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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 90, pp. 1--32.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/90\hfil
Initial-boundary value problems in a plane corner]
{Initial-boundary value problems in a plane corner for the heat equation}

\author[B. V. Bazaliy, N. Vasylyeva\hfil EJDE-2010/90\hfilneg]
{Borys V. Bazaliy, Nataliya Vasylyeva}  % in alphabetical order

\address{Borys V. Bazaliy \newline
Institute of Applied Mathematics and Mechanics of NAS of Ukraine,
R. Luxemburg, str. 74, 83114 Donetsk, Ukraine}
\email{bazaliy@iamm.ac.donetsk.ua}

\address{Nataliya Vasylyeva \newline
Institute of Applied Mathematics and Mechanics of NAS of Ukraine,
R. Luxemburg, str. 74, 83114 Donetsk, Ukraine}
\email{vasylyeva@iamm.ac.donetsk.ua}

\thanks{Submitted September 11, 2009. Published June 30, 2010.}
\subjclass[2000]{35K20, 35K365}
\keywords{Initial boundary value problems; nonsmooth domains;
\hfill\break\indent  weighted H\"older spaces; trigonometric series}

\begin{abstract}
 We study the Dirichlet initial problem for the heat equation by
 the Fourier-Bessel method in a plane corner. We prove classical
 solvability for the problem in weighted H\"older spaces.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}

There are various approaches in investigations of initial boundary
value problems for parabolic equations in domains with
singularities. In the works of Grisvard \cite{g3},
Solonnikov \cite{s1}, Amann \cite{a1},
 Garroni, Solonnikov and Vivaldi \cite{g1}, Frolova \cite{f1}, the
existence of solutions and qualitative properties of solutions are
described in the terms of Sobolev or weighted Sobolev spaces. The
similar results in H\"older classes are represented in works
Guidetti \cite{g4}, Colombo, Guidetti, and Lorenzi \cite{c2},
Solonnikov \cite{s1}
(see also references in these works).

Note that in the pointed out works the method of the Green
function or the theory of analytic semigroups were used to
construct some explicit representation of a solution and to obtain
the optimal estimates.

In the present paper we use the classical Fourier method to get a
solution in the form of Fourier-Bessel series in an angular
domain. Then we apply some results on trigonometric series theory,
in particular, Bernstein theorem and Jackson's construction of
approximating trigonometric polynomials to obtain estimates  of
the higher derivatives of the solution to the Dirichlet initial
problem for the heat equation in weighted H\"older spaces.

Sometimes a classical solution of the initial value problem for a
parabolic equation is defined as a function, that has required
higher derivatives in any internal subdomain of a cylindrical
domain. In the one dimensional case a similar result was published
by Chernyatin \cite{c1} where a solution was represented as the sum of
the trigonometric series. But to the best of our knowledge, we
have not found similar results concerning with two-dimensional
case.

The paper is organized as follows: in Section 2, we formulate the
problem, introduce the appropriate functional space, and show the
formal solution to the Dirichlet initial problem for the heat
equation in the form of the sum of the trigonometric series.
Section 3 contains some auxiliary estimates. In Section 4, we
recall some results from the trigonometric series theory, and in
Section 5 we show that the trigonometric series representing the
formal solution converge together with its higher order
derivatives. Section 6 consists of some final remarks to the
existence and uniqueness theorem and some results concerning the
parabolic equation with singular coefficients.

\section{The statement of the problem and main results}

We use the polar coordinate system $(r,\varphi)$ on a plane
$\mathbb{R}^{2}$. Let
\[
G=\{(r,\varphi):r>0,\;0<\varphi<\theta\},\; \theta \in (0,2\pi),
\]
be an infinite angle on $\mathbb{R}^{2}$, $G_{T}=G\times[0,T]$,
$T\in \mathbb{R}_{+}$, and its boundary be
\begin{gather*}
g=g_0\cup g_{1}, g_0=\{(r,\varphi):r>0,\varphi=0\},\\
g_{1}=\{(r,\varphi):r>0,\varphi=\theta\}, g_{iT}=g_{i}\times
[0,T],\quad i=0,1.
\end{gather*}
Let $\alpha \in (0,1)$, $l$ be an integer. We use the weighted
H\"older space $P^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$
of
 functions $u(r,\varphi,t)$ with the finite norm
\begin{align*}
&\|u\|_{P_{s}^{l+\alpha,(l+\alpha)/2}(\overline{G}_{T})}
\equiv |u|^{l+\alpha}_{s,G_{T}}\\
&=\sum _{0\leq
\beta_{1}+\beta_{2}+2a\leq l}\sup_{(r,\varphi,t)\in
\overline{G}_{T}} r^{-s+\beta_{1}+2a}|D_{r}
^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u|\\
&\quad+\sum_{0<l+\alpha-(\beta_{1}+\beta_{2}+2a)<2}
\Big\{\langle D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}
u\rangle_{t;s-\beta_{1}-2a-\alpha,G_{T}}^{(\frac
{l+\alpha-\beta_{1}-\beta_{2}-2a}{2})} \\
&\quad  + [D_{r}^{\beta_{1}}D_{\varphi
}^{\beta_{2}}D_{t}^{a}u]_{r,t;s-\beta_{1}-2a-2\alpha,G_{T}}^{(\alpha
,\frac {l+\alpha-\beta_{1}-\beta_{2}-2a}{2})}
+[D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u]_{\varphi
,t;s-\beta_{1}-2a-\alpha,G_{T}}^{(\alpha,\frac
{l+\alpha-\beta_{1}-\beta_{2}-2a}{2})}\Big\}
\\
&\quad +\sum _{ \beta_{1}+\beta_{2}+2a=
l}\Big\{\langle D_{r}^{\beta_{1}}D_{\varphi
}^{\beta_{2}}D_{t}^{a}u\rangle_{r;s-\beta_{1}-2a-\alpha,G_{T}}^{(\alpha)}
\\
&\quad +\langle D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u\rangle_{\varphi;s-\beta
_{1}-2a,G_{T}}^{(\alpha)}\Big\},
\end{align*}
with the seminorms  defined as follows,
$\overline{r}=\min (\rho,r)$, $\alpha,\gamma\in (0,1)$:
\begin{gather*}
\langle v\rangle
_{r;\mu,G_{T}}^{(\alpha)}=
\sup_{(\rho,\varphi,t),(r,\varphi,t)\in
\overline{G}_{T},\; |\rho-r|\leq\overline{r}/2}
\overline{r}^{-\mu}\frac
{|v(\rho,\varphi,t)-v(r,\varphi,t)|}{|\rho-r|^{\alpha}},
\\
 \langle v\rangle_{\varphi;\mu,G_{T}}^{(\alpha)}
=\sup_{(r,\varphi,t),(r,\psi,t)\in \overline{G}_{T}}
r^{-\mu}\frac{|v(r,\varphi,t)-v(r,\psi,t)|}{|\varphi-\psi|^{\alpha}},
\\
 \langle v\rangle _{t;\mu,G_{T}}^{(\gamma)}
 =\sup_{(\rho,\varphi,t),(r,\varphi,\tau)\in \overline{G}_{T}}
r^{-\mu}\frac{|v(r,\varphi,t)-v(r,\varphi,\tau)|}{|t-\tau|^{\gamma}},
\\
[v]_{r,t;\mu,G_{T}}^{(\alpha,\gamma)}
=\sup_
{\substack{
(\rho,\varphi,t),(r,\varphi,t),\\
(r,\varphi,\tau),(\rho,\varphi,\tau)\in \overline{G}_{T}\\
|\rho-r|\leq\overline{r}/2 
}}
\overline{r}^{-\mu} \frac{|v(r,\varphi,t)-v(r,\varphi,\tau)
-v(\rho,\varphi,t)+v(\rho,\varphi,\tau)|}
{|\rho-r|^{\alpha}|t-\tau|^{\gamma}},
\\
[v]_{\varphi,t;\mu,G_{T}}^{(\alpha,\gamma)}
=\sup_{\substack{(r,\varphi,t),(r,\psi,t),\\
(r,\varphi,\tau),(r,\psi,\tau)\in \overline{G}_{T}}
}
r^{-\mu}\frac{|v(r,\varphi,t)-v(r,\varphi,\tau)-v(r,\psi,t)
+v(r,\psi,\tau)|} {|\varphi-\psi|^{\alpha}|t-\tau|^{\gamma}}.
\end{gather*}
The seminorms $[\cdot]^{(\alpha,\gamma)}$ were introduced in
\cite{s2}. In a similar way we introduce the space
$P^{l+\alpha}_{s}(\overline{G})$ of the functions $u(r,\varphi)$
on $G$. Hereinafter we will use the subspace
$\widehat{P}^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$
of the space
$P^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$ which is
introduced as follows.
Let
\[
\mathbb{R}_{T}=\{(r,t):r>0,t\in (0,T)\},
\]
and function $v(r,\varphi,t)\in
P^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$ be such that
\[
v(r,\varphi,t)=\sum_{k=1}^{\infty}v_{k}(r,t)\sin\lambda_{k}\varphi,
\quad \lambda_{k}=\frac{\pi k}{\theta},
\]
where
\[
v_{k}(r,t)=\frac{2}{\theta}\int_0^{\theta}v(r,\psi,t)\sin(\lambda
_{k}\psi)d\psi.
\]
We will say the function $v(r,\varphi,t)\in
\widehat{P}^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$
if $v(r,\varphi,t)\in
P^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$ and the
following inequality holds:
\begin{align*}
S(v) _{\widehat{P}_{s}^{l+\alpha,(l+\alpha)/2}(\overline{G}_{T})}
&:=\sum_{k=1}^{\infty}\Big( \sum _{0\leq
\beta_{1}+\beta_{2}+2a\leq l}\sup_{(r,t)\in \overline{R}_{T}}
r^{-s+\beta_{1}+2a}\lambda_{k}^{\beta_{2}}|D_{r}
^{\beta_{1}}D_{t}^{a}v_{k}(r,t)|\\
&\quad
+\sum_{0<l+\alpha-(\beta_{1}+\beta_{2}+2a)<2}
\Big\{\langle D_{r}^{\beta_{1}}D_{t}^{a}
v_{k}\rangle _{t;s-\beta_{1}-2a-\alpha,\mathbb{R}_{T}}^{(\frac
{l+\alpha-\beta_{1}-\beta_{2}-2a}{2})}\\
&\quad +[D_{r}^{\beta_{1}}D_{t}^{a}v_{k}]_{r,t;s-\beta_{1}
  -2a-2\alpha,\mathbb{R}_{T}}^{(\alpha,\frac {l+\alpha-\beta_{1}
  -\beta_{2}-2a}{2})}\Big\}\lambda_{k}^{\beta_{2}} \\
&\quad+\sum _{\beta_{1}+\beta_{2}+2a= l}
\lambda_{k}^{\beta_{2}}\langle D_{r}^{\beta_{1}}D_{t}^{a}v_{k}
\rangle_{r;s-\beta_{1}-2a-\alpha,\mathbb{R}_{T}}^{(\alpha)}
\Big)<\infty.
\end{align*} % \label{e2.1)
The subspace $\widehat{P}^{l+\alpha}_{s}(\overline{G})$ of the
function $v(r,\varphi)$ on $G$ is introduced similarly. One can
easily check that if $v(r,\varphi,t)\in
\widehat{P}^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$
or $v(r,\varphi)\in\widehat{ P}^{l+\alpha}_{s}(\overline{G})$ then
there are the constants $c_{i}$ or $\tilde{c}_{i}$, $i=1,2$, such
that
\begin{align*}
&c_{1}\|v\|_{P_{s}^{l+\alpha,\frac{l+\alpha}{2}}(\overline{G}_{T})}\\
&\leq S(v)_{\widehat{P}_{s}^{l+\alpha,(l+\alpha)/2}(\overline{G}_{T})}
+\sum_{ \beta_{1}+\beta_{2}+2a=
l}\langle D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}v
\rangle_{\varphi;s-\beta_{1}-2a,G_{T}}^{(\alpha)}
\\
&\quad +\sum_{0<l+\alpha-(\beta_{1}+\beta_{2}+2a)<2}
[D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}v]_{\varphi
,t;s-\beta_{1}-2a-\alpha,G_{T}}^{(\alpha,\frac
{l+\alpha-\beta_{1}-\beta_{2}-2a}{2})}\\
&\leq c_{2} \|v\| _{P_{s}^{l+\alpha,\frac{l+\alpha}{2}}
(\overline{G}_{T})};
\end{align*}
\begin{equation} \label{e2.2}
\tilde{c}_{1}\|v\| _{P_{s}^{l+\alpha}(\overline{G})}
\leq S(v)_{\widehat{P}_{s}^{l+\alpha}(\overline{G})}
+\sum _{\beta_{1}+\beta_{2}= l}
\langle D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}v
\rangle_{\varphi;s-\beta _{1},G}^{(\alpha)}
\leq \tilde{c}_{2} \|v\|_{P_{s}^{l+\alpha}(\overline{G})}.
\end{equation}
 Along with the spaces
$P^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$,
we will use the usual H\"older classes
$C^{\alpha,\beta}_{x,t}:=C^{\alpha,\beta}_{x,t}(\overline{\Omega}_{T})$
where $\beta \in (0,1)$, $x\in \Omega$, $t\in(0,T)$, and
\begin{gather*}
\|u\|_{C^{\alpha,\beta}_{x,t}(\overline{\Omega}_{T})}
=\sup_{(x,t)\in \overline{\Omega}_{T}}|u(x,t)|+\langle u
\rangle^{(\alpha)}_{x}+\langle u \rangle^{(\beta)}_{t},
\\
\langle u \rangle^{(\alpha)}_{x}=\sup_{(x,t),(y,t)\in
\overline{\Omega}_{T}}\frac{|u(x,t)-u(y,t)|}{|x-y|^{\alpha}},
\\
\langle u \rangle^{(\beta)}_{t}=\sup_{(x,t),(x,\tau)\in
\overline{\Omega}_{T}}\frac{|u(x,t)-u(x,\tau)|}{|t-\tau|^{\beta}}.
\end{gather*}
We are looking for a solution of the Dirichlet initial problem
\begin{equation} \label{e2.3}
\begin{gathered}
\frac{\partial u}{\partial t}-\frac{1}{r}\frac{\partial}{\partial r}
r\frac{\partial u}{\partial r}-\frac{1}{r^{2}}\frac{\partial^{2}u}
{\partial\varphi^{2}}=f(r,\varphi,t),\quad (r,\varphi,t)\in G_{T},
\\
u|_{g_{iT}}=0,\quad  u|_{t=0}=u_0(r,\varphi ),\quad (r,\varphi)\in G.
\end{gathered}
\end{equation}
We suppose that
\begin{equation} \label{e2.4}
f(r,0,t)=f(r,\theta,t)=0.
\end{equation}
As can be seen from further arguments, conditions \eqref{e2.4} are, at
least formally, necessary to get a solution to problem \eqref{e2.3} in
the form of the Fourier-Bessel series in the space
$P^{2+\alpha,(2+\alpha)/2}_{s+2}(\overline{G}_{T})$. Note that,
due to the presence of the seminorms $[\cdot]^{(\alpha,\alpha/2)}$
in the definition of the norm in the space
$P^{2+\alpha,(2+\alpha)/2}_{s+2}(\overline{G}_{T})$, the series
solution is more smooth than the solution from the ordinary
weighted H\"older spaces.

\begin{theorem} \label{thm2.1}
 Let equality \eqref{e2.4} and the consistency conditions of the
first order in problem \eqref{e2.3}
 be fulfilled.  The functions
$f\in\widehat{ P}^{\alpha,\alpha/2}_{s}(\overline{G}_{T})$ and
$u_0\in \widehat{P}^{2+\alpha}_{s+2}(\overline{G})$.
  Then there exists a unique solution
$u\in \widehat{P}^{2+\alpha,(2+\alpha)/2}_{s+2}(\overline{G}_{T})$ with
\begin{equation} \label{e2.5}
\begin{aligned}
&\|u\|_{P_{s+2}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}+S(u)
_{\widehat{P}_{s+2}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}\\
&\leq \textrm{const.}(\|f\|_{P_{s}^{\alpha,\alpha/2}(\overline{G}_{T})}
+\|u_0\|_{P_{s+2}^{2+\alpha}(\overline{G})}
+S(f) _{\widehat{P}_{s}^{\alpha,\alpha/2}(\overline{G}_{T})}
+S(u_0)_{\widehat{P}_{s+2}^{2+\alpha}(\overline{G})}),
\end{aligned}
\end{equation}
where the constant in \eqref{e2.5} is independent of $u$,
$-\pi/\theta+\alpha< s+2<\pi/\theta$, $\alpha\in (0,1)$.
\end{theorem}

Under the proof of Theorem \ref{thm2.1}, we will omit the subindex $G_{T}$
in the notations of the seminorms if it is clearly from the
context. We will assume that the function $f(r,\varphi,0)=0$ and,
hence, can be extended by zero onto $t<0$ with the same norm.

It can be easily seen that one of the factor in the eigenfunctions
to problem \eqref{e2.3} is $\sin(\lambda_{k}\phi)$,
$\lambda_{k}=\frac{\pi k}{\theta}$, $k=1,2,\dots$. So, after the
standard procedure of separation of variables (see Appendix 7.1),
we get the series representation of the solution:
\begin{gather}
u(r,\varphi,t)= R_1 (r,\varphi,t)+ R_2(r,\varphi,t),
 \label{e2.6} \\
\begin{aligned}
 R_1 (r,\varphi,t)
&=\sum_{k}\sin(\lambda_{k}\varphi)\int_{-\infty
}^{t}d\tau\int_0^{\infty}d\rho\frac{\rho}{2(t-\tau)}e^{-\frac
{\rho^{2}+r^{2}}{4(t-\tau)}}I_{\lambda_{k}}(\frac{\rho
r}{2(t-\tau)} )b_{k}(\rho,\tau)\\
&\equiv \sum_{k} R_{1,k}(r,t)\sin(\lambda_{k}\varphi),
\end{aligned} \label{e2.7}
\\
 R_2(r,\varphi,t)=\sum_{k}\sin(\lambda_{k}\varphi)\int_0
^{\infty}d\rho\frac{\rho}{2t}e^{-\frac{\rho^{2}+r^{2}}{4t}}I_{\lambda_{k}
}\big(\frac{\rho r}{2t}\big)u_{0k}(\rho),
 \label{e2.8} \\
u_{0k}(r)=\frac{2}{\theta}\int_0^{\theta}u_0(r,\psi)\sin(\lambda
_{k}\psi)d\psi,\quad
b_{k}(r,t)=\frac{2}{\theta}\int_0^{\theta}f(r,\psi,t)\sin(\lambda
_{k}\psi)d\psi, \label{e2.9}
\end{gather}
where $I_{\mu}(z)$ is a modified Bessel function. Equality \eqref{e2.6}
means that the desired solution is the sum of the volume potential
$ R_1 (r,\phi,t)$ and the potential of the initial data
$ R_2(r,\phi,t)$.

