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

# include "fsolve_be.h"
# include "midpoint.h"

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

void midpoint ( void dydt ( double t, double y[], double dy[] ), 
  double tspan[2], double y0[], int n, int m, double t[], double y[] )

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

    midpoint() uses the midpoint method + fsolve_be() to solve an ODE.

  Licensing:

    This code is distributed under the MIT license.

  Modified:

    09 November 2023

  Author:

    John Burkardt

  Input:

    dydt: a function that evaluates the right hand side of the ODE,
    of the form
      void dydt ( double t, double y[], double dy[] )

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

    double y0[m]: 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 dt;
  double *fm;
  int i;
  int info;
  int j;
  double theta;
  double tm;
  double to;
  double tol;
  double *ym;
  double *yo;

  fm = ( double * ) malloc ( m * sizeof ( double ) );
  yo = ( double * ) malloc ( m * sizeof ( double ) );
  ym = ( double * ) malloc ( m * sizeof ( double ) );

  dt = ( tspan[1] - tspan[0] ) / ( double ) ( n );

  theta = 0.5;
  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
    {
      to = t[i-1];
      for ( j = 0; j < m; j++ )
      {
        yo[j] = y[(i-1)*m+j];
      }

      tm = t[i-1] + theta * dt;
      for ( j = 0; j < m; j++ )
      {
        ym[j] = y[(i-1)*m+j];
      }

      info = fsolve_be ( dydt, m, to, yo, tm, ym, fm, tol );

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

      t[i] = t[i-1] + dt;
      for ( j = 0; j < m; j++ )
      {
        y[i*m+j] = (       1.0 / theta ) * ym[j] 
                 + ( 1.0 - 1.0 / theta ) * y[(i-1)*m+j];
      }
    }
  }

  free ( fm );
  free ( ym );
  free ( yo );

  return;
}