This is a preprint of: “Crossing limit cycles of planar piecewise linear Hamiltonian systems without equilibrium points”, Rebiha Benterki, Jaume Llibre, Mathematics, vol. 8(755), 2020. DOI: [10.3390/math8050755] CROSSING LIMIT CYCLES OF PLANAR PIECEWISE LINEAR HAMILTONIAN SYSTEMS WITHOUT EQUILIBRIUM POINTS REBIHA BENTERKI1 AND JAUME LLIBRE2 Abstract. In this paper we study the existence of limit cycles of planar piece- wise linear Hamiltonian systems without equilibrium points. First we prove that if these systems are separated by a parabola they can have at most two crossing limit cyles, and if they are separated by a hyperbola or an ellipse they can have at most three crossing limit cycles. Additionally we prove that these upper bounds are reached. Second we show that there is an example of two crossing limit cycles when these systems have four zones separated by three straight lines. 1. Introduction The problem of existence of limit cycles is one of the most and difficult problem in the qualitative theory of differential systems in the plane. Limit cycles appear in natural way in many appliations. We recall that a limit cycle is a periodic orbit of a differential system which is isolated in the set of all periodic orbits of the system. Recently the problem of existence and the number of limit cycles has also been studied for discontinuous piecewise linear differential systems, this study goes back to Andronov et all [1], and still have attention by researchers, mainly due to their simplicity and to their applications to a large number of phenomena, such as switches in electronic circuits, see for instance [2, 10, 11]. Lum and Chua [15] conjectured that a continuous planar piecewise linear system with two zones sepa- rated by a straight line can exhibit at most one limit cycle. Freire et al [5] proved this conjecture in 1998. For the planar discontinuous piecewise linear systems, Han and Zhang [7] conjectured that these systems can have at most two crossing limit cycles when we separate them by a straight line, but Huan and Yang [8] gave a numerical example with three limit cycles, this result was proved analytically by Llibre and Ponce [13]. In 2015 Llibre et all [12] proved that if we separate the planar discontinuous piecewise linear differential centers by a straight line we can not have any limit cycle. Recently, in [3, 9, 14] were studied planar discontinous linear differential centers separated by an algebraic curve, such that a conic, or a reducible and irreducible cubic, and it was proved that these differential systems can exhibit at most three crossing limit cycles having two intersection points with 2010 Mathematics Subject Classification. Primary 34A36, 34C25, 34A26. Key words and phrases. Piecewise smooth vector field, Hamiltonian system, crossing limit cycles, conics. 1 2 R.BENTERKI AND J. LLIBRE the conic of separation and the same result is proved if the curve of separation is a cubic. In the literature we find many papers studying pieccewise smooth vector fields with two zones, and few papers for three and four zones. In this paper we consider planar piecewise linear Hamiltonian systems without equilibrium points. Our first objective is to provide the exact maximum number of crossing limit cycles of planar discontinuous piecewise linear Hamiltonian systems without equi- librium points (or simply PHS) and separated by a conic Σ. We follow the Filippov rules for defining the flow of the piecewise differential systems on a line of discon- tinuity, see [4]. We know that any conic takes nine canonical formes, but the four following