9/20/20151 Numerical Solution of Partial Differential Equations Using Compactly Supported Radial Basis Functions C.S. Chen Department of Mathematics University.

04/19/23 1

Numerical Solution of Partial Differential Equations Using Compactly Supported Radial Basis Functions

C.S. Chen

Department of Mathematics

University of Southern Mississippi

04/19/23 2

PURPOSE OF THE LECTUREPURPOSE OF THE LECTUREPURPOSE OF THE LECTUREPURPOSE OF THE LECTURE

TO SHOW HOW COMPACTLY SUPPORTED RADIAL

BASIS FUNCTIONS (CS-RBFs) CAN BE USED TO

EXTEND THE APPLICABILITY OF BOUNDARY

METHODS TO PROVIDE 'MESH FREE' METHODS

FOR THE NUMERICAL SOLUTION OF PARTIAL

DIFFERENTIAL EQUATIONS.

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04/19/23 4

MESH METHODMESHLESS METHOD

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The Method of Particular Solutions

Consider the Poisson’s equation

,),( xxfLu,),( xxgu

Where ,3,2, dRd is a bounded open nonempty domain

with sufficiently regular boundary .Let puuv where pu satisfying )(xfLu p but does not necessary satisfy the boundary condition in (2).

(1)(2)

v satisfies , ,0 xLv

. ),()( xxx pugu

(3)

(4)(5)

04/19/23 6

• Domain integral Domain integral

• Atkinson’s method (C.S. Chen, M.A. Golberg & Y.C. Hon, The MFS and quasi-Monte Carlo method for diffusion equations, Int. J. Num. Meth. Eng. 43,1421-1435, 1998)

(Requires no meshing if = circle or sphere)

• Radial Basis Function Approximation of

• Others

)();,()( dvQfQPGPu p

for solution lfundamenta a is );,( QPG L

ˆ,)();,()( dvPfQPGPu p

f

04/19/23 7

Assume that )(ˆ)( xx ff and that we can obtain an analytical solution up

to

).(ˆˆ xfu p Then

.ˆ pp uu

To approximate f by f we usually require fitting the given

data set xi

N

1 of pairwise distinct centres with the imposed

conditions

.1 ),(ˆ)( Niff xx

The Method of Particular Solutions

(6)

04/19/23 8

The linear system

( ) , ,f a i Ni ii

N

i jx x x

1

1

is well-posed if the interpolation matrix is non-singular

A i ji N

x x1

Once f in (6) has been established,

u ap i ii

N

1

(7)

(8)

where i iand

i i i i x x x x, .

(9)

04/19/23 9

Compactly Supported RBFsLet be a continous function with (0) 0. If let : ,R R i

x

i i ix x x ,

where is the Euclidean norm. Then is called the RBF. i

References• Z. Wu, Multivariate compactly supported positive definite radial functions, Adv. Comput. Math., 4, pp. 283-292, 1995.• R. Schaback, Creating surfaces from scattered data using radial basis functions, in Mathematical Methods for Curves and Surface, eds. M. Dahlen, T. Lyche and L. Schumaker, Vanderbilt Univ. Press, Nashville, pp. 477-496, 1995• W. Wendland, Piecewise polynomial, positive definite and compactly supported RBFs of minumal degree, Adv. Comput. Math., 4, pp 389- 396, 1995.

04/19/23 10

• C.S. Chen, C.A. Brebbia and H. Power, Dual receiprocity method using compactly supported radial basis functions, Communications in Numerical Methods in Engineering, 15, 1999, 137-150.

• C.S. Chen, M. Marcozzi and S. Choi, The method of fundamental solutions and compactly supported radial basis functions - a meshless approach to 3D problems, Boundary Element Methods XXI, eds. C.A. Brebbia, H. Power, WIT Press, Boston, Southampton, pp. 561-570, 1999.

• C.S. Chen, M.A. Golberg, R.S. Schaback, Recent developments of the dual reciprocity method using compactly supported radial basis functions, in: Transformation of Domain Effects to the Boundary, ed. Y.F. Rashed, WIT PRESS, pp183-225, 2003.

