Given the importance of this lemma in the study of positive definite functions in Appendix B theproof is presented in its complete form... It is also apparent that it is not necessary for
Trang 1|a ij (x) − a ij (y) | ≤ n
'max
Truncated Taylor Representation of a Function
We present now a result well known from calculus and optimization In thefirst case, it comes from the ‘theorem of Taylor’ and in the second, it comesfrom what is known as ‘Lagrange’s residual formula’ Given the importance
of this lemma in the study of positive definite functions in Appendix B theproof is presented in its complete form
Lemma A.4 Let f : IR n → IR be a continuous function with continuous
partial derivatives up to at least the second one Then, for each x ∈ IR n , there exists a real number α (1 ≥ α ≥ 0) such that
f (x) = f (0) + ∂f
∂x(0)
T x +1
2x T H(αx)x where H(αx) is the Hessian matrix (that is, its second partial derivative) of
Trang 2[1− t)] ˙y(t) T x dt + y(0) T x (A.7)
Now, using the mean-value theorem for integrals2, and noting that (1−t) ≥
0 for all t between 0 and 1, the integral on the right-hand side of Equation
(A.7) may be written as
Incorporating this in (A.7) we get
1 We recall here the formula:
2 Recall that for functions h(t) and g(t), continuous on the closed interval a ≤ t ≤ b,
and where g(t) ≥ 0 for each t from the interval, there always exists a number c
such that a ≤ c ≤ b and
b a
h(t)g(t) dt = h(c)
b a
g(t) dt
Trang 32y(α)˙ T x + y(0) T x. (A.8)
On the other hand, using the definition of y(t) given in (A.6), we get
f (x) = e x = 1 + x +1
2e
αx x2
where for each x ∈ IR there exists an α (1 ≥ α ≥ 0) Specifically, for
x = 0 ∈ IR any α ∈ [0, 1] applies (indeed, any α ∈ IR) In the case that x
Trang 4−100 −50 0 50 100
0.00
0.25
0.50
0.75
1.00 α
x
Figure A.1.Example A.1: graph of α
A.3 Functional Spaces
A special class of vectorial spaces are the so-called L n
p (pronounce “el/pi:/en”)
where n is a positive integer and p ∈ (0, ∞] The elements of the L n
p spaces are functions with particular properties
The linear spaces denoted by L n
2 and L n ∞, which are defined below, are
often employed in the analysis of interconnected dynamical systems in the theory of input–output stability Formally, this methodology involves the use
of operators that characterize the behavior of the distinct parts of the inter-connected dynamic systems
We present next a set of definitions and properties of spaces of functions that are useful in establishing certain convergence properties of solutions of differential equations
For the purposes of this book, we say that a function f : IR n → IR m is
said to be continuous if
lim
x→x0
f(x) = f(x0) ∀ x0∈ IR n
A necessary condition for a function to be continuous is that it is defined at
every point x ∈ IR n It is also apparent that it is not necessary for a function
to be continuous that the function’s derivative be defined everywhere For
instance the derivative of the continuous function f (x) = |x| is not defined
at the origin, i.e at x = 0 However, if a function’s derivative is defined
everywhere then the function is continuous
The space L n
2 consists in the set of all the continuous functions f : IR+→
IRn such that ∞
0
f(t) T f(t) dt =
∞
0 f(t)2dt < ∞
Trang 5In words, a function f belongs to the L n
∞ space consists of the set of all continuous functions f : IR+→ IR n
such that their Euclidean norms are upperbounded as3,
sup
t ≥0 f(t) < ∞
The symbols L2and L ∞ denote the spaces L1 and L1∞ respectively.
We present next an example to illustrate the above-mentioned definitions
Example A.2 Consider the continuous functions f (t) = e −αt andg(t) = α sin(t) where α > 0 We want to determine whether f and g belong to the spaces of L2 and L ∞.
Consider first the function f (t):
t ≥ 0, hence f ∈ L ∞ We conclude that f (t) is bounded and integrable, i.e f ∈ L ∞ ∩ L2 respectively.
square-Consider next the function g(t) Notice that the integral
∞
0 |g(t)|2 dt = α2 ∞
0
sin2(t) dt
does not converge; consequently g 2 Nevertheless|g(t)| = |α sin(t)|
≤ α < ∞ for all t ≥ 0, and therefore g ∈ L ∞. ♦
A useful observation for analysis of convergence of solutions of differential
equations is that if we consider a function x : IR+ → IR n and a radially
unbounded positive definite function W : IR n → IR+ then, since W (x) is continuous in x the composition w(t) := W (x(t)) satisfies w ∈ L ∞ if and
only if x ∈ L n
∞.
