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
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--

  Dr Anna Kalogirou
  Research Fellow
  School of Mathematics
  University of Leeds

  http://www1.maths.leeds.ac.uk/~matak/