Research Article Structural Stability of Planar Bimodal...

Research ArticleStructural Stability of Planar Bimodal Linear Systems

Josep Ferrer Marta Pentildea and Antoni Susiacuten

Departament de Matematica Aplicada I Escola Tecnica Superior drsquoEnginyeria Industrial de BarcelonaUniversitat Politecnica de Catalunya Diagonal 647 08028 Barcelona Spain

Correspondence should be addressed to Marta Pena martapenyaupcedu

Received 14 July 2014 Revised 21 October 2014 Accepted 9 November 2014 Published 23 December 2014

Academic Editor Do Wan Kim

Copyright copy 2014 Josep Ferrer et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Structural stability ensures that the qualitative behavior of a system is preserved under small perturbations We study it for planarbimodal linear dynamical systems that is systems consisting of two linear dynamics acting on each side of a given hyperplane andassuming continuity along the separating hyperplane We describe which one of these systems is structurally stable when (real)spiral does not appear and when it does we give necessary and sufficient conditions concerning finite periodic orbits and saddleconnections In particular we study the finite periodic orbits and the homoclinic orbits in the saddlespiral case

1 Introduction

Structural stability ensures that the qualitative behavior ofa system is preserved under small perturbations a systemis structurally stable if anyone in some neighborhood isequivalent to it (in particular they have the same dynamicalbehavior) We study this property for a class of piecewiselinear systems Piecewise linear systems have attracted theinterest of the researchers in recent years by their wide rangeof applications as well as the possible theoretical approachesSee for example [1ndash8] In particular bimodal linear sys-tems consist of two subsystems acting on each side of agiven hyperplane assuming continuity along the separatinghyperplane We focus on the planar case Indeed it is verycommonly found in applications (see the above references)

As we have pointed out a definition of structural stabilityinvolves a topology in the set of the considered systems(which defines the ldquosmall perturbationsrdquo) and an equivalencerelation (which defines the ldquopreservation of the behaviorrdquo)For piecewise linear systems the natural topology is theone of the Euclidean space formed by the coefficients ofthe matrices which determine each subsystem Concerningthe equivalence relation there are some different naturaloptions For example for single linear systems those havingpositive trace and positive determinant form a unique 1198620-class whereas they are partitioned in four 1198621-classes (spiralsnodes improper nodes and starred nodes) Anyway when

a topology and an equivalence relation are fixed the struc-tural stability points are those belonging to an open equiva-lence class

Alternative approaches are possible For example in [9]one asks about generic properties which are verified byldquoalmost allrdquo piecewise linear systems From a topologicalpoint of view it is a matter of density instead of opennessIndeed the properties there are both generic and stable AlsoArnoldrsquos techniques [10] can be partially applied becausealthough the equivalence relation is not defined by theaction of a Lie group the equivalence classes are probablydifferentiable manifolds

Here we focus on structural stability in the sense in [11]where a list of necessary and sufficient conditions is givenfor planar piecewise linear systems Our aim is to specifythese criteria in terms of the coefficients of the matricesin the particular case of bimodal linear systems The firststep is collected in Theorem 6 However further specificstudies are necessary in several cases As a second step wetackle (Theorem 7) the existence of homoclinic orbits andfinite periodic orbits in the saddlespiral case It allows us(Corollary 11) to ensure its structural stability for certainvalues of the parametersWe expect that for bimodal systemsa full characterization of the structural stability in terms of thecoefficients of the matrices is possible

Even more we expect that also a systematic study of thebifurcations is possible Bifurcations are the frontier points

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 892948 8 pageshttpdxdoiorg1011552014892948

2 Mathematical Problems in Engineering

of an open class so that they come out of their class bysmall perturbations Again it depends on the consideredequivalence relation For example the improper nodes andthe starred nodes are 1198621-bifurcation between spirals andnodes but not 1198620-bifurcation because all of them are 1198620-equivalent Indeed the 1198621-frontier of spiralsnodes is strat-ified as follows a 1-codimensional manifold formed by theimproper nodes and a 3-codimensional manifold formed bythe starred nodes (Hence improper nodes appear genericallyin 1-parameterized families of linear systems whereas starrednodes appear only in 3-parameterized families) Here threebifurcations are presented in Corollary 11 1-codimensional(two of them) and 2-codimensional (the third one)

In Section 3 we adapt the conditions stated in [11] forpiecewise linear planar dynamical systems to the particularclass of bimodal ones We conclude that if some subsystemis a starred node a center or a degenerate node then thebimodal system is not structurally stable Moreover we listthe remaining possible cases and we ensure that the bimodalsystem is structurally stable if none of the subsystems is a(real) spiral The other cases need further specific analysis

In particular when a (real) spiral appears it is necessaryto study the finite periodic orbits and the homoclinic orbitsIn Section 4 we study the structural stability of bimodalsystems for the saddlespiral case We conclude that thisbimodal system is structurally stable if 0 lt 120574

1lt 120574119867 where 120574

1

is the trace of the spiral matrix and 120574119867is the only value where

a homoclinic orbit appears The study will be continued infuture works (see [12])

Throughout the paper R will denote the set of realnumbers119872

119899times119898(R) the set of matrices having 119899 rows and 119898

columns and entries in R (in the case where 119899 = 119898 we willsimply write 119872

119899(R)) and 119866119897

119899(R) the group of nonsingular

matrices in 119872119899(R) Finally we will denote by 119890

1 119890

119899the

natural basis of the Euclidean space R119899

2 Structurally Stable BLDS Definitions

We consider

(119905) = 1198601119909 (119905) + 1198611 if 119862119909 (119905) le 0

(119905) = 1198602x (119905) + 119861

2if 119862119909 (119905) ge 0

(1)

where 1198601 1198602isin 119872119899(R) 119861

1 1198612isin 119872119899times1(R) 119862 isin 119872

1times119899(R) We

assume that the dynamic is continuous along the separatinghyperplane 119867 = 119909 isin R119899 119862119909 = 0 that is to say bothsubsystems coincide with 119862119909(119905) = 0

By means of a linear change in the state variable 119909(119905) wecan consider 119862 = (1 0 sdot sdot sdot 0) isin 119872

1times119899(R) Hence119867 = 119909 isin

R119899 1199091= 0 and continuity along119867 is equivalent to

1198612= 1198611 119860

2119890119894= 1198601119890119894 2 le 119894 le 119899 (2)

We will write from now on 119861 = 1198611= 1198612

Definition 1 In the above conditions one says that the tripletof matrices (119860

1 1198602 119861) defines a bimodal linear dynamical

system (BLDS)

Table 1 Critical points classification

Spiral 11988621= 0 120582

1 1205822conjugate complex numbers

Saddle 11988621= 0 120582

1sdot 1205822lt 0

Node 11988621= 0 120582

1sdot 1205822gt 0 120582

1= 1205822

Starred node 11988621= 0 120582

1= 1205822= 120582 120582 = 0 120582 isin R

Improper node 11988621= 1 120582

1= 1205822= 120582 120582 = 0 120582 isin R

Degenerate node 11988621= 0 120582

1= 120582 120582

2= 0 120582 = 0 120582 isin R

The placement of the equilibrium points will play asignificative role in the dynamics of a BLDS So one definesthe following

Definition 2 Let one assume that a subsystem of a BLDShas a unique equilibrium point not lying in the separatinghyperplane One says that this equilibrium point is real if itis located in the half-space corresponding to the consideredsubsystem Otherwise one says that the equilibrium point isvirtual

It is clear that not any pair of equilibrium points arecompatible For example two real saddles are not possible(Table 1 lists the compatible pairs excluding centers starrednodes and degenerate nodes)

