Integrability of a linear center perturbed by a fourth degree homogeneous polynomial
Abstract
In this work we study the integrability of a two-dimensional autonomous system in the plane with linear part of center type and non-linear part given by homogeneous polynomials of fourth degree. We give sufficient conditions for integrability in polar coordinates. Finally we establish a conjecture about the independence of the two classes of parameters which appear in the system; if this conjecture is true the integrable cases found will be the only possible ones.
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Publicacions Matem`atiques, Vol 40 (1996), 21–39. INTEGRABILITY OF A LINEAR CENTER PERTURBED BY A FOURTH DEGREE HOMOGENEOUS POLYNOMIAL* Javier Chavarriga and Jaume Gin´ e Abstract In this work we study the integrability of a two-dimensional autonomous system in the plane with linear part of center type and non-linear part given by homogeneous polynomials of fourth degree. We give sufficient conditions for integrability in polar coordinates. Finally we establish a conjecture about the independence of the two classes of parameters which appear in the system; if this conjecture is true the integrable cases found will be the only possible ones. 1. Introduction We consider the system (1.1) ˙x=−y+Xs(x, y), ˙y=x+Ys(x, y), where Xs(x, y) and Ys(x, y) are homogeneous polynomials of degree s, with s≥2. The aim of this paper is to find the integrable cases of system (1.1) when s= 4 (see Theorem 1). The integrable cases for quadratic systems, s= 2, and cubic homogeneous systems, s= 3, have been studied by several authors: Bautin [1], Chavarriga [2], Coppel [5], Frommer [6], Kapteyn [7], Lloyd [8], Lunkevich and Sibirskii [9], Schlomiuk [11] and ˙ Zoladek [15]. Poincar´e[10] developed an important technique for the general solution of these problems. It consists in finding a formal power series of the form (1.2) H(x, y)= ∞ n=2 Hn(x, y), *Research partially supported by a University of Lleida Project/94.
22 J. Chavarriga, J. Gin´ e where H2(x, y)=(x2+y2) 2, and for each n,Hn(x, y) is a homogeneous polynomial of degree n, so that the derivative of Halong the solutions of system (1.1) satisfies ˙ H= ∞ k=2 V2k(x2+y2)k, where V2kare real numbers called Lyapunov constants. The vanishing of all Lyapunov constants is a necessary condition for the integrability of the system (1.1); in this case the series H(x, y) would be a first integral of the system if it converges. It is not known if the series converges when the Lyapunov constants vanish. On the other hand, it is not possible in general to express this first integral (if it exists) by means of elementary functions. Lyapunov constants are polynomials in the variables given by the coefficients of the polynomials Xs(x, y) and Ys(x, y). Thus the ideal generated by V2khas a finite number of generators by Hilbert’s Theorem. We denote by M(s) the minimum number of generators of such an ideal. It is well known that the number of small amplitudes limit cycles around the origin is at least M(s) (see for instance [12]). The vanishing of these generators is a sufficient condition for the vanishing of all Lyapunov constants and for the integrability of the system. In Section 2 we give without proof some known results which are used in the proof of Theorem 1. In Section 3 we prove Theorem 1. This theorem characterizes the integrable cases by means of polar coordinates for s= 4. Finally, we give an appendix on the computation of the Lyapunov constants for s=4. 