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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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