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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]  %Headings in Russian
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
     {\large Siberian Electronic Mathematical Reports} \\[1mm]
     {\LARGE\tt{http://semr.math.nsc.ru}}\\[0.5mm]
%     {\small Институт математики им. С.Л.Соболева СО РАН}\\[-1mm]
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     \noindent
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     {\flushleft\it Том 16, стр. 144--144 (2019) \hspace{65mm}{\rm\small УДК \udcs}} %Volume, pages in Russian,
                                                                                    %filled by Editorial board
     \newline
         {\rm\small DOI~10.33048/semi.2019.16.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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\begin{document}
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\title[A MIXED PROBLEM FOR THE THIRD EQUATION ORDER]{A MIXED PROBLEM FOR THE THIRD EQUATION ORDER WITH BRIEF CHARACTERISTICS}
\author{{Yu. P. Apakov, A. Kh. Zhuraev}}%
\address{Yusupjon Pulatovich Apakov
\newline\hphantom{iii} Namangan engineering-construction institute, Institute of Mathematics Namangan refional department Uzbekistan Academy of Sciences.
\newline\hphantom{iii} Ziyekor st, 8,
\newline\hphantom{iii} 160103, Namangan, Uzbekistan}%
\email{yusupjonapakov@gmail.com}%
\address{Abdulla Hatamovich Juraev
\newline\hphantom{iii} Namangan engineering-construction institute.
\newline\hphantom{iii} I.Karimov st, 12,
\newline\hphantom{iii} 160103, Namangan, Uzbekistan}%
\email{juraevabdulla@gmail.com}%
\thanks{\sc Yu. P. Apakov, A. Kh. Zhuraev.,
A MIXED PROBLEM FOR THE THIRD EQUATION ORDER WITH BRIEF CHARACTERISTICS}
\thanks{\copyright \ 2020 Yu. P. Apakov, A. Kh. Zhuraev}
\thanks{\it Received   2020, published  }%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } The unique solvability of the mixed problem in a rectangular domain is proven for a third-order equation with multiple characteristics containing the second derivative of time.
\medskip

\noindent{\bf Keywords:} third-order equations with multiple characteristics, mixed problem, uniqueness, existence, series, regular convergence.
 \end{quote}
}


\section{Introduction}
In working [1], taking into account the properties of viscosity and thermal conductivity of a gas, a third-order equation with multiple characteristics containing second time derivatives was obtained from the Navier-Stokes system
$${u_{xxx}} + {u_{yy}} - \frac{\nu }{y}{u_y} = {u_x}{u_{xx}},\,\,\,\,\,\nu  = const.\eqno(1)$$
Equation (1) under $\nu  = 1$ describes an axisymmetric flow, and under $\nu  = 0$ describes a plane-parallel flow [2].

Third-order partial differential equations are encountered in the solution of problems of the theory of nonlinear acoustics, hydrodynamic theory of cosmic plasmas, and filtration of liquids in porous media. In the collection of all third-order equations, an important place is occupied by so-called equations with multiple characteristics.\\
The first results on a third-order equation with multiple characteristics were obtained by H. Block [3],  E. Del Vecchio [4].
Catabriga [5] constructed a fundamental solution of the equation  $D_x^{2n + 1}u - D_y^2u = 0$ in the form of a double improper integral and studied the properties of the potential. In [6, 7], the fundamental solutions expressed via hypergeometric functions were constructed, their properties were studied, and the estimates as  $\left| t \right| \to \infty $  were established. We also mention the works [8–19] in which boundary-value problems for third-order equations were considered.
