 Open Access
 Total Downloads : 666
 Authors : B.M. Yisa, R.B. Adeniyi
 Paper ID : IJERTV1IS6014
 Volume & Issue : Volume 01, Issue 06 (August 2012)
 Published (First Online): 30082012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Generalization of Canonical Polynomials for Overdetermined M – Th Order Ordinary Differential Equations(ODEs)
GENERALIZATION OF CANONICAL POLYNOMIALS FOR OVERDETERMINED m th ORDER ORDINARY DIFFERENTIAL EQUATIONS(ODEs)
B.M. YISA AND R.B. ADENIYI
DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILORIN, ILORIN, NIGERIA
KEY WORDS: Overdetermination, Canonical Polynomials, Differential Operator.
Canonical polynomials play a remarkable roles in Lanczos Recursive formulation of the tau method. Meanwhile, their construction are done for individual cases, and the problems of indeterminate ones are most of the time overwhelming, if not impossible for overdetermined cases. In this paper, we shall present a derived formula for a general class of mth order overdetermined ODEs. As their derivatives are of equal level of imporatnce, a general formula for that is also reported in this paper. The principle of mathematical induction is employed to establish the validity of the two formulae.
Ortiz [8] gave a stepbystep account of Lanczos [6] Tau method and its applications in solving both initial value problems (IVPs) and boundary value problems (BVPs). The essential of the Tau method (Lanczos [6] and Ortiz [8]) is to perturb the given differential problem in such a way that its exact solution becomes a polynomial. To achieve this, a polynomial pertur bation term is added to the right hand side of the differential equation. The derived Tau approximation is written in terms of a special polynomial ba sis, uniquely associated with the given differential operator L (see Ortiz [8]) which defines the given problem. Such basis does not depend on the degree of approximation. The order of the approximation can be increased by just adding one or more canonical polynomials to those already generated and updating the coefficients affecting them.

In this paper, we intend to obtain a general formula for the canonical polynomials and the derivatives of such polynomials for overdetermined m th order initial value problems (IVPs)
Ly(x) :=
Ly(x) :=
m
r=0
m
r=0
Nr
k=0 Nr
k=0
Prkxk
Prkxk
y(r)(x) =
y(r)(x) =
F
r=0
F
r=0
frxr (2.1a)
frxr (2.1a)
(r)
m1
Ly(xrk) := arky (xrk) = k, k = 1(1)m (2.1b)
r=0
where Nr, F are given nonnegative integers and ark, xrk, k, fr, Prk are given real numbers by seeking an approximant
n
yn(x) = arxr, n < + (2.2)
r=0
which is the exact solution of the corresponding perturbed problem
F
Lyn(x) = frxr + Hn(x) (2.3a)
r=0
Lyn(xrk) = k, k = 1(1)m (2.3b)
where
Hn(x) =
m+s1 r=0
m+sr Tnm+r+1(x) (2.4)
is the perturbation term. The parameters r, r = 1(1)m + s, are to be determined,
Tr(x) = Cos
rCos1
2x a b b a
r
(
k=0
(r)
C
k
xk (2.5)
is the Chebyshev polynomial valid in the interval [a, b] (assuming that (2.1) is defined in this interval) and
s = max {Nr r 0 r m} (2.6)

THE GENERALIZED CANONICAL POLYNOMIAL FOR OVERDETERMINED m th ORDER ODEs
The canonical polynomials for the initial value problems (2.1) will be ob tained in this section for cases m = 1, 2,3 and 4 before the general formula is obtained. Since we shall be considering overdetermined cases, the formulae for s = 1,2 and 3 will be presented before that of general s (s is the number of overdetermination).