3For those readers not familiar with the sup of a function f (t), it corresponds
to the smallest possible number which is larger than f (t) for all t ≥ 0 For
instance sup| tanh(t)| = 1 but | tanh(t)| has no maximal value since tanh(t) is
ever increasing and tends to 1 as t → ∞.
Trang 6We remark that a continuous function f belonging to the space L n
2 may not
have a limit We present next a result from the functional analysis literature
which provides sufficient conditions for functions belonging to the L n
2 space
to have a limit at zero This result is very often used in the literature ofmotion control of robot manipulators and in general, in the adaptive controlliterature
Lemma A.5 Consider a once continuously differentiable function f : IR+→
IR n Suppose that f and its time derivative satisfy the following
Then, necessarily lim t →∞ f(t) = 0 ∈ IR n
Proof It follows by contradiction4 Specifically we show that if the conclusion
of the lemma does not hold then the hypothesis that f ∈ L n
2 is violated.
To that end we first need to establish a convenient bound for the function
f(t)2= f (t) T f(t) Its total time derivative is 2f(t) T f(t) and is continuous˙
by assumption so we may invoke the mean value theorem (see Theorem A.2)
to conclude that for any pair of numbers t, t1 ∈ IR+ there exists a number s laying on the line segment that joins t and t1, such that
f(t)2− f(t1)2 ≤ 2f(s) T f(s) |t − t˙ 1|
On the other hand, since f , ˙ f ∈ L n
∞ it follows that there exists k > 0 such
that f(t)2− f(t1)2 ≤ k |t − t
1| ∀ t, t1∈ IR+. (A.9)
Next, notice that
f(t)2=f(t)2− f(t1)2+f(t1)2for all t, t1∈ IR+ Now we use the inequality|a + b| ≥ |a| − |b| which holds for all a, b ∈ IR, with a = f(t1)2and b =
f(t)2− f(t1)2
to see that
4 Proof “by contradiction” or, “by reductio ad absurdum”, is a technique widely used in mathematics to prove theorems and other truths To illustrate the method consider a series of logical statements denoted A, B, C, etc and their negations, denoted A, B, C, etc Then, to prove by contradiction the claim, “A and B =⇒
C”, we proceed as follows Assume that A and B hold but not C Then, we seek for a series of implications that lead to a negation of A and B, i.e we look for other statements D, E, etc such that C = ⇒ D =⇒ E =⇒ A and B So we conclude that C = ⇒ A and B However, in view of the fact that A and B must hold, this contradicts the initial hypothesis of the proof that C does not hold (i.e C) Notice that A and B = A or B.
Trang 7f(t)2 ≥ f(t1)2− f(t)2− f(t1)2
for all t, t1∈ IR+ Then, we use (A.9) to obtain
f(t)2≥ f(t1)2− k |t − t1| (A.10)Assume now that the conclusion of the lemma does not hold i.e, eitherlimt →∞
each T ≥ 0 there exists an infinite unbounded sequence {t1, t2, }, denoted {t n } ∈ IR+ with t n → ∞ as n → ∞, and a constant ε > 0 such that
f(t i)2> ε ∀ t i ≥ T (A.11)
To better see this, we recall that if limt →∞ f(t) exists and is zero then,
for any ε there exists T (ε) such that for all t ≥ T we have f(t)2 ≤ ε Furthermore, without loss of generality, defining δ := ε
2k, we may assumethat for all i ≤ n, t i+1− t i ≥ δ —indeed, if this does not hold, we may always
extract another infinite unbounded subsequence {t
i } such that t
i+1− t
i ≥ δ for all i.
Now, since Inequality (A.10) holds for any t and t1 it also holds for any
element of{t n } Then, in view of (A.11) we have, for each t ibelonging to{t n } and for all t ∈ IR+,
Notice that in the integrals above, t ∈ [t i , t i + δ] therefore, −k|t − t i | ≥ −kδ.
