Instability of a system subject to periodic perturbation (cont'd)

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This is a follow-up to this question.

Consider the following 2-dimensional system
$$
dotx(t) = A(t)x(t) quad x(0)inmathbbR^2,
$$
where $A(t)$ is a 2-dimensional time-varying matrix. Suppose that the origin of the above system is an unstable equilibrium.

Now consider the following "perturbed" system
$$
dotz(t) = (A(t)+delta(t)C)z(t) quad z(0)inmathbbR^2,
$$
where $C$ is a constant $2times 2$ matrix and $delta(t)$ is a zero-mean periodic function of $t$.

My question: Is the origin of the "perturbed" system unstable for every choice of $C$ and $delta(t)$ as above?

As in my previous question, my feeling is that the answer is in the negative; finding an explicit counterexample does not seem easy though.

differential-equations analysis dynamical-systems control-theory stability-theory

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asked Jul 29 at 0:00

Ludwig

737613

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up vote
2
down vote

favorite

1

This is a follow-up to this question.

Consider the following 2-dimensional system
$$
dotx(t) = A(t)x(t) quad x(0)inmathbbR^2,
$$
where $A(t)$ is a 2-dimensional time-varying matrix. Suppose that the origin of the above system is an unstable equilibrium.

Now consider the following "perturbed" system
$$
dotz(t) = (A(t)+delta(t)C)z(t) quad z(0)inmathbbR^2,
$$
where $C$ is a constant $2times 2$ matrix and $delta(t)$ is a zero-mean periodic function of $t$.

My question: Is the origin of the "perturbed" system unstable for every choice of $C$ and $delta(t)$ as above?

As in my previous question, my feeling is that the answer is in the negative; finding an explicit counterexample does not seem easy though.

differential-equations analysis dynamical-systems control-theory stability-theory

share|cite|improve this question

asked Jul 29 at 0:00

Ludwig

737613

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

This is a follow-up to this question.

Consider the following 2-dimensional system
$$
dotx(t) = A(t)x(t) quad x(0)inmathbbR^2,
$$
where $A(t)$ is a 2-dimensional time-varying matrix. Suppose that the origin of the above system is an unstable equilibrium.

Now consider the following "perturbed" system
$$
dotz(t) = (A(t)+delta(t)C)z(t) quad z(0)inmathbbR^2,
$$
where $C$ is a constant $2times 2$ matrix and $delta(t)$ is a zero-mean periodic function of $t$.

My question: Is the origin of the "perturbed" system unstable for every choice of $C$ and $delta(t)$ as above?

As in my previous question, my feeling is that the answer is in the negative; finding an explicit counterexample does not seem easy though.

differential-equations analysis dynamical-systems control-theory stability-theory

share|cite|improve this question

asked Jul 29 at 0:00

Ludwig

737613

This is a follow-up to this question.

Consider the following 2-dimensional system
$$
dotx(t) = A(t)x(t) quad x(0)inmathbbR^2,
$$
where $A(t)$ is a 2-dimensional time-varying matrix. Suppose that the origin of the above system is an unstable equilibrium.

Now consider the following "perturbed" system
$$
dotz(t) = (A(t)+delta(t)C)z(t) quad z(0)inmathbbR^2,
$$
where $C$ is a constant $2times 2$ matrix and $delta(t)$ is a zero-mean periodic function of $t$.

My question: Is the origin of the "perturbed" system unstable for every choice of $C$ and $delta(t)$ as above?

As in my previous question, my feeling is that the answer is in the negative; finding an explicit counterexample does not seem easy though.

differential-equations analysis dynamical-systems control-theory stability-theory

share|cite|improve this question

asked Jul 29 at 0:00

Ludwig

737613

share|cite|improve this question

share|cite|improve this question

asked Jul 29 at 0:00

Ludwig

737613

asked Jul 29 at 0:00

Ludwig

737613

asked Jul 29 at 0:00

Ludwig

737613

737613

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
4
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accepted

The answer would be no, namely a known counter example would be a Kapitza pendulum. This is an inverted pendulum whose base is oscillating vertically. When linearizing at the origin the dynamics can be written as

$$
beginbmatrix
dottheta \ ddottheta
endbmatrix = left(
beginbmatrix
0 & 1 \ alpha & -beta
endbmatrix +
sin(omega,t)
beginbmatrix
0 & 0 \ gamma & 0
endbmatrix
right)
beginbmatrix
theta \ dottheta
endbmatrix,
$$

where $alpha,beta>0$ thus $A(t)$, while actually not time varying, is unstable. For certain choices for $omega$ and $gamma$ this system can be made stable. For example the system is stable when using $beta=omega=gamma=1$ and $alpha=0.2$.

