Chapter 6 f Inter

8/13/2019 Chapter 6 f Inter

1/34

A. J. Clark School of Engineering Department of Civil and Environmental Engineering

by

Dr. Ibrahim A. AssakkafSpring 2001

ENCE 203 - Computation Methods in Civil Engineering IIDepartment of Civil and Environmental Engineering

University of Maryland, College Park

CHAPTER 6f.

NUMERICAL

INTERPOLATION

Assakkaf

Slide No. 164


ENCE 203 CHAPTER 6f. NUMERICAL INTERPOLATION

Interpolation Using Splines

Generally, the higher-order polynomial

gives higher accuracy.

However, this is not true in some

situations, especially when the data

points include local abrupt changes in

f(x) values for steady changes inx

values.

In these situations, the accuracy

decreases.


2/34

Assakkaf

Slide No. 165




In these cases, the function can lead to

erroneous results because of round-off

error and overshoot.

An alternative approach is to use a

lower-order polynomial to subsets of the

data points.

Such connecting polynomials are calledsplines functions.

Assakkaf

Slide No. 166




The concept of spline came from the

drafting technique of using thin flexible

strip, called spline, to draw smooth

curves through a set of points as shown

in the figure of the viewgraph.

The drafter places paper over woodenboard and hammer nails into the paper

and the board at the locations of the

data points.


3/34

Assakkaf

Slide No. 167




Origin of Spline Concept

Assakkaf

Slide No. 168




Need for Splines

f(x)

x

Abrupt change indicates oscillation in interpolating polynomial.


4/34

Assakkaf

Slide No. 169




Need for Splines

f(x)

x


Assakkaf

Slide No. 170




Need for Splines

f(x)

x



5/34

Assakkaf

Slide No. 171




Need for Splines

f(x)

x

The spline provides a much more acceptable approximation.

Assakkaf

Slide No. 172




Types of Splines

1. Linear (simplest),

2. Quadratic, and

3. Cubic (most popular)


6/34

Assakkaf

Slide No. 173




Linear or First-order Splines

f(x)

x

S1S2 S3

Assakkaf

Slide No. 174




Quadratic or Second-order Splines

f(x)

x

S1

S2

S3


7/34

Assakkaf

Slide No. 175




Cubic or Third-order Splines

f(x)

x

S1S2

S3

Assakkaf

Slide No. 176




Linear Splines

Linear or first-order splines are the

simplest to construct.

They are straight lines connecting two pair

of the data points.

Therefore, an algorithm for straight lines

that connect each pair of the data points

can be developed very easily.


8/34

Assakkaf

Slide No. 177




Linear Splines The interpolation function can be expressed as

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

( ) nnnnn

nnnn

iii

ii

iiii

xxxxxxx

xfxfxfxf

xxxxxxx

xfxfxfxf

xxxxxxx

xfxfxfxf

xxxxxxx

xfxfxfxf

+=

+=

+=

+=

+

+

+

11

1

111

1

1

1

322

23

2322

211

12

1211

for

for

for

for

!

(1)

Assakkaf

Slide No. 178




Linear Splines

Requirements for Linear Splines:

The linear spline functions as presented by the

previous equations are to satisfy the following

condition:

( ) ( ) 1,,2,1For1 == + nixfxf iiii "

(2)


9/34

Assakkaf

Slide No. 179




Example 1 (Linear Splines)

Fit the following set of data with first-order

(linear) splines. Evaluate the function atx

= 5. Also plot the spline functions.

x 3 4.5 7 9

f(x) 2.5 1 2.5 0.5

Assakkaf

Slide No. 180




Example 1 - Linear Splines

For the first pairs of data:

( ) ( ) ( ) ( )

( )

( ) ( )( )

( ) 5.43For5.5

)3(5.2

335.4

5.21

5.2

1

1

1

12

1211

=

=

+=

+=

xxxf

xxf

xxf

xxxx

xfxfxfxf 14.5

2.53

f(x)x


10/34

Assakkaf

Slide No. 181




Example 1 (contd) - Linear Splines

For the second pairs of data:

( ) ( ) ( ) ( )

( )

( ) ( )

( )

( ) 75.4For6.07.1

)5.4(6.01

5.45.47

15.21

1

2

2

23

2322

+=

+=

+=

+=

xxxf

xxf

xxf

xxxx

xfxfxfxf

2.57

14.5

f(x)x

Assakkaf

Slide No. 182





For the third pairs of data:

