This is a preprint of: “Crossing limit cycles for a class of piecewise linear differential centers separated by a conic”, Jeidy Johana Jimenez Ruiz, Jaume Llibre, Joao Carlos Medrado, Electron. J. Differential Equations, vol. 41, 1–36, 2020. CROSSING LIMIT CYCLES FOR A CLASS OF PIECEWISE LINEAR DIFFERENTIAL CENTERS SEPARATED BY A CONIC JOHANA JIMENEZ1 , JAUME LLIBRE2 AND JOÃO C. MEDRADO3 Abstract. These last years the study of the version of Hilbert's 16th problem for piece- wise linear dierential systems in the plane, has increased strongly and there are many papers studying the maximum number of crossing limit cycles when the dierential system is dened in two zones separated by a straight line, in particular in [11, 13] it was proved that piecewise linear dierential centers separated by a straight line have no crossing limit cycles, but in the papers [14, 15] it was shown that the maximum number of crossing limit cycles of piecewise linear dierential centers, can change depending of the shape of the discontinuity curve. In this work we study the maximum number of crossing limit cycles of piecewise linear dierential centers separated by a conic. 1. Introduction and statement of the main results The study of the discontinuous piecewise linear dierential systems in the plane started with Andronov, Vitt and Khaikin in [1], and after these systems have been a topic of great interest in the mathematical community due to their applications in various areas, because they are used for modeling real phenomena and dierent modern devices, see for instance the books [4, 21] and references quoted therein. In the qualitative theory of dierential systems in the plane a limit cycle is a periodic orbit which is isolated in the set of all periodic orbits of the system. This concept was dened by Poincaré [17, 18]. In several papers as [3, 10, 22] it was shown that the limits cycles model many phenomena of the real world. After these works the non-existence, existence, the maximum number and other properties of the limit cycles were extensively studied by mathematicians and physicists and more recently, by biologists, economist and engineers, see for instance [4, 23]. As for the general case of planar dierential systems one of the main problems for the particular case of the piecewise linear dierential centers is to determine the existence and the maximum number of crossing limits cycles that these systems can exhibit. In this paper we study the crossing limit cycles which are periodic orbits isolated in the set of all periodic orbits of the piecewise linear dierential centers, which only have isolated points of intersection with the discontinuity curve. To establish an upper bound for the number of crossing limit cycles for the family of piecewise linear dierential systems in the plane separated by a straight line has been the subject of many recent papers, see for instance [2, 5, 7, 20]. In 1990 Lum and Chua [16] conjectured that the continuous piecewise linear systems in the plane separated by one straight line have at most one limit cycle, in 1998 this conjecture was proved by Freire et al [6]. Afterwards in 2010 Han and Zhang [8] conjectured that discontinuous piecewise linear dierential systems in the plane separated by a straight line have at most two crossing limit cycles but in 2012 Huan and Yang [9] gave a negative answer to this conjecture through a numerical example with three crossing limit cycles, later on Llibre and Ponce in [12] proved 2010 Mathematics Subject Classication. Primary 34C05, 34C07, 37G15. Key words and phrases. discontinuous piecewise linear dierential centers, limit cycles, conics. 1 2 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO the existence of these three limit cycles analytically, but it is still an open problem to know if 3 is the maximum number of crossing limit cycles that this class of systems can have. In the paper [11] it was extended the problem of Lum and Chua to the class of discon- tinuous piecewise linear dierential systems in the plane separated by a straight line, in particular it was proved that the class of planar discontinuous piecewise linear dierential centers, has no crossing limit cycles. However, recently in [14, 15] were studied planar discontinuous piecewise linear dierential centers where the curve of discontinuity is not a straight line, and it was shown that the number of crossing limit cycles in these systems is non-zero. For this reason it is interesting to study the role which plays the shape of the dis- continuity curve in the number of crossing limit cycles that planar discontinuous piecewise linear dierential centers can have. In this paper we provide an upper bound for the maximum number of crossing limit cycles of the planar discontinuous piecewise linear dierential centers separated by a conic Σ. Using an ane change of coordinates, any conic can be written in one of following nine canonical forms: (p): x2 + y 2 = 0 two complex straight lines intersecting at a real point; (CL): x2 + 1 = 0 two complex parallel straight lines; (CE): x2 + y 2 + 1 = 0 complex ellipse; (DL): x2 = 0 one double real straight line; (PL): x2 − 1 = 0 two real parallel straight lines; (LV): xy = 0 two real straight lines intersecting at a real point; (E): x2 +y 2 −1 = 0 ellipse; (H): x2 −y 2 −1 = 0, hyperbola; (P): y − x2 = 0 parabola. We do not consider conics of type (p), (CL) or (CE) because they do not separate the plane in connected regions. We observe that we have two options for crossing limit cycles of discontinuous piecewise linear dierential centers separated by a conic Σ, rst we have the crossing limit cycles such that intersect the discontinuity curve in exactly two points and second we have the crossing limit cycles such that intersect the discontinuity curve Σ in four points, we study these two cases in the following sections. 1.1. Crossing limit cycles intersecting the discontinuity curve Σ in two points. The maximum number of crossing limit cycles of piecewise linear dierential centers sep- arated by a conic Σ such that intersect Σ in exactly two points is given in the following theorems Theorem 1. Consider a planar discontinuous piecewise linear dierential centers where Σ is a conic. If Σ is of the type (a) (LV), (PL) or (DL), then there are no crossing limit cycles. (b) (E), then the maximum number of crossing limit cycles intersecting Σ in two points is two. (a) (P), then the maximum number of crossing limit cycles intersecting Σ in two points is three. Theorem 1 is proved in section 2. In the cases studied up to now, there is no a result determining the maximum number of crossing limit cycles for discontinuous piecewise linear dierential centers when Σ is a hyperbola (H), we determine it in the following Theorem Theorem 2. Consider F0 the family of planar discontinuous piecewise linear dierential centers where Σ is a hyperbola (H). Then the following statement hold. (a) There are systems in F0 without crossing limit cycles. 3 (b) There are systems in F0 having exactly one crossing limit cycle that intersects Σ in two points, see Figure 3. (c) There are systems in F0 having exactly two crossing limit cycles that intersect Σ in two points, see Figure 4. (d) For this family of systems F0 we have that the maximum number of crossing limit cycles that intersect Σ in two points is two. Theorem 2 is proved in section 3. 1.2. Crossing limit cycles intersecting the discontinuity curve Σ in four points. Here we do not consider the case where the discontinuity curve is the conic (DL), because rst in [11, 13] it was proved that discontinuous piecewise linear dierential systems separated by a straight line have no crossing limit cycles and second because the crossing limit cycles of these discontinuous piecewise linear centers cannot have four points on the discontinuity curve. In the following theorems we analyze the maximum number of crossing limit cycles for pla- nar discontinuous piecewise linear dierential centers with four points on discontinuity curve, where the plane is divided by the curve of discontinuity Σ ∈ {(P L), (LV ), (P ), (E), (H)}. Theorem 3. Let F1 be the family of planar discontinuous piecewise linear dierential sys- tems formed by three linear centers and with Σ of type (PL). Then for this family the maximum number of crossing limit cycles that intersect Σ in four points is one. Moreover there are systems in this class having one crossing limit cycle. Theorem 3 for a particular linear center between the two parallel straight lines was done in [13], in section 4 we prove it for any linear center. If the discontinuity curve Σ is of the type (LV), then we have the following regions in the plane: R1 = {(x, y) ∈ R2 : x > 0 and y > 0}, R2 = {(x, y) ∈ R2 : x < 0 and y > 0}, R3 = {(x, y) ∈ R2 : x < 0 and y < 0} and R4 = {(x, y) ∈ R2 : x > 0 and y < 0}. Moreover, 1 ∪ Γ1 ∪ Γ2 ∪ Γ2 , where Γ1 = {(x, y) ∈ R : x = 0, y ≥ 0}, Γ1 = {(x, y) ∈ R : x = − − − Σ = Γ+ + + 2 2 0, y ≤ 0}, Γ2 = {(x, y) ∈ R : y = 0, x ≥ 0} and Γ2 = {(x, y) ∈ R : y = 0, x ≤ 0}. In + 2 − 2 this case we have two types of crossing limit cycles, namely crossing limit cycles of type 1 which intersect only two branches of Σ in exactly two points in each branch, and crossing limit cycles of type 2 which intersect in a unique point each branch of the set Σ. Theorem 4. Let F2 be the family of planar discontinuous piecewise linear dierential sys- tems formed by four linear centers and with Σ of the type (LV). The maximum number of crossing limit cycles type 1 is one. Moreover there are systems in this class having one crossing limit cycle. Theorem 4 is proved in Section 5. Theorem 5. Consider the family of planar discontinuous piecewise linear dierential cen- ters F2 . Then the following statement hold. (a) There are systems in F2 with exactly one crossing limit cycle of type 2, see Figure 8. (b) There are systems in F2 with exactly two crossing limit cycles of type 2, see Figure 9. (c) There are systems in F2 with exactly three crossing limit cycles of type 2, see Figure 10 Theorem 5 is proved in Section 6. 