# Journal of Nonlinear Mathematical Physics

Volume 26, Issue 2, March 2019, Pages 169 - 187

# Liouvillian integrability of a general Rayleigh-Duffing oscillator

Authors
Jaume Giné
Departament de Matemàtica, Inspires Research Centre, Universitat de Lleida Avda. Jaume II, 69; 25001 Lleida, Catalonia, Spain,gine@matematica.udl.cat
Claudia Valls
Departamento de Matemática, Instituto Superior Técnico, Av. Rovisco Pais 1049-001, Lisboa, Portugal,cvalls@math.ist.utl.pt
Received 4 May 2018, Accepted 15 October 2018, Available Online 6 January 2021.
DOI
10.1080/14029251.2019.1591710How to use a DOI?
Keywords
Liouvillian integrability; Rayleigh-Duffing oscillator; first integrals
Abstract

We give a complete description of the Darboux and Liouville integrability of a general Rayleigh-Duffing oscillator through the characterization of its polynomial first integrals, Darboux polynomials and exponential factors.

© 2019 The Authors. Published by Atlantis and Taylor & Francis
Open Access
This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).

## 1. Introduction and statement of the main results

We will consider the polynomial differential system

x˙=y,y˙=ax+by+cx3+dy3,(1.1)
where a, b, c, d ∈ ℝ are arbitrary parameters and c2 + d2 ≠ 0 (otherwise the system is linear). When c = d = −1 this system is the well-known Rayleigh-Duffing oscillator introduced by Rayleigh in [15] (in that case, x represents the displacement, a is related with the stiffness and b with the damping) and commonly studied from the dynamical point of view in the references [3, 7, 8]. Note that system (1.1) is invariant by the symmetry (x, y) ↦ (−x, −y).

For a deterministic system, the existence of constants or integrals of motion are responsible for the regular evolution of the phase-space trajectories of the system in well-defined regions of the phase space. For a planar differential system the existence of integrals of motion (called below first integrals) imply the regular evolution of the phase-space trajectories of the system in well-defined regions of the phase space and so the system can be integrated, at least qualitatively, since it can be determined theoretically its phase portrait, see [1]. However, for a differential system it is a difficult problem to determine the existence of integrals of motion (or first integrals). We recall that a function H(x, y) is a first integral of a planar differential if it is a C1 function defined on a full Lebesgue measure subset U ⊆ ℝ2 which is constant along each orbit in U of that system but it is not locally constant on any positive Lebesgue measure subset of U.

When the differential system is polynomial, we can use the Darboux theory of integrability because it provides conditions (that are sufficient) to obtain integrability inside the family of Liouvillian functions (i.e., functions obtained from complex rational functions by a finite process of integrations, exponentiations and algebraic operations). This will be our main tool in the paper. The Darboux theory of integrability can be applied to real or complex polynomial differential systems, however the study of complex algebraic solutions is necessary for obtaining the real first integrals of any real polynomial differential system. In section 2 we have included all the results related with the Darboux theory of integrability that are needed in the paper.

The main result in the work is the following:

### Theorem 1.1.

The following holds for system (1.1) with c2 + d2 ≠ 0:

1. (a)

If d = 0 and c ≠ 0 it is Liouville integrable if and only if either b = 0 or a = −2b2/9.

2. (b)

If d ≠ 0 and c = 0 it is Liouville integrable if and only if a = 0.

3. (c)

If cd ≠ 0 it is not Liouville integrable.

The proof of statements (a) and (b) of Theorem 1.1 is given in section 3 and the proof of Theorem 1.1 (c) is given in section 4. In section 2 we have included all the auxiliary results that will be needed to prove Theorem 1.1. The proof of statement (c) of Theorem 1.1 corrects statement (b) and some gaps in the proof of Theorem 1 in [6].

## 2. Preliminary results

Consider a polynomial differential system of degree d ∈ ℕ

x˙=P(x),x=(x1,x2)2,(2.1)
where P(x) = (P1(x), P2(x)), Pi ∈ ℂ[x], d = max{degP1, degP2} is the degree of system (2.1) and the dot denotes derivative with respect to the independent variable t.

Let 𝒳 be the vector field associated with system (2.1), i.e.,

𝒳=P(mathbfx)=(P1(x1,x2),P2(x1,x2)).

If H is a first integral, then

𝒳(H)=P1Hx1+P2Hx2=0.

A Darboux polynomial of system (2.1) is a polynomial f ∈ ℂ[x] such that

𝒳(f)=P1fx1+P1fx2=Kf,(2.2)
where x = (x1, x2), K ∈ ℂ[x] and has degree at most d − 1. The polynomial K is said to be the cofactor. An invariant algebraic curve is a curve given by f = 0 such that f ∈ ℂ[x].

An exponential factor of system (2.1) is a function E = exp(g/f), with f, g ∈ ℂ[x] being coprime, such that

𝒳(E)=P1Ex1+P2Ex2=LE,(2.3)
where L ∈ ℂ[x] and has degree at most d − 1. It is known that either f is constant, or f is a Darboux polynomial of system (2.1) and 𝒳(g) = Kg + Lf, where K is the cofactor of f.

An inverse integrating factor of system (2.1) is a function V such that

𝒳(V)=P1Vx1+P2Vx2=(P1x1+P2x2)V.

When V ∈ ℂ[x1, x2], it is a Darboux polynomial whose cofactor is the divergence of the system.

We use the following result in [5] for finding Liouville first integrals.