Case m = 1, s = 1
(
F
(P0,0 + P0,1x) y(x) + P1,0 + P1,1x + P1,2×2 y'(x) = frxr, F n (2.7)
r=0
L (P
1,0
+ P1,1
x + P
1,2
x2 d dx
+ (P0,0
+ P0,1x)
(
Lxr = P1,0 + P1,1x + P1,2×2 rxr1 + (P0,0 + P0,1x) xr Lxr = rP1,0xr1 + (rP1,1 + P0,0) xr + (rP1,2 + P0,1) xr+1
Lxr = rP1,0LQr1(x) + (rP1,1 + P0,0) LQr(x) + (rP1,2 + P0,1) LQr+1(x) Lxr = L (rP1,0Qr1(x) + (rP1,1 + P0,0) Qr(x) + (rP1,2 + P0,1) Qr+1(x))
Due to the existence of L1 as a result of linearity of L,
xr = rP1,0Qr1(x) + (rP1,1 + P0,0) Qr(x) + (rP1,2 + P0,1) Qr+1(x)
From where Qr+1(x) is obtained as
xr rP1,0Qr+1(x) (rP1,1 + P0,0) Qr(x)
Qr+1(x) =
rP1,2
+ P0,1
, r 0 (2.8)
when r = 0,
1 P0,0
Q (x) = Q (x)
r = 1,
1 P0,1 P0,1 0
x P1,1 + P0,0
Q (x) =
2 P1,2 + P0,1 P0,1 (P1,2 + P0,1)
P0,0 (P1,1 + P0,0) P0,1P1,0
+ Q (x)
r = 2,
x2
2P1,1 + P0,0
P0,1 (P1,2 + P0,1) 0
Q3(x) = 2P
1,2
+ P0,1
(P1,2
+ P0,1
) (2P
1,2
x
+ P0,1)
+ (2P1,1 + P0,0) (P1,1 + P0,0) P1,0 (P1,2 + P0,1)
P0,1 (P1,2 + P0,1) (2P1,2 + P0,1)
P0,0 (2P1,1 + P0,0) (P1,1 + P0,0) P0,1P1,0 (2P1,1 + P0,0) Q (x)
P0,1 (P1,2 + P0,1) (2P1,2 + P0,1) 1
P1,0P0,0
+ Q (x)
Case m = 2, s = 1
P0,1 (2P1,2 + P0,1) 0
(P2,0 + P2,1x + P2,2×2 + P2,3×3 y''(x) + (P1,0 + P1,1x + P1,2×2 y'(x)
r=0
+ (P0,0 + P0,1x) y(x) = frxr (2.9)
F
Following the same procedure as in the case m = 1, we have
xr r (r 1) P2,0Qr2(x) [r (r 1) P2,1 + rP1,0]Qr1(x)
Qr+1(x) =
r (r 1) P
2,3
+ rP1,2
+ P0,1
Case m = 3, s = 1
[r (r 1) P2,2 + rP1,1 + P0,0]Qr(x) , r 0 (2.10)
r (r 1) P2,3 + rP1,2 + P0,1
(
P3,0 + P3,1x + P3,2×2 + P3,3×3 + P3,4×4 y'''(x)
+ (P2,0 + P2,1x + P2,2×2 + P2,3×3 y''(x) + (P1,0 + P1,1x + P1,2×2 y'(x)
r=0
+ (P0,0 + P0,1x) y(x) = frxr (2.11)
F
The Qr+1(x) for this case is obtained as
xr [r (r 1) (r 2) P3,0]Qr3(x)
Qr+1(x) = P
0,1
+ rP1,2
+ r (r 1) P
2,3
+ r (r 1) (r 2) P
3,4
[r (r 1) (r 2) P3,1 + r (r 1) P2,0]Qr2(x) P0,1 + rP1,2 + r (r 1) P2,3 + r (r 1) (r 2) P3,4 [r (r 1) (r 2) P3,2 + r (r 1 )P2,1 + rP1,0]Qr1(x)
P0,1 + rP1,2 + r (r 1) P2,3 + r (r 1) (r 2) P3,4
[r (r 1) (r 2) P3,3 + r (r 1) P2,2 rP1,1 P0,0]Qr(x)
P0,1 + rP1,2 + r (r 1) P2,3 + r (r 1) (r 2) P3,4
(2.12)
Studying the pattern of Qr+1(x) for m = 1, 2, and 3 above, we arrived at general formula for case s = 1 as:
xr m j=k j!(r Pj,jk Qrk(x)
k=0
k,k+1
Qr+1(x) =
k=1 m
m
k=0
j
(
k
k!(r P
j
m j=o m
j! r Pj,j Qr(x) k!(r P
, r 0 (2.13)
k=0
Now for s = 2 cases, Qr+2(x) we obtained:
k k,k+1
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
Qr+2(x) =
m k=0
k!(r P
j
k
( j! P Q (x)j,j1 r+1
k,k+1
1
k,k+1
k=0
m j=0
k=0
We equally obtained for case s = 3:
m
r j
k
k!(r P
, r 0 (2.14)