• M.A. Golberg, C.S. Chen, M. Ganesh, Particular solutions of the 3D modified Helmholtz equation using compactly supported radial basis functions, Engineering Analysis with Boundary Elements, 24, pp. 539-547, 2000.

References

04/19/23 11

Wendland’s CS-RBFs

( )( ) , ,

, .1

1 0 1

0 1

rr r

rn

nDefine

For d=1,

( )

( ) ( )

( ) (8 )

1

1 3 1

1 5 1

0

3 2

5 2 4

r C

r r C

r r r C

For

For d=2, 3,

( )

( ) ( )

( ) ( )

( ) ( )

1

1 4 1

1 35 18 3

1 32 25 8 1

2 0

4 2

6 2 4

8 3 2 6

r C

r r C

r r r C

r r r r C

(10)

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Globally Supported RBFs

φ=1+r r c2 2

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Compactly Supported RBFs

( )1 2r ( ) ( )1 4 14r r

04/19/23 14

Compact support cut-off parameter (scaling factor)

04/19/23 15

Analytic Particular Solutions L= in 2D

r

r rr

r

1 4 1

0

4

, ,

, .

1, in 2D,

d dr

r dr dr

4

7 6 5 4 2

5 4 3 2

= 1 4 1

4 5 4 5, .

7 2 2 2

d r r rr r dr

dr

r r r r rA r

04/19/23 16

6 5 4 3

15 4 3 2

7 6 5 4 2

5 4 3 2

4 5 4 5=

7 2 2 2

4 5 4 5, .

49 12 5 8 4

r r r r r rC dr

r r r r r AB r

r

where C and D are to be determined by matching and ' at .r

Choose A= B = 0, we have

7 6 5 4 2

5 4 3 2

4 5 4 5, ,

= (*)49 12 5 8 4ln( ) , ,

r r r r rrr

C r D r

04/19/23 17

6 5 4 3

5 4 3 2

4 5 4 5= , ,

7 2 2 2

,

d r r r r r rr

drC

rr

Note that

6 5 4 3

5 4 3 2

2

4 5 4 5For ,

7 2 2 2

14 14

Cr

CC

From (*), we have 2 2529

ln5880 14

D

04/19/23 18

7 6 5 4 2

5 4 3 2

2 2

4 5 4 5, ,

49 12 5 8 4=

529ln , ,

5880 14

r r r r rr

r

rr

04/19/23 19

Example in 2D

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25( , ) sin( ) cos , ( , ) ,

4 2

( , ) sin( ) cos , ( , ) .2

yu x y x x y

yu x y x x y

Example 2D

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

10,000 uniform grid points

04/19/23 23Sparse matrix for uniform grid points.

04/19/23 24

CS-RBF used:

4

1 4 1r r

MFS: Fictitious circle with r = 4.

Interpolation points: 100x100 uniform grid points

α RMSE NZ C PU (sec)

0.10 3.76E-4 561,013 1.65

0.20 1.55E-5 2,171,057 3.36

0.25 8.43E-6 3,376,025 5.32

04/19/23 25

Analytic Particular Solutions L= in 3D

r

r rr

r

1 4 1

0

4

, ,

, .

Recall

Since

12

2

r

d

drr

d

dr, in 3D,

we have

(11)

r

r r r r rr

rr

2 4

2

5

3

5

4

7

5

2 36 2

2

3

5

16 14

14 42

,

, .

04/19/23 26

Numerical Example in 3D

Consider the following Poisson’s problem

u x y z x y z x y z

u x y z x y z x y z

( , , ) cos( ) cos( ) cos( ), ( , , ) ,

( , , ) cos( ) cos( ) cos( ), ( , , ) .

3

-1

0

1-0.5-0.2500.250.5-0.5

-0.25

0

0.25

0.5

-1

0

1

-0.5

-0.25

0

0.25

0.5

Physical Domain

04/19/23 27

-1.0 -0.5 0.0 0.5 1.0 1.50

1

2

3

4

5

6

7

8

= 1.4

= 1.0

= 0.7

Rel

ativ

e E

rror

s (

%)

X Axis

The effect of various scaling factor α

We choose 1 4 14r r( ) to approximate the forcing term.

