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fix: stop divide by zero warning in LU solvers #790

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merged 3 commits into from
Nov 29, 2021
Merged

fix: stop divide by zero warning in LU solvers #790

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f813825
fix: stop divide by zero warning in LU solvers
wd15 Mar 22, 2021
3682f7c
Merge branch 'master' into lu-divide-by-zero
wd15 Sep 13, 2021
a61c1d4
ping the build
guyer Nov 29, 2021
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18 changes: 9 additions & 9 deletions fipy/solvers/petsc/linearLUSolver.py
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Original file line number Diff line number Diff line change
Expand Up @@ -18,13 +18,13 @@ class LinearLUSolver(PETScSolver):
A direct solve is performed.

"""

def __init__(self, tolerance=1e-10, iterations=10, precon="lu"):
"""
:Parameters:
- `tolerance`: The required error tolerance.
- `iterations`: The maximum number of iterative steps to perform.
- `precon`: *Ignored*.
- `precon`: *Ignored*.

"""
PETScSolver.__init__(self, tolerance=tolerance,
Expand All @@ -38,26 +38,26 @@ def _solve_(self, L, x, b):
# TODO: SuperLU invoked with PCFactorSetMatSolverType(pc, MATSOLVERSUPERLU)
# see: http://www.mcs.anl.gov/petsc/petsc-dev/src/ksp/ksp/examples/tutorials/ex52.c.html
# PETSc.PC().setFactorSolverType("superlu")

L.assemble()
ksp.setOperators(L)
ksp.setFromOptions()

for iteration in range(self.iterations):
errorVector = L * x - b
tol = errorVector.norm()

if iteration == 0:
tol0 = tol
if (tol / tol0) <= self.tolerance:

if tol <= self.tolerance * tol0:
break

xError = x.copy()

ksp.solve(errorVector, xError)
x -= xError

if 'FIPY_VERBOSE_SOLVER' in os.environ:
from fipy.tools.debug import PRINT
# L.view()
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2 changes: 1 addition & 1 deletion fipy/solvers/pysparse/linearLUSolver.py
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Expand Up @@ -69,7 +69,7 @@ def _solve_(self, L, x, b):
for iteration in range(self.iterations):
errorVector = L * x - b

if (numerix.sqrt(numerix.sum(errorVector**2)) / error0) <= self.tolerance:
if numerix.sqrt(numerix.sum(errorVector**2)) <= self.tolerance * error0:
break

xError = numerix.zeros(len(b), 'd')
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2 changes: 1 addition & 1 deletion fipy/solvers/scipy/linearLUSolver.py
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Expand Up @@ -38,7 +38,7 @@ def _solve_(self, L, x, b):
for iteration in range(min(self.iterations, 10)):
errorVector = L * x - b

if (numerix.sqrt(numerix.sum(errorVector**2)) / error0) <= self.tolerance:
if numerix.sqrt(numerix.sum(errorVector**2)) <= self.tolerance * error0:
break

xError = LU.solve(errorVector)
Expand Down