formes: x2 + 1 = 0, x2 + y 2 = 0, and x2 + y 2 + 1 = 0 do not sepatate the plane in connected regions, then we omit them. We do not study the crossing limit cycles separated by the conic x2 − 1 = 0, because in [6] it was proved that PHS with three zones which separated by two parallel straight lines have at most one crossing limit cycle. The second objective of this paper is to study the crossing limit cycles of piecewise smooth differential systems such that in each piece the differential system is linear, Hamiltonian and without equilibrium points. Then easy computations show a such differential system in each piece must have a vector field of the form Xi (x, y) = (−λi bi x + bi y + γi , −λ2i bi x + λi bi y + δi ), δi 6= λi γi and bi 6= 0, with i = 1 . . . 4. Their corresponding Hamiltonian function is Hi (x, y) = (−λ2i bi /2)x2 + λi bi xy − (bi /2)y 2 + δi x − γi y. For more details see [6]. 1.1. Crossing limit cycles for planar piecewise linear Hamiltonian systems without equilibrium points separated by a conic. In this subsection we give the upper bound of crossing limit cycles of PHS separated by a parabola, P: y−x2 = 0, by a hyperbola H: x2 − y 2 − 1 = 0 or by an ellipse E: x2 + y 2 − 1 = 0. We consider only the crossing limit cycles that intersect the conics in exactly two points, for this reason we will not study the crossing limit cycles separated by two intersecting straight lines xy = 0. Our first main result is the following. Theorem 1. The following statements hold. (a) The maximum number of crossing limit cycles of PHS intersecting the parabola P in two points is at most two, and this maximum is reached, see Figure 1. (b) The maximum number of crossing limit cycles of PHS intersecting the hy- perbola H in two points is at most three, and this maximum is reached, see Figure 2. 3 (c) The maximum number of crossing limit cycles of PHS intersecting the el- lipse E in two points is at most three, and this maximum is reached, see Figure 3. The proof of Theorem 1 is given in section 2. 1.2. Crossing limit cycles for planar piecewise linear Hamiltonian systems without equilibrium points with four zones. In this subsection we study the existence of crossing limit cycles of the planar piecewise linear Hamiltonian systems without equilibrium points with four zones    X1 (x, y), x ≤ −1,  X2 (x, y), −1 ≤ x ≤ 0, (1) X(x, y) =   X3 (x, y), 0 ≤ x ≤ 1,  X4 (x, y), x ≥ 1. satisfying the condition: C. The vector fields X1 , X2 , X3 and X4 are linear and Hamiltonian without equilibrium points. Our second results are the following. Theorem 2. Continuous planar piecewise Hamiltonian systems without equilib- rium points with four zones satisfying C, have no crossing limit cycles. Theorem 3. There are discontinuous planar piecewise Hamiltonian systems with- out equilibrium points with four zones satisfying C, exhibiting exactly two crossing limit cycles. The proofs of Theorems 2 and 3 are given in section 3. 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 -1.0 -0.5 0.0 0.5 1.0 Figure 1. Two crossing limit cycles of PHS intersecting the parabola in two points. 2. Proof of Theorem 1 Proof of statement (a) of Theorem 1. In the region R1 = {(x, y) : y − x2 ≥ 0} we consider the planar discontinuous piecewise linear Hamiltonian systems without 4 R.BENTERKI AND J. LLIBRE 4 2 0 -2 -4 -6 -4 -2 0 2 4 6 Figure 2. Three crossing limit cycles of PHS intersecting the hy- perbola in two points. 