Lecture 6: 2-D conservative systems and centers, closed ...dima/YFX1520/LectureNotes_6.pdf · We have encountered a new type of dynamics represented by the heteroclinic orbit. One

Lecture notes 6 Nonlinear Dynamics YFX1520

Lecture 6 2-D conservative systems and centers closedorbits and limit-cycles null-cline heteroclinic orbitDulacrsquos criterion Poincareacute-Bendixson theorem

Contents

1 Properties of conservative systems 211 Nonlinear centers in 2-D conservative systems 212 Example Mathematical pendulum 2

121 Model and its fixed points 2122 Proof of conserved quantity 3123 Linear analysis 4

2 Limit-cycles 5

3 Testing for closed orbits 631 Dulacrsquos criterion 6

311 Example 1 6312 Example 2 (home assignment) 7

32 Proof of Dulacrsquos criterion 833 Poincareacute-Bendixson theorem 8

331 Example 1 9332 Example 2 Glycolysis 11

Material teaching aids Phase portrait of mathematical pendulum periodic phase portrait of mathematicalpendulum

D Kartofelev 115 1003211 As of February 14 2020


1 Properties of conservative systems

11 Nonlinear centers in 2-D conservative systems

Figure 1 Isolated fixed point where 983171 ≪ 1 No other fixed points exist in close vicinity to the fixedpoint shown with the dashed circle

Slide 3

Centers and conservative systems

Theorem Suppose ~x = ~f(~x) is conservative and ~f is continuously

dicrarrerentiable in ~x 2 R2 E(~x) is a conserved quantity and ~x

is an

isolated fixed point If that fixed point is a local minimum or

maximum of E(~x) then that isolated fixed point ~xis a center ie

all trajectories close to ~xare closed orbits

DKartofelev YFX1520 3 23

Theorem Suppose x = 983187f(983187x) is conservative and 983187f is continuously differentiable in 983187x isin R2 E(983187x)is a conserved quantity and 983187xlowast is an isolated fixed point shown in Fig 1 If that fixed point is alocal minimum or maximum of E(983187x) shown in Fig 2 then that isolated fixed point 983187xlowast is a centerie all trajectories close to 983187xlowast are closed orbits

Figure 2 Closed trajectories close to the local minimum or maximum of the conserved quantity E(983187x)Closed orbits are shown with the red lines

12 Example Mathematical pendulum

121 Model and its fixed points



Slide 4

Mathematical pendulum

Mathematical pendulum1is given in the form

+ sin = 0 (1)

where is the angular displacement For angular velocity = we

write ( =

= sin (2)

1See Mathematica nb file uploaded to course webpage


Normalised and dimensionless model of mathematical pendulum is given by

θ + sin θ = 0 (1)

where θ is the angular displacement shown in Fig 3 For angular velocity ω = θ we rewrite the Eq (1)as follows 983083

θ = ω

ω = minus sin θ(2)

Figure 3 Mathematical pendulum where θ is the downwards angular displacement

Notice that Eq (1) is explicitly independent of both θ and t hence there is no damping or friction of anykind and there are no time-dependent external driving forces This indicates that the system is conservativeLetrsquos study the dynamics of the pendulum model (1) for minus2π le θ le 2π Fixed points (θlowastωlowast) of the systemare the following

(minus2π 0) (minusπ 0) (0 0) (π 0) (2π 0) (3)

122 Proof of conserved quantity

We use Eq (1) to prove the existence of conserved quantity which in this case is energy (see Lecture 5)

θ + sin θ = 0 | middot θ

θθ + sin(θ)θ = 0 (4)

equivalently and for ω = θ we writed

dt

983061ω2

2minus cos θ

983062= 0 (5)

where the first term corresponds to the kinetic energy and the second to the potential energy The potentialenergy V follows directly from Eq (1) Since force F is defined as follows

