October 20, 2020 20:22 ”Continuous piecewise isochronous centers” This is a preprint of: “Limit cycles of continuous piecewise differential systems formed by linear and quadratic isochronous centers I”, Bilal Ghermoul, Jaume Llibre, Tayeb Salhi, Internat. J. Bifur. Chaos Appl. Sci. Engrg., vol. 32(1), 2250003 (28 pages), 2022. DOI: [10.1142/S0218127422500031] Limit cycles of continuous piecewise differential systems formed by linear and quadratic isochronous centers I B. Ghermoul1 , J. Llibre2 and T. Salhi1 1 Department of Mathematics, University Mohamed El Bachir El Ibrahimi Bordj Bou Arreridj 34265, El–Anasser, Algeria bilal.ghermoul@univ-bba.dz, t.salhi@univ-bba.dz 2 Departament de Matemàtiques, Universitat Autònoma de Barcelona 08193 Bellaterra, Barcelona, Catalonia, Spain jllibre@mat.uab.cat Received (to be inserted by publisher) First we study the planar continuous piecewise differential systems separated by the straight line x = 0 formed by a linear isochronous center in x > 0 and an isochronous quadratic center in x < 0. We prove that these piecewise differential systems cannot have crossing periodic orbits, and consequently they do not have crossing limit cycles. Second we study the crossing periodic orbits and limit cycles of the planar continuous piecewise differential systems separated by the straight line x = 0 having in x > 0 the general quadratic isochronous center ẋ = −y + x2 − y 2 , ẏ = x(1 + 2y) after an affine transformation, and in x < 0 an arbitrary quadratic isochronous center. For these kind of continuous piecewise differential systems the maximum number of crossing limit cycles is one, and there are examples having one crossing limit cycles In short for these families of continuous piecewise differential systems we have solved the exten- sion of the 16th Hilbert problem to themselves. Keywords: Limit cycles, isochronous quadratic centers, continuous piecewise linear differential systems, first integrals 1. Introduction In the qualitative theory of planar differential systems a limit cycle is an isolated periodic solution in the set of all periodic solutions, which remained the most sought solutions when modeling physical systems in the plane. As far as we known the notion of limit cycle appeared in the year 1885 in the work of Poincaré [Poincaré, 1928]. Most of the early examples in the theory of limit cycles in planar differential systems were commonly related to practical problems with mechanical and electronic systems, but periodic behavior appears in all branches of the sciences. To determine the existence of non–existence of limit cycles is one of the more difficult objects in the qualitative theory of planar differential equations. A large amount of references deals with the subject of limit cycles, many of them motivated for the famous second part of the Hilbert’s 16th problem, which ask for the maximum number of limit cycles that the planar polynomial differential systems of a given degree can exhibit, see for details [Hilbert, 1900; Ilyashenko, 2002; Li, 2003]. Since 1930’s the study of the limit cycles also became important in the continuous and discontinuous piecewise differential systems separated by a straight line, due to their applications to mechanics, electrical 1 October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 2 B. Ghermoul, J. Llibre, T. Salhi circuits, ... see for instance the books [Andronov et al., 1966; di Bernardo et al., 2008; Simpson, 2010] and the references therein. As usual a center p of a planar differential system is a singular point for which there is a neighborhood U such that U \ {p} is filled with periodic orbits. When all the periodic orbits surrounding a center have the same period this center is called isochronous. The centers started to be studied by Poincaré [Poincaré, 1881] and Dulac [Dulac, 1908], but the notion of isochronocity goes back to Huygens [Huygens, 2002] in 1673. In this paper we consider continuous piecewise differential systems separated by the straight line x = 0 having in x ≤ 0 and in x ≥ 0 linear or quadratic isochronous centers, and we want to study the non– existence, and the existence of crossing periodic orbits and of crossing limit cycles, and in this last case we want also to know the maximum number of crossing limit cycles for these systems. Here a crossing periodic orbit or a crossing limit cycle is a periodic orbit or a limit cycle which intersects exactly in two points the discontinuity line x = 0. The continuity of a piecewise differential system separated by the straight line x = 0 formed by two centers means that the vector fields defined by these two centers (linear or quadratic) coincide on the line of discontinuity x = 0. So a continuous piecewise differential system is a continuous differential system in R2 and is an analytic differential system in R2 \ {x = 0}. 1.1. Linear centers It is well known that the linear differential centers are isochronous and that the general expression of such centers is as follows, see for a proof [Llibre & Teixeira, 2018]. Lemma 1. A linear differential system having a center can be written in the form 4β 2 + ω 2 ẋ = −βx − y + δ, ẏ = αx + βy + γ, (1) 4α with α > 0 and ω > 0. The linear differential system (1) has the first integral HL (x, y) = 4(αx + βy)2 + 8α(γx − δy) + ω 2 y 2 . 1.2. Quadratic isochronous centers We consider the quadratic polynomial differential systems having an isochronous center. This kind of centers were classified by Loud in the paper [Loud, 1964]. Those systems after an affine change of coordinates become one of the following four systems: ẋ = − y + x2 − y 2 , ẏ =x(1 + 2y), (2) 2 ẋ = − y + x , ẏ =x(1 + y), (3)   4x2 16 ẋ = − y − , ẏ =x 1 − y , (4) 3 3   16 4 8 ẋ = − y + x2 − y 2 , ẏ =x 1 + y . (5) 3 3 3 We are interested in the general expressions of the quadratic isochronous centers. So we transform their normal forms (2), (3), (4) and (5) through the following general affine change of variables (x, y) → (a1 x + b1 y + c1 , a2 x + b2 y + c2 ), (6) with a1 b2 − a2 b1 6= 0. (7) October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 3 Generalized isochronous system (2). Using the change of variables (6) the quadratic system (2) becomes ẋ = ( − b2 c21 + b1 c1 + 2b1 c2 c1 + b2 c22 + b2 c2 + (2a2 b1 c1 + 2a1 b1 c2 − 2a1 b2 c1 + 2a2 b2 c2 + a1 b1 + a2 b2 ) x   + 2b21 c2 + 2b22 c2 + b21 + b22 y + 2a2 b21 + 2a2 b22 xy    + a21 (−b2 ) + 2a2 a1 b1 + a22 b2 x2 + b32 + b21 b2 y 2 /(a2 b1 − a1 b2 ),  (8) ẏ = ( a2 c21 − a1 c1 − 2a1 c2 c1 − a2 c22 − a2 c2 + −2a21 c2 − 2a22 c2 − a21 − a22 x + (2a2 b1 c1 − 2a1 b1 c2 − 2a1 b2 c1 − 2a2 b2 c2 − a1 b1 − a2 b2 ) y   + −2a21 b2 − 2a22 b2 xy + −a32 − a21 a2 x2   + a2 b21 − 2a1 b2 b1 − a2 b22 y 2 /(a2 b1 − a1 b2 ). Since (x2 + y 2 )/(2y + 1) is a first integral of system (2), doing to it the change of variables (6) we get the following first integral of the generalized isochronous quadratic system (8) (a1 x + b1 y + c1 ) 2 + (a2 x + b2 y + c2 ) 2 H2 (x, y) = . 2 (a2 x + b2 y + c2 ) + 1 Generalized isochronous system (3). System (3) is equivalent to the following generalized isochronous system after the linear change of variables (6) ẋ = ( b2 c21 − b1 c1 − b1 c2 c1 − b2 c2 + (−a2 b1 c1 − a1 b1 c2 + 2a1 b2 c1 − a1 b1 − a2 b2 ) x   + b21 (−c2 ) + b2 b1 c1 − b21 − b22 y + a1 b1 b2 − a2 b21 xy   + a21 b2 − a1 a2 b1 x2 /(a1 b2 − a2 b1 ), (9)  ẏ = ( a2 c21 − a1 c1 − a1 c2 c1 − a2 c2 + a21 (−c2 ) + a2 a1 c1 − − a21 a22 x + (2a2 b1 c1 − a1 b1 c2 − a1 b2 c1 − a1 b1 − a2 b2 ) y    + a1 a2 b1 − a21 b2 xy + a2 b21 − a1 b1 b2 y 2 /(a1 b2 − a2 b1 ). The quadratic system (3) has the first integral (x2 + y 2 )/(y + 1)2 . Therefore a first integral of system (9) is (a1 x + b1 y + c1 )2 + (a2 x + b2 y + c2 )2 H3 (x, y) = . (a2 x + b2 y + c2 + 1)2 Generalized isochronous system (4). The quadratic system (4) is equivalent to the following general- ized quadratic system after the linear change of variables (6) ẋ = ( 4b2 c21 + 3b1 c1 − 16b1 c2 c1 + 3b2 c2 + (−16a2 b1 c1 − 16a1 b1 c2 + 8a1 b2 c1 + 3a1 b1 + 3a2 b2 ) x   + −16b21 c2 − 8b2 b1 c1 + 3b21 + 3b22 y + −16a2 b21 − 8a1 b2 b1 xy   + 4a21 b2 − 16a1 a2 b1 x2 − 12b21 b2 y 2 /3 (a2 b1 − a1 b2 ) , ẏ = ( − 4a2 c21 − 3a1 c1 + 16a1 c2 c1 − 3a2 c2 (10)  + 16a21 c2 + 8a2 a1 c1 − 3a21 − 3a22 x + (−8a2 b1 c1 + 16a1 b1 c2 + 16a1 b2 c1 − 3a1 b1 − 3a2 b2 ) y  + 16a21 b2 + 8a2 a1 b1 xy + 12a21 a2 x2   + 16a1 b1 b2 − 4a2 b21 y 2 /3 (a2 b1 − a1 b2 ) . October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 4 B. Ghermoul, J. Llibre, T. Salhi Since (32x2 − 24y + 9)2 /(3 − 16y) is a first integral of system (10), then a first integral of system (10) is 2 32 (a1 x + b1 y + c1 ) 2 − 24 (a2 x + b2 y + c2 ) + 9 H4 (x, y) = . 3 − 16 (a2 x + b2 y + c2 ) Generalized isochronous system (5). Doing the affine change of variables (6) the generalized isochronous system for the quadratic system (5) is ẋ = ( − 16b2 c21 + 3b1 c1 + 8b1 c2 c1 + 4b2 c22 + 3b2 c2 + (8a2 b1 c1 + 8a1 b1 c2 − 32a1 b2 c1 + 8a2 b2 c2 + 3a1 b1 + 3a2 b2 ) x  + 8b21 c2 − 24b2 b1 c1 + 8b22 c2 + 3b21 + 3b22 y   + 8a2 b21 − 24a1 b2 b1 + 8a2 b22 xy + −16a21 b2 + 8a2 a1 b1 + 4a22 b2 x2   + 4b32 − 8b21 b2 y 2 /3 (a2 b1 − a1 b2 ) , (11) ẏ = ( 16a2 c21 − 3a1 c1 − 8a1 c2 c1 − 4a2 c22 − 3a2 c2  + −8a21 c2 + 24a2 a1 c1 − 8a22 c2 − 3a21 − 3a22 x + (32a2 b1 c1 − 8a1 b1 c2 − 8a1 b2 c1 − 8a2 b2 c2 − 3a1 b1 − 3a2 b2 ) y   + −8a21 b2 + 24a2 a1 b1 − 8a22 b2 xy + 8a21 a2 − 4a32 x2   + 16a2 b21 − 8a1 b2 b1 − 4a2 b22 y 2 /3 (a2 b1 − a1 b2 ) . The quadratic system (5) has the first integral (−256x2 + 128y 2 + 96y + 9)/(8y + 3)4 , which gives the following first integral for system (11) −256 (a1 x + b1 y + c1 ) 2 + 128 (a2 x + b2 y + c2 )2 + 96 (a2 x + b2 y + c2 ) + 9 H5 (x, y) = . (8 (a2 x + b2 y + c2 ) + 3)4 1.3. Statement of the main results It has been proved in [Llibre & Teixeira, 2018] that continuous piecewise linear differential systems separated by one straight line formed by two linear centers have no crossing limit cycles. Now we shall see that this result extends to continuous piecewise linear differential systems separated by one straight line formed by one linear center and one quadratic isochronous center have no crossing limit cycles. Theorem 1. The continuous piecewise differential systems formed by a linear differential center (which is isochronous) and an isochronous quadratic center separated by the straight line x = 0 have no crossing periodic orbits, and consequently no crossing limit cycles. Theorem 1 is proved in section 2. In what follows we characterize the existence and non-existence of crossing periodic orbits and crossing limit cycles for continuous piecewise linear differential systems separated by one straight line formed by two quadratic isochronous centers. Theorem 2. The following statements hold for the continuous piecewise differential systems formed by two generalized isochronous quadratic centers separated by the straight line x = 0. (a) If the generalized centers are (8) and (8), then the piecewise differential systems can have crossing periodic orbits but they cannot have crossing limit cycles. (b) If the generalized centers are (8) and (9), then the piecewise differential systems can have crossing periodic orbits but they cannot have crossing limit cycles. (c) If the generalized centers are (8) and (10), then the piecewise differential systems has at most one crossing limit cycle and there are systems realizing this limit cycle. (d) If the generalized centers are (8) and (11), then the piecewise differential systems has at most one crossing limit cycle and there are systems realizing this limit cycle. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 5 Theorem 2 is proved in section 3. So in particular Theorems 1 and 2 solve the 16th Hilbert problem extended to these piecewise differential systems. This a relatively long paper for this reason we have left the study on the maximum number of crossing limit cycles of the continuous piecewise differential systems of the type (i)-(j) with (i) and (j) ∈ {(9), (10), (11)} for a future new article. 2. Proof of Theorem 1 The first objective is to study periodic orbits and limit cycles of continuous piecewise differential systems formed by a linear center (1) and a generalized quadratic system (2), (3), (4) or (5). To find the crossing periodic orbits and the crossing limit cycles of such piecewise differential systems, we must solve the following algebraic system Hk (0, y1 ) − Hk (0, y2 ) = 0, HL (0, y1 ) − HL (0, y2 ) = 0, (12) where Hk (x, y) and HL (x, y) are the first integrals of the quadratic isochronous center and of the linear center respectively, and (0, y1 ) and (0, y2 ) with y1 6= y2 are the two intersection points of the crossing periodic orbits with the straight line x = 0. In order that we have a continuous piecewise differential system (1)-(8), we impose that both systems coincide on x = 0, and then both systems must verify the following algebraic system ẋk − ẋL |x=0 = 0, ẏk − ẏL |x=0 = 0, (13) where ẋL , ẏL , ẋk and ẏk are the derivatives with respect to time t of x and y for linear system and quadratic system, respectively. Thus we get the following algebraic system −a2 b1 δ + a1 b2 δ − b2 c21 + b1 c1 + 2b1 c2 c1 + b2 c22 + b2 c2 = 0, 4a2 b1 β 2 − 4a1 b2 β 2 + a2 b1 ω 2 − a1 b2 ω 2 + 4αb21 + 4αb22 + 8αb21 c2 + 8αb22 c2 = 0,  −b2 b21 + b22 = 0, −a2 b1 γ + a1 b2 γ + a2 c21 − a1 c1 − 2a1 c2 c1 − a2 c22 − a2 c2 = 0, −a2 βb1 + a1 βb2 + 2a2 b1 c1 − 2a1 b1 c2 − 2a1 b2 c1 − 2a2 b2 c2 − a1 b1 − a2 b2 = 0, a2 b21 − 2a1 b2 b1 − a2 b22 = 0. Since when we evaluate a2 b1 − a1 b2 in all real solutions (See supplementary section 5.1) for this alge- braic system we obtain zero, we get a contradiction with (7). Therefore there are no continuous piecewise differential systems (1)-(8). In order that the piecewise differential systems (1)-(9) be continuous they must coincide on x = 0, then the algebraic system (13) must be satisfied, which gives the following algebraic system −a2 b1 δ + a1 b2 δ − b2 c21 + b1 c1 + b1 c2 c1 + b2 c2 = 0, 4a2 b1 β 2 − 4a1 b2 β 2 + a2 b1 ω 2 − a1 b2 ω 2 + 4αb21 + 4αb22 − 4αb1 b2 c1 + 4αb21 c2 = 0, −a2 b1 γ + a1 b2 γ + a2 c21 − a1 c1 − a1 c2 c1 − a2 c2 , = 0, −a2 βb1 + a1 βb2 + 2a2 b1 c1 − a1 b1 c2 − a1 b2 c1 − a1 b1 − a2 b2 = 0, b1 = 0. All solutions (See supplementary section 5.2) of this algebraic system give a2 b1 − a1 b2 = 0, except the 4a2 c2 +8a a c +a2 ω 2 +4a22 −a c2 +a c +a c c +a c c2 −c following solution {α = 1 1 2 4a1 11b2 1 , β = a1 ca11+a2 , γ = 2 1 1 a1 1 b21 2 1 2 2 , δ = 1a1 2 , b1 = 0} for which a2 b1 − a1 b2 = −a1 b2 . Solving the algebraic system (12) for the values of the solution s4 we get y1 = y2 . So the continuous piecewise differential system(1)-(9) with the values of s4 has no crossing periodic orbits. