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Research article

Spatial twisted central configuration for Newtonian (2N+1)-body problem

  • Received: 04 November 2023 Revised: 11 March 2024 Accepted: 10 April 2024 Published: 20 May 2024
  • 70F10, 70F15

  • For a spatial twisted central configuration of the Newtonian (2N+1)-body problem where 2N masses are at the vertices of two paralleled regular N-polygons with distance h>0, and the twist angle between the two regular N-polygons is 0θ<2π, we study the sufficient and necessary conditions for the existence of the spatial twisted central configuration. Additionally, we obtain the uniqueness of the spatial twisted central configuration.

    Citation: Liang Ding, Jinrong Wang, Jinlong Wei. Spatial twisted central configuration for Newtonian (2N+1)-body problem[J]. Communications in Analysis and Mechanics, 2024, 16(2): 388-415. doi: 10.3934/cam.2024018

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  • For a spatial twisted central configuration of the Newtonian (2N+1)-body problem where 2N masses are at the vertices of two paralleled regular N-polygons with distance h>0, and the twist angle between the two regular N-polygons is 0θ<2π, we study the sufficient and necessary conditions for the existence of the spatial twisted central configuration. Additionally, we obtain the uniqueness of the spatial twisted central configuration.



    The refinement schemes can be viewed as a class of iterative algorithms. The main feature of these schemes is that they use initially the rough sketch of any curve called polygon to finally produce smooth sketch also called refined sketch. Initially, a refinement scheme was introduced by de Rahm [3] in 1956. Later on, Chaikin [1] introduced the scheme for curve generation in 1974. Deslauriers and Dubuc presented the interpolating schemes for curve modeling in 1989. This triggered off the modern era for the investigation of the refinement schemes. These refinement schemes have been used for curve and surface modeling for a long time by designers, engineers, computer scientists and mathematicians. The details of these applications are presented by Farin [7].

    Initially, the binary refinement schemes were introduced. These schemes have two rules to refine each edge of the polygon. These rules are same for each edge of the polygon except the initial and final edges. These rules are classified as even and odd rules. These schemes take initial sketch as an input and return the refined sketch as an output. In the next step, these schemes take previously refined sketch as an input and produce more refined sketch. Repeated application of this procedure give the smooth curve/shape. Nowadays, there are many generalizations of the binary schemes. The class of ternary refinement schemes is one of the generalization of the class of binary schemes. In this class of the schemes, three rules are used instead of two rules to refine each edge of the polygon.

    The analysis of the schemes is the crucial issue. Initially, Dyn et al. [4], introduced the divided difference (DD) technique to analyze the schemes in 1987. Later on, Dyn et al. [5], extended this technique for the analysis of the class of binary schemes in 1991. The study of the limiting curves produced by the binary refinement schemes was also presented by Micchelli and Prautzch [11] in 1989. Sabin [15] in 2010, gave further insight in this technique. The further extension of Dyn et al.'s work was done by Mustafa and Zahid [12] in 2013. The cross-differences of directional DD techniques was introduced by Qu [14] in 1991. The further generalization of the DD technique has not been done so for.

    Another technique for the smoothness analysis of the schemes was introduced by Dyn [6] in 2002. This technique is known as Laurent polynomial technique. This technique was used by Hassan and Dodgson [8], Hassan et al. [9], Mustafa et al. [13]. Khan and Mustafa [10], Siddiqi and Rehan [16], Zheng et al. [17] and many others to analyze their binary and ternary refinement schemes. Nowadays many authors are also using this technique. However, this technique also has some limitations and need improvement. In this technique, first a sequence of coefficients used in the refinement rules of the schemes is converted into the polynomial. Secondly, the polynomial has been factorized. In this technique, multiplication, division, factorization of the polynomials are involved. Further, the computation of inequalities and the comparison of the terms are also involved. The technique, we are going to introduce is the generalization of DD technique. In this technique, the computation of inequalities and the comparison of three terms are involved. There are three simple algebraic expressions in each inequality. Further these algebraic expressions contain only the coefficients used in the refinement rules of the schemes.

    The rest of this paper is structured as follows. In Section 2, a general ternary refinement scheme and its DD schemes are introduced. Their convergence is also presented in this section. In Section 3, the deviation between successive levels of polygons produced by the ternary and its DD refinement schemes are presented. The inequalities for the analysis of ternary schemes are presented in Section 4. Applications of these inequalities are also presented in this section. Summary of the work and comparison are presented in Sections 5 and 6 respectively.

    We first introduce the ternary and its first, second and third order divided differences (DD) refinement schemes. Then we present their convergence in this section.

    Let fk={(i/3k,fki)RN, iZ, N2, k0}, be a polygon at kth refinement level then the (k+1)th polygon fk+1, can be obtained by the following three refinement rules

    {fk+13i=mj=0ρjfki+j,fk+13i+1=mj=0ϱjfki+j,fk+13i+2=mj=0σjfki+j. (2.1)

    whereas m>0 while if tki=i/3k then (tk+13i=tki, fk+13i), (tk+13i+1=13(2tki+tki+1), fk+13i+1)

    and (tk+13i+2=13(tki+2tki+1), fk+13i+2) are the points of the polygon fk+1.

    The above rules are the affine combinations of fki's so

    mν=0ρν=mν=0ϱν=mν=0σν=1. (2.2)

    The one step of the refinement procedure is shown in Figure 1. The repeated application of these rules is known as ternary refinement scheme.

