This is a preprint of: “The extended 16th Hilbert problem for a class of discontinuous piecewise differential systems”, Meriem Barkat, Rebiha Benterki, Jaume Llibre, Nonlinear Dynam., vol. 111, 1475–1484, 2023. DOI: [10.1007/s11071-022-07891-9] THE EXTENDED 16TH HILBERT PROBLEM FOR A CLASS OF DISCONTINUOUS PIECEWISE DIFFERENTIAL SYSTEMS MERIEM BARKAT 1 , REBIHA BENTERKI1 AND JAUME LLIBRE2 Abstract. This paper solves the extended 16th Hilbert problem for a family of discontinuous planar differential systems with two regions separated by the straight line x = 0. By using the first integrals, we prove that the maximum number of crossing limit cycles in the family of systems formed by a linear center and a class of Hamiltonian isochronous global centers with a polynomial first integral of degree 2n is 5. 1. Introduction In this paper we are interesting is studying discontinuous piecewise differential systems defined by  − F (x) = (F1− (x), F2− (x))T x ∈ Σ− (1) ẋ = F(x) = + + + T F (x) = (F1 (x), F2 (x)) x ∈ Σ+ . with x = (x, y). Where Σ− and Σ+ are two regions in the plane defined by Σ− = {(x, y) : x ≤ 0}, Σ+ = {(x, y) : x ≥ 0}, and Σ = {(x, y) : x = 0} is the separation line of the plane. For this kind of systems the problem of analysing the existence and the maximum number of limit cycles remains in general open, and to find an upper bound for the maximum number of limit cycles in this class of systems it is known as the extended 16th Hilbert problem. The original 16th Hilbert problem consists in finding an upper bound for the maximum number of limit cycles that the class of polynomial differential systems of a given degree can exhibit, see [10, 11, 12]. The dynamics over the line of discontinuity x = 0 is defined following the Fil- ippov’s convention, see [8]. If F1− (0, y)F1+ (0, y) > 0, then both vector fields have the same direction, the point (0, y) is called a crossing point, and the cross region is defined as follows Σc = {(0, y) ∈ Σ | F1− (0, y)F1+ (0, y) > 0}. Around 1920 Andronov, Vitt and Khaikin [1] started the study of the piecewise differential systems separated by a straight line and until nowadays such systems have deserved the attention of the researchers. Thus these differential systems are 2010 Mathematics Subject Classification. Primary 34C29, 34C25, 47H11. Key words and phrases. Limit cycles, discontinuous piecewise linear differential systems, linear Hamiltonian systems, irreducible cubic curves. 1 2 M. BARKAT, R. BENTERKI AND J. LLIBRE widely used to model processes appearing in mechanics, electronics, economy, etc . . . , see for instance [3, 16, 19]. In the last years many authors worked on the extended 16th Hilbert’s problem for distinct classes of discontinuous piecewise linear differential systems separated by a straight line, where they determine or bound how many limit cycles can appear in these classes of difFerential systems, see for instance [2, 4, 7, 9]. In what follows we deal with the following two classes of isochronous centers. First the class of linear differential systems. The the following lemma provides a normal form for an arbitrary linear differential center. Lemma 1. Through a linear change of variables and a rescaling of the independent variable every linear differential center in R2 can be written a2 + ω 2 (2) ẋ = −ax − y + b, d ẏ = dx + ay + c, with ω > 0 d > 0. The first integral of this linear differential system is (3) H1 (x, y) = (ay + dx)2 + 2d(cx − by) + ω 2 y 2 . We consider the class of Hamiltonian isochronous global centers of the form (4) ẋ = −(δnxy n−1 + δ 2 ny 2n−1 + y), ẏ = δy n + x, with n ∈ N and δ ̸= 0. The first intgral of system (4) is given by 1 H(x, y) = ((δy n + x)2 + y 2 ). 