Nonlinear Systems of Equations 1
3.10 Nonlinear Systems of Equations
1. For each of the following nonlinear systems, write out the vector-valued function
Fassociated with the system and compute the Jacobian of F.
(a) x1−x2−x3
1= 0
x1+x2−x3
2= 0
(b) 1 + x2−e−x1= 0
x3
1−x2= 0
(c)
2x1−3x2+x3−4 = 0
2x1+x2−x3+ 4 = 0
x2
1+x2
2+x2
3−4 = 0
(a) Define
f1(x1, x2) = x1−x2−x3
1and f2(x1, x2) = x1+x2−x3
2
(b) Define
f1(x1, x2) = 1 + x2−e−x1and f2(x1, x2) = x3
1−x2
2Section 3.10
(c) Define
f1(x1, x2, x3) = 2x1−3x2+x3−4,
2. For each of the nonlinear systems in Exercise 1, carry out two iterations of
Newton’s method. Use the initial vector indicated below.
(a) x(0) =1
2
1
2T
(b) x(0) =1 1 T
(c) x(0) =−1
2−3
2
3
2T
(a) From Exercise 1a, we know that
and
Nonlinear Systems of Equations 3
Solving the linear system J(x(0))v(0) =−F(x(0))yields the update vector
For the second iteration, we calculate
Thus,
(b) From Exercise 1b, we know that
and
4Section 3.10
For the second iteration, we calculate
(c) From Exercise 1c, we know that
and
2−3 1
For the second iteration, we calculate
F(x(1)) = 0 0 29
400 T
,
Nonlinear Systems of Equations 5
3. For each of the nonlinear systems in Exercise 1, carry out two iterations of
Broyden’s method. Use the initial vectors indicated in Exercise 2.
(a) From Exercise 1a, we know that
where x=x1x2T. With x(0) =1
2
1
2T
, we find
For the next iteration, we start by computing
6Section 3.10
and also noting that ∆=v(0). To compute A−1
1according to equation (4),
we will need the intermediate results
(b) From Exercise 1b, we know that
F(x) = 1 + x2−e−x1
x3
1−x2,
Nonlinear Systems of Equations 7
and also noting that ∆=v(0). To compute A−1
1according to equation (4),
we will need the intermediate results
(c) From Exercise 1c, we know that
2x1−3x2+x3−4
where x=x1x2x3T. With x(0) =−1
2−3
2
3
2T
, we find
8Section 3.10
For the next iteration, we start by computing
0y= 0.070325. Then
0.01030927835052 0.60824742268041 0.20618556701031
and
In Exercises 4 – 10, solve the indicated nonlinear system of equations using both
Newton’s method and Broyden’s method. Use the indicated initial vector, and
terminate the iteration process when the maximum norm of the difference be-
tween successive iterates falls below 5×10−6. Compare the number of iterations
required by the two methods to achieve convergence.
Nonlinear Systems of Equations 9
4. 5 cos x+ 6 cos(x+y)−10 = 0
5 sin x+ 6 sin(x+y)−4 = 0 x(0) =0.7 0.7T
Define
With x(0) =0.7 0.7Tand a convergence tolerance of 5×10−6, Newton’s
method converges in seven iterations:
nx(n)T
1−0.598549 1.833946
40.147912 0.425999
With the same initial vector and convergence tolerance, Broyden’s method requires
thirteen iterations to achieve convergence:
nx(n)T
1−0.598549 1.833946
60.203952 0.330526
5.
x2
1+x2
2+x2
3−1 = 0
x2
1+x2
3−0.25 = 0
x2
1+x2
2−4x3= 0
x(0) =111T
10 Section 3.10
Define
f1(x1, x2, x3) = x2
1+x2
2+x2
3−1,
x2
and
2x12x22x3
With x(0) =111Tand a convergence tolerance of 5×10−6, Newton’s
method converges in six iterations:
nx(n)T
10.791667 0.875000 0.333333
20.523653 0.866071 0.238095
nx(n)T
10.791667 0.875000 0.333333
20.586953 0.865584 0.244054
6.
x2
1+x2
2+x2
3−10 = 0
x1+ 2x2−2 = 0
x1+ 3x3−9 = 0
x(0) =202T
Nonlinear Systems of Equations 11
Define
f1(x1, x2, x3) = x2
1+x2
2+x2
3−10,
x2
1+x2
2+x2
3−10
nx(n)T
12.250000 −0.125000 2.250000
With the same initial vector and convergence tolerance, Broyden’s method requires
five iterations to achieve convergence:
nx(n)T
12.250000 −0.125000 2.250000
7. x3+ 10x−y−5 = 0
x+y3−10y+ 1 = 0 x(0) =1 0 T
Define
and let x=x y T. Then