December 8, 2021 12:24 CarGonLlib2021IJBCFinal This is a preprint of: “Limit cycles on piecewise smooth vector fields with coupled rigid centers”, Tiago de Carvalho, Luiz Fernando Gonçalves, Jaume Llibre, Internat. J. Bifur. Chaos Appl. Sci. Engrg., vol. 31(15), 2150224, 2021. DOI: [10.1142/S0218127421502242] Limit cycles on piecewise smooth vector fields with coupled rigid centers Tiago Carvalho Departamento de Computação e Matemática Faculdade de Filosofia, Ciências e Letras de Ribeirão Preto Universidade de São Paulo Av. Bandeirantes, 3900, zip-code 14040-901, Ribeirão Preto, SP, Brazil tiagocarvalho@usp.br Luiz Fernando Gonçalves* Instituto de Matemática e Estatı́stica Universidade Federal de Goiás Campus Samambaia, CEP 74690-900, Goiânia, GO, Brazil luiz.goncalves@ufg.br Jaume Llibre Departament de Matemàtiques, Facultat de Ciències Universitat Autònoma de Barcelona 08193 Bellaterra, Barcelona, Spain jllibre@mat.uab.cat Received (to be inserted by publisher) We provide an upper bound for the maximum number of limit cycles of the classes of discontin- uous piecewise differential systems formed by two differential systems separated by a straight line presenting rigid centers. These two rigid centers are polynomial differential systems with a linear part and a nonlinear homogeneous part. We study the maximum number of limit cycles that such a class of piecewise differential systems can exhibit. Keywords: Piecewise smooth vector field; rigid centers; limit cycle. 1. Introduction and main results The search for limit cycles is one of the most important studies in the qualitative theory of the planar ordinary differential equations. Such importance is evidenced by the 16th Hilbert’s problem (see [Hilbert, 1902]) that seeks the determination of an upper bound for the number of limit cycles for the class of planar polynomial vector fields of degree n, a problem that remains unsolved for n ≥ 2. Recently the study of the limit cycles also is relevant in the discontinuous piecewise differential systems. In the last decades the study of discontinuous piecewise vector fields has had a great growth in the mathematical community, since such vector fields can be used as important models in applied science. ∗ Corresponding author 1 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 2 Indeed, several models used in applied problems are described by systems that are not completely differen- tiable, but in different parts, where a law of evolution is suddenly interrupted by another law of evolution that will begin to govern such system. The modeling of such systems consists of different vector fields defined in distinct regions separated by a switching manifold and are known as piecewise smooth vector fields, discontinuous piecewise systems, or Filippov systems. Pioneering studies initiated by Andronov [Andronov et al., 1966] and Filippov [Filippov, 1988] led to a theoretical foundation for this kind of differential systems. Nowadays a vast literature on these vector fields is available. See for instance [di Bernardo et al., 2008] for the main theory and some applications, [Chua et al., 1986] for applications in electrical circuits, [Brogliato, 1999; Leine & Nijmeijer, 2004] for applications in mechanical models, [di Bernardo et al., 2001; Jacquemard & Tonon, 2012] for applications in relay systems, among others. As in the regular case the study of the existence and location of limit cycles in piecewise smooth vector fields is also of great importance, because in addition to the smooth dynamic elements, there are new ones which did not exist in the smooth world. Let p ∈ R2 be a singularity of an analytic differential system in the plane. The singularity p is a center if there exists an open neighborhood U of p such that all the solutions in U \ {p} are periodic. Without loss of generality we may assume that the equilibrium point is at the origin. Denote by Tq the period of the periodic orbit through q ∈ U \ {p}. We say that p is an isochronous center if Tq is constant for all q ∈ U \ {p}. An isochronous center is uniform or rigid if the angular velocity of the vector field is the same for all periodic orbits in U \ {p}, that is if in polar coordinates (r, θ) defined by x = r cos θ, y = r sin θ it can be written as ṙ = G(r, θ), θ̇ = k, where k ̸= 0 is a constant. See [Chavarriga & Sabatini, 1999; Conti, 1994] for details. After scaling the time (if necessary) it is always possible to consider θ̇ = 1 in the previous expression. Isochronicity in the regular case has been extensively studied in the last decades, see for instance [Algaba & Reyes, 2003, 2009; Dias & Mello, 2012; Gasull et al., 2005; Gasull & Torregrosa, 2005; Han & Romanovski, 2012; Llibre & Rabanal, 2015] and references therein. Such importance is due to its applica- tions in physical phenomena and its relation with the famous center-focus problem, a classical open problem in the qualitative theory of planar differential equations to distinguish an equilibrium point between a focus and a center. In recent years isochronicity has also been explored for the discontinuous piecewise differ- ential systems, see for instance [Benterki & Llibre, 2020; Itikawa et al., 2017; Llibre & Teixeira, 2018], by considering the coupling of two or more rigid systems and investigating their dynamics. The main goal of this paper is to provide the maximum number of limit cycles that can bifurcate from a discontinuous piecewise differential systems formed by the coupling of two rigid centers whose switching manifold is the straight line {x = 0}. More precisely, in one half-plane we consider a rigid center of degree n of the form n−1 X n−1 X n−i−1 i ẋ = −y + x si x y, ẏ = x + y si xn−i−1 y i , (1) i=0 i=0 and in the other half-plane we consider either an arbitrary linear rigid center, that is, ẋ = −y, ẏ = x, (2) or a rigid center analogous to (1) but with degree m. The linear-linear case was studied in [Itikawa & Llibre, 2015] and has no limit cycles. Our first main result is the following one. Theorem 1. The discontinuous piecewise differential systems separated by the straight line x = 0 having the linear rigid center (2) in x ≤ 0 and the nonlinear rigid center (1) in x ≥ 0, have no limit cycles. Our second main result is the next one. Theorem 2. Consider the discontinuous piecewise differential systems formed by system (2) in x ≤ 0 and system (1) in x ≥ 0 after an affine change of variables. Then such discontinuous piecewise differential systems have at most one limit cycle when the degree of system (1) is 2 or 3. Furthermore there are examples of such systems with one limit cycle. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 3 The third main result is the next. Theorem 3. Consider the discontinuous piecewise differential systems formed by system (1), of degree n, in x ≥ 0 and system (1), of degree m, in x ≤ 0 after an affine change of variables. Then such discontinuous piecewise differential systems have at most one limit cycle when the degree of systems are 2 or 3, i.e., m, n ∈ {2, 3}. Furthermore there are examples of such systems with one limit cycle. The paper is organized as follows. In Section 2 we recall the basic theory of the piecewise smooth vector fields that we need for proving our results. Section 3 brings some considerations about the rigid centers considered in this work besides the proof of Theorem 1. Theorem 2 is proved in Section 4. Theorem 3 is proved in Section 5. Finally Section 6 closes the paper with concluding remarks. 2. Preliminary definitions and results In this section we present the basic results of the theory on piecewise smooth vector fields that we need. A piecewise smooth vector field on an open set U ⊂ R2 is a pair of Cr -vector fields X and Y with r ≥ 1, defined on U separated by a smooth codimension one manifold Σ. The switching manifold Σ is obtained by considering Σ = h−1 (0), where h : U ⊂ R2 −→ R is a differentiable function having 0 as a regular value. Note that Σ is the separating boundary of the regions Σ+ = {(x, y) ∈ U | h(x, y) > 0} and Σ− = {(x, y) ∈ U | h(x, y) < 0}. So a piecewise smooth vector field is provided by  X(x, y), h(x, y) ≥ 0, Z(x, y) = (3) Y (x, y), h(x, y) ≤ 0. As usual system (3) is denoted by Z = (X, Y, Σ) or simply by Z = (X, Y ), when the separation line Σ is well understood. In order to establish a definition for the trajectories of Z and investigate its behavior we need a criterion for the transition of the orbits between Σ+ and Σ− across Σ. The contact between the vector field X (or Y ) and the switching manifold Σ is characterized by the derivative of h in the direction of the vector field X (also known as the Lie derivative of h with respect to X), that is by the expression Xh(p) = ⟨∇h(p), X(p)⟩ , and for i ≥ 2 we define X i h(p) = ∇X i−1 h(p), X(p) , where ⟨., .⟩ is the usual inner product in R2 . The basic results of the discontinuous piecewise differential systems in this context were stated by Filippov [Filippov, 1988]. We can divide the switching manifold Σ in the following sets: (a) Crossing set: Σc : {p ∈ Σ : Xh(p) · Y h(p) > 0}; (b) Escaping set: Σe : {p ∈ Σ : Xh(p) > 0 and Y h(p) < 0}; (c) Sliding set: Σs : {p ∈ Σ : Xh(p) < 0 and Y h(p) > 0}. The escaping Σe or sliding Σs regions are respectively defined on points of Σ where both vector fields X and Y simultaneously point outwards or inwards from Σ while the interior of its complement in Σ defines the crossing region Σc (see Figure 1). The complementary of the union of these regions is the set formed by the tangency points between X or Y with Σ. Σ Fig. 1. Crossing, sliding and escaping regions, respectively. A point p ∈ Σ is called a tangency point of X (resp. Y ) if it satisfies Xh(p) = 0 (resp. Y h(p) = 0). A tangency point is called a fold point of X if X 2 h(p) ̸= 0. Moreover, p ∈ Σ is a visible (resp. invisible) fold point of X if X 2 h(p) > 0 (resp. X 2 h(p) < 0). December 8, 2021 12:24 CarGonLlib2021IJBCFinal 4 In order to define a trajectory of a discontinuous piecewise differential system passing through a crossing point, it is enough to concatenate the trajectories of X and Y through that point. However in the sliding and escaping sets we need to define an auxiliary vector field. So we consider the Filippov’s convention (see [Filippov, 1988]) and a new vector field is defined on Σs ∪ Σe . This new vector field, called sliding vector field, is a convex linear combination of X(p) and Y (p) in a way that Z s is tangent to Σ in the cone generated by X(p) and Y (p). Furthermore given a discontinuous piecewise vector field Z = (X, Y ) we say that an equilibrium point p of X is real if p ∈ Σ+ and it is virtual if p ∈ Σ− . In this scenario the trajectories of Z are considered as a concatenation of trajectories of X, Y and Z s . Given a vector field F (x, y) = (F1 (x, y), F2 (x, y)), defined on an open set U ⊂ R2 , we consider the corresponding ordinary differential equations dx dy ẋ = = F1 (x, y), ẏ = = F2 (x, y), (4) dt dt where the independent variable t is called the time. Denote the flow of (4) by φF or simply by φ when there is no danger of confusion. Also denote by φF (t, p) or by φ(t, p) the solution of system (4) by the point p such that φ(0, p) = p. When the trajectory of the vector field X through p ∈ Σ returns to Σ (by the first time) after a positive time t1 (p), called X–flight time, we define the half return map associated with X by πX (p) = ϕX (t1 (p), p) = p1 ∈ Σ. When the trajectory of Y through p1 ∈ Σ returns to Σ (by the first time) after a positive time t2 (p1 ), called Y –flight time, we define the half return map associated with Y by πY (p1 ) = ϕY (t2 (p1 ), p1 ) ∈ Σ. The first return map associated with Z = (X, Y ) is defined by the composition of these two transition maps, that is, πZ (p) = πY ◦ πX (p) = ϕY (t2 (p1 ), ϕX (t1 (p), p)) (5) or the reverse, applying first the flow of Y and after the flow of X. See Figure 2. ϕX (t, p) πZ (p) p p1 Σ ϕY (t, p1 ) Fig. 2. First return map of a discontinuous piecewise differential system. When the vector fields X and Y associated to Z = (X, Y ) have a first integral (see [Arnold, 1992]), the solution curves of the respective differential equations are contained in the level sets of the first integrals. In this scenario the first return map can be handily computed by seeking for points in Σ that are on the same level curves of these first integral functions. In this case we avoid working with flight times. 3. Proof of Theorem 1 The paper [Llibre & Teixeira, 2018] provides the normal form 4b2 + ω 2 ẋ = −bx − y + d, ẏ = ax + by + c, (6) 4a with a ̸= 0 and ω > 0, for an arbitrary planar linear differential system having a linear center. Let us consider the equilibrium point of system (6) at the origin, that is, consider c = 0 and d = 0. Under these conditions. in polar coordinates (r, θ) defined by x = r cos θ and y = r sin θ, system (6) gets the form  ! sin θ cos θ 4a2 − 4b2 − ω 2 ṙ = r − b cos(2θ) θ 4a  2 (7) 2 2 4b + ω sin θ θ̇ = a cos2 θ + 2b sin θ cos(θ) + 4a December 8, 2021 12:24 CarGonLlib2021IJBCFinal 5 So we have that system (6) is a rigid center if and only if b = 0 and ω = 2a. Since a is nonzero we can take a = 1 rescaling the time. Therefore system (6) becomes system (2), that is system ẋ = −y, ẏ = x. We call X(x, y) = (−y, x) the vector field associated with system (2). Equivalently in polar coordinates system (2) can be rewritten as ṙ = 0, θ̇ = 1. We may notice that system (2) has the first integral p F (x, y) = x2 + y 2 or, in polar coordinates F (r, θ) = r. (8) Now we consider a polynomial differential system in R2 , ẋ = P (x, y), ẏ = Q(x, y) of degree n. Assume that it has a center at the origin of coordinates. Then it is well known, see for instance [Conti, 1994; Itikawa & Llibre, 2015], that this center is a uniform isochronous if and only if by doing a linear change of variables and a rescaling of time it can be written as ẋ = −y + xf (x, y), ẏ = x + yf (x, y) (9) where f (x, y) is a polynomial of degree n − 1 in the variables x and y, and f (0, 0) = 0. Consider system (9). Conti [Conti, 1994] proved the following result. Pn−1 n−i−1 y i be a homogeneous polynomial of degree n − 1. Then the Proposition 1. Let f (x, y) = i=0 ai x origin is a uniform isochronous center of system (9) if either n is even, or n is odd and X Z 2π n−1 n−1 X n−i−1 i ai cos θ sin θ dθ = ai In−i−1,i = 0. (10) i=0 0 i=0 Pn−1 When n is even, i=0 ai In−i−1,i = 0 always holds, because f (cos(θ + π), sin(θ + π)) = −f (cos θ, sin θ), where n−1 X f (cos θ, sin θ) = ai cosn−i−1 θ sini θ. i=0 In the sections 2.511 and 2.512 of [Gradshteyn & Ryzhik, 2007] we obtain the following expressions for integrals of power of trigonometric functions: Z t1 " m−1 #!x=t1 sinp+1 x X (2m − 1)(2m − 3) . . . (2m − 2k + 1) cos2m−2k−1 x sinp x cos2m xdx = cos2m−1 x + t0 2m + p (2m + p − 2)(2m + p − 4) . . . (2m + p − 2k) k=1 x=t0 Z t1 (2m − 1)!! + sinp xdx, (2m + p)(2m + p − 2) . . . (p + 2) t0 for arbitrary real p, except for the negative even integers −2, −4, . . ., −2n. We also have, Z t1 " l−1 #!x=t1 cos x X (2l − 1)(2l − 3) . . . (2l − 2k + 1) sin2l−2k−1 x sin2l xdx = − sin2l−1 x + t0 2l 2k (l − 1)(l − 2) . . . (l − k) k=1 x=t0 (2l − 1)!! + (t1 − t0 ) 2l l! and Z " l−1 k+1 #!x=t1 t1 cos x X 2 l(l − 1) . . . (l − k) sin2l−2k−2 x 2l+1 2l sin xdx = − sin x + . t0 2l + 1 (2l − 1)(2l − 3) . . . (2l − 2k − 1) k=0 x=t0 For arbitrary real p, except for the negative odd integers −1, −3, . . ., −(2n + 1), we have Z t1 " m #!x=t1 sinp+1 x X 2k m(m − 1) . . . (m − k + 1) cos2m−2k x sinp x cos2m+1 xdx = cos2m x + t0 2m + p + 1 (2m + p − 1)(2m + p − 3) . . . (2m + p − 2k + 1) k=1 x=t0 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 6 Also, for real p, except for the negative even integers −2, −4, . . ., −2n, we have Z t1 " m−1 #!x=t1 cos p+1 x X (2m − 1)(2m − 3) . . . (2m − 2k + 1) sin2m−2k−1 x cosp x sin2m xdx = − sin2m−1 x + t0 2m + p (2m + p − 2)(2m + p − 4) . . . (2m + p − 2k) k=1 x=t0 Z t1 (2m − 1)!! + cosp xdx. (2m + p)(2m + p − 2) . . . (p + 2) t0 Moreover, Z " l−1 #!x=t1 t1 sin x X (2l − 1)(2l − 3) . . . (2l − 2k + 1) cos2l−2k−1 x cos2l xdx = cos2l−1 x + t0 2l 2k (l − 1)(l − 2) . . . (l − k) k=1 x=t0 (2l − 1)!! + (t1 − t0 ) 2l l! and Z " l−1 k+1 #!x=t1 t1 sin x X 2 l(l − 1) . . . (l − k) cos2l−2k−2 x cos2l+1 xdx = cos2l x + . t0 2l + 1 (2l − 1)(2l − 3) . . . (2l − 2k − 1) k=0 x=t0 For arbitrary real p, except for the negative odd integers −1, −3, . . ., −(2n + 1) we have Z t1 " m #!x=t1 cos p+1 x X 2 k m(m − 1) . . . (m − k + 1) sin2m−2k x cosp x sin2m+1 xdx = − sin2m x + . t0 2m + p + 1 (2m + p − 1)(2m + p − 3) . . . (2m + p − 2k + 1) k=1 x=t0 If n odd, from the formulas above we get that In−i−1,i = 0 when i odd, and consequently, " # X Z 2π n−1 2π(n − 2)!! n−i−1 i ai cos θ sin θ dθ = a0 n−1  + n−1 i=0 0 2 2 2 ! n−3 X 2 a2k 2π(2k − 1)!!(n − 2k − 2)!! +  (11) n−2k−1 n−2k−1 (n − 1)(n − 3) · · · (n − 2k + 1) 2 k=1 2 2 ! " # 2π(n − 2)!! + an−1 n−1  , 2 2 n−1 2 ! where (2p)!! = 2p(2p − 2)(2p − 4) · · · 2 and (2p + 1)!! = (2p + 1)(2p − 1) · · · 3 · 1. Thus, hypothesis (10) is equivalent to supposing    n−3   2 n−1 n−1 !X 2  (2j − 1)!!(n − 2j − 2)!!  2 2   an−1 = −a0 − a2j  . (12) (n − 2)!!  