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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 50, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/50\hfil Gevrey problem for parabolic equations]
{Gevrey problem for parabolic equations with changing time direction}
\author[I. S. Pulkin\hfil EJDE-2006/50\hfilneg]
{Igor S. Pulkin}  

\address{Igor S. Pulkin \newline
Moscow Technical University of Radioengineering,
Electronics, and Automation, Moscow, Russia}
\email{igor492@yandex.ru}


\date{}
\thanks{Submitted March 13, 2006. Published April 18, 2006.}
\subjclass[2000]{35K05, 35K20, 35K99, 35M99}
\keywords{Gevrey problem; equation with changing time direction}

\begin{abstract}
 This article concerns  parabolic equations with changing time
 direction  and Gevrey's boundary condition. Using expansion
 series and biorthogonal systems, we prove the existence of
 classical solutions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theroem}[section]
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction and Main Result}

This paper is devoted to the study of existence of classical
solution of the parabolic problem
\begin{equation} \label{eq3}
\frac{\partial{u}}{\partial{t}}-\mathop{\rm sign}(x)
\frac{\partial^2{u}}{\partial{x}^2}=0
\end{equation}
in a domain $\Omega=\{(x,t): x \in (-1;0) \cup (0;1), t \in (0;T)\}$
with Gevrey boundary conditions  (see \cite{g1,t1})
\begin{equation}\label{GU3}
\begin{gathered}
 u(-1;t)=u(1;t)=0,\\
 {u(x;0)=u_{0}(x), \quad  t\in(0;1)}, \\
 {u(x;T)=u_{T}(x), \quad  t\in(-1;0)};
\end{gathered}
\end{equation}
Also with sewing conditions (see \cite{k1})
\begin{equation} \label{US3}
\begin{gathered}
 u(-0; t) = u(+0; t),  \\
\frac{\partial{u}}{\partial{x}}(-0;t)
=-\frac{\partial{u}}{\partial{x}}(+0;t).
\end{gathered}
\end{equation}
Note that \eqref{eq3} is a parabolic equation with changing time
direction. The boundary-value problems with such sewing
conditions appear when modelling, for example, process of
interaction between two reciprocal flows with mutual permeating,
or when designing  certain heat exchangers. Furthermore, the
results obtained in the linear case, may be used for investigating
nonlinear problems with changing time direction. Such problems
arise in supersonic dynamics, boundary layer theory \cite{l1}.

The main aim of the paper is to prove the existence of a classical
solution. We call a function $u: [-1,1]\times[0;T]
\to \mathbb{R}$ a classical solution of the problem
\eqref{eq3}-\eqref{GU3}-\eqref{US3} if $u \in C^2(\Omega)$ and
satisfies to equation \eqref{eq3} in $\Omega$, Gevrey condition
\eqref{GU3} and sewing conditions \eqref{US3} at 0.


To state our main result, we need some preliminary
constructions.
First using separation of variables technique we obtain the
following problem for $X(x)$, where $u(x,t)=X(x)T(t)$,
\begin{gather}
\mathop{\rm sign}(x) \frac{X''}{X}=\frac{T'}{T}=\mu,
\\  \label{obykn3}
\mathop{\rm sign}(x) X''=\mu X.
\\  \label{obyGU3}
\left. \begin{gathered}
 X(-1)=X(1)=0,  \\
 X|_{x=+0} = X|_{x=-0},  \\
 \frac{\partial{X}}{\partial{x}}|_{x=+0} = -
\frac{\partial{X}}{\partial{x}}|_{x=-0}.
\end{gathered}\right\}
 \end{gather}

