p. venkataraman rochester institute of technologygraduate seminar, january, 7, 2010 bezier functions...

P. VenkataramanP. Venkataraman

Rochester Institute of Technology Graduate Seminar, January, 7, 2010

Bezier Functions : From Airfoils to the Inverse Problem

Mechanical Engineering

Bezier Functions From Airfoils to the Inverse Problem

P. Venkataraman





One Dimensional Example

0 50 100 150 200 2501.2

1.25

1.3

1.35

1.4

1.45x 10

4

Ori

gin

al D

ata,

Fit

ted

Dat

a

points

Closing DJIA between Aug and Dec 2007

0 50 100 150 200 2501.2

1.25

1.3

1.35

1.4

1.45x 10

4

Ori

gin

al D

ata,

Fit

ted

Dat

a

points

DJIA - Adjust Close 17 Sep - Dec 18

A Bezier function over all the data

Order of function = 20

Mean original data = 13172.432

Mean Bezier data = 13172.423

Avg. Error = 98.34

Maximum Data = 14164.53

Std. Dev (original) = 530.19

Std. Dev. (Bezier) = 514.68

1





What is a Bezier Function ?

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

x: independent variable

y: d

ep

end

en

t va

ria

ble

s

[a1,b1][a1,b1]

[a2,b2]

[a3,b3]

[a4,b4]

[a5,b5]

Convex hullBezier VerticesBezier Curve: order 4

,0

( ) ( ) ( ) , 0 1

n

i n ii

Bx p y p J p p

1, ( ) ( )i n in i

nJ p p p

i

p : parameter

Bernstein basis

Number of vertices: 5

Order of the function : 4

A Bezier function is a Bezier curve that behaves like a function

The Bezier curve is defined using a parameter

Instead of y=f(x);

both x and y depend on the same parameter value; x = x(p) and y = y(p)

2





Matrix Description of Bezier Function

0 1 2 3 4 50

0.5

1

1.5

2

2.5

3

3.5

4

x: independent variable

y: d

ep

end

en

t va

ria

ble

s

[a1,b1][a1,b1]

[a2,b2]

[a3,b3]

[a4,b4]

[a5,b5]

Convex hullBezier VerticesBezier Curve: order 4

[ ( ) ( )] [ ][ ][ ]x p y p P N B

4 3 2[ ] [ 1];

1 -4 6 -4 1 0 0

-4 12 -12 4 0 1 3

[ ] 6 -12 6 0 0 [ ] 2 1

-4 4 0 0 0 3 2

1 0 0 0 0 5

P p p p p

N B

0

This allows the use of Array Processing for shorter computer time

3





For a selected order of the Bezier function (n) Given a set of (m) vector data ya,i , or [Y], find the coefficient matrix, [B] so that the corresponding data set yb,i , [YB ] produces the least sum of the squared error

2

, ,

m

a i b ii

E y y T T

B B A AE Y Y Y Y Y P NB Y P NB

0E

B

1[ ] [ ] [ ]T TA A AB P P P Y

Minimize

FOC:

The Best Bezier Function to fit the Data

Once the coefficient matrix is known, all other information can be generated using array processing

4

0 50 100 150 200 2501.2

1.25

1.3

1.35

1.4

1.45x 10

4

Ori

gin

al D

ata,

Fit

ted

Dat

a

points

DJIA - Adjust Close 17 Sep - Dec 18





Bezier Airfoils 6

x/c

y/c

01

curve 1 curve 2

curve 3 curve 4

There are 2 curves for the top surface

There are 2 curves for the bottom surface

All curves are 6 th order

Slope continuity is enforced at all curve junctions (except off course the leading and trailing edge)

Properties

Second derivative continuity is enforced between the forward and rear curves

Second derivative direction continuity is enforced at the leading edge

Any past/contemporary/ single element airfoil, low speed or transonic, can be constructed using the Bezier Curves shown above.





