# Fixed-Point Iteration for Nonlinear System of Equations

#### irony of truth

Hello:

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.

if you put your system in the form

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

$$|GraF|<1$$ gra=gradient of the function....

#### irony of truth

$$F(x,y,z)=0$$

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.

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving