ROBUST STABILITY VIA H∞ METHODSc⃝ Alberto Isidori , 2011

1 The L2 gain of a stable linear system

Consider a linear system described by equations of the form

x = Ax+Buy = Cx+Du

(1)

with state x ∈ Rn, input u ∈ Rm, output y ∈ Rp.Let γ > 0 be a fixed number. Suppose there exists a positive definite symmetrix n × n

matrix P such that the quadratic function V (x) = xTPx satisfies, some ϵ > 0, the inequality1

∂V

∂x(Ax+Bu) ≤ −ϵ∥x∥2 + γ2∥u∥2 − ∥Cx+Du∥2 (2)

for all x ∈ Rn and all u ∈ Rm. Note, to begin with, that this property can only hold if thesystem is asymptotically stable. For u = 0 this reduces in fact to

∂V

∂xAx ≤ −ϵ∥x∥2

which implies asymptotic stability.Suppose the input u(·) of (1) is a function in L2[0,∞).2 Integration of the inequality (2)

on the interval [0, T ] yields, for any initial state x(0),

V (x(T )) ≤ V (x(0)) + γ2∫ T

0∥u(t)∥2dt−

∫ T

0∥y(t)∥2dt

≤ V (x(0)) + γ2∫ ∞

0∥u(t)∥qdt ,

from which it is deduced that the response x(t) of the system is defined for all t ∈ [0,∞) andbounded. Now, suppose x(0) = 0 and observe that the previous inequality yields

V (x(T )) ≤ γ2∫ T

0∥u(t)∥2dt−

∫ T

0∥y(t)∥2dt

1A inequality of this kind is often called a dissipation inequality.2The space L2[0,∞) is the space of all piecewise-continuous inputs defined on [0,∞) satisfying∫ ∞

0

∥u(t)∥2dt < ∞

The nonnegative number

∥u(·)∥2 :=

(∫ ∞

0

∥u(t)∥2dt) 1

2

is the L2 norm of u(·).

1

for any T > 0. Since V (x(T )) ≥ 0, we deduce that∫ T

0∥y(t)∥2dt ≤ γ2

∫ T

0∥u(t)∥qdt ≤ γ2

[∥u(·)∥2

]2for any T > 0 and therefore [

∥y(·)∥2]2

≤ γ2[∥u(·)∥2

]2,

i.e.∥y(·)∥2 ≤ γ∥u(·)∥2 .

In other words, for any u(·) ∈ L2[0,∞), the response of the system from the initial statex(0) = 0 is defined for all t ≥ 0, produces an output y(·) which is a function in L2[0,∞), andthe ratio between the L2 norm of the output and the L2 norm of the input is bounded by γ.For this reason, the system is said to have a finite L2 gain, bounded above by the number γ.

Note that functions in L2[0,∞) represent signals having finite energy over the infinitetime interval [0,∞), and therefore the number γ in the inequality (2) can be given theinterpretation of (an upper bound of the) ratio between the energies of output and input.Another interpretation, which does not necessarily require the consideration of inputs havingfinite energy, is possible. Suppose the input is a periodic function of time, with period T , i.e.that

u(t+ kT ) = u◦(t), for all t ∈ [0, T ), k ≥ 0

for some piecewise continuous function u◦(t), defined on [0, T ). Also, suppose that, for somesuitable initial state x(0) = x◦, the state response x◦(t) of the system is defined for allt ∈ [0, T ] and satisfies

x◦(T ) = x◦ .

Then, it is obvious that x◦(t) exists for all t ≥ 0, and is a periodic function, having the sameperiod T of the input, namely

x◦(t+ kT ) = x◦(t), for all t ∈ [0, T ), k ≥ 0

and so is the corresponding output response y(t).For the triplet {u(t), x◦(t), y(t)} thus defined, integration of the inequality (2) over an

interval [t0, t0 + T ], with arbitrary t0 ≥ 0, yields

V (x◦(t0 + T ))− V (x◦(t0)) ≤ γ2∫ t0+T

t0

∥u(s)∥2ds−∫ t0+T

t0

∥y(s)∥2ds ,

i.e., since V (x◦(t0 + T )) = V (x◦(t0)),∫ t0+T

t0

∥y(s)∥2ds ≤ γ2∫ t0+T

t0

∥u(s)∥2ds . (3)

Observe that the integrals on both sides of this inequality are independent of t0, becausethe integrands are periodic functions having period T , and recall that the root mean square

2

value of any (possibly vector-valued) periodic function f(t) (which is usually abbreviated asr.m.s. and characterizes the average power of the signal represented by f(t) ) is defined as

∥f(·)∥r.m.s. =( 1

T

∫ t0+T

t0

∥f(s)∥2ds) 1

2.

With this in mind, (3) yields

∥y(·)∥r.m.s. ≤ γ∥u(·)∥r.m.s. . (4)

In other words, in a system having finite L2 gain system, the number γ (which appears inthe inequality (2)) happens to be also an upper bound for the ratio between the r.m.s. valueof the output and the r.m.s value of the input, whenever a periodic input is producing (froman appropriate initial state) a periodic (state and output) response.

We can therefore conclude that, in a system which satisfies a inequality of the form (2)the number γ can be given these two interpretations. If the input represents a signal whoseenergy over the infinite interval [0,∞) is finite, then the corresponding output from theinitial state x(0) = 0 is a function having finite energy over the interval [0,∞) and the ratiobetween the energies of output and input is bounded from above by the number γ. On theother hand, if the input is a periodic function which produces, from some appropriate initialstate x(0) = x◦, a periodic state and output response, the number γ provides a bound forthe ratio between the average powers of the output and input.

We will see in the next section that the fulfillment of an inequality of the form (2) isequivalent to the fulfillment of a linear matrix inequality involving the system data A,B,C,D,the number γ, and the matrix P .

2 An LMI characterization of the L2 gain

In this section, we derive alternative characterizations of the inequality (2).

Lemma 1 Let V (x) = xTPx, with P a positive definite symmetric matrix and let γ > 0 bea fixed number. There exists a number γ < γ and a number ϵ > 0 such that

∂V

∂x(Ax+Bu) ≤ −ϵ∥x∥2 + γ2∥u∥2 − ∥Cx+Du∥2 (5)

for all x ∈ Rn and all u ∈ Rm if and only if

DTD − γ2I < 0 (6)

PA+ATP + CTC + [PB + CTD][γ2I −DTD]−1[PB + CTD]T < 0 (7)

Proof. . Necessity . Recall that∂V

∂x= 2xTP

and observe that the inequality (5) becomes

2xTP [Ax+Bu] + ϵxTx− γ2uTu+ xTCTCx+ 2xTCDu+ uTDTDu ≤ 0 (8)

3

for all x ∈ Rn and all u ∈ Rm. For x = 0 this implies, in particular,

−γ2I +DTD ≤ 0 .

