Polar Coordinates¶

By assembling general linear combinations of differential operators with variable coefficients in findiff, you can use vector calculus operators in coordinates other than cartesian. Here we show an example for using polar coordinates in 2D.

import numpy as np
from findiff import FinDiff, Coefficient, Laplacian

Sample Problem¶

In this example, we calculate the Laplacian of the 2D paraboloid

$f(x, y) = x^2 + y^2$

once in cartesian coordinates and once in polar coordinates. Let’s start with cartesian.

The Laplacian is obviously

$\nabla^2 f = 2 + 2 = 4$

for all points $(x, y)$ .

Now let’s do this with findiff. First we define the grid and the function.

x, y = [np.linspace(-5, 5, 100)] * 2
dx, dy = x[1] - x[0], y[1] - y[0]
X, Y = np.meshgrid(x, y, indexing='ij')
f = X**2 + Y**2

Then we define and apply the Laplacian in cartesian coordinates:

laplace = Laplacian(h=[dx, dy])
laplace_f = laplace(f)

We get

laplace_f

array([[4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.],
       ...,
       [4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.]])

Polar Coordinates¶

In polar coordinates the paraboloid is simply

$f(r, \varphi) = r^2$

We define our grid in polar coordinates and calculate the $f$ in polar coordinates:

r = np.linspace(0.1, 10, 100)
phi = np.linspace(0, 2*np.pi, 100, endpoint=False)
dr, dphi = r[1] - r[0], phi[1] - phi[0]
R, Phi = np.meshgrid(r, phi, indexing='ij')
f_polar = R**2

Of course, the two arrays f and f_polar are completely different:

f - f_polar

array([[ 49.99      ,  48.99010203,  48.01061014, ...,  48.01061014,
         48.99010203,  49.99      ],
       [ 48.96010203,  47.96020406,  46.98071217, ...,  46.98071217,
         47.96020406,  48.96010203],
       [ 47.93061014,  46.93071217,  45.95122028, ...,  45.95122028,
         46.93071217,  47.93061014],
       ...,
       [-48.01938986, -49.01928783, -49.99877972, ..., -49.99877972,
        -49.01928783, -48.01938986],
       [-49.00989797, -50.00979594, -50.98928783, ..., -50.98928783,
        -50.00979594, -49.00989797],
       [-50.        , -50.99989797, -51.97938986, ..., -51.97938986,
        -50.99989797, -50.        ]])

The Laplacian in polar coordinates is

$\nabla^2 = \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \varphi^2}$

or in findiff:

laplace_polar = FinDiff(0, dr, 2) + Coefficient(1/R) * FinDiff(0, dr) + Coefficient(1/R**2) * FinDiff(1, dphi, 2)
result = laplace_polar(f_polar)

And we get the same result

result

array([[4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.],
       ...,
       [4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.],
       [4., 4., 4., ..., 4., 4., 4.]])