Abstract This paper deals with the integrability and linearizability problem of threedimensional systems
x = x(1 + ax + by + cz),
y = −y + dx2 + exy + f xz + gyz + hy2 + kz2,
z = z(1 + �x + my + pz).
More precisely, we give a complete set of necessary conditions for integrability andlinearizability and then prove their sufficiency using Darboux method and Darbouxinverse Jacobi multiplier, power series argument and a solution of a Riccati equation.
The problem of finding a complete set of first integrals for an ordinary differentialequations is one of the most classical problems in the theory of differential equations.This paper continuous the work started in [1], which studied the integrability andlinearizability problem for three dimensional Lotka–Volterra equations with (λ : μ :ν)-resonance at the origin of the form
x = x(λ+ax+by+cz), y = y(μ+dx+ey+ f z), z = z(ν+gx+hy+kz), (1)
W. Aziz (B)Department of Mathematics, College of Science, Salahaddin University, Arbil, Iraqe-mail: [email protected]
W. Aziz
More precisely, necessary and sufficient conditions for (1 : −1 : 1), (2 : −1 : 1) and(1 : −2 : 1)–resonance are given.
A natural way to generalize Lotka–Volterra equations in (1) is the following threedimensional system
x = P = x(λ+ ax + by + cz),
y = Q = μy + dx2 + exy + f xz + gyz + hy2 + kz2,
z = R = z(ν + �x + my + pz), (2)
where λ,μ, ν �= 0. This case is closer to the general three dimensional system but stilltractable. We are interested in working with eigenvalues in Siegel domain, that is theconvex hull of the eigenvalues contains the origin inside or on the boundary [10], andhave two independent (λ : μ) and (μ : ν)-resonances. This means that, without loss ofgenerality and a possible rescaling of time, we can take λ, μ and ν as coprime integernumbers with λ, ν > 0 and μ < 0. We say that the origin has (λ : μ : ν)-resonance.
Similar problems for two dimensional systems have been the focus of research formany years. Zoładek in [14], considered a complex polynomial vector fields with ap : −q resonant singularity and gave a set of sufficient conditions of the existence of afirst integral. Fronville et al. [8], investigated quadratic system with 1 : −2 resonanceand they found necessary and sufficient conditions to have a local analytic first integralat the origin. In [7], the authors considered complex quadratic vector fields having asaddle or saddle-node type of singularity with 1 : −λ ratio of eigenvalues. They usedthe Darboux method and other tools for showing normalizability, integrability andlinearizability. Investigations of three dimensional systems has been done by Basov andRomanovski [2], who study the existence of a first integral with one zero eigenvalue.Additional results for three dimensional Lotka–Volterra equations has been given in[4], where the authors studied the center variety of a singular point away from theaxes by different mechanisms and the number of limit cycles which can appear afterperturbation. One can consult [12] for a survey on results regarding to the existenceof a first integral of two-dimensional system and for polynomial vector fields in R
n orC
n .Our goal in this paper is, under the assumption of (1 : −1 : 1)-resonance, to provide
the integrability and linearizability conditions and prove their sufficiency through theDarboux method with inverse Jacobi multiplier and by power series arguments. Alsoshould mention that in some cases the solution of a Ricatti equation comes into playto linearize the third variable. The main result of the paper is Theorem 2, see Sect. 3.
2 Tools and Basic Definitions
In this section we recall some basic definitions and tools relating to integrability andlinearizability problem. The definitions can be find in [1,11–13].
We denote by X , the corresponding vector field of system (2), namely
X = P∂
∂x+ Q
∂
∂y+ R
∂
∂z.
Integrability and Linearizability of Three Dimensional Vector Fields
Definition 1 We say that a non-constant analytic function H = H(x, y, z) is a firstintegral of system (2) if it is constant on all solutions (trajectories). That means, thefunction H is a first integral of system (2) if and only if
X H = P∂H
∂x+ Q
∂H
∂y+ R
∂H
∂z= 0.
Definition 2 The algebraic surface F = 0 is an invariant algebraic surface of thepolynomial differential system if it satisfies
F = X F = P∂F
∂x+ Q
∂F
∂y+ R
∂F
∂z= C F, (3)
for some polynomial C ∈ C[x, y, z]. Such a polynomial C is called a cofactor of theinvariant algebraic surface F = 0. One can note that from Eq. (3) that any cofactor hasat most degree one since the polynomial vector field has degree two (independentlyof the degree of F).
We also use the exponential factor which plays the same role as the invariantalgebraic surface to obtain first integrals of the polynomial differential systems (2).
Definition 3 Let E(x, y, z) = exp( f (x, y, z)/g(x, y, z))where f, g ∈ C[x, y, z] andassume they are relatively prime. Then E in an exponential factor of (2) if X E = E L E
for some polynomial L E of degree at most one. The polynomial L E is called thecof actor of E .
