\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_{00,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_{0-1$, i.e. for$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 $$\label{e3.9} \lim_{t\to 0} \Delta_{-2,s}=0.$$ 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 $$\label{e3.10} \lim_{t\to 0} R_2(r,\varphi,t)=u_0(r,\varphi)$$ 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}=zthen \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 $$\label{e3.11} \sum_{k} \lim_{t\to 0}d_{k}=0.$$ 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*} whereD_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 forr^{-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} $$\label{e4.1} E_{n}(f)=O(1/n^{\alpha})$$ 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$$$\label{e4.2} |u_{n}(x)-f(x)|\leq6\omega(1/n).$$ \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 $$\label{e4.3} \max_{x,q} |f|\geq \textrm{const.}{\sum_{k}}|b_{k}(q)|.$$ 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 $$\label{e4.4} v(x,q)=Af(x,q)={\sum_{k}}\mu_{k}(b_{k}(q))\sin kx,\quad |\mu_{k}(b_{k}(q))|\leq M\max_q |b_{k}(q)|,$$ 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 $$\label{e4.5} T_{n}(x,q)-v(x,q)=c(n){\int_0^{\pi/2}} A\big\{f(x+2z,q)+f(x-2z,q)-2f(x,q)\big\} K(2z)dz.$$ The definition of the operatorA$together with the properties of the function$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 $$\label{e4.6} \langle Af(x,q)\rangle _{x}^{(\alpha)}\leq \textrm{const.}\langle f(x,q)\rangle _{x}^{(\alpha)}$$ 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 $$\label{e4.7} {\sum_{k}}(\underset{t}{\max} |b_{k}(t)|+\langle b_{k} \rangle ^{\beta}_{t})< \infty.$$ 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 $$\label{e4.8} \max_{x,t_{1},t_{2}} \frac{|T_{n}(x,t_{1})-T_{n}(x,t_{2})|} {|t_{1}-t_{2}|^{\beta}}\leq \textrm{const.}\max_{x,t_{1},t_{2}} \frac{|f(x,t_{1})-f(x,t_{2})|}{|t_{1}-t_{2}|^{\beta}}.$$ If one passes to a limit in \eqref{e4.8} asn\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 $$\label{e4.9} \langle Af\rangle _{x}^{(\alpha)}\leq \textrm{const.}\langle f(x,t)\rangle _{x}^{(\alpha)},\quad\langle Af\rangle _{t}^{(\beta)}\leq \textrm{const.}\langle f(x,t)\rangle _{t}^{(\beta )}.$$ \end{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 $$\label{e5.1} \langle \frac{\partial^{2} R_1 }{\partial\varphi^{2}}\rangle _{\varphi;s+2,G_{T}}^{(\alpha)} +\sum_{k}\lambda_{k}^{2}\max_{\bar{\mathbb{R}}_{T}} r^{-s-2}|R_{1,k}(r,t)|\leq \textrm{const.}\langle f\rangle _{\varphi;s,G_{T}}^{(\alpha)}.$$ \end{lemma} \begin{proof} After a formal differentiation with respect to$\varphi$one can obtain $$\label{e5.2} \frac{\partial^{2} R_1 }{\partial\varphi^{2}}=- {\sum_{k}} \lambda_{k}^{2}\sin(\lambda_{k}\varphi)\int_0^{t}d\tau\int _0^{\infty}L_{k}(\rho,r,t-\tau)b_{k}(\rho,\tau)d\rho$$ 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}, $$\label{e5.3} |B_{k}|\leq \textrm{const.