1
$\begingroup$

I was curious if someone could share their opinion on this matter. I have noticed that some people in literature normalize their Legendre polynomials, i.e. divide or multiply the polynomial by $$\sqrt{\frac{2n+1}{2}},$$ where $n$ is the order of the polynomial. I am quite new to DG, but I am experiencing better results when I normalize my Legendre polynomial than when I don't. Is there a particular reason why people normalize?

Density at $T = 0.3$ with normalized Legendre polynomial. enter image description here

Density at $T = 0.3$ with unnormalized Legendre Polynomial enter image description here

In terms of the resolution, I am running on a $50\times 50$ grid points.

Polynomial subroutine:

function legendre (x,n)
integer :: n
real(kind=8) :: x
real(kind=8) :: legendre
x = min(max(x,-1.0),1.0)
select case(n)
case(0)
 legendre = 1.0
case(1)
 legendre = x
case(2)
 legendre = 0.5*(3*x**2-1)
case(3)
 legendre = 0.5*(5.0*x**3-3.0*x)
case(4)
 legendre = 0.125*(35.0*x**4-30.0*x**2+3.0)
case(5)
 legendre = 0.125*(63.0*x**5-70.0*x**3+15.0*x)
case(6)
 legendre = 1.0/16.0*(231.0*x**6-315.0*x**4+105.0*x**2-5.0)
end select
legendre = sqrt((2.0*dble(n)+1.0)/2.0)*legendre
return
end function legendre

function legendre_prime (x,n)
integer :: n
real(kind=8) :: x
real(kind=8) :: legendre_prime
x = min(max(x,-1.0),1.0)

select case(n)
case(0)
  legendre_prime = 0.0
case(1)
  legendre_prime = 1.0
case(2)
  legendre_prime = 3.0*x
case(3)
  legendre_prime = 0.5*(15.0*x**2-3.0)
case(4)
  legendre_prime = 0.125*(140.0*x**3-60.0*x)
case(5)
  legendre_prime = 0.125*(315.0*x**4-210.0*x**2+15.0)
case(6)
  legendre_prime = 1.0/16.0*(1386.0*x**5-1260.0*x**3+210.0*x)
end select
legendre_prime = sqrt((2.0*dble(n)+1.0)/2.0)*legendre_prime
return
end function legendre_prime
Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Are you using explicit or implicit schemes ? For explicit schemes, it does not matter. For implicit schemes, the matrix conditioning may depend on this scaling. Can you elaborate in what way your results are worse/better and for what problem ? $\endgroup$
    – cfdlab
    Feb 13, 2019 at 11:22
  • 1
    $\begingroup$ What problem are you solving? Are you doing modal or nodal DG? $\endgroup$
    – Vikram
    Feb 13, 2019 at 16:20
  • $\begingroup$ Sorry for the late replies. I am solving the 2D compressible Euler equations for an ideal gas. In this case I am solving a 4 shock Riemann problem as a means to benchmark my code. In terms of the time integration, I am using a 3rd-order SSP Runge-Kutta method. I have edited my post to include solutions at T = 0.3 of the normalized and unnormalized polynomials. $\endgroup$
    – Simon
    Feb 19, 2019 at 2:54
  • 1
    $\begingroup$ Are you using a limiter ? If yes, then you have to be careful with its implementation since it depends on the normalization used. $\endgroup$
    – cfdlab
    Feb 19, 2019 at 5:12
  • $\begingroup$ GIven I am dealing with a Riemann problem, as far as I know I have to use a limiter to ensure stability. The type of limiter I am using is the TVB Minmod limiter. In terms of the normalization, does it matter if its multiplicative or not? I have edited my post to share the way I am doing the normalization process. $\endgroup$
    – Simon
    Feb 21, 2019 at 18:07

1 Answer 1

Reset to default
2
$\begingroup$

You have to careful while applying TVD limiter and account for the normalization of your basis functions. With reference to [1], if your solution is as $$ u_h(x,y) = \bar{u} + u_x \phi_i(x) + u_y \psi_j(y) $$ where $$ \phi_i(x) = \frac{x - x_i}{\Delta x_i/2}, \qquad \psi_j(y) = \frac{y - y_j}{\Delta y_j/2} $$ Here $\bar{u}, u_x, u_y$ are the dofs or solution variables and $\bar{u}$ is the cell average value. You limit the slope as $$ u_x = minmod(u_x, \bar{u}_{i,j} - \bar{u}_{i-1,j}, \bar{u}_{i+1,j} - \bar{u}_{i,j}) $$ Due to your normalization, your solution is of the form $$ u_h(x,y) = \bar{u} \sqrt{1/2} + u_x \sqrt{3/2} \phi_i(x) + u_y \sqrt{3/2} \psi_j(y) $$ Now $\bar{u}$ is not the cell average value. You have to account for these differences and properly modify the limiter step.

[1] Bernardo Cockburn and Chi-Wang Shu, The Runge–Kutta Discontinuous Galerkin Method for Conservation Laws V, JCP, 141, 1998 Multidimensional Systems

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.