# include <cstdlib>
# include <iostream>
# include <iomanip>
# include <ctime>

using namespace std;

int main ( );
void assemble ( double adiag[], double aleft[], double arite[], double f[], 
  double h[], int indx[], int nl, int node[], int nu, int nquad, int nsub, 
  double ul, double ur, double xn[], double xquad[] );
double ff ( double x );
void geometry ( double h[], int ibc, int indx[], int nl, int node[], int nsub, 
  int *nu, double xl, double xn[], double xquad[], double xr );
void init ( int *ibc, int *nquad, double *ul, double *ur, double *xl, 
  double *xr );
void output ( double f[], int ibc, int indx[], int nsub, int nu, double ul, 
  double ur, double xn[] );
void phi ( int il, double x, double *phii, double *phiix, double xleft, 
  double xrite );
double pp ( double x );
void prsys ( double adiag[], double aleft[], double arite[], double f[], 
  int nu );
double qq ( double x );
void solve ( double adiag[], double aleft[], double arite[], double f[], 
  int nu );
void timestamp ( );

//****************************************************************************80

int main ( )

//****************************************************************************80
//
//  Purpose:
//
//    MAIN is the main program for FEM1D.
//
//  Discussion:
//
//    FEM1D solves a one dimensional ODE using the finite element method.
//
//    The differential equation solved is
//
//      - d/dX (P dU/dX) + Q U  =  F
//
//    The finite-element method uses piecewise linear basis functions.
//
//    Here U is an unknown scalar function of X defined on the
//    interval [XL,XR], and P, Q and F are given functions of X.
//
//    The values of U or U' at XL and XR are also specified.
//
//
//    The interval [XL,XR] is "meshed" with NSUB+1 points,
//
//    XN(0) = XL, XN(1)=XL+H, XN(2)=XL+2*H, ..., XN(NSUB)=XR.
//
//    This creates NSUB subintervals, with interval number 1
//    having endpoints XN(0) and XN(1), and so on up to interval
//    NSUB, which has endpoints XN(NSUB-1) and XN(NSUB).
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameters:
//
//    double ADIAG(NU), the "diagonal" coefficients.  That is, ADIAG(I) is
//    the coefficient of the I-th unknown in the I-th equation.
//
//    double ALEFT(NU), the "left hand" coefficients.  That is, ALEFT(I) 
//    is the coefficient of the (I-1)-th unknown in the I-th equation.
//    There is no value in ALEFT(1), since the first equation
//    does not refer to a "0-th" unknown.
//
//    double ARITE(NU).
//    ARITE(I) is the "right hand" coefficient of the I-th
//    equation in the linear system.  ARITE(I) is the coefficient
//    of the (I+1)-th unknown in the I-th equation.  There is
//    no value in ARITE(NU) because the NU-th equation does not
//    refer to an "NU+1"-th unknown.
//
//    double F(NU).
//    ASSEMBLE stores into F the right hand side of the linear
//    equations.
//    SOLVE replaces those values of F by the solution of the
//    linear equations.
//
//    double H(NSUB)
//    H(I) is the length of subinterval I.  This code uses
//    equal spacing for all the subintervals.
//
//    int 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.
//
//    int INDX[NSUB+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 NSUB, which will be unknown NSUB or NSUB+1,
//    depending on whether there was an unknown at node 0.
//
//    int NL.
//    The number of basis functions used in a single
//    subinterval.  (NL-1) is the degree of the polynomials
//    used.  For this code, NL is fixed at 2, meaning that
//    piecewise linear functions are used as the basis.
//
//    int NODE[NL*NSUB].
//    For each subinterval I:
//    NODE[0+I*2] is the number of the left node, and
//    NODE[1+I*2] is the number of the right node.
//
//    int NQUAD.
//    The number of quadrature points used in a subinterval.
//    This code uses NQUAD = 1.
//
//    int NSUB.
//    The number of subintervals into which the interval [XL,XR] is broken.
//
//    int NU.
//    NU is the number of unknowns in the linear system.
//    Depending on the value of IBC, there will be NSUB-1,
//    NSUB, or NSUB+1 unknown values, which are the coefficients
//    of basis functions.
//
//    double 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.
//
//    double 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.
//
//    double XL.
//    XL is the left endpoint of the interval over which the
//    differential equation is being solved.
//
//    double XN(0:NSUB).
//    XN(I) is the location of the I-th node.  XN(0) is XL,
//    and XN(NSUB) is XR.
//
//    double XQUAD(NSUB)
//    XQUAD(I) is the location of the single quadrature point
//    in interval I.
//
//    double XR.
//    XR is the right endpoint of the interval over which the
//    differential equation is being solved.
//
{
# define NSUB 5
# define NL 2

