Add solver variants

This commit is contained in:
Jan Eitzinger 2023-06-18 10:08:56 +02:00
parent 1d99310b2e
commit 9c8585c7bc
6 changed files with 217 additions and 157 deletions

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@ -428,6 +428,7 @@ int solve(Solver* solver)
double dy2 = solver->dy * solver->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
// identical to 1/((2/dx2)+(2/dy2))
double factor = solver->omega * 0.5 * (dx2 * dy2) / (dx2 + dy2);
double* p = solver->p;
double* rhs = solver->rhs;

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@ -247,68 +247,6 @@ void solveRB(Solver* solver)
#endif
}
void solveRBA(Solver* solver)
{
int imax = solver->imax;
int jmax = solver->jmax;
double eps = solver->eps;
int itermax = solver->itermax;
double dx2 = solver->dx * solver->dx;
double dy2 = solver->dy * solver->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double factor = 0.5 * (dx2 * dy2) / (dx2 + dy2);
double omega = solver->omega;
double* p = solver->p;
double* rhs = solver->rhs;
double epssq = eps * eps;
int it = 0;
double anorm = 1.0, anormf = 0.0;
int pass, jsw, isw;
while ((anorm >= eps * anormf) && (it < itermax)) {
anorm = 0.0;
jsw = 1;
for (pass = 0; pass < 2; pass++) {
isw = jsw;
for (int j = 1; j < jmax + 1; j++) {
for (int i = isw; i < imax + 1; i += 2) {
double r = RHS(i, j) -
((P(i + 1, j) - 2.0 * P(i, j) + P(i - 1, j)) * idx2 +
(P(i, j + 1) - 2.0 * P(i, j) + P(i, j - 1)) * idy2);
anorm += fabs(r);
P(i, j) -= (omega * factor * r);
}
isw = 3 - isw;
}
jsw = 3 - jsw;
}
for (int i = 1; i < imax + 1; i++) {
P(i, 0) = P(i, 1);
P(i, jmax + 1) = P(i, jmax);
}
for (int j = 1; j < jmax + 1; j++) {
P(0, j) = P(1, j);
P(imax + 1, j) = P(imax, j);
}
#ifdef DEBUG
printf("%d Residuum: %e\n", it, res);
#endif
it++;
}
#ifdef VERBOSE
printf("Solver took %d iterations to reach %f\n", it, anorm);
#endif
}
static double maxElement(Solver* solver, double* m)
{
int size = (solver->imax + 2) * (solver->jmax + 2);

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@ -16,6 +16,7 @@ jmax 200 # number of interior cells in y-direction
itermax 10000 # maximal number of pressure iteration in one time step
eps 0.000001 # stopping tolerance for pressure iteration
omg 1.9 # relaxation parameter for SOR iteration
# omg 0.9999 # relaxation parameter for SOR iteration
omg 1.93 # relaxation parameter for SOR iteration
#===============================================================================

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@ -4,18 +4,19 @@
* Use of this source code is governed by a MIT-style
* license that can be found in the LICENSE file.
*/
#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <limits.h>
#include <float.h>
#include <limits.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include "parameter.h"
#include "solver.h"
#include "timing.h"
int main(int argc, char** argv)
{
double startTime, endTime;
Parameter params;
Solver solver;
initParameter(&params);
@ -29,7 +30,10 @@ int main (int argc, char** argv)
printParameter(&params);
initSolver(&solver, &params, 2);
startTime = getTimeStamp();
solve(&solver);
endTime = getTimeStamp();
printf("Solution took %.2fs\n", endTime - startTime);
writeResult(&solver);
return EXIT_SUCCESS;

