Piecewise complementary Lidstone interpolation and error inequalities

Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

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Piecewise complementary Lidstone interpolation and error inequalities Ravi P. Agarwal a,b,∗ , Patricia J.Y. Wong c a

Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901-6975, USA

b

Mathematics and Statistics Department, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia

c

School of Electrical and Electronic Engineering, Nanyang Technological University, 50 Nanyang Avenue, Singapore 639798, Singapore

article

info

Article history: Received 15 January 2010 Received in revised form 26 March 2010 MSC: 41A05 65D05

abstract The purpose of this paper is to develop piecewise complementary Lidstone interpolation in one and two variables and establish explicit error bounds for the derivatives in L∞ and L2 norms. © 2010 Elsevier B.V. All rights reserved.

Keywords: Complementary Lidstone interpolating polynomial Piecewise interpolation Error analysis

1. Introduction Polynomial interpolation has several attractive features, however, polynomial interpolation of a given function often has the drawback of producing approximations that may be wildly oscillatory. To overcome this difficulty, we divide the interval of interest into small subintervals and in each subinterval we construct polynomials of relatively low degree and finally ‘piece together’ these polynomials — therefore the term ‘piecewise interpolation’. This subject has steadily developed over the past sixty years, and at present there are thousands of research papers on piecewise polynomial interpolation and their applications, see [1–4]. Complementary Lidstone interpolation was very recently introduced in [5] and drawn on by Agarwal, Pinelas and Wong in [6], where they considered an odd order differential equation

(−1)m x(2m+1) (t ) = f (t , x, x0 , . . . , x(q) ),

t ∈ (0, 1), m ≥ 1

(1.1)

(0 ≤ q ≤ 2m is fixed) together with boundary data at the odd order derivatives x(0) = α0 ,

x(2k−1) (0) = αk ,

x(2k−1) (1) = βk ,

1 ≤ k ≤ m.

(1.2)

The boundary conditions (1.2) are known as complementary Lidstone boundary conditions. In the field of approximation theory, complementary Lidstone interpolating polynomial of degree 2m satisfies complementary Lidstone conditions P (0) = x(0),

∗

P (2k−1) (0) = x(2k−1) (0),

P (2k−1) (1) = x(2k−1) (1),

1 ≤ k ≤ m.

Corresponding author at: Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901-6975, USA. E-mail addresses: [email protected] (R.P. Agarwal), [email protected] (P.J.Y. Wong).

0377-0427/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2010.03.029

(1.3)

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Complementary Lidstone interpolation naturally complements Lidstone interpolation [7–9] that involves even order derivatives. To be precise, the Lidstone interpolating polynomial of degree (2m − 1) satisfies the Lidstone conditions P (2k) (0) = x(2k) (0), P (2k) (1) = x(2k) (1), 0 ≤ k ≤ m − 1. Further, the complementary Lidstone boundary value problem (1.1), (1.2) complements the Lidstone boundary value problem that consists of the even order differential equation x(2m) (t ) = f (t , x, x0 , . . . , x(q) ) (0 ≤ q ≤ 2m − 1 is fixed) and the Lidstone boundary conditions x(2k) (0) = αk , x(2k) (1) = βk , 0 ≤ k ≤ m − 1. Lidstone interpolation has a long history since 1929 when Lidstone [10] introduced a generalization of Taylor’s series that approximates a given function in the neighborhood of two points instead of one. In terms of completely continuous functions it has been characterized in [11–18], and others. Further, the Lidstone boundary value problem and several of its particular cases have been the subject matter of several investigations [19,7,20–32]. Other work on piecewise Lidstone interpolation and Lidstone splines can be found in [33,21,34,5,35–38]. Motivated by all these related works, in this paper we shall provide explicit representations of piecewise complementary Lidstone interpolates in one and two independent variables, and obtain the error bounds for the derivatives in L∞ and L2 norms. The outline of the paper is as follows. In Section 2, we list the notations used and state some preliminary results from [6]. One-variable piecewise complementary Lidstone interpolation is developed in Section 3 and thereafter we establish the error inequalities in L∞ and L2 norms. Finally, in Section 4 we discuss two-variable piecewise complementary Lidstone interpolation and derive the related error inequalities in L∞ and L2 norms. 2. Preliminaries We shall list the notations used in this paper. Let −∞ < a < b < ∞ and −∞ < c < d < ∞. We let ∆ : a = t0 < t1 < · · · < tN +1 = b and ∆0 : c = u0 < u1 < · · · < uM +1 = d denote uniform partitions of [a, b] a c and [c , d] with step sizes h = Nb− and ` = Md− , respectively. Further, we let ρ = ∆ × ∆0 be a rectangular partition of +1 +1

[a, b] × [c , d]. For the functions x(t ) and f (t , u) and each positive integer r, we shall denote Dr x = ddt rx , Drt f = ∂dt rf and ∂r f Dru f = ∂ ur . For each positive integer r and for each p, 1 ≤ p ≤ ∞, we will let PC r ,p [a, b] be the set of all real-valued functions x(t ) such that (i) x(t ) is (r − 1) times continuously differentiable on [a, b], (ii) there exist si , 0 ≤ i ≤ L + 1 with a = s0 < s1 < · · · < sL+1 = b such that on each open subinterval (si , si+1 ), 0 ≤ i ≤ L, Dr −1 x is continuously differentiable, r

r

and (iii) the Lp -norm of Dr x is finite, i.e., r

kD x kp =

L Z X

si+1

! 1p |D x(t )| dt r

p

< ∞,

1≤p<∞

si

i =0

and when p = ∞,

kDr xk∞ = max

sup

0≤i≤L t ∈(s ,s i i+1 )

|Dr x(t )| < ∞.

Similarly, for each positive integer r and for each p, 1 ≤ p ≤ ∞, we will let PC r ,p ([a, b] × [c , d]) be the set of all real-valued µ functions f (t , u) such that (i) f (t , u) is (r − 1) times continuously differentiable, i.e., Dt Dνu f , 0 ≤ µ + ν ≤ r − 1 exists and is continuous on [a, b] × [c , d], (ii) there exist si , 0 ≤ i ≤ L + 1 and zj , 0 ≤ j ≤ R + 1 with a = s0 < s1 < · · · < sL+1 = b and c = z0 < z1 < · · · < zR+1 = d such that on each open subrectangle (si , si+1 ) × (zj , zj+1 ), 0 ≤ i ≤ L, 0 ≤ j ≤ R µ and for all 0 ≤ µ ≤ r − 1, 0 ≤ ν ≤ r − 1 such that µ + ν = r − 1, Dt Dνu f is continuously differentiable, and (iii) for all µ ν 0 ≤ µ ≤ r , 0 ≤ ν ≤ r such that µ + ν = r the Lp -norm of Dt Du f is finite, i.e., µ ν

kDt Du f kp =

L X R Z X i=0 j=0

si+1

zj+1

Z

si

µ ν

! 1p

|Dt Du f (t , u)| dudt p

< ∞,

1≤p<∞

zj

and when p = ∞, µ

kDt Dνu f k∞ = max 0≤i≤L 0≤j≤R

sup

(t ,u)∈(si ,si+1 )×(zj ,zj+1 )

µ

|Dt Dνu f (t , u)| < ∞. µ

We will also need the set PC r1 ,r2 ,p ([a, b] × [c , d]) of all real-valued functions f (t , u) such that (i) Dt Dνu f , 0 ≤ µ ≤ r1 − 1, 0 ≤ ν ≤ r2 − 1 exists and is continuous on [a, b]×[c , d], (ii) on each open subrectangle (si , si+1 )×(zj , zj+1 ), 0 ≤ i ≤ L, 0 ≤ j ≤ R µ and for all 0 ≤ µ ≤ r1 , 0 ≤ ν ≤ r2 , Dt Dνu f exists and is continuous, and (iii) for all 0 ≤ µ ≤ r1 , 0 ≤ ν ≤ r2 the Lp -norm of µ ν Dt Du f is finite. We shall also need the following fundamental results. Lemma 2.1 (Wirtinger’s Inequality [33]). If x(t ) ∈ PC 1,2 [a, b] and x(a) = x(b) = 0, then b

Z

x2 (t )dt ≤ a

(b − a)2 π2

b

Z

[Dx(t )]2 dt . a

Moreover, in the above relation equality holds if and only if x(t ) = c sin[π (x − a)/(b − a)], where c is an arbitrary constant.