2 1 0 -1 -2 -3 -4 -2 -1 0 1 2 3 4 Figure 3. Three crossing limit cycles of PHS intersecting the el- lipse in two points. equilibrium points (2) ẋ = −λ1 b1 x + b1 y + γ1 , ẏ = −λ21 b1 x + λ1 b1 y + δ1 , with b1 6= 0 and δ1 6= λ1 γ1 . Its corresponding Hamiltonian function is (3) H1 (x, y) = −(λ21 b1 /2)x2 + λ1 b1 xy − (b1 /2)y 2 + δ1 x − γ1 y. In the region R2 = {(x, y) : y − x2 ≤ 0} we consider (4) ẋ = −λ2 b2 x + b2 y + γ2 , ẏ = −λ22 b2 x + λ2 b2 y + δ2 , with b2 6= 0 and δ2 6= λ2 γ2 . Its corresponding Hamiltonian function is (5) H2 (x, y) = −(λ22 b2 /2)x2 + λ2 b2 xy − (b2 /2)y 2 + δ2 x − γ2 y. In order to have a crossing limit cycle which intersects the parabola y − x2 = 0 in the points (xi , yi ) and (xk , yk ), these points must satisfy the following system H1 (xi , yi ) − H1 (xk , yk ) = 0, H2 (xi , yi ) − H2 (xk , yk ) = 0, (6) yi − x2i = 0, yk − x2k = 0. 5 We suppose that the two systems (2) and (4) have three crossing limit cycles. Then system (6) must have three pairs of points as solutions, namely pi = (ri , ri2 ) and qi = (si , s2i ), with i = 1, 2, 3. Due to the fact that these points satisfy system (6) and if we consider the points p1 = (r1 , r12 ) and q1 = (s1 , s21 ), from (6) we obtain that the parameters γ1 and γ2 must be 1 γ1 = (−r1 r13 − b1 r12 s1 − b1 r1 s21 − b1 s31 + 2δ1 + 2b1 r12 λ1 + 2b1 r1 s1 λ1 2(r1 + s1 ) +2b1 s21 λ1 − b1 r1 λ21 − b1 s1 λ21 ), and γ2 has the same expression that γ1 changing (b1 , λ1 , δ1 ) by (b2 , λ2 , δ2 ). If the second points p2 = (r2 , r22 ) and q2 = (s2 , s22 ) satisfy system (6), then the parameters δ1 and δ2 must be b1 δ1 = (−r13 r2 − r1 r23 + r12 r2 s1 − r23 s1 + r1 r2 s21 + r2 s31 + r13 s2 2(r1 − r2 + s1 − s2 ) −r1 r22 s2 + r12 s1 s2 − r22 s1 s2 + r1 s21 s2 + s31 s2 − r1 r2 s22 − r2 s1 s22 − r1 s32 − s1 s32 −2r12 r2 λ1 + 2r1 r22 λ1 − 2r1 r2 s1 λ1 + 2r22 s1 λ1 − 2r2 s21 λ1 − 2r12 s2 λ1 + 2r1 r2 s2 λ1 −2r1 s1 s2 λ1 + 2r2 s1 s2 λ1 − 2s21 s2 λ1 + 2r1 s22 λ1 + 2s1 s22 λ1 ), and δ2 has the same expression that δ1 changing (b1 , λ1 ) by (b2 , λ2 ). Finally, we suppose that the points p3 = (r3 , r32 ) and q3 = (s3 , s23 ) satisfy system (6), then the parameters λ1 and λ2 must be λ1 = A/B where A = r13 (r2 − r3 + s2 − s3 ) + r12 s1 (r2 − r3 + s2 − s3 ) + r23 (r3 − s1 + s3 ) + r22 s2 (r3 − s1 +s3 ) + r1 (−r23 + r33 − r3 s21 − r22 s2 + s21 s2 − s32 + r2 (s21 − s22 ) + r32 s3 − s21 s3 + r3 s23 +s33 ) + (s1 − s2 )(r33 + r32 s3 + (s1 − s3 )(s2 − s3 )(s1 + s2 + s3 ) − r3 (s21 + s1 s2 + s22 −s23 )) − r2 (r33 − s31 + s1 s22 + r32 s3 − s22 s3 + s33 + r3 (−s22 + s23 )), B = 2((s1 − s2 )(r32 + (s1 − s3 )(s2 − s3 ) − r3 (s1 + s2 − s3 )) + r12 (r2 − r3 + s2 − s3 ) +r22 (r3 − s1 + s3 ) + r1 (−r22 + r32 − r3 s1 + r2 (s1 − s2 ) + s1 s2 − s22 + r3 s3 − s1 s3 +s23 ) − r2 (r32 + r3 (−s2 + s3 ) − (s1 − s3 )(s1 − s2 + s3 )). And λ2 has the same expression that λ1 changing b1 by b2 . We replace γ1 , λ1 and δ1 in the expression of H1 (x, y), and γ2 , λ2 and δ2 in the expression of H2 (x, y) and we obtain H1 (x, y) = H2 (x, y). So the piecewise linear differential system becomes a linear differential system, which does not have limit cycles. So the maximum number of crossing limit cycles in this case is two. Example with two limit cycles. Consider the planar discontinuous piecewise linear Hamiltonian system without equilibrium points separated by the parabola P: (7) ẋ = 5.5x − 0.5y + 3, ẏ = 60.5x − 5.5y + 0.2, in the region R1 , its corresponding Hamiltonian function is H1 (x, y) = 30.25x2 − 5.5xy + 0.2x + 0.25y 2 − 3y. The second system is (8) ẋ = 0.2x − 0.1y − 0.778814, ẏ = 0.4x − 0.2y + 0.00727332, in the region R2 , its corresponding Hamiltonian function is H2 (x, y) = 0.2x2 − 0.2xy + 0.00727332x + 0.05y 2 + 0.778814y. 6 R.BENTERKI AND J. LLIBRE These PHS have the limit cycles shown in Figure 1. This completes the proof of statement (a) of Theorem 1.  