Relationship between time-instant number and precision of ZeaD formulas with proofs

Abstract

Recently, zeroing dynamics (ZD) method has shown wonderful effect in solving time-varying problems. Generally, the continuous-time ZD model should be discretized as a discrete-time algorithm for numerical computation. A good choice is to apply Zhang et al. discretization (ZeaD) formulas, which are effective time-discretization formulas. Actually, ZeaD formulas can also be applied to obtaining various discrete-time algorithms, not only the ZD algorithms. In the previous work, some piecemeal ZeaD formulas have been proposed and investigated. However, the relationship between the time-instant number and precision of ZeaD formulas is not found, which is emphatically investigated in this paper. Specifically, the ZeaD formulas from two to nine instants are investigated, and the general ZeaD formula groups are studied. Two-instant and three-instant ZeaD formula groups have linear precision at most. Four-instant ZeaD formula group has quadratic precision at most. Five-instant and six-instant ZeaD formula groups have cubic precision at most. Seven-instant and eight-instant ZeaD formula groups have quartic precision at most. Nine-instant ZeaD formula group has quintic precision at most. Theoretical analyses are presented to substantiate the relationship. Moreover, the ZeaD formulas as well as ZD method are applied to solving time-varying quadratic optimization problem, and the numerical results verify the effectiveness of ZeaD formulas.

Appendices

Appendix: 1. Proof of three-instant ZeaD formula group

Evidently, when m = 3, three-instant ZeaD formula can be presented as

$$ \dot{x}_{k}=a_{2}\frac{x_{k+1}}{\tau}+a_{1}\frac{x_{k}}{\tau}+a_{0}\frac{x_{k-1}}{\tau}+O(\tau^{p}). $$

According to (2), if p = 2, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{2}+a_{1}+a_{0}=0,\\ a_{2}-a_{0}=1,\\ a_{2}+a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{2}=1/2,\\ a_{1}=0,\\ a_{0}=-1/2. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ P_{2}(\gamma)=\frac{1}{2}\gamma^{2}-\frac{1}{2}=0, $$

whose roots are γ₀ = 1 and γ₁ = − 1. Hence, the synthesized formula is not enough 0-stable [32], i.e., three-instant ZeaD formula cannot have O(τ²) precision.

Meanwhile, if p = 1, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{2}+a_{1}+a_{0}=0,\\ a_{2}-a_{0}=1, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{2}=a_{0}+1,\\ a_{1}=-2a_{0}-1. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ P_{2}(\gamma)=(a_{0}+1)\gamma^{2}-(2a_{0}+1)\gamma+a_{0}=0. $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets

$$ P_{2}(\omega)=2\omega+4a_{0}+2=0. $$

From Routh stability criterion [33, 34], the 0-stable condition is a₀ > − 1/2. Besides, a₀≠ 0 should be satisfied. The proof is thus completed.

Appendix: 2. Proof of four-instant ZeaD formula group

When m = 4, four-instant ZeaD formula can be presented as

$$ \dot{x}_{k}=a_{3}\frac{x_{k+1}}{\tau}+a_{2}\frac{x_{k}}{\tau}+a_{1}\frac{x_{k-1}}{\tau}+a_{0}\frac{x_{k-2}}{\tau}+O(\tau^{p}). $$

According to (2), if p = 3, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{3}-a_{1}-2a_{0}=1,\\ a_{3}+a_{1}+4a_{0}=0,\\ a_{3}-a_{1}-8a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{3}=1/3,\\ a_{2}=1/2,\\ a_{1}=-1,\\ a_{0}=1/6. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ P_{3}(\gamma)=\frac{1}{3}\gamma^{3}+\frac{1}{2}\gamma^{2}-\gamma+\frac{1}{6}=0, $$

whose roots are γ₀ = 1, $\gamma _{1}=(\sqrt {33}-5)/4$, and $\gamma _{2}=(-\sqrt {33}-5)/4$. Evidently, |γ₂| > 1. Hence, the synthesized formula is not 0-stable [32], i.e., four-instant ZeaD formula cannot have O(τ³) precision.

