\section{Introduction}\label{sec:intr} 

In~\cite{KmRe} we developed an analytical approach
for proving Fredholmness property for a class of linear hyperbolic problems. 
This subject is related to the following two topics, known
results on which are far from being conclusive:
\begin{itemize}
\item
Correct posedness of initial-boundary value problems for hyperbolic PDEs;
\item
Local investigation of nonlinear 
hyperbolic equations, as smooth continuation and local bifurcations
(the Fredholmness property is crucial here).
\end{itemize}

The Fredholmness in the context of hyperbolic PDEs is related to propagating 
singularities along special curves called characteristics. This entails two main
complications:
\begin{itemize}
\item
Inverses to hyperbolic operators do not improve the regularity of right-hand sides of
differential equations so much as inverse elliptic or parabolic operators do it.
This causes a complication with using the well-known classical  fact that any Fredholm operator is bijective
modulo a compact operator.  Namely,  
to prove the desired compactness property, we would need more regularity than  the hyperbolicity
gives us;
\item
To ensure an optimal regularity trade-off (needed for  Fredholmness) between the spaces of
solutions and the spaces of right hand sides, one hardly could speak about the regularity of
solutions separately in each independent variable. This conclusion is based on the fact that the 
hyperbolicity does not ensure uniform regularity of solutions in both characteristic and 
non-characteristic  directions.
\end{itemize}

To overcome these complications for a certain class of hyperbolic PDEs, 
we here prove a new Fredholmness 
criterion for linear operators in Banach
spaces (Section~2, Theorem~\ref{thm:criter}) and construct spaces of solutions by means
of the corresponding graph norms as suggested in~\cite{KmRe} (Section~3 of the present 
paper). In the case of $n=1$ Theorem~\ref{thm:criter} gives a well-known classical 
result. If $n=2$, it is a part of our result from~\cite{KmRe},
where Fredholm Alternative is proved for periodic-Dirichlet problem for linear 
one-dimensional hyperbolic systems. 

A novelty of the paper  consists in generalization of the Fredholmness criterion 
obtained in \cite{KmRe} from $n=2$ to an arbitrary $n$. Another result is a
construction of multidimensional hyperbolic partial differential operators 
for which the generalized criterion applies with $n=3$ and does not with  $n<3$. 




\section{Fredholmness criterion}


We here generalize the Fredholmness criterion proved in~\cite{KmRe}.

\begin{thm}\label{thm:criter}
Let $W$ be a Banach space, $I$ be the identity in $W$, and $K\in{\cal L}(W)$ with $K^n$ being
compact for some $n\in\mathbb{N}$.  Then $I-K$ is Fredholm.
\end{thm}


\begin{proof}
Since 
$$
I-K^n=\sum\limits_{i=0}^{n-1}K^i(I-K)
$$
and $K^n$ is compact, we have
\begin{equation}
\label{eq:endl}
\mathop{\rm dim}\mathop{\rm ker}(I-K)\le \mathop{\rm dim}\mathop{\rm ker}(I-K^n)<\infty.
\end{equation}
Similarly $\mathop{\rm dim}\mathop{\rm ker}(I-K)^*<\infty$, hence
$
\mathop{\rm codim}\overline{\mathop{\rm im}(I-K)}<\infty.
$
It remains to show that $\mathop{\rm im}(I-K)$ is closed.

Take a sequence $(w_j)_{j\in\mathbb{N}}\subset W$ and an element $w\in W$ such that
\begin{equation}\label{eq:4.1}
(I-K)w_j\to w.
\end{equation}
We have to show that $w \in \mathop{\rm im}(I-K)$.

Because of (\ref{eq:endl}) there exists a closed subspace $V$ of $W$ such that 
\begin{equation}
\label{eq:dirSum} 
W=\mathop{\rm ker}(I-K)\oplus V,
\end{equation}
Using the decomposition
$$
w_{j}=u_j+v_j, \; u_j\in\mathop{\rm ker}(I-K), \; v_j \in V,
$$
we get from (\ref{eq:4.1})
\begin{equation}\label{eq:v}
(I-K)v_j\to w.
\end{equation}