Definition 4 A (multi-valued) function of the form
f λ11 · · · f
λpp
(exp
(g1h1))μ1 · · ·
(exp
(gqhq))μq
,
where λi , μ j ∈ C, fi are irreducible invariant algebraic curves and exp(
g jh j)
are expo-nential factors for i = 1, . . . , p and for j = 1, . . . , q, is said to be a Darboux function.
Definition 5 System (2) is integrable at the origin if and only if there is a change ofcoordinates
(X,Y, Z) = (x + O(x, y, z), y + O(x, y, z), z + O(x, y, z)
),
which transforms the system (2) into
X = Xζ, Y = −Y ζ, Z = Zζ. (4)
where ζ = ζ(X,Y, Z) = 1 + O(X,Y, Z).
Therefore first integrals of system (4) are
φ = X−μY λ and ψ = Y ν Z−μ, (5)
W. Aziz
which pull back to first integrals of system (2)
φ1 = x−μ(yλ + O(x, y, z)
), and φ2 = z−μ(
yν + O(x, y, z)).
Conversely, given two independent first integrals of the form φ1 and φ2, it is easilyseen that there exits a change of coordinates X , Y , Z so that these first integrals canbe written in the form (5) and the new system is of the form (4) for some ζ .
Definition 6 System (2) is linearizable at the origin if and only if the change ofcoordinates can be chosen to make ζ ≡ 1.
This can be shown to be equivalent to asking that all the coefficients in the normalform vanish.
Following [3], we generalize the concept of inverse integrating factor from two-dimensional systems to the concept of inverse Jacobi multipliers for three or higherdimensional systems.
Definition 7 A function M is an inverse Jacobi multiplier for the vector field X if itsatisfies the equation X (M) = MdivX ⇐⇒ div(X /M) = 0.
Recall that the system is completely integrable if there exists two independent firstintegrals [5,6,9]. In general, this is equivalent to finding one first integral together withan inverse Jacobi multiplier.
The following theorem generalizes Theorem 1 in [1] to our context. As the proofis very similar, we do not give all the details. In the following we use the notationX I = xi y j zk .
Theorem 1 Consider the analytic vector field
x
⎛⎝λ+
∑|I |>0
Ax I X I
⎞⎠ ∂
∂x+
⎛⎝μy+
∑|I |>1
AyI X I
⎞⎠ ∂
∂y+ z
⎛⎝ν+
∑|I |>0
AzI X I
⎞⎠ ∂
∂z,
(6)
with λ, ν > 0 andμ < 0. Suppose that it has a first integral = xαzγ (1+O(x, y, z))with at leastα or γ �= 0 and an inverse Jacobi multiplier I J M = xr zt (1+O(x, y, z)).Suppose the cross product of (r − i − 1,− j − 1, t − k − 1) and (α, 0, γ ) is boundedaway from zero for any integers i, k ≥ 0 and j ≥ −1, then the system has a secondanalytic first integral of the form � = x1−r y z1−t (1 + O(x, y, z)), and hence isintegrable.
Proof Without loss of generality, we can rewrite system (6) as:
x
⎛⎝λ+
∑|I ∗|>0
Ax I ∗ X I ∗⎞⎠ ∂
∂x+y
⎛⎝μ+
∑|I ∗|>0
AyI ∗ X I ∗⎞⎠ ∂
∂y+z
⎛⎝ν+
∑|I ∗|>0
AzI ∗ X I ∗⎞⎠ ∂
∂z,
where I ∗ = (i1, i2, i3) with i1, i2 + 1, i3 ≥ 0, Ax I ∗ = AzI ∗ = 0 for i2 = −1,Ay(1,−1,0) = Ay(0,−1,1) = Ay(0,−1,0) = 0 and (λ, μ, ν) = A(0,0,0). We now assume
Integrability and Linearizability of Three Dimensional Vector Fields
that α > 0. After an analytic change of coordinates of the form (x, y, z) → (x(1 +O(x, y, z)), y(1 + O(x, y, z)), z(1 + O(x, y, z))), which will not alter the form ofthe vector field or the inverse Jacobi multiplier, we can assume that φ = X δ is a firstintegral where δ = (α, β, γ ) and let the inverse Jacobi multiplier M to be X θ , whereθ = (r, s, t). We take AI ∗ = (Ax I ∗ , AyI ∗ , AzI ∗).
