e^{\lambda t}
 +\Big[(\frac{\partial}{\partial t}+\lambda)e^{-\lambda t}\Big]\Theta(t)
  \frac{e^{-\frac{|x|^2}{4t}}}{(2\sqrt{\pi t})^n} \\
&=\delta(x,t)e^{-\lambda t}-\lambda e^{-\lambda t} \Theta(t)
 \frac{e^{-\frac{|x|^2}{4t}}}{(2\sqrt{\pi t})^n}
 +\lambda e^{-\lambda t}\Theta(t)\frac{e^{-\frac{|x|^2}{4t}}}{(2\sqrt{\pi t})^n} \\
&=\delta (x,t) e^{-\lambda t}=\delta(x,t).
\end{align*}
\end{proof}

\begin{lemma} \label{lem2.2}
The fundamental solution of the Helmholtz equation \eqref{e1.2} can be
represented as
\begin{equation} \label{e2.2}
\varepsilon^-_{n+1}(\overline{x},\lambda)
= \frac{1}{(n-1)\omega_{n+1}|\overline{x}|^{n-1}}
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})}
  \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|),
\end{equation}
where
$|\overline{x}|^2=x^2_1+\dots+x_n^2+t^2$,
$$
K_{\nu}(z)=\frac{1}{2}\big(\frac{z}{2}\big)^{\nu}
 \int_{0}^{\infty}\xi^{-\nu-1}e^{-\xi -\frac{z^2}{4\xi}}d\xi
$$
is the Macdonald function \cite[p. 183]{W8} and
$$
\omega_{n+1}=\frac{2(\sqrt{\pi})^{n+1}}{\Gamma(\frac{n+1}{2})}
$$
is the area of a unit sphere in $R^{n+1}$.
\end{lemma}

\begin{proof}
We note that when $n=1,2,3$ the fundamental solutions of the Helmholtz 
equation are given in \cite[p. 203-205]{V9}:
\begin{gather*}
\varepsilon^{-}_{1}(x,\lambda)=-\frac{1}{2ik}e^{-i\sqrt{\lambda}|x|}, \\
\varepsilon^{-}_{2}(x_{1},x_{2},\lambda)=\pm\frac{i}{4}H_{0}^{1,2}
(\sqrt{\lambda}|x|), \quad |x|=(x_{1}^{2}+x_{2}^{2})^{\frac{1}{2}}, \\
\varepsilon^{-}_{3}(x_{1},x_{2},x_{3},\lambda)
=\pm\frac{e^{\pm i\sqrt{\lambda}|x|}}{4\pi |x|}, \quad
|x|=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{\frac{1}{2}}.
\end{gather*}

The proof for $n\geq3$ is based on the descent method for the fundamental 
solution of the heat equation with a scalar parameter
\begin{equation*}
\big(\frac{\partial}{\partial \eta}-\Delta_{\overline{x}}+\lambda\big)
\varepsilon^+_{n+2} (\overline{x},\eta,\lambda)=\delta(\overline{x},\eta), \quad
\overline{x} \in R^{n+1},\; \eta>0,
\end{equation*}
where $\overline{x}=(x_1,x_2,\dots,x_n,t)$.

Putting the function
\begin{equation*}
\varepsilon^+_{n+2}(\overline{x},\eta,\lambda)=\Theta(\eta) 
\frac{e^{-\frac{|\overline{x}|^2}{4\eta}}}{(2\sqrt{\pi \eta})^{n+1}}
e^{-\lambda \eta}
\end{equation*}
in the formula of the descent method
\begin{equation} \label{e2.3}
\varepsilon^-_{n+1}(\overline{x},\lambda)
=\int_{0}^{\infty}\varepsilon^+_{n+2} (\overline{x},\eta,\lambda)d\eta
\end{equation}
and after replacing
\begin{equation*}
\xi=\frac{|\overline{x}|^2}{4\eta}, \quad
\eta=\frac{|\overline{x}|^2}{4\xi}, \quad 
d\eta=-\frac{|\overline{x}|^2}{4\xi^2}d\xi,
\end{equation*}
we find that
\begin{align*}
\varepsilon^-_{n+1}(\overline{x},\lambda)
=&\frac{1}{(2\sqrt{\pi})^{n+1}}
 \int_{0}^{\infty}e^{-\xi -\lambda \frac{|\overline{x}|^2}{4\xi}}
 \Big(\frac{|\overline{x}|^2}{4\xi}\Big)^{-\frac{n+1}{2}}
 \frac{|\overline{x}|^2}{4\xi^2}d\xi \\
=&\frac{4^{\frac{n-1}{2}}|\overline{x}|^{1-n}}{(2\sqrt{\pi})^{n+1}}
 \int_{0}^{\infty}e^{-\xi -\lambda \frac{|\overline{x}|^2}{4\xi}}
 \xi^{\frac{n}{2}+\frac{1}{2}-2}d\xi.
\end{align*}
The obtained fundamental solution $\varepsilon^-_{n+1} (\overline{x},\lambda)$ 
we will be expressed in terms of the MacDonald function
\begin{align*}
&\varepsilon^-_{n+1}(\overline{x},\lambda)\\
&=\frac{1}{2(\sqrt{\pi})^{n+1}|\overline{x}|^{n-1}}
  \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 \frac{1}{2}\Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{1-n}{2}}
 \int_{0}^{\infty}   e^{-\xi-\frac{(\sqrt{\lambda}|\overline{x}|)^2}{4\xi}}
 \xi^{-(\frac{1-n}{2})-1} d\xi \\
&=\frac{1}{2(\sqrt{\pi})^{n+1}|\overline{x}|^{n-1}}
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|), \quad
\operatorname{Re}\sqrt{\lambda}\geq0.\\
\end{align*}
Then we will express it with the fundamental solution 
$\varepsilon_{\Delta}^{n+1}(\overline{x})
=\frac{1}{(n-1)\omega_{n+1}|\overline{x}|^{n-1}}$ 
of the Laplace equation
\begin{align*}
\varepsilon^-_{n+1}(\overline{x},\lambda)
=&\frac{1}{2(\sqrt{\pi})^{n+1}|\overline{x}|^{n-1}} 
\frac{\omega_{n+1}(n-1)}{\omega_{n+1}(n-1)} 
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|)=\\
=&\frac{1}{(n-1)\omega_{n+1}|\overline{x}|^{n-1}}
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})}
  \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|)=\\
=&\varepsilon_{\Delta}^{n+1} (\overline{x}) \frac{(n-1)}{\Gamma(\frac{n+1}{2})} 
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|).
\end{align*}

We can show that when $\lambda\to 0$ we obtain the fundamental solution 
of the Laplace equation.

