On the existence of limit cycles and invariant surfaces of sewing piecewise linear differential systems on \mathbb{R}^3
OON THE EXISTENCE OF LIMIT CYCLES AND INVARIANTSURFACES OF SEWING PIECEWISE LINEAR DIFFERENTIALSYSTEMS ON R BRUNO R. DE FREITAS AND JO ˜AO C. MEDRADO
Abstract.
We consider a class of discontinuous piecewise linear differential sys-tems in R with two pieces separated by a plane. In this class we show thatthere exist differential systems having: ( i ) a unique limit cycle, ( ii ) a unique one-parameter family of periodic orbits, ( iii ) scrolls, ( iv ) invariant cylinders foliatedby orbits which can be periodic or no. Introduction and Main Results
The discontinuous piecewise linear differential systems plays an important roleinside the theory of nonlinear dynamical systems. In the models of physical problemsor processes is natural to use the piecewise-smooth dynamical systems when theirmotion is characterized by smooth flow and eventually, interrupted by instantaneousevents (see [1], [4], [8]). There are many non-smooth processes in this context,for example, impact, switching, sliding and other discrete state transitions. Theyare used also in nonlinear engineering models, where certain devices are accuratelymodeled by them, see for instance [5], [15], [19] and, references quoted in these.There are several papers about the existence of limit cycles for discontinuouspiecewise linear differential systems on the plane. The study of this class of dif-ferential systems on R is beginning. In [3], [17] and [18], the authors consider afamily of continuous piecewise linear systems in R and characterize limit cycles andcones foliated by periodic orbits. In [13] is proved the existence of limit cycles andinvariant cylinders for a class of discontinuous vector field in dimension 2 n .In this work, we consider a class of discontinuous piecewise linear differentialsystems in R with two pieces separated by a plane and we investigate the existenceof limit cycles and invariant surfaces. In this way, we give conditions for the existenceof differential systems having: ( i ) a unique limit cycle (Figure 6); ( ii ) a uniqueone-parameter family of periodic orbits (Figure 7); ( iii ) scrolls (Figure 1.( a )); ( iv )invariant cylinders foliated by orbits periodic or no (Figure 1.( b ) and 1.( c )).We note that the existence of the one-parameter family of periodic orbits is ananalogous result to the Lyapunov Center Theorem related to smooth vector fields.See, for instance [2], [9], [10] and, [12].This work is an extension of [16] and in our approach we use essentially theTheorem of Rolle for dynamical systems (see [11]) to address the problem of toshow the existence of limit cycles or invariant surfaces to find zeroes of intersectionof algebraic curves. Mathematics Subject Classification.
Primary: 34C25, 34C07, 37C27.
Key words and phrases.
Uniqueness of limit cycle, surface of periodic orbits, piecewise sewinglinear differential system. a r X i v : . [ m a t h . D S ] A ug B.R. DE FREITAS AND J. C. MEDRADO
Let X ± (x) = A ± (x) + B ± be polynomial vector fields on R of degree one wherex = ( x, y, z ), A ± = ( a ± ij ) × , B = ( b ± , b ± , b ± ) and h the real function given by h ( x, y, z ) = z . We consider the piecewise linear vector field Z = ( X + , X − ) definedby Z ( x, y, z ) = (cid:26) X + ( x, y, z ) , if ( x, y, z ) ∈ Σ + ,X − ( x, y, z ) , if ( x, y, z ) ∈ Σ − , (1)where Σ = h − (0) and Σ ± = ± h >
0. We observe that R = Σ ∪ Σ + ∪ Σ − .On Σ + (Σ − ) the orbit ϕ Z ( t, p ) of the piecewise vector field Z is given by the orbit ϕ + X ( t, p )( ϕ − X ( t, p )) of X + ( X − ), respectively. At points of Σ the orbit of Z is bivaluedon the sewing region given by Σ S = { p ∈ Σ; X + h ( p ) X − h ( p ) > } or in the tangencyset Σ T = { p ∈ Σ; X + h ( p ) = 0 or X − h ( p ) = 0 } . If a point p belongs to the Σ S orΣ T , the orbit of Z through this point is given by the union of the orbits of X + and X − through this same point.We say that an orbit γ of vector field Z = ( X + , X − ) is a Σ − sewing periodic orbit or simply, Σ − periodic orbit , if γ is closed and γ ∩ Σ S (cid:54) = ∅ . In Σ \ (Σ S ∪ Σ T ) thedefinition of the orbit of Z is given following the convetion of Filippov in [6] .The tangency set Σ T of Z is formed by the tangency straight lines L X + and L X − ,i.e., Σ T = L X + ∪ L X − , where L X ± = { p ∈ Σ : X ± h ( p ) = 0 } . If p ∈ Σ T it can be of fold or cusp of piecewise vector field Z . In the first case we say that p is a visible(invisible) fold of Z if sgn(( ± X ± ) h ( p ) > < p ∈ Σ T is not afold of Z but ( X + ) h ( p ) (cid:54) = 0 or ( X − ) h ( p ) (cid:54) = 0 then p is a cusp of Z .In this work, we deal with piecewise linear vector field Z = ( X + , X − ) expressedby (1) with Σ = Σ S ∪ Σ T , whereΣ T ⊂ Σ T = { p ∈ Σ T : p is an invisible fold point of X + or X − } . So, we get the canonical form for to the vector field Z = ( X + , X − ) expressed by X + = ( a + x + b + z, c + y + d + z − , y ) ,X − = ( a − x + b − z + m, c − y + d − z + 1 , y ) . (2)The eigenvalues of X + and X − are λ ± = a ± , λ ± = ( c ± + (cid:112) ( c ± ) + 4 d ± ) / λ ± = ( c ± − (cid:112) ( c ± ) + 4 d ± / . We observe that Z = ( X + , X − ) has distinct dynamics depending of values ofthese eigenvalues. In order to analyze all possible dynamic types of X ± and becomeeasier the presentation of our results, we define seven types:( i ) Sa : If λ ± λ ± < ii ) No : If λ ± , λ ± ∈ R and λ ± λ ± > iii ) Nd : If λ ± = λ ± .( iv ) Fo : If λ ± , λ ± ∈ C and Re( λ ± ) Im( λ ± ) (cid:54) = 0.( v ) Ce : If λ ± , λ ± ∈ C , Re( λ ± ) = 0 and, Im( λ ± ) (cid:54) = 0.( vi ) D : If λ ± λ ± = 0 and ( λ ± ) + ( λ ± ) (cid:54) = 0.( vii ) D : If ( λ ± ) + ( λ ± ) = 0.We remark that in the types D and D , X ± does not have equilibrium points. Definition 1.
