Systems & Control Letters 57 (2008) 546–553 www.elsevier.com/locate/sysconle Stability radii of differential algebraic equations with structured perturbations Nguyen Huu Du ∗ Faculty of Mathematics, Informatics and Mechanics, Vietnam National University, 334 Nguyen Trai, Thanh Xuan, Hanoi, Viet Nam Received March 2006; received in revised form 14 December 2007; accepted 14 December 2007 Available online February 2008 Abstract This paper deals with a formula for computing stability radii of a differential algebraic equation of the form AX (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 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 f (X (t), X (t)) = 0, (1.1) where the leading term X cannot be explicitly solved from X (t) The linear form of this equation is AX (t) − B X (t) = 0, (1.2) where A and B are two constant matrices (see [2,5]) ∗ Corresponding address: Faculty of Mathematics-Mechanics and Informatics, Hanoi National University, 334 Nguyen Trai, Thanh Xuan, Hanoi, Viet Nam Fax: +84 8588817 E-mail address: nhdu2001@yahoo.com 0167-6911/$ - see front matter c 2008 Elsevier B.V All rights reserved doi:10.1016/j.sysconle.2007.12.001 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 socalled 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 radius ρ, 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 547 N.H Du / Systems & Control Letters 57 (2008) 546–553 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−1 is positive and M = −A−1 B 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 4.4 and 4.7) It seems that this is a difficult problem in general, and so far we 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 iR, 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 deals with a formula for computing the complex stability radius of the system (1.1) where the structured perturbations are considered The Section is concerned with a special class of {A, B} where the complex and the real stability radii are equal Consider the differential algebraic equation: (2.1) 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) ≡ 0) and the index of {A, B} = k (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 = W diag(Ir , U )T −1 , B = W diag(B1 , Im−r )T −1 , (2.2) 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 Q = T diag(0r , Im−r )T −1 ; P = Im − Q = T diag(Ir , 0m−r )T −1 ( P X ) (t) − P A−1 B( P X )(t) = 0, Q A−1 AX (t) − Q A−1 1 B X (t) = (2.4) Also by the decomposition (2.2) and the definition (2.3) of Q −1 −1 is a we see that Q A−1 A = T diag(0, U (U − Im−r ) )T −1 −1 nilpotent matrix and Q A−1 B = T diag(0, (U − Im−r ) )T Write T −1 X (t) = Y (t) Z (t) where Z (t) ∈ Rm−r , we obtain U Z (t) − Z (t) = It is easy to see that this equation has a unique solution Z (t) ≡ Therefore, if X (t) is a solution of (2.1) then Q X (t) = T diag(0, Im−r )T −1 X (t) Y (t) = T diag(0, Im−r ) =0 for any t This relation implies that the initial condition X (0) = P x0 must be satisfied Thus, we have the following definition of the stability Definition 2.1 The trivial solution X ≡ of (2.1) is said to be asymptotically