This is a preprint of: “Isochronicity for trivial quintic and septic planar polynomial Hamiltonian systems”, Francisco Braun, Jaume Llibre, Ana Cristina Mereu, Discrete Contin. Dyn. Syst., vol. 36(10), 5245–5255, 2016. DOI: [10.3934/dcds.2016029] ISOCHRONICITY FOR TRIVIAL QUINTIC AND SEPTIC PLANAR POLYNOMIAL HAMILTONIAN SYSTEMS FRANCISCO BRAUN, JAUME LLIBRE AND ANA C. MEREU Abstract. In this paper we completely characterize trivial isochronous centers of degrees 5 and 7. Precisely, we provide formulas, up to linear change of coordinates, ( for ) the Hamiltonian H of the isochronous centers such that H = f12 + f22 /2 has degrees 6 and 8, and f = (f1 , f2 ) : R2 → R2 is a polynomial map with det Df = 1 and f (0, 0) = (0, 0). 1. Introduction Let P (x, y) and Q(x, y) be real polynomials in the variables x and y. We say that a polynomial vector field X = (P, Q) has degree n when max{deg P, deg Q} = n. Given a polynomial Hamiltonian H : R2 → R of degree n + 1, the associated polynomial Hamiltonian system of degree n is (1) x˙ = −Hy (x, y), y˙ = Hx (x, y). System (1) has a center at (0, 0) if there is a neighbourhood of the origin filled of periodic orbits except the origin. The maximum connected set filled of periodic orbits having in its inner boundary the origin is called the period annulus of the center localized at the origin. If the period annulus is R2 \{(0, 0)}, we call the center global. We say that a polynomial Hamiltonian system has an isochronous center at the origin if (0, 0) is a center of (1) and all the orbits in the period annulus of the center have the same period. The following characterization of the polynomial Hamiltonian systems possessing an isochronous center at the origin was given in [6]. The polyno- mial Hamiltonian system (1) has an isochronous center of period 2π at the origin if and only if f1 (x, y)2 + f2 (x, y)2 (2) H(x, y) = , 2 for all (x, y) in a neighborhood N0 of the origin, where f = (f1 , f2 ) : N0 → R2 is an analytic map with Jacobian determinant det Df constant and equal to 1, and f (0, 0) = (0, 0). We observe that this characterization still holds for analytic Hamiltonians. When f can be taken polynomial, we say that 2010 Mathematics Subject Classification. Primary 34A34 ; Secondary: 34C25, 37C37, 14R15. Key words and phrases. Isochronous centers, polynomial Hamiltonian systems, Jaco- bian conjecture. 1 2 F. BRAUN, J. LLIBRE AND A. C. MEREU the polynomial Hamiltonian isochronous center is trivial. In this case, it is clear that f will be defined in all R2 . From [8], when the center is trivial, it is a global center if and only if f is globally injective. Thus the problem of knowing whether a trivial polynomial Hamiltonian isochronous center is global or not is equivalent to the Jacobian conjecture in R2 , which stands that a polynomial map f : R2 → R2 with constant Jacobian determinant is globally injective. We mention here that if the degrees of f1 and f2 are less than or equal to 101 then f is globally injective (see [7] and, for other results on the Jacobian conjecture, see [3]). Thus all the trivial polynomial Hamiltonian isochronous centers of degree less than or equal to 201 are global ones. In [2] it was proved that all isochronous centers of cubic polynomial Hamil- tonian