This is a preprint of: “Limit cycles created by piecewise linear centers”, Jaume Llibre, Xiang Zhang, Chaos, vol. 29(5), 053116:1–14, 2019. DOI: [10.1063/1.5086018] LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS JAUME LLIBRE1 AND XIANG ZHANG2 Abstract. In this paper we study the limit cycles of the discon- tinuous piecewise linear differential systems in the plane R2 formed by three arbitrary linear centers separated by the set Σ = {(x, y) ∈ R2 : y = 0 or x = 0 and y ≥ 0}. We prove that such discontinuous piecewise linear differential sys- tems can have 1, 2 or 3 limit cycles which intersect in a unique point each branch of the set Σ, and that they cannot have more than 3 of such limit cycles. 1. Introduction and statement of the main result For a differential system in the plane R2 a periodic orbit which is isolated in the set of all periodic orbits is a limit cycle. At the end of the 19th century started the studies of the limit cycles of the differential systems with Poincaré [15]. Limit cycles modelize many phenomena of the real world, see the Belousov–Zhabotinskii reaction [2, 21], or the van der Pol oscillator [16, 17], or the motion of the galaxies [5], and many examples can be found in the survey [13], or in the book [3]. In the book of Andronov, Vitt and Khaikin [1] appeared some of the first studies on the discontinuous piecewise linear differential systems in the plane R2 separated by straight lines. The work on such differential systems has continued up today, mainly due to their applications, for instance in mechanics, economy, electrical circuits, etc, see the surveys [13, 20] and the books [3, 19]. There are two types of limit cycles in the planar discontinuous piece- wise linear differential systems, the crossing and sliding ones. The sliding limit cycles contain some arc of the lines of discontinuity which 2010 Mathematics Subject Classification. Primary 34A30, 34C05, 34C25, 34C07, 37G15. Key words and phrases. limit cycles, linear centers, discontinuous piecewise dif- ferential systems, first integrals. 1 2 JAUME LLIBRE AND XIANG ZHANG separate the different linear differential systems (more precise definition can be found in [14]). The crossing limit cycles only contain isolated points of the lines of discontinuity. In this paper we only consider the crossing limit cycles of some planar discontinuous piecewise linear d- ifferential systems separated by pieces of straight lines. From now on we only shall work with crossing limit cycles, but we simply call them limit cycles instead of crossing limit cycles. The easiest discontinuous piecewise linear differential systems in the plane are the discontinuous piecewise linear differential systems sep- arated by a unique straight line. It is known that such differential systems can have 3 limit cycles, see [4, 6, 7, 8, 9, 11]. But at this mo- ment it is an open problem to know if 3 is the maximum number of limit cycles that such discontinuous piecewise linear differential systems can have. Here our objective is to study the number of limit cycles which can exhibit the planar discontinuous piecewise linear differential systems separated by pieces of straight lines such that all their linear differen- tial systems are formed by centers. There are some results on these piecewise linear differential systems. Thus in Theorem 4 of [10] it is proved: Theorem 1. A discontinuous piecewise linear differential system sep- arated by one straight line formed by two linear centers has no limit cycles. Figure 1. The limit cycle of the discontinuous piecewise linear differential system (1). The two straight lines of sepa- ration between the differential systems are drawn in red col- or. But in [12] it is proved the existence of discontinuous piecewise linear differential system separated by two parallel straight lines and