stable if there are positive constants α, c such that the initial value problem Preliminary AX (t) − B X (t) = 0, −1 and P A−1 B = P A1 B P Multiplying both sides of (2.1) by −1 P A−1 and Q A1 respectively we obtain (2.3) It is known that for any α ∈ R such that (α A + B) is nonsingular, one has Rm = ker[(α A + B)−1 A]k ⊕ im[(α A + B)−1 A]k , and Q is the projection onto ker[(α A+B)−1 A]k along the space im[(α A + B)−1 A]k (see [2]) In particular, Q does not depend on the choice of W and T Let A1 = A − B Q = W diag(Ir , U − Im−r )T −1 Since U is a nilpotent matrix, it is clear that A1 is invertible Further, by using (2.2) and (2.3) it follows that P A−1 A = P AX (t) − B X (t) = 0, X (0) = P x0 has a unique solution, and the estimate X (t) c P x0 e−αt , t 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) = 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 where σ (A, B) = ∅, (2.1) has a unique solution X (t) ≡ and we consider (2.1) to be asymptotically stable 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 AX (t) − (B + E∆F)X (t) = 0, (3.1) where ∆ ∈ K p×q The matrix E∆F is called a structured perturbation of (2.1) Write VK = {∆ ∈ K p×q : (3.1) is either irregular or unstable} 548 N.H Du / Systems & Control Letters 57 (2008) 546–553 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{ ∆ : ∆ ∈ VK } ∆G(s0 )u = G(s0 ) −1 uy ∗ G(s0 )u = G(s0 ) −1 u G(s0 ) = u (3.5) Since u = 0, 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)−1 E = F T diag (s Ir − B1 )−1 , (sU − Im−r )−1 W −1 E, for s and (3.2) ∆ G(s0 ) −1 Combining these inequalities we obtain ∆ = G(s0 ) −1 Further, from (3.4) and (3.5) it follows that (s0 A − B − E∆F)x = 0, i.e., s0 ∈ σ (A, B + E∆F) which implies that the system AX (t) − (B + E∆F)X (t) = is either unstable or irregular This means that ∆ ∈ VC Further, −1 Theorem 3.1 dC ∆ = G(s0 ) −1 (a) The complex stability radius of the system (2.1) is given by −1 dC = sup G(s) (3.3) 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 = corresponding to s0 , i.e., (s0 A − B)x = E∆F x This relation implies that F x = F(s0 A − B)−1 E∆F x = G(s0 )∆F x Since F x = 0, −1 [ G(s0 ) ] −1 sup G(s) s∈C+ Therefore, −1 sup G(s) s∈C+ Conversely, take an ε > and an s0 ∈ C+ such that −1 G(s0 ) −1 sups∈C+ G(s) + ε Following the same argument as in [6,7], we find a vector u ∈ C p satisfying u = and G(s0 )u = G(s0 ) Let y ∗ be a linear functional defined on Cq such that y ∗ = and y ∗ (G(s0 )u) = G(s0 )u = G(s0 ) Put ∆ = G(s0 ) −1 uy ∗ ∈ C p×q , x = (s0 A − B)−1 Eu (3.4) It is clear that ∆ G(s0 ) sups∈C+ G(s) Thus, −1 (b) There exists a “bad” matrix ∆ ∈ VK such that ∆ = dC if and only if G(s) reaches its maximum value on iR (c) When E = F = I (unstructured perturbations), dC > if and only if ind (A, B) dC −1 Since ε is arbitrary, dC s∈i R ∆ + ε sup G(s) s∈C+ dC = sup G(s) s∈C+ Note that the function G(s) is analytic on C+ By the maximum principle, max G(·) is reached only on the boundary of C+ , i.e., on {∞} ∪ iR Therefore, −1 dC = sup G(s) s∈{∞}∪i R On the other hand, it is clear from (3.2) that lims→∞ G(s) exists Thus, sups∈{∞}∪i R