Proof of statements (b) of Theorem 1. In the region R1 = {(x, y) : x2 − y 2 − 1 ≥ 0} we consider the PHS given in (2). Its corresponding Hamiltonian function is given by equation (3). In the region R2 = {(x, y) : x2 − y 2 − 1 ≤ 0} we consider the PHS given in (2). Its corresponding Hamiltonian function is given by equation (5). In order that to have a crossing limit cycle which intersects the hyperbola x2 − 2 y − 1 = 0 in the points (xi , yi ) and (xk , yk ), these points must satisfy the system H1 (xi , yi ) − H1 (xk , yk ) = 0, H2 (xi , yi ) − H2 (xk , yk ) = 0, (9) x2i − yi2 = 1, x2k − yk2 = 1. We assume that the two systems (2) and (4) have four crossing limit cycles. So sys- tem (9) must have four pairs of points pi = (cosh ri , sinh ri ) and qi = (cosh si , sinh si ) for i = 1, 2, 3, 4 as solutions. Since these points satisfy system (9), we consider the points p1 = (cosh r1 , sinh r1 ) and q1 = (cosh s1 , sinh s1 ), and from (9) we obtain that the parameters γ1 and γ2 must be 1 γ1 = (2δ1 cosh r1 − b1 λ21 cosh2 r1 + b1 λ21 cosh2 s1 − 2 cosh s1 (δ1 2(sinh r1 − sinh s1 ) +b1 λ1 sinh s1 ) + b1 (− sinh2 r1 + λ1 sinh(2r1 ) + sinh2 s1 ). If we change (b1 , λ1 , δ1 ) by (b2 , λ2 , δ2 ) in the expression of γ1 we get the expression of γ2 . We suppose that the second points p2 = (cosh r2 , sinh r2 ) and q2 = (cosh s2 , sinh s2 ) satisfy system (9), then the parameters δ1 and δ2 must be 1 r1 − s1 δ1 =   r − 2r + s   r + s − 2s  (b1 csch( )(−λ21 cosh2 r1 1 2 1 1 1 2 2 4 cosh − cosh 22 2 sinh r2 + λ1 cosh s1 sinh r2 − sinh2 r1 sinh r2 + λ1 sinh(2r1 ) sinh r2 + sinh r1 sinh2 r2 2 −λ1 sinh r1 sinh(2r2 ) + λ21 cosh2 r2 (sinh r1 − sinh s1 ) − sinh2 r2 sinh s1 + λ1 sinh(2r2 ) sinh s1 + sinh r2 sinh2 s1 + λ21 cosh2 s2 (− sinh r1 + sinh s1 ) − λ1 sinh r2 sinh(2s1 ) +λ21 cosh2 r1 sinh s2 − λ21 cosh2 s1 sinh s2 + sinh2 r1 sinh s2 − λ1 sinh(2r1 ) sinh s2 − sinh2 s1 sinh s2 + λ1 sinh(2s1 ) sinh s2 − sinh r1 sinh2 s2 + sinhs1 sinh2 s2 +λ1 (sinh r1 − sinh s1 ) sinh(2s2 ))). If we change (b1 , λ1 ) by (b2 , λ2 ) in the expression of δ1 we obtain δ2 . Now we suppose that points p3 = (cosh r3 , sinh r3 ) and q3 = (cosh s3 , sinh s3 ) (1) (2) satisfy system (9), then we obtain two values of λ1 we name them λ1 and λ1 (1) (2) and two values of λ2 we name them λ2 and λ2 . The first value of λ1 is given by 7 (1) √ (2) √ λ1 = (A − (1/2) B)/C and λ1 = (A + (1/2) B)/C, where  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 A = − sinh + sinh  r − r − 3r 2+ s − s − s   r − 3r − r 2+ s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 − sinh + sinh  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 − sinh + sinh  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 − sinh + sinh  3r − r + r 2+ s − s + s   r − r + 3r 2+ s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 + sinh − sinh 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 + sinh − sinh , 2 2   r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 B = −4 cosh − cosh 2  r − r − 3r + s − s − s   r − 3r − r + s − s − s 2 1 2 3 1 2 3 1 2 3 1 2 3 + cosh − cosh 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cosh + cosh 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cosh + cosh  3r − r + r 2+ s − s + s   r − r + 3r 2+ s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 + cosh − cosh  r − r + r +2 3s − s + s   r − r + r 2+ s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 + cosh − cosh2 )  r − r − r 2+ s − s − 3s   r − r − r +2 s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 +4(sinh − sinh  r − r − 3r +2s − s − s   r − 3r − r +2s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 + sinh − sinh 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 + sinh − sinh 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 + sinh − sinh 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 − sinh + sinh 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 − sinh + sinh2 , 2 2 and the expression of C is  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 C = cosh − cosh  r − r − 3r2 + s − s − s   r − 3r − r2 + s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 + cosh − cosh  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cosh + cosh  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cosh + cosh  3r − r + r 2+ s − s + s   r − r + 3r 2+ s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 + cosh − cosh 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 + cosh − cosh . 2 2 (1) (2) We get the expression of λ2 and λ2 by changing b1 by b2 in the expression of (1) (2) λ1 and λ1 , respectively. (i) (i) We replace γ1 , δ1 and λ1 in the expression of H1 (x, y), and γ2 , δ2 and λ2 in the expression of H2 (x, y), and we obtain H1 (x, y) = H2 (x, y), for i = 1, 2. Hence 8 R.BENTERKI AND J. LLIBRE in these cases the piecewise linear differential system becomes a linear differential system, which does not have limit cycles. So the maximum number of crossing limit cycles in this case is two. (2) (1) (1) (2) Now we consider either λ1 and λ2 , or λ1 and λ2 , by replacing the expres- (2) (1) sions of γ1 , δ1 and λ1 (resp. λ1 ) in the expression of H1 (x, y), and γ2 , δ2 and (1) (2) λ2 (resp. λ2 ) in the expression of H2 (x, y) we have H1 (x, y) 6= H2 (x, y). Then we assume that points p4 = (cosh r4 , sinh r4 ) and q4 = (cosh s4 , sinh s4 ) satisfy system (9), then we obtain b1 = 0 and b2 = 0. This is a contradiction because by the assumptions they are not zero. Then we proved that the maximum number of crossing limit cycles for PHS separated by a hyperbola is at most three. Example with three limit cycles. We consider a PHS separated by the hyperbola H: 1 (10) ẋ = −0.14..x + 1.4y + , ẏ = −0.014..x + 0.14y + 1.9, 5 in the region R1 = {(x, y) : x2 − y 2 − 1 ≤ 0}. It has the Hamiltonian function y H1 (x, y) = −0.007..x2 + 0.14xy + 1.9x − 0.7y 2 − . 5 Now we consider the second PHS y (11) ẋ = 5x − − 7.14286.., ẏ = 50x − 5y − 67.8571.., 2 in the region R2 = {(x, y) : x2 − y 2 − 1 ≥ 0}. This differential system has the Hamiltonian function y2 H2 (x, y) = 25x2 − 5xy − 67.8571..x + + 7.14286..y. 4 The PHS (10)–(11) has exactly three crossing limit cycles, because the system of equations H1 (α, β) − H1 (γ, δ) = 0, H2 (α, β) − H2 (γ, δ) = 0, (12) α2 − β 2 − 1 = 0, γ 2 − δ 2 − 1 = 0, has three real solutions (α1 , β1 , γ1 , δ1 ) = (3.99376.., 3.86653.., 3.31341.., −3.1589..), (α2 , β2 , γ2 , δ2 ) = (3.43842.., 3.28979.., 2.86513.., −2.68496..) and (α3 , β3 , γ3 , δ3 ) = (2.64219.., 2.44565.., 2.2285.., −1.99154..), see Figure 2.  Proof of statement (c) of Theorem 1. We consider the PHS given in (2) in the re- gion R1 = {(x, y) : x2 + y 2 − 1 ≥ 0}, with its corresponding Hamiltonian function (3). We consider the PHS given in (2) in the region R2 = {(x, y) : x2 + y 2 − 1 ≤ 0}, with its corresponding Hamiltonian function (5). 