Maximum number of limit cycles for certain piecewise linear dynamical systems

This paper deals with the question of the determinacy of the maximum number of limit cycles of some classes of planar discontinuous piecewise linear differential systems defined in two half-planes separated by a straight line Σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Sigma $$\end{document}. We restrict ourselves to the non-sliding limit cycles case, i.e., limit cycles that do not contain any sliding segment. Among all cases treated here, it is proved that the maximum number of limit cycles is at most 2 if one of the two linear differential systems of the discontinuous piecewise linear differential system has a focus in Σ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Sigma $$\end{document}, a center, or a weak saddle. We use the theory of Chebyshev systems for establishing sharp upper bounds for the number of limit cycles. Some normal forms are also provided for these systems.


Introduction and statement of the main results
Non-smooth dynamical systems emerge in a natural way modeling many real processes and phenomena, for instance, recently piecewise linear differential equations appeared as idealized models of cell activity, see [10,37,38]. Due to that, in these last years, the mathematical community became very interested in understanding the dynamics of these kind of systems. In general, some of the main source of motivation to study non-smooth systems can be found in control theory [4], impact and friction mechanics [5,8,27], nonlinear oscillations [1,34], economics [19,24], and biology [3,26]. See for more details the book [11] and the references therein. In this paper, we are interested in discontinuous piecewise linear differential systems. The study of this particular class of non-smooth dynamical systems has started with Andronov and coworkers [1].
We start with a historical fact. Lum and Chua [33] conjectured that a continuous piecewise linear vector field in the plane with two zones separated by a straight line, which is the easiest example of this kind of system, has at most one limit cycle. This conjecture was proved by Freire et al. [13]. Even this relatively easy case demanded a hard work to show the existence of at most one limit cycle.
In this paper, we address the problem of Lum and Chua, for non-sliding limit cycles, extended to the class of discontinuous piecewise linear differential systems in the plane with two zones separated by a straight line. Here, a non-sliding limit cycle is a limit cycle that does not contain any sliding segment in . This problem is very related to the Hilbert's 16th problem [23].
Limit cycles of discontinuous piecewise linear differential systems with two zones separated by a straight line have been studied recently by several authors, see among others [2,7,9,17,18,[20][21][22]28,[30][31][32]. Nevertheless, the problem of Lum and Chua remains open for this class of differential equations. In this work, we give a partial solution for this problem. We note that in [12] the authors proved that if one of the two linear systems has its singular point on the discontinuity straight line, then the number of limit cycles of such a system is at most 4. Our results reduce this upper bound to 2, and, additionally, we prove that it is reached.
Our point of interest in the Lum and Chua problem is aligned with two directions which face serious technical difficulties. First, while solutions in each linear region are easy to find, the times of passage along the regions are not simple to achieve. It means that matching solutions across regions is a very difficult task. Second, to control all possible configurations one must generally consider a large number of parameters.
It was conjectured in [18] that a planar piecewise linear differential systems with two zones separated by a straight line have at most 2 non-sliding limit cycles. A negative answer for this conjecture was provided in [20] via a numerical example having 3 nonsliding limit cycles. Analytical proofs for the existence of these 3 limit cycles were given in [15,31]. Finally, in [16] it was studied general conditions to obtain 3 nonsliding limit cycles in planar piecewise linear differential systems with two zones separated by a straight line. Recently, perturbative techniques (see [29,30]) were used together with newly developed tools on Chebyshev systems (see [36]) to obtain 3 limit cycles in such systems when they are near to non-smooth centers.
