Cubic X-Spline Interpolatory Functions

CUBIC X-SPLINE INTERPOLATORY

FUNCTIONS

Manprit Kaur and Arun Kumar

([email protected], [email protected])

Department of Mathematics and Computer Science, R. D. University, Jabalpur, INDIA.

——————————————————————————————

Abstract:

Cubic X-spline functions was introduced to generalize cubic spline

functions. Here we study existence and error of approximation of cubic spline interpolation having interpolatory condition area matching in each

sub-intervals.

Keywords: Cubic X-spline functions, Cubic spline functions, Inter-polation.

AMS Classification: MCS(2010) 41A15

——————————————————————————————-1. DEFINITIONS AND NOTATIONS :

Let ∆ : 0 = x0<x1 <x2 <...<xn = 1, be a knot sequence in the

interval [0,1]. We denote the sub-intervals of the partition by Ii = [xi−1, xi], i = 1,2,...,n. Let α={αi}ni=1−1 be a parameter of n-1 real numbers.

A cubic polynomial function s ∈ C1 _{[0,1] is termed as a Cubic}

X-Spline on the interval [0,1] with knot sequence ∆, parameterα={αi}n−1 i=1,

if it satisfies the following three conditions:

(i) s is a Cubic polynomial on each sub-interval [ xi−1, xi] , i=1,2,...,n and

it is denoted by si.

(ii)s∈C1_{[0,1], that is, for i = 1,2,...,n, s}

i(xi) = si+1(xi), s 0

i(xi) = s 0 i+1(xi).

and

(iii)s00_i+1(xi) - s 00

i(xi) =αi[ s 000

i+1(xi) - s 000

i (xi) ], i = 1,2,...,n.

The X-Spline was considered to generalize the conventional spline. Here the second derivative is allowed to possess discontinuities at the knots.

The magnitudes of these discontinuities are related to those of the discon-tinuities in the third derivative by means of a simple relationship which

introduces one free parameter to each sub-interval. This gives a choice of the parameter for meeting certain requirements, for example to choose it to

adjust error.

The importance of spline function have been considered in (e.g. [1],[3] and [4]). An error analysis of finding the optimal estimates for the

uniform norm of approximation of continuous function by interpolating cu-bic and quadratic X-splines with arbitrary spaced knots result was obtained

in [2] and [5]. It is interesting to have a result on Cubic X-splines, so that it includes the corresponding result of Cubic Spline as a particular case.

Subsequently, the result concerning X-Spline was obtained by Das [4]. The error of approximation was also studied.

In this paper we consider X-Spline the same as considered by Das [4]

and replacing the condition of point interpolation by Area Matching. We first prove the following theorem and its error of approximation is obtained

subsequently.

THEOREM : Let f be a 1- periodic bounded continuous function and let

{xi}n

i=0 be knot points of the partition ∆ with xi - xi−1 = h for all i. For

a given α= {αi}ni=1−1 the set of non-negative real numbers, there exists a

unique 1-periodic Cubic X-Splinesssatisfying the interpolating condition :

(1) Rxi

xi−1s(x)dx= Rxi

xi−1f(x)dx,

(2) ifh≤ 4αiαi−1δ(6δ+δ

2_{+ 6)}

(12αi−1(1 +δ) +αi−1δ2+ 12αi(1 + 2δ) +αiδ2(13 +δ)) .

Proof of the theorem : We writes0(xi) =Mi , i = 0,1,...,n-1, and let

si(xi) =yi, i = 1,2,...,n value of the spline defined atxion the i-thinterval i.e., in [xi−1, xi].

It can be seen that such splines can be written on the interval

[xi−1, xi], i = 1,2,...,n as

IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 7, September 2014.

(3) h3s(x) =Mi−1h(x−xi−1)(xi−x)2

−Mih(x−xi−1)2(xi−x)

+yi−1(xi−x)2[2(x−xi−1) +h]

+yi(x−xi−1)2[2(xi−x) +h] .

We have

(4) h3_s0_{(x) =Mi}

−1h[(xi−x)2−2(xi−x)(x−xi−1)]

−Mih[2(x−xi−1)(xi−x)−(x−xi−1)2]

+2yi−1[(xi−x)2−(xi−x){2(x−xi−1) +h}]

+2yi[−(x−xi−1)2+ (x−xi−1){2(xi−x) +h}] .

We find thats0_i(xi) =s 0

i+1(xi) =Mi, i= 0,1, ..., n−1. This shows

that s0(x) is continuous at the knots,i.e.,s(x)∈C1_.

Further, we see that

(5) h3s000_i (x) = 6Mi−1h+ 6Mih+ 12yi−1−12yi.

Thus splines considered here are X-splines. i.e., we have

s00_i+1(xi)−s00_i(xi) =αi[ s000_i+1(xi)−s000_i (xi)], fori= 1,2, ..., n.