  double adiag[NSUB+1];
  double aleft[NSUB+1];
  double arite[NSUB+1];
  double f[NSUB+1];
  double h[NSUB];
  int ibc;
  int indx[NSUB+1];
  int node[NL*NSUB];
  int nquad;
  int nu;
  double ul;
  double ur;
  double xl;
  double xn[NSUB+1];
  double xquad[NSUB];
  double xr;

  timestamp ( );

  cout << "\n";
  cout << "FEM1D\n";
  cout << "  C++ version\n";
  cout << "\n";
  cout << "  Solve the two-point boundary value problem\n";
  cout << "\n";
  cout << "  - d/dX (P dU/dX) + Q U  =  F\n";
  cout << "\n";
  cout << "  on the interval [XL,XR], specifying\n";
  cout << "  the value of U or U' at each end.\n";
  cout << "\n";
  cout << "  The interval [XL,XR] is broken into NSUB = " 
       << NSUB << " subintervals\n";
  cout << "  Number of basis functions per element is NL = " 
       << NL << "\n";
//
//  Initialize the data that defines the problem.
//
  init ( &ibc, &nquad, &ul, &ur, &xl, &xr );
//
//  Compute the quantities which define the geometry of the
//  problem.
//
  geometry ( h, ibc, indx, NL, node, NSUB, &nu, xl, xn, xquad, xr );
//
//  Assemble the linear system.
//
  assemble ( adiag, aleft, arite, f, h, indx, NL, node, nu, nquad, 
    NSUB, ul, ur, xn, xquad );
//
//  Print out the linear system.
//
  prsys ( adiag, aleft, arite, f, nu );
//
//  Solve the linear system.
//
  solve ( adiag, aleft, arite, f, nu );
//
//  Print out the solution.
//
  output ( f, ibc, indx, NSUB, nu, ul, ur, xn );
//
//  Terminate.
//
  cout << "\n";
  cout << "FEM1D:\n";
  cout << "  Normal end of execution.\n";

  cout << "\n";
  timestamp ( );

  return 0;
# undef NL
# undef NSUB
}
//****************************************************************************80

void assemble ( double adiag[], double aleft[], double arite[], double f[], 
  double h[], int indx[], int nl, int node[], int nu, int nquad, int nsub, 
  double ul, double ur, double xn[], double xquad[] )