When a general curve of discontinuity is considered instead of a straight line, there is no upper bound for the maximum number of non-sliding limit cycles that a system of this family can have. It is a consequence of a conjecture stated by Braga and Mello in [6] and then proved by Novaes and Ponce in [35].
In this paper, we deal with planar vector fields Z expressed asż = F(z) + sign(x)G(z), where z = (x, y) ∈ R 2 , and F and G are linear vector fields in R 2 or, equivalently, where X (z) = F(z) + G(z) and Y (z) = F(z) − G(z). The line = {x = 0} is called discontinuity set. Our main goal is to study the maximum number of nonsliding limit cycles that the discontinuous piecewise linear differential system (1) can have. The systemsż = X (z) andż = Y (z) are called lateral linear differential systems (or just lateral systems) and more specifically right system and left system, respectively. A linear differential system is called degenerate if its determinant is zero; otherwise, it is called nondegenerate. From now on in this paper, we only consider non-degenerate linear differential systems.
System (1) can be classified according to the singularities of the lateral linear differential systems. A nondegenerate linear differential system can have the following singularities: saddle (S), node (N ), focus (F), and center (C). Among the above classes of singularities, we shall also distinguish the following ones: a weak saddle, i.e., a saddle such that the sum of its eigenvalues is zero (S 0 ); a diagonalizable node with distinct eigenvalues (N ); star node, i.e., a diagonalizable node with equal eigenvalues (N * ); and an improper node, i.e., a non-diagonalizable node (i N). We say that the discontinuous differential system (1) is an L R-system with L , R ∈ {S, S 0 , N , N * , i N, F, C}, when the left system has a singularity of type L and the right system has a singularity of type R.
We define subclasses of L R-systems according to the position of the singularity of each lateral system. The right system can have a virtual singularity (R v ), i.e., a singularity p = ( p x , p y ) with p x < 0; a boundary singularity (R b ), i.e., a singularity p = ( p x , p y ) with p x = 0; or a real singularity (R r ), i.e., a singularity p = ( p x , p y ) with p x > 0. Accordingly, the left system can have a virtual singularity (L v ), i.e., a singularity p = ( p x , p y ) with p x > 0; a boundary singularity (L b ), i.e., a singularity p = ( p x , p y ) with p x = 0; or a real singularity (L r ), i.e., a singularity p = ( p x , p y ) with p x < 0.
We denote by N (L , R) the maximum number of non-sliding limit cycles that an L R-system can have. Clearly, N (L , R) = N (R, L).
In this paper, we compute the exact value of N (L , R) always when one of the lateral systems is a saddle of kind S v , S b , S 0 , a node of kind N r , N b , N * , i N r , i N b , a focus of kind F b , and a center C. Particularly, we obtain that N (L , R) ≤ 2 in all the above cases.
It is easy to see that if one of the lateral linear differential systems is of type S v , S b , N r , N b , N * , i N r , or i N b , then the first return map on the straight line x = 0 of system (1) is not defined. Consequently, system (1) does not admit non-sliding limit cycles in all these cases. So N (R, L) = 0 for the systems having one of these kind of equilibria.
It remains to study the cases when one of the lateral system is F b , C or S 0 r . For these cases, we shall prove the following theorems.
, and N (S 0 r , S r ) are equal to 1, and all numbers N (S 0 r , C) and N (S 0 r , S 0 r ) are equal to 0. We shall see that the next result can be obtained as an immediately corollary of the proofs of Theorems 1 and 2.
and N (C b , S r ) are equal to 1, and all numbers N (C b , C) and N (C b , S 0 r ) are equal to 0. The equalities of Corollary 3 can be extended for all linear centers.
and N (C, S r ) are equal to 1, and all numbers N (C, C) and N (C, S 0 r ) are equal to 0. Theorems 1, 2, and 4, and Corollary 3 are proved in Sect. 3.
Our results give sufficient conditions in order to guarantee that system (1) has at most 2, 1, or 0 limit cycles. We study the non-degenerate cases for which the expression of the time that a trajectory starting in p ∈ remains in the region x > 0 (or x < 0) is known. The remaining cases are those ones whose this associated time is not explicitly determined for both regions.
The systems studied in [15,16,20,29,31], possessing 3 limit cycles, have in one side a real focus, and in the other side either a real focus or a linear system with trace distinct from zero. Thus, they do not satisfy the conditions of our theorems.

