mathematics Article Crossing Limit Cycles of Planar Piecewise Linear Hamiltonian Systems without Equilibrium Points Rebiha Benterki 1 and Jaume LLibre 2, * 1 Département de Mathématiques, Université Mohamed El Bachir El Ibrahimi, Bordj Bou Arréridj 34265, El Anasser, Algeria; r.benterki@univ-bba.dz 2 Departament de Matematiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Spain * Correspondence: jllibre@mat.uab.cat  Received: 20 April 2020; Accepted: 6 May 2020; Published: 10 May 2020  Abstract: In this paper, we study the existence of limit cycles of planar piecewise linear Hamiltonian systems without equilibrium points. Firstly, we prove that if these systems are separated by a parabola, they can have at most two crossing limit cycles, 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. Secondly, we show that there is an example of two crossing limit cycles when these systems have four zones separated by three straight lines. Keywords: piecewise smooth vector field; Hamiltonian system; crossing limit cycles; conics MSC: Primary 34A36; 34C25; 34A26 1. Introduction The problem of the existence of limit cycles and mainly the problem of controlling their maximum number are two of the most difficult problems in the qualitative theory of differential systems in the plane. We solve these two problems for the class of discontinuous piecewise differential systems here considered. 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. Limit cycles appear in a natural way in many applications. Thus, recently, the problem of the existence and the number of limit cycles has also been studied for discontinuous piecewise linear differential systems; this study goes back to Andronov et al. [1] and still has been given attention by researchers, mainly due to its simplicity and to its applications to a large number of phenomena, such as switches in electronic circuits, mechanical systems, etc.; see for instance [2–4], the books [5,6], and the hundreds of references quoted therein. Lum and Chua [7] conjectured that a continuous planar piecewise linear system with two zones separated by a straight line can exhibit at most one limit cycle. Freire et al. [8] proved this conjecture in 1998. For the planar discontinuous piecewise linear systems, Han and Zhang [9] conjectured that these systems can have at most two crossing limit cycles when we separate them by a straight line, but Huan and Yang [10] gave a numerical example with three limit cycles; this result was proven analytically by Llibre and Ponce [11]. In 2015, Llibre et al. [12] proved that if we separate the planar discontinuous piecewise linear differential centers by a straight line, we cannot have any limit cycle. Recently, in the works [13–16], planar discontinuous linear differential centers separated by an algebraic curve, such as a conic, or a Mathematics 2020, 8, 755; doi:10.3390/math8050755 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 755 2 of 14 reducible and irreducible cubic, were studied, and it was proven that these differential systems can exhibit at most three crossing limit cycles having two intersection points with the conic of separation; the same result was proven if the curve of separation was a cubic. In the literature, we find many papers studying piecewise 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 equilibrium points (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 discontinuity; see [17]. We know that any conic takes nine canonical forms, but the four following forms: x2 + 1 = 0, x + y2 = 0, and x2 + y2 + 1 = 0 do not separate the plane in connected regions, then we omit them. 2 We do not study the crossing limit cycles separated by the conic x2 − 1 = 0, because in [18], it was proven that PHS with three zones 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 that 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)y2 + δi x − γi y. For more details, see [18]. 