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The application of a least square procedure to the solution of a homogeneous linear integral equation problem by polynomial approximation

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(1)

TiiE APTLICA1.1.'ICiJ OF A L;i,AST S~UAEE f .ci.CC~DUM TO THE SOLUTIC~~ OF A FC1•:CGE1s.iCUS LIN~Al-i :iN1'ZG.R.AL i;UA'l'ION

P.R.OBIEE BY POLYEOllIAL Af FB.G:rii·iAl'ION

1

1

homas

11I.

Creese

Submi

tt'3d in partial fulfillinent

of

the requirements

for the decree of Bachelor of Science

at the

HASSACi-iUSJT'IS INSTITUTE OF TEC.h"}IOI.DGY

Nay

21

,

1956

Signature redacted

-

~ ., ,.,

.

...,

__

. .

.

.

.

. .

.

.

.

. .

r).

-

#

-.♦\

·,.

J

,

c. ..

.

.

;l(o.1i

.

,

.

Z.t;ICfS6

.

.

. .

Signature of Author

,l"I

Certified by

1 I'\ Department

Signature redacted

.

. .

.

.

. .

.

.

.

.

.

Thesis Advisor

Date

(2)

Table

of

Contents

I

Introduction

1

II Outline of the Method 2

III The Problem

5

IV The Solution

7

V The Results 114

VI The Accuracy 16

VII The Summary 19

References 21

(3)

1

I Introduction

It is desired to demonstrate the use of the least square procedure

in a polynomial approximation' in connection with a homogeneous linear

integral equation problem. For the sake of comparison a relatively

simple problem is chosen for which exact closed form results are known and the polynomial anproximation solution is available. In this way the method is demonstrated and compared with exact results and a previous method.2 The problem chosen is Problem la.1

1 The polynomial approximation is demonstrated in Reference 1, the

least square procedure in Reference 2, and so important a part do these two references play in this paper that items, e.g. equations and tables, contained therein will be referred to (with suffixes to distinguish between them) directly by number: equation (6a) and table (3a) occur in

Reference 1; equation (2b) and table (3b) occur in Reference 2.

2 Except where noted "(exact)", the data given in this paper are obtained by omitting all but ten figures in operations on a ten bank

(4)

II Outline of the Method

In section 2a it is shown that values of

a

polynomial

y(x)

=

a.+ aix + a2x2 + ... * a2nx2n (la)

over an interval divided into 2n equal parts by 2n

+

1 evenly spaced

points, may be given exactly by equation (6a),

y -n

X7-n*

..

0+K+K(~e

KlAY+ ..

+Kn

E

(1)

where

x 0

is the midpoint of the interval,

h

is the spacing of the points,

AX=x-x

, yj y(xi)

h

and values of

K.(A)

are given in tables (3a) and (6b) for n

=

2, the

value we will be using. The requirement that (1) exactly satisfy the

solution of our problem at five points implies

a

polynomial approximation

to the solution, which is given in Section 3a.

In Section 2b it is shown how

a

least square procedure may be

applied to this. If equation (1) is inverted we find

n

2Fy

I (x)

(x)

(lb )

i

=

-n

Placing the restriction

xn -n 2

2L

yiIi(x)(- (xjdx

=

minimum,

(2b)

x-n i= -n

replacing

the exact integral by an approximate one, and taking the

(5)

m

rn

12:

S Dk

L

Yii

xk)- k(x minimum, (3b)

k

=1

i n

from which the vanishing of the partial derivatives finally gives

n m y Dk ixk pk) - Dk kp(xk) (2)

i

=

-n

k=

k-(p=1, ... , 2n 4 1) where (' (xk) k '

or, in matrix formulation,

n m

E IID k] (k) ' ik) Yi - D kV kk k3

2. - n k -1

p= row index k

=

row index (p

1,

, 2n -t- 1)

k

=

column index i

=

column index

For ease in notation the indices will be related to the matrices

by a dash and the range noted after, e.g., in (3),

p - row (l,,2n`+1) k - row (1, m)

k - column (1, m) i - column(-n, n)

