Ukrainian Mathematical Journal, Vol. 53, No. 4, 2001
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING POLYNOMIALS M. M. Rizk
UDC 517.53
For the case of Hermitian interpolation, we consider the approximation-iterative method introduced by Dzyadyk. We construct a practical algorithm.
1. Introduction The mathematical formulation of physical phenomena in many situations often lead to one or several initialvalue problems of the form y ′ ( x ) = f [ x, y (x ) ],
y ( x0 ) = y0 ,
x ∈ [ x0 , x0 + h ] , y ∈ R,
h > 0.
(1)
In 1980 – 1984, Dzyadyk [1, 2] introduced a new technique, in which he constructed an algorithm based on both approximation and iteration. In this algorithm, the function f ( x, y ) in (1) is assumed to be analytic in a certain domain G ⊃ [ x0 , x0 + h ] or sufficiently smooth on [ x0 , x0 + h ] . We call this technique the Dzyadyk approximationiterative method [1 – 5]. The following theorem is true (it is called the Picard theorem and its proof can be found in [5] and [6]): Theorem 1 (Picard theorem). If f (x, y ) is defined and continuous on D = [ x0 , x0 + h ] × [ y0 – H , y0 + H ], H > 0, satisfies the Lipschitz condition with respect to y with constant A, and, in addition, satisfies the inequality q
f
−1 s C D A ∫ e ds < H,
q = Ah,
0
then the unique solution y ( x ) of the initial-value problem (1) is given by
y ( x ) = y0 +
∞
∑ [ y[µ +1]( x) − y[µ]] ,
µ=0
where [0]
y ( x ) ≡ y0 ,
x
[µ ]
y ( x ) = y0 +
∫ f [σ, y
x0
Furthermore,
y − y[µ] ≤
f
CD e
A x−a
Aµ x − a
µ +1
[µ − 1]
]
(σ) dσ .
/(µ + 1)!.
Cairo University, Giza, Egypt. Published in Ukrains’kyi Matematychnyi Zhurnal, Vol. 53, No. 4, pp. 501 – 512, April, 2001. Original article submitted June 21, 1999. 0041–5995/01/5304–0569 $25.00
© 2001 Plenum Publishing Corporation
569
570
M. M. RIZK
Within the framework of the Dzyadyk approximation-iterative method (technique), Dzyadyk used the following projection operator for the approximation of f (ξ, y (ξ)) : An ( f ; ξ) = L n ( f ; ξ) =
n
∑ f (ξi , y (ξi )) ln, i (ξ),
ξ, ξ i ∈ [ x0 , x0 + h ] , i = 0, n,
i=0
where ln, i (ξ) are the fundamental Lagrange interpolation polynomials given by ln, i (ξ) =
n
ξ − ξj
∏ j=0 ξ j ≠i
i
− ξj
.
In this paper, instead of L n , we use a projection operator depending on Hermite interpolation polynomials H 2n + 1 for the approximation of f (ξ, y (ξ)) . Namely, An ( f ; ξ) = H 2n + 1( f ; ξ) =
n
∑ f (ξi , y (ξi ))h2n +1, i (ξ)
n
+
i=0
∑ f ′ (ξi , y (ξi )) h2n +1, i (ξ),
i=0
where
[
]
h2n + 1, i (ξ) = 1 − 2ln′, i (ξi )(ξ − ξi ) ln2, i (ξ) ,
h2n + 1, i (ξ) = (ξ − ξi ) ln2, i (ξ) .