4 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO By the calculations made for this case and the illustrated examples in Theorem 5 we get the following conjecture. Conjecture 1. For the family of systems F2 the maximum number of crossing limit cycles of type 2 is three. Theorem 6. Let F3 be the family of planar discontinuous piecewise linear dierential sys- tems formed by two linear centers and with Σ of type (P). Then for this family the maximum number of crossing limit cycles that intersect Σ in four points is one. Moreover there are systems in this class having one crossing limit cycle. Theorem 6 is proved in Section 7. Theorem 7. Let F4 be the family of planar discontinuous piecewise linear dierential sys- tems formed by two linear centers and with Σ of type (E). Then for this family the maximum number of crossing limit cycles that intersect Σ in four points is one. Moreover there are systems in this class having one crossing limit cycle. Theorem 7 is proved in Section 8. Theorem 8. Let F5 be the family of planar discontinuous piecewise linear dierential sys- tems formed by three linear centers and with Σ of type (H). Then for this family the maximum number of crossing limit cycles that intersect Σ in four points is one. Moreover there are systems in this class having one crossing limit cycle. Theorem 8 is proved in Section 9. 1.3. Crossing limit cycles with four and with two points on the discontinuity curve Σ simultaneously. Here we study the maximum number of crossing limit cycles of planar discontinuous piecewise linear dierential centers that intersect the discontinuity curve Σ in two and in four points simultaneously. We do not consider planar discontinuous piecewise linear dierential centers with discon- tinuity curve a conic of type (DL), (PL) and (LV) because as in the proof of Theorem 1 they do not have crossing limit cycles that intersect the discontinuity curve in two points. Then we study the maximum number of crossing limit cycles with two and with four points in Σ simultaneously by the families F3 , F4 and F5 . Theorem 9. The following statements hold. (a) The planar discontinuous piecewise linear dierential centers that belong to the fam- ily F3 , can have simultaneous one crossing limit cycle that intersects (P) in two points and one crossing limit cycle that intersects (P) in four points. (b) The planar discontinuous piecewise linear dierential centers that belong to the fam- ily F4 , can have simultaneous one crossing limit cycle that intersects (E) in two points and one crossing limit cycle that intersects (E) in four points. (c) The planar discontinuous piecewise linear dierential centers that belong to the fam- ily F5 , can have simultaneous one crossing limit cycle that intersects (H) in two points and one crossing limit cycle that intersects (H) in four points. Theorem 9 is proved in Section 10. In Subsection 1.3 we do not consider the planar discontinuous piecewise linear dierential centers in the family F2 , because they do not have crossing limit cycles that intersect the discontinuity curve (LV) in two points. However in this family there are two types of crossing limit cycles like it was dened in Subsection 1.2. 5 1.4. Crossing limit cycles of types 1 and 2 simultaneously for planar discontinuous piecewise linear dierential centers in F2 . In this case we study the maximum number of crossing limit cycles of types 1 and 2 that planar discontinuous piecewise linear dierential centers in the family F2 can have simultaneously. Theorem 10. The following statements hold. (a) There are planar discontinuous piecewise linear dierential centers that belong to the family F2 such that have one crossing limit cycle of type 1 and one crossing limit cycle of type 2 simultaneously. (b) There are planar discontinuous piecewise linear dierential centers that belong to the family F2 such that have one crossing limit cycle of type 1 and two crossing limit cycles of type 2 simultaneously. Theorem 10 is proved in Section 11. By the illustrated examples in Theorem 10 we get the following conjecture. Conjecture 2. The planar discontinuous piecewise linear dierential centers that belong to the family F2 can have at most one crossing limit cycle of type 1 and two crossing limit cycles of type 2 simultaneously. 2. Proof of Theorem 1 Analyzing the case of discontinuous piecewise linear dierential centers with discontinuity curve a conic of the type (LV), (PL) or (DL) the maximum number of crossing limit cycles is equal to the maximum number of crossing limit cycles in discontinuous piecewise linear dierential centers in the plane separated by a single straight line which was studied in [11, 13]. In these papers it was proved that the discontinuous piecewise linear dierential centers separated by one straight line have no crossing limit cycles. This proves the statement (a) of Theorem 1. In the paper [15] the authors considered discontinuous piecewise linear dierential centers separated by the parabola y = x2 and proved that they have at most three crossing limit cycles that intersect Σ in two points, i.e. statement (b) of Theorem 1. Moreover they provide the following system. Figure 1. The three limit cycles of the discontinuous piecewise linear dif- ferential of Example 1. Example 1. Consider the discontinuous piecewise linear dierential system in R2 separated by the parabola (P) when both linear dierential systems are centers and are dened as 6 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO follows: 18 99 ẋ = − − y, ẏ = x − , in R1 = {(x, y) : y ≥ x2 }, 5 50 189 38221 720042289 41 189 3296401 ẋ = − x+ y− , ẏ = − x + y− , in R2 = {(x, y) : y ≤ x2 }. 100 20500 205000000 20 100 1000000 This discontinuous piecewise linear dierential system has the three crossing limit cycles of Figure 1. With regard to the discontinuous piecewise linear dierential systems separated by an ellipse, in the paper [14] the authors shown that the class of planar discontinuous piecewise linear dierential centers separated by the circle S1 has at most two crossing limit cycles. Moreover, there are discontinuous piecewise linear dierential centers which reach the upper bound of 2 crossing limit cycles, see Example 2. Then we have the statement (c) of Theorem 1. Example 2. We consider the discontinuous piecewise linear dierential system in R2 sep- arated by the ellipse (E) and both linear dierential centers are dened as follows: √ √ ẋ = −2x − 2y − 2 − 1, ẏ = 4x + 2y + 2, in the unbounded region limited by the ellipse (E), and in the bounded region with boundary the ellipse (E) we have the linear dierential center 5 1 1 1 ẋ = x − y + √ + , ẏ = x − y − √ . 4 2 8 2 This discontinuous piecewise dierential system has exactly two crossing limit cycles, see Figure 2. Figure 2. The two limit cycles of the discontinuous piecewise linear dier- ential of Example 2. 3. Proof of Theorem 2 For the systems of the class F0 we have following regions in the plane R1 = {(x, y) ∈ R2 : x2 − y 2 > 1}, which is a region that consist of two connected components. And the region R2 = {(x, y) ∈ R2 : x2 − y 2 < 1}. 7 In order to have a crossing limit cycle, which intersects the hyperbola x2 − y 2 = 1 in two dierent points p = (x1 , y1 ) and q = (x2 , y2 ), these points must satisfy the closing equations H1 (x1 , y1 ) = H1 (x2 , y2 ), H2 (x2 , y2 ) = H2 (x1 , y1 ), (1) x21 − y12 = 1, x22 − y22 = 1. Proof of statement (a) of Theorem 2. We consider a discontinuous piecewise linear dieren- tial system which has the linear center (2) ẋ = −y, ẏ = x, in the region R2 , the orbits of this center intersect the hyperbola in two or in four points, when it intersects the hyperbola in exactly two points these are (±1, 0), which are points of tangency between the hyperbola and the solution curves of the center (2), then it is impossible that there are crossing periodic orbits independent of the linear dierential center that can be considered in the region R1 . So the orbits which can produce a crossing limit cycle intersect the hyperbola in four points and clearly these orbits cannot be crossing limit cycles with exactly two points on the discontinuity curve (H).  Proof of statement (b) of Theorem 2. In the region R1 we consider the linear dierential center √ (3) ẋ = 27 − 5 − 25y, ẏ = −2 + x, √ this system has the rst integral H1 (x, y) = 4(−4 + x)x + 4y(−54 + 2 5 + 25y). In the region R2 we have the linear dierential center √ 3 5 y 3 (4) ẋ = 2 − − , ẏ = − + x, 4 4 2 √ which has the rst integral H2 (x, y) = 4(−3 + x)x + y(−16 + 6 5 + y). This discontinuous Figure 3. The crossing limit cycle of the discontinuous piecewise linear dierential system formed by the centers (3) and (4). piecewise dierential system formed by the linear dierential centers (3) and (4) has one √ real solution (p, q) with p 6= q of the closing equations crossing limit cycle, because the unique given in (1), is p = (1, 0) and q = ( 5, 2). See the crossing limit cycle of this system in Figure 3.  