# Integral Aspects of $(1:-3:1)$ Resonance Lotka–Volterra Equations with
Nonstandard Analysis

Chiman Qadir1, Waleed Aziz1, Ibrahim Hamad1 1 Department of Mathematics,
College of Science, Salahaddin University–Erbil, Kurdistan region–Iraq
<EMAIL_ADDRESS>waleed.aziz<EMAIL_ADDRESS>
###### Abstract.

In this paper the problems of integrable and linearizable Lotka-Volterra
equations with $(\delta:-3\delta:\delta)$-resonance are studied. The necessary
conditions for both problems are obtained in the case when $\delta=1$ and it’s
sufficiency are proved. It is also shown that non-standard analysis has an
important role in proving the sufficient integrable conditions in some cases.
The non-standard analysis approaches have been used for both perturbed and
unperturbed cases for integral aspects of the given system.

###### Key words and phrases:

Darboux integrability, linearizability, perturbation technique and nonstandard
analysis.

###### 2010 Mathematics Subject Classification:

Primary 37K10 Secondary 34A05, 34E18

##  1\. Introduction

The first integral aspects of differential equations are one of interesting
topics in the qualitative theory of ordinary differential equations. The
knowledge of this kind of integral gives important information about the
behavior of orbits near the origin. We mean, here, this kind of problem
behaves as a local integrability problem of differential systems.

One useful method to study the integrability problem, which particularly plays
an important role in the study of the dynamics of polynomial differential
systems is Darboux theory of integrability. This method is based on using
invariant algebraic surfaces to construct first integrals.

For systems of dimension $n$, the existence of less than $n-1$ functionally
independent first integrals means that the system is partially integrable, and
the existence of $n-1$ functionally independent first integrals means that the
system is completely integrable.

In this paper the problems of complete integrability and linearizability at
the origin for a three dimensional Lotka-Volterra system

$\displaystyle\begin{split}\dot{x}&=P=x(\delta+ax+by+cz),\\\
\dot{y}&=Q=y(-3\delta+dx+ey+fz),\\\ \dot{z}&=R=z(\delta+gx+hy+kz),\end{split}$
(1)

are investigated where variables. The parameters are considered as nonstandard
hyperreals where $\delta$ is limited and bounded away from zero. The system
(1) and its associated vector field $\chi$

$\chi=P\frac{\partial}{\partial x}+Q\frac{\partial}{\partial
y}+R\frac{\partial}{\partial z},$ (2)

are analytic around the origin.

Many models in the nature can be related to Lotka–Volterra equations
introduced individually by Alfred Lotka (1925) and Vito Volterra (1926). These
kind of equations are emphasized in studying of population dynamics and in
applied sciences as well. More recently, Lotka–Volterra equations are based in
models of COVID-19 [28, 29]. Particularly, Lotka–Volterra system was studied
by several researchers. In two dimensional systems, it was considered by
Christopher and Rousseau [8], Laurant and Llibre [23], Liu [24] and many
others to find conditions for the existence of a local first integrals in a
neighborhood of the origin. Many authors generalized a similar concept from
two dimensional systems into dimension three. An investigation of necessary
and sufficient conditions for integrability and linearizability for a three
dimensional Lotka–Volterra systems was done by Laurant [22, 23] and for the
resonance $(1:-1:1)$, $(2:-1:1)$ and $(1:-2:1)$ by Aziz and Christopher [1]
and $(1:3:-1)$ in [2]. The authors in [4, 5, 13], give necessary and
sufficient conditions for a family of three dimensional quadratic
nonlinearities.

Nonstandard analysis is considered in many different fields in mathematics and
particularity in differential equations. Moreover, nonstandard analysis tools
enable us to approach the non-exact solution more precisely than that with
classical methods. That is, with nonstandard there is no mathematical
restrictions to find solutions near singularity and regarding it as an
alternative to the exact solution or as a shadow of the exact solution. Some
previous and recent works of this technique are done by Sari [27, 32] who gave
a nonstandard model for perturbation of differential equations, Lorby and Sari
[25] about uniform infinitesimal boundedness of singular perturbation of
differential equation. Hamad and Pirdawood [17] investigated nonstandard
solution in the monad of singularity of differential. Hamad, in [18, 19] gives
nonstandard generalization of Sierpinski Theorem and introduced generalized
curvature and torsion in nonstandard analysis with some other concepts in
differential geometry.

The aim of this paper is to study the integrability problem of Lotka-Volterra
system in two different aspects. We are first interested in determining the
necessary and sufficient conditions for both integrability and linearizability
problems for Lotka–Volterra system with $(1:-3:1)$ resonant singular point at
the origin. The second is to extend the studied problem to hyperreal. We
consider the system (1) with $(\delta:-3\delta:\delta)$ resonant so that
$\delta=1+\varepsilon$ where $\varepsilon$ is infinitesimal or $\delta$ is
itself infinitesimal, we find for both eigenvalues $(1:-3:1)$ and
$(1+\varepsilon:-3(1+\varepsilon):1+\varepsilon)$ in which the vector fields
are infinitely close to each other and have two independent first integrals
which are infinitely close to standard two first integrals where the variables
are near standard points in a perturbation case. The perturbation of vector
fields was proved as a deformation in nonstandard [32]. In [15, 16], Gao
showed that the homogeneous case $\delta=0$ will reduce to a two dimensional
Lotka–Volterra equations. Other nonstandard transform use for unperturbation
case of the system, which leads to the same integrability and linearizability
conditions as in standard form. On the other hand, making an infinitesimal
perturbation with infinitesimal $\delta$ leads to new different results. If we
conditionally treated parameter $a$, then the obtained two independent first
integrals of the system (1) for $\delta=1+\varepsilon$, where $\varepsilon$ is
infinitesimal, belong to monad(0), galaxy(0) or monad($\omega$), where
$\omega$ is unlimited. However, if $\delta$ itself is infinitesimal, then the
obtained two independent first integrals of the system (1) belong to monad(0)
or galaxy(0).

##  2\. Basic Definitions and Known Results

In this section, some basic notions regarding integrability, linearizability
problems and mathematical analysis are introduced. The first part is devoted
to the emphasis of basic concepts and known results of integrability problem,
and in the second part some background of mathematical analysis is described.

We recall that a non-constant analytic function $\phi(x,y,z)$ is a first
integral of the differential system (1) if it is constant on all its solutions
that satisfies

$\chi(\phi)=P\frac{\partial\phi}{\partial x}+Q\frac{\partial\phi}{\partial
y}+R\frac{\partial\phi}{\partial z}=0.$

It is well known that the equation $\phi(x,y,z)=c$ for some constant
$c\in\mathbb{R}$ provides an implicit set of trajectories of the given system.

The system (1) is said to be integrable at the origin if there exists a change
of coordinates

$X=x(1+O(x,y,z)),\quad Y=y(1+O(x,y,z)),\quad Z=z(1+O(x,y,z)),$

bringing (1) to a system orbitally equivalent to

$\displaystyle\dot{X}=\delta X\kappa,\quad\dot{Y}=-3\delta
Y\kappa,\quad\dot{Z}=\delta Z\kappa$ (3)

where $\kappa=\kappa(X,Y,Z)=1+O(X,Y,Z)$.
Thus two functionally independent first integrals of system (3) are

$\Psi_{1}=X^{3}Y\quad\mbox{and}\quad\Psi_{2}=YZ^{3},$

and pull back to first integrals of system (1), we obtain the required
functionally independent first integrals

$\phi_{1}=x^{3}y(1+O(x,y,z))\quad\mbox{and}\quad\phi_{2}=yz^{3}(1+O(x,y,z)).$

Sometimes we can find an analytic change of variables that reduces the
nonlinear system into a linear system. Then the reduced system is said to be
linearizable. If the change of coordinates can be selected so that $\kappa=1$
in equation (3) then we say the system is linearizable.

Let $l\in\mathbb{C}[x,y,z]$ be a polynomial, then a surface $l=0$ is called an
invariant algebraic surface of the system (1), if $l$ satisfies the linear
partial differential equation

$\chi(l)=P\frac{\partial l}{\partial x}+Q\frac{\partial l}{\partial
y}+R\frac{\partial l}{\partial z}=lC_{l},$

for some polynomial $C_{l}\in\mathbb{C}[x,y,z]$ are called the cofactor of the
invariant algebraic surface $l=0$. It is well known that the degree of the
cofactor is at most one.

A function of the form $E(x,y,z)=\exp(\frac{f}{g})$ where
$f,g\in\mathbb{C}[x,y,z]$ is an exponential factor if it satisfies

$\chi(E)=C_{E}E,$

for some polynomial $C_{E}$ of degree at most one. The polynomial $C_{E}$ is
called the cofactor of $E$.

If $l_{i}$ are invariant algebraic surfaces and $E=\exp(\frac{f}{g})$ is an
exponential factor such that there exist
$\lambda_{0},\lambda_{i}\in\mathbb{C}$ for $i=1,\ldots,n$, then the function
of the form

$\Phi=\prod l_{i}^{\lambda^{i}}E^{\lambda_{0}f/g},$

is a Darboux function. Given a Darboux function $\Phi$, we can compute

$\chi(\Phi)=\Phi(\sum\lambda_{i}C_{l_{i}}+\lambda_{0}C_{E}),$

where $C_{l_{i}}$ and $C_{E}$ are respective cofactors of $l_{i}$ and $E$.
Obviously, the function $\Phi$ is a non-trivial first integral of the system
if and only if the cofactors $C_{l_{i}}$ and $C_{E}$ are linearly dependent.

Note that the function $\mu$ is an inverse Jacobi multiplier for the vector
field (2) if it satisfies the equation

$\chi(\mu)=\mu\>{\mbox{div}}(\chi).$

In three dimensional system, if we construct one first integral $\Phi$ and an
inverse Jacobi multiplier $\mu$, then we can find a second independent first
integral.

######  Theorem 2.1.

[3] Suppose the analytic vector fields

$(\lambda+\sum\nolimits_{|I|>0}A_{xI}X^{I})\frac{\partial}{\partial
x}+y(\mu+\sum\nolimits_{|I|>0}A_{yI}X^{I})\frac{\partial}{\partial
y}+(v+\sum\nolimits_{|I|>0}A_{zI}X^{I})\frac{\partial}{\partial z},$

has a first integral $\phi=x^{\alpha}y^{\beta}z^{\gamma}(1+O(x,y,z))$ with at
least one of $\alpha,\beta,\gamma\neq 0$ and an inverse Jacobi multiplier
$M=x^{r}y^{s}z^{t}(1+O(x,y,z))$ and suppose that the cross product of
$(r-i-1,s-j-1,t-k-1)$ and $(\alpha,\beta,\gamma)$ is bounded away from zero
for any integers $i,j,k\geq 0$, then the system has a second analytic first
integral of the form $\psi=x^{1-r}y^{1-s}z^{1-t}(1+O(x,y,z))$, and hence the
system (1) is integrable.

######  Theorem 2.2.

[1] If the system (1) with $\delta=1$ is integrable and there exists a
function $\zeta=x^{\alpha}y^{\beta}z^{\gamma}(1+O(x,y,z))$ such that
$\chi(\zeta)=k\zeta$ for some constant $k=\alpha\lambda+\beta\mu+\gamma\nu$,
then the system is linearizable.

######  Theorem 2.3.

[7] For $\lambda\in\mathbb{N}-\\{1\\}$ the Lotka–Volterra system

$\displaystyle\dot{x}=x(1+c_{20}x+c_{11}y),\quad\dot{y}=y(-\lambda+d_{11}x+d_{02}y),$
(4)

has a linearizable saddle at the origin if and only if one of the following
conditions is satisfied

  1. i.

$mc_{20}+d_{11}=0,m=0,\ldots,\lambda-2$.

  2. ii.

$c_{11}=d_{02}=0$.

  3. iii.

$c_{02}-d_{11}=c_{11}-d_{02}=0$.

  4. iv.

$c_{11}=(\lambda-1)c_{20}+d_{11}=0.$

##  3\. Poincaré-Dulac Normal Form Theory

The theory of normal form of a nonlinear differential systems first introduced
in the work of Poincaré. This theory is crucial in studying the integral
aspects of differential systems and gives a clue of general prospective of the
form of their integrals. So, here, we give main facts of normal forms and some
notations as well.

Let us now consider an analytic system

$\dot{x}=Ax+F(x),$ (5)

where $A$ is an $n\times n$ matrix with eigenvalues
$\alpha_{1},\ldots,\alpha_{n}$, $x\in\mathbb{R}^{n}$ and $F(x)=O(2)$ is an
analytic vector-valued function without constant or linear term in
$(\mathbb{R}^{n},0)$.

######  Definition 3.1.

[6] If the convex hull of the spectrum of matrix $\sigma(A)$ does not contain
the origin, then $\sigma(A)$ is said to be in the Poincaré domain. If the
origin lies inside the convex hull of $\sigma(A)$, then we say $\sigma(A)$ is
in the Siegel domain.

######  Remark 1.

A singular point whose spectrum lies in the Poincaré domain can be brought to
normal form by an analytic change of coordinates. Particularly, a node with
two analytic separatrices can have no resonant terms in its normal form and so
must be analytically linearizable [1].

######  Theorem 3.2.

[6] If the spectrum of matrix is in the Poincaré domain, then there are at
most finitely many resonant monomials.

######  Definition 3.3.

An n-tuple $\alpha=(\alpha_{1},\ldots,\alpha_{n})$ of eigenvalues of $A$ is
called resonant, if it satisfies the resonant identity

$\alpha_{i}=\langle\ell,\alpha\rangle=\sum^{n}_{i=1}\ell_{i}\alpha_{i},\quad|\ell|\geq
2,$ (6)

for non-negative integers $\ell_{i}\in\mathbb{N}_{0}=\\{0,1,2,\ldots\\},$ and
the natural numbers $|\ell|=\sum^{n}_{i=1}\ell_{i}$ is the order of the
resonance. The coefficient $X^{\ell}_{i}$ of the monomial $x$ is called a
resonant coefficient and the corresponding term is called a resonant term.