It is well known that Macdonald defined the function $K_{\nu}(z)$ for arbitrary 
numbers $\nu$ on the basis of equality
$$
K_{\nu}(z)=\frac{\pi}{2\sin\nu\pi}\left(I_{-\nu}(z)-I_{\nu}(z)\right),
$$
where
$$
I_{\nu}(z)=\sum_{m=0}^{\infty}\frac{(\frac{z}{2})^{\nu+2m}}{m!\Gamma(\nu+m+1)}.
$$
So, when $\nu=(1-n)/2$ and $m=0$ we get
\begin{align*}
&\lim_{\lambda\to 0}\varepsilon^-_{n+1} (\overline{x},\lambda) \\
&=\lim_{\lambda\to 0}\varepsilon_{\Delta}^{n+1}(\overline{x})
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})} 
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|) \\
&=\lim_{\lambda\longrightarrow0}\varepsilon_{\Delta}^{n+1}(\overline{x})
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})} 
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 \frac{\pi}{2\sin\frac{1-n}{2}\pi}
 \Big(\sum_{m=0}^{\infty}\frac{(\frac{\sqrt{\lambda}|\overline{x}|}{2}
 )^{-\frac{1-n}{2}+2m}}{m!\Gamma(-\frac{1-n}{2}+m+1)} \\
&\quad -\sum_{m=0}^{\infty}\frac{(\frac{\sqrt{\lambda}|\overline{x}|}{2}
 )^{\frac{1-n}{2}+2m}}{m!\Gamma(\frac{1-n}{2}+m+1)}\Big) \\
&=\lim_{\lambda\longrightarrow0}\varepsilon_{\Delta}^{n+1}(\overline{x})
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})} 
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
 \frac{\pi}{2\sin\frac{1-n}{2}\pi}
 \Big(\frac{(\frac{\sqrt{\lambda}|\overline{x}|}{2})^{-\frac{1-n}{2}}}
 {\Gamma(-\frac{1-n}{2}+1)}
 -\frac{(\frac{\sqrt{\lambda}|\overline{x}|}{2})^{\frac{1-n}{2}}}
 {\Gamma(\frac{1-n}{2}+1)}\Big) \\
&=\lim_{\lambda\longrightarrow0}\varepsilon_{\Delta}^{n+1}
 (\overline{x})\frac{(n-1)}{\Gamma(\frac{n+1}{2})} 
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{n-1}
 \frac{\pi}{2\sin\frac{1-n}{2}\pi}\frac{1}{\Gamma(-\frac{1-n}{2}+1)} \\
&\quad -\varepsilon_{\Delta}^{n+1}(\overline{x})
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})} 
 \frac{\pi}{2\sin\frac{1-n}{2}\pi}\frac{1}{\Gamma(\frac{1-n}{2}+1)}\\
&=J_{1}-J_{2}=0-J_{2}.
\end{align*}

If we use the the formula of the Gamma function 
$\frac{\pi}{sin\pi z}=\Gamma(z)\Gamma(1-z)$ and $\Gamma(z+1)=z\Gamma(z)$,
\begin{align*}
-J_{2}&=-\varepsilon_{\Delta}^{n+1}(\overline{x})
 \frac{(n-1)}{2\Gamma(\frac{n+1}{2})} 
 \frac{\pi}{sin\frac{1-n}{2}\pi}\frac{1}{\Gamma(\frac{1-n}{2}+1)} \\
&=-\varepsilon_{\Delta}^{n+1}(\overline{x})
 \frac{(n-1)}{2\Gamma(\frac{n+1}{2})} 
 \Gamma(\frac{1-n}{2})\Gamma(1-\frac{1-n}{2})\frac{1}{\Gamma(\frac{1-n}{2}+1)} \\
&=-\varepsilon_{\Delta}^{n+1}(\overline{x})\frac{(n-1)}{2\Gamma(\frac{n+1}{2})} 
 \Gamma(\frac{1-n}{2})\Gamma(\frac{1+n}{2})\frac{1}
 {\Gamma(\frac{1-n}{2})(\frac{1-n}{2})} \\
&=\varepsilon_{\Delta}^{n+1}(\overline{x})=\frac{1}{(n-1)\omega_{n+1}
 |\overline{x}|^{n-1}}.
\end{align*}
It should be noted that the function
$$
\widetilde{K}_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|)
=\frac{(n-1)}{\Gamma(\frac{n+1}{2})}
 \Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|)
$$
does not have singularity.

By the property of analytic continuation we obtained for all complex $\lambda$,
 \begin{equation} \label{e2.4}
  \varepsilon^-_{n+1} (\overline{x},\lambda)
= \frac{1}{(n-1)\omega_{n+1}|\overline{x}|^{n-1}}
 \frac{(n-1)}{\Gamma(\frac{n+1}{2})}
\Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|).
 \end{equation}
This completes the proof.
 \end{proof}

\begin{lemma} \label{lem2.3}
 Let $f(x,t) \in C^{\alpha}(\overline{D})$. Then
 \begin{equation} \label{e2.5}
  u(x,t) = (L^{-1}_B f)(x,t) \in C^{\alpha}(\overline{D})\cap C^{2+\alpha}
(\overline{D^-}) \cap C_{x,t}^{2+\alpha,1+\alpha}(\overline{D^+}).
 \end{equation}
\end{lemma}

