This is a preprint of: “Rational limit cycles on Abel equations”, Jaume Llibre, Clàudia Valls, Commun. Pure Appl. Anal., vol. 20(3), 2021. DOI: [10.3934/cpaa.2021007] RATIONAL LIMIT CYCLES ON ABEL EQUATIONS JAUME LLIBRE1 AND CLÀUDIA VALLS2 Abstract. In this paper we deal with Abel equations dy/dx = A(x)y 2 + B(x)y 3 , where A(x) and B(x) are real polynomials. We prove that these Abel equations can have at most three rational limit cycles and we characterize when this happens. Moreover, we provide examples of these Abel equations with three nontrivial rational limit cycles. We also prove that in this case the limit cycles cannot be hyperbolic. 1. Introduction and statement of the results We study the Abel equations dy (1) = A(x)y 2 + B(x)y 3 , dx where x, y are real variables and A(x) and B(x) are polynomials. The limit cycles of these equations have been intensively investigated mainly when the functions A(x) and B(x) are periodic (see for instance [1, 2, 3, 4, 5, 6, 7, 9, 12, 13, 15, 16, 17, 18, 19, 21, 22, 23, 24]), and also when A(x) and B(x) are polynomial (see for instance [8, 10, 11, 14, 20]). Here we are interested in the rational limit cycles of equation (1) when the functions A(x) and B(x) are polynomial. A periodic solution of equation (1) is a solution y(x) defined in the closed interval [0, 1] such that y(0) = y(1). We say that a limit cycle is a periodic solution isolated in the set of periodic solutions of a differential equation (1). Without loss of generality we will assume that the period is 1. The limit cycle is called a polynomial limit cycle if the periodic solution y(x) is a polynomial in the variable x. In particular the authors of [14] proved that any polynomial limit cycle of system (1) is of the form y = c with c ∈ R, and that if a polynomial limit cycle exists with c 6= 0, then no other polynomial limit cycles can exist. In this paper we want to consider the existence of rational limit cycles for system (1), i.e. we want to consider limit cycles of the form y(x) = q(x)/p(x) 2010 Mathematics Subject Classification. 34C05, 34C07, 34C08. Key words and phrases. Algebraic limit cycles, Rational limit cycles, Abel equations, Hyperbolic limit cycles. 