Note that if condition (6) does not hold, then $\alpha_{1},\ldots,\alpha_{n}$
are called non-resonant and then the differential equation can be formally
linearized (i.e. a formal, but in general, non-convergent transformation
series exists). For more details consult [20, 31].

Let $\mathcal{H}_{s}$ denote the vector space of functions from
$\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ each of whose components is a homogeneous
polynomial of degree $s$.

The linear operator (homological operator) $\mathcal{L}$ on $\mathcal{H}_{s}$
is defined by

$\mathcal{L}h(y)=d\,h(y)\,Ay-Ah(y).$

By performing a sequence of transformations of the form

$x=H(y)=y+h^{s}(y),\qquad h^{s}\in\mathcal{H}_{s},$

we can remove all non-resonant terms of system (5) and is formally equivalent
to

$\dot{y}=Ay+G(y).$ (7)

The problem of when an analytic system is analytically equivalent to its
normal form is classical and is an open problem.

######  Theorem 3.4.

[31] Let $(\alpha_{1},\ldots,\alpha_{n})$ be the eigenvalues of $A$ in (5). If
the equality (6) does not holds for all $i\in{1,\ldots,n}$ and for all
$\alpha\in\mathbb{N}_{0}$ in which $|\alpha|\geq 0$, then (5) and (6) are
formally equivalent for all $F$ and $G$.

Let

$\mathcal{M}_{\alpha}=\Big{\\{}\ell=(\ell_{1},\ldots,\ell_{n}):\langle\ell,\alpha\rangle=\sum^{n}_{i=1}\ell_{i}\alpha_{i}=0,\quad|\ell|\geq
1\Big{\\}},$

be a set of vectors and let $R_{\alpha}$ be the rank of vectors in the set
$\mathcal{M}_{\alpha}$. It is obvious $R_{\alpha}\leq n-1$.

######  Theorem 3.5.

[35] Suppose that the origin of system (5) is non-degenerate. Then system (5)
has $n-1$ linearly independent analytic first integrals if and only if
$R_{\lambda}=n-1$, and it is analytically equivalent to its normal form

$\dot{y}_{i}=\lambda_{i}y_{i}(1+G(y)),\quad i=1,\ldots,n,$

by an analytic normalization, where $G(x)$ has no constant term and is an
analytic function of $y^{\ell}$ with $\ell\in\mathcal{M}$ and
$\ell=(\ell_{1},\ldots,\ell_{n})=1$.

For more detail on the formal and analytic transformations to the normal form
using Poincaré-Dulac normal form, one can consult [14, 34].

##  4\. Mathematical Analysis and Nonstandard backgrounds

We briefly recall some mathematical analysis concepts and results related to
the problem in this paper.
A subset $S$ of a metric space $(\mathsf{X},d)$ is compact if every sequence
from $S$ has a sub sequence which converges to an element of $S$. If
$\mathsf{X}$ is compact, we say the metric space itself is compact.

######  Theorem 4.1 (Heine-Borel Theorem).

[30] Every closed and bounded subset of $\mathbb{R}^{n}$ is compact.

Hyperreal numbers are an extension of the real numbers, which contain
infinitesimals and unlimited numbers. The set of hyperreal numbers is denoted
by ${}^{*}\mathbf{R}$ and for simplicity we use $\mathbf{R}$ without the star.
A number $x\in\mathbf{R}$ is unlimited or infinite if $|x|>n$ for all standard
$n\in\mathbb{N}$, $x$ is limited or finite if $|x|<n$ for some standard
$n\in\mathbb{N}$ and $x$ is infinitesimal if $|x|<\frac{1}{n}$ for all
standard $n\in\mathbb{N}$. Note that the number zero is the only standard
infinitesimal. A real number is appreciable if it is neither unlimited nor
infinitesimal. A real number $x$ is called near-standard if $x$ is infinitely
close to $y$ for some standard real number $y$, and denoted by NS.

The two real numbers $x$ and $y$ are infinitely close if $x-y$ is
infinitesimal and is written as $x\simeq y$.

A number $x$ is limited in $\mathbf{R}$, then it is infinitely close to a
unique standard real number; this unique number is called the standard part of
$x$. It is also well known that

  1. (1)

$\mbox{monad}(x)=\\{y\in\mathbf{R}:x\simeq y\\}$ for limited $x$, and is
dented by $\mbox{m}(x)$.

  2. (2)

$\alpha$-monad $(x)=\\{y\in\mathbf{R}:\frac{y-x}{\alpha}\simeq 0\\}$, and
denoted by $\alpha$-m(x).

  3. (3)

galaxy$(x)=\\{y\in\mathbf{R}:x-y\ \ is\ \ \mbox{limited}\\}$ for limited $x$,
and is dented by $\mbox{gal}(x)$.

  4. (4)

$\alpha$-galaxy$(x)=\\{y\in\mathbf{R}:\frac{y-x}{\alpha}\ \ is\ \
\mbox{limited}\\}$, and dented by $\alpha$-gal(x).

  5. (5)

$\alpha$-micromonad $(x)=\\{y\in\mathbf{R}:y\leq{\alpha}^{n},\mbox{for all
standard $n$}\\}$ .(see [11, 12]).

######  Definition 4.2.

[32] Let $F:D\to P$ and $F_{0}:D_{0}\to P$ be mappings from the open subsets
$D$ and $D_{0}$ of the standard topological space $\mathsf{X}$ to the uniform
space $P$, $F_{0}$ standard. The mapping $F$ is said to be perturbation of the
mapping $F_{0}$, which is denoted by $F\simeq F_{0}$, if ${}^{NS}D_{0}\in D$
and $F(x)\simeq F_{0}(x)$ for all $x\in{}^{NS}D_{0}$, where $x\in{}^{NS}D_{0}$
is $\\{x\in\mathsf{X}:\exists^{st}x_{0}\in D_{0},x\simeq x_{0}\\}$.

This definition make a sense, because $D_{0}$ is being a standard open subset
of $\mathsf{X}$, ${}^{NS}D_{0}\subset D$, so $F(x)$ and $F_{0}(x)$ are both
defined for all $x\in{}^{NS}D_{0}$. Let $\Omega_{\mathsf{X},P}$ be the set of
mappings defined on an open subsets of $\mathsf{X}$ to $P$, that is

$\Omega_{\mathsf{X},P}=\\{(F,D):D\>\mbox{open subset
of}\>\mathsf{X}\>\mbox{and}\>F:D\to P\\}.$

The topology of this set is defined as follows. Let
$(F_{0},D_{0})\in\Omega_{\mathsf{X},P}$. The family of sets of the form

$\\{(F,D)\in\Omega_{\mathsf{X},P}:K\subset D\quad\forall x\in
K\quad(F(x),F_{0}(x))\in U\\}$

where $K$ is a compact subset of $D_{0}$ and $U$ is an entourage of the
uniform space $P$ is a basis of the set of neighborhood of $(F_{0},D_{0})$.
This topology is known as topology of uniform convergence on compacta. If all
the mappings are defined on the same open set $D$, this topology is the usual
topology of uniform convergence on compacta on the set of functions on $D$ to
$P$. For more details, see [11, 21, 26, 32, 33].

######  Proposition 1.

[32] Assume $\mathsf{X}$ is locally compact. The mapping $F$ is a perturbation
of the standard mapping $F_{0}$ if and only if $F$ is infinitely close to
$F_{0}$ for the topology of uniform convergence on compacta.

##  5\. Computation of necessary conditions of integrability and
linearizability

The considering problem in this section is the problem of integrals. This
problem and other closely related problems, here, can be reduced to
investigate the polynomial ideals and its associate varieties.

In this section, we study the integrability and linearizability conditions at
the origin of system (1) with $\delta=1$ and the variables and parameters are
real numbers. The necessary conditions were found by computing the conditions
of the existence of two independent first integrals up to a degree.

Here, we are looking for two functionally independent first integrals

$\phi=x^{3}y+\ldots,$ (8)

and

$\psi=z^{3}y+\ldots,$ (9)

which satisfies respectively,

$\chi\phi=\sum\mathcal{R}_{\phi}X,$ (10)

$\chi\psi=\sum\mathcal{R}_{\psi}Y,$ (11)

where $X$ and $Y$ are monomial resonance and $\mathcal{R}_{\phi}$ and
$\mathcal{R}_{\psi}$ are polynomials in the coefficients of the given system,
which are obstacle of the existence of first integrals (8) and (9). The
vanishing of these obstacles gives necessary and sufficient conditions for
integrability conditions. Hence, the first integrals $\phi$ and $\psi$ exist
whenever all $\mathcal{R}_{\phi}$ and $\mathcal{R}_{\psi}$ are zero. Moreover,
let $\mathcal{B}$ be the ideal generated by $\mathcal{R}_{\phi}$ and
$\mathcal{R}_{\psi}$. We compute the variety of the ideal $\mathcal{B}$, and
let us denote it by $V(\mathcal{B})$, and then the irreducible decomposition
of $V(\mathcal{B})$, variety of integral needs to be computed. Note that the
set of $\mathcal{R}_{\phi}$ and $\mathcal{R}_{\psi}$ involving infinitely many
polynomials and finding $\mathcal{B}$, in this case, is impossible. In fact,
by the aid of Hilbert Basic Theorem every ideal in the polynomial ring is
finitely generated. Hence, the ideal $\mathcal{B}$ will stabilize at some
finite order. The degrees 16 of resonant monomials were used using computer
system Maple. The irreducible decomposition were found by using the minAssGTZ
in computer algebra system Singular [9, 10].

For linearizing, we use the same method of computing the conditions for the
existence of a linearizing change of coordinates up to some finite order to
find necessary conditions, and exhibiting a linearizing change of variables
for sufficiency.

Hence, this gives the following theorem.

Consider the three dimensional Lotka-Volterra system

$\displaystyle
F_{0}=\left\\{\begin{array}[]{rc}\dot{x}&=F_{01}=x(1+ax+by+cz),\\\
\dot{y}&=F_{02}=y(-3+dx+ey+fz),\\\
\dot{z}&=F_{03}=z(1+gx+hy+kz),\end{array}\right.$ (15)

where $F_{0}=(F_{01},F_{02},F_{03})$, and all variables and parameters are
standard numbers.

######  Theorem 5.1.

The three dimensional Lotka-Volterra system (15) is integrable at origin if
and only if the system satisfies each of the following conditions

$\displaystyle 1)$ $\displaystyle g=f=d=c=0$ $\displaystyle 2)$ $\displaystyle
f=d=c=g+a=0$ $\displaystyle 2^{*})$ $\displaystyle g=f=d=c+k=0$ $\displaystyle
3)$ $\displaystyle g=f=d=a=0$ $\displaystyle 3^{*})$ $\displaystyle f=d=c=k=0$
$\displaystyle 4)$ $\displaystyle h=g=f=b-e=a-d=0$ $\displaystyle 4^{*})$
$\displaystyle d=c=b=f-k=e-h=0$ $\displaystyle 5)$ $\displaystyle
f=e+h=d+g=c=b+h=a+g=0$ $\displaystyle 5^{*})$ $\displaystyle
g=f-k=e-h=d=c+k=b+h=0$ $\displaystyle 6)$ $\displaystyle
f=2e+h=2d+g=c=2b+h=2a+g=0$ $\displaystyle 6^{*})$ $\displaystyle
g=d=f-k=e-h=c+2k=b+2h=0$ $\displaystyle 7)$ $\displaystyle f=d=b-h=0$
$\displaystyle 8)$ $\displaystyle f=5e-3h=d+g=c=5b+h=3a-g=0$ $\displaystyle
9)$ $\displaystyle f=d=c-k=a-g=0$ $\displaystyle 10)$ $\displaystyle
f=8e-3h=4d+3g=c=8b+h=4a-g=0$ $\displaystyle 11)$ $\displaystyle
f=8e-3h=4d+3g=c+2k=8b+h=4a-g=0$ $\displaystyle 12)$ $\displaystyle
k=f=c=b=2a+d=0$ $\displaystyle 12^{*})$ $\displaystyle h=g=f+2k=d=a=0$
$\displaystyle 13)$ $\displaystyle h=g=f=b=c+k=a+d=0$ $\displaystyle 13^{*})$
$\displaystyle h=f+k=d=c=b=a+g=0$ $\displaystyle 14)$ $\displaystyle
h=f=c=b=2a+d=0$ $\displaystyle 14^{*})$ $\displaystyle h=g=d=b=f+2k=0$
$\displaystyle 15)$ $\displaystyle f=d+g=c=b=2a-g=0$ $\displaystyle 15^{*})$
$\displaystyle h=g=d=f+2k=c-2k=0$ $\displaystyle 16)$ $\displaystyle
f=c=b=3d+2g=3a-g=0$ $\displaystyle 16^{*})$ $\displaystyle h=g=d=c-3k=f+2k=0$
$\displaystyle 17)$ $\displaystyle f=3d+2g=c+k=b=3a-g=0$ $\displaystyle
17^{*})$ $\displaystyle h=f+2k=d=c-3k=a+g=0$ $\displaystyle 18)$
$\displaystyle g=f=c=a+d=0$ $\displaystyle 18^{*})$ $\displaystyle
g=f+k=d=c=0$ $\displaystyle 19)$ $\displaystyle f=k=c=a+d=0$ $\displaystyle
19^{*})$ $\displaystyle g=d=a=f+k=0$ $\displaystyle 20)$ $\displaystyle
f=c=b-h=a+d=0$ $\displaystyle 20^{*})$ $\displaystyle g=f+k=d=b-h=0$
$\displaystyle 21)$ $\displaystyle f=2d+g=c=2a-g=0$ $\displaystyle 21^{*})$
$\displaystyle g=f+k=d=c-2k=0$ $\displaystyle 22)$ $\displaystyle
f=e+h=3d+g=2c-k=b=3a-g=0$ $\displaystyle 22^{*})$ $\displaystyle
h=f+k=c-3k=b+e=a-2g=d=0$ $\displaystyle 23)$ $\displaystyle
f+k=e+h=d+g=c-3k=b+h=a+g=0$ $\displaystyle 23^{*})$ $\displaystyle
f-k=e-h=3d+g=c+k=b+h=3a-g=0$ $\displaystyle 24)$ $\displaystyle
f+k=5e-3h=d+g=c+k=5b+h=3a-g=0$ $\displaystyle 25)$ $\displaystyle
h=g=f+k=c=b=a+d=0$ $\displaystyle 26)$ $\displaystyle h=g=f+2k=c-2k=a+d=0$
$\displaystyle 26^{*})$ $\displaystyle f+k=d+g=c=b=2a-g=0$ $\displaystyle 27)$
$\displaystyle h=f+k=3d+2g=c=b=3a-g=0$ $\displaystyle 27*)$ $\displaystyle
h=g=b=f+2k=c-3k=a+d=0$ $\displaystyle 28)$ $\displaystyle
h=f+k=3d+2g=2c-k=b=3a-g=0$ $\displaystyle 28^{*})$ $\displaystyle
h=f+2k=d+2g=c-3k=b=a-2g=0$ $\displaystyle 29)$ $\displaystyle
h=g=f+k=b-e=a-d=0$ $\displaystyle 29^{*})$ $\displaystyle f-k=e-h=c=b=a+d=0$
$\displaystyle 30)$ $\displaystyle h=g=f+2k=b-e=a-d=0$ $\displaystyle 30^{*})$
$\displaystyle f-k=e-h=c=b=2a+d=0$ $\displaystyle 31)$ $\displaystyle
f+k=e=3d+g=c-3k=b+h=3a-g=0$ $\displaystyle 32)$ $\displaystyle cd-2dk-
gk=f+k=b-h=a+d=0$ $\displaystyle 33)$ $\displaystyle f+k=d+g=c-k=a-g=0$
$\displaystyle 34)$ $\displaystyle h=e=b$ $\displaystyle 35)$ $\displaystyle
cd-2dk-2gk=h=f+2k=b=2a+d=0$ $\displaystyle 36)$ $\displaystyle
h=f+2k=3d+2g=c-3k=b=3a-g=0$ $\displaystyle 37)$ $\displaystyle
g=f+3k=e+3h=d=c-3k=b+5h=0$ $\displaystyle 38)$ $\displaystyle
g=f+3k=e+3h=d=c-4k=b+8h=0$ $\displaystyle 39)$ $\displaystyle
f+3k=e+3h=d=c-4k=b+8h=2a+g=0$ $\displaystyle 40)$ $\displaystyle
f+3k=e+3h=d-g=c-3k=b+5h=a+g=0$ $\displaystyle 41)$ $\displaystyle
ef+2ek-3hk=cd+3cg-2dk+fg-3gk=bk-ce+ek-hk$ $\displaystyle=bf+2bk-3ch-
fh+hk=bd+3bg-de-dh-2eg=af+3ak-dk-3gk$ $\displaystyle=ae+3ah-de-3eg=ac-2ak-
cd-3cg+2dk-fg+4gk=ab+2ah-bd-3bg+dh=0.$