systems are trivial (and hence global) and after a linear change of coordinates the Hamiltonian can be written as ( )2 H(x, y) = (k1 x)2 + k2 y + k3 x + k4 x2 , where k1 , k2 , k3 , k4 ∈ R and k1 k2 ̸= 0. We also mention that there are no polynomial Hamiltonian isochronous centers of degree 4, see [4]. Moreover, it was proved in [5] that there are no polynomial Hamiltonian isochronous centers of even degree for which the analytical function f of the Hamiltonian (2) is defined in the whole plane. In particular, there are no trivial polyno- mial Hamiltonian isochronous centers of even degree. On the other hand, there are examples of non-trivial polynomial Hamiltonian isochronous cen- ters of degree 6k + 1 for all k ≥ 1, see section 3. We point out that in these examples the map f is defined in the whole plane. The following are thus natural questions. Open question 1: Are there non-trivial quintic polynomial Hamiltonian isochronous centers? Open question 2: Are there non-trivial polynomial Hamiltonian isochro- nous centers with Hamiltonian (2) such that f is not analytical in the whole plane R2 ? We observe that if the open question 2 has a negative answer, then by [5] there are no polynomial Hamiltonian isochronous centers of even degree. Our main result is the characterization of the quintic and the septic trivial polynomial Hamiltonian isochronous centers, see theorems 4 and 5 respec- tively, where we provide formulas for the Hamiltonian of these systems. We also give an alternative formula for the Hamiltonian of the cubic polyno- mial isochronous centers (with a trivial proof), using that these centers are trivial, see Proposition 3. 2. Trivial polynomial Hamiltonian isochronous centers We shall use the following technical result. Lemma 1. Let p, q : R2 → R be homogeneous polynomials of degree m and n respectively such that det D(p, q) ≡ 0. Let also d = gcd(m, n), we define ISOCHRONICITY FOR QUINTIC AND SEPTIC HAMILTONIAN SYSTEMS 3 m′ = m/d and n′ = n/d. Then there exists a homogeneous polynomial ′ r : R2 → R of degree d and constants cp , cq ∈ R such that p = cp rm and ′ q = cq rn . Proof. It is enough to prove that the rational function f = pn /q m is constant. In order to do that, it is enough to show that fx = fy = 0. We have pn−1 q m−1 pn−1 q m−1 fx = (nqp x − mpq x ) = y (px qy − qx py ) = 0, q 2m q 2m where in the second equality above we used the Euler’s Theorem for homo- geneous maps. The proof that fy = 0 is analogous.  We address the reader to [1] for a much more general version of Lemma 1. In the proofs of the results of this section, we will have to solve partial differential equations of the form px + βpy = h, where p, h : R2 → R are homogeneous polynomials of degrees k and k − 1, respectively, and β ∈ R. By defining q : R2 → R by (3) p(x, y) = q(x, y − βx) = q(x1 , y1 ), the original equation turns to qx1 = h(x1 , y1 + βx1 ), ∫x and hence there is c ∈ R such that q(x1 , y1 ) = 0 1 h(s, y1 + βs)ds + cy1k , and p is given by (3). For further references, we enunciate this procedure in the following result. Lemma 2. Let p, h : R2 → R be homogeneous polynomials of degrees k and k − 1, respectively, satisfying px + βpy = h, with β ∫∈ R. Then there is c ∈ R x such that p(x, y) = q(x, y − βx) where