(2)
For any n ∈ N, Rizk [7] expanded the fundamental Hermite polynomials h2n + 1, i (ξ) and h2n + 1, i (ξ) , i = 0, n, built by either the nodes ξ*j = – cos ( j π / n ) or the nodes ξ 0j = – cos (( 2j + 1) π / (2n + 2)) , j = 0, n, on [–1, 1] according to the following formulas: 1. If ξ j = : ξ*j and cj = : cos ( j π / n ) , then, for i = 1, 2, … , n – 1,
h*2n + 1, i (ξ) =
1 n2
1 1 − 2n + 1 T1(ξ) + ci n − 1 − c2i 1 − c2i 2n − 2
+
∑ (−1)k (2n − k ) cik Tk (ξ)
k =2
+
h2*n + 1, i (ξ) =
1 2 n2
1 – ci 1 + T2n − 1(ξ) ( − c ) 2 1 2i
ci 1 T2n (ξ) – T2n + 1(ξ) , 2 (1 − c2i ) 1 − c2i
ci − c2i T1(ξ) +
(3a)
2n − 2
∑ (−1)k [ci (k +1) − ci (k −1) ] Tk (ξ)
k =2
1 1 + c2i − T2n − 1(ξ) – ci T2n (ξ) + T2n + 1(ξ) 2 2
(4a)
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS
571
and, for i = 0 and i = n, h*2n + 1, i (ξ) =
1 48 n 2
2 2 4 n + 6n − 1 – 2 2n + 24n − 11 ci T1(ξ)
(
)
(
)
2n − 2
+ 24
∑ (−1)k (2n − k ) cik Tk (ξ)
k =2
)
– 4 n 2 − 1 T2n (ξ) + 2n 2 + 1 ci T2n + 1(ξ) ,
(3b)
1 c − T (ξ) + 1 T 1 2 n − 1(ξ) − ci T2 n (ξ) + T2 n + 1(ξ) . i 2 2
(4b)
(
1 h2*n + 1, i (ξ) = 2 8n
(
+ 2n 2 − 23 ci T2n − 1(ξ)
)
(
)
2. If ξ j = : ξ 0j and S j, k : = cos (( 2j + 1) π / (2n + 2)) , then, for i = 0, 1, … , n, h20n + 1, i (ξ) =
2n + 1 1 k 1 + 2 ∑ (−1) n +1 k =1
h20n + 1, i (ξ) =
1 2 (n + 1)2
k 1 − 2n + 2 Si, k Tk (ξ) ,
(5)
2n + 1
∑ (−1)k [Si, k +1 − Si, k −1] Tk (ξ).
(6)
k =1
2. Analytic Method Let n be an arbitrary natural number, let {ξ j } j = 0, n , – 1 ≤ ξ0 < ξ1 < … < ξn ≤ 1, be nodes on [–1, 1] , and let h > 0 be a real number. Using the linear transformations ξ = –1 +
2 ( x − x0 ) : [ x0 , x0 + h ] → [–1, 1] h
or (7)
h x = x0 + (1 + ξ) : [–1, 1] → [ x0 , x0 + h ] 2 on the segment [ x0 , x0 + h ] .
we construct nodes {x j } j = 0, n ˜ , f˜ , l˜ , h˜ , and h˜ are the transfers of ω, f, l, h, and h , respectively, from [–1, 1] to [ x0 , x0 + h ] , If ω then it is easy to show that f˜ ( x ) = f ( x ),
2 n +1 ˜ ω n + 1(ξ) = ω n + 1( x ) , h
–1 2 ˜ ˜ d 2 ω n + 1(ξ) dω n + 1(ξ) –1 hd ω n + 1 ( x ) dω n + 1 ( x ) = , 2 dx dξ dx 2 dξ 2
h2n + 1, i (ξ) = h˜ 2n + 1, i ( x ),
h2n + 1, i (ξ) =
2 ˜ h2n + 1, i ( x ). h
(8)
(9)
(10)
572
M. M. RIZK
The algorithm consists of the following steps: Step 1. Construct the nodes {ξ j }nj = 0 , – 1 ≤ ξ0 < ξ1 < … < ξn ≤ 1, on [–1, 1] .
[ ]i, j = 0, n
Step 2. Construct matrices aij
[ ]i, j = 0, n
and bij
of order (n + 1) × (n + 1) as follows:
ξj
aij = aij (n, ξ j ) =
ξj
∫ h2n +1, i (ξ) dξ ,
−1
bij = bij (n, ξ j ) =
∫ h2n +1, i (ξ) dξ .