Our goal is to characterize the planar BLDS which arestructurally stable in the sense of [11]

Definition 3 A triplet of matrices (1198601 1198602 119861) defining a

BLDS is said to be (regularly) structurally stable if ithas a neighborhood 119881(119860

1 1198602 119861) such that for every

(1198601015840

1 1198601015840

2 1198611015840) isin 119881(119860

1 1198602 119861) there is a homeomorphismofR2

preserving the hyperplane119867 which maps the oriented orbitsof (11986010158401 1198601015840

2 1198611015840) into those of (119860

1 1198602 119861) and it is differentiable

when restricted to finite periodic orbits

A natural tool in the study of BLDS is simplifying thematrices 119860

1 1198602 119861 by means of changes in the variables 119909(119905)

which preserve the qualitative behavior of the system (inparticular the condition of structural stability) See [5] forsome partial results and [13] for a systematic obtention ofreduced forms So we consider linear changes in the statevariables space preserving the hyperplanes 119909

1(119905) = 119896 which

will be called admissible basis changes Thus they are basischanges given by a matrix 119878 isin 119866119897

119899(R)

119878 = (1 0

119880 119879) 119879 isin 119866119897

119899minus1(R) 119880 isin 119872

119899minus1times1(R) (3)

Also translations parallel to the hyperplane 119867 areallowed

3 Structurally Stable Planar BLDSGeneral Criteria

Let us consider a planar BLDS For each subsystem wefollow the terminology in [14] according to its Jordan reducedform except for the ldquofocusrdquo which we have substituted bythe denomination ldquostarred noderdquo Here we reproduce thisclassification

Mathematical Problems in Engineering 3

If we denote by ( 1205821 0119886211205822

) the reduced matrix then weidentify critical points classification as shown in Table 1

For the particular case of BLDS the general conditionsin [11] in order to be structurally stable can be simplified asfollows

Corollary 4 A planar BLDS is structurally stable if and onlyif the following conditions hold

(1) Singularities conditions

(a) all its singularities at infinity are disjoint from theseparating axis

(b) all its singularities at infinity are hyperbolic(c) all its finite singularities are disjoint from the

separating axis(d) all its finite singularities are hyperbolic(e) all its tangencies with the separating axis are

isolated

(2) Periodic orbits conditions

(a) all its finite periodic orbits are not tangent to theseparating axis

(b) all its finite periodic orbits are hyperbolic(c) the infinite periodic orbit at infinity is hyperbolic

(3) There are no finite orbits which joint either

(a) two different saddle points (ldquosaddle-saddleorbitsrdquo)

(b) a saddle point with itself (ldquosaddle-loop orbitsrdquo orldquohomoclinic orbitsrdquo)

(c) a saddle point and a tangency (ldquosaddle-tangencyorbitsrdquo)

We will specify these conditions for a triplet of matricesdefining a planar BLDS We begin with conditions (1)(a) and(1)(c) in Corollary 4

Lemma 5 The triplets of matrices representing a structurallystable planar BLDS can be reduced to the form

1198601= (

11988611

11988620) 119860

2= (

12057411

12057420)

119861 = (0

1198872

) 1198872

= 0

(4)

Proof Given a planar BLDS defined by a triplet (1198601 1198602 119861)

such as

1198601= (

11988611198863

11988621198864

) 1198602= (

12057411198863

12057421198864

) 119861 = (1198871

1198872

) (5)

the condition (1)(a) in Corollary 4 is equivalent to

(11988611198863

11988621198864

)(0

1) = 120582 (

0

1) (6)

which gives 1198863

= 0 As it is proved in [13] when 1198863

= 0 bymeans of a suitable admissible basis change the triplet can bereduced to

(11988611

11988620) (

12057411

12057420) (

1198871

1198872

) (7)

Moreover by means of the translation 1199092= 1199092+1198871 we obtain

(11988611

11988620) (

12057411

12057420) (

0

1198872

) (8)

Then the condition (1)(c) in Corollary 4 is equivalent to

(11988611

11988620)(

0

1199092119890

) + (0

1198872

) = (0

0) (9)

which gives 1198872

= 0

Now we apply the remaining conditions in Corollary 4

Theorem 6 Let one consider a planar BLDS as in Lemma 5

(1) The only tangency (ie 1(0 1199092) = 0) is the origin

(0 0)(2) If one of the subsystems is a center a degenerate node or

a starred node then the BLDS is not structurally stableMore in general the only BLDS verifying (1)(a) (1)(c)and (1)(d) in Corollary 4 are those in Table 2

(3) The cases 1 2 4 5 6 8 9 10 12 13 14 and 16(those where none of the subsystems is a real spiral) arestructurally stable

(4) In case 3 it is structurally stable if and only if

(a) the finite periodic orbits are hyperbolic(b) there are no saddle-loop orbits(c) there are no finite orbits connecting a saddle and

a tangency point

(5) In the cases 7 11 and 15 the BLDS is structurally stableif and only if the above condition (4)(a) holds

Proof We proceed with the same order for the proof

(1) Clearly 1(0 1199092) = 0 implies 119909

2= 0

(2) Starred nodes have been excluded by (1)(a) The con-dition (1)(d) excludes degenerate nodes and centersthat is to say for 119894 = 1 2

det119860119894= 0

if det119860119894gt 0 then trace119860

119894= 0

(10)

Therefore the subsystems have a unique equilibriumpoint

(minus1198872

1198862

1198861

1198872

1198862

) (minus1198872

1205742

1205741

1198872

1205742

) (11)


Table 2 Enumeration of the possible cases

Subsystem 1Subsystem 2 Virtual saddle Real node Real spiral Real improper nodeReal saddle 1 (119887

2gt 0) 2 (119887

2gt 0) 3 (119887

2gt 0) 4 (119887

2gt 0)

Virtual node 5 (1198872lt 0) 6 (119887

2gt 0) 7 (119887

2gt 0) 8 (119887

2gt 0)

Virtual spiral 9 (1198872lt 0) 10 (119887

2lt 0) 11 (119887

2gt 0) 12 (119887

2gt 0)

Virtual improper node 13 (1198872lt 0) 14 (119887

2lt 0) 15 (119887

2lt 0) 16 (119887

2gt 0)

But not all combinations (realvirtual) are possibleFor example let us assume a real saddle as subsystem1 (ie 119886

2gt 0 and 119887

2gt 0) Then the subsystem 2 must

be for 1205742gt 0 a virtual saddle and for 120574

2lt 0 a real

node a real improper node or a real spiral They arethe cases 1 2 4 and 3 respectively Analogously forthe remainder cases of the table

(3) As starred nodes and degenerate nodes are excludedall the BLDS in Table 2 satisfy condition (1)(b)in Corollary 4 Moreover the feature that the onlytangency point is the origin implies the conditions(1)(e) and (2)(a) (we notice that centers are excluded)Concerning (2)(c) it is clear that only in the case 11there exists infinite periodic orbit at infinity and thatit is hyperbolic (its character of attractingrepellingdepends on the signs of the real and imaginary part ofthe complex eigenvalues) Obviously condition (3)(a)makes no sense in Table 2Again the remaining conditions (2)(b) (3)(b) and(3)(c) make no sense in the cases listed in point 3 ofTheorem 6 so that they are structurally stable

(4) On the other hand they must be verified in case 3 weadapt them to (a) (b) and (c) in point 4 ofTheorem 6

(5) Finally in the cases in point 5 only (a) must beverified because again (b) and (c) make no sense