We say that the piecewise vector field Z = ( X + , X − ) is of type ( T + , T − ) for T ± ∈ { Sa , No , Nd , Fo , Ce , D , D } , if X ± is of type T ± . We observe that the type ( T + , T − ) is equal to ( T − , T + ) i.e., there is an equivalencebetween ( T + , T − ) and ( T − , T + ), for details see [7]. IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R In this paper we prove the following main theorems.
Theorem 2.
Let Z = ( X + , X − ) be a piecewise linear vector field of type ( T + , T − ) .The following statements hold.(1) The vector field Z has scrolls (see Figure 1 ( a ) ) if only if is true one of thefollowing conditions:(a) T + = Sa and T − ∈ { Sa , No , Nd , Fo , Ce , D } ;(b) T + = No and T − ∈ { No , Nd , Fo , Ce , D , D } ;(c) T + = Nd and T − ∈ { Nd , Fo , D , D } ;(d) T + = Fo and T − = D ;(e) T + = D and T − ∈ { D , D } ;with κ + λ (cid:54) = 0 and κλ ≥ , or α λ/κ ≤ and κλ < .(2) The vector field Z has at most a unique invariant cylinder (see Figure 1 ( b ) )if only if is true one of the following conditions:(a) T + = Sa and T − ∈ { Sa , No , Nd , Fo , Ce , D , D } ;(b) T + = No and T − ∈ { No , Nd , Fo , Ce , D } ;(c) T + = Nd and T − ∈ { Nd , Fo , Ce , D } ;(d) T + = Fo and T − ∈ { Fo , Ce , D , D } ;(e) T + = Ce and T − = D ;with κλ < and α λ/κ > .(3) The vector field Z has infinitely many invariant cylinders (see Figure 1 ( c ) )if only if κ = λ = 0 and is true one of the following conditions:(a) T + = Sa and T − ∈ { Sa , No , Nd , Fo , Ce , D } ;(b) T + = No and T − ∈ { No , Nd , Fo , Ce , D } ;(c) T + = Nd and T − ∈ { Nd , Fo , D } ;(d) T + = Ce and T − ∈ { Ce , D } ;(e) T + = D and T − = D ;(f ) T + = D and T − = D .The parameters κ and λ depend on the parameters a ± , b ± , c ± , d ± and m of Z . Theseparameters are given in the Tables 1 and 2. ( b ) ( c )( a ) Figure 1.
Invariant surfaces of (2): ( a ) Scroll. ( b ) A unique invariantcylinder. ( c ) Infinitely many invariant cylinders. B.R. DE FREITAS AND J. C. MEDRADO
Table 1.
We present the expressions of κ and λ that depend of pa-rameters of the vector field Z . We note that α and β are given inLemmas 9 and 10, respectively.( T + , T − ) κ and λ (Sa,Sa) κ = α ( αc − + βc + − c + − c − )( αβc − − βc + − βc − + c + ) ,λ = ( αβc + − αc + − αc − + c − )( αβc + + αβc − − αc + − βc − ) . (Sa,No) κ = − ( βc + − c + + ( − α ) c − )(( c + + ( − α + 1) c − ) β − c + ) α ,λ = ((( c + + c − ) β − c + ) α − βc − )(( βc + − c + − c − ) α + c − ) . (Sa,Nd) κ = α (( − α + 1) β + c + ) ,λ = − (( β + c + ) α − β ) . (Sa,Fo) κ = 4 α (( β + 1 / − α ) ( c − ) − c + ( − α ) c − + ( c + ) ) ,λ = − (4( β + 1 / − α ) ( c − ) − c + α ( − α ) c − − c + ) α . (Sa,Ce) κ = (( − α ) β + ( c + ) ) α ,λ = − ( − α ) β − ( c + ) α . (Sa,D ) κ = − α (( − α + 1) c − + c + ) ,λ = ( c + + c − ) α − c − . (No,No) κ = − ( βc + − c + + ( − α − c − )(( c + + (1 + α ) c − ) β − c + ) α ,λ = (( βc + − c + − c − ) α − c − )((( c + + c − ) β − c + ) α + βc − ) . (No,Nd) κ = ((1 + α ) β + c + ) α ,λ = − (( β + c + ) α + β ) . (No,Fo) κ = (4((1 + α ) ( β + 1 / c − ) + c + (1 + α ) c − + ( c + ) )) α ,λ = − α ) ( β + 1 / c − ) − c + α (1 + α ) c − − c + ) α . (No,Ce) κ = α ((1 + α ) β + ( c + ) ) ,λ = ( − β − c + ) ) α − αβ − β . (No,D ) κ = α ((1 + α ) c − + c + ) ,λ = ( − c + − c − ) α − c − . (Nd,Nd) κ = − ( α + β ) ,λ = 2 β. (Nd,Fo) κ = − (4 β ( c − ) + 4 α + 4 αc − + ( c − ) ) ,λ = (8 β + 2)( c − ) + 4 αc − . (Nd,D ) κ = − ( αβ + 1) ,λ = 1 . (D ,Fo) κ = 4(1 + c − α ) + ( c − ) α (4 β + 1) ,λ = − ( c − ) α (4 β + 1) . (D ,D ) κ = λ = α + β. Table 2.