formed LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 3 by three linear centers, which exhibit one limit cycle. As far as we know this discontinuous system was the first example that only centers in a piecewise linear differential system can create limit cycles. More precisely, consider the discontinuous piecewise linear differential system with three pieces separated by the two parallel straight lines x = ±1 defined by ẋ = 2(−4 + 4x + 5y), ẏ = −8(1 + x + y), if x > 1, (1) ẋ = 1 + 2y, ẏ = 1 − 2x, if −1 < x < 1, ẋ = (16 + 32x + 65y)/2, ẏ = −8(x + 2y), if x < −1. These three linear differential systems are centers, and the discontin- uous piecewise linear differential system has the limit cycle of Figure 1. Figure 2. The unique limit cycle of the discontinuous piecewise linear differential system (2). Our main result is the following: Theorem 2. Consider three linear differential systems formed by cen- ters and separated by the set Σ = {(x, y) : y = 0 or x = 0 and y ≥ 0}. Such discontinuous piecewise linear differential systems can have at most 3 limit cycles intersecting the three branches of Σ \ {(0, 0)} in one point. Moreover, inside this class of discontinuous piecewise linear differential systems there are systems with exactly either 1,or 2, or 3 limit cycles. Theorem 2 is proved in section 2. 4 JAUME LLIBRE AND XIANG ZHANG We remark that in general it is not easy to provide an explicit upper bound for the maximum number of limit cycles in a class of differential systems, and such that this bound be reached. Figure 3. The two limit cycles of the discontinuous piece- wise linear differential system (3). The three components of R2 \ Σ are the positive or first quadrant Q1 = {(x, y) ∈ R2 : x > 0 and y > 0}, the second quadrant Q2 = {(x, y) ∈ R2 : x < 0 and y > 0}, and the half-plane H = {(x, y) ∈ R2 : y < 0}. Proposition 3. Consider the discontinuous piecewise linear differen- tial system with three pieces separated by the set Σ defined by 1 ẋ = − y − 1, ẏ = x − 1, in Q1 , 4 1 4 (2) ẋ = − y, ẏ = x + , in Q2 , 4 3 1 ẋ = − y, ẏ = x + 1, in H. 4 These three linear differential systems are centers, and the discontinu- ous piecewise linear differential system has exactly one limit cycle, see it in Figure 2. LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 5 Figure 4. The three limit cycles of the discontinuous piece- wise linear differential system (4). Proposition 4. Consider the discontinuous piecewise linear differen- tial system with three pieces separated by the set Σ defined by 1 1 ẋ = −y + , ẏ = x + , in Q1 , 2 2 479 97 1 31 (3) ẋ = − y+ , ẏ = x − , in Q2 , 1000 400 2 200 1 1 ẋ = − y, ẏ = 2x + , in H. 8 10 These three linear differential systems are centers, and the discontinu- ous piecewise linear differential system has exactly two limit cycles, see them in Figure 3. Proposition 5. Consider the discontinuous piecewise linear differen- tial system with three pieces separated by the set Σ defined by (4) 2 379 237 ẋ = − y− , ẏ = 2x − , in Q1 , 1565 1565 313 √ 4 11566 4 4430533 ẋ = − y− , ẏ = 8x + , in Q2 , 1565 10955 2191 √  2 4430533 − 1299 ẋ = −2y, ẏ = 8x + , in H. 2191 These three linear differential systems are centers, and the discontin- uous piecewise linear differential system has exactly three limit cycles, see them in Figure 4. Propositions 3, 4 and 5 are proved in section 3. 6 JAUME LLIBRE AND XIANG ZHANG 2. Proof of Theorem 2 It is known that a normal form for a linear differential system having a center is given in the next result. Since the proof is short we provide it. Lemma 6. The equations of a linear differential system with a center are 4b2 + ω 2 ẋ = −bx − y + d, ẏ = ax + by + c, 4a with a > 0 and ω > 0. Proof. Consider an arbitrary linear differential system ẋ = Ax + By + d, ẏ = ax + by + c, in the plane and suppose that it has a center. Then the eigenvalues of this system are p A + b ± 4aB + (A − b)2 . 