Variational Equation or Continuous Dependence on

Initial Condition or Trajectory Sensitivity

& Floquet Theory & Poincaré Map

1. General idea of trajectory sensitivity .........................................................................2 2. Homogeneous Linear Time Invariant System ..........................................................3 3. Non - Homogeneous Linear Time Invariant System ................................................3 Example 1 (LTI): ...........................................................................................................4 4. Homogeneous Linear Time Varying System.............................................................5 Example 2 (LTV):..........................................................................................................5 5. Non – Homogeneous Linear Time Varying System..................................................6 6. Nonlinear system .......................................................................................................6 Example 3 (NL): ............................................................................................................6 Summary ........................................................................................................................8 7. Stability Analysis .......................................................................................................9 Summary:.....................................................................................................................10 8. Floquet Theory.........................................................................................................10 Summary ......................................................................................................................11 9. Poincaré Map ...........................................................................................................12 Overall Summary:........................................................................................................13 Example: Poincaré Map vs Floquet Theory.................................................................14 1. Initial numerical analysis .........................................................................................14 2. Polar Coordinates.....................................................................................................16 2.1 Equilibria and solution of system in polar coordinates..........................................17 2.2 Poincare map and Jacobian from polar coordinates ..............................................17 3. Cartesian coordinates ...............................................................................................19 3.1 Solution in Cartesian coordinates ..........................................................................19 3.2 Jacobian in Cartesian coordinates..........................................................................20 3.3 Perturbation calculation using the Jacobian...........................................................21 Floquet Theory.............................................................................................................23 4. Jacobian around the periodic orbit...........................................................................23 5. Solution of Matrix Differential Equation.................................................................24 5.1 Calculation of Monodromy matrix ........................................................................24 5.2 Perturbation calculation using the Monodromy matrix .........................................26

1/27

1. General idea of trajectory sensitivity

• Assume a solution )00 ,, xtt(φ to an IVP

• How will the orbit behave if we start very close to it?

( )0000 ,, xttx φ=

( )00 ,, xttφ( )000

0

,, xttx

+=+δφ

δ

( )00,, xtt +δφ

( )001 ,, xtt +δφ

( )001 ,, xtt +δφ

Figure 1

• We define:

a. ( ) ( ) ( )000000000 ,,,,,, xttxttxtt φφδφδ Δ=−+=

b. ( ) ( ) ( )000000 ,,,,,, xttxttxtt φδφφ −+=Δ

• Now we have to express ( )00 ,, xttφΔ as a function of the original perturbation:

( ) ( ) ( )

( ) ( )( )

( )⇔

∂∂

=−+

⇔∂

∂+=+

Δ

δφ

φδφ

δφ

φδφ

φ 0

00

,,0000

0

000000

,,,,,,

,,,,,,

00x

xttxttxtt

xxtt

xttxtt

xtt

TS

4444 34444 21

( ) ( ) ( 0000

0000 ,,,,,, xtt

xxttxtt φ

φφ Δ

∂∂

=Δ ) (1)

This is the variational equation.

For we have that 0tt =

( ) ( ) ( ) ( )nnI

xxttxtt

xxttxtt 00 ,, ×=

∂∂

⇔Δ∂

∂=Δ

0

000000

0

0000

,,,,,, φφ

φφ

2/27

So if we want to see the sensitivity to the ICs we need to find ( )

0

00 ,,x

xtt∂

∂φ

2. Homogeneous Linear Time Invariant System

If we have a linear system ( ) ( )tAxtx =& then the solution is:

( ) ( ) ( )0000,, txextt ttA −=φ . Then of course:

( ) ( 0

0

00 ,, ttAex

xtt −=∂

)∂φ (2)

Notice that through that last equation we can give another definition for the

STM (in this exponential matrix): the partial derivative of the solution wrt the

initial condition: ( )0

00 ,,x

xtt∂

∂φ

Thus: ( ) ( ) ( )00000 ,,,, 0 xttextt ttA φφ Δ=Δ −

3. Non - Homogeneous Linear Time Invariant System

If we have a linear system ( ) ( ) ( )tBUtAxtx +=& then the solution is:

( ) ( ) ( ) ( ) ( )∫ −− +=t

t

tAttA dBUetxextt0

0000 ,, ττφ τ

Then of course: ( ) ( 0

0

00 ,, ttAex

xtt −=∂

)∂φ (3)

Thus: ( ) ( ) ( )00000 ,,,, 0 xttextt ttA φφ Δ=Δ −

(see example 1 in the following page)