}r^{2+s}\underset{r,t}{\max}r^{-s}|b_{k}|$$ 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$r. Moreover, \begin{gather} \label{e5.4} \sum_{k}\lambda_{k}^{2}\langle R_{1,k}\rangle _{t;s+2-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}\leq \textrm{const.}\langle f\rangle _{t;s-\alpha,G_{T} }^{(\alpha/2)}, \\ \label{e5.5} \sum_{k}\lambda_{k}^{2}\langle R_{1,k}\rangle _{r;s+2-\alpha,\mathbb{R}_{T}}^{(\alpha)}\leq \textrm{const.}\langle f\rangle _{r;s-\alpha,G_{T} }^{(\alpha)}, \\ \label{e5.6} \begin{aligned} &[\frac{\partial^{2} R_1 }{\partial\varphi^{2}} ]_{\varphi,t;s+2-\alpha,G_{T}}^{(\alpha,\alpha/2)} +\sum_{k}\lambda_{k}^{2}[ R_{1,k}]_{r,t;s+2-2\alpha,\mathbb{R}_{T}}^{(\alpha,\alpha/2)}\\ & \leq \textrm{const.}([f]_{\varphi,t;s-\alpha,G_{T}}^{(\alpha,\alpha/2)} +[f]_{r,t;s-2\alpha,G_{T}}^{(\alpha,\alpha/2)}). \end{aligned} \end{gather} \end{lemma} \begin{proof} The proof of estimates \eqref{e5.4} and \eqref{e5.5} follows from the properties of the functionf(r,\varphi,t)$, 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 $$\label{e5.7} v_{k}(r,t)=\frac{\partial}{\partial t}\int_0^{t}d\tau \int_0^{\infty}d\rho L_{k}(\rho,r,\tau)\int_0^{\theta}\frac {2}{\theta}f(\rho,\psi,t-\tau)\sin(\lambda_{k}\psi)d\psi.$$ 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 $$\label{e5.8} v_{k}(r,t)=\int_0^{t}d\tau\int_0^{\infty}d\rho\frac{\partial L_{k}}{\partial\tau}(\rho,r,\tau)b_{k}(\rho,t-\tau)+\underset{\varepsilon\to 0}{\lim}\int_0^{\infty}d\rho L_{k}(\rho,r,\varepsilon)b_{k}(\rho,t).$$ 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 \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} \begin{lemma} \label{lem5.3} The following estimate holds $$\label{e5.13} \sum_{k}\max_{\overline{\mathbb{R}}_{T}} r^{-s}|\frac{\partial R_{1,k}}{\partial t}|\leq \textrm{const.}\|f\| _{P_{s}^{\alpha,\alpha/2}(G_{T})}\,.$$ \end{lemma} \begin{proof} First we justify the estimate $$\label{e5.14} |\frac{\partial R_{1,k}}{\partial t}| =\big|\int ^{t}_{-\infty}d\tau \int^{\infty}_0d\rho \frac{\partial L_{k}}{\partial t}(\rho,\varphi,t-\tau)[b_{k}(\rho,\tau)-b_{k}(\rho,t)]\big| \leq \textrm{const.}\langle b_{k}\rangle^{(\alpha/2)}_{t,s-\alpha}r^{s}$$ 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 functionI_{\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 \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} 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 $$\label{e5.16} |M_{1}|\leq r^{s}\frac{1}{\lambda_{k}^{\alpha/2-\mu}},\quad \mu<\alpha/2.$$ 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 $$\label{e5.17} \int ^{\infty}_0d\tau \int _0^{\infty}d\rho \rho^{s-\alpha}\tau^{\alpha/2}(m_{1}(\rho,r,\tau)+m_{3}(\rho,r,\tau))\leq \textrm{const.} r^{s}\frac{1}{\lambda_{k}^{\alpha/2-\mu}}.$$ 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 $$\label{e5.18} \int ^{\infty}_0d\tau \int _0^{\infty}d\rho \rho^{s-\alpha}\tau^{\alpha/2}(m_{2}(\rho,r,\tau)-m_{4}(\rho,r,\tau)) \leq \textrm{const.} r^{s}\frac{1}{\lambda_{k}^{\alpha/2-\mu}}.