# Proof by qingqing19771029

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```									MA557/MA578/CS557
Lecture 24

Spring 2003
Prof. Tim Warburton
timwar@math.unm.edu

1
2D Advection Equation
• Equation:
C    C    C
u    v    0
t    x    y
• Scheme (assuming s’th order time integration):

Find C  P p T   P s  dt intervals  such that   P p T 

 C         C     C                    
  , t  u x  v y     ,  2         C    0
                      T  
 



T
where    u  n  ,    C  C  C
2
Recall Stability
• We established stability (neglecting boundary terms):

d                                      u n
             C  L2 e  0
2

2
C               
dt     L2
Tj         unique     2
                  edges e
   j      

3
Consistency

• We will now establish consistency.

• We will use a truncation analysis to determine
consistency and directly convergence.

• We will also need two inverse inequalities, a polynomial
approximation result and a trace inequality.

4
Polynomial Inverse Trace Inequality
• We now wish to bound the surface semi-norm of a
polynomial function by the volume norm of the
polynomial function.

5
Polynomial Inverse Trace Inequality
• Check out:
• http://www.dam.brown.edu/scicomp/publications/Reports/Y20
02/BrownSC-2002-19.pdf

• We will establish the following relationship for an arbitrary
planar triangle T:

u  P p T  , u



 p  1 p  2  T


L2  T                            u   L2 T 
          2         T 
                      
• Where |T| is the area of T and |dt| is the length of the
perimeter of T.

• This is a relatively new result.
6
Inverse Polynomial Trace Theorem
• For an arbitrary planar triangle T, equipped with a
p’th order polynomial space the following holds:

u  P T  , u
p



 p  1 p  2  T


L2  T                            u   L2 T 
          2         T 
                      

7
Proof: Step 1
• For the purpose of this proof we consider the
reference triangle That and discretize the polynomial
space using the PKDO orthonormalized basis:

2i  1 2i  2 j  2 0,0      1  b  2i 1,0
i

 ij   r , s                      Pi  a          Pj      b 
2         2                2 
where:
1 a 
r       1  b   1
 2 
sb
ˆ 
then P p T   ij 
0 i  j  p
8
Proof: Step 2
• We now construct the mass matrix for the s=-1 face
of the reference triangle

   ,   
ij    kl        ˆ
T1

 2i  1  2i  2 j  2   2k  1  2k  2l  2  2i 1,0                                2
                                              Pj       1 Pl 2 k 1,0  1  ik
 2            2         2           2                                             2i  1

 2i  2 j  2   2i  2l  2  2i 1,0
                              Pj       1 Pl 2i1,0  1  ik
      2              2      


        2i  2 j  2  
        2i  2l  2  

  1                  1                  ik
j                         l


             2        
             2      
9
Proof: Step 3
• We now consider the semi-norm of the function over
the s=-1 edge of the triangle:
1
u L2  Tˆ    u  r , 1 dr
2                                      2
1
1

 n M              mM               
1
    un n  r , 1   um m  r , 1  dr                             {Expand in PKDO}
1  n 1              m1               
i  p j  p i k  p l  p  k                   1
          u u      r , 1    r , 1 dr
ij       kl        ij      kl
i 0    j 0 k 0 l 0                           1

i  p j  p i k  p l  p  k                  
        2i  2 j  2  
        2i  2l  2  

                              u ij u kl   1                  1                  ik
j                         l

i 0    j 0 k 0 l 0                          
             2        
             2      
 j  p i  l  p  i                                                          
                 1              
i p
      u ij uil   1 i  j  1                                     i  l 1 
j                                          l

i 0  j 0  l 0                                                                            10
Proof: Step 3 cont
• This looks very complicated:
 j  pi  l  p i                                    
                   1              
i p
u L2  Tˆ        u ij uil   1 i  j  1             i  l 1 
2                                         j                  l

i 0  j 0  l 0                                           
1

 u t Mu
ˆ ˆ

• But: The matrix Mtilde is a block diagonal matrix:

i=0

M
i=1

i=2               11
Proof: Step 4
• Because the matrix Mtilde is block diagonal we can
find its spectral radius by considering maximum of
the spectral radii of all the diagonal blocks.

• The i’th block matrix has the entries (0<=j,i<=p-i)


Mijl   1
j

i  j 1    1
l

i  l 1

• Thus the i’th block matrix is the outer product of a
vector with itself.