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 6 B. Ghermoul, J. Llibre, T. Salhi The piecewise differential systems (1)-(10) is continuous if and only if system (13) is verified, i.e. −3a2 b1 δ + 3a1 b2 δ + 4b2 c21 + 3b1 c1 − 16b1 c2 c1 + 3b2 c2 = 0, 12a2 b1 β 2 − 12a1 b2 β 2 + 3a2 b1 ω 2 − 3a1 b2 ω 2 + 12αb21 + 12αb22 − 32αb1 b2 c1 − 64αb21 c2 2c2 = 0, −3a2 βb1 + 3a1 βb2 − 8a2 b1 c1 + 16a1 b1 c2 + 16a1 b2 c1 − 3a1 b1 − 3a2 b2 = 0, −3a2 b1 γ + 3a1 b2 γ − 4a2 c21 − 3a1 c1 + 16a1 c2 c1 − 3a2 c2 = 0, −4b1 (a2 b1 − 4a1 b2 ) = 0, −4b21 b2 = 0. But all real solutions (See supplementary section 5.3) of this algebraic system give a2 b1 − a1 b2 = 0, 1024a21 c21 −384a2 a1 c1 +9a21 ω 2 +36a22 4a c2 +3a c −16a c c +3a c except the solution {α = 36a1 b2 , β = 3a2 −16a 3a1 1 c1 , γ = 2 1 1 13a1 b2 1 2 1 2 2 , δ = −4c21 −3c2 3a1 , b1 = 0} which gives a2 b1 − a1 b2 = −a1 b2 . By solving system (12) for the values of this solution we get y1 = y2 . Therefore the continuous piecewise differential systems (1)-(10) cannot have crossing periodic orbits. To get a continuous piecewise differential system we impose that both systems (1)-(11) coincide on x = 0 by using (13), and we obtain the following algebraic system −3a2 b1 δ + 3a1 b2 δ − 16b2 c21 + 3b1 c1 + 8b1 c2 c1 + 4b2 c22 + 3b2 c2 = 0, 12a2 b1 β 2 − 12a1 b2 β 2 + 3a2 b1 ω 2 − 3a1 b2 ω 2 + 12αb21 + 12αb22 − 96αb1 b2 c1 + 32αb21 c2 + 32αb22 c2 = 0, −3a2 b1 γ + 3a1 b2 γ + 16a2 c21 − 3a1 c1 − 8a1 c2 c1 − 4a2 c22 − 3a2 c2 = 0, −3a2 βb1 + 3a1 βb2 + 32a2 b1 c1 − 8a1 b1 c2 − 8a1 b2 c1 − 8a2 b2 c2 − 3a1 b1 − 3a2 b2 = 0,  4 4a2 b21 − 2a1 b2 b1 − a2 b22 = 0,  −4b2 b22 − 2b21 = 0. Since all solutions (See supplementary section 5.4) of this algebraic system give a2 b1 − a1 b2 = 0, then there are no continuous piecewise differential systems (1)-(11). In summary Theorem 1 is proved. 3. Proof of Theorem 2 In what follows we consider in one side of the straight line x = 0 the generalized isochronous systems of (2) and in the other side a generalized isochronous system (2), (3), (4) and (5). In the first generalized systems (8) we rename the parameters a2 , b2 and c2 by α1 , β1 and γ1 , respectively; and in the second generalized systems (j) for (j=8, 9, 10 or 11) we rename the parameters a1 , b1 , c1 , a2 , b2 and c2 by a2 , b2 , c2 , α2 , β2 and γ2 , respectively. Doing this condition (7) becomes α1 b1 − a1 β1 6= 0 and α2 b2 − a2 β2 6= 0. (14) In order to study the crossing periodic orbits and limit cycles of a piecewise differential system (8)- (j) formed by two generalized isochronous systems with (j) ∈ {(8, 9, 10, 11}, we must solve the following algebraic system H2 (0, y1 ) − H2 (0, y2 ) = 0, Hj (0, y1 ) − Hj (0, y2 ) = 0, (15) where H2 and Hj are the first integral of the generalized systems (8) and (j). Proof. [Proof of statement (a) of Theorem 2] In order that the piecewise differential systems (8)-(8) be October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 7 continuous, they must coincide on x = 0, which means the following algebraic system must be satisfied. −2a2 b1 β2 γ1 c1 + 2a1 b2 β1 c2 γ2 − a2 b1 β2 c1 + a1 b2 β1 c2 − a2 β1 β2 γ12 + a1 β1 β2 γ22 − a2 β1 β2 γ1 +a1 β1 β2 γ2 + a2 β1 β2 c21 − a1 β1 β2 c22 + α1 b2 β1 γ12 − α1 b1 β2 γ22 + α1 b2 β1 γ1 − α1 b1 β2 γ2 +α1 (−b2 ) β1 c21 + α1 b1 β2 c22 + 2α1 b1 b2 γ1 c1 − 2α1 b1 b2 c2 γ2 + α1 b1 b2 c1 − α1 b1 b2 c2 = 0, −2a2 β2 b21 γ1 + 2a1 b22 β1 γ2 − a2 β2 b21 + a1 b22 β1 − 2a2 β12 β2 γ1 + 2a1 β1 β22 γ2 + a1 β1 β22 − a2 β12 β2 −2α1 β22 b1 γ2 + 2α1 b2 β12 γ1 − α1 β22 b1 + α1 b2 β12 + 2α1 b2 b21 γ1 − 2α1 b22 b1 γ2 + α1 b2 b21 −α1 b22 b1 = 0, −a2 b21 β2 β1 + a1 b22 β2 β1 − a2 β2 β13 + a1 β23 β1 + α1 b2 β13 + α1 b21 b2 β1 − α1 b1 β23 − α1 b1 b22 β2 = 0, a2 α1 β2 γ12 − a1 α1 β1 γ22 + a2 α1 β2 γ1 − a1 α1 β1 γ2 − 2a1 α1 b2 γ1 c1 + 2a2 α1 b1 c2 γ2 − a1 α1 b2 c1 (16) +a2 α1 b1 c2 − a2 α1 β2 c21 + a1 α1 β1 c22 + 2a1 a2 β2 γ1 c1 − 2a1 a2 β1 c2 γ2 + a1 a2 β2 c1 − a1 a2 β1 c2 −α12 b2 γ12 + α12 b1 γ22 − α12 b2 γ1 + α12 b1 γ2 + α12 b2 c21 − α12 b1 c22 = 0, 2a2 α1 β1 β2 γ1 − 2a1 α1 β1 β2 γ2 − a1 α1 β1 β2 + a2 α1 β1 β2 − 2a1 α1 b1 b2 γ1 + 2a2 α1 b1 b2 γ2 −a1 α1 b1 b2 + a2 α1 b1 b2 + 2a1 a2 b1 β2 γ1 − 2a1 a2 b2 β1 γ2 − a1 a2 b2 β1 + a1 a2 b1 β2 −2a1 α1 b2 β1 c1 + 2a1 α1 b2 β1 c2 − 2a2 α1 b1 β2 c1 + 2a2 α1 b1 β2 c2 + 2a1 a2 β1 β2 c1 −2a1 a2 β1 β2 c2 − 2α12 b2 β1 γ1 + 2α12 b1 β2 γ2 − α12 b2 β1 + α12 b1 β2 + 2α12 b1 b2 c1 − 2α12 b1 b2 c2 = 0, −a1 α1 β1 β22 + a2 α1 β12 β2 − a2 α1 β2 b21 − 2a1 α1 b2 β1 b1 + 2a2 α1 b2 β2 b1 + a1 α1 b22 β1 +2a1 a2 β1 β2 b1 − 2a1 a2 b2 β1 β2 + α12 β22 b1 − α12 b2 β12 + α12 b2 b21 − α12 b22 b1 = 0, together with conditions (14). From the sixth equation of (16) we get   α1 a2 β2 −b21 + 2b2 b1 + β12 + α2 b1 β22 − α2 b2 β12 + b1 (b2 − b1 ) a1 =  , (17) β1 2a2 (b2 − b1 ) β2 + α2 β22 + α2 (2b1 − b2 ) b2 if the denominator of a1 is non-zero. If it is zero then β1 = 0 or 2a2 (b2 − b1 ) β2 +α2 β22 +α2 (2b1 − b2 ) b2 = 0. This last equation is equivalent to the following solution −2a2 β2 b2 − α2 β22 + α2 b22 d4 = {b1 = }. 2 (α2 b2 − a2 β2 ) by neglecting all solutions (See supplementary section 5.5) which do not satisfying conditions (14). So it remains to study this solution. If the denominator of (17) is non-zero and α1 = 0, then from (17) we obtain that a1 = 0, and consequently α1 b1 − a1 β1 = 0, which is a contradiction with first condition in (14). In summary for solving system (16) we consider the following cases. Case 1: Suppose that b1 6= (−2a2 β2 b2 − α2 β22 + α2 b22 )/2 (α2 b2 − a2 β2 ), β1 6= 0 and α1 6= 0. Since a1 is defined in (17). From the fifth equation of (16), we get c1 = (b1 (2γ1 + 1)(α2 b2 − a2 β2 ) + b2 (2α2 β1 c2 − a2 (2β1 γ2 − 4β2 γ1 + β1 − 2β2 )) + β2 (α2 (−2β1 γ2 +2β2 γ1 − β1 + β2 ) − 2a2 β1 c2 ) + b22 (−(2α2 γ1 + α2 )))/2β1 (α2 b2 − a2 β2 ). Now we solve simultaneously the second and the third equations of (16), and we obtain the only solution α β2 (b22 +β22 ) {β1 = 2a (b −b )β 1+α β 2 +α (2b −b )b , 2 2 1 2 2 2 2 1 2 2 2b2 (2γ1 +1)(a2 β2 +α2 b1 )+β2 (β2 (2α2 γ1 −α1 +α2 )−2a2 (2b1 γ1 +b1 ))+b22 (−(2α2 γ1 +α1 +α2 )) γ2 = 2α1 (b22 +β22 ) }, among all real solutions (See supplementary section 5.6) for which b1 6= (−2a2 β2 b2 − α2 β22 + α2 b22 )/2 (α2 b2 − a2 β2 ), β1 6= 0, α1 6= 0 and conditions (14) are satisfied. Solving now simultaneously the first and the fourth equations of (16). The only real solutions (See supplementary section 5.7) for which b1 6= (−2a2 β2 b2 − α2 β22 + α2 b22 )/2(α2 b2 − a2 β2 ), β1 6= 0, α1 6= 0 and conditions (14) are satisfied are solutions s1 = {b1 = b2 , α1 = −α2 } and s2 = {b1 = b2 , α1 = α2 }. Then we must discuss these two solutions s1 and s2 in the following subcases to detect whether the continuous piecewise differential systems (8)-(8) have or do not have limit cycles. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 8 B. Ghermoul, J. Llibre, T. Salhi Subcase 1.1: Consider s1 . Solving system (15) we get y1 = y2 , or the following solution  2b1 (2γ1 c2 + c2 ) + b21 (2γ1 y2 + y2 ) + β2 2c22 − 2γ1 (γ1 + 1) + y2 (2β2 γ1 + β2 ) y1 =  . b21 + β22 (−2γ1 + 2β2 y2 − 1) Since in order to have a crossing periodic orbit we must have y1 6= y2 . Therefore these continuous piecewise differential systems have a continuum of crossing periodic orbits and then no limit cycles. Subcase 1.2: Now we consider s2 , so system (15) gives y1 = y2 , or  2b1 (2γ1 c2 + c2 ) + b21 (2γ1 y2 + y2 ) + β2 −2c22 + 2γ1 (γ1 + 1) + y2 (2β2 γ1 + β2 ) y1 = −  . b21 + β22 (2γ1 + 2β2 y2 + 1) As in the previous subcase no limit cycles. Case 2: We suppose that b1 = (−2a2 β2 b2 − α2 β22 + α2 b22 )/2(α2 b2 − a2 β2 ) and β1 6= 0. With this value for b1 the sixth equation of (16) gives the only following four real solutions s1 = {α1 = 0}, s2 = {a2 = 0, α2 = 0}, s3 = {b2 = 0, β2 = 0} and s4 = {α2 = 0, β2 = 0}. The only solution that satisfying conditions (14) is s1 , which is equivalent to α1 = 0. Now from the third equation of (16), regarding all the real solutions (See supplementary section 5.8), the only solution verifying (14) is 4a2 α2 b2 β23 +2β22 (2a22 (b22 +β12 )−α22 b22 )−4a2 α2 b2 β2 (b22 +2β12 )+α22 β24 +α22 b22 (b22 +4β12 ) a1 = 4β2 (b22 +β22 )(a2 β2 −α2 b2 ) . By solving the fifth equation of (16) we obtain from all real solutions (See supplementary section 5.9) the following solutions s1 , s2 and s8 2a b (2β γ −β +β )+α β (2β γ −2β +β )+4a β1 β2 (c1 −c2 )+b22 (−(2α2 γ1 +α2 ))+4α2 β1 b2 (c2 −c1 ) s1 = {γ2 = 2 2 2 1 1 2 2 2 2 1 1 4β21 (a2 b22 +α 2 β2 ) }, s2 = {a2 = −α2 β2 /b2 , c2 = 2b2 γ1 +b 2 +4β1 c1 4β1 }, and s3 = {b2 = 0, c2 = c1 , α2 = 0}, which satisfying (14). Then we divide case 2 into the following three subcases. Subcase 2.1: We consider the solution s1 . Solving the second equation of (16) we get the following real solutions u1 = {γ1 = − 2β1 (c1 α−c2b)(α 2 b2 −a2 β2 ) − 21 }, u2 = {c2 = c1 , α2 = 0} and u3 = {α2 = 0, β1 = 0}. 2 ( 2 +β2 ) 2 2 But u1 and u2 are the only solutions satisfying (14) which must be studied separately into two subcases. Subcase 2.1.1: Consider u1 . Solving simultaneously the first and the fourth equations of (16), we get the only real solution β1 = 0, in contradiction with the hypothesis of Case 2. Subcase 2.1.2: Consider u2 . Then all equations of (16) are verified except the first one, which becomes   a2 β2 β22 − β12 −4b2 β1 c1 (2γ1 + 1) + (2b2 γ1 + b2 ) 2 + β12 4c21 + 1 = 0. Solving this equation we obtain the following set of real solutions v1 = {a2 = 0}, v2 = {β2 = 0}, v3 = {β2 = −β1 } and v4 = {β2 = β1 }. The solutions for which conditions (14) are verified are v3 and v4 . Now we must discuss these two subcases as follows. Subcase 2.1.2.1: Consider v3 . Then we have a continuous piecewise differential system. We prove that there are a continuum of crossing periodic orbits and no limit cycles, because solving the algebraic system (15) we get y1 = y2 , or the solution  2b2 (2γ1 c1 + c1 ) + b22 (2γ1 y2 + y2 ) + β1 −2c21 + 2γ1 (γ1 + 1) + y2 (2β1 γ1 + β1 ) y1 = −  . b22 + β12 (2γ1 + 2β1 y2 + 1) Subcase 2.1.2.2: Consider v4 . Then we get a continuous piecewise differential system. Solving the algebraic system (15) we obtain y1 = y2 , or  2b2 (2γ1 c1 + c1 ) + b22 (2γ1 y2 + y2 ) + β1 −2c21 + 2γ1 (γ1 + 1) + y2 (2β1 γ1 + β1 ) y1 = −  . b22 + β12 (2γ1 + 2β1 y2 + 1) This gives a continuum of crossing periodic orbits. Subcase 2.2: Assume that solution s2 holds. Solving the second equation of (16) we get the following sets of solutions {α2 = 0}, {γ2 = (2β2 γ1 − β1 + β2 )/(2β1 )} and {β1 = 0, γ1 = −1/2}. The only solution October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 9 for which conditions (14) are satisfied is γ2 = (2β2 γ1 − β1 + β2 )/(2β1 ). The remaining unsolved equations of (16) are the first and the fourth equations, we solve them simultaneously and we get only one real solution α2 = 0, but this solution gives a contradiction with (14). Then there are no continuous piecewise differential systems (8)-(8) in this case. Subcase 2.3: Consider the solution s3 . Then the remaining unsolved equations of system (16) are the first, the second and the fourth equations. Solving these three equations simultaneously we get one of the following sets of real solutions t1 = {a2 = 0}, t2 = {β1 = 0}, t3 = {β2 = −β1 , γ2 = −γ1 − 1} and t4 = {β2 = β1 , γ2 = γ1 }. The sets of solutions t3 and t4 are the only for which conditions (14) are verified. We now have a continuous piecewise differential system (8)-(8). Consider these two cases t3 and t4 separately. Subcase 2.3.1: Consider solution t3 . Then for showing whether the piecewise differential system (8)-(8) has a limit cycle or not, we solve the algebraic system (15). Here we get y1 = y2 , or 2c21 − 2γ1 (γ1 + 1) − y2 (2β1 γ1 + β1 ) y1 = . β1 (2γ1 + 2β1 y2 + 1) We have then a continuum of crossing periodic orbits. Subcase 2.3.2: Consider the solution t4 . I In a similar way solving the algebraic system (15). We get y1 = y2 , or 2c21 − 2γ1 (γ1 + 1) − y2 (2β1 γ1 + β1 ) y1 = . β1 (2γ1 + 2β1 y2 + 1) Then we get a continuum of crossing periodic orbits. Case 3: Assume that b1 6= (−2a2 β2 b2 − α2 β22 + α2 b22 )/2 (α2 b2 − a2 β2 ), β1 = 0. The third equation of (16) becomes −α1 b1 β2 b22 + β22 = 0, and since β1 = 0 we must have β2 = 0, because b1 = 0 or α1 = 0 gives b1 α1 − a1 β1 = 0. Solving the five remaining equations of (16), the only solutions from all real solutions (See supplementary section 5.10) for which (14) and b1 6= (−2a2 β2 b2 − α2 β22 + α2 b22 )/(2 (α2 b2 − a2 β2 )) are verified are s1 , s2 , s3 , s4 and s5 given as follows s1 = {b1 = b2 , c1 = c2 , γ1 = −1/2, γ2 = −1/2}, s2 = {a2 = a1 , b1 = b2 , c1 = c2 , α1 = −α2 , γ1 = −γ2 − 1}, s3 = {b1 = b2 , c1 = c2 , α1 = −α2 , γ1 = −1/2, γ2 = −1/2}, s4 = {a2 = a1 , b1 = b2 , c1 = c2 , α1 = α2 , γ1 = γ2 } and s5 = {b1 = b2 , c1 = c2 , α1 = α2 , γ1 = −1/2, γ2 = −1/2}. For these five cases we have a continuous piecewise differential systems (8)-(8). Now we discuss these five subcases. Case 3.1: Consider the solution s1 . Then the first integrals of systems (8)-(8) are H1 (x, y) = [8a1 b1 xy + 8a1 c1 x + 4a21 x2 + 8b1 c1 y + 4b21 y 2 + 4c21 + 4α12 x2 − 4α1 x + 1]/[8α1 x], and H1 (x, y) = [8a2 b1 xy + 8a2 c1 x + 4a22 x2 + 8b1 c1 y + 4b21 y 2 + 4c21 + 4α22 x2 − 4α2 x + 1]/[8α2 x], respectively. So they are not defined on the y-axis, and consequently we cannot have continuous piecewise differential systems. Case 3.2: Consider the solution s2 . Then solving the algebraic system (14) we obtain y1 = y2 , or y1 = −y2 − 2c1 /b1 . We conclude that there is a continuum of crossing periodic orbits. Case 3.3: Consider the solution s3 . Then the first integrals of systems (8) are H1 (x, y) = −[8a1 b2 xy + 8a1 c2 x + 4a21 x2 + 8b2 c2 y + 4b22 y 2 + 4c22 + 4α22 x2 + 4α2 x + 1]/[8α2 x], and H1 (x, y) = [8a2 b2 xy + 8a2 c2 x + 4a22 x2 + 8b2 c2 y + 4b22 y 2 + 4c22 + 4α22 x2 − 4α2 x + 1]/[8α2 x]. The same conclusion than in case 3.1. Case 3.4: Consider the solution s4 . Now solving the algebraic system (14) we obtain y1 = y2 , or y1 = −y2 − 2c2 /b2 . There are a continuum of crossing periodic orbits. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 10 B. Ghermoul, J. Llibre, T. Salhi Case 3.5: Consider the solution s5 . Then the first integrals of systems (8) are H1 (x, y) = [8a1 b2 xy + 8a1 c2 x + 4a21 x2 + 8b2 c2 y + 4b22 y 2 + 4c22 + 4α22 x2 − 4α2 x + 1]/[8α2 x], and H1 (x, y) = [8a2 b2 xy + 8a2 c2 x + 4a22 x2 + 8b2 c2 y + 4b22 y 2 + 4c22 + 4α22 x2 − 4α2 x + 1]/[8α2 x]. Again as case 3.1. Case 4: Assume that b1 = (−2a2 β2 b2 −α2 β22 +α2 b22 )/(2 (α2 b2 − a2 β2 )) and β1 = 0. Then the sixth equation of (16) becomes 2 α1 α2 β2 2 + 2a2 β2 b2 − α2 b2 2 = 0. 4(a2 β2 − α2 b2 ) Since we have β1 = 0 we cannot take α1 = 0 otherwise we get a contradiction with the first condition in (14). So we must take α2 β2 2 + 2a2 β2 b2 − α2 b2 2 = 0. By solving this equation, we obtain a2 = (b2 2 α2 − α2 β2 2 )/2b2 β2 , and by replacing a2 in b1 = (−2a2 β2 b2 − α2 β22 + α2 b22 )/(2 (α2 b2 − a2 β2 )) we get b1 = 0, which is a contradiction with the first condition in (14). Then we cannot have continuous piecewise differential system int this case.  