Since γ > γ we see that (5) implies

DTD < γ2I ,

which is condition (6) and

2xTP [Ax+Bu] + ϵxTx− γ2uTu+ xTCTCx+ 2xTCTDu+ uTDTDu ≤ 0 (9)

for all x ∈ Rn and all u ∈ Rm.Now, observe that, for each fixed x, the right-hand side of (9) is a quadratic form in u,

expressible asM(x) +N(x)u− uTWu . (10)

in whichW = γ2I −DTD

andM(x) = xT(PA+ATP + CTC + ϵI)x , N(x) = 2xT(PB + CTD) .

Since W is positive definite, this form has a unique maximum point, at

u∗(x) =1

2W−1N(x)T .

Hence, (9) holds if and only if the value of the form (10) at u = u∗(x) is non positive, whichyields

M(x) +1

4N(x)W−1N(x)T ≤ 0 .

Using the expressions of M(x), N(x),W , this reads as

xT(PA+ATP + CTC + ϵI + [PB + CTD][γ2I −DTD]−1[PB + CTD]T)x

and this, since ϵ > 0, implies condition (7).

Sufficiency. Observe that the left-hand sides of (6) and (7), which are negative definite,by continuity remain negative definite if γ is replaced by a number γ < γ such that thedifference γ − γ is small enough. Thus, for some ϵ > 0, the matrix P satisfies

PA+ATP + CTC − ϵI + [PB + CTD][γ2I −DTD]−1[PB + CTD] < 0.

As a consequence, the form M(x)+N(x)u−uTWu, in which W = γ2I−DTD, is non positiveand (5) holds. ▹

Lemma 2 Let γ be a fixed number. There exists a positive definite symmetric matrix Psatisfying (6) and (7) if and only if there exists a positive definite symmetric matrix Xsatisfying ATX +XA XB CT

BTX −γI DT

C D −γI

< 0 . (11)

4

Proof. Consider the matrix inequality(ATX +XA+ 1

γCTC XB + 1

γCTD

XTP + 1γD

TC −γI + 1γD

TD

)< 0 . (12)

This inequality holds if and only if the lower right block

−γI +1

γDTD

is negative definite, which is equivalent to condition (6), and so is its the Schur’s complement3

ATX +XA+1

γCTC − [XB +

1

γCTD][−γI +

1

γDTD]−1[XB +

1

γCTD] .

This, having replaced X by 1γP , is identical to condition (7).

Rewrite now (12) as(ATX +XA XB

BTX −γI

)+

(CT

DT

)1

γ(C D ) < 0

and use again (backward) Schur’s complement to arrive at (11). ▹

3 The H∞ norm of a transfer function

We have seen in Section 1 that, in a system which satisfies a inequality of the form (2), if theinput is a periodic function which produces, from some appropriate initial state x(0) = x◦,a periodic state and output response, the number γ provides a bound for the ratio betweenthe average powers of the output and input. This observation can be further pursued in thefollowing terms. Let the input u(·) be a sinusoidal function of time, of period T = 2π/ω0,e.g.

u(t) = Uu0 cos(ω0t) (15)

where U > 0 and u0 ∈ Rm has unitary norm. It is well-know that system (1), which isasymptotically stable, exhibits a well-defined steady-state response, which is itself a sinusoidal

3Schur’s complement Lemma. The symmetric matrix(Q SST R

)(13)

is negative definite if and only if

R < 0 and Q− SR−1ST < 0 . (14)

In fact, observe that a necessary condition for (13) to be negative definite is R < 0. Hence R is nonsingularand (13) can be transformed, by congruence, as(

I 0−R−1ST I

)T (Q SST R

)(I 0

−R−1ST I

)=

(Q− SR−1ST 0

0 R

).

from which the conditions (14) follow.

5

function of period t. The response in question can be easily characterized by means of a simplegeometric construction. Observe that the input thus defined can be viewed as generated byan autonomous system of the form w ∈ R2,

w =

(0 ω0

−ω0 0

)w := Sw

u = u0 ( 1 0 )w := Qw(16)

with w ∈ R2, from in the initial state

w(0) =

(10

)U .

Since A has all eivenvalues with negative real part and S has eigenvalues on the imaginaryaxis, the Sylvester equation

ΠS = AΠ+BQ (17)

has a unique solution Π. The composite system(wx

)=

(S 0BQ A

)(wx

)possesses two complementary invariant subspaces: a stable subspace and a center subspace,respectively spanned by the columns of the matrices

V s =

(0I

), V c =

(IΠ

).

The latter, in particular, shows that the center subspace is the set of all pairs (w, x) suchthat

x = Πw .

Consider now the change of variables

x = x−Πw

which, after a simple calculation using (17), yields

w = Sw˙x = Ax .

We see from this that, for any initial condition,

limt→∞

x(t) = 0 ,

which is to say that the (unique) projection of the trajectory along the stable subspaceasymptotically tends to zero. In the original coordinates, this reads as

limt→∞

[x(t)−Πw(t)] = 0 ,

6

from which we see that the steady-state response of system (1) to any input generated by(16) can be expressed as

xss = Πw(t) . (18)

The calculation of the solution Π of the Sylvester equation (17) is straightforward. Set

Π = (Π1 Π2 )

and observe that the equation in question reduces to

Π

(0 ω0

−ω0 0

)= AΠ+Bu0 ( 1 0 ) .

An elementary calculation (multiply first both sides on the right by the vector ( 1 j )T)yields

Π1 + jΠ2 = (jω0I −A)−1Bu0 ,

i.e.Π = (Re[(jω0I −A)−1B]u0 Im[(jω0I −A)−1B]u0 ) .

As shown above, the steady state response has the form (18). Hence, in particular, theperiodic input

u(t) = u0 cos(ω0t)

produces the periodic state response

xss(t) = Πw(t) = Π1 cos(ω0t)−Π2 sin(ω0t) , (19)

and the periodic output response

yss(t) = Cx◦(t) +Du0 cos(ω0t)

= Re[T (jω0)]u0 cos(ω0t)− Im[T (jω0)]u0 sin(ω0t) ,(20)

in whichT (jω) = C(jωI −A)−1B +D .