4 Structurally Stable Planar BLDSSpecific Studies

Theorem 6 collects the conclusions of applying to planarBLDS the general criteria in [11] for a piecewise linear systemto be structurally stable Nevertheless cases 3 7 11 and 15need additional specific studies For example see in [15 16]partial results concerning case 7 In this section we focus onconditions (a) and (b) of case 3 for divergent spirals leavingthe remaining cases for future works

Thus let us assume a BLDS as in Lemma 5 verifying thefollowing

(i) The left subsystem is a real saddle that is 1198862

gt

0 1198872

gt 0 In particular its equilibrium point is(minus11988721198862 1198861(11988721198862)) and the invariant manifold cuts

the separating line at (0 minus11988721205822) and (0 minus119887

21205821)

where 1205822lt 0 lt 120582

1are the eigenvalues of 119860

1

(Consider 1205821+ 1205822= 1198861 12058211205822= minus1198862)

(ii) The right subsystem is a real divergent spiral thatis 1205741gt 0 120574

2lt 0 and 1205742

1lt minus4120574

2 In particular

its equilibrium point is (minus11988721205742 1205741(11988721205742)) We write

120572 plusmn 119894120573 120573 gt 0 the eigenvalues of 1198602 (Consider

2120572 = 1205741 1205722 + 1205732 = minus120574

2)

Theorem 7 As above let one assume

1198872gt 0 119886

2gt 0 120574

1gt 0

1205742lt 0 120574

2

1lt minus4120574

2

(12)

and let

1205822lt 0 lt 120582

1119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 119860

1

120572 plusmn 119894120573 120573 gt 0 119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 1198602

(13)

In addition let119872 gt 0 and 0 lt 120593 lt 120587 defined by

119872 cos (120593) = 120572 minus1205722+ 1205732

1205822

119872 sin (120593) = 120573 (14)

Then consider the following

(1)

(a) If 1198861gt 0 then there is no homoclinic orbit

(b) If 1198861= 0 then there is a homoclinic orbit only for

1205741= 0 which is not a considered case

(c) If 1198861lt 0 the only homoclinic (ie saddle-loop)

orbit appears for the value 120574119867of 1205741verifying

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (15)

being

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(16)

Moreover 120574119867gt 119886112057421198862

(2)

(a) If 1198861gt 0 then there are no finite periodic orbits

(b) If 1198861= 0 then there are finite periodic orbits (all

of them) only for 1205741= 0 which is not a considered

case(c) If 119886

1lt 0 at least a finite periodic orbit appears

for 0 lt 1205741lt 120574119867 all of the finite periodic orbits

being hyperbolic and disjoint from the tangencypoints No saddle-tangency orbits appear


The proof is based on the following lemmas

Lemma 8 A spiral cuts 1199091= 0 in 119909

21and 119909

22 if and only if

exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(17)

where 120583 = 120572 + 119894120573

Proof The solution of the system for the spiral is

(1199091(119905)

1199092(119905)) = (

120583 120583

12057421205742

)(exp (120583119905) 0

0 exp (120583119905))(120583 120583

12057421205742

)

minus1

times(

1199091(0) +

1198872

1205742

1199092(0) minus 120574

1

1198872

1205742

)+(

minus1198872

1205742

1205741

1198872

1205742

)

(18)

where 120583 is the conjugate of the eigenvalue 120583Considering that the starting and final point have 119909

1= 0

and denoting 1199092(0) = 119909

21and 119909

2(119905) = 119909

22 we get

1205742(120583 minus 120583)(

1198872

1205742

11990922minus 1205741

1198872

1205742

)

= (120583 exp (120583119905) 120583 exp (120583119905)1205742exp (120583119905) 120574

2exp (120583119905))(

1198872minus 120583(119909

21minus 1205741

1198872

1205742

)

minus1198872+ 120583(119909

21minus 1205741

1198872

1205742

)

)

(19)

Multiplying both sides of the system by (1205742 minus120583) we obtain

exp (120583119905) =1198872minus 12058311990922+ 1205831205741(11988721205742)

1198872minus 12058311990921+ 1205831205741(11988721205742)

(20)

which is equivalent to

exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(21)

Lemma 9 Let one consider the saddle-spiral orbit passingthrough (0 minus119887

21205822) Then its first intersection with the sepa-

rating hyperplane (if it exists) is determined by

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(22)

Proof Using Lemma 8 imposing that 11990921= minus11988721205822 we get

11990922=

1198872

1205722 + 1205732[(119872 exp (120572119905) cos (120573119905 minus 120593) minus 120572)

+ 119894 (119872 exp (120572119905) sin (120573119905 minus 120593) + 120573)] (23)

Lemma 10 Let one assume that a finite periodic orbit existsThen

119860+1205741= minus119860minus1198861 (24)

where119860+ and119860minus are the enclosed areas in the right and the leftside respectively

Proof An analogous result is proved in [16] by means ofGreenrsquos formula Alternatively here we follow the approachin [17] Let us consider the following continuous energyfunction

119864minus=1

2(11988621199091+ 1198872)2minus1

21199092

2if 1199091le 0

119864+=1

2(12057421199091+ 1198872)2minus1

21199092

2if 1199091ge 0

(25)

If we consider the following energy piecewise functionthe change in energy Δ119864 along a periodic orbit must be nullHence

0 = Δ119864 = 119864+(0 119909119898

2) minus 119864+(0 119909119872

2)

+ 119864minus(0 119909119872

2) minus 119864minus(0 119909119898

2)

= int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092+ int

119909119872

2

119909119898

2

119889119864minus

1198891199092

1198891199092

(26)

where 1199091198722

and 1199091198982are the top and the bottom intersections

with the separating hyperplane respectivelyBut (25) implies

119889119864+

1198891199092

= (12057421199091+ 1198872)1198891199091

1198891199092

minus 1199092

(27)

and from the equations of the bimodal system we have

1198891199091

1198891199092

=12057411199091+ 1199092

12057421199091+ 1198872

(28)

so

119889119864+

1198891199092

= 12057411199091 (29)

And the first integral can be computed by

int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092= minus1205741119860+ (30)

where 119860+ is the enclosed areas in the right side And

analogously for the other integralThus Δ119864 = 0 if and only if

119860+1205741= minus119860minus1198861 (31)

Proof of Theorem 7 (1)(b) (2)(b) For 1198861= 1205741= 0 it is

obvious that we have a saddlecenter (119886 not considered case)being both subsystems symmetric with regard to the axis


1199092

= 0 Then we have a homoclinic orbit (the firstintersection with 119909

1= 0 of the orbit passing through

(0 minus11988721205822) is just (0 minus119887

21205821)) and all the orbits inside it are

finite periodic orbitsIt is also clear that if 120574

1increases (being 119886

1= 0) then the

orbit passing through (0 minus11988721205822) cuts the axis 119909

1= 0 below

(0 minus11988721205821) (notice that the spirals become divergent and the

equilibrium point descends) So the homoclinic and the finiteperiodic orbits disappear

(1)(a) (2)(a) If in addition 1198861increases then the point

(0 minus11988721205821) ascends

(1)(c) Using Lemma 9 imposing that its first intersectionwith the separating hyperplane cuts at (0 minus119887

21205821)

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(32)

being

1198872

1205722 + 1205732(119872 exp (120572119905) cos (120573119905 minus 120593) minus 120572) = minus

1198872

1205821

(33)

we get

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (34)

Moreover for the existence of the homoclinic orbit it mustbe verified that

1205822

2

1205822

1

1205822

1minus 1205741198671205821minus 1205742

1205822

2minus 1205741198671205822minus 1205742

gt 1 (35)


