view doc/interpreter/quad.txi @ 4167:aae05d51353c

[project @ 2002-11-12 02:52:50 by jwe]
author jwe
date Tue, 12 Nov 2002 02:52:51 +0000
parents a4cd1e9d9962
children 6ab0a8767780
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@c Copyright (C) 1996, 1997 John W. Eaton
@c This is part of the Octave manual.
@c For copying conditions, see the file gpl.texi.

@node Quadrature
@chapter Quadrature

@menu
* Functions of One Variable::   
* Orthogonal Collocation::      
@end menu

@node Functions of One Variable
@section Functions of One Variable

@DOCSTRING(quad)

@DOCSTRING(quad_options)

Here is an example of using @code{quad} to integrate the function
@iftex
@tex
$$
 f(x) = x \sin (1/x) \sqrt {|1 - x|}
$$
from $x = 0$ to $x = 3$.
@end tex
@end iftex
@ifinfo

@example
  @var{f}(@var{x}) = @var{x} * sin (1/@var{x}) * sqrt (abs (1 - @var{x}))
@end example

@noindent
from @var{x} = 0 to @var{x} = 3.
@end ifinfo

This is a fairly difficult integration (plot the function over the range
of integration to see why).

The first step is to define the function:

@example
@group
function y = f (x)
  y = x .* sin (1 ./ x) .* sqrt (abs (1 - x));
endfunction
@end group
@end example

Note the use of the `dot' forms of the operators.  This is not necessary
for the call to @code{quad}, but it makes it much easier to generate a
set of points for plotting (because it makes it possible to call the
function with a vector argument to produce a vector result).

Then we simply call quad:

@example
@group
[v, ier, nfun, err] = quad ("f", 0, 3)
     @result{} 1.9819
     @result{} 1
     @result{} 5061
     @result{} 1.1522e-07
@end group
@end example

Although @code{quad} returns a nonzero value for @var{ier}, the result
is reasonably accurate (to see why, examine what happens to the result
if you move the lower bound to 0.1, then 0.01, then 0.001, etc.).

@node Orthogonal Collocation
@section Orthogonal Collocation

@DOCSTRING(colloc)

Here is an example of using @code{colloc} to generate weight matrices
for solving the second order differential equation
@iftex
@tex
$u^\prime - \alpha u^{\prime\prime} = 0$ with the boundary conditions
$u(0) = 0$ and $u(1) = 1$.
@end tex
@end iftex
@ifinfo
@var{u}' - @var{alpha} * @var{u}'' = 0 with the boundary conditions
@var{u}(0) = 0 and @var{u}(1) = 1.
@end ifinfo

First, we can generate the weight matrices for @var{n} points (including
the endpoints of the interval), and incorporate the boundary conditions
in the right hand side (for a specific value of
@iftex
@tex
$\alpha$).
@end tex
@end iftex
@ifinfo
@var{alpha}).
@end ifinfo

@example
@group
n = 7;
alpha = 0.1;
[r, a, b] = colloc (n-2, "left", "right");
at = a(2:n-1,2:n-1);
bt = b(2:n-1,2:n-1);
rhs = alpha * b(2:n-1,n) - a(2:n-1,n);
@end group
@end example

Then the solution at the roots @var{r} is

@example
u = [ 0; (at - alpha * bt) \ rhs; 1]
     @result{} [ 0.00; 0.004; 0.01 0.00; 0.12; 0.62; 1.00 ]
@end example