To evaluate particular solutions, we choose 300 quasi-random pointsin a box [-1.5,1.5]x[-0.5,0.5]x[-0.5,0.5]. The numerical results are compute along the x-axis with y=z=0.

04/19/23 28

Modified Helmholtz Equation in 3D

2 22

1 ( ), 0 ,1

0, .

nr

p r rd dr

r dr drr

Let j ( )( )

, , ( ) lim , , , .rw r

rr

d

dr

w

rj

r

j

j

0 0 0 1 2

0

1 12

22

2r

d

drr

d

dr r

d w

dr

(16)

(15)

(17)

04/19/23 29

Hence, (15) is equivalent to

d w

drw r

rp

rr

r

n2

22 1 0

0

, ,

, .

The general solution of (18) is of the form

w rAe Be q r r

Ce De r

r r

r r( )( ), ,

, .

0

q(r) can be obtained by the method of undetermined coefficients,or by symbolic ODE solver (MATHEMATICA or MAPLE).A,B,C and D in (19) are to be chosen so that in (16) is twice differentiable at r = 0 and

(18)

(19)

04/19/23 30

Theorem 1. Let w be a solution of (18) with w(0)=0. Then defined by (16) is twice continuously differentiable at 0 with (0)=w'(0), '(0)=0, ''(0)=[s2 w'(0)+p(0)]/3.Furthermore, (r) satisfies (15) as limr→0+

A B q

Ae Be q Ce De

A e B e q C e D e

( )

( )

' ( )

0 0

Choose D = 0. Consequently, we get

A B q

Be q q

C B e q e q

[ ( )]( ' ( ) ( )]

( ) ( ) ( )

0

21 02

(20)

(21)

04/19/23 31

Hence, the particular solution Φ is given by

( )

( ( ) ' ( ), ,( )

, ,

, .

r

s B q q rAe Be q r

rr

Ce

rr

sr sr

sr

2 0 0 0

0

Notice that for r > α and large wave number λ

Ce

r

e q q

r

q e q e

rr

r r r r

( ) ( )' ( ) ( ) ( )

.2

0

0 for

(22)

04/19/23 32

d w

drw r

rr

2

22

2

1 0

,

Example

The general solution of

is given by

w rr r r r

Ae Ber r( )

4 6 22 2 2 2 3 2

4 2

q rr r r r

( ) 4 6 22 2 2 2 3 2

4 2

Hence,

A,B,C can be obtain from (21).

04/19/23 33

For

( ) ,rr r

1 4 1

4

q r r

r r

r r r

( )

480 2880 60 1800 1

240 1400 10 300

20 120 15 4

6 3 8 5 4 2 6 4 2

24 3 6 5

32 2 4 4

42 3 4 5 2 4

52 5

6

04/19/23 34

Numerical Results

400I u f

u g

, ,

, .

in

on

( , , ) : ( , , )x y z R H x y z3 1 with

H x y z x x y z( , , ) min ,

3

4

3

4

2 2

2 2

Consider (23)(24)

04/19/23 35

We choose f and g such that the u(x,y,z)=ex+y+z /400 is theexact solution of (23)-(24).To evaluate particular solutions, we choose N=400 quasi-random points . We choose the CS-RBFs

( )rr

12

04/19/23 36

Error estimates for various compact support cut-off parameter

||u-uN || CPU-time (sec.)

0.2 2.2864E-02 01.40

0.4 1.6194E-02 04.43

0.6 9.7795E-03 07.48

0.8 6.4514E-03 10.63

1.0 4.5061E-03 13.71

1.2 3.8208E-03 16.98

1.4 3.3006E-03 20.53

1.6 2.9089E-03 24.08

1.8 2.6375E-03 27.60

04/19/23 37

Multilevel Interpolation

Given a set of X scattered points, we decompose X into a nestedsequence

X X X XM1 2 M subsets X x x X k Mk

kNk

k 1 1( ) ( ), , , .

