Chapter III Stability of Linear Systemsaksikas/math603-cha4.pdf · conditions for the stability of linear systems, but also pave the way to deriving Lyapunov’s linearization method,

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Chapter III

Stability of Linear Systems

1. Stability and state transition matrix

2. Time-varying (non-autonomous) systems

3. Time-invariant systems

1 STABILITY AND STATE TRANSITION MATRIX 2'

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In this chapter, we study the Lyapunov stability of systemsdescribed by linear vector differential equations. The resultspresented here not only enable us to obtain necessary and sufficientconditions for the stability of linear systems, but also pave the wayto deriving Lyapunov’s linearization method, which is presented inthe next chapter.

1 Stability and state transition matrix

Consider a system described by the linear vector differentialequation

x(t) = A(t)x(t), t ≥ 0 (1)

The system (1) is autonomous is A(·) is constant as a function oftime ; otherwise it is non-autonomous. It is clear that 0 is always anequilibrium of the system (1). Further, 0 is an isolated equilibriumif A(t) is nonsingular for some t ≥ 0.


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The general solution of (1) is given by

x(t) = Φ(t, t0)x(t0) (2)

where Φ(·, ·) is the state transition matrix associated with A(·) andis the unique solution of the equation

d

dtΦ(t, t0) = A(t)Φ(t, t0), ∀t ≥ t0 ≥ 0 (3)

Φ(t0, t0) = I, ∀t0 ≥ 0 (4)

With the aid of this explicit characterization of the solutions of (1),it is possible to derive some useful conditions for the stability of theequilibrium 0. Since these conditions involve the state transitionmatrix Φ, they are not of much computational value, because ingeneral it is impossible to derive an analytical expression for Φ.Nevertheless, they are of conceptual value, enabling one tounderstand the mechanisms of stability and instability in linearsystems.


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Theorem 1.1. The equilibrium 0 is stable if and only if for eacht0 ≥ 0 it is true that

supt≥t0‖Φ(t, t0)‖i := m(t0) <∞ (5)

where ‖ · ‖i denotes the induced norm of matrix.

Proof : Assume that (5) is true, and let ε > 0, t0 ≥ 0 be specified.If we define δ(ε, t0) as ε/m(t0), then

‖x(t0)‖ < δ ⇒ ‖x(t)‖ ≤ ‖Φ(t, t0)‖i‖x(t0)‖ < m(t0)δ = ε (6)

This shows that the equilibrium 0 is stable.

Assume that (5) is false, so that ‖Φ(t, t0)‖i is an unboundedfunction of t for some t0 ≥ 0. To show that 0 is an unstableequilibrium, let ε > 0 be any positive number, and let δ be anarbitrary positive number. It is shown that one can choose an x(t0)in the ball Bδ such that the resulting solution x(t) satisfies‖x(t)‖ ≥ ε for some t ≥ t0.


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Select a δ1 in the open interval (0, δ). Since ‖Φ(t, t0)‖i isunbounded as a function of t, there exists a t ≥ t0 such that

‖Φ(t, t0)‖i >ε

δ1(7)

Next, select a vector v of norm one such that

‖Φ(t, t0)v‖ = ‖Φ(t, t0)‖i (8)

This is possible in view of the definition of the induced norm.Finally, let x(t0) = δ1v. Then x ∈ Bδ. Moreover,

‖x(t)‖ = ‖Φ(t, t0)x(t0)‖ = ‖δ1Φ(t, t0)v‖ = δ1‖Φ(t, t0)‖i > ε (9)

Hence, the equilibrium 0 is unstable.

Remark 1.1. Note that, in the case of linear systems, theinstability does indeed imply that some solution trajectories actuallyblow-up. This is in contrast to the case of nonlinear systems, whereinstability of 0 can be accompanied by the boundedness of allsolutions.


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Necessary and sufficient conditions for uniform stability are givennext :

Theorem 1.2. The equilibrium 0 is uniformly stable if and only if

supt0≥0

supt≥t0‖Φ(t, t0)‖i := m0 <∞ (10)

Proof : Suppose m0 is finite, then for any ε and any t0 ≥ 0, thereexists a δ = ε/m0 such that

‖x0‖ ≤ δ, t0 ≥ 0 ⇒ ‖x(t, t0, x0)‖ < ε, ∀t ≥ t0.

