Chapter 11 polar coordinate

Speaker: Lung-Sheng Chien

Book: Lloyd N. Trefethen, Spectral Methods in MATLAB

2u u

Consider eigen-value problem on unit disk

with boundary condition 1, 0u r

22

1 1rr ru u u u u

r r

We adopt polar coordinate cos

sin

x r

y r

, then

, on 0,1r and 0,2

Discretization on unit disk

0,1r1

, and non-periodic Chebyshev grid in

rUsually we take periodic Fourier grid in

Chebyshev grid in

1,1x

1

2

xr

Chebyshev grid in

0,1r

Observation: nodes are clustered near origin 0r , for time evolution problem,

we need smaller time-step to maintain numerical stability. 1,1r 2

cos cos

sin sin

x r r

y r r

, 0,0x y ,r 1 to 2 mapping

Asymptotic behavior of spectrum of Chebyshev diff. matrix

In chapter 10, we have showed that spectrum of Chebyshev differential matrix (second order) approximates

, 1 1, B.C. 1 0xxu u x u with eigenmode 2

2

4k k

Eigenvalue of

2ND

2ND is negative (real number) and 4

max 0.048N

Large eigenmode of 2ND does not approximate to

22

4k k

Since ppw is too small such that

resolution is not enough

1

2

Mode N is spurious and localized near boundaries 1x

1,1r 2

Grid distribution

0,1r1

Consider 1N Chebyshev node on 1,1 , cosj

jx

N

for 0,1,2, ,j N

Preliminary: Chebyshev node and diff. matrix [1]

x

0x1x2x3x/ 2NxNx

1 1

6N

0 1x 1x2x3 0x 4x5x61 x x

0 1x 1x2x03x4x51 x

x

5N

Even case:

Odd case:

Uniform division in arc

Preliminary: Chebyshev node and diff. matrix [2]

Given

1N Chebyshev nodes , cosj

jx

N

and corresponding function value jv

We can construct a unique polynomial of degree

N , called

0

N

j jj

p x v S x

1

0 j k

k jS x

k j

is a basis.

1

0 0

N NN

i j j i ij jj j

p x v S x D v

where differential matrix

NN ijD D is expressed as

2

00

2 1,

6N N

D

22 1

,6

NNN

ND

2

,2 1

jNjj

j

xD

x

for

1,2, , 1j N

1,

i j

N iij

j i j

cD

c x x

for

, , 0,1, 2, ,i j i j N where2 0,

1 otherwisei

i Nc

with identity 0,

NN Nii ij

j j i

D D

Second derivative matrix is 2N N ND D D

Preliminary: Chebyshev node and diff. matrix [3]

Let p x be the unique polynomial of degree

with

N

define 0 j N

, then impose B.C.

1 0p and j jp x v

j jw p x and j jz p x for

We abbreviate 2Nw D v 1 0p ,that is,

0 0Nv v

In order to keep solvability, we neglect

0 Nw w 2 2 1: 1,1: 1N ND D N N ,that is,

0w

1w

2w

1Nw

Nw

0v

1v

2v

1Nv

Nv

2ND

zero

zero

neglect

neglect

Similarly, we also modify differential matrix as 1: 1,1: 1N ND D N N

Preliminary: DFT [1]

Definition: band-limit interpolant of

function , is periodic sinc function

NS x

sin / sin1

2 2 / tan / 2 2 tan / 2

mikx

Nk m

x h mxhS x P e

h x m x

If we write 1

N

j k j kk

v v v

, then 1

1ˆ

2

m Nikx

k k N kk m k

p x P e v v S x x

Also derivative is according to 1

1

N

j j k N j kk

w p x v S x x

Given a set of data point 1 2, , , NNv v v R with

2N m is even,

2h

N

Then DFT formula for jv

1

2ˆ exp

N

k j jj

v v ikxN

for

, 1, , 1,k m m m m

1

1ˆ exp

2

m

j k jk m

v v ikx

for

1,2, ,j N

Preliminary: DFT [2]

Direct computation of derivative of

sin /

2 / tan / 2N

x hS x

h x

, we have

1

0 0 mod

11 cot , 0 mod

2 2

jN j

j N

S x jhj N

1 1 1 15 5 5 5

1 1 1 15 5 5 5

1 1 1 1 15 5 5 5 5 5

1 1 1 15 5 5 5

1 1 1 15 5 5 5

0 2 3 4

0 2 3

2 0 2

3 2 0

4 3 2 0

j k

S h S h S h S h

S h S h S h S h

D S x x S h S h S h S h

S h S h S h S h

S h S h S h S h

Example:

is a Toeplitz matrix.