8 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO Proof of statement (c) of Theorem 2. In the region R1 we consider the linear dierential center √ √ √ 289 − 48 2 + 289 3 − 305 6 x 49 ẋ = + √ − y, 768  192 (5) √ √  p √ 8 3 32 3 − 289 2 1 + 2 + 3 y ẏ = +x− √ , 768 8 3 which has the rst integral    q   1 2 √ √  √ √ H1 (x, y) = 384x + x 32 3 − 289 2 1 + 2 + 3 − 32 3y 96 r  !! √  √ + y 98y − 3 85057 − 9248 6 + 305 6 − 289 . In the region R2 we have the linear dierential center 1 √ √ √  x y 1 √ √  y (6) ẋ = −3 + 8 2 + 3 − 6 − − , ẏ = −1 − 5 2 − 3 + x + , 8 2 2 8 2 this system has the rst integral  √ √   √ √ √  H2 (x, y) = 4x2 − x 1 + 5 2 + 3 − 4y + y 3 − 8 2 − 3 + 6 + 2y . This discontinuous piecewise dierential system formed by the linear dierential centers (5) Figure 4. The two limit cycles of the discontinuous piecewise linear dier- ential system formed by the centers (5) and (6). and (6) has √ two crossing √ limit√cycles, √ because the unique real solutions (p, q) of system (1) are 1, 0, 2, 1 and 2, −1, 3, 2 , therefore the intersection √  points √ ofthe√two√crossing  limit cycles with the hyperbola are the pairs (1, 0), 2, 1 and 2, −1 , 3, 2 . See these two crossing limit cycles in Figure 4.  We will use the following lemma which provides a normal form for an arbitrary linear dierential center, for a proof see [13]. Lemma 11. Through a linear change of variables and a rescaling of the independent variable every center in R2 can be written 4b2 + ω 2 (7) ẋ = −bx − y + d, ẏ = ax + by + c, 4a with a 6= 0 and ω > 0. This system has the rst integral (8) H1 (x, y) = 4(ax + by)2 + 8a(cx − dy) + y 2 ω 2 . 9 Proof of statement (d) of Theorem 2. In the region R1 we consider the arbitrary linear dif- ferential center (7) which has rst integral (8). In the region R2 we consider the arbitrary linear dierential center 4B 2 + Ω2 (9) ẋ = −Bx − y + D, ẏ = Ax + By + C, 4A with A 6= 0 and Ω > 0. Which has the rst integral H2 (x, y) = 4(Ax + By)2 + 8A(Cx − Dy) + y 2 Ω2 . It is possible to do a rescaling of time in the two above systems. Suppose τ = at in R1 and s = At in R2 . These two rescaling change the velocity in which the orbits of systems (7) and (9) travel, nevertheless they do not change the orbits, therefore they will not change the crossing limit cycles that the discontinuous piecewise linear dierential system can have. After these rescalings of the time we can assume without loss of generality that a = A = 1, and the dot in system (7) (resp. (9)) denotes derivative with respect to the new time τ (resp. s). We assume that the discontinuous piecewise linear dierential system formed by the two linear dierential centers (7) and (9) has three crossing periodic solutions. For this we must impose that the system of equations (1) has three pairs of points as solution, namely (pi , qi ), i = 1, 2, 3, since these solutions provide crossing periodic solutions. We consider (10) pi = (cosh ri , sinh ri ) and qi = (cosh si , sinh si ), for i = 1, 2, 3. These points are the points where the three crossing periodic solutions intersect the hyper- bola (H). Now we consider that the point (p1 , q1 ) satises system (1) and with this condition we obtain the following expression 1 d= 4 cosh2 r1 − 4 cosh2 s1 + 8 cosh r1 (c + b sinh r1 ) 8(sinh r1 − sinh s1 )  − 8 cosh s1 (c + b sinh s1 ) + (4b2 + ω 2 )(sinh2 r1 − sinh2 s1 ) , and D has the same expression that d changing (b, c, ω) by (B, C, Ω). We assume that the point (p2 , q2 ) satises system (1) and we get the expression −1 c= ((sinh r2 − sinh s2 ) 8(sinh(r1 − r2 ) + sinh(r2 − s1 ) − sinh(r1 − s2 ) + sinh(s1 − s2 ))  4 cosh2 s1 + 4b sinh(2s1 ) − 4 cosh2 r1 − 4b sinh(2r1 ) + (sinh r1 − sinh s1 ) 4 cosh2 r2 − 4 cosh2 s2 + 8b cosh r2 sinh r2 − 8b cosh s2 sinh s2 + (4b2 + ω 2 )(sinh r2 − sinh s2 ) (− sinh r1 + sinh r2 − sinh s1 + sinh s2 ))) , and C has the same expression that c changing (b, ω) by (B, Ω). 10 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO Finally we are going to impose that the point (p3 , q3 ) satises system (1) and we get an expression for ω 2 . In this case ω 2 = K/L, where the expression of K is        r1 − r2 + s1 − s2 r3 − s3 r1 − r2 − r3 + s1 − s2 − 3s3 4 (1 + b )csch 2 sinh cosh 2 2 2     r1 − r2 − r3 + s1 − 3s2 − s3 r1 − r2 − 3r3 + s1 − s2 − s3 − cosh + cosh 2 2     r1 − 3r2 − r3 + s1 − s2 − s3 3r1 + r2 − r3 + s1 + s2 − s3 − cosh − cosh 2 2     r1 + 3r2 − r3 + s1 + s2 − s3 r1 + r2 − r3 + 3s1 + s2 − s3 + cosh − cosh 2 2      r1 + r2 − r3 + s1 + 3s2 − s3 2 3r1 − r2 + r3 + s1 − s2 + s3 + cosh + b cosh 2 2      r1 − r2 + 3r3 + s1 − s2 + s3 r1 − r2 − r3 + s1 − s2 − 3s3 − cosh − 2b sinh 2 2     r1 − r2 − r3 + s1 − 3s2 − s3 2r1 − r2 + 2r3 + s1 − s2 + s3 − sinh − sinh 2 2       r1 − r3 2r1 + 2r2 − r3 + s1 + s2 − s3 r1 − r2 sinh + 2 cosh sinh 2 2 2       2r1 − r2 + 2r3 + s1 − s2 + s3 r1 − r3 r2 − r3 − 2 cosh sinh + 2 sinh 2 2 2     r1 − 2r2 − 2r3 + s1 − s2 − s3 r1 + r2 − r3 + 2(s1 + s2 ) − s3 cosh + 2 cosh 2 2       s1 − s2 r1 − r2 + r3 + 2s1 − s2 s1 − s3 sinh − 2 cosh + s3 sinh 2 2 2     r1 − r2 + r3 + 2s1 − s2 s1 − s3 + 2(1 + b2 ) sinh + s3 sinh 2 2 and the expression of L is       r1 − r2 + s1 − s2 r3 − s3 r1 − r2 − r3 + s1 − s2 − 3s3 csch sinh − cosh 2 2 2     r1 − r2 − r3 + s1 − 3s2 − s3 r1 − r2 − 3r3 + s1 − s2 − s3 + cosh − cosh 2 2     r1 − 3r2 − r3 + s1 − s2 − s3 3r1 + r2 − r3 + s1 + s2 − s3 + cosh + cosh 2 2     r1 + 3r2 − r3 + s1 + s2 − s3 r1 + r2 − r3 + 3s1 + s2 − s3 − cosh + cosh 2 2     r1 + r2 − r3 + s1 + 3s2 − s3 3r1 − r2 + r3 + s1 − s2 + s3 − cosh − cosh 2 2     r1 − r2 + 3r3 + s1 − s2 + s3 r1 − r2 + r3 + 2s1 − s2 + cosh − sinh + s3 2 2   s1 − s3 2 sinh , 2 and the expression of Ω2 is the same than the expression of ω 2 changing b by B . 11 Now we replace d, c, ω 2 in the expression of the rst integral H1 (x, y) and we have (11) H1 (x, y) = 4(x2 − y 2 ) + h(x, y, r1 , r2 , r3 , s1 , s2 , s3 )b, and analogously we have (12) H2 (x, y) = 4(x2 − y 2 ) + h(x, y, r1 , r2 , r3 , s1 , s2 , s3 )B. Now we are going to analyze if the discontinuous piecewise linear dierential system formed by (7) and (9) has more crossing periodic solutions than the three supposed in (10). Taking into account (11) and (12) the closing equations (1) becomes h(x1 , y1 , r1 , r2 , r3 , s1 , s2 , s3 ) = h(x2 , y2 , r1 , r2 , r3 , s1 , s2 , s3 ), (13) x21 − y12 = 1, x22 − y22 = 1. This is, we must solve a system with three equations and four unknowns x1 , y1 , x2 , y2 , that we know that have at least the three solutions (10), so system (13) has a continuum of solutions which produce a continuum of crossing periodic solutions, so such systems cannot have crossing limit cycles. Since in statement (c), we have proved that there are systems in F0 with two crossing limit cycles, it follows that the maximum number of crossing limit cycles that intersect Σ in two points is two. This completes the proof of Theorem 2.  4. Proof of Theorem 3 In the case where Σ is of the type (PL), we have following three regions in the plane: R1 = {(x, y) ∈ R2 : x < −1}, R2 = {(x, y) ∈ R2 : −1 < x < 1} and R3 = {(x, y) ∈ R2 : x > 1}. We consider a planar discontinuous piecewise dierential system separated by two parallel straight lines and formed by three arbitrary linear centers. By Lemma 11, we have that these linear centers can be as follows 4b2 + ω 2 ẋ = −bx − y + d, ẏ = ax + by + c, in R1 , 4a 4B 2 + Ω2 (14) ẋ = −Bx − y + D, ẏ = Ax + By + C, in R2 , 4A 4β 2 + λ2 ẋ = −βx − y + δ, ẏ = αx + βy + γ, in R3 . 4α These linear centers have the rst integrals H1 (x, y) = 4(ax + by)2 + 8a(cx − dy) + y 2 ω 2 , H2 (x, y) = 4(Ax + By)2 + 8A(Cx − Dy) + y 2 Ω2 , H3 (x, y) = 4(αx + βy)2 + 8α(γx − δy) + y 2 λ2 , respectively. We are going to analyze if the discontinuous piecewise linear dierential center (14) has crossing periodic solutions. Since the orbits in each region Ri , for i = 1, 2, 3, are ellipses or pieces of one ellipse, we have that if there is a crossing limit cycle this must intersect each straight line x = ±1 in exactly two points, namely (1, y1 ), (1, y2 ) and (−1, y3 ), (−1, y4 ), with 12 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO y1 > y2 and y3 > y4 . Therefore we must study the solutions of the system H3 ( 1, y2 ) = H3 ( 1, y1 ), H2 ( 1, y1 ) = H2 (−1, y3 ), H1 (−1, y3 ) = H1 (−1, y4 ), H2 (−1, y4 ) = H2 ( 1, y2 ), or equivalently, we have the system −(y1 − y2 )(8β − 8δ + (4β 2 + λ2 )(y1 + y2 ) = 0, 16C − 8D(y1 − y3 ) + 8B(y1 + y3 ) + (4B 2 + Ω2 )(y12 − y32 ) = 0, (15) (y3 − y4 )(−8b − 8d + (4b2 + ω 2 )(y3 + y4 ) = 0, −16C + 8D(y2 − y4 ) − 8B(y2 + y4 ) − (4B 2 + Ω2 )(y22 − y42 ) = 0. By hypothesis y1 > y2 and y3 > y4 and therefore system (15) is equivalently to the system γ3 − δ3 + l3 (y1 + y2 ) = 0, η − δ2 (y1 − y3 ) + γ2 (y1 + y3 ) + l2 (y12 − y32 ) = 0, (16) −γ1 − δ1 + l1 (y3 + y4 ) = 0, −η + δ2 (y2 − y4 ) − γ2 (y2 + y4 ) − l2 (y22 − y42 ) = 0, where γ1 = 8b, γ2 = 8B, γ3 = 8β, δ1 = 8d, δ2 = 8D, δ3 = 8δ, l1 = 4b2 +ω 2 , l2 = 4B 2 +Ω2 , l3 = 4β 2 + λ2 and η = 16C. As l1 6= 0 and l3 6= 0, we can isolated y1 and y4 of