### Theorem 2.1.

If system (2.1) of degree d has p Darboux polynomials fi with cofactors Ki, i = 1,..., p, and q exponential factors Ej = exp(gj/hj) with Lj, j = 1,...,q, then there exist αi, βj ∈ ℂ not all zero such that

i=1pαiKi+j=1qβjLj=div(𝒳)
if and only if the function
f1α1fpαpE1β1Eqβq,(2.4)
(called Darboux function) is an inverse integrating factor of 𝒳. Here div(𝒳) stands for the divergence of the system.

To prove the results related with the Liouville integrability we will use the following result that it is proved in [14].

### Theorem 2.2.

System (2.1) has a Liouville first integral if and only if it has an integrating factor which is a Darboux function (see (2.4)).

Note that the previous theorem states that with the Darboux method one can find all Liouville first integrals.

Let J be the Jacobian matrix of 𝒳. The following proposition was proved in [12,13] we state it as we will use it. We recall that a polynomial first integral is a first integral that is a polynomial.

### Theorem 2.3.

Assume that the eigenvalues of the Jacobian matrix of 𝒳 at some singularity (x¯1,x¯2) satisfy k1λ1 + k2λ2 ≠ 0 for k1, k2 ∈ 𝕑+ and k1 + k2 ≥ 1. Then 𝒳 has no polynomial first integral.

The following theorem is due to Li et al. in [9, 10].

### Theorem 2.4.

Assume that the eigenvalues of the Jacobian matrix of 𝒳 at some singularity (x¯1,x¯2) satisfy λ1 = 0 and λ2 ≠ 0. Then 𝒳 can have a polynomial first integral if and only if (x¯1,x¯2) is non-isolated.

The singularities appearing in Theorems 2.3 and 2.4 can be real or complex, but system (2.1) is always real.

Let τ: ℂ2[x1, x2] → ℂ2[x1, x2] be the automorphism defined by

τ(x1,x2)=(x1,x2)(2.5)
and for a polynomial p(x1, x2) let τ*(p) = p(τ(x1, x2)).

### Proposition 2.1.

If g is an irreducible Darboux polynomial of degree d of system (2.1) with cofactor K(x1, x2), then f = g · τ*(g) is also a Darboux polynomial invariant by τ with a cofactor of the form K + τ*(K) = K(x1, x2) + K(−x1, −x2).

### Proof.

The proposition follows easily if we show that τ*(g) is a Darboux polynomial with cofactor τ*(K). To do it, note that applying τ* to (2.2) we get

τ*(𝒳(g))=τ*(Kg).(2.6)

Moreover, using that τ−1 = τ we obtain

τ*(𝒳(g))=𝒳(gτ1)=𝒳(gτ)=𝒳(τ*(g)).(2.7)

On the other hand,

τ*(Kg)=(Kg)τ1=(Kg)τ=(Kτ)(gτ)=τ*(K)τ*(g).(2.8)

Using (2.6), (2.7) and (2.8) we conclude that

𝒳(τ*(g))=τ*(K)τ*(g)
as we wanted to prove.

The following result was proved in [2] (see Theorems 13 and 14). We recall that two real numbers ω1, ω2 are said to be rationally independent if none of them can be written as a linear combination of the other with rational coefficients, or in other words, if the only integers k1, k2 such that k1ω1 + k2ω2 = 0 is k1 = k2 = 0.

### Proposition 2.2.

Let (x¯1,x¯2) be a singularity of system (2.1) and λ1, λ2 be the eigenvalues of the Jacobian matrix at (x¯1,x¯2). Assume that f(x1, x2) = 0 is an irreducible invariant algebraic curve with cofactor K(x1, x2). If f(x¯1,x¯2)0 then K(x¯1,x¯2)=0. Moreover if f(x¯1,x¯2)=0, λ2 ≠ 0 and either λ1 and λ2 are rationally independent, or λ1λ2 < 0, then K(x¯1,x¯2){λ1,λ2,λ1+λ2}.

A polynomial f(x1, x2) is said to be a weight homogeneous polynomial if there exist s = (s1, s2) ∈ ℕ2 and m ∈ ℕ such that for all α ∈ ℝ \ {0},

f(αs1x1,αs2x2)=αmf(x1,x2),
where ℝ denotes the set of real numbers, and ℕ the set of positive integers. We shall refer to s = (s1, s2) as the weight of f, m as the weight degree, and x = (x1, x2) ↦(αs1x1, αs2x2) as the weight change of variables.

## 3. Proof of statements (a) and (b) of Theorem 1.1

Statement (a) of Theorem 1.1 is a consequence of Theorem 1.7 in [4]. In [4] are used the Puiseux series (see [4] for its definition) to find the Liouville integrability. In fact we have checked her results using the methods described in the present work and we have found the same results for the case d = 0 and c ≠ 0. Moreover the case d = b = 0 with c ≠ 0 corresponds to the Hamiltonian system x˙=y, y˙=ax+cx3 which has the polynomial first integral H = −2ax2cx4 + 2y2 and the case d = 0, a = −2b2/9 with c ≠ 0 is Liouville integrable because it has the inverse integrating factor

V=(2b2x29cx412bxy+18y2)3/4.

Now we prove Theorem 1.1 (b). First note that if a = 0 then system (1.1) becomes

x˙=y,y˙=bydy3.

This system has the inverse integrating factor V = by + dy3 and in view of Theorems 2.1 and 2.2 it is Liouvillian integrable.

Now we consider a ≠ 0. Statement (b) of Theorem 1.1 will now be an immediate consequence of the following proposition.

