A Bound on the Real Stability Radius of Continuous-Time Linear Infinite-Dimensional Systems

Computational Mathematics and Modeling, Vol. 12, No. 4, 2001

A BOUND ON THE REAL STABILITY RADIUS OF CONTINUOUS-TIME LINEAR INFINITE-DIMENSIONAL SYSTEMS N. A. Bobylev and A. V. Bulatov The article examines bounds on disturbance norms under which a linear infinite-dimensional dynamical system remains stable.

1. Introduction Issues of robustness of various systems arise in numerous problems of control theory. Mathematical formalization of these issues leads to stability problems not for specific systems, but for whole families of systems (absolute stability problems, analysis of the Hurwitz property for families of polynomials, etc.). The robustness theory has been developed in sufficient detail for concentrated-parameter systems [1–15]. Basic theoretical results are available and huge experimental material has been accumulated. At the same time, the robustness theory for distributed-parameter systems is still in an embryonic stage, very far from completion. In this article we present bounds on disturbances of linear bounded stable operators acting in a real separable Hilbert space H that leave the operators in their stability class. The results are effective and can be applied to investigate the robustness of various distributed-parameter systems (for instance, systems described by differential, integral, and integro-differential equations). 2. Statement of the Problem Let a linear bounded operator A be given in a separable Hilbert space H. Assume that the operator A is stable, i.e., the spectrum of this operator is in the half-plane {Re z < 0}. It is required to find a positive number ∆ with the following property: for every linear bounded operator B satisfying the inequality BF < ∆, the operator A + B is stable. Here  F is the norm in the space of linear operators to be defined below (it is an analogue of the Frobenius matrix norm). 3. Main Results Below we use standard notation: L(H) is the set of all linear bounded operators acting in the space H, ρ(A) is the resolvent set of the operator A ∈ L(H), and σ(A) is the spectrum of the operator A. Let µ(A) = max Re λ. λ∈σ(A)

Take an arbitrary orthonormal basis {ek }∞ 1 and define    ∞ (Bek , el )2 . BF =  k,l=1

Translated from Nelineinaya Dinamika i Upravlenie, Issue 1, pp. 77–86, 2001. c 2001 1046–283X/01/1204–0359 $ 25.00


Plenum Publishing Corporation

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N. A. B OBYLEV AND A. V. B ULATOV

In what follows, this norm is used to bound the disturbances of operator systems. We know [16] that BF < ∞ if and only if B is a Hilbert–Schmidt operator. Therefore the proposed approach somewhat restricts the class of admissible disturbances. In practice, however, this restriction is quite acceptable. Theorem 1. For every operator A ∈ L(H) we have the equality µ(A) =

inf

sup (T −1 AT x, x).

(1)

T ∈L(H) x=1 0∈ρ(T )

Proof is given in the Appendix. Take an arbitrary number εA > 0 and choose the operator T0 so that µ(A)
sup (T0−1 AT0 x, x)
µ(A) + εA . x=1

Let C = (T0−1 )∗ T0−1 ,

m = inf (Cx, x), x=1

M = sup (Cx, x), x=1

κ=

M . m

Theorem 2. Let √ 2(µ(A) + εA ) κ . ∆=− 1+κ

(2)

Then the inequality BF < ∆ implies stability of the operator A + B. Proof

is given in the Appendix.

The bound of the disturbance norm given by Theorem 2 suffers from an essential shortcoming. If the minimum in (1) is attained for T = T∗ , we may take T0 = T∗ , εA = 0. In most cases, however, the minimum in (1) is not attained for any T . Examples show that in such cases we should not make εA too small, because κ is large for small εA . Let us accordingly consider the important case when the minimum in (1) is not attained. The eigenvectors corresponding to the eigenvalues λ1 , λ2 , . . ., µ1 ± iν1 , µ2 ± iν2 , . . . of the operator A are denoted by x1 , x2 , . . ., y1 ± iz1 , y2 ± iz2 , . . ., respectively. Use the vectors xk , yl , zm to form the system {sk }∞ 1 , indexing it in an arbitrary order. Let yk = sαk (k  1), zk = sβk (k  1), W is the set of all natural numbers that are not elements of the sequences {αk } and {βk }. Recall that the system {ξk } is called a Riesz basis of the space H if for some T ∈ L(H), 0 ∈ ρ(T ) the system {T ξk } is an orthonormal basis in H. Theorem 3. If the system {sk } is a Riesz basis in H, then µ(A) = sup (T AT −1 x, x), x=1

where T is the operator entering the definition of Riesz basis.

