Three coefficients of a polynomial can determine its instability

Transcription

1 Linear Algebra and its Applications 338 (2001) Three coefficients of a polynomial can determine its instability Alberto Borobia a,,1, Sebastián Dormido b,2 a Departamento de Matemáticas, UNED, Senda del Rey s.n., Madrid, Spain b Departamento de Informática y Automática, UNED, Senda del Rey s.n., Madrid, Spain Received 25 January 2001; accepted 25 April 2001 Submitted by R.A. Brualdi Abstract We will prove that, in some cases, if we now only three coefficients of a polynomial with positive coefficients and without any restriction on the magnitude of its degree, we can conclude that the polynomial is unstable. Namely, if P(x) = i=0 a i x i is a polynomial with positive coefficients and for some q {1,...,n 1} it is satisfied that a 2q < ( q n )a(n q)/n 0 a q/n,thenp(x)is unstable Elsevier Science Inc. All rights reserved. Keywords: Stable or Hurwitz polynomials; Robust control theory 1. Introduction The research in structured real parametric uncertainty saw a breathrough with the seminal wor of Kharitonov [1] in 1978 regarding the Hurwitz stability of a family of interval polynomials. This result was so surprising and elegant that it has been the tae off point of a renewed interest in robust control theory involving structured real parametric uncertainty (see [2 4], and references therein). In fact, when the control clothing is eliminated, many of the basic problems which we study can be formulated in the frame of a very old question: How are the roots of a polynomial related with its coefficients? Corresponding author. addresses: aborobia@mat.uned.es (A. Borobia), sdormido@dia.uned.es (S. Dormido). 1 Partially supported by DGICYT PB C Partially supported by CICYT TAP C /01/$ - see front matter 2001 Elsevier Science Inc. All rights reserved. PII:S (01)

2 68 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) From the practical point of view the uncertainty structures which arise in typical applications are more complicated than those which we have analyzed in the interval or polytopic framewor. Dealing with complicated uncertainty structures in a robustness context is an open problem. In this sense the present paper gives a new result that can open new perspectives in order to deal with the problem of the uncertainty structure. In some cases nowing only three coefficients of a polynomial with positive coefficients it is possible to conclude that the polynomial is unstable. The result is quite surprising because it is independent of the degree of the polynomial. Furthermore, the relationship between the three coefficients is very simple. To be exact, we demonstrate the following result: If P(x) = i=0 a i x i is a polynomial with positive coefficients and for some q {1,...,n 1} it is satisfied that ( n a 2q < a q) (n q)/n 0 a q/n, then P(x)is unstable. The paper has been organized in the following way. In Section 2 we present an introductory example: we will prove the result on the previous paragraph for the case of fourth order polynomials by using Routh s test. This line of tacle is impossible to pursue for polynomials of higher order. In Section 3 we will prove that if we now only two coefficients of a polynomial of positive coefficients, we cannot conclude that the polynomial is unstable. Then we will state our main result: that this is no longer true when we now three coefficients. In Section 4 we will present several auxiliary lemmas which are necessary in order to prove the main result. This proof will be the content of Section Introductory example A complex polynomial of degree n P(x) = a 0 x n + a 1 x n 1 + +a n 1 x 1 + a n is said to be a stable or a Hurwitz polynomial if and only if all its roots lie in the open left half of the complex plane. It is well nown that all coefficients of a real Hurwitz polynomial have the same sign. Based on this property we will restrict ourselves to the study of polynomials with positive coefficients. Consider now the fourth order uncertain polynomial P(x) = a 0 x x 3 + a 2 x x + a 4, where a 0,a 2,a 4 are fixed with a i >0, and 1, 3 are variable parameters with i >0. We will show that there are triples (a 0,a 2,a 4 ) such that for all 1, 3 > 0 the polynomial P(x)is not a Hurwitz polynomial.

