khalil nonlinear systems third edition chapter 14 and appendix

Chapter 14

Nonlinear Design Tools

The complexity of nonlinear feedback control challenges us to come up with system-atic design procedures to meet control objectives and design specifications. Facedwith such challenge, it is clear that we cannot expect one particular procedure to ap-ply to all nonlinear systems. It is also unlikely that the whole design of a nonlinearfeedback controller can be based on one particular tool. What a control engineerneeds is a set of analysis and design tools that cover a wide range of situations.When working with a particular application, the engineer will need to employ thetools that are most appropriate for the problem in hand. We have already coveredseveral such tools in the earlier chapters. In this chapter, we assemble five nonlineardesign tools, which are simple enough to be presented in an introductory textbook,and practical enough to have been used in real-world problems.1

In the first two sections, we deal with robust control under the matching condi-tion; that is, when uncertain terms enter the state equation at the same point as thecontrol input. In the sliding mode control of Section 14.1, trajectories are forced toreach a sliding manifold in finite time and to stay on the manifold for all future time.Motion on the manifold is independent of matched uncertainties. By using a lowerorder model, the sliding manifold is designed to achieve the control objective. TheLyapunov redesign of Section 14.2 uses a Lyapunov function of a nominal systemto design an additional control component that makes the design robust to largematched uncertainties. Both sliding mode control and Lyapunov redesign producediscontinuous controllers, which could suffer from chattering in the presence of de-lays or unmodeled high-frequency dynamics. Therefore, we develop “continuous”versions of the controllers. In Section 14.1, we use a second-order example to mo-tivate the main elements of the sliding-mode-control technique, then we proceedto present stabilization, tracking, and integral control results. In Section 14.2, weshow how Lyapunov redesign can be used to achieve stabilization, and introducenonlinear damping— a technique that guarantees boundedness of trajectories even

1For more nonlinear design tools, see [88], [89], [103], [124], [153], [167], and [172].

551

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552 CHAPTER 14. NONLINEAR DESIGN TOOLS

when no upper bound on the uncertainty is known.The matching condition can be relaxed via the backstepping technique, intro-

duced in Section 14.3. Backstepping is a recursive procedure that interlaces thechoice of a Lyapunov function with the design of feedback control. It breaks a de-sign problem for the full system into a sequence of design problems for lower order(even scalar) subsystems. By exploiting the extra flexibility that exists with lowerorder and scalar subsystems, backstepping can often solve stabilization, tracking,and robust control problems under conditions less restrictive than those encounteredin other methods.

Passivity-based control exploits passivity of the open-loop system in the designof feedback control. Stabilizing a passive system at an equilibrium point amountsto damping injection. In Section 14.4, we describe the basic idea of passivity-based control. We also describe feedback passivation–a technique that uses feedbackcontrol to convert a nonpassive system into a passive one.

Most of the design tools presented in Chapters 12 through 14 require statefeedback. In Section 14.5, we introduce high-gain observers, which allow us toextend many of those tools to output feedback for a particular class of nonlinearsystems.2 The main idea of Section 14.5 is that the performance under globallybounded state feedback control can be recovered by output feedback control whenthe observer gain is sufficiently high.

14.1 Sliding Mode Control

14.1.1 Motivating Example

Consider the second-order system

x1 = x2

x2 = h(x) + g(x)u

where h and g are unknown nonlinear functions and g(x) ≥ g0 > 0 for all x. Wewant to design a state feedback control law to stabilize the origin. Suppose wecan design a control law that constrains the motion of the system to the manifold(or surface) s = a1x1 + x2 = 0. On this manifold, the motion is governed byx1 = −a1x1. Choosing a1 > 0 guarantees that x(t) tends to zero as t tends toinfinity and the rate of convergence can be controlled by choice of a1. The motionon the manifold s = 0 is independent of h and g. How can we bring the trajectoryto the manifold s = 0 and maintain it there? The variable s satisfies the equation

s = a1x1 + x2 = a1x2 + h(x) + g(x)u

Suppose h and g satisfy the inequality∣∣∣∣a1x2 + h(x)g(x)

∣∣∣∣ ≤ �(x), ∀ x ∈ R2

2Other output feedback control tools are described in Exercises 14.47 through 14.49.


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14.1. SLIDING MODE CONTROL 553

for some known function �(x). With V = (1/2)s2 as a Lyapunov function candidatefor s = a1x2 + h(x) + g(x)u, we have

V = ss = s[a1x2 + h(x)] + g(x)su ≤ g(x)|s|�(x) + g(x)su

Takingu = −β(x) sgn(s)

where β(x) ≥ �(x) + β0, β0 > 0, and

sgn(s) =

⎧⎨⎩

1, s > 00, s = 0

−1, s < 0

yields

V ≤ g(x)|s|�(x)− g(x)[�(x) + β0]s sgn(s) = −g(x)β0|s| ≤ −g0β0|s|

Thus, W =√

V = |s| satisfies the differential inequality

D+W ≤ −g0β0

and the comparison lemma shows that

W (s(t)) ≤ W (s(0))− g0β0t

Therefore, the trajectory reaches the manifold s = 0 in finite time and, once onthe manifold, it cannot leave it, as seen from the inequality V ≤ −2g0β0|s|. Insummary, the motion consists of a reaching phase during which trajectories startingoff the manifold s = 0 move toward it and reach it in finite time, followed by asliding phase during which the motion is confined to the manifold s = 0 and thedynamics of the system are represented by the reduced-order model x1 = −a1x1. Asketch of the phase portrait is shown in Figure 14.1. The manifold s = 0 is calledthe sliding manifold and the control law u = −β(x) sgn(s) is called sliding modecontrol. The striking feature of sliding mode control is its robustness with respectto h and g. We only need to know the upper bound �(x) and during the slidingphase, the motion is completely independent of h and g.

The sliding mode controller simplifies if, in some domain of interest, h and gsatisfy the inequality ∣∣∣∣a1x2 + h(x)

g(x)

∣∣∣∣ ≤ k1

for some known nonnegative constant k1. In this case, we can take

u = −k sgn(s), k > k1


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s=0

Figure 14.1: A typical phase portrait under sliding mode control.

which takes the form of a simple relay. This form, however, usually leads to a finiteregion of attraction, which can be estimated as follows: The condition ss ≤ 0 in theset {|s| ≤ c} makes it positively invariant. From the equation

x1 = x2 = −a1x1 + s

and the function V1 = (1/2)x21, we have

V1 = x1x1 = −a1x21 + x1s ≤ −a1x

21 + |x1|c ≤ 0, ∀ |x1| ≥ c

a1

Thus,|x1(0)| ≤ c

a1⇒ |x1(t)| ≤ c

a1, ∀ t ≥ 0

and the set

Ω ={|x1| ≤ c

a1, |s| ≤ c

}sketched in Figure 14.2, is positively invariant if∣∣∣∣a1x2 + h(x)

g(x)

∣∣∣∣ ≤ k1, ∀ x ∈ Ω

Moreover, every trajectory starting in Ω approaches the origin as t tends to infinity.By choosing c large enough, any compact set in the plane can be made a subsetof Ω. Therefore, if k can be chosen arbitrarily large, the foregoing control law canachieve semiglobal stabilization.

Because of imperfections in switching devices and delays, sliding mode controlsuffers from chattering. The sketch of Figure 14.3 shows how delays can causechattering. It depicts a trajectory in the region s > 0 heading toward the sliding


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�

�

��

��

x1

x2s = 0

c/a1

c

Figure 14.2: Estimate of the region of attraction.

��

��

��

��

��

��

��

��

��

Sliding manifold

a

s < 0 s > 0

Figure 14.3: Chattering due to delay in control switching.

manifold s = 0. It first hits the manifold at point a. In ideal sliding mode control,the trajectory should start sliding on the manifold from point a. In reality, therewill be a delay between the time the sign of s changes and the time the controlswitches. During this delay period, the trajectory crosses the manifold into theregion s < 0. When the control switches, the trajectory reverses its direction andheads again toward the manifold. Once again it crosses the manifold, and repetitionof this process creates the “zig-zag” motion (oscillation) shown in the sketch, whichis known as chattering. Chattering results in low control accuracy, high heat lossesin electrical power circuits, and high wear of moving mechanical parts. It may alsoexcite unmodeled high-frequency dynamics, which degrades the performance of thesystem and may even lead to instability.

To get a better feel for chattering, we simulate the sliding mode control of the


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0 2 4 6 80

0.5

1

1.5

2

Time

θ

0 0.05 0.1 0.15 0.2−2

−1.5

−1

−0.5

0

0.5

Time

s

Figure 14.4: Ideal sliding mode control.

pendulum equation

x1 = x2

x2 = −(g0/�) sin(x1 + δ1)− (k0/m)x2 + (1/m�2)uu = −k sgn(s) = −k sgn(a1x1 + x2)

to stabilize the pendulum at δ1 = π/2, where x1 = θ−δ1 and x2 = θ. The constantsm, �, k0, and g0 are the mass, length, coefficient of friction, and acceleration due togravity, respectively. We take a1 = 1 and k = 4. The gain k = 4 is chosen by using∣∣∣∣a1x2 + h(x)

g

∣∣∣∣ =∣∣�2(m− k0)x2 −mg0� cos(x1)

∣∣≤ �2|m− k0|(2π) + mg0� ≤ 3.68

where the bound is calculated over the set {|x1| ≤ π, |x1 + x2| ≤ π} for 0.05 ≤m ≤ 0.2, 0.9 ≤ � ≤ 1.1, and 0 ≤ k0 ≤ 0.05. The simulation is performed by usingm = 0.1, � = 1, and k0 = 0.02. Figure 14.4 shows ideal sliding mode control,while Figure 14.5 shows a nonideal case where switching is delayed by unmodeledactuator dynamics having the transfer function 1/(0.01s + 1)2.

We will present two ideas for reducing or eliminating chattering. The first ideais to divide the control into continuous and switching components so as to reducethe amplitude of the switching one. Let h(x) and g(x) be nominal models of h(x)and g(x), respectively. Taking

u = − [a1x2 + h(x)]g(x)

+ v

results in

s = a1

[1− g(x)

g(x)

]x2 + h(x)− g(x)

g(x)h(x) + g(x)v def= δ(x) + g(x)v


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0 2 4 6 8−0.5

0

0.5

1

1.5

2

Time

θ

2 2.1 2.2 2.3 2.4−4

−2

0

2

4

Time

u

Figure 14.5: Sliding mode control with unmodeled actuator dynamics.

If the perturbation term δ(x) satisfies the inequality∣∣∣∣ δ(x)g(x)

∣∣∣∣ ≤ �(x)

we can takev = −β(x) sgn(s)

where β(x) ≥ �(x)+β0. Because � is an upper bound on the perturbation term, it islikely to be smaller than an upper bound on the whole function. Consequently, theamplitude of the switching component would be smaller. For example, returningto the pendulum equation and taking m = 0.125, � = 1, k0 = 0.025 to be nominalvalues of m, �, k0, we have∣∣∣∣δ(x)

g

∣∣∣∣ =∣∣∣(a1m�2 − a1m�2 − k0�

2 + k0�2)

x2 − g0(m�− m�) cos x1

∣∣∣ ≤ 1.83

where the bound is calculated over the same set as before. The modified slidingmode control is taken as

u = −0.1x2 + 1.2263 cos x1 − 2 sgn(s)

which shows a reduction in the switching term amplitude from 4 to 2. Figure 14.6shows simulation of this modified control in the presence of unmodeled actuatordynamics. The reduction in the amplitude of chattering is clear.

The second idea to eliminate chattering is to replace the signum function by ahigh-slope saturation function; that is, the control law is taken as

u = −β(x) sat(s

ε

)


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0 2 4 6 80

0.5

1

1.5

2

Time

θ

2 2.1 2.2 2.3 2.4−1

0

1

2

3

Time

u

Figure 14.6: Modified sliding mode control with unmodeled actuator dynamics.

�

�

−1

1

y

sgn(y)

�

�

−1

1

yε

sat(y/ε)

Figure 14.7: The signum nonlinearity and its saturation function approximation.

where sat(·) is the saturation function defined by

sat(y) =

⎧⎨⎩

y, if |y| ≤ 1

sgn(y), if |y| > 1

and ε is a positive constant. The signum and saturation functions are shown inFigure 14.7. The slope of the linear portion of sat(s/ε) is 1/ε. Good approximationrequires the use of small ε. In the limit, as ε → 0, the saturation function sat(s/ε)approaches the signum function sgn(s). To analyze the performance of the “contin-uous” sliding mode controller, we examine the reaching phase by using the functionV = (1/2)s2 whose derivative satisfies the inequality

V ≤ −g0β0|s|


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when |s| ≥ ε; that is, outside the boundary layer {|s| ≤ ε}. Therefore, whenever|s(0)| > ε, |s(t)| will be strictly decreasing, until it reaches the set {|s| ≤ ε} in finitetime and remains inside thereafter. Inside the boundary layer, we have

x1 = −a1x1 + s

where |s| ≤ ε. The derivative of V1 = (1/2)x21 satisfies

V1 = −a1x21 + x1s ≤ −a1x

21 + |x1|ε ≤ −(1− θ1)a1x

21, ∀ |x1| ≥ ε

a1θ1

where 0 < θ1 < 1. Thus, the trajectory reaches the set Ωε = {|x1| ≤ ε/(a1θ1), |s| ≤ε} in finite time. In general, we do not stabilize the origin, but we achieve ultimateboundedness with an ultimate bound that can be reduced by decreasing ε. Whathappens inside Ωε is problem dependent. Let us consider again the pendulumequation and see what happens inside Ωε in that case. Inside the boundary layer{|s| ≤ ε}, the control reduces to the linear feedback law u = −ks/ε, and the closed-loop system

x1 = x2

x2 = −(g0/�) sin(x1 + δ1)− (k0/m)x2 − (k/m�2ε)(a1x1 + x2)

has a unique equilibrium point at (x1, 0), where x1 satisfies the equation

εmg0� sin(x1 + δ1) + ka1x1 = 0

and can be approximated for small ε by x1 ≈ −(εmg0�/ka1) sin δ1. Shifting theequilibrium point to the origin by the change of variables

y1 = x1 − x1, y2 = x2

results in

y1 = y2

y2 = −σ(y1)−(

k0

m+

k

m�2ε

)y2

where

σ(y1) = (g0/�)[sin(y1 + x1 + δ1)− sin(x1 + δ1)] + (ka1/m�2ε)y1

Using

V =∫ y1

0σ(s) ds + (1/2)y2

2

as a Lyapunov function candidate, it can be verified that

V ≥ −(g0/2�)y21 + (ka1/2m�2ε)y2

1 + (1/2)y22


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0 2 4 6 80

0.5

1

1.5

2

Time

θ

ε = 0.3

0 0.5 1 1.5 2 2.5

0

1

2

3

4

Time

u

ε = 0.3

0 2 4 6 80

0.5

1

1.5

2

Time

θ

ε = 0.03

0 0.5 1 1.5 2 2.5

0

1

2

3

4

ε = 0.03

Time

u

Figure 14.8: “Continuous” sliding mode control.

is positive definite for (k/ε) > (m�g0/a1) and its derivative satisfies

˙V = −(

k0

m+

k

m�2ε

)y22

Application of the invariance principle shows that the equilibrium point (x1, 0) isasymptotically stable and attracts every trajectory in Ωε.

For better accuracy, we need to choose ε as small as possible, but a too smallvalue of ε will induce chattering in the presence of time delays or unmodeled fastdynamics. Figure 14.8 shows the performance of the “continuous” sliding modecontroller when applied to the pendulum equation for two different values of ε.Figure 14.9 shows the performance in the presence of the unmodeled actuator1/(0.01s + 1)2. The striking observation here is that, while reducing ε improvesaccuracy in the ideal case, it may not have the same effect in the presence of delaysdue to chattering.

One special case where we can stabilize the origin without pushing ε too smallarises when h(0) = 0. In this case, the system, represented inside the boundary


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0 2 4 6 80

0.5

1

1.5

2

Time

θ

ε = 0.3

0 2 4 6 8−3

−2

−1

0

1

2

3

4

Time

u

ε = 0.3

0 2 4 6 80

0.5

1

1.5

2

Time

θ

ε = 0.03

2 2.1 2.2 2.3 2.4−4

−2

0

2

4ε = 0.03

Time

u

Figure 14.9: “Continuous” sliding mode control with unmodeled actuator dynamics.

layer by

x1 = x2

x2 = h(x)−[g(x)k

ε

](a1x1 + x2)

has an equilibrium point at the origin. We need to choose ε small enough to stabilizethe origin and make Ωε a subset of its region of attraction. For the pendulumequation with δ1 = π, repeating the foregoing stability analysis confirms that wecan achieve our goal if k/ε > mg0�/a1. For � ≤ 1.1, m ≤ 0.2, k = 4 and a1 = 1, weneed ε < 1.8534. Figure 14.10 shows simulation results for ε = 1.

If δ1 is any angle other than 0 or π (the open-loop equilibrium points), thesystem will stabilize at an equilibrium point other than the origin, leading to asteady-state error, which was approximated earlier by (εmg0�/ka1) sin δ1. We canstill achieve zero steady-state error by using integral action. Let x0 =

∫x1 and


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0 2 4 6 80

0.5

1

1.5

2

2.5

3

3.5

Time

θ

ε = 1

0 2 4 60

0.5

1

1.5

2

2.5

3

Time

u

ε = 1

Figure 14.10: “Continuous” sliding mode control when δ1 = π.

write the augmented system as

x0 = x1

x1 = x2

x2 = −(g0/�) sin(x1 + δ1)− (k0/m)x2 + (1/m�2)u

Take s = a0x0 + a1x1 + x2, where the matrix

A0 =[

0 1−a0 −a1

]

is Hurwitz. If

m�2 |a0x1 + a1x2 − (g0/�) sin(x1 + δ1)− (k0/m)x2| ≤ k1

over the domain of interest, we can take the “continuous” sliding mode control as

u = −k sat(s

ε

), k > k1

which ensures that s will reach the boundary-layer {|s| ≤ ε} in finite time, since

ss ≤ −(k − k1)|s|, for |s| ≥ ε

Inside the boundary layer, the system is represented by

η = A0η + B0s, where η =[

x0x1

], B0 =

[01

]

Taking V1 = ηT P0η, where P0 is the solution of the Lyapunov equation P0A0 +A0P

T0 = −I, it can be verified that

V1 = −ηT η + 2ηT P0B0s ≤ −(1− θ1)‖η‖22, ∀ ‖η‖2 ≥ 2‖P0B0‖2ε/θ1


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0 5 10 150

0.5

1

1.5

2

Time

θ

ε = 1

0 2 4 6 8−0.5

0

0.5

1

1.5

2

Time

u

ε = 1

Figure 14.11: “Continuous” sliding mode control with integral action when δ1 = π/2 .

where 0 < θ1 < 1. Thus, all trajectories reach the set

Ωε ={

V1(η) ≤ 4‖P0B0‖22ε2‖P0‖2θ21

, |s| ≤ ε

}

in finite time. Inside Ωε, the system

x0 = x1

x1 = x2

x2 = −(g0/�) sin(x1 + δ1)− (k0/m)x2 − (k/m�2ε)(a0x0 + a1x1 + x2)

has a unique equilibrium point at x = [−(εmg0�/ka0) sin δ1, 0, 0]T . Repeatingthe previous stability analysis, it can be shown that for sufficiently small ε theequilibrium point x is asymptotically stable and every trajectory in Ωε convergesto x as t tends to infinity. Hence, θ converges to the desired position δ1. Simulationresults for m = 0.1, � = 1, k0 = 0.02, δ1 = π/2, a0 = a1 = 1, k = 4, and ε = 1 areshown in Figure 14.11.

14.1.2 Stabilization

Consider the system

x = f(x) + B(x)[G(x)E(x)u + δ(t, x, u)] (14.1)

where x ∈ Rn is the state, u ∈ Rp is the control input, and f , B, G, and Eare sufficiently smooth functions in a domain D ⊂ Rn that contains the origin.The function δ is piecewise continuous in t and sufficiently smooth in (x, u) for(t, x, u) ∈ [0,∞) ×D × Rp. We assume that f , B, and E are known, while G and


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δ could be uncertain. Furthermore, we assume that E(x) is a nonsingular matrixand G(x) is a diagonal matrix whose elements are positive and bounded away fromzero; that is, gi(x) ≥ g0 > 0, for all x ∈ D.3 Suppose f(0) = 0 so that, in theabsence of δ, the origin is an open-loop equilibrium point. Our goal is to design astate feedback control law to stabilize the origin for all uncertainties in G and δ.

Let T : D → Rn be a diffeomorphism such that

∂T

∂xB(x) =

[0I

](14.2)

where I is the p× p identity matrix.4 The change of variables[ηξ

]= T (x), η ∈ Rn−p, ξ ∈ Rp (14.3)

transforms the system into the form

η = fa(η, ξ) (14.4)ξ = fb(η, ξ) + G(x)E(x)u + δ(t, x, u) (14.5)

The form (14.4)–(14.5) is usually referred to as the regular form. To design thesliding mode control, we start by designing the sliding manifold s = ξ − φ(η) = 0such that, when motion is restricted to the manifold, the reduced-order model

η = fa(η, φ(η)) (14.6)

has an asymptotically stable equilibrium point at the origin. The design of φ(η)amounts to solving a stabilization problem for the system

η = fa(η, ξ)

with ξ viewed as the control input. This stabilization problem may be solved byusing the techniques of linearization or feedback linearization presented in the pre-vious two chapters or some nonlinear design tools that will be introduced later inthis chapter, such as backstepping or passivity-based control. We assume that wecan find a stabilizing continuously differentiable function φ(η) with φ(0) = 0. Next,we design u to bring s to zero in finite time and maintain it there for all futuretime. Toward that end, let us write the s-equation:

s = fb(η, ξ)− ∂φ

∂ηfa(η, ξ) + G(x)E(x)u + δ(t, x, u) (14.7)

As we saw in the introductory example, we can design u as a pure switching compo-nent or it may contain an additional continuous component that cancels the known

3The method can be extended to the case where G is not diagonal by including the off-diagonalelements in δ. Since the dependence of δ on u will be restricted, the method works for cases whereG is diagonally dominant.

4The existence of T is explored in Exercise 14.9.


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terms on the right-hand side of (14.7).5 If G(x) is a nominal model of G(x), thecontinuous component of u will be −E−1G−1[fb−(∂φ/∂η)fa]. In the absence of un-certainty; that is, when δ = 0 and G is known, taking u = −E−1G−1[fb−(∂φ/∂η)fa]results in s = 0, which ensures that the condition s = 0 can be maintained for allfuture time. To analyze both cases simultaneously, we write the control u as

u = E−1(x){−L(x)

[fb(η, ξ)− ∂φ

∂ηfa(η, ξ)

]+ v

}(14.8)

where L(x) = G−1(x), if the known terms are cancelled, and L = 0, otherwise.Substituting (14.8) into (14.7) yields

si = gi(x)vi + Δi(t, x, v), 1 ≤ i ≤ p (14.9)

where Δi is the ith component of

Δ(t, x, v) = δ(t, x,−E−1(x)L(x)(fb(η, ξ)− (∂φ/∂η)fa(η, ξ)) + E−1(x)v)+ [I −G(x)L(x)] [fb(η, ξ)− (∂φ/∂η)fa(η, ξ)]

and gi is the ith diagonal element of G. We assume that the ratio Δi/gi satisfiesthe inequality∣∣∣∣Δi(t, x, v)

gi(x)

∣∣∣∣ ≤ �(x)+κ0‖v‖∞, ∀ (t, x, v) ∈ [0,∞)×D×Rp, ∀ 1 ≤ i ≤ p (14.10)

where �(x) ≥ 0 (a continuous function) and κ0 ∈ [0, 1) are known. Using theestimate (14.10), we proceed to design v to force s toward the manifold s = 0.Utilizing Vi = (1/2)s2

i as a Lyapunov function candidate for (14.9), we obtain

Vi = sisi = sigi(x)vi + siΔi(t, x, v) ≤ gi(x){sivi + |si|[�(x) + κ0‖v‖∞]}Take6

vi = −β(x) sgn(si), 1 ≤ i ≤ p (14.11)

where

β(x) ≥ �(x)1− κ0

+ β0, ∀ x ∈ D (14.12)

and β0 > 0. Then,

Vi ≤ gi(x)[−β(x) + �(x) + κ0β(x)]|si| = gi(x)[−(1− κ0)β(x) + �(x)]|si|≤ gi(x)[−�(x)− (1− κ0)β0 + �(x)]|si| ≤ −g0β0(1− κ0)|si|

The inequality Vi ≤ −g0β0(1 − κ0)|si| ensures that all trajectories starting off themanifold s = 0 reach it in finite time and those on the manifold cannot leave it.

The procedure for designing a sliding mode stabilizing controller can be sum-marized by the following steps:

5The continuous component is usually referred to as the equivalent control.6For convenience, we take the coefficient of the signum function to be the same for all control

components. See Exercise 14.12 for a relaxation of this restriction.


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• Design the sliding manifold ξ = φ(η) to stabilize the reduced-order system(14.6).

• Take the control u as u = E−1{−G−1[fb − (∂φ/∂η)fa] + v} or u = E−1v.

• Estimate �(x) and κ0 in (14.10), where Δ depends on the choice made in theprevious step.

• Choose β(x) that satisfies (14.12) and take the switching (discontinuous) con-trol v as given by (14.11).

This procedure exhibits model-order reduction because the main design task is per-formed on the reduced-order system (14.6). The key feature of sliding mode controlis its robustness to matched uncertainties. During the reaching phase, the taskof forcing the trajectories toward the sliding manifold and maintaining them thereis achieved by the switching control (14.11), provided β(x) satisfies the inequality(14.12). From (14.10), we see that �(x) is a measure of the size of the uncertainty.Since we do not require �(x) to be small, the switching controller can handle fairlylarge uncertainties, limited only by practical constraints on the amplitude of thecontrol signals. During the sliding phase, the motion of the system, as determinedby (14.6), is independent of the matched uncertain terms G and δ.

The sliding mode controller contains the discontinuous signum function sgn(si),which raises some theoretical as well as practical issues. Theoretical issues likeexistence and uniqueness of solutions and validity of Lyapunov analysis will have tobe examined in a framework that does not require the state equation to have locallyLipschitz right-hand-side functions.7 There is also the practical issue of chatteringdue to imperfections in switching devices and delays, which was illustrated in theintroductory example. To eliminate chattering, we use a continuous approximationof the signum function.8 By using a continuous approximation, we also avoid thetheoretical difficulties associated with discontinuous controllers.9 We approximatethe signum function sgn(si) by the high-slope saturation function sat(si/ε);10 thatis,

vi = −β(x) sat(si

ε

), 1 ≤ i ≤ p (14.13)

where β(x) satisfies (14.12). To analyze the performance of the “continuous” slidingmode controller, we examine the reaching phase by using the Lyapunov function

7To read about differential equations with discontinuous right-hand side, consult [58], [147],[173], and [198].

8Other approaches to eliminate chattering include the use of observers [197] and extendingthe dynamics of the system by using integrators [177]. It should be noted that the continuousapproximation approach cannot be used in applications where actuators have to be used in anon–off operation mode, like thyristors, for example.

9While we do not pursue rigorous analysis of the discontinuous sliding mode controller, thereader is encouraged to use simulations to examine the performance of both the discontinuouscontroller and its continuous approximation.

10Smooth approximations are discussed in Exercises 14.11.


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Vi = (1/2)s2i . The derivative of Vi satisfies the inequality

Vi ≤ gi(x)[−β(x)si sat

(si

ε

)+ �(x)|si|+ κ0β(x)|si|

]In the region |si| ≥ ε, we have

Vi ≤ gi(x)[−(1− κ0)β(x) + �(x)]|si| ≤ −g0β0(1− κ0)|si|which shows that whenever |si(0)| > ε, |si(t)| will decrease until it reaches the set{|si| ≤ ε} in finite time and remains inside thereafter. The set {|si| ≤ ε, 1 ≤ i ≤ p}is called the boundary layer. To study the behavior of η, we assume that, togetherwith the sliding-manifold design ξ = φ(η), there is a (continuously differentiable)Lyapunov function V (η) that satisfies the inequalities

α1(‖η‖) ≤ V (η) ≤ α2(‖η‖) (14.14)

∂V

∂ηfa(η, φ(η) + s) ≤ −α3(‖η‖), ∀ ‖η‖ ≥ γ(‖s‖) (14.15)

for all (η, ξ) ∈ T (D), where α1, α2, α3, and γ are class K functions.11 Noting that

|si| ≤ c, for 1 ≤ i ≤ p ⇒ ‖s‖ ≤ k1c⇒ V ≤ −α3(‖η‖), for ‖η‖ ≥ γ(k1c)

for some positive constant k1,12 we define a class K function α by

α(r) = α2(γ(k1r))

Then,

V (η) ≥ α(c) ⇒ V (η) ≥ α2(γ(k1c)) ⇒ α2(‖η‖) ≥ α2(γ(k1c))⇒ ‖η‖ ≥ γ(k1c) ⇒ V ≤ −α3(‖η‖) ≤ −α3(γ(k1c))

which shows that the set {V (η) ≤ c0} with c0 ≥ α(c) is positively invariant becauseV is negative on the boundary V (η) = c0. (See Figure 14.12.) It follows that theset

Ω = {V (η) ≤ c0} × {|si| ≤ c, 1 ≤ i ≤ p}, with c0 ≥ α(c) (14.16)

is positively invariant whenever c > ε and Ω ⊂ T (D). Choose ε, c > ε, and c0 ≥ α(c)such that Ω ⊂ T (D). The compact set Ω serves as our estimate of the “region ofattraction.” For all initial states in Ω, the trajectories will be bounded for all t ≥ 0.After some finite time, we have |si(t)| ≤ ε. It follows from the foregoing analysisthat V ≤ −α3(γ(k1ε)) for all V (η) ≥ α(ε). Hence, the trajectories will reach thepositively invariant set

Ωε = {V (η) ≤ α(ε)} × {|si| ≤ ε, 1 ≤ i ≤ p} (14.17)11Inequality (14.15) implies local input-to-state stability of the system η = fa(η, φ(η)+ s) when

s is viewed as the input. (See Exercise 4.60.)12The constant k1 depends on the type of norm used in the analysis.


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α(.)

α(ε)

α(c)

c0

ε c

V

|s|

Figure 14.12: A representation of the set Ω for a scalar s. V < 0 above the α(·)-curve.

in finite time. The set Ωε can be made arbitrarily small by choosing ε small enough.In the limit, as ε → 0, Ωε shrinks to the origin, which shows that the “continuous”sliding mode controller recovers the performance of its discontinuous counterpart.Finally, we note that if all the assumptions hold globally and V (η) is radially un-bounded, we can choose Ω arbitrarily large to include any initial state. We summa-rize our conclusions in the next theorem.

Theorem 14.1 Consider the system (14.4)–(14.5). Suppose there exist φ(η), V (η),�(x), and κ0, which satisfy (14.10), (14.14), and (14.15). Let u and v be given by(14.8) and (14.11), respectively. Suppose ε, c > ε, and c0 ≥ α(c) are chosen suchthat the set Ω, defined by (14.16), is contained in T (D). Then, for all (η(0), ξ(0)) ∈Ω, the trajectory (η(t), ξ(t)) is bounded for all t ≥ 0 and reaches the positivelyinvariant set Ωε, defined by (14.17), in finite time. Moreover, if the assumptionshold globally and V (η) is radially unbounded, the foregoing conclusion holds for anyinitial state. �

The theorem shows that the “continuous” sliding mode controller achieves ul-timate boundedness with an ultimate bound that can be controlled by the designparameter ε. It also gives conditions for global ultimate boundedness. Since theuncertainty δ could be nonvanishing at x = 0, ultimate boundedness is the best wecan expect, in general. If, however, δ vanishes at the origin, then we may be ableto show asymptotic stability of the origin, as we do in the next theorem.

Theorem 14.2 Suppose all the assumptions of Theorem 14.1 are satisfied with�(0) = 0 and κ0 = 0. Suppose further that the origin of η = fa(η, φ(η)) is expo-nentially stale. Then, there exits ε∗ > 0 such that for all 0 < ε < ε∗, the originof the closed-loop system is exponentially stable and Ω is a subset of its region ofattraction. Moreover, if the assumptions hold globally, the origin will be globallyuniformly asymptotically stable. �


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Proof: For inequality (14.10) to hold with κ0 = 0, Δi must be independent ofv. Therefore, we write Δi = Δi(t, x). Theorem 14.1 confirms that all trajectoriesstarting in Ω enter Ωε in finite time. Inside Ωε, the closed-loop system is given by

η = fa(η, φ(η) + s)

si = Δi(t, x)− gi(x)β(x)ε

si, 1 ≤ i ≤ p

By (the converse Lyapunov) Theorem 4.14, there exists a Lyapunov function V0(η)that satisfies

c1‖η‖22 ≤ V0(η) ≤ c2‖η‖22∂V0

∂ηfa(η, φ(η)) ≤ −c3‖η‖22∥∥∥∥∂V0

∂η

∥∥∥∥2≤ c4‖η‖2

in some neighborhood Nη of η = 0. By the smoothness of fa and Δ and the factthat |Δi(t, x)| ≤ gi(x)�(x) with �(0) = 0, we have

‖fa(η, φ(η) + s)− fa(η, φ(η))‖2 ≤ k1‖s‖2‖Δ‖2 ≤ k2‖η‖2 + k3‖s‖2

in some neighborhood N of (η, ξ) = (0, 0). Choose ε small enough that Ωε ⊂ Nη

and Ωε ⊂ N . Using the Lyapunov function candidate

W = V0(η) + 12

p∑i=1

s2i

it can be shown that

W ≤ −c3‖η‖22 + c4k1‖η‖2‖s‖2 + k2‖η‖2‖s‖2 + k3‖s‖22 −β0g0

ε‖s‖22

The right-hand side can be made negative definite in Ωε by choosing ε small enough.The rest of the proof is straightforward. �

The basic idea of the foregoing proof is that, inside the boundary layer, thecontrol vi = −β(x)si/ε acts as high-gain feedback control for small ε. By choosingε small enough, the high-gain feedback stabilizes the origin. We could have usedthe high-gain feedback control throughout the space, but the control amplitude willbe too high when s is far from zero.

We have emphasized the robustness of sliding mode control with respect tomatched uncertainties. What about unmatched uncertainties? Suppose equation(14.1) is modified to

x = f(x) + B(x)[G(x)E(x)u + δ(t, x, u)] + δ1(x) (14.18)


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The change of variables (14.3) transforms the system into

η = fa(η, ξ) + δa(η, ξ)ξ = fb(η, ξ) + G(x)E(x)u + δ(t, x, u) + δb(η, ξ)

where [δa

δb

]=

∂T

∂xδ1

The term δb is added to the matched uncertainty δ. Its only effect is to change theupper bound on Δi/gi. The term δa, on the other hand, is unmatched. It changesthe reduced-order model on the sliding manifold to

η = fa(η, φ(η)) + δa(η, φ(η))

The design of φ will have to guarantee asymptotic stability of the origin η = 0 inthe presence of the uncertainty δa. This is a robust stabilization problem that maybe approached by other robust stabilization techniques such as high-gain feedback.The difference between matched and unmatched uncertainties is that sliding modecontrol guarantees robustness for any matched uncertainty provided an upper boundis known and the needed control effort can be provided. There is no such guaranteefor unmatched uncertainties. We may have to restrict its size to robustly stabilizethe system on the sliding manifold. The next two examples illustrate these points.

Example 14.1 Consider the second-order system

x1 = x2 + θ1x1 sin x2

x2 = θ2x22 + x1 + u

where θ1 and θ2 are unknown parameters that satisfy |θ1| ≤ a and |θ2| ≤ b for someknown bounds a and b. The system is already in the regular form with η = x1 andξ = x2. Uncertainty due to θ2 is matched, while uncertainty due to θ1 is unmatched.We consider the system

x1 = x2 + θ1x1 sin x2

and design x2 to robustly stabilize the origin x1 = 0. This can be achieved withx2 = −kx1, k > a, because

x1x1 = −kx21 + θ1x

21 sin(−kx1) ≤ −(k − a)x2

1

The sliding manifold is s = x2 + kx1 = 0 and

s = θ2x22 + x1 + u + k(x2 + θ1x1 sin x2)

To cancel the known term on the right-hand side, we take

u = −x1 − kx2 + v


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to obtains = v + Δ(x)

where Δ(x) = θ2x22 + kθ1x1 sin x2. Because

|Δ(x)| ≤ ak|x1|+ bx22

we takeβ(x) = ak|x1|+ bx2

2 + β0, β0 > 0

andu = −x1 − kx2 − β(x) sgn(s)

This controller, or its continuous approximation with sufficiently small ε, globallystabilizes the origin.

In the foregoing example, we were able to use high-gain feedback to robustly stabilizethe reduced-order model for unmatched uncertainties that satisfy |θ1| ≤ a, withouthaving to restrict a. In general, this may not be possible, as illustrated by the nextexample.

Example 14.2 Consider the second-order system

x1 = x1 + (1− θ1)x2

x2 = θ2x22 + x1 + u

where θ1 and θ2 are unknown parameters that satisfy |θ1| ≤ a and |θ2| ≤ b. Weconsider the system

x1 = x1 + (1− θ1)x2

and design x2 to robustly stabilize the origin x1 = 0. We note that the systemis not stabilizable at θ1 = 1. Hence, we must limit a to be less than one. Usingx2 = −kx1, we obtain

x1x1 = x21 − k(1− θ1)x2

1 ≤ −[k(1− a)− 1]x21

Hence, the origin x1 = 0 can be stabilized by taking k > 1/(1 − a). The slidingmanifold is s = x2 + kx1 = 0. Proceeding as in the previous example, we end upwith the sliding mode control

u = −(1 + k)x1 − kx2 − β(x) sgn(s)

where β(x) = bx22 + ak|x2|+ β0 with β0 > 0.


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14.1.3 Tracking

Consider the single-input–single-output system

x = f(x) + δ1(x) + g(x)[u + δ(t, x, u)] (14.19)y = h(x) (14.20)

where x, u, and y are the state, control input, and controlled output, respectively.We assume that f , g, h, and δ1 are sufficiently smooth in a domain D ⊂ Rn

and δ is piecewise continuous in t and sufficiently smooth in (x, u) for (t, x, u) ∈[0,∞)×D×R. Furthermore, we assume that f and h are known, while g, δ, and δ1could be uncertain. For all possible uncertainties in g, we assume that the system

x = f(x) + g(x)u (14.21)y = h(x) (14.22)

has relative degree ρ in D; that is,

Lgh(x) = · · · = LgLρ−2f h(x) = 0, LgL

ρ−1f h(x) ≥ a > 0

for all x ∈ D.13 Our goal is to design a state feedback control law such that theoutput y asymptotically tracks a reference signal r(t), where

• r(t) and its derivatives up to r(ρ)(t) are bounded for all t ≥ 0, and the ρthderivative r(ρ)(t) is a piecewise continuous function of t;

• the signals r,. . . ,r(ρ) are available on line.

We know from our study of input–output linearization (Section 13.2) that thesystem (14.21)–(14.22) can be transformed into the normal form by the change ofvariables

⎡⎣ η−−−

ξ

⎤⎦ =

⎡⎣ φ(x)−−−ψ(x)

⎤⎦ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

φ1(x)...

φn−ρ(x)−−−h(x)

...Lρ−1

f h(x)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= T (x) (14.23)

where φ1 to φn−ρ satisfy the partial differential equations

∂φi

∂xg(x) = 0, for 1 ≤ i ≤ n− ρ, ∀ x ∈ D

13Without loss of generality, we assume that LgLρ−1f

h is positive. If it is negative, we can

substitute u = −u and proceed to design u. Thus, by solving the problem for a positive LgLρ−1f

h,

we can cover both signs by multiplying the control by sign(LgLρ−1f

h).


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We assume that T (x) is a diffeomorphism over D. Since f and h are known, whileg could be uncertain, the function ψ is known, while φ could be unknown. Wewould like to restrict the perturbations δ and δ1 such that when the change ofvariables (14.23) is applied to the perturbed system (14.19)–(14.20) it preserves thenormal form structure. From the relative-degree condition, it is clear that the stateequation for η will be independent of u. Let us calculate the state equation for ξ:

ξ1 =∂h

∂x[f + δ1 + g(u + δ)] =

∂h

∂x(f + δ1)

If δ1 belongs to the null space of [∂h/∂x]; that is, [∂h/∂x]δ1(x) = 0, for all x ∈ D,we have

ξ1 = Lfh(x) = ξ2

Similarly,

ξ2 =∂(Lfh)

∂x[f + δ1 + g(u + δ)] =

∂(Lfh)∂x

(f + δ1)

If δ1 belongs to the null space of [∂(Lfh)/∂x] for all x ∈ D, we have

ξ2 = L2fh(x) = ξ2

Continuing in the same manner, it can be seen that if

∂(Lifh)

∂xδ1(x) = 0, for 1 ≤ i ≤ ρ− 2, ∀ x ∈ D (14.24)

then the change of variables (14.23) produces the normal form

η = f0(η, ξ)ξ1 = ξ2

......

ξρ−1 = ξρ

ξρ = Lρfh(x) + Lδ1L

ρ−1f h(x) + LgL

ρ−1f h(x)[u + δ(t, x, u)]

y = ξ1

Let

R =

⎡⎢⎣

r...

r(ρ−1)

⎤⎥⎦ , e =

⎡⎢⎣

ξ1 − r...

ξρ − r(ρ−1)

⎤⎥⎦ = ξ −R

The change of variables e = ξ −R yields

η = f0(η, ξ)e1 = e2

......

eρ−1 = eρ

eρ = Lρfh(x) + Lδ1L

ρ−1f h(x) + LgL

ρ−1f h(x)[u + δ(t, x, u)]− rρ(t)


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Asymptotic tracking will be achieved if we design a state feedback control law toensure that e(t) is bounded and converges to zero as t tends to infinity. Boundednessof e will ensure boundedness of ξ because R(t) is bounded. We need also to ensureboundedness of η. For that, we restrict our analysis to the case where the system

η = f0(η, ξ)

is bounded-input–bounded-state stable. This will be the case for any boundedinput ξ and any initial state η(0) if the system η = f0(η, ξ) is input-to-state stable.So, from this point on, we concentrate our attention on showing boundedness andconvergence of e. The e-equation takes the regular form of (14.4)–(14.5) with η =[e1, . . . , eρ−1]T and ξ = eρ. We start with the system

e1 = e2

......

eρ−1 = eρ

where eρ is viewed as the control input. We want to design eρ to stabilize the origin.For this linear system (in the controllable canonical form), we can achieve this taskby the linear control

eρ = −(k1e1 + · · ·+ kρ−1eρ−1)

where k1 to kρ−1 are chosen such that the polynomial

sρ−1 + kρ−1sρ−2 + · · ·+ k1

is Hurwitz. Then, the sliding manifold is

s = (k1e1 + · · ·+ kρ−1eρ−1) + eρ = 0

and

s = k1e2 + · · ·+kρ−1eρ +Lρfh(x)+Lδ1L

ρ−1f h(x)+LgL

ρ−1f h(x)[u+ δ(t, x, u)]− rρ(t)

We can proceed by designing u = v as a pure switching component, or we can take

u = − 1LgL

ρ−1f h(x)

[k1e2 + · · ·+ kρ−1eρ + Lρ

fh(x)− rρ(t)]

+ v

to cancel the known terms on the right-hand side, where g(x) is a nominal modelof g(x). Note that when g is known; that is, when g = g, the term

− 1LgL

ρ−1f h(x)

[Lρ

fh(x)− rρ(t)]

is the feedback linearizing term we used in Section 13.4.2. In either case, the s-equation can be written as

s = LgLρ−1f h(x)v + Δ(t, x, v)


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Suppose ∣∣∣∣∣ Δ(t, x, v)LgL

ρ−1f h(x)

∣∣∣∣∣ ≤ �(x) + κ0|v|, 0 ≤ κ0 < 1

for all (t, x, v) ∈ [0,∞)×D ×R, where � and κ0 are known. Then,

v = −β(x) sgn(s)

where β(x) ≥ �(x)/(1− κ0) + β0 with β0 > 0, and its continuous approximation isobtained by replacing sgn(s) by sat(s/ε). We leave it to the reader (Exercise 14.13)to show that with the “continuous” sliding mode controller, there exits a finitetime T1, possibly dependent on ε and the initial states, and a positive constant k,independent of ε and the initial states, such that |y(t)− r(t)| ≤ kε for all t ≥ T1.

