Piecewise linear differential systems without equilibria produce limit cycles?

In this article, we study the planar piecewise differential systems formed by two linear differential systems separated by a straight line, such that both linear differential have no equilibria, neither real nor virtual. When the piecewise differential system is continuous, we show that the system has no limit cycles. But when the piecewise differential system is discontinuous, we show that it can have at most one limit cycle.

community nowadays, mainly because these systems are widely used to model processes appearing in electronics, mechanics, economy, etc. See, for instance, the books of di Bernardo et al. [5] and Simpson [28], the survey of Makarenkov and Lamb [26], and the hundreds of references which appear in these last three cited works.
The simplest possible piecewise linear differential systems are the ones formed by two linear differential systems separated by a straight line. We note that for these apparently simple systems, when they are continuous, some serious work is necessary for proving that they have at most one limit cycle, see [7] and [21]. This solved the conjecture of Lum and Chua [25] done in 1990 that such continuous differential systems can have at most one limit cycle.
We consider planar piecewise differential systems formed by two linear differential systems separated by the straight line x = 0, such that both linear differential have no equilibria, neither real nor virtual. We say that an equilibrium point (x 0 , y 0 ) of the linear differential system defined in the half-plane x ≥ 0 is real if it x 0 ≥ 0, otherwise it is called virtual.
Theorem 1 Continuous planar piecewise differential systems formed by two linear differential systems sep- Theorem 2 Discontinuous planar piecewise differential systems formed by two linear differential systems separated by a straight line, such that both linear differential have no equilibria, neither real nor virtual, have at most one limit cycle, and the next example shows that there are piecewise linear differential systems with one limit cycle: The piecewise linear differential systeṁ has the limit cycle of Fig. 2.

Remark 3
Concerning Theorem 2, we stress that a closed curve, consisting of union of branches of trajectories of both systems, one in x > 0 and the other in x < 0, passing through points of the closure of the sliding region is not considered a limit cycle. For a definition of sliding region, see for instance [5]. We consider in the half-plane x > 0 the following general linear differential system without equilibriȧ and in the half-plane x < 0 another general linear differential system without equilibriȧ Here the dot denotes derivative with respect to the independent variable t usually called the time. Changing the sign of the time (if necessary), we can assume without loss of generality that In order that both linear differential systems define in the whole plane R 2 a continuous differential system, we must take We denote the divergence of system ( j) by Div j ; thus, we have Div 1 = α + λb, Div 2 = a + λb.
We note that in order that the continuous piecewise linear differential systems in R 2 formed by systems (1) and (2) can have periodic orbits it is necessary that either Div 1 Div 2 < 0, or Div 1 = Div 2 = 0. This follows from the following claim: If either Div 1 Div 2 > 0, or Div i = 0 and Div j = 0 for i = j, then the continuous piecewise linear differential systems in R 2 formed by systems (1) and (2) have no periodic solutions. Now we prove the claim. Clearly, if there is a limit cycle, this must cross the straight line of separation x = 0 between the two linear differential systems because linear differential systems have no limit cycles. The vector field associated with our differential system is We shall use the following well-known result, the Green's theorem; for a proof, see for instance [27].
where the integration path along γ is in counterclockwise sense.
The divergence of a C 1 differential systeṁ Let γ be a periodic solution of the vector field X (x, y), and let R be the bounded region limited by γ . In order to apply the Green's theorem to this curve γ and to the region R, we shall split such an integral as limit of two integrals as follows. We add to the periodic orbit γ the segment S of the y-axis contained in the region bounded by γ ; now we split this segment as limit of two parallel segments S − (ε) and S + (ε) contained in x < 0 and x > 0 and at a distance ε > 0 of S, respectively, and such that a piece γ − (ε) of γ contained in x < 0 together with S − (ε) forms an oval O − (ε). Similarly, we consider a piece γ + (ε) of γ contained in x > 0 such that together with S + (ε) forms another oval O + (ε), in such a way that the union of these ovals tends to γ ∪ S when ε → 0 (see Fig. 1).
Clearly, the two integrals are well defined, and the integral is the limit when ε → 0 of Applying the Green's theorem (Theorem 4) to both integrals of (5), we obtain that where R ±(ε) are the open regions bounded by the ovals O ±(ε) . Now, from (5) and (6) Since γ is a periodic solution of the vector field X (x, y), we have that P dy − Q dx = 0, so which is a contradiction with the assumptions that either Div 1 Div 2 > 0, or Div i = 0 and Div j = 0 for i = j. Hence, the claim is proved. The proof of this claim has been inspired in ref. [23]. Now in order to prove Theorem 1, it remains to show that when either Div 1 = Div 2 = 0, or Div 1 Div 2 < 0 the vector field X (x, y) has no periodic solution, we distinguish two cases. Case 1: Div 1 = Div 2 = 0. Then, the piecewise differential systems (1) and (2) become the linear differential system which clearly has no periodic solutions.
Case 2: Div 1 Div 2 < 0. Without loss of generality, we can assume that Div 1 = α + bλ > 0 and Div 2 = a + bλ < 0, otherwise we reverse the sign of the independent variable in the differential system. Then, the solution (x + (t), y + (t)) of system (1) with the initial condition (x + (0), y + (0)) = (0, y 0 ) is and the corresponding solution (x − (t), y − (t)) of system (2) with the same initial condition is Let t + be the time that the solution (x + (t), y + (t)), starting at the point (0, y 0 ) when t = 0, enters in forward time in the half-plane x > 0 and reaches by first time the straight line x = 0, in case that such solution exists. Similarly, let −t − be the time that the solution (x − (t), y − (t)), starting at the point (0, y 0 ) when t = 0, enters in backward time in the half-plane x < 0 and reaches by first time the straight line x = 0, in case that such solution exists. Therefore, the piecewise linear differential systems (1) and (2) have limit cycles if the system has isolated solutions. We have three equations and three unknowns t + , t − and y 0 .
From equation x + (t + ) = 0, we get that e t + (α+bλ) We note that if some of the denominators in the expressions of the previous both exponentials is zero, it follows easily that system (8) has no solutions. Substituting these two exponentials in equation This is in contradiction with the facts that t + > 0, t − > 0 and d − λc = 0, otherwise system (7) would have a straight line of equilibria. Then system (8) has no solutions, so the piecewise linear differential system has no limit cycles. This completes the proof of Theorem 1.