### Proposition 3.1.

System (1.1) with c = 0 and a ≠ 0 is not Liouvillian integrable.

### Proof.

Let c = 0 and a ≠ 0. We recall that since c = 0 then d ≠ 0. With the change of variables

x1dBx1,yBy1,sB2dt,B={(a/d2)1/4,a>0,(a/d2)1/4,a<0
we get
x1=y1,y1=±x1+by1+y13,(3.1)
where b = b/B and the prime denotes derivative in s. Let τ be the automorphism defined by (2.5) and g be a Darboux polynomial with cofactor K=a00+a10x1+a01y1+a20x12+a11x1y1+a02y12 with aij ∈ ℂ for i, j ∈ {0, 1, 2}. In view of Proposition 2.1 we have that f = g · τ(g) is a Darboux polynomial invariant by τ and with cofactor K=2a00+2a20x12+2a11x1y1+2a02y12. Note that f(x1, y1) = f (−x1, −y1).

To prove Proposition 3.1 we state and prove the following auxiliary lemma.

### Lemma 3.1.

System (3.1) has no Darboux polynomials.

### Proof of Lemma 9.

First, we introduce the weight-change of variables of the form

x1=α3X,y1=α1Y,s=α2r.

This type of weight-change of variables to find invariant Darboux polynomial were first introduced in [11]. In this form, system (3.1) becomes

X=α4Y,Y=±X+bα2Y+Y3,(3.2)
where now the prime denotes derivative in the new time r. Now we set
F(X,Y)=αf(α3X,α1Y)
and
K=2a00+2a20α4X2+2a11α3XY+2a02α2Y2
where is the highest weight degree in the weight homogeneous components of f in X, Y with weight (3, 1).

We note that F = 0 is an invariant algebraic curve of system (3.2) with cofactor K. Indeed

dFdr=α+2dfds=α+2(2a00+2a20α4X2+2a11α3XY+2a02α2Y2)f=(2a02Y2+2a11α1XY+2a20α2X2+2α2a00)F=KF.

If F=i=0nFi where Fi is a weight homogeneous polynomial in X, Y with weight degree i for i = 0,...,n and n, we get

f=F|α=1.

From the definition of invariant algebraic curve we have

α4Yi=0nαiFiX+(±X+α2bY+Y3)i=0nαiFiY=(2a02Y2+2a11α1XY+2a20α2X2+2α2a00)i=0nαiFi.(3.3)

Equating the terms with α−2 in (3.3) we get a20 = 0 and the terms with α−1 we get a11 = 0. Moreover, equating the terms with α0 in (3.3) we obtain

(±X+Y3)F0Y=2a02Y2F0.

Since F0 ≠ 0 (otherwise f would be constant) we must have

F0=β0Xnl(±X+Y3)l,a02=3l2,
where β0 ≠ 0 (otherwise f would be constant which is not possible). Note that = 3n.

Equating now the terms in (3.3) with α1 we obtain

(±X+Y3)F1Y=3lY2F1.

Solving it, we get

F1=F˜1(X)(±X+Y3)l,
where F˜1 is a function in X. Since F1 must be a polynomial of degree − 1 = 3n − 1 and X has weight-degree 3, we must have F˜1=0 and so F1 = 0.

Equating now the terms in (3.3) with α2 we get

(±X+Y3)F2Y+bYF0Y=3lY2F2+2a00F0.

Introducing F0 we conclude that

F2=F˜2(X)(±X+Y3)l+β0Xnl(±X+Y3)l(bl±X+Y3(2a00bl)3X2/3arctan(X1/32Y3X1/3)+2a00bl3X2/3log((X1/3±Y)2X2/3X1/3Y+Y2)).

Since F2 must be a polynomial with degree − 2 = 3n − 2 and X has weight-degree 3, we must have F˜2=0 and a00 = bl/2 (because β0 ≠ 0). Hence,

fn2=bβ0Xnly(±X+Y3)l1.

Computing the terms in (3.3) with α3 we get

(±X+Y3)F3Y=3lY2F3.

Solving this linear differential equation using that F3 has degree − 3 = 3n − 3 we get

F3=β3Xnl1(±X+Y3)l.

Computing the terms in (3.3) with α4 we obtain

(±X+Y3)F4Y+bYF2Y+YF0X=3lY2F4+bl2F2.

Using F2 and F0 and solving this linear differential equation we obtain

F4=F˜4(X)(±X+Y3)l+β018Xnl(±X+Y3)l(54b2(1+l)lXY2(±X+Y3)26l(1+6b2l)Y2±X+Y3+23(6b2l2+3nl(2+3bX2/3))X1/3arctan(X1/32Y3X1/3)+6b2l2+3nl(2+3bX2/3)X1/3log((X1/3±y)2X2/3X1/3Y+Y2)).

Since F4 must be a polynomial of degree − 4 = 3n − 4 we must have F˜4=0 and 6b2l2 + 3nl(2 + 3bX2/3) = 0. Since b ≠ 0 we get l = n = 0 which is not possible. Hence, there are no Darboux polynomials invariant by τ and consequently there are no Darboux polynomials.

Now we continue with the proof of Proposition 3.1. Since by Lemma 3.1 system (1.1) with c = 0 and a ≠ 0 has no Darboux polynomials, the unique exponential factor of that system may have is of the form E = exp(g) where g ∈ ℂ[x, y] and with cofactor L ∈ ℂ[x, y] of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 with bij ∈ ℂ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

ygxE+(ax+by+dy3)gyE=LE,
and after simplifying by E = exp(g), we conclude that the polynomial g must satisfy
ygx+(ax+by+dy3)gy=L.

