Integrability of the Bakirov system:\\a zero-curvature representation
\textsc{S.~Yu.~Sakovich}\bigskip
For the Bakirov system, which is known to possess only one higher-order local generalized symmetry, a zero-curvature representation containing an essential parameter is found.
nlin0206034.tex
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\title{\textbf{Integrability of the Bakirov system:}\\\textbf{a zero-curvature
representation}}
\author{\textsc{S.~Yu.~Sakovich}\bigskip\\{\footnotesize Institute of Physics,
National Academy of Sciences, 220072 Minsk, Belarus}\\{\footnotesize E-mail:
saks@pisem.net}}
\date{}
\maketitle

\begin{abstract}
For the Bakirov system, which is known to possess only one higher-order local
generalized symmetry, a zero-curvature representation containing an essential
parameter is found.
\end{abstract}

\begin{center}
\rule{48pt}{1pt}\medskip
\end{center}

Recently, Sergyeyev \cite{Ser} gave the final proof of that the Bakirov system%
\begin{equation}
u_{t}=u_{xxxx}+v^{2},\qquad v_{t}=\tfrac{1}{5}v_{xxxx} \label{bak}%
\end{equation}
possesses only one higher-order local generalized symmetry, namely, the
sixth-order one.

In the present paper, we find a linear spectral problem associated with the
Bakirov system (\ref{bak}), i.e.~a zero-curvature representation with an
essential parameter. We hope that the obtained result will be useful in future
studies of the relation between Lax pairs and higher symmetries.

A zero-curvature representation (ZCR) of a system of PDEs is the compatibility
condition%
\begin{equation}
D_{t}X=D_{x}T-\left[  X,T\right]  \label{zcr}%
\end{equation}
of the over-determined linear problem%
\begin{equation}
\Psi_{x}=X\Psi,\qquad\Psi_{t}=T\Psi, \label{lin}%
\end{equation}
where $D_{t}$ and $D_{x}$ are the total derivatives, $X$ and $T$ are $n\times
n$ matrix functions of independent and dependent variables and finite-order
derivatives of dependent variables, the square brackets denote the matrix
commutator, $\Psi$ is a column of $n$ functions of independent variables, and
the ZCR (\ref{zcr}) is satisfied by any solution of the represented system of
PDEs. Two ZCRs are equivalent if they are related by a gauge transformation%
\begin{align}
X^{\prime}  &  =GXG^{-1}+\left(  D_{x}G\right)  G^{-1},\nonumber\\
T^{\prime}  &  =GTG^{-1}+\left(  D_{t}G\right)  G^{-1},\nonumber\\
\Psi^{\prime}  &  =G\Psi,\qquad\det G\neq0, \label{gtr}%
\end{align}
where $G$ is an $n\times n$ matrix function of independent and dependent
variables and finite-order derivatives of dependent variables.

Our aim is to find a ZCR (\ref{zcr}) of the Bakirov system (\ref{bak}).
Assuming for simplicity that $X=X\left(  u,v\right)  $ and $T=T\left(
u,v,u_{x},v_{x},u_{xx},v_{xx},u_{xxx},v_{xxx}\right)  $, and using
(\ref{bak}), we rewrite (\ref{zcr}) in the equivalent form%
\begin{equation}
\left(  u_{xxxx}+v^{2}\right)  \frac{\partial X}{\partial u}+\frac{1}%
{5}v_{xxxx}\frac{\partial X}{\partial v}-D_{x}T+\left[  X,T\right]  =0.
\label{ide}%
\end{equation}
Since (\ref{ide}) cannot be a system of ODEs restricting solutions of
(\ref{bak}), it must be an identity with respect to $u$ and $v$, and therefore
$u$, $v$ and all the derivatives of $u$ and $v$ are mutually independent
variables in (\ref{ide}). This allows us to solve (\ref{ide}) and obtain the
following expressions for the matrices $X$ and $T$:%
\begin{align}
X  &  =Pu+Qv+R,\nonumber\\
T  &  =Pu_{xxx}+\tfrac{1}{5}Qv_{xxx}+\left[  R,P\right]  u_{xx}+\tfrac{1}%
{5}\left[  R,Q\right]  v_{xx}+\left[  R,\left[  R,P\right]  \right]
u_{x}\nonumber\\
&  +\tfrac{1}{5}\left[  R,\left[  R,Q\right]  \right]  v_{x}+\left[  R,\left[
R,\left[  R,P\right]  \right]  \right]  u+\tfrac{1}{5}\left[  R,\left[
R,\left[  R,Q\right]  \right]  \right]  v+S, \label{fxt}%
\end{align}
where $P$, $Q$, $R$ and $S$ are any $n\times n$ constant matrices satisfying
the following commutator relations:%
\begin{gather}
P=-\tfrac{1}{5}\left[  Q,\left[  R,\left[  R,\left[  R,Q\right]  \right]
\right]  \right]  ,\qquad\left[  P,Q\right]  =0,\nonumber\\
\left[  P,\left[  Q,R\right]  \right]  =0,\qquad\left[  P,\left[  R,P\right]
\right]  =0,\qquad\left[  Q,\left[  R,Q\right]  \right]  =0,\nonumber\\
\left[  \left[  R,P\right]  ,\left[  R,Q\right]  \right]  =0,\qquad\left[
P,\left[  R,\left[  R,\left[  R,P\right]  \right]  \right]  \right]
=0,\nonumber\\
\left[  \left[  R,P\right]  ,\left[  R,\left[  R,Q\right]  \right]  \right]
=0,\qquad\left[  P,S\right]  +\left[  R,\left[  R,\left[  R,\left[
R,P\right]  \right]  \right]  \right]  =0,\nonumber\\
\left[  Q,S\right]  +\tfrac{1}{5}\left[  R,\left[  R,\left[  R,\left[
R,Q\right]  \right]  \right]  \right]  =0,\qquad\left[  R,S\right]  =0.
\label{rel}%
\end{gather}