\begin{proof}
 Taking into account the conditions $u(x,0-)=u(x,0+)=\tau(x)$, from the formula
$$
u(x,t)=\int_{D^-} \varepsilon^-_{n+1}(x- \xi,t-\eta,\lambda)f^-(\xi,\eta)dD^-,
\quad t<0
$$
it follows that
\begin{gather} \label{e2.6}
  \tau(x) = u(x,0-) = \int_{D^-} \varepsilon^-_{n+1} (x-\xi,-\eta) 
f^-(\xi,\eta)d\xi d\eta \quad \in C^{2+\alpha} (\overline{\Omega}), \\
\label{e2.7}
  \| \tau(x)\|_{C^{2+\alpha} (\overline{\Omega})} 
\leq d_1 \| f^-(x,t)\|_{C^{\alpha} (\overline{D^-})}.
 \end{gather}
From this and by properties of the heat potential and surface heat potential 
we obtain
\begin{equation} \label{e2.8}
\begin{aligned}
 \| (L^{-1}_B f)(x,t)\|_{C_{x,t}^{2+\alpha,1+\alpha}(\overline{D^+})} 
&=\| \int_{0}^{t}d\eta \int_{\Omega} \varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda)
 f^+(\xi,\eta) d\xi \\
&\quad + \int_{\Omega} \varepsilon^+_{n+1} (x-\xi,t,\lambda)
 \tau(\xi)d\xi \|_{C_{x,t}^{2+\alpha,1+\alpha}(\overline{D^+})} \\
& \leq d_2 \Big( \| f^+(x,t)\|_{C^{\alpha}(\overline{D^+})}
 +\| \tau(x)\|_{C^{2+\alpha}(\overline{\Omega})} \Big) \\
& \leq d_3 \Big( \| f^+(x,t)\|_{C^{\alpha}(\overline{D^+})}
 +\| f^-(x,t)\|_{C^{\alpha}(\overline{D^-})} \Big).
\end{aligned}
\end{equation}
Using the properties of the Newton's potential we obtain
\begin{equation} \label{e2.9}
\begin{aligned}
\| (L^{-1}_Bf)(x,t) \|_{C^{2+\alpha}(\overline{D^-})}  
& =\| \int_{D^-} \varepsilon^-_{n+1} (x-\xi,t-\eta,\lambda) f^-(\xi,\eta) d\xi d\eta
\|_{C^{2+\alpha}(\overline{D^-})} \\
&\leq d_4 \| f^-(x,t) \|_{C^{\alpha}(\overline{D^-})}.
\end{aligned}
\end{equation}
Comparing  inequalities \eqref{e2.8}-\eqref{e2.9} we have
\begin{equation*}
 u(x,t) = (L^{-1}_Bf)(x,t) \in C^{\alpha}(\overline{D})
\cap C^{2+ \alpha}(\overline{D^-}) \cap C_{x,t}^{2+\alpha, 1 + \alpha}(\overline{D^+})
\end{equation*}
and
\begin{equation} \label{e2.10}
\begin{aligned}
 \| u(x,t)\| &= \| (L^{-1}_Bf)(x,t)\|_{ C^{\alpha}(\overline{D})
 \cap C^{2+ \alpha}(\overline{D^-}) \cap C_{x,t}^{2+\alpha,
 1 + \alpha}(\overline{D^+}) } \\
&\leq d_5 \| f(x,t)\|_{ C^{\alpha}(\overline{D})}.
\end{aligned}
\end{equation}
The proof is complete.
\end{proof}

As in \cite{KT2,KA3}, it can be proved that the volume heat potential is
\begin{equation} \label{e2.11}
 u_f (x,t) = \int_{D^+} \varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda) 
f^+(\xi,\eta) d\xi d\eta
\end{equation}
which satisfies the inhomogeneous heat equation with a scalar parameter
\begin{equation} \label{e2.12}
\diamondsuit u_{f} (x,t)=\frac{\partial u_{f} (x,t)}{\partial t} 
- \Delta_x u_{f} (x,t) + \lambda u_{f} (x,t) = f^+ (x,t), \quad x \in R^n,\; t>0
\end{equation}
and satisfies the homogeneous initial condition
\begin{equation} \label{e2.13}
u_{f} (x,t)|_{t=0}=0\,.
\end{equation}
Also the surface heat potential is
\begin{equation}
u_{\tau} (x,t) = \int_{\Omega}\varepsilon^+_{n+1}(x-\xi,t,\lambda) \tau(\xi)d\xi
\end{equation} \label{e2.14}
which satisfies the homogeneous heat equation with a scalar parameter
\begin{equation} \label{e2.15}
\diamondsuit u_{\tau} (x,t)=\frac{\partial u_{\tau}(x,t)}{\partial t}
-\Delta_x u_{\tau}(x,t)+ \lambda u_{\tau} (x,t)=0, \quad x \in R^n,\; t>0
\end{equation}
and satisfies the nonhomogeneous initial condition
\begin{equation} \label{e2.16}
u_{\tau} (x,t)|_{t=0}=u_{0} (x)
\end{equation}
satisfy the same lateral potential boundary condition, i.e. the condition
\begin{equation} \label{e2.17}
\begin{aligned}
&-\frac{u_f(x,t)+u_{\tau}(x,t)}{2}
 +\int_{0}^{t}d\eta \int_{\partial \Omega}
\Big( \frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}{\partial n_{xi}}
 (u_f+u_{\tau})(\xi,\eta) \\
& - \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)\frac{\partial
 (u_f+u_{\tau})(\xi,\eta)}{\partial n_{\xi}} \Big)d\xi=0,
\end{aligned}
\end{equation}
for all $(x,t)\in\partial \Omega\cap(0,T)$, where
$\frac{\partial }{\partial n_{\xi}}$ is the normal derivative.

\begin{lemma} \label{lem2.4}
For any function $f(x,t)\in C^{\alpha}(\overline{D})$ the volume heat 
potential \eqref{e2.11} satisfies the inhomogeneous heat equation with 
a scalar parameter \eqref{e2.12}, the homogeneous initial condition \eqref{e2.13}, 
and the lateral boundary condition
\begin{equation} \label{e2.18}
\begin{aligned}
&-\frac{u_f(x,t)}{2}
 +\int_{0}^{t}d\eta \int_{\partial \Omega}
 \Big( \frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}
  u_f(\xi,\eta) \\
&- \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)\frac{\partial u_f(\xi,\eta)}
 {\partial n_{\xi}} \Big)d\xi=0,
\end{aligned}
\end{equation}
for all $(x,t)\in\partial \Omega\cap(0,T)$.

