This is a preprint of: “Limit cycles of planar discontinuous piecewise linear Hamiltonian systems without equilibria separated by nonregular curves”, Johana Jimenez, Jaume Llibre, Clàudia Valls, Internat. J. Bifur. Chaos Appl. Sci. Engrg., vol. 32(12), 2250184, 2022. DOI: [10.1142/S021812742250184X] LIMIT CYCLES OF PLANAR DISCONTINUOUS PIECEWISE LINEAR HAMILTONIAN SYSTEMS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES JOHANA JIMENEZ1 , JAUME LLIBRE2 AND CLAUDIA VALLS3 Abstract. The problem of determining the existence, the maximum number and the position of the limit cycles of the planar discontinuous piecewise linear dierential systems is an important problem in the qualitative theory of the dierential systems. This is due mainly to the fact that these piecewise dierential systems have many applications in mechanical systems, electrical circuits, control theory, ... In this paper we study two families of piecewise linear Hamiltonian systems without equilibria in R2 separated by a non-regular curve. We provide the maximum number of crossing limit cycles that each family can have and show that this maximum is reached. In this way we are solving for each family the extended 16th Hilbert problem. 1. Introduction and statement of the main results A discontinuous piecewise dierential system on R2 is a pair of Cr (with r ≥ 1) dier- ential systems in R2 separated by a smooth curve Σ. The line of discontinuity Σ of the discontinuous piecewise dierential system is given by Σ = h−1 (0), where h : R2 −→ R is a C 1 function having 0 as a regular value. Observe that Σ is the boundary between the regions Σ+ = {(x, y) ∈ R2 | h(x, y) > 0} and Σ− = {(x, y) ∈ R2 | h(x, y) < 0}. Hence  X(x, y), if h(x, y) ≥ 0, (1) Z(x, y) = Y (x, y), if h(x, y) ≤ 0, is the vector eld corresponding to a piecewise dierential system with line of discontinuity Σ. When the vector elds X and Y coincide on the line Σ we have a continuous piecewise dierential system on R2 , that in general it will not be smooth on Σ. The vector eld (1) usually is denoted by Z = (X, Y, Σ) or simply by Z = (X, Y ), if the separation line Σ is known. In order to establish a denition for the trajectories of Z , we must have a criterion for the transition of the trajectories between Σ+ and Σ− across the curve of discontinuity Σ. The contact between the curve of discontinuity Σ and the vector eld X (or Y ) is described by the directional derivative of h with respect to the vector eld X , i.e. Xh(p) = ⟨∇h(p), X(p)⟩ . Here ⟨., .⟩ denotes the usual inner product of the plane R2 . Filippov in [8] stated the main results of the discontinuous piecewise dierential systems. The curve of discontinuity Σ is divided into the three following sets: (a) Σc : {p ∈ Σ : Xh(x) · Y h(x) > 0}, the Crossing set. (b) Σe : {p ∈ Σ : Xh(x) > 0 and Y h(x) < 0}, the Escaping set. (c) Σs : {p ∈ Σ : Xh(x) < 0 and Y h(x) > 0}, the Sliding set. 2010 Mathematics Subject Classication. Primary 34C29, 34C25, 47H11. Key words and phrases. Crossing limit cycles, discontinuous piecewise linear Hamiltonian systems, non- regular curve. 1 2 J. JIMENEZ, J. LLIBRE AND C. VALLS The points of Σ where both vector elds X and Y simultaneously point outwards or inwards dene the escaping Σe or sliding Σs regions, while the interior in Σ of their com- plement denes the crossing region Σc (see Figure 1). The points of Σ with are not in Σc ∪ Σe ∪ Σs are the tangency points between X or Y with Σ. Σ Figure 1. Crossing, sliding and escaping regions, respectively. A limit cycle of a dierential system is an isolated periodic orbit in the set of all periodic orbits of the system. The limit cycles play a main role in the qualitative theory of the planar dierential equations. It is well known that if the vector elds X and Y are linear they cannot have limit cycles, but the piecewise vector eld Z = (X, Y, Σ) can have limit cycles. If the limit cycle only contains isolated points of Σ we say that it is a crossing limit cycle. In this paper we only will study crossing limit cycles, that frequently we only mention them as limit cycles. The study of the piecewise linear dierential systems is a problem that started with Andronov, Vitt and Khainkin [1] in the 1920's, and nowadays is an important problem in the qualitative theory of the dierential systems mainly due to its applications to many physical phenomena, see for instance the books of [7, 16, 17, 23] and the survey [22], and the hundred of papers cited