How to obtain the 9-point Laplacian formula? I'm studying LeVeque's Finite Difference Methods for Ordinary and Parial Differential Equations. In page 64 he states the formula for the 9-point Laplacian
$$ \nabla_9^2u_{ij}=\frac{1}{6h^2}[4u_{i-1,j}+4u_{i+1,j}+4u_{i,j-1}+4u_{i,j+1}+u_{i-1,j-1}+u_{i-1,j+1}+u_{i+1,j-1}+u_{i+1,j+1}-20u_{ij}] $$
I don't understand how he got to this formula. In previous pages he shows how to get the formula for the 5-point stencil by adding centered finite differences. I tried a similar approach to get the 9-point formula, but had no success.
So, I'm deeply curious to know how to get to this formula by adding centered finite differences or something like that.
 A: I was searching for the explanation and found a relevant paper here.
Here is what I got from it:
The "stencil" you mention is actually not trivial to derive and not based on basic algebric calculation (like the classic 5-point stencil).
Basically, you can consider it as a weigthed sum of two 5-points stencils : the first one for the Horizontal-Vertical (HV) samples, the second one for diagonal samples :

where the weigts are $(1-\alpha)$ for the HV stencil, and $\frac{\alpha}{2}$ for the diagonal stencil. The second stencil kinda makes sense to look like this if you consider "rotating" the axis 45°. The only difference is that the distance between 2 adjacent samples is $\sqrt{2}$, not 1 as in the HV stencil. Another advantage is that both stencils are "self-averaged" (the sum of their coefficients is $0$ wathever $\alpha$).
You can get the classic 5-points stencil taking $\alpha=0$.
Now the 9-points laplacian you mention is obtained for $\alpha=1/3$, which is based on a complicated derivation to express the error of the approximation for a specific PDE equation :

where it can be shown that the solution can be expanded as (where is the 2D distance $R=\sqrt{x^2+y^2}$):

and since we can express the actual Laplacian

the optimal value for $\alpha$ that minimizes the approximation error is $1/3$ (which cancels the $R^{-2}$ term in the expansion.
A: The reasoning behind the 9-point stencil follows from the logic that the "diagonal" points offer additional information that can possibly used to construct a better approximation to the derivative.
Instead of the diagonal neighbors, you could also look at the points with an offset of $2h$ to you, also leading to 9-point stencil. This is discussed in this exercise sheet, leading also to a higher order approximation.
In any case, however, you have to get your hands dirty and do the Taylor-Expansion to get the exact coefficients. What you can see for both kinds of $9$-point stencils is the fact that closer points receive more weight what makes intuitively sense.
