UMFPACK reports that the matrix being solved is singular

Question

I created a divided domain with the stokes-equation in the first subdomain and the mixed-poisson-equation (darcy) in the second subdomain. I work with the UnitSquare and the subdomain 1 should be the interval from 0 to 0,5 and the subdomain 2 from 0,5 to 1.

But now i get the following error:

Solving linear variational problem. UMFPACK problem related to call to numeric * Warning: UMFPACK reports that the matrix being solved is singular. UMFPACK problem related to call to solve * Warning: UMFPACK reports that the matrix being solved is singular. assert vmax>=vmin, "empty range, please specify vmin and/or vmax" Assertion error: empty range, please specify vmin and/or vmax

Can anyone help? Thanks!

Here is the code:

#-*- coding: utf-8 -*-

from dolfin import *

import numpy as np

# Define mesh
mesh = UnitSquare(32,32)

#Subdomain 1
# Gitter übergeben
subdomains = CellFunction("uint", mesh)

# Klasse des Teilgebiets
class Domain_1(SubDomain):
   def inside(self, x, on_boundary):
       return between(x[0], (0, 0.5)) # Koordinatenangabe des Teilgebiets

# Objekt der Klasse erstellen
sub_domain1 = Domain_1()
sub_domain1.mark(subdomains,0)

# Definition Funktionenräume
U = FunctionSpace(mesh, "CG", 2)
V = FunctionSpace(mesh, "CG", 1)
W = U*V

# Definition Trial- und Testfunktion
(u, p) = TrialFunctions(W)
(v, q) = TestFunctions(W)

# boundary condition
p_in = 1
p_out = 0
noslip = DirichletBC(W.sub(0), (0),

                      "on_boundary && \

                      (x[1] <= DOLFIN_EPS | x[1] >= 0.5-DOLFIN_EPS)")

inflow = DirichletBC(W.sub(1), p_in, "x[0] <= 0.0 + DOLFIN_EPS*1000")
outflow = DirichletBC(W.sub(1), p_out, "x[0] >= 0.5 - DOLFIN_EPS*1000")

bcp = [noslip,inflow, outflow]

# Definition f
f = Expression("0")

# Variationsformulierung
a = inner(grad(u), grad(v))*dx + div(v)*p*dx(0) + q*div(u)*dx(0)
L = inner(f,v)*dx(0)

# Lösung berechnen
w = Function(W)
problem = LinearVariationalProblem(a, L, w, bcp)
solver = LinearVariationalSolver(problem)
solver.solve()
(u, p) = w.split()

# Subdomain 2
# Gitter übergeben
subdomains = CellFunction("uint", mesh)

# Klasse des Teilgebiets
class Domain_2(SubDomain):
   def inside(self,x,on_boundary):
       return between(x[0], (0.5,1.0)) # Koordinatenangabe des Teilgebiets

# Objekt der Klasse erstellen
sub_domain2 = Domain_2()
sub_domain2.mark(subdomains,1)

# Define function spaces and mixed (product) space
BDM = FunctionSpace(mesh, "BDM", 1)
DG = FunctionSpace(mesh, "DG", 0)
CG = FunctionSpace(mesh, "CG", 1)
W = MixedFunctionSpace([BDM, DG, CG])

# Define trial and test functions
(sigma, u, p) = TrialFunctions(W)
(tau, v, q) = TestFunctions(W)

#Define pressure boundary condition
p_in = 1
p_out = 0
noslip = DirichletBC(W.sub(1), (0),
                      "on_boundary && \
                      (x[1] <= 0.5 + DOLFIN_EPS | x[1] >= 1.0-DOLFIN_EPS)")

inflow = DirichletBC(W.sub(2), p_in, "x[0] <= 0.5 + DOLFIN_EPS*1000")
outflow = DirichletBC(W.sub(2), p_out, "x[0] >= 1.0 - DOLFIN_EPS*1000")
bcp = [noslip,inflow, outflow]

# Define f
#f = Expression("0")
f = Expression("10*exp(-(pow(x[0] - 0.5, 2) + pow(x[1] - 0.5, 2)) / 0.02)")

# Define variational form
a = (dot(sigma, tau) + div(tau)*u + div(sigma)*v)*dx(1) + inner(p,q)*dx(1) + u*q*dx(1)
L = f*v*dx(1)

# Compute solution
w = Function(W)
problem = LinearVariationalProblem(a, L, w, bcp)
solver = LinearVariationalSolver(problem)
solver.solve()
(sigma, u, p) = w.split()

# plot

plot(u, axes = True, interactive=True, title = "u")
plot(p, axes = True, interactive=True, title = "p")

Comments

You really need to provide a detailed description of what you want to solve. It is rather tedious to interprete code.

Answer 1

You matrix will be singular because you have no terms in the bilinear form involving $p$ over parts of the domain. That means that any p is a solution in parts of your domain.

Comments

@Garth, there is a p in both linear forms.

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p is only integrated over the subdomain dx(0) in the first form. Hence p is absent from the integral over parts of the domain, which will leave it undefined.

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OK, I see. You are right.

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Can you explain this?
I´m a bit confused. I didn´t see the difference between the two subdomains.
Thanks!