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

# include "fsolve.h"
# include "rk1_implicit.h"

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

void rk1_implicit ( void dydt ( double t, double y[], double dy[] ), 
  double tspan[2], double y0[], int n, int m, double t[], double y[] )

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

    rk1_implicit() uses Runge-Kutta order 1 implicit method + fsolve_be() to solve an ODE.

  Licensing:

    This code is distributed under the MIT license.

  Modified:

    13 November 2024

  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 *fn;
  int i;
  int info;
  int j;
  double tn;
  double to;
  double tol;
  double *yn;
  double *yo;

  fn = ( double * ) malloc ( m * sizeof ( double ) );
  yo = ( double * ) malloc ( m * sizeof ( double ) );
  yn = ( 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
    {
      to = t[i-1];
      for ( j = 0; j < m; j++ )
      {
        yo[j] = y[(i-1)*m+j];
      }

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

      info = fsolve_be ( dydt, m, to, yo, tn, yn, fn, tol );

      if ( info != 1 )
      {
        printf ( "\n" );
        printf ( "rk1_implicit(): Fatal error!\n" );
        printf ( "  info = %d\n", info );
        exit ( 1 );
      }

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

  free ( fn );
  free ( yn );
  free ( yo );

  return;
}