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@ -4,13 +4,13 @@
* Use of this source code is governed by a MIT style
* license that can be found in the LICENSE file.
*/
#include "stdlib.h"
#include "stdio.h"
#include "math.h"
#include "stdio.h"
#include "stdlib.h"
#include "solver.h"
#include "parameter.h"
#include "allocate.h"
#include "parameter.h"
#include "solver.h"
#define PI 3.14159265358979323846
#define P(i, j) p[(j) * (imax + 2) + (i)]
@ -58,35 +58,31 @@ void initSolver(Solver *solver, Parameter *params, int problem)
}
}
int solve(Solver *solver)
void solve(Solver* solver)
{
double r;
int it = 0;
double res;
int imax = solver->imax;
int jmax = solver->jmax;
double eps = solver->eps;
double omega = solver->omega;
int itermax = solver->itermax;
double dx2 = solver->dx * solver->dx;
double dy2 = solver->dy * solver->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double factor = omega * 0.5 * (dx2*dy2) / (dx2+dy2);
double factor = solver->omega * 0.5 * (dx2 * dy2) / (dx2 + dy2);
double* p = solver->p;
double* rhs = solver->rhs;
double epssq = eps * eps;
int it = 0;
double res = 1.0;
res = eps + 1.0;
while((res >= eps) && (it < itermax)) {
while ((res >= epssq) && (it < itermax)) {
res = 0.0;
for (int j = 1; j < jmax + 1; j++) {
for (int i = 1; i < imax + 1; i++) {
r = RHS(i,j) - ((P(i-1,j) - 2.0 * P(i,j) + P(i+1,j)) * idx2 +
double r = RHS(i, j) -
((P(i - 1, j) - 2.0 * P(i, j) + P(i + 1, j)) * idx2 +
(P(i, j - 1) - 2.0 * P(i, j) + P(i, j + 1)) * idy2);
P(i, j) -= (factor * r);
@ -104,21 +100,139 @@ int solve(Solver *solver)
P(imax + 1, j) = P(imax, j);
}
it++;
//Residuum
res = sqrt(res / (imax*jmax));
res = res / (double)(imax * jmax);
#ifdef DEBUG
printf("%d Residuum: %e\n", it, res);
#endif
it++;
}
printf("Solver took %d iterations\n",it);
if( res < eps ){
return 1;
} else{
return 0;
printf("Solver took %d iterations to reach %f\n", it, sqrt(res));
}
void solveRB(Solver* solver)
{
int imax = solver->imax;
int jmax = solver->jmax;
double eps = solver->eps;
int itermax = solver->itermax;
double dx2 = solver->dx * solver->dx;
double dy2 = solver->dy * solver->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double factor = solver->omega * 0.5 * (dx2 * dy2) / (dx2 + dy2);
double* p = solver->p;
double* rhs = solver->rhs;
double epssq = eps * eps;
int it = 0;
double res = 1.0;
int pass, jsw, isw;
while ((res >= epssq) && (it < itermax)) {
res = 0.0;
jsw = 1;
for (pass = 0; pass < 2; pass++) {
isw = jsw;
for (int j = 1; j < jmax + 1; j++) {
for (int i = isw; i < imax + 1; i += 2) {
double r = RHS(i, j) -
((P(i + 1, j) - 2.0 * P(i, j) + P(i - 1, j)) * idx2 +
(P(i, j + 1) - 2.0 * P(i, j) + P(i, j - 1)) * idy2);
P(i, j) -= (factor * r);
res += (r * r);
}
isw = 3 - isw;
}
jsw = 3 - jsw;
}
for (int i = 1; i < imax + 1; i++) {
P(i, 0) = P(i, 1);
P(i, jmax + 1) = P(i, jmax);
}
for (int j = 1; j < jmax + 1; j++) {
P(0, j) = P(1, j);
P(imax + 1, j) = P(imax, j);
}
res = res / (double)(imax * jmax);
#ifdef DEBUG
printf("%d Residuum: %e\n", it, res);
#endif
it++;
}
printf("Solver took %d iterations to reach %f\n", it, sqrt(res));
}
void solveRBA(Solver* solver)
{
int imax = solver->imax;
int jmax = solver->jmax;
double eps = solver->eps;
int itermax = solver->itermax;
double dx2 = solver->dx * solver->dx;
double dy2 = solver->dy * solver->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double factor = 0.5 * (dx2 * dy2) / (dx2 + dy2);
double rho = solver->omega;
double* p = solver->p;
double* rhs = solver->rhs;
double epssq = eps * eps;
int it = 0;
double res = 1.0;
int pass, jsw, isw;
double omega = 1.0;
while ((res >= epssq) && (it < itermax)) {
res = 0.0;
jsw = 1;
for (pass = 0; pass < 2; pass++) {
isw = jsw;
for (int j = 1; j < jmax + 1; j++) {
for (int i = isw; i < imax + 1; i += 2) {
double r = RHS(i, j) -
((P(i + 1, j) - 2.0 * P(i, j) + P(i - 1, j)) * idx2 +
(P(i, j + 1) - 2.0 * P(i, j) + P(i, j - 1)) * idy2);
P(i, j) -= (omega * factor * r);
res += (r * r);
}
isw = 3 - isw;
}
jsw = 3 - jsw;
omega = (it == 0 && pass == 0 ? 1.0 / (1.0 - 0.5 * rho * rho)
: 1.0 / (1.0 - 0.25 * rho * rho * omega));
}
for (int i = 1; i < imax + 1; i++) {
P(i, 0) = P(i, 1);
P(i, jmax + 1) = P(i, jmax);
}
for (int j = 1; j < jmax + 1; j++) {
P(0, j) = P(1, j);
P(imax + 1, j) = P(imax, j);
}
res = res / (double)(imax * jmax);
#ifdef DEBUG
printf("%d Residuum: %e Omega: %e\n", it, res, omega);
#endif
it++;
}
printf("Final omega: %f\n", omega);
printf("Solver took %d iterations to reach %f\n", it, sqrt(res));
}
void writeResult(Solver* solver)

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@ -16,7 +16,9 @@ typedef struct {
int itermax;
} Solver;
void initSolver(Solver*, Parameter*, int problem);
void writeResult(Solver*);
int solve(Solver*);
extern void initSolver(Solver*, Parameter*, int problem);
extern void writeResult(Solver*);
extern void solve(Solver*);
extern void solveRB(Solver*);
extern void solveRBA(Solver*);
#endif