fsolve

find a zero of a system of n nonlinear functions

Calling Sequence

[x [,v [,info]]]=fsolve(x0,fct [,fjac] [,tol])

Arguments

:x0 real vector (initial value of function argument). : :fct external (i.e function or list or string). : :fjac external (i.e function or list or string). : :tol real scalar. precision tolerance: termination occurs when the

algorithm estimates that the relative error between x and the solution is at most tol. ( tol=1.d-10 is the default value).

: :x : real vector (final value of function argument, estimated zero). : :v : real vector (value of function at x). : :info termination indicator

:0 improper input parameters. : :1 algorithm estimates that the relative error between x and the

solution is at most tol.

: :2 number of calls to fcn reached : :3 tol is too small. No further improvement in the approximate

solution x is possible.

: :4 iteration is not making good progress. :

:

Description

find a zero of a system of n nonlinear functions in n variables by a modification of the powell hybrid method. Jacobian may be provided.

0 = fct(x) w.r.t x.

fct is an “external”. This external returns v=fct(x) given x.

The simplest calling sequence for fct is:

[v]=fct(x).

If fct is a character string, it refers to a C or Fortran routine which must be linked to Scilab. Fortran calling sequence must be

fct(n,x,v,iflag)
integer n,iflag
`double`_ precision x(n),v(n)

and C Calling sequence must be

fct(`int`_ *n, `double`_ x[],`double`_ v[],`int`_ *iflag)

Incremental link is possible (help link).

jac is an “external”. This external returns v=d(fct)/dx (x) given x.

The simplest calling sequence for jac is:

[v]=jac(x).

If jac is a character string, it refers to a to a C or Fortran routine which must be linked to Scilab calling sequences are the same as those for fct. Note however that v must be a nxn array.

Examples

// A simple example with fsolve
a=[1,7;2,8];
b=[10;11];

function y=fsol1(x)
  y=a*x+b
endfunction
function y=fsolj1(x)
  y=a
endfunction

[xres]=fsolve([100;100],fsol1);
a*xres+b

[xres]=fsolve([100;100],fsol1,fsolj1);
a*xres+b

// See SCI/modules/optimization/sci_gateway/fortran/Ex-fsolve.f
[xres]=fsolve([100;100],'fsol1','fsolj1',1.e-7);
a*xres+b

For some starting points and some equations system, the fsolve method can fail. The fsolve method is a local search method. So, to have a good chance to find a solution to your equations system, you must ship, a good starting point to fsolve.

Here is an example on which fsolve can fail:

// Another example with fsolve
function F=feuler(x, r)
  F=x-r-dt*(x^2-x^3);
endfunction
function J=dFdx(x)  //definition de la derivee de F
   J=1-dt*(2*x-3*x^2);
endfunction

r = 0.04257794928862307 ;
dt = 10;

[x,v,info]=fsolve(r,`list`_(feuler,r),dFdx); // fsolve do not find the solution
`disp`_(v); // The residual
`disp`_(info); // The termination indicator

[x,v,info]=fsolve(1,`list`_(feuler,r),dFdx); // fsolve find the solution
`disp`_(v); // The residual
`disp`_(info); // The termination indicator

`clf`_();
x=`linspace`_(0,1,1000);
`plot`_(x,feuler(x))
a=`gca`_();
a.grid=[5 5];

So, each time you use fsolve, be sure to check the termination indicator and the residual value to see if fsolve has converged.

See Also

• external Scilab Object, external function or routine
• qpsolve linear quadratic programming solver
• optim non-linear optimization routine