Journal of Nonlinear Mathematical Physics

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

Liouvillian integrability of a general Rayleigh-Duffing oscillator

Authors
Jaume GinΓ©
Departament de MatemΓ tica, Inspires Research Centre, Universitat de Lleida Avda. Jaume II, 69; 25001 Lleida, Catalonia, Spain,gine@matematica.udl.cat
Claudia Valls
Departamento de MatemΓ‘tica, Instituto Superior TΓ©cnico, 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.

Copyright
Β© 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

xΛ™=y,yΛ™=ax+by+cx3+dy3,(1.1)
where a, b, c, d ∈ ℝ are arbitrary parameters and c2 + d2 β‰  0 (otherwise the system is linear). When c = d = βˆ’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) ↦ (βˆ’x, βˆ’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 βŠ† 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 β‰  0:

  1. (a)

    If d = 0 and c β‰  0 it is Liouville integrable if and only if either b = 0 or a = βˆ’2b2/9.

  2. (b)

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

  3. (c)

    If cd β‰  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 ∈ β„•

xΛ™=P(x),   x=(x1,x2)βˆˆβ„2,(2.1)
where P(x) = (P1(x), P2(x)), Pi ∈ β„‚[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)=P1βˆ‚Hβˆ‚x1+P2βˆ‚Hβˆ‚x2=0.

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

𝒳(f)=P1βˆ‚fβˆ‚x1+P1βˆ‚fβˆ‚x2=Kf,(2.2)
where x = (x1, x2), K ∈ β„‚[x] and has degree at most d βˆ’ 1. The polynomial K is said to be the cofactor. An invariant algebraic curve is a curve given by f = 0 such that f ∈ β„‚[x].

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

𝒳(E)=P1βˆ‚Eβˆ‚x1+P2βˆ‚Eβˆ‚x2=LE,(2.3)
where L ∈ β„‚[x] and has degree at most d βˆ’ 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)=P1βˆ‚Vβˆ‚x1+P2βˆ‚Vβˆ‚x2=(βˆ‚P1βˆ‚x1+βˆ‚P2βˆ‚x2)V.

When V ∈ β„‚[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 Ξ±i, Ξ²j ∈ β„‚ not all zero such that

βˆ‘i=1pΞ±iKi+βˆ‘j=1qΞ²jLj=div(𝒳)
if and only if the function
f1Ξ±1β‹―fpΞ±pE1Ξ²1β‹―EqΞ²q,(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 (xΒ―1,xΒ―2) satisfy k1Ξ»1 + k2Ξ»2 β‰  0 for k1, k2 ∈ 𝕑+ and k1 + k2 β‰₯ 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 (xΒ―1,xΒ―2) satisfy Ξ»1 = 0 and Ξ»2 β‰  0. Then 𝒳 can have a polynomial first integral if and only if (xΒ―1,xΒ―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 Ο„: β„‚2[x1, x2] β†’ β„‚2[x1, x2] be the automorphism defined by

Ο„(x1,x2)=(βˆ’x1,βˆ’x2)(2.5)
and for a polynomial p(x1, x2) let Ο„*(p) = p(Ο„(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 Β· Ο„*(g) is also a Darboux polynomial invariant by Ο„ with a cofactor of the form K + Ο„*(K) = K(x1, x2) + K(βˆ’x1, βˆ’x2).

Proof.

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

Ο„*(𝒳(g))=Ο„*(Kg).(2.6)

Moreover, using that Ο„βˆ’1 = Ο„ we obtain

Ο„*(𝒳(g))=𝒳(gβˆ˜Ο„βˆ’1)=𝒳(gβˆ˜Ο„)=𝒳(Ο„*(g)).(2.7)

On the other hand,

Ο„*(Kg)=(Kg)βˆ˜Ο„βˆ’1=(Kg)βˆ˜Ο„=(Kβˆ˜Ο„)(gβˆ˜Ο„)=Ο„*(K)Ο„*(g).(2.8)

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

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

The following result was proved in [2] (see Theorems 13 and 14). We recall that two real numbers Ο‰1, Ο‰2 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 k1Ο‰1 + k2Ο‰2 = 0 is k1 = k2 = 0.

Proposition 2.2.

Let (xΒ―1,xΒ―2) be a singularity of system (2.1) and Ξ»1, Ξ»2 be the eigenvalues of the Jacobian matrix at (xΒ―1,xΒ―2). Assume that f(x1, x2) = 0 is an irreducible invariant algebraic curve with cofactor K(x1, x2). If f(xΒ―1,xΒ―2)β‰ 0 then K(xΒ―1,xΒ―2)=0. Moreover if f(xΒ―1,xΒ―2)=0, Ξ»2 β‰  0 and either Ξ»1 and Ξ»2 are rationally independent, or Ξ»1Ξ»2 < 0, then K(xΒ―1,xΒ―2)∈{Ξ»1,Ξ»2,Ξ»1+Ξ»2}.

