This is a preprint of: “Limit cycles for discontinuous planar piecewise linear differential systems separated by an algebraic curve”, Jaume Llibre, Xiang Zhang, Internat. J. Bifur. Chaos Appl. Sci. Engrg., vol. 29(2), 1950017 (17 pages), 2019. LIMIT CYCLES FOR DISCONTINUOUS PLANAR PIECEWISE LINEAR DIFFERENTIAL SYSTEMS SEPARATED BY AN ALGEBRAIC CURVE JAUME LLIBRE1 AND XIANG ZHANG2 Abstract. We study how can change the maximum number of limit cycles of the discontinuous piecewise linear differential systems with only two pieces in function of the degree of the discontinuity algebraic curve between the two linear differential systems. These discontinuous differential systems appear frequently in applied sciences. 1. Introduction and statement of the main result The dynamics of the piecewise linear differential systems started to be studied around 1930, mainly in the book of Andronov et al [1]. Many researchers from different fields considered this kind of differential systems, because they are widely used to model phenomena appearing in mechanics, electronics, economy, neuro- science, ..., see for more details and applications the books of Bernardo et al [3] and of Simpson [33], the surveys of Makarenkov and Lamb [29], and Teixeira [36], the articles [19, 20, 21, 31, 35], and the references which appear in all these works. A periodic orbit of a differential system which is isolated in the set of all periodic orbits of the system is a limit cycle. One of the main problems in the dynamics of the differential systems in the plane is to control the existence and the number of their limit cycles. This problem restricted to polynomial differential systems is the famous 16th Hilbert’s problem, see more details in [14, 18, 22]. This last years many authors have studied the limit cycles of discontinuous piece- wise linear differential systems in R2 . Of course, the simplest piecewise linear dif- ferential systems in R2 are the ones having only two pieces separated by a curve, and when this curve is a straight line. Thus the limit cycles of these last class of discontinuous piecewise linear differential systems has been intensively studied, see [2, 4, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 23, 24, 25, 26, 27, 28, 32] and the references quoted in these papers. Up to know the results of all these papers only provide ex- amples that the discontinuous piecewise linear differential systems in R2 separated by a straight line can have 3 crossing limit cycles (they are defined in what follows), but it is an open question to know if 3 is the maximum number of crossing limit cycles that such discontinuous differential systems can have. 2010 Mathematics Subject Classification. Primary 34A30, 34C05, 34C25, 34C07, 37G15. Key words and phrases. non–smooth differential system, limit cycle, discontinuous piecewise linear differential system. 1 http://www.gsd.uab.cat 2 J. LLIBRE AND X. ZHANG In the papers [5, 30] the authors considered discontinuous piecewise linear dif- ferential systems with only two pieces, but now the curve of discontinuity between the two pieces is not a straight line, and they shown that the number of limit cycles can increase arbitrarily with the number of oscillations of that curve. We note that the discontinuity curves considered in these two papers is not algebraic. The main objective of this paper is to study how can change the maximum number of limit cycles of the discontinuous piecewise linear differential systems with only two pieces in function of the degree of the discontinuity curve when this curve is algebraic. Due to the big amount of computations necessary for studying this maximum number of limit cycles we restrict our analysis as follows: first to the class of discontinuous piecewise linear differential systems separated by the algebraic curve y = xn where n > 1 is an arbitrary positive integer; second to the class of discontinuous piecewise linear differential systems formed by two linear differential systems having only centers; and third we only consider crossing limit cycles, they are defined in what follows. We consider discontinuous piecewise differential equations in R2 of the form ! ( ẋ X − (x, y) if (x, y) ∈ L = {(x, y) : y ≤ xn }, (1) = ẏ X + (x, y) if (x, y) ∈ U = {(x, y) : y ≥ xn }, where the vector fields X − and X + are linear. The dynamics over the curve of discontinuity y = xn is defined