This is a preprint of: “Limit cycles of 3-dimensional discontinuous piecewise differential systems formed by linear centers”, Jaume Llibre, Jaime R. de Moraes, Sao Paulo J. Math. Sci., vol. 15(2), 858–874, 2021. DOI: [10.1007/s40863-021-00237-0] LIMIT CYCLES OF 3-DIMENSIONAL DISCONTINUOUS PIECEWISE DIFFERENTIAL SYSTEMS FORMED BY LINEAR CENTERS JAUME LLIBRE1 AND JAIME R. DE MORAES2 Abstract. In this paper we deal with 3-dimensional discontinuous piecewise differential systems formed by linear centers and separated by one plane or two parallel planes. We prove that these systems separated by one plane have no limit cycles, besides the systems separated by two parallel planes have at most one limit cycle, and that there are systems having such a limit cycle. So we solve the extension of the 16th Hilbert problem to this class of differential systems. 1. Introduction and Statement of the Main Results One of the main goals in the qualitative theory is to study the number of limit cycles of the differential systems. In part this problem was motivated by the 16-th Hilbert problem (1900), see [8, 16] for more details. Limit cycles play a main role for understanding the dynamics of many systems, see for instance [1, 2, 6, 9, 10]. On the other hand there are many problems that are modeled using discontinu- ous piecewise differential systems. These systems appear in various situations like mechanical systems and control theory, see for instance [3, 4]. In particular the study of discontinuous piecewise linear differential systems in the plane separated by straight lines is also an important class of differential systems which appear in models of mechanics and electrical circuits among others, see for instance [5, 13, 15]. Following the Filippov’s convention [7] the discontinuous piecewise linear dif- ferential systems can have sliding or crossing limit cycles. A sliding limit cycle contains sliding segments on the line of discontinuity, whereas crossing limit cycles contain only crossing points. In this paper we work with crossing limit cycles, or simply limit cycles. In [11] it was proved that discontinuous piecewise linear differential systems in the plane separated by one straight line and formed by two linear centers have no limit cycles, besides discontinuous piecewise linear differential systems in the plane separated by two parallel straight lines and formed by three linear centers have at most one limit cycle, and that there are systems having such a limit cycle. The main goal of this paper is to extend these previous results to 3-dimensional discontinuous piecewise differential systems in the space separated by one plane or 2010 Mathematics Subject Classification. 34C07, 34C23, 34C25, 37C27, 37G15. Key words and phrases. discontinuous piecewise differential systems, periodic orbits, linear centers, first integrals, limit cycles. 1 2 J. LLIBRE AND J.R. DE MORAES two parallel planes and formed by linear centers. More precisely, we consider the differential system (1) ẋ = −y, ẏ = x, ż = 0. This differential system has the z−axis filled of singular points, and at every plane z = z0 , with z0 a constant, there exists a linear center at (0, 0, z0 ), i.e. all the orbits in this plane are periodic formed by circles centered at (0, 0, z0 ), of course with the exception of the singular point (0, 0, z0 ). We shall pass from the coordinates (x, y, z) to (X, Y, Z) through the affine change of coordinates X = a0 + a1 x + a2 y + a3 z, Y = b0 + b1 x + b2 y + b3 z, Z = c0 + c1 x + c2 y + c3 z, assuming that the determinant of matrix of the change of coordinates D = D(a, b, c) = −a1 b2 c3 + a1 b3 c2 + a2 b1 c3 − a2 b3 c1 − a3 