A polynomial f(x1, x2) is said to be a weight homogeneous polynomial if there exist s = (s1, s2) ∈ β„•2 and m ∈ β„• such that for all Ξ± ∈ ℝ \ {0},

f(Ξ±s1x1,Ξ±s2x2)=Ξ±mf(x1,x2),
where ℝ denotes the set of real numbers, and β„• 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) ↦(Ξ±s1x1, Ξ±s2x2) 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 β‰  0. Moreover the case d = b = 0 with c β‰  0 corresponds to the Hamiltonian system xΛ™=y, yΛ™=ax+cx3 which has the polynomial first integral H = βˆ’2ax2 βˆ’ cx4 + 2y2 and the case d = 0, a = βˆ’2b2/9 with c β‰  0 is Liouville integrable because it has the inverse integrating factor

V=(2b2x2βˆ’9cx4βˆ’12bxy+18y2)3/4.

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

xΛ™=y,   yΛ™=βˆ’byβˆ’dy3.

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 β‰  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 β‰  0 is not Liouvillian integrable.

Proof.

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

xβ†’1dBx1,   yβ†’By1,   sβ†’B2dt,   B={(a/d2)1/4,a>0,(βˆ’a/d2)1/4,a<0
we get
x1β€²=y1,   y1β€²=Β±x1+by1+y13,(3.1)
where b = b/B and the prime denotes derivative in s. Let Ο„ be the automorphism defined by (2.5) and g be a Darboux polynomial with cofactor K=a00+a10x1+a01y1+a20x12+a11x1y1+a02y12 with aij ∈ β„‚ for i, j ∈ {0, 1, 2}. In view of Proposition 2.1 we have that f = g Β· Ο„(g) is a Darboux polynomial invariant by Ο„ and with cofactor K=2a00+2a20x12+2a11x1y1+2a02y12. Note that f(x1, y1) = f (βˆ’x1, βˆ’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=Ξ±βˆ’3X,   y1=Ξ±βˆ’1Y,   s=Ξ±2r.

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

Xβ€²=Ξ±4Y,   Yβ€²=Β±X+bΞ±2Y+Y3,(3.2)
where now the prime denotes derivative in the new time r. Now we set
F(X,Y)=Ξ±β„“f(Ξ±βˆ’3X,Ξ±βˆ’1Y)
and
K=2a00+2a20Ξ±βˆ’4X2+2a11Ξ±βˆ’3XY+2a02Ξ±βˆ’2Y2
where β„“ 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=Ξ±β„“+2dfds=Ξ±β„“+2(2a00+2a20Ξ±βˆ’4X2+2a11Ξ±βˆ’3XY+2a02Ξ±βˆ’2Y2)f=(2a02Y2+2a11Ξ±βˆ’1XY+2a20Ξ±βˆ’2X2+2Ξ±2a00)F=KF.

If F=βˆ‘i=0nFi where Fi is a weight homogeneous polynomial in X, Y with weight degree β„“ βˆ’ i for i = 0,...,n and β„“ β‰₯ n, we get

f=F|Ξ±=1.

From the definition of invariant algebraic curve we have

Ξ±4Yβˆ‘i=0nΞ±iβˆ‚Fiβˆ‚X+(Β±X+Ξ±2bY+Y3)βˆ‘i=0nΞ±iβˆ‚Fiβˆ‚Y=(2a02Y2+2a11Ξ±βˆ’1XY+2a20Ξ±βˆ’2X2+2Ξ±2a00)βˆ‘i=0nΞ±iFi.(3.3)

Equating the terms with Ξ±βˆ’2 in (3.3) we get a20 = 0 and the terms with Ξ±βˆ’1 we get a11 = 0. Moreover, equating the terms with Ξ±0 in (3.3) we obtain

(Β±X+Y3)βˆ‚F0βˆ‚Y=2a02Y2F0.

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

F0=Ξ²0Xnβˆ’l(Β±X+Y3)l,   a02=3l2,
where Ξ²0 β‰  0 (otherwise f would be constant which is not possible). Note that β„“ = 3n.

Equating now the terms in (3.3) with Ξ±1 we obtain

(Β±X+Y3)βˆ‚F1βˆ‚Y=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 β„“ βˆ’ 1 = 3n βˆ’ 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 Ξ±2 we get

(Β±X+Y3)βˆ‚F2βˆ‚Y+bYβˆ‚F0βˆ‚Y=3lY2F2+2a00F0.

Introducing F0 we conclude that

F2=F˜2(X)(Β±X+Y3)l+Ξ²0Xnβˆ’l(Β±X+Y3)l(blΒ±X+Y3βˆ’(2a00βˆ’bl)3X2/3arctan(X1/3βˆ“2Y3X1/3)+2a00βˆ’bl3X2/3log((X1/3Β±Y)2X2/3βˆ“X1/3Y+Y2)).

