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I'm studying mixed finite elements. The problem is a classical saddle-point one: we seek for $(u,p)$ in $V \times Q$: $$A u + B^t p = f$$ $$Bu = g$$ where $A: V \rightarrow V', B:V \rightarrow Q'$ and $V,Q$ are Hilbert spaces. For this problem, we have the inf-sup conditions on $A$ and $B$ that essentially tell us that the problem is well posed.

As far as I understand, the issue one the discrete level (in contrast to what happens for Poisson equation, and other coercive problems) is that if we consider $V_h \subset V$ and $Q_h \subset Q$ as finite dimensional spaces in which we seek the solution, then we can no more apply the continuous properties to the discrete problem because for instance $$\ker(B_h) \not \subset \ker(B)$$ in general.

So far so good. Then, while doing the analysis, my professor wrote the following diagriam

enter image description here and he said that we would like to join with dotted arrow and we would like that $B_h$ maps entire $V_h$ into the entire $Q_h'$ and we would like this diagram to commute. He then said that it's trivial if $\ker(B_h) \subset \ker(B)$.

I really can't understand why the inclusion of the kernels implies the commutativity of the diagram. Any help is highly appreciated, and even an example where $B$ is the usual divergence operator is perfectly fine to me.


EDIT: Maybe I got the way to go: the diagram commuting means that $$ \Pi_{Q_h'} \circ B = B_h \circ \Pi_{V_h}$$

Now, if the $\ker(B_h) \not \subset \ker(B)$, this means there exists a function $v_h \in V_h \subset V$ such that $$B_h v_h = 0$$ and $$B v_h \ne 0$$

If it commutes, it should hold:$$ B_h \Pi_{V_h} v_h=\underbrace{B_h v_h}_{0} = \Pi_{Q_h'} \underbrace{B v_h}_{\ne 0} \ne 0$$

which is clearly false. In the first equality I used the fact that $v_h$ is of course not affected by the projection. What do you think ? I think that something more can surely be said. Please feel free to add more things about this

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  • $\begingroup$ Nice question -- but I don't understand this either. What does it mean this arrow exists. Maybe that $B_h$ together with $V_h$ and $Q_h$ fulfill an inf-sup condition with a constant independent of $h$? (That would be the standard requirement for inf-sup stable finite elements) $\endgroup$ – Jan 2 days ago
  • $\begingroup$ @Jan I edited the question, maybe now it's more clear. Basically, he said that the condition on the kernels implies that the diagram commutes. It seems to be trivial, but I really don't see how $\endgroup$ – bobinthebox 2 days ago

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