The divergence of system (1.1) with c = 0 and a ≠ 0 is b + 3dy2 and in view of Theorem 2.1 we must have L = b + 3dy2. So, g must satisfy

ygx+(ax+by+dy3)gy=L=b+3dy2.(3.4)

Evaluating it on x = y = 0 we get b = 0. Now we expand g as a polynomial in its homogeneous parts as

g(x,y)=j=0ngj(x,y),
where gj are homogeneous polynomials in (x, y), n ≥ 1 and gn ≠ 0. The homogeneous part gn for n ≥ 3 satisfies
dy3gny=0andsogn=Bnxn,
where Bn ∈ ℂ. The homogeneous part gn−1 must satisfy
dy3gn1y=0andsogn1=Bn1xn1,
where Bn−1 ∈ ℂ. Furthermore, the homogeneous part gn−2 must satisfy
dy3gn2y==Bnnxn1y
and so
gn2=Bn2xn2Bnndyxn1,
which yields Bn = 0. Hence gn = 0. This means that g ∈ ℂ[x, y] has degree at most two. We write it as
g(x,y)=λ00+λ10x+λ01y+λ20x2+λ11xy+λ02y2,
where λij ∈ ℂ for i, j ∈ {0, 1, 2}. Solving (3.4) with such g we get a contradiction. So, in view of Theorems 2.1 and 2.2 it is not Liouvillian integrable. This completes the proof of Proposition 3.1.

## 4. Proof of Theorem 1.1 (c)

First note that if dc ≠ 0 by the change of coordinates and reparametrization of time of the form

x1C1/6d1/3X,yC1/6d2/3Y,τC2/6d1/3t,C={c,c>0,c,c<0,
we can write system (1.1) as
X˙=Y,Y˙=a˜X+b˜Y+δX3+Y3,δ{1;1},(4.1)
where ã = ad2/3/c2/3 and b˜=bd1/3/c1/3. For convenience in the notation of the paper we will rename system (4.1) as
x˙=y,y˙=ax+by+δx3+y3,δ{1;1}.(4.2)

The proof of statement (c) of Theorem 1.1 will be a consequence of different results that we will state and prove below.

### Lemma 4.1.

The unique irreducible Darboux polynomials of system (4.2) of degrees one or two are:

1. (i)

f(x, y) = δx + y with cofactor K(x, y) = b0 + x2δxy + y2 if a = δb0 and b = b0δ with b0 ∈ ℝ;

2. (ii)

f(x, y) = 1 + δx2 + xy with cofactor K(x, y) = x2δxy + y2 if a = 1 and b = −3δ;

3. (iii)

f(x, y) = −δx2 + xyδy2 with cofactor K(x, y) = δx2 + δxy + 2y2 if a = −1 and b = δ.

Proof of Lemma 4.1. It follows by direct computations.

### Proposition 4.1.

The unique irreducible Darboux polynomials of system (4.2) with degree n ≥ 1 are the ones given in the statement of Lemma 4.1.

### Proof.

Let τ be the automorphism defined by (2.5) and g be a Darboux polynomial with cofactor K = a00 + a10x + a01y + a20x2 + a11xy + a02y2 with aij ∈ ℂ for i, j ∈ {0, 1, 2}. In view of Proposition 2.1 we have that f = g · τ(g) is a Darboux polynomial invariant by τ with cofactor K = 2a00 + 2a20x2 + 2a11xy + 2a02y2. Note that f(x, y) = f (−x, −y) and so f has no odd degree terms.

First we want to study the Darboux polynomials of system (4.2) invariant by τ and after that we will obtain all the Darboux polynomials. Note that if for some values of the parameters there are no Darboux polynomials invariant by τ then by Proposition 2.1 there are also no Darboux polynomials which may not be invariant by τ. Furthermore, if for some values of the parameters there is a unique irreducible Darboux polynomial invariant by τ then there are no irreducible Darboux polynomials not invariant by τ since otherwise in view of Proposition 2.1 the Darboux polynomial not invariant by τ would be a factor of the Darboux polynomial invariant by τ. But this is not possible since the latter is irreducible. We will see below that the unique irreducible Darboux polynomials invariant by τ are the ones given in the statement of Lemma 4.1.

By the definition of Darboux polynomial invariant by τ we have that f and K satisfy

yfx+(ax+by+δx3+y3)fy=(a00+a20x2+a11xy+a02y2)f.(4.3)

First we expand f as a polynomial in the variable y as

f=j=0mfj(x)yj.

Computing the terms with ym+2 in (4.3) we get

mfm(x)=a02fm(x)thatisfm(x)(a02m)=0,
which yields a02 = m, that is, a02 is an integer which is the greatest power of f in the variable y.

Now we expand f as a polynomial in its homogeneous parts as

f(x,y)=j=0mfj(x,y),
where fj are homogeneous polynomials in (x, y), n ≥ 1 and fn ≠ 0. It follows from (4.3) that the homogeneous part fn satisfies
(δx3+y3)fny=(a20x2+a11xy+a02y2)fn.(4.4)

We will work with δ = 1. The case δ = −1 can be done in a similar manner.