Preliminary results
A linear change of variables in the plane preserving the vertical lines will be called a vertical lines-preserving linear change of variables.
is a (a) S-system then after a vertical lines-preserving linear change of variables and a time-rescaling system (2) becomes (ẋ,ẏ) T = M 1 (x, y) T ; (b) N -system then after a vertical lines-preserving linear change of variables and a time-rescaling system (2) becomes (ẋ,ẏ) T = M 2 (x, y) T ; (c) F-system (C-system) then after a vertical linespreserving linear change of variables and a time-  Since we are assuming that we have a saddle at the origin and in the expression of its eigenvalues appears 4m 12 m 21 + (m 11 − m 22 ) 2 , we must assume that 4m 12  Then, we can rescale the time by Denoting a=(m 11 +m 22 )/ 4m 12 m 21 + (m 11 − m 22 ) 2 system (2) becomes where now the prime denotes the derivative with respect to the new time variable τ . Computing the eigenvalues of the above system {−1 + a, 1 + a}, we conclude that |a| < 1, because this system is a saddle, i.e., the eigenvalues have different sign. Therefore, we have proved statement (a).

Claim 2 The statement (b) holds.
The proof of statement (b) follows similarly to the proof of statement (a). Nevertheless, we conclude that |a| > 1, because in this case the system is a diagonalizable node, i.e., the eigenvalue have the same sign. Thus, we have proved statement (b). where now the prime denotes the derivative with respect to the new time variable τ . Computing the eigenvalues of the above system {−i + a, i + a}, we conclude that when a = 0 this system has a focus and a center when a = 0. Hence, statement (c) is proved.

Claim 4 The statement (d) holds.
One of the entries m 12 or m 21 is distinct of zero. Indeed, suppose that m 12 = 0, so {m 11 , m 22 } are the eigenvalues of the matrix M. Since system (2) is a nondiagonalizable node, we have that m 11 = m 22 which implies that m 21 = 0, in other way the matrix M would be diagonalizable. On the other hand, supposing that m 21 = 0 we obtain m 12 = 0. From here we assume, without loss of generality, that m 12 = 0.
We also have that m 11 + m 22 = 0, we prove this by reduction to the absurd. Suppose that m 11 + m 22 = 0, then ± m 2 11 + m 12 m 21 are the eigenvalues of the matrix M. Since system (2) is a non-diagonalizable node, we have that the matrix M has only one eigenvalue with multiplicity 2. This implies that the eigenvalues are zero, which is a contradiction with the fact that we are working with non-degenerate linear differential systems. In short, we have proved that m 11 + m 22 = 0.
From the expression of the eigenvalues, it is also easy to see that 4m 12  So we can rescale the time by τ = 1 2 |m 11 + m 22 | t system, (2) becomes where λ = ±1, and now the prime denotes the derivative with respect to the new time variable τ . This completes the proof of statement (c).
A limit cycle of our piecewise linear differential system (1) expends a time t R in the region x > 0 and a time t L in the region x < 0. As we shall see later on we know explicitly the time t L , and we do not know explicitly the time t R . The next lemma will help us to work with one of the intersection points of the limit cycle with the discontinuity straight line instead of the unknown time t R .

Lemma 6
We consider the functions The following statements hold.
To prove statement (b), we compute where To prove statement (c), we compute where r (t) = e t − t − 1 and r (t) = e t − 1. Since r (0) = 0 and r (t) ≶ 0 for t ≶ 0, we conclude that r (t) > 0 and consequently H (t) > 0, for t = 0. It implies that H is an monotonic increasing function for t > 0. The proof of statement (c) follows by noting that lim t→0 H (t) = −1.
Some important tools we shall use to prove our main results lie in the theory of Chebyshev systems (for more details, see, for instance, the book of Karlin and Studden [25]). In the sequel, the concept of Chebyshev systems is introduced.
Consider an ordered set of smooth real functions F = ( f 0 , f 1 , . . . , f n ) defined on a interval I . The maximum number of zeros counting multiplicity admitted by any non-trivial linear combination of functions in F is denoted as Z (F).

Definition 1
We say that F is an Extended Chebyshev system or ET-system on I if and only if Z (F) ≤ n.
We say that F is an Extended Complete Chebyshev system or an ECT-system on I if and only if for any k, The next proposition relates the property of an ordered set of functions ( f 0 , f 1 , . . . , f k ) being an ECTsystem with the nonvanishing property of their Wronskians .
The next result has been recently proved by Novaes and Torregrosa in [36].
Assume that all the Wronskians are nonvanishing except W n (x) which have ≥ 0 zeros on (a, b) and all these zeros are simple. Then, Z (F) = n when = 0, and n + 1 ≤ Z (F) ≤ n + when = 0.

Now consider the functions
We define the ordered sets of functions 2 ) for i = 4, 5, 6. The next two technical lemmas together with Definition 1 and Propositions 7 and 8 will be used later on in the proofs of Theorems 1, 2, and 4 to establish sharp upper bounds for the maximum numbers of non-sliding limit cycles that system (1) can have.