The general case of $f(r,\varphi,t)$; i.e., $f(r,\varphi,0)\neq 0$,
can be reduced to mention above with the following procedure. Let
in problem \eqref{e2.3}, \eqref{e2.4}, $f(r,\varphi,0)\in
\widehat{P}^{\alpha}_{s}(\overline{G})$, and the function
$\widehat{w}(r,\varphi)$ be a solution of the problem
\[
\Delta \widehat{w}=-f(r,\varphi,0) \quad G,\quad
\widehat{w}|_{g}=0,
\]
and $\widehat{w}(r,\varphi)\in
\widehat{P}^{2+\alpha}_{s+2}(\overline{G})$ (see \cite{v1}). Then we
consider the functions $w(r,\varphi,t)=\widehat{w}(r,\varphi)\cos
t$ and $v(r,\varphi,t)=u(r,\varphi,t)-w(r,\varphi,t)$ such that
\begin{equation} \label{e2.10}
\begin{gathered}
\begin{aligned}
\frac{\partial v}{\partial t}-\Delta v
&=\frac{\partial u}{\partial
t}-\Delta u+\widehat{w}\sin t+\Delta\widehat{w}\cos t\\
&=f(r,\varphi,t)+\widehat{w}(r,\varphi)\sin t
-f(r,\varphi,0)\cos t \equiv F(r,\varphi,t) \quad \text{in }
 G_{T},
\end{aligned}\\
v|_{g_{iT}}=0,\quad v|_{t=0}=u_0(r,\varphi)-\widehat{w}(r,\varphi).
\end{gathered}
\end{equation}
One can see that the consistency conditions of the first order are
fulfilled in problem \eqref{e2.10}; the function $F(r,\varphi,t)\in
\widehat{P}^{\alpha,\alpha/2}_{s}(\overline{G}_{T})$, satisfies
condition \eqref{e2.4}, and $F(r,\varphi,0)=0$.

Thus, the investigation of problem \eqref{e2.3} can be reduced to study
of problem \eqref{e2.10} with the homogeneous right part in the equation
if $t\leq 0$.

Note that, to prove inequality \eqref{e2.5} and
$u\in \widehat{P}^{2+\alpha,(2+\alpha)/2}_{s+2}(\overline{G}_{T})$ in
Theorem \ref{thm2.1}, it is sufficient to show the following estimate (due to
the first of inequalities in \eqref{e2.2})
\begin{equation} \label{e2.11}
\begin{aligned}
&S(u)_{\widehat{P}_{s+2}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}
+\sum _{ \beta_{1}+\beta_{2}+2a=
2}\langle D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u
\rangle_{\varphi;s+2-\beta
_{1}-2a,G_{T}}^{(\alpha)}
\\
&+\sum_{0<2+\alpha-(\beta_{1}+\beta_{2}+2a)<2}
[D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u]_{\varphi
,t;s+2-\beta_{1}-2a-\alpha,G_{T}}^{(\alpha,\frac
{2+\alpha-\beta_{1}-\beta_{2}-2a}{2})}
\\
&\leq \textrm{const.}\big(\|f\|
_{P_{s}^{\alpha,\alpha/2}(\overline{G}_{T})}+\|u_0\|
_{P_{s+2}^{2+\alpha}(\overline{G})}\Big).
\end{aligned}
\end{equation}
In the all following inequalities, the constants do not depend on
$k$.

\section{Convergence of series \eqref{e2.7} and \eqref{e2.8}}

Let us denote as
\begin{equation} \label{e3.1}
\Delta_{s}=\int_0^{t}d\tau\int_0^{\infty}d\rho\frac
{\rho^{1+s}}{2\tau}e^{-\frac{\rho^{2}+r^{2}}{4\tau}}I_{\lambda_{k}}
(\frac{\rho r}{2\tau}).
\end{equation}

\begin{lemma} \label{lem3.1}
The following estimate holds
\begin{equation} \label{e3.2}
\Delta_{s}\leq \frac{r^{2+s}}{\lambda_{k}^{2}-(s+2)^{2}},\quad
\text{if } s+2<\pi/\theta.
\end{equation}
\end{lemma}

\begin{proof}
The successive changes of variables: $ \frac{\rho r}{2\tau}=x$ and
$x^{2}\tau/r^{2}=z$ leads to
\begin{align*}
\Delta_{s}
&=r^{s}\int_0 ^{t}d\tau\int_0^{\infty}2^{1+s}(
\frac{x\tau}{r^{2}})^{1+s}e^{-\frac{r^{2}}{4\tau}
-\tau\frac{x^{2}}{r^{2}}}I_{\lambda_{k}}(x)dx\\
&=2^{1+s}\int_0^{\infty}\frac{r^{2+s}}{x^{3+s}}I_{\lambda_{k}}(x)dx
\int_0^{t\frac{x^{2}}{r^{2}}}z^{1+s}e^{-z-\frac{x^{2}}{4z}}dz\\
&\leq 2^{1+s}\int_0^{\infty}\frac{r^{2+s}}{x^{3+s}}I_{\lambda_{k}
}(x)dx\int_0^{\infty}z^{1+s}e^{-z-\frac{x^{2}}{4z}}dz.
\end{align*}
The internal integral in the above inequality can be calculated,
\cite[3.471 (9)]{g2},
$$
\int_0^{\infty}z^{1+s}e^{-z-\frac{x^{2}}{4z}}dz
=2(\frac{x^{2}}{4})^{1+s/2}K_{2+s}(x)
$$
where $K_{\mu}(z)$ is a modified Bessel function of the second
kind. Hence,
\begin{equation} \label{e3.3}
\Delta_{s}\leq\int_0^{\infty}\frac{r^{2+s}}{x}I_{\lambda_{k}
}(x)K_{2+s}(x)dx.
\end{equation}
The condition $s+2<\pi/\theta$ is sufficient to obtain the
boundedness of the right part in \eqref{e3.3} for all $k$. Really, we
take into account the tabular integral in the right part of \eqref{e3.3}
\cite[6.576(5)]{g2}, so that
\begin{equation} \label{e3.4}
\begin{aligned}
\int_0^{\infty}\frac{1}{x}I_{\lambda_{k}}(x)K_{2+s}(x)dx
&=\frac{1}{4}\frac{\Gamma((\lambda_{k}+s+2)/2)
\Gamma((\lambda_{k}-s-2)/2)}{\Gamma(\lambda_{k}+1)}\\
&\quad\times F((\lambda_{k}+s+2)/2,(\lambda_{k}-s-2)/2; \lambda_{k}+1;1)\\
&=\frac{1}{4}\frac{1}{\frac{\lambda_{k}^{2}}{4}-(1+\frac{s}{2})^{2}},
\end{aligned}
\end{equation}
here we employed the definition of the function
$F(\alpha,\beta;\gamma;z)$ \cite[9.111]{g2}. Inequality \eqref{e3.3}
together with \eqref{e3.4} complete the proof of Lemma \ref{lem3.1}.
\end{proof}

 Similar arguments lead to forllowing remark.
 
\begin{remark} \label{rmk1} \rm
If $s+2<\pi/\theta$, then
\[
\int_0^{\infty}d\tau\int_0^{\infty}d\rho
\frac{\rho^{1+s}}{2\tau}e^{-\frac{\rho^{2}+r^{2}}{4\tau}}I_{\lambda_{k}}
(\frac{\rho r}{2\tau})
=\textrm{const.} \frac{r^{2+s}}{\lambda_k^2-(s+2)^2}\,.
\]
\end{remark}

Note that we take advantage of some tabular integrals in order to
obtain the sharp estimates of the weight in the statement of
Lemma \ref{lem3.1}. It is possible to apply  simpler arguments to derive only the
asymptotic $\Delta_{s}$ with respect to $\lambda_{k}$.

Hereinafter we will use the following properties of the Bessel
functions
\begin{equation} \label{e3.5}
\begin{gathered}
I_{\mu}(z)\sim \textrm{const.}\frac{(z/2)^{\mu}}{\Gamma(\mu+1)},\quad
\text{for  small  values  of $z$},\\
I_{\mu}(z)\sim e^{z}/\sqrt{2\pi z}+\frac{C(\mu)}{z^{3/2}}, \quad
\text{for  large  values  of $z$}
\end{gathered}
\end{equation}
where $C(\mu)$ is some function,
\begin{equation} \label{e3.6}
\begin{gathered}
 K_{\mu}(z)\sim \textrm{const.} z^{-\mu}\quad
\text{for $|\mu|\leq \textrm{const.}$  and  small  values of $z$},
\\
K_{\mu}(z)\sim e^{-z}/\sqrt{2\pi z} \quad
\text{for $|\mu|\leq \textrm{const.}$ and  large  values of $z$}.
\end{gathered}
\end{equation}

\begin{lemma} \label{lem3.2}
The following estimate holds for $s<\pi/\theta$:
\begin{equation} \label{e3.7}
D_{s}:=D_{s}(r,t)=\int_0^{\infty}d\rho\frac{\rho^{1+s}}{2t}e^{-\frac
{\rho^{2}+r^{2}}{4t}}I_{\lambda_{k}}\big(\frac{\rho r}{2t}\big)\leq
\textrm{const.}r^{s}.
\end{equation}
\end{lemma}

\begin{proof}
Let us consider the problem
\begin{gather*}
\frac{\partial u}{\partial t}-\Delta u=0 \quad in \quad G_{T},\\
u|_{t=0}=r^{s}\sin \lambda_{k}\varphi,\quad u|_{g_{iT}}=0.
\end{gather*}
Denote by $w(r,\varphi)=r^{s}\sin \lambda_{k}\varphi$ and
introduce the function
$v(r,\varphi,t)=u(r,\varphi,t)-w(r,\varphi)$. The function
$w(r,\varphi)$ satisfies the equation
\[
\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial w}{\partial
r}+\frac{1}{r^{2}}\frac{\partial^{2} w}{\partial
\varphi^{2}}=r^{s-2}(s^{2}-\lambda_{k}^{2})\sin
\lambda_{k}\varphi,
\]
and $v(r,\varphi,t)$ is a solution of the problem
\begin{gather*}
\frac{\partial v}{\partial t}-\Delta
v=r^{s-2}(s^{2}-\lambda_{k}^{2})\sin \lambda_{k}\varphi \quad
\text{in } \quad G_{T}, \\
v|_{t=0}=0,\quad v|_{g_{iT}}=0.
\end{gather*}
Hence, by \eqref{e2.7}
\[
v(r,\varphi,t)=(s^{2}-\lambda_{k}^{2})\sin(\lambda_{k}\varphi)\int_{0
}^{t}d\tau\int_0^{\infty}d\rho\frac{\rho^{1+s-2}}{2(t-\tau)}e^{-\frac
{\rho^{2}+r^{2}}{4(t-\tau)}}I_{\lambda_{k}}(\frac{\rho
r}{2(t-\tau)} ),
\]
and due to Lemma \ref{lem3.1}
$|v(r,\varphi,t)|\leq \textrm{const.} r^{s}$,
so that
\[
|u(r,\varphi,t)|\leq \textrm{const.} r^{s}.
\]
On the other hand the solution $u(r,\varphi,t)$ of the initial
problem can be represented by using \eqref{e2.8}
\[
u(r,\varphi,t)=\sin(\lambda_{k}\varphi)\int_0^{\infty}d\rho\frac{\rho^{1+s}}{2t}e^{-\frac
{\rho^{2}+r^{2}}{4t}}I_{\lambda_{k}}(\frac{\rho r}{2t} ).
\]
If we take here $\varphi=\frac{\theta}{2k}$, we will obtain
inequality \eqref{e3.7}.
\end{proof}

\begin{corollary} \label{coro3.1}
The inequality
\[
e^{-z}z^{1/2}I_{\lambda_{k}}(z)\leq \textrm{const.}, \quad z\in(0,\infty)
\]
is valid for any $k$ with a constant is independent of $k$.
\end{corollary}

The proof of this Corollary is given in Appendix
(see subsection 7.2).

\begin{lemma} \label{lem3.3}
The following equality holds if $s=0$,
\begin{equation} \label{e3.8}
\lim_{t\to 0}D_0=1
\end{equation}
for every $\lambda_{k}$.
\end{lemma}

\begin{proof}
First of all we will prove the following fact. Let
\[
\Delta_{-2,s}:=\int_0^{t}d\tau\int_0^{\infty}
\frac{1}{2\tau\rho^{1-s}}e^{-\frac{r^{2}+\rho^{2}}{4\tau}}
I_{\lambda_{k}}(\frac {r\rho}{2\tau})d\rho
\]
where $s$ will be chosen below.
We show that $\lim_{t\to 0} \Delta_{-2,s}=0$. In fact,
using the changes of variables $\frac{\rho r}{2\tau}=x$ and
$\tau\frac{x^{2}}{r^{2}}=z$,
\begin{align*}
\Delta_{-2,s} &  =\int_0^{\infty}\Big(\frac{1}{2x}\Big)  ^{1-s}r^{-s}I_{\lambda_{k}}
(x)dx\int_0^{t}\frac{1}{\tau^{1-s}}e^{-\frac{r^{2}}{4\tau}-\tau
\frac{x^{2}}{r^{2}}}d\tau\\
&  =\Big(\frac{1}{2}\Big)  ^{1-s}\int_0^{\infty}\frac{r^{s}
}{x^{1+s}}I_{\lambda_{k}}(x)dx\int_0^{t\frac{x^{2}}{r^{2}}}
z^{-1+s}e^{-z-\frac{x^{2}}{4z}}dz\\
&  \leq \textrm{const.}\int_0^{\infty}\frac{r^{s}}{x^{1+s}}I_{\lambda_{k}
}(x) \Big(t\frac{x^{2}}{r^{2}}\Big)
^{\alpha}dx\int_0^{\infty
}z^{-1+s-\alpha}e^{-z-\frac{x^{2}}{4z}}dz\\
&  \leq \textrm{const.}t^{\alpha}r^{s-2\alpha}\int_0^{\infty}\frac
{1}{x^{1-\alpha}}I_{\lambda_{k}}(x)K_{-\alpha+s}(x)dx.
\end{align*}
To estimate the inner integral in the next to last inequality, we
used the integral representation of the function $K_{\nu}(y)$
\cite[8.432(6)]{g2}. The convergence of the integral in the right part
as $x\to  0$ is ensured (see \eqref{e3.5},\eqref{e3.6}) if
$-1+2\alpha+\lambda_{k}-s>-1$, i.e. for $s<2\alpha+\lambda_{k}\leq
2\alpha+\pi/\theta$. The convergence of the integral as
$x\to  \infty$ follows from the second expressions in \eqref{e3.5}
and \eqref{e3.6}. That is why
\begin{equation} \label{e3.9}
\lim_{t\to 0} \Delta_{-2,s}=0.
\end{equation}
The function $u(r,\varphi)=\sin (\overline{\lambda_{k}}\varphi)$
where $\overline{\lambda_{k}}$ is some fixed number from the set
$\{\lambda_{k}\}$ is the solution of the problem
\begin{gather*}
\frac{\partial u}{\partial t}-\frac{1}{r}\frac{\partial}{\partial r}
r\frac{\partial u}{\partial r}-\frac{1}{r^{2}}\frac{\partial^{2}u}
{\partial\varphi^{2}}=\frac{\overline{\lambda_{k}}^{2}}{r^{2}}\sin
(\overline{\lambda_{k}}\varphi),
\\
u|_{t=0}=\sin(\overline{\lambda_{k}}\varphi),\quad
u|_{\varphi=0,\theta}=0,
\end{gather*}
and, hence, there is in view of \eqref{e2.6}-\eqref{e2.8}
for the solution
\begin{align*}
&\sin(\overline{\lambda_{k}}\varphi)\\
& =\sin(\overline{\lambda_{k}}
\varphi)\int_0^{t}d\tau\int_0^{\infty}d\rho\frac
{\overline{\lambda_{k}}^{2}}{2\tau\rho}e^{-\frac{\rho^{2}+r^{2}}{4\tau}
}I_{\overline{\lambda_{k}}}(\frac{\rho r}{2\tau})
+\sin(\overline{\lambda_{k}}\varphi)\int_0^{\infty}d\rho\frac{\rho
}{2t}e^{-\frac{\rho^{2}+r^{2}}{4t}}I_{\overline{\lambda_{k}}}(\frac{\rho
r}{2t}).
\end{align*}
After that, \eqref{e3.8} follows from \eqref{e3.9}.
\end{proof}

As an application of Lemma \ref{lem3.3} is the next result.

\begin{lemma} \label{lem3.4}
The equality
\begin{equation} \label{e3.10}
\lim_{t\to 0} R_2(r,\varphi,t)=u_0(r,\varphi)
\end{equation}
is true for the function $ R_2(r,\varphi,t)$ from \eqref{e2.6}.
\end{lemma}

\begin{proof}
Let us denote
\[
L_{k}(\rho,r,t)=\frac{\rho}{2t}e^{-\frac{\rho^{2}+r^{2}}{4t}}I_{
\lambda_{k}}\big(\frac{\rho r}{2t}\big).
\]
To prove the lemma, it suffices to show that the first term in the
right part of the following equality (which follows from Lemma
\ref{lem3.3})
$$
\lim_{t\to 0} \int_0^{\infty}L_{k}(\rho,r,t)u_{0k}
(\rho)d\rho =\lim_{t\to 0}\int_0^{\infty
}L_{k}(\rho,r,t)[u_{0k}(\rho)-u_{0k}(r)]d\rho +u_{0k}(r)=0.
$$
Let
$$
\int_0^{\infty
}L_{k}(\rho,r,t)[u_{0k}(\rho)-u_{0k}(r)]d\rho \equiv d_{k}.
$$
We apply the mean value theorem, Corollary \ref{coro3.1}, and take into
account that $u_0(r,\varphi)\in P^{2+\alpha}_{s+2}(G)$. We have
\[
u_{0k}(\rho)-u_{0k}(r)=(\rho-r)\frac{du_{0k}}{d\rho}(\overline{r}),\quad
\overline{r}\in[r,\rho],
\]
so that
\begin{align*}
|d_{k}|
&\leq \textrm{const.}\overline{r}^{s+1}\int_0^{\infty}L_{k}(\rho,r,t)|\rho
-r|\underset{\overline{r}}{\max}\overline{r}^{-s-1}|
\frac{d u_{0k}(\overline{r})}{d\overline{r}}| d\rho \\
&\leq \textrm{const.}\overline{r}^{s+1}
\max_r r^{-s-1}|\frac{d u_{0k}(r)}
{dr}|\int_0^{\infty}\frac{\rho}{2t}e^{-\frac{\rho^{2}+r^{2}}{4t}+\frac{\rho r}{2t}
}(\frac{2t}{\rho r})^{1/2}|\rho-r|d\rho\\
&\leq \textrm{const.}\frac{\overline{r}^{s+1}}{r^{1/2}}
\max_r r^{-s-1}|\frac{d u_{0k}(r)}{dr}|\int_0^{\infty}
\frac{\rho^{1/2}}{t^{1/2}}e^{-\frac{(\rho-r)^{2}}{4t}}|\rho
-r|d\rho.
\end{align*}
Denote $(\rho-r)/2\sqrt{t}=z$ then
\begin{align*}
|d_{k}|&\leq \textrm{const.}\frac{\overline{r}^{s+1}}{r^{1/2}}
\max_r r^{-s-1}|\frac{d u_{0k}(r)}
{dr}|\int_{-\infty}^{\infty}\frac{|zt^{1/2}
+r|^{1/2}}{t^{1/2}}e^{-z^{2}}zt\,dz\\
&\leq \textrm{const.}t^{1/2}(t^{1/4}+r^{1/2})
\frac{\overline{r}^{s+1}}{r^{1/2}}\max_r r^{-s-1}|\frac{d u_{0k}(r)}
{dr}|.
\end{align*}
Thus,
$\lim_{t\to 0}d_{k}=0$
for every fixed $r$ and all $k$.