12057411986711988621205822minus 12057421205822

2gt 12057411986711988621205821minus 12057421205822

1(36)

and from it we get

120574119867gt11988611205742

1198862

(37)

(2)(c) For 0 lt 1205741lt 120574119867 we claim that there is at least a stable

limit cycleThe first step is to show that a periodic solution exists

We use the classical argument of Poincare (as eg in [17])Consider a trajectory that starts at height 119909

2on the right

side of the separating hyperplane crosses to the left sideand intersects the left side of the same hyperplane at somenew height 119875(119909

2) The mapping from 119909

2to 119875(119909

2) is called

the Poincare map It tells us how the height of a trajectorychanges after one lap If we can show that there is a point119909lowast

2such that 119875(119909lowast

2) = 119909

lowast

2 then the corresponding trajectory

will be a periodic orbit and it is stable if |1198751015840(1199092)| lt 1 Being

linear both subsystems 119875(1199092) can be easily computed See

for example Figure 1 In order to show that such a 119909lowast2must

exist it is sufficient to know what the graph of 119875(1199092) looks

like roughlyLet us consider 0 le 119909

2le minus119887

21205822 For 119909

2= 0

we have the unique tangency trajectory so that the first

0 01 02 03 04 050

01

02

03

04

05

Figure 1 Distances between original and Poincare image pointsIntersection with the bisectrix corresponds to the stable limit cycleParameter values are the ones used for Example (1)

1

05

0

minus05

minus1

minus15

minus2

minus1 minus08 minus06 minus04 minus02 0 02 04 0806

Figure 2 Appearance of a finite periodic orbit in case 3 1198861= minus1

1198862= 1 120574

1= 01 120574

2= minus5 and 119887

2= 1

intersection with the separating line is under (0 0) and finally119875(0) gt 0 On the other hand for 119909

2= minus11988721205822 the first

intersection is upper (0 minus11988721205821) (recall 120574

1lt 120574119867) so that

119875(minus11988721205822) lt minus119887

21205822 Furthermore 119875(119909

2) is a continuous

function (from the theorem about the dependence of thesolutions on initial conditions) and indeed it is a smooth andmonotonic function (if not two trajectories would cross) Soby the intermediate value theorem the graph of 119875(119909

2) must

cross the bisectrix somewhere that intersection is our desired119909lowast

2We must exclude the possibility that 119875(119909

2) equiv 119909

2on

some interval in which case there would be a band ofinfinitelymany closed orbits If it happens Lemma 10 ensuresthat 119860+119860minus is constant (= minus119886

11205741) in this interval but

being119860+ 119860minus analytic functions the quotient will be constanteverywhere which is obviously false (eg when 119909

2rarr 0)

Finally as 119875(119909lowast2) = 119909

lowast

2is an isolated crossing and 119875(119909lowast

2)

is monotonic increasing then 0 lt 1198751015840(1199092) lt 1 so that the

periodic orbit is an attractor orbit


1

05

0

minus05

minus1

minus15

minus2


Figure 3 Appearance of a homoclinic orbit in case 3 1198861= minus1 119886

2=

1 1205741= 120574119867= 0742 120574


2= 1

Other values 1199092verifying 119875(119909

2) = 1199092can appear But the

above reasoning shows that the possibility that 119875(1199092) equiv 1199092on

some interval is excluded so that the corresponding periodicorbit is again hyperbolic

Corollary 11 The systems in Theorem 7 with 1198861lt 0 and 0 lt

1205741lt 120574119867are structurally stable

Bifurcations appear for

(i) 1198861lt 0 120574

1= 120574119867 homoclinic orbit

(ii) 1198861lt 0 120574

1= 0 nonhyperbolic finite periodic orbits

(iii) 1198861= 0 1205741= 0 both kinds of orbits

Proof The existence of a periodic orbit implies that (3)(c)in Corollary 4 is verified Hence this case verifies all theconditions in that corollary

Examples (1)We show the structurally stable case 1198861= minus1

1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We

plot the phase portrait corresponding to the Poincare mapon the section 119909 = 0 for different initial points for each ofthem the orbits are integrated until the next oriented cutThe continuous lines correspond to inward spiraling orbitsand the discontinuous lines to outward spiraling ones Ahyperbolic finite periodic orbit exists between them

(2) Bifurcations are as follows

(i) homoclinic orbit (Figure 3) 1198861= minus1 119886

2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1

(ii) nonhyperbolic periodic orbits (Figure 4) 1198861= minus1

1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1

(iii) both kinds of orbits (Figure 5) 1198861= 0 119886

2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1

Remark 12 In [12] some partial results for 1205741lt 0 and 120574

1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06

Figure 4 Appearance of nonhyperbolic periodic orbits in case 31198861= minus1 119886

2= 1 120574

1= 0 120574


2= 1

minus1 minus08 minus06 minus04 minus02 0 02 04 06 08

15

1

05

0

minus05

minus1

minus15

Figure 5 Appearance of both kinds of singularities in case 3 1198861= 0

1198862= 1 120574

1= 0 120574


2= 1

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors thank Professor Rafael Ramirez for many helpfuldiscussions during the preparation of the paper This paperis supported by DGICYTMTM2011-23892 (Josep Ferrer andMarta Pena) and TIN2013-47137-C2-1-P (Antoni Susın)

References

[1] J C Artes J Llibre J C Medrado and M A TeixeiraldquoPiecewise linear differential systems with two real saddlesrdquoMathematics and Computers in Simulation vol 95 pp 13ndash222014


[2] K Camlibel M Heemels and H Schumacher ldquoStability andcontrollability of planar bimodal linear complementarity sys-temsrdquo in Proceedings of the 42nd IEEE Conference on Decisionand Control pp 1651ndash1656 December 2003

[3] K Camlibel M Heemels and H Schumacher ldquoOn the con-trollability of bimodal piecewise linear systemsrdquo in HybridSystems Computation and Control vol 2993 of Lecture Notesin Computer Science pp 250ndash264 Springer Berlin Germany2004

[4] M K Camlibel W P M H Heemels and J M SchumacherldquoA full characterization of stabilizability of bimodal piecewiselinear systems with scalar inputsrdquoAutomatica vol 44 no 5 pp1261ndash1267 2008

[5] V Carmona E Freire E Ponce and F Torres ldquoOn simplifyingand classifying piecewise-linear systemsrdquo IEEE Transactions onCircuits and Systems I Fundamental Theory and Applicationsvol 49 no 5 pp 609ndash620 2002

[6] M di Bernardo D J Pagano and E Ponce ldquoNonhyperbolicboundary equilibrium bifurcations in planar Filippov systemsa case study approachrdquo International Journal of Bifurcation andChaos vol 18 no 5 pp 1377ndash1392 2008

[7] J Ferrer M D Magret andM Pena ldquoBimodal piecewise lineardynamical systems Reduced formsrdquo International Journal ofBifurcation and Chaos vol 20 no 9 pp 2795ndash2808 2010

[8] J Llibre M Ordonez and E Ponce ldquoOn the existence anduniqueness of limit cycles in planar continuous piecewise linearsystems without symmetryrdquo Nonlinear Analysis Real WorldApplications vol 14 no 5 pp 2002ndash2012 2013

[9] R Lum and L O Chua ldquoGeneric properties of continuouspiecewise-linear vector fields in R2rdquo IEEE Transactions onCircuits and Systems vol 38 no 9 pp 1043ndash1066 1991

[10] V I Arnold ldquoOn matrices depending on parametersrdquo UspekhiMatematicheskikh Nauk vol 26 pp 101ndash114 1971

[11] J Sotomayor and R Garcia ldquoStructural stability of piecewise-linear vector fieldsrdquo Journal of Differential Equations vol 192no 2 pp 553ndash565 2003