The system (3) may be given in matrix formulation:

DkI p • ixk) •

Iy1

IkI

p - row (1, 2n+ 1) k - row (1, m) i- row (-n, n)

k - column (1, m) i - column

(-n,

n)

D JI

(xkf

k4)~kj

p - row (1, 2n + 1) k -row (1, m)

k - column (1, m)

whereDthe

Dk

are the integration coefficients for the least Equare

processand the l and 1 are coefficients arising in the

(6)

the k, introduce the particular problem into the method. I.(xk) k) , pi- n•

The work to be don. consists of secifying I and k

(7)

III The Problem

Problem la: "A beam of constant flexural rigidity EI and length L

is hinged at the ends and subjected to a compressive force P. Find the

critical value of

P

and the first mode

of buckling."

The bending moment 14(x) for

a

beam is proportional to the

displace-ment

s(x), and, for curvature

((x),

M(x)

=

(x)

=

(5)

EI

dx

.

Figure

1

If, therefore, we consider the deflection

ds

due to a bending moment

apied

atx

¶over

an interval

O

il, ith

x, , , = + X < x6)

(8)

is

xkL -

f )

M() , x

<

dy

EIL

and note that

ds is continuous.

d5

The usual notation transforms

(7):

ds

=

G(x,

$)M

()

(8)

df

where G(x,

)is

called the Green's function

3

and

x(L- ) X

EIL

G (x,

)

I

L (9)

EIL

Now, noting that

Ps(s)

(10)

and

integrating

over

,

L

s(x)

P

G(x

)s()d.

(ll)

0

Equation (11) is the complete exoression of the problem.

3 A

more detailed derivation of the Green's function for this

problem

is

given

in

Reference

1.

(9)

IV The Solution

We shall use a polynomial of degree four for our approximation, and divide the interval into four segments of length h, that is, take:

n = 2

(see (4) )

h

L

14

2)

x.=

(i + 2)h

1

(see (1)

)

0

A

(12)

For the least square integration, divide the interval yet again

as in Figure 2.

m

=

4n

+ 1 = 9

(see

(4)

)

Figure 2

i= -2

I..

I I

-1l

p= 1

k

1

2 0

2

3

3

1I

a I I 14

6

7

I I I I

8

.9

j

1

23 45

6

7

8

9

1

11 12 13

4

15 16

17

For IDk we will use the Simpson's rule coefficients:

(i

=

-2, ... ,

2)

k

2

(10)

L m I f~gdf=~L;;

D -f(

)

O k = 1k D=D' I h .1= 1

9

32 L 12 D= D' D'= D

2

4

6

8

1

I / I D /= D, D/= 2_

3

5

12 D :12 D

kk

To get the inversion coefficients, , we note

2 i -2

Substituting in

(11)

PI G(x, )s()df = s(x)= yGx) G(X §) Ki (A)y5cly 00 i - -2

(14)

or

2 L

2=

y

P

G(x,

)K.()d.

(15)

i-2 0

Making an approxinate (Simpson's rule) integration, we get

2

17

yx)

yiPL

D

G(xk,,'

)K(A.)

L

(16)

1, 17

D= 24 D 1

j

2,s 4,

... ,

1

0,16

(11)

Since

2

(x ) = e<jI (x

k

j~2 ikk

and in this problm

k k '

we have

2

yk .

Yi.

i

L xk

i

- -2

and identify

17

I (xk) k

=

PL D G(xk' j - 1 This is equivalent to I (xk) PLI G(xk

j)

D!Ki(> )

k -

row (1,

9)

k - row (1, 9)

j

- row (1, 17)

i - coluin (-2,2)

j

- column (1, 17) i - column (-2, 2)

D.' as in (16)

3

For accuracy (and also ease in tabulation), 128 EI G(xk, 9j)

L

been tabulated in an appendix (see Table #1), and DK.