From this and (A.13) it follows that
Trang 8We see that on one hand, the term on the left-hand side of Inequality (A.15)
is bounded by assumption (since f ∈ L n
2 ) and on the other hand, since{t n }
is infinite and (A.14) holds for each t i the term on the right-hand side ofInequality (A.16) is unbounded From this contradiction we conclude that itmust hold that limt →∞ f(t) = 0 which completes the proof.
♦♦♦
As an application of Lemma A.5 we present below the proof of Lemma 2.2used extensively in Parts II and III of this text
Proof of Lemma 2.2 Since V (t, x, z, h) ≥ 0 and ˙V (t, x, z, h) ≤ 0 for all
x, z and h then these inequalities also hold for x(τ), z(τ ) and h(τ ) and all
τ ≥ 0 Integrating on both sides of ˙V (τ, x(τ), z(τ), h(τ)) ≤ 0 from 0 to t we
obtain5
V (0, x(0), z(0), h(0)) ≥ V (t, x(t), z(t), h(t)) ≥ 0 ∀ t ≥ 0
Now, since P (t) is positive definite for all t ≥ 0 we may invoke the theorem
of Rayleigh–Ritz which establishes that x T Kx ≥ λmin{K}x T x where K is
any symmetric matrix and λmin{K} denotes the smallest eigenvalue of K, to
conclude that there exists6 p
m > 0 such that y T P (t)y ≥ p m {P }y2 for all
y ∈ IR n +m and all t ∈ IR+ Furthermore, with an abuse of notation, we will
denote such constant by λ min {P } It follows that
5 One should not confuse V (t, N, z, h) with V (t, x(t), z(t), h(t)) as often happens
in the literature The first denotes a function of four variables while the ond is a functional In other words, the second corresponds to the function
sec-V (t, x, z, h) evaluated on certain trajectories which depend on time Therefore,
V (t, x(t), z(t), h(t)) is a function of time.
6 In general, for such a bound to exist it may not be sufficient that P is positive definite for each t but we shall not deal with such issues here and rather, we assume that P is such that the bound exists See also Remark 2.1 on page 25.
Trang 9which, using the fact that V (0, x(0), z(0), h(0)) ≥ V (T, x(T ), z(T ), h(T )) ≥ 0
yields the inequality
so using that Q is positive definite we obtain7 x T Q(t)x ≥ λmin{Q}x2 for
all x ∈ IR n and t ∈ IR+ therefore
Finally, since by assumption ˙x ∈ L n
∞, invoking Lemma A.5 we may
con-clude that limt →∞ x(t) = 0.
Another useful observation is the following
Lemma A.7 Consider the two functions f : IR+ → IR n and h : IR+ → IR with the following characteristics:
Trang 10Proof According to the hypothesis made, there exist finite constants k f > 0 and k h > 0 such that
the output and A ∈ IR m ×m , B ∈ IR m ×n and C ∈ IR n ×m are matrices havingconstant real coefficients The transfer matrix function H(s) of the system is then defined as H(s) = C(sI − A) −1 B where s is a complex number (s ∈ C) The following result allows one to draw conclusions on whether y and ˙ y
belong to L n
2 or L n ∞ depending on whether u belongs to L n2 or L n ∞.
Lemma A.8 Consider the square matrix function of dimension n, H(s) ∈
IR n ×n (s) whose elements are rational strictly proper8 functions of the complex
variable s Assume that the denominators of all its elements have all their roots on the left half of the complex plane (i.e they have negative real parts).
Trang 11numer-Corollary A.2 For the transfer matrix function H(s) ∈ IR n ×n (s), let u and
y denote its inputs and outputs respectively and let the assumptions of Lemma A.8 hold If u ∈ L n
Lemma A.2 appears in
• Marcus M., Minc H., 1965, “Introduction to linear algebra”, Dover
Publi-cations, p 207
• Horn R A., Johnson C R., 1985, “Matrix analysis”, Cambridge University
Press, p 346
Theorem A.1 on partitioned matrices is taken from
• Horn R A., Johnson C R., 1985, “Matrix analysis”, Cambridge University
The definition of L p spaces are clearly exposed in Chapter 6 of
• Vidyasagar M., 1993, “Nonlinear systems analysis”, Prentice-Hall, New
Jersey
The proof of Lemma A.5 is based on the proof of the so-called Barb˘alat’slemma originally reported in
Trang 12• Barb˘alat B., 1959, “Syst`emes d’´equations diff´erentielles d’oscillations
non-lin´eaires”, Revue de math´ ematiques pures et appliqu´ ees, Vol 4, No 2, pp.
