Modeling of curves by a design-control approximating refinement scheme

Abstract

This paper introduces a modified method for construction of a new design-control 6-point approximating refinement scheme. The construction of the new scheme is based on translation of the points of 4-point approximating refinement scheme to the new position according to the linear combination of certain displacement vectors. The initial and terminal points of these vectors are the refinement points of two 4-point approximating binary schemes. The new scheme contains three design-control parameters. These parameters increase the efficiency and flexibility of the new scheme. The mathematical and graphical analysis of the refinement scheme show that this scheme is good choice for curve modeling.

Keywords:

• Approximating refinement scheme
• curve
• design-control parameter
• Gibbs oscillation

MSC[2010]:

• 65D07
• 65D10
• 68U07
• 65D17

1. Introduction

Computer aided geometric design (CAGD) is considered as an emerging research field of computational mathematics hat has been rapidly growing since the past two decades due to a wide range of applications (Garg, 2020). In CAGD different mathematical algorithms and refinement rules are used to obtain better shapes of the objects. The CAGD field works on the same principle that was used by Archimedes to approximate a circle by increasing the number of vertices of a regular polygon. This field concerns with modeling and designing different complex objects with the help of elegant mathematical algorithms. Therefore, CAGD is the field which is mostly used by designers for shaping or fitting curves and surfaces with non-uniform shapes or discrete set of control points.

Refinement schemes have been a well-liked techniques in CAGD to produce curves and surfaces. Refinement schemes have various applications in graphics, image processing, engineering etc and thus have become an important area of study (Livesu, 2021; Livesu, Pitzalis, & Cherchi, 2021; Meng et al., 2020; Ning et al., 2020; Zhang et al., 2020). These schemes succeeded to get a great attention due to their high efficiency and clarity. A subdivision process refines the initial polygon recursively to a set of refined polygons which converge to a smooth limiting curve. Also, the refinement schemes were designed mainly for getting the desirable shapes. In addition, by different refinement schemes we can create smooth curves and surfaces by using a sequence of successive refinements. It starts by choosing a rough shape, and then converges finally to a smooth shape (Mahalingam & Koneru, 2020).

The most important classes of refinement schemes are interpolatory and approximating refinement schemes which interpolate and approximate the given data respectively. Both are considered as a non-parametric binary approximating refinement schemes. The construction and analysis of the non-parametric binary approximating refinement schemes are discussed in (Hameed & Mustafa, 2017; Mustafa & Rehman, 2010; Siddiqi & Younis, 2013). Also, the parametric binary approximating schemes and their inspections are presented in (Asghar & Mustafa, 2019; Daniel & Shunmugaraj, 2010; Rehan & Siddiqi, 2015; Siddiqi & Rehan, 2010; 2011).

In this paper, we present a novel method which merges two non-parametric binary refinement schemes to get a new parametric binary approximating refinement schemes called the design-control binary approximating refinement scheme. The schemes which are used in the construction process have equal complexity, but have different geometric behavior. Hence, the new scheme gives the geometric flexibility. Moreover, it contains the geometric behavior of both parent schemes as those are the special cases of the new design-control scheme.

The remainder of this research is organized as follows. In Section 2, we present some basic notations and results. In Section 3, the framework for the construction of the design-control binary refinement scheme is presented. In Section 4, we study the local control property of the design-control refinement scheme. We discuss the smoothness analysis of new refinement scheme in Section 5. And then, we give the Gibbs phenomena analysis of our refinement scheme in Section 6. Finally, summary and conclusions are given in Section 7.

2. Basic notation and results

A uni-variate linear binary refinement scheme $S_a$ is based on repeated (successive) application of the refinement rules. Which are used to map a polygon $P^k={\left\{P_i^k\right\}}_{i\in \mathbb{Z}}\in l(\mathbb{Z})$ to a refined polygon $P^{k+1}={\left\{P_i^{k+1}\right\}}_{i\in \mathbb{Z}}\in l(\mathbb{Z}).$ The general compact form of these refinement rules is defined as: (2.1) $$\begin{array}{cc}{P}_{2i+\xi }^{k+1}\hfill & =\sum _{j\in \mathbb{Z}}{a}_{2j+\xi }{P}_{i+j}^k, \xi =0,1,\hfill \end{array}$$(2.1) where $l(\mathbb{Z})$ denotes the space of scaler-valued sequences. The sequence $a={\left\{a_j\right\}}_{j\in \mathbb{Z}}$ is called the refinement mask. Therefore, the polynomial which uses this mask as coefficients is called the Laurent polynomial. In addition, the Laurent polynomial corresponding to refinement scheme (2.1)(2.1) $$\begin{array}{cc}{P}_{2i+\xi }^{k+1}\hfill & =\sum _{j\in \mathbb{Z}}{a}_{2j+\xi }{P}_{i+j}^k, \xi =0,1,\hfill \end{array}$$(2.1) is: $$a(z)=\sum _{j\in \mathbb{Z}}{a}_{2j}{z}^{2j}+\sum _{j\in \mathbb{Z}}{a}_{2j+1}{z}^{2j+1}.$$

The necessary condition for the convergence of a binary refinement scheme are: (2.2) $$\sum _{j\in \mathbb{Z}}{a}_{2j}=1,\hspace{1em}\sum _{j\in \mathbb{Z}}{a}_{2j+1}=1.$$(2.2)

Which is equivalent to the following relations; (2.3) $$a(1)=2 \text{and} a(-1)=0.$$(2.3)

Definition 2.1.

A linear refinement scheme is called approximating refinement scheme if it can be written as: $$\begin{array}{c}{P}_{2i}^{k+1}=\sum _{i\in \mathbb{Z}}{a}_{2j}{P}_{i-j}^k,\\ P_{2i+1}^{k+1}=\sum _{i\in \mathbb{Z}}{a}_{2j+1}{P}_{i-j}^k.\end{array}$$

This type of refinement schemes generates the limiting curves in which the control points $P_j^k$ of k-th polygon level are not included in the points of the $(k+1)$-th polygon level.

Definition 2.2.

Convergence analysis by Eigenvalues method

We seek the complete demonstration about the necessary and sufficient conditions for uniform convergence of the refinement scheme. This analysis was firstly introduced by Doo and Sabin (Doo & Sabin, 1978), and then developed through the years (see for example (Xumin, Xiaojun, Xianpeng, & Cailing, 2013; Thorne, 2021)). As a matter of fact, the matrix formalism is the technique for derive the necessary conditions for a scheme to be $C^k,$ it based mainly on the eigenvalues of the subdivision matrix.

Let the eigenvalues of the $n\times n$ subdivision matrix be $\{{\lambda }_i:i=0,1,2,\dots n-1\},$ where ${\lambda }_0=1$ and $\left|{\lambda }_i\right|\ge \left|{\lambda }_{i+1}\right|,\forall i\in \mathbb{N}.$ And we have the necessary conditions for the following properties:

• 1: $\left|{\lambda }_1\right|=\left|{\lambda }_2\right|$ shows kink, i.e. not $C^1.$

• 2: ${\lambda }_1^2<{\lambda }_2$ shows unbounded curvature.

• 3: ${\lambda }_1^2=\left|{\lambda }_2\right|=\left|{\lambda }_3\right|$ shows mildly diverging curvature.

• 4: ${\lambda }_1^2=\left|{\lambda }_2\right|>\left|{\lambda }_3\right|$ shows curvature bounded.

• 5: ${\lambda }_1^2>\left|{\lambda }_2\right|$ shows curvature to zero.

Theorem 2.1.

Let the matrix formalism to derive the necessary conditions for a scheme to be $C^k$ based on the eigenvalues of the subdivision matrix. If the limiting curve is $C^2$ continuity, the eigenvalues $\left\{{\lambda }_i\right\}$ satisfy (Hassan, Ivrissimitzis, Dodgson, & Sabin, 2002): $$\begin{array}{cc}{\lambda }_0\hfill & =1,\left|{\lambda }_i\right|>\left|{\lambda }_{i+1}\right|,\forall i\in \mathbb{N},\hfill \end{array}$$ and $$\begin{array}{cc}{\lambda }_1^2\hfill & =\left|{\lambda }_2\right|>\left|{\lambda }_3\right|.\hfill \end{array}$$

And a scheme will be uniformly convergent if and only if there is an integer $L\ge 1$ such that ${\Vert \frac{1}{2}{S}_{a_m}^L\Vert }_{\infty }$ $<1.$

Theorem 2.2.

Let the refinement scheme $S_a$ with mask $a=\left\{a_i\right\}$ where $i\in \mathbb{Z}$ satisfying the condition (Dyn & Levin, 2002) (2.4) $$\sum _{i\in \mathbb{Z}}{a}_{2i}=\sum _{i\in \mathbb{Z}}{a}_{2i+1}=1.$$(2.4)

If there exist an integer $L\ge 1$, such that $\Vert {(\frac{1}{2}{S}_{a_m})}^L{\Vert }_{\infty }\le 1$, then the refinement scheme $S_a$ is $C^n$-continuous, where $$\begin{array}{c}{\Vert {\left(\frac{1}{2}{S}_{a_m}\right)}^L\Vert }_{\infty }=\max \left\{\sum _{j\in \mathbb{Z}}\left|b_{i-2^{L}j}^{[m,L]}\right|:0\le i\le 2^L-1\right\},\end{array}$$ and Laurent polynomial $b^{[m,L]}(z)=b(z)b(z^2)\dots b(z^{2^{L-1}})$, so, $b(z)=\frac{1}{2}{a}_m(z)$. When $L=1,$ $$\begin{array}{c}{\Vert \left(\frac{1}{2}{S}_{a_m}\right)\Vert }_{\infty }=\frac{1}{2}\max \left\{\Vert \sum _{i\in \mathbb{Z}}{a}_{2i}\Vert ,\Vert \sum _{i\in \mathbb{Z}}{a}_{2i+1}\Vert \right\}.\end{array}$$

Let $a_m(z)={(\frac{2z}{1+z})}^{m}a(z)$ with $S_b$ is contractive. Then $S_a$ is also convergent with $C^n$-continuous.

Theorem 2.3.

Given $0\le \psi \le h$, let g be a function defined by (Amat, Ruiz, & Trillo, 2018) $$\begin{array}{c}g(x)=g_{-}(x),g_{-}\in C^n(]-\infty ,\psi ]),\forall x\le \psi ,\\ g(x)=g_{-}(x),g_{+}\in C^n(]\psi ,+\infty ]),\forall x\ge \psi ,\end{array}$$ with $n\ge 2$ and $g_{-}(\psi )>g_{+}(\psi )$. Let $S_a$ be a uni-variate stationary refinement scheme with condition: $${\zeta }_l{(i)}^{\left[k\right]}=\{\begin{array}{cc}\sum _{\tau \le i}{a}_{2^{\left[k\right]}\tau +l}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{a}_{2^{\left[k\right]}\tau +l}\hfill & i>0,\hfill \end{array}$$ where $a^{[k]}$ define as: $$\begin{array}{cc}{a}_j^{\left[k\right]}\hfill & =\sum _{i\in \mathbb{Z}}{a}_i^{[k-1]}{a}_{j-2i},\hfill \end{array}$$ and $0\le l<2^k$. Then, if ${\zeta }_l(i)\ge 0,\forall i,k$; and if h is sufficiently small, we have the following two conditions:

P 1. If $\left|x\right|\ge \max \left\{\right|\frac{M-1}{2}|,|\frac{M+N}{2}+1\left|\right\}h$, then $$\begin{array}{cc}|g(x)-(S_a^{\infty }{g}_h)(x)|\hfill & =O(h^n),\hfill \end{array}$$ with $n\ge 2.$

P 2. If $\left|x\right|\le \max \left\{\right|\frac{M-1}{2}|,|\frac{M+N}{2}+1\left|\right\}h$, there exists ${\beta }_h=O(h)$ such that: $$\begin{array}{c}{g}_{1,h}-{\beta }_h\le g_{+}(h)-{\beta }_h\le (S_a^{\infty }{g}_h)(x)\le g_{-}(0)+{\beta }_h=g_{0,h}+{\beta }_h.\hfill \end{array}$$

3. Construction structure of the new 6-point design-control refinement scheme

In this section, we describe the construction process of the new design-control binary approximating refinement scheme. The new scheme contains three design-control parameters which control the behavior of the limit curves. Therefore, the step by step procedure of the construction is given below:

3.1. Step 1

In this step, we take two already published binary approximating refinement schemes.