Second derivative is

2

22

2

1 0 mod

6 3

1, 0 mod

2sin / 2

jN j

j Nh

S x

j Njh

Preliminary: DFT [3]

2

22

2

1 0 mod

6 3

1, 0 mod

2sin / 2

jN j

j Nh

S x

j Njh

For second derivative operation

2 2

1 1

,N N

j j k N j k N kk k

w p x v S x x D j k v

second diff. matrix is explicitly defined by using Toeplitz matrix (command in MATLAB)

2

22:2

22 22 2

2

2

2

csc 2 / 2 / 2

csc / 2 / 211

,16 3 2sin 1: 1 / 2

6 3csc / 2 / 2

csc 2 / 2 / 2

N

N N j k

h

h

D S x x toeplitzh N h

hh

h

r

01 r

1r

2r

3r

4r

51rN

r r

0 1 2 3 4 5 60 2N

0r x

y

12

3

4 5

6

r coordinatex y coordinate

Fornberg’s idea : extend radius to negative image [1]

5rN (odd): to avoid singularity of coordinate transformation 0r

6N (even): to keep symmetry condition , , mod 2u r u r

1

2

, x y unit disk 1,1r 0,2 and

r

01 r

1r

kr

1kr

1rNr

1rN

r

0 1 2 m 1m

0 2N

0r

r coordinate

2r

1N

x

y

1

2

m

1m

x y coordinate

N

1N

In general 2 1rN k is odd, and 2N m is even, then

Fornberg’s idea : extend radius to negative image [2]

: non-unifr2

N

and

1 : 1 1i rr i r i N

: 1j j j N Active variable , total number is 1rN N

r

01 r

1r

2r

3r

4r

51rN

r r

0 1 2 3 4 5 60 2N

x

y

12

3

4 5

6

x

y

12

3

4 5

6

0r

x yr coordinate coordinate

, , , mod 2 ,r r x y 2 to 1 mapping

redundant

Redundancy in coordinate transformation [1]

r

01 r

1r

2r

3r

4r

51rN

r r

0 1 2 3 4 5 60 2N

x

y

12

3

4 5

6

x

y

12

3

4 5

6

0r

x yr coordinate coordinate

, , , mod 2 ,r r x y 2 to 1 mapping

redundant

Redundancy in coordinate transformation [2]

8.5

2.6180

-0.7236

0.3820

-0.2764

0.5

-10.4721

-1.1708

2

-0.8944

0.6180

-1.1056

2.8944

-2

-0.1708

1.618

-0.8944

1.5279

-1.5279

0.8944

-1.6180

0.1708

2

-2.8944

1.1056

-0.618

0.8944

-2

1.1708

10.4721

-0.5

0.2764

-0.382

0.7236

-2.618

-8.5

rND

0r

0r

1r

1r

2r

2r

3r

3r

4r

4r

5r

5r

Redundancy in coordinate transformation [3]

2 1 5rN k is odd, then Chebyshev differential matrix rN

D is expressed as

rND

-1.1708

2

-0.8944

0.6180

-2

-0.1708

1.618

-0.8944

0.8944

-1.6180

0.1708

2

-0.618

0.8944

-2

1.1708

neglect

Redundancy in coordinate transformation [4]

, ,N Ni j N i N j

D D

Symmetry property of Chebyshev differential matrix :

-1.1708

2

-0.8944

0.6180

-2

-0.1708

1.618

-0.8944

0.8944

-1.6180

0.1708

2

-0.618

0.8944

-2

1.1708

rND

-1.1708

2

-0.8944

0.6180

-2

-0.1708

1.618

-0.8944

0.8944

-1.6180

0.1708

2

-0.618

0.8944

-2

1.1708

Permute column by

r

TND P

1,2,4,3P

-1.1708

2

-2

-0.1708is symmetric

0.8944

-1.6180

-0.618

0.8944is NOT symmetric

1E

2E

1

21 2

3

4

u

uE E

u

u

is faster ?

