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I am trying to solve the following recurrence relations: $$c(n_1-N,n_2)=c(n_1,n_2)$$ and $$c(n_1,n_2-N)=c(n_1,n_2)\mathrm{e}^{-\frac{2\pi i Mn_1}{N^2}-\frac{\alpha_xMi}{N}},$$ where $N$ and $M$ are integers and $\alpha_x$ is a constant.

Now, from the first one, it tells me that I can express it a Fourier series with period $N$ i.e, $$c(n_1,n_2)=f(n_2)\sum_{m\in\mathbb{Z}}\tilde{c}(m)\mathrm{e}^{2\pi in1m/N},$$

But then there is an issue with the $n_2$, since putting this into the bottom condition, leaves an $n_1$ variable and is not a function of $n_2$ only.

Looking at the bottom condition yields $$c(n_1,n_2)=b(n_1,n_2)\mathrm{e}^{\frac{2\pi i Mn_1}{N^3}+\frac{\alpha_xMi}{N^2}},$$ with $b(n_1+N,n_2+N)=b(n_1,n_2)$, but again, this only satisfies one of the equations. I have tried to look for a solution of the form $$c(n_1,n_2)=b(n_1,n_2)\mathrm{e}^{f(n_1,n_2)},$$ where $b(n_1+N,n_2+N)=b(n_1,n_2)$. This leads to: $$f(n_1-N,n_2)=f(n_1,n_2)$$ and $$f(n_1,n_2-N)=f(n_1,n_2)-\frac{2\pi i Mn_1}{N^2}-\frac{\alpha_xMi}{N}.$$ But I am unable to solve this, Mathematica can't do it either...

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Your equations just say that

$$c(n_1,n_2)$$ takes free values in the square $[0,N-1]\times[0,N-1]$, and you can obtain them elsewhere by the formulas

$$c(n_1-N,n_2)=c(n_1,n_2),$$ and by induction

$$c(n_1-kN,n_2)=c(n_1,n_2),$$

and

$$c(n_1,n_2-N)=c(n_1,n_2)e^{an_1+b}$$

and by induction

$$c(n_1,n_2-k'N)=c(n_1,n_2)e^{k'(an_1+b)}.$$

Of course,

$$c(n_1-kN,n_2-k'N)=c(n_1,n_2)e^{k'(an_1+b)}.$$

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  • $\begingroup$ But what $c(n_1,n_2)$ satisfies this? $\endgroup$ – Lewis Proctor Jul 22 at 12:46
  • $\begingroup$ @LewisProctor: any ! $\endgroup$ – Yves Daoust Jul 22 at 12:47
  • $\begingroup$ Say $c(n_1,n_2)=n_1n_2$. From the first condition $(n_1-N)n_2=n_1n_2-Nn_2\neq n_1n_2$, unless $N=0$ or $n_2=0$. I need a general condition which works for all $n_1,n_2$ and $N$. $\endgroup$ – Lewis Proctor Jul 22 at 12:59

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