n−2j−1   Y j  j=1 2 2 n−2j−1 ! (n − 2i + 1)  2 i=1 So in this paper under the conditions of Proposition 1 we shall work with rigid centers of the form (1), i.e. n−1 X n−1 X ẋ = −y + x ai xn−i−1 y i , ẏ = x + y ai xn−i−1 y i . i=0 i=0 We denote by n−1 n−1 ! X X n−i−1 i n−i−1 i Yn (x, y) = −y + x ai x y ,x + y ai x y i=0 i=0 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 7 the vector field associated to system (1). Equivalently in polar coordinates system (1) writes n−1 X ṙ = rn ai cosn−i−1 θ sini θ i=0 (13) θ̇ = 1. System (13) has the first integral X n−1 Z 1 Hn (r, θ) = − ai cosn−i−1 θ sini θdθ. (1 − n)rn−1 i=0 So in Cartesian coordinates the first integral of system (1) becomes n−1 X Z 1 xn−i y i Hn (x, y) = n−1 − ai n+1 dy. (1 − n)(x2 + y 2 ) 2 i=0 (x2 + y 2 ) 2 By coupling the two rigid centers (2) and (1) we can consider the piecewise smooth vector field  Yn (x, y), x ≥ 0, Z(x, y) = (14) X(x, y), x ≤ 0. Observe that the switching manifold is the straight line Σ = h−1 (0), where h(x, y) = x. Since the vector fields X and Yn associated with the piecewise smooth vector field Z = (Yn , X) are integrable, then the solution curves of the respective differential equation are contained in the level sets of their respective first integrals. Thus the first return map associated with the discontinuous piecewise vector field Z can be computed by seeking for points in the switching manifold Σ that are on the same level curves of these first integrals. In this way for a limit cycle of the discontinuous piecewise differential system associated with Z given by (14) which has two intersecting points (0, y1 ) and (0, y2 ) with the line of discontinuity Σ = {x = 0}, its coordinates y1 and y2 must satisfy the set of equations ( (   F (0, y1 ) = F (0, y2 ), F r0 , π2 = F R0 , 3π 2 , or, in polar coordinates,   (15) Hn (0, y1 ) = Hn (0, y2 ), Hn r0 , π2 = Hn R0 , 3π2 , where (r0 , π/2) and (R0 , 3π/2) are the respective intersecting points (0, y1 ) and (0, y2 ) in polar coordinates. The first equation of (15) produces r0 = R0 . Substituting r0 = R0 in the second equation, we get n−1 X Z π n−1 X Z 3π 2 2 n−i−1 i ai cos θ sin θdθ = ai cosn−i−1 θ sini θdθ. (16) i=0 0 i=0 0 Using the formulas of [Gradshteyn & Ryzhik, 2007] described above for the power of trigonometric functions we conclude that (i) If ai = 0 for all i then the equality (16) holds for every r0 , i.e. we have a continuum of period orbits. Therefore there are no limit cycles and the origin is a piecewise rigid center. (ii) If ai ̸= 0 for some i then, from equality (16) we obtain an expression of the form P (a) = k0 a0 + k1 a1 + . . . + kn−1 an−1 where a = (a0 , . . . , an−1 ) and ki ∈ R. In this way the vectors a such that P (a) = 0 correspond to values of ai such that the equality (16) holds, for all r0 > 0. Therefore we also get a continuum of period orbits. The vectors a such that P (a) ̸= 0 correspond to values of ai such that the equality (16) does not hold. Therefore the origin behave like a focus. So there are no periodic orbits and then no limit cycles. In summary there are no limit cycles in the piecewise smooth vector field Z and Theorem 1 is proved. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 8 Example 3.1. We present a numerical example of Theorem 1. Consider the discontinuous piecewise vector field ( (W1 (x, y), W2 (x, y)) , x ≥ 0, W (x, y) = (17) (−y, x) , x ≤ 0, where  W1 (x, y) = −y + x a0 x3 + a1 x2 y + a2 xy 2 + a3 y 3 and  W2 (x, y) = x + y a0 x3 + a1 x2 y + a2 xy 2 + a3 y 3 . A first integral of the linear vector field is given by F (r, θ) = r. So F (r0 , π/2) = F (R0 , 3π/2) wields r0 = R0 . A first integral of the nonlinear system in polar coordinates is   1 4 3 3 H(r, θ) = − 3 − 4a2 sin θ − a0 (9 sin θ + sin(3θ)) + 4a1 cos θ + 9a3 cos θ − a3 cos(3θ) . (18) 12 r Then we obtain that 2 P (a) = H(r0 , π/2) − H(r0 , 3π/2) = − (2a0 + a2 ) . 3 Therefore if a2 = −2a0 we have a continuum of periodic orbits around the singularity at the origin, that is the singularity is a center. If a2 ̸= −2a0 we have no periodic orbits and the origin is a repulsive focus. See Figure 3. (a) (b) Fig. 3. In (a) we have the phase portrait of the discontinuous piecewise vector field W near the origin when a0 = 1, a1 = 1, a2 = −2 and a3 = 1. In (b) we have the phase portrait of the discontinuous piecewise vector field W when a0 = 1, a1 = 1, a2 = 1 and a3 = 1. 4. Systems (2) and (1) after an affine change of variables The discontinuous piecewise vector field (14) has the centers of both smooth vector fields placed at the origin of coordinates. Now, we give the expression of the differential Systems (2) and (1) and their first integrals after the respective general affine change of variables (X, Y ) = (b1 x + b2 y + d1 , b3 x + b4 y + d2 ), bi , dj ∈ R, i = 1, 2, 3, 4 and j = 1, 2, (19) and (X, Y ) = (c1 x + c2 y + M1 , c3 x + c4 y + M2 ), ci , Mj ∈ R, i = 1, 2, 3, 4 and j = 1, 2. (20) We want to investigate the number of limit cycles of the discontinuous piecewise vector field (14) after this change of variables. However after this change of variables we still want that the centers of each smooth December 8, 2021 12:24 CarGonLlib2021IJBCFinal 9 vector field are rigid. First we consider the change of variables (19) with d1 = d2 = 0. Since X = b1 x + b2 y and Y = b3 x + b4 y we have that the linear system in the new variables is of the form  (−b1 b3 − b2 b4 ) X b21 + b22 Y Ẋ = + , b2 b3 − b1 b4 b2 b3 − b1 b4  (21) −b23 − b24 X (b1 b3 + b2 b4 ) Y Ẏ = + . b2 b3 − b1 b4 b2 b3 − b1 b4 Equivalently in polar coordinates we obtain   r b21 sin(2θ) + b22 sin(2θ) − b23 + b24 sin(2θ) − 2b3 b1 cos(2θ) − 2b2 b4 cos(2θ) ṙ = , 2b2 b3 − 2b1 b4  (22) −b21 sin2 (θ) − b22 sin2 (θ) + b1 b3 sin(2θ) + b2 b4 sin(2θ) − b23 + b24 cos2 (θ) θ̇ = . b2 b3 − b1 b4 Using polar coordinates we conclude that (21) has a rigid center at the origin if and only if b1 = b4 and b2 = −b3 and (21) writes Ẋ = −Y Ẏ = X. (23) Now with (d1 , d2 ) ̸= (0, 0) we obtain    ṙ = d2 cos θ − d1 sin θ, Ẋ = −Y + d2 , or, in polar coordinates, (24) Ẏ = X − d1 ,  θ̇ = r − d2 sin θ − d1 cos θ . r Observe that a first integral of (24) is p F1 (X, Y ) = (X − d1 )2 + (Y − d2 )2 , (25) or, in polar coordinates, p F1 (r, θ) = (r cos θ − d1 )2 + (r sin θ − d2 )2 . On the other hand we consider in x ≥ 0 system (1) with the affine change of variables (20) given by (X, Y ) = (c1 x + c2 y + M1 , c3 x + c4 y + M2 ). First we consider M1 = M2 = 0. So we get  c21 + c22 Y − (c1 c3 + c2 c4 ) X Ẋ = + X (c2 c3 − c1 c4 ) 1−n · S, c2 c3 − c1 c4  (26) (c1 c3 + c2 c4 ) Y − c23 + c24 X Ẏ = + Y (c2 c3 − c1 c4 ) 1−n · S, c2 c3 − c1 c4 where n−1 X S= ai (c3 X − c1 Y ) i (c2 Y − c4 X) n−i−1 , i=0 or equivalently in polar coordinates   r c21 + c22 − c23 − c24 sin(2θ) − 2 (c1 c3 + c2 c4 ) cos(2θ) + 2 (c2 c3 − c1 c4 ) 2−n S̄ ṙ = , 2c2 c3 − 2c1 c4 (27) c2 sin2 θ − c3 c1 sin(2θ) + c23 cos2 θ + (c2 sin θ − c4 cos θ) 2 θ̇ = − 1 , c2 c3 − c1 c4 where n−1 X S̄ = ai (c3 r cos θ − c1 r sin θ) i (c2 r sin θ − c4 r cos θ) n−i−1 . i=0 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 10 Again using polar coordinates we conclude that this system has a rigid center at the origin if and only if c4 = c1 and c3 = −c2 . Under these conditions system (26) writes  1−n n−1 X Ẋ = −Y + X c21 + c22 ai (c2 X + c1 Y ) i (c1 X − c2 Y ) n−i−1 , i=0 ! (28) n−1 X  Ẏ = X + Y c21 + c22 1−n ai (c2 X + c1 Y ) i (c1 X − c2 Y ) n−i−1 . i=0 or equivalently n−1 X n  ṙ = r c21 + c22 1−n ai (c1 sin θ + c2 cos θ) i (c1 cos θ − c2 sin(θ)) n−i−1 , i=0 (29) θ̇ = 1. Note that n−1 n−1 Z r1−n c21 + c22 X Hn (r, θ) = − ai (c1 sin θ + c2 cos θ) i (c1 cos θ − c2 sin(θ)) n−i−1 dθ (30) 1−n i=0 is a first integral of (29). Then a first integral of system (28) has the expression   1−n n−1 Z c21 + c22 n−1 X 2 + Y 2 2 X (c1 Y + c2 X)i (c1 X − c2 Y )n−i−1 Hn (X, Y ) = − ai n+1 X dY (31) 1−n (X 2 + Y 2 ) 2 i=0 Now if we apply the change of variables (X , Y) = (X + M1 , Y + M2 ) to system (28) with (M1 , M2 ) ̸= (0, 0), and rewriting it again in the variables X and Y we get  Ẋ = M2 − Y + c21 + c22 1−n (X − M1 ) · SM ,  (32) Ẏ = X − M1 + c21 + c22 1−n (Y − M2 ) · SM , where n−1 X SM = ai (c2 (X − M1 ) + c1 (Y − M2 )) i (c1 (X − M1 ) − c2 (Y − M2 )) n−i−1 . i=0 whose first integral is  n−1  1−n n−1 Z c21 + c22 (M1 − X) 2 + (M2 − Y ) 2 2 X Hn (x, y) = − (X − M1 ) ai T (X, Y ) dY (33) 1−n i=0 where T (X, Y ) = [c1 (Y − M2 ) + c2 (X − M1 )] i [c1 (X − M1 ) − c2 (Y − M2 )] n−i−1 ·  1 · (X − M1 ) 2 + (Y − M2 ) 2 2 (−n−1) In polar coordinates, the corresponding first integral is   1−n n−1 Z c21 + c22 n−1 (M2 − r sin θ) 2 + (M1 − r cos(θ)) 2 2 X Hn (r, θ) = − ai T (r, θ) dθ, (34) 1−n i=0 where r sec θ (r cos θ − M1 ) [c1 (r sin θ − M2 ) + c2 (r cos θ − M1 )] i T (r, θ) = n+1 · [(M2 − r sin θ) 2 + (M1 − r cos(θ)) 2 ] 2 · [c1 (r cos θ − M1 ) − c2 (r sin θ − M2 )]n−i−1 Notice that the above expressions coincide with the expression for the first integral when M1 = M2 = 0. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 11 4.1. Case linear-quadratic We investigate the number of limit cycles of the discontinuous piecewise vector field formed by the translated linear system (24) and system (32) when n = 2, that is we consider the consider   Y2 (x, y) = Y21 (x, y), Y22 (x, y) , x ≥ 0, Z2 (x, y) = (35) X(x, y) = (−y + d2 , x − d1 ) , x ≤ 0, where (x − M1 ) · S2 Y21 (x, y) = −y + M2 + , (36) c21 + c22 (y − M2 ) · S2 Y22 (x, y) = x − M1 + . (37) c21 + c22 and S2 = a1 (c2 (x − M1 ) + c1 (y − M2 )) + a0 (c1 (x − M1 ) − c2 (y − M2 )) . p From equation (25) we have that F1 (x, y) = (x − d1 )2 + (y − d2 )2 is a first integral of X, and from (33) we get that a0 (c2 (M1 − x) + c1 (M2 − y)) + a1 (c1 (x − M1 ) + c2 (M2 − y)) − c21 − c22 H2 (x, y) = p (38) (x − M1 ) 2 + (y − M2 ) 2 is a first integral of Y2 . So the limit cycles for the discontinuous piecewise vector field Z2 are given by the solutions of the following system of equations F1 (0, y1 ) = F1 (0, y2 ), H2 (0, y1 ) = H2 (0, y2 ). (39) From the first equation of (39) we obtain that y2 = −y1 +2d2 , for every y1 , y2 ∈ R with y1 ̸= y2 . Substituting this expression into the second equation of (39) we have that the solutions of H2 (0, y1 ) = H2 (0, −y1 + 2d2 ) √ are given by y1 = d2 and y1 = d2 ± A where A = υ1 υ2 υ3 υ4 /υ5 with υ1 = −a0 c1 − a1 c2 , υ2 = −a1 c1 M1 + c2 (a0 M1 − c2 ) − c21 , υ3 = c1 (a0 (d2 − M2 ) + a1 M1 ) + c2 (a1 (d2 − M2 ) − a0 M1 + c2 ) + c21 , υ4 = −c1 M1 (a1 (M2 − d2 ) + a0 M1 ) + c21 (d2 − M2 ) + c2 (c2 (d2 − M2 ) − M1 (a0 (d2 − M2 ) + a1 M1 )) , and  υ5 = (a0 c1 + a1 c2 ) M1 (a1 c1 − a0 c2 ) + c21 + c22 . Observe that the solution y1 = d2 corresponds to the periodic orbit of X that is tangent to the separation line given by the y-axis. Therefore this solution does not correspond to a limit cycle. Moreover the solutions y11 = d2 + A and y12 = d2 − A are on the same level curve of the first integral. Therefore they correspond to the same limit cycle, because each limit cycle corresponds to a pair of intersections between the curves given by (39) and the switching manifold. As a consequence, we conclude that there are at most one limit cycle associated to the piecewise smooth vector field Z2 . Example 4.1. We now present a numerical example. Consider the discontinuous piecewise vector field    17y 2 1   − + 5 + x (x + 3y − 7) , 3x + 2(−8 + x)y + 6y 2 , x ≥ 0, Q(x, y) = 5 5 5 (40)   (−y + 2, x + 2) , x ≤ 0. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 12 Observe that (40) corresponds to system (35) when d1 = −2, d2 = 2, c2 = −1, c1 = 2, M1 = 2, M2 = 1, a0 = 2 and a1 = 2. By solving the first equation of the respective system (39) we get y2 = 4 − y1 . Then we have that 6x − 2y − 15 H2 (x, y) = p x − 4x + y 2 − 2y + 5 2 is a first integral of the nonlinear system of the vector field Q. So the solutions of H2 (0, y1 )  = H2 (0, 4−y1 ) are  p p given by y1 = 2 and y1 = 2±3 19/34. Therefore there is a limit cycle through the points 0, 2 − 3 19/34  p  and 0, 2 + 3 19/34 , see Figure 4. Fig. 4. The limit cycle of the discontinuous piecewise vector field Q in red. In blue we have the periodic orbit of the vector field Q that is tangent to the separation line and corresponds to the solution y1 = d2 . 4.2. Case linear-cubic Now we investigate the number of limit cycles of the discontinuous piecewise differential system formed by the translated linear system (24) and system (32) when n = 3, i.e. of the discontinuous piecewise vector field   Y3 (x, y) = Y31 (x, y), Y32 (x, y) , x ≥ 0, Z3 (x, y) = (41) X(x, y) = (−y + d2 , x − d1 ) , x ≤ 0, where Y31 (x, y) = −y + M2 + (x − M1 )P (x, y), Y32 (x, y) = x − M1 + (y − M2 )P (x, y), and a2 (c2 (x − M1 ) + c1 (y − M2 )) 2 − a2 (c1 (x − M1 ) − c2 (y − M2 )) 2 P (x, y) =  + c21 + c22 2 a1 (c1 (x − M1 ) − c2 (y − M2 )) (c2 (x − M1 ) + c1 (y − M2 ))  . c21 + c22 2 Note that the vector field Y3 takes into account hypothesis (10) of Proposition 1 to have a rigid center. In this case hypothesis (10) becomes π(ap0 + a2 ) = 0. Therefore we are assuming that a0 = −a2 . Again Equation (25) assures that F1 (x, y) = (x − d1 )2 + (y − d2 )2 is a first integral of X. From (33) we get December 8, 2021 12:24 CarGonLlib2021IJBCFinal 13 that  a1 (x − M1 ) c21 (x − M1 ) + c22 (M1 − x) + 2c2 c1 (M2 − y) H3 (x, y) =  + 2 −2M1 x − 2M2 y + M12 + M22 + x2 + y 2   2a2 (x − M1 ) 2c2 c1 (x − M1 ) + c21 (y − M2 ) + c22 (M2 − y) − c21 + c22 2  2 −2M1 x − 2M2 y + M12 + M22 + x2 + y 2 is a first integral of Y3 . So the limit cycles of the discontinuous piecewise vector field Z3 are given by the solutions of the system of equations F1 (0, y1 ) = F1 (0, y2 ), H3 (0, y1 ) = H3 (0, y2 ). (42) From the first equation of (42) we obtain that y2 = −y1 +2d2 , for every y1 , y2 ∈ R with y1 ̸= y2 . Substituting this expression in the second equation of (42) we have that the solutions of H3 (0, y1 ) = H3 (0, −y1 + 2d2 ) are given by y1 = d2 and y1 = d2 ± B where q  M1 a1 c1 c2 + a2 c22 − c21 B1 B=  M1 a1 c1 c2 + a2 c22 − c21 with   B1 = c21 M1 a1 M1 (d2 − M2 ) − a2 (d2 − M2 ) 2 − M12 + 2c22 (M2 − d2 ) +   + c2 c1 M1 a1 (d2 − M2 ) 2 − M12 + 4a2 M1 (d2 − M2 ) + c42 (M2 − d2 )   + c22 M1 a2 (d2 − M2 ) 2 − M12 + a1 M1 (M2 − d2 ) + c41 (M2 − d2 ) . Also the solutions d2 ± B are on the same level of the first integral H3 because  M1 a1 c1 c2 + a2 c22 − c21 H3 (0, d2 + B) = H3 (0, d2 − B) = . 2 (d2 − M2 ) Since the solution y1 = d2 corresponds to the periodic orbit of X that is tangent to the separation line given by the y-axis and each limit cycle corresponds to a pair of intersections between the curves given by (42) and the switching manifold, we conclude that there is at most one limit cycle associated to the piecewise smooth vector field Z3 . Example 4.2. Consider the discontinuous piecewise vector field (  Y31 (x, y), Y32 (x, y) , x ≥ 0, Q3 (x, y) = (43) (−y + 2, x + 2) , x ≤ 0. where 2x3 22x2 y 19x2 2xy 2 26xy 9x y 2 7y 3 Y31 (x, y) = + − − + − − − + 25 25 25 25 25 10 25 10 4 2x2 y 2x2 22xy 2 42xy 9x 2y 3 17y 2 11y Y32 (x, y) = − + − + − + − + 1. 25 25 25 25 5 25 25 10 Observe that (43) corresponds to system (41) when d1 = −2, d2 = 2, c2 = −1, c1 = 2, M1 = −1/2, M2 = 1, a0 = 2, a1 = 2, and a2 = −2. By solving the first equation of the respective system (42) we get y2 = 4 − y1 . In this case, we have that 44x2 + x(52 − 8y) − 4y − 35 H3 (x, y) = 4x2 + 4x + 4y 2 − 8y + 5 is a first integral of the right-hand side system of the vector field Q3 . Then the solutions of H3 (0, y1 ) = H3 (0, 4 − y1 ) are given by y1 = 2, y1 = −5/2, and y1 = 13/2. Note that H3 (0, −5/2) = H3 (0, 13/2) = −1/2. Therefore there exists a limit cycle through the points (0, −5/2) and (0, 13/2), see Figure 5. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 14 Fig. 5. The limit cycle of the discontinuous piecewise vector field Q3 in red. In blue we have the periodic orbit of the vector field Q that is tangent to the separation line and corresponds to the solution y1 = d2 . The black dots represent the equilibrium points of the two centers of Q3 . 5. Study of cases where none vector field is linear In order to obtain a complete characterization of the n ≤ 3 case, in this section, we analyze the maximum number of limit cycles that can bifurcate from the piecewise smooth vector field ( Un (x, y), x ≥ 0, Zn,m (x, y) = (44) Vm (x, y), x ≤ 0, with Un and Vm , n, m ∈ {2, 3}, being rigid centers of the form (1) after the respective change of variables (19) and (20), with ai = si for the vector field Un . We will call these cases (n, m) with n, m ∈ {2, 3}. In other words, we have that Vm is of the vector field associated with the differential system (32) and Un is the vector field associated with the differential system  Ẋ = d2 − Y + b21 + b22 1−n (X − d1 ) · Sd ,  (45) Ẏ = X − d1 + b21 + b22 1−n (Y − d2 ) · Sd , where n−1 X Sd = si (b2 (X − d1 ) + b1 (Y − d2 )) i (b1 (X − d1 ) − b2 (Y − d2 )) n−i−1 . i=0 Moreover, we found examples that ensure that the upper bound is reached in each case. 5.1. Case (2,2) Consider the piecewise smooth vector field ( U2 (x, y), x ≥ 0, Z2,2 (x, y) = (46) V2 (x, y), x ≤ 0,  where V2 (x, y) = Y21 (x, y), Y22 (x, y) is given by Equations (36) and (37), and U2 (x, y) = (U21 (x, y), U22 (x, y)) has components (x − d1 ) S22 U21 (x, y) = −y + d2 + , b21 + b22 and (y − d2 ) S22 U22 (x, y) = x − d1 + , b21 + b22 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 15 with S22 = s0 (b1 (x − d1 ) − b2 (y − d2 )) + s1 (b2 (x − d1 ) + b1 (y − d2 )) . From (33) we get that H2 given by (38) is a first integral of V2 and b1 (s1 (x − d1 ) + s0 (d2 − y)) − b2 (b2 + s0 (x − d1 ) + s1 (y − d2 )) − b21 F2 (x, y) = p (x − d1 ) 2 + (y − d2 ) 2 is a first integral of U2 . So the limit cycles for the discontinuous piecewise vector field Z2,2 are given by the solutions of the following system of equations F2 (0, y0 ) = F2 (0, y1 ), H2 (0, y0 ) = H2 (0, y1 ). (47) From the second equation of (47) we obtain y1 = y0 and η1 + η2 + η3 + 2a0 (η4 + η5 ) y1 = y1e = (48) η6 + η7 + η8 where   2 η1 = 2a1 c21 + c22 c1 M1 (2M2 − y0 ) + c2 M2 (y0 − M2 ) + M12 + c21 + c22 (2M2 − y0 ) ,   η2 = a20 M1 c21 M1 (y0 − 2M2 ) − 2c2 c1 M2 (y0 − M2 ) + M12 + c22 M1 (2M2 − y0 ) ,   η3 = a21 M1 c21 M1 (2M2 − y0 ) + 2c2 c1 M2 (y0 − M2 ) + M12 + c22 M1 (y0 − 2M2 ) ,   η4 = c22 M1 c2 (y0 − 2M2 ) − a1 M2 (y0 − M2 ) + M12 + c31 M2 (y0 − M2 ) + M12 ,   η5 = c21 M1 a1 M2 (y0 − M2 ) + M12 + c2 (y0 − 2M2 ) +  + c2 c1 2a1 M12 (y0 − 2M2 ) + c2 M2 (y0 − M2 ) + M12 ,   η6 = c22 −2c2 (a1 (M2 − y0 ) + a0 M1 ) + M1 2a1 a0 (M2 − y0 ) + a20 M1 − a21 M1 + c22 + c41 , η7 = 2c31 (a0 (y0 − M2 ) + a1 M1 ) −  − 2c1 c2 a0 (M2 − y0 ) (c2 − a0 M1 ) + a1 M1 (2a0 M1 − c2 ) + a21 M1 (M2 − y0 ) ,   η8 = c21 (−2c2 (a1 (M2 − y0 ) + a0 M1 )) + c21 M1 2a1 a0 (y0 − M2 ) + a20 (−M1 ) + a21 M1 + 2c22 . From the first equation of (47) we obtain y1 = y0 and µ1 + 2b1 b2 (µ2 + µ3 ) + b22 (µ4 + µ5 ) + b21 (µ6 + µ7 ) y1 = y1d = (49) µ8 + µ9 + µ10  where µ1 = 2b31 d1 s1 (2d2 − y0 ) + d2 s0 (y0 − d2 ) + d21 s0 + b41 (2d2 − y0 ) ,   µ2 = d1 2d1 s0 s1 (y0 − 2d2 ) + d2 s20 − s21 (d2 − y0 ) + d21 s21 − s20 ,  µ3 = b2 d1 s1 (2d2 − y0 ) + d2 s0 (y0 − d2 ) + d21 s0 ,   µ4 = d1 d1 s20 − s21 (2d2 − y0 ) + 2d2 s0 s1 (d2 − y0 ) − 2d21 s0 s1 ,  µ5 = 2b2 d1 s0 (y0 − 2d2 ) + d2 s1 (y0 − d2 ) + d21 s1 + b22 (2d2 − y0 ) ,  µ6 = 2b2 d1 s0 (y0 − 2d2 ) + d2 s1 (y0 − d2 ) + d21 s1 − 2b22 (y0 − 2d2 ) ,   µ7 = d1 d1 s20 − s21 (y0 − 2d2 ) + 2d2 s0 s1 (y0 − d2 ) + 2d21 s0 s1 , December 8, 2021 12:24 CarGonLlib2021IJBCFinal 16  µ8 = b22 −2b2 (s1 (d2 − y0 ) + d1 s0 ) + b22 + d1 2s0 s1 (d2 − y0 ) + d1 s20 − s21 + b41 ,  µ9 = 2b31 (s0 (y0 − d2 ) + d1 s1 ) + b21 −2b2 (s1 (d2 − y0 ) + d1 s0 ) + 2b22 + d1 2s0 s1 (y0 − d2 ) + d1 s21 − s20 ,  µ10 = 2b1 b2 b2 (s0 (y0 − d2 ) + d1 s1 ) + d1 s20 (d2 − y0 ) + s21 (y0 − d2 ) − 2d1 s1 s0 . Consider D2,2 the numerator of the difference y1e − y1d . We have that D2,2 has two zeros in the variable y0 whose expressions are complicated and will be omitted. Therefore we conclude that the piecewise smooth vector field Z2,2 given by (46) has at most one crossing limit cycle. Example 5.1. Consider the discontinuous piecewise vector field (  U21 (x, y), U22 (x, y) , x ≥ 0, Q2,2 (x, y) =  (50) V21 (x, y), V22 (x, y) , x ≤ 0. where 1 U21 (x, y) = −y + 1 + (x − 1)(3(x + 2(y − 1) − 1) − 2(x − 1) + y − 1), 5 1 U22 (x, y) = x − 1 + (y − 1)(3(x + 2(y − 1) − 1) − 2(x − 1) + y − 1), 5         1 1 1 1 1 V2 (x, y) = 1 − y + x− 2 −x + 2(y − 1) + +2 2 x− +y−1 , (51) 5 2 2 2 and        1 1 1 1 V22 (x, y) = x − + (y − 1) 2 −x + 2(y − 1) + +2 2 x− +y−1 . (52) 2 5 2 2 Observe that (50) corresponds to system (46) when b1 = 2, b2 = 1, d1 = 1, d2 = 1, s0 = −1, s1 = 3, c2 = −1, c1 = 2, M1 = 1/2, M2 = 1, a0 = 2, and a1 = 2. By solving the first and second equations of the respective system (47) we get 286 − 119y0 102 − 31y0 y1d = and y1e = . 24y0 + 119 32y0 + 31 √  Solving y1e = y1d we get the solutions y0 = 419 ± 7 386 /383 which are the intersections of the limit cycle with the ordinate axis. See Figure 6. Fig. 6. The limit cycle of the discontinuous piecewise vector field Q2,2 in red. The black dots represent the equilibrium points of the two centers of Q2,2 . December 8, 2021 12:24 CarGonLlib2021IJBCFinal 17 5.2. Case (2,3) Consider the piecewise smooth vector field ( U3 (x, y), x ≥ 0, Z2,3 (x, y) = (53) V2 (x, y), x ≤ 0,  where V2 (x, y) = Y21 (x, y), Y22 (x, y) is given by Equations (36) and (37), and U3 (x, y) = (U31 (x, y), U32 (x, y)) has components (x − d1 ) S23 U31 (x, y) = −y + d2 +  , (54) b21 + b22 2 and (y − d2 ) S23 U32 (x, y) = x − d1 +  , (55) b21 + b22 2 with S23 = −s0 (b2 (x − d1 ) + b1 (y − d2 )) 2 + s1 (b1 (x − d1 ) − b2 (y − d2 )) · · (b2 (x − d1 ) + b1 (y − d2 )) + s0 (b1 (x − d1 ) − b2 (y − d2 )) 2 . Note that the vector field U3 is considering the hypothesis s2 = −s0 given by Equation (12) so that we have a rigid center. From (33) we get that H2 given by (38) is a first integral of V2 and  b21 2b22 + (x − d1 ) (s1 (d1 − x) + 2s0 (y − d2 )) F3 (x, y) = −  + 2 −2d1 x − 2d2 y + d21 + d22 + x2 + y 2 2b2 b1 (x − d1 ) (2s0 (x − d1 ) + s1 (y − d2 )) +  + (56) 2 −2d1 x − 2d2 y + d21 + d22 + x2 + y 2 b22 (x − d1 ) (s1 (x − d1 ) + 2s0 (d2 − y)) + b41 + b42 +  2 −2d1 x − 2d2 y + d21 + d22 + x2 + y 2 is a first integral of U3 . So the limit cycles for the discontinuous piecewise vector field Z2,3 are given by the solutions of the following system of equations F3 (0, y0 ) = F3 (0, y1 ), H2 (0, y0 ) = H2 (0, y1 ). (57) From the second equation of (57) we obtain y1 = y0 and y1 = y1e given by Equation (48). From the first equation of (57) we obtain y1 = y0 and ω1 + ω2 + ω3 y1 = y3d = (58) ω4   where ω1 = b22 d1 2s0 d2 (y0 − d2 ) + d21 − d1 s1 (y0 − 2d2 ) + b41 (− (y0 − 2d2 )) − b42 (y0 − 2d2 ),  ω2 = 2b1 b2 d1 −2d1 s0 (y0 − 2d2 ) − s1 d2 (y0 − d2 ) + d21 ,   ω3 = b21 d1 d1 s1 (y0 − 2d2 ) − 2s0 d2 (y0 − d2 ) + d21 − 2b22 (y0 − 2d2 ) , and  ω4 = b21 2b22 − d1 (2s0 (y0 − d2 ) + d1 s1 ) + 2b2 b1 d1 (s1 (d2 − y0 ) + 2d1 s0 ) + + b22 d1 (2s0 (y0 − d2 ) + d1 s1 ) + b41 + b42 . Consider D2,3 the numerator of the difference y1e − y3d . We have that D2,3 has two zeros in the variable y0 whose expressions are complicated and will be omitted. Therefore we conclude that the piecewise smooth vector field Z2,3 given by (53) has at most one crossing limit cycle. The case (3,2) can be treated similarly. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 18 Example 5.2. Consider the discontinuous piecewise vector field (  U31 (x, y), U32 (x, y) , x ≥ 0, Q2,3 (x, y) =  (59) V21 (x, y), V22 (x, y) , x ≤ 0.  where V2 (x, y) = V21 (x, y), V22 (x, y) is given by equations (51) and (52), 16 U31 (x, y) = −y + 1 + (x + 1) (Se ) , (60) 289 and 16 U32 (x, y) = x + 1 + (y − 1) (Se ) , (61) 289 with  2     2 y−1 x+1 y−1 x+1 Se = − 2(x + 1) + + 2(y − 1) − 2(x + 1) − + 2(y − 1) . 2 2 2 2 Observe that (59) corresponds to system (53) when b1 = 1/2, b2 = −2, d1 = −1, d2 = 1, s0 = −1, s1 = 1, s2 = −s0 = 1, c2 = −1, c1 = 2, M1 = 1/2, M2 = 1, a0 = 2, and a1 = 2. By solving the first and second equations of the respective system (57) we get 826 − 325y0 102 − 31y0 y3d = and y1e = . 88y0 + 325 32y0 + 31 √  Solving y1e = y3d we get the solutions y0 = 1091 ± 2 71486 /959 which are the intersections of the limit cycle with the ordinate axis. See Figure 7. Fig. 7. The limit cycle of the discontinuous piecewise vector field Q2,3 in red. The black dots represent the equilibrium points of the two centers of Q2,3 . 5.3. Case (3,3) Consider the piecewise smooth vector field ( U3 (x, y), x ≥ 0, Z3,3 (x, y) = (62) V3 (x, y), x ≤ 0,  where U3 (x, y) = U31 (x, y), U32 (x, y) is given by Equations (54) and (55), and V3 (x, y) = 1 2 (V3 (x, y), V3 (x, y)) has components (x − M1 ) (S33 ) V31 (x, y) = −y + M2 +  , c21 + c22 2 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 19 (y − M2 ) (S33 ) V32 (x, y) = x − M1 +  , c21 + c22 2 with S33 = −a0 (c2 (x − M1 ) + c1 (y − M2 )) 2 + a0 (c1 (x − M1 ) − c2 (y − M2 )) 2 + a1 (c1 (x − M1 ) − c2 (y − M2 )) (c2 (x − M1 ) + c1 (y − M2 )) . Note that the vector field V3 is considering the hypothesis a2 = −a0 given by Equation (12) so that we have a rigid center. From (33) we get that  −a1 (x − M1 ) c21 (x − M1 ) + c22 (M1 − x) + 2c2 c1 (M2 − y) H3 (x, y) = −  2 −2M1 x − 2M2 y + M12 + M22 + x2 + y 2  2a0 (x − M1 ) 2c2 c1 (x − M1 ) + c21 (y − M2 ) + c22 (M2 − y) −  (63) 2 −2M1 x − 2M2 y + M12 + M22 + x2 + y 2  c21 + c22 2 −  2 −2M1 x − 2M2 y + M12 + M22 + x2 + y 2 is a first integral of V3 and F3 given by (56) is a first integral of U3 . The limit cycles for the discontinuous piecewise vector field Z3,3 are given by the solutions of the following system of equations F3 (0, y0 ) = F3 (0, y1 ), H3 (0, y0 ) = H3 (0, y1 ). (64) From the first equation of (64) we obtain y1 = y0 and y1 = y3d given by Equation (58). From the second equation of (64) we obtain y1 = y0 and ν1 − ν2 y1 = y3e = (65) ν3 where    ν1 = a1 M1 c21 M1 (y0 − 2M2 ) − 2c2 c1 M2 (y0 − M2 ) + M12 + c22 M1 (2M2 − y0 ) + c21 + c22 2 (2M2 − y0 ) ,    ν2 = 2a0 M1 c21 M2 (y0 − M2 ) + M12 + 2c2 c1 M1 (y0 − 2M2 ) − 2a0 M1 c22 M2 (y0 − M2 ) + M12 , and  ν3 = c21 2c22 − M1 (2a0 (y0 − M2 ) + a1 M1 ) + 2c2 c1 M1 (a1 (M2 − y0 ) + 2a0 M1 ) + c22 M1 (2a0 (y0 − M2 ) + a1 M1 ) + c41 + c42 . Consider D3,3 the numerator of the difference y3e − y3d . We have that D3,3 has two zeros in the variable y0 whose expressions are complicated and will be omitted. Therefore we conclude that the piecewise smooth vector field Z3,3 given by (62) has at most one crossing limit cycle. Example 5.3. Consider the discontinuous piecewise vector field (  U31 (x, y), U32 (x, y) , x ≥ 0, Q3,3 (x, y) =  (66) V31 (x, y), V32 (x, y) , x ≤ 0. where U31 (x, y) and U32 (x, y) are given by Equations (60) and (61),   1 1 1 V3 (x, y) = −y + 1 + x+ (S33e ) , 25 2 and 1 1 V32 (x, y) = x + + (y − 1) (S33e ) , 2 25 December 8, 2021 12:24 CarGonLlib2021IJBCFinal 20 with  2         2 1 1 1 1 S33e = −2 −x + 2(y − 1) − −3 2 x + +y−1 −x + 2(y − 1) − +2 2 x + +y−1 . 2 2 2 2 Note that (66) corresponds to system (62) when b1 = 1/2, b2 = −2, d1 = −1, d2 = 1, s0 = −1, s1 = 1, s2 = −s0 = 1, c2 = −1, c1 = 2, M1 = −1/2, M2 = 1, a0 = 2, a1 = −3, and a2 = −a0 = −2. By solving the first and second equations of the respective system (64) we get 826 − 325y0 50 − 15y0 y3d = and y3e = . 88y0 + 325 16y0 + 15 √  Solving y3e = y3d we get the solutions y0 = 1102 + 278354 /970 which are the intersections of the limit cycle with the ordinate axis. See Figure 8. Fig. 8. The limit cycle of the discontinuous piecewise vector field Q3,3 in red. The black dots represent the equilibrium points of the two centers of Q3,3 . 6. Conclusions In this paper we have studied the upper bound for the maximum number of limit cycles of discontinuous piecewise differential systems formed by two differential systems separated by the straight line x = 0. We assume that each one of the differential systems has a rigid center formed of a linear part with a homogeneous polynomial nonlinear part of degree m and n respectively. If both centers are localized at the origin of coordinates and n = 1 then for all positive integer m we have proved that such discontinuous piecewise differential systems have no limit cycles, see Theorem 1. If the centers are not at the origin of coordinates for a subclass of these discontinuous piecewise differential systems with n, m = 2, 3 we have proved that they can have at most one limit cycle, see for more details the statement of Theorems 2 and 3. Acknowledgments Tiago Carvalho is partially supported by São Paulo Research Foundation (FAPESP grants 2019/10450- 0 and 2019/10269-3) and Conselho Nacional de Desenvolvimento Cientı́fico e Tecnológico (CNPq Grant 304809/2017-9). J. Llibre is partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER), the Agència de Gestió d’Ajuts Universitaris i de Recerca grant 2017SGR1617, and the H2020 European Research Council grant MSCA- RISE-2017-777911. L. F. Gonçalves thanks Faculdade de Filosofia, Ciências e Letras de Ribeirão Preto of Universidade de São Paulo. December 8, 2021 12:24 CarGonLlib2021IJBCFinal 21 References Algaba, A. & Reyes, M. 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