Consider the set of twice differentiable functions for
$x\in(-1,0)$ and $x\in(0,1)$ satisfying boundary conditions
\eqref{GU3} and sewing conditions\eqref{US3}. Denote by $W_x$ the
completion of this set with respect to the norm
$$
\|h\|^{2} = \int_{-1}^{0} \Big(h^{2} + h_{x}^{2} + h_{xx}^{2} \Big)dx
+ \int_{0}^{1} \Big(h^{2} + h_{x}^{2} + h_{xx}^{2}
\Big)dx.
$$
It can be shown that $W_x$ is a Hilbert space. Now introduce
the  operator $L_x:  W_x\to L_2(-1,1) $ defined by
$L_xX= \mathop{\rm sign}(x) X''$ for every $X \in W_x$. It is easy
to see that $L_x$ is symmetric. Hence there exists denumerable set
of eigenvalues, both positive and negative. The corresponding
eigenfunctions are:
For $\mu =\lambda^{2}>0$,
\begin{equation}\label{sf1}
X_{\lambda}=\begin{cases}
\frac{\sin{\lambda(x+1)}}{\sin{\lambda}},&   x<0 \\
\frac{\sinh{\lambda(1-x)}}{\sinh{\lambda}}, &   x>0,
\end{cases}
\end{equation}
For $\mu =-\lambda^{2}<0$,
\begin{equation}\label{sf2}
 X_{\lambda}=\begin{cases}
\frac{\sinh{\lambda(x+1)}}{\sinh{\lambda}}, &   x<0 \\
\frac{\sin{\lambda(1-x)}}{\sin{\lambda}}, &  x>0,
\end{cases}
\end{equation}
In both cases eigenvalues are the solutions of the equation
\begin{equation}\label{tg}
\tan{\lambda} = \tanh{\lambda}.
\end{equation}
In addition, there exists an eigenfunction for
$\mu =0$,
\begin{equation}\label{sf0} X_{0}=\begin{cases}
\sqrt{\frac{3}{2}}(x+1), &  x<0 \\
\sqrt{\frac{3}{2}}(1-x), &  x>0.
\end{cases} \end{equation}
Functions
 \eqref{sf1}--\eqref{sf0} form orthogonal complete basis in
$W_{x}$.
As it follows from \eqref{tg}, the negative eigenvalues
$\mu_{k} = -\lambda_{k}^{2}$ satisfy asymptotic relations
$$
\lambda_{n} \sim -\frac{\pi}{4} - \pi n, \quad n \in \mathbb{N}
$$
and the positive eigenvalues $\mu_{k} = \lambda_{k}^{2}$  satisfy
asymptotic relations
$$
\lambda_{n} \sim \frac{\pi}{4} + \pi n,\quad n \in \mathbb{N}\,.
$$
We will seek solutions of the problem \eqref{eq3}--\eqref{US3}
in the form
\begin{equation}\label{usum1}
u(x,t) = \sum_{k=0}^{\infty}\Big{(}A_{k}e^{-\lambda_{k}^{2}t}
\frac{\sinh\lambda_{k}(x+1)}{\sinh\lambda_{k}}
+B_{k}e^{-\lambda_{k}^{2}(T-t)}
\frac{\sin\lambda_{k}(x+1)}{\sin\lambda_{k}}\Big{)} + C(x + 1),
\end{equation}
for $x<0$, and
\begin{equation}\label{usum2}
u(x,t) = \sum_{k=0}^{\infty}\Big{(}A_{k}e^{-\lambda_{k}^{2}t}
\frac{\sin\lambda_{k}(x-1)}{\sin\lambda_{k}}
+B_{k}e^{-\lambda_{k}^{2}(T-t)}
\frac{\sinh\lambda_{k}(x-1)}{\sinh\lambda_{k}}\Big{)} + C(1 - x),\
\end{equation}
for $x>0$.

