DSpace at VNU: Stability radii of differential algebraic equations with structured perturbations

Systems & Control Letters 57 2008 546–553www.elsevier.com/locate/sysconle Stability radii of differential algebraic equations with structured perturbations Faculty of Mathematics, Inform

Systems & Control Letters 57 (2008) 546–553

www.elsevier.com/locate/sysconle

Stability radii of differential algebraic equations with

structured perturbations

Faculty of Mathematics, Informatics and Mechanics, Vietnam National University, 334 Nguyen Trai, Thanh Xuan, Hanoi, Viet Nam

Received 3 March 2006; received in revised form 14 December 2007; accepted 14 December 2007

Available online 4 February 2008

Abstract

This paper deals with a formula for computing stability radii of a differential algebraic equation of the form A X0(t) − B X(t) = 0, where

A, B are constant matrices A computable formula for the complex stability radius is given and a key difference between the ordinary differential equation (ODEs for short) and the differential algebraic equation (DAEs for short) is pointed out A special case where the real stability radius and the complex one are equal is considered

c 2008 Elsevier B.V All rights reserved

Keywords: Differential algebraic equation; Index of the pencil of matrices; Stability radii

1 Introduction

In the last decade, there has been an extensive work on the

study of robustness measures, where one of the most powerful

ideas is the concept of the stability radius, introduced by

Hinrichsen and Pritchart [6] The stability radius is defined

as the smallest value ρ of the norm of the real (or complex)

perturbations destabilizing the system A detailed analysis on

the stability radii for ordinary differential equations can be

found in [1,6–8,11]

On the other hand, in the recent years, several technical

problems in electronic circuit theory and robotic designs lead to

the problem of investigating the differential algebraic equation

where the leading term X0 cannot be explicitly solved from

X(t) The linear form of this equation is

where A and B are two constant matrices (see [2,5])

∗ Corresponding address: Faculty of Mathematics-Mechanics and

Infor-matics, Hanoi National University, 334 Nguyen Trai, Thanh Xuan, Hanoi,

Viet Nam Fax: +84 4 8588817.

E-mail address: nhdu2001@yahoo.com

This paper studies the stability radius of the Eq.(1.2)above Note that if A is a nonsingular matrix, then Eq (1.2) can easily be reduced to an ordinary differential equation and this situation has been extensively investigated in the literature (see for example [1,6,7]) Thus we are mainly interested in the case when A is singular In this case, it is known that every solution

of(1.2)contains simultaneously a differential component and

an algebraic one According to [2,9], the investigation of the so-called index of the pencil of matrices { A, B} is necessary but the situation becomes more complicated Indeed, it is known that in the ODEs case, by the continuity of the spectrum in parameters, if the original Eq.(1.2)is stable then it is still stable under a sufficiently small perturbation In general, this property

is no longer valid for differential algebraic equation since the structure of the solutions of a DAEs depends strongly on the index of { A, B} (see [9,3–5]) and the solutions of(1.2)contain several components, which are related by an algebraic equation Under perturbations, the index of the perturbed system might change which leads to a change of the algebraic relations and several eigenvalues may “disappear” This explains that

in the ODEs case, it is possible to point out a perturbation whose norm equals the stability radiusρ, which destabilizes our system, while such a perturbation in a DAEs may not exist (see Theorem 3.1) Moreover, the stability radius of a stable linear ODE is always positive while the stability radius of a DAEs may be zero

0167-6911/$ - see front matter c 2008 Elsevier B.V All rights reserved.

doi:10.1016/j.sysconle.2007.12.001

Another natural problem in the study of the stability radii of

a dynamical system is to determine when the complex stability

radius and the real stability radius coincide In the case of the

ordinary differential equation, it is known that if A−1is positive

and M = − A−1B is a Metlez matrix, i.e., all entries of M

except possibly those on the principal diagonal, are positive,

then the real stability radius and the complex stability radius

are the same (see [7,8])

For DAEs, we are able to provide the solution for this

problem under an, admittedly, very strict assumption (see

Theorems 4.4and4.7) It seems that this is a difficult problem

in general, and so far we do not know whether this property

remains true However, we shall point out that if the resolvent

of the Eq.(1.2)cannot reach its maximum value on i R, the real

and complex stability radii are equal This is a key difference

between the ODEs and the DAEs in studying the stability radii

This article is organized as follows: In the Section 2, we

recall some basic properties of the linear differential algebraic

equation The Section 3 deals with a formula for computing

the complex stability radius of the system (1.1) where the

structured perturbations are considered The Section 4 is

concerned with a special class of { A, B} where the complex

and the real stability radii are equal

2 Preliminary

Consider the differential algebraic equation:

where X(t) ∈ Rm, A and B are m × m-constant matrices whose

entries are in a field K; the underlying field K is either real or

complex We assume that { A, B} is regular (that is, f (λ) =

det(λA − B) 6≡ 0) and the index of {A, B} = k > 1 (see [2])