2. Some preliminary results In the study of this problem we use polar coordinates. In Lemma 1 we give the expression of system (1.1) in polar coordinates. In Proposition 1 we give the expression of series (1.2) in these coordinates. Lemma 1. In polar coordinates x=rcos(ϕ),y=rsin(ϕ)we can write system (1.1) as (2.1) ˙r=Ps(ϕ)rs, ˙ϕ=1+Qs(ϕ)rs−1,
Integrability of a perturbed linear center 23 where Ps(ϕ)and Qs(ϕ)are trigonometric polynomials of the form Ps(ϕ)=Rs+1 cos ((s+1)ϕ+ϕs+1)+Rs−1cos ((s−1)ϕ+ϕs−1) +···+R1cos(ϕ+ϕ1)if sis even; R0if sis odd; Qs(ϕ)=−Rs+1 sin ((s+1)ϕ+ϕs+1)+rs−1sin (s−1)ϕ+ϕs−1 +···+r1sin(ϕ+ϕ1)if sis even; r0if sis odd; being Rs,rs,ϕs,ϕsarbitrary coefficients. Proposition 1. In polar coordinates series (1.2) for system (1.1) is H(r, ϕ)= ∞ m=0 Hm(ϕ)rm(s−1)+2 where H0(ϕ)=1 2and Hm(ϕ), m=0,1,..., are homogeneous trigonometric polynomials of degree m(s−1)+2, satisfying the differential equations (2.2) dHm+1 dϕ +(m(s−1)+2)HmPs(ϕ)+dHm dϕ Qs(ϕ) =0if (m+ 1)(s−1)+2is odd, V(m+1)(s−1)+2 if (m+ 1)(s−1)+2is even, where V(m+1)(s−1)+2,m=0,1,..., are the Lyapunov constants. Lemma 1 and Proposition 1 are proved in [2]. In particular, for s= 4 system (1.1) takes the form (2.3) ˙r=P4(ϕ)r4, ˙ϕ=1+Q4(ϕ)r3, where (2.4) P4(ϕ)=R5cos(5ϕ+ϕ5)+R3cos(3ϕ+ϕ3)+R1cos(ϕ+ϕ1), Q4(ϕ)=−R5sin(5ϕ+ϕ5)+r3sin(3ϕ+ϕ3)+r1sin(ϕ+ϕ1). In this case the evaluation of ˙ H(r, ϕ) from system (2.3) yields dHm+1 dϕ +(3m+2) HmP4(ϕ)+dHm dϕ Q4(ϕ)=Vm+1 =0if3(m+1)+2isodd; V3(m+1)+2 if 3(m+ 1) + 2 is even; for m=0,1,..., with H0(ϕ)=1 2and Hm(ϕ)=H3m+2(ϕ). In Proposition 2 we will give the general form of the symmetric integrable systems. In Proposition 3 we will give a class of integrable systems which have an integrant factor given by a quadratic polynomial in the variable rs−1whose coefficients are functions in ϕ.
24 J. Chavarriga, J. Gin´ e Proposition 2. In the following two cases system (2.1) is integrable (in the sense that all its Lyapunov constants vanish): (i) (s+1)Ps(ϕ)+dQs(ϕ) dϕ =0. (ii) Ps(ϕ)and Qs(ϕ)are of the form Ps(ϕ)=Rs+1 sin(s+1)ω+Rs−1sin(s−1)ω +···+R1sin ωif sis even; R2sin 2ωif sis odd; Qs(ϕ)=Rs+1 cos(s+1)ω+rs−1cos(s−1)ω +···+r1cos ωif sis even; r2cos 2ω+r0if sis odd; where ω=ϕ+ϕ0and ϕ0and the coefficients Rjand rjare arbitrary. Systems satisfaying (i) have null divergence and those satisfying (ii) have a certain resonance between the angular parameters ϕj,ϕj. Proposition 3. For s∈Nwith s≥2and arbitrary k1,k2,ϕ0∈R system (2.1) with (2.5) Ps(ϕ)=2−k1coss−2(ϕ+ϕ0) sin3(ϕ+ϕ0) +k2sins−2(ϕ+ϕ0) cos3(ϕ+ϕ0), Qs(ϕ)=k1coss−1(ϕ+ϕ0)−k2sins−1(ϕ+ϕ0)cos 2(ϕ+ϕ0), is integrable. In cartesian coordinates x=rcos(ϕ+ϕ0) and y=rsin(ϕ+ϕ0)we can write system (2.5) in the form ˙x=−y−k1xs−1y+k2ys−2(2x2−y2),(2.6) ˙y=x+k1xs−2(x2−2y2)+k2xys−1, with s≥2. We note that the origin is a center for system (2.4). Proposition 2 is proved in [2] and Proposition 3 is proved in [3]. In order to establish the cases of Theorem 1, we have used a certain simplification expressed in the form of a conjecture. This conjecture simplifies the great number of factors that appear in the different Lyapunov constants. This assumption, which we think to be always satisfied, is stated as Conjecture 1.
