Differentiating in one direction using FFT2

Starting from a univariateL-periodic function u, sampled at evenly-spaced points in , and stored in the vector u, the following function approximates the first derivative , using FFT

function du = FFTDiff(u,n,L)
  
  % Frequency vector
  k = (2*pi/L) * [0:n/2-1, -n/2:-1]';

  % Compute FFT
  uHat = fft(u);

  % Compute the derivative in the frequency domain, and back to physical domain
  du = ifft( 1i * k .* uHat, 'symmetric');

end

First of all, if you see any alternative ways on how to write the function above, your feedback is welcome. I am now in the process of writing a function that does the following: starting from a bivariate, periodic function u, sampled at evenly spaced points in , I want to write a function that approximates (hence the partial derivative with respect to x), using FFT2. The main problem I have is that I don't quite know how FFT2 outputs the wavenumbers and . It's a bit tricky to understand from the documentation, and I wonder if you have some ideas on how to do that.

[Matt J]

Perhaps you have your reasons, but I thought I should point out that differentiating with FFTs is a highly inefficient thing to do. You should use diff() or gradient().

There might be cases where it makes sense if you are using some large and exotic differencing kernel, but that doesn't appear to be the case in the code you've shown.

[Paul]

Hi Daniele,

Can you show how this code works with a simple input? I'm running into an error.

Is the code supposed to work for L even and L odd?

L = 20;   % period of function
l = 0:L;  % n + 1 equally spaced points in [0,L]
n = numel(l) - 1;      % n
u = sin(2*pi/20*l).';  % u
du = FFTDiff(u,n,L);

Arrays have incompatible sizes for this operation.

Error in solution>FFTDiff (line 16)
du = ifft( 1i * k .* uHat, 'symmetric');

figure
plot(l,du,'-o')

function du = FFTDiff(u,n,L)
  
  % Frequency vector
  k = (2*pi/L) * [0:n/2-1, -n/2:-1]';
  % Compute FFT
  uHat = fft(u);
  % Compute the derivative in the frequency domain, and back to physical domain
  du = ifft( 1i * k .* uHat, 'symmetric');
end

[Walter Roberson]

k = (2*pi/L) * [0:n/2-1, -n/2:-1]';

should probably be

k = (2*pi/L) * [0:n/2-1, -n/2:-1/2]';

Answer 1

[Matt J]

fft2(u) is equivalent to fft( fft(u,[],1) ,[],2), if that helps at all.

Truthfully, it doesn't appear that you need to use fft2. Assuming the x-direction is column-oriented and the y-direction is row-oriented, you can just use your original function as is to differentiate along x.

[Daniele Avitabile]

Not much, but thanks for answering. At the moment I got this code, but I am not sure this is optimal, really

function dudx = FFTDiff2D(u, nx, ny, Lx, Ly)

    % Frequency vectors
    kx = (2 * pi / Lx) * [0:nx/2-1, -nx/2:-1]';  % Wavenumbers in x direction
    ky = (2 * pi / Ly) * [0:ny/2-1, -ny/2:-1];    % Wavenumbers in y direction

    % Create a meshgrid for the wavenumbers
    [KX, KY] = meshgrid(kx, ky);

    % Compute 2D FFT of the input function
    uHat = fft2(u);

    % Compute the derivative in the frequency domain
    dudx_hat = 1i * KX .* uHat;

    % Back to physical domain using inverse FFT
    dudx = ifft2(dudx_hat, 'symmetric');
end

[Matt J]

Like I said, if you are only differentiating along the columns, FFTDiff() should already be sufficient. There is no need for fft2. If you are differentiating along the rows, then you can still use FFTDiff, but just pre-transpose the input.

Answer 2

[Catalytic]

How about this? It generalizes your original function to let you differentiate along any specified dimension dim for any nD array u -

function du = FFTDiff(u,L,dim)
arguments
   u double {mustBeNonempty}
   L (1,1) double   
   dim (1,1) double = find(size(u)>1,1)  %default to first non-singleton dimension
end

  n=size(u,dim);
 
  % Frequency vector
  k = (2*pi/L) * ifftshift( (0:n-1)-ceil((n-1)/2) );
  e=ones(1,ndims(u)); e(dim)=n;
  k=reshape(k,e);
  
  % Compute FFT
  uHat = fft(u,[],dim);

  % Compute the derivative in the frequency domain, and back to physical domain
  du = ifft( 1i * k .* uHat, [],dim,'symmetric');

end

[Walter Roberson]

I think your code might possibly have some trouble of u is empty. In that case, size(u, find(size(u)>1,1)) becomes size(u,[]) which returns [] . But [] cannot be stored in something size (1,1)

If you are successful in storing the [] into dim, then n=size(u,dim) would return empty, and I suspect that would lead to problems.

[Catalytic]

Yes, probably. I fixed it.