First, we show that  the sequence $(v_j)_{j \in \mathbb{N}}$ is bounded. 
If not, without loss of generality we can assume that
\begin{equation}\label{eq:4.11}
\lim\limits_{j\to \infty}\|v_j\|=\infty.
\end{equation}
From (\ref{eq:v}) and (\ref{eq:4.11}) we get 
\begin{equation}
\label{eq:1Null}
(I-K)\frac{v_j}{\|v_j\|} \to 0,
\end{equation}
hence
\begin{equation}
\label{eq:Null}
(I-K^n)\frac{v_j}{\|v_j\|}  \to 0.
\end{equation}
On the other side, because $K^n$ is compact, there exist $v \in W$ and a subsequence 
$(v_{j_k})_{k \in \mathbb{N}}$ such that
\begin{equation}
\label{eq:conv}
K^n\frac{v_{j_k}}{\|v_{j_k}\|}  \to v.
\end{equation}
Inserting (\ref{eq:conv}) into (\ref{eq:Null}), we get 
\begin{equation}
\label{eq:minus}
\frac{v_{j_k}}{\|v_{j_k}\|} \to v \in V.
\end{equation}
Combining (\ref{eq:minus}) with (\ref{eq:1Null}), we get $(I-K)v=0$,
i.e. $v \in V \cap \mathop{\rm ker} (I-K)$ and $\|v\|=1$. However, this contradicts (\ref{eq:dirSum}).

Now we use the boundedness of $(v_j)_{j\in\mathbb{N}}$  to show that $w\in\mathop{\rm im}(I-K)$.
As
$K^n$ is compact, there exists $v\in W$ and a subsequence $(v_{j_k})_{k \in \mathbb{N}}$ such that
$
K^nv_{j_k}\to v.
$
On the other hand, (\ref{eq:v}) yields $(I-K^n)v_j=\sum\limits_{i=0}^{n-1}K^i(I-K)v_j\to  
\sum\limits_{i=0}^{n-1}K^iw$. 
Hence
$$
\lim\limits_{k\to \infty}v_{j_k}=\sum\limits_{i=0}^{n-1}K^iw+v
$$
and, therefore,
$$
w=\lim\limits_{k\to \infty}(I-K)v_{j_k}=(I-K)\left(\sum\limits_{i=0}^{n-1}K^iw+v\right)\in\mathop{\rm im}(I-K)
$$
as desired.
\end{proof}





\section{Application to two-dimensional \\
hyperbolic PDEs}



We here prove a well-posedness result (specifically, the Fredholmness) for two-dimensional mixed
hyperbolic problems of periodic-Dirichlet  type.
Consider the following first-order hyperbolic system
\begin{equation}\label{eq:10}
\begin{array}{rr}
\displaystyle
\sum\limits_{j=1}^na_{ij}\left(
\alpha_i\partial_tu_j  + \partial_xu_j + \beta_i\partial_yu_j + \gamma_i(x,y,t)u_j
\right) + \sum\limits_{j=1}^nb_{ij}(x,y,t)u_j  = f_i(x,y,t),\\
i\le n,\quad (x,y,t)\in(0,1)\times\mathbb{R}\times\mathbb{R},
\end{array}
\end{equation}
supplemented with the  periodic conditions in $y$ and $t$
\begin{equation}\label{eq:11}
\begin{array}{l}
u_i\left(x,y+Y,t+T\right) = u_i(x,y,t),\qquad
i\le n,\quad (x,y,t)\in[0,1]\times\mathbb{R}\times\mathbb{R},
\end{array}
\end{equation}
and the trivial boundary conditions in $x$
\begin{equation}\label{eq:12}
\begin{array}{l}
u_i(0,y,t) = 0,\qquad
i\le k,\quad (y,t)\in\mathbb{R}^2,\\
u_j(1,y,t) = 0,\qquad
k+1\le j\le n,\quad (y,t)\in\mathbb{R}^2.
\end{array}
\end{equation}
We will make the following assumptions: 
$n\ge 3$, $k$ is an integer in the range $2\le k<n$, the coefficients $a_{ij}$, 
 $\alpha_i$, and $\beta_i$, are fixed real constants, the coefficients
$\gamma_i$, $b_{ij}: [0,1]\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ and the right hand sides $f_i: [0,1]\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ 
are known functions.
Given $l\in\mathbb{N}$ in the range $1\le l\le k-1$, we will suppose also that the matrix $A=\left(a_{ij}\right)_{i,j=1}^n$
has the following diagonal-block structure
 \begin{equation}\label{eq:blockA}
  A=\left(
  \begin{array}{cccccc}
    A_1&0 & 0\\
    0& A_2& 0\\
    0& 0 & A_3
  \end{array}
  \right)
 \end{equation}
where $A_1$, $A_2$, and $A_3$ are $l\times l$, $(k-l)\times(k-l)$, and $(n-k)\times(n-k)$-matrices,
respectively, while $0$ denotes the null matrices of respective sizes. Moreover, the matrix
$B=\left(b_{ij}\right)_{i,j=1}^n$ is assumed to be one of the following two kinds:
\begin{equation}\label{eq:blockB}
  A=\left(
  \begin{array}{cccccc}
    0&0 & B_1\\
    B_2& 0& 0\\
    0& B_3 & 0
  \end{array}
  \right)
 \end{equation}
or
\begin{equation}\label{eq:blockBm}
  A=\left(
  \begin{array}{cccccc}
    0&\tilde B_1 & 0\\
    0& 0& \tilde B_2\\
    \tilde B_3& 0 & 0
  \end{array}
  \right),
 \end{equation} 
where $B_1, B_2, B_3, \tilde B_1, \tilde B_2$, and $\tilde B_3$ are, respectively,
$l\times(n-k)$, $(k-l)\times l$, $(n-k)\times(k-l)$, $l\times(k-l)$, $(k-l)\times(n-k)$, and
$(n-k)\times l$-matrices. Without loss of generality we can consider the matrix (\ref{eq:blockB})
(similar argument works also for (\ref{eq:blockBm})).