14.1.4 Regulation via Integral Control


x = f(x) + δ1(x, w) + g(x, w)[u + δ(x, u, w)] (14.25)y = h(x) (14.26)

where x ∈ Rn is the state, u ∈ R is the control input, y ∈ R is the controlledoutput, and w ∈ Rl is a vector of unknown constant parameters and disturbances.The functions f , g, h, δ, and δ1 are sufficiently smooth in (x, u) and continuous inw for x ∈ D ⊂ Rn, u ∈ R, and w ∈ Dw ⊂ Rl, where D and Dw are open connectedsets. We assume that the system

x = f(x) + g(x, w)u (14.27)y = h(x) (14.28)

has relative degree ρ in D uniformly in w; that is,

Lgh(x, w) = · · · = LgLρ−2f h(x, w) = 0, LgL

ρ−1f h(x, w) ≥ a > 0

for all (x, w) ∈ D×Dw. Our goal is to design a state feedback control law such thatthe output y asymptotically tracks a constant reference r ∈ Dr ⊂ R, where Dr is anopen connected set. This is a special case of the tracking problem of the previoussection, where the reference is constant and the uncertainty is parameterized by w.Therefore, we can use the sliding mode controller of the previous section. Whenthe signum function sgn(s) is approximated by the saturation function sat(s/ε),the regulation error will be ultimately bounded by a constant kε for some k > 0.This is the best we can achieve in a general tracking problem, but in a regulationproblem, we can use integral control to achieve zero steady-state error. Proceedingas in Section 12.3, we augment the integral of the regulation error y − r with thesystem and design a feedback controller that stabilizes the augmented system at an


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equilibrium point, where y = r. Toward that end, we assume that, for each pair(r, w) ∈ Dr × Dw, there is a unique pair (xss, uss) that depends continuously on(r, w) and satisfies the equations

0 = f(xss) + δ1(xss, w) + g(xss, w)[uss + δ(xss, uss, w)]r = h(xss)

so that xss is the desired equilibrium point and uss is the steady-state control thatis needed to maintain equilibrium at xss. Assuming that

∂(Lifh)

∂xδ1(x, w) = 0, for 1 ≤ i ≤ ρ− 2, ∀ (x, w) ∈ D ×Dw

the change of variables (14.23) transforms the system (14.25)–(14.26) into the nor-mal form

η = f0(η, ξ, w)ξ1 = ξ2

......

ξρ−1 = ξρ

ξρ = Lρfh(x) + Lδ1L

ρ−1f h(x, w) + LgL

ρ−1f h(x, w)[u + δ(x, u, w)]

y = ξ1

and maps the equilibrium point xss into (ηss, ξss) where ξss = [r, 0, . . . , 0]T . Aug-menting the integrator

e0 = y − r

with the foregoing equation and applying the change of variables

z = η − ηss, e =

⎡⎢⎢⎢⎣

e1e2...eρ

⎤⎥⎥⎥⎦ =

⎡⎢⎢⎢⎣

ξ1 − rξ2...ξρ

⎤⎥⎥⎥⎦

we obtain the augmented system

z = f0(z + ηss, ξ, w) def= f0(z, e, w, r)e0 = e1

e1 = e2

......

eρ−1 = eρ

eρ = Lρfh(x) + Lδ1L


ρ−1f h(x, w)[u + δ(x, u, w)]


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which preserves the normal form structure with a chain of ρ+1 integrators. There-fore, the design of sliding mode control can proceed as in the previous section. Inparticular, we can take

s = k0e0 + k1e1 + · · ·+ kρ−1eρ−1 + eρ

where k0 to kρ−1 are chosen such that the polynomial

sρ + kρ−1sρ−1 + · · ·+ k1s + k0

is Hurwitz. Then,

s = k0e1 + · · ·+ kρ−1eρ + Lρfh(x) + Lδ1L


ρ−1f h(x, w)[u + δ(x, u, w)]

The control u can be taken as

u = v or u = − 1LgL

ρ−1f h(x)

[k0e1 + · · ·+ kρ−1eρ + Lρ

f (x)]

+ v

where g(x) is a nominal model of g(x, w), to obtain

s = LgLρ−1f h(x, w)v + Δ(x, v, w, r)

If ∣∣∣∣∣ Δ(x, v, w, r)LgL

ρ−1f h(x, w)

∣∣∣∣∣ ≤ �(x) + κ0|v|, 0 ≤ κ0 < 1

for all (x, v, w, r) ∈ D ×R×Dw ×Dr, where � and κ0 are known, we can take

v = −β(x) sat(s

ε

)where β(x) ≥ �(x)/(1 − κ0) + β0 with β0 > 0. The closed-loop system has anequilibrium point at (z, e0, e) = (0, e0, 0). Showing convergence to this equilibriumpoint can be done by repeating the analysis of Section 14.1.2. In particular, if thereis a Lyapunov function V1(z, w, r) for the system z = f0(z, e, w, r) that satisfies theinequalities

α1(‖z‖) ≤ V1(z, w, r) ≤ α2(‖z‖)∂V1

∂zf0(z, e, w, r) ≤ −α3(‖z‖), ∀ ‖z‖ ≥ γ(‖e‖)

uniformly in (w, r) for some class K functions α1, α2, α3, and γ, we can show thatthere are two compact positively invariant sets Ω and Ωε such that every trajectorystarting in Ω enters Ωε in finite time. The construction of the sets Ω and Ωε is donein three steps. We write the closed-loop system in the form

z = f0(z, e, w, r)ζ = Aζ + Bs

s = −(LgLρ−1f h)β sat

(s

ε

)+ Δ


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where ζ = [e0, . . . , eρ−1]T and A is Hurwitz, and use the inequality

ss ≤ −aβ0(1− κ0)|s|, for |s| ≥ ε

to show that the set {|s| ≤ c}, with c > ε, is positively invariant. In the secondstep, we use the Lyapunov function V2(ζ) = ζT Pζ, where P is the solution of theLyapunov equation PA + AT P = −I, and the inequality

V2 ≤ −ζT ζ + 2‖ζ‖ ‖PB‖ |s|to show that the set {|s| ≤ c}∩{V2 ≤ c2ρ1} is positively invariant, for some ρ1 > 0.Inside this set, we have ‖e‖ ≤ cρ2, for some ρ2 > 0. Finally, we use the inequality

V1 ≤ −α3(‖z‖), ∀ ‖z‖ ≥ γ(cρ2)

to show thatΩ = {|s| ≤ c} ∩ {V2 ≤ c2ρ1} ∩ {V1 ≤ c0}

is positively invariant for any c0 ≥ α2(γ(cρ2)). Similarly, it can be shown that

Ωε = {|s| ≤ ε} ∩ {V2 ≤ ε2ρ1} ∩ {V1 ≤ α2(γ(ερ2))}is positively invariant and every trajectory starting in Ω enters Ωε in finite time.

If z = 0 is an exponentially stable equilibrium point of z = f0(z, e, w, r), we canrepeat the proof of Theorem 14.2 to show that every trajectory in Ωε approachesthe desired equilibrium point as t → ∞. In particular, if V3(z, w, r) is a converseLyapunov function for the exponentially stable origin z = 0, provided by Theo-rem 4.14, P is the solution of the Lyapunov equation PA + AT P = −I, and (ζ, s)are deviations of (ζ, s) from their equilibrium values, then it can be shown that thederivative of

V0 = V3 + λζT P ζ + 12 s2

with λ > 0, satisfies the inequality

V0 ≤ −⎡⎣ ‖z‖‖ζ‖2|s|

⎤⎦

T ⎡⎣ k1 −k3 −k4−k3 λ −(λk5 + k6)−k4 −(λk5 + k6) (k2/ε)− k7

⎤⎦⎡⎣ ‖z‖‖ζ‖2|s|

⎤⎦

where k1 and k2 are positive constants, while k3 through k7 are nonnegative con-stants. The derivative V0 can be made negative definite by choosing λ > k2

3/k1 andthen choosing ε small enough to make the 3× 3 matrix positive definite.

In the special case when β = k (a constant) and u = v, the “continuous” slidingmode controller is given by

u = −k sat(

k0e0 + k1e1 + · · ·+ kρ−1eρ−1 + eρ

ε

)(14.29)

When ρ = 1, the controller (14.29) is a classical PI controller followed by saturation(Figure 14.13), and when ρ = 2, it is a classical PID controller followed by saturation(Figure 14.14). This is an interesting connection between the “continuous” slidingmode controller and these classical controllers.


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14.2. LYAPUNOV REDESIGN 579

� �

� �1s

KI

�

� KP

�� Plant �

�

r y��

��

k

k

+

−+

Figure 14.13: The “continuous” sliding mode controller (14.29) for relative-degree-onesystems: a PI controller with KI = kk0/ε and KP = k/ε, followed by saturation.

� �

� �1s

KI

�� KP�

� �s KD

�� Plant �

�

r y��

��

k

k

+

−+

Figure 14.14: The “continuous” sliding mode controller (14.29) for relative-degree-twosystems: a PID controller with KI = kk0/ε, KP = kk1/ε, and KD = k/ε, followed bysaturation.

14.2 Lyapunov Redesign


Consider the system

x = f(t, x) + G(t, x)[u + δ(t, x, u)] (14.30)

where x ∈ Rn is the state and u ∈ Rp is the control input. The functions f , G,and δ are defined for (t, x, u) ∈ [0,∞) ×D × Rp, where D ⊂ Rn is a domain thatcontains the origin. We assume that f , G, and δ are piecewise continuous in t andlocally Lipschitz in x and u. The functions f and G are known precisely, while thefunction δ is an unknown function that lumps together various uncertain terms dueto model simplification, parameter uncertainty, and so on. The uncertain term δsatisfies the matching condition. A nominal model of the system can be taken as

x = f(t, x) + G(t, x)u (14.31)

We proceed to design a stabilizing state feedback controller by using this nominalmodel. Suppose we have succeeded to design a feedback control law u = ψ(t, x)such that the origin of the nominal closed-loop system

x = f(t, x) + G(t, x)ψ(t, x) (14.32)


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is uniformly asymptotically stable. Suppose further that we know a Lyapunovfunction for (14.32); that is, we have a continuously differentiable function V (t, x)that satisfies the inequalities

α1(‖x‖) ≤ V (t, x) ≤ α2(‖x‖) (14.33)

∂V

∂t+

∂V

∂x[f(t, x) + G(t, x)ψ(t, x)] ≤ −α3(‖x‖) (14.34)

for all (t, x) ∈ [0,∞)×D, where α1, α2, and α3 are class K functions. Assume that,with u = ψ(t, x) + v, the uncertain term δ satisfies the inequality

‖δ(t, x, ψ(t, x) + v)‖ ≤ ρ(t, x) + κ0‖v‖, 0 ≤ κ0 < 1 (14.35)

where ρ : [0,∞) × D → R is a nonnegative continuous function. The estimate(14.35) is the only information we need to know about the uncertain term δ. Thefunction ρ is a measure of the size of the uncertainty. It is important to emphasizethat we will not require ρ to be small. We will only require it to be known. Ourgoal in this section is to show that with the knowledge of the Lyapunov functionV , the function ρ, and the constant κ0 in (14.35), we can design an additionalfeedback control v such that the overall control u = ψ(t, x) + v stabilizes the actualsystem (14.30) in the presence of the uncertainty. The design of v is called Lyapunovredesign.

Before we carry on with the Lyapunov redesign technique, let us illustrate thatthe feedback linearization problem of the previous chapter fits into the frameworkof the current problem.

Example 14.3 Consider the feedback linearizable system

x = f(x) + G(x)u

where f : D → Rn and G : D → Rn×p are smooth functions on a domain D ⊂ Rn

and there is a diffeomorphism T : D → Rn such that Dz = T (D) contains the originand T (x) satisfies the partial differential equations

∂T

∂xf(x) = AT (x)−Bγ(x)α(x)

∂T

∂xG(x) = Bγ(x)

where (A, B) is controllable and γ(x) is nonsingular for all x ∈ D. The change ofvariables z = T (x) transforms the system into the form

z = Az + Bγ(x)[u− α(x)]

Consider also the perturbed system

x = f(x) + Δf (x) + [G(x) + ΔG(x)]u


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with smooth perturbations that satisfy the conditions

∂T

∂xΔf (x) = Bγ(x)Δ1(x), ΔG(x) = G(x)Δ2(x)

on D.14 The perturbed system can be represented in the form (14.30); that is,

z = Az −Bγ(x)α(x) + Bγ(x)[u + δ(x, u)]

where δ(x, u) = Δ1(x)+Δ2(x)u. Since the nominal system is feedback linearizable,we can take the nominal stabilizing feedback control as

ψ(x) = α(x)− γ−1(x)Kz = α(x)− γ−1(x)KT (x)

where K is chosen such that (A − BK) is Hurwitz. A Lyapunov function for thenominal closed-loop system

z = (A−BK)z

can be taken as V (z) = zT Pz, where P is the solution of the Lyapunov equation

P (A−BK) + (A−BK)T P = −I

With u = ψ(x) + v, the uncertain term δ(x, u) satisfies the inequality

‖δ(x, ψ(x) + v)‖ ≤ ‖Δ1(x) + Δ2(x)α(x)−Δ2(x)γ−1(x)Kz‖+ ‖Δ2(x)‖ ‖v‖

Thus, to satisfy (14.35), we need the inequalities

‖Δ2(x)‖ ≤ κ0 < 1 (14.36)

and‖Δ1(x) + Δ2(x)α(x)−Δ2(x)γ−1(x)KT (x)‖ ≤ ρ(x) (14.37)

to hold over a domain that contains the origin for some continuous function ρ(x).Inequality (14.36) is restrictive because it puts a definite limit on the perturbationΔ2. Inequality (14.37), on the other hand, is not restrictive, because we do notrequire ρ to be small. It is basically a choice of a function ρ to estimate the growthof the left-hand side of (14.37).

Consider now the system (14.30) and apply the control u = ψ(t, x) + v. Theclosed-loop system

x = f(t, x) + G(t, x)ψ(t, x) + G(t, x)[v + δ(t, x, ψ(t, x) + v)] (14.38)

14It can be easily seen that the perturbed system is still feedback linearizable with the samediffeomorphism T (x), provided I + Δ2 is nonsingular. Condition (14.36) implies that I + Δ2 isnonsingular.


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is a perturbation of the nominal closed-loop system (14.32). Let us calculate thederivative of V (t, x) along the trajectories of (14.38). For convenience, we will notwrite the argument of the various functions. We have

V =∂V

∂t+

∂V

∂x(f + Gψ) +

∂V

∂xG(v + δ) ≤ −α3(‖x‖) +

∂V

∂xG(v + δ)

Set wT = [∂V/∂x]G and rewrite the last inequality as

V ≤ −α3(‖x‖) + wT v + wT δ

The first term on the right-hand side is due to the nominal closed-loop system. Thesecond and third terms represent, respectively, the effect of the control v and theuncertain term δ on V . Due to the matching condition, the uncertain term δ appearson the right-hand side exactly at the same point where v appears. Consequently,it is possible to choose v to cancel the (destabilizing) effect of δ on V . We willnow explore two different methods for choosing v so that wT v + wT δ ≤ 0. Supposeinequality (14.35) is satisfied with ‖ · ‖2; that is,

‖δ(t, x, ψ(t, x) + v)‖2 ≤ ρ(t, x) + κ0‖v‖2, 0 ≤ κ0 < 1

We have

wT v + wT δ ≤ wT v + ‖w‖2 ‖δ‖2 ≤ wT v + ‖w‖2[ρ(t, x) + κ0‖v‖2]Taking

v = −η(t, x) · w

‖w‖2 (14.39)

with a nonnegative function η, we obtain

wT v + wT δ ≤ −η‖w‖2 + ρ‖w‖2 + κ0η‖w‖2 = −η(1− κ0)‖w‖2 + ρ‖w‖2Choosing η(t, x) ≥ ρ(t, x)/(1− κ0) for all (t, x) ∈ [0,∞)×D yields

wT v + wT δ ≤ −ρ‖w‖2 + ρ‖w‖2 = 0

Hence, with the control (14.39), the derivative of V (t, x) along the trajectories ofthe closed-loop system (14.38) is negative definite.

As an alternative idea, suppose (14.35) is satisfied with ‖ · ‖∞; that is,

‖δ(t, x, ψ(t, x) + v)‖∞ ≤ ρ(t, x) + κ0‖v‖∞, 0 ≤ κ0 < 1

We have

wT v + wT δ ≤ wT v + ‖w‖1 ‖δ‖∞ ≤ wT v + ‖w‖1[ρ(t, x) + κ0‖v‖∞]

Choosev = −η(t, x) sgn(w) (14.40)


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where η(t, x) ≥ ρ(t, x)/(1 − κ0) for all (t, x) ∈ [0,∞) × D and sgn(w) is a p-dimensional vector whose ith component is sgn(wi). Then,

wT v + wT δ ≤ −η‖w‖1 + ρ‖w‖1 + κ0η‖w‖1= −η(1− κ0)‖w‖1 + ρ‖w‖1≤ −ρ‖w‖1 + ρ‖w‖1 = 0

Hence, with the control (14.40), the derivative of V (t, x) along the trajectories ofthe closed-loop system (14.38) is negative definite. Notice that the control lawsgiven by (14.39) and (14.40) coincide for single-input systems (p = 1).

The control laws given by (14.39) and (14.40) are discontinuous functions of thestate x. This discontinuity causes some theoretical as well as practical problems.Theoretically, we have to change the definition of the control law to avoid divisionby zero. We also have to examine the question of existence and uniqueness of so-lutions more carefully, since the feedback functions are not locally Lipschitz in x.Practically, the implementation of such discontinuous controllers is characterized bythe phenomenon of chattering, where, due to imperfections in switching devices orcomputational delays, the control has fast switching fluctuations across the switch-ing surface.15 Instead of trying to work out all these problems, we will choose theeasy and more practical route of approximating the discontinuous control law bya continuous one. The development of such approximation is similar for both con-trol laws. Therefore, we continue with the control (14.39) and leave the paralleldevelopment of a continuous approximation of (14.40) to Exercises 14.21 and 14.22.

Consider the feedback control law

v =

⎧⎪⎨⎪⎩−η(t, x)(w/‖w‖2), if η(t, x)‖w‖2 ≥ ε

−η2(t, x)(w/ε), if η(t, x)‖w‖2 < ε

(14.41)

With (14.41), the derivative of V along the trajectories of the closed-loop system(14.38) will be negative definite whenever η(t, x)‖w‖2 ≥ ε. We only need to checkV when η(t, x)‖w‖2 < ε. In this case,

V ≤ −α3(‖x‖2) + wT[−η2 · w

ε+ δ]

≤ −α3(‖x‖2)− η2

ε‖w‖22 + ρ‖w‖2 + κ0‖w‖2‖v‖2

= −α3(‖x‖2)− η2

ε‖w‖22 + ρ‖w‖2 +

κ0η2

ε‖w‖22

≤ −α3(‖x‖2) + (1− κ0)(− η2

ε‖w‖22 + η‖w‖2

)

15See Section 14.1 for further discussion of chattering.


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The term

− η2

ε‖w‖22 + η‖w‖2

attains a maximum value ε/4 at η‖w‖2 = ε/2. Therefore,

V ≤ −α3(‖x‖2) +ε(1− κ0)

4

whenever η(t, x)‖w‖2 < ε. On the other hand, when η(t, x)‖w‖2 ≥ ε, V satisfies

V ≤ −α3(‖x‖2) ≤ −α3(‖x‖2) +ε(1− κ0)

4

Thus, the inequality

V ≤ −α3(‖x‖2) +ε(1− κ0)

4

is satisfied irrespective of the value of η(t, x)‖w‖2. Take r > 0 such that Br ⊂ D,choose ε < 2α3(α−1

2 (α1(r)))/(1− κ0), and set μ = α−13 (ε(1− κ0)/2) < α−1

2 (α1(r)).Then,

V ≤ − 12α3(‖x‖2), ∀ μ ≤ ‖x‖2 < r

Application of Theorem 4.18 results in the following theorem, which shows that thesolutions of the closed-loop system are uniformly ultimately bounded by a class Kfunction of ε.

Theorem 14.3 Consider the system (14.30). Let D ⊂ Rn be a domain that con-tains the origin and Br = {‖x‖2 ≤ r} ⊂ D. Let ψ(t, x) be a stabilizing feedbackcontrol law for the nominal system (14.31) with a Lyapunov function V (t, x) thatsatisfies (14.33) and (14.34) in 2-norm for all t ≥ 0 and all x ∈ D, with someclass K functions α1, α2, and α3. Suppose the uncertain term δ satisfies (14.35)in 2-norm for all t ≥ 0 and all x ∈ D. Let v be given by (14.41) and chooseε < 2α3(α−1

2 (α1(r)))/(1 − κ0). Then, for any ‖x(t0)‖2 < α−12 (α1(r)), there exists

a finite time t1 such that the solution of the closed-loop system (14.38) satisfies

‖x(t)‖2 ≤ β(‖x(t0)‖2, t− t0), ∀ t0 ≤ t < t1 (14.42)

‖x(t)‖2 ≤ b(ε), ∀ t ≥ t1 (14.43)

where β is a class KL function and b is a class K function defined by

b(ε) = α−11 (α2(μ)) = α−1

1 (α2(α−13 (ε(1− κ0)/2)))

If all the assumptions hold globally and α1 belongs to class K∞, then (14.42) and(14.43) hold for any initial state x(t0). �


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Thus, in general, the continuous Lyapunov redesign given by (14.41) does not stabi-lize the origin as its discontinuous counterpart (14.39) does. Nevertheless, it guar-antees uniform ultimate boundedness of the solutions. Since the ultimate boundb(ε) is a class K function of ε, it can be made arbitrarily small by choosing ε smallenough. In the limit, as ε → 0, we recover the performance of the discontinuous con-troller. Notice that there is no analytical reason to require ε to be very small. Theonly analytical restriction on ε is the requirement ε < 2α3(α−1

2 (α1(r)))/(1 − κ0).This requirement is satisfied for any ε when the assumptions hold globally and αi

(i = 1, 2, 3) are class K∞ functions. Of course, from a practical viewpoint, we wouldlike to make ε as small as feasible, because we would like to drive the state of thesystem to a neighborhood of the origin that is as small as it could be. Exploiting thesmallness of ε in the analysis, we can arrive at a sharper result when the uncertaintyδ vanishes at the origin. Suppose there is a ball Ba = {‖x‖2 ≤ a}, a ≤ r, such thatthe following inequalities are satisfied for all x ∈ Ba:

α3(‖x‖2) ≥ φ2(x) (14.44)η(t, x) ≥ η0 > 0 (14.45)ρ(t, x) ≤ ρ1φ(x) (14.46)

Here, φ : Rn → R is a positive definite function of x. Choosing ε < b−1(a) ensuresthat the trajectories of the closed-loop systems will be confined to Ba after a finitetime. When η(t, x)‖w‖2 < ε, the derivative V satisfies

V ≤ −α3(‖x‖2)− η2(1− κ0)ε

‖w‖22 + ρ‖w‖2

≤ − 12α3(‖x‖2)− 1

2φ2(x)− η2

0(1− κ0)ε

‖w‖22 + ρ1φ(x)‖w‖2

≤ − 12α3(‖x‖2)− 1

2

[φ(x)‖w‖2

]T [1 −ρ1−ρ1 2η2

0(1− κ0)/ε

] [φ(x)‖w‖2

]

The matrix of the quadratic form will be positive definite if ε < 2η20(1 − κ0)/ρ2

1.Thus, choosing ε < 2η2

0(1 − κ0)/ρ21, we have V ≤ −α3(‖x‖2)/2. Since V ≤

−α3(‖x‖2) ≤ −α3(‖x‖2)/2 when η(t, x)‖w‖2 ≥ ε, we conclude that

V ≤ − 12α3(‖x‖2)

which shows that the origin is uniformly asymptotically stable.

Corollary 14.1 Assume the inequalities (14.44) through (14.46) are satisfied, inaddition to all the assumptions of Theorem 14.3. Then, for all ε < min{2η2

0(1 −κ0)/ρ2

1, b−1(a)}, the origin of the closed-loop system (14.38) is uniformly asymp-totically stable. If αi(r) = kir

c, then the origin is exponentially stable. �

Corollary 14.1 is particularly useful when the origin of the nominal closed-loopsystem (14.32) is exponentially stable and the perturbation δ(t, x, u) is Lipschitz in


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x and u and vanishes at (x = 0, u = 0). In this case, φ(x) is proportional to ‖x‖2and the uncertain term satisfies (14.35) with ρ(x) that satisfies (14.46). In general,the condition (14.46) may require more than just a vanishing perturbation at theorigin. For example if, in a scalar case, φ(x) = |x|3, then a perturbation term xcannot be bounded by ρ1φ(x).

The stabilization result of Corollary 14.1 is dependent on the choice of η tosatisfy (14.45). It can be shown (Exercise 14.20) that if η does not satisfy (14.45),the feedback control may fail to stabilize the origin. When η satisfies (14.45), thefeedback control law (14.41) acts in the region η‖w‖2 < ε as a high-gain feedbackcontrol v = −kw with k ≥ η2

0/ε. Such high-gain feedback control law can stabilizethe origin when (14.44) through (14.46) are satisfied (Exercise 14.24).

Example 14.4 Let us continue Example 14.3 on feedback linearizable systems.Suppose inequality (14.36) is satisfied in ‖ · ‖2. Suppose further that

‖Δ1(x) + Δ2(x)α(x) + Δ2(x)γ−1(x)Kz‖2 ≤ ρ1‖z‖2for all z ∈ Br ⊂ Dz. Then, (14.37) is satisfied with ρ = ρ1‖z‖2. We take the controlv as in (14.41) with

η = 1 +ρ1‖z‖2(1− κ0)

, wT = 2zT PB, ε < min{

2(1− κ0)ρ21

,2r2λmin(P )

(1− κ0)λmax(P )

}

It can be verified that all the assumptions of Theorem 14.3 and Corollary 14.1 aresatisfied with α1(r) = λmin(P )r2, α2(r) = λmax(P )r2, α3(r) = r2, φ(z) = ‖z‖2,and a = r. Thus, with the overall feedback control u = ψ(x) + v, the origin ofthe perturbed closed-loop system is exponentially stable. If all the assumptionshold globally and T (x) is a global diffeomorphism, the origin x = 0 will be globallyasymptotically stable.16

Example 14.5 Reconsider the pendulum equation of Example 13.18, with δ1 = π,

x1 = x2

x2 = a sin x1 − bx2 + cu

We want to stabilize the pendulum at the open-loop equilibrium point x = 0. Thissystem is feedback linearizable with T (x) = x. A nominal stabilizing feedbackcontrol can be taken as

ψ(x) = −(

a

c

)sin x1 −

(1c

)(k1x1 + k2x2)

16The origin z = 0 will be globally exponentially stable, but we cannot conclude that x = 0will be globally exponentially stable, unless the linear growth conditions ‖T (x)‖ ≤ L1‖x‖ and‖T−1(z)‖ ≤ L2‖z‖ hold globally.


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where a and c are the nominal values of a and c, respectively, and k1 and k2 arechosen such that

A−BK =[

0 1−k1 −(k2 + b)

]

is Hurwitz. With u = ψ(x) + v, the uncertainty term δ is given by

δ =1c

[(ac− ac

c

)sin x1 −

(c− c

c

)(k1x1 + k2x2)

]+(

c− c

c

)v

Hence,|δ| ≤ ρ1‖x‖2 + κ0|v|, ∀ x ∈ R2 ∀ v ∈ R

where

κ0 =∣∣∣∣c− c

c

∣∣∣∣ , ρ1 =k

c, and k =

∣∣∣∣ ac− ac

c

∣∣∣∣+∣∣∣∣c− c

c

∣∣∣∣√

k21 + k2

2

Assuming κ0 < 1 and taking v as in the previous example, we find that the controllaw u = ψ(x)+ v makes the origin globally exponentially stable. In Example 13.18,we analyzed the same system under the control u = ψ(x). Comparing the resultsshows exactly the contribution of the additional control component v. In Exam-ple 13.18, we were able to show that the control u = ψ(x) stabilizes the systemwhen k is restricted to satisfy

k <1

2√

p212 + p2

22

This restriction has now been completely removed, provided we know k.

Example 14.6 Once again, consider the pendulum equation of the previous ex-ample. This time, suppose the suspension point of the pendulum is subjected toa time-varying, bounded, horizontal acceleration. For simplicity, neglect friction(b = 0). The state equation is given by

x1 = x2

x2 = a sin x1 + cu + h(t) cos x1

where h(t) is the (normalized) horizontal acceleration of the suspension point. Wehave |h(t)| ≤ H for all t ≥ 0. The nominal model and the nominal stabilizingcontrol can be taken as in the previous example (with b = 0). The uncertain termδ satisfies

|δ| ≤ ρ1 ‖x‖2 + κ0 |v|+ H/c

where ρ1 and κ0 are the same as in that example. This time, we have ρ(x) =ρ1‖x‖2 + H/c, which does not vanish at x = 0. The choice of η in the control law(14.41) must satisfy η ≥ (ρ1‖x‖2 +H/c)/(1−κ0). We take η(x) = η0 +ρ1‖x‖2/(1−


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κ0) with η0 ≥ H/c(1 − κ0). In the previous example, we arbitrarily set η(0) = 1.This is the only modification we have to make to accommodate the nonvanishingdisturbance term h(t) cos x1. Since ρ(0) �= 0, Corollary 14.1 does not apply. We canonly conclude, by Theorem 14.3, that the solutions of the closed-loop system areuniformly ultimately bounded and the ultimate bound is proportional to

√ε.

14.2.2 Nonlinear Damping

Reconsider the system (14.30) with δ(t, x, u) = Γ(t, x)δ0(t, x, u); that is,

x = f(t, x) + G(t, x)[u + Γ(t, x)δ0(t, x, u)] (14.47)

As before, we assume that f and G are known, while δ0(t, x, u) is an uncertainterm. The function Γ(t, x) is known. We assume that f , G, Γ, and δ0 are piecewisecontinuous in t and locally Lipschitz in x and u for all (t, x, u) ∈ [0,∞)×Rn ×Rp.We assume also that δ0 is uniformly bounded for all (t, x, u). Let ψ(t, x) be a nom-inal stabilizing feedback control law such that the origin of the nominal closed-loopsystem (14.32) is globally uniformly asymptotically stable and there is a known Lya-punov function V (t, x) that satisfies (14.33) and (14.34) for all (t, x) ∈ [0,∞)×Rn

with class K∞ functions α1, α2, and α3. If an upper bound on ‖δ0(t, x, u)‖ is known,we can design the control component v, as before, to ensure robust global stabi-lization. In this section, we show that even when no upper bound on δ0 is known,we can design the control component v to ensure boundedness of the trajectoriesof the closed-loop system. Toward that end, let u = ψ(t, x) + v and recall that thederivative of V along the trajectories of the closed-loop system satisfies

V =∂V

∂t+

∂V

∂x(f + Gψ) +

∂V

∂xG(v + Γδ0) ≤ −α3(‖x‖) + wT (v + Γδ0)

where wT = [∂V/∂x]G. Taking

v = −kw‖Γ(t, x)‖22, k > 0 (14.48)

we obtainV ≤ −α3(‖x‖)− k‖w‖22‖Γ‖22 + ‖w‖2‖Γ‖2k0

in which k0 is an (unknown) upper bound on ‖δ0‖. The term

−k‖w‖22‖Γ‖22 + ‖w‖2‖Γ‖2k0

attains a maximum value k20/4k at ‖w‖2‖Γ‖2 = k0/2k. Therefore,

V ≤ −α3(‖x‖2) +k20

4k

Since α3 is class K∞, it is always true that V is negative outside some ball. It followsfrom Theorem 4.18 that for any initial state x(t0), the solution of the closed-loopsystem is uniformly bounded. The Lyapunov redesign (14.48) is called nonlineardamping. We summarize our conclusion in the next lemma.


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14.3. BACKSTEPPING 589

Lemma 14.1 Consider the system (14.47) and let ψ(t, x) be a stabilizing feedbackcontrol for the nominal system (14.31) with a Lyapunov function V (t, x) that sat-isfies (14.33) and (14.34) for all t ≥ 0 and all x ∈ Rn, with some class K∞ func-tions α1, α2, and α3. Suppose the uncertain term δ0 is uniformly bounded for(t, x, u) ∈ [0,∞)×Rn×Rp. Let v be given by (14.48) and take u = ψ(t, x)+v. Then,for any x(t0) ∈ Rn, the solution of the closed-loop system is uniformly bounded. �

Example 14.7 Consider the scalar system

x = x2 + u + xδ0(t)

where δ0(t) is a bounded function of t. With the nominal stabilizing control ψ(x) =−x2 − x, the Lyapunov function V (x) = x2 satisfies (14.33) and (14.34) globallywith α1(r) = α2(r) = α3(r) = r2. The nonlinear damping component (14.48), withk = 1, is given by v = −2x3. The closed-loop system

x = −x− 2x3 + xδ0(t)

has a bounded solution no matter how large the bounded disturbance δ0 is, thanksto the nonlinear damping term −2x3.

14.3 Backstepping

We start with the special case of integrator backstepping. Consider the system

η = f(η) + g(η)ξ (14.49)ξ = u (14.50)

where [ηT , ξ]T ∈ Rn+1 is the state and u ∈ R is the control input. The functionsf : D → Rn and g : D → Rn are smooth17 in a domain D ⊂ Rn that containsη = 0 and f(0) = 0. We want to design a state feedback control law to stabilize theorigin (η = 0, ξ = 0). We assume that both f and g are known. This system canbe viewed as a cascade connection of two components, as shown in Figure 14.15(a);the first component is (14.49), with ξ as input, and the second component is theintegrator (14.50). Suppose the component (14.49) can be stabilized by a smoothstate feedback control law ξ = φ(η), with φ(0) = 0; that is, the origin of

η = f(η) + g(η)φ(η)

is asymptotically stable. Suppose further that we know a (smooth, positive definite)Lyapunov function V (η) that satisfies the inequality

∂V

∂η[f(η) + g(η)φ(η)] ≤ −W (η), ∀ η ∈ D (14.51)

17We require smoothness of all functions for convenience. It will become clear, however, that ina particular problem, we only need existence of derivatives up to a certain order.


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where W (η) is positive definite. By adding and subtracting g(η)φ(η) on the right-hand side of (14.49), we obtain the equivalent representation

η = [f(η) + g(η)φ(η)] + g(η)[ξ − φ(η)]ξ = u

which is shown in Figure 14.15(b). The change of variables

z = ξ − φ(η)

results in the system

η = [f(η) + g(η)φ(η)] + g(η)zz = u− φ

which is shown in Figure 14.15(c). Going from Figure 14.15(b) to Figure 14.15(c)can be viewed as “backstepping” −φ(η) through the integrator. Since f , g, and φare known, the derivative φ can be computed by using the expression

φ =∂φ

∂η[f(η) + g(η)ξ]

Taking v = u− φ reduces the system to the cascade connection

η = [f(η) + g(η)φ(η)] + g(η)zz = v

which is similar to the system we started from, except that now the first componenthas an asymptotically stable origin when the input is zero. This feature will beexploited in the design of v to stabilize the overall system. Using

Vc(η, ξ) = V (η) + 12z2

as a Lyapunov function candidate, we obtain

Vc =∂V

∂η[f(η) + g(η)φ(η)] +

∂V

∂ηg(η)z + zv

≤ −W (η) +∂V

∂ηg(η)z + zv

Choosing

v = − ∂V

∂ηg(η)− kz, k > 0

yieldsVc ≤ −W (η)− kz2


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�∫

� g(η) � �+ �∫

�

�f(·)

�

u ξ η

f(η)(a)

�∫

� �+�

� g(η) � �+ �∫

�

�f(·) + g(·)φ(·)

�

u ξ η

−φ(η)(b)

� �+�

�∫

� g(η) � �+ �∫

�

�f(·) + g(·)φ(·)

�

u z η

−φ(c)

Figure 14.15: (a) The block diagram of system (14.49)–(14.50); (b) introducing φ(η);(c) “backstepping” −φ(η) through the integrator.


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which shows that the origin (η = 0, z = 0) is asymptotically stable. Since φ(0) = 0,we conclude that the origin (η = 0, ξ = 0) is asymptotically stable. Substitutingfor v, z, and φ, we obtain the state feedback control law

u =∂φ

∂η[f(η) + g(η)ξ]− ∂V

∂ηg(η)− k[ξ − φ(η)] (14.52)

If all the assumptions hold globally and V (η) is radially unbounded, we can concludethat the origin is globally asymptotically stable. We summarize our conclusions inthe next lemma.

Lemma 14.2 Consider the system (14.49)–(14.50). Let φ(η) be a stabilizing statefeedback control law for (14.49) with φ(0) = 0, and V (η) be a Lyapunov function thatsatisfies (14.51) with some positive definite function W (η). Then, the state feedbackcontrol law (14.52) stabilizes the origin of (14.49)–(14.50), with V (η)+[ξ−φ(η)]2/2as a Lyapunov function. Moreover, if all the assumptions hold globally and V (η) isradially unbounded, the origin will be globally asymptotically stable. �

Example 14.8 Consider the system

x1 = x21 − x3

1 + x2

x2 = u

which takes the form (14.49)–(14.50) with η = x1 and ξ = x2. We start with thescalar system

x1 = x21 − x3

1 + x2

with x2 viewed as the input and proceed to design the feedback control x2 = φ(x1)to stabilize the origin x1 = 0. With

x2 = φ(x1) = −x21 − x1

we cancel the nonlinear term x21 to obtain18

x1 = −x1 − x31

and V (x1) = x21/2 satisfies

V = −x21 − x4

1 ≤ −x21, ∀ x1 ∈ R

Hence, the origin of x1 = −x1 − x31 is globally exponentially stable. To backstep,

we use the change of variables

z2 = x2 − φ(x1) = x2 + x1 + x21

18We do not cancel −x31 because it provides nonlinear damping. (See Example13.19.)


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to transform the system into the form

x1 = −x1 − x31 + z2

z2 = u + (1 + 2x1)(−x1 − x31 + z2)

TakingVc(x) = 1

2x21 + 1

2z22

as a composite Lyapunov function, we obtain

Vc = x1(−x1 − x31 + z2) + z2[u + (1 + 2x1)(−x1 − x3

1 + z2)]= −x2

1 − x41 + z2[x1 + (1 + 2x1)(−x1 − x3

1 + z2) + u]

Takingu = −x1 − (1 + 2x1)(−x1 − x3

1 + z2)− z2

yieldsVc = −x2

1 − x41 − z2

2

Hence, the origin is globally asymptotically stable.

Application of integrator backstepping in the preceding example is straightfor-ward due to the simplicity of scalar designs. For higher order systems, we may retainthis simplicity via recursive application of integrator backstepping, as illustrated bythe next example.

Example 14.9 The third-order system

x1 = x21 − x3

1 + x2

x2 = x3

x3 = u

is composed of the second-order system of the previous example with an additionalintegrator at the input side. We proceed to apply integrator backstepping as in theprevious example. After one step of backstepping, we know that the second-ordersystem

x1 = x21 − x3

1 + x2

x2 = x3

with x3 as input, can be globally stabilized by the control

x3 = −x1 − (1 + 2x1)(x21 − x3

1 + x2)− (x2 + x1 + x21)

def= φ(x1, x2)

andV (x1, x2) = 1

2x21 + 1

2 (x2 + x1 + x21)

2


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is the corresponding Lyapunov function. To backstep, we apply the change ofvariables

z3 = x3 − φ(x1, x2)

to obtain

x1 = x21 − x3

1 + x2

x2 = φ(x1, x2) + z3

z3 = u− ∂φ

∂x1(x2

1 − x31 + x2)− ∂φ

∂x2(φ + z3)

Using Vc = V + z23/2 as a composite Lyapunov function, we obtain

Vc =∂V

∂x1(x2

1 − x31 + x2) +

∂V

∂x2(z3 + φ)

+ z3

[u− ∂φ

∂x1(x2

1 − x31 + x2)− ∂φ

∂x2(z3 + φ)

]= −x2

1 − x41 − (x2 + x1 + x2

1)2

+ z3

[∂V

∂x2− ∂φ

∂x1(x2

1 − x31 + x2)− ∂φ

∂x2(z3 + φ) + u

]

Taking

u = − ∂V

∂x2+

∂φ

∂x1(x2

1 − x31 + x2) +

∂φ

∂x2(z3 + φ)− z3

yieldsVc = −x2

1 − x41 − (x2 + x1 + x2

1)2 − z2

3

Hence, the origin is globally asymptotically stable. Let us move now from (14.49)–(14.50) to the more general system

η = f(η) + g(η)ξ (14.53)ξ = fa(η, ξ) + ga(η, ξ)u (14.54)

where fa and ga are smooth. If ga(η, ξ) �= 0 over the domain of interest, the inputtransformation

u =1

ga(η, ξ)[ua − fa(η, ξ)] (14.55)

will reduce (14.54) to the integrator ξ = ua. Therefore, if a stabilizing state feedbackcontrol law φ(η) and a Lyapunov function V (η) exist such that the conditions ofLemma 14.2 are satisfied for (14.53), then the lemma and (14.55) yield

u = φc(η, ξ)

=1

ga(η, ξ)

{∂φ

∂η[f(η) + g(η)ξ]− ∂V

∂ηg(η)− k[ξ − φ(η)]− fa(η, ξ)

}(14.56)


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for some k > 0 andVc(η, ξ) = V (η) + 1

2 [ξ − φ(η)]2 (14.57)

as a stabilizing state feedback control law and a Lyapunov function, respectively,for the overall system (14.53)–(14.54). By recursive application of backstepping, wecan stabilize strict-feedback systems of the form

x = f0(x) + g0(x)z1

z1 = f1(x, z1) + g1(x, z1)z2

z2 = f2(x, z1, z2) + g2(x, z1, z2)z3

...zk−1 = fk−1(x, z1, . . . , zk−1) + gk−1(x, z1, . . . , zk−1)zk

zk = fk(x, z1, . . . , zk) + gk(x, z1, . . . , zk)u

where x ∈ Rn, z1 to zk are scalars, and f0 to fk vanish at the origin. The reasonfor referring to such systems as “strict feedback” is that the nonlinearities fi and gi

in the zi-equation (i = 1, . . . , k) depend only on x, z1, . . . , zi; that is, on the statevariables that are “fed back.” We assume that

gi(x, z1, . . . , zi) �= 0 for 1 ≤ i ≤ k

over the domain of interest. The recursive procedure starts with the system

x = f0(x) + g0(x)z1

where z1 is viewed as the control input. We assume that it is possible to determinea stabilizing state feedback control law z1 = φ0(x), with φ0(0) = 0, and a Lyapunovfunction V0(x) such that

∂V0

∂x[f0(x) + g0(x)φ0(x)] ≤ −W (x)

over the domain of interest for some positive definite function W (x). In many appli-cations of backstepping, the variable x is scalar, which simplifies this stabilizationproblem. With φ0(x) and V0(x) in hand, we proceed to apply backstepping in asystematic way. First, we consider the system

x = f0(x) + g0(x)z1

z1 = f1(x, z1) + g1(x, z1)z2

as a special case of (14.53)–(14.54) with

η = x, ξ = z1, u = z2, f = f0, g = g0, fa = f1, ga = g1


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We use (14.56) and (14.57) to obtain the stabilizing state feedback control law andthe Lyapunov function as

φ1(x, z1) =1g1

[∂φ0

∂x(f0 + g0z1)− ∂V0

∂xg0 − k1(z1 − φ)− f1

], k1 > 0

V1(x, z1) = V0(x) + 12 [z1 − φ(x)]2

Next, we consider the system

x = f0(x) + g0(x)z1

z1 = f1(x, z1) + g1(x, z1)z2

z2 = f2(x, z1, z2) + g2(x, z1, z2)z3

as a special case of (14.53)–(14.54) with

η =[

xz1

], ξ = z2, u = z3, f =

[f0 + g0z1

f1

], g =

[0g1

], fa = f2, ga = g2

Using (14.56) and (14.57), we obtain the stabilizing state feedback control law andthe Lyapunov function as

φ2(x, z1, z2) =1g2

[∂φ1

∂x(f0 + g0z1) +

∂φ1

∂z1(f1 + g1z2)− ∂V1

∂z1g1 − k2(z2 − φ1)− f2

]

for some k2 > 0 and

V2(x, z1, z2) = V1(x, z1) + 12 [z2 − φ2(x, z1)]2

This process is repeated k times to obtain the overall stabilizing state feedbackcontrol law u = φk(x, z1, . . . , zk) and the Lyapunov function Vk(x, z1, . . . , zk).