From the hypothesis, we can take K > 0 such that for all I ∗,
|(θ − I ∗ − 1) × δ| > K . (7)
Furthermore, since φ is a first integral, then Xφ = 0 gives for all I ∗
δ · A∗I = 0. (8)
Since M is a Jacobi multiplier, we have X (M) = div(X)M , and writing
div(X) = ∂P
∂x+ ∂Q
∂y+ ∂R
∂z= 1 · �+
∑I ∗(I ∗ + 1)Ax I ∗ X I ∗
,
where 1 = (1, 1, 1), we see that, for all I ∗,
(θ − I ∗ − 1) · AI ∗ = 0. (9)
By hypothesis, (θ− I ∗ −1) and δ are linearly independent, and so (8) and (9) implythat there exists some constants kI ∗ such that
AI = kI ∗ (θ − I ∗ − 1) × δ. (10)
For ease of calculation we work with the associated 2-form � = P dy ∧ dz +Q dz ∧ dx + R dx ∧ dy rather than X . In this case, a function φ is a first integral ifand only if dφ ∧� = 0. Now
�
M= P dy ∧ dz + Q dz ∧ dx + R dx ∧ dy
X θ
=∑I ∗
(kI ∗ (θ − I ∗ − 1) × δ
) ·(
dydz
yz,
dzdx
zx,
dxdy
xy
)X I ∗−θ+1
=∑I ∗
kI ∗((θ − I ∗ − 1) ·
(dx
x,
dy
y,
dz
z
))∧
(δ ·
(dx
x,
dy
y,
dz
z
))X I ∗−θ+1
=∑
I
kI
((θ − I ∗ − 1
) ·(
dx
x,
dy
y,
dz
z
)X I ∗−θ+1
)∧ dφ
φ
= d
(∑I ∗
kI ∗ X I ∗−θ+1
)∧ dφ
φ. (11)
W. Aziz
Thus, we have a formal first integral of the form
ψ =∑I ∗
kI ∗ X I ∗−θ+1.
From (7) and (10), we must have that |kI ∗ | < K |AI ∗ | for all I ∗, and so ψ is in factanalytic. �
3 Mechanism to Find Integrability and Linearizability Conditions
In this section, we briefly explain the mechanism to find integrability and linearizabilityconditions for system (2) with (λ : μ : ν) = (1 : −1 : 1).
Firstly we seek two analytic first integrals of the form
φ1 = x
(y +
∑i+ j+k>0
ai jk xi y j zk)
and φ2 = z
(y +
∑i+ j+k>0
bi jk xi y j zk).
Then compute resonant terms which are correspond to the obstacle to the existence ofφ1 and φ2. This will be done by calculating the successive terms in the power seriesexpansion, up to degree 12, of the supposed first integrals, that is Xφ1 = 0 and Xφ2 =0. Having calculated a number of these quantities, we then solve them simultaneouslyby computing a factorized Gröbner basis to obtain necessary integrability conditions.The calculations were performed in computer algebra system Maple and Reduce.Finally the minAssGTZ algorithm in Singular was used to check that the conditionsfound were irreducible. We need finally to prove sufficiency of these conditions byexhibiting two independent first integrals via the Darboux method together with inverseJacobi multipliers or some other techniques, for instance, if we can choose a coordinatesystem so that two of the variables decouple to give a linearizable node at the origin,then it just remains to find a linearizing transformation for the third variable via somesimple power series arguments or a solution of the Riccati equation.
For linearizability, we computed the conditions up to degree 12 for the existence ofa linearizing change of coordinates and then for sufficiency, we provided a linearizingchange of coordinates.
Theorem 2 Consider the system (2). The origin is integrable if and only if one of thefollowing conditions are satisfied:
1) a + � = b + m = c − 3p = d = e + � = g + p = h + m = k = 0
1∗) 3a − � = b + m = c + p = d = 3e + � = g − p = h − m = k = 0
2) a = b + m = c = d = e = g = k = � = p = 0
3) a = b + m = c = d = e = g = h − 2m = � = p = 0
3∗) a = b + m = c = e = g = h + 2m = k = � = p = 0
4) ab − eh = ac − 2ap + �p = ag + ap − ep − �p = ah + am − eh − h�
= be + b�− eh − em = bg − cm − gm + mp = bp − ch + hp − mp
Integrability and Linearizability of Three Dimensional Vector Fields
ce + c�− 2ep + g�− �p = d = f = gh − mp = k = 0
5) 4a − � = 2b + m = c = d = 4e + � = g = 2h − m = k = p = 0
5∗) a = b + 2m = c − 4p = d = e = g + p = h + m = k = � = 0
6) a − 2� = b − 2m = c = d = e − 2� = f = g = h − 2m = p = 0
6∗) a = 2b − m = 2c − p = e = f = g − p = h − m = k = � = 0
7) 5a − 2� = b + 2m = c = d = 5e + 2� = f = g = h − 2m = p = 0
7∗) a = 2b + m = 2c − 5p = e = f = g + p = h + m = k = � = 0
8) a = c = d = e = f = g = h − 2m = � = p = 0
8∗) a = 2b − h = c = e = f = g = k = � = p = 0
9) a = b − h + m = c = d = e = g = k = � = p = 0
10) a − e = b − h = d = g = � = m = 0
10∗) b = c = e = g − p = h − m = k = 0
11) 2a − � = b + h = c�− 2 f h = d = 2e + � = g = m = 0
11∗) b = c − 2p = e = f m + �p = g + p = h + m = k = 0
12) b = c = e = f = g = k = p = 0
12∗) a = d = e = f = g = � = m = 0
13) b = c = e = f = g = h − 2m = p = 0
13∗) a = 2b − h = e = f = g = � = m = 0
14) b = h = m = 0
15) b = e = g = m = 0
16) a = b − m = c = e = g = h − 2m = � = p = 0
17) a − � = b − m = c − p = e + 2� = g + 2p = 2h + m = 0
Moreover, the system is linearizable if and only if either one of the conditions (2), (3),(6), (8), (9), (10), (13), (16), (17) or one of the following holds:
1.1) a + � = b + m = c = d = e + � = g = h + m = k = p = 0
1.1∗) a = b + m = c + p = d = e = g − p = h − m = k = � = 0
4.1) a = c = d = e = f = g = k = � = p = 0
4.2) a − e = b − h = c = d = em − h� = f = g = k = 0
4.2∗) a = bp − cm = d = e = f = g − p = h − m = k = � = 0.