2 For more details on the Hamiltonian system (4) see Theorem B of [17] and Theorem 2.3 of [18]. The main result of this work is to study the upper bounds of crossing limit cycles for discontinuous piecewise differential systems separated by the straight line x = 0, and formed by a linear center (2) and the class of Hamiltonian isochronous global center (4) after an arbitrary affine change of variables. Theorem 2. Consider discontinuous piecewise differential systems separated by the straight line x = 0 and formed by two differential systems, when these differential systems are linear centers (2) or the class of Hamiltonian isochronous global centers (4) after an arbitrary affine change of variables. The maximum number of limit cycles of these discontinuous piecewise differential systems is (a) no limit cycles if n = 1; (b) at most one if n = 2, and this maximum is reached, see Fig. 1; (c) at most two if n = 3, and this maximum is reached, see Fig. 2; (d) at most three if n = 4, and this maximum is reached, see Fig. 3; (e) at most four if n = 5, and this maximum is reached, see Fig. 4; (f ) at most five if n ≥ 6, and this maximum is reached, see Fig. 5. Theorem 2 is proved in section 3. The phase portraits of the piecewise differential systems (2)-(4) could be studied using the techniques of [5, 6]. 3 2. The class of Hamiltonian isochronous global center (4) after an arbitrary affine change of variables We present the expression of the class of Hamiltonian isochronous global centers (4) and their first integrals after an arbitrary affine change of variables (x, y) → (a1 X + b1 Y + c1 , α1 X + β1 Y + γ1 ), with b1 α1 − a1 β1 ̸= 0. Thus, after this affine change of variables the differential system (4) becomes −1 Ẋ = − b1 (c1 + a1 x + b1 y + (xα1 + yβ1 + γ1 )n δ) (b1 α1 − a1 β1 ) 1 − β1 (xα1 + yβ1 + γ1 )2 + n(c1 + a1 x + b1 y)(xα1 (xα1 + yβ1 + γ1 )  +yβ1 + γ1 )n δ + n(xα1 + yβ1 + γ1 )2n δ 2 , (5) −1 Ẏ = a1 (c1 + a1 x + b1 y + (xα1 + yβ1 + γ1 )n δ) (b1 α1 − a1 β1 ) 1 + α1 (xα1 + yβ1 + γ1 )2 + n(c1 + a1 x + b1 y)(xα1 (xα1 + yβ1 + γ1 )  +yβ1 + γ1 )n δ + n(xα1 + yβ1 + γ1 )2n δ 2 , with its first integral 1 H2 (x, y) = ((xα1 + yβ1 + γ1 )2 + (c1 + a1 x + b1 y + (xα1 + yβ1 + γ1 )n δ)2 ). 2 2.6 0.8 2.4 0.6 2.2 0.4 0.2 2.0 0.0 1.8 -0.2 1.6 -0.4 -0.6 -0.4 -0.2 0.0 0.2 -0.4 -0.3 -0.2 -0.1 0.0 0.1 Figure 1. The unique Figure 2. The two crossing limit cycle of crossing limit cycles the discontinuous piece- of the discontinuous wise differential system piecewise differential (7)–(8) system (10)–(11) 3. Proof of Theorem 2 The statement (a) of Theorem 2 has already been proved in Theorem 4 of [15] and in Theorem 3 of [13]. 4 M. BARKAT, R. BENTERKI AND J. LLIBRE 1.0 0.5 0.8 0.6 0.0 0.4 0.2 -0.5 0.0 -0.2 -0.4 -1.0 0.0 0.5 1.0 -1.0 -0.5 0.0 0.5 1.0 Figure 4. The four Figure 3. The three crossing limit cycles of crossing limit cycles of the discontinuous piece- the discontinuous piece- wise differential system wise differential system (16)–(17) (13)–(14) 0.5 0.0 -0.5 -1.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 Figure 5. The five crossing limit cycles of the discontinuous piece- wise differential systems (20)–(19) Proof of statement (b) of Theorem 2. For n = 2 and in the half plane R2 = {(x, y) : x < 0} we consider the Hamiltonian isochronous global center (5) with its first 5 integral H2 (x, y), and in the half plane R1 = {(x, y) : x > 0} we consider the planar linear differential center (2) with its first integral H1 (x, y). Our goal is to prove that the discontinuous piecewise differential systems formed by system (5) and system (2) have at most one crossing limit cycle intersecting the line of discontinuity x = 0 in two different points (0, y) and (0, Y ). We know that these two points must satisfy the system of equations H1 (0, y) − H1 (0, Y ) = (y − Y )P1 (y, Y ) = 0, (6) 1 H2 (0, y) − H2 (0, Y ) = (y − Y )Q1 (y, Y ) = 0, 2 where Q1 (y, Y ) and P1 (y, Y ) are polynomials of degrees one and three, respectively, and their expressions are P1 (y, Y ) = −2bd + a2 y + ω 2 y + a2 Y + ω 2 Y,  Q1 (y, Y ) = b21 (y  + Y ) + 2b1 c1 + δ γ12 + β12 y 2 + yY + Y 2 + 2β1 γ1 (y  +Y ) + β1 (2γ1 + β1(y + Y )) 2c1 δ + δ 2 2γ12 + β12 y 2 + Y 2 +2β1 γ1 (y + Y ) + 1 . From P1 (y, Y ) = 0 we know that y = f (Y ). Substituting the expression of y in Q1 (y, Y ) = 0, we obtain a quadratic equation in the variable Y , where this equation has at most two real solutions Y1 and Y2 . Therefore system (6) has at most two real solutions. We can easily show that the two solutions verify the symmetry (Y1 , f (Y1 )) = (f (Y2 ), Y2 ), which means that both solutions provide the same limit cycle for the discontinuous piecewise differential system (2)–(5). So statement (b) is proved if we provide an exemple of a discontinuous piecewise differential system (2)–(5) having one limit cycle for n = 2. In the half plane R1 , we consider the linear differential center (7) ẋ = 4 + x − 2.1473579210218228..y, ẏ = 2.5 + 7.916705377141101..x − y, with its first integral H1 (x, y) = (7.91671..x − y)2 + 15.8334..