Lemma 9
The following statements hold.
From here, it is easy to see that for each a = 0 the Wronskians W 1 2 , W 1 3 , and W 4 2 do not vanish at any point of the intervals (0, π) and (−π, 0); for each a / ∈ {0, ±1} the Wronskians W 2 2 , W 2 3 , and W 5 2 do not vanish at any point of R + , and the Wronskians W 3 2 , W 3 3 , and W 6 2 do not vanish at any point of R + . So statements (a)-( f ) are proved.
To see statement (a ), we compute the Wronskians Again it is easy to see that for each a > 0 (resp. a < 0) the Wronskian W 1 2 does not vanish at any point of the interval (π, 2π) and (resp. (−2π, −π)).
Finally, statement (d ) follows by showing that the Wronskian W 4 2 (t) has exactly one zero in each one of the intervals (π, 2π) and (−2π, −π). Indeed, Since csc(t) cosh(at) is nonvanishing for every a ∈ R, it is sufficient to study the zeros of P a (t) in order to study the zeros of W 4 So, for a = 0, there exist t a ∈ (π, 2π) and t a ∈ (−2π, −π) such that P a (t a ) = P a (t a ) = 0. Indeed, function P a (t) is continuous on the intervals (π, 2π) and (−2π, −π). Computing P a (t) = csc 2 (t) tanh(at) − a cot(t)sech 2 (at), we see that P a (t) = 0 for every a = 0 and t ∈ (π, 2π) ∪ (−2π, −π), which implies that P a (t) has at most one zero in each one of these intervals. This proof ends by applying Proposition 8 for n = = 1.
Lemma 9 was stated assuming a = 0. For a = 0, we define the sets of functions G i = {ξ 1 , ξ i 2 } for i = 1, 2 and we prove the next lemma.
Proof Assuming a = 0 and proceeding analogously to the proof of Lemma 9, we compute the Wronskians.
From here, it is easy to see that the Wronskian W 1 2 does not vanish at any point of the interval (0, π), (−π, 0), (π, 2π), and (−2π, −π) and that the Wronskian W 2 2 does not vanish at any point of R + .

Proof of Theorems 1, 2 and 4, and Corollary 3
The proofs of Theorem 1 and Corollary 3 will be immediate consequences of Propositions 11-16; the proof of Theorem 2 will be an immediate consequence of Propositions 16-21; and the proof of Theorem 4 will be an immediate consequence of Propositions 22-25 and Corollary 3.
We note that some of the partial results contained in this section could be obtained using different approaches. Particularly, the results in [14] may lead to the Propositions 11, 12, and 13. For sake of completeness, we shall prove all propositions using the same technique.
Using Proposition 5, the matrix which defines the right system X of (1) is transformed into one of the matrices of the statements (a)-(d), namely A = (a i j ) i, j . Of course, the transformation is applied to the whole system (1), so the matrix which defines the left system Y is also transformed into a (general) matrix B = (b i j ) i, j . Then, system (1), after this transformation, reads The solution of (4) can be easily computed, because it is a piecewise linear differential system. So let x, y) be the solution of (4) for x > 0 such that ϕ + (0, x, y) = (x, y).
Assuming that t + (y) > 0 and t − (y) < 0 are defined then there exists a limit cycle passing through the point (0, y) with y ∈ J * = Dom(t + ) ∩ Dom(t − ) if and only if ϕ + 2 (t + (y), 0, y) = ϕ − 2 (t − (y), 0, y). Thus, in this case, we must study the zeros y * of the function on the domain J * . Equivalently, if t + (y) < 0 and t − (y) > 0 are defined, then there exists a limit cycle passing through 0, y). Thus, in this case, we must study the zeros y * of the function on the domain J * .
Since the vectors fields X and Y are linear, then a limit cycle passing through a point (x 0 , y 0 ) must contain points of kind (0, y * ) and (0, y * ) such that y * ∈ J * and y * ∈ J * . Therefore, detecting all the zeros of (5) or (6) we must detect all non-sliding limit cycles of (4).
Let X = (X 1 , X 2 ) and Y = (Y 1 , Y 2 ). We say that a point (0, y) is an (a) invisible fold point for the right system when X 1 (0, y) = 0 and ∂ X 1 ∂ y (0, y)X 2 (0, y) < 0; (c) invisible fold point for the left system when An affine (linear) change of variables in the plane preserving the straight line x = 0 will be called in what follows a -preserving affine (linear) change of variables, and a -preserving affine (linear) change of variables which also preserves the semiplane x > 0 will be called in what follows a + -preserving affine (linear) change of variables. Clearly, a + -preserving affine (linear) change of variables also preserves the semiplane x < 0.
The case when the left system has a focus or a center on will be studied in Sect. 3.1, the case when the left system has a weak saddle will be studied in Sect. 3.2, and the case when the left system has a virtual or real center will be studied in Sect. 3.3.

