The improvement of the reduced coordinate iterative procedure for simultaneous linear equations
by Timothy Mark Hodges
A thesis submitted in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE
in Mechanical Engineering
Montana State University
© Copyright by Timothy Mark Hodges (1982)
Abstract:
An iterative method for solving simultaneous linear equations is considered. Previously, the trial
solution generation routine depended on knowledge of the physical system. A method is introduced to
generate the trial solutions based on the system of equations. The new method is used to solve several
differential equations with various boundary conditions, so as to show that the method is independent
from the physical system. Examples of two steady-state problems and a transient problem, with 1000
unknowns, demonstrate the performance as compared to existing techniques. Results indicate that the
method is superior in cases where the boundary conditions are more complicated, and the method is
comparable in all other cases. IHE IMPROVEMENT OF THE REDUCED COORDINATE ITERATIVE
PROCEDURE F O R SIMULTANEOUS LINEAR EQUATIONS
by
Timothy Mark Hodges
A thesis submitted in partial fulfillment
of the requirements for the degree
of
MASTER OF SCIENCE
in
Mechanical Engineering
MONTANA STATE UNIVERSITY
Bozeman, Montana
December 1982
MAiN u b .
n1
37?
(pkb ^
dap' ^ l
H
ii
APPROVAL
of a thesis submitted by
Timothy Mark Hodges
This thesis has been read by each m e mber of the thesis committee
and has been found to be satisfactory regarding content, English usage,
format, citations, bibliographic style, and consistency, and is ready
for submission to the College of Graduate Studies.
XvU-CDate
Z 7
/<?
A
Chairperson, Graduate Committee
Approved for the Major Department
42 2-1
>
^ .
Date
Head, Major Department
Approved for tl
i'Z
Date
"
7 J ' t 2'
Graduate Dean
iii
STATEMENT OF PERMISSION TO USE
In presenting
this
thesis
in partial
ments for a master's degree at Montana
fulfillment
of the require­
State University,
I agree
that
the L i b r a r y shall m a k e it a v a i l a b l e to b o r r o w e r s u n d e r rules of the
library.
Brief quotations
special permission,
f r o m this
thesis
are a l l o w a b l e w i t h o u t
provided that accurate acknowledgement of source is
made.
P e r m i s s i o n for e x t e n s i v e q u o t a t i o n from or reproduction of this
thesis m a y be granted by m y major professor,
the Director of Libraries when,
or in his/her absence,
in the opinion of either,
use of the m a t e r i a l is for s c h o l a r l y purposes.
the m a t e r i a l
Date
the proposed
A n y co p y i n g or use of
in this thes i s for f i n a n c i a l g a i n shall not be a l l o w e d
without m y written permission.
Signature.
by
V
ACKNOWLEDGEMENTS
The aut h o r w i s h e s to th a n k Dr. D. 0. B l a c k k e t t e r and Dr. Ri 0.
W a r r i n g t o n for t h e i r h e l p and guid a n c e
study.
The
auth o r
in the p e r f o r m a n c e
of this
also w i s h e s to t h a n k Dr. R. J. Cona n t for all his
help in advising over the past year.
all her support and understanding.
And finally thanks go to Lisa for
vi
TABLE OF CONTENTS
Page
L I S T S .........................
List of T a b l e s.......... . .................................... . ...
List of F i g u r e s ..........
ABSTRACT.................................. ............. , ..... .
1.
2.
3
INTRODUCTION.... ...........................
DESCRIPTION OF METHOD .............. ................ ....... . . .
vii
vii
ix
xi
I
3
. DESCRIPTION OF THE NEW TRIAL SOLUTION GENERATION..............
6
4.
APPLICATIONS AND R E S U L T S ........................ ...............
. 10
5.
DEVELOPMENT OF RCDP WITH APPLICATIONS AND RESULTS .......
47
6.
CONCLUSIONS....... .....
58
REFERENCES C I T E D ..................
60
APPENDICES,... ...............................
62
Appendix A— Illustrative Example of RCIP
.................
Appendix B- S o l u t i o n of a S i m p l e B o u n d a r y Va l u e P r o b l e m
Using RCIP ..........................
Appendix C- Flow Chart of RCIP Computer Program .............
Appendix D- S o l u t i o n of a S i m p l e B o u n d a r y Value P r o b l e m
Using R C D P .........................................
63
68
72
74
vii
LIST OF TABLES
Table
1
2
3
4
5
6
7
8
Page
Results of I-D Ordinary Differential Equation Problem I,
with Derivative Boundary Conditions, for Various Numbers
of Trial Solutions.........................................
14
Results of I-D Ordinary Differential Equation Problem 2,
w i t h Composite Boundary Conditions, For Various Numbers
of Trial Solutions ........................................
16
Results of I-D Ordinary Differential Equation Problem 3,
with Mixed Boundary Conditions, For Various Numbers of
Trial Solutions .....
17
Results of I-D Ordinary Differential Equation Problem 3,
with Mixed Boundary Conditions, for Various SOR Factors
18
Results of I-D Ordinary Differential Equation Problem 4,
w i t h Dirichlet Boundary Conditions, for Various Numbers
of Trial Solutions .......................
21
Results of I-D Ordinary Differential Equation Problem 4,
w i t h D i r i c h l e t B o u n d a r y C onditions, for V a r i o u s SOR
Factors ......................
22
2 -D L a p l a c e s E q u a t i o n R e s u l t s F o r Va r i o u s N u m b e r s
Trial Solutions ............
of
25
2-D Laplaces Equation Results for Various SOR Factors
..
26
Heat Condu c t i o n
P r o b l e m , C u b e - C u b e G e o m e t r y , F o r Va r i o u s N u m b e r s of
Trial Solutions ...........................................
33
Results of the 3-D Steady-State Heat Conduction Problem,
Cube-Cube Geometry, For Various SOR Factors.............
34
Results of 3-D Transient Heat Conduction Problem, Cube
Geometry, Using RCIP ......................................
38
Results of 3-D Transient Heat Conduction Problem, Cube
Geometry, Using ADEP ......................................
40
9 R e s u l t s for the 3-D S t e a d y - S t a t e
10
11
12
viii
LIST OF TABLES (continued)
Table
13
Page
Results of RCDP for the Ordinary Differential Equation,
with Derivative Boundary Conditions, For Various Numbers
of Equations..,........... .................................
50
14v R e s u l t s of G a u s s i a n - E l i m ! n a t i o n for the O r d i n a r y
!./
Differential Equation, w i t h Derivative Boundary
’
Conditions, For Various Numbers of Equations............
50
V
*
15
Results of RCDP for the Ordinary Differential Equation,
with Composite Boundary Conditions, for Various Numbers
of E q u ations........................
52
R e s u l t s of G a u s s i a n - E l i m i n a t i o n for the O r d i n a r y
Differential Equation,
with Composite Boundary
Conditions, For Various Numbers of Equations ...........
52
Results of RCDP for the Ordinary Differential Equation,
with Mixed Boundary Conditions
Conditions, for Various
Number of Equations .......................................
54
I S ^ R e s u l t s of G a u s s i a n - E l i m i n a t i on for the O r d i n a r y
Differential Equation, with M i x e d B o u n d a r y Conditions,
for Various Numbers of Equations ........................
54
16
17
19
20
Results of RCDP for the Ordinary Differential Equation,
with Dirichlet Boundary Conditions, for Various Numbers
of Equations ...............................................
R e s u l t s of G a u s s i a n - E l i m i n a t i o n for the O r d i n a r y
Differential Equation,
with Dirichlet Boundary
C o n d i t i o n s , for V a r i o u s N u m b e r s of E q u a t i o n s ........
56
56
ix
LIST OF FIGURES
Fig.
1
2
3
4
5
6
7
8
9
10
11
Page
T r i a l S o l u t i o n s and D i f ferent Relaxation Factors Versus
U s e r E x e c u t i o n T i m e for 1—D O r d i n a r y D i f f e r e n t i a l
Equation Problem 3, With Mixed Boundary Conditions......
19
Trial Solutions and Different R e l a x a t i o n F a c t o r s V e r s u s
U s e r E x e c u t i o n T i m e F or I-D O r d i n a r y D i f f e r e n t i a l
Equation Problem 4, with Dirichlet Boundary Conditions...
23
T r i a l S o l u t i o n s and D i f f e r e n t Relaxation Factors Versus
User Execution Time for the 2-D Laplaces Equation........
27
C u b e — C u b e G e o m e t r y f o r the 3 — D S t e a d y — S t a t e H e a t
Conduction P r o blem..........................................
30
Tr i a l S o l u t i o n s and D i f f e r e n t Relaxation Factors Versus
User Execution Time for 3—D Steady— State Heat Conduction
Problem, Cube-Cube Geometry......... ...... .............. .
35
U s e r E x e c u t i o n T i m e V e r s u s A v e r a g e P e r c e n t Error. R C I P
Versus ADEP for the 3-D Transient Heat Transfer Problem
with Elapsed Time T=0.1 .....................................