Furthermore, the conditions $(1)-(7),(9),(12)-(23),(25)-(36)$ are satisfied if
and only if the system is linearizable or one of the following holds

$\displaystyle 41.1$ $\displaystyle dh-eg=k=f=c=b-e=a-d=0$ $\displaystyle
41.1^{*}$ $\displaystyle bk-ch=g=f-k=e-h=d=a=0$ $\displaystyle 41.2$
$\displaystyle f-k=e-h=d-g=c-k=b-h=a-g=0.$

###### Proof.

The dual cases are obtained under the transformation $(x,z)$$\longrightarrow$
$(z,x)$ and we do not consider them further. The other cases are considered
below.

Case 1: In this case, the system takes the form

$\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3+ey),\qquad\dot{z}=z(1+hy+kz).$ (16)

The change of coordinate $Y=\frac{y}{1-\frac{e}{3}y}$ linearizes the second
equation to $\dot{Y}=-3Y$. To linearize the third equation, we seek an
invariant algebraic surface of the form $\alpha(Y)+\beta(Y)z=0$ with cofactor
$hy+kz$, where $\alpha$ and $\beta$ are analytic so that $\alpha(0)=1$. Thus,
$\alpha$ and $\beta$ must satisfy the equations

$-3Y\dot{\alpha}=hy\alpha,$ (17) $3Y\dot{\beta}-\beta=-k\alpha.$

The function $\alpha=(1+\frac{e}{3}Y)^{(-\frac{h}{e})}=\sum\nolimits_{n\geq
0}\eta Y^{n}$ with $|\frac{e}{3}Y|<1$ where
$\eta=\frac{(-1)^{n}(\frac{e}{3})^{n}\prod_{i=-1}^{n-1}(\frac{h}{e}+i)}{n!(\frac{h}{e}-1)}$
satisfied equation (17). It can also be seen that $\beta=\sum\nolimits_{n\geq
0}\frac{-\eta k}{3n-1}Y^{n-\frac{1}{3}}+c_{1}Y^{\frac{1}{3}}$. Similarly to
linearize the first equation of (16), we proceed as before by seeking an
invariant algebraic surface of the form
$\tilde{\alpha}(Y)+\tilde{\beta}(Y)x=0$ with cofactor $ax+by$. Finally, the
change of variables
$(X,Z)=(\frac{x}{\tilde{\alpha}(Y)+\tilde{\beta}(Y)x},\frac{z}{\alpha(Y)+\beta(Y)z})$,
will linearize the first and third equations.

Case 2: The system reduces to

$\dot{x}=x(1+ax+by-kz),\qquad\dot{y}=y(-3+ey),\qquad\dot{z}=z(1+hy+kz).$ (18)

To linearize the system, the second and third equations are the same as
equations in previous case, so it suffices to linearize the first equation
only. We investigate the existence of an invariant algebraic surface
$A(Y,Z)+B(Y,Z)x=0$ with cofactor $ax+by-kz$, where $A$ and $B$ are analytic
with $A(0,0)=1$. The change of coordinate $X=\frac{x}{A(Y,Z)+B(Y,Z)x}$ will
linearize the first equation. To find such $A$ and $B$, they must satisfy the
equations

$\dot{A}=(by-kz)A,$ (19) $\dot{B}+B=aA.$ (20)

We write $A=\exp(\alpha(Y,Z))$ and put in equation(19), we get
$\dot{\alpha}(Y,Z)=by(Y,Z)-kz(Y,Z)$ and let $\alpha=\sum\nolimits_{i+j\geq
0}\alpha_{ij}Y^{i}Z^{j}$.
Then $\dot{\alpha}(Y,Z)=\sum\nolimits_{i+j\geq
0}\alpha_{ij}(j-3i)Y^{i}Z^{j}=by(Y,Z)-kz(Y,Z)=\sum\nolimits_{i+j\geq
0}\gamma_{ij}Y^{i}Z^{j}$, therefore $\alpha_{ij}=\frac{\gamma_{ij}}{j-3i}$ for
$i+j>0$ and the convergence of $\sum\nolimits_{i+j\geq
0}\gamma_{ij}Y^{i}Z^{j}$ imply the convergence of $\alpha$ and hence of $A$.
If $A=\sum\nolimits_{i+j\geq 0}a_{ij}Y^{i}Z^{j}$ and $B=\sum\nolimits_{i+j\geq
0}b_{ij}Y^{i}Z^{j}$ then from equation (20), we get $B=\sum\nolimits_{i+j\geq
0}a\ a_{ij}/(j-3i-1)Y^{i}Z^{j}$ which is convergent if $j\neq 3i+1$, then the
system is linearizable.

Case 3: Under the conditions of integrability, the system reduces to

$\dot{x}=x(1+by+cz),\qquad\dot{y}=y(-3+ey),\qquad\dot{z}=z(1+hy+kz).$

In this case, we try to linearize first equation and the rest are similar as
in Case 1. The change of variable $X=x\exp(-A)$ will give $\dot{X}=X$ whenever
$A$ satisfies

$\dot{A}(Y,Z)=by(Y)+cz(Y,Z).$ (21)

We put $A=\sum\nolimits_{i+j\geq 0}a_{ij}Y^{i}Z^{j}$. Then
$\dot{A}(Y,Z)=\sum\nolimits_{i+j\geq 0}a_{ij}(3i-j)Y^{i}Z^{j}$, and it is
obvious equation (21) can be solved analytically for $A$ with $z(Y,Z)$
contains no term of the type $(YZ^{3})^{n}$. Since
$z(Y,Z)=\frac{Z\alpha(Y)}{1-Z\beta(Y)}=\sum\alpha\beta^{i-1}Z^{i}$.
If we let $i-1=3m$, then it is enough to show that $\alpha\beta^{3m-1}$
contains no term in $Y^{m}$. Simple calculation shows

$\alpha\beta^{3m-1}=\sum\nolimits_{n\geq 0}\eta Y^{n}\sum\nolimits_{p\geq
0}C_{p}^{3m-1}(c_{1})^{-(p+3m-1)}(\sum\nolimits_{n\geq 0}-\frac{\eta
k}{3n-1}Y^{n})^{p}Y^{m-\frac{p-1}{3}},$

which does not contain any term of $Y^{m}$ and we have established the
existence of linearizing change of coordinates.

Case 4: The system has the form

$\dot{x}=x(1+dx+ey+cz),\qquad\dot{y}=y(-3+dx+ey),\qquad\dot{z}=z(1+kz).$

The change of coordinates $X=\frac{x}{1+dx-\frac{e}{3}y}$ and
$Y=\frac{y}{1+dx-\frac{e}{3}y}$ gives

$\dot{X}=X(1-cz+dczX),\qquad\dot{Y}=Y(-3-dezX),\qquad\dot{z}=z(1+kz).$

The first and third equations are linearizable node and the change of
coordinates are $\tilde{X}=\tilde{X}(X,z)$ and $Z=Z(z)$. For linearizing the
second equation, it is enough to find $\Gamma(\tilde{X},Z)$ such that
$\dot{\Gamma}=d\>e\>z\>X$ and the change of coordinate $\tilde{Y}=Ye^{\Gamma}$
gives $\dot{\tilde{Y}}=-3\tilde{Y}$. Let $\Gamma=\sum\nolimits_{i+j\geq
0}\gamma_{ij}\tilde{X}^{i}Z^{j}$ and
$X(\tilde{X},Z)z(Z)=\sum\nolimits_{i+j\geq 0}\eta_{i}\tilde{X}^{i}Z^{j}$.
We find that

$\dot{\Gamma}(\tilde{X},Z)=\sum\nolimits_{i+j\geq
0}(i+j)\gamma_{ij}\tilde{X}^{i}Z^{j}=\sum\nolimits_{i+j\geq
0}de\eta_{i}\tilde{X}^{i}Z^{j},$

Therefore, we see that $\gamma_{ij}=\frac{de\eta_{ij}}{i+j}$, which guaranties
a convergent for $\Gamma$.

Case 5: The conditions in this case yield the system

$\dot{x}=x(1-gx-hy),\qquad\dot{y}=y(-3-gx-hy),\qquad\dot{z}=z(1+gx+hy+kz).$

The system admits invariant algebraic surfaces $\l_{1}=1+kz+\frac{1}{2}kz(hy-
gx)=0$ and $\l_{2}=1-gx+\frac{1}{3}hy=0$ with respective cofactors $kz$ and
$-gx-hy$. Then the transformation $X=x\hskip 2.84544ptl^{-1}_{2},Y=y\hskip
2.84544ptl^{-1}_{2},Z=z\hskip 2.84544ptl^{-1}_{1}\hskip 2.84544ptl_{2}$
linearizes the system to $\dot{X}=X,\dot{Y}=-3Y,\dot{Z}=Z$.

Case 6: The system becomes

$\dot{x}=x(1+dx+ey),\qquad\dot{y}=y(-3+dx+ey),\qquad\dot{z}=z(1-2dx-2ey+kz).$

In this case we have two invariant algebraic surfaces
$\l_{1}=1+dx-\frac{1}{3}ey=0$ with cofactor $dx+ey$ and $\l_{2}=1+kz(1+dx-
ey+\frac{1}{2}(dx)^{2}+\frac{1}{5}(ey)^{2})=0$ with cofactor $kz$, which allow
us to linearize the system by the substitution $(X,Y,Z)=(x\hskip
2.84544ptl^{-1}_{1},y\hskip 2.84544ptl^{-1}_{1},z\hskip
2.84544ptl^{2}_{1}\hskip 2.84544ptl_{2}^{-1})$.

