Systems of Linear Algebraic Equations - Numerical Methods in Engineering with MATLAB - page 93

valueof n ):
⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

⎡
⎣

⎤
⎦

2 − 1 00 ... 0001 x 1

x 2
x 3
.
x n − 2
x n − 1
x n

0
0
0
.
0
0
1

− 12 − 1 0 ... 0000

0 − 12 − 1 ... 0000

. . . . . . . .

=

0000 ... − 12 − 1 0

0000 ... 0 − 12 − 1

1 000 ... 00 − 12

Run the programwith n = 20. The exact solution can be shown to be x i =− n / 4 + i / 2,

i = 1 , 2 ,..., n .

Solution In thiscase the iterativeformulas in Eq. (2.35) are

x 1 = ω ( x 2 − x n ) / 2 + (1 − ω ) x 1

x i = ω ( x i − 1 + x i + 1 ) / 2 + (1 − ω ) x i , i = 2 , 3 ,..., n − 1 (a)

x n = ω (1 − x 1 + x n − 1 ) / 2 + (1 − ω ) x n

which areevaluatedbythe following function:

functionx=fex2 17(x,omega)

% Iteration formula Eq. (2.35) for Example 2.17.

_

n = length(x);
x(1) = omega*(x(2) - x(n))/2 + (1-omega)*x(1);
fori=2:n-1
x(i) = omega*(x(i-1) + x(i+1))/2 + (1-omega)*x(i);

end
x(n) = omega *(1 - x(1) + x(n-1))/2 + (1-omega)*x(n);

=

The solution can beobtainedwith a single command (note that x 0 is the

starting vector):

>> [x,numIter,omega] = gaussSeidel(@fex2

_

17,zeros(20,1))

resulting in

x=
-4.5000
-4.0000
-3.5000
-3.0000