Proof. [Proof of statement (b) of Theorem 2] In order that the piecewise differential system (8)-(9) be continuous they must coincide on x = 0, so the following algebraic system must be satisfied −2a2 β2 b1 c1 γ1 + a1 β1 b1 c2 γ2 + a1 β1 b1 c2 − a2 β2 b1 c1 − a2 β1 β2 γ12 − a2 β1 β2 γ1 +a1 β1 β2 γ2 + a2 β1 β2 c21 − a1 β1 β2 c22 + α2 β1 b1 γ12 + α2 β1 b1 γ1 − α1 β2 b1 γ2 −α2 β1 b1 c21 + α1 β2 b1 c22 + 2α2 b21 c1 γ1 − α1 b21 c2 γ2 + α2 b21 c1 − α1 b21 c2 = 0, −2a2 β2 b21 γ1 + a1 β1 b21 γ2 + a1 β1 b21 − a2 β2 b21 − a1 β1 β2 b1 c2 − 2a2 β12 β2 γ1 + a1 β1 β22 −a2 β12 β2 + 2α2 β12 b1 γ1 + α2 β12 b1 − α1 β22 b1 + 2α2 b31 γ1 − α1 b31 γ2 + α2 b31 −α1 b31 + α1 β2 b21 c2 = 0,  −β1 b21 + β12 = 0, 2a2 α1 β1 β2 γ1 + a2 α1 β1 β2 − a1 α2 β1 β2 + a2 α1 b1 b2 γ2 − 2a1 α2 b1 b2 γ1 + a2 α1 b1 b2 (18) −a1 α2 b1 b2 + 2a1 a2 b1 β2 γ1 − a1 a2 b2 β1 γ2 − a1 a2 b2 β1 + a1 a2 b1 β2 − 2a2 α1 b1 β2 c1 +a2 α1 b1 β2 c2 − 2a1 α2 b2 β1 c1 + 2a1 α2 b2 β1 c2 + 2a1 a2 β1 β2 c1 − a1 a2 β1 β2 c2 −2α2 α1 b2 β1 γ1 − α2 α1 b2 β1 + α2 α1 b1 β2 + 2α2 α1 b1 b2 c1 − 2α2 α1 b1 b2 c2 = 0, a2 α1 β2γ12 + a2 α1 β2 γ1 − a1 α2 β1 γ2 − 2a1 α2 b2 γ1 c1 + a2 α1 b1 c2 γ2 − a1 α2 b2 c1 +a2 α1 b1 c2 − a2 α1 β2 c21 + a1 α2 β1 c22 + 2a1 a2 β2 γ1 c1 − a1 a2 β1 c2 γ2 + a1 a2 β2 c1 −a1 a2 β1 c2 − α1 α2 b2 γ12 − α1 α2 b2 γ1 + α1 α2 b1 γ2 + α1 α2 b2 c21 − α1 α2 b1 c22 = 0, −2a1 β1 b1 + a1 b2 β1 − α1 β12 + α1 b21 − α1 b2 b1 = 0, with α1 b1 − a1 β1 6= 0 and α2 b2 − a2 β2 6= 0. We see easily that a necessary condition for a continuous piecewise system is β1 = 0 and this implies that α1 6= 0 and b1 6= 0, otherwise we have a contradiction with the condition α1 b1 − a1 β1 6= 0. Substituting β1 = 0 into the last equation of (18) we obtain b1 (b1 − b2 )α1 = 0, and consequently b2 = b1 , because b1 and α1 cannot be zero. Now in all cases of this proof we take β1 = 0, b2 = b1 , α1 6= 0 and b1 6= 0. Case 1: b1 c2 + β2 6= 0, β2 6= 0 and γ1 (γ1 + 1) 6= 0. Then from the first equation of (18) we get −2a2 β2 c1 γ1 − a2 β2 c1 + 2α2 b1 c1 γ1 − α1 b1 c2 + α2 b1 c1 + α1 β2 c22 γ2 = . α1 (b1 c2 + β2 ) From the second equation of (18) we obtain b1 (2γ1 + 1) (b1 (c2 − c1 ) + β2 ) (α2 b1 − a2 β2 ) α1 =  . β2 b21 + β22 October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 11 Using the fifth equation of (18) we have a1 = −b1 (−a2 b1 β2 − 2a2 b1 β2 c22 + 4a2 b1 β2 c1 c2 − 2a2 b1 β2 c12 − α2 β22 + a2 β22 c1 − a2 β22 c2 −α2 b1 β2 c1 + α2 b1 β2 c2 + 2α2 b21 c21 + 2α2 b21 c22 − 4α2 b21 c1 c2 )/(β2 (b21 + β22 )). Finally from the fourth equation of (18) we get b1 (c1 − c2 )(c21 − 2c2 c1 + c22 + γ12 + γ1 + 1) β2 = . γ1 (γ1 + 1) With these values of the parameters the piecewise differential systems (8)-(9) are continuous. Now we must solve the system H2 (0, y1 ) − H2 (0, y2 ) = 0, H3 (0, y1 ) − H3 (0, y2 ) = 0, (19) for studying the crossing periodic orbits and the crossing limit cycles of these continuous piecewise differ- ential systems. The first equation of (19) in this case is b1 (y1 − y2 )(2c1 + b1 y1 + b1 y2 ) = 0. (20) 1 + 2γ1 Since in order to have a crossing periodic orbit we must have y1 6= y2 , we obtain that y2 = −(2c1 +b1 y1 )/b1 . Substituting y2 into the second equation of (19) we get that y1 = y2 = −c1 /b1 . Therefore this piecewise differential systems has no crossing periodic orbits. Case 2: Assume that b1 c2 + β2 6= 0, β2 6= 0 and γ1 (1 + γ1 ) = 0. Then from the case 1 we have the same values for γ2 , α1 and a1 substituting in them γ1 = 0 or γ1 = −1. Then from the fourth equation of (18) we should have c2 = c1 otherwise we have non–continuous piecewise differential system. With these values the piecewise differential systems (8)-(9) are continuous. Again from (20) we obtain y2 = −(2c1 + b1 y1 )/b1 . Substituting y2 into the second equation of (19) we get that y1 = y2 = −c1 /b1 . So no crossing periodic orbits. Case 3: b1 c2 + β2 6= 0 and β2 = 0. Then c2 6= 0. Now as in the case 1, from first equation of system (18) we obtain γ2 = (c1 α2 (1 + 2γ1 ) − c2 α1 )/(c2 α1 ). Then the second equation of (18) becomes b31 (c2 − c1 )α2 (1 + 2γ1 ) − = 0. (21) c2 Which α2 and b1 cannot be zero from the condition of (14). So in this case we have three subcases satisfy (c2 − c1 )(1 + 2γ1 ) = 0. Subcase 3.1: 1 + 2γ1 6= 0. Then c2 = c1 to verify equation (21). The fifth equation of (18) becomes −(a1 − a2 )b21 α2(1 + 2γ1 ) = 0, so we must take a2 = a1 , otherwise the piecewise differential system cannot be continuous. Finally from the fourth equation of (18) we get α2 = (α1 (1 + γ1 + γ12 ))/(1 + 2γ1 ). With these values of the parameters the piecewise differential systems (8)-(9) are continuous. Now the first and second equation of (19) become b1 (y1 − y2 )(2c1 + b1 y1 + b1 y2 ) b1 (y1 − y2 )(2c1 + b1 y1 + b1 y2 ) = 0, = 0, 1 + 2γ1 (1 + γ1 + γ12 )2 respectively. Since the numerator of both equations coincide we have a continuum of crossing periodic orbits in this subcase. Subcase 3.2: c2 6= c1 . Then we must take 1 + 2γ1 = 0 to satisfy the equation (21). Using the fifth equation of (18) we obtain 2b21 (c1 − c2 )α1 α2 = 0, we see that cannot verify this equation under the conditions of this subcase. Then we cannot have continuous piecewise differential system in subcase. Subcase 3.3: 1 + 2γ1 = c2 − c1 = 0. Then the fifth equation of (18) is satisfied, and the fourth equation of (18) becomes −3b1 α1 α2 /4 = 0, which cannot be satisfied otherwise we have a contradiction with (14). So in this subcase the piecewise differential system cannot be continuous. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 12 B. Ghermoul, J. Llibre, T. Salhi Case 4: b1 c2 +β2 = 0 and c2 6= 0. The first equation of (18) gives α1 = (a2 c1 c2 +c1 α2 )(1+2γ1 )/(c2 (1+c22 )). The second equation of (18) becomes b31 (a2 c2 + α2 )(1 + 2γ1 )(c2 + c32 − c1 (1 + 2c22 + γ2 )) = 0. c2 (1 + c22 ) From the condition b1 c2 + β2 = 0 of this case we get c2 = −β2 /b1 and from the starting of the proof of this statement we have b1 = b2 , by substituting in a2 c2 + α2 = 0 gives a2 β2 − b2 α2 = 0; which is a contradiction with second condition in (14). If 1 + 2γ1 = 0, from the above formula of α1 we get α1 = 0 and since β1 = 0 we obtain a1 β1 − b1 α1 = 0; which is a contradiction with first condition in (14). So a2 c2 + α2 and 1 + 2γ1 cannot be zero, we must take c2 + c32 − c1 (1 + 2c22 + γ2 ) = 0. (22) Subcase 4.1: c1 = 0. Then from the second equation of (18) we have b31 (a2 c2 + α2 )(1 + 2γ1 ) = 0, Since a2 c2 + α2 and 1 + 2γ1 cannot be zero as we have seen above this last equation cannot be satisfy, and consequently the piecewise differential system cannot be continuous. Subcase 4.2: c1 6= 0. Then from (22) we get γ2 = (−c1 + c2 − 2c1 c22 + c32 )/c1 . The fifth equation of (18) is b21 (a2 c2 + α2 )(a2 c2 (1 + 2c21 − 3c1 c2 + c22 ) − a1 (c2 + c32 ) + c1 (2c1 − 3c2 )α2 )(1 + 2γ1 ) = 0. c2 (1 + c22 ) As we have seen in the previous case a2 c2 + α2 and 1 + 2γ1 cannot be zero. So we should have a2 c2 (1 + 2c21 − 3c1 c2 + c22 ) − a1 (c2 + c32 ) + c1 (2c1 − 3c2 )α2 = 0. Solving this equation we get a2 c2 + 2a2 c21 c2 − 3a2 c1 c22 + a2 c32 + 2c21 α2 − 3c1 c2 α2 a1 = . c2 (1 + c22 ) And solving the fourth equation of (18) we obtain p −c1 ± −c1 (3c1 + 4c31 − 4c2 − 12c21 c2 + 12c1 c22 − 4c32 ) γ1 = . 2c1 We assume that the squareroot which appears in the expression of γ1 is real, otherwise the piecewise differential system is not continuous. With these values of the parameters we have satisfied all equations of system (18). Then the piecewise differential systems (8)-(9) are continuous. Now the first equation of (19) becomes b1 c1 (y1 − y2 )(2c1 + b1 y1 + b1 y2 ) −p = 0. −c1 (3c1 + 4c31 − 4c2 − 12c21 c2 + 12c1 c22 − 4c32 ) In order to get crossing periodic orbits we need that 2c1 + b1 y1 + b1 y2 = 0, or equivalently y2 = (−2c1 − b1 y1 )/b1 . Substituting y2 into the second equation of (19) we get 4c31 (1 + c22 )(c1 + b1 y1 )3 = 0. c22 (−1 + 2c1 c2 − c22 + b1 c1 y1 )2 (1 + 2c21 − 2c1 c2 + c22 + b1 c1 y1 )2 By solving this equation we get y1 = −c1 /b1 and yield to y2 = −c1 /b1 . Then in this case the piecewise differential system cannot have crossing periodic orbits. Case 5: b1 c2 + β2 = 0 and c2 = 0. Then β2 = 0, and the first equation of system (18) becomes b21 c1 α2 (1 + 2γ1 ) = 0. Since b1 and α2 cannot be zero, so we have the cases verify c1 (1 + 2γ1 ) = 0, otherwise we have non-continuous piecewise differential system piecewise. Then we have the two following subcases. Subcase 5.1: c1 = 0 and 1 + 2γ1 6= 0. Then the second equation of system (18) becomes b31 (α2 + 2α2 γ1 − α1 (1+γ2 )) = 0. Since b1 cannot be zero we have that α2 +2α2 γ1 −α1 (1+γ2 ) = 0. By solving this equation we October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 13 get γ2 = (−α1 + α2 + 2α2 γ1 )/α1 . Thus system (18) is satisfied in this subcase and the piecewise differential system is continuous. Now the first and second equation of (19) become 2 b21 (y1 − y2 )(y1 + y2 ) b21 (y1 − y2 )(y1 + y2 ) =0 and =0 1 + 2γ1 1 + γ1 + γ12 respectively. This system has the two solutions y1 = y2 and y1 = −y2 . The last solution means that the piecewise differential system has a continuum of periodic orbits. Then no crossing limit cycles. Subcase 5.2: c1 6= 0 and 1 + 2γ1 = 0. The second equation of system (18) becomes √ −b31 α1 (1 + γ2 ) = 0. Since b1 and α1 cannot be zero, we have that γ2 = −1. The fifth equation becomes 3b21 α1 α2 = 0, but this is a contradiction because all the parameters which appear in it are nonzero. Therefore the piecewise differential system cannot be continuous. Subcase 5.3: c1 = 1 + 2γ1 = 0. The same proof and result as in subcase 5.2.  Proof. [Proof of statement (c) of Theorem 2] For studying the maximum number of limit cycles of the piecewise differential systems (8)-(10), we must solve system (15). Solving the first equation of (15) with respect to y1 we obtain y1 = y2 and  2b1 (2γ1 c1 + c1 ) + b21 (2γ1 y2 + y2 ) + β1 −2c21 + 2γ1 (γ1 + 1) + y2 (2β1 γ1 + β1 ) y1 = −  . (23) b21 + β12 (2γ1 + 2β1 y2 + 1) In order that the piecewise differential systems (8)-(10) be continuous they must coincide on x = 0, which means the following algebraic system must be satisfied. −6a2 b1 β2 γ1 c1 − 16a1 b2 β1 c2 γ2 − 3a2 b1 β2 c1 + 3a1 b2 β1 c2 − 3a2 β1 β2 γ12 − 3a2 β1 β2 γ1 +3a1 β1 β2 γ2 + 3a2 β1 β2 c21 + 4a1 β1 β2 c22 + 3α2 b2 β1 γ12 + 3α2 b2 β1 γ1 − 3α1 b1 β2 γ2 −3α2 b2 β1 c21 − 4α1 b1 β2 c22 + 6α2 b1 b2 γ1 c1 + 16α1 b1 b2 c2 γ2 + 3α2 b1 b2 c1 − 3α1 b1 b2 c2 = 0, −6a2 β2 b21 γ1 − 16a1 b22 β1 γ2 − 3a2 β2 b21 + 3a1 b22 β1 − 8a1 b2 β1 β2 c2 − 6a2 β12 β2 γ1 + 3a1 β1 β22 −3a2 β12 β2 + 6α2 b2 β12 γ1 − 3α1 β22 b1 + 3α2 b2 β12 + 6α2 b2 b21 γ1 + 16α1 b22 b1 γ2 + 3α2 b2 b21 −3α1 b22 b1 + 8α1 b2 β2 b1 c2 = 0, a2 b21 β2 β1 + 4a1 b22 β2 β1 + a2 β2 β13 − α2 b2 β13 − α2 b21 b2 β1 − 4α1 b1 b22 β2 = 0, 3a2 α1 β2 γ12 + 3a2 α1 β2 γ1 − 3a1 α2 β1 γ2 − 6a1 α2 b2 γ1 c1 − 16a2 α1 b1 c2 γ2 − 3a1 α2 b2 c1 (24) +3a2 α1 b1 c2 − 3a2 α1 β2 c21 − 4a1 α2 β1 c22 + 6a1 a2 β2 γ1 c1 + 16a1 a2 β1 c2 γ2 + 3a1 a2 β2 c1 −3a1 a2 β1 c2 − 3α1 α2 b2 γ12 − 3α1 α2 b2 γ1 + 3α1 α2 b1 γ2 + 3α1 α2 b2 c21 + 4α1 α2 b1 c22 = 0, 6a2 α1 β1 β2 γ1 + 3a2 α1 β1 β2 − 3a1 α2 β1 β2 − 16a2 α1 b1 b2 γ2 − 6a1 α2 b1 b2 γ1 + 3a2 α1 b1 b2 −3a1 α2 b1 b2 + 6a1 a2 b1 β2 γ1 + 16a1 a2 b2 β1 γ2 − 3a1 a2 b2 β1 + 3a1 a2 b1 β2 − 6a2 α1 b1 β2 c1 −16a2 α1 b1 β2 c2 − 6a1 α2 b2 β1 c1 − 8a1 α2 b2 β1 c2 + 6a1 a2 β1 β2 c1 + 16a1 a2 β1 β2 c2 −6α2 α1 b2 β1 γ1 − 3α2 α1 b2 β1 + 3α2 α1 b1 β2 + 6α2 α1 b1 b2 c1 + 8α2 α1 b1 b2 c2 = 0, 3a2 α1 β12 β2 − 3a2 α1 β2 b21 − 6a1 α2 b2 β1 b1 − 16a2 α1 b2 β2 b1 − 4a1 α2 b22 β1 + 6a1 a2 β1 β2 b1 +16a1 a2 b2 β1 β2 − 3α1 α2 b2 β12 + 3α1 α2 b2 b21 + 4α1 α2 b22 b1 = 0, together with conditions (14). Solving the sixth equation of (24) we get   a2 α1 β2 −3b21 − 16b2 b1 + 3β12 + α1 α2 b2 3b21 + 4b2 b1 − 3β12 a1 = , (25) 2α2 b2 (3b1 + 2b2 ) β1 − 2a2 (3b1 + 8b2 ) β1 β2 except when denominator 2α2 b2 (3b1 + 2b2 ) β1 − 2a2 (3b1 + 8b2 ) β1 β2 = 0. This denominator vanishes if and October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 14 B. Ghermoul, J. Llibre, T. Salhi only if one of the following solutions holds: d1 = {β1 = 0}, d2 = {β2 = αa22b(3b 2 (3b1 +2b2 ) 1 +8b2 ) }, d3 = {a2 = 0, b1 = −2b2 /3}, d4 = {a2 = 0, α2 = 0}, (26) d5 = {b1 = −8b2 /3, α2 = 0}, d6 = {a2 = 0, b2 = 0}, d7 = {b1 = 0, b2 = 0}. To continue solving the algebraic system (24) we must discuss all cases when 2α2 b2 (3b1 + 2b2 ) β1 − 2a2 (3b1 + 8b2 ) β1 β2 = 0 and 2α2 b2 (3b1 + 2b2 ) β1 − 2a2 (3b1 + 8b2 ) β1 β2 6= 0. Case 1: Assume that 2α2 b2 (3b1 + 2b2 ) β1 − 2a2 (3b1 + 8b2 ) β1 β2 6= 0. (27) Using a1 given by (25) we solve the third equation of the algebraic system (24) and we obtain the following solution α1 = a2 (3b1 +8b2 )β1 β6b2 −α 2β 2 b2 (3b1 +2b2 )β1 , 2 2 the only allowed solution (See supplementary section 5.11) for which conditions (27) and (14) are verified, then we take this value of α1 and solve the second equation of (24) and we get the solution (See supple- mentary section 5.12) 3b2 (8β γ +β +4β2 )−8β1 β2 b2 c2 +3β1 β22 γ2 = 2 2 1 1 16b 2β , 2 1 which does not contradict conditions (27) or (14), so for this value of γ2 solving the fifth equation of system (24) we obtain 2 −6β1 c1 −8β1 c2 ) β2 = b2 (8b2 γ1 +b1 (6γ1 +3)+4b 3β1 , All real solutions (See supplementary section 5.13) are in contradiction with (27) or (14) except the so- lution v1 . Replacing β2 in the remaining equations of the algebraic system (24) and by solving the first equation of this system we obtain the following allowed solutions (See supplementary section 5.14) w1 and w2 √ 3S 2 (15b1 +56b2 )β1 (2γ1 +1)− b2 2 −30β1 c1 1 +β1 w1 = {c2 = 112β12 } and √ 3S+β1 (b21 +β12 )(15b1 (2γ1 +1)+56b2 (2γ1 +1)−30β1 c1 ) w2 = {c2 = 112β12 (b21 +β12 ) }, whith S = β12 (b21 + β12 )(36β1 b31 c1 (2γ1 + 1) + b21 (448(2b2 γ1 + b2 )2 − 3β12 (12c21 + 12γ1 (γ1 + 1) + 31)) + 4β1 b1 c1 (2γ1 + 1)(9β12 − 448b22 ) + 448b22 β12 (4c21 + 1) − 9b41 (2γ1 + 1)2 − 12β14 (3c21 + 7)). Without solving the fourth equation of the algebraic system (24) to get a piecewise continuous differential systems (8)-(10). We solve the algebraic system (15). In these two cases replacing the value of y1 given by (23) into the second equation of (15) its left hand side is equivalent to a long polynomial of second degree with respect to y2 . Subcase 1.1: If we consider w1 . Then the polynomial of second degree with respect to y2 , after equating it to zero (See supplementary section 5.15), we compute the two roots of this polynomial we get two different values y21 and y22 . By replacing y21 we get y11 , and replacing y22 we get y12 . Since we obtain that y11 = y22 and y12 = y21 , it follows that in this subcase the system has at most one limit cycle. Subcase 1.2: If we consider w2 . Then the polynomial of second degree with respect to y2 , after equating it to zero (See supplementary section 5.16), we compute the two roots of this polynomial we get two different values y21 and y22 . By replacing y21 we get y11 , and replacing y22 we get y12 . Due to the facts that y11 = y22 and y12 = y21 , the system in this subcase has at most one limit cycle. Case 2: Assume that 2α2 b2 (3b1 + 2b2 ) β1 − 2a2 (3b1 + 8b2 ) β1 β2 = 0. Here we consider the subcases d1 , d2 , d3 , d5 and d7 defined by (26) except d4 and d6 because are in contradiction with (14). Subcase 2.1: Consider the set d1 of (26). In this case we have β1 = 0. The third equation (24) becomes 4b22 β2 = 0, this gives b2 = 0 or β2 = 0. b2 and β2 cannot vanish at the same time otherwise we get a October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 15 contradiction with (14). If b2 = 0 and β2 6= 0, then the sixth equation of (24) becomes −3a2 b1 β2 = 0 which gives a2 = 0 or b1 = 0, since β1 = 0 and b2 = 0 we get a contradiction with (14). Then we have no continuous piecewise differential systems (8)-(10) in this case. Finally if we have β2 = 0 and b2 6= 0, then the sixth equation of (24) becomes b2 (3b1 + 4b2 )α2 = 0, which gives b2 = 0, α2 = 0 or 3b1 + 4b2 = 0, and since β2 = 0, all these parameters give a contradiction with (14) except b2 = −3b1 /4. Solving the four remaining equation of (24) we get the only solutions (See supplementary section 5.17) s1 and s2 that verify (14) as follows s1 = {c1 = −4c2 /3, γ1 = −1/2, γ2 = 3/16} and α22 s2 = {a2 = −3a1 /4, c1 = −4c2 /3, γ1 = − 2α 3α1 − 1/2, γ2 = 3/16 − 3α21 }. 