Transcript Numerical Methods for Partial Differential Equations

Numerical Methods for Partial
Differential Equations
CAAM 452
Spring 2005
Lecture 15
Instructor: Tim Warburton
CAAM 452 Spring 2005
Two-Dimensional Advection
• Recall in one-dimensions we chose an arbitrary
section of a pipe.
• We monitored the flux of fluid into and out of the ends
of the pipe.
• The conservation equation in 1D we derived was:
b
d
C  u  b, t  C  b, t   u  a, t  C  a, t 

dt a
• Now we start with a 2D domain and consider any
simply connected sub-region:
CAAM 452 Spring 2005
Our 2D Domain 

CAAM 452 Spring 2005
Our Sub-region 


CAAM 452 Spring 2005
Outward Normal n

n

CAAM 452 Spring 2005
Conservation
• Suppose the fluid has a mean velocity u everywhere.
• This time u is a two-vector.
n=outward pointing normal

u
• The flux of fluid through the boundary is the integral of the
concentration C being advected normal to the surface of w:
flux     u  n  CdS

CAAM 452 Spring 2005
Conservation Law
• Now we know how much of the concentrate is
being advected through the boundary of our subregion.
d
CdV     u  n  CdS

dt 


• The conservation law is now:
d
CdV     u  n  CdS

dt 

     uC  dV

• We apply the divergence theorem (beware this
relies on smoothness arguments):
CAAM 452 Spring 2005
In Component Form
• Assuming ubar is constant this becomes:
u 
u 
v 
  
 x 
 
  
 x 
d
CdV     u   CdV

dt 

 
 
    u  v  CdV
y 
  x
CAAM 452 Spring 2005
Finalize
• For all sub-regions:
 
d
 
CdV    u  v  CdV

dt 
x
y 

 C
C
C 
 
u
v
 dV  0
t
x
y 

• Assuming enough continuity we obtain:
C
C
C
u
v
0
t
x
y
CAAM 452 Spring 2005
Basic Technology For Meshes in 2D
• Our target is to be able to use triangulations of a
domain in 2D.
• For example here we have a 5 triangle mesh of an
L-shape domain:
CAAM 452 Spring 2005
Global Vertex Numbering
• In order to represent the topology of the triangles
we create a global numbering of the unique
vertices in the mesh: 3
5
4
6
7
1
2
• Notice – the allocation of number to vertex is
arbitrary.
CAAM 452 Spring 2005
Triangle as Ordered Triplet
• We next label each triangle in the mesh (numbers
in triangles):
3
5
1
7
4
2
4
5
3
6
1
2
• Again the labeling of the triangles is in an arbitrary
order.
CAAM 452 Spring 2005
Triangle as Ordered Triplet cont
We can represent each triangle as a
triplet of integers which are the global
vertex numbers of the triangle in a
counter-clockwise ordering:
Triangle 1:
Triangle 2:
Triangle 3:
Triangle 4:
Triangle 5:
7,5,3
2,5,7
4,6,1
1,7,4
2,7,1
3
5
1
7
4
2
4
5
3
6
1
2
CAAM 452 Spring 2005
Class Exercise Part 1
Given this element to vertex representation of the mesh,
can you determine determine the topology of the triangle (i.e. what
might the mesh look like)?
Triangle 1:
Triangle 2:
Triangle 3:
Triangle 4:
2,3,6
5,3,2
6,3,1
2,6,4
What is the cost of your algorithm?
CAAM 452 Spring 2005
Class Exercise Part 2
Given this element to vertex representation of the mesh,
can you determine which triangles share an edge:
Triangle 1:
Triangle 2:
Triangle 3:
Triangle 4:
Triangle 5:
Triangle 6:
Triangle 7:
1,2,9
2,3,9
3,4,5
3,5,6
3,6,9
9,6,7
8,9,7
What is the cost of your algorithm?
CAAM 452 Spring 2005
Naïve Algorithm
• We could naively try a brute force search.
• Two triangles will share an edge if they both contain
two vertices in common.
a
b
• So the obvious algorithm is:
CAAM 452 Spring 2005
Alogithm 1
Brute force – test to see which out of
all the edges connects to a given
edge.
Cost approx. 9*K^2
We can trim the number of tests
by ½ easily (loop elmt2=elmt1+1:K
and store result in (elmt1,edge1)
and (elmt2,edge2).
Very loopy so Matlab will be very
slow.
CAAM 452 Spring 2005
Alogithm 2
Set up the edge to node matrix
using Matlab’s sparse matrix facility.
Thinking about the connectivity
matrices as sparse matrices allows
us to easily find other connectivities:
NodeToEdge =
transpose(EdgeToNode)
…
CAAM 452 Spring 2005
Topology Determined
• So with either approach we now know the topology of the
triangles (i.e. which triangles share an edge).
• Consider the advection equation: q
 c q  0
t
or it's conservation law version:
d
qdA    c  nqds

