5.3.3 The matlab routine function fsolve

The MATLAB routine function fsolve is used to solve sets of nonlinear algebraic equations using a quasi-Newton method.

Syntax

x = fsolve(fun,x0,options) solves the equations with the optimization options specified in the structure options. Use function optimset to set these options.

Input Arguments

Function Arguments contains general descriptions of arguments passed into fsolve:

fun fun is a function that accepts a vector x and returns a vector F, the nonlinear equations evaluated at x.

If the Jacobian can also be computed and the Jacobian option is 'on', set by options = optimset('Jacobian','on') the function fun must return, in a second output argument, the Jacobian value J, a matrix, at x;

x0 Initial point for x;

options Options structure created with function optimset.

Output Arguments

[x,fval] = fsolve(fun,x0) returns the value of the objective function fun at the solution x.

[x,fval,exitflag] = fsolve(...) returns a value exitflag that describes the exit condition.

[x,fval,exitflag,output] = fsolve(...) returns a structure output that contains information about the optimization.

[x,fval,exitflag,output,jacobian] = fsolve(...) returns the Jacobian of fun at the solution x.

Example 5.2. This example solves the system of two equations and two unknowns:

Rewrite the equations in the form F(x) = 0:

Start your search for a solution at x0 = [-5 -5]. First, write a file that computes F, the values of the equations at x.

function F = myfun(x)

F = [2*x(1) - x(2) - exp(-x(1));

-x(1) + 2*x(2) - exp(-x(2))];

Save this function file as myfun.m somewhere on your MATLAB path. Next, set up the initial point and options and call fsolve:

>>x0 = [-5; -5]; % Make a starting guess at the solution

>>options=optimset('Display','iter'); % Option to display output

>>[x,fval] = fsolve(@myfun,x0,options) % Call solver

After several iterations, fsolve finds an answer:

Norm of First-order Trust-region

Iteration Func-count f(x) step optimality radius

0 3 23535.6 2.29e+004 1

1 6 6001.72 1 5.75e+003 1

2 9 1573.51 1 1.47e+003 1

3 12 427.226 1 388 1

4 15 119.763 1 107 1

5 18 33.5206 1 30.8 1

6 21 8.35208 1 9.05 1

7 24 1.21394 1 2.26 1

8 27 0.016329 0.759511 0.206 2.5

9 30 3.51575e-006 0.111927 0.00294 2.5

10 33 1.64763e-013 0.00169132 6.36e-007 2.5

Equation solved. fsolve completed because the vector of function values is near zero as measured by the default value of the function tolerance, and the problem appears regular as measured by the gradient.

x =

0.5671

0.5671

fval =

1.0e-006 *

-0.4059

-0.4059

Example 5.3. This example solves the system of the equations:

Write a file that computes F and Jacobian, the values of the equations at x.

function [y,j]=funscj(x)

y=[x(1)+x(2)-sin(pi*x(1));x(1)-x(2)-cos(pi*x(2))];

if nargout>1

j=[1-pi*cos(pi*x(1)) 1;1 -1+pi*sin(pi*x(2))];

end;

A call to the function fsolve:

>>x0 = [1;0.5]; % Make a starting guess at the solution

>>options=optimset('Jacobian','on'); %

>>[x,ff,e_flag,inform]=fsolve(@funscj, x0,options)

x = 0.5000 0.5000

ff = 1.0e-008 *

0.1653

0.0000

e_flag = 1

inform = iterations: 6

funcCount: 7

algorithm: 'trust-region dogleg'

firstorderopt: 1.6525e-009

message: 'Optimization terminated: first-order optimality is less than options.TolFun.'