(2.5x − 4y) + 16y 2 . In the half plane R2 , we consider the Hamiltonian isochronous global center Ẋ = 0.0344828..(9((3x + 0.4y − 5.5)2 + 5x − 9y − 6) − (0.4 2(3x  1 +0.4y − 5.5)2 + (3x + 0.4y − 5.5)2 , (8) (3x + 0.4y− 5.5) 2 Ẏ = 0.0344828.. 9 (3x + 0.4y − 5.5) + 5x − 9y − 6 + 3 2(3x 1 +0.4y − 5.5)2 + (3x + 0.4y − 5.5)2 )) , (3x + 0.4y − 5.5) with its first integral 1 H2 (x, y) = ((3x + 0.4y − 5.5)2 + ((3x + 0.4y − 5.5)2 + 5x − 9y − 6)2 ). 2 The unique real solution of system (6) is (1.6508469135171708.., 2.074661499255111..). So, the discontinuous piecewise differential system (7)–(8) has one limit cycle as in Fig. 1. □ Proof of statement (c) of Theorem 2. For n = 3 in the half plane R1 we consider the Hamiltonian isochronous global center (5) with its first integral H2 (x, y), and in 6 M. BARKAT, R. BENTERKI AND J. LLIBRE the half plane R2 we consider the linear differential center (2) with its first integral H1 (x, y). In order that the discontinuous piecewise differential system (2)–(5) has crossing limit cycle, it must intersect the line of discontinuit x = 0 in two different points (0, y) and (0, Y ), and these two points must satisfy the system of equations H1 (0, y) − H1 (0, Y ) = (y − Y )P1 (y, Y ) = 0, (9) 1 H2 (0, y) − H2 (0, Y ) = (y − Y )Q2 (y, Y ) = 0. 2 where Q2 (y, Y ) is a polynomial of degree five given by  Q2 (y, Y ) = b21 (y + Y ) +  2b1 c1 + δ γ13 +β13 (y + Y ) y 2 + Y 2 + 3β12 γ1 y 2  +yY + Y 2 + 3β 2 2 2 2  1 γ1 (y + Y2) + β1 2c1 δ 3γ1 + 2β1 y 2 + 2yY + Y 2 +3β1γ1 (y + Y ) + 2γ1 + δ (2γ1 + β1 (y + Y )) γ1 + β1 y − yY  +Y 2 + β1 γ1 (y + Y ) 3γ12 + β12 y 2 + yY + Y 2 + 3β1 γ1 (y + Y ) +β1 (y + Y ) . From P1 (y, Y ) = 0 we obtain y = f (Y ). By Substituting the value of Y in Q2 (y, Y ) = 0, we obtain a quartic equation in the variable y. This equation has at most four real solutions, symmetric in the sense stated in the proof of statment (b). Therefore system (9) has at most two real solutions (y, Y ). Hence the discontinuous piecewise differential system (2) + (5) with n = 3 has at most two limit cycles. Now we prove that our result is reached for n = 3, by giving an example with ex- actly two limit cycles. In the half plane R1 we consider the Hamiltonian isochronous global center  Ẋ = 14.6449.. 2 − (0.5x − 0.232929..y + 1.68341..)3 + 4x − 2y + 2 1 + 0.232929.. 3(0.5x − 0.232929..y (0.5x − 0.232929..y + 1.68341..) +1.68341..) − 3(4x − 2y + 2)(0.5x− 0.232929..y + 1.68341..)3 6 +(0.5x − 0.232929..y + 1.68341..)2 ,  (10) Ẏ = 14.6449.. 4 − (0.5x − 0.232929..y + 1.68341..)3 + 4x − 2y + 2 1 + 0.5 3(0.5x − 0.232929..y (0.5x − 0.232929..y + 1.68341..) +1.68341..)6 − 3(4x − 2y + 2)(0.5x− 0.232929..y + 1.68341..)3 +(0.5x − 0.232929..y + 1.68341..)2 , with the first integral 1 H2 (x, y) = 2 (0.5x − 0.232929..y + 1.68341..)2 + − (0.5x − 0.232929..y 2  +1.68341..)3 + 4x − 2y + 2 . In the half plane R2 , we consider the linear differential center 37y (11) ẋ = 2 − x − , ẏ = −1 + 4x + y, 4 with the first integral H1 (x, y) = 8(−x − 2y) + 36y 2 + (4x + y)2 . 2 √ 2 √   16  The two real solutions of system (9) are (4−3 10), 4−3 10) and 0, . 37 37 37 Then the two crossing limit cycles of these systems are shown in Fig.2. □ 7 Proof of statement (d) of Theorem 2. For n = 4 and in the half plane R1 we con- sider the Hamiltonian isochronous global center (5) with its first integral H2 (x, y), and in the half plane R2 we consider the linear differential center (2) with its first integral H1 (x, y). We have to prove that the piecewise differential systems formed by system (5) and system (2) have at most three crossing limit cycles, each one of them intersects the line of discontinuity x = 0 in two different points (0, y) and (0, Y ). We know that these two points must satisfy the system of equations H1 (0, y) − H1 (0, Y ) = (y − Y )P1 (y, Y ) = 0, (12) H2 (0, y) − H2 (0, Y ) = 12 (y − Y )Q3 (y, Y ) = 0. where Q3 (y, Y ) is a polynomial of degree seven given by 1  Q3 (y, Y ) = 2 b21 (y + Y) + 2b1 c1 + δ γ14 + 4β13 γ1 (y + Y ) y 2+ Y 2 + 6β12 γ12 y2 +yY + Y 2 + β14 y 4 + y 3 Y + y 2 Y 2 + yY 3 + Y 4 + 4β1 γ13 (y +  Y) +β1 (2γ1 + β1 (y + Y )) δ 2γ12 + β12 y 2 + Y 2 + 2β1 γ1 (y + Y ) 2c1 +δ 2γ14 + β14 y 4+ Y 4+ 4β13 γ1 y 3 + Y 3 + 6β12 γ12 y 2 + Y 2 +4β1 γ13 (y + Y ) + 1 . From P1 (y, Y ) = 0 we obtain y = f (Y ). By Substituting the value of Y in Q3 (y, Y ) = 0, we obtain an equation of degree seven in the variable y, which has at most seven real solutions (y, Y ), which are again symmetric. So these solutions provide at most three limit cycles of the discontinous piecewise differential systems (2)-(5) with n = 4. Now we give an example of a discontinuous piecewise differential system (2)– (5) with n = 4 having three limit cycles. In the half plane R1 , we consider the Hamiltonian isochronous global center Ẋ = 0.0526154.. 0.2 − 0.00991602..(0.5x − 4.72646..y − 1)4 − 4x −0.2y − 1.3326 1 + 4.72646.. 0.00039331..(0.5x − 4.72646..y (0.5x − 4.72646..y − 1) 8 −1) − 0.0396641..(−4x − 0.2y − 1.3326..)(0.5x − 4.72646..y − 1)4 2 (13) +(0.5x − 4.72646..y − 1) , Ẏ = 0.0526154.. 4 − 0.00991602..(0.5x − 4.72646..y − 1)4 − 4x − 0.2y  1 −1.3326.. − 0.5 0.00039331..(0.5x− (0.5x − 4.72646..y − 1) 4.72646..y − 1)8 − 0.0396641..(−4x − 0.2y −  1.3326..)(0.5x −4.7264..6y − 1)4 + (0.5x − 4.72646..y − 1)2 , with the first integral 1 2 H2 (x, y) = 2 − 0.00991602..(0.5x −4.72646..y − 1)4 − 4x − 0.2y − 1.3326.. +(0.5x − 4.72646..y − 1)2 . In the half plane R2 , we consider the linear differential center (14) ẋ = −2 − 3x − 9y, ẏ = −1 + 5x + 3y, with the first integral H1 (x, y) = 36y 2 + 10(−x + 2y) + (5x + 3y)2 . 8 M. BARKAT, R. BENTERKI AND J. LLIBRE 1 √ 1 √ The three real solutions of system (12) are ( (−10 − 1045), (−10 − 1045)), 45 45 2 √ 2 √ 1 √ 1 √ ( (−5−4 10), (−5+4 10)) and ( (−10− 235), (−10+ 235)). Then 45 45 45 45 the three crossing limit cycles of the discontinuous piecewise differential system (13)–(14) are shown in Fig. 3. □ Proof of statement (e) of Theorem 2. For n = 5 and in the half plane R2 , we con- sider the Hamiltonian isochronous global center (5) with its first integral H2 (x, y). In the half plane R1 we consider the linear differential center (2) with its first integral H1 (x, y). If the discontinuous piecewise differential system (2)–(5) has a crossing limit cycle intersecting the line of discontinuity x = 0 in the points (0, y) and (0, Y ), these points must satisfy the system of equations H1 (0, y) − H1 (0, Y ) = (y − Y )P1 (y, Y ), (15) 1 H2 (0, y) − H2 (0, Y ) = (y − Y )Q4 (y, Y ) = 0. 2 Where Q4 (y, Y ) is a polynomial of degree nine given by 1  Q4 (y, Y ) = 2 b21 y + b21 Y + 2δ b1 γ15 + 10β13 γ12 b1 (y + Y ) y 2 + Y 2 + c1 y 2 +yY + Y 2 + 10β12γ13 b1 y 2 + yY + Y 2 + c1 (y + Y ) + β15 b1 (y +Y ) y 2 + yY + Y 2 y 2 − yY + Y 2 + c1 y 4 + y 3Y + y 2 Y 2 + yY 3 +Y  + 5β14 γ1 b1 y 4 + y 3 Y + y2 Y 2 + yY 3 + Y 4 + c1 (y + Y ) y 2 4 +Y 2 + 5β1 γ14 (b1 (y + Y ) + c1 ) + 2b1 c1 + 2β1 γ1 + β1 δ 2 (2γ1 +β1 (y + Y )) γ14 + β13 γ1 (y + Y ) 3y 2 − 4yY + 3Y 2 + 2β12 γ12 2y 2  −yY + 2Y 2 + β14 y 4 − y 3 Y + y 2 Y 2 − yY 3 + Y 4 + 2β1γ13 (y + Y ) 5γ14 + 5β13 γ1 (y + Y ) y 2 +Y 2 + 10β12 γ12 y 2 + 2  yY 2+ Y 2+ β 4 4 1 y 3 2 2 3 4 3 +y Y + y Y + yY + Y + 10β1 γ1 (y + Y ) + β1 y + β1 Y . By Bezout Theorem (see for instance [20]), system (15) has at most nine real solutions (y, Y ) with y different from Y, by symmetry the maximum number of crossing limit cycles of these systems is at most four. We provide an example with exactly four limit cycles of the discontinuous piece- wise differential system (2)–(5) for n = 5. So in the half plane R1 we consider the linear differential center 29y (16) ẋ = 3 + 2x − , ẏ = 1 + 3x − 2y, 3 with the first integral H1 (x, y) = 6(x − 3y) + (3x − 2y)2 + 25y 2 . In the half plane R2 we consider the Hamiltonian isochronous global center Ẋ = 1.93362.. 1.03433..  − 0.977806..(0.5x − 1.)5 + 3x − 1.03433..y −0.656806.. , Ẏ = 1.93362.. 3 − 0.977806(0.5x − 1.)5 + 3x − 1.03433..y (17)  1 −0.656806.. + 0.5 − 4.88903..