function [ ibc, imax, nprint, nquad, problem, ul, ur, xl, xr ] = init  ( dummy )

%% INIT initializes variables that define the problem.
%
%  Modified:
%
%    01 November 2006
%
%  Parameters:
%
%    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, integer IMAX.
%    The number of Newton iterations to carry out.
%
%    Output, integer NPRINT.
%    The number of points at which the computed solution
%    should be printed out when compared to the exact solution.
%
%    Output, integer NQUAD.
%    The number of quadrature points used in a subinterval.
%    This code uses NQUAD = 1.
%
%    Output, integer PROBLEM, indicates which problem to be solved.
%    * 1, u = x, p=1, q=0, f=x, u(0)=0, u'(1)=1.
%    * 2, u = 2*(1-cos(0.5*pi*x))/pi, p=1, q=0,
%    f = -0.5*pi*cos(0.5*pi*x) + 2*sin(0.5*pi*x)*(1-cos(0.5*pi*x)/pi
%    u(0) = 0, u'(1)=1.
%
%    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 XR.
%    XR is the right endpoint of the interval over which the
%    differential equation is being solved.
%
  ibc = 1;
  imax = 20;
  nprint = 9;
  nquad = 1;
  problem = 1;
  ul = 0.0;
  ur = 1.0;
  xl = 0.0;
  xr = 1.0;
%
%  Print out the values that have been set.
%
  fprintf ( 1, '\n' );
  fprintf ( 1, 'The equation is to be solved for\n' );
  fprintf ( 1, 'X greater than XL  =  ', xl );
  fprintf ( 1, ' and less than XR = ', 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

  if ( problem == 1 )
    fprintf ( 1, '\n' );
    fprintf ( 1, '  This is test problem #1:\n' );
    fprintf ( 1, '\n' );
    fprintf ( 1, '  P(X) = 1, Q(X) = 0, F(X) = X.\n' );
    fprintf ( 1, '  Boundary conditions: U(0) = 0, U''(1) = 1.\n' );
    fprintf ( 1, '\n' );
    fprintf ( 1, '  The exact solution is U(X) = X\n' );
  elseif ( problem == 2 )
    fprintf ( 1, '\n' );
    fprintf ( 1, '  This is test problem #2:\n' );
    fprintf ( 1, '\n' );
    fprintf ( 1, '  P(X) = 1, Q(X) = 0, \n' );
    fprintf ( 1, '  F(X) = -0.5*pi*cos(0.5*pi*X)\n' );
    fprintf ( 1, '        + 2*sin(0.5*pi*X)*(1-cos(0.5*pi*X)/pi.\n' );
    fprintf ( 1, '  Boundary conditions: U(0) = 0, U''(1) = 1.\n' );
    fprintf ( 1, '\n' );
    fprintf ( 1, '  The exact solution is U(X) = 2*(1-cos(pi*x/2))/pi\n' );
  end

  fprintf ( 1, '\n' );
  fprintf ( 1, 'Number of quadrature points per element is %d\n', nquad );
  fprintf ( 1, 'Number of iterations is %d\n', imax );