inside these references. Many of the study developed on the piecewise linear dierential systems were done con- sidering piecewise linear dierential systems with only two zones and separated by a straight line. Few studies have been done with more zones or considering discontinuity curves dif- ferent to a straight line. In 1990 Lum and Chua[20, 21] conjectured that the planar continuous piecewise dieren- tial systems separated by a straight line have at most one limit cycle. In 1998 Freire, Ponce, Rodrigo and Torres [10] gave a proof for this conjecture, a shorter proof was given later on by Llibre, Ordóñez and Ponce in [18]. While that for discontinuous piecewise dierential systems separated by a straight line Han and Zhang [11] in 2010 conjectured that these sys- tems can exhibit at most two limit cycles. A numerical counterexample to this conjecture was given by Huan and Yang [13] in 2012 providing a discontinuous piecewise dierential systems separated by a straight line with three limit cycles. In 2012 Llibre and Ponce [19] proved analytically the existence of these three limit cycles, and later on several authors also nd discontinuous piecewise dierential systems separated by a straight line with three limit cycles. But we still do not know if three is the maximum number of limit cycles for this class of piecewise dierential systems. The famous and unsolved 16th Hilbert problem asked for an upper bound on the max- imum number of limit cycles that polynomial dierential systems of a given degree can exhibit, see [12]. These last years several authors have extended this problem to dierent classes of dierential systems in particular to the discontinuous piecewise dierential systems formed by distinct linear dierential systems. Thus recently in [3, 4, 14, 15] the authors studied the extension of the 16th Hilbert problem for discontinuous piecewise dierential systems formed by linear centers which have either two or more zones and they are sepa- rated by either conics, or reducible cubics, or irreducible cubics. In [2, 5, 6, 9] the authors considered discontinuous piecewise linear Hamiltonian systems without equilibrium points where such systems are separated by either two parallel straight lines, or conics, or reducible LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 3 cubics, or irreducible cubics, in each of these classes of piecewise dierential systems the authors determined the maximum number of crossing limit cycles that these piecewise linear systems can exhibit. The results that were obtained in the mentioned papers shown that the shape of the discontinuity curve plays an important role in the number of limit cycles that the piecewise dierential systems can have. The objective of this paper is to study the maximum number of crossing limit cycles for two families of discontinuous piecewise linear Hamiltonian systems without equilibrium points in R2 separated by a non-regular curve. First in subsection 1.1 we present our results on the limit cycles of discontinuous piecewise linear Hamiltonian systems without equilibrium points in R2 where the discontinuity curve is the non-regular line  ΣA = (x, y) ∈ R2 : xy = 0 and y ≥ 0, x ≥ 0 . After in subsection 1.2 we present our results on the limit cycles of discontinuous piecewise linear Hamiltonian systems without equilibrium points in R2 where the discontinuity curve is the non-regular line  ΣB = (x, y) ∈ R2 : xy = 0 and y ≥ 0 . 1.1. Crossing limit cycles intersecting the discontinuity curve ΣA . We observed that ΣA = Σ+x ∪ Σy , where +   Σ+ 2 x = (x, y) ∈ R : y = 0, x ≥ 0 and Σ+ 2 y = (x, y) ∈ R : x = 0, y ≥ 0. The discontinuity curve ΣA separates the plane R2 into the following two pieces 1 RA ={(x, y) ∈ R2 : x > 0, y > 0}, RA 2 = R2 \ (RA 1 ∪ ΣA ). We denote by FA the class of discontinuous piecewise linear dierential systems formed by two linear Hamiltonian systems without equilibrium points and separated by ΣA . We shall study the following class of limit cycles. A crossing limit cycle of type A1 is a limit cycle that intersects Σ+ x and Σy in exactly one + point. A crossing limit cycle of type A2 is a limit cycle that intersects Σ+ x and Σy in exactly two + point. We observed that it is not possible to have limit cycles of systems in FA that intersect only either Σ+x , or Σy , see for instance [9]. + Theorem 1. The following statements hold for the discontinuous piecewise linear Hamil- tonian systems without equilibria when