Example 14.10 Consider a single-input–single output system in the special nor-mal form

x = f0(x) + g0(x)z1

z1 = z2

...zr−1 = zr

zr = γ(x, z)[u− α(x, z)]y = z1

where x ∈ Rn−r, z1 to zr are scalars, and γ(x, z) �= 0 for all (x, z). This represen-tation is a special case of the normal form (13.16)–(13.18) because the x-equationtakes the form f0(x) + g0(x)z1, instead of the more general form f0(x, z). The


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system is in the strict feedback form. The origin can be globally stabilized by re-cursive backstepping if we can find a smooth function φ0(x) and a smooth radiallyunbounded Lyapunov function V0(x) such that

∂V0

∂x[f0(x) + g0(x)φ0(x)] ≤ −W (x), ∀ x ∈ Rn

for some positive definite function W (x). If the system is minimum phase, the originof x = f0(x) is globally asymptotically stable, and we know a Lyapunov functionV0(x) that satisfies

∂V0

∂xf0(x) ≤ −W (x), ∀ x ∈ Rn

for some positive definite function W (x), we can simply take φ0(x) = 0. Otherwise,we have to search for φ0(x) and V0(x). This shows that backstepping allows usto stabilize nonminimum phase systems, provided the stabilization problem for thezero dynamics is solvable. Example 14.11 The second-order system

η = −η + η2ξ

ξ = u

was considered in Example 13.16, where it was shown that u = −kξ, with sufficientlylarge k > 0, can achieve semiglobal stabilization. In this example, we achieve globalstabilization via backstepping. Starting with the system

η = −η + η2ξ

it is clear that ξ = 0 and V0(η) = η2/2 result in

∂V0

∂η(−η) = −η2, ∀ η ∈ R

Using V = V0 + ξ2/2 = (η2 + ξ2)/2, we obtain

V = η(−η + η2ξ) + ξu = −η2 + ξ(η3 + u)

Thus,u = −η3 − kξ, k > 0

globally stabilizes the origin. Example 14.12 As a variation from the previous example, consider the system19

η = η2 − ηξ

ξ = u

19With the output y = ξ, the system is nonminimum phase, because the origin of the zero-dynamics equation η = η2 is unstable.


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This time, η = η2 − ηξ cannot be stabilized by ξ = 0. It is easy, however, to seethat ξ = η + η2 and V0(η) = η2/2 result in

∂V0

∂η[η2 − η(η + η2)] = −η4, ∀ η ∈ R

Using V = V0 + (ξ − η − η2)2/2, we obtain

V = η(η2 − ηξ) + (ξ − η − η2)[u− (1 + 2η)(η2 − ηξ)]= −η4 + (ξ − η − η2)[−η2 + u− (1 + 2η)(η2 − ηξ)]

The control

u = (1 + 2η)(η2 − ηξ) + η2 − k(ξ − η − η2), k > 0

yieldsV = −η4 − k(ξ − η − η2)2

which shows that the origin is globally asymptotically stable.

In the previous two sections, we saw how to use sliding mode control and Lya-punov redesign to robustly stabilize an uncertain system when the uncertainty sat-isfies the matching condition. Backstepping can be used to relax the matchingcondition. To illustrate the idea, let us consider the single-input system

η = f(η) + g(η)ξ + δη(η, ξ) (14.58)

ξ = fa(η, ξ) + ga(η, ξ)u + δξ(η, ξ) (14.59)

defined on a domain D ⊂ Rn+1 that contains the origin (η = 0, ξ = 0), whereη ∈ Rn and ξ ∈ R. Suppose ga(η, ξ) �= 0 and all functions are smooth for all(η, ξ) ∈ D. Let f , g, fa, and ga be known and δη and δξ be uncertain terms.We assume that f and fa vanish at the origin and the uncertain terms satisfy theinequalities

‖δη(η, ξ)‖2 ≤ a1‖η‖2 (14.60)|δξ(η, ξ)| ≤ a2‖η‖2 + a3|ξ| (14.61)

for all (η, ξ) ∈ D. Inequality (14.60) restricts the class of uncertainties, because itrestricts the upper bound on δη(η, ξ) to depend only on η. Nevertheless, it is lessrestrictive than the matching condition that would have required δη = 0. Startingwith the system (14.58), suppose we can find a stabilizing state feedback controllaw ξ = φ(η) with φ(0) = 0 and a (smooth, positive definite) Lyapunov functionV (η) such that

∂V

∂η[f(η) + g(η)φ(η) + δη(η, ξ)] ≤ −b‖η‖22 (14.62)


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for all (η, ξ) ∈ D for some positive constant b. Inequality (14.62) shows that η = 0is an asymptotically stable equilibrium point of the system

η = f(η) + g(η)φ(η) + δη(η, ξ)

Suppose further that φ(η) satisfies the inequalities

|φ(η)| ≤ a4‖η‖2,∥∥∥∥∂φ

∂η

∥∥∥∥2≤ a5 (14.63)

over D. Consider now the Lyapunov function candidate

Vc(η, ξ) = V (η) + 12 [ξ − φ(η)]2 (14.64)

The derivative of Vc along the trajectories of (14.58)–(14.59) is given by

Vc =∂V

∂η(f + gφ + δη) +

∂V

∂ηg (ξ − φ)

+ (ξ − φ)[fa + gau + δξ − ∂φ

∂η(f + gξ + δη)

]

Taking

u =1ga

[∂φ

∂η(f + gξ)− ∂V

∂ηg − fa − k(ξ − φ)

], k > 0 (14.65)

and using (14.62), we obtain

Vc ≤ −b‖η‖22 + (ξ − φ)[δξ − ∂φ

∂ηδη

]− k(ξ − φ)2

By using (14.60), (14.61), and (14.63), it can be shown that

Vc ≤ −b‖η‖22 + 2a6‖η‖2|ξ − φ| − (k − a3)(ξ − φ)2

= −[ ‖η‖2|ξ − φ|

]T [b −a6−a6 (k − a3)

] [ ‖η‖2|ξ − φ|

]

for some a6 ≥ 0. Choosing

k > a3 +a26

b

yieldsVc ≤ −σ[‖η‖22 + |ξ − φ|2]

for some σ > 0. Thus, we have completed the proof of the next lemma.


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Lemma 14.3 Consider the system (14.58)–(14.59), where the uncertainty satisfiesinequalities (14.60) and (14.61). Let φ(η) be a stabilizing state feedback control lawfor (14.58) that satisfies (14.63), and V (η) be a Lyapunov function that satisfies(14.62). Then, the state feedback control law (14.65), with k sufficiently large, sta-bilizes the origin of (14.58)–(14.59). Moreover, if all the assumptions hold globallyand V (η) is radially unbounded, the origin will be globally asymptotically stable. �

As an application of backstepping, consider the robust stabilization of the system

xi = xi+1 + δi(x), 1 ≤ i ≤ n− 1xn = γ(x)[u− α(x)] + δn(x)

}(14.66)

defined on a domain D ⊂ Rn that contains the origin x = 0, where x = [x1, . . . , xn]T .Suppose γ(x) �= 0 and all functions are smooth for all x ∈ D. Let α and γ be known,and δi for 1 ≤ i ≤ n be uncertain terms. The nominal system is feedback lineariz-able. We assume that the uncertain terms satisfy the inequalities

|δi(x)| ≤ ai

i∑k=1

|xk|, for 1 ≤ i ≤ n (14.67)

for all x ∈ D, where the nonnegative constants a1 to an are known. Inequalities(14.67) restrict the class of uncertainties for 1 ≤ i ≤ n − 1 because they restrictthe upper bound on δi(x) to depend only on x1 to xi; that is, the upper boundsappear in a strict feedback form. Nevertheless, they are less restrictive than thematching condition that would have required δi = 0 for 1 ≤ i ≤ n − 1. To applythe backstepping recursive design procedure, we start with the scalar system

x1 = x2 + δ1(x)

where x2 is viewed as the control input and δ1(x) satisfies the inequality |δ1(x)| ≤a1|x1|. In this scalar system, the uncertainty satisfies the matching condition. Theorigin x1 = 0 can be robustly stabilized by the high-gain feedback control x2 =−k1x1 where k1 > 0 is chosen sufficiently large. In particular, let V1(x1) = x2

1/2 bea Lyapunov function candidate. Then,

V1 = x1[−k1x1 + δ1(x)] ≤ −(k1 − a1)x21

and the origin is stabilized for all k1 > a1. From this point on, backstepping andLemma 14.3 are applied recursively to derive the stabilizing state feedback control.The procedure is illustrated by the following example.

Example 14.13 We want to design a state feedback control law to stabilize thesecond-order system

x1 = x2 + θ1x1 sin x2

x2 = θ2x22 + x1 + u


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where θ1 and θ2 are unknown parameters that satisfy |θ1| ≤ a and |θ2| ≤ b for someknown bounds a and b. The system takes the form (14.66) with δ1 = θ1x1 sin x2and δ2 = θ2x

22. The function δ1 satisfies the inequality |δ1| ≤ a|x1| globally. The

function δ2 satisfies the inequality |δ2| ≤ bρ|x2| on the set |x2| ≤ ρ. Starting withthe system

x1 = x2 + θ1x1 sin x2

we take x2 = φ1(x1) = −k1x1 and V1(x1) = x21/2 to obtain

V1 = x1φ1(x1) + θ1x21 sin x2 ≤ −(k1 − a)x2

1

We choose k1 = 1 + a. To backstep, we apply the change of variables z2 = x2 +(1 + a)x1 and rewrite the system as

x1 = −(1 + a)x1 + θ1x1 sin x2 + z2

z2 = ψ1(x) + ψ2(x, θ) + u

whereψ1 = x1 + (1 + a)x2, ψ2 = (1 + a)θ1x1 sin x2 + θ2x

22

Using Vc = (x21 + z2

2)/2 as a composite Lyapunov function, we obtain

Vc ≤ −x21 + z2[x1 + ψ1(x) + ψ2(x, θ) + u]

Takingu = −x1 − ψ1(x)− kz2

yields

Vc ≤ −x21 + z2ψ2(x, θ)− kz2

2

≤ −x21 + a(1 + a)|x1| |z2|+ bx2

2|z2| − kz22

On the setΩc = {x ∈ R2 | Vc(x) ≤ c}

we have |x2| ≤ ρ for some ρ dependent on c.20 Restricting our analysis to Ωc, weobtain

Vc ≤ −x21 + a(1 + a)|x1| |z2|+ bρ|z2 − (1 + a)x1| |z2| − kz2

2

≤ −x21 + (1 + a)(a + bρ)|x1| |z2| − (k − bρ)z2

2

Choosingk > bρ + (1 + a)2(a + bρ)2/4

ensures that the origin is exponentially stable21 with Ωc contained in the region ofattraction. Since the preceding inequality can be satisfied for any c > 0 by choosingk sufficiently large, the feedback control can achieve semiglobal stabilization.

20ρ can be estimated by√

2c(1 + k21).

21Note that we conclude exponential stability, rather than asymptotic stability as guaranteedby Lemma 14.3. Why?


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Example 14.14 Consider again the system

x1 = x2 + θ1x1 sin x2

x2 = θ2x22 + x1 + u

from the previous example, where |θ1| ≤ a and |θ2| ≤ b. In that example, we limitedthe analysis to a compact set in order to satisfy the linear growth inequalities (14.67).Consequently, we could only achieve semiglobal stabilization. In this example, weachieve global stabilization by combining backstepping with Lyapunov redesign. Weproceed as in the previous example until we reach the point

Vc ≤ −x21 + z2[x1 + ψ1(x) + ψ2(x, θ) + u]

In Example 14.13, we used u = −x1 − ψ1(x)− kz2 and relied on the high gain k todeal with the uncertainty. This required the nonlinear term x2

2 to be bounded bythe linear term ρ|x2|. Here, we take

u = −x1 − ψ1(x)− kz2 + v, k > 0

where the control component v is to be designed. Then,

Vc ≤ −x21 − kz2

2 + z2[ψ2(x, θ) + v]

Noting that|ψ2(x, θ)| ≤ a(1 + a)|x1|+ bx2

2

we take

v =

⎧⎨⎩−η(x) sgn(z2), if η(x)|z2| ≥ ε

−η2(x)z2/ε, if η(x)|z2| < ε

where η(x) = η0 +a(1+a)|x1|+ bx22 for some η0 > 0 and ε > 0. When η(x)|z2| ≥ ε,

z2[ψ2(x, θ) + v] ≤ |ψ2||z2| − η|z2| ≤ 0

and when η(x)|z2| < ε,

z2[ψ2(x, θ) + v] ≤ |η||z2| − η2z22

ε≤ ε

4

Thus,Vc ≤ −x2

1 − kz22 +

ε

4This inequality shows that within a finite time interval, the state x enters a ball Br

of radius r = k0√

ε for some k0 > 0. Inside Br, we have

|ψ2(x, θ)| ≤ a(1 + a)|x1|+ br|x2| ≤ ρ1(|x1|+ |z2|)


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for some ρ1 > 0 independent of ε. In the intersection of Br and the boundary layer{η(x)|z2| < ε}, we have

Vc ≤ −x21 − kz2

2 + ρ1|x1||z2|+ ρ1|z2|2 − η20z2

2

ε

= − 12x2

1 − kz22 −

[12x2

1 − ρ1|x1||z2|+(

η20

ε− ρ1

)z22

]

The bracketed term can be made nonnegative by choosing ε small enough. Hence,

Vc ≤ −12x2

1 − kz22

and the origin is globally asymptotically stable. We conclude the section by noting that backstepping can be applied to multiin-

put systems in what is known as block backstepping, provided certain nonsingularityconditions are satisfied. Consider the system

η = f(η) + G(η)ξ (14.68)ξ = fa(η, ξ) + Ga(η, ξ)u (14.69)

where η ∈ Rn, ξ ∈ Rm, and u ∈ Rm, in which m could be greater than one. Supposef , fa, G, and Ga are (known) smooth functions over the domain of interest, f andfa vanish at the origin, and the m×m matrix Ga is nonsingular. Suppose furtherthat the component (14.68) can be stabilized by a smooth state feedback controllaw ξ = φ(η) with φ(0) = 0, and we know a (smooth, positive definite) Lyapunovfunction V (η) that satisfies the inequality

∂V

∂η[f(η) + G(η)φ(η)] ≤ −W (η)

for some positive definite function W (η). Using

Vc = V (η) + 12 [ξ − φ(η)]T [ξ − φ(η)]

as a Lyapunov function candidate for the overall system, we obtain

Vc =∂V

∂η(f + Gφ) +

∂V

∂ηG (ξ − φ) + [ξ − φ]T

[fa + Gau− ∂φ

∂η(f + Gξ)

]

Taking

u = G−1a

[∂φ

∂η(f + Gξ)−

(∂V

∂ηG

)T

− fa − k(ξ − φ)

], k > 0

results in

Vc =∂V

∂η(f + Gφ)− k[ξ − φ(η)]T [ξ − φ(η)] ≤ −W (η)− k[ξ − φ(η)]T [ξ − φ(η)]

which shows that the origin (η = 0, ξ = 0) is asymptotically stable.


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14.4 Passivity-Based Control

In Chapter 6, we introduced the notion of passivity and studied its role in ana-lyzing the stability of feedback connections. The ideas of passivity-based controlthat we are going to introduce here are straightforward applications of the resultsof Chapter 6. However, we do not need the elaborate details of Chapter 6 to under-stand these ideas. It is enough to recall the definitions of passivity and zero-stateobservability.

We consider the p-input–p-output system

x = f(x, u) (14.70)y = h(x) (14.71)

where f is locally Lipschitz in (x, u) and h is continuous in x, for all x ∈ Rn

and u ∈ Rm. We assume that f(0, 0) = 0, so that the origin x = 0 is an open-loop equilibrium point, and h(0) = 0. We recall that the system (14.70)–(14.71)is passive if there exists a continuously differentiable positive semidefinite functionV (x) (called the storage function) such that

uT y ≥ V =∂V

∂xf(x, u), ∀ (x, u) ∈ Rn ×Rm (14.72)

The system is zero-state observable if no solution of x = f(x, 0) can stay identicallyin the set {h(x) = 0} other than the trivial solution x(t) ≡ 0. Throughout thissection we will require the storage function to be positive definite. The basic ideaof passivity-based control is illustrated in the next theorem.

Theorem 14.4 If the system (14.70)–(14.71) is

(1) passive with a radially unbounded positive definite storage function and

(2) zero-state observable,

then the origin x = 0 can be globally stabilized by u = −φ(y), where φ is any locallyLipschitz function such that φ(0) = 0 and yT φ(y) > 0 for all y �= 0.

Proof: Use the storage function V (x) as a Lyapunov function candidate for theclosed-loop system

x = f(x,−φ(y))

The derivative of V is given by

V =∂V

∂xf(x,−φ(y)) ≤ −yT φ(y) ≤ 0

Hence, V is negative semidefinite and V = 0 if and only if y = 0. By zero-stateobservability,

y(t) ≡ 0 ⇒ u(t) ≡ 0 ⇒ x(t) ≡ 0


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14.4. PASSIVITY-BASED CONTROL 605

Therefore, by the invariance principle, the origin is globally asymptotically stable.�

The intuition behind the theorem becomes clear when we think of the storagefunction as the energy of the system. A passive system has a stable origin. Allthat is needed to stabilize the origin is the injection of damping so that energy willdissipate whenever x(t) is not identically zero. The required damping is injected bythe function φ. There is great freedom in the choice of φ. We can choose it to meetany constraint on the magnitude of u. For example, if u is constrained to |ui| ≤ ki

for 1 ≤ i ≤ p, we can choose φi(y) = ki sat(yi) or φi(y) = (2ki/π) tan−1(yi).The utility of Theorem 14.4 can be increased by transforming nonpassive systems

into passive ones. Consider, for example, a special case of (14.70), where

x = f(x) + G(x)u (14.73)

Suppose a radially unbounded, positive definite, continuously differentiable functionV (x) exists such that

∂V

∂xf(x) ≤ 0, ∀ x

Take

y = h(x) def=[∂V

∂xG(x)

]T

Then the system with input u and output y is passive. If it is also zero-stateobservable, we can apply Theorem 14.4.


x1 = x2

x2 = −x31 + u

Let V (x) = x41/4 + x2

2/2. Then

V = x31x2 − x2x

31 + x2u = x2u

Set y = x2 and note that, with u = 0, y(t) ≡ 0 implies that x(t) ≡ 0. Thus, all theconditions of Theorem 14.4 are satisfied and a globally stabilizing state feedbackcontrol can be taken as u = −kx2 or u = −(2k/π) tan−1(x2) with any k > 0.

Allowing ourselves the freedom to choose the output function is useful, but weare still limited to state equations for which the origin is open-loop stable. We cancover a wider class of systems if we use feedback to achieve passivity. Consideragain the system (14.73). If a feedback control

u = α(x) + β(x)v (14.74)


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and an output function h(x) exist such that the system

x = f(x) + G(x)α(x) + G(x)β(x)v (14.75)y = h(x) (14.76)

with input v and output y, satisfies the conditions of Theorem 14.4, we can globallystabilize the origin by using v = −φ(y). The use of feedback to convert a nonpassivesystem into a passive one is known as feedback passivation.22

Example 14.16 The nonlinear dynamic equations for an m-link robot take theform

M(q)q + C(q, q)q + Dq + g(q) = u

where q is an m-dimensional vector of generalized coordinates representing jointpositions, u is an m-dimensional control input, and M(q) is a symmetric inertiamatrix that is positive definite for all q ∈ Rm. The term C(q, q)q accounts forcentrifugal and Coriolis forces. The matrix C has the property that M − 2C is askew-symmetric matrix for all q, q ∈ Rm, where M is the total derivative of M(q)with respect to t. The term Dq, where D is a positive semidefinite symmetricmatrix, accounts for viscous damping. The term g(q), which accounts for gravityforces, is given by g(q) = [∂P (q)/∂q]T , where P (q) is the total potential energyof the links due to gravity. Consider the regulation problem of designing a statefeedback control law so that q asymptotically tracks a constant reference qr. Lete = q − qr. Then, e satisfies the differential equation

M(q)e + C(q, q)e + De + g(q) = u

Our goal is to stabilize the system at (e = 0, e = 0), but this point is not anopen-loop equilibrium point. Let

u = g(q)−Kpe + v

where Kp is a positive definite symmetric matrix and v is an additional controlcomponent to be chosen. Substituting u into the e-equation yields

M(q)e + C(q, q)e + De + Kpe = v

TakeV = 1

2 eT M(q)e + 12eT Kpe

as a storage function candidate. The function V is positive definite and its derivativesatisfies

V = eT Me + 12 eT M e + eT Kpe

= 12 eT (M − 2C)e− eT De− eT Kpe + eT v + eT Kpe

≤ eT v22For the system (14.73) with output y = h(x), it is shown in [31] that the system is

locally feedback equivalent to a passive system with a positive definite storage function ifrank{[∂h/∂x](0)G(0)} = p and the zero dynamics have a stable equilibrium point at the originwith a positive definite Lyapunov function.


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Defining the output as y = e, we see that the system with input v and output yis passive with V as the storage function. It is interesting to note that the role ofthe passifying feedback component g(q)−Kpe is to reshape the potential energy to(1/2)eT Kpe, which has a unique minimum at e = 0. The sum of the kinetic energyand the reshaped potential energy is the storage function. With v = 0,

y(t) ≡ 0 ⇔ e(t) ≡ 0 ⇒ e(t) ≡ 0 ⇒ Kpe(t) ≡ 0 ⇒ e(t) ≡ 0

which shows that the system is zero-state observable. Hence, it can be globallystabilized by the control v = −φ(e) with any function φ such that φ(0) = 0 andyT φ(y) > 0 for all y �= 0. The choice v = −Kde with a positive definite symmetricmatrix Kd results in the control

u = g(q)−Kp(q − qr)−Kdq

which takes the form of a classical PD controller plus a gravity compensation term.

One class of systems that is amenable to feedback passivation is the cascadeconnection of a passive system with a system whose unforced dynamics have astable equilibrium point at the origin. Consider the system

z = fa(z) + F (z, y)y (14.77)x = f(x) + G(x)u (14.78)y = h(x) (14.79)

where fa(0) = 0, f(0) = 0, and h(0) = 0. The functions fa, F , f , and G are locallyLipschitz and h is continuous. We view the system as a cascade connection of thedriving system (14.78)–(14.79) and the driven system (14.77).23 We assume thatthe representation (14.77)–(14.79) is valid globally, the driving system is passivewith a radially unbounded positive definite storage function V (x), the origin ofz = fa(z) is stable, and we know a radially unbounded Lyapunov function W (z)for z = fa(z), which satisfies

∂W

∂zfa(z) ≤ 0, ∀ z

Using U(z, x) = W (z) + V (x) as a storage function candidate for the full system(14.77)–(14.79), we obtain

U =∂W

∂zfa(z) +

∂W

∂zF (z, y)y +

∂V

∂xf(x) +

∂V

∂xG(x)u

23A driven system of the form z = f0(z, y) with sufficiently smooth f0 can be represented in theform (14.77) by taking

fa(z) = f0(z, 0) and F (z, y) =

∫ 1

0

∂f0

∂y(z, sy) ds


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≤ ∂W

∂zF (z, y)y + yT u = yT

[u +

(∂W

∂zF (z, y)

)T]

The feedback control

u = −(

∂W

∂zF (z, y)

)T

+ v

results inU ≤ yT v

Hence, the system

z = fa(z) + F (z, y)y (14.80)

x = f(x)−G(x)(

∂W

∂zF (z, y)

)T

+ G(x)v (14.81)

y = h(x) (14.82)

with input v and output y is passive with U as the storage function. If the system(14.80)–(14.82) is zero-state observable, we can apply Theorem 14.4 to globallystabilize the origin.

Example 14.17 The rotational motion of a rigid body subject to three indepen-dent scalar control torques can be modeled by24

ρ = 12 [I3 + S(ρ) + ρρT ]ω

Mω = −S(ω)Mω + u

where ω ∈ R3 is the velocity vector and ρ ∈ R3 is a particular choice of the kinematicparameters, which leads to a three-dimensional representation of the rotation group.The matrix S(x) is a skew-symmetric matrix, defined by

S(x) =

⎡⎣ 0 −x3 x2

x3 0 −x1−x2 x1 0

⎤⎦

M is a positive definite symmetric inertia matrix, and I3 is the 3 × 3 identitymatrix. Taking y = ω, it can be seen that the system takes the form of the cascadeconnection (14.77)–(14.79) with

Mω = −S(ω)Mω + u, y = ω

as the driving system and

ρ = 12 [I3 + S(ρ) + ρρT ]ω

24See [97] and [151] for the derivation of the model. If ε ∈ R3 and η ∈ R are the Eulerparameters, then ρ = ε/η.


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as the driven system. Taking V (ω) = (1/2)ωT Mω, it can be seen that

V = ωT Mω = −ωT S(ω)Mω + ωT u = yT u

where we used the property ωT S(ω) = 0. Hence, the driving system is passive.The unforced driven system ρ = 0 has a stable equilibrium point at ρ = 0 and anyradially unbounded, positive definite, continuously differentiable function W (ρ) willqualify as a Lyapunov function. Thus, all our assumptions are satisfied and thesystem can be made passive by the control

u = −{

∂W

∂ρ12 [I3 + S(ρ) + ρρT ]

}T

+ v

Taking W (ρ) = k ln(1 + ρT ρ), with k > 0, yields

u = −{

kρT

1 + ρT ρ[I3 + S(ρ) + ρρT ]

}T

+ v = −kρ + v

where we used the property ρT S(ρ) = 0. We need to check zero-state observabilityof the passive system

ρ = 12 [I3 + S(ρ) + ρρT ]ω

Mω = −S(ω)Mω − kρ + v

y = ω

With v = 0,y(t) ≡ 0 ⇔ ω(t) ≡ 0 ⇒ ω(t) ≡ 0 ⇒ ρ(t) ≡ 0

Hence, the system is zero-state observable and can be globally stabilized by

u = −kρ− φ(ω)

with any locally Lipschitz function φ such that φ(0) = 0 and yT φ(y) > 0 for ally �= 0.

Checking zero-state observability of the full system (14.80)–(14.82) can be avoidedif we strengthen the assumption on W (z) to

∂W

∂zfa(z) < 0, ∀ z �= 0, and

∂W

∂z(0) = 0 (14.83)

which implies that the origin of z = fa(z) is globally asymptotically stable. Taking

u = −(

∂W

∂zF (z, y)

)T

− φ(y) (14.84)


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where φ is any locally Lipschitz function such that φ(0) = 0 and yT φ(y) > 0 for ally �= 0, and using U as a Lyapunov function candidate for the closed-loop system,we obtain

U ≤ ∂W

∂zfa(z)− yT φ(y) ≤ 0

Moreover, U = 0 implies that z = 0 and y = 0, which in turn imply that u = 0. Ifthe driving system (14.78)–(14.79) is zero-state observable, the conditions u(t) ≡ 0and y(t) ≡ 0 imply that x(t) ≡ 0. Hence, by the invariance principle, the origin(z = 0, x = 0) is globally asymptotically stable. We summarize this conclusion inthe next theorem.

Theorem 14.5 Suppose the system (14.78)–(14.79) is zero-state observable andpassive with a radially unbounded, positive definite storage function. Suppose theorigin of z = fa(z) is globally asymptotically stable and let W (z) be a radiallyunbounded, positive definite Lyapunov function that satisfies (14.83). Then, thecontrol (14.84) globally stabilizes the origin (z = 0, x = 0). �


η = −η + η2ξ

ξ = u

which was studied in Examples 13.16 and 14.11. With y = ξ as the output, thesystem takes the form (14.77)–(14.79). The system ξ = u, y = ξ is passive withthe storage function V (ξ) = ξ2/2. It is clearly zero-state observable, since y = ξ.The origin of η = −η is globally exponentially stable with the Lyapunov functionW (η) = η2/2. Thus, all the conditions of Theorem 14.5 are satisfied and a globallystabilizing state feedback control can be taken as

u = −η3 − kζ, k > 0

which is the same control we derived by using backstepping

14.5 High-Gain Observers

The nonlinear design techniques discussed in this chapter, and the previous one,assume state feedback; that is, measurements of all state variables are available. Inmany practical problems we cannot measure all state variables, or we may choosenot to measure some of them due to technical or economic reasons. Therefore, it isimportant to extend these techniques to output feedback. In some special cases, wecan modify the technique to produce an output feedback controller. Examples areexplored in Exercises 14.47 and 14.48. The first exercise shows that for minimum-phase, relative-degree-one systems, we can design sliding mode control as outputfeedback. The second exercise shows passivity-based control for a system that has a


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14.5. HIGH-GAIN OBSERVERS 611

passive map from the input to the derivative of the output. In more general cases,we have to use dynamic compensation to extend state feedback designs to outputfeedback. One form of dynamic compensation is to use observers that asymptoti-cally estimate the state from output measurements. For some nonlinear systems,the design of such observers could be as easy as in linear systems. Suppose, forexample, that a nonlinear system can be transformed into the form25

x = Ax + g(y, u) (14.85)y = Cx (14.86)

where (A, C) is observable. This form is special because the nonlinear function gdepends only on the output y and the control input u. Taking the observer as

˙x = Ax + g(y, u) + H(y − Cx) (14.87)

it can be easily seen that the estimation error x = x− x satisfies the linear equation

˙x = (A−HC)x

Hence, designing C such that A − HC is Hurwitz guarantees asymptotic errorconvergence, that is, limt→∞ x(t) = 0. Exercise 14.49 explores the use of the ob-server (14.87) in output feedback control. Aside from the fact that the observer(14.87) works only for a special class of nonlinear systems, its main drawback is theassumption that the nonlinear function g is perfectly known. Any error in modelingg will be reflected in the estimation error equation. In particular, if the observer isimplemented as

˙x = Ax + g0(y, u) + H(y − Cx)

where g0 is a nominal model of g, the ˙x-equation becomes

˙x = (A−HC)x + g(y, u)− g0(y, u)

It is no longer obvious that a Hurwitz A −HC can handle the perturbation termg− g0. In this section, we give a special design of the observer gain that makes theobserver robust to uncertainties in modeling the nonlinear functions. The technique,known as high-gain observers, works for a wide class of nonlinear systems andguarantees that the output feedback controller recovers the performance of the statefeedback controller when the observer gain is sufficiently high. In Section 14.5.1,we use a second-order example to motivate the idea of high-gain observers. InSection 14.5.2, we use the observer in output feedback stabilization. The main resultof that section is a separation principle that allows us to separate the design intotwo tasks. First, we design a state feedback controller that stabilizes the system and

25Necessary and sufficient conditions for a nonlinear system to be equivalent to the form (14.85)–(14.86) are given in [124, Chapter 5]. The same reference gives a different approach for introducingdynamic compensation through filtered transformations.


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meets other design specifications. Then, an output feedback controller is obtainedby replacing the state x by its estimate x provided by the high-gain observer. Akey property that makes this separation possible is the design of the state feedbackcontroller to be globally bounded in x. High-gain observers can be used in a widerange of control problems.26 As an example, we show in Section 14.5.3 the use ofhigh-gain observers in output feedback integral regulators.

14.5.1 Motivating Example

Consider the second-order nonlinear system

x1 = x2 (14.88)x2 = φ(x, u) (14.89)y = x1 (14.90)

where x = [x1, x2]T . Suppose u = γ(x) is a locally Lipschitz state feedback controllaw that stabilizes the origin x = 0 of the closed-loop system

x1 = x2 (14.91)x2 = φ(x, γ(x)) (14.92)

To implement this feedback control using only measurements of the output y, weuse the observer

˙x1 = x2 + h1(y − x1) (14.93)˙x2 = φ0(x, u) + h2(y − x1) (14.94)

where φ0(x, u) is a nominal model of the nonlinear function φ(x, u). The estimationerror

x =[

x1x2

]=[

x1 − x1x2 − x2

]

satisfies the equation

˙x1 = −h1x1 + x2 (14.95)˙x2 = −h2x1 + δ(x, x) (14.96)

where δ(x, x) = φ(x, γ(x)) − φ0(x, γ(x)). We want to design the observer gainH = [h1, h2]T such that limt→∞ x(t) = 0. In the absence of the disturbance termδ, asymptotic error convergence is achieved by designing H such that

Ao =[ −h1 1−h2 0

]26See [100] for a survey of the use of high-gain observers in various control problem formulations.


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is Hurwitz. For this second-order system, Ao is Hurwitz for any positive constantsh1 and h2. In the presence of δ, we need to design H with the additional goal ofrejecting the effect of δ on x. This is ideally achieved, for any δ, if the transferfunction

Go(s) =1

s2 + h1s + h2

[1

s + h1

]

from δ to x is identically zero. While this is not possible, we can make supω∈R |Go(jω)|arbitrarily small by choosing h2 � h1 � 1. In particular, taking

h1 =α1

ε, h2 =

α2

ε2 (14.97)

for some positive constants α1, α2, and ε, with ε � 1, it can be shown that

Go(s) =ε

(εs)2 + α1εs + α2

[ε

εs + α1

]

Hence, limε→0 Go(s) = 0. The disturbance rejection property of the high-gainobserver can be also seen in the time domain by representing the error equa-tion (14.95)–(14.96) in the singularly perturbed form. Towards that end, definethe scaled estimation errors

η1 =x1

ε, η2 = x2 (14.98)

The newly defined variables satisfy the singularly perturbed equation

εη1 = −α1η1 + η2 (14.99)εη2 = −α2η1 + εδ(x, x) (14.100)

This equation shows clearly that reducing ε diminishes the effect δ. It shows alsothat, for small ε, the scaled-estimation error η will be much faster than x. Notice,however, that η1(0) will be O(1/ε) whenever x1(0) �= x1(0). Consequently, thesolution of (14.99)–(14.100) will contain a term of the form (1/ε)e−at/ε for somea > 0. While this exponential mode decays rapidly, it exhibits an impulsive-likebehavior where the transient peaks to O(1/ε) values before it decays rapidly towardszero. In fact, the function (a/ε)e−at/ε approaches an impulse function as ε tends tozero. This behavior is known as the peaking phenomenon. It is important to realizethat the peaking phenomenon is not a consequence of using the change of variables(14.98) to represent the error dynamics in the singularly perturbed form. It is anintrinsic feature of any high-gain-observer with h2 � h1 � 1. This point can beseen by calculating the transition matrix exp(Aot) and noting that its (2,1) elementis given by

−2h2√4h2 − h2

1

e−h1t/2 sin

(t√

4h2 − h21

2

)


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when 4h2 > h21 and

−h2√h2

1 − 4h2

{exp

[−(

h1 −√

h21 − 4h2

2

)t

]− exp

[−(

h1 +√

h21 − 4h2

2

)t

]}

when 4h2 < h21. The amplitude of the exponential mode is greater than

√h2 in the

first case and h2/h1 in the second one. Thus, as we increase h1 and h2/h1, we drivethe amplitude toward infinity.

To get a better feel for the peaking phenomenon, let us simulate the system

x1 = x2

x2 = x32 + u

y = x1

which can be globally stabilized by the state feedback controller

u = −x32 − x1 − x2

The output feedback controller is taken as

u = −x32 − x1 − x2

˙x1 = x2 + (2/ε)(y − x1)˙x2 = (1/ε2)(y − x1)

where the observer gain assigns the eigenvalues of Ao at −1/ε and −1/ε. Fig-ure 14.16 shows the performance of the closed-loop system under state and outputfeedback. Output feedback is simulated for three different values of ε. The initialconditions are x1(0) = 0.1, x2(0) = x1(0) = x2(0) = 0. Peaking is induced by[x1(0)− x1(0)]/ε = 0.1/ε when ε is sufficiently small. Figure 14.16 shows a counterintuitive behavior as ε decreases. Since decreasing ε causes the estimation errorto decay faster toward zero, one would expect the response under output feedbackto approach the response under state feedback as ε decreases. Figure 14.16 showsthe opposite behavior, where the response under output feedback deviates from theresponse under state feedback as ε decreases. This is the impact of the peakingphenomenon. The same figure shows the control u on a much shorter time intervalto exhibit peaking. This control peaking is transmitted to the plant causing itsstate to peak. If peaking of the state takes it outside the region of attraction, itcould destabilize the system. Figure 14.17 shows that this is exactly what happensin the current case as we decrease ε to 0.004 where the system has a finite escapetime shortly after t = 0.07.

Fortunately, we can overcome the peaking phenomenon by saturating the controloutside a compact region of interest in order to create a buffer that protects theplant from peaking. Suppose the control is saturated as

u = sat(−x32 − x1 − x2)


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0 1 2 3 4 5 6 7 8 9 10−2

−1.5

−1

−0.5

0

0.5x 1

SFBOFB ε = 0.1OFB ε = 0.01OFB ε = 0.005

0 1 2 3 4 5 6 7 8 9 10−3

−2

−1

0

1

x 2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1−400

−300

−200

−100

0

u

t

Figure 14.16: Performance under state (SFB) and output (OFB) feedback.

Figure 14.18 shows the performance of the closed-loop system under saturated stateand output feedback. The control u is shown on a shorter time interval that exhibitscontrol saturation during peaking. The peaking period decreases with ε. The statesx1 and x2 exhibit the intuitive behavior we expected earlier; namely, the responseunder output feedback approaches the response under state feedback as ε decreases.Note that we decrease ε to 0.001, beyond the value 0.004 where instability wasdetected in the unsaturated case. Not only does the system remain stable, butthe response under output feedback is almost indistinguishable from the responseunder state feedback. What is more interesting is that the region of attraction underoutput feedback approaches the region of attraction under state feedback as ε tendsto zero. This is shown in Figures 14.19 and 14.20. The first figure shows the phaseportrait of the closed-loop system under u = sat(−x3

2 − x1 − x2). It has a boundedregion of attraction enclosed by a limit cycle. The second figure shows that theintersection of the boundary of the region of attraction under u = sat(−x3

2−x1−x2)with the x1–x2 plane approaches the limit cycle as ε tends to zero.

The behavior we saw in Figures 14.18 and 14.20 will be realized with any glob-ally bounded stabilizing function γ(x). During the peaking period, the control γ(x)saturates. Since the peaking period shrinks to zero as ε tends to zero, for sufficientlysmall ε, the peaking period becomes so small that the state of the plant x remainsclose to its initial value. After the peaking period, the estimation error becomesO(ε) and the feedback control γ(x) becomes close to γ(x). Consequently, the tra-jectories of the closed-loop system under output feedback asymptotically approach


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0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08−0.6

−0.4

−0.2

0

0.2

x 1

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08−600

−400

−200

0

x 2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08−1000

0

1000

2000

u

t

Figure 14.17: Instability induced by peaking at ε = 0.004.

its trajectories under state feedback as ε tends to zero. This leads to recovery of theperformance achieved under state feedback. The global boundedness of γ(x) canbe always achieved by saturating the state feedback control, or the state estimates,outside a compact region of interest.

The analysis of the closed-loop system under output feedback proceeds as fol-lows: The system is represented in the singularly perturbed form

x1 = x2

x2 = φ(x, γ(x))εη1 = −α1η1 + η2

εη2 = −α2η1 + εδ(x, x)

where x1 = x1 − εη1 and x2 = x2 − η2. The motion of the slow variables (x1, x2)can be approximated by a slow model obtained by setting ε = 0. Since ε = 0 yieldsη = 0, the slow model is the closed-loop system under state feedback, given by(14.91)–(14.92). The fast motion of (η1, η2) can be approximated by the fast model

εη =[ −α1 1−α2 0

]η

def= A0η

obtained by neglecting εδ. Let V (x) be a Lyapunov function for the slow modeland W (η) = ηT P0η be a Lyapunov function for the fast model, where P0 is the


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0 1 2 3 4 5 6 7 8 9 10−0.05

0

0.05

0.1

0.15

x 1

SFBOFB ε = 0.1OFB ε = 0.01OFB ε = 0.001

0 1 2 3 4 5 6 7 8 9 10

−0.1

−0.05

0

0.05

x 2

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

−1

−0.5

0

u

t

Figure 14.18: Performance under state (SFB) and output (OFB) feedback with satu-ration.

solution of the Lyapunov equation P0A0 + AT0 PT

0 = −I. Define the sets Ωc andΣ by Ωc = {V (x) ≤ c} and Σ = {W (η) ≤ ρε2}, where c > 0 is chosen such thatΩc is in the interior of the region of attraction of (14.91)–(14.92). The analysis canbe divided in two basic steps. In the first step we show that for sufficiently largeρ, there is ε∗1 > 0 such that, for every 0 < ε ≤ ε∗1, the origin of the closed-loopsystem is asymptotically stable and the set Ωc × Σ is a positively invariant subsetof the region of attraction. The proof makes use of the fact that in Ωc × Σ, η isO(ε). In the second step, we show that for any bounded x(0) and any x(0) ∈ Ωb,where 0 < b < c, there exists ε∗2 > 0 such that, for every 0 < ε ≤ ε∗2, the trajectoryenters the set Ωc × Σ in finite time. The proof makes use of the fact that Ωb is inthe interior of Ωc and γ(x) is globally bounded. Hence, there exits a time T1 > 0,independent of ε, such that any trajectory starting in Ωb will remain in Ωc for allt ∈ [0, T1]. Using the fact that η decays faster than an exponential mode of the form(1/ε)e−at/ε, we can show that the trajectory enters the set Ωc × Σ within a timeinterval [0, T (ε)], where limε→0 T (ε) = 0. Thus, by choosing ε small enough, we canensure that T (ε) < T1. Figure 14.21 gives a sketch that illustrates this behavior.

The full-order observer (14.93)–(14.94) provides estimates (x1, x2) that are usedto replace (x1, x2) in the feedback control law. Since y = x1 is measured, we canuse x1 in the control law and only replace x2 by x2. This approach does not change


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−3 −2 −1 0 1 2 3−2

−1

0

1

2

x1

x 2

Figure 14.19: Phase portrait of the closed-loop system under u = sat(−x32−x1−x2).

the analysis of the closed-loop system, and we obtain the same results as before.On the other hand, we can use the reduced-order observer

w = − h (w + hy) + φo(x, u) (14.101)x2 = w + hy (14.102)

where h = α/ε for some positive constants α and ε with ε � 1, to estimate x2. Itis not difficult to see that the high-gain reduced-order observer (14.101)–(14.102)exhibits the peaking phenomenon, and global boundedness of the state feedbackcontrol plays the same role as in the full-order observer case.

The high-gain observer is basically an approximate differentiator. This pointcan be easily seen in the special case when the nominal function φ0 is chosen to bezero, for then the observer is linear. For the full-order observer (14.93)–(14.94), thetransfer function from y to x is

α2

(εs)2 + α1εs + α2

[1 + (εα1/α2)s

s

]→

[1s

]as ε → 0

and for the reduced-order observer (14.101)–(14.102), the transfer function from yto x2 is

s

(ε/α)s + 1→ s as ε → 0

Thus, on a compact frequency interval, the high-gain observer approximates y forsufficiently small ε.

Realizing that the high-gain observer is basically an approximate differentiator,we can see that measurement noise and unmodeled high-frequency sensor dynamicswill put a practical limit on how small ε could be. Despite this limitation, thereare interesting applications where the range of permissible values of ε allows forsuccessful application of high-gain observers.27

27Examples of application to induction motors and mechanical systems are given in [3], [47],and [186].