In the present paper, we study the third-order boundary-value problem for a third-order equation with multiple characteristics in a rectangular domain. Consider an equation
$${{\text{U}}_{xxx}} - {U_{yy}} + A{U_{xx}} + B{U_x} + C{U_y} + DU = 0$$
  where
$$A,\,B,C,D \in R$$
note that by replacing variables
$$U\left( {x,y} \right) = u\left( {x,y} \right){e^{ - \frac{A}{3}x + \frac{C}{2}y}}$$
this equation is reduced to the form:
$${u_{xxx}} - {u_{yy}} + a{u_x} + cu = 0,\eqno(2)$$
 where
$$a =  - \frac{{{A^2}}}{3} + B,\,\,\,\,\,\,\,\,\,c = \frac{{2{A^3}}}{{27}} + \frac{{{C^2}}}{4} - \frac{{AB}}{3} + D.\,\,\,\,$$
In what follows, we assume that $c > 0,\,\,\,a > 0.$\\
In the domain $D = \left\{ {\left( {x,y} \right):0 < x,y < 1} \right\}$  we consider the following problem:

Problem${B_3}$.   In the domain $D$, to find the solution of Eq. (2) from the class\\
$C_{x,y}^{3,2}\left( D \right) \cap C_{x,y}^{2,1}\left( {\overline D } \right),$ satisfying the boundary condition:
$$\left\{ \begin{gathered}
  \alpha u(x,0) + \beta {u_y}(x,0) = 0 \hfill \\
  \gamma u(x,1) + \delta {u_y}(x,1) = 0 \hfill \\
\end{gathered}  \right.\,\,,\,\,\,\,\,\,\,0 \leqslant x \leqslant 1\,\,\,\,\eqno(3)$$
\[u(0,y) = {\varphi _1}(y),\,\,\,\,\,{u_x}(1,y) = {\varphi _2}(y),\,\,\,\,\,\,{u_{xx}}(1,y) = {\varphi _3}(y)\eqno(4)\]
where\\
${\varphi _i}(y) \in {C^3}[0,1],\,\,\,\,i = 1,3,\,$ are given functions and, moreover:
\[\beta \,{\varphi '_i}(0) = 0,\,\,\,\,\,\,\,\gamma \,{\varphi _i}(1) + \delta \,{\varphi '_i}(1) = 0,\,\,\,\,\,\varphi ''(0) = \varphi ''(1) = 0.\]

Here  $\alpha ,\beta ,\gamma ,\delta $ - are constants, and ${\alpha ^2} + {\beta ^2} \ne 0,\,\,{\gamma ^2} + {\delta ^2} \ne 0.$\\
Note that, similar problems were investigated in [20] for $a = c = 0.$

\section{Uniqueness of results}

\begin{theorem}
If $\alpha \beta  \leqslant 0,\,\,\,\,\,\,\delta \gamma  \geqslant 0,$  then problem  $B_{3}$ is uniquely solvable.
\end{theorem}

\begin{proof}
Assume the contrary, let $u_{1}(x,y)$  and $u_{2}(x,y)$  be solutions of $B_{3}$. Then  $u(x,y)=u_{1}(x,y)-u_{2}(x,y)$  is a solution of homogenous problem   $B_{3}$  in domain $D$ .
Consider the identity
$$u\left( {{u_{xxx}} - {u_{yy}} + a{u_x} + cu} \right) = 0,$$
 or
\[\frac{\partial }{{\partial x}}\left( {u{u_{xx}}} \right) - \frac{1}{2}\frac{\partial }{{\partial x}}\left( {u_x^2} \right) - \frac{\partial }{{\partial y}}\left( {u{u_y}} \right) + u_y^2 + \frac{a}{2}\frac{\partial }{{\partial x}}\left( {{u^2}} \right) + c{u^2} = 0.\]
Integrating this identity over the domain $D$, we get
\[\iint\limits_D {\frac{\partial }{{\partial x}}\left( {u{u_{xx}}} \right)}\,dxdy - \frac{1}{2}\iint\limits_D {\frac{\partial }{{\partial x}}\left( {u_x^2} \right)}\,dxdy - \iint\limits_D {\frac{\partial }{{\partial y}}\left( {u{u_y}} \right)}\,dxdy + \iint\limits_D {u_y^2}dxdy + \]
\[ + \frac{a}{2}\iint\limits_D {\frac{\partial }{{\partial x}}\left( {{u^2}} \right)dxdy + c\iint\limits_D {{u^2}dxdy}} = 0.\]
Using the Green formula, we obtain
\[\,\,\int\limits_0^1 {\left[ {u\left( {1,y} \right){u_{xx}}\left( {1,y} \right) - u\left( {0,y} \right){u_{xx}}\left( {0,y} \right)} \right]dy - \frac{1}{2}} \int\limits_0^1 {\left[ {u_x^2\left( {1,y} \right) - u_x^2\left( {0,y} \right)} \right]dy - } \]
\[ - \int\limits_0^1 {\left[ {u\left( {x,1} \right){u_y}\left( {x,1} \right) - u\left( {x,0} \right){u_y}\left( {x,0} \right)} \right]} dx + \iint\limits_D {u_y^2\left( {x,y} \right)dxdy} + \]
\[ + \frac{a}{2}\int\limits_0^1 {\left[ {{u^2}\left( {1,y} \right) - {u^2}\left( {0,y} \right)} \right]} dy + c\iint\limits_D {{u^2}\left( {x,y} \right)dxdy = 0}.\]

In view of the boundary conditions (3), under the assumption that $\alpha  \ne 0,\,\,\,\gamma  \ne 0$  we obtain
\[\frac{\delta }{\gamma }\int\limits_0^1 {u_y^2\left( {x,1} \right)} dx - \frac{\beta }{\alpha }\int\limits_0^1 {u_y^2\left( {x,0} \right)\,} dx + \frac{a}{2}\int\limits_0^1 {u_{}^2\left( {1,y} \right)\,} dy + \]
\[ + \frac{1}{2}\int\limits_0^1 {u_x^2\left( {0,y} \right)dy + \iint\limits_D {u_y^2\left( {x,y} \right)dxdy + c\iint\limits_D {{u^2}\left( {x,y} \right)dxdy} = 0}} .\]
Takin into account the conditions of theorem and $a > 0,\,\,\,\,c > 0$ ,  we obtain $u\left( {x,y} \right) \equiv 0$, in $\left( {x,y} \right) \in \overline D $.\\
In the cases$\beta  \ne 0,\,\,\,\,\delta  \ne 0;\,\,\,\,\,\,\alpha  \ne 0,\,\,\delta  \ne 0;\,\,\,\,\,\gamma  \ne 0,\,\,\beta  \ne 0\,\,$ in a similar way, we obtain the equality $u\left( {x,y} \right) \equiv 0$ , in  $\left( {x,y} \right) \in \overline D $.