42
U s e r E x e c u t i o n t i m e V e r s u s A v e r a g e Pe r c e n t E r r o r . R C I P
Versus ADEP for the 3-D Transient Heat Transfer Problem
with Elapsed Time T = 0 . 5 ....................................
43
U s e r E x e c u t i o n T i m e V e r s u s A v e r a g e Pe r c e n t Error. R C l P
Versus ADEP for the 3—D Transient Heat Transfer Problem
with an Elapsed Time T=I .0...................... .'..........
44
User Execution Time Versus Average Percent Error for RCDP
Versus Gaussian-Elimination of I-D Ordinary Differential
Equation Problem I, Derivative Boundary Conditions.... ,..
51
User Execution Time Versus Average Percent Error of RCDP
Versus Gaussian-Elimination for I-D Ordinary Differential
Equation Problem 2, Composite Boundary Conditions........
53
User Execution Time Versus Average Percent Error of RCDP
Versus Gaussian-Elimination for I-D Ordinary Differential
Equation Problem 3, Mixed Boundary Conditions............
55
X
LIST OF FIGURES (continued)
Fig.
12
Page
User Execution Time Versus Average Percent Error of RCDP
Versus Gaussian-Elimination for I-D Ordinary Differential
Equation Problem 4, Dirichlet Boundary Conditions..... ..
57
xi
ABSTRACT
An iterative method for solving simultaneous linear equations is
considered.
Previously, the trial solution generation routine depended
on k n o w l e d g e of the p h y s i c a l system.
A m e t h o d is i n t r o d u c e d to
generate the trial solutions based on the system of equations. The new
m e t h o d is u s e d to solve seve r al d i f f e r e n t i a l e q u a t i o n s w i t h va r i o u s
boundary conditions, so as to show that the method is independent from
the p h y s i c a l system.
E x a m p l e s of t w o s t e a d y - s t a t e p r o b l e m s and a
transient problem, w i t h 1000 unknowns, demonstrate the performance as
compared to existing techniques.
Results indicate that the method is
superior in cases where the boundary conditions are more complicated,
and the method is comparable in all other cases.
(I)
CHAPTER I
INTRODUCTION
Many numerical methods have been developed to accurately solve a
large s y s t e m of s i m u l t a n e o u s e q u a t i o n s .
that has the possibility of being
A m e t h o d is p r e s e n t e d here
faster and more
accurate
than these
methods.
The Reduced Coordinate Iterative Procedure (RCIP)[1] uses a linear
c o m b i n a t i o n of trial s o l u t i o n s , w i t h each trial s o l u t i o n having an
unknown weighting
coefficient.
mined by minimizing
E a c h weighting coefficient is deter­
an error function.
This minimizing
process pro­
duces a smaller set of equations which can be solved directly for these
w e i g h t i n g co e f f i c i e n t s .
tions,
Therefore,
in solving a large set of e q u a ­
the RCIP method reduces the number of equations to be solved.
The m e t h o d of least squares can be r elated to the R C I P method,
because
of
method.[3]
passes
the
minimization
of
the
s quared
residual
in the
RCIP
The least squares method finds the equation of a curve that
through scattered points.
This
curve
is found
so that the sum
of the square distances from each point to the curve is a minimum.
the
RCIP method
equation set,
used
for
the
s u m of the
the variance,
curve
equations as RCIP
fitting
is.
squares
is minimized.
and
not
for
the
of the r e s i d u a l s
The least
solu t i o n
squares
of
from
method
In
the
is
simultaneous
(2)
Another method that RCIP is related to is the Conjugate Direction
M e t h o d [4,8].
The R C I P and C o n j u g a t e D i r e c t i o n m e t h o d s b o t h use the
minimization of an error function in the solution of the problem.
Conjugate
Direction
direc t i o n s .
The
Method
minimizes
the
error
along
Conjugate Direction Method will
iterations than there are unknowns.
The
o r t hogonal
converge
in less
The methods have been shown to be
different [I].
To
evaluate
techniques were
using
the p e r f o r m a n c e
used:
successive
of the R C I P m ethod,
The first was
a Gauss-Seidel
o v e r - r e l a x a t i o n (SOR)
[5],
iterative
scheme
and the second w as the
Alternating Direction Explicit Procedure (ADEP)[6],
methods
t w o existing
Both are excellent
for c o m p a r a t ion, b e c a u s e of the Ir speed of c o n v e r g e n c e and
their accuracy.
The RCIP method previously has had a systematic procedure for the
generation of trial solutions.
was
dependent
on knowledge
However,
the trial solution generation
of the physical
system which the equation
represented. The trial solution generation should be independent of the
knowledge of the physical system.
It
is
the
purpose
of
this
paper
to p resent
a trial
solu t i o n
generation scheme that is independent of the knowledge of the physical
system while continuing to produce rapid convergence and be competitive
with existing methods.
(3)
CHAPTER II
DESCRIPTION OF THE METHOD
The RCIP method uses a linear
tors
with
each
efficient.
duced set
trial
combination of trial
solution having
solution vec­
an unknown scalar weighting
co­
The w e i g h t i n g c o e f f i c i e n t s are o b t a i n e d b y solving a r e ­
of equations which are obtained by minimizing the variance.
The r e d u c e d set is s o l v e d usi n g an exact or direct method.
The trial
s o l u t i o n s are d i f f e r e n t at each i t e r a t i o n and the it e r a t i v e process
continues
until
the
desired
convergence
is
reached
[2].
The
formulation of RCIP proceeds as follows:
Let
AX=F
(I)
be a system of n equations in n unknowns.
(m is the number of trial solutions,
Initially choose m vectors
G(k,m)Q n in length,
the subscript
0 indicates the first iteration)
G(k,I)q ,'G(k, 2 )q ,....,G(k,m) q
k=l
(2 )
where m < n, and set
m
k=l
X0 = Z ctj G(k,j)0
(3)
J=I
where
the
scalars
(a^'Qg"
o m ) are to be determined.
Let
,am )= < A X 0- F , A X 0-F> = | IA X q- F I I2
(4)
(4)
be
the
inner
(R q = A X q -F),
product
and
the
respectively.
square
F i n d the
of
the
norm
of
the
residual
s o l u t i o n (<Zq ^,....,aQ m ) to the
system
p
I ,..., m .
(5)
Designate
V 0 V ( a 0 , l ,,--,<z0 , m )
and let
X(l,l)= 2
k -!#•••# He
UQ j G(k,j)0
j=l
Then
...,am )= l IAX1-Fl I
solve the system
p !#•••>m
and if Ca1
m ) is the solution, de s i g n a t e
(10)
V 1= V 1 (O1 1 1 --^al m )
and let
m
X(2,l)= ^
O1 . G(Icj)1
j=l
k = l ,...,n
(11)
k Ii111 >u
(12)
At the ith stage.
X(i,D— 2 ai—i 4 ®'Ck, j )^_i
3=1
’J
Select v e c t o r s G(k,2)
X. =
G(k,m)
m
2 Oj G ( ^ j ) i
J=I
and let
k = l , 11.,n
(13)
(5)
Then
Vi U
1, . . . , <zmM
IAXi - F l I 2 .
. (14)
Again solve the system
9V.
P=I,..,m.
(15)
If this s o l u t i o n is (^i !,•••,^ i m ), then
V i = V i (ai>1,.,•,ai>m) .
(16)
The algorithm continues with
X(i+l,l)=
U i j G(k,j)i
k=l,...,n ,
(17)
j=l
s e l e c t i o n of G(k,2) i + 1 ,...,G(k,m)i+1
,
X i+1™ 2
“ j G (k,j)i+1
j=l
k = l ,...,n ,
(18)
and a subsequent minimization of
V i + l (al ’---’a m ) = ^ A X i+l_ F ^ 2 -
(19)
The solution of the original system is derived from
X = L i m i_ >+eoG(i,l).
(20)
T o i l l u s t r a t e the t h e o r y of the above s e c t i o n an e x a m p l e p r o b l e m
presented
in Appendix A.
is
(6)
CHAPTER III
DESCRIPTION OF NEW TRIAL SOLUTION GENERATION
P r e v i o u s l y the trial s o l u t i o n g e n e r a t i o n p r o c e s s w a s b a s e d on
knowledge
of the physical
system
represented.
The new
trial solution
selection process successfully makes the RCIP method independent of the
k n o w l e d g e of the p h y s i c a l system.
system
of
equations
and
The n e w m e t h o d u t i l i z e s only the
information
resulting
from
other
trial
solutions to obtain sets of trial solutions.
At the beginning of the first iteration,
set is selected.
an initial trial solution
O n s u c c e s s i v e i t e r a t i o n s the trial s o l u t i o n set is
c o m p o s e d of the last b e s t solution, and a sequence of special G a u s s —
Seidel solutions.
E q u a t i o n s to be solv e d b y G a u s s - S e i d e l for trial
solutions are formed from the same
coefficient matrix as the original
p r o b l e m and a n e w right h a n d side.
F o r the second trial solution the
n e w right h a n d side is the r e sid u a l f r o m the e q u a t i o n set using the
last
best
solution.
s o l u t i o n is zero.
o b t a i n trial
The
initial
Three Gauss-Seidel
s o l u t i o n two.
trial solution two.
is f o r m e d f r o m
guess
used
for
iterations
the
Gauss-Seidel
are p e r f o r m e d
to
The third trial s o l u t i o n is s i m i l a r to
The difference is that another new' right hand side
the r e s i d u a l
of the seco n d trial solution.