9 In order that systems (2) and (4) have crossing limit cycles intersect the ellipse x2 + x2 − 1 = 0 in the points (xi , yi ) and (xk , yk ), they must satisfy the system H1 (xi , yi ) − H1 (xk , yk ) = 0, H2 (xi , yi ) − H2 (xk , yk ) = 0, (13) x2i + yi2 = 1, x2k + yk2 = 1. We suppose that systems (2) and (4) have four crossing limit cycles. So system (13) must have four pairs of points pi = (cos ri , sin ri ) and qi = (cos si , sin si ) for i = 1, 2, 3, 4 as solutions. So if we consider the points p1 = (cos r1 , sin r1 ) and q1 = (cos s1 , sin s1 ), from (13) we obtain that the parameters γ1 and γ2 must be 1 γ1 = (4δ1 cos r1 + b1 cos(2r1 ) − b1 λ21 cos(2r1 ) − 4δ1 cos s1 − b1 cos(2s1 ) 4(sin r1 − sin s1 ) +b1 λ21 cos(2s1 ) + 2b1 λ1 sin(2r1 ) − 2b1 λ1 sin(2s1 )). If we change (b1 , λ1 , δ1 ) by (b2 , λ2 , δ2 ) in the expression of γ1 we get the expression of γ2 . Now if the second points p2 = (cos r2 , sin r2 ) and q2 = (cos s2 , sin s2 ) satisfy system (13), then the parameters δ1 and δ2 take the values r1 cos((r1 + s1 )/2) csc((r2 − s2 )/2) csc((r1 − r2 + s1 − s2 )/2) 2 δ1 =   (λ1 cos2 r2 sin r1 4 sin r1 − sin s1 −λ21 cos2 s2 sin r1 − 2λ1 cos r2 sin r1 sin r2 + sin r1 sin2 r2 + 2λ1 cos(r1 + s1 ) sin r2 sin(r1 − s1 ) − λ21 cos2 r2 sin s1 + λ21 cos2 s2 sin s1 + 2λ1 cos r2 sin r2 sin s1 − sin2 r2 sin s1 − sin r2 sin(r1 − s1 ) sin(r1 + s1 ) + λ21 sin r2 sin(r1 − s1 ) sin(r1 + s1 ) −2λ1 cos(r1 + s1 ) sin(r1 − s1 ) sin s2 + sin(r1 − s1 ) sin(r1 + s1 ) sin s2 − λ21 sin(r1 − s1 ) sin(r1 + s1 ) sin s2 − sin r1 sin2 s2 + sin s1 sin2 s2 + λ1 sin r1 sin(2s2 ) −λ1 sin s1 sin(2s2 )). If we change (b1 , λ1 ) by (b2 , λ2 ) in the expression of δ1 we obtain δ2 . If we assume that the points p3 = (cos r3 , sin r3 ) and q3 = (cos s3 , sin s3 ) satisfy (1) (2) system (13), then we obtain two values of λ1 namely λ1 and λ1 and two values (1) (2) (1) √ (2) √ of λ2 namely λ2 and λ2 , such that λ1 = (A + B)/C and λ1 = (A − B)/C, where  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 A= − sin + sin  r − r − 3r 2+ s − s − s   r − 3r − r 2+ s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 − sin + sin  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 − sin + sin  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 − sin + sin 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 + sin − sin 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 + sin − sin , 2 2 10 R.BENTERKI AND J. LLIBRE  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 B = cos − cos  r − r − 3r2 + s − s − s   r − 3r − r2 + s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 + cos − cos 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cos + cos 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cos + cos 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 + cos − cos  r − r + r +2 3s − s + s   r − r + r 2+ s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 + cos − cos2  2   r − r − r +2 s − 3s − s  1 2 3 1 2 3 +(sin r1 − r2 − r3 + s1 − s2 − 3s3 − sin  r2 − r − 3r + s − s − s   r − 3r − r +2s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 + sin − sin  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 + sin − sin 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 + sin − sin 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 − sin + sin 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 − sin + sin2 , 2 2 and the expression of C is  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 C = cos − cos  r − r − 3r2 + s − s − s   r − 3r − r2 + s − s − s  1 2 3 1 2 3 1 2 3 1 2 3 + cos − cos 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cos + cos 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  1 2 3 1 2 3 1 2 3 1 2 3 − cos + cos 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 1 2 3 1 2 3 + cos − cos 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  1 2 3 1 2 3 1 2 3 1 2 3 + cos − cos . 