MA557/MA578/CS557
Lecture 8
Spring 2003
Prof. Tim Warburton
[email protected]
1
Discontinuous Galerkin Scheme
• Recall that under sufficient conditions (i.e. smoothness)
on an interval we obtained the density advection PDE
q
q
u
0
t
x
• We again divide the pipe into sections divided by
x1,x2,…,xN .
• On each section we are going to solve a weak version
of the advection equation.
Find i  P p  xi , xi 1  such that v  P p  xi , xi 1 
xi 1

xi
 
 
v  i  u i dx  v  xi  u  i 1  xi   i  xi  
x 
 t
2
1
3
2
x1 x2 x3
 N 1
xN-1 xN
In the i’th cell we will calculate an approximation of q,
denoted by i . Typically, this will be a p’th order
polynomial on the i’th interval. We will solve the
following weak equation for i i=1,..,N-1
Find i  P p  xi , xi 1  such that v  P p  xi , xi 1 
xi 1

xi
i 
 i
v
u
dx  v  xi  u  i 1  xi   i  xi  
x 
 t
3
Discretization of Pp
• In order to work with the DG scheme we need to choose a
basis for Pp and formulate the scheme as a finite
dimensional problem
Find i  P p  xi , xi 1  such that v  P p  xi , xi 1 
i 
 i
v

u
dx  v  xi  u  i 1  xi   i  xi  
x  t
x 
i
• We expand the density in terms of the Legendre
m p
polynomials
xi 1
i   i ,m Lm  x 
m 0
where:
i ,m is the m'th coefficient of the Legendre
expansion of the density in the i'th cell.
2 x  xi  xi 1
x
xi 1  xi
4
Discretization of Pp
• The scheme:
Find i  P p  xi , xi 1  such that v  P p  xi , xi 1 
xi 1

xi
i 
 i
v
u
dx  v  xi  u  i 1  xi   i  xi  
x 
 t
• Becomes:
Find i ,m m  0,..., p such that
 dx 1
 d i ,m 
  Ln Lm dx 


m 0 
 dt 
 dx 1
m p
m p
m p
 dx 1 dLm 

u    Ln
dx  i ,m   uLn  1  i 1,m Lm 1  uLn  1  i ,m Lm  1
dx
dx
m 0 
m 0
m 0
 1


m p
5
Simplify, simplify
• Starting with:
Find i ,m m  0,..., p such that
 dx 1
 d i ,m 
  Ln Lm dx 


m 0 
 dt 
 dx 1
m p
m p
m p
 dx 1 dLm 

u    Ln
dx  i ,m   uLn  1  i 1,m Lm 1  uLn  1  i ,m Lm  1
dx
dx
m 0 
m 0
m 0
 1


m p
• We define two matrices:
1
M nm
dx
  Ln  x  Lm  x  dx
dx 1
Dnm
dx
d
  Ln  x  Lm  x  dx
dx 1
dx
1
1
dx dx
d

Ln  x  Lm  x  dx

dx dx 1
dx
1
  Ln  x 
1
d
Lm  x  dx
dx
6
Then:
Find i ,m m  0,..., p such that
 dx 1
 d i ,m 
  Ln Lm dx 


m 0 
 dt 
 dx 1
m p
m p
m p
 dx 1 dLm 

u    Ln
dx  i ,m   uLn  1  i 1,m Lm 1  uLn  1  i ,m Lm  1
dx
dx
m 0 
m 0
m 0
 1


m p
Becomes:
Find i ,m m  0,..., p such that
m p
m p
m p
d i ,m 

 M nm
  u   Dnm i ,m   uLn  1  i 1,m Lm 1  uLn  1  i ,m Lm  1

dt 
m 0 
m 0
m 0
m 0
m p
7
Questions
• This raises the following questions how can
we compute Mnm and Dnm
• We will need some of the following results.
8
The Legendre Polynomials
(used to discretize Pp)
• Let’s introduce the Legendre polynomials.
• The Legendre Lm  x  , m  0,1,... polynomials
are defined as eigenfunctions of the singular
Sturm-Liouville problem:
d 
dLm

 x    m(m  1) Lm  x   0
 1  x 1  x 
dx 
dx

9
Closed Form Formula for Lm
• If we choose Lm(1) =1 then:
1 l  floor ( m / 2)
l  m  2m  2l  m2 l
Lm  x   m   1  
x

2
l 0
 l  m 
• Examples:
L0  x   1
1
0  1  2  1 
L1  x     1    x   x
2
 0  1  
1
3x 2  1
0  2  4  2
1  2  2  0 
L2  x   2   1    x   1    x  
2 
2
 0  2 
 1  2  
• Warning: this is not a stable way to compute
the value of a Legendre polynomial
(unstable as m increases)
10
Stable Formula for Computing Lm
• We can also use the following recurrence
relation to compute Lm
2m  1
m
Lm1  x  
xLm  x  
Lm1  x 
m 1
m 1
• Starting with L0=1 L1=x we obtain:
3
1
1
2
L2  x   xL1  x   L0  x    3x  1
2
2
2
• This is a more stable way to compute the
Legendre polynomials at some x.
11
The Rodriguez Formula
• We can also obtain the m’th order Legendre
polynomial using the Rodriguez formula:
m
1 dm 2
Lm  x   m
x  1
m 
2 m! dx
12
Orthogonality of the Legendre Polynomials
• The Legendre polynomials are orthogonal to
each other:
1
p
m
1 dp 2
1 dm 2
1 Lp  x  Lm  x  dx  1 2 p p! dx p  x  1 2m m! dx m  x  1 dx
1
m
p d
m
1
1 1 dp 2
2
 p
x