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

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Corollary 2.2 ([33]). If x(t ) ∈ PC 2,2 [a, b] and x(a) = x(b) = 0, then b

Z

x2 (t )dt ≤ a

(b − a)4 π4

Z

b

[D2 x(t )]2 dt .

a

Moreover, in the above relation equality holds if and only if x(t ) = c sin[π (x − a)/(b − a)], where c is an arbitrary constant. Lemma 2.3 (Jensen’s Inequality [33]). Let 0 < q < p and let ai , 1 ≤ i ≤ n be nonnegative numbers. Then n X

! 1p p ai

n X

≤

i =1

! 1q q ai

.

i=1

Definition 2.1. For a given function x : C (2m+1) [0, 1] → R, let P2m (t ) be the interpolating polynomial of degree 2m satisfying the complementary Lidstone conditions (2k−1)

P2m (0) = x(0),

P2m

(2k−1)

(0) = x(2k−1) (0),

P2m

(1) = x(2k−1) (1),

1 ≤ k ≤ m.

(2.1)

Then, P2m (t ) is known as the complementary Lidstone interpolating polynomial of x. Theorem 2.4 ([5,6]). Let x ∈ C (2m+1) [0, 1]. Then, x(t ) = P2m (t ) + R(t ),

(2.2)

where P2m (t ) is the complementary Lidstone interpolating polynomial of degree 2m and R(t ) is the residue term given respectively by P2m (t ) = x(0) +

m X 

x(2k−1) (0)(vk (1) − vk (1 − t )) + x(2k−1) (1)(vk (t ) − vk (0))



(2.3)

k=1

and R(t ) =

1

Z

hm (t , s)x(2m+1) (s)ds.

(2.4)

0

Here

 m X (1 − s)2m−2k+1   (vk (t ) − vk (0)) ,  Z t − (2m − 2k + 1)! k =1 hm ( t , s ) = gm (τ , s)dτ = m X  s2m s2m−2k+1 0   + (vk (1 − t ) − vk (1)) ,  (2m)! k=1 (2m − 2k + 1)!

t ≤s (2.5) s≤t

and

vk (t ) = Λ0k (t ),

k≥0

(2.6)

where gm (t , s) and Λk (t ) are given by g 1 ( t , s) =

 (t − 1)s, (s − 1)t ,

s≤t t ≤s

gm (t , s) =

1

Z

g1 (t , τ )gm−1 (τ , s)dτ ,

m≥2

0

Λ0 ( t ) = t , Λ00k (t ) = Λk−1 (t ), Λk (0) = Λk (1) = 0, k ≥ 1 Z 1 Z 1 Λk (t ) = gk (t , s)sds, Λk (1 − t ) = gk (t , s)(1 − s)ds, k ≥ 1. 0

Remark 2.1. From (2.6) it is clear that v0 (t ) = 1; vk0 (t ) = Λk−1 (t ), k ≥ 1; 0, k ≥ 2; vk (t ) = 0

v0 (t ) = 1,

(2.7)

0

Rt 0

vk−1 (s)ds, k ≥ 1; and v1 (t ) =

t2 2

1

− , 6

v2 (t ) =

t4 24

−

t2 12

+

7 360

R1 0

vk (s)ds = 0, k ≥ 1; vk0 (0) = 0, k ≥ 1; vk0 (1) =

.

Theorem 2.5 ([6]). Let x ∈ C (2m+1) [0, 1]. Then, (k) |x(k) (t ) − P2m (t )| ≤ C2m+1,k max |x(2m+1) (t )|, t ∈[0,1]

0 ≤ k ≤ 2m

(2.8)

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where




C2m+1,0 = (−1)

m

4 22m+2 − 1

B2m+2

(2m + 2)! (−1)m−k+1 E2m−2k+2 C2m+1,2k−1 = 2m−2k+2 , 1≤k≤m 2 (2m − 2k + 2)!
2 22m−2k+2 − 1 C2m+1,2k = (−1)m−k B2m−2k+2 , 1 ≤ k ≤ m. (2m − 2k + 2)! 1 1 (Note Bj is the jth Bernoulli number and Ej is the jth Euler number, C3,0 = 12 , C3,1 = 81 , C3,2 = 12 , C5,0 = 120 , C5,1 = 5 1 1 1 , C = , C = , C = ). Moreover, the constants C in (2.8) are the best possible, as equalities hold for the 5 ,2 5,3 5,4 2m+1,k 384 24 8 2 function x(t ) = E2m+1 (t ) − E2m+1 (0) (where E2m+1 (t ) is the Euler polynomial of degree (2m + 1)) whose complementary Lidstone interpolating polynomial P2m (t ) ≡ 0, and only for this function up to a constant factor.

3. Piecewise complementary Lidstone interpolation For a fixed ∆, we define the set Lm (∆) = {p(t ) ∈ C [a, b] : p(t ) is a polynomial of degree at most 2m in each subinterval [ti , ti+1 ], 0 ≤ i ≤ N }. It is clear that Lm (∆) is of dimension [(2m + 1)(N + 1) − N ]. Definition 3.1. For a given function x(t ) ∈ C (2m−1) [a, b], we say L∆ m x(t ) is the Lm (∆)-interpolate of x(t ), also known as the complementary Lidstone interpolate of x(t ) if L∆ x ( t ) ∈ L ( ∆ ) with m m L∆ 0 ≤ j ≤ N; m x(tj ) = x(tj ) = xj , (2k−1) (2k−1) D2k−1 L∆ , x ( t ) = x ( t i ) = xi i m

1 ≤ k ≤ m, 0 ≤ i ≤ N + 1.

In view of Theorem 2.4, L∆ m x(t ) exists uniquely and can be expressed explicitly in the subinterval [ti , ti+1 ] as L∆ m x( t ) = xi +

m  X

(2k−1)

xi



vk (1) − vk



ti+1 − t h

k=1



    t − ti + x(i+2k1−1) vk − vk (0) h2k−1 , h

t ∈ [ti , ti+1 ], 0 ≤ i ≤ N .

(3.1)

Therefore, it follows that L∆ m x( t ) =

N X

xi sm,i (t ) +

N +1 X m X

i=0

(2j−1)

xi

rm,i,j (t ),

(3.2)

i=0 j=1

where sm,i (t ), 0 ≤ i ≤ N and rm,i,j (t ), 0 ≤ i ≤ N + 1, 1 ≤ j ≤ m satisfy D2ν−1 sm,i (t ) = 0,

1 ≤ ν ≤ m, t ∈ [tµ , tµ+1 ], 0 ≤ µ ≤ N

(3.3)

and D2ν−1 rm,i,j (tµ ) = δiµ δν j h2j−1 ,

1 ≤ ν ≤ m, 0 ≤ µ ≤ N + 1;

(3.4)

and appear as sm,i (t ) = 1,

= 0,

t ∈ [ti , ti+1 ], 0 ≤ i ≤ N otherwise

(3.5)

and

   ti+1 − t vj (1) − vj h2j−1 , h     t − ti−1 = vj − vj (0) h2j−1 ,

rm,i,j (t ) =

h

= 0,

t ∈ [ti , ti+1 ], 0 ≤ i ≤ N t ∈ [ti−1 , ti ], 1 ≤ i ≤ N + 1

otherwise.

(3.6) (2m−1)

It is clear from Definition 3.1 as well as the representation (3.2) that the function x(t ) need not be in C [a, b], rather (2k−1) 2k−1 ∆ (2k−1) it is sufficient that for the function x(t ), L∆ x ( t ) = x ( t ) = x , 0 ≤ i ≤ N and D L x ( t ) = x ( t ) = x ,1 ≤ k ≤ i i i i i m m i m, 0 ≤ i ≤ N + 1 exist.

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

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Theorem 3.1. Let x(t ) ∈ PC 2m+1,∞ [a, b]. Then 2m+1−k kDk (x − L∆ kD2m+1 xk∞ , m x)k∞ ≤ C2m+1,k h

Proof. This follows from Theorem 2.5.

0 ≤ k ≤ 2m.

(3.7)



Remark 3.1. It is clear from Theorem 2.5 that the constants C2m+1,k are the best possible in the inequalities (3.7). Theorem 3.2. Let x(t ) ∈ PC 2m−1,∞ [a, b]. Then 2m−1−k kDk (x − L∆ max m x)k∞ ≤ C2m−1,k h

sup

0≤i≤N t ∈(t ,t i i+1

    2m−1 ti+1 − t t − ti (2m−1) (2m−1) D x ( t ) − x − x i i +1 h h )

≤ 2C2m−1,k h2m−1−k kD2m−1 xk∞ ,

0 ≤ k ≤ 2m − 1.

(3.8) (3.9)

Proof. Without loss of generality, we let a = 0, b = 1 and h = 1 so that (3.1) gives L∆ m x(t ) = x(0) +

m X 

x(2k−1) (0)(vk (1) − vk (1 − t )) + x(2k−1) (1)(vk (t ) − vk (0)) .



k=1

Therefore, from Theorem 2.4 for any x(t ) ∈ PC 2m+1,1 [0, 1] it follows that ∆

x(t ) − Lm x(t ) =

1

Z

hm (t , s)D2m+1 x(s)ds.