2 Subcase 2.1.1: Consider the set s1 . In this case we have c1 = −4c2 /3, γ1 = −1/2 and γ2 = 3/16. Now the algebraic system (24) is verified. But the first integrals of systems (8) and (10) are H1 (x, y) = [72a1 b1 xy − 96a1 c2 x + 36a21 x2 − 96b1 c2 y + 36b21 y 2 + 64c22 + 36α12 x2 − 36α1 x + 9]/[72α1 x], and 2 H3 (x, y) = −[ −96a2 b1 xy + 128a2 c2 x + 64a22 x2 − 96b1 c2 y + 36b21 y 2 + 64c22 − 48α2 x + 9 ]/[64α2 x], respectively, note that they are not defined on the y-axis, which means that there are no continuous piecewise differential systems formed by (8) and (10). Subcase 2.1.2: Consider the set s2 . Which gives the parameters a2 = −3a1 /4, c1 = −4c2 /3, γ1 = −2α2 /(3α1 ) − 1/2 and γ2 = 3/16 − α22 /3α12 . With these parameters we get continuous piecewise differential systems (8)-(10). Solving the algebraic system (15) we obtain y1 = y2 or y1 = 8c2 /(3b1 ) − y2 , this provided a continuum of periodic orbits. Subcase 2.2: Consider the set d2 of (26). In this case we have β2 = α2 b2 (3b1 + 2b2 )/(a2 (3b1 + 8b2 )). Since b2 = 0 or α2 = 0 is in contradiction with a2 β2 − b2 α2 6= 0, then we have either β2 = 0 and b2 = −3b1 /2, (28) or a2 (3b1 + 8b2 ) β2 α2 = . (29) b2 (3b1 + 2b2 )   Using (28) the third and the sixth equations of (24) become −β1 b21 + β12 = 0 and −α1 b21 + β12 = 0, respectively. Therefore either β1 = α1 = 0 or b1 = β1 = 0 and this contradicts (14). Then in this case we cannot have a continuous piecewise differential system.  Now using the value of α2 from (29). The sixth equation of (24) becomes −α1 b21 + β12 = 0. Since b1 = β1 = 0 contradicts (14) we must take α1 = 0. Solving now the fifth equation of (24), all solutions (See supplementary section 5.18) give a contradiction with (14) except s1 and s2 given as follows 2b2 β1 (−2β2 b2 (9c1 +8c2 )+b22 (16γ2 −3)−12β22 )+3b1 (b22 (β1 (16γ2 −3)−6β2 )+8β1 β2 b2 c2 −3β1 β22 ) s1 = {γ1 = 36b1 b22 β2 } and (30) 1 b2 (16γ2 −3) 12β2 s2 = {b1 = 0, c1 = 18 ( β2 − b2 − 16c2 )}. Then we must study these two cases separately. Subcase 2.2.1: Consider the set s1 of (30). By solving the third equation of (24), all solutions (See supplementary section 5.19) give a contradiction with (14) except u1 and u2 √ 2a1 b2 (3b1 +2b2 )−3a2 b21 u1 = {β1 = − √ √ 3 a2 } and √ 2 (31) 2a1 b2 (3b1 +2b2 )−3a2 b1 u2 = {β1 = √ √ 3 a }. 2 Then we must study these two cases separately. Subcase 2.2.1.1: Consider the set u1 of (31). Solving the second equation of (24) we get from the sets of all real solutions (See supplementary section 5.20) the only allowed solution is 3b1 (8β2 b2 c2 + b22 (16γ2 − 3) − 3β22 ) + 4b2 (−16β2 b2 c2 + b22 (16γ2 − 3) − 12β22 ) c1 = . 72b22 β2 October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 16 B. Ghermoul, J. Llibre, T. Salhi The first equation of (24) gives the only allowed solution among the sets of real solutions (See supplemen- tary section 5.21) which is given by 4b2 a2 (32β2 b32 c2 (3 − 16γ2 ) + β22 b22 (256c22 − 384γ2 + 153) + 384β23 b2 c2 + b42 (3 − 16γ2 )2 + 144β24 ) 2 b1 = − b2 . 9a1 (32β2 b32 c2 (3 − 16γ2 ) + 2β22 b22 (128c22 + 24γ2 + 9) − 48β23 b2 c2 + b42 (3 − 16γ2 )2 + 9β24 ) 3 Solving the fourth equation of (24) gives from all real solutions (See supplementary section 5.22), the only allowed solution is   3a1 32β2 b32 c2 (3 − 16γ2 ) + 8β22 b22 32c22 + 6γ2 + 9 − 48β23 b2 c2 + b42 (3 − 16γ2 ) 2 + 63β24 a2 = −  . 4 32β2 b32 c2 (3 − 16γ2 ) + β22 b22 256c22 − 384γ2 + 153 + 384β23 b2 c2 + b42 (3 − 16γ2 ) 2 + 144β24 Now we have a continuous piecewise differential systems (8)-(10). We solve the algebraic system (15) with respect to √ y1 and y2 , we get one of the following √ sets of solutions R+8β2 b2 c2 +b22 (16γ2 −3)−3β22 R+8β2 b2 c2 +b22 (16γ2 −3)−3β22 s0 = {y1 = − 24b22 β2 , y2 = − 24b22 β2 }, √ √ R−8β2 b2 c2 −16b22 γ2 +3b22 +3β22 R−8β2 b2 c2 −16b22 γ2 +3b22 +3β22 s1 = {y1 = 24b22 β2 , y2 = 24b22 β2 }, 3 2 4 2 √ √ 2 3 5 27β2 b2 +β2 b2 (128c2 +96γ2 −27)− 3 T −144β2 b2 c2 +8b2 c2 (3−16γ2 ) s2 = {y1 = 8b42 (−16β2 b2 c2 +b22 (16γ2 −3)+6β22 ) , 3 2 4 2 √ √ 2 3 5 27β2 b2 +β2 b2 (128c2 +96γ2 −27)+ 3 T −144β2 b2 c2 +8b2 c2 (3−16γ2 ) y2 = 8b42 (−16β2 b2 c2 +b22 (16γ2 −3)+6β22 ) }, √ √ 27β23 b22 +β2 b42 (128c22 +96γ2 −27)+ 3 T −144β22 b32 c2 +8b52 c2 (3−16γ2 ) s3 = {y1 = 8b42 (−16β2 b2 c2 +b22 (16γ2 −3)+6β22 ) , 3 2 4 2 √ √ 2 3 5 27β2 b2 +β2 b2 (128c2 +96γ2 −27)− 3 T −144β2 b2 c2 +8b2 c2 (3−16γ2 ) y2 = 8b42 (−16β2 b2 c2 +b22 (16γ2 −3)+6β22 ) }. Where R = (8β2 b2 c2 + b22 (16γ2 − 3) − 3β22 )2 − 48b22 β2 (b2 c2 (16γ2 − 3) − β2 (4c22 + 3γ2 )) and T = b42 (−16β2 b2 c2 + 2b22 (8γ2 − 3) + 3β22 )(32β2 b32 c2 (3 − 16γ2 ) + 2β22 b22 (128c22 + 120γ2 − 27) − 240β23 b2 c2 + b42 (3 − 16γ2 )2 + 45β24 ). Here s0 and s1 give y1 = y2 ; y1 of s2 is equal to y2 of s3 , and y2 of s2 is equal to y1 of s3 . Then the continuous piecewise differential systems (8)-(10) can have at most one limit cycle. Numerical example. 4395 370 a1 = −1, a2 = 19252 , b1 = 231 , b2 = − 21 , c1 = 4022 11 465 231 , c2 = − 2 , α1 = 0, α2 = − 4813 , β1 =  q  q 1 − 77 4 5541 5 1 , β2 = 1, γ1 = − 154 233 1847 1 15 − 2 and γ2 = 10. Figure 1 shows the crossing limit cycle in this case. Here the points of intersection with the y−axis are y1 = −9.83073.. and y2 = −15.1693... Fig. 1. Crossing limit cycle formed by systems (8) and (10). October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 17 Subcase 2.2.1.2: Consider the set u2 of (31). We follow the same steps as in Subcase 2.2.1.1 (See supplementary section 5.23). We get a continuous piecewise differential systems (8)-(10). We solve the algebraic system (15) with respect to y1 and y2 , and we get one of the same sets of solutions s0 , s1 , s2 and s3 as in Subcase 2.2.1.1. Then the continuous piecewise differential systems (8)-(10) can have at most one limit cycle. Subcase 2.2.2: Consider the set s2 of (30). Solving the second equation of (24) we get a2 = 4a1 b22 /(3β12 ), then we solve the third equation of (24) and we get, among all sets of solutions (See supplementary section 5.24), the only allowed solution 3b22 (8β2 γ1 + β1 + 4β2 ) − 8β1 β2 b2 c2 + 3β1 β22 γ2 = . 16b22 β1 Then for this value of γ2 solving simultaneously the first and the fourth equation of (24) we get the only one real solution β2 = 0, which gives a contradiction with (14). Then there are no continuous piecewise differential systems (8)-(10) in this case. Subcase 2.3: Consider the set d3 of (26). In this case we have a2 = 0 and b1 = −2b2 /3. With these parameters the sixth equation of (24) becomes α1 4b22 + 9β12 = 0. Solving now the third and the fourth equations of (24) we conclude that all sets of real solutions (See supplementary section 5.25) give a contradiction with (14) except the following one α2 (4b22 +9β12 ) 6b c γ +3b c +4β c2 {β2 = 36a1 b2 , γ2 = − 2 1 1 3β21 1 1 2 }. Then for these values of β2 and γ2 solving the first, the second, and the fifth equations of (24) all obtained solutions are complex. Then we have no continuous piecewise differential systems (8)-(10) in this case. Subcase 2.4: Consider the set 2 2  d5 of (26). Here we have α2 = 0 and b1 = −8b2 /3. The sixth equation of (24) becomes α1 64b2 + 9β1 = 0. Since α2 = 0 we must have b2 6= 0, otherwise we get a contradiction with (14), which yields to α1 = 0. Solving now the third and the fourth equations of (24) simultaneously, all real solutions (See supplementary section 5.26) give a contradiction with (14) except the following one 1 9β12 3 c1 (2β2 γ1 +β2 ) s1 = {a1 = 36 a2 (− b22 − 64), γ2 = 16 (1 − β 1 c2 )}, 1 9β12 s2 = {a1 = 36 a2 (− b2 − 64), c2 = 0, γ1 = −1/2} and (32) 2 1 9β12 s3 = {a1 = 36 a2 (− b22 − 64), c1 = 0, c2 = 0}, which must be discussed separately in the following subcases. Subcase 2.4.1: Consider the set s1 of (32). We solve now the remaining equations, the first, the second and the fifth equations of (24). All real solutions (See supplementary section 5.27) give a contradiction with (14) except the following two solutions q b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 3β12 (16b22 +9β12 ) β 1 c2 u1 = {c1 = −8c2 /3, 128b22 β1 −9β13 , γ1 = − q + b2 − 21 } and 8 b22 β12 (128b22 −9β12 )(16b22 +9β12 ) q (33) b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 3β12 (16b22 +9β12 ) β 1 c2 1 u2 = {c1 = −8c2 /3, β2 = 9β13 −128b22 β1 , γ1 = q + b2 − 2 }. 8 b22 β12 (128b22 −9β12 )(16b22 +9β12 ) Then this case is divided into two other subcases. Subcase 2.4.1.1: Consider the set u1 of (33). We have now a continuous piecewise differential systems (8)-(10). Solving the first equation of (15) with respect to y1 and y2 we get y2 = y1 which does not give limit cycles for the piecewise differential systems (8)-(11) in this subcase (See supplementary section 5.28). Subcase 2.4.1.2: Consider the set u2 of (33). We have now a continuous piecewise differential systems (8)- (10). As in Subcase 2.4.1.2 we get y2 = y1 which does not give limit cycles for the piecewise differential systems (8)-(11) in this subcase (See supplementary section 5.29). Subcase 2.4.2: Consider the set s2 of (32). Solving now the remaining equations, the first, the second and the fifth equations of (24) simultaneously. All solutions (See supplementary section 5.30) give a contradiction with (14). Then we have no continuous piecewise differential systems in this case. Subcase 2.4.3: Consider the set s3 of (32). We solve the remaining equations, the first, the second and the fifth equations of (24). All sets of solutions (See supplementary section 5.31) give a contradiction with October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 18 B. Ghermoul, J. Llibre, T. Salhi (14) except the sets u1 and u2 . q 16β1 b42 +9β13 b22 −36β12 b22 +3 b22 β12 (1008β12 b22 +2048b42 −81β14 )+512b42 u1 = {β2 = q , γ1 = − 1024b42 −72b22 β12 , b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 54b22 9 γ2 = 2 −128b2 + 16 }, 9βq 1 2 q (34) b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 36β12 b22 +3 b22 β12 (128b22 −9β12 )(16b22 +9β12 )−512b42 u2 = {β2 = 9β13 −128b22 β1 , γ1 = 8b22 (128b22 −9β12 ) , 54b22 9 γ2 = 9β1 −128b22 2 + 16 }. Then two other subcases bifurcate from this case. Subcase 2.4.3.1: Consider the set u1 of (34). Now we have a continuous piecewise differential systems (8)- (10). Solving the algebraic system (15) we get y1 = y2 , then there are no periodic orbits, and consequently no limit cycles. Subcase 2.4.3.2: Consider the set u2 of (34). By solving the algebraic system (15) we get y1 = y2 , then there are no periodic orbits in this case. Subcase 2.5: Consider the set d7 of (26), which means b1 = b2 = 0. By solving the algebraic system (24). All sets of real solutions (See supplementary section 5.32) give no continuous piecewise differential systems can be found, because all solutions give a contradiction with (14).  Proof. [Proof of statement (d) of Theorem 2] In order that the piecewise differential systems (8)-(11) be continuous they must coincide on x = 0, so these systems must satisfy the following algebraic system −6a2 b1 β2 γ1 c1 + 8a1 b2 β1 c2 γ2 − 3a2 b1 β2 c1 + 3a1 b2 β1 c2 − 3a2 β1 β2 γ12 + 4a1 β1 β2 γ22 −3a2 β1 β2 γ1 + 3a1 β1 β2 γ2 + 3a2 β1 β2 c21 − 16a1 β1 β2 c22 + 3α2 b2 β1 γ12 − 4α1 b1 β2 γ22 +3α2 b2 β1 γ1 − 3α1 b1 β2 γ2 − 3α2 b2 β1 c21 + 16α1 b1 β2 c22 + 6α2 b1 b2 γ1 c1 − 8α1 b1 b2 c2 γ2 +3α2 b1 b2 c1 − 3α1 b1 b2 c2 = 0, −6a2 β2 b21 γ1 + 8a1 b22 β1 γ2 − 3a2 β2 b21 + 3a1 b22 β1 − 24a1 b2 β1 β2 c2 − 6a2 β12 β2 γ1 + 8a1 β1 β22 γ2 +3a1 β1 β22 − 3a2 β12 β2 − 8α1 β22 b1 γ2 + 6α2 b2 β12 γ1 − 3α1 β22 b1 + 3α2 b2 β12 + 6α2 b2 b21 γ1 −8α1 b22 b1 γ2 + 3α2 b2 b21 − 3α1 b22 b1 + 24α1 b2 β2 b1 c2 = 0, −3a2 b21 β2 β1 − 8a1 b22 β2 β1 − 3a2 β2 β13 + 4a1 β23 β1 + 3α2 b2 β13 + 3α2 b21 b2 β1 − 4α1 b1 β23 +8α1 b1 b22 β2 = 0, 3a2 α1 β2 γ12 − 4a1 α2 β1 γ22 + 3a2 α1 β2 γ1 − 3a1 α2 β1 γ2 − 6a1 α2 b2 γ1 c1 + 8a2 α1 b1 c2 γ2 (35) −3a1 α2 b2 c1 + 3a2 α1 b1 c2 − 3a2 α1 β2 c21 + 16a1 α2 β1 c22 + 6a1 a2 β2 γ1 c1 − 8a1 a2 β1 c2 γ2 +3a1 a2 β2 c1 − 3a1 a2 β1 c2 − 3α1 α2 b2 γ12 + 4α1 α2 b1 γ22 − 3α1 α2 b2 γ1 + 3α1 α2 b1 γ2 +3α1 α2 b2 c21 − 16α1 α2 b1 c22 = 0, 6a2 α1 β1 β2 γ1 − 8a1 α2 β1 β2 γ2 + 3a2 α1 β1 β2 − 3a1 α2 β1 β2 + 8a2 α1 b1 b2 γ2 − 6a1 α2 b1 b2 γ1 +3a2 α1 b1 b2 − 3a1 α2 b1 b2 + 6a1 a2 b1 β2 γ1 − 8a1 a2 b2 β1 γ2 − 3a1 a2 b2 β1 + 3a1 a2 b1 β2 −6a2 α1 b1 β2 c1 + 8a2 α1 b1 β2 c2 − 6a1 α2 b2 β1 c1 + 32a1 α2 b2 β1 c2 + 6a1 a2 β1 β2 c1 − 8a1 a2 β1 β2 c2 −6α2 α1 b2 β1 γ1 + 8α2 α1 b1 β2 γ2 − 3α2 α1 b2 β1 + 3α2 α1 b1 β2 + 6α2 α1 b1 b2 c1 − 32α2 α1 b1 b2 c2 = 0, −4a1 α2 β1 β22 + 3a2 α1 β12 β2 − 3a2 α1 β2 b21 − 6a1 α2 b2 β1 b1 + 8a2 α1 b2 β2 b1 + 16a1 α2 b22 β1 +6a1 a2 β1 β2 b1 − 8a1 a2 b2 β1 β2 + 4α1 α2 β22 b1 − 3α1 α2 b2 β12 + 3α1 α2 b2 b21 − 16α1 α2 b22 b1 = 0, together with conditions (14). Solving the second and the fifth equation of (35). Regarding the sets of all real solutions (See supple- October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 19 mentary section 5.33), the only allowed solutions which do not contradict (14) are the following s1 = {c1 = [−3b21 (2γ1 + 1)(4α2 b2 − a2 β2 ) + b1 (9a1 β2 b2 (2γ1 + 1) + α1 β22 (8γ2 + 3) +4α1 b22 (8γ2 + 3)) − β1 (3(2γ1 + 1)(−a2 β2 β1 + 4α2 b2 β1 − 3α1 b2 β2 ) +a1 (8γ2 + 3)(4b22 + β22 ))]/[18b2 β2 (α1 b1 − a1 β1 )] if b2 β2 6= 0, (−3b21 (2γ1 +1)(α2 b2 −a2 β2 )−β1 (3β1 (2γ1 +1)(α2 b2 −a2 β2 )+a1 (8γ2 +3)(b22 +β22 ))+α1 b1 (8γ2 +3)(b22 +β22 ) c2 = (24b2 β2 (α1 b1 −a1 β1 ) }, a1 β2 b1 (6γ1 +3)−3α2 β1 (2β1 γ1 +β1 )+β1 β2 (−8a1 c2 +6α1 γ1 +3α1 )−3b21 (2α2 γ1 +α2 )+8α1 β2 b1 c2 s2 = {b2 = 0, c1 = , (36) 6β2 (α1 b1 −a1 β1 ) −3a2 (2γ1 +1)(b21 +β12 ) γ2 = 8β2 (α1 b1 −a1 β1 ) − 3/8} and −3a2 (2γ1 +1)(b21 +β12 )+a1 b2 (b1 (6γ1 +3)−32β1 c2 )+α1 b2 (32b1 c2 +6β1 γ1 +3β1 ) s3 = {c1 = 6b2 (α1 b1 −a1 β1 ) , β2 = 0, 3(β1 (α2 β1 (2γ1 +1)+a1 b2 )+b21 (2α2 γ1 +α2 )−α1 b1 b2 ) γ2 = 8b2 (α1 b1 −a1 β1 ) }, which are studied in the following three cases. Case 1: Consider the set s1 of (36). Solving the third equation of (35) we get one of the following sets of allowed real solutions (See supplementary section 5.34)  β2 4β22 (α1 b1 − a1 β1 ) + 3a2 b21 β1 + 8a1 b22 β1 + 3a2 β13 − 8α1 b1 b22 u1 = {α2 =  } if b2 β1 6= 0, 3b2 β1 b21 + β12 β2 u2 = {b2 = − √ , β1 = 0} and (37) 2 β2 u3 = {b2 = √ , β1 = 0}, 2 which do not contradict (14) and b2 β2 6= 0. Subcase 1.1: Consider the set u1 of (37). We solve the sixth equation of (35) to get one of the following allowed real solution (See supplementary section 5.35)  4β22 (a1 β1 − α1 b1 ) + b2 6a1 β1 b1 − 16a1 b2 β1 + 3α1 β12 − 3α1 b21 + 16α1 b2 b1 a2 =  }, (38) 3β1 b21 + β12 which does not contradicts (14), b2 β2 6= 0 and b2 β1 6= 0. We solve the first equation of (35) to get one of the following real solutions (See supplementary section 5.36) (β1 (8γ2 +3)−4β2 ) z1 = {γ1 = 8β2 9(−b22 β12 β24 (9b21 +32b22 +9β12 −16β22 )(β22 −2b22 )(4β22 (9b21 −32b22 +9β12 )+b22 (9b21 −256b22 +9β12 )−16β24 ))1/2 − 8β22 (β22 −2b22 )(−4β22 (9b21 −32b22 +9β12 )+b22 (−9b21 +256b22 −9β12 )+16β24 ) }, (β1 (8γ2 +3)−4β2 ) z2 = {γ1 = 8β2 9(−b22 β12 β24 (9b21 +32b22 +9β12 −16β22 )(β22 −2b22 )(4β22 (9b21 −32b22 +9β12 )+b22 (9b21 −256b22 +9β12 )−16β24 ))1/2 + 2 2 2 2 2 2 2 2 2 2 2 4 }, q 8β2 (β2 −2b2 )(−4β2 (9b1 −32b2 +9β1 )+b2 (−9b1 +256b2 −9β1 )+16β2 ) 32β22 β2 (39) z3 = {b1 = − 31 2 3 − 9β1 , b2 = − 6 }, √ q 32β22 β2 z4 = {b1 = 13 − 9β12 , b2 = − √ }, q3 6 32β2 2 β2 z5 = {b1 = − 13 2 3 − 9β1 , b2 = 6 } and √ q 32β22 β2 z6 = {b1 = 13 2 3 − 9β1 , b2 = 6 }. √ The only allowed solutions which do not contradict (14), b2 β2 6= 0 and b2 β1 6= 0. Again these solutions gives six other different cases. Subcase 1.1.1: Consider the set z1 of (39). Now the remaining fourth equation of (35) becomes     β2 b22 + β22 (α1 b1 − a1 β1 ) 2 12b1 β22 − 6b22 + b2 128b22 + 9β12 − 32β22 + 9b2 b21 = 0,  since b22 + β22 (α1 b1 − a1 β1 ) 6= 0, otherwise we have a contradiction with (14), and from b2 β2 6= 0, we have β2 6= 0. This last equation is equivalent to the following one   12b1 β22 − 6b22 + b2 128b22 + 9β12 − 32β22 + 9b2 b21 = 0. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 20 B. Ghermoul, J. Llibre, T. Salhi From b2 β2 6= 0 we have b2 6= 0, then we can get the solution for this last equation by solving it with respect to b1 , and hence we get the solutions for the fourth equation of (35). In fact we have the two solutions q  ± − 9β12 + 16β22 b22 + 16b42 + 4β24 + 12b22 − 2β22 b1 = . 3b2 With every one of these two values for b1 we have a continuous piecewise differential systems (8)-(11). Moreover, using all the parameters which provide the continuity of the system except the final value b1 , we solve the algebraic system (15), we get only one pair (y1 , y2 ) excluding y1 = y2 and permutation between y1 and y2 . Then the continuous piecewise differential systems (8)-(11) can have at most one limit cycle. Numerical example. √ √  a1 = 1, a2 = 173549 260215428 13039+147712067 , b1 = 4501 2308 − 13039 , b2 = 3/2, √ q √  q √  19( 13039−2308) 1 6995470812 13039+10285238052791 3 254(5052 13039+1718561) c1 = − 5040 − 8 76047982270 , c 2 = 112 95 − 14 299401505 , √  q √  7(34487 13039+7413985) 9 10(5052 13039+1718561) α1 = 1, α2 = 477061618 , β 1 = 11/10, β2 = 7/5, γ 1 = 112 17 − 231 7604798227 and γ2 = 2. The crossing limit cycle is shown in Figure 2. Fig. 2. Crossing limit cycle formed by systems (8)-(11). Zooming in the graph near the points of intersection y1 = −0.493173 and y2 = 0.58115 with the y−axis, we get the graph as shown in Figure 3. Fig. 3. Zooming in the graph of crossing limit cycle formed by systems (8)-(11). October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 21 Subcase 1.1.2: Consider the set z2 of (39). The remaining fourth equation of (35) becomes     β2 b22 + β22 12b1 β22 − 6b22 + b2 128b22 + 9β12 − 32β22 + 9b2 b21 (α1 b1 − a1 β1 ) 2 = 0, This last equation is equivalent to the following one   12b1 β22 − 6b22 + b2 128b22 + 9β12 − 32β22 + 9b2 b21 = 0. From b2 β2 6= 0 we have b2 6= 0, then we can get all solution for this last equation by solving it with respect to b1 . Hence we get the two solutions for the fourth equation of (35) q  ± − 9β12 + 16β22 b22 + 16b42 + 4β24 + 12b22 − 2β22 b1 = . 3b2 With every one of these two values for b1 we have a continuous piecewise differential systems (8)-(11). Again using all parameters which provide the continuity except the final value b1 , we solve the algebraic system (15), and we get only one pair (y1 , y2 ) excluding y1 = y2 and permutation between y1 and y2 . Then the continuous piecewise differential systems (8)-(11) can have at most one limit cycle. Subcase 1.1.3: Consider the set z3 of (39). Here we do not need to solve the fourth equation of (35) to get a continuous piecewise differential systems (8)-(11), because solving the algebraic system (15) we get only one pair (y1 , y2 ) except a permutation between y1 and y2 which are y1 = [β15 (−(704γ2 (4γ2 + 3) + 639))(800γ2 (4γ2 + 3) + 693) + 16β2 β14 (2γ1 + 1)(8γ2 + 3)(31264γ2 (4γ2 + 3) + 27549) −√ 672β13 (3904γ 2 2 3 √2 (4γ2 + 3) + 2601)(2β2 γ1 + β2 ) + 374272β1 (8γ2 2+ 3)(2β2 γ1 + β2 ) − 311296β1 (2β2 γ1 + 4 β2 ) +12 6(2γ1 +1) R]/[24β1 (32β√1 β√ 2 (β1 (8γ2 +3)−4(2β2 γ1 +β2 ))(β1 (164γ2 (4γ2 +3)+153)−82β2 β1 (2γ1 + 2 1)(8γ2 + 3) + 164(2β2 γ1 + β2 ) ) − 6 R)] and √ √ y2 = [ 6 R + 8β1 β2 (−4β13 (8γ2 + 3)(164γ2 (4γ2 + 3) + 153) + 9β2 β12 (2γ1 + 1)(640γ2 (4γ2 + 3) + 387) − 912β1 (8γ2 +3)(2β2 γ1 +β2 )2 +512(2β2 γ1 +β2 )3 )]/[48β12 β22 (β12 (704γ2 (4γ2 +3)+639)−352β2 β1 (2γ1 +1)(8γ2 + 3) + 704(2β2 γ1 + β2 )2 )], where R = −β12 β22 (β12 (224γ2 (4γ2 + 3) + 207) − 112β2 β1 (2γ1 + 1)(8γ2 + 3) + 224(2β2 γ1 + β2 )2 )(β12 (512γ2 (4γ2 + 3) + 531) − 256β2 β1 (2γ1 + 1)(8γ2 + 3) + 512(2β2√ γ1 + β2 )2 )2 . And since from z10 of (39) we have b1 = − S/3, with S = 32β22 /3 − 9β12 . This provided that S ≥ 0 and R ≥ 0 which are verified if and only if β1 = 0, this value of β1 is not allowed because we have already from (38)  4β22 (a1 β1 − α1 b1 ) + b2 6a1 β1 b1 − 16a1 b2 β1 + 3α1 β12 − 3α1 b21 + 16α1 b2 b1 a2 =  . 3β1 b21 + β12 This proves that the possible continuous piecewise differential systems (8)-(11) have no limit cycles. Subcase 1.1.4: Consider the set z4 of (39). This case follows in a similar results to Subcase 1.1.3. Then we have no limit cycles for the possible continuous piecewise differential systems (8)-(11). Subcase 1.1.5: Consider the set z5 of (39). Again this case gives similar results to Subcase 1.1.3. Subcase 1.1.6: Consider the set z6 of (39). Also this case gives similar results to Subcase 1.1.3. Subcase 1.2: Consider the set u2 of (37). Before solving the remaining equations of (35) to get a continuous piecewise differential systems (8)-(11), we solve the algebraic system (15) by using only one parameters β1 = 0 to get √four pairs (y1 , y2√) as follows S0 = {y1 = − R16b 1 +16c1 , y2 = R16b 1 −16c1 }, √ 1 √ 1 R1 −16c1 R1 +16c1 S1 = {y1 = , y 2 = − }, √16b √1 16b√ √ 1 R2 +16c1 R2 −16c1 S2 = {y1 = − 2 16b , y2 = 2 16b }, √ √ 1 √ √ 1 2 R2 −16c1 2 R2 +16c1 S3 = {y1 = 16b1 , y2 = − 16b1 }, where R1 = [2048c31 (β22 − 2b22 )β24 + 256b1 c21 (4(8γ2 + 3)b22 + 16c2 β2 b2 − 3β22 (8γ2 + 3))β23 + 16b21 c1 (β2 (256c22 + 96γ2 (4γ2 + 3) + 63) − 128b2 c2 (8γ2 + 3))β23 − 2b31 (8γ2 + 3)((256c22 + 32γ2 (4γ2 + 3) + 27)β22 − 96b2 c2 (8γ2 + 3)β2 + October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 22 B. Ghermoul, J. Llibre, T. Salhi 4b22 (8γ2 + 3)2 )β2 − 2b1 [β22 (b1 (8γ2 + 3) − 8c1 β2 )2 (512c21 (2(32γ2 (4γ2 + 3) + 9)b22 − 64c2 β2 (8γ2 + 3)b2 + (256c22 + 9)β22 )β24 − 128b1 c1 (8γ2 + 3)(2(32γ2 (4γ2 + 3) + 9)b22 − 64c2 β2 (8γ2 + 3)b2 + (256c22 + 9)β22 )β23 + b21 ((256c22 + 9)(256c22 + 128γ2 (4γ2 + 3) + 81)β24 − 128b2 c2 (8γ2 + 3)(256c22 + 64γ2 (4γ2 + 3) + 45)β23 + 8b22 (8γ2 + 3)2 (768c22 + (16γ2 + 3)(16γ2 + 9))β22 − 512b32 c2 (8γ2 + 3)3 β2 + 16b42 (8γ2 + 3)4 ))]1/2 ]/[β23 (8c1 β2 (β22 − 2b22 ) + b1 (4(8γ2 + 3)b22 − 16c2 β2 b2 − β22 (8γ2 + 3)))] and R2 = [1024c31 (β22 −2b22 )β24 +8b21 c1 (β2 (256c22 +96γ2 (4γ2 +3)+63)−128b2 c2 (8γ2 +3))β23 −b31 (8γ2 +3)((256c22 + 32γ2 (4γ2 +3)+27)β22 −96b2 c2 (8γ2 +3)β2 +4b22 (8γ2 +3)2 )β2 +(128c21 (4(8γ2 +3)b22 +16c2 β2 b2 −3β22 (8γ2 +3))β23 + [β22 (b1 (8γ2 +3)−8c1 β2 )2 (512c21 (2(32γ2 (4γ2 +3)+9)b22 −64c2 β2 (8γ2 +3)b2 +(256c22 +9)β22 )β24 −128b1 c1 (8γ2 + 3)(2(32γ2 (4γ2 + 3) + 9)b22 − 64c2 β2 (8γ2 + 3)b2 + (256c22 + 9)β22 )β23 + b21 ((256c22 + 9)(256c22 + 128γ2 (4γ2 + 3) + 81)β24 − 128b2 c2 (8γ2 + 3)(256c22 + 64γ2 (4γ2 + 3) + 45)β23 + 8b22 (8γ2 + 3)2 (768c22 + (16γ2 + 3)(16γ2 + 9))β22 − 512b32 c2 (8γ2 +3)3 β2 +16b42 (8γ2 +3)4 ))]1/2 )b1 ]/[β23 (8c1 β2 (β22 −2b22 )+b1 (4(8γ2 +3)b22 −16c2 β2 b2 −β22 (8γ2 +3)))]. Since y1 (y2 ) of S0 is equal y2 (y1 ) of S1 , and y1 (y2 ) of S2 is equal to y2 (y1 ) of S3 , then we can choose two pairs S0 and S2 . This proves that the possible continuous piecewise differential systems (8)-(11) can have at most two limit cycles in this case, but in what follows we will prove that they cannot have limit cycles because R1 < 0 and R2 < 0. Return now to the continuity condition. Solving the sixth √ equation of (35) we get one allowed solution which does not contradict (14), then we have β2 = −3b1 /(4 2) (See supplementary section 5.37). Now the first equation of (35) becomes equivalent to √ √ √ b21 (448a2 α2 (2γ1 +1)2 +216a1 (2γ1 +1)2 ( 2a2 +α2 )−64 2(2a2 γ1 +a2 )2 + 2(256(2α2 γ1 +α2 )2 −81α12 )) = 0, and since β1 = 0 we must have b1 6= 0, otherwise they contradict (14). Solving this last equation with respect to α1 we get two following values r  √  2 √ α1 = ± (2γ1 + 1) 2 56 2α2 a2 + 27 2a1 α2 − 16a22 + 54a1 a2 + 64α22 . (40) 9 Without solving the fourth remaining equation of (35), we use all parameters we always obtain R1 < 0 and R2 < 0. Subcase 1.3: Consider the set u3 of (37). Working as in Subcase 1.2 the piecewise differential system has no limit cycles in this subcase. Case 2: Consider the set s2 of (36). Solving the third and the sixth equation of (35) we get the only one solution {a1 = 3a2 (b21 + β12 )/(4β22 ) + α1 b1 /β1 , α2 = 3a2 b1 /(2β2 ) + α1 β2 /β1 } which verify the condition (14) (See supplementary section 5.38). Then solving the first and the fourth equation of (35) simultaneously, The solutions for which the condition (14) is verified are two (See supplementary section 5.39) p p u1 = {b1 = 0, β1 = −4βp 2 2 2 64c2 + 9/(3 p 256c2 + 9)} and 2 2 (41) u2 = {b1 = 0, β1 = 4β2 64c2 + 9/(3 256c2 + 9)}. For any of these solutions we have now a continuous piecewise differential systems (8)-(11). Subcase 2.1: Consider the set u1 of (41). Here we consider the first integrals for (8)-(11) which are given respectively as follows (a1 x + b1 y + c1 ) 2 + (γ1 + α1 x + β1 y) 2 = k1 , (42) 2 (γ1 + α1 x + β1 y) + 1 and −256 (a2 x + b2 y + c2 ) 2 + 128 (γ2 + α2 x + β2 y) 2 + 96 (γ2 + α2 x + β2 y) + 9 = k2 , (43) (8 (γ2 + α2 x + β2 y) + 3) 4 where k1 and k2 are two arbitrary constants. The points of intersections of orbits of system (8) with the y−axis is given by solving equation (42) with respect(to y, then we get the set of solutions as follows √ √ q 3β2 64c22 +9 256c22 +9(k1 −γ1 )+ β22 (−64c22 −9)(256c22 +9)(16c22 −9k1 (k1 +1)) Y1= y =− 4β22 (64c22 +9) , October 20, 2020 20:22 ”Continuous piecewise isochronous centers” Limit cycles of continuous piecewise differential systems 23 √ √ q ) 3β2 64c22 +9 256c22 +9(γ1 −k1 )+ β22 (−64c22 −9)(256c22 +9)(16c22 −9k1 (k1 +1)) y= 4β22 (64c22 +9) } . To obtain the points of intersections of orbits of system (9) with the y−axis, we solve equation (43) with respect(to y, then we have the following r √ solutions √ 2 3 256c2 +9(2γ1 +1) 1− −256c22 k2 −9k2 +1 Y2= y = √ 2 − 18 β22 k2 , 8β2 64c2 +9 √ r √ ! 1 3 256c22 +9(2γ1 +1) 1− −256c22 k2 −9k2 +1 y= √ + , 8 β2 64c22 +9 β22 k2 √ r√ 3 256c22 +9(2γ1 +1) 1 −256c22 k2 −9k2 +1+1 y= √ − , 8β2 64c22 +9 8 β22 k2 √ r√ !) 1 3 256c22 +9(2γ1 +1) −256c22 k2 −9k2 +1+1 y= √ + . 8 β2 64c22 +9 β22 k2 All possible cases for which the orbits of (8) coincide with the orbits of (11) on the y− axis to forme a closed orbits are given by solving, with respect to k1 and k2 , the algebraic system formed by connecting the two points of Y 1 together with any possible two points of Y 2. All such cases give the following different pairs (k n1 , k2 ) as follows o K1 = k1 = − 21 , k2 = − 256c32 +9 , n p 2 o K2 = k1 = 61 (− 64c22 + 9 − 3), k2 = 256c12 +9 and n p 2 o K3 = k1 = 16 ( 64c22 + 9 − 3), k2 = 256c12 +9 . 2 Using all these three pairs, Y 1 becomes: For K1: ( Y 1 isqgiven by √ √ β22 (−(64c22 +9)2 )(256c22 +9)−3β2 64c22 +9 256c22 +9(2γ1 +1) Y1= y =− 8β22 (64c22 +9) , √ √ q ) 3β2 64c22 +9 256c22 +9(2γ1 +1)+ β22 (−(64c22 +9)2 )(256c22 +9) y= 8β22 (64c22 +9) , which are complex. For K2: ( Y 1√is given by ) √ 2  √ √ 256c22 +9 64c2 +9+6γ1 +3 256c22 +9 64c22 +9+6γ1 +3 Y1= y = √ 2 ,y = √ 2 , 8β2 64c2 +9 8β2 64c2 +9 which gives only one element. For K3: ( Y 1 is√given by √ )  √ √ 256c22 +9 64c22 +9−6γ1 −3 256c22 +9 64c22 +9−6γ1 −3 Y1= y =− √ 2 ,y = − √ 2 , 8β2 64c2 +9 8β2 64c2 +9 which gives only one element. This provided no limit cycle for the piecewise differential system (8)-(11) in this subcase. Subcase 2.2: Consider the set u2 of (41). Working as in Subcase 2.1. Again this provided no limit cycles. Case 3: Consider the set s3 of (36). Solving the third and the sixth equation of (35) we get the following set of real solutions {b2 = 3b1 /16, β1 = 0} for which the condition (14) is verified (See supplementary section 5.40). Solving the first equation of (35) we obtain one of the following sets of real solutions (See supplementary section 5.41) u1 = {a2 = 3a1 /16} and (44) u2 = {γ1 = −1/2}, for which the condition (14) is verified. For any of these solutions we have now a continuous piecewise differential systems (8)-(11). October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 24 B. Ghermoul, J. Llibre, T. Salhi Subcase 3.1: Consider the set u1 of (44). We shall solve the fourth equation of (35) which now is  α2 b1 256 (2α2 γ1 + α2 ) 2 − 9α12 γ12 + γ1 + 1 = 0. Since b1 6= 0 and α2 6= 0 (because β1 = β2 = 0), this equation has two solutions p ∓ −9α12 + 1024α22 ∓ 3 3072α12 α22 − 27α14 γ1 = . 18α12 − 2048α22 Subcase 3.1.1: For γ1− we have a continuous piecewise differential systems (8)-(11). We solve the algebraic system (15). The first two equations become (3α1 − 32α2 ) (3α1 + 32α2 ) b1 (y1 − y2 ) (3b1 y1 + 3b1 y2 + 32c2 ) √ q  = 0, 9 3 −α12 9α12 − 1024α22 (3α1 − 32α2 ) 2 (3α1 + 32α2 ) 2 b1 (y1 − y2 ) (3b1 y1 + 3b1 y2 + 32c2 ) = 0. 15925248α24 From these equations we get y1 = y2 or y1 = − 32c 3b1 − y2 . Then this provided a continuum of periodic orbits 2 and then no limit cycles. Subcase 3.1.2: For γ1+ . We again have a continuous piecewise differential systems (8)-(11). We solve the algebraic system (15) its two first equations become (3α1 − 32α2 ) (3α1 + 32α2 ) b1 (y1 − y2 ) (3b1 y1 + 3b1 y2 + 32c2 ) √ q 2 2 2  = 0, 9 3 −α1 9α1 − 1024α2 (3α1 − 32α2 ) 2 (3α1 + 32α2 ) 2 b1 (y1 − y2 ) (3b1 y1 + 3b1 y2 + 32c2 ) = 0. 15925248α24 From these equations we get y1 = y2 or y1 = − 32c 3b1 − y2 . Then this provided a continuum of periodic orbits 2 and then no limit cycles. Subcase 3.2: Consider the set u2 of (41). Then the fourth equation of (35) becomes −27α1 α2 b1 = 0. This equation has α1 = 0, α2 = 0 or b1 = 0 but all these solutions are not allowed because they contradict (14). Then in this subcase there are no continuous piecewise differential systems (8)-(11).  4. Conclusion First we have studied the planar continuous piecewise differential systems formed by a quadratic isochronous center and an isochronous linear center separated by the straight line x = 0, and we have proved that they cannot neither crossing periodic orbits, nor limit cycles, see Theorem 1. Second we study the crossing periodic orbits and limit cycles of the planar continuous piecewise dif- ferential systems separated by the straight line x = 0 having in x > 0 the quadratic isochronous center (8), and in x < 0 an arbitrary quadratic isochronous center, i.e. one of the systems (8), (9), (10) or (11). We remark that these last four quadratic polynomial differential systems are all the quadratic polynomial differential systems having an isochronous center. For these four families of continuous piecewise differential systems the maximum number of crossing limit cycles is one, and there are examples having one crossing limit cycles. See Theorem 2. In short for the eight different families of planar continuous piecewise differential systems here studied we have proved the extension of the 16th Hilbert problem to them. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” REFERENCES 25 Acknowledgments The first author is partially supported by the university Mohammed El Bachir El Ibrahimi, Bordj Bou Arreridj, Algerian Ministry of Higher Education and Scientific Research and Direction Générale de la Recherche Scientifique et du Développement Technologique, DGRSDT. MESRS Algeria, and Research project number: PRFU C00L03UN190120180007. The second author is partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agen- cia Estatal de Investigación grants MTM2016-77278-P (FEDER) and PID2019-104658GB-I00 (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. The third author is partially supported by the university Mohammed El Bachir El Ibrahimi, Bordj Bou Arreridj, Algerian Ministry of Higher Education and Scientific Research. References Andronov, A. A., Vitt, A. A. & Khaikin, S. E. [1966] “Theory of oscillators”, Translated from the Russian by F. Immirzi; translation edited and abridged by W. Fishwick (Pergamon Press, Oxford). di Bernardo, M., Budd, C. J., Champneys, A. R. & Kowalczyk, P. [2008] “Piecewise-smooth dynamical systems”, Applied Mathematical Sciences, Vol. 163 (Springer-Verlag London Ltd., London), theory and applications. Dulac, H. [1908] “Détermination et integration d’une certaine classe d’équations différentielle ayant par point singulier un centre”, Bull. Sci. Math. Sér.(2) 32, 230–252. Hilbert, D. [1900] “Mathematische Probleme”, Lecture, Second Internat. Congr. Math.8, 437–479; Bull. (New Series) Amer. Math. Soc. 37 (2000), 407–436. Huygens, C. “Horologium Oscillatorium”, translated by Ian Bruce (July 2007), (PDF). 17thcentury- maths.com. Retrieved 2009-03-01. Ilyashenko, Y. [2002] “Centennial history of Hilbert’s 16th problem”, Bull. (New Series) Amer. Math. Soc. 39, 301–354. Li, J. [2003] “Hilbert’s 16th problem and bifurcations of planar polynomial vector fields”, Internat. J. Bifur. Chaos Appl. Sci. Engrg. 13, 47–106. Llibre, J., Novaes, D. D. & Teixeira, M. A. [2015] “Maximum number of limit cycles for certain piecewise linear dynamical systems”, Nonlinear Dyn. 82, 1159–1175. Llibre, J. & Teixeira, M. A. [2018] “Piecewise linear differential systems with only centers can create limit cycles?”, Nonlinear Dyn. 91 (2018), 249–255. Li, J. [1964] “Behavior of the periodic of solutions of certain plane autonomous systems near centers”, Contributions to Differential Equations 3, 21–36. Poincaré, H. [1881] “Mémoire sur les courbes définies par les équations différentielles”, Journal de Mathématiques 37, 375–422; Oeuvres de Henri Poincaré, vol. I, Gauthier-Villars, Paris, 1951, pp 3–84. Poincaré, H. [1928] “Sur les courbes définies par une équation différentielle”, Oevres complètes 1. Simpson, D. J. W. [2010] “Bifurcations in piecewise–smooth continuous systems”, World Scientific Series on Nonlinear Science A 69. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 5. SUPPLEMENTARY INFORMATION FOR REVIEW ONLY 5.1. SECTION All real solutions s1 = {a2 = 0, b2 = 0, c2 = −1/2}, q (2a1 c2 +a1 )2 +4a2 (a2 c22 +a2 c2 )+2a1 c2 +a1 − s2 = {b1 = 0, b2 = 0, c1 = 2a2 }, q (2a1 c2 +a1 ) +4a2 (a2 c2 +a2 c2 )+2a1 c2 +a1 2 2 s3 = {b1 = 0, b2 = 0, c1 = 2a2 }, s4 = {a1 = 0, a2 = 0, b1 = 0, b2 = 0}, s5 = {a2 = 0, b1 = 0, b2 = 0, c1 = 0}, s6 = {a2 = 0, b1 = 0, b2 = 0, c2 = −1/2} and s7 = {α = 0, a1 = 0, a2 = 0, b2 = 0, c1 = 0}. 5.2. SECTION The algebraic system has the following real solutions −a1 ) s1 = {b1 = 0, b2 = 0, c2 = c1a(a12c1c1+a 2 }, s2 = {a1 = 0, a2 = 0, b1 = 0, b2 = 0}, s3 = {a2 = 0, b1 = 0, b2 = 0, c1 = 0}, 4a2 c2 +8a a c +a2 ω 2 +4a22 −a c2 +a c +a c c +a c c21 −c2 s4 = {α = 1 1 2 4a1 11b2 1 , β = a1 ca11+a2 , γ = 2 1 1 a1 1 b21 2 1 2 2 , δ = a1 , b1 = 0} and s5 = {α = 0, a1 = 0, a2 = 0, b1 = 0, c2 = c21 }. 5.3. SECTION The algebraic system has the following real solutions s1 = {b1 = 0, b2 = 0, c2 = c116a (4a2 c1 +3a1 ) 1 c1 −3a2 }, s2 = {a2 = 0, b2 = 0, c2 = 3/16}, s3 = {a1 = 0, a2 = 0, b1 = 0, b2 = 0}, s4 = {a2 = 0, b1 = 0, b2 = 0, c1 = 0}, s5 = {a2 = 0, b2 = 0, c1 = 0, c2 = 3/16}, 1024a21 c21 −384a2 a1 c1 +9a21 ω 2 +36a22 4a2 c21 +3a1 c1 −16a1 c2 c1 +3a2 c2 −4c21 −3c2 s6 = {α = 36a1 b2 , β = 3a2 −16a 3a1 1 c1 ,γ = 3a1 b2 ,δ = 3a1 , b1 = 0}, s7 = {α = 0, a1 = 0, a2 = 0, b1 = 0, c2 = −4c21 /3}, s8 = {α = 0, a1 = 0, a2 = 0, b2 = 0, c1 = 0} and s9 = {a1 = 0, a2 = 0, b2 = 0, c1 = 0, c2 = 3/16}. 5.4. SECTION The algebraic system has the following real solutions s1 = {a2 = 0, b2 = 0, c2 = −3/8}, s2 = {a1 = 0, a2 = 0, b2 = 0, c2 = −3/8}, s3 = {a1 = 0, a2 = 0, b1 = 0, b2 = 0}, s4 = {a2 = 0, b1 = 0, b2 = 0, c1 = 0}, s5 = {a2 = 0, b1 = 0, b2 = 0, c2 = −3/8}, s6 = {α = 0, a1 = 0, a2 = 0,qb2 = 0, c1 = 0}, − (8a1 c2 +3a1 )2 +64a2 (4a2 c22 +3a2 c2 )+8a1 c2 +3a1 s7 = {b1 = 0, b2 = 0, c1 = 32a2 }, q (8a1 c2 +3a1 )2 +64a2 (4a2 c22 +3a2 c2 )+8a1 c2 +3a1 s8 = {b1 = 0, b2 = 0, c1 = 32a2 p }, √ 2 √ √ s9 = {α = 0, a1 = 0, a2 = 0, b1 = −b2 / 2, c1 = (−p 288c2 + 216c2 + 9/2 − 4 2c2 − 3/ 2)/32}, √ √ √ s10 = {α = 0, a1 = 0, a2 = 0, b1 = −b2 / 2, c1 = ( 288c22 + 216c2 + 9/2 − 4 2c2 − 3/ 2)/32}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” √ p √ √ s11 = {α = 0, a1 = 0, a2 = 0, b1 = b2 / 2, c1 = (− 288c 2 + 216c + 9/2 + 4 2c + 3/ 2)/32} and √ p 2 2 √ 2 √ s12 = {α = 0, a1 = 0, a2 = 0, b1 = b2 / 2, c1 = ( 288c22 + 216c2 + 9/2 + 4 2c2 + 3/ 2)/32}. 5.5. SECTION d1 = {a2 = 0, α2 = 0}, d2 = {b2 = 0, β2 = 0}, d3 = {α2 = 0, β2 = 0}, −2a2 β2 b2 −α2 β22 +α2 b22 d4 = {b1 = 2(α2 b2 −a2 β2 ) }. 5.6. SECTION All real solutions 2a β2 b2 +α2 β22 −α2 b22 r1 = {b1 = 2 2a 2 β2 −2α2 b2 , α1 = 0}, r2 = {b2 = 0, β2 = 0}, r3 = {α1 = 0, β1 = 0, γ1 = −1/2}, r4 = {a2 = 0, α1 = 0, α2 = 0}, r5 = {b1 = b2 /2, β2 = 0, γ2 = −1/2}, r6 = {b1 = b2 /2, α1 = 0, β2 = 0}, r7 = {α1 = 0, α2 = 0, β2 = 0}, r9 = {β1 = 0, β2 = 0, γ2 = 2α2 b1 (2γ1 +1)−b 2 (2α2 γ1 +α1 +α2 ) 2α1 b2 }, α1 β2 (b22 +β22 ) r10 = {β1 = 2a2 (b2 −b1 )β2 +α2 β22 +α2 (2b1 −b2 )b2 , 2b2 (2γ1 +1)(a2 β2 +α2 b1 )+β2 (β2 (2α2 γ1 −α1 +α2 )−2a2 (2b1 γ1 +b1 ))+b22 (−(2α2 γ1 +α1 +α2 )) γ2 = 2α1 (b22 +β22 ) }. 5.7. SECTION All real solutions p − α22 (−b21 + b22 + β22 )/β2 }, s1 = {a2 = α2 b2 /β2 , α1 = p s2 = {a2 = α2 b2 /β2 , α1 = α22 (−b21 + b22 + β22 )/β2 }, α β 2 −α b2 +2α b b s3 = {a2 = 2 22b1 β22 −2b 2 2 1 2 2 β2 , α1 = 0}, s4 = {b1 = b2 , α1 = −α2 }, s5 = {b1 = b2 , α1 = α2 }, s6 = {b2 = 0, α1 = 0}, s7 = {b2 = 0, β2 = 0}, s8 = {b1 = b2 /2, β2 = 0}, s9 = {b1 = b2 , β2 = 0}, s10 = {b2 = 0, β2 = 0}, s11 = {α2 = 0, β2 = 0}, s12 = {β2 = 0, γ1 = −1/2}, s13 = {α1 = 0, γ1 = −1/2}, p s14 = {a2 = 0, b2 = 0, α1 = p − α22 (β22 − b21 )/β2 }, s15 = {a2 = 0, b2 = 0, α1 = α22 (β22 − b21 )/β2 }, s16 = {b1 = b2 , α1 = 0, α2 = 0} and s17 = {α1 = 0, α2 = 0, γ1 = −1/2}. 5.8. SECTION All real solutions r1 = {β1 = 0}, r2 = {a2 = 0, α2 = 0}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” r3 = {b2 = 0, β2 = 0}, r4 = {α2 = 0, β2 = 0} and 4a2 α2 b2 β23 +2β22 (2a22 (b22 +β12 )−α22 b22 )−4a2 α2 b2 β2 (b22 +2β12 )+α22 β24 +α22 b22 (b22 +4β12 ) r5 = {a1 = 4β2 (b22 +β22 )(a2 β2 −α2 b2 ) }. 5.9. SECTION All real solutions 2a b (2β γ −β +β )+α β (2β γ −2β +β )+4a β1 β2 (c1 −c2 )+b22 (−(2α2 γ1 +α2 ))+4α2 β1 b2 (c2 −c1 ) s1 = {γ2 = 2 2 2 1 1 2 2 2 2 1 1 4β21 (a2 b22 +α 2 β2 ) }, s2 = {a2 = −α2 β2 /b2 , c2 = 2b2 γ1 +b 2 +4β1 c1 4β1 }, s3 = {a2 = 0, α2 = 0}, s4 = {a2 = α2 (b2 −β 2 )(b2 +β2 ) 2b2 β2 , β1 = 0}, s5 = {β1 = 0, γ1 = −1/2}, s6 = {b2 = 0, β2 = 0}, s7 = {a2 = 0, b2 = 0, α2 = 0}, s8 = {b2 = 0, c2 = c1 , α2 = 0}, s9 = {b2 = 0, α2 = 0, β1 = 0}, s10 = {b2 = 0, β1 = 0, γ1 = −1/2} and s11 = {α2 = 0, β1 = 0, β2 = 0}. 5.10. SECTION α1 (c21 −γ1 (γ1 +1)) s1 = {a1 = 2γ1 c1 +c1 , b1 = 0}, α2 (c22 −γ2 (γ2 +1)) s2 = {a2 = 2γ2 c2 +c2 , b2 = 0}, s3 = {b1 = 0, b2 = 0}, s4 = {b2 = 0, α1 = 0}, s5 = {α1 = 0, α2 = 0}, s6 = {α1 = 0, γ1 = −1/2}, s7 = {b1 = 0, α2 = 0}, s8 = {α1 = 0, α2 = 0}, s9 = {α2 = 0, γ2 = −1/2}, s10 = {b1 = 0, c1 = 0, γ1 = −1}, s11 = {b1 = 0, c1 = 0, γ1 = 0}, s12 = {b2 = 0, c2 = 0, γ2 = −1}, s13 = {b2 = 0, c2 = 0, γ2 = 0}, s14 = {b1 = 0, c1 = 0, α1 = 0}, s15 = {b1 = 0, α1 = 0, γ1 = −1/2}, s16 = {b1 = b2 , α1 = 0, γ1 = −1/2}, s17 = {b1 = 0, b2 = 0, α1 = 0}, s18 = {b1 = 0, α1 = 0, α2 = 0}, s19 = {b1 = b2 , α2 = 0, γ2 = −1/2}, s20 = {b1 = 0, b2 = 0, α2 = 0}, s21 = {b2 = 0, c2 = 0, α2 = 0}, s22 = {b2 = 0, α2 = 0, γ2 = −1/2}, s23 = {b1 = b2 , α1 = 0, α2 = 0}, s24 = {b1 = b2 , c1 = c2 , γ1 = −1/2, γ2 = −1/2}, s25 = {a2 = α2 (2a2α 1 γ1 +a1 +2α1 c2 ) 1 γ2 +α1 2α1 c2 − 2γ1 +1 , b1 = 0, b2 = 0, c1 = α1 c2 (2γ2 +1) 2α2 γ1 +α2 }, s26 = {a1 = 0, b1 = 0, c1 = 0, α1 = 0}, s27 = {b1 = 0, b2 = 0, α1 = 0, γ1 = −1/2}, s28 = {b1 = 0, b2 = 0, α2 = 0, γ2 = −1/2}, s29 = {a2 = 0, b2 = 0, c2 = 0, α2 = 0}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” s30 = {a2 = a12α α2 (2γ1 +1) 1 γ2 +α1 , b1 = 0, b2 = 0, c1 = 0, c2 = 0}, 2α2 c1 s31 = {a2 = − 2γ2 +1 , b1 = 0, b2 = 0, c2 = 0, γ1 = −1/2}, 2α1 c2 s32 = {a1 = − 2γ 1 +1 , b1 = 0, b2 = 0, c1 = 0, γ2 = −1/2}, s33 = {b1 = 0, b2 = 0, c1 = c2 , γ1 = −1/2, γ2 = −1/2}, s34 = {a1 = 0, b1 = 0, b2 = 0, c1 = 0, α1 = 0}, s35 = {a2 = a1 , b1 = b2 , c1 = c2 , α1 = −α2 , γ1 = −γ2 − 1}, s36 = {b1 = b2 , c1 = c2 , α1 = −α2 , γ1 = −1/2, γ2 = −1/2}, +a1 −2α2 c2 s37 2α2 c2 = {a2 = 2γ 1 +1 − 2a1 γ12γ 2 +1 , b1 = 0, b2 = 0, c1 = − 2γ2γ2 c2 +c2 1 +1 , α1 = −α2 }, s38 = {a2 = a1 , b1 = b2 , c1 = c2 , α1 = α2 , γ1 = γ2 }, s39 = {b1 = b2 , c1 = c2 , α1 = α2 , γ1 = −1/2, γ2 = −1/2}, s40 = {a2 = 2a1 γ12γ +a1 +2α2 c2 2 +1 − 2γ2α2 c2 1 +1 , b1 = 0, b2 = 0, c1 = 2γ2γ 2 c2 +c2 1 +1 , α1 = α2 }, s41 = {a2 = 0, b1 = 0, b2 = 0, c2 = 0, α2 = 0}, s42 = {a2 = − 2a2γ 1 γ1 +a1 2 +1 , b1 = 0, b2 = 0, c1 = 0, c2 = 0, α1 = −α2 }, 2α2 c1 s43 = {a2 = − 2γ2 +1 ), b1 = 0, b2 = 0, c2 = 0, α1 = −α2 , γ1 = −1/2}, 2α2 c2 s44 = {a1 = 2γ 1 +1 , b1 = 0, b2 = 0, c1 = 0, α1 = −α2 , γ2 = −1/2}, s45 = {b1 = 0, b2 = 0, c1 = c2 , α1 = −α2 , γ1 = −1/2, γ2 = −1/2}, s46 = {a2 = 2a2γ1 γ1 +a1 2 +1 , b1 = 0, b2 = 0, c1 = 0, c2 = 0, α1 = α2 }, 2α2 c1 s47 = {a2 = − 2γ2 +1 , b1 = 0, b2 = 0, c2 = 0, α1 = α2 , γ1 = −1/2}, 2α2 c2 s48 = {a1 = − 2γ 1 +1 , b1 = 0, b2 = 0, c1 = 0, α1 = α2 , γ2 = −1/2}, s49 = {b1 = 0, b2 = 0, c1 = c2 , α1 = α2 , γ1 = −1/2, γ2 = −1/2}. 5.11. SECTION All real solutions are r1 = {α1 = a2 (3b1 +8b2 )β1 β6b2 −α 2β 2 b2 (3b1 +2b2 )β1 }, 2 2 r2 = {b1 = 0, b2 = 0}, r3 = {b2 = 0, β1 = 0}, r4 = {b2 = 0, β2 = 0}, r5 = {β1 = 0, β2 = 0}, r6 = {b1 = −2b2 /3, β2 = 0}, r7 = {α2 = 0, β2 = 0} and r8 = {a2 = 0, b2 = 0}. 5.12. SECTION All real solutions are 3b2 (8β γ +β +4β2 )−8β1 β2 b2 c2 +3β1 β22 s1 = {γ2 = 2 2 1 1 16b 2β }, 2 1 s2 = {b2 = 0, β1 = 0}, s3 = {b2 = 0, β2 = 0}, s4 = {β1 = 0, γ1 = −1/2} and s5 = {β1 = 0, β2 = 0}. 5.13. SECTION All sets of real solutions 2 −6β1 c1 −8β1 c2 ) v1 = {β2 = b2 (8b2 γ1 +b1 (6γ1 +3)+4b 3β1 }, v2 = {b2 = 0, β1 = 0}, v3 = {b1 = −4b2 /3, β1 = 0} and v4 = {β1 = 0, γ1 = −1/2}. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 5.14. SECTION All sets of real solutions √ 3S 2 (15b1 +56b2 )β1 (2γ1 +1)− b2 2 −30β1 c1 1 +β1 w1 = {c2 = 112β12 }, √ 2 2 3S+β1 (b1 +β1 )(15b1 (2γ1 +1)+56b2 (2γ1 +1)−30β1 c1 ) w2 = {c2 = 112β12 (b21 +β12 ) }, a2 (8b2 γ1 +b1 (6γ1 +3)+4b2 −6β1 c1 −8β1 c2 ) w3 = {α2 = 3β1 }, w4 = {a2 = 0, β1 = 0}, w5 = {b1 = 0, β1 = 0}, w6 = {b2 = −3b1 /4, β1 = 0}, w7 = {b2 = −3b1 /16, β1 = 0} and w8 = {β1 = 0, γ1 = −1/2}. whith S = β12 (b21 + β12 )(36β1 b31 c1 (2γ1 + 1) + b21 (448(2b2 γ1 + b2 )2 − 3β12 (12c21 + 12γ1 (γ1 + 1) + 31)) + 4β1 b1 c1 (2γ1 + 1)(9β12 − 448b22 ) + 448b22 β12 (4c21 + 1) − 9b41 (2γ1 + 1)2 − 12β14 (3c21 + 7)). 5.15. SECTION The polynomial of degree two of the Subcase 1.1 in the proof of statement (c) of Theorem 2 is 648c31 β18 + 19656c1 β18 + 1296c31 γ1 β18 + 39312c1 γ1 β18 − 15120b2 c41 β17 − 972b1 c21 β17 − 504b2 c21 β17 − 3888b1 c21 γ12 β17 − 15120b2 c21 γ12 β17 − 39312b1 γ12 β17 +7056b2 γ12 β17 −9828b1 β17 +3528b2 β17 −3888b1 c21 γ1 β17 −15120b2 c21 γ1 β17 −39312b1 γ1 β17 +7056b2 γ1 β17 +1296b21 c31 β16 − 334656b22 c31 β16 + 30240b1 b2 c31 β16 + 3888b21 c1 γ13 β16 + 30240b1 b2 c1 γ13 β16 + 5832b21 c1 γ12 β16 + 45360b1 b2 c1 γ12 β16 + 39798b21 c1 β16 − 83664b22 c1 β16 + 504b1 b2 c1 β16 + 2592b21 c31 γ1 β16 − 669312b22 c31 γ1 β16 + 60480b1 b2 c31 γ1 β16 + 81540b21 c1 γ1 β16 − 167328b22 c1 γ1 β16 + 16128b1 b2 c1 γ1 β16 − 150528b32 c41 β15 − 15120b21 b2 c41 β15 − 1296b31 γ14 β15 − 15120b21 b2 γ14 β15 − 19737b31 β15 − 18816b32 β15 − 2592b31 γ13 β15 − 30240b21 b2 γ13 β15 + 41832b1 b22 β15 −1944b31 c21 β15 −112896b32 c21 β15 +501984b1 b22 c21 β15 −26964b21 b2 c21 β15 −80568b31 γ12 β15 −37632b32 γ12 β15 +167328b1 b22 γ12 β15 − 7776b31 c21 γ12 β15 − 150528b32 c21 γ12 β15 + 2007936b1 b22 c21 γ12 β15 − 120960b21 b2 c21 γ12 β15 − 5292b21 b2 γ12 β15 + 5166b21 b2 β15 − 79272b31 γ1 β15 − 37632b32 γ1 β15 + 167328b1 b22 γ1 β15 − 7776b31 c21 γ1 β15 − 150528b32 c21 γ1 β15 + 2007936b1 b22 c21 γ1 β15 − 120960b21 b2 c21 γ1 β15 + 9828b21 b2 γ1 β15 + 648b41 c31 β14 + 301056b1 b32 c31 β 4 √1 − √334656b 2 2 3 4 3 3 4 4 3 4 3 3 4 1 b2 c1 β1 + 30240b1 b2 c1 β1 + 7776b1 c1 γ1 β1 + 301056b1 b2 c1 γ1 β1 − 2007936b1 b2 c1 γ1 β1 + 2 2 3 4 3 3 4 2 4 4 2 4 3 2 4 2 2 2 4 3 2 4 √ √1 b2 c1 γ1 β1 +√1332√ 3 R1 c1 β1 + 11664b1 c1 γ1 β1 + 451584b1 b2 c1 γ1 β1 − 120960b 3011904b1 b2 c1 γ1 β1 + 181440b1 b2 c1 γ1 β1 − 252 3 R1 β14 − 504 3γ1 R1 β14 + 20628b41 c1 β14 + 112896b1 b32 c1 β14√− √ 334656b21 b22 c1 β14 + 11844b31 b2 c1 β14 + 1296b41 c31 γ1 β14 + 602112b1 b32 c31 γ1 β14 − 669312b21 b22 c31 γ1 β14 + 60480b31 b2 c31 γ1 β14 + 2664 3 R1 c21 γ1 β14 + 45144b41 c1 γ1 β14 + 376320b1 b32 c1 γ1 β14 − 1673280b21 b22 c1 γ1 β14 + √ 84168b √ 3 4 5 3 5 4 3 2 3 4 3 1 b2 c1 γ1 β1 − 9990b1 β1 − 2592b1 γ1 β1 − 150528b1 b2 γ1 β1 + 669312b1 b2 γ1 β1 − 45360b1 b2 γ1 β1 − 3 2 4 3 4 4 3 2 3 3 3 3 5 3 3 2 3 3 3 3 2 3 3 4 3 3 3 2 3 37632b1 b2 β1 + 1848 3 R1 b2 c1 β1 − 5184b1 γ1 β1 − 301056b1 b2 γ1 β1 + 1338624b1 b2 γ1 β1 − 90720b1 b2 γ1 β1 + 83664b1 b2 β1 − 972b51 c21 β13 − 263424b21 b32 c21 β13 + 501984b31 b22 c21 β13 − 26460b41 b2 c21 β13 − 43200b51 γ12 β13 − 338688b21 b32 γ12 β13 + 1171296b √ √ 3 2 2 3 1 b2 γ 1 β 1 − 5 2 2 3 2 3 2 2 3 3 2 2 2 3 4 2 2 3 4 2 3 2 3 3888b√1 c1 γ1 β1 √ − 1053696b1 b2 c√ γ β 1 1 1 √ + 2007936b b c 1 2 1√1 1 γ β − 105840b √ b c 1 2 1 1 1γ β − 57708b 1√b γ 2 1 β √1 + 1848 3 R 1 2 c1 γ1 β1 − b 5328 3b1 c1 γ1 R1 β1 − 1332 3b1 c1 R1 β1 − 5328 3b1 c1 γ1 R1 β1 − 252b1 b2 β1 + 924 3 R1 b2 c1 β1 − 40608b1 γ1 β13 − 2 3 3 3 4 3 3 5 188160b21 b32 γ1 β13 + 501984b 3 2 3 √ √ 1 b2 γ1 β1 − 3888b1 c√ 5 2 1 γ1√ β13 − 1053696b21 b32 c21 γ1 β13 + 2007936b31 b22 c21 γ1 β13 − 105840b41 b2 c21 γ1 β13 − 12348b41 b2 γ1 β13 + 1848 √3 √R1 b2 c1 γ1 β13 + √ 2664 √ 3 R1 b21 γ13 β12 √ 6 3 2 √ 12c12γ12β1 + 903168b + 3888b 3 3 3 2 √ √1 b2 c21 γ21 β21 − 2007936b 4 2 3 2 √ √1 b2 c21 γ21 β21 + 5 3 2 2 2 2 2 90720b1 b2 c1 γ1 β1 + 81 3 R1 b1 β1 + 952 3 R1 b2 β1 + 1332 3 R1 b1 c1 β1 + 3808 √ 3 R31√ b2 c1 β1 + 3996√ 3 R1 b√ 1 γ1 β1 + 5832b√61 c1 γ12 β12 +1354752b 3 3 1 b 2√c1 γ1 2 2 β 1 −3011904b 4 2 1 b 2 c √ 1 γ 2 2 1 β 1 +136080b 