# Normalized legendre and quadrature basis for discontinous Galerkin method

I've successfully implemented a 1D DG code with non-normalized Legendre basis and I've now moved onto developing a 2D code using tensor products. For my 2D code I've chosen to have normalized Legendre polynomials,i.e.

$$L_{0}(x) = \sqrt{\frac{1}{2}}\\ L_{1}(x) = \sqrt{\frac{3}{2}} x.$$

My question is: If we normalize the Legendre polynomials, do I also have to normalized my Gauss-Legendre weights as well? Or do we only normalized the polynomials in terms of the basis?

subroutine GaussQuad (xq,wq,n)
use parameters
implicit none
integer :: n
real(kind=8),dimension(n) :: xq,wq

integer :: i,iter
real(kind=8) :: xx
real(kind=8) :: legendre,dlegendre

do i=1,n
xx = cos(dpi*(i - 0.25d0)/(n + 0.5d0))

do iter=1,500
xx = xx - legendre(xx,n)/dlegendre(xx,n)
end do

xq(i) = xx
wq(i) = (2.0*dble(n) + 1.0)*2.0d0/((1.0d0-xx**2.0)*dlegendre(xx,n)**2.0)
end do


In order for my 2D code to work I need to use the normalized Legendre polynomials in the Gauss quadrature routine along with the $$2n + 1$$ normalization on the weights (see how its been added to wq(i)). However, I only got this to work due to an ad hoc guess. I would like to avoid this as I don't personally understand why this is required for my solver to work.

EDIT 4/1/2020:

1. In my code, I do all my operations in the reference element. In fact, double checked and my Gauss-Legendre points values are within $$[-1,+1]$$ and my weights also sum to 2.
2. To map from physical space to the reference space, I use the following $$X(\eta) = \frac{x_{i} - x_{i-1}}{2} \eta + \frac{x_{i} + x_{i-1}}{2}\\ Y(\zeta) = \frac{y_{j} - y_{j-1}}{2} \zeta + \frac{y_{j} + y_{j-1}}{2}$$
3. To integrate the volume fluxes I use a tensor product Gauss-Legendre scheme with $$M^{2}$$ points

do ix=1,nx
do iy=1,ny
do i=1,mx
do j=1,my
do inode=1,mx
do jnode=1,my
call Flux(un(:,ix,iy,inode,jnode),FFlux,GFlux)

flux_vol1(:,ix,iy,i,j) = flux_vol1(:,ix,iy,i,j) + &
& 0.5*FFlux(:)* &
& dlegendre(xg(inode),i-1)*&
& legendre(xg(jnode),j-1)*&
& wg(inode)*&
& wg(jnode)

flux_vol2(:,ix,iy,i,j) = flux_vol2(:,ix,iy,i,j) + &
& 0.5*GFlux(:)* &
& dlegendre(xg(jnode),j-1)*&
& legendre(xg(inode),i-1)*&
& wg(inode)* &
& wg(jnode)
end do
end do
end do
end do
end do
end do


Above is a snippet in how I perform the tensor product quadrature. I'm a little confused on why removing the normalizing of the Gauss quadrature is causing such severe results.
Solution:

The standard way of computing tensor products is $$u_{h}(x,y) = \sum_{k = 0}^{N} \sum_{l = 0}^{M} u_{ij}(t) \phi_{k}(x) \phi_{l}(y) (1),$$ The tensor product formula adapted in the last two papers are $$U_{h}(x,y) = \sum_{k=0}^{\frac{1}{2}(k + 1)(k + 2)} u_{ij}^{k}(t) \phi_{k}(x,y) (2)$$ Is there a reason to use formula (1) over formula (2)? Mathematically, they seem both correct.

• Quadrature weights/nodes do not depend on the normalisation of the basis functions. They depend on your reference interval. Usually it is [-1,1] since Legendre are defined on this interval and the weights then sum to 2. But if you use the reference interval as [0,1], then you must map the nodes to lie in this interval and weights must be scaled to sum to unity. Apr 1, 2020 at 6:45
• That makes sense. I included some more info to maybe help illustrate my problem. Its quite perplexing. BTW the PDE i'm solving is 2D linear advection with a Gaussian pulse (very smooth). Apr 1, 2020 at 16:02
• My answer was only about using Legendre as basis in which case normalisation does not matter. I see that you are trying to compute the weights yourself. In this case, the weights do depend on the normalisation. The formulae e.g., here en.wikipedia.org/wiki/Gaussian_quadrature#Gauss–Legendre_quadrature are without normalisation. The weights are also given in this site, you can directly use them as they are exact. Apr 2, 2020 at 3:02
• Hi CFD Lab. I managed to fix my issue. I have included what I did along with a side question. Apr 8, 2020 at 16:05
• Both types can be used in DG since there are no continuity requirements. Tensor product polynomials are normally used on quadrilateral/hexahedral elements. The other type are used on any type of elements. It is better to ask this as a new question, maybe somebody else will give more insight. Apr 9, 2020 at 5:00

So I managed to fix my problem. First, off I was missing a factor of 0.5 when calculating the surface integral. Additionally, I am using the following normalization $$\phi(x) = \sqrt{2m + 1} L(\xi)$$.