q(x1 , y1 ) = 0 1 h(s, y1 + βs)ds + cy1k . We begin with Proposition 3, where we give an alternative characteriza- tion of the cubic polynomial Hamiltonian isochronous centers using the fact that they are trivial (recall the mentioned result of [2]). Right after the proof, we relate the formula of Proposition 3 and the one presented in [2]. Proposition 3. Assume that the polynomial Hamiltonian system (1) has degree 3 and has an isochronous center of period 2π at the origin. Then up to a linear change of variables, the Hamiltonian can be written as P 2 + (y + λP )2 H= , 2 with P = x + c2 y 2 , c2 , λ ∈ R and c2 ̸= 0. 4 F. BRAUN, J. LLIBRE AND A. C. MEREU Proof. Since the cubic polynomial Hamiltonian isochronous centers are triv- ial, there exists f = (f1 , f2 ) : R2 → R2 a polynomial ( map ) of degree 2 with det Df = 1 and f (0, 0) = (0, 0) such that H = f12 + f22 /2. After a linear change of variables, it is clear that we can write f1 = x + p2 , f 2 = y + q2 , with p2 and q2 homogeneous polynomials of degree 2. Without loss of gen- erality we can assume that p2 ̸= 0 (otherwise we change the roles of f1 and f2 and of x and y). The hypothesis det Df = 1 also gives that (4) p2 x + q2 y = 0, p2 x q2 y − q2 x p2 y = 0. The second equation of (4) gives from Lemma 1 that there exists λ ̸= 0 such that q2 = λp2 . Substituting this in the first equation of (4) we obtain p2 x + λp2 y = 0, which solved for a homogeneous polynomial of degree 2 (by Lemma 2) determines c2 ∈ R such that p2 = c2 (y − λx)2 , with c2 ̸= 0. We then apply the change of variables (x, y) 7→ (x, y − λx), finishing the proof.  √ √ √ √ We observe √ that changing √ x to√y and taking √ 2k1 = 1/ 1 + λ2 , 2k2 = 1 + λ , 2k3 = λ/ 1 + λ and 2k4 = c2 1 + λ2 , the formula of Proposi- 2 2 tion 3 satisfies the mentioned formula for the cubic polynomial Hamiltonian isochronous√ centers of [2]. On the other hand, the change of coordinates (x, y) 7→ ( 2 (k2 y + k3 x, k1)x) transforms the mentioned √ formula of [2] in H(x, y) = x + c2 y 2 )2 + y 2 /2, with c2 = k4 /( 2k1 ), which is the formula of Proposition 3 with λ = 0. We observe that in [2] it was not assumed that the isochronous center has period exactly 2π. For trivial polynomial Hamiltonian isochronous centers of degrees 5 and 7 we have similar formulas to the Hamiltonians, see the following theorems 4 and 5. Theorem 4. Assume that the polynomial Hamiltonian system (1) has degree 5 and has a trivial isochronous center of period 2π at the origin. Then up to a linear change of variables, the Hamiltonian can be written as P 2 + (y + λP )2 H= , 2 with P = x + c2 y 2 + c3 y 3 , c2 , c3 , λ ∈ R and c3 ̸= 0. ( ) Proof. We consider H = f12 + f22 /2, with f = (f1 , f2 ) a polynomial map of degree 3 satisfying f (0, 0) = (0, 0) and det Df (x, y) = 1 for all (x, y) ∈ R2 . It is clear that after a linear change of variables we can write f1 = x + p2 + p3 , f 2 = y + q2 + q3 , ISOCHRONICITY FOR QUINTIC AND SEPTIC HAMILTONIAN SYSTEMS 5 where pi and qi are homogeneous polynomials of degree i, i = 1, 2. Without loss of generality we can suppose that p3 ̸= 0. The assumption det Df = 1 gives p 2 x + q2 y = 0, p 3 x + q3 y + p 2 x q2 y − q2 x p 2 y = 0, (5) p 3 x q2 y − q2 x p 3 y + p 2 x q3 y − q3 x p 2 y = 0, p 3 x q3 y − q3 x p 