How to solve a linear PDEs with trigonometric function as coefficients

Is a general method for solving a system of linear partial differential equation with trigonometric function as coefficients exist ?

For example something like that: $q$ is the unknown function, $2 \pi$ periodic, from $\mathfrak{R}^2\times [0,T]$ to $\mathfrak{C}$ and $A$, $B$, $C$ function of $\mathfrak{R}^2$ to $\mathfrak{C}$ are the coefficient of the system. $i$ is an integer varying from $1$ to $4$. The equation look like, using Einstein summation convention: $$\partial_t q_i(y,z)= A_j(y,z)\partial_{j} q_{i}(y,z) + q_j(y,z) B_{ij}(y,z) + C_{i}(y,z)$$ $A$ is of the from: $$A_j(y,z)=a^{(1)}_j \sin(2\pi y) + a^{(2)}_j \sin(2\pi z) + a^{(3)}_j \cos(2\pi y) + a^{(4)}_j \cos(2\pi z)+a_j^{(5)}$$ with $a_{j}^{(k)}$ in $\mathfrak{C}$. Functions $B_{ij}$ and $C_{j}$ are define in the same way that $A_{j}$. The initial condition $q(y,z,t=0)$ is given.

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Thanks for your comments. I use Einstein summation convention in the equation for the right end side. The explanation for $A_j$ was totally mest up with typographyc error. I fix it, hope it more understandable. – aberration Apr 30 '12 at 16:30
Are you prescribing initial data? Are you interested in $2\pi$ periodic solutions, or are you interested in general solutions? – Willie Wong May 1 '12 at 8:09
Yes I prescibing initial data. And I'm looking for a $2 \pi$ periodic condition. Thanks. – aberration May 1 '12 at 16:29
Have you tried taking the Fourier Transform/Fourier Series of both sides of the equation? Using the elementary properties of the Fourier transform, you should end up with an "ODE" with constant coefficients in $\ell^2$. If you are lucky it is diagonalisable. – Willie Wong May 2 '12 at 7:55
I try to take the Fourier transform in $y$, $z$ and $t$ but after this I get a functional equation. $t$ is not really a problem in fact. I see the $sin(2\pi z)$ multiplier as a modulation and I end up with some contribution in $\hat a(kz-2\pi)$. – aberration May 2 '12 at 14:23