[12] J Ferrer M Pena and A Susin ldquoTangency-saddle singularitiesof planar bimodal linear systemsrdquo in Proceedings of the Inter-national Conference on Mathematical Models and Methods inApplied Sciences Saint Petersburg Russia September 2014

[13] J Ferrer M Magret and M Pena ldquoDifferentiable families ofplanar bimodal linear control systemsrdquoMathematical Problemsin Engineering vol 2014 Article ID 292813 9 pages 2014

[14] M W Hirsch and S Smale Differential Equations DynamicalSystems and Linear Algebra Academic Press London UK 1974

[15] E Freire E Ponce F Rodrigo and F Torres ldquoBifurcationsets of continuous piecewise linear systems with two zonesrdquoInternational Journal of Bifurcation and Chaos vol 8 no 11 pp2073ndash2097 1998

[16] J Llibre and J Sotomayor ldquoPhase portraits of planar controlsystemsrdquoNonlinear Analysis Theory Methods and Applicationsvol 27 no 10 pp 1177ndash1197 1996

[17] S H Strogatz Nonlinear Dynamics and Chaos Perseus Books2000

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


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Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


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Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


of an open class so that they come out of their class bysmall perturbations Again it depends on the consideredequivalence relation For example the improper nodes andthe starred nodes are 1198621-bifurcation between spirals andnodes but not 1198620-bifurcation because all of them are 1198620-equivalent Indeed the 1198621-frontier of spiralsnodes is strat-ified as follows a 1-codimensional manifold formed by theimproper nodes and a 3-codimensional manifold formed bythe starred nodes (Hence improper nodes appear genericallyin 1-parameterized families of linear systems whereas starrednodes appear only in 3-parameterized families) Here threebifurcations are presented in Corollary 11 1-codimensional(two of them) and 2-codimensional (the third one)

In Section 3 we adapt the conditions stated in [11] forpiecewise linear planar dynamical systems to the particularclass of bimodal ones We conclude that if some subsystemis a starred node a center or a degenerate node then thebimodal system is not structurally stable Moreover we listthe remaining possible cases and we ensure that the bimodalsystem is structurally stable if none of the subsystems is a(real) spiral The other cases need further specific analysis

In particular when a (real) spiral appears it is necessaryto study the finite periodic orbits and the homoclinic orbitsIn Section 4 we study the structural stability of bimodalsystems for the saddlespiral case We conclude that thisbimodal system is structurally stable if 0 lt 120574

1lt 120574119867 where 120574

1

is the trace of the spiral matrix and 120574119867is the only value where

a homoclinic orbit appears The study will be continued infuture works (see [12])

Throughout the paper R will denote the set of realnumbers119872

119899times119898(R) the set of matrices having 119899 rows and 119898

columns and entries in R (in the case where 119899 = 119898 we willsimply write 119872

119899(R)) and 119866119897

119899(R) the group of nonsingular

matrices in 119872119899(R) Finally we will denote by 119890

1 119890

119899the

natural basis of the Euclidean space R119899

2 Structurally Stable BLDS Definitions

We consider

(119905) = 1198601119909 (119905) + 1198611 if 119862119909 (119905) le 0

(119905) = 1198602x (119905) + 119861

2if 119862119909 (119905) ge 0

(1)

where 1198601 1198602isin 119872119899(R) 119861

1 1198612isin 119872119899times1(R) 119862 isin 119872

1times119899(R) We

assume that the dynamic is continuous along the separatinghyperplane 119867 = 119909 isin R119899 119862119909 = 0 that is to say bothsubsystems coincide with 119862119909(119905) = 0

By means of a linear change in the state variable 119909(119905) wecan consider 119862 = (1 0 sdot sdot sdot 0) isin 119872

1times119899(R) Hence119867 = 119909 isin

R119899 1199091= 0 and continuity along119867 is equivalent to

1198612= 1198611 119860

2119890119894= 1198601119890119894 2 le 119894 le 119899 (2)

We will write from now on 119861 = 1198611= 1198612

Definition 1 In the above conditions one says that the tripletof matrices (119860

1 1198602 119861) defines a bimodal linear dynamical

system (BLDS)

Table 1 Critical points classification

Spiral 11988621= 0 120582

1 1205822conjugate complex numbers

Saddle 11988621= 0 120582

1sdot 1205822lt 0

Node 11988621= 0 120582

1sdot 1205822gt 0 120582

1= 1205822

Starred node 11988621= 0 120582

1= 1205822= 120582 120582 = 0 120582 isin R

Improper node 11988621= 1 120582

1= 1205822= 120582 120582 = 0 120582 isin R

Degenerate node 11988621= 0 120582

1= 120582 120582

2= 0 120582 = 0 120582 isin R

The placement of the equilibrium points will play asignificative role in the dynamics of a BLDS So one definesthe following

Definition 2 Let one assume that a subsystem of a BLDShas a unique equilibrium point not lying in the separatinghyperplane One says that this equilibrium point is real if itis located in the half-space corresponding to the consideredsubsystem Otherwise one says that the equilibrium point isvirtual

It is clear that not any pair of equilibrium points arecompatible For example two real saddles are not possible(Table 1 lists the compatible pairs excluding centers starrednodes and degenerate nodes)

Our goal is to characterize the planar BLDS which arestructurally stable in the sense of [11]

Definition 3 A triplet of matrices (1198601 1198602 119861) defining a

BLDS is said to be (regularly) structurally stable if ithas a neighborhood 119881(119860

1 1198602 119861) such that for every

(1198601015840

1 1198601015840

2 1198611015840) isin 119881(119860

1 1198602 119861) there is a homeomorphismofR2

preserving the hyperplane119867 which maps the oriented orbitsof (11986010158401 1198601015840

2 1198611015840) into those of (119860

1 1198602 119861) and it is differentiable

when restricted to finite periodic orbits

A natural tool in the study of BLDS is simplifying thematrices 119860

1 1198602 119861 by means of changes in the variables 119909(119905)

which preserve the qualitative behavior of the system (inparticular the condition of structural stability) See [5] forsome partial results and [13] for a systematic obtention ofreduced forms So we consider linear changes in the statevariables space preserving the hyperplanes 119909

1(119905) = 119896 which

will be called admissible basis changes Thus they are basischanges given by a matrix 119878 isin 119866119897

119899(R)

119878 = (1 0

119880 119879) 119879 isin 119866119897

119899minus1(R) 119880 isin 119872

119899minus1times1(R) (3)

Also translations parallel to the hyperplane 119867 areallowed

3 Structurally Stable Planar BLDSGeneral Criteria

Let us consider a planar BLDS For each subsystem wefollow the terminology in [14] according to its Jordan reducedform except for the ldquofocusrdquo which we have substituted bythe denomination ldquostarred noderdquo Here we reproduce thisclassification


If we denote by ( 1205821 0119886211205822








isolated










1198601= (

11988611

11988620) 119860

2= (

12057411

12057420)

119861 = (0

1198872

) 1198872

= 0

(4)


such as

1198601= (

11988611198863

11988621198864

) 1198602= (

12057411198863

12057421198864

) 119861 = (1198871

1198872

) (5)


(11988611198863

11988621198864

)(0

1) = 120582 (

0

1) (6)

which gives 1198863



(11988611

11988620) (

12057411

12057420) (

1198871

1198872

) (7)


(11988611

11988620) (

12057411

12057420) (

0

1198872

) (8)