the discontinuity curve is ΣA . (i) The maximum number of crossing limit cycles of type A1 is two. See Figure 2a. (ii) The maximum number of limit cycles of type A2 is one. See Figure 2b. (iii) There are examples of discontinuous piecewise linear Hamiltonian systems having simultaneously one limit cycle of type A1 and one limit cycle of type A2 . See Figure 2c. Note that although the maximum number of possible simultaneous crossing limit cycles of types A1 and A2 are two and one respectively we could not found an example realizing this upper bound. To obtain a limit cycle of a given type it is not enough to obtain the points where the candidate to be a limit cycle intersects the line of discontinuity in the prescribed number, the orbits through these points must close must be well oriented and forming a 4 J. JIMENEZ, J. LLIBRE AND C. VALLS 5 6 4 0 4 3 -5 2 2 -10 0 1 -15 0 -2 -20 -15 -10 -5 0 5 -2 0 2 4 6 -1 0 1 2 3 (a) Two limit cycles of type (c) One limit cycle of type A1 A1 . (b) One limit cycle of type A2 . and one limit cycle of type A2 . Figure 2. Crossing limit cycles in the family FA . periodic orbit, and this periodic orbit must be isolated in order to dene a limit cycle. This implies that the search of examples can be very dicult and that the upper bound obtained by our method which does not take into account all these matters of orientation of the orbits, isolateness and pieces of orbits in the prescribed regions could not be optimal. Theorem 1 is proved in section 3. 1.2. Crossing limit cycles intersecting the discontinuity curve ΣB . We observed that ΣB = Σx ∪ Σ+ y , where Σx = Σx ∪ Σx , being + −  Σ− 2 x = (x, y) ∈ R : y = 0, x ≤ 0 . The discontinuity curve ΣB separates the plane R2 into the following three pieces 1 = {(x, y) ∈ R2 : x > 0, y > 0}, R2 = {(x, y) ∈ R2 : x < 0, y > 0}, RB B 3 = {(x, y) ∈ R2 : y < 0}. RB We denote by FB the class of discontinuous piecewise linear dierential systems formed by three linear Hamiltonian systems without equilibrium points and separated by ΣB . We shall study the following types of limit cycles. A crossing limit cycle of type B1 is a limit cycle that intersects Σ+ x , Σx and Σy in exactly − + one point. A crossing limit cycle of type B2 is a limit cycle that intersects either Σ+x and Σy , or Σx + − and Σ+ y in exactly two points. Theorem 2. The following statements hold for the discontinuous piecewise linear Hamil- tonian systems without equilibria when the discontinuity curve is ΣB . (i) The maximum number of crossing limit cycles of type B1 is three. See Figure 3a. (ii) The maximum number of crossing limit cycles of type B2 is one. See Figure 3b. (iii) There are examples of discontinuous piecewise linear Hamiltonian systems having simultaneously two limit cycles of type B1 and one limit cycle of type B2 . See Figure 3c. Note that although the maximum number of possible simultaneous crossing limit cycles of types B1 and B2 are three and one respectively we could not found an example realizing this upper bound. Theorem 2 is proved in section 4. LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 5 6 10 6 4 5 5 2 4 3 0 0 2 -2 -5 1 -4 0 -10 -4 -2 0 2 4 -1 -4 -2 0 2 4 6 8 (a) Three limit cycles of type -2 0 2 4 (c) Two limit cycles of type B1 B1 . (b) One limit cycle of type B2 . and one limit cycle of type B2 . Figure 3. Crossing limit cycles in the family FB . 2. Preliminary result In this paper we consider the normal form for an arbitrary linear dierential Hamiltonian system in R2 without equilibrium points provided in [9]. Lemma 1. The normal form for a linear dierential Hamiltonian system in R2 without equilibria is (2) ẋ = −λbx + by + γ, ẏ = −λ2 bx + λby + δ, with δ ̸= λγ, and b ̸= 0. This linear dierential system has the rst integral 1 b (3) H(x, y) = − λ2 bx2 + λbxy − y 2 + δx − γy. 2 2 Note that in the dierential system (2) we can assume that b = 1 doing a rescaling of the time. 3. Proof of Theorem 1 Proof of statement (i) of Theorem 1. A piecewise linear dierential system in the family FA is formed by one linear dierential Hamiltonian without equilibrium points in each region RA1 and R2 . Hence by Lemma 1 the general equations for this class of piecewise dierential A systems are ẋ = −λ1 x + y + γ1 , ẏ = −λ21 x + λ1 y + δ1 , in RA 1, ẋ = −λ2 x + y + γ2 , ẏ = −λ22 x + λ2 y + δ2 , in RA 2. Moreover by Lemma 1 we have that 1 y2 Hi (x, y) = − λ2i x2 + λi xy − + δi x − γi y, i = 1, 2, 2 2 is a rst integral of this system in the