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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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>
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