The Kronecker decomposition of the pencil of matrices { A, B}

indicates that there exists a pair of nonsingular matrices W, T

such that

A = Wdiag(Ir, U)T−1,

where Is is the unit matrix in Ks×s and B1 is a matrix in

Kr ×r Further, U ∈ K(m−r)×(m−r)is a nilpotent matrix whose

nilpotency degree is exactly k Denote

b

Q = Tdiag(0r, Im−r)T−1;

b

P = Im−Q = Tb diag(Ir, 0m−r)T−1 (2.3)

It is known that for any α ∈ R such that (α A + B) is

nonsingular, one has

Rm=ker[(α A + B)−1A]k⊕im[(α A + B)−1A]k,

and bQis the projection onto ker[(α A+B)−1A]kalong the space

im[(α A + B)−1A]k(see [2]) In particular, bQdoes not depend

on the choice of W and T Let

A1= A − B bQ = Wdiag(Ir, U − Im−r)T−1

Since U is a nilpotent matrix, it is clear that A1is invertible

Further, by using(2.2)and(2.3)it follows that bP A−1A = bP

and bP A−11 B = bP A−11 B bP Multiplying both sides of(2.1)by b

P A−11 and bQ A−11 respectively we obtain

(P Xb )0(t) −bP A−11 B(P Xb )(t) = 0, b

Q A−11 A X0(t) −Q Ab −11 B X(t) = 0 (2.4) Also by the decomposition(2.2)and the definition(2.3)of bQ

we see that bQ A−11 A = Tdiag(0, U(U − Im−r)−1)T−1 is a nilpotent matrix and bQ A−11 B = Tdiag(0, (U − Im−r)−1)T−1 Write T−1X(t) =Y (t)

Z (t)

 where Z(t) ∈ Rm−r, we obtain

U Z0(t) − Z(t) = 0

It is easy to see that this equation has a unique solution Z(t) ≡

0 Therefore, if X(t) is a solution of(2.1)then b

Q X(t) = T diag(0, Im−r)T−1X(t)

=Tdiag(0, Im−r) Y(t)

0



=0

for any t > 0 This relation implies that the initial condition

X(0) = bP x0 must be satisfied Thus, we have the following definition of the stability

Definition 2.1 The trivial solution X ≡ 0 of (2.1)is said to

be asymptotically stable if there are positive constantsα, c such that the initial value problem

A X0(t) − B X (t) = 0, X(0) = bP x0

has a unique solution, and the estimate

kX(t)k 6 ckbP x0ke−αt, t > 0 holds

Let C+ = {z ∈ C : Rz > 0} and C− = C \ C+, where C is the complex plane We denote by σ (C, D) the spectrum of the pencil {C, D}, i.e., the set of all solutions of the equation det(λC − D) = 0 When C = I , we write simply σ (D) for σ (I, D) It is known that the equation Eq (2.1)is asymptotically stable iff all finite eigenvalues of { A, B} lie within C−(see [9]) In the case whereσ (A, B) = ∅,(2.1) has a unique solution X(t) ≡ 0 and we consider (2.1)to be asymptotically stable

3 Structured perturbations

As what is done in the ODE’s case (see [1,6–8]), fix a pencil of matrices { A, B} Assume that {A, B} is regular and asymptotically stable (i.e., (2.1)is asymptotically stable); let

a pair of the matrices E ∈ Km× p, F ∈ Kq×m be given We consider the perturbed system

where ∆ ∈ Kp×q The matrix E ∆F is called a structured perturbation of(2.1) Write

VK = {∆ ∈ Kp×q:(3.1)is either irregular or unstable}

We think of VK as the set of “bad” perturbations Throughout

this paper, all matrix norms are deduced from the vector norm

of the corresponding normed linear space Let

dK=inf{k∆k : ∆ ∈ VK}

We call dK the structured stability radius of the quadruple

{A, B, E, F} If K = C, we have a definition of the complex

stability radius and if K = R, dK is called the real stability

radius

First, we investigate the complex stability radius of the

system(2.1), i.e., K = C Denote

G(s) = F(s A − B)−1E

=F Tdiag(sIr −B1)−1, (sU − Im−r)−1W−1E, (3.2)

for s> 0

Theorem 3.1

(a) The complex stability radius of the system(2.1)is given by

dC=



sup

s∈iR

kG(s)k

−1

(b) There exists a “bad” matrix ∆ ∈ VK such that k∆k = dC

if and only if kG(s)k reaches its maximum value on iR

(c) When E = F = I (unstructured perturbations), dC> 0 if

and only ifind(A, B) 6 1

Proof (a) We follow similar arguments given in [6,7] Let

∆ ∈ VC Then, either the pencil of matrices { A, B + E∆F}

is irregular or it is regular but is not asymptotically stable In

both cases, we can always choose an eigenvalue s0∈σ (A, B +

E∆F)∩C+and an eigenvector x 6= 0 corresponding to s0, i.e.,

(s0A − B)x = E∆Fx This relation implies that

F x = F(s0A − B)−1E∆F x = G(s0)∆Fx

Since F x 6= 0,

k∆k > [kG(s0)k]−1

>

"

