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Consider the 1D advection equation in its strong and weak forms

$$\ u_t + a u_x = 0 $$ $$\ \int_{x_{j-0.5}}^{x_{j+0.5}} w u_t \ dx - a \int_{x_{j-0.5}}^{x_{j+0.5}} w_x u \ dx + a [w(x)\hat{u}]_{x_{j-0.5}} ^{x_{j+0.5}} = 0 $$

Having introduced the piecewise constant Lagrange polynomials or the first two Legendre Polynomials as our test functions, the system can be written as :

$$ \begin{bmatrix} \int (\phi_1)^2 dx \ \int \phi_1 \phi_2 dx \\ \int \phi_2 \phi_1 dx \ \int (\phi_2)^2 dx \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}_t - a \begin{bmatrix} \int (\phi_1)_x \phi_1 dx \ \int (\phi_1)_x \phi_2 dx \\ \int (\phi_2)_x \phi_1 dx \ \int (\phi_2)_x \phi_2 dx \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + a \begin{bmatrix} \phi_1(x_{j-0.5}) \ \phi_1(x_{j+0.5}) \\ \phi_2(x_{j-0.5}) \ \phi_2(x_{j+0.5}) \end{bmatrix} \begin{bmatrix} -\hat{u}_{x_{j-0.5}} \\ \hat{u}_{x_{j+0.5}} \end{bmatrix} = 0 $$

When solving through the use of the Lagrange polynomial basis, the values $ u_1 $ and $ u_2 $ that are being solved for are the values of the function $ u(x,t) $ at the nodes. In the case of the Legendre Polynomials, what are we solving for? What is the physical significance of $ u_1 $ and $ u_2 $ and how do they relate to the function values at the nodes, cell average and flux ?

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    $\begingroup$ are you asking about nodal vs. modal DG? if so, then the DG book by Hesthaven and Warburton or these slides may be of help to you. $\endgroup$
    – GoHokies
    Mar 13, 2017 at 10:24
  • $\begingroup$ I think you have to place the nodes depending on which order you use. For instance, if you used 0th order Legendre basis, you would be solving for 1 node placed at the center of the element. I've been looking for an answer to this question since a few weeks now, and I'm not sure this is the correct answer. $\endgroup$
    – mfnx
    Nov 30, 2020 at 11:47

1 Answer 1

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To illustrate, consider the 1D case with linear approximations on the elements, i.e., polynomials of degree 1. Furthermore, given the formulas you wrote down, I assume that your reference element lies on the $[0, 1]$ interval.

Lagrange polynomials

You already noted that for the Lagrange Base the approximation on the $i$'th element corresponds to esentially interpolating between the nodes, i.e., $$u_i(t,x) = \sum_{j=0}^{N} u_{i,j}(t) l_j(x) \tag{1} \label{1}$$ (here $N=1$) where $$ l_j(x) := \prod_{m = 0, m \neq j}^N \frac{x - x_m}{x_j - x_m}.\tag{2}$$ To keep the notatin relatively compact, I will suppress the temporal dependency of the $u_{i,j}$ for now.

Clearly, the Lagrange polynomials $l_j(x)$ fulfill a Kronecker property at the interpolation nodes $x_m$ and thus also, the finite element approximation $u_i(x)$ fulfills $u_i(x_m) = u_{i,m}(t)$ since $$u_i(x_m) = \sum_{j=0}^{N} u_{i,j} l_j(x_m) = \sum_{j=0}^{N} u_{i,j} \delta_{jm} = u_{i,m}.\tag{3}$$

I would like to stress that until now, we have not decided where to place the nodes $x_j$! These nodes are in fact usually determined from the choice of the quadrature rule used for approximating the integrals in the weak form. Typical choices include Gauss-Lobatto or Gauss-Legendre. The reason to choose the same nodes $x_m$ for interpolation and quadrature (collocation) becomes appearant when you take a look at the weak formulation $$ \int_0^1 u_t(t,x) w(x) dx + \int_0^1 a u(t,x)_x w(x) dx = 0 \quad \forall w \in W^1 \tag{4} \label{4} $$ where here, it makes sense to choose the Lagrange polynomials as a basis for the testspace $W^1$. To get the idea, it suffices to consider the first term in \eqref{4}, for a detailed discussion you can take a look at this documentation. Alright, let's substitute \eqref{1} into $\int_0^1 u_t(t,x) w(x) dx$ and test with the basis of functions $l_k(x)$ of $W^1$: \begin{align} &\int_0^1 \sum_{j=0}^{N} u_{i,j}(t)' l_j(x) l_k(x) d x, &k = 0, 1 \tag{5}\\ \overset{\text{Quadrature of sufficient order}}{=}& \sum_{l=0}^N w_l \sum_{j=0}^{N} u_{i,j}(t)' l_j(x_l) l_k(x_l) &k = 0, 1 \tag{6} \\ \overset{\text{Same nodes for interpolation and quadrature}}{=}& \sum_{l=0}^N w_l \sum_{j=0}^{N} u_{i,j}(t)' \delta_{jl} \delta_{kl} &k = 0, 1 \tag{7}\\ =& w_k u_{i,k}(t)' &k = 0, 1 \tag{8} \end{align}

Which is a drastic, yet correct simplification of the integration which motivates the use of identical nodes for interpolation and quadrature.