In this table are given κ and λ for the cases that they are constant.( T + , T − ) κ and λ (Sa,D ), (Ce,Nd), (Ce,Fo), (Ce,D ) and (D ,Fo) κ = 1 and λ = − ), (Nd,D ) and (D ,D ) κ = λ = 1.(Ce,Ce), (D ,Ce) and (D ,D ) κ = λ = 0. IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R Theorem 3.
Let Z = ( X + , X − ) be a piecewise linear vector field with X + and X − defined in (2) . The following statements hold.(1) If ( a + ) + ( a − ) = 0 or Z has no invariant cylinder then there are not limitcycles.(2) If ( a + ) + ( a − ) (cid:54) = 0 and Z has at most a unique invariant cylinder then Z has at most a unique limit cycle in this cylinder.(3) If ( a + ) + ( a − ) (cid:54) = 0 , a + a − ≥ and Z has infinitely many invariant cylin-ders then there is an invariant surface formed of periodic orbits, where eachperiodic orbit is contained in an invariant cylinder. The statement (3) of Theorem 3 is similar a Lyapunov Center Theorem relatedto smooth vector fields.In the Section 2 we obtain the canonical form (2) and characterize the tangencialsets (straight lines of tangency) formed by orbits of Z and Σ. The Theorems 2 and 3will be proved in Section 3.2. The tangency straight lines and canonical forms of Z The purpose of this section is to characterize the tangency straight lines L X ± andto obtain a canonical form to the piecewise vector field (1) such that Σ = Σ S ∪ Σ T with Σ T = L X + ∪ L X − , where L X ± = { p ∈ Σ : X ± h ( p ) = 0 and ( ± X ± ) h ( p ) < } and the sets Σ T , L X + and L X − are not empty. As L X + and L X − are not empty weget that ( a ± ) + ( a ± ) (cid:54) = 0. In the sequel, we will assume these hypothesis. Lemma 4.
Let Z = ( X + , X − ) be defined in (1) with L X ± the tangency straightlines of X ± . If X + h ( p ) X − h ( p ) ≥ , for all p ∈ Σ then the tangency straight linesare the same, i.e., L X + ≡ L X − .Proof. Each tangency straight line L X + or L X − separate Σ in two regions where thesignals of X ± ( p ) are opposite. So, by the hypothesis we get L X + ≡ L X − and withoutloss of generality we can assume that the tangency straight line contains the origin.Moreover, we get that L ± X in Σ is given by y = 0. (cid:3) Using the Lemma 4 and rescaling the time, we have that the vector fields X + and X − can be written by X + ( x, y, z ) = ( a +11 x + a +12 y + a +13 z + b +1 , a +21 x + a +22 y + a +23 z + b +2 , y + a +33 z ) ,X − ( x, y, z ) = ( a − x + a − y + a − z + b − , a − x + a − y + a − z + b − , y + a − z ) . (3) Lemma 5.
Let X + be as given in (3) . The tangency straight line L X + is charac-terized by one of the following statements:(1) If a +21 = 0 then for all p ∈ L X + , p is a visible or invisible fold when b +2 > or b +2 < , respectively;(2) If a +21 = b +2 = 0 then L X + is invariant by X + ;(3) If a +21 (cid:54) = 0 then there is a unique point p ∈ L X + such that it is a cusp (if a +21 b +1 − a +11 b +2 (cid:54) = 0 ) or a singular point of X + (if a +21 b +1 − a +11 b +2 = 0 ).Proof. From (3) we get X + h ( x, y,
0) = y that implies L X + = { y = z = 0 } and( X + ) h ( x, ,
0) = a +21 x + b +2 . If a +21 = 0 we get two possibilities: ( i ) when b +2 (cid:54) = 0,the point ( x, ,
0) is a fold for all x ∈ R (statement 1); ( ii ) when b +2 = 0, we havethat L X + is invariant by the vector field X + (statement 2). Finally, for a (cid:54) = 0, B.R. DE FREITAS AND J. C. MEDRADO follows from X h ( x, ,
0) = 0 that ( X + ) h ( − b +2 /a +21 , ,
0) = − a +11 b +2 + b +1 a +21 . Thus,if − a +11 b +2 + b +1 a +21 (cid:54) = 0 the point ( − b +2 /a +21 , ,
0) is a cusp, otherwise is a singularpoint. (cid:3)
Proposition 6.
Let Z = ( X + , X − ) the vector field defined as (1) . Suppose that X + hX − h ( p ) ≥ and that L X ± ≡ L X ± . Then the vectors field X + and X − can bewritten as (2) .Proof. From Lemmas 4 and 5 we get L X ± = { y = z = 0 } , a ± = 0, and thatsgn( ± b ± ) <
0. Doing the twin change of variables given by ϕ + ( x, y, z ) = ( x − ( b +1 /b +2 ) y − (( b +1 a +11 − b +1 a +22 + a +12 b +2 ) /b +2 ) z, y + a +33 z, − b +2 z ) on Σ + , ϕ − ( x, y, z ) = ( x − ( b +1 /b +2 ) y − (( b +2 a − − b +1 a − + b +1 a − ) /b +2 ) z, y + a − z, b − z ) on Σ − ,and rescaling the time t → − t/b +2 on Σ + and t → , t/b − on Σ − we obtain the canonicalform (2). (cid:3) Proposition 7.