following the Filippov’s convention, see [8]. Thus, on the curve of discontinuity y = xn we distinguish generically two different kind of points: the crossing points {p = (x, xn ) : X − (p)X + (p) > 0}, and the sliding points {p = (x, xn ) : X − (p)X + (p) < 0}. We assume that each one of the linear differential systems of the discontinuous piecewise linear differential systems (1) has a center. We want to find upper bound for the maximum number of crossing limit cycles that such discontinuous piecewise linear differential systems can exhibit. A crossing limit cycle for such discontinuous differential systems is formed by two pieces of orbits, one piece of orbit of the linear differential system contained in y < xn and the other piece of orbit of the linear differential system contained in y > xn , furthermore both pieces connect in two crossing points over the curve y = xn . We must mention that when n = 1, i.e. when the discontinuous curve is a straight line, and both linear differential systems have a center the discontinuous differential system (1) has no crossing limit cycles, see for details [28]. This proves the result of the next Theorem 2 for the particular case n = 1. An equilibrium p of a linear differential system of a discontinuous piecewise linear differential systems in R2 separated by the algebraic curve y = xn is called real when p belongs to the closure of the component of R2 \ {(x, y) : y = xn } where is defined the mentioned linear differential system, otherwise p is called virtual. The following normal form for the discontinuous piecewise linear differential systems in R2 separated by the algebraic curve y = xn when both linear differential systems have a center will help us to prove our manin result, the theorem which follows the next proposition. DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 3 Proposition 1. After a linear change of variables and a rescaling of the indepen- dent variable any discontinuous piecewise linear differential systems in R2 separated by the algebraic curve f (x, y) = 0 when both linear differential systems have a center can be written as 4b2 + ω 2 ẋ = −bx − y + d, ẏ = ax + by + c, in L = {(x, y) : y ≤ xn }, (2) 4a ẋ = −y − α, ẏ = x − β, in U = {(x, y) : y ≥ xn }, with a > 0 and ω > 0. Figure 1. The three limit cycles of a discontinuous piecewise linear differential system (2) for n = 2. Theorem 2. Let n be a positive integer. (a) An upper bound for the maximum number of crossing limit cycles that a discontinuous piecewise linear differential system in R2 separated by the algebraic curve y = xn can have when both linear differential systems have a center, real or virtual, is dn equal to 2n2 − n − 1 2n2 − n if n is odd, and if n is even. 2 2 (b) For n = 2 there are discontinuous piecewise linear differential systems which reach the upper bound of 3 limit cycles of statement (a), see the 3 limit cycles in Figure 1. (c) For n = 2, 3 we provide the explicit polynomial of degree 2dn whose real roots x provide the points (x, xn ) where the crossing limit cycles of the discontinuous piecewise linear differential system (2) intersect the curve y = xn . Theorem 2 is proved in section 3. We remark that in statement (a) of Theorem 2 we provide upper bound for the maximum number of crossing limit cycles that the discontinuous piecewise linear differential systems studied in that theorem can have. The major part of the 4 J. LLIBRE AND X. ZHANG results mentioned previously on the limit cycles of discontinuous piecewise linear differential systems studied until now only provide lower bounds for that maximum number. The unique other three papers that we know providing upper bounds are when the curve of discontinuity is a straight line, and either one of the linear differential systems has an equilibrium on the the discontinuity straight line [7, 24], or one of the linear differential systems has a center [28]. It remains the following open question. Open question. Are the upper bounds given in the statement of Theorem 2 reached for values of n > 2? 