The
so-called Canonical polynomials introduced by Lanczos(A) have hitherto been
used in application to the Tau method for the solution of ordinary differential
equation via Legendre polynomials and Chebyshev polynomials.
In
this paper, we described how canonical polynomials can easily be constructed as
basis to the solution of first order differential equations. From a
computational point of view, the canonical polynomials are attractive, easily
generated, uing a simple recursive relation and its associated conditional of
the given problem via Legendre and Chebyshev polynomials is of great
importance.
The
paper of Oritz(B) gives an account of the theory of the Tau method which it
subsequently uses in the problems considered to illustrate the effectiveness
and superiority of Legendre polynomials to Chebyshev polynomials.
THE
METHOD USED
IN
THIS SECTION, WE GENERATE CANONICAL POLYNOMIALS FOLLOWING LANCZOS(A), WE DEFINE
CANONICAL POLYNOMIALS QK(X); K = 0 WHICH ARE UNIQUELY ASSOCIATED
WITH THE OPERATOR. CONSIDER A LINEAR DIFFERENTIAL EQUATION
Y'-Y=0,
Y(0)=1 .................. 1
GENERATING
THE CANONICAL POLYNOMIALS,
L=D/DX
- 1
LXK=
KXK-1 - XK
THUS,
LXK
- KLQK-1(X) - LQK(X) - LQK(X)=0
FROM
THE LINEARITY OF L, AND THE EXISTENCE OFL- ', WE HAVE
QK(X)
+ XK = KQ K-1(X)
SINCE
DQK(X) = XK
FROM
THE BOUNDARY CONDITION
DY(X)
= 0 => XK = 0
QK(X)
= KQK-1(X)
IT
FOLLOWS THAT
QK(X)
= K!SK(X)
FOR
THE DIFFERENTIAL EQUATION CONSIDER
CHEBYSHEV
POLYNOMIALS
WE
RECALL SOME WELL-KNOWN PROPERTIES OF THE CHEBYSHEVPOLYNOMIALS:
TN*(X)
= F(-N,N,1/2,X) TN*(X) = COS N(COS-1(X), -1 = X = 1 WHERE X
= COS 0. TO EVALUATE THE FIRST FEW POLYNOMIALS, WE FOLLOW
T0(X)
= T0(COS 0 ) = 1
T1(X)
= T1(COS 0) = X
WE
NOW MAKE USE OF THE RECURSIVE RELATION
TN+1(X)
= 2XTN(X) - TN-1(X)
TO
GENERATE OTHERS FOR N=1,2,3,...
LEGENDRE
POLYNOMIALS
LEGENDRE
POLYNOMIALS PN*(X), DEFINED BY THE HYPERGEOMETRIC SERIES
PN*(X)
= F(-N, N+1, 1:X) = F(α,β,δ: X)
=>
PN*(X) = 1 + αβX
+ α(α +1) β(β
+1)X2 + +
δ δ
(δ+1)
α (α+1)( a+2)
.+( α+N)
β
(β+1)
( β+N)X
δ (δ+1)( d+2) +
..+
( δ+N)
WHERE
N=0
P0*(X)
= 1
WHEN
N=1 => P1*(X) = 1-2X
WHEN
N=2 => 1 - 6X + 12X2 ETC.
THE
TAU METHOD
ORITZ
(B) GIVES AN ACCOUNT OF THE THEORY OF TAU METHOD; SUCH IS APPLIED TO THE
FOLLOWING BASIC PROBLEM.
LY(X)=PM(X)Y(M)(X)+.......+P0(X)Y(X)
= F(X); A=X=B; YM(X) STANDS FOR THE DERIVATIVE OF ORDER M OF Y(X)
AND Y(X)=YN(X)= S A X = S A Q (X)
WHERE
QK(X) IS THE CANONICAL POLYNOMIAL. HERE, WE NEED A SMALL PERTURB
TERM WHICH LEADS TO THE CHOICE OF CHEBYSHEV POLYNOMIALS WHICH OSCILLATES WITH
EQUAL AMPLITUDE IN THE RANGE CONSIDERED.