Consider the boundary value problem ˙x = Y ± (x) = P x + Q ± with x = ( x (0) , y (0) , z (0)) = ( x , y , , P = γ δ σ ψ and Q ± = M ± , where γ, δ, σ, ψ, M ∈ R .Let ϕ ± ( t, ( x , y , be the solutions of ˙x = Y ± (x) and consider the straight line r = { ( x, y, z ) ∈ R : y = y , z = 0 } . Let τ ± ∈ R / { } such that z ( τ ± ) = 0 then ϕ ± ( τ ± , r ) is a straight line parallel to r given by r = { ( x, y, z ) ∈ R : y = y , z =0 } .Proof. The general solution ϕ ± ( t, ( x , y , P t x + e P t (cid:90) t e − P η
Qdη.
Observe that the matrix e
P t has zeroes at positions (2 ,
1) and (3 , ϕ ± ( t, ( x , y , x ± ( t ) , y ± ( t ) , z ± ( t )) where x ± ( t ) = e γt x + f ± y + f ± ,y ± ( t ) = f ± y + f ± ,z ± ( t ) = f ± y + f ± , (4)where f ± ij = f ± ij ( t, γ, δ, σ, ψ, M ) , for i, j = 1 , , τ ± ( y ) such that z ± ( τ ± ( y )) = 0. Then y = y ± ( τ ± ( y )) depends only of y . This implies thatall orbits of Y ± with origin at r intersect Σ = { z = 0 } after time τ ± ( y ), i.e., ϕ ± ( τ ± ( y ) , r ) is the straight line r . (cid:3) Follows from of Proposition 7 that if Z has a periodic orbit then it is in a invariantcylinder. So, all periodic orbit are 1 − periodic. Corollary 8.
Consider the boundary value problems ( A ) : ˙x = X + (x) , ( x (0) , y (0) , z (0)) = ( x , y , , ( x ( τ ) , y ( τ ) , z ( τ )) = ( x , y , , ( B ) : ˙x = X − (x) , ( x (0) , y (0) , z (0)) = ( (cid:101) x , (cid:101) y , , ( x ( τ ) , y ( τ ) , z ( τ )) = ( (cid:101) x , (cid:101) y , . IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R where X + and X − are given by (2) . If y = (cid:101) y and y = (cid:101) y then there is an invariantcylinder for the vector field Z = ( X + , X − ) .Proof. Let ϕ ± ( t, p ) be the solutions of ( A ) and ( B ) respectively and the straightlines r = { ( x, y, z ) ∈ R : y = y , z = 0 } and r = { ( x, y, z ) ∈ R : y = y , z = 0 } .From Proposition 7, we have that ϕ + ( r , τ ) = r and ϕ − ( r , τ ) = r . So, we obtainan invariant cylinder (See Figure 2). (cid:3) y y y y X + X − Figure 2.
Superior half cylinder and inferior half cylinder.In the next section we will determine a maximum quota for the number of invariantcylinders and in the sequel a maximum quota for the number of cycles in eachcylinder. 3.
Proof of Theorems
The proofs of Theorems 2 and 3 is done in the subsections 3.1 and 3.2, respectively,except when Z is of type (Fo,Fo) which is proved in the subsection 3.3.In the subsections 3.1 and 3.2 we present a complete proof considering Z of type(Sa,Sa) under hypothesis given in the statements (1), (2) and (3) of Theorems 2and 3 because for all other cases, except (Fo,Fo), the proof is done in similar way.We present in the Tables 1 and 2 the distinguished elements used in the proof.Finally, we conclude the proofs of these theorems in the subsection 3.3 when Z is ofthe type (Fo,Fo).In order to prove these theorems, when the return applications is defined, wemake a substitution of variables given by Propositions 9 and 10 and we address theproof to determine intersection points of curves which are associated to existenceof invariant cylinders. For to determine the number of intersection points betweenthese curves, we use also Theorem 11 proved by Kovanskii([11]). Lemma 9.
Consider the boundary value problem x (cid:48) = a + x + b + z,y (cid:48) = c + y + d + z − ,z (cid:48) = y, ( x (0) , y (0) , z (0)) = ( x , y , , ( x ( τ ) , y ( τ ) , z ( τ )) = ( x , y , , (5) where a + (cid:54) = 0 . Then the next statements are true. B.R. DE FREITAS AND J. C. MEDRADO ( i ) If X + is of the type Sa doing ( ρ, v, w ) = (e a + τ , e − ( c + + s ) τ/ , e ( c + − s ) τ/ ) thenwe get v = ρ α , w = ρ α and w = v α , where α = − ( c + + s ) / a + , α =( c + − s ) / a + , s = (cid:112) ( c + ) + 4 d + , α = ( s − c + ) / ( s + c + ) . Moreover, w = c + y − sy − c + y − sy − , v = c + y + sy − c + y + sy − ,ρ = 4( a + ) x − a + ) c + x + a + ( c + ) x − a + s x + 4 a + b + y − b + a + ) x − a + ) c + x + a + ( c + ) x − a + s x + 4 a + b + y − b + . (6)( ii ) If X + is of the type No , doing ( ρ, v, w ) = (e a + τ , e ( c + + s ) τ/ , e ( c + − s ) τ/ ) thenwe get v = ρ α , w = ρ α and w = v α , where α = ( c + + s ) / a + , α =( c + − s ) / a + , s = (cid:112) ( c + ) + 4 d + , α = ( c + − s ) / ( s + c + ) . Moreover, w = c + y − sy − c + y − sy − , v = c + y + sy − c + y + sy − ,ρ = 4( a + ) x − a + ) c + x + a + ( c + ) x − a + s x + 4 a + b + y − b + a + ) x − a + ) c + x + a + ( c + ) x − a + s x + 4 a + b + y − b + . ( iii ) If X + is of the type Nd , doing ( ρ, v, w ) = (e