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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 310, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/310\hfil Cyclic factorizable derivations]
{Fields of rational constants of cyclic factorizable derivations}

\author[J. Zieli\'nski \hfil EJDE-2015/310\hfilneg]
{Janusz Zieli\'nski}

\address{Janusz Zieli\'nski \newline
Faculty of Mathematics and Computer Science,
N. Copernicus University, ul. Chopina 12/18,
87-100 Toru\'n, Poland}
\email{ubukrool@mat.uni.torun.pl}

\thanks{Submitted November 12, 2014. Published December 21, 2015.}
\subjclass[2010]{34A34, 13N15, 12H05, 92D25}
\keywords{Lotka-Volterra derivation; factorizable derivation;
\hfill\break\indent rational constant; rational first integral}

\begin{abstract}
 We describe all rational constants of a large family of four-variable
 cyclic factorizable derivations. Thus, we determine all rational
 first integrals of their corresponding systems of differential equations.
 Moreover, we give a characteristic of all four-variable Lotka-Volterra 
 derivations  with a nontrivial rational constant.
 All considerations are over an arbitrary field of characteristic zero.
 Our main tool is the investigation of the cofactors of strict Darboux 
 polynomials.  Factorizable derivations are important in derivation theory.
 Namely, we may associate the factorizable  derivation with any given
 derivation of a polynomial ring and that construction
 helps to determine rational constants of arbitrary derivations.
 Besides, Lotka-Volterra systems play a significant role in population 
 biology, laser physics and plasma physics.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

One of the main results of the paper is Theorem~\ref{gljedno}, which
gives the description of the fields of rational constants
of a family of four-variable Lotka-Volterra derivations.
This is a generalization of Theorem \ref{gl},
which describes the rings of polynomial constants.
As an important consequence we obtain Corollary \ref{exist},
which characterizes all four-variable Lotka-Volterra derivations
with a nontrivial rational constant. Such a problem for
three variables was studied by Moulin Ollagnier in \cite{Mou}.
We extend the results of Theorem~\ref{gljedno} and
Corollary~\ref{exist} to cyclic factorizable
derivations via diagonal automorphisms.
All our considerations are over an arbitrary
field $k$ of characteristic zero.

Recall that if $R$ is a commutative $k$-algebra, then
a $k$-linear mapping $d:R\to R$ is
called a~\emph{derivation} of $R$ if for all $a,b\in R$
$$
d(ab)=ad(b)+d(a)b.
$$
We call $R^d=\ker d$
the \emph{ring of constants} of the derivation $d$.
Then $k\subseteq R^d$ and
a \emph{nontrivial} constant of $d$ is
an element of the set $R^d\setminus k$.

Let us fix some notation:
$\mathbb{Q}_+$ - the set of positive rationals,
$\mathbb{N}$ - the set of nonnegative integers,
$\mathbb{N}_+$ - the set of positive integers,
$n$ - an integer $\geq 3$,
$k[X]:=k[x_1,\ldots ,x_n]$, the ring of polynomials in $n$ variables,
$k(X):=k(x_1,\ldots ,x_n)$, the field of rational functions in $n$ variables.

If $f_1,\ldots ,f_n\in k[X]$, then there exists exactly one
derivation $d:k[X]\to k[X]$ such that $d(x_1)=f_1,
\ldots ,d(x_n)=f_n$. A derivation $d:k[X]\to k[X]$
is called \emph{factorizable} if $d(x_i)=x_if_i,$
where the polynomials $f_i$ are of degree $1$ for $i=1,\ldots ,n$.
We may associate the factorizable derivation with any given derivation
of $k[X]$ and that construction helps to
obtain new facts on constants, especially rational constants,
of the initial derivation (see, for instance, \cite{Mac}, \cite{NZ}).
A derivation $d:k[X]\to k[X]$ is said to be
\emph{cyclic factorizable} if $d(x_i)=x_i(A_ix_{i-1}+B_ix_{i+1}),$
where $A_i, B_i\in k$ for $i=1,\ldots ,n$
(we adopt the convention that $x_{n+1}=x_1$ and
$x_0=x_n$). Special cases of cyclic factorizable derivations
are Lotka-Volterra derivations (see Section 2).

