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Assume we want to solve the Poisson eq. with the FEM on some Domain $\Omega$, i.e. $$\begin{cases} -\Delta u = f, \; \Omega\\ u = 0, \; \partial \Omega \end{cases}$$ For the sake of the discussion let $\Omega = [0,1] \subset \mathbb {R}$. Let $\Omega_h := \{x_0 =0, \dots, x_N =1 \}$ be some vertices of the discrete mesh and let the cells be the respective intervals.

Now, we seek some $u_h \in V_h$ such that for all $v \in V_h$ it holds $$\int_\Omega u_h'v' \mathrm{d}x = \int_\Omega fv \,\mathrm{d} x.$$

We can decompose the functions now into a linear combination of the finite element basis $\{\varphi_i\}$, $$u_h = \sum_i \alpha_i \varphi_i,$$ whereby the $\alpha_i$ are called the DOFs. The $\alpha_i$ can be calculated by solving a linear system. So far, so good.

In literature (for example [1]), it is written that a DOF is a linear form $\alpha_i$, and in the case of nodal Lagrange elements, for example, it is just a point-wise evaluation, $$\alpha_i (u) := u(x_i).$$ Obviously, in this case, I get an explicit expression for $u$ and even the values of u on the vertices $x_i$ directly. However, there are more complicated DOFs. Consider for example the modal Lagrange elements, where $$\alpha_i (u) := \frac{1}{|\Omega|} \int_{\Omega} u \varphi_i \mathrm{d} x.$$ Inserting this into the decomposition gives $$u_h = \sum_i \frac{1}{|\Omega|} \int_{\Omega} u_h \varphi_i \mathrm{d} x \;\varphi_i.$$

I don't understand how I can recover my FEM solution from this expression. Do I ignore the exact definition of $\alpha_i$ and just take the numerical value as the basis coefficients? If so, what is the whole point of introducing them in the first place?

(PS: Disclaimer: I know, I have been a little bit sloppy with the distinction of the whole domain $\Omega$ and the physical elements and mappings from the reference element, but I did this for the sake of simplicity and I hope that it doesn't change much regarding my question.)

[1] Finite Elements I, Alexandre Ern, Jean-Luc Guermond.

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I think the symbols $\varphi_i$ in your notation are overloaded, which may be contributing to your confusion.

The choice of degrees of freedom and choice of basis are directly linked. It is perhaps easiest to think of the degrees of freedom as the coefficients of the expansion in a basis. So, starting with a basis $\{ \varphi_i \}$ for the finite element space $V_h$, an element $u_h \in V_h$ can be written $$ u_h(x) = \sum_i \alpha_i \varphi_i(x), $$ which gives the degrees of freedom $\alpha_i$.

Going the other direction, we can start with a basis for the dual space. In this case, we define $\{ \alpha_i \}$ to be a collection of linearly independent linear functionals. These linear functionals induce a basis $\{ \varphi_i \}$ satisfying the duality property $$ \alpha_i ( \varphi_j ) = \delta_{ij} $$ (where $\delta_{ij}$ is the Kronecker delta).

In the first example you give, the degrees of freedom of the nodal Lagrange elements correspond to evaluation at nodal points $\{x_i\}$, i.e. $$ \alpha_i(v) = v(x_i). $$ These degrees of freedom induce basis functions that satisfy the Kronecker delta property above. These basis functions on each element are given by the Lagrange polynomials corresponding to the nodal points.

In your example of a modal basis, the degrees of freedom are defined as the moments against certain test functions $\xi_i$, i.e. $$ \alpha_i(v) = \frac{1}{|K|} \int_K v \xi_i \, dx. $$ This choice of degrees of freedom induces a basis $\{ \varphi_i \}$. In the case that the test functions $\xi_i$ are orthogonal and scaled appropriately, then $\varphi_i = \xi_i$, but this will not hold in general.

So, in order to perform computations on a function $u_h$ given these degrees of freedom, you need to know what the basis functions $\varphi_i$ are. You can compute these in terms of some known basis using a Vandermonde-type matrix. If you have a known basis $\{ \psi_i \}$ (e.g. Lagrange polynomials, monomials, etc.), then you can form the matrix $V$ with entries $$ V_{ij} = \alpha_i(\psi_j). $$ The entries of $V^{-1}$ will give you the coefficients of each function $\varphi_i$ expanded in the known basis $\{ \psi_j \}$. In some special cases, you may know a priori the basis functions $\varphi_i$ (e.g. when they come from integrating against orthogonal polynomials), and this computation is unnecessary.

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  • $\begingroup$ Thank you very much! I think I'm beginning to understand. So is the following correct? I want to expand the solution of my PDE in term of some basis $\{\varphi_i \}$, i.e. $u = \sum_i \beta_i \varphi_i$. I order to do this, I need to construct a basis $\{\varphi_i \}$. In order to construct this basis, I start with a dual basis $\{\alpha_i \}$ and some known basis $\{\psi_i \}$ and expand each basis function $\varphi_i$ in term of the known basis $\{\varphi_j \}$. The coefficients I get by $V_{ij} =\alpha_i(\psi_j)$. [Part 1] $\endgroup$
    – welahi
    Dec 14, 2022 at 23:07
  • $\begingroup$ However, doesn't this give me an unnecessarily complicated basis transformation on a finite-dimensional space? Why just not work directly with this the known basis $\{\varphi_i \}$? [Part 2] $\endgroup$
    – welahi
    Dec 14, 2022 at 23:09
  • $\begingroup$ It depends. If your "known basis" is, for example, the monomials, it isn't really suitable for use with finite elements (poorly conditioned, unnatural to enforce continuity, etc.). In many standard cases, we know the induced basis functions already (e.g. Lagrange polynomials for nodal degrees of freedom), so you don't need to use the Vandermonde. Note that the coefficients are the entries of the inverse of the Vandermonde. $\endgroup$
    – Will P.
    Dec 15, 2022 at 4:18
  • $\begingroup$ Thanks again. Is the term "DOF" maybe also overloaded? On one hand, it is used to describe the coefficients of the "wanted" basis $\{\varphi_i\}$ in terms of the "known" basis $\{\psi_i\}$, and on the other hand it is the name of the coefficients of my PDE solution in terms of the basis $\{\varphi_i\}$, correct? $\endgroup$
    – welahi
    Dec 16, 2022 at 19:07
  • $\begingroup$ Usually "DOF" will refer to the coefficients $\alpha_i$ in the expansion $u_h = \sum_i \alpha_i \varphi_i$ (or to the interpretation of $\alpha_i$ as a linear functional). It is not usually used to describe the coefficients of one basis in terms of another. $\endgroup$
    – Will P.
    Dec 16, 2022 at 21:09

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