41.6

21.2859

-1.8472

0.7141

-0.9528

-8

-68.3607

-31.5331

7.3167

-1.9056

2.2111

17.5724

40.8276

12.6833

-10.0669

5.7889

-3.6944

-23.6393

-23.6393

-3.6944

5.7889

-10.0669

12.6833

40.8276

17.5724

2.2111

-1.9056

7.3167

-31.5331

-68.3607

-8

-0.9528

0.7141

-1.8472

21.2859

41.6

2

r r rN N ND D D

0r

0r

1r

1r

2r

2r

3r

3r

4r

4r

5r

5r

Redundancy in coordinate transformation [5]

2 1 5rN k is odd, then Chebyshev differential matrix 2

rND is expressed as

2

rND

neglect

-31.5331

7.3167

-1.9056

2.2111

12.6833

-10.0669

5.7889

-3.6944

-3.6944

5.7889

-10.0669

12.6833

2.2111

-1.9056

7.3167

-31.5331

Redundancy in coordinate transformation [6]

2

rND

-31.5331

7.3167

-1.9056

2.2111

12.6833

-10.0669

5.7889

-3.6944

-3.6944

5.7889

-10.0669

12.6833

2.2111

-1.9056

7.3167

-31.5331

2

r

TND P

-31.5331

7.3167

-1.9056

2.2111

12.6833

-10.0669

5.7889

-3.6944

-3.6944

5.7889

-10.0669

12.6833

2.2111

-1.9056

7.3167

-31.5331

Permute column by 1,2,4,3P

2 2

, ,N Ni j N i N jD D

Symmetry property of Chebyshev differential matrix :

-31.5331

7.3167

12.6833

-10.06691D is NOT sym.

-3.6944

5.7889

2.2111

-1.90562D is NOT sym.

r

01 r

1r

2r

3r

4r

51rN

r r

0 1 2 3 4 5 60 2N

0r

Row-major indexing: remove redundancy [1]

redundant

1 2 3 4 5 6

7 8 9 10 11 12

11

12

13

u

u

u

14

15

16

u

u

u

21

22

23

u

u

u

24

25

26

u

u

u

1u

2u

1

2

3

4

5

6

7

8

9

10

11

12

,ij i ju u r

Index order

Index order

,1 ,2 , , 1 , 2 ,, , , , , , ,T

i i i i m i m i m i Nu u u u u u u

Define active variable for 1 1ri N and 1 j N

total number of active variables is

2 6 12k N

NOT

1 4 6 24rN N

0 1r 1rkr0

1kr 1rNr 1

rNr

rk jr 1k jr

1 0k kr r

1 0k j k jr r

1 1 0rN

r r

2 1rN k

2N m

is odd, and

is even, and

Row-major indexing: remove redundancy [2]

suppose 0 j k , then 1 0k j k jr r

cos : 0i rr

ir i N

N

2 11

1cos cos cosrN k

k j k jr r r

k j k j k jr r

N N N

since

2,

N m

: 1j j j N

Hence forj m j 1 ,j m and mod 2m j j

Row-major indexing: remove redundancy [3]

0 i k for

1 1, , mod 2k i j k i ju r u r From symmetry condition, we have

0 j m 1 , ,k i j k i m ju r u r

and

1 , ,k i m j k i ju r u r , symmetry condition implies

1, , 1,

2, 1, 2,

1, 1, ,

1

1

1r

k j k m j m j

k j k m j m j

N j m j k m j

u u u

u u u

u u u

1, , 1,

2, 1, 2,

1, 1, ,

1

1

1r

k m j k j j

k m j k j j

N m j j k j

u u u

u u u

u u u

Therefore, we have two important relationships

1

2

r

01 r

1r

2r

31 r

00 1 2 3

Kronecker product [1]