(0.5x − 1.)5 (3x (0.5x − 1) −1.03433..y − 0.656806..) + 4.78052..(0.5x − 1.)10  2 +(0.5x − 1) , 9 with its first integral 1 H2 (x, y) = ((−0.977806..(0.5x−1.)5 +3x−1.03433..y −0.656806..)2 +(0.5x−1)2 ). 2 The discontinuous piecewise differential system (16)–(17) has exactly four crossing 1 limit cycles intersecting the straight line x = 0 in the following points: (9 −  1  1 29 √ 1 √ √ 1 √ √ 1 313), (9 + 313) , (9 − 226), (9 + 226) , (9 − 134), (9 + √ 29 1 √ 29 1 29 √  29 29 134) , and (9 − 2 13), (9 + 2 13) . These limit cycles are shown in Fig. 29 29 4. □ We shall need to use the following technical Proposition in the proof of statement (f) of Theorem 2. Proposition 3. Let f0 , ..., fn be analytic functions defined on an open interval I ⊂ R. If f0 , ..., fn are linearly independent, then thereP exists s1 , ..., sn ∈ I and n λ0 , . . . , λn ∈ R such that for every j ∈ {1, ..., n} we have i=0 λi fi (sj ) = 0. For a proof of this result see for instance Proposition 1 of [14]. Proof of statement (f ) of Theorem 2. For n = 6 we consider the linear differential center (2) with its first integral H1 (x, y) in the half plane R2 , and the Hamiltonian isochronous global center (5) with its first integral H2 (x, y) in the half plane R1 . In order that the discontinuous piecewise differential system (2)–(5) has a crossing limit cycle, it must intersect the discontinuous line x = 0 in two different points (0, y) and (0, Y ). These points satisfy the equations H1 (0, y) − H1 (0, Y ) = (y − Y )P1 (y, Y ), (18) 1 H2 (0, y) − H2 (0, Y ) = (y − Y )Q5 (y, Y ) = 0. 2 Where Q5 (y, Y ) is a polynomial of degree eleven. From P1 (x, y) = 0 we have y = f (Y ). Substituting the expression of y in Q5 (y, Y )/(y − Y ) = 0, we obtain the polynomial K(Y ) = A0 + A1 Y + A2 Y 2 + A3 Y 3 + A4 Y 4 + A5 Y 5 + A6 Y 6 + A7 Y 7 + A8 Y 8 +A9 Y 9 + A10 Y 10 , of degree 10, where the Ai ’s for i = 0, . . . , 10, are given in the appendix. Since the rank of the Jacobian matrix of the function A = (A0 , . . . , A10 ) with re- spect to its twelve parameters: a, b, c, d, α1 , β1 , γ1 , c1 , a1 , b1 , δ, ω, is six of the eleven functions Ai , with i = 0, . . . , 10, which are linearly independent. according Propo- sition 3 it follows that the polynomial K(Y ) can have at most five real solutions. We provide an example with exactly five limit cycles of the discontinuous piece- wise differential system (2)-(5) for n = 6. So in the half plane R2 we consider the linear differential center 17y (19) ẋ = −x − − 1, ẏ = 4x + y − 2, 4 with the first integral H1 (x, y) = (4x + y)2 + 8(y − 2x) + 16y 2 . 10 M. BARKAT, R. BENTERKI AND J. LLIBRE In the half plane R1 , we consider the Hamiltonian isochronous global center 0.15486.. Ẋ = − (1.61436..((0.379849.. + 0.02x (0.379849.. + 0.02x + 1.61436..y) +1.61436..y)2 + 2.96641..(−0.399469.. + 4x)(0.379849.. + 0.02x +1.61436..y)6 + 1.46659..(0.379849.. + 0.02x + 1.61436..y)1 2)) , (20) Ẏ = 0.15486.. (4(−0.399469.. + 4x + 0.494401.. (0.379849.. + 0.02x + 1.61436..y) 6 (0.379849.. + 0.02x + 1.61436..y) ) + (0.02((0.379849.. + 0.02x +1.61436..y)2 + 2.96641..(−0.399469.. + 4x)(0.379849.. + 0.02x +1.61436..y)6 + 1.46659..(0.379849.. + 0.02x + 1.61436..y)1 2)) , with its first integral 1 H2 (x, y) = 2 ((0.379849..+ 0.02x + 1.61436..y)2 + (−0.399469.. + 4x + 0.494401.. (0.379849.. + 0.02x + 1.61436..y)6 )2 ). The discontinuous piecewise differential system (19)–(20) has exactly five cross- 1 ing limit cycles intersecting the straight line x = 0 in the following points: (−4−     17 √ 1 √ 1 √ 1 √ 1 √ 1 237), (−4+ 237)) , (−4− 186), (−4+ 186) , (−4−3 15), (−4+ 17 17 17 17 17 √  2 √ 2 √  1 √ 1 √  3 15) , (−2 − 21), (−2 + 21) and (−4 − 33), (−4 + 33) . 