$$ 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, \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} 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 $$\label{e5.20} r^{-s}\frac{\partial R_1}{\partial t}(r,\varphi,t)\in C_{\varphi}^{\alpha },\quad\langle \frac{\partial R_1}{\partial t}\rangle _{\varphi;s,G_{T}}^{(\alpha)}\leq \textrm{const.}[f]_{\varphi,t;s-\alpha,G_{T}} ^{(\alpha,\alpha/2)}.$$ 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$\bigtriangleup t=t_{2}-t_{1}. We have \label{e5.21} \begin{aligned} &\frac{\partial R_1}{\partial t}(r,\varphi,t_{2}) -\frac{\partial R_1}{\partial t}(r,\varphi,t_{1})\\ &= {\sum_{k}} \sin(\lambda_{k}\varphi) \int_{2t_{1}-t_{2}}^{t_{2}}d\tau\int _0^{\infty}d\rho\frac{\partial L_{k}}{\partial\tau}(\rho,r,t_{2} -\tau)[b_{k}(\rho,\tau)-b_{k}(\rho,t_{2})]\\ &\quad - {\sum_{k}} \sin(\lambda_{k}\varphi)\int_{2t_{1}-t_{2}}^{t_{1}}d\tau \int _0^{\infty}d\rho\frac{\partial L_{k}}{\partial\tau}(\rho,r,t_{1} -\tau)[b_{k}(\rho,\tau)-b_{k}(\rho,t_{1})]\\ &\quad + {\sum_{k}} \sin(\lambda_{k}\varphi)\int_{-\infty}^{2t_{1}-t_{2}}d\tau \int_0^{\infty}d\rho[ b_{k}(\rho,\tau)-b_{k}(\rho ,t_{1})]\\ &\quad\times [ \frac{\partial L_{k}}{\partial\tau}(\rho,r,t_{2}-\tau )-\frac{\partial L_{k}}{\partial\tau}(\rho,r,t_{1}-\tau)] \\ &\quad + {\sum_{k}} \sin(\lambda_{k}\varphi)\int_{-\infty}^{2t_{1}-t_{2}}d\tau\int_0^{\infty }d\rho[ b_{k}(\rho,t_{1})-b_{k}(\rho,t_{2})] \frac{\partial L_{k}} {\partial\tau}(\rho,r,t_{2}-\tau) \\ &= {\sum_{i=1}^{4}}{\sum_{k}} \sin(\lambda_{k}\varphi){\sum_{j=1}^{2}}A_{j,k}^{(i)}, \end{aligned} whereA_{1,k}^{(i)}$,$i=\overline{1,4}$, 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 \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} The estimate ofA_{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$\overline{t}-2t_{1}+t_{2}\geq\Delta t, thus, Lemma \ref{lem3.2} gives \begin{align*} A_{1,k}^{(3,1)} & \leq(\Delta t)\langle b_{k}\rangle _{t,s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}\int_{-\infty}^{2t_{1}-t_{2}}d\tau \int_0^{\infty}d\rho\frac{\rho^{s+1-\alpha}}{(\overline{t} -\tau)^{3-\alpha/2}}e^{-\frac{\rho^{2}+r^{2}}{4(\overline{t}-\tau)} }I_{\lambda_{k}}(\frac{\rho r}{2(\overline{t}-\tau)})\\ & \leq \textrm{const.}r^{s-\alpha}(\Delta t)^{\alpha/2}\langle b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}. \end{align*} The estimate ofA_{j,k}^{(4)}$in \eqref{e5.21} is obtained simultaneously for$j=1$and$j=2. We have \begin{align*} |{\sum_{j=1}^{2}} A_{j,k}^{(4)}| & =\big|\int_{-\infty}^{2t_{1}-t_{2}} d\tau\int_0^{\infty}d\rho[ b_{k}(\rho,t_{1})-b_{k}(\rho ,t_{2})]\frac{\partial L_{k}}{\partial\tau}(\rho,r,t_{2}-\tau)\big| \\ & =\big|\int_0^{\infty}d\rho[ b_{k}(\rho,t_{1})-b_{k} (\rho,t_{2})]\int_{-\infty}^{2t_{1}-t_{2}}\frac{\partial L_{k}} {\partial\tau}(\rho,r,t_{2}-\tau)d\tau\big| \\ & =\big|\int_0^{\infty}d\rho[ b_{k}(\rho,t_{1})-b_{k} (\rho,t_{2})]\frac{\rho}{2\Delta t}e^{-\frac{\rho^{2}+r^{2}}{4\Delta t} }I_{\lambda_{k}}(\frac{\rho r}{2\Delta t})\big| \\ &\leq \textrm{const.