Case 7: The integrability conditions imply the system

$\dot{x}=x(1+ax+hy+cz),\qquad\dot{y}=y(-3+ey),\qquad\dot{z}=z(1+gx+hy+kz).$

The change of variables $X=x\ n(y)$ and $Z=z\ n(y)$ for some analytic function
$n(y)$ with $n(0)=1$, then the system takes the form

$\dot{X}=X(1+aXn^{-1}+cZn^{-1}+hy+\frac{\dot{n}}{n}y(-3+ey)),\\\ $
$\displaystyle\dot{y}=y(-3+ey),$ (22)
$\dot{Z}=Z(1+gXn^{-1}+kZn^{-1}+hy+\frac{\dot{n}}{n}y(-3+ey)).$

We can select $n$ such that $1+hy+\frac{\dot{n}}{n}y(-3+ey)=\frac{1}{n}$ with
$n(0)=1$.
This equation, clearly, is an ordinary first order linear differential
equation

$\dot{n}-\frac{1}{3}\frac{1+hy}{y(1-\frac{e}{3}y)}n=-\frac{1}{3y(1-\frac{e}{3}y)}.$

Therefor, its solution is

$n=-\frac{1}{3}e^{\int-\frac{1+hy}{3y(1-\frac{e}{3}y)}dy}(\int
e^{\int\frac{1+hy}{3y(1-\frac{e}{3}y)}dy}y^{-1}(1-\frac{e}{3}y)^{-1}dy+C).$

Manipulating some calculations gives

$\displaystyle n$
$\displaystyle=-\frac{1}{3}y^{-\frac{1}{3}}(1-\frac{e}{3})^{\frac{h}{e}+\frac{1}{3}}\int
y^{-\frac{2}{3}}(1-\frac{e}{3}y)^{\frac{h}{e}+\frac{4}{3}}dy$
$\displaystyle=-\frac{1}{3}(1-\frac{e}{3}y)^{\frac{h}{e}+\frac{1}{3}}\sum_{i=0}^{\infty}(-1)^{i}\frac{\prod_{j=0}^{i}(\frac{4}{3}+\frac{h}{e}-(j-1))}{i!(\frac{4}{3}+\frac{h}{e}+1)(i+\frac{1}{3}))}(-\frac{e}{3}y)^{i},$

which can be seen to be analytic in $y$ with $n(0)=1$. Then rescaling it by
$\frac{1}{n(y)}$, the system (22) becomes

${X^{\prime}}=X(1+aX+cZ),\qquad{y^{\prime}}=y(-3+ey)n(y),\qquad{Z^{\prime}}=Z(1+gX+kZ).$

where ${}^{\prime}=\frac{d}{d\tau}$ and $t=n\tau$ .The first and third
equations gives a linearizable node. For linearizing the second equation, we
substitute $Y=m(y)$, such that

$y(-3+ey)n(y)\frac{dm(y)}{dy}=-3m(y),\quad m(0)=0,\quad m^{\prime}(0)=1,$

put $n(y)$ in equation above then it is analytically solvable .

Case 8: The system takes the form

$\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3-3ax-3by),\qquad\dot{z}=z(1+3ax-5by+kz),$

which has two independent first integrals $\phi_{1}=x^{3}y$ and
$\phi_{2}=yz^{3}l^{-3}$, where $\l=(1+ax+by)^{2}+kz(1+\frac{1}{2}ax)-by=0$ is
an invariant algebraic surface with cofactor $2ax-6by+kz$.

Case 9: Here the corresponding system is

$\displaystyle\dot{x}=x(1+ax+by+kz),\quad\hskip
2.84544pt\dot{y}=y(-3+ey),\quad\hskip 2.84544pt\dot{z}=z(1+ax+hy+kz).$ (23)

The system has an invariant algebraic curve $\l=1-\frac{1}{3}ey=0$ with
cofactor $ey$. Then there is a first integral $\Phi=xz^{-1}l^{\frac{h-b}{e}}$
and an inverse Jacobi multiplier
$x^{2}y^{\frac{4}{3}}zl^{(\frac{2}{3}-{\frac{b}{e}})}$. Thus, the first
integral will be of the form $\Psi=x^{-1}y^{\frac{-1}{3}}(1+O(x,y,z))$. The
desired first integrals are $\phi_{1}=\Psi^{-3}=x^{3}y(1+O(x,y,z)$ and
$\phi_{2}=(\Phi\Psi)^{-3}=yz^{3}(1+O(x,y,z)$. However, if
$\xi=y(1-\frac{e}{3}y)^{-1}$ satisfies $\dot{\xi}=-3\xi$, then the system (23)
must also be linearizable by Theorem 2.2.

Case 10: The system of this case is written as

$\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3-3ax-3by),\qquad\dot{z}=z(1+4ax-8by+kz).$

We construct two independent first integrals $\phi_{1}=x^{3}y$ and
$\phi_{2}=yz^{3}l^{-3}$, where $\l=(1+ax+by)^{3}-3ab^{2}xy^{2}+kz(1+ax-
by-\frac{1}{3}(ax)^{2}-2abxy-\frac{1}{5}(by)^{2})=0$ is invariant algebraic
surface with cofactor $3ax-9by+kz$.

Case 11: The system has the form

$\dot{x}=x(1+ax+by-2kz),\qquad\dot{y}=y(-3-3ax-3by),\qquad\dot{z}=z(1+4ax-8by+kz).$

This case has an invariant algebraic surface
$\l=(1+ax+by)^{3}+kz(3ax+2(ax)^{2}-6abxy+akxz)=0$ with cofactor $3ax-9by$
which yields the Darboux integral $\Phi=xyz^{2}l^{-2}$ and an inverse Jacobi
multiplier $xy^{\frac{4}{3}}z^{2}l^{\frac{2}{3}}$. Therefor the other first
integral is of the form $\Psi=y^{\frac{-1}{3}}z^{-1}(1+O(x,y,z))$ by Theorem
2.1. The two independent first integrals will be desired as
$\phi_{1}=\Phi^{3}\Psi^{6}=x^{3}y(1+\cdots)$ and
$\phi_{2}=(\Psi)^{-3}=yz^{3}(1+\cdots)$.

Case 12: The system in this case is

$\ \dot{x}=x(1+ax),\qquad\dot{y}=y(-3-2ax+ey),\qquad\dot{z}=z(1+gx+hy).$

The linearizing change is given by $X=xl^{-1}_{1}$, $Y=yl^{3}_{1}l^{-1}_{2}$
and $Z=zl^{({\frac{h}{e}}-\frac{g}{a})}_{1}l^{-\frac{h}{e}}_{2}$, where
$\l_{1}=1+ax=0$ with cofactor $ax$ and $\l_{2}=1+ax-\frac{1}{3}ey=0$ with
cofactor $ax+ey$ are invariant algebraic surfaces.

Case 13: The reduced system is

$\dot{x}=x(1+ax-kz),\qquad\dot{y}=y(-3-ax+ey),\qquad\dot{z}=z(1+kz).$

The change of coordinates $(X,Y,Z)=(x\hskip 2.84544ptl^{-1}_{1}l_{3},y\hskip
2.84544ptl_{1}l^{-1}_{2}l_{3},z\hskip 2.84544ptl^{-1}_{3})$ will linearize the
system where $\l_{1}=1+ax+\frac{1}{2}akxz$,
$\l_{2}=1-\frac{1}{3}ey+kz+\frac{1}{6}aexy-\frac{1}{3}ekyz$ and $l_{3}=1+kz$.

Case 14: The system can be written in the form

$\dot{x}=x(1+ax),\qquad\dot{y}=y(-3-2ax+ey),\qquad\dot{z}=z(1+gx+kz).$

The system has invariant surfaces $\l_{1}=1+ax=0$,
$\l_{2}=1+ax-\frac{1}{3}ey=0$ with cofactors $ax$ and $ax+ey$, respectively.
The first and third equations are decoupled in the variables $x$ and $z$ which
give linearizable node and by the change of variables $X=x(1+O(x,z))$ and
$Z=z(1+O(x,z))$ will linearize $\dot{X}=X$, $\dot{Z}=Z$. The second equation
is linearized by the substitution $Y=yl^{3}_{1}l^{-1}_{2}$ which gives
$\dot{Y}=-3Y$.

Case 15: The system is of the form

$\dot{x}=x(1+ax),\qquad\dot{y}=y(-3-2ax+ey),\qquad\dot{z}=z(1+2ax+hy+kz).$

This system will be transformed to

$\dot{X}=X,\qquad\dot{Y}=Y(-3+eY),\qquad\dot{Z}=Z(1+hY+kZ),$ (24)

where $X=\frac{x}{1+ax}$, $Y=\frac{y}{1+ax}$ and $Z=\frac{z}{1+ax}$.
The second equation of (24) will be linearized by using the substitution
$\hat{Y}=\frac{Y}{1-\frac{1}{3}eY}$. For linearizing third equation in (24),
the idea is the same in Case 1.

Case 16: The system is of the form

$\displaystyle\dot{x}=x(1+ax),\qquad\dot{y}=y(-3-2ax+ey),\qquad\dot{z}.=z(1+3ax+hy+kz).$

We have two invariant algebraic surfaces $l_{1}=1+ax=0$ and
$l_{2}=1+ax-\frac{1}{3}ey=0$ with respective cofactors $ax$ and $ax+ey$,
respectively. The substitution
$(X,Y)=(\frac{x}{1+ax},\frac{y(1+ax)^{3}}{1+ax-\frac{1}{3}ey})$ will linearize
the first and second equations to $\dot{X}=X$ and $\dot{Y}=-3Y$. For
linearizing the third equation, we seek an algebraic surface of the form
$l=\gamma+\gamma_{1}z=0$ with cofactor $3ax+hy+kz$ with
$\gamma_{1}=\gamma_{1}(X,Y)$ and $\gamma=\gamma(X,Y)$ such that
$\gamma(0,0)=1$. The form $Z=\frac{z}{(\gamma+\gamma_{1}z)}$ will linearize
the third one. To find each of $\gamma$ and $\gamma_{1}$ we have to solve the
following two differential equations

$\dot{\gamma_{1}}-\gamma_{1}=k\gamma,$ (25) $\dot{\gamma}=(3ax+hy)\gamma.$
(26)

We write $\gamma=e^{\eta(X,Y)}$ and from equation (26) we obtain

$\dot{\eta}(X,Y)=3ax(X,Y)+hy(X,Y).$

Some calculations show

$y=\sum\nolimits_{n\geq 1}(-\frac{e}{3})^{n-1}Y^{n}(1-aX)^{3n+2},$

which contain no terms of the form $(X^{3}Y)^{n}$ because always $3n+2>3n$.

Case 17: The system takes the form

$\dot{x}=x(1+ax-kz),\qquad\dot{y}=y(-3-2ax+ey),\qquad\dot{z}=z(1+3ax+hy+kz).$

In this case we can neither linearize the system nor obtain two independent
first integrals for the system in classical way. So, if we use nonstandard
approach like when $a\simeq 0$ implies that $d\simeq 0$ and $g\simeq 0$, then
the system becomes

$\dot{x}\simeq x(1-kz),\qquad\dot{y}\simeq y(-3+ey),\qquad\dot{z}\simeq
z(1+hy+kz).$

The change of coordinate of $X=xz$ gives

$\dot{X}\simeq X(2+hy),\qquad\dot{y}\simeq y(-3+ey),\qquad\dot{z}\simeq
z(1+hy+kz).$

The system involvs an algebraic curve $l=1-\frac{e}{3}y=0$ with cofactor $ey$,
then the substitution $(\hat{X},Y)=(Xl^{\frac{-h}{e}},yl^{-1})$ will linearize
first and second equations. For linearizing third equation, it is the same as
in Case 1 as considered before.

Case 18: The system reduces to

$\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3-ax+ey),\qquad\dot{z}=z(1+hy+kz).$

We can see that the first and second equations are linearizable saddle by
Theorem 2.3 part i. For linearizing the third equation, we seek an invariant
algebraic surface of the form $A+Bz=0$ with cofactor $hy+kz$ with $A=A(X,Y)$
and $B=B(X,Y)$ such that $A(0,0)=1$. Then the linearizable change is given by
$Z=\frac{z}{(A+Bz)}$. To find such $A$ and $B$, we have to solve

$\dot{A}=Ahy,$ (27) $\dot{B}+B=kA.$

We now write $A=\exp(\eta(X,Y))$ and from equation (27) we obtain
$\dot{\eta}=hy$.
To solve this equation, let $\eta=\sum\nolimits_{i+j>0}a_{ij}X^{i}Y^{j}$.
Then
$\dot{\eta}(X,Y)=\sum\nolimits_{i+j>0}(i-3j)a_{ij}X^{i}Y^{j}=hy(X,Y)=b_{0,1}hY+\sum\nolimits_{i+j>1}b_{ij}X^{i}Y^{j}$,
for some $b_{i,j}$. Note that $a_{1,0}=0,a_{0,1}=\frac{-h}{3}$ and
$a_{ij}=\frac{b_{ij}}{(i-3j)}$ for $i+j>1$. The convergence of
$\sum\nolimits_{i+j>0}b_{ij}X^{i}Y^{j}$ guarantees the convergence of $\eta$
and hence of $A$. It remains to show that the inverse transformation
$z=\frac{AZ}{1-BZ}=\sum\nolimits AB^{n}Z^{n+1}$ does not involve the term
$X^{3}Y$.
Suppose $n+1=3m$ for some $m$. It suffices to show that $AB^{3m-1}$ has no
term $Y^{m}$. Direct calculation shows

$\dot{B}=\frac{dB}{dt}=-3Y\frac{dB}{dY},$
$\frac{-3Y}{3m}\frac{dB^{3m}}{dY}-B^{3m}=kAB^{3m-1}.$

Clearly, the coefficient in $Y^{n}$ in $B^{3m}$ vanishes on the right hand
side so that either $k=0$, that mean $B\equiv 0$, or the coefficients of
$Y^{n}$ in $B^{3m-1}$ vanishes. Therefore $Z=\frac{z}{A+Bz}$ contains no term
like $(X^{3}Y)^{n}$. Thus, we found the existence of a linearizing
transformation.

Case 19: The system in this case is

$\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3-ax+ey),\qquad\dot{z}=z(1+gx+hy).$

The first two equations give a linearizable saddle by Theorem 2.3, part i and
hence $\dot{X}=X,\quad\dot{Y}=-3Y$.
Using the substitution $Z=z\exp(-\gamma(X,Y))$ to linearize the remaining
equation. Now, it suffices to find $\gamma(X,Y)$ such that

$\dot{\gamma}=gx(X,Y)+hy(X,Y).$ (28)

Let

$\gamma(X,Y)=\sum\nolimits_{i+j>0}\alpha_{ij}X^{i}Y^{j},$ (29)

then

$gx(X,Y)+hy(X,Y)=\sum\nolimits_{i+j>0}\beta_{ij}X^{i}Y^{j}.$

After differentiating (29) and equating with (28) we obtain

$\displaystyle\sum\nolimits_{i+j>0}(i-3j)\alpha_{ij}X^{i}Y^{j}=\sum\nolimits_{i+j>0}\beta_{ij}X^{i}Y^{j}$

Then we can choose $\alpha_{ij}=\frac{\beta_{ij}}{(i-3j)}$ for $i\neq 3j$ and
the convergence follows from the convergence of
$\sum\nolimits_{i+j>0}\beta_{ij}X^{i}Y^{j}$. Thus the system is linearizable.

