1D Discontinuous Galerkin - Lagrange vs Legendre Basis

Question

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 ?

Answer 1

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}.