\end{proof}

\section{Solution existence}
We seek the solution of the problem by the method of separation of variables:
$$u(x,y) = X(x)Y(y).\eqno(5)$$
Thus, equation (2) implies that
$$\frac{{X''' + aX'}}{X} - \frac{{Y''}}{Y} =  - c,\,\,\,\,\,\,\,\,\,\,\,\,\, - c =  - v + \mu .$$
This yields
\[X''' + aX' + \nu X  = 0,\eqno(6)\]                                                                                                       \[Y'' + \mu \,Y = 0.\,\eqno(7)\]
To determine  we obtain problems of the Sturm-Louvilly type
\[\left\{ \begin{gathered}
  Y'' + \mu \,Y = 0, \hfill \\
  \alpha \,Y(0) + \beta \,Y'(0) = 0, \hfill \\
  \gamma \,Y(1) + \delta \,Y'(1) = 0. \hfill \\
\end{gathered}  \right.\eqno(8)\]
Problem (8) has eigenvalues [20] in the form of positive roots of the transcendental
\[ctg\sqrt \mu   = \frac{{\alpha \gamma  + \delta \beta \mu }}{{\sqrt \mu  \left( {\gamma \beta  - \alpha \delta } \right)}}\eqno(9)\]
Which implies that   $\sqrt {{\mu _n}}  = \pi n + {\varepsilon _n}$, where $\mathop {\lim }\limits_{n \to \infty } {\varepsilon _n} = 0,$  or${\mu _n} = {\rm O}\left( {{n^2}} \right),\,\,\,\,n \to \infty .$\\
The corresponding eigenfunctions take the form
$${Y_n}(y) = \left( {\alpha \sin \sqrt {{\mu _n}} y - \beta \sqrt {{\mu _n}} \cos \sqrt {{\mu _n}} y} \right).\eqno(10)$$
The orthogonality of the system of functions (10) is proved as in [20].\\
We now find the norm of the eigenfunctions ${Y_n}\left( y \right)$ in ${L_2}\left[ {0,1} \right]$ :
\[\,{\left\| {{Y_n}\left( y \right)} \right\|^2} = \int\limits_0^1 {Y_n^2\left( y \right)} dy = \frac{1}{2}\left( {{\alpha ^2} + {\beta ^2}{\mu _n} - \alpha \beta } \right) + \]
$$ + \left( {\frac{{{\beta ^2}\sqrt {{\mu _n}} }}{4} - \frac{{{\alpha ^2}}}{{4\sqrt {{\mu _n}} }}} \right)\sin 2\sqrt {{\mu _n}}  + \frac{{\alpha \beta }}{2}\cos 2\sqrt {{\mu _n}}  = \frac{1}{2}{\beta ^2}{\mu _n}\left( {1 + {\text{O}}\left( {\frac{1}{{\sqrt {{\mu _n}} }}} \right)} \right).\eqno(11)$$
The characteristic equation (7) takes the form:
 $${m^3} + am + \nu  = 0\eqno(12)$$
Equation (12)  has the roots [21]:
$${m_1} =  - 2\tau \,,\,\,\,\,\,\,\,\,\,\,{m_{2,3}} = \tau  \pm i\rho ,\,\,\,\,\,\,\,\,\tau  =  - \frac{1}{2}\left( {u + \upsilon } \right),\,\,\,\,\,\,\,\,\,\,\rho  = \frac{{\sqrt 3 }}{2}\left( {u - \upsilon } \right),$$
where
$$u = \sqrt[3]{{ - \frac{\nu }{2} + \sqrt {\frac{{{\nu ^2}}}{4} + \frac{{{a^3}}}{{27}}} }}\,,\,\,\,\,\,\,\,\,\,\upsilon  = \sqrt[3]{{ - \frac{\nu }{2} - \sqrt {\frac{{{\nu ^2}}}{4} + \frac{{{a^3}}}{{27}}} }},$$
since  $a > 0,\,\,\,\,\,\nu  > 0$ то $\frac{{{\nu ^2}}}{4} + \frac{{{a^3}}}{{27}} > 0$.\\
We write the general solution of Eq. (6) in the form
