# Tag Info

5

The triangle inequality is your friend. Let's ignore the issue of boundary approximation for a moment, then you are computing a solution with inexact linear solver. Let's call it $u_h$. We will call the exact solution of the PDE $u$, and the exact finite element solution $u_\text{FEM}$; neither of these can be computed exactly (in the case of $u_\text{FEM}$ ...

4

It turns out that I have just the right paper for you on this subject: https://www.math.colostate.edu/~bangerth/publications/2013-pattern.pdf

3

The face integrals don't show up when using continuous basis functions, and the weak form that you wrote down indeed assumes that the basis functions are continuous. The finite volume method is a special case of the finite element method when you allow discontinuous basis functions instead of only continuous ones. For a discontinuous Galerkin (or DG) ...

3

I think that errors 1 and 2 classify as variational crimes, as also classifies approximating the domain by a mesh (but that's not your case). That being said, I am not aware of how big these errors are in comparison with the approximation error. Now, regarding rounding errors I remember Nick's Trefethen quote  If rounding errors vanished, numerical ...

2

When one learn about functional analysis methods for PDEs, usually starts from common theorems like the Riesz representation and the Lax-Milgram lemmas, which work quite good with linear PDEs. When dealing with non-linear PDEs the story is far more complicated. There are some results from functional analysis that may come in hand, and usually these take the ...

2

The virtual strain energy should be \begin{equation} \delta U = \int\limits_V {\delta {{\bf{\varepsilon }}^T}{\bf{C}\left(\varepsilon - {\bf{\bar \varepsilon }} \right)dV}} \end{equation} where \begin{equation} {\bf{\bar \varepsilon }} = \alpha \Delta T\left\{ \begin{array}{l} 1\\ 1\\ 1\\ 0\\ 0\\ 0 \end{array} \right\} \end{equation} ...

1

There are many variations of the idea of Taylor-Hood elements that remain stable. The Taylor-Hood element on quadrilateral and hexahedral elements are generally understood (though historically incorrect) to be the space $Q_2\times Q_1$ for velocity and pressure. But this is not optimal: We can make the velocity space larger and instead use $Q_2 \times P_{-1}... 1 Have you looked at the Bauer-Fike theorem? See for instance https://reader.elsevier.com/reader/sd/pii/S0377042712000787?token=EADDECFFF960A9B46EBB6438D0F431B804609AE7B8926B13BE768A77B616FF71096792A06899155B21E6062C13A55175&originRegion=us-east-1&originCreation=20211021031529 1 The solution$u\in H^1(\Omega)for the Poisson problem is continuous-differentiable and therefore we can apply the div-theorem such that we end up with the weak form \begin{align} -\int_\Omega \Delta u v dx= \int_\Omega \nabla u \nabla v \ dx - \underbrace{\int_{\Gamma} \nabla u \cdot n v \ ds}_{=0\text{ since } u=0 \text{ on } \Gamma} =\int_\Omega f v \ ... 1 The residual error estimater can help in your case. Letu_h\in V_h$be the FE solution given at a time$t$and assume you know$\frac{\partial^2 u_h}{\partial t^2}$at the same time$t$then you can define the residual at an element$Tas: \begin{align} \Delta u_h|_T - \frac{\partial^2 u_h}{\partial t^2}|_T \end{align} As you mentioned\Delta u_h\$ is not ...

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