Case 20: The system with integrability conditions has the form

$\displaystyle\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3-ax+ey),\qquad\dot{z}=z(1+gx+hy+kz).$

The first and second equations are linearizable saddle by Theorem 2.3, part i.
The third equation can be linearized in the similar way of Case 16.

Case 21: The system is written as

$\displaystyle\dot{x}=x(1+ax+by),\qquad\dot{y}=y(-3-ax+ey),\qquad\dot{z}=z(1+2ax+hy+kz).$

The first two equations are linearizable saddle by Theorem 2.3, part i. The
linearization of the third equation is similar to the Case 16.

Case 22: The corresponding system in this case is

$\dot{x}=x(1+ax+cz),\qquad\dot{y}=y(-3-ax-hy),\qquad\dot{z}=z(1+3ax+hy+2cz),$

which has an invariant algebraic surface $\l=(1+ax)^{2}+cz(2+hy)=0$ with
cofactor $2(ax+cz)$. We construct a first integral of the form
$\Phi=x^{2}yzl^{-2}$ with an inverse Jacobi multiplier
$IJM=x^{4}y^{2}zl^{\frac{-1}{2}}$, then the other first integral is
$\Psi=x^{-3}y^{-1}l^{\frac{-3}{2}}$ by Theorem 2.1. Now, the required first
integrals are $\phi_{1}=\Phi^{3}\Psi^{2}=yz^{3}(1+\cdots)$ and
$\phi_{2}=(\Psi)^{-1}=x^{-1}y(1+\cdots)$. Since $\xi=xl^{-\frac{1}{2}}$
satisfies $\dot{\xi}=\xi$, then the system in this case must also be
linearizable by Theorem 2.2.

Case 23: The system

$\displaystyle\dot{x}=x(1-gx-hy+3kz),\hskip 8.5359pt\dot{y}=y(-3-gx-hy-
kz),\hskip 8.5359pt\dot{z}=z(1+gx+hy+kz),$

has an invariant algebraic surface $\l=(1+kz)^{2}+kz(hy-gx)=0$ with cofactor
$2kz$. We find a first integral $\Phi=xyz^{2}l^{-2}$ and an inverse Jacobi
multiplier $IJM=xy^{1/2}z^{-1/2}l$ and the other first integral
$\Psi=y^{\frac{1}{2}}z^{\frac{3}{2}}(1+...)$ is guaranteed by Theorem 2.1. The
two first integrals in standard form are
$\phi_{1}=\Phi^{3}\Psi^{-4}=x^{3}y(1+\cdots)$ and
$\phi_{2}=(\Psi)^{2}=yz^{3}(1+\cdots)$. The substitution
$(X,Y,Z)=(\frac{\Phi}{yz^{3}},\frac{\phi_{2}}{z^{3}},\frac{\Phi}{xy})$
linearizes the system.

Case 24: Under these conditions the corresponding system has the form

$\displaystyle\dot{x}=x(1+ax+by-kz),\qquad\dot{y}=y(-3-3ax-3by-
kz),\qquad\dot{z}=z(1+3ax-5by+kz).$

The system has two invariant algebraic surfaces $\l_{1}=(1+ax+by)^{2}+kz(ax-
by)=0$ and $\l_{2}=(1+ax+by+kz)^{2}-4bkyz=0$ with respective cofactors
$2ax-6by$ and $2ax-6by+2kz$. Then the system has two independent first
integrals $\phi_{1}=x^{3}y(l_{1}l_{2})^{2})$ and
$\phi_{2}=yz^{3}l_{1}^{2}l_{2}^{-1}$.

Case 25: The corresponding system is

$\displaystyle\dot{x}=x(1-dx),\qquad\dot{y}=y(-3+dx+ey-
kz),\qquad\dot{z}=z(1+kz).$

The surfaces $\l_{1}=1+kz=0$, $\l_{2}=1-dx=0$ and
$\l_{3}=(1+kz)^{2}-\frac{1}{3}ey(1+\frac{1}{2}dx-\frac{3}{2}kz)=0$ are
invariant algebraic surfaces with cofactors $kz$, $-dx$ and $ey+2kz$,
respectively.
By the change of coordinates $X=xl^{-1}_{2}$, $Y=yl_{1}^{3}l_{2}l^{-1}_{3}$
and $Z=zl^{-1}_{1}$, we obtain $\dot{X}=X,\dot{Y}=-3Y,\dot{Z}=Z$.

Case 26: In this case the system is

$\displaystyle\dot{x}=x(1+ax+by+2kz),\qquad\dot{y}=y(-3-ax+ey-2kz),\qquad\dot{z}=z(1+kz).$

By using $X=\frac{x}{1+kz}$, $Y=\frac{y}{1+kz}$ and $Z=\frac{z}{1+kz}$, the
system transforms to

$\displaystyle\dot{X}=X(1+aX+bY),\hskip 14.22636pt\dot{Y}=Y(-3-aX+eY),\hskip
14.22636pt\dot{Z}=Z.$

where the first two equations are linearizable saddle by Theorem 2.3, part i
and then the system is linearizable.

Case 27: The system takes the form

$\displaystyle\dot{x}=x(1+ax),\qquad\dot{y}=y(-3-2ax+ey-
kz),\qquad\dot{z}=z(1+3ax+kz).$

The change of variable $Y=yl_{1}l_{2}l_{3}$ linearzies the second equation
where $\l_{1}=(1+ax)^{2}+kz(1+\frac{1}{2}ax)=0$,
$\l_{2}=1+ax-\frac{1}{3}ey(1+kz)=0$ and $\l_{3}=1+ax=0$ are invariant
algebraic surfaces with respective cofactors $2ax+kz$, $ax+ey$ and $ax$. The
other two equations are linearizable node and the change of coordinates
$X=x(1+O(x,z))$ and $Z=z(1+O(x,z)$) gives $\dot{X}=X,\dot{Z}=Z$.

Case 28: Under the conditions of this case, the system (15) is written as

$\dot{x}=x(1+ax+cz),\qquad\dot{y}=y(-3+2ax+ey-2cz),\quad\dot{z}=z(1+3ax+2cz).$

The system has two invariant algebraic surfaces $\l_{1}=(1+ax)^{2}+2cz)=0$ and
$\l_{2}=1+\frac{1}{3}ey(-1+ax+cz-(ax)^{2})=0$ with cofactors $2(ax+cz)$ and
$ey$, respectively.
The second equation of the system is linearized by $Y=yl_{1}l_{2}^{-1}$ to
$\dot{Y}=-3Y$ and the two other equations of the system are linearizable node
with the change of variables $X=x(1+O(x,z))$ and $Z=z(1+O(x,z)$) so that
$\dot{X}=X,\>\dot{Z}=Z$.

Case 29: The system reduces to

$\dot{x}=x(1+ax+ey+cz),\qquad\dot{y}=y(-3+ax+ey-kz),\qquad\dot{z}=z(1+kz).$

The system has an invariant algebraic curve $l=1+kz=0$ with cofactor $kz$.
Then we find a first integral $\Phi=x^{-1}yz^{4}l^{(\frac{c}{k}-3)}$ with an
inverse Jacobi multiplier $xy^{2}z^{4}l^{-1}$. The other independent first
integral is $\Psi=y^{-1}z^{-3}(1+O(x,y,z))$. Therefore the two first integrals
in desired forms are $\phi_{1}=\Phi^{-3}\Psi^{-4}$ and $\phi_{2}=\Psi^{-1}$.

Case 30: The system takes the form

$\dot{x}=x(1+ax+ey+cz),\qquad\dot{y}=y(-3+ax+ey-2kz),\quad\dot{z}=z(1+kz).$

This case has an invariant algebraic curve $l=1+kz=0$ with cofactor $kz$. We
find the first integral $\Phi=x^{-1}yz^{4}l^{(\frac{c}{k}-2)}$ and the inverse
Jacobi multiplier is $xy^{2}z^{4}$. Theorem 2.1 guarantees that the existence
of the other independent first integral is $\Psi=y^{-1}z^{-3}$. Thus, the two
first integrals in desired form are $\phi_{1}=\Phi^{-3}\Psi^{-4}$ and
$\phi_{2}=\Psi^{-1}$. Since $\xi=z(1+kz)^{-1}$ satisfies $\dot{\xi}=\xi$, then
the system must also be linearizable by Theorem 2.2.

Case 31: The system becomes

$\dot{x}=x(1+ax-hy+3kz),\qquad\dot{y}=y(-3-ax-
kz),\qquad\dot{z}=z(1+3ax+hy+kz).$

We use nonstandard approach again as in Case 17. Assuming $a\simeq 0$ implies
that $d\simeq 0$ and $g\simeq 0$, then the system becomes

$\dot{x}\simeq x(1-hy+3kz),\qquad\dot{y}\simeq y(-3-kz),\qquad\dot{z}\simeq
z(1+hy+kz).$

The change of coordinates $(X,Y,Z)=(xy^{3},y,yz)$ transforms the system to

$\dot{X}\simeq X(-8-hY),\qquad\dot{Y}\simeq-3Y-kZ,\qquad\dot{Z}\simeq
Z(-2+hY).$

The critical point at the origin of this system is in the Poincaré domain and
hence is linearizable via an analytic change of coordinates which can be
chosen as $(\hat{X},\hat{Y},\hat{Z})\simeq(X(1+O(1)),Y(1+O(1)),Z(1+O(1)))$.
The two first integrals are $\hat{\phi}=\hat{X}^{-3}\hat{Y}^{2}$ and
$\hat{\psi}=\hat{Y}^{2}\hat{X}^{-3}$, then pulling back these two first
integrals, we see that the original system has two independent first integrals
of required form $\phi_{1}=\hat{\phi}^{-1}=x^{3}y(1+\cdots)$ and
$\phi_{2}=\hat{\psi}^{-1}=yz^{3}(1+\cdots)$.

Case 32: When $k=0$ we have two subcases

i) $c=0,f=0,b=h,d=-a$, then the system takes the form

$\displaystyle\dot{x}=x(1+ax+hy),\qquad\dot{y}=y(-3-ax+ey),\qquad\dot{z}=z(1+gx+hy).$

is the same proof as in Case 19.
ii) $d=0,a=0,f=0,h=b$ then the system in this case is

$\dot{x}=x(1+by+cz),\quad\hskip 28.45274pt\dot{y}=y(-3+ey),\quad\hskip
28.45274pt\dot{z}=z(1+gx+by)$

is the same proof of Case 7.
If $k\neq 0$ then the system takes

$\dot{x}=x(1-dx+hy+cz),\qquad\dot{y}=y(-3+dx+ey-
kz),\qquad\dot{z}=z(1+\frac{(-2k+c)}{k}dx+hy+kz).$

The substitution $(X,y,Z)=(dx-kz,y,yz)$ the system reduces to

$\dot{X}=X(1-X+hy),\qquad\dot{y}=y(-3+X+ey),\qquad\dot{Z}=Z(-2+\frac{(c-k)}{k}dX+(e+h)y).$

The first two equations give linearizable saddle by Theorem 2.3, part i and
clearly $\dot{\hat{X}}=\hat{X}$ and $\dot{Y}=-3Y$.
The transformation $\hat{Z}=Z\exp(-\gamma(\hat{X},Y))$ will linearize the
third equation. It just remains to find $\gamma(\hat{X},Y)$ such that

$\dot{\gamma}=\frac{(c-k)}{k}dX(\hat{X},Y)+(e+h)y(\hat{X},Y).$ (30)

To see this, let

$\gamma(\hat{X},Y)=\sum\nolimits_{i+j>0}\alpha_{ij}\hat{X}^{i}Y^{j}.$ (31)

Also assume $X(\hat{X},Y)=\sum\nolimits_{i+j>0}g_{ij}\hat{X}^{i}Y^{j}$ and
$y(\hat{X},Y)=\sum\nolimits_{i+j>0}h_{ij}\hat{X}^{i}Y^{j}$.
Differentiating (31) and equating with (30), we get

$\sum\nolimits_{i+j>0}(i-3j)\alpha_{ij}X^{i}Y^{j}=\sum\nolimits_{i+j>0}(\frac{c-k}{k})dg_{ij}\hat{X}^{i}Y^{j}+\sum\nolimits_{i+j>0}(e+h)h_{ij}\hat{X}^{i}Y^{j},$

then we can choose
$\alpha_{ij}$=$\frac{(\frac{c-k}{k})dg_{ij}+(e+h)h_{ij}}{i-3j}$ for $i\neq 3j$
or otherwise $\alpha_{ij}=0$. Therefore, the system is linearizable.

Case 33: The system takes the form

$\displaystyle\dot{x}=x(1-gx+by+kz),\qquad\dot{y}=y(-3-gx+ey-
kz),\qquad\dot{z}=z(1+gx+hy+kz).$

The change of coordinates $(X,Y,Z)=(xy,y,yz)$ reduce the system into

$\displaystyle\dot{X}=X(-2+(b+e)Y),\qquad\dot{y}=Y(-3+eY)-gX-
kZ,\qquad\dot{Z}=Z(-2+(e+h)Y)$

The critical point at the origin of system is in the Poincaré domain and hence
is linearizable by the analytic change of variables which can be chosen to be
of the form
$(\hat{X},\hat{Y},\hat{Z})=(X(1+O(1)),Y-gX-kZ+O(2),Z(1+O(1)))$.
The two first integrals $\hat{\phi}=\hat{X}^{-3}\hat{Y}^{2}$ and
$\hat{\psi}=\hat{Y}^{2}\hat{Z}^{-3}$ pull back these two first integrals, we
see that the initial system has two independent analytic first integrals of
desired form $\phi_{1}=\hat{\phi}^{-1}=x^{3}y(1+\cdots)$ and
$\phi_{2}=\hat{\psi}^{-1}=yz^{3}(1+\cdots)$.