There is no general procedure for determining
all constants of a derivation. Even for a given
derivation the problem may be difficult, see for
instance counterexamples to Hilbert's fourteenth problem
(all of them are of the form $k[X]^d$, however it took more than
a half century to find at least one of them, for more details
we refer the reader to \cite{Now,Kur}) or Jouanolou derivations
(where the rings and fields of constants are trivial, see
\cite{Mac,Now}).

The main motivations of our study are the following:
\begin{itemize}
  \item Lagutinskii's procedure of association of the factorizable
derivation with any given derivation (for instance, \cite{Mac}, \cite{NZ});
  \item applications of Lotka-Volterra systems
in population biology, laser physics and plasma physics
(see, among many others, \cite{Al}, \cite{Bo}, \cite{Cai});
  \item links to invariant theory,
mainly to connected algebraic groups (see \cite{Now})
\end{itemize}

If $\delta $ is a derivation of $k(X)$ such that $\delta (x_i)=f_i$
for $i=1,\ldots ,n$, then
the set $k(X)^{\delta}\setminus k$ coincides with the set of all
rational first integrals of a system of ordinary differential
equations
$$\frac{dx_i(t)}{dt}=f_i(x_1(t),\ldots ,x_n(t)),$$
where $i=1,\ldots ,n$ (for more details
we refer the reader to \cite{Now}). Therefore, we
describe both: all rational constants of a
derivation and all rational first integrals of
its corresponding system of differential equations.

\section{Lotka-Volterra derivations and polynomial constants}

Let $d:k[X]\to k[X]$ be
a cyclic factorizable derivation of the form $d(x_i)=x_i(A_ix_{i-1}+B_ix_{i+1})$
where $A_i, B_i\in k$ for $i=1,\ldots ,n$. Suppose that $A_i\neq 0$ for all $i$.
Consider an automorphism $\sigma :k[X]\to k[X]$ defined by
$\sigma (x_i)=A_{i+1}^{-1} x_i$ for $i=1,\ldots ,n$.
Then $\Delta=\sigma d \sigma^{-1}$ is also a derivation of the ring $k[X]$.
Moreover, $f$ is a nontrivial polynomial (respectively: rational, see Section 3)
constant of a derivation $d$ if and only if $\sigma (f)$ is a nontrivial
polynomial (respectively: rational) constant of a derivation $\Delta$.
Clearly $\sigma^{-1}(x_i)=A_{i+1} x_i$ and a short
computation shows that $\Delta (x_i)=x_i(x_{i-1}-C_ix_{i+1})$ for
$C_i=-B_iA_{i+2}^{-1}$ (we allow $C_i=0$) and $i=1,\ldots ,n$.
We can proceed similarly if $A_i=0$ for some $i$ but $B_i\neq 0$
for all $i$.

Let $C_1,\ldots ,C_n\in k$.
From now on,
$d:k[X]\to k[X]$ is a derivation of the form
$$
d(x_i)=x_i(x_{i-1}-C_ix_{i+1})
$$
for $i=1,\ldots ,n$
(we still adhere to the convention that $x_{n+1}=x_1$ and
$x_0=x_n$).
We call $d$ a \emph{Lotka-Volterra derivation} with
parameters $C_1,\ldots ,C_n$.