3 y = 0. The last equation gives by Lemma 1 that there exists λ ∈ R such that (6) q3 = λp3 . Substituting this in the third equation in (5), we obtain (7) p3 x (q2 − λp2 )y − (q2 − λp2 )x p3 y = 0. Here we have two possibilities: either q2 ̸= λp2 or q2 = λp2 . In the first one, equation (7) gives by Lemma 1 that there exist a, b, c2 , c3 ∈ R, with c2 c3 (a2 + b2 ) ̸= 0 such that (8) q2 = λp2 + c2 (ax + by)2 , p3 = c3 (ax + by)3 . Using then the first equation of (8) and the first one of (5), we obtain p2 x + λp2 y = −2bc2 (ax + by) . From Lemma 2, we obtain d2 ∈ R such that p2 = −bc2 ((a + bλ)x + 2b(y − λx)) x + d2 (y − λx)2 . Then we substitute p2 , (8) and (6) in the second equation of (5) and, after dividing by ax + by and equating the coefficients of x and y to 0, we obtain the system ( )( ) ( ) a λ ( 3 (a + bλ) c3 ) 0 = . b −1 4 b3 c2 + (a + bλ)d2 c2 0 If a + bλ = 0, it follows that b = 0 (since c2 ̸= 0), and hence a = 0, a contradiction. On the other hand if a + bλ ̸= 0, it follows that c3 = 0, which is again a contradiction. If now q2 = λp2 , it follows from (6) and from the first and the second equations in (5) that p2 x + λp2 y = 0, p3 x + λq3 y = 0. Solving these equations for homogeneous polynomials of degrees 2 and 3 (see Lemma 2), respectively, we obtain c2 , c3 ∈ R such that p2 = c2 (y − λx)2 , p3 = c3 (y − λx)3 , with c3 ̸= 0. So f1 = x + c2 (y − λx)2 + c3 (y − λx)3 , ( ) f2 = y + λ c2 (y − λx)2 + c3 (y − λx)3 . 6 F. BRAUN, J. LLIBRE AND A. C. MEREU Applying the linear change of variables (x, y) 7→ (x, y − λx), we end the proof of the theorem.  The following is a characterization of trivial polynomial Hamiltonian iso- chronous centers of degree 7. Theorem 5. Assume that the polynomial Hamiltonian system (1) has degree 7 and has a trivial isochronous center of period 2π at the origin. Then up to a linear change of variables the Hamiltonian H has one of the following two forms: P12 + (y + λP1 )2 H= , P1 =x + c2 y 2 + c3 y 3 + c4 y 4 , 2 P 2 + (Q + λP2 )2 H= 2 , P2 =x + β1 Q + β2 Q2 , 2 where Q = y + Γx2 , c2 , c3 , c4 , β1 , β2 , λ, Γ ∈ R and c4 β2 Γ ̸= 0. Proof. Let f1 and f2 be polynomials of degree 4 such that det D(f1 , f2 ) = 1 and f1 (0, 0) = f2 (0, 0) = 0. After a linear change of variables, it is clear we can write (9) f1 = x + p2 + p3 + p4 , f 2 = y + q2 + q3 + q4 , where pi and qi are homogeneous polynomials of degree i for i = 2, 3, 4. Moreover, since the homogeneous terms of positive degrees of the Jacobian determinant of (f1 , f2 ) are zero, we obtain the following equations: (10) p2 x + q2 y = 0, (11) p3 x + q3 y + p2 x q2 y − q2 x p2 y = 0, (12) p4 x + q4 y + p2 x q3 y − q3 x p2 y + p3 x q2 y − q2 x p3 y = 0, (13) p2 x q4 y − q4 x p2 y + p4 x q2 y − q2 x p4 y + p3 x q3 y − q3 x p3 y = 0, (14) p3 x q4 y − q4 x p3 y + p4 x q3 y − q3 x p4 y = 0, (15) p4 x q4 y − q4 x p4 y = 0. Equation (15) and Lemma 1 give that (16) q4 = λp4 . Substituting this in equation (14) yields p4 x (q3 − λp3 )y − (q3 − λp3 )x p4 y = 0. We have two possibilities, either (17) q3 = λp3 or, from Lemma 1, there exist a, b, c3 , c4 ∈ R such that (18) p4 = c4 (ax + by)4 , q3 = λp3 + c3 (ax + by)3 , with (a2 + b2 )c3 c4 ̸= 0. ISOCHRONICITY FOR QUINTIC AND SEPTIC HAMILTONIAN SYSTEMS 7 Assuming (17), we obtain from (13) that