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−1.5 −1 −0.5 0 0.5 1 1.5

−1

−0.5

0

0.5

1

x1

x 2

Figure 14.20: The region of attraction under state feedback (solid) and intersectionof the region of attraction under output feedback with the x1-x2 plane for ε = 0.1(dashed) and ε = 0.05 (dash-dot).


Consider the multi-input–multi-output nonlinear system

x = Ax + Bφ(x, z, u) (14.103)z = ψ(x, z, u) (14.104)y = Cx (14.105)ζ = q(x, z) (14.106)

where u ∈ Rp is the control input, y ∈ Rm and ζ ∈ Rs are measured outputs, andx ∈ Rρ and z ∈ R� constitute the state vector. The ρ × ρ matrix A, the ρ × mmatrix B, and the m× ρ matrix C, given by

A = block diag[A1, . . . , Am], Ai =

⎡⎢⎢⎢⎢⎢⎣

0 1 · · · · · · 00 0 1 · · · 0...

...0 · · · · · · 0 10 · · · · · · · · · 0

⎤⎥⎥⎥⎥⎥⎦

ρi×ρi

B = block diag[B1, . . . , Bm], Bi =

⎡⎢⎢⎢⎢⎢⎣

00...01

⎤⎥⎥⎥⎥⎥⎦

ρi×1


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�

�

x

ηO(1/ε)

O(ε)

Ωc

Ωb��

��

��

��

��

Figure 14.21: Illustration of fast convergence to the set Ωc × Σ.

C = block diag[C1, . . . , Cm], Ci =[

1 0 · · · · · · 0]1×ρi

where 1 ≤ i ≤ m and ρ = ρ1 + · · · + ρm, represent m chains of integrators. Thefunctions φ, ψ, and q are locally Lipschitz in their arguments for (x, z, u) ∈ Dx ×Dz × Rp, where Dx ⊂ Rρ and Dz ⊂ Rs are domains that contain their respectiveorigins. Moreover, φ(0, 0, 0) = 0, ψ(0, 0, 0) = 0, and q(0, 0) = 0. Our goal is todesign an output feedback controller to stabilize the origin.

The two main sources for the model (14.103)–(14.106) are the normal form ofinput–output linearizable systems and models of mechanical and electromechanicalsystems, where displacement variables are measured while their derivatives (veloc-ities, accelerations, etc.) are not measured. The normal form for a single-input–single-output system is given by (13.16)–(13.18). It is easy to see that the equationtakes the form (14.103)–(14.105) with x = ξ and z = η.28 If y is the only measuredvariable, we can drop equation (14.106). However, in many problems, we can mea-sure some state variables in addition to those at the end of the chains of integrators.For example, the magnetic suspension system of Exercise 1.18 is modeled by

x1 = x2

x2 = g − k

mx2 − L0ax2

3

2m(a + x1)2

x3 =1

L(x1)

[−Rx3 +

L0ax2x3

(a + x1)2+ u

]

where x1 is the ball position, x2 is its velocity, and x3 is the electromagnet cur-rent. Typically, we measure the ball position x1 and the current x3. The modelfits the form (14.103)–(14.106) with (x1, x2) as the x component and x3 as the z

28See [88, Section 5.1] for the multivariable normal form.


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component. The measured outputs are y = x1 and ζ = x3. Another source for themodel (14.103)–(14.106) where (14.106) is significant arises in system in which thedynamics are extended by adding integrators. In Example 13.8, we saw a systemrepresented by an nth order differential equation. The dynamics were extendedby adding m integrators at the input side. Inspection of the resulting state modelshows that it fits the form (14.103)–(14.106) with z as the states of the m integra-tors and x as the output and its derivatives up to y(n−1). In this case, the wholevector z is measured, and equation (14.106) takes the form ζ = z.

We use a two-step approach to design the output feedback controller. First, apartial state feedback controller that uses measurements of x and ζ is designed toasymptotically stabilize the origin. Then, a high-gain observer is used to estimatex from y. The state feedback controller is allowed to be a dynamical system of theform

ϑ = Γ(ϑ, x, ζ) (14.107)u = γ(ϑ, x, ζ) (14.108)

where γ and Γ are locally Lipschitz functions in their arguments over the domainof interest and globally bounded functions of x. Moreover, γ(0, 0, 0) = 0 andΓ(0, 0, 0) = 0. A static state feedback controller u = γ(x, ζ) will be viewed as aspecial case of the foregoing equation by dropping the ϑ-equation. For convenience,we write the closed-loop system under state feedback as

X = f(X ) (14.109)

where X = (x, z, ϑ). The output feedback controller is taken as

ϑ = Γ(ϑ, x, ζ) (14.110)u = γ(ϑ, x, ζ) (14.111)

where x is generated by the high-gain observer

˙x = Ax + Bφ0(x, ζ, u) + H(y − Cx) (14.112)

The observer gain H is chosen as

H = block diag[H1, . . . , Hm], Hi =

⎡⎢⎢⎢⎢⎢⎣

αi1/ε

αi2/ε2

...αi

ρi−1/ερi−1

αiρi

/ερi

⎤⎥⎥⎥⎥⎥⎦

ρi×1

(14.113)

where ε is a positive constant to be specified and the positive constants αij are

chosen such that the roots of

sρi + αi1s

ρi−1 + · · ·+ αiρi−1s + αi

ρi= 0


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are in the open left-half plane, for all i = 1, . . . , m. The function φ0(x, ζ, u) is anominal model of φ(x, z, u), which is required to be locally Lipschitz in its argumentsover the domain of interest and globally bounded in x. Moreover, φ0(0, 0, 0) = 0.

Theorem 14.6 Consider the closed-loop system of the plant (14.103)–(14.106) andthe output feedback controller (14.110)–(14.112). Suppose the origin of (14.109) isasymptotically stable and R is its region of attraction. Let S be any compact set inthe interior of R and Q be any compact subset of Rρ. Then,

• there exists ε∗1 > 0 such that, for every 0 < ε ≤ ε∗1, the solutions (X (t), x(t))of the closed-loop system, starting in S ×Q, are bounded for all t ≥ 0.

• given any μ > 0, there exist ε∗2 > 0 and T2 > 0, both dependent on μ, suchthat, for every 0 < ε ≤ ε∗2, the solutions of the closed-loop system, starting inS ×Q, satisfy

‖X (t)‖ ≤ μ and ‖x(t)‖ ≤ μ, ∀ t ≥ T2 (14.114)

• given any μ > 0, there exists ε∗3 > 0, dependent on μ, such that, for every0 < ε ≤ ε∗3, the solutions of the closed-loop system, starting in S ×Q, satisfy

‖X (t)−Xr(t)‖ ≤ μ, ∀ t ≥ 0 (14.115)

where Xr is the solution of (14.109), starting at X (0).

• if the origin of (14.109) is exponentially stable, then there exists ε∗4 > 0 suchthat, for every 0 < ε ≤ ε∗4, the origin of the closed-loop system is exponentiallystable and S ×Q is a subset of its region of attraction.

�

Proof: See Appendix C.23.

The theorem shows that the output feedback controller recovers the performanceof the state feedback controller for sufficiently small ε. The performance recoverymanifests itself in three points. First, recovery of exponential stability. Second, re-covery of the region of attraction in the sense that we can recover any compact set inits interior. Third, the solution X (t) under output feedback approaches the solutionunder state feedback as ε tends to zero. For convenience, recovery of asymptoticstability is shown only for the exponentially stable case.29 Notice, however, that thefirst three bullets of the theorem, which show boundedness, ultimate boundedness,and trajectory convergence, are valid without the exponential stability assumption.

As a corollary of the theorem, it is clear that if the state feedback controllerachieves global or semiglobal stabilization with local exponential stability, then forsufficiently small ε, the output feedback controller achieves semiglobal stabilizationwith local exponential stability.

29See [16] for the more general case when the origin is asymptotically, but not exponentially,stable.


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Example 14.19 In Section 14.1, we designed the continuous sliding mode statefeedback controller

u = −k sat

(a1(θ − π) + θ

μ

)

with a1 = 1, k = 4, and μ = 1, to stabilize the pendulum equation

m�2θ + mg0� sin θ + k0�2θ = u

at (θ = π, θ = 0). Suppose now that we only measure θ. An output feedbackcontroller can be taken as

u = −k sat

(a1(θ − π) + ω

μ

)

where θ and ω are the estimates of θ and ω = θ provided by the high-gain observer

˙θ = ω + (2/ε)(θ − θ)˙ω = φ0(θ, u) + (1/ε2)(θ − θ)

where φ0 = −a sin θ + cu is a nominal model of φ = −(g0/�) sin θ − (k0/m)θ +(1/m�2)u in which a and c are nominal values of (g0/�) and (1/m�2), respectively,while the nominal value of the coefficient of friction k0 is taken to be zero. Theobserver is designed to have multiple poles at −1/ε. In Figure 14.22, we compare theperformance of the state and output feedback controllers for ε = 0.05 and ε = 0.01.The pendulum parameters are m = 0.15, � = 1.05, and k0 = 0.02 and the initialconditions are θ(0) = π/4 and ω(0) = θ(0) = ω(0) = 0. We consider three casesfor the observer. In the first case, the observer uses the nominal values a = 9.81and c = 10, which correspond to the nominal parameters m = 0.1 and � = 1. Inthe second case, we use a = 9.3429 and c = 6.0469, which correspond to the actualparameters; that is, m = m = 0.15 and � = � = 1.05. In the third case, we use alinear observer by setting a = c = 0. In all cases, we see that the response underoutput feedback approaches the response under state feedback as ε decreases. Whenε is relatively large, we see an advantage for including φ0 in the observer when it isa good model of φ. However, if the model is not that good, a linear observer mayperform better. The important thing to notice here is that the differences betweenthe three observers diminish as ε decreases. This is expected, because decreasing εrejects the effect of the uncertainty in modeling φ.

14.5.3 Regulation via Integral Control


x = f(x, w) + g(x, w)[u + δ(x, u, w)]y = h(x, w)


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0 2 4 6 8 100.5

1

1.5

2

2.5

3

3.5

θ

(a)

0 2 4 6 8 10−2

−1

0

1

2

ω

(b)

0 2 4 6 8 100.5

1

1.5

2

2.5

3

3.5

θ

Time

(c)

0 2 4 6 8 100

0.5

1

1.5

2

2.5

3

3.5

θ

(d)

Time

SFBOFB ε = 0.05OFB ε = 0.01

Figure 14.22: Comparison of state feedback (SFB) and output feedback (OFB) forExample 14.19. Figures (a) and (b) show θ and ω = θ for a nonlinear high-gainobserver with nominal m and �. Figure (c) shows θ for a nonlinear high-gain observerwith actual m and �. Figure (d) shows θ for a linear high-gain observer.

where x ∈ Rn is the state, u ∈ R is the control input, y ∈ R is the controlled aswell as measured output, and w ∈ Rl is a vector of unknown constant parametersand disturbances. The functions f , g, h, and δ are sufficiently smooth in (x, u) andcontinuous in w for x ∈ D ⊂ Rn, u ∈ R, and w ∈ Dw ⊂ Rl, where D and Dw areopen connected sets. We assume that the system

x = f(x, w) + g(x, w)uy = h(x, w)

has relative degree ρ in D uniformly in w; that is,

Lgh(x, w) = · · · = LgLρ−2f h(x, w) = 0, LgL

ρ−1f h(x, w) ≥ a > 0

for all (x, w) ∈ D ×Dw. Our goal is to design an output feedback controller suchthat the output y asymptotically tracks a constant reference r ∈ Dr ⊂ R, where Dr


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14.6. EXERCISES 625

is an open connected set.This is the output feedback version of the problem we studied in Section 14.1.4.

The only difference is that we allow f and h to depend on w, while in Section 14.1.4,they were restricted to be independent of w. The restriction was needed there be-cause the variables h, Lfh, . . . , Lρ−1

f h were used to compute the state feedback con-trol. In the output feedback case, these variables are calculated from the measuredsignal y by using a high-gain observer. Because we allow f to depend on w, theterm δ1 has been absorbed in f . We are not going to repeat the assumptions andderivations of Section 14.1.4. Let us only recall the sliding mode state feedbackcontroller (14.29), namely,

e0 = e1

u = −k sat(

k0e0 + k1e1 + k2e2 + · · ·+ kρ−1eρ−1 + eρ

μ

)

where e1 = y−r is the regulation error and e2 to eρ are the derivatives of e1.30 Thecontrol is globally bounded and the signal e1 is available on-line. To implement thiscontroller using output feedback, we estimate e2 to eρ by utilizing a linear high-gainobserver. Thus, the output feedback controller can be taken as

e0 = e1

u = −k sat(

k0e0 + k1e1 + k2e2 + · · ·+ kρ−1eρ−1 + eρ

μ

)˙ei = ei+1 +

(αi

εi

)(e1 − e1), 1 ≤ i ≤ ρ− 1

˙eρ =(αρ

ερ

)(e1 − e1)

where the positive constants α1 to αρ are chosen such the roots of

sρ + α1sρ−1 + · · ·+ αρ−1s + αρ = 0

have negative real parts. For relative-degree-one systems (ρ = 1), the high-gain ob-server is not used. Under the assumptions of Section 14.1.4, the closed-loop systemunder state feedback has an exponentially stable equilibrium point at (z, e0, e) =(0, e0, 0). We leave it to the reader (Exercise 14.50) to verify that, for sufficientlysmall ε, the output feedback controller recovers the performance of the state feed-back controller.

14.6 Exercises

14.1 Consider the system

x1 = x2 + sin x1, x2 = θ1x21 + (1 + θ2)u, y = x1

30We have written the sliding mode boundary-layer parameter as μ to reserve ε for the high-gainobserver parameter.


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where |θ1| ≤ 2 and |θ2| ≤ 1/2. Using sliding mode control,

(a) design a continuous state feedback controller to stabilize the origin.

(b) design a continuous state feedback controller so that the output y(t) asymp-totically tracks a reference signal r(t). Assume that r, r, and r are continuousand bounded.

14.2 A simplified model of an underwater vehicle in yaw is given by [60]

ψ + aψ|ψ| = τ

where ψ is the heading angle, τ is a normalized torque input, and a is a positiveparameter. It is desired that ψ tracks a desired trajectory ψr(t), where ψr(t), ψr(t),and ψr(t) are bounded functions of t. Let a = 1 be the nominal value of a.

(a) Using x1 = ψ and x2 = ψ as state variables, u = τ as control input and y = ψas output, find the state model.

(b) Show that the system is input–output linearizable.

(c) Assuming that a = a = 1, use feedback linearization to design a state feedbackcontroller that achieves global asymptotic tracking.

(d) Assuming that |a − a| ≤ 0.01 and ψr(t) = sin 2t, show that the controllerdesigned in part (c) will achieve asymptotic tracking with tolerance |ψ(t) −ψr(t)| ≤ δ1 and estimate δ1. Is this tolerance achievable for all initial states?

(e) Assuming that |a−a| ≤ k, where k is known, design a state feedback controllerto achieve global asymptotic tracking with tolerance |ψ(t)− ψr(t)| ≤ 0.01.

14.3 ([176]) Consider the controlled van der Pol equation

x1 = x2, x2 = −ω2x1 + εω(1− μ2x21)x2u

where ω, ε, and μ are positive constants and u is the control input.

(a) Show that for u = 1, there is a stable limit cycle outside the surface x21+x2

2/ω2 =1/μ2 and for u = −1, there is an unstable limit cycle outside the same surface.

(b) Let s = x21 +x2

2/ω2−r2, with r < 1/μ. Show that restricting the motion of thesystem to the surface s = 0 (that is, s(t) ≡ 0) results in a harmonic oscillator

x1 = x2, x2 = −ω2x1

which produces a sinusoidal oscillation of frequency ω and amplitude r.

(c) Design a state feedback sliding mode controller to drive all trajectories in theband |x1| < 1/μ to the manifold s = 0 and have them slide on that manifold.


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14.6. EXERCISES 627

(d) Simulate the response of this system for the ideal sliding mode controller andfor a continuous approximation. Use ω = μ = ε = 1.


x1 = x2 + ax1 sin x1, x2 = bx1x2 + u

where a and b are unknown constants, but we know the bounds |a − 1| ≤ 1 and|b − 1| ≤ 2. Using sliding mode control, design a continuous globally stabilizingstate feedback controller.

14.5 The equation of motion for a pendulum whose suspension point is subjectedto a time-varying, bounded, horizontal acceleration is given by

m�θ + mg sin θ + k�θ = T/� + mh(t) cos θ

where h is the horizontal acceleration, T is the torque input, and the other variablesare defined in Section 1.2.1. Assume that

0.9 ≤ � ≤ 1.1, 0.5 ≤ m ≤ 1.5, 0 ≤ k ≤ 0.2, |h(t)| ≤ 1

and g = 9.81. It is desired to stabilize the pendulum at θ = 0 for arbitrary initialconditions θ(0) and θ(0). Design a continuous sliding mode state feedback controllerto achieve ultimate boundedness with |θ| ≤ 0.01 and |θ| ≤ 0.01.

14.6 ([108]) Consider the system

x1 = x1x2, x2 = x1 + u

(a) Using sliding mode control, design a continuous globally stabilizing state feed-back controller.

(b) Can you globally stabilize the origin via feedback linearization?


x1 = −x1 + tanh(x2), x2 = x2 + x3, x3 = u + δ(x)

where δ(x) is an uncertain function that satisfies |δ(x)| ≤ �(x) for all x, for someknown function �. Design a continuous sliding mode state feedback controller suchthat, for all ‖x(0)‖∞ ≤ k, x(t) is bounded and |x1(t)| is ultimately bounded by0.01.

14.8 The tank system of Example 12.5 (with integrator included) is given by

y =1

A(y)(u− c

√y), σ = y − r


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where r is a desired set point. Let c and A(y) be nominal models of c and A(y),respectively, and suppose that you know positive constants �1 > 0, �2 > 0, �3 ≥ 0,and 0 ≤ �4 < 1 such that

�1 ≤ A(y) ≤ �2, |c− c| ≤ �3, and

∣∣∣∣∣A(y)− A(y)A(y)

∣∣∣∣∣ ≤ �4

Using sliding mode control, design a continuous state feedback controller such thatall state variables are bounded and |y(t)− r| converges to zero as t →∞.

14.9 Consider the system (14.1).

(a) Let B be a constant matrix of rank p. Show that there is an n×n nonsingularmatrix M such that MB = [0, Ip]T , where Ip is the p × p identity matrix.Verify that T (x) = Mx satisfies (14.2).

(b) Let B(x) be a smooth function of x and assume that B has rank p for allx in a domain D ⊂ Rn. Let Δ = span{b1, . . . , bp}, where b1, . . . , bp arethe columns of B. Suppose Δ is involutive. Show that for every x0 ∈ D,there exist smooth functions φ1(x), . . . , φn−p(x), with linearly independentdifferentials ∂φ1/∂x, . . . , ∂φn−p/∂x at x0, such that [∂φi/∂x]B(x) = 0 for1 ≤ i ≤ n− p. Show that we can find smooth functions φn−p+1(x), . . . , φn(x)such that T (x) = [φ1(x), . . . , φn(x)]T is a diffeomorphism in a neighborhoodof x0 and satisfies (14.2).Hint: Apply Frobenius theorem.

14.10 Consider the nonautonomous regular form

η = fa(t, η, ξ) + δa(t, η, ξ)ξ = fb(t, η, ξ) + G(t, x)E(t, x)u + δ(t, x, u)

where, for all (t, x) ∈ [0,∞)×D, E is a known nonsingular matrix with a boundedinverse and G is a positive diagonal matrix whose elements are bounded away fromzero. Suppose there is a continuously differentiable function φ(t, η), with φ(t, 0) = 0,such that the origin of η = fa(t, η, φ(t, η)) + δη(t, η, φ(t, η)) is uniformly asymptoti-cally stable. Let

u = E−1[−L

(fb − ∂φ

∂t− ∂φ

∂ηfa

)+ v

]

where L = G−1 or L = 0 and G(t, x) is a nominal model of G(t, x). Let

Δ = (I −GL)(

fb − ∂φ

∂t− ∂φ

∂ηfa

)+ δ − ∂φ

∂ηδa

Suppose Δi satisfies the inequality (14.10) with � = �(t, x). Taking s = ξ − φ(t, η),design a sliding mode controller to stabilize the origin. State and prove a theoremsimilar to Theorem 14.1.


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14.6. EXERCISES 629

14.11 Suppose (14.13) is replaced by

vi = − β(x) σ(si

ε

)where σ : R → R is a continuously differentiable, odd, monotonically increasingfunction with the properties

σ(0) = 0, limy→∞σ(y) = 1, and yσ(y) ≥ σ(1)y2, ∀ |y| ≤ 1

(a) Verify that σ(y) = tanh(y), σ(y) = (2/π) tan−1(πy/2), and σ(y) = y/(1 + |y|)satisfy the foregoing properties.

(b) Show that if (14.10) is satisfied with κ0 < σ(1) and β is chosen to satisfy

β(x) ≥ �(x)σ(1)− κ0

+ β0, β0 > 0

thensisi ≤ −g0β0[σ(1)− κ0]|si|, for |si| ≥ ε

(c) Prove Theorems 14.1 and 14.2 for this sliding mode control.

14.12 Replace inequality (14.10) by

Δi

gi≤ �i(x) +

p∑j=1

κij |vj |, ∀ 1 ≤ i ≤ p

LetK be the p×p matrix whose elements are κij and suppose I−K is an M -matrix.31

Recall from the properties of M -matrices that the following three conditions areequivalent:32

(i) I −K is an M -matrix.

(ii) I −K is nonsingular and all elements of (I −K)−1 are nonnegative.

(iii) There is a vector w whose elements are all positive such that the elements ofb = (I −K)w are all positive.

Let �i(x) ≥ �i(x) for all 1 ≤ i ≤ p and σ(x) = (I −K)−1[�1(x), . . . , �p(x)]T .

(a) Show that vi = −βi(x) sgn(si) with βi(x) = σi(x) + wi yields sisi ≤ −big0|si|for 1 ≤ i ≤ p.

(b) Suppose∑p

j=1 κij ≤ κ0 < 1 and �i(x) = �(x) for 1 ≤ i ≤ p. Show that I−K isan M -matrix and �i(x) and w can be chosen to produce the control (14.11).

31See Lemma 9.7 for the definition of M -matrices.32See [57].


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14.13 With reference to the “continuous” sliding mode controller of Section 14.1.3,show that there exists a finite time T , possibly dependent on ε and the initialstates, and a positive constant k, independent of ε and the initial states, such that|y(t)− r(t)| ≤ kε for all t ≥ T .

14.14 Repeat Exercise 14.5 using Lyapunov redesign.

14.15 Repeat Exercise 14.8 using Lyapunov redesign.

14.16 Use the Numerical data of the pendulum equation in Section 14.1.1 tosimulate the Lyapunov redesign of Example 14.5. Choose the design parametersof the Lyapunov redesign to obtain the same control level as in the sliding modedesign. Compare the performance of the two controllers.

14.17 For each of the scalar systems that follow, use the nonlinear damping toolto design a state feedback controller that guarantees boundedness of the state x(t)as well as uniform ultimate boundedness by an ultimate bound μ. The functionδ(t) is bounded for all t ≥ 0, but we do not know an upper bound on |δ(t)|.

(a) x = −x + x2[u + δ(t)], (b) x = x2[1 + δ(t)]− xu

14.18 Consider a single-input–single-output system in the normal form (13.16)–(13.18) and suppose (13.16) is input-to-state stable. Let α(x) and γ(x) be nominalmodels of the functions α(x) and γ(x), respectively, and suppose the modeling errorssatisfy the inequalities

|γ(x)[α(x)− α(x)]| ≤ ρ0(x),∣∣∣∣γ(x)− γ(x)

γ(x)

∣∣∣∣ ≤ k < 1

where the function ρ0(x) and the constant k are known. Let r(t) be a referencesignal and suppose r and its derivatives up to r(ρ) are continuous and bounded.Using Lyapunov redesign, design a continuous state feedback controller such that theoutput y asymptotically tracks r with prespecified tolerance μ; that is, |y(t)−r(t)| ≤μ for all t ≥ T for some finite time T .

14.19 Repeat the previous exercise for the case of constant reference, using integralcontrol. Verify that the regulation error converges to zero.


x1 = x2, x2 = u + δ(x)

where δ is unknown, but we know an estimate ρ1 such that |δ(x)| ≤ ρ1‖x‖2. Letu = ψ(x) = −x1 − x2 be the nominal stabilizing control, and

v =

⎧⎨⎩−ρ1‖x‖2(w/‖w‖2), if ρ1‖x‖2‖w‖2 ≥ ε

−ρ21‖x‖22(w/ε), if ρ1‖x‖2‖w‖2 < ε


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14.6. EXERCISES 631

where wT = 2xT PB and V (x) = xT Px is a Lyapunov function for the nominalclosed-loop system. Apply the control u = −x1 − x2 + v.

(a) Verify that all the assumptions of Corollary 14.1 are satisfied, except (14.45),which holds with η0 = 0.

(b) Show that when δ(x) = 2(x1 + x2) and ρ1 = 2√

2, the origin is unstable.

14.21 Assume that (14.33) through (14.35) are satisfied with ‖ · ‖∞. Consider thefollowing continuous approximation of the discontinuous control (14.40):

vi =

⎧⎨⎩−η(t, x) sgn(wi), if η(t, x)|wi| ≥ ε

−η2(t, x)(wi/ε), if η(t, x)|wi| < ε

for i = 1, 2, . . . , p, where wT = [∂V/∂x]G(t, x).

(a) Show that

V ≤ −α3(‖x‖∞) +∑i∈I

[η(t, x)|wi| − η2(t, x)|wi|2

ε

]

where i ∈ I if η(t, x)|wi| < ε.

(b) State and prove a theorem similar to Theorem 14.3 for the current controller.

14.22 Consider the controller of Exercise 14.21. We want to prove a result similarto Corollary 14.1. Assume that α3(‖x‖∞) ≥ φ2(x), η(t, x) ≥ η0 > 0, and

|δi| ≤ ρ1φ(x) + κ0|vi|, 0 ≤ κ0 < 1

for i = 1, 2, . . . , p. These inequalities imply that (14.35) holds with ‖ · ‖∞, but theyare more restrictive because the upper bound on |δi| depends only on |vi|.(a) Show that

V ≤ −φ2(x) +∑i∈I

{−(1− κ0)η2

0|wi|2

ε+ ρ1φ(x)|wi|

}

(b) State and prove a result similar to Corollary 14.1.

14.23 Suppose the inequality (14.35) takes the form

‖δ(t, x, ψ(t, x) + v)‖2 ≤ ρ0 + ρ1φ(x) + κ0‖v‖2, 0 ≤ κ0 < 1

where φ(x) =√

α3(‖x‖2). Let η(x) = η0 + η1φ(x), where η0 ≥ ρ0/(1 − κ0) andη1 ≥ ρ1/(1− κ0). Consider the feedback control

v =

⎧⎨⎩−[η0 + η1φ(x)](w/‖w‖2), if ‖w‖2 ≥ ε

−[η0 + η1φ(x)](w/ε), if ‖w‖2 < ε


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(a) Show that the derivative of V along the trajectories of the closed-loop system(14.38) satisfies

V ≤ −φ2(x) +ε

4[ρ0 + ρ1φ(x)] ≤ − 1

2α3(‖x‖2) +

ε2ρ21

32+

ερ0

4

(b) Apply Theorem 4.18 to arrive at a conclusion similar to Theorem 14.3.

(c) Compare this controller with (14.41).

14.24 Consider the problem treated in Section 14.2 and assume that (14.35) issatisfied with 2-norm. Suppose further that (14.44) and (14.46) are satisfied. Showthat the control law

u = ψ(t, x)− γw; wT =∂V

∂xG(t, x)

with sufficiently large γ will stabilize the origin.

14.25 Consider the problem treated in Section 14.2. Show that, instead of usingthe control law u = ψ(t, x)+v, we can simply use u = v, where ρ(t, x) is taken fromthe inequality

‖δ(t, x, u)− ψ(t, x)‖2 ≤ ρ(t, x) + κ0‖u‖2, 0 ≤ κ0 < 1

14.26 Suppose that, in addition to the matched uncertainty δ, the system (14.30)also has unmatched uncertainty Δ; that is,

x = f(t, x) + Δ(t, x) + G(t, x)[u + δ(t, x, u)]

Suppose that, over a domain D ⊂ Rn, all the assumptions of Theorem 14.3 aresatisfied, the inequalities (14.44) through (14.46) are satisfied, and the unmatcheduncertainty satisfies ‖[∂V/∂x]Δ(t, x)‖2 ≤ μφ2(x), for some μ ≥ 0. Let u = ψ(t, x)+v, where v is determined by (14.41). Show that if μ < 1, then the feedback controllaw will stabilize the origin of the closed-loop system provided that the ε chosen issmall enough to satisfy ε < 4(1− μ)(1− κ0)η2

0/ρ21.

14.27 Consider the system x = f(x) + G(x)[u + δ(x, u)], and suppose there areknown smooth functions ψ(x), V (x), and ρ(x), all vanishing at x = 0, and a knownconstant k such that

c1‖x‖2 ≤ V (x) ≤ c2‖x‖2, ∂V

∂x[f(x) + G(x)ψ(x)] ≤ −c3‖x‖2

‖δ(x, ψ(x) + v)‖ ≤ ρ(x) + κ0‖v‖, 0 ≤ κ0 < 1, ∀ x ∈ Rn, ∀ v ∈ Rp

where c1 to c3 are positive constants.


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14.6. EXERCISES 633

(a) Show that it is possible to design a continuous state feedback controller u =γ(x) such that the origin of

x = f(x) + G(x)[γ(x) + δ(x, γ(x))]

is globally exponentially stable.

(b) Apply the result of part (a) to the system

x1 = x2, x2 = (1 + a1)(x31 + x3

2) + (1 + a2)u

where a1 and a2 are unknown constants that satisfy |a1| ≤ 1 and |a2| ≤ 1/2.

14.28 Repeat exercise 14.1 using backstepping.



14.31 Using backstepping, design a state feedback controller to globally stabilizethe system

x1 = x2 + a + (x1 − a1/3)3, x2 = x1 + u

where a is a known constant.

14.32 ([108]) Consider the system

x1 = −x2 − 32x2

1 − 12x3

1, x2 = u

(a) Using backstepping, design a linear state feedback controller to globally sta-bilize the origin.Hint: Avoid cancellation of nonlinear terms.

(b) Design a globally stabilizing state feedback controller by using feedback lin-earization.

(c) Compare the two designs. Using computer simulations, compare their perfor-mance and the control effort used in each case.


x1 = x2, x2 = x1 − x31 + u

(a) Find a smooth state feedback controller u = ψ(x) such that the origin is globallyexponentially stable.


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(b) Extend the dynamics of the system by connecting an integrator in series withthe input, to obtain

x1 = x2, x2 = x1 − x31 + z, z = v

Using backstepping, find a smooth state feedback controller v = φ(x, z) suchthat the origin is globally asymptotically stable.

14.34 Consider the system of Exercise 13.17.

(a) Starting with the x1-equation and stepping back to the x2-equation, design astate feedback controller u = ψ(x) such that the origin (x1 = 0, x2 = 0) ofthe first two equations is globally exponentially stable.

(b) Show that, under the feedback controller of part (a), the origin x = 0 of thefull system is globally asymptotically stable.Hint: Use input-to-state properties of the third equation.


x1 = x2 + θx21, x2 = x3 + u, x3 = x1 − x3, y = x1

where θ ∈ [0, 2]. Using backstepping, design a state feedback controller such that|y − a sin t| is ultimately bounded by μ, where μ is a design parameter that can bechosen arbitrarily small. Assume that |a| ≤ 1 and ‖x(0)‖∞ ≤ 1.

14.36 Repeat exercise 14.4 using a combination of backstepping and Lyapunovredesign.


x1 = −x1 + x1x2, x2 = x2 + x3, x3 = x21 + δ(x) + u

where δ(x) is an unknown (locally Lipschitz) function of x that satisfies |δ(x)| ≤k‖x‖2 for all x, with a known constant k. Design a globally stabilizing state feedbackcontroller.

14.38 Consider again Exercise 14.7.

(a) With δ = 0, use backstepping to design a globally stabilizing state feedbackcontroller.

(b) Use the stabilizing controller of part (a) and Lyapunov redesign to design astate feedback controller such that, for all ‖x(0)‖∞ ≤ k, x(t) is bounded and|x1(t)| is ultimately bounded by 0.01.


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14.39 Consider the magnetic suspension system of Exercise 1.18 and suppose theball is subject to a vertical disturbance force d(t); that is,

my = −ky + mg + F (y, i) + d(t)

Suppose further that |d(t)| ≤ d0 for all t ≥ 0, where the upper bound d0 is known.

(a) Viewing the force F as the control input, use Lyapunov redesign to design astate feedback control F = γ(y, y) such that |y − r| is ultimately boundedby μ, where μ is a design parameter that can be chosen arbitrarily small.Design your control such that γ is a continuously differentiable function of itsarguments.

(b) Using backstepping, design a state feedback control law for the voltage inputu that will ensure that |y − r| is ultimately bounded by μ.


x1 = −x1 + x21[x2 + δ(t)], x2 = u

where δ(t) is a bounded function of t for all t ≥ 0, but we do not know an upperbound on |δ(t)|. By combining backstepping and the nonlinear damping tools,design a state feedback controller that ensures global boundedness of the state xfor all initial states x(0) ∈ R2.

14.41 Repeat Exercise 14.40 for the system

x1 = −x1x2 + x21[1 + δ(t)], x2 = u

14.42 Consider the linear system x = Ax+Bu and suppose there exists a positivedefinite symmetric matrix P such that PA + AT P ≤ 0 and the pair (A, BT P ) isobservable. Design a globally stabilizing state feedback control law u = −ψ(x) suchthat ‖ψ(x)‖ ≤ k for all x, where k is a given positive constant.


x1 = x2, x2 = −x31 + ψ(u)

where ψ is a locally Lipschitz function that satisfies ψ(0) = 0 and uψ(u) > 0 for allu �= 0. Design a globally stabilizing state feedback control.

14.44 Consider a relative-degree-one, single-input–single-output system with aglobally defined normal form

η = f0(η, y), y = b(η, y) + a(η, y)u

where f0(0, 0) = 0 and a(η, y) ≥ a0 > 0. Suppose there is a (known) radially un-bounded Lyapunov function W (η), with [∂W/∂η](0) = 0, such that [∂W/∂η]f0(η, 0) <0 for all η �= 0. Design a globally stabilizing state feedback controller.


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14.45 Design a passivity-based globally stabilizing state feedback control for thesystem of Example 13.17.

14.46 Design a globally stabilizing state feedback control law for the system

x1 = −(1 + x3)x31, x2 = x3, x3 = x2

2 − 1 + u

14.47 Consider a relative-degree-one, single-input–single-output system that isrepresented, in some domain that contains the origin, by the normal form

η = f0(η, y), y = b(η, y) + a(η, y)u

where f0, a, and b are sufficiently smooth, a(η, y) ≥ a0 > 0, f0(0, 0) = 0, andb(0, 0) = 0. Suppose the origin of η = f0(η, 0) is asymptotically stable and there isa Lyapunov function V (η) such that

α1(‖η‖) ≤ V (η) ≤ α2(‖η‖) and∂V

∂ηf0(η, y) ≤ −α3(‖η‖), ∀ ‖η‖ ≥ γ(|y|)

for some class K functions α1, α2, α3, and γ. Let a(y) and b(y) be smooth nominalmodels of a(η, y) and b(η, y), respectively, such that a(y) ≥ a0 > 0 and∣∣∣∣∣ b(η, y)

a(η, y)− b(y)

a(y)

∣∣∣∣∣ ≤ �(y) (14.116)

over the domain of interest, where �(y) is known. The choice a = 1, b = 0 ispossible.

(a) Show that a continuous stabilizing sliding mode controller can be taken as

u = − b(y)a(y)

− β(y) sat(y

ε

)

where β(y) ≥ �(y) + β0, for some positive constants ε and β0. In particular,show that there are compact positively invariant sets Ω and Ωε = {V (η) ≤α(ε), |y| ≤ ε} ⊂ Ω, for some class K function α, such that every trajectorystarting in Ω enters Ωε in finite time.

(b) Show that if the origin of η = f0(η, 0) is exponentially stable, then for suffi-ciently small ε the origin of the closed-loop system is exponentially stable andΩ is a subset of its region of attraction.

(c) Show that (14.116) can be satisfied on any compact set with a constant �.

(d) Under what conditions will this controller achieve semiglobal stabilization?


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14.6. EXERCISES 637

(e) Design a stabilizing output feedback controller for the system

x1 = −x1 − x32, x2 = −x2 − x3

3 − u, x3 = x21 − x3 + u, y = x2

14.48 Consider the p-input–p-output system

x = f(x, u), y = h(x)

where f is locally Lipschitz, h is continuously differentiable, f(0, 0) = 0, and h(0) =0. Suppose the system

x = f(x, u), y =∂h

∂xf(x, u) def= h(x, u)

with output y, is passive with a radially unbounded, positive definite storage func-tion V (x); that is, V ≤ uT y, and is zero-state observable. Let zi be the outputof the linear transfer function bis/(s + ai) whose input is yi, where ai and bi arepositive constants.

(a) Using V (x)+∑p

i=1(ki/2bi)z2i as a Lyapunov function candidate, show that the

output feedback control ui = −kizi, 1 ≤ i ≤ p, ki > 0, globally stabilizes theorigin.

(b) Using V (x)+∑p

i=1(1/bi)∫ zi

0 φi(σ) dσ as a Lyapunov function candidate, whereφi is a locally Lipschitz function such that φi(0) = 0 and σφi(σ) > 0 forall σ �= 0, show that the output feedback control ui = −φi(zi), 1 ≤ i ≤ p,stabilizes the origin. Under what conditions on φi will this control law achieveglobal stabilization?

(c) Apply the result of part (a) to globally stabilize the pendulum equation

m�θ + mg sin θ = u

at the angle θ = δ1 by using feedback from θ, but not from θ.

14.49 Consider the system (14.85)–(14.86) and suppose u = γ(x) is a locallyLipschitz state feedback control that globally stabilizes the origin. Let x be thestate estimate provided by the observer (14.87). Show that the output feedbackcontrol u = γ(x) globally stabilizes the origin (x = 0, x = 0) of the closed-loopsystem if the system

x = Ax + g(Cx, γ(x− v))

with input v, is input-to-state stable.

14.50 Verify that the output feedback controller of Section 14.5.3 recovers theperformance of the state feedback controller for sufficiently small ε. In particular,show that the closed-loop system under output feedback has an exponentially stableequilibrium point at (z, e0, e, e) = (0, e0, 0, 0).


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In the next seven exercises, we present a few case studies.

14.51 A current-controlled induction motor can be modeled by33

Jω = kt(λaib − λbia)− TL

λa = − Rr

Lrλa − pωλb +

RrM

Lria

λb = − Rr

Lrλb + pωλa +

RrM

Lrib

where ω is the rotor speed, TL is the load torque, λa and λb are the componentsof the rotor flux vector, ia and ib are the components of the stator current vector,Lr and M are the rotor and mutual inductances, Rr is the rotor resistance, J isthe moment of inertia, p is the number of pole pairs, and kt is a positive constant.The load torque TL can be taken as TL = To + φ(ω), where φ ∈ [0,∞] is a locallyLipschitz function that models the load due to friction, while To models the speed-independent load. The currents ia and ib are the control variables. It is desiredto design feedback control so that the speed ω tracks a reference speed ωr in thepresence of unknown load torque. In this exercise, we will pursue the design bytransforming the equations into field-oriented coordinates. This transformationdecouples the control problem into two separate control problems, one for the speedand the other for the rotor flux. The transformation should be applied on-line andrequires measurement of the rotor flux. We will start by assuming that the rotor fluxis available. Then we will use an observer to estimate the rotor flux and analyze theperformance of the controller when the rotor flux is replaced by its estimate. Theanalysis will take into consideration uncertainty in the rotor resistance Rr, whichmay change significantly with temperature during the operation of the motor.

(a) Let ρ be the angle of the rotor flux vector and λd be its magnitude; that is,ρ = tan−1(λb/λa) and λd =

√λ2

a + λ2b . Replacing λa and λb by λd and ρ as

state variables and transforming ia and ib into id and iq, defined by[idiq

]=[

cos ρ sin ρ− sin ρ cos ρ

] [iaib

](14.117)

show that the motor can be represented by the state model

Jω = kt λdiq − TL, λd = − Rr

Lrλd +

RrM

Lrid, ρ = pω +

RrMiqLrλd

provided λd > 0.

(b) The first two equations of the foregoing model are independent of ρ. Therefore,to design the feedback control laws for id and iq, we can drop the state equationfor ρ. However, we still need ρ to calculate ia and ib by using the inverse ofthe transformation given by (14.117). Show that for constant ω, iq, and λd, ρwill be unbounded. Explain why an unbounded ρ does not cause a problem.

33See, for example, [50, Appendix C] or [117].


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14.6. EXERCISES 639

(c) Note that the torque generated by the motor is proportional to λdiq. Therefore,we can control the speed by first regulating λd to a desired constant flux λr

using the control input id; then we can design iq assuming λd = λr. Thisassumption will be justified if the dynamics of the flux loop are much fasterthan the dynamics of the speed loop.34 In this part of the exercise, we designthe flux control. Show that id = (λr/M) − k(λd − λr), with k ≥ 0, achievesthe desired regulation.

(d) Since the currents ia and ib must be limited to certain maximum values, inour design, we would like to assume that id and iq are limited to Id and Iq,respectively. Show that if Id > λr/M and 0 < λd(0) < λr, then under thesaturated control

id = Id sat(

λr/M − k(λd − λr)Id

)(14.118)

λd(t) changes monotonically from λd(0) to λr. Estimate the settling time.

(e) For speed control, let us assume that ωr(t), ωr(t), and To(t) are bounded.Design a sliding mode controller of the form iq = −Iq sat(s/ε) with appro-priately chosen s and ε. Give conditions under which the controller will workand estimate the ultimate bound on the tracking error.

(f) Suppose, in addition to the foregoing assumption, ωr(t) satisfies limt→∞ ωr(t) =ωr and limt→∞ ωr(t) = 0 and we want to achieve zero steady-state error whenTo is constant. Using integral control, design a sliding mode controller of theform iq = −Iq sat(s/ε) with appropriately chosen s and ε. Give conditionsunder which the controller will work and verify that it achieves zero steady-state error.

(g) The assumption that the rotor flux can be measured is not a practical one. Inapplications, it is common to estimate the flux by using the observer

˙λa = − Rr

Lrλa − pωλb +

RrM

Lria,

˙λb = − Rr

Lrλb + pωλa +

RrM

Lrib

where Rr is a nominal value (or estimate) of Rr. The field orientation angleρ and the flux magnitude λd are calculated by using ρ = tan−1(λb/λa) and

λd =√

λ2a + λ2

b ; id and iq are defined by using (14.117) with the newly definedρ. To write down a state model for the overall system, we define the fluxestimation errors ed and eq by[

ed

eq

]=[

cos ρ sin ρ− sin ρ cos ρ

] [λa − λa

λb − λb

](14.119)

34This can be justified by the singular perturbation theory of Chapter 11.


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Using ω, λd, ρ, ed, and eq as state variables, show that the overall system canbe represented by the state model

Jω = kt(λdiq + eqid − ediq)− TL

λd = − Rr

Lrλd +

RrM

Lrid

ρ = pω +RrMiqLrλd

ed = − Rr

Lred +

RrMiqLrλd

eq +

(Rr −Rr

Lr

)(Mid − λd)

eq = − RrMiqLrλd

ed − Rr

Lreq +

(Rr −Rr

Lr

)Miq

(h) Verify that the flux control (14.118) still regulates λd to λr, provided Id > λr/Mand 0 < λd(0) < λr. Furthermore, show that

|Mid(t)− λd(t)| ≤ (1 + kM)[MId − λd(0)] exp[−(Rr/Lr)t]

(i) Using V = (1/2)(e2d+e2

q) and the comparison lemma, show that ‖e‖ =√

e2d + e2

q

satisfies the bound

‖e(t)‖ ≤ k1e−γt + k2

∣∣∣∣∣ Rr −Rr

Rr

∣∣∣∣∣MIq

for some positive constants γ, k1, and k2.

