Hi guys I am trying to prove the following theorem and would like if possible to get comments on my work and maybe some guidance for on of the directions of the proof.
Let $y''+Q(t)y=0$ where $Q$ is a periodic function $Q(t)=Q(t+a)$. $p^2-Dp+1=0$ is the the characteristic polynomial of the Monodromy matrix, with roots $p_1,p_2$ which are the Floquet multipliers. The claim is that there exists a solution $y(t)$ with period $ka$, $k\in \mathbb{Z}$ if and only if $D=2cos(\frac{2\pi j}{k})$ for some integer $j$.
My approach was to show one inclusion first where I assume that $D=2cos(\frac{2\pi j}{k})$ and aim to show there is a solution with period $ka$.
Case 1. Say $k=1$ then $D=2$ thus $p^2-2p+1=0$ yells the solutions two solutions $p_1=1$ and $p_2=1$. Note that $1=e^{2\pi t}$. Then by some work in the class we know every solution is of the form $y(t)=c_1q_1(t)p_1(t)+c_2q_2(t)p_2(t)$ where $q_1, q_2$ are $a$ periodic functions. Thus for us the solution is of the form $y(t)=c_1q_1(t)e^{2\pi t}+c_2q_2(t)e^{2\pi t}$. Which is clearly $1$ periodic. ( I am worried that because it is a double root the solution may have to have an extra t in front of $c_2q_2(t)$)
Case 2: Let $k=2$ now we need to consider $D=2cos(\pi j)$ if $j$ is even we get $D=2$ back to the previous case. If $j$ is odd then $D=-2$ then we see that $p^2+2p+1=0$ gives the roots $p_1=-1$ and $p_2=-1$ then similarly to the other case we have $y(t)=+c_1q_1(t)e^{\pi t}+c_2q_2(t)e^{\pi t}$
Case 3: Let $k>2$ then we have the following $p^2-2cos(\frac{2\pi j}{k})p+1=0$ by the quadratic formula we get that $p_1= 2cos(\frac{2\pi j}{k}) +i2sin(\frac{2\pi j}{k})$ and $p_2$ is the conjugate. I see how this is $ka$ periodic but I am not sure how this becomes a function of $t$ and we in put it in the solution. What I was thinking was that $y(t)=c_1q_1 2(cos(\frac{2\pi j}{k}t) +i2sin(\frac{2\pi j}{k}t))+c_2q_2( 2cos(\frac{2\pi j}{k}t) -i2sin(\frac{2\pi j}{k}t))$. Can someone please comment on what I am doing and possibly guide me to doing the other direction show that $ka$ periodic solution has to have $D=2cos(\frac{2\pi j}{k})$. Thank you for the time.
The statement is true if and only if the $k$th power of one of the eigenvalues is $1$. Since the eigenvalues are a complex conjugate pair, both have to be complex-conjugate $k$th unit roots, $p=e^{i\frac{2\pi j}k}$, so that $D=2Re(p)=2\cos\frac{2\pi j}k$.