\begin{theorem} \label{thm1.1}
 Assume $u_{0}, u_{T} \in L_{2}(0;1)$. Then there exists unique
classical solution $u(x,t)$ of the problem
\eqref{eq3}-\eqref{GU3}-\eqref{US3}.
Furthermore, there exists unique collection of coefficients
$A_{k}, B_{k} \in l_{2}$  such that the solution $u(x,t)$ on the
set $\Omega$ express by the series expansion \eqref{usum1} and
\eqref{usum2}, respectively. If $x < 0, t = T$ then the sum of
series \eqref{usum1} is $u(x,T) = u_{T}$, if $x > 0, t = 0$ then
the sum of series \eqref{usum2} is $u(x,0) = u_{0}$. Moreover, if
$u_{0}, u_{T} \in C_{0}^{2}[0;1]$, then \eqref{usum1} and
\eqref{usum2} converge absolutely in rectangular $\Omega = (-1;0)
\times (0;T) \cup (0;1) \times (0;T)$.
\end{theorem}

\section{Proof of the main result}

First, we substitute initial data \eqref{obyGU3} into
\eqref{usum1} and \eqref{usum2}. Hence we obtain that
for $t=T, x<0$,
$$
u(x,T)=\sum_{k=0}^{\infty}\Big(A_{k}e^{-\lambda_{k}^{2}T}
\frac{\sinh\lambda_{k}(x+1)}{\sinh\lambda_{k}}+
B_{k}\frac{\sin\lambda_{k}(x+1)}{\sin\lambda_{k}}\Big),
$$
and for $t=0, x>0$
$$
u(x,0)=\sum_{k=0}^{\infty}\Big(A_{k}
\frac{\sin\lambda_{k}(1-x)}{\sin\lambda_{k}}+
B_{k}e^{-\lambda_{k}^{2}T}\frac{\sinh\lambda_{k}(1-x)}
{\sinh\lambda_{k}}\Big).
$$
Now we make the change of variables setting $x=y-1$ for $x<0$ and
$x=1-y$ for $x>0$. Then the foregoing equations will take the form
\begin{gather}\label{sis1}
u_{0}(y)= \sum_{k=1}^{\infty}\Big(
A_{k}\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}+
B_{k}e^{-\lambda_{k}^{2}T}
\frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big) + Cy,
\\  \label{sis2}
u_{T}(y)=
\sum_{k=1}^{\infty}\Big(
A_{k}e^{-\lambda_{k}^{2}T}
\frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}+
B_{k}\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}\Big) + Cy,
\end{gather}
where
\begin{gather*}
u_{0}(y)=u_{0}(1-x),\\
u_{T}(y)=u_{T}(1+x).
\end{gather*}
Adding and subtracting \eqref{sis1} and \eqref{sis2} term by term
we obtain
\begin{gather}\label{sist1}
u_{0} + u_{T} - 2Cy = \sum_{k=1}^{\infty}\big
(A_{k} + B_{k}\big)\Big(\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
+ e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big),
\\  \label{sist2}
u_{0} - u_{T} =
\sum_{k=1}^{\infty}
\big( A_{k} - B_{k} \big)
\Big(\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
 -e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big).
\end{gather}
It is easy to see that both these equations represent the
expansion of given functions with respect to
\begin{equation}\label{funct1}
\alpha_{k} = \Big( \frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
+ e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big)
\end{equation}
and
\begin{equation}\label{funct2}
\beta_{k} = \Big( \frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
- e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big).
\end{equation}

Suppose that there exist biorthogonal systems $ \{\psi_{n}\}$,
$\{\omega_{n}\}$, $n = 1, \dots \infty$ in $L_2(0,1)$, such that
\begin{gather*}
(\alpha_{k} ; \psi_{n}) = \delta_{kn}, \\
(\beta_{k} ; \omega_{n}) = \delta_{kn},
\end{gather*}
where $\delta_{kn}$ is the delta Kronecker,  and $(\cdot ; \cdot)$
 is a scalar product in  $L_2(0,1)$. Then coefficients $A_{k}$ and
$B_{k}$ could be found for arbitrary left sides of \eqref{sist1}
and \eqref{sist2}. It means that the following equalities will
satisfy:
\begin{gather*}
\int_{0}^{1} \psi_{n} \Big(\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
+ e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big) dy
=  \delta_{kn}, \\
\int_{0}^{1} \omega_{n} \Big(\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
- e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big) dy
=  \delta_{kn}.
\end{gather*}