Observe now that ∫ 2πω0

0∥u(s)∥2ds = π

ω0∥u0∥2

and that, in view of the specific form of Π,∫ 2πω0

0∥yss(s)∥2ds =

π

ω0∥T (jω0)u0∥2 .

In other words∥u(·)∥2r.m.s. = 1

2∥u0∥2

∥yss(·)∥2r.m.s. = 12∥T (jω0)u0∥2 .

Thus, from the interpretation illustrated above one can conclude that, if the systemsatisfies (2), then

∥T (jω0)u0∥2 = 2∥yss(·)∥2r.m.s. ≤ γ22∥u(·)∥2r.m.s. = γ2∥u0∥2

7

i.e.∥T (jω0)u0∥ ≤ γ∥u0∥ .

Observing that both u0 and ω0 are arbitrary, it is seen from this that

supω∈R

max∥u∥=1

∥T (jω)u∥ ≤ γ . (21)

The quantity on the left-hand side is, by definition, the so-called H∞ norm of the matrixT (jω). Therefore, one can conclude that a linear system that satisfies (2) is asymptoticallystable and the H∞ norm of its frequency response matrix is bounded from above by the numberγ.

4 The Bounded Real Lemma

We have shown above that, for a system of the form (1), if (and only if) there exists anumber γ > 0 and a symmetric positive definite matrix X satisfying (11), then there existsa number γ and a positive definite matrix P satisfying (2) for some ϵ. This, in view of theinterpretation provided above, proves that the fulfillment of (11) for some γ (with X positivedefinite) implies that the system is asymptotically stable and that:

(i) its L2 gain is strictly less than γ,

(ii) the H∞ norm of its transfer function is strictly less than γ.

However, put in these terms, we have only said that (2) implies both (i) and (ii) andwe have not investigated yet whether the reverse implications might hold. In this section,we complete the analysis, by showing that the two properties (i) and (ii) are, in fact, twodifferent characterizations of the same property and both imply (11).

This will be done by means of a circular proof involving another equivalent version ofthe property that γ is an upper bound for the H∞ norm of the transfer function matrix ofthe system, which is very useful for practical purposes, since it can be rather easily verified.More precisely, the fact that γ is an upper bound for the H∞ norm of the transfer functionmatrix of the system can be checked by looking at the spectrum of a matrix of the form

H =

(A0 R0

−Q0 −AT0

), (22)

in which R0 and Q0 are symmetric matrices which, together with A0, depend on the matricesA,B,C,D which characterize the system and on the number γ.4

Theorem 1 Consider the linear system (1) and let γ > 0 be a fixed number. The followingare equivalent:

(i) there exists γ < γ, ϵ > 0 and a symmetric positive definite matrix P such that (5) holdsfor V (x) = xTPx ,

4A matrix with this structure is called an Hamiltonian matrix and has the property that its spectrum issymmetric with respect to the imaginary axis (see Section 7). Thus, if λ is an eigenvalue of H, so is also −λ.Since the entries of H are real numbers, this shows that if a+ jb is an eigenvalue of H, so is also −a+ jb.

8

(ii) all the eigenvalues of A have negative real part and the frequency response matrix of thesystem T (jω) = C(jωI −A)−1B +D satisfies

supω∈R

max∥u∥=1

∥T (jω)u∥ < γ , (23)

(iii) all the eigenvalues of A have negative real part, the matrix W = γ2I −DTD is positivedefinite, and the Hamiltonian matrix

H =

(A+BW−1DTC BW−1BT

−CTC − CTDW−1DTC −AT − CTDW−1BT

)(24)

has no eigenvalues on the imaginary axis,

(iv) there exists a positive definite symmetric matrix X satisfyingATX +XA XB CT

BTX −γI DT

C D −γI

< 0 . (25)

Proof. We have already shown, in the previous sections, that (i) implies that (1) is anasymptotically stable system, with a frequency response matrix satisfying

supω∈R

max∥u∥=1

∥T (jω)u∥ ≤ γ .

Thus, (i) ⇒ (ii).To show that (ii) ⇒ (iii), first of all that note, since

limω→∞

T (jω) = D

it necessarily follows that ∥Du∥ < γ for all u with ∥u∥ = 1 and this implies γ2I > DTD, i.e.the matrix W is positive definite.

Now observe that the Hamiltonian matrix (24) can be expressed in the form

H = L+MN

for

L =

(A 0

−CTC −AT

), M =

(B

−CTD

),

N = (W−1DTC W−1BT ) .

Suppose, by contradiction, that the matrix H has eigenvalues on the imaginary axis. Bydefinition, there exist a 2n-dimensional vector x0 and a number ω0 ∈ R such that

(jω0I − L)x0 = MNx0.

Observe now that the matrix L has no eigenvalues on the imaginary axis, because itseigenvalues coincide with those of A and −AT, and A is by hypothesis stable. Thus (jω0I−L)is nonsingular. Observe also that the vector u0 = Nx0 is nonzero because otherwise x0 would

9

be an eigenvector of L associated with an eigenvalue at jω0, which is a contradiction. Asimple manipulation yields

u0 = N(jω0I − L)−1Mu0. (26)

It is easy to check that

N(jω0I − L)−1M = W−1[TT(−jω0)T (jω0)−DTD] (27)

where T (s) = C(sI −A)−1B +D. In fact, it suffices to compute the transfer function of

x = Lx+Mu

y = Nx

and observe that N(sI − L)−1M = W−1[TT(−s)T (s)−DTD].Multiply (27) on the left by uT0 W and on the right by u0, and use (26), to obtain

uT0 Wu0 = uT0 [TT(−jω0)T (jω0)−DTD]u0 ,

which in turn, in view of the definition of W , yields

γ2∥u0∥2 = ∥T (jω0)u0∥2

which contradicts (ii), thus completing the proof.To show that (iii) ⇒ (iv), set

F = (A+BW−1DTC)T

Q = −BW−1BT

GGT = CT(I +DW−1DT)C

(the latter is indeed possible because I +DW−1DT is a positive definite matrix: in fact, itis a sum of the positive definite matrix I and of the positive semidefinite matrix DW−1DT;hence there exists a nonsingular matrix M such that I +DW−1DT = MTM ; in view of this,the previous expression holds with GT = MC).

The pair (F,GGT) thus defined is stabilizable. In fact, suppose that this is not the case.Then, there is a vector x = 0 such that

xT (F − λI GGT ) = 0

for some λ with non-negative real part. Then,

0 =

(A+BW−1DTC − λI

CTMTMC

)x .

