Applying Floquet's Theorem - Floquet Multiplier of $-1$ implies solution of at least period $2omega$
I have the linear system $y' = A(t)y$ where $A(t)$ is a continuous n-by-n matrix of period $omega$. I want to show that if $-1$ is a multiplier of this system, then there is a solution of at least period $2omega$. I am stuck on figuring out how to begin.
Here is the relevant material: 
So Step 1: What happens when $-1$ is a multiplier of the system? This means $-1$ is an eigenvalue of the nonsingular matrix $exp(omega R)$ where $R$ is a matrix. Here I do not know how to proceed.
differential-equations
asked Dec 3 '18 at 9:18
Nalt
686
|
show 2 more comments
I have the linear system $y' = A(t)y$ where $A(t)$ is a continuous n-by-n matrix of period $omega$. I want to show that if $-1$ is a multiplier of this system, then there is a solution of at least period $2omega$. I am stuck on figuring out how to begin.
Here is the relevant material:
So Step 1: What happens when $-1$ is a multiplier of the system? This means $-1$ is an eigenvalue of the nonsingular matrix $exp(omega R)$ where $R$ is a matrix. Here I do not know how to proceed.
differential-equations
asked Dec 3 '18 at 9:18
Nalt
686
1
So, $-1$ is an eigenvalue of $exp{(omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(omega)$ equals? What $y(2omega)$ equals?
– Evgeny
Dec 3 '18 at 12:00
Dear Nalt: Would you consider undeleting you post math.stackexchange.com/questions/3038615/… ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers!
– Robert Lewis
Dec 14 '18 at 17:32
@RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose.
– Nalt
Dec 14 '18 at 17:35
Thank you! By the way, what was the "similar post" if you don't mind?
– Robert Lewis
Dec 14 '18 at 17:37
This: math.stackexchange.com/a/1374462/597047
– Nalt
Dec 14 '18 at 17:39
|
show 2 more comments
I have the linear system $y' = A(t)y$ where $A(t)$ is a continuous n-by-n matrix of period $omega$. I want to show that if $-1$ is a multiplier of this system, then there is a solution of at least period $2omega$. I am stuck on figuring out how to begin.
Here is the relevant material:
So Step 1: What happens when $-1$ is a multiplier of the system? This means $-1$ is an eigenvalue of the nonsingular matrix $exp(omega R)$ where $R$ is a matrix. Here I do not know how to proceed.
differential-equations
asked Dec 3 '18 at 9:18
Nalt
686
I have the linear system $y' = A(t)y$ where $A(t)$ is a continuous n-by-n matrix of period $omega$. I want to show that if $-1$ is a multiplier of this system, then there is a solution of at least period $2omega$. I am stuck on figuring out how to begin.
Here is the relevant material:
So Step 1: What happens when $-1$ is a multiplier of the system? This means $-1$ is an eigenvalue of the nonsingular matrix $exp(omega R)$ where $R$ is a matrix. Here I do not know how to proceed.
differential-equations
differential-equations
asked Dec 3 '18 at 9:18
Nalt
686
asked Dec 3 '18 at 9:18
Nalt
686
asked Dec 3 '18 at 9:18
Nalt
686
asked Dec 3 '18 at 9:18
Nalt
686
asked Dec 3 '18 at 9:18
Nalt
686
686
1
So, $-1$ is an eigenvalue of $exp{(omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(omega)$ equals? What $y(2omega)$ equals?
– Evgeny
Dec 3 '18 at 12:00
Dear Nalt: Would you consider undeleting you post math.stackexchange.com/questions/3038615/… ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers!
– Robert Lewis
Dec 14 '18 at 17:32
@RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose.
– Nalt
Dec 14 '18 at 17:35
Thank you! By the way, what was the "similar post" if you don't mind?
– Robert Lewis
Dec 14 '18 at 17:37
This: math.stackexchange.com/a/1374462/597047
– Nalt
Dec 14 '18 at 17:39
|
show 2 more comments
1
So, $-1$ is an eigenvalue of $exp{(omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(omega)$ equals? What $y(2omega)$ equals?