(A)

3 i

j

.24 D.K.(A) (see Table #2) , and in terms of these, we have

3 1

4Since, as will be seen in (20), for

j

= 1, 17, and i

=

-2, 2

DK(A)is multilied by zero only, (see (20) and Table

#3),

these

values have been omitted from Table #2 and dashes inserted in their

place.

(lb

)

('7)

.(18)

-I

(19)

has

(i

=

-2,

...

,

2)

5j )K i( )

(12)

X124•128

2 Ii(xk) k - row (1, 9) i - column (-2, 2) 128 El G(xk,'•jj

k

-

row

(1,

9)

j

-

column

(1,

17)

I (xk) is tabulated in Table #3. S(xk+

k)

j

- row (1, 17) i - column (-2,

2)

(20)

(p

=

1,

*..

,

5),

(k

=1,

..

,9)

(21)

(i

=

-2, ...

,

2)

Having now assembled all the components of

(h),

(the D's in (13),

the I s

in

(20) and (21)

),

we

lack

only, (k

= 1,

..

,9),

since

yi

=yk

2

(k=

1,

3,

5,

7,

9)

(i

=-2,

...

,

2)

Setting

/pk

l2I?iIp(xk)

(see Table #4 for values), and

pi

=1

12

X20f, p - row (1,

5

) i - column (-2, 2) 112 X/pk I k - column (1, 9) *I i xk) I

(23)

i - column (-2, 2)

is

tabulated

in

Table

#5.

Substituting all this,

(h)

becomes

(22)

I

p - row (1,

5

)

k - row (1, 9 )

I IX'

(xk) I

(13)

12 1 p - row (1,

5)

i- row (-2, 2) i - column (-2, 2) 11 / -k I 1 k

12'

p -

row(1,

)

k

-

row (1,9)

k- column (1, 9)

2

9

~~Y

<.7= x

~~

7k 1

(-

=

1,

,

5)

.

(25)

We can eliminate y

2

,

y4

y

6

, and y8 by means of (1) in the following

manner:

y =K()y- K1 1 4y.

0( 0 K1()y

2XY -2 -1(7-1+ Ko(\)yo + K2 )Y2

7-2 = 2=

(26)

so

Yk

-1

-

7- 70

Y,#

KI(kX1

Substituting

(27) into

(25)

gives

1

1

7i

i

=-l

i2.1 -1

(k

=

2,

4, 6,3

)

,

(27)

(p 1, ••• ,

5)

where

L~i

Z= N

=

3072

Z

=

E.I

I

(28)

(29)

11

or

(24)

'k

=

2,

4,

6, 8

kT

(14)

12 For increased accuracy, N1 k was calculated first and tabulated

in Table #6;Y is in Table

#7.

Now, combining

in

(28):

1i

-

=

Since

0

for p

=1,5,

we may change subscripts:

(30)

q

=

p - 1 ,

(p

=

2, 3,

)

.

Now we have four unknowns, the y's and Z, three linear equations,

and a secular equation:

- Z )yi

=

0 (q

=

1, 2, 3)

det

-

Z

0

(q

=

1,

2,

3)

(i

-1, 0, 1)

.1

31)

By the symmetry of theproblem

yi =Y-1

and, substituting numbers from Tables# 6 and #7 into the above, (31)

becomes (167 865.578 1 - 1 938

960 Z)yI

(loo

051.753

9

-

787 620

Z)y

0

=

0

(200 103.507

8

-

1 633 976

Z)y,

(130 172.462

9

-

1

526

250

Z)yO

:

0

C2)

along with a corresponding secular equation, from which one gets solu-tions for the first two symmetric modes of buckling.

(15)

i)

From (29) and the solution of (32),

Pi

EI .l1

1

,

E=

9,869 649 837

L

.101 320 717 2

L

72 P

yl

=.7082L6

2

44

33,

=

y_

.

I

(16)

14

V The itesults

The results from (33), rounded down to two error digits or five decimal places (whichever is greater) are given below in Table A.