Again a
Gauss-Seidel solution scheme is used to solve for the solution to this
new
set.
An
initial
guess
of zero
is used,
w i t h three
iterations
(7)
x
performed.
The rest of the trial solutions are calculated in the same
fashion until m number of trial solutions are reached. An example, of m
equal to three trial solutions,
is shown below.
Let
A*X=F
(21)
be an nxn system of equations.
For the first iteration the first trial
solution is
G(i,I) = initial estimate
i=l,...,n.
Now find the residual vector for thefirst
R^T.S.
(22)
trial solution (or T.S. I)
I) = A*G(i,l) - F i
i=l,...,n.
(23)
i=l,...,n .
(24)
Form the new equation set for trial solution two
A*G(i,2) = R i CT-S. I)
Solve using Gauss-Seidel with an initial
iterations
to o b t a i n G d , 2).
Now
guess of zero.
Perform three
find the residual v e c t o r for the
second trial solution
R i (T.S. 2)= A*G(i,2) - R ^ T . S . I)
i=l,...,n.
(25)
Form the new equation set for trial solution three
A*G(i,3) = R i (T.S. 2)
Solve
this
Perform
set
three
using
Gauss-Seidel
iterations
i=l,...,n.
with
to obtain G d , 3).
for the first i t e r a t i o n have b e e n found.
an
initial
The
three
solution.
guess
of zero.
trial
solutions
For the s e c o n d th r o u g h the
nth iterations the only change is that the first trial
last best
(26)
solution is the
(8)
The ba s i s for this trial s o l u t i o n s e l e c t i o n p r o c e s s is b a s e d on
the following:
A*G(i,l) - F -
subtract
R i CT-S. D = O
(27a)
A*G(i,2) - R i CT-S. I) - R i CT-S- 2) = 0
(27b)
A*G(i,3) - R i CT-S- 2) = 0
(27c)
(27c)
from (27b)
A*(G(i,2) - G ( i,3)) - R i CT-S- D = O
subtract
(28) from
A*(G(i,D
(28)
(27a)
- G(i,2) + G(i,3)) - F
= 0.
(29)
N o t e , if G d , 3) is an exact solution to equation (27c) and
X i = Gd,I)
then X i is an exact
- G d , 2) + G d , 3).
s o l u t i o n to e q u a t i o n
(30)
(21). Note
that
G d , 3) is
approximate therefore, X i is approximate.
The numerical algorithm for the generation of the trial solutions
is as follows. The first trial solution for the first iteration is
Gd,I)
= initial estimate
i=l,...,n.
(31)
The first trial solution for the second through the nth iterations is
Gd,!)
= X.
i=l,...,n.
(32)
(9)
The second through the m t h trial solution is calculated as follows
i=l,...,n
i=l,...,n
k k = 2 ,m
p = l , ..,3
(33)
(34a)
R i (kk)
= (A(i,k,)*G(k,kk-l)) - R i(Rk-I)
(34b)
G(i,kk)
= 0
(34c)
G(i,kk)p = R i (Rk) / A(i,i)
n
- Z
A(i,k)*G(k,kk)
k=l
(34d)
/ A(i,i)
.
The above a l g o r i t h m s w i l l c a l c u l a t e m n u m b e r of trial s o l u t i o n s . A n
illustrative problem is solved and included in Appendix B.
(10)
CHAPTER IV
APPLICATIONS AND RESULTS
This c h a p t e r is d e v o t e d to the s o l u t i o n of p r o b l e m s b y the R C I P
method
and a c o m p a r i s o n w i t h
some
particular the SOR and ADEP methods
The
RCIP
equations.
method
is
Therefore,
i n t e n d e d to
of the exis t i n g
techniques.
In
are compared with the RCIP method.
solve
linear
systems
of algebraic
to solve the following differential equations a
finite difference technique is used to model each differential equation
as a system of algebraic equations.
In each p r o b l e m
the c o n v e r g e n c e lim i t used is a l w a y s that the
v a r i a n c e be less than or equal
to some
convergence
criterion.
The
v a r i a n c e of the e q u a t i o n set is d e f i n e d as the s um of the residuals
squared,
for all the equations.
average percent
W h e n an exact s o l u t i o n is k n o w n an
error can be obtained which
of the d i f f e r e n c e s b e t w e e n the exact
is defined as the average
s o l u t i o n and the
approximate
solution divided by the exact solution.
PROBLEMS ONE THROUGH FOUR
The
following
four ordinary differential
show the generality of the new method.
equations
are
solved to
Various boundary conditions are
(11)
employed
to
illustrate
a spectrum of problems.
Problem
one
follows:
Solve
A
— -
+
y = 0
(35)
dx
for the region 0 <= x <=
y' (0) = 0
To
illustrate
differential
,
tt/2
with the derivative boundary conditions
y' (tt/2) = I
(35a)
the finite difference
technique
e q u a t i o n and b o u n d a r y c o n d i t i o n s
a l g e b r a i c form.
the above governing
will
be
modeled
in
The cent r a l d i f f e r e n c e a p p r o x i m a t i o n of the second
order derivative is
d2J
y i+1 ~ 2 Ji + Yi-!
---= --------------------dx^
Ax^
(36a)
equation (35) becomes
?i+l - % ?i + y i-l
Ax
+
2
Yi
(36b)
0
The central difference approximation of the first order derivative is
iI dx
equations
y i+l ~ y i~l
(36c)
2Ax
(35a) become
y i+l(0) = y i_ 1 (0)
y i +l(n/2) - y i_ 1 (n/2) = 2Ax
The
above
finite
(36d)
.
(36e)
d i f f e r e n c e m o d e l approximates the differential
equation as a system of linear equations.
i
(12)
The exact solution for problem one is
Tesact = " C0S x
•
<36f)
The finite d i f f e r e n c e m o d e l for p r o b l e m I is u s e d to illustrate
the SOR algorithm used.
The algorithm follows:
Y j + i(r+1) + Y . * (r+1)
Y. (r+1) = (I-X) *Y.(r) + X * -------------------------2 - AX2
the
variable
indicates
r
indicates
the p r e s e n t
nodal point.
the
previous
iteration,
iteration,
and the subscript
The variable X is the SOR factor,
the
val u e
of
X=1.0
is
just
the
r+1
i indicates the
which is between I and
2 for over r e l a x a t i o n and b e t w e e n 0 and I for under relaxation.
va l u e
(37)
G a u s s - S e i d e I routine.
The
p r o b l e m s w e r e m o d e l e d in the same m a n n e r as p r o b l e m I.
The
remaining
And the S OR
factor used in the following problems was the same as above.
[5]
Problem two follows: Solve
d2y
--dx2
+
y = 0
(38)
for the region 0 <= x <= rc/2 with the composite boundary conditions
y ' (0) + y(0) = I
y' (n/2) - y(jr/2) = -I
(39)
.
(40)
The exact solution for equations (38-40)
Yexact = (SIN x + C0S x)/2
Problem three follows:
x
2 d2y
--. dx2
•
is
,
(41)
Solve
dy
o
— 2x — + 2y = x
dx
(42)
(13)
for the region I <= x <= 2 with the mixed boundary conditions
y ( I) = 0
,
y ' (2) = 0.
(43)
The exact solution for equations (42-43)
4
Yexact “ " x
3
5 ’5 2
---x
+
3
"
is
1 3
" x
2
•
<4 4 >
Problem four follows: Solve
2 d2y
(1—x )
dx2
—
dy
6x —
dx
— 4y = 0
(45)
for the region 0 <= x <= 0.5 with the Dirichlet boundary conditions
y(0) = -4.5 , y ( 0 .5) = 2
.
(46)
The exact solution for equations (45-46)
27
0
2 2
Yexact ~ —
(3x-x /(1-x ) )
6
S t a n d a r d fini t e d i f f e r e n c e
approximate the above problems.
is
-
27
2 2
—
(l/(l-x ) )
6
.
(47)
t e c h n i q u e s are used to algebraically
In each of the first
nodal n e t w o r k is c h o s e n so that there is
four problems
a
50 spaces for each region,
which requires a different number of equations
for each problem.
The
d i f f e r e n t n u m b e r of e q u a t i o n s r e s u l t e d f r o m the d i f f e r e n t b o u n d a r y
conditions.
A u n i f o r m initial e s t i m a t e of 0.5 is c h o s e n to start the
methods.
In problem one the convergence limit was that the variance in the
51 e q u a t i o n set be less t h a n or equal to 0.1.
The r e s u l t s of the R C I P
m e t h o d for v a r i o u s n u m b e r s of trial solutions are s h o w n in Table I.
The SOR method,
obtain
the
with over and then under relaxation factors,
convergence
criterion
did
not
in 10,000 iter a t i o n s w i t h a user
Table I. Results of I-D Ordinary Differential Equation Problem I , with Derivative
Boundary Conditions, for Various Numbers of Trial Solutions.