\[{X_n}\left( x \right) = {C_{1n}}{e^{ - 2{\tau _n}x}} + {e^{{\tau _n}x}}\left( {{C_{2n}}\cos {\rho _n}x + {C_{3n}}\sin {\rho _n}x} \right),\]
where    are arbitrary constants.  According to (5), the functions
\[{u_n}\left( {x,y} \right) = \left[ {{C_{1n}}{e^{ - 2{\tau _n}x}} + {e^{{\tau _n}x}}\left( {{C_{2n}}\cos {\rho _n}x + {C_{3n}}\sin {\rho _n}x} \right)} \right]{Y_n}\left( y \right),\]
are particular solutions of Eq. (2) satisfying conditions (3).\\
In view of the linearity and homogeneity of Eq. (2), the sum of particular solutions is also a particular solution

of Eq. (2). This is also true for the series
\[u\left( {x,y} \right) = \sum\limits_{n = 1}^\infty  {\left[ {{C_{1n}}{e^{ - 2{\tau _n}x}} + {C_{2n}}{e^{{\tau _n}x}}\cos {\rho _n}x + {C_{3n}}{e^{{\tau _n}x}}\sin {\rho _n}x} \right]{Y_n}\left( y \right)}. \eqno(13)\]
In view of the fact that the function u(x, y) satisfies the boundary conditions (4), we arrive at the system of Equations
\[\left\{ \begin{gathered}
  {A_{1n}} = {C_{1n}} + {C_{2n}}\,, \hfill \\
  {A_{2n}} =  - 2{\tau _n}{e^{ - 2{\tau _n}}}{C_{1n}} + {e^{{\tau _n}}}\left( {{\tau _n}cos{\rho _n} - {\rho _n}\sin {\rho _n}} \right){C_{2n}} + {e^{{\tau _n}}}\left( {{\tau _n}\sin {\rho _n} + {\rho _n}cos{\rho _n}} \right){C_{3n}}\,, \hfill \\
  {A_{3n}} = 4\tau _n^2{e^{ - 2{\tau _n}}}{C_{1n}} + {e^{{\tau _n}}}\left[ {\left( {\tau _n^2 - \rho _n^2} \right)cos{\rho _n} - 2{\tau _n}{\rho _n}\sin {\rho _n}} \right]{C_{2n}} +  \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {e^{{\tau _n}}}\left[ {\left( {\tau _n^2 - \rho _n^2} \right)\sin {\rho _n} + 2{\tau _n}{\rho _n}cos{\rho _n}} \right]{C_{3n}}, \hfill \\
\end{gathered}  \right.\eqno(14)\]
 where
 \[{A_{in}} = \frac{1}{{{{\left\| {{Y_n}} \right\|}^2}}}\int\limits_0^1 {{\varphi _i}\left( \eta  \right)} {Y_n}\left( \eta  \right)d\eta ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,i = 1,2,3.\]
This system has the determinant
\[\Delta  = {\tau _n}{e^{2{\tau _n}}}\left\{ {\left( {\frac{1}{2}\tau _n^2 - \frac{3}{2}\rho _n^2} \right)\sin 2{\rho _n} + {\rho _n}\left( {\frac{3}{2}{\tau _n} - {\rho _n}} \right)\cos 2{\rho _n} + } \right.\]
\[ + 2{e^{ - 3{\tau _n}}}\left. {\left[ {\left( {\tau _n^2 - \rho _n^2} \right)\sin {\rho _n} + 2{\tau _n}{\rho _n}\cos {\rho _n}} \right]} \right\},\]
Assume that $\Delta  \ne 0$. We prove the following lemma:
\begin{lemma}
The boundary-value problem
$$\left\{ \begin{gathered}
  X''' + aX' + \nu X = 0\,, \hfill \\
  X\left( 0 \right) = X'\left( 1 \right) = X''\left( 1 \right) = 0\,, \hfill \\
\end{gathered}  \right.\eqno(16)$$
possesses solely the trivial solution.
