1

Spring 2003

Prof. Tim Warburtontimwar@math.unm.edu

MA557/MA578/CS557Lecture 24

2

2D Advection Equation

• Equation:

• Scheme (assuming s’th order time integration):

0C C C

u vt x y

Find dt intervals such that

, , 02

where ,

p s p

TT

C P T P P T

C C Cu v C

t x y

C C C

u n

3

Recall Stability

• We established stability (neglecting boundary terms):

22

22

e

02

jj

L eL T unique

edges

dC C

dt

u n

4

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.

5

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.

6

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:

• Where |T| is the area of T and |dt| is the length of the perimeter of T.

• This is a relatively new result.

2 2

1 2,

2p

L T L T

p p Tu P T u u

T

7

Inverse Polynomial Trace Theorem

• For an arbitrary planar triangle T, equipped with a p’th order polynomial space the following holds:

2 2

1 2,

2p

L T L T

p p Tu P T u u

T

8

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:

0,0 2 1,0

0

2 1 2 2 2 1,

2 2 2

where:

11 1

2

ˆthen

ii

i jij

pij

i j p

i i j br s P a P b

ar b

s b

P T

9

Proof: Step 2

• We now construct the mass matrix for the s=-1 face of the reference triangle

1̂

2 1,0 2 1,0

2 1,0 2 1,0

,

2 1 2 2 2 2 1 2 2 2 21 1

2 2 2 2 2 1

2 2 2 2 2 21 1

2 2

2 2 2 2 2 21 1

2 2

ij klT

i kj l ik

i ij l ik

j l

ik

i i j k k lP P

i

i j i lP P

i j i l

10

Proof: Step 3

• We now consider the semi-norm of the function over the s=-1 edge of the triangle:

21

12 2

ˆ

1

1

1 11

1

0 0 0 0 1

, 1

, 1 , 1

, 1 , 1

2 2 2 2 2 21 1

2 2

L T

n M m M

n n m mn m

i p j p i k p l p k

ij kl ij kli j k l

j l

ikij kll

u u r dr

u r u r dr

u u r r dr

i j i lu u

0 0 0 0

0 0 0

1 1 1 1

i p j p i k p l p k

i j k

i p j p i l p ij l

ij ili j l

u u i j i l

{Expand in PKDO}

11

Proof: Step 3 cont

• This looks very complicated:

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

21

2

ˆ0 0 0

1 1 1 1

ˆ ˆ

i p j p i l p ij l

ij ilL Ti j l

u u u i j i l

tu Mu

i=0

i=1

i=2

M

12

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)

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

1 1 1 1j li

jl i j i l M

13

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:

• Taking the maximum over the p+1 blocks we find the maximum eigenvalue of Mtilde as:

2

0

0

1 1

1

11 1

22 1

2

j p iji

j

j p i

j

i j

i j

p i p ii p i

p i p i

0 2 1

2

p p

14

Proof: Step 5

• We now know that the spectral radius of Mtilde is:

• So returning to the edge 1 norm:

• But here’s the trick – we can rewrite the rhs term using the orthonormality of the PKDO basis:

0 2 1

2

p p

21

2

ˆ

0

ˆ ˆ

ˆ ˆ

L Tu

t

t

u Mu

u u

2 21

2 20 2 0 2 2 0ˆ ˆ

0 0 ˆ

ˆ ˆn M n M

n n nL T L Tn n T

u u u drds u

15

• So we have established the following result for the semi-norm on the s=-1 edge:

• We use a scaling argument for an arbitrary planar triangle:

Proof: Step 6

2 21

2 2

ˆ ˆ

1 2

2L T L T

p pu u

2 21 1

2

2

2

2 21ˆ

21ˆ

21

21

.2

1 2.

2 21 2 2

.2 2

1 2

2

L T L T

L T

L T

L T

Tu u

p pTu

p pTu

T

p p Tu

T

16

Proof: Stage 7

• Since we have just established:

• 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:

2 21

2 211 2

2L T L T

p p Tu u

T

2 2

2

2

32 2

1

321

1

2

1 2

2

1 2

2

e

e

L T L Te

e

L Te

L T

u u

p p Tu

T

p p Tu

T

QED

17

Derivative Inverse Inequality

• We now wish to bound the norm of the spatial derivative of a polynomial function by the norm of the polynomial.

18

Inverse Inequality 2

• The following holds:

• See: Ch. Schwab “p- and hp- Finite Element Methods: Theory and Applications in Solid and Fluid Mechanics”, Oxford University Press, 1998.

2

2

2

For the following inequality holds:

max

for a universal constant c independent of T and p

p

e

L TeL T

u P T

Tucp u

x T

19

Trace Theorem

• We now wish to bound the surface norm of the larger class of functions which has derivatives with bounded L2-norm.

