Piecewise Bounded Quadratic Systems in the Plane

In this paper we study the sliding mode of piecewise bounded quadratic systems in the plane given by a non-smooth vector field Z=(X,Y)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Z=(X,Y)$$\end{document}. Analyzing the singular, crossing and sliding sets, we get the conditions which ensure that any solution, including the sliding one, is bounded.

where f : ⊂ R 2 −→ R and X, Y : ⊂ R 2 −→ R 2 are of class C r (r 1) and 0 is a regular value of f . The dynamics of the system on the switching manifold given by the points x such that f (x) = 0 is also studied. We use the notation Z (x) = (X (x), Y (x)) or simply Z = (X, Y ) and assume that Z is bi-valuated on . The flow of Z = (X, Y ) is defined respectively by the vector fields X and Y as usual. The convention adopted by Filippov, in the pioneering work [4], for the transition of orbits (trajectories) between the two regions separated by and its permanence in is presented in the second section.
In this work we consider that X and Y are bounded quadratic vector fields in the plane. More precisely we study X and Y as one of the normal forms of the bounded quadratic vector fields, studied in [6]. A piecewise differential system is said bounded when all its trajectories are bounded for t 0. A natural question is The following proposition says that in general the answer is negative.

Proposition 1 There exist bound quadratic vector fields X and Y such that Z = (X, Y ) is not bounded.
The proof of this proposition is presented in the second and third subsections of the second section of two forms: the first gives us a vector field Z unbounded in a sliding region and the second in a crossing region. For details about this regions see the second subsection of the second section.
In this paper we consider the following: Problem Let Z = (X, Y ) be a non-smooth vector field with X and Y bounded quadratic.

Is it possible to give sufficient conditions on the coefficients of the linear parts of X and Y which ensures that Z is bounded?
The paper is organized as follows. In the second section we present preliminaries results and state our main results. In the third section we prove the main results. In the fourth section we present an important application of the piecewise bounded quadratic system: the massspring system. This model is extremely important for the study of the natural phenomena. It is used as a good approximation for small amplitude oscillations. The oscillatory systems can be studied by second order differential equations, from the application of laws physical, such as Newton's laws, Hooke's law and hypothesis such as damping proportional to the speed etc. We analyzed the mass-springer system under the action of a force with different intensities and the switching manifold is interpreted as the interrupt of each one of the forces to start the other. For more details see the fourth section.

Bounded Quadratic Systems in the Plane
Consider an autonomous quadratic differential system in the plane with an equilibrium point at (0, 0) given byẋ where a, b, c, d, e, f are real constants.
According to Markus (see [6]) if ϕ(x) = 0, then it is linearly equivalent to one of the following forms: We say that a quadratic system is bounded if all its trajectories are bounded for t 0. Dickson and Perko (see [3]) determined what of those systems (1) are bounded. More preciselly they proved the following.
Theorem 2 Considerẋ = Ax + ϕ(x) an autonomous quadratic differential system in the plane with an equilibrium point at (0, 0), as above, and initial condition 2 , x 2 2 ) T , then the system has an unbounded trajectory (as t → ∞) Theorem 2 classifies the quadratic systems in the plane that are bounded. If a quadratic system is linearly equivalent to a system with the quadratic part given by (0, The complete study of the phase portraits of the bounded quadratic systems having limit cycles can be found in [1] and their perturbation inside all quadratic systems in [5]. Moreover we refer [2] as a survey of the known results for this kind of systems.

Piecewise Smooth Dynamical Systems in
We denote χ r ( ) the set of the vector fields X : ⊂ R 2 −→ R 2 of C r class with r 1. Consider X, Y ∈ χ r ( ). A vector field like the following is said to be a non-smooth vector field with switching manifold , and it is denoted by Z = (X, Y ). The set of these vector fields is denoted for χ r ( , f ).
for any x ∈ , where ·, · is a usual inner product. For illustration of the non-smooth vector field Z see Fig. 1.
We classify the points on according the product X f (x) · Y f (x). By the Filippov convention, the points x ∈ such that X f (x) · Y f (x) > 0 determine a crossing region (Fig.  2a, b) and the points x ∈ such X f (x) · Y f (x) < 0 determine a sliding region (Fig. 2c, d).
The points x ∈ such that X f (x) · Y f (x) = 0 are called tangent points of Z (Fig. 3). Thus In the sliding region, we define the Filippov vector field, which is given by for any is a usual inner product. See Fig. 4.
The set of the regular points is denoted by reg and the set of the singular points is denoted for sin . Thus reg = c ∪ ( s \ e ) and sin = t ∪ e .