Now we prove the existence of biorthogonal systems.

\begin{proposition} \label{prop2.1}
Let $\{h_{n}\}$, be a biogthogonal system with  $\{\sigma_n\}$,
$n = 1, \dots \infty$ and $ \{h_{n}\}$ uniformly
bounded. Then there  exists a biorthogonal system for
$\{\sigma_{n} + \tau_{n}\}$, provided that for all $k$,
\begin{equation}\label{uslo1}
\sum_{n=1}^{\infty}|(\tau_{k}, h_{n})|^2 < \delta < 1;
\end{equation}
and
\begin{equation}
\label{uslo2}
\sum_{n=1}^{\infty}|\tau_{n}| < \infty.
\end{equation}
\end{proposition}

\begin{proof}
We look for a biorthogonal system in the form
$$
\tilde{h}_{n} = h_{n} + \sum_{i=1}^{\infty}b_{in}h_{i}.
$$
Then
$$
\delta_{kn} = (\sigma_{k} + \tau_{k}; h_{n} + \sum_{i=1}^{\infty}b_{in}h_{i}) =
\delta_{kn} + (\tau_{k}; h_{n}) + b_{kn} + \sum_{i=1}^{\infty}b_{in}(\tau_{k}; h_{i}).$$
Let us denote $A$ the matrix whose elements are
$(\tau_{k}; h_{n})$ and $B$ the matrix with elements
 $b_{kn}$. Then we obtain the equation
$$
A + B + BA = 0, \quad\text{or}\quad B(E + A) = - A
$$
where $E$ is an identity matrix.
Note that under condition \eqref{uslo1} this equation is resolvable
as there exists an inverse matrix
$$
(E + A)^{-1} = E - A + A^{2} - A^{3} + \dots
$$
Hence,
$$
B = - A + A^{2} - A^{3} + \dots,
$$
Furthermore,
$$
\|B\| \leq \frac{\|A\|}{1 - \|A\|}.
$$
Thus  $B$ is the matrix of certain linear bounded operator on the space
 $l_{2}$, that is why  the collection $\{b_{ik} \}$,
$i = 1,2,\dots$
belongs $l_{2}$ for all $k$. As $|h_{i}|$ is uniformly bounded and
\eqref{uslo2} is fulfilled, then all series
$\sum_{i=1}^{\infty}b_{in}h_{i}$ converge.
\end{proof}

\begin{proposition} \label{prop2.2}
There exists  $T_0>0$ such that for
arbitrary $T>T_0$ there exists biorthogonal systems for systems of
functions $\alpha_{k}$ and $\beta_{k}$ .
\end{proposition}

\begin{proof} As follows from \cite{m1}, for the system
\begin{equation}\label{moiseev}
\sigma_{n} = \frac{\sin(\frac{\pi}{4} + \pi n)x}{\sin(\frac{\pi}{4}
+ \pi n)}
\end{equation}
there exists a biorthogonal system
 $ \{h_{n}\}$; moreover
the functions $ \{h_{n}\} $ are uniformly bounded, for example by
the constant 10.