(11988611

11988620)(

0

1199092119890

) + (0

1198872

) = (0

0) (9)

which gives 1198872

= 0









a tangency point



(1) Clearly 1(0 1199092) = 0 implies 119909

2= 0


det119860119894= 0


119894= 0

(10)


(minus1198872

1198862

1198861

1198872

1198862

) (minus1198872

1205742

1205741

1198872

1205742

) (11)




2gt 0) 2 (119887

2gt 0) 3 (119887

2gt 0) 4 (119887

2gt 0)

Virtual node 5 (1198872lt 0) 6 (119887

2gt 0) 7 (119887

2gt 0) 8 (119887

2gt 0)


2lt 0) 11 (119887

2gt 0) 12 (119887

2gt 0)


2lt 0) 15 (119887

2lt 0) 16 (119887

2gt 0)


2gt 0 and 119887



2lt 0 a real









gt

0 1198872



21205821)

where 1205822lt 0 lt 120582


1

(Consider 1205821+ 1205822= 1198861 12058211205822= minus1198862)


2lt 0 and 1205742

1lt minus4120574

2 In particular



2120572 = 1205741 1205722 + 1205732 = minus120574

2)


1198872gt 0 119886

2gt 0 120574

1gt 0

1205742lt 0 120574

2

1lt minus4120574

2

(12)

and let

1205822lt 0 lt 120582

1119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 119860

1

120572 plusmn 119894120573 120573 gt 0 119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 1198602

(13)


119872 cos (120593) = 120572 minus1205722+ 1205732

1205822

119872 sin (120593) = 120573 (14)


(1)






119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (15)

being

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(16)

Moreover 120574119867gt 119886112057421198862

(2)




case(c) If 119886







21and 119909

22 if and only if

exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(17)

where 120583 = 120572 + 119894120573


(1199091(119905)

1199092(119905)) = (

120583 120583

12057421205742

)(exp (120583119905) 0

0 exp (120583119905))(120583 120583

12057421205742

)

minus1

times(

1199091(0) +

1198872

1205742

1199092(0) minus 120574

1

1198872

1205742

)+(

minus1198872

1205742

1205741

1198872

1205742

)

(18)


1= 0

and denoting 1199092(0) = 119909

21and 119909

2(119905) = 119909

22 we get

1205742(120583 minus 120583)(

1198872

1205742

11990922minus 1205741

1198872

1205742

)

= (120583 exp (120583119905) 120583 exp (120583119905)1205742exp (120583119905) 120574

2exp (120583119905))(

1198872minus 120583(119909

21minus 1205741

1198872

1205742

)

minus1198872+ 120583(119909

21minus 1205741

1198872

1205742

)

)

(19)


exp (120583119905) =1198872minus 12058311990922+ 1205831205741(11988721205742)

1198872minus 12058311990921+ 1205831205741(11988721205742)

(20)


exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(21)




exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(22)


11990922=

1198872


+ 119894 (119872 exp (120572119905) sin (120573119905 minus 120593) + 120573)] (23)


119860+1205741= minus119860minus1198861 (24)



119864minus=1

2(11988621199091+ 1198872)2minus1

21199092

2if 1199091le 0

119864+=1

2(12057421199091+ 1198872)2minus1

21199092

2if 1199091ge 0

(25)


0 = Δ119864 = 119864+(0 119909119898

2) minus 119864+(0 119909119872

2)

+ 119864minus(0 119909119872

2) minus 119864minus(0 119909119898

2)

= int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092+ int

119909119872

2

119909119898

2

119889119864minus

1198891199092

1198891199092

(26)

where 1199091198722



119889119864+

1198891199092

= (12057421199091+ 1198872)1198891199091

1198891199092

minus 1199092

(27)


1198891199091

1198891199092

=12057411199091+ 1199092

12057421199091+ 1198872

(28)

so

119889119864+

1198891199092

= 12057411199091 (29)


int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092= minus1205741119860+ (30)



119860+1205741= minus119860minus1198861 (31)




1199092







1= 0) then the


1= 0 below




(0 minus11988721205821) ascends


21205821)

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(32)

being

1198872


1198872

1205821

(33)

we get

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (34)


1205822

2

1205822

1

1205822

1minus 1205741198671205821minus 1205742

1205822

2minus 1205741198671205822minus 1205742

gt 1 (35)


12057411986711988621205822minus 12057421205822

2gt 12057411986711988621205821minus 12057421205822

1(36)

and from it we get

120574119867gt11988611205742

1198862

(37)




2on the right



2to 119875(119909

2) is called



2) = 119909

lowast







2le minus119887

21205822 For 119909

2= 0


0 01 02 03 04 050

01

02

03

04

05


1

05

0

minus05

minus1

minus15

minus2



1198862= 1 120574

1= 01 120574


2= 1




1lt 120574119867) so that

119875(minus11988721205822) lt minus119887

21205822 Furthermore 119875(119909

2) is a continuous


2) must



2) equiv 119909

2on




2rarr 0)


lowast


2)




1

05

0

minus05

minus1

minus15

minus2



2=

1 1205741= 120574119867= 0742 120574


2= 1








(i) 1198861lt 0 120574


(ii) 1198861lt 0 120574





1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We




2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1


1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1


2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1


1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06


2= 1 120574

1= 0 120574


2= 1


15

1

05

0

minus05

minus1

minus15


1198862= 1 120574

1= 0 120574


2= 1



Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










If we denote by ( 1205821 0119886211205822








isolated










1198601= (

11988611

11988620) 119860

2= (

12057411

12057420)

119861 = (0

1198872

) 1198872

= 0

(4)


such as

1198601= (

11988611198863

11988621198864

) 1198602= (

12057411198863

12057421198864

) 119861 = (1198871

1198872

) (5)


(11988611198863

11988621198864

)(0

1) = 120582 (

0

1) (6)

which gives 1198863



(11988611

11988620) (

12057411

12057420) (

1198871

1198872

) (7)


(11988611

11988620) (

12057411

12057420) (

0

1198872

) (8)


(11988611

11988620)(

0

1199092119890

) + (0

1198872

) = (0

0) (9)

which gives 1198872

= 0









a tangency point



(1) Clearly 1(0 1199092) = 0 implies 119909

2= 0


det119860119894= 0


119894= 0

(10)


(minus1198872

1198862

1198861

1198872

1198862

) (minus1198872

1205742

1205741

1198872

1205742

) (11)




2gt 0) 2 (119887

2gt 0) 3 (119887

2gt 0) 4 (119887

2gt 0)

Virtual node 5 (1198872lt 0) 6 (119887

2gt 0) 7 (119887

2gt 0) 8 (119887

2gt 0)


2lt 0) 11 (119887

2gt 0) 12 (119887

2gt 0)


2lt 0) 15 (119887

2lt 0) 16 (119887

2gt 0)


2gt 0 and 119887



2lt 0 a real









gt

0 1198872



21205821)

where 1205822lt 0 lt 120582


1

(Consider 1205821+ 1205822= 1198861 12058211205822= minus1198862)


2lt 0 and 1205742

1lt minus4120574

2 In particular



2120572 = 1205741 1205722 + 1205732 = minus120574

2)


1198872gt 0 119886

2gt 0 120574

1gt 0

1205742lt 0 120574

2

1lt minus4120574

2

(12)

and let

1205822lt 0 lt 120582

1119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 119860

1

120572 plusmn 119894120573 120573 gt 0 119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 1198602

(13)


119872 cos (120593) = 120572 minus1205722+ 1205732

1205822

119872 sin (120593) = 120573 (14)


(1)






119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (15)

being

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(16)

Moreover 120574119867gt 119886112057421198862

(2)




case(c) If 119886







21and 119909

22 if and only if

exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(17)

where 120583 = 120572 + 119894120573


(1199091(119905)