(11)
−1
From (3a,b), (4a,b), and (11), we get the proof of the following lemma: Lemma 1. If ξj = ξ*j = – cos ( j π / n ) , then the matrices aij and bij can be expressed in the following explicit form: For i = 1, 2, … , n – 1,
(
)
aij = aij* = aij n, ξ*j =
1 4n 2 2 (1 − c j ) n − 2 (4n − 1)(1 − c2i ) n +
ci (1 − c2 j ) n2 (2n − 1) − 2 4 (n − 1)(1 − c2i ) 2n − 1
+
(
bij = bij* = bij n, ξ*j
)
=
(2n − k ) cik c(k − 1) j c(k + 1) j 2 − − 2 , 2 k + 1 k − 1 k −1 k =2
∑
1 1 4n 2 1 − c2 j 1 ( − ) + c c c2i + 2 i j 2 2 4 2n 4n − 1 n − 1
–
1 2
2n − 1
c(k − 1) j c(k + 1) j 2 − − 2 c(k − 1) i − c(k + 1) i , k −1 k +1 k −1 k =2
[
∑
and, for i = 0 and i = n,
(
)
aij = aij* = aij n, ξ*j =
2 n2 − 1 1 1 ( − ) + − + c n n 1 6 1 j 4n 2 3 4n 2 − 1 +
ci (1 − c2 j ) 2 2n 2 + 1 n + n − + 2 24 11 24 n 2 − 1 2n − 1
+
c(k − 1) j c(k + 1) j 2 − − , k −1 k + 1 k 2 − 1
∑ (2n − k ) cik
k =2
]
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS
(
bij = bij* = bij n, ξ*j
)
=
ci (1 − c j ) 2
2 (4n − 1)
+
1 − c2j 32 (n 2 − 1)
573
,
where ck = cos ( k π / n ) . From (5), (6), and (11), we get the proof of the following lemma: Lemma 2. If ξj = ξ 0j = – cos (( 2j + 1) π / (2n + 2)) , j = 0, n, then the matrices a i j a n d b i j can be expressed in the following explicit form for i = 0, 1, … , n :
(
aij = aij0 = aij n, ξ 0j
)
=
(1 − S j, 2 )(2n + 1) Si,1 1 1 − S j,1 + n +1 4 (n + 1) 2n + 1
+
k
S j, (k − 1) S j, (k + 1) 2 − − 2 , k −1 k +1 k − 1
∑ 1 − 2 (n + 1) Si, k
k =2
(
)
bij = bij0 = bij n, ξ 0j =
{
1 1 (Si, 2 − 1)(1 − S j, 2 ) 2 4 (n + 1) 2 2n + 1
+
S j, (k − 1) S j, (k + 1) 2 − − 2 , k −1 k +1 k − 1
∑ (Si, k +1 − Si, k −1)
k =2
where Si, j = cos (( 2i + 1) j π / (2n + 2)) .
[ ]i, j = 0, n
Note that aij
[ ]i, j = 0, n
and bij
depend on the natural number n and the choice of the nodes
{ξ j } j = 0, n ,
but does not depend on our problem (1). Step 3. Construct a sequence
{x j } j = 0, n
( x0 ≤ x0 < x1 < … < xn ≤ x0 + h) on [ x0 , x0 + h ] as follows:
x j = x0 +
h (1 + ξ j ) , j = 0, 1, … , n. 2
Step 4. Using the matrices in Step 2, we write the following nonlinear system of ( n + 1 ) equations in ( n + 1 ) unknowns yj , j = 0, n : yj = y0 +
h n h2 aij f ( xi , yi ) + ∑ 2 i=0 4
n
∑ bij [ fx′ + f fy′ ]( xi , yi ),
j = 0, n.
(12)
i=0
By solving system (12) (using simple successive iteration or Newton iteration), we determine the values of yj , j = 0, n. Theorem 2. The values y j given by (12) represent the values of the following polynomial (of degree not exceeding 2n + 2 ) at the points x = x j ∈ [ x0 , x0 + h ] :
574
M. M. RIZK x
P2n + 2 ( x ) = y0 +
∫ H˜ 2n +1[ f (⋅; P2n + 2 (⋅)); σ] dσ
x0
= y0 +
n
x
i=0
x0
∑ f ( xi , P2n + 2 ( xi )) ∫ h˜2n +1, i (σ) dσ +
n
∑[
i=0
x
]
˜ fx′ + f fy′ ( xi , P2n + 2 ( xi )) ∫ h2n + 1, i (σ) dσ ,
(13)
x0
i.e., P2n + 2 ( x j ) = yj , j = 0, n. Proof. By using transformation (7) and Eqs. (10), we get xj
ξj
∫ h˜2n +1, i (σ) dσ =
∫ h2n +1, i (ξ) 2 dξ
x0 xj
˜ ∫ h2n +1, i (σ) dσ =
x0
h
−1
=
h aij , 2
ξj
h h h2 h d bij , ( ξ ) ξ = 2 + 1 n , i ∫ 2 4 −1 2
which obviously yields the required result. Step 5. By using Theorem 2 and relation (12), we now get the following polynomial of degree not exceeding ( 2n + 2 ) : P2n + 2 ( x ) = y0 +
h n h2 f ( xi , yi ) Π 2n + 2, i (ξ) + ∑ 2 i=0 4
n
∑ [ fx′ + f fy′ ]( xi , yi ) Π2n + 2, i (ξ) ,
(14)
i=0
where ξ = – 1 + 2 ( x − x0 )/ h, ξ
Π 2n + 2, i (ξ) =
∫ h2n +1, i (t) dt ,
−1
ξ
and
Π2n + 2, i (ξ) =
∫ h2n +1, i (t) dt .