1
x

1
dx


m
p 
m 


1
2 p ! 2 m! dx
dx
m 1
p d
m
1
1 1 d p 1 2
2
 p
x

1
x

1
dx


m
p 1 
m 1 


1
2 p ! 2 m! dx
dx
  1
p
m p
p d
m
1
1 1 2
2
x  1
x  1 dx

p
m
m p 


1
2 p ! 2 m!
dx
0
• Since 2m < m+p
13
Miscellaneous Relations
Lm  x   1,
L m  1   1
-1  x  1
m
m  m  1
dLm
,
 x 
dx
2
-1  x  1
dLm
m m  m  1
 1   1
dx
2
1
2
2
1 Lm  x  dx  2m  1
14
Differentiation
• Suppose we have a Legendre expansion for
the density in a cell:
m p
i   i ,m Lm  x 
m 0
• We can find the coefficients of the Legendre
expansion for the derivative of the function
m p
as:
d i
x 
 (1) L x
dx
where:
  
i ,m
m
 
m 0
i(1)
, m   2m  1
j p

i , j
j  m 1
j  m odd
15
Coefficient Differentiation Matrix

(1)
i ,m
j p
2

 2m  1  i , j
xi 1  xi
j  m 1
j  m odd
(2m  1) if j>m and j+m odd
ˆ
Dmj  
0 otherwise

j p
2
(1)
i ,m 
Dˆ mj i , j

xi 1  xi j 0
or ρi(1) 
2
ˆ in matrix-vector notation
Dρ
i
xi 1  xi
i.e. if we create a vector of the Legendre coefficients for
rho in the I’th cell then we can compute the Legendre
coefficients of the derivative of rho in the I’th cell by
pre-multiplying the vector of coefficients with the matrix:
2
ˆ
D
xi 1  xi
16
The D Matrix
1
dx

Ln  x  Lm  x  dx
dx 1
• We can use a simple trick to compute D
M nm
• Recall:
dx
d
Dnm 
L
x
Lm  x  dx


n

dx 1
dx
1
1
dx
d
Dnm   Ln  x  Lm  x  dx
dx 1
dx
simple change of variables
1
• We use:
dx dx
d

L
x
Lm  x  dx


n

dx dx 1
dx
1
  Ln  x 
1
d
Lm  x  dx
dx
j p
  M nj Dˆ jm
j 0

2 ˆ
Dnm
2n  1
17
The Vandermonde Matrix
• It will be convenient for future applications to be able
to transform between data at a set of nodes and the
Legendre coefficients of the p’th order polynomial
which interpolates the data at these nodes.
• Consider the I’th cell again, but imagine that there are
p+1 nodes in that cell then we have:
m p
i  xn    i ,m Lm  xn  n=0,...,p
m 0
• Notice that the Vandermonde matrix defined by Vnm  Lm  xn 
is square and we can obtain the coefficients by
multiplying both sides by the inverse of V
m p
i  xn    Vnm i ,m
m 0
n p
 i ,m   V 1  i  xn 
n 0
mn
18
Vandermonde
• Reiterate:
– that the inverse of the Vandermonde matrix is
very useful for transforming nodal data to
Legendre coefficients
– that the Vandermonde matrix is very useful for
transforming Legendre coefficients to nodal
data
19
Results So Far
• 0<=n,j,m<=p
Mass matrix:
x x
 i 1 i  Ln  x Lm  x  dx  0
2 1
1
M nm
(n  m)
 xi 1  xi 2
n=m 


=  2
2n  1
 0
(n  m)
Coeff. Differentiation
(2m  1) if  j>m  and  j+m odd 
Dˆ mj  
0 otherwise

1
Dnm   Ln  x 
Stiffness matrix
Vandermonde matrix
1
dLm
 x  dx
dx
 2  ˆ

 Dnm
 2n  1 
Vnm  Lm  xn 
20
Homework 3
Q1) Code up a function (LEGENDRE) which
takes a vector x and m as arguments and
computes the m’th order Legendre
polynomial at x using the stable recurrence
relation.
Q2) Create a function (LEGdiff) which takes p
as argument and returns the coefficient
differentiation matrix D
Q3) Create a function (LEGmass) which takes p
as argument and returns the mass matrix M
21
Homework 3 cont
Q4) Create a function LEGvdm which returns the following matrix:
Vij  L j  xi  0  i,j  p
Q5) Test them with the following matlab routine:
3)
Compute the Chebychev nodes
4)
Build the Vandermonde matrix using the
Chebychev nodes and Legendre polynomials
7)
Build the coefficient differentiation matrix
9) Build the Legendre mass matrix
12) Evaluate x^m at the Chebychev nodes
14) Compute the Legendre coefficients
16) Multiply by Dhat (i.e. differentiate in coeff space)
18) Evaluate at the nodes
20) Compute the error in the derivative
22
Homework 3 cont
Q5cont) Explain the output from this test case
23
Next Lecture (9)
• We will put everything together to build a 1D
DG scheme for advection…
24