sup

s∈C +

kG(s)k

#−1

Therefore,

dC>

"

sup

s∈C +

kG(s)k

#−1

Conversely, take an ε > 0 and an s0 ∈ C+ such that

kG(s0)k−1

6 hsups∈C+kG(s)ki−1+ε Following the same

argument as in [6,7], we find a vector u ∈ Cp satisfying

kuk = 1 and kG(s0)uk = kG(s0)k Let y∗ be a linear

functional defined on Cqsuch that ky∗k =1 and y∗(G(s0)u) =

kG(s0)uk = kG(s0)k Put

∆ = kG(s0)k−1uy∗∈ Cp×q, x =(s0A − B)−1E u (3.4)

It is clear that

k∆k 6 kG(s0)k−1kukky∗k = kG(s0)k−1

and

∆G(s0)u = kG(s0)k−1uy∗G(s0)u

= kG(s0)k−1u kG(s0)k = u (3.5) Since u 6= 0,

k∆k > kG(s0)k−1 Combining these inequalities we obtain

k∆k = kG(s0)k−1 Further, from (3.4) and (3.5) it follows that (s0A − B − E∆F)x = 0, i.e., s0 ∈ σ (A, B + E∆F) which implies that the system

A X0(t) − (B + E∆F)X (t) = 0

is either unstable or irregular This means that ∆ ∈ VC Further,

dC6 k∆k = kG(s0)−1k6

"

sup

s∈C +

kG(s)k

#−1

+ε

Sinceε is arbitrary, dC6hsups∈C+kG(s)ki−1 Thus,

dC=

"

sup

s∈C +

kG(s)k

#−1

Note that the function G(s) is analytic on C+ By the maximum principle, max kG(·)k is reached only on the boundary of C+, i.e., on {∞} ∪ i R Therefore,

dC=

"

sup

s∈{∞}∪iR

kG(s)k

#−1

On the other hand, it is clear from(3.2)that lims→∞kG(s)k exists Thus, sups∈{∞}∪iRkG(s)k = sups∈iRkG(s)k, so (3.3) follows

(b) From the above argument, we see that if there exists an

s0 ∈ C+ such that kG(s0)k = [sups∈C+kG(s)k], then the matrix ∆ defined in(3.4)is a “bad” matrix and dC= k∆k When max kG(s)k does not reach its maximum on iR, i.e., kG(t)k < sups∈iRkG(s)k for any t ∈ iR, we show that no matrix ∆ exists such that dC = k∆k and the system

A X0−(B + E∆F)X = 0 is unstable Suppose, on the contrary, there was such a matrix ∆ Let s0 ∈ σ (A, B + E∆F) ∩ C+

and x 6= 0 be an eigenvector corresponding to s0, i.e., s0Ax − (B + E∆F)x = 0 Similar as above, we can prove that

k∆k > kG(s0)k−1 > [sups∈C+kG(s0)k]−1 = dC This is a contradiction

(c) Let E = F = I It is proved in part (a) that

dC=

 sup

s∈iR

kG(s)k

−1

,

where G(s) = (s A − B)−1 Suppose that ind(A, B) = k > 1 From(3.2)we get

G(s) = T (diag(s Ir, sU) − diag(B1, Im−r))−1W−1

=Tdiag (s Ir −B1)−1, −

k−1

X (sU)i

!

W−1

=Tdiag(sIr −B1)−1, 0m−r



W−1 +Tdiag 0r, −

k−1

X

i =0

(sU)i

!

W−1

It is easy to prove that lims→∞kTdiag((s Ir −B1)−1, 0)W−1k

=0 Therefore, kG(s)k tends to ∞ as s → ∞ which implies

that dC=0 The proof is complete 

Remark 3.2 Suppose that kG(s)k does not have a maximum

value on C+ Let (sn) ⊂ C+ be a sequence such that

limn→∞sn = ∞ and limn→∞kG(sn)k = sups∈C+kG(s)k

Let ∆n be given by(3.4)associated to sn Suppose, by using

a subsequence if necessary, limn→∞k∆nk = k∆0k = dC

From part (b) ofTheorem 3.1, we see that the system A X0−

(B + E∆0F)X = 0 is stable Since the set of matrices ∆ for

which { A, B + E∆F} is index-1 tractable is open, the index of

{A, B + E∆0F }must be higher than 1

We need the following simple lemma:

Lemma 3.3 Suppose that the bounded linear operator triplet:

M : X → Y, P : Y → Z , N : Z → X is given, where X, Y, Z

are Banach spaces Then the operator I − MPN is invertible if

and only if f I − PNM is invertible

Proof Suppose that I − MPN is invertible By direct

calculation, it is easy to verify that the inverse of I − PNM

is

(I − PNM)−1=I + PN(I − MPN)−1M

Furthermore, if (I − MPN)−1 is bounded then so is (I −

PNM)−1 The converse is proved similarly 

Suppose that (2.1) is index-1 tractable This assumption

implies that(A − BQb)−1A = bP and bQ = − bQ(A − BQb)−1B

(see [9]) Putting bB = bP(A − BQb)−1B = bP(A − BQb)−1B bP

we obtain

(A − BQ)b−1(λA − B) = λ(A − BQ)b−1A −(A − BQ)b−1B

=λP − bb Q(A − BQb)−1B − bP(A − BQb)−1B

=λP + bb Q − bB

=(P + bb Q/λ)(λI − (P + bb Q/λ)−1

b

B)

=(P + bb Q/λ)(λI − (P +b λQb)bB)

=(P + bb Q/λ)(λI −bB)

Hence,

(λA − B)−1 =(bP + bQ/λ)(λI −bB)−1

(A − BQb)−1

=(λI −bB)−1(P +b λQb)(A − BQb)−1

=(λI −bB)−1

b

P(A − BQb)−1+λ(λI −bB)−1

×Qb(A − BQb)−1

Since(λI −bB)Q =b λQ, we haveb λ(λI −bB)−1

b

Q = bQ and thus

(λA − B)−1 =(λI −bB)−1

b

P(A − BQb)−1

From the relation lim

λ→∞k(λI −bB)−1k =0,

it follows that the limit lim

λ→∞G(λ) = FQb(A − BQb)−1E := G(∞) exists

Theorem 3.4 Suppose that(2.1)is index-1 tractable

(1) If ∆ ∈ Cp×qwith k∆k< kG(∞)k−1then(3.1)is

index-1 tractable

(2) There exists ∆ ∈ VC with k∆k = kG(∞)k−1 such that {A, B + F∆F} is not index-1

Proof (1) If ∆ ∈ Cp×q satisfies k∆k < kG(∞)k−1 then k∆k · kF bQ(A − BQb)−1E k < 1 Using Banach’s lemma, it follows that the matrix I − F bQ(A − BQb)−1E∆ is invertible Moreover,

A −(B + E∆F)Q =b (A − BQb)

×



I −(A − BQ)b −1E∆F bQ Applying Lemma 3.3 with M = F bQ, P = I and N = (A − BQb)−1E∆ we see that I −(A − BQb)−1E∆F bQis also invertible This implies that A −(B + E∆F)Qbis invertible as well, i.e.,(3.1)is index-1 tractable

(2) Eq (3.1) is not index-1 tractable if I − ∆F bQ(A −

B bQ)−1E is singular Let u ∈ Cp satisfy kuk = 1 and

kF bQ(A − BQb)−1E uk = kF bQ(A − BQb)−1E k According

to the Hahn–Banach lemma, we can find a linear functional

x∗ defined on Cq such that kx∗k = 1 and x∗(FQb(A −

B bQ)−1E u) = kFQb(A − BQb)−1E uk Let ∆ = kF bQ(A −

B bQ)−1E k−1ux∗ It is clear that (I − ∆FQb(A − BQb)−1E)u

=(I − kFQb(A − BQb)−1E k−1ux∗(FQb(A − BQb)−1E))u

=u − kF bQ(A − BQb)−1E k−1ux∗(FQb(A − BQb)−1E u)

=u − u =0

This means that I − ∆F bQ(A − BQ)b−1E is singular On the other hand,

k∆k 6 kF bQ(A − BQb)−1E k−1kuk.kx∗k

6 kF bQ(A − BQb)−1E k−1, and

k∆(FQ(A − Bb Q)b−1E u)k = kFQ(A − Bb Q)b−1E k−1ku.x∗

×(FQb(A − BQb)−1E u)k = 1, which implies that

k∆k > kF bQ(A − BQb)−1E uk−1= kF bQ(A − BQb)−1E k−1 Thus,

k∆k = kF bQ(A − BQb)−1E k−1 The proof is complete 

Corollary 3.5 If ∆ ∈ Cp×q satisfies k∆k< dCthen(3.1)is

asymptotically stable and index-1 tractable

Example 1 Let us calculate the stability radius of the equation

under the structured perturbations A X0(t)−(B + E∆F)X (t) =

0 where ∆ is a perturbation and

A =













,

E =















It is easy to verify that ind(A, B) = 2 and σ (A, B) = −1

3 Therefore, the pencil { A, B} is asymptotically stable By the

direct computation, we obtain

G(s) = F(s A − B)−1E

=









s

3s + 1

s 3s + 1

s 3s + 1

s +1

3s + 1

s +1 3s + 1

s +1 3s + 1

s

s 3s + 1







 , Rs > 0

Let k · k3be the maximum norm of C3 We have kG(s)k∞ =

3 max{|3s+1s |, |s+1

3s+1|, | s

3s+1|}, where k · k∞ is the operator’s norm induced by k · k3 It is easy to see that kG(s)k∞attains