The difference $\tau_{k}$ of $\alpha_{k}$ and \eqref{moiseev}
(note that for $\beta_{k}$ all reasonings are the same) consists
of two terms
\begin{gather*}
\tau^{(1)}_{k} =
\frac{\sin(\frac{\pi}{4} + \pi k)x}{\sin(\frac{\pi}{4} + \pi k)} -
\frac{\sin\lambda_{k}x}{\sin\lambda_{k}} ; \\
\tau^{(2)}_{k} =
e^{-\lambda_{k}^{2}T} \frac{\sinh\lambda_{k}x}{\sinh\lambda_{k}} .
\end{gather*}
By virtue of Lagrange theorem for $\tau^{(1)}_{k}$,
$$
|\tau^{(1)}_{k}| \leq
\sqrt{2}\cos\lambda_{k}(\lambda_{k} - \frac{\pi}{4} - \pi k).
$$
Taking into account that $\lambda_{k}$ is a root of the equation
$\tan\lambda = \tanh\lambda $, we obtain
\begin{align*}
|\tau^{(1)}_{k}| &\leq \sqrt{2}\cos\lambda_{k}|\tanh\lambda_{k} - 1|
\cdot\frac{1}{\cos^{2}\lambda_{k}} \\
&\leq 2|\tanh\lambda_{k} - 1| =
\frac{2}{\cosh\lambda_{k}(\cosh\lambda_{k} + \sinh\lambda_{k})},
\end{align*}
hence, norms $\tau^{(1)}_{k}$ are rapidly decreasing
(at the rate $e^{-Ck^{2}}$).
Thus, for $\tau^{(1)}_{k}$ condition \eqref{uslo2} holds. In order to prove
 \eqref{uslo1} we need to estimate sum of the series.
Direct calculation shows that even first root $\lambda_{1} \approx
3,9266$ differ from  $\frac{5\pi}{4} \approx 3,9270$ less than by
$0,5\cdot10^{-3}$, then the sum of squares  $\tau^{(1)}_{k}$ can
be bounded above by geometric progression
$$\sum_{k=1}^{\infty} 10^{-6} e^{-2\pi k} =
\frac{10^{-6} }{1 - e^{-2\pi}} \leq 10^{-5}.$$

The validity of \eqref{uslo2}  for $\tau^{(2)}_{k}$ is obvious for
all positive  $T$, and in order to estimate the sum of squared norms
we calculate at first
$$
\int_{0}^{1}
\Big( \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}} \Big)^{2} dy =
\frac{1}{2\lambda_{k}\tan\lambda_{k}} - \frac{1}{2\sinh^{2}\lambda_{k}}.
$$
Hence the series of sums of the squared norms is majorized by
$$
\sum_{k=1}^{\infty}\frac{e^{10^{-3}-(\frac{\pi}{2}
+ 2\pi k)T^2}}{\lambda_{k}}.
$$
Direct calculations show that the sum of this series for
$T = 0,15$ is $1,25 \cdot 10^{-3}$.
  From the estimate
\begin{align*}
\sum_{k=1}^{\infty}|(\tau_{k}, h_{n})|^2
&\leq \sum_{k=1}^{\infty}|\tau_{k}|^2 \cdot |h_{n}|^2 \\
&\leq \sum_{k=1}^{\infty}100|\tau^{(1)}_{k} + \tau^{(2)}_{k}|^2 \\
&\leq \sum_{k=1}^{\infty}200(|\tau^{(1)}_{k}|^2 + |\tau^{(2)}_{k}|^2)
< 0,3
\end{align*}
we can see that if $T_0 = 0,15$ it is sufficient for concluding
the proof.
\end{proof}

\begin{proposition}
The following two systems
\begin{gather}\label{osw1}
\Big\{  X^{+}_{0} = \sqrt{3}y ,\
X^{+}_{k}=\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
+\frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big\},
\\  \label{osw2}
\Big\{X^{-}_{k}=\frac{\sin\lambda_{k}y}{\sin\lambda_{k}}
-\frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}\Big\},
\end{gather}
$k=1\dots\infty$, are complete and orthogonal in $L_{2}(0;1)$.
\end{proposition}

