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Sep 19, 2020 at 7:56 comment added Vefhug Oh, I see now. I'm having difficulties to understand how it's done the $(i,j)$ entry of the matrix: is it this one $$\int a(\sum_j u_k \phi_k) \phi_i{'} \phi_j{'}$$ ? I've never seen such an assembly, so I don't know how to move
Sep 19, 2020 at 5:12 comment added cfdlab Its a nonlinear problem. You have to integrate with $a(u_h)$ inside the integral. The matrix depends on the solution, so you have to recompute it every time $u_h$ changes.
Sep 18, 2020 at 13:31 comment added Vefhug My bad, sorry! I have a question about that term. When I discretize, $a(u_h)$ is a vector. So I have, after calling $A$ the matrix with $(A)_{ij} = \int \phi_i' \phi_j' dx$ $$\vec{a} A$$ where the multiplication is entrywise. Should this be written in a program as $\text{diag(a(uh)) * A$?
Sep 18, 2020 at 13:09 comment added cfdlab You must know the function $a(u)$. So $a(u_h)$ just means evaluate it at the numerical solution.
Sep 18, 2020 at 7:22 comment added Vefhug Also, what is, explicitely, $a(u_h)$?
Sep 18, 2020 at 7:21 comment added Vefhug could you expand a bit your answer? It's not clear to me how you got the $B_{ij}$, i.e. a matrix and not a tensor, as Wolfgang wrote in the linked question
Sep 18, 2020 at 3:23 history answered cfdlab CC BY-SA 4.0