1 2 J. LLIBRE AND C. VALLS where p, q ∈ R[x] and (p(x), q(x)) = 1. As usual R[x] denotes the set of all real polynomials in the variable x. We will study only the rational limit cycles that are not polynomial limit cycles. In this scenario we will distinguish between trivial limit cycles (the polynomial ones) and non-trivial limit cycles (the rational limit cycles that are not polynomials). In [17] the authors provide examples of differential equations (1) having one or two rational limit cycles, and that these limit cycles are hyperbolic. Our main theorem is the following one. Theorem 1. System (1) has at most three rational limit cycles, and when it has three rational limit cycles 1/yi (x) for i = 1, 2, 3, then there exist a polynomial S(x) and two different constants c1 , c2 ∈ R \ {0} such that S(0) = S(1) and S(x) 6= 0, S(x) + c1 6= 0 and S(x) + c2 6= 0 for x ∈ [0, 1]. Moreover (2) B(x) = S(x)S 0 (x)(S(x) + c1 )(S(x) + c2 ), and the three limit cycles and satisfy y1 (x) = S(x)(S(x) + c1 ), (3) y2 (x) = S(x)(S(x) + c2 ), y3 (x) = (S(x) + c1 )(S(x) + c2 ). We provide an example of system (1) with three rational limit cycles. The proof of Theorem 1 and the example are given in section 2. Denote by y(x, x0 ) the solution of equation (1) such that y(0, x0 ) = x0 . Clearly a zero of the function φ(x0 ) = y(1, x0 ) − x0 implies that y(x, x0 ) is a periodic solution of (1), and an isolated zero of φ(x0 ) implies that y(x, x0 ) is a limit cycle of system (1). When we have a simple isolated zero of φ(x0 ), i.e. φ(x0 ) = 0 and φ0 (x0 ) 6= 0, then we say that y(x, x0 ) is a hyperbolic limit cycle. Theorem 2. When system (1) has three limit cycles, they are not hyper- bolic. Theorem 2 is proved in section 3. 2. Proof of Theorem 1 It was proved in Lemma 2 of [17] the following result, but for completeness we provide it here. Lemma 3. The rational function y = q(x)/p(x) with p(x) non-constant is a periodic solution of system (1) if and only if q(x) = c ∈ R \ {0}, p(0) = p(1) and p(x) has no zero in [0, 1] and p(x)p0 (x) (4) cB(x) + + p(x)A(x) = 0. c RATIONAL LIMIT CYCLES 3 Proof. For the reverse implication, we note that if q(x) = c ∈ R \ {0}, p(0) = p(1), p(x) has no zero in [0, 1] and equality (4) holds then it is clear that the rational function y = c/p(x) is a periodic solution of system (1). For the direct implication, we note that if y(x) = q(x)/p(x) is a periodic solution of system (1) then p(x) 6= 0 for x ∈ [0, 1]. Let g(x, y) = p(x)y−q(x). Then dg(x, y) dy 0= |g(x,y)=0 = p0 (x)y + p(x) − q 0 (x) dx dx = p0 (x)y + p(x)(A(x)y 2 + B(x)y 3 ) − q 0 (x). Note that g(x, y) is irreducible, so there exists a polynomial k(x, y) so that (5) p0 (x)y + p(x)(A(x)y 2 + B(x)y 3 ) − q 0 (x) = k(x, y)g(x, y). Since the highest degree in y in the left-hand side is 3 and the highest degree in y in g(x, y) is 1 we get that the highest degree in y in k(x, y) is 2 and so it can