We have to find a solution of (\ref{rel}), nontrivial in the following sense:
$X$ contains both $u$ and $v$, i.e.~(\ref{zcr}) gives expressions for both
equations of (\ref{bak}); $\left[  X,T\right]  \neq0$, because commutative
ZCRs are simply matrices of conservation laws (for this reason, and without
loss of generality, the matrices $P$, $Q$, $R$ and $S$ are set to be
traceless); $X$ contains a parameter (`essential' or `spectral') which cannot
be removed by gauge transformations (\ref{gtr}). We solve (\ref{rel}), using
the \textit{Mathematica} computer algebra system \cite{Wol}, successively
taking $Q$ in all possible Jordan forms, suppressing the excessive
arbitrariness of solutions by transformations (\ref{gtr}) with constant $G$,
and increasing the matrix dimension $n$. The cases of $2\times2$ and
$3\times3$ matrices contain no nontrivial solutions of (\ref{rel}), while the
case $4\times4$ gives us the following:%
\begin{equation}
P=%
\begin{pmatrix}
0 & 0 & \frac{8}{5}z\left(  -3+6z-11z^{2}\right)  \alpha^{3} & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
,\label{map}%
\end{equation}%
\begin{equation}
Q=%
\begin{pmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{pmatrix}
,\label{maq}%
\end{equation}%
\begin{equation}
R=%
\begin{pmatrix}
\alpha & 0 & 0 & 0\\
0 & z\alpha & 0 & \alpha\\
0 & 0 & \left(  -1+2z\right)  \alpha & 0\\
0 & \left(  -3+6z-11z^{2}\right)  \alpha & 0 & -3z\alpha
\end{pmatrix}
,\label{mar}%
\end{equation}%
\begin{equation}
S=%
\begin{pmatrix}
S_{11} & 0 & 0 & 0\\
0 & S_{22} & 0 & S_{24}\\
0 & 0 & S_{33} & 0\\
0 & S_{42} & 0 & S_{44}%
\end{pmatrix}
\label{mas}%
\end{equation}
with%
\begin{align}
S_{11} &  =\tfrac{8}{5}\left(  2-12z+21z^{2}-18z^{3}+3z^{4}\right)  \alpha
^{4},\nonumber\\
S_{22} &  =\tfrac{8}{5}\left(  3-10z+15z^{2}-4z^{4}\right)  \alpha
^{4},\nonumber\\
S_{24} &  =\tfrac{8}{5}\left(  -1+3z+z^{2}-3z^{3}\right)  \alpha
^{4},\nonumber\\
S_{33} &  =\tfrac{8}{5}\left(  -8+28z-39z^{2}+22z^{3}-7z^{4}\right)
\alpha^{4},\nonumber\\
S_{42} &  =\tfrac{8}{5}\left(  3-15z+26z^{2}-18z^{3}-29z^{4}+33z^{5}\right)
\alpha^{4},\nonumber\\
S_{44} &  =\tfrac{8}{5}\left(  3-6z+3z^{2}-4z^{3}+8z^{4}\right)  \alpha
^{4},\label{sss}%
\end{align}
where $\alpha$ is an arbitrary parameter, and $z$ is any of the four roots%
\begin{align}
z_{1,2} &  =\frac{1}{20}\left(  9+\mathrm{i}\sqrt{39}\pm\sqrt{-138-2\mathrm{i}%
\sqrt{39}}\right)  ,\nonumber\\
z_{3,4} &  =\frac{1}{20}\left(  9-\mathrm{i}\sqrt{39}\pm\sqrt{-138+2\mathrm{i}%
\sqrt{39}}\right)  \label{zzz}%
\end{align}
of the algebraic equation%
\begin{equation}
3-12z+21z^{2}-18z^{3}+10z^{4}=0.\label{eqz}%
\end{equation}