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 − y2 − 1 = 0, or by an ellipse E: x2 + y2 − 1 = 0. We consider only the crossing limit cycles that intersect the conics in exactly two points, and 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 at 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 hyperbola H at two points is at most three, and this maximum is reached; see Figure 2. (c) The maximum number of crossing limit cycles of PHS intersecting the ellipse E at two points is at most three, and this maximum is reached; see Figure 3. Mathematics 2020, 8, 755 3 of 14 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 planar discontinuous piecewise linear Hamiltonian systems (PHS) intersecting the parabola at two points. 4 2 0 -2 -4 -6 -4 -2 0 2 4 6 Figure 2. Three crossing limit cycles of PHS intersecting the hyperbola at 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 ellipse at two points. The proof of Theorem 1 is given in Section 2. Mathematics 2020, 8, 755 4 of 14 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,   X ( x, y), 2 −1 ≤ x ≤ 0, X ( x, y) = (1)   X 3 ( 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 equilibrium points with four zones satisfying C have no crossing limit cycles. Theorem 3. There are discontinuous planar piecewise Hamiltonian systems without equilibrium points with four zones satisfying C, exhibiting exactly two crossing limit cycles; see Figure 4. 8 6 4 2 0 -2 -4 -6 -10 -5 0 Figure 4. Two crossing limit cycles of PHS with four zones. The proofs of Theorems 2 and 3 are given in Section 3. 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 equilibrium points: ẋ = −λ1 b1 x + b1 y + γ1 , ẏ = −λ21 b1 x + λ1 b1 y + δ1 , (2) with b1 6= 0 and δ1 6= λ1 γ1 . The corresponding Hamiltonian function is: H1 ( x, y) = −(λ21 b1 /2) x2 + λ1 b1 xy − (b1 /2)y2 + δ1 x − γ1 y. (3) In the region R2 = {( x, y) : y − x2 ≤ 0}, we consider: Mathematics 2020, 8, 755 5 of 14 ẋ = −λ2 b2 x + b2 y + γ2 , ẏ = −λ22 b2 x + λ2 b2 y + δ2 , (4) with b2 6= 0 and δ2 6= λ2 γ2 . The corresponding Hamiltonian function is: H2 ( x, y) = −(λ22 b2 /2) x2 + λ2 b2 xy − (b2 /2)y2 + δ2 x − γ2 y. (5) In order to have a crossing limit cycle that intersects the parabola y − x2 = 0 at 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 − xi2 = 0, yk − xk2 = 0. 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 ), solving the first two equations of (6) with respect to the parameters γ1 and γ2 , we get: 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 changes (b1 , λ1 , δ1 ) by (b2 , λ2 , δ2 ). If the second points p2 = (r2 , r22 ) and q2 = (s2 , s22 ) satisfy System (6), then solving the first two equations of (6) with respect to the parameters δ1 and δ2 , we get: 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 changes (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 )). λ2 has the same expression that λ1 changes 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). Therefore, the piecewise linear differential system becomes a linear Mathematics 2020, 8, 755 6 of 14 differential system, which does not have limit cycles. Therefore, 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: 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 + 0.25y2 − 3y. The second system is: 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 + 0.05y2 + 0.778814y. Now system (6) has the two solutions (α1 , β 1 , γ1 , δ1 ) = (0.191567.., 0.0366978.., −0.191502.., 0.036673..) and (α2 , β 2 , γ2 , δ2 ) = (0.395114.., 0.156115.., −0.372692.., 0.138899..). Which provide the two limit cycles shown in Figure 1. This completes the proof of Statement (a) of Theorem 1. Proof of Statement (b) of Theorem 1. In the region R1 = {( x, y) : x2 − y2 − 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 − y2 − 1 ≤ 0}, we consider the PHS given in (2). Its corresponding Hamiltonian function is given by Equation (5). In order to have a crossing limit cycle that intersects the hyperbola x2 − y2 − 1 = 0 at 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, (7) xi2 − y2i = 1, xk2 − y2k = 1. We assume that the two systems (2) and (4) have four crossing limit cycles. Therefore, System (7) 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 (7), we consider the points p1 = (cosh r1 , sinh r1 ) and q1 = (cosh s1 , sinh s1 ), and from (7), 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 . Mathematics 2020, 8, 755 7 of 14 We suppose that the second points p2 = (cosh r2 , sinh r2 ) and q2 = (cosh s2 , sinh s2 ) satisfy System (7), then the parameters δ1 and δ2 must be: 1 r1 − s1 δ1 =  r − 2r + s   r + s − 2s  (b1 csch( )(−λ21 cosh2 r1 2  1 2 1 1 1 2 4 cosh − cosh 2 2 sinh r2 + λ1 cosh s1 sinh r2 − sinh2 r1 sinh r2 + λ1 sinh(2r1 ) sinh r2 + sinh r1 sinh2 r2 2 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 ) satisfy System (7), (1) (2) (1) then we obtain two values of λ1 (we name them λ1 and λ1 ) and two values of λ2 (we name them λ2 (2) (1) √ (2) √ and λ2 ). The first value of λ1 is given by 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 2 3 2 3 A= − sinh + sinh 1 1  r − r − 3r 2+ s − s − s   r − 3r − r 2+ s − s − s  2 3 2 3 2 3 2 3 − sinh 1 1 + sinh 1 1 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 2 3 2 3 − sinh + sinh 1 1 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  2 3 2 3 2 3 2 3 − sinh 1 1 + sinh 1 1 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 2 3 2 3 + sinh − sinh 1 1 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  2 3 2 3 2 3 2 3 + sinh 1 1 − sinh 1 1 , 2 2   r − r − r + s − s − 3s   r − r − r + s − 3s − s  2 3 2 3 2 3 2 3 B= −4 cosh 1 1 − cosh 1 1  r − r − 3r + s2 − s − s   r − 3r − r + s2 − s − s  2 3 2 3 2 3 2 3 + cosh 1 1 − cosh 1 1  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 2 3 2 3 − cosh + cosh 1 1  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  2 3 2 3 2 3 2 3 − cosh 1 1 + cosh 1 1 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 2 3 2 3 + cosh − cosh 1 1 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  2 3 2 3 2 3 2 3 + cosh 1 1 − cosh2 1 1 ) 2  r − r − r + s − s − 3s  2  r − r − r + s − 3s − s  2 3 2 3 2 3 2 3 +4(sinh 1 1 − sinh 1 1 2  r − r − 3r + s − s − s  2  r − 3r − r + s − s − s  2 3 2 3 2 3 2 3 + sinh 1 1 − sinh 1 1 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 2 3 2 3 + sinh − sinh 1 1  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  2 3 2 3 2 3 2 3 + sinh 1 1 − sinh 1 1  3r − r + r 2+ s − s + s   r − r + 3r 2+ s − s + s  1 2 3 1 2 3 2 3 2 3 − sinh + sinh 1 1  r − r + r +2 3s − s + s   r − r + r 2+ s − s + 3s  2 3 2 3 2 3 2 3 − sinh 1 1 + sinh2 1 1 , 2 2 Mathematics 2020, 8, 755 8 of 14 and the expression of C is:  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 2 3 2 3 C= cosh − cosh 1 1  r − r − 3r2 + s − s − s   r − 3r − 2r + s − s − s  2 3 2 3 2 3 2 3 + cosh 1 1 − cosh 1 1 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 2 3 2 3 − cosh + cosh 1 1 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  2 3 2 3 2 3 2 3 − cosh 1 1 + cosh 1 1 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 2 3 2 3 + cosh − cosh 1 1 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  2 3 2 3 2 3 2 3 + cosh 1 1 − cosh 1 1 . 