Our aim is to prove the Fredholmness of the operator generated by problem (\ref{eq:10})--(\ref{eq:12}) 
between the two spaces of continuous functions, namely, between the space of
the right-hand sides 
$$
W:=\left({\mbox{C}}_{per}\left([0,1]\times\mathbb{R}\times\mathbb{R}\right)\right)^n
$$ 
and the space of solutions
\begin{equation}\label{eq:13}
\begin{array}{cc}
\displaystyle
V:= \Bigl\{u\in W \,:\,u_i(0,y,t)=0 \mbox{ for } i\le k, u_i(1,y,t)=0 \mbox{ for } k+1\le i\le n,\nonumber\\
\displaystyle
\sum\limits_{j=1}^na_{ij}\left(\alpha_i\partial_tu_j+\partial_xu_j+\beta_i\partial_yu_j\right)\in W \mbox{ for } i\le n
\Bigr\}.\nonumber
\end{array}
\end{equation}
Here $u=(u_1,\dots,u_n)$, ${\mbox{C}}_{per}\left([0,1]\times\mathbb{R}\times\mathbb{R}\right)$ denotes
the space of continuous on $[0,1]\times\mathbb{R}\times\mathbb{R}$ functions being $Y$-periodic 
in $y$ and $T$-periodic in $t$.
The function spaces $W$ and $V$ are endowed with the norms
$$
\|u\|_{W}:=\sum\limits_{i=1}^n\max\limits_{[0,1]\times[0,Y]\times[0,T]}
\left|u_i\right|
$$
and 
\begin{equation}\label{eq:14}
\begin{array}{cc}
\displaystyle
\|u\|_{V}:=\|u\|_{W} + \left\|\sum\limits_{i=1}^na_{ij}\left(\alpha_i\partial_tu_j+\partial_xu_j+\beta_i\partial_yu_j\right)
\right\|_{W}.
\end{array}
\end{equation}