5 1 b 2 c 1 γ 2 2 1 β √ 1 −1848 3b 1 b 2 γ 1 R 1 β 1 2 −2772 3b 1 b 2 c2 1 R1 β12 − 2√ 2 √ 2 2 √ 2 √ 2 6 2 2772 3b1 b2 γ1 R1 β1 − 462 3b1 b2 R1 β1 − 5544 3b1 b2 c1 γ1 R1 β√ 1 −√1848 3b1 b2 γ1 R1 β1 + 486b1 c1√ β1 √+ 112896b1 b2 c1 β12 − 3 3 4 2 2 5 2 2 2 250992b √ √ b c 1 2 1 1 β + √ √ 11340b b c 1 2 1 1 β + 1494 3 R b γ 1 1 1 1 β + 1904 3 R1 b22 γ1 β12 + 2 2 2664 3 R1 b1 c1 γ1 β1 + 7616 3 R1 b2 c1 γ1 β1 + 2916b1 c1 γ1 β1 + 677376b1 b2 c1 γ1 β1 − 1505952b1 b2 c1 γ1 β1 + 68040b51 b2 c1 γ1 β12 − 2 2 2 2 6 2 3 3 2 4 2 2 81b71 β1 − 1296b71 γ14 β1 − 301056b41 b32 γ14 β1 + 669312b51 b22 γ14 β1 − 30240b61 b2 γ14 β1 − 18816b41 b32 β1 − 2592b71 γ13 β1 − 602112b41 b32 γ13 β1 + 5 2 3 6 3 5 2 7 2 4 3 2 5 2 2 6 √ √1 b2 γ1 β1 − 260480b1 b2 γ√ 1338624b 1 β1 + 41832b √ 1 b2 β1 − 1944b √ 1 γ1 β1 − 451584b √ 1 b2 γ1 β1√+ 1003968b √ 1 b2 γ1 β1 − √45360b1√ b2 γ12 β1 + 7392√3 R1 b21√ b2 c1 γ1 β1 − 5328√ 3b31 c1 γ12 √ R1 β1 − 15232 3b1 b22 c1 γ12 R1√β1√− 1332 3b31 c1 R1 β1 − 3808 3b1 b22 c1 R1 β1 − 5328 3b31 c1 γ1 R1 β1 − 15232 3b1 b22 c1 γ1 √R√ 6 1 β1 − 1890b1 b2 β1 + 1848 √ √3 R 2 1 b1 b2 c1 β√1 − √ 648b 7 1 γ1 β1 − 150528b 4 3 √ √ 1 b22γ21 β31 + 5 2 6 2 4 4 3 √ √1 b2 γ1 β1 − 15120b1 b2 γ1 β1 + 7392 3 R1 b1 b2 c1 γ1 β1 + 333 3 R1 b1 + 2664 3 R1 b1 γ1 + 7616 3 R1 b1 b2 γ1 + 334656b 952 3 R1 b21 b22 + (−15120b2 c21 β19 + 7056b2 β19 + 15120b1 b2 c1 β18 + 30240b1 b2 c1 γ1 β18 − 37632b32 β17 − 150528b32 c21 β17 − 30240b21 b2 c21 β17 − 15120b21 b2 γ12 β17 + 10332b21 b2 β17 − 15120b21 b2 γ1 β17 + 150528b1 b32 c1 β16 + 30240b31 b2 c1 β16 + 301056b1 b32 c1 γ1 β16 + 3 60480b√ 1√ b2 c1 γ1 β16 − 75264b21 b32 β15 − 150528b21 b32 c21 β15 − 15120b41 b√2 c21 β15 √ − 150528b21 b32 γ1√ 2 5 β1 − 30240b 4 2 5 √ 1 b42 γ1 β1 − 504b 4 5 1 b2 β1 + 1848 3 R1 b2 c1 β1 − 150528b1 b2 γ1 β1 − 30240b1 b2 γ1 β1 − 924 3b1 b2 R1 β1 − 1848 3b1 b2 γ1 R1 β1 + 150528b1 b2 c1 β14 + 5 2 3 5 4 5 4 3 3 5 15120b√ 1√ b2 c1 β14 + 301056b31 b32 c1 γ1 β14 + 30240b51 b2 c1 γ1 β14 − 37632b 4 3 3 √ 13b2 β√ 1 − 150528b1 b√ 4 3 2 3 2 γ1 β1 − √ 15120b61 b2 γ12 β13 −√ 3780b 6 √ 1 b24β12 + 3 1848 √ 3 √ R1 b21 b2 c1 β13 − 150528b 4 3 b √ 1 2 1 1√γ β 3 − 15120b 6 b γ 1 2√1 1 β 3 − 924 3b √ 1 2 b R β 1√ 1 2 − 1848 √ 3b 3 b γ R 1 2 1 √ 1 1 2√ β 2 2 )y + 3996 3 R 1 1 1+ b γ 11424 3 R1 b21 b22 γ12 − 3696 3b31 b2 γ13 R1 − 5544 3b31 b2 γ12 R1 − 462 3b31 b2 R1 − 2772 3b31 b2 γ1 R1 + (1296c31 β19 + 39312c1 β19 − 1944b1 c21 β18 − 15120b2 c21 β18 − 19656b1 β18 + 7056b2 β18 − 3888b1 c21 γ1 β18 − 30240b2 c21 γ1 β18 − 39312b1 γ1 β18 + October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 14112b2 γ1 β18 + 2592b21 c31 β17 − 669312b22 c31 β17 + 3888b21 c1 γ12 β17 + 60480b1 b2 c1 γ12 β17 + 79596b21 c1 β17 − 167328b22 c1 β17 + 15120b1 b2 c1 β17 + 3888b21 c1 γ1 β17 + 60480b1 b2 c1 γ1 β17 − 39474b31 β16 − 37632b32 β16 − 1296b31 γ13 β16 − 30240b21 b2 γ13 β16 + 83664b1 b22 β16 − 3888b31 c21 β16 − 150528b32 c21 β16 + 1003968b1 b22 c21 β16 − 30240b21 b2 c21 β16 − 1944b31 γ12 β16 − 45360b21 b2 γ12 β16 + 10332b21 b2 β16 − 79596b31 γ1 β16 − 75264b32 γ1 β16 + 167328b1 b22 γ1 β16 − 7776b 3 2 √ 1√ c1 γ1 β16 − 301056b32 c21 γ1 β16 + 2007936b1 b22 c21 γ1 β16 − 60480b21 b2 c21 γ1 β16 + 2 6 4 3 5 2 2 3 5 2 5 4 2 5 3 2 5 2 2 2 5 5544b1 b2 γ1 β1 + 1296b1 c1√ β1 −√669312b 1 b2 c1 β1 + 2664 3 R1 c1 β1 + 7776b1 c1 γ1 β1 + 602112b1 b2 c1 γ1 β1 − 2007936b1 b2 c1 γ1 β1 + 3 2 5 120960b1 b2 c1 γ1 β1 − 504 3 R1 β1 + 41256b1 c1 β1 + 150528b1 b2 c1 β1 − 669312b1 b2 c1 β1 + 30240b1 b2 c1 β1 + 7776b1 c1 γ1 β15 + 5 4 5 3 5 2 2 5 3 5 4 602112b1 b32 c1 γ1 β15 − 2007936b21 b22 c1 γ1 β15 + 120960b31 b2 c1 γ1 β15 − 19980b51 β14 − 75264b21 b32 β14 − 2592b51 γ13 β14 − 301056b21 b32 γ13 β14 + 669312b31 b22 γ13 β14 − 60480b41 b2 γ13 β14 + 167328b31 b22 β14 − 1944b51 c21 β14 − 150528b21 b32 c21√ β14 + 1003968b √ 3 2 2 4 1 b2 c1 β1 √ − 15120b41√b2 c21 β14 − 5 2 4 2 3 2 4 3 2 2 4 4 2 4 4 4 3888b1 γ1 β1 − 451584b √ √ 1 b2 γ1 β14 + 1003968b 1 b2 γ1 β1 − 90720b1 b2 γ1 β1 − 2664 3b1 c1 R1 β1 − 5328 3b1 c1 γ1 R1 β1 − 4 4 5 4 2 3 4 3 2 4 5 2 4 2 3 2 504b1 b2 β1 + 1848 3 R1 b2 c1 β1 − 41256b1 γ1 β1 − 301056b1 b2 γ1 β1√+√ 669312b1 b2 γ1 β1 − 3888b√ √1 c1 γ12β13 − 301056b √ 1√ b2 c1 γ1 β14 + 3 2 2 4 4 2 4 4 4 4 2 3 √ √1 b2 c1 γ1 β1 − 30240b 2007936b √ 1√b2 c1 γ21 β21 3− 31248b √1 b√2 γ1 β1 + 3696 3 R1 b2 c1 γ1 β1 + 162 3 R1 b1 β1 + 1904 3 R1 b2 β1 + 2664 3 R1 b21 c21 β13 + 7616 √ 3 R b 1 √ c 2 1 1β + 2664 √3 R b 1√ 2 2 3 γ 1 1 1β + 3888b 6 c γ 2 3 1√1 1 1 √β + 602112b 3 3 b c γ 1 2 1 1 1 2 3 β − 2007936b 4 2 2 3 1 2 1 1 β1 + b c γ 5 2 3 2 3 3 60480b1 b2 c1 γ1 β1 − 3696 3b1 b2 γ1 R1 β1 − √ 3b2 1 b2 3 R1 β1 − 3696 √ 924 3b1 b2 γ1 R1 β1 + 972b1 c1 β1 + 150528b1 b2 c1 β13 − 3 6 3 3 3 501984b1 b2 c1 β1 + 15120b1 b2 c1 β1 + 2664 3 R1 b1 γ1 β1 + 3888b1 c1 γ1 β13 + 602112b31 b32 c1 γ1 β13 − 2007936b41 b22 c1 γ1 β13 + 4 2 3 5 3 6 60480b51 b2 c1 γ1 β13 − 162b71 β12 − 37632b41 b32 β12 − 1296b71 γ13 β12 − 301056b41 b32 γ13 β12 + √669312b 5 2 3 2 1 b2 γ1 β1 − √ 30240b61 b2√γ13 β12 + 5 2 2 7 2 2 4 3 2 2 5 2 2 2 6 2 2 3 √ 2 83664b b β √ 1 32 1 √ −1944b γ β 1 1 1 −451584b b √ 1 22 1 √ γ β 1 +1003968b b γ 1 2 1 1 β −45360b 1√b γ 2 √ β 1 1 −2664 3b c 1 1 R β 1 1 −7616 3b1 b22 c1 R1 β12 − 2 2 6 2 2 2 7 2 4 3 2 5328 3b1 c1 γ1 R1 β1 − 15232 3b1 b2 c1 γ√ 1 √ R1 β1 − 3780b1 b2 β1 +√1848 √ 34 R1 b1 b2 c1√ β1 √− 972b1 γ1 β1 − 225792b √ √ 1 b24γ12β1 + 5 2 2 6 2 2 2 2 2 501984b √ √ 1 b2 γ1 β1 − 22680b1 b2 γ1 β √1 +3 3696 3 R1 b1 b2 c1 γ1√β1 + 666 3 R1 b1 β1 +√1904 3 √R1 b1 b2 β1 + 2664√ 3√ R1 b1 γ1 β1 + 2√ 3 √ 7616√3√R1 b21 b22 γ12 β1 − 3696 √3b√ 1 b2 γ1 R1 β1 −√924 √ 3b 1 b 2 R1 β1 − 3696 3b31 b2 γ1 R1 β1 + 2664 3 R1 b41 γ1 β1 + 7616 3 R1 b21 b22 γ1 β1 )y2 + 1998 3 R1 b41 γ1 + 5712 3 R1 b21 b22 γ1 = 0, where R1 = β12 (b21 + β12 )(−9(2γ1 + 1)2 b41 + 36c1 β1 (2γ1 + 1)b31 + (448(2γ1 b2 + b2 )2 − 3β12 (12c21 + 12γ1 (γ1 + 1) + 31))b21 + 4c1 β1 (9β12 − 448b22 )(2γ1 + 1)b1 − 12(3c21 + 7)β14 + 448b22 (4c21 + 1)β12 ). 5.16. SECTION The polynomial of degree two of the Subcase 1.2 in the proof of statement (c) of Theorem 2 is −648c31 β18 −19656c1 β18 −1296c31 γ1 β18 −39312c1 γ1 β18 +15120b2 c41 β17 +972b1 c21 β17 +504b2 c21 β17 +3888b1 c21 γ12 β17 +15120b2 c21 γ12 β17 + 39312b1 γ12 β17 −7056b2 γ12 β17 +9828b1 β17 −3528b2 β17 +3888b1 c21 γ1 β17 +15120b2 c21 γ1 β17 +39312b1 γ1 β17 −7056b2 γ1 β17 −1296b21 c31 β16 + 334656b22 c31 β16 − 30240b1 b2 c31 β16 − 3888b21 c1 γ13 β16 − 30240b1 b2 c1 γ13 β16 − 5832b21 c1 γ12 β16 − 45360b1 b2 c1 γ12 β16 − 39798b21 c1 β16 + 83664b22 c1 β16 − 504b1 b2 c1 β16 − 2592b21 c31 γ1 β16 + 669312b22 c31 γ1 β16 − 60480b1 b2 c31 γ1 β16 − 81540b21 c1 γ1 β16 + 167328b22 c1 γ1 β16 − 16128b1 b2 c1 γ1 β16 + 150528b32 c41 β15 + 15120b21 b2 c41 β15 + 1296b31 γ14 β15 + 15120b21 b2 γ14 β15 + 19737b31 β15 + 18816b32 β15 + 2592b31 γ13 β15 + 30240b21 b2 γ13 β15 − 41832b1 b22 β15 +1944b31 c21 β15 +112896b32 c21 β15 −501984b1 b22 c21 β15 +26964b21 b2 c21 β15 +80568b31 γ12 β15 +37632b32 γ12 β15 −167328b1 b22 γ12 β15 + 7776b31 c21 γ12 β15 + 150528b32 c21 γ12 β15 − 2007936b1 b22 c21 γ12 β15 + 120960b21 b2 c21 γ12 β15 + 5292b21 b2 γ12 β15 − 5166b21 b2 β15 + 79272b31 γ1 β15 + 37632b32 γ1 β15 − 167328b1 b22 γ1 β15 + 7776b31 c21 γ1 β15 + 150528b32 c21 γ1 β15 − 2007936b1 b22 c21 γ1 β15 + 120960b21 b2 c21 γ1 β15 − 9828b21 b2 γ1 β15 − 648b41 c31 β14 − 301056b1 b32 c31 β 4 √1 + √334656b 2 2 3 4 3 3 4 4 3 4 3 1 b2 c1 β1 − 30240b1 b2 c1 β1 − 7776b1 c1 γ1 β1 − 301056b1 b2 c1 γ1 β1 + 2007936b1 b2 c1 γ1 β1 − 3 4 2 2 3 4 3 3 4 2 4 4 2 4 3 2 4 2 2 2 4 3 2 4 √ √1 b2 c1 γ1 β1 +√1332√ 3 R2 c1 β1 − 11664b1 c1 γ1 β1 − 451584b1 b2 c1 γ1 β1 + 120960b 3011904b1 b2 c1 γ1 β1 − 181440b1 b2 c1 γ1 β1 − 252 3 R2 β14 − 504 3γ1 R2 β14 − 20628b41 c1 β14 − 112896b1 b32 c1 β14√+ √ 334656b21 b22 c1 β14 − 11844b31 b2 c1 β14 − 1296b41 c31 γ1 β14 − 602112b1 b32 c31 γ1 β14 + 669312b21 b22 c31 γ1 β14 − 60480b31 b2 c31 γ1 β14 + 2664 3 R2 c21 γ1 β14 − 45144b41 c1 γ1 β14 − 376320b1 b32 c1 γ1 β14 + 1673280b21 b22 c1 γ1 β14 − √ 84168b √ 3 4 5 3 5 4 3 2 3 4 3 1 b2 c1 γ1 β1 + 9990b1 β1 + 2592b1 γ1 β1 + 150528b1 b2 γ1 β1 − 669312b1 b2 γ1 β1 + 45360b1 b2 γ1 β1 + 3 2 4 3 4 4 3 2 3 3 3 3 5 3 3 2 3 3 3 3 2 3 3 4 3 3 3 2 3 37632b1 b2 β1 + 1848 3 R2 b2 c1 β1 + 5184b1 γ1 β1 + 301056b1 b2 γ1 β1 − 1338624b1 b2 γ1 β1 + 90720b1 b2 γ1 β1 − 83664b1 b2 β1 + 972b51 c21 β13 + 263424b21 b32 c21 β13 − 501984b31 b22 c21 β13 + 26460b41 b2 c21 β13 + 43200b51 γ12 β13 + 338688b21 b32 γ12 β13 − 1171296b √ √ 3 2 2 3 1 b2 γ 1 β 1 + 5 2 2 3 2 3 2 2 3 3 2 2 2 3 4 2 2 3 4 2 3 2 3 3888b√1 c1 γ1 β1 √ + 1053696b1 b2 c√ γ β 1 1 1 √ − 2007936b b c 1 2 1√1 1 γ β + 105840b √ 1 b 2 c γ 1 1 1 β + 57708b b 1√ 12 γ β √1 + 1848 3 R 2 b 2 c1 γ1 β1 − 5328 3b1 c1 γ1 R2 β1 − 1332 3b1 c1 R2 β1 − 5328 3b1 c1 γ1 R2 β1 + 252b1 b2 β1 + 924 3 R2 b2 c1 β1 + 40608b1 γ1 β13 + 2 3 3 3 4 3 3 5 188160b21 b32 γ1 β13 − 501984b 3 2 3 √ √ 1 b2 γ1 β1 + 3888b1 c√ 5 2 1 γ1√ β13 + 1053696b21 b32 c21 γ1 β13 − 2007936b31 b22 c21 γ1 β13 + 105840b41 b2 c21 γ1 β13 + 12348b41 b2 γ1 β13 + 1848 √3 √R2 b2 c1 γ1 β13 + √ 2664 √ 3 R2 b21 γ13 β12 √ − 3888b 6 3 2 √ 12c12γ12β1 − 903168b 3 3 3 2 √ √1 b2 c21 γ21 β21 + 2007936b 4 2 3 2 √ √1 b2 c21 γ21 β21 − 5 3 2 2 2 2 2 90720b1 b2 c1 γ1 β1 + 81 3 R2 b1 β1 + 952 3 R2 b2 β1 + 1332 3 R2 b1 c1 β1 + 3808 √ 3 R32√ b2 c1 β1 + 3996√ 3 R2 b√ 1 γ1 β1 − 5832b√61 c1 γ12 β12 −1354752b 3 3 1 b 2√c 1 γ 2 2 1 β1 +3011904b 4 2 1 b 2 c √ 1 γ 2 2 1 β 1 −136080b 5 1 b 2 c 1 γ 2 2 1 β √ 1 −1848 3b 1 b 2 γ 1 R 2 β 2 1 −2772 3b 1 b 2 c 2 1 R2 β12 − 2√ 2 √ 2 2 √ 2 √ 2 6 2 3 3 2 2772 3b1 b2 γ1 R2 β1 − 462 3b1 b2 R2√ β1 −√ 5544 3b1 b2 c1 γ1 √R√ 2 β1 − 1848 3b1 b2 γ √1 √R2 β1 − 486b1 c1 β1 −√112896b √ 1 b2 c1 β1 + 250992b41 b22 c1 β12 − 11340b51 b2 c1 β12 + 1494 3 R2 b21 γ1 β12 + 1904 3 R2 b22 γ1 β12 + 2664 3 R2 b21 c21 γ1 β12 + 7616 3 R2 b22 c21 γ1 β12 − 2916b61 c1 γ1 β12 − 677376b31 b32 c1 γ1 β12 + 1505952b41 b22 c1 γ1 β12 − 68040b51 b2 c1 γ1 β12 + 81b71 β1 + 1296b71 γ14 β1 + 301056b41 b32 γ14 β1 − 669312b51 b22 γ14 β1 + 30240b61 b2 γ14 β1 + 18816b41 b32 β1 + 2592b71 γ13 β1 + 602112b41 b32 γ13 β1 −√1338624b √ 5 2 3 1 b2 γ 1 β 1 + 6 3 5 2 7 2 4 3 2 5 2 2 6 2 2 60480b √ 1 b32 γ1 β21√ − 41832b1 b2 β1 +√ 1944b1 γ1 β1√ + 451584b1 b2 γ1√ β1 − 1003968b 1 b 2 γ 1 β 1 + √ 45360b 1 b 2 γ 1 β 1 + 7392 3 √ R 2 b 1 b 2√c1 γ12 β1 − 2 2 3 √ 2 √ 3 5328 √ 3b1 c1 γ1 R√ β 2 1 − 15232 3b b c 1 2 1 1 γ R β 2 1√ √ − 1332 3b c 1 1 R β 2 1 − 3808 3b b 1 2 1c R β 2 1 − 5328 3b 1 1 1 R2 β1 − c γ 15232 3b1 b22 c1 γ1 R2 β√ 1 + √ 1890b 6 1 b 2 β 1 + 1848 √ √ 3 R 2 b 2 1 b 2 c 1 β 1 + 648b √ √ 14 3 7 γ 1 β 1 + 150528b 4 3 b √ √1 2 2 2 3 γ 1 β 1 − 334656b √ √ 5 2 1 b2 γ1 β1 + 6 15120b1 b2 γ1 β1 + 7392 3 R2 b1 b2 c1 γ1 β1 + 333 3 R2 b1 + 2664 3 R2 b1 γ1 + 7616 3 R2 b1 b2 γ1 + 952 3 R2 b21 b22 + 2 4 (15120b2 c21 β19 − 7056b2 β19 − 15120b1 b2 c1 β18 − 30240b1 b2 c1 γ1 β18 + 37632b32 β17 + 150528b32 c21 β17 + 30240b21 b2 c21 β17 + 15120b21 b2 γ12 β17 − 10332b21 b2 β17 + 15120b21 b2 γ1 β17 − 150528b1 b32 c1 β16 − 30240b31 b2 c1 β16 − 301056b1 b32 c1 γ1 β16 −√ 60480b 3 6 √ 1 b2 c1 γ51 β1 + 75264b 2 3 5 1 b2 β1 + 2 3 2 5 4 2 5 2 3 2 5 4 2 5 4 5 2 3 5 150528b1 b2 c1 β1 + 15120b b 2 c √ 1 1√ 1 4 β + 150528b 1√b γ 2 1 1 β + 30240b √ 1 b 2 γ β 1 1 + 504b 1 b 2 β 1 + 1848 3 R 2 b 2 c1 β1 + 150528b 1 2 γ1 β1 + b 4 5 4 3 3 4 5 4 3 3 4 30240b1 b2 γ1 β1 − 924 3b1 b2 R2 β1 − 1848 3b1 b2 γ1 R2 β1 − 150528b1 b2 c1 β1 − 15120b1 b2 c1 β1 − 301056b1 b2 c1 γ1 β1 − October 20, 2020 20:22 ”Continuous piecewise isochronous centers” √ √ 30240b51 b2 c1 γ1 β14 + 37632b √ 4 3 3 1 b√2 β1 + 150528b√ 4 3 2 3 1 b2 γ1 β1 + √ 15120b61 b2 γ12 β13 + √ 3780b √ 6 3 2 3 1 b2 β1 + 1848√ 3√ R2 b1 b2 c1 β1 + 150528b 4 3 √ 3 1 b23γ√1 β1 + 3 6 3 3 2 3 2 2 4 2 2 2 2 15120b √ 1 b2 γ1 β1√ −924 3b1 b√ 2 R2 β1 √−1848 3b1 b√ 2 γ1 R2 β√ 1 )y2 +3996 3 R2 b1 γ1 +11424 3 R2 b1 b2 γ1 −3696 3b1 b2 γ1 R2 − 5544 3b31 b2 γ12 R2 − 462 3b31 b2 R2 − 2772 3b31 b2 γ1 R2 + (−1296c31 β19 − 39312c1 β19 + 1944b1 c21 β18 + 15120b2 c21 β18 + 19656b1 β18 − 7056b2 β18 + 3888b1 c21 γ1 β18 + 30240b2 c21 γ1 β18 + 39312b1 γ1 β18 − 14112b2 γ1 β18 − 2592b21 c31 β17 + 669312b22 c31 β17 − 3888b21 c1 γ12 β17 − 60480b1 b2 c1 γ12 β17 − 79596b21 c1 β17 + 167328b22 c1 β17 − 15120b1 b2 c1 β17 − 3888b21 c1 γ1 β17 − 60480b1 b2 c1 γ1 β17 + 39474b31 β16 + 37632b32 β16 + 1296b31 γ13 β16 + 30240b21 b2 γ13 β16 − 83664b1 b22 β16 + 3888b31 c21 β16 + 150528b32 c21 β16 − 1003968b1 b22 c21 β16 + 30240b21 b2 c21 β16 + 1944b31 γ12 β16 + 45360b21 b2 γ12 β16 − 10332b21 b2 β16 + 79596b31 γ1 β16 + 75264b32 γ1 β16 − 167328b1 b22 γ1 β16 + 7776b√31 c√ 2 6 3 2 6 2 2 6 2 2 6 2 6 1 γ1 β1 + 301056b2 c1 γ1 β1 − 2007936b1 b2 c1 γ1 β1 + 60480b1 b2 c1 γ1 β1 − 5544b1 b2 γ1 β1 − 1296b1 c1 β1 + 669312b 4 3 5 √ √ 2 2 3 5 1 b2 c1 β1 + 2664 3 R2 c1 β1 − 7776b1 c1 γ1 β1 − 602112b1 b2 c1 γ1 β1 + 2007936b1 b2 c1 γ1 β1 − 120960b1 b2 c1 γ1 β1 − 504 3 R2 β15 − 2 5 4 2 5 3 2 5 2 2 2 5 3 2 5 41256b41 c1 β15 − 150528b1 b32 c1 β15 + 669312b21 b22 c1 β15 − 30240b31 b2 c1 β15 − 7776b41 c1 γ1 β15 − 602112b1 b32 c1 γ1 β15 + 2007936b21 b22 c1 γ1 β15 − 120960b31 b2 c1 γ1 β15 + 19980b51 β14 + 75264b21 b32 β14 + 2592b51 γ13 β14 + 301056b21 b32 γ13 β14 − 669312b31 b22 γ13 β14 + 60480b41 b2 γ13 β14 − 167328b31 b22 β14 + 1944b51 c21 β14 + 150528b21 b32 c21√β14 − 1003968b √ 3 2 2 4 1 b2 c1 β1√+ 15120b 4 2 4 √1 b2 c14β1 + 3888b 5 2 4 1 γ1 β1 + 451584b 2 3 2 4 √ √ 1 b2 γ1 β41 − 3 2 2 4 4 2 4 4 4 4 1003968b1 b2 γ1 β1 + 90720b1 b2 γ1 β1 − 2664 3b1 c1 R2 β1 − 5328 3b1 c1 γ1 R2 β1 + 504b1 b2 β1 + 1848 3 R2 b2 c1 β1 + 41256b51 γ1 β14 +301056b21√b32√ γ1 β14 −669312b31 b22 γ1 β 4 √1 +3888b √ 5 2 4 2 3 2 4 √ √ 1 b22 c31 γ1 β1 −2007936b 1 c1 γ1 β1 +301056b √ √ 3 2 2 4 1 b2 c1 γ1 β1 +30240b 4 2 4 √ √1 b2 c21 γ21 β31 + 4 4 4 2 3 2 2 3 31248b √ 1√ b2 γ1 β1 + 3696 3 R2 b2 c1 γ1 β1 + 162 3 R2 b1 β1 + 1904 3 R2 b2 β1 + 2664 3 R2 b1 c1 β1 + 7616 √ 3 R22√ b2 c1 β1 + 2 2 3 6 2 3 3 3 2 3 4 2 2 3 5 2 3 3 2664 √ 3 R√ 2 b1 γ1 β1 − 3888b1 c1 γ1 √β1 − 602112b √ 1 b2 c1 γ1 β1 + 2007936b1 b2 c1 γ1 β1 − 60480b1 b2 c1 γ1 β1 − 3696 3b1 b2 γ1 R2 β1 − 3 3 6 3 3 3 3 4 2 3 924 3b1 b2 R2 β1 −√ √3696 3b1 b2 γ1 R2 β1 − 972b1 c1 β1 − 150528b1 b2 c1 β1 + 501984b1 b2 c1 β1 − 15120b51 b2 c1 β13 + 2664 3 R2 b21 γ1 β13 − 3888b61 c1 γ1 β13 − 602112b31 b32 c1 γ1 β13 + 2007936b41 b22 c1 γ1 β13 − 60480b51 b2 c1 γ1 β13 + 162b71 β12 + 37632b41 b32 β12 + 1296b71 γ13 β12 + 301056b41 b32 γ13 β12 − 669312b √ 3 1√ 5 2 3 2 b2 γ1 β1 + 30240b 6 3 2 1 − 83664b1 b2 β √ 1 b2 γ21 β√ 5 2 2 √1 +3 1944b 7 2 2 √1 γ1 β12 + 4 3 2 2 5 2 2 2 6 2 2 2 2 451584b b γ √1 2 1 1 √ β − 1003968b b γ 1 2 1 1 β + 45360b b 1 2 1γ β √ 1√ − 2664 3b c 1 1 R β 2 1 − 7616 3b b c 1 2 1 R β 2 1 − 5328 3b 1 1 1 R2 β1 − c γ 15232 3b1 b22 c1 γ1 R2√β12√+ 3780b61 b2 β12 + 1848 √ √ 3 R 2 b 2 1 b 2 c 1 β√ 2 1 √+ 972b 7 1 γ 1 β 2 1 + 225792b 4 3 √ √ 14 22 1 b γ 1 β 2 − 501984b 5 2 2 √ √ 1 b22 γ21 β21 + 6 2 2 2 4 2 2 22680b √ 1 b2 γ1 β1√ +3696 3 R2 b√ 1 b2 c1 γ1 √β1 +666 3 R2√ b1 β1 +1904 √ 3 R2 b1 b2 β1√ +2664 √ 3 R2 b1 γ1 β1 +7616 √ √ 3 R2 b1 b2 γ1 β1 − 3696√3b√31 b2 γ12 R2 β1 −√924 √ 3b31 b2 R2 β1 − 3696 3b31 b2 γ1 R2 β1 + 2664 3 R2 b41 γ1 β1 + 7616 3 R2 b21 b22 γ1 β1 )y2 + 1998 3 R2 b41 γ1 + 5712 3 R2 b21 b22 γ1 = 0. where R2 = β12 (b21 + β12 )(−9(2γ1 + 1)2 b41 + 36c1 β1 (2γ1 + 1)b31 + (448(2γ1 b2 + b2 )2 − 3β12 (12c21 + 12γ1 (γ1 + 1) + 31))b21 + 4c1 β1 (9β12 − 448b22 )(2γ1 + 1)b1 − 12(3c21 + 7)β14 + 448b22 (4c21 + 1)β12 ). 