( ) ( ) ( ) ( )

( )

( ) ( )( )

( ) 97For5.9

)7(5.2

779

5.25.0

5.2

1

2

3

34

3433

=

=

+=

+=

xxxf

xxf

xxf

xxxx

xfxfxfxf

0.59

2.57

f(x)x


11/34

Assakkaf

Slide No. 183





Finding the value of the function whenx = 5

Ifx = 5, then the following spline applies:

Thus,

( ) 75.4For6.07.1 += xxxf

( )

( )3.1

56.07.1

6.07.15

=+=

+= xf

Assakkaf

Slide No. 184





Plot of the linear spline functions:

0

0.5

1

1.5

2

2.5

3

0 2 4 6 8 10

x

f(x)

f1f2

f3


12/34

Assakkaf

Slide No. 185




Quadratic Splines

The quadratic splines provide a quadratic

equation connecting any two adjacent data

points within the data set of the type

( )[ ] nixfx ii ,,3,2,1for, "=

f(xn)f(x4)f(x3)f(x2)f(x1)f(x)

xnx4x3x2x1x

n4321i

Assakkaf

Slide No. 186




Quadratic Splines

The general form of a quadratic equation

between point [xi,f(xi)] and [xi+1,f(xi+1)] is

( ) 1,,2,1For2 =++= nicxbxaxfiiii

"

(3)


13/34

Assakkaf

Slide No. 187




Quadratic Splines

Every two adjacent data points have an

interpolation equation given by Eq. 3, with

three constants ai, bi, andci.

As a result, there are 3(n 1)unknowns

that need to be determined using the data

set, requiring 3(n 1) conditions

Assakkaf

Slide No. 188




Quadratic Splines

Conditions for Quadratic Splines:

1. The splines must pas through the data points.

For the ith spline, this condition can be

expressed as

( ) ( )

( ) ( ) 1,,3,2,1for

1,,3,2,1for

11

2

11

2

==++=

==++=

++++ nixfcxbxaxf

nixfcxbxaxf

iiiiiiii

iiiiiiii

"

" (4a)

(4b)


14/34

Assakkaf

Slide No. 189




Quadratic Splines


2. The splines must be continuous at the interior

data points. This condition can be expressed

using the first derivatives of the quadratic

splines as

2,,3,2,1for22 1111 =+=+ ++++ nibxabxa iiiiii " (5)

Assakkaf

Slide No. 190




Quadratic Splines


3. The last condition can be arbitrary. The

second derivative for the spline between the

first two data points can be set equal to zero.

Since the second derivative for the first splineis 2a1, this condition can be written as

002 11 == aa (6)


15/34

Assakkaf

Slide No. 191




Quadratic Splines

Equations 4 to 6 provide the needed

3(n 1) conditions to solve for the 3(n 1)

unknowns ai, bi, and ci, i = 1,2,, n 1.

The resulting quadratic splines provide the

desired continuity while passing through

the data points.

Assakkaf

Slide No. 192




Example 2 Quadratic Splines

Fit the following set of data with second-

order (quadratic) splines. Evaluate the

function atx = 5. Also plot the spline

functions.

0.52.512.5f(x)

974.53x

4321i


16/34

Assakkaf

Slide No. 193




Example 2 (contd) Quadratic Splines

For the first data pairs, Eqs. 4a and 4b

apply as follows:

( )

( ) 1,,3,2,1for

1,,3,2,1for

11

2

1

2

==++

==++

+++ nixfcxbxa

nixfcxbxa

iiiiii

iiiiii

"

"

( ) ( )( ) ( ) 15.45.4

5.23311

2

1

11

2

1

=++

=++

cbacba

2

1

i

14.5

2.53

f(x)x

(7)

Assakkaf

Slide No. 194





For the second data pairs, Eqs. 4a and 4b

apply as follows:

( )

( ) 1,,3,2,1for

1,,3,2,1for

112

1

2

==++

==++

+++ nixfcxbxa

nixfcxbxa

iiiiii

iiiiii

"

"

( ) ( )

( ) ( ) 5.277

15.45.4

22

2

2

22

2

2

=++

=++

cba

cba

3

2

i

2.57

14.5

f(x)x

(8)