    Figure 1.  The part of polygons at kth and (k+1)th levels are shown by solid and dash lines respectively.

    The first, second and third order divided difference (DD) ternary refinement schemes (TRS) are presented in this section.

    Lemma 2.1. Let fk={(i/3k,fki)} be a polygon of TRS at kth refinement level and d[1]kj=3k(fkj+1fkj) be the first divided difference then the first order DD TRS is defined as

    {d[1](k+1)3i=3m1μ=0{μν=0(ρμνϱμν)d[1]ki+μ},d[1](k+1)3i+1=3m1μ=0{μν=0(ϱμνσμν)d[1]ki+μ},d[1](k+1)3i+2=3mμ=0{μν=0(σμνρμν1)d[1]ki+μ}. (2.3)

    Proof. Since the first order DD is defined as

    d[1]kν=3k(fkν+1fkν), (2.4)

    therefore by (2.1), we get

    d[1](k+1)3i=3k+1{(ϱ0ρ0)fki+(ϱ1ρ1)fki+1+(ϱ2ρ2)fki+2++(ϱm1ρm1)fki+m1+(ϱmρm)fki+m}. (2.5)

    Now consider the linear combination

    d[1](k+1)3i=y0d[1]ki+y1d[1]ki+1+y2d[1]ki+2++ym1d[1]ki+(m1). (2.6)

    The goal is to find the unknown y0,,ym1. It can be written as

    d[1](k+1)3i=3k{y0fki+(y0y1)fki+1+(y1y2)fki+2++(ym3ym2)fkm2+(ym2ym1)fki+(m1)+ym1fki+m}. (2.7)

    Comparing (2.5) and (2.7), we get

    y0=3(ρ0ϱ0),y1=3(ρ0ϱ0)+3(ρ1ϱ1),y2=3(ρ0ϱ0)+3(ρ1ϱ1)+3(ρ2ϱ2),ym1=3(ρ0ϱ0)+3(ρ1ϱ1)++3(ρm1ϱm1).

    Substituting in (2.6), we get

    d[1](k+1)3i=3[(ρ0ϱ0)d[1]ki+{(ρ0ϱ0)+(ρ1ϱ1)}d[1]ki+1+{(ρ0ϱ0)+(ρ1ϱ1)+(ρ2ϱ2)}d[1]ki+2++{(ρ0ϱ0)+(ρ1ϱ1)++(ρm1ϱm1)}d[1]ki+(m1)].

    This implies

    d[1](k+1)3i=3m1μ=0{μν=0(ρμνϱμν)d[1]ki+μ}.

    Similarly, we get

    d[1](k+1)3i+1=3m1μ=0{μν=0(ϱμνσμν)d[1]ki+μ},

    and

    d[1](k+1)3i+2=3mμ=0{μν=0(σμνρμν1)d[1]ki+μ}.

    This completes the proof. By adopting the similar procedure, we get the following results.

    Lemma 2.2. Let fk={(i/3k,fki)} be a polygon of TRS at kth refinement level and

    d[2]kj=32k(2!)1(fkj12fkj+fkj+1)

    be the second order divided difference then the second order DD TRS is defined as

    {d[2](k+1)3i=32m1μ=0[μν=0{(ν+1)σμν+νϱμν2νρμν}d[2]ki+μ],d[2](k+1)3i+1=32m2μ=0[μν=0{(ν+1)(ρμν+σμν2ϱμν)}d[2]ki+μ+1],d[2](k+1)3i+2=32m1μ=0[μν=0{νρμν+(ν+1)ϱμν(2ν+2)σμν}d[2]ki+μ+1]. (2.8)

    Lemma 2.3. Let fk={(i/3k,fki)} be a polygon of TRS at kth refinement level and

    d[3]k)j=33k(3!)1(fkj1+3fkj3fkj+1+fkj+2)

    be the third order divided difference then the third order DD TRS is defined as

    {d[3](k+1)3i=33m2μ=0[μν=0{(ν+1)σμν3ν(ν+1)2(ρμνϱμν)}d[3]ki+μ],d[3](k+1)3i+1=33m2μ=0[μν=0{(ν+1)ρμν3(ν+1)(ν+2)2(ϱμνσμν)}d[3]ki+μ+1],d[3](k+1)3i+2=33m2μ=0[μν=0{3ν(ν+1)2ρμν+(ν+1)ϱμν3(ν+1)(ν+2)2σμν}d[3]ki+μ+1]. (2.9)

    The convergence of first, second and third order divided difference (DD) ternary refinement schemes (TRS) is presented in this section. Throughout the paper, π[0,n] denotes the set of continuous functions on the closed and bounded interval [0,n].

    Lemma 2.4. Let fk={(i/3k,fki)} be a polygon of TRS at kth refinement level and d[1]k={(i/3k,d[1]ki)} be a polygon of first order DD scheme. If limkd[1]k=d[1]π[0,n], and f is the limiting curve/shape produced by TRS then d[1]=f.

    Proof. Bernstein polynomial for x[0,n] with data values fki is

    Bk(x)=si=0(si)(xn)i(1xn)sifki.

    Its first derivative is

    Bk(x)=si=0(si)in(xn)i1(1xn)sifki+si=0(si)(xn)i(si)(1n)(1xn)si1fki. (2.10)

    This implies

    Bk(x)=sns1i=0(s1i)(xn)i(1xn)si1(fki+1fki).

    For s=3kn, we get

    Bk(x)=s1i=0(s1i)(xn)i(1xn)si13k(fki+1fki).