region RAi for i = 1, 2. Now if there is a crossing limit of type A1 , this intersects the discontinuity curve ΣA in two points (X, 0) and (0, Y ) with X > 0 and Y > 0. Of course these two points must satisfy the following closing equations e1 = H1 (X, 0) − H1 (0, Y ) = Y 2 + 2Y γ1 + 2Xδ1 − X 2 λ21 = 0, (4) e2 = H2 (X, 0) − H2 (0, Y ) = Y 2 + 2Y γ2 + 2Xδ2 − X 2 λ22 = 0. It follows from Bézout theorem that system (4) has at most four real solutions, but since one solution is (0, 0), which cannot produce a limit cycle of type A1 , we conclude that system 6 J. JIMENEZ, J. LLIBRE AND C. VALLS (4) has at most three real solutions that without loss of generality we can assume that they are (Xi , 0), (0, Yi ) with i = 1, 2, 3 satisfying 0 < X1 < X2 < X3 and 0 < Y1 < Y2 < Y3 , (otherwise the solutions would intersect which is not possible by the uniqueness of solutions of a dierential system). We consider two dierent cases. If λ1 = λ2 = 0 then e1 = 0 is a parabola symmetric with respect to some horizontal straight line and passing through the origin. Moreover, E2 = e1 − e2 = 0 become E2 = 2(δ1 − δ2 )X + 2(γ1 − γ2 )Y = 0, which is a straight line. Since a parabola and a straight line intersect at most in two points we have that the system e1 = E2 = 0 intersect at most in two points where one of these two is the origin, then we have at most one point satisfying 0 < X1 and 0 < Y1 , therefore there is at most one limit cycle. If λ1 λ2 ̸= 0 it follows from e1 = e2 = 0 that E1 = λ22 e1 − λ21 e2 = 0 and E2 = e1 − e2 = 0 become E1 := 2(λ22 δ1 − λ21 δ2 )X + 2(λ22 γ1 − λ21 γ2 )Y + (λ22 − λ21 )Y 2 = 0, E2 := 2(δ1 − δ2 )X + 2(γ1 − γ2 )Y + (λ22 − λ21 )X 2 = 0. If λ22 = λ21 (i.e., λ2 = ±λ1 ) then equations E1 = E2 = 0 have at most one solution and so there is at most one limit cycle. Assume now that λ2 ̸= ±λ1 . If λ22 δ1 − λ21 δ2 = 0 then E1 = 0 reduces to either one horizontal straight line, or two horizontal parallel straight lines passing one of these two straight lines through the origin. The equation E2 = 0 is either a parabola symmetric with respect to some vertical straight line, or one vertical straight line, or two vertical parallel straight lines passing one of these two straight lines through the origin. Since E1 = E2 = 0 pass through the origin, there are at most two intersection points satisfying 0 < X1 < X2 , but 0 < Y1 = Y2 and so there is at most one limit cycle. If γ1 − γ2 = 0 then E2 = 0 reduces to either one vertical straight line, or two vertical parallel straight lines passing one of these two straight lines through the origin. The equation E1 = 0 is either a parabola symmetric with respect to some horizontal straight line, or one horizontal straight line, or two horizontal parallel straight lines passing one of these two straight lines through the origin. Since E1 = E2 = 0 pass through the origin, there are at most two intersection points satisfying 0 < Y1 < Y2 , but 0 < X1 = X2 and so there is at most one limit cycle. Finally, assume that λ22 δ1 − λ21 δ2 ̸= 0 and γ1 − γ2 ̸= 0. In this case, E1 = 0 is a parabola symmetric with respect to some horizontal straight line and E2 = 0 is a parabola symmetric with respect to some vertical line. Since both parabolas intersect at the origin, the maximum number of intersections points is three. When we have three intersection points they satisfy 0 < X1 < X2 < X3 and 0 < Y3 < Y2 < Y1 , and then there cannot be three limit cycles. Hence there are at most two intersection points satisfying 0 < X1 < X2 and 0 < Y1 < Y2 , and so at most two limit cycles. Now we verify that this found upper bound is reached. We provide a discontinuous piecewise linear dierential system in FA having exactly two limit cycles of type A1 . In the LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 7 region RA1 we consider the linear Hamiltonian system 271 1837341 73441 271 118235781 ẋ = x+y− , ẏ = − x− y+ , 100 95000 10000 100 3800000 which has the rst integral 73441 2 271 118235781 1 1837341 H1 (x, y) = − x − xy + x − y2 + y; 20000 100 3800000 2 95000 in region RA2 we consider the Hamiltonian system 1207 1001997 85697 1207 14426523 ẋ = x − 34y + , ẏ = x− y+ 50 47500 5000 50 1900000 with the rst integral 85697 2 1207 14426523 1001997 H2 (x, y) = − x + xy − x − 17y 2 + y. 10000 50 1900000 47500 With these two systems we obtain that system (4) has exactly two reals solutions (Xj , Yj ), with Xj > 0 and Yj > 0, for j = 1, 2, namely     3 21 39 15 (X1 , Y1 ) = , and (X2 , Y2 ) = , , 2 10 10 4 which generated the two crossing limit cycles of type A1 in Figure 2a. □ Proof of statement (ii) of Theorem 1. If there exists a periodic solution candidate to be a limit cycle of type A2 , then this periodic solution has four intersection points on the dis- continuity line ΣA of the form (x1 , 0), (x2 , 0), (0, y1 ) and (0, y2 ), satisfying 0 < x1 < x2 , 0 < y1 < y2 . Hence the following equations must be satised f1 = H1 (x1 , 0) − H1 (0, y1 ) = 2δ1 x1 + 2γ1 y1 − λ21 x21 + y12 = 0, f2 = H1 (x2 , 0) − H1 (0, y2 ) = 2δ1 x2 + 2γ1 y2 − λ21 x22 + y22 = 0, (5) f3 = H2 (x1 , 0) − H2 (x2 , 0) = (x1 − x2 )(−2δ2 + λ22 x1 + λ22 x2 ) = 0, f4 = H2 (0, y1 ) − H2 (0, y2 ) = (y2 − y1 )(2γ2 + y1 + y2 ) = 0. We consider two dierent cases. Case 1: λ2 = 0. In this case from f3 = 0 we get δ2 = 0. This is not possible, since by Lemma 1 we have that δ2 ̸= λ2 γ2 . Case 2: λ2 ̸= 0. From f3 = 0 and f4 = 0, we get 2δ2 (6) x2 = − x1 , y2 = −2γ2 − y1 . λ22 Substituting these last expressions in f1 and f2 and then setting F3 = f1 −f2 = 0, we obtain that system (5) reduces to  2 δ12 2δ2 λ21 −δ1 λ22 f2 = −γ12 + λ21 − λ 1 x1 − + ((2γ2 − γ1 ) + y1 )2 = 0, λ1 λ22 (7)  2 δ2 (δ2 λ21 −δ1 λ22 ) δ2 λ21 F3 = γ1 γ2 − γ2 + λ42 + δ1 − 2 x1 + (γ1 − γ2 )y1 = 0. λ2 We observed that equation F3 = 0 is a straight line and equation f2 = 0 is a hyperbola. By Bézout theorem system (7) has at most two real solutions. If they are two real solutions (x±1 , y1 ) from (6) we obtain the correspondent (x2 , y2 ). Let (x1 , x2 , y1 , y2 ) be ± ± ± + + + + the four intersection points of a rst limit cycle of type A2 with 0 < x+ 1 < x2 , 0 < y1 < y2 , + + + and (x1 , x2 , y1 , y2 ) be the four intersection points of a second limit cycle of type A2 with − − − − 0 < x−1 < x2 and 0 < y1 < y2 . If there are two limit cycles of type A2 , we can suppose − − − 8 J. JIMENEZ, J. LLIBRE AND C. VALLS without loss of generality that the rst limit cycle is located in the interior region enclosed by the second limit cycle. Then the intersection points must satisfy (8) 0 < x− + + 1 < x1 < x2 < x2 − and 0 < y1− < y1+ < y2+ < y2− . We claim that it is not possible to have two solutions (x± 1 , y1 ) of system (7) which satisfy ± the conditions in (8). If γ1 −γ2 = 0 it follows from F3 = 0 that x1 = δ2 /λ22 , but then from (6) we get x2 = δ2 /λ22 which is not possible. Hence, γ1 − γ2 ̸= 0 and solving F3 = 0 in the variable y1 we obtain   −δ22 λ21 + λ42 −γ1 γ2 + γ22 − δ1 x1 + δ2 λ22 δ1 + λ21 x1 y1 = . λ42 (γ1 − γ2 ) Now we introduce y1 into f2 = 0 and we get P (x1 ) = C0 + C1 x1 + C2 x21 = 0, being    γ2 λ42 (γ1 − γ2 ) − δ1 δ2 λ22 + δ22 λ21 λ42 − 2γ12 − 3γ1 γ2 + γ22 − δ1 δ2 λ22 + δ22 λ21 C0 = , λ82 (γ1 − γ2 )2   2δ2 λ42 λ21 (γ1 − γ2 )2 − δ12 + 2δ1 δ2 λ21 λ22 − δ22 λ41 C1 = , λ62 (γ1 − γ2 )2  λ42 δ12 − λ21 (γ1 − γ2 )2 − 2δ1 δ2 λ21 λ22 + δ22 λ41 C2 = . λ42 (γ1 − γ2 )2 The polynomial P (x1 ) is quadratic in the variable x1 , whose roots x1,± , are p δ2 λ22 ∆/4 x1,± = 2± , λ2 C2 where  4   ∆= 8 δ2 λ21 − λ22 λ1 (γ1 − γ2 ) + δ1 ) δ2 λ21 − λ22 (λ1 (γ2 − γ1 ) + δ1 ) λ2 (γ1 − γ2 )2   γ2 λ42 (2γ1 − γ2 ) − 2δ1 δ2 λ22 + δ22 λ21 . Note that from (6) we get p 2δ2 δ2 λ22 ∆/4 x2,± = 2 − x1,± = 2 ∓ = x1,∓ . λ2 λ2 C2 The argument of the claim follows. Therefore the maximum number of limit cycles of type A2 that our discontinuous piecewise linear dierential system can have is one. Now we verify that the upper bound found is reached. We provide a discontinuous piecewise linear dierential system in FA having exactly one limit cycle of type A2 . In the region RA1 we consider the Hamiltonian system 3 12647 9 3 327 ẋ = − x − y + , ẏ = x + y − 2 8960 4 2 128 this system has the rst integral 9 3 327 1 12647 H1 (x, y) = − x2 − xy + x − y2 + y; 8 2 128 2 8960 and in the region RA2 we consider the Hamiltonian system 5 14 25 5 45 ẋ = x − y + , ẏ = x − y − 8 5 64 8 64 LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 9 which has the rst integral 25 2 5 45 1 14 H2 (x, y) = − x + xy + x − y 2 + y; 128 8 64 2 5 With these systems in each region RA we have that the system of closing equations (5) has i exactly one real solutions (x1 , x2 , y1 , y2 ) which satisfy 0 < x1 < x2 and 0 < y1 < y2 , namely   1 31 7 22 (x1 , x2 , y1 , y2 ) = , , , . 