– Evgeny
Dec 3 '18 at 12:00
Dear Nalt: Would you consider undeleting you post math.stackexchange.com/questions/3038615/… ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers!
– Robert Lewis
Dec 14 '18 at 17:32
@RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose.
– Nalt
Dec 14 '18 at 17:35
Thank you! By the way, what was the "similar post" if you don't mind?
– Robert Lewis
Dec 14 '18 at 17:37
This: math.stackexchange.com/a/1374462/597047
– Nalt
Dec 14 '18 at 17:39
1
1
So, $-1$ is an eigenvalue of $exp{(omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(omega)$ equals? What $y(2omega)$ equals?
– Evgeny
Dec 3 '18 at 12:00
So, $-1$ is an eigenvalue of $exp{(omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(omega)$ equals? What $y(2omega)$ equals?
– Evgeny
Dec 3 '18 at 12:00
Dear Nalt: Would you consider undeleting you post math.stackexchange.com/questions/3038615/… ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers!
– Robert Lewis
Dec 14 '18 at 17:32
Dear Nalt: Would you consider undeleting you post math.stackexchange.com/questions/3038615/… ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers!
– Robert Lewis
Dec 14 '18 at 17:32
@RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose.
– Nalt
Dec 14 '18 at 17:35
@RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose.
– Nalt
Dec 14 '18 at 17:35
Thank you! By the way, what was the "similar post" if you don't mind?
– Robert Lewis
Dec 14 '18 at 17:37
Thank you! By the way, what was the "similar post" if you don't mind?
– Robert Lewis
Dec 14 '18 at 17:37
This: math.stackexchange.com/a/1374462/597047
– Nalt
Dec 14 '18 at 17:39
This: math.stackexchange.com/a/1374462/597047
– Nalt
Dec 14 '18 at 17:39
|
show 2 more comments
1 Answer
1
active
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The situation is that you know from the fundamental matrix solution $Phi$ to your ODE system,
$$
Φ'(t)=A(t)Φ(t),~~Φ(0)=I
$$
that $Φ(ω)$ has an eigenvalue $-1$. Additionally you know from the periodicity of $A$ that $Φ(ω+t)=Φ(t)Φ(ω)$.
Let $v$ be an eigenvector, then $y(t)=Φ(t)v$ is a solution with initial value $y(0)=v$ and
$$
y(ω)=Φ(ω)v=-v,~~y(2ω)=Φ(ω)^2v=v.
$$
which means that the trajectory of that solution returns to the initial point.
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
add a comment |
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The situation is that you know from the fundamental matrix solution $Phi$ to your ODE system,
$$
Φ'(t)=A(t)Φ(t),~~Φ(0)=I
$$
that $Φ(ω)$ has an eigenvalue $-1$. Additionally you know from the periodicity of $A$ that $Φ(ω+t)=Φ(t)Φ(ω)$.
Let $v$ be an eigenvector, then $y(t)=Φ(t)v$ is a solution with initial value $y(0)=v$ and
$$
y(ω)=Φ(ω)v=-v,~~y(2ω)=Φ(ω)^2v=v.
$$
which means that the trajectory of that solution returns to the initial point.
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
add a comment |
The situation is that you know from the fundamental matrix solution $Phi$ to your ODE system,
$$
Φ'(t)=A(t)Φ(t),~~Φ(0)=I
$$
that $Φ(ω)$ has an eigenvalue $-1$. Additionally you know from the periodicity of $A$ that $Φ(ω+t)=Φ(t)Φ(ω)$.
Let $v$ be an eigenvector, then $y(t)=Φ(t)v$ is a solution with initial value $y(0)=v$ and
$$
y(ω)=Φ(ω)v=-v,~~y(2ω)=Φ(ω)^2v=v.
$$
which means that the trajectory of that solution returns to the initial point.
