# Properties of the Set of Hadamardized Hurwitz Polynomials

Authors: Baltazar Aguirre-Hernández; Edgar-Cristian Díaz-González; Carlos-Arturo Loredo-Villalobos; Faustino-Ricardo García-Sosa
Journal: Mathematical Problems in Engineering
 (2015)

---

## Abstract

We say that a Hurwitz polynomialpt is a Hadamardized polynomial if there are two Hurwitz polynomials
ft and gt such that f∗g=p, where f∗g is the Hadamard product of f and g. In this paper, we prove that the set of all Hadamardized Hurwitz polynomials is an open, unbounded,
nonconvex, and arc-connected set. Furthermore, we give a result so that a fourth-degree Hurwitz interval
polynomial is a Hadamardized polynomial family and we discuss an approach of differential topology in the
study of the set of Hadamardized Hurwitz polynomials.


---

## Body

## 1. Introduction

The study of Hurwitz polynomials is important for the following: if the characteristic polynomial of a matrixA is a Hurwitz polynomial, then the system x˙=Ax is stable. Such study began when Maxwell posed the problem of finding necessary and sufficient conditions for a polynomial has all of its roots with negative real part. Since then, a lot information has been generated and part of it can be found in [1–4]. On the other hand, in the study of Hurwitz polynomials, topological approaches have recently been reported in [5–9]. Now, in this paper we present some topological and geometric results about the set of Hurwitz polynomials that admit a Hadamard factorization; such set will be called the set of Hadamardized polynomials. First, we give some definitions. Given two polynomials ft=antn+an-1tn-1+⋯+a0 and g(t)=bntn+bn-1tn-1+⋯+b0, we say that (1)f∗gt=anbntn+an-1bn-1tn-1+⋯+a0b0is the Hadamard product of f and g. Garloff and Wagner have shown that the set of Hurwitz polynomials is closed under the Hadamard product (see [10]) and preserves total nonnegativity (see [11]). And, conversely, we say that a Hurwitz polynomial p(t) admits a Hadamard factorization if there are two Hurwitz polynomials f(t) and g(t) such that f∗g(t)=p(t). In this case, we say that p(t) is a Hadamardized Hurwitz polynomial, we denote by H(Had) the set of all Hadamardized Hurwitz polynomials, and we denote by H(Had)n the set of all n-degree Hadamardized Hurwitz polynomials. It is shown in [12] that there are Hurwitz polynomials of degree equal to four that do not have a Hadamard factorization. In [13] necessary conditions for a Hurwitz polynomial to have a Hadamard factorization were obtained, and in [14, 15] necessary and sufficient conditions for a Hurwitz polynomial of degree equal to four to admit a Hadamard factorization were obtained. Now in this paper we prove that H(Had) is an unbounded, nonconvex open set and arc-connected. Additionally, in Section 3 we give a result about Hadamard factorization of a fourth-degree Hurwitz interval polynomials. Finally, in Section 4 we present an approach of differential topology for studying the set of Hadamardized Hurwitz polynomials.


## 2. Main Results

### 2.1. The Set of Hadamardized Hurwitz Polynomials Is an Open Set

The fact that the set of Hadamardized Hurwitz polynomials of degree equal to four is an open set is an immediate consequence of TheoremA.3; since q(t)=α4t4+α3t3+α2t2+α1t+α0 is a Hadamardized Hurwitz polynomial, then(2)α0α3α1<α1α4α3-α22,and this inequality holds for small perturbations of the coefficients. Since it is not known necessary and sufficient conditions (in terms of its coefficients) for an n-degree Hurwitz polynomial to have a Hadamard factorization, we need a result that allows us to prove the case for polynomials of arbitrary degree. Such a result is the Open Mapping Theorem.Theorem 1.
H
(
H
a
d
) is an open set.Proof.
Letp(t)=antn+an-1tn-1+⋯+a1t+a0 be a Hadamardized Hurwitz polynomial. Then, there are two Hurwitz polynomials f(t)=bntn+bn-1tn-1+⋯+b0 and g(t)=cntn+cn-1tn-1+⋯+c0 such that (f∗g)(t)=p(t). Consider the operator Tf:Pn→Pn, where Pn is the set of n-degree polynomials, defined by Tf(h(t))=(f∗h)(t). T is a linear operator and if h(t)=dntn+dn-1tn-1+⋯+d1t+d0, then(3)Tfht=bndn,…,b1d1,b0d0≤bn,…,b1,b0dn,…,d1,d0.That is, Tf is a bounded linear operator. It is clear that Tf is a surjective mapping because if q(t)=qntn+qn-1tn-1+⋯+q1t+q0 is an arbitrary polynomial, then(4)Tfqnbntn+qn-1bn-1tn-1+⋯+q1b1t+q0b0=qt.On the other hand, since Hn (n-degree Hurwitz polynomials set) is an open set, let U be an open set such that g∈U⊂Hn. Then, Tf[U]⊂Hn (see [10]), and, by the Open Mapping Theorem [16], Tf[U] is an open set. Since p(t)∈Tf[U], we have that p(t)∈Tf[U]⊂H(Had), and H(Had) is an open set.