Conversely, if $u(x,t)\in W^{2,1}_{2}(D)$ is a solution of the inhomogeneous 
heat equation with a scalar parameter \eqref{e2.12}, which satisfies the
 homogeneous initial condition \eqref{e2.13} and the lateral boundary 
condition \eqref{e2.18}, then it coincides with the volume heat potential 
\eqref{e2.11}.
\end{lemma}

\begin{proof}
We consider the heat potential
\begin{equation*}
 u_f (x,t) = \int_{D^+} \varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda) f^+(\xi,\eta) d\xi d\eta=
\end{equation*}
\begin{equation*}
=\int_{D^+} \frac{e^{-\frac{|x-\xi|^2}{4(t-\eta)}}e^{-\lambda (t-\eta)}}{(2\sqrt{\pi (t-\eta)})^n}f^+(\xi,\eta) d\xi d\eta=\int_{D^+} \frac{e^{-\frac{|x-\xi|^2}{4(t-\eta)}}e^{-\lambda (t-\eta)}}{(2\sqrt{\pi (t-\eta)})^n} \diamondsuit u_{f} (\xi,\eta)d\xi d\eta.
\end{equation*}

Since the integral
\begin{equation*}
\int_{0}^{t}d\eta\int_{\Omega} \varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda) \diamondsuit u_{f} (\xi,\eta)d\xi
\end{equation*}
the improper integral as an integral of the function
 $\varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda)$
with a singularity at $t = \eta$ then we understand it as 
\begin{equation*}
 \lim_{\delta\to 0}u_{\delta} (x,t)
=\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\Omega} 
\varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda) \diamondsuit u_{f} (\xi,\eta)d\xi.
\end{equation*}
By a direct calculation and integration by parts for $x\in\Omega$ and 
$0<\delta<t$ it can be verified that
\begin{equation} \label{e2.19}
\begin{aligned}
&u_{f} (x,t) \\
&=\lim_{\delta\to 0}u_{\delta} (x,t) \\
&=\lim_{\delta\to 0}
 \int_{0}^{t-\delta}d\eta\int_{\Omega} \varepsilon^+_{n+1} 
 (x-\xi,t-\eta,\lambda) (\frac{\partial}{\partial\eta}-\Delta_{\xi}+\lambda) 
 u_{f} (\xi,\eta)d\xi \\
&=\lim_{\delta\to 0}\int_{\Omega} \varepsilon^+_{n+1} (x-\xi,\delta,\lambda)u_{f} 
 (\xi,t-\delta)d\xi \\
&\quad -\lim_{\delta\to 0}\int_{\Omega} \varepsilon^+_{n+1} (x-\xi,t,\lambda)u_{f}
  (\xi,0)d\xi \\
&\quad -\lim_{\delta\to 0}\int_{0}^{t-\delta} d\eta
\int_{\partial\Omega}\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)\frac{\partial u_{f}(\xi,\eta)}{\partial n_{\xi}}d\xi \\
&\quad +\lim_{\delta\to 0}\int_{0}^{t-\delta} d\eta
\int_{\partial\Omega}\frac{\partial \varepsilon^+_{n+1}(x-\xi,
 t-\eta,\lambda)}{\partial n_{\xi}}u_{f}(\xi,\eta)d\xi \\
&\quad +\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\Omega}u_{f}(\xi,\eta)
(-\frac{\partial}{\partial\eta}-\Delta_{\xi}+\lambda)\varepsilon^+_{n+1} 
 (x-\xi,t-\eta,\lambda)d\xi \\
&=I_{1}+I_{2}+I_{3}+I_{4}+I_{5},
\end{aligned}
\end{equation}
for all $(x,t)\in\Omega\times(0,T)$.

To calculate the value of the integral $I_{1}$, we use the explicit form 
of the fundamental solution and the change of variables
\begin{align*}
I_{1}&=\lim_{\delta\to 0}\int_{\Omega} \varepsilon^+_{n+1} 
 (x-\xi,\delta,\lambda)u_{f} (\xi,t-\delta)d\xi \\
&=\lim_{\delta\to 0}\int_{\Omega}\frac{1}{(2\sqrt{\pi\delta})^{n}}
e^{-\frac{|x-\xi|^2}{4\delta}}e^{-\lambda \delta}
u_{f} (\xi,t-\delta)d\xi \\
&=\lim_{\delta\to 0}\int_{R^{n}}\frac{1}{(2\sqrt{\pi\delta})^{n}}
e^{-\frac{|x-\xi|^2}{4\delta}}e^{-\lambda \delta}\widehat{u}_{f} 
 (\xi,t-\delta)d\xi=|\frac{|x-\xi|}{2\sqrt{\delta}}=z,
\end{align*} 
\begin{align*}
d\xi&=-2\sqrt{\delta}dz|
=\lim_{\delta\to 0}\frac{1}{(\sqrt{\pi})^{n}}
\int_{\frac{|x-a|}{2\sqrt{\delta}}}^{\frac{|x+a|}{2\sqrt{\delta}}}
{u}_{f} (x-2\sqrt{\delta}z,t-\delta)e^{-z^{2}}e^{-\lambda \delta}dz \\
&=\lim_{\delta\to 0}\frac{1}{(\sqrt{\pi})^{n}}
{u}_{f} (x-2\sqrt{\delta}z,t-\delta)e^{-\lambda \delta}
\int_{\frac{|x-a|}{2\sqrt{\delta}}}^{\frac{|x+a|}{2\sqrt{\delta}}}
e^{-z^{2}}dz \\
&=u(x,t)\frac{1}{(\sqrt{\pi})^{n}} \int_{-\infty}^{+\infty}
e^{-z^{2}}dz=u(x,t),
\end{align*}
where $\widehat{u}(x,t)$ is the extension by a zero the function $u(x,t)$ 
to the cube $-a<\xi<a$ from the $R^{n}$ containing the domain $\Omega$.

Since $u(x,0)=0$ then the integral is $I_{2}=0$.
The integrals $I_{3}$ and $I_{4}$ have a limit at $\delta\to 0$ and equals to
\begin{gather*}
I_{3}=\int_{0}^{t}d\eta\int_{\partial\Omega}\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)\frac{\partial u_{f}(\xi,\eta)}{\partial n_{\xi}}d\xi, \\
I_{4}=\int_{0}^{t}d\eta\int_{\partial\Omega}
\frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}u_{f}
(\xi,\eta)d\xi.
\end{gather*}
In that $\eta\leq t-\delta<t$, then
\[
(-\frac{\partial}{\partial\eta}-\Delta_{\xi}+\lambda)\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)\equiv0,
\]
therefore $I_{5}=0$.

