# [SOLVED]Floquet

#### dwsmith

##### Well-known member
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
If x is one dimensional then by separation of variables you get:

$$x(t)=x_0 e^{\int a(t) dt}$$

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
$$p(t)=e^{a'(0)t^2/2+a''(0)t^3/6+...}$$

$$p'(t)=p(t) (a(t)-a(0))=p(t)(a(t)-a(T))$$

plug t-> t+T to get:
$$p'(t+T)=0$$
i.e $$p(t+T)=const$$

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

#### dwsmith

##### Well-known member
If x is one dimensional then by separation of variables you get:

$$x(t)=x_0 e^{\int a(t) dt}$$

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
$$p(t)=e^{a'(0)t^2/2+a''(0)t^3/6+...}$$

$$p'(t)=p(t) (a(t)-a(0))=p(t)(a(t)-a(T))$$

plug t-> t+T to get:
$$p'(t+T)=0$$
i.e $$p(t+T)=const$$

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 $\int\frac{\dot{x}}{x}dx = \int a(t)dt$.
$$x = C\exp\left[\int a(t)dt\right].$$
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
The solution is
$$x(t)=x_0 e^{a(T)t} e^{a'(0)t^2/2+a''(0)t^3/6+...}$$

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

$$p'(t)=p(t)(a(t)-a(0))$$
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
The solution is
$$x(t)=x_0 e^{a(T)t} e^{a'(0)t^2/2+a''(0)t^3/6+...}$$

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

$$p'(t)=p(t)(a(t)-a(0))$$
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
You have $$x(t)=x_0 e^{\int a(t)dt}$$

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
You have $$x(t)=x_0 e^{\int a(t)dt}$$

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
$$p'(t)=p(t)(a(t)-a(0))$$
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
$$p'(t)=p(t) (a(t)-a(0))=p(t)(a(t)-a(T))$$

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

#### dwsmith

##### Well-known member
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
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

#### Raluca

##### New member
do you think i could apply this floquet theory in economics?

#### dwsmith

##### Well-known member
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.