its maximum at s0 =0 and kG(0)k∞=3 Hence, dC=1/3

With u = (1, 1, 1)> we see that kuk3 = 1; kuk3 = 1 and

kG(0)uk3= kG(0)k∞=3 Further, let y∗=(0 1 0) and

∆ = kG(0)k−1uy∗=







,

then det(s A − B − E∆F) = 2s = 0 for s = 0 which implies

that ∆ ∈ VKand k∆k = 1/3

Example 2 Let us consider the equation A X0(t) − B X (t) = 0

where A = −12 −42and B = −21 −20 It is clear that ind

(A, B) = 1; σ (A, B) = −1 and G(s) = (s A − B)−1 =

s/(s + 1) 1/2

1/2 1/4



Let k · k2be the maximum norm of C2 We have

kG(s)k∞=max{|s/(s + 1)| + 1/2, 1/2 + 1/4}, where k · k∞

is the operator’s norm induced by k · k2 We see that kG(s)k∞

cannot reach its maximum on i R and lims→∞kG(s)k∞=3/2,

i.e., dC = 2/3 If we choose un = (n − i)/ p

n 2+1 1

 with

n ∈ N then kunk2 = 1 and kG(in)unk2 = kG(in)k∞ =

n/√n2+1 + 1/2 when n > 1 Thus, with y∗ = 1 0
we

have y∗(G(in)un) = kG(in)k∞and

∆n= kG(in)k−1

∞uny∗ converges to

2/3 0



as n → ∞ For any n ∈ N, it follows that n ∈ σ (A, B + ∆) which

implies that the system A X0(t) − (B + ∆n)X (t) = 0 is

not asymptotically stable However, it is easy to verify that

det(s A − B − ∆) = −8/3 for all s Hence, σ(A, B + ∆) = ∅ and the equation A X0(t) − (B + ∆)X (t) = 0; that is the system







−x01+2x20 −5

3x1+2x2=0, 2x10 −4x20 +4

3x1=0, has a unique solution x1 = 0; x2 =0, which is considered to

be asymptotically stable (seeRemark 3.2)

4 The equality of real and complex stability radii of DAEs

In this section, we consider the problem when the real and complex stability radii are equal It seems that this is a difficult problem in the DAEs because in this case, the positive cone Rm+

is no longer invariant under the action of the pencil of matrices {A, B}; even when both A and B are positive (see the definition below) However, we are able to answer this question under

a very strict assumption Firstly, we have the following result which provides a difference between ODEs and DAEs

We consider a differential algebraic equation with structured perturbations

where A, B are constant matrices in Rm×m, the pencil { A, B}

is regular; E ∈ Rm× p;F ∈ Rq×m Theorem 4.1 If kG(s)k does not reach its maximum on iR then dC=dR

Proof It is clear that dC 6 dR Therefore, it is sufficient to prove that there exists a sequence of disturbances(∆n) ⊂ VR

such that limn→∞k∆nk 6 dC Since ind(A, B) = k > 1, by using(3.2)we have

G(s) = FT diag(sIr−B1)−1, (sU − I )−1W−1E

= F Tdiag (s Ir−B1)−1, −

k−1

X

i =0

(sU)i

!

W−1E

= F Tdiag(sIr−B1)−1, −IW−1E

−F Tdiag 0r,

k−1

X

i =1

(sU)i

!

It is clear that lim

s→∞kF Tdiag(sIr−B1)−1, −IW−1E k

= kF Tdiag(0r, −I ) W−1E k, and

kF Tdiag 0r,k−1X

i =1

(sU)i

!

W−1E k

= |s| F Tdiag 0r,

k−1

X

si −1Ui

!

W−1E

Then, from(4.2)it follows that the limit lims→∞kG(s)k exists

and it is a finite number if and only if

kF Tdiag 0r,

k−1

X

i =1

(sU)i

!