(j) Find conditions under which the sliding mode controller with integral action,designed in part (f), will still achieve zero steady-state error.

14.52 The nonlinear dynamic equations of an m-link robot are given by

M(q)q + C(q, q)q + Dq + g(q) = u (14.120)

where all variables are defined in Exercise 1.4. We assume that M , C, and g arecontinuous functions of their arguments and

0 < λmyT y ≤ yT M(q)y ≤ λMyT y, ∀ q, y ∈ Rm, y �= 0

for some positive constants λm and λM . We want to design a state feedback controllaw such that q(t) asymptotically tracks a reference trajectory qr(t), where qr(t),qr(t), and qr(t) are continuous and bounded. In this exercise, we design a sliding


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14.6. EXERCISES 641

mode controller. Take the sliding surface as s = Λe + e = 0, where Λ is a positivediagonal matrix, and let

u = M(q)v + L[C(q, q)q + g(q) + M(q)qr − M(q)Λe]

where M , C, and g are nominal models of M , C, and g, respectively, and L is eitherzero or identity, leading to two different control laws.

(a) Show that s satisfies an equation of the form

s = v + Δ(q, q, qr, qr, qr, v)

and give expressions for Δ when L = 0 and L = I.

(b) Assuming that

‖M−1(q)M(q)− I‖∞ ≤ κ0 < 1, ∀ q ∈ Rm (14.121)

show that Δi satisfies the inequality

|Δi| ≤ ρ(·) + κ0‖v‖∞, for 1 ≤ i ≤ m

where ρ may depend on (q, q, qr, qr, qr).

(c) Letvi = −β(·) sat(si/ε), ε > 0, for 1 ≤ i ≤ m

where β may depend on (q, q, qr, qr, qr). Show how to choose β to ensure thatthe error e is globally uniformly ultimately bounded and give an estimate ofthe ultimate bound in terms of ε.

(d) What properties can you prove for the sliding mode controller when β is takenas a constant?

14.53 Consider the m-link robot of the previous exercise. In the current exercise,we derive a different sliding mode controller [180] that uses the skew symmetricproperty of (M − 2C) and avoids the condition (14.121).

(a) Taking s as in the previous exercise and W = (1/2)sT M(q)s as a Lyapunovfunction candidate for the s-equation, show that

W = sT [MΛe + C(Λe− qr)−Dq − g −Mqr + u]

(b) Letu = v + L[−M(q)Λe− C(q, q)(Λe− qr) + g(q) + M(q)qr]

where L = 0 or L = I, leading to two different control laws. Show that

W = sT [v + Δ(q, q, qr, qr, qr)]

and give expressions for Δ when L = 0 and L = I.


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(c) Let

v = −β(·)ϕ(s/ε), ε > 0, ϕ(y) =

⎧⎨⎩

y/‖y‖2, for ‖y‖2 ≥ 1

y, for ‖y‖2 < 1

where β may depend on (q, q, qr, qr, qr). Show how to choose β to ensure thatthe error e is globally uniformly ultimately bounded and give an estimate ofthe ultimate bound in terms of ε.

(d) What properties can you prove for the sliding mode controller when β is takenas a constant?

��

��

��

��

q1

q2

Load

Figure 14.23: Two-link robot.

14.54 The two-link robot, shown in Figure 14.23, can be modeled [171] by equa-tion (14.120) with

M =[

a1 + 2a4 cos q2 a2 + a4 cos q2a2 + a4 cos q2 a3

], C = a4 sin q2

[ −q2 −(q1 + q2)q1 0

]

g =[

b1 cos q1 + b2 cos(q1 + q2)b2 cos(q1 + q2)

]

where a1 through a4, b1, and b2 are positive constants which depend on masses,moments of inertial, and lengths of the two links, as well as the acceleration dueto gravity. We neglect damping and take D = 0. Let the nominal values of theparameters be

a1 = 200.01, a2 = 23.5, a3 = 122.5, a4 = 25, b1 = 784.8, b2 = 245.25


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14.6. EXERCISES 643

To move the arm from the initial position (q1 = 0, q2 = 0) to the terminal position(q1 = π/2, q2 = π/2), we take the reference trajectory as

qr1(t) = qr2(t) = (π/2)[1− exp(−5t)(1 + 5t)]

We assume that the control inputs are constrained to |u1| ≤ 6000 Nm and |u2| ≤5000 Nm. In the previous two exercises, we designed four different sliding modecontrol laws, given by

u = −Mβ sat(s/ε)u = −Mβ sat(s/ε) + Cq + g + M qr − MΛe

u = −β ϕ(s/ε)u = −β ϕ(s/ε) + C(qr − Λe) + g + M qr − MΛe

For convenience, we take β to be constant.

(a) Using simulation, choose the design parameters Λ, β, and ε for each of thefour controllers. Include limiters in your simulation to impose the controlconstraints.

(b) Compare the performance of the four controllers when, due to an unknownload, the actual system parameters are perturbed to

a1 = 259.7, a2 = 58.19, a3 = 157.19, a4 = 56.25, b1 = 1030.1, b2 = 551.8125

(c) Suppose we only measure the angles q1 and q2. Design a high-gain observerto implement the state feedback controllers. Using simulation, compare theperformance of the output and state feedback controllers, for any one of thefour control laws.

14.55 Reconsider the two-link robot of the previous exercise.

(a) Following Example 14.16, design a passivity-based controller to regulate theangles (q1, q2) from (0, 0) to (π/2, π/2). Use simulation to choose the designparameters Kp and Kd and compare with the sliding mode controllers of theprevious exercise.

(b) Suppose we only measure the angles q1 and q2. Design a high-gain observerto implement the state feedback controller. Using simulation, compare theperformance of the output and state feedback controllers.

14.56 Consider the TORA system of Exercise 1.16. In this exercise, we design apassivity-based control law [146] to globally stabilize the origin.

(a) Using the sum of the potential energy (1/2)kx2c and kinetic energy (1/2)vT D(θ)v,

where v = [θ, xc]T , as the storage function, show that the system with inputu and output θ is passive. Is it zero-state observable?


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(b) Let u = −φ1(θ) + w where φ1 is locally Lipschitz, φ1(0) = 0, yφ1(y) > 0 for ally �= 0, and lim|y|→∞

∫ y

0 φ1(λ) dλ = ∞. Using

V =12vT D(θ)v +

12kx2

c +∫ θ

0φ1(λ) dλ, v =

[θxc

]

as the storage function, show that the system with input w and output θ ispassive. Show also that it is zero-state observable.

(c) Let φ2 be any locally Lipschitz function such that φ2(0) = 0 and yφ2(y) > 0for all y �= 0. Show that u = −φ1(θ)− φ2(θ) globally stabilizes the origin.

(d) Let M = 1.3608 Kg, m = 0.096 Kg, L = 0.0592 m, I = 0.0002175 Kg/m2, andk = 186.3 N/m. Verify that

u = −Up sat(Kpθ)− Uv sat(Kv θ)

will be globally stabilizing for any positive constants Up, Uv, Kp, and Kv.Choose Up and Uv such that Up + Uv ≤ 0.1 to guarantee that u satisfiesthe constraint |u| ≤ 0.1. We want to design these four positive constants toreduce the settling time. We will use simulation and local analysis to choosethe constants. By linearizing the closed-loop system about the origin, showthat the closed-loop characteristic equation is given by

1 + β0(s2 + β2)(s + β1)

s2(s2 + β3)= 0

where

β0 =UvKv(m + M)

Δ0, β1 =

UpKp

UvKv, β2 =

k

m + M, β3 =

k(I + mL2)Δ0

and Δ0 = Δ(0). Verify that the characteristic polynomial is Hurwitz andconstruct the root locus as β0 varies from zero to infinity.

(e) By simulating the closed-loop system with the initial states θ(0) = π, θ(0) = 0,xc(0) = 0.025, and xc(0) = 0, and by using the root locus analysis of part (d),choose the constants Up, Uv, Kp, and Kv to make the settling time as smallas you can. You should be able to achieve a settling time of about 30 sec.

(f) Suppose now that we can only measure θ. Using Exercise 14.48, show that theorigin is globally stabilized by the output feedback controller

u = −Up sat(Kpθ)− Uv sat(Kvz)

where z is the output of the transfer function s/(εs+1) driven by θ and ε is anypositive constant. By viewing this transfer function as the transfer function


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14.6. EXERCISES 645

of a reduced-order high-gain observer, use the analysis of Section 14.5 to showthat the output feedback controller recovers the performance of the statefeedback controller as ε tends to zero. Simulate the closed-loop system fordifferent values of ε and compare with the performance under state feedback.

14.57 Consider the TORA system of Exercise 1.16. In this exercise, we design asliding mode controller35 to stabilize the origin.

(a) Show that the change of variables

η1 = xc +mL sin θ

m + M, η2 = xc +

mLθ cos θ

m + M, η3 = θ, ξ = θ

transforms the system into the regular form (14.4)–(14.5).

(b) Using

V0(η) =(m + M)k1

2mL

(η1 − mL

m + Msin η3

)2

+(m + M)2k1

2kmLη22 +

k2

2η23

with positive constants k1 and k2, show that

ξ = φ(η) def= k1

(η1 − mL

m + Msin η3

)cos η3 − k2η3

globally stabilizes the origin of

η = η2, η2 = − k

m + M

(η1 − mL

m + Msin η3

), η3 = ξ

Verify that the sliding surface can be taken as

s = θ + k2θ − k1xc cos θ = 0

and note that s is independent of the system parameters.

(c) Choose β(x) such that u = −β(x) sat(s/μ) globally stabilizes the origin forsufficiently small μ.

(d) The expression of β(x) in the previous part could be complicated. To simplifythe control law, we take β as a positive constant and write the control law as

u = −β sat

(θ + k2θ − k1xc cos θ

μ

)

where the positive constants k1, k2, β and μ are the design parameters. Showthat, for sufficiently small μ, this control law stabilizes the origin and guaran-tees that the region of attraction includes a compact set around the origin.

35The design uses ideas from the passivity-based design of [172].


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(e) Let the system parameters be as in part (d) of exercise 14.56. We will usesimulation and local analysis to choose the design parameters to reduce thesettling time. By linearizing the closed-loop system about the origin, showthat the closed-loop characteristic equation is given by

1 + γ0s3 + γ1s

2 + γ2s + γ3

s2(s2 + γ4)= 0

where

γ0 =β(m + M)

μΔ0, γ1 = k2 +

mLk1

m + M, γ2 =

k

m + M

γ3 =kk2

m + M, γ4 =

k(I + mL2)Δ0

, Δ0 = Δ(0)

Verify that the characteristic polynomial is Hurwitz and construct the rootlocus as γ0 varies from zero to infinity. Compare the root locus with theone obtained in exercise 14.56 and comment on the role played by the term−k1xc cos θ in the choice of s.

(f) By simulating the closed-loop system with the initial states θ(0) = π, θ(0) = 0,xc(0) = 0.025, and xc(0) = 0, and by using the root locus analysis of part (e),choose the constants k1, k2, β, and μ to make the settling time as small as youcan. You should be able to achieve a settling time of about 4 sec. Comparewith exercise 14.56.

(g) Suppose now that we can only measure θ and xc. Using the analysis of Sec-tion 14.5, show that, for sufficiently small μ and ε, the origin is stabilized bythe output feedback controller

u = −β sat(

z + k2θ − k1xc cos θ

μ

)

where z is the output of the transfer function s/(εs + 1) driven by θ, whichcorresponds to a reduced-order high-gain observer. Verify that the outputfeedback controller recovers the performance of the state feedback controlleras ε tends to zero. Simulate the closed-loop system for different values of εand compare with the performance under state feedback.


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Appendix A

Mathematical Review

Euclidean Space

The set of all n-dimensional vectors x = [x1, . . . , xn]T , where x1, . . . , xn arereal numbers, defines the n-dimensional Euclidean space denoted by Rn. The one-dimensional Euclidean space consists of all real numbers and is denoted by R.Vectors in Rn can be added by adding their corresponding components. They canbe multiplied by a scalar by multiplying each component by the scalar. The innerproduct of two vectors x and y is xT y =

∑ni=1 xiyi.

Vector and Matrix Norms

The norm ‖x‖ of a vector x is a real-valued function with the properties

• ‖x‖ ≥ 0 for all x ∈ Rn, with ‖x‖ = 0 if and only if x = 0.

• ‖x + y‖ ≤ ‖x‖+ ‖y‖, for all x, y ∈ Rn.

• ‖αx‖ = |α| ‖x‖, for all α ∈ R and x ∈ Rn.

The second property is the triangle inequality. We consider the class of p-norms,defined by

‖x‖p = (|x1|p + · · ·+ |xn|p)1/p, 1 ≤ p < ∞

and‖x‖∞ = max

i|xi|

The three most commonly used norms are ‖x‖1, ‖x‖∞, and the Euclidean norm

‖x‖2 =(|x1|2 + · · ·+ |xn|2

)1/2=(xT x

)1/2

647


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648 APPENDIX A. MATHEMATICAL REVIEW

All p-norms are equivalent in the sense that if ‖ · ‖α and ‖ · ‖β are two differentp-norms, then there exist positive constants c1 and c2 such that

c1‖x‖α ≤ ‖x‖β ≤ c2‖x‖α

for all x ∈ Rn. For the 1-, 2-, and ∞-norms, these inequalities take the form

‖x‖2 ≤ ‖x‖1 ≤√

n ‖x‖2, ‖x‖∞ ≤ ‖x‖2 ≤√

n ‖x‖∞, ‖x‖∞ ≤ ‖x‖1 ≤ n ‖x‖∞An important result concerning p-norms is the Holder inequality

|xT y| ≤ ‖x‖p‖y‖q,1p

+1q

= 1

for all x ∈ Rn, y ∈ Rn. Quite often when we use norms, we only use propertiesdeduced from the three basic properties satisfied by any norm. In those cases, thesubscript p is dropped, indicating that the norm can be any p-norm.

An m× n matrix A of real elements defines a linear mapping y = Ax from Rn

into Rm. The induced p-norm of A is defined by1

‖A‖p = supx�=0

‖Ax‖p

‖x‖p= max‖x‖p=1

‖Ax‖p

which for p = 1, 2, and ∞ is given by

‖A‖1 = maxj

m∑i=1

|aij |, ‖A‖2 =[λmax(AT A)

]1/2, and ‖A‖∞ = max

i

n∑j=1

|aij |

where λmax(AT A) is the maximum eigenvalue of AT A. Some useful properties ofinduced matrix norms for real matrices A and B of dimensions m × n and n × �,respectively, are as follows:

1√n‖A‖∞ ≤ ‖A‖2 ≤

√m ‖A‖∞,

1√m‖A‖1 ≤ ‖A‖2 ≤

√n ‖A‖1

‖A‖2 ≤√‖A‖1 ‖A‖∞, ‖AB‖p ≤ ‖A‖p ‖B‖p

Topological Concepts in Rn

Convergence of Sequences: A sequence of vectors x0, x1, . . ., xk, . . . in Rn,denoted by {xk}, is said to converge to a limit vector x if

‖xk − x‖ → 0 as k →∞which is equivalent to saying that, given any ε > 0, there is an integer N such that

‖xk − x‖ < ε, ∀ k ≥ N

1sup denotes supremum, the least upper bound; inf denotes infimum, the greatest lower bound.


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APPENDIX A. MATHEMATICAL REVIEW 649

The symbol “∀” reads “for all.” A vector x is an accumulation point of a sequence{xk} if there is a subsequence of {xk} that converges to x; that is, if there is aninfinite subset K of the nonnegative integers such that {xk}k∈K converges to x.A bounded sequence {xk} in Rn has at least one accumulation point in Rn. Asequence of real numbers {rk} is said to be increasing (monotonically increasingor nondecreasing) if rk ≤ rk+1 ∀ k. If rk < rk+1, it is said to be strictly increas-ing. Decreasing (monotonically decreasing or nonincreasing) and strictly decreasingsequences are defined similarly with rk ≥ rk+1. An increasing sequence of real num-bers that is bounded from above converges to a real number. Similarly, a decreasingsequence of real numbers that is bounded from below converges to a real number.

Sets: A subset S ⊂ Rn is said to be open if, for every vector x ∈ S, one canfind an ε-neighborhood of x

N(x, ε) = {z ∈ Rn | ‖z − x‖ < ε}such that N(x, ε) ⊂ S. A set S is closed if and only if its complement in Rn isopen. Equivalently, S is closed if and only if every convergent sequence {xk} withelements in S converges to a point in S. A set S is bounded if there is r > 0 suchthat ‖x‖ ≤ r for all x ∈ S. A set S is compact if it is closed and bounded. A pointp is a boundary point of a set S if every neighborhood of p contains at least onepoint of S and one point not belonging to S. The set of all boundary points of S,denoted by ∂S, is called the boundary of S. A closed set contains all its boundarypoints. An open set contains none of its boundary points. The interior of a set S isS − ∂S. An open set is equal to its interior. The closure of a set S, denoted by S,is the union of S and its boundary. A closed set is equal to its closure. An open setS is connected if every pair of points in S can be joined by an arc lying in S. A setS is called a region if it is the union of an open connected set with some, none, orall of its boundary points. If none of the boundary points are included, the regionis called an open region or domain. A set S is convex if, for every x, y ∈ S andevery real number θ, 0 < θ < 1, the point θx + (1 − θ)y ∈ S. If x ∈ X ⊂ Rn andy ∈ Y ⊂ Rm, we say that (x, y) belongs to the product set X × Y ⊂ Rn ×Rm.

Continuous Functions: A function f mapping a set S1 into a set S2 is denotedby f : S1 → S2. A function f : Rn → Rm is said to be continuous at a point x iff(xk) → f(x) whenever xk → x. Equivalently, f is continuous at x if, given ε > 0,there is δ > 0 such that

‖x− y‖ < δ ⇒ ‖f(x)− f(y)‖ < ε

The symbol “⇒” reads “implies.” A function f is continuous on a set S if it iscontinuous at every point of S, and it is uniformly continuous on S if, given ε > 0there is δ > 0 (dependent only on ε) such that the inequality holds for all x, y ∈ S.Note that uniform continuity is defined on a set, while continuity is defined at apoint. For uniform continuity, the same constant δ works for all points in the set.


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Clearly, if f is uniformly continuous on a set S, then it is continuous on S. Theopposite statement is not true in general. However, if S is a compact set, thencontinuity and uniform continuity on S are equivalent. The function

(a1f1 + a2f2)(·) = a1f1(·) + a2f2(·)is continuous for any two scalars a1 and a2 and any two continuous functions f1 andf2. If S1, S2, and S3 are any sets and f1 : S1 → S2 and f2 : S2 → S3 are functions,then the function f2 ◦ f1 : S1 → S3, defined by

(f2 ◦ f1)(·) = f2(f1(·))is called the composition of f1 and f2. The composition of two continuous functionsis continuous. If S ⊂ Rn and f : S → Rm, then the set of f(x) such that x ∈ S iscalled the image of S under f and is denoted by f(S). If f is a continuous functiondefined on a compact set S, then f(S) is compact; hence, continuous functions oncompact sets are bounded. Moreover, if f is real valued, that is, f : S → R, thenthere are points p and q in the compact set S such that f(x) ≤ f(p) and f(x) ≥ f(q)for all x ∈ S. If f is a continuous function defined on a connected set S, then f(S)is connected. A function f defined on a set S is said to be one to one on S ifwhenever x, y ∈ S, and x �= y, then f(x) �= f(y). If f : S → Rm is a continuous,one-to-one function on a compact set S ⊂ Rn, then f has a continuous inverse f−1

on f(S). The composition of f and f−1 is identity; that is, f−1(f(x)) = x. Afunction f : R → Rn is said to be piecewise continuous on an interval J ⊂ R if forevery bounded subinterval J0 ⊂ J , f is continuous for all x ∈ J0, except, possibly,at a finite number of points where f may have discontinuities. Moreover, at eachpoint of discontinuity x0, the right-side limit limh→0 f(x0+h) and the left-side limitlimh→0 f(x0 − h) exist; that is, the function has a finite jump at x0.

Differentiable functions: A function f : R → R is said to be differentiable atx if the limit

f ′(x) = limh→0

f(x + h)− f(x)h

exists. The limit f ′(x) is called the derivative of f at x. A function f : Rn → Rm issaid to be continuously differentiable at a point x0 if the partial derivatives ∂fi/∂xj

exist and are continuous at x0 for 1 ≤ i ≤ m, 1 ≤ j ≤ n. A function f is continuouslydifferentiable on a set S if it is continuously differentiable at every point of S. Fora continuously differentiable function f : Rn → R, the row vector ∂f/∂x is definedby

∂f

∂x=[

∂f

∂x1, . . . ,

∂f

∂xn

]The gradient vector, denoted by ∇f(x), is

∇f(x) =[∂f

∂x

]T


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APPENDIX A. MATHEMATICAL REVIEW 651

For a continuously differentiable function f : Rn → Rm, the Jacobian matrix[∂f/∂x] is an m×n matrix whose element in the ith row and jth column is ∂fi/∂xj .Suppose S ⊂ Rn is open, f maps S into Rm, f is continuously differentiable atx0 ∈ S, g maps an open set containing f(S) into Rk, and g is continuously differ-entiable at f(x0). Then the mapping h of S into Rk, defined by h(x) = g(f(x)), iscontinuously differentiable at x0 and its Jacobian matrix is given by the chain rule

∂h

∂x

∣∣∣∣x=x0

=∂g

∂f

∣∣∣∣f=f(x0)

∂f

∂x

∣∣∣∣x=x0

Mean Value and Implicit Function Theorems

If x and y are two distinct points in Rn, then the line segment L(x, y) joining xand y is

L(x, y) = {z | z = θx + (1− θ)y, 0 < θ < 1}Mean Value Theorem

Assume that f : Rn → R is continuously differentiable at each point x of anopen set S ⊂ Rn. Let x and y be two points of S such that the line segmentL(x, y) ⊂ S. Then there exists a point z of L(x, y) such that

f(y)− f(x) =∂f

∂x

∣∣∣∣x=z

(y − x)

Implicit Function Theorem

Assume that f : Rn×Rm → Rn is continuously differentiable at each point (x, y)of an open set S ⊂ Rn × Rm. Let (x0, y0) be a point in S for which f(x0, y0) = 0and for which the Jacobian matrix [∂f/∂x](x0, y0) is nonsingular. Then there existneighborhoods U ⊂ Rn of x0 and V ⊂ Rm of y0 such that for each y ∈ V theequation f(x, y) = 0 has a unique solution x ∈ U . Moreover, this solution can begiven as x = g(y), where g is continuously differentiable at y = y0.

The proof of these two theorems, as well as the other facts stated earlier inthis appendix, can be found in any textbook on advanced calculus or mathematicalanalysis.2

Gronwall–Bellman Inequality

Lemma A.1 Let λ : [a, b] → R be continuous and μ : [a, b] → R be continuous andnonnegative. If a continuous function y : [a, b] → R satisfies

y(t) ≤ λ(t) +∫ t

a

μ(s)y(s) ds

2See, for example, [10].


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for a ≤ t ≤ b, then on the same interval

y(t) ≤ λ(t) +∫ t

a

λ(s)μ(s) exp[∫ t

s

μ(τ) dτ

]ds

In particular, if λ(t) ≡ λ is a constant, then

y(t) ≤ λ exp[∫ t

a

μ(τ) dτ

]

If, in addition, μ(t) ≡ μ ≥ 0 is a constant, then

y(t) ≤ λ exp[μ(t− a)]

�

Proof: Let z(t) =∫ t

aμ(s)y(s) ds and v(t) = z(t) + λ(t) − y(t) ≥ 0. Then, z is

differentiable and

z = μ(t)y(t) = μ(t)z(t) + μ(t)λ(t)− μ(t)v(t)

This is a scalar linear state equation with the state transition function

φ(t, s) = exp[∫ t

s

μ(τ) dτ

]

Since z(a) = 0, we have

z(t) =∫ t

a

φ(t, s)[μ(s)λ(s)− μ(s)v(s)] ds

The term ∫ t

a

φ(t, s)μ(s)v(s) ds

is nonnegative. Therefore,

z(t) ≤∫ t

a

exp[∫ t

s

μ(τ) dτ

]μ(s)λ(s) ds

Since y(t) ≤ λ(t) + z(t), this completes the proof in the general case. In the specialcase when λ(t) ≡ λ, we have∫ t

a

μ(s) exp[∫ t

s

μ(τ) dτ

]ds = −

∫ t

a

d

ds

{exp

[∫ t

s

μ(τ) dτ

]}ds

= −{

exp[∫ t

s

μ(τ) dτ

]}∣∣∣∣s=t

s=a

= −1 + exp[∫ t

a

μ(τ) dτ

]

which proves the lemma when λ is a constant. The proof when both λ and μ areconstants follows by integration. �


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Appendix B

Contraction Mapping

Consider an equation of the form x = T (x). A solution x∗ to this equation is saidto be a fixed point of the mapping T , since T leaves x∗ invariant. A classical ideafor finding a fixed point is the successive approximation method. We begin withan initial trial vector x1 and compute x2 = T (x1). Continuing in this manner it-eratively, we compute successive vectors xk+1 = T (xk). The contraction mappingtheorem gives sufficient conditions under which there is a fixed point x∗ of x = T (x)and the sequence {xk} converges to x∗. It is a powerful analysis tool for provingthe existence of a solution of an equation of the form x = T (x). The theorem isvalid, not only when T is a mapping from one Euclidean space into another, butalso when T is a mapping between Banach spaces. We will use the contractionmapping theorem in this general setting. We start by introducing Banach spaces.1

Vector Spaces: A linear vector space X over the field R is a set of elementsx, y, z, . . . called vectors such that for any two vectors x, y ∈ X , the sum x + y isdefined, x + y ∈ X , x + y = y + x, (x + y) + z = x + (y + z), and there is a zerovector 0 ∈ X such that x + 0 = x for all x ∈ X . Also for any numbers α, β ∈ R,the scalar multiplication αx is defined, αx ∈ X , 1 ·x = x, 0 ·x = 0, (αβ)x = α(βx),α(x + y) = αx + αy, and (α + β)x = αx + βx, for all x, y ∈ X .

Normed Linear Space: A linear space X is a normed linear space if, to eachvector x ∈ X , there is a real-valued norm ‖x‖ that satisfies

• ‖x‖ ≥ 0 for all x ∈ X , with ‖x‖ = 0 if and only if x = 0.

• ‖x + y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X .

• ‖αx‖ = |α| ‖x‖ for all α ∈ R and x ∈ X .

1For a complete treatment of Banach spaces, consult any textbook on functional analysis. Alucid treatment can be found in [121, Chapter 2].

653


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654 APPENDIX B. CONTRACTION MAPPING

If it is not clear from the context whether ‖ · ‖ is a norm on X or a norm on Rn,we will write ‖ · ‖X for the norm on X .

Convergence: A sequence {xk} ∈ X , a normed linear space, converges tox ∈ X if

‖xk − x‖ → 0 as k →∞Closed Set: A set S ⊂ X is closed if and only if every convergent sequence

with elements in S has its limit in S.Cauchy Sequence: A sequence {xk} ∈ X is said to be a Cauchy sequence if

‖xk − xm‖ → 0 as k, m→∞Every convergent sequence is Cauchy, but not vice versa.

Banach Space: A normed linear space X is complete if every Cauchy sequencein X converges to a vector in X . A complete normed linear space is a Banach space.

Example B.1 Consider the set of all continuous functions f : [a, b] → Rn, denotedby C[a, b]. This set forms a vector space on R. The sum x + y is defined by(x + y)(t) = x(t) + y(t). The scalar multiplication is defined by (αx)(t) = αx(t).The zero vector is the function that is identically zero on [a, b]. We define a normby

‖x‖C = maxt∈[a,b]

‖x(t)‖

where the right-hand side norm is any p-norm on Rn. Clearly ‖x‖C ≥ 0 and is zeroonly for the zero vector. The triangle inequality follows from

max ‖x(t) + y(t)‖ ≤ max[‖x(t)‖+ ‖y(t)‖] ≤ max ‖x(t)‖+ max ‖y(t)‖Also,

max ‖αx(t)‖ = max |α| ‖x(t)‖ = |α|max ‖x(t)‖where all maxima are taken over [a, b]. Hence, C[a, b], together with the norm ‖·‖C ,is a normed linear space. It is also a Banach space. To prove this claim, we need toshow that every Cauchy sequence in C[a, b] converges to a vector in C[a, b]. Supposethat {xk} is a Cauchy sequence in C[a, b]. For each fixed t ∈ [a, b],

‖xk(t)− xm(t)‖ ≤ ‖xk − xm‖C → 0 as k, m→∞So {xk(t)} is a Cauchy sequence in Rn. But Rn with any p-norm is complete becauseconvergence implies componentwise convergence and R is complete. Therefore,there is a real vector x(t) to which the sequence converges: xk(t) → x(t). Thisproves pointwise convergence. We prove next that the convergence is uniform int ∈ [a, b]. Given ε > 0, choose N such that ‖xk − xm‖C < ε/2 for k, m > N . Thenfor k > N

‖xk(t)− x(t)‖ ≤ ‖xk(t)− xm(t)‖+ ‖xm(t)− x(t)‖≤ ‖xk − xm‖C + ‖xm(t)− x(t)‖


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APPENDIX B. CONTRACTION MAPPING 655

By choosing m sufficiently large (which may depend on t), each term on the right-hand side can be made smaller than ε/2; so ‖xk(t) − x(t)‖ < ε for k > N . Hence,{xk} converges to x, uniformly in t ∈ [a, b]. To complete the proof, we need to showthat x(t) is continuous and {xk} converges to x in the norm of C[a, b]. To provecontinuity, consider

‖x(t + δ)− x(t)‖ ≤ ‖x(t + δ)− xk(t + δ)‖+ ‖xk(t + δ)− xk(t)‖+ ‖xk(t)− x(t)‖

Since {xk} converges uniformly to x, given any ε > 0, we can choose k large enoughto make both the first and third terms on the right-hand side less than ε/3. Becausexk(t) is continuous, we can choose δ small enough to make the second term less thanε/3. Therefore, x(t) is continuous. The convergence of xk to x in ‖ · ‖C is a directconsequence of the uniform convergence.

Theorem B.1 (Contraction Mapping) Let S be a closed subset of a Banachspace X and let T be a mapping that maps S into S. Suppose that

‖T (x)− T (y)‖ ≤ ρ‖x− y‖, ∀ x, y ∈ S, 0 ≤ ρ < 1

then

• there exists a unique vector x∗ ∈ S satisfying x∗ = T (x∗).

• x∗ can be obtained by the method of successive approximation, starting fromany arbitrary initial vector in S. �

Proof: Select an arbitrary x1 ∈ S and define the sequence {xk} by the formulaxk+1 = T (xk). Since T maps S into S, xk ∈ S for all k ≥ 1. The first step of theproof is to show that {xk} is Cauchy. We have

‖xk+1 − xk‖ = ‖T (xk)− T (xk−1)‖≤ ρ‖xk − xk−1‖ ≤ ρ2‖xk−1 − xk−2‖ ≤ · · · ≤ ρk−1‖x2 − x1‖

It follows that

‖xk+r − xk‖ ≤ ‖xk+r − xk+r−1‖+ ‖xk+r−1 − xk+r−2‖+ · · ·+ ‖xk+1 − xk‖≤ [

ρk+r−2 + ρk+r−3 + · · ·+ ρk−1] ‖x2 − x1‖

≤ ρk−1∞∑

i=0

ρi ‖x2 − x1‖ =ρk−1

1− ρ‖x2 − x1‖

The right-hand side tends to zero as k →∞. Thus, the sequence is Cauchy. BecauseX is a Banach space,

xk → x∗ ∈ X as k →∞


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656 APPENDIX B. CONTRACTION MAPPING

Moreover, since S is closed, x∗ ∈ S. We now show that x∗ = T (x∗). For anyxk = T (xk−1), we have

‖x∗ − T (x∗)‖ ≤ ‖x∗ − xk‖+ ‖xk − T (x∗)‖ ≤ ‖x∗ − xk‖+ ρ‖xk−1 − x∗‖

By choosing k large enough, the right-hand side of the inequality can be madearbitrarily small. Thus, ‖x∗ − T (x∗)‖ = 0; that is, x∗ = T (x∗). It remains to showthat x∗ is the unique fixed point of T in S. Suppose that x∗ and y∗ are fixed points.Then,

‖x∗ − y∗‖ = ‖T (x∗)− T (y∗)‖ ≤ ρ‖x∗ − y∗‖Since ρ < 1, we have x∗ = y∗. �


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Appendix C

Proofs

C.1 Proof of Theorems 3.1 and 3.2

Proof of Theorem 3.1: We start by noting that if x(t) is a solution of

x = f(t, x), x(t0) = x0 (C.1)

then, by integration, we have

x(t) = x0 +∫ t

t0

f(s, x(s)) ds (C.2)

Conversely, if x(t) satisfies (C.2), then x(t) satisfies (C.1). Thus, the study of exis-tence and uniqueness of the solution of the differential equation (C.1) is equivalentto the study of existence and uniqueness of the solution of the integral equation(C.2). We proceed with (C.2). Viewing the right-hand side of (C.2) as a mappingof the continuous function x : [t0, t1] → Rn, and denoting it by (Px)(t), we canrewrite (C.2) as

x(t) = (Px)(t) (C.3)

Note that (Px)(t) is continuous in t. A solution of (C.3) is a fixed point of the map-ping P that maps x into Px. Existence of a fixed point of (C.3) can be establishedby using the contraction mapping theorem. This requires defining a Banach spaceX and a closed set S ⊂ X such that P maps S into S and is a contraction over S.Let

X = C[t0, t0 + δ], with norm ‖x‖C = maxt∈[t0,t0+δ]

‖x(t)‖

andS = {x ∈ X | ‖x− x0‖C ≤ r}

where r is the radius of the ball B and δ is a positive constant to be chosen. Wewill restrict the choice of δ to satisfy δ ≤ t1− t0 so that [t0, t0 + δ] ⊂ [t0, t1]. Notice

657


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658 APPENDIX C. PROOFS

that ‖x(t)‖ denotes a norm on Rn, while ‖x‖C denotes a norm on X . Also, B is aball in Rn, while S is a ball in X . By definition, P maps X into X . To show thatit maps S into S, write

(Px)(t)− x0 =∫ t

t0

f(s, x(s)) ds =∫ t

t0

[f(s, x(s))− f(s, x0) + f(s, x0)] ds

By piecewise continuity of f , we know that f(t, x0) is bounded on [t0, t1]. Let

h = maxt∈[t0,t1]

‖f(t, x0)‖

Using the Lipschitz condition (3.2) and the fact that for each x ∈ S,

‖x(t)− x0‖ ≤ r, ∀ t ∈ [t0, t0 + δ]

we obtain

‖(Px)(t)− x0‖ ≤∫ t

t0

[‖f(s, x(s))− f(s, x0)‖+ ‖f(s, x0)‖] ds

≤∫ t

t0

[L‖x(s)− x0‖+ h] ds ≤∫ t

t0

(Lr + h) ds

= (t− t0)(Lr + h) ≤ δ(Lr + h)

and‖Px− x0‖C = max

t∈[t0,t0+δ]‖(Px)(t)− x0‖ ≤ δ(Lr + h)

Hence, choosing δ ≤ r/(Lr + h) ensures that P maps S into S. To show that P isa contraction mapping over S, let x and y ∈ S and consider

‖(Px)(t)− (Py)(t)‖ =∥∥∥∥∫ t

t0

[f(s, x(s))− f(s, y(s))] ds

∥∥∥∥≤

∫ t

t0

‖f(s, x(s))− f(s, y(s))‖ ds

≤∫ t

t0

L‖x(s)− y(s)‖ ds ≤∫ t

t0

ds L‖x− y‖C

Therefore,

‖Px− Py‖C ≤ Lδ‖x− y‖C ≤ ρ‖x− y‖C for δ ≤ ρ

L

Thus, choosing ρ < 1 and δ ≤ ρ/L ensures that P is a contraction mapping over S.By the contraction mapping theorem, we can conclude that if δ is chosen to satisfy

δ ≤ min{

t1 − t0,r

Lr + h,ρ

L

}for ρ < 1 (C.4)


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C.2. PROOF OF LEMMA 3.4 659

then (C.2) will have a unique solution in S. This is not the end of the proof,though, since we are interested in establishing uniqueness of the solution among allcontinuous functions x(t), that is, uniqueness in X . It turns out that any solutionof (C.2) in X will lie in S. To see this, note that since x(t0) = x0 is inside theball B, any continuous solution x(t) must lie inside B for some interval of time.Suppose that x(t) leaves the ball B and let t0 + μ be the first time x(t) intersectsthe boundary of B. Then,

‖x(t0 + μ)− x0‖ = r

On the other hand, for all t ≤ t0 + μ,

‖x(t)− x0‖ ≤∫ t

t0

[‖f(s, x(s))− f(s, x0)‖+ ‖f(s, x0)‖] ds

≤∫ t

t0

[L‖x(s)− x0‖+ h] ds ≤∫ t

t0

(Lr + h) ds

Therefore,

r = ‖x(t0 + μ)− x0‖ ≤ (Lr + h)μ ⇒ μ ≥ r

Lr + h≥ δ

Hence, the solution x(t) cannot leave the set B within the time interval [t0, t0 + δ],which implies that any solution in X lies in S. Consequently, uniqueness of thesolution in S implies uniqueness in X .

Proof of Theorem 3.2: The key point of the proof is to show that the constantδ of Theorem 3.1 can be made independent of the initial state x0. From (C.4), wesee that the dependence of δ on the initial state comes through the constant h in theterm r/(Lr + h). Since in the current case the Lipschitz condition holds globally,we can choose r arbitrarily large. Therefore, for any finite h, we can choose r largeenough so that r/(Lr + h) > ρ/L. This reduces (C.4) to the requirement

δ ≤ min{

t1 − t0,ρ

L

}for ρ < 1

If t1 − t0 ≤ ρ/L, we could choose δ = t1 − t0 and be done. Otherwise, we choose δto satisfy δ ≤ ρ/L. Now, divide [t0, t1] into a finite number of subintervals of lengthδ ≤ ρ/L, and apply Theorem 3.1 repeatedly.1 �

C.2 Proof of Lemma 3.4

The upper right-hand derivative D+v(t) is defined by

D+v(t) = lim suph→0+

v(t + h)− v(t)h

1Note that the initial state of each subinterval x1, say, will satisfy ‖f(t, x1)‖ ≤ h1 for somefinite h1.


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where lim supn→∞ (the limit superior) of a sequence of real numbers {xn} is a realnumber y satisfying the following two conditions:

• for every ε > 0, there exists an integer N such that n > N implies xn < y + ε;

• given ε > 0 and m > 0, there exists an integer n > m such that xn > y − ε.

The first statement means that ultimately all terms of the sequence are less thany + ε, and the second one means that infinitely many terms are greater than y − ε.One of the properties of lim sup2 is that if zn ≤ xn for each n = 1, 2 . . ., thenlim supn→∞ zn ≤ lim supn→∞ xn. From this property, we see that if |v(t + h) −v(t)|/h ≤ g(t, h), ∀ h ∈ (0, b] and limh→0+ g(t, h) = g0(t), then D+v(t) ≤ g0(t).

To prove lemma 3.4, consider the differential equation

z = f(t, z) + λ, z(t0) = u0 (C.5)

where λ is a positive constant. On any compact interval [t0, t1], we conclude fromTheorem 3.5 that for any ε > 0, there is δ > 0 such that if λ < δ, then (C.5) has aunique solution z(t, λ) defined on [t0, t1] and

|z(t, λ)− u(t)| < ε, ∀ t ∈ [t0, t1] (C.6)

Claim 1: v(t) ≤ z(t, λ) for all t ∈ [t0, t1].This claim can be shown by contradiction, for if it were not true, there would betimes a, b ∈ (t0, t1] such that v(a) = z(a, λ) and v(t) > z(t, λ) for a < t ≤ b.Consequently,

v(t)− v(a) > z(t, λ)− z(a, λ), ∀ t ∈ (a, b]

which implies

D+v(a) ≥ z(a, λ) = f(a, z(a, λ)) + λ > f(a, v(a))

which contradicts the inequality D+v(t) ≤ f(t, v(t)).Claim 2: v(t) ≤ u(t) for all t ∈ [t0, t1].

Again, this claim can be shown by contradiction, for if it were not true, there wouldexist a ∈ (t0, t1] such that v(a) > u(a). Taking ε = [v(a)− u(a)]/2 and using (C.6),we obtain

v(a)− z(a, λ) = v(a)− u(a) + u(a)− z(a, λ) ≥ ε

which contradicts the statement of Claim 1.Thus, we have shown that v(t) ≤ u(t) for all t ∈ [t0, t1]. Since this is true on everycompact interval, we conclude that it holds for all t ≥ t0. If it was not the case,let T < ∞ be the first time the inequality is violated. We have v(t) ≤ u(t) forall t ∈ [t0, T ) and, by continuity, v(T ) = u(T ). Consequently, we can extend theinequality to the interval [T, T + Δ] for some Δ > 0, which contradicts the claimthat T is the first time the inequality is violated.

2See [10, Theorem 12-4].


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Since x(t) is bounded, by the Bolzano–Weierstrass theorem,3 it has an accumulationpoint as t → ∞; hence the positive limit set L+ is nonempty. For each y ∈ L+,there is a sequence ti with ti →∞ as i→∞ such that x(ti) → y as i→∞. Becausex(ti) is bounded uniformly in i, the limit y is bounded; that is, L+ is bounded. Toshow that L+ is closed, let {yi} ∈ L+ be some sequence such that yi → y as i→∞and prove that y ∈ L+. For every i, there is a sequence {tij} with tij → ∞ asj → ∞ such that x(tij) → yi as j → ∞. We will construct a particular sequence{τi}. Given the sequences tij , choose τ2 > t12 such that ‖x(τ2)− y2‖ < 1/2; chooseτ3 > t13 such that ‖x(τ3)−y3‖ < 1/3; and so on for i = 4, 5, . . .. Of course, τi →∞as i → ∞ and ‖x(τi) − yi‖ < 1/i for every i. Now, given ε > 0, there are positiveintegers N1 and N2 such that

‖x(τi)− yi‖ <ε

2, ∀ i > N1 and ‖yi − y‖ <

ε

2, ∀ i > N2

The first inequality follows from ‖x(τi) − yi‖ < 1/i and the second one from thelimit yi → y. Thus,

‖x(τi)− y‖ < ε, ∀ i > N = max{N1, N2}

which shows that x(τi) → y as i → ∞. Hence, L+ is closed. This proves that theset L+ is compact, because it is closed and bounded.

To show that L+ is an invariant set, let y ∈ L+ and φ(t; y) be the solution of(4.1) that passes through y at t = 0; that is, φ(0; y) = y, and show that φ(t; y) ∈ L+,∀ t ∈ R. There is a sequence {ti} with ti → ∞ as i → ∞ such that x(ti) → y asi → ∞. Write x(ti) = φ(ti; x0), where x0 is the initial state of x(t) at t = 0. Byuniqueness of the solution,

φ(t + ti; x0) = φ(t; φ(ti; x0)) = φ(t; x(ti))

where, for sufficiently large i, t + ti ≥ 0. By continuity,

limi→∞

φ(t + ti; x0) = limi→∞

φ(t; x(ti)) = φ(t; y)

which shows that φ(t; y) ∈ L+.Finally, to show that x(t) → L+ as t → ∞, use a contradiction argument.

Suppose this is not the case; then, there is an ε > 0 and a sequence {ti} withti → ∞ as i → ∞ such that dist(x(ti), L+) > ε. Since the sequence x(ti) isbounded, it contains a convergent subsequence x(t

′i) → x∗ as i→∞. The point x∗

must belong to L+ and at the same time be at a distance ε from L+, which is acontradiction.

3See [10].