#### 2.1.1. Application

Consider the systemx˙=A(λ)x where A(λ)∈Mn×n(R) and λ∈Rn is a parameter. Let pλ(t)=an(λ)tn+an-1(λ)tn-1+⋯+a1(λ)+a0(λ) be the characteristic polynomial of the matrix A(λ).Suppose thatPλ0 is a Hadamardized Hurwitz polynomial. By Theorem 1, there is an open ball Bϵ(pλ0) such that every element of Bϵ(pλ0) is a Hadamardized Hurwitz polynomial. Then, we can find two Hurwitz polynomials:(5)fλt=bnλtn+bn-1λtn-1+⋯+b1λt+b0λ,gλt=cnλtn+cn-1λtn-1+⋯+c1λt+c0λsuch that fλ∗gλ=Pλ for λ≈λ0. The idea is to obtain a Hadamard factorization of Pλ(t) where the coefficients of fλ(t) and gλ(t) are simpler than those of Pλ(t). To illustrate this idea, we give the following example.Example 2.
Consider the systemx˙=Ax, where the entries aij(λ) of A are given by(6)a11=-λ2-8λ-15,a12=-λ2-11λ-29,a13=-λ2-12λ-36,a14=2λ+7,a21=0,a22=0,a23=1,a24=0,a31=0,a32=0,a33=0,a34=1,a41=-λ2-8λ-15,a42=-λ2-11λ-30,a43=-λ2-12λ-36,a44=λ+7.The characteristic polynomial is Pλ(t)=t4+(λ2+6λ+8)t3+(λ2+12λ+36)t2 + (λ2+11λ+30)t+(λ2+8λ+15). For λ=0, P0(t)=t4+8t3+36t2+30t+15 is a Hadamardized Hurwitz polynomial by Theorem A.4.By Theorem1, we can find two Hurwitz polynomials such that the Hadamard product of them is equal to Pλ(t) for small enough λ. Consider the Hurwitz polynomials:(7)F0t=t4+4t3+6t2+6t+5,G0t=t4+2t3+6t2+5t+3.Note that F0∗G0=t4+8t3+36t2+30t+15. The families of polynomials(8)Fλt=t4+λ+4t3+λ+6t2+λ+6t+λ+5,Gλt=t4+λ+2t3+λ+6t2+λ+5t+λ+3are Hurwitz stable for small enough λ, the Hadamard product Fλ∗Gλ satisfies Fλ∗Gλ=Pλ(t), and the coefficients of Fλ and Gλ are simpler than those of Pλ.


### 2.2. The Set of Hadamardized Hurwitz Polynomials Is a Nonconvex Set

We have thatH(Had)⊆H (see [13]), where H(Had)≠H and H is the set of Hurwitz polynomials, and it is known that the set of Hurwitz polynomials is a nonconvex set (see [17, 18]). However, this does not imply that the set of Hadamardized Hurwitz polynomials is also a nonconvex set. Thus, the nonconvexity of H(Had) has to be proven.Considered the ray of polynomials:(9)Fkt=t4+t3+6.9t2+2t+3+kt3+t2+2t+1;that is(10)Fkt=t4+1+kt3+6.9+kt2+2+2kt+3+k.Then, Fk(t) is Hurwitz if and only if(11)1+k>0,6.9+k>0,2+2k>0,3+k>0,1+k6.9+k2+2k-2+2k2-1+k23+k>0.This inequality holds if and only if k>-1, and Fk(t) is Hadamardized if and only if(12)3+k1+k2+2k<2+2k1+k-6.9+k2.Thus, Fk(t) is Hurwitz and Hadamardized if and only if k∈(-1,1.2-3.2)∪(1.2+3.2,∞), and Fk(t) is Hurwitz but not Hadamardized if and only if k∈[1.2-3.2,1.2+3.2].Theorem 3.
The set of Hadamardized Hurwitz polynomials is a nonconvex set.Proof.
In the discussion above, takek=-0.5 and k=2.9. Then we get the Hadamardized Hurwitz polynomials r(t)=t4+t3+6.9t2+2t+3-0.5(t3+t2+2t+1) and h(t)=t4+t3+6.9t2+2t+3+2.9(t3+t2+2t+1). Now consider the convex combination λr(t)+(1-λ)h(t). For λ=1/2, we get the polynomial t4+2.2t3+8.1t2+4.4t+4.2, which is not Hadamardized.


### 2.3. The Set of Hadamardized Hurwitz Polynomials Is an Unbounded Set

Theorem 4.
There are rays of polynomials contained in the set of Hadamardized Hurwitz polynomials. Hence, it is an unbounded set.Let(13)Gkt=t4+t3+7t2+2t+3+kt3+t2+2t+1=t4+1+kt3+7+kt2+2+2kt+3+k.Then, Gk(t) is Hurwitz if and only if(14)1+k>0,7+k>0,2+2k>0,3+k>0,1+k7+k2+2k-2+2k2-1+k23+k>0,and Gk(t) is Hadamardized if and only if(15)3+k1+k2+2k<2+2k1+k-7+k2.Consequently, Gk(t) is a Hadamardized Hurwitz polynomial for k>-1 and H(Had) is a unbounded set.The existence of rays of Hadamardized Hurwitz polynomials allows us to show an application in the designing of stabilizing controls.