//****************************************************************************80
//
//  Purpose:
//
//    ASSEMBLE assembles the matrix and right-hand-side of the linear system.
//
//  Discussion:
//
//    The linear system has the form:
//
//      K * C = F
//
//    that is to be solved for the coefficients C.
//
//    Numerical integration is used to compute the entries of K and F.
//
//    Note that a 1 point quadrature rule, which is sometimes used to
//    assemble the matrix and right hand side, is just barely accurate
//    enough for simple problems.  If you want better results, you
//    should use a quadrature rule that is more accurate.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt.
//
//  Parameters:
//
//    Output, double ADIAG(NU), the "diagonal" coefficients.  That is, 
//    ADIAG(I) is the coefficient of the I-th unknown in the I-th equation.
//
//    Output, double ALEFT(NU), the "left hand" coefficients.  That is, 
//    ALEFT(I) is the coefficient of the (I-1)-th unknown in the I-th equation.
//    There is no value in ALEFT(1), since the first equation
//    does not refer to a "0-th" unknown.
//
//    Output, double ARITE(NU).
//    ARITE(I) is the "right hand" coefficient of the I-th
//    equation in the linear system.  ARITE(I) is the coefficient
//    of the (I+1)-th unknown in the I-th equation.  There is
//    no value in ARITE(NU) because the NU-th equation does not
//    refer to an "NU+1"-th unknown.
//
//    Output, double F(NU).
//    ASSEMBLE stores into F the right hand side of the linear
//    equations.
//    SOLVE replaces those values of F by the solution of the
//    linear equations.
//
//    Input, double H(NSUB)
//    H(I) is the length of subinterval I.  This code uses
//    equal spacing for all the subintervals.
//
//    Input, int INDX[NSUB+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 NSUB, which will be unknown NSUB or NSUB+1,
//    depending on whether there was an unknown at node 0.
//
//    Input, int NL.
//    The number of basis functions used in a single
//    subinterval.  (NL-1) is the degree of the polynomials
//    used.  For this code, NL is fixed at 2, meaning that
//    piecewise linear functions are used as the basis.
//
//    Input, int NODE[NL*NSUB].
//    For each subinterval I:
//    NODE[0+I*2] is the number of the left node, and
//    NODE[1+I*2] is the number of the right node.
//
//    Input, int NU.
//    NU is the number of unknowns in the linear system.
//    Depending on the value of IBC, there will be NSUB-1,
//    NSUB, or NSUB+1 unknown values, which are the coefficients
//    of basis functions.
//
//    Input, int NQUAD.
//    The number of quadrature points used in a subinterval.
//    This code uses NQUAD = 1.
//
//    Input, int NSUB.
//    The number of subintervals into which the interval [XL,XR] is broken.
//
//    Input, double 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, double 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, double XL.
//    XL is the left endpoint of the interval over which the
//    differential equation is being solved.
//
//    Input, double XR.
//    XR is the right endpoint of the interval over which the
//    differential equation is being solved.
//
{
  double aij;
  double he;
  int i;
  int ie;
  int ig;
  int il;
  int iq;
  int iu;
  int jg;
  int jl;
  int ju;
  double phii;
  double phiix;
  double phij;
  double phijx;
  double x;
  double xleft;
  double xquade;
  double xrite;
//
//  Zero out the arrays that hold the coefficients of the matrix
//  and the right hand side.
//
  for ( i = 0; i < nu; i++ )
  {
    f[i] = 0.0;
  }
  for ( i = 0; i < nu; i++ )
  {
    adiag[i] = 0.0;
  }
  for ( i = 0; i < nu; i++ )
  {
    aleft[i] = 0.0;
  }
  for ( i = 0; i < nu; i++ )
  {
    arite[i] = 0.0;
  }
//
//  For interval number IE,
//
  for ( ie = 0; ie < nsub; ie++ )
  {
    he = h[ie];
    xleft = xn[node[0+ie*2]];
    xrite = xn[node[1+ie*2]];
//
//  consider each quadrature point IQ,
//
    for ( iq = 0; iq < nquad; iq++ )
    {
      xquade = xquad[ie];
//
//  and evaluate the integrals associated with the basis functions
//  for the left, and for the right nodes.
//
      for ( il = 1; il <= nl; il++ )
      {
        ig = node[il-1+ie*2];
        iu = indx[ig] - 1;

        if ( 0 <= iu )
        {
          phi ( il, xquade, &phii, &phiix, xleft, xrite );
          f[iu] = f[iu] + he * ff ( xquade ) * phii;
//
//  Take care of boundary nodes at which U' was specified.
//
          if ( ig == 0 )
          {
            x = 0.0;
            f[iu] = f[iu] - pp ( x ) * ul;
          }
          else if ( ig == nsub )
          {
            x = 1.0;
            f[iu] = f[iu] + pp ( x ) * ur;
          }
//
//  Evaluate the integrals that take a product of the basis
//  function times itself, or times the other basis function
//  that is nonzero in this interval.
//
          for ( jl = 1; jl <= nl; jl++ )
          {
            jg = node[jl-1+ie*2];
            ju = indx[jg] - 1;

            phi ( jl, xquade, &phij, &phijx, xleft, xrite );

            aij = he * ( pp ( xquade ) * phiix * phijx 
                       + qq ( xquade ) * phii  * phij   );
//
//  If there is no variable associated with the node, then it's
//  a specified boundary value, so we multiply the coefficient
//  times the specified boundary value and subtract it from the
//  right hand side.
//
            if ( ju < 0 )
            {
              if ( jg == 0 )
              {
                f[iu] = f[iu] - aij * ul;
              }
              else if ( jg == nsub )
              {               
                f[iu] = f[iu] - aij * ur;
              }
            }
//
//  Otherwise, we add the coefficient we've just computed to the
//  diagonal, or left or right entries of row IU of the matrix.
//
            else
            {
              if ( iu == ju )
              {
                adiag[iu] = adiag[iu] + aij;
              }
              else if ( ju < iu )
              {
                aleft[iu] = aleft[iu] + aij;
              }
              else
              {
                arite[iu] = arite[iu] + aij;
              }
            }
          }
        }
      }
    }
  }
  return;
}
//****************************************************************************80

double ff ( double x )

