# Open Problems Related to the Hurwitz Stability of Polynomials Segments

Authors: Baltazar Aguirre-Hernández; Faustino Ricardo García-Sosa; Carlos Arturo Loredo-Villalobos; Raúl Villafuerte-Segura; Eric Campos-Cantón
Journal: Mathematical Problems in Engineering
 (2018)

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## Abstract

In the framework of robust stability analysis of linear systems, the development of techniques and methods that help to obtain necessary and sufficient conditions to determine stability of convex combinations of polynomials is paramount. In this paper, knowing that Hurwitz polynomials set is not a convex set, a brief overview of some results and open problems concerning the stability of the convex combinations of Hurwitz polynomials is then provided.


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## Body

## 1. Introduction

The design of control systems arises from the need of human beings to manipulate real plants with some degree of confidence and accuracy. However, this design requires mathematical models which are often very complex if all the dynamics that involve a real plant are taken into account. Thus, a complex model requires a complex controller design. Robust control theory intends to analyze a complex model by studying its linear approximation and always assuming that this approximation will incur with some degree of modeling errors. This error is regarded as uncertainty which is modeled and bounded to determine stability conditions of a system and control laws are obtained.Problems that arise from robust control theory containing one uncertain parameter, such as gain or constants time, and the system stability should be determined for the entire uncertainty range. The study of criteria for deciding whether a convex combination of polynomials is stable can be applied to resolution of a wide variety of uncertainty systems. The problem to be raised is as follows: find conditions about the stable polynomialsp0(t) and p1(t) such that the convex combination described by p(t,λ)=λP0(t)+(1-λ)P1(t) is stable (Hurwitz), for all λ∈[0,1].The first result, where necessary and sufficient conditions were presented, is known as the Bialas Theorem [1–3]. A different approach in terms of the frequency domain is known as theSegment Lemma and it was found by Bhattacharyya et al. [4, 5]. This result shows that the stability of p(t,λ) is equivalent to certain conditions that must be met by even and odd part of the polynomials p0(t) and p1(t). In complex polynomials case, Bose developed a method to determine the stability of the segment of complex polynomials [6]. In regard to sufficient conditions, in [7, 8], the well-known Rantzer’s conditions were obtained.Based on the above criteria computer algorithms have been developed to resolve this problem. The Segment Lemma has been used to design an algorithm in [9]. On this same path, a method to verify the stability of convex combinations of polynomials in a finite number of calculus is obtained in [10]. In [11] an algorithm to determine the stability of segment of complex polynomials is proposed; it is based on the work of Bose [12]. In the present paper, we will discuss some of these criteria and propose some open problems in these topics. Applications of Hurwitz stability criterion to Engineering problems are huge, for example, for the stability of wave propagation in strain gradient [13] or mixture [14].The paper is organized as follows. In Section2, the problem statement is presented. Some results with necessary and sufficient condition on stability of segment of Hurwitz polynomials and its open problems are presented in Section 3. In Section 4, some open problems with sufficient condition and using Rantzer’s conditions are postulated. In Section 5, a related problem with segments of stable polynomials is the calculating of the Minimum Left Extreme, which was studied by Bialas. In Section 6 some results and open problems related to Hadamardized Hurwitz polynomials are postulated. Finally, in Section 7, some concluding remarks are provided.


## 2. Problem Statement

In this section some definitions and one motivation example in the framework of the stability of polynomials are given.Definition 1.
A polynomial with real coefficients,f(t)=b0tn+b1tn-1+⋯+bn-1t+bn, is Hurwitz if all its roots have negative real part.Definition 2.
Suppose thatp0(t) and p1(t) are polynomials (real or complex) of degree n. Let(1)Pt,λ=λp0t+1-λp1t,and consider the following one parameter family of polynomials:(2)p0t,p1t=Pt,λ:λ∈0,1.This family will be referred to as a segment of polynomials. We shall say that the segment is stable if and only if every polynomial on the segment is stable.The following example shows that a segment of polynomials is not necessarily a stable segment although the two extremes are Hurwitz polynomials.Example 3.
LetP0(t)=t3+6t2+11t+6=(t+1)(t+2)(t+3) and P1(t)=6t2+5t+216 be Hurwitz; nevertheless(3)λP0t+1-λP1t=λt3+6t2+11t+6t+1-λ6t2+5t+216is not Hurwitz for all λ≥0:(i)
forλ∈0,5-2/7 it is Hurwitz;(ii)
forλ∈5-2/7,5+2/7 it is not Hurwitz;(iii)
forλ∈5+2/7,1 it is Hurwitz.Now, there are several criteria to determine the stability of the segment of polynomials; we present some of them below.


