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Fixed-Point Iteration for Nonlinear System of Equations

by irony of truth
Tags: equations, fixedpoint, iteration, nonlinear
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irony of truth
Dec20-06, 07:43 AM
P: 91

I am solving for the fixed point of this nonlinear system:
x^2 - x + 2y^2 + yz - 10 = 0
5x - 6y + z = 0
-x^2 - y^2 + z = 0

Somehow, I got stuck with my function for g, g(x) = x. I ran this in a program applying the Newton's method and I got its solution easily. However, I find it difficult using fixed-point method.

My function g should look something like this: My first expression shall be "equated to x", but the equal sign should not appear. My second expression shall be "equated to y" but no equal sign should appear (and so forth). I have tried several forms of g but the iteration would simply diverge from the fixed point.

For example,
x = x^2 + 2y^2 + yz - 10; others: (x^2 + x+ 2y^2 + yz - 10)/2
y = (5x + y + z) /7; others: (5x + z)/6; (5x + 2y + z) /8
z = x^2 + y^2

So, my function would be:
(x^2 + 2y^2 + yz - 10)
g = ((5x + y + z) /7 )
( x^2 + y^2 )
(but for any initial guess, the fixed point can't be found (as it diverges)).

Is this possible to converge to a fixed point?

Any help will be appreciated.
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Dec27-06, 05:12 PM
P: 134
if you put your system in the form

[tex] F(x,y,z)=0 [/tex] i think the condition for convergence (no t pretty sure) is:

[tex] |GraF|<1 [/tex] gra=gradient of the function....
irony of truth
Dec27-06, 10:06 PM
P: 91
Hi Karlisbad:

[tex] F(x,y,z)=0 [/tex]

I believe that this will be effective for the Newton's Method, which I got one of its solutions using a scilab program. It's pretty difficult for me to use the fixed point iteration, but I am hoping that somehow, I can get that fixed point.

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