    This again implies

    Bk(x)=s1i=0(s1i)(xn)i(1xn)si1d[1]ki.

    Since the Bernstien polynomials are uniformly convergent therefore limkBk=f and limkBk=d[1] on [0,n]. So f=d[1]π[0,n]. Hence the proof.

    Lemma 2.5. Let fk={(i/3k,fki)} and d[2]k={(i/3k,d[2]ki)} be the polygons of TRS and its second order DD schemes respectively. If limkd[2]k=d[2]π[0,n], and f is the limiting curve produced by TRS then d[2]=f.

    Proof. Take the derivative of (2.10)

    Bk(x)=1n2{s(s1)si=2(s2)!(i2)!(si)! (xn)i2(1xn)sifki2s(s1)s1i=1(s2)!(i1)!(si1)! (xn)i1(1xn)si1fki +s(s1)s2i=0(s2)!i!(si2)!(xn)i(1xn)si2fki}. (2.11)

    This implies

    Bk(x)=s(s1)n2{s2i=0(s2)!(i)!(si2)! (xn)i(1xn)si2×(fki+22fi+1+fki)}.

    Since s=3kn therefore s(s1)n2=32k21, so we get

    Bk(x)=32k21{s2i=0(s2)!(i)!(si2)! (xn)i(1xn)si2×(fki+22fi+1+fki)}.

    This implies

    Bk(x)=s2i=0{(s2)!(i)!(si2)! (xn)i(1xn)si2}d[2]ki+1,

    where d[2]ki+1 = 32k(2!)1(fki+22fki+1+fki).

    Again the uniform convergence of the Bernstien polynomials implies limkBk=d[2] and limkBk=f. Therefore when k then d[2] converges uniformly to f on [0,n]. This concludes d[2]=fπ[0,n]. Hence the proof.

    Lemma 2.6. Let fk={(i/3k,fki)} and d[3]k)={(i/3k,d[3]ki)} be the polygons of TRS and its third order DD schemes respectively. If limkd[3]k=d[3]π[0,n] then d[3]=f.

    Proof. Now take the derivative of (2.11), then

    Bk(x)=1n3si=3s(s1)(s2)(s3)!(i3)!(si)!(xn)i3(1xn)sifki3n3s1i=2s(s1)(s2)(s3)!(i2)!(si1)!(xn)i2(1xn)si1fki+3n3s2i=1s(s1)(s2)(s3)!(i1)!(si2)!(xn)i1(1xn)si2fki1n3s3i=0s(s1)(s2)(s3)!i!(si3)!(xn)i(1xn)si3fki.

    Since, s=3kn, then

    Bk(x)=33k(3!)1s3i=0(s3i)(xn)i(1xn)si3×(fki+3fki+13fki+2+fki+3).

    This implies

    Bk(x)=s3i=0(s3i)(xn)i(1xn)si3d[3]ki+1,

    where

    d[3]ki+1=33k(3!)1(fki+3fki+13fki+2+fki+3).

    This implies Bkd[3] and Bkf. This implies that for k the d[3] converges uniformly to f on [0,n]. This concludes that d[3]=fπ[0,n]. Hence the proof.

    We first introduce the inequalities to compute the deviation between two consecutive points at the same refinement level then we introduce the inequalities to compute the deviation between successive levels of polygons produced by ternary and its DD refinement schemes.

    Lemma 3.1. If f0={(i/30,f0i)} is the initial polygon and fk+1={(i/3k+1,fk+1i)} is the polygon obtained by TRS at (k+1)th refinement level. If ϖ<1 then the deviation between two consecutive points at (k+1)th level is

    maxifk+1i+1fk+1i(ϖ)k+1maxif0i+1f0i, (3.1)

    where

    ϖ=max{ϖ1,ϖ2,ϖ3}<1, (3.2)
    {ϖ1=m1μ=0|μν=0(ρμνϱμν)|,ϖ2=m1μ=0|μν=0(ϱμνσμν)|,ϖ3=mμ=0|μν=0(σμνρμν1)|. (3.3)

    Proof. From (2.1), we get

    fk+13i+1fk+13i=mν=0ϱνfki+νmν=0ρνfki+ν=mν=0(ϱνρν)fki+ν.

    This further implies

    fk+13i+1fk+13i=[(ρ0ϱ0)(fki+1fki)+{(ρ0ϱ0)+(ρ1ϱ1)}(fki+2fki+1)+{(ρ0ϱ0)+(ρ1ϱ1)+(ρ2ϱ2)}(fki+3fki+2)++{(ρ0ϱ0)+(ρ1ϱ1)++(ρm1ϱm1)}(fki+mfki+m1)+{(ϱ0a0)+(ϱ1ρ1)++(ϱmρm)}fki+m].

    This leads to

    fk+13i+1fk+13i=m1μ=0{μν=0(ρμνϱμν)}(fki+μ+1fki+μ)+mν=0(ϱνρν)fki+m.

    Since by (2.2), mν=0(ϱνρν)=0, then

    fk+13i+1fk+13i=m1μ=0{μν=0(ρμνϱμν)(fki+μ+1fki+μ)}.

    By taking maximum norm, we get

    fk+13i+1fk+13iϖ1maxifki+1fki, (3.4)

    where

    ϖ1=m1μ=0|μν=0(ρμνϱμν)|.

