Using meshgrid in MATLAB when the function has matrix operations - matlab

Question

I need to plot level surfaces of a 3 variable function. The variables are in a column vector X = [x, y, z]^t. The function is f(X) = X^t * A * X. Where ^t means transpose and A is a 3x3 constant matrix. I know for a fact that A is symmetric and therefore diagonalizable, i.e. A = V * D * V^t. Just in case it turns out to be useful.
I intend to use isosurface to get the points of the function where it equals a certain level and then use patch to plot.
However I can't figure out how to compute the value of the function for every point in the grid. Ideally I'd like to do
x = linspace(-1, 1,10); y=x; z = x;
[XX,YY,ZZ]=meshgrid(x,y,z);
f = [XX YY ZZ]'*A*[XX YY ZZ];
level = 1;
s = isosurface(XX,YY,ZZ,f,level);
patch(s, 'EdgeColor','none','FaceColor','blue');
but this won't work obviously because of the sizes of XX and A. What I've done so far is do the math myself to obtain the function as a polynomial of XX, YY and ZZ but it's incredibly ugly and not practical.
Anybody knows how to do this? Thanks!

Answer 1

If I understand correctly, this does that you want. In the following explanation I will use 10x10x10 points as given in your example (although the code works for any number of points). Also, I define a random 3x3 matrix A.
Once XX, YY and ZZ have been generated as 10x10x10 arrays (step 1), the key is to build a 1000x3 matrix in which the first column is x coordinate, the second is y and the third is z. This is variable XYZ in the code below (step 2).
Since XYZ is a matrix, not a vector, the function f can't be computed using matrix multiplication. But it can be obtained efficiently with bsxfun. First compute an intermediate 1000x3x3 variable (XYZ2) with all 3x3 products of coordinates for each of the 1000 points (step 3). Then reshape it into a 1000x9 matrix and multiply by the 9x1 vector obtained from linearizing A (step 4).
The f thus obtained has size 1000x1. You need to reshape it into a 10x10x10 array to match XX, YY and ZZ (step 5). Then you can use isosurface as per your code (step 6).
A = rand(3);
x = linspace(-1, 1,10); y = x; z = x;
[XX,YY,ZZ] = meshgrid(x,y,z); %// step 1
XYZ = [XX(:) YY(:) ZZ(:)]; %// step 2
XYZ2 = bsxfun(#times, XYZ, permute(XYZ, [1 3 2])); %// step 3
f = reshape(XYZ2,[],numel(A))*A(:); %// step 4
f = reshape(f, size(XX)); %// step 5
level = 1;
s = isosurface(XX,YY,ZZ,f,level);
patch(s, 'EdgeColor','none','FaceColor','blue'); %// step 6

Answer 2

I suspect that you could explicitly define your function as
ffun = #(x,y,z) [x y z]*A*[x; y; z];
then use arrayfun to apply this function to each element of your coordinate vectors:
f = arrayfun(ffun,XX,YY,ZZ);
This will return f(i)=ffun(XX(i),YY(i),ZZ(i)) for each i in 1:numel(XX), which I think is what you are after. The output f will have the same shape as your input arrays, which seems perfectly fine to use with isosurface later.