b1 c2 + a3 b2 c1 6= 0, where a = (a1 , a2 , a3 ), b = (b1 , b2 , b3 ), c = (c1 , c2 , c3 ). In the new coordinates (X, Y, Z) the differential system (1) writes Ẋ = P0 + P1 X + P2 Y + P3 Z, (2) Ẏ = Q0 + Q1 X + Q2 Y + Q3 Z, Ż = R0 + R1 X + R2 Y + R3 Z, where P0 = −D−1 (a1 (−a0 b1 c3 + a0 b3 c1 − a3 b0 c1 + a3 b1 c0 ) +a2 (−a0 b2 c3 + a0 b3 c2 + a2 b0 c3 − a2 b3 c0 − a3 b0 c2 + a3 b2 c0 ) +a21 (b0 c3 − b3 c0 )), P1 = D−1 (−a1 b1 c3 + a1 b3 c1 − a2 b2 c3 + a2 b3 c2 ),  P2 = D−1 (c3 a21 + a22 − a3 (a1 c1 + a2 c2 )),  P3 = −D−1 (b3 a21 + a22 − a3 (a1 b1 + a2 b2 )),   Q0 = D−1 (c3 a0 b21 + b22 − a1 b0 b1 − a2 b0 b2 + b3 (−a0 (b1 c1 + b2 c2 )  +a1 b1 c0 + a2 b2 c0 ) + a3 b0 b1 c1 + b2 (b0 c2 − b2 c0 ) − b21 c0 ),  Q1 = −D−1 (c3 b21 + b22 − b3 (b1 c1 + b2 c2 )), Q2 = D−1 (c3 (a1 b1 + a2 b2 ) − a3 (b1 c1 + b2 c2 )),  Q3 = −D−1 (b3 (a1 b1 + a2 b2 ) − a3 b21 + b22 ), R0 = −D−1 (−a0 b1 c1 c3 − a0 b2 c2 c3 + a0 b3 c21 + a0 b3 c22 + a1 c1 (b0 c3 − b3 c0 )   +a2 b0 c2 c3 − a2 b3 c0 c2 + a3 −b0 c21 + c22 + b1 c0 c1 + b2 c0 c2 ),  R1 = −D−1 (c3 (b1 c1 + b2 c2 ) − b3 c21 + c22 ),  R2 = −D−1 (a3 c21 + c22 − c3 (a1 c1 + a2 c2 )), R3 = D−1 (−a1 b3 c1 − a2 b3 c2 + a3 b1 c1 + a3 b2 c2 ). For simplicity we rename the variables X, Y and Z by x, y and z, respectively. Thus system (2) writes (ẋ, ẏ, ż)T = V + M · (x, y, z)T , LIMIT CYCLES OF DISCONTINUOUS PIECEWISE LINEAR SYSTEMS 3 where   P1 P2 P3 V (a, b, c) = (P0 , Q0 , R0 )T and M (a, b, c) = Q1 Q2 Q3  . R1 R2 R3 Now we consider the 3-dimensional discontinuous piecewise differential system separated by plane x = 0 (ẋ, ẏ, ż)T = V (a, b, c) + M (a, b, c) · (x, y, z)T , in x > 0, (3) T T (ẋ, ẏ, ż) = V (α, β, γ) + M (α, β, γ) · (x, y, z) , in x < 0, and the one separated by the parallel planes x = −1 and x = 1 (ẋ, ẏ, ż)T = V (a, b, c) + M (a, b, c) · (x, y, z)T , in x > 1, (4) (ẋ, ẏ, ż)T = V (A, B, C) + M (A, B, C) · (x, y, z)T , in |x| < 1, (ẋ, ẏ, ż)T = V (α, β, γ) + M (α, β, γ) · (x, y, z)T , in x < −1, where we assume that D(a, b, c) 6= 0, D(α, β, γ) 6= 0 and D(A, B, C) 6= 0. In what follows we state the main results of this paper. Theorem 1. The discontinuous piecewise linear differential systems (3) have no limit cycles. Theorem 2. The discontinuous piecewise linear differential systems (4) have at most one limit cycle. Moreover there are systems in this class having one limit cycle. Theorems 1 and 2 are proved in sections 2 and 3, respectively. The main tool for proving these theorems is the use of the first integrals of the differential sys- tems which form the discontinuous piecewise differential systems, this technique for studying the limit cycles already was used in [12]. Proposition 1. The discontinuous piecewise linear differential system separated by the two parallel planes x = −1 and x = 1 ẋ = p0 , ẏ = y − z, ż = p0 , in x > 1, (5) ẋ = p1 , ẏ = p2 , ż = p3 , in |x| < 1, ẋ = −x + 2y − 2z, ẏ = −x + 2y − 2z, ż = −x + 2y − z, in x < −1, where 1 x z 13x 11y 21z 5 p0 = − − + 2y − , p1 = − + − + , 2 2 2 16 8 8 16 5x y 3z 3 33x 7y 9z 7 p2 = + − + , p3 = − + + , 8 4 4 8 32 16 16 32 has one limit cycle, the one of Figure 1. Moreover this limit cycle is stable. Proposition 1 is proved at the end of section 3. 4 J. LLIBRE AND J.R. DE MORAES Figure 1. The limit cycle of discontinuous piecewise linear differential system (5). 