Jacobi method#Weighted Jacobi method

Let $A\backslash mathbf\; x\; =\; \backslash mathbf\; b$ be a square system of n linear equations, where:

When $A$ and $\backslash mathbf\; b$ are known, and $\backslash mathbf\; x$ is unknown, we can use the Jacobi method to approximate $\backslash mathbf\; x$. The vector $\backslash mathbf\; x^\{(0)\}$ denotes our initial guess for $\backslash mathbf\; x$ (often $\backslash mathbf\; x^\{(0)\}\_i=0$ for $i=1,2,...,n$). We denote $\backslash mathbf\{x\}^\{(k)\}$ as the k-th approximation or iteration of $\backslash mathbf\{x\}$, and $\backslash mathbf\{x\}^\{(k+1)\}$ is the next (or k+1) iteration of $\backslash mathbf\{x\}$.

Then A can be decomposed into a diagonal component D, a lower triangular part L and an upper triangular part U:The solution is then obtained iteratively via

:$\backslash mathbf\{x\}^\{(k+1)\}\; =\; D^\{-1\}\; (\backslash mathbf\{b\}\; -\; (L+U)\; \backslash mathbf\{x\}^\{(k)\}).$

The element-based formula for each row $i$ is thus:$$x^\{(k+1)\}\_i\; =\; \backslash frac\{1\}\{a\_\{ii\}\}\; \backslash left(b\_i\; -\backslash sum\_\{j\backslash ne\; i\}a\_\{ij\}x^\{(k)\}\_j\backslash right),\backslash quad\; i=1,2,\backslash ldots,n.$$The computation of $x\_i^\{(k+1)\}$ requires each element in $\backslash mathbf\{x\}^\{(k)\}$ except itself. Unlike the Gauss–Seidel method, we cannot overwrite $x\_i^\{(k)\}$ with $x\_i^\{(k+1)\}$, as that value will be needed by the rest of the computation. The minimum amount of storage is two vectors of size n.

Input: {{nowrap|initial guess x⁽⁰⁾ to the solution}}, (diagonal dominant) matrix A, right-hand side vector b, convergence criterion

Output: {{nowrap|solution when convergence is reached}}

Comments: pseudocode based on the element-based formula above

{{nowrap|1=k = 0}}

while convergence not reached do

for i := 1 step until n do

{{nowrap|1=σ = 0}}

for j := 1 step until n do

if j ≠ i then

{{nowrap|1=σ = σ + a_ij x_j^(k)}}

end

end

{{nowrap|1=x_i^(k+1) = (b_i − σ) / a_ii}}

end

increment k

end

The standard convergence condition (for any iterative method) is when the spectral radius of the iteration matrix is less than 1:

:

A sufficient (but not necessary) condition for the method to converge is that the matrix A is strictly or irreducibly diagonally dominant. Strict row diagonal dominance means that for each row, the absolute value of the diagonal term is greater than the sum of absolute values of other terms:

:$\backslash left\; |\; a\_\{ii\}\; \backslash right\; |\; >\; \backslash sum\_\{j\; \backslash ne\; i\}\; \{\backslash left\; |\; a\_\{ij\}\; \backslash right\; |\}.$

The Jacobi method sometimes converges even if these conditions are not satisfied.

Note that the Jacobi method does not converge for every symmetric positive-definite matrix. For example,

A =

\begin{pmatrix}

29 & 2 & 1\\

2 & 6 & 1\\

1 & 1 & \frac{1}{5}

\end{pmatrix}

\quad \Rightarrow \quad

D^{-1} (L+U) =

\begin{pmatrix}

0 & \frac{2}{29} & \frac{1}{29}\\

\frac{1}{3} & 0 & \frac{1}{6}\\

5 & 5 & 0

\end{pmatrix}

\quad \Rightarrow \quad

\rho(D^{-1}(L+U)) \approx 1.0661 \,.

A linear system of the form $Ax=b$ with initial estimate $x^\{(0)\}$ is given by

:$A=$

\begin{bmatrix}

2 & 1 \\

5 & 7 \\

\end{bmatrix},

\ b=

\begin{bmatrix}

11 \\

13 \\

\end{bmatrix}

\quad \text{and} \quad x^{(0)} =

\begin{bmatrix}

1 \\

1 \\

\end{bmatrix} .