\begin{proof} The first system is a complete system of
eigenfunctions of operator $L_{4}u = u^{IV}$ in the space of
smooth functions, satisfying boundary conditions
$$
u(0) = u''(0) = u''(1) = u'''(1) = 0.
$$
It is easy to see if we integrate by parts that this operator is
 symmetric in this space, then eigenfunctions are orthogonal.
As this system consists of all eigenfunctions then it is complete.
Indeed, if we denote $a^{4}$,  an eigenvalue of
$L_{4}$, then all eigenfunctions, corresponding $a^{4}$,  take the form
$$
A\sin{ax} + B\cos{ax} + C\sinh{ax} + D\cosh{ax}.
$$
Since $u(0) = u''(0) = 0$, it follows that $B = D = 0$.
 From the boundary condition $u''(1) = 0$ we obtain the equation
$$
-Aa^{2}\sin{a} + Ca^{2}\sinh{a} = 0.
$$
It turned out that
$$
A = \frac{Z}{\sin{a}},\quad  C = \frac{Z}{\sinh{a}},
$$
where $Z$ is certain nontrivial parameter and without loss of
generality we may let $Z=1$.
It follows from $u'''(1) = 0$, that
$$
-Aa^{3}\cos{a} + Ca^{3}\cosh{a} = 0.
$$
Substituting the expression obtained for $A$ and $C$, into this
equation, we have the relation
$$  \tan{a} = \tanh{a},
$$
which must be valid for all eigenvalues. Using precisely the same
reasoning, we prove the statement for second system; the only
difference is  that the boundary conditions are in the form
$$
u(0) = u(1) = u'(1) = u''(0) = 0.
$$
It should be noted that
\begin{equation}\label{norv}
\mu_{0}^{2} = | X^{+}_{0} | = 1;\quad
\mu_{k}^{2} = | X^{+}_{k} | = | X^{-}_{k} | =
\frac{1}{2\sin^{2}\lambda_{k}} - \frac{1}{2\sinh^{2}\lambda_{k}}.
\end{equation}
By  \eqref{tg} and the asymptotic behavior of eigenvalues,
the norms of the vectors tend to unit as $k\to \infty$.
\end{proof}

\begin{proposition} \label{prop2.4}
The biorthogonal systems for \eqref{funct1}--
\eqref{funct2}, are unique.
\end{proposition}


\begin{proof} The validity of this statement will follow
immediately from the completeness of systems $\alpha_{k}$ and
$\beta_{k}$. We prove the completeness only for the first system
as the proof for the second is similar. Assume that the system
$\alpha_{k}$ is not complete.Then there exists a vector $Z$ such
that for all $k$
$$
(\alpha_{k},Z) = 0.
$$
Since the system of vectors $X_{k}^{+}$ is complete, we may expand vector
$Z$ with respect to this system:
$$
Z = \sum_{i = 1}^{\infty}z_{i}X_{i}.
$$
We denote here
$$
X_{k} = \frac{X_{k}^{+}}{\mu_{k}},
$$
which form an orthonormal system. Also denote
\begin{gather*}
\alpha_{k}^{*} = \frac{\alpha_{k}}{\mu_{k}},\\
\gamma_{k} = X_{k} - \alpha_{k}^{*} = \frac{1}{\mu_{k}}
(1 - e^{-\lambda_{k}^{2}T}) \frac{\sinh\lambda_{k}y}{\sinh\lambda_{k}}.
\end{gather*}
Then for all $k$,
\begin{gather*}
(X_{k} - \gamma_{k} ; \sum_{i = 1}^{\infty}z_{i}X_{i}) = 0,\\
z_{k} - \sum_{i = 1}^{\infty}z_{i}(\gamma_{k} ; X_{i}) = 0.
\end{gather*}
Hence $\{z_{k}\}$ are  eigenvectors of the matrix with elements
$M_{ki} = (\gamma_{k} ; X_{i})$, and corresponding eigenvalue $+1$.
The operator associated with this matrix is
$$
M : X_{k} \longmapsto \gamma_{k}.
$$
Then the operator
$$E - M : X_{k} \longmapsto \alpha_{k}^{*},$$
and  $\{z_{k}\}$ is and eigenvector of this operator,
corresponding to eigenvalue $0$.
 In other words,
$$\sum_{k = 1}^{\infty}z_{k}\alpha_{k}^{*} = 0.
$$
Existence of such liner combination with not only trivial
$z_{k}$ contradicts with the fact of linearly independence
of $\alpha_{k}$.
\end{proof}

\begin{proposition} \label{prop2.5}
Assume $u_{0}, u_{T}$ are in $C_{0}^{2}[0;1]$. Then the coefficients
$A_{k}$ and $B_{k}$ belong $ l_{1}$.
\end{proposition}