Let $n=4$. For arbitrary $C_1, C_2, C_3, C_4\in k$
we may consider the four sentences:
\begin{align*}
s_1:&\quad C_1C_2C_3C_4=1.\\
s_2:&\quad C_1,C_3\in \mathbb{Q}_+ \text{ and } C_1C_3=1.\\
s_3:&\quad C_2,C_4\in \mathbb{Q}_+ \text{ and } C_2C_4=1.\\
s_4:&\quad C_1C_2C_3C_4=-1\quad \text{and $C_i=1$ for two consecutive indices $i$}.
\end{align*}
In case $s_2$ let $C_1=\frac{p}{q}$, where
$p,q\in \mathbb{N}_+$
and $\gcd (p,q)=1$.
In case $s_3$ let $C_2=\frac{r}{t}$, where
$r,t\in \mathbb{N}_+$
and $\gcd (r,t)=1$. In case $s_4$ we define the polynomial $f_4$,
namely for $C_1=C_2=1$ let
$$
f_4=x_1^2+x_2^2+x_3^2+C_3^2x_4^2+2x_1x_2-2x_1x_3
 -2C_3x_1x_4+2x_2x_3-2C_3x_2x_4+2C_3x_3x_4,
$$
for the other possibilities one has to rotate the indices appropriately.

Obviously sentences $s_1$ and $s_4$ are mutually exclusive.
Note also that if $s_2\wedge s_3$,
then $s_1$. This means that we have ten cases to consider,
depending on the truth values of the sentences $s_1, s_2, s_3, s_4$.
Denote by $\neg s$ the negation of the sentence $s$.

\begin{theorem}[{\cite[Theorem 2]{PHJZ}}]  \label{gl}
Let $d:k[X]\to k[X]$ be a derivation of the form
$$
d=\sum_{i=1}^4 x_i(x_{i-1}-C_ix_{i+1})\frac{\partial}{\partial x_i},
$$
where $C_1,C_2,C_3,C_4\in k$.
Then the ring of constants of $d$ is always finitely generated over $k$
with at most three generators. In each case it is a polynomial
ring, more precisely:
\begin{itemize}

\item[(1)] if $\neg s_1\wedge \neg s_2\wedge \neg s_3\wedge \neg s_4$,
then $k[X]^d=k$,

\item[(2)] if $s_1\wedge \neg s_2\wedge \neg s_3$,
then $k[X]^d=k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4]$,

\item[(3)] if $\neg s_1\wedge \neg s_2\wedge \neg s_3\wedge s_4$,
then $k[X]^d=k[f_4]$,

\item[(4)] if $\neg s_1\wedge \neg s_2\wedge s_3\wedge \neg s_4$,
then $k[X]^d=k[x_2^tx_4^r],$\smallskip

\item[(5)] if $\neg s_1\wedge s_2\wedge \neg s_3\wedge \neg s_4$,
then $k[X]^d=k[x_1^qx_3^p]$,

\item[(6)] if $\neg s_1\wedge \neg s_2\wedge s_3\wedge s_4$,
then $k[X]^d=k[f_4, x_2^tx_4^r]$,

\item[(7)] if $\neg s_1\wedge s_2\wedge \neg s_3\wedge s_4$,
then $k[X]^d=k[f_4, x_1^qx_3^p]$,

\item[(8)] if $s_1\wedge \neg s_2\wedge s_3$,
then $k[X]^d=k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_2^tx_4^r]$,

\item[(9)] if $s_1\wedge s_2\wedge \neg s_3$,
then $k[X]^d=k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_1^qx_3^p]$,

\item[(10)] if $s_2\wedge s_3$,
then $k[X]^d=k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_1^qx_3^p,\, x_2^tx_4^r]$.

\end{itemize}
\end{theorem}

\section{Darboux polynomials and rational constants}

A polynomial $g\in k[X]$ is said to be \emph{strict}
if it is homogeneous and not divisible by the variables
$x_1,\ldots,x_n$.
For $\alpha=(\alpha_1,\ldots ,\alpha_n)\in \mathbb{N}^n$,
we denote by $X^{\alpha }$ the monomial
$x_1^{\alpha_1}\ldots x_n^{\alpha_n}\in k[X]$.
Every nonzero homogeneous polynomial
$f\in k[X]$ has a unique representation $f=X^{\alpha}g$, where
$X^{\alpha}$ is a monomial and $g$ is strict.

We call a nonzero polynomial $f\in k[X]$ a \emph{Darboux polynomial}
(or an \emph{integral element})
of a derivation $\delta :k[X]\to k[X]$
if $\delta(f)=\Lambda f$ for some $\Lambda \in k[X]$. We will call
$\Lambda $ a \emph{cofactor} of $f$. Since $d$ is a homogeneous
derivation of degree $1$, the cofactor of each homogeneous form
is a linear form.
Denote by $k[X]_{(m)}$ the homogeneous component
of $k[X]$ of degree $m$.

\begin{lemma}[{\cite[Lemma 3.2]{Czech}}] \label{lem21}
Let $n=4$. Let $g\in k[X]_{(m)}$ be a Darboux
polynomial of $d$ with the cofactor $\lambda_1x_1+\ldots +\lambda_4x_4$.
Let $i\in \{1,2,3,4\}$. If $g$ is not divisible by $x_i$,
then $\lambda_{i+1}\in \mathbb{N}$. More precisely, if
$g(x_1,\ldots, x_{i-1},0,x_{i+1},\ldots,x_4)=x_{i+2}^{\beta_{i+2}}
\overline{G}$ and $x_{i+2}\not \; \mid \overline{G}$, then
$\lambda_{i+1}=\beta_{i+2}$ and $\lambda_{i+3}=-C_{i+2}\lambda_{i+1}$.
\end{lemma}

\begin{corollary}[{\cite[Corollary 3.3]{Czech}}] \label{calk}
Let $n=4$. If $g\in k[X]$ is a strict Darboux polynomial,
then its cofactor is a linear form with coefficients in $\mathbb{N}$.
\end{corollary}

For any derivation $\delta:k[X]\to k[X]$
there exists exactly one derivation $\bar{\delta}:k(X)\to k(X)$
such that $\bar{\delta}_{\mid k[X]}=\delta$.
By a \emph{rational constant} of the derivation $\delta:k[X]\to k[X]$
we mean the constant of its corresponding derivation $\bar{\delta}:k(X)\to k(X)$.
The rational constants of $\delta$ form a field.
For simplicity, we write $\delta$ instead of $\bar{\delta}$.

\begin{lemma}[{\cite[Lemma 2]{JZ5}}] \label{mono}
Let $n=4$. The field $k(X)^d$ contains a nontrivial rational monomial constant
if and only if at least one of the following
two conditions is fulfilled:
\begin{enumerate}

\item $C_1, C_3\in \mathbb{Q}$ and $C_1C_3=1$,

\item $C_2, C_4\in \mathbb{Q}$ and $C_2C_4=1$.
\end{enumerate}
\end{lemma}

\begin{proposition}[{\cite[Prop. 2.2.2]{Now}}]\label{N1}
Let $\delta:k[X]\to k[X]$ be a derivation and let $f$ and $g$
be nonzero relatively prime polynomials from $k[X]$. Then
$\delta(\frac{f}{g}) = 0$ if and only if $f$ and $g$ are Darboux
polynomials of $\delta$ with the same cofactor.
\end{proposition}

\begin{proposition}[{\cite[Prop. 2.2.3]{Now}}] \label{jednor}
Let $\delta$ be a homogeneous derivation of $k[X]$ and let $f\in k[X]$
be a Darboux polynomial of $\delta$ with the cofactor $\Lambda \ink[X]$.
Then $\Lambda $ is homogeneous and each homogeneous
component of $f$ is also a Darboux polynomial of $\delta$ with the same cofactor
$\Lambda $.
\end{proposition}

\begin{proposition}[{\cite[Prop 2.2.1]{Now}}] \label{czyn}
Let $\delta$ be a derivation of $k[X]$. Then $f\in k[X]$ is a Darboux
polynomial of $\delta$ if and only if all factors of $f$ are Darboux
polynomials of $\delta$. Moreover, if $f=f_1f_2$ is a Darboux
polynomial, then sum of the cofactors of $f_1$ and $f_2$
equals the cofactor of $f$.
\end{proposition}