The following theorems and will be used in the proof of Theorem 2.
Lemma 1 Consider the system
X = λX, y = y(−μ+ A(X, Z)+ ε y), Z = νZ ,
W. Aziz
where λ,μ, ν ∈ Z+, ε is a parameter and A(X, Z) is a function of X and Z. The
following statement holds
(i) ε = 0. The system is linearizable if there exists a function ζ(X, Z) such thatζ (X, Z) = A(X, Z).
(ii) ε �= 0. The system is linearizable if there exists functions γ (X, Z) and ξ(X, Z)such that γ (0, 0) = 1 and satisfying
ξ − μξ = ε γ, γ = γ A. (12)
Proof We distinguish the following casesCase i: When α = 0, we only need to show that there exists a function ζ(X, Z)
such that ζ (X, Z) = A(X, Z). The transformation Y = y e−ζ will linearize the secondequation and gives Y = −μY .
To find such a function γ , we write A(X, Z) = ∑i+ j>0 ai j Xi Z j and ζ(X, Z) =∑
i+ j>0 bi j Xi Z j . Clearly we require bi j = ai jiλ+ jν which gives a convergent expres-
sion for ζ .
Case ii: When ε �= 0, we seek an invariant algebraic surface γ (X, Z)+ξ(X, Z)y =0 with cofactor A(X, Z) + εy. Hence the change of variable Y = y/(γ + ξ y) willlinearize the second equation and gives Y = −μY . To find such functions γ (X, Z)and ξ(X, Z), we need to solve the following differential equations
ξ − μξ = ε γ, γ = γ A.
To find γ , we write γ = eθ where θ = θ(X, Z) and solve θ = A. Letθ(X, Z) = ∑
i+ j>0 ci j Xi Z j and A(X, Z) = ∑i+ j>0 ai j Xi Z j . Then
∑i+ j>0(iλ+
jν)ci j Xi Z j = ∑i+ j>0 ai j Xi Z j . Hence clearly ci j = ai j
iλ+ jν for i + j > 0 and the
convergent of∑
i+ j>0 ai j Xi Z j guarantees the convergent of θ and hence γ .
Now writing γ = ∑i+ j>0 bi j Xi Z j , and we can find that ξ = ∑
i+ j>0ε bi j
iλ+ jν−μgives a convergent expression.
Lemma 2 Consider the system
X = λX, y = y(−μ+ A(X, Z))+ B(X, Z), Z = νZ ,
where λ,μ, ν ∈ Z+. The system is linearizable if there exist functions α and γ such
that α+γ B = −μα and γ +γ A = 0, where A(X, Z) and B(X, Z) are functionsof X and Z.
Proof The change of variable Y = α+γ y will gives Y = −μY . Then this differentialequation gives the following equations
α + γ B = −μα, γ + γ A = 0. (13)
To find γ , we write γ = e−ζ and solve ζ = A(X, Z). Writing ζ = ∑i+ j>0 ci j Xi Z j
and A = ∑i+ j>0 ai j Xi Z j , then
∑i+ j>0 (iλ+ jν)ci j Xi Z j = ∑
i+ j>0 ai j Xi Z j ,
Integrability and Linearizability of Three Dimensional Vector Fields
and hence ci j = ai jiλ+ jν for i + j > 0. The convergence of A(X, Z), guarantees the
convergence of ζ(X, Z) and hence γ . Again we write α(X, Z) = ∑i+ j>0 di j Xi Z j ,
γ (X, Z) B(X, Z) = ∑i+ j>0 bi j Xi Z j . From first equation in (13) we see di j =
− bi jiλ+ jν+μ for i + j > 0. Again the convergence of γ B, guarantees the convergence
of α.
Lemma 3 Consider the system
X = λX, y = −μy + A(X, Z)+ ε y2, Z = νZ ,
where λ,μ, ν ∈ Z+ and such that A(X, Z) is a function of X, Z and ε is a parameter
different from zero. The system is linearizable if the Riccati equation y = −μy +A(X, Z)+ε y2 has a particular solution y1(X, Z) and if there exists functions γ (X, Z)and ξ(X, Z) such that γ (0, 0) = 1 and satisfying
ξ − μξ = ε γ, γ = 2ε y1γ. (14)
Proof Assume that y1 is a particular solution of the Riccati equation then, y1 =−μy1 + A(X, Z)+ αy2
1 . A change of coordinates y = Y + y1 then gives
Y = Y (−μ+ 2ε y1 + ε Y ).