Also included are ordinates for intermediate points obtained by means of equation (1) and Table (3a), and, for comparison, both the exact solution, sin(7tfx/L), and values obtained in Reference 1 by

(17)

9 6 0 609*6

-4-_ d

60

9L 160.

5

1T6 *

09

L96*

00

000i1

9!°6

17L7

f409 69

6

=

0

L5r

911

69,

61

00

0

60

T69

L26

0000-1

66

6L

10

691

00

011

0 1

0

0011

'0

'j5

0. C---'

oz1*

C.-Li

0

-[

(-f/x.~ut

I

W/(Xi~

~

;

[

rf/x

C JUT-pon"-i 0 Opo-IT, q.s.ITL'T ;DLU i - k

(18)

1U

VI The Accuracy

It is instructive to take a very close look at Table A, page 15.

For convenience, values of

[y(x)

-psin(f x/L /A, and

[z(x) -,sin(iTx/L)//, and., in addition, the squares of these two

quantities, are given in Table B to five decimal places.

a) One notes first that the least square procedure in this

particular case apparently reduces by over one third the integral of the square of the error. We would expect a reduction larger still in problems where the polynomial approximation oscillates highly, for the

strength of the least square procedure is particularly useful in that

kind of situation.

b) The accuracy in the value of Pi is increased nearly three decimal places, a significant improvement.

c) An odd feature of this solution is that x, falls very close to the point of maximum error. In other words, the best polynomial for approximating the sine curve has its maximum error very near to the point in which we are most interested.

d) None of the three curves may be distinguished from the others

on even a full-page graph. A graph of the error is included in the Appendix instead (see Figure 3).

e) There are a few remarks about the third (second symmetric) mode of buckling in Reference 1 giving results,z 3(x), obtained by

polynomial approximation. Table C contains these, and the corresponding results from the present method. Sin(3ix/L) is the exact solution.

(19)

~U~T~Ofljo

Gp097-. LT qsjiZaqq u 1

7O9

99~

-e 0

i,9901,c

0

0

0

0

0Z

0791

'0

ZIL1000 -06 0007 9*1 56* '5i0 061 69C IC ( l 06 *'01

0069

0'CCU

06

91

99

0000~-696

99007

00

06 000

*

1

09- 1C

I~ LI

000.

69L

lTo

10 701

5L 52

91 000

z

0-1

5007

0L

'09.oc

CE

000

69 Oki~

~

J 91 tr07 I~ 11000

~j

0 0 0 0 0 0101 *0

V)

z V

rj/x

!

-r-(ri/y, 4)

u svr-

(X) z

(20)

Table C x/L

y

3

(x)/"

sin(3vx/L)

z3 x)/A

50

0

1.00 1.00 1.00

.25, .75

1.0

-

.62

-

.71

-

.9

0, 1.00 2.0 0 0 0 I? PL2 1.h 2i

=

l.00 2 9192 _9f2

The Third Mode of Buckling

Here, too, the increase in accuracy is substantial as far as the critical load is concerned, and apreciable in the buckling deflection, as might be expected from the greater strength of the present method.

(21)

Vill The Summary

Since it is not the problem but the application of a method which is concerned in this paper, it is not upon the solution but upon the accuracy and efficiency.which attention should be focused. The efficiency is no longer questioned.5

As is evident from VI, the accuracy of this method is a consider-able improvement over that of the straight polynomial approximation with the same order polynomial. Accuracy in the latter is increased

by either increasing the order or using the least square procedure.

Increasing the order involves one very soon in machine error, while it will be noted that in the present case the earliest machine

error appeared in the final equations (32). Again, increased order

means an increased order in the secular equation, with corresponding

increased error in the solution.

Use of the least square procedure, on the other hand, substitutes an additional approximate integration for this machine error, but, if this integration is reasonably successful, the new solution cannot be less accurate, in a least square sense, than the original polynomial

approximation.

5

See References

1

and (particularly) 2.

