# include <cmath>
# include <iomanip>
# include <iostream>

using namespace std;

# include "fsolve.hpp"
# include "rk1_implicit.hpp"

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

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

//****************************************************************************80
//
//  Purpose:
//
//    rk1_implicit() uses Runge-Kutta order 1 implicit + 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 = new double[m];
  yn = new double[m];
  yo = new double[m];

  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 )
      {
        cout << "\n";
        cout << "rk1_implicit(): Fatal error!\n";
        cout << "  info = " << info << "\n";
        exit ( 1 );
      }

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

  delete [] fn;
  delete [] yn;
  delete [] yo;

  return;
}