2. Normal form Now we shall prove the normal form for the discontinuous piecewise linear dif- ferential systems that we study. Proof of Proposition 1. Doing the linear change of variables which writes the matrix of the linear differential system in U , having a real or virtual center, in its real normal form, and rescaling the independent variable of the differential system (if necessary) we can write such a linear differential system as ẋ = −y − α, ẏ = x − β. Now the linear differential system in the region L is ẋ = Ax + By + d, ẏ = ax + by + c, assuming that it has a center. Since the eigenvalues of this system are p A + b ± 4aB + (A − b)2 , 2 in order to have a center we must have that A + b = 0 and 4aB + (A − b)2 = −ω 2 for some ω > 0 and aB < 0, or equivalently A = −b, B = −(4b2 + ω 2 )/(4a) and a > 0. This completes the proof of the proposition.  It is easy to check that the differential linear system in the region L has the first integral HL = 4(ax + by)2 + 8a(cx − dy) + ω 2 y 2 , while the first integral of the differential linear system in the region U is HU = (y + α)2 + (x − β)2 . 3. Proofs of Theorem 2 In order that the piecewise linear differential system (2) has a crossing limit cycle Γ, it must intersect the discontinuous curve y = xn in two points. Let (xL , xnL ) and (xR , xnR ) be two intersecting points. Then, taking into account that HL and HR are first integrals these two points satisfy the equations eL := HL (xL , xnL ) − HL (xR , xnR ) = 0, (3) eU := HU (xL , xnL ) − HU (xR , xnR ) = 0. DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 5 Note that eL and eU are two polynomials of degree 2n in the variables xL and xR . Clearly xL − xR is a factor of both eL and eR . Some direct calculations show that EL = eL /(xL − xR ) and ER = eR /(xL − xR ) are given by   EL = 4b2 + ω 2 x2n−1 R + xL x2n−2 R + . . . + xn−3 L xR n+2 + xn−2 n+1 L xR   + (4b2 + ω 2 )xn−1 L + 8ab xnR + (4b2 + ω 2 )xnL + 8abxL − 8ad xn−1 R  n−2 + (4b2 + ω 2 )xn+1 L + 8abx 2 L − 8adx L x R  2 + . . . + (4b2 + ω 2 )x2n−3 L + 8abx n−2 L − 8adxn−3 L xR 2 2 2n−2 n−1 n−2 2  + (4b + ω )xL + 8abxL − 8adxL + 4a xR + (4b2 + ω 2 )x2n−1 L n−1 + 8abxnL − 8adxL + 4a2 xL + 8ac, and EU = x2n−1 R + xL x2n−2 R + . . . + xn−3 L xR n+2 + xn−2 L xR n+1 + xn−1 L xR n n−1 n−2  + (xnL + 2α) xR + xL xR + . . . + xn−3 L xR 2 n−2  + (xnL + 2α) xL + 1 xR + (xnL + 2α) xn−1 L + xL − 2β. Note that EL and EU are polynomials in the variables xL and xR of degree 2n − 1. To study the existence and the number of limit cycles of system (2), it is necessary to compute the common zeros (xL , xR ) of EL and ER . For doing so we calculate the resultants Resultant[EL , ER , xR ] and Resultant[EL , ER , xL ] of EL and ER with respect to xR and xL , respectively. By the symmetry of EL and ER with respect to xL and xR , we know that Resultant[EL , ER , xR ] and Resultant[EL , ER , xL ] have the same expression. So we only need to compute Resultant[EL , ER , xR ]. In what follows we denoted by RL (respectively RR ) for simplifying the nota- tion the Resultant[EL , ER , xR ] (respectively Resultant[EL , ER , xL ]) because this resultant is a polynomial in the variable xL (respectively xR ). The next result provides the degree of the resultant RL , which is the most technical lemma in the proof of Theorem 2. We remark that in general the degree of the resultant of two arbitrary polynomials of degree 2n − 1 is (2n − 1)2 , but since the polynomials EL and ER are special the degree in this case is lower, and to compute this degree is the more difficult part of this paper. Lemma 3. The maximum degree of the resultant RL is 2n2 − n − 1 if n is odd, and 2n2 − n if n is even. Proof. Let S0 = 4b2 + ω 2 , S1 = −8ad, S2 = 4a2 , S3 = 8ac and A = 8ab. And for simplifying notations we replace xL by x. Then by the definition of the resultant of two polynomials RL is the determinant of a (4n − 2) × (4n − 2) matrix, more precisely C1 01 ··· 0n−1 ··· 02n−2 C2 01 ··· 0n−1 ··· 02n−2 C1 C2 .. .. . . RL = , C1 C2 .. .. . . C1 C2 6 J. LLIBRE AND X. ZHANG where 0k is a k–dimensional column vector for k ∈ {1, . . . , n − 2}, the entries which are empty are zeros, and C1 and C2 are 2n-dimensional column vectors given respectively as     S0 1  S0 x   x       S0 x2   x2       ..   ..   .   .       S0 xn−3   xn−3       S0 xn−2   xn−2      C1 =   S0 xn−1 + A  , C2 =    xn−1 .   S0 xn + Ax + S1   n x + 2α       S0 xn+1 + Ax2 + S1 x   xn+1 + 2αx       ..   ..   .   .       