Dear Prof. Bangerth,
Thank you very much for your clear explanations.
So according to that, is it still correct if I sum up the equations in
discretized form? (Considering the fact that they're still containing trial
and test functions, though in discretized form).
Also could you please point me out to an example or a universal approach to
implement equations in matrix-vector forms (linear systems)? Mixed Laplace
seems to be serving as a typical example in this regard but not sure as it
also involves some considerations on preconditioning and solver usage.
Thank you,
Ali
On Tue, Oct 19, 2021, 8:31 PM Wolfgang Bangerth <bange...@colostate.edu>
wrote:
>
> Ali,
>
> > As my first question (which is an elementary math level question) and
> > considering step-20 as an example, I was wondering how two equations
> > (bilinear forms) containing two unknowns can be summed up to a single
> > equation? While for example in the case of a system of two linear
> equations it
> > is not allowed. (my guess: bilinear forms are still in integral form)
> > I noticed this lack of understanding when I was trying to implement a
> vector
> > valued problem. It is of nonlinear type and composed of two equations. I
> > linearized and discretized equations separately. Now I have a 2*2 matrix
> > (consider A, B | A' , B') multiplied by {del.u ,del. v} (as unknowns)
> and on
> > the right hand side I have {f1, f2}.
>
> Take the mixed Laplace equation as an example:
> u + grad p = 0
> -div u = f
> If you multiply the first equation by a test function v and the second by
> a
> test function q, then you will get the following two weak forms after
> integration by parts (ignoring boundary conditions for a moment):
> (v,u) - (div v, p) = 0 \forall v
> -(q, div u) = (q,f) \forall q
> We probably agree on this, right?
>
> Now, we generally add these equations together and then have
> (v,u) - (div v, p) - (q, div u) = (q,f) \forall v,q
> If I understand your question right, then you are asking about why we can
> add
> these equations together? That is because it is true *for all v,q*, and
> consequently it is in particular true if we choose q=0. But if we choose
> q=0,
> then the previous equation simplifies to
> (v,u) - (div v, p) = 0 \forall v
> On the other hand, if we had chosen v=0, then we will get that
> -(q, div u) = (q,f) \forall q
>
> In other words, the presence of the test functions and the statement "for
> all
> test functions" allows us to recover the original equations from the
> summed-up
> equation.
>
> Does that make sense?
>
> Best
> Wolfgang
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth email: bange...@colostate.edu
> www: http://www.math.colostate.edu/~bangerth/
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