}r^{s-\alpha}(\Delta t)^{\alpha/2}\langle b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}, \end{align*} where Lemma \ref{lem3.2} has been applied. The coefficientsA^{(i)}_{j,k}$,$i=\overline{1,3}$,$j=2$, are evaluated similarly. Thus, the above gives an estimate for all$i=\overline{1,4}$, and$j=1,2$$|A_{j,k}^{(i)}|\leq \textrm{const.}r^{s-\alpha}(\Delta t)^{\alpha/2}\langle b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}.$ This inequality together with the convergence of$ {\sum_{k}} \langle b_{k}\rangle _{t;s-\alpha,\mathbb{R}_{T}}^{(\alpha/2)}$lead to $$\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)}$$ 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 ash>0\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} Letr_{h}=r+h$. One can easy estimate the coefficients$V_{1,k}^{j}$,$j=1,2with 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 estimateV_{1,k}^{3}in \eqref{e5.24}, we apply the mean value theorem. We have \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} 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, \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} since the onlyh\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 $$\label{e5.27} \sum_{k}\langle \frac{\partial R_{1,k}}{\partial t}\rangle _{r;s-\alpha ,\mathbb{R}_{T}}^{(\alpha)}\leq \textrm{const.}\langle f\rangle_{r,t;s-\alpha,G_{T} }^{(\alpha,\alpha/2)}.$$ \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 off(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 valueA_{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 $$\label{e5.28} [\frac{\partial R_1 }{\partial t}]_{\varphi,t;s-\alpha,G_{T}} ^{(\alpha,\alpha/2)}\leq \textrm{const.}[f]_{\varphi,t;s-\alpha,G_{T}}^{(\alpha ,\alpha/2)}.$$ \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 $$\label{e6.3} \frac{\partial u}{\partial t}-\Big( \frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial u}{\partial r}+\frac{b}{r}\frac{\partial u}{\partial r}\Big)-\frac{1}{r^{2}}\Big( \frac{\partial^{2} u}{\partial \varphi^{2}}+b\frac{\cos \varphi}{\sin \varphi}\frac{\partial u}{\partial \varphi}\Big)=f(r,\varphi,t),\quad (r,\varphi,t)\in G_{T}.$$ 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$$v_{1}(\varphi)=J_0(\lambda_{k}\varphi),\quad v_{2}(\varphi)=N_0(\lambda_{k}\varphi),$ where$J_{\nu}(x)$and$N_{\nu}(x)$are the Bessel functions of the first and second kind. The Bessel functions$J_{\nu}(x)$has the power series representation $J_{\nu}(x)=\frac{x^{\nu}}{2^{\nu}}\sum^{\infty}_{k=0}(-1)^{k}\frac{x^{2k}}{2^{2k}k!\Gamma(\nu+k+1)}.$ In view of this expansion the eigenfunctions$v_{2}(\phi)$for$b\neq 1$and$v_{1}(\phi)$for$b= 1$are appropriate for our purpose. They have the bounded second derivative and satisfy the first boundary condition in \eqref{e6.5}. To satisfy the second one, we define$\lambda=\lambda_{k}$as the solutions of the equation$J_{-q/2}(\lambda_{k}\theta)=0$,$k=1,2,\dots$. We will say about the case$b\neq 1$, the case$b=1$can be studied similarly. The formal solution of problem \eqref{e6.1}, \eqref{e6.2} is represented as $$\label{e6.6} u(r,\varphi,t)= R_{1b}(r,\varphi,t) + R_{2b}(r,\varphi,t),$$ 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 $$\label{e6.7} f(r,\theta,t)=0$$ is necessary for the solvability of the problem in$P^{2+\alpha,(2+\alpha)/2}_{s,b}(\overline{G}_{T})$. \begin{theorem} \label{thm6.1} Assume the consistency conditions of the first order and condition \eqref{e6.7} are fulfilled. The functions$f\in\widehat{ P}^{\alpha,\alpha/2}_{s,b}(\overline{G}_{T})$and$u_0\in \widehat{P}^{2+\alpha}_{s+2,b}(\overline{G})$Then there exists a unique solution$u(r,\varphi,t)\in \widehat{P}^{2+\alpha,\frac{2+\alpha}{2}}_{s,b}(\overline{G}_{T})and \label{e6.8} \begin{aligned} &\|u\|_{P_{s+2,b}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}+S(u) _{\widehat{P}_{s+2,b}^{2+\alpha,(2+\alpha)/2}(\overline{G}_{T})}\\ &\leq \textrm{const.}(\|f\| _{P_{s,b}^{\alpha,\alpha/2}(\overline{G}_{T})}+\| u_0\| _{P_{s+2,b}^{2+\alpha}(\overline{G})} +S(f) _{\widehat{P}_{s,b}^{\alpha,\alpha/2}(\overline{G}_{T})}+S(u_0) _{\widehat{P}_{s+2,b}^{2+\alpha}(\overline{G})}), \end{aligned} where the constant in \eqref{e6.8} is independent ofu(r,\varphi,t)$,$\alpha\in (0,1)$and$s+2<(\lambda_{1}^{2}+b^{2}/4)^{1/2}$,$\lambda_{1}\theta$is the smallest root of the equation$J_{-q/2}(\lambda_{k}\theta)=0$. \end{theorem} In general, the proof of Theorem \ref{thm6.1} repeats our arguments from the proof of Theorem \ref{thm2.1}. We note only that if$k>>1$, $J_{-q/2}(\lambda_{k}\varphi)\sim \sqrt{\frac{1}{\lambda_{k}\varphi}} \cos(\lambda_{k}\varphi+ \pi(q-1)/4),\quad \lambda_{k}\theta\sim (k-(q+1)/4)\pi+O(1/k),$ that gives the possibility to apply here the theorems from the trigonometric series theory. \section{Appendix} \subsection{Formal representation of the solution to \eqref{e2.3}} To obtain the formal solution of \eqref{e2.3}, we applied the method of the separation of the variables. In detail, one consists in the following. Let us consider case of$u_0(r,\varphi)\equiv 0$(this case corresponds to$ R_2(r,\varphi,t)=0$in \eqref{e2.4}). We look for the solution$u(r,\varphi,t)$of the problem $$\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}$$ as $$\label{e7.2} u(r,\varphi,t)=\sum_{k}V_{k}(r,t)\Phi_{k}(\varphi).$$ After the substitution of the function$V_{k}(r,t)\Phi_{k}(\varphi)$into the homogenous equation and boundary condition from \eqref{e7.1}, we obtain \begin{gather} \label{e7.3} r^{2}\frac{\frac{\partial V_{k}}{\partial t}-\frac{1}{r}\frac{\partial}{\partial r}r\frac{\partial V_{k}}{\partial r}}{V_{k}} =\frac{\frac{\partial ^{2}\Phi_{k}}{\partial \varphi^{2}}}{\Phi_{k}}\equiv -\lambda_{k}^{2}, \\ \label{e7.4} \Phi_{k}(0)=\Phi_{k}(\theta)=0. \end{gather} Conditions \eqref{e7.3} and \eqref{e7.4} lead to the function$\Phi_{k}$being the solution of the problem $$\label{e7.5} \begin{gathered} \Phi''_{k}(\varphi)+\lambda_{k}^{2}\Phi_{k}=0,\\ \Phi_{k}(0)=\Phi_{k}(\theta)=0. \end{gathered}$$ The solution of \eqref{e7.5} is the function $$\label{e7.6} \Phi_{k}=\sin\lambda_{k}\varphi,\quad \lambda_{k}=\pi k/\theta,\; k=1,2\dots.