How to get the laplacian of a scalar field?

[cvex]

Hi,

I am trying to calculate the laplacian of a scalar field but I might actually need something else. So basically I am applying reaction diffusion on a 2d image. I am reading the neighbours, multiplying them with these weights and then add them.

This works great. I don't know if what I am doing is called laplacian but I was told that it is.

Now I am trying to do the same thing in 3d volumes:
http://www.openvdb.org/

But the problem is I don't know what weights to use to sum the neighbours. If I use the same weights on a 2d volume, it works just the same. But in 3d, I am not sure what the weights would be.

I also tried using actual laplacian (OpenVDB has tools to calculate it), but after 7-8 iterations the values in the volume go to infinity. Maybe what I need is not really laplacian. But whatever is coming from the neighbours has to be in balance so just like in the 2d image version, the values don't get bigger and bigger while nothing is getting smaller. That's my understanding.

I tried calculating the laplacian manually also by using the actual definition, and got different result that the build-in tools. Not sure if I am missing something:
http://paste.ofcode.org/QYUec5tu63AM6n2Bmax6C7

Do you guys know what I need?Cheers :)

 

[jedishrfu]

Are the weights tied to the number of surrounding cells? So that for 3d you'd have 26 neighbors instead of 8 for 2d.

Or tied the number of lines through the point? so for 2d it would be 4 and for 3d it would be 4+9 or 13.

This is just my guess. It would be best if you knew how the weights are calculated.

 

[cvex]

Thanks a lot for replying. It's tied to the number of surrounding cells. So for 2d, it's 8 and for 3d it's 26 just like you said.

Also the positive and negative weights have to cancel out, which is why in 2d the original cell has a weight of -1 and the surrounding cells have a total weight of 1.

So in a way the 3d volume is like a stack of 2d images.

 

[the_wolfman]

I think the wikipedia page on the discrete Laplace operator has what you are looking for:
http://en.wikipedia.org/wiki/Discrete_Laplace_operator

Its the last series of equations in the Image processing section.

 

[cvex]

Thanks wolfman, I think that works. I tried it before I think but didn't divide the result by 6 :( Now it seems to work at least.

Thanks a lot :)