3/27

Example 1 (LTI):

% initial clc, clear, close syms tau t; t0=0; A=[1 -4; 2 3]; B=[1;1]; U=sin(t);Utau=sin(tau); % nominal orbit x0=[0.01;0.01]; xsol=expm(A*(t-t0))*x0+int(expm(A*(t-tau))*B*Utau,tau,0,t); cnt=1; t2=0:0.01:0.5; for t1=t2; xsol1(cnt,1:2)=subs(xsol,t,t1); cnt=cnt+1; end % perturbed orbit x0=[0.01;0.01]+0.005; xsol=expm(A*(t-t0))*x0+int(expm(A*(t-tau))*B*Utau,tau,0,t); cnt=1; t2=0:0.01:0.5; for t1=t2; xsol2(cnt,1:2)=subs(xsol,t,t1); cnt=cnt+1; end % plot there difference subplot(2,1,1) plot(xsol2(:,1)-xsol1(:,1),xsol2(:,2)-xsol1(:,2)) xlim([min(xsol2(:,1)-xsol1(:,1)), max(xsol2(:,1)-xsol1(:,1))]) % Calculate the diff Dx0=[0.005; 0.005]; J=expm(A*(t-t0)); cnt=1; t2=0:0.01:0.5; for t1=t2; Dx(cnt,1:2)=subs(J,t,t1)*Dx0; cnt=cnt+1; end subplot(2,1,2) plot(Dx(:,1),Dx(:,2),'r') xlim([min(Dx(:,1)), max(Dx(:,1))])

-20 -15 -10 -5 0x 10

-3

0

0.01

0.02

x1

x 2

-20 -15 -10 -5 0 5x 10

-3

0

0.01

0.02

x1

x 2

4/27

4. Homogeneous Linear Time Varying System

If we have a linear system ( ) ( ) ( )txtAtx =&

Then the solution is ( ) ( ) 0000 ,,, xttxtt STMΦ=φ and hence of course

( ) ( )0,tt0

00 ,,x

xttSTMΦ=

∂∂φ

(4)

To find now this STM we have to numerically solve:

( ) ( ) ( ) ( ) nnSTMSTMSTM ItttttAttdtd

×=ΦΦ=Φ 0000 ,,,,

Notice that through that last equation we can give another definition for the STM (in this case: the partial derivative of the solution wrt the initial condition:

( )0

00 ,,x

xtt∂

∂φ

Example 2 (LTV):

Figure 2

5/27

5. Non – Homogeneous Linear Time Varying System

The analysis here is exactly the same as before so no examples are given.

6. Nonlinear system

If we have a nonlinear system:

( ) ( )( )

( ) ( )( )

( ) ( )( )( )

( )

( )

( )

( )( )( )

( )

( )( ) ( )( )

( )00,,

00

0

00

00

00

0

00

0 0

00

00

00

0

00

0 0

00

0

00

000000

,,,,,

,,,,

,,,,,

,,,,

,,,,,

,,,,,

,,,,,

000

0xtt

xtxfxtt

dtd

xxt

xtttxttf

xxtt

dtd

dx

xtxtxtfI

xxtt

dx

xtfIx

xtt

dxtfxxtt

xxttx

x

JacobiantY

t

t

t

φφ

τφφφφ

ττφτφ

ττφφ

τττφφ

τττφφ

φ=∂∂

=

⇔∂

∂∂

∂=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

∂∂

⇔∂

∂∂

∂+=

∂∂

⇔∂

∂+=

∂∂

⇔+=

∫

∫

∫

44 344 2143421

Now, the RHS must be equal to the LHS for 0tt = which means that

( ) nnx Ixtt ×=000 ,,0

φ . Also, in a linear system ( ) ( )tAx

txf=

∂∂ , which means again that

( )00 ,,0

xttxφ is nothing more than the STM.