the rst and the third equations of system (16), respectively. Then, we obtain −l3 y2 + γ3 − δ3 −l1 y3 + γ1 + δ1 y1 = , y4 = . l3 l1 Now replacing these expressions of y1 and y4 in the second and fourth equations of (16), we have the system of two equations l2 (l3 (y2 − y3 ) + ψ3 )(l3 (y2 + y3 ) + ψ3 ) + l3 (l3 (η + (y3 − y2 )γ2 + (y2 + y3 )δ2 ) − ψ2 ψ3 ) E1 = , l32 l2 ψ12 − l1 ψ1 (2l2 y3 + γ2 + δ2 ) − l12 (η + (y2 − y3 )(l2 (y2 + y3 ) + γ2 ) − (y2 + y3 )δ2 ) E2 = . l12 Doing the Groebner basis of the two polynomials E1 and E2 with respect to the variables y2 and y3 , we obtain the equations (17) m0 + m1 y3 + m2 y32 = 0, k0 + k1 y3 + k2 y2 = 0, with 1 m0 = 2l13 l33 ψ22 (l3 ψ1 (γ2 + δ2 ) + l1 (2l3 η − ψ2 ψ3 )) − l12 l2 l32 (l32 ψ12 (γ22 − 6γ2 δ2 + δ22 ) + 4l1 l3 ψ2 l14 l32  (2l1 η + ψ1 (γ2 + δ2 ))ψ3 − 5l12 ψ22 ψ32 + 2l1 l22 l3 (−l33 ψ13 (γ2 + δ2 ) + 2l1 l32 ψ12 ψ2 ψ3 − 2l13 ψ2 ψ33 )  + l12 l3 (2l1 η + ψ1 (γ2 + δ2 ))ψ32 +l23 (l3 ψ1 + l1 ψ3 )2 (l3 ψ1 − l1 ψ3 )2 , 4l2 ψ1 (−2l1 l3 ψ1 + l1 ψ3 ))(2l1 l3 δ2 − l2 (l3 ψ1 − l1 ψ3 )) m1 = , l13 4l2 (l2 (l3 ψ1 + l1 ψ3 − 2l1 l3 γ2 ))(l2 (l3 ψ1 − l1 ψ3 − 2l1 l2 δ2 )) m2 = , l12 (l1 l3 (l3 ψ1 (γ2 + δ2 )ψ12 + l12 ψ32 )) (2l3 (l2 ψ1 − l1 (γ2 + δ2 )) k0 = 2 , k1 = , k2 = 2(l3 ψ2 − l2 ψ3 ), l1 l3 l1 where ψ1 = γ1 + δ1 , ψ2 = γ2 − δ2 and ψ3 = γ3 − δ3 . We recall that Bézout Theorem states that if a polynomial dierential system of equations has nitely many solutions, then the number of its solutions is at most the product of the degrees of the polynomials which appear in the system, see [19]. Then by Bézout Theorem 13 in this case, we have that system (17) has at most two solutions. Moreover, from these two solutions (y21 , y31 ) and (y22 , y32 ) of (17), we will have two solutions of (16) which are of the form (y11 , y21 , y31 , y41 ) and (y12 , y22 , y32 , y42 ), but analyzing system (16) we have that if (y11 , y21 , y31 , y41 ) is a solution, then (y21 , y11 , y41 , y31 ) is another solution. Finally due to the fact that y1 > y2 and y3 > y4 , at most one of these two solutions will be satisfactory. Therefore we have proved that the planar discontinuous piecewise dierential systems of the family F1 , can have at most one crossing limit cycle. Now we verify that this upper bound is reached, for this we present a discontinuous piecewise linear dierential system that belongs to the family F1 and has exactly one crossing limit cycle. We consider the discontinuous piecewise linear dierential centers 27 x 5 1 y ẋ = − − − y, ẏ = +x+ , in R1 , 64 2 16 16 2 43 x 5 31 y (18) ẋ = − − y, ẏ = − + x + , in R2 , 64 4 16 128 4 1 x 13 1 y ẋ = − − y, ẏ = + x + , in R3 . 3 3 36 7 3 These systems have the rst integrals   21 1 H1 (x, y) = 4x + x + 4y + y(27 + 10y), 2 8   31 1 H2 (x, y) = 4x2 + x − + 2y + y(−43 + 10y), 16 8 8 8 4 H3 (x, y) = x − y + y 2 + (3x + y)2 , 7 3 9 respectively. Then the discontinuous piecewise dierential system formed by the linear Figure 5. The crossing limit cycle of the discontinuous piecewise linear dierential system (18) with three centers separated by the conic (PL). dierential centers (18) has one crossing limit cycle that intersects (PL) in four points, because the unique real solution (y1 , y2 , y3 , y4 ) with y1 > y2 and y3 > y4 of system (15) is the point (y1 , y2 , y3 , y4 ) = (1.5, −1.5, 1, −0.5). See the crossing limit cycle of this system in Figure 5. This completes the proof of Theorem 3. 5. Proof of Theorem 4 We consider a planar discontinuous piecewise linear dierential system with four zones separated by (LV) and formed by four arbitrary linear centers. By Lemma 11 this piecewise 14 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO linear dierential system can be as follows 4b21 + ω12 ẋ = −b1 x − y + d1 , ẏ = a1 x + b1 y + c1 , in R1 , 4a1 4b22 + ω22 ẋ = −b2 x − y + d2 , ẏ = a2 x + b2 y + c2 , in R2 , 4a2 (19) 4b23 + ω32 ẋ = −b3 x − y + d3 , ẏ = a3 x + b3 y + c3 , in R3 , 4a3 4b24 + ω42 ẋ = −b4 x − y + d4 , ẏ = a4 x + b4 y + c4 , in R4 , 4a4 with ai 6= 0 and ωi > 0 for i = 1, 2, 3, 4. The regions Ri for i = 1, 2, 3, 4 are dened just before the statement of Theorem 4. These linear dierential centers have the rst integrals H1 , H2 , H3 and H4 respectively, where (20) Hi (x, y) = 4(ai x + bi y)2 + 8ai (ci x − di y) + y 2 ωi2 , for i = 1, 2, 3, 4. If the discontinuous piecewise linear center (19) has two crossing limit cycles of type 1, these two crossing limit cycles should be some of Figure 6. We observe that the cases of Figures (a) (b) (c) (d) Figure 6. Possible cases of two crossing limit cycles of type 1 of discontin- uous piecewise linear center (19). 6b, 6c and 6d are not possible because in these cases the pieces of the ellipses of linear dierential centers in the regions R4 , R1 and R2 , respectively would not be nested which contradicts that the linear dierential systems in each of these regions are linear centers. Therefore if the discontinuous piecewise linear center (19) has two crossing limit cycles of type 1 these could be as in Figure 6a. Now we going to study the conditions in order that the piecewise linear dierential center (19) has crossing limit cycles of type 1 and we will show that the maximum number of crossing limits cycles of type 1 is one. Without loss of generality we assume that the crossing limit cycles intersect the branches Γ+ 1 and Γ2 in the points (0, y1 ), (0, y2 ) and (x1 , 0), (x2 , 0), + respectively, where 0 < y1 < y2 and 0 < x1 < x2 . Then taking into account the rst integrals (20) for each linear center, these points must satisfy the following equations H1 (x2 , 0) = H1 (0, y2 ), H2 (0, y2 ) = H2 (0, y1 ), H1 (0, y1 ) = H1 (x1 , 0), H4 (x1 , 0) = H4 (x2 , 0), 15 equivalently we have 4a21 x22 + 8a1 (c1 x2 + d1 y2 ) − y22 l1 = 0, −(y1 − y2 )(−8a2 d2 + (y1 + y2 )l2 ) = 0, (21) −4a21 x21 − 8a1 (c1 x1 + d1 y1 ) + y12 l1 = 0, 4a4 (x1 − x2 )(2c4 + a4 (x1 + x2 )) = 0, where l1 = 4b21 + ω12 , l2 = 4b22 + ω22 and η = (a4 c1 − a1 c4 ). Moreover, by hypothesis x1 < x2 and y1 < y2 , then from the second and the fourth equations of (21), we have 8a2 d2 − l2 y2 2c4 + a4 x1 y1 = , x2 = − . l2 a4 Substituting these expressions of y1 and x2 in the rst and third equations of (21) we obtain the two equations 4a21 (2c4 + a4 x1 )2 − 8a1 a4 (2c1 c4 + a4 c1 x1 − a4 d1 y2 ) − a24 y22 l1 E1 = , a24   l1 (y2 l2 − 8a2 d2 )2 8a2 d1 d2 E2 = 4a21 x21 − − 8a 1 d y 1 2 − − c x 1 1 . l22 l2 Doing the Groebner basis of the two polynomials E1 and E2 with respect to the variables x1 and y2 we get the two equations (22) α0 + α1 y2 + α2 y22 = 0, β0 + β1 x1 + β2 y2 = 0, where  a1 c4 η 2 (−2a4 c1 + a1 c4 ) 16a32 a24 d32 l1 (a2 d2 l1 − 2a1 d1 l1 l2 ) α0 =4 + + (8a2 d2 (a1 η 2 d1 l2 a24 l24  2 2 2 2 1 + a2 d2 (2a1 a4 d1 − η l1 )) 2 , l2 8a2 d2 α1 = −32a22 a24 b21 d22 (2b21 + ω12 ) − 4a22 a24 d22 ω14 + 8a1 a2 a24 d1 d2 l1 l2 + a24 c21 l22 l12 l23  − 2a1 a4 c1 c4 l1 l22 + a21 (−4a24 d21 + c24 l1 l2 ,   2 2 2 2 2 2 4a22 d22 l12 8a1 a2 d1 d2 l1 α2 =2a1 a4 c1 c4 l1 − a1 c4 l1 + a4 −c1 l1 + 4a1 d1 + − , l22 l2 a1 c4 η 4a22 a4 d22 l1 4a1 a2 a4 d1 d2 a2 a4 d2 l1 β0 = − + 2 − , β1 = −a1 η, β2 = a1 a4 d1 − . a4 l2 l2 l2 The Bézout Theorem (see [19]) applied to system (22) says that this system has at most two isolated solutions. Therefore system (21) has two solutions which are of the form (x11 , x12 , y11 , y21 ) and (x21 , x22 , y12 , y22 ), but it is possible to prove that if (x1 , x2 , y1 , y2 ) is a solution of system (21), then (x2 , x1 , y2 , y1 ) is also a solution of this system. Since we must have that x1 < x2 and y1 < y2 , then system (21) has a unique solution, and therefore the discontinuous piecewise linear dierential center (19) that belongs to the family F2 can have at most one crossing limit cycle of type 1 intersecting Γ+ 1 and Γ2 . + Now we verify that this upper bound is reached. That is, that there are piecewise linear dierential centers in the family F2 having one crossing limit cycle of type 1. We consider 16 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO the following discontinuous piecewise linear dierential center 12049 325 2113 ẋ = − 3x − y, ẏ = − + 3x + 3y, in R1 , 2160 108 540 153 x 17 9 x y ẋ = + − y, ẏ = + − , in R2 , (23) 64 2 16 16 4 2 ẋ = −y − β, ẏ = x + α, in R3 , 577 577 20 10 ẋ = − + 2x − y, ẏ = − + x − 2y, in R4 . 192 480 3 3 In the region R3 we can consider any linear dierential center, because the crossing limit cycle will be formed by parts of the orbits of the centers of the regions R1 , R2 and R4 . The Figure 7. The crossing limit cycle of type 1 of the discontinuous piecewise linear dierential system (23) separated by the conic (LV). centers in (23) have the rst integrals   2 4226 1 H1 (x, y) = 36x + x − + 72y + y(−12049 + 3250y), 45 90 1  H2 (x, y) = 8x2 + 4x (9 − 8y) + 17y(−9 + 2y) , 32 H3 (x, y) = (x + α)2 + (y + β)2 , 1  H4 (x, y) = 1600x2 + 577y(5 + y) − 640x(10 + 3y) , 36 in Ri , i = 1, 2, 3, 4, respectively. Then for the discontinuous piecewise linear dierential center (23) system (21) becomes 1  −8452x2 + 3240x22 + (12049 − 3250y2 )y2 = 0, 90 17 − (y1 − y2 )(−9 + 2y1 + 2y2 ) = 0, (24) 4226 32 1 x1 − 36x21 + y1 (−12049 + 3250y1 ) = 0, 45 90 400 (x1 − x2 )(−4 + x1 + x2 ) = 0. 9 Taking into account that the solutions (x1 , x2 , y1 , y2 ) must satisfy x1 < x2 and y1 < y2 , we have that the unique solution of system (24) is the point (x1 , x2 , y1 , y2 ) = (1, 3, 0.5, 4). See the crossing limit cycle of type 1 of the piecewise linear dierential centers (23) in Figure 7. This completes the proof of Theorem 4. 