Solving (4.4) with δ = 1 we get

fn=Gn(x)(x+y)(a11a20a02)/3(x2xy+y2)(a20a112a02)/6(3xi(2yx)3x+i(2yx))3(a20+a11)i/6,
where Gn is any function in the variable x. Note that since fn must be a polynomial and a02 is an integer, we must have a20 + a11 = 0, i.e., a11 = −a20 but then we get
fn=Gn(x)(x+y)(2a20+a02)/3(x2xy+y2)(a02a20)/3.

Again, since fn is a homogeneous polynomial of degree n we have a20 = l1l2 and a02 = 2l2 + l1 for some l1, l2 ∈ ℕ (here ℕ denotes the set of nonnegative integers) and

fn=Anxnl12l2(x+y)l1(x2xy+y2)l2,
for some constant An ≠ 0. Without loss of generality we can take An = 1. Since fn(x, y) = τ(fn(x, y)) = fn(−x, −y) we must have n even.

The homogeneous part fn−1 must have degree n − 1 which is odd and so fn−1 = 0 (because f must be invariant by τ, see the observation at the beginning of the proof of Proposition 4.1). The homogeneous part fn−2 must have degree n − 2 and satisfies

(x3+y3)fn2y+hn2=(a20x2+a11xy+a02y2)fn2,
where
hn2=yfnx(ax+2by)fnya00fn.

Solving it we get

fn2=(x+y)l11(x2xy+y2)l21[Gn2(x)(x+y)(x2xy+y2)rn2(x,y)s1xnl12l22(x+y)(x2xy+y2)arctan(2yx3x)s2xnl12l22(x+y)(x2xy+y2)log(x+yx2xy+y2)],
where Gn−2 is any function in the variable x, rn−2(x, y) is a polynomial in the variables x, y of degree nl1 − 2l2 + 1 and
s1=3a00+(1+a4b)l1+(5a2b)l23n,s2=a00+(a1)l1(1+a+2b)l2+n.

Since fn−2 must be a homogeneous polynomial of degree n−2 and An ≠ 0 we must have Gn−2(x) = An−2xnl1−2l2−2 for some constant An−2 and s1 = s2 = 0. Instead of solving s1 = s2 = 0 we only save the values of s1, s2 and we will solve it later and consider the solution

fn2=(x+y)l11(x2xy+y2)l21(rn2(x,y)+An2xnl12l22(x+y)(x2xy+y2)).

The homogeneous part fn−3 must have degree n − 3 which is odd and so fn−3 = 0 (because f must be invariant by τ and fn−1 = 0). The homogeneous part fn−4 must have degree n − 4 and satisfies

(x3+y3)fn4y+hn4=(a20x2+a11xy+a02y2)fn4,
where
hn4=yfn2x(ax+2by)fn2ya00fn2.

Solving it we get

fn4=(x+y)l12(x2xy+y2)l22[Gn4(x)(x+y)2(x2xy+y2)2rn4(x,y)s3xnl12l24(x+y)2(x2xy+y2)2arctan(2yx3x)s4xnl12l24(x+y)2(x2xy+y2)2log(x+yx2xy+y2)],
where Gn−4 is any function in the variable x, rn−4(x, y) is a polynomial in the variables x, y of degree nl1 − 2l2 + 2 and s3, s4 are very big and we do not write them here. Since fn−4 must be a homogeneous polynomial of degree n − 4 we must have s3 = s4 = 0 and then we write it as
fn4=(x+y)l12(x2xy+y2)l22(rn4(x,y)+An4xnl12l24(x+y)2+(x2xy+y2)2),
for some constant An−4. The homogeneous part fn−5 must have degree n − 5 which is odd and so fn−5 = 0 (because f must be invariant by τ and fn−3 = 0). The homogeneous part fn−6 must have degree n − 6 and satisfies
(x3+y3)fn6y+hn6=(a20x2+a11xy+a02y2)fn6,
where
hn6=yfn4x(ax+2by)fn4ya00fn4.

Solving it we get

fn6=(x+y)l13(x2xy+y2)l23[Gn6(x)(x+y)3(x2xy+y2)3rn6(x,y)s5xnl12l26(x+y)3(x2xy+y2)2arctan(2yx3x)s6xnl12l26(x+y)3(x2xy+y2)3log(x+yx2xy+y2)],
where Gn−6 is any function in the variable x, rn−6(x, y) is a polynomial in the variables x, y of degree nl1 − 2l2 + 3 and s5, s6 are very big and we do not write them here. Since fn−6 must be a homogeneous polynomial of degree n − 6 we must have s5 = s6 = 0. Now we solve the system
s1=s2=s3=s4=s5=s6=0
with n ≥ 1 (an integer) and l12+l220 (the case l1 = l2 = 0 yields n = 0 which is not possible) yield the solutions:
1. (i)

a = −1, b = 1, l1 = 0, n = 2l2 and a00 = l2;

2. (ii)

a = 5/2, b = −9/2, l2 = 0, n = 3l1 and a00 = l1/2;

3. (iii)

a = 1, b = −3, l2 = 0, n = 2l1 and a00 = 0;

4. (iv)

b = a − 1, l2 = 0, n = l1 and a00 = al1.

Recall that we are only interested in solutions with a and b real. Note that the case in which the cofactor is zero we have, in particular, that l1 = l2 = 0 and so it is not possible because this implies n = 0. Now we study each of the cases (i)–(iv) separately in different lemmas.

### Lemma 4.2.

The unique irreducible Darboux polynomial satisfying (i) is the one given in Lemma 4.1 (iii).