share|cite|improve this answer

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
  – Ludwig
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
  – Kwin van der Veen
  Jul 29 at 19:03
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

The answer would be no, namely a known counter example would be a Kapitza pendulum. This is an inverted pendulum whose base is oscillating vertically. When linearizing at the origin the dynamics can be written as

$$
beginbmatrix
dottheta \ ddottheta
endbmatrix = left(
beginbmatrix
0 & 1 \ alpha & -beta
endbmatrix +
sin(omega,t)
beginbmatrix
0 & 0 \ gamma & 0
endbmatrix
right)
beginbmatrix
theta \ dottheta
endbmatrix,
$$

where $alpha,beta>0$ thus $A(t)$, while actually not time varying, is unstable. For certain choices for $omega$ and $gamma$ this system can be made stable. For example the system is stable when using $beta=omega=gamma=1$ and $alpha=0.2$.

share|cite|improve this answer

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
  – Ludwig
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
  – Kwin van der Veen
  Jul 29 at 19:03
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

The answer would be no, namely a known counter example would be a Kapitza pendulum. This is an inverted pendulum whose base is oscillating vertically. When linearizing at the origin the dynamics can be written as

$$
beginbmatrix
dottheta \ ddottheta
endbmatrix = left(
beginbmatrix
0 & 1 \ alpha & -beta
endbmatrix +
sin(omega,t)
beginbmatrix
0 & 0 \ gamma & 0
endbmatrix
right)
beginbmatrix
theta \ dottheta
endbmatrix,
$$

where $alpha,beta>0$ thus $A(t)$, while actually not time varying, is unstable. For certain choices for $omega$ and $gamma$ this system can be made stable. For example the system is stable when using $beta=omega=gamma=1$ and $alpha=0.2$.

share|cite|improve this answer

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
  – Ludwig
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
  – Kwin van der Veen
  Jul 29 at 19:03
  

  
  
  
  

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

The answer would be no, namely a known counter example would be a Kapitza pendulum. This is an inverted pendulum whose base is oscillating vertically. When linearizing at the origin the dynamics can be written as

$$
beginbmatrix
dottheta \ ddottheta
endbmatrix = left(
beginbmatrix
0 & 1 \ alpha & -beta
endbmatrix +
sin(omega,t)
beginbmatrix
0 & 0 \ gamma & 0
endbmatrix
right)
beginbmatrix
theta \ dottheta
endbmatrix,
$$

where $alpha,beta>0$ thus $A(t)$, while actually not time varying, is unstable. For certain choices for $omega$ and $gamma$ this system can be made stable. For example the system is stable when using $beta=omega=gamma=1$ and $alpha=0.2$.

share|cite|improve this answer

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

The answer would be no, namely a known counter example would be a Kapitza pendulum. This is an inverted pendulum whose base is oscillating vertically. When linearizing at the origin the dynamics can be written as

$$
beginbmatrix
dottheta \ ddottheta
endbmatrix = left(
beginbmatrix
0 & 1 \ alpha & -beta
endbmatrix +
sin(omega,t)
beginbmatrix
0 & 0 \ gamma & 0
endbmatrix
right)
beginbmatrix
theta \ dottheta
endbmatrix,
$$

where $alpha,beta>0$ thus $A(t)$, while actually not time varying, is unstable. For certain choices for $omega$ and $gamma$ this system can be made stable. For example the system is stable when using $beta=omega=gamma=1$ and $alpha=0.2$.

share|cite|improve this answer

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

answered Jul 29 at 12:36

Kwin van der Veen

4,3292826

4,3292826

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
  – Ludwig
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
  – Kwin van der Veen
  Jul 29 at 19:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
  – Ludwig
  Jul 29 at 16:03
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
  – Kwin van der Veen
  Jul 29 at 19:03
  

  
  
  
  

Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
– Ludwig
Jul 29 at 16:03

Thanks! Out of curiosity, does there exist an analytic way to check for which values of $alpha$, $beta$, $gamma$ and $omega$ the origin of the system is unstable?
– Ludwig
Jul 29 at 16:03

1

1

@Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
– Kwin van der Veen
Jul 29 at 19:03

@Ludwig I am not aware of a general analytical method. When I analyzed the stability of this system I numerically evaluated the state transition matrix after one period, after which the dynamics would repeat, so that result could be considered as discrete LTI system. So the stability could be analyzed by looking whether its eigenvalues inside the unit circle.
– Kwin van der Veen
Jul 29 at 19:03

add a comment | 

 

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