17 6. Proof of Theorem 5 Proof of statement (a) of Theorem 5. In the region R1 we consider the linear dierential center 13 x y y (25) ẋ = − − − , ẏ = 1 + x + , 4 2 2 2 this system has the rst integral H1 (x, y) = 2(2x2 + 2x(2 + y) + y(13 + y)). In the region R2 we have the linear dierential center 851 x 181 3 y (26) ẋ = − − − y, ẏ = +x+ , 3600 3 900 2 3 which has the rst integral H2 (x, y) = 4x2 + 4x(9 + 2y)/3 + y(851 + 362y)/450. In the region R3 we have the linear dierential center 43 x 5 1 y (27) ẋ = − + − y, ẏ = − + x − , 32 4 16 2 4 which has the rst integral H3 (x, y) = 4x2 − 3x(2 + y) + y(−43 + 5y)/4. And in the region R4 we have the linear dierential center 137 x 25 3 y (28) ẋ = + − y, ẏ = +x− , 72 3 144 2 3 which has the rst integral H4 (x, y) = 4x(3 + x) − (137 + 24x)y/9 + 25y 2 /36. In order to have a crossing limit cycle of type 2, which intersects the discontinuity conic (LV) in four dierent points p1 = (x1 , 0), q1 = (0, y1 ), p2 = (x2 , 0) and q2 = (0, y2 ), with x1 , y1 > 0 and x2 , y2 < 0, these points must satisfy the closing equations e1 = H1 (x1 , 0) − H1 (0, y1 ) = 0, e2 = H2 (0, y1 ) − H2 (x2 , 0) = 0, (29) e3 = H3 (x2 , 0) − H3 (0, y2 ) = 0, e4 = H4 (0, y2 ) − H4 (x1 , 0) = 0. Considering the four above linear dierential centers (25), (26), (27) and (28) and their respective rst integrals Hi (x, y), i = 1, 2, 3, 4, we have the following equivalent system 4x1 (2 + x1 ) − 2y1 (13 + y1 ) = 0, 1 −4x2 (3 + x2 ) + y1 (851 + 361y1 ) = 0, 450 (30) 1 4(x2 − 1)x2 + (43 − 5y2 )y2 = 0, 4 1 −4x1 (x1 + 3) + y2 (−548 + 25y2 ) = 0, 36 the unique real solution (p1 , q1 , p2 , q2 ) of (30) is p1 = (3, 0), q1 = (0, 2), p2 = (−7/2, 0) and q2 = (0, −4), therefore the piecewise dierential system formed by the linear dierential centers (25), (26), (27) and (28) has exactly one crossing limit cycle of type 2. See the crossing limit cycle of this system in Figure 8.  Proof of statement (b) of Theorem 5. In the region R1 we consider the linear dierential center 7 x y 7 y (31) ẋ = − − , ẏ = +x+ , 24 2 2 8 2 18 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO Figure 8. The crossing limit cycle of type 2 of the discontinuous piecewise linear dierential system formed by the linear centers (25), (26), (27) and (28) separated by (LV). which has the rst integral H1 (x, y) = x (7 + 4x) − 7y/3 + 4xy + 2y 2 . In the region R2 we consider the linear dierential center 251 109 293 (32) ẋ = − −x− y, ẏ = − + x + y, 400 100 200  this system has the rst integral H2 (x, y) = 200x2 + y (251 + 218y) + x (−586 + 400y) /50. In the region R3 we have the linear dierential center 5 x 5 23 y (33) ẋ = + − y, ẏ = +x− , 96 4 16 24 4 this system has the rst integral H3 (x, y) = 4x2 + x (23/3 − 2y) + 5y (−1 + 3y) /12. In the region R4 we have the linear dierential center 137 x 25 53 y (34) ẋ = − + − y, ẏ = − +x− , 288 3 144 48 3  this system has the rst integral H4 (x, y) = 144x2 − 6x (53 + 16y) + y (137 + 25y) /36. This discontinuous piecewise dierential system formed by the linear dierential centers (31), (32),(33) and (34) has two crossing limit cycles of type 2, because the unique real solutions (pi1 , q1i , pi2 , q2i ), with i = 1, 2 of system (29) are p11 = (1, 0), q11 = (0, 3), p12 = (−5/2, 0) and q21 = (0, −2) and p21 = (2, 0), q12 = (0, 9/2), p22 = (−4, 0) and q22 = (0, −5). See these two crossing limit cycles in Figure 9.  Proof of statement (c) of Theorem 5. In the region R1 we consider the linear dierential center 1 x 4 1 y (35) ẋ = − − y, ẏ = +x+ , 3 2 9 2 2  which has the rst integral H1 (x, y) = 4 9x(1 + x) − 6y + 9xy + 4y 2 /9. In the region R2 we have the linear dierential center 1 x 1 y (36) ẋ = + − y, ẏ = − + x − , 4 2 4 2 this system has the rst integral H2 (x, y) = 4x2 + 2y (−1 + 2y) − 2x(1 + 2y). In the region R3 we have the linear dierential center 216 2 47 15 2 (37) ẋ = − + x − y, ẏ = + x − y, 160 3 80 8 3 19 Figure 9. The two crossing limit cycle of type 2 of the discontinuous piece- wise linear dierential system formed by the linear centers (31), (32), (33) and (34) separated by (LV). this system has the rst integral H3 (x, y) = 4x2 + x (15 − 16y/3) + y (219 + 47y) /20. In the region R4 we have the linear dierential center 7 x y 3 y (38) ẋ = − − − , ẏ = − + x + , 10 4 5 2 4 this system has the rst integral H4 (x, y) = 4x + 2x (−6 + y) + 4y(7 + y)/5. This dis- 2 continuous piecewise dierential system formed by the linear dierential centers (35), (36), (37) and (38) has three crossing limit cycles of type 2, because the unique real solutions (pi1 , q1i , pi2 , q2i ), with i = 1, 2, 3 of system (29) are p11 = (1, 0), q11 = (0, 3), p12 = (−5/2, 0) and q21 = (0, −2); p21 = (2, 0), q12 = (0, 9/2), p22 = (−4, 0) and q22 = (0, −5); and p31 = (3, 0), q13 = (0, 6), p32 = (−11/2, 0) and q23 = (0, −7). See these three crossing limit cycles of type 2 in Figure 10.  Figure 10. The three crossing limit cycle of type 2 of the discontinuous piecewise linear dierential system formed by the centers (35), (36),(37) and (38) separated by (LV). 7. Proof of Theorem 6 If the discontinuity curve Σ is of the type (P), we have following two regions in the plane: R1 = {(x, y) ∈ R2 : x2 < y}, and R2 = {(x, y) ∈ R2 : x2 > y}. 20 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO We consider a planar discontinuous piecewise linear dierential system formed by two linear arbitrary centers. By Lemma 11 these piecewise linear dierential centers can be as follows 4b2 + ω 2 ẋ = −bx − y + d, ẏ = ax + by + c, in R1 , 4a (39) 4β 2 + ω 2 ẋ = −βx − y + δ, ẏ = αx + βy + γ, in R2 . 4α These linear dierential centers have the rst integrals H1 (x, y) = 4(ax + by)2 + 8a(cx − dy) + y 2 ω 2 , (40) H2 (x, y) = 4(αx + βy)2 + 8α(γx − δy) + y 2 Ω2 , respectively. After two rescalings of time as in the proof Theorem 2 we can assume without loss of generality that a = α = 1. In order that the piecewise linear dierential centers (39) has crossing limit cycles with four point on (P). We must study the solutions of the system: e1 = H1 (x1 , x21 ) − H1 (x2 , x22 ) = 0, e2 = H2 (x2 , x22 ) − H2 (x3 , x23 ) = 0, (41) e3 = H1 (x3 , x23 ) − H1 (x4 , x24 ) = 0, e4 = H2 (x4 , x24 ) − H2 (x1 , x21 ) = 0, or equivalently (42) e1 = 4x21 (1 + bx1 )2 − 4x22 (1 + bx2 )2 + 8x1 (c − dx1 ) + 8x2 (dx2 − c) + (x41 − x42 )ω 2 = 0, e2 = 4x22 (1 + βx2 )2 + 8x2 (γ − δx2 )2 − 4x23 (1 + βx3 )2 + 8x3 (δx3 − γ) + (x42 − x43 )Ω = 0, e3 = 4x23 (1 + bx3 )2 − 4x24 (1 + bx4 )2 + 8x3 (c − dx3 ) + 8x4 (dx4 − c) + (x43 − x44 )ω 2 = 0, e4 = 4x24 (1 + βx4 )2 + 8x1 (δx1 − γ) − 4x21 (1 + βx1 )2 + 8x4 (γ − δx4 ) + (x44 − x41 )Ω2 = 0. We assume that the discontinuous piecewise linear dierential centers (39) has two crossing periodic solutions. For this we must have that system of equations (42) has two real solu- tions, namely (p1 , p2 , p3 , p4 ) and (q1 , q2 , q3 , q4 ), where pi = (ki , ki2 ) and qi = (Li , L2i ), with i = 1, 2, 3, 4. These points are the points where the two crossing periodic solution intersect discontinuity curve (P). If the point (p1 , p2 , p3 , p4 ) satises system (42), by the equation e1 of (42) we obtain the following expression 8c + 4(k1 + k2 )(1 + b(k1 + k2 )) + 4b(k12 + k22 ) + (k1 + k2 )(k12 + k22 )l1 d= , 8(k1 + k2 ) by the equation e2 of (42) we get the expression 8γ + 4(k2 + k3 )(1 + β(k2 + k3 )) + 4β(k22 + k32 ) + (k2 + k3 )(k22 + k32 )l2 δ= , 8(k2 + k3 ) by the equation e3 of (42) we obtain the expression   2  k1 + k2 (k1 + k22 )l1 k 2 + k22 c= (k42 − k32 ) + 1 + (1 + b(k1 + k2 )) + b 1 2(k1 + k2 − k3 − k4 )(k3 − k4 ) 4 k1 + k2  + 2b(k43 − k33 ) + (k44 − k34 )l1 and by the equation e4 of (42) we obtain the expression 1 γ= 8β(k1 + k4 )k4 − l2 (k22 + k32 − k42 )(k2 + k3 )(k1 + k4 ) − 2(k22 + k2 k3 8(k1 − k2 − k3 + k4 )  + k32 ) + (k2 + k3 )(k13 l2 + k12 (8β + k4 l2 ) , 21 here we consider l1 = 4b2 + ω 2 and l2 = 4β 2 + Ω2 . We assume that the point (q1 , q2 , q3 , q4 ) satises system (42), then we can obtain the remaining parameters of discontinuous piecewise linear dierential center (39). By the equation e1 of (42) we obtain ω 2 = S/T , where −4b(L1 − L2 ) S= (bk1 + (bk2 + 2))(k3 + k4 − L1 − L2 )k12 + k1 −bk33 + k2 (bk2 + 2) k1 + k2 − k3 − k4  (k3 + k4 − L1 − L2 ) −(k32 + k3 k4 + k42 )(bk4 + 2) + bL32 + (L21 + L1 L2 + L22 )(bL1 + 2) + k2 (bk2 + 2)(k3 + k4 − L1 − L2 ) + k2 (bL1 + 2)(L21 + L1 L2 + L22 ) − (k32 + k3 k4 + k42 )  (bk4 + 2) − bk33 + bk2 L32 + (L2 + L1 ) bk33 + (k3 + k4 )(k4 − L1 )(b(k4 + L1 ) + 2) + k32 (bk4 + 2) ) − L22 (k3 + k4 )((bL1 + 2) − bL2 ) ) , and L1 − L2 T = (k13 + k12 k2 + k1 k22 + k23 )(k3 + k4 − L1 − L2 ) + (k1 + k2 ) ((L1 + L2 ) k1 + k2 − k3 − k4   (L21 + L22 ) − (k3 + k4 )(k32 + k42 ) + (k3 + k4 )(L1 + L2 ) k32 + k42 − L21 − L22 , by the equation e2 of (42) we obtain Ω2 = V /W , where −4β(L2 − L3 ) V = k1 (βk1 + (βk4 + 2))(k2 + k3 − L2 − L3 ) − 2k1 k22 + k2 (k3 − k4 ) k1 − k2 − k3 + k4   +(k3 − L2 )(k3 − k4 + L2 ) + L3 (k4 − L2 ) − L23 + βk1 −k23 − k22 k3 + k2 k42 − k32  −k33 + k3 k42 + (L2 + L3 ) −k42 + L22 + L23 + βk23 (−k4 + L2 + L3 ) − k22 (βk3 + 2)  (k4 − L2 − L3 ) − βk2 k32 (k4 − L2 − L3 ) − k43 + (L2 + L3 )(L22 + L23 ) − 2k2 L22 − k42  + k3 (k4 − L2 − L3 ) + L2 L3 + L23 − (k3 − k4 ) (βk32 + k3 (βk4 + 2))(k4 − L2 − L3 )   + β(L2 + L3 ) −k42 + L22 + L23 + 2 −k4 (L2 + L3 ) + L22 + L2 L3 + L23 , and L2 − L3 W = k 2 (k2 + k3 − L2 − L3 )(k1 + k4 ) + k1 k2 (k42 − k32 ) − k23 − k22 k3 − k33 k1 − k2 − k3 + k4 1    + k3 k42 + (L2 + L3 ) −k42 + L22 + L23 + L2 (k2 + k3 ) k22 + k32 − k43 − k4 (k2 + k3 )   k22 + k32 − k42 + L3 k23 + k22 k3 + k2 k32 − L22 (k2 + k3 − k4 ) + k33 − k43 − (k2 + k3 − k4 ) (L32 + L2 L23 + L33 ) ) , by the equations e3 and e4 we get that b = β = 0. This implies that the linear dierential systems in R1 and in R2 are of the form 1 ẋ = , ẏ = x, 2 which is a contradiction because by hypothesis each of the linear dierential systems con- sidered is a center. Therefore we have proved that the maximum number of crossing limit cycles of the discontinuous piecewise linear dierential centers in F3 is one. Now we verify that this upper bound is reached. That is, that there are piecewise lin- ear dierential centers in the family F3 having one crossing limit cycle. We consider the discontinuous piecewise linear dierential system formed by the following linear dierential centers 22 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO 1694273 15 321 975799 15y (43) ẋ = + x− y, ẏ = +x− in R1 , 153600 16 256 153600 16 146901 39047 (44) ẋ = + x − 2y, ẏ = − +x−y in R2 . 