Airfoil Optimization 7

x/c

y/c

01

curve 1 curve 2

curve 3 curve 4

Single and Multipoint Airfoil Design

Single-Point design: cruise Multi-Point (Two-Point) design –

cruise and takeoff.

The airfoil geometry is parameterized using Bezier CurvesThe aerodynamic information is obtained using the XFOIL program(Professor Drela MIT)

Airfoils can be designed for geometry• Area • Maximum thickness • Maximum thickness for top and bottom• Location of maximum thickness • Disparate locations of maximum thickness

Airfoils can be designed for performance• Maximum CL

• Minimum CD

• Maximum CL/CD

• Maximum CL3/2 /CD





8Differential Equations - ODE

A disk of radius R is rotating with the angular speed ω in still fluid. The flow is steady, incompressible, has constant property, and is axisymmetric. The fluid at the disk has to satisfy the no slip condition. The centrifugal effects cause the fluid to leave the disk radially near the disk. The flow above the disk must replace this airflow through a downward spiraling flow. A cylindrical coordinate system (r, θ, z) is used for description. Vr, Vθ, Vz, are the velocity components. p is the pressure, ν, the dynamic viscosity. The continuity and the Navier-Stokes equations are

2 2 2

2 2

2 2

2 2

2 2

2 2

0

1

1 1

zr r

zr r r rr z

rr z

z z z z zr z

VV Vr r z

VVV p V V VV V

r r z r r r r z

V V V V V V VV V

r r z r r r z

V V V V VpV V

r z z r r r z

0: 0; ; 0;

: 0; 0;r z

r

z V V r V

z V V

Navier-Stokes equation

Boundary Conditions :

Flow Over a Rotating Disk





9Differential Equations - ODE

Transformation Relations


Flow Over a Rotating Disk

Transformed Navier-Stokes equation

; ( ); ( );

( ); ( ) ( );

r

z

zZ z V r F Z V r G Z

V H Z p z P Z

2 2

2 0

0

2 0

0

F H

F FH G F

FG HG G

P HH H

0; 0; 1; 0; 0

( 6); 0; 0

Z F G H P

Z F G





10Differential Equations - ODE Flow Over a Rotating Disk

Bezier Solution :

Three Bezier functions will be used to identify the functions F, G, and H. This is now a coupled set of nonlinear differential equations.

Optimization Problem :

22 2 2

1

1002

1

2

2

pn

i i i i i i ii

i i i i ii

f F H F F H G F

FG H G G

Minimize :

Subject to : 1 1 1 1 1 1

1 1 1 1

, [0,0]; , [0,1]; [ , ] [0,0]

, [0,0]; , [0,0];m m m m

Z F Z G Z H

Z F Z G

6 5 4 3 2

6 5 4 3 2

6 5 4 3

( ) -0.36800 - 0.23400 1.1100 0.68000 1.4400 3.3720

( ) 1.5270 - 4.4280 2.7150 3.4200 - 4.9500 1.7160

( ) - 0.56300 3.1740 - 6.7500 6.1000 - 0.8

Z p p p p p p p

F p p p p p p p

G p p p p p

2

6 5 4 3 2

0 1

8500 - 2.0760 1

( ) 1.7270 - 3.8640 - 0.69000 7.7600 - 5.7750

p

p p

H p p p p p p

Solution :





11Differential Equations - ODE Flow Over a Rotating Disk

Bezier Solution :

0 1 2 3 4 5 6-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Z

(F,

G,

H)

U - velocity

Curve:U - velocity

V - velocityCurve:V - velocity

W - Velocity

Curve:W - Velocity

-1 -0.5 0 0.5 10

1

2

3

4

5

6Laminar Flow near a Spinning Disk

Z

F, G, H

F (Vr)

F-Bezier

G (V)

G - Bezier

H (Vz)