17 17 17 17 These limit cycles are shown in Fig. 5. This completes the proof of statement (f) of theorem 2 for n = 6. In the general case for n > 6 we consider the linear differential center (2) with its first integral H1 (x, y) in the half plane R2 , and the Hamiltonian isochronous global center (5) with n > 6 with its first integral H2 (x, y), in the half plane R1 . If there exists a limit cycle of the discontinuous piecewise differential systems (2) –(5) it must intersect the discontinuity line x = 0 in two different points (0, y) and (0, Y ). These two points must satisfy the system H1 (0, y) − H1 (0, Y ) = (y − Y )P1 (y, Y ) = 0, (21) H2 (0, y) − H2 (0, Y ) = (y − Y )Qn (y, Y ) = 0. where Qn (y, Y ) is a polynomial of degree less than or equal 2n−1. From P1 (y, Y ) = 0 we get the expression of y. By replacing y = f (Y ) in the equation Qn (y, Y ) = 0, we obtain again a polynomial D(Y ) of degree at most 2n − 1 in the variable Y. Assume that the degree of D(Y ) is 2n − 1, it the degree is smaller we can use the same arguments for proving that the discontinupus piecewise differential system has at most five limit cycles. So we write D(Y ) = C0 + C1 Y + C2 Y 2 + · · · + C2n−1 Y 2n−1 . Let M(2n)×12 be the Jacobian matrix of the function C = (C0 , . . . , C2n−1 ) with respect to its twelve parameters. We know that the rank of the matrix of order (2n) × 12 for n > 6 is at most 12. Hence the maximum number of real roots of the polynomial D(Y ) is at most eleven, and due to the symmetry of its real roots we know that the maximum number of real solutions of system (21) is at most 5. Then the maximum number of crossing limit cycles of the discontinuous piecewise differential system (2)–(5) for n > 6 is at most 5. □ 11 4. The Appendix Here we provide the values Ai , with i = 0, . . . , 10. 1 A0 = 11 (b1 c1 a22 + β1 γ1 a22 + 11b1 c1 ω 2 a20 + bdβ12 a20 + bb21 da20 + 11ω 2 (a2 + ω2 β1 a20 + 55b1 c1 ω 4 a18 + 10bb21 dω 2 a18 + 10bdω 2 β12 a18 + 55ω 4 β1 γ1 a18 + 165b1 c1 ω 6 a16 + 45bb21 dω 4 a16 + 45bdω 4 β12 a16 + 165ω 6 β1 γ1 a16 + 330b1 c1 ω 8 a14 +120bb21 dω 6 a14 + 120bdω 6 β12 a14 + 330ω 8 β1 γ1 a14 + 462b1 c1 ω 10 a12 + 210b b21 dω 8 a12 + 210bdω 8 β12 a12 + 462ω 10 β1 γ1 a12 + 462b1 c1 ω 12 a10 + 252bb21 dω 10 a10 + 252bdω 10 β12 a10 + 462ω 12 β1 γ1 a10 + 330b1 c1 ω 14 a8 + 210bb21 dω 12 a8 +210bdω 12 β12 a8 + 330ω 14 β1 γ1 a8 + 165b1 c1 ω 16 a6 + 120bb21 dω 14 a6 + 120bd ω 14 + 165ω 16 β1 γ1 a6 + 55b1 c1 ω 18 a4 + 45bb21 dω 16 a4 + 45bdω 16 β12 a4 + 55ω 18 β1 γ1 a4 + 11b1 c1 ω 20 a2 + 10bb21 dω 18 a2 + 10bdω 18 β12 a2 + 11ω 20 β1 γ1 a2 + b1 c1 ω 22 + bb21 dω 20 + bdω 20 β12 + 2β1 (512b11 d11 β111 + 3072b10 d10 (a2 + ω 2 )γ1 β110 +8448b9 d9 (a2 + ω 2 )2 γ12 β19 + 14080b8 d8 (a2 + ω 2 )3 γ13 β18 + 15840b7 d7 (a2 +ω 2 )4 γ14 β17 + 12672b6 d6 (a2 + ω 2 )5 γ15 β16 + 7392b5 d5 (a2 + ω 2 )6 γ16 β15 + 3168b4 d4 (a2 + ω 2 )7 γ17 β14 + 990b3 d3 (a2 + ω 2 )8 γ18 β13 + 220b2 d2 (a2 + ω 2 )9 γ19 β12 +33bd(a2 + ω 2 )10 γ110 β1 + 3(a2 + ω 2 )11 γ111 )δ 2 + ω 22 β1 γ1 + (a2 + ω 2 )5 (32b5 d5 (2bb1 d + c1 (a2 + ω 2 ))β16 + 96b4 d4 (a2 + ω 2 )(2bb1 d + c1 (a2 + ω 2 ))γ1 β15 +120b3 d3 (a2 + ω 2 )2 (2bb1 d + c1 (a2 + ω 2 ))γ12 β14 + 80b2 d2 (a2 + ω 2 )3 (2bb1 d +c1 (a2 + ω 2 ))γ13 β13 + 30bd(a2 + ω 2 )4 (2bb1 d + c1 (a2 + ω 2 ))γ14 β12 + 6(a2 + ω 2 )5 (2bb1 d + c1 (a2 + ω 2 ))γ15 β1 + b1 (a2 + ω 2 )6 γ16 )δ), 1 A1 = − 10 2bdβ12 δ(5γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )a18 + 5γ12 (9γ1 (22β1 δγ16 (a2 + ω2 +3b1 γ1 + 4c1 β1 )ω 2 + 2bdβ1 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 ))a16 + 4γ1 (45γ12 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 4 + 20bdβ1 γ1 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 2 + 9b2 d2 β12 (5b1 γ1 + 2β1 (66δγ16 + c1 )))a14 + 4(105γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 6 +70bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 4 + 63b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 +c1 ))ω 2 + 8b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 ))a12 + 2(315γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 8 + 280bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 6 + 378b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 4 + 96b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 2 + 40b4 d4 β14 (396β1 δγ15 + b1 ))a10 + 10(63γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 10 + 70bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 8 + 126b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 6 +48b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 4 + 40b4 d4 β14 (396β1 δγ15 + b1 )ω 2 +4752b5 d5 β16 γ14 δ)a8 + 20(21γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 