Hello, I encountered the following system of linear first-order differential equations:


$y'(z)=A(z) y(z)$


where


$y(z): R \rightarrow R^2$ and
$A(z)=\begin{pmatrix} 
0 & B Cos(\alpha z + \Phi\_b) \cr 
A Cos(\alpha z + \Phi\_a) & 0 \end{pmatrix}$


All quantities are real and constant if not explicitly variable. (Probably - if anyone wants to talk about approximations, which would be ok to some extent -) $A$,$B$ are $<1$, while $\alpha>10^3$ and $z<10^{-2}$.


I found Floquet's theorem, which only gives my the structure of the solution but isn't of much help for me. 
In approximation, I thought about using the Magnus expansion.


I added math.SP, as I got the suggestion to transform the equation to a Schrodinger equation with $Y(z)=A\_{1,2}(z)^{1/2} y(z)$ but that did not produce any new ideas for a solution.


I am at the end my (physicist's) knowledge, hope anyone has an idea.



Homogeneous linear differential equation system with simple periodical coefficient matrix
OK. First of all, change $(y\_1,y\_2)$ to $(y\_1,\sqrt{B/A}y\_2)$ and the time $z$ to $t=\alpha z+\frac{\Phi\_A+\Phi\_B}{2}$. Then we'll get the system with the matrix 
$$
A(t)=\begin{pmatrix} 
0 & C\cos (t + \Phi) \cr 
C \cos(t- \Phi) & 0 \end{pmatrix}
$$
where $C=\alpha^{-1}\sqrt{AB}$ and $\Phi=(\Phi\_B-\Phi\_A)/2$. We want our approximation to be decent on $[\alpha z\_{\min} ,\alpha z\_{\max}]\subset [-\alpha,\alpha]$. Noting that $\cos(t-\Phi)=\cos(\Phi-t)$, we see that $A(t)=A^\*(2\pi-t)$, which immediately tells us that the monodromy matrix $M$ from $0$ to $2\pi$ is self-adjoint. Also, denoting $\psi(t)=C\cos(\Phi+t)$=ce^{it}+\bar c{e^{-it}}$ with $c=\frac 12Ce^{i\Phi}, we see that the fundamental matrix $M(t)$ of the solution on $[0,2\pi]$ can be obtained (by the standard Piquard iterations) as the sum
$$
M(t)=\begin{pmatrix} 
1 & 0 \cr 
0 & 1 \end{pmatrix}+
\begin{pmatrix} 
0 & \psi\_1(t) \cr 
-\psi\_1(-t) & 0 \end{pmatrix}
$$
$$+
\begin{pmatrix} 
 -\psi\_2(t) & 0 \cr 
0 & -\psi\_2(-t) \end{pmatrix}+
\begin{pmatrix} 
0 & -\psi\_3(t) \cr 
\psi\_3(-t) & 0 \end{pmatrix}+
O(C^4)
$$ 
Where $\psi\_0(t)=1$ and $\psi\_{k+1}(t)=\int\_0^t\psi(s)\psi\_k(-s)ds$.


You can write a long series but I want to convince you that 4 first terms are enough for your problem. We can find the first two $\psi$'s:


$$
\psi\_1=\frac 1i[(ce^{it}-\bar c{e^{-it}})-(c-\bar c)
$$


$$
\psi\_2=-\frac{c-\bar c}{i}\psi\_1+\frac 1i(c^2-\bar c^2)t+\frac{|c|^2}2(e^{2it}+e^{-2it})-|c|^2
$$


and the linear term in $\psi\_3$, which is
$$
(c^2-\bar c^2)t[(c-\bar c)+ce^{it}-\bar ce^{-it}]
$$


Plugging in $t=2\pi$, we see that the monodromy matrix is $\begin{pmatrix} 
1-2\pi v & 4\pi vs \cr 
4\pi vs & 1+2\pi v \end{pmatrix}+\begin{pmatrix} 
O(C^4) & O(C^5) \cr 
O(C^5) & O(C^4)\end{pmatrix}$
with $v=2\Im (c^2)=\frac 12 C^2\sin 2\Phi= \frac 12 \alpha^{-2}AB\sin(\Phi\_B-\Phi\_A)$, $s=c-\bar c=C\sin\Phi=\alpha^{-1}\sqrt{AB}\sin\frac{\Phi\_B-\Phi\_A}2$. 


Now, the life is easy: the growth/decay part is essentially given by the matrix 
$$
G(t)=\begin{pmatrix} 
e^{-vt} & 0 \cr 
0 & e^{vt} \end{pmatrix}
$$ 
the rotation part is essentially given by 
$$
T=\begin{pmatrix} 
1 & -s \cr 
s & 1 \end{pmatrix}
$$
and the oscillation is essentially given by 
$$
H(t)=\begin{pmatrix} 
1-\widetilde\psi\_2(t) & \psi\_1(t) \cr 
-\psi\_1(-t) & 1-\widetilde\psi\_2(-t) \end{pmatrix}
$$
where $\widetilde \psi\_2$ is $\psi\_2$ with the term $\frac 1i(c^2-\bar c^2)t$ removed. 
So, $M(t)\approx H(t)T^{-1}G(t)T$ (well, $T^{-1}GT$ and $H$ don't commute but the commutator effect is of size $C^3$).


This should work just fine letting you to see just enough in your range. 


P.S. My original answer had an error in that I neglected the rotation of eigenvectors, but now it should be fine even for fairly large $C$ like $0.1$. Check agaist your numerics and see if it works well enough for you.