NUMBER
OF TRIAL
SOLUTIONS
2,3,4
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
DID NOT CONVERGE AT 100 ITERATIONS.
11
0.3208
12.97
6
11
1.1182
15.59
7
8
1.1586
13.40
8
6
0.7707
11.63
9
5
0.2624
11.13
The SOR Method Did Not Converge at 10,000 Iterations.
(14)
5
(15)
execution time of 660 seconds.
to S O R in that R C I P o b t a i n s
Therefore,
the RCIP method is superior
a s o l u t i o n and SO R does not.
W h e n the
number of trial solutions was less than 5 the convergence criterion was
not o b t a i n e d w i t h the R C I P method,
it did not conv e r g e at 100 it e r a ­
tions w i t h a u s e r t i m e of 240 seconds.
This is b e c a u s e the s m a l l e r
number of trial solutions does not contain enough information to obtain
rapid convergence.
Problem
2
is v e r y
similar
to p r o b l e m
I.
s o l u t i o n of 51 e q u a t i o n s are s h o w n in Table 2.
trial
solutions
is
less
than
8 the
convergence
The
r esults
of the
W h e n the n u m b e r of
c r i t e r i o n w as
not
obtained w i t h the RCIP method , this was because of the smaller number
of trial
s o l u t i o n s not
containing
enough information to obtain rapid
convergence. The trial solutions 2—7 did not converge at 100 iterations
with
a time
of a p p r o x i m a t e l y 240 seconds.
The S O R m e t h o d did not
o b t a i n the c o n v e r g e n c e c r i t e r i o n at 10,000 iterations w i t h a time of
660 seconds.
Several
under relaxation,
again,
relaxation factors
were utilized,
with no convergence being obtained.
produces a solution and SOR does not.
over and then
The RCIP method
The limit of convergence
for this problem is that the variance be less than or equal to 0.001.
Problem 3 was
solved and the results are shown in Table 3 for the
R C I P m e t h o d and in T a b l e 4 for the S O R method.
execution
method,
I).
time
versus
the n u m b e r
of trial
A g r a p h of the u s e r
so l u t i o n s
for
the R C I P
and over relaxation factor for the SOR method is included (Fig.
Clearly,
as
the
graph
shows,
the
conver g e n c e
to
a
solution
r e q u i r e s less t i m e for the R C l P m e t h o d than for the S O R method.
The
Table 2. Results of I-D Ordinary Differential Equation Problem 2, with Composite Boundary
Conditions, for Various Numbers of Trial Solutions.
NUMBER
OF TRIAL
SOLUTIONS
2-7
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
DID NOT CONVERGE AT 100 ITERATIONS.
22
.2306
41.29
9
10
.1739
21.69
10
4
.2227
9.93
11
4
.1333
10.95
The SOR Method did not Converge at 10,000 Iterations.
(16)
8
Table 3. Results of I-D Ordinary Differential Equation Problem 3, with Mixed
Boundary Conditions, for Various Numbers of Trial Solutions.
NUMBER
OF TRIAL
SOLUTIONS
2
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
DID NOT CONVERGE AT 100 ITERATIONS.
20
.0761
17.77
6
8
.0686
10.99
8
6
.0088
11.19
10
5
.0181
11.84
12
4
.0324
11.54
(17)
4
Table 4. Results of I-D Ordinary Differential Equation Problem 3, with Mixed
Boundary Conditions, for Various SOR Factors.
SOR
FACTORS
I.0-1.3
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
DID NOT CONVERGE AT 1000 ITERATIONS.
837
.1216
52.21
1.4
347
.1594
21.92
1.5
648
.0629
40.50
1.6
680
.0634
42.52
(18)
1.35
(19)
SOR FACTOR
DSER EXECUTION TIME (SEC)
1.45
4
6
8
□
RCIP METHOD
A
SOR METHOD
10
12
NUMBER OF TRIAL SOLUTIONS
Figure I. Trial Solutions and Different Relaxation Factors Versus User
Execution Time for I-D Ordinary Differential Equation Problem
3, with Mixed Boundary Conditions
(20)
g r a p h of the trial s o l u t i o n s v e r s u s the e x e c u t i o n t i m e for the R C I P
method is flat indicating about the same execution time for any number
of trial solutions,
time
where the graph of the SOR factor versus execution
for the S O R m e t h o d has a dip i n d i c a t i n g the i m p o r t a n c e of the
optimum SOR factor.
a variance
The solution to problem 3 is for 50 equations and
less than or equal to 0.001.
Problem 4 was solved and the results are found in Table 5 for the
RCIP
method
execution
method,
and T a b l e
time
and
(Fig. 2).
versus
6 for
the S O R method.
the n u m b e r
of
trial
A graph
so l u t i o n s
of the user
for
the R C I P
the over relaxation factors for the SOR method is included
In F i g u r e 2, the S O R and the R C I P m e t h o d b o t h have a p p r o x i ­
mately the same results.
The RCIP method was faster, by approximately
10 seconds, in the e x e c u t i o n t i m e t h a n the S OR m ethod, until the S O R
method
was
approaches
its optimum
faster. F o r this p r o b l e m
methods
relaxation factor where
the t w o m e t h o d s
are
the SOR method
c omparable.
The
solved 49 equations for a variance less than or equal to 0.001.
PROBLEM FIVE
Laplace's equation was solved in a rectanglar region of length one
in the y d i r e c t i o n
and t w o
in the
x
direction.
Each
side
of
the
rectangle is maintained at a different temperature.
The problem is formally stated as follows: Solve
d2T
d2T
--- -f — — = O
9x2
(48)
3y2
for the region
0
<= x <= 2 , 0 <= y <= I
(49)
Table 5. Results of I-D Ordinary Differential Equations Problem 4, with Dirichlet
Boundary Conditions, for Various Numbers of Trial Solutions.
NUMBER
OF TRIAL
SOLUTIONS
2-3
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
DID NOT CONVERGE AT 100 ITERATIONS.
51
7.5369
42.83
6
20
7.5286
25.89
8
10
7.5297
17.77
10
6
7.5252
13.58
11
5
7.5354
12.58
(21)
4
Table 6. Results of I-D Ordinary Differential Equation Problem 4, with Dirichlet
Boundary Conditions, for Various SOR Factors.
SOR
FACTORS
I.0-1.3
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
DID NOT CONVERGE AT 1000 ITERATIONS •
925
7.5231
55.71
1.5
709
7.5234
42.78
1.6
517
7.5237
31.25
1.7
342
7.5248
20.84
1.8
181
7.5463
11.22
1.9
148
7.5367
9.23
(22)
1.4
(23)
SOR FACTOR
USER EXECUTION TIME (SEC)
1.4
30
1.5
1.6
1.7
1.8
1.9
A
SOR FACTOR
□
RCIP METHOD
-
NUMBER OF TRIAL SOLUTIONS
Figure 2. Trial Solutions and Different Relaxation Factors Versus User
Execution Time for I-D Ordinary Differential Equation
Problem 4, with Dirichlet Boundary Conditions
(24)
with the boundary conditions
T(x,0) = 200
(50a)
T(x,2) = 400
(50b)
T(0,y) = 300
(51a)
T(l,y) = 100
(51b)
A standard finite difference technique was used to approximate the
equation in (48).
A nodal network of 7 nodes in the y direction and 15»
nodes in the x direction,
was selected. This results in Ax and Ay being
0.125.
All solution attempts began with an initial estimate of a uniform
temperature
distribution equal
the analytical solution [7].
to unity,
the
results
are
compared
to
The convergence limit for the solution of
the 105 e q u a t i o n s w a s set at the v a r i a n c e b e i n g less t h a n or equal to
0.000001.
D a t a f r o m the s o l u tions us i n g d i f f e r e n t n u m b e r s of trial
s o l u t i o n s is p r e s e n t e d in T a b l e 7 for the R C I P m ethod,
the
solutions using different relaxation factors
and data f r o m
is presented in Table
8 for the SOR method.
The results for the RCIP method are compared to the results of the
S O R meth o d . A g r a p h of e x e c u t i o n t i m e v e r s u s trial s o l u t i o n s for the
RCIP method,
and SOR factor for the SOR method will indicate the effec­
tiveness of the RCIP method in speed and accuracy.(Fig. 3)
F i g u r e 3 i n d i c a t e s that the S O R m e t h o d has a s t e a d y decrease in
execution time for each successive SOR factor until
of
1.53
is
reached.
After
the
optimum
is
the optimum value
reached
an
increase
is
Table 7. 2-D Laplace's Equation Results for Various Numbers of Trial Solutions
NUMBER
OF TRIAL
SOLUTIONS
NUMBER
OF
ITERATIONS
2
23
7.0529
1.3430
3
9
. 7:0528
.9422
4
6
7.0527
.8967
5
4
7.0528
.7912
6
3
7.0527
.7387
7
3
7.0527
.8490
AVERAGE %
ERROR
USER
EXECUTION
TIME (MIN)
(25)
Table 8. 2-D Laplace's Equation Results for Various SOR Factors
SOR
FACTOR
NUMBER
OF
ITERATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (MIN)
7.0529
1.6250
1.2
84
7.0529
1.0895
1.4
51
7.0529
.6958
1.5
35
7.0528
.5023
1.6
35
7.0527
.5010
1.7
48
7.0527
.6588
1.8
75
7.0527
.9808
(26)
129
1.0
(Gauss-Seidel)
(27)
SOR FACTOR
RCIP METHOD
A
SOR METHOD
USER EXECUTION TIME (MIN)
□
1
2
3
4
5
6
7
8
NUMBER OF TRIAL SOLUTIONS
Figure 3. Trial Solutions and Different Relaxation Factors Versus User
Execution Time for 2-D Laplace's Equation
(28)
observed.
