A family of univariate ternary subdivision schemes with 5 parameters

Abstract

In this paper, we propose an efficient strategy for the construction of a family of ternary subdivision schemes with five parameters. These five parameters allow us to increase the choices when drawing and designing different models from the same initial control points. Moreover, by choosing appropriate values of parameters, we can get many existing ternary approximating subdivision schemes. Some of the existing ternary interpolatory and many new ternary approximating/interpolatory subdivision schemes with arbitrary order of regularities can also be obtained.

Appendices

Appendix A

1.1 A.1 Proof of Proposition 8

We shall prove it by mathematical induction on m. When $m=1$,

$$\begin{aligned} b(z)= & {} \alpha _{4}z^{6}+(\alpha _{4}+\alpha _{3})z^{5}+(\alpha _{4}+a_{3}+\alpha _{2})z^{4}+(\alpha _{3}+\alpha _{2}+\alpha _{1})z^{3} +(\alpha _{2}+\alpha _{1}\\&+\,\alpha _{0})z^{2}+(\alpha _{1}+\alpha _{0})z+\alpha _{0}. \end{aligned}$$

The sum of the coefficients of the terms having the powers of z: 3p, $3p+1$ and $3p+2$ in b(z) are

$$\begin{aligned} \alpha _{4}+(\alpha _{3}+\alpha _{2}+\alpha _{1})+\alpha _{0}=\sum \limits _{i=0}^{4}\alpha _{i}=3^{0}\sum \limits _{i=0}^{4}\alpha _{i}=3^{m-1}\sum \limits _{i=0}^{4}\alpha _{i},\\ (\alpha _{4}+\alpha _{3})+(\alpha _{2}+\alpha _{1}+\alpha _{0})=\sum \limits _{i=0}^{4}\alpha _{i}=3^{0}\sum \limits _{i=0}^{4}\alpha _{i}=3^{m-1}\sum \limits _{i=0}^{4}\alpha _{i}, \end{aligned}$$

and

$$\begin{aligned} (\alpha _{4}+\alpha _{3}+\alpha _{2})+(\alpha _{1}+\alpha _{0})=\sum \limits _{i=0}^{4}\alpha _{i}=3^{0}\sum \limits _{i=0}^{4}\alpha _{i}=3^{m-1}\sum \limits _{i=0}^{4}\alpha _{i}. \end{aligned}$$

Now assume that when $m=k$, the result holds up i.e.

$$\begin{aligned} b(z)= & {} (\alpha _{4}z^{4}+\alpha _{3}z^{3}+\alpha _{2}z^{2}+\alpha _{1}z+\alpha _{0})(1+z+z^{2})^{k} =\beta _{4+2k}z^{4+2k}+\beta _{3+2k}z^{3+2k}\\&+\,\beta _{2+2k}z^{2+2k}+\dots \beta _{4+k}z^{4+k}+\beta _{3+k}z^{3+k}+\beta _{2+k}z^{2+k}+\cdots +\beta _{2}z^{2}+\beta _{1}z+\beta _{0}. \end{aligned}$$

If we assume that $4+2k=3p$, then we have

$$\begin{aligned}&\beta _{4+2k}+\beta _{1+2k}+\cdots +\beta _{4+k}+\beta _{1+k}+\cdots +\beta _{3}+\beta _{0}= \beta _{3+2k}+\beta _{2k}+\cdots \\&\quad +\beta _{3+k}+\beta _{k}+\cdots +\beta _{4}+\beta _{1}=\beta _{2+2k}+\beta _{2k-1}+\cdots +\beta _{2+k}+\beta _{k-1}+\cdots \\&\quad +\beta _{5} +\beta _{2}=3^{k-1}\sum \limits _{i=0}^{4}\alpha _{i}. \end{aligned}$$

Now we shall prove that for $m=k+1$, the result holds up.