G(s) = sups∈i R G(s) , so (3.3) follows (b) From the above argument, we see that if there exists an s0 ∈ C+ such that G(s0 ) = [sups∈C+ G(s) ], then the matrix ∆ defined in (3.4) is a “bad” matrix and dC = ∆ When max G(s) does not reach its maximum on iR, i.e., G(t) < sups∈i R G(s) for any t ∈ iR, we show that no matrix ∆ exists such that dC = ∆ and the system AX − (B + E∆F)X = is unstable Suppose, on the contrary, there was such a matrix ∆ Let s0 ∈ σ (A, B + E∆F) ∩ C+ and x = be an eigenvector corresponding to s0 , i.e., s0 Ax − (B + E∆F)x = Similar as above, we can prove that ∆ G(s0 ) −1 > [sups∈C+ G(s0 ) ]−1 = dC This is a contradiction (c) Let E = F = I It is proved in part (a) that −1 dC = sup G(s) , s∈i R where G(s) = (s A − B)−1 Suppose that ind(A, B) = k > From (3.2) we get G(s) = T (diag(s Ir , sU ) − diag(B1 , Im−r ))−1 W −1 k−1 −1 u y ∗ = G(s0 ) −1 = T diag (s Ir − B1 )−1 , − (sU )i i=0 W −1 549 N.H Du / Systems & Control Letters 57 (2008) 546–553 = T diag (s Ir − B1 )−1 , 0m−r W −1 From the relation lim k−1 + T diag 0r , − (sU ) i W −1 λ→∞ it follows that the limit i=0 )−1 , 0)W −1 It is easy to prove that lims→∞ T diag((s Ir − B1 = Therefore, G(s) tends to ∞ as s → ∞ which implies that dC = The proof is complete Remark 3.2 Suppose that G(s) does not have a maximum value on C+ Let (sn ) ⊂ C+ be a sequence such that limn→∞ sn = ∞ and limn→∞ G(sn ) = sups∈C+ G(s) Let ∆n be given by (3.4) associated to sn Suppose, by using a subsequence if necessary, limn→∞ ∆n = ∆0 = dC From part (b) of Theorem 3.1, we see that the system AX − (B + E∆0 F)X = 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∆0 F} must be higher than 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)−1 M 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 − B Q)−1 A = P and Q = − Q(A − B Q)−1 B (see [9]) Putting B = P(A − B Q)−1 B = P(A − B Q)−1 B P we obtain (A − B Q)−1 (λA − B) = λ(A − B Q)−1 A − (A − B Q)−1 B = λ P − Q(A − B Q) −1 (λI − B)−1 = 0, −1 B − P(A − B Q) B lim G(λ) = F Q(A − B Q)−1 E := G(∞) λ→∞ exists Theorem 3.4 Suppose that (2.1) is index-1 tractable (1) If ∆ ∈ C p×q with ∆ < G(∞) −1 then (3.1) is index1 tractable (2) There exists ∆ ∈ VC with ∆ = G(∞) −1 such that {A, B + F∆F} is not index-1 Proof (1) If ∆ ∈ C p×q satisfies ∆ < G(∞) −1 then ∆ · F Q(A − B Q)−1 E < Using Banach’s lemma, it follows that the matrix I − F Q(A − B Q)−1 E∆ is invertible Moreover, A − (B + E∆F) Q = (A − B Q) × I − (A − B Q)−1 E∆F Q Applying Lemma 3.3 with M = F Q, P = I and N = (A − B Q)−1 E∆ we see that I − (A − B Q)−1 E∆F Q is also invertible This implies that A − (B + E∆F) Q is invertible as well, i.e., (3.1) is index-1 tractable (2) Eq (3.1) is not index-1 tractable if I − ∆F Q(A − B Q)−1 E is singular Let u ∈ C p satisfy u = and F Q(A − B Q)−1 Eu = F Q(A − B Q)−1 E According to the Hahn–Banach lemma, we can find a linear functional x ∗ defined on Cq such that x ∗ = and x ∗ (F Q(A − B Q)−1 Eu) = F Q(A − B Q)−1 Eu Let ∆ = F Q(A − B Q)−1 E −1 ux ∗ It is clear that (I − ∆F Q(A − B Q)−1 E)u = (I − F Q(A − B Q)−1 E = u − F Q(A − B Q)−1 E −1 −1 ux ∗ (F Q(A − B Q)−1 