\end{lemma}
\begin{proof}
Assume the contrary, let   $X\left( x \right) \ne 0$.

Consider the identity
$$X\left( {X''' + aX' + \nu X} \right) = 0,$$
or
$${\left( {XX'' - \frac{1}{2}{{\left( {X'} \right)}^2} + \frac{a}{2}{X^2}} \right)^\prime } + \nu {X^2} = 0,$$
Integrating over the domain $\left( {0 < x < 1} \right)$,   we get
\[\int\limits_0^1 {{{\left( {XX'' - \frac{1}{2}{{\left( {X'} \right)}^2} + \frac{a}{2}{X^2}} \right)}^\prime }dx + \nu \int\limits_0^1 {{X^2}} } dx = 0,\]
\[X\left( 1 \right)X''\left( 1 \right) - X\left( 0 \right)X''\left( 0 \right) - \frac{1}{2}{\left( {X'\left( 1 \right)} \right)^2} + \frac{1}{2}{\left( {X'\left( 0 \right)} \right)^2} + \]
\[ + \frac{a}{2}{X^2}\left( 1 \right) - \frac{a}{2}{X^2}\left( 0 \right) + \nu \int\limits_0^1 {{X^2}dx}  = 0.\]
Given the boundary conditions, we obtain
\[\frac{1}{2}{\left( {X'\left( 0 \right)} \right)^2} + \frac{a}{2}{X^2}\left( 1 \right) + \nu \int\limits_0^1 {{X^2}dx}  = 0,\]
since $a > 0,\,\,\,\,\nu  > 0$,   then   $X\left( x \right) \equiv 0$.
\end{proof}
Assume that $\Delta  = 0$,  then there exist constants   $C_1^ * ,C_{^2}^ * ,C_{^3}^ * $   that are not simultaneously equal to zero and satisfy the following system of  equations:
\[\left\{ \begin{gathered}
  0 = C_{_1}^* + C_{_2}^*\,, \hfill \\
  0 =  - 2{\tau _n}{e^{ - 2{\tau _n}}}C_{_1}^* + {e^{{\tau _n}}}\left( {{\tau _n}cos{\rho _n} - {\rho _n}\sin {\rho _n}} \right)C_{_2}^* + {e^{{\tau _n}}}\left( {{\tau _n}\sin {\rho _n} + {\rho _n}cos{\rho _n}} \right)C_{_3}^*\,,\,\,\,\,\, \hfill \\
  0 = 4\tau _n^2{e^{ - 2{\tau _n}}}C_{_1}^* + {e^{{\tau _n}}}\left[ {\left( {\tau _n^2 - \rho _n^2} \right)cos{\rho _n} - 2{\tau _n}{\rho _n}\sin {\rho _n}} \right]C_{_2}^* +  \hfill \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {e^{{\tau _n}}}\left[ {\left( {\tau _n^2 - \rho _n^2} \right)\sin {\rho _n} + 2{\tau _n}{\rho _n}cos{\rho _n}} \right]C_{_3}^*, \hfill \\
\end{gathered}  \right.\]
This means that the function
\[{V_n}\left( x \right) = C_{^1}^ * {e^{ - 2{\tau _n}x}} + {e^{{\tau _n}x}}\left( {C_{^2}^ * \cos {\rho _n}x + C_{^3}^ * \sin {\rho _n}x} \right),\]
is a solution of the boundary-value problem (16), however, by the proved lemma:
\[C_{^1}^ * {e^{ - 2{\tau _n}x}} + C_{^2}^ * {e^{{\tau _n}x}}\cos {\rho _n}x + C_{^3}^ * {e^{{\tau _n}x}}\sin {\rho _n}x \equiv 0,\]
which is impossible in view of the linear independence of the functions
\[{e^{ - 2{\tau _n}x}},\,\,\,\,\,{e^{{\tau _n}x}}\cos {\rho _n}x,\,\,\,\,\,{e^{{\tau _n}x}}\sin {\rho _n}x.\]
So  $\Delta  \ne 0$.\\The system of equations (14) possesses a unique solution of the form
\[{C_{1n}} = \frac{1}{\Delta }\left[ {{e^{2{\tau _n}}}\left( {2\tau _n^2{\rho _n} - \rho _n^3} \right){A_{1n}} - {e^{{\tau _n}}}\left[ {\left( {\tau _n^2 - \rho _n^2} \right)\sin {\rho _n} + 2{\tau _n}{\rho _n}cos{\rho _n}} \right]{A_{2n}} + } \right.\]
\[\left. { + {e^{{\tau _n}}}\left( {{\tau _n}\sin {\rho _n} + {\lambda _n}cos{\rho _n}} \right){A_{3n}}} \right],\]
