NonlinearBook10pt November 20, 2007
626 CHAPTER 10
Specifically, with (10.118), (10.134), and (10.141), the Hamiltonian has the
form
H(x, u) = L
1
(x) + L
2
(x)u + u
T
R
2
(x)u + V
′
(x)(f(x) + G(x)u)
+
1
4γ
2
V
′
(x)J
1
(x)J
T
1
(x)V
′T
(x) + [h(x) + J(x)u]
T
[h(x) + J(x)u].
Now, the proof follows as the proof of Corollary 10.5.
Note that since Γ(x, φ(x)) ≥ 0, x ∈ R
n
, (10.139) implies that
˙
V (x)
△
= V
′
(x)[f(x) + G(x)φ(x)] < 0, x ∈ R
n
, x 6= 0, (10.149)
with φ(x) given by (10.143). Furthermore, (10.136), (10.138), and (10.149)
ensure th at V (·) is a Lyapunov function for the undisturbed closed-loop
system (10.142). In addition, with L
1
(x) given by (10.146) and φ(x) given
by (10.143), L(x, u) + Γ(x, u) can be expressed as
L(x, u) + Γ(x, u)
= [u − φ(x)]
T
R
2a
(x)[u − φ(x)] −V
′
(x)[f(x) + G(x)u]
= [u +
1
2
R
−1
2a
(x)(L
T
2
(x) + 2J
T
(x)h(x)]
T
R
2a
(x)
[u +
1
2
R
−1
2a
(x)(L
T
2
(x) + 2J
T
(x)h(x)]
−V
′
(x)[f(x) + G(x)φ(x)] −
1
4
V
′
(x)G(x)R
−1
2a
(x)G
T
(x)V
′T
(x).
(10.150)
Since R
2a
(x) ≥ R
2
(x) > 0 for all x ∈ R
n
the first term of the right-hand
side of (10.150) is nonnegative, while (10.149) implies that the second term
is nonnegative. Thus, we have
L(x, u) + Γ(x, u) ≥ −
1
4
V
′
(x)G(x)R
−1
2a
(x)G
T
(x)V
′T
(x), (10.151)
which shows that L(x, u) + Γ(x, u) may be negative. As a result, there
may exist a control in put u for which the auxiliary performance functional
J(x
0
, u) is negative. However, if the disturbance rejection control u is a
regulation controller, that is, u ∈ S(x
0
), th en it f ollows from (10.144) and
(10.147) that
J(x
0
, u(·)) ≥ V (x
0
) ≥ 0, x
0
∈ R
n
, u(·) ∈ S(x
0
).
Furth ermore, in this case substituting u = φ(x) into (10.150) yields
L(x, φ(x)) + Γ(x, φ(x)) = −V
′
(x)[f(x) + G(x)φ(x)],
which, by (10.149), is positive.
Next, we specialize Theorem 10.4 to the p assivity case. Specifically,
we consider th e case where p = d and r(z, w) = 2z
T
w. For the following
result we consider performance variables
z(t) = h(x(t)) + J(x(t))u(t) + J
2
(x(t))w(t), (10.152)