Let $m=k+1$, then

$$\begin{aligned} b(z)= & {} (\alpha _{4}z^{4}+\alpha _{3}z^{3}+\alpha _{2}z^{2}+\alpha _{1}z+\alpha _{0})(1+z+z^{2})^{k+1}= (\beta _{4+2k}z^{4+2k}+\beta _{3+2k}z^{3+2k}\\&+\,\beta _{2+2k}z^{2+2k}+\cdots \beta _{4+k}z^{4+k}+\beta _{3+k}z^{3+k}+\beta _{2+k}z^{2+k}+\cdots +\beta _{2}z^{2}+\beta _{1}z+\beta _{0}) \\&\times \, (1+z+z^{2})=\beta _{4+2k}z^{6+2k}+(\beta _{4+2k}+\beta _{3+2k})z^{5+2k}+(\beta _{4+2k}+\beta _{3+2k}+\beta _{2+2k}) \\&\times \, z^{4+2k}+(\beta _{3+2k}+\beta _{2+2k}+\beta _{1+2k})z^{3+2k}+\cdots +(\beta _{3}+\beta _{2}+\beta _{1})z^{3}+(\beta _{2}+\beta _{1}\\&+\,\beta _{0})z^{2}+(\beta _{1}+\beta _{0})z +\beta _{0}. \end{aligned}$$

So the sum of the coefficients of the terms with the powers of z: 3p, $3p+1$ and $3p+2$ in b(z) are

$$\begin{aligned}&\beta _{4+2k}+\beta _{3+2k}+\cdots +\beta _{4+k}+\beta _{3+k}+\cdots +\beta _{2}+\beta _{1}+\beta _{0}=\beta _{4+2k}+\beta _{3+2k}+\beta _{2+2k}\\&\quad +\cdots +\beta _{4+k}+\beta _{3+k}+\cdots +\beta _{2}+\beta _{1}+\beta _{0}= \beta _{4+2k}+\beta _{3+2k}+\beta _{2+2k}+\cdots +\beta _{4+k}\\&\quad +\beta _{3+k}+\cdots +\beta _{2}+\beta _{1}+\beta _{0}=3^{k}\sum \limits _{i=0}^{4}\alpha _{i}=3^{m-1}\sum \limits _{i=0}^{4}\alpha _{i}. \end{aligned}$$

Result holds up for $m=k+1$. Therefore, the Proposition 8 holds up.

Appendix B

1.1 B.1 Proof of Proposition 9

We shall prove it by mathematical induction on m.

Let $m=1$,

$$\begin{aligned} b(z)= & {} (\alpha _{4}z^{4}+\alpha _{3}z^{3}+\alpha _{2}z^{2}+\alpha _{1}z+\alpha _{0})(1+z+z^2) =\alpha _{4}z^{6}+(\alpha _{4}+\alpha _{3})z^{5}\\&+(\alpha _{4}+a_{3}+\alpha _{2})z^{4}+(\alpha _{3}+\alpha _{2}+\alpha _{1})z^{3} +(\alpha _{2}+\alpha _{1}+\alpha _{0})z^{2}+(\alpha _{1}\\&+\alpha _{0})z+\alpha _{0}. \end{aligned}$$

The generating polynomial of the difference scheme b(z) is corresponding to the following $4 \times 4$ order difference matrix $B_{1}$

$$\begin{aligned} B_{1}= \begin{bmatrix} \alpha _{4}&\quad \alpha _{3}+\alpha _{2}+\alpha _{1}&\quad \alpha _{0}&\quad 0 \\ 0&\quad \alpha _{4}+\alpha _{3}&\quad \alpha _{2}+\alpha _{1}+\alpha _{0}&\quad 0\\ 0&\quad \alpha _{4}+\alpha _{3}+\alpha _{2}&\quad \alpha _{1}+\alpha _{0}&\quad 0\\ 0&\quad \alpha _{4}&\quad \alpha _{3}+\alpha _{2}+\alpha _{1}&\quad \alpha _{0} \end{bmatrix}_{4 \times 4}. \end{aligned}$$

Eigenvalues of $B_{1}$ are roots of the equation $|B_{1}-\lambda I|=0$, where I is the $4\times 4$ order unit matrix. Therefore we have

$$\begin{aligned} |B_{1}-\lambda I|= \begin{vmatrix} \alpha _{4}-\lambda&\quad \alpha _{3}+\alpha _{2}+\alpha _{1}&\quad \alpha _{0}&\quad 0 \\ 0&\quad \alpha _{4}+\alpha _{3}-\lambda&\quad \alpha _{2}+\alpha _{1}+\alpha _{0}&\quad 0\\ 0&\quad \alpha _{4}+\alpha _{3}+\alpha _{2}&\quad \alpha _{1}+\alpha _{0}-\lambda&\quad 0\\ 0&\quad \alpha _{4}&\quad \alpha _{3}+\alpha _{2}+\alpha _{1}&\quad \alpha _{0}-\lambda \end{vmatrix}_{4 \times 4}=0. \end{aligned}$$

