Limit cycles of 3-dimensional discontinuous piecewise differential systems formed by linear centers

In this paper we deal with 3-dimensional discontinuous piecewise differential systems formed by linear centers and separated by one plane or two parallel planes. We prove that these systems separated by one plane have no limit cycles, besides the systems separated by two parallel planes have at most one limit cycle, and that there are systems having such a limit cycle. So we solve the extension of the 16th Hilbert problem to this class of differential systems.


Introduction and statement of the main results
One of the main goals in the qualitative theory is to study the number of limit cycles of the differential systems. In part this problem was motivated by the 16-th Hilbert problem (1900), see [8,16] for more details. Limit cycles play a main role for understanding the dynamics of many systems, see for instance [1,2,4,9,10].
On the other hand there are many problems that are modeled using discontinuous piecewise differential systems. These systems appear in various situations Communicated by Marco Antonio Teixeira. like mechanical systems and control theory, see for instance [3,5]. In particular the study of discontinuous piecewise linear differential systems in the plane separated by straight lines is also an important class of differential systems which appear in models of mechanics and electrical circuits among others, see for instance [6,13,15].
Following the Filippov's convention [7] the discontinuous piecewise linear differential systems can have sliding or crossing limit cycles. A sliding limit cycle contains sliding segments on the line of discontinuity, whereas crossing limit cycles contain only crossing points. In this paper we work with crossing limit cycles, or simply limit cycles.
In [11] it was proved that discontinuous piecewise linear differential systems in the plane separated by one straight line and formed by two linear centers have no limit cycles, besides discontinuous piecewise linear differential systems in the plane separated by two parallel straight lines and formed by three linear centers have at most one limit cycle, and that there are systems having such a limit cycle.
The main goal of this paper is to extend these previous results to 3-dimensional discontinuous piecewise differential systems in the space separated by one plane or two parallel planes and formed by linear centers. More precisely, we consider the differential system This differential system has the z-axis filled of singular points, and at every plane z = z 0 , with z 0 a constant, there exists a linear center at (0, 0, z 0 ) , i.e. all the orbits in this plane are periodic formed by circles centered at (0, 0, z 0 ) , of course with the exception of the singular point (0, 0, z 0 ).
We shall pass from the coordinates (x, y, z) to (X, Y, Z) through the affine change of coordinates assuming that the determinant of matrix of the change of coordinates For simplicity we rename the variables X, Y and Z by x, y and z, respectively. Thus system (2) writes where Now we consider the 3-dimensional discontinuous piecewise differential system separated by plane x = 0 and the one separated by the parallel planes x = −1 and x = 1 In what follows we state the main results of this paper.

Theorem 1
The discontinuous piecewise linear differential systems (3) have no limit cycles.

Theorem 2
The discontinuous piecewise linear differential systems (4) have at most one limit cycle. Moreover there are systems in this class having one limit cycle.
Theorems 1 and 2 are proved in Sects. 2 and 3, respectively. The main tool for proving these theorems is the use of the first integrals of the differential systems which form the discontinuous piecewise differential systems, this technique for studying the limit cycles already was used in [12].

Proposition 1
The discontinuous piecewise linear differential system separated by the two parallel planes x = −1 and x = 1 where has one limit cycle, the one of Fig. 1. Moreover this limit cycle is stable. Proposition 1 is proved at the end of Sect. 3.

Proof of Theorem 1
The differential system in x > 0 of the discontinuous linear piecewise differential system (3) has the two independent first integrals and the differential system in x < 0 has the two independent first integrals f 1 and f 2 obtained respectively from h 1 and h 2 , changing the parameters a i , b i and c i by i , i and i respectively, for i = 0, 1, 2, 3.
A limit cycle for discontinuous piecewise differential system (3) must intersect the plane x = 0 in two distinct points, denoted by (0, y 1 , z 1 ) and (0, y 2 , z 2 ) , and such two points must satisfy the system of equations Taking the change of variables y 1 = y + y 2 and z 1 = z + z 2 we obtain The limit cycle of discontinuous piecewise linear differential system (5) In order that the system e 2 = e 4 = 0 has non-trivial solutions we need that the following determinant be zero However now there are only three independent equations e 1 = e 2 = e 3 = 0 and four unknowns variables y 1 , y 2 , z 1 and z 2 . Thus always we have a continuum of periodic solutions and no limit cycles. Therefore Theorem 1 is proved.