1199092(119905)) = (

120583 120583

12057421205742

)(exp (120583119905) 0

0 exp (120583119905))(120583 120583

12057421205742

)

minus1

times(

1199091(0) +

1198872

1205742

1199092(0) minus 120574

1

1198872

1205742

)+(

minus1198872

1205742

1205741

1198872

1205742

)

(18)


1= 0

and denoting 1199092(0) = 119909

21and 119909

2(119905) = 119909

22 we get

1205742(120583 minus 120583)(

1198872

1205742

11990922minus 1205741

1198872

1205742

)

= (120583 exp (120583119905) 120583 exp (120583119905)1205742exp (120583119905) 120574

2exp (120583119905))(

1198872minus 120583(119909

21minus 1205741

1198872

1205742

)

minus1198872+ 120583(119909

21minus 1205741

1198872

1205742

)

)

(19)


exp (120583119905) =1198872minus 12058311990922+ 1205831205741(11988721205742)

1198872minus 12058311990921+ 1205831205741(11988721205742)

(20)


exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(21)




exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(22)


11990922=

1198872


+ 119894 (119872 exp (120572119905) sin (120573119905 minus 120593) + 120573)] (23)


119860+1205741= minus119860minus1198861 (24)



119864minus=1

2(11988621199091+ 1198872)2minus1

21199092

2if 1199091le 0

119864+=1

2(12057421199091+ 1198872)2minus1

21199092

2if 1199091ge 0

(25)


0 = Δ119864 = 119864+(0 119909119898

2) minus 119864+(0 119909119872

2)

+ 119864minus(0 119909119872

2) minus 119864minus(0 119909119898

2)

= int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092+ int

119909119872

2

119909119898

2

119889119864minus

1198891199092

1198891199092

(26)

where 1199091198722



119889119864+

1198891199092

= (12057421199091+ 1198872)1198891199091

1198891199092

minus 1199092

(27)


1198891199091

1198891199092

=12057411199091+ 1199092

12057421199091+ 1198872

(28)

so

119889119864+

1198891199092

= 12057411199091 (29)


int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092= minus1205741119860+ (30)



119860+1205741= minus119860minus1198861 (31)




1199092







1= 0) then the


1= 0 below




(0 minus11988721205821) ascends


21205821)

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(32)

being

1198872


1198872

1205821

(33)

we get

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (34)


1205822

2

1205822

1

1205822

1minus 1205741198671205821minus 1205742

1205822

2minus 1205741198671205822minus 1205742

gt 1 (35)


12057411986711988621205822minus 12057421205822

2gt 12057411986711988621205821minus 12057421205822

1(36)

and from it we get

120574119867gt11988611205742

1198862

(37)




2on the right



2to 119875(119909

2) is called



2) = 119909

lowast







2le minus119887

21205822 For 119909

2= 0


0 01 02 03 04 050

01

02

03

04

05


1

05

0

minus05

minus1

minus15

minus2



1198862= 1 120574

1= 01 120574


2= 1




1lt 120574119867) so that

119875(minus11988721205822) lt minus119887

21205822 Furthermore 119875(119909

2) is a continuous


2) must



2) equiv 119909

2on




2rarr 0)


lowast


2)




1

05

0

minus05

minus1

minus15

minus2



2=

1 1205741= 120574119867= 0742 120574


2= 1








(i) 1198861lt 0 120574


(ii) 1198861lt 0 120574





1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We




2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1


1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1


2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1


1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06


2= 1 120574

1= 0 120574


2= 1


15

1

05

0

minus05

minus1

minus15


1198862= 1 120574

1= 0 120574


2= 1



Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












2gt 0) 2 (119887

2gt 0) 3 (119887

2gt 0) 4 (119887

2gt 0)

Virtual node 5 (1198872lt 0) 6 (119887

2gt 0) 7 (119887

2gt 0) 8 (119887

2gt 0)


2lt 0) 11 (119887

2gt 0) 12 (119887

2gt 0)


2lt 0) 15 (119887

2lt 0) 16 (119887

2gt 0)


2gt 0 and 119887



2lt 0 a real









gt

0 1198872



21205821)

where 1205822lt 0 lt 120582


1

(Consider 1205821+ 1205822= 1198861 12058211205822= minus1198862)


2lt 0 and 1205742

1lt minus4120574

2 In particular



2120572 = 1205741 1205722 + 1205732 = minus120574

2)


1198872gt 0 119886

2gt 0 120574

1gt 0

1205742lt 0 120574

2

1lt minus4120574

2

(12)

and let

1205822lt 0 lt 120582

1119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 119860

1

120572 plusmn 119894120573 120573 gt 0 119905ℎ119890 119890119894119892119890119899V119886119897119906119890119904 119900119891 1198602

(13)


119872 cos (120593) = 120572 minus1205722+ 1205732

1205822

119872 sin (120593) = 120573 (14)


(1)






119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (15)

being

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(16)

Moreover 120574119867gt 119886112057421198862

(2)




case(c) If 119886







21and 119909

22 if and only if

exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(17)

where 120583 = 120572 + 119894120573


(1199091(119905)

1199092(119905)) = (

120583 120583

12057421205742

)(exp (120583119905) 0

0 exp (120583119905))(120583 120583

12057421205742

)

minus1

times(

1199091(0) +

1198872

1205742

1199092(0) minus 120574

1

1198872

1205742

)+(

minus1198872

1205742

1205741

1198872

1205742

)

(18)


1= 0

and denoting 1199092(0) = 119909

21and 119909

2(119905) = 119909

22 we get

1205742(120583 minus 120583)(

1198872

1205742

11990922minus 1205741

1198872

1205742

)

= (120583 exp (120583119905) 120583 exp (120583119905)1205742exp (120583119905) 120574

2exp (120583119905))(

1198872minus 120583(119909

21minus 1205741

1198872

1205742

)

minus1198872+ 120583(119909

21minus 1205741

1198872

1205742

)

)

(19)


exp (120583119905) =1198872minus 12058311990922+ 1205831205741(11988721205742)

1198872minus 12058311990921+ 1205831205741(11988721205742)

(20)


exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(21)




exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(22)


11990922=

1198872


+ 119894 (119872 exp (120572119905) sin (120573119905 minus 120593) + 120573)] (23)


119860+1205741= minus119860minus1198861 (24)



119864minus=1

2(11988621199091+ 1198872)2minus1

21199092

2if 1199091le 0

119864+=1

2(12057421199091+ 1198872)2minus1

21199092

2if 1199091ge 0

(25)


0 = Δ119864 = 119864+(0 119909119898

2) minus 119864+(0 119909119872

2)

+ 119864minus(0 119909119872

2) minus 119864minus(0 119909119898

2)

= int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092+ int

119909119872

2

119909119898

2

119889119864minus

1198891199092

1198891199092

(26)

where 1199091198722



119889119864+

1198891199092

= (12057421199091+ 1198872)1198891199091

1198891199092

minus 1199092

(27)


1198891199091

1198891199092

=12057411199091+ 1199092

12057421199091+ 1198872

(28)

so

119889119864+

1198891199092

= 12057411199091 (29)


int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092= minus1205741119860+ (30)



119860+1205741= minus119860minus1198861 (31)




1199092







1= 0) then the


1= 0 below




(0 minus11988721205821) ascends


21205821)

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(32)

being

1198872


1198872

1205821

(33)

we get

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (34)


1205822

2

1205822

1

1205822

1minus 1205741198671205821minus 1205742

1205822

2minus 1205741198671205822minus 1205742

gt 1 (35)


