1
$\begingroup$

Please consider the following theorem:

Theorem (Floquet): Consider the equation $\dot{x} = A(t)x$ with $A(t)$ a continuous $T$-periodic $n\times n$-matrix. Each fundamental matrix $\Phi(t)$ of the equation $\dot{x} = A(t)x$ can be written as the product of two $n\times n$-matrices $$ \Phi(t) = P(t)e^{Bt} $$ with $P(t)$ $T$-periodic and $B$ a constant $n\times n$-matrix.

I want to solve the following exercise:

Consider the equation $\dot{x} = A(t)x$ with $A(t)$ a smooth $T$-periodic $n\times n$-matrix, $x\in\mathbb{R}^n$, $f(t)$ a smooth scalar $T$-periodic function. Consider the case when $n = 1, A(t) = f(t)$. Determine $P(t)$ and $B$ in the Floquet theorem.

What I've tried: If $n = 1, A(t) = f(t)$ the fundamental matrix $\Phi(t) = x(t)$ is a scalar function and the solution to the equation $\dot{x} = f(t)x$. According to the Floquet theoren we can write $\Phi(t) = x(t) = P(t)e^{Bt} = p(t)e^{bt}$ where $p(t)$ is a $T$-periodic scalar function and $b$ a scalar. If I substitute this into the original equation I get the following $$ \dot{x} = f(t)p(t)e^{bt} $$ Unfortunately I don't really know how to proceed from here..

Question: How should I solve this exercise?

$\endgroup$
2
$\begingroup$

Given that $x(t)=p(t)e^{bt}$ we get by substituting into the original equation $\frac{dx}{dt}=f(t)x$ that $p(t)$ must satisfy:

$$\dot{p}+(b-f(t))p=0\Rightarrow e^{\int^{t} (b-f(t'))dt'}p(t)=C\Rightarrow p(t)=Ce^{-bt}e^{\int^{t}f(t')dt'}$$

and thus the most general solution to the problem is:

$$x(t)=Ce^{\int^t f(t')dt'}=x(t_0)\exp\int_{t_0}^{t}f(t')dt'$$

for some arbitrary time $t_0$. Note that there is some arbitrariness in the way we can define b in this simple problem. To finish we should choose it's value such that $p(t)$ is periodic. We have that

$$\frac{d}{dt}\ln\frac{p(t+T)}{p(t)}=\frac{d}{dt}\int_t^{t+T}(f(t')-b)dt'=f(t+T)-f(t)=0$$

and therefore for every t we get that:

$$p(t+T)=p(t)\exp\int_{t_0}^{t_0+T}(f(t')-b)dt', A\in\mathbb{R}^+$$

However from mean value theorem we are guaranteed that there exists a $\xi\in(0,T)$ (the mean value of f in that interval) s.t:

$$\frac{1}{T}\int_{t_0}^{t_0+T}f(t')dt'=f(\xi)$$

If we pick $b=f(\xi)$ we get a periodic $p(t)$ and the Floquet theorem requirements are fulfilled.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.