\section{Fields of rational constants of LV derivations}

From now on, $n=4$.
For $C_1, C_2, C_3, C_4\in k$ consider the sentences:
\begin{align*}
\tilde{s}_2:&\quad  C_1,C_3\in \mathbb{Q}\text{ and }C_1C_3=1.\\
\tilde{s}_3:&\quad C_2,C_4\in \mathbb{Q}\text{ and }C_2C_4=1.
\end{align*}
In case $\tilde{s}_2$ let $C_1=\frac{p}{q}$, where
$p,q\in \mathbb{Z}$, $q\neq 0$
and $\gcd (p,q)=1$.
In case $\tilde{s}_3$ let $C_2=\frac{r}{t}$, where
$r,t\in \mathbb{Z}$, $t\neq 0$
and $\gcd (r,t)=1$. Note that these presentations of $C_i$
are unique up to sign. Sentences $s_1$, $s_2$, $s_3$, $s_4$
and polynomial $f_4$ are as in Section~2.

\begin{theorem} \label{gljedno}
Let $d:k(X)\to k(X)$ be a four-variable Lotka-Volterra
derivation with parameters $C_1,C_2,C_3,C_4\in k$.
Then:
\begin{itemize}

\item[(1)] if $\neg s_1\wedge \neg \tilde{s}_2\wedge \neg \tilde{s}_3\wedge \neg s_4$,
then $k(X)^d=k$,

\item[(2)] if $s_1\wedge \neg \tilde{s}_2\wedge \neg \tilde{s}_3$,
then $k(X)^d=k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4)$,

\item[(3)] if $\neg s_1\wedge \neg \tilde{s}_2\wedge \neg \tilde{s}_3\wedge s_4$,
then $k(X)^d=k(f_4)$,

\item[(4)] if $\neg s_1\wedge \neg \tilde{s}_2\wedge s_3\wedge \neg s_4$,
then $k(X)^d=k(x_2^tx_4^r)$,

\item[(5)] if $\neg s_1\wedge s_2\wedge \neg \tilde{s}_3\wedge \neg s_4$,
then $k(X)^d=k(x_1^qx_3^p)$,

\item[(6)] if $\neg s_1\wedge \neg \tilde{s}_2\wedge s_3\wedge s_4$,
then $k(X)^d=k(f_4, x_2^tx_4^r)$,

\item[(7)] if $\neg s_1\wedge s_2\wedge \neg \tilde{s}_3\wedge s_4$,
then $k(X)^d=k(f_4, x_1^qx_3^p)$,

\item[(8)] if $s_1\wedge \neg \tilde{s}_2\wedge s_3$,
then $k(X)^d=k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_2^tx_4^r)$,

\item[(9)] if $s_1\wedge s_2\wedge \neg \tilde{s}_3$,
then $k(X)^d=k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_1^qx_3^p)$,

\item[(10)] if $s_2\wedge s_3$,
then $k(X)^d=k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,
x_1^qx_3^p, x_2^tx_4^r)$.
\end{itemize}
\end{theorem}

\begin{proof}
All inclusions of the form $\supseteq$ follow from
Theorem \ref{gl}.
Next we show the inclusions of the form $\subseteq$.

Let $\psi =\frac{f}{g}\in k(X)^d$, where $f, g\in k[X]\setminus
\{0\}$ and $\gcd (f,g)=1$.
By Proposition \ref{N1} we have
$d(f)=\Lambda f$ and $d(g)=\Lambda g$ for some $\Lambda \in k[X]$.
Let $f=\sum f_j$ and $g=\sum g_j$, where
$f_j$ and $g_j$ are homogeneous polynomials of degree $j$.
By Proposition \ref{jednor}, since $d$ is homogeneous,
we have $d(f_j)=\Lambda f_j$ and
$d(g_j)=\Lambda g_j$ for all $j\in \mathbb{N}$.
Then, by Proposition \ref{N1} again, we have
$d(\frac{f_j}{g_i})=0$
and $d(\frac{g_j}{g_i})=0$ for
all $i$ and $j$. Moreover, obviously
 $$
\frac{f}{g}= \frac{\sum_j \frac{f_j}{g_i}}{\sum_j \frac{g_j}{g_i}}
$$ 
for some fixed $i$. Therefore it suffices to prove
the assertion of Theorem \ref{gljedno} for
homogeneous $f$ and $g$.

