# Discrete Laplacian on a triangle mesh

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Perry Mason on 30 Sep 2021
Commented: Alec Jacobson on 4 Oct 2021
Hi everybody,
I have a triangle mesh with N nodes and T triangles given by the following matrices
n(N,3): which defines the nodes coordinates
t(T,3): which is the connectivity matrix
If I note L as the Discrete Laplacian on my triangle mesh, and I code this
function1= n(:,1).^2;
function2= L*function1;
function2 should be an array of all twos, shouldn't it?
Unfortunately, although I have found some Discrete Laplacian matlab functions in Matlabcentral none of them retrieve a result where function2 should be an array of all twos.
I would appreciate any tip in this matter about how to compute the Discrete Laplacian
Cheers.

Alec Jacobson on 30 Sep 2021
L discretizes the integrated discrete Laplacian in the sense that:
x' * L * x discretizes Dirichlet energy: ∫ ‖∇x‖² dx. Via Green's idenity this energy is related to the Laplacian as:
∫ ‖∇x‖² dx = -∫ x∆x dx + boundary terms
So, there are at least 3 reasons you won't get all twos:
1) If n,t is a curved surface (as in, all vertices are not on the same plane) then L also encodes the non-Euclidean metric,
2) if n,t has a boundary, then for vertices along the boundary you'll get the integrate Laplacian + boundary terms
3) L computes integrated values, to convert to intergral average, pointwise values you can "unintegrate" (i.e., divide by local area) by multiplying with the inverse of the mass matrix: M \ L * x

Perry Mason on 4 Oct 2021
In order to understand what you said I produced the following uniform spherical (to evoid boundaries) mesh (of 0.9 m radius) with N=3840 nodes given the vector n(N:,3) and 1280 triangular elements, where the connectivity matrix is t(T,3).
By using gptoolbox, I try to compute the Laplacian ( ) over the spherical surface of the function with the following code (as you suggested in your previous answer)
L = -cotmatrix(n,t);
M = massmatrix(n,t);
f1=n(:,1).^2;
f2=M\L*f1;
I was expecting to be an array of all twos, but this is what I get Once more I would appreciate any comment or correction about what I am doing.
Cheers
Alec Jacobson on 4 Oct 2021
See #1 in the answer above. If your surface is curved, then you don't have a flat metric and won't get just 2s.
Try this in the Euclidean domain via:
[V,F] = create_regular_grid(10,10);
L = cotmatrix(V,F);
M = massmatrix(V,F);
M\(L*f1)
everything is 2.0 except the boundary.