    Using (2.1) and similar procedure as above, we get

    fk+13i+2fk+13i+1ϖ2maxifki+1fki, (3.5)

    where

    ϖ2=m1μ=0|μν=0(ϱμνσμν)|.

    and

    fk+13i+3fk+13i+2ϖ3maxifki+1fki, (3.6)

    where

    ϖ3=mμ=0|μν=0(σμνρμν1)|.

    Combining (3.4), (3.5) and (3.6), it proceeds as

    maxifk+1i+1fk+1iϖmaxifki+1fki,

    where

    ϖ=max{ϖ1,ϖ2,ϖ3}.

    This concludes

    maxifk+1i+1fk+1i(ϖ)k+1maxif0i+1f0i.

    This completes the proof.

    Theorem 3.2. If f0={(i/30,f0i)} is the initial polygon and fk+1={(i/3k+1,fk+1i)} is the polygon obtained by TRS at (k+1)th refinement level. If ϖ<1 then the deviation between two successive polygons at kth and (k+1)th levels is

    fk+1fkϖmaxifki+1fki, (3.7)

    where

    ϖ=max{ϖ4,ϖ5,ϖ6}, (3.8)
    {ϖ4=m1μ=0|1μν=0ρμν|,ϖ5=|23ϱ0|+m1μ=1|1μν=0ϱμν|,ϖ6=|13σ0|+m1μ=1|1μν=0σμν|. (3.9)

    Proof. Since the maximum deviation between fk+1 and fk occurs at the diadic values tk+13i=tki, tk+13i+1=13(2tki+tki+1) and tk+13i+2=13(tki+2tki+1) respectively, therefore

    fk+1fkmax{Z1k,Z2k,Z3k}, (3.10)

    where

    {Z1k=maxifk+13ifki,Z2k=maxifk+13i+113(2fki+fki+1),Z3k=maxifk+13i+213(fki+2fki+1). (3.11)

    From (2.1), we obtain

    fk+13ifki=mν=0ρνfki+νfki.

    Since by (2.1), mν=0ρν1=0, then

    fk+13ifki=m1μ=0(1μν=0ρμν)(fki+μ+1fki+μ).

    Taking norm, we get

    fk+13ifkiϖ4maxifki+1fki, (3.12)

    where

    ϖ4=m1μ=0|1μν=0ρμν|.

    Using (2.1) and similar procedure as above, we get

    fk+13i+1(23fki+13fki+1)ϖ5maxifki+1fki, (3.13)

    where

    ϖ5=|23ϱ0|+m1μ=1|1μν=0ϱμν|.
    fk+13i+2(13fki+23fki+1)ϖ6maxifki+1fki, (3.14)

    where

    ϖ6=|13σ0|+m1μ=1|1μν=0σμν|.

    From (3.10)-(3.14), we get

    fk+1fkϖmaxi|fki+1fki|,

    where

    ϖ=max{ϖ4,ϖ5,ϖ6}.

    Hence the proof.

    Lemma 3.3. If d[1]0={(i/30,d[1]0i)} is the initial polygon and d[1](k+1)=

    {(i/3k+1,d[1](k+1)i)} is the polygon obtained by first order DD TRS at (k+1)th refinement level. If ϖ<1 and

    {c1=m1ν=0(mν)(2ϱνσνρν)=0,c2=mν=0{(2ν+1)σmν(mν)(ρν+ϱν)}=0,c3=mν=0{2νρmννϱmν(ν+1)σmν}=0, (3.15)

    then the deviation between two consecutive points at (k+1)th level is

    maxid[1](k+1)i+1d[1](k+1)i(ϖ)k+1maxid[1]0i+1d[1]0i, (3.16)

    where

    ϖ=max{ϖ1,ϖ2,ϖ3}, (3.17)
    {ϖ1=3m2μ=0|μν=0(ν+1)(ρμν2ϱμν+σμν)|,ϖ2=3m1μ=0|μν=0{(ν+1)ϱμν(2ν+2)σμν+νρμν}|,ϖ3=3m1μ=0|μν=0{(ν+1)σμν2νρμν+νϱμν}|. (3.18)

    Proof. Using (2.3) and proceeding similarly as in Lemma 3.1, we get

    If, c1=m1ν=0(mν)(2ϱνσνρν)=0, then

    d[1](k+1)3i+1d[1](k+1)3iϖ1maxid[1]ki+1d[1]ki, (3.19)

    where

    ϖ1=3m2μ=0|μν=0(ν+1)(ρμν2ϱμν+σμν)|.

    Using (2.3) and similar procedure as above, we get

    c2=mν=0{(2ν+1)σmν(mν)(ρν+ϱν)}=0,

    d[1](k+1)3i+2d[1](k+1)3i+1ϖ2maxid[1]ki+1d[1]ki. (3.20)

    where

    ϖ2=3m1μ=0|μν=0{(ν+1)ϱμν(2ν+2)σμν+νρμν}|.

    And

    c3=mν=0{2νρmννϱmν(ν+1)σmν}=0,

    d[1](k+1)3i+3d[1](k+1)3i+2ϖ3maxid[1]ki+1d[1]ki, (3.21)

    where

    ϖ3=3m1μ=0|μν=0{(ν+1)σμν2νρμν+νϱμν}|.

    Combining Eqs (3.19), (3.20) and (3.21), we get

    maxid[1](k+1)i+1d[1](k+1)iϖmaxid[1]ki+1d[1]ki,

    whereas

    ϖ=max{ϖ1,ϖ2,ϖ3}.

