Split red-balack and standard sor solvers

This commit is contained in:
Jan Eitzinger 2024-03-06 07:27:28 +01:00
parent a834a57bf8
commit b9bf4d7b63
4 changed files with 81 additions and 65 deletions

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@ -1,8 +1,8 @@
# Supported: GCC, CLANG, ICC
TAG ?= CLANG
ENABLE_OPENMP ?= false
# Supported: sor, mg
SOLVER ?= sor
# Supported: sor, rb, mg
SOLVER ?= rb
# Run in debug settings
DEBUG ?= false

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@ -26,8 +26,8 @@ p_init 0.0 # initial value for pressure
xlength 1.0 # domain size in x-direction
ylength 1.0 # domain size in y-direction
imax 100 # number of interior cells in x-direction
jmax 100 # number of interior cells in y-direction
imax 128 # number of interior cells in x-direction
jmax 128 # number of interior cells in y-direction
# Time Data:
# ---------

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@ -0,0 +1,76 @@
/*
* Copyright (C) NHR@FAU, University Erlangen-Nuremberg.
* All rights reserved. This file is part of nusif-solver.
* Use of this source code is governed by a MIT style
* license that can be found in the LICENSE file.
*/
#include "solver.h"
#include "util.h"
void initSolver(Solver* s, Discretization* d, Parameter* p)
{
s->grid = &d->grid;
s->itermax = p->itermax;
s->eps = p->eps;
s->omega = p->omg;
}
void solve(Solver* solver, double* p, double* rhs)
{
int imax = solver->grid->imax;
int jmax = solver->grid->jmax;
double eps = solver->eps;
int itermax = solver->itermax;
double dx2 = solver->grid->dx * solver->grid->dx;
double dy2 = solver->grid->dy * solver->grid->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double factor = solver->omega * 0.5 * (dx2 * dy2) / (dx2 + dy2);
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++;
}
#ifdef VERBOSE
printf("Solver took %d iterations to reach %f\n", it, sqrt(res));
#endif
}

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@ -15,7 +15,7 @@ void initSolver(Solver* s, Discretization* d, Parameter* p)
s->omega = p->omg;
}
void solveSOR(Solver* solver, double* p, double* rhs)
void solve(Solver* solver, double* p, double* rhs)
{
int imax = solver->grid->imax;
int jmax = solver->grid->jmax;
@ -66,63 +66,3 @@ void solveSOR(Solver* solver, double* p, double* rhs)
printf("Solver took %d iterations to reach %f\n", it, sqrt(res));
#endif
}
void solve(Solver* solver, double* p, double* rhs)
{
int imax = solver->grid->imax;
int jmax = solver->grid->jmax;
double eps = solver->eps;
int itermax = solver->itermax;
double dx2 = solver->grid->dx * solver->grid->dx;
double dy2 = solver->grid->dy * solver->grid->dy;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double factor = solver->omega * 0.5 * (dx2 * dy2) / (dx2 + dy2);
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++;
}
#ifdef VERBOSE
printf("Solver took %d iterations to reach %f\n", it, sqrt(res));
#endif
}