2 If this system has a center then A + b = 0 and 4aB + (A − b)2 = −ω 2 for some ω > 0 and aB < 0, i.e. if A = −b, B = −(4b2 + ω 2 )/(4a) and a > 0. Therefore the lemma is proved.  We remark that the normal form in Lemma 6 is independent of the change of coordinates, so in the next proof of Theorem 2 we can use this normal form in each of the three regions Q1 , Q2 and H for the three centers. Now we start the proof of Theorem 2. Suppose that we have a discontinuous piecewise linear differential system separated by the set Σ formed by three linear centers. By Lemma 6 we can write such a discontinuous piecewise linear differential system as 4β 2 + ω 2 ẋ = −βx − y + δ, ẏ = αx + βy + γ, in Q1 , 4α 4b2 + w2 (5) ẋ = −bx − y + d, ẏ = ax + by + c, in Q2 , 4a 4B 2 + W 2 ẋ = −Bx − y + D, ẏ = Ax + By + C, in H. 4A LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 7 The linear centers in Q1 , Q2 and H have the first integrals H1 , H2 and H3 , respectively, where H1 (x, y) = 4(αx + βy)2 + 8α(γx − δy) + ω 2 y 2 , H2 (x, y) = 4(ax + by)2 + 8a(cx − dy) + w2 y 2 , H3 (x, y) = 4(Ax + By)2 + 8A(Cx − Dy) + W 2 y 2 . Suppose that this discontinuous piecewise differential system with three linear centers has some limit cycle intersecting each branch of Σ \ {(0, 0)} in one point, namely (x+ , 0) with x+ > 0, (0, y+ ) with y+ > 0, and (x− , 0) with x− < 0. Then the first integrals H1 , H2 and H3 must satisfy the following three equations H1 (x+ , 0) − H1 (0, y+ ) = 0, H2 (0, y+ ) − H2 (x− , 0) = 0, H3 (x− , 0) − H3 (x+ , 0) = 0, or equivalently 4α2 x2+ − (4β 2 + ω 2 ) y+ 2 + 8αγx+ + 8αδy+ = 0, (6) 4a2 x2− − (4b2 + w2 ) y+ 2 + 8acx− + 8ady+ = 0, −4A(x+ − x− )(Ax+ + Ax− + 2C) = 0. Since x+ − x− > 0, by Bezout Theorem (see for instance [18]), the system of the three polynomial equations (6) of degrees 2, 2 and 1 have at most 4 real solutions (x+ , y+ , x− ). So the discontinuous piecewise linear differential system (2) can have at most 4 limit cycles. Since x+ > 0, x− < 0 and A 6= 0 the third equation of system (6) reduces to Ax+ + Ax− + 2C = 0. Isolating x− from this equation and substituting it into the second equation of system (6), we get the system 4α2 x2+ − (4β 2 + ω 2 ) y+ 2 + 8αγx+ + 8αδy+ = 0, 8a(2aC − Ac) (7) 4a2 x2+ − (4b2 + w2 ) y+ 2 + x+ A 16aC(aC − Ac) +8ady+ + = 0. A2 Note that we can assume C(aC − Ac) 6= 0, otherwise the polynomial system (7) has the solution (0, 0) which cannot contribute with a limit cycle having a unique point in each branch of Σ \ {(0, 0)}. So, in this case system (7) only can have at most 3 solutions which can provide 8 JAUME LLIBRE AND XIANG ZHANG at most 3 limit cycles for the discontinuous piecewise linear differential system (2), and the theorem is proved if C(aC − Ac) = 0. In order that the polynomial system (6) can have eventually four real k solutions, (xk+ , y+ , xk− ) for k = 1, 2, 3, 4, producing four limit cycles for the discontinuous piecewise linear differential system (2) it is necessary that x1+ , x2+ , x3+ and x4+ have the same order as that of y+ 1 2 , y+ 3 , y+ 4 and y+ . For instance, (8) x1+ < x2+ < x3+ < x4+ 1 and y+ 2 < y+ 3 < y+ 4 < y+ , otherwise the solutions of the piecewise linear differential system (2) k connecting the points (xk+ , 0) and (0, y+ ) in the quadrant Q1 would in- tersect, in contradiction with the Uniqueness Theorem for the solutions of an ordinary differential equation. In what follows, but only inside the rest of this section, in order to simplify the notation we shall write xk and yk instead of xk+ and y+ k , respectively. We consider the following two conics F1 (x, y) = ax2 + by 2 + cx + dy = 0, (9) F2 (x, y) = Ax2 + By 2 + Cx + Dy + E = 0. where abABE 6= 0. We claim that the polynomial systems (9) cannot have 4 solutions (xk , yk ) with x1 , x2 , x3 , x4 and y1 , y2 , y3 , y4 having the same order as illustrated in (8). Since the systems (9) contain the systems (7), it follows that systems (7) cannot have 4 solutions (xk , yk ) with x1 , x2 , x3 , x4 and y1 , y2 , y3 , y4 having the same order. Hence the discontinuous piecewise linear differential system (2) cannot have 4 limit cycles, and this completes the proof of the theorem after proving Propositions 3, 4 and 5. For proving the claim we consider that the two conics of (9) share four points (xk , yk ) with x1 , x2 , x3 , x4 and y1 , y2 , y3 , y4 having the same order, and we shall arrive to a contradiction. Since the first conic (9) has the four distinct points (xk , yk ), these points satisfy the system ax21 + by12 + cx1 + dy1 = 0, ax22 + by22 + cx2 + dy2 = 0, (10) ax23 + by32 + cx3 + dy3 = 0, ax24 + by42 + cx4 + dy4 = 0. LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 9 The determinant of this linear system in the variables a, b, c and d is the determinant of the matrix  2 2  x1 y 1 x1 y1  2 2   x2 y 2 x2 y2  (11) M =  x2 y 2 x ,  3 3 3 y3   x24 y42 x4 y4 which is zero, otherwise the coefficients of the first conic of (9) will be all zero, in contradiction with the fact that ab 6= 0. So the rank of the matrix M is at most 3. Since the second conic (9) also has the four points (xk , yk ), these points satisfy the system Ax21 + By12 + Cx1 + Dy1 = −E, Ax22 + By22 + Cx2 + Dy2 = −E, (12) Ax23 + By32 + Cx3 + Dy3 = −E, Ax24 + By42 + Cx4 + Dy4 = −E. Case 1. Assume that the rank of the matrix M is 3. So a row of this matrix is a linear combination of the other three, without loss of generality assume that it is the fourth arrow. Then there exist λi for i = 1, 2, 3 not all zero such that x24 = λ1 x21 + λ2 x22 + λ3 x23 , y42 = λ1 y12 + λ2 y22 + λ3 y32 , x4 = λ1 x1 + λ2 x2 + λ3 x3 , y4 = λ1 y1 + λ2 y2 + λ3 y3 . We define K = x21 x2 y32 − x21 x3 y22 − x1 x22 y32 + x1 x23 y22 + x22 x3 y12 − x2 x23 y12 , and we consider two subcases. Subcase 1.1: K 6= 0. Under this assumption we can solve the first three equations of system (12) with respect the variables A, B and C obtaining A1 D + A2 E B1 D + B2 E C1 D + C2 E A= , B= , C= , K K K 10 JAUME LLIBRE AND XIANG ZHANG where A1 = −x1 y22 y3 + x1 y2 y32 + x2 y12 y3 − x2 y1 y32 − x3 y12 y2 + x3 y1 y22 , A2 = −x1 y22 + x1 y32 + x2 y12 − x2 y32 − x3 y12 + x3 y22 , B1 = −x21 x2 y3 + x21 x3 y2 + x1 x22 y3 − x1 x23 y2 − x22 x3 y1 + x2 x23 y1 , B2 = (x2 − x1 )(x1 − x3 )(x2 − x3 ), C1 = x21 y22 y3 − x21 y2 y32 − x22 y12 y3 + x22 y1 y32 + x23 y12 y2 − x23 y1 y22 , C2 = x21 y22 − x21 y32 − x22 y12 + x22 y32 + x23 y12 − x23 y22 . Substituting A, B and C in the fourth equation of (12) we obtain E(λ1 + λ2 + λ3 − 1) = 0. Since E 6= 0 we have λ1 + λ2 + λ3 = 1. Now we solve the system λ1 + λ2 + λ3 − 1 = 0, (13) (λ1 x1 + λ2 x2 + λ3 x3 )2 − λ1 x21 − λ2 x22 − λ3 x23 = 0, (λ1 y1 + λ2 y2 + λ3 y3 )2 − λ1 y12 − λ2 y22 − λ3 y32 = 0. with respect to the variables λ1 , λ2 and λ3 , and we get 1 λ1 = (−x1 y2 + x1 y3 − x2 y1 + 2x2 y2 − x2 y3 + x3 y1 − x3 y2 ) L (−x1 y2 + x1 y3 − x2 y1 + x2 y3 + x3 y1 + x3 y2 − 2x3 y3 ), 1 λ2 = − (−2x1 y1 + x1 y2 + x1 y3 + x2 y1 − x2 y3 + x3 y1 − x3 y2 ) L (−x1 y2 + x1 y3 − x2 y1 + x2 y3 + x3 y1 + x3 y2 − 2x3 y3 ), λ3 = 1 − λ1 − λ2 , provided that L = (x1 (y3 − y2 ) + x2 (y1 − y3 ) + x3 (y2 − y1 ))2 6= 0. We note that if L = 0 then taken −x1 y2 + x1 y3 + x2 y1 − x2 y3 x3 = , y1 − y2 and solving the system (13) we get −2λ3 y1 + y1 − y2 + 2λ3 y3 + R λ1 = , 2y1 − 2y2 −2λ3 y1 + y1 − y2 + 2λ3 y3 − R λ2 = , 2y1 − 2y2 p where R = ± y12 − 2y1 (−2λ3 y2 + y2 + 2λ3 y3 ) + y22 − 4λ3 y2 y3 + 4λ3 y32 . If R is complex we are done. If R is real then λ1 and λ2 depend on λ3 , and the rank of the matrix M is not 3 as we are considering, so L cannot be zero. LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 11 In short, we have that (14) F2 = D(d1 x2 + d2 