## 3. Necessary and Sufficient Conditions

In this section, we present two of the most known necessary and sufficient conditions on the stability of segment of Hurwitz polynomials.


### 3.1. The Bialas Theorem

Theorem 4 (see [2]).
Ifp0 is Hurwitz and deg⁡p0>deg⁡p1, then P(t,λ)=λp0(t)+(1-λ)p1(t) is Hurwitz for all λ∈[0,1], if and only if the matrix H-1(p0)H(p1) has eigenvalues not in (-∞,0), where H-1(p0) and H(p1) are matrices Hurwitz of p0 and p1, respectively.Open Problem 1.
Is it possible to find necessary and sufficient conditions in a similar way of the Bialas Theorem fordeg⁡p0=deg⁡p1 case?
In order to establish the Segment Lemma we give some definitions.
Given a polynomialp(t)=a0+a1t+a2t2+⋯+antn, we define (4)pevens=a0+a2t2+⋯,podds=a1t+a3t3+⋯,then p(t)=peven(t)+podd(t) and p(iω) can be written as p(iω)=pe(ω)+iωpo(ω), where (5)peω=a0-a2ω2+a4ω4+⋯,poω=a1-a3ω2+a5ω4+⋯.
In the Bialas Theorem, it is necessary that the two extremes of the segment of polynomials have different degree. Now we present the Segment Lemma, where the two extremes of the segment of polynomials have equal degrees.


### 3.2. The Segment Lemma

Lemma 5 (see [4, 5]).
Letp1(t) and p2(t)   be n-degree Hurwitz polynomials with leading coefficients of the same sign. Then the segment of polynomials [p1(t),p2(t)] is Hurwitz, if and only if there are not ω>0 that satisfies the three conditions: (1)
p1e(ω)p2o(ω)-p2e(ω)p1o(ω)=0,(2)
p1e(ω)p2e(ω)≤0,(3)
p1o(ω)p2o(ω)≤0.Open Problem 2.
Is it possible to get a result, based on the approach of the Segment Lemma, to be applicable when polynomialsp0 and p1 have different degrees?
After we have discussed two necessary and sufficient conditions, now in the next section, we expose two sufficient conditions: Rantzer’s conditions and one approach of matrix inequalities.


## 3.1. The Bialas Theorem

Theorem 4 (see [2]).
Ifp0 is Hurwitz and deg⁡p0>deg⁡p1, then P(t,λ)=λp0(t)+(1-λ)p1(t) is Hurwitz for all λ∈[0,1], if and only if the matrix H-1(p0)H(p1) has eigenvalues not in (-∞,0), where H-1(p0) and H(p1) are matrices Hurwitz of p0 and p1, respectively.Open Problem 1.
Is it possible to find necessary and sufficient conditions in a similar way of the Bialas Theorem fordeg⁡p0=deg⁡p1 case?
In order to establish the Segment Lemma we give some definitions.
Given a polynomialp(t)=a0+a1t+a2t2+⋯+antn, we define (4)pevens=a0+a2t2+⋯,podds=a1t+a3t3+⋯,then p(t)=peven(t)+podd(t) and p(iω) can be written as p(iω)=pe(ω)+iωpo(ω), where (5)peω=a0-a2ω2+a4ω4+⋯,poω=a1-a3ω2+a5ω4+⋯.
In the Bialas Theorem, it is necessary that the two extremes of the segment of polynomials have different degree. Now we present the Segment Lemma, where the two extremes of the segment of polynomials have equal degrees.


## 3.2. The Segment Lemma

Lemma 5 (see [4, 5]).
Letp1(t) and p2(t)   be n-degree Hurwitz polynomials with leading coefficients of the same sign. Then the segment of polynomials [p1(t),p2(t)] is Hurwitz, if and only if there are not ω>0 that satisfies the three conditions: (1)
p1e(ω)p2o(ω)-p2e(ω)p1o(ω)=0,(2)
p1e(ω)p2e(ω)≤0,(3)
p1o(ω)p2o(ω)≤0.Open Problem 2.
Is it possible to get a result, based on the approach of the Segment Lemma, to be applicable when polynomialsp0 and p1 have different degrees?
After we have discussed two necessary and sufficient conditions, now in the next section, we expose two sufficient conditions: Rantzer’s conditions and one approach of matrix inequalities.