Taking into account  \eqref{e2.19}, we get that for all $(x,t)\in\Omega\times(0,T)$,
\begin{align*}
I_{u}(x,t)&=\int_{0}^{t}d\eta\int_{\partial\Omega}
 \Big(\frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}
 {\partial n_{\xi}}u_{f}(\xi,\eta) \\
&\quad -\varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda)
 \frac{\partial u_{f}(\xi,\eta)}{\partial n_{\xi}}\Big)d\xi=0.
\end{align*}
When $x\to\partial\Omega$, using the properties of the double layer potential,
 we obtain
\begin{equation} \label{e2.20}
\begin{aligned}
I_{u}(x,t) &=  -\frac{u_{f}(x,t)}{2}+  \int_{0}^{t}  
 d\eta\int_{\partial\Omega}   \Big(\frac{\partial \varepsilon^+_{n+1}
(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}u_{f}(\xi,\eta) \\
&\quad -\varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda)\frac{\partial u_{f}(\xi,\eta)}
{\partial n_{\xi}}\Big)  d\xi=0.
\end{aligned}
\end{equation}
for all $(x,t)\in\partial\Omega\times(0,T)$.

When $x\neq\xi$ and $~t\neq\eta$, we have
\begin{gather*}
(\frac{\partial}{\partial t}-\Delta_{x}+\lambda)\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)\equiv0, \\
(\frac{\partial}{\partial t}-\Delta_{x}
 +\lambda)\frac{\partial}{\partial n_{\xi}}\varepsilon^+_{n+1}
 (x-\xi,t-\eta,\lambda)\equiv0,
\end{gather*}
so we obtain
\begin{equation} \label{e2.21}
(\frac{\partial}{\partial t}-\Delta_{x}+\lambda)I_{u}(x,t)\equiv0.
\end{equation}
Since $I_{u}(x,t)$ is the solution of the homogeneous heat equation \eqref{e2.21},
  by  the uniqueness of the mixed Cauchy problem, the identity
$$
I_{u}(x,t)\equiv0
$$
is equivalent to \eqref{e2.20}, i.e.
$$
I_{u}(x,t)|_{x\in\partial\Omega}=0
$$
is the lateral boundary condition of the volume heat potential \eqref{e2.11}
 for the heat equation with a scalar parameter.

Now we prove the converse statement.
If $u_{1}(x,t)$ is an arbitrary solution of the inhomogeneous heat equation 
with a scalar parameter \eqref{e2.12}, which satisfies the homogeneous 
initial condition \eqref{e2.13} and the lateral boundary condition 
\eqref{e2.18}, then it coincides with the volume heat potential $u_{f}(x,t)$,
 i.e. $u_{1}(x,t)=u_{f}(x,t)$.

If not, then the function
$$
\vartheta(x,t)=u_{1}(x,t)-u_{f}(x,t)
$$
satisfies the homogeneous heat equation with a scalar parameter
$$
\lozenge\vartheta(x,t)=\lozenge u_{1}(x,t)-\lozenge u_{f}(x,t)=0
$$
and the homogeneous initial condition
$$
\vartheta(x,0)=u_{1}(x,0)-u_{f}(x,0)=0
$$
and the lateral boundary condition
$$
I_{\vartheta}(x,t)=I_{u_{1}}(x,t)-I_{u_{f}}(x,t)=0.
$$
As above, by direct calculation we obtain
\begin{align*}
0&=\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\Omega} \varepsilon^+_{n+1}
 (x-\xi,t-\eta,\lambda) \lozenge\vartheta(\xi,\eta)d\xi \\
&=\vartheta(x,t)+\int_{0}^{t}d\eta\int_{\partial\Omega}
 \Big(\frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}
 {\partial n_{\xi}}\vartheta(\xi,\eta)\\
&\quad -\varepsilon^+_{n+1} 
 (x-\xi,t-\eta,\lambda)
 \frac{\partial \vartheta(\xi,\eta)}{\partial n_{\xi}}\Big)d\xi
\end{align*}
for all $(x,t)\in\Omega\times(0,T)$.
Therefore,
\begin{gather*}
\left(\vartheta(x,t)+I_{\vartheta}(x,t)\right)|_{x\in\partial\Omega}=0,\\
\vartheta(x,t)|_{x\in\partial\Omega}=0.
\end{gather*}
By the uniqueness of the mixed Cauchy problem for the homogeneous heat 
equation with a scalar parameter, according to the maximum principle, 
we have 
\begin{gather*}
\vartheta(x,t)\equiv 0, \\
u_{1}(x,t)=u_{f}(x,t)
\end{gather*}
for all $(x,t)\in\Omega\times(0,T)$.

Thus, the lateral boundary condition \eqref{e2.18} and the initial 
condition \eqref{e2.13} for the heat equation with a scalar parameter 
\eqref{e2.12} generates a volume heat potential uniquely.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
For any function $u_{0}(x)\in W^{2}_{2}(\Omega)$ the surface heat 
potential \eqref{e2.14} satisfies the homogeneous heat equation with a 
scalar parameter \eqref{e2.15}, the nonhomogeneous initial condition 
\eqref{e2.16}, and the following lateral boundary condition:
\begin{equation} \label{2.22}
\begin{aligned}
&-\frac{u_\tau(x,t)}{2}
 +\int_{0}^{t}d\eta \int_{\partial \Omega} \Big( \frac{\partial
\varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}} u_\tau(\xi,\eta) \\
& - \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)
 \frac{\partial u_\tau(\xi,\eta)}{\partial n_{\xi}} \Big)d\xi=0,
\end{aligned}
\end{equation}
for all $(x,t)\in\partial \Omega\cap(0,T)$.