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Define ψ(s) byψ(s) = inf

s≤‖x‖≤rV (x) for 0 ≤ s ≤ r

The function ψ(·) is continuous, positive definite, and increasing. Moreover, V (x) ≥ψ(‖x‖) for 0 ≤ ‖x‖ ≤ r. But ψ(·) is not necessarily strictly increasing. Let α1(s)be a class K function such that α1(s) ≤ kψ(s) with 0 < k < 1. Then,

V (x) ≥ ψ(‖x‖) ≥ α1(‖x‖) for ‖x‖ ≤ r

On the other hand, define φ(s) by

φ(s) = sup‖x‖≤s

V (x) for 0 ≤ s ≤ r

The function φ(·) is continuous, positive definite, and increasing (not necessarilystrictly increasing). Moreover, V (x) ≤ φ(‖x‖) for ‖x‖ ≤ r. Let α2(s) be a class Kfunction such that α2(s) ≥ kφ(s) with k > 1. Then

V (x) ≤ φ(‖x‖) ≤ α2(‖x‖) for ‖x‖ ≤ r

If D = Rn, the definitions of ψ(s) and φ(s) are changed to

ψ(s) = inf‖x‖≥s

V (x), φ(s) = sup‖x‖≤s

V (x), for s ≥ 0

The functions ψ and φ are continuous, positive definite, increasing, and

ψ(‖x‖) ≤ V (x) ≤ φ(‖x‖), ∀ x ∈ Rn

The functions α1 and α2 can be chosen as before. If V (x) is radially unbounded,ψ(s) and φ(s) tend to infinity as s→∞. Hence, we can choose α1 and α2 to belongto class K∞.


Since α(·) is locally Lipschitz, the equation has a unique solution for every initialstate y0 ≥ 0. Because y(t) < 0 whenever y(t) > 0, the solution has the propertythat y(t) ≤ y0 for all t ≥ t0. Therefore, the solution is bounded and can be extendedfor all t ≥ t0. By integration, we have

−∫ y

y0

dx

α(x)=∫ t

t0

dτ

Let b be any positive number less than a, and define

η(y) = −∫ y

b

dx

α(x)


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The function η(y) is a strictly decreasing differentiable function on (0, a). Moreover,limy→0 η(y) = ∞. This limit follows from two facts. First, the solution of thedifferential equation y(t) → 0 as t →∞, since y(t) < 0 whenever y(t) > 0. Second,the limit y(t) → 0 can happen only asymptotically as t → ∞; it cannot happenin finite time due to the uniqueness of the solution. Let c = − limy→a η(y) (c maybe ∞). The range of the function η is (−c,∞). Since η is strictly decreasing, itsinverse η−1 is defined on (−c,∞). For any y0 > 0, the solution y(t) satisfies

η(y(t))− η(y0) = t− t0

Hence,y(t) = η−1(η(y0) + t− t0)

On the other hand, if y0 = 0, then y(t) ≡ 0, because y = 0 is an equilibrium point.Define a function σ(r, s) by

σ(r, s) ={

η−1(η(r) + s), r > 00, r = 0

Then, y(t) = σ(y0, t − t0) for all t ≥ t0 and y0 ≥ 0. The function σ is continuous,since both η and η−1 are continuous in their domains and limx→∞ η−1(x) = 0. Itis strictly increasing in r for each fixed s because

∂

∂rσ(r, s) =

α(σ(r, s))α(r)

> 0

and strictly decreasing in s for each fixed r because

∂

∂sσ(r, s) = −α(σ(r, s)) < 0

Furthermore, σ(r, s) → 0 as s →∞. Thus, σ is a class KL function.


Uniform Stability: Suppose there is a class K function α such that

‖x(t)‖ ≤ α(‖x(t0)‖), ∀ t ≥ t0 ≥ 0, ∀ ‖x(t0)‖ < c

Given ε > 0, let δ = min{c, α−1(ε)}. Then, for ‖x(t0)‖ < δ, we have

‖x(t)‖ ≤ α(‖x(t0)‖) < α(δ) ≤ α(α−1(ε)) = ε

Now assume that given ε > 0, there is δ = δ(ε) > 0 such that

‖x(t0)‖ < δ ⇒ ‖x(t)‖ < ε, ∀ t ≥ t0


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For a fixed ε, let δ(ε) be the supremum of all applicable δ(ε). The function δ(ε)is positive and nondecreasing, but not necessarily continuous. We choose a classK function ζ(r) such that ζ(r) ≤ kδ(r), with 0 < k < 1. Let α(r) = ζ−1(r).Then, α(r) is class K. Let c = limr→∞ ζ(r). Given x(t0), with ‖x(t0)‖ < c, letε = α(‖x(t0)‖). Then, ‖x(t0)‖ < δ(ε) and

‖x(t)‖ < ε = α(‖x(t0)‖) (C.7)

Uniform Asymptotic Stability: Suppose there is a class KL function β(r, s)such that (4.20) is satisfied. Then,

‖x(t)‖ ≤ β(‖x(t0)‖, 0)

which implies that x = 0 is uniformly stable. Moreover, for ‖x(t0)‖ < c, the solutionsatisfies

‖x(t)‖ ≤ β(c, t− t0)

which shows that x(t) → 0 as t → ∞, uniformly in t0. Suppose now that x = 0 isuniformly stable and x(t) → 0 as t → ∞, uniformly in t0, and show that there isa class KL function β(r, s) for which (4.20) is satisfied. Due to uniform stability,there is a constant c > 0 and a class K function α such that for any r ∈ (0, c], thesolution x(t) satisfies

‖x(t)‖ ≤ α(‖x(t0)‖) < α(r), ∀ t ≥ t0, ∀ ‖x(t0)‖ < r (C.8)

Moreover, given η > 0, there exists T = T (η, r) ≥ 0 (dependent on η and r, butindependent of t0) such that

‖x(t)‖ < η, ∀ t ≥ t0 + T (η, r)

Let T (η, r) be the infimum of all applicable T (η, r). The function T (η, r) is nonneg-ative and nonincreasing in η, nondecreasing in r, and T (η, r) = 0 for all η ≥ α(r).Let

Wr(η) =2η

∫ η

η/2T (s, r) ds +

r

η≥ T (η, r) +

r

η

The function Wr(η) is positive and has the following properties:

• for each fixed r, Wr(η) is continuous, strictly decreasing, and Wr(η) → 0 asη →∞, and

• for each fixed η, Wr(η) is strictly increasing in r.

Take Ur = W−1r . Then, Ur inherits the foregoing two properties of Wr and

T (Ur(s), r) < Wr(Ur(s)) = s. Therefore,

‖x(t)‖ ≤ Ur(t− t0), ∀ t ≥ t0, ∀ ‖x(t0‖ < r (C.9)


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C.7. PROOF OF THEOREM 4.16 665

From (C.8) and (C.9), it is clear that

‖x(t)‖ ≤√

α(‖x(t0)‖)Uc(t− t0), ∀ t ≥ t0, ∀ ‖x(t0)‖ < c

Thus, inequality (4.20) is satisfied with β(r, s) =√

α(r)Uc(s).

Global Uniform Asymptotic Stability: If (4.20) holds for all x(t0) ∈ Rn,then it can be easily seen, as in the previous case, that the origin is globally uni-formly asymptotically stable. To prove the opposite direction, notice that in thecurrent case the function δ(ε) has the additional property δ(ε) → ∞ as ε → ∞.Consequently, the class K function α can be chosen to belong to class K∞, andinequality (C.7) holds for all x(t0) ∈ Rn. Moreover, inequality (C.9) holds for anyr > 0. Let

ψ(r, s) = min{

α(r), infρ∈(r,∞)

Uρ(s)}

Then,‖x(t)‖ ≤ ψ(‖x(t0)‖, t− t0), ∀ t ≥ t0, ∀ x(t0) ∈ Rn

If ψ would be class KL, we would be done. This may not be the case, so we definethe function

φ(r, s) =∫ r+1

r

ψ(λ, s) dλ +r

(r + 1)(s + 1)

which is positive and has the following properties:

• for each fixed s ≥ 0, φ(r, s) is continuous and strictly increasing in r,

• for each fixed r ≥ 0, φ(r, s) is strictly decreasing in s and tends to zero ass →∞, and

• φ(r, s) ≥ ψ(r, s).

Thus,‖x(t)‖ ≤ φ(‖x(t0)‖, t− t0), ∀ t ≥ t0, ∀ x(t0) ∈ Rn (C.10)

From (C.10) and the global version of (C.7), we see that

‖x(t)‖ ≤√

α(‖x(t0)‖)φ(‖x(t0)‖, t− t0), ∀ t ≥ t0, ∀ x(t0) ∈ Rn

Thus, the inequality (4.20) is satisfied globally with β(r, s) =√

α(r)φ(r, s).

C.7 Proof of Theorem 4.16

The construction of a Lyapunov function is accomplished by using a lemma, knownas Massera’s lemma. We start by stating and proving Massera’s lemma.


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Lemma C.1 Let g : [0,∞) → R be a positive, continuous, strictly decreasing func-tion with g(t) → 0 as t → ∞. Let h : [0,∞) → R be a positive, continuous,nondecreasing function. Then, there exists a function G(t) such that

• G(t) and its derivative G′(t) are class K functions defined for all t ≥ 0.

• For any continuous function u(t) that satisfies 0 ≤ u(t) ≤ g(t) for all t ≥ 0,there exist positive constants k1 and k2, independent of u, such that∫ ∞

0G(u(t)) dt ≤ k1;

∫ ∞

0G′(u(t))h(t) dt ≤ k2

�

Proof of Lemma C.1: Since g(t) is strictly decreasing, we can choose a sequencetn such that

g(tn) ≤ 1n + 1

, n = 1, 2, . . .

We use this sequence to define a function η(t) as follows:

(a) η(tn) = 1/n.

(b) Between tn and tn+1, η(t) is linear.

(c) In the interval 0 < t ≤ t1, η(t) = (t1/t)p, where p is a positive integer chosenso large that the derivative η′(t) has a positive jump at t1, η′(t−1 ) < η′(t+1 ).

The function η(t) is strictly decreasing, and for t ≥ t1, we have g(t) < η(t). Ast → 0+, η(t) grows unbounded. The inverse of η(t), denoted by η−1(s), is a strictlydecreasing function that grows unbounded as s→ 0+. Obviously,

η−1(u(t)) ≥ η−1(g(t)) > η−1(η(t)) = t, ∀ t ≥ t1

for any nonnegative function u(t) ≤ g(t). Define

H(s) =exp[−η−1(s)]

h(η−1(s)), s ≥ 0

Since η−1 is continuous and h is positive, H(s) is continuous on 0 < s < ∞, whileη−1(s) →∞ as s → 0+. Hence, H(s) defines a class K function on [0,∞). It followsthat the integral

G(r) =∫ r

0H(s) ds

exists and both G(r) and G′(r) = H(r) are class K functions on [0,∞). Now, letu(t) be a continuous nonnegative function such that u(t) ≤ g(t). We have

G′(u(t)) =exp[−η−1(u(t))]

h(η−1(u(t)))≤ e−t

h(t), ∀ t ≥ t1


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Thus, ∫ ∞

t1

G′(u(t))h(t) dt ≤∫ ∞

t1

e−t ≤ 1

and ∫ ∞

0G′(u(t))h(t) dt ≤

∫ t1

0G′(g(t))h(t) dt + 1 ≤ k2

which shows that the second integral in the lemma is bounded. For the first integral,we have∫ ∞

t1

G(u(t)) dt =∫ ∞

t1

∫ u(t)

0

exp[−η−1(s)]h(η−1(s))

ds dt ≤∫ ∞

t1

∫ η(t)

0

exp[−η−1(s)]h(0)

ds dt

For 0 ≤ s ≤ η(t), we have−η−1(s) ≤ −t

Hence, ∫ η(t)

0

exp[−η−1(s)]h(0)

ds ≤∫ η(t)

0

e−t

h(0)ds =

e−t

h(0)η(t) ≤ e−t

h(0)

for t ≥ t1. Consequently,∫ ∞

0G(u(t)) dt ≤

∫ t1

0G(g(t)) dt +

∫ ∞

t1

e−t

h(0)dt ≤ k1

Therefore, the first integral of the lemma is bounded. The proof of the lemma iscomplete. �

To prove the theorem, let

V (t, x) =∫ ∞

t

G(‖φ(τ ; t, x)‖2) dτ

where φ(τ ; t, x) is the solution that starts at (t, x) and G is a class K function tobe selected using Lemma C.1. To see how to choose G, let us start by checking theupper bound on

∂V

∂x=∫ ∞

t

G′(‖φ‖2) φT

‖φ‖2 φx dτ

We saw in the proof of Theorem 4.14 that the assumption ‖∂f/∂x‖2 ≤ L, uniformlyin t, implies that ‖φx(τ ; t, x)‖2 ≤ exp[L(τ − t)]. Therefore,∥∥∥∥∂V

∂x

∥∥∥∥2≤

∫ ∞

t

G′(‖φ(τ ; t, x)‖2) exp[L(τ − t)] dτ

≤∫ ∞

t

G′(β(‖x‖2, τ − t)) exp[L(τ − t)] dτ

≤∫ ∞

0G′(β(‖x‖2, s)) exp(Ls) ds


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With β(r0, s) and exp(Ls) as the functions g and h of Lemma C.1, we take G asthe class K function provided by the lemma. Hence, the integral∫ ∞

0G′(β(‖x‖2, s)) exp(Ls) ds

def= α4(‖x‖2)

is bounded for all ‖x‖2 ≤ r0, uniformly in x. Moreover, it is a continuous andstrictly increasing function of ‖x‖2, because β(‖x‖2, s) is a class K function in ‖x‖2for every fixed s. Thus, α4 is a class K function, which proves the last inequality inthe theorem statement. Consider now

V (t, x) =∫ ∞

t

G(‖φ(τ ; t, x)‖2) dτ

≤∫ ∞

t

G(β(‖x‖2, τ − t)) dτ =∫ ∞

0G(β(‖x‖2, s)) ds

def= α2(‖x‖2)

By Lemma C.1, the last integral is bounded for all ‖x‖2 ≤ r0. The function α2is a class K function. Recall from the proof of Theorem 4.14 that the assumption‖∂f/∂x‖2 ≤ L, uniformly in t, implies that ‖φ(τ ; t, x)‖2 ≥ ‖x‖2 exp[−L(τ − t)].Therefore,

V (t, x) ≥∫ ∞

t

G(‖x‖2e−L(τ−t)) dτ =∫ ∞

0G(‖x‖2e−Ls) ds

≥∫ (ln 2)/L

0G( 1

2‖x‖2) ds =ln 2L

G( 12‖x‖2)

def= α1(‖x‖2)

Clearly, α1(‖x‖2) is a class K function. Hence, V satisfies the inequality

α1(‖x‖2) ≤ V (t, x) ≤ α2(‖x‖2)for all ‖x‖2 ≤ r0. Finally, the derivative of V along the trajectories of the systemis given by

∂V

∂t+

∂V

∂xf(t, x) =

−G(‖x‖2) +∫ ∞

t

G′(‖φ‖2) φT

‖φ‖2 [φt(τ ; t, x) + φx(τ ; t, x)f(t, x)] dτ

Sinceφt(τ ; t, x) + φx(τ ; t, x)f(t, x) ≡ 0, ∀ τ ≥ t

we have∂V

∂t+

∂V

∂xf(t, x) = −G(‖x‖2)

Consequently, the three inequalities of the theorem statement are satisfied for all‖x‖2 ≤ r0. Notice that, due to the equivalence of norms, we can state the inequali-ties in any p-norm. If the system is autonomous, the solution depends only on τ− t;


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that is, φ(τ ; t, x) = ψ(τ − t; x). Therefore,

V =∫ ∞

t

G(‖ψ(τ − t; x)‖2) dτ =∫ ∞

0G(‖ψ(s; x)‖2) ds

which is independent of t.


For any given solution x(t) of

x = f(x), x(0) = x0 ∈ RA (C.11)

the change of time variable from t to τ =∫ t

0 (1+‖f(x(s))‖) ds results in the system

dx

dτ=

11 + ‖f(x)‖f(x) def= f(x), x(0) = x0 (C.12)

where x(τ) = x(t) with t substituted in terms of τ . The origin is an asymptoticallystable equilibrium point of (C.12), and RA is its region of attraction. If V (x) is aLyapunov function for (C.12) that satisfies

∂V

∂xf(x) ≤ −W (x)

for some positive definite function W (x), then

∂V

∂xf(x) = (1 + ‖f(x)‖)∂V

∂xf(x) ≤ −(1 + ‖f(x)‖)W (x) ≤ −W (x)

because 1+ ‖f(x)‖ ≥ 1. Therefore, it is sufficient to construct a Lyapunov functionfor (C.12). It is easier to work with (C.12) because the property ‖f‖ ≤ 1 impliesthat there is no finite escape time for t < 0. For the rest of the proof, we work with(C.12), which we rewrite as

x = f(x) (C.13)

by dropping the bar from x and using x to denote the derivative with respect to τ .We know from Lemma 8.1 that RA is an open set. When RA �= Rn, let F be

the complement of RA in Rn. For every x ∈ RA define

ω(x) = max{‖x‖, 1

dist(x, F )− 2

dist(0, F )

}(C.14)

if RA �= Rn and ω(x) = ‖x‖ if RA = Rn. It is easy to verify that ω(x) is positivedefinite and locally Lipschitz. Since dist(x, F ) → 0 as x approaches ∂RA, ω(x) →∞as x approaches ∂RA. Moreover, for r0 = (1/2)dist(0, F ), we have

infy∈F

{‖x− y‖} ≥ infy∈F

{‖y‖ − ‖x‖} ≥ infy∈F

{‖y‖ − r0}, ∀ ‖x‖ ≤ r0


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Hence,

dist(x, F ) ≥ dist(0, F )− 12dist(0, F ) = 1

2dist(0, F ), ∀ ‖x‖ ≤ r0

Therefore, ω(x) = ‖x‖ for all ‖x‖ ≤ r0.

Lemma C.2 The solution of (C.13) satisfies

ω(x(t)) ≤ β(ω(x(0)), t), ∀ t ≥ 0, ∀ x(0) ∈ RA (C.15)

where β(r, s) is a class KL function defined for all r ≥ 0 and s ≥ 0, and β(r, 0) isa class K∞ function. �

Proof of Lemma C.2: We start by showing that for any constant r > 0, there isa constant b = b(r) > 0 such that the solution of (C.13) with ω(x(0)) ≤ r satisfiesω(x(t)) ≤ b for all t ≥ 0. Suppose to the contrary that this is not the case, then thereis a sequence of solutions x(i)(t) of (C.13) and constants ti such that ω(x(i)(0)) ≤ rfor i = 1, 2, 3, · · · and ω(x(i)(ti) > i for i = 1, 2, 3, · · ·. Let T ∗ be the supremum ofall T ≥ 0 such that x(i)(t), for i = 1, 2, 3, · · ·, are finite for all t ∈ [0, T ]; that is,

T ∗ = sup{T ≥ 0 | lim supi→∞

{ max0≤t≤T

{ω(x(i)(t))} < ∞} (C.16)

We consider the cases T ∗ < ∞ and T ∗ = ∞ separately. When T ∗ < ∞, let τi bea sequence of positive constants such that for every i ≥ 1 the solution of (C.13)satisfies

ω(x(0)) ≤ i ⇒ ω(x(t)) ≤ i + 1, ∀ 0 ≤ t ≤ 2τi (C.17)

This sequence is not empty because of continuity of x(t). We can always chooseτi such that T ∗ > τ1 > τ2 > · · · and limi→∞ τi = 0. Let x(1,i)(t) be the sequenceof all functions x(i)(t) such that ω(x(i)(T ∗ − τ1)) > 1. The functions x(1,i)(t) forman infinite sequence. To see this fact, let I1 be the set of indices i such that x(i)(t)does not belong to the sequence x(1,i)(t). The choice of τ1 implies that

ω(x(i)(t)) ≤ 2 for T ∗ − τ1 ≤ t ≤ T ∗ + τ1, ∀ i ∈ I (C.18)

If there was only a finite number of functions x(i)(t) in the sequence x(1,i)(t), (C.18)would imply that

lim supi→∞

{ max0≤t≤T ∗+τ1

{ω(x(i)(t))}} <∞ (C.19)

which contradicts the definition (C.16) of T ∗. Thus, the sequence x(1,i)(t) is infinite.Let x(2,i)(t) be the sequence of all functions x(1,i)(t) such that ω(x(1,i)(T ∗−τ2)) > 2.We can repeat the foregoing argument to show that the sequence x(2,i)(t) is infinite.Proceeding in the same manner, we can construct a family of subsequences until weend up with a sequence x(i)(t) such that

ω(x(i)(T ∗ − τj)) ≥ j, ∀ j = 1, 2, 3, · · · , ∀ i = 1, 2, 3, · · · (C.20)


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Since ‖x(i)(0)‖ ≤ ω(x(i)(0)) ≤ r and ‖f(x)‖ ≤ 1, the solutions x(i)(t) belong to thecompact set {‖x‖ ≤ r + T} for any 0 < T < T ∗. Thus, the sequence ω(x(i)(t))is bounded on the interval t ∈ [0, T ], uniformly in i. We can select from thesequence x(i)(t) a subsequence x(i)(t) that converges uniformly on the interval [0, T ],0 < T < T ∗, to a solution x(t), defined for t ∈ [0, T ∗). From (C.20) and the fact thatlimi→∞ τi = 0, we conclude that limt→T ∗ ω(x(t)) = ∞. Similarly, when T ∗ = ∞,the solutions x(i)(t) belong to the compact set {‖x‖ ≤ r+T} for any T > 0. Hence,we can select a subsequence x(i)(t) that converges uniformly on the interval [0, T ]to a solution x(t), defined for t ∈ [0,∞), with limt→∞ ω(x(t)) = ∞. Thus, wehave shown that there is a constant T ∗, 0 < T ∗ ≤ ∞, and a solution x(t) suchthat ω(x(0)) ≤ r and limt→T ∗ ω(x(t)) = ∞. However, this is impossible, becausex(0) ∈ RA. Therefore, we conclude that for any r > 0, there is b = b(r) > 0 suchthat the solution of (C.13) with ω(x(0)) ≤ r satisfies ω(x(t)) ≤ b for all t ≥ 0. Theconstant b(r) can be chosen as an increasing function of r and, in view of the factthat the origin is stable, b(r) → 0 as r → 0. Moreover, b(r) → ∞ as r → ∞, sinceb(r) ≥ r. We can find a class K∞ function α(r) such that b(r) ≤ α(r) for all r ≥ 0.Thus, the solution of (C.13) satisfies

ω(x(t)) ≤ α(ω(x(0))), ∀ t ≥ 0, ∀ x(0) ∈ RA (C.21)

On the other hand, given any positive constants r and η, we can show that there isT = T (η, r) > 0 such that

ω(x(0)) < r ⇒ ω(x(t)) < η, ∀ t ≥ T (C.22)

If this were not the case, there would exist a sequence of solutions x(i)(t) of (C.13)and constants τi such that

limi→∞

τi = ∞, ω(x(i)(0)) ≤ r, and ω(x(i)(τi)) ≥ η

However, from (C.21), we see that for any positive constant δ < α−1(η), everysolution of (C.13) with ω(x(τ)) ≤ δ, satisfies ω(x(t)) < η for all t ≥ τ . Hence,ω(x(i)(t)) ≥ δ for 0 ≤ t ≤ τi. Because ω(x(i)(t)) ≤ α(r) for all t ≥ 0, we select fromthe sequence x(i)(t) a subsequence x(i)(t) that converges uniformly in every interval[0, T ] with 0 < T < ∞. The function x(t) = limi→∞ x(i)(t) is a solution of (C.13)for which ω(x(t)) ≥ δ for all t ≥ 0. This, however, is impossible since x(0) ∈ RA.Thus, T (η, r) exists. By repeating the steps of the proof of Lemma 4.5 (the globaluniform asymptotic stability case), we can use (C.21) and (C.22) to show that thereis a class KL function β(r, s) with β(r, 0) ≥ α(r) such that (C.15) is satisfied. �

Let φ(t; x) denote the solution of (C.13) that starts from x at t = 0. Because‖f(x)‖ is bounded, φ(t; x) is defined for all t ≤ 0 . Moreover, since RA is an invariantset (Lemma 8.1), φ(t; x) ∈ RA for all t ≤ 0. Define g : RA → R by

g(x) = inft≤0{ω(φ(t; x))} (C.23)


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It is clear from the definition that

g(φ(t; x)) ≤ g(x), ∀ t ≥ 0, ∀ x ∈ RA (C.24)

α−1(ω(x)) ≤ g(x) ≤ ω(x), ∀ x ∈ RA (C.25)

The first inequality of (C.25) holds because, from (C.21), ω(x) ≤ α(ω(φ(t; x))),∀ t ≤ 0. Let us show that g(x) is locally Lipschitz for x ∈ RA, x �= 0. Equivalently,we show that g(x) is Lipschitz on the compact set H = {x ∈ RA | c1 ≤ ω(x) ≤ c2},with c2 > c1 > 0. Inequality (C.15) implies that

c1 ≤ ω(x) ≤ β(ω(φ(t; x)),−t), ∀ t < 0, ∀ x ∈ H

Let T1 satisfy β(2c2, T1) = c1. Then, for all t ≤ −T1,

β(2c2, T1) = c1 ≤ β(ω(φ(t; x)),−t) ≤ β(ω(φ(t; x)), T1)

Hence,ω(φ(t; x)) ≥ 2c2 ≥ 2ω(x) ≥ 2g(x), ∀ t ≤ −T1, ∀ x ∈ H

This shows that, for all x ∈ H, the infimum defining g(x) is reached within theinterval [−T1, 0]. Using the facts that φ(t; x) is Lipschitz in x on H for any compacttime interval (Theorem 3.4) and ω is locally Lipschitz on RA, we conclude that g(x)is Lipschitz on H. Notice that this argument does not hold for c1 = 0 and thatis why we do not show that g(x) is locally Lipschitz at x = 0. However, g(x) iscontinuous for all x ∈ RA. This is so because g(0) = 0, g(x) ≤ ω(x) (from (C.25)),and ω(x) is continuous.

Define the function V : RA → R by

V (x) = supt≥0

{g(φ(t; x))

1 + 2t

1 + t

}(C.26)

Using (C.24) and (C.25), it is easy to verify that

α−1(ω(x)) ≤ g(x) ≤ V (x) ≤ 2g(x) ≤ 2ω(x) (C.27)

Let us show that V (x) is locally Lipschitz for x ∈ RA, x �= 0, by showing that V (x)is Lipschitz on H = {x ∈ RA | c1 ≤ ω(x) ≤ c2} with 0 < c1 < c2. Using (C.15) and(C.25), we have

g(φ(t; x))1 + 2t

1 + t≤ 2ω(φ(t; x)) ≤ 2β(ω(x), t) ≤ 2β(c2, t)

for all x ∈ H. Let T2 > 0 satisfy 4β(c2, T2) = α−1(c1). Then, for all t ≥ T2,

g(φ(t; x))1 + 2t

1 + t≤ 1

2α−1(c1) ≤ 12α−1(ω(x)) ≤ 1

2 V (x)


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Thus, the supremum defining V (x) is reached within the interval [0, T2]. By repeat-ing the argument used with g(x), we can show that V (x) is Lipschitz on H. SinceV (0) = 0, it follows from (C.27) that V (x) is continuous for all x ∈ RA.

Next, we show that V (x(t)) is decreasing along the solution of (C.13). BecauseV (x) is only locally Lipschitz, its derivative along the solution of (C.13) is calculatedby using

˙V (x) = lim suph→0+

1h

[V (φ(h; x))− V (x)] (C.28)

For x �= 0, take r > ω(x). From the properties of class KL functions, we canfind a function γr(ρ), defined for 0 < ρ < ∞ and 0 < r < ∞, such that foreach fixed r, γr(ρ) is continuous and decreasing in ρ, for each fixed ρ, γr(ρ) isincreasing in r, and 4β(r, γr(ρ)) ≤ α−1(ρ/2) for all 0 < ρ <∞. Let h0 be such thatω(φ(t; x)) ≥ (1/2)ω(x) for all t ∈ [0, h0] and pick h ∈ [0, h0]. Then,

V (φ(h; x)) = supt≥0

{g(φ(t; φ(h; x)))

1 + 2t

1 + t

}= sup

t≥0

{g(φ(t + h; x))

1 + 2t

1 + t

}

Using

g(φ(t + h; x))1 + 2t

1 + t≤ 2ω(φ(t + h; x)) ≤ 2β(ω(x), t + h) < 2β(r, t + h)

we see that for all t + h ≥ γr(ω(x)),

g(φ(t + h; x))1 + 2t

1 + t≤ 1

2α−1 ( 12ω(x)

) ≤ 12α−1(ω(φ(h; x))) ≤ 1

2 V (φ(h; x))

Hence, the supremum defining V (φ(h; x)) is reached at some time t′ such thatt′ + h ≤ γr(ω(x)). Therefore,

V (φ(h; x)) = g(φ(t′ + h; x))1 + 2t′

1 + t′

= g(φ(t′ + h; x))1 + 2t′ + 2h

1 + t′ + h

[1− h

(1 + 2t′ + 2h)(1 + t′)

]

≤ V (x)[1− h

2[1 + γr(ω(x)]2

]

Setting

ηr(s) =α−1(s)

2[1 + γr(s)]2

when s > 0 and ηr(0) = 0, it can be easily verified that ηr(s) is a class K∞ functionand

˙V (x) ≤ −ηr(ω(x))


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Since the preceding inequality holds for all r > ω(x), it will hold with η(s) =supr>s ηr(s); that is, ˙V (x) ≤ −η(ω(x)) for all x �= 0. Define

η(s) =∫ 2s+1

2s

ηr(s) dr

when s > 0 and η(0) = 0. Then η(s) is a continuous, positive definite function on[0,∞) and η(s) ≤ η(s). It follows that

˙V (x) ≤ −η(ω(x)), ∀ x ∈ RA, x �= 0 (C.29)

The function V (x) satisfies all the conditions stated in Theorem 4.17, except thatit is not smooth. To finish the proof, we smooth out V (x) by using the next twolemmas, which we quote without proof.

Lemma C.3 Let D be an open subset of Rn, and suppose there are locally Lipschitzfunctions Φ : D → R and g : D → Rn and a continuous function ψ : D → R suchthat the derivative of Φ(x) along the trajectories of x = g(x) satisfies Φ(x) ≤ ψ(x)for all x ∈ D. Then, given any continuous functions μ : D → (0,∞) and ν : D →(0,∞), there is a smooth function Ψ : D → R such that |Φ(x)−Ψ(x)| ≤ μ(x) andΨ(x) ≤ ψ(x) + ν(x) for all x ∈ D. �

Proof of Lemma C.3: See [118, Theorem B.1].

Lemma C.4 Let D ⊂ Rn be a domain that contains the origin and Φ : D → [0,∞)be a locally Lipschitz, positive definite function such that Φ(x) is smooth for x �= 0.Then, there exists a class K∞ function σ, smooth on (0,∞), such that σ(i)(r) → 0as r → 0+ for each i = 0, 1, · · ·, σ′(r) > 0 for all r > 0, and Ψ(x) = σ(Φ(x)) issmooth on D. �

Proof of Lemma C.4: See [118, Lemma 4.3] (with the domain restricted to Dinstead of Rn).

Apply Lemma C.3 with D = RA − {0}, Φ(x) = V (x), g(x) = f(x), ψ(x) =−η(ω(x)), μ(x) = (1/2)α−1(ω(x)), and ν(x) = (1/2)η(ω(x)) to find a functionV (x), smooth on RA − {0}, such that

α1(ω(x)) ≤ V (x) ≤ α2(ω(x)) and ˙V (x) ≤ −α3(ω(x))

where α1(r) = (1/2)α−1(r) and α2(r) = 2r + (1/2)α−1(r) are class K∞ functionsand α3(r) = (1/2)η(r) is continuous and positive definite on [0,∞). Now applyLemma C.4 with D = RA and Φ = V to find a class K∞ function σ such thatV (x) = σ(V (x)) is smooth on RA. It is easy to verify that αi(r) = σ(αi(r)), i = 1, 2,are class K∞ functions,

α3(r) = α3(r) mint∈[α1(r),α2(r)]

σ′(t)


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is continuous and positive definite on [0,∞),

α1(ω(x)) ≤ V (x) ≤ α2(ω(x))

andV (x) = σ′(V (x)) ˙

V (x) ≤ −σ′(V (x))α3(ω(x)) ≤ −α3(ω(x))

The function V satisfies all the conditions stated in Theorem 4.17. The fact thatfor any c > 0 the set {V (x) ≤ c} is a compact subset of RA follows from {V (x) ≤c} ⊂ {ω(x) ≤ α−1

1 (c)}.


The statement of this theorem reduces to that of Theorem 4.9 when μ = 0. There-fore, the current proof shares some ideas and terminology with the proof of Theo-rem 4.9. Let ρ = α1(r). Then α2(μ) < ρ and α2(‖x(t0)‖) ≤ ρ. Let η = α2(μ) anddefine Ωt,η = {x ∈ Br | V (t, x) ≤ η} and Ωt,ρ = {x ∈ Br | V (t, x) ≤ ρ}. Then

Bμ ⊂ Ωt,η ⊂ {α1(‖x‖) ≤ η} ⊂ {α1(‖x‖) ≤ ρ} = Br ⊂ D

andΩt,η ⊂ Ωt,ρ ⊂ Br ⊂ D

The sets Ωt,ρ and Ωt,η have the property that a solution starting in either set cannotleave it because V (t, x) is negative on the boundary. Since

α2(‖x(t0)‖) ≤ ρ ⇒ x(t0) ∈ Ωt0,ρ

we conclude that x(t) ∈ Ωt,ρ for all t ≥ t0. A solution starting in Ωt,ρ must enterΩt,η in finite time because in the set {Ωt,ρ − Ωt,η}, V satisfies

V (t, x) ≤ −k < 0

where k = min{W3(x)} over the set {μ ≤ ‖x‖ ≤ r}, which contains {Ωt,ρ − Ωt,η}.The foregoing inequality implies that

V (t, x(t)) ≤ V (t0, x(t0))− k(t− t0) ≤ ρ− k(t− t0)

which shows that V (t, x(t)) reduces to η within the time interval [t0, t0 +(ρ−η)/k].For a solution starting inside Ωt,η, inequality (4.43) is satisfied for all t ≥ t0, becauseΩt,η ⊂ {α1(‖x‖) ≤ α2(μ)}. For a solution starting inside Ωt,ρ, but outside Ωt,η, lett0 + T be the first time it enters Ωt,η. For all t ∈ [t0, t0 + T ]

V ≤ −W3(x) ≤ −α3(‖x‖) ≤ −α3(α−1

2 (V )) def= −α(V )


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where α3 and α are class K functions. The existence of α3 follows from Lemma 4.3.Similar to the proof of Theorem 4.9, we can show that there is a class KL functionσ such that

V (t, x(t)) ≤ σ(V (t0, x(t0)), t− t0), ∀ t ∈ [t0, t0 + T ]

Defining β(r, s) = α−11 (σ(α2(r), s)), we obtain

‖x(t)‖ ≤ β(‖x(t0)‖, t− t0), ∀ t ∈ [t0, t0 + T ]

If D = Rn, α3, and consequently β, can be chosen independent of ρ. If α1 belongsto class K∞, so does α2 and the bound α−1

2 (ρ) can be made arbitrarily large bychoosing ρ large enough. Hence, any initial state x(t0) can be included in the set{‖x‖ ≤ α−1

2 (ρ)}.


We complete the proof of Theorem 5.4 by showing that the L2 gain is equal tosupω∈R ‖G(jω)‖2. Let c1 be the L2 gain and c2 = supω∈R ‖G(jω)‖2. We knowthat c1 ≤ c2. Suppose c1 < c2 and set ε = (c2 − c1)/3. Then, for any u ∈ L2with ‖u‖L2 ≤ 1, we have ‖y‖L2 ≤ c2 − 3ε. We will establish a contradictionby finding a signal u with ‖u‖L2 ≤ 1 such that ‖y‖L2 ≥ c2 − 2ε. It is easierto construct such signal if we define signals on the whole real line R. There isno loss of generality in doing so due to the fact (shown in Exercise 5.19) thatthe L2 gain is the same whether signals are defined on [0,∞) or on R. Now,select ω0 ∈ R such that ‖G(jω0)‖2 ≥ c2 − ε. Let v ∈ Cm be the normalizedeigenvector (v∗v = 1) corresponding to the maximum eigenvalue of the Hermitianmatrix GT (−jω0)G(jω0). Hence, v∗GT (−jω0)G(jω0)v = ‖G(jω0)‖22. Write v as

v =[α1e

jθ1 , α2ejθ2 , . . . , αmejθm

]Twhere αi ∈ R is such that θi ∈ (−π, 0]. Take 0 ≤ βi ≤ ∞ such that θi =−2 tan−1(ω0/βi), with βi = ∞ if θi = 0. Define the m × 1 transfer function H(s)by

H(s) =[α1

β1 − s

β1 + s, α2

β2 − s

β2 + s, . . . , αm

βm − s

βm + s

]T

with 1 replacing (βi − s)/(βi + s) if θi = 0. It can be easily seen that H(jω0) = vand HT (−jω)H(jω) =

∑mi=1 α2

i = v∗v = 1 for all ω ∈ R. Take uσ(t) as the outputof H(s) driven by the scalar function

zσ(t) =(

11 + e−ω2

0σ/2

)1/2( 8πσ

)1/4

e−t2/σ cos(ω0t), σ > 0, t ∈ R


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It can be verified that zσ ∈ L2 and ‖zσ‖L2 = 1.4 Consequently, uσ ∈ L2 and‖uσ‖L2 ≤ 1. The Fourier transform of zσ(t) is given by

Zσ(jω) =(

11 + e−ω2

0σ/2

)1/2 (πσ

2

)1/4 [e−(ω−ω0)2σ/4 + e−(ω+ω0)2σ/4

]

Let yσ(t) be the output of G(s) when the input is uσ(t). The Fourier transformof yσ(t) is given by Yσ(jω) = G(jω)Uσ(jω) = G(jω)H(jω)Zσ(jω). By Parseval’stheorem,

‖yσ‖2L2=

12π

∫ ∞

−∞ZT

σ (−jω)HT (−jω)GT (−jω)G(jω)H(jω)Zσ(jω) dω

=12π

∫ ∞

−∞HT (−jω)GT (−jω)G(jω)H(jω) |Zσ(jω)|2 dω

Using

|Zσ(jω)|2 ≥ 11 + e−ω2

0σ/2

(πσ

2

)1/2 [e−(ω−ω0)2σ/2 + e−(ω−ω0)2σ/2

]def= ψσ(ω)

we obtain

‖yσ‖2L2≥ 1

2π

∫ ∞

−∞HT (−jω)GT (−jω)G(jω)H(jω)ψσ(ω) dω

By letting σ → ∞, one can concentrate the frequency spectrum ψσ(ω) around thethe two frequencies ω = ±ω0.5 Hence, the right-hand side of the last inequalityapproaches

HT (−jω0)GT (−jω0)G(jω0)H(jω0) = ‖G(jω0)‖22 ≥ (c2 − ε)2

as σ →∞. Therefore, we can choose a finite σ large enough such that ‖yσ‖L2 ≥ c2−2ε. This, however, contradicts the inequality ‖yσ‖L2 ≤ c2 − 3ε. The contradictionshows that c1 = c2.


Sufficiency: Suppose the conditions of the lemma are satisfied. Since G(s) isHurwitz, there exist positive constants δ and μ∗ such that poles of all elements ofG(s− μ) have real parts less than −δ, for all μ < μ∗. To show that G(s) is strictlypositive real, it is sufficient to show that G(jω − μ) + GT (−jω − μ) is positive

4For the purpose of this discussion, the L2-norm is defined by ‖z‖2L2=∫∞−∞ zT (t)z(t) dt.

5ψσ(ω) approaches π[δ(ω − ω0) + δ(ω + ω0)] as σ →∞, where δ(·) is the impulse function.


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semidefinite for all ω ∈ R. Let {A, B, C, D} be a minimal realization of G(s).Then,

G(s− μ) = D + C(sI − μI −A)−1B

= D + C(sI −A)−1(sI −A)(sI − μI −A)−1B

= D + C(sI −A)−1(μI + sI − μI −A)(sI − μI −A)−1B

= G(s) + μN(s) (C.30)

whereN(s) = C(sI −A)−1(sI − μI −A)−1B

Since A and (A + μI) are Hurwitz, uniformly in μ, there is k0 > 0 such that

σmax[N(jω) + NT (−jω)

] ≤ k0, ∀ ω ∈ R (C.31)

Moreover, limω→∞ ω2N(jω) exists. Hence, there exist k1 > 0 and ω1 > 0 such that

ω2σmax[N(jω) + NT (−jω)

] ≤ k1, ∀ |ω| ≥ ω1 (C.32)

If G(∞) + GT (∞) is positive definite, there is σ0 > 0 such that

σmin[G(jω) + GT (−jω)

] ≥ σ0, ∀ ω ∈ R (C.33)

From (C.30), (C.31), and (C.33),

σmin[G(jω − μ) + GT (−jω − μ)

] ≥ σ0 − μk0, ∀ ω ∈ R

Choosing μ < σ0/k0 ensures that G(jω − μ) + GT (−jω − μ) is positive definite forall ω ∈ R. If G(∞) + GT (∞) is singular, the third condition of the lemma ensuresthat G(jω) + GT (−jω) has q singular values with limω→∞ σi(ω) > 0 and (p − q)singular values with limω→∞ σi(ω) = 0 and limω→∞ ω2σi(ω) > 0. Therefore, thereexist σ1 > 0 and ω2 > 0 such that

ω2σmin[G(jω) + GT (−jω)

] ≥ σ1, ∀ |ω| ≥ ω2 (C.34)

From (C.30), (C.32), and (C.34),

ω2σmin[G(jω − μ) + GT (−jω − μ)

] ≥ σ1 − μk1, ∀ |ω| ≥ ω3 (C.35)

where ω3 = max{ω1, ω2}. On the compact frequency interval [−ω3, ω3], we have

σmin[G(jω) + GT (−jω)

] ≥ σ2 > 0 (C.36)

Hence, from (C.30), (C.31), and (C.36),

σmin[G(jω − μ) + GT (−jω − μ)

] ≥ σ2 − μk0, ∀ |ω| ≤ ω3 (C.37)


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Choosing μ < min{σ1/k1, σ2/k0} ensures that G(jω−μ)+GT (−jω−μ) is positivedefinite for all ω ∈ R.

Necessity: Suppose G(s) is strictly positive real. There exists μ > 0 such thatG(s− μ) is positive real. It follows that G(s) is Hurwitz and positive real. Conse-quently,

G(jω) + GT (−jω) ≥ 0, ∀ ω ∈ R

Therefore,G(∞) + GT (∞) ≥ 0

Let {A, B, C, D} be a minimal realization of G(s). By Lemma 6.3, there exist P ,L, W , and ε that satisfy (6.14) through (6.16). Let Φ(s) = (sI −A)−1. We have

G(s) + GT (−s) = D + DT + CΦ(s)B + BT ΦT (−s)CT

Substitute for C by using (6.15) and for D + DT by using (6.16). Then,

G(s) + GT (−s) = WT W + (BT P + WT L)Φ(s)B + BT ΦT (−s)(PB + LT W )= WT W + WT LΦ(s)B + BT ΦT (−s)LT W

+ BT ΦT (−s)[−AT P − PA]Φ(s)B

Using (6.14) yields

G(s) + GT (−s) =[WT + BT ΦT (−s)LT

][W + LΦ(s)B] + εBT ΦT (−s)PΦ(s)B

From the last equation, it can be seen that G(jω) + GT (−jω) is positive definitefor all ω ∈ R, for if it were singular at some frequency ω, there would exist x ∈ Cp,x �= 0, such that

(x∗)T [G(jω) + GT (−jω)]x = 0 ⇒ (x∗)T BT ΦT (−jω)PΦ(jω)Bx = 0 ⇒ Bx = 0

Also,

(x∗)T [G(jω) + GT (−jω)]x = 0 ⇒ (x∗)T [W + LΦ(−jω)B]T [W + LΦ(jω)B] x = 0

Since Bx = 0, the preceding equation implies that Wx = 0. Hence,

(x∗)T [G(s) + GT (−s)]x ≡ 0, ∀ s

which contradicts the assumption that det[G(s) + GT (−s)] is not identically zero.Now if G(∞) + GT (∞) is positive definite, we are done. Otherwise, let M be anyp× (p− q) full rank matrix such that MT (D + DT )M = MT WT WM = 0. Then,WM = 0 and

MT [G(jω) + GT (−jω)]M = MT BT ΦT (−jω)(LT L + εP )Φ(jω)BM


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We note that BM must have full column rank; otherwise, there is x �= 0 such thatBMx = 0. Taking y = Mx yields

yT [G(jω) + GT (−jω)]y = 0, ∀ ω ∈ R

which contradicts the positive definiteness of G(jω) + GT (−jω). Now

limω→∞ω2MT [G(jω) + GT (−jω)]M = MT BT (LT L + εP )BM

The fact that BM has full column rank implies that MT BT (LT L + εP )BM ispositive definite.