PN(X)
= tTN*(X) WHERE TN*(X)
IS THE SHIFTED CHEBYSHEV POLYNOMIAL WHICH ARE OFTEN USED WITH THE TAU METHOD
AND
N
TN*(X)
=∑CKNXK
WHERE CKN ARE COEFFICIENTS OF XK. WE
K=0
ASSUME
HERE THAT A TRANSFORMATION HAS BEEN MADE SUCH THAT A=0 AND B=1 TO SIMPLIFY
MATTER FURTHER IN ORDER TO GET THE SHIFTED CHEBYSHEV POLYNOMIAL I.E
T0*(X)=1,
T1* (X)=X=(1-2q), T2*(X)=1- 8q
+ 8q2=1-8X+8X2
RESULT AND DISCUSSION
CONSIDER THE DIFFERENTIAL EQUATION
Y'-Y
= 0, Y(0) = 1 ...... 1
WHICH
DEFINES THE EXPONENTIAL FUNCTION.
Y(X)
= EX = 1+X+X2/2+X3/3+..2
WHICH
CONVERGES IN THE ENTIRE COMPLEX PLAIN. IF WE TRUNCATE THE TAYLOR SERIES
YN(X)
= 1+X+X2/2!+.........+XN/N!+.......3
THIS
FUNCTION SATISFIES THE DIFFERENTIAL EQUATION
Y'N-YN
= XN/N! 4
SUPPOSE
WE ARE SOLVING 1 IN THE RANGE OF (0,1).
NOW
BY CHOSING CHEBYSHEV POLYNOMIALS TN*(X) DEFINED WITH THE HELP OF THE
HYPERGEOMETRIC SERIES TN*(X) = F(-N, N, 1/2; X) AS THE ERROR TERM ON
THE RIGHT HAND SIDE OF (1) WE THEREFORE SOLVE THE DIFFERNTIAL EQUATION
Y'N-YN
= δTN*(X) ...............5
BY
INTRODUCING CANONICAL POLYNOMIAL QK.QK(X) IS DEFINED BY
Q'K
- QK(X) = XK
=>
QK(X) = -K!SK(X) ............. 6
IF
WE DENOTE ITS PARTIAL SUM OF THE FIRST K+1 TERMS OF THE TAYLOR SERIES BY SK(X) SUCH THAT
SK(X)
= 1+X+X2/2!+.......+XN/N!
...7 WRITING OUT
POLYNOMIALS T*N(X) EXPLICITLY AS
N
TN*(X)
= CN°+CN1X+CN2X2+.........+CNNXN
=∑CKNXK
.8
K=0
BY
SUPERPOSITION OF LINEAR OPERATION WE HAVE
N
YN(X)
= -τΣCKNK!SK(X)
..........9
K=0
SATISFY
THE BOUNDARY CONDITION
YN(0)
= 1, WILL YIELDS
N
-
τΣCKNK!SK(0)
= 1
K=0
1
-
τ= N
ΣCKNK!
K=0
THE
FINAL SOLUTION BECOMES
N
∑CKNK!SK(X)
YN(X)
= K=0 .....10
N
∑CKNK!
K=0
WHEN
N = 4
T4*(X)
= 1-32X+160X2-256X3+128X4
N
∑CK4K!SK(X)
Y4(X)
= K=0
N
∑ CK4K!
K=0
WHERE
4
∑CK4K!SK(X)
=
K=0
C040!S0(X)+C141!S1(X)+C242!S2(X)+C343!S3(X)+C444!S4(X)
4
∑CK4K!SK(X)
= C040!+C141!+C242!+C343!+C444!
K=0
N
SK(X)
= 1+X+X2/2!+..........+XN/N! = ∑XK/K!