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Proof is given in the Appendix. In practice, instead of checking that the real and imaginary parts of the eigenvectors form a Riesz basis in H it is better to check that the full eigenvectors form a Riesz basis in the complexified space Hc = {x+iy|x, y ∈ H}. Let us state some associated propositions. Proposition 1. Let {ξk } be an orthonormal basis in Hc . Then the operator B uniquely defined by the equalities Bξαk = ξαk + iξβk

(k  1),

Bξβk = ξαk − iξβk

(k  1),

(k ∈ W )

Bξk = ξk is bounded and 0 ∈ ρ(B).

Proposition 2. Assume that the system {tk } defined by the equalities tαk = sαk + isβk

(k  1),

tβk = sαk − isβk

(k  1),

tk = sk

(k ∈ W )

forms a Riesz basis in Hc . Then the system {sk } is a Riesz basis in H, and the operator T entering the definition of Riesz basis has the form U BTc , where Tc is the operator entering the definition of the Riesz basis for the system {tk }, B is the operator constructed from the orthonormal basis {Tc tk } in accordance with the condition of Proposition 1, and U is the unitary operator transforming the orthonormal basis {Tc tk } into some real orthonormal basis. Appendix Proof of Theorem 1. We will first prove a number of lemmas. Lemma A.1 . Let A ∈ L(H), m = inf (Ax, x), x=1

M = sup (Ax, x). x=1

If λ ∈ σ(A), then Re λ ∈ [m, M ].

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Proof. In the complex Hilbert space Hc = {x + iy|x, y ∈ H} define the operator Ac that acts according to the rule Ac (x + iy) = Ax + iAy. By definition, σ(Ac ) = σ(A). Direct calculations show that inf

z=1 z∈Hc

Re (Ac z, z) = m,

sup Re (Ac z, z) = M. z=1 z∈Hc

Take an arbitrary ε > 0. We will show that all the points in the half-plane Q = {λ ∈ C | Re λ > M + ε} are contained in ρ(Ac ). To prove this assertion, we use the method of continuation in a parameter [17]. Since λ ∈ ρ(Ac ) at |λ| > Ac , it suffices to prove the inequality (Ac − λE)z  γ(ε)z

(z ∈ Hc , λ ∈ Q),

(A.1)

where γ(ε) is a positive constant. Let λ = λ1 + iλ2 , w = w1 + iλ2 (λ1 , λ2 , w1 ∈ R). Then (Ac − λE)z2 = Ac z − wz2 − 2 Re (Ac z − wz, (λ − w)z) + |λ − w|2 z2  (λ1 − w1 )(−2 Re (Ac z, z) + (λ1 + w1 )z2 ). Setting w1 = M + ε/2, we obtain 3 (Ac − λE)z2  ε2 z2 . 4 We have proved the bound (A.1). We can use the same method to prove that all the points of the half-plane {λ ∈ C | Re λ < m − ε} are regular points of the operator Ac . Q.E.D. Corollary. For every operator A ∈ L(H), µ(A)


inf

sup (T −1 AT x, x).

T ∈L(H) x=1 0∈ρ(T )

Lemma A.2 . If the scalar product (, )∗ defined in H induces a norm equivalent to the original norm, then there exists an operator T ∈ L(H) such that 0 ∈ ρ(T ) and (x, y)∗ = (T x, T y) (x, y ∈ H). Proof. For every fixed x, the functional f (y) = (y, x)∗ is a linear bounded functional on H. By the Riesz theorem, the functional f is representable in the form f (y) = (y, H(x)). A direct check√will show that the operator H is linear, continuous, selfadjoint and positive definite. Then the operator T = H satisfies the conditions of Lemma A.2. Q.E.D.