In
the
execution
time
RCIP
from
2
method
trial
there
is
s o l u tions
a
large
decrease
to 3 trial
in
the
solutions,
but
thereafter until 7 trial solutions the execution time decreases by only
9%.
The decrease in execution time from 3 to 4 trial solutions is only
0.046 minutes,
where
the average decrease
is 0.068 minutes per trial solution.
from 3 to 6 trial
solutions
This small decrease is due to the
next to last iterations variance being just above the convergence limit
this
causes
the
method
decrease in the variance,
The
SOR methods
to p e r f o r m
one
more
iteration
with
0.2X10
below the convergence limit.
s o l u t i o n had
a s m a l l e r e x e c u t i o n time
a 98%
at its
o p t i m u m v a l u e of 1.53 t h a n any of the s o l u tions of the R C I P method.
The
RCIP
method
solutions,
This
is
flat,
in its solutions
shows
that
exception of 2,
any cho i c e
o nly
9%
difference
for different numbers
of trial
solutions,
from
of
3
to
trial
with
6
trial
solutions.
the possible
is going to take approximately the same amount of time
to obtain a solution. Where the SOR method has a large difference , 69%
from Gauss-Seidel
to the optimum SOR factor,
in its execution time.
A
accurate solution with acceptable execution time is only obtained close
to the optimum value.
PROBLEM SIX
The elliptical partial differential equation to be solved is
d2T
where T, x, y,
S2T
9 2T
and z are non-dimensional variables for temperature and
three spatial coordinates.
The g e o m e t r y is a cube of face l e ngth 0.4
(29)
l o c a t e d c o n c e n t r i c a l l y in a large cube of face l e n g t h one.
cube
The inner
is maintained at a temperature of unity while the surface of the
ou t e r cube is set at zero t e m p e rature.
problem
(Fig.
By using the s y m m e t r y of the
o n l y a s i x t e e n t h of the c o m p o s i t e cube n e e d be considered.
4)
The problem is formally stated as: Solve
9 2T
d 2T
B2T
dx2
By2
8z2
0
(53)
for the region
0 <= x <= z
,
0 <= y <= I
,
0 <= z <= I
(54)
with the boundary conditions
T(x,y,l)
= 0
(55a)
T(x,l,z)
= 0
(55b)
T(x,y,0.4) = I
(55c)
T(x,0.4,z) = I
(55d)
BT
— (z,y,z) = 0
Bx
,
II
O
N
N
I
BT
dz
BT
— (x,0,z) = 0
Sy
(55e)
BT
— (0,y,z) = 0
Bx
(55f)
A standard finite difference technique was used to approximate the
second
order
derivatives
approximation was
(55f).
in e q u a t i o n
u s e d for
the
(53) .
derivatives
A
central
difference
in e q u a t i o n s
A nodal network of 10 nodes per side was
(55e) and
selected resulting in
10 nodal spaces, therefore, the number of equations is 475 with Ax = Ay
Az = 0.1.
(30)
F i g u r e 4.
C u b e - C u b e G e o m e t r y for 3 -D S t e a d y - S t a t e Heat
Problem
Conduction
(31)
To d e t e r m i n e
fluxes were
cube
a measure
of the a c c u r a c y of the solutions, heat
calculated for the inner cube
surfaces.
and for the outer
Tb n o n - d i m e n s i o n a l ize these quantities,
divided by the heat flux of concentric
conditions
surfaces,
(T^ inner cube,
spheres with the
T q outer cube),
they w e r e
same boundary
and radii equal to the face
l e n g t h s of the inner (r%) and out e r (r^) cubes.
The v a l u e T 2
is the
b o u n d a r y c o n d i t i o n acti n g over the area A. The va l u e T^ is the inner
parallel nodal point associated with T 2 , with a distance of Ax between
them.
This non-dimensional heat flux (q') is as follows
-A(T2- T 1 )(ro-r.)
(56)
q' =
Ax 4jr T iXo (Ti-To )
Ti = I , T0
Ax = .1
0 , Ti =
(57a)
.4
(57b)
, A = .1 X .1
- O 1IS(T2-T1 )
(58)
4 TT
D i f f e r e n t n u m b e r s of trial solutions, for the R C I P method, w e r e
u s e d to o b t a i n s o l u t i o n s to this p r o b l e m . A n a c c e p t a b l e s o l u t i o n w a s
reached,
trial
with
a variance
solution set.
The
less
than or equal to 0.0000001,
solutions using
two
or three
for each
trial
solutions
w e r e s l o w e r to c o n v e r g e t h a n the so l u t i o n s using 4, 5, 6, or 7 trial
solutions,
which all
average rate increase
trial solution.
take
from
approximately the same
amount
of time.
5 to 7 trial solutions is 0.045 seconds per
The increase of 1.51 seconds in execution time from 5
to 6 trial solutions is higher than the average rate increase.
due
to the next
The
to
last
i t e rations
variance
being
just
This is
above
the
(32)
c o n v e r g e n c e limit.
iteration.
0.1X10
This
T h i s causes the m e t h o d to cont i n u e w i t h an o t h e r
continuation
to
the last
iteration gives
a variance,
100% lower than the convergence limit.
The r e s u l t s f r o m the R C I P m e t h o d (Table 9) w e r e c o m p a r e d to the
results obtained by solving the same problem using the SOR method.
same convergence limit was used as for the RCIP method,
than
or
equal
to
0.0000001,
to
obtain
comparable
variance less
results.
relaxation factors ranging from 1.0 (which makes the routine
to G a u s s - S e i d e l )
to a v a l u e
obtain solutions.
of 1.8 w e r e
The optimum value
used
in the
The
Over
identical
SOR method
to
of the over relaxation factor was
f o u n d to be 1.6.
A uniform temperature distribution of 0.5 was used as the initial
e s t i m a t e in e a c h meth o d .
s h o w n in T a b l e 10.
The re s u l t s f r o m the v a r i o u s S O R runs are
The r e s ults are also s h o w n on a g r a p h (Fig. 5) of
execution time versus the number of trial solutions in the RCIP method,
and the relaxation factor in the SOR method.
of various SOR factors and execution time
rate
of 1.89 seconds per 0.1 increase
decreases
seconds
Thereafter.,
per
trial
solution
the curve is flat
at an approximate
in the SO R factor.
graph of the RCIP methods results decreases
2.6
The graph of the results
from
Whi l e the
at an approximate
2
to
4
trial
rate of
solutions.
with a 2% increase for trial
solutions
4, 5, 6, and 7.
The t w o curves cross w i t h the S O R h a v i n g the s m a l l e s t e x e c u t i o n
time at the optimum.
If the optimum SOR factor is known then the SOR
method has a significant advantage.
The disadvantage of the SOR method
Table 9. Results for the 3-D Steady-State Heat Conduction Problem, Cube-Cube
Geometry, for Various Numbers of Trial Solutions.
NUMBER
OF TRIAL
SOLUTIONS
NUMBER
OF
ITERATIONS
HEAT FLUX
OUTER
INNER
CUBE
CUBE
% DIFFERENCE
IN HEAT
BALANCE
USER
EXECUTION
TIME (SEC)
18
.09527
.09530
.03845
12.29
3
7
.09527
.09529
.02601
8.55
4
4
.09527
.09527
.00304
7.09
5
3
.09527
.09529
.01795
7.15
6
3
.09527
.09527
.00110
8.66
7
2
.09527
.09528
.00943
7.24
(33)
2
Table 10. Results of the 3-D Steady-State Heat Conduction Problem, Cube-Cube
Geometry, for Various SOR Factors.
SOR
FACTOR
NUMBER
OF
ITERATIONS
HEAT FLUX
OUTER
INNER
CUBE
CUBE
% DIFFERENCE
IN HEAT
BALANCE
USER
EXECUTION
TIME (SEC)
.09526
.09531
.05595
16.55
1.2
69
.09526
.09531
.05081
11.44
1.4
44
.09526
.09530
.03516
7.86
1.5
33
.09527
.09529
.02392
6.18
1.6
(optimum)
26
.09527
.09527
.00129
5.23
1.7
33
.09527
.09527
.00141
6.19
I i8
51
.09527
.09527
.00073
8.82
(34)
104
1.0
(Gauss-Seidel)
(35)
SOR FACTOR
RCIP METHOD
A
SOR METHOD
USER EXECUTION TIME (SEC)
□
1
2
3
4
5
6
7
8
NUMBER OF TRIAL SOLUTIONS
Figure 5.