Sufficiency: Suppose there exist P = PT > 0, L, and W that satisfy (6.11)through (6.13). Using V (x) = xT Px as a Lyapunov function for x = Ax, (6.11)shows that the origin of x = Ax is stable. Hence, A has no eigenvalues in Re[s] > 0.Let Φ(s) = (sI −A)−1. We have

G(s) + GT (s∗) = D + DT + CΦ(s)B + BT ΦT (s∗)CT

Substitute for C by using (6.12) and for D + DT by using (6.13). Then,

G(s) + GT (s∗) = WT W + (BT P + WT L)Φ(s)B + BT ΦT (s∗)(PB + LT W )= WT W + WT LΦ(s)B + BT ΦT (s∗)LT W

+ BT ΦT (s∗)[(s + s∗)P −AT P − PA]Φ(s)B

Using (6.11) yields

G(s) + GT (s∗) =[WT + BT ΦT (s∗)LT

][W + LΦ(s)B]

+ (s + s∗)BT ΦT (s∗)PΦ(s)B (C.38)

which shows that for all s in Re[s] ≥ 0, G(s) + GT (s∗) ≥ 0. It follows that, for allω for which jω is not a pole of any element of G(s), G(jω) + GT (−jω) is positivesemidefinite. It remains to show that G(s) satisfies the third condition of Defini-tion 6.4. Suppose jω0 is a pole of order m of any element of G(s). Then, for anyp-dimensional complex vector x, the values taken by (x∗)T G(s)x on a semicirculararc of arbitrarily small radius ρ, centered at jω0, are

(x∗)T G(s)x ≈ (x∗)T K0xρ−me−jmθ, −π

2≤ θ ≤ π

2

Therefore,

ρmRe[(x∗)T G(s)x] ≈ Re[(x∗)T K0x] cos mθ + Im[(x∗)T K0x] sinmθ


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For m > 1, this expression could have either sign, while (C.38) implies that it isnonnegative. Hence, m must be limited to one. For m = 1, choosing θ near −π/2,0, and π/2 shows that Im[(x∗)T K0x] = 0 and Re[(x∗)T K0x] ≥ 0. Hence, K0 ispositive semidefinite Hermitian.

Necessity: We prove necessity first for the special case when A is Hurwitz. Thenwe extend the proof to the general case when A may have eigenvalues on the imag-inary axis.Special case: The proof uses a spectral factorization result, which we quote with-out proof.

Lemma C.5 Let the p×p proper rational transfer function matrix U(s) be positivereal and Hurwitz. Then, there exists an r × p proper rational Hurwitz transferfunction matrix V (s) such that

U(s) + UT (−s) = V T (−s)V (s) (C.39)

where r is the normal rank of U(s) + UT (−s), that is, the rank over the field ofrational functions of s. Furthermore, rank V (s) = r for Re[s] > 0. Proof: See [214, Theorem 2].

Suppose now that G(s) is positive real and Hurwitz, and recall that {A, B, C, D}is a minimal realization of G(s). From Lemma C.5, there exists an r × p transferfunction matrix V (s) such that (C.39) is satisfied. Let {F, G, H, J} be a minimalrealization of V (s). The matrix F is Hurwitz, since V (s) is Hurwitz. It can be easilyseen that {−FT , HT ,−GT , JT } is a minimal realization of V T (−s). Therefore,

{A1,B1, C1,D1} ={ [

F 0HT H −FT

],

[G

HT J

],[

JT H −GT], JT J

}

is a realization of the cascade connection V T (−s)V (s). By checking controllabilityand observability and by using the property that rank V (s) = r for Re[s] > 0, itcan be seen that this realization is minimal. Let us show the controllability test;6

the observability test is similar. By writing⎡⎣ I 0

H(sI − F )−1 I

⎤⎦⎡⎣ sI − F G

−H J

⎤⎦ =

⎡⎣ sI − F G

0 H(sI − F )−1G + J

⎤⎦

=

⎡⎣ sI − F G

0 V (s)

⎤⎦

6The controllability argument uses the fact that V T (−s) has no transmission zeros at the polesof V (s).


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it can be seen that

rank V (s) = r, ∀ Re[s] > 0 ⇔ rank

⎡⎣ sI − F G

−H J

⎤⎦ = nF + r, ∀ Re[s] > 0

where nF is the dimension of F . Now, show controllability of (A1,B1) by contra-diction. Suppose (A1,B1) is not controllable. Then, there are a complex number λand a vector w ∈ CnF +r, partitioned into nF and r subvectors, such that

(w∗1)T F + (w∗2)T HT H = λ(w∗1)T (C.40)−(w∗2)T FT = λ(w∗2)T (C.41)

(w∗1)T G + (w∗2)T HT J = 0 (C.42)

Equation (C.41) shows that Re[λ] > 0, since F is Hurwitz, and (C.40) and (C.42)show that

[(w∗1)T (w∗2)T HT

] ⎡⎣ λI − F G

−H J

⎤⎦ = 0 ⇒ rank V (λ) < r

which contradicts the fact that rank V (s) = r for Re[s] > 0. Thus, (A1,B1) iscontrollable.

Consider the Lyapunov equation

KF + FT K = −HT H

Because the pair (F, H) is observable, there is a unique positive definite solutionK. This fact is shown in Exercise 4.22. Using the similarity transformation[

I 0K I

]

we obtain the following alternative minimal realization of V T (−s)V (s):

{A2,B2, C2,D2} ={ [F 00 −FT

],

[G

KG + HT J

],[

JT H + GT K −GT], JT J

}

On the other hand, {−AT , CT ,−BT , DT } is a minimal realization of UT (−s).Therefore,

{A3,B3, C3,D3} ={ [

A 00 −AT

],

[BCT

],[

C −BT], D + DT

}

is a realization of the parallel connection U(s) + UT (−s). Since the eigenvaluesof A are in the open left-half plane, while the eigenvalues of −AT are in the open


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right-half plane, it can be easily seen that this realization is minimal. Thus, due to(C.39), {A2,B2, C2,D2} and {A3,B3, C3,D3} are equivalent minimal realizations ofthe same transfer function. Therefore, they have the same dimension and there isa nonsingular matrix T such that7

A2 = TA3T−1, B2 = TB3, C2 = C3T−1, JT J = D +DT

The matrix T must be a block-diagonal matrix. To see this point, partition Tcompatibly as

T =[

T11 T12T21 T22

]Then, the matrix T12 satisfies the equation

FT12 + T12AT = 0

Premultiplying by exp(Ft) and postmultiplying by exp(AT t), we obtain

0 = exp(Ft)[FT12 + T12AT ] exp(AT t) =

d

dt[exp(Ft)T12 exp(AT t)]

Hence, exp(Ft)T12 exp(AT t) is constant for all t ≥ 0. In particular, since exp(0) = I,

T12 = exp(Ft)T12 exp(AT t) → 0 as t →∞Therefore, T12 = 0. Similarly, we can show that T21 = 0. Consequently, the matrixT11 is nonsingular and

F = T11AT−111 , G = T11B, JT H + GT K = CT−1

11

DefineP = TT

11KT11, L = HT11, W = J

It can be easily verified that P , L, and W satisfy the equations

PA + AT P = −LT L, PB = CT − LT W, WT W = D + DT

which completes the proof for the special case.General case: Suppose now that A has eigenvalues on the imaginary axis. Thereis a nonsingular matrix Q such that

QAQ−1 =[

A0 00 An

], QB =

[B0Bn

], CQ−1 =

[C0 Cn

]where A0 has eigenvalues on the imaginary axis and An has eigenvalues with nega-tive real parts. The transfer function G(s) can be written as G(s) = G0(s)+Gn(s),where G0(s) = C0(sI −A0)−1B0 has all poles on the imaginary axis, and Gn(s) =

7See [35, Theorem 5-20].


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Cn(sI − An)−1bn + D has all poles in the open left-half plane. It follows thatthe poles of G0(s) are simple and the corresponding residue matrices are positivesemidefinite Hermitian. Due to this property, we can choose Q such that

A0 + AT0 = 0, C0 = BT

0 (C.43)

To see this fact, notice that G0(s) can be written as

G0(s) =1sF0 +

m∑i=1

1s2 + ω2

i

(Fis + Hi) =1sF0 +

m∑i=1

[1

s− jωiRi +

1s + jωi

R∗i

]

where F0 is positive semidefinite symmetric and Ri is positive semidefinite Hermi-tian. If, for each term in this summation, we can find a minimal realization thathas the property (C.43), the parallel connection will be a minimal realization ofG0(s) that has the same property. It is sufficient to consider the terms (1/s)F0and [1/(s2 + ω2

i )](Fis + Hi). If r0 = rank F0, (1/s)F0 has a minimal realiza-tion of dimension r0, given by {0, N0, N

T0 }, where F0 = NT

0 N0. If ri = rank Ri,[1/(s2 + ω2

i )](Fis + Hi) has a minimal realization of dimension 2ri, given by

Ai =[

0 ωiI−ωiI 0

], Bi =

[Mi1Mi2

], Ci =

[MT

i1 MTi2]

whereMi1 =

1√2(Ni + N∗

i ), Mi2 =j√2(Ni −N∗

i ), Ri = (N∗i )T Ni

It is clear that {Ai, Bi, Ci} has the property (C.43).Since Gn(s) is positive real and Hurwitz, from the special case treated earlier,

there exist matrices Pn = Pn > 0, Ln, and W such that

PnAn + ATnPn = −LT

nLn, PnBn = CTn − LT

nW, WT W = D + DT

It can be easily verified that

P = QT

[I 00 Pn

]Q, L =

[0 Ln

]Q

and W satisfy equations (6.11) through (6.13).


We start by writing a time-domain version of the infinite-dimensional equation(7.20). Consider the space S of all half-wave symmetric periodic signals of funda-mental frequency ω, which have finite energy on any finite interval. A signal y ∈ Scan be represented by its Fourier series

y(t) =∑

k odd

ak exp(jkωt),∑

k odd

|ak|2 <∞


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Define a norm on S by

‖y‖2 =ω

π

∫ 2π/ω

0y2(t) dt = 2

∑k odd

|ak|2

With this norm, S is a Banach space. Define gk(t− τ) by

gk(t− τ) =ω

π{G(jkω) exp[jkω(t− τ)] + G(−jkω) exp[−jkω(t− τ)]}

For odd integers m and k > 0, we have

∫ π/ω

0gk(t− τ) exp(jmωτ) dτ =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

G(jkω) exp(jkωt), if m = k

G(−jkω) exp(−jkωt), if m = −k

0, if |m| �= k

(C.44)

Define the linear mapping g and the nonlinear mapping gψ on S by

gy =∫ π/ω

0

∑k odd; k>0

gk(t− τ)y(τ) dτ

gψy =∫ π/ω

0

∑k odd; k>0

gk(t− τ)ψ(y(τ)) dτ

wherey(t) =

∑k odd

ak exp(jkωt) and ψ(y(t)) =∑

k odd

ck exp(jkωt)

Using (C.44), it can be seen that

gy =∑

k odd

G(jkω)ak exp(jkωt)

gψy =∑

k odd

G(jkω)ck exp(jkωt)

With these definitions, the condition for existence of half-wave symmetric periodicoscillation can be written as

y = −gψy (C.45)

Equation (C.45) is equivalent to (7.20). In order to separate the effect of the higherharmonics from that of the first harmonic, define a mapping P1 by

P1y = y1 = a1 exp(jωt) + a1 exp(−jωt) = 2Re[a1 exp(jωt)]

and a mapping Ph by

Phy = yh = y − y1 =∑

k odd; |k|�=1

ak exp(jkωt)


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Without loss of generality, we take a1 = a/2j so that y1(t) = a sin ωt. Solving(C.45) is equivalent to solving both (C.46) and (C.47):

yh = −Phgψ(y1 + yh) (C.46)y1 = −P1gψ(y1 + yh) (C.47)

By evaluating the right-hand side of (C.47), it can be seen that this equation isequivalent to (7.35). The error term δΨ, defined after (7.36), satisfies

P1gψy1 − P1gψ(y1 + yh) = 2Re[G(jω)a1δΨ exp(jωt)] (C.48)

Thus, to obtain a bound on δΨ, we need to find a bound on yh, which we willfind by using the contraction mapping theorem, without solving (C.46). By adding[Phg(β + α)/2]yh to both sides of (C.46), we can rewrite it as(

I + Phgβ + α

2

)yh = −Phg

[ψ(y1 + yh)− β + α

2yh

](C.49)

Let us consider the linear mapping K = I + Phg(β + α)/2, which appears on theleft-hand side of (C.49). It maps S into S. Given any z ∈ S defined by

z(t) =∑

k odd

bk exp(jkωt)

we consider the linear equation Kx = z and seek a solution x in S. Representing xas

x(t) =∑

k odd

dk exp(jkωt)

we have(I + Phg

β + α

2

)x = x1 +

∑k odd; |k|�=1

[1 +

β + α

2G(jkω)

]dk exp(jkωt)

Hence, the linear equation Kx = z has a unique solution if

infk odd; |k|�=1

∣∣∣∣1 +β + α

2G(jkω)

∣∣∣∣ �= 0 (C.50)

In other words, condition (C.50) guarantees that the linear mapping K has aninverse. This condition is always satisfied if ω ∈ Ω, because the left-hand side of(C.50) can vanish only if ρ(ω) = 0. Denote the inverse of K by K−1 and rewrite(C.49) as

yh = −K−1Phg

[ψ(y1 + yh)− β + α

2(y1 + yh)

]def= Tyh


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where we have used the fact that Phgy1 = 0. We would like to apply the contractionmapping theorem to the equation yh = Tyh. Clearly, T maps S into S. We needto verify that T is a contraction on S. To that end, consider

Ty(2) − Ty(1) = K−1Phg[ψT (y1 + y(2))− ψT (y1 + y(1))

]where

ψT (y) = ψ(y)− β + α

2y

LetψT (y1 + y(2))− ψT (y1 + y(1)) =

∑k odd; |k|�=1

ek exp(jkωt)

Then,

∥∥∥Ty(2) − Ty(1)∥∥∥2

= 2∑

k odd; |k|�=1

∣∣∣∣ G(jkω)1 + [(β + α)/2]G(jkω)

∣∣∣∣2

|ek|2

≤{

supk odd; |k|�=1

∣∣∣∣ G(jkω)1 + [(β + α)/2]G(jkω)

∣∣∣∣}2

×∥∥∥ψT (y1 + y(2))− ψT (y1 + y(1))

∥∥∥2

Due to the slope restriction on ψ, we have

∣∣∣ψT (y1 + y(2))− ψT (y1 + y(1))∣∣∣ ≤ (β − α

2

) ∣∣∣y(2) − y(1)∣∣∣

Moreover,

supk odd; |k|�=1

∣∣∣∣ G(jkω)1 + [(β + α)/2]G(jkω)

∣∣∣∣ ≤ 1ρ(ω)

where ρ(ω) is defined by (7.38). Hence,

∥∥∥Ty(2) − Ty(1)∥∥∥ ≤ 1

ρ(ω)

(β − α

2

)∥∥∥y(2) − y(1)∥∥∥

Since1

ρ(ω)

(β − α

2

)< 1, ∀ ω ∈ Ω

we conclude that as long as ω ∈ Ω, T is a contraction mapping. Thus, by thecontraction mapping theorem, the equation yh = Tyh has a unique solution. Notingthat T (−y1) = 0, we rewrite the equation yh = Tyh as

yh = Tyh − T (−y1)


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and conclude that

‖yh‖ ≤ 1ρ(ω)

(β − α

2

)‖yh‖+

1ρ(ω)

(β − α

2

)a

Therefore,

‖yh‖ ≤ a[(β − α)/2] /ρ(ω)1− [(β − α)/2] /ρ(ω)

=a[(β − α)/2]

ρ(ω)− [(β − α)/2]=

2σ(ω)aβ − α

which proves (7.40). To prove (7.41), consider (C.47) rewritten as

y1 + P1gψy1 = P1g[ψy1 − ψ(y1 + yh)] (C.51)

by adding P1gψy1 to both sides. Taking norms on both sides of (C.51) gives

|1 + G(jω)Ψ(a)|a ≤ |G(jω)|(

β − α

2

)‖yh‖ ≤ |G(jω)|σ(ω)a

from which, we can calculate the bound

|δΨ| =∣∣∣∣ 1G(jω)

+ Ψ(a)∣∣∣∣ =

∣∣∣∣1 + G(jω)Ψ(a)G(jω)

∣∣∣∣ ≤ σ(ω)

which completes the proof of the lemma.


If, in the proof of Lemma 7.1, we had defined P1 = 0 and Ph = I, then the mappingT would still be a contraction mapping if ω ∈ Ω. Therefore, y = yh = 0 would bethe unique solution of yh = Tyh. This proves that there is no half-wave symmetricperiodic solutions with fundamental frequency ω ∈ Ω. The necessity of the condition∣∣∣∣ 1

G(jω)+ Ψ(a)

∣∣∣∣ ≤ σ(ω)

shows that there would be no half-wave symmetric periodic solutions with funda-mental frequency ω ∈ Ω′ if the corresponding error circle does not intersect the−Ψ(a) locus. Thus, we are left with the third part of the theorem where we wouldlike to show that for each complete intersection defining Γ, there is a half-wavesymmetric periodic solutions with (ω, a) ∈ Γ. The proof of this part uses a resultfrom degree theory, so let us explain that result first.

Suppose we are given a continuously differentiable function φ : D → Rn, whereD ⊂ Rn is open and bounded. Let p ∈ Rn be a point such that φ(x) = p for somex inside D, but φ(x) �= p on the boundary ∂D of D. We are interested in showingthat φ(x) = p has a solution in D, where φ(x) is a perturbation of φ(x). Degree


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theory achieves this by ensuring that no solution leaves D as φ is perturbed to φ;that is why no solutions were allowed on the boundary ∂D. Assume that at everysolution xi ∈ D of φ(x) = p, the Jacobian matrix [∂φ/∂x] is nonsingular. Definethe degree of φ at p relative to D by

d(φ, D, p) =∑

xi=φ−1(p)

sgn{

det[∂φ

∂x(xi)

]}

Notice that if φ(x) �= p ∀ x ∈ D, the degree is zero. The two basic properties of thedegree are as follows:8

• If d(φ, D, p) �= 0, then φ(x) = p has at least one solution in D.

• If η : D × [0, 1] → Rn is continuous and η(x, μ) �= 0 for all x ∈ ∂D and allμ ∈ [0, 1], then d[η(·, μ), D, p] is the same for all μ ∈ [0, 1].

The second property is known as the homotopic invariance of d.Let us get back to our problem and define

φ(ω, a) = Ψ(a) +1

G(jω)

on Γ; φ is a complex variable. By taking the real and imaginary parts of φ ascomponents of a second-order vector, we can view φ as a mapping from Γ into R2.By assumption, the equation φ(w, a) = 0 has a unique solution (ωs, as) in Γ. TheJacobian of φ with respect to (ω, a) at (ωs, as) is given by

⎡⎣

ddaΨ(a)

∣∣a=as

−Ψ2(as){

ddω Re[G(jω)]

}ω=ωs

0 −Ψ2(as){

ddω Im[G(jω)]

}ω=ωs

⎤⎦

The assumptions

d

daΨ(a)

∣∣∣∣a=as

�= 0;d

dωIm[G(jω)]

∣∣∣∣ω=ωs

�= 0

guarantee that the Jacobian is nonsingular. Thus,

d(φ,Γ, 0) = ±1

We are interested in showing that

φ(ω, a) def=1

G(jω)+ Ψ(a)− δΨ(ω, a) = 0 (C.52)

8See [26] for the proof of these properties.


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has a solution in Γ. This will be the case if we can show that

d(φ,Γ, 0) �= 0

To that end, define

η(ω, a, μ) = (1− μ)φ(ω, a) + μφ(ω, a) = φ(ω, a)− μδΨ(ω, a)

for μ ∈ [0, 1], so that η = φ at μ = 0 and η = φ at μ = 1. It can be verified that∣∣∣∣Ψ(a) +

1G(jω)

∣∣∣∣ ≥ σ(ω), ∀ (ω, a) ∈ ∂Γ (C.53)

For example, if we take the boundary a = a1, then, with reference to Figure 7.20,the left-hand side of (C.53) is the length of the line connecting the point on thelocus of −Ψ(a) corresponding to a = a1 to the a point on the locus of 1/G(jω)with ω1 ≤ ω ≤ ω2. By construction, the former point is outside (or on) the errorcircle centered at the latter one. Therefore, the length of the line connecting thetwo points must be greater than (or equal to) the radius of the error circle, that is,σ(ω). Using (C.53), we have

|η(ω, a, μ)| ≥ |φ(ω, a)| − μ|δΨ(ω, a)|=

∣∣∣∣Ψ(a) +1

G(jω)

∣∣∣∣− μ|δΨ(ω, a)| ≥ σ(ω)− μσ(ω)

where we have used the bound (7.41). Thus, for all 0 ≤ μ < 1, the right-handside of the last inequality is positive, which implies that η(ω, a, μ) �= 0 on ∂Γ whenμ < 1. There is no loss of generality in assuming that η(ω, a, 1) �= 0 on ∂Γ, becauseequality would imply we had found a solution as required. Thus, by the homotopicinvariance property of d, we have

d(φ,Γ, 0) = d(φ,Γ, 0) �= 0

Therefore, (C.52) has a solution in Γ, which concludes the proof of the theorem.

C.15 Proof of Theorems 8.1 and 8.3

The main element in the proofs of these two theorems is a contraction mappingargument, which is used almost identically in the two proofs. To avoid repetition,we will state and prove a lemma that captures the needed contraction mappingargument and then use it to prove the two theorems. The statement of the lemmaappears to be very similar to the statement of Theorem 8.1, but it has an additionalclaim that is needed in Theorem 8.3.


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C.15. PROOF OF THEOREMS 8.1 AND 8.3 691

Lemma C.6 Consider the system

y = Ay + f(y, z) (C.54)z = Bz + g(y, z) (C.55)

where y ∈ Rk, z ∈ Rm, the eigenvalues of A have zero real parts, the eigenvaluesof B have negative real parts, and f and g are twice continuously differentiablefunctions that vanish together with their first derivatives at the origin. Then, thereexist δ > 0 and a continuously differentiable function η(y), defined for all ‖y‖ < δ,such that z = η(y) is a center manifold for (C.54)–(C.55). Moreover, if ‖g(y, 0)‖ ≤k‖y‖p for all ‖y‖ ≤ r, where p > 1 and r > 0, then there is c > 0 such that‖η(y)‖ ≤ c‖y‖p. �

Proof: It is more convenient to show the existence of a center manifold whensolutions in the manifold are defined for all t ∈ R. A center manifold for (C.54)–(C.55) may, in general, be only local; that is, a solution in the manifold may bedefined only on an interval [0, t1) ⊂ R. Therefore, the following idea is used inthe proof: We consider a modified equation that is identical to (C.54)–(C.55) inthe neighborhood of the origin, but that has some desired global properties whichensure that a solution in a center manifold will be defined for all t. We prove theexistence of a center manifold for the modified equation. Since the two equationsagree in the neighborhood of the origin, this proves the existence of a (local) centermanifold for the original equation.

Let ψ : Rk → [0, 1] be a smooth (continuously differentiable infinitely manytimes) function9 with ψ(y) = 1 when ‖y‖ ≤ 1 and ψ(y) = 0 when ‖y‖ ≥ 2. Forε > 0, define F and G by

F (y, z) = f(yψ(y

ε

), z)

; G(y, z) = g(yψ(y

ε

), z)

The functions F and G are twice continuously differentiable, and they, togetherwith their first partial derivatives, are globally bounded in y; that is, whenever‖z‖ ≤ k1, the function is bounded for all y ∈ Rk. Consider the modified system

y = Ay + F (y, z) (C.56)z = Bz + G(y, z) (C.57)

9An example of such a function in the scalar (k = 1) case is ψ(y) = 1 for |y| ≤ 1, ψ(y) = 0 for|y| ≥ 2, and

ψ(y) = 1− 1b

∫ |y|

1

exp( −1

x− 1

)exp( −1

2− x

)dx, for 1 < |y| < 2

where

b =

∫ 2

1

exp( −1

x− 1

)exp( −1

2− x

)dx


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We prove the existence of a center manifold for (C.56)–(C.57). Let X denote the setof all globally bounded, continuous functions η : Rk → Rm. With supy∈Rk ‖η(y)‖as a norm, X is a Banach space. For c1 > 0, c2 > 0, and c3 > 0, let S ⊂ X be theset of all continuously differential functions η : Rk → Rm such that

η(0) = 0,∂η

∂y(0) = 0, ‖η(y)‖ ≤ c1,

∥∥∥∥∂η

∂y(y)∥∥∥∥ ≤ c2,

∥∥∥∥∂η

∂y(y)− ∂η

∂y(x)∥∥∥∥ ≤ c3‖y − x‖

for all x, y ∈ Rk. To show that S is closed, let ηi(y) be a convergent sequence in Sand show that η(y) = limi→∞ ηi(y) belongs to S. The key step is to show that η(y)is continuously differentiable. The rest can be shown by contradiction. Becausecontinuous differentiability can be shown component wise, we do it for a scalar η.Let v be an arbitrary vector in Rk, with ‖v‖ = 1, and μ be a positive constant. Bythe mean value theorem,

ηi(y + μv)− ηi(y) =∂ηi

∂y(y + αiμv)μv

ηj(y + μv)− ηj(y) =∂ηj

∂y(y + αjμv)μv

where 0 < αi < 1 and 0 < αj < 1. By adding and subtracting terms, we can writethe following equation:[

∂ηi

∂y(y)− ∂ηj

∂y(y)]

μv =[∂ηi

∂y(y)− ∂ηi

∂y(y + αiμv)

]μv

−[∂ηj

∂y(y)− ∂ηj

∂y(y + αjμv)

]μv

+ [ηi(y + μv)− ηj(y + μv)]− [ηi(y)− ηj(y)]

Given ε > 0, find integers i0 and j0 large enough that

supy∈Rk

‖ηi(y)− ηj(y)‖ <ε2

16c3

for all i > i0 and j > j0, which is possible because {ηi} is convergent. Using thepreceding inequality and∥∥∥∥

[∂η�

∂y(y)− ∂η�

∂y(y + α�μv)

]μv

∥∥∥∥ ≤ c3α�μ2‖v‖2 < c3μ

2 for � = i or j

it can be shown that ∥∥∥∥[∂ηi

∂y(y)− ∂ηj

∂y(y)]

v

∥∥∥∥ < 2c3μ +ε2

8c3μ

Taking μ = ε/(4c3) yields ∥∥∥∥∂ηi

∂y(y)− ∂ηj

∂y(y)∥∥∥∥ < ε


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Hence, ∂ηi/∂y is a Cauchy sequence in the Banach space of globally bounded,continuous functions from Rk into Rk. Consequently, it converges to a continuousfunction J(y). It follows that η(y) is differentiable and ∂η/∂y = J(y).10

For a given η ∈ S, consider the system

y = Ay + F (y, η(y)) (C.58)z = Bz + G(y, η(y)) (C.59)

Due to the boundedness of η(y) and [∂η/∂y], the right-hand side of (C.58) is globallyLipschitz in y. Therefore, for every initial state y0 ∈ Rk, (C.58) has a uniquesolution defined for all t. We denote this solution by y(t) = π(t; y0, η), whereπ(0; y0, η) = y0; the solution is parameterized in the fixed function η. Equation(C.59) is linear in z; therefore, its solution is given by

z(t) = exp[B(t− τ)]z(τ)

+∫ t

τ

exp[B(t− λ)]G(π(λ− τ ; y(τ), η), η(π(λ− τ ; y(τ), η))) dλ

Multiply through by exp[−B(t− τ)], move the integral term to the other side, andchange the integration variable from λ to s = λ− τ to obtain

z(τ) = exp[−B(t− τ)]z(t)

+∫ 0

t−τ

exp(−Bs)G(π(s; y(τ), η), η(π(s; y(τ), η))) ds

This expression is valid for any t ∈ R. Taking the limit t → −∞ results in

z(τ) =∫ 0

−∞exp(−Bs)G(π(s; y(τ), η), η(π(s; y(τ), η))) ds (C.60)

Let us rewrite the right-hand side of (C.60) with y(τ) replaced by y and denote itby (Pη)(y).

(Pη)(y) =∫ 0

−∞exp(−Bs)G(π(s; y, η), η(π(s; y, η))) ds (C.61)

With this definition, (C.60) says that z(τ) = (Pη)(y(τ)) for all τ . Hence, z =(Pη)(y) defines an invariant manifold for (C.58)–(C.59) parameterized in η.Consider (C.56)–(C.57). If η(y) is a fixed point of the mapping (Pη)(y), that is,

η(y) = (Pη)(y)

then z = η(y) is a center manifold for (C.56)–(C.57). This fact can be seen asfollows: First, using the properties of η ∈ S and the fact that y = 0 is an equilibriumpoint of (C.58), it can be seen from (C.61) that

(Pη)(0) = 0;∂(Pη)

∂y(0) = 0

10See [111, Theorem 9.1].


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Second, since z = (Pη)(y) is an invariant manifold for (C.58)–(C.59), (Pη)(y)satisfies the partial differential equation

∂

∂y(Pη)(y)[Ay + F (y, (Pη)(y))] = B(Pη)(y) + G(y, (Pη)(y))

If η(y) = (Pη)(y), then clearly η(y) satisfies the same partial differential equation;hence, it is a center manifold for (C.56)–(C.57). It remains now to show thatthe mapping (Pη) has a fixed point, which will be done by an application of thecontraction mapping theorem. We want to show that the mapping Pη maps S intoitself and is a contraction mapping on S. From the definitions of F and G, there isa nonnegative continuous function ρ(ε) with ρ(0) = 0 such that

‖F (y, z)‖ ≤ ερ(ε);∥∥∥∥∂F

∂y(y, z)

∥∥∥∥ ≤ ρ(ε);∥∥∥∥∂F

∂z(y, z)

∥∥∥∥ ≤ ρ(ε) (C.62)

‖G(y, z)‖ ≤ ερ(ε);∥∥∥∥∂G

∂y(y, z)

∥∥∥∥ ≤ ρ(ε);∥∥∥∥∂G

∂z(y, z)

∥∥∥∥ ≤ ρ(ε) (C.63)

for all y ∈ Rk and all z ∈ Rm with ‖z‖ < ε. Because the eigenvalues of B havenegative real parts, there exist positive constants β and C such that for s ≤ 0,

‖ exp(−Bs)‖ ≤ C exp(βs) (C.64)

Since the eigenvalues of A have zero real parts, for each α > 0, there is a positiveconstant M(α) (which may tend to ∞ as α → 0) such that for s ∈ R,

‖ exp(As)‖ ≤M(α) exp(α|s|) (C.65)

To show that Pη maps S into itself, we need to show that there are positive constantsc1, c2, and c3 such that if η(y) is continuously differentiable and satisfies

‖η(y)‖ ≤ c1;∥∥∥∥∂η

∂y(y)∥∥∥∥ ≤ c2;

∥∥∥∥∂η

∂y(y)− ∂η

∂y(x)∥∥∥∥ ≤ c3‖y − x‖

for all x, y ∈ Rk, then (Pη)(y) is continuously differentiable and satisfies the sameinequalities. Continuous differentiability of (Pη)(y) is obvious from (C.61). Toverify the inequalities, we need to use the estimates on F and G provided by (C.62)and (C.63); therefore, we choose c1 to satisfy 0.5ε < c1 < ε. Using (C.64) and theestimates on G and η, we have from (C.61),

‖(Pη)(y)‖ ≤∫ 0

−∞‖ exp(−Bs)‖ ‖G‖ ds ≤

∫ 0

−∞C exp(βs) ερ(ε) ds =

Cερ(ε)β

The upper bound on (Pη)(y) will be less than c1 for sufficiently small ε. Letπy(t; y, η) denote the Jacobian of π(t; y, η) with respect to y. It satisfies the varia-tional equation

πy =[A +

(∂F

∂y

)+(

∂F

∂z

) (∂η

∂y

)]πy, πy(0; y, η) = I


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where (∂F

∂y

)=(

∂F

∂y

)(π(t; y, η), η((π(t; y, η)))

with similar expressions for ∂F/∂z and ∂η/∂y. Hence, for t ≤ 0,

πy(t; y, η) = exp(At)−∫ 0

t

exp[A(t− s)][(

∂F

∂y

)+(

∂F

∂z

) (∂η

∂y

)]πy(s; y, η) ds

Using (C.65) and the estimates on F and η, we obtain

‖πy(t; y, η)‖ ≤ M(α) exp(−αt) +∫ 0

t

M(α) exp[α(s− t)](1 + c2)ρ(ε)‖πy(s; y, η)‖ ds

Multiply through by exp(αt) and apply the Gronwall–Bellman inequality to showthat11

‖πy(t; y, η)‖ ≤ M(α) exp(−γt)

where γ = α + M(α)(1 + c2)ρ(ε). Using this bound, as well as (C.64) and the esti-mates on G and η, we proceed to calculate a bound on the Jacobian [∂(Pη)(y)/∂y].From (C.61),

∂(Pη)(y)∂y

=∫ 0

−∞exp(−Bs)

[(∂G

∂y

)+(

∂G

∂z

) (∂η

∂y

)]πy(s; y, η) ds

Thus, ∥∥∥∥∂(Pη)(y)∂y

∥∥∥∥ ≤∫ 0

−∞C exp(βs)(1 + c2)ρ(ε)M(α) exp(−γs) ds

=C(1 + c2)ρ(ε)M(α)

β − γ

provided ε and α are small enough that β > γ. This bound on the Jacobian of(Pη)(y) will be less than c2 for sufficiently small ε. To show that the Jacobian[∂(Pη)(y)/∂y] is Lipschitz with a Lipschitz constant c3, we note that the Jacobianmatrices [∂F/∂y], [∂F/∂z], [∂G/∂y], and [∂G/∂z] satisfy Lipschitz inequalities ofthe form ∥∥∥∥∂F

∂y(y, z)− ∂F

∂y(x, w)

∥∥∥∥ ≤ L[‖y − x‖+ ‖z − w‖]

11Because t appears in the lower limit of the integral, you actually apply a version of theGronwall–Bellman inequality that says: If

y(t) ≤ λ(t) +

∫ a

t

μ(s)y(s) ds

then

y(t) ≤ λ(t) +

∫ a

t

λ(s)μ(s) exp

[∫ s

t

μ(τ) dτ

]ds


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for all x, y ∈ Rk and z, w ∈ Bε. By working with ε < ε∗ for some ε∗ > 0, we canchoose the Lipschitz constant L independent of ε. Moreover, we can use the same Lfor all four Jacobian matrices. Using these inequalities and the Gronwall–Bellmaninequality, we can repeat the foregoing derivations to show that

‖πy(t; y, η)− πy(t; x, η)‖ ≤ L1 exp(−2γt)‖y − x‖for all x, y ∈ Rk and t ≤ 0, where L1 = [(1 + c2)2L + ρ(ε)c3]M3(α)/γ. The lastinequality can be used to show that∥∥∥∥∂(Pη)

∂y(y)− ∂(Pη)

∂y(x)∥∥∥∥ ≤ CL1(2γ − α)

M(β − 2γ)‖y − x‖

provided β − 2γ > 0. Choose α and ε small enough that β − 2γ > β/2. Then,∥∥∥∥∂(Pη)∂y

(y)− ∂(Pη)∂y

(x)∥∥∥∥ ≤ [L2 + ρ(ε)L3c3]‖y − x‖

where L2 and L3 are independent of ε and c3. Choosing c3 > L2, we can choose εsmall enough that L2 + ρ(ε)L3c3 < c3. Thus, we have shown that, for sufficientlysmall c1 and sufficiently large c3, the mapping Pη maps S into itself. To show thatit is a contraction on S, let η1(y) and η2(y) be two functions in S. Let π1(t) andπ2(t) be the corresponding solutions of (C.58), which start at y; that is,

πi(t) = π(t; y, ηi), i = 1, 2

Using the estimates given by (C.62) and (C.63), it can be shown that

‖F (π2, η2(π2))− F (π1, η1(π1))‖ ≤ (1 + c2)ρ(ε)‖π2 − π1‖+ ρ(ε) supy∈Rk

‖η2 − η1‖

‖G(π2, η2(π2))−G(π1, η1(π1))‖ ≤ (1 + c2)ρ(ε)‖π2 − π1‖+ ρ(ε) supy∈Rk

‖η2 − η1‖

From (C.58), ‖π2 − π1‖ satisfies

‖π2(t)− π1(t)‖ ≤∫ 0

t

M(α) exp[α(s− t)][ρ(ε) supy∈Rk

‖η2 − η1‖

+ (1 + c2)ρ(ε)‖π2(s)− π1(s)‖] ds

≤ 1α

M(α)ρ(ε) supy∈Rk

‖η2 − η1‖ exp(−αt)

+∫ 0

t

(γ − α) exp[α(s− t)]‖π2(s)− π1(s)‖ ds

where we used γ = α+M(α)(1+ c2)ρ(ε) and the fact that π1(0) = π2(0). Multiplythrough by exp(αt) and apply the Gronwall–Bellman inequality to show that

‖π2(t)− π1(t)‖ ≤ 1α

M(α)ρ(ε) supy∈Rk

‖η2 − η1‖ exp(−γt)


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Using this inequality in

(Pη2)(y)− (Pη1)(y) =∫ 0

−∞exp(−Bs) [G(π2, η2(π2))−G(π1, η1(π1))] ds

we obtain

‖(Pη2)(y)− (Pη1)(y)‖ ≤∫ 0

−∞Ceβs[(1 + c2)ρ(ε)‖π2(s)− π1(s)‖

+ ρ(ε) supy∈Rk

‖η2 − η1‖ ] ds

≤ Cρ(ε) supy∈Rk

‖η2 − η1‖[

1β

+∫ 0

−∞eβs(1 + c2)

1α

M(α)ρ(ε)e−γs ds

]≤ b sup

y∈Rk

‖η2 − η1‖

where

b = Cρ(ε)[

1β

+γ − α

α(β − γ)

]

By choosing ε small enough, we can ensure that b < 1; hence, Pη is a contractionmapping on S. Thus, by the contraction mapping theorem, the mapping Pη has afixed point in S.

Suppose now that ‖g(y, 0)‖ ≤ k‖y‖p. The function G(y, 0) satisfies the samebound. Consider the closed subset

Y = {η ∈ S | ‖η(y)‖ ≤ c4‖y‖p}

where c4 is a positive constant to be chosen. To complete the proof of the lemma,we want to show that the fixed point of the mapping Pη is in Y , which will be thecase if we can show that c4 can be chosen such that Pη maps Y into itself. Usingthe estimate on G provided by (C.63), we have

‖G(y, η(y))‖ ≤ ‖G(y, 0)‖+ ‖G(y, η(y))−G(y, 0)‖ ≤ k‖y‖p + ρ(ε)‖η(y)‖

Since, in the set Y , ‖η(y)‖ ≤ c4‖y‖p,

‖G(y, η(y))‖ ≤ [k + c4ρ(ε)]‖y‖p

Using this estimate in (C.61) yields

‖(Pη)(y)‖ ≤∫ 0

−∞C exp(βs) [k + c4ρ(ε)]‖π(s; y, η)‖p ds


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Because π(t; 0, η) = 0 and ‖πy(t; y, η)‖ ≤ M(α) exp(−γt), it can be shown, as inthe proof of Lemma 3.1, that

‖π(t; y, η)‖ ≤M(α) exp(−γt)‖y‖for t ≤ 0. Thus,

‖(Pη)(y)‖ ≤ C[k + c4ρ(ε)]Mp(α)β − pγ

‖y‖p def= c5‖y‖p

provided ε and α are small enough that β − pγ > 0. By choosing c4 large enoughand ε small enough, we have c5 < c4. Therefore, (Pη) maps Y into itself, whichcompletes the proof of the lemma. �

Proof of Theorem 8.1

It follows from Lemma C.6 with A = A1, B = A2, f = g1, and g = g2. �

Proof of Theorem 8.3

Define μ(y) = h(y)−φ(y). Using the fact thatN (h(y)) = 0 andN (φ(y)) = O(‖y‖p),where N (h(y)) is defined by (8.11), we can show that μ(y) satisfies the partialdifferential equation

∂μ

∂y(y)[A1 + N(y, μ(y))]−A2μ(y)−Q(y, μ(y)) = 0 (C.66)

whereN(y, z) = g1(y, φ(y) + z)

and

Q(y, z) = g2(y, φ(y) + z)− g2(y, φ(y)) +N (φ(y))

− ∂φ

∂y(y)[g1(y, φ(y) + z)− g1(y, φ(y))]

A function μ(y) satisfying (C.66) is a center manifold for an equation of the form(C.54)–(C.55) with A = A1, B = A2, f = N , and g = Q. Furthermore, in this case,

Q(y, 0) = N (φ(y)) = O(‖y‖p)

Hence, by Lemma C.6, there exists a continuously differentiable function μ(y) =O(‖y‖p) that satisfies (C.66). Therefore, h(y) − φ(y) = O(‖y‖p). The reducedsystem is given by

y = A1y + g1(y, h(y))= A1y + g1(y, φ(y)) + g1(y, h(y))− g1(y, φ(y))


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Since g1 is twice continuously differentiable and its first partial derivatives vanishat the origin, we have ∥∥∥∥∂g1

∂z(y, z)

∥∥∥∥ ≤ k1‖y‖+ k2‖z‖

in the neighborhood of the origin. By the mean value theorem,

g1i(y, h(y))− g1i(y, φ(y)) =∂g1i

∂z(y, ζ(y))[h(y)− φ(y)]

where‖ζ(y)‖ ≤ ‖μ(y)‖+ ‖φ(y)‖ ≤ k3‖y‖p ≤ k3‖y‖

for ‖y‖ < 1. Therefore,

‖g1(y, h(y))− g1(y, φ(y))‖ ≤ k4‖y‖ ‖μ(y)‖ = O(‖y‖p+1)

which completes the proof of the theorem. �


To show that RA is invariant, we need to show that

x ∈ RA ⇒ x(s) def= φ(s; x) ∈ RA, ∀ s ∈ R

Sinceφ(t; φ(s; x)) = φ(t + s; x)

it is clear that limt→∞ φ(t; x(s)) = 0 for all s ∈ R. Hence, RA is invariant. Toshow that RA is open, take any point p ∈ RA and show that every point in aneighborhood of p belongs to RA. To that end, let T > 0 be large enough that‖φ(T ; p)‖ < a/2, where a is chosen so small that the domain ‖x‖ < a is containedin RA. Consider the neighborhood ‖x− p‖ < b of p. By continuous dependence ofthe solution on initial states, we can choose b small enough to ensure that for anypoint q in the neighborhood ‖x− p‖ < b, the solution at time T satisfies

‖φ(T ; p)− φ(T ; q)‖ <a

2

Then,‖φ(T ; q)‖ ≤ ‖φ(T ; q)− φ(T ; p)‖+ ‖φ(T ; p)‖ < a

This shows that the point φ(T ; q) is inside RA. Hence, the solution starting at qapproaches the origin as t →∞. Thus, q ∈ RA and the set RA is open. We leave itto the reader (Exercise 8.13) to show that RA is connected. The statement aboutthe boundary of RA follows from the next lemma.

Lemma C.7 The boundary of an open invariant set is an invariant set. Hence, itis formed of trajectories.