(3.10)

0

Using integration by parts in the right side of (3.10), and noting from (2.5) that hm (t , 0) = hm (t , 1) = 0, we get x(t ) − L∆ m x(t ) = −

1

Z

h¯ m (t , s)D2m x(s)ds,

(3.11)

0

where

 m X (1 − s)2m−2k   , (v ( t ) − v ( 0 ))  k k (2m − 2k)! ∂ hm (t , s)  k=1 h¯ m (t , s) = = m X  s2m−2k s2m−1 ∂s   + (vk (1 − t ) − vk (1)) ,  (2m − 1)! k=1 (2m − 2k)!

t ≤s (3.12) s ≤ t.

It can be checked that h¯ m (t , 0) = vm (1 − t ) − vm (1),

∂ h¯ m (t , s) = hm−1 (t , s). ∂s

h¯ m (t , 1) = vm (t ) − vm (0),

(3.13)

Moreover, in view of (2.5)–(2.7) we find h¯ m (t , 0) = vm (1 − t ) − vm (1) = Λ0m (1 − t ) − Λ0m (1) = −

1

Z

Λ00m (s)ds 1 −t

1

Z

Λm−1 (s)ds = −

=− 1−t 1

Z

t

Z

Λm−1 (1 − s)ds = −

Z t Z

0 t

Z



gm−1 (s, τ )ds (1 − τ )dτ = −

=− 0

0

gm−1 (s, τ )(1 − τ )dτ

 ds

0

1

Z

0

1

hm−1 (t , τ )(1 − τ )dτ

(3.14)

0

and h¯ m (t , 1) = vm (t ) − vm (0) = Λ0m (t ) − Λ0m (0) =

t

Z

Λ00m (s)ds 0

t

Z

Λm−1 (s)ds =

= 0

0 1

Z

t

Z

1

gm−1 (s, τ )τ dτ

0

 ds

0



gm−1 (s, τ )ds τ dτ =

= 0

Z t Z

1

Z

hm−1 (t , τ )τ dτ . 0

Now, integrating by parts the right side of (3.11) and also noting (3.13)–(3.15) gives

(3.15)

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∂ h¯ m (t , s) 2m−1 D x(s)ds ∂s 0 Z 1 hm−1 (t , s)D2m−1 x(s)ds = −(vm (t ) − vm (0))x(2m−1) (1) + (vm (1 − t ) − vm (1))x(2m−1) (0) +

(2m−1) ¯ x(t ) − L∆ (1) + h¯ m (t , 0)x(2m−1) (0) + m x(t ) = −hm (t , 1)x

1

Z

0 1

Z =

  hm−1 (t , s) −sx(2m−1) (1) − (1 − s)x(2m−1) (0) + D2m−1 x(s) ds.

0

Therefore, we have for t ∈ [0, 1] and 0 ≤ k ≤ 2m − 1,

k ∂ hm−1 (t , s) ds · sup D2m−1 x(t ) − tx(2m−1) (1) − (1 − t )x(2m−1) (0) k ∂t t ∈(0,1) 0 Z 1 k ∂ hm−1 (t , s) ds · sup D2m−1 x(t ) [1 + t + (1 − t )] ≤ k ∂ t t ∈(0,1) 0 Z 1 k ∂ hm−1 (t , s) ds · sup D2m−1 x(t ) . =2 k ∂t t ∈(0,1) 0 Z

|Dk (x(t ) − L∆ m x(t ))| ≤

1

(3.16)

(3.17)

Note that

Z max

t ∈[0,1]

0

1

k Z ∂ hm−1 (t , s) ds = max t ∈[0,1] ∂tk

1 0

k−1 ∂ gm−1 (t , s) ds = C2m−1,k , ∂ t k−1

0 ≤ k ≤ 2m − 1

where the last equality follows from [7]. The proof is complete by substituting (3.18) into (3.16) and (3.17).

(3.18) 

Remark 3.2. For each k the inequality (3.8) is the best possible, as equality holds for the function x(t ) = E2m−1 (t )− E2m−1 (0) for a = 0, b = 1 and h = 1, whose complementary Lidstone interpolate is L∆ m (E2m−1 (t ) − E2m−1 (0)) = vm (1) − vm (1 − t ) + vm (t ) − vm (0), and only for this function up to a constant factor. We also note that the inequality connecting the right sides of (3.8) and (3.9) is also the best possible. For this, it suffices to note that for the continuous function x(t ) =

  −1 + 4t ,  3 − 4t ,

0≤t ≤ 1 2

1 2

≤t≤1

the equality maxt ∈[0,1] |x(t )−(1 − t )x(0)− tx(1)| = 2 maxt ∈[0,1] |x(t )| holds. However, as such the sharpness of (3.9) remains undecided. Theorem 3.3. Let x(t ) ∈ PC 2m,∞ [a, b], 1 ≤ m ≤ 3. Then 2m−k kDk (x − L∆ kD2m xk∞ , m x)k∞ ≤ ζ2m,k h

0 ≤ k ≤ 2m − 1

(3.19)

where the constants ζ2m,k are given in the following table. k

m=1

m=2

m=3

0

1 4

7817 600 000

1 ∗ 240

1

1 2

1 24

121 3 51 840

√

2 3

2 3 27 1 2

√ √

√

1

2(3+ 30)(225−30 30) 2 10 125 1 24

√

4

2 3 27

5

1 2

All ζ2m,k are exact maxima except ζ6,0 which is indicated with ∗ . Proof. Without loss of generality, we let a = 0, b = 1 and h = 1. Then, it follows from (3.11) that

|Dk (x − L∆ m x)(t )| ≤

1

Z 0

k  ∂ h¯ m (t , s) ds kD2m xk∞ , ∂tk

0 ≤ k ≤ 2m − 1

(3.20)

¯ where hm (t , s) is defined in (3.12). The constants ζ2m,k are obtained by maximizing (if possible) or giving an upper bound of R 1 ∂ k h¯ m (t ,s) ds over t ∈ [0, 1]. Indeed, all ζ2m,k given in the table are exact maxima except ζ6,0 . 0 ∂tk

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

Case m = 1: From (3.12) and the explicit expressions of some vk (t ) in Remark 2.1, we obtain h¯ 1 (t , s) =

 2  t , 2

t ≤s 1

 s + [(1 − t )2 − 1],

s≤t

2

and then

 2 t   ,   2 1 |h¯ 1 (t , s)| = −s − [(1 − t )2 − 1],  2    s + 1 [(1 − t )2 − 1],

t ≤s≤1 0 ≤ s ≤ t1 t1 ≤ s ≤ t

2

[1 − (1 − t )2 ] ∈ [0, t ] for all t ∈ [0, 1]. Subsequently, we find   Z 1 Z t1  Z t Z 1 1 2 2 ¯ −s − [(1 − t ) − 1] ds + |h1 (t , s)|ds = s + [(1 − t ) − 1] ds +

where t1 =

1 2

2

0

0

=

t 2 (t − 2)2 4

2

t

= p2,0 (t ).

Clearly, maxt ∈[0,1] p2,0 (t ) = p2,0 (1) =

 ∂ h¯ 1 (t , s) t, = −(1 − t ), ∂t

t1

= ζ2,0 . Next, we have

1 4

t
Hence,

Z t Z 1 ∂ h¯ 1 (t , s) ds = (1 − t )ds + tds = 2t (1 − t ) = p2,1 (t ) ∂t 0 0 t
and maxt ∈[0,1] p2,1 (t ) = p2,1 12 = 21 = ζ2,1 . Case m = 2: We have  2 t4 t    [3(1 − s)2 − 1] + = A, 24 h¯ 2 (t , s) = 12 2 2  s (1 − t ) 1   (4s − 6) + (3s2 − 1) + [(1 − t )4 + 1] = B, Z

1

24

12

24

For t ≤ s, we find

   A,        |h¯ 2 (t , s)| = −A,          −A, where t2 = 1 −



2−t 2 6

 12

   B,        |h¯ 2 (t , s)| = B,          −B,

" s ≤ t2 , t ∈

0,

" s ≥ t2 , t ∈

" t ∈

0,

√ 6−

8

7

√ #

6−

8

7

√ #

6−

8

7

# ,1

h √ i ∈ [t , 1] for t ∈ 0, 6−7 8 . For s ≤ t, we obtain " √ # 6− 8 t ∈ 0, 7

" s ≤ t3 , t ∈

8

7

" s ≥ t3 , t ∈

√ 6−

√ 6− 7

8

# ,1 # ,1

1

t ≤s s ≤ t.