y 2 + d3 x + d4 y) + E(e1 x2 + e2 y 2 + e3 x + e4 y + e5 ), where d1 = −x1 y22 y3 + x1 y2 y32 + x2 y12 y3 − x2 y1 y32 − x3 y12 y2 + x3 y1 y22 , d2 = −x21 x2 y3 + x21 x3 y2 + x1 x22 y3 − x1 x23 y2 − x22 x3 y1 + x2 x23 y1 , d3 = x21 y22 y3 − x21 y2 y32 − x22 y12 y3 + x22 y1 y32 + x23 y12 y2 − x23 y1 y22 , d4 = x21 x2 y32 − x21 x3 y22 − x1 x22 y32 + x1 x23 y22 + x22 x3 y12 − x2 x23 y12 , e1 = −x1 y22 + x1 y32 + x2 y12 − x2 y32 − x3 y12 + x3 y22 , e2 = (x2 − x1 )(x1 − x3 )(x2 − x3 ), e3 = x21 y22 − x21 y32 − x22 y12 + x22 y32 + x23 y12 − x23 y22 , e4 = 0, e5 = x21 x2 y32 − x21 x3 y22 − x1 x22 y32 + x1 x23 y22 + x22 x3 y12 − x2 x23 y12 . From the computations for obtaining the expression of F2 given in (14) it follows immediately the expression of F1 , i.e. (15) F1 = d(d1 x2 + d2 y 2 + d3 x + d4 y). Now solving the system F1 (x, y) = 0 and F2 (x, y) = 0, we obtain the four points (xk , yk ) for k = 1, 2, 3, 4 with 2(x1 − x3 )(x3 − x2 )(y1 − y2 ) x4 = − + x1 + x2 − x3 , (16) N 2(y1 − y3 )(y3 − y2 )(x1 − x2 ) y4 = + y1 + y2 − y3 , N where N = x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 ) 6= 0 because L = N 2 and we have proved that L 6= 0. We can label the points (xk , yk ) in order that (17) x1 < x2 < x3 and y1 < y2 < y3 . Now we claim that the following four systems of inequalities do not hold: x1 < x2 < x3 < x4 and y1 < y2 < y3 < y4 , x1 < x2 < x4 < x3 and y1 < y2 < y4 < y3 , (18) x1 < x4 < x2 < x3 and y1 < y4 < y2 < y3 , x4 < x1 < x2 < x3 and y4 < y1 < y2 < y3 . 12 JAUME LLIBRE AND XIANG ZHANG We prove this claim. We can assume without loss of generality that N > 0. Using the expressions of x4 and y4 given in (16) we obtain x2 − x3 x 4 − x1 = (2(x1 − x3 )(y1 − y2 ) (19) N + x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 )) Since x2 > x1 we have 2(x1 − x3 )(y1 − y2 ) + x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 ) > 2(x1 − x3 )(y1 − y2 ) + x1 (y2 − y1 ) + x3 (y1 − y2 ) = (x1 − x3 )(y1 − y2 ) > 0. Then from (17) and (19) it follows that x4 − x1 < 0. Now note that y2 − y3 y4 − y1 = (−2(x1 − x2 )(y1 − y3 ) (20) N + x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 )). Since x3 > x1 we have −2(x1 − x2 )(y1 − y3 ) + x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 ) < −2(x1 − x2 )(y1 − y3 ) + x1 (y1 − y3 ) + x2 (y3 − y1 ) = −(x1 − x2 )(y1 − y3 ) < 0. Therefore from (17) and (20) we obtain y4 − y1 > 0. Doing similar computations for x4 − x2 and y4 − y2 we get x1 − x3 x4 − x2 = (2(x2 − x3 )(y1 − y2 ) N + x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 )) (x1 − x3 )(x2 − x3 )(y1 − y2 ) < < 0, (21) N y1 − y3 y4 − y2 = (−2(x1 − x2 )(y2 − y3 ) N + x1 (y2 − y3 ) + x2 (y3 − y1 ) + x3 (y1 − y2 )) (−2(x1 − x2 )(y1 − y3 )(y2 − y3 ) > > 0. N LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 13 Finally 2(x1 − x3 )(x3 − x2 )(y1 − y2 ) x4 − x3 = − + x1 + x2 − 2x3  N  (x1 − x3 )(x2 − x3 )(y1 − y2 ) = + x1 − x3 N   (x1 − x3 )(x2 − x3 )(y1 − y2 ) + + x2 − x3 N <0 + 0 = 0, (22) 2(y1 − y3 )(y3 − y2 )(x1 − x2 ) y4 − y3 = + y1 + y2 − 2y3  N  (y1 − y3 )(y3 − y2 )(x1 − x2 ) = + y1 − y3 N   (y1 − y3 )(y3 − y2 )(x1 − x2 ) + + y2 − y3 N >0 + 0 = 0. Summarizing these results we obtain (x4 − x1 )(y4 − y1 ) < 0, (x4 − x2 )(y4 − y2 ) < 0, (x4 − x3 )(y4 − y3 ) < 0. This proves the claim. Since at least one of the four inequalities in (18) must hold in order that the piecewise linear differential system (2) has 4 limit cycles, it follows that in this subcase the system has at most 3 limit cycles. Subcase 1.2: K = 0. Then solving the system F1 (x, y) = 0 and F2 (x, y) = 0 we get that y = ((y1 − y2 )x − x2 y1 + x1 y2 )/(x1 − x2 ), So we have a continuum of solutions, and consequently no limit cycles. Case 2: Assume that the rank of the matrix M is 2. Now two rows of the matrix M are linear combination of the other two rows. Without loss of generality we can suppose that