p4 x (q2 − λp2 )y − p4 y (q2 − λp2 )x = 0. We have then another two possibilities, either (19) q2 = λp2 or, from Lemma 1 there exist a, b, c, c2 , c4 ∈ R such that ( )2 ( ) (20) p4 = c4 ax2 + 2bxy + cy 2 , q2 = λp2 + c2 ax2 + 2bxy + cy 2 , with (a2 + b2 + c2 )c2 c4 ̸= 0. Assuming (19), it follows from (16) and (17) that equations (10), (11) and (12) turn to p2 x + λp2 y = 0, p3 x + λp3 y = 0, p4 x + λp4 y = 0, respectively. Solving these equations for homogeneous polynomials of de- grees 2, 3 and 4, we obtain that p2 = c2 (y − λx)2 , p3 = c3 (y − λx)3 , p4 = c4 (y − λx)4 . By applying the linear change of coordinates (x, y) 7→ (x, y − λx) in (9), we obtain the first Hamiltonian of the theorem. Now if we assume (17) and (20), equation (10) turns to p2 x + λp2 y + 2c2 (bx + cy) = 0. Using Lemma 2, we obtain d2 ∈ R such that ( ) (21) p2 = −c2 bx + cy + c(y − λx) x + d2 (y − λx)2 . Substituting this in (11), using (17) and (20), we obtain a partial differential equation of the form p3 x + λp3 y = h, with h(x, y + λx) = 4(b2 − ac)c22 x2 + 4c2 (bcc2 + ad2 + bd2 λ + Lλ)xy + 4c2 Ly 2 , where (22) L = c2 c2 + d2 (b + cλ). Then from Lemma 2, we obtain d3 ∈ R such that 4 p3 = c22 (b2 − ac)x3 + 2c2 (bcc2 + ad2 + bd2 λ + Lλ)x2 (y − λx) (23) 3 +4c2 Lx(y − λx)2 + d3 (y − λx)3 . We finally substitute (21) and (23), and also (16), (17) and (20) in equation (12), obtaining an identically zero homogeneous polynomial of degree 3. Recalling that (a2 + b2 + c2 )c2 c4 ̸= 0, we will divide the analysis in two cases: either b = −cλ or b + cλ ̸= 0. In the first possibility, the coefficient of y 3 of the polynomial is 8cc22 L. Since L = c2 c2 , it follows that c = 0 (and thus L = 0). Then the coefficient of xy 2 turns to −6ac2 d3 , which gives that d3 = 0, and hence the coefficient 8 F. BRAUN, J. LLIBRE AND A. C. MEREU of x3 gives that c4 = c22 d2 . In particular, d2 ̸= 0. With these information, it follows from (21), (23) and (20) that p2 = d2 (y − λx)2 , p3 = 2ac2 d2 x2 (y − λx), p4 = a2 c22 d2 x4 , and q2 , q3 and q4 are given by (20), (17) and (16), respectively. Then by applying the change of coordinates (x, y) 7→ (x, y − λx) in (9), it follows that in the new variables f1 = x + d2 (y 2 + 2ac2 x2 y + a2 c22 x4 ), ( ) f2 = y + λ x + d2 (y 2 + 2ac2 x2 y + a2 c22 x4 ) + ac2 x2 . By defining β2 = d2 , Γ = ac2 and Q = y + Γx2 , we clearly obtain the second Hamiltonian of the theorem with β1 = 0. Now we analyze the second possibility b + cλ ̸= 0. The coefficients of y 3 and xy 2 of the polynomial defined by equation (12) give the following linear system (recall that L is given by (22)) ( ) ( ) c4 ( −8cc22 L ) (24) A = , d3 8c22 (b2 − ac)d2 + 3cλL where ( ) ( 4c (b + cλ) ) −6c ( 2 (b + cλ) ) A= . 4 3b(b + cλ) + ac − b2 −6c2 a − bλ − 2cλ2 The determinant of A is 48c2 (b + cλ)3 ̸= 0. Thus c4 and d3 are given by inverting A in (24). We substitute them in the coefficients of x3 and x2 y and obtain (using (22)), respectively, that 4c32 (b2 − ac)2 (2b2 + ac + 3bcλ) 12cc32 (b2 − ac)2 = 0, = 0, (b + cλ)3 (b + cλ)2 which gives that c ̸= 0 and a = b2 /c. We finally obtain b2 c22 L 2cc2 L a= , c4 = , d3 = , c b + cλ b + cλ with Lc(b + cλ) ̸= 0. Now from (23) and from (20) (after some calculations) 2c2 L p3 = (bx + cy)2 (y − λx), c(b + cλ) (25) c2 