Example 3 (NL):

Consider the system ( )

( ) ⎪⎪⎭

⎪⎪⎬

⎫

−−+=

−−+−=

22

22

1

1

yxyxdtdy

yxxydtdx

. The numerical solution for x(0)=1

and y(0)=0 is:

6/27

-1 -0.5 0 0.5 10

0.2

0.4

0.6

0.8

1

Figure 3

The Jacobian is: ( )⎥⎦

⎤⎢⎣

⎡

−−−−−−−

=∂

∂22

22

31212131

yxxyxyyx

xxf which needs to be evaluated

along the orbit ( 00 ,, xtt )φ in order to find the STM:

( )( ) ( )( )

( )00,,

00 ,,,,,0

000

xttx

txfxttdtd

xxttx

x φφφ=∂

∂= , ( ) nnx Ixtt ×=000 ,,

0φ

The Jacobian is:

Figure 4

Hence, after 2s the STM has the value:

>> [DPA' DPB'] ans = -0.00762199518287 -0.90929742682579 0.01665436331215 -0.41614683654700

7/27

So let’s add a small perturbation and I expct the perturbation

after 2s to be:

( ) ⎥⎦

⎤⎢⎣

⎡=Δ

002.0001.0

,, 000 xttφ

>> [DPA' DPB']*[0.001; 0.002] ans = -0.00182621684883 -0.00081563930978

x

y

original2

x

y

original1

e2

To Workspace1

e1

To Workspace

>> [e1; e2] ans = -0.00182360306785 -0.00081662773015

Summary

8/27

So to summarise: • To find how the perturbations evolve we use:

( ) ( ) ( )0000

0000 ,,

,,,, xtt

xxtt

xtt φφ

φ Δ∂

∂=Δ

• Thus we need ( )0

00 ,,x

xtt∂

∂φ which is the STM.

• Hence to find the STM:

o For LTI systems ( ) ( )0

0

00 ,, ttAex

xtt −=∂

∂φ

o For LTV systems ( ) ( ) ( )0,tt , nI ×=0, tAttdtd

STMSTM Φ=Φ ( ) nSTM ttΦ 00 ,

o For NL systems again solve the previous equation but now

( ) ( )( )00 ,,

,

xttxxtxftA

φ=∂∂

= .

7. Stability Analysis

Now, let’s assume that we have a periodic orbit then we can still apply the previous

method and we can try to find the perturbations using:

( ) ( ) ( )0000

0000 ,,,,,, xtt

xxttxtt φ

φφ Δ

∂∂

=Δ or ( ) ( ) ( 0000000 ,,,,,, xttxttxtt )φφ ΔΦ=Δ

Notice that we used instead of ( 00 ,, xttΦ ) ( )0, ttΦ as in the general nonlinear case

( )0

00 ,,x

xtt∂

∂φ may still be a function of x0.

This matrix now can be found by solving

( ) ( )( )

( )0,,

0 ,,,00

ttx

txfttdtd

STMxttx

STM Φ∂

∂=Φ

=φ, ( ) nnSTM Itt ×=Φ 00 ,

Let’s evaluate the above perturbations at t=t0+T:

( ) ( ) ( )000000000 ,,,,,, xttxttTxtTt φφ Δ+Φ=+Δ

Then we call the matrix the Monodromy Matrix (MM) of the limit

cycle.

( 000 ,, xttT +Φ )

9/27

Now,( ) ( ) ( )( ) ( ) ( ) (( ) ( ) ( )000000

2000

000000000000

000000000

,,,,,,2

,,,,,,,,2,,,,,,2

xttxttTxttT

xttxttTxttTxttTxttTxttTxttT

φφ

φφ )φφ

Δ+Φ=+Δ

⇔Δ+Φ+Φ=+Δ⇔+Δ+Φ=+Δ

And also, ( ) ( ) ( )000000000 ,,,,2,,2 xttxttTxttT φφ Δ+Φ=+Δ

Thus we immediately see that ( ) ( )0002

000 ,,,,2 xttTxttT +Φ=+Φ

( )000 ,, xtt+

( )

which implies

that: or that ( )000 ,, TxttkT kΦ=+Φ

( ) ( )00000 ,,,, xttxt φΔ 0000 ,, tTxttkT k +Φ=+φΔ

Using eigenvalue decomposition: ( ) ( )0001

000 ,,,, xttUUxttkT k φφ ΔΛ=+Δ −

Which implies that if the all eigenvalues of the monodromy matrix have absolute

value less than 1 the orbit is stable.

Summary:

So to summarise:

• ( ) ( ) ( )0000

0000 ,,

,,,, xtt

xxtt

xtt φφ

φ Δ∂

∂=Δ

• Thus we need ( )0

00 ,,x

xtt∂

∂φ which is the STM.

• Hence to find the STM:

o For LTI systems ( ) ( )0ttA −

0

00 ,,e

xxtt

=∂

∂φ

o For LTV systems ( ) ( ) ( )0,tt , nI ×=0, tAttdtd

STMSTM Φ=Φ ( ) nSTM ttΦ 00 ,

o For NL systems again solve the previous equation but now

( ) ( )( )00 ,,

,

xttxxtxftA

φ=∂∂

= .