the rows three and four are combination of the rows one and two, i.e. (23) x3 = λ1 x1 +λ2 x2 , y3 = λ1 y1 +λ2 y2 , x23 = λ1 x21 +λ2 x22 , y32 = λ1 y12 +λ2 y22 . A similar expression we have for x4 , y4 , x24 and y42 but with µ’s instead of λ’s. 14 JAUME LLIBRE AND XIANG ZHANG Again we assume that the conics F1 = 0 and F2 = 0 share four points (xk , yk ) for k = 1, 2, 3, 4. Substituting them in F2 we obtain Ax21 + By12 + Cx1 + Dy1 + E = 0, Ax22 + By22 + Cx2 + Dy2 + E = 0, (24) E(1 − λ1 − λ2 ) = 0, E(1 − µ1 − µ2 ) = 0. respectively. From (23) and (16) we get that either (λ1 , λ2 ) = (1, 0), or (λ1 , λ2 ) = (0, 1). Therefore, the third row is either equal to the first row, or two the second. In contradiction with the fact that the four points (xk , yk ) are distinct. Case 3: The rank of the matrix M is one. But then the four points (xk , yk ) are collinear in contradiction with the fact that the two hyper- bolas (7) cannot intersect in four collinear points. In summary, the discontinuous piecewise linear differential system (2) has at most 3 limit cycles. So the proof of Theorem 2 is completed, modulo the proof of the Propositions 3, 4 and 5. 3. Proofs of Propositions 3, 4 and 5 Proof of Proposition 3. We shall prove that the discontinuous piecewise linear differential system (2) having a center in each of the pieces Q1 , Q2 and H has exactly 1 limit cycle. Since ±4i are the eigenvalues of the matrices of the three linear differential systems of (2), these systems are centers. The three linear differential systems of (3) have the following first integrals H1 (x, y) = 4x2 + y 2 − 8x − 4y, 32 H2 (x, y) = 4x2 + y 2 + x, 3 2 2 H3 (x, y) = 4x + y + 8x, LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 15 in Q1 , Q2 and H, respectively. Then for the discontinuous piecewise linear differential system (2) the system (6) becomes 4x2+ − y+ 2 − 8x+ + 4y+ = 0, 32 4x2− − y+ 2 + x− = 0, 3 (x+ − x− )(2 + x− + x+ ) = 0. Taking into account that we are only interested in the solutions (x+ , y+ , x− ) satisfying x+ > 0, x− < 0 and y+ > 0, the unique solution of the pre- 1 vious system is (x1+ , y+ , x1− ) = (1, 2, −3). The first linear differential system of (2) has the solution x1 (t) = sin2 (4t) + cos2 (4t) − cos(4t),  y1 (t) = 2 sin2 (4t) + cos2 (4t) − sin(4t) , satisfying the initial conditions x1 (0) = 1 and y1 (0) = 0. The second linear differential system of (2) has the solution 1  x2 (t) = − 4 sin2 (4t) + 4 cos2 (4t) + 5 cos(4t) , 3 10 y2 (t) = sin(4t), 3 satisfying the initial conditions x2 (0) = 0 and y2 (0) = 2. And the third linear differential system of (2) has the solution x3 (t) = 2 cos(4t) − sin2 (4t) − cos2 (4t), y3 (t) = −4 sin(4t), satisfying the initial conditions x3 (0) = −3 and y3 (0) = 0. The time that the solution (x1 (t), y1 (t)) contained in Q1 needs to reach the point (0, 2) is t1 = 0.3926990954490209.. The time that the solution (x2 (t), y2 (t)) contained in Q2 needs to reach the point (−3, 0) is t2 = 0.624522886199127.. The time that the solution (x3 (t), y3 (t)) con- tained in H needs to reach the point (1, 0) is t3 = 0.785398163397448.. Drawing the the orbit (xk (t), yk (t)) for the times t ∈ [0, tk ] for k = 1, 2, 3, we obtain the limit cycle of Figure 2.  Proof of Proposition 4. We shall prove that the discontinuous piecewise linear differential system (3) having a center in each of the pieces Q1 , Q2 and H has exactly 2 limit cycles. 16 JAUME LLIBRE AND XIANG ZHANG r 1 479 1 Since ±i, ± i and ± i are the eigenvalues of the matrices of 20 5 2 the three linear differential systems of (3), these systems are centers. The three linear differential systems of (3) have the following first integrals H1 (x, y) = x2 + y 2 + x − y, 479 2 31 97 H2 (x, y) = x2 + y − x− y, 500 50 100 8 H3 (x, y) = 16x2 + +y 2 + x, 5 in Q1 , Q2 and H, respectively. Then for the discontinuous piecewise linear differential system (3) the system (6) becomes (x+ − y+ + 1)(x+ + y+ ) = 0, (25) 2 900x2− − 977y+ − 1410x− + 955y+ = 0, (x+ − x− )(1 − 10x+ − 10x− ) = 0. Taking into account that we are only interested in the solutions (x+ , y+ , x− ) satisfying x+ > 0, x− < 0 and y+ > 0, the unique two solutions of the previous system are     1 1 1 19 2 2 2 9 (x+ , y+ , x− ) = 2, 3, − and (x+ , y+ , x− ) = 1, 2, − . 