F = minusdVdθ (6)



and Eq (1) is a balance of forces we derive the potential

V = minus983133

(minus sin θ)dθ =

983133sin θ dθ = minus cos θ + C (7)

where the integration constant C = 0 From result (5) it is clear that the sum of kinetic and potentialenergies is indeed conserved in time The conserved energy or more specifically the Hamiltonian has theform

E(θω) =ω2

2983167983166983165983168Kin en

minus cos θ983167 983166983165 983168Pot en

= const (8)

Slide 5 shows the energy values as plotted against angle θ and angular velocity ω The following numericalfile contains the code that produced the plot

Numerics cdf1 nb1Mathematical pendulum and heteroclinic orbit Integrated numerical solution

Slide 5


Figure Hamiltonian energy surface


123 Linear analysis

Jacobian matrix of Sys (2) is

J =

983091

983109983109983109983107

partθ

partθ

partθ

partωpartω

partθ

partω

partω

983092

983110983110983110983108=

9830610 1

minus cos θ 0

983062 (9)

Evaluation of the matrix about the system fixed points (θlowastωlowast) = (0 0) (plusmnπ 0) (plusmn2π 0) is the following

J983055983055(θlowastωlowast)

= J983055983055(00)

= J983055983055(plusmn2π0)

=

9830610 1minus1 0

983062 (10)

here the trace τ = 0 and the determinant ∆ = 1 According to the linear fixed point classification we havea linear center and according to the theorem presented in Sec 11 this center is also a true nonlinearcenter because it is located at a local minimum of the Hamiltonian (conserved quantity) (8) see Slide 5


= J983055983055(plusmnπ0)

=

9830610 11 0

983062 (11)



here the matrix determinant ∆ = minus1 and we have a saddle Now we have all the information required tosketch the phase portrait Slide 6 shows the phase portrait The qualitatively accurate phase portrait andthe interactive numerical solution of Eq (1) or Sys (2) are presented in the following numerical file


Slide 6


Figure Phase portrait showing five fixed points () = (2 0)( 0) (0 0) ( 0) (2 0) Heteroclinic orbit shown with the red line


We have encountered a new type of dynamics represented by the heteroclinic orbit One could alsoargue that this orbit is homoclinic since the pendulum is 2π-periodic Fixed points (plusmn2π 0) and (0 0)or (minusπ 0) and (π 0) are actually the same fixed points

2 Limit-cycles

A limit-cycle is an isolated closed trajectory Isolated means that neighbouring trajectories are not closedthey spiral either toward or away from the limit-cycle see Fig 4

Figure 4 Limit-cycles are shown with the closed continuous or dashed red lines Continuous lines indicatethe stable and dashed the unstable limit-cycles (Left) Stable or attracting limit-cycle (Middle) Unstableor repelling limit-cycle (Right) Half-stable limit-cycle that is stable from the inside and unstable from theoutside

If all neighbouring trajectories approach a limit-cycle we say that this limit-cycle is stable or attractingOtherwise the limit-cycle is unstable or repelling or in exceptional cases half-stable Half-stable limit-cycle can be stable from inside and unstable from outside as shown in Fig 4 (Right) of vice versa

Stable limit-cycles are very important scientificallymdashthey model systems that exhibit self-sustained os-cillations In other words these systems oscillate even in the absence of external periodic forcing Of thecountless examples that could be given we mention only a few the beating of a heart the periodic fir-ing of a pacemaker neurone daily rhythms in human body temperature and hormone secretion chemical



reactions that oscillate spontaneously and dangerous self-excited vibrations in bridges and airplane wingsthat can lead to structure damage and failure In each case there is a standardnominal oscillation ofsome preferred period waveform and amplitude If the system is perturbed slightly it always returns tothe nominal cycle Limit-cycles are inherently nonlinear phenomena they can not occur in linearsystems (by linear system we mean x = A983187x where system matrix A has constant and real valued coeffi-cients)