S0 x2n−3 + Axn−2 + S1 xn−3   x2n−3 + 2αxn−3       S0 x2n−2 + Axn−1 + S1 xn−2 + S2   2n−2 x + 2αxn−2 + 1  S0 x2n−1 + Axn + S1 xn−1 + S2 x + S3 x 2n−1 + 2αxn−1 + x − 2β For computing the determinant RL , we first apply the next algorithm to RL : • We add the (2n − 1 + j)-th column multiplied by −(4b2 + ω 2 ) to the j-th column for j = 1, . . . , 2n − 1. Then • We expand the resulting determinant of a (4n−2)×(4n−2) matrix through its first row. • We expand the new resulting determinant of a (4n − 3) × (4n − 3) matrix through its first row. • We expand this second new resulting determinant of a (4n − 4) × (4n − 4) matrix again through its first row. • In total we repeat this process n − 1 times, and at the end we have that RL is the determinant of a (3n − 1) × (3n − 1) matrix. After all these computations we achieve that RL = (−1)n−1 RL1 with D1 01 ··· 0n−2 0n−1 0n ··· 02n−2 C2 01 ··· 0n−2 0n−1 D1 C2 .. .. . . D1 C2 RL 1 = , D1 C2 D1 .. . D1 DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 7 where the entries without letters are zeros, and D1 is the n + 1–dimensional column vector given by   A  Ax + B     Ax 2 + Bx     . ..  D1 =  ,    Ax n−2 + Bx n−3     Axn−1 + Bxn−2 + C  Axn + Bxn−1 + Cx + D    where B = −8ad−2α 4b2 + ω 2 , C = 4a2 − 4b2 + ω 2 and D = 8ac+2β 4b2 + ω 2 . We separate the rest of the proof in three cases, first n > 3, after n = 2 and finally n = 3. We do the next computations with RL1 when n > 3: • We add the first row multiplied by −x` to the (`+1)-th row for ` = 1, . . . , n. • In the new resulting determinant we add the second row multiplied by −x` to the (` + 2)-th row for ` = 1, . . . , n − 1. • In this second new resulting determinant we add the third row multiplied by −x` to the (` + 3)-th row for ` = 1, . . . , n − 2. • In this third new resulting determinant we add the fourth row multiplied by −x` to the (` + 3)-th row for ` = 1, . . . , n − 3. • In total we repeat this process n times. Now RL1 is the determinant of a (3n − 1) × (3n − 1) matrix of the form M1 M2 RL1 = M3 M4 where M1 , M2 , M3 and M4 are respectively the (n + 1) × (n + 1), (n + 1) × (2n − 2), (2n − 2) × (n + 1) and (2n − 2) × (2n − 2) matrices given by   A  B A     ..   B .     . . . .  M1 =  . . ,   .. ..   . .     .   C .. A  D C B A  T  0n−2 1 0  1     .. . ..   .  M2 =  ,  1 0     1  0Tn−2 2α 0 8 J. LLIBRE AND X. ZHANG   02n−2 fn (x) gn−1 (x) hn−2 (x) hn−3 (x) · · · ··· h1 (x)  .. .. ..   02n−3 fn (x) gn−1 (x) hn−2 (x) . . .     .. .. .. .. ..   02n−4 . . . . .     .. .. ..   . . . hn−3 (x)  M3 =  ,  .. ..   . . hn−2 (x)     ..   . gn−1 (x)     fn (x)  0n−2 where fn (x) = Axn + Bxn−1 + Cx + D, gn−1 (x) = Axn−1 + Bxn−2 + C, hj (x) = Axj + Bxj−1 for j = n − 2, . . . , 1, and M4 is equal to   A rn+1 (x) rn (x) xn−1 ··· ··· x3 x2  .. ..   h1 (x) A rn+2 (x) rn+1 (x) rn (x) . . x4 x3     . . .. .. .. .. .. .. .. ..   .. .. . . . . . . . .     . . .. .. .. .. .. .. ..   .. .. . . r2n−3 (x) . . . . x n−1 .     .. .. .. .. .. .. .. ..   . . . . A q2n−2 (x) . . . . rn (x) x n−1     . .. .. .. .. .. ..   .. . . . . . .   hn−2 (x) h1 (x) p2n−1 (x) rn+1 (x) rn (x)   .. .. .. ,  .. .. .. .. .. ..   gn−1 (x) hn−2 (x) . . . . . . . . .   .. .. ..   .. .. .. .. ..   fn (x) gn−1 (x) . . . . . . . .     .. .. .. .. .. ..   fn (x) . . . . . r2n−3 (x) .     .. .. ..   . . hn−2 (x) . q2n−2 (x) r2n−3 (x)     ..   . gn−1 (x) p2n−1 (x) q2n−2 (x)  fn (x) p2n−1 (x) where  p2n−1 (x) = x2n−1 + 2αxn−2 + 1 x − 2β, q2n−2 (x) = x2n−2 + 2αxn−2 + 1, rj (x) = xj + 2αxj−n for j = 2n − 3, . . . , n with j ≥ 3. Since det M1 6= 0 because ab 6= 0, doing some linear algebra we get that det RL1 = det M1 det(M4 − M3 M1−1 M2 ) = (det A)n+1 det(M4 − M3 M1−1 M2 ). DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 9 Direct calculations show that   w1  w2 w1     w3 w2 w1     ..   . w3 w2 w1     .. .. .. .. ..   . . . . .  M1−1 =    .. . . .. .. .. ..   . . . . . .     . .. .. .. .. ..   wn−1 .. . . . . .     .. .. .. ..   u w . . . . w2 w1  n n−1 vn+1 un wn−1 ··· ··· ··· w3 w2 w1 where wk = (−1)k−1 A−k B k−1 for k = 1, 2, . . . , n − 1, −n un = (−1) n−1 A B n−1 − A−2 C, vn+1 = (−1)n A−(n+1) B n + 2A−3 BC − A−2 D. Furthermore,   0Tn−2 w1  0Tn−2 w2 w1     0Tn−2 w3 w2 w1     .. ..   . . w3 w2 w1     .. .. .. .. .. ..   . . . . . .  M1−1 M2 =  ,  .. .. .. .. .. .. ..   . . . . . . .     .. .. .. .. .. ..   0T wn−1 . . . . . .   n−2   .. .. .. ..   