dt Tk
Tk
• We can use this as the foundation of an upwind finite
volume solver:
k ,e
k ,e

c

n

c

n
dqk
1
   
dt
Tk e1 edge e 
2
where ke  ElmtToElmt(k , e)
e3

  qke  qk 

CAAM 452 Spring 2005
Geometric Factors
• With Tk being the k’th triangle we need to be able to
compute Tk and the outward facing unit normals on
each edge n k ,e
 c  n k ,e  c  n k ,e 
dqk
1 e 3

 

dt
Tk e1 edge e 
2
where ke  ElmtToElmt(k , e)
v
n k ,3
v k ,1
k ,2
k ,1


v

v
1
y
y
k ,1
n  k ,1 
where Lks ,1  v k ,2  v k ,1

k
,2
k
,1
Ls    vx  vx  
k ,3
n k ,2
n k ,1
  qke  qk 


v
k ,2
n k ,2
k ,3
k ,2
1  vy  vy 
 k ,2 
where Lks ,2  v k ,3  v k ,2

k
,3
k
,2
Ls    vx  vx  
n k ,3
k ,1
k ,3
1  vy  vy 
 k ,3 
where Lks ,3  v k ,1  v k ,3

k
,1
k
,2
Ls    vx  vx  
CAAM 452 Spring 2005
Computing the Area of a Triangle
• We can use a simple geometric argument to
determine the area of a triangle:
1 k ,2
v  v k ,1    v k ,3  v k ,1 

2
1
  vxk ,2  vxk ,1  v yk ,3  v yk ,1    v yk ,2  v yk ,1  vxk ,3  vxk ,1 
2
Tk 
CAAM 452 Spring 2005
cont
•
We make some approximations:
1) that q is in fact constant over the element
2) Euler-forward in time
and we obtain:
k ,e
k ,e

 n
c

n

c

n
dt
n1
n
qk  qk   Tk ,e 
qke  qkn 




Tk e1
2


where Tk ,e is the length of the e'th edge on triangle k
e3
CAAM 452 Spring 2005
Notes on the Geometric Factors
We use the formula: area of triangle = ½*(base*perpendicular height)
Lk ,2
h k ,1
hk ,2
h k ,3
Lk ,1
Lk ,1
Lk ,1
2


1 k ,1 k ,1 h k ,1
Tk
h L
2
Lk ,3
Lk ,2
Lk ,2
2


1 k ,2 k ,2 h k ,2
Tk
h L
2
Lk ,3
Lk ,3
2


1 k ,3 k ,3 h k ,3
Tk
h L
2
The ratio of the edge length to triangle area gives 2 divided by the perpendicular height
from the edge to the vertex not on the edge.
CAAM 452 Spring 2005
cont
• Simplifying:
k ,e
k ,e

c

n

c

n
qkn1  qkn  dt  
k ,e

h
e1 
e3
 n
n
  qke  qk 

• This looks very much like a one-dimensional
scheme applied at each edge – which it is.
• At each edge we use the component of the
advection velocity in the direction of the outwards
facing normal.
CAAM 452 Spring 2005
Project 1- Overview
• I grabbed a polygonal description of part of coast line
of the UK and Eire:
• I then used the Triangle
program by Shewchuk
to generate a triangle
mesh.
• Your task: solve the linearized
Euler equations with a pressure
pulse set off near the coast.
http://www.mar.dfo-mpo.gc.ca/science/ocean/coastal_hydrodynamics/resolute/resolute.html
CAAM 452 Spring 2005
Project 1 - Equations
• The linearized Euler equations we will consider are:
u
 p
0
     
v 
0 
p  0, BC: unx  vny  0
t   x   y  
 p
u
v
• where p is the pressure and u,v are the x and y
components of the water velocity.
• The boundary condition states that at the boundary
the velocity field is tangential to the boundary.
CAAM 452 Spring 2005
cont
• We write a first order finite-volume scheme for the
linearized Euler equations using upwind fluxes for:
u
 p
0
     
v 
0 
p  0, BC: unx  vny  0





t   x   y  
 p
u
v
• as (after board demo of characteristic treatment):
k ,e
 ukn1   ukn 

n
x 
e 3
1  k ,e  k , e n
 n1   n 
n
k ,e
n
n
n
n
v

v

dt
n
n
u

u

n
v

v

p

p






x
ke
k
y
ke
k
ke
k 
k ,e  y 
 k   k 
e 1 h
 p n1   p n 
 1 
k
k

  