−1
For specific nodes, by using (3a,b), (4a,b), (5), and (6) and performing certain transformations, we determine the explicit form of each Π 2n + 2, i (ξ) and Π2n + 2, i (ξ) in the following two lemmas: Lemma 3. If ξj = ξ*j = – cos ( j π / n ) and c i = cos ( j π / n ) , j = 0, n , then Π 2n + 2, i (ξ) a n d Π2n + 2, i (ξ) can be expressed in the following explicit form: I. For i = 1, 2, … , n – 1, Π 2n + 2, i (ξ) = Π*2n + 2, i (ξ) =
ci 1 1 n− (1 + ξ) + 2 2 1 − c2i n 2n − 1
+
∑
k =2
(−1)k
1 2 (2n − 1) − 1 − c 1 − ξ 2i
[
(2n − k ) cik Tk + 1(ξ) Tk − 1(ξ) 2 (−1)k k + 1 − k − 1 − k2 − 1 2
]
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS
–
ci T2n + 2 (ξ) T2n − 2 (ξ) 1 − + 2 4 (1 − c2i ) 2n + 2 2n − 2 n − 1
+
T2n + 1(ξ) T2n − 1(ξ) 1 2 − − 2 , 2 (1 − c2i ) 2n + 1 2n − 1 4n − 1
575
1 2 2 4ci (1 + ξ) + 2c2i (1 − ξ ) 8n
Π2n + 2, i (ξ) = Π2*n + 2, i (ξ) =
2n − 1
Tk + 1(ξ) Tk − 1(ξ) 2 (−1)k − − 2 k +1 k −1 k − 1
∑ (−1)k (c(k +1) i − c(k −1) i )
+ 2
k =2
T2n + 2 (ξ) T2n − 2 (ξ) T2n + 1(ξ) T2n − 1(ξ) 2 1 – 2ci − − + − 2 − 2 . 2n − 1 2n − 2 4n − 1 n −1 2n + 2 2n + 1 II. For i = 0 and i = n, Π 2n + 2, i (ξ) = Π*2n + 2, i (ξ) =
1 48n 2
2 2 2 4(n + 6n − 1)(1 + ξ) + ci (2n + 24n − 11)(1 − ξ ) 2n − 1
+ 12
Tk + 1(ξ) Tk − 1(ξ) 2 (−1)k − − 2 k +1 k −1 k − 1
∑ (−1)k (2n − k ) cik
k =2
(2n 2 + 1) ci 2
+
T2n + 2 (ξ) T2n − 2 (ξ) 1 2n + 2 − 2n − 2 + n 2 − 1
T2n + 1(ξ) T2n − 1(ξ) 2 – 2(n 2 − 1) − − 2 , 2n − 1 4n − 1 2n + 1 Π2n + 2, i (ξ) = Π2*n + 2, i (ξ) =
1 32 n 2
2 4ci (1 + ξ) + 2 (1 − ξ )
T2n + 2 (ξ) T2n − 2 (ξ) T2n + 1(ξ) T2n − 1(ξ) 2 1 + – 2ci − + 2 − − 2 . 2n − 1 2n – 2 4n − 1 n −1 2n + 2 2n + 1 Lemma 4. If ξj = ξ 0j = – cos (( 2j + 1) π / (2n + 2)) , j = 0, n, then Π 2n + 2, i (ξ) and Π2n + 2, i (ξ) can be expressed in the following explicit form. If Si, k = cos (( 2i + 1) k π / (2n + 2)) , then, for i = 0, … , n, Π 2n + 2, i (ξ) = Π 20n + 2, i (ξ) =
2n + 1 1 Si,1 1 − ξ 2 1 + ξ + n +1 2n + 2
[
]
2n + 1
+
k Tk + 1(ξ) Tk − 1(ξ) 2 (−1)k k (− 1 ) S 1 − − − 2 , ∑ i, k k −1 k − 1 2n + 2 k + 1 k =2
(15)
576
M. M. RIZK
Π2n + 2, i (ξ) = Π20n + 2, i (ξ) =
1 2 Si, 2 − 1 2 1− ξ 4 (n + 1)
[
2n + 1
+
][
]
Tk + 1(ξ) Tk − 1(ξ) 2 (−1)k k (− 1 ) S − S − − 2 . ∑ i, k + 1 i, k − 1 k −1 k − 1 k +1 k =2
[
]
(16)
Using the Picard iteration (Theorem 1) for the solution of system (12), we now obtain the following approximation polynomial [polynomial (14)] depending on µ = 0, 1, … : P2[n0]+ 2 ≡ y0, P2[nµ+] 2 ( x ) = y0 +
h n h2 f xi , yi[µ − 1] Π 2n + 2, i (ξ) + ∑ 2 i=0 4
(
)
n
∑ [ fx′ + f fy′]( xi , yi[µ −1] ) Π2n + 2, i (ξ),
i=0
where ξ = – 1 + 2 ( x − x0 )/ h . Theorem 3 (error estimation). If f ( x, y) ∈ C(2n + 2)[ x0 , x0 + h] , there exists a finite positive constant H and the following conditions are satisfied: (i) (ii)
f ( x, y) defined and continuous on D : = [ x0 , x0 + h] × [ y0 − H, y0 + H ] ; f ( x, y) − f ( x, z) ≤ A y − z ∀ x ∈ [ x0 , x0 + h] , ∀ y, z ∈ [ y0 − H, y0 + H ] ;
(iii) q : = Ah < 1. Then there exists a finite positive constant M2n + 2 such that, for µ = 1, 2, … , the following inequalities are satisfied: I. If ξ j = ξ*j = – cos ( j π / n ), j = 0, n, then
P2[nµ+] 2 − y
C[ x0 , x0 + h ]
≤
M2n + 2 h 2n + 2 1 − q µ + 2 4n − 1(2n + 2)! 1 − q
f
q CD e
qµ . (µ + 1)!