### Proof of Lemma 4.2.

The origin is always a singular point of system (4.2) whose Jacobian matrix at that point has eigenvalues (b±b2+4a)/2. Under the assumptions a = −1, b = 1 the origin is a focus and the eigenvalues of the Jacobian matrix at that point are (1±3i)/2. Note that they are rationally independent. Moreover, the cofactor of the irreducible Darboux polynomial (if it exists) is of the form

K=K(x,y)=l2(1x2+xy+2y2).

In particular, K(0, 0) = l2 ≠ 0 because n = 2l2 ≠ 0. So, by Proposition 2.2 we must have

l2{1,1+3i2,13i2}.

Taking into account that l2 is an integer we conclude that l2 = 1 and then n = 2. We thus conclude that the unique Darboux polynomial invariant by τ satisfying (i) must have degree two and so it must be the one in Lemma 4.1 (iii). By the observation at the beginning of the proof of Proposition 4.1 we conclude that if there are irreducible Darboux polynomials that are not invariant by τ must be divisors of the Darboux polynomial −δx2 + xyδy2 but this is not possible because they would have degree one and there are none for these values of the parameters. Hence, the only irreducible Darboux polynomial must be invariant by τ and so is the one given in Lemma 4.1 (iii).

### Lemma 4.3.

There are no irreducible Darboux polynomials satisfying (ii).

### Proof of Lemma 4.3.

The origin is always a singular point of system (4.2) whose Jacobian matrix at that point has eigenvalues (b±b2+4a)/2. Under the assumptions a = 5/2, b = −9/2 the origin is a saddle and the eigenvalues of the Jacobian matrix at that point are −5, 1/2. Moreover, the cofactor of the irreducible Darboux polynomial (if it exists) is of the form

K=K(x,y)=l1(12+x2xy+y2).

In particular, K(0, 0) = l1/2 ≠ 0 because n = 3l1 ≠ 0. So, by Proposition 2.2 we must have

l12{12,5,92}.

Taking into account that l1 is a positive integer we conclude that l1 = 1 and then n = 3. We thus conclude that the unique irreducible Darboux polynomial satisfying (iii) must have degree three. Computing it we see that it does not exist. Hence, there are no Darboux polynomials invariant by τ. By the observation at the beginning of the proof of Proposition 4.1 we conclude that in fact there are no Darboux polynomials in this case.

### Lemma 4.4.

The unique irreducible Darboux polynomial satisfying (iii) is the one given in Lemma 4.1 (ii).

### Proof of Lemma 4.4.

We proceed by contradiction. If we denote by ϕ(x, y) the irreducible Darboux polynomial of Lemma 4.1 (ii), that is, ϕ(x, y) = 1−x2 +xy and set Φ(x, y) = ϕl1 then Φ is a Darboux polynomial of system (4.2) with cofactor K = l1(x2xy + y2). Moreover Φ has degree n = 2l1 and if we expand Φ in its homogeneous terms we get

Φ(x,y)=xl1(x+y)l1+j=02l12Φj(x,y)
where each Φj is a homogeneous polynomial of degree j (we recall that Φ cannot have odd degree terms). Now define T = Φ(x, y) − f(x, y) where f is the Darboux polynomial satisfying (iii) with degree 2l1. Clearly T is a Darboux polynomial of system (4.2) with cofactor K = l1(x2xy + y2). If we expand T in its homogeneous terms we get that T2l1 = 0 and
T=j=02l12Tj(x,y),
where again each Tj is a homogeneous polynomial of degree j. Using that T is a Darboux polynomial we get that
T2l12=A2l12xl12(x+y)l1,A2l12.

Note that the terms T2l1−4 satisfy

(x3+y3)T2l14y+yT2l12x+xT2l12y3yT2l12y=l1(x2xy+y2)T2l14.

Solving it we get

T2l14=A2l14x2l13(x+y)l1A2l12xl14(x+y)l11[l1+33(x+y)arctan(2yx3x)+(x+y)log(x2xy+y2(x+y2))].

Since T2l1−4 must be a polynomial we get A2l1−2 = 0 and so T2l1−2 = 0. Now proceeding inductively we get T2l1−2j = 0 for j = 0,...,l1 which yields T = 0. In short, f(x, y) = Φ(x, y) = ϕl1. Note that ϕ is irreducible. Hence, since f is also irreducible we must have l1 = 1 and then n = 2. We thus conclude that the unique irreducible Darboux polynomial invariant by τ satisfying (iii) must have degree two and so it must be the one in Lemma 4.1 (i). By the observation at the beginning of the proof of Proposition 4.1 we conclude that the only irreducible Darboux polynomial must be invariant by τ and so is the one given in Lemma 4.1 (ii).

### Lemma 4.5.

The unique irreducible Darboux polynomial satisfying (iv) is the one given in Lemma 4.1 (i).

### Proof of Lemma 4.5.

The proof is similar to the proof of Lemma 4.4. We also proceed by contradiction. Let Φ(x, y) = (x + y)l1. Note that Φ is a Darboux polynomial of system (4.2) with cofactor K = l1(b0 + x2xy + y2). Now let T = Φ(x, y) − f(x, y) where f is the Darboux polynomial satisfying (iv) with degree l1. Clearly, T is a Darboux polynomial of system (4.2) with cofactor K = l1(b0 + x2xy + y2). If we expand T in its homogeneous terms we get Tl1 = 0 and

T=j=0l11Tj(x,y),
where again each Tj is a homogeneous polynomial of degree j. Using that T is a Darboux polynomial we get
Tl11=Al11(x+y)l11,Al11.