    This implies

    maxid[1](k+1)i+1d[1](k+1)i(ϖ)k+1maxid[1]0i+1d[1]0i.

    This completes the proof.

    Theorem 3.4. If d[1]0={(i/30,d[1]0i)} is the initial polygon and d[1](k+1)=

    {(i/3k+1,d[1](k+1)i)} is the polygon obtained by first order DD TRS at (k+1)th refinement level. If ϖ<1 and

    {c4=13+mν=1ν(ρmνϱmν)=0,c5=13+mν=1ν(ϱmνσmν)=0,c6=13+mν=0{(ν+1)σmννρmν}=0, (3.22)

    then the deviation between two successive polygons at kth and (k+1)th levels is

    d[1](k+1)d[1]kϖmaxid[1]ki+1d[1]ki, (3.23)

    where

    ϖ=max{ϖ4,ϖ5,ϖ6}, (3.24)
    {ϖ4=3m2μ=0|13+μν=0{(ν+1)(ϱμνρμν)}|,ϖ5=3|232ϱ0+σ0|+3m2μ=1|13+μν=0{(ν+1)(σμνϱμν)}|,ϖ6=3|132σ0|+3m1μ=1|13+μν=0(νρμν(ν+1)σμν)|. (3.25)

    Proof. Since the maximum deviation between d[1](k+1) and d[1]k occurs at the diadic values tk+13i=tki, tk+13i+1=13(2tki+tki+1) and tk+13i+2=13(tki+2tki+1) respectively, therefore

    d[1](k+1)d[1]kmax{Z4k,Z5k,Z6k}, (3.26)

    where

    {Z4k=maxid[1](k+1)3id[1]ki,Z5k=maxid[1](k+1)3i+113(2d[1]ki+d[1]ki+1),Z6k=maxid[1](k+1)3i+213(d[1]ki+2d[1]ki+1). (3.27)

    By (2.3), we get

    d[1](k+1)3id[1]ki=3m1μ=0{μν=0(ρμνϱμν)d[1]ki+μ}d[1]ki.

    This leads to

    d[1](k+1)3id[1]ki=3[m2μ=0{13+μν=0(ν+1)(ϱμνρμν)(d[1]ki+μ+1d[1]ki+μ+1)}+{13+mν=1ν(ρmνϱmν)}d[1]ki+m1].

    If, c4=13+mν=1ν(ρmνϱmν)=0, taking norm we get

    d[1](k+1)3id[1]kiϖ4maxid[1]ki+1d[1]ki, (3.28)

    where,

    ϖ4=3m2μ=0|13+μν=0{(ν+1)(ϱμνρμν)}|.

    Using (2.3) and similar procedure as above, we get

    c5=13+mν=1ν(ϱmνσmν)=0,

    d[1](k+1)3i+1(23d[1]ki+13d[1]ki+1)ϖ5maxid[1]ki+1d[1]ki, (3.29)

    where

    ϖ5=3|232ϱ0+σ0|+3m2μ=1|13+μν=0{(ν+1)(σμνϱμν)}|.

    And

    c6=13+mν=0{(ν+1)σmννρmν}=0,

    maxiˆdk+13i+2(13ˆdki+23ˆdki+1)ϖ6maxid[1]ki+1d[1]ki, (3.30)

    where

    ϖ6=3|132σ0|+3m1μ=1|13+μν=0(νρμν(ν+1)σμν)|.

    From (3.26)-(3.30)

    d[1](k+1)d[1]kϖmaxid[1]ki+1d[1]ki,

    whereas

    ϖ=max{ϖ4,ϖ5,ϖ6}.

    Hence the proof.

    Lemma 3.5. If d[2]0={(i/30,d[2]0i)} is the initial polygon and d[2](k+1)=

    {(i/3k+1,d[2](k+1)i)} is the polygon obtained by second order DD TRS at (k+1)th refinement level. If ϑ<1 and

    {χ1=m1ν=0{3ν(ν+1)2(ρmν1ϱmν1)(ν+1)σmν1}=0,χ2=m1ν=0{(ν+1)(3ν+2)2ϱm1ν(ν+1)(3ν+4)2σm1ν}=0,χ3=m1ν=0{3(ν+1)(ν+2)2σm1ν(ν+1)ϱm1ν3ν(ν+1)2ρm1ν}=0, (3.31)

    then the deviation between two consecutive points at (k+1)th level is

    maxid[2](k+1)i+1d[2](k+1)i(ϑ)k+1maxid[2]0i+1d[2]0i, (3.32)

    where

    ϑ=max{ϑ1,ϑ2,ϑ3}, (3.33)
    {ϑ1=32m2μ=0|μν=0{(ν+1)σμν+3ν(ν+1)2(ϱμνρμν)}|,ϑ2=32m2μ=0|μν=0{(ν+1)ρμν+3(ν+1)(ν+2)2(σμνϱμν)}|,ϑ3=32m2μ=0|μν=0{3ν(ν+1)2ρμν+(ν+1)ϱμν3(ν+1)(ν+2)2σμν}|. (3.34)

    Theorem 3.6. If d[2]0={(i/30,d[2]0i)} is the initial polygon and d[2](k+1)=

    {(i/3k+1,d[2](k+1)i)} is the polygon obtained by second order DD TRS at (k+1)th refinement level. If ϑ<1 and