Proof of Theorem 2
We note that the differential system in x > 1 of the discontinuous piecewise differential system (4) has the same two independent first integrals h 1 and h 2 given in the proof of Theorem 1. The differential system in |x| < 1 has the two independent first integrals f 1 and f 2 obtained respectively from h 1 and h 2 replacing the parameters a i , b i and c i by A i , B i and C i , respectively for i = 0, 1, 2, 3 , and the differential system in x < −1 has the two independent first integrals g 1 and g 2 obtained respectively from h 1 and h 2 , changing the parameters a i , b i and c i by i , i and i respectively for i = 0, 1, 2, 3.
Proof Since K 6 ≠ 0 and K 24 ≠ 0 , we solve E 2 = 0 in the variable y 1 and E 6 = 0 in the variable y 3 , and we obtain Replacing y 1 and y 3 in each equation E i = 0 for i ∈ {1, 3, 4, 5, 7, 8} , in particular we obtain that Since K 15 ≠ 0 we solve E 4 = 0 in the variable y 2 and we substitute it in each E i = 0 for i ∈ {1, 3, 5, 7, 8} . Thus E 8 becomes where T 1 and T 3 are polynomials of degree 1 and T 2 and T 4 are polynomials of degree 2. We do not explicit them here due to their lengths. If z 3 = z 4 then from (8) we get y 3 = y 4 and we have no limit cycles. Thus we must study the zeros of the system we get the new equation such that T 5 is a polynomial of degree 1. In short, we have four polynomials and the product of their degrees is 2.
Thus if system (6) has finitely many solutions by Bezout Theorem (see for instance [14]) it has at most 2 solutions. By Remark 1 these two solutions correspond to the same limit cycle. So the discontinuous piecewise linear differential system (4) has at most one limit cycle. ◻ Lemma 2 If K 6 ≠ 0 , K 24 = 0 , K 15 ≠ 0 and K 16 K 6 − K 15 K 7 ≠ 0 , then system (4) has at most one limit cycle.
Proof Since K 6 ≠ 0 we solve E 2 = 0 in the variable y 1 and we obtain (7). Taking K 24 = 0 and replacing y 1 given in (7) in each E i = 0 , for i ∈ {1, 3, .., 8} , in particular we get E 6 = K 23 (z 3 − z 4 ) = 0 . So z 3 = z 4 , otherwise K 23 = 0 and we do not have finitely many solutions and therefore we have no limit cycles. Then we have that Solving E 8 = 0 in the variable y 2 we obtain Substituting y 2 in each E i = 0 for i ∈ {1, 3, 4, 5, 7} , in particular we have Solving E 4 = 0 in the variable y 3 and substituting it in each E i = 0 for i ∈ {1, 3, 5, 7} we obtain the following four polynomial equations in the variables z 1 , z 2 , y 4 and z 4 where T 1 and T 3 are polynomials of degree 1 and T 2 and T 4 are polynomials of degree 2. If z 1 = z 2 then from (7) we have y 1 = y 2 and we have no limit cycles. Thus we must study the number of the zeros of the system T 1 = T 2 = T 3 = T 4 = 0 . Taking E 3,7 = E 3 + E 7 we get the new equation such that T 5 is a polynomial of degree 1. In short we have four polynomials and the product of their degrees is 2. Thus as in the proof of the previous lemma we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle. ◻ Lemma 3 If K 6 ≠ 0 , K 24 ≠ 0 , K 15 = 0 and K 16 K 6 − K 15 K 7 ≠ 0 , then system (4) has at most one limit cycle.
Proof Consider K 15 = 0 . So K 16 ≠ 0 . Since K 6 ≠ 0 and K 24 ≠ 0 we substituted y 1 and y 3 given in (7) and (8)  Solving E 4 = 0 and E 8 = 0 in the variables z 2 and z 1 respectively, we obtain Substituting them in each equation E i = 0 for i ∈ {1, 3, 5, 7} , we obtain the following four polynomial equations in the variables y 2 , y 3 , z 3 and z 4 : where T 1 and T 3 are polynomials of degree 1 and T 2 and T 4 are polynomials of degree 2. As in the proof of Lemma 1 we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle. ◻