\begin{lemma}\label{lem:2.2}
The space $V$ is complete.
\end{lemma}

\begin{proof}
Let $(u^{m})_{m\in\mathbb{N}}$ be a fundamental sequence
in $V$.
Then $(u^{m})_{m\in\mathbb{N}}$ and 
$$
\left(\sum\limits_{j=1}^na_{1j}(\alpha_1\d_tu_j^{m}+\d_xu_j^{m}+\beta_1\d_yu_j^{m}),\dots,
\sum\limits_{j=1}^na_{nj}(\alpha_n\d_tu_j^{m}+\d_xu_j^{m}+\beta_n\d_yu_j^{m})\right)_{m\in\mathbb{N}}
$$
are fundamental sequences in $W$. Due to completeness of $W$,
there exist
$u,v\in W$  such that
$$
u^{m}\to u \; \mbox{ and } \;
\left(\sum\limits_{j=1}^na_{ij}(\alpha_i\d_tu^{m}+\d_xu^{m}+\beta_i\d_yu^{m})\right)_{i=1}^n\to  v
$$
in $W$ as $m\to\infty$.
It remains to show that, given $i\le n$,
$
\sum_{j=1}^na_{ij}(\alpha_i\d_tu_j+\d_xu_j+\beta_i\d_yu_j)=v_i
$
in the sense of generalized derivatives, where $v_i$ are the components of vector $v$. 
But this is obvious: Take a smooth
function
$\varphi: (0,1)\times\left(0,Y\right)\times\left(0,T\right)\to\mathbb{R}$ with compact
support. Then
\begin{eqnarray*}
&\displaystyle
\int\limits_{0}^1\int\limits_{0}^{Y}\int\limits_{0}^T\sum\limits_{j=1}^na_{ij}
u_j(\alpha_i\d_t+\d_x+\beta_i\d_y)\varphi\,
dx\,dy\,dt&\\
&\displaystyle
=
\lim\limits_{m\to\infty}\int\limits_{0}^1\int\limits_{0}^{Y}\int\limits_{0}^T
\sum\limits_{j=1}^na_{ij}u_j^{m}(\alpha_i\d_t+\d_x+\beta_i\d_y)\varphi\,dx\,dy\,dt&\\
&\displaystyle
=-\lim\limits_{m\to\infty}\int\limits_{0}^1\int\limits_{0}^{Y}\int\limits_{0}^T
\sum\limits_{j=1}^na_{ij}(\alpha_i\d_t+\d_x+\beta_i\d_y)u_j^{m}\varphi\,dx\,dy\,dt=-
\int\limits_{0}^1\int\limits_{0}^{Y}\int\limits_{0}^T
v_i\varphi\,dx\,dy\,dt&
\end{eqnarray*}
as desired. Since $i$ is an arbitrary natural number in the range $i\le n$, the
proof is complete
\end{proof}


To formulate our result let us introduce linear
operators: Given $a_{ij},\alpha_i,\beta_i\in\mathbb{R}$ and $\gamma_i, b_{ij}\in {\mbox{C}}_{per}\left([0,1]\times\mathbb{R}\times\mathbb{R}\right)$, we define 
$$
Cu
:=
\left(\sum\limits_{j=1}^na_{ij}\left(
\alpha_i\partial_tu_j+\partial_xu_j+\beta_i\partial_yu_j+\gamma_i(x,y,t)u_j\right)\right)_{i=1}^n,
$$
$$
Du
:=
\left(\sum\limits_{j=1}^n
b_{ij}(x,y,t)u_j\right)_{i=1}^n.
$$
Moreover, we will need a modified notation for $\alpha_i$ and $\beta_i$, namely,
$$
\tilde\alpha_i=
\cases{0
&if
$b_{ij}\equiv 0$ for all $1\le j\le n$,
 \cr \alpha_i
&otherwise
, \cr}
$$
$$
\tilde\beta_i=
\cases{0
&if
$b_{ij}\equiv 0$ for all $1\le j\le n$,
 \cr \beta_i
&otherwise
. \cr}
$$

\begin{thm}\label{thm:fredh}
Assume that $\gamma_i, b_{ij}\in{\mbox{C}}_{per}^1\left([0,1]\times\mathbb{R}\times\mathbb{R}\right)$,
$\det(a_{ij})_{i,j=1}^n\ne 0$, and
\begin{equation}\label{eq:ab}
(\tilde\beta_i-\tilde\beta_j)(\tilde\alpha_j-\tilde\alpha_s)-
(\tilde\beta_j-\tilde\beta_s)(\tilde\alpha_i-\tilde\alpha_j)\ne 0
\end{equation}
for all $i,j,s\in\{1,\dots,n\}$ with $1\le i\le l$, $l+1\le j\le k$, $k+1\le s\le n$
unless $\tilde\alpha_i=\tilde\alpha_j=0$.
Then the following is true:

(i) The operator $C$ is an isomorphism from $V$ onto $W$.