0T un wn−1 . . . . w2 w1  n−2 0Tn−2 vn+1 + 2αw1 un wn−1 ··· ··· ··· w3 w2 and   0n×(n−2) Mn×n M3 M1−1 M2 = , Mn×n = (Fk` )k,`=1,...,n , 0(n−2)×(n−2) 0(n−2)×n with 0i×j the zero i × j matrix, and n+1−k X Fk` = hk−1+s wn+3−`−s , s=1 where for simplifying the notation we have used wj = 0 for j ≤ 0, hn = fn , hn−1 = gn−1 , wn = un and wn+1 = vn+1 + 2αw1 . Since det A = 8ab 6= 0, the computation of RL1 has been reduced to calculate the determinant of the matrix M4 − M3 M1−1 M2 . Note that M4 and M3 M1−1 M2 are (2n − 2) × (2n − 2) matrices. Now we work with these two matrices according to the following algorithm: 10 J. LLIBRE AND X. ZHANG • we add the i-th row multiplied by −x to the (i + 1)-th row for i = n − 3, n − 2, . . . , 1, successively. At the end of this process the matrix M4 has been reduced to the matrix M4∗ given by   A xn+1 + 2αx xn + 2α xn−1 ··· ··· ··· x3 x2  B A 2α   .. ..   B . .     .. .. ..   . . .     .. .. ..   . . .     ..   . A 1 2α     B −2β 1 2α     C pe2n −2β 1 ,    ..   D C pe2n −2β .     e .. .. ..   fn+1 D . pe2n . .     .. .. .. .. ..   fen+1 . . . . .   .. .. .. ..   . . . .   C 1   .. ..   . D C . −2β 1  fen+1 D pe2n −2β where fen+1 = −xfn (x) and pe2n = −xp2n−1 (x) are polynomials of degrees n + 1 and 2n respectively. Applying the same algorithm to the matrix M3 M1−1 M2 , this has been reduced to the matrix M3∗ given by   F11 F12 F13 F14 ··· ··· ··· F1,n−2 F1,n−1 F1,n  E2 (x) E1 (x) w0 Cw1   E3 (x) E2 (x) E1 (x) w0 Cw1     .. .. .. .. .. ..   01 . . . . . .     .. .. .. .. .. .. ..   . . . . . . .     .. .. .. .. .. .. .. ..   . . . . . . . .   ,  .. .. .. .. .. .. .. .. ..   . . . . . . . . .     En−2 (x) En−3 (x) En−4 (x) ··· ··· ··· E2 (x) E1 (x) w0 Cw1     En−1 (x) En−2 (x) En−3 (x) ··· ··· ··· ··· E2 (x) E1 (x) w0   Fn (x) Fn−1 (x) Fn−2 (x) ··· ··· ··· ··· F3 (x) F2 (x) F1 (x)     −xfn wn+1 −xfn wn −xfn wn−1 ··· ··· ··· ··· −xfn w4 −xfn w3 −xfn w2    02 03 DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 11 where 01 = 0(n+1)×(n−2) , 02 = 0(n−3)×(n−2) , 03 = 0(n−3)×n , w0 = Dw1 + Cw2 , and Ei (x) = −xfn (x)wi + Dwi+1 + Cwi+2 for i = 1, . . . , n − 1, Fi (x) = −xfn (x)wi + Dwi+1 , for i = 1, . . . , n, F11 = −A−1 Bxn + (A−2 BC − A−1 D + 2α)x, + (−1)n A−n−1 B n+1 + 2A−3 B 2 C − 2A−2 BD + 2αA−1 D, F12 = xn + (−1)n−1 A−n B n − A−2 BC + A−1 D, F13 = xn−1 + (−1)n−2 A−(n−1) B n−1 + A−1 C, F1` = xn+2−` + (−1)n+1−` A−(n+2−`) B n+2−` , for ` = 4, . . . , n.  Now det M4 − M3 M1−1 M2 = det (M4∗ − M3∗ ), and this last determinant is the determinant of the matrix  A G1 σ2 σ3 σ4 ··· ··· σn−1 σn  B A e2 E e1 E δ −cw1    .. ..   B . e3 E Ee2 e1 E δ .   . . .   .. .. . . . .. .. ..   . . . . . . . .   .. . . .. .. .. ..   . en−3 . . . . . .   A E . .   B A en−2 E en−3 E Een−4 ··· ··· Ee1 δ −cw1   en−1 en−2 en−3 en−4 e1   B E E E E ··· ··· E δ   C e2n,1 p en−1 F Fen−2 Fen−3 ··· ··· e2 F e1 F ,    D C xfn wn+1 e2n,2 p fen+1,1 fen+1,2 xfn wn−3 ··· xfn w3 xfn w2     fen+1 ..   D . e2n p −2β 1   .. .. .. ..   fen+1 . . . .   e2n p   .. .. .. ..   . . C . . 1     .. ..  . D C . −2β 1 fen+1 D e2n p −2β where the entries without anything are all zeros, and G1 = G1 (x) = xn+1 + A−1 Bxn + (A−2 BC − A−1 D)x − (−1)n A−n−1 B n+1 − 2A−3 B 2 C + 2A−2 BD − 2αA−1 D, σ2 = 2α + (−1)n A−n B n + A−2 BC − A−1 D, σ3 = (−1)n−1 A−(n−1) B n−1 − A−1 C, σ` = (−1)n+2−` A−(n+2−`) B n+2−` , ` = 4, . . . , n, δ = 2α − Dw1 − Cw2 , Een−2 = 1 − En−2 , Een−1 = −2β − En−1 , Eej = −Ej , j = 1, . . . , n − 3 e Fn−1 = −2β − Fn−1 , e Fn−2 = 1 − Fn−2 , e Fj = −Fj , j = 1, . . . , n − 3 fen+1 = −xfn , fen+1,1 = −2β + xfn wn−1 , fen+1,2 = 1 + xfn wn−2 pe2n = −xp2n−1 , pe2n,1 = −xp2n−1 − Fn , pe2n,2 = −xp2n−1 + xfn wn . Denote this last (2n−2)×(2n−2) matrix as M2n−2 . Note that A, B, C, D, δ, −cw1 , −2β, σi are constants, pe2n,1 , pe2n,2 and pe2n are polynomials of degree 2n, all the other letters denote polynomials of degree n + 1. 12 J. LLIBRE AND X. ZHANG Since by definition the determinant of a matrix (aij ) of order m is X (−1)τ (j1 ,...,jm ) a1j1 . . . amjm , σ(j1 ,...,jm ) where σ is a permutation of {1, . . . , m} and the summation is taken over all these possible permutations, and τ evaluated on a permutation of {1, . . . , m} is the min- imum number of transpositions for passing the permutation to the identity, for calculating the degree of this determinant, we must first compute the highest de- gree terms a1j1 . . . amjm in the summation. In order to realize this aim we must choose the aiji ’s in such a way that the product has the highest possible degree. Applying this idea to the determinant of the matrix M2n−2 , in the row and col- umn where pe2n,1 , pe2n,2 , pe2n is located, we must choose either one of these three polynomials or the two ones