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
1homas
11I.Creese
Submi
tt'3d in partial fulfillinentof
the requirementsfor the decree of Bachelor of Science
at the
HASSACi-iUSJT'IS INSTITUTE OF TEC.h"}IOI.DGY
Nay
21
,
1956
Signature redacted
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Signature of Author
,l"ICertified by
1 I'\ DepartmentSignature redacted
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Thesis Advisor
Date
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
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
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(~eKlAY+ ..
+KnE
(1)
where
x 0is 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
m
rn
12:
S Dk
L
Yii
xk)- k(x minimum, (3b)k
=1
i nfrom 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 (p1,
, 2n -t- 1)k
=
column index i=
column indexFor 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 Equareprocessand the l and 1 are coefficients arising in the
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
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 momentapied
atx
¶over
an intervalO
il, ithx, , , = + X < x6)
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
3and
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.
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= -2I..
I I-1l
p= 1
k
1
2 02
3
3
1I
a I I 146
7
I I I I8
.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
L m I f~gdf=~L;;
D -f(
)
O k = 1k D=D' I h .1= 19
32 L 12 D= D' D'= D2
4
6
8
1
I / I D /= D, D/= 2_3
5
12 D :12 Dkk
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 L2=
y
P
G(x,
)K.()d.
(15)
i-2 0Making 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
Since
2
(x ) = e<jI (x
k
j~2 ikkand in this problm
k k '
we have
2
yk .Yi.
iL xk
i
- -2
and identify17
I (xk) k=
PL D G(xk' j - 1 This is equivalent to I (xk) PLI G(xkj)
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),
thesevalues have been omitted from Table #2 and dashes inserted in their
place.
(lb
)
('7)
.(18)
-I
(19)
has
(i
=
-2,
...
,
2)
5j )K i( )X124•128
2 Ii(xk) k - row (1, 9) i - column (-2, 2) 128 El G(xk,'•jjk
-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) I12 1 p - row (1,
5)
i- row (-2, 2) i - column (-2, 2) 11 / -k I 1 k12'
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- 70Y,#
KI(kX1Substituting
(27) into
(25)
gives
1
1
7i
i=-l
i2.1 -1(k
=
2,
4, 6,3
)
,
(27)
(p 1, ••• ,5)
whereL~i
Z= N=
3072
Z=
E.II
(28)
(29)
11or
(24)'k
=
2,
4,
6, 8
kT12 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.
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
4433,
=
y_
.
I
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
9 6 0 609*6
-4-_ d
60
9L 160.
5
1T6 *
09
L96*
00
000i1
9!°6
17L7f409 69
6
=
0
L5r
911
69,
61
00
0
60
T69
L26
0000-1
66
6L
10
691
00
011
0° 0 10
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 - k1U
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.
~U~T~Ofljo
Gp097-. LT qsjiZaqq u 17O9
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 *'010069
0'CCU
0691
99
0000~-696
99007
00
06 000
*
1
09- 1C
I~ LI000.
69L
lTo10 701
5L 5291 000
z
0-1
5007
0L
'09.oc
CE000
69 Oki~~
J 91 tr07 I~ 11000~j
0 0 0 0 0 0101 *0V)
z Vrj/x
!-r-(ri/y, 4)
u svr-
(X) z
Table C x/L
y
3(x)/"
sin(3vx/L)
z3 x)/A50
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 _9f2The 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.
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.
'-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 equation21
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.
Calculated Values (exact): 128 EI G(xk, ) , L D K. ()
X
i(xk)
Transposed
Transposed
Table #1 Table #2 Table #3 Table #4Table ,5
Table #6 Table # 7Iransposed
TV.
Graph of Ay , Az as a function of x Pgnk pkiabiefrl
128 EI G(xk j) L(k=1,
*.. ,9)
(j
=1,
..
P 17)
Transposed 128 EI G(xksyj)= L3
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
108
6
42
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) A2kj
- )(9 - k)(j
1) 2k 1(exact)
2x
03
6
9
12
15
18
-21
24
27
30
25
20
15
10
5
0
9
0 24
6
8
10
1214
16
18
20 22 2418
126
0 i i i i0
1
2
3
4
5
6
7
8
9
10
12
13
14
7
0
0 0 0 0 0 00
0 0 0 0 0 0 00
0
0
f j (L- Xk) -1)(17- j),
-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
.703125000
.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$390 250 0-
.21434
3750
-.15625 0000
-.2).69
3750
0.687005
00
1.50390
6250
1.00000 0000
2.28515 625o
1.09375 0000
1.50390
625o
6
7
8
9
1011
12
13
114
16
17
2 4 (i=-2,
.. ,' 2)
(j
-p
1,
..
j
17)
Table #3
128
*24-El Ii(xk)
PT2(exact)
(i=
-21, ... , 2)(k
=
1, ...
, 9 )
-1 0 I I0
90.343 75
132.87-5
129.093 75
102.5
74.593 75
52.7$
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
030.843 75
52.8075
74.593
75
102.5
129.093 75
132.875
90.343 75
0
128•24*
E li(xk) PL2 Ivalues 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
25128
EI1 G(xkt1j) ,/*
L -- Lf
3pk=12
XkIpxk)=
12• 24-128
31'kspxk) (exact)PI4L
(k 1, ... , 9)(p
1,
,5) Transposedp=l
2
3
4
5
kl
0
0 0 0 02
0180.687
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 06
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
iabio
#5
(exact)
O p,=
1222123
•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
20
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
20
4
5
0
0
=
27pable
#6
(k= 1, ... , 9)(p= 1,
...
,5)
Transnosed
p=
12
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 = Z2/
3pk
= P 2NZtpk
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
432.03016.0
152.0
192.0
072.0
0
"Illpk /9-2p-#k EI-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.
1526
250.
787
620.
0
1 0111117.
816988.
827
78b.
values of Npok from Table #6
values of Kj(Ak) from table (3a)