Suppose m(t0) is unbounded as a function of t0. Then at least onecomponent of Φ(·, ·), say the ij-component, has the property that

supt≥t0|φij(t, t0)| is unbounded as a function of t0 (11)

Let x0 = ej , the elementary vector. Then (11) implies that thequantity ‖x(t)‖/‖x0‖ = ‖Φ(t, t0)x0‖/‖x0‖ cannot be boundedindependent of t0. Hence, 0 is not uniformly stable.


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The next theorem characterizes uniform asymptotic stability.

Theorem 1.3. The equilibrium 0 is (globally) uniformlyasymptotically stable if and only if

supt0≥0

supt≥t0‖Φ(t, t0)‖i := m0 <∞ (12)

‖Φ(t+ t0, t0)‖i → 0 as t→∞, uniformly in t0 (13)

Proof : ⇒ Assume (12) holds, then 0 is uniformly stable. Also, if(13) holds, then the ratio

‖Φ(t, t0)x0‖/‖x0‖ approaches 0 uniformly in t0

so 0 is uniformly attractive. Hence, 0 is uniformly asymptoticallystable.

⇐ left as an exercice.


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The next theorem shows that, for linear systems, uniformasymptotic stability is equivalent to exponential stability.

Theorem 1.4. The equilibrium 0 is uniformly stable if and only ifthere exist constants m,λ > 0 such that

‖Φ(t, t0)‖i ≤ m exp[−λ(t− t0)], ∀t ≥ t0 (14)

Proof : Assume (14) is satisfied. Then clearly (12) and (13) are alsotrue, whence 0 is uniformly asymptotically stable by Theorem 1.3.

Conversely, suppose (12) and (13) are true. Then there exist finiteconstants µ and T such that

‖Φ(t, t0)‖i ≤ µ,∀t ≥ t0 ≥ 0 (15)

‖Φ(t+ t0, t0)‖i ≤ 1/2, ∀t ≥ T, ∀t0 ≥ 0. (16)

In particular, (16) implies that

‖Φ(T + t0, t0)‖i ≤ 1/2, ∀t0 ≥ 0. (17)

Now, given any t0 and any t ≥ t0, pick an integer k such that


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t0 + kT ≤ t ≤ t0 + (k + 1)T . Then

Φ(t, t0) = Φ(t, t0 + kT )Φ(t0 + kT, t0 + kT −T ) . . .Φ(t0 +T, t0) (18)

Hence

‖Φ(t, t0)‖i = ‖Φ(t, t0 + kT )‖iΠkj=1‖Φ(t0 + jT, t0 + jT − T )‖i

≤ µ2−k ≤ 2µ2−(t−t0)/T

Hence (14) is satisfied if we define

m = 2µ and λ =log2

T.

This completes the proof.

This section contains several results that relate the stabilityproperties of a linear system to its state transition matrix. Sincethese results require an explicit expression of the state transitionmatrix, they are not of much use for testing purposes. Nevertheless,they do provide some insight.

2 TIME-INVARIANT SYSTEMS 10'

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2 Time-invariant systems

Throughout this section, attention is restricted to lineartime-invariant systems of the form

x(t) = Ax(t) (19)

In this special case, Lyapunov theory is very complete, as we shallsee.

Theorem 2.1.

(1)The equilibrium 0 of (19) is exponentially stable if and only if allthe eigenvalues of A have negative real parts.

(2) The equilibrium 0 of (19) is stable if and only if all theeigenvalues of A have nonpositive real parts.

Proof : The state transition matrix Φ(t, t0) of the system (19) isgiven by

Φ(t, t0) = exp[A(t− t0)] (20)


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where exp(·) is the matrix exponential. Furthermore, exp(At) canbe expressed as

exp(At) =

r∑i=1

mi∑j=1

pij(A)tj−1 exp(λit) (21)

where r is the number of distinct eigenvalues of A ; λ1, . . . λr arethe distinct eigenvalues ; mi is the multiplicity of the eigenvalue λiand pij are interpolating polynomials. The stated conditions forstability and for asymptotic stability now follow readily fromTheorem 1.2 and 1.3.

Example 2.1.

1.

A =

[1 −10 2

]The eigenvalues are λ1 = 1 and λ2 = 2. Then the equilibrium 0is unstable.