Firstly, the 4-point approximating refinement scheme which was presented by (Deslauriers & Dubuc, 1989; Mustafa & Rehman, 2010) is (3.1) $$\{\begin{array}{c}{Q}_{2i}^{k+1}=-\frac{7}{128}{P}_{i-1}^k+\frac{105}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{5}{128}{P}_{i+2}^k,\\ \\ Q_{2i+1}^{k+1}=-\frac{5}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k+\frac{105}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k,\end{array}$$(3.1) where $\{Q_j^{k+1}:j=0,1,\dots ,2n+1\}$ are the refined points of the subdivided polygon by scheme (3.1)(3.1) $$\{\begin{array}{c}{Q}_{2i}^{k+1}=-\frac{7}{128}{P}_{i-1}^k+\frac{105}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{5}{128}{P}_{i+2}^k,\\ \\ Q_{2i+1}^{k+1}=-\frac{5}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k+\frac{105}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k,\end{array}$$(3.1) if $\{P_j^k:j=0,1,\dots ,n\}$ are the given points of the given polygon.

Secondly, the 4-point approximating B-spline refinement scheme which is constructed by B-spline basis function of degree-6 is (3.2) $$\{\begin{array}{c}{R}_{2i}^{k+1}=\frac{7}{64}{P}_{i-1}^k+\frac{35}{64}{P}_i^k+\frac{21}{64}{P}_{i+1}^k+\frac{1}{64}{P}_{i+2}^k,\\ \\ R_{2i+1}^{k+1}=\frac{1}{64}{P}_{i-1}^k+\frac{21}{64}{P}_i^k+\frac{35}{64}{P}_{i+1}^k+\frac{7}{64}{P}_{i+2}^k,\end{array}$$(3.2) where $\{R_j^{k+1}:j=0,1,\dots ,2n+1\}$ are the refined points of the subdivided polygon by scheme (3.2)(3.2) $$\{\begin{array}{c}{R}_{2i}^{k+1}=\frac{7}{64}{P}_{i-1}^k+\frac{35}{64}{P}_i^k+\frac{21}{64}{P}_{i+1}^k+\frac{1}{64}{P}_{i+2}^k,\\ \\ R_{2i+1}^{k+1}=\frac{1}{64}{P}_{i-1}^k+\frac{21}{64}{P}_i^k+\frac{35}{64}{P}_{i+1}^k+\frac{7}{64}{P}_{i+2}^k,\end{array}$$(3.2) if $\{P_j^k:j=0,1,\dots ,n\}$ are the given points of the given polygon.

3.2. Step 2

In this step, firstly we define two displacement vectors. The first displacement vector is denoted by $D_{2i}^{k+1}$ and is shown in Figure 1(a). We get this vector by subtracting refinement point $R_{2i}^{k+1}$ defined in (3.2)(3.2) $$\{\begin{array}{c}{R}_{2i}^{k+1}=\frac{7}{64}{P}_{i-1}^k+\frac{35}{64}{P}_i^k+\frac{21}{64}{P}_{i+1}^k+\frac{1}{64}{P}_{i+2}^k,\\ \\ R_{2i+1}^{k+1}=\frac{1}{64}{P}_{i-1}^k+\frac{21}{64}{P}_i^k+\frac{35}{64}{P}_{i+1}^k+\frac{7}{64}{P}_{i+2}^k,\end{array}$$(3.2) from the refinement point $Q_{2i}^{k+1}$ defined in (3.1)(3.1) $$\{\begin{array}{c}{Q}_{2i}^{k+1}=-\frac{7}{128}{P}_{i-1}^k+\frac{105}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{5}{128}{P}_{i+2}^k,\\ \\ Q_{2i+1}^{k+1}=-\frac{5}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k+\frac{105}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k,\end{array}$$(3.1) . That is $$\begin{array}{cc}{D}_{2i}^{k+1}\hfill & =Q_{2i}^{k+1}-R_{2i}^{k+1}.\hfill \end{array}$$

Figure 1. Vectors $D_{2i}^{k+1}$ and $D_{2i+1}^{k+1}.$

Similarly, we get the second displacement vector $D_{2i+1}^{k+1}$ which is shown in Figure 1(b) by subtracting refitment point $R_{2i+1}^{k+1}$ defined in (3.2)(3.2) $$\{\begin{array}{c}{R}_{2i}^{k+1}=\frac{7}{64}{P}_{i-1}^k+\frac{35}{64}{P}_i^k+\frac{21}{64}{P}_{i+1}^k+\frac{1}{64}{P}_{i+2}^k,\\ \\ R_{2i+1}^{k+1}=\frac{1}{64}{P}_{i-1}^k+\frac{21}{64}{P}_i^k+\frac{35}{64}{P}_{i+1}^k+\frac{7}{64}{P}_{i+2}^k,\end{array}$$(3.2) from the refinement point $Q_{2i+1}^{k+1}$ defined in (3.1)(3.1) $$\{\begin{array}{c}{Q}_{2i}^{k+1}=-\frac{7}{128}{P}_{i-1}^k+\frac{105}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{5}{128}{P}_{i+2}^k,\\ \\ Q_{2i+1}^{k+1}=-\frac{5}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k+\frac{105}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k,\end{array}$$(3.1) $$\begin{array}{cc}{D}_{2i+1}^{k+1}\hfill & =Q_{2i+1}^{k+1}-R_{2i+1}^{k+1}.\hfill \end{array}$$

Hence the displacement vectors can be expressed in the linear combination of the given control points $\{P_j^k:j=0,1,\dots ,n\}.$ That is (3.3) $$\{\begin{array}{c}{D}_{2i}^{k+1}=-\frac{21}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k-\frac{7}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k,\\ \\ D_{2i+1}^{k+1}=-\frac{7}{128}{P}_{i-1}^k-\frac{7}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{21}{128}{P}_{i+2}^k.\end{array}$$(3.3)

If we increase/decrease the indices of refinement equations defined in (3.1)–(3.2) and define four more displacement vectors in same manner, we get (3.4) $$\begin{array}{c}\{\begin{array}{c}{D}_{2i-2}^{k+1}=-\frac{21}{128}{P}_{i-2}^k+\frac{35}{128}{P}_{i-1}^k-\frac{7}{128}{P}_i^k-\frac{7}{128}{P}_{i+1}^k,\\ \\ D_{2i-1}^{k+1}=-\frac{7}{128}{P}_{i-2}^k-\frac{7}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k-\frac{21}{128}{P}_{i+1}^k,\\ \\ D_{2i+2}^{k+1}=-\frac{21}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k-\frac{7}{128}{P}_{i+3}^k,\\ \\ D_{2i+3}^{k+1}=-\frac{7}{128}{P}_i^k-\frac{7}{128}{P}_{i+1}^k+\frac{35}{128}{P}_{i+2}^k-\frac{21}{128}{P}_{i+3}^k.\end{array}\hfill \end{array}$$(3.4)

3.3. Step 3

In this step, we use the properties of “vector addition” and “scalar multiplication” to get the resultant vectors. We use the six vectors defined in (3.3)–(3.4) along with three scalars $\mu$ $\nu$ and $\omega$ $\in$ $\mathbb{R}.$ Hence we get two resultant vectors denote by: $D_E$ and $D_O$ and they defined below: (3.5) $$\{\begin{array}{c}{Q}_E=\mu D_{2i-2}^{k+1}+\nu D_{2i}^{k+1}+\omega D_{2i+2}^{k+1},\\ \\ D_O=\omega D_{2i-1}^{k+1}+\nu D_{2i+1}^{k+1}+\mu D_{2i+3}^{k+1}.\end{array}$$(3.5)

3.4. Step 4

In this step, we calculate refinement rules of the new 6-point approximating refinement scheme by translating the refinement points of scheme (3.1)(3.1) $$\{\begin{array}{c}{Q}_{2i}^{k+1}=-\frac{7}{128}{P}_{i-1}^k+\frac{105}{128}{P}_i^k+\frac{35}{128}{P}_{i+1}^k-\frac{5}{128}{P}_{i+2}^k,\\ \\ Q_{2i+1}^{k+1}=-\frac{5}{128}{P}_{i-1}^k+\frac{35}{128}{P}_i^k+\frac{105}{128}{P}_{i+1}^k-\frac{7}{128}{P}_{i+2}^k,\end{array}$$(3.1) . For the translation of these points we use the resultant vectors $D_E$ and $D_O$ which are defined in (3.5)(3.5) $$\{\begin{array}{c}{Q}_E=\mu D_{2i-2}^{k+1}+\nu D_{2i}^{k+1}+\omega D_{2i+2}^{k+1},\\ \\ D_O=\omega D_{2i-1}^{k+1}+\nu D_{2i+1}^{k+1}+\mu D_{2i+3}^{k+1}.\end{array}$$(3.5) . Hence the first refinement rule is: $$P_{2i}^{k+1}=Q_{2i}^{k+1}-Q_E.$$

Similarly, the second refinement rule is: $$P_{2i+1}^{k+1}=Q_{2i+1}^{k+1}-Q_O.$$

Thus, the new design-control approximating scheme is: (3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) where, $$\begin{array}{c}{a}_{-5}=a_4=\frac{21}{128}\mu , a_{-3}=a_2=-\frac{7}{128}-\frac{35}{128}\mu +\frac{21}{128}\nu ,\\ a_{-1}=a_0=\frac{105}{128}+\frac{7}{128}\mu -\frac{35}{128}\nu +\frac{21}{128}\omega ,\\ a_1=a_{-2}=\frac{35}{128}+\frac{7}{128}\mu +\frac{7}{128}\nu -\frac{35}{128}\omega ,\\ a_3=a_{-4}=-\frac{5}{128}+\frac{7}{128}\nu +\frac{7}{128}\omega , a_5=a_{-6}=\frac{7}{128}\omega .\end{array}$$

Here, $P_{2i}^{k+1}$ and $P_{2i+1}^{k+1}$ are points at $(k+1)$-th refinement level, whereas $\mu ,$ $\nu$ and $\omega$ are the scalars used as design-control parameters. Thus the sequence $\{\dots ,0,0,a_{-6},a_{-5},a_{-4},a_{-3},a_{-2},$ $a_{-1},a_0,a_1,a_2,a_3,a_4,0,0,\dots \}$ is the mask of the scheme (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) . The new 6-point approximating refinement scheme is denoted by $S_{a,\mu ,\nu ,\omega }.$