## 4. Sufficient Conditions

In this section, we present two sufficient conditions for the stability of segment that are known in the literature: Rantzer’s conditions and one interesting approach with matrix inequalities.


### 4.1. Rantzer’s Conditions

Supposing thatp0(t) is a Hurwitz polynomial and P1(t) is semistable (their roots have real part 0), then the segment of polynomials [p0(t),p1(t)] consists of Hurwitz polynomials if you have one of the following four conditions.(i) The differenced=p1-p0 satisfies (6)∂arg⁡diω∂ω<0,ω∈w>0∣diw≠0.(ii) Each of the polynomialsp0 and p1 has at least one root in C- and(7)∂arg⁡diω∂ω<sin⁡2arg⁡diω2ω,ω∈w>0∣diw≠0.(iii) Each of the polynomialsp0 and p1 has at least one root in C- and(8)∂arg⁡diω∂ω≤0,ω∈w>0∣diw≠0.(iv) Each of the polynomialsp0 and p1 has at least two roots in C- and(9)∂arg⁡diω∂ω≤sin⁡2arg⁡diω2ω,ω∈w>0∣diw≠0.The previous conditions are known as Rantzer’s conditions; see [8].Open Problem 3.
Is there a mechanism to describe Rantzer’s conditions in terms of the polynomial coefficients?Open Problem 4.
What should we add to Rantzer’s conditions in order to obtain necessary and sufficient conditions?In the next subsection, we present an interesting sufficient condition for checking the stability of segment of polynomials based on matrix inequalities.


### 4.2. A Matrix Inequality

Given a polynomialp0(t)=tn+a1tn-1+⋯+an, we define the matrix E(n,n)∈M(n+1)×(n+1) as(10)En,n=1000⋯00-a2a1-10⋯00a4-a3a2-a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an.Theorem 6.
Letp1=c1tn+c2tn-1+⋯+cn+1 be a polynomial of degree n. If the polynomials p0(t) and p1(t) are Hurwitz and vector c=(c1,c2,…,cn+1)T⪰0 satisfies the system of linear inequalities,(11)En,nc⪶0,then λp0(t)+(1-λ)p1(t) is Hurwitz for all λ∈[0,1]. Here the symbol ⪰0 (⪯0) means that all components of a given vector are ≥0 (≤0) and the symbol ⪶0 means that all components of a given vector are ≥0 but there is at least one component which is >0. See [15, 16].Open Problem 5.
With these approach matrix inequalities, what should we add to obtain necessary and sufficient conditions?


#### 4.2.1. Case for Gradep1=n-1

A similar case is obtained ifdeg⁡p1t=n-1. In this case, the matrix E(n,n-1)∈M(n+1)×n is defined by(12)En,n-1=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an,and the corresponding inequality is(13)En,n-1c⪶0.There is a relation between segments and rays of polynomials, which let us give applications in control theory. We explain it in the next two subsections.