#### 2.3.1. Application

The following example illustrates the designing of a state feedback control that stabilizes the system for each positive value of the parameter. Consider the system(16)x˙1x˙2x˙3x˙4=010000100001-3-2-7-1x1x2x3x4+0001u;and the state feedback control(17)ux=k-1,-2,-1,-1x1x2x3x4,where k∈R is a parameter. Then, the controlled system is(18)x˙1x˙2x˙3x˙4=010000100001-3-k-2-2k-7-k-1-kx1x2x3x4.The characteristic polynomial Gk(t)=t4+t3+7t2+2t+3+k(t3+t2+2t+1) is a Hadamardized Hurwitz polynomial for all k>-1, and consequently the system is stable for all k>-1. Additionally, Gk(t) is Hadamardized for all k>-1.


### 2.4. The Set of Hurwitz Hadamardized Polynomials with Positive Coefficients Is Arc-Connected

To prove the main result of this section, we need the following lemma.Lemma 5.
H
n
+ is an arc-connected set.Proof.
LetP1(t) and P2(t) be two Hurwitz polynomials with positive coefficients. Suppose that n=2m. Then, P1(t) and P2(t) can be written as(19)P1t=b0t2+b1t+b2·t2+b3t+b4⋯t2+b2m-1t+b2m,P2t=a0t2+a1t+a2·t2+a3t+a4⋯t2+a2m-1t+a2m,where bi>0 and ai>0 for all i=0,1,…,2m. Define(20)Hλt=λa0+1-λb0t2+λa1+1-λb1t+λa2+1-λb2t2+λa3+1-λb3t+λa4+1-λb4⋯t2+λa2m-1+1-λb2m-1t+λa2m+1-λb2mwith λ∈[0,1]. We have that H0(t)=P1(t), H1(t)=P2(t), and Hλ(t) is a Hurwitz polynomial for all λ∈[0,1]. Therefore, Hn+ is arc-connected set.Theorem 6.
The set of Hadamardized Hurwitz polynomials with positive coefficients is arc-connected.Proof.
LetP1(t)=antn+an-1tn-1+⋯+a0 and P2(t)=bntn+bn-1tn-1+⋯+b0 Hadamardized Hurwitz polynomials with positive coefficients; then, there are(21)f1t=αntn+αn-1tn-1+⋯+α0,f2t=βntn+βn-1tn-1+⋯+β0,g1t=γntn+γn-1tn-1+⋯+γ0,g2t=δntn+δn-1tn-1+⋯+δ0such that f1, f2, g1, and g2 are Hurwitz polynomials with positive coefficients and (f1∗g1)(t)=P1(t) and (f2∗g2)(t)=P2(t); that is,(22)αnγntn+αn-1γn-1tn-1+⋯+α0γ0=antn+an-1tn-1+⋯+a0,βnδntn+βn-1δn-1tn-1+⋯+β0δ0=bntn+bn-1tn-1+⋯+b0.Since Hn+ is arc-connected set, then there exist(23)G1,λt=cnλtn+cn-1λtn-1+⋯+c0λ,G2,λt=dnλtn+dn-1λtn-1+⋯+d0λsuch that(24)G1,0t=cn0tn+cn-10tn-1+⋯+c00=f1t,G2,0t=dn0tn+dn-10tn-1+⋯+d00=g1t,G1,1t=cn1tn+cn-11tn-1+⋯+c01=f2t,G2,1t=dn1tn+dn-11tn-1+⋯+d01=g2tand each Gi,λ(t)∈Hn+ for all λ∈[0,1]. Now, consider(25)G1,λ∗G2,λt=cnλdnλtn+cn-1λdn-1λtn-1+⋯+c0λd0λand then (G1,λ∗G2,λ)(t) is a Hadamardized Hurwitz polynomial for all λ∈[0,1] (see [10]) and(26)G1,0∗G2,0t=f1∗g1t=P1t,G1,1∗G2,1t=f2∗g2t=P2t.Consequently, the set of Hadamardized Hurwitz polynomials whit positive coefficients is arc-connected.


## 2.1. The Set of Hadamardized Hurwitz Polynomials Is an Open Set

The fact that the set of Hadamardized Hurwitz polynomials of degree equal to four is an open set is an immediate consequence of TheoremA.3; since q(t)=α4t4+α3t3+α2t2+α1t+α0 is a Hadamardized Hurwitz polynomial, then(2)α0α3α1<α1α4α3-α22,and this inequality holds for small perturbations of the coefficients. Since it is not known necessary and sufficient conditions (in terms of its coefficients) for an n-degree Hurwitz polynomial to have a Hadamard factorization, we need a result that allows us to prove the case for polynomials of arbitrary degree. Such a result is the Open Mapping Theorem.Theorem 1.
H
(
H
a
d
) is an open set.Proof.
Letp(t)=antn+an-1tn-1+⋯+a1t+a0 be a Hadamardized Hurwitz polynomial. Then, there are two Hurwitz polynomials f(t)=bntn+bn-1tn-1+⋯+b0 and g(t)=cntn+cn-1tn-1+⋯+c0 such that (f∗g)(t)=p(t). Consider the operator Tf:Pn→Pn, where Pn is the set of n-degree polynomials, defined by Tf(h(t))=(f∗h)(t). T is a linear operator and if h(t)=dntn+dn-1tn-1+⋯+d1t+d0, then(3)Tfht=bndn,…,b1d1,b0d0≤bn,…,b1,b0dn,…,d1,d0.That is, Tf is a bounded linear operator. It is clear that Tf is a surjective mapping because if q(t)=qntn+qn-1tn-1+⋯+q1t+q0 is an arbitrary polynomial, then(4)Tfqnbntn+qn-1bn-1tn-1+⋯+q1b1t+q0b0=qt.On the other hand, since Hn (n-degree Hurwitz polynomials set) is an open set, let U be an open set such that g∈U⊂Hn. Then, Tf[U]⊂Hn (see [10]), and, by the Open Mapping Theorem [16], Tf[U] is an open set. Since p(t)∈Tf[U], we have that p(t)∈Tf[U]⊂H(Had), and H(Had) is an open set.