H - Bezier

Comparison of Bezier Solution with Numerical Solution





Differential Equations - PDE Flow in a Channel

A steady, two-dimensional, constant property flow takes place in a two dimensional channel. The x-velocity (u) at the inlet is constant with the value U0. There is no y-velocity (v) at the inlet. The no slip conditions apply on both wall

x

y

L1

L2

U0 u = 0, v =0

u = 0, v =0

v = 0Navier-Stokes Equations :

2 2

2 2

2 2

2 2

0

1

1

continuity:

x-momentum:

y-momentum:

u v

x y

u u p u uu v

x y x x y

v v p v vu v

x y y x y


0

2 2 2

0 0 0 0 0

0 0 0 0 0

0 0

; ( , ) ( , ) ; ( , ) ( , ) ; ( , ) ( , ) ;

; ( , ) ( , ) ; ( , ) ( , ) ;

; ( , ) ( , ) ; ( , ) ( , ) ;

x u x y u y U v x y v y p x y p y c

y u x y u x v x y v x

y L u x y u x L v x y v x L

In the above, ρ is the fluid density and ν is the fluid kinematic viscosity. L1 is the length of the channel. L2 is the width of the channel. The domain is called the entering region of the flow as the viscous effects through the walls will shape the velocity profile in the channel as the flow proceeds left to right.

12






x

y

L1

L2

U0 u = 0, v =0

u = 0, v =0

v = 0

The nonlinear BVP problem will be solved using Bezier functions.

Here the solution will be represented by three surfaces in the solution domain.

The first is the solution for the velocity in the x-direction u(x, y), the second is the solution for the velocity in the y-direction v(x, y), and the third one is the solution for the pressure p(x ,y).

The Optimization Problem :

2

1 1

22 2

2 21 1

22 2

2 21 1

1

1

, ,

, ,, ,

, , , ,

( )p q

i j i j i j

p q

i j i j i ji j i j

p q

i j i j i j i j i j

u vMinimize F X

x y

u u p u uu v

x y x x y

v v p v vu v

x y y x y

0

2 2 2

0 0 0 0

0 0 0 0 0 0 0

0 0 0 0

; ( , ) ; ( , ) ;

; ( , ) ; ( , ) ;

; ( , ) ; ( , ) ;

x u y U v y

y u x v x

y L u x L v x L

Boundary Conditions:

13






x

y

L1

L2

U0 u = 0, v =0

u = 0, v =0

v = 0

Bezier Solution :

The solution presented corresponds to m = 9 and n = 6

0.33

0.661

0.991

1.32

1.65

1.98

2.31

2.64

2.97

3.3

3.63

3.96

4.294.62

4.95

5.28

5.61

5.95

6.28

6.61

4.62

4.95

y

x

Contour for u

0 10 20 30 400

0.2

0.4

0.6

0.8

1-0.0416-0.0328-0.024-0.0152

-0.00636

-0.00197-0.00197

0.00243

-0.00636

0.00683

0.0112

-0.0108

0.0156

-0.0152-0.0196

0.02

-0.024

0.0244

-0.0284

0.0288

-0.0328

0.0332

-0.0372

0.0376

-0.00197

0.00243

-0.0416

0.042 -0.00197

y

x

Contour for v

0 10 20 30 400

0.2

0.4

0.6

0.8

1

0 50

0.2

0.4

0.6

0.8

1

y

u

Profile for u

x =0

x =2.25

x =4.5

x =6.75

x =11.25

x =18

x =22.5

x =31.5

x =36

x =45-0.05 0 0.050

0.2

0.4

0.6

0.8

1

y

v

Profile for v

x =0

x =2.25

x =4.5

x =6.75

x =11.25

x =18

x =22.5

x =31.5

x =36

x =45

u velocity v velocity

14






x

y

L1

L2

U0 u = 0, v =0

u = 0, v =0

v = 0

Bezier Solution :

The solution presented corresponds to m = 9 and n = 6

p - solution

-1.41

1.3

4

6.71

9.42

12.1

14.8

17.5

20.2

22.9

All of the solutions can be represented by explicit polynomials in two parameters– which has not be done before