• Evaluate the STM at Ttt += 0

8. Floquet Theory

Another way to see the above is to linearise around the limit cycle:

10/27

( ) ( )( )

( ) ( )( ) ( ) ( )( ) ( )( ) ( )( )( ) ( )

( ) ( ) ( )( )( ) ( )

( ) ( )( )( ) ( )txtx

ttxftxdtd

txtx

ttxftxtx

txtx

ttxfttxfttxtxfttxftx

ttxftx

p

p

p

xx

xxp

xxpp

pp

Δ∂

∂=Δ

⇔Δ∂

∂=−

⇔Δ∂

∂+=Δ+==

=

=

=

=

,

,

,,,,

,

&&

&

&

Which in the previous notation this can be written as:

( ) ( )( )( ) ( )

( 00,,

00 ,,,,,00

xtttx

ttxfxttdtd

xttx

φφφ

Δ∂

∂=Δ

=

) and thus the solution is:

( ) ( ) ( )0000000 ,,,,,, xttxttxtt STM φφ Δ×Φ=Δ which again implies that

( ) ( )( )( ) ( )

( 00,,

00 ,,,,,00

xtttx

ttxfxttdtd

STMxttx

STM Φ∂

∂=Φ

=φ00

,,00 ,,,,,

00

xtttx

ttxfxttdtd

STMxttx

STM Φ∂

∂=Φ

=φ

) and that and that ( ) IxttSTM =( ) IxttSTM =Φ 00 ,, .

Hence we have a similar equation to the one before and hence we get the same result.

Thus to find the stability properties we have to find the monodromy matrix and each

eigenvalues.

Summary: • Linearise the system around the periodic orbit and hence express the

perturbations through a LTV differential equation • To solve that we need the STM which can be found by solving a matrix

differential equation:

• ( ) ( )( )( ) ( )

( )00,,

00 ,,,,,00

xtttx

ttxfxttdtd

STMxttx

STM Φ∂

∂=Φ

=φ

, ( ) IxttSTM =Φ 00 ,,

• The stability is found by evaluating ( )00 ,, xttSTMΦ after 0tTt += .

11/27

9. Poincaré Map

Going back to the general solution of a nonlinear (it can also be linear) system:

( ) ( )( )∫+=t

dxtfxxtt0

00000 ,,,,, τττφφ

Now, the stroboscopic Poincare map is nothing more than:

( ) ( )( ) ( ) ( )( )∫∫+

+

+

+==⇔+=+00

001

0000000 ,,,,,,,,

tT

nnnn

tT

dxtfxxPxdxtfxxttT τττφτττφφ

( )( ) ( )( ) 0,,,,,,00

00

00 =⇔=+ ∫∫

++ tT

FPFP

tT

FPFP dxtfxdxtfx τττφτττφThe fixed point is

The stability analysis is similar to the previous case:

( ) ( )( )( )

( )∫+

∂∂

∂∂

+=∂

∂ 0

0 0

00

00

00 ,,,,

,,,tT

FP

FP dx

xtxtxtfI

xxP ττφ

τφττφ

Hence the monodromy matrix is nothing more than the Jacobian of the Poincaré

map!

Unfortunately again in the general case we have to numerically solve the matrix

differential equation:

( )( ) ( )( )

( )FPxxttx

FPx xttx

txfxttdtd

FPFP

,,,,, 0,,

0 00

φφφ=∂

∂= for [ ]Tt ,0∈

Summary:

• Evaluate the solution after T time: ( )000 ,, xttT +φ .

• The stability can be found from ( )0

00 ,,x

xtt∂

∂φ .

• To calculate this, again we have to solve a matrix differential equation.

12/27

Overall Summary: 1. In all cases we start from assuming that we have the solution ( )00 ,, xttφ .