Left system has a focus or a center on
In this case v 1 = 0, 4b 12 b 21 +(b 11 − b 22 ) 2 ≤ 0 and the point (0, −v 2 ) is a singularity of focus or center type.
In order to fix the clockwise orientation of the flow of system (4), we assume that Proof From Proposition 5(c), we can assume that a 11 = a 22 = a with a ∈ R, a 12 = −a 21 = 1, and by a + -preserving translation we can take u 2 = 0. Moreover, u 1 > 0 because the right system has a focus or a center which is virtual for system (4).
It is easy to see that the point (0, −a u 1 ) ∈ is an invisible fold point for the right system. So the function t + (y) > 0 is defined for every y > −a u 1 (see Fig. 1). Moreover, its image is the interval (0, π). Indeed, given y > −a u 1 consider the line (y) passing through the focus point (−u 1 , 0) and (0, y). The trajectory of the left system starting at (0, y) returns to the line (y) at t = π , so it must return to in a time t < π. Thus, t + (y) ∈ (0, π) for every y > −a u 1 .
The left system has a center if and only if b 22 = −b 11 and b 2 11 + b 12 b 21 < 0. In this case, δ = 1, k 1 = 2v 2 , k 3 = −k 2 = −u 1 , so the function (7) becomes Multiplying g 4 by a parameter, if needed, we see that k 1 and k 2 can be chosen freely. Hence, applying Lemma 9(d) we conclude that N (C b , F v ) = 1. Finally, the lateral systems are centers if and only if a = 0, b 22 = −b 11 and b 2 11 + b 12 b 21 < 0. In this case, the function (7) becomes g 1 (t) = k 1 . So if k 1 = 0, that is v 2 = 0, then there is no solutions for the equation g 1 (t) = 0; if k 1 = 0, that is v 2 = 0, then g 1 = 0, which implies that all the solutions of system (4) passing through (0, y) for y > Y M are periodic solutions, in other words there are no limit cycles. Hence, we conclude that N (C b , C v ) = 0.

Proposition 12 The equalities
Proof From Proposition 5(c) and by a + -preserving translation, we can assume that a 11 = a 22 = a with a ∈ R, a 12 = −a 21 = 1, u 2 = 0, and u 1 < 0 because the right system has a focus which is real for system (4).
In the case that a < 0 it is easy to see that the focus (−u 1 , 0) is an attractor singularity and that the point (0, −au 1 ) ∈ is a visible fold point for the right system. So the function t + (y) < 0 is defined for Fig. 2 Left: real focus for the right system when a < 0. In this case, the shaded line represents the domain of the definition of the function t + (y) < 0. Right: real focus for the right system when a > 0. In this case, the shaded line represents the domain of the definition of the function t + (y) > 0 x y every y < −au 1 . Moreover, its image is the interval (−τ, −π), where τ = −t + (−au 1 ) so π < τ < 2π . Indeed, given y < −au 1 consider the line (y) passing through the focus point (−u 1 , 0) and (0, y). The trajectory of the left system starting at (0, y) returns to the line (y) at t = −π , so it must return to for −2π < −τ < t < −π . Thus, t + (y) ∈ (−τ, −π) for every y < −au 1 (see Fig. 2

left).
In the other case, a > 0 the focus (−u 1 , 0) is a repulsive singularity. Considering now the function t + (y) > 0 defined for every y > −au 1 , the same analysis can be done (see Fig. 2 right).
From now on in this proof we assume, without loss of generality, that a < 0.
The equalities N (F b , C r ) = 1 and N (C b , C r ) = 0 follow in a similar way to the proof of Proposition 11.
In this case, the shaded line represents the domain of the definition of the function t + (y) > 0. Right: virtual diagonalizable node for the right system when a > 1. In this case, the shaded line represents the domain of the definition of the function x y and (0, y + (t 2 )). It concludes the proof of this proposition.

Proposition 13 The equalities
Proof Here u 1 = 0, because the right system has its focus on the line . From Proposition 5(c) and by a + -preserving translation, we can assume that a 11 = a 22 = a with a ∈ R, a 12 = −a 21 = 1, and u 2 = 0. The function t + (y) > 0 is defined for every y > 0, because the point (0, 0) is a focus for the right system. Moreover, we compute t + (y) = π .
Let Y M = max{0, −v 2 }, so computing the zeros of the function (5) for y > Y M is equivalent to compute the zeros of the linear function (8). From here, the equalities N (F b , C b ) = 1 and N (C b , C b ) = 0 follow similarly to the proof of Proposition 11. It concludes the proof of this proposition. N (F b , N