This implies in particular 0 = xTCTMTMCx = ∥MCx∥2 and hence Cx = 0, because Mis nonsingular. This in turn implies Ax = λx, and this is a contradiction because all theeigenvalues of A have negative real part.

Moreover, it is easy to check that

HT =

(F −GGT

−Q −FT

)10

and this matrix by hypothesis has no eigenvalues on the imaginary axis.Thus5 there is a unique solution Y − of

Y −F + FTY − − Y −GGTY − +Q = 0 , σ(F −GGTY −) ⊂ C− . (28)

Moreover6, the set of solutions Y of the inequality

Y F + FTY − Y GGTY +Q > 0 (29)

is nonempty and any Y in this set is such that Y < Y −.Observe now that

0 = Y −F + FTY − − Y −GGTY − +Q

= Y −(AT + CTDW−1BT) + (A+BW−1DTC)Y −

− Y −CT(I +DW−1DT)CY − −BW−1BT

= Y −AT +AY − − [Y −CTD −B]W−1[DTCY − −BT]− Y −CTCY −,

which yieldsY −AT +AY − ≥ 0 .

Setting V (x) = xTY −x, this inequality shows that, along the trajectories of

x = ATx , (30)

the function V (x(t)) is non-decreasing, i.e. V (x(t)) ≥ V (x(0)) for any x(0) and any t ≥ 0.On the other hand, system (30) is by hypothesis asymptotically stable, i.e. limt→∞ x(t) = 0.Therefore, necessarily, V (x(0)) ≤ 0, i.e the matrix Y − is negative semi-definite. From this,it is concluded that any solution Y of (29), that is of the inequality

Y AT +AY − [Y CTD −B]W−1[DTCY −BT]− Y CTCY > 0 , (31)

which necessarily satisfies Y < Y − ≤ 0, is a negative definite matrix.Take any of the solutions Y of (31) and consider P = −Y −1. By construction, this matrix

is a positive definite solution of the inequality in (7). Thus, by Lemma 2, (iv) holds.The proof that (iv) ⇒ (i) is provided by Lemma 2 and 1. ▹

5 Small gain theorem and robust stability

We discuss in this section the stability properties of feedback interconnected systems. Moreprecisely, consider two systems Σ1 and Σ2, described by equations of the form

xi = Aixi +Biui

yi = Cixi +Diui(32)

5See Appendice C, Proposition 16See Appendice C, Proposition 2

11

with i = 1, 2, in which we assume that

dim(u2) = dim(y1)

dim(u1) = dim(y2) .

Suppose that the matrices D1 and D2 are such that the constraint

u2 = y1

u1 = y2(33)

makes sense. In other words, suppose that for each x1, x2 there is a unique pair u1, u2satisfying

u2 = C1x1 +D1u1

u1 = C2x2 +D2u2 .

This is the case if the system of equations

u2 = C1x1 +D1u1

u1 = C2x2 +D2u2

as a unique solution u1, u2, and this occurs if and only if the matrix(−D1 II −D2

)is invertible, i.e. the matrix I −D2D1 is nonsingular.

The system defined by (33) is the pure feedback interconnection of Σ1 and Σ2.Suppose now there exist two positive definite matrices P1, P2 and a two real numbers

γ1, γ2 such that Σ1 and Σ2 satisfy inequalities of the form (2), namely

∂Vi

∂xi(Aixi +Biui) ≤ −ϵi∥xi∥2 + γ2i ∥ui∥2 − ∥Cixi +Diui∥2 (34)

in which Vi(xi) = xTi Pixi and ϵi > 0.Consider, for the pure feedback interconnection of Σ1 and Σ2 the candidate Lyapunov

functionW (x1, x2) = V (x1) + aV (x2)

in which a > 0 is a number to be determined. A simple calculation shows that

∂W

∂x1x1 +

∂W

∂x2x2 ≤ −ϵ1∥x1∥2 − ϵ2∥x1∥2 + γ21∥u1∥2 − ∥y1∥2 + aγ22∥u2∥2 − a∥y2∥2

≤ −ϵ1∥x1∥2 − ϵ2∥x1∥2 + γ21∥y2∥2 − ∥y1∥2 + aγ22∥y1∥2 − a∥y2∥2

= −ϵ1∥x1∥2 − ϵ2∥x1∥2 + ( yT1 yT2 )

((−1 + aγ22)I 0

0 (γ21 − a)I

)(y1y2

).

If ((−1 + aγ22)I 0

0 (γ21 − a)I

)≤ 0 , (35)

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we have∂W

∂x1x1 +

∂W

∂x2x2 ≤ −α1(∥x1∥)− aα2(∥x2∥)

and hence the interconnected system is asymptotically stable. Condition (35), on the otherhand, is can be fulfilled for some a > 0 if (and only if)

−1 + aγ22 ≤ 0, γ21 − a ≤ 0

i.e. if

γ21 ≤ a ≤ 1

γ22.

All of the above can be summarized in the following important result.

Theorem 2 Consider a pair of systems (32) and suppose that the matrix I −D2D1 is non-singular. Suppose (32) satisfy inequalities of the form (34), with P1, P2 positive definite and

γ1γ2 ≤ 1 .

Then, the pure feedback interconnection of Σ1 and Σ2 is asymptotically stable.

Note also that, in view of the Bounded Real Lemma, the Theorem above can be rephrasedin terms of H∞ norms of the transfer functions of the two component subsystems, as follows.

Theorem 3 Consider a pair of systems (32) and suppose that the matrix I −D2D1 is non-singular. Suppose both systems are asymptotically stable. Let

Ti(s) = Ci(sI −Ai)−1Bi +Di

denote the respective transfer functions and suppose

∥T1∥∞ < γ1 , ∥T2∥∞ < γ2 .

Ifγ1γ2 ≤ 1

the pure feedback interconnection of Σ1 and Σ2 is asymptotically stable.

This theorem is the point of departure for the study of robust stability via H∞ methods.