267–270
See also Lemma 2.12 in
• Narendra K., Annaswamy A., 1989, Stable adaptive systems, Prentice-Hall,
p 85
Lemma A.8 is taken from
• Desoer C., Vidyasagar M., 1975, “Feedback systems: Input–output ties”, Academic Press, New York, p 59.
1
f (t)
Figure A.2. Problem 1
Hint: Notice that f2(t) ≤ h2(t) where
Trang 14Support to Lyapunov Theory
B.1 Conditions for Positive Definiteness of Functions
The interest of Lemma A.4 in this textbook resides in that it may be used
to derive sufficient conditions for a function to be positive definite (locally orglobally) We present such conditions in the statement of the following lemma
Lemma B.1 Let f : IR n → IR be a continuously differentiable function with continuous partial derivatives up to at least second order Assume that
• f(0) = 0 ∈ IR
• ∂x ∂f(0) = 0∈ IR n
Furthermore,
• if the Hessian matrix satisfies H(0) > 0, then f(x) is a positive definite
function (at least locally).
• If the Hessian matrix H(x) > 0 for all x ∈ IR n , then f (x) is a globally
positive definite function.
Proof Considering Lemma A.4 and the hypothesis made on the function f (x)
we see that for each x ∈ IR n there exists an α (1 ≥ α ≥ 0) such that
f (x) = 1
2x T H(αx)x
Under the hypothesis of continuity up to the second partial derivative, if
the Hessian matrix evaluated at x = 0 is positive definite (H(0) > 0), then the Hessian matrix is also positive definite in a neighborhood of x = 0 ∈ IR n,
e.g for all x ∈ IR n such thatx ≤ ε and for some ε > 0, i.e.
H(x) > 0 ∀ x ∈ IR n:x ≤ ε
Trang 15Of course, H(αx) > 0 for all x ∈ IR n such that x ≤ ε and for any α
(1≥ α ≥ 0) Since for all x ∈ IR n there exists an α (1 ≥ α ≥ 0) and
f (x) = 1
2x T H(αx)x , then f (x) > 0 for all x n such thatx ≤ ε Furthermore, since by hypothesis f (0) = 0, it follows that f (x) is positive definite at least locally.
On the other hand, if the Hessian matrix H(x) is positive definite for all
x ∈ IR n , it follows that so is H(αx) and this, not only for 1 ≥ α ≥ 0 but for
any real α Therefore, f (x) > 0 for all x n and, since we assumed
that f (0) = 0 we conclude that f (x) is globally positive definite.
♦♦♦
Next, we present some examples to illustrate the application of the ous lemma
previ-Example B.1 Consider the following function f : IR2→ IR used in the
study of stability of the origin of the differential equation that modelsthe behavior of an ideal pendulum, that is,
f (x1, x2) = mgl[1 − cos(x1)] + J x
2 2
and is positive definite at x = 0 ∈ IR2 Hence, according to Lemma
B.1 the function f (x1, x2) is positive definite at least locally Notice
that this function is not globally positive definite since cos(x1) = 0
for all x1 = nπ2 with n = 1, 2, 3 and cos(x1) < 0 for all x1 ∈
Trang 16func-Example B.2 Consider the function f : IR n → IR defined as
f (˜ q) = U(qd − ˜q) − U(q d ) + g(q d)T q +˜ 1
ε q˜T K p˜q
where K p = K T
p > 0, q d ∈ IR n is a constant vector, ε is a real positive
constant andU(q) stands for the potential energy of the robot Here,
we assume that all the joints of the robot are revolute
The objective of this example is to show that if K p is selected sothat1
λmin{K p } > ε2k g then f (˜ q) is a globally positive definite function.
To prove the latter we use Lemma B.1 Notice first that f (0) = 0.
The gradient of f (˜ q) with respect to ˜q is
Clearly the gradient of f (˜ q) is zero at ˜q = 0 ∈ IR n
On the other hand, the (symmetric) Hessian matrix H(˜ q) of f(˜q),