Integrability of a perturbed linear center 25 Conjecture 1. A necessary condition for all the Lyapunov constants of system (2.3) are zero is that the angular parameters (ϕ5,ϕ 3, ϕ3,ϕ 1, ϕ1) and the radial ones (R5,R 3,r 3,R 1,r 1)must be independent. With reference to the number of small amplitude limit cycles around the origin that appear when we perturbed system (2.3) inside the same class of systems, we note that in the third integrable case of Theorem 1 (Case 10 in its proof) the number of relations of parameters is seven. So if Conjecture 1 is true, we think that the number of small amplitude limit cycles is at least seven. The Lyapunov constants were obtained using the computer algebra system Mathematica. 3. The main result Theorem 1. System (2.3) is integrable in the following cases: (i) ϕ1=ϕ1,ϕ3=ϕ3,5R3+3r3=0and 5R1+r1=0. (ii) ϕ1=ϕ1,ϕ3=ϕ3,ϕ5=5ϕ1and ϕ3=3ϕ1. (iii) ϕ1=ϕ1,ϕ3=ϕ3,ϕ5=2ϕ1+ϕ3,r3=3R3,r1=2R1,R5=R3, and |R1|=2|R3|. (iv) ϕ1=ϕ1,ϕ3=ϕ3,R5=0and (iv.1) R3r1−3r3R1=0, (iv.2) r1=3R1and r3=−3R3, (iv.3) r1=R1=|3R3|and r3=−3R3. This theorem is independent on Conjecture 1, but if Conjecture 1 is true then system (2.3) will be integrable only in the cases given by Theorem 1. Proof of Theorem 1: The first not zero Lyapunov constant is V8=−1 2(R1r1sin(ϕ1−ϕ1)+R3r3sin(ϕ3−ϕ3)) . In particular the previous constant vanishes when ϕ1=ϕ1and ϕ3= ϕ3. On the other hand, if Conjecture 1 is true, we arrive at the same condition as above. With this assumption the next not zero Lyapunov constant is V14 given by 40V14 =5((R3r1−3r3R1)(r1−3R1)(r1−R1)) sin(3ϕ1−ϕ3) +3(5r2 1−6r1R1−11R2 1)r3+ (11r2 1−74r1R1+75R2 1)R3 R5sin(2ϕ1+ϕ3−ϕ5) +(15r1−51R1)r2 3+ 22(r1−R1)r3R3−(21r1−25R1)R2 3 R5sin(ϕ1−2ϕ3+ϕ5).
26 J. Chavarriga, J. Gin´ e The simultaneous vanishing of three factors of V14 respect to the radials parameters with arbitrary angular parameters leads to the following cases: 1. 5R1+r1= 0 and 5R3+3r3=0; 2. R1=r1=0; 3. R3=r3=0; 4. R5=0,R3r1−3r3R1=0,R2 1+r2 1= 0 and R2 3+r2 3=0; 5. R5= 0 and r1−3R1=0; 6. R5= 0 and r1−R1=0. If we impose that the values of the angular parameters are not arbitrary, the possible dependence relations between them are the following: 7. ϕ3−3ϕ1= 0 and ϕ5−5ϕ1=0; 8. ϕ3−3ϕ1=0; 9. ϕ1−2ϕ3+ϕ5=0; 10. 2ϕ1+ϕ3−ϕ5=0; 11. ϕ5−5ϕ1=0; 12. ϕ5−3ϕ3+4ϕ1=0. Case 1: In this case it is easy to see that 5P4+Q 4= 0, where =d dϕ , and the divergence of the vector field defined by system (2.3) is zero. So the system is integrable and all Lyapunov constants are zero (see Proposition 2). Case 2: If r1=R1= 0 then the first not zero Lyapunov constant after V14 is V26, that is 8400V26 =((5R3+3r3)(3R3−r3)(3R3−5r3)(R3−3r3)(29R3−89r3)) R3 5sin(5ϕ3−3ϕ5). The vanishing of the first factor corresponds to Case 1. The situation when R3= 0 and r3= 0 corresponds to a degenerate case of zero divergence. The vanishing of the rest of the factors of V26 allows to express r3in function of R3. If we introduce these relations in the next non-zero constant V32 (see Appendix 2), we can find R5as a function of R3. Finally by substituting them in the next non-zero Lyapunov constant V38 (see Appendix 2), we can see that this constant never vanishes. Therefore there is no possible integrable cases. The vanishing of R5will be seen later. The vanishing of sin(5ϕ3−3ϕ5) corresponds to Case 7. Case 3: If r3=R3= 0 then the first non-zero Lyapunov constant after V14 is V20, that is 320V20 =((r1+5R1)(5r1−3R1)(2r1−3R1)(r1−3R1)(r1−6R1)) R5sin(5ϕ1−ϕ5).