L p4 = 2 2 (bx + cy)4 , c (b + cλ) c2 q2 = λp2 + (bx + cy)2 . c ( ) We consider the change of variables (x, y) 7→( (bx + cy)/(b + cλ), y − λx , ) inverse is the transformation (x, y) 7→ x − cy/(b + cλ), λx + by/(b + whose cλ) . Observe that the determinant of the change is 1. By applying the ISOCHRONICITY FOR QUINTIC AND SEPTIC HAMILTONIAN SYSTEMS 9 transformation in (21) and in (25), we obtain that L p2 = −c2 (b + cλ)x2 + y2, b + cλ 2c2 (b + cλ)L 2 p3 = x y, (26) c c2 (b + cλ)3 L 4 p4 = 2 x , c2 c2 (b + cλ)2 2 q2 = λp2 + x . c Therefore ( )2 L c2 (b + cλ)2 2 (27) p2 + p3 + p4 = −c2 (b + cλ)x2 + y+ x b + cλ c = −c2 (b + cλ)x2 + β2 Q2 , with β2 = L/(b + cλ) and Q = y + Γx2 , where Γ = c2 (b + cλ)2 /c. Then from (9), in the new variables, c f1 = x − y − c2 (b + cλ)x2 + β2 Q2 b + cλ c =x− Q + β2 Q2 . b + cλ Similarly, substituting the last equation of (26) and equations (17), (16) and (27) in (9), we get that in the new variables b c2 (b + cλ)2 2 ( ) f2 = y+ x + λ x − c2 (b + cλ)x2 + β2 Q2 b + cλ c ( ) c =Q+λ x− Q + β2 Q2 . b + cλ By defining β1 = −c/(b + cλ), we obtain that the above f1 and f2 satisfy the second Hamiltonian of the theorem (now with β1 ̸= 0). We have yet to analyze possibility (18). The remaining part of the proof will be to show that if we assume this possibility we get a contradiction. We will treat this analyzing two cases: a = −bλ and a + bλ ̸= 0. In the first case, we consider new c3 and c4 not zero in order that (18) turn to (28) p4 = c4 (y − λx)4 , q3 = λp3 + c3 (y − λx)3 . We take the change of coordinates (x, y) 7→ (x, y − λx). Then using (28) we write equations (10), (11), (12) and (13) in these new variables, where we denote pi (x, y) = pi (x, y − λx) and qi (x, y) = qi (x, y − λx), i = 2, 3, 4: p2 x + (q 2 − λp2 )y = 0, p3 x + p2 x q 2 y − q 2 x p2 y + 3c3 y 2 = 0, (29) p3 x (q 2 − λp2 )y − (q 2 − λp2 )x p3 y + 3c3 p2 x y 2 = 0, (−4c4 (q 2 − λp2 )x y + 3c3 p3 x ) y 2 = 0. 10 F. BRAUN, J. LLIBRE AND A. C. MEREU Integrating the fourth equation of (29) in y, we obtain d3 ∈ R such that 4c4 p3 = (q − λp2 )y + d3 y 3 . 3c3 2 On the other hand, defining p2 = a1 x2 + 2a2 xy + a3 y 2 and integrating in y the first equation of (29), we obtain a4 ∈ R such that q 2 = a4 x2 + 2 (a2 λ − a1 ) xy + (a3 λ − a2 ) y 2 . Then we substitute the above q 2 and p3 in the second and in the third equations of (29), obtaining two identically zero homogeneous polynomials of degrees 2 and 3, respectively. We denote this by h2 = 0 and h3 = 0, respectively. The coefficient of y 3 of h3 is −8c4 (a4 − a1 λ)2 /(3c3 ). Since c4 ̸= 0, it follows that a4 = a1 λ. Then the coefficient of x2 of h2 turns to −4a21 , and hence a1 = 0. Finally, the coefficients of y 2 and y 3 of h2 and h3 , respectively, are −4a22 + 3c3 and 6a2 c3 , implying that a2 = c3 = 0, which is a contradiction. ( In the second ) case we consider the change of variables (x, y) → ax + by, y − λx and write pi and q i , i = 2, 3, 4 the maps pi and qi in these new variables, i.e. pi (x, y) = pi (ax + by, y − λx), qi (x, y) = qi (ax + by, y − λx). Then denoting the new variables by (x, y) again, it follows that p4 = c4 x4 , q 4 = λp4 and q 3 = λp3 + c3 x3 , and equations (10), (11), (12) and (13) turn, respectively, to ( ap2 x − λp2 y +)bq 2 x + q 2 y = 0, (a + bλ) p3 x + p2 x q 2 y − q 2 x p2 y + 3bc3 x2 = 0, ( (30) (a + bλ) p3 x (q 2 − λp2 )y − (q 2 − λp2 )x p3 y + 4c4 x3 ) −3c3 p2y x2 = 0, ( ) (a + bλ) 4c4 (q 2 − λp2 )y x − 3c3 p3 y x2 = 0. The second and the fourth equations of (30) give, respectively, that 3bc3 2 4c4 (31) p3 x = −(p2 x q 2 y − q 2 x p2 y ) − x , p3 y = (q − λp2 )y x. a + bλ 3c3 2 Substituting (31) in the third equation of (30), we obtain that ( ) 4c4 3bc3 2 p2 x q 2 y − p2 y q 2 x + (q − λp2 )x x + x (q 2 − λp2 )y (32) 3c3 2 a + bλ −4c4 x3 + 3c3 p2 y x2 = 0. Then defining p2 = a1 x2 + 2a2 xy + a3 y 2 the first equation of (30) will be a differential equation in q 2 . Solving it for a homogeneous polynomial of degree 2 (now it is similar to Lemma 2, but we apply the change of variables x 7→ x − by instead and integrate in y), it follows that there exists a4 ∈ R such that q 2 = a4 (x − by)2 − 2(aa1 − a2 λ)(x − by)y − (a(a2 + a1 b) − (a3 + a2 b)λ) y 2 . ISOCHRONICITY FOR QUINTIC AND SEPTIC HAMILTONIAN SYSTEMS 11 Substituting the above p2 and q2 in (32) we obtain an identically zero ho- mogeneous polynomial of degree 3. The coefficient of x3 of this polynomial gives that ( ) 2 8a4 c4 9bc3 −4c4 − (aa1 + a4 b − a2 λ) + = 0. 3 c3 a + bλ Now we substitute p2 and q2 in the equation p3 xy − p3 yx = 0 given by (31) and obtain an identically zero homogeneous polynomial of degree 1. The coefficient of x of this polynomial gives that 16c4 (aa1 + a4 b − a2 λ) = 0. 3c3 Combining the last two equations, we obtain that c4 = 0, a contradiction.  We observe that the two hamiltonians that appear in Theorem 5 can not be transformed in each other using a linear change of coordinates. This is so because applying the linear change (x, y) 7→ (ax + by, cx + dy) in the second Hamiltonian, the only way to make the monomial x8 disappear is to take a = 0. Then to make the monomial x4 disappear, we have to take c = 0. But then we no longer have a change of coordinates. We also observe that our results give formulas (up to linear change of coordinates) for all the polynomial maps f = (f1 , f2 ) : R2 → R2 such that det Df = 1, f (0, 0) = (0, 0) and degree of f is 2, 3 or 4. Using these formulas it is very simple to see that such maps are injective. 3. Examples of polynomial Hamiltonian isochronous centers The following example shows that there exist non trivial polynomial Hamiltonian isochronous centers of degree 6k + 1, for all k ∈ {1, 2 . . .}. Example 6. Let λ ∈ R and k ∈ {1, 2 . . .}. Let f = (f1 , f2 ) : R2 → R2 be defined by ( )2 ( )2 x + λy k x + λy k + 1 + (x + λy k )2 y f1 = √ , f2 = √ . 1 + (x + λy k )2 1 + (x + λy k )2 It follows that the Jacobian ( determinant ) of f is constant and equal to 1. Moreover, taking H = f12 + f22 /2, it is simple to see that 2H is the poly- nomial ( ) ( )3 (x + λy k )2 + 2y(x + λy k )2 1 + (x + λy k )2 + y 2 1 + (x + λy k )2 , which clearly has degree 6k + 2 if λ ̸= 0 (and degree 8 if λ = 0). Thus system (1) with Hamiltonian H has an isochronous center of degree 6k + 1 at the origin if λ ̸= 0 (degree 7 if λ = 0). Lemma 7. The isochronous center presented in Example 6 is non-trivial. 