We use the equation $x^\{(k+1)\}=D^\{-1\}(b\; -\; (L+U)x^\{(k)\})$, described above, to estimate $x$. First, we rewrite the equation in a more convenient form $D^\{-1\}(b\; -\; (L+U)x^\{(k)\})\; =\; Tx^\{(k)\}\; +\; C$, where $T=-D^\{-1\}(L+U)$ and $C\; =\; D^\{-1\}b$. From the known values

$$D^\{-1\}=$$

\begin{bmatrix}

1/2 & 0 \\

0 & 1/7 \\

\end{bmatrix},

\ L=

\begin{bmatrix}

0 & 0 \\

5 & 0 \\

\end{bmatrix}

\quad \text{and} \quad U =

\begin{bmatrix}

0 & 1 \\

0 & 0 \\

\end{bmatrix} .

we determine $T=-D^\{-1\}(L+U)$ as

$$T=$$

\begin{bmatrix}

1/2 & 0 \\

0 & 1/7 \\

\end{bmatrix}

\left\{

\begin{bmatrix}

0 & 0 \\

-5 & 0 \\

\end{bmatrix}

+

\begin{bmatrix}

0 & -1 \\

0 & 0 \\

\end{bmatrix}\right\}

=

\begin{bmatrix}

0 & -1/2 \\

-5/7 & 0 \\

\end{bmatrix} .

Further, $C$ is found as

$$C\; =$$

\begin{bmatrix}

1/2 & 0 \\

0 & 1/7 \\

\end{bmatrix}

\begin{bmatrix}

11 \\

13 \\

\end{bmatrix}

=

\begin{bmatrix}

11/2 \\

13/7 \\

\end{bmatrix}.

With $T$ and $C$ calculated, we estimate $x$ as $x^\{(1)\}=\; Tx^\{(0)\}+C$:

$$x^\{(1)\}=$$

\begin{bmatrix}

0 & -1/2 \\

-5/7 & 0 \\

\end{bmatrix}

\begin{bmatrix}

1 \\

1 \\

\end{bmatrix}

+

\begin{bmatrix}

11/2 \\

13/7 \\

\end{bmatrix}

=

\begin{bmatrix}

5.0 \\

8/7 \\

\end{bmatrix}

\approx

\begin{bmatrix}

5 \\

1.143 \\

\end{bmatrix} .

The next iteration yields

$$x^\{(2)\}=$$

\begin{bmatrix}

0 & -1/2 \\

-5/7 & 0 \\

\end{bmatrix}

\begin{bmatrix}

5.0 \\

8/7 \\

\end{bmatrix}

+

\begin{bmatrix}

11/2 \\

13/7 \\

\end{bmatrix}

=

\begin{bmatrix}

69/14 \\

-12/7 \\

\end{bmatrix}

\approx

\begin{bmatrix}

4.929 \\

-1.714 \\

\end{bmatrix} .

This process is repeated until convergence (i.e., until $\backslash |Ax^\{(n)\}\; -\; b\backslash |$ is small). The solution after 25 iterations is

:$x=\backslash begin\{bmatrix\}$

7.111\\

-3.222

\end{bmatrix}

.

Suppose we are given the following linear system:

:$$

\begin{align}

10x_1 - x_2 + 2x_3 & = 6, \\

-x_1 + 11x_2 - x_3 + 3x_4 & = 25, \\

2x_1- x_2+ 10x_3 - x_4 & = -11, \\

3x_2 - x_3 + 8x_4 & = 15.

\end{align}

If we choose {{math|(0, 0, 0, 0)}} as the initial approximation, then the first approximate solution is given by

\begin{align}

x_1 & = (6 + 0 - (2 * 0)) / 10 = 0.6, \\

x_2 & = (25 + 0 + 0 - (3 * 0)) / 11 = 25/11 = 2.2727, \\

x_3 & = (-11 - (2 * 0) + 0 + 0) / 10 = -1.1,\\

x_4 & = (15 - (3 * 0) + 0) / 8 = 1.875.

\end{align}

Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after five iterations.