Due to $u_0(r,\varphi)\in P^{2+\alpha}_{s+2}(G)$, we have
$r^{-s-2}u_{0k}(r)\sim\frac{1}{k^{2+\alpha}}$ and
$r^{-s-1}\frac{du_{0k}(r)}{dr}\sim\frac{1}{k^{1+\alpha}}$, and
the all written
above gives
\begin{equation} \label{e3.11}
\sum_{k} \lim_{t\to 0}d_{k}=0.
\end{equation}
Let us represent $ R_2(r,\varphi,t)$ as
\begin{align*}
& R_2(r,\varphi,t)\\
&=\sum_{k}
\sin(\lambda_{k}\varphi)\int_0^{\infty
}L_{k}(\rho,r,t)[u_{0k}(\rho)-u_{0k}(r)]d\rho
+\sum_{k} \sin(\lambda_{k}\varphi)u_{0k}(r)D_0(r,t)
\end{align*}
where $D_0(r,t)$ was introduced in Lemma \ref{lem3.2}. After passing on
to the limit in this representation and taking into account \eqref{e3.8}
and \eqref{e3.11}, we obtain
\begin{align*}
\lim_{t\to 0} R_2(r,\varphi,t)
&=\lim_{t\to 0}\sum_{k} \sin(\lambda_{k}\varphi)\int_0^{\infty
}L_{k}(\rho,r,t)[u_{0k}(\rho)-u_{0k}(r)]d\rho\\
&\quad +\lim_{t\to 0}\sum_{k}
\sin(\lambda_{k}\varphi)u_{0k}(r)D_0(r,t)=u_0(r,\varphi).
\end{align*}
\end{proof}

As a some preliminary result we note that Lemma \ref{lem3.1} gives the
order of the decreasing to the coefficients of the trigonometric
series for $r^{-s-2} R_1 (r,\varphi,t)$. If one takes into account
that the Fourier coefficients of functions from H\"older classes
$C^{\alpha}$ have the order $1/k^{\alpha}$, Lemma \ref{lem3.1} will lead
the Fourier coefficients of $r^{-s-2} R_1 (r,\varphi,t)$ have the
order $1/k^{2+\alpha}$. Therefore, the function
$r^{-s-2} R_1 (r,\varphi,t)$ can be differentiated with respect to
$\varphi$ in the case $r$ and $t$ are fixed. We will show that the
function will be differentiated twice with respect to $\varphi$.
If the function $r^{-s-2}u_0(r,\varphi)$ from
$ R_2(r,\varphi,t)$ has the second derivative with respect to
$\varphi$ for the fixed $r$ which belongs to classes $C^{\alpha}$,
Lemma \ref{lem3.2} asserts that the Fourier coefficients of the function
$r^{-s-2} R_2(r,\varphi,t)$ have also the order
$1/k^{2+\alpha}$.

\section{Some facts from the trigonometric series theory}

Let $f(x)$ be a $2\pi$-periodic function with the corresponding
trigonometric series
\[
f(x)\sim\frac{a_0}{2}+
\sum_{n=1}^{\infty} (a_{n}\cos nx+b_{n}\sin nx)=S[f].
\]
 Note that the series $S[f]$ converges to $f(x)$ in the point
$x$ due to Dini's test for $f\in C^{\alpha}$.
 Let $f(x)$ be a continuous function, and $T_{n}(x)$ be any
 trigonometric polynomial of the order not higher then $n$,
\[
\Delta(T_{n}):=\max_{x\in[0,2\pi]} |f(x)-T_{n}(x)|,
\quad
E_{n}(f):=\inf\Delta(T_{n})
\]
where the infimum is considered throughout the set of the
polynomials $T_{n}(x)$. The value of $E_{n}(f)$ is called the best
approximation of the order $n$ to the function $f(x)$
(see \cite[Ch.3, n.13]{z1}).

\begin{theorem}[{Bernstein's Theorem
\cite[Appendix to Ch.4, n.7]{b1}}])  \label{thm4.1}
\begin{equation} \label{e4.1}
E_{n}(f)=O(1/n^{\alpha})
\end{equation}
 if and only if $f(x)\in C^{\alpha}$, $\alpha\in(0,1)$. Moreover, if
 $$
 E_{n}(f)\leq A\frac{1}{n^{\alpha}},
 $$
 then
 $ \langle f \rangle^{(\alpha)}_{x}\leq \textrm{const.} A$.
\end{theorem}

The proof can be found in \cite[Ch.4, n.2]{n1}. The next theorem
contains the method of the building of the approximating
trigonometric polynomial (Jackson's construction
\cite[Ch.4, n.2]{n1}).

\begin{theorem} \label{thm4.2}
Let a $2\pi$-periodic function $f(x)\in C^{\alpha}([0,2\pi])$
 and have the module of continuity
$\omega(\delta)$. Define
\[
u_{n}(x)=c(n) \int_{-\pi}^{\pi} f(l)K(l-x)dl,\quad
c(n)=\frac{3}{2\pi n(2n^{2}+1)},\quad
K(z)=\Big( \frac{\sin(nz/2)}{\sin(z/2)}\Big)^{4}.
\]
Then the following statements hold
\begin{enumerate}
 \item The function $u_{n}(x)$ has the form
 \[
u_{n}(x)=A+
\sum_{k=1}^{2n-2} (a_{k}\cos kx+b_{k}\sin kx);
\]
i.e., $u_{n}(x)$ is a trigonometric polynomial of the $(2n-2)$
order.
\item If ${\int_{-\pi}^{\pi}}f(x)dx=0$, then
    $A=0$.
\item The following estimates holds for all $x$
    \begin{equation} \label{e4.2}
|u_{n}(x)-f(x)|\leq6\omega(1/n).
\end{equation}
\end{enumerate}
\end{theorem}

We apply these theorems in the following case. Let one have the
function $f(x,q)={\sum_{k}} b_{k}(q)\sin kx$
where $q\in \Omega\subset \mathbb{R}^{1}$, and
$b_{k}(q)=\frac{2}{\pi}{\int_{-\pi}^{\pi}}
f(x,q)\sin kx dx$, $f(x,q)$ is continuous with respect to $x$ and
$q$, $f(x,q)\in C^{\alpha}_{x}([0,2\pi])$ with $\alpha \in (0,1)$,
uniformly with respect to $q$,  and $\omega(\delta)$ be the module
of continuity to the function $f(x,q)$ with respect to $x$ which
is uniform with respect to $q$,
\[
{\sum_{k}} \max_q |b_{k}(q)|<\infty.
\]
This inequality implies
\begin{equation} \label{e4.3}
\max_{x,q} |f|\geq \textrm{const.}{\sum_{k}}|b_{k}(q)|.
\end{equation}
Indeed, because the series ${\sum_{k}}\max_q  |b_{k}(q)|$
converges it is possible to choose $N$ so as
\[
{\sum_{k=N+1}^{\infty}} \max_q
|b_{k}(q)|\leq\frac{1}{2}\max_{x,q} |f|.
\]
On the other hand the suitable constant can be searched such that
\[
\frac{1}{2}\max_{x,q} |f|\geq \textrm{const.}
{\sum_{k=1}^{N}}
\max_q |b_{k}(q)|,
\]
that completes the proof of \eqref{e4.3}.

 Let us introduce the linear operator
$A:C\to  C$ which acts by the following rule
\begin{equation} \label{e4.4}
v(x,q)=Af(x,q)={\sum_{k}}\mu_{k}(b_{k}(q))\sin kx,\quad
|\mu_{k}(b_{k}(q))|\leq
M\max_q |b_{k}(q)|,
\end{equation}
where $M$ is an independent constant of $k$ and $q$. The question
arises, does $v(x,q)$ have the same module of continuity as
$f(x,q)$. In accordance with Bernstein's theorem this question can
be reformulated as is it possible to construct the approximating
polynomial to $v(x,q)$ with the same approximation like \eqref{e4.1}.
Denote $T_{n}(x,q)=Au_{n}(x,q)$ where $u_{n}(x,q)$ is the
approximating trigonometric polynomial to the function $f(x,q)$,
then
\[
v(x,q)-T_{n}(x,q)=Af(x,q)-Au_{n}(x,q).
\]
Following the proof of Theorem \ref{thm4.2}, we can write
\begin{align*}
Au_{n}(x,q)
& =A\big\{  c(n) {\int_0^{\pi/2}}
[f(x+2z,q)+f(x-2z,q)]K(2z)dz\big\}  \\
& =c(n){\int_0^{\pi/2}}
A\big\{f(x+2z,q)+f(x-2z,q)\big\}  K(2z)dz.
\end{align*}
After that we use the equality
\[
2c(n){\int_0^{\pi/2}}
\Big(  \frac{\sin(nz)}{\sin(z)}\Big)  ^{4}dz=2c(n)
{\int_0^{\pi/2}}K(2z)dz=1,
\]
and obtain
\begin{equation} \label{e4.5}
T_{n}(x,q)-v(x,q)=c(n){\int_0^{\pi/2}}
A\big\{f(x+2z,q)+f(x-2z,q)-2f(x,q)\big\}  K(2z)dz.
\end{equation}
The definition of the operator $A$ together with the properties of
the function $f(x,q)$ lead to the estimates:
\[
\max_{x,q} |A(f(x,q))|\leq M {\sum_{k}}
\max_q |b_{k}(q)|\leq
\textrm{const.}\max_{x,q} |f(x,q)|,
\]
and that is why
\[
\max_{x,q} |A\{ f(x+2z,q)+f(x-2z,q)-2f(x,q)\}|
\leq \textrm{const.}\omega(2z).
\]
After that the estimate of the right part can be finished like the
proof of \cite[Theorem 4.2]{n1}. We have
\[
|T_{n}(x,q)-v(x,q)|\leq \textrm{const.}\omega(1/n)
\leq \textrm{const.}\frac{1}{n^{\alpha}}.
\]
Bernstein's theorem leads to $Af(x,q)\in C^{\alpha}([0,2\pi])$.
Moreover, the estimate
\begin{equation} \label{e4.6}
\langle Af(x,q)\rangle _{x}^{(\alpha)}\leq
\textrm{const.}\langle f(x,q)\rangle _{x}^{(\alpha)}
\end{equation}
follows from the proof of Bernstein's theorem. Thus we obtained
the following fact.

\begin{lemma} \label{lem4.1}
Let $f(x,q)$ be continuous with respect to $x,q$ and
$f(x,q)\in C_{x}^{\alpha}$ uniformly with respect to $q$
and $\alpha \in (0,1)$,
\[
{\sum_{k}}\max_q  |b_{k}(q)|<
\infty,
\]
then $Af\in C_{x}^{\alpha}$ and estimate \eqref{e4.6} holds.
\end{lemma}

Assume that $f(x,t)$ is a $2\pi$-periodical function with respect
to $x$ and $f(x,t)\in C_{x,t}^{\alpha,\beta}$, $\alpha,\beta \in
(0,1)$, and
\begin{equation} \label{e4.7}
{\sum_{k}}(\underset{t}{\max} |b_{k}(t)|+\langle
b_{k} \rangle ^{\beta}_{t})< \infty.
\end{equation}
Let
\[
u_{n}(x,t)=c(n){\int_{-\pi}^{\pi}} f(s,t)K(x-s)ds
\]
be the trigonometric polynomial from Theorem \ref{thm4.2} which
approximates the function $f(x,t)$. We have
\[
u_{n}(x,t_{1})-u_{n}(x,t_{2})=c(n){\int_{-\pi}^{\pi}}
[f(s,t_{1})-f(s,t_{2})]K(x-s)ds.
\]
The properties of the kernel $K(x-s)$ ensure the inequality
\[
\max_{x,t_{1},t_{2}} \frac{|u_{n}(x,t_{1})-u_{n}(x,t_{2})|}
{|t_{1}-t_{2}|^{\beta}}\leq \textrm{const.}\langle f(x,t)\rangle
_{t}^{(\beta)};
\]
i.e., the trigonometric polynomial approximating $f(x,t)$ has a
uniformly bounded H\"older constant with respect to $t$. Let, as
before, $T_{n}(x,t)=Au_{n}(x,t)$. Then
\begin{align*}
|T_{n}(x,t_{1})-T_{n}(x,t_{2})|
& =|c(n) {\int_{-\pi}^{\pi}} A\{f(s,t_{1})-f(s,t_{2})\}K(x-s)ds|  \\
& \leq \textrm{const.}\max_{x,t_{1},t_{2}} |f(x,t_{1})-f(x,t_{2})|.
\end{align*}
It leads to
\begin{equation} \label{e4.8}
\max_{x,t_{1},t_{2}} \frac{|T_{n}(x,t_{1})-T_{n}(x,t_{2})|}
{|t_{1}-t_{2}|^{\beta}}\leq \textrm{const.}\max_{x,t_{1},t_{2}}
\frac{|f(x,t_{1})-f(x,t_{2})|}{|t_{1}-t_{2}|^{\beta}}.
\end{equation}
If one passes to a limit in \eqref{e4.8} as $n\to  \infty$ (here
we keep in mind that $T_{n}(x,t_{k})\to  Af(x,t_{k}),\quad
k=1,2$) then
\[
\max_{x,t_{1},t_{2}} \frac{|Af(x,t_{1})-Af(x,t_{2})|}{|t_{1}
-t_{2}|^{\beta}}\leq \textrm{const.}\langle f(x,t)\rangle
_{t}^{(\beta)}.
\]

\begin{lemma} \label{lem4.2}
Let the function $f(x,t)$ be a $2\pi$-periodical function with
respect to $x$, and $f(x,t)\in C_{x,t}^{\alpha,\beta}$,
$\alpha,\beta \in (0,1)$ and \eqref{e4.7} holds. Then $Af(x,t)\in
C_{x,t}^{\alpha,\beta}$ and
\begin{equation} \label{e4.9}
\langle Af\rangle _{x}^{(\alpha)}\leq
\textrm{const.}\langle f(x,t)\rangle
_{x}^{(\alpha)},\quad\langle Af\rangle
_{t}^{(\beta)}\leq \textrm{const.}\langle f(x,t)\rangle
_{t}^{(\beta )}.
\end{equation}
\end{lemma}

\begin{remark} \label{rmk4.1} \rm
Lemmas \ref{lem4.1} and \ref{lem4.2} will hold if we change the functions $f(x,q)\in
C^{\alpha}_{x}$ and $f(x,t)\in C^{\alpha,\beta}_{x,t}$ onto
$f(x,q_{1},\dots q_{n})\in C^{\alpha}_{x}$ uniformly with respect to
$q_{1}\dots q_{n}$ in Lemma \ref{lem4.1}, and $f(x,t_{1},\dots t_{n})\in
C^{\alpha,\beta_{1},\dots\beta_{n}}_{x,t_{1},\dots t_{n}}$ with
$0<\beta_{i}< 1$, $i=\overline{1,n}$  in Lemma \ref{lem4.2},
correspondingly, and the inequality like \eqref{e4.7} holds.
\end{remark}

\section{Estimates of the higher seminorms of the solution}

\subsection{Estimate for
$\frac{\partial^{2} R_1 }{\partial\varphi^{2}}$}

\begin{lemma} \label{lem5.1}
The function $\frac{\partial^{2} R_1 }{\partial\varphi^{2}}$
meets the H\"older condition with respect to $\varphi$ and
\begin{equation} \label{e5.1}
\langle
\frac{\partial^{2} R_1 }{\partial\varphi^{2}}\rangle
_{\varphi;s+2,G_{T}}^{(\alpha)}
+\sum_{k}\lambda_{k}^{2}\max_{\bar{\mathbb{R}}_{T}} r^{-s-2}|R_{1,k}(r,t)|\leq
\textrm{const.}\langle f\rangle _{\varphi;s,G_{T}}^{(\alpha)}.
\end{equation}
\end{lemma}

\begin{proof}
After a formal differentiation with respect to $\varphi$ one can
obtain
\begin{equation} \label{e5.2}
\frac{\partial^{2} R_1 }{\partial\varphi^{2}}=-
{\sum_{k}}
\lambda_{k}^{2}\sin(\lambda_{k}\varphi)\int_0^{t}d\tau\int
_0^{\infty}L_{k}(\rho,r,t-\tau)b_{k}(\rho,\tau)d\rho
\end{equation}
where $b_{k}(r,t)$ are the Fourier coefficients of the function
$f(r,\varphi,t)$. The function $f(r,\varphi,t)$ is continued odd
onto the interval $(-\theta,0)$, and $f(r,\varphi,t)=0$ if
$\varphi=0,\theta$ or $t<0$. In the case of a $2\theta-$
periodical function, the change of variables allows keeping the
mentioned above argumentations regarding to use of the
approximating trigonometric polynomial to a $2\pi-$ periodical
function. Let us denote by
\[
B_{k}=-\lambda_{k}^{2}\int_0^{t}d\tau
\int_0^{\infty}L_{k}(\rho,r,t-\tau)b_{k}(\rho,\tau)d\rho,
\]
in view of Lemma \ref{lem3.1},
\begin{equation} \label{e5.3}
|B_{k}|\leq \textrm{const.}r^{2+s}\underset{r,t}{\max}r^{-s}|b_{k}|
\end{equation}
with the constant is independent of $k$. After that,  we put in
\eqref{e4.4}: $x=\varphi$,  $f(x,q):= f(r,\varphi,t)$,
$b_{k}(q):=r^{-s}b_{k}(r,t)$, $\mu_{k}(b_{k}(q)):=r^{-2-s} B_{k}$,
$Af(x,q):=r^{-2-s}\frac{\partial^{2} R_1 }{\partial\varphi^{2}}$.
Then Lemma \ref{lem4.1} together with the properties of the function
$f(r,\varphi,t)$ (namely, $f\in
\widehat{P}_{s}^{\alpha,\alpha/2}(\overline{G}_{T})$, i.e.
$r^{-s}f\in C^{\alpha}_{\varphi}([0,\theta])$ uniformly with
respect to $t$ and $r$, inequality like \eqref{e4.3} holds) lead to
estimate \eqref{e5.1}.
\end{proof}

\begin{lemma} \label{lem5.2}
The function
$\frac{\partial^{2} R_1 }{\partial\varphi^{2}}(r,\varphi,t)$
satisfies the H\"older conditions with respect to $t$ and $r$.
Moreover,
\begin{gather} \label{e5.4}
\sum_{k}\lambda_{k}^{2}\langle  R_{1,k}\rangle
_{t;s+2-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}\leq \textrm{const.}\langle
f\rangle _{t;s-\alpha,G_{T} }^{(\alpha/2)}, \\
 \label{e5.5}
\sum_{k}\lambda_{k}^{2}\langle  R_{1,k}\rangle
_{r;s+2-\alpha,\mathbb{R}_{T}}^{(\alpha)}\leq \textrm{const.}\langle
f\rangle _{r;s-\alpha,G_{T} }^{(\alpha)}, \\
 \label{e5.6} \begin{aligned}
&[\frac{\partial^{2} R_1 }{\partial\varphi^{2}}
]_{\varphi,t;s+2-\alpha,G_{T}}^{(\alpha,\alpha/2)}
+\sum_{k}\lambda_{k}^{2}[
 R_{1,k}]_{r,t;s+2-2\alpha,\mathbb{R}_{T}}^{(\alpha,\alpha/2)}\\
& \leq
\textrm{const.}([f]_{\varphi,t;s-\alpha,G_{T}}^{(\alpha,\alpha/2)}
+[f]_{r,t;s-2\alpha,G_{T}}^{(\alpha,\alpha/2)}).
\end{aligned}
\end{gather}
\end{lemma}

\begin{proof}
The proof of estimates \eqref{e5.4} and \eqref{e5.5} follows from
the properties of the function $f(r,\varphi,t)$,
Lemma \ref{lem3.1} and Lemma \ref{lem4.2}.
Regarding inequality \eqref{e5.6}, it is obtained if one applies
Lemmas \ref{lem4.2} and \ref{lem5.1} to the function
$[\frac{\partial^{2} R_1 }{\partial\varphi^{2}}(r,\varphi,
t_{2})-\frac{\partial^{2} R_1 }{\partial\varphi^{2}}(r,\varphi,
t_{1})]$.
\end{proof}