Boundary condition:
ukne  ukn , vkne  vkn , pkne  pkn at the domain boundary
CAAM 452 Spring 2005
Project Tasks
Q1) Write code to read in the mesh from a .neu mesh
file (more about that next time)
Q2) Write start up code to compute:
– The area of each triangle
– The length of each edge of each triangle
– The outwards facing unit normal at each edge of
each triangle
Q3) Implement the first order finite volume scheme
for the linearized acoustics. Use a dt which obeys:
0  dt  C min  h
k ,e
k ,e

1
with safety factor C such that 0  C 
2
CAAM 452 Spring 2005
Project Tasks cont
Q4) By generating a sequence of increasingly refined
meshes for a square [-1,1]x[-1,1] find the solution
order of accuracy of the scheme.
A suitable initial condition is:
0
 u  x, y,0   

 v x, y,0   

0



 

 p  x, y,0    cos  x  cos  y  

 

The solution for all time is:




 sin  x  cos  y  sin  t 2 / 2 
 u  x, y , t   

 v x, y, t    cos  x sin  y sin  t 2 / 2 
     
 

 p  x, y , t   

  cos  x  cos  y  cos  t 2 




CAAM 452 Spring 2005
Project Tasks cont
Q5) Use the ocean mesh and model a pressure pulse
(Gaussian profile) off the coast of England
• Repeat with 3 increasingly refined meshes to
visually display and compare results.
• Plot snap shots at 4 time intervals showing the
pulse spreading and interacting with the coast line.
• I will now demo how to build meshes with some
in house software based on Shewchuk’s
triangle mesh generator.
CAAM 452 Spring 2005
Project Details
• You may work in groups of 1 or 2.
• No groups of 3 or more will be permitted.
• The project is due: 03/29/05
• Be prepared to present your results (i.e. make a 5
minute PowerPoint presentation).
• Prepare a report – and I strongly suggest using the
Latex stylesheet available from the website.
CAAM 452 Spring 2005
More Advanced Than FV
• As you will find out the finite volume method is
robust, but not very accurate.
• To increase the solution order of accuracy we are
going to use the discontinuous Galerkin method
(DGM or DG)
• The idea: on each triangle we create a p’th order
polynomial expansion local to the triangle.
• We then use a flux type formulation to exchange
information between triangles.
CAAM 452 Spring 2005
Basics
• First we need to discuss how to create local
polynomial approximations on each triangle.
• i.e. we need a robustly computable basis for
P p T  = x i y j
T
0i+j p
• In finite volume and finite element methods it is
common to perform calculus type operations
(integration, differentiation and interpolation) on a
standard element and map the results to a physical
element.
CAAM 452 Spring 2005
Reference Triangle
• The following will be our reference triangle:
(-1,1)
s
r
(-1,-1)
(1,-1)
• The k’th triangle is the image of this triangle under
the map:
k ,1
k ,2
k ,3





v
v
v
x
 
rs x
 1 r  x
 1 s  x 
 y     2   v k ,1    2   v k ,2    2   v k ,3 

 y  
 y  
 y 
 
CAAM 452 Spring 2005
Reference Triangle Mapped
To Physical Triangle
• A typical map will have the following action:
v
s
(-1,1)
k ,3
x
, v yk ,3 
r
v
(-1,-1)
k ,1
x
,v
k ,1
y

v
k ,2
x
, vyk ,2 
(1,-1)
k ,1
k ,2
k ,3
 x
 r  s   vx   1  r   vx   1  s   vx 
 y     2   v k ,1    2   v k ,2    2   v k ,3 

 y  
 y  
 y 
 
CAAM 452 Spring 2005
Orthonormal Basis for the Triangle
• Fortunately a well behaved, orthonormal basis for the
triangle has been discovered (and rediscovered multiple
times).
• First we need to know some details about the Jacobi
polynomials. These polynomials are parameterized by two
reals:  ,  and their integer orders n,m such that they satisfy
the orthogonality relationship (for integer alpha,beta):
n   ! n   !