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Lemma A.3 . If for some B ∈ L(H) we have the inequality r(B) < 1, then there exists an operator T ∈ L(H) such that 0 ∈ ρ(T ) and T BT −1  < 1. Proof.

By Gel’fand’s formula (see, e.g., [16]), r(B) = lim

n→∞

 n B n .

Therefore the power series 

B n tn

n
0

converges in the disk {|t| < 1/r(B)}. In particular, the series 

B n ,

n
0



B n 2

n
0

are convergent. We denote by M 2 the sum of the last series. In the space H define the scalar product (x, y)B =

∞ 

(B n x, B n y)

(x, y ∈ H).

n=0

A direct calculation shows that x
xB
M x

(x ∈ H).

Therefore B2B = sup (Bx, Bx)B = sup ((x, x)B − (x, x)) xB =1

=1−

xB =1

inf (x, x)
1 −

xB =1

1 < 1. M2

Lemma A.2 implies the existence of an operator T ∈ L(H) such that 0 ∈ ρ(T ) and (x, y)B = (T x, T y) (x, y ∈ H). Then T BT −1 2 = sup (T BT −1 y, T BT −1 y) (y,y)=1

=

sup (T x,T x)=1

Q.E.D.

(T Bx, T Bx) =

sup (x,x)B =1

(Bx, Bx)B = B2B < 1.

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N. A. B OBYLEV AND A. V. B ULATOV

Lemma A.4 . If µ(A) < 0, then there exists an operator T ∈ L(H) such that 0 ∈ ρ(T ) and (T AT −1 x, x)
0 for all x ∈ H. Proof. Since 1 ∈ σ(A), the operator B = (I − A)−1 (I + A) is defined, where I is the identity operator. The spectrum of the operator B is contained in the disk {|z| < 1} (see, e.g., [19]). Here A = (B − I)(I + B)−1 . By Lemma A.3, there exists an operator T ∈ L(H) such that 0 ∈ ρ(T ) and T BT −1  < 1. Let G = T BT −1 . We have T AT −1 = T (B − I)T −1 T (I + B)−1 T −1 = (G − I)(I + G)−1 . Therefore, for x = 0, (T AT −1 x, x) = ((G − I)(I + G)−1 x, x) = ((G − I)y, (I + G)y)   = (Gy − y, y + Gy) = (Gy, Gy) − (y, y)
G2 − 1 y
0 (here y = (I + G)−1 x). Q.E.D. Let us now proceed to prove Theorem 1. Let m = µ(A). For all ε > 0 we have µ(A − (m + ε)I) < 0. By Lemma A.4, for every ε > 0 there exists an operator Tε such that sup (Tε (A − (m + ε)I)Tε−1 x, x)
0.

x=1

Hence sup (Tε ATε−1 x, x)
m + ε.

x=1

(A.2)

Since ε in inequality (A.2) is arbitrarily small, equality (1) follows by Lemma A.1. Q.E.D. Proof of Theorem 2. By Theorem 1, µ(A + B)
sup (T0−1 (A + B)T0 x, x) x=1


sup (T0−1 AT0 x, x) + sup (T0−1 BT0 x, x) x=1

x=1


µ(A) + εA + sup (T0−1 BT0 x, x). x=1

(A.3)

Let ν(ε) be the maximum in the following constrained maximization problem:  (T0−1 BT0 x, x) → max,      BF
ε,      x = 1.

(A.4)

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From (A.3) it follows that for B∗
ε, µ(A + B)
µ(A) + εA + ν(ε).