\[{C_{2n}} = \frac{1}{\Delta }\left[ {2{\tau _n}{e^{ - {\tau _n}}}\left[ {\left( {3\tau _{_n}^2 - \rho _n^2} \right)\sin {\rho _n} + 4{\tau _n}{\rho _n}\cos {\rho _n}} \right]{A_{1n}} + } \right.\]
\[ + \left. {{e^{{\tau _n}}}\left[ {\left( {\tau _{_n}^2 - \rho _n^2} \right)\sin {\rho _n} + 2{\tau _n}{\rho _n}\cos {\rho _n}} \right]{A_{2n}} - {e^{{\tau _n}}}\left( {{\tau _n}\sin {\rho _n} + {\rho _n}cos{\rho _n}} \right){A_{3n}}} \right],\]
\[{C_{3n}} = \frac{1}{\Delta }\left[ { - 2{\tau _n}{e^{ - {\tau _n}}}\left( {\left( {3\tau _n^2 - \rho _n^2} \right)\cos {\rho _n} - 4{\tau _n}{\rho _n}\sin {\rho _n}} \right){A_{1n}} + } \right.\]
\[ + \left( {4\tau _n^2{e^{ - 2{\tau _n}}} - {e^{{\tau _n}}}\left[ {\left( {\tau _n^2 - \rho _n^2} \right)cos{\rho _n} - 2{\tau _n}{\rho _n}\sin {\rho _n}} \right]} \right){A_{2n}}\, + \]
\[\left. { + \left[ {{e^{{\tau _n}}}\left( {{\tau _n}cos{\rho _n} - {\rho _n}\sin {\rho _n}} \right) + 2{\tau _n}{e^{ - 2{\tau _n}}}} \right]{A_{3n}}} \right]\,.\]
If series (13) and its derivatives ${u_{xxx}},\,\,{u_{yy\,}}$ are uniformly convergent in both variables in the domain D, then this series is the classical solution of problem ${B_3}$.
We now prove the uniform convergence of series (13) in the domain D.\\
By using (13) we get
\[\left| {u\left( {x,y} \right)} \right| \leqslant \sum\limits_{n = 1}^\infty  {\left[ {\left| {{C_{1n}}} \right|{e^{ - 2{\tau _n}x}} + \left| {{C_{2n}}} \right|{e^{{\tau _n}x}} + \left| {{C_{3n}}} \right|{e^{{\tau _n}x}}} \right]}  \cdot \left| {Y\left( y \right)} \right|.\eqno(17)\]
Given that
$$\Delta  = \beta {e^{2{\alpha _n}}}\bar \Delta ,\,\,\,\,\,\,\,\,\,\,\,\bar \Delta  = 1 - {e^{ - 3{\alpha _n}}}\left( {\frac{{3{\alpha _n}}}{{{\beta _n}}}\sin {\beta _n} + \cos {\beta _n}} \right),$$
  we have
$$\mathop {\lim }\limits_{n \to \infty } {\bar \Delta _n} = 1.$$
Estimate the first term
$$\left| {{C_{1n}}} \right|{e^{ - 2{\tau _n}x}} \leqslant \frac{{{e^{ - 2{\tau _n}x}}{e^{2{\tau _n}}}}}{{{\tau _n}{e^{2{\tau _n}}}\left| {\overline \Delta  } \right|}}\left\{ {\left| {2\tau _n^2 - \rho _n^3} \right|\left| {{A_{1n}}} \right| + {e^{ - {\tau _n}}}\left( {\left| {\tau _n^2 - \rho _n^2} \right| + 2\left| {{\tau _n}{\rho _n}} \right|} \right)\left| {{A_{2n}}} \right|} \right. + $$
$$\left. { + {e^{ - {\tau _n}}}\left( {\left| {{\tau _n}} \right| + \left| {{\rho _n}} \right|} \right)\left| {{A_{3n}}} \right|} \right\} \leqslant \left( {\left| {{A_{1n}}} \right| + \left| {{A_{2n}}} \right| + \left| {{A_{3n}}} \right|} \right){M_1},$$
similarly
$$\left| {{C_{2n}}} \right|{e^{{\tau _n}x}} \leqslant \left( {\left| {{A_{1n}}} \right| + \left| {{A_{2n}}} \right| + \left| {{A_{3n}}} \right|} \right)M_2^{},$$
$$\left| {{C_{3n}}} \right|{e^{{\tau _n}x}} \leqslant \left( {\left| {{A_{1n}}} \right| + \left| {{A_{2n}}} \right| + \left| {{A_{3n}}} \right|} \right){M_3}.$$
Equation (17) yields