$$ Now we return to problem \eqref{e7.1} and represent$f(r,\varphi,t)$as $$\label{e7.7} f(r,\varphi,t)=\sum_{k}b_{k}(r,t)\sin\lambda_{k}\varphi,$$ with $$\label{e7.8} b_{k}(r,t)=\frac{2}{\theta}\int^{\theta}_0f(r,\psi,t)\sin\lambda_{k}\psi d\psi.$$ After that we substitute \eqref{e7.2}, \eqref{e7.7} and \eqref{e7.8} to the equation and the initial condition of \eqref{e7.1} and have $$\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}$$ Here we use that the function$\Phi_{k}(\varphi)$satisfies the equation in \eqref{e7.5}. Let us denote the Hankel transformation (see, for example, \cite{g2,p1,s3} for discussion) with respect to$r$of the functions$V_{k}(r,t)$and$b_{k}(r,t)$by$\widehat{V}_{k}(\mu,t)$and$\widehat{b}_{k}(\mu,t)$, respectively,$\mu$is the parameter under the transformation: $$\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}$$ where$J_{\lambda_{k}}(\mu r)$is the Bessel function \cite{g2}. Expressions \eqref{e7.9} and \eqref{e7.10} lead to the following problem for the function$\widehat{V}_{k}(\mu,t)$, $$\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}$$ It is easy to check that the function $$\label{e7.12} \widehat{V}_{k}(\mu,t)=\int^{t}_0e^{-\mu^2(t-\tau)} \widehat{b}_{k}(\mu,\tau)d\tau$$ gives the solution of problem \eqref{e7.11}. After applying the inverse Hankel transformation in \eqref{e7.12}, we obtain $$\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.$$ Then the formal solution \eqref{e7.1} follows from \eqref{e7.2}, \eqref{e7.6} and \eqref{e7.13}, so, $$\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.$$ To obtain formula \eqref{e2.7}, we transform \eqref{e7.14} applying formula \cite[6.633(2)]{g2}: $\int_0^{\infty}\mu e^{-\mu^2(t-\tau)} J_{\lambda_{k}}(\mu r)J_{\lambda_{k}}(\mu \rho)d\mu=\frac{1}{2(t-\tau)}I_{\lambda_{k}}\Big(\frac{r\rho}{2(t-\tau)}\Big) \exp\Big(-\frac{r^{2}+\rho^{2}}{4(t-\tau)}\Big)$ where$I_{\lambda_{k}}(x)is the modified Bessel function. Thus, \begin{align*} &u(r,\varphi,t)\\ &=\sum_{k=1}\sin\lambda_{k}\varphi\int^{t}_0d\tau\int^{\infty}_0d \rho{b}_{k}(\rho,\tau)\rho \int^{\infty}_0 \mu J_{\lambda_{k}}(\mu r)e^{-\mu^2(t-\tau)}J_{\lambda_{k}}(\mu \rho) d\mu\\ &=\sum_{k=1}\sin\lambda_{k}\varphi\int^{t}_0d\tau \int^{\infty}_0d \rho{b}_{k}(\rho,\tau) \frac{\rho}{2(t-\tau)}I_{\lambda_{k}} \Big(\frac{r\rho}{2(t-\tau)}\Big)\exp\Big(-\frac{r^{2}+\rho^{2}}{4(t-\tau)}\Big). \end{align*} That gives \eqref{e2.6}, \eqref{e2.7} with R_2\equiv 0$(due to$u_0\equiv 0$). To obtain the complete formula \eqref{e2.6}; i.e., with$ R_2\neq 0$, it is enough to consider the problem $$\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}$$ and apply all reasoning mentioned above to this problem. After that, the solution of \eqref{e2.3} is represented as $$\label{e7.16} u(r,\varphi,t)= R_1 (r,\varphi,t)+ R_2(r,\varphi,t)$$ where$ R_1 (r,\varphi,t)$and$ R_2(r,\varphi,t)$are the solutions of \eqref{e7.1} and \eqref{e7.15}, correspondingly. $$\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}$$ $$\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.