    {χ4=132+m1ν=0[(ν+1)2{νϱm1ν+(ν+2)σm1ν2νρm1ν}]=0,χ5=132+m1ν=1[ν(ν+1)2{ρmν1+σmν12ϱmν1}]=0,χ6=132+mν=1{ν(ν+1)2ϱmνν(ν+1)σmν+ν(ν1)2ρmν}=0, (3.35)

    then the deviation between two successive polygons at kth and (k+1)th levels is

    d[2](k+1)d[2]kϑmaxid[2]ki+1d[2]ki, (3.36)

    where

    ϑ=max{ϑ4,ϑ5,ϑ6}, (3.37)
    {ϑ4=32m2μ=0|132+μν=0{ν(ν+1)ρμνν(ν+1)2ϱμν(ν+1)(ν+2)2σμν}|,ϑ5=32|233|+32m2μ=1|19+μν=1{ν(ν+1)2(2ϱμνρμνσμν)}|,ϑ6=32|133|+32m1μ=1|132+μν=1{ν(ν+1)σμνν(ν+1)2ϱμνν(ν1)2ρμν}|. (3.38)

    Lemma 3.7. If d[3]0={(i/30,d[3]0i)} is the initial polygon and d[3](k+1)=

    {(i/3k+1,d[3](k+1)i)} is the polygon obtained by third order DD TRS at (k+1)th refinement level. If ϑ<1 and

    {χ7=m2ν=0[(ν+1)(ν+2)2{(ν+1)ρmν2(2ν+3)ϱmν2+(ν+2)σmν2}]=0,χ8=m2ν=0[(ν+1)(ν+2){(ν1)2ρmν2+(ν+4)2ϱmν2(ν+3)σmν2}]=0,χ9=m2ν=0[(ν+1)(ν+2){νρmν2+(ν1)2ϱmν2+(ν+4)2σmν2}]=0, (3.39)

    then the deviation between two consecutive points at (k+1)th level is

    maxid[3](k+1)i+1d[3](k+1)i(ϑ)k+1maxid[3]0i+1d[3]0i, (3.40)

    where

    ϑ=max{ϑ7,ϑ8,ϑ9}, (3.41)
    {ϑ7=33m2μ=0|μν=0{ν(ν+1)(ν+2)ϱμνν(ν+1)(ν+3)2ρμν(ν1)(ν+1)(ν+2)2σμν}|,ϑ8=33m3μ=0|μν=0{(ν1)(ν+1)(ν+2)2ρμν(ν+1)(ν+2)(ν+4)2ϱμν+(ν+1)(ν+2)(ν+3)σμν}|,ϑ9=33m3μ=0|μν=0{ν(ν+1)(ν+2)ρμν(ν1)(ν+1)(ν+2)2ϱμν(ν+1)(ν+2)(ν+4)2σμν}|. (3.42)

    Theorem 3.8. If d[3]0={(i/30,d[3]0i)} is the initial polygon and d[3](k+1)={(i/3k+1,d[3](k+1)i)} is the polygon obtained by third order DD TRS at (k+1)th refinement level. If ϑ<1 and

    {χ10=133+m2ν=0[(ν+1)(ν+2)2{σm2ν+νϱm2ννρm2ν}]=0,χ11=133+m2ν=0[(ν+1)(ν+2)2{ρmν2(ν+3)ϱmν2+(ν+3)σmν2}]=0,χ12=133+m2ν=0[(ν+1)(ν+2)2{ϱmν2(ν+3)σmν2+νρmν2}]=0, (3.43)

    then the deviation between two successive polygons at kth and (k+1)th levels is

    d[3](k+1)d[3]kϑmaxid[3]ki+1d[3]ki, (3.44)

    where

    ϑ=max{ϑ10,ϑ11,ϑ12}, (3.45)
    {ϑ10=33m3μ=0|133+μν=0{(ν+1)(ν+2)2(νρμννϱμνσμν)}|,ϑ11=33|234|+33m2μ=1|133+μν=1{(ν+1)2{νρμν+(ν+2)ϱμν(ν+2)σμν}}|,ϑ12=33|134|+33m2μ=1|{133+μν=1ν(ν+1)2{(ν+2)σμνϱμν(ν1)ρμν}}|. (3.46)

    Now we present the analysis of the ternary refinement schemes. A function is called Cm-continuous if its mth order derivative is continuous. The common class of continuous function is C=C0. It is natural to think of a Cm function as being a little bit rough, but the graph of a C3 function looks smooth. So we discuss the analysis of the schemes up to C3-continuity. The results can be extended similarly. Let {πm[0,n], m=0,1,2,3} denotes the set of Cm-continuous functions on the closed and bounded interval [0,n].

    Theorem 4.1. If f0={(i/30,f0i)} is the initial polygon and fk+1={(i/3k+1,fk+1i)} is the polygon obtained by TRS at (k+1)th refinement level. If ϖ, ϖ<1 then limkfk=fπ[0,n]=π0[0,n].

    Proof. The (3.7), gives

    fk+1fkϖmaxifki+1fki,

    and (3.1), gives

    fk+1fkϖ(ϖ)kmaxif0i+1f0i,

    Since ϖ, ϖ<1 therefore {fk}k=0 is a Cauchy sequence on closed and bounded interval [0,n]. Hence it is convergent. That is

    limkfk=fπ[0,n]=π0[0,n].

    Hence the proof.

    Lemma 4.2. If d[1]0={(i/30,d[1]0i)} is the initial polygon and d[1](k+1)=

    {(i/3k+1,d[1](k+1)i)} is the polygon obtained by first order DD TRS at (k+1)th refinement level. If ϖ, ϖ<1 and c1,c2,c3,c4,c5,c6=0, then limkd[1]k=d[1]π[0,n]=π0[0,n].