### 2.1.1. Application

Consider the systemx˙=A(λ)x where A(λ)∈Mn×n(R) and λ∈Rn is a parameter. Let pλ(t)=an(λ)tn+an-1(λ)tn-1+⋯+a1(λ)+a0(λ) be the characteristic polynomial of the matrix A(λ).Suppose thatPλ0 is a Hadamardized Hurwitz polynomial. By Theorem 1, there is an open ball Bϵ(pλ0) such that every element of Bϵ(pλ0) is a Hadamardized Hurwitz polynomial. Then, we can find two Hurwitz polynomials:(5)fλt=bnλtn+bn-1λtn-1+⋯+b1λt+b0λ,gλt=cnλtn+cn-1λtn-1+⋯+c1λt+c0λsuch that fλ∗gλ=Pλ for λ≈λ0. The idea is to obtain a Hadamard factorization of Pλ(t) where the coefficients of fλ(t) and gλ(t) are simpler than those of Pλ(t). To illustrate this idea, we give the following example.Example 2.
Consider the systemx˙=Ax, where the entries aij(λ) of A are given by(6)a11=-λ2-8λ-15,a12=-λ2-11λ-29,a13=-λ2-12λ-36,a14=2λ+7,a21=0,a22=0,a23=1,a24=0,a31=0,a32=0,a33=0,a34=1,a41=-λ2-8λ-15,a42=-λ2-11λ-30,a43=-λ2-12λ-36,a44=λ+7.The characteristic polynomial is Pλ(t)=t4+(λ2+6λ+8)t3+(λ2+12λ+36)t2 + (λ2+11λ+30)t+(λ2+8λ+15). For λ=0, P0(t)=t4+8t3+36t2+30t+15 is a Hadamardized Hurwitz polynomial by Theorem A.4.By Theorem1, we can find two Hurwitz polynomials such that the Hadamard product of them is equal to Pλ(t) for small enough λ. Consider the Hurwitz polynomials:(7)F0t=t4+4t3+6t2+6t+5,G0t=t4+2t3+6t2+5t+3.Note that F0∗G0=t4+8t3+36t2+30t+15. The families of polynomials(8)Fλt=t4+λ+4t3+λ+6t2+λ+6t+λ+5,Gλt=t4+λ+2t3+λ+6t2+λ+5t+λ+3are Hurwitz stable for small enough λ, the Hadamard product Fλ∗Gλ satisfies Fλ∗Gλ=Pλ(t), and the coefficients of Fλ and Gλ are simpler than those of Pλ.


## 2.1.1. Application

Consider the systemx˙=A(λ)x where A(λ)∈Mn×n(R) and λ∈Rn is a parameter. Let pλ(t)=an(λ)tn+an-1(λ)tn-1+⋯+a1(λ)+a0(λ) be the characteristic polynomial of the matrix A(λ).Suppose thatPλ0 is a Hadamardized Hurwitz polynomial. By Theorem 1, there is an open ball Bϵ(pλ0) such that every element of Bϵ(pλ0) is a Hadamardized Hurwitz polynomial. Then, we can find two Hurwitz polynomials:(5)fλt=bnλtn+bn-1λtn-1+⋯+b1λt+b0λ,gλt=cnλtn+cn-1λtn-1+⋯+c1λt+c0λsuch that fλ∗gλ=Pλ for λ≈λ0. The idea is to obtain a Hadamard factorization of Pλ(t) where the coefficients of fλ(t) and gλ(t) are simpler than those of Pλ(t). To illustrate this idea, we give the following example.Example 2.
Consider the systemx˙=Ax, where the entries aij(λ) of A are given by(6)a11=-λ2-8λ-15,a12=-λ2-11λ-29,a13=-λ2-12λ-36,a14=2λ+7,a21=0,a22=0,a23=1,a24=0,a31=0,a32=0,a33=0,a34=1,a41=-λ2-8λ-15,a42=-λ2-11λ-30,a43=-λ2-12λ-36,a44=λ+7.The characteristic polynomial is Pλ(t)=t4+(λ2+6λ+8)t3+(λ2+12λ+36)t2 + (λ2+11λ+30)t+(λ2+8λ+15). For λ=0, P0(t)=t4+8t3+36t2+30t+15 is a Hadamardized Hurwitz polynomial by Theorem A.4.By Theorem1, we can find two Hurwitz polynomials such that the Hadamard product of them is equal to Pλ(t) for small enough λ. Consider the Hurwitz polynomials:(7)F0t=t4+4t3+6t2+6t+5,G0t=t4+2t3+6t2+5t+3.Note that F0∗G0=t4+8t3+36t2+30t+15. The families of polynomials(8)Fλt=t4+λ+4t3+λ+6t2+λ+6t+λ+5,Gλt=t4+λ+2t3+λ+6t2+λ+5t+λ+3are Hurwitz stable for small enough λ, the Hadamard product Fλ∗Gλ satisfies Fλ∗Gλ=Pλ(t), and the coefficients of Fλ and Gλ are simpler than those of Pλ.