Trial Solutions and Different Relaxation Factors Versus User
Execution time for 3-d Steady-State Heat Conduction Problem
Cube-Cube Geometry
(36)
was that a trial and error process must be used to obtain the optimum
value of the relaxation factor.
The RCIP method will be more efficient
for the trial solutions 4, 5, 6, and 7, because of the small variation
in the e x e c u t i o n times.
as
efficient
T h e choice of 2 or 3 trial so l u t i o n s w a s not
as 4 to 7 trial
solutions,
but
2 or
3 still
p r oduce
excellent results.
PROBLEM SEVEN
The parabolic partial differential equation to be solved is
9T
S2T
92T
92T
9r
9x2
9y2
9z2
for the cube r e g i o n 0 <= x <= I, 0 <= y <= I, and 0 <= z <= I, wh e r e T,
t
,
x,
y,
and
z
are
t e m p e r a t u r e , time,
all
non-dimensional
variables
and three spatial coordinates,
representing
respectively. The
boundary conditions for (59) are:
T = 0 at x=l or y=l or z=l
(60a)
9T
— = 0 at x=0
9x
(60b)
9T
— = 0 at y=0
9y
(60c)
9T
— = 0 at z=0
9z
(60d)
and the initial condition is
T = I at T = 0
.
(61)
The a l g e b r a i c a p p r o x i m a t i o n of the p a r a b o l i c e q u a t i o n (59) w a s
a c c o m p l i s h e d by u s i n g the C r a n k N i c o l s o n m e t h o d
[5],
and a central
(37)
difference was used to approximate
60d).
the first order derivative
The cube geometry is approximated by a nodal network of 10X10X10
nodes which gives 1000 equations and 1000 unknowns.
spaces
0.1.
in (60b-
on each face
This results in 10
of the c u b e , w i t h a m e s h size of A x = Ay = Az =
Results were determined for three elapsed times,
t =0.1,
.0.5,
and
1.0. The effe c t of u s i n g d i f f e r e n t v a l u e s of the t i m e i n c r e m e n t w a s
also studied.
The ADEP method was
compared to the RCIP method for this problem.
A unif o r m temperature distribution,
equal to unity, w a s used as the
initial e s t i m a t e to start the s o l u t i o n pr o c e s s for b o t h the R C I P and
A D E P m e t h o d s . T a b l e s 11 and 12, and F i g u r e s 6, 7, and 8 illustrate the
relation between the execution time and accuracy of each method.
A value of 4 trial solutions was used to obtain the results
in T a b l e 11 for the R C I P method.
chosen from
the results
w h i c h 4 trial
of
shown
T he n u m b e r of trial solutions w a s
the previous
3-D steady-state
problem,
in
s o l u t i o n s p r o d u c e d the s m a l l e s t e x e c u t i o n time (7.09
seconds).
F o r the e l a p s e d t i m e of T =0.1 T a b l e 12 and Fig. 6 s h o w s that A D E P
p r o d u c e s an a v e r a g e p e r c e n t e r r o r of
steps
except
for At=O.01,
m o r e than .26% for all the time
which proved too large to maintain accuracy.
The average percent error increases from .26% to . 2 % as the time
gets
smaller.
This
increasing
error
step
is due to the large n u m b e r of
iterations and the error in the algebraic approximation.
Table 11. Results of 3-D Transient Heat Conduction Problem, Cube Geometry,
Using R C I P .
NUMBER OF
TIME STEPS
ELAPSED
TIME, -c
0.02
5
0.1
6.2355 .
.1739E-07
.4635
0.0125
8
0.1
.6703
.197 8E-10
.6825
0.01
10
0.1
.4398
.87 91E-06
.7838
0.005
20
0.1
.2852
.6876E-09
.8332
0.004
25
0.1
.2811
.4648E-10
1.0180
0.0025
40
0.1
.2808
.8269E-13
1.5627
0.002
50
0.1
.2814
.3033E-14
1.9273
0.025
20
0.5
2.4292
,9586E-10
1.6240
0.02
25
0.5
.5827
.6116E-06
1.8158
.40
0.5
.1233
.1627E-07
2.1403
50
0.5
.0303
.2478E-08
2.2113
0.005
100
0.5
.1010
.2145E-11
3.7417
0.004
125
0.5
.1164
.1476E-12
4.6422
0.002
250
0.5
.1368
.9767E-17
9.2708
TIME STEP
At
0.01
VARIANCE
USER EXECUTION
TIME (MIN)
(38)
0.0125
AVERAGE %
ERROR
Table 11
(continued)
0.05
20
1.0
1540.0572
.1800E-08
1.9290
0.04
25
1.0
237.1156
.1974E-06
2.0658
0.025
40
1.0
1.7589
.1723E-08
2.4108
0.02
50
1.0
.4782
.3728E-09
2.6940
0.0125
80
1.0
.3681
.1043E-10
3.5928
0.01
100
1.0
.5620
.1529E-11
4.0158
0.005
200
1.0
.8218
.1343E-14
7.3598
VO
Table 12. Results of 3-D Transient Heat Conduction Problem, Cube Geometry,
Using A D E P .
NUMBER OF
TIME STEPS
ELAPSED
TIME, T
10
0.1
24.2707
.1210
0.001
100
0.1
.2672
.3162
0.0005
200
0.1
.2605
.5322
0.0002
500
0.1
.2784
1.1833
0.0001
1000
0.1
.2818
2.2705
500
0.5
1.2731
1.1302
0.0005
1000
0.5
.4254
2.2125
0.00025
2000
0.5
.2140
4.3562
0.0002
2500
0.5
.1887
5.4307
0.000125
4000
0.5
.1612
8.6435
0.0001
5000
0.5
.1549
10.7778
500
1.0
11.1638
1.1108
TIME STEP
At
0.01
0.002
USER EXECUTION
TIME (MIN)
(40)
0.001
AVERAGE %
ERROR
Table 12.
(Continued)
0.001
1000
1.0
3.4006
2.1768
0.0005
2000
1.0
1.5261
4.3140
0.00025
4000
1.0
1.0616
8.5410
H
A
ADEP METHOD
□
RCIP METHOD
(42)
AVERAGE % ERROR
I
I
2
USER EXECUTION TIME(MIN)
Figure 6. User Execution Time Versus Average Percent Error. RCIP Versus ADEP for the
3-D T r a n s i e n t Heat T r a n s f e r P r o b l e m w i t h an El a p s e d T i m e t =0.I.
A
ADEP METHOD
(43)
AVERAGE % ERROR
RCIP METHOD
USER EXECUTION TIME(MIN)
Figure 7. User Execution Time Versus Average Percent Error. RCIP Versus ADEP for the
3-D T r a n s i e n t Heat T r a n s f e r P r o b l e m w i t h an Elapsed T i m e x=0.5.
A
ADEP METHOD
(44)
AVERAGE % ERROR
RCIP METHOD
USER EXECUTION TIME(MIN)
Figure 8. User Execution Time Versus Average Percent Error. RCIP Versus ADEP for the
3-D Transient Heat Transfer Problem with an Elapsed Time t=1.0.
(45)
The
graph
method (Fig.
of the
execution time
6) indicates
v e rsus
that when the
accuracy
smallest percent
for the R C I P
error (0.28%)
w a s r e a c h e d an o p t i m u m va l u e of the t i m e step h ad also b e e n reached.
Therefore,
further refinement of the time
step would
amount of time to produce the same accuracy.
to r e a c h the a c c u r a c y of the A D E P me t h o d ,
execution time for a specified
require a larger
The RCIP method was able
but A D E P w a s
s u p e r i o r in
accuracy over the RCIP method.
due to the s m a l l e l a p s e d t i m e T=0.1.
B e c a u s e of this sm a l l time the
solution process resembles that of the steady-state problem.
noting that as the elapsed time is increased larger time
used.
This is
However,
steps can.be
The ability of RCIP to use larger time steps, as compared to the
smaller
time
steps
that A D E P requires,
e n ables R C I P to o b t a i n the
required accuracy in less time.
For
the
indicate
that
elapsed
time
of
T=0.5,
the RCIP method produces
time than the ADEP method.
Tables
more
11,
12
and
Figure
accurate, results
7,
in less
The RCIP method obtained an average percent
error equal to 0.0303% in 2.1403 minutes of execution time, with a time
step of 0.0125 and 40 iterations.
results
time
T he A D E P m e t h o d r e a c h e d its best
(0.1612%) at At=O.000125 and 4000 iterations,
of 8.6435 m i n u t e s .
with an execution
The graph. Fig. 7, indicates that w h e n the
s m a l l e s t p e r c e n t e r r o r 0^0303% occurs the R C I P m e t h o d has reached an
optimum
time
step.
The further refinement
of the time
poorer accuracy (0.1164%) and larger execution times
F o r the e l a p s e d t i m e of
t
step produced
(4.6422 minutes).
=1.0. T a b l e 11, 12, and Fig. 8, the graph
of the results of the RCIP method is entirely below the graph of ADEP's
(46)
results.