20

Trace Theorem

• Suppose:

• Then:

2 2

1 ,L T L T

u uu H T u

x y

2 2 2 2

2 21L T L T L T L T

u c u u uh

21

Polynomial Approximation

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

22

Polynomial Approximation on T

• We now introduce an estimate for polynomial approximation of a function f . For there exists a constant C dependent on r and the geometry of T, but independent of f, h, p such that:

• For the L2 projection:

• Here:

2

where min , 1 , 0

rq

q

p r q H TH T

hf P f c f

p

r p q

, 0rf H T r

: r ppP H T P T

20 0

r

i jj r ii r

i jH Ti j L T

ff

x y

See: http://www.math.ubc.ca/~feldman/m606/sobolev.pdf for info on Sobolev norms

23

Summary of Theorems:

• We will now use the

2 2

2

2

2 2 2 2

2

2 21

2

1 21) For ,

2

2) For , max

13) For ,

4) For ,

where min

rq

p

L T L T

ep

L TeL T

L T L T L T L T

qr

p r q H TH T

p p Tu P T u u

T

Tuu P T cp u

x T

u H T u c u u uh

hu H T u P u c u

p

r

, 1 , 0 qp

24

Convergence Theorem

• Assume that a solution exists 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:

• Where c (lower case) is independent of h and p.

, 2rC H r

21

1

2 0,1

, , , ,

, ,0 max , ,

k

r rk k

k K

k L T

k K

r rH T H Ts tk

C x y t C x y t

h hc C x y t C x y s

p p

25

Proof Step 1:

• Recall the DG scheme:

• We define the Truncation error TC as

Find dt intervals such that

, , 02

where ,

p s p

TT

C P T P P T

C C Cu v C

t x y

C C C

u n

CFind T dt intervals such that

, , ,2

p s p

p p pCpT

T T

P T P P T

P C P C P CT u v P C

t x y

26

Proof Step 2:Three Important Equations

• We now consider three equations:

1) Numerical:

2) Truncation – projection of exact solution in numerical scheme:

3) Exact equation, with exact solution:

, 0p

T

C C CP u v

t x y

, , 02

TT

C C Cu v C

t x y

, , ,2

p p p Cp T

TT

P C P C P Cu v P C T

t x y

27

Proof Step 4: Truncation Error

• Subtract equation 3 from equation 2:

• Substitute phi=Tc and apply Cauchy-Schwarz:

, ,

,2

p p pCp p pT

T

p

T

P C P C P CC C CT P u P v P

t t x x y y

P C

2

2,

,2

p p pC Cp p pL T

T

Cp

T

P C P C P CC C CT T P u P v P

t t x x y y

T P C

28

Proof Step 4: cont

• Substituting phi=Tc

2

2

2 2 2

2

2,

,2

2

p p pC Cp p pL T

T

Cp

T

p p pCp p pL T

L T L T L T

C

L T

P C P C P CC C CT T P u P v P

t t x x y y

T P C

P C P C P CC C CT P u P v P

t t x x y y

T

2

p

L T

P C

29

Individual Volume Terms:

• Assume exact time discretization:

• Using the polynomial approximation theorem:

2

0 assuming time discretization is exactpp

L T

P C CP

t t

2 2 2

1

1

2

p p pp p p p p

L T L T L T

r H T

P C P C P CC C CP P P P P

x x x x x x

hc Cp

2 2 2

1

1

2

p p pp p p p p

L T L T L T

r H T

P C P C P CC C CP P P P P

x x x x x x

hc Cp

30

Surface Term

• We can add the jump in C due to continuity of C:

2 2

2

2

2 2

by assumed continuity of C

2

2

p p

L T L T

p

L T

p

L T

P C P C C

P C C

P C C

31

Jump Terms

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

2 2

2 2

2 2 22

2

2 2

1 2 12

2 2

1

2

1

considering the - terms:

1

r r r

r

p pL T L T

p pL T L T

p p p pL T L T L TL T

r r rH T H T H T

p r H TL T

P C P C C

P C C P C C

P C C c P C C P C C P C Ch

h h hc C C Cp p p

hP C C c C

p

32

Plugging In These Results

• Using the previous results and the polynomial inverse trace inequality:

2 2

2 2 2

2

2

2

2

1 1

2 2

0.5

1

2

0

1 2

2 2

r r

r

p p pC Cp p pL T L T

L T L T L T

CpL T

L T

Cr rH T H TL T

r H T

P C P C P CC C CT T P u P v P

t t x x y y

T P C

h hT u c C v c C

p p

p p T hc C

T p

33

Simplifying

• We can now bound the truncation error:

2

1

2 rC

r H TL T

hT c C

p

34

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