## 2.2. The Set of Hadamardized Hurwitz Polynomials Is a Nonconvex Set

We have thatH(Had)⊆H (see [13]), where H(Had)≠H and H is the set of Hurwitz polynomials, and it is known that the set of Hurwitz polynomials is a nonconvex set (see [17, 18]). However, this does not imply that the set of Hadamardized Hurwitz polynomials is also a nonconvex set. Thus, the nonconvexity of H(Had) has to be proven.Considered the ray of polynomials:(9)Fkt=t4+t3+6.9t2+2t+3+kt3+t2+2t+1;that is(10)Fkt=t4+1+kt3+6.9+kt2+2+2kt+3+k.Then, Fk(t) is Hurwitz if and only if(11)1+k>0,6.9+k>0,2+2k>0,3+k>0,1+k6.9+k2+2k-2+2k2-1+k23+k>0.This inequality holds if and only if k>-1, and Fk(t) is Hadamardized if and only if(12)3+k1+k2+2k<2+2k1+k-6.9+k2.Thus, Fk(t) is Hurwitz and Hadamardized if and only if k∈(-1,1.2-3.2)∪(1.2+3.2,∞), and Fk(t) is Hurwitz but not Hadamardized if and only if k∈[1.2-3.2,1.2+3.2].Theorem 3.
The set of Hadamardized Hurwitz polynomials is a nonconvex set.Proof.
In the discussion above, takek=-0.5 and k=2.9. Then we get the Hadamardized Hurwitz polynomials r(t)=t4+t3+6.9t2+2t+3-0.5(t3+t2+2t+1) and h(t)=t4+t3+6.9t2+2t+3+2.9(t3+t2+2t+1). Now consider the convex combination λr(t)+(1-λ)h(t). For λ=1/2, we get the polynomial t4+2.2t3+8.1t2+4.4t+4.2, which is not Hadamardized.


## 2.3. The Set of Hadamardized Hurwitz Polynomials Is an Unbounded Set

Theorem 4.
There are rays of polynomials contained in the set of Hadamardized Hurwitz polynomials. Hence, it is an unbounded set.Let(13)Gkt=t4+t3+7t2+2t+3+kt3+t2+2t+1=t4+1+kt3+7+kt2+2+2kt+3+k.Then, Gk(t) is Hurwitz if and only if(14)1+k>0,7+k>0,2+2k>0,3+k>0,1+k7+k2+2k-2+2k2-1+k23+k>0,and Gk(t) is Hadamardized if and only if(15)3+k1+k2+2k<2+2k1+k-7+k2.Consequently, Gk(t) is a Hadamardized Hurwitz polynomial for k>-1 and H(Had) is a unbounded set.The existence of rays of Hadamardized Hurwitz polynomials allows us to show an application in the designing of stabilizing controls.


### 2.3.1. Application

The following example illustrates the designing of a state feedback control that stabilizes the system for each positive value of the parameter. Consider the system(16)x˙1x˙2x˙3x˙4=010000100001-3-2-7-1x1x2x3x4+0001u;and the state feedback control(17)ux=k-1,-2,-1,-1x1x2x3x4,where k∈R is a parameter. Then, the controlled system is(18)x˙1x˙2x˙3x˙4=010000100001-3-k-2-2k-7-k-1-kx1x2x3x4.The characteristic polynomial Gk(t)=t4+t3+7t2+2t+3+k(t3+t2+2t+1) is a Hadamardized Hurwitz polynomial for all k>-1, and consequently the system is stable for all k>-1. Additionally, Gk(t) is Hadamardized for all k>-1.


## 2.3.1. Application

The following example illustrates the designing of a state feedback control that stabilizes the system for each positive value of the parameter. Consider the system(16)x˙1x˙2x˙3x˙4=010000100001-3-2-7-1x1x2x3x4+0001u;and the state feedback control(17)ux=k-1,-2,-1,-1x1x2x3x4,where k∈R is a parameter. Then, the controlled system is(18)x˙1x˙2x˙3x˙4=010000100001-3-k-2-2k-7-k-1-kx1x2x3x4.The characteristic polynomial Gk(t)=t4+t3+7t2+2t+3+k(t3+t2+2t+1) is a Hadamardized Hurwitz polynomial for all k>-1, and consequently the system is stable for all k>-1. Additionally, Gk(t) is Hadamardized for all k>-1.