At each time step the RCIP method is much faster.
the larger time
steps results
The use of
in the RCIP method having the advantages
of less i t e r a t i o n s w i t h less error, and a faster e x e c u t i o n time.
larger
elapsed
times
are
used ADEP will perform
numerous
If
iterations,
therefore the ADEP method will be slower than the RCIP method and less
accurate.
All computer runs were accomplished on the Honeywell CP-6 computer
in double precision.
in Appendix C.
A flow chart of the computer program is included
(47)
CHAPTER V
D E V E L O P M E N T OF R C D P W I T H
APPLICATION AND RESULTS
M a n y dir e c t m e t h o d s have b e e n d e v e l o p e d to solve sets of l i near
equations.
These methods have been restricted to solving small sets of
e q u a t i o n s u s u a l l y less t h a n 40 [5]. A n e w m e t h o d
is p r e s e n t e d here
w h i c h cou l d pr o v e to be m u c h f a ster for the s o l u t i o n of large sets of
tridiagonal equations.
T h i s n e w meth o d ,
(RCDP), uses
two
c a l l e d the R e d u c e d Coordinate Direct Procedure
trial
solutions
coefficients in the RCIP method.
of one trial
to
for
t he
weighting
The trial solutions generated consist
s o l u t i o n due to the
coefficient
s o l u t i o n due to the b o u n d a r y conditions.
successfully solve the problem
solve
matrix
and one
trial
These t w o trial solutions
in one iteration.
An attempt
was made
to e x p a n d this m e t h o d to a general routine to solve n o n - t r i d i a g o n a l
systems.
A problem was
d i s c o v e r e d w i t h this expansion.
tridiagonal systems more
solve the problem.
the
subsequent
than t w o trial solutions are n e c e s s a r y to
In the development of the extra trial solutions and
reduced matrix
the problem was found.
the
to solve
for the weighting coefficients
When more than two trial solutions is required
reduced matrix becomes
closer and closer to being singular as the
n u m b e r of trial s o l u t i o n s is increased.
give
erroneous
For n o n ­
results.
T h e r e f o r e the m e t h o d w o u l d
(48)
The a l g o r i t h m s to f o r m the trial s o l u t i o n s for R C D P are as
follows.
Let
A*X = F
(62)
be a s y s t e m of n e q u a t i o n s
and n u n k n owns.
To c a l c u l a t e
the trial
solution due to the boundary conditions set
G(i,m) = 0
i=l,...,m-1
(I)
then
G(i+m-l,m) = F(i)/A(i,i+m— I)
i+m-2
Z
A(i,j)*G(j,m)/A(i,i+m-1)
(2)
J=I
i=l,...,n-m+1
To calculate the coefficient matrix trial solution set
G(i,i) = I
G ( j ,i) = 0
j = l , ...,m-1 , j /= i ,
i=l,...,m-1
(3)
i=l, —
(4)
,m-1
then
G(i+m-1,k)= -
i+m—2
^
A(i,j)*G(j,k)/A(i,i+m-1)
(5)
J=I
i=l,...,n-m+1
The
algorithms
are
written
,
k=l,...,m-l
in
general
t r i d i a g o n a l s y s t e m m is equal to two.
existing
technique,
(6)
form,
although
for
a
A c o m p a r i s i o n of R C D P w i t h an
Gaussian-Elimination
the v a l i d i t y of the method.
.
[5],
were
made
An i l l u s t r a t i v e p r o b l e m
to
determine
is solved and
included in Appendix D.
The
four one-dimensional problems
and 12 are to be s o l v e d b y each method.
stated in
Chapter IV pages 11
A n u m b e r of d i f f e r e n t m e s h
sizes were chosen to obtain results for small to large equation sets.
(49)
The
results
of the ECDP and the Gaussian-Elim!nation methods for
various numbers of equations are
shown in the following
eight tables.
C o m p a r i s o n s are m a d e for each p r o b l e m on a graph of e x e c u t i o n time
versus average percent error.
The r e s u l t s of the R C D P m e t h o d and the results of the G a u s s i a n Elimination method show that at larger mesh sizes the two methods are
comparable.
the
RCDP
As the m e s h size gets smaller, or larger e q u a t i o n sets,
method
produces
a c cu r a t e
r esults
in
less
time
than
the
Gaussian-Elimination method. - This is due to the number of calculations
performed by each method.
The Gaussian-Elimination method requires n
calculations
a
to
produce
solution,
and
the
RCDP
requires
2 8
n '
calculation to produce a solution.
In p r o b l e m
4
the
graph Figure
p r o d u c e s m u c h m o r e a c c u r a t e r esults
method in less time.
12
sh o w s
that
the R C D P m e t h o d
t h a n the G a u s s i a n - E l i m i n a t i o n
(50)
Table 13. Results of RCDP for the Ordinary Differential Equation,
with Derivative Boundary Conditions, for Various Numbers
of Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
11
.3099
.18
21
.0772
.30
51
.0123
.80
101
.0031
2.20
201
.0008
7.53
USER
EXECUTION
TIME (SEC)
Table 14. Results of Gaussian-Elimination for the Ordinary Differential
Equ a t i o n , w i t h D e r i v a t i v e B o u n d a r y Conditions, for Various
Numbers of Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
11
.3099
.16
21
.0772
.30
51
.0123
1.53
101
.0031
8.94
201
.0008
66.02
A
Gaussian-Elimination
D
RCDP Method
(51)
AVERAGE % ERROR
.07
USER EXECUTION TIME (SEC)
Figure 9.
User Execution Time Versus Average Percent Error for RCDP Versus GaussianElimination of I-D Ordinary Differential Equation Problem I, Derivative
Boundary Conditions.
(52)
Table 15. Results of RCDP for the Ordinary Differential Equation,
with Composite Boundary Conditions, for Various Numbers
of Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
11
.1246
.17
21
.0318
.29
51
.0052
.80
101
.0013
2.19
201
.0003
7.40
Table 16. Results of Gaussian-Elimination for the Ordinary Differential
Equation, with Composite Boundary Conditions, for Various
Numbers of Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
11
.1246
.16
21
.0318
.30
51
.0052
1.53
101
.0013
8.86
201
.0003
65.39
Gaussian-Elimination
□
RCDP Method
(53)
AVERAGE % ERROR
A
USER EXECUTION TIME(SEC)
Figure 10. User Execution Time Versus Average Percent Error of RCDP Versus GaussianElimination for I-D Ordinary Differential Equation Problem 2, Composite
Boundary Conditions.
(54)
Table 17. Results of RCDP for the Ordinary Differential Equation,
with Mixed Boundary Conditions, for Various Numbers of
Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
10
.7485
.17
20
.1857
.28
50
.0296
.77
100
.0074
2.18
200
.0018
7.59
Table 18. Results of Gaussian-Elimination for the Ordinary Differential
Equation, with Mixed Boundary Conditions, for Various Numbers
of Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
10
.7485
.15
20
.1857
.28
50
.0296
1.46
100
.0074
8.77
200
.0018
67.00
Gaussian-Elimination
D
RCDP Method
(55)
AVERAGE % ERROR
A
USER EXECUTION TIME (SEC)
Figure 11. User Execution Time Versus Average Percent Error for RCDP Versus GaussianElimination of I-D Ordinary Differential Equation Problem I, Mixed
Boundary Conditions.
(56)
Table 19. Results of RCDP for the Ordinary Differential Equation,
with Dirichlet Boundary Conditions, for Various Numbers
of Equations.
NUMBER
OF
EQUATIONS
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
9
2.6465
.15
19
5.9274
.26
49
7.2303
.74
99
4.4752
2.13
199
.3946
7.28
Table 20. Results of Gaussian-Elimination for the Ordinary Differential
Equation, with Dirichlet
Boundary
Conditions, for Various
Numbers of Equations.
NUMBER
OF
EQUATION
AVERAGE %
ERROR
USER
EXECUTION
TIME (SEC)
9
19.0485
.14
19
94.2719
.26
49
123.7314
1.37
99
62.9538
8.38
199
27.2497
63.46
Gaussian-Elimination
□
RCDP Method
(57)
AVERAGE % ERROR
A
USER EXECUTION TIME(SEC)
Figure 12. User Execution Time Versus Average Percent Error of RCDP Versus GaussianElimination for I-D Ordinary Differential Equation Problem 4, Dirichlet
Boundary Conditions
(58)
CHAPTER VI
CONCLUSIONS
The r e s u l t s s h o w that the R C I P m e t h o d is an e x c e l l e n t solu t i o n
r o u t i n e for line a r e q u a t i o n sets.
T he R C I P m e t h o d p r o v e d to be m o s t
useful on problems with complicated boundary conditions.
The first two
I-D problems having mixed and composite boundary conditions were solved
by
the
RCIP
solution.
method.
The
Conclusion,
try
SOR
method
however,
the RCIP method
did
when
not
pr o d u c e
convergence
a
problems
are encou n t e r e d . The R C I P m e t h o d and S O R m e t h o d w i t h o p t i m u m o v e r ­
relaxation factor solution produces
3-D steady-state problems.
comparable results for the 2-D and
If the optimum SOR factor is not known then
a trial and error search must be accomplished to determine the optimum
r e l a x a t i o n factor.
