Consider the Stokes problem and the usual divergence operator $B:V \rightarrow Q'$, $\langle Bv, q\rangle = b(v,q)=(\operatorname{div} v,q)$ and its discrete versione $B_h : V_h \rightarrow Q_h'$.

In the lecture notes, I read the following consideration:

It must be $\dim(V_h) \geq \dim(Q_h)$. If not, then $\ker(B_h)= \{0\}$ and the only solution to $Au + B^t p =f$ is is $u=0$.

Question: Why does $\dim(V_h) < \dim(Q_h)$ implies $\ker(B_h) =\{0\}$?

For the inf-sup condition on $B_h$ to be satistfied, I need $B_h$ to be surjective, BUT if I have a linear application (to make an example) from $\mathbb{R}^3 \rightarrow \mathbb{R}^5$, then by the rank theorem: we have indeed $$3 = \dim(\text{Im}) + \dim(\ker)$$ and this implies that it cannot be surjective. I'd say that the only way for that remark to make sense is when I have equality in the dimensions: in that case, being surjective is equivalent to be injective. Therefore, $B_hu_h=0$ only for $u_h=0$.

But I feel link I'm missing something: consider the following discussion, found on Brezzi-Boffi-Fortin: Mixed finite elements

enter image description here

He says essentially that, even if you don't have equality in the dimensions, you're going to have $\ker(B_h)=\{0\}$, and I really can't figure out why.

  • $\begingroup$ Is that book really making such a strong statement? It's just that P1-P0 happens to not work. Inequality of dimensions is required, but not necessarily sufficient condition, so I don't see a conflict here. Or have I misunderstood this part of the question? $\endgroup$ Jul 21 at 23:31
  • $\begingroup$ Yes, I attached the screenshot. Do you mean that even if the function cannot be surjective, it may be injective (and hence have locking, right?) $\endgroup$ Jul 21 at 23:48
  • $\begingroup$ Yes I saw it, I'm just not sure about the strong conclusion "you're going to have ...". I'm not smart enough to give a proper answer for this question, just that $\dim(V_h) \geq \dim(B_h)$ is necessary but not sufficient, and I don't see the book section contradicting this. It even includes the qualifier "likely", as it depends on the mesh and boundary conditions. (b.t.w I found this text helpful: researchgate.net/publication/… ) $\endgroup$ Jul 22 at 1:34

I have a friend who is amazing on FEM and mixed FEM (our research group works on HDG -which is pointwise divergence-free- for incompressible flows), but unfortunately, he is not on SE. I will see if he has time to look at some of your questions. My naive (but somewhat educated) opinion is that, this is a specific example for some meshes.

(Before I begin, just to make sure that we are on the same page, $\text{dim}(V_h)$ would be the dimensionality of the physical space (for example, in 2-D, 2) times the number of interior velocity DoFs (N) -since the velocity vanishes on the boundary- times the polynomial degree used (in this case, 1), so 2*N. $\text{dim}(Q_h)$ would be equal to the number of triangles (t) minus 1. I am not sure if this is confusion on my side, but wanted to make sure.)

First, say you have a full column-rank matrix $A\in\mathbb{R}^{m\times n}$ with $m>n$ and you want to solve $Ax=0$. It is clear that the unique solution to this problem is $x=0$. Since $B_h$ is a linear operator, it has a matrix representation, say $\mathbb{B}$, and $\mathbb{B}$ is $\text{dim}(Q_h)$-by-$\text{dim}(V_h)$. If $\text{dim}(Q_h)>\text{dim}(V_h)$ then $\mathbb{B}u = 0_{\text{dim}(Q_h)}$ has only one solution $u=0_{\text{dim}(V_h)}$.

In case of $P_1-P_0$ elements, if you were to construct a piecewise linear continuous divergence-free velocity field, it may end up being zero almost everywhere, since it has to vanish on boundaries and its divergence has to be zero. We can show that by counting the constraints(Brezzi-Fortin-Doffi appeals to the famous Euler's identity for planar graphs, also check this math.SE answer). According to their argument, $u_h$ is over-constrained($\text{dim}(Q_h)>\text{dim}(V_h)$), hence it must be that $u_h\equiv 0$. If you also check example 8.10.2 (criss-cross mesh for $P_1-P_0$ elements) on pg.508 (in the version I have), where the row rank of the matrix representation of $B_h$ is less than its number of columns, hence $u_h$ is not over-constrained. Though, apparently, there are other problems.

  • $\begingroup$ Thanks a lot for your answer. Indeed, the screenshot I took is precisely the one you mentioned in the third paragraph (as you see, they're counting the constraints). However, I want to make sure I understood why "$u_h$ is overconstrained, hence it must be $0$". Is this only because using the matrix representation of $\mathcal{B}$ we have $ \mathcal{B} u_h=0_{Q_h}$ implies $u_h=0_{V_h}$, since $\mathcal{B}$ has full column rank? $\endgroup$ Jul 22 at 7:37
  • $\begingroup$ Another thing. I would say that the condition $\text{dim}(Q_h)>\text{dim}(V_h)$ is bad also because it implies that $B_h$ cannot be a surjective linear map. Therefore, the infsup condition on $B_h$ doesn't hold, because to make it possible, we need to require the surjectivity of $B_h$. @AbdullahAliSivas $\endgroup$ Jul 22 at 7:45
  • $\begingroup$ Yes, at least in the context of that finite element method, it is because $\mathbb{B}$ has full column rank. They actually define what locking is on pg. 330 (Section 5.6.2), and they describe two cases: total locking happens if $\mathbb{B}u=0$ implies $u=0$, and that happens under a lot of conditions too. Check also the partial locking phenomenon, which is another way of over-constraining the solution. In that case, $u_h\in \text{Ker}B_h$ is not necessarily zero a.e. but since $\text{Ker}B_h$ is not large enough, $u_h$ won't be a good approximation to the infinite dimensional $u$. $\endgroup$ Jul 22 at 15:42
  • $\begingroup$ Is your affermative answer referred to both of my questions ? :-) Btw, thanks a lot, I missed that section. @AbdullahAliSivas $\endgroup$ Jul 22 at 17:03
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    $\begingroup$ Long Chen teaches this course 226 at UCI, you can find most of their teaching material at math.uci.edu/~chenlong/teaching.html . I am particularly referring to math.uci.edu/~chenlong/226/FEMStokes.pdf . Regarding the BBF book, it is definitely dense and difficult to read, but I know some really smart people who can explain everything in that book. You can become one of them if you persist :) I just learned what I needed for my research (preconditioning) and then stopped, didn't try hard enough to understand everything :) $\endgroup$ Jul 23 at 18:50

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