t2 2

ds

2549

2550

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

h

√

i , 1 . Taking into account all the above gives "  √ # 6− 8   1 t 2 (2 − t 2 )(12 − 6t 2 ) 21 = p  t ∈ 0, 4,0,a (t ),  Z 1  108 7 ¯ " # |h2 (t , s)|ds = √ Z Z t t3  0  6− 8  1 t 3 (1 − t )(4 − 3t ) +  Bds = p4,0,b (t ), t ∈ Bds − ,1 

where t3 is the root of the equation B(s) = 0 in [0, t ] for t ∈

24

≤

6− 8 7

6−

√ ! 8

7

7

t3

0

    p   4,0,a

" = 0.0112,

0,

t ∈

"     p4,0,b (1) = ζ4,0 ,

t ∈

6−

√ # 8

7

√ 6−

8

7

# ,1

≤ ζ4,0 . Using a similar technique, we get

Z



 2 2 1    t (1 − t )(1 + t )(3 − 3t ) 2 = p4,1,a (t ), 1 ¯ ∂ h2 (t , s) 27 ∂ t ds =  2 1 0   t (t − 1)(t − 2)(6t − 3t 2 ) 2 = p4,1,b (t ),

t ∈ 0,

 t ∈

27

   1   p = ζ4,1 , 4 , 1 , a  2   ≤  1  p4,1,b = ζ4,1 , 2



t ∈ 0,

 t ∈

1 2

1

1 2

1



2

 ,1



2



,1

= ζ4,1 . Next, we find

" r #  1 2 2 2 21   − (−1 + 3t )(3 − 9t ) = p4,2,a (t ), t ∈ 0, 1 −   27 3           4 4 2 2 2 1 1   − + t − t 2 (−6 + 18t − 9t 2 ) 2 + Z 1 2¯ − t (3 − 9t 2 ) 2 = p4,2,b (t ),  ∂ h2 (t , s) 27 " 9 r9 r # 27 9 ∂ t 2 ds =  1 1 0   t ∈ 1− ,   3 3   # "r     1 1 2 2 12  − (4 − 12t + 6t )(−6 + 18t − 9t ) = p4,2,c (t ), t ∈ ,1 27

3

   p4,2,a (0) = ζ4,2 ,      r !   1 ≤ p4,2,b = 0.0406,  3         p4,2,c (1) = ζ4,2 ,

r #

"

1

0, 1 −

t ∈

" t ∈

r 1−

"r t ∈

1 3

3 1 3

r # ,

1 3

# ,1

≤ ζ4,2 . ∂3

Finally, we have ∂ t 3 h¯ 2 (t , s) = ∂∂t h¯ 1 (t , s) and so ζ4,3 = ζ2,1 . Case m = 3: From (3.12) we get

 2 t (1 − s)4 t 2 (t 2 − 2)(1 − s)2 t 2 (t 4 − 5t 2 + 7)   + + = Q,    48 720  5 48 4 2 2 4 2 s s [(1 − t ) − 1] s [(1 − t ) − 2(1 − t ) + 1] h¯ 3 (t , s) = + +  120 48 48   6 4 2    + (1 − t ) − 5(1 − t ) + 7(1 − t ) − 3 = R, s ≤ t . 720

t ≤s

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

2551

For t ≤ s, we find

( −Q , |h¯ 3 (t , s)| = Q , Q,

s ≤ t4 , t ∈ [0, α] s ≥ t4 , t ∈ [0, α] t ∈ [α, 1]



q

 12

∈ [t , 1] for t ∈ [0, α], and α = 0.4897 is a root of the equation R1 t − t4 = 0. The above enables us to obtain explicit expression of t |h¯ 3 (t , s)|ds. However, for s ≤ t, the expression of h¯ 3 (t , s) is a quintic polynomial in s, and R(s) =R 0 cannot be solved for s analytically. t Hence, we can only provide an upper bound (but not the explicit expression) for 0 |h¯ 3 (t , s)|ds. Here, it is found that ∂ R(t , s) ≥ 0 for s ∈ [0, t ], subsequently R(t , s) is nondecreasing in s and therefore ∂s where t4 = 1 − 1 − 21 t 2 −

1 2

11 4 t 15

− 83 t 2 +

32 15

max |R(t , s)| = max{|R(t , 0)|, |R(t , t )|}.

s∈[0,t ]

Using the above relation, we get t

Z

|h¯ 3 (t , s)|ds ≤ t · max{|R(t , 0)|, |R(t , t )|}. 0

Note that

|R(t , 0)| = −R(t , 0),

t ∈ [0, 1]



and |R(t , t )| =

−R(t , t ), R(t , t ),

t ∈ [0, α] t ∈ [α, 1].

All the above leads to 1

Z 0

 Z t4 Z 1   Q ds + Q ds, t ∈ [0, α] t · max{−R(t , 0), −R(t , t )} − t4 |h¯ 3 (t , s)|ds ≤ Z 1t   t · max{−R(t , 0), R(t , t )} + Q ds, t ∈ [α, 1] t  Z t4  Z 1    −α · R (α, 0 ) + − Q ds + Q ds = 0.0017, t ∈ [0, α]  t t4 t =α ≤ Z 1     Q ds = ζ6,0 , t ∈ [α, 1] −R(1, 0) + t

t =1

≤ ζ6,0 . Next, using a similar technique as in case 2, we find

Z

   p6,1,a (t ), 1 ¯ ∂ h3 (t , s) ds = ∂t  0  p6,1,b (t ),



t ∈ 0,

 t ∈

1 2

   1   p = ζ6,1 ,  6,1,a 2   ≤  1  p6,1,b = ζ6,1 , 2

1



2



,1



t ∈ 0,

 t ∈

1 2

1



2



,1

= ζ6,1 where p6,1,a (t ) =

p6,1,b (t ) =

1

p

2



15(1 − t )(2t + 1)(6t 2 − 7t + 7)t (t − 1) 50 625 × (36t 7 − 84t 6 − 44t 5 + 56t 4 + 89t 3 − 91t 2 − 56t − 56), 450t − 225t 2 −

p

 12

3015t (t − 2)(3t 2 − 6t + 1)

10125  p × 3t 2 − 6t + 6 + 15t (t − 2)(3t 2 − 6t + 1) .

t (t − 1)(t − 2)

2552

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

We also obtain

    p6,2,a (t ),         Z 1 2¯  ∂ h3 (t , s) ds = p6,2,b (t ), ∂t2  0           p6,2,c (t ),

√

 t ∈ 0, 1 −

2 30

1−

15

√

 t ∈ 1 −

1−

2 30

t ∈ 1−

2 30

√

! 12 , 1−

15

√



! 12  

! 12

2 30 15

! 12  

 , 1

15

√



    p6,2,a (0) = ζ6,2 ,         √ ! 12    2 30  = 0.0010, ≤ p6,2,b  1 −  15           p6,2,c (1) = ζ6,2 ,

t ∈ 0, 1 −

1−

2 30 15

√

 t ∈ 1 −

2 30

1−

t ∈ 1−

2 30

√

! 12 , 1−

15

√



! 21  

! 12

2 30 15

! 12  

 , 1

15

≤ ζ6,2 where 2

p6,2,a (t ) = −

1

[45t 4 + (3t 2 − 1)β1 (t ) − 3][225 − 675t 2 − 30β1 (t )] 2 , 10 125 p6,2,b (t ) = β2 (t ) + β2 (1 − t ),    p6,2,c (t ) = β2 (t ) −

1

120

1

(1 − t ) − (1 − t ) 5

3

6

1

2

2

t −

1

6

− (1 − t )

1

24

4

t −

1 12

2

t +



7 360

;

here 1

β1 (t ) = (450t 4 − 225t 2 + 30) 2 ,   1 5 1 1 1 1 5 β2 (t ) = − t − t3 (1 − t )2 − + [β3 (t )] 2 120 6 2 6 h 60  i 1 p 1 1 1 7 1 3 2 (1 − t ) − + 2 β3 (t ) − t (1 − t )4 − (1 − t )2 + , + [β3 (t )] 2 3

2

β3 (t ) = 1 − 3(1 − t )2 −

1

6

24



32(1 − t )4 − 16(1 − t )2 +

2

32 15

 12

12

360

.

j+2 j Finally, we have ∂∂t j+2 h¯ 3 (t , s) = ∂∂t j h¯ 2 (t , s), j = 1, 2, 3, therefore ζ6,j+2 = ζ4,j , j = 1, 2, 3. This completes the proof.



Remark 3.3. The sharpness of the inequalities (3.19) remains undecided. However, these inequalities cannot be improved further by using the same technique. Now we shall obtain error bounds for the derivatives of the interpolation error x(t )−L∆ m x(t ) in L2 norm. Given the partition ∆, let h·ip (1 ≤ p < ∞) be defined as

" hxip =

N −1 Z X i=0

ti+1

# 1p |x(s)| ds p

.

ti

Clearly, hxip ≤ kxkp =

hP

N i=0

R ti+1 ti

|x(s)|p ds

i 1p

.