Adding all the columns in the second last column and expanding the resulting determinant according to the last column and then according to the first column, we get

$$\begin{aligned} (\alpha _{4}-\lambda )(\alpha _{0}-\lambda ) \begin{vmatrix} \alpha _{4}+\alpha _{3}-\lambda&\sum \limits _{i=0}^{4}\alpha _{i}-\lambda \\ \alpha _{4}+\alpha _{3}+\alpha _{2}&\sum \limits _{i=0}^{4}\alpha _{i}-\lambda \end{vmatrix}_{2 \times 2}=0. \end{aligned}$$

Now extract the common factor $\sum \nolimits _{i=0}^{4}\alpha _{i}-\lambda$ from the last column and then subtracting last row from the first row, we get

$$\begin{aligned} (\alpha _{4}-\lambda )(\alpha _{0}-\lambda )(\sum \limits _{i=0}^{4}\alpha _{i}-\lambda ) \begin{vmatrix} -\alpha _{2}-\lambda&0\\ \alpha _{4}+\alpha _{3}+\alpha _{2}&1 \end{vmatrix}_{2 \times 2}=0. \end{aligned}$$

Therefore, eigenvalues of the difference matrix $B_{1}$ are $\alpha _{4}$, $\alpha _{0}$, $\sum \nolimits _{i=0}^{4}\alpha _{i}$ and $-\alpha _{2}$.

Suppose that the result holds up for $m=k$, we have to prove that the result also holds up for $m=k+1$.

When $m=k$, the generating polynomial b(z) of difference scheme is

$$\begin{aligned} b(z)= & {} b_{k}(z)=(\alpha _{4}z^{4}+\alpha _{3}z^{3}+\alpha _{2}z^{2}+\alpha _{1}z+\alpha _{0})(1+z+z^2)^{k} =\beta _{4+2k}z^{4+2k}\\&+\,\beta _{3+2k}z^{3+2k}+\beta _{2+2k}z^{2+2k}+\cdots \beta _{4+k}z^{4+k}+\beta _{3+k}z^{3+k}+\beta _{2+k}z^{2+k}+\cdots \\&+\,\beta _{2}z^{2}+\beta _{1}z+\beta _{0}. \end{aligned}$$

If we assume that $4+2k=3p$, then by the Proposition 8, we have

$$\begin{aligned}&\beta _{4+2k}+\beta _{1+2k}+\cdots +\beta _{4+k}+\beta _{1+k}+\cdots +\beta _{3}+\beta _{0}=\beta _{3+2k}+\beta _{2k}+\cdots +\beta _{3+k}+\beta _{k}\\&\quad +\cdots +\beta _{4}+\beta _{1}=\beta _{2+2k}+\beta _{2k-1}+\cdots +\beta _{2+k}+\beta _{k-1}+\cdots +\beta _{5}+\beta _{2}= 3^{k-1}\sum \limits _{i=0}^{4}\alpha _{i}. \end{aligned}$$

Let $B_{k}$ of order $(3+2k) \times (3+2k)$ be the subdivision matrix corresponding to the generating polynomial of difference scheme $b_{k}(z)$, then

$$\begin{aligned} B_{k}= \begin{bmatrix} \beta _{2k+4}&\quad \beta _{2k+1}&\quad \beta _{2k-2}&\quad \beta _{2k-5}&\quad \dots&\quad 0&\quad 0&\quad 0 \\ 0&\quad \beta _{2k+3}&\quad \beta _{2k}&\quad \beta _{2k-3}&\quad \dots&\quad 0&\quad 0&\quad 0\\ 0&\quad \beta _{2k+2}&\quad \beta _{2k-1}&\quad \beta _{2k-4}&\quad \dots&\quad 0&\quad 0&\quad 0\\ 0&\quad \beta _{2k+4}&\quad \beta _{2k+1}&\quad \beta _{2k-2}&\quad \dots&\quad 0&\quad 0&\quad 0\\ \vdots&\quad \vdots&\quad \vdots&\quad \vdots&\quad \vdots&\quad \vdots&\quad \vdots&\quad \vdots \\ 0&\quad 0&\quad 0&\quad 0&\quad \dots&\quad \beta _{5}&\quad \beta _{2}&\quad 0\\ 0&\quad 0&\quad 0&\quad 0&\quad \dots&\quad \beta _{4}&\quad \beta _{1}&\quad 0 \\ 0&\quad 0&\quad 0&\quad 0&\quad \dots&\quad \beta _{6}&\quad \beta _{3}&\quad \beta _{0} \end{bmatrix}, \end{aligned}$$

where $4+2k=3p$.