3 Testing for closed orbits

31 Dulacrsquos criterion

Dulacrsquos criterion is a negative criterion used to rule out limit-cycles

Slide 7

Dulacrsquos criterion

Let ~x = ~f(~x) be a continuously dicrarrerentiable vector field defined on

a simply connected subset R of a plane If there exists a continuously

dicrarrerentiable real valued function g(~x) such that

div(g~x) = r middot (g~x) (3)

has one sign throughout R then there are no closed orbits lying

entirely in R


In a simply connected region it is possible to shrink the circumference or perimeter of the region to beinfinitely small (not a technical definition) Figure 5 shows a comparison between a simply connected regionand not simply connected region

Figure 5 (Left) Simply connected region R having no holes (Right) Not simply connected region

311 Example 1

Show that system 983083x = x(2minus xminus y)

y = y(4xminus x2 minus 3)(12)

has no closed orbits in simply connected region R where x gt 0 and y gt 0 shown in Fig 6We have to come up with function g(983187x) Letrsquos pick (educated guess)

g =1

xy (13)



Figure 6 Region of interest in 2-D plain

Now we need to study the sign of

div(gx) = nabla middot (gx) =983061

part

partxpart

party

983062middot983061gxgy

983062=

part

partxgx+

part

partygy =

part

partx

983063x(2minus xminus y)

xy

983064+

part

party

983063y(4xminus x2 minus 3)

xy

983064=

part

partx

9830612minus xminus y

y

983062+

part

party

9830614xminus x2 minus 3)

x

983062= minus1

ylt 0 (14)

in region R because y gt 0 Since the sign is strictly negative we do not have closed orbits in the region RWe can confirm this conclusion using a computer Numerical file linked below shows the phase portrait ofSys (12)

Numerics cdf2 nb2Dulacrsquos criterion and limit-cycles A numerical example integrated solution and phase portraitPhase portrait of Sys (12) shown below confirms that there are no closed orbits in region R

0 1 2 3 4

0

1

2

3

4

x(t)

y(t)

Figure 7 Phase portrait of Sys (12) featuring three unstable fixed points shown with the empty bullets

312 Example 2 (home assignment)

Show that system 983083x = y

y = minusxminus y + x2 + y2(15)

has no closed orbits in simply connected region R isin R2We have to come up with function g(983187x) Letrsquos pick (educated guess)

g = eminus2x (16)


div(gx) = nabla middot (gx) = part

partx

983043eminus2xy

983044+

part

party

983045eminus2x

983043minusxminus y + x2 + y2

983044983046=

minus2eminus2xy minus eminus2x +2eminus2xy = minuseminus2x lt 0 (17)

in region R Since the sign is strictly negative we do not have closed orbits in the region R isin R2



Numerics cdf3 nb3Dulacrsquos criterion and limit-cycles Numerical example integrated solution and phase portrait

Phase portrait of Sys (15) shown below confirms that there are no closed orbits in region R isin R2

-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)

Figure 8 Phase portrait of Sys (15) featuring a stable spiral and the an unstable saddle node

32 Proof of Dulacrsquos criterion

Slide 10

Proof by contradiction Dulacrsquos criterionLet C be a closed orbit in subset R

and let A be the region inside C

Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)

Therefore there is no closed orbit C in R


Reminder The dot product of orthogonal vectors is 0

33 Poincareacute-Bendixson theorem

Now that we know how to rule out closed orbits we turn to the opposite task finding a method toestablish that closed orbits exist in particular systems The following theorem is one of the few results inthis direction



Slide 11

Poincare-Bendixson theorem

Suppose that

1 R is a closed bounded subset in R2 called the trapping region

2 ~x = ~f(~x) is a continuously dicrarrerentiable vector field on an open

set containing R

3 R does not contain any fixed points (P ) and

4 there exists a trajectory C that is ldquoconfinedrdquo in R in the sense

that it starts in R and stays in R for all future time

Then either C is a closed orbit or it spirals toward a closed orbit as

t 1 In either case R contains a closed orbit (shown as a heavy

curve in the above figure


Trapping region R usually has an annular shape Figure 9 shows a relatively simple trapping regionTrapping regions are not simply connected subsets of phase plane

