Two different recurrence relations, same solution Say, I want to solve two recurrence relations in 2D given by
$$\begin{cases}\alpha_{i,j}=\frac{1}{4}\left(\alpha_{i+1,j}+\alpha_{i-1,j}+\alpha_{i,j+1}+\alpha_{i,j-1} \right)\\\alpha_{i,j}=\frac{1}{4}\left(  \alpha_{i+1,j+1} + \alpha_{i-1,j+1} + \alpha_{i-1,j-1} +\alpha_{i+1,j-1} \right)\end{cases}$$
where $\alpha_{i,j}$ is the solution to both equations simultaneously for the same choice of boundary values $\alpha_{i,j}^B$ for each equation. Now I'm wonderng, if there can exist such solutions. My intuition is that there shouldn't, especially as the equations are not linear superpositions of each other. I'm not sure if my question is already ruled out by uniqueness of the solution.
 A: If you have a mesh of $n \times m$ points, you have

*

*$mn$ unknowns

*$2(m+n-1)$ values at the boundary.

*$2(m-2)(n-2)$ equations at the center

The total number of equations you have is
$$
2(m-2)(n-2) + 2(m+n-1) = 2mn -2m -2n +6
$$
As you have $mn$ values to find, this solution is over-determined for
$$
\left(2mn -2m -2n +6\right) - mn < 0
$$
$$
m(n-2)< 2n-6
$$
$$
m<2-\dfrac{2}{n-2} < 2
$$
So, if there's at least a solution to this problem, there are $(mn-2m-2n+6)$ equations that are linear-dependent on the others.
Your solution depends highly on the boundary conditions. It's different if you have only one equation (instead of two) because you can put whatever you want for $\alpha^{B}$ and you still solve the system.
For example: For a mesh $3 \times 3$, $8$ values are at BC and there's only $1$ unknown. But that means the $8$ values on the boundary must satisfy
$$
\alpha_{00} + \alpha_{20} + \alpha_{22} + \alpha_{02} = \alpha_{10} + \alpha_{01} + \alpha_{12} + \alpha_{21} 
$$
So, instead of $8$ values on the boundary, you can put only $7$ values and get the same solution.
For a mesh of $n \times m$ points, there are

*

*$nm$ unknowns

*$2(m+n-1)$ values at the boundary

*$2(n-2)(m-2)$ equations at the center.

In fact, the DOFs (Degrees of Freedom) on your BC is only
$$
mn - 2\left(m-2\right)(n-2) = -mn+4m+4n-8
$$
instead of $2(m+n-1)$. And if you have $m$ and $n$ very big, this value becomes negative and only one possibility is allowed:
$$
\alpha_{ij} = \text{const}
$$