(ii) The operator $C+D$ is Fredholm 
from $V$ into $W$.
\end{thm}


\begin{proof}
To prove (i), we have to show that, given $f\in W$, there exists a unique $u\in V$ such that
\begin{equation}\label{eq:15}
\sum\limits_{j=1}^na_{ij}\left(
\alpha_i\partial_tu_j  + \partial_xu_j + \beta_i\partial_yu_j + \gamma_i(x,y,t)u_j
\right) = f_i(x,y,t),\quad i\le n.
\end{equation}
Rewrite (\ref{eq:15}) as
$$
\left(\frac{d}{d\xi}+\gamma_i\right)\left[\sum\limits_{j=1}^na_{ij}u_j\left(\xi,
y+\beta_i(\xi-x),t+\alpha_i(\xi-x)\right)\right]\bigg|_{\xi=x} = f_i(x,y,t), \quad i\le n.
$$
Set
$$
  A_0=\left(
  \begin{array}{cccccc}
    A_1&0 \\
    0& A_2
  \end{array}
  \right).
$$
On the account of (\ref{eq:blockA}) and the assumption that $\det(a_{ij})_{i,j=1}^n\ne 0$, system (\ref{eq:15}) 
has a unique solution in $V$ explicitely given by the formula
\begin{eqnarray}
\label{16}
\lefteqn{
u_i(x,y,t) = \frac{1}{\det A_0}\sum\limits_{j=1}^k\left(A_{ji}^0\right)^{ad}}\nonumber\\
\displaystyle
&&
\times
\int\limits_0^x\exp\left\{\int\limits_x^\xi\gamma_j(\xi_1,y+\beta_j(\xi_1-x),t+\alpha_j(\xi_1-x))\,d\xi_1\right\}\nonumber\\
\displaystyle
&&
\times
f_j(\xi,y+\beta_j(\xi-x),t+\alpha_j(\xi-x))\,d\xi,\qquad i\le k;\nonumber\\
\lefteqn{
u_i(x,y,t) = \frac{1}{\det A_3}\sum\limits_{j=k+1}^n\left(A_{ji}^3\right)^{ad}}\\
\displaystyle
&&
\times
\int\limits_x^1\exp\left\{\int\limits_x^\xi\gamma_j(\xi_1,y+\beta_j(\xi_1-x),t+\alpha_j(\xi_1-x))\,d\xi_1\right\}\nonumber\\
\displaystyle
&&
\times
f_j(\xi,y+\beta_j(\xi-x),t+\alpha_j(\xi-x))\,d\xi,\qquad k+1\le i\le n.\nonumber
\end{eqnarray}
Here and subsequently, $\left\{(A_{ij}^s)^{ad}\right\}_{i,j}$ stands for the
adjoint matrix to $A_s$.
Assertion (i) is therewith proved.

To prove Assertion (ii), note that $C+D$ is Fredholm from $V$ into $W$ iff $I+DC^{-1}$ is Fredholm
from $W$ into $W$, where $I$ is the identity in $W$. Set $K=DC^{-1}$ and apply Theorem~\ref{thm:criter}
with $n=3$. Our task is therefore reduced to showing that $K^3$ is compact.