Proposition 14 The equalities
Proof From Proposition 5(b) and by a + -preserving translation, we can assume that a 11 = a 22 = a with |a| > 1, a 12 = a 21 = 1, u 2 = 0, and u 1 > 0, because the right system is a diagonalizable node, which is virtual for system (4).
It is easy to see that the point (0, −a u 1 ) ∈ is an invisible fold point for the right system.
In the case a < −1, the node (−u 1 , 0) is an attractor singularity. The stable manifold and the strong stable manifold of the node intersect at the points (0, y s ) and (0, y ss ), respectively, where y s = u 1 < −a u 1 and y ss = −u 1 < u 1 . So the function t + (y) > 0 is defined for every y > −au 1 (see Fig. 3

left).
In the other case a > 1, the node (−u 1 , 0) is an repulsive singularity. The stable manifold and the strong stable manifold of the node intersect at the points (0, y s ) and (0, y ss ), respectively, where y s = −u 1 > −a u 1 and y ss = u 1 > −u 1 . So the function t + (y) < 0 is defined for every y < −au 1 (see Fig. 3  right).
From now on in this proof we assume, without loss of generality, that a < −1.
Hence, taking y + (t) = u 1 G(t) for t ∈ R + we have that y + t + (y) = y for every y > −au 1 .
The image of the function t + is R + . Indeed, computing implicitly the derivative in the variable y of the identity y + t + (y) = y we obtain a + coth(θ )) .
From here, the equality N (C b , N v ) = 1 follows similarly to the proof of Proposition 11 but now applying Lemma 9(e) to the function g 2 (t) = k 1 ξ 1 − 2k 2 ξ 5 2 . It completes the proof of this proposition. N (F b , i N Fig. 4 Virtual non-diagonalizable node for the right system when λ = 1. In this case, the shaded line represents the domain of the definition of the function t + (y) < 0

Proposition 15 The equalities
Proof From Proposition 5(d) and by a + -preserving translation, we can assume that a 11 = a 12 = a 22 = λ with λ = ±1, a 21 = 0, u 2 = 0, and u 1 > 0, because the right system is a non-diagonalizable node, which is virtual for system (4).
It is easy to see that for λ = ±1 the point (0, −u 1 ) ∈ is a invisible fold point for the right system and that the invariant manifold of the node intersects at the origin (0, 0) (see Fig. 4). In order to fix the clockwise orientation of the flow of system (4), we assume that λ = 1; otherwise, the first return map would not be defined and there would not exist limit cycles. In this case, the function t + (y) < 0 is defined for every y < −u 1 .
From here, the equality N (C b , i N v ) = 1 follows similarly to the proof of Proposition 11 but now applying Lemma 9( f ) to the function g 3 (t) = k 1 ξ 1 −2k 2 ξ 6 2 . It concludes the proof of proposition.