Figure 9 Trapping region R with annular donut-like shape

Practical tip for constructing a trapping region Find an annulus such that vector field points into iton its boundaries According to Poincareacute-Bendixson theorem a closed orbit will be trapped in that annulusAlso find and prefer annuli that have minimal area Figure 10 shows an example of such annular region

Figure 10 Annulus R with vector field vectors pointing into it on its boundaries

331 Example 1

Consider the following system given in polar coordinates983083r = r(1minus r2) + micror cos θ

θ = 1(18)

where micro is the control parameter Show that closed orbits exist for small positive micro ie 0 lt micro ≪ 1If micro = 0 then the system takes the form

r = r(1minus r2)

θ = 1(19)

This system is decoupled Angular velocity θ is constant and positive The behaviour of a trajectory inthe radial direction is described by the first equation of this system This equation is similar to previously



introduced Logistic equation 1-D phase portrait of the first equation is shown in Fig 11 We can sketchthe phase portrait corresponding to Sys (19) by combining the above observations Figure 12 shows theresulting phase portrait and the stable limit-cycle associated with the carrying capacity rlowast = K = 1 of thefirst equation present in Sys (19)

Figure 11 Phase portrait of 1-D equation featured in (19) where the quantity similar the carrying capacityK of the Logistic equation is rlowast = K = 1

Figure 12 Phase portrait shown in polar coordinates corresponding to Sys (19) Stable limit-cycle is shownwith the red closed trajectory

Letrsquos consider the full system where micro ∕= 0 According to Poincareacute-Bendixson theorem we need to constructan annular trapping region Figure 13 shows a promising candidate We seek two concentric circles withradii r = rmin and r = rmax such that r lt 0 on the outer circle and r gt 0 on the inner circle Then theannulus 0 lt rmin le r le rmax will be our trapping region Note that there are no fixed points in the annulussince θ gt 0 ∕= 0 hence if rmin and rmax can be found then Poincareacute-Bendixson theorem will imply theexistence of a closed orbit

Figure 13 Annular trapping region shown in polar coordinates Radial distance r = rmin is the innerboundary and r = rmax is the outer boundary of the proposed annulus

Letrsquos show that vector field defined by (18) flows into the selected annulus on its boundaries Firstly weconsider the inner boundary r = rmin where it must hold that r gt 0 for all θ

r = r(1minus r2) + micror cos θ = r(1minus r2 + micro cos θ) gt 0 (20)

Since minus1 le cos θ le 1 a sufficient condition for rmin is

r(1minus r2 minus micro) gt 0 | divide r (21)

1minus r2 minus micro gt 0 (22)

r2 lt 1minus micro (23)



r lt983155

1minus micro (24)

any such r will do as long as micro lt 1 which is fine since in our case micro ≪ 1Secondly we consider the outer boundary r = rmax where it must hold that r lt 0 for all θ

r = r(1minus r2 + micro cos θ) lt 0 (25)

Since minus1 le cos θ le 1 a sufficient condition for rmax is

r(1minus r2 + micro) lt 0 | divide r (26)

1minus r2 + micro lt 0 (27)

r2 gt 1 + micro (28)

r gt983155

1 + micro (29)

Now that rmin and rmax have been found we have proven the existence of a limit-cycle Obtained result canbe confirmed using a computer

Numerics cdf4 nb4Poincareacute-Bendixson theorem Phase portrait of a nonlinear system in polar coordinatesNumerically obtained results agree with the math derived above

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Full equation μ lt 1

Figure 14 Phase portraits of Sys (18) featuring a stable limit-cycle shown with the closed red trajectoryAnnular trapping region is shown with the black concentric circles

Note The limit-cycle will also persist for micro gt 1 cf (24)