Let $f=X^{\alpha}h$, where $X^{\alpha}$ is a monomial and $h$ is strict
(analogously we proceed for $g$). By
Proposition \ref{czyn} both $X^{\alpha}$ and $h$
are Darboux polynomials of $d$.
Let $\lambda =\lambda_1x_1+\ldots +\lambda_4x_4$
be the cofactor of $h$.
By Lemma~\ref{lem21} we have
\begin{equation}\label{lambdy}
\lambda_{i+3}=-C_{i+2}\lambda_{i+1}
\end{equation}
for all $i$ in the cyclic sense.
Moreover, Corollary~\ref{calk}
gives $\lambda_i\in \mathbb{N}$ for $i=1,\ldots ,4$.
\smallskip

\noindent\textbf{Cases (1)--(3).}
Suppose that $\lambda_1\neq 0$. Then \eqref{lambdy} for $i=2$
implies that also $\lambda_3\neq 0$ and
$C_4=-\frac{\lambda_1}{\lambda_3}\in \mathbb{Q}$.
Likewise, \eqref{lambdy} for $i=4$ gives
$C_2=-\frac{\lambda_3}{\lambda_1}\in \mathbb{Q}$.
Therefore $C_2C_4=1$, which is a contradiction to
$\neg \tilde{s}_3$. This proves that $\lambda_1=0$.
Analogously we proceed for $\lambda_2$, $\lambda_3$
and $\lambda_4$.
Hence we have $\lambda_1=\ldots =\lambda_4=0$ and the only strict Darboux
polynomials of $d$ are constants of $d$.
Note that $s_2\Rightarrow \tilde{s}_2$. Hence
$\neg \tilde{s}_2\Rightarrow \neg s_2$. The same for
$s_3$ and $\tilde{s}_3$.
Thus, in view of Theorem~\ref{gl}, we have $h\in k$ or
$h\in k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4]$ or $h\in k[f_4]$, respectively.
Furthermore, by Proposition \ref{czyn}, the cofactor of $X^{\alpha}$
is equal to $\Lambda $, since the cofactor of $h$ equals $0$.

Similarly, $g=X^{\beta}l$, where $l\in k[X]^d$
and $X^{\beta}$ is a Darboux monomial with the cofactor $\Lambda $.
Then $\frac{X^{\alpha}}{X^{\beta}}\in k(X)^d$,
by Proposition \ref{N1}.
In view of Lemma \ref{mono}, $\frac{X^{\alpha}}{X^{\beta}}\in k$.
Hence $\psi =c\frac{h}{l}$, where $c\in k$ and $h, l\in k[X]^d$.
Thus $\psi \in k$ or $\psi \in k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4)$ 
or $\psi \in k(f_4)$, respectively.
\smallskip