12 + 28bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 10 + 63b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 8 +32b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 6 + 40b4 d4 β14 (396β1 δγ15 + b1 )ω 4 +9504b5 d5 β16 γ14 δω 2 + 2464b6 d6 β17 γ13 δ)a6 + 4(45γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 ) ω 14 + 70bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 12 + 189b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 10 + 120b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 8 + 200b4 d4 β14 (396 β1 δγ15 + b1 )ω 6 + 71280b5 d5 β16 γ14 δω 4 + 36960b6 d6 β17 γ13 δω 2 + 8448b7 d7 β18 γ12 δ) a4 + (45γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 16 + 80bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 12 M. BARKAT, R. BENTERKI AND J. LLIBRE β1 )ω 14 + 252b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 12 + 192b3 d3 β13 (462β1 δγ16 +6b1 γ1 + c1 β1 )ω 10 + 400b4 d4 β14 (396β1 δγ15 + b1 )ω 8 + 190080b5 d5 β16 γ14 δω 6 +147840b6 d6 β17 γ13 δω 4 + 67584b7 d7 β18 γ12 δω 2 + 13824b8 d8 β19 γ1 δ)a2 + 2560b9 d9 β110 δ + 47520b5 d5 ω 8 β16 γ14 δ + 49280b6 d6 ω 6 β17 γ13 δ + 33792b7 d7 ω 4 β18 γ12 δ +13824b8 d8 ω 2 β19 γ1 δ + 80b4 d4 ω 10 β14 (396β1 δγ15 + b1 ) + 5ω 18 γ13 (22β1 δγ16 +3b1 γ1 + 4c1 β1 ) + 10bdω 16 β1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 ) + 32b3 d3 ω 12 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 ) + 36b2 d2 ω 14 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))), 1 A2 = β 2 δ(5γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )a18 + 5γ12 (9γ1 (22β1 δγ16 + 3b1 γ1 (a2 + ω 2 )9 1 +4c1 β1 )ω 2 + 2bdβ1 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 ))a16 + 4γ1 (45γ12 (22β1 δγ16 + 3b1 γ1 +4c1 β1 )ω 4 + 20bdβ1 γ1 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 2 + 12b2 d2 β12 (5b1 γ1 + 2β1 (66δγ16 + c1 )))a14 + 28(15γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 6 + 10bdβ1 γ12 (99β1 δγ16 +8b1 γ1 + 6c1 β1 )ω 4 + 12b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 2 + 2b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 ))a12 + 2(315γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 8 + 280bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 6 + 504b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 4 +168b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 2 + 88b4 d4 β14 (396β1 δγ15 + b1 ))a10 +10(63γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 10 + 70bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 8 +168b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 6 + 84b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 4 + 88b4 d4 β14 (396β1 δγ15 + b1 )ω 2 + 12672b5 d5 β16 γ14 δ)a8 + 20(21γ13 (22β1 δγ16 +3b1 γ1 + 4c1 β1 )ω 12 + 28bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 10 + 84b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 8 + 56b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 6 + 88b4 d4 β14 (396β1 δγ15 + b1 )ω 4 + 25344b5 d5 β16 γ14 δω 2 + 7744b6 d6 β17 γ13 δ)a6 + 4(45γ13 (22β1 δγ16 +3b1 γ1 + 4c1 β1 )ω 14 + 70bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 + 6c1 β1 )ω 12 + 252b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 10 + 210b3 d3 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 )ω 8 + 440b4 d4 β14 (396β1 δγ15 + b1 )ω 6 + 190080b5 d5 β16 γ14 δω 4 + 116160b6 d6 β17 γ13 δω 2 + 30624b7 d7 β18 γ12 δ)a4 + (45γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 )ω 16 + 80bdβ1 γ12 (99β1 δγ16 + 8b1 γ1 +6c1 β1 )ω 14 + 336b2 d2 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 + c1 ))ω 12 + 336b3 d3 β13 (462β1 δγ16 +6b1 γ1 + c1 β1 )ω 10 + 880b4 d4 β14 (396β1 δγ15 + b1 )ω 8 + 506880b5 d5 β16 γ14 δω 6 + 464640b6 d6 β17 γ13 δω 4 + 244992b7 d7 β18 γ12 δω 2 + 56832b8 d8 β19 γ1 δ)a2 + 11776b9 d9 β110 δ + 126720b5 d5 ω 8 β16 γ14 δ + 154880b6 d6 ω 6 β17 γ13 δ + 122496b7 d7 ω 4 β18 γ12 δ + 56832b8 d8 ω 2 β19 γ1 δ + 176b4 d4 ω 10 β14 (396β1 δγ15 + b1 ) + 5ω 18 γ13 (22β1 δγ16 + 3b1 γ1 + 4c1 β1 ) + 10bdω 16 β1 γ12 (99β1 δγ16 + 8b1 γ1 +6c1 β1 ) + 56b3 d3 ω 12 β13 (462β1 δγ16 + 6b1 γ1 + c1 β1 ) + 48b2 d2 ω 14 β12 γ1 (5b1 γ1 + 2β1 (66δγ16 +c1 ))), 13 1 A3 = − 8 (4bβ14 dδ(15a14 b1 γ12 + 6a14 β1 c1 γ1 + 36a12 bb1 