Case 34: The system (15) can be written as

$\dot{x}=x(1+ax+cz),\qquad\dot{y}=y(-3+dx+fz),\qquad\dot{z}=z(1+gx+kz).$

The first and third equations are linearizable node, with the change of
coordinates $X=x(1+O(x,z))$ and $Z=z(1+O(x,z)$) such that
$\dot{X}=X,\dot{Z}=Z$. The second equation can be linearized by the
transformation $Y=ye^{-\eta}$ if $\eta$ can be selected so that

$\dot{{\eta}}(X,Z)=dx(X,Z)+fz(X,Z).$

Let $\eta(x,Z)=\sum\nolimits_{i,j\geq 0}a_{ij}X^{i}Z^{j}$. Then

$\dot{{\eta}}=\sum\nolimits_{i,j\geq
0}(i+j)a_{ij}X^{i}Z^{j}=dx(X,Z)+fz(X,Z)=\sum\nolimits_{i,j\geq
0}b_{ij}X^{i}Z^{j}.$

Setting $a_{ij}=\frac{b_{ij}}{i+j}$ for $i+j>0$ which yields the convergence
of $\eta$.

Case 35: When $a\neq 0$, then the system can be written as

$\dot{x}=x(1+ax+(2k-\frac{gk}{a})z),\qquad\dot{y}=y(-3-2ax+ey-2kz),\qquad\dot{z}=z(1+gx+kz).$

We can see that the first and third equations of the system are lineaizable
node and the change of variable $Y=yl_{1}^{3}l_{2}^{-1}$ will linearize the
second equation, where $\l_{1}=1+ax+kz=0$ and
$\l_{2}=1+ax-\frac{1}{3}ey+kz=0$.
When $a=0$, we have two sub-cases.

  1. i)

$d=g=b=h=f+2k=0$, then the system becomes

$\dot{x}=x(1+cz),\qquad\dot{y}=y(-3+ey-2kz),\qquad\dot{z}=z(1+kz).$

The transformation
$(X,Y,Z)=(xl_{1}^{-\frac{c}{k}},y(l_{1}l_{2})^{-1},zl_{1}^{-1})$ will
linearize the system where $l_{1}=1+kz$ and $l_{2}=1-\frac{e}{3}y+kz$ .

  2. ii)

$d=k=b=h=f=0$, then the system takes the form

$\dot{x}=x(1+cz),\qquad\dot{y}=y(-3+ey),\qquad\dot{z}=z(1+gx).$

The system has an invariant algebraic curve $l=1-\frac{e}{3}y=0$ with cofactor
$ey$. The second equation linearized by the substitution $Y=yl^{-1}$ gives
$\dot{Y}=-3Y$. The remained equations are in the variables $x$ and $z$ which
give linearizable node and by the change of coordinates $X=x(1+O(x,z))$ and
$Z=z(1+O(x,z))$ will linearize so that $\dot{X}=X$, $\dot{Z}=Z$.

Case 36: The system can be written as

$\dot{x}=x(1+ax+3kz),\qquad\dot{y}=y(-3-2ax+ey-2kz),\qquad\dot{z}=z(1+3ax+kz).$

This case have invariant algebraic surfaces
$\l_{1}=1-\frac{1}{3}ey+\frac{1}{3}eaxy+\frac{1}{3}ekyz-\frac{1}{3}ea^{2}yx^{2}+\frac{2}{3}keaxyz-\frac{1}{3}ek^{2}yz^{2}=0$
and $\l_{2}=1+2ax2kz-2kaxz+(ax)^{2}+(kz)^{2}=0$ with cofactors $ey$ and
$2(ax+kz)$, respectively. The transformation $Y=yl_{1}^{-1}l_{2}$ linearizes
the second equation. Note also that the first and third equations of the
system are lineaizable node.

Case 37: The integrability conditions in this case takes the form

$\dot{x}=x(1+ax-5hy+3kz),\qquad\dot{y}=y(-3-3hy-3kz),\qquad\dot{z}=z(1+hy+kz).$

The system has an invariant algebraic surface
$l=(1+hy+kz)^{2}+\frac{1}{2}ax(2-hy+kz)=0$ with cofactor $ax-6hy+2kz$, which
gives two independent first integrals $\phi_{1}=x^{3}yl^{-3}$ and
$\phi_{2}=yz^{3}$.

Case 38: The system has the form

$\dot{x}=x(1+ax-8hy+4kz),\qquad\dot{y}=-3y(1+hy+kz),\qquad\dot{z}=z(1+hy+kz).$

It is obvious we have a first integral $\Phi=yz^{3}$. We can also find an
inverse Jacobi multiplier $x^{2}y^{1/3}z^{-2}l^{-2}$ which yields the second
first integral $\Psi=x^{-1}y^{\frac{1}{3}}z^{3}l^{-2}$. We write the
independent first integrals in required forms
$\phi_{1}=\Phi^{3}\Psi^{-3}=x^{3}y(1+\cdots)$ and $\phi_{2}=yz^{3}$.

Case 39: The system becomes

$\dot{x}=x(1+ax-8hy+4kz),\qquad\dot{y}=y(-3+hy+kz),\qquad\dot{z}=z(1-2ax+hy+kz).$

In this case, the system has an invariant algebraic surface
$\l=(1+hy+kz)^{3}+akxz(3+ax-6hy+2kz)=0$ with cofactor $-9hy+3kz$ which lets us
deduce a first integral $\Phi=x^{2}yzl^{-2}$ and an inverse Jacobi multiplier
$x^{2}y^{4/3}zl^{-2/3}$. We can obtain a second integral
$\Psi=x^{-1}y^{\frac{-1}{3}}(1+\cdots)$ by Theorem 2.1. We write these first
integrals in desired forms $\phi_{1}=\Psi^{-3}=x^{3}y(1+cdots)$ and
$\phi_{2}=(\Psi)^{6}\Phi^{3}=yz^{3}(1+\cdots)$.

Case 40: The system reduces to

$\dot{x}=x(1-gx-5hy+3kz),\qquad\dot{y}=y(-3+gx-3hy-3kz),\qquad\dot{z}=z(1+gx+hy+kz).$

Invariant algebraic surfaces in this case are $l_{1}=(1+hy+kz)+gx(hy-kz)=0$
and $l_{2}=(1-gx+hy+kz)^{2}+4ghxy=0$ with respective cofactors $-2(3hy-kz)$
and $-2(gx+3hy-kz)$ yielding two independently first integrals as follows
$\phi_{1}=x^{3}yl_{1}^{-2}l_{2}^{-1}$ and
$\phi_{2}=yz^{3}l_{1}^{-2}l_{2}^{2}$.

Case 41: If $e\neq 0$, then we have two subcases.

  1. 1.

The integral conditions appears as

$k(b+e-h)-ce=a(3b-2e)-de=k(2e-3h)+fe=a(b-e-h)+ge=0.$

The system has an invariant algebraic surface $l=1+ax-\frac{1}{3}ey+kz=0$ with
cofactor $ax+ey+kz$ and produces two first integrals
$\phi_{1}=x^{3}yl^{-1-\frac{3b}{e}}$ and $\phi_{2}=yz^{3}l^{-1+\frac{3h}{e}}$.

  2. 2.

With the conditions $a=g=-\frac{d}{3},b=h=-\frac{e}{3}$ and
$c=k=-\frac{f}{3}$, the system (15) reduces to

$\dot{x}=x(1-\frac{d}{3}x-\frac{e}{3}y-\frac{f}{3}z),\qquad\dot{y}=y(-3+dx+ey+fz),\qquad\dot{z}=z(1-\frac{d}{3}x-\frac{e}{3}y-\frac{f}{3}z)$

Therefore, the system has two independent first integrals of the form
$\phi_{1}=x^{3}y$ and $\phi_{2}=yz^{3}$.

We now consider the case when $e=0$. Again we have some subcases.

  1. i.

$h=k=0$.
If $b=0$, $a=0$ and $d=\frac{-g(f+3c)}{c}$, the system has exponential factor
$E_{1}=\exp(-gx+cz)$ with cofactor $-gx+cz$, which yields two first integrals
$\phi_{1}=x^{3}yE_{1}^{-3-\frac{f}{c}}$ and
$\phi_{2}=yz^{3}E_{1}^{\frac{-f}{c}}$.

If $b\neq 0$ , $a=0$ $f=0$ and $d=-3g$, then the system admits an exponential
factor $E_{2}=\exp(3gx+by-3cz)$ with cofactor $3(gx-by-cz)$, and we construct
two first integrals $\phi_{1}=x^{3}yE_{2}$ and $\phi_{2}=yz^{3}$.

  2. ii.

If $h\neq 0$, $k=0$, $a=0$ ,$b=\frac{h(f+3c)}{f}$ and $d=-\frac{g(f+3c)}{c}$,
we have an exponential factor $E_{3}=\exp(-fgx-hey+fcz)$ with cofactor
$-fgx+3hey+fcz$, with two first integrals
$\phi_{1}=x^{3}yE_{3}^{\frac{-(3c+f)}{fc}}$ and
$\phi_{2}=yz^{3}E_{3}^{\frac{-1}{c}}$.

  3. iii.

If $k\neq 0$, $h=0$ $b=0$, $c=\frac{k(2a-g)}{a}$and $f=-\frac{k(3a-d-3g)}{a}$,
the system has an algebraic surface $\l=1+ax+kz=0$ with cofactor $ax+kz$, with
two first integrals $\phi_{1}=x^{3}yl^{-\frac{(3a+d)}{a}}$ and
$\phi_{2}=yz^{3}l^{-\frac{(d+3g)}{a}}$.

Case 41.1: The system is of the form

$\dot{x}=x(1+dx+ey),\qquad\dot{y}=y(-3+dx+ey),\qquad\dot{z}=z(1+gx+\frac{eg}{d}y).$

We have two subcases depending on the parameter $d$.

  1. 1.

When $d\neq 0$, the change of variables $X=xl^{-1}$, $Y=yl^{-1}$ and
$Z=zl^{-\frac{g}{d}}$ where $l=1+dx-\frac{e}{3}y=0$ linearizes the system.

  2. 2.

When $d=0$, $e\neq 0$ and $g=0$, the change of variables $X=xl^{-1}$,
$Y=yl^{-1}$ and $Z=zl^{-\frac{h}{e}}$ linearizes the system.
Now, if $e=0$ and $g\neq 0$, the change of variables $X=x$, $Y=y$ and $Z=zE$
where $E=\exp(gx-\frac{h}{3}y)$ linearizes the system.

Case 41.2: The system in this case has the form

$\dot{x}=x(1+gx+hy+kz),\qquad\dot{y}=y(-3+gx+hy+kz),\qquad\dot{z}=z(1+gx+hy+kz).$

The system will be linearized by the change of coordinates
$(X,Y,Z)=(xl^{-1},yl^{-1},zl^{-1})$ where $l$ is an invariant algebraic
surface of the form $l=1+gx+hy+kz=0$ with cofactor $gx+hy+kz$.
∎

##  6\. Nonstandard results

This section motivated two different approaches to construct two independent
first integrals. The first motivation is infinitely close to standard first
integrals by using nonstandard perturbation theory of differential equations.
The other motivation uses the unperturbation nonstandard transformation to
find infinitesimal, limited or unlimited first integrals.
We study two types of system (1) depends on eigenvalues which are
$\delta=(1+\varepsilon)$, $\varepsilon$ is infinitesimal, in these cases,
either $\delta$ is appreciable or $\delta$ is itself infinitesimal.

Now, the change of variables $x_{1}=\delta^{-1}x$, $y_{1}=\delta^{-1}y$,
$z_{1}=\delta^{-1}z$ and $\tau=\delta t$ transform the system (1) with respect
to $\tau$ to

$\displaystyle{x^{\prime}_{1}}$ $\displaystyle=$ $\displaystyle
x_{1}(1+ax_{1}+by_{1}+cz_{1}),$ (32) $\displaystyle{y^{\prime}_{1}}$
$\displaystyle=$ $\displaystyle y_{1}(-3+dx_{1}+ey_{1}+fz_{1}),$
$\displaystyle{z^{\prime}_{1}}$ $\displaystyle=$ $\displaystyle
z_{1}(1+gx_{1}+hy_{1}+kz_{1}),$

where ′ stands for $\frac{d}{d\tau}$ and in general the variables and
parameters are hyperreals.

###  6.1. Perturbation Case

In this subsection, the study of the system with $\delta=(1+\varepsilon)$
satisfies the perturbation technique to find first integrals and linearization
of differential system, besides that, the $\delta$ is infinitesimal type of
system and it does not satisfy the perturbation technique.
Where $\delta=(1+\varepsilon)$ the system becomes

$\displaystyle
F=\left\\{\begin{array}[]{rc}\dot{x}&=F_{1}=x((1+\varepsilon)+ax+by+cz),\\\
\dot{y}&=F_{2}=y(-3(1+\varepsilon)+dx+ey+fz),\\\
\dot{z}&=F_{3}=z((1+\varepsilon)+gx+hy+kz),\end{array}\right.$ (36)

where $F=(F_{1},F_{2},F_{3})$, where variables are near standard numbers and
parameters are standard.
Let $\mathbb{R}^{3}$ be a standard topological space. A point
$p=(x,y,z)\in\mathbb{R}^{3}$ is said to be infinitely close to a standard
point $p_{0}=(x_{0},y_{0},z_{0})\in\mathbb{R}^{3}$ which is denoted by
$p\simeq p_{0}$, if $p$ is in any standard neighborhood of $p_{0}$. Let
$D_{0}$ be a subset of $\mathbb{R}^{3}$. A point $p\in\mathbb{R}^{3}$ is said
to be nearstandard in $D_{0}$ if there is a standard $p_{0}$ $\in$$D_{0}$ such
that $p\simeq p_{0}$ . Let us denote by

$\displaystyle{}^{NS}D_{0}=\\{{p\in\mathbb{R}^{3}:\exists^{st}p_{0}\in
D_{0},,p\simeq p_{0}}\\},$

the external-set of nearstandard points in $D_{0}$.
The reformulation of Theorem 5.1 to standard case by using the definition of
standard with $\varepsilon=0$ and its nonstandard formulation proof is the
same as the classical proof.