2 2 (1) (2) (1) (2) We get the expression of λ2 and λ2 by changing b1 by b2 in the expression of 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, in these cases, the piecewise linear differential system becomes a linear differential system, which does not have limit cycles. Therefore, the maximum number of crossing limit cycles in this case is two. (2) (1) (1) (2) (2) Now, we consider either λ1 and λ2 or λ1 and λ2 , by replacing the expressions of γ1 , δ1 , and λ1 (1) (1) (2) (resp. λ1 ) in the expression of H1 ( x, y) and γ2 , δ2 , and 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 (7), 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 ẋ = −0.14..x + 1.4y + , ẏ = −0.014..x + 0.14y + 1.9, (8) 5 in the region R1 = {( x, y) : x2 − y2 − 1 ≤ 0}. It has the Hamiltonian function: y H1 ( x, y) = −0.007..x2 + 0.14xy + 1.9x − 0.7y2 − . 5 Now, we consider the second PHS: y ẋ = 5x − − 7.14286.., ẏ = 50x − 5y − 67.8571.., (9) 2 in the region R2 = {( x, y) : x2 − y2 − 1 ≥ 0, x > 1}. This differential system has the Hamiltonian function: y2 H2 ( x, y) = 25x2 − 5xy − 67.8571..x + + 7.14286..y. 4 The PHS (8)–(9) 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, (10) α2 − β2 − 1 = 0, γ2 − δ2 − 1 = 0, Mathematics 2020, 8, 755 9 of 14 has three real solutions (α1 , β 1 , γ1 , δ1 ) = (3.99376..., 3.86653..., 3.31341..., −3.1589..), (α2 , β 2 , γ2 , δ2 ) = (3.43842..., 3.28979..., 2.86513..., −2.68496..), and (α3 , β 3 , 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 region R1 = {( x, y) : x2 + y2 − 1 ≥ 0}, with its corresponding Hamiltonian function (3). We consider the PHS given in (2) in the region R2 = {( x, y) : x2 + y2 − 1 ≤ 0}, with its corresponding Hamiltonian function (5). In order that Systems (2) and (4) have crossing limit cycles intersecting the ellipse x2 + x2 − 1 = 0 at 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, (11) xi2 + y2i = 1, xk2 + y2k = 1. We suppose that Systems (2) and (4) have four crossing limit cycles. Therefore, System (11) 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. Therefore, if we consider the points p1 = (cos r1 , sin r1 ) and q1 = (cos s1 , sin s1 ), from (11), 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 (11), 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 System (11), then we (1) (2) (1) (2) obtain two values of λ1 , namely λ1 and λ1 , and two values of λ2 , namely λ2 and λ2 , such that (1) √ (2) √ λ1 = ( A + B)/C and λ1 = ( A − B)/C, where: Mathematics 2020, 8, 755 10 of 14  r − r − r + s − s − 3s   r − r − r + s − 3s − s  1 2 3 1 2 3 2 3 2 3 A= − sin + sin 1 1 2  r − r − 3r + s − s − s  2  r − 3r − r + s − s − s  2 3 2 3 2 3 2 3 − sin 1 1 + sin 1 1 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 2 3 2 3 − sin + sin 1 1  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  2 3 2 3 2 3 2 3 − sin 1 1 + sin 1 1  3r − r + r 2+ s − s + s   r − r + 3r 2+ s − s + s  1 2 3 1 2 3 2 3 2 3 + sin − sin 1 1  r − r + r +2 3s − s + s   r − r + r +2 s − s + 3s  2 3 2 3 2 3 2 3 + sin 1 1 − sin 1 1 , 2 2  r − r − r + s − s − 3s   r − r − r + s − 3s − s  2 3 2 3 2 3 2 3 B= cos 1 1 − cos 1 1 2  r − r − 3r + s − s − s  2  r − 3r − r + s − s − s  2 3 2 3 2 3 2 3 + cos 1 1 − cos 1 1 2  3r + r − r + s + s − s  2  r + 3r − r + s + s − s  1 2 3 1 2 3 2 3 2 3 − cos + cos 1 1 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  2 3 2 3 2 3 2 3 − cos 1 1 + cos 1 1 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 2 3 2 3 + cos − cos 1 1 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  2 3 2 3 2 3 2 3 + cos 1 1 − cos2 1 1 r − r − r + 2 2 2 3 s1 − s2 − 3s3   r − r2 − r3 + s1 − 3s2 − s3  +(sin 1 − sin 1  r − r − 3r + 2 2 2 3 s1 − s2 − s3   r − 3r − r + 2 3 s1 − s2 − s3  + sin 1 − sin 1  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 2 3 2 3 + sin − sin 1 1  r + r − r +2 3s + s − s   r + r − r +2 s + 3s − s  2 3 2 3 2 3 2 3 + sin 1 1 − sin 1 1 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 2 3 2 3 − sin + sin 1 1 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  2 3 2 3 2 3 2 3 − sin 1 1 + sin2 1 1 , 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 2 3 2 3 C= cos − cos 1 1  r − r − 3r2 + s − s − s   r − 3r − 2r + s − s − s  2 3 2 3 2 3 2 3 + cos 1 1 − cos 1 1  3r + r − r 2+ s + s − s   r + 3r − r 2+ s + s − s  1 2 3 1 2 3 2 3 2 3 − cos + cos 1 1 2  r + r − r + 3s + s − s  2  r + r − r + s + 3s − s  2 3 2 3 2 3 2 3 − cos 1 1 + cos 1 1 2  3r − r + r + s − s + s  2  r − r + 3r + s − s + s  1 2 3 1 2 3 2 3 2 3 + cos − cos 1 1 2  r − r + r + 3s − s + s  2  r − r + r + s − s + 3s  2 3 2 3 2 3 2 3 + cos 1 1 − cos 1 1 . 2 2 (1) (2) (1) (2) The expressions of λ2 and λ2 are the same as 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. Therefore, the maximum number of crossing limit cycles in these cases is two. Mathematics 2020, 8, 755 11 of 14 (2) (1) (1) (2) (2) Now, we consider either λ1 and λ2 or λ1 and λ2 , by replacing the expressions of γ1 , δ1 , and λ1 (1) (1) (2) (resp. λ1 ) in the expression of H1 ( x, y) and γ2 , δ2 and 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 (11), 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. Example with three limit cycles. In the region R1 = {( x, y) : x2 + y2 − 1 ≥ 0}, we consider the linear PHS: ẋ = 2.53..x + 1.1..y − 0.6.., ẏ = −5.819..x − 2.53..y − 0.4..; (12) its Hamiltonian function is: H1 ( x, y) = −2.9095..x2 − 2.53..xy − 0.4..x − 0.55..y2 + 0...6y. In the region R2 = {( x, y) : x2 + y2 − 1 ≤ 0}, we consider the linear PHS: ẋ = −0.308696..x + 0.71..y + 0.0732085.., ẏ = −0.134216..x + 0.308696..y + 0.0488056.. (13) Its Hamiltonian function is: H2 ( x, y) = −0.0671078..x2 + 0.308696..xy + 0.0488056..x − 0.355..y2 − 0.0732085..y. The linear PHS (12)–(13) 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, (14) α2 + β2 − 1 = 0, γ2 + δ2 − 1 = 0, has three real solutions (α1 , β 1 , γ1 , δ1 ) = (−0.0450412..., −0.998985..., 0.730814..., −0.682576...), (α2 , β 2 , γ2 , δ2 ) = (−0.40163..., −0.915802..., 0.92153..., −0.388307...), and (α3 , β 3 , γ3 , δ3 ) = (−0.760814.., −0.64897.., 0.99956.., −0.0296781..) We mention that the proof of Theorem 1 can be also analyzed using the results of [19]. 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: 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. Mathematics 2020, 8, 755 12 of 14 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 ) satisfying: H1 (−1, y0 ) = H1 (−1, y5 ), H2 (−1, y0 ) = H2 (0, y1 ), H2 (−1, y5 ) = H2 (0, y4 ), (15) H3 ((0, y1 ) = H3 (1, y2 ), H3 ((0, y4 ) = H3 (1, y3 ), H4 ((1, y2 ) = H4 (1, y3 ), or equivalently: (y0 − y5 )(2b1 λ1 + b1 y0 + b1 y5 + 2γ1 ) = 0, (16) −b2 λ22 − b2 y20 − 2b2 λ2 y0 + b2 y21 − 2δ2 − 2γ2 y0 + 2γ2 y1 = 0, (17) −b2 λ22 + b2 y24 − b2 y25 − 2b2 λ2 y5 − 2δ2 + 2γ2 y4 − 2γ2 y5 = 0, (18) b3 λ23 − b3 y21 + b3 y22 − 2b3 λ3 y2 − 2δ3 − 2γ3 y1 + 2γ3 y2 = 0, (19) b3 λ23 + b3 y23 − 2b3 λ3 y3 − b3 y24 − 2δ3 + 2γ3 y3 − 2γ3 y4 = 0, (20) (y2 − y3 )(−2b4 λ4 + b4 y2 + b4 y3 + 2γ4 ) = 0. (21) As y0 6= y5 and y2 6= y3 , we can solve Equation (16) for y5 , as well as we can solve Equation (21) for y3 . Substituting the obtained values of y5 and y3 into Equations (18) and (20), respectively, we obtain the following two equations: 2γ1 1 γ2 ( + 2λ1 + y0 + y4 ) − δ2 − 2 (b2 (b1 (2λ1 − λ2 + y0 − y4 ) + 2γ1 ) b1 2b1 (22) (b1 (2λ1 − λ2 + y0 + y4 ) + 2γ1 )) = 0, and: b3 (b4 (λ3 − 2λ4 + y2 − y4 ) + 2γ4 )(b4 (λ3 − 2λ4 + y2 + y4 ) + 2γ4 ) (23) −2b4 (b4 (δ3 + γ3 (−2λ4 + y2 + y4 )) + 2γ3 γ4 ) = 0. First, we solve Equation (17) for y0 , and we get: q y0 = (1/b2 )(−b2 λ2 − γ2 ± b22 y21 + 2b2 γ2 λ2 − 2b2 δ2 + 2b2 γ2 y1 + γ22 ); (24) then, we solve Equation (19) for y2 , and we get: q y2 = (1/b3 )(+b3 λ3 − γ3 ± b32 y21 − 2b3 γ3 λ3 + 2b3 δ3 + 2b3 γ3 y1 + γ32 ). (25) Substituting (24) into (22), we obtain two equations f 1,2 (y1 , y4 ) = 0 depending on y1 and y4 . Then, substituting (25) into (23), we obtain two equations g1,2 (y1 , y4 ) = 0 depending on y1 and y4 . Therefore, we compute the product F (y1 , y2 ) = f 1 (y1 , y2 ) f 2 (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 . Mathematics 2020, 8, 755 13 of 14 By using Bézout’s theorem, we obtain that the number of solutions of the system: F (y1 , y4 ) = 0, G (y1 , y4 ) = 0. (26) 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. Therefore, we obtain that the number of different solutions of System (26) is at most eight, which is an upper bound for the maximum number of limit cycles that can have the PHS (15). 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 − y2 + 3y, 16 2 H2 ( x, y) = x2 − 2xy + 30x + y2 − 2y, H3 ( x, y) = −4x2 − 4xy + 13x − y2 − 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 following 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. 4. Conclusions We considered four classes of discontinuous piecewise differential systems formed by linear Hamiltonian systems without equilibrium points in the plane separated either by a parabola, a hyperbola, an ellipse, or three parallel lines. For each class, we provided the maximum number of crossing limit cycles that the differential systems of the class can exhibit. Furthermore, we provided examples exhibiting the maximum number of limit cycles for each class. We characterized the maximum number of crossing limit cycles for classes of discontinuous piecewise differential systems formed by linear Hamiltonian systems without equilibrium points separated by conics, but it remains to study these maximum numbers when the separation is done by cubics, or more general algebraic curves. Author Contributions: Conceptualization, R.B. and J.L.; methodology, J.L.; software, R.B.; validation, R.B. and J.L.; formal analysis, R.B. and J.L.; investigation, R.B. and J.L.; writing–original draft, R.B. and J.L. All authors have read and agreed to the published version of the manuscript. Mathematics 2020, 8, 755 14 of 14 Funding: This work is supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación Grant 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. Conflicts of Interest: The authors declare no conflict of interest. References 1. Andronov, A.; Vitt, A.; Khaikin, S. Theory of Oscillations; Pergamon Press: Oxford, UK, 1966. 2. Banerjee, S.; Verghese, G. Nonlinear Phenomena in Power Electronics; Attractors, bifurcations chaos and nonlinear control; Wiley-IEEE Press: New York, NY, USA, 2001. 3. 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