function output ( u, ibc, indx, nsub, nu, ul, ur, xn )

%% OUTPUT prints out the computed solution at the nodes.
%
%  Modified:
%
%    01 November 2006
%
%  Parameters:
%
%    Input, real U(NU), the solution of the linear equations.
%
%    Input, 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.
%
%    Input, integer INDX(1:N+1).
%    For a node I, INDX(I) is the index of the unknown
%    associated with node I.
%    If INDX(I) is equal to -1, then no unknown is associated
%    with the node, because a boundary condition fixing the
%    value of U has been applied at the node instead.
%    Unknowns are numbered beginning with 1.
%    If IBC is 2 or 4, then there is an unknown value of U
%    at node 0, which will be unknown number 1.  Otherwise,
%    unknown number 1 will be associated with node 1.
%    If IBC is 1 or 4, then there is an unknown value of U
%    at node N, which will be unknown N or N+1,
%    depending on whether there was an unknown at node 0.
%
%    integer NSUB.
%    The number of subintervals into which the interval
%    [XL,XR] is broken.
%
%    Input, integer NU.
%    NU is the number of unknowns in the linear system.
%    Depending on the value of IBC, there will be N-1,
%    N, or N+1 unknown values, which are the coefficients
%    of basis functions.
%
%    Input, 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.
%
%    Input, 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.
%
%    Input, real XN(1:N+1).
%    XN(I) is the location of the I-th node.  XN(1) is XL,
%    and XN(N+1) is XR.
%
  fprintf ( 1, '\n' );
  fprintf ( 1, 'Computed solution:\n' );
  fprintf ( 1, '\n' );
  fprintf ( 1, '    Node        X(I)          U(X(I))\n' );
  fprintf ( 1, '\n' );

  for i = 0 : nsub

    if ( i == 0 )
      if ( ibc == 1 | ibc == 3 )
        value = ul;
      else
        value = u(indx(i+1));
      end
    elseif ( i == nsub )
      if ( ibc == 2 | ibc == 3 )
        value = ur;
      else
        value = u(indx(i+1));
      end
    else
      value = u(indx(i+1));
    end

    fprintf ( 1, '  %6d  %12f  %12f\n', i, xn(i+1), value );

  end