Systems
of
Nonlinear Equations
A system of nonlinear algebraic equations may arise
when one is dealing with problems involving
optimization and numerical integration (Gauss
quadratures).
Generally, the system of equations may not be of the
polynomial variety.
Therefore a system of n equations in n unknowns is
called nonlinear if one or more of the equations in
the systems is/are nonlinear.
The numerical methods we discussed so far have
been concerned with finding a root of a nonlinear
algebraic equation with one independent variable.
We now consider methods for solving systems of
nonlinear algebraic equations in which each
equation is a function of a specified number of
variables.
Consider the system of two nonlinear equations with
two variables
 f1 ( x, y )  0

 f 2 ( x, y )  0

The problem can be stated as follows:
f1 ( x, y)  0
Given the continuous functions ----------------f 2 ( x, y)  0 , find the values x = α and y = β
and ---------------such that
f1 ( ,  )  0 and ----------------------f 2 ( ,  )  0
------------------

f1 ( x, y)
The function -----------------and
----------------f 2 ( x, y)
may be algebraic equations, transcendental or any
nonlinear relationship between the input x and y
f1 ( x, y)  0
f 2 ( x, y)  0
and the output -----------------and
------------------.

The solutions to ---------are
the intersections of the
--------------------------------------------------,--------------
f1 ( x, y)  f 2 ( x, y)  0
This problem is considerably more complicated then
solution of a single nonlinear equation.
The one-point iterative method discussed in the
previous Section (Secant Method ) for the solution
of a single equation may be extended to system.
So to solve the system of nonlinear equations we have
many methods but we will use the Newton’s method.
 f1 ( x, y )  4 x 3  y  6  0

 f 2 ( x, y )  x 2 y  1  0
Newton’s Method
Consider the two nonlinear equations specified by
the equations
 f1 ( x, y )  0

 f 2 ( x, y )  0

( xn , yn ) an approximation to a
Suppose that ------------is
root (α, β), then by using the Taylor’s theorem for
f1 ( x, y) and --------functions of two variables for -----------( xn , yn )
f 2 ( x, y)
-------------expanding
about -----------,
Newton’s Method
we have
and
Newton’s Method
f 2 ( ,  )  0 these
Since --------------f1 ( ,  )  0 and -----------------,
equations, with x = α and y = β, give
Newton’s Method
The Newton’s method has a condition that initial
( xn , yn )
approximation -------------should
sufficiently close
to exact root (α, β), therefore, the higher order terms
may be neglected to obtain

Newton’s Method
We see that this represents a system of two linear
algebraic equations for α and β.
Of course, since the higher order terms are omitted in
the derivation of these equations, their solution (α, β) is
no longer an exact root of --------and
------.


However, it will usually be a better approximation
than ( xn , yn )
Newton’s Method
( xn 1 , yn 1 ) -------------------, so replacing (α, β) by -----------------in
----and
------,

 gives the iterative scheme
Then writing in the matrix form, we have
Newton’s Method

f1 , f 2
where ---------and
their partial derivatives--------------f1 , f 2
are evaluated at-------------.
( xn , yn )
x
x
Newton’s Method
Hence
We call the following matrix J a Jacobian matrix
Newton’s Method
Note that -------can
 be written in the simplified form as
follows
Newton’s Method
where h and k can be evaluated as
and

Newton’s Method
where all functions are to be evaluated at (x, y).
The Newton’s method for a pair of equations in two
unknowns is therefore
where (h, k) are given by ------( xn , yn )
 evaluated at -------------.
Newton’s Method
At a starting approximation ---------------,
( x0 , y0 ) the
f 2 y are
functions f1 , f1x , f1y , f 2 , f 2x and -------,
evaluated.
The linear equations are then solved for -------------and
( x1, y1 )
whole process is repeated until convergence is
obtained.
Newton’s Method
By comparison of the
and
shows that the above procedure is indeed an extension
of the Newton’s method in one variable, where division
by f ′ generalized to pre-multiplication by J−1.
Newton’s Method
Procedure (Newton’s Method for Two Nonlinear Equations)
1. Choose the initial guess for the roots of the system,
so that the determinant of the Jacobian matrix is
not zero.
2. Establish Tolerance ϵ(> 0).
3. Evaluate the Jacobian at initial approximations
and then find inverse of Jacobian.
Newton’s Method
Procedure (Newton’s Method for Two Nonlinear Equations)
1. Choose the initial guess for the roots of the system,
so that the determinant of the Jacobian matrix is
not zero.
2. Establish Tolerance ϵ(> 0).
3. Evaluate the Jacobian at initial approximations
and then find inverse of Jacobian.
Newton’s Method
Procedure (Newton’s Method for Two Nonlinear Equations)
4. Compute new approximation to the roots by using iterative
h

