I am trying to find the laplacian of a point in 3D but the major issue is that the distances between my points not constant. For example in 1D : Illustration, the problem is the same for each axis.
In the case of a uniform distance between each points, I have this formula : $$ \nabla^2f(x,y,z) \approx \frac{f(x+\Delta_x,y,z)+f(x-\Delta,y,z)-2f(x,y,z)}{\Delta_x^2} + \frac{f(x,y+\Delta_y,z)+f(x,y-\Delta_y,z)-2f(x,y,z)}{\Delta_y^2} + \frac{f(x,y,z+\Delta_z)+f(x,y,z-\Delta_z)-2f(x,y,z)}{\Delta_z^2} $$ I tried to find a new one by considering 2 distances on the $x$ axis, $h$ and $a*h$ where $a$ is a real $<0$ but I can't simplify the result as the first formula. Maybe I can still use this by considering a point close to the others to have a constant distance and approximate the value as there is not a huge gap between each values but I am not sure about the approach.
Do you know if there is a good way to solve my problem ?
Thanks
The nonuniform grid estimate for 1D: $$ f''(0)\approx\frac{f(h)-(1-a^{-1})f(0)-a^{-1}f(ah)}{\frac12(1-a)h^2} $$ can be obtained if you try to fit a quadratic $f(x)=\alpha x^2+\beta x+\gamma$ and fiddle with the coefficients to get $2\alpha$. When $a=-1$, this reduces to the symmetric case $f''(0)\approx h^{-2}[f(h)-2f(0)+f(-h)]$.