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

using namespace std;

# include "fsolve_be.hpp"
# include "midpoint.hpp"

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

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

//****************************************************************************80
//
//  Purpose:
//
//    midpoint() uses the midpoint method + fsolve_be() to solve an ODE.
//
//  Licensing:
//
//    This code is distributed under the MIT license.
//
//  Modified:
//
//    10 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 = new double[m];
  ym = new double[m];
  yo = new double[m];

  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 )
      {
        cout << "\n";
        cout << "midpoint(): Fatal error!\n";
        cout << "  info = " << info << "\n";
        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];
      }
    }
  }

  delete [] fm;
  delete [] ym;
  delete [] yo;

  return;
}