\subsection{Estimates of the derivative of the function
$ R_1 (r,\varphi,t)$ with respect to $t$}

First we obtain the representation of
$\partial  R_1 /\partial t$.
Let
\begin{equation} \label{e5.7}
v_{k}(r,t)=\frac{\partial}{\partial t}\int_0^{t}d\tau
\int_0^{\infty}d\rho
L_{k}(\rho,r,\tau)\int_0^{\theta}\frac
{2}{\theta}f(\rho,\psi,t-\tau)\sin(\lambda_{k}\psi)d\psi.
\end{equation}
Assume from the beginning that $f(r,\varphi,t)$ is differentiated
with respect to $t$. Then differentiation under the integral sign
acts on $f(r,\varphi,t)$, and integrating by parts gives
\begin{equation} \label{e5.8}
v_{k}(r,t)=\int_0^{t}d\tau\int_0^{\infty}d\rho\frac{\partial
L_{k}}{\partial\tau}(\rho,r,\tau)b_{k}(\rho,t-\tau)+\underset{\varepsilon\to
0}{\lim}\int_0^{\infty}d\rho
L_{k}(\rho,r,\varepsilon)b_{k}(\rho,t).
\end{equation}
 Note that the derivative of the function $f(r,\varphi,t)$ is
not required in \eqref{e5.8}. Using the relation above, we obtain the
following representation
\begin{align*}
\frac{\partial  R_1 }{\partial t}(r,\varphi,t)
&  ={\sum_{k}}\sin(\lambda_{k}\varphi)\int_0^{t}d\tau\int_0^{\infty}
d\rho\frac{\partial
L_{k}}{\partial\tau}(\rho,r,\tau)\\
&\quad\times \int_0^{\theta}\frac{2}{\theta}[f(\rho,\psi,t-\tau)
 -f(\rho,\psi,t)]\sin(\lambda_{k}\psi)d\psi\\
&\quad +
{\sum_{k}} \sin(\lambda_{k}\varphi)\int_0^{\infty}d\rho
L_{k}(\rho,r,t)
\int_0^{\theta}\frac{2}{\theta}f(\rho,\psi,t)\sin(\lambda
_{k}\psi)d\psi\\
&\equiv A_{1}+A_{2}.
\end{align*}
Straight away, we obtain another useful representation of
$\frac{\partial  R_1 }{\partial t}(r,\varphi,t)$. Let
\begin{gather*}
v_{1k}(\rho,t)=\int_{-\infty}^{t}d\tau\int_0^{\infty}d\rho
L_{k}(\rho,r,t-\tau)b_{k}(\rho,\tau),\\
v_{1k}^{h}(\rho,t)=\int_{-\infty}^{t-h}d\tau\int_0^{\infty
}d\rho L_{k}(\rho,r,t-\tau)b_{k}(\rho,\tau).
\end{gather*}
The derivative of $\partial  v_{1k}/\partial t$ is
$\lim_{h\to 0} \frac{\partial v_{1k}^{h}}{\partial
t}$.   Non-complicated calculations and Corollary \ref{coro3.1} give
\[
\lim_{t\to  +\infty}{\lim}L_{k}(\rho,r,t)=0,
\]
and then
\begin{gather} \label{e5.10}
\frac{\partial v_{1k}}{\partial t}=\int_{-\infty}^{t}d\tau
\int_0^{\infty}\frac{\partial L_{k}}{\partial
t}(\rho,r,t-\tau )[b_{k}(\rho,\tau)-b_{k}(\rho,t)]d\rho,
\\
 \begin{aligned}
\frac{\partial R_1}{\partial t}(r,\varphi,t)
&  = {\sum_{k}}
\sin(\lambda_{k}\varphi)\int_{-\infty}^{t}d\tau\int_0^{\infty}
d\rho\frac{\partial L_{k}}{\partial t}(\rho,r,t-\tau) \\
&\quad \times\int_0^{\theta}\frac{2}{\theta}[f(\rho,\psi,\tau
)-f(\rho,\psi,t)]\sin(\lambda_{k}\psi)d\psi.
\end{aligned}\label{e5.11}
\end{gather}
We will use the next representation
\begin{equation} \label{e5.12}
\begin{aligned}
\frac{\partial L_{k}}{\partial t}(\rho,r,t)
&  =-\frac{\rho}{2t^{2}}
e^{-\frac{\rho^{2}+r^{2}}{4t}}I_{\lambda_{k}}(\frac{\rho
r}{2t})+\frac
{\rho}{2t}\frac{\partial}{\partial t}\{e^{-\frac{\rho^{2}+r^{2}}{4t}
}I_{\lambda_{k}}\big(\frac{\rho r}{2t}\big)\} \\
&  =i_{1k}(\rho,r,t)+i_{2k}(\rho,r,t).
\end{aligned}
\end{equation}

\begin{lemma} \label{lem5.3}
The following estimate holds
\begin{equation} \label{e5.13}
\sum_{k}\max_{\overline{\mathbb{R}}_{T}} r^{-s}|\frac{\partial
 R_{1,k}}{\partial t}|\leq \textrm{const.}\|f\|
_{P_{s}^{\alpha,\alpha/2}(G_{T})}\,.
\end{equation}
\end{lemma}

\begin{proof}
First  we justify the estimate
\begin{equation} \label{e5.14}
|\frac{\partial R_{1,k}}{\partial t}|
=\big|\int
^{t}_{-\infty}d\tau \int^{\infty}_0d\rho \frac{\partial
L_{k}}{\partial t}(\rho,\varphi,t-\tau)[b_{k}(\rho,\tau)-b_{k}(\rho,t)]\big|
\leq
\textrm{const.}\langle b_{k}\rangle^{(\alpha/2)}_{t,s-\alpha}r^{s}
\end{equation}
where
\begin{align*}
\frac{\partial}{\partial t}L_{k}(\rho,r,t)
&=-\frac{1}{t}L_{k}(\rho,r,t)+\frac{\rho(\rho^{2}+r^{2})}{8t^{3}}I_{\lambda_{k}}(\frac{r\rho}{2t})
e^{-\frac{\rho^{2}+r^{2}}{4t}}
\\
&\quad -\frac{r\rho^{2}}{4t^{3}}e^{-\frac{\rho^{2}+r^{2}}{4t}}\frac{d}{d
x}I_{\lambda_{k}}(x),\quad x=\frac{r\rho}{2t}.
\end{align*}
 From the representation of the function $I_{\lambda_{k}}(x)$ (see
\cite[8.431(1)]{g2})
\[
I_{\lambda_{k}}(x)=
\frac{(x/2)^{\lambda_{k}}}{\Gamma(\lambda_{k}+1/2)
\Gamma(1/2)}\int^{1}_{-1} e^{xy}(1-y^{2})^{\lambda_{k}-1/2}dy,
\]
it follows
\begin{align*}
\frac{d I_{\lambda_{k}}(x)}{d x}
&=\frac{\lambda_{k} }{x}I_{\lambda_{k}}(x)
 +\frac{(x/2)^{\lambda_{k}}}{\Gamma(\lambda_{k}+1/2)\Gamma(1/2)}\int^{1}_{-1}
ye^{xy}(1-y^{2})^{\lambda_{k}-1/2}dy \\
&\equiv \frac{\lambda_{k}}{x}I_{\lambda_{k}}(x)+Q_{\lambda_{k}}(x).
\end{align*}
On the other hand,(see \cite[8.486(4)]{g2})
\[
x\frac{d I_{\lambda_{k}}(x)}{d
x}=\lambda_{k}I_{\lambda_{k}}(x)+xI_{\lambda_{k}+1}(x),
\]
and, hence,
$xQ_{\lambda_{k}}(x)=xI_{\lambda_{k}+1}(x)$.
 From this equation and the definition of $Q_{\lambda_{k}}(x)$, we
obtain
\[
xQ_{\lambda_{k}}(x)\leq \textrm{const.}
\begin{cases}
 \frac{x^{\lambda_{k}+2}}{\Gamma(\lambda_{k}+1)},
&\text{for } x\leq 1,\\
xI_{\lambda_{k}}(x), &\text{for } x> 1.
\end{cases}
\]
Returning to $\frac{\partial L_{k}}{\partial t}(\rho,r,t)$, we
have
\begin{align*}
\frac{\partial L_{k}}{\partial t}(\rho,r,t)
&=-\frac{1}{t}L_{k}+\frac{1}{t}L_{k}\frac{\rho^{2}+r^{2}}{4t}
-\frac{\lambda_{k}}{t}L_{k}
-\frac{\rho}{2t^{2}}xQ_{\lambda_{k}}(x)e^{-\frac{\rho^{2}+r^{2}}{4t}}
\\
&\equiv -m_{1}(\rho,r,t)+m_{2}(\rho,r,t)-m_{3}
 (\rho,r,t)-m_{4}(\rho,r,t),\quad x=\frac{r\rho}{2t}.
\end{align*}
Let
\[
M(r,t)=\int  ^{t}_{-\infty}d\tau\int
_0^{\infty}d\rho \rho^{s-\alpha}(t-\tau)^{\alpha/2}L_{k}(\rho,r,t-\tau).
\]
Since
\[
M(r,t)=\int  ^{\infty}_0dz\int  _0^{\infty}d\rho
\rho^{s-\alpha}z^{\alpha/2}L_{k}(\rho,r,z),
\]
then
$\frac{\partial M}{\partial t}(r,t)=0$.
Due to Lemma \ref{lem3.2},
\[
\lim_{t\to 0} t^{\alpha/2} \int
_0^{\infty}d\rho \rho^{s-\alpha}L_{k}(\rho,r,t)=0,
\]
and therefore
\begin{equation} \label{e5.15}
\begin{aligned}
\frac{\partial M}{\partial t}(r,t)
&=\int ^{t}_{-\infty}d\tau\int  _0^{\infty}d\rho
[\frac{\partial}{\partial t}(t-\tau)^{\alpha/2}]
\rho^{s-\alpha}L_{k}(\rho,r,t-\tau) \\
&\quad +\int  ^{t}_{-\infty}d\tau\int  _0^{\infty}d\rho
(t-\tau)^{\alpha/2} \rho^{s-\alpha}\frac{\partial}{\partial t}
L_{k}(\rho,r,t-\tau).
\end{aligned}
\end{equation}
Let us consider the integral
\[
M_{1}=(1+\lambda_{k})\int  ^{\infty}_0dt\int
_0^{\infty}d\rho \rho^{s-\alpha}t^{-1+\alpha/2}L_{k}(\rho,r,t)
\]
corresponding to  $(m_{1}(\rho,r,t)+m_{3}(\rho,r,t))$ in
the representation of $\frac{\partial L_{k}}{\partial
t}(\rho,r,t)$. Then the following estimate holds
\begin{equation} \label{e5.16}
|M_{1}|\leq r^{s}\frac{1}{\lambda_{k}^{\alpha/2-\mu}},\quad
\mu<\alpha/2.
\end{equation}
We represent its proof in the Appendix (see Subsection 7.3).
Now, using the integral representations of $I_{\lambda_{k}}(x)$
and $Q_{\lambda_{k}}(x)$, we have
\begin{align*}
& m_{2}(\rho,r,t)-m_{4}(\rho,r,t) \\
&=\frac{\rho}{8t^{3}}e^{-\frac{\rho^{2}+r^{2}}{4t}}\{(\rho^{2}+
r^{2})I_{\lambda_{k}}(r\rho/2t)-2r\rho
Q_{\lambda_{k}}(r\rho/2t)\}\\
&=\frac{\rho}{8t^{3}}e^{-\frac{\rho^{2}+r^{2}}{4t}}\{(\rho^{2}-2r\rho
+r^{2})I_{\lambda_{k}}(r\rho/2t)\\
&\quad +2r\rho\frac{(r\rho/4t)^{\lambda_{k}}}{\Gamma(\lambda_{k}+1/2)\Gamma(1/2)}\int^{1}_{-1}
(1-y)e^{xy}(1-y^{2})^{\lambda_{k}-1/2}dy\}\geq 0.
\end{align*}
Estimate \eqref{e5.16} implies
\begin{equation} \label{e5.17}
\int  ^{\infty}_0d\tau \int  _0^{\infty}d\rho
\rho^{s-\alpha}\tau^{\alpha/2}(m_{1}(\rho,r,\tau)+m_{3}(\rho,r,\tau))\leq
\textrm{const.} r^{s}\frac{1}{\lambda_{k}^{\alpha/2-\mu}}.
\end{equation}
Note that the estimate of the first term at the right part of
\eqref{e5.15} is contained in \eqref{e5.16}, thus, by the equation
$\frac{\partial M}{\partial t}(r,t)=0$, we have
\begin{equation} \label{e5.18}
\int  ^{\infty}_0d\tau \int  _0^{\infty}d\rho
\rho^{s-\alpha}\tau^{\alpha/2}(m_{2}(\rho,r,\tau)-m_{4}(\rho,r,\tau))
\leq \textrm{const.} r^{s}\frac{1}{\lambda_{k}^{\alpha/2-\mu}}.
\end{equation}
At last, we are ready  with \eqref{e5.17} and \eqref{e5.18} to prove inequality
\eqref{e5.14}:
\begin{align*}
|\frac{\partial R_{1k}}{\partial t}|
&\leq \langle b_{k}\rangle^{(\alpha/2)}_{t,
 s-\alpha,\mathbb{R}_{T}}r^{s}\int^{\infty}_0
d\tau\int^{\infty}_0d\rho|\frac{\partial
L_{k}}{\partial
t}(\rho,\varphi,\tau)|\rho^{s-\alpha}\tau^{\frac{\alpha}{2}}\\
&=\langle b_{k}\rangle^{(\alpha/2)}_{t,s-\alpha,\mathbb{R}_{T}}r^{s}
\int^{t}_{-\infty}\int^{\infty}_0\rho^{s-\alpha}\tau^{\frac{\alpha}{2}}
(m_{1}(\rho,r,\tau)+m_{3}(\rho,r,\tau)\\
&\quad +m_{2}(\rho,r,\tau)-m_{4}(\rho,r,\tau))\\
&\leq \textrm{const.}\langle
b_{k}\rangle^{(\alpha/2)}_{t,s-\alpha,\mathbb{R}_{T}}\frac
{r^{s}}{\lambda_{k}^{\alpha/2-\mu}}.
\end{align*}
As the series $\sum_{k}\langle
b_{k}\rangle^{(\alpha/2)}_{t,s-\alpha,\mathbb{R}_{T}}$ converges,
we arrive at \eqref{e5.13} which completes the proof.
\end{proof}

\subsection{H\"older constant for  $\frac{\partial
 R_1}{\partial t}(r,\varphi,t)$ with respect to $\varphi$}
We apply Theorem \ref{thm4.2} to estimate the H\"older constant of the
function $\frac{\partial  R_1}{\partial t}(r,\varphi,t)$
with respect to $\varphi$.
Let us consider the  function
\[
F(x,t,\tau)=g(x,t-\tau)-g(x,t),\quad g(x,t)\in
C_{x,t}^{\alpha,\alpha/2}(\Omega_{T})
\]
where
$\Omega_{T}:=[0,2\pi]\times [0,T]$, $\alpha\in(0,1)$,
and the function $g(x,t)$ satisfies \eqref{e4.7}.
The approximating polynomial to $F(x,t,\tau)$ is
\[
u_{n}(x,t,\tau)=c(n){\int_0^{\pi/2}}
[F(x+2l,t,\tau)+F(x-2l,t,\tau)]K(2l)dl.
\]
Let
\[
T_{n}(x,t,\tau)=\overline{A}u_{n}(x,t,\tau)
\]
where the operator $\overline{A}$, on the one hand, is the
operator like $A$ from \eqref{e4.4} with $f(x,q):=F(x,t,\tau)$,
and, on the other hand, models the operator from the right
hand side in \eqref{e5.11}. In the same way as above,
\begin{equation} \label{e5.19}
\begin{aligned}
& T_{n}(x,t,\tau)-\overline{A}F(x,t,\tau)\\
&=c(n) {\int_0^{\pi/2}} \overline{A}\{F(x+2l,t,\tau)+F(x-2l,t,\tau)
-2F(x,t,\tau)\}K(2l)dl.
\end{aligned}
\end{equation}
After applying the operator $\overline{A}$ and following the
proof of Lemma \ref{lem5.3}, we have
\[
\max_{x,t,\tau} |\overline{A}F(x,t,\tau)|\leq
\textrm{const.}\max_{x,t,\tau} \{ \frac{|F(x,t,\tau
)|}{\tau^{\alpha/2}}\}.
\]
We apply this estimate to the integrand in \eqref{e5.19} and obtain
\begin{align*}
&|T_{n}(x,t,\tau)-\overline{A}F(x,t,\tau)|\\
&\leq \textrm{const.}c(n){\int_0^{\pi/2}} K(2l)
\max_{x,t,\tau} \big\{
\frac{|F(x+2l,t,\tau)+F(x-2l,t,\tau)-2F(x,t,\tau)|}{\tau
^{\alpha/2}} \big\}dl.
\end{align*}
It is obvious that
\[
\max_{x,t,\tau} \frac{|F(x+2l,t,\tau)+F(x-2l,t,\tau)
-2F(x,t,\tau)|}{\tau^{\alpha/2}}\leq
\textrm{const.}[g]_{x,t;\Omega_{T}}^{(\alpha,\alpha/2)}l^{\alpha}.
\]
That is why following  the proof of Theorem \ref{thm4.2}, we obtain that
the studied function $\overline{A}F(x,t,\tau)$ belongs to
$C^{\alpha}_{x}[0,2\pi])$.