//****************************************************************************80
//
//  Purpose:
//
//    FF evaluates the right hand side function.
//
//  Discussion:
//
//    This routine evaluates the function F(X) in the differential equation.
//
//      -d/dx (p du/dx) + q u  =  f
//
//    at the point X.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    John Burkardt
//
//  Parameters:
//
//    Input, double X, the argument of the function.
//
//    Output, double FF, the value of the function.
//
{
  double value;

  value = 0.0;

  return value;
}
//****************************************************************************80

void geometry ( double h[], int ibc, int indx[], int nl, int node[], int nsub, 
  int *nu, double xl, double xn[], double xquad[], double xr )

//****************************************************************************80
//
//  Purpose: 
//
//    GEOMETRY sets up the geometry for the interval [XL,XR].
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameters:
//
//    Output, double H(NSUB)
//    H(I) is the length of subinterval I.  This code uses
//    equal spacing for all the subintervals.
//
//    Input, int 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, int INDX[NSUB+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 NSUB, which will be unknown NSUB or NSUB+1,
//    depending on whether there was an unknown at node 0.
//
//    Input, int NL.
//    The number of basis functions used in a single
//    subinterval.  (NL-1) is the degree of the polynomials
//    used.  For this code, NL is fixed at 2, meaning that
//    piecewise linear functions are used as the basis.
//
//    Output, int NODE[NL*NSUB].
//    For each subinterval I:
//    NODE[0+I*2] is the number of the left node, and
//    NODE[1+I*2] is the number of the right node.
//
//    Input, int NSUB.
//    The number of subintervals into which the interval [XL,XR] is broken.
//
//    Output, int *NU.
//    NU is the number of unknowns in the linear system.
//    Depending on the value of IBC, there will be NSUB-1,
//    NSUB, or NSUB+1 unknown values, which are the coefficients
//    of basis functions.
//
//    Input, double XL.
//    XL is the left endpoint of the interval over which the
//    differential equation is being solved.
//
//    Output, double XN(0:NSUB).
//    XN(I) is the location of the I-th node.  XN(0) is XL,
//    and XN(NSUB) is XR.
//
//    Output, double XQUAD(NSUB)
//    XQUAD(I) is the location of the single quadrature point
//    in interval I.
//
//    Input, double XR.
//    XR is the right endpoint of the interval over which the
//    differential equation is being solved.
//
{
  int i;
//
//  Set the value of XN, the locations of the nodes.
//
  cout << "\n";
  cout << "  Node      Location\n";
  cout << "\n";
  for ( i = 0; i <= nsub; i++ )
  {
    xn[i]  =  ( ( double ) ( nsub - i ) * xl 
              + ( double )          i   * xr ) 
              / ( double ) ( nsub );
    cout << "  " << setw(8) << i
         << "  " << setw(14) << xn[i] << "\n";
  }
//
//  Set the lengths of each subinterval.
//
  cout << "\n";
  cout << "Subint    Length\n";
  cout << "\n";
  for ( i = 0; i < nsub; i++ )
  {
    h[i] = xn[i+1] - xn[i];
    cout << "  " << setw(8) << i+1
         << "  " << setw(14) << h[i] << "\n";
  }
//
//  Set the quadrature points, each of which is the midpoint
//  of its subinterval.
//
  cout << "\n";
  cout << "Subint    Quadrature point\n";
  cout << "\n";
  for ( i = 0; i < nsub; i++ )
  {
    xquad[i] = 0.5 * ( xn[i] + xn[i+1] );
    cout << "  " << setw(8) << i+1
         << "  " << setw(14) << xquad[i] << "\n";
  }
//
//  Set the value of NODE, which records, for each interval,
//  the node numbers at the left and right.
//
  cout << "\n";
  cout << "Subint  Left Node  Right Node\n";
  cout << "\n";
  for ( i = 0; i < nsub; i++ )
  {
    node[0+i*2] = i;
    node[1+i*2] = i + 1;
    cout << "  " << setw(8) << i+1
         << "  " << setw(8) << node[0+i*2] 
         << "  " << setw(8) << node[1+i*2] << "\n";
  }
//
//  Starting with node 0, see if an unknown is associated with
//  the node.  If so, give it an index.
//
  *nu = 0;
//
//  Handle first node.
//
  i = 0;
  if ( ibc == 1 || ibc == 3 )
  {
    indx[i] = -1;
  }
  else
  {
    *nu = *nu + 1;
    indx[i] = *nu;
  }
//
//  Handle nodes 1 through nsub-1
//
  for ( i = 1; i < nsub; i++ )
  {
    *nu = *nu + 1;
    indx[i] = *nu;
  }
//
//  Handle the last node.
//
  i = nsub;