Proposition 16 The equalities
Proof From Proposition 5(a) and by a + -preserving translation, we can assume that a 11 = a 22 = a with |a| < 1, a 12 = a 21 = 1, u 2 = 0, and u 1 < 0, because the right system is a saddle, which is real for system (4). It is easy to see that the point (0, −au 1 ) ∈ is an invisible fold point for the right system and that the stable and unstable invariant manifolds of the saddle intersect at the points (0, y s ) and (0, y u ), respectively, where y s = −u 1 and y s = u 1 . So the function t + (y) > 0 is defined for every −au 1 < y < −u 1 .
The right system has a saddle with trace equal to 0 if and only if a = 0, in this case ξ 2 2 (t) = ξ 2 3 (t) = coth(t) − csch(t). So the equality N (F b , S 0 r ) = 1 follows applying lemma 10(b) to the function g 2 (t) = k 1 ξ 1 (t)+2k 2 ξ 2 2 . From here, the equalities N (C b , S r ) = 1 and N (C b , S 0 r ) = 0 follow similar to the proof of Proposition 11 but now by applying Lemma 9(e) to the function g 2 (t) = k 1 ξ 1 (t) − 2k 2 ξ 5 2 (t). It concludes the proof of this proposition. , and let y s be the y-coordinate of the intersection between the stable manifold with . We compute 12 and In order to fix the clockwise orientation of the flow of system (4), we assume that y s < y s , which is equivalent to assume that b 12 > 0.
The left system has an invisible fold point (0,y) given by For y s < y < y y , we define So t − (y) = t * (y) < 0 fory < y < y u and t − (y) = t * (y) > 0 for y s < y <y.
Proof From Proposition 5(c), we can assume that a 11 = a 22 = a with a ∈ R, a 12 = −a 21 = 1, and by a + -preserving translation we can take u 2 = 0. Moreover, u 1 > 0 because the right system has a focus which is virtual for system (4). From the proof of Proposition 11, we know that the function t + : (−a u 1 , ∞) → (0, π), such that ϕ + (t + (y), 0, y) = 0 for y > −a u 1 , is invertible with inverse y + : (0, π) → (−a u 1 , ∞) given by y Let Y M = max −a u 1 ,y , so computing the zeros of the function (5) for Y M < y < y u is equivalent to compute the zeros of the function g 4 (t) = f (y + (t)) = k 1 ξ 1 + k 2 ξ 4 2 (11) for t ∈ I ⊂ (0, π), where k 1 = 2(b 11 v 1 + b 12 v 2 )/b 12 and k 2 = −2u 1 , and here I = t + ((Y M , y u )). Multiplying the function g 4 by a parameter, if necessary, we see that k 1 and k 2 can be chosen freely. So applying Lemma 9(d), we conclude that N (S 0 r , F v ) = 1. The right system has a center if and only if a = 0. In this case, ξ 4 2 = 0 and the function (11) becomes then there are no solutions for the equation g 4 (t) = 0; and if k 1 = 0, that is b 11 v 1 = −b 12 v 2 , then g 4 = 0, that is system (4) is a center. Hence, we conclude that N (S 0 r , C v ) = 0.
Proof From Proposition 5(c), we can assume that a 11 = a 22 = a with a ∈ R, a 12 = −a 21 = 1, and by a + -preserving translation we can take u 2 = 0.
Let Y m = min{−au 1 ,y}, so computing the zeros of the function (6) for y s < y < Y m is also equivalent to compute the zeros of the function (11) now for t ∈ I ⊂ (−τ, −π), where I = t + ((y s , Y m )).
The equality N (S 0 r , C r ) = 0 follows similarly to the proof of Proposition 17. It concludes the proof of this proposition.

Proposition 19
The equality N (S 0 r , N v ) = 1 holds. Proof From Proposition 5(b) and by a + -preserving translation, we can assume that a 11 = a 22 = a with |a| > 1, a 12 = a 21 = 1, u 2 = 0, and u 1 > 0, because the right system has a diagonalizable node, which is virtual for system (4).
Following the proof of Proposition 14, the function t + : (−au 1 , ∞) → R + is invertible with inverse y + : R + → (−au 1 , ∞) given by y + (t) = u 1 G(t). Here, as we have done in the proof of Proposition 13 we are assuming, without loss of generality, that a < 1.
Let Y M = max{−a u 1 ,y}, so computing the zeros of the function (5) for Y M < y < y u is equivalent to compute the zeros of the function g 5 (t) = f (y + (t)) = k 1 ξ 1 + k 2 ξ 5 2 (12) for t ∈ I ⊂ R + , where k 1 = 2(b 11 v 1 + b 12 v 2 )/b 12 , k 2 = −2u 1 , and I = y + ((Y M , y u )). Multiplying the function g 5 (t) by a parameter, if necessary, we see that the parameters k 1 and k 2 can be chosen freely. So applying Lemma 9(e), we conclude that N (S 0 r , N v ) = 1.

Proposition 20
The equality N (S 0 r , i N v ) = 1 holds.
Proof From Proposition 5(b) and by a + -preserving translation, we can assume that a 11 = a 12 = a 22 = λ with λ = ±1, a 21 = 0, u 2 = 0, and u 1 > 0, because the right system has a non-diagonalizable node, which is virtual for system (4). Following the proof of Proposition 15, the function t + : (−u 1 , ∞) → R + is invertible with inverse y + : u 1 H (t). Here, as we have done in the proof of Proposition 15 we are assuming, without loss of generality, that λ = 1.
Let Y M = max{−u 1 ,y}, so computing the zeros of the function (5) for Y M < y < y u is equivalent to compute the zeros of the function 12 , k 2 = −2u 1 , and I = y + ((Y M , y u )). Multiplying the function g 6 (t) by a parameter, if necessary, we see that k 1 and k 2 can be chosen freely. So applying Lemma 9( f ), we conclude that N (S 0 r , i N v ) = 1.
Let Y M = max{−a u 1 ,y} and Y m = min{u 1 , y u }, so computing the zeros of the function (5) for Y M < y < Y m is equivalent to compute the zeros of the function (12) for t ∈ I ⊂ R + , where k 1 = 2(b 11 v 1 + b 12 v 2 )/b 12 and k 2 = −2u 1 . Multiplying the function (12) by a parameter, if necessary, we see that k 1 and k 2 can be chosen freely. So applying Lemma 9(e) we conclude that N (S 0 r , S r ) = 1. The right system has a saddle with trace equal to 0 if and only if a = 0. In this case, ξ 5 2 = 0 and the function (12) becomes g 5 (t) = k 1 . So if k 1 = 0, that is b 11 v 1 = 0, then there are no solutions for the equation g 5 (t) = 0. If k 1 = 0, that is b 11 v 1 = 0, then g 5 = 0, which implies that all the solutions of system (4) passing through (0, y) for Y M < y < Y m are periodic solutions; in other words, there are no limit cycles. Hence, we conclude that N (S 0 r , S 0 r ) = 0.