On substitution in the above X-spline equation we find that

(6) 2hMi+1(h+ 3αi)−8h2Mi−2hMi−1(h−3αi) =

6[(h+ 2αi )(yi+1−yi) + (h−2αi)(yi−yi−1)].

Now applying area matching interpolatory condition (1), we have

Rxi

xi−1f(x)dx=

Mi−1h−2R

xi

xi−1 (x−xi−1)(xi−x)

2_dx

−Mih−2Rxi

xi−1 (x−xi−1)

2_(xi−x)dx

+yi−1h−3R

xi

xi−1{(xi−x)

2_[2(x−xi

+yih−3Rxi

xi−1{(x−xi−1)

2_{[2(xi−x) +h]}dx,}

i.e.,

(7) Rxi

xi−1f(x)dx=Mi−1

h2

12−Mi

h2

12+

h

2yi−1+

h

2yi =Fi,

say. This yields,

(8) Fi+1−Fi=h

2

6 Mi−

h2

12Mi+1−

h2

12Mi−1

+h

2(yi+1−yi) +

h

2(yi−yi−1).

After eliminatingy0_isin the above equation from (6) and (7), we have

(9) Wi=pMi−2+qMi−1+rMi+sMi+1, i= 1,2, ..., n,

where

Wi= 12h−2[αi−1(h+ 2αi)(Fi+1−Fi) +αi(h−2αi−1)(Fi−Fi−1)],

p=hαi−4αiαi−1,

q=hαi−1−12αiαi−1+ 10hαi,

r= 10hαi−1+ 12αiαi−1+hαi,

s=hαi−1+ 4αiαi−1.

The system of equations have unique solution if the coefficient matrix of the above system of equations is non-singular. In order to establish its

non-singularity, we decompose the matrix as the following :

B=            

1 β 0 . . . 0 0

0 1 β . . . 0 0

0 0 1 β . . 0 0

. . . . . . . . . . . . β 0 0 0 . . 0 1

           

, C=

               

p λ s

β 0 0 . . . 0 0

0 p λ s

β 0 . . . 0 0

0 0 p λ s

β 0 . . 0 0 . . . . . . . . . . . .

λ s

β 0 0 . . . . 0 p

                ,

IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 7, September 2014.

whereβ andλare determined by the set of the following equations :

λ+pβ=q

and

s

β +λβ=r.

Substituting the values ofλ,sandrwe see that

s

β +λβ−r= 0.

The values ofβ are given by the equation

(αi−1+ 10αi−αiβ)β2+αi−1−(10αi−1+αi)βh

−12αiαi−1β+ 4αiαi−1+ (−12αiαi−1+ 4αiαi−1β)β2= 0.

Let us consider β = −1 +δ,for small positive δ, i.e., δ < 1

2. The value of the above expression is

(12αi−1(1−δ) +αi−1δ2+ 12αi(1−2δ) +αiδ2(13−δ))h

+24αiαi−1δ(1−δ) + 4αiαi−1δ,

which is positive. And atβ =−1−δ,it becomes

12αi−1(1 +δ) +αi−1δ2+ 12αi(1 + 2δ) +αiδ2(13 +δ)h

−4αiαi−1δ(6δ+δ2+ 6),

which is negative if

(10) h≤ 4αiαi−1δ(6δ+δ

2_{+ 6)}

(12αi−1(1 +δ) +αi−1δ2+ 12αi(1 + 2δ) +αiδ2(13 +δ)) .

Thus, one of the root ofβ lies in [−1−δ,−1 +δ] under the condition onδ.

Now we show that the matrix is diagonally dominant with the

diag-onal elementλ. We have

λ= (αi−1+ 10αi−αiβ)h+ 4αiαi−1(β−3)

and this is negative if

h≤ 4αiαi−1(3−β)

This is true forβ∈[−1−δ,−1 +δ] if

(11) h≤ 12αiαi−1−4αiαi−1(−1 +δ)

αi−1+ 10αi−αi(−1−δ) .

Next we see that from the hypothesis (2)pis also negative.

Obviouslysis positive.

We have

(|λ| − |p| − |s

β| ≥(|λ| − |p| − s

−1 +δ)

therefore

(12) D=−−2αi−1+αi−1δ−10αi+ 11αiδ−αiδ

2

−1 +δ h

−8αiαi−1−16αiαi−1δ+ 4αiαi−1δ

2

−1 +δ .

This is positive if

(13) h < 4αiαi−1(2−4δ+δ 2₎ αi(10 +δ2−11δ) + (2−δ)αi−1

.