17/34

Assakkaf

Slide No. 195





For the third data pairs, Eqs. 4a and 4b

apply as follows:

( )

( ) 1,,3,2,1for

1,,3,2,1for

11

2

1

2

==++

==++

+++ nixfcxbxa

nixfcxbxa

iiiiii

iiiiii

"

"

( ) ( )( ) ( ) 5.099

5.27733

2

3

33

2

3

=++

=++

cbacba

4

3

i

0.59

2.57

f(x)x

(9)

Assakkaf

Slide No. 196





For the first and second data pairs, Eqs. 5

applies as follows:

( ) ( )

( ) ( ) 3322

2211

7272

5.425.42

baba

baba

+=+

+=+

(10)

2,,3,2,1for22 1111 =+=+ ++++ nibxabxa iiiiii "

3

2

i

2.57

14.5

f(x)x


18/34

Assakkaf

Slide No. 197





The last condition comes from Eq. 6 as

Since a1 = 0, the resulting system of 3(n-1)

- 1= 3(4 1)-1 = 8 equations can be

obtained as follows:

01=a

Assakkaf

Slide No. 198





01414

09

5.0981

5.2749

5.2749

15.425.20

15.4

5.23

3322

221

333

333

222

222

11

11

=+

=

=++

=++

=++

=++

=+

=+

baba

bab

cba

cba

cba

cba

cb

cb

(11)


19/34

Assakkaf

Slide No. 199





Eq. 11 can be rewritten in a matrix form as

=

0

0

5.0

5.2

5.2

1

1

5.2

0114011400

00001901

198100000

174900000

000174900

00015.425.2000

00000015.4

00000013

3

3

3

2

2

2

1

1

c

b

a

c

b

a

c

b

(12)

Assakkaf

Slide No. 200





Therefore,

=

3.91

6.24

6.1

46.18

76.6

64.0

5.5

1

3

3

3

2

2

2

1

1

c

b

a

c

b

a

c

b( )

( )

( ) 97for3.916.246.1

75.4for46.1876.664.0

5.43for5.5

2

3

2

2

1

+=

+=

+=

xxxxf

xxxxf

xxxf (13a)

(13b)

(13c)


20/34

Assakkaf

Slide No. 201




Example 1 (contd) - Quadratic Splines

Finding the value of the function whenx = 5

Ifx = 5, then the following spline of Eq. 13b

applies:

Thus,( ) ( ) ( )

66.0

46.18576.6564.052

=

+=f

( ) 75.4for46.1876.664.0 22 += xxxxf

Assakkaf

Slide No. 202




Example 2 (contd) - Quadratic Splines

Plot of the quadratic spline functions:

0

0.5

1

1.5

2

2.5

3

3.5

0 2 4 6 8 10

x

f(x) f1 f2f3


21/34

Assakkaf

Slide No. 203




Cubic Splines

The cubic splines provide a cubic equation

connecting any two adjacent data points

within the data set of the type

( )[ ] nixfx ii ,,3,2,1for, "=

f(xn)f(x4)f(x3)f(x2)f(x1)f(x)

xnx4x3x2x1x

n4321i

Assakkaf

Slide No. 204




Cubic Splines

The general form of a cubic equation

between point [xi,f(xi)] and [xi+1,f(xi+1)] is

( ) 1,,2,1For23 =+++= nidxcxbxaxf iiiii "

(14)


22/34

Assakkaf

Slide No. 205




Cubic Splines

Every two adjacent data points have an

interpolation equation given by Eq. 14, with

three constants ai, bi, ci,anddi.

As a result, there are 4(n 1)unknowns

that need to be determined using the data

set, requiring 4(n 1) conditions

Assakkaf

Slide No. 206




Cubic Splines

Conditions for Cubic Splines:

1. The splines must pas through the data points.

For the ith spline, this condition can be

expressed as

( ) ( )

( ) ( ) 1,,3,2,1for

1,,3,2,1for

1

2

1

3

11

23

==+++=

==+++=

+=++ nixfdxcxbxaxf

nixfdxcxbxaxf

iiiiiiiiii

iiiiiiiiii

"

" (15a)

(15b)


23/34

Assakkaf

Slide No. 207




Cubic Splines


2. The splines must be continuous at the interior

data points. This condition can be expressed

using the first derivatives of the cubic splines

as

2,,3,2,1for

2323 1112

111

2

1

=

++=++ +++++++

ni

cxbxacxbxa iiiiiiiiii

"

(16)

Assakkaf

Slide No. 208




Cubic Splines


3. The splines must satisfy the second

derivative continuity at the interior points.