f
q CD e
qµ . (µ + 1)!
II. If ξj = ξ 0j = – cos (( 2j + 1) π / (2n + 2)) , j = 0, n, then
P2[nµ+] 2 − y
C[ x0 , x0 + h ]
≤
M2n + 2 h 2n + 2 1 − q µ + 2 4n + 1(2n + 2)! 1 − q
Proof. If y[µ]( x ) , µ = 0, 1, 2, … , is the Picard sequence of functions given by Theorem 1, then, from (13), Theorem 1, and condition (ii), we get
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS
P2[nµ+] 2 − y[µ]( x ) =
∫ {H˜ 2n +1[ f (⋅; P2n + 2 (⋅)); σ] − f (σ, y x
[µ − 1]
µ −1
x
∫
x0
≤
≤
)}
(σ) dσ
x0
≤
H˜ 2n + 1[ f ( ⋅; P2[nµ+−21](⋅)); σ] − f (σ, P2[nµ+−21](σ)) dσ +
f (2 n + 2 ) (2n + 2)!
x
∫
˜ 2n + 1(σ) dσ + A ω
x0
x
∫
577
x
[µ − 1] µ −1 ∫ f (σ ; P2n + 2 (⋅)) − f (σ, y (σ))
dσ
x0
P2[nµ+−21](⋅) − y µ − 1(σ) dσ
x0
M2n + 2 h 2n + 2 2 ω 2n + 1 + Ah P2[nµ+−21] − y[µ − 1] . (2n + 2)! 2
(17)
Thus, if ∆ µ = P2[nµ+] 2 − y[µ] , then we get ∆ µ ≤ B + q∆ µ −1 ≤ B + qB + q 2 ∆ µ − 2 ≤ … ≤ B
1 − qµ , 1− q
(18)
where B =
2 M2n + 2 h 2n + 2 2 ω n +1 , (2n + 2)! 2
q = Ah.
Therefore, from the triangle inequality and relations (17) and (18), we get P2[nµ+] 2 − y
C[ x0 , x0 + h ]
≤
2 M2n + 2 h 2n + 2 2 1 − qµ ω + n +1 1− q (2n + 2)! 2
f
q CD e
qµ . (µ + 1)!
(19)
If ξj = ξ*j or ξj = ξ 0j , then we have
(
)(
) (
)
ω *n + 1(ξ) = ξ − ξ*0 ξ − ξ1* … ξ − ξ*n = Cn 1 − ξ 2 sin (n arccos ξ) ,
(20)
Cn = − 2(1 − n) ,
(
)(
) (
)
ω 0n + 1(ξ) = ξ − ξ 00 ξ − ξ10 … ξ − ξ 0n = Cn0 sin ((n + 1) arccos ξ) ,
Cn0 = 2 − n .
(21)
From (19) and (20), we now get assertion I and, from (19) and (21), we get assertion II. Theorem 3 is thus proved. 3. One Step Method In this section, we present a one step method based on the use of polynomials (14) and give a relation between its parameters. Let [ x0 , x0 + H ] be any interval, let M, n, N ∈ N be any natural numbers, let h = H / M, and let x N = x0 + Nh, N = 0, 1, 2, … , M, and
{ξ j }nj = 0
be nodes on [–1, 1] .
