1D Discontinuous Galerkin - Lagrange vs Legendre Basis

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 ?

• 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. – GoHokies Mar 13 '17 at 10:24
• 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. – mfnx Nov 30 '20 at 11:47