Discontinuous piecewise differential system: Proof of Theorem 2
Again we consider the planar piecewise differential systems formed by two linear differential systems separated by a straight line, such that both linear differential have no equilibria, neither real nor virtual, defined by (1) and (2). But now we assume that this piecewise differential systems are discontinuous, i.e. we do not consider the conditions (4). Using the notation of the proof of Theorem 1, we have Div 1 = α + λβ, Div 2 = a + μb.
Here, we must separate the proof of Theorem 2 in three cases.
From equation x + (t + ) = 0, we get that and from equation x − (−t − ) = 0, we obtain that We note that the denominator of t + cannot be zero, otherwise t + will be infinity, and we cannot have a periodic solution solution satisfying (10). Also the denominator of y 0 cannot be zero, otherwise b = 0 and therefore x − (t) = ct. So if c = 0 never holds x − (−t − ) = 0; and if c = 0 we have that x − (t) = 0 for all t, so in this case we cannot have periodic solutions of the ones here consider with a piece in the half-space x > 0 and another piece in the half-space x < 0. Substituting t + and y 0 in equation y Hence, if this last equality does not hold, then system (10) has no solutions, and consequently, the system has no periodic solutions. If c/b − γ /β = 0, then solving the second equation from (10) with respect to t − > 0 we get at most one solution, which substituted in (12) provides at most one periodic solution. So the theorem is proved in Case 1.
From equation x + (t + ) = 0, we get that t + is given in (11), and from equation x − (−t − ) = 0, we obtain that Again we note that if some of the denominators in the expressions of t + or y 0 is zero, it follows easily that system (8) has no solutions. Substituting t + and y 0 in equation y + (t + ) − y(−t − ) = 0, we have that Since the function z coth(z) for z > 0 is strictly increasing, and at z = 0 takes the zero value, it follows that Eq. (14) has at most one solution for z = 1 2 (a + bμ)t − (i.e. for t − ) and consequently a unique solution for y 0 and t + from (13) and (11), respectively. Obtaining in this way at most one limit cycle for the discontinuous piecewise linear differential systems (1) and (2), hence Theorem 2 is proved under the assumptions of Case 2.
We note that the discontinuous piecewise linear differential system which provides the limit cycle of Fig. 2 is a particular system of the ones studied in this Case 2.
Defining t + and t − as before the piecewise linear differential system (1) and (2) has limit cycles if system (8) has isolated solutions for the unknowns t + , t − and y 0 .
We change the parameters α and a by u 2 and v 2 defined as α = u 2 − βλ and a = −v 2 − bμ, respectively. From equation x + (t + ) = 0, we get that We note that if some of the denominators in the expressions of the previous both exponentials is zero, the present study can be done in a similar and easier way. Substituting these two exponentials in equation y + (t + ) − y(−t − ) = 0, we have that From this last equation, we obtain that Isolating y 0 from equation x − (−t − ) = 0, substituting t + and y 0 in equation x + (t + ) = 0, and changing the variable t − by the variable z through we obtain the equation where and f 0 (z) = e z − 1 e k − 1 , We note that in the particular case that k = 1 Eq. (17) becomes which has no solutions for z > 0. Now we shall prove that the functions f 0 and f 1 form an extended complete Chebyshev system if k = 1. So, if k = 1, Eq. (17) can have at most one zero, and this can be reached. See for more details on Chebyshev system the appendix. In short, if Eq. (17) as at most one solution t − , from it we get a unique value for y 0 and t + , and consequently at most a unique limit cycle for the discontinuous piecewise linear differential system (1) and (2) under the assumptions of Case 3. This completes the proof of Theorem 2.
Only remains to show that the functions f 0 and f 1 form an extended complete Chebyshev system if k = 1. Indeed, from Eq. (15) and since t + and t − are positive (by definition), it follows that (d − cμ)(γ λ − δ) > 0. Therefore, from (16) we get that z > 0 and k > 0. Now we change the variable z by the new variable t through z = log t with t > 1. So, in function of the new variable t we have f 0 (t) = (t − 1)(t k − 1) > 0, Hence, from Theorem 5 of the appendix, in order to see that the functions f 0 and f 1 form an extended complete Chebyshev system for t > 1, we only need to show that the Wronskian does not vanish for k > 0 and k = 1. Since we have that W (t) < 0 if k ∈ (0, 1), W (t) = 0 if k = 1, and W (t) > 0 if k > 1 (see also Figs. 3 and 4, the functions f 0 and f 1 form an extended complete Chebyshev system if k = 1.