Split red-balack and standard sor solvers
This commit is contained in:
parent
a834a57bf8
commit
b9bf4d7b63
@ -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
|
||||
|
||||
|
@ -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:
|
||||
# ---------
|
||||
|
76
BasicSolver/2D-seq/src/solver-rb.c
Normal file
76
BasicSolver/2D-seq/src/solver-rb.c
Normal file
@ -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
|
||||
}
|
@ -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
|
||||
}
|
||||
|
Loading…
Reference in New Issue
Block a user