Take a bounded set $N\subset W$ and let $M$ be its image under $(DC^{-1})^3$. To show that $M$ is precompact in 
$W$, we use Arzela-Ascoli precompactness criterion in ${\mbox{C}}\left([0,1]\times[0,Y]\times[0,T]\right)$.
As $(DC^{-1})^3$ is a bounded operator on $W$, the set $M$ is uniformly bounded in $W$. It remains to
check the equicontinuity property of $M$ in $W$. Given $\overline u\in W$, set 
$\tilde u:=DC^{-1}D\overline u$. On the account of the representation (\ref{eq:15}) for $C^{-1}$, we get
\begin{eqnarray}
\label{eq:16}
\lefteqn{
\tilde u_i(x,y,t) = 
\sum\limits_{j=k+1}^nb_{ij}(x,y,t)\frac{1}{\det A_3}
\sum\limits_{r=k+1}^n\left(A_{rj}^3\right)^{ad}}\nonumber\\
\displaystyle
&&\times
\int\limits_{x}^1\exp\left\{\int\limits_{x}^\xi\gamma_r
(\xi_1,y+\beta_r(\xi_1-x),t+\alpha_r(\xi_1-x))\,d\xi_1\right\}\nonumber\\
\displaystyle
&&\times
\sum\limits_{q=l+1}^k(b_{rq}\overline u_q)(\xi,y+\beta_r(\xi-x),t+\alpha_r(\xi-x))\,d\xi,\qquad i\le l;\nonumber\\
\displaystyle
\lefteqn{
\tilde u_i(x,y,t) = 
\sum\limits_{j=1}^lb_{ij}(x,y,t)\frac{1}{\det A_1}
\sum\limits_{r=1}^l\left(A_{rj}^1\right)^{ad}}\nonumber\\
\displaystyle
&&\times
\int\limits_0^x\exp\left\{\int\limits_{x}^\xi\gamma_r
(\xi_1,y+\beta_r(\xi_1-x),t+\alpha_r(\xi_1-x))\,d\xi_1\right\}\\
\displaystyle
&&\times
\sum\limits_{q=k+1}^n(b_{rq}\overline u_q)(\xi,y+\beta_r(\xi-x),t+\alpha_r(\xi-x))\,d\xi,\qquad l+1\le i\le k;\nonumber\\
\displaystyle
\lefteqn{
\tilde u_i(x,y,t) = 
\sum\limits_{j=l+1}^kb_{ij}(x,y,t)\frac{1}{\det A_2}
\sum\limits_{r=l+1}^k\left(A_{rj}^2\right)^{ad}}\nonumber\\
\displaystyle
&&\times
\int\limits_0^x\exp\left\{\int\limits_{x}^\xi\gamma_r
(\xi_1,y+\beta_r(\xi_1-x),t+\alpha_r(\xi_1-x))\,d\xi_1\right\}\nonumber\\
\displaystyle
&&\times
\sum\limits_{q=1}^l(b_{rq}\overline u_q)(\xi,y+\beta_r(\xi-x),t+\alpha_r(\xi-x))\,d\xi,\qquad k+1\le i\le n.
\nonumber
\end{eqnarray}
Obviously, given $f\in W$, we have $\tilde u=\left(DC^{-1}\right)^3f$ iff $\tilde u$ is 
defined by (\ref{eq:16}) with $\overline u=C^{-1}DC^{-1}f$ explicitely given by
\begin{eqnarray*}
\lefteqn{
\overline u_i(x,y,t) = 
\frac{1}{\det A_1}
\sum\limits_{j=1}^l\left(A_{ji}^1\right)^{ad}}\\
\displaystyle
&&\times
\int\limits_{0}^x\exp\left\{\int\limits_{x}^\xi\gamma_j
(\xi_1,y+\beta_j(\xi_1-x),t+\alpha_j(\xi_1-x))\,d\xi_1\right\}\\
\displaystyle
&&\times
\sum\limits_{r=k+1}^nb_{jr}(\xi,y+\beta_j(\xi-x),t+\alpha_j(\xi-x))\,d\xi
\frac{1}{\det A_3}
\sum\limits_{q=k+1}^n\left(A_{qr}^3\right)^{ad}\\
\displaystyle
&&\times
\int\limits_{\xi}^1\exp\left\{\int\limits_{\xi}^{\xi_1}\gamma_q
(\xi_2,y+\beta_j(\xi-x)+\beta_q(\xi_2-\xi),t+\alpha_j(\xi-x)+\alpha_q(\xi_2-\xi))\,d\xi_2\right\}\\
\displaystyle
&&\times
f_q(\xi_1,y+\beta_j(\xi-x)+\beta_q(\xi_1-\xi),t+\alpha_j(\xi-x)+\alpha_q(\xi_1-\xi))\,d\xi_1,\quad
i\le l,\\
\lefteqn{
\overline u_i(x,y,t) = 
\frac{1}{\det A_2}
\sum\limits_{j=l+1}^k\left(A_{ji}^2\right)^{ad}}\\
\displaystyle
&&\times
\int\limits_{0}^x\exp\left\{\int\limits_{x}^\xi\gamma_j