Thus, similar considerations as in the case of the function
$\frac{\partial  R_1}{\partial t}(r,\varphi,t)$ lead to
\begin{equation} \label{e5.20}
r^{-s}\frac{\partial  R_1}{\partial t}(r,\varphi,t)\in
C_{\varphi}^{\alpha },\quad\langle \frac{\partial
 R_1}{\partial t}\rangle
_{\varphi;s,G_{T}}^{(\alpha)}\leq
\textrm{const.}[f]_{\varphi,t;s-\alpha,G_{T}} ^{(\alpha,\alpha/2)}.
\end{equation}
This is the place where the additional smoothness of the function
$f(r,\varphi,t)$; i.e., the boundedness of the seminorm
$[f]^{(\alpha,\alpha/2)}_{\phi,t;s-\alpha,G_{T}}$, is used. That,
of course, is stipulated by the approach to the investigation of
the problem.

\subsection{H\"older constant for $\frac{\partial
 R_1 }{\partial t}(r,\varphi,t)$ with respect to $t$}

In this section we make use representation \eqref{e5.11} of the function
$\frac{\partial  R_1}{\partial t}(r,\varphi,t)$. Let
$t_{2}>t_{1}$, and $\bigtriangleup t=t_{2}-t_{1}$. We have
\begin{equation} \label{e5.21}
\begin{aligned}
&\frac{\partial  R_1}{\partial t}(r,\varphi,t_{2})
-\frac{\partial  R_1}{\partial t}(r,\varphi,t_{1})\\
&= {\sum_{k}} \sin(\lambda_{k}\varphi)
\int_{2t_{1}-t_{2}}^{t_{2}}d\tau\int _0^{\infty}d\rho\frac{\partial
L_{k}}{\partial\tau}(\rho,r,t_{2}
-\tau)[b_{k}(\rho,\tau)-b_{k}(\rho,t_{2})]\\
&\quad - {\sum_{k}}
\sin(\lambda_{k}\varphi)\int_{2t_{1}-t_{2}}^{t_{1}}d\tau
\int _0^{\infty}d\rho\frac{\partial
L_{k}}{\partial\tau}(\rho,r,t_{1}
-\tau)[b_{k}(\rho,\tau)-b_{k}(\rho,t_{1})]\\
&\quad + {\sum_{k}}
\sin(\lambda_{k}\varphi)\int_{-\infty}^{2t_{1}-t_{2}}d\tau
\int_0^{\infty}d\rho[ b_{k}(\rho,\tau)-b_{k}(\rho
,t_{1})]\\
&\quad\times [  \frac{\partial
L_{k}}{\partial\tau}(\rho,r,t_{2}-\tau )-\frac{\partial
L_{k}}{\partial\tau}(\rho,r,t_{1}-\tau)] \\
&\quad + {\sum_{k}}
\sin(\lambda_{k}\varphi)\int_{-\infty}^{2t_{1}-t_{2}}d\tau\int_0^{\infty
}d\rho[ b_{k}(\rho,t_{1})-b_{k}(\rho,t_{2})] \frac{\partial
L_{k}} {\partial\tau}(\rho,r,t_{2}-\tau) \\
&= {\sum_{i=1}^{4}}{\sum_{k}}
\sin(\lambda_{k}\varphi){\sum_{j=1}^{2}}A_{j,k}^{(i)},
\end{aligned}
\end{equation}
where $A_{1,k}^{(i)}$, $i=\overline{1,4}$, correspond to $i_{1k}$
in representation \eqref{e5.12} for the function
 $\partial L_{k}/\partial t$ and $A_{2,k}^{(i)}$,
$i=\overline{1,4}$, do to $i_{2k}$. By the
definition
\[
A_{1,k}^{(1)}=-\int_{2t_{1}-t_{2}}^{t_{2}}d\tau\int_0^{\infty
}d\rho\frac{\rho}{2(t_{2}-\tau)^{2}}e^{-\frac{\rho^{2}+r^{2}}{4(t_{2}-\tau
)}}I_{\lambda_{k}}(\frac{\rho r}{2(t_{2}-\tau)})[b_{k}(\rho,\tau)-b_{k}
(\rho,t_{2})],
\]
so that the inequality
\[
|A_{1,k}^{(1)}|\leq \textrm{const.}\int_{2t_{1}-t_{2}}^{t_{2}}d\tau
\int_0^{\infty}d\rho\frac{\rho^{s+1-\alpha}}{(t_{2}-\tau
)^{2-\alpha/2}}e^{-\frac{\rho^{2}+r^{2}}{4(t_{2}-\tau)}}I_{\lambda_{k}
}(\frac{\rho r}{2(t_{2}-\tau)})\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}
\]
is valid. After applying Lemma \ref{lem3.2}, we obtain
\begin{equation} \label{e5.22}
\begin{aligned}
|A_{1,k}^{(1)}|
&\leq \textrm{const.}r^{s-\alpha}\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}
\int_{2t_{1}-t_{2}}^{t_{2}}\frac{d\tau}
{(t_{2}-\tau)^{1-\alpha/2}}\\
&\leq \textrm{const.}r^{s-\alpha}(\Delta t)^{\alpha /2}\langle
b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}.
\end{aligned}
\end{equation}
The estimate of $A_{1,k}^{(2)}$ has been done  the same way.
To estimate
\[
A_{1,k}^{(3)}=\int_{-\infty}^{2t_{1}-t_{2}}d\tau\int_0
^{\infty}d\rho[
b_{k}(\rho,\tau)-b_{k}(\rho,t_{1})][i_{1k}(\rho
,r,t_{2}-\tau)-i_{1k}(\rho,r,t_{1}-\tau)],
\]
we apply the mean value theorem. To this end we calculate
\begin{align*}
\frac{\partial}{\partial t}\big\{
\frac{\rho}{2t^{2}}e^{-\frac{\rho
^{2}+r^{2}}{4t}}I_{\lambda_{k}}\big(\frac{\rho r}{2t}\big)\big\}
&=-\frac{\partial i_{1k}}{\partial t}\\
&=-\frac{\rho}{t^{3}}e^{-\frac{\rho
^{2}+r^{2}}{4t}}I_{\lambda_{k}}\big(\frac{\rho r}{2t}\big)
+\frac{\rho}{2t^{2}}
\frac{\partial}{\partial t}\big\{  e^{-\frac{\rho^{2}+r^{2}}{4t}}
I_{\lambda_{k}}\big(\frac{\rho r}{2t}\big)\big\} \\
&=J_{1k}+J_{2k}.
\end{align*}
Let $\overline{t}\in(t_{1},t_{2})$ and
\[
A_{1,k}^{(3,1)}=\int_{-\infty}^{2t_{1}-t_{2}}d\tau\int_0
^{\infty}d\rho[
b_{k}(\rho,\tau)-b_{k}(\rho,t_{1})]\frac{\rho
(t_{2}-t_{1})}{(\overline{t}-\tau)^{3}}e^{-\frac{\rho^{2}+r^{2}}
{4(\overline{t}-\tau)}}I_{\lambda_{k}}(\frac{\rho
r}{2(\overline{t}-\tau)}).
\]
We restrict ourself only by the estimate of $A_{1,k}^{(3,1)}$,
that is the part of $A_{1,k}^{(3)}$ corresponding to $J_{1k}$. The
rest estimates are proved with the same way.

Note that $\overline{t}-\tau\geq t_{1}-\tau$ and
$\overline{t}-2t_{1}+t_{2}\geq\Delta t$, thus, Lemma \ref{lem3.2} gives
\begin{align*}
A_{1,k}^{(3,1)}
&  \leq(\Delta t)\langle b_{k}\rangle
_{t,s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}\int_{-\infty}^{2t_{1}-t_{2}}d\tau
\int_0^{\infty}d\rho\frac{\rho^{s+1-\alpha}}{(\overline{t}
-\tau)^{3-\alpha/2}}e^{-\frac{\rho^{2}+r^{2}}{4(\overline{t}-\tau)}
}I_{\lambda_{k}}(\frac{\rho r}{2(\overline{t}-\tau)})\\
&  \leq \textrm{const.}r^{s-\alpha}(\Delta t)^{\alpha/2}\langle
b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}.
\end{align*}
The estimate of $A_{j,k}^{(4)}$ in \eqref{e5.21} is obtained
 simultaneously for $j=1$ and $j=2$. We have
\begin{align*}
|{\sum_{j=1}^{2}} A_{j,k}^{(4)}|
&  =\big|\int_{-\infty}^{2t_{1}-t_{2}}
d\tau\int_0^{\infty}d\rho[
b_{k}(\rho,t_{1})-b_{k}(\rho ,t_{2})]\frac{\partial
L_{k}}{\partial\tau}(\rho,r,t_{2}-\tau)\big| \\
&  =\big|\int_0^{\infty}d\rho[ b_{k}(\rho,t_{1})-b_{k}
(\rho,t_{2})]\int_{-\infty}^{2t_{1}-t_{2}}\frac{\partial L_{k}}
{\partial\tau}(\rho,r,t_{2}-\tau)d\tau\big| \\
&  =\big|\int_0^{\infty}d\rho[ b_{k}(\rho,t_{1})-b_{k}
(\rho,t_{2})]\frac{\rho}{2\Delta t}e^{-\frac{\rho^{2}+r^{2}}{4\Delta t}
}I_{\lambda_{k}}(\frac{\rho r}{2\Delta t})\big| \\
&\leq \textrm{const.}r^{s-\alpha}(\Delta t)^{\alpha/2}\langle
b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)},
\end{align*}
where Lemma \ref{lem3.2} has been applied.

 The coefficients $A^{(i)}_{j,k}$, $i=\overline{1,3}$, $j=2$, are
evaluated similarly. Thus, the  above gives an
estimate for all $i=\overline{1,4}$, and $j=1,2$
\[
|A_{j,k}^{(i)}|\leq \textrm{const.}r^{s-\alpha}(\Delta
t)^{\alpha/2}\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}.
\]
This inequality together with the convergence of $
{\sum_{k}} \langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}$ lead to
\begin{equation} \label{e5.23}
\sum_{k}\langle \frac{\partial  R_{1,k}}{\partial t}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}
\leq \textrm{const.}\langle f\rangle _{t;s-\alpha,G_{T} }^{(\alpha/2)}
\end{equation}
as it was to be proved.

\subsection{H\"older constant of the function $\frac{\partial
 R_1}{\partial t}(r,\varphi,t)$ with respect to $r$}

We change  the variables in the representation $\frac{\partial
 R_1}{\partial t}(r,\varphi,t)$ from \eqref{e5.11}:
$t-\tau\to  \tau$, and consider as the example, the part of
one which corresponds to $i_{1k}(\rho,r,t)$ in \eqref{e5.12}. Let
\begin{align*}
V_{1}(r,\varphi,t)
&={\sum_{k}}\sin(\lambda_{k}\varphi)\int_0^{\infty}d\tau\int_0^{\infty
}d\rho i_{1k}(\rho,r,\tau)[b_{k}(\rho,t-\tau)-b_{k}(\rho,t)] \\
&\equiv {\sum_{k}} \sin(\lambda_{k}\varphi) V_{1,k}(r,t).
\end{align*}
Consider the difference as $h>0$
\begin{equation} \label{e5.24}
\begin{aligned}
&V_{1}(r+h,\varphi,t)-V_{1}(r,\varphi,t) \\
&={\sum_{k}} \sin(\lambda_{k}\varphi)\int_0^{h^{2}}d\tau\int_0^{\infty
}d\rho i_{1k}(\rho,r+h,\tau)[b_{k}(\rho,t-\tau)-b_{k}(\rho,t)] \\
&\quad -{\sum_{k}} \sin(\lambda_{k}\varphi)\int_0^{h^{2}}d\tau
 \int_0^{\infty }d\rho i_{1k}(\rho,r,\tau) [b_{k}(\rho,t-\tau)
 -b_{k}(\rho,t)]\\
&\quad + {\sum_{k}} \sin(\lambda_{k}\varphi)
\int_{h^{2}}^{\infty}d\tau\int_0
^{\infty}d\rho[ i_{1k}(\rho,r+h,\tau)-i_{1k}(\rho,r,\tau)]\\
&\quad\times [ b_{k}(\rho,t-\tau)-b_{k}(\rho,t)]
\equiv {\sum_{j=1}^{3}}{\sum_{k}}
V_{1,k}^{j}\sin(\lambda_{k}\varphi).
\end{aligned}
\end{equation}
Let $r_{h}=r+h$. One can easy estimate the coefficients
$V_{1,k}^{j}$, $j=1,2$ with Lemma \ref{lem3.2}. For instance,
\begin{align*}
|V_{1,k}^{1}|
&\leq\langle b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}
^{(\alpha/2)}\int_0^{h^{2}}d\tau\int_0^{\infty}d\rho
\frac{\rho^{s+1-\alpha}}{2\tau^{2}}\tau^{\alpha/2}e^{-\frac{\rho^{2}
+r_{h}^{2}}{4\tau}}I_{\lambda_{k}}(\frac{\rho r_{h}}{2\tau})\\
&\leq \textrm{const.}r^{s-\alpha}\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}\int_0^{h^{2}}
 d\tau\frac{1}{\tau^{1-\alpha/2}}\\
&\leq \textrm{const.}r^{s-\alpha}\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha /2)}h^{\alpha}.
\end{align*}
To estimate $V_{1,k}^{3}$ in \eqref{e5.24}, we apply the mean value
theorem. We have
\begin{equation} \label{e5.25}
\begin{aligned}
\frac{\partial i_{1k}(\rho,r,t)}{\partial r}
&  =\frac{r\rho}{4t^{3}
}e^{-\frac{\rho^{2}+r^{2}}{4t}}I_{\lambda_{k}}(\frac{\rho r}
{2t})-\frac{\rho^{2}}{4t^{3}}e^{-\frac{\rho^{2}+r^{2}}{4t}}\frac{d}{dx}I_{\lambda_{k}}(x) \\
&  =\frac{\rho}{2t^{2}}\left\{
\frac{r}{2t}I_{\lambda_{k}}(x)-\frac{\rho
}{2t}\frac{d}{dx}I_{\lambda_{k}}(x)\right\}  e^{-\frac{\rho^{2}+r^{2}
}{4t}} \\
&  =\frac{\rho}{2t^{2}} \frac{r-\rho}{2t}I_{\lambda_{k}}
(x)e^{-\frac{\rho^{2}+r^{2}}{4t}}-\frac{\rho}{2t^{2}}\frac{\rho}{2t}[
\frac{d}{dx}I_{\lambda_{k}}(x)
 -I_{\lambda_{k} }(x) ] e^{-\frac{\rho^{2}+r^{2}}{4t}}\\
& =j_{1k}+j_{2k}
\end{aligned}
\end{equation}
where $\rho r/2t=x$. In compliance with \eqref{e5.25} the Fourier
coefficients $V_{1,k}^{3}$ can be represented as
$V_{1,k}^{3}=V_{1,k}^{3,1}+V_{1,k}^{3,2}$. First  we estimate
$V_{1,k}^{3,1}$
\[
V_{1,k}^{3,1}=h\int_{h^{2}}^{\infty}d\tau\int_0^{\infty}
d\rho\frac{\rho(\overline{r}-\rho)}{4\tau^{3}}e^{-\frac{\rho^{2}
+\overline{r}^{2}}{4\tau}}I_{\lambda_{k}}(\frac{\rho\overline{r}}{2\tau
})[b_{k}(\rho,t-\tau)-b_{k}(\rho,t)]
\]
where  $\overline{r}\in (r,r+h)$.
 We have by properties of the function $b_{k}(\rho,t)$
\[
|V_{1,k}^{3,1}|\leq\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha
/2)}h\int_{h^{2}}^{\infty}d\tau\int_0^{\infty}d\rho\frac
{\rho^{1+s-\alpha}|\overline{r}-\rho|}{4\tau^{3}}\tau^{\alpha/2}
e^{-\frac{\rho^{2}+\overline{r}^{2}}{4\tau}}I_{\lambda_{k}}(\frac
{\rho\overline{r}}{2\tau}),
\]
and as it follows from Subsection 7.4 in the appendix,
\[
\int_0^{\infty}d\rho\frac{\rho^{1+s-\alpha}|\overline{r}-\rho
|}{4t^{3/2}}e^{-\frac{\rho^{2}+\overline{r}^{2}}{4t}}I_{\lambda_{k}}
(\frac{\rho\overline{r}}{2t})\leq \textrm{const.}\overline{r}^{s-\alpha}.
\]
Therefore,
\begin{equation} \label{e5.26}
\begin{aligned}
|V_{1,k}^{3,1}|
& \leq \textrm{const.}\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}
}^{(\alpha/2)}h\overline{r}^{s-\alpha}\int_{h^{2}}^{\infty}\tau
^{\alpha/2-3/2}d\tau \leq \textrm{const.}\langle b_{k}\rangle
_{t;s-\alpha,\mathbb{R}_{T}
}^{(\alpha/2)}h^{\alpha}\overline{r}^{s-\alpha}\\
& \leq \textrm{const.}\langle b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}
}^{(\alpha/2)}h^{\alpha}(r+h)^{s-\alpha}\leq \textrm{const.}\langle
b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}
}^{(\alpha/2)}h^{\alpha}r^{s-\alpha},
\end{aligned}
\end{equation}
since the only  $h\leq r$ should be considered, to obtain the
H\"older constant for $\partial  R_1 /\partial t$ with respect
to $r$.