The R C I P m e t h o d does not require a search for a
optimum
trial
number
solutions,
of
except one,
solutions,
r e s ults
choos i n g
any n u m b e r
in fast accurate
solutions.
of
trial
In the
s o l u t i o n to the t r a n s i e n t p r o b l e m R C I P has some d e f i n i t e a d vantages
over A D E P , which has the ablity to operate with larger time steps and
rapid convergence of the RCIP method.
The number of trial solutions to
p i c k if a single s o l u t i o n a t t e m p t is sought is six trial solutions.
This number' of trial solutions produced excellent results.
,i
(59)
The R G D P m e t h o d and G a ussia n - E l i m i n a t i o n are comparable methods
for the solution of small equation sets.
producing
accurate
The RCDP has an advantage of
r e s u l t s m u c h faster than the Gaussian-Elimination
routine for equation sets larger than 40.
(60)
REFERENCES CITED
(61)
1.
Blackfcetter, D.O., V/arrington, R.O., Henry, M.S., and Garner,
E.R., 'A N e w
I t e r a t i v e M e t h o d for So l v i n g S i m u l t a n e o u s L i n e a r
"Equations',Advances, in C o m p u t e r M ethods for P a r t i a l D i f f e r e n t i a l
Equations, Vol III> June 1979, pp 70-72;
2.
B l a c k k e t t e r , D.O., V/arrington, R.O., and Horning, R., ' E v a l u a t i o n
' of a N e w I t e r a t i v e P r o c e d u r e f o r C o n d u c t i o n H e a t T r a n s f e r
Problems', P r e s e n t e d at the 2 nd N a t i o n a l C o n f e r e n c e on N u m e r i c a l
Methods in Heat Tranfer, Univ. of Maryland, Sept. 1981.
3.
Golub, G., ' N umerical M e t h o d s for S olving L i n e a r L e a s t
Problems,' Numerische Mathematik 7, p. 206-216, 1965.
Squares
4.
Hesteries, M.R., and E. Stiefel, 'Methods of Conjugate Gradients for
Solving Linear Systems,' Journal of Research of the National Bureau
of Standards, Vol. 49, No. 6, Dec., 1952.
5.
H o r n be c k , R.W., N u m e r i c a l M e t h o d s , N e w York: Q u a n t u m P u b l i s h e r s
I n c ., 1975.
6.
H o r ning, R.P., 'A N e w I t e r a t i v e M e t h o d for So l v i n g S i m u l t a n e o u s
Linear Equations w i t h Direct Applications to Three-Dimensional Heat
Conduction Problems', Masters Thesis, Montana State Univ., August
1980.
•
.
V1 -
7.
K a r l e k a r , B.V., and D e s m o n d , R.M., 'E n g i n e e r i n g Heat T r a n s f e r ,New
York: West Publishing Co., 1977.
8.
Stewart, G.W., 'Conjugate Direction Methods for Solving Systems of
Linear Equations,' Numerische Mathematik 21, 285—297, 1972.
(62)
APPENDICES
-
(63)
APPENDIX A
ILLUSTRATIVE EXAMPLE OF RCIP
(64)
Consider the following example.
1 1 0
2
C 2I
0
Li.
t-M
, L x3 3
O
I
rH-i
For an initial estimate select
IT
G ( 0 ,I) =
thus n=3 and m=2.
Ol
O
O
and
G (0,2) =
I
.IJ
•
Then equation (3) becomes
Gr(Ojl) 4" (%2 G(0,2) —z (&2
X1
q
and
1 1 0
A X 0 -F
1 2
0
0
0
Oi +
1
©2 — 2
Oj + 202 _ 3
O2 - 1
<A X 0-F, AX q - F > = (a1+a2-2)2 + (a1+2a2-3)2 + (O2-I)2
w h i c h is equal to the V 0 (a^,a2 ) in e q u a t i o n (4).
derivatives according to equation (5) yields
av0
—
=
2c^ + 3o2 - 5
F o r m i n g the p artial
(65)
and
av0
—
= «2 + ^a2 “ 3 .
9a2
Thus
dV0
—
= O
and
9V0
—
= O
Sa2
yeilds the system
2a^ + Sa2 = 5
a^ + 202 = 3
w i t h the solution
a^ = I
,
a2 = I
T h a t is (a(0,l),a(0,2)) = (1,1) and thus
f
X 0 = G(0,1) + G(0,2) = / I
For the second iteration,
equation (7) becomes
G(l,I) = X r
select arbitrarily
G(l,2)
Therefore following equations
(8)-(12) we have
(66)
O1 +
X 1 = O1GCL,!) + o2G(l,2)
o2
O1 + 2 o 2
O1 + 2o 2
2o1 + 3o2 — 2
A X 1-F
So1 + So2 _ 3
O1 + 2 o 2 — I
^aI'a2 ^ ~ ^AX1-FiAX1-F>
(2a1 + 3o2 - 2)^ + (Sa1 + So2 - 3)^
+(O1 + 2a2 — I)^ ,
14a1 + 23O2 - 14
23a1 + 3 Sa2 - 23
and the equations to be solved are
14o1 + 23o2 = 14
23O1 + 3 Sq2 = 23
with solution
O1 = I , o2 = O .
Thus
,
(67)
the algorithm terminates w i t h the solution
X
I
(68)
APPENDIX B
SOLUTION OF A SIMPLE BOUNDARY VALUE PROBLEM USING RGIP
(69)
The problem is as follows
d2u
U +
0
X
on
0 < x < I ,
with
u(0) = 0
Use
a c e n tral
and u ( I) = 0 , (U exact = x - sinh(x)/sinh(I)) .
difference
approximation,
and a step
size
of
obtain the following equation set to be solved,
-201
100
100 -201
0
0
0
. . . .
0
100
. . . .
0
0
. . .
100
-201
ITERATION I
We have n = 9 and pick m = 3.
For the
initial estimate a constant value
r -.088
/ -.131
I -.139
) -.142
< -.174
) -.257
[ -.272
I -.207
\-.103
V 0 = .849
A v g . % error = 4.763
Uj = .0695
»2 = -.0663
a3 = .0483
(70)
ITERATION 2
G(l,I)
^ O 193s
.0326
.0429
.0499
.0530 >,
.0520
.0450
.0330
Ix OlSSy
/ . o o 4 i Sn
.00385
.00303
.00209
G(l,2) =, .00105 \,
\ -.000237
-.00361
-.00691
-.00604 .
V 1 = .00612
A v g . % error = 1.218
Ct1 = 1.04
a2 = -1.01
«3 = 1.28
ITERATION 3
At this iteration the
V 2 = .0000873
and
A v g . % error = .399 .
ITERATION 5
At this iteration the
V 5 = .000000000722
and the
A v g . % error = .0762
Z
^
-.000864
-.00127
-.00122
) -.000544
G d , 3) =S .00079
.00218
.00260
.00204
X. 0 0 1 0 2 /
(71)
The m e t h o d has c o n v e r g e d to the s p e c i f i e d li m i t that the v a r i a n c e be
less
than or equal
to .00000001.
The
solution to the p r o b l e m
follows:
^014 7 5 N'
.02866
.04085
.05044
X 5 = <(.05655 >
.05822
.05447
.04426
^ 0 2650y
.
is as
APPENDIX C
FLOW CHART OF RCIP COMPUTER.PROGRAM
(73)
START
DETERMINE
EQUATION SET
INPUT
INITIAL ESTIMATE
DETERMINE
TRIAL SOLUTIONS
DETERMINE
ERROR MATRIX
SOLVE FOR WEIGHTING
COEFFICIENTS
TIME STEP
DETERMINE NEW
F MATRIX
OBTAIN APPROX.
SOLUTION
DETERMINE
VARIANCE V
VARIANCE
S V <= E / /
TRANSIENT
ONLY __
SOLUTION
E = CONVERGENCE LIMIT
(74)
APPENDIX D
SOLUTION OF A SIMPLE BOUNDARY VALUE PROBLEM USING RCDP
(75)
The same p r o b l e m
RCDP m e t h o d .
d2u
— dx2
s o l v e d in A p p e n d i x B is s o lved here using the
The problem is as follows
- u + x = 0
on
0 < x < I
,
with
u(0) = 0
and
u(l) = 0 , (Ue2act = x - sinh(x)/sinh(I)).
The coefficient trial solution is
and the boundary condition trial solution is
0 .0000’
0.0010
-
-0.0040
-
0.0101
0.0202
-0.0356
-0.0573
-0.0866
1-0.1247
The weighting coefficients are
U1 = 0.0147
«2
=
1.00
,
and the variance and average percent error are
Variance = 0.22E— 32
Average % Error = 0 . 0 7 5 6
(76)
The solution is
ui
0.0148X
0.0287 V
0.0408
0.0505 /
0.0566
j
\
0.OSSlC
0.0545 \
0.0443
0.0265/
J
MONTANA STATE UN]V E B S I^
BOZEMAN
3 1762 10636202 1
N378
H666
cop.2
DATE
Hodges, T. M.
The improvement of the
reduced coordinate
iterative procedure ...
ISSUED TO
&
/(>7
p