The eigenvalues of the matrix $B_{k}$ are the roots of the characteristic equation $|B_{k}-\lambda I|=0$.

When $m=k+1$, the generating polynomial of the difference scheme become

$$\begin{aligned} b_{k+1}(z)= & {} b_{k}(z)(1+z+z^2)=\beta _{4+2k}z^{6+2k}+(\beta _{4+2k}+\beta _{3+2k})z^{5+2k}+(\beta _{4+2k}+\beta _{3+2k}\\&+\,\beta _{2+2k})z^{4+2k}+(\beta _{3+2k}+\beta _{2+2k}+\beta _{1+2k})z^{3+2k}+\cdots +(\beta _{3}+\beta _{2}+\beta _{1})z^{3}\\&+\,(\beta _{2}+\beta _{1}+\beta _{0})z^{2}+(\beta _{1}+\beta _{0})z+\beta _{0}. \end{aligned}$$

Then subdivision matrix $B_{k+1}$, which is corresponding to the generating polynomial of difference scheme $b_{k+1}(z)$ is

$$\begin{aligned} B_{k+1} = \left[ \begin{array}{ccccccccccccccccc} \beta _{2k+4} &{} \beta _{2k+3}+\beta _{2k+2}+\beta _{2k+1} &{} \beta _{2k}+\beta _{2k-1}+\beta _{2k-2} \\ 0 &{} \beta _{2k+4}+\beta _{2k+3} &{} \beta _{2k+2}+\beta _{2k+1}+\beta _{2k} \\ 0 &{} \beta _{2k+4}+\beta _{2k+3}+\beta _{2k+2} &{} \beta _{2k+1}+\beta _{2k}+\beta _{2k-1}\beta _{2k-1} \\ 0 &{} \beta _{2k+4} &{} \beta _{2k+3}+\beta _{2k+2}+\beta _{2k+1} \\ \vdots &{} \vdots &{} \vdots \\ \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \end{array} \right. \\ \left. \begin{array}{ccccccccccccccccc} \beta _{2k-3}+\beta _{2k-4}+\beta _{2k-5} &{} \dots &{} 0 &{} 0 &{} 0 \\ \beta _{2k-1}+\beta _{2k-2}+\beta _{2k-3} &{} \dots &{} 0 &{} 0 &{} 0\\ \beta _{2k-2}+\beta _{2k-3}+\beta _{2k-4} &{} \dots &{} 0 &{} 0 &{} 0\\ \beta _{2k}+\beta _{2k-1}+\beta _{2k-2} &{} \dots &{} 0 &{} 0 &{} 0\\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots \\ 0 &{} \dots &{} \beta _{5}+\beta _{4}+\beta _{3} &{} \beta _{2}+\beta _{1}+\beta _{0} &{} 0\\ 0 &{} \dots &{} \beta _{4}+\beta _{3}+\beta _{2} &{} \beta _{1}+\beta _{0} &{}0 \\ 0 &{} \dots &{} \beta _{6}+\beta _{5}+\beta _{4} &{} \beta _{3}+\beta _{2}+\beta _{1} &{} \beta _{0} \end{array} \right] . \end{aligned}$$