Meanwhile, if p = 2, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{3}-a_{1}-2a_{0}=1,\\ a_{3}+a_{1}+4a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{3}=-a_{0}+1/2,\\ a_{2}=3a_{0},\\ a_{1}=-3a_{0}-1/2. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ P_{3}(\gamma)=\left( -a_{0}+\frac{1}{2}\right)\gamma^{3}+3a_{0}\gamma^{2}-\left( 3a_{0}+\frac{1}{2}\right)\gamma+a_{0}=0. $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets

$$ 2\omega^{2}+2\omega-8a_{0}=0. $$

From Routh stability criterion [33, 34], the 0-stable condition is a₀ < 0. The general four-instant ZeaD formula (5) is obtained. The proof is thus completed.

Appendix: 3. Proof of five-instant ZeaD formula group

When m = 5, five-instant ZeaD formula can be presented as

$$\dot{x}_{k}=a_{4}\frac{x_{k+1}}{\tau}+a_{3}\frac{x_{k}}{\tau}+a_{2}\frac{x_{k-1}}{\tau}+a_{1}\frac{x_{k-2}}{\tau}+a_{0}\frac{x_{k-3}}{\tau}+O(\tau^{p}). $$

According to (2), if p = 4, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{4}-a_{2}-2a_{1}-3a_{0}=1,\\ a_{4}+a_{2}+4a_{1}+9a_{0}=0,\\ a_{4}-a_{2}-8a_{1}-27a_{0}=0,\\ a_{4}+a_{2}+16a_{1}+81a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{4}=1/4,\\ a_{3}=5/6,\\ a_{2}=-3/2,\\ a_{1}=1/2,\\ a_{0}=-1/12. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ P_{4}(\gamma)=\frac{1}{4}\gamma^{4}+\frac{5}{6}\gamma^{3}-\frac{3}{2}\gamma^{2}+\frac{1}{2}\gamma-\frac{1}{12}=0. $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets the transformed characteristic equation

$$ 2\omega^{3}+4\omega^{2}+\frac{2}{3}\omega-\frac{8}{3}=0. $$

Because there is a negative coefficient − 8/3, the synthesized formula is not 0-stable [33, 34]. That is, five-instant ZeaD formula cannot have O(τ⁴) precision.

Meanwhile, if p = 3, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{4}-a_{2}-2a_{1}-3a_{0}=1,\\ a_{4}+a_{2}+4a_{1}+9a_{0}=0,\\ a_{4}-a_{2}-8a_{1}-27a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{4}=a_{0}+1/3,\\ a_{3}=-4a_{0}+1/2,\\ a_{2}=6a_{0}-1,\\ a_{1}=-4a_{0}+1/6. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ P_{4}(\gamma)=\left( a_{0}+\frac{1}{3}\right)\gamma^{4}-\left( 4a_{0}-\frac{1}{2}\right)\gamma^{3}+\left( 6a_{0}-1\right)\gamma^{2}-\left( 4a_{0}-\frac{1}{6}\right)\gamma+a_{0}=0. $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets

$$ 2\omega^{3}+4\omega^{2}+\frac{2}{3}\omega+16a_{0}-\frac{4}{3}=0. $$

From Routh stability criterion [33, 34], the 0-stable condition is 1/12 < a₀ < 1/6. The general five-instant ZeaD formula (6) is obtained. The proof is thus completed.

Appendix: 4. Proof of six-instant ZeaD formula group

When m = 6, six-instant ZeaD formula can be presented as

$$\dot{x}_{k}=a_{5}\frac{x_{k+1}}{\tau}+a_{4}\frac{x_{k}}{\tau}+a_{3}\frac{x_{k-1}}{\tau}+a_{2}\frac{x_{k-2}}{\tau}+a_{1}\frac{x_{k-3}}{\tau}+a_{0}\frac{x_{k-4}}{\tau}+O(\tau^{p}). $$