W−1E k =0 Since kG(s)k does not reach its maximum on iR,

dC−1= sup

s∈iR

kG(s)k = lim

s→∞

If lims→∞kG(s)k = +∞ then dC=0 and we can choose ∆ =

0 to see that dC=dR=0 Suppose that lims→∞kG(s)k < ∞

By virtue of(4.3)it follows that dC−1 =limn∈N;n→∞kG(n)k

For any n ∈ N, let un ∈ Rp be a vector with kunk =

1; kG(n)unk = kG(n)k; let y∗

n be a linear functional defined

on Rq with kyn∗k = 1 and yn∗(G(n)un) = kG(n)unk as in

Section3 By denoting ∆n = kG(n)k−1un· yn∗ we see that

n is an eigenvalue of the pencil { A, B + E∆nF } with the

corresponding eigenvector xn = (n A − B)−1E un Therefore,

∆nis in VR Further, k∆nk = kG(n)k−1kun.y∗

nk6 kG(n)k−1

and k∆n(G(n)un)k = kG(n)k−1kun.y∗

n(G(n)un)k = kunk =

1 which implies that k∆nk = kG(n)k−1and limn→∞k∆nk =

limn→∞kG(n)k−1=dC This relation says that dR6 dC The

proof is complete 

When kG(s)k attains its absolute maximum at a finite point

in i R, we need further assumptions A matrix H = (αi j) ∈

Rm×m is said to be positive, written as H > 0, if αi j > 0 for

any i, j We define a partial order relation in Rm×mby

M > N ⇔ M − N > 0

We call the absolute value of a matrix M = (mi j) is the

matrix |M| = (|mi j|); similarly, for a vector x, we put |x| =

(|x1|, |x2|, , |xm|) Let µ(C, D) be the abscissa spectrum of

the pencil of matrices {C, D}, i.e., µ(C, D) := max{Rλ : λ ∈

σ (C, D)}

Hypothesis 4.2

(i) Assume that

(ii) There exists a sequence(tn); tn> 0; tn→ ∞such that

(tnA − B)−1

(iii) The system(4.1)is asymptotically stable

We remark that the above conditions ensure the positivity of

(4.1)in ODEs Indeed, suppose that A = I and the inequality

(4.5)holds for a sequence tn → ∞ From the relations

(tnI − B)−1 =tn−1(I − B/tn)−1

=tn−1(I + B/tn+o(1/tn)) > 0,

it follows that B is a Metlez matrix, i.e., all entries of B are

positive, except possibly those on the principle diagonal Thus

(4.1)is positive (see [7])

Lemma 4.3 Suppose that the system (4.1) satisfies the Hypothesis 4.2 For any λ ∈ C with Rλ > µ(A, B), the inequality

holds for any x ∈ Rm Proof Letλ = t + iω with t > µ(A, B) We show that

|[(t + iω)A − B]−1x |6 (t A − B)−1|x | for all x ∈ Rm Suppose that tn−t> µ(A, B) for any n > n0

By a simple calculation we have ((t + iω)A − B)−1=(tnA − B)−1

×hI −(tn−t − iω)A(tnA − B)−1i−1 Denoting R(tn) = (tnA − B)−1, we obtain [((t + iω)A − B)]−1=R(tn) [I − (tn−t − iω)AR(tn)]−1

=R(tn)

∞

X

k=0

(tn−t − iω)k(AR(tn))k (4.7)

The above series is absolutely convergent because

|(tn−t − iω)r(AR(tn))| < 1, for sufficiently n large, where r(M) denotes the spectral radius

of the matrix M Indeed, since limtn→∞(tn− |tn−t − iω|) = t,

we see that tn− |tn−t − iω| > t > µ(A, B) which implies that tn −µ(A, B) > |tn −t − iω| for n sufficiently large Furthermore, by theHypothesis 4.2 it follows that A R(tn) is

a positive matrix and then by the Perron–Frobenius theorem we have r(AR(tn)) = µ(AR(tn)) ∈ σ (AR(tn)) This means that det[r(AR(tn))I − AR(tn)] = 0 Hence,

det[r(AR(tn))I − AR(tn)] = 0

⇔det[(tnA − B) − A/r(AR(tn))] = 0

⇔det[(tn−1/r(AR(tn)))A − B] = 0

Thus, tn− 1

r (AR(t n )) ∈ σ(A, B) which implies that 1

r (AR(t n )) >

tn−µ(A, B), or equivalently, |tn−t − iω|r(AR(tn)) < 1 for

nsufficiently large

Since A> 0 and R(tn) > 0, from(4.7)it follows that

|((t + iω)A − B)−1x |

6 R(tn)

∞

X

k=0

|(tn−t − iω)|k(AR(tn))k|x |

=R(tn) [I − |tn−t − iω|AR(tn)]−1|x |

= [(tn− |tn−t − iω|)A − B]−1|x | for any x ∈ Rm Letting tn→ ∞, we obtain

|[(t + iω)A − B]−1x |6 (t A − B)−1|x | TheLemma 4.3is proved 

Suppose that the system(4.1)satisfies theHypothesis 4.2

We choose a monotone norm in Cm That is, if |x| 6 |y| then kxk 6 kyk (see [10, Chapter 5 Section 11]) Because

E > 0; F > 0, from(4.6)we get

|G(λ)x| = |F(λA − B)−1E x |6 F|(λA − B)−1(Ex)|

by (4.6)