If we put $i=-2,-1,0,1,2$ in the pair of refinement equation (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , then we get a system of refinement equations. Which can be written in a matrix form as: $$\left(\begin{array}{c}{P}_{-4}^{k+1}\\ P_{-3}^{k+1}\\ P_{-2}^{k+1}\\ P_{-1}^{k+1}\\ P_0^{k+1}\\ P_1^{k+1}\\ P_2^{k+1}\\ P_3^{k+1}\\ P_4^{k+1}\\ P_5^{k+1}\end{array}\right)\mathrm{ }=\left(\begin{array}{cccccccccc}{a}_{-5}& a_{-3}& a_{-1}& a_1& a_3& a_5& 0& 0& 0& 0\\ a_{-6}& a_{-4}& a_{-2}& a_0& a_2& a_4& 0& 0& 0& 0\\ 0& a_{-5}& a_{-3}& a_{-1}& a_1& a_3& a_5& 0& 0& 0\\ 0& a_{-6}& a_{-4}& a_{-2}& a_0& a_2& a_4& 0& 0& 0\\ 0& 0& a_{-5}& a_{-3}& a_{-1}& a_1& a_3& a_5& 0& 0\\ 0& 0& a_{-6}& a_{-4}& a_{-2}& a_0& a_2& a_4& 0& 0\\ 0& 0& 0& a_{-5}& a_{-3}& a_{-2}& a_1& a_3& a_5& 0\\ 0& 0& 0& a_{-6}& a_{-4}& a_{-2}& a_0& a_2& a_4& 0\\ 0& 0& 0& 0& a_{-5}& a_{-3}& a_{-1}& a_1& a_3& a_5\\ 0& 0& 0& 0& a_{-6}& a_{-4}& a_{-2}& a_0& a_2& a_4\end{array}\right)\left(\begin{array}{c}{P}_{-4}^k\\ P_{-3}^k\\ P_{-2}^k\\ P_{-1}^k\\ P_0^k\\ P_1^k\\ P_2^k\\ P_3^k\\ P_4^k\\ P_5^k\end{array}\right).$$

Then, the mask of refinement scheme $S_{a,\mu ,\nu ,\omega }$ defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) is $$\begin{array}{cc}a=\{\dots ,0,0,\frac{7}{128}\omega ,\frac{21}{128}\mu ,-\frac{5}{128}+\frac{7}{128}\nu +\frac{7}{128}\omega ,-\frac{7}{128}-\frac{35}{128}\mu +\frac{21}{128}\nu ,\frac{35}{128}+\frac{7}{128}\mu +\frac{7}{128}\nu -\frac{35}{128}\omega ,\frac{105}{128}+\frac{7}{128}\mu -\frac{35}{128}\nu +\frac{21}{128}\omega ,\frac{105}{128}+\frac{7}{128}\mu -\frac{35}{128}\nu +\frac{21}{128}\omega ,\frac{35}{128}+\frac{7}{128}\mu +\frac{7}{128}\nu -\frac{35}{128}\omega ,-\frac{7}{128}-\frac{35}{128}\mu +\frac{21}{128}\nu ,-\frac{5}{128}+\frac{7}{128}\nu +\frac{7}{128}\omega ,\frac{21}{128}\mu ,\frac{7}{128}\omega ,0,0,\dots \}.\hfill & \hfill \end{array}$$

Remark 3.1.

In this article, we use the “sub-scheme” word for the refinement scheme which is the special case of our refinement scheme $S_{a,\mu ,\nu ,\omega }$ defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) .

3.5. Sub-scheme $S_{t,\nu }$ of the refinement scheme $S_{a,\mu ,\nu ,\omega }$

In this section, we present the sub-scheme, with design-control parameter $\nu ,$ of scheme $S_{a,\mu ,\nu ,\omega }$ that is defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) . We set $\mu =\omega =0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) to get the following sub-scheme of scheme $S_{a,\mu ,\nu ,\omega }$ (3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) where $$\begin{array}{c}{t}_{-3}=t_2=-\frac{7}{128}+\frac{21}{128}\nu , t_{-1}=t_0=\frac{105}{128}-\frac{35}{128}\nu ,\\ t_1=t_{-2}=\frac{35}{128}+\frac{7}{128}\nu , t_3=t_{-4}=-\frac{5}{128}+\frac{7}{128}\nu .\end{array}$$

Where, the mask of above scheme is $\{\dots ,0,0,t_{-4},t_{-3},t_{-2},t_{-1},t_0,t_1,t_2,t_3,0,0,\dots \}.$ Hence, the mask of corresponding refinement scheme $S_{t,\nu }$ is, $$\begin{array}{c}t=\{\dots ,0,0,-\frac{5}{128}+\frac{7}{128}\nu ,-\frac{7}{128}+\frac{21}{128}\nu ,\frac{35}{128}+\frac{7}{128}\nu ,\frac{105}{128}-\frac{35}{128}\nu ,\\ \frac{105}{128}-\frac{35}{128}\nu ,\frac{35}{128}+\frac{7}{128}\nu ,-\frac{7}{128}+\frac{21}{128}\nu ,-\frac{5}{128}+\frac{7}{128}\nu ,0,0,\dots \}.\end{array}$$

Remark 3.2.

For the scheme $S_{t,\nu },$ the vectors defined in (3.5)(3.5) $$\{\begin{array}{c}{Q}_E=\mu D_{2i-2}^{k+1}+\nu D_{2i}^{k+1}+\omega D_{2i+2}^{k+1},\\ \\ D_O=\omega D_{2i-1}^{k+1}+\nu D_{2i+1}^{k+1}+\mu D_{2i+3}^{k+1}.\end{array}$$(3.5) reduce to $$Q_E=\nu D_{2i}^{k+1} \text{and} D_O=\nu D_{2i+1}^{k+1}.$$

3.6. Sub-scheme $S_{\mathrm{\ell ,\mu }}$ of the refinement scheme $S_{\mathrm{a,\mu ,\nu ,\omega }}$

In this section, we give the sub-scheme with design-control parameter $\mu$ of scheme $S_{a,\mu ,\nu ,\omega }$ defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) . We set $\mu =\omega$ and $\nu =-2\mu +1$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) then, to get the following scheme denotes by $S_{\ell ,\mu }$ (3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) where, $$\begin{array}{c}{\ell }_{-5}={\ell }_4=\frac{21}{128}\mu , {\ell }_{-3}={\ell }_2=\frac{7}{64}-\frac{77}{128}\mu , {\ell }_{-1}={\ell }_0=\frac{35}{64}+\frac{49}{64}\mu ,\\ {\ell }_1={\ell }_{-2}=\frac{21}{64}-\frac{21}{64}\mu , {\ell }_3={\ell }_{-4}=\frac{1}{64}-\frac{7}{128}\mu , {\ell }_5={\ell }_{-6}=\frac{7}{128}\mu ,\end{array}$$ with $\{\dots ,0,0,{\ell }_{-6},{\ell }_{-5},{\ell }_{-4},{\ell }_{-3},{\ell }_{-2},{\ell }_{-1},{\ell }_0,{\ell }_1,{\ell }_2,{\ell }_3,{\ell }_4,{\ell }_5,0,0,\dots \}$ is the mask of this refinement scheme. Thus, the mask of corresponding refinement scheme $S_{\ell ,\mu }$ is, $$\begin{array}{cc}\ell \hfill & =\{\dots ,0,0,\frac{7}{128}\mu ,\frac{21}{128}\mu ,\frac{1}{64}-\frac{7}{128}\mu ,\frac{7}{64}-\frac{77}{128}\mu ,\frac{21}{64}-\frac{21}{64}\mu ,\frac{35}{64}+\frac{49}{64}\mu ,\frac{35}{64}+\frac{49}{64}\mu ,\hfill \\ \frac{21}{64}-\frac{21}{64}\mu ,\frac{7}{64}-\frac{77}{128}\mu ,\frac{1}{64}-\frac{7}{128}\mu ,\frac{21}{128}\mu ,\frac{7}{128}\mu ,0,0,\dots \}.\hfill & \hfill \end{array}$$

Remark 3.3.

For the scheme $S_{\ell ,\mu },$ the vectors which are defined in (3.5)(3.5) $$\{\begin{array}{c}{Q}_E=\mu D_{2i-2}^{k+1}+\nu D_{2i}^{k+1}+\omega D_{2i+2}^{k+1},\\ \\ D_O=\omega D_{2i-1}^{k+1}+\nu D_{2i+1}^{k+1}+\mu D_{2i+3}^{k+1}.\end{array}$$(3.5) take the following form: $$\begin{array}{c}{Q}_E=D_{2i}^{k+1}+\mu \left[(D_{2i-2}^{k+1}-D_{2i}^{k+1})+(D_{2i+2}^{k+1}-D_{2i}^{k+1})\right],\\ \\ D_O=D_{2i+1}^{k+1}+\mu \left[(D_{2i-1}^{k+1}-D_{2i+1}^{k+1})+(D_{2i+3}^{k+1}-D_{2i+1}^{k+1})\right].\end{array}$$

3.7. Refinement rules for boundary points when the given polygon is open

Here, we discuss the refinement rules for the open polygon of our new approximating refinement scheme $S_{a,\mu ,\nu ,\omega }.$ Whenever, we have to smooth an open polygon, we use a specific relations which replace the unknowns point with known points to get the refinement rules for boundary points of our refinement scheme.

Let the indices of the given points be $P_j^k:$ $j=0,1,\dots ,n$ as shown in Figure 2. Hence, the point $P_j^k:$ $j<0$ and $P_j^k:$ $j>n$ are unknowns. So, first we use the following relation to find out the unknown points $P_j^k:$ $j<0.$ (3.9) $$\begin{array}{c}{P}_{-m}^k=2P_0^k-P_m^k, \text{where}\mathrm{ } m=1,2.\end{array}$$(3.9)

Figure 2. Sketch of open polygon.

Then, we use the following relation to replace the unknown points $P_j^k:$ $j>n$ with the known points $P_j^k:$ $j=0,1,\dots ,n.$ And then, (3.10) $$\begin{array}{c}{P}_{n+m}^k=2P_n^k-P_{n-m}^k, \mathrm{ }\text{where} m=1,2,3.\end{array}$$(3.10)