10 10 The first linear differential system of (3) has the solution 1 x1 (t) = ((2u + 1) cos t + sin t − 1), 2 1 y1 (t) = ((2u + 1) sin t − cos t + 1), 2 satisfying the initial conditions x1 (0) = u and y1 (0) = 0. The second linear differential system of (3) has the solution r ! r ! 485 − 958v 1 479 31 1 479 31 x2 (t) = √ sin t − cos t + , 20 2395 20 5 100 20 5 100   r ! r ! 485 1 479 31 1 479 485 y2 (t) = v − cos t − √ sin t + , 958 20 5 2 2395 20 5 958 LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 17 satisfying the initial conditions x2 (0) = 0 and y2 (0) = v. And the third linear differential system of (3) has the solution     1 t x3 (t) = (20w + 1) cos −1 , 20 2   1 t y3 (t) = (20w + 1) sin , 5 2 satisfying the initial conditions x3 (0) = w and y3 (0) = 0. Now we consider the solution (x1k (t), yk1 (t)) for k = 1, 2, 3 of the dis- continuous piecewise linear differential system (3) corresponding to the solution (x1+ , y+1 , x1− ) = (u, v, w) = (2, 3, −19/10) of system (25). Then the time that the solution (x11 (t), y11 (t)) contained in Q1 needs to reach the point (0, v) is t1 = 1.5707963267948966.. The time that the so- lution (x12 (t), y21 (t)) contained in Q2 needs to reach the point (w, 0) is t2 = 3.3659397308004984.. The time that the solution (x13 (t), y31 (t)) con- tained in H needs to reach the point (u, 0) is t3 = 6.283185307179586.. Let (x2k (t), yk2 (t)) for k = 1, 2, 3 be the solution of the discontinu- ous piecewise linear differential system (3) corresponding to the solu- 2 tion (x2+ , y+ , x2− ) = (u, v, w) = (1, 2, −9/10) of system (25). Then the time that the solution (x21 (t), y12 (t)) contained in Q+ needs to reach the point (0, v) is T1 = 1.5707963267948966.. The time that the so- lution (x22 (t), y22 (t)) contained in Q2 needs to reach the point (w, 0) is T2 = 3.473334629546722.. The time that the solution (x23 (t), y32 (t)) con- tained in H needs to reach the point (u, 0) is T3 = 6.283185307179586.. Drawing the two orbits (xjk (t), ykj (t)) for j = 1, 2 and for the times t ∈ [0, tk ] and t ∈ [0, Tk ] for k = 1, 2, 3, respectively, we obtain the 2 limit cycles of Figure 3.  Proof of Proposition 5. We shall prove that the discontinuous piecewise linear differential system (4) having a center in each of the pieces Q1 , Q2 and H has exactly 3 limit cycles. r 2i 2 Since ± √ , ±4 i and ±4i are the eigenvalues of the ma- 1565 1565 trices of the three linear differential systems of (4), these systems are centers. 18 JAUME LLIBRE AND XIANG ZHANG The three linear differential systems of (4) have the following first integrals H1 (x, y) = 1565x2 + y 2 − 1185x + 379y, √ H2 (x, y) = 21910x2 + 7y 2 + 10 4430533x + 5783y, √ H3 (x, y) = 8764x2 + 2191y 2 + (2 4430533 − 2598)x, in Q1 , Q2 and H, respectively. Then for the discontinuous piecewise linear differential system (4) the system (6) becomes 2 1565x2+ − y+ − 1185x+ − 379y+ = 0, 2 2 √ (26) 21910x− − 7y+ + 10 4430533x− − 5783y+ = 0, √  (x+ − x− ) 4382x+ + 4382x− + 4430533 − 1299 = 0. Taking into account that we are only interested in the solutions (x+ , y+ , x− ) satisfying x+ > 0, x− < 0 and y+ > 0, the unique three solutions of the previous system are √ ! −3083 − 4430533 (x1+ , y+ 1 , x1− ) = 1, ,1 , 4382 √ ! −7465 − 4430533 (x2+ , y+ 2 , x2− ) = 2, , 10 , 4382 √ ! −11847 − 4430533 (x3+ , y+ 3 , x3− ) = 3, , 26 . 