2
 1  x   1  x   ,
 ,
1 2   2  Pn  x  Pm  x dx   nm 2n      1 n! n     !
1


 ,
Pn
1
n   !


n! !
Pn ,   x    1 Pn ,   x 
n
CAAM 452 Spring 2005
Jacobi Polynomial Recurrence Relation
We can generate the n’th order Jacobi polynomials at a given x in [-1,1] through the
following recurrence relationship:
P0 ,  x   1
P1 ,  x  
1
2   1      2  x  1 

2
2  n  1 n      1 2n      Pn1,  x  

2n      2 !  ,

2
2
 2n      1      x
 Pn  x 
 2n      1! 

 2  n    n    2n      2  Pn1,  x 
See:
http://www.caam.rice.edu/~timwar/MA578S03/MatlabScripts/JACOBI1D.m
CAAM 452 Spring 2005
Orthonormal Basis for the Triangle
• The following basis is due to Koornwinder (later
revived by Proriol, Dubiner, Owens,….)
 1  b  2 n1,0
 n ,m   r , s   Pn0,0  a  
b 
 Pm
 2 
where:
n
2(1  r )
1
(1  s )
bs
a
• It satisfies the following orthogonality condition:
1 s
2
2
1 1i, j  k ,l drds   ik 2i  1 jl 2 j  2i  2
CAAM 452 Spring 2005
cont
• Notice that the (n,m) basis member is an n+m’th
order polynomial.
 1  b  2 n1,0
 n ,m   r , s   Pn0,0  a  
b 
 Pm
 2 
where:
n
2(1  r )
1
(1  s )
bs
a
• Moreover, the orthogonality property allows us to
determine that the set of basis members with:
• n+m<=p supports all bivariate polynomials with
maximum total degree p.
CAAM 452 Spring 2005
Basis Ordering
In order to simplify the notation we can order the
polynomials with a single index (this is not a unique
ordering):
 1  0,0 
 p 2  0,1
 2  1,0 
 p 3  1,1
 3   2,0 
 p 1   
 M  0,p 
 2 p 2   p 1,1
p ,0 
where M=(p+1)(p+2)/2 and we have normalized each of the { i }
CAAM 452 Spring 2005
cont
• Next class we will discuss how to use this basis to
interpolate, differentiate and integrate data at points
in the triangle.
• We continue here to discuss the question of which
points to use on the triangle for polynomial
interpolation.
CAAM 452 Spring 2005
Interpolation Using Generalized
Vandermonde Matrix
• Given a set of nodes lying in the triangle we use V to
construct an interpolating polynomial for a function who’s
iM
values we know at the nodes:
f  r , s    i  r , s  fˆi
• The interpolation condition yields:
i 1
iM
f  rj , s j    i  rj , s j  fˆi
i 1
iM
 f  rj , s j    V ji fˆi where V ji   i  rj , s j 
i 1
 fˆi 
j M
1
V
   f  rj , s j 
j 1
ij
CAAM 452 Spring 2005
Differentiation
• Suppose we wish to find the derivative of a p’th order
f  P p T  = x i y j
polynomial:
T 0i+j p
• First we note that the approximation becomes equality:
iM
f  r , s    i  r , s  fˆi
i 1
• And interpolation allows us to find the polynomial
m M
coefficients:
f  rn , sn    Vnm fˆm
m 1
• So differentiation requires us to compute:
m M
 m
f
 r, s   
 r , s  fˆm
r
m 1 r
m M
 m
f
 r, s   
 r , s  fˆm
s
m 1 s
CAAM 452 Spring 2005
Differentiation cont
• So we need to be able to compute:
 m
 m
and
r
s
• Recall the definition of the basis functions:
  1  s  2 n1,0
0,0  2(1  r )
 nm  r , s   Pn 
 1 
s
 Pm
 (1  s)
 2 
n
• R-derivative:
 nm
 2  dP  2(1  r )   1  s  2 n1,0
 1 
 r, s   
s

 Pm

r
 1  s  dx  (1  s)
 2 
0,0
n
n
CAAM 452 Spring 2005
Quick Jacobi Polynomial Identity
• We will make extensive use of the following:
dPn ,
 n      1   1, 1
 x  
 x
 Pn1
dx
2


CAAM 452 Spring 2005
r-Derivative
• Ok we need to calculate:
 nm
0,0
 2  dPn  2(1  r )   1  s  2 n1,0
 1 
 r, s   
s