As for $V_{1,k}^{3,2}$ corresponding to $j_{2k}$ in \eqref{e5.25},
it can be represented as ($x=r\rho/2t$)
\[
j_{2k}=\frac{-\rho^{2}}{4t^{3}}[
\frac{d}{dx}I_{\lambda_{k}}(x)-I_{\lambda_{k} }(x)]
e^{-\frac{\rho^{2}+r^{2}}{4t}}=\frac{\rho^{2}}{4t^{3}}e^{-\frac{\rho^{2}+r^{2}}{4t}}
[I_{\lambda_{k}}(x)-\frac{\lambda_{k}}{x}I_{\lambda_{k}
}(x)-Q_{\lambda_{k}}(x)]
\]
and
\[
j_{2k}\leq
\frac{\rho\lambda_{k}}{2rt^{2}}I_{\lambda_{k}}(r\rho/2t)e^{-\frac{\rho^{2}+r^{2}}{4t}}+
\frac{\rho^{2}}{4t^{3}}e^{-\frac{\rho^{2}+r^{2}}{4t}}
[I_{\lambda_{k}}(r\rho/2t)-I_{\lambda_{k}+1}(r\rho/2t)].
\]
Note that the first term in the right part of the last inequality
is estimated in the proof of Lemma \ref{lem5.3} (see \eqref{e5.16}), as for the
second term one is evaluated like $V^{3,1}_{1,k}$. If we take into
account that from the equation
$I_{\lambda_{k}+1}(x)=Q_{\lambda_{k}}(x)$, we have
$I_{\lambda_{k}+1}(x)\leq \textrm{const.} I_{\lambda_{k}}(x)$, and, hence,
by Corollary \ref{coro3.1}, $x^{1/2}e^{-x}I_{\lambda_{k}+1}(x)\leq const$.
uniformly in $k$.  From here it follows that
$I_{\lambda_{k}}(x)-I_{\lambda_{k}+1}(x)\sim \textrm{const.} x^{-3/2}$ for
large value of $x$ where the constant in independent of $k$. Using
this fact, we can repeat the arguments from Subsection 7.4.
 Thus, the
estimate like \eqref{e5.26} holds for $V_{1,k}^{3,2}$. On account of
convergence of ${\sum_{k}} \langle
b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T} }^{(\alpha/2)}$ we have
\[
\sum_{k}\langle V_{1,k}\rangle _{r;s-\alpha
,\mathbb{R}_{T}}^{(\alpha)}\leq \textrm{const.}\langle
f\rangle_{r,t;s-\alpha,G_{T} }^{(\alpha,\alpha/2)}.
\]
Finally, we note that the analogous methods are applied to treat
the function (which corresponds to $i_{2k}$ from \eqref{e5.12})
\begin{align*}
V_{2}(r,\varphi,t)
&={\sum_{k}} \sin(\lambda_{k}\varphi)\int_0^{\infty}d\tau\int_0^{\infty
}d\rho i_{2k}(\rho,r,\tau)[b_{k}(\rho,t-\tau)-b_{k}(\rho,t)]\\
&\equiv \sum_{k}V_{2,k}(r,t)\sin(\lambda_{k}\varphi),
\end{align*}
and the following is true
\[
\sum_{k}\langle V_{2,k}\rangle _{r;s-\alpha
,\mathbb{R}_{T}}^{(\alpha)}\leq \textrm{const.}\langle
f\rangle_{r,t;s-\alpha,G_{T} }^{(\alpha,\alpha/2)}.
\]
The above estimates lead to
\begin{equation} \label{e5.27}
\sum_{k}\langle \frac{\partial  R_{1,k}}{\partial
t}\rangle _{r;s-\alpha ,\mathbb{R}_{T}}^{(\alpha)}\leq
\textrm{const.}\langle f\rangle_{r,t;s-\alpha,G_{T}
}^{(\alpha,\alpha/2)}.
\end{equation}


\begin{remark} \label{rmk5.1} \rm
Note that the estimate of $\sum _{k}[\frac{\partial
 R_{1,k}}{\partial t}]^{\alpha,
\alpha/2}_{r,t;s-2\alpha,\mathbb{R}_{T}}$
will be obtained in the same way if we apply the arguments above
to the difference $[\frac{\partial  R_1 }{\partial
t}(r,\varphi,t_{2})-\frac{\partial  R_1 }{\partial t}(r,\varphi$,
$t_{1})]$.
\end{remark}

\subsection{Estimate for the seminorm $[\frac{\partial
 R_1 }{\partial
t}]_{\varphi,t;s-\alpha,G_{T}}^{(\alpha,\alpha/2)}$}

We will use representation \eqref{e5.21} to the difference
of $\frac{\partial  R_1 }{\partial t}
(r,\varphi,t_{2})-\frac{\partial  R_1
}{\partial t}(r,\varphi,t_{1})$ to obtain the desired estimate.
Let us consider the item
$A^{(1)}_{j,k}=A^{(1)}_{j,k}(r,\varphi,t_{1},t_{2})$. Let $A_{1}$
be the operator corresponding to $A^{(1)}_{j,k}$; i.e.,
\[
A_{1}(f(r,\varphi,\tau)-f(r,\varphi,t_{2}))
=A^{(1)}_{j,k}(r,\varphi,t_{1},t_{2})=v(r,\varphi),
\]
and
\begin{align*}
u_{n}(f(r,\varphi,\tau)-f(r,\varphi,t_{2}))
&  =c(n) {\int_0^{\pi/2}}
[f(r,\varphi+2l\theta/\pi,\tau)-f(r,\varphi-2l\theta/\pi,\tau)\\
&\quad -f(r,\varphi+2l\theta/\pi,t_{2})
+f(r,\varphi-2l\theta/\pi,t_{2})]K(2l)dl
\end{align*}
be the approximating trigonometric polynomial of
$f(r,\varphi,\tau)-f(r,\varphi,t_{2})$. After that, we introduce
the approximating trigonometric polynomial of
$A_{1}(f(r,\varphi,\tau)-f(r,\varphi$, $t_{2}))$ as
$T_{n}(r,\varphi,\tau,t_{2})=A_{1}u_{n}(f(r,\varphi,\tau)-f(r,\varphi,t_{2}))$.
Then, as before,
\begin{align*}
v-T_{n}
& =c(n) {\int_0^{\pi/2}} A_{1}
\big\{f(r,\varphi+2l\theta/\pi,\tau)-f(r,\varphi-2l\theta/\pi,\tau)\\
&\quad -f(r,\varphi+2l\theta/\pi,t_{2})+f(r,\varphi-2l\theta/\pi,t_{2})\\
&\quad -2[f(r,\varphi,\tau)-f(r,\varphi,t_{2})]\}K(2l)dl.
\end{align*}
Estimate \eqref{e5.22} ensured that the value $A_{1}\{\dots\}$ where
$\{\dots\}$ is the expression in the braces in the integrand can be
evaluated as
\[
|A_{1}\{\dots\}|\leq \textrm{const.}l^{\alpha}r^{s-\alpha}|\Delta t|^{\alpha
/2}[f]_{\varphi,t;s-\alpha,G_{T}}^{(\alpha,\alpha/2)}.
\]
After that, ending the estimate as well as  the proof of Theorem
\ref{thm4.2} and applying Theorem \ref{thm4.1}, we obtain
$\frac{A^{(1)}_{j,k}(r,\varphi,t_{1},t_{2} )}{r^{s-\alpha}|\Delta
t|^{\alpha/2}}\in C_{\varphi}^{\alpha}$ uniformly with respect to
the rest variables. The same arguments are true in the case of
other terms in \eqref{e5.21}. This implies
\begin{equation} \label{e5.28}
[\frac{\partial  R_1 }{\partial t}]_{\varphi,t;s-\alpha,G_{T}}
^{(\alpha,\alpha/2)}\leq
\textrm{const.}[f]_{\varphi,t;s-\alpha,G_{T}}^{(\alpha ,\alpha/2)}.
\end{equation}

\section{Proof of Theorem \ref{thm2.1} and applications}

To complete the proof of Theorem \ref{thm2.1}, we note the following.  The exact
representation of the solution in \eqref{e2.6} has been got. We have
shown the proof of the estimates to the higher derivatives of the
solution with respect to $\varphi$ and $t$. After that the
derivatives of the solution with respect to $r$ are evaluated with
these estimates and the equation. We have given the estimates of
the solution corresponding  to the bulk potential, and the
estimates of the potential corresponding to the initial data are
done with the same way. This proves estimate \eqref{e2.11}. A uniqueness
of the solution in the wider class has been proved in \cite{s1}. Thus,
Theorem \ref{thm2.1} has been proved.

\begin{remark} \label{rmk6.1} \rm
Problem \eqref{e2.3} with not uniform boundary conditions can be studied
with reduction one to the problem with uniformly boundary value
problem if the boundary functions are extended into the domain
$G_{T}$ (see \cite{s1}).
\end{remark}

\begin{remark} \label{rmk6.2} \rm
The described method makes possible to consider the homogeneous
Dirichlet initial problem in an arbitrary domain in $\mathbb{R}^{2}$ with
an corner point on the boundary.
\end{remark}

In this section we  formulate only  results relating to the
problem for the parabolic equation with singular coefficients of
the form
\begin{gather} \label{e6.1}
\frac{\partial u}{\partial
t}-\Big(\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial
u}{\partial r}+\frac{b}{r}\frac{\partial u}{\partial
r}\Big)-\frac{1}{r^{2}}\Big(\frac{\partial^{2} u}{\partial
\varphi^{2}}+\frac{b}{\varphi}\frac{\partial u}{\partial
\varphi}\Big)=f(r,\varphi,t),\quad (r,\varphi,t)\in G_{T},
\\
\label{e6.2}
\frac{\partial u}{\partial \varphi}|_{\varphi=0}=0, \quad
u|_{\varphi=\theta}=0, \quad u|_{t=0}=u_0(r,\varphi),
\end{gather}
where $b=\textrm{const.}>0$.

Equation \eqref{e6.1} is the main part of the parabolic equation with the
Bessel operator
\[
\frac{\partial u}{\partial t}-\frac{\partial ^{2}u}{\partial
x^{2}}-\Big(\frac{\partial ^{2}u}{\partial y^{2}}+\frac
{b}{y}\frac{\partial u}{\partial y}\Big)=f(x,y,t),\quad
(x,y,t)\in G_{T},
\]
which can be also rewritten in the form
\begin{equation} \label{e6.3}
\frac{\partial u}{\partial t}-\Big(
\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial u}{\partial
r}+\frac{b}{r}\frac{\partial u}{\partial
r}\Big)-\frac{1}{r^{2}}\Big( \frac{\partial^{2} u}{\partial
\varphi^{2}}+b\frac{\cos \varphi}{\sin \varphi}\frac{\partial
u}{\partial \varphi}\Big)=f(r,\varphi,t),\quad (r,\varphi,t)\in
G_{T}.
\end{equation}
If $b=0$, we get the problem for the heat equation.

We shall use the representation of a solution to problem \eqref{e6.1},
\eqref{e6.2} in the form of the Fourier series by using eigenfunctions of
the problem
\begin{gather} \label{e6.4}
\frac{\partial^{2}v}{\partial
\varphi^{2}}+\frac{b}{\varphi}\frac{\partial v}{\partial
\varphi}=-\lambda^{2}v \quad \varphi\in (0,\theta),
\\
\label{e6.5}
\frac{\partial v}{\partial \varphi}|_{\varphi=0}=0,\quad
v|_{\varphi=\theta}=0.
\end{gather}
Equation \eqref{e6.4} has the two linearly independent solutions:
\[
v_{1}(\varphi)=\varphi^{q/2}J_{q/2}(\lambda_{k}\varphi),\quad
v_{2}(\varphi)=\varphi^{q/2}J_{-q/2}(\lambda_{k}\varphi),\quad
q=1-b,b\neq1,
\]
and if $b=1$
\[
v_{1}(\varphi)=J_0(\lambda_{k}\varphi),\quad
v_{2}(\varphi)=N_0(\lambda_{k}\varphi),
\]
where $J_{\nu}(x)$ and $N_{\nu}(x)$ are the Bessel functions of
the first and second kind. The Bessel functions $J_{\nu}(x)$ has
the power series representation
\[
J_{\nu}(x)=\frac{x^{\nu}}{2^{\nu}}\sum^{\infty}_{k=0}(-1)^{k}\frac{x^{2k}}{2^{2k}k!\Gamma(\nu+k+1)}.
\]
In view of this expansion the eigenfunctions $v_{2}(\phi)$ for
$b\neq 1$ and $v_{1}(\phi)$ for $b= 1$ are appropriate for our
purpose. They have the bounded second derivative and satisfy the
first boundary condition in \eqref{e6.5}. To satisfy the second one, we
define $\lambda=\lambda_{k}$ as the solutions of the equation
$J_{-q/2}(\lambda_{k}\theta)=0$, $k=1,2,\dots$. We will say about
the case $b\neq 1$, the case $b=1$ can be studied similarly.

The formal solution of problem \eqref{e6.1}, \eqref{e6.2}
is represented as
\begin{equation} \label{e6.6}
u(r,\varphi,t)= R_{1b}(r,\varphi,t)
+ R_{2b}(r,\varphi,t),
\end{equation}
where $ R_{1b}(r,\varphi,t)$ is the volume potential
\begin{align*}
& R_{1b}(r,\varphi,t)\\
&=\sum_{k}\varphi^{q/2}J_{-q/2}(\lambda_{k}\varphi)
\int_0^{t}d\tau\int_0^{\infty}\Big(\frac{\rho}{r}\Big)^{b/2}
\frac{\rho}{2(t-\tau)}e^{-\frac{\rho^{2}+r^{2}}{4(t-\tau)}}
I_{\nu_{k}}\Big( \frac{\rho r}{2(t-\tau)}\Big)a_{k}d\rho
\end{align*}
with
\[
a_{k}=\Big( \frac
{\theta^{2}}{2}J^{2}_{1-q/2}(\lambda_{k}\theta)\Big)^{-1}
\int_0^{\theta}\psi^{1-q/2}J_{-q/2}(\lambda_{k}\psi)f(\rho,\psi,\tau)d\psi,
\]
and $ R_{2b}(r,\varphi,t)$ is the initial data potential
\[
 R_{2b}(r,\varphi,t)
=\sum_{k}\varphi^{q/2}J_{-q/2}(\lambda_{k}\varphi)
\int_0^{\infty}\Big(\frac{\rho}{r}\Big)^{b/2}
\frac{\rho}{2t}e^{-\frac{\rho^{2}+r^{2}}{4t}}
 I_{\nu_{k}}\Big( \frac{\rho r}{2t} \Big)a_{0k}d\rho
\]
with
\[
a_{0k}=\Big(\frac
{\theta^{2}}{2}J^{2}_{1-q/2}(\lambda_{k}\theta)\Big)^{-1}
\int_0^{\theta}\psi^{1-q/2}J_{-q/2}(\lambda_{k}\psi)u_0(\rho,\psi)d\psi,
\]
and $\nu _{k}^{2}=\lambda_{k}^{2}+b^{2}/4$.

It turns out that the natural space for solutions of problem
\eqref{e6.1}, \eqref{e6.2} is the space
$P_{s,b}^{l+\alpha,(l+\alpha)/2}(\overline{G}_{T})$ of the
functions with the finite norm ($l$ is an integer, $\alpha
\in(0,1)$)
\begin{align*}
 \| u\| _{P_{s,b}^{l+\alpha,(l+\alpha)/2}(\overline{G}_{T})}
& = \sum _{0\leq \beta_{1}+\beta_{2}+2a\leq
l}\sup_{(r,\phi,t)\in \overline{G}_{T}}
r^{-s+\beta_{1}+2a}\varphi^{b/2}|D_{r}
^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u| \\
&\quad +\sum _{0< l+\alpha-(\beta_{1}+\beta_{2}+2a)<2
}\Big\{\langle \varphi^{b/2}D_{r}^{\beta_{1}}D_{\varphi}
 ^{\beta_{2}}D_{t}^{a}u\rangle_{t;s-\beta_{1}-2a-\alpha,G_{T}}^{(\frac
{l+\alpha-\beta_{1}-\beta_{2}-2a}{2})} \\
&\quad +[\varphi^{b/2}D_{r}^{\beta_{1}}D_{\varphi
}^{\beta_{2}}D_{t}^{a}u]_{r,t;s-\beta_{1}-2a-2\alpha,G_{T}}^{(\alpha
,\frac {l+\alpha-\beta_{1}-\beta_{2}-2a}{2})} \\
&\quad +[\varphi^{b/2}D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}D_{t}^{a}u]_{\varphi
,t;s-\beta_{1}-2a-\alpha,G_{T}}^{(\alpha,\frac
{l+\alpha-\beta_{1}-\beta_{2}-2a}{2})}\Big\}\\
&\quad +\sum _{
\beta_{1}+\beta_{2}+2a=
l}\big\{\langle \varphi^{b/2}D_{r}^{\beta_{1}}D_{\varphi
}^{\beta_{2}}D_{t}^{a}u\rangle_{r;
 s-\beta_{1}-2a-\alpha,G_{T}}^{(\alpha)} \\
&\quad +\langle \varphi^{b/2}D_{r}^{\beta_{1}}D_{\varphi}^{\beta_{2}}
D_{t}^{a}u\rangle _{\varphi;s-\beta _{1}-2a,G_{T}}^{(\alpha)}\big\}.
\end{align*}

We introduce the subspace
$\widehat{P}^{l+\alpha,\frac{l+\alpha}{2}}_{s,b}(\overline{G}_{T})$
($\widehat{P}^{l+\alpha}_{s,b}(\overline{G})$) of the space
$P^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$
($P^{l+\alpha}_{s}(\overline{G})$) like the definition of
$\widehat{P}^{l+\alpha,\frac{l+\alpha}{2}}_{s}(\overline{G}_{T})$
($\widehat{P}^{l+\alpha}_{s}(\overline{G})$). We are looking for
the solution to the problem in the form of the series and waiting
that these series converge in $\overline{G}_{T}$. All their terms
are equal to zero at $\varphi=0$, thus, the condition
\begin{equation} \label{e6.7}
f(r,\theta,t)=0
\end{equation}
is necessary for the solvability of the problem in
$P^{2+\alpha,(2+\alpha)/2}_{s,b}(\overline{G}_{T})$.

\begin{theorem} \label{thm6.1}
Assume  the consistency conditions of the first order and
condition \eqref{e6.7} are fulfilled.  The functions $f\in\widehat{
P}^{\alpha,\alpha/2}_{s,b}(\overline{G}_{T})$ and $u_0\in
\widehat{P}^{2+\alpha}_{s+2,b}(\overline{G})$ Then there exists a
unique solution $u(r,\varphi,t)\in$ $
\widehat{P}^{2+\alpha,\frac{2+\alpha}{2}}_{s,b}$
$(\overline{G}_{T})$ and
\begin{equation} \label{e6.8}
\begin{aligned}
&\|u\|_{P_{s+2,b}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}+S(u)
_{\widehat{P}_{s+2,b}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}\\
&\leq \textrm{const.}(\|f\|
_{P_{s,b}^{\alpha,\alpha/2}(\overline{G}_{T})}+\|
u_0\| _{P_{s+2,b}^{2+\alpha}(\overline{G})}
+S(f) _{\widehat{P}_{s,b}^{\alpha,\alpha/2}(\overline{G}_{T})}+S(u_0)
_{\widehat{P}_{s+2,b}^{2+\alpha}(\overline{G})}),
\end{aligned}
\end{equation}
where the constant in \eqref{e6.8} is independent of $u(r,\varphi,t)$,
$\alpha\in (0,1)$ and $s+2<(\lambda_{1}^{2}+b^{2}/4)^{1/2}$,
$\lambda_{1}\theta$ is the smallest root of the equation
$J_{-q/2}(\lambda_{k}\theta)=0$.
\end{theorem}

In general, the proof of Theorem \ref{thm6.1} repeats our arguments from
the proof of Theorem \ref{thm2.1}. We note only that if $k>>1$,
\[
J_{-q/2}(\lambda_{k}\varphi)\sim
\sqrt{\frac{1}{\lambda_{k}\varphi}} \cos(\lambda_{k}\varphi+
\pi(q-1)/4),\quad
\lambda_{k}\theta\sim (k-(q+1)/4)\pi+O(1/k),
\]
that gives the possibility to apply here the theorems from the
trigonometric series theory.