  x 
x 



n1

n

n




X

formula X ----------------;
--------------,
-----------,------------

Z X
 y  Z    X    .
n 1
n 1

n 1
n

n
k 
n
 yn 
) (xn−1,
( xn1, yyn−1)∥
5. Check tolerance limit. If ∥(xn,( xyn)
n , yn −
n1 )   ≤ ϵ, for n ≥ 0,
then end; otherwise, go back to step 3, and repeat the process.
‫سوف نستعرض حساب مقلوب مصفوفة في الحالة‬
‫على المثال التالي‪:‬‬
‫‪0 0 1 ‬‬
‫‪‬‬
‫‪‬‬
‫‪A  2  1 3‬‬
‫‪1 1 4‬‬
‫‪n=3‬‬
------------det( A) ‫نحسب قيمة‬
0 0 1

det( A)  2  1 3
1 1 4
0 0

2  1
1 1 
det( A)  ((0 * 1* 4)  (0 * 3 *1)  (1* 2 *1))
 ((0 * 2 * 4)  (0 * 3 *1)  (1* 1*1))
det( A)  (0  0  2)  ((0)  (0)  (1))
det( A)  2  1
det( A)  3  0
 c11 c12
adj ( A)  C T  c21 c22
c31 c32
c13 
c23 
c33 
T
------------adj ( A) ‫نحسب قيمة‬
0 0 1 
c11  (1)11 2  1 3  (1)11 (( 1* 4)  (3 *1))  (4  3)  7
1 1 4
0 0 1 
c12  (1)1 2 2  1 3  (1)1 2 (( 2 * 4)  (3 *1))  (8  3)  5
1 1 4
0 0 1 
c13  (1)13 2  1 3  (1)13 (( 2 *1)  (1*1))  (2  1)  3
1 1 4
0 0 1 
c21  (1) 21 2  1 3  (1) 21 ((0 * 4)  (1*1))  (0  1)  1
1 1 4
0 0 1 
c22  (1) 2 2 2  1 3  (1) 2 2 ((0 * 4)  (1*1))  (0  1)  1
1 1 4
0 0 1 
c23  (1) 23 2  1 3  (1) 23 ((0 *1)  (0 *1))  (0  0)  0
1 1 4
0 0 1 
c31  (1)31 2  1 3  (1)31 ((0 * 3)  (1* 1))  (0  1)  1
1 1 4
0 0 1 
c32  (1)3 2 2  1 3  (1)3 2 ((0 * 3)  (1* 2))  (0  2)  2
1 1 4
0 0 1 
c33  (1)33 2  1 3  (1)33 ((0 * 1)  (0 * 2))  (0  0)  0
1 1 4
 7  5 3
C   1  1 0
 1
2 0
 7 1 1 
C T    5  1 2
 3
0 0
7 3
A1  5 3
 1
1
3
1
3
0

2 
3

0 
1
3
0 0 1 7 3
AA1  2  1 3 5 3
1 1 4  1
 0  0 1

1
AA   14 3  5 3  3
 7 3  5 3  4
1
3
1
3
0

2 
3

0 
1
3
000
000 
2  1 0
2  2  0
3
3
3
3

1  1 0
1  2  0
3
3
3
3

1 0 0 


1
AA  0 1 0  I
0 0 1 
a b ------------n=2 ‫في الحالة‬
A

c d 
det( A)  ad  bc
T
 d  c
 d  b
adj ( A)  C  




 b a 
 c a 
T
d

ad bc
1
A   c
 ad bc
b
ad bc
a
ad bc