15





Bezier Function in 3D

A 3D Bezier function will be a surface in 2D. Bezier surface can be described as a vector-valued function of two parameters r and s

[ ( , ) ( , ) ( , )]; 0 , 1x r s y r s u r s r s

, ,

, , ,0 0

( ) (1 ) ; 0 1; ( ) (1 ) ; 0 1

[ ( , ) ( , ) ( , )] ( , ) ( ) ( )

i m i j n j

m i n j

m n

i j m i n ji j

m nJ r r r r K s s s s

i j

x r s y r s u r s Q r s B J r K s

2 2

4 3 2

3 4 4

3 2 3

3 2 2 3 2

( , ) -1 3 3

( , ) - 4 8 - 6 8

( , ) (-1 48 - 78 12 )

(4 -112 180 - 24 )

(-6 60 - 90 ) (-8 12 8)

0 1 0 1

x r s r r r

y r s s s s s

u r s r r r s

r r r s

r r s r r s

r s

16





Matrix Form of Bezier Function in 3D

[ ( , ) ( , ) ( , )]; 0 , 1x r s y r s u r s r s

[ ( , ) ( , ) ( , )] ( , )T

x r s y r s u r s Q r s R M B N S

4 3 21[ ] [ ... ] 1TTn nS s s s s s s s

1 3 2 1[ ...1] 1m mR r r r r r

1 0

1 2

1 0

0

11 1 ... 1

0 0 1 1

11 1 ... 0

0 1 1 2

[ ] . . . 0

11 1 ... 0

0 1 1 0

1 0 ... 00 0

m m

m m

m m m m m m m

m m m m

m m m m

m m

M

m m m m

m m

[ ]M

-1 3 -3 1

3 -6 3 0

-3 3 0 0

1 0 0 0

[ ]N

1 -4 6 -4 1

-4 12 -12 4 0

6 -12 6 0 0

-4 4 0 0 0

1 0 0 0 0

[ ]B

[0 0 0] [0 2 2] [0 3 3] [0 5 4] [0 6 5]

[1 0 0] [1 2 2] [1 3 3] [1 5 2] [1 6 1]

[3 0 0] [3 2 3] [3 3 0] [3 5 3] [3 6 5]

[5 0 0] [5 2 3] [5 3 0] [5 5 3] [5 6 5]

18





Minimize

FOC:

Bezier Filter for 3D Data

Once the coefficient matrix is known, all other information can be generated using array processing

For the filter, the best order is chosen on minimum absolute error

Given a set of array data [U], assuming an order for each dimension (m, n),

find the Bezier function coefficient matrix, [BU] so that the corresponding approximate data [UB] generates the least value for the sum of the squared error over the data array

2B

i j

E U U TTB A U A A U AU R M B N S F B G

0U

E

B

1T TU A A A A AB G IF F G IF U[ ] [ ] [ ]

18





Three Dimensional Bezier Function – Smooth Datay

x

Original Data

5 10 15 20 25 30 35 40 45 50

5

10

15

20

25

30

35

40

45

50

-6

-4

-2

0

2

4

6

Original Data about 2600 points based on MATLAB Peaks function

3D View of the Data

010

2030

4050

60

0

10

20

30

40

50

60-8

-6

-4

-2

0

2

4

6

8

10

x

Original Data

y

Ori

gin

al

-6

-4

-2

0

2

4

6

8

y

x

m =12, n =15 ,Least Sum of Absolute Error :179.8217

5 10 15 20 25 30 35 40 45 50

5

10

15

20

25

30

35

40

45

50

-6

-4

-2

0

2

4

6

Using the Bezier Filter

010

2030

4050

60

0

10

20

30

40

50

60-8

-6

-4

-2

0

2

4

6

8

x

Bezier Data

y

Bez

ier

-6

-4

-2

0

2

4

6 Contour Plot

3D Plot

original Bezier

mean 0.317 0.312

std. dev. 1.116 1.086

maximum 8.042 7.360

minimum -6.521 -6.405

average error: 6.91e-02

19





Three Dimensional Bezier Function – Rough Data

Same peaks function but randomly perturbed on both sides

y

x

Original Data

5 10 15 20 25 30 35 40 45 50

5

10

15

20

25

30

35

40

45

50

-6

-4

-2

0

2

4

6

Less dominant peaks diffused3D plot

010

2030

4050

60

0

10

20

30

40

50

60-8

-6

-4

-2

0

2

4

6

8

10

x

Original Data

y

Ori

gin

al

-6

-4

-2

0

2

4

6

8 Bezier FilterContour plot

y

x

m =12, n =12 ,Least Sum of Absolute Error :1702.726

5 10 15 20 25 30 35 40 45 50

5

10

15

20

25

30

35

40

45

50

-6

-4

-2

0

2

4

6

3D plot

010

2030

4050

60

0

10

20

30

40

50

60-8

-6

-4

-2

0

2

4

6

8

x

Bezier Data

y

Bez

ier

-6

-4

-2

0

2

4

6

average error: 6.54e-01

original Bezier

mean 0.322 0.325

std. dev. 0.859 1.035

maximum 8.253 7.481

minimum -7.651 -6.565

20





Bezier Function in Image Handling

The original image is 960 x 1280 pixels of size 671 KB

200 400 600 800 1000 1200

100

200

300

400

500

600

700

800

900

True image processing in MATLAB

Bezier filter applied to Red, Green and Blue color separately and combined

Highly nonlinear color distribution

21





Single Bezier Functions for the Image

200 400 600 800 1000 1200

100

200

300

400

500

600

700

800

900

200 400 600 800 1000 1200

100

200

300

400

500

600

700

800

900

Size = 671 KB

Bezier function representation

Function order 20 x 20

Coefficient storage = 11 KB (3 color streams)

Original image

22





Bezier Function in Four Quadrants

Original Image 671 KB

200 400 600 800 1000 1200

100

200

300

400

500

600

700

800

900

Four quads

Bezier function representation

200 400 600 800 1000 1200

100

200

300

400

500

600

700

800

900

Function order 20 x 20

Coefficient storage = 4*11 KB (3 color streams) = 44 KB

23





The Inverse ODE Problem

The inverse problem in this paper is very direct :

find the differential equation and the boundary conditions if the discrete solution is known everywhere

If [xi, yi], i = 1,2, .. p is known as the solution to

0( ) 0; (0)f D y y y

Then find f(D) and y0

OR

f(D) may be a linear or a nonlinear operatorThe ODE is homogenous

the forward or the direct boundary value problem is the determination of the solution everywhere if the differential equation is known and the boundary conditions are given

after all

24





The Solution Process

The procedure involves two steps:

Step 1: A “best” Bezier function is fitted to the data

This function, which is also the solution to the ODE, will satisfy the differential equation and identify the boundary condition

Step 2: The specific form of the differential equation is determined

This form is established from a generic representation of the ODE using a set of exponent and coefficient values

25





Why a Bezier Function?

Bezier functions can provide explicit solutions to the forward boundary value problem very effectively

The author’s papers in previous CIE conferences have shown Bezier functions can solve linear or nonlinear, single or multi variable, ordinary or partial differential equations, with initial and/or boundary values

Bezier functions are parametric curves based on Bernstein polynomial basis functions

“the Bernstein polynomial approximation to a continuous function mimics the gross features of the function remarkably well” - Gordon and Riesenfeld

As the order of the polynomial is increased, this approximation converges uniformly to the function and its derivatives where they exist

The Bezier curve delivers, at the minimum, the same smoothness as the primitive function it is trying to emulate