\noindent\textbf{Cases (4), (6).}
 As above $\lambda_2=\lambda_4=0$.
If, contrary to our claim,
$\lambda_2\neq 0$, then by \eqref{lambdy} we have $\lambda_4\neq 0$,
$C_1=-\frac{\lambda_2}{\lambda_4}\in \mathbb{Q}$ and
$C_3=-\frac{\lambda_4}{\lambda_2}\in \mathbb{Q}$,
in contradiction with $\neg \tilde{s}_2$.
By \eqref{lambdy} for $i=4$:
\begin{equation}
\label{lambdy2}
\lambda_3=-C_2\lambda_1.
\end{equation}
However, by Corollary~\ref{calk} and by $s_3$,
the left-hand side of \eqref{lambdy2} is nonnegative,
whereas the right-hand side of \eqref{lambdy2} is nonpositive.
Therefore
$\lambda_3=0$ and, since $C_2>0$, we have
$\lambda_1=0$.
Then $h\in k[X]^d$. Since $\neg s_1\wedge 
\neg \tilde{s}_2\wedge s_3\wedge \neg s_4
\Rightarrow \neg s_1\wedge \neg s_2\wedge s_3\wedge \neg s_4$
and $\neg s_1\wedge \neg \tilde{s}_2\wedge s_3\wedge s_4
\Rightarrow \neg s_1\wedge \neg s_2\wedge s_3\wedge s_4$,
we have case (4) or (6) of
Theorem~\ref{gl}, respectively.
Therefore we have $h\in k[x_2^tx_4^r]$ or $h\in k[f_4, x_2^tx_4^r]$,
respectively. Moreover, the cofactor of $X^{\alpha}$
is equal to $\Lambda $.

Analogously, $g=X^{\beta}l$, where $l\in k[x_2^tx_4^r]$
and $X^{\beta}$ is a Darboux monomial with the cofactor $\Lambda $.
Then $\frac{X^{\alpha}}{X^{\beta}}\in k(X)^d$,
by Proposition \ref{N1} again.
Let $\frac{X^{\alpha}}{X^{\beta}}=x_1^ax_2^bx_3^cx_4^e$,
where $a,b,c,e\in \mathbb{Z}$. Then
$$
d\left(\frac{X^{\alpha}}{X^{\beta}}\right)
=x_1^ax_2^bx_3^cx_4^e((b-eC_4)x_1+(c-aC_1)x_2+
(e-bC_2)x_3+(a-cC_3)x_4).
$$
Since $d(\frac{X^{\alpha}}{X^{\beta}})=0$, we have two
systems of linear equations:
\begin{equation}\label{be}
\begin{gathered}
b-eC_4=0\\
e-bC_2=0
\end{gathered}
\end{equation}
and
\begin{equation} \label{ca}
\begin{gathered}
c-aC_1=0\\
a-cC_3=0.
\end{gathered}
\end{equation}
Since $\neg \tilde{s}_2$, we obtain $a=c=0$.
Moreover, $e-b\frac{r}{t}=0$ implies $et=br$.
Since $\gcd (r,t)=1$, we have $r\mid e$, and thus
$e=jr$ for some $j\in \mathbb{Z}$. Therefore
$br=jrt$, and since $r\neq 0$, we have $b=jt$.
Consequently, 
$$
\frac{X^{\alpha}}{X^{\beta}}=
x_2^{jt}x_4^{jr}=(x_2^tx_4^r)^j\in k(x_2^tx_4^r).
$$
Thus $\psi =\frac{X^{\alpha}}{X^{\beta}}\frac{h}{l}$ belongs
to $k(x_2^tx_4^r)$ or $k(f_4, x_2^tx_4^r)$, respectively.
\smallskip

\noindent\textbf{Cases (5), (7).} 
These two cases are completely analogous to cases (4)
and (6), respectively.
\smallskip

\noindent\textbf{Case (8).} Similarly to case (4) we show that
$\lambda_1=\ldots =\lambda_4=0$.
Therefore $h\in k[X]^d$. Since $s_1\wedge \neg \tilde{s}_2\wedge s_3
\Rightarrow s_1\wedge \neg s_2\wedge s_3$, we have case (8) of
Theorem~\ref{gl}. Hence, 
$h\in k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\, x_2^tx_4^r]$ 
and the cofactor of $X^{\alpha}$
is equal to $\Lambda $. Likewise, $g=X^{\beta}l$,
where $l\in k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4, \, x_2^tx_4^r]$
and the cofactor of $X^{\beta}$ equals $\Lambda $.
In the same way as in case (4) we show that
$\frac{X^{\alpha}}{X^{\beta}}\in k(x_2^tx_4^r)$.
Finally, $\psi \in k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\, x_2^tx_4^r)$.
\smallskip