K=0
=>
S0(X) = 1, S1(X) = 1+X, S2(X) = 1+X+X2/2!
S3(X)
= 1+X+X2/2!+X3/3!, S4(X) = 1+X+X2/2!+X3/3!+X4/4!
HENCE
Y4(X)
= 1325+1824X+928X2+256X3+128X4 .11
1825
THE
ABOVE SOLUTION LOOKS LIKE WEIGHTED AVERAGE OF THE PARTIAL SUMS SK(X).
THIS WEIGHTING IS VERY EFFICIENT IF X = 1 WE OBTAIN
Y4(1)
= 4961/1825 = 2.718356..12
THE
EXACT VALUE
Y4(1)
= E1 = 2.7182818284 .13
HENCE
ERROR = EXACT VALUE - APPROXIMATE VALUE.
ERROR
= -7.4*10-5
WHEREAS
THE UNWEIGHTED PARTIAL SUM S4(1) GIVES
65/24
= 2.70832
WITH
ERROR = 1.0 * 10-2
HERE,
WE SEE THE GREAT INCREASED CONVERGENCE THUS OBTAINED.
HOWEVER,
THE RANGE (0, 1) IS ACCIDENTAL NOW TESTING WITH ANALYTIC FUNCTIONS WHICH ARE
DEFINED AT ALL POINTS OF THE COMPLEX PLANE EXCEPT FOR SINGULAR POINTS.
HENCE,OUR AIM WILL BE TO OBTAIN Y(Z) WHERE Z MAY BE CHOSEN AS ANY NON-SINGULAR
COMPLEX POINT.
IN
VIEW OF THIS, WE CHOOSE OUR ERROR POLYNOMIAL IN THE FORM TN*(X/Z)
AND SOLVE THE GIVEN DIFFERENTIAL EQUATION ALONG THE COMPLEX RAY WHICH CONNECTS
THE POINT X=0 WITH THE POINT X=Z. THEN SOLVING THE DIFFERENTIAL EQUATION
DYN(X)
= jTN*(X/Z) .............. 14
BY
CONSIDERING Z MERELY AS A GIVEN CONSTANT, WE FINALLY SUBSTITUTE FOR X THE
END-POINT X=Z OF THE RANGE IN WHICH TN*(X/Z) IS USABLE.
HENCE,
N
TN*(X/Z)
= ∑CKNXK
K=0 ZK
.............. .15
WE
OBTAIN
N
∑ CKNSK(Z)K!
YN(Z)
= K=0 ZK
TN*(-1/Z)
..............................
16
THE
PREVIOUS APPROXIMATIONS HAVE NOW TURNED INTO RATIONAL APPROXIMATIONS GIVING THE
SUCCESSIVE APPROXIMATES AS THE RATIO OF TWO POLYNOMIALS OF ORDER N.
WHEN
N=4, WE HAVE
N
∑ CK4SK(Z)K!
Y4(X)
= K=0 ZK
N
.........17
∑CK4K!
K=0 ZK
=
3072+1536Z+320Z2+32Z3+Z4
3072-1536Z+320Z2-32Z3+Z4
..17A
NOW
REPLACING THE COEFFICIENT CKN OF THE CHEBYSHEV POLYNOMIAL
BY THE CORRESPONDING COEFFICIENT OF THE LEGENDRE POLYNMIAL PN* (X)
DEFINED THE HYPERGEOMETRIC SERIES PN* (X) = F(-N, N+1, 1; X)
HENCE
N
∑ PKNSK(Z)
YNP(Z)
= K=0 ZK
N ....18
∑ PKN
K=0 ZK
WHEN
N = 4
N
∑ PK4SK(Z)
Y4P(Z)
= K=0 ZK
N
.18A
∑ PK4
K=0 ZK
P4*(X)
= 1-20X+180X2-840X3+1680X4
WE
NOW HAVE Y4P(Z) = 1680+840Z+180Z2+20Z3+Z4
1680-840Z+180Z2-20Z3+Z4
PUTTING
Z = 1, WE OBTAIN
Y4P(1)
= 2721/1001 = 2.71828172...........19
WHEREAS
THE EXACT VALUE = 2.7182818284
HENCE
THE ERROR =V = 1.1*10-7
COMPARING
THE RESULT OF CHEBYSHEV WITH LEGENDRE WE DISCOVER THAT LEGENDRE SOLUTION GIVE
MUCH CLOSER E-VALUE THEN THE VALUES OBTAINED BY THE CHEBYSHEV WEIGHTING.
