# include <stdbool.h>
# include <stdio.h>
# include <stdlib.h>

# include "bdf2.h"
# include "fsolve.h"

/******************************************************************************/

void bdf2 ( void dydt ( double x, double y[], double yp[] ), double tspan[2], 
  double y0[], int n, int m, double t[], double y[] )

/******************************************************************************/
/*
  Purpose:

    bdf2() approximates the solution to an ODE using the BDF2 method.

  Licensing:

    This code is distributed under the MIT license.

  Modified:

    01 December 2023

  Author:

    John Burkardt

  Input:

    dydt: a function that evaluates the right hand side of the ODE.

    double tspan[2]: contains the initial and final times.

    double y0[m]: a column vector containing the initial condition.

    int n: the number of steps to take.

    int m: the number of variables.

  Output:

    double t[n+1], y[m*(n+1)]: the times and solution values.
*/
{
  double *dely;
  double dt;
  double *fn;
  int i;
  int info;
  int j;
  double t1;
  double t2;
  double t3;
  double tol;
  double *y1;
  double *y2;
  double *y3;

  dely = ( double * ) malloc ( m * sizeof ( double ) );
  fn = ( double * ) malloc ( m * sizeof ( double ) );
  y1 = ( double * ) malloc ( m * sizeof ( double ) );
  y2 = ( double * ) malloc ( m * sizeof ( double ) );
  y3 = ( double * ) malloc ( m * sizeof ( double ) );

  dt = ( tspan[1] - tspan[0] ) / ( double ) ( n );
  tol = 1.0E-05;

  for ( i = 0; i <= n; i++ )
  {
    if ( i == 0 )
    {
      t[i] = tspan[0];
      for ( j = 0; j < m; j++ )
      {
        y[i*m+j] = y0[j];
      }
    }
    else if ( i == 1 )
    {
      t1 = t[i-1];
      for ( j = 0; j < m; j++ )
      {
        y1[j] = y[(i-1)*m+j];
      }

      t2 = t1 + dt;
      dydt ( t1, y1, dely );
      for ( j = 0; j < m; j++ )
      {
        y2[j] = y1[j] + dt * dely[j];
      }
      info = fsolve_be ( dydt, m, t1, y1, t2, y2, fn, tol );
      if ( info != 1 )
      {
        printf ( "\n" );
        printf ( "bdf2(): Fatal error!\n" );
        printf ( "  info = %d\n", info );
        exit ( 1 );
      }

      t[i] = t2;
      for ( j = 0; j < m; j++ )
      {
        y[i*m+j] = y2[j];
      }
    }
    else
    {
      t1 = t[i-2];
      for ( j = 0; j < m; j++ )
      {
        y1[j] = y[(i-2)*m+j];
      }
      t2 = t1 + dt;
      for ( j = 0; j < m; j++ )
      {
        y2[j] = y[(i-1)*m+j];
      }
      t3 = t2 + dt;

      dydt ( t2, y2, dely );
      for ( j = 0; j < m; j++ )
      {
        y3[j] = y2[j] + dt * dely[j];
      }

      info = fsolve_bdf2 ( dydt, m, t1, y1, t2, y2, t3, y3, fn, tol );

      if ( info != 1 )
      {
        printf ( "\n" );
        printf ( "bdf2(): Fatal error!\n" );
        printf ( "  info = %d\n", info );
        exit ( 1 );
      }

      t[i] = t3;
      for ( j = 0; j < m; j++ )
      {
        y[i*m+j] = y3[j];
      }
    }
  }

  free ( dely );
  free ( fn );
  free ( y1 );
  free ( y2 );
  free ( y3 );

  return;
}