A problem of robust stabilization can be cast, in rather general terms, as follows. A plantwith control input u and measurement output y whose model is uncertain can be, from arather general viewpoint, thought of as the interconnection of a nominal system modeled byequations of the form

x = Ax+B1v +B2uz = C1x+D11v +D12uy = C2x+D21v

(36)

in which the “additional” input v and the “additional” output z are seen seen as output and,respectively, input of a system

xp = Apxp +Bpzv = Cpxp +Dpz

(37)

13

whose parameters are uncertain .7

This setup includes the special case of a plant of fixed dimension whose parameters areuncertain, that is a plant modeled as

x = (A0 + δA)x+ (B0 + δB)uy = (C0 + δC)

(38)

in which A0, B0, C0 represent nominal values and δA, δB, δC uncertain perturbations. Infact, the latter can be seen as interconnection of a system of the form (36) in which

A = A0 B1 = ( I 0 ) B2 = B0

C1 =

(I0

)D11 =

(00

)D12 =

(0I

)C2 = C0 D21 = ( 0 I )

with a system of the form

v =

(δA δBδC 0

)z

which is indeed a special case of a system of the form (37). The interconnection (36) –(37) of a nominal model and a of dynamic perturbation is more general, though, because itaccommodates for perturbations which possibly include unmodelled dynamics. In this setup,all modeling uncertainties are confined in the model (37), including the dimension itself ofxp.

Suppose now that (37) Ap is a Hurwitz matrix and that the transfer function

Tp(s) = Cp(sI −Ap)−1Bp +Dp

has an H∞ norm which is bounded by a known number γp. That is, assume that, no matterwhat the perturbations are, the perturbing system (37) is a stable system satisfying

∥Tp∥∞ < γp . (39)

for some γp.Let

xc = Acxc +Bcyu = Ccxc +Dcy

(40)

be a controller for the nominal plant (36) yielding a closed loop system(xξ

)=

(A+B2DcC2 B2Cc

BcC2 Ac

)(xxc

)+

(B1 +B2DcD21

BcD21

)w

z = (C1 +DcC2 D12Cc )

(xxc

)+ (D11 +DcD21)w

(41)

7Note that, for the interconnection (36) – (37) to be well-defined, the matrix I −DpD11 is required to beinvertible.

14

which is asymptotically stable and whose transfer function, between the input w and outputz has an H∞ norm bounded by a number γ satisfying

γγp < 1 . (42)

If this is the case, thanks to the small gain theorem, it can be concluded that the controller(44) stabilizes any of the perturbed plants (36) – (37), so long as the perturbation is such that(37) is asymptotically stable and the bound (39) holds.

In this way, the problem of robust stabilization is reduced to a problem of stabilizing anominal plant and to simultaneously enforce an H∞ bound.

6 The coupled LMIs approach to the problem of γ-suboptimalH∞ feedback design

We summarize in this Section some relevant results about the so-called problem of γ-suboptimalH∞ feedback design. Consider a linear system described by equations of the form

x = Ax+B1v +B2uz = C1x+D11v +D12uy = C2x+D21v .

(43)

Let γ > 0 be a fixed number. The problem of γ-suboptimal H∞ feedback design consistsin finding a controller

ξ = Acξ +Bcyu = Ccξ +Dcy

(44)

yielding a closed loop system(xξ

)=

(A+B2DcC2 B2Cc

BcC2 Ac

)(xξ

)+

(B1 +B2DcD21

BcD21

)v

z = (C1 +D12DcC2 D12Cc )

(xξ

)+ (D11 +D12DcD21) v

(45)

which is asymptotically stable and whose transfer function, between the input v and outputz has an H∞ norm strictly less than γ

Rewrite system (45) asx = Ax+ Bvz = Cx+Dv

where

A =

(A+B2DcC2 B2Cc

BcC2 Ac

), B =

(B1 +B2DcD21

BcD21

)C = (C1 +D12DcC2 D12Cc ) , D = D11 +D12DcD21 .

In view of the Bounded Real Lemma, the properties in question are achieved if and onlyif there exists a symmetric matrix X satisfying

X > 0 (46)

15

ATX + XA XB CT

BTX −γI DT

C D −γI

< 0 . (47)

Thus, the problem is to try to find a quadruplet {Ac, Bc, Cc, Dc} such that (46) and (47)hold for some symmetric X .

The basic inequality (47) will be now transformed as follows. Let

x ∈ Rn, ξ ∈ Rk, v ∈ Rm1 , u ∈ Rm2 , z ∈ Rp1 , y ∈ Rp2 .

Set

A0 =

(A 00 0

), B0 =

(B1

0

), C0 = (C1 0 ) , (48)

ΨX =

AT0 X + XA0 XB0 CT

0

BT0 X −γI DT

11

C0 D11 −γI

, (49)

P =

(0 Ik 0 0BT

2 0 0m2×m1 DT12

), (50)

Q =

(0 Ik 0 0C2 0 D21 0p2×p1

), (51)

and

ΞX =

X 0 00 Im1 00 0 Ip1

.

Collecting the parameters of the controller (44) in the (n+m2)× (n+ p2) matrix

K =

(Ac Bc

Cc Dc

)(52)

the inequality (47) can be rewritten as

ΨX +QTKT(PΞX ) + (PΞX )TKQ < 0 . (53)

Thus, the problem of γ-suboptimal H∞ feedback design can be cast as the problem offinding a symmetric matrix X > 0 and a matrix K such that (53) holds. In this context, thefollowing result is very useful.8

Lemma 3 Given a symmetric m ×m matrix Ψ and two matrices P,Q having m columns,consider the problem of finding some matrix K of compatible dimensions such that

Ψ+QTKP + PTKTQ < 0 . (54)

Let WP be a matrix such that Im(WP ) = Ker(P ) and let WQ be a matrix such thatIm(WQ) = Ker(Q). Then (54) is solvable in K if and only if

WTQΨWQ < 0

WTP ΨWP < 0 .

(55)

8For a proof, see: P.Gahinet and P. Apkarian, A Linear Matrix Inequality Approach to H∞ Control, in Int.J. Robust and Nonlinear Control, vol. 4, pp 421-448 (1994).

16

This Lemma can be used to eliminate K from (53) and obtain existence conditions de-pending only on X and on the parameters of the plant (43). Let WPX be a matrix whosecolumns span Ker(PΞX ) and let WQ be a matrix whose columns span Ker(Q). Accordingto Lemma 3, there exists K for which (53) holds if and only if

WTQΨXWQ < 0

WTPX

ΨXWPX < 0 .(56)

The second of these two inequalities can be further manipulated observing that if WP is amatrix whose columns span Ker(P), the columns of the matrix Ξ−1

X WP span Ker(PΞX ).Thus, having set

ΦX =

A0X−1 + X−1AT0 B0 X−1CT

0

BT0 −γI DT

11

C0X−1 D11 −γI

, (57)

the second of (56) can be rewritten as

WTP Ξ−1

X ΨXΞ−1X WP = WT

PΦXWP < 0 .