(22)

'-The problem given here illustrates the ease with which the least square procedure is adapted to a homogeneous6 linear integral equation problem with reasonably satisfactory results.

6

Note that Reference 2 contains only non-homogeneous equation

(23)

21

References

1. Crout, P.D., "An Application of Polynomial Approximation to the

Solution of Integral Equations Arising in Physical Problems,

Journal

of Mathematics

and

Physics, Vol. XIX, No. 1, January

1940.

2. Hildebrand, F.B., and Crout, P.D., "A Least Square Procedure for

Solving Integral Equations by Polynomial Approximation,"

Journal of Mathematics and Physics, Vol. XX, No. 3, August

1941.

(24)

Calculated Values (exact): 128 EI G(xk, ) , L D K. ()

X

i(xk)

Transposed

Transposed

Table #1 Table #2 Table #3 Table #4

Table ,5

Table #6 Table # 7

Iransposed

TV.

Graph of Ay , Az as a function of x Pgnk pk

(25)

iabiefrl

128 EI G(xk j) L

(k=1,

*.. ,

9)

(j

=1,

..

P 17)

Transposed 128 EI G(xksyj)= L

3

2

3

14

5

6

7

8

9

10

11

12

13

114

16

17

0

7

14

13

12

11

10

9

8

7

6

5

14

3

2

1

0

4

0

6

12

18

214

22

20

18

16

14

12

10

8

6

4

2

0

128

L

(k

6

7

0

5

10

15

20

25

30

27

214

21

18

12

9

6

3

0

(L

8

0

4

8

12

16

20

24

28

32

28

214

20

16

12

8

14

0

-j)

, x ( 5

(j

4 1) A2k

j

- )(9 - k)

(j

1) 2k 1

(exact)

2

x

0

3

6

9

12

15

18

-21

24

27

30

25

20

15

10

5

0

9

0 2

4

6

8

10

12

14

16

18

20 22 24

18

12

6

0 i i i i

0

1

2

3

4

5

6

7

8

9

10

12

13

14

7

0

0 0 0 0 0 0

0

0 0 0 0 0 0 0

0

0

0

f j (L- Xk) -

1)(17- j),

(26)

-Table #2 2# I D i i~j (4 214 -1 1.50390 6250 1.09375 0000

1.00000000

150390

6250

.6875

0000

.41015 6250

0

-

.24609

37150

- .15625 0000 -

.21484 1375C

0

.75390 625o

.21375 0000

.4o15650

0

1

3

14

1 -

.96679

6375

-

.5,687 S000o

-

.60$556 6375

.75195 3125

.70312

5000

1.85470 3125

1.0000 0000

1.8470 3125

.70312

5000

.75195 3125

0

-

.685h 6375

-

.54687

000

-

.96679o6875

values of Ki(A ) from tables (3a),(6b)

Ag = j- 9

2,

j

1, 17

3,5,...

,

13, 15

j

2,1,...

,

14, 16

(exact)

2

i

=

-2

.)ioi5

6250

.2175 0000

.2$390 250 0

-

.21434

3750

-

.15625 0000

-

.2).69

3750

0

.687005

00

1.50390

6250

1.00000 0000

2.28515 625o

1.09375 0000

1.50390

625o

6

7

8

9

10

11

12

13

114

16

17

2 4 (i

=-2,

.. ,

' 2)

(j

-p

1,

..

j

17)

(27)

Table #3

128

*

24-El Ii(xk)

PT2

(exact)

(i

=

-21, ... , 2)

(k

=

1, ...