Hence, the boundary refinement points for the open polygons are: $$\begin{array}{cc}{P}_0^{k+1}\hfill & =\frac{1}{128}[(-21\mu +7\nu +21\omega +91)P_0^k+(42+42\mu -14\nu -35\omega )P_1^k\hfill \\ +(-21\mu +7\nu +7\omega -5)P_2^k+7\omega P_3^k],\hfill & \hfill \end{array}$$ $$\begin{array}{cc}{P}_1^{k+1}\hfill & =\frac{1}{128}[(7\mu +21\nu -7\omega +25)P_0^k+(7\mu -42\nu +14\omega +110)P_1^k\hfill \\ +(-35\mu +21\nu -7\omega -7)P_2^k+21\mu P_3^k],\hfill & \hfill \end{array}$$ $$\begin{array}{cc}{P}_2^{k+1}\hfill & =\frac{1}{128}[(7\mu +21\nu -7)P_0^k+(-14\mu -35\nu +21\omega +105)P_1^k\hfill \\ +(7\mu +7\nu -35\omega +35)P_2^k+(7\nu +7\omega -5)P_3^k+7\omega P_4^k],\hfill & \hfill \end{array}$$ $$\begin{array}{ccc}{P}_3^{k+1}\hfill & =\frac{1}{128}[(7\nu +21\omega -5)P_0^k+(7\mu +7\nu -42\omega +35)P_1^k+(7\mu -35\nu +21\omega +105)P_2^k+(-35\mu +21\nu -7)P_3^k+21\mu P_4^k],\hfill & \hfill \end{array}$$ $$\begin{array}{ccc}{P}_{2n-4}^{k+1}\hfill & =\frac{1}{128}[21\mu P_{n-4}^k+(-35\mu +21\nu -7)P_{n-3}^k+(7\mu -35\nu +21\omega +105)P_{n-2}^k+(7\mu +7\nu -42\omega +35)P_{n-1}^k+(7\nu +7\omega -5)P_n^k],\hfill & \hfill \end{array}$$ $$\begin{array}{cc}{P}_{2n-3}^{k+1}\hfill & =\frac{1}{128}[7\omega P_{n-4}^k+(7\nu +7\omega -5)P_{n-3}^k+(7\mu +7\nu -35\omega +35)\hfill \\ \times P_{n-2}^k+(14\mu -35\nu +21\omega +105)P_{n-1}^k+(7\mu +21\nu -7)P_n^k],\hfill & \hfill \end{array}$$ $$\begin{array}{ccc}{P}_{2n-2}^{k+1}\hfill & =\frac{1}{128}[21\mu P_{n-3}^k+(-35\mu +21\nu -7\omega -7)P_{n-2}^k+(7\mu -42\nu +14\omega +110)P_{n-1}^k+(7\mu +21\nu -7\omega +25)P_n^k],\hfill & \hfill \end{array}$$ $$\begin{array}{ccc}{P}_{2n-1}^{k+1}\hfill & =\frac{1}{128}[7\omega P_{n-3}^k+(-21\mu +7\nu +7\omega -5)P_{n-2}^k+(42\mu +14\nu -35\omega +14)P_{n-1}^k+(-21\mu +7\nu +21\omega +91)P_n^k],\hfill & \hfill \end{array}$$ $$\begin{array}{ccc}{P}_{2n}^{k+1}\hfill & =\frac{1}{128}[-7\mu P_{n-3}^k+(21\mu -7\nu -7\omega +5)P_{n-2}^k+(-42\mu -42\nu +35\omega -42)P_{n-1}^k+(21\mu -7\nu -21\omega +165)P_n^k],\hfill & \hfill \end{array}$$ $$\begin{array}{ccc}{P}_{2n+1}^{k+1}\hfill & =\frac{1}{128}[-21\mu P_{n-3}^k+(35\mu -21\nu +7\omega +7)P_{n-2}^k+(-7\mu +42\nu -14\omega -110)P_{n-1}^k+(-7\mu -21\nu +7\omega +231)P_n^k].\hfill & \hfill \end{array}$$

4. Support of design-control refinement scheme

In this section, we analyze the support of our design-control refinement scheme. The support of our refinement scheme shows that how better it locally controls the limiting curves. The support of our refinement scheme is small, hence our scheme has better local control on shapes. We calculate the support of our refinement scheme $S_{a,\mu ,\nu ,\omega }$ and its sub-schemes $S_{t,\nu }$ and $S_{\ell ,\mu }$ which are defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) and (3.8)(3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) respectively. The support of these schemes is analyzed theoretically and graphically as follows.

Lemma 4.1.

If we use refinement scheme $S_{a,\mu ,\nu ,\omega }$ on initial data (4.1) $$P_i^0=\{\begin{array}{cc}& \begin{array}{cc}1\hfill & i=0,\hfill \\ 0\hfill & i\ne 0.\hfill \end{array}\end{array}$$(4.1)

Then, after first subdivision step the non-zero points are $P_{-6}^1,$ $P_{-5}^1,$ $\dots ,$ $P_4^1,$ $P_5^1.$

Proof.

If we use $i=-4$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_{-8}^1\hfill & =a_{-5}{P}_{-6}^0+a_{-3}{P}_{-5}^0+a_{-1}{P}_{-4}^0+a_1{P}_{-3}^0+a_3{P}_{-2}^0+a_5{P}_{-1}^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(0)+a_{-1}(0)+a_1(0)+a_3(0)+a_5(0)=0,\hfill \end{array}$$ and $$\begin{array}{cc}{P}_{-7}^1\hfill & =a_{-6}{P}_{-6}^0+a_{-4}{P}_{-5}^0+a_{-2}{P}_{-4}^0+a_0{P}_{-3}^0+a_2{P}_{-2}^0+a_4{P}_{-1}^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(0)+a_{-2}(0)+a_0(0)+a_2(0)+a_4(0)=0.\hfill \end{array}$$

If we put $i=-3$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_{-6}^1\hfill & =a_{-5}{P}_{-5}^0+a_{-3}{P}_{-4}^0+a_{-1}{P}_{-3}^0+a_1{P}_{-2}^0+a_3{P}_{-1}^0+a_5{P}_0^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(0)+a_{-1}(0)+a_1(0)+a_3(0)+a_5(1)=a_5,\hfill \\ =\frac{7}{128}\omega ,\hfill & \hfill \end{array}$$ and $$\begin{array}{cc}{P}_{-5}^1\hfill & =a_{-6}{P}_{-5}^0+a_{-4}{P}_{-4}^0+a_{-2}{P}_{-3}^0+a_0{P}_{-2}^0+a_2{P}_{-1}^0+a_4{P}_0^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(0)+a_{-2}(0)+a_0(0)+a_2(0)+a_4(1)=a_4,\hfill \\ =\frac{21}{128}\mu .\hfill & \hfill \end{array}$$

If we substitute (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) by $i=-2$ and $k=0,$ we get $$\begin{array}{cc}{P}_{-4}^1\hfill & =a_{-5}{P}_{-4}^0+a_{-3}{P}_{-3}^0+a_{-1}{P}_{-2}^0+a_1{P}_{-1}^0+a_3{P}_0^0+a_5{P}_1^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(0)+a_{-1}(0)+a_1(0)+a_3(1)+a_5(0)=a_3,\hfill \\ =-\frac{5}{128}+\frac{7}{128}\nu +\frac{7}{128}\omega ,\hfill & \hfill \end{array}$$ and $$\begin{array}{cc}{P}_{-3}^1\hfill & =a_{-6}{P}_{-4}^0+a_{-4}{P}_{-3}^0+a_{-2}{P}_{-2}^0+a_0{P}_{-1}^0+a_2{P}_0^0+a_4{P}_1^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(0)+a_{-2}(0)+a_0(0)+a_2(1)+a_4(0)=a_2,\hfill \\ =-\frac{7}{128}-\frac{35}{128}\mu +\frac{21}{128}\nu .\hfill & \hfill \end{array}$$

For $i=-1$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_{-2}^1\hfill & =a_{-5}{P}_{-3}^0+a_{-3}{P}_{-2}^0+a_{-1}{P}_{-1}^0+a_1{P}_0^0+a_3{P}_1^0+a_5{P}_2^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(0)+a_{-1}(0)+a_1(1)+a_3(0)+a_5(0)=a_1,\hfill \\ =\frac{35}{128}+\frac{7}{128}\mu +\frac{7}{128}\nu -\frac{35}{128}\omega ,\hfill & \hfill \end{array}$$ and $$\begin{array}{cc}{P}_{-1}^1\hfill & =a_{-6}{P}_{-3}^0+a_{-4}{P}_{-2}^0+a_{-2}{P}_{-1}^0+a_0{P}_0^0+a_2{P}_1^0+a_4{P}_2^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(0)+a_{-2}(0)+a_0(1)+a_2(0)+a_4(0)=a_0,\hfill \\ =\frac{105}{128}+\frac{7}{128}\mu -\frac{35}{128}\nu +\frac{21}{128}\omega .\hfill & \hfill \end{array}$$

If we put $i=0$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_0^1\hfill & =a_{-5}{P}_{-2}^0+a_{-3}{P}_{-1}^0+a_{-1}{P}_0^0+a_1{P}_1^0+a_3{P}_2^0+a_5{P}_3^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(0)+a_{-1}(1)+a_1(0)+a_3(0)+a_5(0)=a_{-1},\hfill \\ =\frac{105}{128}+\frac{7}{128}\mu -\frac{35}{128}\nu +\frac{21}{128}\omega ,\hfill & \hfill \end{array}$$ and $$\begin{array}{cc}{P}_1^1\hfill & =a_{-6}{P}_{-2}^0+a_{-4}{P}_{-1}^0+a_{-2}{P}_0^0+a_0{P}_1^0+a_2{P}_2^0+a_4{P}_3^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(0)+a_{-2}(1)+a_0(0)+a_2(0)+a_4(0)=a_{-2},\hfill \\ =\frac{35}{128}+\frac{7}{128}\mu +\frac{7}{128}\nu -\frac{35}{128}\omega .\hfill & \hfill \end{array}$$

If we put $i=1$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_2^1\hfill & =a_{-5}{P}_{-1}^0+a_{-3}{P}_0^0+a_{-1}{P}_1^0+a_1{P}_2^0+a_3{P}_3^0+a_5{P}_4^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(1)+a_{-1}(0)+a_1(0)+a_3(0)+a_5(0)=a_{-3},\hfill \\ =-\frac{7}{128}-\frac{35}{128}\mu +\frac{21}{128}\nu ,\hfill & \hfill \end{array}$$ and $$\begin{array}{cc}{P}_3^1\hfill & =a_{-6}{P}_{-1}^0+a_{-4}{P}_0^0+a_{-2}{P}_1^0+a_0{P}_2^0+a_2{P}_3^0+a_4{P}_4^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(1)+a_{-2}(0)+a_0(0)+a_2(0)+a_4(0)=a_{-4},\hfill \\ =-\frac{5}{128}+\frac{7}{128}\nu +\frac{7}{128}\omega .\hfill & \hfill \end{array}$$

If we use $i=2$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_4^1\hfill & =a_{-5}{P}_0^0+a_{-3}{P}_1^0+a_{-1}{P}_2^0+a_1{P}_3^0+a_3{P}_4^0+a_5{P}_5^0,\hfill \\ \hfill & =a_{-5}(1)+a_{-3}(0)+a_{-1}(0)+a_1(0)+a_3(0)+a_5(0)=a_{-5},\hfill \\ =\frac{21}{128}\mu ,\hfill & \hfill \end{array}$$ and $$\begin{array}{cc}{P}_5^1\hfill & =a_{-6}{P}_0^0+a_{-4}{P}_1^0+a_{-2}{P}_2^0+a_0{P}_3^0+a_2{P}_4^0+a_4{P}_5^0,\hfill \\ \hfill & =a_{-6}(1)+a_{-4}(0)+a_{-2}(0)+a_0(0)+a_2(0)+a_4(0)=a_{-6},\hfill \\ =\frac{7}{128}\omega .\hfill & \hfill \end{array}$$

If we use $i=3$ and $k=0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) , we get $$\begin{array}{cc}{P}_6^1\hfill & =a_{-5}{P}_1^0+a_{-3}{P}_2^0+a_{-1}{P}_3^0+a_1{P}_4^0+a_3{P}_5^0+a_5{P}_6^0,\hfill \\ \hfill & =a_{-5}(0)+a_{-3}(0)+a_{-1}(0)+a_1(0)+a_3(0)+a_5(0)=0,\hfill \end{array}$$ and $$\begin{array}{cc}{P}_7^1\hfill & =a_{-6}{P}_1^0+a_{-4}{P}_2^0+a_{-2}{P}_3^0+a_0{P}_4^0+a_2{P}_5^0+a_4{P}_6^0,\hfill \\ \hfill & =a_{-6}(0)+a_{-4}(0)+a_{-2}(0)+a_0(0)+a_2(0)+a_4(0)=0.\hfill \end{array}$$

Hence, if we use the design-control refinement scheme $S_{a,\mu ,\nu ,\omega }$ defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) on the given initial data (4.1)(4.1) $$P_i^0=\{\begin{array}{cc}& \begin{array}{cc}1\hfill & i=0,\hfill \\ 0\hfill & i\ne 0.\hfill \end{array}\end{array}$$(4.1) , then after first subdivision step the non-zero points are $P_{-6}^1,$ $P_{-5}^1,$ …, $P_4^1$ and $P_5^1.$ □

Lemma 4.2.