We summarize this fact as follows.

Proposition 1 There exists a k-dimensional controller that solves the problem of γ-suboptimalH∞ feedback design if and only if there exists a (n+ k)× (n+ k) symmetric matrix X suchthat

WTQΨXWQ < 0

WTPΦXWP < 0 ,

(58)

in which ΨX and ΦX are the matrices defined in (49) and (57).

The two inequalities (58) thus found can be further simplified. To this end, it is convenientto partition X and X−1 as

X =

(S NNT ∗

), X−1 =

(R MMT ∗

)(59)

in which R and S are n × n and N and M are n × k. With the partition (59), the matrix(49) becomes

ΨX =

SA+ATS ATN SB1 CT

1

NTA 0 NTB1 0BT

1 S BT1 N −γI DT

11

C1 0 D11 −γI

(60)

and the matrix (57) becomes

ΦX =

AR+RAT AM B1 RCT

1

MTAT 0 0 MTCT1

BT1 0 −γI DT

11

C1R C1M D11 −γI

. (61)

17

From the definition of Q, it is readily seen that a matrix WQ whose columns span Ker(Q)can be expressed as

WP =

Z1 00 0Z2 00 Ip1

in which (

Z1

Z2

)is a matrix whose columns span

Ker (C2 D21 ) .

Then, an easy calculation shows that the first inequality in (58) can be rewritten as

(ZT1 ZT

2 00 0 I

)ATS + SAT SB1 CT1

BT1 S −γI DT

11

C1 D11 −γI

Z1 0Z2 00 I

< 0 .

Likewise, from the definition of P, it is readily seen that a matrix WP whose columnsspan Ker(P) can be expressed as

WP =

V1 00 00 Im1

V2 0

in which (

V1

V2

)is a matrix whose columns span

Ker (BT2 DT

12 ) .

Then, an easy calculation shows that the second inequality in (58) can be rewritten as

(V T1 V T

2 00 0 I

)AR+RAT RCT1 B1

C1R −γI D11

BT1 DT

11 −γI

V1 0V2 00 I

< 0 .

In this way, the two inequalities (58) have been transformed into two inequalities involvingsubmatrix S of X and the submatrix R of X−1. To complete the analysis, it remains toconnect these matrices to each other. This is achieved via the following Lemma.

Lemma 4 Let S and R be symmetric n × n matrices. There exist a n × k matrix N andk × k symmetric matrix Z a such that(

S NNT Z

)> 0 (62)

18

and (S NNT Z

)−1

=

(R ∗∗ ∗

)(63)

if and only ifrank(I − SR) ≤ k (64)

and (S II R

)≥ 0 . (65)

Proof. To prove necessity, write

X =

(S NNT Z

)and X−1 =

(R MMT ∗

).

Then, by definition we haveSR+NMT = I

NTR+ ZMT = 0 .(66)

Thus I − SR = NMT and this implies (64), because N has k columns. Now, set

T =

(I R0 MT

).

Using (66), the first of which implies MNT = I−RS because S and R are symmetric, observethat

T TXT =

(S II R

).

Pick y ∈ R2n and define x ∈ Rn+k as x = T y. Then, it is seen that

yT(S II R

)y = xTXx ≥ 0

because the matrix X is positive definite by assumption. This concludes the proof of thenecessity.

To prove sufficiency, let k ≤ k be the rank of I − SR and let N , M two n× k matrices ofrank k such that

I − SR = NMT. (67)

Using the property (67) it is possible to show that the equation

NTR+ ZMT = 0 (68)

has a solution Z. In fact, observe that

MNTR−RNMT = (I −RS)R−R(I − SR) = 0 .

Let L be any matrix such that MTL = −Ik (which exists because the k rows of MT areindependent) and, from the identity above, obtain

MNTRL = −RN .

19

This shows that the matrix Z = (NTRL)T satisfies (68). The matrix Z is symmetric. Infact, note that

0 = M(NTR+ ZMT) = (I −RS)R+ MZMT = R−RSR+ MZMT .

and hence MZMT is symmetric. This yields

MZMT − MZTMT = M(Z − ZT)MT = 0

from which it is seen that Z = ZT because MT has independent rows.As a consequence of (67) and(68), the symmetric matrix

X =

(S NNT Z

)(69)

is a solution of (S INT 0

)= X

(I R0 MT

). (70)

The symmetric matrix X thus found is invertible because, otherwise, the independence ofthe rows of the matrix on the left-hand side of (70) would be contradicted. It is easily seenthat

X−1 =

(R ∗MT ∗

). (71)

Moreover, it is also possible to prove that X > 0. In fact, letting

T =

(I R0 MT

)observe that

T TX T =

(S II R

).

Assumption (65) implies that X ≥ 0. In fact, suppose xTXx < 0 for some x. Using the factthat the rows of T are independent, find y such that x = T y. This would make

yT(S II R

)y < 0 ,

a contradiction. But we have shown before that X is nonsingular and hence, knowing thatthis matrix is positive semidefinite, we conclude it is positive definite.

If k = k the sufficiency is proven. Otherwise, set ℓ = k − k and

X =

S N 0NT Z 00 0 Iℓ×ℓ

and observe that X is positive definite and

X−1 =

R ∗ 0MT ∗ 00 0 Iℓ×ℓ

20

has the required structure. ▹

Remark. Note that condition (65), alone, implies S > 0 and R > 0. In fact, in the proof ofsufficiency, using this condition we have shown that (69) is positive definite. Hence P > 0.Its inverse has the structure indicated in (71) and hence also R > 0. Finally, note that thecondition (64) is immaterial if k ≥ n because the matrix on the right is n× n. ▹

This Lemma establishes a coupling condition between the two submatrices S and R iden-tified in the previous analysis, that determines the positivity of the matrix X . Using thisLemma it is therefore possible to arrive at the following conclusion.