\section{Appendix}
\subsection{Formal representation of the solution to \eqref{e2.3}}
To obtain the formal solution of  \eqref{e2.3}, we applied the
method of the separation of the variables. In detail, one consists
in the following. Let us consider case of
$u_0(r,\varphi)\equiv 0$ (this case corresponds to
$ R_2(r,\varphi,t)=0$ in \eqref{e2.4}).
 We look for the solution $u(r,\varphi,t)$ of the problem
\begin{equation} \label{e7.1}
\begin{gathered}
\frac{\partial u}{\partial t}-\frac{1}{r}\frac{\partial}{\partial r}
r\frac{\partial u}{\partial r}-\frac{1}{r^{2}}\frac{\partial^{2}u}
{\partial\varphi^{2}}=f(r,\varphi,t),\quad (r,\varphi,t)\in G_{T},
\\
u|_{g_{iT}}=0,\quad  u|_{t=0}=0,\quad (r,\varphi)\in  G,
\end{gathered}
\end{equation}
 as
 \begin{equation} \label{e7.2}
 u(r,\varphi,t)=\sum_{k}V_{k}(r,t)\Phi_{k}(\varphi).
 \end{equation}
After the substitution of the function
$V_{k}(r,t)\Phi_{k}(\varphi)$ into the homogenous equation  and
boundary condition from \eqref{e7.1}, we obtain
\begin{gather} \label{e7.3}
r^{2}\frac{\frac{\partial V_{k}}{\partial
t}-\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial
V_{k}}{\partial r}}{V_{k}} =\frac{\frac{\partial
^{2}\Phi_{k}}{\partial \varphi^{2}}}{\Phi_{k}}\equiv
-\lambda_{k}^{2},
\\
\label{e7.4}
\Phi_{k}(0)=\Phi_{k}(\theta)=0.
\end{gather}
Conditions \eqref{e7.3} and \eqref{e7.4} lead to the function
$\Phi_{k}$ being the solution of the problem
\begin{equation} \label{e7.5}
\begin{gathered}
\Phi''_{k}(\varphi)+\lambda_{k}^{2}\Phi_{k}=0,\\
\Phi_{k}(0)=\Phi_{k}(\theta)=0.
\end{gathered}
\end{equation}
The solution of \eqref{e7.5} is the function
\begin{equation} \label{e7.6}
\Phi_{k}=\sin\lambda_{k}\varphi,\quad  \lambda_{k}=\pi k/\theta,\;
k=1,2\dots.
\end{equation}
Now we return to problem \eqref{e7.1} and represent $f(r,\varphi,t)$ as
\begin{equation} \label{e7.7}
f(r,\varphi,t)=\sum_{k}b_{k}(r,t)\sin\lambda_{k}\varphi,
\end{equation}
with
\begin{equation} \label{e7.8}
b_{k}(r,t)=\frac{2}{\theta}\int^{\theta}_0f(r,\psi,t)\sin\lambda_{k}\psi
d\psi.
\end{equation}
After that we substitute \eqref{e7.2}, \eqref{e7.7} and \eqref{e7.8}
to the equation and the initial condition of \eqref{e7.1} and have
\begin{equation} \label{e7.9}
\begin{gathered}
\frac{\partial V_{k}}{\partial
t}-\frac{1}{r}\frac{\partial}{\partial r} r\frac{\partial
V_{k}}{\partial r}+\lambda_{k}^{2}\frac{V_{k}}{r^{2}}=b_{k}(r,t),
\\
V_{k}(r,0)=0.
\end{gathered}
\end{equation}
Here we use that the function $\Phi_{k}(\varphi)$ satisfies  the
equation in \eqref{e7.5}.

Let us denote the Hankel transformation (see, for example,
\cite{g2,p1,s3} for discussion)
with respect to $r$ of the functions
$V_{k}(r,t)$ and $b_{k}(r,t)$ by $\widehat{V}_{k}(\mu,t)$ and
$\widehat{b}_{k}(\mu,t)$, respectively, $\mu$ is the parameter
under the transformation:
\begin{equation} \label{e7.10}
\begin{gathered}
\widehat{V}_{k}(\mu,t)=\int^{\infty}_0V_{k}(r,t)rJ_{\lambda_{k}}(\mu
r)dr;\\
\widehat{b}_{k}(\mu,t)=\int^{\infty}_0b_{k}(r,t)rJ_{\lambda_{k}}(\mu
r)dr,
\end{gathered}
\end{equation}
where $J_{\lambda_{k}}(\mu r)$ is the Bessel function \cite{g2}.

Expressions \eqref{e7.9} and \eqref{e7.10} lead to the following
problem for the function $\widehat{V}_{k}(\mu,t)$,
\begin{equation} \label{e7.11}
\begin{gathered}
\frac{d \widehat{V}_{k}}{d
t}+\mu^{2}\widehat{V}_{k}=\widehat{b}_{k}(\mu,t), \\
\widehat{V}_{k}(\mu,0)=0.
\end{gathered}
\end{equation}
It is easy to check that the function
\begin{equation} \label{e7.12}
\widehat{V}_{k}(\mu,t)=\int^{t}_0e^{-\mu^2(t-\tau)}
\widehat{b}_{k}(\mu,\tau)d\tau
\end{equation}
gives the solution of problem \eqref{e7.11}.

After applying the inverse
Hankel transformation in \eqref{e7.12}, we obtain
\begin{equation} \label{e7.13}
V_{k}(r,t)=\int^{\infty}_0\mu J_{\lambda_{k}}(\mu
r)\int^{t}_0e^{-\mu^2(t-\tau)}\widehat{b}_{k}(\mu,\tau)d\tau d
\mu.
\end{equation}
Then the formal solution \eqref{e7.1} follows from
\eqref{e7.2}, \eqref{e7.6} and \eqref{e7.13}, so,
\begin{equation} \label{e7.14}
u(r,\varphi,t)=\sum_{k=1}\sin\lambda_{k}\varphi\int^{\infty}_0\mu
J_{\lambda_{k}}(\mu
r)\int^{t}_0e^{-\mu^2(t-\tau)}\widehat{b}_{k}(\mu,\tau)d\tau d
\mu.
\end{equation}
To obtain formula \eqref{e2.7}, we transform \eqref{e7.14}
applying formula \cite[6.633(2)]{g2}:
\[
\int_0^{\infty}\mu e^{-\mu^2(t-\tau)} J_{\lambda_{k}}(\mu
r)J_{\lambda_{k}}(\mu
\rho)d\mu=\frac{1}{2(t-\tau)}I_{\lambda_{k}}\Big(\frac{r\rho}{2(t-\tau)}\Big)
\exp\Big(-\frac{r^{2}+\rho^{2}}{4(t-\tau)}\Big)
\]
where $I_{\lambda_{k}}(x)$ is the modified Bessel function.
Thus,
\begin{align*}
&u(r,\varphi,t)\\
&=\sum_{k=1}\sin\lambda_{k}\varphi\int^{t}_0d\tau\int^{\infty}_0d
\rho{b}_{k}(\rho,\tau)\rho \int^{\infty}_0 \mu
J_{\lambda_{k}}(\mu r)e^{-\mu^2(t-\tau)}J_{\lambda_{k}}(\mu \rho)
d\mu\\
&=\sum_{k=1}\sin\lambda_{k}\varphi\int^{t}_0d\tau
\int^{\infty}_0d \rho{b}_{k}(\rho,\tau)
\frac{\rho}{2(t-\tau)}I_{\lambda_{k}}
\Big(\frac{r\rho}{2(t-\tau)}\Big)\exp\Big(-\frac{r^{2}+\rho^{2}}{4(t-\tau)}\Big).
\end{align*}
That  gives \eqref{e2.6}, \eqref{e2.7} with $ R_2\equiv 0$
(due to $u_0\equiv 0$). To obtain the complete formula \eqref{e2.6};
 i.e., with $ R_2\neq 0$, it is enough to consider
the problem
\begin{equation} \label{e7.15}
\begin{gathered}
\frac{\partial u}{\partial t}-\frac{1}{r}\frac{\partial}{\partial r}
r\frac{\partial u}{\partial r}-\frac{1}{r^{2}}\frac{\partial^{2}u}
{\partial\varphi^{2}}=0,\quad (r,\varphi,t)\in G_{T},
\\
u|_{g_{iT}}=0,\quad  u|_{t=0}=u_0(r,\varphi),(r,\varphi)\in G,
\end{gathered}
\end{equation}
and apply all reasoning mentioned above to this problem.

After that, the solution of  \eqref{e2.3} is represented as
\begin{equation} \label{e7.16}
u(r,\varphi,t)= R_1 (r,\varphi,t)+ R_2(r,\varphi,t)
\end{equation}
where $ R_1 (r,\varphi,t)$ and $ R_2(r,\varphi,t)$
are the solutions of  \eqref{e7.1} and \eqref{e7.15}, correspondingly.
\begin{equation} \label{e7.17}
\begin{gathered}
 R_1 (r,\varphi,t)=\sum_{k}\sin(\lambda_{k}\varphi)\int_{0
}^{t}d\tau\int_0^{\infty}d\rho\frac{\rho}{2(t-\tau)}e^{-\frac
{\rho^{2}+r^{2}}{4(t-\tau)}}I_{\lambda_{k}}
\Big(\frac{\rho r}{2(t-\tau)}\Big)b_{k}(\rho,\tau),
\\
 R_2(r,\varphi,t)=\sum_{k}\sin(\lambda_{k}\varphi)\int_0
^{\infty}d\rho\frac{\rho}{2t}e^{-\frac{\rho^{2}+r^{2}}{4t}}I_{\lambda_{k}
}\Big(\frac{\rho r}{2t}\Big)u_{0k}(\rho),
\end{gathered}
\end{equation}
\begin{equation} \label{e7.18}
u_{0k}(r)=\frac{2}{\theta}\int_0^{\theta}u_0(r,\psi)\sin(\lambda
_{k}\psi)d\psi,\quad
b_{k}(r,t)=\frac{2}{\theta}\int_0^{\theta}f(r,\psi,t)\sin(\lambda
_{k}\psi)d\psi.
\end{equation}
Equation \eqref{e2.6} implies that the desired solution is
the sum of the volume potential $ R_1 (r,\phi,t)$ and the
potential of the initial data $ R_2(r,\phi,t)$.

The representation for $ R_1 (r,\varphi,t)$ can be rewritten also
as
\begin{equation} \label{e7.19}
 R_1 (r,\varphi,t)=\sum_{k}\sin(\lambda_{k}\varphi)\int_{-\infty
}^{t}d\tau\int_0^{\infty}d\rho\frac{\rho}{2(t-\tau)}e^{-\frac
{\rho^{2}+r^{2}}{4(t-\tau)}}I_{\lambda_{k}}
\Big(\frac{\rho r}{2(t-\tau)}\Big)b_{k}(\rho,\tau),
\end{equation}
if $f(r,\varphi,t)=0$ for $t<0$, that was assumed.
Thus representations \eqref{e7.16}-\eqref{e7.19}
give \eqref{e2.6}-\eqref{e2.9}.

\subsection{Proof of Corollary \ref{coro3.1}}
 In the integral from \eqref{e3.7}, we change the variable
$\frac{\rho r}{2t}=x$ and then $x^{2}=y$,
\begin{align*}
D_{s}  & =\int_0^{\infty}\Big(\frac{2xt}{r}\Big)  ^{1+s}
\frac{2t}{r}\frac{1}{2t}e^{-\frac{r^{2}}
{4t}-t\frac{x^{2}}{r^{2}}}I_{\lambda_{k}}(x)dx \\
& =\Big(\frac{2t}{r}\Big)  ^{1+s}\frac{1}{2r}e^{-\frac{r^{2}}{4t}}
\int_0^{\infty}y^{s/2}e^{-t\frac{y}{r^{2}}}I_{\lambda_{k}}
(y^{1/2})dy.
\end{align*}
Using tabular integral \cite[6.643(2)]{g2}, we obtain
\[
D_{s}=2^{1+s}r^{s}(t/r^{2})^{\frac{1+s}{2}}e^{-\frac{r^{2}}{8t}}\frac
{\Gamma(\frac{\lambda_{k}+s+2}{2})}{\Gamma(\lambda_{k}+1)}M_{-\frac{1+s}
{2},\frac{\lambda_{k}}{2}}(r^{2}/4t).
\]
In our case (see \cite[9.221]{g2})
\[
M_{-\frac{1+s}{2},\frac{\lambda_{k}}{2}}(r^{2}/4t)
=\frac{(r^{2}/4t)^{\frac
{\lambda_{k}+1}{2}}}{2^{\lambda_{k}}B(\frac{\lambda_{k}-s}{2},\frac
{\lambda_{k}+s+2}{2})}N\Big(\lambda_{k},\frac{r^{2}}{8t}\Big),
\]
where $B(x,y)=\frac{\Gamma (x)\Gamma (y)}{\Gamma(x+y)}$, $\Gamma
(x)$ is the Gamma function, and
\[
N\Big(\lambda_k,\frac{r^{2}}{8t}\Big)  
=\int_{-1}^{1}
(1+z)^{\frac{\lambda_{k}+s}{2}}(1-z)^{\frac{\lambda_{k}-2-s}{2}}
e^{z\frac{r^{2}}{8t}}dz.
\]
By the substitution $1+z=x$ we go to
\[
N\Big(\lambda_k,\frac{r^{2}}{8t}\Big)  =\int_0
^{2}e^{-\frac{r^{2}}{8t}}x^{\frac{\lambda_{k}+s}{2}}(2-x)^{\frac{\lambda
_{k}-2-s}{2}}e^{x\frac{r^{2}}{8t}}dx.
\]
In this equality, we put $s=-1$ and use tabular integral
\cite[3.383(2)]{g2}, then
\[
N\Big(\lambda_{k},\frac{r^{2}}{8t}\Big) 
=\sqrt{\pi}\Big(\frac{16t}{r^{2}}\Big) ^{\lambda_{k}/2}
\Gamma\Big(\frac{\lambda_{k}+1}{2}\Big) 
I_{\lambda_{k}/2}\Big(\frac{r^{2}}{8t}\Big).
\]
Finally, we gather our calculations and obtain
\[
D_{-1}= \textrm{const.}r^{-1}e^{-z} z^{\frac{1}{2}}I_{\frac{\lambda_{k}}{2}
}(z),\quad z=r^{2}/8t.
\]
Lemma \ref{lem3.2} leads to
\[
e^{-z} z^{\frac{1}{2}}I_{\frac{\lambda_{k}}{2}}(z)\leq \textrm{const.},
\]
where the constant does not depend on $k$. Recall that
$\lambda_{k}=\frac{\pi}{\theta}k$, so, if we take $k=2n$,
$n=1,2,\dots$, we will obtain our assertion. This ends the proof of
Corollary \ref{coro3.1}.

\subsection{Estimate for the integral
$M_{1}=(1+\lambda_{k})\int^{\infty}_0dt\int^{\infty}_0d\rho\frac{\rho^{s-\alpha}}
{t^{1-\alpha/2}}L_{k}(\rho,r,t)$}