2. Proof of Theorem 1 The differential system in x > 0 of the discontinuous linear piecewise differential system (3) has the two independent first integrals h1 (x, y, z) = (−a3 b1 c0 + a1 b3 c0 + a3 b0 c1 − a0 b3 c1 − a1 b0 c3 + a0 b1 c3 +b3 c1 x − b1 c3 x − a3 c1 y + a1 c3 y + a3 b1 z − a1 b3 z)2 +(a3 b2 c0 − a2 b3 c0 − a3 b0 c2 + a0 b3 c2 + a2 b0 c3 − a0 b2 c3 −b3 c2 x + b2 c3 x + a3 c2 y − a2 c3 y − a3 b2 z + a2 b3 z)2 , h2 (x, y, z) = −a2 b1 c0 + a1 b2 c0 + a2 b0 c1 − a0 b2 c1 − a1 b0 c2 + a0 b1 c2 +(b2 c1 − b1 c2 )x + (−a2 c1 + a1 c2 )y + (a2 b1 − a1 b2 )z, and the differential system in x < 0 has the two independent first integrals f1 and f2 obtained respectively from h1 and h2 , changing the parameters ai , bi and ci by αi , βi and γi respectively, for i = 0, 1, 2, 3. A limit cycle for discontinuous piecewise differential system (3) must intersect the plane x = 0 in two distinct points, denoted by (0, y1 , z1 ) and (0, y2 , z2 ), and such two points must satisfy the system of equations e1 = h1 (0, y2 , z2 ) − h1 (0, y1 , z1 ) = 0, e2 = h2 (0, y2 , z2 ) − h2 (0, y1 , z1 ) = 0, e3 = f1 (0, y2 , z2 ) − f1 (0, y1 , z1 ) = 0, e4 = f2 (0, y2 , z2 ) − f2 (0, y1 , z1 ) = 0. LIMIT CYCLES OF DISCONTINUOUS PIECEWISE LINEAR SYSTEMS 5 Taking the change of variables y1 = y + y2 and z1 = z + z2 we obtain e2 = (a2 c1 − a1 c2 )y + (−a2 b1 + a1 b2 )z, e4 = (α2 γ1 − α1 γ2 )y + (−α2 β1 + α1 β2 )z. In order that the system e2 = e4 = 0 has non-trivial solutions we need that the following determinant be zero ∆ = −a2 α2 β1 c1 + α1 a2 β2 c1 + a2 α2 b1 γ1 − a1 α2 b2 γ1 + a1 α2 β1 c2 −a1 α1 β2 c2 − α1 a2 b1 γ2 + a1 α1 b2 γ2 . However now there are only three independent equations e1 = e2 = e3 = 0 and four unknowns variables y1 , y2 , z1 and z2 . Thus always we have a continuum of periodic solutions and no limit cycles. Therefore Theorem 1 is proved. 3. Proof of Theorem 2 We note that the differential system in x > 1 of the discontinuous piecewise differential system (4) has the same two independent first integrals h1 and h2 given in the proof of Theorem 1. The differential system in |x| < 1 has the two inde- pendent first integrals f1 and f2 obtained respectively from h1 and h2 replacing the parameters ai , bi and ci by Ai , Bi and Ci , respectively for i = 0, 1, 2, 3, and the differential system in x < −1 has the two independent first integrals g1 and g2 obtained respectively from h1 and h2 , changing the parameters ai , bi and ci by αi , βi and γi respectively for i = 0, 1, 2, 3. A limit cycle of the discontinuous piecewise differential system (4) must intersect each plane x = 1 and x = −1 in two distinct points, denoted by (1, y1 , z1 ) and (1, y2 , z2 ), and (−1, y3 , z3 ) and (−1, y4 , z4 ), respectively. Such four points must satisfy the system of equations (6) E1 = h1 (1, y2 , z2 ) − h1 (1, y1 , z1 ) = 0, E2 = h2 (1, y2 , z2 ) − h2 (1, y1 , z1 ) = 0, E3 = f1 (1, y2 , z2 ) − f1 (−1, y3 , z3 ) = 0, E4 = f2 (1, y2 , z2 ) − f2 (−1, y3 , z3 ) = 0, E5 = g1 (−1, y3 , z3 ) − g1 (−1, y4 , z4 ) = 0, E6 = g2 (−1, y3 , z3 ) − g2 (−1, y4 , z4 ) = 0, E7 = f1 (−1, y4 , z4 ) − f1 (1, y1 , z1 ) = 0, E8 = f2 (−1, y4 , z4 ) − f2 (1, y1 , z1 ) = 0. Applying the change of parameters K0 = b3 c1 − b1 c3 , K1 = a3 c1 − a1 c3 , K2 = a3 b1 − a1 b3 , K3 = b3 c2 − b2 c3 , K4 = a3 c2 − a2 c3 , K5 = a3 b2 − a2 b3 , K6 = a2 c1 − a1 c2 , K7 = a2 b1 − a1 b2 , K8 = B3 C1 − B1 C3 , K9 = A3 C1 − A1 C3 , K10 = A3 B1 − A1 B3 , K11 = B3 C2 − B2 C3 , K12 = A3 C2 − A2 C3 , K13 = A3 B2 − A2 B3 , K14 = B2 C1 − B1 C2 , K15 = A2 C1 − A1 C2 , K16 = A2 B1 − A1 B2 , K17 = α3 β1 − α1 β, K18 = α3 γ1 − α1 γ3 , K19 = α3 β2 − α2 β3 , K20 = β1 γ3 − β3 γ1 , K21 = α3 γ2 − α2 γ3 , K22 = β2 γ3 − β3 γ2 , K23 = α2 β1 − α1 β2 , K24 = α2 γ1 − α1 γ2 , 6 J. LLIBRE AND J.R. DE MORAES the polynomials Ei for i = 1, ..., 8 write E1 = −((a0 − 1)K0 + K1 (y1 − b0 ) + K2 (c0 − z1 ))2 + ((a0 − 1)K0 +K1 (y2 − b0 ) + K2 (c0 − z2 ))2 − ((a0 − 1)K3 + K4 (y1 − b0 ) +K5 (c0 − z1 ))2 + ((a0 − 1)K3 + K4 (y2 − b0 ) + K5 (c0 − z2 ))2 , E2 = K6 (y1 − y2 ) − K7 (z1 − z2 ), E3 = ((A0 − 1)K8 + K9 (y2 − B0 ) + K10 (C0 − z2 ))2 − ((A0 + 1)K8 +K9 (y3 − B0 ) + K10 (C0 − z3 ))2 + ((A0 − 1)K11 + K12 (y2 − B0 ) +K13 (C0 − z2 ))2 − ((A0 + 1)K11 + K12 (y3 − B0 ) + K13 (C0 − z3 ))2 , E4 = 2K14 − K15 (y2 − y3 ) + K16 (z2 − z3 ), E5 = (K17 (z3 − γ0 ) + K18 (β0 − y3 ) + (α0 + 1)K20 )2 − (K17 (z4 − γ0 ) +K18 (β0 − y4 ) + (α0 + 1)K20 )2 + (K19 (z3 − γ0 ) + K21 (β0 − y3 ) +(α0 + 1)K22 )2 − (K19 (z4 − γ0 ) + K21 (β0 − y4 ) + (α0 + 1)K22 )2 , E6 = K23 (z3 − z4 ) − K24 (y3 − y4 ), E7 = −((A0 − 1)K8 + K9 (y1 − B0 ) + K10 (C0 − z1 ))2 + ((A0 + 1)K8 +K9 (y4 − B0 ) + K10 (C0 − z4 ))2 − ((A0 − 1)K11 + K12 (y1 − B0 ) +K13 (C0 − z1 ))2 + ((A0 + 1)K11 + K12 (y4 − B0 ) + K13 (C0 − z4 ))2 , E8 = −2K14 + K15 (y1 − y4 ) − K16 (z1 − z4 ). Remark 1. If (y1 , y2 , y3 , y4 , z1 , z2 , z3 , z4 ) is a solution of system (6), then (y2 , y1 , y4 , y3 , z2 , z1 , z4 , z3 ) is also a solution. In what follows we provide 14 lemmas. Lemma 1. If K6 6= 0, K24 6= 0, K15 6= 0 and K16 K6 − K15 K7 6= 0, then system (4) has at most one limit cycle. Proof. Since K6 6= 0 and K24 6= 0, we solve E2 = 0 in the variable y1 and E6 = 0 in the variable y3 , and we obtain K7 (7) y1 = y2 + (z1 − z2 ), K6 K23 (8) y3 = y4 + (z3 − z4 ). K24 Replacing y1 and y3 in each equation Ei = 0 for i ∈ {1, 3, 4, 5, 7, 8}, in particular we obtain that K15 K23 (9) E4 = 2K14 + K16 (z2 − z3 ) + K15 (y4 − y2 ) + (z3 − z4 ) = 0. K24 Since K15 =6 0 we solve E4 = 0 in the variable y2 and we substitute it in each Ei = 0 for i ∈ {1, 3, 5, 7, 8}. Thus E8 becomes K7 K15 K15 K23 E8 = K16 (−z1 + z2 − z3 + z4 ) + (z1 − z2 ) + (z3 − z4 ) = 0. K6 K24 Since K16 K6 −K15 K7 6= 0 we solve E8 = 0 in the variable z1 and we substitute it in each equation Ei = 0 for i ∈ {1, 3, 5, 7}. Now we have the following four polynomial LIMIT CYCLES OF DISCONTINUOUS PIECEWISE LINEAR SYSTEMS 7 equations in the variables z2 , z3 , z4 and y4 : (K15 K23 − K16 K24 ) E1 = 2 (K K − K K )2 (z3 − z4 )T1 = 0, K15 K24 16 6 15 7 1 E3 = − 2 T2 = 0, K15 K24 1 E5 = − 2 (z3 − z4 )T3 = 0, K24 1 E7 = 2 2 T4 = 0, K15 K24 (K16 K6 − K15 K7 )2 where T1 and T3 are polynomials of degree 1 and T2 and T4 are polynomials of degree 2. We do not explicit them here due to their lengths. If z3 = z4 then from (8) we get y3 = y4 and we have no limit cycles. Thus we must study the zeros of the system T1 = T2 = T3 = T4 = 0. Taking E3,7 = E3 + E7 we get the new equation 1 E3,7 = 2 (K K − K K )2 (z3 − z4 )T5 = 0, K15 K24 16 6 15 7 such that T5 is a polynomial of degree 1. In short, we have four polynomials and the product of their degrees is 2. Thus if system (6) has finitely many solutions by Bezout Theorem (see for in- stance [14]) it has at most 2 solutions. By Remark 1 these two solutions correspond to the same limit cycle. So the discontinuous piecewise linear differential system (4) has at most one limit cycle.  Lemma 2. If K6 6= 0, K24 = 0, K15 6= 0 and K16 K6 − K15 K7 6= 0, then system (4) has at most one limit cycle. Proof. Since K6 6= 0 we solve E2 = 0 in the variable y1 and we obtain (7). Taking K24 = 0 and replacing y1 given in (7) in each Ei = 0, for i ∈ {1, 3, .., 8}, in particular we get E6 = K23 (z3 − z4 ) = 0. So z3 = z4 , otherwise K23 = 0 and we do not have finitely many solutions and therefore we have no limit cycles. Then we have that   1 E8 = −2K14 + K16 z4 + K15 (y2 − y4 ) − (K16 K6 − K15 K7 )z1 + K15 K7 z2 . K6 Solving E8 = 0 in the variable y2 we obtain 2K14 K6 + K16 K6 z1 − K15 K7 z1 + K15 K7 z2 − K16 K6 z4 y2 = y4 + . K15 K6 Substituting y2 in each Ei = 0 for i ∈ {1, 3, 4, 5, 7}, in particular we have 1 E4 = K15 (y3 − y4 ) − (K16 K6 − K15 K7 )(z1 − z2 ) = 0. K6 8 J. LLIBRE AND J.R. DE