If we use refinement scheme $S_{a,\mu ,\nu ,\omega }$ on initial data $$\begin{array}{cc}{P}_i^0\hfill & =\{\begin{array}{cc}& \begin{array}{cc}1\hfill & i=0,\hfill \\ 0\hfill & i\ne 0.\hfill \end{array}\end{array}\hfill \end{array}$$

Then, after second subdivision step, the non-zero points are $P_{-18}^2,$ $P_{-17}^2,$ $\dots ,$ $P_{14}^2,$ $P_{15}^2.$

Lemma 4.3.

If we use refinement scheme $S_{a,\mu ,\nu ,\omega }$ on initial data $$\begin{array}{cc}{P}_i^0\hfill & =\{\begin{array}{cc}& \begin{array}{cc}1\hfill & i=0,\hfill \\ 0\hfill & i\ne 0.\hfill \end{array}\end{array}\hfill \end{array}$$

Then, after third subdivision step, the non-zero points are $P_{-42}^3,$ $P_{-41}^3,$ $\dots ,$ $P_{34}^3,$ $P_{35}^3.$

Theorem 4.4.

The support width of our binary refinement scheme $S_{a,\mu ,\nu ,\omega }$ defined in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) is 11. So, it vanishes outside the closed interval [−5.5,5.5].

Proof.

In order to prove the above result, we use the results of Lemmas 4.1, 4.2 and 4.3. Let us define a set $$\begin{array}{cc}{F}_k\hfill & =\left\{\frac{j}{2^k}: \forall j\in \mathbb{Z}\right\},\hfill \end{array}$$ such that $$\begin{array}{cc}\varphi \left(\frac{j}{2^k}\right)\hfill & =P_j^k \forall j\in \mathbb{Z}.\hfill \end{array}$$

By Lemma 4.1, we get that: if we use one subdivision step on initial data by our design-control refinement scheme, the leftmost non-zero point is $$\begin{array}{cc}{P}_{-6}^1\hfill & =\varphi \left(-\frac{6}{2}\right),\hfill \end{array}$$ and the rightmost non-zero point is $$\begin{array}{cc}{P}_5^1\hfill & =\varphi \left(\frac{5}{2}\right).\hfill \end{array}$$

By Lemma 4.2, we get that: if we use two subdivision steps on initial data by refinement scheme $S_{a,\mu ,\nu ,\omega },$ the leftmost non-zero point is $$\begin{array}{cc}{P}_{-18}^2=P_{-6(1+2)}^3\hfill & =\varphi \left(-\frac{6(1+2)}{2^2}\right),\hfill \end{array}$$ and the rightmost non-zero point is $$\begin{array}{cc}{P}_{15}^2=P_{5(1+2)}^3\hfill & =\varphi \left(\frac{5(1+2)}{2^2}\right).\hfill \end{array}$$

Similarly, from Lemma 4.3, we get that: if we use three subdivision steps on initial data by scheme $S_{a,\mu ,\nu ,\omega },$ the leftmost non-zero point is $$\begin{array}{cc}{P}_{-42}^3=P_{-6(1+2+2^2)}^3\hfill & =\varphi \left(-\frac{6(1+2+2^2)}{2^3}\right),\hfill \end{array}$$ and the rightmost non-zero point is $$\begin{array}{cc}{P}_{35}^3=P_{5(1+2+2^2)}^3\hfill & =\varphi \left(\frac{5(1+2+2^2)}{2^3}\right).\hfill \end{array}$$

If we continue this process, then after k subdivision steps, the leftmost non-zero point is $$\begin{array}{cc}{P}_{-6(1+2+2^2{\dots 2}^{k-1})}^k\hfill & =\varphi \left(-\frac{6(1+2+2^2{\dots 2}^{k-1})}{2^k}\right),\hfill \end{array}$$ and the rightmost non-zero point is $$\begin{array}{cc}{P}_{5(1+2+2^2{\dots 2}^{k-1})}^k\hfill & =\varphi \left(\frac{5(1+2+2^2{\dots 2}^{k-1})}{2^k}\right).\hfill \end{array}$$

The difference between left and right non-zero points at k-th subdivision steps is $$\begin{array}{cc}\mathit{Difference}=\hfill & \left[\frac{5(1+2+2^2{\dots 2}^{k-1})}{2^k}-\frac{-6(1+2+2^2{\dots 2}^{k-1})}{2^k}\right]\hfill \\ =\left[\frac{5(1+2+2^2{\dots 2}^{k-1})}{2^k}+\frac{6(1+2+2^2{\dots 2}^{k-1})}{2^k}\right]\hfill & \hfill \\ =\hfill & \left[(5+6)\left(\frac{1+2+2^2{\dots 2}^{k-1}}{2^k}\right)\right]\hfill \\ =\left[11\left(\frac{1}{2}+\frac{1}{2^2}+\dots ..+\frac{1}{2^k}\right)\right].\hfill & \hfill \end{array}$$

If $k\to \infty ,$ we get the support width of refinement scheme $S_{a,\mu ,\nu ,\omega }$ $$\begin{array}{cc}\text{Support} \text{Width}\hfill & =\underset{k\to \infty }{\lim }\left[11\left(\frac{1}{2}+\frac{1}{2^2}+\dots ..+\frac{1}{2^k}\right)\right]\hfill \\ =11\left[\frac{\frac{1}{2}}{1-\frac{1}{2}}\right]\hfill \\ =11.\hfill & \hfill \end{array}$$

Hence, the support width of our refinement scheme $S_{a,\mu ,\nu ,\omega }$ is 11. So, it vanishes outside the closed interval [−5.5, 5.5]. □

Corollary 4.5.

The support width of our refinement scheme $S_{t,\nu }$ is 7. So, it vanishes outside the closed interval [−3.5, 3.5].

Proof.

The proof of this corollary is trivial, because $S_{t,\nu }$ is a sub-scheme of scheme $S_{a,\mu ,\nu ,\omega }$ and it is obtained by putting $\mu =\omega =0$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) . □

Corollary 4.6.

The support width of our refinement scheme $S_{\ell ,\mu }$ is 11. So, it vanishes outside the closed interval [−5.5, 5.5].

Proof.

The proof of this corollary is also trivial, because $S_{\ell ,\mu }$ is a sub-scheme of scheme $S_{a,\mu ,\nu ,\omega }$ and it is obtained by putting $\mu =\omega$ and $\nu =1-2\mu$ in (3.6)(3.6) $$\{\begin{array}{c}{P}_{2i}^{k+1}=a_{-5}{P}_{i-2}^k+a_{-3}{P}_{i-1}^k+a_{-1}{P}_i^k+a_1{P}_{i+1}^k+a_3{P}_{i+2}^k+a_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}=a_{-6}{P}_{i-2}^k+a_{-4}{P}_{i-1}^k+a_{-2}{P}_i^k+a_0{P}_{i+1}^k+a_2{P}_{i+2}^k+a_4{P}_{i+3}^k,\end{array}$$(3.6) . □

Now, we analyze the support of our refinement scheme graphically.

Experiment 4.1.

In this experiment, we plot an initial sketch by using data points $(-7,0),$ $(-6,0),$ $(-5,0),$ $(-4,0),$ $(-3,0),$ $(-2,0),$ $(-1,0),$ $(0,1),$ $(1,0),$ $(2,0),$ $(3,0),$ $(4,0),$ $(5,0),$ $(6,0),$ $(7,0)$. The curves generated by our refinement scheme $S_{a,\mu ,\nu ,\omega }$ are shown in Figure 3. Figures 3(a–c) show the curves fitted by our refinement scheme after first, second and third refinement steps respectively. Here, the values of parameters are $(\mu ,\nu ,\omega )=(-0.1,1.2,-0.1).$

Figure 3. Black solid lines and red bullets represent initial control polygons and initial control points respectively. Blue solid lines represent curves fitted by our refinement scheme after first, second and third subdivision steps respectively.

Experiment 4.2.

In this experiment, we draw initial sketch by using the following initial control points $(5,15),$ $(5,7),$ $(2,7),$ $(2,6),$ $(1,6),$ $(1,4),$ $(2,4),$ $(2,3),$ $(5,3),$ $(5,4),$ $(6,4),$ $(6,11),$ $(11,11),$ $(11,16),$ $(8,6),$ $(8,5),$ $(7,5),$ $(7,3),$ $(8,3),$ $(8,2),$ $(11,2),$ $(11,3),$ $(12,3),$ $(12,15),$ $(5,15)$. The black dashed-dotted lines in Figure 4(a) represent the initial sketch. While, red curve in Figure 4(a) is the curve fitted by our scheme after three refinement steps. Now we draw another initial sketch by changing the position of one point of the sketch which is shown in Figure 4(a). The new sketch is shown in Figure 4(b) by black dashed-dotted lines. Which is simply obtained by moving point $(8.5,11)$ of previous sketch to the new position $(8.5,7)$. Figure 4(b) shows the curve fitted by our scheme after three refinement steps. Here, the values of parameters are: $(\mu ,\nu ,\omega )=(0,1,0)$. By comparing Figures 4(a and b), we conclude that by moving one control point of the initial sketch, only a very small portion of limit curve is affected. This property shows that our refinement scheme has a good local control.

Figure 4. Local control property of new design-control refinement scheme.

5. Continuity analysis of new design control refinement scheme

In this section, we present the convergence and smoothness analysis of the sub-schemes $S_{t,\nu }$ and $S_{\ell ,\mu }$ of new design-control refinement scheme. For the analysis, we use the theoretical results of (Hassan et al., 2002) and (Dyn & Levin, 2002) which are given in Theorem 2.1 and Theorem 2.2 respectively.

5.1. Continuity analysis of refinement scheme $S_{t,\nu }$

Here, we analyzed the convergence of binary approximating refinement scheme $S_{t,\nu }$ by using Eigenvalues of the subdivision matrix. We also check the level of smoothness of our proposed scheme by using the Laurent polynomial method for $L=1$ and $L=2.$

Theorem 5.1.

The refinement scheme (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) satisfies the necessary conditions for $C^2$ continuity if the range of the tension parameter is $0.7142857143<\nu <1.857142857.$

Proof.

To find the convergence of the refinement scheme $S_{t,\nu }$ which is defined in (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) , consider the local subdivision matrix T of the scheme $S_{t,\nu }.$ If we put $i=-1,0,1$ in the pair of refinement equations (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) , then we get a system of refinement equations. Which can be written in matrix form as: $$\begin{array}{cc}{\overline{P}}^{k+1}\hfill & =T{\overline{P}}^k,\hfill \end{array}$$ where, ${\overline{P}}^{k+1}=[P_{-2}^{k+1}$ $P_{-1}^{k+1}$ $P_0^{k+1}$ $P_1^{k+1}$ $P_2^{k+1}{]}^T,$ ${\overline{P}}^k=[P_{-2}^k$ $P_{-1}^k$ $P_0^k$ $P_1^k$ $P_2^k{]}^T$ and $$\begin{array}{cc}T\hfill & =\left(\begin{array}{cccccc}{t}_{-3}& t_{-1}& t_1& t_3& 0& 0\\ t_{-4}& t_{-2}& t_0& t_2& 0& 0\\ 0& t_{-3}& t_{-1}& t_1& t_3& 0\\ 0& t_{-4}& t_{-2}& t_0& t_2& 0\\ 0& 0& t_{-3}& t_{-1}& t_1& t_3\\ 0& 0& t_{-4}& t_{-2}& t_0& t_2\end{array}\right),\hfill \end{array}$$ the eigenvalues of the subdivision matrix T are: $${\lambda }_0=1,{\lambda }_1=\frac{1}{2},{\lambda }_2=\frac{1}{4},{\lambda }_3=\frac{1}{8},{\lambda }_4=\frac{1}{16},{\lambda }_5=\frac{9}{64}-\frac{7}{64}\nu .$$

From Theorem 2.1, the necessary conditions for $C^2$-continuity are: (5.1) $${\lambda }_0=1, \left|{\lambda }_i\right|\ge \left|{\lambda }_{i+1}\right| \forall i\in \mathbb{N} \text{and} |{\lambda }_1{|}^2=|{\lambda }_2|>|{\lambda }_3|.$$(5.1)

Now, by putting $i=4$ in (5.5)(5.5) $${\lambda }_0=1, \left|{\lambda }_i\right|\ge \left|{\lambda }_{i+1}\right| \forall i\in \mathbb{N} \text{and} |{\lambda }_1{|}^2=|{\lambda }_2|>|{\lambda }_3|.$$(5.5) , and compare the values $$\left|\frac{1}{16}\right|>\left|\frac{9}{64}-\frac{7}{64}\nu \right|.$$

After simplification we get: $$\frac{5}{7}<\nu <\frac{13}{7}.$$

Thus, we have $$0.7142857143<\nu <1.857142857.$$

Which, implies that the given scheme is convergent in this interval $0.7142857143<\nu <1.857142857.$ □

Theorem 5.2.