332 Example 2 Glycolysis

This example is less contrived compared to the previous one Letrsquos consider a simplified dimensionlessmodel of glycolysis given by 983083

x = minusx+ ay + x2y

y = bminus ay minus x2y(30)

where a and b are the kinetic parameters x and y are the concentrations of ADP (adenosine diphosphate)and F6P (fructose-6-phosphate) molecules respectively Here x y a b gt 0 Using Poincareacute-Bendixsontheorem show that chemical oscillations are possible Determine the values of a and b that lead to oscillatingreaction

A useful tools for studying phase portraits are null-clines The null-clines are curves on the 2-D phaseportrait corresponding to x = 0 and y = 0 In the case x = 0 and for the first equation in Sys (30) we write

x = 0 rArr minusx+ ay + x2y = 0 rArr yx=0(x) =x

a+ x2 (31)



In the case y = 0 and for the second equation in Sys (30) we write

y = 0 rArr bminus ay minus x2y = 0 rArr yy=0(x) =b

a+ x2 (32)

The point where null-clines intersect each other corresponds to the fixed point Null-clines (31) and (32)are sketched by hand (explained during the lecture) in Fig 15 along with some representative vectors Aneasy way to determine the flow direction of the vector field defined by Sys (30) is to estimate or calculateone vector and deduce the others by relying on the fact that the field must be continuous We estimatethat for x ≫ 1 vector components x asymp x2y gt 0 and y asymp minusx2y lt 0 This vector shown in the upper-rightcorner of Fig 15 is used as a starting point to populate the phase portrait with other vectors

Figure 15 Null-clines (31) and (32) Flow direction of the vector field is shown with the arrows Fixedpoint shown with the hollow bullet is assumed to be unstable

Slide 14 shows a promising trapping region If we can show that vector field defined by Sys (30) flowsinto the trapping region then we have proven the existence of a closed orbit

Slide 14

Glycolysis trapping region

Figure Annular trapping region shown with the red lines Local vector

field flow directions are shown with the arrows


Null-clines (31) and (32) are shown with the continuous black lines

First we focus on the outer boundary It is evident that the flow direction on outer boundaries shownwith the continuous red lines on Slide 14 is indeed pointing into the annulus But it is not so clear withthe upper-right slanted part of the boundary shown with the dashed line How was point x = b selectedand why the slope dydx = minus1 was selected

As shown above the slope of the field vectors for x ≫ 1 can be easily approximated from the algebraicallydominant parts of the field (30) x asymp x2y and y asymp minusx2y The slope dydx = yx asymp minus1 This result is not



accurate enough We should compare the sizes of x and minusy more precisely

xminus (minusy) (33)

minus x+ay +x2y + (bminusay minus

x2y ) (34)

bminus x rArr x = b (35)

Henceminus y gt x if x gt b (36)

This inequality implies that the vector field points inward on the slanted line shown on Slide 14 becausedydx is more negative than minus1 and therefore the vectors are steeper than the slanted line Also thequestion regarding the selection of point x = b is answered here

Now letrsquos focus on the vector field flow through the inner boundary of the trapping region

Slides 15ndash19

Glycolysis inner boundary of trapping region

Secondly we focus on the inner boundary of the proposed trapping

region We need to find and show that the fixed point

(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)

is unstable ie it repels the vector field



Letrsquos analyse the fixed point (10) using linear analysis The Jacobian

of Sys (9) has the form

J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)



Jacobian evaluated at the fixed point (10) has the form

J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)

The determinant = det J |(xy) = a+ b2 gt 0 is positive because

a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)



In order to ensure repelling unstable fixed points for gt 0 trace has to be positive The dividing line between repelling unstable fixed

points and stable ones is = 0 Solving

= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)

This result defines a line in the parameter space of Sys (9) For

parameters a and b in the region corresponding to gt 0 we are

guaranteed that Sys (9) has a closed orbitmdashoscillating reaction


In order to ensure repelling unstable fixed points for ∆ gt 0 trace τ has to be positive Figure 16shows the positions of repelling fixed points as they appear on the classification graph for the linearfixed points The result shown on Slide 18 defines a line in the parameter space of Sys (30) shown onSlide 19 For parameters a and b in the region corresponding to τ gt 0 we are guaranteed that Sys (30)has a closed orbitmdashoscillating reaction