Conversely, if $u(x,t)\in W^{2,1}_{2}(D^{+})$ is a solution of the 
homogeneous heat equation with a scalar parameter \eqref{e2.15},
 which satisfies the nonhomogeneous initial condition \eqref{e2.16} and 
the lateral boundary condition \eqref{2.22}, then it coincides with the 
surface heat potential \eqref{e2.11}.
\end{lemma}

\begin{proof}
In this case formula \eqref{e2.19} becomes
\begin{equation} \label{e2.23}
\begin{aligned}
0&=\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\Omega} 
 \varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda) 
 (\frac{\partial}{\partial\eta}-\Delta_{\xi}+\lambda) u_{\tau} (\xi,\eta)d\xi \\
&=\lim_{\delta\to 0}\int_{\Omega} \varepsilon^+_{n+1} (x-\xi,\delta,\lambda)u_{\tau} 
(\xi,t-\delta)d\xi \\
&\quad -\lim_{\delta\to 0}\int_{\Omega} \varepsilon^+_{n+1} (x-\xi,t,\lambda)u_{\tau} 
 (\xi,0)d\xi \\
&\quad -\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\partial\Omega}
 \varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda)
 \frac{\partial u_{\tau}(\xi,\eta)}{\partial n_{\xi}}d\xi \\
&\quad +\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\partial\Omega}
 \frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}u_{\tau}
 (\xi,\eta)d\xi \\
&\quad +\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\Omega}u_{\tau}(\xi,\eta)
(-\frac{\partial}{\partial\eta}-\Delta_{\xi}+\lambda)\varepsilon^+_{n+1} 
 (x-\xi,t-\eta,\lambda)d\xi \\
&=I_{1}+I_{2}+I_{3}+I_{4}+I_{5}
\end{aligned}
\end{equation}
for all $(x,t)\in\Omega\times(0,T)$.

The integral $I_{1}$ is calculated as in the case of the heat potential and 
$I_{1}=u_{\tau}(x,t)$. 
By definition of the surface heat potential the integral $I_{2}$ coincides 
with the surface heat potential
\begin{equation} \label{e2.24}
u_{\tau}(x,t)=\int_{\Omega}\varepsilon^+_{n+1}(x-\xi,t,\lambda)\tau (\xi)d\xi=
\int_{\Omega}\varepsilon^+_{n+1}(x-\xi,t,\lambda)u_{\tau}(\xi,0)d\xi=I_{2}.
\end{equation}
As above
\begin{gather*}
I_{3}=\int_{0}^{t}d\eta\int_{\partial\Omega}\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)\frac{\partial u_{\tau}(\xi,\eta)}{\partial n_{\xi}}d\xi,
\\
I_{4}=\int_{0}^{t}d\eta\int_{\partial\Omega}\frac{\partial \varepsilon^+_{n+1}
(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}u_{\tau}(\xi,\eta)d\xi,
\\
I_{5}=\lim_{\delta\to 0}\int_{0}^{t-\delta}d\eta\int_{\Omega}u_{\tau}(\xi,\eta)
(-\frac{\partial}{\partial\eta}-\Delta_{\xi}+\lambda)\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)d\xi=0.
\end{gather*}
We obtain that for all $(x,t)\in\Omega\times(0,T)$,
\begin{align*}
I_{u}(x,t)&=\int_{0}^{t}d\eta\int_{\partial\Omega}
\Big(\frac{\partial \varepsilon^+_{n+1}(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}
u_{\tau}(\xi,\eta) \\
&\quad -\varepsilon^+_{n+1} (x-\xi,t-\eta,\lambda)\frac{\partial u_{\tau}
(\xi,\eta)}{\partial n_{\xi}}\Big)d\xi=0.
\end{align*}

When $x\to\partial\Omega$, once again using the properties of the double 
layer potential, we obtain the lateral boundary condition for the surface 
heat potential
\begin{equation} \label{e2.25}
\begin{aligned}
I_{u}(x,t) &=  -\frac{u_{\tau}(x,t)}{2}+  \int_{0}^{t}    d\eta 
 \int_{\partial\Omega}   \Big(\frac{\partial \varepsilon^+_{n+1}
(x-\xi,t-\eta,\lambda)}{\partial n_{\xi}}u_{\tau}(\xi,\eta)\\
&\quad -\varepsilon^+_{n+1} 
(x-\xi,t-\eta,\lambda)\frac{\partial u_{\tau}(\xi,\eta)}{\partial n_{\xi}}\Big)  d\xi
 =0.
\end{aligned}
\end{equation}
for all $(x,t)\in\partial\Omega\times(0,T)$ and
\begin{equation} \label{e2.26}
(\frac{\partial}{\partial t}-\Delta_{x}+\lambda)I_{u}(x,t)\equiv0.
\end{equation}

As above, since $I_{u}(x,t)$ is the solution of the homogeneous heat equation
 \eqref{e2.26}, then, by virtue of the uniqueness of the mixed Cauchy problem, 
the identity
$$
I_{u}(x,t)\equiv 0
$$
is equivalent to \eqref{e2.25}, i.e.
$$
I_{u}(x,t)|_{x\in\partial\Omega}=0
$$
is the lateral boundary condition of the surface heat potential \eqref{e2.14} 
for the homogeneous heat equation with a scalar parameter.

The converse statement is proved as in the case of the volume heat potential.
This completes the proof.
\end{proof}

As in \cite{KS1} we can show that Newton's potential (volume potential)
\begin{equation} \label{e2.27}
u(x,t)=(L^{-1}_Bf)(x,t)=\int_{D^-} \varepsilon^-_{n+1}
(x- \xi,t-\eta,\lambda)f^-(\xi,\eta)dD^-,\; t<0,
\end{equation}
satisfies Helmholtz equation
\begin{equation} \label{e2.28}
-\frac{\partial^2 u(x,t)}{\partial t^2}-\Delta_x u(x,t)+\lambda u(x,t)
=f^- (x,t), \quad x\in R^n,\; t<0
\end{equation}
satisfies the potential boundary condition
\begin{equation} \label{e2.29}
\begin{aligned}
&-\frac{u(x,t)}{2}+\int_{\partial D^{-}}
  \Big( \frac{\partial \varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda)}
{\partial n_{\xi}}u(\xi,\eta)\\
&-\varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda) 
\frac{\partial u(\xi,\eta)}{\partial n_{\xi}}\Big)dD^{-}=0,
\end{aligned}
\end{equation}
for all $(x,t)\in\partial D^{-}=\sigma\cup\Omega$.