Since F2 must be a polynomial with degree β„“ βˆ’ 2 = 3n βˆ’ 2 and X has weight-degree 3, we must have F˜2=0 and a00 = bl/2 (because Ξ²0 β‰  0). Hence,

fnβˆ’2=bΞ²0Xnβˆ’ly(Β±X+Y3)lβˆ’1.

Computing the terms in (3.3) with Ξ±3 we get

(Β±X+Y3)βˆ‚F3βˆ‚Y=3lY2F3.

Solving this linear differential equation using that F3 has degree β„“ βˆ’ 3 = 3n βˆ’ 3 we get

F3=Ξ²3Xnβˆ’lβˆ’1(Β±X+Y3)l.

Computing the terms in (3.3) with Ξ±4 we obtain

(Β±X+Y3)βˆ‚F4βˆ‚Y+bYβˆ‚F2βˆ‚Y+Yβˆ‚F0βˆ‚X=3lY2F4+bl2F2.

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

F4=F˜4(X)(Β±X+Y3)l+Ξ²018Xnβˆ’l(Β±X+Y3)l(54b2(βˆ’1+l)lXY2(Β±X+Y3)2βˆ’6l(1+6b2l)Y2Β±X+Y3+23(6b2l2+3nβˆ’l(2+3bX2/3))X1/3arctan(X1/3βˆ“2Y3X1/3)+6b2l2+3nβˆ’l(2+3bX2/3)X1/3log((X1/3Β±y)2X2/3βˆ“X1/3Y+Y2)).

Since F4 must be a polynomial of degree β„“ βˆ’ 4 = 3n βˆ’ 4 we must have F˜4=0 and 6b2l2 + 3n βˆ’ l(2 + 3bX2/3) = 0. Since b β‰  0 we get l = n = 0 which is not possible. Hence, there are no Darboux polynomials invariant by Ο„ 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 β‰  0 has no Darboux polynomials, the unique exponential factor of that system may have is of the form E = exp(g) where g ∈ β„‚[x, y] and with cofactor L ∈ β„‚[x, y] of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 with bij ∈ β„‚ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

yβˆ‚gβˆ‚xE+(ax+by+dy3)βˆ‚gβˆ‚yE=LE,
and after simplifying by E = exp(g), we conclude that the polynomial g must satisfy
yβˆ‚gβˆ‚x+(ax+by+dy3)βˆ‚gβˆ‚y=L.

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

yβˆ‚gβˆ‚x+(ax+by+dy3)βˆ‚gβˆ‚y=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)=βˆ‘j=0ngj(x,y),
where gj are homogeneous polynomials in (x, y), n β‰₯ 1 and gn β‰  0. The homogeneous part gn for n β‰₯ 3 satisfies
dy3βˆ‚gnβˆ‚y=0   and so  gn=Bnxn,
where Bn ∈ β„‚. The homogeneous part gnβˆ’1 must satisfy
dy3βˆ‚gnβˆ’1βˆ‚y=0   and so  gnβˆ’1=Bnβˆ’1xnβˆ’1,
where Bnβˆ’1 ∈ β„‚. Furthermore, the homogeneous part gnβˆ’2 must satisfy
dy3βˆ‚gnβˆ’2βˆ‚y==Bnnxnβˆ’1y
and so
gnβˆ’2=Bnβˆ’2xnβˆ’2βˆ’Bnndyxnβˆ’1,
which yields Bn = 0. Hence gn = 0. This means that g ∈ β„‚[x, y] has degree at most two. We write it as
g(x,y)=Ξ»00+Ξ»10x+Ξ»01y+Ξ»20x2+Ξ»11xy+Ξ»02y2,
where Ξ»ij ∈ β„‚ 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 β‰  0 by the change of coordinates and reparametrization of time of the form

xβ†’1C1/6d1/3X,   yβ†’C1/6d2/3Y,   τ→C2/6d1/3t,   C={c,c>0,βˆ’c,c<0,
we can write system (1.1) as
XΛ™=Y,   YΛ™=a˜X+b˜Y+Ξ΄X3+Y3,β€‰β€‰β€‰Ξ΄βˆˆ{βˆ’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
xΛ™=y,   yΛ™=ax+by+Ξ΄x3+y3,β€‰β€‰β€‰Ξ΄βˆˆ{βˆ’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) = Ξ΄x + y with cofactor K(x, y) = b0 + x2 βˆ’ Ξ΄xy + y2 if a = Ξ΄b0 and b = b0 βˆ’ Ξ΄ with b0 ∈ ℝ;

  2. (ii)

    f(x, y) = 1 + Ξ΄x2 + xy with cofactor K(x, y) = x2 βˆ’ Ξ΄xy + y2 if a = 1 and b = βˆ’3Ξ΄;

  3. (iii)

    f(x, y) = βˆ’Ξ΄x2 + xy βˆ’ Ξ΄y2 with cofactor K(x, y) = Ξ΄ βˆ’ x2 + Ξ΄xy + 2y2 if a = βˆ’1 and b = Ξ΄.