1 2 3

4 5 6

11

12

13

u

u

u

21

22

23

u

u

u

1u

2u

1

2

3

4

5

6

,ij i ju u r Define active variable

for 1 2i and 1 3j

Separation of variable: assume matrix A acts on r-dir and matrix B acts on

1 1,11 12

2,21 222

,

,

j j

jj

r ua aAu

ua ar

dir

is independent of j

1 11 12 13 ,1

2 21 22 23 ,2

3 31 32 33 ,3

,

,

,

i i

i i

i i

r b b b u

Bu r b b b u

r b b b u

is independent of i

Let 1 6

2

uX R

u

be row-major

index of active variable

Kronecker product [2]

11 12 6 6

21 22

a B a BA B R

a B a B

Kronecker product is defined by

Case 1:

A I

11 12 1 11 1 12 2

21 22 2 21 1 22 2

a B a B u a Bu a BuA B X

a B a B u a Bu a Bu

1

2

BuI B X

Bu

1 11 12 13 ,1

2 21 22 23 ,2

3 31 32 33 ,3

,

,

,

i i

i i

i i

r b b b u

Bu r b b b u

r b b b u

Case 2:

B I

11 21

11 12 12 22

13 2311 1 12 2

21 1 22 2 11 21

21 12 22 22

13 23

u u

a u a u

u ua u a uA I X

a u a u u u

a u a u

u u

1 1,11 12

2,21 221

,

,

j j

jj

r ua aAu

ua ar

Kronecker product [3]

Case 3: permutation, if permute

2 3

r

01 r

1r

2r

31 r

00 1 2 3

1 2 3

4 5 6

r

01 r

1r

2r

31 r

00 1 2 3

1 2 3

4 5 6

11

12

13

u

u

u

21

22

23

u

u

u

1

2

uX

u

11

13 1

12

u

u P u

u

21

23 2

22

u

u P u

u

1,3,2P 1

2

uX

u

1

2

P uI P X X

P u

Kronecker product [4]

Case 4:

1 2,A diag a a

1 1

2 2

a BuA B X

a Bu

1 1 11 12 13 1,1

1 1 2 1 21 22 23 1,2

1 3 31 32 33 1,3

,

,

,

r b b b u

a Bu r a b b b u

r b b b u

2 1 11 12 13 2,1

2 2 2 2 21 22 23 2,2

2 3 31 32 33 2,3

,

,

,

r b b b u

a Bu r a b b b u

r b b b u

2

2

1 1 rr ru u u u

r r

2,2 2 2

1 2

1 1 1, , , N

k

diag Dr r r

1i

2i

22

1 1rr ru u u u

r r , on 1,1r and 0,2

2 22

1 1r rN N r ND u D u D u u

r r

2 1rN k 2N m is odd, and is even, then

cos : 1ir

ir i k

N

: 1j j j N

Active variable is

Total number is ,kN NOT

1rN N

Non-active variable active variable [1]

, that is 0, 0 2i jr

x

y

12

3

4 5

6 1 0u r

Note that differential matrix rN

D is of dimension

0,

1,

1,

,

r

r

j

j

N j

N j

u

u

u

u

Neglect due to B.C.,acts on

1rN

However 2

rND

rNDand act on

1,

2,

,

j

j

k j

u

u

u

1,

2,

1,r

k j

k j

N j

u

u

u

NOT active variable, how to deal with?

Non-active variable active variable [2]

1, , 1,

2, 1, 2,

1, 1, ,

: 1:1

r

k j k m j m j

k j k m j m j

N j m j k m j

u u u

u u uk

u u u

1, , 1,

2, 1, 2,

1, 1, ,

: 1:1

r

k m j k j j

k m j k j j

N m j j k j

u u u

u u uk

u u u

: 1:1k where is a permutation matrix

and

From previous discussion, we have following relationships which can solve this problem

Non-active variable active variable [3]

,ij i ju u r Recall for 1 1ri N and 1 j N

, ,, row of ,r rr N i j r N jD u r i D u

Consider Chebyshev differential matrix , rr ND acts on

u and evaluate at

,i jr

We write in matrix notation

, ,1 ,2 , , 1 , 2 , 1,r rr N i j i i i k i k i k i ND u r d d d d d d

0r 0r

1,

2,

,

j

j

k j

u

u

u

1,

2,

1,r

k j

k j

N j

u

u

u

0r

0r

Question: How about if we arrange equations on , i jr i when fixed j

Non-active variable active variable [4]