12
Proof: Step 4 cont
• Thus the i’th block matrix is the outer product of a
vector with itself  the rank of the i’th block matrix is
one and the one non-zero eigenvalue is:

  1                  
j  p i                      2
                       i  j 1
i                   j

j 0
j  p i
    i  j  1
j 0

  i  1 p  i  1   
 p  i  p  i  1
2


 p  i  2  p  i  1
2
• Taking the maximum over the p+1 blocks we find the
maximum eigenvalue of Mtilde as:
 
0   p  2  p  1
13
2
Proof: Step 5
• We now know that the spectral radius of Mtilde is:

 
0   p  2  p  1
2

• So returning to the edge 1 norm:

u L2  Tˆ   u t Mu
2
ˆ ˆ
1

  0u t u
ˆ ˆ

• But here’s the trick – we can rewrite the rhs term
using the orthonormality of the PKDO basis:
n M               nM
u L2  Tˆ    0  un   0  un  n drds   0 u L2 Tˆ 
2                                                     2
ˆ2         ˆ2 2
1
n 0       n 0   ˆ     T                    14
Proof: Step 6
• So we have established the following result for the semi-
norm on the s=-1 edge:              p  1 p  2 
u L2  Tˆ  
2                             2
u L2 Tˆ 
1
2
• We use a scaling argument for an arbitrary planar triangle:
T1

2                           2
u   L2  T1 
. u L2  Tˆ 
2             1

T1  p  1 p  2  2
    .                  u L2 Tˆ 
2              2
T1  p  1 p  2  2

2
.                    u L2 T 
2              2     T


 p  1 p  2       T1
u
2
L2 T 
2              T                                15
Proof: Stage 7
• Since we have just established:

u
2

 p  1 p  2            T1
u
2
L2  T1                                              L2 T 
2              T

• We can use the fact that the assignment of the r,s coordinate
system was arbitrary to generalize this for each of the edges.
• We now sum over all faces and:
e3
 u
2                         2
u   L2  T                  L2  Te 
e1
e 3
 p  1 p  2        T1

2
u   L2 T 
e1               2            T                         QED
 p  1 p  2        T

2
u   L2 T            16
2              T
Derivative Inverse Inequality
• We now wish to bound the norm of the spatial
derivative of a polynomial function by the norm of the
polynomial.

17
Inverse Inequality 2
• The following holds:

For u  P p T  the following inequality holds:
u                    Te 
 cp max 
2
 u L2 T 
x  L2 T 
e
 T 
for a universal constant c independent of T and p

• See: Ch. Schwab “p- and hp- Finite Element
Methods: Theory and Applications in Solid and Fluid
Mechanics”, Oxford University Press, 1998.
18
Trace Theorem
• We now wish to bound the surface norm of the larger
class of functions which has derivatives with bounded
L2-norm.

19
Trace Theorem

• Suppose:
 u
                               u                

u  H T   u
1
,                 
 x
                     L2 T    y   L2 T       


• Then:                                                  1                    
 c u             u L2 T   u
2                                                       2
L2 T  
u   L2  T             L2 T 
                           h                    

20
Polynomial Approximation
• We now introduce a theorem providing an estimate
for a projection to the polynomial space.

21
Polynomial Approximation on T
• We now introduce an estimate for polynomial
approximation of a function f . For f  H T  , r  0
r

there exists a constant C dependent on r and the
geometry of T, but independent of f, h, p such that:
h q
f  Pp f        H q T 
 c r 2 q f         H r T 
p
where   min  r , p  1 , 0  q  
• For the L2 projection: Pp : H r T   P p T 

i  r j  r i
i j f
• Here:         f   H r T 
 
i 0     j 0    xi y j   L2 T 
22
See: http://www.math.ubc.ca/~feldman/m606/sobolev.pdf for info on Sobolev norms
Summary of Theorems:
• We will now use the

1) For u  P T  ,
p


u L2  T   
 p  1 p  2            T   

u      L2 T 
              2                  T    
                                      

u                          Te 
2) For u  P T  ,
p
 cp max 
2
u                L2 T 
x        L2 T 
e
 T 

                                     1             
3) For u  H 1 T  , u              c u                u           
2                                                        2
L2 T  
L2  T              L2 T         L2 T 
u
                                     h             

h q
4) For u  H T  ,
r
u  Ppu q  c r 2 q u                      H r T 
H T  p
where   min  r , p  1 , 0  q                                                                23
Convergence Theorem
• Assume that a solution C  H    , r  2 exists
r

in the domain Omega. Then the numerical solution
Ctilde to the semidiscrete approximation from the DG
scheme converges to the exact solution, and the
global error is bounded as:
k K

 C  x, y , t   C  x , y , t 
k 1                                 L2 Tk 

 h
k K
h 1                               
 c   r C  x, y,0         H r Tk 
 t r 2 max C  x, y, s    H r Tk  
k 1  p                               p s0,t                           
• Where c (lower case) is independent of h and p.