E))u ux ∗ (F Q(A − B Q)−1 Eu) = u − u = = λP + Q − B This means that I − ∆F Q(A − B Q)−1 E is singular On the other hand, = ( P + Q/λ)(λI − ( P + Q/λ)−1 B) = ( P + Q/λ)(λI − ( P + λ Q) B) ∆ = ( P + Q/λ)(λI − B) Hence, F Q(A − B Q)−1 E −1 F Q(A − B Q)−1 E −1 u x∗ , and (λA − B)−1 = ( P + Q/λ)(λI − B) −1 (A − B Q)−1 ∆(F Q(A − B Q)−1 Eu) = F Q(A − B Q)−1 E = (λI − B)−1 ( P + λ Q)(A − B Q)−1 = (λI − B) −1 P(A − B Q) −1 × Q(A − B Q) −1 −1 which implies that Since (λI − B) Q = λ Q, we have λ(λI − thus ∆ = Q and F Q(A − B Q)−1 Eu (3.6) −1 Thus, ∆ = F Q(A − B Q)−1 E (λA − B)−1 = (λI − B)−1 P(A − B Q)−1 + Q(A − B Q)−1 u.x ∗ × (F Q(A − B Q)−1 Eu) = 1, + λ(λI − B) B)−1 Q −1 The proof is complete −1 = F Q(A − B Q)−1 E −1 550 N.H Du / Systems & Control Letters 57 (2008) 546–553 Corollary 3.5 If ∆ ∈ C p×q satisfies ∆ < dC then (3.1) is asymptotically stable and index-1 tractable Example Let us calculate the stability radius of the equation under the structured perturbations AX (t)−(B + E∆F)X (t) = where ∆ is a perturbation and 1 −2 −1 B = −1 −1 , A = 0 1 , 0 −1 −1 1 1 0 E = 1 1 , F = 0 0 0 0 It is easy to verify that ind(A, B) = and σ (A, B) = − 31 Therefore, the pencil {A, B} is asymptotically stable By the direct computation, we obtain G(s) = F(s A − B)−1 E s s 3s + 3s + s+1 s+1 = 3s + 3s + s s − − 3s + 3s + s 3s + s+1 , 3s + s − 3s + Rs Let · be the maximum norm of C3 We have G(s) ∞ = s s+1 s max{| 3s+1 |, | 3s+1 |, | 3s+1 |}, where · ∞ is the operator’s norm induced by · It is easy to see that G(s) ∞ attains its maximum at s0 = and G(0) ∞ = Hence, dC = 1/3 With u = (1, 1, 1) we see that u = 1; u = and G(0)u = G(0) ∞ = Further, let y ∗ = (0 0) and 1/3 ∆ = G(0) −1 uy ∗ = 0 1/3 0 , 1/3 then det(s A − B − E∆F) = 2s = for s = which implies that ∆ ∈ VK and ∆ = 1/3 det(s A − B − ∆) = −8/3 for all s Hence, σ (A, B + ∆) = ∅ and the equation AX (t) − (B + ∆)X (t) = 0; that is the system −x1 + 2x2 − x1 + 2x2 = 0, 2x1 − 4x2 + x1 = 0, has a unique solution x1 = 0; x2 = 0, which is considered to be asymptotically stable (see Remark 3.2) 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 AX (t) − (B + E∆F)X (t) = 0, 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 then dC = dR G(s) = F T diag (s Ir − B1 )−1 , (sU − I )−1 W −1 E k−1 = F T diag (s Ir − B1 )−1 , − 2/3 ∆ := 2/3 0 −1 ∗ ∞ un y converges to as n → ∞ For any n ∈ N, it follows that n ∈ σ (A, B + ∆) which implies that the system AX (t) − (B + ∆n )X (t) = is not asymptotically stable However, it is easy to verify that (sU )i W −1 E i=0 (A, B) = 1; ∆n = G(in) G(s) does not reach its maximum on iR Proof It is clear that dC dR Therefore, it is sufficient to prove that there exists a sequence of disturbances (∆n ) ⊂ VR such that limn→∞ ∆n dC Since ind(A, B) = k 1, by using (3.2) we have Example Let us consider the equation AX (t) − B X (t) = −2 where A = −1 −4 and B = −2 It is clear that ind σ (A, B) = −1 and G(s) = (s A − B)−1 = s/(s + 1) 1/2 1/2 1/4 Let · be the maximum norm of C We have G(s) ∞ = max{|s/(s + 1)| + 1/2, 1/2 + 1/4}, where · ∞ is the operator’s norm induced by · We see that G(s) ∞ cannot reach its maximum on iR and lims→∞ G(s) ∞ = 3/2, i.e., dC = 2/3 If we choose u n = (n − i)/1 n + with n √ ∈ N then u n = and G(in)u n = G(in) ∞ = n/ n + + 1/2 when n > Thus, with y ∗ = we have y ∗ (G(in)u n ) = G(in) ∞ and (4.1) = F T diag (s Ir − B1 ) −1 , −I W −1 E k−1 − F T diag 0r , (sU )i W −1 E i=1 It is clear that lim s→∞ F T diag (s Ir − B1 )−1 , −I W −1 E = F T diag (0r , −I ) W −1 E , and k−1 F T diag 0r , (sU )i W −1 E i=1 k−1 = |s| F T diag 0r , s i−1 U i i=1 W −1 E (4.2) 551 N.H Du / Systems & Control Letters 57 (2008) 546–553 Then, from (4.2) it follows that the limit lims→∞ G(s) exists and it is a finite number if and only if k−1 F T diag 0r , (sU )i W −1 E = |(λA − B)−1 x| i=1 −1 = sup G(s) = lim G(s) dC (4.3) s→∞ (RλA − B)−1 |x| (4.6) holds for any x ∈ Rm Since G(s) does not reach its maximum on iR, s∈i R 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 Proof Let λ = t + iω with t > µ(A, B) We show that |[(t + iω)A − B]−1 x| (t A − B)−1 |x| If lims→∞ G(s) = +∞ then dC = and we can choose ∆ = to see that dC = dR = Suppose that lims→∞ G(s) < ∞ −1 = limn∈N;n→∞ G(n) By virtue of (4.3) it follows that dC p For any n ∈ N, let u n ∈ R be a vector with u n = 1; G(n)u n = G(n) ; let yn∗ be a linear functional defined on Rq with yn∗ = and yn∗ (G(n)u n ) = G(n)u n as in Section By denoting ∆n = G(n) −1 u n · yn∗ we see that n is an eigenvalue of the pencil {A, B + E∆n F} with the corresponding eigenvector xn = (n A − B)−1 Eu n Therefore, ∆n is in VR Further, ∆n = G(n) −1 u n yn∗ G(n) −1 −1 ∗ and ∆n (G(n)u n ) = G(n) u n yn (G(n)u n ) = u n = which implies that ∆n = G(n) −1 and limn→∞ ∆n = limn→∞ G(n) −1 = dC This relation says that dR dC The proof is complete for all x ∈ Rm Suppose that tn − t > µ(A, B) for any n By a simple calculation we have When G(s) attains its absolute maximum at a finite point in iR, we need further assumptions A matrix H = (αi j ) ∈ Rm×m is said to be positive, written as H 0, if αi j for m×m any i, j We define a partial order relation in R by 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 the Hypothesis 4.2 it follows that A R(tn ) is a positive matrix and then by the Perron–Frobenius theorem we have r (A R(tn )) = µ(A R(tn )) ∈ σ (A R(tn )) This means that det[r (A R(tn ))I − A R(tn )] = Hence, M N ⇔M−N We call the absolute value of a matrix M = (m i j ) is the matrix |M| = (|m i 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 A 0, 0, E F (4.4) (ii) There exists a sequence (tn ); tn > 0; tn → ∞ such that (tn A − B)−1 for all n (4.5) n0 ((t + iω)A − B)−1 = (tn A − B)−1 × I − (tn − t − iω)A(tn A − B)−1 −1 Denoting R(tn ) = (tn A − B)−1 , we obtain [((t + iω)A − B)]−1 = R(tn ) [I − (tn − t − iω)A R(tn )]−1 ∞ = R(tn ) (tn − t − iω)k (A R(tn ))k (4.7) k=0 The above series is absolutely convergent because |(tn − t − iω)r (A R(tn ))| < 1, det[r (A R(tn ))I − A R(tn )] = ⇔ det[(tn A − B) − A/r (A R(tn ))] = ⇔ det[(tn − 1/r (A R(tn )))A − B] = 1 ∈ σ (A, B) which implies that r (A R(t Thus, tn − r (A R(t n )) n )) tn − µ(A, B), or equivalently, |tn − t − iω|r (A R(tn )) < for n sufficiently large Since