Theorem 4 Consider a plant of modelled by equations of the form (43). Let V1, V2, Z1, Z2

be matrices such that

Im(Z1

Z2

)= Ker (C2 D21 ) , Im

(V1

V2

)= Ker (BT

2 DT12 ) .

The problem of γ-suboptimal H∞ feedback design has a solution if and only if there existsymmetric matrices S and R satisfying the following system of linear matrix inequalities(

ZT1 ZT

2 00 0 I

)ATS + SAT SB1 CT1

BT1 S −γI DT

11

C1 D11 −γI

Z1 0Z2 00 I

< 0 (72)

(V T1 V T

2 00 0 I

)AR+RAT RCT1 B1

C1R −γI D11

BT1 DT

11 −γI

V1 0V2 00 I

< 0 (73)

(S II R

)≥ 0 . (74)

In particular, there exists a solution of dimension k < n if and only if there exist R and Ssatisfying (72), (73), (74) and, in addition,

rank(I −RS) ≤ k . (75)

The result above describes necessary conditions for the existence of a controller that solvesthe problem of problem of γ-suboptimal H∞ feedback design. For the actual construction of acontroller, one may proceed as follows. Assuming that S and R are positive definite symmetricmatrices satisfying the system of linear matrix inequalities (72), (73), (74), construct a matrixX as indicated in the proof of Lemma 4, that is, find two matrices n× k matrices N and Msuch that

I − SR = NMT

with k = rank(I − SR), and solve for X the linear equation(S INT 0

)= X

(I R0 MT

).

By construction, the matrix in question satisfies (58) and their equivalent versions (56).Thus, according to Lemma 3, there exists a matrix K satisfying (53). This is a linear matrix

21

inequality in the only unknown K. The solution of such inequality provides the requiredcontroller as

K =

(Ac Bc

Cc Dc

).

22

A Hamiltonian Matrices and Algebraic Riccati Equations

In this section, we will study Riccati equations of the form

ATX +XA+Q+XRX = 0 (76)

with R and Q symmetric matrices, which can also be rewritten in the equivalent form

(X −I )

(A R−Q −AT

)(IX

)= 0

From either one of these expressions, it is easy to deduce the following identity(A R−Q −AT

)(IX

)=

(IX

)(A+RX) (77)

and to conclude that X is a solution of the Riccati equation (76) if and only if the subspace

V = Im

(IX

)(78)

is an (n-dimensional) invariant subspace of the matrix

H =

(A R−Q −AT

)(79)

In particular, (77) also shows that if X is a solution of (76), the matrix A+RX characterizesthe restriction of H to its invariant subspace (78). A matrix of the form (79), with real entriesand in which R andQ are symmetric matrices, is called an Hamiltonian matrix. Some relevantfeatures of the Hamiltonian matrix (79) and their relationships with the Riccati equation (76)are described in the following Lemmas.

Lemma 5 The spectrum of the Hamiltonian matrix (79) is symmetric with respect to theimaginary axis.

Proof. Set

J =

(0 −II 0

)and note that

J−1HJ =

(−AT Q−R A

)= −HT .

HenceH and -HT are similar. As a consequence, if λ is an eigenvalue ofH so is also −λ. Sincethe entries of H are real numbers, and therefore the spectrum of this matrix is symmetricwith respect to the real axis, the result follows. ▹

Suppose now that the matrix (79) has no eigenvalues on the imaginary axis. Then, thematrix in question has exactly n eigenvalues in C− and n eigenvalues in C+. As a conse-quence, there exist two complementary n-dimensional invariant subspaces of H: a subspaceV − on which the restriction of H has all eigenvalues with strictly negative real part, which is

23

called the stable eigenspace and a subspace V + on which the restriction of H has all eigen-values with strictly positive real part, which is called the unstable eigenspace. A situation ofspecial interest in the subsequent analysis is the one in which the stable eigenspace (respec-tively, the unstable eigenspace) of the matrix (79) can be given the form (78); in this case infact, as observed before, it is possible to associate with this subspace a particular solution ofthe Riccati equation (76). Observing that an n-dimensional subspace V of R2n can be giventhe form (78) if and only if

V ∩ Im

(0I

)= {0} , (80)

it is easy to arrive at the following conclusions.

Lemma 6 The following are equivalent:

(i) the Hamiltonian matrix (79) has no eigenvalue on the imaginary axis and its stableeigenspace V − (respectively its unstable eigenspace V +) satisfies the condition (80),

(ii) there exists a unique symmetric solution X − (respectively, X +) of the Riccati equation(76) such that A+RX − has all eigenvalues in C− (A+RX + has all eigenvalues in C+).

Proof. The identity (77) already shows the equivalence between (i) and the existence of asolutionX of (76) such that A+RX has all eigenvalues either in C− or in C+. The uniquenessof such a solution follows from the uniqueness of the expression of an n-dimensional invariantsubspace of (79) in the special form (78). Finally, to prove that such a solution is symmetric,multiply (77) on the left by ( I XT ) J to obtain

( I XT ) JH

(IX

)= (XT −X)(A+RX)

and, taking the transpose,

( I XT )HTJT

(IX

)= (A+RX)T(X −XT) .

Since JH = HTJT, these yield

(XT −X)(A+RX) + (A+RX)T(XT −X) = 0 .

This is a Sylvester equation in (XT−X), which has a unique solution because (A+RX)and −(A+RX)T have no common eigenvalues. This solution is necessarily the trivial solution(XT −X) = 0 and the result follows. ▹

The solutions X − and X + of the Riccati equation (76) thus characterized, whenever theyexist, are called the stabilizing solution and, respectively, the antistabilizing solution of theequation in question. We describe a special cases in which these particular solutions exist.To this end all we need, in view of the previous Lemma, is to establish conditions underwhich the Hamiltonian matrix (79) has no eigenvalues on the imaginary axis and its stableeigenspace V − (respectively its unstable eigenspace V +) satisfies the condition (80).

24

Lemma 7 Suppose the Hamiltonian matrix (79) has no eigenvalues on the imaginary axisand R is a (either positive or negative) semidefinite matrix. If the pair (A,R) is stabilizable,the stable eigenspace V − of (79) satisfies the condition (80). If the pair (A,R) is antistabi-lizable, the unstable eigenspace V + of (79) satisfies the condition (80).

Proof. Express V − in the form

V − = Im

(X1

X2

)and note that, by definition,

H

(X1

X2

)=

(X1

X2

)S

where S is a stable matrix. Multiplying this relation on thee left by

(XT1 XT

2 ) J = (XT2 −XT

1 )

one obtains

(XT1 XT

2 ) JH

(X1

X2

)= (XT

2 X1 −XT1 X2)S

Since the matrix on the left-hand side is symmetric, so is also (XT2 X1 − XT

1 X2)S and theconclusion

XT2 X1 = XT

1 X2 (81)

follows by the same type of arguments used in the proof of the previous Lemma.We prove now that Ker(X1) is invariant under S. To this end, note that

( I 0 )H

(X1

X2

)= ( I 0 )

(X1

X2

)S

yieldsAX1 +RX2 = X1S . (82)

Let x ∈ Ker(X1). Then, using (81), we obtain

(X2x)TR(X2x) = xTXT

2 (AX1 +RX2)x = xTXT2 X1Sx = xTXT

1 X2Sx = 0 .

Since R is sign semidefinite by hypothesis, then RX2x = 0 and (82) yields X1Sx = 0 i.e.Sx ∈ Ker(X1).

