# Tag Info

6

The other answer has everything you already need, but it's also worth pointing out that $u_h$ is computable whereas $I_hu$ is not: The latter requires you to know the exact solution, which in general we of course don't know (and if we did, we didn't need to compute a Galerkin approximation $u_h$).

6

Let's assume $u$ is the solution to the variational problem and $u_h$ is the Galerkin approximation of the solution on the subspace $V_h \subset V$. By Cea's lemma you have: $$||u-u_h||_V \leq C\inf\limits_{v_h \in V_h}||u-v_h||_V$$ for some positive constant $C$. Now you define the projection $I_h:V \rightarrow V_h$. As you ...

4

One approach to convert this into an ODE is with index reduction methods. These allow you to convert high-index DAEs into low-index DAEs or ODEs. See section VII.2 of "Solving Ordinary Differential Equations II" by Hairer and Wanner. Consider a generic, Hessenberg index-2 DAE \begin{align} y' &= f(y, z) \\ 0 &= g(y) \end{align} ...

4

You normally don't assign material properties to a node. They are part of a subdomain and, hence, "assigned" to an element. They appear in the integrals you compute, e.g. $$\int\limits_{\Omega_e} k \nabla u \nabla w d\Omega\, ,$$ here $\Omega_e$ represents the element identified as $e$. Nevertheless, due to interpolation the computed integrals are ...

2

The statement as given is indeed correct (i.e., left and right hand side are different) for almost any function $f(x)$. What it says, once you write out what these norms are, is that $$\sum_m \sqrt{\int_{T_m} f(x)^2 } \neq \sqrt{ \sum_m \int_{T_m} f(x)^2 }.$$

1

MPI Q1: In a naive implementation yes, you will end up with a copy of the mesh on each process. You can reduce this somewhat by only duplicating the parts of the mesh which contain the surface, since you are only doing a surface integral and not a volume integral. This in theory will reduce how much memory is duplicated per process ($O(n^2)$ vs. $O(n^3)$). I ...

1

We have recently proposed a mixed u-p formulation that results in a global symmetric matrix irrespective of the volumetric energy function (for hyperelasticity). The paper contains sufficient details on the linearisation (2nd derivative).

1

Disclaimer: I'm outside my area of expertise here. Due to continuity constraints, when solving electromagnetic simulations with FEM, one uses edge based vector basis functions. After solving, I would expect you to construct your solution field $$\vec{H}(x) = \sum_i H_i \vec{N}_i(x)$$ or, element-wise \vec{H}(x) = \sum_i H^e_i \vec{N}^e_i(x) \quad\forall x \...

1

He is only going to the reference triangle to have the constant independent of $h$, as required. But yes, he is using the fact that having the bubble function inside the norm double bars does not stop it from being a norm.

1

I think $u - \Pi u$ is zero on a part of the boundary. Then you can use Poincare inequality to bound $\| u - \Pi u \|_0 \leq C| u - \Pi u|_1$. It could also be possible to look at the case $m = 0$ and argue the $L^2$ part is of higher order and, hence, smaller in the asymptotic limit.

1

Split $[0,1]$ into two elements $[0,1/2]$ and $[1/2,1]$. Consider the function $f$ satisfying $f(x)=0$ for $x < 1/2$ and $f(x)=1$ otherwise. For this function the $H^1$ norm is infinity but if you calculate the $H^1$ norm over the two elements separately you get 0 and $1/2$, respectively. In particular, the derivative of the Heaviside function is a delta ...

1

Brezzi, Fortin, and Marini showed that a version of the MINI element is stable for 2D elasticity. The stress and velocity spaces are $\Sigma_h = (\mathscr{L}_1^1 + \mathscr{B}_3)_s^{2\times 2}$, $V_h = (\mathscr{L}_1^1)^2$ where $\mathscr{L}_p^k$ is the space of polynomials of degree $p$ with $k$ continuous derivatives across triangle boundaries and \$\...

1

I have a paper on this which, as long as your boundary is flat and you don't expecct it to pass vortexes, should do you well: https://doi.org/10.1002/fld.1427. The gist of it is that as long as the boundary is flat and you enforce no tangential flow, the normal traction is equal to the pressure in an incompressible flow. This is an OK BC for Poiseuille-like ...

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