2 10 5 5 This real solution generated the crossing limit cycle of type A2 in the family FA in Figure 2b. □ Proof statement (iii) of Theorem 1. Here we provide a piecewise linear Hamiltonian system in FA which has exactly one limit cycle of type A1 and one limit cycle of type A2 . In region RA1 we consider the Hamiltonian system r r 6 1351 13682 48636 6 1351 85596 ẋ = − x−y+ , ẏ = x+ y− , 13 79 5135 13351 13 79 13351 which has the rst integral r 24318 2 6 1351 85596 1 13682 H1 (x, y) = − x − xy + x − y2 + y; 13351 13 79 13351 2 5135 in region RA2 we consider the Hamiltonian system r r 31 14 124 31 372 ẋ =2 x − y + , ẏ = x−2 y− , 39 5 39 39 65 this Hamiltonian system has the rst integral r 62 31 372 1 14 H2 (x, y) = x2 + 2 xy + x − y 2 + y. 39 39 65 2 5 With these Hamiltonian systems we have that system (4) has exactly the one real solution   39 33 (X, Y ) = , , 10 5 and system (5) has exactly the one real solution   1 31 7 22 (x1 , x2 , y1 , y2 ) = , , , . 2 10 5 5 These reals solutions generated one crossing limit cycle of type A1 and one crossing limit cycle of type A2 in FA in Figure 2c. This completes the proof of Theorem 1. □ 4. Proof of Theorem 2 Proof of statement (i) of Theorem 2. We have that a piecewise linear dierential system in the family FB is formed by one linear dierential Hamiltonian without equilibrium points in each region RBi for i = 1, 2, 3. Hence by Lemma 1 the general equations for this class of piecewise dierential systems are ẋ = −λ1 x + y + γ1 , ẏ = −λ21 x + λ1 y + δ1 , in RB 1, (9) ẋ = −λ2 x + y + γ2 , ẏ = −λ22 x + λ2 y + δ2 , in RB 2, ẋ = −λ3 x + y + γ3 , ẏ = −λ23 x + λ3 y + δ3 , in RB 3, 10 J. JIMENEZ, J. LLIBRE AND C. VALLS and 1 y2 Hi (x, y) = − λ2i x2 + λi xy − + δi x − γi y, i = 1, 2, 3 2 2 is a rst integral of the system in the region RBi for i = 1, 2, 3. In order to have a crossing limit of type B1 , it must intersects the discontinuity curve ΣB in three points (X, 0), (0, Y ) and (x, 0) with x < 0 < X and Y > 0. Of course these three points must satisfy the following closing equations ẽ1 = H1 (X, 0) − H1 (0, Y ) = Y 2 + 2Y γ1 + 2Xδ1 − X 2 λ21 = 0, (10) ẽ2 = H2 (x, 0) − H2 (0, Y ) = Y 2 + 2Y γ2 + 2xδ2 − x2 λ22 = 0, ẽ3 = H3 (X, 0) − H3 (x, 0) = (X − x)((x + X)λ23 − 2δ3 ) = 0. Since X − x ̸= 0 equation ẽ3 reduces to ẽ3 = (x + X)λ23 − 2δ3 = 0. If λ3 = 0, then δ3 = 0 which is not possible because by Lemma 1 δ3 ̸= λ3 γ3 . Therefore we have that λ3 ̸= 0. Isolating x and substituting in equation ẽ2 system (10) reduces to system ẽ1 = Y 2 + 2γ1 Y + 2δ1 X − λ21 X 2 = 0, (11) 2  δ3  ẽ2 = Y 2 + 2γ2 Y + 2 2δ3 λ22 − δ2 λ23 X − λ22 X 2 + 4 4 δ2 λ23 − δ3 λ22 = 0. λ3 λ3 It follows from Bézout theorem that system (11) has at most four real solutions, (Xi , 0), (0, Yi ) with i = 1, 2, 3, 4 where (12) 0 < X1 < X2 < X3 < X4 and 0 < Y1 < Y2 < Y3 < Y4 . We consider dierent cases. If either δ3 = 0, or δ2 λ23 − δ3 λ22 = 0, then equation ẽ1 = ẽ2 = 0 are hyperbolas passing through the origin. Since both hyperbolas intersect at most in four points being one of these the origin, we have that there is at most three points satisfying condition (12). Therefore there are at most three limit cycles. We assume that δ3 ̸= 0 and δ2 λ23 − δ3 λ22 ̸= 0. If λ1 = λ2 = 0, system (9) has at most two limit cycles because the resultant of the polynomial ẽ1 and ẽ2 with respect to the variable X is 8δ1 δ2 δ3 2Y 2 (δ1 + δ2 ) + 4Y (γ2 δ1 + γ1 δ2 ) + , λ23 which has at most two positive solutions in the variable Y and equation ẽ2 = 0 reduces to 4δ2 δ3 Y 2 + 2γ2 Y − 2δ2 X + = 0. λ3 2 Since δ2 ̸= 0, for each solution Y there is at most one solution X of ẽ2 = 0 and so there are at most two limit cycles. Consider now that λ1 = 0, λ2 ̸= 0 (λ1 ̸= 0, λ2 = 0), we have that ẽ1 = 0 reduces to either a parabola symmetric with respect to some horizontal straight line, or one horizontal straight line, or two horizontal parallel straight lines passing one of these two straight lines through the origin (hyperbola passing through the origin). Moreover, ẽ2 = 0 is a hyperbola (is either a parabola symmetric with respect to some horizontal straight line, or one horizontal straight line, or two horizontal parallel straight lines). In both cases we have that there are at most three intersection points satisfying condition (12) and so at most three limit cycles. LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 11 Assume now that λ1 λ2 ̸= 0. From ẽ1 = 0 = ẽ2 it follows that Ẽ1 = λ22 ẽ1 − λ21 ẽ2 = 0 and Ẽ2 = ẽ1 − ẽ2 = 0 become    δ3 λ21 (δ3 λ22 − δ2 λ23 ) 2 2 2δ3 λ21 Ẽ1 :=4 + 2 λ1 δ2 + λ2 δ1 − X + 2(λ22 γ1 − λ21 γ2 )Y λ43 λ23 + (λ22 − λ21 )Y 2 = 0,   δ3 (δ3 λ22 − δ2 λ23 ) 2δ3 λ22 Ẽ2 :=4 + 2 δ1 + δ2 − X + 2(γ1 − γ2 )Y + (λ22 − λ21 )X 2 = 0. λ43 λ23 If λ22 = λ21 (i.e., λ2 = ±λ1 ) then equations Ẽ1 = Ẽ2 = 0 are straight lines, and so they have at most one solution implying that there is at most one limit cycle. Assume now that λ2 ̸= ±λ1 . If γ1 − γ2 = 0 then Ẽ2 = 0 reduces to either one vertical straight line, or two vertical parallel straight lines. The equation Ẽ1 = 0 is either a parabola symmetric with respect to some horizontal straight line, or one horizontal straight line, or two horizontal parallel straight lines. Then Ẽ1 = Ẽ2 = 0, have at most four intersection points but only two intersection points satisfying 0 < X1 < X2 and 0 < Y1 < Y2 , and so at most two limit cycles.   