From Case (ii) of Theorem 1, implies that the system is linearizable since equationsin (14) are just equations in (12) required.
Proof of Theorem 2 Cases 1∗, 3∗, 5∗, 6∗, 7∗, 8∗, 10∗, 11∗, 12∗, 13∗, 1.1∗ and 4.2∗ aredual to Cases 1, 3, 5, 6, 7, 8, 10, 11, 12, 13, 1.1 and 4.2 under the transformation(x, y, z) → (z, y, x), and do not need to be considered separately. The other casesare considered below.
Case 1: The system has an invariant algebraic surface
� = 1+2ax−2by+ 2c
3z+a2x2+2abxy−
(2
3ac + b f
)xz+b2 y2− 2bc
3yz+ c2
9z2 = 0,
with cofactor C� = 2ax + 2by + 2c3 z giving a first integral φ = x−1z� and inverse
Jacobi multiplier I J M = x76 z− 1
6 �13 . Theorem 1 now guarantees the existence of
the second first integral ψ = x− 16 yz
76 (1 + · · · ). We can construct two independent
first integrals of the desired form as φ1 = φ− 76ψ = x(y + · · · ) and φ2 = φ− 1
6ψ =z(y + · · · ).
Case 1.1: The system has invariant algebraic surfaces �1 = 1 + 2ax − 2by +a2x2 + 2abxy + b f xz + b2 y2 = 0 and �2 = b2x + z + 2axz − 2byz + a2x2z +2abxyz + b f xz2 + b2 y2z = 0 with cofactors C�1 = 2ax + 2by and C�2 = 1 +ax + by. Therefore, a linearizing change of coordinates is given by (X,Y, Z) =(
x �− 1
21 ,�
− 12
1 �2, z �
121
).
W. Aziz
Case 2: System (2) has the form
x = x(1 + by), y = −y + f xz + hy2, z = z(1 − by). (15)
When h �= 0, the system has invariant algebraic surface � = 1−2hy+ f hxz+h2 y2 =0 with cofactor C� = 2hy and hence the transformation (X, Z) = (x �
− b2h , z �
b2h )
linearizes the first and third equations, that is X = X, Z = Z . One can note that thesecond equation
y = −y + f xz + hy2 = −y + f X Z + hy2,
is a Riccati equation. We seek a solution of the Riccati equation of the form y =G(X Z) = G(xz), then
G ′(X Z) = f
2− 1
2X ZG(X Z)+ h
2X ZG2(X Z),
or
G ′(xz) = f
2− 1
2xzG(xz)+ h
2xzG2(xz),
which has a particular solution y1 = G(xz) = sin(√
f hxz)−cos(√
f hxz)√
f hxz)h sin(
√f hxz)
. Note
that y1 is analytic with leading term f3 xz. From Theorem 3, we see that a change of
variables y = Y + y1 transforms the second equation in (15) to
Y = Y (−1 + 2hy1 + hY ).
Furthermore, there exists an invariant algebraic surface of the form α(X, Z) +β(X, Z)Y = 0 with cofactor 2hy1 + hY and α(0, 0) = 1, so that the transforma-
tion Y = Yα+βY will gives ˙Y = −Y . To find such α and β, we have to solve
β − β = α h, α = 2 h α y1. (16)
Write α = eψ and we have to solve ψ = 2 h y1. Supposeψ = ∑i+ j>0 ci j Xi Z j , then
∑i+ j>0
(i + j) ci j Xi Z j = 2 h y1(X, Z) = 2
3f h X Z +
∑i+ j>2
ai j Xi Z j .
Note that c10 = c01 = 0, c11 = 23 f h and ci j = ai j
i+ j for i + j > 2. Clearly
the convergence of∑
i+ j>2 ai j Xi Z j , guarantees the convergence of ψ and so α.
Moreover, in fact, α does not contain X and Z term. Now write β = ∑bi j Xi Z j , then
from first equation of (16) we see that b11 = 23 f h2 and bi j = h
i+ j−1 ai j for i + j > 2and the convergence is clear.
Integrability and Linearizability of Three Dimensional Vector Fields
When h = 0, the change of variables
(X,Y, Z) =(
xeb(y− f xz), y − f
3xz, ze−b(y− f xz)
),
gives X = X (1 − b f X Z), Y = −Y, Z = Z(1 + b f X Z). Clearly, the first andthird equations give a linearizable node. Hence there exists a change of coordinates
X = X (1 + o(X, Z)) and Z = Z(1 + o(X, Z)) such that ˙X = X , ˙Z = Z .Case 3: The system (2) reduces to
x = x(1 + by), y = −y + f xz − 2by2 + kz2, z = z(1 − by). (17)
Performing the change of variables (X,Y, Z) = (x2, x2 y, xz), the system (17)becomes
X = 2X + 2bY, Y = Y + f X Z + k Z2, Z = 2Z .