## 2.4. The Set of Hurwitz Hadamardized Polynomials with Positive Coefficients Is Arc-Connected

To prove the main result of this section, we need the following lemma.Lemma 5.
H
n
+ is an arc-connected set.Proof.
LetP1(t) and P2(t) be two Hurwitz polynomials with positive coefficients. Suppose that n=2m. Then, P1(t) and P2(t) can be written as(19)P1t=b0t2+b1t+b2·t2+b3t+b4⋯t2+b2m-1t+b2m,P2t=a0t2+a1t+a2·t2+a3t+a4⋯t2+a2m-1t+a2m,where bi>0 and ai>0 for all i=0,1,…,2m. Define(20)Hλt=λa0+1-λb0t2+λa1+1-λb1t+λa2+1-λb2t2+λa3+1-λb3t+λa4+1-λb4⋯t2+λa2m-1+1-λb2m-1t+λa2m+1-λb2mwith λ∈[0,1]. We have that H0(t)=P1(t), H1(t)=P2(t), and Hλ(t) is a Hurwitz polynomial for all λ∈[0,1]. Therefore, Hn+ is arc-connected set.Theorem 6.
The set of Hadamardized Hurwitz polynomials with positive coefficients is arc-connected.Proof.
LetP1(t)=antn+an-1tn-1+⋯+a0 and P2(t)=bntn+bn-1tn-1+⋯+b0 Hadamardized Hurwitz polynomials with positive coefficients; then, there are(21)f1t=αntn+αn-1tn-1+⋯+α0,f2t=βntn+βn-1tn-1+⋯+β0,g1t=γntn+γn-1tn-1+⋯+γ0,g2t=δntn+δn-1tn-1+⋯+δ0such that f1, f2, g1, and g2 are Hurwitz polynomials with positive coefficients and (f1∗g1)(t)=P1(t) and (f2∗g2)(t)=P2(t); that is,(22)αnγntn+αn-1γn-1tn-1+⋯+α0γ0=antn+an-1tn-1+⋯+a0,βnδntn+βn-1δn-1tn-1+⋯+β0δ0=bntn+bn-1tn-1+⋯+b0.Since Hn+ is arc-connected set, then there exist(23)G1,λt=cnλtn+cn-1λtn-1+⋯+c0λ,G2,λt=dnλtn+dn-1λtn-1+⋯+d0λsuch that(24)G1,0t=cn0tn+cn-10tn-1+⋯+c00=f1t,G2,0t=dn0tn+dn-10tn-1+⋯+d00=g1t,G1,1t=cn1tn+cn-11tn-1+⋯+c01=f2t,G2,1t=dn1tn+dn-11tn-1+⋯+d01=g2tand each Gi,λ(t)∈Hn+ for all λ∈[0,1]. Now, consider(25)G1,λ∗G2,λt=cnλdnλtn+cn-1λdn-1λtn-1+⋯+c0λd0λand then (G1,λ∗G2,λ)(t) is a Hadamardized Hurwitz polynomial for all λ∈[0,1] (see [10]) and(26)G1,0∗G2,0t=f1∗g1t=P1t,G1,1∗G2,1t=f2∗g2t=P2t.Consequently, the set of Hadamardized Hurwitz polynomials whit positive coefficients is arc-connected.


## 3. The Kharitonov Theorem and the Set of Hadamardized Hurwitz Polynomials

The Kharitonov theorem (see [19]) has motivated a lot of research on the subject of stable polynomials. Many papers have been published since Kharitonov’s theorem was reported (see [20–24]). We now address the following question. Suppose we have an interval family of Hurwitz polynomials. If the four Kharitonov polynomials are Hadamardized polynomials, is each element of the family a Hadamardized polynomial? We will give the answer in the case of 4-degree polynomials. First, we need the following result. Although the proof is simple, we have decided to include it.Theorem 7.
The Hurwitz polynomialf(t)=α4t4+α3t3+α2t2+α1t+α0 is Hadamardized if and only if g(t)=α4t4+α3t3+α2t2+α1t+α0 is a Hurwitz polynomial.Proof.
The necessary condition is immediate. To prove sufficient condition, we suppose thatf(t) is Hadamardized, and then (see [14, 15])(27)α0α3α1<α1α4α3-α22.Note that α2α3-α1α4>0 (since f is Hurwitz), and then(28)α2-α1α4α3>0.Consequently, if f(t) is Hadamardized, then(29)α0α3α1<α2-α1α4α3.Whence, 0<α1α2α3-α1α4-α0α3, which implies that g(t) is a Hurwitz polynomial.The last theorem let us establish the following result.Theorem 8.
Consider the interval familyF(t)=[a4,b4]t4+[a3,b3]t3+[a2,b2]t2+[a1,b1]t+[a0,b0] which consists of Hurwitz polynomials. If the four Kharitonov polynomials are Hadamardized polynomials, then each element of the family is a Hadamardized polynomial.Proof.
The four Kharitonov polynomials are(30)k1t=a4t4+b3t3+b2t2+a1t+a0,k2t=a4t4+a3t3+b2t2+b1t+a0,k3t=b4t4+b3t3+a2t2+a1t+b0,k4t=b4t4+a3t3+a2t2+b1t+b0.Each one is a Hurwitz polynomial. If the four Kharitonov polynomials are Hadamardized, then, applying the previous theorem, the polynomials(31)g1t=a4t4+b3t3+b2t2+a1t+a0,g2t=a4t4+a3t3+b2t2+b1t+a0,g1t=b4t4+b3t3+a2t2+a1t+b0,g1t=b4t4+a3t3+a2t2+b1t+b0are Hurwitz. Now, for Kharitonov’s theorem, the polynomial(32)Gt=a4,b4t4+a3,b3t3+a2,b2t2+a1,b1t+a0,b0is also Hurwitz. Let f(t)=c4t4+c3t3+c2t2+c1t+c0 be an arbitrary element of F(t), where ci∈[ai,bi] for i=1,2,3,4, and consider the polynomial in G(t)(33)gt=c4t4+c3t3+c2t2+c1t+c0and then g(t) is Hurwitz. Hence,(34)ft=gt∗gtand the theorem is proved.Remark 9.
The discussion above allows us to propose the following open problem.Open Problem 1.
Consider the Hurwitz polynomialf(t)=αntn+⋯+α1t+α0 with n≥5; if f(t) is a Hadamardized polynomial, then g(t)=αntn+⋯+α1t+α0 is a Hurwitz polynomial.Remark 10.
Note that, by repeated application of the square root, the polynomialαn2mtn+⋯+α12mt+α02m converges to the polynomial tn+tn-1+⋯+t+1 when m→∞, which has equally spaced roots on the unit circle and consequently tn+tn-1+⋯+1 is not a Hurwitz polynomial. It let us establish the following problem.Open Problem 2.
Letαntn+⋯+α1t+α0 and αntn+⋯+α1t+α0 be Hurwitz polynomials; what is the maximum m such that αn2mtn+⋯+α12mt+α02m is a Hurwitz polynomial?Remark 11.
Since forn degree greater than or equal to 5 we do not have a similar result to Theorem 8, then we can propose the following two open problems.Open Problem 3.
Consider the family of Hurwitz polynomials with degreen≥5: [an,bn]tn+⋯+[a1,b1]t+[a0,b0]. If the four Kharitonov polynomials are Hadamardized polynomials, every element is a Hadamardized polynomial.Open Problem 4.
If all of the vertices of an interval family of polynomials with degreen≥5 are Hadamardized Hurwitz polynomials, does the family consist of Hadamardized Hurwitz polynomials?