######  Proposition 2.

Let $F:D\to\mathbb{R}^{3}$ and $F_{0}:D_{0}\to\mathbb{R}^{3}$ be vector
fields, $p=(x,y,z)\in D$ and $p_{0}=(x_{0},y_{0},z_{0})\in D_{0}$ and $p_{0}$
is standard. Then vector field $F$ is perturbation to the vector field $F_{0}$
and system (36) has the same integrability conditions of system (15), if
$p\simeq p_{0}$ and $F\simeq F_{0}$, for topology of uniform convergence on
compacta.

###### Proof.

According to Heine Borel’s theorem, $\mathbb{R}^{3}$ is locally compact (any
compact set is locally compact), we assume $p=(x,y,z)\in{}^{NS}D_{0}$, then
there exists a standard $p_{0}=(x_{0},y_{0},z_{0})\in D_{0}$ such that
$p\simeq p_{0}$.
To prove $F$ is infinitely close to $F_{0}$, that means

$\displaystyle F\simeq F_{0}.$

Since $\mathbb{R}^{3}$ is locally compact, there exists a standard compact
neighborhood $K$ of $p_{0}$ such that $p\simeq p_{0}$. Hence $p\in K\subset D$
and $p=(x,y,z)$ belongs to standard set $D_{0}$.
Now, $F_{1}(p)=x(1+\varepsilon+ax+by+cz)$ and $F_{01}(p)=x(1+ax+by+cz)$, so

$F_{1}(p)-F_{01}(p)=x(1+\varepsilon+ax+by+cz)-x(1+ax+by+cz)=x\varepsilon.$

Then

$F_{1}(p)\simeq F_{01}(p).$

We repeat the processes for the two other $F_{2}(p)\simeq F_{02}(p)$ and
$F_{3}(p)\simeq F_{03}(p)$.
Thus, $F$ is infinitely close to $F_{0}$, then $F$ is perturbation to $F_{0}$.
Using the same transformation as in the system (32) into the system (36), we
get the same integrability conditions as in Theorem 5.1.
∎

######  Proposition 3.

Let a vector field $F:D\to\mathbb{R}^{3}$ is a perturbation of a standard
vector field $F_{0}:D_{0}\to\mathbb{R}^{3}$ and system (36) has two first
integrals of the form $H_{1}(x,y,z)=x^{3}y(1+O(x,y,z))$ and
$H_{2}(x,y,z)=yz^{3}(1+O(x,y,z))$ where $(x,y,z)\in{}^{NS}D_{0}$, then
$H_{1}(x,y,z)\simeq\phi_{1}(x,y,z)$ and $H_{2}(x,y,z)\simeq\phi_{2}(x,y,z)$.

###### Proof.

By Proposition 1, the vector field $F=(F_{1},F_{2},F_{3})$ in (36) is a
perturbation to the standard vector field $F_{0}=(F_{01},F_{02},F_{03})$ in
system (15).

Since $(x,y,z)\in{}^{NS}D_{0}$, we can set $x=x_{0}+\varepsilon_{1}$,
$y=y_{0}+\varepsilon_{2}$ and $z=z_{0}+\varepsilon_{3}$, where
$\varepsilon_{1},\varepsilon_{2},\varepsilon_{3}$ are infinitesimal and
$(x_{0},y_{0},z_{0})\in D_{0}$.

$\displaystyle
H_{1}(x_{0}+\varepsilon_{1},y_{0}+\varepsilon_{2},z_{0}+\varepsilon_{3})-\phi_{1}(x_{0},y_{0},z_{0})$
$\displaystyle=(x_{0}+\varepsilon_{1})^{3}(y_{0}+\varepsilon_{2})(1+O(x_{0}+\varepsilon_{1},y_{0}+\varepsilon_{2},z_{0}$
$\displaystyle+\varepsilon_{3}))-x_{0}^{3}y_{0}(1+O(x_{0},y_{0},z_{0})).$ (37)

Performing some calculations, we can simplify equation (6.1) to

$x_{0}^{3}y_{0}(1+O(x_{0},y_{0},z_{0})+\tau_{1})-x_{0}^{3}y_{0}(1+O(x_{0},y_{0},z_{0}))=x_{0}^{3}y_{0}\tau_{1},$

where $\tau_{1}$ is infinitesimal and this implies $H_{1}\simeq\phi_{1}$ for
all near standard point of $D_{0}$
In similar way we can obtain $H_{2}\simeq\phi_{2}$.
∎

######  Proposition 4.

If the vector field $E_{1}:D\to\mathbb{R}^{3}$ is a perturbation of a standard
vector field $E_{0}:D_{0}\to\mathbb{R}^{3}$, with $p\in{}^{NS}D_{0}$ and
$p_{0}\in D_{0}$, then the linearizablity of the two vector fields (36) and
(15) are $E_{1}\simeq E_{0}$ and $p\simeq p_{0}$, where

$E_{1}=\left\\{\begin{array}[]{rcc}\dot{x}&=E_{11}=(1+\varepsilon)x,\\\
\dot{y}&=E_{12}=-3(1+\varepsilon)y),\\\
\dot{z}&=E_{13}=(1+\varepsilon)z,\end{array}\right.$

and

$E_{0}=\left\\{\begin{array}[]{rccc}\dot{x}&=E_{01}=x,\\\
\dot{y}&=E_{02}=-3y,\\\ \dot{z}&=E_{03}=z,\end{array}\right.$

where $E_{1}=({E_{11},E_{12},E_{13}})$ and $E_{0}=({E_{01},E_{02},E_{03}})$

###### Proof.

: Using Proposition 3, the vector field $E$ is a perturbation to the standard
vector field $E_{0}$. If we let
$p=(x_{0}+\varepsilon_{1},y_{0}+\varepsilon_{2},z_{0}+\varepsilon_{3})$ and
$p_{0}=(x_{0},y_{0},z_{0})$ where $\varepsilon_{1},\varepsilon_{2}$ and
$\varepsilon_{3}$ are infinitesimals, then directly from the vector fields
$E_{1}$ and $E_{0}$, we conclude that $p\simeq p_{0}$ and $E_{1}\simeq E_{0}.$
∎

######  Remark 2.

In system (1), when $\delta$ is infinitesimal, then the Proposition 1 for
perturbation does not work because the difference of the two vector fields is
appreciable not infinitesimal.

###  6.2. Unperturbation Case

In this section we analyze the two independent first integrals in the
perspective view of nonstandard analysis.

######  Proposition 5.

If the conditions of Theorem 5.1 are satisfied, then the system (1) with
$\delta=1+\varepsilon$ has two independent first integrals of the form
$\psi_{1}={(1+\varepsilon)}^{-4}x^{3}y(1+O({(1+\varepsilon)}^{-1}x$,
${(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$ and
$\psi_{2}={(1+\varepsilon)}^{-4}yz^{3}(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
where $(x,y,z)\in\textbf{R}^{3}$ where parameters are standard, and the two
first integrals are limited or unlimited.

###### Proof.

The transformation of (32), reduces the system (36) with
$\delta=1+\varepsilon$ to

$\displaystyle{x^{\prime}_{1}}$ $\displaystyle=$ $\displaystyle
x_{1}(1+ax_{1}+by_{1}+cz_{1}),$ $\displaystyle{y^{\prime}_{1}}$
$\displaystyle=$ $\displaystyle y_{1}(-3+dx_{1}+ey_{1}+fz_{1}),$
$\displaystyle{z^{\prime}_{1}}$ $\displaystyle=$ $\displaystyle
z_{1}(1+gx_{1}+hy_{1}+kz_{1}).$

Using the arguments similar in Theorem 5.1, the system has two independent
first integrals of the form
$\phi_{1}=x_{1}^{3}y_{1}(1+O(x_{1},y_{1},z_{1}))=x_{1}^{3}y_{1}p_{1}$ and
$\phi_{2}=y_{1}z_{1}^{3}(1+O(x_{1},y_{1},z_{1}))=y_{1}z_{1}^{3}p_{2}$.
then the first integrals for the original system are

$\psi_{1}={(1+\varepsilon)}^{-4}x^{3}y(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))={(1+\varepsilon)}^{-4}x^{3}yp_{1}$

and

$\psi_{2}={(1+\varepsilon)}^{-4}yz^{3}(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))={(1+\varepsilon)}^{-4}yz^{3}p_{2}$

Since the variables are nonstandard and parameters are standard, then we have
different types of first integrals.

  1. I.

The two first integrals $\psi_{1},\psi_{2}$ are appreciable .
Case 1: If $x,y,z\in\mbox{gal}(0)$, then the two first integrals are obtained
directly from the substitution of $x,y,z$ in $\psi_{1}$ and $\psi_{2}$.

Case 2: If $y\in\mbox{m}(\omega)$, wher $\omega$ is unlimited and
$(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
contain $y$, then by assuming
$y(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))=y_{1}$
and since the product of two unlimited is unlimited therefore $y_{1}$ is
unlimited. Let $x=\frac{\alpha}{\sqrt[3]{y_{1}}}$ and
$z=\frac{\beta}{\sqrt[3]{y_{2}}}$ where $\alpha,\beta\in\mbox{gal}(0)$. Then
we deduce that the two first integrals are appreciable.

  2. II.

The two first integrals $\psi_{1},\psi_{2}$ are infinitesimal
Case 1: If $x,y,z\in\mbox{m}(0)$, then the infinitesimal of the two first
integrals are directly obtained by the substitution of variables in
$\psi_{1}$, $\psi_{2}$.

Case 2: If $y\in\mbox{m}(0)$ and $x,z\in\mbox{gal}(0)$, the multiplication of
$x^{3}$ and $(1+O({(1+\varepsilon)}^{-1}x,\\\
{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$ stay limited. When we
multiply the results by $y$, we see that the two first integrals are
infinitesimal.

Case 3: If $y\in\mbox{m}(\omega)$ and $(1+O(x,y,z))$ contain $y$, then suppose
$y(1+O({(1+\varepsilon)}^{-1}x,\\\
{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))=y_{1}$ and the product of
two unlimited is unlimited, hence $y_{1}$ is unlimited. If we take
$x=\frac{\alpha}{\sqrt[3]{y_{1}^{2}}}$ and
$z=\frac{\beta}{\sqrt[3]{y_{2}^{2}}}$, where $\alpha,\beta\in\mbox{gal}(0)$,
then the two first integrals are infinitesimal.

  3. III.

The two first integrals $\psi_{1},\psi_{2}$ are unlimited
If $y\in\mbox{m}(\omega)$ and
$(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
contain $y$, then suppose
$y(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))=y_{1}$
and the product of two unlimited is unlimited, therefore $y_{1}$ is unlimited.
If we take $x=\frac{\alpha}{\sqrt[6]{y_{1}}}$ and
$z=\frac{\beta}{\sqrt[6]{y_{2}}}$ where $\alpha,\beta\in\mbox{gal}(0)$, thus
the results stay unlimited.

∎

######  Proposition 6.

Let the system (1) with $\delta=1+\varepsilon$ and all parameters are standard
except $a$. If the conditions of Theorem 5.1 are satisfied, then it has two
independent first integrals of the form
$\psi_{1}={(1+\varepsilon)}^{-4}x^{3}y(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
and
$\psi_{2}={(1+\varepsilon)}^{-4}yz^{3}(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$,
which are either limited or unlimited.

###### Proof.

From the proof of the first part of Proposition 5 with $\delta=1+\varepsilon$,
the system has two independent first integrals of the form
$\phi_{1}=x_{1}^{3}y_{1}(1+O(x_{1},y_{1},z_{1}))$ and
$\phi_{2}=y_{1}z_{1}^{3}(1+O(x_{1},y_{1},z_{1}))$. Then the two first
integrals of system(1) are

$\psi_{1}={(1+\varepsilon)}^{-4}x^{3}y(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$

and

$\psi_{2}={(1+\varepsilon)}^{-4}yz^{3}(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z)).$

The condition will be on the parameter $a$ and other parameters are standard.
Let $G_{1}(G_{2})$ be the set of coefficients $x,y,z$ in
$O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z)$.
We have different types of first integrals which are

  1. I.

The two first integrals $\psi_{1},\psi_{2}$ are appreciable.
Case 1: If $x(z)$, $y$ and $a$ are appreciable, then we have two appreciable
first integrals directly from the substitution of variables in the first
integrals.
Case 2: If $x(z)$ and $y$ are appreciable, $a$ is unlimited and $a^{-1}\in
G_{1}(G_{2})$, then the term
$(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
stays appreciable and then is multiplied by $x^{3}y$. Therefore, the result is
unchanged.
Case 3: If $x(z)$ and $y$ are appreciable when $a\not\in G_{1}(G_{2})$, the
result stays appreciable.
Case 4: If $x(z)$ and $y$ are limited with $G_{1}(G_{2})=${0}, then obviously
the two first integrals are appreciable since variables are appreciable.