The critical point at the origin is then in Poincaré domain and is linearizable via ananalytic change of coordinates of the form (X , Y , Z) = (X + 2bY + O(2), Y +O(2), Z). The first integrals of the linear system are φ = X− 1
2 Y and ψ = X− 32 Y Z .
Pulling back first integrals to the original coordinates, we get first integrals of desiredform φ1 = x(y + · · · ) and φ2 = z(y + · · · ).
Case 4: If h �= 0, then the system has an invariant algebraic surface � = 1 +ax − hy + pz = 0 with cofactor C� = ax + hy + pz. It is not difficult to see that
φ = xz−1�
m−bh is a first integral and I J M = x �
2+ mh is an inverse Jacobi multiplier.
Now, Theorem 1 gives a second first integral ψ = y z(1 + · · · ). The desired firstintegrals are φ1 = φ ψ = x(y + · · · ), and φ2 = ψ = z(y + · · · ). When h = 0, wehave a = p = 0, bm �= 0 (clearly when b = 0 or m = 0, we are in Case 15) withbg − cm − gm = 0 and be + b� − em = 0. The system has an exponential factorE = exp(emx − bmy + bgz) with cofactor CE = emx + bmy + bgz which allows
us to construct two independent first integrals φ1 = x y E− 1m and φ2 = y z E− 1
b .Case 4.1: The transformation
(X,Y, Z) = (x(1 − hy)−
bh , y(1 − hy)−1, z(1 − hy)−
mh),
linearizes the equations. When h = 0, we replace (1 − hy)− bh and (1 − hy)− m
h byexp(−by) and exp(−my) respectively.
Case 4.2: When b �= 0, a linearizing change of coordinates is given by
(X,Y, Z) =( x
1 + ax − by,
x
1 + ax − by,
x
(1 + ax − by)mb
).
When b = 0, then we have two subcases:
W. Aziz
(i) If a = 0, m �= 0, then the change of coordinates
(X,Y, Z) = (x, y, z exp(−�x) exp(my)
)
linearizes the system.(ii) If m = 0, a �= 0, the transformation
(X,Y, Z) =(
x
1 + ax,
y
(1 + ax)ea,
z
(1 + ax)�a
),
linearizes the system.
When a = m = 0, we get sub cases which already have been taken into account.Case 4.3: In this case the change of coordinates
(X,Y, Z) =(
x
1 + ax − by + cz,
y
1 + ax − by + cz,
z
1 + ax − by + cz
),
linearizes the system.
Case 5: In this case, the system has invariant algebraic surface
� = (1 + ax + by)3 − 1
2ab f z(3ax + 3by + 2a2x2 + 6abxy − b f xz) = 0,
with cofactor 3ax − 3by. One can find a first integral φ = xz−1� and inverse Jacobi
multiplier I J M = z�13 and hence Theorem 1 guarantees the existence of second
first integral ψ = xy(1 + · · · ). The desired first integrals are φ1 = ψ = x(y + · · · )and ψ2 = φ−1ψ = z(y + · · · ). When a = 0, φ = xy − f
3 x2z and ψ = xz−1�−3
are first integrals where � = 1 + by − f b2 xz. Then the desired first integrals are
Case 6: A linearizing change of coordinates is given by
(X,Y, Z) = (x�
−1, (−3y + kz2)�−1, z�− 12)
where � = 1 + 2�x − 2my + kmz2.
Case 7: The system is
x = x
(1 + 2
5�x − 2my
), y = −y−2
5�xy+2my2+kz2, z = z(1+�x+my). (18)
The change of variables(w, Z
) = (x(1 + 2
5�x − 2my), xz2)
transforms the system(18) to
x = w, w =(
1 + 4
5�x
)w − 2mk Z , Z = 3Z
(1 + 4
5�x
). (19)
Integrability and Linearizability of Three Dimensional Vector Fields
Now we apply the transformation (X , W , Z) =(
xL ,
wL2 ,
ZL3
)where L = 1+ 4
5�x and
after rescaling the system by L we get
˙X = W
(1 − 4
5�X
),
˙W = W − 2mk Z − 8
5�W 2,
˙Z = 3Z
(1 − 4
5�W
). (20)
Clearly the second and third equation of (20) gives a linearizable node and hence thereexists a change of coordinates
(W , Z
) = (W + mk Z + O(W , Z), Z(1 + O(W , Z))
)
such that ˙W = W and ˙Z = 3Z . Then one first integral is given by φ = W 3 Z−1. Pullback to first integrals of the original system, we get φ = x2z−2
(1 + O(2)
). System
(18) has inverse Jacobi multiplier I J M = x− 23 z
53 and Theorem 1 therefore guarantees
the existence of a second first integral of the form ψ = x5/3 y z−2/3(1 + O(1)).From these two integrals it is easy to construct integrals of the form required asφ1 = φ−1/3ψ = x(y + · · · ) and φ2 = φ−5/6ψ = z(y + · · · ).