a + τ , e c + τ/ , c + τ / then we get v = ρ α , w = ln( ρ α ) and w = ln( v ) , where α = c + / a + , α = c + / , w = 2 c + ( y − y )( c + y − cy − , v = cy − cy − ,ρ = 4( a + ) x − a + ) c + x + a + ( c + ) x + 4 a + b + y − b + a + ) x − a + ) c + x + a + ( c + ) x + 4 a + b + y − b + . ( iv ) If X + is of the type Fo , doing ( ρ, v, w ) = (e a + τ , e c + τ , tan( sτ / then weget v = ρ α , w = tan(ln( ρ α )) and w = tan(ln( v α )) where α = c + /a + ,α = s/ a + , s = (cid:112) − (( c + ) + 4 d + ) , α = s/ c + , w = 2 s ( − y + y )(( c + ) + s ) y − c + ) y − c + y + 4 , v = 4 − c + y + (( c + ) + s ) y − c + y + (( c + ) + s ) y ρ = 4 x ( a + ) − x ( a + ) c + + ((( c + ) + s ) x + 4 b + y + 1) a − b + a + ) x − c + ( a + ) x + ((( c + ) + s ) x + 4 b + y ) a + − b + . ( v ) If X + is of the type Ce , doing ( ρ, v, w ) = (e a + τ , cos( ατ ) , sin( ατ )) then we get v = cos(ln( ρ α/a + )) , w = sin(ln( ρ α/a + )) and w + v = 1 , where α = √− d + , w = α ( y − y ) α y + 1 , v = α y y + 1 α y + 1 ,ρ = a x + aα x + aby − ba x + aα x + aby − b . ( vi ) If X + is of the type D , doing ( ρ, v, w ) = (e a + τ , e c + τ , c + τ ) then we get v = ρ α , w = ln( ρ α ) and w = ln( v ) , where α = c + /a + , α = 1 /c + , w = y − y , v = c + y − c + y − ,ρ = ( a + ) x − ( a + ) c + x + a + b + y − b + ( a + ) x − ( a + ) c + x + a + b + y − b + . IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R ( vii ) If X + is of the type D , doing ( ρ, v, w ) = (e a + τ , τ, τ ) then we get v =ln( ρ α ) , w = (ln( ρ α )) and w = v , where α = 1 /a + , w = 2 y ( y − y ) , v = y − y ,ρ = ( a + ) x + a + b + y − b + ( a + ) x + a + b + y − b + . Proof. ( i ) By the Proposition 7 the solution ( x ( t ) , y ( t ) , z ( t )) is(e λ t x + ηy + µ, λ e λ t − λ e λ t s y ± e λ t − e λ t s , e λ t − e λ t s y ∓ e λ t − e λ t − sλ λ ) , where λ = a + , λ , = ( c + ± s ) / η, µ ) obtained replacing ( δ, M, σ, ψ, ±
1) =( b + , , c + , d + , − { x ( τ ) = x , y ( τ ) = y , z ( τ ) = 0 } in { ρ, v, w } where ( ρ, v, w ) = (e a + τ , e − ( c + + s ) τ/ , e ( c + − s ) τ/ ) and we obtain (6).The remaining cases are obtained analogously. (cid:3) In similar way we obtain analogous result associated to X − . Lemma 10.
Consider the boundary value problem x (cid:48) = a − x + b − z + m,y (cid:48) = c − y + d − z + 1 ,z (cid:48) = y, ( x (0) , y (0) , z (0)) = ( (cid:101) x , (cid:101) y , , ( x ( τ ) , y ( τ ) , z ( τ )) = ( (cid:101) x , (cid:101) y , , (7) where a − (cid:54) = 0 . Then the next statements are true. ( i ) If X − is of the type Sa and No , doing ( ξ, V, W ) = (e a − τ , e − ( c − + S ) τ/ , e − ( − c − + S ) τ/ ) then we get V = ξ β , W = ξ β and W = V β , where β = − ( c − + S ) / a − , β = ( c − − S ) / a − , S = (cid:112) ( c − ) + 4 d − , β = ( S − c − ) / ( S + c − ) . Moreover, W = S (cid:101) y − c − (cid:101) y − S (cid:101) y − c − (cid:101) y − , V = S (cid:101) y + c − (cid:101) y + 2 S (cid:101) y + c − (cid:101) y + 2 ,ξ = ( S + 4 a − ( c − − a − )) m + ( S + c − (4 a − − c − ) − a − ) ) a − (cid:101) x − b − ( a − (cid:101) y + 1)( S + 4 a − ( c − − a − )) m + ( S + c − (4 a − − c − ) − a − ) ) a − (cid:101) x − b − ( a − (cid:101) y + 1) . (8)( ii ) If X − is of the type Nd , doing ( ξ, V, W ) = (e a − τ , e c − τ/ , c − τ / then we get V = ξ β , W = ln( ξ β ) and W = ln( V ) , where β = c − / a − , β = c − / , W = 4 c − ( (cid:101) y − (cid:101) y )2( c − (cid:101) y + 2)( c − (cid:101) y + 2) , V = c − (cid:101) y + 2 c − (cid:101) y + 2 ,ξ = (( c − ) − c − a − + 4( a − ) ) m + (( c − ) a − − c − ( a − ) + 4( a − ) ) (cid:101) x + 4 b − ( a − (cid:101) y + 1)(( c − ) − c − a − + 4( a − ) ) m + (( c − ) a − − c − ( a − ) + 4( a − ) ) (cid:101) x + 4 b − ( a − (cid:101) y + 1) . ( iii ) If X − is of the type Fo , doing ( ξ, V, W ) = (e a − τ , e c − τ , tan( Sτ / then we get V = ξ β , W = tan(ln( ξ β )) and W = tan(ln( V β )) where β = c − /a − , β = S/ a − , S = (cid:112) − (( c − ) + 4 d − ) , β = S/ c − , W = 2 S ( (cid:101) y − (cid:101) y ) S (cid:101) y (cid:101) y + ( c − ) (cid:101) y (cid:101) y + 2 c − (cid:101) y + 2 c − (cid:101) y + 4 , V = S (cid:101) y + ( c − ) (cid:101) y + 4 c − (cid:101) y + 4 S (cid:101) y + ( c − ) (cid:101) y + 4 c − (cid:101) y + 4 ξ = ( S + ( c − ) + 4 a − ) m + (( S + ( c − ) ) a − + 4( a − ) ( a − − c − )) (cid:101) x + 4 b − ( a − (cid:101) y + 1)( S + ( c − ) + 4 a − ) m + (( S + ( c − ) ) a − + 4( a − ) ( a − − c − )) (cid:101) x + 4 b − ( a − (cid:101) y + 1) . ( iv ) If X − is of the type Ce , doing ( ξ, V, W ) = (e a − τ , cos( βτ ) , sin( βτ )) then weget V = cos(ln( ξ βa − )) , W = sin(ln( ξ βa − )) and W + V = 1 , where β = √− d − and V = − β (cid:101) y (cid:101) y − − β (cid:101) y − , W = β ( (cid:101) y − (cid:101) y ) − β (cid:101) y − ,ξ = β a − (cid:101) x + ( a − ) (cid:101) x + β m + b − a − (cid:101) y + m ( a − ) + b − β a − (cid:101) x + ( a − ) (cid:101) x + β m + b − a − (cid:101) y + m ( a − ) + b − . ( v ) If X − is of the type D , doing ( ξ, V, W ) = (e a − τ , e c − τ , c − τ ) then we get V = ξ β , W = ln( ξ β ) and W = ln( V ) , where β = c − /a − , β = 1 /c − , W = c − ( (cid:101) y − (cid:101) y ) , V = c − (cid:101) y + 1 c − (cid:101) y + 1 ,ξ = − c − ( a − ) (cid:101) x + ( a − ) (cid:101) x + b − a − (cid:101) y − c − ma − + m ( a − ) + b − − c − ( a − ) (cid:101) x + ( a − ) (cid:101) x + b − a − (cid:101) y − c − ma − + m ( a − ) + b − . ( vi ) If X − is of the type D , doing ( ξ, V, W ) = (e a − τ , τ , τ ) then we get V =ln( ξ β ) , W = (ln( ξ β )) and W = V where β = 1 /a − , W = 2 (cid:101) y ( (cid:101) y − (cid:101) y ) , V = (cid:101) y − (cid:101) y ,ξ = ( a − ) (cid:101) x + b − a − (cid:101) y + m ( a − ) + b − ( a − ) (cid:101) x + b − a − (cid:101) y + m ( a − ) + b − . Proof.
The proof is analogous to the proof of Lemma 9. (cid:3)
Theorem 11. (Kovanskii, [11] ) Let X be a C planar vector field without singularpoints in an open region Ω ⊂ R . If a C curve, γ ⊂ Ω , intersects an integral curveof X at two points then in between these points, there exists a point of tangencybetween γ and X . Proof of Theorem 2.
Considering the boundary value problem (5) and state-ment ( i ) of Lemma 9 follows that α >
1, 0 < v, w < w = v α . Expliciting x , y , y in (6) we get y = − ( − α )( αv − vw − α + v ) c + ( vw − α , y = − ( − α )( αvw − αw − w + 1) c + ( vw − α , and x = ρx + B , where B = 4 b + ( a + ρy − a + y − ρ + 1) a + (4( a + ) − a + c + + ( c + ) − s ) . From statement ( i ) of Lemma 10 and expliciting (cid:101) x in (8), we get W = S (cid:101) y − c − (cid:101) y − S (cid:101) y − c − (cid:101) y − , V = S (cid:101) y + c − (cid:101) y + 2 S (cid:101) y + c − (cid:101) y + 2 and (cid:101) x = 1 ξ (cid:101) x + C, where C = 4 a − b − ( ξ (cid:101) y − (cid:101) y ) + ( ξ − m (4( a − ) − c − a − − ( S − ( c − ) )) + 4( ξ − b − a − ξ ( c − − a − + S )( − c − + 2 a − + S ) . Substituting (cid:101) y = y and (cid:101) y = y in the expressions of V, W , we consider in theregion ∆ = (0 , × (0 ,
1) contained in the plane vw , the curves C f = { ( v, w ) ∈ ∆; f ( v, w ) = w − v α = 0 } ,C F = { ( v, w ) ∈ ∆; F ( v, w ) = W − V β = 0 } , (9) IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R where V = ( α c − + αβc + − αc + − αc − ) vw + ( − α c − + c − ) w − αβc + + αc + + αc − − c − ( αβc + − αc + − αc − + c − ) vw + ( α c − − c − ) v − α c − − αβc + + αc + + αc − ,W = ( c − β (1 − α ) + c + α (1 − β )) vw + ( α − c − βv + α ( c + ( β −
1) + c − β (1 − α ))( c − αβ ( α −
1) + c + α (1 − β )) vw + (1 − α ) c − βw + αc + ( β −
1) + c − β ( α − . Thus, for each point of intersection of the curves C f and C F , the piecewise linearvector field Z has an invariant cylinder. Note that the curve C F does not dependof a + or a − . Consequently, when a + a − = 0 the number of invariant cilinders is thesame.When v → − we have that w → , y → y →
0. From this fact, followsthat V → W → C f and C F pass through the point (1 , . Let (cid:101) X = ( v, αw ) be a vector field defined in ∆. So, C f is an integral curve of (cid:101) X .Consider the following system { F ( v, w ) = 0 , ∇ F ( v, w ) · (cid:101) X = 0 } . (10)We get ∇ F ( v, w ) · (cid:101) X = f ( v, w ) f ( v, w ) where f ( v, w ) = y (cid:98) y y (cid:98) y ( β + 1)( α + 1) β ( c − ) (cid:98) D ,f ( v, w ) = κw ( v − + λv ( w − , with κ = α ( αc − + βc + − c + − c − )( αβc − − βc + − βc − + c + ) ,λ = ( αβc + − αc + − αc − + c − )( αβc + + αβc − − αc + − βc − ) . We denote by (cid:98) y , (cid:98) y and (cid:98) D the denominators of y , y and ∇ F ( v, w ) · (cid:101) X respectively.Since that y , y and ∇ F ( v, w ) · (cid:101) X are well defined we obtain (cid:98) y (cid:98) y (cid:98) D (cid:54) = 0. Note thatthe system (10) is equivalent the { F ( v, w ) = 0 , f ( v, w ) = 0 } . Let C f be the curve { ( v, w ) ∈ ∆; f ( v, w ) = 0 } . Now, we get that ( v, w ) ∈ ∆ ∩ C f ∩ C f if and only if( v, w ) satisfies { w = v α , κw ( v − + λv ( w − = 0 } , or equivalently λv ( v α − κv α ( v − + 1 = 0 . This equation admits one zero for v ∈ (0 ,
1) if 1 + α λ/κ >
0, otherwise it does notadmit zeros in (0 , p p vw (1 , C F C f Figure 3.