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2.

A =

[−1 10 −2

]The eigenvalues are λ1 = −1 and λ2 = −2. Then theequilibrium 0 is exponentially stable.

3.

A =

[1 10 −2

]The eigenvalues are λ1 = 1 and λ2 = −2. Then the equilibrium0 is unstable.

4.

A =

[0 −11 0

]The eigenvalues are λ1 = i and λ2 = −i. Then the equilibrium0 is stable.


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5.

A =

[1 −11 1

]The eigenvalues are λ1 = 1 + i and λ2 = 1− i. Then theequilibrium 0 is unstable.

Thus in the case of linear time-invariant systems of the form (19),the stability status of the equilibrium 0 can be ascertained bystudying the eigenvalues of A. However it is possible to formulatean entirely different approach to the problem, based on the use ofquadratic Lyapunov functions. This theory is of interest in itself,and is also useful in studying non-linear systems using linearizationmethods.


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Given the system (19), the idea is to choose a Lyapunov functioncandidate of the form

V (x) = x′Px (22)

where P is a real symmetric matrix. Then V is given by

V (x) = x′Px+ x′Px = −x′Qx (23)

whereA′P + PA = −Q (24)

Equation (24) is commonly known as the Lyapunov MatrixEquation. By means of this equation, it is possible to study thestability properties of the equilibrium 0 of the system (19). Forexample, if a pair of matrices (P,Q) satisfying (24) can be found

such that both P and Q are positive definite, then both V and −Vare positive definite functions, and V is radially unbounded. Hence,by Theorem 3.4 in Chapter 3, 0 is globally exponentially stable. Onthe other hand, if a pair (P,Q) can be found such that Q is positive

definite and P has at least one nonpositive eigenvalue, then −V ispositive definite, and V assumes nonpositive values arbitrarily closeto the origin. Hence, by Theorem 88, the origin is unstable.


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There are two ways in which (24) can be tackled :

1) Given a matrix A, one can pick a particular matrix P and studythe properties of the matrix Q resulting from (24).

2) Given a matrix A, one can pick a particular matrix Q and studythe matrix P resulting from (24).


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One difficulty with selecting Q and trying to find the correspondingP is that, depending on the matrix A, (24) may not have a uniquesolution for P . The next result gives necessary and sufficientconditions under which (24) has a unique solution corresponding toeach Q.

Lemma 2.1. Let A ∈ Rn×n, and let {λ1, λ2, . . . , λn} denote the(not necessarily distinct) eigenvalues of A. Then (24) has a uniquesolution for P corresponding to each Q ∈ Rn×n if and only if

λi + λj 6= 0, ∀i, j (25)

On the basis of Lemma 2.1, one can state the following corollary :

Corollary 2.1. If for some choice of Q ∈ Rn×n, Equation (24)does not have a unique solution P , then the origin is notasymptotically stable.

Proof : If all the eigenvalues of A have negative real parts, then(25) is satisfied.


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The following lemma provides an alternate characterization of thesolutions of (24). Note that a matrix A is called Hurwitz if all ofits eigenvalues have negative real parts.

Lemma 2.2. Let A be a Hurwitz matrix. Then, for eachQ ∈ Rn×n, the corresponding unique solution of (24) is given by

P =

∫ ∞0

eA′tQeAtdt (26)

Proof : If A is Hurwitz, then the condition (25) is satisfied, and(24) has a unique solution for P corresponding to each Q ∈ Rn×n.Moreover, if A is Hurwitz, then the integral on the right side of(26) is well-defined.

Let M denote this integral. It is now shown that

A′M +MA = −Q (27)


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By the uniqueness of solutions to (24), it then follows that P = M .To prove (27), observe that

A′M +MA =

∫ ∞0

[A′eA′tQeAt + eA

′tQeAtA]dt

=

∫ ∞0

d[eA′tQeAt] =

[eA

′tQeAt]∞0

= −QThis completes the proof.

Remark 2.1. Note that the above lemma also provides aconvenient way to compute infinite integrals of the form (26).


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We can now state one of the main results of the Lyapunov matrixequation :

Theorem 2.2. Given a matrix A ∈ Rn×n, the following threestatements are equivalent :

(1) A is a Hurwitz matrix.