( ) ( ) ( )0000

0000 ,,,,,, xtt

xxttxtt φφφ Δ

∂∂

=Δ 2. Express the perturbations as:

( ) ( )0

0000

,,,,x

xttxttSTM ∂∂

=Φφ solve the matrix DE: ( ) ( ) ( )0000 ,,,, xtttAxtt

dtd

STMSTM Φ=Φ , ( ) nnSTM Ixtt ×=Φ 000 ,, 3. To find

4. The solution of the above equation will give ( )00 ,, xttSTMΦ and hence we can get the MM: ( )000 ,, xttTSTM +Φ . The eigenvalues of this will

determine the stability as ( ) ( ) ( )0000

00000 ,,,,,, xtt

xxttxtkTt

k

φφφ Δ⎟⎟⎠

⎞⎜⎜⎝

⎛∂

∂=+Δ

5. For the Floquet Theory we must first derive ( ) ( )( )( ) ( )

( )00,,

00 ,,,,,00

xtttx

ttxfxttdtd

xttx

φφφ

Δ∂

∂=Δ

=

which is LTV and hence we need the STM to

express ( ) ( ) ( )0000

0000 ,,,,,, xtt

xxttxtt φφφ Δ

∂∂

=Δ .

6. For the Poincare map it is exactly the same as before but now we write directly: ( ) ( ) ( )0000

000000 ,,,,,, xtt

xxtTtxtTt φφφ Δ

∂+∂

=+Δ .

Thus • The Floquet Theory needs an extra step than the Trajectory Sensitivity • The Poincaré map is just a special case of the Trajectory Sensitivity

13/27

Example: Poincaré Map vs Floquet Theory

1. Initial numerical analysis

A system is given by

( )( ) ⎪⎪⎭

⎪⎪⎬

⎫

−−+=

−−+−=

22

22

1

1

yxyxdtdy

yxxydtdx

The EPs are:

( )( ) ⎪⎭

⎪⎬⎫

−−+=

−−+−=22

22

10

10

yxyx

yxxy

From the 1st eqn: ( ) ( )2222 101 yxxyyxxy −−=⇔=−−+− And by replacing that at the 2nd eqn:

( )( )( ) ( )⎪⎩

⎪⎨⎧

=−−+

=⇒=⇒−−+=

−−−−+=

nt inconsiste is which 011

0010

110

222

222

2222

yx

yxyxxx

yxyxxx

Hence there is only one equilibrium point at the origin. The system has a stable limit cycle as it can be seen by its numerical simulation.

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

To study the system we can use polar coordinates:

( )( )θθ

sincos

ryrx

==

which will give: and 222 yxr += ⎟⎠⎞

⎜⎝⎛= −

xy1tanθ

14/27

0 2 4 6 8 10 12 14 16 18 200.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

0 2 4 6 8 10 12 14 16 18 20-4

-3

-2

-1

0

1

2

3

4

15/27

2. Polar Coordinates

Lets calculate the model in polar coordinates:

( )

( )( ) (( )( )2222

222

111

1

222

yxyxyyxxyx )rdt

drdtdyy

dtdxx

rdtdr

dtdyy

dtdxx

dtdrr

yxdtd

dtdr

−−++−−+−=

⎟⎠⎞

⎜⎝⎛ +=

+=

+=

Also, ( )( )θθ

sincos

ryrx

==

, so , Hence 222 ryx =+ ( )( ) (( ))( )22 111 ryxyrxyxrdt

dr−++−+−=

( ) ( )( )( ) ( )( )

( ) ( )( )( )( )( )( )( )22

222

2222

2222

2222

11

11

111

111

111

rrrdt

dr

ryxrdt

dr

ryrxrdt

dr

ryyxrxxyrdt

dr

ryyxrxxyrdt

dr

−=

−+=

−+−=

−++−+−=

−++−+−=

( )21 rrdtdr

−=

And

( )( ) ( )( )( )

( )( ) ( )( )( )

( )( ) ( )( )( )

( ) ( )( )

( ) 111

111

111

111

111

1

1

1tan

22

222

22222

22222

222

222222

22

22

2

221

==+

=−−+−+

=−+−−−+

=−+−−−+

=+

−−+−−−−+

=+

⎟⎠⎞

⎜⎝⎛ −

=+

−=

⎟⎠⎞

⎜⎝⎛+

⎟⎠⎞

⎜⎝⎛=⇒⎟

⎠⎞

⎜⎝⎛= −

rr

ryx

rrxyyryxx

rrxyyryxx

ryrxyxryx

yxyyxxyxyxyx

yxy

dtdxx

dtdy

yxx

x

ydtdxx

dtdy

xyx

ydtd

dtd

xy θθ

16/27

So the system in polar coordinates is:

( )1

1 2

=

−=

dtd

rrdtdr

θ , Simulink validated this as well.