A and R(tn ) 0, from (4.7) it follows that |((t + iω)A − B)−1 x| ∞ R(tn ) |(tn − t − iω)|k (A R(tn ))k |x| k=0 (iii) The system (4.1) is asymptotically stable = R(tn ) [I − |tn − t − iω|A R(tn )]−1 |x| 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 = [(tn − |tn − t − iω|)A − B]−1 |x| (tn I − 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]) for any x ∈ Rm Letting tn → ∞, we obtain |[(t + iω)A − B]−1 x| (t A − B)−1 |x| The Lemma 4.3 is proved Suppose that the system (4.1) satisfies the Hypothesis 4.2 We choose a monotone norm in Cm That is, if |x| |y| then x y (see [10, Chapter Section 11]) Because E 0; F 0, from (4.6) we get 552 N.H Du / Systems & Control Letters 57 (2008) 546–553 F|(λA − B)−1 (E x)| |G(λ)x| = |F(λA − B)−1 E x| by (4.6) F(RλA − B)−1 |E x| = G(Rλ)|x| as t → Suppose that pi j = From (4.10) we have F(RλA − B)−1 E|x| (4.8) for any λ ∈ C with Rλ > µ(A, B) and x ∈ Rm The relation (4.8) implies sup G(λ) = x∈Rm ; x =1 sup x∈Rm ; x =1 G(λ)x Rm G(Rλ)|x| Rm G(Rλ) , λ∈i R On the other hand, µ(A, B) < then by using (4.8) with λ ∈ iR we see that |G(λ)x| G(0)|x| = −F B −1 E|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, u = such that G(0)u = G(0) By virtue of the Hahn–Banach theorem, there exists a positive linear functional y ∗ satisfying y ∗ (G(0)u) = G(0)u = G(0) and y ∗ = Let ∆ = G(0) −1 uy ∗ , it is easy to see that ∆ is a “bad” matrix and ∆ = 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 dC and the real stability radius dR are equal and dR = ( F B −1 E )−1 As is mentioned above, the assumption of the positivity of (tn A − B)−1 for 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} = and let Q be 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) with P(X (0) − x0 ) = satisfies the condition X (t) for all t > From (2.4), if X (t) is a solution of (4.1) then Q X = which implies that X (t) = P X (t) for any t Therefore, (2.1) can be rewritten X (0) = P x0 , (4.11) By dividing both sides of (4.11) by t and letting t → we obtain bi j Conversely, if B is a P-Metlez matrix, then we can find α such that α P + B In noting that the projection P and the matrix B are commutative we have Since α P + B or sup G(λ) = sup G(λ) = G(0) = F B −1 E X − B X = 0; tbi j + o(t)i j exp(t B) P = exp(−αt P + t ( B + α P)) P = exp(−αt P) exp(t ( B + α P)) P provided Rλ > µ(A, B) Therefore, λ∈C+ (4.10) (4.9) and P X (t) = exp(t B) P x0 0, it follows that exp(t B) P 0, 0, ∀ x0 The proof is complete Lemma 4.6 If B is a P-Metlez matrix then (t I − B)−1 P 0, for any t > Proof Since B is a P-Metlez matrix, we can find α > such that α P + B By the calculation we see that the matrix (α + t) P + t Q is invertible and ((α + t) P + t Q)−1 = P/(α + t) + Q/t Therefore, (t I − B)−1 P = (t + α) P + t Q − (α P + B) (t + α) P + t Q = × (α P + B) I − (t + α) P + t Q −1 P −1 −1 P = I − (t + α) P + t Q × (t + α) P + t Q −1 −1 (α P + B) P −1 = I− = α+t P Q + t +α t (α P + B) t +α I− Since α P + B 0, P Q + t +α t (α P + B) I− P −1 P t+α (α P −1 + B) Hence, where B = P(A − B Q)−1 B P (t I − Theorem 4.5 The system (4.1) is positive if and only if