Legendre Polynomials

Now, let's consider Legendre polynomials. The first two Legendre polynomials on interval $[-1, 1]$ are given by $P_0(x) = 1, P_1(x) = x$. From this definition it is clear that the $P_j(x)$ (in particular $P_0(x)$) can not fulfill a Kronecker property at any choice of nodes. Instead, the Legendre polynomials are orthogal with respect to the $L^2$ scalar product: $$ \int_{-1}^1 P_i(x) P_j(x) d x = \frac{2}{2k+1} \delta_{ij}.\tag{9}$$ This is a big qualitative difference compared to the Lagrange basis!

Let's map the interval $[0,1]$ onto $[-1, 1]$ through $$ \xi \mapsto 2\xi -1, [0, 1] \rightarrow [-1, 1] \tag{10}$$ and define the basis functions $\phi_k(x), k = 0,1$ as $$\phi_k(x) := \sqrt{2k+1} P_k(2x - 1) \tag{11}$$ which posess the Kronecker property $$\int_0^1 \phi_i(x) \phi_j(x) dx = \delta_{ij}, \tag{12}\label{12}$$ i.e., they form a proper basis of $W^1:=\{w \in L^2([0, 1]) : w \in \mathcal{P}^1\}$ where $\mathcal{P}^p$ denotes the vector space of polynomials of degree $p$.

Again, we expand the approximation $u_i(t, x)$ on the $i$'th interval in these base functions as

$$u_i(x) = \sum_{j=0}^{N} u_{i,j} \phi_j(x). \tag{13} \label{13}$$ Now, let's consider again the first term in the weak formulation \eqref{4} and test with the $\phi_l(x)$ (since they form a basis of our test-space $W^1$): \begin{align} & \int_0^1 \sum_{j=0}^{N} u_{i,j}' \phi_j(x) \phi_l(x) d x & l = 0, 1 \tag{14}\\ =& \sum_{j=0}^{N} u_{i,j}' \int_0^1 \phi_j(x) \phi_l(x) d x & l = 0, 1 \tag{15}\\ \overset{\eqref{12}}{=}& \sum_{j=0}^{N} u_{i,j}' \delta_{jl} & l = 0, 1 \tag{16} \\ =& u_{i,l}' &l = 0, 1 \tag{16} \end{align}

Answer

So to finally answer your question, the $u_{i,j}$ can be seen as a weight that quantifies the influence of the $j$'th basis function $\phi_j(x)$ when tested with $u_i(t, x)$. In contrast to Lagrange, the $u_{i,j}$ provide sort of a "global" statement on $u_i$, whereas in the Lagrange case, this requires the integration weights $w_j$.

Values at nodes, interfaces and average can then be simply computed from \eqref{13}.

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    $\begingroup$ That's a nice answer, however a more concrete answer would be: (1) In case of Legendre polynomials $u_1$ and $u_2$ are the constant and linear contributions to the final solution. (2) $u_1$ is the cell average. (3) There is no simple connection between the modal values and the physical/numerical flux. $\endgroup$
    – ConvexHull
    Oct 26, 2022 at 17:30
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    $\begingroup$ (1) I elaborated that much to show the difference between both approaches. Of course, you could just say that $u_0, u_1$ are the coefficients of the constant and linear part, but that gives about as much information as saying $u_0, u_1$ are the contributions of the first and second Lagrange polynomial. (2) Correct, I might add that to my answer. (3) Well I mean that's what approximate Riemann solvers are for, right? You provide trace values and then they take over the business. $\endgroup$
    – Dan Doe
    Oct 26, 2022 at 18:13
  • $\begingroup$ My intention was merely to complement your very good, detailed answer, especially with regard to the original very specific question. ;) $\endgroup$
    – ConvexHull
    Oct 26, 2022 at 18:36
  • $\begingroup$ I didnt mean to sound that rude, the character limit got me :D $\endgroup$
    – Dan Doe
    Oct 26, 2022 at 19:47

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