 Pm

r
 1  s  dx  (1  s)
 2 
 nm 
n
 2  n  1  1,1  2(1  r )   1  s  2 n1,0
 1 
 r, s   
s n  1

 Pn1 
 Pm
r
 1  s  2 
 (1  s)
 2 
n
 0 when n  0 (since P0 ,  x   1)
• We can compute these using the definition of the
Jacobi polynomials.
• Watch out for s=1 (top vertex) – the r-derivative of
all the basis is functions is zero at r=1,s=-1
CAAM 452 Spring 2005
s-Derivative
We use the chain and product rule to obtain:
n
0,0


2
1

r




d
P
2(1

r
)
1

s

 2 n1,0
n

1
 r , s   
s

 Pm


2
s


 2 
 1  s  dx  (1  s )
 nm 
n 1

  n  1 s  
 2 n1,0
0,0  2(1  r )
 Pn 
 1  
s
  Pm
 (1  s )

 2 2  


  1  s   dPm2 n1,0
0 ,0  2(1  r )
 Pn 
 1 
 s 
 
 (1  s )
  2   dx

n
CAAM 452 Spring 2005
s-Derivative
From which:
n


 2 1  r   n  1  1,1  2(1  r )
  1  s  2 n1,0
 1  
 r, s   
s
 Pn1 
 Pm
2 
s
 (1  s )


 1  s   2 
 2 
 nm 

 n  1 s 
0,0  2(1  r )
 Pn 
 1   

(1

s
)
2
2





n 1

 2 n1,0
s
 Pm


  1  s   m  2n  2  2 n 2,1

0,0  2(1  r )
 Pn 
 1 
P
s



 
 m1
(1

s
)
2
2
 




n
CAAM 452 Spring 2005
Special Cases
 0 m 
 1,0
 r , s   Pm  s 
s
s
Don’t worry about all those denominators having (1-s)
since the functions are just polynomials and not
singular functions…
CAAM 452 Spring 2005
Recap
 nm 
 2  n  1  1,1  2(1  r )   1  s  2 n1,0
 1 
 r, s   
s n  1

 Pn1 
 Pm
r
 1  s  2 
 (1  s)
 2 
n
 0 when n  0 (since P0 ,  x   1)
 2  n  1  1,1  2(1  r )    1  s  n 2 n1,0
 1  
 r, s   
s
 Pn1 
 Pm
2 
s
 (1  s )
   2 
 1  s   2 
 nm 
  n  1  s 
0,0  2(1  r )
 Pn 
 1  

(1

s
)
2
2



 
n 1
 2 n1,0
s
 Pm

 2(1  r )   1  s   m  2n  2  2 n2,1 
 Pn0,0 
 1 
 
 Pm1  s  
2


 (1  s )
  2  
n
CAAM 452 Spring 2005
Derivative matrices
• Given data at M=(p+1)(p+2)/2 points we can
directly r and s derivatives with:
fˆm 
m M
1
V
 
m1
mj
f  rj , s j 
m M
m
f
r ˆ
r
ˆ
ˆ
 rn , sn    Dnm f m where Dnm 
 rn , sn 
r
r
m1
m M
m
f
s ˆ
s
ˆ
ˆ
r
,
s

D
f
where
D

 n n   nm m
 rn , sn 
nm
s
s
m1
CAAM 452 Spring 2005
One-Stage Differentiation
• Given a vector of values of f at a set of nodes we
can obtain a vector of the r and s derivatives at the
nodes by:
f
ˆ r V 1
 Dr f where Dr  D
r
f
s
s
s 1
ˆ
 D f where D  D V
s
CAAM 452 Spring 2005
Inner Product Matrices
• We will we need to compute:
 f , g T  
fgdV
T
  x, y 
   f  r, s  g  r, s 
drds
  r, s 
1 1
1 s
• It is not obvious how to do this given the value
of f and g at a set of M points.
• Same old trick – construct the polynomial
coefficients with the Vandermonde matrix and
use the orthogonality relationship..
CAAM 452 Spring 2005
Inner-Product
 f , g T
  x, y  1  s