\begin{proof} To prove this statement note that
\begin{gather*}
\alpha_{k} = \frac{1}{2}((X^{+}_{k} + X^{-}_{k}) +
e^{-\lambda_{k}^{2}T}(X^{+}_{k} - X^{-}_{k})), \\
\beta_{k} = \frac{1}{2}((X^{+}_{k} + X^{-}_{k}) -
e^{-\lambda_{k}^{2}T}(X^{+}_{k} - X^{-}_{k})).
\end{gather*}
The outcome of these formulas is that if both series of coefficients when
expanding with respect to $X^{+}_{k}$ and $X^{-}_{k}$ converge
absolutely, then  both series of coefficients when expanding with
respect to $\alpha_{k}$ and $\beta_{k}$ also converge absolutely.

Let a $f$ be a function in $C_{0}^{2}[0;1]$. Then there exists an
expansion with respect to $X^{-}_{k}$,
$$
f = \sum_{k = 1}^{\infty}c_{k}X^{-}_{k},
$$
where the coefficients can be calculated using the orthogonality
by formulas:
$$
c_{k} = \frac{1}{\mu_{k}^{2}}\int_{0}^{1}f \cdot X^{-}_{k}dx.
$$
Since the second derivative of $f$ is square integrable, there exists
similar expansion with respect to $X^{+}_{k}$,
$$
f'' = \sum_{k = 0}^{\infty}d_{k}X^{+}_{k}
$$
where the coefficients are
$$
d_{k} = \frac{1}{\mu_{k}^{2}}\int_{0}^{1}f'' \cdot X^{+}_{k}dx.
$$
Note that if we differentiate the system $X^{-}_{k}$ twice,
 it will turn into the system
$-\lambda_{k}^{2}X^{+}_{k}$. So that, if we integrate twice and
take into account that
$f$ and $f'$ vanish in the boundary points of segment $[0; 1]$,
we derive the relation
$$
d_{k} = - \lambda_{k}^{2}c_{k},
$$
which implies the inclusion $c_{k} \in l_{1}$.

The coefficient $C$ of \eqref{sist1} may be found from the
boundary condition if $y = 1$,
$$
C = \frac{1}{2} (u_{0} + u_{T} -
\sum_{k=1}^{\infty}(A_{k} + B_{k})(1 + e^{-\lambda_{k}^{2}T})).
$$
To complete the proof of the theorem we need only to
show that functional series \eqref{usum1}--\eqref{usum2} and
the series of their first and second derivatives with respect to
$x$ and $t$ converge uniformly. For $0 < t_{1} < t_{2} < T$
convergence on a segment $[t_{1}; t_{2}]$ is a consequence of
majoring criterion for convergence. Since $[t_{1}$ and $t_{2}]$
are arbitrary the series converge uniformly for every interior
point of the interval $(0;T)$.
The proof of the theorem is complete.
\end{proof}

\begin{thebibliography}{00}

\bibitem{g1} Gevrey M.; \emph{Sur les equations aux derivees partielles
du type parabolique}.  J. Math. Appl., 1913, t.9, Sec.6, p. 305-475.

\bibitem{k1} Kislov N. V.; \emph{Pulkin I.S--On existence and uniqueness
of a week solution to Gevrey problem with generalized sewing conditions},
 Vestnik MEI, N.6, 2002, p. 88--92.

\bibitem{l1} Larkin N. A., Novikov V. A., Ianenko N. N.;
\emph{Nonlinear alternating kind equations}, Novosibirsk, Nauka, 1983.

\bibitem{m1} Moiseev E. I.; \emph{On basis property of sine and cosine
 systems}.  Doklady Acad. Nauk SSSR, 1984. V. 275, N. 4., p. 794--798.

\bibitem{t1} Tersenov S. A.; \emph{Parabolic equations with the changing
time direction}, Novosibirsk, 1985.

\end{thebibliography}
\end{document}