(\xi_1,y+\beta_j(\xi_1-x),t+\alpha_j(\xi_1-x))\,d\xi_1\right\}\\
\displaystyle
&&\times
\sum\limits_{r=1}^lb_{jr}(\xi,y+\beta_j(\xi-x),t+\alpha_j(\xi-x))\,d\xi
\frac{1}{\det A_1}
\sum\limits_{q=1}^l\left(A_{qr}^1\right)^{ad}\\
\displaystyle
&&\times
\int\limits_{0}^\xi\exp\left\{\int\limits_{\xi}^{\xi_1}\gamma_q
(\xi_2,y+\beta_j(\xi-x)+\beta_q(\xi_2-\xi),t+\alpha_j(\xi-x)+\alpha_q(\xi_2-\xi))\,d\xi_2\right\}\\
\displaystyle
&&\times
f_q(\xi_1,y+\beta_j(\xi-x)+\beta_q(\xi_1-\xi),t+\alpha_j(\xi-x)+\alpha_q(\xi_1-\xi))\,d\xi_1,\quad
l+1\le i\le k,\\
\lefteqn{
\overline u_i(x,y,t) = 
\frac{1}{\det A_3}
\sum\limits_{j=k+1}^n\left(A_{ji}^3\right)^{ad}}\\
\displaystyle
&&\times
\int\limits_{x}^1\exp\left\{\int\limits_{x}^\xi\gamma_j
(\xi_1,y+\beta_j(\xi_1-x),t+\alpha_j(\xi_1-x))\,d\xi_1\right\}\\
\displaystyle
&&\times
\sum\limits_{r=l+1}^kb_{jr}(\xi,y+\beta_j(\xi-x),t+\alpha_j(\xi-x))\,d\xi
\frac{1}{\det A_2}
\sum\limits_{q=l+1}^k\left(A_{qr}^2\right)^{ad}\\
\displaystyle
&&\times
\int\limits_{0}^\xi\exp\left\{\int\limits_{\xi}^{\xi_1}\gamma_q
(\xi_2,y+\beta_j(\xi-x)+\beta_q(\xi_2-\xi),t+\alpha_j(\xi-x)+\alpha_q(\xi_2-\xi))\,d\xi_2\right\}\\
\displaystyle
&&\times
f_q(\xi_1,y+\beta_j(\xi-x)+\beta_q(\xi_1-\xi),t+\alpha_j(\xi-x)+\alpha_q(\xi_1-\xi))\,d\xi_1,\quad
k+1\le i\le n.
\end{eqnarray*}
 To prove the desired equicontinuity property we have
to show the existence of a function $\alpha: \mathbb{R}_+\to\mathbb{R}$ with $\alpha(p)\to 0$ as $p\to 0$
for which we have
\begin{equation}\label{eq:18}
\left\|\tilde u(x+h_1,y+h_2,t+h_3)-\tilde u(x,y,t)\right\|\le\alpha\left(|h|\right)\|f\|_W,
\end{equation}
the estimate being uniform in $f\in W$ and $h=(h_1,h_2,h_3)\in\mathbb{R}^3$. To achieve this property 
we transform the expression for $\tilde u$ to a form appropriate for our purposes. We make
calculations only for one summand contributing into $\tilde u$ 
(similar argument works for all other summands as well).
Specifically, we will consider the following summand (up to a multiplicative  constant):
\begin{eqnarray}
\label{eq:*}
\lefteqn{
b_{ij}(x,y,t)\int\limits_{0}^x\exp\left\{\int\limits_{x}^\xi\gamma_r
(\xi_1,y+\beta_r(\xi_1-x),t+\alpha_r(\xi_1-x))\,d\xi_1\right\}}\nonumber\\
\displaystyle
&&\times b_{rq}(\xi,y+\beta_r(\xi-x),t+\alpha_r(\xi-x))\,d\xi\nonumber\\
\displaystyle
&&\times\int\limits_{\xi}^1\exp\left\{\int\limits_{\xi}^{\xi_1}\gamma_p
(\xi_2,y+\beta_r(\xi-x)+\beta_p(\xi_2-\xi),t+\alpha_r(\xi-x)+\alpha_p(\xi_2-\xi))\,d\xi_2\right\}\nonumber\\
\displaystyle
&&\times  b_{ps}(\xi_1,y+\beta_r(\xi-x)+\beta_p(\xi_1-\xi),t+\alpha_r(\xi-x)+\alpha_p(\xi_1-\xi))\,d\xi_1\\
\displaystyle
&&\times\int\limits_{0}^{\xi_1}\exp\biggl\{\int\limits_{\xi_1}^{\xi_2}\gamma_m
(\xi_3,y+\beta_r(\xi-x)+\beta_p(\xi_1-\xi)+\beta_m(\xi_3-\xi_1),\nonumber\\
\displaystyle
&&t+\alpha_r(\xi-x)+\alpha_p(\xi_1-\xi)+\alpha_m(\xi_3-\xi_1))\,d\xi_3\biggr\}\nonumber\\
\displaystyle
&&\times  f_m(\xi_2,y+\beta_r(\xi-x)+\beta_p(\xi_1-\xi)+\beta_m(\xi_2-\xi_1),\nonumber\\
\displaystyle