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Proof: Let M be an open invariant set and x be a boundary point of M . Thereexists a sequence xn ∈ M that converges to x. Since M is an invariant set, thesolution φ(t; xn) ∈ M for all t ∈ R. The sequence φ(t; xn) converges to φ(t; x) forall t ∈ R. Thus, φ(t; x) is an accumulation point of M for all t. On the otherhand, φ(t; x) �∈M , since x is a boundary point of M . Therefore, the solution φ(t; x)belongs to the boundary of M for all t. �


We will work with the full problem given by (11.10) and (11.11) in the (x, y) vari-ables. The error estimate for z will then follow from the change of variables (11.9).Let y belong to the domain Dy, where (11.9) maps Dx × Dy into Dz. When weanalyze (11.12) with the slowly varying t and x, we want to use the uniform expo-nential stability property (11.15) of the boundary-layer model. Inequality (11.15) isvalid only when x ∈ Dx, so to use it, we need to confirm that the slowly varying xwill always be in Dx. We anticipate that this will be true because the solution x ofthe reduced problem (11.8) belongs to S, a compact subset of Dx, and we anticipatethat the error ‖x(t, ε) − x(t)‖ will be O(ε). Then, for sufficiently small ε, x willbelong to Dx. However, the estimate ‖x(t, ε) − x(t)‖ = O(ε) has not been provenyet, so we cannot start by using it. We use a special technique to get around thisdifficulty.12 If Dx �= Rn, let E be the complement of Dx in Rn and define

k = 12 inf {‖x− y‖ | x ∈ S, y ∈ E} > 0

If Dx = Rn, take k to be any positive constant. The sets

S1 = {x ∈ Rn | dist(x, S) ≤ k/2} and S2 = {x ∈ Rn | dist(x, S) ≤ k}are compact subsets of Dx and S ⊂ S1 ⊂ S2. Let ψ : Rn → [0, 1] be a smooth(continuously differentiable infinitely many times) function with ψ(x) = 1 when xbelongs to S1 and ψ(x) = 0 when x is outside S2.13 We define F and G by

F (t, x, y, ε) = f(t, ϕ(x), y + h(t, ϕ(x)), ε) (C.67)

G(t, x, y, ε) = g(t, ϕ(x), y + h(t, ϕ(x)), ε)− ε∂h

∂t(t, ϕ(x))

− ε∂h

∂x(t, ϕ(x))f(t, ϕ(x), y + h(t, ϕ(x)), ε) (C.68)

where ϕ(x) = (x − ξ0)ψ(x) + ξ0. It can easily seen that, for all x ∈ Rn, ϕ(x) isbounded and belongs to Dx, since Dx is convex. When x ∈ S1, we have ϕ(x) = x;hence, the functions F and f are identical. The same is true for the functions Gand g − ε[(∂h/∂t) + (∂h/∂x)f ]. It can be verified that for all (t, x, y, ε) ∈ [0, t1] ×Rn × Ω1 × [0, ε0], where Ω1 is any compact subset of Dy, we have the following:

12The same technique is used in the proof of the center manifold theorem. (See Appendix C.15.)13The existence of ψ is shown in Lemma 6.2 of Chapter 23 of [111].


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• F and G and their first partial derivatives with respect to ε are continuousand bounded.

• F (t, x, y, 0) has bounded first partial derivatives with respect to (x, y).

• G(t, x, y, 0) and [∂G(t, x, y, 0)/∂y] have bounded first partial derivatives withrespect to (t, x, y).

Consider the modified singular perturbation problem

x = F (t, x, y, ε), x(t0) = ξ(ε) (C.69)εy = G(t, x, y, ε), y(t0) = η(ε)− h(t0, ξ(ε)) (C.70)

The modified problem of (C.69) and (C.70) is identical to the original problem of(11.10) and (11.11) when x ∈ S1. The set S1 has been chosen in anticipation thatthe solution x(t, ε) will be confined to S1, which is based on the fact that x(t) ∈ S.The boundary-layer model

dy

dτ= G(t, x, y, 0) (C.71)

has an equilibrium point at y = 0. Since

G(t, x, y, 0) = g(t, ϕ(x), y + h(t, ϕ(x)), 0)

for any fixed x ∈ Rn, the boundary-layer model (C.71) can be represented as aboundary-layer model of the form (11.14) with ϕ(x) ∈ Dx as the frozen parameter.Since inequality (11.15) holds uniformly in the frozen parameter, it is clear that thesolutions of (C.71) satisfy the same inequality for all x ∈ Rn; that is,

‖y(τ)‖ ≤ k‖y(0)‖ exp(−γτ), ∀ ‖y(0)‖ < ρ0, ∀ (t, x) ∈ [0, t1]×Rn, ∀ τ ≥ 0 (C.72)

The reduced problem for (C.69) and (C.70) is

x = F (t, x, 0, 0), x(t0) = ξ0 (C.73)

This problem is identical to the reduced problem (11.8) whenever x ∈ S1. Since(11.8) has a unique solution x(t) defined for all t ∈ [t0, t1] and x(t) ∈ S, it followsthat x(t) is the unique solution of (C.73) for t ∈ [t0, t1]. We proceed to provethe theorem for the modified singular perturbation problem given by (C.69) and(C.70). Upon completion of this task, we will show that, for sufficiently small ε,the solution x(t, ε) of (C.69)–(C.70) belongs to S1. This will establish that theoriginal and modified problems have the same solution and proves the theorem forthe original problem given by (11.10) and (11.11).

Consider the boundary-layer model (C.71). Since [∂G/∂y] has bounded first par-tial derivatives with respect to (t, x) and G(t, x, 0, 0) = 0 for all (t, x), the Jacobianmatrices [∂G/∂t] and [∂G/∂x] satisfy∥∥∥∥∂G

∂t

∥∥∥∥ ≤ L1‖y‖;∥∥∥∥∂G

∂x

∥∥∥∥ ≤ L2‖y‖


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Using these estimates and (C.72), we conclude from Lemma 9.8 that there is aLyapunov function V1(t, x, y) which satisfies

c1‖y‖2 ≤ V1(t, x, y) ≤ c2‖y‖2 (C.74)

∂V1

∂yG(t, x, y, 0) ≤ −c3‖y‖2 (C.75)∥∥∥∥∂V1

∂y

∥∥∥∥ ≤ c4‖y‖;∥∥∥∥∂V1

∂t

∥∥∥∥ ≤ c5‖y‖2;∥∥∥∥∂V1

∂x

∥∥∥∥ ≤ c6‖y‖2 (C.76)

for all y ∈ {‖y‖ < ρ0} and all (t, x) ∈ [0, t1] × Rn. The derivative of V1 along thetrajectories of the full system (C.69)–(C.70) is given by

V1 =1ε

∂V1

∂yG(t, x, y, ε) +

∂V1

∂t+

∂V1

∂xF (t, x, y, ε)

=1ε

∂V1

∂yG(t, x, y, 0) +

1ε

∂V1

∂y[G(t, x, y, ε)−G(t, x, y, 0)]

+∂V1

∂t+

∂V1

∂xF (t, x, y, ε)

Using (C.75) and (C.76) and the estimates

‖F (t, x, y, ε)‖ ≤ k0; ‖G(t, x, y, ε)−G(t, x, y, 0)‖ ≤ εL3

we obtain

V1 ≤ − c3

ε‖y‖2 + c4L3‖y‖+ c5‖y‖2 + c6k0‖y‖2

≤ − c3

2ε‖y‖2 + c4L3‖y‖, for ε ≤ c3

2c5 + 2c6k0

Thus, if at some time t∗ ≥ t0, ‖y(t∗, ε)‖ < ρ0√

c1/c2def= μ, the solution y(t, ε) of

the full problem will satisfy the exponentially decaying bound

‖y(t, ε)‖ ≤ μ√

c2/c1 exp[−α(t− t∗)

ε

]+ εδ, ∀ t ≥ t∗ (C.77)

where α = c3/4c2 and δ = 2c2c4L3/c1c3. On the other hand, y(t0, ε) = η(ε) −h(t0, ξ(ε)) = η0−h(t0, ξ0)+O(ε) and η0−h(t0, ξ0) belongs to Ωy, a compact subsetof the region of attraction of the boundary-layer model

dy

dτ= G(t0, ξ0, y, 0) = g(t0, ξ0, y + h(t0, ξ0), 0) (C.78)

We recall from (the converse Lyapunov) Theorem 4.17 that there is a Lyapunovfunction V0(y) such that

∂V0

∂yg(t0, ξ0, y + h(t0, ξ0), 0) ≤ −W0(y)


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over the region of attraction, where W0(y) is positive definite and {V0(y) ≤ c} is acompact subset of the region of attraction for any c > 0 . Choose c0 such that Ωy

is in the interior of {V0(y) ≤ c0}. The derivative of V0 along the trajectories of thefull system (C.69)–(C.70) is given by

V0 =1ε

∂V0

∂yG(t, x, y, ε)

=1ε

∂V0

∂yG(t0, ξ0, y, 0) +

1ε

∂V0

∂y[G(t, x, y, ε)−G(t0, ξ0, y, 0)]

It can be verified that, for all (t, y) ∈ [t0, t0 + εT ]× {V0(y) ≤ c0},

V0 ≤ 1ε

[−W0(y) + a0ε(1 + T )]

for some a0 > 0. Application of Theorem 4.18 shows that there is ε∗1 > 0 such that,for 0 < ε < ε∗1, y(t, ε) satisfies the inequality

‖y(t, ε)‖ ≤ β(μ2, (t− t0)/ε) + �(ε(1 + T ))

over the interval [t0, t0 +εT ], where β is a class KL function, � is a class K function,and μ2 is a positive constant. Choose T large enough that β(μ2, T ) < μ/2; thenchoose ε∗ < ε∗1 small enough that �(ε∗(1 + T )) < μ/2. It follows that, for ε < ε∗,y(t, ε) satisfies

‖y(t, ε)‖ < μ1 + μ/2 for t ∈ [t0, t0 + εT ] and ‖y(t0 + εT, ε)‖ < μ (C.79)

with μ1 = β(μ2, 0). Then, (C.77) and (C.79) yield

‖y(t, ε)‖ ≤ k1 exp[−α(t− t0)

ε

]+ εδ, ∀ t ≥ t0 (C.80)

for some k1 > 0.Consider (C.69). By rewriting the right-hand side as

F (t, x, y, ε) = F (t, x, 0, 0) + [F (t, x, y, ε)− F (t, x, 0, 0)]

we view (C.69) as a perturbation of the reduced system (C.73). The bracketedperturbation term satisfies

‖F (t, x, y, ε)− F (t, x, 0, 0)‖ ≤ ‖F (t, x, y, ε)− F (t, x, y, 0)‖+ ‖F (t, x, y, 0)− F (t, x, 0, 0)‖

≤ L4ε + L5‖y‖≤ θ1ε + θ2 exp

[−α(t− t0)ε

]


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where θ1 = L4 + L5δ and θ2 = L5k1. Define

u(t, ε) = x(t, ε)− x(t)

Then,

u(t, ε) = ξ(ε)− ξ(0) +∫ t

t0

[F (s, x(s, ε), y(s, ε), ε)− F (s, x(s), 0, 0)] ds

= ξ(ε)− ξ(0) +∫ t

t0

[F (s, x(s, ε), y(s, ε), ε)− F (s, x(s, ε), 0, 0)] ds

+∫ t

t0

[F (s, x(s, ε), 0, 0)− F (s, x(s), 0, 0)] ds

and

‖u(t, ε)‖ ≤ k2ε +∫ t

t0

{θ1ε + θ2 exp

[−α(s− t0)ε

]}ds +

∫ t

t0

L6‖u(s, ε)‖ ds

≤ k2ε +[θ1ε(t1 − t0) +

θ2ε

α

]+∫ t

t0

L6‖u(s, ε)‖ ds

By the Gronwall–Bellman lemma, we arrive at the estimate

‖x(t, ε)− x(t)‖ ≤ εk3[1 + t1 − t0] exp[L6(t1 − t0)] (C.81)

which proves the error estimate for x. We can also conclude that, for sufficientlysmall ε, the solution x(t, ε) is defined for all t ∈ [t0, t1].

To prove the error estimate for y, consider (C.70), which, for convenience, iswritten in the τ time scale as

dy

dτ= G(t0 + ετ, x(t0 + ετ, ε), y, ε)

Let y(τ) denote the solution of the boundary-layer model

dy

dτ= G(t0, ξ0, y, 0), y(0) = η0 − h(t0, ξ0)

and setv(τ, ε) = y(τ, ε)− y(τ)

By differentiating both sides with respect to τ and substituting for the derivativesof y and y, we obtain

dv

dτ= G(t0 + ετ, x(t0 + ετ, ε), y(τ, ε), ε)−G(t0, ξ0, y(τ), 0)

We add and subtract G(t0 + ετ, x(t0 + ετ, ε), v, 0) to obtain

dv

dτ= G(t, x, v, 0) + ΔG (C.82)


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where t = t0 + ετ , x = x(t0 + ετ, ε), ΔG = Δ1 + Δ2 + Δ3, and

Δ1 = G(t, x, y, 0)−G(t, x, y, 0)−G(t, x, v, 0)Δ2 = G(t, x, y, ε)−G(t, x, y, 0)Δ3 = G(t, x, y, 0)−G(t0, ξ0, y, 0)

It can be verified that

‖Δ1‖ ≤ k4‖v‖2 + k5‖v‖ ‖y‖, ‖Δ2‖ ≤ εL3

‖Δ3‖ ≤ L1|t− t0| ‖y‖+ L2‖x− ξ0‖ ‖y‖ ≤ (L1ετ + L2εa + L2ετk0)‖y‖

for some nonnegative constants k4, k5, and a. Repeating the derivation that led to(C.80), it can be shown that

‖y(τ)‖ ≤ k1e−ατ , ∀ τ ≥ 0 (C.83)

Hence,

‖ΔG‖ ≤ k4‖v‖2 + k5k1‖v‖ e−ατ + εL3 + εa1k1(1 + τ)e−ατ

≤ k4‖v‖2 + k5k1‖v‖e−ατ + εa2 (C.84)

where a1 = max{L2a, L1 + L2k0} and a2 = L3 + a1k1 max{1, 1/α}. We have usedthe fact that (1 + τ)e−ατ ≤ max{1, 1/α}. Equation (C.82) can be viewed as aperturbation of

dv

dτ= G(t, x, v, 0) (C.85)

which, by Lemma 9.8, has a Lyapunov function V1(t, x, v) that satisfies (C.74)through (C.76). Calculating the derivative of V1 along the trajectories of (C.82)and using the estimate (C.84), we obtain

V1 =∂V1

∂t+

∂V1

∂xF +

1ε

∂V1

∂v[G(t, x, v, 0) + ΔG]

≤ c5‖v‖2 + c6k0‖v‖2 − c3

ε‖v‖2 +

c4

ε‖v‖ (k4‖v‖2 + k5k1‖v‖e−ατ + εa2

)For ‖v‖ ≤ c3/4c4k4 and 0 < ε < c3/4(c5 + c6k0), we have

V1 ≤ − c3

2ε‖v‖2 +

c4k5k1

ε‖v‖2e−ατ + c4a2‖v‖

≤ − 2ε

(ka − kbe

−ατ)V1 + 2kc

√V1

where ka = c3/4c2, kb = c4k5k1/2c1, and kc = c4a2/2√

c1. Taking W =√

V1 yields

D+W (τ) ≤ − (ka − kbe−ατ

)W + εkc


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where D+W (τ) is the upper right-hand derivative of W with respect to τ . By thecomparison principle (Lemma 3.4), we conclude that

W (τ) ≤ φ(τ, 0)W (0) + ε

∫ τ

0φ(τ, σ)kc dσ

where

φ(τ, σ) = exp[−∫ τ

σ

(ka − kbe

−αλ)

dλ

]and |φ(τ, σ)| ≤ kge

−αg(τ−σ)

for some kg, αg > 0. Using the fact that v(0) = O(ε), we conclude that v(τ) = O(ε)for all τ ≥ 0. This completes the proof that the solution of (C.69)–(C.70) satisfies

x(t, ε)− x(t) = O(ε), y(t, ε)− y

(t

ε

)= O(ε)

∀ t ∈ [t0, t1] for sufficiently small ε. Since x(t) ∈ S, there is ε∗2 > 0 small enoughsuch that x(t, ε) ∈ S1 for all t ∈ [t0, t1] and all ε < ε∗2. Hence, x(t, ε) and y(t, ε) arethe solutions of (11.10)–(11.11). From (11.9), we have

z(t, ε)− h(t, x(t))− y

(t

ε

)= y(t, ε)− y

(t

ε

)+ h(t, x(t, ε))− h(t, x(t)) = O(ε)

where we have used the fact that h is Lipschitz in x. Finally, since y(τ) satisfies(C.83) and

exp[−α(t− t0)

ε

]≤ ε, ∀ α(t− t0) ≥ ε ln

(1ε

)

the term y(t/ε) will be O(ε) uniformly on [tb, t1] if ε is small enough to satisfy

ε ln(

1ε

)≤ α(tb − t0)

The proof of Theorem 11.1 is now complete.


The proof follows closely that of Theorem 11.1. We will only point out two maindifferences, one in showing that x belongs to Dx and the other in analyzing theerror x − x. The first point makes use of the Lyapunov function V of the reducedsystem, while the second one makes use of its stability properties.

Using the Lyapunov function V , we will argue that x belongs to the compactset {W1(x) ≤ c} for all t ≥ t0. Therefore, we do not need to truncate x by using thefunction ψ(x), as we did in the proof of Theorem 11.1. The functions F and G arestill defined by (C.67) and (C.68), but with ϕ(x) replaced by x. They have the same


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properties as before for all (t, x, y, ε) ∈ [0,∞)×{W1(x) ≤ c}×Ω1×[0, ε0]. Moreover,[∂F/∂x](t, x, 0, 0) is Lipschitz in x, uniformly in t. For all x ∈ {W1(x) ≤ c}, we canrepeat the earlier derivation to show that y(t, ε) satisfies (C.80). Hence,

V ≤ −W3(x) + k6ε + k7 exp[−α(t− t0)

ε

]

Using the fact that ξ0 ∈ {W2(x) ≤ ρc}, we can see that there is a time T1 > 0,independent of ε, such that, for sufficiently small ε, x(t, ε) ∈ {W2(x) ≤ c} for allt ∈ [t0, t0 + T1]. For t ≥ t0 + T1, the exponential term exp [−α(t− t0)/ε] is O(ε).Thus,

V ≤ −W3(x) + k8ε

Using this inequality, we can show that V is negative on the boundary V (t, x) = c.Therefore, x(t, ε) ∈ {W1(x) ≤ c} for all t ≥ t0.

To analyze the approximation error u(t, ε) = x(t, ε)− x(t), we view (C.69) as aperturbation of the reduced system (C.73). Instead of using the Gronwall–Bellmanlemma to derive an estimate of u, we employ Lyapunov analysis that exploits theexponential stability of the origin of (C.73). The Lyapunov analysis is very similarto the boundary-layer analysis of Theorem 11.1’s proof. Therefore, we will describeit only briefly. The error u satisfies the equation

u = F (t, u, 0, 0) + ΔF (C.86)

where

ΔF = [F (t, x + u, 0, 0)− F (t, x, 0, 0)− F (t, u, 0, 0)] + [F (t, x, y, ε)− F (t, x, 0, 0)]

It can be verified that

‖ΔF‖ ≤ k4‖u‖2 + k5‖u‖ ‖x‖+ k6 exp[−α(t− t0)

ε

]+ εk7

The system (C.86) is viewed as a perturbation of

u = F (t, u, 0, 0) (C.87)

Because the origin of (C.87) is exponentially stable, we can obtain a Lyapunovfunction V (t, u) for it by using Theorem 4.14. Using this function with (C.86), weobtain

˙V =∂V

∂t+

∂V

∂uF (t, u, 0, 0) +

∂V

∂uΔF

≤ −c3‖u‖2 + c4‖u‖{

k4‖u‖2 + k5‖u‖ ‖x‖+ k6 exp[−α(t− t0)

ε

]+ εk7

}

For ‖u‖ ≤ c3/2c4k4, we have

˙V ≤ −2[ka − kbe

−α(t−t0)]V + 2

{εkc + kd exp

[−α(t− t0)ε

]}√V


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for some ka, α > 0 and kb, kc, kd ≥ 0. Applying the comparison principle yields

W (t) ≤ φ(t, t0)W (0) +∫ t

t0

φ(t, s){

εkc + kd exp[−α(s− t0)

ε

]}ds

where W =√

V and

|φ(t, s)| ≤ kge−σ(t−t0), σ > 0, kg > 0

Since u(t0) = O(ε) and∫ t

t0

exp[−σ(t− s)] exp[−α(s− t0)

ε

]ds = O(ε)

we can show that W (t) = O(ε) and, subsequently, u(t, ε) = O(ε). The rest of theproof proceeds exactly as in Theorem 11.1. Notice that the boundary-layer analysisin that proof is valid for all τ ≥ 0.


We analyze the closed-loop system (12.45) as a slowly varying system by using theresults of Section 9.6. Since dependence on w does not play a role in the proof, wewrite g(X , ρ, w) as g(X , ρ). It can be verified that g(X , ρ) is continuously differen-tiable in a domain DX ×Dρ and Xss(α) and Ams(α) are continuously differentiablein Dρ. Because Ams(α) is Hurwitz for all α ∈ Dρ, it is Hurwitz uniformly in α forall α ∈ S (a compact subset of Dρ). Hence, Ams satisfies all the assumptions ofLemma 9.9 for α ∈ S. Let Pms = Pms(α) be the solution of the Lyapunov equationPmsAms + AT

msPms = −I. We use V (Xδ, α) = X Tδ PmsXδ as a Lyapunov function

candidate for the frozen system Xδ = g(Xδ + Xss(α), α). Lemma 9.9 shows thatV (Xδ, α) satisfies (9.41), (9.43), and (9.44). We only need to verify (9.42). Thefrozen system can be rewritten as

Xδ = Ams(α)Xδ + Δg(Xδ, α)

where‖Δg(Xδ, α)‖2 = ‖g(Xδ + Xss(α), α)−Ams(α)Xδ‖2 ≤ k1‖Xδ‖22

in some domain {‖Xδ‖2 < r1}. Thus, the derivative of V along the trajectories ofXδ = g(Xδ + Xss(α), α) satisfies

V ≤ −‖Xδ‖22 + 2c2k1‖Xδ‖32 ≤ − 12‖Xδ‖22

for ‖Xδ‖2 < 1/(4c2k1). Therefore, there exists r > 0 such that V (Xδ, α) satis-fies (9.41) through (9.44) for all (Xδ, α) ∈ {‖Xδ‖2 < r} × S. The conclusions ofTheorem 12.1 follow from Theorem 9.3.


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To analyze the closed-loop system (12.58)–(12.59), we combine stability analysis ofslowly varying systems from Section 9.6 with stability analysis of singularly per-turbed systems from Section 11.5. Since dependence on w does not play a role inthe proof, we write (12.58)–(12.59) as

X = g(X , ρ) + N(ρ)[ϑ− φ(X , ρ)] (C.88)εϑ = −ϑ + φ(X , ρ) (C.89)

When ε = 0, we obtain the reduced system X = g(X , ρ), which was analyzed in theproof of Theorem 12.1 by using the quadratic Lyapunov function X T

δ PmsXδ. It canbe verified that φ(X , ρ) is continuously differentiable in a domain DX × Dρ. Thechange of variables

Y = X − Xss(ρ), Z = ϑ− φ(X , ρ)

transforms the system (C.88)–(C.89) into

Y = g(Y + Xss(ρ), ρ) + N(ρ)Z − ∂Xss

∂ρρ (C.90)

εZ = −Z − ε∂φ

∂X [g(Y + Xss(ρ), ρ) + N(ρ)Z]− ε∂φ

∂ρρ (C.91)

Using V = YT PmsY + (1/2)ZTZ as a Lyapunov function for (C.90)–(C.91), weobtain

V = −YTY + 2YT Pms

[g(Y + Xss(ρ), ρ)−Ams(ρ)Y + N(ρ)Z − ∂Xss

∂ρρ

]

+ YT

[d

dtPms(ρ)

]Y

− 1εZTZ − ZT

{∂φ

∂X [g(Y + Xss(ρ), ρ) + N(ρ)Z] +∂φ

∂ρρ

}

≤ −‖Y‖22 −1ε‖Z‖22 + c1‖Y‖32 + c2‖Y‖2 ‖Z‖2 + c3‖Y‖2‖ρ‖2

+ c4‖Y‖22‖ρ‖2 + c5‖Z‖22 + c6‖Z‖2‖ρ‖2in some neighborhood of the origin, for some positive constants ci. Limiting ouranalysis to a neighborhood where ‖Y‖2 ≤ c7 ≤ 1/(4c1), we arrive at the inequality

V ≤ − 12‖Y‖22 −

12ε‖Z‖22 + (c3‖Y‖2 + c4c7‖Y‖2 + c6‖Z‖2)‖ρ‖2

−[ ‖Y‖2‖Z‖2

]T [1/4 − c2/2− c2/2 1/(2ε)− c5

] [ ‖Y‖2‖Z‖2

]


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Choosing ε∗ small enough that the 2×2 matrix is positive definite for all 0 < ε < ε∗,we end up with

V ≤ −2αV + 2β√V‖ρ‖2

for some positive constants α and β. Hence, W =√V satisfies the inequality

D+W ≤ −αW + β‖ρ‖2Applying the comparison lemma concludes the proof.


The proof uses the notions of Lie brackets and involutive distributions, introducedin Section 14.3, as well as the notion of complete integrability. A nonsingulardistribution Δ on D, generated by f1, . . . , fk, is completely integrable if for eachx0 ∈ D, there exists a neighborhood N of x0 and n−k real-valued smooth functionsh1(x), . . . , hn−k(x) such that

∂hj

∂xfi(x) = 0, ∀ 1 ≤ i ≤ k and 1 ≤ j ≤ n− k

and the row vectors dh1(x), . . . , dn−k(x) are linearly independent for all x ∈ D,where

dh(x) =∂h

∂x=[

∂h

∂x1, . . . ,

∂h

∂xn

]is called the differential of h. A key result from differential geometry is Frobeniustheorem,14 which states that a nonsingular distribution is completely integrable ifand only if it is involutive.

We start by stating and proving two preliminary lemmas.

Lemma C.8 For all x ∈ D and all integers k and j such that k ≥ 0 and 0 ≤ j ≤ρ− k − 1, we have

LadjfgL

kfh(x) =

⎧⎨⎩

0, 0 ≤ j + k < ρ− 1

(−1)jLgLρ−1f h(x) �= 0, j + k = ρ− 1

(C.92)

�

Proof: Prove it by induction of j. For j = 0, (C.92) holds by the definition ofrelative degree. Assume now that (C.92) holds for some j and prove it for j + 1.Recall from the Jacobi identity (Exercise 13.8) that

L[f,β]λ(x) = LfLβλ(x)− LβLfλ(x)

14See [88] for the proof of Frobenius theorem.


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for any real-valued function λ and any vector fields f and β. Taking λ = Lkfh and

β = adjfg, we obtain

Ladj+1f

gLkfh(x) = L[f,adj

fg]L

kfh(x) = LfLadj

fgL

kfh(x)− Ladj

fgL

k+1f h(x)

We note that the first term on the right-hand side vanishes since

j + k + 1 ≤ ρ− 1 ⇒ j + k < ρ− 1 ⇒ LfLadjfgL

kfh(x) = 0

Moreover,

LadjfgL

k+1f h(x) =

⎧⎨⎩

0, 0 ≤ j + k + 1 < ρ− 1

(−1)jLgLρ−1f h(x) �= 0, j + k + 1 = ρ− 1

by the assumption that (C.92) holds for j. Thus,

Ladj+1f

gLkfh(x) =

⎧⎨⎩

0, 0 ≤ j + k + 1 < ρ− 1

(−1)j+1LgLρ−1f h(x) �= 0, j + k + 1 = ρ− 1

which completes the proof of the lemma. �

Lemma C.9 For all x ∈ D,

• the row vectors dh(x), dLfh(x), . . . , dLρ−1f h(x) are linearly independent;

• the column vectors g(x), adfg(x), . . . , adρ−1f g(x) are linearly independent.

�

Proof: We have⎡⎢⎣

dh(x)...

dLρ−1f h(x)

⎤⎥⎦ [ g(x) . . . adρ−1

f g(x)]

=

⎡⎢⎢⎢⎢⎣

Lgh(x) Ladf gh(x) . . . . . . Ladρ−1f

gh(x)

LgLfh(x) Ladρ−2f

gLfh(x) ∗...

...LgL

ρ−1f h(x) ∗ . . . ∗

⎤⎥⎥⎥⎥⎦


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From the previous lemma, the right-hand side matrix takes the form⎡⎢⎢⎢⎢⎢⎣

0 . . . . . . 0 �0 � ∗...

...0 � ∗� ∗ . . . ∗

⎤⎥⎥⎥⎥⎥⎦

where � denotes a nonzero element. Thus, the matrix is nonsingular, which provesthe lemma, for if any of the two matrices on the left-hand side has rank less thanρ, their product must be singular. �

Lemma C.9 shows that ρ ≤ n. We are now ready to prove Theorem 13.1. Theproof for the case ρ = n follows from Lemma C.9, whose first statement shows that[∂T/∂x] is nonsingular. Consider the case ρ < n. The distribution Δ = span{g}is nonsingular, involutive, and has dimension one.15 By Frobenius theorem, Δ iscompletely integrable. Hence, for every x0 ∈ D, there exist a neighborhood N1of x0 and n − 1 smooth functions φ1(x), . . . , φn−1(x), with linearly independentdifferentials such that

Lgφi(x) = 0, for 1 ≤ i ≤ n− 1, ∀ x ∈ N1

BecauseLgL

ifh(x) = 0, for 0 ≤ i ≤ ρ− 2

and dh(x), . . . , dLρ−2f h(x) are linearly independent, we can use h, . . . , Lρ−2

f h aspart of these n − 1 functions. In particular, we take them as φn−ρ+1, . . . , φn−1.Since LgL

ρ−1f h(x) �= 0, the row vector dLρ−1

f h(x0) is linearly independent of therow vectors dφ1(x0), . . . , dφn−1(x0). Therefore,

rank[∂T

∂x(x0)

]= n ⇒ ∂T

∂x(x0) is nonsingular

and there is a neighborhood N2 of x0 such that T (x), restricted to N2, is a diffeo-morphism on N2. Taking N = N1 ∩N2 completes the proof of the theorem. �


The systemx = f(x) + g(x)u

15Note that any nonsingular distribution of dimension one is automatically involutive.


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is feedback linearizable if and only if a sufficiently smooth function h(x) exists suchthat the system

x = f(x) + g(x)u, y = h(x)

has relative degree n in D0 ⊂ D; that is, h(x) satisfies

LgLifh(x) = 0, for 0 ≤ i ≤ n− 2 and LgL

n−1f h(x) �= 0, ∀ x ∈ D0 (C.93)

Thus, to prove the theorem, we need to show that the existence of h(x) satisfying(C.93) is equivalent to conditions 1 and 2.Necessity: Suppose there is h(x) satisfying (C.93). Lemma C.9 shows that rank G =n. Then, D is nonsingular and has dimension n− 1. From (C.92), with k = 0 andρ = n, we have

Lgh(x) = Ladf gh(x) = · · · = Ladn−2f

gh(x) = 0

which can be written as

dh(x)[g(x), adfg(x), . . . , adn−2f g(x)] = 0

This equation implies that D is completely integrable and it follows from Frobeniustheorem that D is involutive.Sufficiency: Suppose conditions 1 and 2 are satisfied. Then, D is nonsingular andhas dimension n− 1. By Frobenius theorem, there exists h(x) satisfying

Lgh(x) = Ladf gh(x) = · · · = Ladn−2f

gh(x) = 0

Using the Jacobi identity (Exercise 13.8), it can be verified that

Lgh(x) = LgLfh(x) = · · · = LgLn−2f h(x) = 0

Furthermore,

dh(x)G(x) = dh(x)[g(x), adfg(x), . . . , adn−1f g(x)] = [0, . . . , 0, Ladn−1

fgh(x)]

Since rank G = n and dh(x) �= 0, it must be true that Ladn−1f

gh(x) �= 0. Using the

Jacobi identity, it can be verified that LgLn−1f h(x) �= 0, which completes the proof

of the theorem.


For the purpose of analysis, the observer dynamics are replaced by the equivalentdynamics of the scaled estimation error

ηij =xij − xij

ερi−j


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for 1 ≤ i ≤ m and 1 ≤ j ≤ ρi. Hence, we have x = x−D(ε)η, where

η = [η11, . . . , η1ρ1 , . . . , ηm1, . . . , ηmρm]T ,

D(ε) = block diag[D1, . . . , Dm]Di = diag[ερi−1, . . . , 1]ρi×ρi

The closed-loop system can be represented by

x = Ax + Bφ(x, z, γ(ϑ, x−D(ε)η, ζ))z = ψ(x, z, γ(ϑ, x−D(ε)η, ζ))ϑ = Γ(ϑ, x−D(ε)η, ζ)

εη = A0η + εBδ(x, z, ϑ, D(ε)η)

whereδ(x, z, ϑ, D(ε)η) = φ(x, z, γ(ϑ, x, ζ))− φ0(x, ζ, γ(ϑ, x, ζ))

and (1/ε)A0 = D−1(ε)(A−HC)D(ε) is a ρ× ρ Hurwitz matrix. For convenience,we rewrite the system in the compact singularly perturbed form

X = F (X , D(ε)η) (C.94)εη = A0η + εBΔ(X , D(ε)η) (C.95)

where F (X , 0) = f(X ). The initial states are X (0) = (x(0), z(0), ϑ(0)) ∈ S andx(0) ∈ Q. Thus, we have η(0) = D−1(ε)[x(0)− x(0)]. Setting ε = 0 in (C.95) yieldsη = 0 and the reduced system

X = f(X ) (C.96)

is nothing, but the closed–loop system under state feedback. The boundary-layermodel, obtained by applying the change of time variable τ = t/ε then setting ε = 0,is given by

dη

dτ= A0η

Since the origin of (C.96) is asymptotically stable and R is its region of attraction,by (the converse Lyapunov) Theorem 4.17, there is a smooth, positive definitefunction V (X ) and a continuous, positive definite function U(X ), both defined forall X ∈ R, such that

V (X ) →∞ as X → ∂R∂V

∂X f(X ) ≤ −U(X ), ∀ X ∈ Rand for any c > 0, {V (X ) ≤ c} is a compact subset of R. Let S be any compactset in the interior of R. Choose positive constants b and c such that c > b >maxX∈S V (X ). Then

S ⊂ Ωb = {V (X ) ≤ b} ⊂ Ωc = {V (X ) ≤ c} ⊂ R


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For the boundary-layer system, the Lyapunov function W (η) = ηT P0η, where P0 isthe positive definite solution of the Lyapunov equation P0A0 +AT

0 P0 = −I, satisfies

λmin(P0)‖η‖2 ≤ W (η) ≤ λmax(P0)‖η‖2

∂W

∂ηA0η ≤ −‖η‖2

where throughout the proof we use ‖ · ‖ = ‖ · ‖2. Due to equivalence of norms, itis sufficient to prove (14.114) and (14.115) in the 2-norm. Let Σ = {W (η) ≤ �ε2}and Λ = Ωc × Σ. Due to the global boundedness of F and Δ in x, for all X ∈ Ωc

and η ∈ Rρ, we have

‖F (X , D(ε)η)‖ ≤ k1, ‖Δ(X , D(ε)η)‖ ≤ k2

where k1 and k2 are positive constants independent of ε. Moreover, for any 0 < ε <1, there is L1, independent of ε, such that for all (X , η) ∈ Λ and every 0 < ε ≤ ε,we have

‖F (X , D(ε)η)− F (X , 0)‖ ≤ L1‖η‖We will always consider ε ≤ ε. We start by showing that there exist positiveconstants � and ε1 (dependent on �) such that the compact set Λ is positivelyinvariant for every 0 < ε ≤ ε1. This can be done by verifying that

V ≤ −U(X ) + εk3

andW ≤ − 1

ε‖η‖2 + 2‖η‖‖P0‖‖B‖k2 ≤ − 1

ε‖η‖2 + 2‖η‖‖P0‖k2

for all (X , η) ∈ Λ, where k3 = L1L2√

�/λmin(P0), ‖P0‖ = λmax(P0), ‖B‖ = 1, andL2 is an upper bound for ‖∂V/∂X‖ over Ωc. Taking � = 16k2

2‖P0‖3 and ε1 = β/k3,where β = minX∈∂Ωc U(X ), it can be shown that, for every 0 < ε ≤ ε1, we haveV ≤ 0 for all (X , η) ∈ {V (X ) = c} × Σ and W ≤ 0 for all (X , η) ∈ Ωc × {W (η) =�ε2}. Hence, Λ is positively invariant.

Now we consider the initial state (X (0), x(0)) ∈ S×Q. It can be verified that thecorresponding initial error η(0) satisfies ‖η(0)‖ ≤ k/ε(ρmax−1) for some nonnegativeconstant k dependent on S and Q, where ρmax = max {ρ1, ..., ρm}. Since X (0) is inthe interior of Ωc, it can be shown that

‖X (t)−X (0)‖ ≤ k1t (C.97)

as long as X (t) ∈ Ωc. Thus, there exists a finite time T0, independent of ε, suchthat X (t) ∈ Ωc for all t ∈ [0, T0]. During this time interval, we have

W ≤ − 12ε‖η‖2 − 1

2ε‖η‖2 + 2k2‖P0‖‖η‖ ≤ − 1

2ε‖η‖2, for W (η) ≥ �ε2


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sinceW (η) ≥ �ε2 ⇒ ‖P0‖‖η‖2 ≥ 16k2

2‖P0‖3ε2 ⇔ ‖η‖ ≥ 4k2‖P0‖εTherefore,

W (η(t)) ≤ σ2

ε2(ρmax−1) exp (−σ1t/ε) (C.98)

where σ1 = 1/(2‖P0‖) and σ2 = k2‖P0‖. Choose ε2 > 0 small enough that

T (ε) def=ε

σ1ln (

σ2

�ε2ρmax) ≤ 1

2T0

for all 0 < ε ≤ ε2. We note that ε2 exists, since T (ε) tends to zero as ε tends to zero.It follows that W (η(T (ε))) ≤ �ε2 for every 0 < ε ≤ ε2. Taking ε∗1 = min {ε, ε1, ε2}guarantees that, for every 0 < ε ≤ ε∗1, the trajectory (X (t), η(t)) enters Λ during theinterval [0, T (ε)] and remains there for all t ≥ T (ε). Consequently, the trajectoryis bounded for all t ≥ T (ε). On the other hand, for t ∈ [0, T (ε)], the trajectory isbounded by virtue of inequalities (C.97) and (C.98).

Next, we show (14.114). We know that, for every 0 < ε ≤ ε∗1, the solutions areinside the set Λ for all t ≥ T (ε), where Λ is O(ε) in the direction of the variable η.Thus, we can find ε3 = ε3(μ) ≤ ε∗1 such that, for every 0 < ε ≤ ε3, we have

‖η(t)‖ ≤ μ/2, ∀ t ≥ T (ε3) = T (μ) (C.99)

Using the fact that V ≤ −U(x) + εk3 for all (X , η) ∈ Λ, we conclude that

V ≤ −12U(x), for X �∈ {U(X ) ≤ 2k3ε

def= ν(ε)} (C.100)

Because U(X ) is positive definite and continuous, the set {U(X ) ≤ ν(ε)} is acompact set for sufficiently small ε. Let c0(ε) = maxU(X )≤ν(ε){V (X )}; c0(ε) isnondecreasing and limε→0 c0(ε) = 0. Consider the compact set {V (X ) ≤ c0(ε)}.We have {U(X ) ≤ ν(ε)} ⊂ {V (X ) ≤ c0(ε)}. Choose ε4 = ε4(μ) ≤ ε∗1 small enoughsuch that, for all ε ≤ ε4, the set {U(X ) ≤ ν(ε)} is compact, the set {V (X ) ≤ c0(ε)}is in the interior of Ωc, and

{V (X ) ≤ c0(ε)} ⊂ {‖X‖ ≤ μ/2} (C.101)

Then, for all X ∈ Ωc, but X �∈ {V (X ) ≤ c0(ε)}, we have an inequality similarto (C.100). Therefore, we conclude that the set {V (X ) ≤ c0(ε)} × Σ is positivelyinvariant and every trajectory in Ωc×Σ reaches {V (X ) ≤ c0(ε)}×Σ in finite time.In other words, given (C.101), there exists a finite time T = T (μ) such that forevery 0 < ε ≤ ε4

‖X (t)‖ ≤ μ/2. ∀ t ≥ T (C.102)

Take ε∗2 = ε∗2(μ) = min{ε3, ε4} and T2 = T2(μ) = max{T , T}. Then, (14.114)follows from (C.99), (C.102), x = x−D(ε)η, and ‖D(ε)‖ = 1.

To show (14.115), we divide the interval [0,∞) into three intervals [0, T (ε)],[T (ε), T3], and [T3,∞) and show (14.115) for each interval. From the ultimate


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boundedness of X (t), shown in (14.114), and the asymptotic stability of the originof (C.96), we conclude that there exists a finite time T3 ≥ T (ε), independent of ε,such that, for every 0 < ε ≤ ε∗2, we have

‖X (t)−Xr(t)‖ ≤ μ, ∀ t ≥ T3 (C.103)

From (C.97), we know that

‖X (t)−X (0)‖ ≤ k1t

during the interval [0, T (ε)]. Similarly, it can be shown that

‖Xr(t)−X (0)‖ ≤ k1t

during the same interval. Hence,

‖X (t)−Xr(t)‖ ≤ 2k1T (ε), ∀ t ∈ [0, T (ε)]

Since T (ε) → 0 as ε → 0, there exists 0 < ε5 ≤ ε∗2 such that, for every 0 < ε ≤ ε5,we have

‖X (t)−Xr(t)‖ ≤ μ, ∀ t ∈ [0, T (ε)] (C.104)

Over the interval [T (ε), T3], the solution X (t) satisfies

X = F (X , D(ε)η(t)), with ‖X (T (ε))−Xr(T (ε))‖ ≤ δ1(ε)

where D(ε)η is O(ε) and δ1(ε) → 0 as ε → 0. Thus, by Theorem 3.5, we concludethat there exists 0 < ε6 ≤ ε∗2 such that, for every 0 < ε ≤ ε6, we have

‖X (t)−Xr(t)‖ ≤ μ, ∀ t ∈ [T (ε), T3] (C.105)

Take ε∗3 = min{ε5, ε6}. Then, (14.115) follows from (C.103) through (C.105).Finally, assuming that the origin of (C.96) is exponentially stable, it follows from

(the converse Lyapunov) Theorem 4.14 that there exists a continuously differentiableLyapunov function V1(X ) which satisfies the inequalities

b1‖X‖2 ≤ V1(X ) ≤ b2‖X‖2, ∂V1

∂X F (X , 0) ≤ −b3‖X‖2,∥∥∥∥∂V1

∂X∥∥∥∥ ≤ b4‖X‖

over the ball Br ⊂ R for some positive constants r, b1, b2, b3, and b4. Using thelocal Lipschitz property of F and Δ and the fact that F (0, 0) = 0 and Δ(0, 0) = 0,it can be shown that the composite Lyapunov function V2(X , η) = V1(X ) + W (η)satisfies

V2 ≤ −YT QYwhere

Q =[

b3 −β1−β1 (1/ε)− β2

], Y =

[ ‖X‖‖η‖

]


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for some nonnegative constants β1 and β2. The matrix Q will be positive definitefor sufficiently small ε. Hence, there is a neighborhood N of the origin, independentof ε, and ε7 > 0 such that for every 0 < ε ≤ ε7, the origin is exponentially stableand every trajectory in N converges to the origin as t → ∞. By (14.114), thereexists ε8 > 0 such that for every 0 < ε ≤ ε8, the solutions starting in S×Q enter Nin finite time. Hence, for every 0 < ε ≤ ε∗4 = min{ε7, ε8}, the origin is exponentiallystable and S ×Q is a subset of the region of attraction.


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khalil nonlinear systems third edition chapter 14 and appendix

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