Case 8: The change of coordinates (X,Y, Z) = (x�
− b2m , (−3y+kz2)�−1, z�− 1
2)
linearizes the systems where � = 1 − 2my + kmz2. When m = 0, we are in Case13*, which is a dual with Case 13.
Case 9: The change of variables (X,Y, Z) = (x�
mh −1, (y − f
3 xz)�−1, z�− mh)
where � = 1−hy + f h2 xz, linearizes the system. When h = 0, the linearizing change
is (X,Y, Z) = (x E−m, y − f3 xz, z Em) where E = exp(y − f
2 xy).
Case 10: The transformation (X,Y ) =(
x1+ax−by+ζ z ,
y1+ax−by+ζ z
)where ζ =
12 (p + √
p2 − 4bk) yields a new system
X = X (1 + (c + ζ )z − η Xz),
Y = Y(
− 1 − ζ z − η Xz + bkz2
1 + ζ z
)+ f Xz + (1 − aX)
kz2
1 + ζ z,
z = z(1 + pz), (21)
where η = ac − b f − 2aζ . The first and third equation gives a linearizable nodeand then there exists a change of coordinates X = X(X, z) = x(1 + o(1)) and
Z = Z(z) = z(1 + o(1)) such that ˙X = X , ˙Z = Z . It remains to linearize thesecond equation in (21). Applying Theorem 2, we see that there exists the change
of variable of the form Y = A(X , Z)Y + B(X , Z) with ˙Y = −Y if the followingequation can be satisfied
A − A(ζ z + ηXz − bkz2
1 + ζ z
)= 0, B + A
(( f Xz + (1 − aX)
bkz2
1 + ζ z
)= −B.
It is easy to show that this can be done with no obstructions.
W. Aziz
Case 11: If b = 0, we obtain sub-cases of Case 15. If b �= 0, the system takes theform
x = x(1+ax +by+cz), y = −y−axy− ac
bxz−by2+kz2, z = z(1+2ax + pz).
which has invariant algebraic surfaces �1 = 1 + ax + by + 12 (p +√
p2 + 4kb)z = 0
and �2 = 1 + ax + by + 12 (p − √
p2 + 4kb)z = 0 with cofactors C�1 = ax − by +12 (p + √
p2 + 4kb)z and C�2 = ax − by + 12 (p − √
p2 + 4kb)z which producing afirst integral
φ = x z−1�
− 2c−p−√
p2+4bk
2√
p2+4bk1 �
2c−p+√
p2+4bk
2√
p2+4bk2
and inverse Jacobi multiplier I J M = z �
2c+p+√
p2+4bk
2√
p2+4bk1 �
−2c−p+√
p2+4bk
2√
p2+4bk2 . Now
Theorem 1 guarantees the existence of a second first integralψ = xy(1+· · · ). There-fore, the desired first integrals areφ1 = ψ = x(y+· · · ) andφ2 = φ−1ψ = z(y+· · · ).
When√
p2 + 4bk = 0 we replace �
− 2c−p−√
p2+4bk
2√
p2+4bk1 and �
2c+p+√
p2+4bk
2√
p2+4bk1 by (1+ax +
by + 12 p) as well as �
2c−p+√
p2+4bk
2√
p2+4bk2 by �
12 p−c3 and �
−2c−p+√
p2+4bk
2√
p2+4bk2 by �
12 p+c3 where
�3 = exp
(z
1+ax+by+ 12 pz
).
Case 12: When ah �= 0, the system has invariant algebraic surfaces �1 = 1+ax =0, �2 = 1+ 1
2 (a+√a2 − 4hd)x−hy = 0 and �3 = 1+ 1
2 (a−√a2 − 4hd)x−hy = 0
with cofactors C�1 = ax , C�2 = 12 (a + √
a2 − 4hd)x + hy and C�3 = 12 (a −√
a2 − 4hd)x +hy. A linearizing change of coordinates of the first and third equationis given by the transformation
(X, Z) =(
x �−11 , z �
m2h − l
a1 (�2 �3)
− m2h
).
Note that
y1 = (1 + ax)−√
a2−4hda
(1 + 1
2 (a + √a2 − 4hd)x
) − 1 − 12 (a − √
a2 − 4hd)x
h((1 + ax)−
√a2−4hd
a − 1)
= 1
3d x2(1 + · · · ),
is a particular solution of the equation y = −y + dx2 + hy2 (or the corresponding
Riccati equation dydx = − y
x(1+ax)+ d x1+ax + h y2
x(1+ax) ). The change of variable y = Y +y1
transform the Riccati equation to Y = Y (−1 + 2hy1 + hY ). Note that this equation isexactly as Case 2 and hence is linearizable using Theorem 3.
