Calculate derivatives in a non-uniform grid

I have a non-uniform grid (non-equal intervals between nodes). [x,y]

I also have the data (U) calculated on this grid.

I want to calculate the derivatives of U.

How can I do this?

Answer 1

[Star Strider]

The gradient function is an option, specifically:

dydx = gradient(y) ./ gradient(x);

for vectors, or more generally for matrices:

[dxr,dxc] = gradient(x);
[dyr,dyc] = gradient(y);
dc = dyc./dxc;                                      % Column Derivatives
dr = dyr./dxr;                                      % Row Derivatives

Try that to see if it gives you an acceptable result.

[moreza7]

Sorry, I don't understand!

What does your code exactly do?!

I have a non uniform grid ([x,y]) with it's data (U(x,y)).

I want to calculate dU/dy and dU/dx. How can I use this code to do so?

[Star Strider]

I am guessing that your ‘non-uniform grid’ (that I call ‘G’ here) is a matrix, as is ‘U’.

I would do something like this:

[dxr,dxc] = gradient(G);
[dyr,dyc] = gradient(U);
dc = dyc./dxc;                                      % Column Derivatives
dr = dyr./dxr;                                      % Row Derivatives

So ‘dc’ takes the derivatives along the columns, and ‘dr’ along the rows. See the documentation for the gradient function (that I linked to in my Answer) for details.

[Walter Roberson]

[dux, duy] = gradient(U, x, y);
duy ./ dux

perhaps?

Answer 2

[Alessandro Mura]

Hi,

the solution is using the Jacobian matrix.

This is used to transform gradients between the coordinate system of (row,column) to (x,y).

First calculate the gradients of U,x and y

[dxi,dxj]=gradient(x);
[dyi,dyj]=gradient(y);
[dui,duj]=gradient(u);

then the Jacobian of the transformation is

[dxi dxj
dyi dyj]

the determinant is

DET=dxi.*dyj- dxj.*dyi;

the inverse of the Jacobian has these 4 coefficients:

JAC11=DET.*dxi;
JAC21=DET.*dyi;
JAC12=DET.*dxj;
JAC22=DET.*dyj;

Then to transform the gradient [dui,duj] into dux, duy you just do:

dux=dui.*JAC11+duj.*JAC12;
duy=dui.*JAC21+duj.*JAC22;

Answer 3

[Justino Martinez]

I don't know if I have missing something in the notation, but the inverse of the Jacobian for these 4 coefficients shold be

JAC11 = dyj./DET

JAC21 = -dxj./DET

JAC12 = -dyi./DET

JAC22 = dxi./DET

isn't it?