Closed orbits

Stable fp

005 010 015a

0204060810b

Figure Parameter space defining the parameter values corresponding to

unstable fixed point (10)


Figure 16 ∆ vs τ fixed point classification graph The marked region is populated by repellers unstablespirals unstable centers unstable stars and unstable degenerate nodes

The result shown on Slide 19 is calculated using the following numerical file

Numerics cdf5 nb5Glycolysis phase portrait and null-clines Numerical solution of glycolysis model

We conclude that the annulus shown on Slide 14 is indeed the trapping region and a closed orbit exists in itNumerical integration shows that closed orbit is a stable limit-cycle Slides 20 and 21 show the phase portraitand numerically integrated time-domain results for a typical case where a = 007 b = 053

Slides 20 21

Glycolysis3 limit-cycle

00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)



Glycolysis time-domain result

05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP


Numerical file that was used to generate the above figures is shown below

Numerics cdf5 nb5Glycolysis phase portrait and null-clines Numerical solution of glycolysis modelPhase portrait and null-clines for a typical case where a = 007 b = 053



00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

Figure 17 Phase portrait and null-clines of Sys (30)

Poincareacute-Bendixson theorem is one of the central results of nonlinear dynamics It says that the dynam-ical possibilities in the phase plane are very limited if a trajectory is confined to a closed bounded re-gion that contains no fixed points then the trajectory must eventually approach a closed orbit Nothingmore complicated is possible This result depends crucially on the two-dimensionality of the plane Inhigher-dimensional systems (n ge 3) Poincareacute-Bendixson theorem no longer applies The theorem alsoimplies that chaos can never occur in the phase plane

Reading suggestion

Link File name CitationPaper1 paper0pdf Evgeni E Selrsquokov ldquoSelf-oscillations in glycolysis 1 A simple kinetic modelrdquo

European Journal of Biochemistry 4(1) pp 79ndash86 (1968)doi101111j1432-10331968tb00175x

Revision questions

1 Expand on the connection between 2-D conservative systems and centers2 Sketch a heteroclinic orbit3 What is limit-cycle4 Sketch a stable limit-cycle5 Sketch an unstable limit-cycle6 Sketch a half-stable (stable from outside) limit-cycle7 Sketch a half-stable (stable from inside) limit-cycle8 Define and sketch a null-cline9 What is Dulacrsquos criterion

10 State Poincareacute-Bendixson theorem11 Does Poincareacute-Bendixson theorem apply to 3-D systems12 Can chaos occur in 2-D systems



1 Properties of conservative systems

11 Nonlinear centers in 2-D conservative systems

Figure 1 Isolated fixed point where 983171 ≪ 1 No other fixed points exist in close vicinity to the fixedpoint shown with the dashed circle

Slide 3

Centers and conservative systems

Theorem Suppose ~x = ~f(~x) is conservative and ~f is continuously

dicrarrerentiable in ~x 2 R2 E(~x) is a conserved quantity and ~x

is an

isolated fixed point If that fixed point is a local minimum or

maximum of E(~x) then that isolated fixed point ~xis a center ie

all trajectories close to ~xare closed orbits


Theorem Suppose x = 983187f(983187x) is conservative and 983187f is continuously differentiable in 983187x isin R2 E(983187x)is a conserved quantity and 983187xlowast is an isolated fixed point shown in Fig 1 If that fixed point is alocal minimum or maximum of E(983187x) shown in Fig 2 then that isolated fixed point 983187xlowast is a centerie all trajectories close to 983187xlowast are closed orbits

Figure 2 Closed trajectories close to the local minimum or maximum of the conserved quantity E(983187x)Closed orbits are shown with the red lines

12 Example Mathematical pendulum

121 Model and its fixed points



Slide 4



+ sin = 0 (1)


write ( =

= sin (2)




