Systems of Nonlinear Equations

CH EN 2450 - Numerical Methods

Prof. Tony Saad (www.tsaad.net)
Department of Chemical Engineering
University of Utah


Example 1

A system of nonlinear equations consists of several nonlinear functions - as many as there are unknowns. Solving a system of nonlinear equations means funding those points where the functions intersect each other. Consider for example the following system of equations \begin{equation} y = 4x - 0.5 x^3 \end{equation} \begin{equation} y = \sin(x)e^{-x} \end{equation}

The first step is to write these in residual form \begin{equation} f_1 = y - 4x - 0.5 x^3,\\ f_2 = y - \sin(x)e^{-x} \end{equation}

In [21]:
import numpy as np
from numpy import cos, sin, pi, exp
%matplotlib inline
%config InlineBackend.figure_format = 'svg'
import matplotlib.pyplot as plt
from scipy.optimize import fsolve
In [22]:
y1 = lambda x: 4 * x - 0.5 * x**3
y2 = lambda x: sin(x)*exp(-x)
x = np.linspace(-3.5,4,100)
plt.ylim(-8,6)
plt.plot(x,y1(x), 'k')
plt.plot(x,y2(x), 'r')
plt.grid()
plt.savefig('example1.pdf')
In [26]:
def newton_solver(F, J, x, tol):
    F_value = F(x)
    err = np.linalg.norm(F_value, ord=2)  # l2 norm of vector
    err = tol + 100
    niter = 0
    while abs(err) > tol and niter < 100:
        J_value = J(x)
        delta = np.linalg.solve(J_value, -F_value)
        x = x + delta
        F_value = F(x)
        err = np.linalg.norm(F_value, ord=2)
        niter += 1

    # Here, either a solution is found, or too many iterations
    if abs(err) > tol:
        niter = -1
    return x, niter, err
In [27]:
def F(xval):
    x = xval[0]
    y = xval[1]
    f1 = 0.5 * x**3 + y - 4*x
    f2 = y - sin(x)*exp(-x)
    return np.array([f1,f2])


def J(xval):
    x = xval[0]
    y = xval[1]
    return np.array([[1.5 * x**2 - 4, 1],
                     [-cos(x)*exp(-x)+sin(x)*exp(-x), 1]])

def Jdiff(F,xval):
    n = len(xval)
    J = np.zeros((n,n))
    col = 0
    for x0 in xval:
        delta = np.zeros(n)
        delta[col] = 1e-4 * x0 + 1e-12
        diff = (F(xval + delta) - F(xval))/delta[col]
        J[:,col] = diff
        col += 1
    return J
In [28]:
tol = 1e-8
xguess = np.array([-2,-4])
x, n, err = newton_solver(F, J, xguess, tol)
print (n, x)
print ('Error Norm =',err)
4 [-1.46110592 -4.28481694]
Error Norm = 1.523439979143842e-11
In [29]:
fsolve(F,(-2,-4))
Out[29]:
array([-1.46110592, -4.28481694])

Example 2

Find the roots of the following system of equations \begin{equation} x^2 + y^2 = 1, \\ y = x^3 - x + 1 \end{equation} First we assign $x_1 \equiv x$ and $x_2 \equiv y$ and rewrite the system in residual form \begin{equation} f_1(x_1,x_2) = x_1^2 + x_2^2 - 1, \\ f_2(x_1,x_2) = x_1^3 - x_1 - x_2 + 1 \end{equation}

In [39]:
x = np.linspace(-1,1)
y1 = lambda x: x**3 - x + 1
y2 = lambda x: np.sqrt(1 - x**2)
plt.plot(x,y1(x), 'k')
plt.plot(x,y2(x), 'r')
plt.grid()
In [33]:
def F(xval):
    x1 = xval[0]
    x2 = xval[1]
    f1 = x1**2 + x2**2 - 1
    f2 = x1**3 - x1 + 1 - x2
    return np.array([f1,f2])


def J(xval):
    x1 = xval[0]
    x2 = xval[1]
    return np.array([[2.0*x1,       2.0*x2],
                     [3.0*x1**2 -1, -1]])
In [31]:
tol = 1e-8
xguess = np.array([0.5,0.5])
x, n, err = newton_solver(F, J, xguess, tol)
print (n, x)
print ('Error Norm =',err)
6 [0.74419654 0.66796071]
Error Norm = 4.965068306494546e-16
In [32]:
fsolve(F,(0.5,0.5))
Out[32]:
array([0.74419654, 0.66796071])
In [40]:
import urllib
import requests
from IPython.core.display import HTML
def css_styling():
    styles = requests.get("https://raw.githubusercontent.com/saadtony/NumericalMethods/master/styles/custom.css")
    return HTML(styles.text)
css_styling()
Out[40]:
CSS style adapted from https://github.com/barbagroup/CFDPython. Copyright (c) Barba group