### 4.3. Relationship between Rays and Segment of Polynomials

Ifp0(t)+kp1(t) is Hurwitz for all k≥0, then 1/1+kp0(t)+k/1+kp1(t) is Hurwitz for all k≥0, from which if p1 is Hurwitz, then, the stability of ray p0(t)+kp1(t) is equivalent to the stability of the segment p0(t),p1(t). An analysis in terms of ray polynomials was made in [15].Fordeg⁡p1t=n, D(n,n)∈Mn×(n+1) is the matrix:(14)Dn,n=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯-an-2an-30000⋯an-an-1;for deg⁡p1t=n-1, D(n,n-1)∈Mn×n is the matrix: (15)Dn,n-1=1000⋯00-a2a1-10⋯00a4-a3a2-a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-2-an-30000⋯-anan-1;for deg⁡p1t=n-2, D(n,n-2)∈M(n-1)×(n-1) is the matrix: (16)Dn,n-2=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-2-an-30000⋯-anan-1.Rewriting the results shown in [15, 16] in terms of segment polynomials yields the following result.Theorem 7.
Consider the Hurwitz polynomialp0(t)=tn+a1tn-1+⋯+an. If p1(t) is Hurwitz with deg⁡p1t=n,n-1 or n-2 and the vector of coefficients c satisfies the system of linear inequalities (17)Dc⪶0;then the polynomial λp0(t)+(1-λ)p1(t) is Hurwitz for all λ∈[0,1]; where the matrix D is determined by the degree of p1(t) and is one of the matrices D(n,n),D(n,n-1), or D(n,n-2).Example 8.
Consider the polynomial Hurwitzp0(t)=t3+1/12t2+5t+1/12. The vector of coefficients c of polynomial p1(t)=t3+37t2+11/2t+1 is solution of system of linear inequalities (11):(18)E3,3c=1000-5112-100-1125-1120001121371121=18172112;therefore, the segment polynomial p0,p1 is Hurwitz. However, c is not a solution of (17):(19)D3,3c=12-100-125-1210012-51371121=-7327314-94,and Rantzer’s conditions are not verified: to the difference d=p1-p0=73/2t2+1/12t+1/12 has(20)∂arg⁡diω∂ω=1/14+73/4ω21/12-73/2ω22+1/12ω2>0,sin2arg⁡diω2ω=1/14-73/4ω21/12-73/2ω22+1/12ω2.Open Problem 6.
Will there be a stable segment of polynomial that satisfies Rantzer’s conditions but does not satisfy (11) or (17)?
The importance of the segments and rays of Hurwitz polynomials with the design of stabilizing controls is pointed out in the next subsection.


### 4.4. Applications to Control Systems

Consider the systemx˙=Ax+bu, where(21)A=010⋯0001⋯0⋮⋮⋮⋮000⋯1-an-an-1-an-2⋯-a1,b=00⋮01.For the uncontrolled system x˙=Ax, the characteristic polynomial is p0(t)=tn+a1tn-1+a2tn-2+⋯+an. If into the controlled system x˙=Ax+bu we choose a control u as u(x)=-kcTx=-k(cn,cn-1,…,c1)x, where k>0 and c∈Rn, then the controlled system takes the form:(22)x˙=A-kbcTx,and its characteristic polynomial is (23)pt=tn+a1tn-1+a2tn-2+⋯+an+kc1tn-1+c2tn-2+⋯+cn,that is,(24)pt=p0t+kp1t,where p1(t)=c1tn-1+c2tn-2+⋯+cn.Example 9.
Consider the system(25)x˙=010001-1-132-2x+001-2k,-4k,-2kx.In this case, the matrix E is given by(26)E=2-10-132-2001,then we have the system(27)2-10-132-2001242=002.Therefore, the system is stable ∀k≥0. But this cannot be verified with the results presented by Aguirre et al. in [15].(28)D=100-322-10-132,100-322-10-132242=23-1.The Minimum Left Extreme is a subject very related to the study of segment of Hurwitz polynomials; hence in the next sections we present some ideas about the Minimum Left Extreme in order to propose some open problems.


## 4.1. Rantzer’s Conditions

Supposing thatp0(t) is a Hurwitz polynomial and P1(t) is semistable (their roots have real part 0), then the segment of polynomials [p0(t),p1(t)] consists of Hurwitz polynomials if you have one of the following four conditions.(i) The differenced=p1-p0 satisfies (6)∂arg⁡diω∂ω<0,ω∈w>0∣diw≠0.(ii) Each of the polynomialsp0 and p1 has at least one root in C- and(7)∂arg⁡diω∂ω<sin⁡2arg⁡diω2ω,ω∈w>0∣diw≠0.(iii) Each of the polynomialsp0 and p1 has at least one root in C- and(8)∂arg⁡diω∂ω≤0,ω∈w>0∣diw≠0.(iv) Each of the polynomialsp0 and p1 has at least two roots in C- and(9)∂arg⁡diω∂ω≤sin⁡2arg⁡diω2ω,ω∈w>0∣diw≠0.The previous conditions are known as Rantzer’s conditions; see [8].Open Problem 3.
Is there a mechanism to describe Rantzer’s conditions in terms of the polynomial coefficients?Open Problem 4.
What should we add to Rantzer’s conditions in order to obtain necessary and sufficient conditions?In the next subsection, we present an interesting sufficient condition for checking the stability of segment of polynomials based on matrix inequalities.