Curves C f and C F .Now, we will prove the statement (1) of Theorem 2. So, we suppose that κ + λ (cid:54) =0 and κλ = 0. Without loss of generality we can suppose that κ (cid:54) = 0 and λ = 0 (the proof when κ = 0 and λ (cid:54) = 0 is analogous). In this case f = κw ( v − . Now, weassume that C f and C F intersect at a point p in ∆ (see Figure 3). Follows fromTheorem 11 that there is p ∈ ∆ and this point is a solution of { F = 0 , f = 0 } ,but this is a contradiction since f (cid:54) = 0 in ∆. Therefore there are no invariantcylinders and consequently there are no periodic orbits, i.e., the differential systemhas a scroll.The case κ + λ (cid:54) = 0 and κλ > f (cid:54) = 0 in ∆.Suppose now that 1 + α λ/κ ≤ κλ <
0. The curves C f , C F and C f pass vw C f C f C F p p vw (1 , C f C f C F ( a ) ( b ) Figure 4.
Curves C f , C f and C F .through the point (1 ,
1) and we have that ∇ F (1 ,
1) = ( F v (1 , , F w (1 , F v (1 , > F w (1 , < (cid:101) XF (1 ,
1) = (cid:101) X F (1 ,
1) = 0 and (cid:101) X F (1 , >
0. Weillustrate the relative positions between these curves in the Figure 4 ( a ). Assume that C f and C F intersect at a point p in ∆ (see Figure 4 ( b )). Follows from Theorem 11that there is p ∈ ∆ and this point is a solution of { F = 0 , f = 0 } , i.e., p ∈ C f ,but this is a contradiction.Now, in order to prove the statement (2) we assume that κλ < α λ/κ > ∇ F (1 ,
1) = ( F v (1 , , F w (1 , F v (1 , > F w (1 , < (cid:101) XF (1 ,
1) = (cid:101) X F (1 ,
1) = 0 and (cid:101) X F (1 , < a )).Assume that C f and C F intersect at two points p and p in ∆ (see Figure 5( b ) ). Follows from Theorem 11 that there are p , p ∈ ∆ which are solutionsof { F = 0 , f = 0 } , i.e., p , p ∈ C f , but this is a contradiction since that C f intersects C f at most at one point.If the intersection points p or p of C f ∩ C F are tangent with odd (or even)multiplicity the proof is analogous.Finally, if κ = λ = 0, then ∇ F ( v, w ) · (cid:101) X ≡ C f and C F arecoincident. In this case there is a continuous of invariant cylinders.The proof for all the other cases is similar to this one done for the case (Sa , Sa). Inthe Tables 1 and 2, we present the distinguished elements necessary to analyze thenumber of intersections points between curves C f and C F , consequently the numberof invariant cylinders. Thus, we conclude the proof of Theorem 2.3.2. Proof the Theorem 3.
We present the complete proof about existence ofisolated periodic orbits when the vector field Z is of type (Sa,Sa). So, under thishypotesis, we start observing that if Z has no invariant cylinder then do not existlimit cycles. As the curve C F given in (9) does not depend of a + or a − , the number IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R vw C F p p p p C f vw (1 , C f C f C F ( a ) ( b ) Figure 5.
Curves C f , C f and C F .of invariant cylinders remains the same, independently of the configuration of a + and a − .Suppose that ( a + ) + ( a − ) = 0. From boundary value problems (5) and (7)and from (4) of the Proposition 7, we obtain respectively x = x + ηy + µ and (cid:101) x = (cid:101) x + (cid:101) η (cid:101) y + (cid:101) µ . Here, we obtain ( η, µ ) and ( (cid:101) η, (cid:101) µ ) are obtained directly from Propo-sition 7 replacing ( γ, δ, M, σ, ψ, ±
1) = ( a + , b + , , c + , d + , −
1) and ( γ, δ, M, σ, ψ, ±
1) =( a − , b − , m, c − , d − , (cid:101) y = y and (cid:101) y = y ,we get that the return times τ and τ are also fixed. Thus, in this invariant cylinder,the number of limit cycles is given by the intersections of the straight lines r ± givenby r + : x = x + ηy + µ with r − : (cid:101) x = (cid:101) x − ( (cid:101) ηy + (cid:101) µ ) , where (cid:101) x = x and (cid:101) x = x . Doing ( B, C ) = ( ηy + µ, − ( (cid:101) ηy + (cid:101) µ )) we obtain thateither all solutions are closed in the cylinder, if B = C , or there is no closed solutionsin this cylinder when B (cid:54) = C .Suppose that ( a + ) + ( a − ) (cid:54) = 0 and Z at most one invariant cylinder. Fixedthe invariant cylinder, the number of limit cycles is given by the intersections of x = ρx + B if a + (cid:54) = 0 (or x = x + B if a + = 0) and x = x /ξ + C if a − (cid:54) = 0 (or x = x + C if a − = 0), where ρ = e a + τ , ξ = e a − τ , ( B, C ) are obtained as above and(
B, C ) obtained in the proof of Theorem 2. Thus, there is at most one limit cycle.Suppose that ( a + ) + ( a − ) (cid:54) = 0 , a + a − ≥ Z has infinitely many invariantcylinders. Suppose initially a + a − >
0. We will show that in each invariant cylinderthere is a unique isolated periodic orbit. Indeed, in each cylinder, the orbit periodicis given by the intersection of the straight lines r ± given by r + : x = ρx + B and r − : x = 1 ξ x + C. These straight lines has a unique intersection point provided that ρ (cid:54) = 1 /ξ . Notethat ρ = 1 /ξ ⇔ τ = − a − τ /a + , where τ and τ are positives. With the hypothesisof that a + a − >
0, the relation τ = − a − τ /a + can not be satisfied and thus ρ (cid:54) = 1 /ξ .The intersection point in each cylinder is given by x = C − B ξ − ρ e x = ρ C − B ξ − ρ + B. Varying continuously the cylinders, the terms x and x also range continuously,and we obtain one invariant surface formed of periodc orbits, where each orbit is aninvariant cylinder. If a + = 0 and a − (cid:54) = 0, the periodic orbit in each cylinder is givenby intersection of stright lines x = x + B and x = 1 ξ x + C. The case a + (cid:54) = 0 and a − = 0 follows analogously. Example 12.
