Jacobi Method

The iterative methods depend on the rearrangement of the equations in this manner:

$\displaystyle \boxed{x_i=\frac{b_i}{a_{ii}}- \sum_{j=1,j\neq i}^n \frac{a_{ij}}... ...,\ldots, n,\mapsto x_1=\frac{11}{6}-\left(\frac{-2}{6}x_2+\frac{1}{6}x_3\right)$

(6.4)

Each equation now solved for the variables in succession:

$\begin{indisplay}\begin{array}{l} x_1 = 1.8333 + 0.3333x_2 - 0.1667x_3 \\ x_2 =... ....2857x_3 \\ x_3 = 0.2000 + 0.2000x_1 + 0.4000x_2 \\ \end{array}\end{indisplay}$

We begin with some initial approximation to the value of the variables.

Say initial values are; $x_1=0,x_2=0,x_3=0$

. Each component might be taken equal to zero if no better initial estimates are at hand.

The new values are substituted in the right-hand sides to generate a second approximation,

and the process is repeated until successive values of each of the variables are sufficiently alike.

Now, general form

$\begin{indisplay}\begin{array}{l} x_1^{(n+ 1)} = 1.8333 + 0.3333x_2^{(n)}- 0.166... ...+ 1)} = 0.2000 + 0.2000x_1^{(n)} + 0.4000x_2^{(n)}\\ \end{array}\end{indisplay}$

(6.5)

Starting with an initial vector of $x^{(0)}=(0,0,0,)$ , we obtain Table 6.1

Table 6.1: Successive estimates of solution (Jacobi method)

First	Second	Third	Fourth	Fifth	Sixth	$\ldots$	Ninth
0	1.833	2.038	2.085	2.004	1.994	$\ldots$	2.000
0	0.714	1.181	1.053	1.001	0.990	$\ldots$	1.000
0	0.200	0.852	1.080	1.038	1.001	$\ldots$	1.000

Rewrite in matrix notation; let $A = L + D + U$

$\begin{indisplay} Ax=b \Longrightarrow \left[ \begin{array}{rrr} 6 &-2 & 1 \\ ... ...=\left[ \begin{array}{r} 11\\ 5\\ -1\\ \end{array}\right] \end{indisplay}$

$\begin{indisplay} L=\left[ \begin{array}{rrr} 0 & 0 & 0 \\ -2 & 0 & 0 \\ 1 ... ...0 &-2 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \\ \end{array} \right] \end{indisplay}$

$\displaystyle \begin{array}{l} Ax = (L + D + U)x= b \\ Dx = -(L + U)x + b\\ x = -D^{-1}(L + U)x + D^{-1}b\\ \end{array}$

From this we have, identifying $x$

on the left as the new iterate,

$\displaystyle x^{(n+1)} = -D^{-1}(L + U)x^{(n)} + D^{-1}b$

In Eqn. 6.5,

$\begin{indisplay} b'=D^{-1}b=\left[ \begin{array}{l} 1.8333 \\ 0.7143 \\ 0.2000 \\ \end{array} \right] \end{indisplay}$

$\begin{indisplay} D^{-1}(L + U)=\left[ \begin{array}{rrr} 0 &-0.3333 & 0.1667 \... ... & 0 & 0.2857 \\ -0.2000 & -0.4000 & 0 \\ \end{array} \right] \end{indisplay}$

This procedure is known as the Jacobi method, also called ”the method of simultaneous displacements”,

because each of the equations is simultaneously changed by using the most recent set of $x$

-values (see Table 6.1).

Example py-file: The Jacobi approximation to the solution of AX = B. myJacobi.py