10100 5050 These linear dierential centers have the rst integrals   2 975799 15y y(−1694273 + 96300y) H1 (x, y) = 4x + x − + , 19200 2 19200   2 156188 293802y H2 (x, y) = 4x − x + 8y − + 8y 2 , 2525 2525 respectively. The discontinuous piecewise dierential center formed by the linear dieren- Figure 11. The crossing limit cycle of the discontinuous piecewise linear dierential centers formed by (43) and (44) separated by the conic (P). tial centers (43) and (44) has one crossing limit cycle, because the unique real solution (p1 , p2 , p3 , p4 ) of system (42) is p1 = (4, 16), p2 = (−13/5, 169/25), p3 = (−3/2, 9/4), and p4 = (2, 4). See the crossing limit cycle of this system in Figure 11. 8. Proof of Theorem 7 When the discontinuity curve Σ is of the type (E), we have following two regions in the plane: R1 = {(x, y) ∈ R2 : x2 + y 2 < 1}, and R2 = {(x, y) ∈ R2 : x2 + y 2 > 1}. By Lemma 11 a piecewise linear dierential center of family F4 can be consider as (39) where the rst integrals are given in (40). Now we going to study the conditions in order that a piecewise linear dierential center in the family F4 has crossing limit cycles intersecting the discontinuity curve (E) in exactly four points. Taking into account the rst integrals (40) a piecewise linear dierential center in F4 has crossing limit cycles if there are points (xi , yi ) for i = 1, 2, 3, 4 satisfying the equations e1 = H1 (x1 , y1 ) − H1 (x2 , y2 ) = 0, e2 = H2 (x2 , y2 ) − H2 (x3 , y3 ) = 0, e3 = H1 (x3 , y3 ) − H1 (x4 , y4 ) = 0, e4 = H2 (x4 , y4 ) − H2 (x1 , y1 ) = 0, E1 = x21 + y12 − 1 = 0, E2 = x22 + y22 − 1 = 0, E3 = x23 + y32 − 1 = 0, E4 = x24 + y42 − 1 = 0, 23 considering l1 = 4b2 + ω 2 and l2 = 4β 2 + Ω2 , we have the equivalent system (45) e1 = 4(x21 − x22 ) + 8(bx1 y1 − bx2 y2 + c(x1 − x2 ) − dy1 + dy2 ) + l1 (y12 − y22 ) = 0, e2 = 4(x22 − x23 ) + 8x2 (γ + βy2 ) − 8x3 (γ + βy3 ) + (y2 − y3 ) (l2 (y2 + y3 ) − 8δ) = 0, e3 = 4(x23 − x24 ) + 8(bx3 y3 − bx4 y4 + c(x3 − x4 ) − dy3 + dy4 ) + l1 (y32 − y42 ) = 0, e4 = 8δ(y1 − y4 ) + 8x4 (γ + βy4 ) − 4(x21 − x24 ) − 8x1 (γ + βy1 ) + l2 (y42 − y12 ) = 0, E1 = 0, E2 = 0, E3 = 0, E4 = 0. Where we consider without generality a = α = 1 as in the proof Theorem 2. We assume that this piecewise linear dierential center has two crossing periodic solutions. For this we have that system (45) has two pairs of solutions, (p1 , p2 , p3 , p4 ) and (q1 , q2 , q3 , q4 ) with pi 6= pj , and qi 6= qj , for i 6= j and i, j = 1, 2, 3, 4. Since these solution points are on the circle (E), then we can consider them in the following way (46) pi = (ki , λi ), with ki = cos si , λi = sin si and qi = (mi , ni ), with mi = cos ti , ni = sin ti , with si , ti ∈ [0, 2π), for i = 1, 2, 3, 4. Substituting the rst solution (p1 , p2 , p3 , p4 ) with pi as in (46) in (45) we can determine the parameters d, δ, c, γ of the piecewise linear dierential centers (39), and we get (8c(k1 − k2 ) + 4(k1 − k2 + bλ1 − bλ2 )(k1 + k2 + b(λ1 + λ2 )) + (λ21 − λ22 )ω 2 ) d= , 8(λ1 − λ2 ) 4k22 − 4k32 + 8k2 (λ2 β + γ) − 8k3 (λ3 β + γ) + (λ22 − λ23 )l2 δ= , 8(λ2 − λ3 ) 1 c= (4(λ1 − λ2 )(k42 − k32 − 2bλ3 k3 + 2bk4 λ4 ) 8((k3 − k4 )(λ1 − λ2 ) − (k1 − k2 )(λ3 − λ4 )) + (λ3 − λ4 )(4(k12 − k22 ) + 8b(k1 λ1 − k2 λ2 ) + (λ1 − λ2 )(λ1 + λ2 − λ3 − l4 ) l1 )) , 1 γ= 4(λ1 − λ4 )(k32 − k22 + 2k3 λ3 β − 2k2 λ2 ) 8(−(k1 − k4 )(λ2 − λ3 ) + (k2 − k3 )(λ1 − λ4 )) + (λ2 − λ3 )(4k12 − 4k42 + 8k1 λ1 β − 8k4 λ4 β + (λ1 − λ4 )(λ1 − λ2 − λ3 + λ4 ) l2 )) . Analogously, substituting the second solution (q1 , q2 , q3 , q4 ) with qi as in (46) in (45) we get remaining parameters ω, Ω, b, β. Substituting ki , λi , mi , ni like (46) we obtain that b = β = 0, therefore we get that the piecewise linear dierential centers is formed by linear dierential center ẋ = −y, ẏ = x, in the regions R1 and R2 . This is a contradiction because with this linear dierential center is not possible to generate crossing limit cycles. Then we proved that the maximum number of crossing limits cycles for piecewise linear dierential centers in F4 is one. Moreover this maximum number is reached, that is, there are piecewise linear dierential centers in F4 such that have one crossing limit cycle with four points on (E), as we see below Consider the discontinuous piecewise linear dierential centers in the family F4 formed by the following two linear dierential centers √ √ −107 − 89 2 5x 345y −71 + 89 2 5y (47) ẋ = − − , ẏ = +x+ in R1 , 1024 16 256 1024 16 1  √  41y 1  √  (48) ẋ = −7 + 18 2 − x − , ẏ = 43 + 18 2 + x + y in R2 . 64 16 64 24 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO Figure 12. The crossing limit cycle of the discontinuous piecewise linear dierential system formed by the centers (47) and (48) separated by the conic (E). These linear dierential centers have the rst integrals 1  √ √  H1 (x, y) = 512x2 + x(−71 + 89 2 + 320y) + y(107 + 89 2 + 690y) , 128 1 √ √  H2 (x, y) = 32x2 + x(43 + 18 2 + 64y) + y(7 − 18 2 + 82y) , 8 respectively. Then the discontinuous piecewise dierential system formed by the linear dierential centers (47) and (48) has one crossing limit√cycle, because √  the unique real solution (p1 , p2 , p3 , p4 ) of system (45) is p1 = (1, 0), p2 = − 2/2, 1/ 2 , p3 = (−1, 0), and p4 = (0, −1). See the crossing limit cycle of this system in Figure 12. 9. Proof of Theorem 8 If the discontinuity curve Σ is of the type (H) we have following three regions in the plane: R1 = {(x, y) ∈ R2 : x2 − y 2 > 1, x > 0}, R2 = {(x, y) ∈ R2 : x2 − y 2 < 1} and R3 = {(x, y) ∈ R2 : x2 − y 2 > 1, x < 0}. We consider a planar discontinuous piecewise linear dierential system formed by three linear arbitrary centers. By Lemma 11 these linear dierential centers can be as follows 4b21 + ω12 ẋ = −b1 x − y + d1 , ẏ = a1 x + b1 y + c1 , in R1 , 4a1 4b2 + ω22 (49) ẋ = −b2 x − 2 y + d2 , ẏ = a2 x + b2 y + c2 , in R2 , 4a2 4b2 + ω32 ẋ = −b3 x − 3 y + d3 , ẏ = a3 x + b3 y + c3 , in R3 . 4a3 These linear dierential centers have the rst integrals H1 (x, y) = 4(a1 x + b1 y)2 + 8a1 (c1 x − d1 y) + y 2 ω12 , (50) H2 (x, y) = 4(a2 x + b2 y)2 + 8a2 (c2 x − d2 y) + y 2 ω22 , H3 (x, y) = 4(a3 x + b3 y)2 + 8a3 (c3 x − d3 y) + y 2 ω32 , respectively. 25 In order to have a crossing limit cycle, which intersects the discontinuity curve (H) in four dierent points pi = (xi , yi ), i = 1, 2, 3, 4, these points must satisfy the following equations e1 = H1 (x1 , y1 ) − H1 (x2 , y2 ) = 0, e2 = H2 (x2 , y2 ) − H2 (x3 , y3 ) = 0, e3 = H3 (x3 , y3 ) − H3 (x4 , y4 ) = 0, e4 = H2 (x4 , y4 ) − H2 (x1 , y1 ) = 0, (51) E1 = x21 − y12 − 1 = 0, E2 = x22 − y22 − 1 = 0, E3 = x23 − y32 − 1 = 0, E4 = x24 − y42 − 1 = 0, equivalently, we have (52) e1 = 4(x21 − x22 ) + 8(c1 (x1 − x2 ) − d1 y1 + b1 x1 y1 + d1 y2 − b1 x2 y2 ) + (y12 − y22 )l1 = 0, e2 = 4(x22 − x23 ) + 8(c2 (x2 − x3 ) − d2 y2 + b2 x2 y2 + d2 y3 − b2 x3 y3 ) + (y22 − y32 )l2 = 0, e3 = 4(x23 − x24 ) + 8(c3 (x3 − x4 ) − d3 y3 + b3 x3 y3 + d3 y4 − b3 x4 y4 ) + (y32 − y42 )l3 = 0, e4 = 4(x24 − x21 ) + 8(c2 (x4 − x1 ) − d2 y4 − b2 x1 y1 + d2 y1 + b4 x4 y4 ) + (y42 − y12 )l2 = 0, E1 = 0, E2 = 0, E3 = 0, E4 = 0, where li = 4b2i + ωi2 , for i = 1, 2, 3. Here we are taking without generality a1 = a2 = a3 = 1 as in the proofs of the previous theorems. We assume that the discontinuous piecewise linear dierential system formed by the three linear dierential centers in (49) has two crossing periodic solutions. For this we must impose that the system of equations (52) has two of real solution, namely (p1 , p2 , p3 , p4 ) and (q1 , q2 , q3 , q4 ). Since these solutions provide crossing periodic solutions and these points are the points where the crossing periodic solutions intersect the hyperbola (H) we can consider pi = (ki , λi ) = (cosh ri , sinh ri ) and qi = (mi , ni ) = (cosh si , sinh si ), (53) with ri , si ∈ R for i = 1, 2, 3, 4. Now we assume that the point (p1 , p2 , p3 , p4 ) with pi = (ki , λi ), i = 1, 2, 3, 4 satises system (52), and then we obtain the following expressions 8ci (ki − ki+1 ) + 4(ki − ki+1 + bi (λi − λi+1 ))(ki + ki+1 + bi (λi + λi+1 )) + (λ2i − λ2i+1 )ωi2 di = , 8(λi − λi+1 ) for i = 1, 2, 3, and 1 c2 = 4(k32 + 2b2 k3 λ3 − k22 − 2b2 k2 λ2 )(λ4 − λ1 ) 8((k2 − k3 )(λ1 − λ4 ) − (k1 − k4 )(λ2 − λ3 )) + (λ2 − λ3 )(4(k12 − k42 ) + 8b2 (k1 λ1 − k4 λ4 ) + (λ1 − λ4 )(λ1 − λ2 − λ3 + λ4 ) l2 )) . We assume that the point (q1 , q2 , q3 , q4 ) with qi as in (53) satises system (52). By the rst equation in (52) we get that 1 c1 = (λ1 − λ2 )(−4(m21 − m22 ) − 8b1 m1 n1 8((λ1 − λ2 )(m1 − m2 ) − (k1 − k2 )(n1 − n2 )) 4(k1 − k2 + b1 (λ1 − λ2 ))(k1 + k2 + b1 (λ1 + λ2 ))(n1 − n2 ) + + 8b1 m2 n2 + (λ1 + λ2 )ω12 λ1 − λ2  (n1 − n2 ) − (n21 − n22 )l1 ) . 