MORAES Solving E4 = 0 in the variable y3 and substituting it in each Ei = 0 for i ∈ {1, 3, 5, 7} we obtain the following four polynomial equations in the variables z1 , z2 , y4 and z4 1 E1 = − (z2 − z1 )T1 = 0, K15 K62 1 E3 = − 2 T2 = 0, K15 K6 (K16 K6 − K15 K7 ) E5 = 2 K2 (z2 − z1 )T3 = 0, K15 6 1 E7 = 2 T4 = 0, K15 where T1 and T3 are polynomials of degree 1 and T2 and T4 are polynomials of degree 2. If z1 = z2 then from (7) we have y1 = y2 and we have no limit cycles. Thus we must study the number of the zeros of the system T1 = T2 = T3 = T4 = 0. Taking E3,7 = E3 + E7 we get the new equation 1 E3,7 = 2 K (z1 − z2 )T5 = 0, K15 6 such that T5 is a polynomial of degree 1. In short we have four polynomials and the product of their degrees is 2. Thus as in the proof of the previous lemma we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle.  Lemma 3. If K6 6= 0, K24 6= 0, K15 = 0 and K16 K6 − K15 K7 6= 0, then system (4) has at most one limit cycle. Proof. Consider K15 = 0. So K16 6= 0. Since K6 6= 0 and K24 6= 0 we substituted y1 and y3 given in (7) and (8) respectively, in each equation Ei = 0 for i = 1, ..., 8. Thus E4 and E8 become respectively E4 = 2K14 + K16 z2 − K16 z3 , E8 = −2K14 − K16 z1 + K16 z4 . Solving E4 = 0 and E8 = 0 in the variables z2 and z1 respectively, we obtain z1 = (−2K14 + K16 z4 )/K16 , z2 = (−2K14 + K16 z3 )/K16 . Substituting them in each equation Ei = 0 for i ∈ {1, 3, 5, 7}, we obtain the follow- ing four polynomial equations in the variables y2 , y3 , z3 and z4 : 1 E1 = (z3 − z4 )T1 = 0, K16 K62 1 E3 = − 2 2 T2 = 0, K16 K24 1 E5 = − 2 (z3 − z4 )T3 = 0, K24 1 E7 = 2 2 T4 = 0, K16 K6 where T1 and T3 are polynomials of degree 1 and T2 and T4 are polynomials of degree 2. As in the proof of Lemma 1 we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle.  LIMIT CYCLES OF DISCONTINUOUS PIECEWISE LINEAR SYSTEMS 9 Lemma 4. If K6 = 0, K24 6= 0, K15 6= 0 and K16 K6 − K15 K7 6= 0, then system (4) has at most one limit cycle. Proof. Consider K6 = 0. So we have E2 = −K7 (z1 − z2 ) = 0. Since K7 6= 0, we have that z1 = z2 . Taking z1 = z2 and solving E6 = 0 in the variable y3 we obtain (8). Substituting it in each equation Ei = 0 for i ∈ {1, 3, 4, 5, 7, 8}, in particular we obtain E4 = 0 given in (9). Solving E4 = 0 in the variable y2 and substituting it in each equation Ei = 0 for i ∈ {1, 3, 5, 7, 8}, E8 becomes E8 = −2K14 + K15 (y1 − y4 ) + K16 (z4 − z2 ). Solving E8 = 0 in the variable y1 and substituting it in each equation Ei = 0 for i ∈ {1, 3, 5, 7}, we obtain the following four polynomial equations in the variables y4 , z2 , z3 and z4 : K15 K23 − K16 K24 E1 = 2 K2 (z3 − z4 )T1 = 0, K15 24 1 E3 = − 2 T2 = 0, K15 K24 1 E5 = − 2 (z3 − z4 )T3 = 0, K24 1 E7 = 2 T4 = 0, K15 where T1 and T3 are polynomials of degree 1 and T2 and T4 are polynomials of degree 2. As in the proof of Lemma 1 we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle.  Lemma 5. If K6 = 0, K24 = 0, K15 6= 0 and K16 K6 − K15 K7 6= 0, then system (4) has at most one limit cycle. Proof. Consider K6 = 0, K24 = 0 and K15 6= 0. So we have that K7 6= 0. From E2 = 0 and E6 = 0 we obtain that z1 = z2 and z3 = z4 . Solving E4 = 0 and E8 = 0 in the variables y2 and y1 , respectively and substituting them in each Ei = 0 for i ∈ {1, 3, 5, 7} we get the following four polynomial equations in the variables y3 , y4 , z2 and z4 : 1 E1 = (y3 − y4 )T1 = 0, K15 1 E3 = − 2 T2 = 0, K15 E5 = (y4 − y3 )T3 = 0, 1 E7 = 2 T4 = 0, K15 where T1 and T3 are polynomials of degree 1 and T2 and T4 are polynomials of degree 2. Again as in the proof of Lemma 1 we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle.  