3 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) For this simple example we can use Routh s test. To perform the test we must first prepare the following array: Row 1 a 0 a 2 a 4 Row Row 3 a 2 αa 0 a 4 Row 4 1 (α a 4 a 2 αa 0 ) Row 5 a 4 where α = 3 / 1. Imposing that all the elements in the left column of the Routh array are positive, it can be verified the following result: If the coefficients a 0,a 2,a 4 fulfil the relationship a 2 < 2 a 0 a 4, then there are no 1, 3 > 0suchthatP(x)is a Hurwitz polynomial. Remar. Let P(x) = n i=0 a i x n i be a real polynomial. The even and odd parts of P(x)are defined by P even (x) = a n + a n 2 x 2 + a n 4 x 4 +, P odd (x) = a n 1 x + a n 3 x 3 + a n 5 x 5 + The Hermite Bieler Theorem (see Section 1.3 of [4]) states that a necessary condition for the stability of P(x) is that all roots of P even (x) and P odd (x) lie on the imaginary axis. This result implies certain independence of the even and odd parts of a real polynomial when we study its instability. For instance, in our introductory example if the coefficients a 0,a 2,a 4 fulfil the relationship a 2 < 2 a 0 a 4, then the even part of the polynomial will have some roots that do not lie on the imaginary axis and then by the Hermite Bieler Theorem the polynomial will be unstable. Note that the coefficients of the odd part of the polynomial do not play any role on this argument. 3. Main results We introduce the following notation: P n will denote the set of polynomials of order n with positive coefficients, and PE n the set of real stable polynomials of order n with positive coefficients. It is clear that PE n P n. Moreover we will denote by P i 1,...,i t n [α 1,...,α t ] with 0 i 1 < <i t n and α 1,...,α t > 0 to the set of polynomials n i=0 a i x n i P n such that a i1 = α 1,...,a it = α t. We also define the set PE i 1,...,i t n [α 1,...,α t ]=P i 1,...,i t n [α 1,...,α t ] PE n. We will now prove that if we now only two coefficients of a polynomial of P n we cannot conclude that the polynomial is unstable.

4 70 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) Lemma 1. For any α, β > 0 and any r, s N with 0 r<s n it is satisfied that PE r,s n [α, β] /=. Proof. Let P(x) = n i=0 a i x n i PE n be any stable polynomial. Let z 1,z 2,..., z n C be the roots of P(x).For any λ, µ > 0 define the polynomial n P(x,λ,µ) = λµ i a i x n i. i=0 Note that µz 1,µz 2,...,µz n C are the roots of P (x, λ, µ), and therefore P(x,λ,µ) PE n. We can choose λ and µ in such a way that P(x,λ,µ)belongs to PEn r,s [α, β]. For it we only need to solve the system λµ r a r = α, λµ s a s = β. The values that we obtain for λ and µ are ( ) 1/r s αas µ = and λ = β βa r µ s. a s The situation changes when we now three coefficients of a polynomial of positive coefficients. Indeed we will see that there exist polynomials of P (without restrictions on the magnitude of n) for which the value of only three of their coefficients determine its instability. We are going to concentrate in the study of the [1, 1] (that is, the stable polynomials of order n which are monic and such that the product of their roots is equal to 1). One reason for this restriction is the following: the proof of Lemma 1 says that choosing any stable polynomial P(x) = n i=0 a i x n i PE n and taing µ = (a 0 /a n ) 1/n and λ = 1/a 0 we can set PE 0,n n [1, 1]. That is, each stable polynomial has a unique representative in PEn 0,n [1, 1]. Note that to determine PE n and to determine PEn 0,n [1, 1] are equivalent problems. The argument would be the same if we fix any other two coefficients with any values. But when we mae calculations, to fix the first and the last coefficients with value equal to 1 has great advantages. We now state the main result of this wor. construct the polynomial P(x,1/a 0, (a 0 /a n ) 1/n ) that belongs to PE 0,n n Theorem 2. PEn 0,q,n [1,α,1] = if and only if q and n are even numbers and α ( n/2 q/2 ). 4. Auxiliary results Before we give the proof of Theorem 2 we present three auxiliary lemmas.