, 9 )

-1 0 I I

0

90.343 75

132.87-5

129.093 75

102.5

74.593 75

52.7$

30.843 75

0

0

33.359 375

88.687 5

143.184 375

166.25

143.434 375

88.687

5

33.359 375

0

1

0

30.843 75

52.8075

74.593

75

102.5

129.093 75

132.875

90.343 75

0

128•

24*

E li(xk) PL2 I

values from Tables #1 and #2

X

I (xk)

=

i= -2

k= 1

2

3

6

7

8

9

2

0

0

0

0

0

0

0

0

0

25

128

EI1 G(xkt

1j) ,/*

L -- L

(28)

f

3pk

=12

XkIpxk)=

12• 24-

128

31'kspxk) (exact)

PI4L

(k 1, ... , 9)

(p

1,

,5) Transposed

p=l

2

3

4

5

k

l

0

0 0 0 0

2

0

180.687

5

66.71

75

61.637

5

0

3 0

132.875

883.687

5

52.875

0

4

0

258.187 5

286.968

75

149.187

5

0

5

o

102.*

166.25

102.5 0

6

o

19

.187

5

236.968

75

258.137

5

0

7

0

2.G75

38,6375

132.875

0

8

0

61.687

5

5

6.718

75

180.687

5

0

90

0

0

0

0

values of 24 123 EII D(xk)

P71

from Table

#3

2~ Pk

=1,

9

12 D 1 , k 3,

5,

7

L2

,

k

=2,

4, 6, 8

(29)

iabio

#5

(exact)

O p,

=

12

22123

24

EI 2

'

(p= 1

..

,5) (i

=-2,

... , 2)

-1

0

0

0

93

6L3.289

062 5001109

01.753

906

3101100

051.753 906 2501130 172.462 690

250

625

74 222.289

062 oolloo 0571.53 906 250

0

0

1

0

2

0

74

222.289 062

500

0

10051.753 906

250

0

93

643.289 062 000

0

0

12 128

• 2

Pi

12• 24 * 128 El1 Dklp fXk ) I

~

FL

.124.

128

EI

Ii(xk)

values

from Tables #3 and #4

i=

-2

110

2

0

4

5

0

0

=

27

(30)

pable

#6

(k= 1, ... , 9)

(p= 1,

...

,

5)

Transnosed

p

=

1

2

555

31h

458

162

189

0

072.0

192.0

152.0

880.

)

304.0

432.0

504.0

0

3

20h

27?

881

510

881

272

20h

0

96150.0

!488.0o

568.o

720.0

568.o

6488.0

960.0

4a,0

N

=

3072

X=

Z NOgi = Z

2/

3

pk

= P 2

NZtpk

values of/pk from Table

#4

(exac U)

k

1

2

3

h

5

6

7

8

9

189

162

45

314

793

)08

555

0

432.0

3016.0

152.0

192.0

072.0

0

"Illpk /9-2p-#k EI

(31)

-Table#7 2

(exact)

k

=

2, , 6, 8

(p

=

1,

... ,

5)

(i=

-2, ... , 2)

0

827 784.

816 988.

1

l1

176.

0

0

0

787

620.

1

526

250.

787

620.

0

1 0

111117.

816988.

827

78b.

values of Npok from Table #6

values of Kj(Ak) from table (3a)

'pi=

i=

-2

1

2

3

4

0

0

0

0

0

2

0 0 0

0

0

I i 29 -1

- 1.5

(32)

~1

-A FIGURE 3

POSMON.)

17,-H1'

I -4fH

ti 't PA. H+FH -1H+H-H+11H-++I-4111H+-4 H+H+H + r+4-H .,Ax1 3 ~c3 - LU X=L/2

4

21

~H±~I9I1

xl

3 T; 0 -A

.~>

I

2Z -+H+±H-L, - -, : 1 -, I-, . I--lag PO S TI ON1 I

M,1111 i I I-IT

T I I J I I i #fl-H+ I I I I TT i i ....i ...' W i I 1 1 / rfl

I

4

Figure

Table  of  Contents
Table  C  x/L y 3 (x)/&#34; sin(3vx/L) z3  x)/A 50  0  1.00  1.00  1.00 .25,  .75  1.0  - .62  - .71  - .9 0,  1.00  2.0  0  0  0 I?  PL 2 1.h  2i  =  l.00  2 9192  _9f2

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