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 β‰₯ 1 are the ones given in the statement of Lemma 4.1.

Proof.

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

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

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

yβˆ‚fβˆ‚x+(ax+by+Ξ΄x3+y3)βˆ‚fβˆ‚y=(a00+a20x2+a11xy+a02y2)f.(4.3)

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

f=βˆ‘j=0mfj(x)yj.

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

mfm(x)=a02fm(x)   that is   fm(x)(a02βˆ’m)=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)=βˆ‘j=0mfj(x,y),
where fj are homogeneous polynomials in (x, y), n β‰₯ 1 and fn β‰  0. It follows from (4.3) that the homogeneous part fn satisfies
(Ξ΄x3+y3)βˆ‚fnβˆ‚y=(a20x2+a11xy+a02y2)fn.(4.4)

We will work with Ξ΄ = 1. The case Ξ΄ = βˆ’1 can be done in a similar manner.

Solving (4.4) with Ξ΄ = 1 we get

fn=Gn(x)(x+y)βˆ’(a11βˆ’a20βˆ’a02)/3(x2βˆ’xy+y2)βˆ’(a20βˆ’a11βˆ’2a02)/6(3xβˆ’i(2yβˆ’x)3x+i(2yβˆ’x))βˆ’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 = βˆ’a20 but then we get
fn=Gn(x)(x+y)(2a20+a02)/3(x2βˆ’xy+y2)(a02βˆ’a20)/3.

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

fn=Anxnβˆ’l1βˆ’2l2(x+y)l1(x2βˆ’xy+y2)l2,
for some constant An β‰  0. Without loss of generality we can take An = 1. Since fn(x, y) = Ο„(fn(x, y)) = fn(βˆ’x, βˆ’y) we must have n even.

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

(x3+y3)βˆ‚fnβˆ’2βˆ‚y+hnβˆ’2=(a20x2+a11xy+a02y2)fnβˆ’2,
where
hnβˆ’2=yβˆ‚fnβˆ‚xβˆ’(ax+2by)βˆ‚fnβˆ‚yβˆ’a00fn.

Solving it we get

fnβˆ’2=(x+y)l1βˆ’1(x2βˆ’xy+y2)l2βˆ’1[Gnβˆ’2(x)(x+y)(x2βˆ’xy+y2)βˆ’rnβˆ’2(x,y)βˆ’s1xnβˆ’l1βˆ’2l2βˆ’2(x+y)(x2βˆ’xy+y2)arctan(2yβˆ’x3x)βˆ’s2xnβˆ’l1βˆ’2l2βˆ’2(x+y)(x2βˆ’xy+y2)log(x+yx2βˆ’xy+y2)],
where Gnβˆ’2 is any function in the variable x, rnβˆ’2(x, y) is a polynomial in the variables x, y of degree n βˆ’ l1 βˆ’ 2l2 + 1 and
s1=3a00+(1+aβˆ’4b)l1+(5βˆ’aβˆ’2b)l2βˆ’3n,s2=a00+(aβˆ’1)l1βˆ’(1+a+2b)l2+n.

Since fnβˆ’2 must be a homogeneous polynomial of degree nβˆ’2 and An β‰  0 we must have Gnβˆ’2(x) = Anβˆ’2xnβˆ’l1βˆ’2l2βˆ’2 for some constant Anβˆ’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

fnβˆ’2=βˆ’(x+y)l1βˆ’1(x2βˆ’xy+y2)l2βˆ’1(rnβˆ’2(x,y)+Anβˆ’2xnβˆ’l1βˆ’2l2βˆ’2(x+y)(x2βˆ’xy+y2)).

The homogeneous part fnβˆ’3 must have degree n βˆ’ 3 which is odd and so fnβˆ’3 = 0 (because f must be invariant by Ο„ and fnβˆ’1 = 0). The homogeneous part fnβˆ’4 must have degree n βˆ’ 4 and satisfies

(x3+y3)βˆ‚fnβˆ’4βˆ‚y+hnβˆ’4=(a20x2+a11xy+a02y2)fnβˆ’4,
where
hnβˆ’4=yβˆ‚fnβˆ’2βˆ‚xβˆ’(ax+2by)βˆ‚fnβˆ’2βˆ‚yβˆ’a00fnβˆ’2.