Integrability and Linearizability of Three Dimensional Vector Fields
When a = 0, h �= 0, the system has an exponential factor �1 = exp(x) andinvariant algebraic surfaces �2 = 1 + i
√hdx − hy = 0 and �3 = 1 − i
√hdx −
hy = 0 with cofactors C�1 = x , C�2 = i√
hdx + hy and C�3 = −i√
hdx + hy.A linearizing transformation of the first and third equation is given by (X, Z) =(x, z �
−�1 (�2 �3)
− m2h
). In this case the Riccati equation above has a particular solution
y1 = 1
h+ i
√dh
hx coth
(i√
dh x) = 1
3d x2(1 + · · · ),
and in the same way, the transformation y = Y + y1 linearizes the Riccati equation.When h = 0, a �= 0, E = exp(dx − ay) is an exponential factor with cofactordx + ay = 0. The first and third equations linearize by
(X, Z) =(
x �−11 , z �
mda2 − l
a
1 E− ma
).
In this case y1 = da3(−2−2ax)ln(1+ax)+2ax+a2x2
x = 13 d x2(1 + · · · ) is a particular
solution of the linear differential equation y = −y + dx2 (or the corresponding linearequation dy
dx = − yx(1+ax) + d x
1+ax ). The transformation y = Y + y1 linearizes theequation above. Finally when a = h = 0, then the linearizing change is given by
(X,Y, Z) = (x, 3y − dx2, z e−(�x−m y+ 12 m d x2)).
Case 13: The transformation (Y, Z) = (y �−1, z �
− 12 ) where � = 1 − 2my +
kmz2 + 12λx = 0, λ = a + √
a2 − 8md , converts the system to
x = x (1 + ax),
Y = −Y + d x2 (1 + 2mY − km Z2)
1 + λx+ k Z2 − λxY − km(1 − 2�)λxY Z2,
Z = Z(1 + �x − 1
2λx Z − 1
2mk(1 − 2�)λx Z2). (22)
Clearly, the first and third equations give a linearizable node and hence there is alinearizing transformation X = x(1 + O(x, z)) and Z = Z(1 + O(x, Z)) such that˙X = X , ˙Z = Z . To linearize the second equation from (22), as in Case 10, using
Theorem 2 there exists a transformation of the form Y = A(X , Z)Y + B(X , Z) such
that ˙Y = −Y if the following equation can be satisfied
x = x(1+ax +cz), y = −y+dx2 +exy+ f xz+gyz+kz2, z = z(1+�x + pz).
W. Aziz
The first and third equations gives a linearizable node, so there exists a change ofcoordinates X = x(1 + O(x, z)) and Z = z(1 + O(x, z)) such that X = X, Z = Z .
It is suffices to find Y = α + βy where α = α(X, Z) and β = β(X, Z) such thatY = −Y if the following equations can be satisfied
α + β(dx2 + f xz + kz2) = −α and β + β(ex + gz) = 0. (25)
By the same way as Case 10 and 13, apply Theorem 2, we can see that there are noobstructions to the existence of such α and β satisfying equations in (25).
Case 15: The reduces system is given by
x = x(1 + ax + cz), y = −y + dx2 + f xz + hy2 + kz2, z = z(1 + �x + pz).
The first and third equations gives a linearizable node. We denote the linearizingcoordinates by X and Z . In order to linearize the second equation, note that the secondequation is a Riccati equation and suppose that it has a solution of the form y = − U
hUwhere U = X ∂U
∂X + Z ∂U∂Z . This gives
U + U = −hU (dx2 + f xz + kz2). (26)
Write U , x and z as a power series U = ∑ai j Xi Z j , x = ∑
bi j Xi Z j , z =∑ci j Xi Z j . By Eq. (26), we have
∑(i + j +1)(i + j)ai j Xi Z j =−h
∑ai j Xi Z j
(d
( ∑bi j Xi Z j
)2
+f∑
bi j Xi Z j∑
ci j Xi Z j +k
( ∑ci j Xi Z j
)2).
We can now find each ai j uniquely by induction, and the convergence of U canbe obtained by standard majorization techniques. As in Case 2 and Case 13, thetransformation Y = y − U will linearize the second equation.
Case 16:This case has invariant algebraic surfaces �1 = 1−2by +bdx2 + f bxz +kbz2 = 0 and
�2 = 1−6by+3dbx2+3 f bxz+12b2 y2+3kbz2−12b2dx2 y−12b2 f xyz−8b3 y3
−12kb2 yz2+3b2d2x4+6 f db2x3z+3b3dx2 y2+(6b2dk + 3 f 2b2)x2z2
+3b3 f xy2z+6kb2 f xz3+3kb3 y2z2+3b2k2z4 =0,
with cofactors C�1 = 2by and C�2 = 6by. The change of coordinates (X,Y, Z) =(x�
− 12
1 , x�− 1
62 , z�
− 12
1 ) linearize the system.
Case 17: The system has an invariant algebraic surface
� = y − 1
3dx2 + axy − 1
3f xz + 1
2by2 + cyz − 1
3kz2 = 0,
Integrability and Linearizability of Three Dimensional Vector Fields
with cofactor C� = −(1 + ax + by + cz) which allows the contraction of twoindependent first integrals in the desired form φ1 = x � = x (y + · · · ) and φ2 =z � = z (y + · · · ).Acknowledgments I would like to thank Colin Christopher for helpful discussions during the preparationof this work.
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