Lemma 6. If K6 6= 0, K24 = 0, K15 = 0 and K16 K6 − K15 K7 6= 0, then system (4) has no limit cycles. 10 J. LLIBRE AND J.R. DE MORAES Proof. Take K24 = 0 and K15 = 0. So we have that K16 6= 0, E6 = K23 (z3 − z4 ) and E8 = −2K14 − K16 z1 + K16 z4 . As in the proof of Lemma 2 we have that z3 = z4 . Solving E2 = 0 in the variable y1 we obtain (7). Solving E8 = 0 in the variable z4 and substituting it in E4 = 0 we get E4 = −K16 (z1 − z2 ) = 0. If z1 = z2 we have from (7) that y1 = y2 and we have no limit cycles. Therefore system (4) has no limit cycles.  Lemma 7. If K6 6= 0, K24 6= 0, K15 6= 0 and K16 K6 − K15 K7 = 0, then system (4) has no limit cycles. Proof. Consider K6 6= 0, K24 6= 0, K15 6= 0 and K16 K6 − K15 K7 = 0. Solving E2 = 0 and E6 = 0 in the variables y1 and y3 respectively, we obtain y1 and y3 given in (7) and (8). Substituting y1 , y3 and K16 = K15 K7 /K6 in E4 = 0 and E8 = 0 in particular we obtain K15 K7 K15 K23 E4 = 2K14 + K15 (y4 − y2 ) + (z2 − z3 ) + (z3 − z4 ) = 0. K6 K24 Solving E4 = 0 in the variable y2 and substituting it in E8 = 0 we obtain K15 (K23 K6 − K24 K7 )(z3 − z4 ) E8 = = 0. K24 K6 If z3 = z4 , then from (8) we get y3 = y4 and we have no limit cycles. If K23 K6 − K24 K7 = 0, then there are more unknown variables than equations in system (6) and therefore there are no limit cycles.  Lemma 8. If K6 6= 0 and K24 6= 0, K15 = 0 and K16 K6 −K15 K7 = 0, then system (4) has no limit cycles. Proof. Consider K6 6= 0 and K24 6= 0. Taking K15 = 0 we have that K16 = 0. So we obtain E4 = −E8 = 2K14 . Therefore there are more unknown variables than equations in system (6) and therefore there are no limit cycles.  Lemma 9. If K6 6= 0 and K24 = 0, K15 = 0 and K16 K6 −K15 K7 = 0, then system (4) has no limit cycles. Proof. The proof of this lemma is analogous to the one of the previous lemma.  Lemma 10. If K6 = 0, K24 = 0 and K15 = 0, then system (4) has no limit cycles. Proof. Considering K6 = 0, K24 = 0 and K15 = 0 we have E2 = −K7 (z1 − z2 ) and E6 = K23 (z3 − z4 ). So for having limit cycles it is necessary that z1 = z2 and z3 = z4 . However in this case E4 = −E8 = 2K14 + K16 z2 − K16 z4 and therefore there are no limit cycles.  Lemma 11. If K6 6= 0, K24 = 0, K15 6= 0 and K16 K6 − K15 K7 = 0, then system (4) has no limit cycles. Proof. Consider K6 6= 0, K24 = 0, K15 6= 0 and K16 = K15 K7 /K6 . Thus we have that E6 = K23 (z3 − z4 ) = 0. It is necessary that z3 = z4 for having limit cycles. LIMIT CYCLES OF DISCONTINUOUS PIECEWISE LINEAR SYSTEMS 11 Solving E2 = 0 in the variable y1 we get y1 given in (7). Substituting it in E4 = 0 and E8 = 0 we obtain 2K14 K6 − K15 K6 y2 + K15 K6 y3 + K15 K7 z2 − K15 K7 z4 E4 = = 0, K6 2K14 K6 − K15 K6 y2 + K15 K6 y4 + K15 K7 z2 − K15 K7 z4 E8 = − = 0. K6 Solving E8 = 0 in the variable y2 and substituting it in E4 = 0 we have that E4 = K15 (y3 − y4 ) = 0. If y3 = y4 we have no limit cycles because z3 = z4 .  Lemma 12. If K6 = 0, K24 = 0, K15 6= 0 and K16 K6 − K15 K7 = 0, then system (4) has no limit cycles. Proof. Consider K6 = 0, K24 = 0 and K15 6= 0. Since K16 K6 − K15 K7 = 0 we have K7 = 0. Thus we vanish E2 and system (6) has seven equations and eight unknown variables y1 , .., y4 , z1 , ..., z4 . So there are no limit cycles.  Lemma 13. If K6 = 0, K24 6= 0, K15 6= 0 and K16 K6 − K15 K7 = 0, then system (4) has no limit cycles. Proof. The proof of this lemma is analogous to the one of the previous lemma.  