I would like to solve the following algebraic linear system of q-difference functional equations:

\begin{cases} a_{11}\left(x\right)f\left(x\right)+a_{12}\left(x\right)g\left(x\right)=f\left(qx\right)\\ \\ a_{21}\left(x\right)f\left(x\right)+a_{22}\left(x\right)g\left(x\right)=g\left(qx\right) \end{cases}

where all the functions are $\mathbb{R}\rightarrow\mathbb{R}$. In more detail, the functions $a_{ij}\left(x\right)$ are known (moreover they are analytic and bounded between $0$ and $1$), and the purpose is to determine $f\left(x\right)$ and $g\left(x\right)$. The parameter $q$ is smaller than $1$. If we define:

\begin{equation} A\left(x\right)\overset{\mathrm{def}}{=}\left[\begin{array}{cc} a_{11}\left(x\right) & a_{12}\left(x\right)\\ a_{21}\left(x\right) & a_{22}\left(x\right) \end{array}\right] \end{equation}

we have $\det A\left(x\right)\neq0\;\forall x\neq0$ and $\det\left(I-A\left(0\right)\right)\neq0$ (this is due to the special form of the functions $a_{ij}\left(x\right)$ I'm using). From numerical analysis I know that this system has also non-zero solutions.

MY APPROACH

I've tried to solve this problem using the methods developed for q-difference equations (see for example this link). From the system of equations we obtain:

\begin{equation} \left[\begin{array}{c} f\left(x\right)\\ g\left(x\right) \end{array}\right]=A^{-1}\left(x\right)\left[\begin{array}{c} f\left(qx\right)\\ g\left(qx\right) \end{array}\right] \end{equation}

therefore if we iterate $n$-times we obtain:

\begin{equation} \left[\begin{array}{c} f\left(x\right)\\ g\left(x\right) \end{array}\right]=\left[\prod_{i=0}^{n-1}A^{-1}\left(q^{i}x\right)\right]\left[\begin{array}{c} f\left(q^{n}x\right)\\ g\left(q^{n}x\right) \end{array}\right] \end{equation}

where:

\begin{equation} \prod_{i=0}^{n-1}A^{-1}\left(q^{i}x\right)=A^{-1}\left(x\right)A^{-1}\left(qx\right)\cdots A^{-1}\left(q^{n-1}x\right) \end{equation}

Thefore in the limit $n\rightarrow\infty$, since $q<1$, we obtain:

\begin{equation} \left[\begin{array}{c} f\left(x\right)\\ g\left(x\right) \end{array}\right]=\left[\prod_{i=0}^{\infty}A^{-1}\left(q^{i}x\right)\right]\left[\begin{array}{c} f\left(0\right)\\ g\left(0\right) \end{array}\right] \end{equation}

Now, for $x=0$, the system of equations becomes:

\begin{cases} a_{11}\left(0\right)f\left(0\right)+a_{12}\left(0\right)g\left(0\right)=f\left(0\right)\\ \\ a_{21}\left(0\right)f\left(0\right)+a_{22}\left(0\right)g\left(0\right)=g\left(0\right) \end{cases}

therefore we obtain:

\begin{equation} \left[\begin{array}{c} f\left(0\right)\\ g\left(0\right) \end{array}\right]=\left[I-A\left(0\right)\right]^{-1}\left[\begin{array}{c} 0\\ 0 \end{array}\right]=\left[\begin{array}{c} 0\\ 0 \end{array}\right] \end{equation}

since $\det\left(I-A\left(0\right)\right)\neq0$ by hypothesis. Clearly the matrix $\prod_{i=0}^{\infty}A^{-1}\left(q^{i}x\right)$ is singular, since $\det A\left(0\right)=0$ by hypothesis. Therefore we have obtained the indeterminate solution $\left[\begin{array}{c} f\left(x\right)\\ g\left(x\right) \end{array}\right]=\infty\cdot0$, which is not good. Since this method doesn't work, do you know how this system can be solved? Maybe there is an alternative approach. Thanks in advance!

EDIT: simplification of the problem

In order to simplify further the description of this problem, below I write its 1D equivalent case:

\begin{equation} a\left(x\right)f\left(x\right)=f\left(qx\right) \end{equation}

where $a\left(x\right)$ is known and such that $a\left(0\right)=0$. So for example we can consider the functional equation $xf\left(x\right)=f\left(qx\right)$. How can I calculate $f\left(x\right)$, given that $f\left(x\right)=0$ for $x\leq0$?