2δ3 λ21 Ifλ21 δ2 + λ22 δ1 − = 0 then Ẽ1 = 0 reduces to either one horizontal straight line, λ23 or two horizontal parallel straight lines passing. The equation Ẽ2 = 0 is either a parabola symmetric with respect to some vertical straight line, or one vertical straight line, or two vertical parallel straight lines. Therefore Ẽ1 = Ẽ2 = 0, have at most four intersection points but only two intersection points satisfying 0 < X1 < X2 and 0 < Y1 < Y2 , and so at most two limit cycles.   2δ3 λ21 Finally, assume that λ21 δ2 δ1 − + λ22 ̸= 0 and γ1 − γ2 ̸= 0. In this case Ẽ1 = 0 is λ23 a parabola symmetric with respect to some horizontal straight line and Ẽ2 = 0 is a parabola symmetric with respect to some vertical line. Then the maximum number of intersections points is four but we only have three intersection points satisfying 0 < X1 < X2 < X3 and 0 < Y1 < Y2 < Y3 , and so there are at most three limit cycles. Now we provide a discontinuous piecewise linear dierential system in FB which has exactly three crossing limit cycles of type B1 . And this proves that the upper bound found is reached. In the region RB1 we consider the Hamiltonian system r r 483 20003 483 483 37989 ẋ = x+y− , ẏ = − x− y+ , 109 2180 109 109 2180 this system has the rst integral r 483 2 483 37989 20003 H1 (x, y) = − x − xy + x − y2 + y; 218 109 2180 2180 in the region RB2 we consider the Hamiltonian system r r 15 10589 135 15 1983 ẋ = 3 x−y+ , ẏ = x−3 y+ , 157 3140 157 157 628 which has the rst integral r 135 2 15 1983 1 10589 H2 (x, y) = − x +3 xy − x − y2 + y; 314 157 628 2 3140 12 J. JIMENEZ, J. LLIBRE AND C. VALLS and nally in region RB3 we have the Hamiltonian system 22 ẋ = −2x − y + 5, ẏ = 4x + 2y + , 5 this system has the rst integral 22 1 H3 (x, y) = −2x2 − 2xy − x − y 2 + 5y. 5 2 With these system in each region RBi we have that the system of closing equations (10) has exactly three real solutions (Xj , Yj , xj ) such that they satisfy condition (12) and xj < 0, for j = 1, 2, 3, namely       1 7 21 2 5 21 47 3 33 11 (X1 , Y1 , x ) = 2, , − , (X2 , Y2 , x ) = , ,− , (X3 , Y3 , x ) = , 5, − . 2 5 2 5 10 10 2 These three real solutions generated the three crossing limit cycles of type B1 in the family FB in Figure 3a. □ Proof of statement (ii) of Theorem 2. If there exists a periodic solution candidate to be a limit cycle of type B2 , then this periodic solution has four intersection points on the discontinuity curve ΣB of the form (x1 , 0), (x2 , 0), (0, y1 ) and (0, y2 ), satisfying 0 < x1 < x2 , 0 < y1 < y2 . Hence, the following equations must be satised f˜1 = H1 (x2 , 0) − H1 (0, y2 ) = 2δ1 x2 + 2γ1 y2 − λ21 x22 + y22 = 0, f˜2 = H1 (x1 , 0) − H1 (0, y1 ) = 2δ1 x1 + 2γ1 y1 − λ21 x21 + y12 = 0, (13) f˜3 = H2 (0, y2 ) − H2 (0, y1 ) = (y1 − y2 )(2γ2 + y1 + y2 ) = 0, f˜4 = H3 (x2 , 0) − H3 (x1 , 0) = (x1 − x2 )(−2δ3 + λ23 x1 + λ23 x2 ) = 0 We observe that λ3 ̸= 0, since if λ3 = 0 from f˜4 = 0 we get that δ3 = 0 and this is not possible because by Lemma 1 we have that δ3 ̸= λ3 γ3 . From f˜3 = 0 and f˜4 = 0 we get that 2δ3 − λ23 x1 (14) x2 = , y2 = −y1 − 2γ2 . λ23 Using these last expressions we can write the rst and second equations of (13) in terms of x1 and y1 , and we get !   