Consider that a + = , b + = 0 , c + = − , d + = , c − = , d − = , a − = m = b − = 1 . y y Figure 6.
Invariant cylinder with a unique limit cycle.
Example 13.
Consider that c + = c − = m = 0 , a + = b − = d + = − , d − = − , b + =1 and a − = − . In this case we get that the vector field Z has infinitely manyinvariant cylinders. In each invariant cylinder there is a unique limit cycle. Alllimit cycles form a cone. (see Figure 7). Figure 7.
Cone of Periodic Orbits.
IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R The case focus-focus.
As mentioned earlier, the case (Fo,Fo) is not resolvedwith the techniques of Subsection 3.1. We use here [14]. For this, we get a differentcanonical form of (2). Proceeding as in Proposition 6, doing the twin change ofvariables given by ϕ + ( x, y, z ) = ( x − ( b +1 /b +2 ) y + (( b +1 a +11 − b +1 a +22 + a +12 b +2 ) /b +2 ) z, y + a +22 z, z ) on Σ + , ϕ − ( x, y, z ) = ( x − ( b +1 /b +2 ) y + (( b +2 a − − b +1 a − + b +1 a − ) /b +2 ) z, y + a − z, z ) on Σ − ,in vector field X + ( x, y, z ) = ( a +11 x + a +12 y + a +13 z + b +1 , a +22 y + a +23 z + b +2 , − y + a +33 z ) ,X − ( x, y, z ) = ( a − x + a − y + a − z + b − , a − y + a − z + b − , − y + a − z ) , we obtain X + ( x, y, z ) = ( a + x + b + z, D z + a , − y + T z ) ,X − ( x, y, z ) = ( a − x + b − z + m, D z + a , − y + T z ) . (11)To study the case (Fo-Fo), we must assume that a > , a < , T i − D i < T i (cid:54) = 0, with i = 1 ,
2. The Corollary 8 is also satisfied in this case. Thus, todetermine quotas for the number of limit cycles, we will determine first the numberof cylinders invariants. We note that in plane ( y, z ), the vectors field X + and X − are as in Theorem 4 of [14]. As the cylinders are determined by straight lines oftype r = { ( x, y, z ) ∈ R : y = y , z = 0 } and r = { ( x, y, z ) ∈ R : y = y , z = 0 } ,we will consider only the components in the plane ( y, z ). By Theorems 4 and 5 of[14], there is at most one limit cycle for y (cid:48) = D z + a ,z (cid:48) = − y + T z, if z > , y (cid:48) = D z + a ,z (cid:48) = − y + T z, if z < , with a > , a < , T i − D i < T i (cid:54) = 0, with i = 1 , X + and X − as in (11), there is at most one invariant cylinder. Let usconsider the boundary value problems x (cid:48) = a + x + b + z,y (cid:48) = D z + a ,z (cid:48) = − y + T z, ( x (0) , y (0) , z (0)) = ( x , y , , ( x ( τ ) , y ( τ ) , z ( τ )) = ( x , y , , and x (cid:48) = a − x + b − z + m,y (cid:48) = D z + a ,z (cid:48) = − y + T z, ( x (0) , y (0) , z (0)) = ( (cid:101) x , (cid:101) y , , ( x ( τ ) , y ( τ ) , z ( τ )) = ( (cid:101) x , (cid:101) y , . Fixed the invariant cylinder, y = (cid:101) y and y = (cid:101) y the times τ and τ are also fixed.With the boundary value problems above, we can write again x = ρx + B, and (cid:101) x = 1 ξ (cid:101) x + C, where ρ = e a + τ , ξ = e a − τ with B and C obtained as Proposition 7. Thus, doing x = (cid:101) x and x = (cid:101) x , exists in this cylinder at most one limit cycle. The caseswhere a + a − = 0 are studied analogously, noting once again that if a + = 0 we obtain x = x + B and if a − = 0, (cid:101) x = (cid:101) x + C . Acknowledgments
Both authors are partially supported by the FAPEG, by the CNPq grants numbers475623/2013-4 and 306615/2012-6 and, by the CAPES grant numbers PROCAD-88881.068462/2014-01 and by CSF/PVE-88881.030454/2013-01.
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IMIT CYCLES AND INVARIANT SURFACES OF SEWING PWL DIFF. SYSTEMS ON R Instituto de Matem´atica e Estat´ıstica, Universidade Federal de Goi´as, 74001-970Goiˆania, Goi´as, Brazil
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