578
M. M. RIZK
For all i, j = 0, 1, 2, … , n, we consider the following parameters: 1 (1 + ξi ) , 2
ci =
Aij =
1 a ji , 2
1 b ji , 2
Bij =
(22) 1
1 h2n + 1, i (ξ) dξ, 2 −∫1
Bi =
1
Bi =
1 h2n + 1, i (ξ) dξ. 2 −∫1
1. If ξj = ξ*j = – cos ( j π / n ) , then it is obvious that 1 ain 2
Bi = Bi* =
and
Bi =
1 bin . 2
2. If ξj = ξ 0j = – cos (( 2j + 1) π / (2n + 2)) , then, by using (15) and (16), one can easily show that n k 1 1 1 − 2 ∑ Si, 2k 1 − , n +1 n + 1 4k 2 − 1 k =1
Bi = Bi0 =
Bi = Bi0 =
1 2 (n + 1)2
n
Si, 2k − 1 − Si, 2k + 1
k =1
4k 2 − 1
∑
,
Si j = cos
(2i + 1) jπ . 2n + 2
The relation between constants (22) is given by the following theorem: Theorem 4. For any n ∈ N and
{ξ j } ,
j = 0, 1, … , n, the following relations are true:
I. For all j = 0, n and 1 ≤ k ≤ 2n + 2, n
∑ Aji cik −1
=
i=0
k −1 n 1 k cj – ∑ Bji cik − 2 ; 2 i=0 k
(23)
k − 1 n k −2 1 – ∑ c Bi . k 2 i=0 i
(24)
II. For all 1 ≤ k ≤ 2n + 2, n
∑ Bi cik −1 =
i=0
Proof. I. From (22) and (11), we get n
∑
i=0
Aji cik − 1
1 n 1 n = aij cik − 1 = ∑ ∑ 2 i=0 2 i=0 1 = k 2
ξj
ξj
1 ∫ h2n +1, i (ξ) dξ 2 −1
k −1
(1 + ξi )k − 1
n
(1 + ξi )k − 1 h2n + 1, i (ξ) dξ . ∫ i∑ =0
−1
Consider the function g ( ξ ) = ( 1 + ξ ) k – 1, k ≤ 2n + 2. We have
(25)
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS n
∑ (1 + ξi )k −1 h2n +1, i (ξ)
H 2n + 1(g; ξ j ) =
i=0
579
n
+ (k – 1) ∑ (1 + ξi )k − 2 h2n + 1, i (ξ) = (1 + ξi )k − 1.
(26)
i=0
Integrating each side of (40) from – 1 to ξ j , we get ξj
∫
n
n
1 k k 2 cj . k
∑ (1 + ξi )k −1 h2n +1, i (ξ) dξ + (k – 1) ∑ 2 k − 2 cik − 2 bij = i=0
−1 i = 0
(27)
From (25) and (27), we get assertion I of Theorem 4. II. From (22) and (11), we get n
∑ Bi cik −1 =
i=0
1
1 n ∑ 2 i=0
1
∫ h2n +1, i (ξ) dξ 2 k −1 (1 + ξi )
k −1
.
(28)
−1
Integrating each side of (24) from – 1 to 1, we obtain n
∑
i=0
(1 + ξi )k − 1
1
∫
−1
n
h2n + 1, i (ξ) dξ + (k – 1) ∑ (1 + ξi )k − 2 b = i=0
2k . k
From (28) and (29), we get assertion II of Theorem 4. n
Remark 1. In (23), if k = 1 or k = 2, then
∑ Aji = cj
i=0 n
Remark 2. In (24), if k = 1 or k = 2, then
∑ Bi = 1 and
i=0
Remark 3. Let
{ξ j }in= 0
n
n
i=0
i=0
∑ Bji = cj – 2 ∑ Ajici .
and
n
∑ Bi = 1 – 2
i=0
n
∑ Bici .
i=0
be any nodes in [–1, 1] and let G ∈ C [–1, 1] be the function defined as follows:
(ξ + 1) (2ξ − (ξ 0 − 1)) (ξ − ξ 0 ) (ξ 0 + 1) 2 (ξ − ξ i ) 2ξ − ξ i + ξ i + 1 ξ − ξ i + 1 G(ξ) = 2 ξi + 1 − ξi (ξ − ξ n ) (2ξ − (ξ n + 1)) (ξ − 1) (1 − ξ n ) 2
(
(
(
)
)) (
)
if
ξ ∈[–1, ξ 0 ],
if
ξ ∈ ξi, ξi + 1 ,
if
ξ ∈[ξ n, 1] .