  if ( ibc == 2 || ibc == 3 )
  {
    indx[i] = -1;
  }
  else
  {
    *nu = *nu + 1;
    indx[i] = *nu;
  }

  cout << "\n";
  cout << "  Number of unknowns NU = " << *nu << "\n";
  cout << "\n";
  cout << "  Node  Unknown\n";
  cout << "\n";
  for ( i = 0; i <= nsub; i++ )
  {
    cout << "  " << setw(8) << i
         << "  " << setw(8) << indx[i] << "\n";
  }

  return;
}
//****************************************************************************80

void init ( int *ibc, int *nquad, double *ul, double *ur, double *xl, 
  double *xr )

//****************************************************************************80
//
//  Purpose: 
//
//    INIT assigns values to variables which define the problem.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameters:
//
//    Output, int *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, int *NQUAD.
//    The number of quadrature points used in a subinterval.
//    This code uses NQUAD = 1.
//
//    Output, double *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, double *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, double *XL.
//    XL is the left endpoint of the interval over which the
//    differential equation is being solved.
//
//    Output, double *XR.
//    XR is the right endpoint of the interval over which the
//    differential equation is being solved.
//
{
//
//  IBC declares what the boundary conditions are.
//
  *ibc = 1;
//
//  NQUAD is the number of quadrature points per subinterval.
//  The program as currently written cannot handle any value for
//  NQUAD except 1//
//
  *nquad = 1;
//
//  Set the values of U or U' at the endpoints.
//
  *ul = 0.0;
  *ur = 1.0;
//
//  Define the location of the endpoints of the interval.
//
  *xl = 0.0;
  *xr = 1.0;
//
//  Print out the values that have been set.
//
  cout << "\n";
  cout << "  The equation is to be solved for\n";
  cout << "  X greater than XL = " << *xl << "\n";
  cout << "  and less than XR = " << *xr << "\n";
  cout << "\n";
  cout << "  The boundary conditions are:\n";
  cout << "\n";

  if ( *ibc == 1 || *ibc == 3 )
  {
    cout << "  At X = XL, U = " << *ul << "\n";
  }
  else
  {
    cout << "  At X = XL, U' = " << *ul << "\n";
  }

  if ( *ibc == 2 || *ibc == 3 )
  {
    cout << "  At X = XR, U = " << *ur << "\n";
  }
  else
  {
    cout << "  At X = XR, U' = " << *ur << "\n";
  }

  cout << "\n";
  cout << "  Number of quadrature points per element is " << *nquad << "\n";

  return;
}
//****************************************************************************80

void output ( double f[], int ibc, int indx[], int nsub, int nu, double ul, 
  double ur, double xn[] )

//****************************************************************************80
//
//  Purpose:
//
//    OUTPUT prints out the computed solution.
//
//  Discussion:
//
//    We simply print out the solution vector F, except that, for
//    certain boundary conditions, we are going to have to get the
//    value of the solution at XL or XR by using the specified
//    boundary value.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameters:
//
//    Input, double F(NU).
//    ASSEMBLE stores into F the right hand side of the linear
//    equations.
//    SOLVE replaces those values of F by the solution of the
//    linear equations.
//
//    Input, int 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, int INDX[NSUB+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 NSUB, which will be unknown NSUB or NSUB+1,
//    depending on whether there was an unknown at node 0.
//
//    Input, int NSUB.
//    The number of subintervals into which the interval [XL,XR] is broken.
//
//    Input, int NU.
//    NU is the number of unknowns in the linear system.
//    Depending on the value of IBC, there will be NSUB-1,
//    NSUB, or NSUB+1 unknown values, which are the coefficients
//    of basis functions.
//
//    Input, double 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, double 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, double XN(0:NSUB).
//    XN(I) is the location of the I-th node.  XN(0) is XL,
//    and XN(NSUB) is XR.
//
{
  int i;
  double u;