be written as k(x, y) = k0 (x) + k1 (x)y + k2 (x)y 2 , where k0 , k1 , k2 ∈ R[x]. Comparing the coefficients of y 0 , y 1 , y 2 and y 3 in (5) we get q 0 (x) = k0 (x)q(x), p0 (x) = k0 (x)p(x) − k1 (x)q(x), (6) p(x)A(x) = k1 (x)p(x) − k2 (x)q(x), p(x)B(x) = k2 (x)p( x). From the first relation we get that q(x)|q 0 (x). This implies that q(x) is a constant that we denote by c, that is, q(x) = c ∈ R. If c = 0 then y = q(x)/p(x) = 0. This is not possible and so c 6= 0. Moreover, y = q(x)/p(x) = c/p(x) is a periodic solution, then p(0) = p(1). From the second relation we get that k1 (x) = −p0 (x)/c and from the fourth relation we obtain k2 (x) = B(x). Substituting them in the third relation we get (4) and the direct inclusion is proved.  In view of Lemma 3 it is not restrictive to take c = 1 and consider all rational limit cycles of the form y = 1/p(x) with p(x) satisfying p(0) = p(1) with p(x) having no zero in [0, 1] and satisfying (4). From (4) we must have that B(x) is multiple of p(x) and so B(x) = p(x)r(x) for some polynomial r(x). Therefore, (4) becomes Z 0 (7) r(x)+p (x)+A(x) = 0 and so p(x) = κ− (A(s)+r(s)) ds, κ ∈ R. Assume that equation (1) has two rational limit cycles, y(x) = 1/p1 (x) and y(x) = 1/p2 (x) with p1 (x), p2 (x) ∈ R[x] \ R. Denote by q(x) = (p1 (x), p2 (x)), i.e. the maximum common divisor of the polynomials p1 (x) and p2 (x), and consequently (8) p1 (x) = q(x)s1 (x), p2 (x) = q(x)s2 (x) 4 J. LLIBRE AND C. VALLS with q(x), si (x) ∈ R[x] and (s1 (x), s2 (x)) = 1. Note that in view of the above observation we must have that (9) B(x) = q(x)s1 (x)s2 (x)s3 (x) for some s3 (x) ∈ R[x]. Lemma 4. The following equalities hold (10) s3 (x) = q 0 (x) and s1 (x) − s2 (x) = c ∈ R. Proof. Note that in view of (7) we have Z q(x)s1 (x) = κ0 − (A(s) + s2 (s)s3 (s)) ds, (11) Z q(x)s2 (x) = κ1 − (A(s) + s1 (s)s3 (s)) ds, with κ0 , κ1 ∈ R. Hence, q 0 (x)s1 (x) + q(x)s01 (x) = −A(x) − s2 (x)s3 (x), q 0 (x)s2 (x) + q(x)s02 (x) = −A(x) − s1 (x)s3 (x), and so q 0 (x)s1 (x) + q(x)s01 (x) − q 0 (x)s2 (x) + q(x)s02 (x)) = −s2 (x)s3 (x) + s1 (x)s3 (x) = (s1 (x) − s2 (x))s3 (x), which gives q 0 (x)(s1 (x) − s2 (x)) + q(x)(s1 (x) − s2 (x))0 = (s1 (x) − s2 (x))s3 (x), that is q(x)(s1 (x) − s2 (x))0 = (s1 (x) − s2 (x))(s3 (x) − q 0 (x)). Hence (s1 (x) − s2 (x))0 s3 (x) q 0 (x) = − . (s1 (x) − s2 (x)) q(x) q(x) Therefore Z  1 s2 (s) (12) s1 (x) − s2 (x) = κ2 exp ds , q(x) q(s) for some κ2 ∈ R. Since s1 (x) − s2 (x) must be a polynomial we must have s3 (x) = κ3 q 0 (x) for some κ3 ∈ R. Indeed note that we must have Z s3 (s) (13) ds = κ3 log h(x), κ3 ∈ R, h(x) ∈ R[x] \ {0}. q(s) Let H(x) = (q(x), s3 (x)). Then s3 (x) H(x)s̄3 (x) s̄3 (x) h0 (x) = = = κ3 . q(x) H(x)q̄(x) q̄(x) h(x) RATIONAL LIMIT CYCLES 5 Therefore q̄(x) = h(x) and s̄3 (x) = κ3 q̄ 0 (x). So q(x) = H(x)h(x). From (12) and (13) we have   1 q(x) κ3 s1 (x) − s2 (x) = κ2 . q(x) H(x) Since s1 (x) − s2 (x) must be a polynomial and κ3 6= 0, it follows that H(x) can be one. Hence, (14) s3 (x) = κ3 q 0 (x) and s1 (x) − s2 (x) = κ2 q(x)κ3 −1 . On the other hand, doing a change of variables of the form Y = βy where β 2 = sign(κ3 )κ3 , the Abel equation (1) becomes dY A(x) 2 B(x) 3 (15) = Y + Y = A(x)Y 2 + B(x)Y 3 . dx β β2 Since B(x) = q(x)s1 (x)s2 (x)κ3 q 0 (x), then B(x) = ±q(x)s1 (x)s2 (x)q 0 (x). In what follows we shall work with the Abel equation (15). Repeating the previous computations starting with the Abel equation (15) we will arrive to equation (14) which now writes s3 (x) = q 0 (x) and s1 (x) − s2 (x) = κ2 , because κ3 = ±1 and only can be one. This concludes the proof of the lemma.  Note that from (8), (9) and Lemma 4 we have that (16) B(x) = q(x)q 0 (x)s1 (x)s2 (x). Proof of Theorem 1. Assume that equation (1) has three rational limit cy- cles, y = 1/p1 (x) and y = 1/p2 (x) and y3 = 1/p3 (x) with p1 , p2 , p3 ∈ R[x] \ R. Denote by q1 (x) = (p1 (x), p2 (x)), q2 (x) = (p1 (x), p3 (x)) and q3 (x) = (p2 (x), p3 (x)). In view of Lemma 4 we have p1 (x) = q1 (x)s1 (x) = q2 (x)s2 (x), (17) p2 (x) = q1 (x)(s1 (x) + c1 ) = q3 (x)s3 (x), p3 (x) = q2 (x)(s2 (x) + c2 ) = q3 (x)(s3 (x) + c3 ), for some polynomials s1 (x), s2 (x), s3 (x) and constants c1 , c2 , c3 ∈ R \ {0}. Hence, we get p2 (x) − p1 (x) = q1 (x)c1 , c1 ∈ R, p3 (x) − p1 (x) = q2 (x)c2 , c2 ∈ R, p3 (x) − p2 (x) = q3 (x)c3 , c3 ∈ R, and so (18) q2 (x)c2 = q1 (x)c1 + q3 (x)c3 . We consider two situations. 6 J. LLIBRE AND C. VALLS Case 1: q1 (x) and s2 (x) are coprime. Note that from (17) we have that q1 (x)s1 (x) = q2 (x)s2 (x), and then from (18) we get q1 (x)s1 (x)c2 = q2 (x)c2 = q1 (x)c1 + q3 (x)c3 . s2 (x) In particular there exists T (x) ∈ R[x] so that q3 (x) = q1 (x)T (x), and consequently s1 (x)c2 = c1 + T (x)c3 , s2 (x) which yields s2 (x) s1 (x) = (c1 + T (x)c3 ). c2 Therefore from (17) we get s2 (x) q2 (x)s2 (x) = q1 (x)s1 (x) = q1 (x) (c1 + T (x)c3 ), c2 and so c1 + T (x)c3 q2 (x) = q1 (x) . c2 Hence we have s2 (x) p1 (x) = q1 (x)s1 (x) = q1 (x) (c1 + T (x)c3 ), c2   s2 (x) (19) p2 (x) = q1 (x)(s1 (x) + c1 ) = q1 (x) (c1 + T (x)c3 ) + c1 , c2 c1 + T (x)c3 p3 (x) = q2 (x)(s2 (x) + c2 ) = q1 (x) (s2 (x) + c2 ). c2 We consider two subcases. Subcase 1.1: Assume that T (x) and s2 (x) + c2 are coprime. Then the maximum common divisor between p2 (x) and p3 (x) is q1 (x). Indeed, we will show that s2 (x) r1 (x) = (c1 + T (x)c3 ) + c1 c2 and r2 (x) = (c1 + T (x)c3 )(s2 (x) + c2 ) are coprime. Note that if x∗ is a zero of c1 + T (x)c3 then we have that r2 (x∗ ) = 0 but r1 (x∗ ) = c1 6= 0. Moreover, if x̂ is a solution of s2 (x) + c2 = 0 then r2 (x̂) = 0 but r1 (x̂) = −(c1 + T (x̂)c3 ) + c1 = T (x̂)c3 6= 0. Therefore, using p1 (x) and p2 (x) from (16) and (19) we can write   0 s2 (x) s2 (x) B(x) = q1 (x)q1 (x) (c1 + T (x)c3 ) (c1 + T (x)c3 ) + c1 , c2 c2 RATIONAL LIMIT CYCLES 7 and from p1 (x) and p3 (X) we can write   0 s2 (x) c1 + T (x)c3 B(x) = q1 (x)q1 (x) (c1 + T (x)c3 ) + c1 (s2 (x) + c2 ), c2 c2 and so s2 (x) = s2 (x) + c2 which is not possible because c2 6= 0. Subcase 1.2: Assume that T (x) and s2 (x) + c2 are not coprime. Write T (x) = α1 (x)α2 (x), s2 (x) + c2 = α1 (x)α3 (x), where α2 , α3 ∈ R[x] and α1 (x) ∈ R[x] \ R. Then c1 + T (x)c3 p3 (x) = q1 (x)α1 (x)α3 (x) , c2 α1 (x) p2 (x) = q1 (x) (c1 α3 (x) + s2 (x)α2 (x)c3 ). c2 We first note that the maximum common divisor between p2 (x) and p3 (x) is q1 (x)α1 (x). To do so, we will show that r3 (x) = α3 (x)(c1 + T (x)c3 ) and r4 (x) = c1 α3 (x) + s2 (x)α2 (x)c3 are coprime. If x∗ is a zero of α3 (x) then r3 (x∗ ) = 0 but r4 (x∗ ) = s2 (x∗ )α2 (x∗ )c3 = −c2 α2 (x∗ )c3 . Since α2 (x) and α3 (x) are coprime, we get that α2 (x∗ ) 6= 0, and then r4 (x∗ ) 6= 0. Moreover, if c1 + T (x̂)c3 = 0 then r4 (x̂) = c1 6= 0. So r3 (x) and r4 (x) are coprime. From p1 (x), p2 (x), (16) and (19) we get s2 (x) α1 (x) (20) B(x) = q1 (x)q10 (x) (c1 + T (x)c3 ) (c1 α3 (x) + s2 (x)α2 (x)c3 ). c2 c2 Note that from p2 (x), p3 (x), (16) and (19) we have (21) q1 (x) B(x) = α1 (x)(q1 (x)α1 (x))0 α3 (x)(c1 +T (x)c3 )(c1 α3 (x)+s2 (x)α2 (x)c3 ). c22 Comparing (20) with (21) we obtain α3 (x)(q1 (x)α1 (x))0 = q10 (x)(α1 (x)α3 (x) − c2 ), i.e. −c2 q10 (x) = −α3 (x)q1 (x)α10 (x), which is not possible unless either α3 (x) = 0 or α10 (x) = 0, but then q1 (x) would be constant, a contradiction. In short, Case 1 is not possible. Case 2: q1 (x) and s2 (x) are not coprime We write q1 (x) = R1 (x)R2 (x), s2 (x) = R1 (x)R3 (x) with R1 (x), R2 (x), R3 (x) ∈ R[x] and R1 (x) 6∈ R. 8 J. LLIBRE AND C. VALLS We consider two different subcases. Subcase 2.1: R3 (x) = R ∈ R. So s2 (x) = R1 (x)R and q1 (x) = R2 (x)s2 (x)/R. We also consider two cases 2.1.1: R2 (x) = R2 ∈ R. From (8) we have q1 (x)s1 (x) = q2 (x)s2 (x) and so q2 (x) = R2 s1 (x)/R.Then R2 p1 (x) = s1 (x)s2 (x), R R2 p2 (x) = s2 (x)(s1 (x) + c1 ), R R2 p3 (x) = s1 (x)(s2 (x) + c2 ). R From p1 (x), p2 (x), (16) and (19) we get  R 2 2 B(x) = s2 (x)s1 (x)(s1 (x) + c1 )s02 (x), R and from p1 (x), p3 (x), (16) and (19) we obtain  R 2 2 B(x) = s2 (x)s1 (x)s01 (x)(s2 (x) + c2 ), R and so s02 (x)(s1 (x) + c1 ) = s01 (x)(s2 (x) + c2 ), which yields s02 (x) s01 (x) = s2 (x) + c2 s1 (x) + c1 and integrating log(s2 (x) + c2 ) = κ + log(s1 (x) + c1 ), κ∈R and so s2 (x) + c2 = κ1 (s1 (x) + c1 ), κ1 = eκ ∈ R+ . Hence R2 p1 (x) = s1 (x)(κ1 (s1 (x) + c1 ) − c2 ), R R2 (22) p2 (x) = (κ1 (s1 (x) + c1 ) − c2 )(s1 (x) + c1 ), R R2 p3 (x) = s1 (x)κ1 (s1 (x) + c1 ), R and from (16) we obtain  R 2 2 B(x) = κ1 s1 (x)s01 (x)(s1 (x) + c1 )(κ1 (s1 (x) + c1 ) − c2 ). R Doing the rescaling Y = βy, we can assume that the constant (R2 /R)2 κ1 = 1 and the constants R2 /R and κ1 in the expression of pi (x) is one (see the proof of Lemma 4). Note that B(x) is as in the statement of the theorem as well as pi (x) for i = 1, 2, 3. RATIONAL LIMIT CYCLES 9 2.1.2: R2 (x) ∈ R[x] \ R. Since R3 (x) = R we have s2 (x) = R1 (x)R and q1 (x) = R2 (x)s2 (x)/R. From (18) we get R2 (x)s1 (x)c2 = R2 (x)s2 (x)c1 + q3 (x)c3 (x), R and so   R2 (x) c1 s2 (x) q3 (x) = s1 (x)c2 − . c3 R Since q1 (x)s1 (x) = q2 (x)s2 (x) we get q2 (x) = R2 (x)s1 (x)/R. In short R2 (x) p1 (x) = s1 (x)s2 (x), R R2 (x) p2 (x) = s2 (x)(s1 (x) + c1 ), R R2 (x) p3 (x) = s1 (x)(s2 (x) + c2 ). R We consider two cases: 2.1.2.1: s1 (x) and s2 (x) are coprime. In this case the maximum common divisor between p2 (x) and p3 (x) is R2 (x) and so from (16) we get R2 (x) B(x) = s2 (x)(R2 (x)s2 (x))0 s1 (x)(s1 (x) + c1 ) R2 R2 (x) = s2 (x)s1 (x)R20 (x)(s1 (x) + c1 )(s2 (x) + c2 ), R2 and so (R2 (x)s2 (x))0 = R20 (x)(s2 (x) + c2 ), that is R20 (x)s2 (x) + R2 (x)s02 (x) = R20 (x)s2 (x) + R20 (x)c2 , which yields R20 (x) s0 (x) = 2 . R2 (x) c2 Hence R2 (x) = ec2 s2 (x) which is not possible. 2.1.2.2: s1 (x) and s2 (x) are not coprime. In this case we write s1 (x) = κ(x)ŝ1 (x), s2 (x) = κ(x)ŝ2 (x), with κ(x), ŝ1 (x), ŝ2 (x) ∈ R[x] with κ(x) 6∈ R. Then R2 (x) 2 p1 (x) = κ (x)ŝ1 (x)ŝ2 (x), R R2 (x) p2 (x) = κ(x)ŝ2 (x)(κ(x)ŝ1 (x) + c1 ), R R2 (x) p3 (x) = κ(x)ŝ1 (x)(κ(x)ŝ2 (x) + c2 ). R 10 J. LLIBRE AND C. VALLS Then R2 (x) B(x) = κ(x)(R2 (x)κ(x))0 ŝ2 (x)(κ(x)ŝ1 (x) + c1 )ŝ1 (x)(κ(x)ŝ2 (x) + c2 ) R2 R2 (x) = κ(x)ŝ2 (x)(R2 (x)κ(x)ŝ2 (x))0 ŝ1 (x)(κ(x)ŝ1 (x) + c1 )κ(x), R2 and so (R2 (x)κ(x))0 (κ(x)ŝ2 (x) + c2 ) = κ(x)(R2 (x)κ(x)ŝ2 (x))0 , which yields (R2 (x)κ(x))0 c2 = (R2 (x)κ(x))κ(x)ŝ2 (x). This is not possible because the left hand side has less dimension than the right hand side. In summary, Subcase 2.1.2 is not possible. Subcase 2.2: R2 (x) ∈ R[x] \ R. We have q1 (x) = R1 (x)R2 (x) and s2 (x) = R1 (x)R3 (x). Then R2 (x)s1 (x)c2 = R1 (x)R2 (x)c1 + q3 (x)c3 . R3 (x) In particular there exists T (x) ∈ R[x] so that q3 (x) = R2 (x)T (x), and so s1 (x)c2 = R1 (x)c1 + T (x)c3 , R3 (x) which yields s1 (x) = R4 (x)R3 (x). Therefore, from p1 (x) in (8) we get q2 (x)s2 (x) = q1 (x)s1 (x) = R1 (x)R2 (x)R3 (x)R4 (x) = q2 (x)R1 (x)R3 (x) and so q2 (x) = R2 (x)R4 (x). Hence we have p1 (x) = q1 (x)s1 (x) = R1 (x)R2 (x)R3 (x)R4 (x), p2 (x) = q1 (x)(s1 (x) + c1 ) = R1 (x)R2 (x)(R3 (x)R4 (x) + c1 ), p3 (x) = q2 (x)(s2 (x) + c2 ) = R2 (x)R4 (x)(R1 (x)R3 (x) + c2 ). We consider two cases. 2.2.1: R1 (x) and R4 (x) are coprime. We have B(x) = R1 (x)R2 (x)(R1 (x)R2 (x))0 R3 (x)R4 (x)(R3 (x)R4 (x) + c1 ) = R2 (x)R20 (x)R1 (x)R4 (x)(R3 (x)R4 (x) + c1 )(R1 (x)R3 (x) + c2 ), and so (R1 (x)R2 (x))0 R3 (x) = R20 (x)(R1 (x)R3 (x) + c2 ), which yields R10 (x)R2 (x)R3 (x) = c2 R20 (x). This is not possible because the right hand side has less dimension than the left hand side. RATIONAL LIMIT CYCLES 11 2.2.2: R1 (x) and R4 (x) are not coprime. We write R1 (x) = R(x)R̂1 (x), R4 (x) = R(x)R̂4 (x) where R(x), R̂1 (x), R̂4 (x) ∈ R[x] with R(x) 6∈ R. Note that p1 (x) = q1 (x)s1 (x) = R2 (x)R̂1 (x)R2 (x)R3 (x)R̂4 (x), p2 (x) = q1 (x)(s1 (x) + c1 ) = R(x)R̂1 (x)R2 (x)(R3 (x)R(x)R̂4 (x) + c1 ), p3 (x) = q2 (x)(s2 (x) + c2 ) = R2 (x)R(x)R̂4 (x)(R(x)R̂1 (x)R3 (x) + c2 ). Then B(x) = (R(x)R̂1 (x)R2 (x))0 R(x)R̂1 (x)R2 (x)R(x)R3 (x)R̂4 (x)(R3 (x)R(x)R̂4 (x) + c1 ) = R(x)R2 (x)(R(x)R2 (x))0 R̂1 (x)R̂4 (x)(R3 (x)R4 (x) + c1 )(R(x)R̂1 (x)R3 (x) + c2 ), and so (R(x)R̂1 (x)R2 (x))0 R(x)R3 (x) = (R(x)R2 (x))0 (R(x)R̂1 (x)R3 (x) + c2 ), that is (R(x)R2 (x))0 R̂1 (x)R(x)R3 (x) + (R(x)R2 (x))R̂10 (x)R(x)R3 (x) = (R(x)R2 (x))0 (R(x)R̂1 (x)R3 (x) + c2 (R(x)R2 (x))0 , which yields R(x)R2 (x)(R̂1 (x))0 R(x)R3 (x) = c2 (R(x)R2 (x))0 . This is not possible because the right hand side has less dimension than the left hand side. So subcase 2.2 is not possible. In short from (22) there are at most three rational limit cycles and if they exist then there must exist a polynomial S(x) = s1 (x) and two different constants c1 , c2 ∈ R \ {0} such that from Lemma 3 we have S(0) = S(1), and S(x) 6= 0, S(x) + c1 6= 0 and S(x) + c2 6= 0 for x ∈ [0, 1], and B(x) is as in (2). Moreover the three limit cycles can be taken to be p1 (x) = S(x)(S(x) + c1 ), p2 (x) = (S(x) + c1 )(S(x) + c2 ), p3 (x) = S(x)(S(x) + c2 ), which are the ones given in (3) in the statement of the theorem. Note that yi (x) = 1/pi (x) for i = 1, 2, 3, satisfy (4) with c = 1 and that pi (0) = pi (1) and pi (x) 6= 0 for x ∈ [0, 1]. Hence, in view of Lemma 3, the three solutions y1 , y2 , y3 are three periodic solutions of (1). Moreover they are isolated and so they are limit cycles of (1). This concludes the proof of the theorem.  Now we provide an Abel equation (1) with three rational limit cycles. Indeed, taking S(x) = x2 − x + 1, c1 = 1 and c2 = 2 we construct the Abel 12 J. LLIBRE AND C. VALLS equation (1) with A(x) = −3(−1 + 2x) − 3(−1 + 2x)(1 − x + x2 ), B(x) = (−1 + 2x)(1 − x + x2 )(2 − x + x2 )(3 − x + x2 ). Then system (1) has the three rational solutions yi (x) = 1/pi (x) for i = 1, 2, 3 with p1 (x) = (x2 − x + 1)(x2 − x + 2), p2 (x) = (x2 − x + 2)(x2 − x + 3), p3 (x) = (x2 − x + 1)(x2 − x + 3). Note that S(0) = S(1) and S(x) 6= 0 for x ∈ [0, 1]. Moreover, S(x) + 1 = x2 − x + 2 6= 0 for x ∈ [0, 1], and S(x) + 2 = x2 − x + 3 6= 0 for x ∈ [0, 1]. The Abel system that we have constructed has three rational limit cycles. 3. Proof of Theorem 2 To decide whether a periodic solution is a hyperbolic limit cycle we need the following lemma, whose proof can be found in [17]. Lemma 5. If y = 1/p(x) is a periodic solution of system (1) then it is a R1 hyperbolic limit cycle if and only if 0 (B(x)/p2 (x)) dx 6= 0. Now we can prove Theorem 2. In view of Theorem 1 the three limit cycles are yi = 1/pi (x) with pi given in (3) for i = 1, 2, 3. We will prove first that y0 is not a hyperbolic limit cycle. In view of (3) since p1 (x) = S(x)(S(x) + c1 ), we have that B(x) S 0 (x)(S(x) + c2 ) = . p21 (x) S(x)(S(x) + c1 ) Note that we can write S 0 (x)(S(x) + c2 ) c2 S 0 (x) c1 − c2 S 0 (x) = + . S(x)(S(x) + c1 ) c1 S(x) c1 S(x) + c1 Hence Z 1 B(x) c2 1 c1 − c2 1 2 dx = log S(x) 0 + log(S(x) + c1 ) 0 = 0, 0 p1 (x) c1 c1 because S(0) = S(1). So the periodic solution y = 1/p1 (x) is not a hyper- bolic limit cycle. For y = 1/p2 (x) with p2 (x) = S(x)(S(x) + c2 ), we have B(x) S 0 (x)(S(x) + c1 ) = p22 (x) S(x)(S(x) + c2 ) and the proof is the same as for 1/p1 (x) interchanging the roles of c1 and c2 . RATIONAL LIMIT CYCLES 13 Finally, for the case of y = 1/p3 (x) since p3 (x) = (S(x) + c1 )(S(x) + c2 ) it follows that B(x) S 0 (x)S(x) 2 = . p2 (x) (S(x) + c1 )(S(x) + c2 ) Note that we can write S 0 (x)S(x) c1 S 0 (x) c2 S 0 (x) = − . 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