/**********
ExplicitDiffusion2: solves the diffusion equation for the test problem 
using the explicit difference method

This uses an entire array for the solution 
- compare Mathematica code uses a few vectors.

[in]: int M : number of time steps
      int N : number of positive and negative space steps

[out] : array of values in file for solution at given time

This uses static dimensionalization to avoid array class issues

Version 1.0, William Shaw, 2006.

*************/
#include <cmath>
#include <iostream>
#include <fstream>

using namespace std;

void ExplicitDiffusion(int N, int M)
{
       int i, j;
       double dt = 0.1/M ;
       double dx = 2.0/N;;
       const double PI = 3.141592653589793;
       double solvals[500][500] = {0.0};  // stores solution values, hard wired!
       double alpha = 0.0; // key diffusion parameter
       
       alpha = dt/(dx*dx);
       
       //output dt
       cout << "dt = " << dt << "\n";
           //output dt
       cout << "dx = " << dx << "\n";
       // output alpha
       cout << "alpha = " << alpha << "\n";
       
       // initialize initital data
       
       for (j = 0; j <= 2*N; j++)
       {
           solvals[0][j] = sin(PI*(j-N)*dx/2);
       }
             
       // initialize boundary conditions
       
       for (i = 1; i< M; i++)
       {
           solvals[i][0] = 0;
           solvals[i][2*N] = 0;
           }
       // Run the explicit algorithm
       
       for (i= 1; i<= M; i++)
       {
           for (j = 1; j<= 2*N-1; j++)
           {solvals[i][j] = (1-2*alpha)*solvals[i-1][j] + alpha*(solvals[i-1][j-1]+solvals[i-1][j+1]);
           }   
       }
       
    ofstream out("fdoutput.txt");
    for (j=0; j<= 2*N; j++)
    {   out.precision(15);
        out << solvals[M][j] << "\n";
        }
       return ;
}
       
       
#include <iostream>

#include <cmath>
using namespace std;
int main()
{
    double result;
    char name[20];
    int N;
    int M;
    cout << "Explicit Diffusion Example \n ";
    cout << "Enter number of time steps (M) \n ";
    cin >> M;
    cout << M  << " time steps \n" ;
    cout << "Enter space step parameter (N) \n ";
    cin >> N;
    cout << 2*N  << " space steps \n" ;
    cout << " Result being written to fdoutput.txt \n " << endl;
    ExplicitDiffusion(N,M);
    cout << "Hit any key+<RET> to finish \n ";
    cin >> name;
 
    return(0);
}