6 F(RλA − B)−1|E x |6 F(RλA − B)−1E |x |

for anyλ ∈ C with Rλ > µ(A, B) and x ∈ Rm The relation

(4.8)implies

kG(λ)k = sup

x ∈R m ;kx k=1

kG(λ)xkR m

x ∈R m ;kx k=1

kG(Rλ)|x|kR m 6 kG(Rλ)k, provided Rλ > µ(A, B) Therefore,

sup

λ∈C +

kG(λ)k = sup

λ∈iR

kG(λ)k = kG(0)k = kFB−1E k

On the other hand,µ(A, B) < 0 then by using(4.8)withλ ∈

i R we see that 0 6 |G(λ)x| 6 G(0)|x| = −F B−1E |x |, ∀ x ∈

Rm This means that G(0) is a positive matrix Therefore, from

the Perron–Frobenius theorem, we can find u ∈ Rq;u >

0, kuk = 1 such that kG(0)uk = kG(0)k By virtue of the

Hahn–Banach theorem, there exists a positive linear functional

y∗satisfying y∗(G(0)u) = kG(0)uk = kG(0)k and ky∗k =1

Let ∆ = kG(0)k−1uy∗, it is easy to see that ∆ is a “bad” matrix

and k∆k = dC

Summing up, we obtain

Theorem 4.4 Suppose that (4.1)satisfies the Hypothesis 4.2

and a monotone norm in Rm is chosen The complex stability

radius dCand the real stability radius dRare equal and dR=

(kFB−1E k)−1

As is mentioned above, the assumption of the positivity of

(tnA − B)−1for a sequence(tn) tending to ∞ is strong and it

is difficult to verify it We now give a conditions to ensure the

above hypothesis to be satisfied Suppose that ind{ A, B} = 1

and let bQbe the projection on ker A given by(2.3)

It is said that the system(4.1)is positive if for any x0∈ Rm+,

the solution X(t) withPb(X (0)− x0) = 0 satisfies the condition

X(t) > 0 for all t > 0 From(2.4), if X(t) is a solution of(4.1)

then bQ X =0 which implies that X(t) =P Xb (t) for any t > 0

Therefore,(2.1)can be rewritten

where bB = bP(A − BQb)−1B bP

Theorem 4.5 The system(4.1)is positive if and only if bP> 0

and bB is a bP-Metlez matrix, i.e., all entries of bB are

non-negative, except possibly the entries bbi j with bpi j > 0, where

b

P =(bpi j)

Proof It is seen that (4.1) is positive if and only if (4.9) is

positive The general solution of(4.9)is

X(t) = exp(tbB)P xb 0

The positivity condition of(4.9)implies that X(0) = P xb 0> 0

if x0> 0, i.e., bP> 0 On the other hand,

exp(tbB)P =b I +

∞

X (tbB)n

! b

P = bP + t bB + o(t),

Suppose thatbpi j =0 From(4.10)we have

By dividing both sides of (4.11)by t and letting t → 0 we obtain bi j > 0

Conversely, if bB is a bP-Metlez matrix, then we can findα such thatαbP + bB > 0 In noting that the projection bPand the matrix bBare commutative we have

exp(tbB)P =b exp(−αtP + tb (bB +αPb))Pb

=exp(−αtP) exp(t(b bB +αbP))bP

SinceαP + bb B > 0 and bP > 0, it follows that exp(t bB)Pb> 0, or

X(t) = exp(tbB)P xb 0> 0, ∀x0> 0

The proof is complete  Lemma 4.6 If bB is a bP-Metlez matrix then (t I −bB)−1

b

P > 0, for any t > 0

Proof Since B is a bP-Metlez matrix, we can find α > 0 such that αP + bb B > 0 By the calculation we see that the matrix(α + t)bP + t bQis invertible and((α + t)bP + t bQ)−1= b

P/(α + t) +Qb/t Therefore, (t I −bB)−1

b

P = (t + α)P + t bb Q −(αP + bb B)
−1

b P

= (t + α)bP + t bQ
I − (t + α)P + t bb Q
−1

× (αbP + bB) −1Pb

=



I − (t + α)P + t bb Q
−1(αbP + bB)

× (t + α)bP + t bQ
−1Pb

t +α+

b Q t

! (αP + bb B)

!−1

b P

t +α+

b Q t

! b P

α + t



t +α(αP + bb B)

−1

b

P

Since αbP + bB > 0, I −t +α1 (αbP + bB)−1 > 0 Hence, (t I −bB)−1

b

P > 0 The proof is complete  Theorem 4.7 Suppose that the system (4.1) is positive and b

P(A − BQb)−1 > 0, bQ(A − BQb)−1 > 0 If E > 0 and

F> 0 then dC=dR Proof By a similar argument as in the proof ofLemma 4.3we obtain

|(λI −bB)−1

b

P x |6 (RλI − bB)−1

b

for any x ∈ Rm and any λ with Rλ > 0 Indeed, let λ =

t +iω, t > 0 Applying(4.7)with A = I, B = bB, tn =s ∈ R+

and bP x we have ((t + iω) −bB)
−1

b

P x = R(s)

