Variational Equation or Continuous Dependence on … Equation or Continuous Dependence on Initial...
Transcript of Variational Equation or Continuous Dependence on … Equation or Continuous Dependence on Initial...
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
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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
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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:
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-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