## 4.2. A Matrix Inequality

Given a polynomialp0(t)=tn+a1tn-1+⋯+an, we define the matrix E(n,n)∈M(n+1)×(n+1) as(10)En,n=1000⋯00-a2a1-10⋯00a4-a3a2-a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an.Theorem 6.
Letp1=c1tn+c2tn-1+⋯+cn+1 be a polynomial of degree n. If the polynomials p0(t) and p1(t) are Hurwitz and vector c=(c1,c2,…,cn+1)T⪰0 satisfies the system of linear inequalities,(11)En,nc⪶0,then λp0(t)+(1-λ)p1(t) is Hurwitz for all λ∈[0,1]. Here the symbol ⪰0 (⪯0) means that all components of a given vector are ≥0 (≤0) and the symbol ⪶0 means that all components of a given vector are ≥0 but there is at least one component which is >0. See [15, 16].Open Problem 5.
With these approach matrix inequalities, what should we add to obtain necessary and sufficient conditions?


### 4.2.1. Case for Gradep1=n-1

A similar case is obtained ifdeg⁡p1t=n-1. In this case, the matrix E(n,n-1)∈M(n+1)×n is defined by(12)En,n-1=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an,and the corresponding inequality is(13)En,n-1c⪶0.There is a relation between segments and rays of polynomials, which let us give applications in control theory. We explain it in the next two subsections.


## 4.2.1. Case for Gradep1=n-1

A similar case is obtained ifdeg⁡p1t=n-1. In this case, the matrix E(n,n-1)∈M(n+1)×n is defined by(12)En,n-1=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an,and the corresponding inequality is(13)En,n-1c⪶0.There is a relation between segments and rays of polynomials, which let us give applications in control theory. We explain it in the next two subsections.


## 4.3. Relationship between Rays and Segment of Polynomials

Ifp0(t)+kp1(t) is Hurwitz for all k≥0, then 1/1+kp0(t)+k/1+kp1(t) is Hurwitz for all k≥0, from which if p1 is Hurwitz, then, the stability of ray p0(t)+kp1(t) is equivalent to the stability of the segment p0(t),p1(t). An analysis in terms of ray polynomials was made in [15].Fordeg⁡p1t=n, D(n,n)∈Mn×(n+1) is the matrix:(14)Dn,n=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯-an-2an-30000⋯an-an-1;for deg⁡p1t=n-1, D(n,n-1)∈Mn×n is the matrix: (15)Dn,n-1=1000⋯00-a2a1-10⋯00a4-a3a2-a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-2-an-30000⋯-anan-1;for deg⁡p1t=n-2, D(n,n-2)∈M(n-1)×(n-1) is the matrix: (16)Dn,n-2=a1-100⋯00-a3a2-a11⋯00a5-a4a3-a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-2-an-30000⋯-anan-1.Rewriting the results shown in [15, 16] in terms of segment polynomials yields the following result.Theorem 7.
Consider the Hurwitz polynomialp0(t)=tn+a1tn-1+⋯+an. If p1(t) is Hurwitz with deg⁡p1t=n,n-1 or n-2 and the vector of coefficients c satisfies the system of linear inequalities (17)Dc⪶0;then the polynomial λp0(t)+(1-λ)p1(t) is Hurwitz for all λ∈[0,1]; where the matrix D is determined by the degree of p1(t) and is one of the matrices D(n,n),D(n,n-1), or D(n,n-2).Example 8.
Consider the polynomial Hurwitzp0(t)=t3+1/12t2+5t+1/12. The vector of coefficients c of polynomial p1(t)=t3+37t2+11/2t+1 is solution of system of linear inequalities (11):(18)E3,3c=1000-5112-100-1125-1120001121371121=18172112;therefore, the segment polynomial p0,p1 is Hurwitz. However, c is not a solution of (17):(19)D3,3c=12-100-125-1210012-51371121=-7327314-94,and Rantzer’s conditions are not verified: to the difference d=p1-p0=73/2t2+1/12t+1/12 has(20)∂arg⁡diω∂ω=1/14+73/4ω21/12-73/2ω22+1/12ω2>0,sin2arg⁡diω2ω=1/14-73/4ω21/12-73/2ω22+1/12ω2.Open Problem 6.
Will there be a stable segment of polynomial that satisfies Rantzer’s conditions but does not satisfy (11) or (17)?
The importance of the segments and rays of Hurwitz polynomials with the design of stabilizing controls is pointed out in the next subsection.