Eigenvalues of matrix $B_{k+1}$ are roots of the equation $\left| B_{k+1}-\lambda I\right| =0$, therefore

$$\begin{aligned} B_{k+1}-\lambda I= & {} \left[ \begin{array}{ccccccccccccccccc} \beta _{2k+4}-\lambda &{} \beta _{2k+3}+\beta _{2k+2}+\beta _{2k+1} &{} \beta _{2k}+\beta _{2k-1}+\beta _{2k-2} \\ 0 &{} \beta _{2k+4}+\beta _{2k+3}-\lambda &{} \beta _{2k+2}+\beta _{2k+1}+\beta _{2k} \\ 0 &{} \beta _{2k+4}+\beta _{2k+3}+\beta _{2k+2} &{} \beta _{2k+1}+\beta _{2k}+\beta _{2k-1}\beta _{2k-1}-\lambda \\ 0 &{} \beta _{2k+4} &{} \beta _{2k+3}+\beta _{2k+2}+\beta _{2k+1}\\ \vdots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} 0\\ 0 &{} 0 &{} 0 \\ 0 &{} 0 &{} 0 \\ \end{array} \right. \\&\left. \begin{array}{ccccccccccccccccc} \dots &{} 0 &{} 0 &{} 0 \\ \dots &{} 0 &{} 0 &{} 0\\ \dots &{} 0 &{} 0 &{} 0\\ \dots &{} 0 &{} 0 &{} 0\\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ \dots &{} \beta _{5}+\beta _{4}+\beta _{3}-\lambda &{} \beta _{2}+\beta _{1}+\beta _{0} &{} 0\\ \dots &{} \beta _{4}+\beta _{3}+\beta _{2} &{} \beta _{1}+\beta _{0}-\lambda &{}0 \\ \dots &{} \beta _{6}+\beta _{5}+\beta _{4} &{} \beta _{3}+\beta _{2}+\beta _{1} &{} \beta _{0}-\lambda \end{array} \right] . \end{aligned}$$

Now we only need to show that the eigenvalues of the matrix $B_{k+1}$ are the eigenvalues of the matrix $B_{k}$ and $3^{k}\sum \nolimits _{i=0}^{4}\alpha _{i}$. Adding all the columns in the last column and extracting the factor $\sum \nolimits _{i=0}^{4+2k}\beta _{i}-\lambda =3^{k}\sum \nolimits _{i=0}^{4}\alpha _{i}-\lambda$, then all the elements of last columns will be 1. Now by subtracting each previous row from its consecutive next row, we get

$$\begin{aligned} B_{k+1}-\lambda I= & {} \left( 3^{k}\sum \limits _{i=0}^{4}\alpha _{i}-\lambda \right) \left[ \begin{array}{ccccccccccccccccc} \beta _{2k+4}-\lambda &{} \beta _{2k+2}+\beta _{2k+1}-\beta _{2k+4}+\lambda \\ 0 &{} -\beta _{2k+2}-\lambda \\ 0 &{} \beta _{2k+3}+\beta _{2k+2} \\ 0 &{} \beta _{2k+4} \\ \vdots &{} \vdots \\ 0 &{} 0 \\ 0 &{} 0 \\ 0 &{} 0 \\ \end{array} \right. \\&\left. \begin{array}{ccccccccccccccccc} \beta _{2k-1}+\beta _{2k-2}-\beta _{2k+1}-\beta _{2k+2} &{} \dots &{} 0 &{} 0 &{} 0 \\ \beta _{2k+2}-\beta _{2k-1}+\lambda &{} \dots &{} 0 &{} 0 &{} 0\\ \beta _{2k}+\beta _{2k-1}-\beta _{2k+3}-\beta _{2k+2}-\lambda &{} \dots &{} 0 &{} 0 &{} 0\\ \beta _{2k+2}+\beta _{2k+1}-\beta _{2k+4} &{} \dots &{} 0 &{} 0 &{} 0\\ \vdots &{} \vdots &{} \vdots &{} \vdots &{} \vdots \\ 0 &{} \dots &{} \beta _{5}-\beta _{2}-\lambda &{} \beta _{2}+\lambda &{} 0\\ 0 &{} \dots &{} \beta _{3}+\beta _{2}-\beta _{5}-\beta _{6} &{} \beta _{0}-\beta _{2}+\beta _{3}-\lambda &{}0 \\ 0 &{} \dots &{} \beta _{6}+\beta _{5}+\beta _{4} &{} \beta _{3}+\beta _{2}+\beta _{1} &{} 1 \end{array} \right] . \end{aligned}$$

Multiplying the both sides of the above matrix with the transformation matrix

we get

We can write the above matrices in the form of a determinant, therefore the eigenvalues of the matrix $B_{k+1}$ are the eigenvalues of the matrix $B_{k}$ and $3^{k}\sum \nolimits _{i=0}^{4}\alpha _{i}$. Hence the Proposition 9 has proved.