function [ ibc, tol, ul, ur, xl, xn, xr ] = init ( n )

%% INIT initializes some parameters that define the problem.
%
%  Modified:
%
%    06 November 2006
%
%  Parameters:
%
%    Input, integer N
%    The number of subintervals into which the interval
%    [XL,XR] is broken.
%
%    Output, integer IBC.
%    IBC declares what the boundary conditions are.
%    1, at the left endpoint, U has the value UL,
%       at the right endpoint, U' has the value UR.
%    2, at the left endpoint, U' has the value UL,
%       at the right endpoint, U has the value UR.
%    3, at the left endpoint, U has the value UL,
%       and at the right endpoint, U has the value UR.
%    4, at the left endpoint, U' has the value UL,
%       at the right endpoint U' has the value UR.
%
%    Output, real TOL.
%    A tolerance that is used to determine whether the estimated
%    error in an interval is so large that it should be subdivided
%    and the problem solved again.
%
%    Output, real UL.
%    If IBC is 1 or 3, UL is the value that U is required
%    to have at X = XL.
%    If IBC is 2 or 4, UL is the value that U' is required
%    to have at X = XL.
%
%    Output, real UR.
%    If IBC is 2 or 3, UR is the value that U is required
%    to have at X = XR.
%    If IBC is 1 or 4, UR is the value that U' is required
%    to have at X = XR.
%
%    Output, real XL.
%    XL is the left endpoint of the interval over which the
%    differential equation is being solved.
%
%    Output, real XN(0:N).
%    XN(I) is the location of the I-th node.  XN(0) is XL,
%    and XN(N) is XR.
%
%    Output, real XR.
%    XR is the right endpoint of the interval over which the
%    differential equation is being solved.
%
  tol = 0.01;
%
%  Find out which problem we're working on.
%
  problem = get_problem ( 'DUMMY' );
%
%  Set the boundary conditions for the problem, and
%  print out its title.
%
  if ( problem == 1 )

    ibc = 3;
    ul = 0.0;
    ur = 1.0;
    xl = 0.0;
    xr = 1.0;
    fprintf ( 1, '\n' );
    fprintf ( 1, 'Exact solution is U = X\n' );

  elseif ( problem == 2 )

    ibc = 3;
    ul = 0.0;
    ur = 1.0;
    xl = 0.0;
    xr = 1.0;
    fprintf ( 1, '\n' );
    fprintf ( 1, 'Exact solution is U = X*X\n' );

  elseif ( problem == 3 )

    ibc = 3;
    ul = 0.0;
    ur = 1.0;
    xl = 0.0;
    xr = 1.0;
    fprintf ( 1, '\n' );
    fprintf ( 1, 'Exact solution is U = SIN(PI*X/2)\n' );

  elseif ( problem == 4 )

    ibc = 3;
    ul = 1.0;
    ur = 0.0;
    xl = 0.0;
    xr = 1.0;
    fprintf ( 1, '\n' );
    fprintf ( 1, 'Exact solution is U = COS(PI*X/2)\n' );

  elseif ( problem == 5 )

    ibc = 3;
    beta = get_beta ( 'DUMMY' );
    ul = 0.0;
    ur = 1.0 / ( ( beta + 2.0 ) * ( beta + 1.0 ) );
    xl = 0.0;
    xr = 1.0;
    fprintf ( 1, '\n' );
    fprintf ( 1, 'Rheinboldt problem\n' );

  elseif ( problem == 6 )

    ibc = 3;
    alpha = get_alpha ( 'DUMMY' );
    xl = 0.0;
    xr = 1.0;
    ul = uexact ( xl );
    ur = uexact ( xr );
    fprintf ( 1, '\n' );
    fprintf ( 1, 'Arctangent problem\n' );

  end
%
%  The nodes are defined here, and not in the geometry routine.
%  This is because each new iteration chooses the location
%  of the new nodes in a special way.
%
  for i = 0 : n
    xn(i+1) = ( ( n - i ) * xl   ...
              + (     i ) * xr ) ...
              / ( n     );
  end

  fprintf ( 1, 'The equation is to be solved for\n' );
  fprintf ( 1, 'X greater than %f\n', xl );
  fprintf ( 1, ' and less than %f\n', xr );
  fprintf ( 1, '\n' );
  fprintf ( 1, 'The boundary conditions are:\n' );
  fprintf ( 1, '\n' );

  if ( ibc == 1 | ibc == 3 )
    fprintf ( 1, '  At X = XL, U = %f\n', ul );
  else
    fprintf ( 1, '  At X = XL, U'' = %f\n', ul );
  end

  if ( ibc == 2 | ibc == 3 )
    fprintf ( 1, '  At X = XR, U = %f\n', ur );
  else
    fprintf ( 1, '  At X = XR, U'' %f\n= ', ur );
  end