5 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) Lemma 3. Let P(x) = n i=0 a i x n i be a monic polynomial of degree n and real coefficients. Let r 1,...,r s R and z 1, z 1,...,z t, z t C R with s + 2t = n be the roots of P(x).The coefficient a d of x n d in P(x)is equal to ( 1) u+v 2 v r i1 r iu Re z j1 Re z jv z z w 2. u + v + 2w = d 1 i 1 < <i u s 1 j 1 < <j v t 1 1 < < w t j a /= b Remar. If the polynomial P(x)is stable, then all the terms that appear in the sum that defines a d are positive. Proof. We have that P(x)= = n a i x n i = i=0 s (x r i ) i=1 s (x r i ) i=1 t ((x z j )(x z j )) j=1 t (x 2 (2 Rez j )x + z j 2 ). j=1 When we develop this last expression we obtain a polynomial such that the coefficient of x n d is ( r i1 ) ( r iu )( 2 Rez j1 ) ( 2 Rez jv ) z z w 2. u + v + 2w = d 1 i 1 < <i u s 1 j 1 < <j v t 1 1 < < w t j a /= b Lemma 4. Let H = { (x 1,...,x n ) R n : x 1 x 2 x n = 1,x i > 0 i {1,...,n} }. For each r {1,...,n} define the function f r : H R (x 1,...,x n ) 1 i 1 < <i r n x i 1...x ir Then f r (x 1,...,x n ) ( n r ) for all (x 1,...,x n ) H and equality holds only for (1,...,1) H. Proof. For each i, j {1,...,n} with i/= j and each λ R {0} define the function Ti,j λ : H H (a 1,...,a n ) (b 1,...,b n ) with b i = λa i,b j = a j /λ and b = a if /= i, j.

6 72 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) (i) If (a 1,...,a n ) H with a i 1 a j and λ>1, then f r (Ti,j λ (a 1,...,a n )) f r (a 1,...,a n ) ( ( aj )) = (λa i a i ) + λ a j ( = (λ 1) a i a j λ ) 1 i 1 < <i r 1 n i 1,...,i r 1 /= i, j 1 i 1 < <i r 1 n i 1,...,i r 1 /= i, j a i1...a ir 1 a i1...a ir 1 > 0. That is, f r (Ti,j λ (a 1,...,a n )) > f r (a 1,...,a n ). (ii) Let (a 1,...,a n ), (c 1,...,c n ) H be such that it is satisfied for each {1,...,n} that if c 1, then c a 1andifc 1, then c a 1. Suppose that there exists some i {1,...,n} such that c i >a i 1, then also there exists some j {1,...,n} {i} such that c j <a j 1. Let β = min{c i /a i,a j /c j } > 1 and (b 1,...,b n ) = T β i,j (a 1,...,a n ). Then c i b i >a i 1andc j b j <a j 1, and for h/= i, j b h = a h. Therefore, it is satisfied for each {1,...,n} that if c > 1, then c b a 1 and if c < 1, then c b a 1. Moreover (c 1,...,c n ) and (b 1,...,b n ) have one or two more common coordinates that (c 1,...,c n ) and (a 1,...,a n ) since c i = b i or c j = b j, or both coincide. (iii) From (ii) it follows that for any b = (b 1,...,b n ) H there exists a finite sequence of atmost n elements 1,...,a(0) n ), (a(1) 1,...,a(1) n ),...,(a(r) 1,...,a(r) n ) H that satisfies: (1) (a (0) 1,...,a(0) n ) = (1, 1,...,1) and (a (r) 1,...,a(r) n ) = (b 1,...,b n ). (2) For {1,...,n} if b 1, then b = a (r) a (r 1) a (0) = 1. (3) For {1,...,n} if b 1, then b = a (r) a (r 1) a (0) = 1. (4) For h {0, 1,...,r 1} the points (a (h+1) 1,...,a n (h+1) ) and (b 1,...,b n ) have one or two more common coordinates than (a (h) 1,...,a(h) n ) and (b 1,...,b n ). We now explain how to construct the sequence. Suppose (a (h) 1,...,a(h) n ) is different from (b 1,...,b n ). Then there exists some i h {1,...,n} such that b ih >a (h) i h 1, and some j h {1,...,n} {i h } with b jh <a (h) j h 1. Tae λ h = min{b ih /a (h) i h, a (h) j h /b jh } > 1 and define (a (0) (a (h+1) 1,...,a n (h+1) ) = T λ h i h,j h (a (h) 1,...,a(h) n ).