(A.5)

Let us derive an upper bound for ν(ε). Substituting y = T0 x, we reduce the problem (A.4) to the equivalent problem  (CBy, y) → max,      BF
ε,      (Cy, y) = 1. The maximum in (A.6) depends on C. Indeed, substituting z =

√

(A.6)

Cy, we obtain the problem

√  √ CB C −1 z, z) → max, (      BF
ε,      (z, z) = 1. √ √ The continuous dependence of C and C −1 on C is proved, e.g., in [19]. From the spectral expansion theorem for a bounded symmetrical operator [18] we obtain that the operator C can be approximated with any accuracy by an operator of the form N 

λ k Pk ,

k=1

where λk ∈ [m, M ], Pk are pairwise orthogonal orthoprojectors, and N 

Pk = E.

k=1

Let us investigate problem (A.6) for operators C of this form. In this case, there exists an orthonormal basis {fk }∞ 1 that consists of the eigenvectors of the operator C: Cfk = dk fk ,

dk ∈ {λ1 , λ2 , . . ., λN }.

Define the unitary operator U that takes the orthonormal basis {ek } into the orthonormal basis {fk } and the operator D acting by the rule Dek = dk ek (k  1). Then C = U DU ∗ . Introducing the new variables w = U ∗ y,

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N. A. B OBYLEV AND A. V. B ULATOV

V = U ∗ BU, we reduce problem (A.6) to the form  (DV w, w) → max,      V F
ε,      (Dw, w) = 1.

(A.7)

Let Rn be the orthogonal projector on the subspace span (e1 , e2 , . . ., en ). If V F
ε, then Rn V

Rn 2F

n 

=

(V el , em )2
ε2 .

l,m=1

Moreover, lim V − Rn V Rn  = 0

n→∞

(see, e.g., [18]). We can therefore first investigate problem (A.7) subject to the additional constraints V ek ∈ span (e1 , e2 , . . ., en ) (1
k
n), V ek = 0 (k > n) and then find the maximum over n. Let us pass to coordinate notation. Let

w=

∞ 

wk ek ,

k=1

V ek =

n 

vlk el

(1
k
n).

l=1

In this notation, problem (A.7) takes the form  n     dl vlk wk wl → max,     k,l=1      n   |vkl |2
ε2 ,   k,l=1       ∞     dk |wk |2 = 1.   k=1

Since the maximum in problem (A.8) is nonnegative, it is attained for n  k,l=1

|vkl |2 = ε2

(A.8)

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and wn+1 = wn+2 = . . . = 0. Thus, problem (A.8) is equivalent to the problem  n     dl vlk wk wl → max,     k,l=1      n   |vkl |2 = ε2 ,   k,l=1      n      dk |wk |2 = 1.  

(A.9)

k=1

Problem (A.9) has been investigated in [20]. It has been shown that the maximum ν˜(ε) in this problem satisfies the inequality ε(di + dj ) ε(1 + κ0 ) 
, √ 1i, j n 2 di dj 2 κ0

ν˜(ε)
max

(A.10)

where max dk

κ0 =

1kn

min dk

.

1kn

By construction, κ0


M = κ. m

Therefore, the solution of problem (A.4) satisfies the bound ν(ε)


ε(1 + κ) √ . 2 κ)

(A.11)

The assertion of Theorem 2 now follows from (A.5) and (A.11). Indeed, let BF = β < ∆. Then µ(A + B)
µ(A) + εA +

β(1 + κ) ∆(1 + κ) √ √ < µ(A) + εA + = 0. 2 κ 2 κ

Thus, the operator A + B is stable. Q.E.D. Proof of Theorem 3. Lemma A.5 . If the system {sk } forms an orthonormal basis in H, then µ(A) = sup (Ax, x). x=1

(A.12)

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Proof. If the system {sk } is an orthonormal basis in H, then the space H is representable as the direct sum of the subspaces Hk , k  1, dim Hk
2, that are invariant with respect to A, and the spectra of the restrictions Ak of the operator A on Hk consist either of a single real eigenvalue or of a pair of complex conjugate eigenvalues. The equality (A.12) is easily checked for the operators Ak . Then the equality 

sup (Ax, x) = sup k
1

x=1

sup (Ax, x) x=1 x∈Hk

leads to equality (A.12) for the operator A. Q.E.D. To prove Theorem 3, it suffices to apply Lemma A.5 to the operator T AT −1 and use the equality σ(A) = σ(T AT −1 ). This work has been supported by the Russian Foundation for Basic Research, grants 99–01–00883 and 98–01–00586. REFERENCES 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

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