The binary refinement scheme (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) satisfies the sufficient conditions of $C^2$-continuity for $0.7142857143<\nu <1.857142857.$

Proof.

The Laurent polynomial of refinement scheme (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) is (5.2) $$\begin{array}{cc}t(z)\hfill & =\left(-\frac{5}{128}+\frac{7}{128}\nu \right)z^{-3}+\left(-\frac{7}{128}+\frac{21}{128}\nu \right)z^{-2}+\left(\frac{35}{128}+\frac{7}{128}\beta \right)z^{-1}+\left(\frac{105}{128}-\frac{35}{128}\nu \right)z^0+\left(\frac{105}{128}-\frac{35}{128}\nu \right)z^1+\left(\frac{35}{128}+\frac{7}{128}\nu \right)z^2+\left(-\frac{7}{128}+\frac{21}{128}\nu \right)z^3+\left(-\frac{5}{128}+\frac{7}{128}\nu \right)z^4.\hfill \end{array}$$(5.2)

By Theorem 2.2. (5.3) $$\begin{array}{cc}{b}^{[m,L]}(z)\hfill & =\frac{1}{2^L}{t}_m^{\left[L\right]}(z),\hspace{1em}m=1,2,\dots \hfill \end{array}$$(5.3) where, $$\begin{array}{cc}{t}_m(z)\hfill & =\frac{2z}{1+z}{t}_{m-1}(z)={\left(\frac{2z}{1+z}\right)}^{m}t(z),\hfill \end{array}$$ and, $$\begin{array}{cc}{t}_m^{\left[L\right]}(z)\hfill & =\prod _{j=0}^{L-1}{t}_m(z^{2j}).\hfill \end{array}$$ (5.4) $${\Vert {\left(\frac{1}{2}{S}_{t_m}\right)}^L\Vert }_{\infty }=\max \left\{\sum _j\left|b_{\gamma +2^{L}j}^{[m,L]}\right|;\gamma =0,1,\dots ,2^L-1\right\}<1, m=1,2,3.$$(5.4)

Now, by putting $L=1$ and $m=1$ in (5.3)(5.3) $$\begin{array}{cc}{b}^{[m,L]}(z)\hfill & =\frac{1}{2^L}{t}_m^{\left[L\right]}(z),\hspace{1em}m=1,2,\dots \hfill \end{array}$$(5.3) , we get: $$\begin{array}{cc}{b}_1^1(z)\hfill & =\frac{1}{2}{t}_1(z)=\frac{z}{z+1}t(z)=\left(\frac{7}{128}\nu -\frac{5}{128}\right)z^4+\left(\frac{7}{64}\nu -\frac{1}{64}\right)z^3+(-\frac{7}{128}\nu \hfill \\ +\frac{37}{128})z^2+\left(-\frac{7}{32}\nu +\frac{17}{32}\right)z^1+\left(-\frac{7}{128}\nu +\frac{37}{128}\right)z^0+\left(\frac{7}{64}\nu -\frac{1}{64}\right)z^{-1}+\left(\frac{7}{128}\nu -\frac{5}{128}\right)z^{-2}.\hfill & \hfill \end{array}$$

This implies that $$\begin{array}{cc}{\Vert {\left(\frac{1}{2}{S}_{t_1}\right)}^1\Vert }_{\infty }\hfill & =\max \{\left|\frac{7}{128}\nu -\frac{5}{128}\right|+\left|-\frac{7}{128}\nu +\frac{37}{128}\right|+\left|-\frac{7}{128}\nu +\frac{37}{128}\right|\hfill \\ +\left|\frac{7}{128}\nu -\frac{5}{128}\right|,\left|\frac{7}{64}\nu -\frac{1}{64}\right|+\left|-\frac{7}{32}\nu +\frac{17}{32}\right|+\left|\frac{7}{64}\nu -\frac{1}{64}\right|\}<1.\hfill & \hfill \end{array}$$

The above relation is true if $\nu$ lies between 0.7142857143 to 1.857142857. Hence the scheme is $C^0$-continuous. If we put $L=1$ and $m=2$ in (5.3)(5.3) $$\begin{array}{cc}{b}^{[m,L]}(z)\hfill & =\frac{1}{2^L}{t}_m^{\left[L\right]}(z),\hspace{1em}m=1,2,\dots \hfill \end{array}$$(5.3) , then, we obtain $$\begin{array}{cc}{b}_2^1(z)\hfill & =\frac{1}{2}{t}_2(z)=\frac{2z^2}{{(z+1)}^2}t(z)=\left(\frac{7}{64}\nu -\frac{5}{64}\right)z^4+\left(\frac{7}{64}\nu +\frac{3}{64}\right)z^3+(-\frac{7}{32}\nu \hfill \\ +\frac{17}{32})z^2+\left(-\frac{7}{32}\nu +\frac{17}{32}\right)z^1+\left(\frac{7}{64}\nu +\frac{3}{64}\right)z^0+\left(\frac{7}{64}\nu -\frac{5}{64}\right)z^{-1}.\hfill & \hfill \end{array}$$

This implies that $$\begin{array}{cc}{\Vert \frac{1}{2}{S}_{t_2}\Vert }_{\infty }\hfill & =\max \{\left|\frac{7}{64}\nu -\frac{5}{64}\right|+\left|-\frac{7}{32}\nu +\frac{17}{32}\right|+\left|\frac{7}{64}\nu +\frac{3}{64}\right|,\left|-\frac{7}{32}\nu +\frac{17}{32}\right|\hfill \\ +\left|\frac{7}{64}\nu -\frac{5}{64}\right|\}<1.\hfill & \hfill \end{array}$$

Hence, this scheme is $C^1$-continuous for $0.7142857143<\nu <1.857142857.$

Now we substitute $L=1$ and $m=3$ in (5.3)(5.3) $$\begin{array}{cc}{b}^{[m,L]}(z)\hfill & =\frac{1}{2^L}{t}_m^{\left[L\right]}(z),\hspace{1em}m=1,2,\dots \hfill \end{array}$$(5.3) , we get $$\begin{array}{ccc}{b}_3^1(z)\hfill & =\frac{1}{2}{t}_3(z)=\frac{4z^3}{{(z+1)}^3}t(z)=\left(\frac{7}{32}\nu -\frac{5}{32}\right)z^4+\left(\frac{1}{4}\right)z^3+\left(-\frac{7}{16}\nu +\frac{13}{16}\right)z^2+\left(\frac{1}{4}\right)z^1+\left(\frac{7}{32}\nu -\frac{5}{32}\right)z^0.\hfill & \hfill \end{array}$$

This implies that $$\begin{array}{cc}{\Vert \frac{1}{2}{S}_{t_3}\Vert }_{\infty }\hfill & =\max \left\{\left|\frac{7}{32}\nu -\frac{5}{32}\right|+\left|-\frac{7}{16}\nu +\frac{13}{16}\right|+\left|\frac{7}{32}\nu -\frac{5}{32}\right|,\left|\frac{1}{4}\right|+\left|\frac{1}{4}\right|\right\}<1.\hfill \end{array}$$

Thus the scheme is $C^2$ -continuous for $0.7142857143<\nu <1.857142857.$ □

In Table 1, we present the continuity results of our refinement scheme $S_{t,\nu }$ by Laurent polynomial method, that we have obtained by using results of Theorem 2.2.

Table 1. The order of smoothness of the refinement scheme $S_{t,\nu }$ for specific range of parameter.

Order of continuity	For $L=1$	For $L=2$
$C^0$	$-1<\nu <3.5714$	$-1.425464<\nu <5.1397500$
$C^1$	$-1<\nu <3.5714$	$-1.2013<\nu <4.7292$
$C^2$	$0.1429<\nu <2.4286$	$-0.0698<\nu <3.2127$
$C^3$	$0.1429<\nu <2.4286$	$0.1429<\nu <2.4286$
$C^4$	$0.1429<\nu <1.1429$	$0.1429<\nu <1.6388$

5.2. Continuity analysis of the refinement scheme $S_{\ell ,\mu }$

In this section, we analyzed the convergence of binary approximating refinement scheme $S_{\ell ,\mu }$ by analyzing the eigenvalues of the subdivision matrix. We also check the level of smoothness of our proposed scheme by using the Laurent polynomial method for $L=1$ and $L=2.$

Theorem 5.3.

The refinement scheme (3.8)(3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) satisfies the necessary conditions for $C^2$ continuity if the range of the tension parameter is $-0.1428571429<\mu <-0.06578510188.$

Proof.

To find the convergence of the refinement scheme $S_{\ell ,\mu },$ defined in (3.8)(3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) , consider the local subdivision matrix $\overline{T}$ of the scheme $S_{\ell ,\mu }.$ If we put $i=-2,-1,0,1,2$ in the pair of refinement equations (3.8)(3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) , then we get a system of refinement equations. Which can be written in matrix form as: $$\begin{array}{cc}{\widehat{P}}^{k+1}\hfill & =\overline{T}{\widehat{P}}^k,\hfill \end{array}$$ where ${\widehat{P}}^{k+1}=[P_{-3}^{k+1}$ $P_{-2}^{k+1}$ $P_{-1}^{k+1}$ $P_0^{k+1}$ $P_1^{k+1}$ $P_2^{k+1}$ $P_3^{k+1}{]}^T,$ ${\widehat{P}}^k=[P_{-3}^k$ $P_{-2}^k$ $P_{-1}^k$ $P_0^k$ $P_1^k$ $P_2^k$ $P_3^k{]}^T$ and, $$\begin{array}{c}\overline{T}=\left(\begin{array}{cccccccccc}{\ell }_{-5}& {\ell }_{-3}& {\ell }_{-1}& {\ell }_1& {\ell }_3& {\ell }_5& 0& 0& 0& 0\\ {\ell }_{-6}& {\ell }_{-4}& {\ell }_{-2}& {\ell }_0& {\ell }_2& {\ell }_4& 0& 0& 0& 0\\ 0& {\ell }_{-5}& {\ell }_{-3}& {\ell }_{-1}& {\ell }_1& {\ell }_3& {\ell }_5& 0& 0& 0\\ 0& {\ell }_{-6}& {\ell }_{-4}& {\ell }_{-2}& {\ell }_0& {\ell }_2& {\ell }_4& 0& 0& 0\\ 0& 0& {\ell }_{-5}& {\ell }_{-3}& {\ell }_{-1}& {\ell }_1& {\ell }_3& {\ell }_5& 0& 0\\ 0& 0& {\ell }_{-6}& {\ell }_{-4}& {\ell }_{-2}& {\ell }_0& {\ell }_2& {\ell }_4& 0& 0\\ 0& 0& 0& {\ell }_{-5}& {\ell }_{-3}& {\ell }_{-1}& {\ell }_1& {\ell }_3& {\ell }_5& 0\\ 0& 0& 0& {\ell }_{-6}& {\ell }_{-4}& {\ell }_{-2}& {\ell }_0& {\ell }_2& {\ell }_4& 0\\ 0& 0& 0& 0& {\ell }_{-5}& {\ell }_{-3}& {\ell }_{-1}& {\ell }_1& {\ell }_3& {\ell }_5\\ 0& 0& 0& 0& {\ell }_{-6}& {\ell }_{-4}& {\ell }_{-2}& {\ell }_0& {\ell }_2& {\ell }_4\end{array}\right).\hfill \end{array}$$