Theorem 3.4. Let x(t ) ∈ PC 2m,2 [a, b]. Then

kDk (x − L∆ m x)k2 ≤

 2m−k h

π

kD2m xk2 ,

1 ≤ k ≤ 2m

(3.21)

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

2553

and

hx − L∆ m xi2 ≤

 2m h

π

hD2m xi2 .

(3.22)

Proof. Using integration by parts and the interpolating conditions (Definition 3.1) it can be shown that for 0 ≤ i ≤ N , ti+1

Z

2 [D2m L∆ m x(t )] dt +

ti+1

Z

2 [D2m x(t ) − D2m L∆ m x(t )] dt

ti

ti ti+1

Z =

[D2m x(t )]2 dt + 2

ti+1

= ti+1

Z

ti+1 2m ∆ [D2m x(t )]2 dt + 2 D2m−1 [x(t ) − L∆ Lm x(t ) t − 2 m x(t )]D i

ti

=

2m ∆ D2m [L∆ Lm x(t )dt m x(t ) − x(t )]D

ti

ti

Z

ti+1

Z

Z

ti+1

2m+1 ∆ D2m−1 [x(t ) − L∆ Lm x(t )dt m x(t )]D

ti

[D2m x(t )]2 dt .

(3.23)

ti

Summing (3.23) from i = 0 to N and then taking the square root yields 2m kD2m (x − L∆ x k2 , m x)k2 ≤ kD

i.e., (3.21) when k = 2m. Now, for a fixed i such that 0 ≤ i ≤ N, we have D2j−1 [x(tµ ) − L∆ m x(tµ )] = 0 for µ = i, i + 1 and 1 ≤ j ≤ m. Hence, applying Corollary 2.2 and Lemma 2.1 successively it follows that ti+1

Z

2 {D2j−1 [x(t ) − L∆ m x(t )]} dt ≤

 4 Z

ti+1

h

π

ti

2 {D2j+1 [x(t ) − L∆ m x(t )]} dt

ti

 8 Z h

ti+1

2 {D2j+3 [x(t ) − L∆ m x(t )]} dt π ti ······  4(m−j) Z ti+1 h 2 {D2m−1 [x(t ) − L∆ ≤ m x(t )]} dt π ti  2(2m−2j+1) Z ti+1 h 2 {D2m [x(t ) − L∆ ≤ m x(t )]} dt π ti  2(2m−2j+1) Z ti+1 h ≤ [D2m x(t )]2 dt , π ti

≤

(3.24)

where in the last inequality we have used (3.23). Set k = 2j − 1, 1 ≤ j ≤ m.Summing (3.24) from i = 0 to N and then taking the square root leads to (3.21) when k = 1, 3, . . . , 2m − 1. Next, for a fixed i such that 0 ≤ i ≤ N and 1 ≤ j ≤ m − 1, we find, by using integration by parts and the Cauchy–Schwartz inequality, ti+1

Z ti

ti+1 2 2j ∆ 2j−1 {D2j [x(t ) − L∆ [x(t ) − L∆ m x(t )]} dt = D [x(t ) − Lm x(t )]D m x(t )] t i

ti+1

Z −

2j+1 D2j−1 [x(t ) − L∆ [x(t ) − L∆ m x(t )]D m x(t )]dt

ti ti+1

Z =−

2j+1 D2j−1 [x(t ) − L∆ [x(t ) − L∆ m x(t )]D m x(t )]dt

ti ti+1

Z ≤

2 [D2j−1 (x(t ) − L∆ m x(t ))] dt

ti

h

h

π

π

 2(2m−2j) Z =

ti+1

h

π

ti+1

2 [D2j+1 (x(t ) − L∆ m x(t ))] dt

 12

ti

 2m−2j+1  2m−2j−1 Z ≤

 12 Z

ti

ti+1

[D2m x(t )]2 dt

ti

[D2m x(t )]2 dt ,

(3.25)

2554

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

where (3.24) is used in the last inequality. Set k = 2j, 1 ≤ j ≤ m − 1. Summing (3.25) from i = 0 to N and then taking the square root gives (3.21) when k = 2, 4, . . . , 2m − 2. Finally, for a fixed i such that 0 ≤ i ≤ N − 1, we have x(tµ ) − L∆ m x(tµ ) = 0 for µ = i, i + 1. Hence, applying Corollary 2.2 and (3.25) successively we find ti+1

Z

2 [x(t ) − L∆ m x(t )] dt ≤

 4 Z

ti+1

h

π

ti

2 {D2 [x(t ) − L∆ m x(t )]} dt

ti

 4  2(2m−2) Z ≤

h

π  2(2m) Z

=

ti+1

h

π

ti+1

h

π

[D2m x(t )]2 dt

ti

[D2m x(t )]2 dt .

(3.26)

ti

Summing (3.26) from i = 0 to (N − 1) and then taking the square root immediately leads to (3.22).



Corollary 3.5. Let x(t ) ∈ PC 2m,2 [a, b]. Then

hDk (x − L∆ m x)i2 ≤

 2m−k h

π

hD2m xi2 ,

0 ≤ k ≤ 2m.

(3.27)

Proof. In the proof of Theorem 3.4, we sum (3.23)–(3.25) from i = 0 to (N − 1) (instead of N) and then take the square root, this gives (3.27) when 1 ≤ k ≤ 2m. The case k = 0 is simply (3.22).  Theorem 3.6. Let x(t ) ∈ PC 2m+1,2 [a, b]. Then

kDk (x − L∆ m x)k2 ≤

 2m+1−k h

π

kD2m+1 xk2 ,

1 ≤ k ≤ 2m + 1

(3.28)

and

hx − L∆ m xi2 ≤

 2m+1 h

π

hD2m+1 xi2 .

(3.29)

Proof. The inequality (3.28) for k = 2m + 1 is obvious from the fact that L∆ m x(t ) is a polynomial of degree 2m in each subinterval [ti , ti+1 ], 0 ≤ i ≤ N . Now, for a fixed i such that 0 ≤ i ≤ N and 1 ≤ j ≤ m, we apply Corollary 2.2 repeatedly (as in (3.24)) to get ti+1

Z

2 {D2j−1 [x(t ) − L∆ m x(t )]} dt ≤

 4(m−j) Z

ti+1

h

π

ti

 4(m−j+1) Z

ti+1

h

≤

2 {D2m−1 [x(t ) − L∆ m x(t )]} dt

ti

π

2 {D2m+1 [x(t ) − L∆ m x(t )]} dt

ti

 2[2m+1−(2j−1)] Z

ti+1

h

=

π

[D2m+1 x(t )]2 dt .

(3.30)

ti

Set k = 2j − 1, 1 ≤ j ≤ m. Summing (3.30) from i = 0 to N and then taking the square root leads to (3.28) when k = 1, 3, . . . , 2m − 1. Next, for a fixed i such that 0 ≤ i ≤ N and 1 ≤ j ≤ m, we obtain, as in (3.25), ti+1

Z

2 {D2j [x(t ) − L∆ m x(t )]} dt ≤

ti

Z

ti+1

2 [D2j−1 (x(t ) − L∆ m x(t ))] dt

 21 Z

ti

h

h

π

π

 2(2m+1−2j) Z =

ti+1

h

π

2 [D2j+1 (x(t ) − L∆ m x(t ))] dt

 12

ti

 2m+1−(2j−1)  2m+1−(2j+1) Z ≤

ti+1

ti+1

[D2m+1 x(t )]2 dt

ti

[D2m+1 x(t )]2 dt ,

(3.31)

ti

where we have used (3.30) in the last inequality. Set k = 2j, 1 ≤ j ≤ m. Summing (3.31) from i = 0 to N and then taking the square root gives (3.28) when k = 2, 4, . . . , 2m.

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

2555

Finally, for a fixed i such that 0 ≤ i ≤ N − 1, we apply Corollary 2.2 and (3.31) successively to get ti+1

Z

 4 Z

ti+1

h

2 [x(t ) − L∆ m x(t )] dt ≤

π

ti

2 {D2 [x(t ) − L∆ m x(t )]} dt

ti

 4  2(2m+1−2) Z ≤

h

h

π

π

[D2m+1 x(t )]2 dt

ti

 2(2m+1) Z

ti+1

h

=

ti+1

π

[D2m+1 x(t )]2 dt .

(3.32)

ti

Summing (3.32) from i = 0 to (N − 1) and then taking the square root immediately leads to (3.29).



Corollary 3.7. Let x(t ) ∈ PC 2m+1,2 [a, b]. Then ∆

hD (x − Lm x)i2 ≤ k

 2m+1−k h

π

hD2m+1 xi2 ,

0 ≤ k ≤ 2m + 1.