2 2 (1) (2) (1) (2) The expression of λ2 and λ2 are the same than the expressions of λ1 and λ1 , respectively, if we change b1 by b2 . (i) (i) We replace γ1 , δ1 and λ1 in the expression of H1 (x, y), and γ2 , δ2 and λ2 in the expression of H2 (x, y) and we obtain H1 (x, y) = H2 (x, y) for i = 1, 2. So the maximum number of crossing limit cycles in these cases is two. (2) (1) (1) (2) Now we consider either λ1 and λ2 , or λ1 and λ2 , by replacing the expres- (2) (1) sions of γ1 , δ1 and λ1 (resp. λ1 ) in the expression of H1 (x, y), and γ2 , δ2 and (1) (2) λ2 (resp. λ2 ) in the expression of H2 (x, y), and we get two different expressions of the Hamiltonian functions H1 (x, y) and H2 (x, y). Then we assume that points p4 = (cos r4 , sin r4 ) and q4 = (cos s4 , sin s4 ) satisfy system (13), and by solving this system we obtain b1 = 0 and b2 = 0, which is a contradiction. Then we proved that the maximum number of crossing limit cycles for PHS separated by an ellipse is at most three. 11 Example with three limit cycles. In the region R1 = {(x, y) : x2 +y 2 −1 ≥ 0} we consider the linear PHS (14) ẋ = 2.53x + 1.1y − 0.6, ẏ = −5.819x − 2.53y − 0.4, its Hamiltonian function is H1 (x, y) = −2.9095x2 − 2.53xy − 0.4x − 0.55y 2 + 0.6y. In the region R2 = {(x, y) : x2 + y 2 − 1 ≤ 0} we consider the linear PHS (15) ẋ = −0.308696x + 0.71y + 0.0732085, ẏ = −0.134216x + 0.308696y + 0.0488056. Its Hamiltonian function is H2 (x, y) = −0.0671078x2 + 0.308696xy + 0.0488056x − 0.355y 2 − 0.0732085y. The linear PHS (14)–(15) has exactly three crossing limit cycles, because the system of equations H1 (α, β) − H1 (γ, δ) = 0, H2 (α, β) − H2 (γ, δ) = 0, (16) α2 + β 2 − 1 = 0, γ 2 + δ 2 − 1 = 0, has three real solutions (α1 , β1 , γ1 , δ1 ) = (−0.0450412.., −0.998985.., 0.730814.., −0.682576..), (α2 , β2 , γ2 , δ2 ) = (−0.40163.., −0.915802.., 0.92153.., −0.388307..) and (α3 , β3 , γ3 , δ3 ) = (−0.760814.., −0.64897.., 0.99956.., −0.0296781..)  8 6 4 2 0 -2 -4 -6 -10 -5 0 Figure 4. Two crossing limit cycles of PHS with four zones. 3. Proof of Theorems 2 and 3 Proof of Theorem 2. Consider a continuous linear Hamiltonian differential system separated by the straight lines x = −1, x = 0 and x = 1. According to the continuity of the vector field X we obtain 12 R.BENTERKI AND J. LLIBRE X1 (−1, y) = X2 (−1, y), X2 (0, y) = X3 (0, y) and X3 (1, y) = X4 (0, y), ∀y ∈ R. Which imply that b1 = b2 = b3 = b4 = b, δ1 = δ2 = δ3 = δ4 = δ, γ1 = γ2 = γ3 = γ4 = γ, λ1 = λ2 = λ3 = λ4 = λ. Therefore, from system (1) the piecewise vector field becomes the vector field X(x, y) = (−λbx + by + γ, −λ2 bx + λby + δ), δ 6= λγ, b 6= 0. Since this linear differential system has no equilibrium point it has no periodic orbits, then no limit cycles. This completes the proof of Theorem 2.  Proof of Theorem 3. If the PHS with four zones have crossing limit cycles, then there are crossing points (−1, y0 ), (−1, y5 ); (0, y1 ), (0, y4 ); and (1, y2 ), (1, y3 ) satis- fying H1 (−1, y0 ) = H1 (−1, y5 ), H2 (−1, y0 ) = H2 (0, y1 ), H2 (−1, y5 ) = H2 (0, y4 ), (17) H3 ((0, y1 ) = H3 (1, y2 ), H3 ((0, y4 ) = H3 (1, y3 ), H4 ((1, y2 ) = H4 (1, y3 ), or equivalently (18) (y0 − y5 )(2b1 λ1 + b1 y0 + b1 y5 + 2γ1 ) = 0, (19) −b2 λ22 − b2 y02 − 2b2 λ2 y0 + b2 y12 − 2δ2 − 2γ2 y0 + 2γ2 y1 = 0, (20) −b2 λ22 + b2 y42 − b2 y52 − 2b2 λ2 y5 − 2δ2 + 2γ2 y4 − 2γ2 y5 = 0, (21) b3 λ23 − b3 y12 + b3 y22 − 2b3 λ3 y2 − 2δ3 − 2γ3 y1 + 2γ3 y2 = 0, (22) b3 λ23 + b3 y32 − 2b3 λ3 y3 − b3 y42 − 2δ3 + 2γ3 y3 − 2γ3 y4 = 0, (23) (y2 − y3 )(−2b4 λ4 + b4 y2 + b4 y3 + 2γ4 ) = 0. As y0 6= y5 and y2 6= y3 , we can solve equation (18) for y5 as well as we can solve equation (23) for y3 . Substituting the obtained values of