The eigenvalue of the above subdivision matrix are: $$\begin{array}{cc}{\lambda }_0\hfill & =1,{\lambda }_1=\frac{1}{2},{\lambda }_2=\frac{1}{4},{\lambda }_3=\frac{1}{8},{\lambda }_4=\frac{1}{16},{\lambda }_5=\frac{1}{32},{\lambda }_6=\frac{1}{64},{\lambda }_6=\frac{1}{64},{\lambda }_7=-\frac{7}{32}\mu ,\hfill \\ {\lambda }_8\hfill & =\frac{7}{128}\mu +\frac{1}{128}+\frac{1}{128}\sqrt{833{\mu }^2+70\mu +1},{\lambda }_9=\frac{7}{128}\mu +\frac{1}{128}-\frac{1}{128}\sqrt{833{\mu }^2+70\mu +1}.\hfill & \hfill \end{array}$$

From Theorem 2.1, the necessary conditions for $C^2$-continuity are: (5.5) $${\lambda }_0=1, \left|{\lambda }_i\right|\ge \left|{\lambda }_{i+1}\right| \forall i\in \mathbb{N} \text{and} |{\lambda }_1{|}^2=|{\lambda }_2|>|{\lambda }_3|.$$(5.5)

Let $$\left|\frac{1}{64}\right|>\left|-\frac{7}{32}\mu \right|.$$

This implies that $$-\frac{1}{14}<\mu <\frac{1}{14}.$$

Thus, we have (5.6) $$-0.071429<\mu <0.071429.$$(5.6)

Let, $$\left|-\frac{7}{32}\mu \right|>\left|\frac{7}{128}\mu +\frac{1}{128}+\frac{1}{128}\sqrt{833{\mu }^2+70\mu +1}\right|.$$

This, implies that $$-1.290803377<\mu <-0.06578510188 \text{and}-0.01824851156<\mu <0.7193748053.$$

By combining the above we get result as: (5.7) $$-1.290803377<\mu <0.7193748053.$$(5.7)

Let, $$\left|\frac{7}{128}\mu +\frac{1}{128}+\frac{1}{128}\sqrt{833{\mu }^2+70\mu +1}\right|>\left|\frac{7}{128}\mu +\frac{1}{128}-\frac{1}{128}\sqrt{833{\mu }^2+70\mu +1}\right|.$$

This, implies that $$-0.1428571429<\mu <-0.06578510188 \text{and} -0.01824851156<\mu .$$

By combining above we get the result (5.8) $$-0.1428571429<\mu <-0.06578510188.$$(5.8)

If we take the common part of all inequalities (5.6)(5.8) $$-0.1428571429<\mu <-0.06578510188.$$(5.8) , (5.7)(5.7) $$-1.290803377<\mu <0.7193748053.$$(5.7) and (5.8)(5.8) $$-0.1428571429<\mu <-0.06578510188.$$(5.8) , we get the following common interval (5.9) $$-0.1428571429<\mu <-0.06578510188.$$(5.9)

Theorem 5.4.

The refinement scheme (3.8)(3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) satisfies the sufficient conditions for $C^2$-smoothness for $-0.1071428571<\mu <-0.01824851156.$

In Table 2, we summarize the continuity results of our refinement scheme $S_{\ell ,\mu }$ by Laurent polynomial method which we have obtained by using results of Theorem 2.2.

Table 2. The order of smoothness of the refinement scheme $S_{\ell ,\mu }$ for specific range of parameter.

Order of continuity	For $L=1$	For $L=2$
$C^0$	$-0.6429<\mu <0.500$	$-1.58005<\mu <0.7943$
$C^1$	$-0.6429<\mu <0.500$	$-1.15356<\mu <0.69177$
$C^2$	$-0.4082<\mu <0.2449$	$-0.6982410<\mu <0.345555$
$C^3$	$-0.3571<\mu <0.2143$	$-0.48060<\mu <0.2857$
$C^4$	$-0.1429<\mu <0.0714$	$-0.24155<\mu <0.09394$
$C^5$	$-0.1429<\mu <0.0476$	$-0.1429<\mu <0.05055$
$C^6$	$-0.0351<\mu <0$	$-0.07<\mu <0$

Now, we give an experiment to show the smoothness property of our design-control refinement scheme geometrically.

Experiment 5.1.

In this experiment, we draw the initial control polygon by using there initial control points: $(-14,4),$ $(-12,12),$ $(-10,13),$ $(0,13),$ $(16,13),$ $(15,9),$ $(10,9),$ $(6,-7),$ $(5,-11),$ $(5,-12),$ $(7,-13),$ $(12,-13),$ $(11,-16),$ $(4,-16),$ $(3,-15),$ $(2,-13),$ $(2,-11),$ $(4,-1),$ $(6,9),$ $(0,9),$ $(-3,9),$ $(-5,-1),$ $(-8,-16),$ $(-12,-16),$ $(-9,-1),$ $(-7,9),$ $(-9,9),$ $(-10,4)$ and $(-14,4)$. We show the curve generated by our refinement scheme $S_{a,\mu ,\nu ,\omega }$ in Figure 5. Figure 5(a) shows the initial control polygon. Figures 5(b–d) show the curves fitted by our refinement scheme after one, two and three refinement steps respectively. Where $(\mu ,\nu ,\omega )$ = $(0,1,0).$

Figure 5. Smooth curve fitted by our refinement scheme.

6. Gibbs phenomenon analysis

In this section, we theoretically and graphically analyze the Gibbs phenomenon of sub-schemes $S_{t,\nu }$ and $S_{\ell ,\mu }.$ By using the mask of these refinement schemes, we obtain certain conditions to analyze whether there exist a Gibbs phenomenon close to the discontinuous points or not (Table 3).

Table 3. Values of ${\zeta }_0^{[1]}(i)$ and ${\zeta }_1^{[1]}(i)$ for $S_{t,\nu }.$

i	$-2$	$-1$	0	1
${\zeta }_0^{[1]}(i)$	$-\frac{5}{128}+\frac{7}{128}\nu$	$\frac{30}{128}+\frac{14}{128}\nu$	0	$-\frac{7}{128}+\frac{21}{128}\nu$
${\zeta }_1^{[1]}(i)$	$-\frac{7}{128}+\frac{21}{128}\nu$	$\frac{49}{64}-\frac{7}{64}\nu$	0	$-\frac{5}{128}+\frac{7}{128}\nu$

Theorem 6.1.

The binary approximating refinement scheme $S_{t,\nu }$ defined in (3.7)(3.7) $$\{\begin{array}{c}{P}_{2i}^{k+1}=t_{-3}{P}_{i-1}^k+t_{-1}{P}_i^k+t_1{P}_{i+1}^k+t_3{P}_{i+2}^k,\\ \\ P_{2i+1}^{k+1}=t_{-4}{P}_{i-1}^k+t_{-2}{P}_i^k+t_0{P}_{i+1}^k+t_2{P}_{i+2}^k,\end{array}$$(3.7) does not produce Gibbs oscillation in the limiting curves for $0.7142857\le \nu \le 7.$

Proof.

By Theorem 2.3, a stationary scheme is free from Gibbs oscillation near to the discontinuity if (6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) where, (6.2) $$\begin{array}{cc}{\zeta }_{\overline{l}}^{\left[k\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{a}_{2^{\left[k\right]}\tau +\overline{l}}^{\left[k\right]}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{a}_{2^{\left[k\right]}\tau +\overline{l}}^{\left[k\right]}\hfill & i>0,\hfill \end{array}\hfill \end{array}$$(6.2) and, $0⩽\overline{l}<2^k.$

Since, the Laurent polynomial of our scheme $S_{t,\nu }$ is: (6.3) $$\begin{array}{cc}t(z)\hfill & =\sum _{j=-4}^3{t}_j{z}^j,\hfill \end{array}$$(6.3)

Therefore, if we fix $k=1$ and put $a_j^{[1]}=t_j:j=-4,-3,\dots ,3$ in (6.2)(6.2) $$\begin{array}{cc}{\zeta }_{\overline{l}}^{\left[k\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{a}_{2^{\left[k\right]}\tau +\overline{l}}^{\left[k\right]}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{a}_{2^{\left[k\right]}\tau +\overline{l}}^{\left[k\right]}\hfill & i>0,\hfill \end{array}\hfill \end{array}$$(6.2) , we get (6.4) $$\begin{array}{cc}{\zeta }_{\overline{l}}^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau +\overline{l}}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau +\overline{l}}\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.4)

Now, we discuss two different cases corresponding to two values of $\overline{l},$ namely, $\overline{l}=0$ and $\overline{l}=1.$

Case 1:

The value of $\overline{l}$ in this case is fixed, that is $\overline{l}=0.$ Therefore we have (6.5) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau }\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau }\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.5)

If we use $i=-2$ in (6.5)(6.5) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau }\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau }\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.5) , we get (6.6) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(-2)\hfill & =\sum _{\tau \le -2}{t}_{2\tau }=t_{-4}+t_{-6}+t_{-8}+\dots =t_{-4}.\hfill \end{array}$$(6.6)

From (6.1)(6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) , the following relation is required $${\zeta }_0^{\left[1\right]}(-2)\ge 0.$$

This implies that $$\begin{array}{cc}{t}_{-4}\hfill & =\left(-\frac{5}{128}+\frac{7}{128}\nu \right)\ge 0.\hfill \end{array}$$

Thus, we have (6.7) $${\zeta }_0^{\left[1\right]}(-2)\ge 0 \text{for} \nu \ge 0.7142857.$$(6.7)

If we put $i=-1$ in (6.5)(6.5) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau }\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau }\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.5) , we get (6.8) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(-1)\hfill & =\sum _{\tau \le -1}{t}_{2\tau }=t_{-2}+t_{-4}+t_{-6}+\dots =t_{-2}+t_{-4}.\hfill \end{array}$$(6.8)

From (6.1)(6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) , the following relation is required $${\zeta }_0^{\left[1\right]}(-1)\ge 0.$$

This implies that $$\begin{array}{cc}{t}_{-2}+t_{-4}\hfill & =\left(\frac{35}{128}+\frac{7}{128}\nu \right)+\left(-\frac{5}{128}+\frac{7}{128}\nu \right),\hfill \\ \hfill & =\left(\frac{30}{128}+\frac{14}{128}\nu \right)\ge 0.\hfill \end{array}$$

Thus, we have (6.9) $${\zeta }_0^{\left[1\right]}(-1)\ge 0 \text{for} \nu \ge -2.142857143.$$(6.9)

If we put $i=0$ in (6.5)(6.5) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau }\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau }\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.5) , we get (6.10) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(0)\hfill & =0 \text{forall} \nu .\hfill \end{array}$$(6.10)