Solving it we get

fnβˆ’4=(x+y)l1βˆ’2(x2βˆ’xy+y2)l2βˆ’2[Gnβˆ’4(x)(x+y)2(x2βˆ’xy+y2)2βˆ’rnβˆ’4(x,y)βˆ’s3xnβˆ’l1βˆ’2l2βˆ’4(x+y)2(x2βˆ’xy+y2)2arctan(2yβˆ’x3x)βˆ’s4xnβˆ’l1βˆ’2l2βˆ’4(x+y)2(x2βˆ’xy+y2)2log(x+yx2βˆ’xy+y2)],
where Gnβˆ’4 is any function in the variable x, rnβˆ’4(x, y) is a polynomial in the variables x, y of degree n βˆ’ l1 βˆ’ 2l2 + 2 and s3, s4 are very big and we do not write them here. Since fnβˆ’4 must be a homogeneous polynomial of degree n βˆ’ 4 we must have s3 = s4 = 0 and then we write it as
fnβˆ’4=βˆ’(x+y)l1βˆ’2(x2βˆ’xy+y2)l2βˆ’2(rnβˆ’4(x,y)+Anβˆ’4xnβˆ’l1βˆ’2l2βˆ’4(x+y)2+(x2βˆ’xy+y2)2),
for some constant Anβˆ’4. The homogeneous part fnβˆ’5 must have degree n βˆ’ 5 which is odd and so fnβˆ’5 = 0 (because f must be invariant by Ο„ and fnβˆ’3 = 0). The homogeneous part fnβˆ’6 must have degree n βˆ’ 6 and satisfies
(x3+y3)βˆ‚fnβˆ’6βˆ‚y+hnβˆ’6=(a20x2+a11xy+a02y2)fnβˆ’6,
where
hnβˆ’6=yβˆ‚fnβˆ’4βˆ‚xβˆ’(ax+2by)βˆ‚fnβˆ’4βˆ‚yβˆ’a00fnβˆ’4.

Solving it we get

fnβˆ’6=(x+y)l1βˆ’3(x2βˆ’xy+y2)l2βˆ’3[Gnβˆ’6(x)(x+y)3(x2βˆ’xy+y2)3βˆ’rnβˆ’6(x,y)βˆ’s5xnβˆ’l1βˆ’2l2βˆ’6(x+y)3(x2βˆ’xy+y2)2arctan(2yβˆ’x3x)βˆ’s6xnβˆ’l1βˆ’2l2βˆ’6(x+y)3(x2βˆ’xy+y2)3log(x+yx2βˆ’xy+y2)],
where Gnβˆ’6 is any function in the variable x, rnβˆ’6(x, y) is a polynomial in the variables x, y of degree n βˆ’ l1 βˆ’ 2l2 + 3 and s5, s6 are very big and we do not write them here. Since fnβˆ’6 must be a homogeneous polynomial of degree n βˆ’ 6 we must have s5 = s6 = 0. Now we solve the system
s1=s2=s3=s4=s5=s6=0
with n β‰₯ 1 (an integer) and l12+l22β‰ 0 (the case l1 = l2 = 0 yields n = 0 which is not possible) yield the solutions:
  1. (i)

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

  2. (ii)

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

  3. (iii)

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

  4. (iv)

    b = a βˆ’ 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 (bΒ±b2+4a)/2. Under the assumptions a = βˆ’1, b = 1 the origin is a focus and the eigenvalues of the Jacobian matrix at that point are (1Β±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(1βˆ’x2+xy+2y2).

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

l2∈{1,1+3i2,1βˆ’3i2}.

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 Ο„ 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 Ο„ must be divisors of the Darboux polynomial βˆ’Ξ΄x2 + xy βˆ’ Ξ΄y2 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 Ο„ 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 (bΒ±b2+4a)/2. Under the assumptions a = 5/2, b = βˆ’9/2 the origin is a saddle and the eigenvalues of the Jacobian matrix at that point are βˆ’5, 1/2. Moreover, the cofactor of the irreducible Darboux polynomial (if it exists) is of the form

K=K(x,y)=l1(12+x2βˆ’xy+y2).

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

l12∈{12,βˆ’5,βˆ’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 Ο„. 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 Ο•(x, y) the irreducible Darboux polynomial of Lemma 4.1 (ii), that is, Ο•(x, y) = 1βˆ’x2 +xy and set Ξ¦(x, y) = Ο•l1 then Ξ¦ is a Darboux polynomial of system (4.2) with cofactor K = l1(x2 βˆ’ xy + y2). Moreover Ξ¦ has degree n = 2l1 and if we expand Ξ¦ in its homogeneous terms we get

Ξ¦(x,y)=xl1(x+y)l1+βˆ‘j=02l1βˆ’2Ξ¦j(x,y)
where each Ξ¦j is a homogeneous polynomial of degree j (we recall that Ξ¦ cannot have odd degree terms). Now define T = Ξ¦(x, y) βˆ’ 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(x2 βˆ’ xy + y2). If we expand T in its homogeneous terms we get that T2l1 = 0 and
T=βˆ‘j=02l1βˆ’2Tj(x,y),
where again each Tj is a homogeneous polynomial of degree j. Using that T is a Darboux polynomial we get that
T2l1βˆ’2=A2l1βˆ’2xl1βˆ’2(x+y)l1,   A2l1βˆ’2βˆˆβ„‚.