Consequently, a nontrivial ZCR (\ref{zcr}) of the Bakirov system (\ref{bak})
is determined by the following $4\times4$ matrices $X$ and $T$:%
\begin{equation}
X=%
\begin{pmatrix}
\alpha & v & \frac{8}{5}z\left(  -3+6z-11z^{2}\right)  \alpha^{3}u & 0\\
0 & z\alpha & v & \alpha\\
0 & 0 & \left(  -1+2z\right)  \alpha & 0\\
0 & \left(  -3+6z-11z^{2}\right)  \alpha & 0 & -3z\alpha
\end{pmatrix}
\label{max}%
\end{equation}
and%
\begin{equation}
T=%
\begin{pmatrix}
T_{11} & T_{12} & T_{13} & T_{14}\\
0 & T_{22} & T_{23} & T_{24}\\
0 & 0 & T_{33} & 0\\
0 & T_{42} & T_{43} & T_{44}%
\end{pmatrix}
\label{mat}%
\end{equation}
with%
\begin{align}
T_{11}  &  =\tfrac{8}{5}\left(  2-12z+21z^{2}-18z^{3}+3z^{4}\right)
\alpha^{4},\nonumber\\
T_{12}  &  =\tfrac{1}{5}\left(  -4\left(  2-3z+6z^{2}+3z^{3}\right)
\alpha^{3}v-2\left(  1-2z+5z^{2}\right)  \alpha^{2}v_{x}\right. \nonumber\\
&  \left.  +\left(  1-z\right)  \alpha v_{xx}+v_{xxx}\right)  ,\nonumber\\
T_{13}  &  =\tfrac{8}{5}z\left(  -3+6z-11z^{2}\right)  \alpha^{3}\left(
8\left(  1-z\right)  ^{3}\alpha^{3}u+4\left(  1-z\right)  ^{2}\alpha^{2}%
u_{x}\right. \nonumber\\
&  \left.  +2\left(  1-z\right)  \alpha u_{xx}+u_{xxx}\right)  ,\nonumber\\
T_{14}  &  =\tfrac{1}{5}\alpha\left(  4z\left(  -3+z\right)  \alpha
^{2}v-2\left(  1+z\right)  \alpha v_{x}-v_{xx}\right)  ,\nonumber\\
T_{22}  &  =\tfrac{8}{5}\left(  3-10z+15z^{2}-4z^{4}\right)  \alpha
^{4},\nonumber\\
T_{23}  &  =\tfrac{1}{5}\left(  4\left(  -2+9z-18z^{2}+19z^{3}\right)
\alpha^{3}v-2\left(  1-2z+5z^{2}\right)  \alpha^{2}v_{x}\right. \nonumber\\
&  \left.  +\left(  1-z\right)  \alpha v_{xx}+v_{xxx}\right)  ,\nonumber\\
T_{24}  &  =\tfrac{8}{5}\left(  -1+3z+z^{2}-3z^{3}\right)  \alpha
^{4},\nonumber\\
T_{33}  &  =\tfrac{8}{5}\left(  -8+28z-39z^{2}+22z^{3}-7z^{4}\right)
\alpha^{4},\nonumber\\
T_{42}  &  =\tfrac{8}{5}\left(  3-15z+26z^{2}-18z^{3}-29z^{4}+33z^{5}\right)
\alpha^{4},\nonumber\\
T_{43}  &  =\tfrac{1}{5}\left(  -3+6z-11z^{2}\right)  \alpha\left(  4z\left(
-3+5z\right)  \alpha^{2}v\right. \nonumber\\
&  \left.  +\left(  2-6z\right)  \alpha v_{x}+v_{xx}\right)  ,\nonumber\\
T_{44}  &  =\tfrac{8}{5}\left(  3-6z+3z^{2}-4z^{3}+8z^{4}\right)  \alpha^{4}.
\label{ttt}%
\end{align}
Let us remind that, in (\ref{max})--(\ref{ttt}), $z$ is any of the roots
(\ref{zzz}) of the equation (\ref{eqz}), and $\alpha$ is an arbitrary parameter.