$$\left| {u\left( {x,y} \right)} \right| \leqslant \sum\limits_{n = 1}^\infty  {\left( {\left| {{A_{1n}}{Y_n}\left( y \right)} \right| + \left| {{A_{2n}}{Y_n}\left( y \right)} \right| + \left| {{A_{3n}}{Y_n}\left( y \right)} \right|} \right)} {M_4}.\eqno(18)$$
Further, we estimate the quantity $\left| {{A_{in}}{Y_n}\left( y \right)} \right|$
$$\left| {{A_{in}}{Y_n}\left( y \right)} \right| \leqslant \frac{{\left| {{Y_n}\left( y \right)} \right|}}{{{{\left\| {{Y_n}} \right\|}^2}}}\int\limits_0^1 {{\varphi _i}\left( \eta  \right)} {Y_n}\left( \eta  \right)d\eta .$$
By using the unequality
$$\left| {{Y_n}(y)} \right| \leqslant \left| {\alpha \sin \sqrt {{\mu _n}} y - \beta \sqrt {{\mu _n}} \cos \sqrt {{\mu _n}} y} \right| \leqslant \left| \alpha  \right| + \left| \beta  \right|\sqrt {{\mu _n}}, $$
we obtain
$$\left| {{A_{in}}{Y_n}\left( y \right)} \right| \leqslant \frac{{{{\left( {\left| \alpha  \right| + \left| \beta  \right|\sqrt {{\mu _n}} } \right)}^2}}}{{{{\left\| {{Y_n}} \right\|}^2}}}\int\limits_0^1 {{\varphi _i}\left( \eta  \right)} d\eta .$$
The expression  $\frac{{{{\left( {\left| \alpha  \right| + \left| \beta  \right|\sqrt {{\mu _n}} } \right)}^2}}}{{{{\left\| {{Y_n}} \right\|}^2}}}$   is bounded as $n \to \infty $ , i.e. as ${\mu _n} \to \infty $, because it follows from (11) that
\[\frac{{{{\left( {\left| \alpha  \right| + \left| \beta  \right|\sqrt {{\mu _n}} } \right)}^2}}}{{{{\left\| {{Y_n}} \right\|}^2}}} = \frac{{{\alpha ^2} + 2\left| {\alpha \beta } \right|\sqrt {{\mu _n}}  + {\beta ^2}{\mu _n}}}{{{{\left\| {{Y_n}} \right\|}^2}}} \to 2.\]
We conclude that, starting from a certain number n, the following inequality is true:
\[\left| {{A_{in}}} \right|\left| {Y\left( y \right)} \right| < \int\limits_0^1 {\left| {{\varphi _i}\left( \eta  \right)} \right|} d\eta ,\,\,\,\,\,\,\,\,\,i = \overline {1,3} ,\,\,where\,\,\,\,\,B > 2.\]
Integrating by parts and taking into account condition (15), we arrive at the estimates
\[\left| {{\varphi _i}} \right| \leqslant {N_i}\frac{{\left| {{\varphi _{in}}^{\prime \prime \prime }} \right|}}{{{n^3}}},\,\,\,\,\,\,\,i = \overline {1,3} ,\,\,\]
where
\[{\varphi '''_{in}} = \frac{1}{{{{\left\| {{Y_n}} \right\|}^2}}}\int\limits_0^l {{{\varphi '''}_{in}}\left( \eta  \right)} {Y_n}\left( \eta  \right)d\eta ,\,\,\,\,\,\,\,i = \overline {1,3} .\]
Then
\[\left| {{A_{in}}{Y_n}} \right| \leqslant 2{N_i}B\frac{{\left| {{\varphi _{in}}^{\prime \prime \prime }} \right|}}{{{n^3}}},\,\,\,\,\,\,i = \overline {1,3} .\eqno(19)\]
By using (19), we rewrite estimate (18) in the form
\[\left| {u\left( {x,y} \right)} \right| \leqslant B{M_4}{N_i}\sum\limits_{n = 1}^\infty  {\frac{{\left| {{{\varphi '''}_{in}}} \right|}}{{{n^3}}}}  < \infty ,\,\,\,\,\,\,i = \overline {1,3} .\]
This implies that series (13) is absolutely and uniformly convergent. We prove the uniform convergence of  $\frac{{{\partial ^2}u}}{{\partial {y^2}}}$. It follows from (13) that