∞

X (s − t − iω)k(R(s))k

b

P x,

where R(s) = (s I −bB)−1 Since bPand R(s) are commutative,

(R(s))k

b

P = (R(s)bP)k

b

P = Pb(R(s)Pb)k

b

P Therefore, by applying theLemma 4.6we get

((t + iω) −bB)
−1

b

P x

6R(s)bP

∞

X

k=0

|s − t − iω|k

×(R(s)bP)k

b P|x |

=R(s)(I − |s − t − iω|R(s))−1

b P|x |

=((s − |s − t − iω|)I −bB)−1

b P|x | Letting s → ∞ we obtain(4.12)

Further, from(3.6)we see that

G(λ) = F(λA − B)−1E

= F(λI −bB)−1

b P(A − BQ)b−1+Q(A − Bb Q)b−1E

Therefore,

|G(λ)x| = |F(λI −bB)−1

b

P(A − BQb)−1E x +F bQ(A − BQb)−1E x |

6 |F(λI − bB)−1

b

P(A − BQ)b−1E x | + |F bQ(A − BQb)−1E x |

6 F(RλI − bB)−1

b P(A − BQ)b−1E |x | +F bQ(A − BQb)−1E |x |

=G(Rλ)|x|

Hence, maxs∈iRkG(s)k = kG(0)k, i.e., kG(·)k has the

maximum value on i R at 0 Therefore, we can use the same

technique as above to find a “bad” real matrix ∆ with k∆k =

dC This means that dR=dC The theorem is proved 

Example Compute the stability radius of the system A X0(t) −

B X(t) = 0 with

A =















It is seen that ind(A, B) = 1; σ (A, B) = {−3/2 −

√

5/2; −3/2 +√5/2} and

b

P =















Thus bBis bP-Metlez and

b

Q(A − BQb)−1=







> 0;

b

P(A − BQb)−1=







> 0

Then G(s) > 0 for any s > 0 and kG(s)k, s ∈ C+attains the

maximum value at s0=0 with

G(0) =







 ; and kG(0)k = 5

Therefore, dR=dC=1/5 = k∆k where

∆ =









5 Conclusion

In this paper, we present a formula for the computation of the complex stability radius and obtain a characterization of the stability radius formula for the differential algebraic equation (see items (b) and (c) ofTheorems 3.1and4.1) We also provide some sufficient conditions under which the complex stability radius and the real stability radius are the same However, the additional assumptions in Theorem 4.7 are quite strict For example, there are many examples where the system (4.1)is positive while the resolvent (t A − B)−1is not So far we do not know whether the positive condition in (4.1) implies the equality of dCand dR An answer to this problem would be of great interest

Acknowledgment The author wishes to thank the referee for many suggestions which helped to improve the organization of the paper and several proofs

References

[1] A.V Bulatov, P Diamond, Real structural stability radius of infinite-dimensional linear systems: Its estimate, Automat Remote Control 63 (5) (2002) 713–722.

[2] S.L Campbell, Singular Systems of Differential Equations, Pitman Advanced Publishing Program, San Francisco, London, Melbourne, 1982 [3] V Dragan, The asymptotic behavior of the stability radius for a singularly perturbed linear system, Int J Robust Nonlinear Control 8 (1998) 817–829.

[4] M Hanke, E.I Macana, R M¨arz, On asymptotics in case of linear index

2 differential algebraic equations, SIAM J Numer Anal 35 (1) (1998) [5] M Hanke, Asymptotic expansions for regularization methods of linear fully implicit differential algebraic equations, Z Anal Ihre Anw 13 (1994) 513–535.

[6] D Hinrichsen, A.J Pritchard, Stability radii of linear systems, Syst Control Lett 7 (1986) 4–10.

[7] D Hinrichsen, A.J Pritchard, Stability radii for structured perturbations and the algebraic riccati equations linear systems, Syst Control Lett 8 (1986) 105–113.

[8] D Hinrichsen, N.K Son, Stability radii of positive discrete-time systems under affine parameter perturbations, Int J Robust Nonlinear Control 8 (1998) 1169–1188.

[9] R M¨arz, Extra-ordinary differential equation Attempts to an analysis of differential algebraic system, Progr Math 168 (1998) 313–334 [10] H.H Schaefer, Banach Latices and Positive Operators, Springer-Verlag, Berlin, Heidelberg, New York, 1974.

[11] P.J Psarrakos, M.J Tsatsomeros, On the stability radius of matrix polynomials, Linear Multilinear Algebra 50 (1) (2002) 151–165.

[8] D Hinrichsen, N.K Son, Stability radii. .. for the computation of the complex stability radius and obtain a characterization of the stability radius formula for the differential algebraic equation (see items (b) and (c) ofTheorems 3.1and4.1)... organization of the paper and several proofs

References

[1] A.V Bulatov, P Diamond, Real structural stability radius of infinite-dimensional linear systems: Its estimate, Automat