-u(2)+u(1)*(1-u(1)^2-u(2)^2)

u(1)+u(2)*(1-u(1)^2-u(2)^2)

y0

x0

sqrt(u(1)^2+u(2)^2)

atan(u(2)/u(1))

2y

1x

y

x1

x theta

r

Zero-OrderHold

MATLABFunction

MATLAB Fcn3

MATLABFunction

MATLAB Fcn2

MATLABFunction

MATLAB Fcn1

MATLABFunction

MATLAB Fcn

1s

Integrator1

1s

Integrator

x

y

2

2

2

2

2r

2.1 Equilibria and solution of system in polar coordinates

The 2nd equation has no EP but the first has at:

( )⎩⎨⎧

==

⇒=−10

01 2

rr

rr

The Jacobian is ( )⎩⎨⎧

−==

⇒−=−∂∂

2:11:0

31 23

rr

rrrr

. So the origin is unstable and r=1 is

stable. The ODE in polar coordinates can be solved and we have:

( )

( ) tt

er

rtr t

+=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−−

00

21

22

00

,

111,

θθθ

So we can see that the angle θ will continuously increase while the amplitude r will converge to 1. 2.2 Poincare map and Jacobian from polar coordinates

But regarding the original system the cosine/sine of the angle will be periodic of period π2 and as the amplitude will converge to a steady state value then this implies that the system will have a limit cycle which will be a perfect circle of radius 1, centre at the origin and a period of π2 .

17/27

So if we start at r0 then the next point after one period (Poincare map) is:

( ) ( )2

1

42

000 111,2

−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+== ππ e

rrPrr , which can be written as:

( )2

1

421 111

−−

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+== πe

rrPr

nnn

To find the fixed point:

( ) ( ) 1111111

11111111111

24424242422

42

22

1

42

21

42

=⇒=⇔−=−⇔=−+⇔=−+

⇔=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+⇔=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+⇔

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−−−−−

−−

−

−

nnnnnnn

nn

nn

nn

rreererererr

er

rer

rer

r

πππππ

πππ

As the FP is 1 then the Jacobian at that point is:

( ) ( ) 23

42

0

30

42

23

42

342 111111

0

−

−−

=

−

−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

∂∂

⇒⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

∂∂ ππππ e

rre

rrPe

rre

rrP

rrn

n

nn

n

n

n

And hence: ( )

11048.3 64

12 <×==

∂∂ −−

=

πerrP

nrn

n and hence it is stable.

Or that simply the perturbations around the fixed point are given by: ( ) ( )nrenr Δ=+Δ − π41

which has been crosschecked using Simulink. To do that I simulated the original system at r0=1 and the same system with r0=0.1 and I sampled their difference with π2=T . Then I simulated the ( ) ( )nrenr Δ=+Δ − π41 with initial condition 1-0.9=0.1 and I compared the previous difference and that output. The smaller the IC the smaller their difference:

18/27

3. Cartesian coordinates

3.1 Solution in Cartesian coordinates

Now let’s find the solution in the x-y plane:

( )( )

( )

( )

( ) ( ) ( )( ) ( ) ( )

( ) ( )( )

( )

( )⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⇒

⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪

⎬

⎫

⎭⎬⎫

==

⇒⎭⎬⎫

==

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⇒⎭⎬⎫

==

−

−

−

−

−

−

−

−

=

tex

y

tex

x

yrx

ryrx

ter

y

ter

x

ryrx

t

t

t

t

sin111

cos111

0000

0sin000cos00

sin111

cos111

sincos

21

220

21

2202

1

22

0

21

22

000θ

θθ

19/27

This has been validated by Simulink. I.e. I simulated the previous set of equations that give analytical solution and compare that with the numerical that I get from Simulink. The error was 10-14.

If I do not assume that θ =0:

( )( )

( )

( )

( ) ( ) ( )

( ) ( )( )

⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++=

⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++=

⇒

⎪⎪⎪⎪⎪⎪

⎭

⎪⎪⎪⎪⎪⎪

⎬

⎫

⎪⎭

⎪⎬

⎫

⎟⎟⎠

⎞⎜⎜⎝

⎛=

+=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⇒⎭⎬⎫

==

−

−

−

−

−

−

−

−

−

−

−

=

txy

eyx

y

txy

eyx

x

xy

yxr

ter

y

ter

x

ryrx

t

t

t

t

0

012

1

22

02

0

0

012

1

22

02

0

1

22

0

21

22

0

0

21

22

00

tansin111

tancos111

00tan0

000

sin111

cos111

sincos 0

θ

θ

θ

θθ θ

3.2 Jacobian in Cartesian coordinates

Now let’s find the Poincare map in the x-y domain:

20/27

( )

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

=

−

−

−

−

−

−

=

π

π

π

π

π

2tansin111

2tancos111

,

0

012

1

42

02

0

0

012

1

42

02

0200

xy

eyx

xy

eyx

yx tP

The Jacobian of that is rather cumbersome so I will use Matlab

clc, clear syms x0 y0 T r2=x0^2+y0^2; ra=(1/r2)-1; rb=ra*exp(2*T); rb=ra*exp(-2*T); rc=1/sqrt(1+rb); rat=y0/x0; th1=atan(rat)+T; f1=rc*cos(th1); f2=rc*sin(th1); Df11=diff(f1,x0); Df12=diff(f1,y0); Df21=diff(f2,x0); Df22=diff(f2,y0); Df11=subs(Df11,{T,x0,y0},{2*pi,1,0}); Df12=subs(Df12,{T,x0,y0},{2*pi,1,0}); Df21=subs(Df21,{T,x0,y0},{2*pi,1,0}); Df22=subs(Df22,{T,x0,y0},{2*pi,1,0}); DF=[Df11 Df12; Df21 Df22] eig(DF)

DF = 0.00000348734236 0.00000000000000

-0.00000000000000 1.00000000000000

We get 2 eigenvalues: 0.00000348734236 1.00000000000000 The 1st one is the same as the one that we got from the polar coordinates and 1 is from the fact that the system is autonomous. Notice that we could have calculated the Jacobian if we had solved

( )( ) ( )( )

( )FPxxttx

FPx xttx

txfxttdtd

FPFP

,,,,, 0,,

0 00

φφφ=∂

∂= for [ ]T,0t∈

3.3 Perturbation calculation using the Jacobian

21/27

>> DF*[0.001;0.001] ans = 1.0e-003 * 0.00000348734236 1.00000000000000

x0

y0

u(1)+u(2)*(1-u(1)^2-u(2)^2)

-u(2)+u(1)*(1-u(1)^2-u(2)^2)

x0+0.001

y0+0.001

u(1)+u(2)*(1-u(1)^2-u(2)^2)

-u(2)+u(1)*(1-u(1)^2-u(2)^2)

Subtract Scope

MATLABFunction

MATLAB Fcn3

MATLABFunction

MATLAB Fcn2

MATLABFunction

MATLAB Fcn1

MATLABFunction

MATLAB Fcn

1s

Integrator3

1s

Integrator2

1s

Integrator1

1s

Integrator

-4.955e-007

0.000999

Display

2

2

2

2

y

x

2

2

2

2

y

x 2

22

Note: If I did not know the IC for the Poincare map, i.e. its fixed point I have to use a Newton Rapshon method.

22/27

Floquet Theory

4. Jacobian around the periodic orbit

So now I have a nonlinear system with a stable periodic orbit with T=6.28… and also I know its solution. Now I will try to use the Floquet Theory. The nonlinear vector field is:

( ) ( )( ⎥

⎦

⎤⎢⎣

⎡

−−+−−+−

= 22

22

11,

yxyxyxxyyxf ) and hence ( ) ( )yx

dtd ,fXfX

==

So the Jacobian around the orbit is:

( )( )( ) ( )( )

( )( ) ( )( ) =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂−−+∂

∂−−+∂

∂−−+−∂

∂−−+−∂

=∂∂

yyxyx

xyxyx

yyxxy

xyxxy

2222

2222

11

11

XXf

( )( )( ) ( )( )

( )( ) (( ))⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂−−+∂

∂−−+∂

∂−−+−∂

∂−−+−∂

=∂∂

yyyxyx

xyyxyx

yxyxxy

xxyxxy

3232

2323

XXf which will give:

( )

⎥⎦

⎤⎢⎣

⎡

−−−−−−−

=∂∂

22

22

31212131

yxxyxyyx

XXf

Now this must be evaluated around the orbit:

( )

( )⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−−

−−

tex

y

tex

x

t

t

sin111

cos111

21

220

21

220

starting

on the point x0=1, y0=0 as this is a point on the cycle.