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
Qr+3(x) =
m k=0
k!(r P
j
k
( j! P Q (x)j,j+2 r+2
k,k+2
2
k=0
m j=0
m
k=0
r j
k
k!(r P
, r 0 (2.15)
k,k+2
Continuing with these process, we derived for case m = m and s = s the general canonical poynomial
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
j
k,k+s
Qr+s(x) =
m k!(r P
k=0
k
s1 m
j!(r Pj,j+k Qr+k (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+s
, r 0 (2.16)
Let m be the order of the ODE (1.1) and let s be the number of overdeter minations, then the canonical polynomial associated with the DE is
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
j
k,k+s
Qr+s(x) =
m k!(r P
k=0
k
s1 m
j!(r Pj,j+k Qr+k (x)
k=0
j=0
m
j
k!(r P
, r 0 (2.17)
k=0
k k,k+s
We shall employ the principles of mathematical induction over the summa tion variables m and s to establish the validity of (2.17).This will be achieved by varying one of these variables at a time while the other is fixed. Firstly, let s be fixed at one in (2.17) so that
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
Qr+1(x) =
m k=0
k!(r P
j
k
k,k+1
m j=0
j!(r Pj,j+k Qr(x)
j
k
m k=0
k!(r P
k,k+1
, r 0 (2.18)
We use induction on m for fixed s = 1. We shall show that the formula (2.18) holds for m = 1:
x
r 1
k=1
1
j=1
j!(r Pj,jk Qrk (x)
k
Qr+1(x) =
1
j
k=0
k!(r P
k,k+1
j
j=0 j! Pj,j+k r+k
1 (r Q (x)
1
k
k=0
k!(r P
k,k+1
, r 0 (2.19)
xr rP1,0Qr1(x) (P0,0 + rP1,1) Qr(x)
Qr+1(x) =
P0,1
+ rP1,2
(2.20)
which is the same as Qr+1(x) in (2.13). Hence the formula (2.17) is true for
m = 1. Now assume that (1.2) is true for m = n. Thus (2.17) becomes
x
r n
k=1
n j=k
j!(r Pj,jk Qrk (x)
Qr+1(x) =
n k=0
k!(r P
j
k
n
k,k+1
j!(r Pj,j Qr(x)
k=0
k,k+1
j=0
n
j
k
k!(r P
, r 0 (2.21)
We now show that the formula (2.17) holds for m = n + 1.
From our construction of Qr+1(x) in (2.18) for m = 1 up to m = n + 1, we have
x
r n
k=1
n j=k
k,k+1
j!(r Pj,jk Qrk (x)
n+1,n+2
Qr+1(x) = n
k!(r P + P (n + 1)!( r
j
k
n+1
Pn+1,nk+1 (n + 1)!( r
Qrk(x)
k=0
n
n+1
k=0
k
n+1
k!(r Pk,k+1 + Pn+1,n+2 (n + 1)!( r
n j!(r Pj,j Qr(x) + Pn+1,n+1 (n + 1)!( r Qr(x)
k=0
k,k+1
n+1,n+2
j=0 j
n
k!(r P + P
n+1
k
n+1
(n + 1)!( r
(2.22)
k=1
xr n
n j=k
j!(r Pj,jk
k,k+1
Qr+1(x) =
n+1 k!(r P
j
k=0
k
Pn+1,nk+1 (n + 1)!( r
Qrk(x)
n+1
k=0
k
n+1 k!(r Pk,k+1
n
j!(r Pj,j + Pn+1,n+1 (n + 1)!( r Qr(x)
k,k+1
j=0
j
k=0
k
xr n+1 n+1 j!(r Pj,jk Qrk (x)
n+1 k!(r P
n+1
(2.23)
Qr+1(x) =
n+1 k!(r P
k
k=0
k,k+1
k=1
j=k
j
j=0
j
k=0
k
n+1 j!(r Pj,j Qr(x)
k,k+1
n+1 k!(r P (2.24)
Thus, (2.17) holds for m = n + 1 and hence, from the above steps, for all positive integral values of m.