5.17. SECTION All sets of real solutions s1 = {b1 = β1 }, s2 = {α1 = β1 , α2 = β1 }, s3 = {α1 = β1 , γ1 = −1/2}, s4 = {α2 = β1 , γ2 = 3/16}, s5 = {c1 = −4c2 /3, γ1 = −1/2, γ2 = 3/16} and α22 s6 = {a2 = −3a1 /4, c1 = −4c2 /3, γ1 = − 2α2 3α1 − 1/2, γ2 = 3/16 − 3α21 }. 5.18. SECTION All solutions are s1 = {a1 = 0}, 2b2 β1 (−2β2 b2 (9c1 +8c2 )+b22 (16γ2 −3)−12β22 )+3b1 (b22 (β1 (16γ2 −3)−6β2 )+8β1 β2 b2 c2 −3β1 β22 ) s2 = {γ1 = 36b1 b22 β2 }, 1 b2 (16γ2 −3) 12β2 s3 = {b1 = 0, c1 = 18 ( β2 − b2 − 16c2 )}, s5 = {b2 = 0, β1 = 0}, s6 = {b2 = 0, β2 = 0}, s7 = {b1 = 0, β1 = 0}, s8 = {β1 = 0, β2 = 0}, s9 = {b1 = −2b2 /3, β2 = 0} and s10 = {β2 = 0, γ2 = 3/16}. 5.19. SECTION All solutions are u1 = {β1 = 0},√ 1 22a b (3b +2b )−3a b2 u2 = {β1 = − √1 √ 2 2 1 }, 3 a2 √ 2 2a1 b2 (3b1 +2b2 )−3a2 b1 u3 = {β1 = √ √ }, 3 a2 u4 = {a1 = 0, a2 = 0}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” u5 = {a2 = 0, b1 = −2b2 /3} and u6 = {a2 = 0, b2 = 0}. 5.20. SECTION All solutions are v1 = {a1 = 0}, 2a b (3b +2b2 ) v2 = {a2 = 1 2 3b21 }, 1 v3 = {b2 = −3b1 /2}, 3b1 (8β2 b2 c2 +b22 (16γ2 −3)−3β22 )+4b2 (−16β2 b2 c2 +b22 (16γ2 −3)−12β22 ) v4 = {c1 = 72b22 β2 }, v5 = {b1 = −4b2 /3, β2 = 0} and v6 = {β2 = 0, γ2 = 3/16}. 5.21. SECTION All solutions are 2 2(2a1 b2 +3a1 b1 b2 ) w1 = {a2 = 3b2 }, 1 w2 = {a1 = 0, b1 = 0}, w3 = {b1 = 0, b2 = 0}, 4b2 a2 (32β2 b32 c2 (3−16γ2 )+β22 b22 (256c22 −384γ2 +153)+384β23 b2 c2 +b42 (3−16γ2 )2 +144β24 ) w4 = {b1 = 9a1 (32β2 b32 c2 (3−16γ2 )+2β22 b22 (128c22 +24γ2 +9)−48β23 b2 c2 +b42 (3−16γ2 )2 +9β24 ) − 23 b2 }, w5 = {a1 = 0, a2 = 0}, w6 = {a1 = 0, b2 = 0}, w7 = {b2 = 0, β2 = 0} and w8 = {β2 = 0, γ2 = 3/16}. 5.22. SECTION All solutions are 3a1 (32β2 b32 c2 (3−16γ2 )+8β22 b22 (32c22 +6γ2 +9)−48β23 b2 c2 +b42 (3−16γ2 )2 +63β24 ) z1 = {a2 = − }, 4(32β2 b32 c2 (3−16γ2 )+β22 b22 (256c22 −384γ2 +153)+384β23 b2 c2 +b42 (3−16γ2 )2 +144β24 ) z2 = {b2 = 0}, z3 = {b2 = 0, β2 = 0} and z4 = {β2 = 0, γ2 = 3/16}. 5.23. SECTION Solving the second equation of (24) we get one of the following sets of real solutions v1 = {a1 = 0}, 2a b (3b +2b2 ) v2 = {a2 = 1 2 3b21 }, 1 v3 = {b2 = −3b1 /2}, 3b1 (8β2 b2 c2 +b22 (16γ2 −3)−3β22 )+4b2 (−16β2 b2 c2 +b22 (16γ2 −3)−12β22 ) v4 = {c1 = 72b22 β2 }, v5 = {b1 = −4b2 /3, β2 = 0} and v6 = {β2 = 0, γ2 = 3/16}. The only allowed solution is v3 then we have 3b1 (8β2 b2 c2 + b22 (16γ2 − 3) − 3β22 ) + 4b2 (−16β2 b2 c2 + b22 (16γ2 − 3) − 12β22 ) c1 = . 72b22 β2 The first equation of (24) gives one of the following sets of real solutions 2(2a1 b22 +3a1 b1 b2 ) w1 = {a2 = 3b21 }, w2 = {a1 = 0, b1 = 0}, w3 = {b1 = 0, b2 = 0}, 8b2 a2 (32β2 b32 c2 (3−16γ2 )+β22 b22 (256c22 −384γ2 +153)+384β23 b2 c2 +b42 (3−16γ2 )2 +144β24 ) w4 = {b1 = 9a1 (32β2 b32 c2 (3−16γ2 )+2β22 b22 (128c22 +24γ2 +9)−48β23 b2 c2 +b42 (3−16γ2 )2 +9β24 ) − 23 b2 }, w5 = {a1 = 0, a2 = 0}, w6 = {a1 = 0, b2 = 0}, w7 = {b2 = 0, β2 = 0} and October 20, 2020 20:22 ”Continuous piecewise isochronous centers” w8 = {β2 = 0, γ2 = 3/16}, The only allowed solution is w4 for which the value of b1 is given as follows 8b2 a2 (32β2 b32 c2 (3 − 16γ2 ) + β22 b22 (256c22 − 384γ2 + 153) + 384β23 b2 c2 + b42 (3 − 16γ2 )2 + 144β24 ) 2 3 2 2 2 3 4 2 4 − b2 . 9a1 (32β2 b2 c2 (3 − 16γ2 ) + 2β2 b2 (128c2 + 24γ2 + 9) − 48β2 b2 c2 + b2 (3 − 16γ2 ) + 9β2 ) 3 Solving the fourth equation of (24) gives one of the following sets of real solutions 3a1 (32β2 b32 c2 (3−16γ2 )+8β22 b22 (32c22 +6γ2 +9)−48β23 b2 c2 +b42 (3−16γ2 )2 +63β24 ) z1 = {a2 = − }, 4(32β2 b32 c2 (3−16γ2 )+β22 b22 (256c22 −384γ2 +153)+384β23 b2 c2 +b42 (3−16γ2 )2 +144β24 ) z2 = {b2 = 0}, z3 = {b2 = 0, β2 = 0} and z4 = {β2 = 0, γ2 = 3/16}. The only allowed solution is z1 , so the value of a2 is fixed. Now we have a continuous piecewise differential systems (8)-(10). We solve the algebraic system (15) with respect to y1 and y2 , and we get one of the same sets of solutions s0 , s1 , s2 and s3 as in Subcase 2.2.1.1. Then the continuous piecewise differential systems (8)-(10) can have at most one limit cycle. 5.24. SECTION All sets of solutions are s1 = {a1 = 0}, s2 = {b2 = 0}, 3b22 (8β2 γ1 +β1 +4β2 )−8β1 β2 b2 c2 +3β1 β22 s3 = {γ2 = 16b22 β1 }, s4 = {b2 = 0, β1 = 0}, s5 = {b2 = 0, β2 = 0}, s6 = {β1 = 0, γ1 = −1/2} and s7 = {β1 = 0, β2 = 0}. 5.25. SECTION All sets of real solutions s1 = {a1 = a2 , β1 = a2 }, s2 = {b2 = a2 , β1 = a2 }, s3 = {a1 = a2 , α2 = a2 }, s4 = {b2 = a2 , β1 = a2 }, s5 = {c1 = a2 , β1 = a2 }, s6 = {β1 = a2 , γ1 = −1/2}, α2 (4b22 +9β12 ) 6b c γ +3b c +4β1 c22 s7 = {β2 = 36a1 b2 , γ2 = − 2 1 1 3β21 1 }, s8 = {b2 = a2 , α2 = a2 , γ2 = −4c22 /3}, 6b2 c1 γ1 +3b2 c1 +4β1 c22 s9 = {a1 = a2 , α2 = a2 , γ2 = − 3β1 }, α2 b2 s10 = {c1 = a2 , β1 = a2 , β2 = 9a1 }, s11 = {β1 = a2 , β2 = α9a 2 b2 1 , γ1 = −1/2}, s12 = {a1 = a2 , α2 = a2 , β1 = a2 , γ1 = −1/2}, s13 = {a1 = a2 , c1 = a2 , α2 = a2 , β1 = a2 }. 5.26. SECTION All solutions are s1 = {a1 = 0, β1 = 0}, 2 1 9β1 3 c1 (2β2 γ1 +β2 ) s2 = {a1 = 36 a2 (− b22 − 64), γ2 = 16 (1 − β1 c2 )}, s3 = {a1 = 0, a2 = 0}, s4 = {c1 = 0, β1 = 0}, s5 = {β1 = 0, β2 = 0}, s6 = {β1 = 0, γ1 = −1/2}, 3 c (2β γ +β ) s7 = {a2 = 0, b2 = 0, γ2 = 16 (1 − 1 β21 c12 2 )}, s8 = {b2 = 0, c1 = 0, β1 = 0}, s9 = {b2 = 0, β1 = 0, β2 = 0}, s10 = {b2 = 0, β1 = 0, γ1 = −1/2}, 2 1 9β1 s11 = {a1 = 36 a2 (− b22 − 64), c2 = 0, γ1 = −1/2}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 2 1 9β1 s12 = {a1 = 36 a2 (− b22 − 64), c1 = 0, c2 = 0}, 1 9β12 s13 = {a1 = 36 a2 (− b22 − 64), c2 = 0, β2 = 0}, s14 = {a1 = −16a2 /9, β1 = 0, γ1 = −1/2}, s15 = {a1 = −16a2 /9, c1 = 0, β1 = 0}, s16 = {a1 = −16a2 /9, β1 = 0, β2 = 0}, s17 = {a2 = 0, b2 = 0, c1 = 0, c2 = 0}, s18 = {a2 = 0, b2 = 0, c2 = 0, β2 = 0} and s19 = {a2 = 0, b2 = 0, c2 = 0, γ1 = −1/2}. 5.27. SECTION All solutions are u1 = {a2 = 0}, u2 = {β2 = 0}, u3 = {β1 = 0, γ1 = −1/2}, q b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 3β12 (16b22 +9β12 ) β1 c2 u4 = {c1 = −8c2 /3, 128b2 β −9β 3 , γ1 = − q + b2 − 12 }, 2 1 1 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 q b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 3β12 (16b22 +9β12 ) β1 c2 u5 = {c1 = −8c2 /3, β2 = 9β13 −128b22 β1 , γ1 = q + b2 − 21 } and 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 u6 = {c1 = −8c2 /3, β1 = 0, γ1 = −1/2}. 5.28. SECTION Solving the first equation of (15) with respect to y2 after substituting in it only c1 , we get   128b2 (2γ1 c2 + c2 ) + 64b22 (2γ1 y1 + y1 ) + β1 −128c22 + 18γ1 (γ1 + 1) + 9y1 (2β1 γ1 + β1 ) y2 = −  . 64b22 + 9β12 (2γ1 + 2β1 y1 + 1) Replacing this value of y2 into the second equation of (15) and solving it with respect to y2 after invoking the remaining parameters β2 and γ1 , we get a polynomial of degree four which has the four roots √ q 3β1 (16b22 +9β12 ) y1 = − 12 3 3 128b21−9β 2 + q − cb22 , 2 1 8 b22 β12 (128b22 −9β12 )(16b22 +9β12 ) √ q 3β1 (16b22 +9β12 ) y1 = − 21 3 3 128b21−9β 2 + q − cb22 , 2 1 8 b22 β12 (128b22 −9β12 )(16b22 +9β12 ) √ q 3β1 (16b22 +9β12 ) y1 = 32 3 128b21−9β 2 + q − cb22 and 2 1 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 √ q 3β1 (16b22 +9β12 ) y1 = 32 3 128b21−9β 2 + q − cb22 . 2 1 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 Replacing these values of y1 into y2 , all cases give y2 = y1 which does not give limit cycles for the piecewise differential systems (8)-(11) in this subcase. 5.29. SECTION Solving the first equation of (15) with respect to y2 after substituting in it c1 , we get   128b2 (2γ1 c2 + c2 ) + 64b22 (2γ1 y1 + y1 ) + β1 −128c22 + 18γ1 (γ1 + 1) + 9y1 (2β1 γ1 + β1 ) y2 = −  . 64b22 + 9β12 (2γ1 + 2β1 y1 + 1) Replacing this value of y2 into the second equation of (15) and solving it with respect to y2 after invoking the remaining parameters β2 and γ1 , we get a polynomial of degree four which has the four roots √ q 3β1 (16b22 +9β12 ) y1 = − 12 3 3 128b21−9β 2 + q − cb22 , 2 1 8 b22 β12 (128b22 −9β12 )(16b22 +9β12 ) √ q 3β1 (16b22 +9β12 ) y1 = − 21 3 3 128b21−9β 2 + q − cb22 , 2 1 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 √ q 3β1 (16b22 +9β12 ) y1 = 32 3 128b21−9β 2 + q − cb22 and 2 1 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 √ q 3β1 (16b22 +9β12 ) y1 = 32 3 128b21−9β 2 + q − cb22 . 2 1 8 b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 Replacing these values of y1 into y2 , all cases cases give y2 = y1 which does not give limit cycles for the piecewise differential systems (8)-(11) in this subcase. October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 5.30. SECTION All solutions are s1 = {a2 = 0}, s2 = {β1 = 0}, s3 = {β2 = 0, γ2q= 3/16},  s4 = {β1 = − 43 b22 12c21 − 1 , β2 = 0, γ2 = 3/16} and q  s5 = {β1 = 34 b22 12c21 − 1 , β2 = 0, γ2 = 3/16}. 5.31. SECTION All solutions are u1 = {a2 = 0}, u2 = {β1 = 0, γ1 = −1/2}, u3 = {β1 = 0, β2 = 0}, u4 = {β2 = 0, γ2 = 3/16}, u5 = {β1 = 0, β2 = 0, γ2 = −9γ1 (γ1 + 1)/16}, u6 = {β1 = 0, γ1 = −1/2, γ2 = 9/64}, q 16β1 b42 +9β13 b22 −36β12 b22 +3 b22 β12 (1008β12 b22 +2048b42 −81β14 )+512b42 u7 = {β2 = q , γ1 = − 1024b42 −72b22 β12 , b22 β12 (128b22 −9β12 )(16b22 +9β12 ) 54b22 9 γ2 = 9β12 −128b22 + 16 }, q q b2 β1 (128b22 −9β12 )(16b22 +9β12 ) 2 2 36β12 b22 +3 b22 β12 (128b22 −9β12 )(16b22 +9β12 )−512b42 u8 = {β2 = 9β 3 −128b2 β , γ1 = , 1 2 1 8b22 (128b22 −9β12 ) 54b22 9 γ2 = + 16 9β12 −128b22 }, √ u9 = {β1 = −8√ 2b2 /3, β2 = 0, γ1 = 1, γ2 = 3/16} and u10 = {β1 = 8 2b2 /3, β2 = 0, γ1 = 1, γ2 = 3/16}. 5.32. SECTION All solutions are s1 = {a1 = 0, a2 = 0, β1 √ = 0}, √ s2 = {a1 = 0, c1 = − γ√ 1 γ1 + 1, α1 = 0, β1 = 0}, √ s3 = {a1 = 0, c1 = γ1 γ1 + 1, α1 = 0, β1 = 0} and a (6a1 c1 + 16a1 c2 + 6α1 γ1 + 3α1 ) a (−c21 + γ12 + γ1 ) 4c2 s4 = {α2 = 2 , β1 = 0, β2 = 0, γ2 = 2 − 2 }. 3a1 a1 3 5.33. SECTION 2 (8γ +3)(b2 +β 2 ) All solutions are s1 = {a2 = α2 b2 /β2 , c2 = 2 24b2 β2 2 }, s2 = {b1 = 0, β1 = 0}, s3 = {c1 = [−3b21 (2γ1 + 1)(4α2 b2 − a2 β2 ) + b1 (9a1 β2 b2 (2γ1 + 1) + α1 β22 (8γ2 + 3) + 4α1 b22 (8γ2 + 3)) − β1 (3(2γ1 + 1)(−a2 β2 β1 + 4α2 b2 β1 − 3α1 b2 β2 ) + a1 (8γ2 + 3)(4b22 + β22 ))]/[18b2 β2 (α1 b1 − a1 β1 )], (−3b2 (2γ +1)(α b −a β )−β (3β (2γ +1)(α b −a β )+a (8γ2 +3)(b22 +β22 ))+α1 b1 (8γ2 +3)(b22 +β22 ) 1 2 2 2 2 1 1 1 2 2 2 2 1 c2 = 1 (24b2 β2 (α1 b1 −a1 β1 ) }, s4 = {a2 = α2 b2 /β2 , α1 = a1 β1 /b1 }, s5 = {α1 = a1 β1 /b1 , γ1 = −1/2}, s6 = {b2 = 0, β2 = 0}, s7 = {a1 = 0, a2 = α2 b2 /β2 , α1 = 0}, s8 = {a1 = 0, α1 = 0, γ1 = −1/2}, s9 = {a1 = 0, a2 = α2 b2 /β2 , b1 = 0}, s10 = {a1 = 0, b1 = 0, γ1 = −1/2}, s11 = {a2 = 0, b2 = 0, γ2 = −3/8}, s12 = {b1 = 0, b2 = 0, β1 = 0}, s13 = {a2 = 0, b2 = 0, α1 = a1 β1 /b1 }, s14 = {b2 = 0, α1 = a1 β1 /b1 , γ1 = −1/2}, a1 β2 b1 (6γ1 +3)−3α2 β1 (2β1 γ1 +β1 )+β1 β2 (−8a1 c2 +6α1 γ1 +3α1 )−3b21 (2α2 γ1 +α2 )+8α1 β2 b1 c2 s15 = {b2 = 0, c1 = 6β2 (α1 b1 −a1 β1 ) , −3a2 (2γ1 +1)(b21 +β12 ) γ2 = 8β2 (α1 b1 −a1 β1 ) − 3/8}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” (8γ +3)(b2 +β 2 ) 2 s16 = {a2 = 0, c2 = 2 24b2 β2 2 , α2 = 0}, s17 = {b1 = 0, β1 = 0, β2 = 0}, s18 = {α1 = a1 β1 /b1 , α2 = 0, β2 = 0}, s19 = {α1 = a1 β1 /b1 , β2 = 0, γ1 = −1/2}, s20 = {α2 = 0, β2 = 0, γ2 = −3/8}, −3a2 (2γ1 +1)(b21 +β12 )+a1 b2 (b1 (6γ1 +3)−32β1 c2 )+α1 b2 (32b1 c2 +6β1 γ1 +3β1 ) s21 = {c1 = 6b2 (α1 b1 −a1 β1 ) , β2 = 0, 3(β1 (α2 β1 (2γ1 +1)+a1 b2 )+b21 (2α2 γ1 +α2 )−α1 b1 b2 ) γ2 = 8b2 (α1 b1 −a1 β1 ) }, s22 = {a1 = 0, a2 = α2 b2 /β2 , b1 = 0, α1 = 0}, s23 = {a1 = 0, b1 = 0, α1 = 0, γ1 = −1/2}, s25 = {a1 = 0, b2 = 0, α1 = 0, γ1 = −1/2}, s26 = {a1 = 0, a2 = 0, b1 = 0, b2 = 0}, s27 = {a1 = 0, b1 = 0, b2 = 0, γ1 = −1/2}, s28 = {a1 = 0, α1 = 0, α2 = 0, β2 = 0}, s29 = {a1 = 0, α1 = 0, β2 = 0, γ1 = −1/2}, s30 = {a1 = 0, b1 = 0, α2 = 0, β2 = 0}, s31 = {a1 = 0, b1 = 0, β2 = 0, γ1 = −1/2}, s32 = {a1 = 0, a2 = 0, b1 = 0, b2 = 0, α1 = 0}, s33 = {a1 = 0, b1 = 0, b2 = 0, α1 = 0, γ1 = −1/2}, s34 = {a1 = 0, b1 = 0, α1 = 0, α2 = 0, β2 = 0} and s35 = {a1 = 0, b1 = 0, α1 = 0, β2 = 0, γ1 = −1/2}. 5.34. SECTION   β2 4β22 (α1 b1 − a1 β1 ) + 3a2 b21 β1 + 8a1 b22 β1 + 3a2 β13 − 8α1 b1 b22 All solutions are u1 = {α2 =  }, 3b2 β1 b21 + β12   β 3a2 (b21 +β12 ) u2 = {b2 = 0, α1 = 1 4a1 − β22 }, 4b1 u3 = {b1 = 0, β1 = 0}, β u4 = {b2 = − √2 , β1 = 0}, 2 β2 u5 = {b2 = √ , β1 = 0}, 2 u6 = {α1 = 0, β1 = 0}, u7 = {β1 = 0, β2 = 0}, u8 = {b2 = 0, β2 = 0} and 4a β 2 u9 = {a2 = 1 22 , b1 = 0, b2 = 0}. 3β1 5.35. SECTION All solutions are   4β22 (a1 β1 − α1 b1 ) + b2 6a1 β1 b1 − 16a1 b2 β1 + 3α1 β12 − 3α1 b21 + 16α1 b2 b1 v1 = {a2 =  }, 3β1 b21 + β12 α b v2 = {β1 = 1 1 }, a1 v3 = {β2 = 0},√ v4 = {β2 = √− 2b2 }, v5 = {β2 = 2b2 }, v6 = {a1 = 0, b1 = 0}, v7 = {a1 = 0, α1 = 0}, v8 = {b1 = 0, β1 = 0}, v9 = {α1 = 0, β1 = 0},p v10 = {β1 = 0, β2 = − p b2 (16b2 − 3b1 )/2} and v11 = {β1 = 0, β2 = b2 (16b2 − 3b1 )/2}. 5.36. SECTION α1 b1 All solutions are z1 = {β1 = a1 }, (β1 (8γ2 +3)−4β2 ) z2 = {γ1 = 8β2 October 20, 2020 20:22 ”Continuous piecewise isochronous centers” 9(−b22 β12 β24 (9b21 +32b22 +9β12 −16β22 )(β22 −2b22 )(4β22 (9b21 −32b22 +9β12 )+b22 (9b21 −256b22 +9β12 )−16β24 ))1/2 − 8β22 (β22 −2b22 )(−4β22 (9b21 −32b22 +9β12 )+b22 (−9b21 +256b22 −9β12 )+16β24 ) }, (β1 (8γ2 +3)−4β2 ) z3 = {γ1 = 8β2 9(−b22 β12 β24 (9b21 +32b22 +9β12 −16β22 )(β22 −2b22 )(4β22 (9b21 −32b22 +9β12 )+b22 (9b21 −256b22 +9β12 )−16β24 ))1/2 + 8β22 (β22 −2b22 )(−4β22 (9b21 −32b22 +9β12 )+b22 (−9b21 +256b22 −9β12 )+16β24 ) }, z4 = {a1 = 0, b1 = 0}, z5 = {a1 = 0, αq1 = 0}, z6 = {b1 = − 13 4β22 − 9β12 , b2 = 0}, q z7 = {b1 = 13 4β22 − 9β12 , b2 = 0}, β2 z8 = {b2 = − √ , β1 = 0}, 2 β2 z9 = {b2 = √ , β1 = 0}, 2 q 32β22 β2 z10 = {b1 = − 13 2 3 − 9β1 , b2 = − √ }, q 6 1 32β22 β z11 = {b1 = 3 3 − 9β12 , b2 = − √ }, 2 q 6 32β22 1 2 β z12 = {b1 = −3 3 − 9β1 , b2 = 6 }, √ 2 q 32β 2 1 2 β2 z13 = {b1 = 3 3 − 9β1 , b2 = 6 }, 2 √ 2 2 4(4b +β ) z14 = {b1 = − √ 22 2 2 , β1 = 0}, 3 b2 +4β2 4(4b22 +β22 ) z15 = {b1 = √ , β1 = 0}, 3 b22 +4β22 q z16 = {b1 = − 31 256b22 − 9β12 , β2 = 0}, q z17 = {b1 = 31 256b22 − 9β12 , β2 = 0}, z18 = {b2 = 0, β2 = 0}, z19 = {β1 = 0, β2 = 0}, z20 = {β2 = 0, γ2 = − 38 }, q β2 z21 = {b1 = − 41 2 β2 , b2 = − √ , β1 = 0}, q 3 3 6 β2 z22 = {b1 = 43 23 β2 , b2 = − √ , β1 = 0}, q 6 4 2 β z23 = {b1 = − 3 3 β2 , b2 = √2 , β1 = 0} and q 6 β2 z24 = {b1 = 43 23 β2 , b2 = √ , β1 = 0}. 6 5.37. SECTION All solutions are v1 = {b1 = 0}, v2 = {α1 = 0},√ v3 = {α2 = − 2a2 }, v4 = {β2 = 0} and v5 = {β2 = − 3b √1 }. 4 2 5.38. SECTION All solutions are u1 = {β2 = 0}, u2 = {a1 = 3a2 (b21 + β12 )/(4β22 ) + α1 b1 /β1 , α2 = 3a2 b1 /(2β2 ) + α1 β2 /β1 }, u3 = {a1 = α1 b1 /β1 , a2 = 0}, u4 = {b1 = 0, β1 = 0}, u5 = {α1 = 0, β1 = 0}, u6 = {a2 = 0, β2 = 0}, u7 = {a2 = 0, α1 = 0, β1 = 0} and u8 = {a2 = 0, β1 = 0, β2 = 0}. 5.39. SECTION All solutions are u1 = {a2 = 0}, u2 = {β1 = 0}, October 20, 2020 20:22 ”Continuous piecewise isochronous centers” q q u3 = {b1 = 0, β1 = −4β2 64c22 + 9/(3 256c22 + 9)}, q q u4 = {b1 = 0, β1 = 4β2 64c22 + 9/(3 256c22 + 9)}, q q u5 = {b1 = −4β2 64c22 + 9/(3 256c22 + 9), β1 = 0}, q q u6 = {b1 = 4β2 64c22 + 9/(3 256c22 + 9), β1 = 0}, q q u7 = {a2 = 0, β1 = − 16β22 (64c22 + 9) − 9b21 (256c22 + 9)/(3 256c22 + 9)} and q q u8 = {a2 = 0, β1 = 16β22 (64c22 + 9) − 9b21 (256c22 + 9)/(3 256c22 + 9)}. 5.40. SECTION All solutions are u1 = {b2 = 0}, u2 = {α2 = 0}, u3 = {b1 = 0, β1 = 0}, u4 = {b2 = 3b1 /16, β1 = 0} and u5 = {α1 = 0, β1 = 0}. 5.41. SECTION All solutions are u1 = {a2 = 3a1 /16}, u2 = {b1 = 0}, u3 = {α2 = 0} and u4 = {γ1 = −1/2}.