[
]
i = 0, n − 1,
+ – It is obvious that G′ ( – 1 ) = G′ ( ξi ) = G′ ( 1 ) = 1 and G ( – 1 ) = G ( ξi ) = G (1) = 0 ∀ i = 0, 1, … , n and 1
∫ G (t) dt
−1
= 0.
(29)
580
M. M. RIZK
Table 1 c0
A00
A01
…
A0n
B00
B01
…
B0n
c1
A10
A11
…
A1n
B00
B01
…
B0n
M
M
M
O
M
M
M
O
M
cn
An 0
An 1
…
An n
Bn 0
Bn 1
…
Bn n
B0
B1
…
Bn
B0
B1
…
Bn
= G (ξ) ,
n ≥ 1.
Therefore,
H 2n + 1(G; ξ j ) =
n
n
∑ G (ξi ) h2n +1, i (ξ)
∑ G′(ξi ) h2n +1, i (ξ)
+
i=0
i=0
(30)
Integrating each side of (30) from – 1 to 1, we get n
∑ Bi
= 0.
i=0
Thus, from 2, we get n
∑ Bi Ci
i=0
=
1 . 2
In polynomial (14), if we put x = xN and yN ≈ y( x N ), n = 0, 1, 2, … , M, then we can consider the following scheme: Ki = f x N + ci h, YN + h
n
∑ Aij K j
j=0
+
h2 2
n
∑ Bij K j ,
i = 0, n,
j=0
(31) Ki = f ′ x N + ci h, YN + h
h ∑ Aij K j + 2 ∑ Bij K j , j=0 j=0 n
2
n
i = 0, n.
Furthermore, n
YN +1 = YN + h
∑ Bi Ki
j=0
+
h2 2
n
∑ Bi Ki ,
j=0
N = 0, M − 1.
(32)
The parameters of (31) and (32) are presented in Table 1. Definition 1. We define the following properties with respect to the parameters A ij, B ij, B i , Bi , a n d c i , i, j = 0, n :
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS
581
Table 2 h
µ
enJ of Example 1
µ
enJ of Example 2
µ
enJ of Example 3
0.1
06
3.367306E–13
07
8.570922E–13
09
7.371880E–14
0.5
08
1.263820E–08
10
5.537792E–13
15
1.206146E–12
1.0
29
1.582177E–05
14
2.633049E–09
21
3.812061E–12
0.1
07
9.992007E–16
05
5.759837E–13
08
9.237056E–14
0.5
13
3.721246E–12
11
1.506573E–13
13
3.323564E–12
1.0
31
3.055127E–08
18
1.887379E–14
19
1.044498E–12
0.1
07
1.665335E–15
05
1.827427E–13
08
9.769963E–15
0.5
17
1.842970E–14
11
2.252643E–13
14
1.154632E–13
1.0
34
4.580791E–11
17
2.278178E–13
20
5.165646E–12
0.1
06
7.771561E–16
05
3.186340E–14
08
2.398082E–14
0.5
17
3.330667E–16
11
2.333689E–13
13
4.920508E–13
1.0
44
1.565414E–16
16
9.414691E–14
18
2.664535E–13
n
3
5
7
9
B (ξ) :
C (η) :
E (η, ξ) :
n
∑
i, j = 0
B jcir −1 A jicik −1 =
n
∑ Bicik −1
i=0
n
∑ Ajicik −1
i=0
=
=
k − 1 n k −2 1 – ∑ c Bi , 2 i=0 i k
1 ≤ k ≤ ξ,
k −1 n 1 k cj – ∑ Bji cik − 2 , j = 0, n, k 2 i=0 n
1 ≤ k ≤ η,
(33)
(34)
n
1 k −1 k + r −1 B jcir −1B jicik −2 – – ∑ cik +r −2 Bi , 1 ≤ r ≤ η, 1 ≤ k ≤ ξ . (35) k (k + r ) 2 i,∑ 2 k j =0 i =0
Theorem 4 yields the following statement: Theorem 5. The algorithm given by (31) and (32) possesses the properties B ( 2n + 2 ) and C ( 2n + 2 ) for each n. By using (33), (34), and (35), one can easily show that B ( η + ξ ) and C (ξ) yield E (η, ξ) . 4. Examples By using scheme (31), (32), we solved the following five examples (with known analytic solution) on PC jπ Pentium II-MMX 300 Mhz, using the nodes ξ j = ξ*j = − cos , j = 0, n. n
582
M. M. RIZK
Table 3
n
3
5
7
9
h
µ
enN of Example 4
µ
enN of Example 5
0.5
04
1.970673E–10
02
3.039236E–14
2.0
07
7.501384E–10
02
1.385558E–12
4.0
08
8.105921E–07
02
1.108447E–11
30.0
43
7.651603E–07
02
4.678441E–09
0.5
06
1.131317E–13
02
1.149081E–14
2.0
08
4.142714E–09
02
5.533352E–12
4.0
09
4.438719E–08
02
1.567884E–10
30.0
43
3.542338E–07
02
3.368408E–06
0.5
06
9.858780E–14
02
2.428613E–15
2.0
09
2.060452E–10
02
2.178169E–11
4.0
10