    Proof. By (3.23), we have

    d[1](k+1)d[1]kϖmaxid[1]ki+1d[1]ki,

    Using (3.16), we have

    d[1](k+1)d[1]kϖ(ϖ)kmaxid[1]0i+1d[1]0i,

    Since ϖ, ϖ<1, therefore {d[1]k}k=0 defines a Cauchy sequence on [0,n] and

    limkd[1]k=d[1]π[0,n]=π0[0,n].

    This completes the proof.

    Theorem 4.3. Let fπ0[0,n] be the limiting curve produced by the ternary refinement scheme. If ϖ, ϖ<1 and c1,c2,c3,c4,c5,c6=0 then fπ1[0,n].

    Proof. Lemma 4.2 leads to limkd[1]k=d[1]π0[0,n]. Lemma 2.4 gives d[1]=f. Therefore, we have the result, fπ1[0,n].

    Lemma 4.4. If d[2]0={(i/30,d[2]0i)} is the initial polygon and d[2](k+1)=

    {(i/3k+1,d[2](k+1)i)} is the polygon obtained by second order DD TRS at (k+1)th refinement level. If ϑ, ϑ<1 and χ1,χ2,χ3,χ4,χ5,χ6=0 then limkd[2]k=d[2]π0[0,n].

    Proof. By (3.36), we have

    d[2](k+1)d[2]kϑmaxid[2]ki+1d[2]ki,

    Using (3.32), we have

    d[2](k+1)d[2]kϑ(ϑ)kmaxid[2]0i+1d[2]0i,

    Since ϑ, ϑ<1 therefore {d[2]k}k=0 is a Cauchy convergent sequence. So

    limkd[2]k=d[2]π0[0,n].

    Hence the proof.

    Theorem 4.5. Let fπ0[0,n] be the limiting curve produced by the ternary refinement scheme. If ϑ, ϑ<1 and φ1,φ2,φ3,φ4,φ5,φ6=0 then fπ2[0,n].

    Proof. Lemma 4.4 gives limkd[2]k=d[2]π0[0,n]. Lemma 2.5 leads to d[2]=f. This implies fπ2[0,n].

    Lemma 4.6. If d[3]0={(i/30,d[3]0i)} is the initial polygon and d[3](k+1)=

    {(i/3k+1,d[3](k+1)i)} is the polygon obtained by third order DD TRS at (k+1)th refinement level. If ϑ, ϑ<1 and φ7,φ8,φ9,φ10,φ11,φ12=0 then limkd[3]k)=d[3]π0[0,n].

    Proof. By (3.44), we have

    d[3](k+1)d[3]kϑmaxid[3]ki+1d[3]ki,

    Using (3.40), we have

    d[3](k+1)d[3]kϑ(ϑ)kmaxid[3]0i+1d[3]0i,

    Since ϑ, ϑ<1 therefore {d[3]k}k=0 is a Cauchy convergent sequence. Thus

    limkd[3]k=d[3]π0[0,n].

    Hence the proof.

    Theorem 4.7. Let fπ0[0,n] be the limiting curve produced by the ternary refinement scheme. If ϑ, ϑ<1 and φ7,φ8,φ9,φ10,φ11,φ12=0 then fπ3[0,n].

    Proof. Lemma 4.6, implies that limkd[3]k)=d[3]π0[0,n] while Lemma 2.6 leads to d[3]=f. This implies fπ3[0,n].

    Application and authenticity of the results

    We discuss the analysis of the ternary refinement schemes introduced by [8,9,10,16,17]. We conclude that the results obtained by our methods are always equivalent to the one returned by the Laurent polynomial method. We add the Figure 2 for the interest of general readers by the direction of anonymous reviewer of this paper. The Figure 2(d) depicts the smoothness of the limiting curve produced by the ternary scheme. The Figure 2(a), (b) and (c) represent the initial, first and second refinement level of the ternary scheme.

    Figure 2.  Applications of ternary refinement scheme [17]. Here, the value of μ is -154.

    Example 4.1. If the curve is produced by the following 4-point ternary refinement scheme [17]

    {fk+13i=ρ0fki1+ρ1fki+ρ2fki+1+ρ3fki+2,fk+13i+1=ϱ0fki1+ϱ1fki+ϱ2fki+2+ϱ3fki+3,fk+13i+2=σ0fki1+σ1fki+σ2fki+2+σ3fki+3,

    where

    {ρ0=(581+53μ), ρ1=(19273μ), ρ2=(1354+μ), ρ3=(1162+13μ),ϱ0=μ,  ϱ1=(12μ), ϱ2=(12μ), ϱ3=μ,σ0=(1162+13μ),   σ1=(1354+μ),   σ2=(19273μ), σ3=(581+53μ),

    then by Laurent polynomial method the scheme produces C3-continuous curve for μ(127,227). By Theorems 4.1, 4.3, 4.5 and 4.7, we also get the same results.

    Example 4.2. If the curve is produced by the following 4-point ternary refinement scheme [9]

    {fk+13i=ρ1fki,fk+13i+1=ϱ0fki1+ϱ1fki+ϱ2fki+2+ϱ3fki+3,fk+13i+2=σ0fki1+σ1fki+σ2fki+2+σ3fki+3,

    where

    {ρ0=0,ρ1=1,ρ2=0,ρ3=0,ϱ0=11816μ, ϱ1=1318+12μ,  ϱ2=71812μ,  ϱ3=118+16μ,σ0=118+16μ,σ1=71812μ, σ2=1318+12μ, σ3=11816μ,

    then by Laurent polynomial method the scheme produces C2-continuous curve over the interval μ(115,19). By Theorems 4.1, 4.3, and 4.5, we also get the same results.

