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[SOLVED] Floquet

dwsmith

Well-known member
Feb 1, 2012
1,673
For the scalar linear ODE with periodic coefficients,
$$
x' = a(t)x,\quad\quad a(t + T) = a(t),
$$
show that the solution is of the form
$$
x(t) = x_0e^{\mu t}p(t),
$$
where $\mu$ and $x_0$ are constants, and $p(t)$ is a $T$-periodic function.

How can I show the solution is of the form mentioned? Can I just say by Floquet Theory, the solution is of the form $X(t) = \sum\limits_{n=1}^{n}c_nx_n(t)$ where $x_n(t) = e^{\mu_nt}p_n(t)$ but then how do I get the $x_0$ for $c_1$?
By Floquet Theory, we can define $X_k = X(t)v_k$.Suppose $Bv_k = \lambda_kv_k$.
$$
X_k(t + T) = X(t + T)v_k = X(t)Bv_k = X(t)\lambda_kv_k = \lambda_kX(t)v_k = \lambda_kX_k(t)
$$
Let $\underbrace{\lambda_k}_{\text{characteristic multipliers}} = \text{exp}\left[\overbrace{\rho_k}^{\text{characteristic exponents}}T\right]$.
Define $p_k(t) = \frac{X_k(t)}{e^{\rho_k t}}$.
Then
\begin{alignat*}{3}
p_k(t + T) & = & \frac{X_k(t + T)}{\text{exp}\left[\rho_k(t + T)\right]}\\
& = & \frac{\lambda_kX_k(t)}{\text{exp}\left[\rho_k(t + T)\right]}\\
& = & \frac{\lambda_kX_k(t)}{\lambda_k\text{exp}\left[\rho_kt\right]}\\
& = & \frac{X_k(t)}{\text{exp}\left[\rho_kt\right]}\\
& = & p_k(t)
\end{alignat*}
Thus $p(t)$ is T periodic.

Since $x_0 = x(0)$ is an initial condition, can I just say that is why it is constant. What about $\mu$?
 
Last edited:

Alan

Member
Jul 21, 2012
60
If x is one dimensional then by separation of variables you get:

[tex]x(t)=x_0 e^{\int a(t) dt}[/tex]

Now if you can expand a(t) to power series around 0 (which is not given in your premise), then

a(t) = a(0)+ a'(0)t + a''(0)t^2/2!+...

a(T)=a(0)=\mu

you'll get one part of exp(\mu t) and the other part is
[tex] p(t)=e^{a'(0)t^2/2+a''(0)t^3/6+...}[/tex]

[tex]p'(t)=p(t) (a(t)-a(0))=p(t)(a(t)-a(T))[/tex]

plug t-> t+T to get:
[tex]p'(t+T)=0[/tex]
i.e [tex]p(t+T)=const[/tex]

And this I believe finishes the proof, but it's not general without assuming something on our 'a'.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
If x is one dimensional then by separation of variables you get:

[tex]x(t)=x_0 e^{\int a(t) dt}[/tex]

Now if you can expand a(t) to power series around 0 (which is not given in your premise), then

a(t) = a(0)+ a'(0)t + a''(0)t^2/2!+...

a(T)=a(0)=\mu

you'll get one part of exp(\mu t) and the other part is
[tex] p(t)=e^{a'(0)t^2/2+a''(0)t^3/6+...}[/tex]

[tex]p'(t)=p(t) (a(t)-a(0))=p(t)(a(t)-a(T))[/tex]

plug t-> t+T to get:
[tex]p'(t+T)=0[/tex]
i.e [tex]p(t+T)=const[/tex]

And this I believe finishes the proof, but it's not general without assuming something on our 'a'.
By separation of variables, we have that $\int\frac{\dot{x}}{x}dx = \int a(t)dt$.
Solving the integral leads too
$$
x = C\exp\left[\int a(t)dt\right].
$$
Let's expand the the Taylor series of $a(t)$ about 0.
$$
a(0) = a(0) + a'(0)t + \frac{a''(0)t^2}{2} + \cdots
$$
Since $a(t)$ is periodic with period $T$, we have $a(T) = a(0) = \mu$.
By substitution, we have
\begin{alignat*}{3}
x(0) & = & C\exp\left[\int \mu dt\right]\\
& = & Ce^{\mu t}
& = & C = x_0
\end{alignat*}
Therefore, we have $x(t) = x_0e^{\mu t}$

But why is p(t) multiplied by x? How do we go from the x above to saying it is $x(t) = x_0e^{\mu t}p(t)$
 

Alan

Member
Jul 21, 2012
60
The solution is
[tex] x(t)=x_0 e^{a(T)t} e^{a'(0)t^2/2+a''(0)t^3/6+...}[/tex]

Denote by p(t) the third factor, you can see that it satisfies:

[tex]p'(t)=p(t)(a(t)-a(0))[/tex]
So you get that p'(T)=0, and this is true for any m integer that p'(mT)=0, thus p(t+T)=p(t).
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The solution is
[tex] x(t)=x_0 e^{a(T)t} e^{a'(0)t^2/2+a''(0)t^3/6+...}[/tex]

Denote by p(t) the third factor, you can see that it satisfies:

[tex]p'(t)=p(t)(a(t)-a(0))[/tex]
So you get that p'(T)=0, and this is true for any m integer that p'(mT)=0, thus p(t+T)=p(t).
How do you go from
$$
\exp\left[a(t + T)\right]
$$
to
$$
e^{a(T)t} e^{a'(0)t^2/2+a''(0)t^3/6+...}
$$
 

Alan

Member
Jul 21, 2012
60
You have [tex]x(t)=x_0 e^{\int a(t)dt}[/tex]

I assumed that I can expand a(t) by a power series (I don't see how to show this otherwise), so a(t)=a(0)+a'(0)t+a''(0)t^2/2!+... and a(0)=a(T) from T-periodicity of a(t), plug back to the integral to get what I wrote.

If I can't expand a(t) with a powers series then I am not sure how to solve this.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
You have [tex]x(t)=x_0 e^{\int a(t)dt}[/tex]

I assumed that I can expand a(t) by a power series (I don't see how to show this otherwise), so a(t)=a(0)+a'(0)t+a''(0)t^2/2!+... and a(0)=a(T) from T-periodicity of a(t), plug back to the integral to get what I wrote.

If I can't expand a(t) with a powers series then I am not sure how to solve this.
So $p(t) = \exp\left[\int\left(a'(0)t + \frac{a''(0)t^2}{2} + \cdots\right)dt\right]$ is defined this way?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
[tex]p'(t)=p(t)(a(t)-a(0))[/tex]
So you get that p'(T)=0, and this is true for any m integer that p'(mT)=0, thus p(t+T)=p(t).
Ok so I understand p(t) now. Why are you taking the derivative of p(t) to show it is T-periodic?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
[tex]p'(t)=p(t) (a(t)-a(0))=p(t)(a(t)-a(T))[/tex]

plug t-> t+T to get:
[tex]p'(t+T)=0[/tex]
i.e [tex]p(t+T)=const[/tex]
I don't get this part.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
This is all wrong. Is there another way to do this?
What I wrote original comes from Floquet. Is it enough to just say what I was doing in post 1?
 

Raluca

New member
Jul 24, 2013
2
Hello, I am a phd student in Romania and I have to study Floquet Theory. I need some help with documentation and research. Do you have any? Appreciate your help
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Raluca

New member
Jul 24, 2013
2
do you think i could apply this floquet theory in economics?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
do you think i could apply this floquet theory in economics?
This is just the mathematics about it so you should be able to since the ecnomic version is probably built off of the math.