the steady state $u_{m,n}=a$. Therefore, after changing the coordinates (\[eq:COR\]), substituting $u_{m,n}(t)=a+\xi_{m,n}(t)$, and linearizing the obtained equation in small perturbations $\xi_{m,n}(t)$, we obtain $$\begin{aligned}
\begin{aligned}  \label{eq:xi}
\frac{\mathrm{d}}{\mathrm{d}t}\xi_{m,n}(t)  &= \left(\alpha+i\beta-i\Omega-2a^{2}\right)\xi_{m,n}(t)-a^{2}\xi_{m,n}^{\ast}(t) + \\
 &\quad +\frac{C}{2}e^{-i\Omega\tau}\left(e^{ik_{1}}\xi_{m-1,n}(t-\tau)+e^{ik_{2}}\xi_{m,n-1}(t-\tau)\right).
\end{aligned}\end{aligned}$$ The solutions of the obtained linear equation can be found by the ansatz $$\xi_{m,n}(t)=b_{1}e^{\lambda t-iq_{1}m-iq_{2}n}+b_{2}^{\ast}e^{\lambda^{\ast}t+iq_{1}m+iq_{2}n}.\label{eq:bloch}$$ The ansatz (\[eq:bloch\]) can be obtained, e.g. by rewriting the system (\[eq:xi\]) in the real form, diagonalizing it with the discrete Fourier transform (similar to (\[eq:SLld\])), and noticing that the equations for the Fourier components $\tilde{\xi}_{q_{1},q_{2}}$ and $\tilde{\xi}_{-q_{1},-q_{2}}$ are complex conjugate, and hence, they have the same stability properties with the complex conjugate eigenvalues, see also [@Cross2009]. After substituting (\[eq:bloch\]) into (\[eq:xi\]), the coefficients at the two linearly independent functions $e^{\lambda t-iq_{1}m-iq_{2}n}$ and $e^{\lambda^{*}t+iq_{1}m+iq_{2}n}$ should be zero. This leads to the system of two linear equations for unknowns $b_{1}$ and $b_{2}$: $$\begin{aligned}
\left[\alpha+i\beta-i\Omega-\lambda-2a^{2}+\frac{C}{2}e^{-(\lambda+i\Omega)\tau}\left(e^{i(k_{1}+q_{1})}+e^{i(k_{2}+q_{2})}\right)\right]b_{1}-a^{2}b_{2} & =0,\\
-a^{2}b_{1}+\left[\alpha-i\beta+i\Omega-\lambda-2a^{2}+\frac{C}{2}e^{-(\lambda-i\Omega)\tau}\left(e^{-i(k_{1}-q_{1})}+e^{-i(k_{2}-q_{2})}\right)\right]b_{2} & =0.\end{aligned}$$ This system has a nontrivial solution if the determinant of its matrix is zero. We arrive at the characteristic equation $$\begin{aligned}
%\begin{aligned} 
\chi(\lambda,q_{-},q_{+})= &\; \lambda^{2}+2(a^{2}+R\cos k_{\tau})\lambda+R^{2}+2Ra^{2}\cos k_{\tau}+R_{+}R_{-}e^{-2\lambda\tau+2iq_{+}} - \nonumber  \\
 & -\Bigl[(a^{2}+R\cos k_{\tau}+\lambda)\left(R_{+}e^{ik_{\tau}}+R_{-}e^{-ik_{\tau}}\right) - \label{eq: SLpsolCE} \\
 &\quad\,\, -iR\sin k_{\tau}\left(R_{+}e^{ik_{\tau}}-R_{-}e^{-ik_{\tau}}\right)\Bigr]e^{-\lambda\tau+iq_{+}}, \nonumber
%\end{aligned}\end{aligned}$$ where we additionally denote $q_{\pm}:=\frac{1}{2}\left(q_{1}\pm q_{2}\right)$ and $R_{\pm}:=C\cos(k_{-}\pm q_{-})$. Now the 