on the row and column where one of these three poly- nomials is located, because their product has the highest degree. The details are the following: • For pe2n,1 and pe2n,2 we can choose respectively G1 Fe1 and E e1 xfn w1 . But e we can choose only one of them, because F1 and xfn w1 are located on the same column. So in the product we choose G1 Fe1 and pe2n,2 . Of course we can also choose pe2n,1 and E e1 xfn w1 as the elements in the product. • Next in order to get the highest degree in the product, we must choose the B or A in the second row, which is associated to pe2n,1 . We choose A. • Then for the first pe2n from above, we choose fen+1 and E e1 , which are located respectively on the row and column where the first pe2n is located. • For the second pe2n , we can only choose itself, because there are no other two elements on the row and column where the second pe2n is located, whose product has degree larger than 2n. • Correspondingly in the fourth row we can choose either the B or the A, we choose A. • To the third pe2n , we choose fen+1 and Ee1 , which are located respectively on the row and column where the third pe2n is located. • In the row and column where the fourth pe2n are located, we can choose only pe2n itself as an element of the product. • Then in the sixth row we can choose only either the B or the A, and for the fifth pe2n we must choose fen+1 and E e1 as the elements of the product. • According to this algorithm we get the highest degree terms in the summa- tion in the following products – n polynomials of degree n + 1 and (n − 2)/2 polynomials of degree 2n provided that n is even, – n − 1 polynomials of degree n + 1 and (n − 1)/2 polynomials of degree 2n provided that n is odd. Summarizing we have proved that the highest degree of RL is at most n−2 n(n + 1) + 2n = 2n2 − n if n is even, 2 n−1 (n − 1)(n + 1) + 2n = 2n2 − n − 1 if n is odd. 2 DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 13 Note that this last two numbers are both even, and recall that they are denoted 2dn . The above proof provides the maximum degree of the polynomial RL for n > 3. We now study the degree of RL for n = 2, 3, and for these values of n we present the explicit expressions of RL . For n = 2 we work in a similar way as we did for n > 3, and we get that RL = −(8ab)3 det (M4∗ − M3∗ ) , where   p3 (x) q2 (x) M4∗ = , −xp3 (x) −2β   −η1 f2 + (η2 − A−2 D)g1 + (2α + 1)A−1 g1 A−1 f2 − η1 g1 M3∗ = , η1 xf2 + (η2 − A−2 D)D + (2α + 1)A−1 D −A−1 xf2 − η1 D with η1 = A−2 (B +C) and η2 = A−3 (B +C)2 . Since q2 (x)−(A−1 f2 (x)−η1 g1 (x)) = A−2 (B + C)2 − A−1 D + 2α + 1, it follows that RL is of degree 6 with the monomial x6 of higher degree, which comes from the product of the monomials x3 of higher degree in the polynomials p3 (x) and A−1 xf2 (x). For n = 3 similar computations yield RL = (8ab)4 det (M4∗ − M3∗ ) with   A q4 (x) r3 (x) x2  B −2β 1 2α  M4∗ =   C −xp5 (x) , −2β 1  D 0 −xp5 (x) −2β  −2  0 −A Bf3 + u3 g2 + γ4 h1 A−1 f3 − A−2 Bg2 + u3 h1 A−1 g2 − A−2 Bh1  0 A Bxf3 + u3 D + γ4 C −A−1 xf3 − A−2 BD + u3 C −2 A−1 D − A−2 BC  M3∗ =   0 , −u3 xf3 + γ4 D A−2 Bxf3 + u3 D −A−1 xf3 − A−2 BD  0 −γ4 xf3 −u3 xf3 A−2 Bxf3 where γ4 = −A−4 B 3 + 2A−3 BC − A−2 D + 2αA−1 . Note that A−1 g2 − A−2 Bh1 = x2 − A−2 B 2 + A−1 C. Some additional calculations show that polynomial entries m∗i1 for i = 1, . . . , 4, m∗13 and m∗14 of the matrix M4∗ − M3∗ := (m∗ij (x))4×4 have degree zero, and that the degree of the polynomials m∗32 and m∗43 is 6. All the other entries have degree 4. In the determinant of the matrix M4∗ − M3∗ , the product containing the two entries of degree 6 provides only degree 12. Taking one degree 6 entry and two entries of degree 4 we get the maximum degree 14. This completes the proof of the lemma.  Proof of statement (a) Theorem 2. From (3) we have that for each solution (xL , xR ) with xL 6= xR of the polynomial systems EL = 0 and ER = 0, we have a cross- ing limit cycle of the discontinuous piecewise linear differential system (2) which intersect the curve y = xn at the points (xL , xnL ) and (xR , xnR ). Recall that we denote by 2dn the degree of the resultant RL computed in Lemma 3. Of course, the expression of 2dn changes according the parity of n. 14 J. LLIBRE AND X. ZHANG If (xL , xR ) with xL 6= xR is a solution of the polynomial system EL = 0 and ER = 0, the value xL is a root of the resultant RL , and the value xR is a root of the resultant RR . But both resultants RL and RR have the same roots, because these two polynomials are the same, we can pass from one to the other interchanging the variable xL by the variable xR . So the possible values for xL and for xR are the same. Consequently we only have at most 2dn points (xL , xnL ) and (xR , xnR ), all together, which can be points where the crossing limit cycles intersect the curve y = xn . Hence we cannot have more then dn crossing limit cycles. This completes the proof of statement (a) of the theorem.  