26





For a selected order of the Bezier function (n) Given a set of (m) vector data ya,i , or [Y], find the coefficient matrix, [B] so that the corresponding data set yb,i , [YB ] produces the least sum of the squared error

2

, ,

m

a i b ii

E y y

T T

B B A AE Y Y Y Y Y P NB Y P NB

0E

B

1[ ] [ ] [ ]T TA A AB P P P Y

Minimize

FOC:

Step 1:The Best Bezier Function to fit the Data

Once the coefficient matrix is known, all other information, including the derivatives can be generated using array processing

This is Step 1 of the solution process

The best m is determined by the lowest value of E

27





Step 2:The Generic Form of ODE

Many 3rd order ODE generic forms are used in the paper. For example

There are two types of unknowns: the exponents of the derivativesthe coefficients multiplying the terms

The exponents are expected to be integers The coefficients are unrestrictedThe function and its derivatives are known quantities after Step 1

Linear Generic Form

1 2 3

1 2

11

1

3 22 4

1 2 3 3 2

232

1 2 3 2

21 2 3

21 2 3

( )

( )

( )

( ) 0

a a aa

b bb

cc

d

d y d y dye e x e x y

dxdx dx

d y dyf f x f x y

dxdx

dyg g x g x y

dx

h h x h x y

Nonlinear Generic Form

1 1

11

3 22 2

1 2 3 1 2 33 2

2 21 2 3 1 2 3

( ) ( )

( ) ( ) 0

a b

cd

d y d ye e x e x f f x f x

dx dx

dyg g x g x h h x h x y

dx

28





Establishing the Unknowns

A Least Squared Error Technique is used to determine the unknowns

1 2 13 2 1

4 3 2 1

23 2 2

1 2 3 43 2 21

( )

a a ba b cN a b c di i i i

i i ii i i i

Minimize F

d f d f df d f df dfA f A f A f A f

dx dx dxdx dx dx

N: the number of data points

This is the objective for linear constant coefficient form A similar one can be used for the generic nonlinear form

A continuous application of standard optimization technique was unsuccessful because the exponents were not integers

A mixed integer (exponents) – continuous (coefficients) approach was also unsuccessful because the solution will determine trivial values

Solution was only possible through discrete programming

29





Discrete Programming Used in The Paper

Two procedures are considered in this paper

1. Exhaustive Enumerationall of the values for the unknowns are considered in combination before the

optimum is determined

2. Simple Heuristic Programmingsimple heuristic exhaustive enumeration over predetermined number of cycles

(1 billion)

Discrete Programming is incredibly time extensive

For the linear constant coefficient form, allowing 3 values for each unknown required 1.0*105 cpu seconds on a Linux Opteron running MATLAB 2007a

30





Example 1 (Step 1)

1 1.5 2 2.5 3 3.5 4 4.5 5-5

-4

-3

-2

-1

0

1

2

y -

Ori

gin

al D

ata

x0 1 2 3 4 5 6-5

-4

-3

-2

-1

0

1

2

3

y -

Ori

gin

al D

ata,

Fit

ted

Dat

a

x

original data

Bezier approximationBest order of Fit (based on y-data) : 14Number of data points: 200 Sum of Absolute Error (y): 7.27217e-005Sum of Squared Error (y): 3.96250e-011Average Error (y): 3.63608e-007Sum of Absolute Error (x): 4.56362e-007Sum of Squared Error (x): 2.05982e-015Average Error (x): 2.28181e-009

Type original data Bezier datax (initial) 1 1x (final) 5 5y (initial) 1 1y (final) 2 2dy/dx (initial) not given -7.2728dy/dx (final) not given 2.1184d2y/dx2 (initial)

not given 6.5163

The original data is discrete x-y data

The derivatives are those predicted for the data

31





Example 1 (Step 2)

0 1 2 3 4 5 6-5

-4

-3

-2

-1

0

1

2

3

y -

Ori

gin

al D

ata,

Fit

ted

Dat

a

x

original data

Bezier approximation

1 1

11

3 22 2

1 2 3 1 2 33 2

2 21 2 3 1 2 3

( ) ( )