111 12 1, 1, 1 1, 2 1, 1

221 22 2, 2, 1 2, 2 2, 1

,1 ,2 , , 1 , 2 , 1

1,1 1,2 1,1

2

1

,

,

,

,

,

,

r

r

r

r

r

jk k k N

jk k k N

k j k k k k k k k k k N

N

k k k kk j

k j

N j

rd d d d d d

r d d d d d d

r d d d d d dD u

d d d dr

r

r

1, 1 1, 2 1, 1

2,1 2,2

1,1 1,2 1, 1, 1 1, 1

r

r r r r r r

k k k k k N

k k

N N N k N k N N

d d

d d

d d d d d

1,

2,

,

j

j

k j

u

u

u

1,

2,

1,r

k j

k j

N j

u

u

u

,

1,

1,

k m j

k m j

m j

u

u

u

0r

0r

0r

0r

We only keep operations on active variable 0,u r

Later on, we use the same symbol 1 2rND E E

, ju r

That is, only consider equation ,rN i jD u r for 0ir

1 2

3 4rN

E ED

E E

abbreviate

Non-active variable active variable [5]

Define permutation matrix 11: , 1: 1: 1

1

r

I

P k N k

k

k

1,

2,

,

j

j

k j

u

u

u

1,

2,

1,r

k j

k j

N j

u

u

u

, ju r , jP u r

1,

2,

,

j

j

k j

u

u

u

1,

2,

,

m j

m j

k m j

u

u

u

,

1,

1,

k m j

k m j

m j

u

u

u

Active variable with the same indexing in r-direction

Let

Non-active variable active variable [6]

, ,r r

TN j N jD u r D P Pu r

Moreover, we modify differential matrix according to permutation P by

11 12 1, 1, 1 1, 2 1, 1

21 22 2, 2, 1 2, 2 2, 1

1 2

,1 ,2 , , 1 , 2 , 1

r

r

r

r

k N k k

k N k kTN

k k k k k N k k k k

d d d d d d

d d d d d dD P E E

d d d d d d

where

0r 0r

1, 1,

2, 2,

, 1 2

, ,

,r

j m j

j m j

r N j

k j k m j

u u

u uD u r E E

u u

such that for 1 j m

1, 1,

2, 2,

, 1 2

, ,

,r

m j j

m j j

r N m j

k m j k j

u u

u uD u r E E

u u

and

Evaluated at 1 2, , , , , ,j j k jr r r

Non-active variable active variable [7]

1,1 1,2 1, 1, 1 1, 2 1,2

2,1 2,2 2, 2, 1 2, 2 2,2, 1 2 1 2

,1 ,2 , , 1 , 2 ,2

, , , , , ,r

m m m m

m m m mr N m

k k k m k m k m k m

u u u u u u

u u u u u uD u r r r E E

u u u u u u

1, 1 1, 2 1,2 1,1 1,2 1,

2, 1 2, 2 2,2 2,1 2,2 2,, 1 2 2 1 2

, 1 , 2 ,2 ,1 ,2 ,

, , , , , ,r

m m m m

m m m mr N m m m

k m k m k m k k k m

u u u u u u

u u u u u uD u r r r E E

u u u u u u

,1

,2

,

,

i

ii

i m

u

uU

u

define

, 1

, 2

,2

i m

i mi

i m

u

uV

u

and active variable

1 1

2 21 2 1 2, ,

T T

T T

T Tk k

U V

U VX X X X X

U V

, 1 2 1 1 2 2 2 1, , , , | |rr N mD u r r E X X E X X To sum up

Non-active variable active variable [8]

1 1

2 21 2

T T

T T

T Tk k

U V

U VX X X

U V

Note that under row-major indexing, memory storage of

is 1 1 2 2| | | | | | | | |TT T T T T T T T

j j k kmem X U V U V U V U V

but 2 1 1 1 2 2| | | | | | | | | |TT T T T T T T T

j j k kmem X X V U V U V U V U

If we adopt Kronecker products, then 2 1| mk

m

Imem X X I mem X

I

, 1 2 1 2, , , ,r

m mr N m

m m

I ID u r r E E mem X

I I

Definition: Kronecker product is defined by

11 12 1

12 22 2

1 2

n

n

m m mn

a B a B a B

a B a B a BA B

a B a B a B

1 22,

3 4rr N

D DD

D D

Non-active variable active variable [9]

