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On 21/10/15 14:32, Anna Kalogirou wrote:
> But the matrix A has the same form as C, namely it has
nonzero
> variables in a block submatrix only.
Ah, oops, I see.
So because you're using an iterative solver without any
preconditioning, the operator is never being inverted
directly (hence
you never get zero pivot errors in an LU factorisation).
Your
operator just has a gigantic null space. If the right hand
side is
consistent you might get lucky in that the Krylov solver can
just
explore the space orthogonal to the null space (and you end
up with a
solution in that subspace).
I guess what you really want to do is only solve the problem
in the
region where the heaviside function is nonzero, but we don't
currently
have proper support for this. I guess if you want to be
able to
invert the system somehow, we can add the identity matrix to
the zero
block and (presumably) zero the right hand side in the
appropriate rows.
Note that this is equivalent to prescribing a dirichlet
BC everywhere in the 0 block.
Lawrence
>
> On 21/10/15 14:26, Lawrence Mitchell wrote: On 21/10/15
14:14, Anna
> Kalogirou wrote:
>>>> Dear Lawrence,
>>>>
>>>> I believe the code (attached) sets up the
>>>> matrices/vectrors/operators correctly.
However, I have one
>>>> question: Due to the use of a heavyside
function, the
>>>> operator B on the LHS is nonzero in a block
NxN, say, in the
>>>> bottom right corner. For that reason, when
I solved the
>>>> linear system in Matlab I only considered
these nonzero
>>>> values in the block, because otherwise the
LHS would not be
>>>> invertible. How come and it doesn't
complain here?
> So it's clear that the matrix C = Q1*Q2^T is not
invertible,
> however, presumably you're solving for:
>
> B = A + C
>
> where A is full rank and invertible. So I don't
immediately see
> why there would be a problem in trying to invert B
which is what
> the solver does.
>
> Lawrence
>
>>
>> _______________________________________________
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>> list firedrake@imperial.ac.uk
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>
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