\noindent\textbf{Case (9).} It is entirely analogous to case (8).
\smallskip

\noindent\textbf{Case (10).} Since all $C_i$ are positive, the
the left-hand side of \eqref{lambdy} is nonnegative
and the right-hand side of \eqref{lambdy} is nonpositive
for $i=1,\ldots 4$. Thus $\lambda_1=\ldots =\lambda_4=0$
and $h\in k[X]^d$. Hence, by Theorem~\ref{gl}, we have
$h\in k[x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_1^qx_3^p,\, x_2^tx_4^r]$. Moreover,
the cofactor of $X^{\alpha}$
equals $\Lambda $. Analogously, $g=X^{\beta}l$,
where $l\in k[X]^d$ and
the cofactor of $X^{\beta}$ equals $\Lambda $.
Then $\frac{X^{\alpha}}{X^{\beta}}\in k(X)^d$.
If $\frac{X^{\alpha}}{X^{\beta}}=x_1^ax_2^bx_3^cx_4^e$,
where $a,b,c,e\in \mathbb{Z}$, then we again obtain
the systems of linear equations of the form \eqref{be}
and \eqref{ca}. Similarly to case (4) we obtain
$e=jr$, $b=jt$ for some $j\in \mathbb{Z}$
and $a=sq$, $c=sp$ for some $s\in \mathbb{Z}$.
Thus $\frac{X^{\alpha}}{X^{\beta}}\in k(x_1^qx_3^p,\, x_2^tx_4^r)$.
Consequently, $\psi \in k(x_1+C_1x_2+C_1C_2x_3+C_1C_2C_3x_4,\,
x_1^qx_3^p,\, x_2^tx_4^r)$.
\end{proof}


Note that Theorem \ref{gl} covers all the cases.
Theorem \ref{gljedno} does not cover all the cases,
however huge majority of them. Nevertheless,
the following corollary covers all the cases.

\begin{corollary}\label{exist}
If $d$ is a four-variable Lotka-Volterra derivation, then $k(X)^d$
contains a nontrivial rational constant
if and only if at least one of the following
four conditions is fulfilled:
\begin{itemize}
\item[(1)] $C_1C_2C_3C_4=1$,

\item[(2)]  $C_1, C_3\in \mathbb{Q}$ and $C_1C_3=1$,

\item[(3)]  $C_2, C_4\in \mathbb{Q}$ and $C_2C_4=1$,

\item[(4)] $C_1C_2C_3C_4=-1$ and $C_i=1$ for two consecutive indices $i$.
\end{itemize}
\end{corollary}

\begin{proof}
If $\neg s_1\wedge \neg \tilde{s}_2\wedge \neg \tilde{s}_3\wedge \neg s_4$,
then $k(X)^d=k$, by Theorem \ref{gljedno}.
If $\tilde{s}_2$ or $\tilde{s}_3$, then
$k(X)^d\neq k$, by Lemma \ref{mono}.
If $s_1$ or $s_4$, then $k(X)^d\neq k$, by Theorem \ref{gljedno}.
\end{proof}

Note that if $d$ is as in Theorem \ref{gljedno},
then the field of rational constants equals
the field of fractions of the ring of polynomial constants.
Which is not true in general, even $k[X]$ may be trivial,
while $k(X)$ nontrivial.

\begin{example} \rm
Let $k\in \{ \mathbb{R}, \mathbb{C}\}$.
Let $d:k[X]\to k[X]$ be a derivation
defined by
\begin{gather*}
d(x_i)=x_i(x_{i-1}+x_{i+1}),\quad  \text{for } i=1,3,\\
d(x_i)=x_i(x_{i-1}-\Pi x_{i+1}),\quad \text{for } i=2,4.
\end{gather*}
By Theorem \ref{gl},
$k[X]^d=k$. Nevertheless, $\frac{x_1}{x_3}\in k(X)^d$.
\end{example}


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\end{document}