  cout << "\n";
  cout << "  Computed solution coefficients:\n";
  cout << "\n";
  cout << "  Node    X(I)        U(X(I))\n";
  cout << "\n";

  for ( i = 0; i <= nsub; i++ )
  {
//
//  If we're at the first node, check the boundary condition.
//
    if ( i == 0 )
    {
      if ( ibc == 1 || ibc == 3 )
      {
        u = ul;
      }
      else
      {
        u = f[indx[i]-1];
      }
    }
//
//  If we're at the last node, check the boundary condition.
//
    else if ( i == nsub )
    {
      if ( ibc == 2 || ibc == 3 )
      {
        u = ur;
      }
      else
      {
        u = f[indx[i]-1];
      }
    }
//
//  Any other node, we're sure the value is stored in F.
//
    else
    {
      u = f[indx[i]-1];
    }

    cout << "  " << setw(8)  << i
         << "  " << setw(8)  << xn[i]
         << "  " << setw(14) << u << "\n";
  }

  return;
}
//****************************************************************************80

void phi ( int il, double x, double *phii, double *phiix, double xleft, 
  double xrite )

//****************************************************************************80
//
//  Purpose:
//
//    PHI evaluates a linear basis function and its derivative.
//
//  Discussion:
//
//    The evaluation is done at a point X in an interval [XLEFT,XRITE].
//
//    In this interval, there are just two nonzero basis functions.
//    The first basis function is a line which is 1 at the left
//    endpoint and 0 at the right.  The second basis function is 0 at
//    the left endpoint and 1 at the right.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameters:
//
//    Input, int IL, the index of the basis function.
//    1, the function which is 1 at XLEFT and 0 at XRITE.
//    2, the function which is 0 at XLEFT and 1 at XRITE.
//
//    Input, double X, the evaluation point.
//
//    Output, double *PHII, *PHIIX, the value of the
//    basis function and its derivative at X.
//
//    Input, double XLEFT, XRITE, the left and right
//    endpoints of the interval.
//
{
  if ( xleft <= x && x <= xrite )
  {
    if ( il == 1 )
    {
      *phii = ( xrite - x ) / ( xrite - xleft );
      *phiix =         -1.0 / ( xrite - xleft );
    }
    else
    {
      *phii = ( x - xleft ) / ( xrite - xleft );
      *phiix = 1.0          / ( xrite - xleft );
    }
  }
//
//  If X is outside of the interval, just set everything to 0.
//
  else
  {
    *phii  = 0.0;
    *phiix = 0.0;
  }

  return;
}
//****************************************************************************80

double pp ( double x )

//****************************************************************************80
//
//  Purpose:
//
//    PP evaluates the function P in the differential equation.
//
//  Discussion:
//
//    The function P appears in the differential equation as;
//
//      - d/dx (p du/dx) + q u  =  f
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    John Burkardt
//
//  Parameters:
//
//    Input, double X, the argument of the function.
//
//    Output, double PP, the value of the function.
//
{
  double value;

  value = 1.0;

  return value;
}
//****************************************************************************80

void prsys ( double adiag[], double aleft[], double arite[], double f[], 
  int nu )

//****************************************************************************80
//
//  Purpose:
//
//    PRSYS prints out the tridiagonal linear system.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameter:
//
//    Input, double ADIAG(NU), the "diagonal" coefficients.  That is, 
//    ADIAG(I) is the coefficient of the I-th unknown in the I-th equation.
//
//    Input, double ALEFT(NU), the "left hand" coefficients.  That is, ALEFT(I) 
//    is the coefficient of the (I-1)-th unknown in the I-th equation.
//    There is no value in ALEFT(1), since the first equation
//    does not refer to a "0-th" unknown.
//
//    Input, double ARITE(NU).
//    ARITE(I) is the "right hand" coefficient of the I-th
//    equation in the linear system.  ARITE(I) is the coefficient
//    of the (I+1)-th unknown in the I-th equation.  There is
//    no value in ARITE(NU) because the NU-th equation does not
//    refer to an "NU+1"-th unknown.
//
//    Input, double F(NU).
//    ASSEMBLE stores into F the right hand side of the linear
//    equations.
//    SOLVE replaces those values of F by the solution of the
//    linear equations.
//
//    Input, int NU.
//    NU is the number of unknowns in the linear system.
//    Depending on the value of IBC, there will be NSUB-1,
//    NSUB, or NSUB+1 unknown values, which are the coefficients
//    of basis functions.
//
{
  int i;

