Compute divergence of vector field using python

Question

Is there a function that could be used for calculation of the divergence of the vectorial field? (in matlab) I would expect it exists in numpy/scipy but I can not find it us

Answer 1

What Daniel had modified is the right answer, let me explain self defined func divergence further in more detail :

Function np.gradient() defined as : np.gradient( f) = df/dx, df/dy, df/dz +...

but we need define func divergence as : divergence ( f) = dfx/dx + dfy/dy + dfz/dz +... = np.gradient( fx) + np.gradient(fy) + np.gradient(fz) + ...

Let's test, compare with example of divergence in matlab

import numpy as np
import matplotlib.pyplot as plt

NY = 50
ymin = -2.
ymax = 2.
dy = (ymax -ymin )/(NY-1.)

NX = NY
xmin = -2.
xmax = 2.
dx = (xmax -xmin)/(NX-1.)

def divergence(f):
    num_dims = len(f)
    return np.ufunc.reduce(np.add, [np.gradient(f[i], axis=i) for i in range(num_dims)])

y = np.array([ ymin + float(i)*dy for i in range(NY)])
x = np.array([ xmin + float(i)*dx for i in range(NX)])

x, y = np.meshgrid( x, y, indexing = 'ij', sparse = False)

Fx  = np.cos(x + 2*y)
Fy  = np.sin(x - 2*y)

F = [Fx, Fy]
g = divergence(F)

plt.pcolormesh(x, y, g)
plt.colorbar()
plt.savefig( 'Div' + str(NY) +'.png', format = 'png')
plt.show()

Answer 2

Based on @paul_chen answer, and with some additions for Matplotlib 3.3.0 (a shading param needs to be passed, and default colormap I guess has changed)

import numpy as np
import matplotlib.pyplot as plt

NY = 20; ymin = -2.; ymax = 2.
dy = (ymax -ymin )/(NY-1.)
NX = NY
xmin = -2.; xmax = 2.
dx = (xmax -xmin)/(NX-1.)

def divergence(f):
    num_dims = len(f)
    return np.ufunc.reduce(np.add, [np.gradient(f[i], axis=i) for i in range(num_dims)])

y = np.array([ ymin + float(i)*dy for i in range(NY)])
x = np.array([ xmin + float(i)*dx for i in range(NX)])

x, y = np.meshgrid( x, y, indexing = 'ij', sparse = False)

Fx  = np.cos(x + 2*y)
Fy  = np.sin(x - 2*y)


F = [Fx, Fy]
g = divergence(F)

plt.pcolormesh(x, y, g, shading='nearest', cmap=plt.cm.get_cmap('coolwarm'))
plt.colorbar()
plt.quiver(x,y,Fx,Fy)
plt.savefig( 'Div.png', format = 'png')

Answer 3

Just a hint for everybody reading that:

the functions above do not compute the divergence of a vector field. they sum the derivatives of a scalar field A:

result = dA/dx + dA/dy

in contrast to a vector field (with three dimensional example):

result = sum dAi/dxi = dAx/dx + dAy/dy + dAz/dz

Vote down for all! It is mathematically simply wrong.

Cheers!

Answer 4

Based on Juh_'s answer, but modified for the correct divergence of a vector field formula

def divergence(f):
    """
    Computes the divergence of the vector field f, corresponding to dFx/dx + dFy/dy + ...
    :param f: List of ndarrays, where every item of the list is one dimension of the vector field
    :return: Single ndarray of the same shape as each of the items in f, which corresponds to a scalar field
    """
    num_dims = len(f)
    return np.ufunc.reduce(np.add, [np.gradient(f[i], axis=i) for i in range(num_dims)])

Matlab's documentation uses this exact formula (scroll down to Divergence of a Vector Field)

Answer 5

The divergence as a built-in function is included in matlab, but not numpy. This is the sort of thing that it may perhaps be worthwhile to contribute to pylab, an effort to create a viable open-source alternative to matlab.

http://wiki.scipy.org/PyLab

Edit: Now called http://www.scipy.org/stackspec.html

Answer 6

I don't think the answer by @Daniel is correct, especially when the input is in order [Fx, Fy, Fz, ...].

A simple test case

See the MATLAB code:

a = [1 2 3;1 2 3; 1 2 3];
b = [[7 8 9] ;[1 5 8] ;[2 4 7]];
divergence(a,b)

which gives the result:

ans =

   -5.0000   -2.0000         0
   -1.5000   -1.0000         0
    2.0000         0         0

and Daniel's solution:

def divergence(f):
    """
    Daniel's solution
    Computes the divergence of the vector field f, corresponding to dFx/dx + dFy/dy + ...
    :param f: List of ndarrays, where every item of the list is one dimension of the vector field
    :return: Single ndarray of the same shape as each of the items in f, which corresponds to a scalar field
    """
    num_dims = len(f)
    return np.ufunc.reduce(np.add, [np.gradient(f[i], axis=i) for i in range(num_dims)])


if __name__ == '__main__':
    a = np.array([[1, 2, 3]] * 3)
    b = np.array([[7, 8, 9], [1, 5, 8], [2, 4, 7]])
    div = divergence([a, b])
    print(div)
    pass

which gives:

[[1.  1.  1. ]
 [4.  3.5 3. ]
 [2.  2.5 3. ]]

Explanation

The mistake of Daniel's solution is, in Numpy, the x axis is the last axis instead of the first axis. When using np.gradient(x, axis=0), Numpy actually gives the gradient of y direction (when x is a 2d array).

My solution

There is my solution based on Daniel's answer.

def divergence(f):
    """
    Computes the divergence of the vector field f, corresponding to dFx/dx + dFy/dy + ...
    :param f: List of ndarrays, where every item of the list is one dimension of the vector field
    :return: Single ndarray of the same shape as each of the items in f, which corresponds to a scalar field
    """
    num_dims = len(f)
    return np.ufunc.reduce(np.add, [np.gradient(f[num_dims - i - 1], axis=i) for i in range(num_dims)])

which gives the same result as MATLAB divergence in my test case.