IF
WE PROCEED BY PUTTING Z = I, WE OBTAIN SUCCESSIVE APPROXIMATIONS OF
EI
= COS1+ISIN1 = 0.54030231+0.84147098I
IN
THE CASE N = 4 CONSIDERED
Y4P(I)
= 1501+820I
1501-820I
=
1580601+2461640I
2925401
Y4P(I)
= 0.540302338+0.841470964I
ERROR
h
= -3*10-8+2*10-8I
WHEREAS
THE WEIGHTING BY CHEBYSHEV COEFFICIENT YIELDS
Y4C(I)
= 2753+1504I
2753-1504I
=
5316993+8281024I
9841025
Y4C(I)
= 0.5402885+0.8414798I
ERROR
η= 1.4*10-5-0.9*10-5I
SEE
TABLE 1 FOR SOME NUMERICAL RESULTS FOR THE ERROR ESTIMATES BASED ON THE EXAMPLE
1, WHEN X = 1
EXAMPLE 2
Y'(1+X) = 1, Y(0) = 0.
THE EXACT VALUE (SOLUTION) => Y(X) = LOG(1+X)
=> Y(X) = X-X2/2+X3/3
FOLLOWING THE ILLUSTRATION OF EXAMPLE 1 WE HAVE
CANONICAL POLYNOMIAL BECOMES
QK(X) = (-1)K-1SK(X)
THE PERTURBED TERM BECOMES
Y'(1+X) = 1+jT*N(X)
N
YN(X) = j∑CKN(-1)KSK(X)
K=1
WHERE
N
SK(X) = ∑ (-1)K+1XK
K=0 K
N
TN*(X) = ∑CKNXK
K=0
HENCE WE HAVE THE TABLE FOR THE RESULT OF EXAMPLE
CONCLUSIONS:
The polynomials of legendre and chebyshev has been
described. The two method is shown to be accurate efficient and general in
application for sufficiently solution y(x) and for tau polynomial approximation
yn(x).
the
result obtained in the present work demonstrate the effectiveness and
superiority of legendre polynomials to chebyshev polynomials for the solution
of order linear differential equation. The variants of the error estimated
described the case of reciprocal radii in which the point x = 0 becomes a
singular point of our domain legendre polynomial fail to give better value than
the chebyshev polynomials even of the end point x = 1. By excluding, however
the point x = 0 by defining our range as (e,1) which by a simple linear
transformation can then be changed back to the standard range (0,1). The
condition that our domain shall contain no singular points is now satisfied.
in
the vicinity of singularity pn*(x) (i.e. the legendre polynomials)
gives larger errors than the tn*(x) (i.e. chebyshev) for small
values of n. As n increases, the polynomials pn*(x) compete with tn*(x)
with increasing accuracy to the tn*(x) for the purpose of end point
approximation.
Acknowledgement:
The authors wish to thank ayeni e.o and tunde okewale
for performing the numerical experiments. The subject matter of this paper
has been discussed with numerous individuals and the authors are grateful to
them
all. The following deserve particular mention: r.o ayeni (lautech), tejumola,
dosu ojengbede, omosoju, eegunyomi, and opoola t.o (of university
of ilorin)
REFERENCES