Calculate derivatives in a non-uniform grid
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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?
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답변 (3개)
Star Strider
2019년 8월 1일
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.
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Star Strider
2019년 8월 1일
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.
Alessandro Mura
2020년 6월 22일
편집: Alessandro Mura
2020년 6월 22일
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;
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Justino Martinez
2024년 1월 15일
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?
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