(2) There exists some positive definite matrix Q ∈ Rn×n such that(24) has a corresponding unique positive definite solution P .

(3) For every positive definite matrix Q ∈ Rn×n, (24) has a uniquepositive definite solution P .

Proof : “(3)⇒ (2)” Obvious.

“(2)⇒ (1)” Suppose (2) is true for some particular Q. Then we canapply Theorem 3.3 in Chapter 3, with the Lyapunov functioncandidate V (x) = x′Px. Then V (x) = −x′Qx, and one canconclude that 0 is asymptotically stable. By Theorem 2.1, thisimplies that A is Hurwitz.


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“(1)⇒ (3)” Suppose A is Hurwitz and let Q be positive definitebut otherwise arbitrary. By Lemma 2.2, Equation (24) has acorresponding unique solution P given by (26). It only remains toshow that P is positive definite. For this purpose, factor Q in theform M ′M where M is nonsingular. Now it is claimed that P ispositive definite because

x′Px > 0, ∀x 6= 0 (28)

with Q = M ′M , P becomes

P =

∫ ∞0

eA′tM ′MeAtdt (29)

Thus, for any x ∈ Rn,

x′Px =

∫ ∞0

x′eA′tM ′MeAtxdt =

∫ ∞0

‖MeAtx‖22dt ≥ 0 (30)

where ‖ · ‖2 denotes the Euclidean norm. Next, if x′Px = 0, then

MeAtx = 0, ∀t ≥ 0 (31)

Substituting t = 0 gives Mx = 0, which implies x = 0. Hence, P is


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positive definite.

Remark 2.2. Theorem 2.2 is very important in that it enables oneto determine the stability status of the equilibrium 0 in thefollowing manner : Given A ∈ Rn×n, pick Q ∈ Rn×n to be anypositive definite matrix (a logical choice is the identity matrix).Attempt to solve (24) for P .

(a) If (24) has no solution or has non-unique solution, then 0 isnot asymptotically stable.

(b) If P is unique but not positive definite, then once again 0 is notasymptotically stable.

(c) If P is uniquely determined and positive definite, then 0 isasymptotically stable.


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Example 2.2. In this example we demonstrate the necessary stepsrequired in applying the Lyapunov stability test. Consider thefollowing continuous time invariant system represented by x1(t)

x2(t)x3(t)

=

0 1 01 0 1−1 −2 −3

x1(t)x2(t)x3(t)

(32)

It is easy to check by MATLAB function eig that the eigenvalues ofthis system are

λ1 = −2.3247, λ2 = −0.3367 + i0.5623, λ3 = −0.3367− i0.5623

and hence this system is asymptotically stable. In order to apply theLyapunov method, we first choose a positive definite matrix Q. Thestandard initial guess for Q is identity, i.e. Q = I3. With the helpof the MATLAB function lyap (used for solving the algebraicLyapunov equation), we can execute the following statementP = lyap(A′, Q) and obtain the solution P as

P =

2.3 2.1 0.52.1 4.6 1.30.5 1.3 0.6


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Examining the positive definiteness of the matrix P (all eigenvaluesof P must be in the closed right half plane), we get that theeigenvalues of this matrix are given

6.1827, 1.1149, 0.2024

hence P is positive definite and the Lyapunov test indicates that thesystem under consideration is asymptotically stable.

It can be seen from this particular example that the Lyapunovstability test is not numerically very efficient since we have first tosolve the linear algebraic Lyapunov equation and then to test thepositive definiteness of the matrix P , which requires finding itseigenvalues. Of course, we can find the eigenvalue of the matrix Aimmediately and from that information determine the systemstability. It is true that the Lyapunov stability test is not the rightmethod to test the stability of linear systems when the systemmatrix is given by numerical entries. However, it can be used as auseful concept in theoretical considerations, e.g. to prove someother stability results.


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Theorem 2.2 shows that, if A is Hurwitz and Q is positive definite .The next result shows that, under certain conditions, P is positivedefinite even when Q is positive semi-definite.Lemma 2.3. Suppose A ∈ Rn×n and satisfies (25). SupposeC ∈ Rn×n, and that

rank L =

CCA

...CAn−1

= n (33)

Under these conditions, the equation

A′P + PA = −C ′C (34)

has a unique positive definite solution P .