Next, we assume that (2.17) holds for s = n, that is
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
Qr+n(x) =
m k=0
k!(r P
j
k
k,k+n
n1 m
j!(r Pj,j+k Qr+k (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+n
(2.25)
and then show that it holds for s = n + 1, that is
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
Qr+n+1(x) =
m k=0
(k!(r P
j
k
k,k+1
+ k!(r P
k,k+n+1
k
k=0
j=0
n1 n
j!(r Pj,j+k Qr+k (x) Qr+n+1(x)
m
(k!(r Pk,k+1 + k!(r Pk,k+n+1
( j! P Q (x)j,j+n+1 r+n+1
j
k=0
k k
m r
k
k
j=0 j
m k=0
(k!(r P
k,k+1
+ k!(r P
k,k+n+1
(2.26)
x
r m
k=1
m j=k
j!(r Pj,jk Qrk (x)
Qr+n+1(x) =
m k=0
n
k! r P
m
k,k+n+1
j
k
(
j!(r Pj,j+k Qr+k (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+n+1
(2.27)
Thus, (2.17) holds for all m and s.
3.0 THE n th DERIVATIVES OF THE CANONICAL POLYNOMIALS OF m th ORDER OVERDETERMINED ORDINARY DIFFERENTIAL EQUATIONS ODES
The n th derivatives of the canonical polynomials presented in section 2 above are presented in this section. This is achieved by first obtaining the derivatives for individual cases and from that, now seek the general n th derivatives for all cases. As in the previous sections, m shall be the order of the differential equation, s the number of overdetermination and n, the order of the derivatives.
r1
Case m = 1, s = 1, n = 1
r
Q1 (x) =
rxr1 rP1,0Q1
(x) (P0,0 + rP1,1) Q1 (x)
, r 0 (3.1)
r+1
Case m = 1, s = 1, n = 2
P0,1 + rP1,2
Q11
(x) =
r (r 1) xr2 rP1,0Q11
(x) (P0,0 + rP1,1) Q11 (x)
, r 0
r+1
Case m = 1, s = 1, n = 3
P0,1 + rP1,2
r1
r
(3.2)
Q111
r
(x) =
r (r 1) (r 2) xr3 rP1,0Q111 (x) (P0,0 + rP1,1) Q111 (x)
, r 0
r+1
P0,1 + rP1,2
r1
(3.3)
r
If we continue with this process, we shall have for case n = n, Case m = 1, s = 1, n = n
n!(r xrn rP1,0Qn (x) (P0,0 + rP1,1) Qn(x)
Q
n r+1
(x) = n
r1
P0,1 + rP1,2
, r 0
(3.4)
Following the same procedure, we shall obtain the following results for the specific cases.