Finally, we prove that Ker(X1) = {0}. By contradiction, suppose is not. Then thereexists λ ∈ C− (because S is stable), and x ∈ Ker(X1) such that Sx = λx. Consider now

( 0 I )H

(X1

X2

)= ( 0 I )

(X1

X2

)S

which yields−QX1 −ATX2 = X2S .

Multiplication on the right by x yields, in turn,

−ATX2x = X2λx

25

i.e. (λI +AT)X2x = 0. Recalling that RX2x = 0, we obtain

(X2x)T (A+ λI R ) = 0

which, if X2x = 0, contradicts the hypothesis that (A,R) is stabilizable, because −λ ∈ C+.Thus necessarily X2x = 0 and x is such that

0 =

(X1xX2x

)=

(X1

X2

)x ,

but this contradicts the fact that V − has dimension n. Thus x = 0, Ker(X1) = {0}, X1 isinvertible and condition (80) holds. ▹

We can merge the previous two Lemma as follows.

Proposition 2 Suppose the Hamiltonian matrix (79) has no eigenvalues on the imaginaryaxis and R is a (either positive or negative) semidefinite matrix.

If the pair (A,R) is stabilizable, the stable eigenspace V − of (79) can be expressed in theform

V − = Im

(I

X −

)in which X − is a symmetric matrix, the (unique) stabilizing solution of the Riccati equation(76).

If the pair (A,R) is antistabilizable, the unstable eigenspace V + of (79) can be expressedin the form

V + = Im

(I

X +

)in which X − is a symmetric matrix, the (unique) antistabilizing solution of the Riccati equa-tion (76).

We discuss now the relation between solutions of the algebraic Riccati equation (76) andof the algebraic Riccati inequality

ATX +XA+Q+XRX > 0 . (83)

Lemma 8 Suppose R is negative semidefinite. Let X − (respectively X +) be a solution ofthe Riccati equation (76) having the property that σ(A + RX −) ∈ C− (respectively, σ(A +RX +) ∈ C+). Let X be any symmetric matrix satisfying (83). Then X < X − (respectively,X > X +).

Proof. Set R = −BBT and observe that

AT(X −X −) + (X −X −)A−XBBTX +X −BBTX − > 0 . (84)

Set P = X−X − and observe that the rate of change of the quadratic function V (x) = xTPxalong the trajectories of

x = (A+RX −)x = (A−BBTX −)x (85)

26

satisfies

V (x) = 2xT(X−X −)(A+RX −)x = xT[AT(X−X −)+(X−X −)A−2(X−X −)BBTX −]x

Note now that

−2(X −X −)BBTX − = −2(X −X −)BBTX − +XBBTX −XBBTX

= −XBBTX +X −BBTX − + (X −X −)BBT(X −X −)

Hence, letting P denote the left-hand side of (84), we obtain

V (x) = xT[P + xTPBBTP ]x .

Since P is positive definite by assumption, so is also P + PBBTP and hence the functionV (x) is increasing along any nontrivial trajectory of (85). In other words,

V (x(t)) > V (x(0)) (86)

along any nontrivial trajectory of (85). The latter, on the other hand, is an asymptoticallystable system by assumption and limt→∞ x(t) = 0 for any x(0). Therefore, (86) implies thatV (x) is negative for any x = 0.9 This shows that P is negative definite, i.e. X < X −. ▹

Lemma 9 Let X − (respectively, X +) be a solution of the Riccati equation (76) having theproperty that σ(A + RX −) ∈ C− (respectively, σ(A + RX +) ∈ C+). Then, there exist areal number ε0 > 0 and a family of symmetric matrices Xε, defined for ε ∈ (−ε0, ε0) andcontinuously depending on ε, satisfying

ATXε +XεA+Q+XεRXε = εI .

and such that limε→0Xε = X − (respectively, X +).

Proof. Consider the function f : Rn×n × R → Rn×n defined by

f(X, ε) = ATX +XA+Q+XRX − εI.

By definition, this function is such that

f(X −, 0) = 0.

Moreover, it is easy to see that

f(X − + δX, 0) = f(X −, 0) + δX(A+RX −) + (A+RX −)TδX + δXRδX .

The linear term in δX on the right-hand side of this relation can be viewed as the value atδX of the linear mapping L : Rn×n → Rn×n defined by

L(δX) = δX(A+RX −) + (A+RX −)TδX

9In fact, suppose V (x) = c > 0 for some x = 0 and set x(0) = x. Since limt→∞ x(t) = 0, there exists T > 0such that |V (x(T ))| ≤ c/2 and this, in view of (86), would yield c/2 ≥ |V (x(T ))| ≥ V (x(T )) ≥ V (x(0)) = c,i.e. a contradiction.

27

and this mapping is indeed invertible since (A+ RX −) has by hypothesis all eigenvalues inC− and thus the Sylvester equation

δX(A+RX −) + (A+RX −)TδX = Y

has a unique solution for each Y ∈ Rn×n. One can therefore conclude, by the implicit functiontheorem, that there exist a positive number ε0 and a continuous function Xε of ε, defined forall |ε| < ε0, such that f(Xε, ε) = 0, satisfying

ATXε +XεA+Q+XεRXε = εI for all |ε|. (87)

By construction (A R

−Q+ εI −AT

)(IXε

)=

(IXε

)(A+RXε).

Since (A+RX −) has all eigenvalues in C− and limε→0Xε = X −, then also (A+RXε) hasall eigenvalues in C− for small |ε|. Hence Xε is the unique stabilizing solution of (87) and isnecessarily symmetric. ▹

We can merge the previous two Lemma as follows.

Proposition 3 Suppose R is negative semidefinite. Let X − (respectively X +) be a solutionof the Riccati equation (76) having the property that σ(A+RX −) ∈ C− (respectively, σ(A+RX +) ∈ C+). Then, the set of solutions of

ATX +XA+Q+XRX > 0 ,

is not empty and any X in this set satisfies X < X − (respectively, X > X +).

28