β1 γ1 d + 6a12 bβ12 c1 d a2 + ω2 +105a12 b1 γ12 ω 2 + 42a12 β1 c1 γ1 ω 2 + 26a10 b2 b1 β12 d2 + 216a10 bb1 β1 γ1 dω 2 + 36a10 b β12 c1 dω 2 + 315a10 b1 γ12 ω 4 + 126a10 β1 c1 γ1 ω 4 + 130a8 b2 b1 β12 d2 ω 2 + 540a8 bb1 β1 γ1 dω 4 +90a8 bβ12 c1 dω 4 + 525a8 b1 γ12 ω 6 + 210a8 β1 c1 γ1 ω 6 + 260a6 b2 b1 β12 d2 ω 4 + 720a6 bb1 β1 γ1 dω 6 + 120a6 bβ12 c1 dω 6 + 525a6 b1 γ12 ω 8 + 210a6 β1 c1 γ1 ω 8 + 260a4 b2 b1 β12 d2 ω 6 +540a4 bb1 β1 γ1 dω 8 + 90a4 bβ12 c1 dω 8 + 315a4 b1 γ12 ω 10 + 126a4 β1 c1 γ1 ω 10 + 130a2 b2 b1 β12 d2 ω 8 + 4β1 δ(4368b6 β16 γ1 d6 (a2 + ω 2 ) + 8184b5 β15 γ12 d5 (a2 + ω 2 )2 + 8800b4 β14 γ13 d4 (a2 + ω 2 )3 + 5940b3 β13 γ14 d3 (a2 + ω 2 )4 + 2574b2 β12 γ15 d2 (a2 + ω 2 )5 + 693bβ1 γ16 d(a2 +ω 2 )6 + 99γ17 (a2 + ω 2 )7 + 2574 + 1024b7 β17 d7 ) + 216a2 bb1 β1 γ1 dω 10 + 36a2 bβ12 c1 d ω 10 + 105a2 b1 γ12 ω 12 + 42a2 β1 c1 γ1 ω 12 + 26b2 b1 β12 d2 ω 10 + 36bb1 β1 γ1 dω 12 + 6bβ12 c1 dω 12 + 15b1 γ12 ω 14 + 6β1 c1 γ1 ω 14 )), 1 A4 = − (β 4 δ(15a14 b1 γ12 + 6a14 β1 c1 γ1 + 36a12 bb1 β1 γ1 d + 6a12 bβ12 c1 d + 105a12 b1 (a2 + ω 2 )7 1 γ12 ω 2 + 42a12 β1 c1 γ1 ω 2 + 36a10 b2 b1 β12 d2 + 216a10 bb1 β1 γ1 dω 2 + 36a10 bβ12 c1 dω 2 +315a10 b1 γ12 ω 4 + 126a10 β1 c1 γ1 ω 4 + 180a8 b2 b1 β12 d2 ω 2 + 540a8 bb1 β1 γ1 dω 4 + 90a8 bβ12 c1 dω 4 + 525a8 b1 γ12 ω 6 + 210a8 β1 c1 γ1 ω 6 + 360a6 b2 b1 β12 d2 ω 4 + 720a6 bb1 β1 γ1 d ω 6 + 120a6 bβ12 c1 dω 6 + 525a6 b1 γ12 ω 8 + 210a6 β1 c1 γ1 ω 8 + 360a4 b2 b1 β12 d2 ω 6 + 540a4 bb1 β1 γ1 dω 8 + 90a4 bβ12 c1 dω 8 + 315a4 b1 γ12 ω 10 + 126a4 β1 c1 γ1 ω 10 + 180a2 b2 b1 β12 d2 ω 8 +4β1 δ(14208b6 β16 γ1 d6 (a2 + ω 2 ) + 22704b5 β15 γ12 d5 (a2 + ω 2 )2 + 20240b4 β14 γ13 d4 (a2 + ω 2 )3 + 10890b3 β13 γ14 d3 (a2 + ω 2 )4 + 3564b2 β12 γ15 d2 (a2 + ω 2 )5 + 693bβ1 γ16 d(a2 + ω 2 )6 +99γ17 (a2 + ω 2 )7 + 3824b7 β17 d7 ) + 216a2 bb1 β1 γ1 dω 10 + 36a2 bβ12 c1 dω 10 + 105a2 b1 γ12 ω 12 + 42a2 β1 c1 γ1 ω 12 + 36b2 b1 β12 d2 ω 10 + 36bb1 β1 γ1 dω 12 + 6bβ12 c1 dω 12 + 15b1 γ12 ω 14 +6β1 c1 γ1 ω 14 )), 1 A5 = − (2bβ16 dδ(−(a2 + ω 2 )((a2 + ω 2 )(44β1 γ12 δ(340b2 β12 γ1 d2 (a2 + ω 2 ) + 135bβ1 (a + ω 2 )6 2 γ12 d(a2 + ω 2 )2 + 27γ13 (a2 + ω 2 )3 + 480b3 β13 d3 ) + 3b1 (a2 + ω 2 )3 ) + 15936b4 β15 γ1 d4 δ) −5024b5 β16 d5 δ)), 1 A6 = − (β 6 δ((a2 + ω 2 )((a2 + ω 2 )(44β1 γ12 δ(160b2 β12 γ1 d2 (a2 + ω 2 + 45bβ1 γ12 d(a2 (a2 + ω 2 )5 1 +ω 2 )2 + 9γ13 (a2 + ω 2 )3 + 300b3 β13 d3 ) + b1 (a2 + ω 2 )3 ) + 12480b4 β15 γ1 d4 δ) +4736b5 β16 d5 δ)), 1 A7 = − (−80bβ19 dδ 2 (42b2 β12 γ1 d2 (a2 + ω 2 ) + 33bβ1 γ12 d(a2 + ω 2 )2 + 11γ13 (a2 (a2 + ω 2 )4 +ω 2 )3 + 20b3 β13 d3 )), 1 A8 = − (β 9 δ 2 (60b2 β12 γ1 d2 (a2 + ω 2 ) + 33bβ1 γ12 d(a2 + ω 2 )2 + 11γ13 (a2 + ω 2 )3 (a2 + ω 2 )3 1 +38b3 β13 d3 ), )), 1 A9 = − (−60bβ111 dδ 2 (γ1 (a2 + ω 2 ) + bβ1 d), (a2 + ω 2 )2 14 M. BARKAT, R. BENTERKI AND J. LLIBRE 1 A10 = − (−60bβ111 dδ 2 (γ1 (a2 + ω 2 ) + bβ1 d))). (a2 + ω 2 )2 Acknowledgements The third author is partially supported by the Agencia Estatal de Investigación grant PID2019-104658GB-I00, and the H2020 European Research Council grant MSCA-RISE-2017-777911. References [1] A. Andronov, A. Vitt and S. Khaikin, Theory of Oscillations, Pergamon Press, Oxford, 1966. [2] J.C. Artés, J. Llibre, J.C. Medrado and M.A. Teixeira, Piecewise linear differential systems with two real saddles, Math. Comput. Simul. 95, 13–22, (2013). [3] M. di Bernardo, C. J. Budd, A. R. Champneys and P. Kowalczyk, Piecewise-Smooth Dynamical Systems: Theory and Applications, Appl. Math. Sci. Series 163, Springer-Verlag, London, 2008. [4] D.C. Braga and L.F.Mello, Limit cycles in a family of discontinuous piecewise linear differential systems with two zones in the plane, Nonlinear Dynam. 73 1283–1288 (2013). [5] E. Diz-Pita, J. Llibre, M.V. Otero-Espinar andC. 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Walker, Algebraic curves, Reprint of the 1950th edition, Springer-Verlag, New York- Heidelberg, 1978. 1 Département of Mathematics, University Mohamed El Bachir El Ibrahimi of Bordj Bou Arréridj 34000, El Anasser, Algeria Email address: 1 meriem.berkat@univ-bba.dz and r.benterki@univ-bba.dz 2 Departament de Matematiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain Email address: jllibre@mat.uab.cat