## 4. The Stability Test, the Set of Hadamardized Hurwitz Polynomials, and Differential Topology

The stability test is a criterion for verifying whether a given polynomial is Hurwitz. We explain it. Consider then-degree polynomial with positive coefficients:(35)Pt=Antn+An-1tn-1+⋯+A1t+A0.Defining the n-1-degree polynomial(36)Rt=An-12tn-1+An-1An-2-AnAn-3tn-2+An-1An-3tn-3+An-1An-4-AnAn-5tn-4+⋯,then we have the following result.Theorem 12.
P
(
t
) is a Hurwitz polynomial if and only if R(t) is a Hurwitz polynomial.Proof.
See [1].The importance of the stability test was appreciated when Kharitonov used it to obtain his celebrated theorem. Moreover, it is worth mentioning that other important properties of the set of Hurwitz polynomials were shown with the stability test (see [6, 13]). Now we use the stability test for studying the following subset of the Hadamardized Hurwitz polynomial set:(37)HHadn∗=ft=antn+⋯+a0∈Hn+:ht=antn+⋯+a0∈Hn+.As was observed in Open Problem 1 of Section 3, at the moment, we do not know the answer to the following question: H(Had)n∗⫋H(Had)n or H(Had)n∗=H(Had)n?In this section, we get a different test, which is applicable to studying the setH(Had)n∗. This special test let us obtain the following property: H(Had)n∗ is a vector bundle on the set H(Had)n-1∗. There are previous works (see [6, 7]), where some ideas of differential topology have been used to study the space of Hurwitz polynomials. A presentation of the concepts used in this section can be consulted in [6, 7, 25, 26].


### 4.1. A Test forH(Had)n∗

Consider then-degree polynomial with positive coefficients as (35); define the (n-1)-degree polynomial(38)Qt=An-12tn-1+An-1An-2-AnAn-32tn-2+An-1An-3tn-3+An-1An-4-AnAn-52tn-4+⋯and then we have the following result.Theorem 13.
P
(
t
)
∈
H
(
H
a
d
)
n
∗ if and only if Q(t)∈H(Had)n-1∗.Proof.
IfP(t)∈H(Had)n∗, then Antn+An-1tn-1+⋯+A1t+A0∈Hn+, and by stability test(39)An-1tn-1+An-1An-2-AnAn-3tn-2+An-1An-3tn-3+An-1An-4-AnAn-5tn-4+⋯∈Hn-1+.Then, Q(t) is a Hurwitz polynomial (see [10]). Consequently, Q(t)∈H(Had)n-1∗. This proves the sufficient condition. Now, if Q(t)∈H(Had)n-1∗, then we obtain (39) and by the stability test(40)Antn+An-1tn-1+⋯+A1t+A0∈Hn+,where P(t)∈Hn+ (see [10]) and then P(t)∈H(Had)n∗. This completes the proof.