According to (2), if p = 4, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{5}-a_{3}-2a_{2}-3a_{1}-4a_{0}=1,\\ a_{5}+a_{3}+4a_{2}+9a_{1}+16a_{0}=0,\\ a_{5}-a_{3}-8a_{2}-27a_{1}-64a_{0}=0,\\ a_{5}+a_{3}+16a_{2}+81a_{1}+256a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{5}=-a_{0}+1/4,\\ a_{4}=5a_{0}+5/6,\\ a_{3}=-10a_{0}-3/2,\\ a_{2}=10a_{0}+1/2,\\ a_{1}=-5a_{0}-1/12. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ \begin{array}{llll} P_{5}(\gamma)=&\left( -a_{0}+\frac{1}{4}\right)\gamma^{5}+\left( 5a_{0}+\frac{5}{6}\right)\gamma^{4}-\left( 10a_{0}+\frac{3}{2}\right)\gamma^{3}\\ &+\left( 10a_{0}+\frac{1}{2}\right)\gamma^{2}-\left( 5a_{0}+\frac{1}{12}\right)\gamma+a_{0}=0. \end{array} $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets transformed characteristic equation

$$ 2\omega^{4}+6\omega^{3}+\frac{14}{3}\omega^{2}-2\omega-32a_{0}-\frac{8}{3}=0. $$

Because there is a negative coefficient − 2, the synthesized formula is not 0-stable [33, 34]. That is, six-instant ZeaD formula cannot have O(τ⁴) precision.

Meanwhile, if p = 3, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{5}-a_{3}-2a_{2}-3a_{1}-4a_{0}=1,\\ a_{5}+a_{3}+4a_{2}+9a_{1}+16a_{0}=0,\\ a_{5}-a_{3}-8a_{2}-27a_{1}-64a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{5}=a_{1}+4a_{0}+1/3,\\ a_{4}=-4a_{1}-15a_{0}+1/2,\\ a_{3}=6a_{1}+20a_{0}-1,\\ a_{2}=-4a_{1}-10a_{0}+1/6. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ \begin{array}{llll} P_{5}(\gamma)=&\left( a_{1}+4a_{0}+\frac{1}{3}\right)\gamma^{5}-\left( 4a_{1}+15a_{0}-\frac{1}{2}\right)\gamma^{4}+(6a_{1}+20a_{0}-1)\gamma^{3}\\ &-\left( 4a_{1}+10a_{0}-\frac{1}{6}\right)\gamma^{2}+a_{1}\gamma+a_{0}=0. \end{array} $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets

$$ 2\omega^{4}+6\omega^{3}+\frac{14}{3}\omega^{2}+\left( 16a_{1}+80a_{0}-\frac{2}{3}\right)\omega +16a_{1}+48a_{0}-\frac{4}{3}=0. $$

From Routh stability criterion [33, 34], the 0-stable condition is presented as the group form of

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} 12a_{1}+36a_{0}-1>0,\\ 24a_{1}+120a_{0}-1>0,\\ 12a_{1}+60a_{0}-11<0,\\ 36{a_{1}^{2}}+360a_{1}a_{0}+900{a_{0}^{2}}+6a_{1}-51a_{0}-2<0. \end{array} \right. \end{array} $$

By analyzing and plotting, the condition can be reduced to

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} 12a_{1}+36a_{0}-1>0,\\ 36{a_{1}^{2}}+360a_{1}a_{0}+900{a_{0}^{2}}+6a_{1}-51a_{0}-2<0. \end{array} \right. \end{array} $$

The general six-instant ZeaD formula (7) is obtained. The proof is thus completed.

Appendix: 5. Proof of seven-instant ZeaD formula group

When m = 7, seven-instant ZeaD formula can be presented as

$$ \dot{x}_{k}=a_{6}\frac{x_{k+1}}{\tau}+a_{5}\frac{x_{k}}{\tau}+a_{4}\frac{x_{k-1}}{\tau}+a_{3}\frac{x_{k-2}}{\tau}+a_{2}\frac{x_{k-3}}{\tau}+a_{1}\frac{x_{k-4}}{\tau}+a_{0}\frac{x_{k-5}}{\tau}+O(\tau^{p}). $$