## 4.4. Applications to Control Systems

Consider the systemx˙=Ax+bu, where(21)A=010⋯0001⋯0⋮⋮⋮⋮000⋯1-an-an-1-an-2⋯-a1,b=00⋮01.For the uncontrolled system x˙=Ax, the characteristic polynomial is p0(t)=tn+a1tn-1+a2tn-2+⋯+an. If into the controlled system x˙=Ax+bu we choose a control u as u(x)=-kcTx=-k(cn,cn-1,…,c1)x, where k>0 and c∈Rn, then the controlled system takes the form:(22)x˙=A-kbcTx,and its characteristic polynomial is (23)pt=tn+a1tn-1+a2tn-2+⋯+an+kc1tn-1+c2tn-2+⋯+cn,that is,(24)pt=p0t+kp1t,where p1(t)=c1tn-1+c2tn-2+⋯+cn.Example 9.
Consider the system(25)x˙=010001-1-132-2x+001-2k,-4k,-2kx.In this case, the matrix E is given by(26)E=2-10-132-2001,then we have the system(27)2-10-132-2001242=002.Therefore, the system is stable ∀k≥0. But this cannot be verified with the results presented by Aguirre et al. in [15].(28)D=100-322-10-132,100-322-10-132242=23-1.The Minimum Left Extreme is a subject very related to the study of segment of Hurwitz polynomials; hence in the next sections we present some ideas about the Minimum Left Extreme in order to propose some open problems.


## 5. The Minimum Left Extreme

A related problem with segment of Hurwitz polynomials is the calculating of the Minimum Left Extreme, which was studied by Bialas. We begin with some definitions.Letp0(t) be a Hurwitz polynomial that generates a family of polynomials given as follows:(29)pt=p0t+kp1t,where k∈R and p1(t)=∑i=1ncitn-i is a polynomial with coefficient vector c=(c1,c2,…,cn)T⪰0. If E(n,n-1) is the matrix defined in the previous section and c is a solution of the system of linear inequalities E(n,n-1)c⪶0, then p0(t)+kp1(t) is a Hurwitz polynomial ∀k≥0.What is the value ofkmin<0, so that p0(t)+kp1(t) is Hurwitz ∀k≥kmin? From Bialas (see [2]),(30)kmin=1λmin--H-1p0Hp1,where H(p0) and H(p1) are Hurwitz matrices of p0 and p1, respectively, and λmin-[-H-1(p0)H(p1)] is the smallest negative eigenvalue of the matrix (31)-H-1p0Hp1.


