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Spring 2003

Prof. Tim Warburtontimwar@math.unm.edu

MA557/MA578/CS557Lecture 22

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Interpolation on the Triangle

• Recall we are considering a two-dimensional domain.

• We assume that a triangulation of the domain is given.

• In each triangle we are going to create a polynomial approximation of the solution to a PDE.

• We discussed briefly an orthogonal basis for the triangle (which you are currently verifying is indeed actually an orthogonal set of functions).

• We construct a generalized Vandermonde basis using this basis and a set of nodes.

3

Reference Triangle

• The following will be our basic triangles:

• All straight sided triangles are the image of this triangle under the map:

s

r

(-1,-1) (1,-1)

(-1,1)

1 2 3

1 2 3

1 1

2 2 2x x x

y y y

v v vx r s r s

v v vy

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Reference Triangle

• The following will be our basic triangles:

• All straight sided triangles are the image of this triangle under the map:

s

r

(-1,-1) (1,-1)

(-1,1)

1 2 3

1 2 3

1 1

2 2 2x x x

y y y

v v vx r s r s

v v vy

2 2,x yv v 1 1,x yv v

3 3,x yv v

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• Given a set of nodes lying in the triangle we use V to construct an interpolating polynomial for a function who’s values we know at the nodes:

• The interpolation condition yields:

0 0

ˆ, ,i p j p i

ij iji j

f r s r s f

0 0

0 0

1

ˆ, ,

ˆ

ˆ

1 2

2

i p j p i

n n n nij iji j

i p j p i

n ij iji j

m M

nm mm

f r s r s f

f

f

p pM

V

V where we have chosen an ordering of the modes

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Differentiation

• Suppose we wish to find the derivative of a p’th order polynomial

• First we note that the approximation becomes equality:

• And interpolation allows us to find the PKDO coefficients:

• So differentiation requires us to compute:

for a triangle pf P T T

0 0

ˆ, ,i p j p i

ij iji j

f r s r s f

1

ˆ, m M

n n nm mm

f r s f

V

0 0

0 0

ˆ, ,

ˆ, ,

i p j p iij

iji j

i p j p iij

iji j

fr s r s f

r r

fr s r s f

s s

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Differentiation cont

• So we need to be able to compute:

• Recall the definition of the basis functions:

• R-derivative:

and ,ij ij

r sr s

0,0 2 1,02(1 ) 1, 1

(1 ) 2

nn

n mnm

r sr s P P s

s

0,0

2 1,02 2(1 ) 1, 1

1 (1 ) 2

nnm nn

m

dP r sr s P s

r s dx s

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Quick Jacobi Polynomial Identity

• We will make extensive use of the following:

,

1, 11

1

2n

n

dP nx P x

dx

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r-Derivative

• Ok we need to calculate:

• 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

0,0

2 1,02 2(1 ) 1, 1

1 (1 ) 2

nnm nn

m

dP r sr s P s

r s dx s

1,1 2 1,01

,0

2 1 2(1 ) 1, 1 1

1 2 (1 ) 2

0 when 0 (since P 1)

nnm n

n m

n r sr s P P s n

r s s

n x

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s-Derivative

2 1,0

0,0 21

0,0

2

2

1,

10

,

0

,0

0

1,

2

2(1 )1

(1 )

2(1 ) 11

(1 )

2 1 2(1 )1

(1

2 2

)

1

2

1

nnm

n

nm

nn

nn

mn

n

m

dPs

dx

sr s P s

r d

s

rP P s

s

r sP

n s

P

s

r

dx ss

We use the chain and product rule to obtain:

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s-Derivative

1,11

2 1

21

,0

0,0 2 1,0

0,0

2

1

1,

2

2(1 )1

(1 )

2(1 ) 11

(1

2 2

2

2 1

1

2 2

1 2(1 )

) 2

12 (1 )1

nnm

nm

nm

nn

n

n

m

n

n

sr

r n rP

sss P s

s

rP P s

s

r s

n s

nPP

s

m

2,1 s

From which:

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Special Cases

0 1,0,m

mr s P ss s

Don’t worry about all those denominators having (1-s)since the functions are just polynomials and not singular functions…

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Recap

1,1 2 1,012

10,0 2 1,0

0,0 2 2,11

2 1 2(1 ) 1, 1

2 (1 ) 21

2(1 ) 11

(1 ) 2 2

2(1 ) 1 2 21

(1 ) 2 2

nnm n

n m

nn

n m

nn

n m

n r sr s P P s

s ss

r n sP P s

s

r s m nP P

s

s

1,1 2 1,01

,0

2 1 2(1 ) 1, 1 1

1 2 (1 ) 2

0 when 0 (since P 1)

nnm n

n m

n r sr s P P s n

r s s

n x

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Derivative matrices

• Given data at M=(p+1)(p+2)/2 points we can directly r and s derivatives with:

1

1

1

1

ˆ ,

ˆˆ ˆ, where ,

ˆˆ ˆ, where ,

m M

m j jmjm

m Mr r m

n n nm m nm n nm

m Ms s m

n n nm m nm n nm

f f r s

fr s f r s

r r

fr s f r s

s s

V

D D

D D

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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:

1

1

ˆ where

ˆ where

r r r

s s s

r

s

fD f D D V

fD f D D V

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Matlab Scripts

• After class on Friday I will post code which computes the derivative matrices Dr and Ds for an arbitrary set of M nodes.

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Inner Product Matrices

• Recall we need to compute:

• 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 PKDO coefficients and use the orthogonality relationship..

1

1 1

,

,, ,

,

TT

s

f g fgdV

x yf r s g r s drds

r s

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Inner-Product

1

1 1

1

1 11 1

1

1 11 1

1

1 1 1

,, , ,

,

, ˆ ˆ, ,,

, ˆ ˆ, ,,

, ˆ ˆ , ,,

s

T

s m M n M

m m n nm n

s n M m M

m m n nn m

sm M

m n m nm

x yf g f r s g r s drds

r s

x yf r s g r s drds

r s

x yf r s g r s drds

r s

x yf g r s r s drds

r s

1

ij1 1

1

, 2 2ˆ ˆ where C, 2 1 2 2 2

, ˆ ˆ,

n M

n

n M m M

m n nm nn m

n M

n n nn

x yf g C

r s i i j

x yf g C

r s

Since the r,s->x,y is linear

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Nodal Mass Matrix

• Suppose we know the value of f and g at M points then we can compute the PKDO coefficients using the generalized Vandermonde matrix.

• We can then integrate by the previous operations (last slide).

• All these operations can be concatenated into one mass matrix.

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Mass-Matrix

1

, ˆ ˆ,,

, where is a diagonal matrix

,

,

,

, where

,

n M

n n nTn

t

tt

tt

x yf g f g C

r s

x y

r s

x y

r s

x y

r s

1 1

1 1

1 1

V f C V g C

f V C V g

f Mg M V C V

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Nodal Mass Matrix

• Setting

• Where:

• Then:

1 1

km1 1

,

,, ,

,

,

,

,

,

tn m n mT

j Mk M

n j j jk m k kk j

j Mk M

nj jkk j

nm

h h

x yh r s h r s

r s

x y

r s

x y

r s

h Mh

M

M

M

,

,

n

m

f h r s

g h r s

1 2, 1 , , where

2n m m nm

p ph r s n m M M

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DG Matrices

• Recall we need to compute:

• We use the coordinate change, chain rule, linearity of the map T->That (reference triangle) and finally the identities we just found:

,T

gfx

ˆ ˆ

ˆ ˆ

, , ,

, ,, ,

, ,

, ,, ,

, ,

, ,

, ,

T T T

T T

T T

t r t s

g r g s gf f fx x r x s

x y x yr g s gf f

r s x r r s x s

x y x yr g s gf f

r s x r r s x s

x y x yr s

r s x r s x

f MD g f MD g

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Summary

Set: where hn is the n’th Lagrange interpolant defined as the p’th order polynomial in r,s for which:

, ,,

, ,

, ,,

, ,

r smn nm nm

T

r smn nm nm

T

x y x yh r sh

x r s x r s x

x y x yh r sh

y r s y r s y

MD MD

MD MD

, , nr s h r s

1 2, 1 , , where

2n m m nm

p ph r s n m M M

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Progress With The DG Scheme For Advection

• The DG scheme now looks like:

• Where we set: and we can now compute everything in the first three inner-product.

• Next time we will discuss how to compute everything in the surface inner-product.

1 1 1

Find , 1,..., such that:

.....

, 02

where ,

, , ,m M m M m M

m m mn m n m n mT

m m mT T

m

n

T

dC h hh h u h C v h C

dt

C t m M

h C

C C

y

C

x

u n

, , nr s h r s

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Summary of Matrices

, ,,

, ,

, ,,

, ,

,,

,

r smn nm nm

T

r smn nm nm

T

n m T

x y x yh r sh

x r s x r s x

x y x yh r sh

y r s y r s y

x yh h

r s

MD MD

MD MD

M

Recall the factors: are constantover a straightsided triangle.

,, , , ,

,

x yr r s s

x y x y r s

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Next Lecture

• We will use a better set of nodes (improve the condition number of the Vandermonde matrix).

• Time permitting we will prove consistency.