$$ Equation \eqref{e2.6} implies that the desired solution is the sum of the volume potential$ R_1 (r,\phi,t)$and the potential of the initial data$ R_2(r,\phi,t)$. The representation for$ R_1 (r,\varphi,t)$can be rewritten also as $$\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),$$ 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),$ whereB(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, \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} Forq\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 functionF(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;q^{2})$is bounded (see \cite[9.102]{g2}) so that $$\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}.$$ For$q\in (1+\varepsilon,\infty), \label{e7.22} \begin{aligned} d_{3}&=\int^{\infty}_0dz z ^{1-\alpha}J_{\lambda_{k}} (zq)J_{\lambda_{k}}(z)\\ &=\frac{q^{-\lambda_{k}+\alpha-2}\Gamma(\lambda_{k} +1-\alpha/2)}{2^{\alpha-1}\Gamma(\lambda_{k}+1)\Gamma(\alpha/2)} F(\lambda_{k}+1-\alpha/2,1-\alpha/2;\lambda_{k}+1;q^{-2})\\ &\leq \textrm{const.} q^{-\lambda_{k}+\alpha-2}\lambda_{k}^{-\alpha/2}. \end{aligned} Estimates \eqref{e7.21} and \eqref{e7.22} lead to $$\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}.$$ 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 usedq=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 usedq=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 $$\label{e7.24} |M^{(2)}_{1}|=|M^{(2,1)}_{1}+M^{(2,2)}_{1}|\leq \textrm{const.} \lambda_{k}^{-1+\mu-\alpha/2}.$$ Finally, the following inequality follows from \eqref{e7.20}, \eqref{e7.23} and \eqref{e7.24}: $$\label{e7.25} |M_{1}|\leq \textrm{const.}\frac{r^{s}}{\lambda_{k}^{\alpha/2-\mu}}$$ 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$z=uvso that \begin{align*} A(v)&=v^{-\beta-1}\int _0^{\infty} z^{\beta}e^{-\gamma(\frac{z}{v}-v)^{2}}I_{\lambda_{k}}(2z)e^{-2z}dz\\ &=v^{-\beta-1}\int_0^{1} z^{\beta}e^{-\gamma(\frac{z}{v}-v)^{2}}I_{\lambda_{k}}(2z) e^{-2z}dz\\ &\quad +v^{-\beta-1}\int _{1}^{\infty} z^{\beta}e^{-\gamma(\frac{z}{v}-v)^{2}}I_{\lambda_{k}}(2z)e^{-2z}dz\\ &\equiv A_{1}(v)+A_{2}(v). \end{align*} To estimateA_{2}(v)$, 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 $\varphi(\xi,v)=(\xi+v)^{\beta-1/2}e^{-\gamma\xi^{2}/2}\leq \textrm{const.} v^{\beta-1/2}$ underv\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$v<1. Therefore, \begin{align*} A_{2}(v)&\leq \textrm{const.} v^{-1-\beta}\int _{1}^{\infty}z^{\beta-1/2}e^{-\gamma\frac{z^{2}}{2v^{2}}} e^{-\gamma\frac{z^{2}}{2v^{2}}}e^{2z}e^{-\gamma v^{2}}dz\\ &\leq \textrm{const.} v^{-1-\beta}e^{-\frac{\gamma}{2 v^{2}}} \int_{1}^{\infty}z^{\beta-1/2}e^{-\gamma\frac{z^{2}}{2}+2z}dz\\ &\leq \textrm{const.} v^{-1-\beta}e^{-\frac{\gamma}{2 v^{2}}} \leq \textrm{const.} v^{-1+\beta} \end{align*} forv<1$. 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*} forv\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. 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