First proof of Proposition 1 Consider
. Applying Theorem 2, statement (c), we conclude that X and Y are bounded. The manifold , that is formed by the points (x 1 , 0), is a sliding region and the Filippov vector field is F Z (x 1 , 0) = (1, 0). This vector field has no equilibrium point and the trajectory of F Z is unbounded, as shown in Fig. 6. Thus Z = (X, Y ) is unbounded. 2×2 , ϕ and ψ as in (1). In this subsection and in the following we consider

Piecewise Bounded Quadratic Systems in the Plane
Observe that the equilibrium point of X and Y are now the points p and q respectively and = {( Similarly as defined in the first subsection, we say that a piecewise quadratic system in the plane is bounded if all its trajectories are bounded for t 0. In the introduction we mentioned that if X and Y are bounded quadratic vector fields, this does not imply that Z = (X, Y ) is bounded and the example that was presented is unbounded in the sliding region s . The vector field Z = (X, Y ) with X and Y bounded, can have an unbounded trajectory ) passing by crossing points as shown the example below.
Second proof of Proposition 1 Consider the vector field Z ∈ Q r ( , f ) given by The lines x 2 = −1 and x 2 = 1 are lines of equilibrium points of X and Y respectively.
Our main result is the following.
In what follows when we refer the coefficients of the matrix of the vector field X we use a i j and for the vector field Y we use b i j .

Theorem 4 Let Q pq = Q p × Q q be a subset of Q r ( , f ) as described above and
i j is bounded. Moreover the boundness is characterized by the sign of rational functions depending on the parameters (a 11 , a 12 , a 21 , a 22 , b 11 , b 12 , b 21 , b 22 ).
Theorem 4 characterizes the non-smooth vector fields Z = (X, Y ) ∈ Q pq which are bounded. In the third section we prove Theorems 3 and 4. To prove Theorem 4, first of all we compute X f · Y f on . We prove (see Lemma 5) that either X f · Y f ≡ 0 or has 0, 1 or 2 tangent points. If X f · Y f ≡ 0 then Z is bounded. Besides, if has 0, 1 or 2 tangent points then Lemmas 7, 8 and 11, respectively, give the conditions for the boundness.

Proof of Theorems 3 and 4
In this section we proof our main results.

Proof of Theorem 3 (a)
Suppose that = s . Thus the Filippov vector field F Z is an one dimensional vector field. Therefore Z is bounded if and only if F Z (x 1 ) ≤ 0 near x 1 = +∞ and F Z (x 1 ) ≥ 0 near x 1 = −∞. If # t = 1 and s is unbounded again the only possibility of Z be unbounded, occurs in sliding region s because if Z has an unbounded trajectory passing by the crossing region c then there exists at least two distinct points (z, 0) and (w, 0) of c with X f (z, 0), Y f (z, 0) > 0 and X f (w, 0), Y f (w, 0) < 0. Thus exist y t = x t such that X f (y t ) · Y f (y t ) = 0. The last case, = s ∪ c ∪ t with c bounded, just make the study of the signal of X f · Y f . The proof is analogous previous arguments (b). If = c then the only possibility of a trajectory of Z be unbounded occurs if it contains at least two distinct points of c . In this case there exists z such that X f (z, 0) = 0 or Y f (z, 0) = 0 and thus (z, 0) ∈ t . Therefore Z is bounded.