&&t+\alpha_r(\xi-x)+\alpha_p(\xi_1-\xi)+\alpha_m(\xi_2-\xi_1))\,d\xi_2.
\nonumber
\end{eqnarray}
Changing the order of integration, we have
\begin{eqnarray}
\label{eq:19}
\lefteqn{
\int\limits_{0}^xd\xi\int\limits_{\xi}^1d\xi_1\int\limits_{0}^{\xi_1}d\xi_2 =
\int\limits_{0}^xd\xi\left[\int\limits_{0}^{\xi}d\xi_2\int\limits_{\xi}^{1}d\xi_1 +
\int\limits_{\xi}^1d\xi_2\int\limits_{\xi_2}^{1}d\xi_1\right]}\nonumber\\
\displaystyle
&&
=\int\limits_{0}^xd\xi_2\int\limits_{\xi_2}^xd\xi\int\limits_{\xi}^1d\xi_1 +
\int\limits_{0}^xd\xi_2\int\limits_{0}^{\xi_2}d\xi\int\limits_{\xi_2}^1d\xi_1 +
\int\limits_{x}^1d\xi_2\int\limits_{0}^xd\xi\int\limits_{\xi}^1d\xi_1.
\end{eqnarray}
Moreover, we introduce new variables $\mu$ and $\eta$ (instead of $\xi$ and $\xi_1$) by
\begin{equation}\label{eq:23}
\begin{array}{ccc}
\displaystyle
\mu:=y-\beta_rx+\xi(\beta_r-\beta_p)+\xi_1(\beta_p-\beta_m)\\
\displaystyle
\eta:=t-\alpha_rx+\xi(\alpha_r-\alpha_p)+\xi_1(\alpha_p-\alpha_m).
\end{array}
\end{equation}
Owing to (\ref{eq:blockB}), the integers $r, p, m$ belong to three different sets
$\{1,\dots,l\}$, $\{l+1,\dots,k\}$, and $\{k+1,\dots,n\}$. Thus, on the account of
(\ref{eq:ab}), the mapping (\ref{eq:23}) is non-generate.
By means of (\ref{eq:19}) and (\ref{eq:23}) the summand (\ref{eq:*}) can be transformed to
\begin{eqnarray}
\label{eq:**}
\lefteqn{
\int\limits_{0}^xd\xi_2\int\!\!\int_{\triangle_1}F(x,y,t,\xi_2,\mu,\eta)
f_m(\xi_2,\mu,\eta)\,\d\mu\,d\eta}
\nonumber\\
&&
+\int\limits_{0}^xd\xi_2\int\!\!\int_{\Pi}F(x,y,t,\xi_2,\mu,\eta)
f_m(\xi_2,\mu,\eta)\,\d\mu\,d\eta\\
&&
+\int\limits_{x}^1d\xi_2\int\!\!\int_{\triangle_2}F(x,y,t,\xi_2,\mu,\eta)
f_m(\xi_2,\mu,\eta)\,\d\mu\,d\eta,
\nonumber
\end{eqnarray}
the triangles $\triangle_1$ and $\triangle_2$ being the images 
under the mapping (\ref{eq:23}) of the triangles on the plane $(\xi,\xi_1)$ with the 
vertices, respectively, $(\xi_2,\xi_2)$, $(\xi_2,1)$, $(x,1)$ and $(0,0)$, $(0,1)$, $(x,1)$;
similarly, the quadrangle $\Pi$  being the  image
under the mapping (\ref{eq:23}) of the quadrangle  on the plane $(\xi,\xi_1)$ with the 
vertices $(0,\xi_2)$, $(0,1)$, $(\xi_2,1)$, $(\xi_2,\xi_2)$. Moreover, $F$ is a known function
of $b_{ij}, b_{rq}$, and $b_{ps}$ continuously differentiable in $x,y,t$. 
Our task is therefore reduced to obtaining the estimate (\ref{eq:18}) with $\tilde u$ 
replaced by (\ref{eq:**}). This is a straightforward consequence of the fact that the 
lines bounding $\triangle_1$, $\triangle_2$, and $\Pi$ depend linearly on $x,y,t$. The latter 
is true due to linearity of (\ref{eq:23}) in $x,y,t$. 
The proof is therewith complete.
\end{proof}

\begin{rem}
Note that in the course of proving Theorem~2, we applied Theorem~1 with $n=3$. 
This choice is essential: a simple analysis of our argument shows that $n=2$ 
would not work. This is the main motivation of extension of 
our approach from~\cite{KmRe} made in the current paper.
\end{rem}

\subsection*{Acknowledgments}

I thank Lutz Recke for useful discussions.