It should be noted that the boundary $\partial D^{-}$ includes a domain 
$\Omega$, which is an internal subset of the domain $D$.

\begin{lemma} \label{lem2.6}
For any function $f(x,t)\in C^{\alpha}(\overline{D})$ the Newton's potential 
\eqref{e2.27} satisfies the inhomogeneous Helmholtz equation \eqref{e2.28} 
and the potential boundary condition
\begin{equation} \label{e2.30}
\begin{aligned}
&-\frac{u(x,t)}{2}
+\int_{\partial D^{-}} \Big(\frac{\partial \varepsilon^-_{n+1}(x-\xi,
 t-\eta,\lambda)}{\partial n_{\xi}} u(\xi,\eta) \\
&-\varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda)\frac{\partial u(\xi,\eta)}
{\partial n_{\xi}} \Big)dD^{-}=0
\end{aligned}
\end{equation}
for all $(x,t)\in\partial D^{-}=\sigma\cup\Omega$.

Conversely, if $u(x,t)\in W^{2}_{2}(D^{-})$ is a solution of the 
inhomogeneous Helmholtz equation \eqref{e2.28}, which satisfies the 
potential boundary condition \eqref{e2.30}, then it coincides with 
 Newton's potential \eqref{e2.27},
where
\begin{equation} \label{e2.31}
\begin{aligned}
\varepsilon^-_{n+1}(\overline{x},\lambda)
&= \frac{1}{(n-1)\omega_{n+1}|\overline{x}|^{n-1}}
\frac{(n-1)}{\Gamma(\frac{n+1}{2})}
\Big(\frac{\sqrt{\lambda}|\overline{x}|}{2}\Big)^{\frac{n-1}{2}}
K_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|) \\
&=\varepsilon_{\Delta}^{n+1}(\overline{x})
\widetilde{K}_{\frac{1-n}{2}}(\sqrt{\lambda}|\overline{x}|)
\end{aligned}
\end{equation}
is the fundamental solution of the Helmholtz equation.
\end{lemma}

\begin{proof}
Assuming that $u(x,t)\in C^{2}(D^{-})\cap C^{1}(\overline{D^{-}})$, 
by direct calculation and using Green's formula for any $(x,t)\in (D^{-})$, we have
\begin{align*}
u(x,t)&=\int_{D^{-}}\varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda)f(\xi,\eta)dD^{-} \\
&=\int_{D^{-}} \varepsilon^-_{n+1} (x-\xi,t-\eta,\lambda) 
 (-\frac{\partial^{2}}{\partial\eta^{2}}-\Delta_{\xi}+\lambda)u(\xi,\eta)dD^{-} \\
&=\int_{\partial D^{-}}\Big(\frac{\partial \varepsilon^-_{n+1}
 (x-\xi,t-\eta,\lambda)}{\partial n}u(\xi,\eta)-\varepsilon^-_{n+1}
 (x-\xi,t-\eta,\lambda)\frac{\partial u(\xi,\eta)}{\partial n}\Big)dS \\
&\quad - \int_{D^{-}}u(\xi,\eta)
(-\frac{\partial^{2}}{\partial\eta^{2}}-\Delta_{\xi}+\lambda)\varepsilon^-_{n+1} 
(x-\xi,t-\eta,\lambda)dD^{-} \\
&=u(x,t)+\int_{\partial D^{-}}\Big(\frac{\partial \varepsilon^-_{n+1}
(x-\xi,t-\eta,\lambda)}{\partial n}u(\xi,\eta)
-\varepsilon^-_{n+1} (x-\xi,t-\eta,\lambda)
\frac{\partial u(\xi,\eta)}{\partial n}\Big)dS.
\end{align*}
Therefore,
\begin{align*}
I_{u}(x,t)
&=\int_{\partial D^{-}}\Big(\frac{\partial \varepsilon^-_{n+1}
(x-\xi,t-\eta,\lambda)}{\partial n}u(\xi,\eta)\\
&\quad -\varepsilon^-_{n+1}
 (x-\xi,t-\eta,\lambda)\frac{\partial u(\xi,\eta)}{\partial n}\Big)dS=0
\end{align*}
for all $(x,t)\in D^{-}$.

When $x\to\partial\Omega$ using the properties of the double layer 
potential for \eqref{e2.31}, we obtain the potential boundary condition 
of the Newton's potential,
\begin{equation} \label{e2.32}
\begin{aligned}
I_{u}(x,t) &=  -\frac{u(x,t)}{2}+\int_{\partial D^{-}} 
 \Big(\frac{\partial \varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda)}
 {\partial n}u(\xi,\eta) \\
&\quad -\varepsilon^-_{n+1} (x-\xi,t-\eta,\lambda)
 \frac{\partial u(\xi,\eta)}{\partial n}\Big)  dS=0
\end{aligned}
\end{equation}
for all $(x,t)\in\partial D^{-}=\sigma\cup\Omega$.

When $x\neq\xi$ and $t\neq\eta$,
\begin{gather*}
(-\frac{\partial^{2}}{\partial t^{2}}-\Delta_{x}+\lambda)\varepsilon^-_{n+1}
(x-\xi,t-\eta,\lambda)\equiv0, \\
(-\frac{\partial^{2}}{\partial t^{2}}-\Delta_{x}+\lambda)
\frac{\partial}{\partial n}\varepsilon^-_{n+1}(x-\xi,t-\eta,\lambda)\equiv0.
\end{gather*}
Then we have
\begin{equation} \label{e2.33}
(-\frac{\partial^{2}}{\partial t^{2}}-\Delta_{x}+\lambda)I_{u}(x,t)\equiv0.
\end{equation}

Since $I_{u}(x,t)$ is the solution of the homogeneous Helmholtz equation 
\eqref{e2.33},  by the uniqueness of the Dirichle problem, it follows that
$$
I_{u}(x,t)\equiv0
$$
is equivalent to \eqref{e2.32}, i.e.
$$
I_{u}(x,t)|_{(x,t)\in\partial D^{-}}=0
$$
is the potential boundary condition of the Newton's potential \eqref{e2.27} 
for the inhomogeneous Helmholtz equation.