### 5.1. Estimation Technique Proposed

Consider the polynomialp0(t)=tn+a1tn-1+⋯+an whereby the following matrix is defined:(32)Zn,n-1=a1-200⋯000a2-2a12⋯0000a3-2a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-2an-20000⋯0an,with Z(n,n-1)i the ith row of the matrix Z(n·n-1) and a=a1,a2,…,anT.Theorem 10.
Letp0(t)=tn+a1tn-1+⋯+an and E(n,n-1) be a Hurwitz polynomial and a matrix defined above, respectively. If the vector c=(c1,c2,…,cn)T⪰0 is a solution to the system of linear inequalities E(n,n-1)c⪶0 and p1(t)=∑i=1n‍citn-i, then p0(t)+kp1(t) is Hurwitz ∀k>k0, where k0=maxi=1,…,n-Z(n,n-1)ia/E(n,n-1)ic; that is, kmin≤k0.Example 11.
Givenp0(t)=t3+7t2+14t+8 and p1(t)=t2+4t+6, we have that(33)Z3,2a=7-20014-140087148=218464,E3,2c=7-10-814-7008146=3648,k0=max-213,-846,-6448=-143,Hp0=7801140078,Hp1=160040016,H-1p0Hp1=74526450-190114507720-44534.Then σ-H-1(p0)H(p1)={-1/34,-1/15±1/15i}. We have that λmin=-1/34, kmin=-1/43.In the above example, the two approaches lead top0(t)+kp1(t) being Hurwitz ∀k>-1/43.Example 12.
Givenp0(t)=t3+7t2+14t+8 and p1(t)=26t2+137t+90, we have that(34)Z3,2a=7-20014-140087148=218464,E3,2c=7-10-814-70082613790=451080720,k0=max-2145,-841080,-64720=-790=-0.07778,Hp0=7801140078,Hp1=269000137002690,H-1p0Hp1=1824582450-134586990091360-3743720454.Then σ-H-1(p0)H(p1)={-11.25,-4.1399,-9.5601}. Finally we have λmin=-11.25, kmin=-0.088889.Now, in Example12 with the approach proposed in Theorem 10, we obtain that p0(t)+kp1(t) is Hurwitz ∀k>-0.07778 and applying the Bialas approach we obtain that p0(t)+kp1(t) is Hurwitz ∀k>-0.088889. That is, kmin<k0.Example 13.
Consider the following system:(35)x˙=010001-8-14-7x+001-90k,-137k,-26x.Here p0(t)=t3+7t2+14t+8 and p1(t)=26t2+137t+90. Then it holds (36)k0=max⁡-2145,-841080,-64720=-790=-0.07778>-1.Therefore, in addition, we can ensure that the system is stable for all k>0 (the control u is a stabilizing control) and we can also say more, the system is stable for all k>-0.07778 because p0(t)+kp1(t) is Hurwitz ∀k>-0.07778.Open Problem 7.
Describe the cases whenk0=kmin.Remark 14.
An advantage of our approach, in relation to the Bialas approach, is that we can estimate the Minimum Left Extreme when the polynomials have the same degree. Given the polynomialp0(t)=tn+a1tn-1+⋯+an, the matrices Z(n,n) and E(n,n) are defined by(37)Zn,n=1000⋯000a1-20⋯0000a2-2a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-2an-20000⋯0an,En,n=1000⋯00-a2a1-10⋯00a4-a3a2-a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an.If the vector c=(c0,c1,…,cn)T⪰0 is a solution of inequalities system E(n,n)c⪶0 and the polynomial p1(t) is given by p1(t)=∑i=0n‍citn-i, then p0(t)+kp1(t) is a Hurwitz polynomial ∀k>k0, where k0=maxi=1,…,n,n+1-Zia/Eic; that is, kmin≤k0.Open Problem 8.
With this approach find conditions for calculatingk0 as deg⁡p0>deg⁡p1 or as deg⁡p0=deg⁡p1.
If we consider the set of the Hadamardized Hurwitz polynomials, then we can plant some open problems also. It is the motivation to include the next section.