y5 and y3 into equations (20) and (22), respectively, we obtain the following two equations 2γ1 1 γ2 ( + 2λ1 + y0 + y4 ) − δ2 − 2 (b2 (b1 (2λ1 − λ2 + y0 − y4 ) + 2γ1 ) (24) b1 2b1 (b1 (2λ1 − λ2 + y0 + y4 ) + 2γ1 )) = 0, and b3 (b4 (λ3 − 2λ4 + y2 − y4 ) + 2γ4 )(b4 (λ3 − 2λ4 + y2 + y4 ) + 2γ4 ) (25) −2b4 (b4 (δ3 + γ3 (−2λ4 + y2 + y4 )) + 2γ3 γ4 ) = 0. First we solve equation (19) for y0 and we get p (26) y0 = (1/b2 )(−b2 λ2 − γ2 ± b22 y12 + 2b2 γ2 λ2 − 2b2 δ2 + 2b2 γ2 y1 + γ22 ) then equation (21) for y2 and we get p (27) y2 = (1/b3 )(+b3 λ3 − γ3 ± b23 y12 − 2b3 γ3 λ3 + 2b3 δ3 + 2b3 γ3 y1 + γ32 ). 13 Substituting (26) into (24) we obtain two equations f1,2 (y1 , y4 ) = 0 depend on y1 and y4 . Then, substituting (27) into (25) we obtain two equations g1,2 (y1 , y4 ) = 0 depending on y1 and y4 . So we compute the product F (y1 , y2 ) = f1 (y1 , y2 )f2 (y1 , y2 ) = 0 and G(y1 , y2 ) = g1 (y1 , y2 )g2 (y1 , y2 ) = 0, and we obtain two quartic polynomial equations with the variables y1 and y4 . By using Bézout Theorem we obtain that the number of solutions of the system (28) F (y1 , y4 ) = 0, G(y1 , y4 ) = 0. is bounded by the product of the degrees of the polynomials F (y1 , y4 ) and G(y1 , y4 ). If (y1 , y4 ) is a solution of these equations, (y4 , y1 ) is also a solution. So we obtain that the number of different solutions of system (28) is at most 8 which is an upper bound for the maximum number of limit cycles that can have the PHS (17). Due to the higher degree of this system and the number of its parameters we only can give an example with two limit cycles. Example with two limit cycles. Consider the vector fields X = (X1 , X2 , X3 , X4 ) such that x x y X1 (x, y) = (− + 2y − 3, − + + 3), 2 8 2 X2 (x, y) = (2 + 2x − 2y, 2x − 2y + 30), X3 (x, y) = (4 + 4x + 2y, 13 − 8x − 4y), x x y X4 (x, y) = (− + y − 3, − + − 3). 2 4 2 Their corresponding Hamiltonian functions are given, respectively, by x2 xy H1 (x, y) = − + + 3x − y 2 + 3y, 16 2 H2 (x, y) = x2 − 2xy + 30x + y 2 − 2y, H3 (x, y) = −4x2 − 4xy + 13x − y 2 − 4y, x2 xy y2 H4 (x, y) = − + − 3x − + 3y. 8 2 2 The first crossing limit cycle intersects the straight lines of discontinuity in the fol- lowing points: (−1, −5.69679..) and (−1, 8.19679..); (0, −1.11032..) and (0, 7.25999..); and (1, 0.66814..) and (1, 6.33186..). The second crossing limit cycle intersects the straight lines of discontinuity in the points: (−1, −5.35506..) and (−1, 7.85506..); (0, 0.177417..) and (0, 0.177417..); and (1, 1.07357..) and (1, 5.92643..). The crossing limit cycles of X are shown in Figure 4.  Acknowledgements This work is supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911 14 R.BENTERKI AND J. LLIBRE References [1] A. Andronov, A. Vitt and S. Khaikin, Theory of Oscillations, Pergamon Press, Oxford, 1966. [2] S. Banerjee and G. Verghese, Nonlinear phenomena in power electronics, Attractors, bifurcations chaos and nonlinear control, Wiley-IEEE Press, New York, 2001. [3] R. Benterki and J. Llibre, The limit cycles of discontinuous piecewise linear differential systems formed by centers and separated by irreducible cubic curves I, preprint, 2019. [4] A.F. Filippov, Differential Equations With Discontinuous Righthand Sides, Kluwer Aca- demic Publishers Group, Dordrecht, 1998. [5] E. Freire, E. Ponce, F. Rodrigo and F. 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J. of Circuit Theory and Appl. 19(3) (1991), 251–307. 1 Département de Mathématiques, Université Mohamed El Bachir El Ibrahimi, Bordj Bou Arréridj 34265, El Anasser, Algeria Email address: 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