7 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) (iv) From (i) and (iii) it follows f r (1, 1,...,1) = ( n r )<f r(x 1,...,x n ) for all (x 1,...,x n ) H {(1, 1,...,1)}. We will now consider extensions of the sets P n,pe n,p i 1,...,i t n [α 1,...,α t ], and [α 1,...,α t ]. We will denote by P n to the set of polynomials of order n with PE i 1,...,i t n nonnegative coefficients, and by PE n to the set composed of those polynomials of P n such that the real part of their roots are non-positive. We have that PE n P n. Moreover, we will denote by P i 1,...,i t n [α 1,...,α t ] with 0 i 1 < <i t n and α 1,...,α t 0 to the set of polynomials n i=0 a i x n i P n such that a i1 = α 1,...,a it = α t. We also define the set PE i 1,...,i t n [α 1,...,α t ]= P i 1,...,i t n [α 1,...,α t ] PE n. Lemma 5. Let P(x) = n i=0 a i x n i,q(x)= n i=0 b i x n i in PE 0,n n [1, 1]. Let r 1,...,r c R and z 1, z 1,...,z d, z d C R be the roots of P(x)and let s 1,...,s c R and w 1, w 1,...,w d, w d C R be the roots of Q(x) with c + 2d = n. If a q < α<b q, then PE 0,q,n n [1,α,1] /=. Proof. For each i {1,...,d} define any continuous function γ i :[0, 1] C such that γ i (0) = z i,γ i (1) = w i, and Re (γ i (t)) < 0forallt (0, 1). For each j {1,...,c} define any continuous function φ j :[0, 1] R such that φ j (0) = r j,φ j (1) = s j, and φ j (t) < 0forallt (0, 1). For all t [0, 1] we define P t (x) = c d (x φ j (t)) ((x γ i (t))(x γ i (t))). j=1 i=1 Note that for any t (0, 1) we have that P t (x) PE n. Following the proof of Lemma 1, for each t (0, 1) let λ t,µ t > 0 such that the polynomial P t (x, λ t,µ t ) belongs to PEn 0,n [1, 1]. Then define the continuous function f : [0, 1] PE 0,n n [1, 1] 0 f(0) = P(x) 1 f(1) = Q(x) t (0, 1) f(t)= P t (x, λ t,µ t ). We have that f(t) PEn 0,n [1, 1] for all t (0, 1), and by a continuity argument the result follows.