This condition can expressed as

2,,3,2,1for

2626 1111

=

+=+ ++++ni

bxabxa iiiiii

"

(17)


24/34

Assakkaf

Slide No. 209




Cubic Splines


4. The last condition can be set as the second

derivative at the first and last data points to

be zero. This condition is as follows:

026

026

11

11

=+

=+

nnn

i

bxa

bxa(18)

Assakkaf

Slide No. 210




Cubic Splines

Equations 15 to 18 provide the needed

4(n 1) conditions to solve for the 4(n 1)

unknowns ai, bi, ci,and di, i = 1,2,, n 1.

The resulting cubic splines provide the

desired continuity while passing through

the data points.


25/34

Assakkaf

Slide No. 211




Example 3 Cubic Splines

Fit the following set of data with third-order

(cubic) splines. Evaluate the function atx =

5. Also plot the spline functions.

0.52.512.5f(x)

974.53x

4321i

Assakkaf

Slide No. 212




Example 3 (contd) Cubic Splines

For the first data pairs, Eqs. 15a and 15b

apply as follows:

( ) ( ) ( )

( ) ( ) ( ) 15.45.45.4

5.2333

11

2

1

3

1

11

2

1

3

1

=+++

=+++

dcba

dcba

2

1

i

14.5

2.53

f(x)x

(19)

( ) ( )

( ) ( ) 1,,3,2,1for

1,,3,2,1for

12 13 11

23

==+++=

==+++=

++++ nixfdxcxbxaxf

nixfdxcxbxaxf

iiiiiiiiii

iiiiiiiiii

"

"


26/34

Assakkaf

Slide No. 213





For the second data pairs, Eqs. 15a and

15b apply as follows:

( ) ( ) ( )( ) ( ) ( ) 5.2777

15.45.45.422

2

2

3

2

22

2

2

3

2

=+++

=+++

dcbadcba

3

2

i

2.57

14.5

f(x)x

(20)

( ) ( )

( ) ( ) 1,,3,2,1for

1,,3,2,1for

1

2

1

3

11

23

==+++=

==+++=

++++ nixfdxcxbxaxf

nixfdxcxbxaxf

iiiiiiiiii

iiiiiiiiii

"

"

Assakkaf

Slide No. 214





For the third data pairs, Eqs. 15a and 15b

apply as follows:

( ) ( ) ( )

( ) ( ) ( ) 5.0999

5.2777

33

2

3

3

3

33

2

3

3

3

=+++

=+++

dcba

dcba

4

3

i

0.59

2.57

f(x)x

(21)

( ) ( )

( ) ( ) 1,,3,2,1for

1,,3,2,1for

12 13 11

23

==+++=

==+++=

++++ nixfdxcxbxaxf

nixfdxcxbxaxf

iiiiiiiiii

iiiiiiiiii

"

"


27/34

Assakkaf

Slide No. 215





For the second and third data pairs, Eq. 16

applies as follows:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 332

322

2

2

22

2

211

2

1

72737273

5.425.435.425.43

cbacba

cbacba

++=++

++=++

(22)

3

2

i

2.57

14.5

f(x)x

2,,3,2,1for

2323 1112

111

2

1

=

++=++ +++++++

ni

cxbxacxbxa iiiiiiiiii

"

Assakkaf

Slide No. 216





For the second and third data pairs, Eq. 17

applies as follows:

( ) ( )

( ) ( ) 3322

2211

276276

25.4625.46

baba

baba

+=+

+=+

(23)

2,,3,2,1for

2626 1111

=

+=+ ++++

ni

bxabxa iiiiii

" 3

2

i

2.57

14.5

f(x)x


28/34

Assakkaf

Slide No. 217





The last condition comes from Eq. 18 as

Therefore, the 12 equations [4(4-1)] are as

follows:

026

026

11

11

=+

=+

nnn

i

bxa

bxa

( )

( ) 0296

0236

33

11

=+

=+

ba

ba(24)