    Example 4.3. If the curve is produced by the following 3-point ternary refinement scheme [8]

    {fk+13i=ρ0fki1+ρ1fki+ρ2fki+1,fk+13i+1=ϱ1fki,fk+13i+2=σ0fki1+σ1fki+σ2fki+1,

    where

    {ρ0=(ρ+13),  ρ1=(232ρ),ρ2=ρ,ϱ0=0,ϱ1=1, ϱ2=0,σ0=ρ,   σ1=(232ρ),σ2=(ρ+13),

    then by both methods the scheme produces C1-continuous curve for w(19,0).

    Example 4.4. If the curve is produced by the following 6-point ternary refinement scheme [10]

    {fk+13i=ρ2fki,fk+13i+1=ϱ0fki2+ϱ1fki1+ϱ2fki+ϱ3fki+1+ϱ4fki+2+ϱ5fki+3,fk+13i+2=σ0fki2+σ1fki1+σ2fki+σ3fki+1+σ4fki+2+σ5fki+3, (4.1)

    where

    {ρ0=0,ρ1=0,ρ2=1,ρ3=0,ρ4=0,ρ5=0,ϱ0=(1181+13ω), ϱ1=(132751ω),ϱ2=(227+74ω),ϱ3=(748146ω),ϱ4=(527+9ω),   ϱ5=ω,σ0=ω,  σ1=(527+9ω),   σ2=(748146ω),σ3=(227+74ω),σ4=(132751ω),   σ5=(1181+13ω),

    then by both methods the scheme produces C2-continuous curve over the interval w(141215,231944).

    Example 4.5. If the curve is produced by the following 3-point ternary refinement scheme [16]

    {fk+13i=ρ0fki1+ρ1fki+ρ2fki+1,fk+13i+1=ϱ0fki1+ϱ1fki+ϱ2fki+2,fk+13i+2=σ0fki1+σ1fki+σ2fki+2, (4.2)

    where

    {ρ0=(2572+μ),ρ1=(23362μ),ρ2=(172+μ),ϱ0=(18+μ), ϱ1=(342μ), ϱ2=(18+μ),σ0=(172+μ),σ1=(23362μ), σ2=(2572+μ),

    then by both methods the scheme produces C2-continuous curve for μ(172,772).

    Here we present the brief summary of the work done so for in this paper.

    ● If max{ϖ1,ϖ2,ϖ3}<1 where ϖ1,ϖ2,ϖ3 are defined in (3.3) then TRS will generate C0 continuous curve.

    ● If c1, c2, c3, c4, c5, c6 defined in (3.15) and (3.22) are equal to zero and max{ϖ1,ϖ2,ϖ3}<1, where ϖ1,ϖ2,ϖ3 are defined in (3.18) then TRS will generate C1 continuous curve.

    ● If χ1, χ2, χ3, χ4, χ5, χ6 defined in (3.31) and (3.35) are equal to zero and max{ϑ1,ϑ2,ϑ3}<1, where ϑ1,ϑ2,ϑ3 are defined in (3.34) then TRS will generate C2 continuous curve.

    ● If χ7, χ8, χ9 χ10, χ11, χ12 defined in (3.39) and (3.43) are equal to zero and max{ϑ7,ϑ8,ϑ9}<1, where ϑ7,ϑ8,ϑ9 are defined in (3.42) then TRS will generate C3 continuous curve.

    There are two major techniques to analyze the refinement schemes. These are called Laurent polynomial and divided difference (DD) techniques.

    ● Our technique is the generalization of the DD technique for ternary refinement schemes.

    ● We have presented the explicit form of the general inequalities for the analysis of the ternary refinement schemes. These inequalities contain simple algebraic expressions. In these inequalities the polynomial factorization, division and summation are not involved. Simple arithmetic operations such as subtraction and multiplication are involved to evaluate the inequalities.

    ● While in Laurent polynomial technique, the polynomial factorization, division and summation are involved to evaluate the inequalities. In this technique, the explicit form of the general inequalities are also not available.

    ● So it is obvious that the computational complexity of our techniques is less than the complexity of Laurent polynomial technique.

    In this paper, we have introduced an alternative technique to analyze a class of ternary refinement schemes. A comparative study of the proposed technique with other existing technique has been presented to prove the effectiveness of the proposed technique. It has been observed that the alternative technique has less computational cost comparative to the existing Laurent polynomial technique.

    The second author is grate full to HEC Pakistan for awarding the fellowship under the Indigenous Ph.D. Scholarship Scheme. The author Yu-Ming Chu is supported by National Natural Science Foundation of China (Grant Nos. 61673169, 11971142).

    Conceptualization, Ghulam Mustafa and Dumitru Baleanu; Formal analysis, Dumitru Baleanu and Yu-Ming Chu; Methodology, Ghulam Mustafa and Syeda Tehmina Ejaz; Supervision, Ghulam Mustafa; Writing original draft, Syeda Tehmina Ejaz and Ghulam Mustafa; Writing, review and editing, Syeda Tehmina Ejaz and Yu-Ming Chu.

    The authors declare no conflict of interest.



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