Integrability of a perturbed linear center 27 The vanishing of the first factor corresponds to Case 1. The situation when R1= 0 and r1= 0 corresponds to a degenerate case of zero divergence. The vanishing of the rest of the factors of V20 allows to express r1in function of R1. If we introduce these relations in the next non-zero constant V26 (see Appendix 3), we can find R5as a function of R1. Finally by substituting them in the next non-zero Lyapunov constant V32 (see Appendix 3), we can see that this constant does not vanish. Therefore there is no possible integrable cases. The vanishing of R5will be seen later. The vanishing of sin(5ϕ1−ϕ5) corresponds to Case 7. We now consider the case R5= 0. Then V14 =1 8(R3r1−3r3R1)(r1−3R1)(r1−R1) sin(3ϕ1−ϕ3). If we impose that V14 = 0 we have four possibilities. The vanishing of sin(3ϕ1−ϕ3) corresponds to Case 7. The others correspond to Cases 4, 5 and 6 which we will be studied. Case 4: In this case R5= 0 and R3r1−3r3R1=0. Ifr1= 0 and r3= 0, then P4(ϕ)=λQ 4(ϕ) where λ=R1/r1=R3/3r3. System (2.3) takes the form ˙r=λr4Q 4(ϕ), ˙ϕ=1+r3Q(ϕ). The previous system is integrable and its first integral is H(r, ϕ)=1+(1+3λ)Q(ϕ)r3r−(3+ 1 λ). If r1=0orr3= 0 implies ˙ϕ= 1 and system (2.3) is trivially integrable. Case 5: In this case R5= 0 and r1−3R1= 0. We can suppose r1=3R1= 0, because r1=R1= 0 corresponds to a particular case of Case 7. Then 64V20 = 27(R3−r3)2(r3+3R3)R3 1sin(3ϕ1−ϕ3), 17920V26 =35283r2 3−4932r3R3−28943R2 3−245232R2 1 (R3−r3)2(r3+3R3)R3 1sin(3ϕ1−ϕ3). The simultaneous vanishing of the two previous Lyapunov constants give rise to two possible cases either r3+3R3=0orr3−R3= 0. The last one corresponds to a particular case of R3r1−3r3R1= 0 which has been
28 J. Chavarriga, J. Gin´ e studied in Case 4. In the first case, that is r1=3R1and r3=−3R3, system (2.3) reduces to ˙r=r4(R3cos(3ϕ+ϕ3)+R1cos(ϕ+ϕ1)) , ˙ϕ=1+r3(−3R3sin(3ϕ+ϕ3)+3R1sin(ϕ+ϕ1)) . If we make the change R=r3the system takes the form ˙ R=3R2(R3cos(3ϕ+ϕ3)+R1cos(ϕ+ϕ1)) , ˙ϕ=1+3R(−R3sin(3ϕ+ϕ3)+R1sin(ϕ+ϕ1)) , which corresponds to an integrable case of a quadratic system with linear part of center type, see for instance [2]. Case 6: In this case R5= 0 and r1−R1= 0. We can suppose r1= R1= 0, because r1=R1= 0 corresponds to a particular case of Case 7. Then 64V20 =(R3−3r3)(−20r2 1+17r2 3−10R3r3−3R2 3) sin(3ϕ1−ϕ3), 23040V26 =799200r4 1+ (2238104R2 3+ 421824r3R3−699048r2 3)r2 1 +18603r4 3−24450r3 3R3−66784r2 3R2 3+44402r3R3 3+13029R4 3 (R3−3r3)r3 1sin(3ϕ1−ϕ3). If R3=3r3the above constants are zero and Case 6 corresponds to a particular case of R3r1−3r3R1= 0 studied in Case 4. If the second factor of V20 is zero, we have (3.1) 20r2 1−17r2 3+10R3r3+3R2 3=0. By substituting the expression obtained for r1from (3.1) in V26 we get 38400V26 =−(r3+3R3)(3r3−R3)2(1019r2 3−326r3R3−273R2 3) r3 1sin(3ϕ1−ϕ3). From the vanishing of the previous constant we obtain three possible cases: R3=−3r3(already studied), 1019r2 3−326r3R3−273R2 3= 0 and r3=−3R3. Let 1019r2 3−326r3R3−273R2 3= 0. From (3.1) we obtain 6520r2 1+ 4648r2 3+ 1752R2 3= 0, which implies r1=r3=R3= 0. Finally let r3=−3R3. By replacing the value of r3in (3.1) we obtain r2 1=9R2 3. So r1=3|R3|, and system (2.3) reduces to ˙r=r4(R3cos(3ϕ+ϕ3)+3|R3|cos(ϕ+ϕ1)) , ˙ϕ=1+r3(−3R3sin(3ϕ+ϕ3)+3|R3|sin(ϕ+ϕ1)) .
