A Python package for finite difference numerical derivatives and partial differential equations in any number of dimensions.

• Differentiate arrays of any number of dimensions along any axis with any desired accuracy order
• Accurate treatment of grid boundary
• Can handle arbitrary linear combinations of derivatives with constant and variable coefficients
• Fully vectorized for speed
• Matrix representations of arbitrary linear differential operators
• Solve partial differential equations with Dirichlet or Neumann boundary conditions
• New in version 0.10: Symbolic representation of finite difference schemes

pip install --upgrade findiff

You can find the documentation of the code including examples of application at https://findiff.readthedocs.io/en/latest/.

findiff allows to easily define derivative operators that you can apply to numpy arrays of any dimension. The syntax for a simple derivative operator is

FinDiff(axis, spacing, degree)

where spacing is the separation of grid points between neighboring grid points. Consider the 1D case with a first derivative $\displaystyle \frac{\partial}{\partial x}$ along the only axis (0):

import numpy as np

x = np.linspace(0, 1, 100)
f = np.sin(x)  # as an example

# Define the derivative:
d_dx = FinDiff(0, dx, 1)

# Apply it:
df_dx = d_dx(f) 

Similary, you can define partial derivative operators along different axes or of higher degree, for example:

We can also take linear combinations of derivative operators, for example:

$$ 2x \frac{\partial^3}{\partial x^2 \partial z} + 3 \sin(y)z^2 \frac{\partial^3}{\partial x \partial y^2} $$

is

Coef(2*X) * FinDiff((0, dz, 2), (2, dz, 1)) + Coef(3*sin(Y)*Z**2) * FinDiff((0, dx, 1), (1, dy, 2))

where X, Y, Z are numpy arrays with meshed grid points.

Chaining differential operators is also possible, e.g.

$$ \left( \frac{\partial}{\partial x} - \frac{\partial}{\partial y} \right) \cdot \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} \right) = \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} $$

can be written as

(FinDiff(0, dx) - FinDiff(1, dy)) * (FinDiff(0, dx) + FinDiff(1, dy))

and

FinDiff(0, dx, 2) - FinDiff(1, dy, 2)

Of course, standard operators from vector calculus like gradient, divergence and curl are also available as shortcuts.

More examples can be found here and in this blog.

When constructing an instance of FinDiff, you can request the desired accuracy order by setting the keyword argument acc. For example:

d2_dx2 = findiff.FinDiff(0, dy, 2, acc=4)
d2f_dx2 = d2_dx2(f)

If not specified, second order accuracy will be taken by default.

Sometimes you may want to have the raw finite difference coefficients. These can be obtained for any derivative and accuracy order using findiff.coefficients(deriv, acc). For instance,

import findiff
coefs = findiff.coefficients(deriv=3, acc=4, symbolic=True)

gives

{'backward': {'coefficients': [15/8, -13, 307/8, -62, 461/8, -29, 49/8],
              'offsets': [-6, -5, -4, -3, -2, -1, 0]},
 'center': {'coefficients': [1/8, -1, 13/8, 0, -13/8, 1, -1/8],
            'offsets': [-3, -2, -1, 0, 1, 2, 3]},
 'forward': {'coefficients': [-49/8, 29, -461/8, 62, -307/8, 13, -15/8],
             'offsets': [0, 1, 2, 3, 4, 5, 6]}}

If you want to specify the detailed offsets instead of the accuracy order, you can do this by setting the offset keyword argument:

import findiff
coefs = findiff.coefficients(deriv=2, offsets=[-2, 1, 0, 2, 3, 4, 7], symbolic=True)

The resulting accuracy order is computed and part of the output:

{'coefficients': [187/1620, -122/27, 9/7, 103/20, -13/5, 31/54, -19/2835], 
 'offsets': [-2, 1, 0, 2, 3, 4, 7], 
 'accuracy': 5}

For a given FinDiff differential operator, you can get the matrix representation using the matrix(shape) method, e.g. for a small 1D grid of 10 points:

d2_dx2 = FinDiff(0, dx, 2)
mat = d2_dx2.matrix((10,))  # this method returns a scipy sparse matrix
print(mat.toarray())

has the output

[[ 2. -5.  4. -1.  0.  0.  0.]
 [ 1. -2.  1.  0.  0.  0.  0.]
 [ 0.  1. -2.  1.  0.  0.  0.]
 [ 0.  0.  1. -2.  1.  0.  0.]
 [ 0.  0.  0.  1. -2.  1.  0.]
 [ 0.  0.  0.  0.  1. -2.  1.]
 [ 0.  0.  0. -1.  4. -5.  2.]]

findiff uses standard stencils (patterns of grid points) to evaluate the derivative. However, you can design your own stencil. A picture says more than a thousand words, so look at the following example for a standard second order accurate stencil for the 2D Laplacian $\displaystyle \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$:

This can be reproduced by findiff writing

offsets = [(0, 0), (1, 0), (-1, 0), (0, 1), (0, -1)]
stencil = Stencil(offsets, partials={(2, 0): 1, (0, 2): 1}, spacings=(1, 1))

The attribute stencil.values contains the coefficients

{(0, 0): -4.0, (1, 0): 1.0, (-1, 0): 1.0, (0, 1): 1.0, (0, -1): 1.0}

Now for a some more exotic stencil. Consider this one:

With findiff you can get it easily:

offsets = [(0, 0), (1, 1), (-1, -1), (1, -1), (-1, 1)]
stencil = Stencil(offsets, partials={(2, 0): 1, (0, 2): 1}, spacings=(1, 1))
stencil.values

which returns

{(0, 0): -2.0, (1, 1): 0.5, (-1, -1): 0.5, (1, -1): 0.5, (-1, 1): 0.5}

As of version 0.10, findiff can also provide a symbolic representation of finite difference schemes suitable for using in conjunction with sympy. The main use case is to facilitate deriving your own iteration schemes.

from findiff import SymbolicMesh, SymbolicDiff

mesh = SymbolicMesh("x, y")
u = mesh.create_symbol("u")
d2_dx2, d2_dy2 = [SymbolicDiff(mesh, axis=k, degree=2) for k in range(2)]

(
    d2_dx2(u, at=(m, n), offsets=(-1, 0, 1)) + 
    d2_dy2(u, at=(m, n), offsets=(-1, 0, 1))
)

Outputs:

$$ \frac{u_{m,n + 1} + u_{m,n - 1} - 2 u_{m,n}}{\Delta y^2} + \frac{u_{m + 1,n} + u_{m - 1,n} - 2 u_{m,n}}{\Delta x^2} $$

Also see the example notebook.

findiff can be used to easily formulate and solve partial differential equation problems

$$ \mathcal{L}u(\vec{x}) = f(\vec{x}) $$

where $\mathcal{L}$ is a general linear differential operator.

In order to obtain a unique solution, Dirichlet, Neumann or more general boundary conditions can be applied.

Find the solution of

$$ \left( \frac{d^2}{dt^2} - \alpha \frac{d}{dt} + \omega^2 \right)u(t) = \sin{(2t)} $$

subject to the (Dirichlet) boundary conditions

$$ u(0) = 0, \hspace{1em} u(10) = 1 $$

from findiff import FinDiff, Id, PDE

shape = (300, )
t = numpy.linspace(0, 10, shape[0])
dt = t[1]-t[0]

L = FinDiff(0, dt, 2) - FinDiff(0, dt, 1) + Coef(5) * Id()
f = numpy.cos(2*t)

bc = BoundaryConditions(shape)
bc[0] = 0
bc[-1] = 1

pde = PDE(L, f, bc)
u = pde.solve()

Result:

A plate with temperature profile given on one edge and zero heat flux across the other edges, i.e.

$$ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) u(x,y) = f(x,y) $$

with Dirichlet boundary condition

$$ \begin{align*} u(x,0) &= 300 \\ u(1,y) &= 300 - 200y \end{align*} $$

and Neumann boundary conditions

$$ \begin{align*} \frac{\partial u}{\partial x} &= 0, & \text{ for } x = 0 \\ \frac{\partial u}{\partial y} &= 0, & \text{ for } y = 0 \end{align*} $$

shape = (100, 100)
x, y = np.linspace(0, 1, shape[0]), np.linspace(0, 1, shape[1])
dx, dy = x[1]-x[0], y[1]-y[0]
X, Y = np.meshgrid(x, y, indexing='ij')

L = FinDiff(0, dx, 2) + FinDiff(1, dy, 2)
f = np.zeros(shape)

bc = BoundaryConditions(shape)
bc[1,:] = FinDiff(0, dx, 1), 0  # Neumann BC
bc[-1,:] = 300. - 200*Y   # Dirichlet BC
bc[:, 0] = 300.   # Dirichlet BC
bc[1:-1, -1] = FinDiff(1, dy, 1), 0  # Neumann BC

pde = PDE(L, f, bc)
u = pde.solve()

Result:

You have used findiff in a publication? Here is how you can cite it:

M. Baer. findiff software package. URL: https://github.com/maroba/findiff. 2018

BibTeX entry:

@misc{findiff,
  title = {{findiff} Software Package},
  author = {M. Baer},
  url = {https://github.com/maroba/findiff},
  key = {findiff},
  note = {\url{https://github.com/maroba/findiff}},
  year = {2018}
}

• Fork the repository
• Clone your fork to your machine
• Install in development mode:

python setup.py develop

From the console:

python -m unittest discover test