  cout << "\n";
  cout << "Printout of tridiagonal linear system:\n";
  cout << "\n";
  cout << "Equation  ALEFT  ADIAG  ARITE  RHS\n";
  cout << "\n";

  for ( i = 0; i < nu; i++ )
  {
    cout << "  " << setw(8) << i + 1
         << "  " << setw(14) << aleft[i]
         << "  " << setw(14) << adiag[i]
         << "  " << setw(14) << arite[i]
         << "  " << setw(14) << f[i] << "\n";
  }

  return;
}
//****************************************************************************80

double qq ( double x )

//****************************************************************************80
//
//  Purpose: 
//
//    QQ evaluates the function Q in the differential equation.
//
//  Discussion:
//
//    The function Q appears in the differential equation as:
//
//      - d/dx (p du/dx) + q u  =  f
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    John Burkardt
//
//  Parameters:
//
//    Input, double X, the argument of the function.
//
//    Output, double QQ, the value of the function.
//
{
  double value;

  value = 0.0;

  return value;
}
//****************************************************************************80

void solve ( double adiag[], double aleft[], double arite[], double f[], 
  int nu )

//****************************************************************************80
//
//  Purpose: 
//
//    SOLVE solves a tridiagonal matrix system of the form A*x = b.
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    31 October 2006
//
//  Author:
//
//    C++ version by John Burkardt
//
//  Parameters:
//
//    Input/output, double ADIAG(NU), ALEFT(NU), ARITE(NU).
//    On input, ADIAG, ALEFT, and ARITE contain the diagonal,
//    left and right entries of the equations.
//    On output, ADIAG and ARITE have been changed in order
//    to compute the solution.
//    Note that for the first equation, there is no ALEFT
//    coefficient, and for the last, there is no ARITE.
//    So there is no need to store a value in ALEFT(1), nor
//    in ARITE(NU).
//
//    Input/output, double F(NU).
//    On input, F contains the right hand side of the linear
//    system to be solved.
//    On output, F contains the solution of the linear system.
//
//    Input, int NU, the number of equations to be solved.
//
{
  int i;
//
//  Carry out Gauss elimination on the matrix, saving information
//  needed for the backsolve.
//
  arite[0] = arite[0] / adiag[0];

  for ( i = 1; i < nu - 1; i++ )
  {
    adiag[i] = adiag[i] - aleft[i] * arite[i-1];
    arite[i] = arite[i] / adiag[i];
  }
  adiag[nu-1] = adiag[nu-1] - aleft[nu-1] * arite[nu-2];
//
//  Carry out the same elimination steps on F that were done to the
//  matrix.
//
  f[0] = f[0] / adiag[0];
  for ( i = 1; i < nu; i++ )
  {
    f[i] = ( f[i] - aleft[i] * f[i-1] ) / adiag[i];
  }
//
//  And now carry out the steps of "back substitution".
//
  for ( i = nu - 2; 0 <= i; i-- )
  {
    f[i] = f[i] - arite[i] * f[i+1];
  }

  return;
}
//****************************************************************************80

void timestamp ( )

//****************************************************************************80
//
//  Purpose:
//
//    TIMESTAMP prints the current YMDHMS date as a time stamp.
//
//  Example:
//
//    31 May 2001 09:45:54 AM
//
//  Licensing:
//
//    This code is distributed under the MIT license. 
//
//  Modified:
//
//    24 September 2003
//
//  Author:
//
//    John Burkardt
//
//  Parameters:
//
//    None
//
{
# define TIME_SIZE 40

  static char time_buffer[TIME_SIZE];
  const struct tm *tm;
  time_t now;

  now = time ( NULL );
  tm = localtime ( &now );

  strftime ( time_buffer, TIME_SIZE, "%d %B %Y %I:%M:%S %p", tm );

  cout << time_buffer << "\n";

  return;
# undef TIME_SIZE
}