The converse statement is proved as in the case of the volume heat potential 
and surface heat potential.
This completes the proof.
\end{proof}

We have proved that an elliptic-parabolic potential $u(x,t)=(L^{-1}_Bf)(x,t)$ 
\eqref{e1.3} satisfies the boundary conditions \eqref{e2.17} and \eqref{e2.29}, 
now we prove the converse statement.
If
\begin{equation*}
 u(x,t) = (L^{-1}_Bf)(x,t) \in C^{\alpha}(\overline{D})
 \cap C^{2+ \alpha}(\overline{D^-}) \cap C_{x,t}^{2+\alpha, 
1 + \alpha}(\overline{D^+})
\end{equation*}
is a solution of
\begin{equation} \label{e2.34}
\begin{gathered}
 \Big( \frac{\partial}{\partial t} - \Delta_x + \lambda \Big)u(x,t)
=f^+(x,t), \quad (x,t) \in D^+  \\
 \Big( -\frac{\partial^{2}}{\partial t^{2}} - \Delta_x + \lambda \Big)u(x,t)
=f^-(x,t), \quad (x,t) \in D^- 
 \end{gathered}
\end{equation}
and satisfies  conditions \eqref{e2.17} and \eqref{e2.29}, 
then $u(x,t)$ coincides with the elliptic-parabolic potential
 $u(x,t)=(L^{-1}_Bf)(x,t)$ \eqref{e1.3}.

From the continuity of solution $u(x,t)$ when $t =0$ we can find
\begin{equation} \label{e2.35}
 \tau(x) = u(x,0-) = \int_{D^-} \varepsilon^-_{n+1} (x-\xi,t-\eta,\lambda) 
f^-(\xi,\eta) d\xi d\eta.
\end{equation}
The general solution of  \eqref{e1.1} in the domain $D^+$ satisfying 
 condition \eqref{e2.35} and $u(x,t)|_{t=0}=\tau(x)$, so we can represent 
the general solution of\eqref{e1.1} in the following form
\begin{equation} \label{e2.36}
\begin{aligned}
 u(x,t) &= (L^{-1}_Bf)(x,t) \\
&= \int_{D^+} \varepsilon^+_{n+1} 
 (x-\xi,t-\eta,\lambda) f^+(\xi,\eta) d\xi d\eta \\ 
&\quad +\int_{\Omega} \varepsilon^+_{n+1} (x-\xi,t,\lambda) 
\Big(\int_{D^-} \varepsilon^-_{n+1} (\xi-\overline{\xi},t-\eta,\lambda) 
 f^-(\overline{\xi},\eta) d\overline{\xi} d\eta\Big) d\xi.
\end{aligned}
\end{equation}


\begin{theorem} \label{thm2.7}
For any $f(x,t) \in C^{\alpha}(\overline{D})$ the elliptic-parabolic potential
\begin{equation*}
 u(x,t) = (L^{-1}_Bf)(x,t) \in C^{\alpha}(\overline{D}) 
\cap C^{2+ \alpha}(\overline{D^-}) \cap C_{x,t}^{2+\alpha, 
1 + \alpha}(\overline{D^+})
\end{equation*}
which represented as \eqref{e1.3} satisfies the Bitsadze-Samarsky boundary 
conditions \eqref{e2.17} and \eqref{e2.29}.

Conversely, if
\begin{equation*}
 u(x,t) = (L^{-1}_Bf)(x,t) \in C^{\alpha}(\overline{D}) 
\cap C^{2+ \alpha}(\overline{D^-}) \cap C_{x,t}^{2+\alpha, 
1 + \alpha}(\overline{D^+})
\end{equation*}
is a solution of the equations \eqref{e1.1}-\eqref{e1.2} and it 
satisfies the Bitsadze-Samarsky boundary conditions \eqref{e2.17}
 and \eqref{e2.29}, then it coincides with the elliptic-parabolic 
potential \eqref{e1.3}.
\end{theorem}

\subsection*{Acknowledgments}
This research was supported by grants AP05133239
and AP05134615 from the Ministry of Education and Science of Republic
 of Kazakhstan.

\begin{thebibliography}{00}

\bibitem{KS1}
T.~Sh.~Kal'menov, D.~Suragan;
Two spectral problems for the volume potential.
\emph{ Doklady Mathematics}, \textbf{80} (2009), P. 646-649.

\bibitem{KT2} T. Sh. Kal'menov, N.~E.~Tokmagambetov;
\newblock On a nonlocal boundary value problem for the multidimensional heat 
equation in a noncylindrical domain.
\emph{Siber. Math. J.}, \textbf{54} (2013), P. 1023-1028.

\bibitem{KA3} T. Sh. Kal'menov, G. D. Arepova;
\newblock On a boundary condition of the surface heat potential.
\newblock {\em AIP Conf. Proc.}, \textbf{1676} (2015), 020054. P. 1-4.

\bibitem{KA4} T. Sh. Kal'menov, G. D. Arepova;
\newblock On a heat and mass transfer model for the locally
inhomogeneous initial data.
\newblock {\em Bulletin SUSU MMCS}, \textbf{9} (2016), P. 124-129.

\bibitem{KA5} T. Sh. Kal'menov, G. D. Arepova;
\newblock On a boundary conditions for the linear integral operators
\newblock {\em Doklady Adgskoj (Cherkesskoj) mezhdunarodnoj akademii nauk
 [Reports of Adyghe (Circassian) International Academy of Sciences].}, 
\textbf{17} (2015), P. 34-41.

\bibitem{KS6} T. Sh. Kal'menov, D. Suragan;
\newblock A boundary condition and spectral problems for the {N}ewton potential.
\newblock {\em Modern aspects of the theory of partial differential
 equations}, \textbf{216} (2011) of {\em Oper. Theory: Adv. Appl.}, P. 187-210.
Birkh{\"a}user/Springer Basel AG, Basel, 2011.

\bibitem{KAS7} T. Sh. Kal'menov, G. D. Arepova, D. Suragan;
\newblock On the symmetry of the boundary conditions of the volume potential.
\newblock {\em AIP Conf. Proc.}, \textbf{1880} (2017), 040014. P. 1-4.

\bibitem{W8} G. W. Watson;
\emph{A treatise on the theory of Bessel functions}.
London, (1922), 804 p.

\bibitem{V9} V. S. Vladimirov;
\emph{Equations of mathematical physics}.
Moscow, (1981), 512 p.

\end{thebibliography}

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