(3.33)

Proof. In the proof of Theorem 3.6, we sum (3.30) and (3.31) from i = 0 to (N − 1) (instead of N) and then take the square root, this gives (3.33) when 1 ≤ k ≤ 2m. The case k = 2m + 1 is obvious and the case k = 0 is just (3.29).  Theorem 3.8. Let p, q ≥ 1, r = max{p, q, 2}, and x(t ) ∈ PC 2m+j,r [a, b] where j = 0, 1. (a) For 1 ≤ p ≤ 2 ≤ q, we have ∆

kD (x − Lm x)kp ≤ k

 2m+j−k h

π

1

1

(b − a) p − q kD2m+j xkq ,

1 ≤ k ≤ 2m + j

(3.34)

and

 2m+j

∆

h

hx − Lm xip ≤

1

1

(tN − a) p − q hD2m+j xiq .

π

(3.35)

(b) For p ≥ 2, we have

kDk (x − L∆ m x)kp ≤

1 2

1

h2

+ 1p

 2m+j−k−1 h

π

kD2m+j xk2 ,

k = 1, 3, . . . , 2m − 1

(3.36)

and

hx − L∆ m xip ≤

1 2

1

h2

+ 1p

 2m+j−1 h

π

hD2m+j xi2 .

(3.37)

Proof. (a) For any function ω(t ) ∈ PC 0,q [a, b] and q ≥ p ≥ 1, Hölder’s inequality gives 1

1

kωkp ≤ (b − a) p − q kωkq .

(3.38)

Therefore, for 1 ≤ p ≤ 2 ≤ q, using (3.38), (3.21) and (3.28) we get for 1 ≤ k ≤ 2m + j, 1

1

p − 2 kDk (x − L∆ x)k kDk (x − L∆ 2 m x)kp ≤ (b − a) m  2m+j−k 1 1 h ≤ ( b − a) p − 2 kD2m+j xk2 π  2m+j−k 1 1 1 1 h ≤ ( b − a) p − 2 (b − a) 2 − q kD2m+j xkq π  2m+j−k 1 1 h = ( b − a) p − q kD2m+j xkq , π

which is (3.34). The proof of (3.35) is similar, here we use (3.22), (3.29) and an analogy of (3.38) given by hωip ≤ 1

1

(tN − a) p − q hωiq . (b) Let ω(t ) ∈ PC 1,q [a, b] satisfy ω(ti ) = 0, 0 ≤ i ≤ N + 1. Then, we have for t ∈ (ti , ti+1 ), Z t  Z ti+1 1 ω(t ) = Dω(s)ds − Dω(s)ds , 2

ti

t

2556

R.P. Agarwal, P.J.Y. Wong / Journal of Computational and Applied Mathematics 234 (2010) 2543–2561

which implies ti+1

Z

1

|ω(t )| ≤

2

|Dω(s)|ds.

ti

Now an application of Hölder’s inequality provides 1

|ω(t )| ≤ where zi =

R

ti+1 ti

" N Z X i =0

2

h

ti+1

1− 1q

zi ,

|Dω(t )|q dt

 1q

. For p ≥ 1, it follows that

# 1p |ω(t )| dt p

≤

ti

1 2

h

! 1p

N X

1− 1q + 1p

p zi

.

i=0

Applying Lemma 2.3, we obtain

kωkp ≤

1 2

1− 1q + 1p

h

N X

! 1q q zi

1

=

2

i =0

h

1− 1q + 1p

kDωkq ,

1 ≤ q ≤ p.

(3.39)

Now, substitute q = 2 and ω = Dk (x − L∆ m x), k = 1, 3, . . . , 2m − 1 in (3.39), we get for p ≥ 2,

kDk (x − L∆ m x)kp ≤

1 2

1

h2

+ 1p

kDk+1 (x − L∆ m x)k2 ≤

1 2

1

h2

+ 1p

 2m+j−k−1 h

π

kD2m+j xk2 ,

where we have used (3.21) and (3.28) in the last inequality. This proves (3.36). The proof of (3.37) is similar, here let ω(t ) ∈ PC 1,q [a, tN ] satisfy ω(ti ) = 0, 0 ≤ i ≤ N. Then, analogous to (3.39) we have

hωip ≤

1 2

h

N −1 X

1− 1q + 1p

! 1q q zi

=

i =0

1 2

h

1− 1q + 1p

hDωiq ,

1 ≤ q ≤ p.

(3.40)

Now, substitute q = 2 and ω = x − L∆ m x in (3.40), we get for p ≥ 2,

hx − L∆ m xip ≤

1 2

1

h2

+ 1p

hD(x − L∆ m x)i2 ≤

1 2

1

h2

+ 1p

 2m+j−1 h

π

hD2m+j xi2 ,

where we have used Corollaries 3.5 and 3.7 in the last inequality. This completes the proof.



Corollary 3.9. Let p, q ≥ 1, r = max{p, q, 2}, and x(t ) ∈ PC 2m+j,r [a, b] where j = 0, 1. (a) For 1 ≤ p ≤ 2 ≤ q, we have

hDk (x − L∆ m x)ip ≤

 2m+j−k h

π

1

1

(tN − a) p − q hD2m+j xiq ,

0 ≤ k ≤ 2m + j.

(3.41)

(b) For p ≥ 2, we have

hDk (x − L∆ m x)ip ≤

1 2

1

h2

+ 1p

 2m+j−k−1 h

π

hD2m+j xi2 ,

k = 0, 1, 3, . . . , 2m − 1.

(3.42)

Proof. (a) For 1 ≤ k ≤ 2m + j, the proof of (3.41) is similar to that of Theorem 3.8(a) with the modification that we use an 1

−1

analogy of (3.38) given by hωip ≤ (tN − a) p q hωiq , as well as Corollaries 3.5 and 3.7 (instead of (3.38), (3.21) and (3.28)). The case k = 0 is simply (3.35). (b) For k = 1, 3, . . . , 2m − 1, the proof of (3.42) is similar to that of Theorem 3.8(b) with the modification that we use (3.40) and Corollaries 3.5 and 3.7 (instead of (3.39), (3.21) and (3.28)). The case k = 0 is simply (3.37). 

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4. Two-variable piecewise complementary Lidstone interpolation For a fixed ρ , we define the set Lm (ρ) as follows: Lm (ρ) = Lm (∆) ⊕ Lm (∆0 ) (the tensor product)

 N +1 m M +1 m N M = Span rm,i,µ (t ), rm,j,ν (u), sm,k (t ), sm,` (u) i=0 µ=1 j=0 ν=1 k=0 `=0 = {p(t , u) ∈ C ([a, b] × [c , d]) : p(t , u) is a two-variable polynomial of degree at most 2m in each variable and in each subrectangle [ti , ti+1 ] × [uj , uj+1 ], 0 ≤ i ≤ N , 0 ≤ j ≤ M }. Since Lm (ρ) is the tensor product of Lm (∆) and Lm (∆0 ), which are of dimensions [(2m + 1)(N + 1) − N ] and [(2m + 1)(M + 1) − M ] respectively, Lm (ρ) is of dimension [(2m + 1)(N + 1) − N ] × [(2m + 1)(M + 1) − M ]. ρ

Definition 4.1. For a given f (t , u) ∈ C (2m−1,2m−1) ([a, b] × [c , d]), we say Lm f (t , u) is the Lm (ρ)-interpolate of f (t , u), also ρ known as the two-variable complementary Lidstone interpolate of f (t , u) if Lm f (t , u) ∈ Lm (ρ) with Lρm f (ti , uj ) = f (ti , uj ) = fi,j , 2µ−1 2ν−1 ρ Du Dt Lm f

(ti , uj ) =

0 ≤ i ≤ N, 0 ≤ j ≤ M;

2µ−1 2ν−1 Du Dt f

(2µ−1,2ν−1)

(ti , uj ) = fi,j

,

1 ≤ µ, ν ≤ m, 0 ≤ i ≤ N + 1, 0 ≤ j ≤ M + 1.

ρ

For f (t , u) ∈ C (2m−1,2m−1) ([a, b] × [c , d]), it is clear that Lm f (t , u) exists uniquely, and in view of (3.2) can be expressed explicitly as Lρm f (t , u) =

+1 X m m M N +1 X X X i=0 µ=1 j=0 ν=1

(2µ−1,2ν−1)

rm,i,µ (t )rm,j,ν (u)fi,j

m X N XX j=0 ν=1 i=0

N X M X

sm,i (t )sm,j (u)fi,j

i=0 j=0

M +1

+

+

(0,2ν−1)

rm,j,ν (u)sm,i (t )fi,j

N +1 X m X M X

+

i=0 µ=1 j=0

(2µ−1,0)

rm,i,µ (t )sm,j (u)fi,j

.

