# Tag Info

7

First of all, interpolation and approximation are slightly different from each other. Given a sufficiently smooth function $f$ (sufficiently smooth just means that I am covering my bases, there are many theorems in approximation theory considering different function classes, from discontinuous -measure zero- functions to infinitely smooth or analytic or ...

7

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 ...

5

This will implicitly depend on your function, $F$, as well as the method you used to derive your $f$ and $g$. If your $F$ were, say, the constant function $F(y,z)=1$, then you'd be guaranteed to have no error whatever you do. On the other hand, if you have a discontinuous $F$, say $$F(x,y)=\begin{cases} 1&y\leq0,\\ 0& y>0, \end{cases}$$ then ...

5

Since you are trying to find the gradient of the function in the vertices of a triangular mesh. So, we are considering an interpolation over the dual mesh defining the value on each centroid, you can try those area based interpolation function e.g, Wachspress Interpolation function. The benefit of this interpolation function is that it is general enough for ...

4

Since you are on a uniform $x-y-z$ grid, you are in luck that others have had similar issues before. Specifically, I would suggest that you take a look at HDF5 for a low-level way of storing this kind of data as arrays. But in the end, you want to not just store an array, but actually do something with it: interpolate. For this, a higher-level approach would ...

3

This is what Griewank et al. call "Piecewise linearization in secant mode", see for instance https://opus4.kobv.de/opus4-zib/files/6164/newton_secant_approx_paper.pdf. The aim of that research was to capture the kinks of absolute value operations with the same precision a tangent or a secant captures the local behavior of a smooth function, with an ...

3

You can treat each of the interpolation nodes as points in the complex plane, i.e., $z_k = \exp(2\pi i k/ 6)$, then form the Vandermonde matrix $V_{kl} = z_k^l$, for $k, l = 1, 2, ..., 6$, and solve $Vc = f$ for the vector of coefficients $f$. Note that in this case, the Vandermonde matrix obeys $V^HV = 6I$ and the interpolation is equivalent to ...

3

As noticed by @AbdullahAliSivas the set of basis functions is not unisolvent. In 1d case to obtain an polynomial interpolation of degree $N$ you just need of $N+1$ distinct points, $x_i \neq x_j$ if $i \neq j$. Changing the points disposition change the quality of the interpolation, because change the Lebesgue's costant. Think to Runge's phenomenon. Going to ...

3

Let's use the polar coordinates ($\rho,\theta$), then for the six vertices of the regular hexagon defined by $\rho$=1 and polar angles $\theta_i$ there are the following equations $f(\theta_i) = a_0 \cos^2(\theta_i) + a_1 \sin^2(\theta_i) + a_2 \cos(\theta_i) \sin(\theta_i) + \\ + a_3 \cos(\theta_i) + a_4 \sin(\theta_i) + a_5 = f_i$ where $f_i$ is the ...

2

You could find the shape functions proposing a quadratic function of the form $$N_i = a_0 + a_1 r + a_2 s + a_3 rs + a_4 r^2 + a_5 s^2\, ,$$ and enforcing the following condition $$N_i(r_j, s_j) = \delta_{ij}\, ,$$ that is, 1 for each node and 0 for the others. I am using as reference element a triangle with vertices $(0, 0), (0, 1), (1, 0)$, and assuming ...

2

You need either a cubic term or 3D coordinate system to successfully interpolate in a hexagon domain. I do not know how this can be proven mathematically. The following two papers might offer some insights. Kamiński, M. (2005). Hexagonal finite elements in heat conduction. International communications in heat and mass transfer, 32(9), 1143-1151. Yang, C., &...

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 $$\sum_m \sqrt{\int_{T_m} f(x)^2 } \neq \sqrt{ \sum_m \int_{T_m} f(x)^2 }.$$

2

Potentially related and useful (I gave these resources to my students when teaching Intro to Computational Mathematics, kinda useful pedagogically too): "Automatic Source-to-Source Error Compensation of Floating-Point Programs" by Laurent Thévenoux, Philippe Langlois and Matthieu Martel : https://hal.archives-ouvertes.fr/hal-01158399/document ...

2

QUICK utilizes the two upwind nodes, $x_U$ and $x_C$, and the downwind node, $x_D$ for a quadratic interpolation at the control volume face, $x_f$. Utilizing the Lagrange polynomial form of a quadratic: \begin{multline*} \phi ({x_f}) = \left[ { - \frac{{\left( {{x_f} - {x_C}} \right)\left( {{x_D} - {x_f}} \right)}}{{\left( {{x_C} - {x_U}} \right)\left( {{...

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

I did something similar for reporting stress in a linear elasticity finite element program, which approximates stress as a constant quantity across a linear tetrahedral element. It is often desirable to plot stress as a continuous quantity, because a smoothly varying quantity is more believable. I figured this was a good compromise. This approach should be ...

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