I am not familiar with this expression |Q1> <Q2|. What I really want is Q1*Q2^T, where Q1 and Q2 and vectors defined in my previous email. Actually I think it's easier if I just provide you with the actual equation I need to solve, please see attached. Anna. On 19/10/15 14:15, Lawrence Mitchell wrote:
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On 19/10/15 14:00, Anna Kalogirou wrote:
Dear Lawrence,
Yes, I have a term where the test function and the basis function are evaluated in separate integrals, this is why I wrote Q*mu*dx. Can I assemble them separately? E.g.
Q1 = assemble(v*dx) Q2 = assemble(mu*dx) The resulting values aren't unknowns (either test or trial functions) so I still don't quite understand.
Q1 is a Co-Function (living in the dual space of V) and Q2 is a Function in V. You can certainly point-wise multiply the values together, but I don't think that's what you want, since that still doesn't give you an operator mapping from the trial to the test space.
Do you want the rank-1 operator formed by:
|Q1> <Q2|
?
Cheers,
Lawrence -----BEGIN PGP SIGNATURE----- Version: GnuPG v2.0.22 (GNU/Linux)
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-- Dr Anna Kalogirou Research Fellow School of Mathematics University of Leeds http://www1.maths.leeds.ac.uk/~matak/