Chapter 4 solving systems of nonlinear equations

Chapter 4: Solving Systems of Nonlinear Equations
System of Nonlinear Equations
Let f1, f2,..., fn be a nonlinear scalar-valued function with variables
x1, x2,..., xn. We want to ﬁnd x1, x2,..., xn such that fi (x1, x2,..., xn)= 0
i = 1,..., n. That is,
f1(x1, x2,..., xn)= 0
f2(x1, x2,..., xn)= 0
.
.
.
fn(x1, x2,..., xn)= 0
Newton’s Method for System of Nonlinear Equations
Given an initial values x0, y0, each of the approximate solutions in iterations
i = 0, 1, 2,... is given by the formula:






















)
y
,
x
(
f
)
y
,
x
(
f
J
y
x
y
x
i
i
2
i
i
1
1
-
i
i
i
1
i
1
i
or 























y
x
y
x
y
x
i
i
1
i
1
i
where























)
y
,
x
(
2
)
y
,
x
(
2
)
y
,
x
(
1
)
y
,
x
(
1
i
i
i
i
i
i
i
i
i
y
f
x
f
y
f
x
f
J is Jacobian matrix








y
x
can be solved from the linear system
















)
y
,
x
(
f
)
y
,
x
(
f
y
x
J
i
i
2
i
i
1
i
For 3 equation, let











3
2
1
x
x
x
x and











x)
(
f
x)
(
f
x)
(
f
f(x)
3
2
1
)
f(x
J
x
x (i)
-1
i
(i)
1)
(i



or (i)
(i)
1)
(i
x
x
x 



where









































i
i
i
i
i
i
i
i
i
x
3
3
x
2
3
x
1
3
x
3
2
x
2
2
x
1
2
x
3
1
x
2
1
x
1
1
i
x
f
x
f
x
f
x
f
x
f
x
f
x
f
x
f
x
f
J
(i)
x
 can be solved from the linear system )
f(x
x
J (i)
(i)
i 


Example Find the Jacobian of the following vector-valued function f(x).













)
sin(x
x
1
x
e
x
x
2
)
x
,
x
,
f(x
1
2
3
2
x
3
1
3
2
1
2

Example An approximate solution (x, y) of the nonlinear system
e2x+y
- x = 0
x2
- y = 0
can be found by using Newton’s method. Determine the approximate solution
from the ﬁrst two iterations of Newton’s method when the initial values for x
and y are x0 = 0 and y0 = 0, respectively. Compute the absolute error using
Euclidean norm and inﬁnity norm in each iteration.

Example: Approximate the solution of the following nonlinear system by
using 2 iterations of Newton’s method
1
x
x
4x
0
x
4x
1
2
2
1
2
2
2
1




with initial value x(0)
= [0, 1]T
. Compute the absolute error by using
Euclidean norm in the last iteration (4 D.P. Rounding).

Fixed-point Method
Fixed-point iteration consists of two main steps:
(I) Transform the equation by constructing the iteration function g(x) so that
g(x)= x and f(x)= 0 have the same solution.
(II) Let x(0)
be an initial starting guess. The approximate solution in iteration
k = 1, 2,.... from Fixed-point method can be computed from:
xk
= g(xk-1
)
Condition for Convergence
1
x
g
...
x
g
x
g
n
i
2
i
1
i 









Example: Suppose we want to ﬁnd approximate solution x1 > 0, x2 > 0 of the
following nonlinear system by using Fixed-point method.
1
x
x
4x
0
x
4x
1
2
2
1
2
2
2
1




There are many possible iteration functions 






(x)
g
(x)
g
g(x)
2
1
so that the above
system has the same solution as x = g(x) where x =[x1, x2]T
.
Show that the followings can be used as iteration functions for this
nonlinear system.
1
1
2
2
1
2
2
1
2
1
2
2
2
1
1
4x
1
x
g2(x)
,
2
x
g1(x)
(II)
converge)
(not
x
1
x
x
4x
(x)
g
,
x
x
4x
(x)
g
(I)











Example: Find the approximate solution x1 > 0, x2 > 0 of the following
nonlinear system by using 2 iterations of Fixed-point method
1
x
x
4x
0
x
4x
1
2
2
1
2
2
2
1




with initial value x(0) = [1, 1]T
and
1
1
2
4x
1
x
g2(x)
,
2
x
g1(x)



Compute the absolute error in the last iteration by using Euclidean norm ( 4
D.P. Rounding).

Chapter 4 solving systems of nonlinear equations

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