Lemma 14. If K6 = 0, K24 6= 0 and K15 = 0, then system (4) has no limit cycles. Proof. Consider K6 = 0, K24 6= 0 and K15 = 0. We have that E2 = −K7 (z1 −z2 ) = 0. So for having limit cycles it is necessary that z1 = z2 . Thus we obtain E8 = −2K14 − K16 z2 + K16 z4 = 0. Note that if K16 = 0 we have no limit cycles. Solving E6 = 0 and E8 = 0 in the variables y3 and z2 , respectively and substituting them in E4 = 0 and E5 = 0 in particular we obtain E4 = −K16 (z3 − z4 ) = 0. Taking z3 = z4 we vanish E4 and E5 . So there are no limit cycles because system (6) has more unknown variables than equations.  Lemmas 1–14 show that if K16 K6 −K15 K7 = 0, then the discontinuous piecewise differential system (4) has no limit cycles; and if K16 K6 − K15 K7 6= 0, then it has at most one limit cycle, except when K6 6= 0, K24 = 0 and K15 = 0 (in this case there are no limit cycles). In short, we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle. In order to complete the proof of Theorem 2 we must prove Proposition 1. Proof of Proposition 1. We shall prove that discontinuous piecewise linear differen- tial system (5) has one limit cycle. 12 J. LLIBRE AND J.R. DE MORAES The first integrals f1 , f2 , g1 , g2 , h1 and h2 for the discontinuous linear piecewise differential system (5) are 2 f1 = (5 + 6x + 53x2 + 16y − 80xy + 52y 2 − 36z + 132xz − 192yz + 180z 2 ), 9 1 f2 = (−5 − 3x + 6y − 6z), 6 g1 = (−x + 2y − z)2 + z 2 , g2 = −x + y, h1 = (−1 − x + 2y + z)2 + (−2y + 2z)2 . h2 = −1 − x + z. So system (6) becomes 0 = −8y12 + 8y22 + 4z1 − 5z12 + 4y1 (2 + z1 ) − 4z2 + 5z22 − 4y2 (2 + z2 ), 0 = −z1 + z2 , 8 0= (3 + 13y22 − 13y32 + 24z2 + 45z22 − 16y2 (1 + 3z2 ) + 42z3 − 45z32 , 9 +24y3 (−1 + 2z3 )), 0 = −1 + y2 − y3 − z2 + z3 , 0 = z32 + (−1 − 2y3 + z3 )2 − z42 − (−1 − 2y4 + z4 )2 , 0 = y3 − y4 , 8 0 = − (3 + 13y12 − 13y42 + 24z1 + 45z12 − 16y1 (1 + 3z1 ) + 42z4 − 45z42 , 9 +24y4 (−1 + 2z4 )), 0 = 1 − y1 + y4 + z1 − z4 . The unique isolated solution of the previous system satisfying (y1 , z1 ) 6= (y2 , z2 ) and (y3 , z3 ) 6= (y4 , z4 ) is (y1∗ , y2∗ , y3∗ , y4∗ , z1∗ , z2∗ , z3∗ , z4∗ ) = (1, 0, 0, 0, 0, 0, 1, 0). The solution (x1 (t), y1 (t), z1 (t)) of system (5) in x > 1 such that (x1 (0), y1 (0), z1 (0)) = (1, y1∗ , z1∗ ) is x(t) = 2 − cos t + sin t, y(t) = 1 + sin t, z(t) = 1 − cos t + sin t. The solution (x2 (t), y2 (t), z2 (t)) of system (5) in |x| < 1 such that (x2 (0), y2 (0), z2 (0)) = (1, y2∗ , z2∗ ) is 1 x(t) = (−1 + 3 cos t − sin t), 2 y(t) = −1 + cos t + sin t, 1 z(t) = (−1 + cos t + 5 sin t). 4 LIMIT CYCLES OF DISCONTINUOUS PIECEWISE LINEAR SYSTEMS 13 The solution (x3 (t), y3 (t), z3 (t)) of system (5) in x < −1 such that (x3 (0), y3 (0), z3 (0)) = (−1, y3∗ , z3∗ ) is x(t) = −2 + cos t − sin t, y(t) = −1 + cos t − sin t, z(t) = cos t. Finally the solution (x4 (t), y4 (t), z4 (t)) of system of (5) in |x| < 1 such that (x4 (0), y4 (0), z4 (0)) = (−1, y4∗ , z4∗ ) is 1 x(t) = (−1 − 7 cos t + 9 sin t), 8 1 y(t) = (5 − 5 cos t − sin t), 4 13 z(t) = − (−1 + cos t + sin t). 16 The time that the solution (x1 (t), y1 (t), z1 (t)) contained in x > 1 needs to reach the point (1, y2∗ , z2∗ ) is t1 = 3π/2. The time that the solution (x2 (t), y2 (t), z2 (t)) contained in −1 < x < 1 needs to reach the point (−1, y3∗ , z3∗ ) is t2 = π/2. The time that the solution (x3 (t), y3 (t), z3 (t)) contained in x < −1 needs to reach the point (−1, y4∗ , z4∗ ) is t3 = 3π/2. Lastly the time that the solution (x4 (t), y4 (t), z4 (t)) contained in −1 < x < 1 needs to reach the point (1, y1∗ , z1∗ ) is t4 = π/2. The limit cycle of Figure 1 is obtained drawing the orbits (xk (t), yk (t), zk (t)), for t ∈ [0, tk ] and k = 1, 2, 3, 4. Now we shall prove that this limit cycle is stable. This limit cycle starts at the point (1, 0, 0) of the plane x = 1, cross the region |x| < 1 until the point (−1, 0, 1) of the plane x = −1, after travels in the region x < 1 until the point (−1, 0, 0) of the plane x = −1, cross again the region |x| < 1 until the point (1, 1, 0) of the plane x = 1, and finally it travels in the region x > 1 until the initial point (1, 0, 0). Let ε and δ be two small real numbers, then the point (1, ε, δ) is close to the point (1, 0, 0) of the limit cycle. Using the first integrals f1 and f2 we compute where the orbit through the point (1, ε, δ) intersect the plane x = −1 near the point (−1, 0, 1), this intersection takes place at the point (−1, y1 , z1 ) where 1 p  y1 = 441δ 2 + 66δ(4 − 7ε) + 121ε2 − 184ε + 144 − 21δ + 21ε − 12 , 10 1 p  z1 = 441δ 2 + 66δ(4 − 7ε) + 121ε2 − 184ε + 144 − 11δ + 11ε − 2 . 10 Now with the first integrals g1 and g2 we compute where the orbit through the point (−1, y1 , z1 ) travels in the region x < −1 until intersecting the plane x = −1 at the point (−1, y2 , z2 ) near the point (−1, 0, 0), where 1 p  y2 = 441δ 2 + 66δ(4 − 7ε) + ε(121ε − 184) + 144 − 21δ + 21ε − 12 , 10 1 p  z2 = 441δ 2 + 66δ(4 − 7ε) + ε(121ε − 184) + 144 − 31δ + 31ε − 12 . 10 Using again the first integrals f1 and f2 we compute where the orbit through the point (−1, y2 , z2 ) travels in the region |x| < 1 until intersecting the plane x = 1 at 14 J. LLIBRE AND J.R. DE MORAES the point (1, y3 , z3 ) near the point (1, 1, 0), where 1 1 y3 = (21δ − T − 21ε + 17) , z3 = (11δ − T − 11ε + 7) , 10 10 here p T = 2205δ 2 + 6δ(354 − 665ε) − 42R(2δ − 2ε + 1) + ε(1885ε − 2044) + 553, p R = 441δ 2 + 66δ(4 − 7ε) + ε(121ε − 184) + 144. Finally with the first integrals h1 and h2 we compute where the orbit through the point (1, y3 , z3 ) travels in the region x > 1 until intersecting the plane x = 1 at the point (1, y4 , z4 ) near the point (1, 0, 0), where 1 1 y4 = (−31δ + S + 31ε − 7), z4 = (11δ − S − 11ε + 7), 20 10 p where S = 2205δ 2 + 6δ(354 − 665ε) − 42R(2δ − 2ε + 1) + ε(1885ε − 2044) + 553. In summary, the Poincaré map F near the point (1, 0, 0) and the limit cycle is f (1, ε, δ) = (1, y4 , z4 ). Therefore  11  −1  14  Df (1, 0, 0) =  , 25 4 − 7 1 √  and their eigenvalues are −16 ± 235 , both negative. Hence the limit cycles is 7 stable. This completes the proof of the proposition.  Remark 2. We note that for the piecewise differential system of Proposition 1 we have that K6 = 0, K7 = 1, K15 = K16 = −2 and K24 = 1. So this piecewise differential system only satisfies Lemma 4. Acknowledgments The first author is partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER) and PID2019-104658GB-I00 (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA-RISE-2017-777911. The second author is supported by FUNDECT–219/2016. References [1] M. Bey, S. Badi, K. Fernane and A. Makhlouf, The number of limit cycles bifurcating from the periodic orbits of an isochronous center, Math. Methods Appl. Sci. 42 (2019), no. 3, 821–829. [2] G. Blé, V. Castellanos and J. Llibre, Existence of limit cycles in a tritrophic food chain model with Holling functional responses of type II and III, Math. Methods Appl. 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Lamb, Dynamics and bifurcations of nonsmooth systems: a sur- vey, Phys. D 241 (2012), 1826–1844. [14] I. R. Shafarevich, Basic Algebraic Geometry, Springer, Berlin, 1974. [15] D. J. W. Simpson, Bifurcations in Piecewise–Smooth Continuous Systems, World Scientific Series on Nonlinear Science A, vol. 69. World Scientific, Singapore, 2010. [16] S. Smale, Mathematical problems for next century, Math. Intelligencer 20 (1998), 7–15. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bel- laterra, Barcelona, Catalonia, Spain. Email address: jllibre@mat.uab.cat 2 Curso de Matemática – UEMS, Rodovia Dourados–Itaum Km 12, CEP 79804–970, Dourados, Mato Grosso do Sul, Brazil. Email address: jaime@uems.br