Case m = 1, s = 2, n = n
Q
n r+2
(x) =
n
r1
n!(r xrn rP1,0Qn (x)
P0,2 + rP1,3
P0,2 + rP1,3
(P0,0 + rP1,1) Qn(x) (P0,1 + rP1,2) Qn
r
(x)
P0,2
+ rP1,3
r+1
, r 0 (3.5)
Case m = 1, s = 3, n = n
Q
n r+3
(x) =
n
r1
n!(r xrn rP1,0Qn (x)
P0,3 + rP1,4
P0,3 + rP1,4
(P0,0 + rP1,1) Qn(x) (P0,2 + rP1,3) Qn
r
(x)
P0,3
+ rP1,4
r+1
, r 0 (3.6)
So that the first derivative for the case m = m and s = s is
n!(r xrn m
r+s
m
j!(r Pj,jk Q1
(x)
Q1 (x) = n
k=1
k
m k=0
j=k
k!(r P
j
k,k+s
rk
s1 m
j!(r Pj,j+k Q1
(x)
The second derivative:
k=0
j=0
k
m k=0
j
k!(r P
k,k+s
r+k
(3.7)
Q
11
r+s
(x) =
k=1
j=k
r (r 1) xr2 m m
j!(r Pj,jk Q11
(x)
j
rk
s1 m
j!(r Pj,j+k Q11
(x)
The third derivative:
k=0
j=0
k
m k=0
j
k!(r P
k,k+s
r+k
(3.8)
r (r 1) (r 2) xr3 m
r+s
k=0
m
j!(r Pj,jk Q111
(x)
Q111 (x) =
k=1
k
k,k+s
m k!(r P
j=k j
rk
s1 m
j!(r Pj,j+k Q111
(x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+s
r+k
(3.9)
Thus, the n th derivative is obtained as
n!(r xrn m m j!(r Pj,jk Q(n) (x)
r+s
Q(n) (x) = n
k=1
k
m k=0
j=k
k!(r P
j
k,k+s
rk
s1 m
j!(r Pj,j+k Q(n) (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+s
r+k
(3.10)
If the generalized canonical polynomials associated with mth order overde termined ODE is given as
x
r m
k=1
m (r Q (x)
j
k=0
k
j=k j! Pj,jk rk
k,k+s
Qr+s(x) =
m k!(r P
s1 m
j!(r Pj,j+k Qr+k (x)
k,k+s
k=0
k=0
j=0
m
j
k
k!(r P
, r 0 (3.11)
Then its n th derivative is
n!(r xrn m
r+s
m
j!(r Pj,jk Q(n) (x)
Q(n) (x) = n
k=1
k
m k=0
j=k
k!(r P
j
k,k+s
rk
s1 m
j!(r Pj,j+k Q(n) (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+s
r+k
(3.12)
We shall employ here again the principle of mathematical induction over the summation variables m and s to establish the validity of (3.12). We shall vary one of these variables at a time while the other is fixed.
Firstly, let us fix s at one in (3.12) so that
n!(r xrn m
r+1
m
j!(r Pj,jk Q(n) (x)
Q(n) (x) = n
k=1
k
m k=0
j=k
k!(r P
j
k,k+1
rk
m j!(r Pj,j+k Q(n) (x)
k=0
k,k+1
j=0
m
j
k
k!(r P
r+k
(3.13)
We use induction on m for s = 1. We shall show that the formula holds for
m = 1;
n
n!(r xrn 1
1 j!(r Pj,jk Q(n) (x)
Q(n) (x) =
r+1
1 (r
k
k=1
j=k
j
rk
k=0 k! Pk,k+1
1 j!(r Pj,j+k Q(n) (x)
k,k+1
j=0
1
j
k=0
k
k!(r P
r+k
(3.14)
Q
n r+1
(x) =
n
r1
P0,1 + rP1,2
(3.15)
n!(r xrn rP1,0Qn
(x) (P0,0 + rP1,1) Qn(x)
r
which is the same as Qr+1(x) in (3.4). Hence the formula (3.13) is true for
m = 1.
Now assume that is true for m = q. Thus (3.13) becomes
n!(r xrn q
r+1
q
j!(r Pj,jk Q(n) (x)
Q(n) (x) = n
k=1
q k=0
j=k
k
k!(r P
j
k,k+1
rk
q j!(r Pj,j+k Q(n) (x)
k=0
k,k+1
j=0
q
j
k
k!(r P
r+k
(3.16)
We now show that the formula (3.13) holds fpr m = q + 1.