If we substitute $i=1$ in (6.5)(6.5) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau }\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau }\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.5) , we get (6.11) $$\begin{array}{cc}{\zeta }_0^{\left[1\right]}(1)\hfill & =\sum _{\tau \ge 1}{t}_{2\tau }=t_2+t_4+t_6+\dots =t_2.\hfill \end{array}$$(6.11)

From (6.1)(6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) , the following relation is required $${\zeta }_0^{\left[1\right]}(1)\ge 0.$$

This implies that $$\begin{array}{cc}{t}_2\hfill & =\left(-\frac{7}{128}+\frac{21}{128}\nu \right)\ge 0.\hfill \end{array}$$

Thus, we have (6.12) $${\zeta }_0^{\left[1\right]}(1)\ge 0 \text{for} 0.3333333\le \nu .$$(6.12)

By Combining (6.7)(6.12) $${\zeta }_0^{\left[1\right]}(1)\ge 0 \text{for} 0.3333333\le \nu .$$(6.12) , (6.9)(6.9) $${\zeta }_0^{\left[1\right]}(-1)\ge 0 \text{for} \nu \ge -2.142857143.$$(6.9) and (6.12)(6.12) $${\zeta }_0^{\left[1\right]}(1)\ge 0 \text{for} 0.3333333\le \nu .$$(6.12) , we get the following result $${\zeta }_0^{\left[1\right]}(-2),{\zeta }_0^{\left[1\right]}(-1),{\zeta }_0^{\left[1\right]}(0),{\zeta }_0^{\left[1\right]}(1)\ge 0 \text{for} 0.7142857\le \nu \le 7.$$

Case 2:

The value of $\overline{l}$ in this case is fixed, that is $\overline{l}=1.$ Therefore we have (6.13) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau +1}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau +1}\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.13)

If we substitute $i=-2$ in (6.13)(6.13) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau +1}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau +1}\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.13) , we get (6.14) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(-2)\hfill & =\sum _{\tau \le -2}{t}_{2\tau +1}=t_{-3}+t_{-5}+t_{-7}+\dots =t_{-3}.\hfill \end{array}$$(6.14)

From (6.1)(6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) , the following relation is required $${\zeta }_1^{\left[1\right]}(-2)\ge 0.$$

This, implies that $$t_{-3}=\left(-\frac{7}{128}+\frac{21}{128}\nu \right)\ge 0.$$

Thus, we have (6.15) $${\zeta }_1^{\left[1\right]}(-2)\ge 0 \text{for} 0.3333333\le \nu .$$(6.15)

Now, if we put $i=-1$ in (6.13)(6.13) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau +1}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau +1}\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.13) , we get (6.16) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(-1)\hfill & =\sum _{\tau \le -1}{t}_{2\tau +1}=t_{-1}+t_{-3}+t_{-5}+\dots =t_{-1}+t_{-3}.\hfill \end{array}$$(6.16)

From (6.1)(6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) , the following relation is required $${\zeta }_1^{\left[1\right]}(-1)\ge 0.$$

This implies that $$\begin{array}{cc}{t}_{-1}+t_{-3}\hfill & =\left(-\frac{7}{128}+\frac{21}{128}\nu \right)+\left(\frac{105}{128}-\frac{35}{128}\nu \right),\hfill \\ =\hfill & \left(\frac{49}{64}-\frac{7}{64}\nu \right)\ge 0.\hfill \end{array}$$

Thus, we have (6.17) $${\zeta }_1^{\left[1\right]}(-1)\ge 0 \text{for} \nu \le 7.$$(6.17)

Now, we put $i=0$ in (6.13)(6.13) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau +1}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau +1}\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.13) , we get $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(0)\hfill & =0 \text{for} \mu .\hfill \end{array}$$

Now, we put $i=1$ in (6.13)(6.13) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(i)\hfill & =\{\begin{array}{cc}\sum _{\tau \le i}{t}_{2\tau +1}\hfill & i<0,\hfill \\ 0,\hfill & i=0,\hfill \\ \sum _{\tau \ge i}{t}_{2\tau +1}\hfill & i>0.\hfill \end{array}\hfill \end{array}$$(6.13) , we get (6.18) $$\begin{array}{cc}{\zeta }_1^{\left[1\right]}(1)\hfill & =\sum _{\tau \ge 1}{t}_{2\tau +1}=t_3+t_5+t_7+\dots =t_3.\hfill \end{array}$$(6.18)

From (6.1)(6.1) $${\zeta }_{\overline{l}}^{\left[k\right]}(i)\ge 0 \forall i,k,$$(6.1) , the following relation is required $${\zeta }_1^{\left[1\right]}(1)\ge 0.$$

This implies that $$\begin{array}{cc}{t}_3\hfill & =\left(-\frac{5}{128}+\frac{7}{128}\nu \right)\ge 0.\hfill \end{array}$$

Thus, we have (6.19) $${\zeta }_1^{\left[1\right]}(1)\ge 0 \text{for} \nu \ge 0.7142857.$$(6.19)

Now, we combine the interval for $\nu$ from (6.15)(6.19) $${\zeta }_1^{\left[1\right]}(1)\ge 0 \text{for} \nu \ge 0.7142857.$$(6.19) , (6.17)(6.17) $${\zeta }_1^{\left[1\right]}(-1)\ge 0 \text{for} \nu \le 7.$$(6.17) and (6.19)(6.19) $${\zeta }_1^{\left[1\right]}(1)\ge 0 \text{for} \nu \ge 0.7142857.$$(6.19) , we get the following result (6.20) $${\zeta }_1^{\left[1\right]}(-2),{\zeta }_1^{\left[1\right]}(-1),{\zeta }_1^{\left[1\right]}(0),{\zeta }_1^{\left[1\right]}(1)\ge 0 \text{for} 0.7142857\le \nu \le 7.$$(6.20)

By Combining (6) and (6.20)(6.20) $${\zeta }_1^{\left[1\right]}(-2),{\zeta }_1^{\left[1\right]}(-1),{\zeta }_1^{\left[1\right]}(0),{\zeta }_1^{\left[1\right]}(1)\ge 0 \text{for} 0.7142857\le \nu \le 7.$$(6.20) , we get the common interval $0.7142857\le \nu \le 7$ for which our scheme $S_{t,\nu }$ does not produces Gibbs oscillations. □

Theorem 6.2.

The binary approximating refinement scheme $S_{\ell ,\mu }$ defined in (3.8)(3.8) $$\{\begin{array}{c}{P}_{2i}^{k+1}={\ell }_{-5}{P}_{i-2}^k+{\ell }_{-3}{P}_{i-1}^k+{\ell }_{-1}{P}_i^k+{\ell }_1{P}_{i+1}^k+{\ell }_3{P}_{i+2}^k+{\ell }_5{P}_{i+3}^k,\\ \\ P_{2i+1}^{k+1}={\ell }_{-6}{P}_{i-2}^k+{\ell }_{-4}{P}_{i-1}^k+{\ell }_{-2}{P}_i^k+{\ell }_0{P}_{i+1}^k+{\ell }_2{P}_{i+2}^k+{\ell }_4{P}_{i+3}^k,\end{array}$$(3.8) does not produce Gibbs oscillation in limiting shapes for $\mu =0.$

Proof.

We can prove this theorem by adopting the steps of Theorem 6.1. □

Experiment 6.1.

Here, in this experiment, we take the initial data from discontinuous function defined as: (6.21) $$f(x)=\{\begin{array}{c}\hspace{0.17em}\sin \hspace{0.17em}(\pi x),\hspace{1em}x\in [0,0.5],\hfill \\ \hfill \\ -\hspace{0.17em}\sin \hspace{0.17em}(\pi x),\hspace{1em}x\in [0.5,1].\hfill \end{array}$$(6.21)

Function (6.21)(6.21) $$f(x)=\{\begin{array}{c}\hspace{0.17em}\sin \hspace{0.17em}(\pi x),\hspace{1em}x\in [0,0.5],\hfill \\ \hfill \\ -\hspace{0.17em}\sin \hspace{0.17em}(\pi x),\hspace{1em}x\in [0.5,1].\hfill \end{array}$$(6.21) provides us the initial data which is given in the Table 4 and shown in Figure 6(a). Figures 6(b and c) show the Gibbs oscillation free curves fitted by the scheme $S_{t,\nu }$ for $\nu =1$ and $\nu =1.5$ respectively.

Figure 6. (a) shows the initial polygon and initial control points (b) and (c) show the curves fitted by our scheme after three subdivision steps.

Table 4. Data of taken from function defined in (6.21)(6.21) $$f(x)=\{\begin{array}{c}\hspace{0.17em}\sin \hspace{0.17em}(\pi x),\hspace{1em}x\in [0,0.5],\hfill \\ \hfill \\ -\hspace{0.17em}\sin \hspace{0.17em}(\pi x),\hspace{1em}x\in [0.5,1].\hfill \end{array}$$(6.21) .

x	0	0.1	0.2	0.3	0.4	0.5	0.5	0.6	0.7	0.8
$f(x)$	0	0.039	0.587	0.809	0.951	1	$-1$	$-0.95$	$-0.809$	$-0.587$
x	0.9	1
$f(x)$	$-0.309$	0

Experiment 6.2.

In this experiment, we plot an initial sketch by using the data points $(2,0),$ $(5,0),$ $(5,2),$ $(6,2),$ $(6,3),$ $(7,3),$ $(7,4),$ $(8,4),$ $(8,3),$ $(10,3),$ $(10,2),$ $(12,2),$ $(12,4),$ $(11,4),$ $(11,5),$ $(10,5),$ $(10,6),$ $(11,6),$ $(11,7),$ $(12,7),$ $(12,8),$ $(13,8),$ $(13,9),$ $(14,9),$ $(14,10),$ $(15,10),$ $(15,11),$ $(16,11),$ $(16,12),$ $(17,12),$ $(17,13),$ $(18,13),$ $(18,16),$ $(15,16),$ $(15,15),$ $(14,15),$ $(14,14),$ $(13,14),$ $(13,13),$ $(12,13),$ $(12,12),$ $(11,12),$ $(11,11),$ $(10,11),$ $(10,10),$ $(9,10),$ $(9,9),$ $(8,9),$ $(8,8),$ $(7,8),$ $(7,9),$ $(6,9),$ $(6,10),$ $(4,10),$ $(4,8),$ $(5,8),$ $(5,6),$ $(6,6),$ $(6,5),$ $(5,5),$ $(5,4),$ $(4,4),$ $(4,3),$ $(2,3)$ and $(2,0)$. We present the curves generated by our refinement scheme $S_{t,\nu }$ in Figure 7. Figures 7(b–d) show the smooth and oscillation free curves fitted by our refinement scheme when we set $\nu =1.2$ $\nu =1.8$ and $\nu =1$ respectively.

Figure 7. Black dashed-dotted lines represent initial control polygons. Red lines represent curves fitted by our refinement scheme after three subdivision steps.

7. Summary and conclusions

The summary of this research is:

• We construct a binary 6-point approximating refinement scheme by using two existing binary 4-point approximating refinement schemes.

• The support width of our scheme is 11 and its support region is $[-5.5,5.5].$

• Our refinement scheme produces limiting curves up to $C^6$ smoothness.

• If we set the value of parameters in the proved fixed range, our scheme produces curve free from artifacts and Gibbs oscillations.

Hence, we conclude that our scheme is a good choice for curve modeling with small support and high continuity, while the extra benefits can be achieved by using special values of parameters.

Disclosure statement

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Data availability

The data used to support the findings of the study is available within this paper.