Now, we have to prove that $\alpha$ is an essential parameter, i.e.~that
$\alpha$ cannot be removed from the obtained ZCR by a gauge transformation
(\ref{gtr}). We do this, using the method of gauge-invariant description of
ZCRs of evolution equations \cite{Sak} (see also the independent work
\cite{M97}, based on the very general and abstract study of ZCRs \cite{M92}).
Since the matrix $X$ (\ref{max}) does not contain derivatives of $u$ and $v$,
the two characteristic matrices of the obtained ZCR are simply $C_{u}=\partial
X/\partial u=P$ and $C_{v}=\partial X/\partial v=Q$. We take one of them,
$C_{u}=P$ (\ref{map}), introduce the operator $\nabla_{x}$, defined as
$\nabla_{x}M=D_{x}M-\left[  X,M\right]  $ for any $4\times4$ matrix function
$M$, compute $\nabla_{x}C_{u}$, and find that%
\begin{equation}
\nabla_{x}C_{u}+2\left(  1-z\right)  \alpha C_{u}=0. \label{clo}%
\end{equation}
In the terminology of \cite{Sak}, the relation (\ref{clo}) is one of the two
closure equations of the cyclic basis. The scalar coefficient $2\left(
1-z\right)  \alpha$ in (\ref{clo}) is an invariant with respect to the gauge
transformations (\ref{gtr}), because the matrices $C_{u}$ and $\nabla_{x}%
C_{u}$ are transformed as $C_{u}^{\prime}=GC_{u}G^{-1}$ and $\nabla
_{x}^{\prime}C_{u}^{\prime}=G\left(  \nabla_{x}C_{u}\right)  G^{-1}$ (see
\cite{Sak} and \cite{M97}). The explicit dependence of the invariant $2\left(
1-z\right)  \alpha$ on the parameter $\alpha$ shows that this parameter cannot
be `gauged out' from the matrix $X$ (\ref{max}).

We think that the ZCR of the Bakirov system, found in this paper, can be used
in future studies of the relation between Lax pairs, recursion operators and
generalized symmetries. The following problems arise from the obtained result.
Is it possible to derive a recursion operator for the Bakirov system from its
ZCR? If not, why? If there exists a recursion operator of the Bakirov system,
why the operator does not produce infinitely many local generalized symmetries?

\begin{center}
\rule{48pt}{1pt}
\end{center}

The author is grateful to Dr.~A.~Sergyeyev for useful information on
symmetries of the Ibragimov--Shabat and Bakirov systems.

\begin{center}
\rule{48pt}{1pt}
\end{center}

\begin{thebibliography}{9}
\bibitem {Ser}A.~Sergyeyev, Symmetries and integrability: Bakirov system
revisited, \textit{J.~Phys. A: Math. Gen.} \textbf{34} (2001) 4983--4990.

\bibitem {Wol}S.~Wolfram, \textit{The Mathematica Book}, 4th ed. (Cambridge
University Press, Cambridge, 1999).

\bibitem {Sak}S.Yu.~Sakovich, On zero-curvature representations of evolution
equations, \textit{J.~Phys. A: Math. Gen.} \textbf{28} (1995) 2861--2869.

\bibitem {M97}M.~Marvan, A direct procedure to compute zero-curvature
representations. The case $\mathrm{sl}_{2}$, \textit{International Conference
on Secondary Calculus and Cohomological Physics}, Moscow, Russia, August
24--31, 1997, 9 pp. (ELibEMS, www.emis.de/proceedings/SCCP97).

\bibitem {M92}M.~Marvan, On zero-curvature representations of partial
differential equations, \textit{Differential Geometry and Its Applications},
eds. O.~Kowalski and D.~Krupka (Silesian University, Opava, 1993) 103--122.
(ELibEMS, www.emis.de/proceedings/5ICDGA).
\end{thebibliography}

\end{document}
nlin0206034.extracted_bib
\begin{thebibliography}{9}
\bibitem {Ser}A.~Sergyeyev, Symmetries and integrability: Bakirov system
revisited, \textit{J.~Phys. A: Math. Gen.} \textbf{34} (2001) 4983--4990.

\bibitem {Wol}S.~Wolfram, \textit{The Mathematica Book}, 4th ed. (Cambridge
University Press, Cambridge, 1999).

\bibitem {Sak}S.Yu.~Sakovich, On zero-curvature representations of evolution
equations, \textit{J.~Phys. A: Math. Gen.} \textbf{28} (1995) 2861--2869.

\bibitem {M97}M.~Marvan, A direct procedure to compute zero-curvature
representations. The case $\mathrm{sl}_{2}$, \textit{International Conference
on Secondary Calculus and Cohomological Physics}, Moscow, Russia, August
24--31, 1997, 9 pp. (ELibEMS, www.emis.de/proceedings/SCCP97).

\bibitem {M92}M.~Marvan, On zero-curvature representations of partial
differential equations, \textit{Differential Geometry and Its Applications},
eds. O.~Kowalski and D.~Krupka (Silesian University, Opava, 1993) 103--122.
(ELibEMS, www.emis.de/proceedings/5ICDGA).
\end{thebibliography}