Proof : Existence and uniqueness of P follows from Lemma (2.1).To show that P is positive definite, we have

x′Px = 0⇒ CeAtx = 0, ∀t ≥ 0

Let f(t) = CeAtx. Then f as well as all its derivatives are


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identically zero. In particular, Lx = 0, which implies x = 0. Hence,P is positive definite.

Example 2.3. Consider the same system matrix as in Example2.2 with the matrix Q1 obtained from

Q1 = CTC =

001

[ 0 0 1]

=

0 0 00 0 00 0 1

Note that the rank of the matrix L is 3, then L is full rank. Then,the Lyapunov algebraic equation

A′P1 + P1A = −Q1 (35)

has the positive definite solution

P1 =

0.1 0.2 00.2 0.7 0.10 0.1 0.2

which can be confirmed by finding the eigenvalues of P1, so that theconsidered linear system is asymptotically stable.

3 TIME-VARYING SYSTEMS 26'

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3 Time-Varying Systems

Here, we interested in the stability of the following time-varyingsystem

x(t) = A(t)x(t), t ≥ 0 (36)

In the case of linear time-varying systems, the stability status ofthe equilibrium 0 can be ascertained, in principle at least, bystudying the state transition matrix.

Existence of Quadratic Lyapunov Functions

For time-invariant systems, it has been shown that if 0 isexponentially stable then a quadratic Lyapunov function exists. Asimilar result is now proved for time-varying systems, under theassumption that 0 is exponentially stable. This result is based ontwo preliminary lemmas :


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Lemma 3.1. Suppose Q : R+ → Rn×n is continuous and bounded,and that the equilibrium 0 of (36) is uniformly asymptoticallystable. Then, for each t ≥ 0, the matrix

P (t) =

∫ ∞t

Φ′(τ, t)Q(τ)Φ(τ, t)dτ (37)

is well-defined ; moreover, P (t) is bounded as a function of t.

Proof : The assumption of uniform asymptotic stability impliesthat 0 is exponentially stable. Thus, there exist constants m,λ > 0such that

‖Φ(τ, t)‖i ≤ m exp[−λ(τ − t)], ∀τ ≥ t ≥ 0. (38)

The previous bound, together with the boundeness of Q(·), provesthe lemma.


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Lemma 3.2. Suppose that, in addition to the assumption ofLemma (3.1), the following conditions also hold :

(1) Q(t) is symmetric and positive definite for each t ≥ 0 ;moreover, there exists a constant α > 0 such that

αx′x ≤ x′Q(t)x, ∀t ≥ 0,∀x ∈ Rn. (39)

(2) The matrix A(·) is bounded, i.e.

m0 := supt≥0‖A(t)‖i,2 <∞ (40)

Under these conditions, the matrix P (t) is defined in (37) ispositive definite for each t ≥ 0 ; moreover, there exists a constantβ > 0 such that

βx′x ≤ x′P (t)x, ∀t ≥ 0, ∀x ∈ Rn. (41)


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Theorem 3.1. Suppose Q(·) and A(·) satisfy the hypotheses ofLemma 3.2. Then for each function Q(·) satisfying the hypotheses,the function

V (t, x) = x′P (t)x

is a Lyapunov function for establishing the exponential stability ofthe equilibrium 0.

Proof : With V (t, x) defined as above, we have

V (t, x) = x′[P (t) +A′(t)P (t) + P (t)A(t)]x

It is easy to verify by differentiating (37) that

P (t) = −A′(t)P (t)− P (t)A(t)−Q(t) (42)

HenceV (t, x) = −x′Q(t)x

Thus the functions V and V satisfy all the needed conditions.


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Example 3.1. Consider the linear time-varying system{x1(t) = 1

2 (cos t− esin t)x1(t) + sin2 t · x2(t)x2(t) = − sin2 t · x1(t) + 1

2 (cos t− esin t)x2(t)

Then

A(t) =

[12 (cos t− esin t) sin2 t− sin2 t 1

2 (cos t− esin t)

].

Taking Q = I, simple calculation shows that

P (t) =

[e− sin t 0

0 e− sin t

]is a solution of (42). The system is exponentially stable.

Chapter III Stability of Linear Systemsaksikas/math603-cha4.pdf · conditions for the stability of linear systems, but also pave the way to deriving Lyapunov’s linearization method,

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