ch-03-ex-03-solution.cpp
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ch-03-ex-03-solution.cpp
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			/*
			* chapter-03-exercise-03.cpp
			*
			* Implements the backward Euler method to solve:
			* dy / dx = -y, y_0 = 1, 0 <= x <= 1
			*
			* The discretization for N grid points is:
			* - step size: h = 1 / (N - 1)
			* - grid points: x_n = n * h
			* - solution: y_(n) - y_(n-1) / h = - y_n ; n = 1,2,...N-1
			*
			*
			* Created on: Sep 2, 2012
			* Author: Radu Popescu <radu.popescu@epfl.ch>
			*/

			#include <iostream>
			#include <fstream>
			#include <cassert>

			int main(int argc, char* argv[])
			{
			// Read from the command line the number of grid points to use
			int N;
			std::cout << "Please input the number of grid points. N = ";
			std::cin >> N;
			assert(N > 1);

			// Calculate step size h
			double h = 1.0 / (N - 1);
			std::cout << "Computed step size. h = " << h << std::endl;

			// Start computing the grid points and the solution in these points
			// At each step, we output the solution to disk
			std::ofstream output_file("xy.dat");
			if (!output_file.is_open()) {
			std::cout << "Error opening output file." << std::endl;
			return 1;
			}
			output_file.setf(std::ios::scientific);
			output_file.setf(std::ios::showpos);
			output_file.precision(9);

			// x_(n-1) = x_0 = 0
			double x_prev = 0;
			// x_(n) = x_1 = h;
			double x_next = h;
			// y_(n-1) = y_0 = 1;
			double y_prev = 1;
			double y_next;

			output_file << x_prev << " " << y_prev << std::endl;

			while (x_next <= 1.0) {
			y_next = y_prev / (h + 1.0);
			output_file << x_next << " " << y_next << std::endl;
			y_prev = y_next;
			x_prev = x_next;
			x_next = x_prev + h;
			}

			output_file.close();

			return 0;
			}