The exact solution of the system is {{math|(1, 2, −1, 1)}}.

import numpy as np

ITERATION_LIMIT = 1000

1. initialize the matrix

A = np.array([[10., -1., 2., 0.],

[-1., 11., -1., 3.],

[2., -1., 10., -1.],

[0.0, 3., -1., 8.]])

1. initialize the RHS vector

b = np.array([6., 25., -11., 15.])

1. prints the system

print("System:")

for i in range(A.shape[0]):

row = [f"{A[i, j]}*x{j + 1}" for j in range(A.shape[1])]

print(f'{" + ".join(row)} = {b[i]}')

print()

x = np.zeros_like(b)

for it_count in range(ITERATION_LIMIT):

if it_count != 0:

print(f"Iteration {it_count}: {x}")

x_new = np.zeros_like(x)

for i in range(A.shape[0]):

s1 = np.dot(A[i, :i], x[:i])

s2 = np.dot(A[i, i + 1:], x[i + 1:])

x_new[i] = (b[i] - s1 - s2) / A[i, i]

if x_new[i] == x_new[i-1]:

break

if np.allclose(x, x_new, atol=1e-10, rtol=0.):

break

x = x_new

print("Solution: ")

print(x)

error = np.dot(A, x) - b

print("Error:")

print(error)

The weighted Jacobi iteration uses a parameter $\backslash omega$ to compute the iteration as

:$\backslash mathbf\{x\}^\{(k+1)\}\; =\; \backslash omega\; D^\{-1\}\; (\backslash mathbf\{b\}\; -\; (L+U)\; \backslash mathbf\{x\}^\{(k)\})\; +\; \backslash left(1-\backslash omega\backslash right)\backslash mathbf\{x\}^\{(k)\}$

with $\backslash omega\; =\; 2/3$ being the usual choice.{{cite book|last=Saad|first=Yousef|author-link=Yousef Saad|title=Iterative Methods for Sparse Linear Systems|edition=2nd|year=2003|publisher=SIAM|isbn=0898715342|page=[https://archive.org/details/iterativemethods0000saad/page/414 414]|url=https://archive.org/details/iterativemethods0000saad/page/414}}

From the relation $L\; +\; U\; =\; A\; -\; D$, this may also be expressed as

:$\backslash mathbf\{x\}^\{(k+1)\}\; =\; \backslash omega\; D^\{-1\}\; \backslash mathbf\{b\}\; +\; \backslash left(\; I\; -\; \backslash omega\; D^\{-1\}\; A\; \backslash right)\; \backslash mathbf\{x\}^\{(k)\}$.

In case that the system matrix $A$ is of symmetric positive-definite type one can show convergence.

Let $C=C\_\backslash omega\; =\; I-\backslash omega\; D^\{-1\}A$ be the iteration matrix.

Then, convergence is guaranteed for

:$$

\rho(C_\omega) < 1

\quad \Longleftrightarrow \quad

0 < \omega < \frac{2}{\lambda_\text{max} (D^{-1}A)} \,,

where $\backslash lambda\_\backslash text\{max\}$ is the maximal eigenvalue.

The spectral radius can be minimized for a particular choice of $\backslash omega\; =\; \backslash omega\_\backslash text\{opt\}$ as follows

\min_\omega \rho (C_\omega) = \rho (C_{\omega_\text{opt}}) = 1-\frac{2}{\kappa(D^{-1}A)+1}

\quad \text{for} \quad

\omega_\text{opt} := \frac{2}{\lambda_\text{min}(D^{-1}A)+\lambda_\text{max}(D^{-1}A)} \,,

where $\backslash kappa$ is the matrix condition number.

• Gauss–Seidel method
• Successive over-relaxation
• Iterative method § Linear systems
• Gaussian Belief Propagation
• Matrix splitting

{{Reflist}}

• {{CFDWiki|name=Jacobi_method}}
• {{MathWorld|urlname=JacobiMethod|title=Jacobi method|author=Black, Noel|author2=Moore, Shirley|author3= Weisstein, Eric W.|name-list-style=amp}}
• [http://www.math-linux.com/spip.php?article49 Jacobi Method from www.math-linux.com]

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