s(x+1, y) (du(x+1, y) - du(x, y))
- s(x-1, y) (du(x, y) - du(x-1, y))
+ s(x, y+1) (du(x, y+1) - du(x, y))
- - s(x, y-1) (du(x, y) - du(x, y-1)) = C
+ - s(x, y-1) (du(x, y) - du(x, y-1)) = -C
s(x+1, y) du(x+1, y) - s(x+1, y) du(x, y)
- s(x-1, y) du(x, y) + s(x-1, y) du(x-1, y)
+ s(x, y+1) du(x, y+1) - s(x, y+1) du(x, y)
- - s(x, y-1) du(x, y) + s(x, y-1) du(x, y-1) = C
+ - s(x, y-1) du(x, y) + s(x, y-1) du(x, y-1) = -C
- (s(x+1, y) + s(x-1, y) + s(x, y+1) + s(x, y-1)) du(x, y) =
- - s(x+1, y) du(x+1, y) - s(x-1, y) du(x-1, y) - s(x, y+1) du(x, y+1) - s(x, y-1) du(x, y-1) + C
+ - s(x+1, y) du(x+1, y) - s(x-1, y) du(x-1, y) - s(x, y+1) du(x, y+1) - s(x, y-1) du(x, y-1) - C
It is interesting to note that if s = 1 uniformly, which would be the case
without our penalizer Ψ(a²), we would have the familiar discrete Laplacian,