Left system has a virtual or real center
In this case, v 1 = 0, b 22 = −b 11 , b 2 11 + b 12 b 21 < 0 and the point (−v 1 , −v 2 ) is a singularity of center type.
The left system has a fold point (0,y) given by which is visible if v 1 > 0, and invisible if v 1 < 0. In order to fix the clockwise orientation of the flow of system (4), we assume that Y 1 (−v 1 , 1−v 2 ) = b 12 > 0.
Proof In Corollary 3, these equalities have already been proved when the left system has a center in . So we can take v 1 = 0. To obtain N (C, F v ) = 1, we follow the proof of Proposition 11, and then, we compute the solutions of the function (5) for y > Y M = max{y, −a u 1 }. To obtain N (C, F r ) = 2, we follow the proof of Proposition 12 and then we compute the solutions of the function (6) for y < Y m = min{y, −a u 1 }. In both cases, the equations to be solved are equivalent to k 1 + k 2 ξ 4 2 (t) = 0, for t ∈ (0, π) and t ∈ (−τ, −π), respectively. Here, k 1 = (b 11 v 1 + b 12 v 2 )/b 12 and k 2 = −u 1 . So applying statements (d) and (d ) of Lemma 9, we conclude that N (C, F v ) = 1 and N (C, F r ) ≤ 2, respectively. Moreover, since N (C b , F r ) = 2, we actually have the equality N (C, F r ) = 2. The equality N (C, C v ) = N (C, C r ) = 0 follows similarly to the proof of Proposition 17. It concludes the proof of this proposition.

Proposition 23
The equalities N (C, F b ) = 1 and N (C, C b ) = 0 hold. Proof In Corollary 3, these equalities have already been proved when the left system has a center in . So we can take v 1 = 0.
To obtain N (C, F v ) = 1, we follow the proof of Proposition 13, and then, we compute the solutions of the function (5) for y > Y M = max{y, 0}, which is equivalent to compute the zeros of the liner equation k 1 + k 2 y = 0. Here, k 1 = 2(b 11 v 1 + b 12 v 2 )/b 12 and k 2 = (1 − e aπ ). The equalities N (C, F b ) = 1 and N (C, C b ) = 0 follow similarly to the proof of Proposition 11. It concludes the proof of this proposition.
Proof In Corollary 3, these equalities have already been proved when the left system has a center in . So we can take v 1 = 0. To prove the equality N (C, N v ) = 1, we follow the proof of Proposition 14, and then, we compute the solutions of the function (5) for y > Y M = max{y, −a u 1 }. To prove the equality N (C, S r ) = 1, we follow the proof of Proposition 16, and then, we compute the solutions of the function (5) for Y M < y < u 1 . In both cases, the equations to be solved are equivalent to k 1 + k 2 ξ 5 2 = 0, where k 1 = 2(b 11 v 1 + b 12 v 2 )/b 12 and k 2 = −2u 1 . From here, the proofs of the equalities N (C, N v ) = 1 and N (C, S r ) = 1 follow similarly to the proofs of the Propositions 19 and 21, respectively. The equality N (C, S 0 r ) = 0 follows similarly to the proof of Proposition 21. It concludes the proof of this proposition.

Proposition 25
The equality N (C, i N v ) = 1 holds. Proof In Corollary 3, this equality has already been proved when the left system has a center in . So we can take v 1 = 0.
Following the proof of Proposition 15, we compute the solutions of the function (6) for y < Y m = {y, −u 1 }, which is equivalent to compute the zeros of the following equation k 1 + k 2 ξ 6 2 = 0, where k 1 = (b 11 v 1 + b 12 v 2 )/b 12 and k 2 = −u 1 . So analogously to the proof of Proposition 20, we conclude that N (C, i N v ) = 1. It concludes the proof of this proposition.