\[\frac{{{\partial ^2}u}}{{\partial {y^2}}} = \sum\limits_{n = 1}^\infty  {\left[ {{C_{1n}}{e^{ - 2{\tau _n}x}} + {C_{2n}}{e^{{\tau _n}x}}\cos {\rho _n}x + {C_{3n}}{e^{{\tau _n}x}}\sin {\rho _n}x} \right]{{Y''}_n}\left( y \right)}, \]
where
$${Y''_n}(y) \leqslant  - {\mu _n}\left( {\alpha \sin \sqrt {{\mu _n}} y - \beta \sqrt {{\mu _n}} \cos \sqrt {{\mu _n}} y} \right){A_n} =  - {\mu _n}{Y_n}(y).$$
Thus, we get
\[\frac{{{\partial ^2}u}}{{\partial {y^2}}} = \sum\limits_{n = 1}^\infty  {\left[ {{C_{1n}}{e^{ - 2{\tau _n}x}} + {C_{2n}}{e^{{\tau _n}x}}\cos {\rho _n}x + {C_{3n}}{e^{{\tau _n}x}}\sin {\rho _n}x} \right]\left( { - {\mu _n}} \right)} {Y_n}\left( y \right),\]
\[\left| {\frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right| = \sum\limits_{n = 1}^\infty  {\left[ {\left| {{C_{1n}}} \right|{e^{ - 2{\tau _n}x}} + \left| {{C_{2n}}} \right|{e^{{\tau _n}x}} + \left| {{C_{3n}}} \right|{e^{{\tau _n}x}}} \right]\left| {{\mu _n}} \right|} \left| {{Y_n}\left( y \right)} \right| \leqslant \]
\[ \leqslant {M_{_4}}\sum\limits_{n = 1}^\infty  {\left[ {\left| {{A_{1n}}{Y_n}\left( y \right)} \right| + \left| {{A_{2n}}{Y_n}\left( y \right)} \right| + \left| {{A_{3n}}{Y_n}\left( y \right)} \right|} \right]\left| {{\mu _n}} \right|} .\]
Since    ${\mu _n} = {\rm O}\left( {{n^2}} \right)$, we find
$$\left| {\frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right| \leqslant 2B{M_4}{N_i}\sum\limits_{i = 1}^\infty  {\frac{{\left| {{{\varphi '''}_{in}}} \right|}}{n}}. $$
By using the Cauchy–Buniakowski and Bessel inequalities [22], we get
\[\left| {\frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right| \leqslant 2B{M_4}{N_i}\sqrt {\sum\limits_{i = 1}^\infty  {{{\left| {{{\varphi '''}_{in}}} \right|}^2}} } \sqrt {\sum\limits_{i = 1}^\infty  {\frac{1}{{{n^2}}}} }  \leqslant 2B{M_4}{N_i}\sqrt {2{{\left\| {{{\varphi '''}_{in}}} \right\|}^2}} \sqrt {\frac{{{\pi ^2}}}{6}}  = 2{M_4}{N_i}\frac{\pi }{{\sqrt 3 }}\left\| {{{\varphi '''}_{in}}} \right\| < \infty, \]
where
\[\sum\limits_{i = 1}^\infty  {{{\left| {{{\varphi '''}_{in}}} \right|}^2}}  = 2\left\| {{{\varphi '''}_{in}}} \right\|_{{L_2}\left( {0,1} \right)}^2,\,\,i = \overline {1,3} ,\,\,\,\,\sum\limits_{i = 1}^n {\frac{1}{{{n^2}}}}  = \frac{{{\pi ^2}}}{6}.\]
In view of Eq. (1), we obtain a similar estimate for  $\frac{{{\partial ^3}u}}{{\partial {x^3}}}$. This implies that  $\left| {\frac{{{\partial ^3}u}}{{\partial {x^3}}}} \right|$ and  $\left| {\frac{{{\partial ^2}u}}{{\partial {y^2}}}} \right|$ are absolutely and uniformly convergent.\\
The theorem is proved.
\begin{theorem}
 If ${\varphi _i}(y) \in {C^3}[0,1],$  and  $\alpha \,{\varphi _i}(0) + \beta \,{\varphi '_i}(0) = 0,\,\,\,\,\gamma \,{\varphi _i}(1) + \delta \,{\varphi '_i}(1) = 0,$$\varphi ''(0) = \varphi ''(1) = 0$, then the solution to the problem   exists and presented in the form (13).
\end{theorem}



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