r+1
From our construction of Qn

in (3.4) for m = 1 up to m = q + 1, we
have
(n)
Qr+1(x) =
n! r xrn q q
n
k=1
j=k
k=0
k
k,k+1
(
q k!(r P + P
j! r Pj,j k Qn
j
rk
q+1,q+2
q+1
(
(q + 1)!( r
(x)
rk
((Pq+1,qk+1 (q + 1)!) Qn (x)
k=0
k
k,k+1
q+1,q+2
q+1
+ q k!(r P + P (q + 1)!( r
q j!(r Pj,j Qn(x) + Pq+1,q+1 (q + 1)!( r Qn(x)
k=0
k,k+1
q+1,q+2
j=0 j
q
r
k
q+1
k!(r P + P
q+1
(q + 1)!( r
r
(3.17)
n
k=1
j=k
n!(r xrn q q
k=0
k,k+1
j!(r Pj,jk Qn
q+1,q+2
(x)
(n)
k
q+1
Qr+1(x) =
q k!(r P + P (q + 1)!( r
rk
((Pq+1,qk+1 (q + 1)!)) Qn (x)
rk
j
k=0
k
k,k+1
q+1,q+2
q+1
+ q k!(r P + P (q + 1)!( r
q
j!(r Pj,j Qn(x) + Pq+1,q+1 (q + 1)!( r
k=0
k,k+1
q+1,q+2
Qn(x)
j=0 j
n!(r xrn q+1 q+1 j!(r Pj,jk Q(n) (x)
q
r
k
q+1
k!(r P + P
q+1
(q + 1)!( r
r
(3.18)
n
k=1
j=k
j
r+1
q+1 k!(r Pk,k+1
k
j,j
r
Q(n) (x) =
rk
j=0
j
k=0
q+1 j!(r P
Q(n)(x)
k,k+1
k=0
k
q+1 k!(r P
(3.19)
Thus (3.12) holds for m = q + 1 and hence, from the steps above, for all integral values of m.
Next we assume that (3.12) holds for s=q, that is
n!(r xrn m
r+q
m
j!(r Pj,jk Q(n) (x)
Q(n) (x) = n
k=1
k
m k=0
j=k
k!(r P
j
k,k+1
rk
q1 m
j!(r Pj,j+k Q(n) (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+1
r+k
(3.20)
and the show that it holds for s = q + 1, that is,
n!(r xrn m
Q (x) =
m
j!(r Pj,jk Q(n) (x)
(n) n
r+q+1
k=1
k
m k=0
j=k
k!(r P
j
k,k+q+1
rk
j=0
q1 m
k=0
j!(r Pj,j Q(n) (x)
r+q+1
Now, by our construction of Q(n)
m k=0
(x),
k!(r P
k
j
r+k
k,k+q+1
(3.21)
n
n!(r xrn m
m
j!(r Pj,jk Q(n) (x)
k
(n)
Qr+q+1(x) =
m k=0
(k!(r P
k=1
j=k
k,k+1
+ k!(r P
k
j
rk
k,k+q+1
j=0
q1 m
k=0
j!(r Pj,j+k Qn
(x)
m
(k!(r Pk,k+1 + k!(r Pk,k+q+1
k
k
k=0
k
m
k=0
k
j
r+k
j=0
j
r+q+1
k
k
k,k+q+1
j!(r Pj,j+q+1 Q(n)
k,k+1
(x)
+ m
(k!(r P
+ k!(r P
(3.22)
n!(r xrn m
Q (x) =
m
j!(r Pj,jk Q(n) (x)
(n) n
q+r+1
k=1
k
m k=0
j=k
k!(r P
j
k,k+q+1
rk
q m
j!(r Pj,j+k Q(n) (x)
k=0
j=0
k
m k=0
j
k!(r P
k,k+q+1
r+k
(3.23)
Thus, (3.12) holds for all m and s.
The derivation of a general formula for the canonical polynomials asso ciated with mth order overdetermined linear ODE together with its associ ated nth order derivative has been presented.
The recursive nature of the formulae makes for easy determination of particular cases for which m will be specified. The fact that the determi nation of canonical pollynomials is independent of the boundary conditions makes it attractive in the Tau approximation problem to the solution of ODEs, and when Tau approximations of higher degrees are needed, the pro cess of their determination does not begin from scratch.
The polynomial reported above will, in the subsequent work, be incor porated into the Tau method for purpose of generalizing the recursive for mulation of the Tau method itself.

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