  2. II.

The two first integrals $\psi_{1},\psi_{2}$ are infinitesimal.
Case 1: If $x(z)$ and $y$ are infinitesimal, then the term
$O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z)$
is infinitesimal. Therefore,
$(1+O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
is appreciable. When we multiply this appreciable term to ${x}^{3}({z}^{3})y$
which is infinitesimal, then the result is infinitesimal.
Case 2: If $x(z)$ and $y$ are infinitesimal with $a\not\in G_{1}(G_{2})$, and
since the coefficients of
$O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z)$
does not contain $a$, then as the Case 1 (II) above it has the same
illustration.
Case 3: Let $x(z)$ and $y$ are infinitesimal or $x(z)$ is appreciable
(infinitesimal) and $y$ is infinitesimal (appreciable) with
$G_{1}(G_{2})=${0}. Since the coefficients of $O({(1+\varepsilon)}^{-1}x,\\\
{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$ are zero, then the two
first integrals are $\psi_{1}={(1+\varepsilon)}^{-4}{x}^{3}y$ and
$\psi_{2}={(1+\varepsilon)}^{-4}y{z}^{3}$. When the variables are
infinitesimal, it is also true for $\psi_{1}$ and $\psi_{2}$ or when $x(z)$ is
appreciable and $y$ is infinitesimal, the result stays infinitesimal.
Case 4: Let $x(z)$ and $y$ are infinitesimal or $x(z)$ is infinitesimal and
$y$ is appreciable with $G_{1}(G_{2})=${0}. Since the coefficients of
$O({(1+\varepsilon)}^{-1}x,{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$
are zero, then the two first integrals are
$\psi_{1}={(1+\varepsilon)}^{-4}{x}^{3}y$ and
$\psi_{2}={(1+\varepsilon)}^{-4}y{z}^{3}$. When the variables are
infinitesimal, it is also true for $\psi_{1}$ and $\psi_{2}$ or when $x(z)$ is
infinitesimal and $y$ is appreciable , the result stays infinitesimal.

  3. III.

The two first integrals $\psi_{1},\psi_{2}$ are unlimited.

Case 1: If $x(z)$, $y$ and $a$ are unlimited, then it clearly obtains two
unlimited first integrals.
Case 2: If $x(z),y$ are appreciable, $a$ is unlimited and $a\in G_{1}(G_{2})$,
then $O({(1+\varepsilon)}^{-1}x,\\\
{(1+\varepsilon)}^{-1}y,{(1+\varepsilon)}^{-1}z))$ is unlimited and when we
multiply it by any appreciable the result is unlimited.
Case 3: If $y$ is unlimited with $G_{1}(G_{2})=\\{0\\}$, then the two first
integrals become $\psi_{1}={(1+\varepsilon)}^{-4}{x}^{3}y$ and
$\psi_{2}={(1+\varepsilon)}^{-4}{yz}^{3}$, with
$x(z)=\frac{\alpha}{\sqrt[6]{y}}$ $(\frac{\beta}{\sqrt[6]{y}})$ where
$\alpha,\beta\in$ gal(0), and hence the two first integrals stay unlimited.

∎

######  Proposition 7.

Consider the system (1) with $\delta$ is infinitesimal. If the conditions of
Theorem 5.1 are satisfied, then the system (1) has two independent first
integrals of the form
$\psi_{1}={\delta}^{-4}x^{3}y(1+O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z))$
and
$\psi_{2}={\delta}^{-4}yz^{3}(1+O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z))$
which they are either limited or unlimited.

###### Proof.

In a similar way as in the proof of Proposition 5, we have the two independent
first integrals of the form
$\psi_{1}={\delta}^{-4}x^{3}y(1+O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z))$
and
$\psi_{2}={\delta}^{-4}yz^{3}(1+O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z))$
.
Now, if we suppose $x=\delta^{n}(\alpha_{1}+\delta)$,
$y=\delta^{n}(\alpha_{2}+\delta)$ and $z=\delta^{n}(\alpha_{3}+\delta)$ where
$\alpha_{1},\alpha_{2},\alpha_{3}$ are standard reals and $n$ is standard non-
negative integer, then we have the following cases:
Case 1: if $n=0$ implies $x,y,z$ $\in$ gal(0), then the two first integrals
$\psi_{1},\psi_{2}$ are unlimited
Case 2: If $n=1$ implies $x,y,z$ $\in$ m(0), then the two first integrals
$\psi_{1},\psi_{2}$ are appreciable.
Case 3: If $1<n$ implies $x,y,z$ $\in$ $\delta$-micromonad(0), then the two
first integrals $\psi_{1},\psi_{2}$ belong to $\delta$-micromonad(0).
∎

######  Proposition 8.

Suppose the system (1) with $\delta$ is infinitesimal and all parameters are
standard reals except $a$. If the conditions of Theorem 5.1 are satisfied then
it has two independent first integrals of the form
$\psi_{1}={\delta}^{-4}x^{3}y(1+O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z))={\delta}^{-4}x^{3}yp_{1}$
and
$\psi_{2}={\delta}^{-4}yz^{3}(1+O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z))={\delta}^{-4}yz^{3}p_{2}$
and the two first integrals are either limited or unlimited.

###### Proof.

Using the first part of the proof of Proposition 5, then the two first
integrals exist with $a$ either infinitesimal or limited, then the proof is
similar to Proposition 7. Now for the remaining cases with unlimited parameter
$a$, assume $G_{1}(G_{2})$ be the set of coefficients $x,y,z$ in
$O({\delta}^{-1}x,{\delta}^{-1}y,{\delta}^{-1}z)$, then we have different
types of first integrals.
Suppose $x=\delta^{n}(\alpha_{1}+\delta)$, $y=\delta^{n}(\alpha_{2}+\delta)$
and $z=\delta^{n}(\alpha_{3}+\delta)$ for $\alpha_{1},\alpha_{2},\alpha_{3}$
are standard reals and $n$ is standard non-negative integer, then we have the
following cases:
Case 1: If $a$ is unlimited and $a^{-1}\in G_{1}(G_{2})$, then

  1. i.

If $2\leq n$ implies $x,y,z\in\delta-\mbox{micromonad(0)}$, then the two first
integrals are belonging to $\delta-\mbox{micromonad(0)}$.

  2. ii.

If $n=1$ implies that $x,y,z\in\mbox{m}(0)$ and $p_{1}$ is appreciable, then
the two first integrals are appreciable.

Case 2: If $a$ is unlimited and $a\in G_{1}(G_{2})$. Let $a={\delta}^{-1}$, we
have the following cases.

  1. i.

If $1<n$ implies $x,y,z\in\delta-\mbox{micromonad(0)}$, then the results
depend on $p_{1}$ as follows.

    1. 1)

For $n=2$, then $p_{1}$ is appreciable and the two first integrals are
infinitesimal.

    2. 2)

For $2<n$, then $p_{1}$ $\in$ $\delta-\mbox{micromonad(0)}$ and the two first
integrals are belonging to $\delta-\mbox{micromonad(0)}$.

  2. ii.

If $n=1$ implies $x,y,z\in\mbox{m}(0)$ and $p_{1}$ is appreciable, then the
two first integrals are appreciable.

  3. iii.

If $n=0$ implies $x,y,z\in\mbox{gal(0)}$ and $p_{1}$ is unlimited, then the
two first integrals are unlimited.

Case 3: If $a$ is unlimited, $a\not\in G_{1}(G_{2})$, therefore the result of
this case is the same as that of Case 2 in proof of Proposition 7.

∎

## References

  * [1] W. Aziz, C. Christipher. _Local integrability and linearizability of three-dimensional Lotka-Volterra systems,_ Journal of Applied Mathematics and computation 219, (2012), 4067-4081.
  * [2] W. Aziz. _Integrability and linearizability problems of three dimensional Lotka-Volterra equations of rank-2_ , Qualitative Theory of Dynamical systems (2019).
  * [3] W.Aziz. _Analytic and algebraic aspects of integrability for first order partial differentail equations_ , Thesis (2012).
  * [4] W. Aziz, A. Amen and C. Pantazi. _Integrability and linearizability of a family of three-dimensional quadratic systems_. Dynamical Systems 36, Number 2, (2021), 317–331.
  * [5] W. Aziz. _Integrability and Linearizability of Three Dimensional Vector Fields_. Qualitative Theory of Dynamical Systems 12, Number 2, (2014), 197–213.
  * [6] S. Chow, C. Li and D. Wang. _Normal forms and bifurcation of planar vector fields_ , Cambridge University Press, 1994.
  * [7] C. Christopher, P. Mardešić and C. Rousseau. _Normalizable, Integrable, and Linearizable Saddle Points for Complex Quadratic Systems in $\mathbb{C}^{2}$_. Journal of Dynamical and Control Systems 9, no. 3 (2003): 311-363.
  * [8] C. Christopher and C. Rousseau. _Normalizable, integrable and linearizable saddle points in the Lotka-Volterra system_.Qualitative Theory of Dynamical System 5, (2004) 11–61.
  * [9] W. Decker, S. Laplagne, G. Pfister, H. A. Schonemann. _SINGULAR 3-1 library for computing the prime decomposition and radical of ideals_ , primdec. lib, 2010.
  * [10] W. Decker, G. Greuel, G. Pfister and H. Schönemann. __ Singular _4-2-1 — A computer algebra system for polynomial computations_.  http://www.singular.uni-kl.de (2021).
  * [11] F. Diener, M. Diener. _Nonstandard Analysis in practice_ , Springer-Verlag,Berlin,Heildberg, 1996.
  * [12] F. Diener and G. Reeb. _Analys non standard_ , Hermann, 1989.
  * [13] M. Dukarić, R. Oliveira, V. Romanovski. _Local integrability and linearizability of a $(1:-1:-1)$ resonant quadratic system_. J. Dyn. Differ. Equ. 29, (2017) 597–613.
  * [14] F. Dumortier, J. Llibre, and J. Artés. _Qualitative Theory of Planar Differential Systems_ , Springer-Verlag Berlin Heidelberg, 1977. 2256001
  * [15] P. Gao and Z.Liu. _An indirect method of finding integrals for three-dimensional homogeneous systems,_ Phys.Letters A 244(1998) 49–52.
  * [16] P. Gao. _Direct integration method and first integrals for three-dimensional Lotka-Volterra systems_ , Phys.Letters A 255(1999) 253–258.
  * [17] I. Hamad and M. Pirdawood. _Nonstandard Solutions for Ordinary Differential Equations near Singularity_ , $3^{rd}$ International Conference on Recent Trends in Multi-Disciplinary Research (ICRTMDR-20), 2020., (Maldives ).
  * [18] I. Hamad. _On Some Nonstandard Development of Intermediate Value Property_ , Journal of Mathematics and Computer Science, Vol 45(2), (2016) 35–45 (Egypt).
  * [19] I. Hamad. _Generalized Curvature and Torsion in Nonstandard Analysis_ ,thesis, LAP Lambert Academic Publishing , ISBN 978-3-8443-0763-4.
  * [20] Y. Ilyashenko, and S. Yakovenko. _Lectures on Analytic Differential Equations_ , Graduate Studies in Mathematics Vol. 86. American Mathematical Society, Providence, RI, 2008.
  * [21] K. Joshi. _Introduction to general topology_. New Age International, 1983.
  * [22] C. Laurant, J. Llibre. _Darboux integrability for 3D Lotka-Volterra systems,_ J. Phys. A: Math. Gen. 33 (2000) 2395–2406.
  * [23] C. Laurant. _Darboux first integral conditions and integrability of the 3D Lotka-Volterra systems,_ Journal of Nonlinear Math. phys. 7(4), (2000), 511–531.
  * [24] C.Liu, G.Chen and G. Chen. _integrability of Lotka–Volterra systems of degree 4_. J. Nonlinear Math. Phys. 7, (2000) 511–531.
  * [25] C. Lobry and T. Sari. _The peaking Phenomenon and singular Perturbations:an extension of on Tykhonov’s theorem_ ,Rapport de recherche INRIA,vol.4051, (2000) ,32 pages. Math.,vol.942(1982) 113–135.
  * [26] P. Loeb and M. Wolff. _Nonstandard analysis for the working mathematician_. Dordrecht, Boston: Kluwer Academic Publishers, 2000.
  * [27] R. Lutz and T. Sari. _Applications of Nonstandard analysis in boundary value problems in singular perturbation theorey_.Springer Lecture Notes in Math.,vol.942, (1982), 113–135.
  * [28] W. Mohammed, E. Aly, A. Matouk, S. Albosaily and E. Elabbasy. _An analytical study of the dynamic behavior of Lotka-Volterra based models of COVID-19_. Results in Physics  26, (2021), 104432.
  * [29] S. Pidurkar, B. Thakran, S. Raut, L. Thakre and N. Vaidya. _Analysis of Covid-19 in India using mathematical modeling_. Journal of Physics: Conference Series 1913, Number 1, (2021), 012110.
  * [30] C.Pugh. _Real mathematical analysis_ , Charles Chapman Pugh, Springer-Verlag,New York 2002.
  * [31] V. Romanovski, and D.Shafer. _The Center and Cyclicity Problems: A Computational Algebra Approach_ , Birkh$\ddot{u}$ser, Boston, Inc., Boston, MA, 2009.250020310.1007/978-0-8176-4727-8
  * [32] T. Sari. _Nonstandard perturbation theory of differential equations._ In Invited talk at the International research symposium on nonstandard analysis and its applications, ICMS, Edinburgh, pp. 11-17. 1996.
  * [33] N.Vakil. _Real analysis through modern infinitesimals_ , Cambridge University Press, 2011.
  * [34] S. Walcher. _Symmetries and convergence of normal form transformations_. Monografías de la Academia de Ciencias de Zaragona (Proceedings of the 6th Conference on Celestial Mechanics (Spanish) 25 (2004), 251–268.
  * [35] X. Zhang. _Analytic normalization of analytic integrable systems and the embedding flows_ , J. Differential Equations 244 (2008), No. 5, 1080–1092.