12057411986711988621205822minus 12057421205822

2gt 12057411986711988621205821minus 12057421205822

1(36)

and from it we get

120574119867gt11988611205742

1198862

(37)




2on the right



2to 119875(119909

2) is called



2) = 119909

lowast







2le minus119887

21205822 For 119909

2= 0


0 01 02 03 04 050

01

02

03

04

05


1

05

0

minus05

minus1

minus15

minus2



1198862= 1 120574

1= 01 120574


2= 1




1lt 120574119867) so that

119875(minus11988721205822) lt minus119887

21205822 Furthermore 119875(119909

2) is a continuous


2) must



2) equiv 119909

2on




2rarr 0)


lowast


2)




1

05

0

minus05

minus1

minus15

minus2



2=

1 1205741= 120574119867= 0742 120574


2= 1








(i) 1198861lt 0 120574


(ii) 1198861lt 0 120574





1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We




2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1


1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1


2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1


1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06


2= 1 120574

1= 0 120574


2= 1


15

1

05

0

minus05

minus1

minus15


1198862= 1 120574

1= 0 120574


2= 1



Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












21and 119909

22 if and only if

exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(17)

where 120583 = 120572 + 119894120573


(1199091(119905)

1199092(119905)) = (

120583 120583

12057421205742

)(exp (120583119905) 0

0 exp (120583119905))(120583 120583

12057421205742

)

minus1

times(

1199091(0) +

1198872

1205742

1199092(0) minus 120574

1

1198872

1205742

)+(

minus1198872

1205742

1205741

1198872

1205742

)

(18)


1= 0

and denoting 1199092(0) = 119909

21and 119909

2(119905) = 119909

22 we get

1205742(120583 minus 120583)(

1198872

1205742

11990922minus 1205741

1198872

1205742

)

= (120583 exp (120583119905) 120583 exp (120583119905)1205742exp (120583119905) 120574

2exp (120583119905))(

1198872minus 120583(119909

21minus 1205741

1198872

1205742

)

minus1198872+ 120583(119909

21minus 1205741

1198872

1205742

)

)

(19)


exp (120583119905) =1198872minus 12058311990922+ 1205831205741(11988721205742)

1198872minus 12058311990921+ 1205831205741(11988721205742)

(20)


exp (120583119905) =1198872+ 12058311990922

1198872+ 12058311990921

(21)




exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(22)


11990922=

1198872


+ 119894 (119872 exp (120572119905) sin (120573119905 minus 120593) + 120573)] (23)


119860+1205741= minus119860minus1198861 (24)



119864minus=1

2(11988621199091+ 1198872)2minus1

21199092

2if 1199091le 0

119864+=1

2(12057421199091+ 1198872)2minus1

21199092

2if 1199091ge 0

(25)


0 = Δ119864 = 119864+(0 119909119898

2) minus 119864+(0 119909119872

2)

+ 119864minus(0 119909119872

2) minus 119864minus(0 119909119898

2)

= int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092+ int

119909119872

2

119909119898

2

119889119864minus

1198891199092

1198891199092

(26)

where 1199091198722



119889119864+

1198891199092

= (12057421199091+ 1198872)1198891199091

1198891199092

minus 1199092

(27)


1198891199091

1198891199092

=12057411199091+ 1199092

12057421199091+ 1198872

(28)

so

119889119864+

1198891199092

= 12057411199091 (29)


int

119909119898

2

119909119872

2

119889119864+

1198891199092

1198891199092= minus1205741119860+ (30)



119860+1205741= minus119860minus1198861 (31)




1199092







1= 0) then the


1= 0 below




(0 minus11988721205821) ascends


21205821)

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(32)

being

1198872


1198872

1205821

(33)

we get

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (34)


1205822

2

1205822

1

1205822

1minus 1205741198671205821minus 1205742

1205822

2minus 1205741198671205822minus 1205742

gt 1 (35)


12057411986711988621205822minus 12057421205822

2gt 12057411986711988621205821minus 12057421205822

1(36)

and from it we get

120574119867gt11988611205742

1198862

(37)




2on the right



2to 119875(119909

2) is called



2) = 119909

lowast







2le minus119887

21205822 For 119909

2= 0


0 01 02 03 04 050

01

02

03

04

05


1

05

0

minus05

minus1

minus15

minus2



1198862= 1 120574

1= 01 120574


2= 1




1lt 120574119867) so that

119875(minus11988721205822) lt minus119887

21205822 Furthermore 119875(119909

2) is a continuous


2) must



2) equiv 119909

2on




2rarr 0)


lowast


2)




1

05

0

minus05

minus1

minus15

minus2



2=

1 1205741= 120574119867= 0742 120574


2= 1








(i) 1198861lt 0 120574


(ii) 1198861lt 0 120574





1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We




2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1


1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1


2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1


1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06


2= 1 120574

1= 0 120574


2= 1


15

1

05

0

minus05

minus1

minus15


1198862= 1 120574

1= 0 120574


2= 1



Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1199092







1= 0) then the


1= 0 below




(0 minus11988721205821) ascends


21205821)

exp (120572119905) sin (120573119905 minus 120593) +120573

119872= 0 120587 + 120593 le 120573119905 le

3120587

2+ 120593

(32)

being

1198872


1198872

1205821

(33)

we get

119905 =1

1205741

ln(1205822

2

1205822

1

1205822

1minus 12057411205821minus 1205742

1205822

2minus 12057411205822minus 1205742

) (34)


1205822

2

1205822

1

1205822

1minus 1205741198671205821minus 1205742

1205822

2minus 1205741198671205822minus 1205742

gt 1 (35)


12057411986711988621205822minus 12057421205822

2gt 12057411986711988621205821minus 12057421205822

1(36)

and from it we get

120574119867gt11988611205742

1198862

(37)




2on the right



2to 119875(119909

2) is called



2) = 119909

lowast







2le minus119887

21205822 For 119909

2= 0


0 01 02 03 04 050

01

02

03

04

05


1

05

0

minus05

minus1

minus15

minus2



1198862= 1 120574

1= 01 120574


2= 1




1lt 120574119867) so that

119875(minus11988721205822) lt minus119887

21205822 Furthermore 119875(119909

2) is a continuous


2) must



2) equiv 119909

2on




2rarr 0)


lowast


2)




1

05

0

minus05

minus1

minus15

minus2



2=

1 1205741= 120574119867= 0742 120574


2= 1








(i) 1198861lt 0 120574


(ii) 1198861lt 0 120574





1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We




2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1


1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1


2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1


1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06


2= 1 120574

1= 0 120574


2= 1


15

1

05

0

minus05

minus1

minus15


1198862= 1 120574

1= 0 120574


2= 1



Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










1

05

0

minus05

minus1

minus15

minus2



2=

1 1205741= 120574119867= 0742 120574


2= 1








(i) 1198861lt 0 120574


(ii) 1198861lt 0 120574





1198862= 1 120574

1= 01 120574


2= 1 in Figure 2 We




2= 1 120574

1= 120574119867

1205742= minus5 and 119887

2= 1


1198862= 1 1205741= 0 1205742= minus5 and 119887

2= 1


2= 1 120574

1= 0

1205742= minus5 and 119887

2= 1


1gt 120574119867

have been presented

1

08

06

04

02

0

minus02

minus04

minus06

minus08

minus1

minus02 minus01 0 01 02 03 04 05 06


2= 1 120574

1= 0 120574


2= 1


15

1

05

0

minus05

minus1

minus15


1198862= 1 120574

1= 0 120574


2= 1



Acknowledgments


References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra

































Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article Structural Stability of Planar Bimodal...

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