24
Proof Step 1:
• Recall the DG scheme:

Find C  P p T   P s  dt intervals  such that   P p T 

 C    C    C                                 
,  u    v      ,                           C    0
 t    x    y T   2                           
                                  T
where    u  n  ,     C  C  C
• We define the Truncation error TC as
Find T C  P p T   P s  dt intervals  such that   P p T 

Pp C                            
   
 Pp C    Pp C
 , T 
C
T
 ,
   t
u
x
v          , 
y             2
  Pp C  
       
                         T                             T
25
Proof Step 2:Three Important Equations
• We now consider three equations:
1) Numerical:

C                       
   
 C    C
,  u    v      ,               C    0
 
 t    x    y T        2
                      T
2) Truncation – projection of exact solution in numerical
scheme:
Pp C                         
   
 Pp C     Pp C
  , t  u x  v y     ,              Pp C     , T C 
                        T        2                            T
                         T
3) Exact equation, with exact solution:
         C    C    C  
  , Pp     u    v      0
         t    x    y  T                                 26
Proof Step 4: Truncation Error
• Subtract equation 3 from equation 2:

  Pp C      C     Pp C      C     Pp C      C  
 , T 
C
 , 
t
 Pp    u 
t       x
 Pp    v 
x       y
 Pp    
y  T
                                       
T

                      
 ,                 Pp C  
     2                  
                           T
• Substitute phi=Tc and apply Cauchy-Schwarz:

C 2
 C  PpC      C     PpC      C     PpC      C  
T                  T ,       Pp    u        Pp    v        Pp    
L2 T 
     t       t      x       x      y       y  T
                     
 T C ,              PpC  
      2                
                          T                        27
Proof Step 4: cont
• Substituting phi=Tc

C 2
 C  Pp C      C     Pp C      C     PpC      C  
T                T ,        Pp    u         Pp    v        Pp    
L2 T 
     t        t      x        x      y       y  T
                            
 T C ,                    Pp C  
         2                    
                                 T
 P C
 p         C                            Pp C    C                  PpC     C          

 T   C
       Pp                       u           Pp                v        Pp             
 t        t                             x      x                   y      y
L2 T 
                       L2 T                          L2 T                       L T  
2

       
T    C
              Pp C 
L2  T        2              
                        L2  T 

28
Individual Volume Terms:
• Assume exact time discretization:
Pp C            C
 Pp                      0 assuming time discretization is exact
t            t    L2 T 

• Using the polynomial approximation theorem:
Pp C    C                           Pp C    C                             Pp C      C 
 Pp                       Pp         Pp                           Pp         Pp    
x      x         L2 T 
x      x             L2 T           x        x    L2 T 

h 1
 c r 2 C      H 1 T 
p

Pp C          C                     Pp C              C                   P C   C 
 Pp                   Pp             Pp                       Pp  p  Pp    
x            x   L2 T 
x                x   L2 T           x    x        L2 T 

h 1
 c r 2 C      H 1 T 
p                                                                         29
Surface Term
• We can add the jump in C due to continuity of C:
                                        
              Pp C                               Pp C   C 
    2                                   2                     
                        L2  T                                          L2  T 

by assumed continuity of C
       

  Pp C   C   

 
    2       
                                        
L2 T     
       

  Pp C   C   

 
    2       
                                          
L2 T 

30
Jump Terms
• Break the jump terms with triangle inequality and use
the trace theorem:

 Pp C 
                           Pp C   C 
             
L2  T                              L2  T 

  Pp C   C                      PpC   C

L2 T                            
L2 T 

considering the - terms:

                         
Pp C   C 2   c   PpC   C 2   PpC   C 2   PpC   C         1
                    
2                                                                                2

L  T                    L T                                                    L2 T 

                                        L  T  h                            
 h          h 1              h 2 1           
 c  r C H r T  r  2 C H r T   2 r C H r T  
2

p            p                   p                

1


 PpC   C L2 T    c p r 1 C
2
h
H r T                                      31
Plugging In These Results
• Using the previous results and the polynomial inverse
trace inequality:

 P C
 p         C                         Pp C    C                          Pp C    C          

C 2
T                  T       C
       Pp                       u        Pp                        v        Pp             
 t        t                          x      x                           y      y
L2 T                  L2 T 
                       L2 T                               L2 T                       L T  
2

         
T      C
                Pp C                
L2  T        2                
                          L2  T 

h 1                            h 1
T   C
 0  u c r 2 C            H r T 
 v c r 2 C            H r T 
L2 T             p                                p


 p  1 p  2    T   
    
h 0.5
c r 1 C                   H r T 
2            T     2      p

32
Simplifying
• We can now bound the truncation error:

h 1
T C 2  c r 2 C    H r T 
L T  p

33
Proof continued – next lecture
• Next Lecture.

• Also – we will start implementing a 2D DG solver.

• Each student will be given a distinct system of pdes
(e.g. Maxwell’s, Acoustics, Euler, …).

• If two students wish to work together they can share
code and work as a group – but with different pde.s

34

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