4

3

i

0.59

2.57

f(x)x

Assakkaf

Slide No. 218





0254

0218

0242242

0227227

014147141470975.60975.60

5.0981729

5.2749343

5.2749343

15.425.20125.91

15.425.20125.91

5.23927

33

11

3322

2211

333222

222111

3333

3333

2222

2222

1111

1111

=+

=+

=+

=+

=++=++

=+++

=+++

=+++

=+++

=+++

=+++

ba

ba

baba

baba

cbacbacbacba

dcba

dcba

dcba

dcba

dcba

dcba

(25)


29/34

Assakkaf

Slide No. 219





=

0

00

0

0

0

5.0

5.2

5.2

1

1

5.2

2540000000000

000000000021800242002420000

00000022700227

011414701141470000

000001975.6001975.60

198172900000000

174934300000000

000017493430000

000015.425.20125.910000

0000000015.425.20125.91

0000000013927

3

3

3

3

2

2

2

2

1

1

1

1

d

cb

a

d

c

b

a

d

c

b

a

Assakkaf

Slide No. 220





Therefore,

=

813.8

326.0

554.0

0528.0

813.33

941.17

163.3

177.0

020.2

253.3

547.1

172.0

3

3

3

3

2

2

2

2

1

1

1

1

d

c

b

a

d

c

b

a

d

c

b

a

( )

( )

( ) 97for813.8326.0554.00528.0

74.5for813.33941.17163.3177.0

5.43for02.2253.3547.1172.0

23

3

23

2

23

1

++=

++=

++=

xxxxxf

xxxxxf

xxxxxf

( )

058.1

813.33)5(941.17)5(163.3)5(177.05 23

=

++=f


30/34

Assakkaf

Slide No. 221





0

0.5

1

1.5

2

2.5

3

0 2 4 6 8 10

x

f(x)

f1f2 f3

Assakkaf

Slide No. 222




Comparison among Linear, Quadratic, and Cubic

Splines for Examples 1, 2, and 3

Splines

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4 5 6 7 8 9 10

x

f(x)

Linear

Quadratic

Cubic


31/34

Assakkaf

Slide No. 223




Comparison among Linear, Quadratic, and Cubic Splines for

Examples 1, 2, and 3 with Third-order Polynomial

0

0.5

1

1.5

2

2.5

3

3.5

2 3 4 5 6 7 8 9 10

x

f(x)

Linear

Quadratic

Cubic

3rd Degree

Assakkaf

Slide No. 224



Multidimensional Interpolation

In general, the function of interest can

take the following form:

In most engineering applications, two-dimensional interpolation is required.

( )nxxxxf ,,,, 321 "


32/34

Assakkaf

Slide No. 225




The previously introduced methods for

one-dimensional case can be applied to

multidimensional interpolation.

However, with increased computational

difficulties.

To illustrate the concept, linear

interpolation is introduced for two-dimensional interpolation.

Assakkaf

Slide No. 226




Two-dimensional Interpolation

Example 4

The following table gives the values of the

gravitational acceleration, g, in feet per second

squared as a function of altitude L (in degrees)

and elevation E(in feet). Find the value of gwhen L = 12.50 and E= 500 feet.


33/34

Assakkaf

Slide No. 227





Example 4

32.101332.104432.1075L = 20032.092932.096032.0991L = 15

0

32.08632.089732.0928L = 100

32.082932.085932.0890L = 50

32.081632.084732.0877L = 00

E= 2000 ftE= 1000 ftE= 0 ft

Assakkaf

Slide No. 228





Example 4 (contd)

First, we interpolate for Eat L = 10 to obtain

g(100

, 500 feet).

( ) ( )ELgxxf ,, 21 =

09125.320928.320897.32

0928.32

01000

0500

0897.321000

500

0928.320

11

1

=

=

g

g

g

gE


34/34

Assakkaf

Slide No. 229





Example 4 (contd)

Second, we interpolate for Eat L = 15 to obtain

g(150, 500 feet).

09755.320991.320960.32

0991.32

01000

0500

0960.321000

500

0991.320

22

1

=

=

g

g

g

gE

Assakkaf

Slide No. 230





Example 4 (contd)

Using the values of g1 and g2, we interpolate for

L at E= 500 to obtain g(12.50, 500 feet).

Therefore, g(12.50, 500 feet) = g3 = 32.0944 ft/s2

0944.3209125.3209755.32

09125.32

1015

105.12

09755.3215

5.12

09125.3210

3

2

3

=

=

g

g

g

gL

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Chapter 6 f Inter