Algorithm:

S f

S f S

S f S

X X

X X

M X kk

M

XM M

1

2 1

1

1

1 1

2 2

| | ,

| ( )| ,

| ( )| .

S x c x xk jk

k jk

j

N k

( ) ( ) ( )

1

S S S fM X X1 2 | |

04/19/23 38

The basic idea is to set α1 relatively large with few interpolationpoints, and to let the αk decrease as k increases with more points.

Reference:M.S. Floater, A. Iske, Multistep scattered data interpolation using compactlysupported radial basis functions, J. Comp. and Appl. Math., 73, 65-78, 1996.

30 interpolation points 100 interpolation points

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300 interpolation points

04/19/23 40

Numerical Results

u e x y D

u e e y x y D

D

x y

x y x

2 , ( , ) ,

cos , ( , ) ,

where is the Oval of Cassinii.

10.50-0.5-1

0.4

0.2

0-0.2

-0.4

04/19/23 41

x y Numer. Exact Abs. Error

0.00 0.00 1.99994 2.00000 0.000061

-0.63 0.00 3.74097 3.74098 0.000012

1.34 0.00 7.65247 7.65261 0.000028

0.17 0.17 2.17206 2.35207 0.000082

0.00 0.20 1.79712 1.79707 0.000052

-0.37 0.37 1.11578 1.11594 0.000156

-1.34 0.00 0.5228 0.52270 0.000130

α/1.5, 1.0, 0.6, 0.2/ n/30, 60, 200, 400/

04/19/23 42

For time-dependent problems, we consider two approaches to convert problems to Helmholtz

equation • LAPLACE TRANSFORM

• FINITE DIFFERENCES IN TIME

ALSO POSSIBLE

• OPERATOR SPLITTING (RAMACHANDRAN AND BALANKRISHMAN)

04/19/23 43

DPPhPu ),0,()0,(

(HEAT EQUATION)

3,2,),,(),(),( dRDPtPftPutPu dt

0,),,(),( tDPtPgtPu

Consider the BVP

where D is a bounded domain in 2D and 3D.

Let

0

),(),(ˆ dttPuesPu st

(28)Then

(25)(26)(27)

),(ˆ sPu satisfies

DPsPgsPu

sPMPhsPfsPussPu

),,(ˆ),(ˆ

),()(),(ˆ),(ˆ),(ˆ (29)(30)

04/19/23 44

(WAVE EQUATION)

)()0,(),()0,(

0,),,(),(

0,),,(),(

00 PvPuPuPu

tDPtPgtPu

tDPtPutPu

t

tt

Consider BVP

Then ),(ˆ sPu satisfies

DPsPgsPu

PvPsusPussPu

),,(ˆ),(ˆ

)()(),(ˆ),(ˆ 002

Similar approach can be applied to hyperbolic-heat equation and heat equations with memory

(31)

(32)(33)

(34)(35)

04/19/23 45

(HEAT EQUATION)1For .0,Let nnn tttnnt

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

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

1

1

nn

nn

tPutPutPu

tPutPutPu

and/)],(),([),( 1 nnt tPutPutPu

satisfies ),()(Then nn tPuPv )(//)1(//11 Pmfvvvv nnnnnn

Method) s(Rothe' 1For nnnn fvvv //11

Nicholson)-(Crank 1/2For )(2/2/2 11 Pmfvvvv nnnnnn

(36)(37)

(38)

(39)

(40)

(41)

04/19/23 46

(J. Su & B. Tabarrok, A time-marching integral equation method for unsteady

state problems, Comp. Meths. Appl. Mech. Eng. 142, 203-214, 1997) (WAVE EQUATION)

211 /)2( nnntt uuuu

satisfies Then nn uv

12

12

11 )1(/)2(/ nnnnn vvvvv

CONVECTION-DIFFUSION EQUATION

2/)(

field) velocity (

11t

nn

t

uuu

VuuVu

nnnnn vVvvvv

2/)()1( 1111

Similar approach works for non-linear equation

),,(),(),( uuPftPutPu t

(41)

(42)

(43)(44)

(45)

(46)