( ) ( )( ) ( )⎪⎭

⎪⎬⎫

=

=

tty

ttx

p

p

sin

cosso:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ⎥⎥⎦⎤

⎢⎢⎣

⎡

−−−−−−−=

∂∂

=22

22

sin3cos1sincos21sincos21sincos31

tttttttt

PXXXXf

Or:

23/27

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

−−−−−=

∂∂

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−−−−−−=

∂∂

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−−−−−−−−=

∂∂

=

=

=

2

2

2

2

222

222

sin22sin12sin1cos2

sin211sincos21sincos211cos21

sinsin2cos1sincos21sincos21sincoscos21

tttt

tttttt

tttttttttt

P

P

P

XX

XX

XX

XXf

XXf

XXf

For t=0 to t=T=6.28… The elements of the Jacobian matrix are:

5. Solution of Matrix Differential Equation

5.1 Calculation of Monodromy matrix

In order to find the modnoromy matrix I need to to solve the MDE: ( ) ( ) ( )

( ) ( ) ( )00

00

,,

,,

tttdt

ttd

ttdt

ttd

ΦAΦ

ΦXXfΦ

PXX

=

∂∂

==

With the IC = I2 The following model will calculate the matrix A from t=0 to t=T.

24/27

1

x1

1

x0

cos

T rigonometricFunction2

sin

T rigonometricFunction1

sin

T rigonometricFunction

DP1

To Workspace1

DP

To Workspace

MatrixMultiply

Product2

MatrixMultiply

Product1

|u|2

MathFunction1

|u|2

MathFunction

1s

Integrator2

1s

Integrator

-K-

Gain6

-K-

Gain5

-K-

Gain4

-K-

Gain3

-2

Gain2

-2

Gain1

2

GainClock

>> DP=[DP1' DP2'] DP = 0.00000348734236 -0.00000000000004 0.00000000000004 0.99999999999983 Notice that this result is very close to the one that we got from the Poincare Map. Now, that the monodromy matrix is found we can find the Floquet multipliers: >> eig(DP) ans = 0.00000348734236 0.99999999999983 Which are very close to the ones from the PM. Another way to do that is to numerically solve the original system and then to use the

numerical solution to ( )⎥⎦

⎤⎢⎣

⎡

−−−−−−−

=∂∂

22

22

31212131

yxxyxyyx

XXf with

( )

( )⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

−−

−−

tex

y

tex

x

t

t

sin111

cos111

21

220

21

220

.

It is important though to remember that the system must start from x0=1, y0=0.

25/27

set x0, y0

1

x1

1

x0

x

y

original

cos

T rigonom etricFunction2

sin

T rigonom etricFunction1

sin

T rigonom etricFunction

DPB

T o Workspace3

DPA

T o Workspace2

DP2

To Workspace1

DP1

T o Workspace

MatrixM ultiply

Product4

MatrixM ul tiply

Product3

MatrixMultiply

Product2

MatrixMultiply

Product1

x

Polar

In1

In2

Numerical of anali tic

|u|2

M athFunction1

|u|2

MathFunction

MAT LABFunction

MAT LAB Fcn4

MAT LABFunction

MATLAB Fcn3

MAT LABFunction

MATLAB Fcn2

MAT LABFunction

MAT LAB Fcn1

Jacobian of Poincare

1s

Integrator3

1s

Integrator2

1s

Integrator1

1s

Integrator

-K-

Gain9

-K-

Gain8

-K-

Gain7

-K-

Gain6

-K-

Gain5

-K-

Gain4

-K-

Gain3

-2

Gain2

-K-

Gain10

-2

Gain1

2

GainClock

x

x

y

y

t 2t

2,1

1,2

2,2

1,1

[2x2]

[2x2]

[2x2]

[2x2]

[2x2]

[2x2]

[2x2]

[2x2]

[2x2]

[2x2][2x2]

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1,1

2,2

1,2

2,1

[2x2]

[2x2]

[2x2]

2

2

2

2

2

2

2

2

2

2

[2x2]

[2x2]

[2x2]

[2x2]

[2x2][2x2]

[2x2]

[2x2]

>> eig(DPa) ans = 0.00000348734236 1.00000000000004 Hence we get the same results. 5.2 Perturbation calculation using the Monodromy matrix

So I have found ( )0,tΦ , now I will use this matrix to prove that ( ) ( ) ( )00, XΦX Δ=Δ tt by knowing that . So for t=0.1T I have got from the NL system: ( )0ΔX [ 01.0= ] 0.02572402284312 0.01868959658690 And from the product >> E1=[DP1' DP2']*[0.1;0] E1 = 0.02302539573201 0.01672892922346 These are in a very good agreement.

26/27

27/27