(4.1)

ρ

The following result provides a characterization of Lm f (t , u) in terms of one-variable interpolation schemes. Lemma 4.1. If f (t , u) ∈ C (2m−1,2m−1) ([a, b] × [c , d]), then ∆ ∆ ∆ Lρm f (t , u) = L∆ m Lm f (t , u) = Lm Lm f (t , u). 0

0

(4.2)

Proof. Applying (3.2), we get ∆0 ∆

( N +1 X m X

∆0

Lm Lm f (t , u) = Lm

rm,i,µ (t )

2µ−1 f Dt

(ti , u) +

N X

i=0 µ=1

" m N +1 X m XX X j=0 ν=1

+

i=0 µ=1

" M N +1 X m X X j=0

i=0 µ=1

sm,i (t )f (ti , u)

i=0

M +1

=

)

(2µ−1,2ν−1) rm,i,µ (t )fi,j

+

N X

(0,2ν−1) sm,i (t )fi,j

# rm,j,ν (u)

i =0

(2µ−1,0) rm,i,µ (t )fi,j

+

N X

# sm,i (t )fi,j sm,j (u)

i=0

= Lρm f (t , u), ρ

0

∆ where the last equality is obvious from (4.1). The proof of L∆ m Lm f (t , u) = Lm f (t , u) is similar.



Now let f (t , u) ∈ C (2m−1,2m−1) ([a, b] × [c , d]) be an arbitrary function. From Lemma 4.1, we have ∆ ∆ f − Lρm f = (f − L∆ m f ) + Lm (f − Lm f ) 0

∆

∆

∆0

(4.3) ∆0

∆0

= (f − Lm f ) + [Lm (f − Lm f ) − (f − Lm f )] + (f − Lm f ) ∆0

∆0

∆ ∆ = (f − L∆ m f ) + [Lm (f − Lm f ) − (f − Lm f )] + (f − Lm f ).

We shall use the above relations to obtain the following error estimates.

(4.4) (4.5)

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Theorem 4.2. Let f (t , u) ∈ PC n,n,∞ ([a, b] × [c , d]) where n = 2m − 1, 2m, 2m + 1. Then

kDkt (f − Lρm f )k∞ ≤ αn,k hn−k kDnt f k∞ + βn,k hn−k `n kDnu Dnt f k∞ + γn,k `n kDnu Dkt f k∞ ,

(4.6)

where 0 ≤ k ≤ n if n = 2m − 1, and 0 ≤ k ≤ n − 1 if n = 2m, 2m + 1. The constants αn,k , βn,k and γn,k are given in the following table.

αn,k βn,k γn,k

n = 2m − 1 0 ≤ k ≤ 2m − 1

n = 2m (m = 1, 2, 3) 0 ≤ k ≤ 2m − 1

n = 2m + 1 0 ≤ k ≤ 2m

2C2m−1,k 4C2m−1,k C2m−1,0 2C2m−1,0

ζ2m,k ζ2m,k ζ2m,0 ζ2m,0

C2m+1,k C2m+1,k C2m+1,0 C2m+1,0

(The constants Cn,k and ζn,k are given respectively in Theorems 2.5 and 3.3.) Proof. We shall only prove the case n = 2m − 1 as the proof for other cases is similar. It is noted that as a function of 0 2m−1,∞ t , (f − L∆ [a, b], thus from (4.4) and (3.9), we find for 0 ≤ k ≤ 2m − 1, m f ) ∈ PC k ∆ ∆ ∆ k ∆ kDkt (f − Lρm f )k∞ ≤ kDkt (f − L∆ m f )k∞ + kDt [Lm (f − Lm f ) − (f − Lm f )]k∞ + kDt (f − Lm f )k∞ 0

0

0

0

0

−1 k ∆ ≤ 2C2m−1,k h2m−1−k kD2m f k∞ + 2C2m−1,k h2m−1−k kDt2m−1 (f − L∆ t m f )k∞ + kDt (f − Lm f )k∞ . (4.7) ∆ k Note that as a function of u, for each 0 ≤ k ≤ 2m − 1, Dkt f ∈ PC 2m−1,∞ [c , d] and Dkt L∆ m f = Lm Dt f . Thus, we can apply (3.9) again to obtain 0

0

2m−1 −1 k kDkt (f − L∆ kD2m D t f k∞ , m f )k∞ ≤ 2C2m−1,0 ` u

Now using (4.8) in (4.7) yields (4.6) immediately.

0

0 ≤ k ≤ 2m − 1.

(4.8)



The next theorem gives a result on a special case of Theorem 4.2 when n = 2 (i.e., n = 2m and m = 1). Theorem 4.3. Let f (t , u) ∈ PC 2,2,∞ ([a, b] × [c , d]). Then ρ

kf − L1 f k∞ ≤ ζ2,0 h2 kD2t f k∞ + ζ2,0 h`2 kD2u Dt f k∞ + ζ2,0 `2 kD2u f k∞

(4.9)

and ρ

kDt (f − L1 f )k∞ ≤ ζ2,1 hkD2t f k∞ + ζ2,0 `2 kD2u Dt f k∞ .

(4.10)

Proof. Let x(t ) ∈ C (1) [a, b]. From (3.1), we have for t ∈ [ti , ti+1 ], ∆

L1 x ( t ) = x i +

h

( "

2

≤ kxk∞ + = kxk∞ +



0

xi 1 − h 2

"

ti+1 − t

h 2(t − ti ) 2

h

2 # +

h



1−

ti+1 − t

2

 +

h

x0i+1



t − ti

t − ti

2 )

h

2 #

h

kx0 k∞

kx 0 k∞

≤ kxk∞ + hkx0 k∞ .

(4.11)

Applying (4.3), (4.11) and (3.19) successively, we find ρ

0

∆ ∆ kf − L1 f k∞ ≤ kf − L∆ 1 f k∞ + kL1 (f − L1 f )k∞ 0

0

∆ ∆ ≤ kf − L∆ 1 f k∞ + kf − L1 f k∞ + hkDt (f − L1 f )k∞

≤ ζ2,0 h2 kD2t f k∞ + ζ2,0 `2 kD2u f k∞ + ζ2,0 h`2 kD2u Dt f k∞ which is (4.9). Next, from (4.11) we have for t ∈ [ti , ti+1 ], 0 Dt L∆ 1 x(t ) = xi



ti+1 − t h



+ x0i+1



t − ti h



≤ kx0 k∞ .

(4.12)

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2559

Using (4.3), (4.12) and (3.19) successively gives ρ

0

∆ ∆ kDt (f − L1 f )k∞ ≤ kDt (f − L∆ 1 f )k∞ + kDt L1 (f − L1 f )k∞ 0

∆ ≤ kDt (f − L∆ 1 f )k∞ + kDt (f − L1 f )k∞

≤ ζ2,1 hkD2t f k∞ + ζ2,0 `2 kD2u Dt f k∞ which is (4.10).



Remark 4.1. While we cannot compare (4.9) and (4.6)|n=2 when k = 0, it is clear that (4.10) is sharper than (4.6)|n=2 when k = 1. ρ

Now we shall obtain error bounds for the derivatives of the interpolation error f (t , u) − Lm f (t , u) in L2 norm. Given the partition ρ , let h·ip (1 ≤ p < ∞) be defined as

hf ip =

" N −1 M −1 Z X X

uj+1

Z

# 1p |f (s, τ )| dτ ds

.

p

uj

ti

i =0 j =0

Clearly, hf ip ≤ kf kp =

ti+1

hP

N i=0

PM R ti+1 R uj+1 j =0

ti

uj

|f (s, τ )|p dτ ds

i 1p

.

Theorem 4.4. Let f (t , u) ∈ PC n,n,2 ([a, b] × [c , d]) where n = 2m, 2m + 1. Then for 0 ≤ k ≤ n, ρ

h (f − Lm f )i2 ≤ Dkt

 n−k h

h

π

 n−k  n  n h ` ` n n i + hDu Dt f i2 + hDnu Dkt f i2 . π π π

Dnt f 2

(4.13)

Proof. Following the proof of Theorem 4.2 and using Corollaries 3.5 and 3.7, we find for 0 ≤ k ≤ n, k ∆ ∆ ∆ k ∆ hDkt (f − Lρm f )i2 ≤ hDkt (f − L∆ m f )i2 + hDt [Lm (f − Lm f ) − (f − Lm f )]i2 + hDt (f − Lm f )i2   n −k  n−k h h 0 k ∆0 ≤ hDnt f i2 + hDnt (f − L∆ m f )i2 + hDt (f − Lm f )i2 . π π 0

0

0

(4.14)

Next, applying Corollaries 3.5 and 3.7 again, we get 0

hDkt (f − L∆ m f )i2 ≤

 n ` hDnu Dkt f i2 , π

0 ≤ k ≤ n.