4.565695E–09
02
1.622595E–10
30.0
42
1.306392E–07
02
7.262734E–06
0.5
06
3.497203E–14
02
9.492407E–15
2.0
10
5.279666E–12
02
2.804867E–12
4.0
11
1.096299E–11
02
9.636381E–11
30.0
42
3.640602E–07
02
5.373036E–06
Table 2 gives the error enJ = Y1[µ] − y ( x0 + h) examples when using the Jacobi iteration, i.e.,
( µ is the number of iteration) for the first, second, and third
Ki[0] = f ( x0 , y0 ) , Ki[µ + 1] = f x0 + ci , y0 + h Ki[µ + 1] = f ′ x0 + ci , y0 + h
n
Ki[0] = f ′( x0 , y0 ) , n
∑ Aij K [jµ] +
j=0
∑ Aij K [jµ] +
j=0
h2 2
n
h2 2
n
∑ Bij K j[µ] ,
j=0
∑ Bij K j[µ] ,
j=0
µ = 0, 1, … ,
for different values of h ( h = 0.1, 0.5, 1.0) and different values of n ( n = 3, 5, 7, 9) . Table 3 gives the error enN = Y1[µ] − y ( x0 + h) for the stiff fourth and fifth examples when using the Newton – Raphson iteration for different values of h ( h = 0.5, 2, 4, 30) and different values of n ( n = 3, 5, 7, 9) .
DZYADYK’S TECHNIQUE FOR ORDINARY DIFFERENTIAL EQUATIONS USING HERMITIAN INTERPOLATING P OLYNOMIALS
583
Example 1. y ′ = – 2xy , x0 = 0 with the initial condition y( x0 ) = 1 and exact solution y( x ) = (1 + x 2 ) . –1
2
Example 2. y ′ = e x − y , x 0 = 0 with the initial condition y( x0 ) = ln (2) and exact solution y( x ) = x + ln (1 + e − x ) . Example 3. y ′ = 4x
y , x0 = 1 with the initial condition y( x0 ) = 4 and exact solution y( x ) = (1 + x 2 ) . 2
Example 4. y ′ = −1000 ( y − x 3 ) + 3x 2 , x 0 = 0 with the initial condition y( x0 ) = 0 and exact solution 3 y( x ) = x .
(
Example 5. y ′ = 1000 y − (1 + x 2 ) −1
y( x ) = (1 + x 2 ) .
−1
) − 2 xy2 , x0 = 0
with the initial condition y( x0 ) = 1 and exact solution
REFERENCES 1. V. K. Dzyadyk, Approximation-Iterative Method for the Approximate Solution of the Cauchy Problem for Ordinary Differential Equations [in Russian], Preprint No. 27, Institute of Mathematics, Ukrainian Academy of Sciences, Kiev (1984). 2. V. K. Dzyadyk and Yu. I. Romanenko, Approximation-Iterative Method for the Polynomial Approximation of Solutions of a Nonlinear Cauchy Problem for Equations of Hyperbolic Type [in Russian], Preprint No. 63, Institute of Mathematics, Ukrainian Academy of Sciences, Kiev (1986). 3. V. K. Dzyadyk, A. M. Basov, and M. M. Rizk, Theory and Application of the Approximation-Iterative Method and Its Comparison with Methods of the Runge – Kutta Type [in Russian], Preprint No. 39, Institute of Mathematics, Ukrainian Academy of Sciences, Kiev (1991). 4. V. K. Dzyadyk and Ya. F. Vasilenko, Application of the Approximation-Iterative Method to the Solution of Stiff Problems for Ordinary Differential Equation [in Russian], Preprint No. 55, Institute of Mathematics, Ukrainian Academy of Sciences, Kiev (1991). 5. V. K. Dzyadyk, Approximation Methods for Solutions of Differential and Integral Equations, VSP, Netherlands, Japan (1995). 6. L. E. Él’sgol’ts, Differential Equations and Calculus of Variations [in Russian], Mir, Moscow (1970). 7. M. M. Rizk, “Expansions for the fundamental Hermite interpolation polynomials in terms of Chebyshev polynomials,” Ukr. Math. Zh., 53, No. 1, 135–143 (2001).