Proof of Theorem 4
In the following theorems we denote by Q 0 t the set of vector fields Z ∈ Q pq without tangent points, by Q 1 t with exactly one tangent point and by Q 2 t with two tangent points. We use the notation Then Z have at most 4 tangent points. In fact, the tangent points are solutions of equation  p 2 ) and q = (q 1 , q 2 ) are equilibrium points of X and Y respectively. Thus , and . Then 0) is a polynomial of degree less than or equal to 4 depending on the coefficients. Thus it has at most 4 real zeros.
If Z = (X, Y ) ∈ Q pq and [X f · Y f ](x 1 ) = 0 for some x 1 ∈ R, then the non-smooth vector field Z has at most 2 tangent points. In fact it follows immediately since a 1 = a 2 = a 4 = b 1 = b 2 = b 4 = 0 in each one of the cases for Q p i . Now we prove the boundness of Z when X f · Y f ≡ 0 on . In this case we conclude that = t . If a trajectory of (X, Y ) is unbounded then either the trajectory passes by crossing region and returns or the trajectory escapes by sliding region. Since c = s = ∅, it cannot happen.
and A (1) = (a i j ) 2×2 with a 11 < 0, a 12 = 0, a 22 0, B (1) (1) (y) = (0, y 1 y 2 ) T with y = (y 1 , y 2 ). If = s , the expression of the Filippov vector field F Z (seen as one dimensional) is (i) Suppose that b 21 = q 2 and a 21 = p 2 . Thus A 2 = a 11 (b 21 − q 2 ) − b 11 (a 21 − p 2 ) and where the symbol ∞ represents ±∞. Therefore Z is bounded if and only if (ii) Consider now b 21 = q 2 and a 21 = p 2 . Thus A 1 = b 11 a 22 p 2 − a 11 b 22 q 2 and B 0 = a 22 p 2 − b 22 q 2 and we have lim Therefore Z is bounded if and only if is a polynomial of degree 1 or 2. In the first case, we have = s ∪ t ∪ c with s and c unbounded. In the other case, = s ∪ t or = c ∪ t . The only possibility of Z be unbounded, occurs in sliding region s because if Z has an unbounded trajectory passing by the crossing region c then there exists at least two distinct points (z, 0) and (w, The remainder of proof is analogous to proof of the statement (a) of the Lemma 7.
Remark 9 If = c ∪ t we cannot ensure that the system is bounded. See the following example.
Example 10 Consider the vector field Z given by The lines x 2 = −1 and x 2 = 1 are lines of equilibrium points of X and Y respectively. We have t = {3}, s = ∅ and c = {x 1 ∈ R : x 1 = 3}. Fig. 8 shows an unbounded trajectory when = c ∪ t .

and n m. If s is composed only by equilibrium points of F Z then Z is trivially bounded.
Proof To verify that = s ∪ c ∪ t with c bounded or = s ∪ c ∪ t with s bounded, just make the study of the signal of X f · Y f . The proof is analogous arguments to the proof of statement (a) of the Lemma 7.

Remark 12 If
= s ∪ c ∪ t we cannot ensure that the system is bounded. See for instance, the example in the second proof of Proposition 1.
To define the set B L i j , stated in Theorem 4, we consider the set of quadratic systems (X, Y ) ∈ B i j that satisfies the conditions given in the Lemmas 5, 7, 8 and 11. It concludes the proof of Theorem 4.

Application: The Mass-Spring System
Consider the vertical mass-spring system, as shown in Fig. 9. The object of mass m when displaced from its equilibrium position s, experiences a restoring force F 1 proportional to the displacement x. This force is described as F 1 = −k(s + x) where k is a positive constant, known as the spring constant. Suppose that a frictional force (damping), for example the air resistance, is also present. This force F 2 is proportional to the velocity and is described as F 2 = −βẋ, where β is a positive constant. Consider the action of gravity F 3 = mg, where g ∼ = 9, 8m/s 2 . The resultant force is F = F i and using the Newton's second law and observing that −ks = mg the we write mẍ = −kx − βẋ.
We want that the planar system to be quadratic and bounded and for this is sufficient chooseẋ = y 2 − cx with a convenient constant c. Thus we write the planar systeṁ x = −cx + y 2 ,ẏ = mc − β 2m y − mc 2 − βc + k 2m (x/y). Fig. 9 The mass-spring system under the action of two forces F (1) 2 and F (2) 2 given by quadratic term y 2 in the system (6) Imposing the condition mc 2 − βc + k = 0, we find the values of c such that the system (6) is quadratic. These values are We admit β 2 − 4mk > 0 and observe that c 1 , c 2 > 0. Now, consider X (x, y) and Y (x, y) the vector fields given by (6) with the respective constants c 1 and c 2 . Observe that mc 1 − β 2m and mc 2 − β 2m are negative constants. Therefore, by Theorem 2, the quadratic systems given by X and Y are bounded. Let Z = (X, Y ) be a non-smooth quadratic system with f (x, y) = y. Let p = ( p 1 , p 2 ) be the equilibrium point of X and q = (q 1 , q 2 ) of Y . The switched manifold = f −1 (0) is interpreted as the interrupt of each one of the forces (given by y > 0 and y < 0) to start the other. Thus x − q 1 y − q 2 + (y − q 2 ) 2 0 if y 0.
We have Z ∈ B 22 and X (x, 0) · Y f (x, 0) = mc 1 − β 2m If p 2 q 2 = 0 then by Lemma 5, Z is bounded. Now consider p 2 q 2 = 0. In this case no have tangent points ( t = ∅). If p 2 q 2 > 0 then = c ( s = ∅) and by item (b) of the Lemma 7, Z = (X, Y ) is bounded. If p 2 q 2 < 0 then = s ( c = ∅) and by item (a) of the Lemma 7 we conclude that Z = (X, Y ) is bounded since R .