Integrability of a perturbed linear center 29 If we make the same change R=r3, then we can write the system as ˙ R=3R2(R3cos(3ϕ+ϕ3)+3|R3|cos(ϕ+ϕ1)) , ˙ϕ=1+3R(−R3sin(3ϕ+ϕ3)+|R3|sin(ϕ+ϕ1)) , which corresponds to an integrable case of a quadratic system with linear part of center type, see for instance [2]. Case 7: We call this case resonance angles. Then system (2.3) is integrable (see Proposition 2). Case 8: Let ϕ3−3ϕ1= 0. We also assume that 5ϕ1−ϕ5= 0 (the opposite case has been studied in Case 7). Then a term that appears in V26 and which must be zero independently on the rest is 1 8400((5R3+3r3)(3R3−r3)(3R3−5r3)(R3−3r3)(29R3−89r3)) R3 5sin(15ϕ1−3ϕ5). From the vanishing of the factors of the previous expression and their substitution in the constants V14,V20 and the remaining terms of V26, see Appendix 1, we obtain cases already studied. Case 9: Let ϕ1−2ϕ3+ϕ5= 0. We also assume 3ϕ1−ϕ3= 0 (the opposite case has been studied in Case 7). Then a term that appears in V20 and which must be zero independently on the rest is 1 320 ((r1+5R1)(5r1−3R1)(2r1−3R1)(r1−3R1)(r1−6R1)) R5sin(6ϕ1−2ϕ3). From the vanishing of the factors of the previous expression and their substitution into the constants V14,V26 and the remaining terms of V20, see Appendix 1, we obtain cases already studied. Case 10: Let 2ϕ1+ϕ3−ϕ5= 0. Then 40V14 =(5(R3r1−3r3R1)(r1−3R1)(r1−R1)) +(15r1−51R1)r2 3+ 22(r1−R1)r3R3−(21r1−25R1)R2 3R5 sin(3ϕ1−ϕ3).
36 J. Chavarriga, J. Gin´ e −572024722500R4 1r3R2 3+ 20500037550r2 1r3 3R2 3 −49479662700r1R1r3 3R2 3−19379780550R2 1r3 3R2 3+ 14229762525r4 1R3 3 −244978014600r3 1R1R3 3+ 632738742075r2 1R2 1R3 3 +1386330907500r1R3 1R3 3−2417530500000R4 1R3 3−7980247800r2 1r2 3R3 3 −70150101300r1R1r2 3R3 3+151108935300R2 1r2 3R3 3−53409883950r2 1r3R4 3 +127889399550r1R1r3R4 3+ 15480487500R2 1r3R4 3−24927916550r2 1R5 3 +211221387500r1R1R5 3−232617000000R2 1R5 3+ 1453308075r4 1r3R2 5 −42985052190r3 1R1r3R2 5+ 242789114880r2 1R2 1r3R2 5 −171680012730r1R3 1r3R2 5−404427233475R4 1r3R2 5 −1421290800r2 1r3 3R2 5+ 3652068060r1R1r3 3R2 5−3585857040R2 1r3 3R2 5 +1543409055r4 1R3R2 5−36386475210r3 1R1R3R2 5 +338814678060r2 1R2 1R3R2 5−1182414401550r1R3 1R3R2 5 +1446537268125R4 1R3R2 5−2616459570r2 1r2 3R3R2 5 +14166828720r1R1r2 3R3R2 5−26265773610R2 1r2 3R3R2 5 +12524885820r2 1r3R2 3R2 5−5917042980r1R1r3R2 3R2 5 −30720899700R2 1r3R2 3R2 5+ 9427256550r2 1R3 3R2 5 −70208058600r1R1R3 3R2 5+119035248750R2 1R3 3R2 5−1130948460r2 1r3R4 5 +1929611592r1R1r3R4 5+ 2916332532R2 1r3R4 5−804688884r2 1R3R4 5 + 6239285496r1R1R3R4 5−7028745300R2 1R3R4 5sin(2ϕ1+ϕ3−ϕ5) +2475 −104580r4 1r2 3−20131965r3 1R1r2 3+ 59842161r2 1R2 1r2 3 +6015393r1R3 1r2 3−86403753R4 1r2 3−1872885r4 1r3R3 −3885144r3 1R1r3R3+ 77226918r2 1R2 1r3R3−246205968r1R3 1r3R3 +273837735R4 1r3R3−1461101r4 1R2 3+ 14956237r3 1R1R2 3 −75810827r2 1R2 1R2 3+ 235432455r1R3 1R2 3 −201316500R4 1R2 3R2 5sin(4ϕ1+2ϕ3−2ϕ5). Appendix 2 Lyapunov constants V32 and V38 if ϕ1=ϕ1,ϕ3=ϕ3and R1=r1=0: 3386880000V32 =(3r3+5R3)R3 5−1459292625r6 3 +8614578420r5 3R3−3561439455r4 3R2 3−34418277320r3 3R3 3 +41410717145r2 3R4 3−16050420540r3R5 3+ 2034246375R6 3 −2170849005r4 3R2 5+ 6255604188r3 3R3R2 5−6059843478r2 3R2 3R2 5
Integrability of a perturbed linear center 37 + 2315564748r3R3 3R2 5−303704613R4 3R2 5sin(5ϕ3−3ϕ5). 1482030950400000V38 =(3r3+5R3)R3 5−1149145404553125r8 3 +8848194134872050r7 3R3+ 1185862616782200r6 3R2 3 −96387033484384650r5 3R3 3+ 58538665263406650r4 3R4 3 +332980469705167550r3 3R5 3−413020341318321800r2 3R6 3 +161016094351964250r3R7 3−20445404262373125R8 3 +2891909356127550r6 3R2 5−42487871210003790r5 3R3R2 5 +152493752666073780r4 3R2 3R2 5−176828521900465980r3 3R3 3R2 5 +96197776031891430r2 3R4 3R2 5−25839946288107990r3R5 3R2 5 +2766024379211400R6 3R2 5−3621290735420280r4 3R4 5 +14619453338344608r3 3R3R4 5−13917796242684048r2 3R2 3R4 5 + 5015027246100768r3R3 3R4 5−622727965831608R4 3R4 5sin(5ϕ3−3ϕ5). Appendix 3 Lyapunov constants V26 and V32 if ϕ1=ϕ1,ϕ3=ϕ3and R3=r3=0: 268800V26 =(r1+5R1)R5(10500r6 1−128940r5 1R1−430920r4 1R2 1 +9577260r3 1R3 1−34912080r2 1R4 1+ 43228080r1R5 1−15479100R6 1 +72515r4 1R2 5−703782r3 1R1R2 5+ 2395276r2 1R2 1R2 5 −3361098r1R3 1R2 5+ 1585089R4 1R2 5sin(5ϕ1−ϕ5). 2483712000V32 =(r1+5R1)R5(97020000r8 1−1217046600r7 1R1 −19884803400r6 1R2 1+ 277869853800r5 1R3 1+ 222079611600r4 1R4 1 −12326128630200r3 1R5 1+ 46086997987800r2 1R6 1 −57154152825000r1R7 1+ 20495280960000R8 1+ 1020766985r6 1R2 5 +1114175216r5 1R1R2 5−223447537033r4 1R2 1R2 5 +1715463721944r3 1R3 1R2 5−5101631600457r2 1R4 1R2 5 +6576435520200r1R5 1R2 5−2941989081255R6 1R2 5+ 1712647422R4 1R4 5 −17633067984r3 1R1R4 5+ 64359273276r2 1R2 1R4 5 −97959123360r1R3 1R4 5+ 53197740630R4 1R4 5sin(5ϕ1−ϕ5).
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Integrability of a perturbed linear center 39 14. K. S. Sibirskii, On the number of limit cycles in the neighborhood of a singular point, Differential Equations 1(1965), 36–47. 15. H. ˙ Zoladek, On certain generalization of the Bautin’s Theorem, Nonlinearity 7(1994), 233–279. 16. H. ˙ Zoladek, The solution of the center-focus problem, Preprint, Institute of Mathematics, University of Warsaw, 1992. Keywords. center-focus problem, integrable systems in the plane. 1991 Mathematics subject classifications: Primary 34A05; Secondary 34C05. Departament de Matem`atica Escola Universit`aria Polit`ecnica Universitat de Lleida Pla¸ca Victor Siurana 1 25003 Lleida SPAIN Primera versi´o rebuda el 16 de Desembre de 1994, darrera versi´o rebuda el 24 de Novembre de 1995