Note that the terms T2l1βˆ’4 satisfy

(x3+y3)βˆ‚T2l1βˆ’4βˆ‚y+yβˆ‚T2l1βˆ’2βˆ‚x+xβˆ‚T2l1βˆ’2βˆ‚yβˆ’3yβˆ‚T2l1βˆ’2βˆ‚y=l1(x2βˆ’xy+y2)T2l1βˆ’4.

Solving it we get

T2l1βˆ’4=A2l1βˆ’4x2l1βˆ’3(x+y)l1βˆ’A2l1βˆ’2xl1βˆ’4(x+y)l1βˆ’1[l1+33(x+y)arctan(2yβˆ’x3x)+(x+y)log(x2βˆ’xy+y2(x+y2))].

Since T2l1βˆ’4 must be a polynomial we get A2l1βˆ’2 = 0 and so T2l1βˆ’2 = 0. Now proceeding inductively we get T2l1βˆ’2j = 0 for j = 0,...,l1 which yields T = 0. In short, f(x, y) = Ξ¦(x, y) = Ο•l1. Note that Ο• 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 Ο„ 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 Ο„ 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 Ξ¦(x, y) = (x + y)l1. Note that Ξ¦ is a Darboux polynomial of system (4.2) with cofactor K = l1(b0 + x2 βˆ’ xy + y2). Now let T = Ξ¦(x, y) βˆ’ 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 + x2 βˆ’ xy + y2). If we expand T in its homogeneous terms we get Tl1 = 0 and

T=βˆ‘j=0l1βˆ’1Tj(x,y),
where again each Tj is a homogeneous polynomial of degree j. Using that T is a Darboux polynomial we get
Tl1βˆ’1=Al1βˆ’1(x+y)l1βˆ’1,   Al1βˆ’1βˆˆβ„‚.

Note that the terms Tl1βˆ’2 satisfy

Tl1βˆ’2=Al1βˆ’2(x+y)l1βˆ’2,   Al1βˆ’2βˆˆβ„‚.

Moreover, the terms Tl1βˆ’3 satisfy

(x3+y3)βˆ‚Tl1βˆ’3βˆ‚y+yβˆ‚Tl1βˆ’3βˆ‚x+b0xβˆ‚Tl1βˆ’1βˆ‚y+(b01+1)yβˆ‚Tl1βˆ’1βˆ‚yβˆ’b0l1Tl1βˆ’1=l1(x2+xy+y2)Tl1βˆ’3.

Solving it and using that Tl1βˆ’3 is a polynomial we get Al1βˆ’1 = 0 and so Tl1βˆ’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 Ξ΄ = βˆ’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 + Ξ΄x)n) where g ∈ β„‚[x, y], n β‰₯ 0 and g is coprime with y + Ξ΄x whenever n > 0. Moreover, the cofactor L ∈ β„‚[x, y] is of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 where bij ∈ β„‚ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

yβˆ‚gβˆ‚xE+(Ξ΄b0x+(b0βˆ’Ξ΄)y+Ξ΄x3+y3)βˆ‚gβˆ‚xEβˆ’n(b0+x2βˆ’Ξ΄xy+y2)gE=L(y+Ξ΄x)nE,
and after simplifying by E = exp(g/(y + Ξ΄x)n), we conclude that g must satisfy
yβˆ‚gβˆ‚x+(Ξ΄b0x+(b0βˆ’Ξ΄)y+Ξ΄x3+y3)βˆ‚gβˆ‚xβˆ’n(b0+x2βˆ’Ξ΄xy+y2)g=L(y+Ξ΄x)n.

Assume that n > 0 and so g is coprime with y + Ξ΄x. If we denote by gΒ― the restriction of g to y = βˆ’Ξ΄x then gΒ―β‰ 0 and satisfies

βˆ’Ξ΄xdgΒ―dxβˆ’n(b0+3x2)gΒ―=0,
because Ξ΄2 = 1. Solving the above linear differential equation we get
gΒ―(x)=Ξ±eβˆ’3nx2/2xβˆ’nb0/Ξ΄,β€‰β€‰β€‰Ξ±βˆˆβ„‚.