θ + sin θ = 0 (1)


θ = ω




(minus2π 0) (minusπ 0) (0 0) (π 0) (2π 0) (3)




θθ + sin(θ)θ = 0 (4)


dt

983061ω2

2minus cos θ

983062= 0 (5)


F = minusdVdθ (6)




V = minus983133

(minus sin θ)dθ =



E(θω) =ω2

2983167983166983165983168Kin en

minus cos θ983167 983166983165 983168Pot en

= const (8)



Slide 5




123 Linear analysis


J =

983091

983109983109983109983107

partθ

partθ

partθ

partωpartω

partθ

partω

partω

983092

983110983110983110983108=

9830610 1

minus cos θ 0

983062 (9)



= J983055983055(00)

= J983055983055(plusmn2π0)

=

9830610 1minus1 0

983062 (10)



= J983055983055(plusmnπ0)

=

9830610 11 0

983062 (11)





Slide 6





2 Limit-cycles











Slide 7







entirely in R




311 Example 1




g =1

xy (13)






part

partxpart

party


983062=

part

partxgx+

part

partygy =

part

partx


xy

983064+

part

party


xy

983064=

part

partx


y

983062+

part

party


x

983062= minus1

ylt 0 (14)



0 1 2 3 4

0

1

2

3

4

x(t)

y(t)






g = eminus2x (16)



partx

983043eminus2xy

983044+

part

party

983045eminus2x


983044983046=







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions





Slide 4



+ sin = 0 (1)


write ( =

= sin (2)




θ + sin θ = 0 (1)


θ = ω




(minus2π 0) (minusπ 0) (0 0) (π 0) (2π 0) (3)




θθ + sin(θ)θ = 0 (4)


dt

983061ω2

2minus cos θ

983062= 0 (5)


F = minusdVdθ (6)




V = minus983133

(minus sin θ)dθ =



E(θω) =ω2

2983167983166983165983168Kin en

minus cos θ983167 983166983165 983168Pot en

= const (8)



Slide 5




123 Linear analysis


J =

983091

983109983109983109983107

partθ

partθ

partθ

partωpartω

partθ

partω

partω

983092

983110983110983110983108=

9830610 1

minus cos θ 0

983062 (9)



= J983055983055(00)

= J983055983055(plusmn2π0)

=

9830610 1minus1 0

983062 (10)



= J983055983055(plusmnπ0)

=

9830610 11 0

983062 (11)





Slide 6





2 Limit-cycles











Slide 7







entirely in R




311 Example 1




g =1

xy (13)






part

partxpart

party


983062=

part

partxgx+

part

partygy =

part

partx


xy

983064+

part

party


xy

983064=

part

partx


y

983062+

part

party


x

983062= minus1

ylt 0 (14)



0 1 2 3 4

0

1

2

3

4

x(t)

y(t)






g = eminus2x (16)



partx

983043eminus2xy

983044+

part

party

983045eminus2x


983044983046=







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions






V = minus983133

(minus sin θ)dθ =



E(θω) =ω2

2983167983166983165983168Kin en

minus cos θ983167 983166983165 983168Pot en

= const (8)



Slide 5




123 Linear analysis


J =

983091

983109983109983109983107

partθ

partθ

partθ

partωpartω

partθ

partω

partω

983092

983110983110983110983108=

9830610 1

minus cos θ 0

983062 (9)



= J983055983055(00)

= J983055983055(plusmn2π0)

=

9830610 1minus1 0

983062 (10)



= J983055983055(plusmnπ0)

=

9830610 11 0

983062 (11)





Slide 6





2 Limit-cycles











Slide 7







entirely in R




311 Example 1




g =1

xy (13)






part

partxpart

party


983062=

part

partxgx+

part

partygy =

part

partx


xy

983064+

part

party


xy

983064=

part

partx


y

983062+

part

party


x

983062= minus1

ylt 0 (14)



