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heat.c
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heat.c
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/*
** PROGRAM: heat equation solve
**
** PURPOSE: This program will explore use of an explicit
** finite difference method to solve the heat
** equation under a method of manufactured solution (MMS)
** scheme. The solution has been set to be a simple
** function based on exponentials and trig functions.
**
** A finite difference scheme is used on a 1000x1000 cube.
** A total of 0.5 units of time are simulated.
**
** The MMS solution has been adapted from
** G.W. Recktenwald (2011). Finite difference approximations
** to the Heat Equation. Portland State University.
**
**
** USAGE: Run with two arguments:
** First is the number of cells.
** Second is the number of timesteps.
**
** For example, with 100x100 cells and 10 steps:
**
** ./heat 100 10
**
**
** HISTORY: Written by Tom Deakin, Oct 2018
**
*/
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <omp.h>
// Key constants used in this program
#define PI acos(-1.0) // Pi
#define LINE "--------------------\n" // A line for fancy output
// Function definitions
void initial_value(const int n, const double dx, const double length, double * restrict u);
void zero(const int n, double * restrict u);
void solve(const int n, const double alpha, const double dx, const double dt, const double * restrict u, double * restrict u_tmp);
double solution(const double t, const double x, const double y, const double alpha, const double length);
double l2norm(const int n, const double * restrict u, const int nsteps, const double dt, const double alpha, const double dx, const double length);
// Main function
int main(int argc, char *argv[]) {
// Start the total program runtime timer
double start = omp_get_wtime();
// Problem size, forms an nxn grid
int n = 1000;
// Number of timesteps
int nsteps = 10;
// Check for the correct number of arguments
// Print usage and exits if not correct
if (argc == 3) {
// Set problem size from first argument
n = atoi(argv[1]);
if (n < 0) {
fprintf(stderr, "Error: n must be positive\n");
exit(EXIT_FAILURE);
}
// Set number of timesteps from second argument
nsteps = atoi(argv[2]);
if (nsteps < 0) {
fprintf(stderr, "Error: nsteps must be positive\n");
exit(EXIT_FAILURE);
}
}
//
// Set problem definition
//
double alpha = 0.1; // heat equation coefficient
double length = 1000.0; // physical size of domain: length x length square
double dx = length / (n+1); // physical size of each cell (+1 as don't simulate boundaries as they are given)
double dt = 0.5 / nsteps; // time interval (total time of 0.5s)
// Stability requires that dt/(dx^2) <= 0.5,
double r = alpha * dt / (dx * dx);
// Print message detailing runtime configuration
printf("\n");
printf(" MMS heat equation\n\n");
printf(LINE);
printf("Problem input\n\n");
printf(" Grid size: %d x %d\n", n, n);
printf(" Cell width: %E\n", dx);
printf(" Grid length: %lf x %lf\n", length, length);
printf("\n");
printf(" Alpha: %E\n", alpha);
printf("\n");
printf(" Steps: %d\n", nsteps);
printf(" Total time: %E\n", dt*(double)nsteps);
printf(" Time step: %E\n", dt);
printf(LINE);
// Stability check
printf("Stability\n\n");
printf(" r value: %lf\n", r);
if (r > 0.5)
printf(" Warning: unstable\n");
printf(LINE);
// Allocate two nxn grids
double *u = malloc(sizeof(double)*n*n);
double *u_tmp = malloc(sizeof(double)*n*n);
double *tmp;
// Set the initial value of the grid under the MMS scheme
initial_value(n, dx, length, u);
zero(n, u_tmp);
//
// Run through timesteps under the explicit scheme
//
// Start the solve timer
double tic = omp_get_wtime();
for (int t = 0; t < nsteps; ++t) {
// Call the solve kernel
// Computes u_tmp at the next timestep
// given the value of u at the current timestep
solve(n, alpha, dx, dt, u, u_tmp);
// Pointer swap
tmp = u;
u = u_tmp;
u_tmp = tmp;
}
// Stop solve timer
double toc = omp_get_wtime();
//
// Check the L2-norm of the computed solution
// against the *known* solution from the MMS scheme
//
double norm = l2norm(n, u, nsteps, dt, alpha, dx, length);
// Stop total timer
double stop = omp_get_wtime();
// Print results
printf("Results\n\n");
printf("Error (L2norm): %E\n", norm);
printf("Solve time (s): %lf\n", toc-tic);
printf("Total time (s): %lf\n", stop-start);
printf(LINE);
// Free the memory
free(u);
free(u_tmp);
}
// Sets the mesh to an initial value, determined by the MMS scheme
void initial_value(const int n, const double dx, const double length, double * restrict u) {
double y = dx;
for (int j = 0; j < n; ++j) {
double x = dx; // Physical x position
for (int i = 0; i < n; ++i) {
u[i+j*n] = sin(PI * x / length) * sin(PI * y / length);
x += dx;
}
y += dx; // Physical y position
}
}
// Zero the array u
void zero(const int n, double * restrict u) {
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
u[i+j*n] = 0.0;
}
}
}
// Compute the next timestep, given the current timestep
void solve(const int n, const double alpha, const double dx, const double dt, const double * restrict u, double * restrict u_tmp) {
// Finite difference constant multiplier
const double r = alpha * dt / (dx * dx);
const double r2 = 1.0 - 4.0*r;
// Loop over the nxn grid
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
// Update the 5-point stencil, using boundary conditions on the edges of the domain.
// Boundaries are zero because the MMS solution is zero there.
u_tmp[i+j*n] = r2 * u[i+j*n] +
r * ((i < n-1) ? u[i+1+j*n] : 0.0) +
r * ((i > 0) ? u[i-1+j*n] : 0.0) +
r * ((j < n-1) ? u[i+(j+1)*n] : 0.0) +
r * ((j > 0) ? u[i+(j-1)*n] : 0.0);
}
}
}
// True answer given by the manufactured solution
double solution(const double t, const double x, const double y, const double alpha, const double length) {
return exp(-2.0*alpha*PI*PI*t/(length*length)) * sin(PI*x/length) * sin(PI*y/length);
}
// Computes the L2-norm of the computed grid and the MMS known solution
// The known solution is the same as the boundary function.
double l2norm(const int n, const double * restrict u, const int nsteps, const double dt, const double alpha, const double dx, const double length) {
// Final (real) time simulated
double time = dt * (double)nsteps;
// L2-norm error
double l2norm = 0.0;
// Loop over the grid and compute difference of computed and known solutions as an L2-norm
double y = dx;
for (int j = 0; j < n; ++j) {
double x = dx;
for (int i = 0; i < n; ++i) {
double answer = solution(time, x, y, alpha, length);
l2norm += (u[i+j*n] - answer) * (u[i+j*n] - answer);
x += dx;
}
y += dx;
}
return sqrt(l2norm);
}