Note that the terms Tl1−2 satisfy

Tl12=Al12(x+y)l12,Al12.

Moreover, the terms Tl1−3 satisfy

(x3+y3)Tl13y+yTl13x+b0xTl11y+(b01+1)yTl11yb0l1Tl11=l1(x2+xy+y2)Tl13.

Solving it and using that Tl1−3 is a polynomial we get Al1−1 = 0 and so Tl1−1 = 0. Proceeding inductively we obtain T = 0 and so f(x, y) = (x + y)l1. Since f must be irreducible we get l1 = 1 and n = 1 so it must be the one of Lemma 4.1 (i).

Proceeding in a similar manner for the case δ = −1 we also obtain that the unique irreducible Darboux polynomials are the ones of degree one and two given in Lemma 4.1.

### Proposition 4.2.

System (4.2) is not Liouvillian integrable.

### Proof.

We separate the proof of Proposition 4.2 into different lemmas.

### Lemma 4.6.

System (4.2) under the assumptions of Lemma 4.1 (i) is not Liouvillian integrable.

### Proof of Lemma 4.6.

The unique exponential factor that system (4.2) may have is of the form E = exp(g/(y + δx)n) where g ∈ ℂ[x, y], n ≥ 0 and g is coprime with y + δx whenever n > 0. Moreover, the cofactor L ∈ ℂ[x, y] is of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 where bij ∈ ℂ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

ygxE+(δb0x+(b0δ)y+δx3+y3)gxEn(b0+x2δxy+y2)gE=L(y+δx)nE,
and after simplifying by E = exp(g/(y + δx)n), we conclude that g must satisfy
ygx+(δb0x+(b0δ)y+δx3+y3)gxn(b0+x2δxy+y2)g=L(y+δx)n.

Assume that n > 0 and so g is coprime with y + δx. If we denote by g¯ the restriction of g to y = −δx then g¯0 and satisfies

δxdg¯dxn(b0+3x2)g¯=0,
because δ2 = 1. Solving the above linear differential equation we get
g¯(x)=αe3nx2/2xnb0/δ,α.

Since g¯ must be a polynomial and n > 0, we get α = 0 and so g¯=0 which is not possible. So, n = 0 and g satisfies

ygx+(δb0x+(b0δ)y+δx3+y3)gx=L.

Note that the divergence of system (4.2) is b0δ + 3y2 and in view of Theorem 2.1 we must have L = −b0 + δ − 3y2λ(b0 + x2δxy + y2) for some λ ∈ ℂ. Hence, g must satisfy

ygx+(δb0x+(b0δ)y+δx3+y3)gx=(b0+δ3y2λ(b0+x2δxy+y2)).(4.5)

Evaluating (4.5) on x = y = 0 we get δ = b0(1 + λ). Now we expand g as a polynomial in its homogeneous parts as

g(x,y)=j=0mgj(x,y),
where gj are homogeneous polynomials in (x, y), m ≥ 1 and gm ≠ 0. The homogeneous part gm for m ≥ 3 satisfies
(δx3+y3)gmy=0andsogm=Bmxm,
where Bm ∈ ℂ. The homogeneous part gm−1 must satisfy
(δx3+y3)gm1y=0andsogm1=Bm1xm1.

Furthermore, the homogeneous part gm−2 must satisfy

(δx3+y3)gm2y=Bmmxm1y
and so
gm2=Bm2xm2Bmmδ6xm2[23arctan(2yδx3δx)+log(δx2δxy+y2(x+y)2)],
which yields Bm = 0. Hence gm = 0. This means that g ∈ ℂ[x, y] and has degree at most two. We write it as
g(x,y)=λ00+λ10x+λ01y+λ20x2+λ11xy+λ02y2,
with λi,j ∈ ℂ for i, j ∈ {0, 1, 2}. Solving equation (4.5) with g as above we get a contradiction. So, in view of Theorem 2.1 and 2.2 it is not Liouvillian integrable.

### Lemma 4.7.

System (4.2) under the assumptions of Lemma 4.1 (ii) is not Liouvillian integrable.

### Proof of Lemma 4.7.

The unique exponential factor that system (4.2) may have is of the form E = exp(g/(1 + δx2 + xy)n) where g ∈ ℂ[x,y], n ≥ 0 and g is coprime with 1 + δx2 + xy whenever n > 0. Moreover, the cofactor L ∈ ℂ[x, y] is of the form L = b00 +b10x+b01y+b20x2 +b11xy+b02y2 where bij ∈ ℂ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

ygxE+(x3δy+δx3+y3)gyEn(x2δxy+y2)gE=L(1+δx2+xy)nE,
and after simplifying by E = exp(g/(1 + δx2 + xy)n), we conclude that g must satisfy
ygx+(x3δy+δx3+y3)gyn(x2δxy+y2)g=L(1+δx2+xy)n.

Note that the divergence of system (4.2) is −3δ + 3y2 and in view of Theorem 2.1 we must have L = 3δ − 3y2λ(x2δxy + y2) for some λ ∈ ℂ, and so

ygx+(x3δy+δx3+y3)gyn(x2δxy+y2)g=(3δ3y2λ(x2δxy+y2))(1+δx2+xy)n.(4.6)

Evaluating (4.6) on x = y = 0 we get 0 = 3δ which is not possible. Therefore, in this case there is no such an exponential factor. This concludes the proof.

### Lemma 4.8.

System (4.2) under the assumptions of Lemma 4.1 (iii) is not Liouvillian integrable.

