DiffEqOperators.jl
Linear operators for discretizations of differential equations and scientific machine learning (SciML)
Install / Use
/learn @SciML/DiffEqOperators.jlREADME
DiffEqOperators.jl
This Package is in the Process of being Deprecated
Alternatives:
- For automated finite difference discretization of symbolically-defined PDEs, see MethodOfLines.jl.
- For MatrixFreeOperators, and other non-derivative operators, see SciMLOperators.jl.
- For VecJacOperators and JacVecOperators, see SparseDiffTools.jl.
README
DiffEqOperators.jl is a package for finite difference discretization of partial differential equations. It allows building lazy operators for high order non-uniform finite differences in an arbitrary number of dimensions, including vector calculus operators.
For automatic Method of Lines discretization of PDEs, better suited to nonlinear systems of equations and more complex boundary conditions, please see MethodOfLines.jl
For the operators, both centered and
upwind operators are provided,
for domains of any dimension, arbitrarily spaced grids, and for any order of accuracy.
The cases of 1, 2, and 3 dimensions with an evenly spaced grid are optimized with a
convolution routine from NNlib.jl. Care is taken to give efficiency by avoiding
unnecessary allocations, using purpose-built stencil compilers, allowing GPUs
and parallelism, etc. Any operator can be concretized as an Array, a
BandedMatrix or a sparse matrix.
Documentation
For information on using the package, see the stable documentation. Use the in-development documentation for the version of the documentation which contains the unreleased features.
Example 1: Finite Difference Operator Solution for the Heat Equation
using DiffEqOperators, OrdinaryDiffEq
# # Heat Equation
# This example demonstrates how to combine `OrdinaryDiffEq` with `DiffEqOperators` to solve a time-dependent PDE.
# We consider the heat equation on the unit interval, with Dirichlet boundary conditions:
# ∂ₜu = Δu
# u(x=0,t) = a
# u(x=1,t) = b
# u(x, t=0) = u₀(x)
#
# For `a = b = 0` and `u₀(x) = sin(2πx)` a solution is given by:
u_analytic(x, t) = sin(2*π*x) * exp(-t*(2*π)^2)
nknots = 100
h = 1.0/(nknots+1)
knots = range(h, step=h, length=nknots)
ord_deriv = 2
ord_approx = 2
const Δ = CenteredDifference(ord_deriv, ord_approx, h, nknots)
const bc = Dirichlet0BC(Float64)
t0 = 0.0
t1 = 0.03
u0 = u_analytic.(knots, t0)
step(u,p,t) = Δ*bc*u
prob = ODEProblem(step, u0, (t0, t1))
alg = KenCarp4()
sol = solve(prob, alg)
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