f  r , s  g  r , s  drds


  r , s  1 1
Since the r,s->x,y is linear
  x, y  1  s  m  M ˆ
 nM

ˆ

  f mm  r , s    g nn  r , s   drds


  r , s  1 1  m1
 n1

  x, y  1  s  n  M m  M ˆ


   f mm  r , s gˆ nn  r , s   drds


  r , s  1 1  n1 m1

1 s
  x, y  n  M m  M ˆ

f m gˆ n   m  r , s n  r , s  drds


  r , s  n1 m1
1 1
  x, y  n  M m  M ˆ

2
 2 

f m gˆ n nmCn where C ij  




  r , s  n1 m1
2
i

1

  2i  2 j  2 
  x, y  n  M ˆ

f n gˆ nCn

  r , s  n1
CAAM 452 Spring 2005
Summary So Far
• We are confident that given a set of (p+1)(p+2)/2
interpolation points we can use the orthogonal
basis to perform differentiation, integration and
interpolation.
• Now – we should be concerned about how to
choose the points.
• Recall – in 1D interpolation it is tempting to use the
equally spaced points on an interval but this can
cause the Runge effect (wild oscillations in the
interpolated function).
CAAM 452 Spring 2005
Accuracy of Polynomial Interpolation
• We introduce the operator supremum-norm and the
L-infinity norm:
I p  sup
Ipf
f 0
f


where f

 max f  x, y 
 x , y T
for the p’th order interpolation operator: I p
• The interpolation error is bounded below by:
f  x, y   f *  x, y 

 f  x, y   I p f  x, y 

where f * is the optimal polynomial approximation
in the .  norm
CAAM 452 Spring 2005
cont
• This tells us that the interpolating polynomial which
approximates a function can only be less or equal
in accuracy to the best polynomial approximation.
• Finding the optimal polynomial which approximates
a function is an unsolved problem.
• However, there is a theorem by Lebesgue which
allows us to bound the interpolation error in terms
of the error one would attain by choosing the best
possible polynomial approximation.
CAAM 452 Spring 2005
Theorem (Lebesgue)
• Assume that f  C T  and we consider a set of
M interpolation points then:
f Ipf

 1   p  f  f *
where  p  I p  max
 x , y T

iM
 h  x, y  is
i 1
i
termed the Lebesgue constant and
hi  x, y  is the i'th Lagrange Cardinal function
• i.e. the Lebesgue constant gives us an idea of the
optimality of the M points. The smaller  p is the
better.
CAAM 452 Spring 2005
Proof
f Ipf

 f  f*
 ff
 ff

 f*Ipf

triangle inequality
nM
*
 f   f  xn , yn  hn  x, y 
*

*

n 1

nM
definition of interpolant

nM
 f  x , y  h  x, y    f  x , y  h  x , y 
*
n
n 1
n
n
n
n 1
n
n

ok since f *  P p  T 
 ff
 ff
 ff

 1 

*

*

*
n M

n 1


n M
  f  x , y   f  x , y  h  x, y 
*
n
n 1

n M

n 1
n
n
n
n

f *  xn , yn   f  xn , yn  hn  x, y 
 ff

nM
*

 h  x, y 
n 1
n

hn  x, y   f  f *



CAAM 452 Spring 2005
Comments
• Note – the Lebesgue number only depends on the
distribution of nodes. The only condition on the function
being approximated is that it is continuous.
• Here’s the tricky part. It is difficult to compute  p exactly
as it requires us to search the continuum of points in the
triangle.
• However, we can approximate this number by randomly
sampling a large number of points in the triangle. i.e. for
some, large, N compute the following:
iM
 p  max  hi  x j , y j 
j 1,.., N
i 1
for some finite subset
 x , y 
j
j
j 1,.., N
T
CAAM 452 Spring 2005
Open Problem
• It is still an open problem to find the set of nodes for the triangle
which minimize the Lebesgue constant.
• Various attempts have been made:
• Hesthaven:
• http://epubs.siam.org/sam-bin/getfile/SINUM/articles/30587.pdf
(You will need access permission – so download on on campus)
• Wingate and Taylor:
• http://citeseer.nj.nec.com/taylor98fekete.html
• And others.. See comparison tables in Hesthaven paper.
CAAM 452 Spring 2005
Next Class
• We will assume one type of these sets of nodes is being
used and will go on to discuss how we can perform
necessary manipulations (integration, differentiation,
interpolation) with data at these nodes.
• We will then introduce the discontinuous Galerkin method,
and show how we can achieve p’th order accurate
numerical solutions when the actual solution is sufficiently
smooth.
• Furthermore, we will see how boundary conditions are
straight forward to implement – in contrast the finite
difference methods we saw previously.
CAAM 452 Spring 2005