( ) ( ) 0

a b

cd

d y d ye e x e x f f x f x

dx dx

dyg g x g x h h x h x y

dx

The exponents and coefficients are drawn from the set of three except for h1 that will belong to a set of 9 values

Solution : Exhaustive Enumeration

a1 = 0, a2 = 1, a3 = 1, a4 = 1, b1 = 1, b2 = 0, b3 = 0, c1 = 1, c2 = 0, d1 = 1.The solution for the exponents:

The solution for the coefficients:

e1 = 1, e2 = 1, e3 = 1, f1 = 0, f2 = 0, f3 = 1, g1 = 0, g2 = 1, g3 = 0, h1 = -0.25 h2 = 0, h3 = 1.

22 2

2( 0.25) 0

d y dyx x x y

dxdx

The differential equation

This was the same differential equation used to generate the discrete data

32





Example 2 (Step 1)

0 1 2 3 4 5 60

0.5

1

1.5

2

2.5

3

3.5

4

4.5

y -

Ori

gin

al D

ata,

Fit

ted

Dat

a

x0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

3

3.5

4

4.5

y -

Ori

gin

al D

ata,

Fit

ted

Dat

a

x

Original Data

Bezier ApproximationBest order of Fit (based on y-data): 12 No. of data points: 101 Sum of Absolute Error (y): 8.90327e-005 Sum of Squared Error (y): 1.51515e-010 Average Error (y): 8.81512e-007 Sum of Absolute Error (x): 1.23819e-008 Sum of Squared Error (x): 2.26533e-018 Average Error (x): 1.22593e-010

Type original data Bezier datax (initial) 0 1.3234e-013x (final) 6 6.0000y (initial) 0 1.9390e-007dy/dx (initial) 0 -1.3919e-005dy/dx (final) 1 1.0001d2y/dx2 (initial) 0.3326 0.3329

The discrete data is created by numerical integration using derivative information

The Bezier data approximates the derivative nicely

33





Example 2 (Step 2)

Solution : Exhaustive EnumerationThe solution for the exponents:

The solution for the coefficients:

The differential equation

0 1 2 3 4 5 6 70

0.5

1

1.5

2

2.5

3

3.5

4

4.5

y -

Ori

gin

al D

ata,

Fit

ted

Dat

a

x

Original Data

Bezier Approximation

1 2 13 24 3

12 1

3 2 2

1 23 2 2

3 4( ) 0

a a ba ba b

cc d

y yd d dy d y dye y e y

dx dxdx dx dx

dye y e y

dx

A constant nonlinear generic form is used (to reduce time of computation)

a1 = 1, a2 = 1, a3 = 2, a4 = 0, b1 = 2, b2 = 2, b3 = 1, c1 = 0, c2 = 0, d1 = 0

e1 = 1, e2 = 0.5, e3 = 0.5, e4 = 0.51 13 2

3 21

d y d y

dy dy

2dy

dx

21 22

20.5

d y dy

dxdy

1

3 2

3 2

0

0.5 0

y

d y d yy

dy dx

This is the Blasius equation used to generate data

34





Work in Process

The computation time is a serious issue for a broader range of values. Global optimization techniques may provide a relief

Extension tocoupled ODEsingle and coupled PDEnon smooth data

are planned for the future

35





Bezier filter is easy to incorporate and can work for regular, unpredictable data, and images

The Bezier functions have excellent blending and smoothing properties

High order but well behaved polynomial functions can be useful in capturing the data content and underlying behavior

Bezier functions naturally decouples the independent and the dependent variables

Properties of the Bezier Function

Gradient and derivative information of the data are easy to obtain

36

Bezier functions coupled with optimization can solve all kinds of mathematical problems





Questions ??

p. venkataraman rochester institute of technologygraduate seminar, january, 7, 2010 bezier functions...

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