### 4.2. An Approach of Differential Topology in the Study of the Set of Hadamardized Hurwitz Polynomials

Suppose that the(n-1)-degree polynomial(41)Qt=bn-1tn-1+bn-2tn-2+⋯+b1t+b0is an element of H(Had)n-1∗; then,(42)qt=bn-1tn-1+bn-2tn-2+⋯+b1t+b0is an element of Hn-1+. We can calculate the fiber (see [6, 13]) of q(t):(43)Pst=stn+bn-11/4tn-1+bn-2bn-11/4+bn-3bn-11/2stn-2+bn-3bn-11/4tn-3+bn-4bn-11/4+bn-5bn-11/2stn-4+⋯with s>0; then, we have(44)Ps∗Pst=s2tn+bn-1tn-1+bn-2bn-1+2bn-2bn-3bn-13/4s+bn-3bn-1s2tn-2+bn-3bn-1tn-3+⋯.Then, (Ps∗Ps)(t) will be the fiber of Q(t) and we have the following result.Theorem 14.
H
(
H
a
d
)
n
∗ is a vector bundle with base H(Had)n-1∗.Proof.
GivenP(t)=Antn+An-1tn-1+⋯+A1t+A0∈H(Had)n∗, Q(t) is the (n-1)-degree polynomial defined according to (38). Then, we can define φ:H(Had)n∗→H(Had)n-1∗×R as follows: φ(P(t))=(Q(t),ln(bn)).
Now we defineφ-1:H(Had)n-1∗×R→H(Had)n∗; given the polynomial Q(t) defined in (41) and l∈R, φ-1(Q(t),l) is the following polynomial φ-1(Q(t),l)=Pel/2∗Pel/2(t), obtained by replacing s by el/2 in (44).
Then, we obtain the following commutative diagram:(45)where  H is determined by the new test established in Theorem 8; that is, H(P(t))=Q(t). Then, H(Had)n∗ is a vector bundle with base H(Had)n-1∗.Remark 15.
By the last theorem, we can say thatH(Had)n∗≈H(Had)n-1∗×R and it implies that (46)HHadn∗≈HHad4×Rn-4.


## 4.1. A Test forH(Had)n∗

Consider then-degree polynomial with positive coefficients as (35); define the (n-1)-degree polynomial(38)Qt=An-12tn-1+An-1An-2-AnAn-32tn-2+An-1An-3tn-3+An-1An-4-AnAn-52tn-4+⋯and then we have the following result.Theorem 13.
P
(
t
)
∈
H
(
H
a
d
)
n
∗ if and only if Q(t)∈H(Had)n-1∗.Proof.
IfP(t)∈H(Had)n∗, then Antn+An-1tn-1+⋯+A1t+A0∈Hn+, and by stability test(39)An-1tn-1+An-1An-2-AnAn-3tn-2+An-1An-3tn-3+An-1An-4-AnAn-5tn-4+⋯∈Hn-1+.Then, Q(t) is a Hurwitz polynomial (see [10]). Consequently, Q(t)∈H(Had)n-1∗. This proves the sufficient condition. Now, if Q(t)∈H(Had)n-1∗, then we obtain (39) and by the stability test(40)Antn+An-1tn-1+⋯+A1t+A0∈Hn+,where P(t)∈Hn+ (see [10]) and then P(t)∈H(Had)n∗. This completes the proof.


## 4.2. An Approach of Differential Topology in the Study of the Set of Hadamardized Hurwitz Polynomials

Suppose that the(n-1)-degree polynomial(41)Qt=bn-1tn-1+bn-2tn-2+⋯+b1t+b0is an element of H(Had)n-1∗; then,(42)qt=bn-1tn-1+bn-2tn-2+⋯+b1t+b0is an element of Hn-1+. We can calculate the fiber (see [6, 13]) of q(t):(43)Pst=stn+bn-11/4tn-1+bn-2bn-11/4+bn-3bn-11/2stn-2+bn-3bn-11/4tn-3+bn-4bn-11/4+bn-5bn-11/2stn-4+⋯with s>0; then, we have(44)Ps∗Pst=s2tn+bn-1tn-1+bn-2bn-1+2bn-2bn-3bn-13/4s+bn-3bn-1s2tn-2+bn-3bn-1tn-3+⋯.Then, (Ps∗Ps)(t) will be the fiber of Q(t) and we have the following result.Theorem 14.
H
(
H
a
d
)
n
∗ is a vector bundle with base H(Had)n-1∗.Proof.
GivenP(t)=Antn+An-1tn-1+⋯+A1t+A0∈H(Had)n∗, Q(t) is the (n-1)-degree polynomial defined according to (38). Then, we can define φ:H(Had)n∗→H(Had)n-1∗×R as follows: φ(P(t))=(Q(t),ln(bn)).
Now we defineφ-1:H(Had)n-1∗×R→H(Had)n∗; given the polynomial Q(t) defined in (41) and l∈R, φ-1(Q(t),l) is the following polynomial φ-1(Q(t),l)=Pel/2∗Pel/2(t), obtained by replacing s by el/2 in (44).
Then, we obtain the following commutative diagram:(45)where  H is determined by the new test established in Theorem 8; that is, H(P(t))=Q(t). Then, H(Had)n∗ is a vector bundle with base H(Had)n-1∗.Remark 15.
By the last theorem, we can say thatH(Had)n∗≈H(Had)n-1∗×R and it implies that (46)HHadn∗≈HHad4×Rn-4.


## 5. Conclusion

The study of the set of Hadamardized Hurwitz polynomials began with a work by Garloff and Shrinivasan in 1996. Later, Loredo-Villalobos and Aguirre-Hernández obtained other results in 2011. We now prove that the set of Hadamardized Hurwitz polynomials is an open, nonconvex, unbounded, and arc-connected set. Additionally, we have given a condition to check when a family of interval polynomials of fourth degree is a Hadamardized polynomial and we have proposed an approach of differential topology for studying the set of Hadamardized Hurwitz polynomials. We have presented some results that are valid forn-degree polynomials and other results that are valid for 4-degree polynomials and then we have established open problems for polynomials with degree greater than or equal to 5. We hope that some constructive techniques are discovered for solving the open problems.



---

*Source: 695279-2015-05-26.xml*