4382 The first linear differential system of (3) has the solution       237 2t 379 2t 237 x1 (t) = u − cos √ + √ sin √ + , 626 1565 2 1565 1565 626 r     1 5 2t 379 2t 379 y1 (t) = (237 − 626u) sin √ + cos √ − , 2 313 1565 2 1565 2 satisfying the initial conditions x1 (0) = u and y1 (0) = 0. LIMIT CYCLES CREATED BY PIECEWISE LINEAR CENTERS 19 The second linear differential system of (3) has the solution r ! r ! 1 2 √ 2 x2 (t) = sin 2 t 3130(14v + 5783) cos 2 t 21910 1565 1565 r !! √ 2 −10 4430533 sin 2 t , 1565   r ! r r ! 5783 2 1 22152665 2 y2 (t) = v+ cos 4 t − sin 4 t 14 1565 7 626 1565 5783 − , 14 satisfying the initial conditions x2 (0) = 0 and y2 (0) = v. And the third linear differential system of (3) has the solution 1 √   x3 (t) = 8764w cos(4t) + 4430533 − 1299 (cos(4t) − 1) , 8764 √  1299 − 8764w − 4430533 sin(4t) y3 (t) = , 4382 satisfying the initial conditions x3 (0) = w and y3 (0) = 0. Now we consider the solution (x1k (t), yk1 (t)) for k = 1, 2, 3 of the discontinuous piecewise linear differential system (4) corresponding to the solution (x1+ , y+ 1 , x1− ) of system (26). Then the time that the so- lution (x11 (t), y11 (t)) contained in Q1 needs to reach the point (0, v) is t1 = 4.10363864680248.. The time that the solution (x12 (t), y21 (t)) con- tained in Q2 needs to reach the point (w, 0) is t2 = 1.11762450719575.. The time that the solution (x13 (t), y31 (t)) contained in H needs to reach the point (u, 0) is t3 = 0.785398163397448.. Let (x2k (t), yk2 (t)) for k = 1, 2, 3 be the solution of the discontinu- ous piecewise linear differential system (4) corresponding to the so- lution (x2+ , y+ 2 , x2− ) of system (26). Then the time that the solution (x21 (t), y12 (t)) contained in Q+ needs to reach the point (0, v) is r1 = 7.93799227264621.. The time that the solution (x22 (t), y22 (t)) contained in Q2 needs to reach the point (w, 0) is r2 = 2.02943545009903.. The time that the solution (x23 (t), y32 (t)) contained in H needs to reach the point (u, 0) is r3 = 0.785398163397448.. Let (x3k (t), yk3 (t)) for k = 1, 2, 3 be the solution of the discontinu- ous piecewise linear differential system (4) corresponding to the so- lution (x3+ , y+ 3 , x3− ) = of system (26). Then the time that the solu- 3 3 tion (x1 (t), y1 (t)) contained in Q+ needs to reach the point (0, v) is 20 JAUME LLIBRE AND XIANG ZHANG s1 = 11.27688306691738.. The time that the solution (x32 (t), y23 (t)) con- tained in Q2 needs to reach the point (w, 0) is s2 = 2.88219547492608.. The time that the solution (x33 (t), y33 (t)) contained in H needs to reach the point (u, 0) is s3 = 0.785398163397448.. Drawing the three orbits (xjk (t), ykj (t)) for j = 1, 2, 3 and for the times t ∈ [0, tk ], t ∈ [0, rk ] and t ∈ [0, sk ] for k = 1, 2, 3, respectively, we obtain the 3 limit cycles of Figure 4.  Acknowledgments This work is supported by the Ministerio de Economı́a, Industria y Competitividad, Agencia Estatal de Investigación grants MTM2016- 77278-P (FEDER) and MDM-2014-0445, the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 Euro- pean Research Council grant MSCA-RISE-2017-777911. 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Simpson, Bifurcations in Piecewise–Smooth Continuous Systems, World Scientific Series on Nonlinear Science A, vol 69, World Scientific, Sin- gapore, 2010. [20] M.A. Teixeira, Perturbation theory for non-smooth systems, In Mathematics of complexity and dynamical systems. Vols. 1–3, Springer, New York, 2012, pp. 1325–1336. [21] A.M. Zhabotinsky, Periodical oxidation of malonic acid in solution (a study of the Belousov reaction kinetics), Biofizika 9 (1964), 306–311. 1 Departament de Matemàtiques, Universitat Autònoma de Barce- lona, 08193 Bellaterra, Barcelona, Catalonia, Spain E-mail address: jllibre@mat.uab.cat 2 School of Mathematical Sciences, MOE–LSC, Shanghai Jiao Tong University, Shanghai, 200240, P. R. China E-mail address: xzhang@sjtu.edu.cn