P and B is a P-Metlez matrix, i.e., all entries of B are nonnegative, except possibly the entries bi j with pi j > 0, where P = ( pi j ) Theorem 4.7 Suppose that the system (4.1) is positive and P(A − B Q)−1 0, Q(A − B Q)−1 If E and F then dC = dR 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(t B) P x0 The positivity condition of (4.9) implies that X (0) = P x0 if x0 0, i.e., P On the other hand, B)−1 P The proof is complete Proof By a similar argument as in the proof of Lemma 4.3 we obtain (t B)n I+ n=1 P = P + t B + o(t), (4.12) for any x ∈ and any λ with Rλ > Indeed, let λ = t +iω, t > Applying (4.7) with A = I, B = B, tn = s ∈ R+ and P x we have Rm ∞ exp(t B) P = (RλI − B)−1 P|x|, |(λI − B)−1 P x| ((t + iω) − B) −1 ∞ (s − t − iω)k (R(s))k P x, P x = R(s) k=0 553 N.H Du / Systems & Control Letters 57 (2008) 546–553 where R(s) = (s I − B)−1 Since P and R(s) are commutative, (R(s))k P = (R(s) P)k P = P(R(s) P)k P Therefore, by applying the Lemma 4.6 we get ((t + iω) − B) −1 ∞ Px |s − t − iω|k R(s) P k=0 × (R(s) P)k P|x| = R(s)(I − |s − t − iω|R(s))−1 P|x| = ((s − |s − t − iω|)I − B)−1 P|x| Letting s → ∞ we obtain (4.12) Further, from (3.6) we see that G(λ) = F(λA − B)−1 E = F (λI − B)−1 P(A − B Q)−1 + Q(A − B Q)−1 E Therefore, |G(λ)x| = |F(λI − B)−1 P(A − B Q)−1 E x + F Q(A − B Q)−1 E x| |F(λI − B)−1 P(A − B Q)−1 E x| + |F Q(A − B Q)−1 E x| −1 F(RλI − B) P(A − B Q)−1 E|x| + F Q(A − B Q)−1 E|x| = G(Rλ)|x| Hence, maxs∈i R G(s) = G(0) , i.e., G(·) has the maximum value on iR at Therefore, we can use the same technique as above to find a “bad” real matrix ∆ with ∆ = dC This means that dR = dC The theorem is proved Example Compute the stability radius of the system AX (t) − B X (t) = with 0 −2 A = 0 0 , B = −1 0 0 −1 It √ is seen that√ind(A, B) = 1; σ (A, B) = {−3/2 − 5/2; −3/2 + 5/2} and 0 −2 P = 0 0 ; B = −1 0 0 0 0 Thus B is P-Metlez and 0 Q(A − B Q)−1 = 0 0 0 0 P(A − B Q)−1 = 0 1 0 0; Then G(s) for any s > and G(s) , s ∈ C+ attains the maximum value at s0 = with G(0) = 1 2 ; and G(0) = Therefore, dR = dC = 1/5 = ∆ where 1/5 ∆ = 0 1/5 0 1/5 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) of Theorems 3.1 and 4.1) 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)−1 is not So far we not know whether the positive condition in (4.1) implies the equality of dC and 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 infinitedimensional 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 (1998) 817829 [4] M Hanke, E.I Macana, R Măarz, On asymptotics in case of linear index differential algebraic 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151–165 ... 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) of Theorems 3.1... Stability radii for structured perturbations and the algebraic riccati equations linear systems, Syst Control Lett (1986) 105–113 [8] D Hinrichsen, N.K Son, Stability radii of positive discrete-time... regular (that is, f (λ) = det(λA − B) ≡ 0) and the index of {A, B} = k (see [2]) The Kronecker decomposition of the pencil of matrices {A, B} indicates that there exists a pair of nonsingular matrices