We see that the expression on the right hand side of the inequality (10) is smaller than the expressions on the right hand side of the inequalities

(11) and (13). Hence the result is true if

h≤ 4αiαi−1δ(6δ+δ

2_{+ 6)}

(12αi−1(1 +δ) +αi−1δ2+ 12αi(1 + 2δ) +αiδ2(13 +δ)) .

To visualize the above condition we take, for example, δ = 0.1 in this case the above condition becomes

h≤ 2.644αiαi−1

13.21αi−1+ 14.531αi .

ERROR OF APPROXIMATION :

We write e(x) = s(x)−f(x). Since s0(xi) = Mi , i = 1,2,...,n, we get e0_i=Mi−f_i0.

IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 7, September 2014.

From (9) we find

(14) pe0_i−2+qe0_i−1+re0_i+se0_i+1=

Wi−pf_i0₋₂−qf_i0₋₁−rf_i0−sf_i+10 ,i = 1,2,...,n.

We denote the right hand side of the above expression byUi, say. We have

Ui= 12h−2_[αi

−1(h+ 2αi)(Fi+1−Fi)

+αi(h−2αi−1)(Fi−Fi−1)]

−(hαi−4αiαi−1)f 0

i−2−(hαi−1−12αiαi−1+ 10hαi)f 0 i−1

−(10hαi−1+ 12αiαi−1+hαi)f 0

i −(hαi−1+ 4αiαi−1)f 0 i+1.

By Taylor’s series expansion, we see that

Fi+1−Fi=

Rxi+1

xi (x−xi)

f0(ξi+1(x))−f 0 i+1

dx

+Rxi

xi−1(xi−x)

f0(ηi(x))−f_i0dx+h

2 2 f 0 i+1+ h2 2 f 0 i.

Ui=h(12h−2₍₍Rxi+1

xi (x−xi)

f0(ξi+1(x))−f_i+10 dx

+Rxi

xi−1(xi−x)

f0(ηi(x))−f_i0dx)αi−1

+(Rxi

xi−1(x−xi−1)

f0(ξi(x))−f_i0dx

+Rxi−1

xi−2(xi−1−x)

f0(ηi−1(x))−f 0 i−1

dx)αi)

+5αi−1f

0

i+1−4αi−1f

0 i + 5αif

0 i −4αif

0

i−1−αif 0

i−2−αi−1f

0 i−1)

+24h−2_α iαi−1(

Rxi+1

xi (x−xi)

f0(ξi+1(x))−f 0 i+1

dx)

+Rxi

xi−1(xi−x)

f0(ηi(x))−f 0 i

dx

−Rxi

xi−1(x−xi−1)

f0(ξi(x))−f_i0dx

−Rxi−1

xi−2(xi−1−x)

f0(ηi−1(x))−f 0 i−1

dx

+12αiαi−1

f_i+10 −f_i0+ 4αiαi−1

f_i0₋₂−f_i+10 .

|Ui| ≤h





12αi−1ω

f0;h+ 12αiωf0;h+ 4αi−1

f_i+10 −f_i0

+4αif_i0−f_i0₋₁+αi−1

f_i+10 −f_i0₋₁+αif_i0 −f_i0₋₂





+ 72αiαi−1ω

f0;h,

≤18((αi−1+αi)h+ 4αiαi−1)ω

f0;h.

If we takeα= max{αi }

then

|Ui| ≤36α(h+ 2α)ω

f0;h.

Hence

||e0_i||= sup|e0_i| ≤ ||A−1_||||U i||

We have

||e0_i|| ≤18((αi−1+αi)h+ 4αiαi−1)D−1ω

f0;h

whereD is defined in (12).

From the condition (1) there is at least one pointx0_i−1in [xi−1, xi] such that e(x0_i−1) = 0.This gives that

e(x) =Rx

x0_i−1e 0

(x)dx.

Therefore

|e(x)| ≤h 18((αi−1+αi)h+ 4αiαi−1)D−1

ωf0;h.

Reference :

1. Ahlberg, J.H. Nilson, E.N and Walsh, J.L., 1967. The theory of splines

and their applications, London, Academic Press.

2. Bujalska A. and Smarzewski R., Quadratic X-Splines, IMA, J.Numer.

Analysis, (1982), 2, 37-47.

3. Clenshaw C.W.; Negus B., The Cubic X-spline and its application to

interpolation, J. Inst. Math. Appl., 22(1978), 109-119.

4. Das, V.B., Spline interpolation by lower degree polynomials using area

matching condition, Ph.D. Thesis, Rani Durgawati Uni. (2004), Jabalpur.

5. Smarzewski R. and Bujalska A.,Uniform Convergence of Cubic and

Qua-dratic X-spline interpolants, IMA, J. Numer. Analysis, (1983) 3, 353-372.

IJISET - International Journal of Innovative Science, Engineering & Technology, Vol. 1 Issue 7, September 2014.