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
add a comment |
The situation is that you know from the fundamental matrix solution $Phi$ to your ODE system,
$$
Φ'(t)=A(t)Φ(t),~~Φ(0)=I
$$
that $Φ(ω)$ has an eigenvalue $-1$. Additionally you know from the periodicity of $A$ that $Φ(ω+t)=Φ(t)Φ(ω)$.
Let $v$ be an eigenvector, then $y(t)=Φ(t)v$ is a solution with initial value $y(0)=v$ and
$$
y(ω)=Φ(ω)v=-v,~~y(2ω)=Φ(ω)^2v=v.
$$
which means that the trajectory of that solution returns to the initial point.
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
The situation is that you know from the fundamental matrix solution $Phi$ to your ODE system,
$$
Φ'(t)=A(t)Φ(t),~~Φ(0)=I
$$
that $Φ(ω)$ has an eigenvalue $-1$. Additionally you know from the periodicity of $A$ that $Φ(ω+t)=Φ(t)Φ(ω)$.
Let $v$ be an eigenvector, then $y(t)=Φ(t)v$ is a solution with initial value $y(0)=v$ and
$$
y(ω)=Φ(ω)v=-v,~~y(2ω)=Φ(ω)^2v=v.
$$
which means that the trajectory of that solution returns to the initial point.
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
answered Dec 3 '18 at 15:55
LutzL
56.4k42054
56.4k42054
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
add a comment |
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
Why do we have $Phi(omega)^2$ instead of $Phi(2omega)$? I know we have $Phi(2omega + t) = Phi(t)Phi(2omega)$, but don't see how this translates to having $Phi(2omega) = Phi(omega)^2$
– Nalt
Dec 3 '18 at 16:06
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
You also have the same relation with $ω$ instead of $2ω$, as mentioned in the answer. From there $Φ(2ω)=Φ(ω)^2$ directly follows. That $Φ(nω)=Φ(ω)^n$ is the inspiration to look into a factorization that makes $Φ(t)=P(t)Φ(ω)^{t/ω}$ formally correct.
– LutzL
Dec 3 '18 at 16:14
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
I see. My other question: The exercise wanted had the wording "...show there is a solution of at least period $2omega$". What you showed was there is a solution of period $2omega$. So why is there the "at least" wording?
– Nalt
Dec 3 '18 at 16:20
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
This argument only shows that $2ω$ is one period of the solution. There might be a minimal period that is a fraction of that. Like $y(t)=cos(5t)$ has a period $2pi$ with $y(0)=1$ and $y(pi)=-1$, but the minimal period is $2pi/5$. But as $2ω$ definitely is a period of the solution, the "at least" might have another interpretation.
– LutzL
Dec 3 '18 at 16:28
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
Thank you. For my very first question I just had to write $Phi(omega+omega)$ to get what we wanted. Apologies and thank you for your time! You really helped a lot!
– Nalt
Dec 3 '18 at 16:34
add a comment |
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1
So, $-1$ is an eigenvalue of $exp{(omega R)}$. That means that there is also an eigenvector $v$ corresponding to this eigenvalue. Start with $v$ as an initial condition $y(0)$. What $y(omega)$ equals? What $y(2omega)$ equals?
– Evgeny
Dec 3 '18 at 12:00
Dear Nalt: Would you consider undeleting you post math.stackexchange.com/questions/3038615/… ? I was on the verge of posting what I think is a pretty detailed answer when--poof! The question was gone! Thank you. Cheers!
– Robert Lewis
Dec 14 '18 at 17:32
@RobertLewis Done. It was deleted because I found a similar post, but maybe it is okay. If a mod doesn't like it they'll handle it I suppose.
– Nalt
Dec 14 '18 at 17:35
Thank you! By the way, what was the "similar post" if you don't mind?
– Robert Lewis
Dec 14 '18 at 17:37
This: math.stackexchange.com/a/1374462/597047
– Nalt
Dec 14 '18 at 17:39