Question

in order to code the DEL2 matlab function in c++ I need to understand the algorithm. I’ve managed to code the function for elements of the matrix that are not on the borders or the edges.
I’ve seen several topics about it and read the MATLAB code by typing “edit del2” or “type del2” but I don’t understand the calculations that are made to obtain the borders and the edges.

Any help would be appreciated, thanks.

Answer 1

You want to approximate u” knowing only the value of u on the right (or the left) of a point.
In order to have a second order approximation, you need 3 equations (basic taylor expansion):

u(i+1) = u(i) + h u’ + (1/2) h^2 u” + (1/6) h^3 u”’ + O(h^4)

u(i+2) = u(i) + 2 h u’ + (4/2) h^2 u” + (8/6) h^3 u”’ + O(h^4)

u(i+3) = u(i) + 3 h u’ + (9/2) h^2 u” + (27/6) h^3 u”’ + O(h^4)

Solving for u” gives (1):

h^2 u” = -5 u(i+1) + 4 u(i+2) – u(i+3) + 2 u(i) +O(h^4)

To get the laplacian you need to replace the traditional formula with this one on the borders.

For example where “i = 0” you’ll have:

del2(u) (i=0,j) = [-5 u(i+1,j) + 4 u(i+2,j) – u(i+3,j) + 2 u(i,j) + u(i,j+1) + u(i,j-1) – 2u(i,j) ]/h^2

EDIT clarifications:

The laplacian is the sum of the 2nd derivatives in the x and in the y directions. You can calculate the second derivative with the formula (2)

u” = (u(i+1) + u(i-1) – 2u(i))/h^2

if you have both u(i+1) and u(i-1). If i=0 or i=imax you can use the first formula I wrote to compute the derivatives (notice that due to the simmetry of the 2nd derivative, if i = imax you can just replace “i+k” with “i-k”). The same applies for the y (j) direction:

On the edges you can mix up the formulas (1) and (2):

del2(u) (i=imax,j) = [-5 u(i-1,j) + 4 u(i-2,j) – u(i-3,j) + 2 u(i,j) + u(i,j+1) + u(i,j-1) – 2u(i,j) ]/h^2

del2(u) (i,j=0) = [-5 u(i,j+1) + 4 u(i,j+2) – u(i,j+3) + 2 u(i,j) + u(i+1,j) + u(i-1,j) – 2u(i,j) ]/h^2

del2(u) (i,j=jmax) = [-5 u(i,j-1) + 4 u(i,j-2) – u(i,j-3) + 2 u(i,j) + u(i+1,j) + u(i-1,j) – 2u(i,j) ]/h^2

And on the corners you’ll just use (1) two times for both directions.

del2(u) (i=0,j=0) = [-5 u(i,j+1) + 4 u(i,j+2) – u(i,j+3) + 2 u(i,j) + -5 u(i,j+1) + 4 u(i+2,j) – u(i+3,j) + 2 u(i,j)]/h^2

Del2 is the 2nd order discrete laplacian, i.e. it permits to approximate the laplacian of a real continuous function given its values on a square cartesian grid NxN where the distance between two adjacent nodes is h.

h^2 is just a constant dimensional-factor, you can get the matlab implementation from these formulas by setting h^2 = 4.

For example, if you want to compute the real laplacian of u(x,y) on the (0,L) x (0,L) square, what you do is writing down the values of this function on an NxN cartesian grid, i.e. you calculate u(0,0), u(L/(N-1),0), u(2L/(N-1),0) … u( (N-1)L/(N-1) =L,0) … u(0,L/(N-1)), u(L/(N-1),L/(N-1)) etc. and you put down these N^2 values in a matrix A.

Then you’ll have
ans = 4*del2(A)/h^2, where h = L/(N-1).

del2 will return the exact value of the continuous laplacian if your starting function is linear or quadratic (x^2+y^2 fine, x^3 + y^3 not fine). If the function is not linear nor quadratic, the result will be more accurate the more points you use (i.e. in the limit h -> 0)

I hope this is more clear, notice that i used 0-based indices for accessing matrix (C/C++ array style), while matlab uses 1-based.