26 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO By the second equation of (52) we obtain that ω22 = K/S where 4 K= (k32 − k22 )((λ1 − λ4 )ψ2 − (k1 − k4 )ψ1 ) + (λ2 (k1 − k4 )(λ2 − λ3 ) − (k2 − k3 )(λ1 − λ4 ) − λ3 )((k12 + b22 (λ1 − λ4 )(λ1 − λ2 − λ3 + λ4 ) − k42 )ψ2 + k4 ψ2 − b22 (λ2 + λ3 − n2 − n3 ) ψ1 + 2b2 (−λ4 ψ2 + m2 n2 − m3 n3 )) + k1 (−ψ2 + b22 (λ2 + λ3 − n2 − n3 )ψ1 + 2b2 (λ1 ψ2 − m2 n2 + m3 n3 ))) + k2 ((λ1 − λ4 )(ψ2 − b22 (λ1 + λ4 − n2 − n3 )ψ1 ) − (k12 − k42 )ψ1 − 2b2 (λ2 (−λ4 ψ2 − (k1 − k4 )ψ1 ) + λ4 ((m2 − k4 )n2 + (k4 − m3 )n3 )λ1 (λ2 ψ2 + (k1 − m2 )n2 + (m3 − k1 )n3 ))) + k3 ((λ4 − λ1 )ψ2 + (k12 − k42 )ψ1 + b22 (λ1 − λ4 )(λ1 + λ4 − n2 − n3 )ψ1 + 2b2 (λ3 (−λ4 ψ2 − (k1 − k4 )ψ1 ) + λ4 (m2 n2 − k4 ψ1 − m3 n3 ) + λ1 (λ3 ψ2 + k1 ψ1 − m2 n2 + m3 n3 )), 1 S= (λ2 λ4 ψ2 + λ3 λ24 ψ2 + (k1 − k4 )(λ23 n2 − λ3 n22 ) (k1 − k4 )(λ2 − λ3 ) − (k2 − k3 )(λ1 − λ4 ) 3 + (k3 − k2 )(λ24 n2 − λ4 n22 ) + λ21 ((k2 − k3 )ψ1 − (λ2 − λ3 )ψ2 ) + λ22 (λ4 ψ2 + (k1 − k4 )ψ1 ) + ((k4 − k1 )λ23 + (k2 − k3 )λ24 )n3 + (k1 λ3 − k4 λ3 − k2 λ4 + k3 λ4 )n23 + λ1 ((λ22 − λ23 )ψ2 − (k2 − k3 )ψ1 (n2 + n3 )) + λ2 (λ24 ψ2 + (k1 − k4 )ψ1 (n2 + n3 )))), by the third equation we get that −1 c3 = (((λ3 − λ4 )(4(m24 − m23 ) + 8b3 (m4 n4 − m3 n3 ) 8(λ4 ψ4 − λ3 ψ2 + (k3 − k4 )ψ1 )) 4(k3 − k4 + b3 (λ3 − λ4 ))(k3 + k4 + b3 (λ3 + λ4 ))ψ1 + + (λ3 + λ4 )ω32 ψ1 − (n23 − n24 )l3 )), λ3 − λ4 where ψ1 = n2 − n3 , ψ2 = m2 − m3 , ψ3 = n3 − n4 and ψ4 = m3 − m4 . Finally by the fourth equation in (52) we have that b2 = 0. With these expressions for d2 , c2 , ω2 and b2 we obtain that the linear dierential system in the region R2 is ẋ = y, ẏ = x, which is a linear dierential system type saddle. This is a contradiction because we are working with centers in each region Ri for i = 1, 2, 3. Therefore we have proved that the maximum number of crossing limit cycles for systems in F5 is one. Moreover it is possible to show that there are piecewise linear dierential centers in F5 such that have one crossing limit cycle. Indeed, consider the discontinuous piecewise linear dierential system in the family F5 formed by the following linear dierential centers √ √ x 29 −355 + 64 10 + 80 21 y (54) ẋ = − − y + √ , ẏ = x + + 1, in R1 , 2 16 64(6 + 21) 2 x 13y 1 ẋ = − + p √ √ √ √ √ √ √ 3 36 24 81 − 12 35(−20 2 + 26 3 − (14 + 13 3) 5 + (7 + 2 10) 7) √ √ √ √ √ √ (−10920 − 12610 2 − (3500 + 1000 10) 3 + (1092 7 − 2548 + 168 21) 5 √ √ √ (55) + (1274 + 240 6 + 1261 10) 7), y 1 √ ẏ =x − + √ √ √ √ √ √ (−240 + 960 2 3 48(−20 2 + 26 3 − 14 5 + 7 7 − 13 15 + 2 70) √ √ √ √ √ √ + 1274 3 + 474 5 + 387 7 − 637 15 − 288 35 − 96 70), in R2 , √ 3 45 3 √ 337(−3 + 2 3) 3 3 (56) ẋ = − x − y + (−11 + 2 3) + √ , ẏ = x + y − , in R3 . 2 16 4 64 5 2 2 These linear dierential centers have rst integrals 27 Figure 13. The crossing limit cycle of the discontinuous piecewise linear dierential system formed by the centers (54), (55) and (56) with disconti- nuity curve the conic (H). √ √ √ (355 − 64 10 − 80 21)y + 2(6 + 21)(16x(2 + x) + 16xy + 29y 2 ) H1 (x, y) = √ , 8(6 + 21)  y 2 1 H2 (x, y) =4 x − + p √ √ √ √ √ √ √ 3 18 27 − 4 35(26 3 − 20 2 − (14 + 13 3) 5 + (7 + 2 10) 7) q √ √ √ √ √ √ √ (−3 27 − 4 35(240 − 960 2 − 1274 3 − 474 5 − 387 7 + 637 15 + 288 35 √ √ √ √ √ √ + 637 15 + 288 35 + 96 70)x + 2(10500 + 10920 3 + 12610 6 + 3000 10 √ √ √ √ √ √ − 720 14 + 2548 15 − 1274 21 − 504 35 − 1092 105 − 1261 210)y) + y 2 , 1 √ √ √ H3 (x, y) = 4x2 + 12x(−1 + y) + y(2640 − 480 3 + 1011 5 − 674 15 + 450y), 40 respectively. √ The uniquepreal solution (p1 , p2√ , p3 , p4 ) that satises (52)p in this case is p1 = ( 10, −3), p2 = (−5/2, 21/4), p3 = (−4, 15) and p4 = (−7/2, − 45/4). See the crossing limit cycle of this system in Figure 13. 10. Proof of Theorem 9 Proof of statement (a) of Theorem 9. In this case we use the notations given in the proof of Theorem 6, then we consider the planar discontinuous piecewise linear center (39) and the rst integrals (40). In order that the discontinuous piecewise linear center (39) has crossing limit cycles with four points, namely (x1 , x21 ), (x2 , x22 ), (x3 , x23 ), (x4 , x24 ) and one crossing limit cycle with two points, namely (x5 , x25 ), (x6 , x26 ) on (P), we must study the solutions (x1 , x2 , x3 , x4 , x5 , x6 ) of system (41) and the equations e5 = H1 (x5 , x25 ) − H1 (x6 , x26 ) = 0, e6 = H2 (x6 , x26 ) − H2 (x5 , x25 ) = 0, or equivalently systems (42) and (57) e5 = 4x25 (1 + bx5 )2 + 8x5 (c − dx5 ) − 4x26 (1 + bx6 )2 + 8x6 (dx6 − c) + (x45 − x46 )ω 2 = 0, e6 = 4x26 (1 + x6 β)2 − 4x25 (1 + x5 β)2 + 8x5 (x5 δ − γ) + 8x6 (γ − x6 δ) + (x46 − x45 )Ω2 = 0. We assume that systems (42) and (57) have two real solutions where each real solution provides one crossing limit cycle with four points on (P) and one crossing limit cycle whit two 28 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO points on (P), but by Theorem 6 we have that discontinuous piecewise linear center (39) has at most 1 crossing limit cycle with four points on (P), therefore if we have two real solutions of systems (42) and (57) they are of the form (x1 , x2 , x3 , x4 , x5 , x6 ) = (k1 , k2 , k3 , k4 , k5 , k6 ) and (x1 , x2 , x3 , x4 , x5 , x6 ) = (k1 , k2 , k3 , k4 , λ5 , λ6 ), with ki , λ5 , λ6 ∈ R for i = 1, 2, 3, 4, 5, 6. If the point (k1 , k2 , k3 , k4 , k5 , k6 ) satises systems (42) and (57), by the equations e1 , e2 , e3 and e4 of (42) we obtain expressions for the parameters d, δ, c and γ as in the proof of Theorem 6, by the equation e5 of system (57) we obtain an expression for ω 2 = S/T with S and T as in the proof of Theorem 6 changing L1 and L2 by k5 and k6 , respectively. By equation e6 of system (57) we obtain Ω2 = V /W where the expression for V and W are the same expressions that in the proof of Theorem 6 changing L3 by k5 . We assume that the point (k1 , k2 , k3 , k4 , λ5 , λ6 ) satises systems (42) and (57), then we have e1 = e2 = e3 = e4 = 0 and by the equations e5 and e6 of system (57) we obtain b = β = 0. As in the proof of Theorem 6 we can conclude that the two linear centers in (39) became ẋ = 1/2, ẏ = x, which is a contradiction. So systems (42) and (57) have at most one solution and therefore planar discontinuous piecewise linear centers in F3 have at most one crossing limit cycle with four point on (P) and one crossing limit cycle with two points on (P) simultaneously. Moreover this upper bound is reached, this is there are systems in F3 with one crossing limit cycle with four points on (P) and one crossing limit cycle with two points on (P) simultaneously. Figure 14. The two crossing limit cycles of the discontinuous piecewise linear dierential system formed by the centers (58) and (59). We consider the discontinuous piecewise linear dierential system formed by the linear centers 1225 x 310 103 y (58) ẋ = + − y, ẏ = − + x − , in R1 , 229 2 229 229 2 6055 x 170 3359 y (59) ẋ = − − y, ẏ = − + x + , in R2 . 712 4 89 229 4 These linear dierential centers have the rst integrals H1 (x, y) = 229x2 + 10y(−245 + 31y) − x(206 + 229y), H2 (x, y) = 356x2 + 2x(−3359 + 89y) + 5y(−1211 + 136y), respectively. The unique real solution of systems (42) and (57) is (x1 , x2 , x3 , x4 , x5 , x6 ) = (3, −2, −3/2, 1, 2, 12/5), therefore we have one crossing limit cycle that intersects (P) in the points (3, 9), (−2, 4), (−3/2, 9/2) and (1, 1), and one crossing limit cycle that intersects (P) in the points (2, 4) and (12/5, 144/25). See these crossing limit cycles in Figure 14.  29 Proof of statement (b) of Theorem 9. In this case we consider the notations of the proof of Theorem 7 and therefore we consider the planar discontinuous piecewise linear center (39) and the rst integrals (40). In order that the discontinuous piecewise linear center (39) has crossing limit cycles with four points on (E), namely (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), (x4 , y4 ) and one crossing limit cycle with two points on (E), namely (x5 , y5 ), (x6 , y6 ), we must study the solutions (p1 , p2 , p3 , p4 , p5 , p6 ) of systems (45) and (60) e5 = 4(x25 − x26 ) + 8(c(x5 − x6 ) − dy5 + bx5 y5 + dy6 − bx6 y6 ) + (y52 − y62 )l1 = 0, e6 = 4(x26 − x25 ) + 8(βx6 y6 − βx5 y5 + y5 δ − x5 γ + x6 γ − y6 δ) + (y62 − y52 )l2 = 0, (60) E5 = x25 + y52 − 1 = 0, E6 = x26 + y62 − 1 = 0. We assume that systems (45) and (60) have two real solutions where each real solution provides one crossing limit cycle with four points on (E) and one crossing limit cycle with two points on (E), like in Theorem 7 we proved that discontinuous piecewise linear center (39) has at most 1 crossing limit cycle with four points on (E), then we have that if there are two real solutions of systems (45) and (60) they are of the form (p1 , p2 , p3 , p4 , p5 , p6 ) and (p1 , p2 , p3 , p4 , q5 , q6 ), with pi and qj as (46) for i = 1, 2, 3, 4, 5, 6 and j = 5, 6. Substituting the rst solution (p1 , p2 , p3 , p4 , p5 , p6 ) in systems (45) and (60) we obtain from the equations