According to (2), if p = 5, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{6}+a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{6}-a_{4}-2a_{3}-3a_{2}-4a_{1}-5a_{0}=1,\\ a_{6}+a_{4}+4a_{3}+9a_{2}+16a_{1}+25a_{0}=0,\\ a_{6}-a_{4}-8a_{3}-27a_{2}-64a_{1}-125a_{0}=0,\\ a_{6}+a_{4}+16a_{3}+81a_{2}+256a_{1}+625a_{0}=0,\\ a_{6}-a_{4}-32a_{3}-243a_{2}-1024a_{1}-3125a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{6}=a_{0}+1/5,\\ a_{5}=-6a_{0}+13/12,\\ a_{4}=15a_{0}-2,\\ a_{3}=-20a_{0}+1,\\ a_{2}=15a_{0}-1/3,\\ a_{1}=-6a_{0}+1/20. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ \begin{array}{llll} P_{6}(\gamma)=&\left( a_{0}+\frac{1}{5}\right)\gamma^{6}-\left( 6a_{0}-\frac{13}{12}\right)\gamma^{5}+\left( 15a_{0}-2\right)\gamma^{4}-\left( 20a_{0}-1\right)\gamma^{3}\\ &+\left( 15a_{0}-\frac{1}{3}\right)\gamma^{2}-\left( 6a_{0}-\frac{1}{20}\right)\gamma+a_{0}=0. \end{array} $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets transformed characteristic equation

$$ 2\omega^{5}+8\omega^{4}+\frac{32}{3}\omega^{3}+\frac{8}{3}\omega^{2}-\frac{94}{15}\omega+64a_{0}-\frac{64}{15}=0. $$

Because there is a negative coefficient − 94/15, the synthesized formula is not 0-stable [33, 34]. That is, seven-instant ZeaD formula cannot have O(τ⁵) precision.

Meanwhile, if p = 4, one has

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{6}+a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=0,\\ a_{6}-a_{4}-2a_{3}-3a_{2}-4a_{1}-5a_{0}=1,\\ a_{6}+a_{4}+4a_{3}+9a_{2}+16a_{1}+25a_{0}=0,\\ a_{6}-a_{4}-8a_{3}-27a_{2}-64a_{1}-125a_{0}=0,\\ a_{6}+a_{4}+16a_{3}+81a_{2}+256a_{1}+625a_{0}=0, \end{array} \right. \end{array} $$

i.e.,

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} a_{6}=-a_{1}-5a_{0}+1/4,\\ a_{5}=5a_{1}+24a_{0}+5/6,\\ a_{4}=-10a_{1}-45a_{0}-3/2,\\ a_{3}=10a_{1}+40a_{0}+1/2,\\ a_{2}=-5a_{1}-15a_{0}-1/12. \end{array} \right. \end{array} $$

In this case, the characteristic equation is

$$ \begin{array}{llll} P_{6}(\gamma)=&\left( -a_{1}-5a_{0}+\frac{1}{4}\right)\gamma^{6}+\left( 5a_{1}+24a_{0}+\frac{5}{6}\right)\gamma^{5}\\ &-\left( 10a_{1}+45a_{0}+\frac{3}{2}\right)\gamma^{4}+\left( 10a_{1}+40a_{0}+\frac{1}{2}\right)\gamma^{3}\\ &-\left( 5a_{1}+15a_{0}+\frac{1}{12}\right)\gamma^{2}+a_{1}\gamma+a_{0}=0. \end{array} $$

By applying bilinear transformation [33, 34], i.e., γ = (ω + 1)/(ω − 1), one gets

$$ 2\omega^{5}+8\omega^{4}+\frac{32}{3}\omega^{3}+\frac{8}{3}\omega^{2}-\left( 32a_{1}+192a_{0}+\frac{14}{3}\right)\omega-\left( 32a_{1}+128a_{0}+\frac{8}{3}\right)=0. $$

From Routh stability criterion [33, 34], the 0-stable condition is in the group form as follows:

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} 48a_{1}+288a_{0}+7<0,\\ 12a_{1}+48a_{0}+1<0,\\ 36a_{1}+240a_{0}+11>0,\\ 432{a_{1}^{2}}+5760a_{1}a_{0}+19200{a_{0}^{2}}-96a_{1}+160a_{0}-3<0. \end{array} \right. \end{array} $$

By analyzing and plotting, the condition can be reduced to

$$ \begin{array}{@{}rcl@{}} \left\{ \begin{array}{l} 12a_{1}+48a_{0}+1<0,\\ 432{a_{1}^{2}}+5760a_{1}a_{0}+19200{a_{0}^{2}}-96a_{1}+160a_{0}-3<0. \end{array} \right. \end{array} $$

The general seven-instant ZeaD formula (8) is obtained. The proof is thus completed.