102 lines
3.0 KiB
C

/*
* 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;
}
double solve(Solver* s, double* p, double* rhs)
{
int imax = s->grid->imax;
int jmax = s->grid->jmax;
int kmax = s->grid->kmax;
double eps = s->eps;
int itermax = s->itermax;
double dx2 = s->grid->dx * s->grid->dx;
double dy2 = s->grid->dy * s->grid->dy;
double dz2 = s->grid->dz * s->grid->dz;
double idx2 = 1.0 / dx2;
double idy2 = 1.0 / dy2;
double idz2 = 1.0 / dz2;
double factor = s->omega * 0.5 * (dx2 * dy2 * dz2) /
(dy2 * dz2 + dx2 * dz2 + dx2 * dy2);
double epssq = eps * eps;
int it = 0;
double res = 1.0;
int pass, ksw, jsw, isw;
while ((res >= epssq) && (it < itermax)) {
res = 0.0;
ksw = 1;
for (pass = 0; pass < 2; pass++) {
jsw = ksw;
for (int k = 1; k < kmax + 1; k++) {
isw = jsw;
for (int j = 1; j < jmax + 1; j++) {
for (int i = isw; i < imax + 1; i += 2) {
double r =
RHS(i, j, k) -
((P(i + 1, j, k) - 2.0 * P(i, j, k) + P(i - 1, j, k)) * idx2 +
(P(i, j + 1, k) - 2.0 * P(i, j, k) + P(i, j - 1, k)) *
idy2 +
(P(i, j, k + 1) - 2.0 * P(i, j, k) + P(i, j, k - 1)) *
idz2);
P(i, j, k) -= (factor * r);
res += (r * r);
}
isw = 3 - isw;
}
jsw = 3 - jsw;
}
ksw = 3 - ksw;
}
for (int j = 1; j < jmax + 1; j++) {
for (int i = 1; i < imax + 1; i++) {
P(i, j, 0) = P(i, j, 1);
P(i, j, kmax + 1) = P(i, j, kmax);
}
}
for (int k = 1; k < kmax + 1; k++) {
for (int i = 1; i < imax + 1; i++) {
P(i, 0, k) = P(i, 1, k);
P(i, jmax + 1, k) = P(i, jmax, k);
}
}
for (int k = 1; k < kmax + 1; k++) {
for (int j = 1; j < jmax + 1; j++) {
P(0, j, k) = P(1, j, k);
P(imax + 1, j, k) = P(imax, j, k);
}
}
res = res / (double)(imax * jmax * kmax);
#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
return res;
}