On 22 Jan 2015, at 15:38, Eike Mueller <E.Mueller@bath.ac.uk> wrote:

Hi Lawrence,

yes, I use the mixed function spaces only in the preconditioner, so I went for your alternative solution. However, now I get:

   L = ufl.replace(r_p, {TestFunction(self._W3), self._mptest}) \
  File "/Users/eikemueller/PostDocBath/EllipticSolvers/ufl/build/lib/ufl/algorithms/replace.py", line 61, in replace
    mapping2 = dict((k, as_ufl(v)) for (k, v) in iteritems(mapping))
  File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/six.py", line 484, in iteritems
    return iter(getattr(d, _iteritems)(**kw))
AttributeError: 'set' object has no attribute ‘iteritems’

Thanks,

Eike

--

Dr Eike Hermann Mueller
Research Associate (PostDoc)

Department of Mathematical Sciences
University of Bath
Bath BA2 7AY, United Kingdom

+44 1225 38 5803
e.mueller@bath.ac.uk
http://people.bath.ac.uk/em459/

On 22 Jan 2015, at 15:27, Lawrence Mitchell <lawrence.mitchell@imperial.ac.uk> wrote:

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On 22/01/15 15:17, Eike Mueller wrote:

Dear firedrakers,

assume I want to do the following solve of a mixed system:

V1 = FunctionSpace(mesh, 'RT', 1) V2 = FunctionSpace(mesh, 'DG',
0)

W = V1 * V2 lmbda = 1 u, p = TrialFunctions(W) v, q =
TestFunctions(W) f = Function(V1) g = Function(V2)

a = (p*q - q*div(u) + lmbda*inner(v, u) + div(v)*p)*dx

L = (dot(u,f)+p*g)*dx

u = Function(W) solve(a == L, u, solver_parameters=...)

But now I take out the line

L = (dot(u,f)+p*g)*dx

and assume that somewhere I calculate instead:

L1 = dot(TestFunction(V1),f)*dx L2 = TestFunction(V2)*g*dx

Can you define these two forms using the test functions v and q
respectively (rather than building new test functions on the
subspaces)?  Then everything should "just work".

Alternately, if you haven't already assembled L1 and L2, you could (to
get L), do:

v1 = TestFunction(V1)
v2 = TestFunction(V2)

L1 = dot(v1, f)*dx
L2 = v2*g*dx

import ufl
L1p = ufl.replace(L1, {v1, v})
L2p = ufl.replace(L2, {v2, v})

L = L1p + L2p

but that's effectively the same as just defining L1 and L2 using v and
q rather v1 and v2.  I guess you're separately using L1 and L2
somewhere in a non-mixed solve?

Cheers,

Lawrence

Cheers,

Lawrence
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