8 74 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) Proof of Theorem 2 (i) Let P(x) = i=0 a i x i PE 0, [1, 1]. Let r 1,...,r 2s R and z 1, z 1,...,z t, z t C R be the roots of P(x). Note that 2s + 2t =. Let x 1 = r 1 r 2,...,x s = r 2s 1 r 2s,x s+1 = z 1 2,...,x n = z t 2. We employ the expression given in Lemma 3 to describe the coefficient a 2q with q {1,...,n 1}. As P(x) is stable then all terms in the sum that defines a 2q are positive. We consider only part of these terms. Namely, those in which tae part only x 1,...,x n. Thus we will have that a 2q > x i1...x iq. 1 i 1 < <i q n As (x 1,...,x n ) H,it follows from Lemma 4 that a 2q >( n q ). Therefore, PE0,2q, [1,α,1]= if α ( n q ). Remar. The Hermite Bieler Theorem and Lemma 4 provide another approach in order to obtain at least the strict inequality above. Consider the real polynomial P(x) = x + a 1 x 1 + a 2 x 2 + +a 2 x 2 + a 1 x and consider the even part of P(x) P even (x) = x + a 2 x 2 + a 4 x 4 + +a 4 x 4 + a 2 x Define the polynomial P e/2 (x) as the polynomial that we obtain from P even (x) by dividing all its degrees by 2. That is P e/2 (x) = x n + a 2 x n 1 + a 4 x n 2 + +a 4 x 2 + a 2 x + 1. If P(x) is stable, then by the Hermite Bieler Theorem all roots of P even (x) lie on the imaginary axis, or equivalently, all roots of P e/2 (x) are real negative numbers. Then Lemma 4 can be employed to conclude that a 2q ( n q ). (ii) Define the function f : [ π 2, ) PE 0, [1, 1] t [ π 2, 0] (x2 + (2cost)x + 1) n t [0, ) (x + 1) 2 (x + (1 + t))(x + 1+t 1 ). The function f is continuous. Except for t = π/2 the remaining polynomials f(t) belongs to PE 0, [1, 1]. We have f ( π 2 ) = (x 2 + 1) n = n q=0 ( ) n x 2q. q

9 For t> π 2 let A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) f(t)= x + f 1 (t)x 1 + f 2 (t)x 2 + +f 1 (t)x + 1. It is easy to chec that f i (t) when t for all i {1,..., 1}. Arguing by continuity we conclude that PE 0,2q, [1,α,1] /= if α>( n q ) and PE0,2q 1, [1,α,1] /= for all α>0. (iii) On the other hand, we define the function g : (0, ) PE 0,+1 +1 [1, 1] t (x t )n (x + t n ). The function g is continuous. It is easy to chec that when t 0 all even coefficients of g(t) except the independent term (that it is always equal to 1) tend to 0 and all odd coefficients of g(t) except the coefficient of x +1 (that it is always equal to 1) tend to, and that when t all odd coefficients of g(t) except the coefficient of x +1 tend to 0 and all even coefficients of g(t) except the independent term tend to. From Lemma 5 we conclude that PE 0,q,+1 +1 [1,α,1] /= for all α>0. Corollary 6. n 1} such that ( n a 2q < q then P(x)is unstable. Let P(x) = i=0 a i x i P. If there exists some q {1,..., ) a (n q)/n 0 a q/n, Proof. By the proof of Lemma 1 we now that the polynomial ( b i x i = P x, 1 ( ) ) 1/ a0, a 0 i=0 a belongs to PE 0, [1, 1]. Moreover, P(x)is stable if and only if P(x, a 1 0,(a 0 /a ) 1/ ) is stable. We have that b 2q = 1 a 0 ( a0 a ) 2q/, a 2q = a 2q a (n q)/n 0 a q/n. If for some q {1,...,n 1} we have that b 2q <( n q ), then it follows from Theorem 2thatP(x, a 1 0, (a 0 /a ) 1/ ), and therefore P(x)is unstable. Acnowledgement The remars at the end of Section 2 and at the beginning of Section 5 are the consequences of suggestions of the referee.

10 76 A. Borobia, S. Dormido / Linear Algebra and its Applications 338 (2001) References [1] V.L. Kharitonov, Asymptotic stability of an equilibrium position of a family of systems of linear differential equations, Differentsial nye Uraveniya, 14 (1978) [English translation: Differential Equations 14 (1979) ]. [2] J. Acermann, Robust Control: Systems with Uncertain Physical Parameters, Springer, New Yor, [3] B.R. Barmish, New Tools for Robustness of Linear Systems, Macmillan, New Yor, [4] S.P. Bhattacharyya, H. Chapellat, L.H. Keel, Robust Control: the Parametric Approach, Prentice-Hall, Englewood Cliffs, NJ, 1995.