The same reason holds for second derivative operator

2, 1 2 1 2, , , ,

r

m mr N m

m m

I ID u r r D D mem X

I I

2, 1 2rr ND D D

neglect equation on

0r 2, 1 2r

Tr ND P D D

then

, ,

1 1, row of matrix

r r

Tr N i j r N

i

D u r i D Pr r

1 2

1 m m

m m

I IR E E

I Ir r

where

1

2

1 2

1/

1/ 1 1 1, , ,

1/k

k

r

rR diag

r r r

r

collect equations on each :1ir i k

we have

Non-active variable active variable [10]

We write second derivative operator on direction as

2

22

2

1 0 mod

6 3

1, 0 mod

2sin / 2

jN j

j Nh

S x

j Njh

22,N N k jD S x x where

,1

,22 2 2, , ,

,2

,

i

i iN i N N

i

i m

u

u UD u r D D

V

u

implies 2 2

, ,2 2

1 1, i

N i Nii

UD u r D

Vr r

for 1 , 1 2 0ii k j m r

,1

,22 2 2, , ,

,2

, row of row of

i

i iN i j N N

i

i m

u

u UD u r j D j D

V

u

so

1 12 2,2 ,2

1 11 11

22 22,,2 22 2

2, 2 222

22

,2,2

1 1

,11

,1

,11

N N

NNN

k

kN

Nkkk

U UD D

V Vr rr

U UDDr

rD u V Vrr

rU DD rVr

k

k

U

V

Non-active variable active variable [11]

If we adopt Kronecker products, then

1

22 2 2, ,2

,

,1

,

N N

k

r

rD u R D mem X

r

r

Non-active variable active variable [12]

2, 1 2 1 2, , , ,

r

m mr N m

m m

I ID u r r D D mem X

I I

summary

, 1 2 1 2

1 1, , , ,

r

m mr N m

m m

I ID u r r R E E mem X

I Ir r r

1

22 2 2, ,2

,

,1

,

N N

k

r

rD u R D mem X

r

r

1

2

3

Note that

1 1 1 1 2 1 2

2 2 1 2 2 2 21 2 2

1 2 2

, , , ,

, , , ,, , ,

, , , ,

m

mm

k k k k m

r r r r

r r r rr r r

r r r r

so that all three system of equations are of the same order.

Non-active variable active variable [13]

22

1 1rr ru u u u

r r , on 1,1r and 0,2

Discretization on mem X

2L mem X mem X

then

1 2 1 2

2 2,

2 21 1 2 2 ,

m m m m

m m m m

N

m mN

m m

I I I IL D D R E E

I I I I

R D

I ID RE D RE R D

I I

where

1

2

1 2

1/

1/ 1 1 1, , ,

1/k

k

r

rR diag

r r r

r

Example: program 28

2u u with boundary condition 1, 0u r

25rN is odd and 20N is even, let eigen-pair be ,k kV

2 21 1 2 2 ,

m mN

m m

I IL D RE D RE R D

I I

sparse structure of

Dimension :

112 20 240

2rN N

Example: program 28 (mash plot of eigenvector)

Eigenvalue is sorted, monotone increasing and normalized to first eigenvalue

1 2 n

Eigenvector is normalized by supremum norm,k

kk

VV

V

1

2

1 5.7832 2 3 14.682 4 5 26.3746 6 30.4713 7 8 40.7065 9 10 42.2185

Example: program 28 (nodal set)

Exercise 1

2u u with boundary condition 1,2 0u r

25rN is odd and 20N is even, let eigen-pair be ,k kV

1 2r on annulus

1,1x Chebyshev node on Chebyshev node on 1 2r

11

2

xr

2 2 2, , ,r rr N r N N NL D RD I R D

Exercise 1 (mash plot of eigenvector)

Eigenvalue is sorted, monotone increasing and normalized to first eigenvalue

1 2 n

Eigenvector is normalized by supremum norm,k

kk

VV

V

1

2