Since gΒ― must be a polynomial and n > 0, we get Ξ± = 0 and so gΒ―=0 which is not possible. So, n = 0 and g satisfies

yβˆ‚gβˆ‚x+(Ξ΄b0x+(b0βˆ’Ξ΄)y+Ξ΄x3+y3)βˆ‚gβˆ‚x=L.

Note that the divergence of system (4.2) is b0 βˆ’ Ξ΄ + 3y2 and in view of Theorem 2.1 we must have L = βˆ’b0 + Ξ΄ βˆ’ 3y2 βˆ’ Ξ»(b0 + x2 βˆ’ Ξ΄xy + y2) for some Ξ» ∈ β„‚. Hence, g must satisfy

yβˆ‚gβˆ‚x+(Ξ΄b0x+(b0βˆ’Ξ΄)y+Ξ΄x3+y3)βˆ‚gβˆ‚x=βˆ’(βˆ’b0+Ξ΄βˆ’3y2βˆ’Ξ»(b0+x2βˆ’Ξ΄xy+y2)).(4.5)

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

g(x,y)=βˆ‘j=0mgj(x,y),
where gj are homogeneous polynomials in (x, y), m β‰₯ 1 and gm β‰  0. The homogeneous part gm for m β‰₯ 3 satisfies
(Ξ΄x3+y3)βˆ‚gmβˆ‚y=0   and so  gm=Bmxm,
where Bm ∈ β„‚. The homogeneous part gmβˆ’1 must satisfy
(Ξ΄x3+y3)βˆ‚gmβˆ’1βˆ‚y=0   and so  gmβˆ’1=Bmβˆ’1xmβˆ’1.

Furthermore, the homogeneous part gmβˆ’2 must satisfy

(Ξ΄x3+y3)βˆ‚gmβˆ’2βˆ‚y=βˆ’Bmmxmβˆ’1y
and so
gmβˆ’2=Bmβˆ’2xmβˆ’2βˆ’BmmΞ΄6xmβˆ’2[23arctan(2yβˆ’Ξ΄x3Ξ΄x)+log(Ξ΄x2βˆ’Ξ΄xy+y2(x+y)2)],
which yields Bm = 0. Hence gm = 0. This means that g ∈ β„‚[x, y] and has degree at most two. We write it as
g(x,y)=Ξ»00+Ξ»10x+Ξ»01y+Ξ»20x2+Ξ»11xy+Ξ»02y2,
with Ξ»i,j ∈ β„‚ 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 + Ξ΄x2 + xy)n) where g ∈ β„‚[x,y], n β‰₯ 0 and g is coprime with 1 + Ξ΄x2 + xy whenever n > 0. Moreover, the cofactor L ∈ β„‚[x, y] is of the form L = b00 +b10x+b01y+b20x2 +b11xy+b02y2 where bij ∈ β„‚ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

yβˆ‚gβˆ‚xE+(xβˆ’3Ξ΄y+Ξ΄x3+y3)βˆ‚gβˆ‚yEβˆ’n(x2βˆ’Ξ΄xy+y2)gE=L(1+Ξ΄x2+xy)nE,
and after simplifying by E = exp(g/(1 + Ξ΄x2 + xy)n), we conclude that g must satisfy
yβˆ‚gβˆ‚x+(xβˆ’3Ξ΄y+Ξ΄x3+y3)βˆ‚gβˆ‚yβˆ’n(x2βˆ’Ξ΄xy+y2)g=L(1+Ξ΄x2+xy)n.

Note that the divergence of system (4.2) is βˆ’3Ξ΄ + 3y2 and in view of Theorem 2.1 we must have L = 3Ξ΄ βˆ’ 3y2 βˆ’ Ξ»(x2 βˆ’ Ξ΄xy + y2) for some Ξ» ∈ β„‚, and so

yβˆ‚gβˆ‚x+(xβˆ’3Ξ΄y+Ξ΄x3+y3)βˆ‚gβˆ‚yβˆ’n(x2βˆ’Ξ΄xy+y2)g=(3Ξ΄βˆ’3y2Ξ»(x2βˆ’Ξ΄xy+y2))(1+Ξ΄x2+xy)n.(4.6)

Evaluating (4.6) on x = y = 0 we get 0 = 3Ξ΄ 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/(βˆ’Ξ΄x2 + xy βˆ’ Ξ΄y2)n) where g ∈ β„‚[x, y], n β‰₯ 0 and g is coprime with βˆ’Ξ΄x2 + xy βˆ’ Ξ΄y2 whenever n > 0. Moreover, the cofactor L ∈ β„‚[x, y] is of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 where bij ∈ β„‚ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

yβˆ‚gβˆ‚xE+(βˆ’x+Ξ΄y+Ξ΄x3+y3)βˆ‚gβˆ‚yEβˆ’n(Ξ΄βˆ’x2+Ξ΄xy+2y2)gE=L(βˆ’Ξ΄x2+xyβˆ’Ξ΄y2)nE,
and after simplifying by E = exp(g/(βˆ’Ξ΄x2 + xy βˆ’ Ξ΄y2)n), we conclude that g must satisfy
yβˆ‚gβˆ‚x+(βˆ’x+Ξ΄y+Ξ΄x3+y3)βˆ‚gβˆ‚yβˆ’n(Ξ΄βˆ’x2+Ξ΄xy+y2)g=L(βˆ’Ξ΄x2+xyβˆ’Ξ΄y2)n.(4.7)