Proof of statement (b) Theorem 2. We provide an example showing that the upper bound of 3 limit cycles given in Theorem 2 for the maximum number of limit cycles that a discontinuous piecewise linear differential system in R2 separated by the parabola y = x2 when both linear differential systems have a center is reached. Consider the linear differential system 189x 38221y 720042289 ẋ = − + − , 100 20500 205000000 (4) 41x 189y 3296401 ẏ = − + − , 20 100 1000000 in the region L, and the linear differential system 18 ẋ = − y, 5 (5) 99 ẏ = x − , 50 in the region U. Both linear differential systems form a discontinuous piecewise linear differential system in R2 separated by the parabola y = x2 having two centers. This discontinuous piecewise linear differential system has the three limit cycles of the Figure 1. We note that the value of the coordinate x of the six points (x, x2 ) of the parabola which belong to the limit cycles are the six real roots of the polynomial RL equal to 10769750976621134140624876981 − 29736678501017804062959688750 x+ 2515419213211056546308719750 x2 + 25945028877701413073475000000 x3 − 5566633022238812671927500000 x4 − 5065968152418838875000000000 x5 + 1454075658590625000000000000 x6 . The discontinuous piecewise linear differential system (4)-(5) proves the statement (b) of Theorem 2.  Proof of statement (c) Theorem 2. For n = 2 we have that dn = 3, and conse- quently by Theorem 2 the discontinuous piecewise linear differential system (2) will have at most 3 limit cycles. The concrete expression of the resultant RL modulo the coefficient of the highest degree monomial is ∗ RL =x6 + C1 x5 + C2 x4 + C3 x3 + C4 x2 + C5 x + C6 , DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 15 as before we have replaced the variable xL by the variable x for simplicity, and C1 = 2Aη1 , C2 = A−1 D + 2A2 η12 + η3 , C3 = 4Dη1 − 4β, C4 = A−2 D2 − A−1 Dη3 + Dη2 + η32 + Aη1 (Dη1 − 4β) + A2 η12 η3 , C5 = 2A−1 D(Dη1 − β) − 2βη3 − 2ADη1 η2 − 2A2 (β − Dη1 )η12 , C6 = A−3 D3 − 2A−2 D2 η3 + A−1 D(η3 + 1)η3 − Dη1 (6β + Dη1 − η1 η3 ) +4β 2 − A(Dη22 − 2βη1 η3 ) + A3 η13 (2β + Dη1 ), where η3 = 2α + 1. For n = 3 we have that dn = 7, therefore by Theorem 2 the discontinuous piecewise linear differential system (2) will have at most 7 limit cycles. The resultant RL =Resultant[EL (xL , xR ), ER (xL , xR ), xR ] has a huge expression directly using the parameters of the system. Here we choose the previous notations A, B, C, D, ∗ and still replace xL by x for simplicity, we get the next formula RL of RL modulo a multiplicative constant. ∗ RL =C0 x14 + C1 x13 + C2 x12 + C3 x11 + C4 x10 + C5 x9 + C6 x8 + C7 x7 + C8 x6 + C9 x5 + C10 x4 + C11 x3 + C12 x2 + C13 x + C14 , where C0 = −3B, C1 = −6A−1 B 2 , C2 = −4A−2 B 3 − 2A−1 B(C + 2D) − 3D + 4Aα, C3 = −2A−2 B 2 (C + 5D) − 12A−1 BD, C4 = −2A−3 B 3 (3C − 5D) − 18A−2 B 2 D − A−2 BD(8C + D) −2A−1 (3B 2 α + 2CD + D2 ) − 7B + 6Cα + 4Dα − 4Aβ, C5 = 4A−4 B 4 (C − D) − 2A−3 B 2 (4BD − 5C 2 + 10CD + D2 ) +2A−2 B(4B 2 α − 5CD + 7D2 ) − 2A−1 (5B 2 + 6BCα − 3D2 ) +2Bβ + 2Dα + 8Aα2 , C6 = A−4 B 3 (15C 2 − 18CD − D2 ) + 2A−3 B(BCD − 13BD2 + 2C 3 − 4C 2 D − CD2 ) −A−2 (4B 3 − 4B 2 Cα − 8B 2 Dα − 18BD2 − 6CD2 − 3C 2 D) +2A−1 (3B 2 β − BC − 3BD + 3C 2 α + 4CDα + D2 α) − 7D + 6Cβ + 4Dβ + 8Aα, C7 = 4A−5 B 4 C(C − D) + 2A−4 B 2 (8BCD + 8C 3 − 10C 2 D − CD2 − 10D2 ) +2A−3 B(B 3 − 4B 2 Dα − 9BD2 + C 2 D − 12CD2 − D3 ) −2A−2 (B 3 β + 5B 2 D + 9B 2 Dα − 2BC 2 α − 2BCDα − 2BD2 α + 4CD2 + 2D3 ) −4A−1 (BCβ + BDβ + 5BD − 3CDα − D2 α) + 8Dα2 ) − 16Aαβ, C8 = 2A−5 B 3 (2BD + 5C 2 )(C − D) + 4A−4 B 4 (C − D)α − 4A−4 B 3 D2 +2A−4 B 2 (15C 2 D − 20CD2 − D3 ) + A−4 BC 2 (4C 2 − D2 − 6CD) −8A−3 B 4 β + 2A−3 B 3 (5C − 2D − 8Dα) − 2A−3 B 2 (5C 2 + 2CD − D2 )α −2A−3 BD2 (8C + 7D) − 2A−3 CD(C 2 + CD + D2 ) − 4A−2 B 3 α2 +2A−2 B 2 (3Cβ − 11D − 2Dβ) − A−2 B(8CD − 26CDα + D2 + 8D2 α) +2A−2 (2C 3 α + 2C 2 Dα + 3CD2 α − 2D3 ) − 2A−1 B 2 α +2A−1 B(2Cα2 + 4Dα2 + 3Dβ) − 2A−1 (2C 2 β + 3CD + 6CDβ + D2 + D2 α) −5B − 4Bαβ + 8Cα + 4Dα + 4Dα2 + 8A(α3 − β), 16 J. LLIBRE AND X. ZHANG  C9 = 8A−5 B 2 C C 2 + 2BD (C − D) + 2A−4 B 4 (C + D) + 4A−4 B 3 C(C − 3D)α +2A−4 B 2 D2 (5C − 11D) + 4A−4 BCD(3C 2 − 5CD − D2 ) −4A−3 B 3 (2Cβ + D + 2Dβ) + 2A−3 B 2 (3C 2 − 4CD − 4CDα − D2 − 8D2 α) −4A−3 B(C 3 α − 2C 2 Dα − 2CD2 α + 3D3 ) − 4A−3 CD2 (2C + D) − 4A−2 B 3 α +12A−2 B 2 Dβ + 2A−2 B(6C 2 β − 3CD − 10CDβ − 7D2 + D2 β) +2A−2 (10C 2 Dα − 4CD2 α + 3D3 α) − 4A−1 B 2 + 12A−1 BCα −2A−1 (4C 2 α2 − 8CDα2 + 2CDβ + 5D2 + 4D2 β) +4(Bβ + Dα − 2Cαβ − 2Dαβ) + 8A(α2 + β 2 ),  C10 = −2A−5 B(C − D) B 4 − 2B 3 Cα − 5B 2 D2 + 10BC 2 D − C 4 +2A−4 B 4 (2Cβ + D − 4Dβ) + A−4 B 3 (4C 2 + 8CDα − D2 − 16D2 α) −2A−4 B 2 (2C 3 α + 2C 2 Dα − CD2 α + 3D3 ) + A−4 B(6C 2 D2 − 20CD3 − D4 ) −A−4 C 2 D(C 2 + 2D2 ) + 2A−3 B 4 α − 4A−3 B 3 (Cα2 + 2Dβ) +2A−3 B 2 (7C 2 β + 5CD − 8CDβ − 8D2 − 9D2 α + 2D2 β) +2A−3 B(C 3 − C 2 D − CD2 + 4D3 α − C 2 Dα) +2A−3 (C 4 α + 3C 2 D2 α − 4CD3 − D4 ) − A−2 B 3 (1 − 4αβ) −4A−2 B 2 (C + D)α + A−2 B(4C 2 α2 + 8CDα2 − 4CDβ − 21D2 − 4D2 β) −A−2 D(2C 2 + 12C 2 β + 8CD − 22CDα + 2CDβ − D2 + 4D2 α) + 8A−1 B 2 β −2A−1 B(8Cαβ + D − 3Dα) + 4A−1 (C 2 α + 2CDα − 5CDα2 + 4D2 α2 + 2D2 β) −4B(α2 + 2β 2 ) − 8Cβ + 8Cα3 − 5D − 20Dαβ − 4Dβ − 4A(2αβ − 1)α, C11 = 4A−5 B(C − D) B 4 β − B 3 (C − Dα) + B 2 C 2 α + 5BCD2 + 2C 3 D −8A−4 B 4 Dβ + 2A−4 B 3 (3C 2 β + 7CD − 2CDβ + D2 β − 3D2 − 4D2 α) +2A−4 B 2 (2C 3 − C 2 D − 3C 2 Dα − CD2 − 4CD2 α + D3 α) −2A−4 B(4C 4 α − 2C 3 Dα − 2C 2 D2 α + 3CD3 + 3D4 ) − 2A−4 CD2 (2C 2 + D2 ) −8A−3 B 4 αβ − 8A−3 B 3 Dα2 − 2A−3 B 2 D(4Cβ + 7D) +2A−3 B(6C 3 β + 2C 2 D − 2D3 α − 10C 2 Dβ − 6CD2 + 2CD2 α) +2A−3 (6C 3 Dα − 2C 2 D2 α + 4CD3 α − D4 ) + 4A−2 B 3 β +2A−2 B 2 (C + 8Cαβ − D − 2Dα + 4Dαβ) + 2A−2 BD(8Cα2 + 4Dα2 + 7Dβ) −2A−2 (4C 3 α2 − 4C 2 Dα2 + 2C 2 Dβ + 3CD2 + 4CD2 β + 2D3 − 3D3 α) −16A−1 B 2 β 2 − 8A−1 B(Cβ + D + 4Dαβ) + 8A−1 (CD − 2CDβ + D2 − 2D2 α)α +16Bβα2 + 8(Cβ 2 + 2Dα3 ) − 16Aαβ,  C12 = 2A−5 B(C − D) B 3 (2Cβ − D) − B 2 C(C − 4Dα) + B(2C 3 α + 3D3 ) + 6C 2 D2 −A−4 B 5 + 4A−4 B 4 Cα − 4A−4 B 3 (C 2 α2 + 2CDβ + D2 β) +A−4 B 2 (6C 3 β + 16C 2 D − 8CD2 + 2CD2 β − 4CD2 α − D3 − 8D3 α) −A−4 BD(18C 3 α + C 2 D − 8C 2 Dα − 4CD2 α + 3D3 ) +A−4 D2 (2C 3 Dα − 6C 2 D2 − D4 ) − 2A−3 B 4 β + 4A−3 B 3 (C − Cαβ − Dα) −2A−3 B 2 (7C 2 α − 3D2 β) + 4A−3 B(3C 3 α2 + C 2 Dβ − 2CD2 − 4CD2 β − 2D3 + D3 β) −2A−3 (2C 3 Dβ + C 2 D2 − 10C 2 D2 α + 2C 2 D2 β + 4CD3 α − 3D4 α) −4A−2 B 3 β 2 + A−2 B 2 (4Cβ − 5D + 4Dβ) + 2A−2 B(4C 2 αβ − CD + 12CDα) +A−2 (4C 2 Dα − 20C 2 Dα2 + 8CD2 α2 + 2CD2 β − 3D3 ) +2A−1 B 2 α − 2A−1 B(4Cα2 − Dβ + 4Dβ 2 ) +2A−1 (4C 2 α3 − 2C 2 β − CD − 4CDβ − 6CDαβ − 2D2 α − 8D2 αβ) −B + 4Bαβ + 2Cα + 8Cα2 β + 8Dα2 + 4Dβ 2 + 4A(2αβ − 1)β,  C13 = 4A−5 B(C − D) 2B 3 Dβ − B 2 (C 2 β + CD − 2D2 α) + BC 2 Dα + 2CD3 +2A−4 B 5 β − 2A−4 B 4 (2Cβ − D)α − 2A−4 B 3 D(2Cα2 + 3Dβ) −2A−4 B 2 D(C 2 β − 6CD + 4D2 + 3D2 α − 2D2 β) +2A−4 BC(4C 3 β + 2C 2 D − CD2 − D3 ) −2A−4 B(2C 3 Dβ + 9C 2 D2 α + C 2 D2 β − 4CD3 α − 2D4 α) +2A−4 CD3 (Cα − 2D) + 4A−3 B 4 β 2 − 4A−3 B 3 (2C − Dα)β +4A−3 B 2 C(3Cβ − 2D)α + 4A−3 BD(3C 2 α2 + CDβ − 2D2 ) −2A−3 D(2C 3 β + C 2 D + 4C 2 Dβ + 2CD2 − 8CD2 α − D3 + 2D3 α) DISCONTINUOUS PIECEWISE LINEAR DIFFERENTIAL SYSTEMS 17 −2A−2 B 2 (4Cβ 2 − 5Dβ + 4Dβ 2 ) + 2A−2 B(2C 2 β + CD − 14CDβα − D2 + 3D2 α) +4A−2 (2C 3 αβ + C 2 Dα − 2C 2 Dαβ − 5CD2 α2 + 2D3 α2 + D3 β) −4A−1 B 2 αβ + 4A−1 B(2Cα2 β − Dα2 − Dβ 2 ) +4A−1 (2C 2 β 2 − CDβ + 2CDα3 − D2 − 2D2 β − D2 αβ) +2(Bβ − 4Bαβ 2 + Dα − 4Dα2 β) + 8Aβ 2 ),  C14 = 2A−5 B(C − D) B 2 D(4Cβ + D) − BC(2C 2 β + CD − 2D2 α) + D4 −4A−4 B 5 β 2 − A−4 B 4 (2Cβ + D + 8Dαβ) + 2A−4 B 3 (2C 2 β + CD − 2D2 α)α −2A−4 B 2 D2 (9Cβ + D − 4Dβ) +A−4 BD(10C 3 β + 5C 2 D − 4C 2 Dβ − 10CD2 α − 2CD2 + 4CD2 β + D4 + 4D3 α) −A−4 D2 (2C 3 β + C 2 D − 2CD2 α + D3 ) +2A−3 B 3 (10Cβ 2 + Dβ) + 2A−3 B 2 (4C 2 β + 2CD + 12CDαβ − D2 α) −2A−3 B(6C 3 αβ + 3C 2 Dα − 6CD2 α2 − 5D3 β) −2A−3 D2 (2C 2 β + CD + 2CDβ + D2 − 3D2 α + D2 β) −20A−2 B 2 Dβ 2 − 2A−2 B(6C 2 β 2 + 7CDβ + 4CDβ 2 + 2D2 + 14D2 αβ) +2A−2 D(6C 2 αβ − 2C 2 β − CD + 3CDα + 2D2 α − 6D2 α2 ) +8A−1 B 2 αβ 2 + 2A−1 B(2Cβ + D + 8Dαβ)α −2A−1 (4C 2 α2 β + 2CDα2 − 6CDβ 2 − 4D2 α3 − 3D2 β − 4D2 β 2 ) −4Bβ 2 − 2Cβ − 16Cαβ 2 − D − 12Dαβ − 8Aβ 3 , where u3 = A−3 B 2 − A−2 C and η4 = −A−4 B 3 + 2A−3 BC − A−2 D + 2A−1 α. 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