### Proof of Lemma 4.8.

The unique exponential factor that system (4.2) may have is of the form E = exp(g/(−δx2 + xyδy2)n) where g ∈ ℂ[x, y], n ≥ 0 and g is coprime with −δx2 + xyδy2 whenever n > 0. Moreover, the cofactor L ∈ ℂ[x, y] is of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 where bij ∈ ℂ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

ygxE+(x+δy+δx3+y3)gyEn(δx2+δxy+2y2)gE=L(δx2+xyδy2)nE,
and after simplifying by E = exp(g/(−δx2 + xyδy2)n), we conclude that g must satisfy
ygx+(x+δy+δx3+y3)gyn(δx2+δxy+y2)g=L(δx2+xyδy2)n.(4.7)

Note that the divergence of system (4.2) is δ + 3y2 and by Theorem 2.1 we must have L = −δ − 3y2λ(δx2 + δxy + y2) for some λ ∈ ℂ. So, by (4.7) g must satisfy

ygx+(x+δy+δx3+y3)gyn(δx2+δxy+y2)g=(δ+3y2λ(δx2+δxy+y2))(δx2+xyδ2)n.(4.8)

Evaluating (4.8) on y = 0 and x=±δ we get 0 = (−1)n+1δ3n+1 which is not possible. Hence, in this case there is no such an exponential factor. This concludes the proof.

### Lemma 4.9.

System (4.2) under none of the assumptions of Lemma 4.1 is not Liouvillian integrable.

### Proof of Lemma 4.9.

Since system (4.2) has no Darboux polynomial, so the unique exponential factor that system (4.2) may have is of the form E = exp(g) where g ∈ ℂ[x, y] and with cofactor L ∈ ℂ[x, y] of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 where bij ∈ ℂ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

ygxE+(ax+by+δx3+y3)gyE=LE,
and after simplifying by E = exp(g), we conclude that g must satisfy
ygx+(ax+by+δx3+y3)gy=L.(4.9)

Note that the divergence of system (4.2) is b + 3y2 and in view of Theorem 2.1 we must have L = −b − 3y2. So, in view of (4.9), g must satisfy

ygx+(ax+by+δx3+y3)gy=L=b3y2.(4.10)

Evaluating it on x = y = 0 we get b = 0. Now we expand g as a polynomial in its homogeneous parts as g(x,y)=j=0ngj(x,y), where gj are homogeneous polynomials in (x, y), n ≥ 1 and gn ≠ 0.

The homogeneous part gn for n ≥ 3 satisfies

(δx3+y3)gny=0andsogn=Bnxn,
where Bn ∈ ℂ. The homogeneous part gn−1 must satisfy
(δx3+y3)gn1y=0andsogn1=Bn1xn1,
where Bn−1 ∈ ℂ. Furthermore, the homogeneous part gn−2 must satisfy
(δx3+y3)gn2y=Bnnxn1y
and so
gn2=Bn2xn2Bnnδ6xn2[23arctan(2yδx3δx)+log(δx2δxy+y2(x+y)2)],
which yields Bn = 0. Hence gn = 0. This means that g ∈ ℂ[x, y] has degree at most two. We write it as
g(x,y)=λ00+λ10x+λ01y+λ20x2+λ11xy+λ02y2,
with λi,j ∈ ℂ for i, j ∈ {0, 1, 2}. Solving equation (4.10) with g as above we get a contradiction. So, in view of Theorem 2.1 and 2.2 it is not Liouvillian integrable.

Now, the proof of Proposition 4.2 follows directly from Lemmas 4.64.9.

## Acknowledgements

The authors are grateful to the referees for their valuable comments and suggestions to improve this paper. The first author is partially supported by a MINECO/FEDER grant number MTM2017-84383-P and an AGAUR (Generalitat de Catalunya) grant number 2017SGR-1276. The second author is supported by FCT/Portugal through UID/MAT/04459/2013.

## References

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[8]D.W. Smith and P. Jordan, Nonlinear ordinary differential equations – An introduction for scientists and engineers, 4th ed, University Press, Oxford, 2007.
[12]H. Poincaré, Mémoire sur les courbes définies par les équations différentielles, Œuvres de Henri Poincaré, Gauthier–Villars, Paris, Vol. I, 1951, pp. 95-114.
[13]H. Poincaré, Sur l’intégration algébrique des équations différentielles du premier ordre et du premier degrée, Rend. Circ. Mat. Palermo, Vol. 5, 1891, pp. 161-191. 11 (1897), 193–239
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Journal
Journal of Nonlinear Mathematical Physics
Volume-Issue
26 - 2
Pages
169 - 187
Publication Date
2021/01/06
ISSN (Online)
1776-0852
ISSN (Print)
1402-9251
DOI
10.1080/14029251.2019.1591710How to use a DOI?
© 2019 The Authors. Published by Atlantis and Taylor & Francis
Open Access
This is an open access article distributed under the CC BY-NC 4.0 license (http://creativecommons.org/licenses/by-nc/4.0/).

TY  - JOUR
AU  - Jaume Giné
AU  - Claudia Valls
PY  - 2021
DA  - 2021/01/06
TI  - Liouvillian integrability of a general Rayleigh-Duffing oscillator
JO  - Journal of Nonlinear Mathematical Physics
SP  - 169
EP  - 187
VL  - 26
IS  - 2
SN  - 1776-0852
UR  - https://doi.org/10.1080/14029251.2019.1591710
DO  - 10.1080/14029251.2019.1591710
ID  - Giné2021
ER  -