e1 , e2 , e3 and e4 of (45) the same expressions than in the proof of Theorem 7 for d, δ, c, γ , and by the equations e5 and e6 of system (60) we obtain the same expressions than in the proof of Theorem 7 for ω and Ω changing (m1 , n1 ) by (k5 , λ5 ) and (m2 , n2 ) by (k6 , λ6 ), respectively. We assume that the point (p1 , p2 , p3 , p4 , q5 , q6 ) satises systems (45) and (60), then we have e1 = e2 = e3 = e4 = 0 and by the equations e5 and e6 of system (60) we obtain b = β = 0. As in the proof of Theorem 7 we obtain that both linear centers in (39) become ẋ = −y, ẏ = x, in contradiction that they have limit cycles. So we can conclude that systems (45) and (60) have at most one solution and therefore planar discontinuous piecewise linear centers in F4 have at most one crossing limit cycle with four points on (E) and one crossing limit cycle with two points on (E) simultaneously. Now we verify that this upper bound is reached, that is there are systems in F4 with one crossing limit cycle with four points on (E) and one crossing limit cycle with two points on (E) simultaneously. We consider the discontinuous piecewise linear dierential system in F4 formed by the linear centers (61) √ √ √ √ √ √ √ 3 2+ 6−6 x 2( 2 + 3 − 2) 3 2+2 3+ 6−4 y ẋ = √ √ − − √ √ y, ẏ = √ √ + x + , in R1 , 4(2 2 + 3 3 − 3) 2 2 2+3 3−3 4(2 2 + 3 3 − 3) 2 (62) √ √ √ √ √ √ 18 − 93 2 + 4 3 + 33 6 x 31 2 + 37 3 − 11 6 − 236 ẋ = √ − + y, 920(1 + 3) 4 230 √ √ √ 89 2 + 58 3 + 7 6192 y ẏ = √ + x + , in R2 . 920(1 + 3) 4 The unique real solution of systems (45) and (60) in this case is (p1 , p2 , p3 , p4 , p5 , p6 ) with p1 = (cos π/2, sin π/2), p2 = (cos π, sin π), p3 = (cos 3π/2, sin 3π/2), p4 = (cos −π/3, sin −π/3), p5 = (cos π/4, sin π/4) and p6 = (cos 0, sin 0). See these crossing limit cycles in Figure 15.  Proof of statement (c) of Theorem 9. Here we consider the notations of the proof of Theo- rem 8 and therefore we consider the planar discontinuous piecewise linear center (49) and the rst integrals (50). In order that the discontinuous piecewise linear center (49) has 30 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO Figure 15. The two limit cycles of the discontinuous piecewise linear dif- ferential system formed by the centers (61) and (62). crossing limit cycles with four points on (H), namely (x1 , y1 ), (x2 , y2 ), (x3 , y3 ), (x4 , y4 ) and one crossing limit cycle with two points on (H), namely (x5 , y5 ), (x6 , y6 ), we must study the solutions (p1 , p2 , p3 , p4 , p5 , p6 ) of systems (52) and (63) (63) e5 = 4(x25 − x26 ) + 8(c2 (x5 − x6 ) − d2 y5 + b2 x5 y5 + d2 y6 − b2 x6 y6 ) + (y52 − y62 )l1 = 0, e6 = 4(x26 − x25 ) + 8(b1 x6 y6 − b1 x5 y5 + y5 d1 − x5 c1 + x6 c1 − y6 d1 ) + (y62 − y52 )l2 = 0, E5 = x25 − y52 − 1 = 0, E6 = x26 − y62 − 1 = 0. We assume that systems (52) and (63) have two real solutions where each real solution provides one crossing limit cycle with four points on (H) and one crossing limit cycle with two points on (H). By Theorem 8 the discontinuous piecewise linear center (49) has at most 1 crossing limit cycle with four points on (H), then we have that if there are two real solutions of systems (52) and (63) they are of the form (p1 , p2 , p3 , p4 , p5 , p6 ) and (p1 , p2 , p3 , p4 , q5 , q6 ), with pi and qj as (53) for i = 1, 2, 3, 4, 5, 6 and j = 5, 6. Considering the rst solution (p1 , p2 , p3 , p4 , p5 , p6 ) of systems (52) and (63) we obtain the same expressions that in the proof of Theorem 8 for d1 , d2 , d3 , c2 , c1 , ω2 changing (m1 , n1 ) by (k5 , λ5 ) and (m2 , n2 ) by (k6 , λ6 ), respectively. Now we assume that the point (p1 , p2 , p3 , p4 , q5 , q6 ) satises systems (52) and (63), then we have e1 = e2 = e3 = e4 = 0, and by the equation e5 of system (63) we obtain b2 = 0 and with this the linear system in the region R2 becomes a saddle which is a contradiction, because we are working with linear centers in each regions Ri for i = 1, 2, 3. Therefore the discontinuous piecewise linear center (49) has at most one crossing limit cycle with four points on (H) and one crossing limit cycle with two points on (H) simultaneously. Moreover this upper bound is reached, that is there are piecewise linear dierential centers in F5 such that have one crossing limit cycle with four points on (H) and one crossing limit cycle with two points on (H) simultaneously. Indeed consider the piecewise linear dierential system 31 formed by the linear centers √ √ √ √ −1215 − 576 2 + 256 7 + 112 13 − 384 15 x 29 ẋ = √ √ √ − − y, 192(2 7 + 13 − α) 2 16 (64) r −1 675 √ √ 2 √ ẏ = r r ! − − 64 7 − 28 13 + 288 (23 − 4 30) 7 13 4 7 48 −1 + 2 + α α r r r r r ! 7 945 13 26 105 195 y + 945 + + 144 + 192 + 96 + x + , in R1 , α 2 α α α α 2 √ √ √ √ √ 1125 + 432 14 + 189 26 + 207 30 + 160 105 + 70 195 x ẋ = − 6ξ 2 √ √ √ √ √ √ 54 2 + 336 7 + 174 13 + 24 15 + 68 210 + 35 390 − η (65) − y, η √ √ √ √ √ √ 855 2 + 3516 7 + 1797 13 + 315 15 + 644 210 + 329 390 y ẏ = − + x + , in R2 , 6ξ 2 (66) 9 73 3 45 3 3 ẋ = − + √ − x − y, ẏ = − + x + y, in R3 , 2 8 2 2 16 2 2 √ √ √ √ √ √ here α =√ 23 + 4 √30, η =√42 + 48 √14 + 24 √26 + 9 30 + 24 105 + 12 195 and ξ = 87 + 108 14 + 54 26 + 16 30 + 40 105 + 20 195. The unique √ real solution√of systems (51) √ and (63) in this case is (p1 , p2 , p√3 , p4 , p5 , p6 ) with p1 = √ (3, − 8), p2 = (4, 15), p3 = (−3, 8), p4 = (−1, 0), p5 = (7/6, − 13/6) and p6 = (4/3, 7/3). See these crossing limit cycles in Figure 16.  Figure 16. The two limit cycles of the discontinuous piecewise linear dif- ferential system formed by the centers (64), (65) and (66). 11. Proof of Theorem 10 Proof of statement (a) of Theorem 10. In the region R1 we consider the linear dierential center 193 x 58 149 y (67) ẋ = − − − y, ẏ = − +x+ , 134 3 67 134 3 32 J. JIMENEZ, J. LLIBRE AND J.C. MEDRADO Figure 17. The crossing limit cycles of types 1 and 2 of the discontinuous piecewise linear dierential system formed by the linear centers (67), (68), (69) and (70) separated by (LV). this system has the rst integral H1 (x, y) = 201x2 + x(134y − 447) + 3y(58y − 193)). In the region R2 we have the linear dierential center 153 x 17 89 y (68) ẋ = − − − y, ẏ = − +x+ , 16 2 4 32 2 which has the rst integral H2 (x, y) = 4x2 + 17y(2y − 9)/2 + x(4y − 89/4). In the region R3 we have the linear dierential center √ √ 43 3 3 3 (69) ẋ = − − x − y, ẏ = 1 + x − y, 40 4 4 4 √ which has the rst integral H3 (x, y) = 4x(2 + x) + 43/5y − 2 3xy + 3y 2 . And in the region R4 we have the linear dierential center 251 y (70) ẋ = − , ẏ = −2 + x, 512 256 which has the rst integral H4 (x, y) = 4x(x − 4) + y(y − 251)/64. In order to have a crossing limit cycle of type 1 and one crossing limit cycle of type 2, we must study the real solutions (p1 , q1 , p2 , q2 , p3 , q3 , p4 , q4 ), of systems (21) and (29) respectively, where pi = (xi , 0) and qi = (0, yi ), with x1 , x2 , x3 , y1 , y2 , y3 > 0 and x4 , y4 < 0. In this case considering the four above centers (67), (68), (69) and (70) and their re- spective rst integrals we have that the unique real solution for systems (21) and (29) is (p1 , q1 , p2 , q2 , p3 , q3 , p4 , q4 ) with p1 = (1, 0), q1 = (0, 1/2), p2 = (3, 0), q2 = (0, 4), p3 = (5, 0), q3 = (0, 6), p4 = (−4, 0) and q4 = (0, −5). See these crossing limit cycles of types 1 and 2 in Figure 17.  Proof of statement (b) of Theorem 10. In the region R1 we consider the linear dierential center 193 x 58 149 y (71) ẋ = − − y, ẏ = − +x+ , 134 3 67 134 3 this system has the rst integral H1 (x, y) = 201x2 + x(134y − 447) + 3y(58y − 193)). In the region R2 we have the linear dierential center 9 x 1 y (72) ẋ = − − 2y, ẏ = − + x + , 2 2 4 2 33 which has the rst integral H2 (x, y) = 2x(4x − 1) + 4y(x − 9) + 8y 2 . In the region R3 we have the linear dierential center √ √ 11 3 3 11 3 (73) ẋ = − + x − y, ẏ = +x− y, 8 4 4 8 4 √ which has the rst integral H3 (x, y) = 4x2 + y(11 + 3y) + x(11 − 2 3y). And in the region R4 we have the linear dierential center √ √ 51831 − 595 16909 x 6775 − 119 16909 y (74) ẋ = + + y, ẏ = −2 + x − , 35912 2 17956 2 √ which√ has the rst integral H4 (x, y) = 17956(x − x(4 + y)) + y(595 16909 − 51831 + 2 (118 16909−6775)y). The unique real solutions for systems (21) and (29) are (p1 , q1 , p2 , q2 , p3 , q3 , p4 , q4 ) Figure 18. One crossing limit cycle of type 1 and two crossing limit cycles of type 2 of the discontinuous piecewise linear dierential system formed by the linear centers (71), (72), (73) and (74) separated by (LV). with p1 = (1, 0), q1 = (0, 1/2), p2 = (3, 0), q2 = (0, 4), p3 = (5, 0), q3 = (0,√6), p4 = (−4, 0) and q4 = (0, −5). And (p1 , q1 , p2 , q2 , l3 , m3 , l4 , m4 ) with l3 = ((149 + 3 16909)/134, 0), m3 = (0, 5), l4 = (−2, 0) and m4 = (0, −3). See these crossing limit cycles of types 1 and 2 in Figure 18.  Acknowledgements The rst author is partially supported by CAPES grant number 88881.188516/2018-01. 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Ye, Theory of limit cycles, Translations of Mathematical Monographs American Mathematical Society, 1986. 1 Universidade Federal do Oeste da Bahia, 46470000 Bom Jesus da Lapa, Bahia, Brasil Email address : jjohanajimenez@gmail.com 2 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain Email address : jllibre@mat.uab.cat 3 Instituto de Matemática e Estatística, Universidade Federal de Goiás, 74001-970 Goiâ- nia, Goiás, Brazil Email address : medrado@mat.ufg.br