Note that the divergence of system (4.2) is Ξ΄ + 3y2 and by Theorem 2.1 we must have L = βˆ’Ξ΄ βˆ’ 3y2 βˆ’ Ξ»(Ξ΄ βˆ’ x2 + Ξ΄xy + y2) for some Ξ» ∈ β„‚. So, by (4.7) g must satisfy

yβˆ‚gβˆ‚x+(βˆ’x+Ξ΄y+Ξ΄x3+y3)βˆ‚gβˆ‚yβˆ’n(Ξ΄βˆ’x2+Ξ΄xy+y2)g=βˆ’(Ξ΄+3y2βˆ’Ξ»(Ξ΄βˆ’x2+Ξ΄xy+y2))(βˆ’Ξ΄x2+xyβˆ’Ξ΄2)n.(4.8)

Evaluating (4.8) on y = 0 and x=Β±Ξ΄ we get 0 = (βˆ’1)n+1Ξ΄3n+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 ∈ β„‚[x, y] and with cofactor L ∈ β„‚[x, y] of the form L = b00 + b10x + b01y + b20x2 + b11xy + b02y2 where bij ∈ β„‚ for i, j ∈ {0, 1, 2}. Hence, in view of (2.3) we have

yβˆ‚gβˆ‚xE+(ax+by+Ξ΄x3+y3)βˆ‚gβˆ‚yE=LE,
and after simplifying by E = exp(g), we conclude that g must satisfy
yβˆ‚gβˆ‚x+(ax+by+Ξ΄x3+y3)βˆ‚gβˆ‚y=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 = βˆ’b βˆ’ 3y2. So, in view of (4.9), g must satisfy

yβˆ‚gβˆ‚x+(ax+by+Ξ΄x3+y3)βˆ‚gβˆ‚y=L=βˆ’bβˆ’3y2.(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)=βˆ‘j=0ngj(x,y), where gj are homogeneous polynomials in (x, y), n β‰₯ 1 and gn β‰  0.

The homogeneous part gn for n β‰₯ 3 satisfies

(Ξ΄x3+y3)βˆ‚gnβˆ‚y=0   and so  gn=Bnxn,
where Bn ∈ β„‚. The homogeneous part gnβˆ’1 must satisfy
(Ξ΄x3+y3)βˆ‚gnβˆ’1βˆ‚y=0   and so  gnβˆ’1=Bnβˆ’1xnβˆ’1,
where Bnβˆ’1 ∈ β„‚. Furthermore, the homogeneous part gnβˆ’2 must satisfy
(Ξ΄x3+y3)βˆ‚gnβˆ’2βˆ‚y=βˆ’Bnnxnβˆ’1y
and so
gnβˆ’2=Bnβˆ’2xnβˆ’2βˆ’BnnΞ΄6xnβˆ’2[23arctan(2yβˆ’Ξ΄x3Ξ΄x)+log(Ξ΄x2βˆ’Ξ΄xy+y2(x+y)2)],
which yields Bn = 0. Hence gn = 0. This means that g ∈ β„‚[x, y] has degree at most two. We write it as
g(x,y)=Ξ»00+Ξ»10x+Ξ»01y+Ξ»20x2+Ξ»11xy+Ξ»02y2,
with Ξ»i,j ∈ β„‚ 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 4.6–4.9.

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

[5]F. Dumortier, J. Llibre, and J.C. ArtΓ©s, Qualitative Theory of Planar Differential Systems, Springer Verlag, New York, 2006.
[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. PoincarΓ©, MΓ©moire sur les courbes dΓ©finies par les Γ©quations diffΓ©rentielles, Ε’uvres de Henri PoincarΓ©, Gauthier–Villars, Paris, Vol. I, 1951, pp. 95-114.
[13]H. PoincarΓ©, Sur l’intΓ©gration algΓ©brique des Γ©quations diffΓ©rentielles du premier ordre et du premier degrΓ©e, Rend. Circ. Mat. Palermo, Vol. 5, 1891, pp. 161-191. 11 (1897), 193–239
[15]J.M. Strutt, Scientific Papers, Macmillan and Co., London, 1943.
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?
Copyright
Β© 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/).

Cite this article

TY  - JOUR
AU  - Jaume GinΓ©
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  - GinΓ©2021
ER  -