## 5.1. Estimation Technique Proposed

Consider the polynomialp0(t)=tn+a1tn-1+⋯+an whereby the following matrix is defined:(32)Zn,n-1=a1-200⋯000a2-2a12⋯0000a3-2a2⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-2an-20000⋯0an,with Z(n,n-1)i the ith row of the matrix Z(n·n-1) and a=a1,a2,…,anT.Theorem 10.
Letp0(t)=tn+a1tn-1+⋯+an and E(n,n-1) be a Hurwitz polynomial and a matrix defined above, respectively. If the vector c=(c1,c2,…,cn)T⪰0 is a solution to the system of linear inequalities E(n,n-1)c⪶0 and p1(t)=∑i=1n‍citn-i, then p0(t)+kp1(t) is Hurwitz ∀k>k0, where k0=maxi=1,…,n-Z(n,n-1)ia/E(n,n-1)ic; that is, kmin≤k0.Example 11.
Givenp0(t)=t3+7t2+14t+8 and p1(t)=t2+4t+6, we have that(33)Z3,2a=7-20014-140087148=218464,E3,2c=7-10-814-7008146=3648,k0=max-213,-846,-6448=-143,Hp0=7801140078,Hp1=160040016,H-1p0Hp1=74526450-190114507720-44534.Then σ-H-1(p0)H(p1)={-1/34,-1/15±1/15i}. We have that λmin=-1/34, kmin=-1/43.In the above example, the two approaches lead top0(t)+kp1(t) being Hurwitz ∀k>-1/43.Example 12.
Givenp0(t)=t3+7t2+14t+8 and p1(t)=26t2+137t+90, we have that(34)Z3,2a=7-20014-140087148=218464,E3,2c=7-10-814-70082613790=451080720,k0=max-2145,-841080,-64720=-790=-0.07778,Hp0=7801140078,Hp1=269000137002690,H-1p0Hp1=1824582450-134586990091360-3743720454.Then σ-H-1(p0)H(p1)={-11.25,-4.1399,-9.5601}. Finally we have λmin=-11.25, kmin=-0.088889.Now, in Example12 with the approach proposed in Theorem 10, we obtain that p0(t)+kp1(t) is Hurwitz ∀k>-0.07778 and applying the Bialas approach we obtain that p0(t)+kp1(t) is Hurwitz ∀k>-0.088889. That is, kmin<k0.Example 13.
Consider the following system:(35)x˙=010001-8-14-7x+001-90k,-137k,-26x.Here p0(t)=t3+7t2+14t+8 and p1(t)=26t2+137t+90. Then it holds (36)k0=max⁡-2145,-841080,-64720=-790=-0.07778>-1.Therefore, in addition, we can ensure that the system is stable for all k>0 (the control u is a stabilizing control) and we can also say more, the system is stable for all k>-0.07778 because p0(t)+kp1(t) is Hurwitz ∀k>-0.07778.Open Problem 7.
Describe the cases whenk0=kmin.Remark 14.
An advantage of our approach, in relation to the Bialas approach, is that we can estimate the Minimum Left Extreme when the polynomials have the same degree. Given the polynomialp0(t)=tn+a1tn-1+⋯+an, the matrices Z(n,n) and E(n,n) are defined by(37)Zn,n=1000⋯000a1-20⋯0000a2-2a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-2an-20000⋯0an,En,n=1000⋯00-a2a1-10⋯00a4-a3a2-a1⋯00⋯⋯⋯⋯⋯⋯⋯0000⋯an-1-an-20000⋯0an.If the vector c=(c0,c1,…,cn)T⪰0 is a solution of inequalities system E(n,n)c⪶0 and the polynomial p1(t) is given by p1(t)=∑i=0n‍citn-i, then p0(t)+kp1(t) is a Hurwitz polynomial ∀k>k0, where k0=maxi=1,…,n,n+1-Zia/Eic; that is, kmin≤k0.Open Problem 8.
With this approach find conditions for calculatingk0 as deg⁡p0>deg⁡p1 or as deg⁡p0=deg⁡p1.
If we consider the set of the Hadamardized Hurwitz polynomials, then we can plant some open problems also. It is the motivation to include the next section.


## 6. Hadamardized Hurwitz Polynomials

In the present section, some results and open problems related to Hadamardized Hurwitz polynomials are postulated.Another property of Hurwitz polynomials is related to Hadamard product. Consider two real polynomialsp(t) and q(t): (38)pt=antn+an-1tn-1+⋯+a1t+a0,qt=bntn+bn-1tn-1+⋯+b1t+b0.The Hadamard product of p(t) and q(t), denoted by (p∗q)(t), is defined by(39)p∗qt=anbntn+an-1bn-1tn-1+⋯+a1b1t+a0b0t.In [17], it was proved that the set of Hurwitz polynomials is closed under Hadamard product. However in [18] it was shown that there are Hurwitz polynomials of degree 4 that do not have a Hadamard factorization in two Hurwitz polynomials. In [19, 20], some conditions to Hadamard factorization existence were presented. The set of Hurwitz polynomials that admits a Hadamard factorization, we called Hadamardized Hurwitz polynomials, is denoted by H(Had)n. Recently, in [21] some topological properties about this set were shown. But regarding the convex combination, we can propose the following problems.Open Problem 9.
Givenf,g∈H(Had)n, what conditions are required to guarantee that λf+(1-λ)g∈H(Had)n?Open Problem 10.
Is there any property inHn set that does not share with H(Had)n set?


## 7. Conclusions

In this paper, a brief overview of some results concerning the stability of the convex combination of Hurwitz polynomials is provided. Also some criteria for deciding if a segment of polynomials is a Hurwitz polynomial were presented. Furthermore some open problems about these polynomials in order to motivate the study of them were given. As mentioned above, the study and solution of these problems also have their field of opportunity in the framework of robust systems.



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*Source: 2075903-2018-02-15.xml*