12 F. BRAUN, J. LLIBRE AND A. C. MEREU Proof. Suppose on the contrary that the center is trivial. Then there exists a polynomial map g = (g2 , g2 ) : R2 → R2 with det Dg = 1 and g(0, 0) = (0, 0) such that g12 + g22 = 2H. The map h = g ◦ T , with T (x, y) = (x − λy k , y) is also polynomial, det Dh = 1, h(0, 0) = (0, 0) and 2 2 2 2 2 2 2 3 e = h1 + h2 = x + 2yx (1 + x ) + y (1 + x ) H 2 2 is a polynomial of degree 8. Thus system (1) with Hamiltonian H e has a trivial isochronous center at the origin. Since h is globally injective (by the mentioned result of [7] or by Theorem 5), it follows from the mentioned result of [8] that this center is global. This is not possible, because the level curve H = 1/2 is not bounded (it is formed by the curves y = −1/(1 + x2 ) and y = (1 − x2 )/(1 + x2 )2 ).  Example 6 with λ = 0 has already appeared in [2]. The following example provide trivial isochronous centers for all even degrees. Example 8. Let k ∈ {2, 3, . . . , }, λ, c2 , . . . , ck ∈ R, with ck ̸= 0 and f = (f1 , f2 ) : R2 → R2 be defined by f1 = x + c2 y 2 + · · · + ck y k , f2 = y + λf1 . It is clear that f (0, 0) = (0, 0) and det Df = 1. Thus system (1) with the Hamiltonian given by H = (f12 + f22 )/2 has a trivial polynomial Hamiltonian isochronous center of degree 2k − 1 at the origin. Since f is clearly injective, this center is global. Acknowledgments The first author is partially supported by a FAPESP-BRAZIL grant 2014/26149-3. The second author is partially supported by a MINECO grant MTM 2013–40998–P, an AGAUR grant number 2014SGR–568, and FP7-PEOPLE-2012-IRSES-316338 and 318999. The third author is par- tially supported by a FAPESP-BRAZIL grant 2012/18780-0 and a CNPq grant 449655/2014-8. The three authors are also supported by a CAPES CSF–PVE grant 88881.030454/ 2013-01 from the program CSF-PVE References [1] K. Baba and Y. Nakai, A generalization of Magnu’s Theorem, Osaka J. Math. 14, (1977), 403–409. [2] A. Cima, F. Man ˜ osas and J. Villadelprat Isochronicity for Several Classes of Hamiltonian Systems, J. Differential Equations 157 (1999), 373–413. [3] A. van den Essen, Polynomial automorphisms and the Jacobian conjecture, Progress in Mathematics 190. Birkh¨auser Verlag, Basel, 2000. [4] X. Jarque and J. Villadelprat, Nonexistence of isochronous centers in planar polynomial Hamiltonian systems of degree four, J. Differential Equations 180 (2002), 334–373. ISOCHRONICITY FOR QUINTIC AND SEPTIC HAMILTONIAN SYSTEMS 13 [5] J. Llibre and V. G. Romanovski, Isochronicity and linearizability of planar poly- nomial Hamiltonian systems, J. Differential Equations 259 (2015), 1649–1662. [6] F. Man ˜ osas and J. Villadelprat Area-Preserving Normalizations for centers of Planar Hamiltonian Systems, J. Differential Equations 179 (2002), 625–646. [7] T. T. Moh, On the Jacobian conjecture and the configurations of roots, J. Reine Angew. Math. 340 (1983), 140–212. [8] M. Sabatini, A connection between isochronous Hamiltonian centers and the Jaco- bian Conjecture, Nonlinear Anal. 34 (1998), 829–838. ˜ o Carlos, SP, Brazil Department of Mathematics. UFSCar, 13565-905, Sa E-mail address: franciscobraun@dm.ufscar.br Departament de Matema ` tiques, Universitat Auto ` noma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain E-mail address: jllibre@mat.uab.cat Department of Physics, Chemistry and Mathematics. UFSCar. 18052-780, Sorocaba, SP, Brazil E-mail address: anamereu@ufscar.br