TheAlgorithms-C/numerical_methods/ode_midpoint_euler.c

190 lines
5.4 KiB
C

/**
* \file
* \authors [Krishna Vedala](https://github.com/kvedala)
* \brief Solve a multivariable first order [ordinary differential equation
* (ODEs)](https://en.wikipedia.org/wiki/Ordinary_differential_equation) using
* [midpoint Euler
* method](https://en.wikipedia.org/wiki/Midpoint_method)
*
* \details
* The ODE being solved is:
* \f{eqnarray*}{
* \dot{u} &=& v\\
* \dot{v} &=& -\omega^2 u\\
* \omega &=& 1\\
* [x_0, u_0, v_0] &=& [0,1,0]\qquad\ldots\text{(initial values)}
* \f}
* The exact solution for the above problem is:
* \f{eqnarray*}{
* u(x) &=& \cos(x)\\
* v(x) &=& -\sin(x)\\
* \f}
* The computation results are stored to a text file `midpoint_euler.csv` and
* the exact soltuion results in `exact.csv` for comparison. <img
* src="https://raw.githubusercontent.com/TheAlgorithms/C/docs/images/numerical_methods/ode_midpoint_euler.svg"
* alt="Implementation solution"/>
*
* To implement [Van der Pol
* oscillator](https://en.wikipedia.org/wiki/Van_der_Pol_oscillator), change the
* ::problem function to:
* ```cpp
* const double mu = 2.0;
* dy[0] = y[1];
* dy[1] = mu * (1.f - y[0] * y[0]) * y[1] - y[0];
* ```
* \see ode_forward_euler.c, ode_semi_implicit_euler.c
*/
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define order 2 /**< number of dependent variables in ::problem */
/**
* @brief Problem statement for a system with first-order differential
* equations. Updates the system differential variables.
* \note This function can be updated to and ode of any order.
*
* @param[in] x independent variable(s)
* @param[in,out] y dependent variable(s)
* @param[in,out] dy first-derivative of dependent variable(s)
*/
void problem(const double *x, double *y, double *dy)
{
const double omega = 1.F; // some const for the problem
dy[0] = y[1]; // x dot
dy[1] = -omega * omega * y[0]; // y dot
}
/**
* @brief Exact solution of the problem. Used for solution comparison.
*
* @param[in] x independent variable
* @param[in,out] y dependent variable
*/
void exact_solution(const double *x, double *y)
{
y[0] = cos(x[0]);
y[1] = -sin(x[0]);
}
/**
* @brief Compute next step approximation using the midpoint-Euler
* method.
* @f[y_{n+1} = y_n + dx\, f\left(x_n+\frac{1}{2}dx,
* y_n + \frac{1}{2}dx\,f\left(x_n,y_n\right)\right)@f]
* @param[in] dx step size
* @param[in,out] x take @f$x_n@f$ and compute @f$x_{n+1}@f$
* @param[in,out] y take @f$y_n@f$ and compute @f$y_{n+1}@f$
* @param[in,out] dy compute @f$y_n+\frac{1}{2}dx\,f\left(x_n,y_n\right)@f$
*/
void midpoint_euler_step(double dx, double *x, double *y, double *dy)
{
problem(x, y, dy);
double tmp_x = (*x) + 0.5 * dx;
double tmp_y[order];
int o;
for (o = 0; o < order; o++) tmp_y[o] = y[o] + 0.5 * dx * dy[o];
problem(&tmp_x, tmp_y, dy);
for (o = 0; o < order; o++) y[o] += dx * dy[o];
}
/**
* @brief Compute approximation using the midpoint-Euler
* method in the given limits.
* @param[in] dx step size
* @param[in] x0 initial value of independent variable
* @param[in] x_max final value of independent variable
* @param[in,out] y take \f$y_n\f$ and compute \f$y_{n+1}\f$
* @param[in] save_to_file flag to save results to a CSV file (1) or not (0)
* @returns time taken for computation in seconds
*/
double midpoint_euler(double dx, double x0, double x_max, double *y,
char save_to_file)
{
double dy[order];
FILE *fp = NULL;
if (save_to_file)
{
fp = fopen("midpoint_euler.csv", "w+");
if (fp == NULL)
{
perror("Error! ");
return -1;
}
}
/* start integration */
clock_t t1 = clock();
double x = x0;
do // iterate for each step of independent variable
{
if (save_to_file && fp)
fprintf(fp, "%.4g,%.4g,%.4g\n", x, y[0], y[1]); // write to file
midpoint_euler_step(dx, &x, y, dy); // perform integration
x += dx; // update step
} while (x <= x_max); // till upper limit of independent variable
/* end of integration */
clock_t t2 = clock();
if (save_to_file && fp)
fclose(fp);
return (double)(t2 - t1) / CLOCKS_PER_SEC;
}
/**
Main Function
*/
int main(int argc, char *argv[])
{
double X0 = 0.f; /* initial value of x0 */
double X_MAX = 10.F; /* upper limit of integration */
double Y0[] = {1.f, 0.f}; /* initial value Y = y(x = x_0) */
double step_size;
if (argc == 1)
{
printf("\nEnter the step size: ");
scanf("%lg", &step_size);
}
else
// use commandline argument as independent variable step size
step_size = atof(argv[1]);
// get approximate solution
double total_time = midpoint_euler(step_size, X0, X_MAX, Y0, 1);
printf("\tTime = %.6g ms\n", total_time);
/* compute exact solution for comparion */
FILE *fp = fopen("exact.csv", "w+");
if (fp == NULL)
{
perror("Error! ");
return -1;
}
double x = X0;
double *y = &(Y0[0]);
printf("Finding exact solution\n");
clock_t t1 = clock();
do
{
fprintf(fp, "%.4g,%.4g,%.4g\n", x, y[0], y[1]); // write to file
exact_solution(&x, y);
x += step_size;
} while (x <= X_MAX);
clock_t t2 = clock();
total_time = (t2 - t1) / CLOCKS_PER_SEC;
printf("\tTime = %.6g ms\n", total_time);
fclose(fp);
return 0;
}