Hi, professor Wolfgang

Yes, I can write the components of V(u(x_q)) by known expression, for
example V(u(x_q))= (2u(x_q),3u(x_q)) where 2u(x_q) is the x-direction and
3u(x_q) is y-direction. I construct a vector<Tensor> named tmp_vector and
assign the values by tmp_vector[0]=2u(x_q), tmp_vector[1]=3u(x_q). Then I
do the multiplication tmp_vector*fe_values.shape_grad(i,q_point). But when
I ran the program, errors occurred because maybe the operation between
these two terms is not permitted. So I define the V(u(x_q)) by
Point<dim>(2u(x_q),3u(x_q)) and the program runs without errors.

I want to ask whether the way I define V(u(x_q)) is right and are there any
better ways to do this?

Thank you very much!

Best,
Toddy

Wolfgang Bangerth <[email protected]> 于2021年9月17日周五 上午8:33写道:

> On 9/15/21 10:47 PM, Toddy Liu wrote:
> >
> > Thank you very much for your reply and it really helps me. You explained
> > how to deal with the latter case (V'(u) . grad u, phi_i) and I've got
> > the point how to solve this. Addtionally. how about the former case
> > -(V(u), grad phi_i). Specifically, how to construct the vector V(u) and
> > then multiply with grad phi_i or to say fe_values.shape_grad(i,q_point)?
>
> You perform the integral via quadrature, and so all you have to know is
> what
>    V(u)(x_q) = V(u(x_q))
> which you evaluate in the same way as you did
>    sin(u(x_q))
> in step-25 with the only exception that now V(u(x_q)) is a vector
> (actually, a Tensor<1,dim>) instead of a scalar quantity.
>
> Best
>   W.
>
> --
> ------------------------------------------------------------------------
> Wolfgang Bangerth          email:                 [email protected]
>                             www: http://www.math.colostate.edu/~bangerth/
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