δ3 δ1 λ23 − δ3 λ21 4δ3 λ21 f˜1 (x1 , y1 ) =4 γ22 − γ1 γ2 + + − 2δ1 x1 + 2(2γ2 − γ1 )y1 λ43 λ23 − λ21 x21 + y12 = 0, f˜2 (x1 , y1 ) =2δ1 x1 + 2γ1 y1 − λ21 x21 + y12 = 0. Setting F̃3 (x1 , y1 ) = f˜2 (x1 , y1 ) − f˜1 (x1 , y1 ) = 0 we obtain    λ2 δ 2 F̃3 (x1 , y1 ) = 4 λ23 γ1 γ2 − λ23 γ22 − δ1 δ3 + 1 2 3 − (λ21 δ3 − λ23 δ1 )x1 + λ23 (γ1 − γ2 )y1 = 0. λ3 If γ1 − γ2 = 0 it follows from F̃3 (x1 , y1 ) = 0 that x1 = δ3 /λ23 , but then from (14) we get x2 = δ3 /λ23 = x1 which is not possible. Hence γ1 − γ2 ̸= 0 and solving F̃3 (x1 , y1 ) = 0 in the variable y1 we obtain γ2 λ43 (γ2 − γ1 ) + δ1 δ3 λ23 − δ32 λ21 + λ23 (δ3 λ21 − δ1 λ23 )x1 y1 = . (γ1 − γ2 )λ43 Now we substituting y1 into f˜2 (x1 , y1 ) = 0 we have P (x1 ) = C̃0 + C̃1 x1 + C̃2 x21 = 0, LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 13 where    γ2 λ43 (γ1 − γ2 ) − δ1 δ3 λ23 + δ32 λ21 −λ43 2γ12 − 3γ1 γ2 + γ22 − δ1 δ3 λ23 + δ32 λ21 C̃0 = , λ83 (γ1 − γ2 )2   2δ3 λ43 λ21 (γ1 − γ2 )2 − δ12 + 2δ1 δ3 λ21 λ23 − δ32 λ41 C̃1 = , λ63 (γ1 − γ2 )2  λ43 δ12 − λ21 (γ1 − γ2 )2 − 2δ1 δ3 λ21 λ23 + δ32 λ41 C̃2 = . λ43 (γ1 − γ2 )2 The polynomial P (x1 ) is quadratic in the variable x1 , whose roots x± 1 , are  q  ∆ 2 2δ3 −C̃2 δ3 ± 4 λ3 x± 1 = , C̃1 λ43 where 4   ∆= δ3 λ21 − λ23 (λ1 (γ1 − γ2 ) + δ1 ) δ3 λ21 − λ23 (λ1 (γ2 − γ1 ) + δ1 ) λ83 (γ1 − γ2 )2  γ2 λ43 (2γ1 − γ2 ) − 2δ1 δ3 λ23 + δ32 λ21 . Note that from (14) we get  q  2δ3 δ3 C̃2 + λ3 C̃1 ∓ λ3 ∆ 2 2 2δ3 − λ23 x± 4 x± 2 = 1 = = x∓ 1. λ23 C̃1 λ43 Since x2 > x1 , there is at most one solution of the two possible solutions x±1 . Therefore, the maximum number of limit cycles of type B2 that our discontinuous piecewise linear dierential system can have is one. Now we verify that this upper bound found is reached. We provide a discontinuous piecewise linear dierential system in FB having exactly one crossing limit cycle of type B2 . And this proves that the upper bound found for this case is reached. We consider the linear Hamiltonian system 3 561299 9 3 7276597 ẋ = − x − y + , ẏ = x + y − 2 147900 4 2 1183200 in region RB1 , and this system has the rst integral 9 3 7276597 1 561299 H1 (x, y) = − x2 − xy + x − y2 + y; 8 2 1183200 2 147900 in the region RB2 we consider the Hamiltonian system 5 43 25 5 3 ẋ = x − y + , ẏ = x − y + 8 10 64 8 4 which has the rst integral 25 2 5 3 1 43 H2 (x, y) = − x + xy − x − y 2 + y; 128 8 4 2 10 and in the region RB3 we consider the Hamiltonian system 22 484 22 71027 ẋ = x − y + 5, ẏ = x− y− 9 81 9 4050 this system has the rst integral 242 2 22 71027 1 H3 (x, y) = − x + xy + x − y 2 + 5y. 81 9 4050 2 14 J. JIMENEZ, J. LLIBRE AND C. VALLS With these three systems we have that system (13) has exactly one real solution, namely   8 31 427 11 (x1 , y1 , y2 , x2 ) = , , , , 5 10 100 2 which generated the crossing limit cycle of type B2 in Figure 3b. □ Proof statement (iii) of Theorem 2. We provide a piecewise linear Hamiltonian system in FB which has exactly two limit cycles of type B1 and one limit cycle of type B2 . In region RB1 we consider the Hamiltonian system r r 314270 720859 314270 314270 17072429 ẋ = − x−y+ , ẏ = x+ y− , 223671 213020 223671 223671 4473420 which has the rst integral r 157135 2 314270 17072429 1 720859 H1 (x, y) = − x − xy + x − y2 + y; 223671 223671 4473420 2 213020 in region RB2 we consider the Hamiltonian system q √ 5(782031760005−556339 1459934861005) 4123 18 ẋ = x−y+ , 10651 5 q √ √  5(782031760005−556339 1459934861005) 5 782031760005 − 556339 1459934861005 4123 ẏ = x− y 467728791523 ! 10651 √ 556339 1459934861005 − 782031760005 704 − + , 140670313240 133 this Hamiltonian system has the rst integral √  5 782031760005 − 556339 1459934861005 2 1 2 18 H2 (x, y) = x + y − y s 935457583046 2 5 √  5 782031760005 − 556339 1459934861005 4123 − xy 10651 ! √ 556339 1459934861005 − 782031760005 704 + + x; 140670313240 133 and in the region RB3 we consider the Hamiltonian system 22 484 22 71027 ẋ = x − y − 1, ẏ = x− y− , 9 81 9 4050 which has the rst integral 242 2 22 71027 1 H3 (x, y) = − x + xy + x − y 2 − y. 81 9 4050 2 With these Hamiltonian systems we have that system (10) has exactly the two real solutions   1 36 44 133 (X1 , Y1 , x ) = , ,− , 5 5 100 √ ! 371 1459934861005 + 720859 31 (X2 , Y2 , x2 ) = , ,− , 50 213020 20 LIMIT CYCLES OF DPWLHS WITHOUT EQUILIBRIA SEPARATED BY NON-REGULAR CURVES 15 and system (13) has exactly the one real solution   427 11 8 17 (x1 , y1 , x2 , y2 ) = , , , . 100 2 5 10 These reals solutions generated the two crossing limit cycles of type B1 and one crossing limit cycle of type B2 in FB in Figure 3c. 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A, vol. 69, World Scientic, Singapore, 2010. 1 Universidade Federal do Oeste da Bahia, 46470000 Bom Jesus da Lapa, Bahia, Brazil Email address : jjohanajimenez@gmail.com Departament de Matematiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, 2 Barcelona, Catalonia, Spain Email address : jllibre@mat.uab.cat Departamento de Matemática, Instituto Superior Técnico, Universidade de Lisboa, Av. 3 Rovisco Pais 1049001, Lisboa, Portugal Email address : cvalls@math.ist.utl.pt