Using the representation of the function $I_{\lambda_{k}}(x)$
from \cite[7.7.3(25)]{b2}, we obtain
\begin{align*}
M_{1}
&=(1+\lambda_{k})\int^{\infty}_0dt\int^{\infty}_0d\rho
 \frac{\rho^{1+s-\alpha}}
{t^{1-\alpha/2}}\int^{\infty}_0J_{\lambda_{k}}
(\rho\mu)J_{\lambda_{k}}(r\mu)e^{-t\mu^{2}}\mu
d\mu\\
&=(1+\lambda_{k})\int^{\infty}_0dt\int^{\infty}_0d\rho
 \frac{\rho^{-1+s-\alpha}}
{t^{1-\alpha/2}}\int^{\infty}_0J_{\lambda_{k}}(zr/\rho)
 J_{\lambda_{k}}(z)e^{-z^{2}t\rho^{-2}}z dz 
\\
&=(1+\lambda_{k})\int^{\infty}_0dy
y^{-1-s+\alpha}\int^{\infty}_0dz
zJ_{\lambda_{k}}(zry)J_{\lambda_{k}}(z)
\int^{\infty}_0dtt^{-1+\alpha/2}e^{-z^{2}ty^{2}},
\end{align*}
In the first equality above we used $\mu=z/\rho$, and in the second
$\rho=y^{-1}$.
The last integral can be calculated (see \cite[3.381(4)]{g2})
\[
\int^{\infty}_0dt t^{-1+\alpha/2}e^{-z^{2}ty^{2}}=\frac{\Gamma
(\alpha/2)}{z^{\alpha}y^{\alpha}},
\]
so that after the changing of the variable $y=q/r$,
\begin{equation} \label{e7.20}
\begin{aligned}
&M_{1}\\
&=(1+\lambda_{k})r^{s}\Gamma
(\alpha/2)\int^{\infty}_0dq q^{-1-s} \int^{\infty}_0dz
z^{1-\alpha}J_{\lambda_{k}}(zq)J_{\lambda_{k}}(z)\\
&=(1+\lambda_{k})r^{s}\Gamma (\alpha/2)
 \Big\{ \int ^{1-\varepsilon}_0+ \int
_{1-\varepsilon}^{1+\varepsilon}+\int
^{\infty}_{1+\varepsilon}\Big\} dq
q^{-1-s}\int^{\infty}_0dz z
^{1-\alpha}J_{\lambda_{k}}(zq)J_{\lambda_{k}}(z)\\
& \equiv(1+\lambda_{k})r^{s}\Gamma (\alpha/2)(M^{(1)}_{1}
+M^{(2)}_{1}+M^{(3)}_{1}).
\end{aligned}
\end{equation}
For $q\in(0,1-\varepsilon)$ the integral
(see \cite[7.7.4(29)]{b2})
\begin{align*}
d_{1}&=\int^{\infty}_0dz z ^{1-\alpha}J_{\lambda_{k}}
(zq)J_{\lambda_{k}}(z)\\
&=\frac{q^{\lambda_{k}} \Gamma(\lambda_{k}+1-\alpha/2)}{2^{\alpha-1}
\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
 F(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;q^{2}).
\end{align*}
The function
$F(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;q^{2})$
is bounded (see \cite[9.102]{g2})
 so that
\begin{equation} \label{e7.21}
d_{1}\leq
\textrm{const.}\frac{q^{\lambda_{k}}}{2^{\alpha-1}\Gamma(\alpha/2)}\lambda_{k}^{-\alpha/2}\leq
\textrm{const.} q^{\lambda_{k}}\lambda_{k}^{-\alpha/2}.
\end{equation}
For $q\in (1+\varepsilon,\infty)$,
\begin{equation} \label{e7.22}
\begin{aligned}
d_{3}&=\int^{\infty}_0dz z ^{1-\alpha}J_{\lambda_{k}}
 (zq)J_{\lambda_{k}}(z)\\
&=\frac{q^{-\lambda_{k}+\alpha-2}\Gamma(\lambda_{k}
+1-\alpha/2)}{2^{\alpha-1}\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
F(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;q^{-2})\\
&\leq \textrm{const.} q^{-\lambda_{k}+\alpha-2}\lambda_{k}^{-\alpha/2}.
\end{aligned}
\end{equation}
Estimates \eqref{e7.21} and \eqref{e7.22} lead to
\begin{equation} \label{e7.23}
M_{1}^{(1)}\leq \textrm{const.}\lambda_{k}^{-1-\alpha/2},\quad
M^{(3)}_{1}\leq \textrm{const.}\lambda_{k}^{-1-\alpha/2}.
\end{equation}
Now we estimate the integral
\[
M^{(2)}_{1}=\Big\{\int^{1}_{1-\varepsilon}
+\int_{1}^{1+\varepsilon}\Big\}q^{-1-s}dq\int^{\infty}_0
z^{1-\alpha}J_{\lambda_{k}}(z)J_{\lambda_{k}}(qz)dz
=M^{(2,1)}_{1}+M^{(2,2)}_{1},
\]
\begin{align*}
M^{(2,1)}_{1}&=\int^{1}_{1-\varepsilon}q^{-1-s}dq\int^{\infty}_0
z^{1-\alpha}J_{\lambda_{k}}(z)J_{\lambda_{k}}(qz)dz\\
&=\int_0^{\varepsilon}(1-x)^{-1-s}dx \int^{\infty}_0
z^{1-\alpha}J_{\lambda_{k}}(z)J_{\lambda_{k}}((1-x)z)dz\\
&=\int_0^{\varepsilon}(1-x)^{-1-s}d_{21}(x)dx
\end{align*}
(in the second inequality above, we used $q=1-x$),
\[
d_{21}=\frac{(1-x)^{\lambda_{k}}\Gamma(\lambda_{k}
 +1-\alpha/2)}{2^{\alpha-1}\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
F(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;(1-x)^{2}),
\]
\begin{align*}
M^{(2,2)}_{1}&=\int_{1}^{1+\varepsilon}q^{-1-s}dq\int^{\infty}_0
z^{1-\alpha}J_{\lambda_{k}}(z)J_{\lambda_{k}}(qz)dz \\
&=\int_0^{\varepsilon}(1+x)^{-1-s} d_{22}(x)dx
\end{align*}
(in the inequality above, we used $q=1+x$),
\[
d_{22}=\frac{(1+x)^{\alpha-2-\lambda_{k}}\Gamma(\lambda_{k}+1-\alpha/2)}{2^{\alpha-1}\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
F(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;(1+x)^{-2}),
\]
where
\[
F(\alpha,\beta;\gamma;x)=1+\frac{\alpha\beta}{1\cdot\gamma}x
+\frac{\alpha(\alpha+1)\beta(\beta+1)}{1\cdot2\cdot\gamma(\gamma+1)}x^{2}
+\dots
= 1+\sum^{\infty}_{p=1}a_{p}x^{p}.
\]
In our case
\[
a_{1}=\frac{(\lambda_{k}+1-\alpha/2)(1-\alpha/2)}{(\lambda_{k}+1)},
\quad
a_{2}=a_{1}\cdot\frac{(\lambda_{k}+2-\alpha/2)(2-\alpha/2)}{
2\cdot(\lambda_{k}+2)},\dots;
\]
i.e., $a_{p}\leq const$. with respect to $p$ and $\lambda_{k}$.
After that,
\begin{align*}
&M^{(2,1)}_{1}+M^{(2,2)}_{1}\\
&=\frac{\Gamma(\lambda_{k}+1-\alpha/2)}{2^{\alpha-1}
\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
\sum_{p=0}a_{p} \int^{\varepsilon}_0[(1-x)^{\lambda_{k}-1-s+2p}
+(1+x)^{-\lambda_{k}-1-s-2p-2+\alpha}]dx
\\
&=\frac{\Gamma(\lambda_{k}+1-\alpha/2)}{2^{\alpha-1}
\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
\sum_{p=0}a_{p}[(1-x)^{\lambda_{k}-s+2p}(\lambda_{k}-s
+2p)^{-1}\\
&\quad +(1+x)^{-\lambda_{k}-s-2p-2+\alpha}
(-\lambda_{k}-s-2p-2+\alpha)^{-1}]|^{x=\varepsilon}_{x=0}\\
&=\frac{\Gamma(\lambda_{k}+1-\alpha/2)}{2^{\alpha-1}
\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
\sum_{p=0}a_{p}\Big\{[(1-\varepsilon)^{\lambda_{k}
-s+2p}(\lambda_{k}-s+2p)^{-1}\\
&\quad +(1+\varepsilon)^{-\lambda_{k}-s-2p-2+\alpha}
(-\lambda_{k}-s-2p-2+\alpha)^{-1}] \\
&\quad -(\alpha-2-2s)(\lambda_{k}-s+2p)^{-1}
(-\lambda_{k}-s-2p-2+\alpha)^{-1}\Big\}\\
&=\frac{\Gamma(\lambda_{k}+1-\alpha/2)}{2^{\alpha-1}
 \Gamma(\lambda_{k}+1)\Gamma(\alpha/2)}
\Big\{\sum_{p=0}a_{p}[(1-\varepsilon)^{\lambda_{k}-s+2p}
(\lambda_{k}-s+2p)^{-1}\\
&\quad +(1+\varepsilon)^{-\lambda_{k}-s-2p-2+\alpha}
(-\lambda_{k}-s-2p-2+\alpha)^{-1}] \\
&\quad +\sum_{p=0}a_{p}(\alpha-2-2s)(\lambda_{k}-s+2p)^{-1}
(\lambda_{k}+s+2p+2-\alpha)^{-1}\Big\}.
\end{align*}
The first series converges because, for example, for every fixed
$\varepsilon\geq \varepsilon_0>0$
\[
a_{p}(1-\varepsilon)^{\lambda_{k}-s+2p}(\lambda_{k}-s+2p)^{-1}\leq
\frac{\textrm{const.}}{\lambda_{k}}q^{\lambda_{k}},\quad q<1,
\]
so
\begin{align*}
&\big|\sum_{p=0}a_{p}[(1-\varepsilon)^{\lambda_{k}-s+2p}
(\lambda_{k}-s+2p)^{-1}\\
&+(1+\varepsilon)^{-\lambda_{k}-s-2p-2+\alpha}
(-\lambda_{k}-s-2p-2+\alpha)^{-1}]\big|
\leq\frac{\textrm{const.}}{\lambda_{k}}.
\end{align*}
As for the second series,
\[
\big|\sum_{p=0}a_{p}(\alpha-2-2s)(\lambda_{k}-s+2p)^{-1}
(\lambda_{k}+s+2p+2-\alpha)^{-1}\big|
\leq \frac{\textrm{const.}}{\lambda_{k}^{1-\mu}}, \quad \mu>0.
\]
Taking into account that
\[
\frac{\Gamma(\lambda_{k}+1-\alpha/2)}{\Gamma(\lambda_{k}+1)}\approx
\lambda_{k}^{-\alpha/2}
\]
for large $\lambda_{k}$, we have
\begin{equation} \label{e7.24}
|M^{(2)}_{1}|=|M^{(2,1)}_{1}+M^{(2,2)}_{1}|\leq \textrm{const.}
\lambda_{k}^{-1+\mu-\alpha/2}.
\end{equation}
Finally, the following inequality follows from
\eqref{e7.20}, \eqref{e7.23} and \eqref{e7.24}:
\begin{equation} \label{e7.25}
|M_{1}|\leq \textrm{const.}\frac{r^{s}}{\lambda_{k}^{\alpha/2-\mu}}
\end{equation}
with $0<\mu<\alpha/2$.

\subsection{Estimate for the integral $\int
_0^{\infty}\frac{\rho^{1+s-\alpha}}{t^{3/2}}|r-\rho|I_{\lambda_{k}}(r\rho/2t)e^{-\frac{r^{2}+\rho^{2}}{4t}}d\rho$
from Subsection 5.5}

In this subsection we show the estimate
\[
I=\int
_0^{\infty}\frac{\rho^{1+s-\alpha}}{t^{3/2}}|r-\rho|I_{\lambda_{k}}(r\rho/2t)e^{-\frac{r^{2}+\rho^{2}}{4t}}d\rho\leq
\textrm{const.} r^{s-\alpha}.
\]
Denote
\[
u=\frac{\rho}{2t^{1/2}},\quad v=\frac{r}{2t^{1/2}},\quad
2uv=\frac{r\rho}{2t},
\]
and change the integration variable $\rho$ by $u$. We obtain
\begin{align*}
I&=\int_0^{\infty}4t\frac{(2t^{1/2}u)^{1+s-\alpha}}{t^{3/2}}
|v-u|e^{-(u^{2}+v^{2})}I_{\lambda_{k}}(2uv)du\\
&\leq \textrm{const.}\int
_0^{\infty}t^{\frac{s-\alpha}{2}}u^{1+s-\alpha}
e^{-\gamma(u-v)^{2}}I_{\lambda_{k}}(2uv)e^{-2uv}du
\end{align*}
where $\gamma\in (0,1)$.
Next we consider the integral
\[
A(v)=\int _0^{\infty}
u^{\beta}e^{-\gamma(u-v)^{2}}I_{\lambda_{k}}(2uv)e^{-2uv}du.
\]
Introduce the new integration variable $z=uv$ so that
\begin{align*}
A(v)&=v^{-\beta-1}\int _0^{\infty}
z^{\beta}e^{-\gamma(\frac{z}{v}-v)^{2}}I_{\lambda_{k}}(2z)e^{-2z}dz\\
&=v^{-\beta-1}\int_0^{1}
z^{\beta}e^{-\gamma(\frac{z}{v}-v)^{2}}I_{\lambda_{k}}(2z)
 e^{-2z}dz\\
&\quad +v^{-\beta-1}\int _{1}^{\infty}
z^{\beta}e^{-\gamma(\frac{z}{v}-v)^{2}}I_{\lambda_{k}}(2z)e^{-2z}dz\\
&\equiv A_{1}(v)+A_{2}(v).
\end{align*}
To estimate $A_{2}(v)$, we use Corollary \ref{coro3.1}, and obtain
\[
A_{2}(v)\leq \textrm{const.}v^{-\beta-1}\int _{1}^{\infty}
z^{\beta-1/2}e^{-\gamma(\frac{z}{v}-v)^{2}}dz.
\]
Now let $\xi=\frac{z}{v}-v$. Then
\begin{align*}
 A_{2}(v)& \leq \textrm{const.}v^{-\beta-1}\int _{-v+1/v}^{\infty}
v^{\beta+1/2}(\xi+v)^{\beta-1/2}e^{-\gamma\xi^{2}}d\xi\\
&\leq \textrm{const.}v^{-1/2}\int _{-v+1/v}^{\infty}
\underset{\xi}{\max}[(\xi+v)^{\beta-1/2}e^{-\gamma\xi^{2}/2}]
e^{-\gamma\xi^{2}/2}d\xi.
\end{align*}
One can verify that
\[
\varphi(\xi,v)=(\xi+v)^{\beta-1/2}e^{-\gamma\xi^{2}/2}\leq \textrm{const.}
v^{\beta-1/2}
\]
under $v\geq v_0(\beta,\gamma)>0$. It implies
\[
A_{2}(v)\leq \textrm{const.} v^{\beta-1} \quad \text{for } v\geq 1.
\]
To estimate $A_{2}(v)$ for $v<1$, notice that
$e^{-\gamma\frac{z^{2}}{2v^{2}}}\leq e^{\frac{-\gamma}{2v^{2}}}$
if $z\geq 1$ and
$e^{-\gamma\frac{z^{2}}{2v^{2}}}\leq e^{\frac{-\gamma z^{2}}{2}}$
if $v<1$. Therefore,
\begin{align*}
A_{2}(v)&\leq \textrm{const.} v^{-1-\beta}\int
_{1}^{\infty}z^{\beta-1/2}e^{-\gamma\frac{z^{2}}{2v^{2}}}
e^{-\gamma\frac{z^{2}}{2v^{2}}}e^{2z}e^{-\gamma v^{2}}dz\\
&\leq \textrm{const.} v^{-1-\beta}e^{-\frac{\gamma}{2 v^{2}}}
 \int_{1}^{\infty}z^{\beta-1/2}e^{-\gamma\frac{z^{2}}{2}+2z}dz\\
&\leq \textrm{const.} v^{-1-\beta}e^{-\frac{\gamma}{2 v^{2}}}
\leq \textrm{const.} v^{-1+\beta}
\end{align*}
for $v<1$.

After that we evaluate the integral $A_{1}(v)$ for $v\geq 1$. For
$z\leq 1$, we use the estimate
\[
I_{\lambda_{k}}(2z)\leq \textrm{const.}
z^{\lambda_{k}}\leq \textrm{const.}
z^{\lambda_{1}},\quad \lambda_{1}=\pi/\theta.
\]
Then
\begin{align*}
A_{1}(v)&\leq \textrm{const.} v^{-\beta-1}\int _0^{1} z^{\beta
+\lambda_{1}}e^{-\gamma(v^{2}-2z+z^{2}/v^{2})}dz \\
&\leq \textrm{const.}v^{-\beta-1}e^{-\gamma v^{2}}
\leq \textrm{const.} v^{-1+\beta}
\end{align*}
for $v\geq 1$.
At last, for $v<1$,
\begin{align*}
A_{1}(v)&\leq \textrm{const.} v^{-\beta-1}\int _0^{1} z^{\beta
+\lambda_{1}}e^{-\gamma(v^{2}-2z+z^{2}/v^{2})}dz \\
&\leq \textrm{const.}v^{-\beta}\int _0^{1/v}dy
(yv)^{\beta+\lambda_{1}}e^{-\gamma y^{2}}\\
&\leq \textrm{const.}
v^{\lambda_{1}}=\textrm{const.}v^{-1+\beta}v^{\lambda_{1}+1-\beta}\leq
\textrm{const.} v^{\beta-1}
\end{align*}
if $\lambda_{1}+1-\beta\geq 0$.

If we take $\beta=1+s-\alpha$, then the condition
$\lambda_{1}+1-\beta\geq 0$ means $1-\alpha\leq \pi/\theta$
that is fulfilled under conditions of Theorem \ref{thm2.1}. Thus, our
calculations lead to
\[
I\leq \textrm{const.}
t^{(s-\alpha)/2}\Big(\frac{r}{2t^{1/2}}\Big)^{-1+1+s-\alpha}
\leq \textrm{const.} r^{s-\alpha},
\]
that was to be proved.

\subsection*{Acknowledgments}
The authors would like to thank R. M. Trigub for the useful discussions.

\begin{thebibliography}{00}

\bibitem{a1} H. Amann;
\emph{Operator-valued Fourier multipliers, vector-valued Besov spaces,
and applications}, Math. Nachr., 186 (1997), 5-56.

\bibitem{b1} N. K. Bari;
\emph{Trigonometric series} (Russian), GITTL, Moscow, 1961.
[N.K. Bari, A treatise on trigonometric series, v. I,II, The
Macmillan Co, New York, 1964].

\bibitem{b2} H. Bateman, A. Erd\'{e}lyi;
\emph{Higher transcendental functions}, v. II, New York Toronto London,
MC Graw-Hill Company, 1953.

\bibitem{c1} V. A. Chernyatin;
\emph{Existence of a classical solution
of a mixed problem for a second-order one-dimensional parabolic
equation}, Moscow Univ. Math. Bull., \textbf{46}, 6 (1990), 7-11.

\bibitem{c2} F. Colombo, D. Guidetti, and A. Lorenzi;
\emph{Elliptic equations in nonsmooth plane domanis with an
applictaion to a parabolic problem},
Adv. Diff. Eq., \textbf{7}, 6 (2002), 695-716.

\bibitem{f1} E. V. Frolova;
\emph{On a certain nonstationary problem
in a dihedral angle I}, J. Soviet. Math., \textbf{70} (1994),
1828-1840.

\bibitem{g1} M. G. Garroni, V. A. Solonnikov, and M. A. Vivaldi;
\emph{Existence and regularity results for oblique derivative problem for
heat equations in an angle},
 Proceedings of the Royal Society of Edinburgh, \textbf{128 A},
(1998), 47-79.

\bibitem{g2} I. S. Gradshteyn and I. M. Ryzhik;
\emph{Table of integrals, sums, series and products},
5-th ed., Academic Press, London, UK, 1994.

\bibitem{g3} P. Grisvard;
\emph{Elliptic problems in nonsmooth domains},
Pitman, London, 1985.

\bibitem{g4} D. Guidetti;
\emph{The mixed Cauchy-Dirichlet problem
for the heat equation in a plane angle in spaces of H\"older
continuous functions}, Adv. Diff. Eq., \textbf{6}, 8 (2001),
897-930.

\bibitem{n1} I. P. Natanson;
\emph{Constructive function theory} (Russian),
GITTL, Moscow, 1949. [I. P. Natanson, Constructive function theory,
v. I, II, III, Frederic Ungar Publishing Co, New York, 1964].

\bibitem{p1} A. D. Polyanin, and A. V. Manzhirov;
\emph{Handbook of Integral Equations}, CRC Press, Boca Raton, 1998.

\bibitem{s1} V. A. Solonnikov;
\emph{Solvability of classical initial boundary-value problems for
a heat equation in dihedral angle}, J. Soviet. Math., \textbf{32},
(1986), 526-546.

\bibitem{s2} V. A. Solonnikov;
\emph{Solvability of a problem on the motion of a viscous
incompressible fluid bounded by a free surface},
Math. USSR-Izv., \textbf{11}, 6  (1978), 1323-1358.

\bibitem{s3} W. R. Smythe;
\emph{Static and Dynamic Electricity} (3rd ed.),
New York: McGraw-Hill, 1968.

\bibitem{v1} N. Vasylyeva;
\emph{On solvability of the Hele-Shaw problem in weighted
H\"older spaces in a plane domain with a corner point}, UMB,
\textbf{2}, 3 (2005), 317-343.

\bibitem{z1} A. Zygmund;
\emph{Trigonometric series}, 2nd. ed. v. I,II,
Cambridge University Press, New York, 1959.

\end{thebibliography}
\end{document}