Copyright (C) 2010-2020 Luke Olson
Copyright (C) 2020 Andreas Kloeckner
Let's consider solving the discretized 2D Poisson equation with iterative methods.
import numpy as np
import numpy.linalg as la
from matplotlib import pyplot as pt
n=16
T = 2*np.eye(n)-np.diag(np.ones(n-1),1)-np.diag(np.ones(n-1),-1)
A = np.kron(np.eye(n),T)+np.kron(T,np.eye(n))
pt.spy(A)
<matplotlib.image.AxesImage at 0x7f11e955ae48>
Define a right-hand side and solve the resulting system directly.
h = 1/(n-1)
b = h*np.arange(0,n*n)
x = la.solve(A,b)
Split the matrix into its diagonal and strictly lower/upper triangular parts.
d = np.diag(A)
D = np.diag(d)
L = np.tril(A,-1)
U = np.triu(A,1)
la.norm(A-(D+L+U))
0.0
Jacobi iteration proceeds using $$\boldsymbol x^{(i+1)} = \boldsymbol D^{-1}(\boldsymbol b- (\boldsymbol L+\boldsymbol U)\boldsymbol x^{(i)}).$$
def jacobi(niter,x0):
xi = x0
for i in range(niter):
xi = np.diag(1./d)@(b-(L+U)@xi)
return xi
niters = np.asarray(2**np.arange(4,12),dtype=int)
x0 = np.random.random(n*n)
jacobi_results = []
err = []
for niter in niters:
jacobi_results.append(jacobi(niter,x0.copy()))
err.append(la.norm(jacobi_results[-1]-x))
pt.plot(niters,err)
pt.yscale('log')
pt.xscale('log')
def cg(A,b,niter,x0):
rk = b - A @ x0
sk = rk
xk = x0
for i in range(niter):
alpha = np.inner(rk,rk)/np.inner(sk, A @ sk)
xk1 = xk + alpha * sk
rk1 = rk - alpha * A @ sk
beta = np.inner(rk1,rk1)/np.inner(rk,rk)
sk1 = rk1 + beta*sk
rk = rk1
xk = xk1
sk = sk1
return xk
iters = np.asarray(2**np.arange(2,7),dtype=int)
x0 = np.random.random(n*n)
cg_results = []
err = []
for niter in iters:
cg_results.append(cg(A,b,niter,x0.copy()))
err.append(la.norm(cg_results[-1]-x))
pt.plot(iters,err)
pt.yscale('log')
pt.xscale('log')