Using (4.15) in (4.14) gives (4.13) immediately.

(4.15)



Theorem 4.5. Let p, q ≥ 1, r = max{p, q, 2}, and f (t , u) ∈ PC 2m+j,2m+j,r ([a, b] × [c , d]) where j = 0, 1. (a) For 1 ≤ p ≤ 2 ≤ q, we have for 0 ≤ k ≤ 2m + j, ρ

h (f − Lm f )ip ≤ Dkt

 2m+j−k h

π

(tN − a)

1 1 p−q

 2m+j 1 1 ` (uM − c ) p − q hDu2m+j Dkt f iq i + π

2m+j Dt f q

h

 2m+j−k  2m+j−1 1 1 1 1 h ` (tN − a) p − q ` 2 + q hDu2m+j Dt2m+j f i2 . 2 π π (b) For p ≥ 2, we have for k = 0, 1, 3, . . . , 2m − 1,  2m+j−k−1  2m+j−1 1 1 1 h 1 1+1 ` +j p 2 hDkt (f − Lρm f )ip ≤ h 2 + p hD2m f i + ` hDu2m+j Dkt f i2 2 t 2 π 2 π  2m+j−k−1  2m+j 1 1 1 h ` + h2+p hDu2m+j Dt2m+j f i2 . 2 π π +

1

(4.16)

(4.17)

Proof. (a) Using a similar technique as in the proof of Theorem 4.2 and (3.41), we obtain for 0 ≤ k ≤ 2m + j, k ∆ ∆ ∆ k ∆ hDkt (f − Lρm f )ip ≤ hDkt (f − L∆ m f )ip + hDt [Lm (f − Lm f ) − (f − Lm f )]ip + hDt (f − Lm f )ip  2m+j−k  2m+j−k 1 1 1 1 h h 0 +j ≤ (tN − a) p − q hD2m f iq + (tN − a) p − q hDt2m+j (f − L∆ t m f )iq π π  2m+j 1 1 ` +j k + (uM − c ) p − q hD2m Dt f iq . u π 0

0

0

(4.18)

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Since q ≥ 2, we apply (3.42) to get 2m+j Dt

h

∆0

(f − Lm f )iq ≤

1 2

`

1 + 1p 2

 2m+j−1 ` +j 2m+j hD2m Dt f i2 . u π

(4.19)

Substituting (4.19) into (4.18) yields (4.16) immediately. (b) Following the proof of part (a) and using (3.42), we find for k = 0, 1, 3, . . . , 2m − 1, k ∆ ∆ k ∆ ∆ hDkt (f − Lρm f )ip ≤ hDkt (f − L∆ m f )ip + hDt [Lm (f − Lm f ) − (f − Lm f )]ip + hDt (f − Lm f )ip  2m+j−k−1  2m+j−k−1 1 1 1 h 1 1+1 h 0 +j hD2m hDt2m+j (f − L∆ ≤ h2+p f i2 + h 2 p t m f )i2 2 π 2 π  2m+j−1 1 1 1 ` +j k + `2+p hD2m Dt f i2 . u 2 π 0

0

0

(4.20)

Next, applying Corollaries 3.5 and 3.7 we get 2m+j Dt

h

 2m+j ` +j 2m+j (f − Lm f )i2 ≤ hD2m Dt f i2 . u π ∆0

(4.21)

Substituting (4.21) into (4.20) gives (4.17) immediately.

 ρ

ρ

Remark 4.2. We note that similar bounds for kDku (f − Lm f )k∞ and hDku (f − Lm f )i2 can be obtained from all the results in this section by interchanging t and u, and h and `. Acknowledgement The authors would like to thank the referee for the comments which help to improve the paper. References [1] R.P. Agarwal, P.J.Y. Wong, Error Inequalities in Polynomial Interpolation and their Applications, Kluwer, Dordrecht, 1993. [2] G. Birkhoff, M.H. Schultz, R.S. Varga, Piecewise Hermite interpolation in one and two variables with applications to partial differential equations, Numerische Mathematik 11 (1968) 232–256. [3] P.G. Ciarlet, M.H. Schultz, R.S. Varga, Numerical methods of high order accuracy for nonlinear boundary value problems, I. One dimensional problem, Numerische Mathematik 9 (1967) 394–430. [4] P.G. Ciarlet, M.H. Schultz, R.S. Varga, Numerical methods of high order accuracy for nonlinear boundary value problems, II. Nonlinear boundary conditions, Numerische Mathematik 11 (1968) 331–345. [5] F.A. Costabile, F. Dell’Accio, R. Luceri, Explicit polynomial expansions of regular real functions by means of even order Bernoulli polynomials and boundary values, J. Comput. Appl. Math. 175 (2005) 77–99. [6] R.P. Agarwal, S. Pinelas, P.J.Y. Wong, Complementary Lidstone interpolation and boundary value problems, J. Inequal. Appl. (in press). [7] R.P. Agarwal, P.J.Y. Wong, Lidstone polynomials and boundary value problems, Comput. Math. Appl. 17 (1989) 1397–1421. [8] P.J. Davis, Interpolation and Approximation, Blaisdell Publishing Co., Boston, 1961. [9] A.K. Varma, G. Howell, Best error bounds for derivatives in two point Birkhoff interpolation problem, J. Approx. Theory 38 (1983) 258–268. [10] G.J. Lidstone, Notes on the extension of Aitken’s theorem (for polynomial interpolation) to the Everett types, Proc. Edinb. Math. Soc. 2 (1929) 16–19. [11] R.P. Boas, A note on functions of exponential type, Bull. Amer. Math. Soc. 47 (1941) 750–754. [12] R.P. Boas, Representation of functions by Lidstone series, Duke Math. J. 10 (1943) 239–245. [13] H. Poritsky, On certain polynomial and other approximations to analytic functions, Trans. Amer. Math. Soc. 34 (1932) 274–331. [14] I.J. Schoenberg, On certain two-point expansions of integral functions of exponential type, Bull. Amer. Math. Soc. 42 (1936) 284–288. [15] J.M. Whittaker, On Lidstone’s series and two-point expansions of analytic functions, Proc. Lond. Math. Soc. 36 (1933–1934) 451–469. [16] J.M. Whittaker, Interpolatory Function Theory, Cambridge, 1935. [17] D.V. Widder, Functions whose even derivatives have a prescribed sign, Proc. National Acad. Sciences 26 (1940) 657–659. [18] D.V. Widder, Completely convex functions and Lidstone series, Trans. Amer. Math. Soc. 51 (1942) 387–398. [19] R.P. Agarwal, G. Akrivis, Boundary value problems occurring in plate deflection theory, J. Comput. Appl. Math. 8 (1982) 145–154. [20] R.P. Agarwal, P.J.Y. Wong, Quasilinearization and approximate quasilinearization for Lidstone boundary value problems, Int. J. Comput. Math. 42 (1992) 99–116. [21] R.P. Agarwal, P.J.Y. Wong, Error bounds for the derivatives of Lidstone interpolation and applications, in: N.K. Govil, et al. (Eds.), Approximation Theory, in Memory of A.K. Varma, Marcal Dekker, New York, 1998, pp. 1–41. [22] P. Baldwin, Asymptotic estimates of the eigenvalues of a sixth-order boundary-value problem obtained by using global phase-integral method, Phil. Trans. R. Soc. Lond. A 322 (1987) 281–305. [23] P. Baldwin, A localized instability in a Bénard layer, Appl. Anal. 24 (1987) 117–156. [24] A. Boutayeb, E.H. Twizell, Finite-difference methods for twelfth-order boundary value problems, J. Comput. Appl. Math. 35 (1991) 133–138. [25] J.M. Davis, J. Henderson, P.J.Y. Wong, General Lidstone problems: multiplicity and symmetry of solutions, J. Math. Anal. Appl. 251 (2000) 527–548. [26] P.W. Eloe, J. Henderson, H.B. Thompson, Extremal points for impulsive Lidstone boundary value problems, Math. Comput. Modelling 32 (2000) 687–698. [27] P.W. Eloe, M.N. Islam, Monotone methods and fourth order Lidstone boundary value problems with impulse effects, Commun. Appl. Anal. 5 (2001) 113–120. [28] P. Forster, Existenzaussagen und Fehlerabschätzungen bei gewissen nichtlinearen Randwertaufgaben mit gewöhnlichen Differentialgleichungen, Numer. Math. 10 (1967) 410–422. [29] Y. Guo, W. Ge, Twin positive symmetric solutions for Lidstone boundary value problems, Taiwanese J. Math. 8 (2004) 271–283.

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