0 1 2 3 4

0

1

2

3

4

x(t)

y(t)






g = eminus2x (16)



partx

983043eminus2xy

983044+

part

party

983045eminus2x


983044983046=







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions







Slide 6





2 Limit-cycles











Slide 7







entirely in R




311 Example 1




g =1

xy (13)






part

partxpart

party


983062=

part

partxgx+

part

partygy =

part

partx


xy

983064+

part

party


xy

983064=

part

partx


y

983062+

part

party


x

983062= minus1

ylt 0 (14)



0 1 2 3 4

0

1

2

3

4

x(t)

y(t)






g = eminus2x (16)



partx

983043eminus2xy

983044+

part

party

983045eminus2x


983044983046=







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions









Slide 7







entirely in R




311 Example 1




g =1

xy (13)






part

partxpart

party


983062=

part

partxgx+

part

partygy =

part

partx


xy

983064+

part

party


xy

983064=

part

partx


y

983062+

part

party


x

983062= minus1

ylt 0 (14)



0 1 2 3 4

0

1

2

3

4

x(t)

y(t)






g = eminus2x (16)



partx

983043eminus2xy

983044+

part

party

983045eminus2x


983044983046=







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions








part

partxpart

party


983062=

part

partxgx+

part

partygy =

part

partx


xy

983064+

part

party


xy

983064=

part

partx


y

983062+

part

party


x

983062= minus1

ylt 0 (14)



0 1 2 3 4

0

1

2

3

4

x(t)

y(t)






g = eminus2x (16)



partx

983043eminus2xy

983044+

part

party

983045eminus2x


983044983046=







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions







-10-05 00 05 10 15 20

-10

-05

00

05

10

x(t)

y(t)



Slide 10



Greenrsquos theorem

ZZ

A

r middot ~F

dA =

I

C

~F middot ~n

dl (6)

If ~F = g~x thenZZ

A

hr middot (g~x)

i

| z 6=0

has onesign by

assumption

dA =

I

C

g~x middot ~n

| z =0~n~x

dl (7)








Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions





Slide 11


Suppose that



set containing R












331 Example 1


θ = 1(18)


r = r(1minus r2)

θ = 1(19)

















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions


















r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions





r lt983155

1minus micro (24)







r gt983155

1 + micro (29)



-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)

Case where μ = 0

-15 -10 -05 00 05 10 15-15

-10

-05

00

05

10

15

x(t)

y(t)











a+ x2 (31)





a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions







a+ x2 (32)




Slide 14













x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions








x2y ) (34)





Slides 15ndash19




(x = 0

y = 0)

(x + ay + x2y = 0

b ay x2y = 0) (x y) =

b

b

a+ b2

(10)






J =

0

BBB

x

x

x

yy

x

y

y

1

CCCA=

2xy 1 a+ x2

2xy a x2

(11)




J |(xy) =

0

B

2b2

a+ b2 1 a+ b2

2b2

a+ b2a b4

1

CA (12)


a b gt 0 The trace

= tr J |(xy) =2b2

a+ b2 1 a b2 (13)





= 0 ) 2b2

a+ b2 1 a b2 = 0 (14)

for b gives

b(a) =

r1

2

1 2aplusmn

p1 8a

(15)









Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions






Closed orbits

Stable fp

005 010 015a

0204060810b








Slides 20 21


00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)

10 15 20x(t)

10

15

20

25

y(t)




05 10 15 20

05

10

15

20

25

x(t)

y(t)

0 10 20 30 40 50 60 7000

05

10

15

20

25

t

y(t)

F6P

0 10 20 30 40 50 60 7000

05

10

15

20

t

x(t)

ADP






00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions





00 05 10 15 20 25 30

00

05

10

15

20

25

30

ADF x(t)

F6Py(t)



Reading suggestion



Revision questions




Documents

Lecture 6: 2-D conservative systems and centers, closed ...dima/YFX1520/LectureNotes_6.pdf · We have encountered a new type of dynamics represented by the heteroclinic orbit. One