Taking z 1 = z 2 and solving E 6 = 0 in the variable y 3 we obtain (8). Substituting it in each equation E i = 0 for i ∈ {1, 3, 4, 5, 7, 8} , in particular we obtain E 4 = 0 given in (9). Solving E 4 = 0 in the variable y 2 and substituting it in each equation E i = 0 for i ∈ {1, 3, 5, 7, 8} , E 8 becomes Solving E 8 = 0 in the variable y 1 and substituting it in each equation E i = 0 for i ∈ {1, 3, 5, 7} , we obtain the following four polynomial equations in the variables y 4 , z 2 , z 3 and z 4 : where T 1 and T 3 are polynomials of degree 1 and T 2 and T 4 are polynomials of degree 2. As in the proof of Lemma 1 we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle. ◻ Lemma 5 If K 6 = 0 , K 24 = 0 , K 15 ≠ 0 and K 16 K 6 − K 15 K 7 ≠ 0 , then system (4) has at most one limit cycle.
Proof Consider K 6 = 0 , K 24 = 0 and K 15 ≠ 0 . So we have that K 7 ≠ 0 . From E 2 = 0 and E 6 = 0 we obtain that z 1 = z 2 and z 3 = z 4 . Solving E 4 = 0 and E 8 = 0 in the variables y 2 and y 1 , respectively and substituting them in each E i = 0 for i ∈ {1, 3, 5, 7} we get the following four polynomial equations in the variables y 3 , y 4 , z 2 and z 4 : where T 1 and T 3 are polynomials of degree 1 and T 2 and T 4 are polynomials of degree 2. Again as in the proof of Lemma 1 we conclude that the discontinuous piecewise differential system (4) has at most one limit cycle. ◻
Proof Consider K 6 ≠ 0 , K 24 ≠ 0 , K 15 ≠ 0 and K 16 K 6 − K 15 K 7 = 0 . Solving E 2 = 0 and E 6 = 0 in the variables y 1 and y 3 respectively, we obtain y 1 and y 3 given in (7) and (8). Substituting y 1 , y 3 and K 16 = K 15 K 7 ∕K 6 in E 4 = 0 and E 8 = 0 in particular we obtain Solving E 4 = 0 in the variable y 2 and substituting it in E 8 = 0 we obtain If z 3 = z 4 , then from (8) we get y 3 = y 4 and we have no limit cycles. If K 23 K 6 − K 24 K 7 = 0 , then there are more unknown variables than equations in system (6) and therefore there are no limit cycles. ◻ Lemma 8 If K 6 ≠ 0 and K 24 ≠ 0 , K 15 = 0 and K 16 K 6 − K 15 K 7 = 0 , then system (4) has no limit cycles.
Proof Consider K 6 ≠ 0 and K 24 ≠ 0 . Taking K 15 = 0 we have that K 16 = 0 . So we obtain E 4 = −E 8 = 2K 14 . Therefore there are more unknown variables than equations in system (6) and therefore there are no limit cycles. ◻
Proof The proof of this lemma is analogous to the one of the previous lemma. ◻
Now we shall prove that this limit cycle is stable. This limit cycle starts at the point (1, 0, 0) of the plane x = 1 , cross the region |x| < 1 until the point (−1, 0, 1) of the plane x = −1 , after travels in the region x < 1 until the point (−1, 0, 0) of the plane x = −1 , cross again the region |x| < 1 until the point (1, 1, 0) of the plane x = 1 , and finally it travels in the region x > 1 until the initial point (1, 0, 0).
Let and be two small real numbers, then the point (1, , ) is close to the point (1, 0, 0) of the limit cycle. Using the first integrals f 1 and f 2 we compute where the orbit through the point (1, , ) intersect the plane x = −1 near the point (−1, 0, 1) , this intersection takes place at the point (−1, y 1 , z 1 ) where Now with the first integrals g 1 and g 2 we compute where the orbit through the point (−1, y 1 , z 1 ) travels in the region x < −1 until intersecting the plane x = −1 at the point (−1, y 2 , z 2 ) near the point (−1, 0, 0) , where Using again the first integrals f 1 and f 2 we compute where the orbit through the point (−1, y 2 , z 2 ) travels in the region |x| < 1 until intersecting the plane x = 1 at the point