Linearization
x = f (x),
f (0) = 0
f is continuously differentiable over D = cfw_kxk < r
J(x) =
f
(x)
x
h() = f (x) for 0 1,
h(1) h(0) =
Z
h () = J(x)x
1
h () d,
h(0) = f (0) = 0
0
f (x) =
Z
0
Nonlinear Control Lecture # 2 Stability of Equilibrium Points
Exponential Stability
Definition 3.3
The equilibrium point x = 0 of x = f (x) is exponentially
stable if
kx(t)k kkx(0)ket , t 0
k 1, > 0, for all kx(0)k < c
It is globally exponentially stable if the inequality is satisfied
for any initial state x(0)
Expo
Scalar Systems (n = 1)
The behavior of x(t) in the neighborhood of the origin can be
determined by examining the sign of f (x)
The requirement for stability is violated if xf (x) > 0 on
either side of the origin
f(x)
f(x)
x
x
Unstable
f(x)
Unstable
Nonlin
Lyapunovs Theorem (3.3)
If there is V (x) such that
V (0) = 0 and V (x) > 0,
V (x) 0,
x D with x 6= 0
xD
then the origin is a stable
Moreover, if
V (x) < 0,
x D with x 6= 0
then the origin is asymptotically stable
Nonlinear Control Lecture # 2 Stability
Example 4.9
x 1 = x2 ,
x 2 = h(x1 ) x2 + u(t),
h(x1 ) = x1 31 x31
|u(t)| d
V (x) =
2 2
x
3 1
1 T
x
2
Z x1
k k
h(y) dy,
x+
k 1
0
x1 h(x1 ) x21 ,
5 2
x
12 1
Z
0
0<k<1
x1
h(y) dy 12 x21 , |x1 | 1
min(P1 )kxk2 xT P1 x V (x) xT P2 x max (P2 )kxk2
k + 56 k
k+1
Example 3.14
x 1 = x2 ,
x 2 = x1 + (x21 1)x2
f
0 1
=
A=
1 1
x x=0
has eigenvalues (1 j 3)/2. Hence the origin is
asymptotically stable
1.5 0.5
T
Take Q = I, P A + A P = I P =
0.5
1
min (P ) = 0.691
Nonlinear Control Lecture # 3 Stability of Equilibrium P
Theorem 4.6
Let V (x) be a continuously differentiable function
1 (kxk) V (x) 2 (kxk)
V
f (x, u) W3 (x), kxk (kuk) > 0
x
x Rn , u Rm , where 1 , 2 K , K, and W3 (x) is
a continuous positive definite function. Then, the system
x = f (x, u) is ISS with = 1
Terminology: A function V (t, x) is said to be
positive semidefinite if V (t, x) 0
positive definite if V (t, x) W1 (x) for some positive
definite function W1 (x)
radially unbounded if V (t, x) W1 (x) and W1 (x) is
radially unbounded
decrescent if V (t, x
Lemma 4.1
Let 1 and 2 be class K functions on [0, a1 ) and [0, a2 ),
respectively, with a1 limra2 2 (r), and be a class KL
function defined on [0, limra2 2 (r) [0, ) with
a1 limra2 (2 (r), 0). Let 3 and 4 be class K
functions. Denote the inverse of i by i
Definition 5.1
y = h(t, u) is
passive if uT y 0
lossless if uT y = 0
input strictly passive if uT y uT (u) for some function
where uT (u) > 0, u 6= 0
output strictly passive if uT y y T (y) for some function
where y T (y) > 0, y 6= 0
Nonlinear Control L
V
Ax = xT x
x
c3 = 1, c4 = 2 kP k = 2max (P ) = 2 1.513 = 3.026
V (x) = xT P x,
kg(x)k = |x2 |3
g(x) satisfies the bound kg(x)k kxk over compact sets of
x. Consider the compact set
c = cfw_V (x) c = cfw_xT P x c,
c>0
max |x2 | = max 0 1 x
xT P xc
xT P x
Lyapunovs Method
Let V (x) be a continuously differentiable function defined in a
domain D Rn ; 0 D. The derivative of V along the
trajectories of x = f (x) is
V (x) =
n
X
V
i=1
=
=
xi
V
,
x1
x i =
n
X
V
i=1
V
,
x2
V
f (x)
x
Nonlinear Control Lecture # 2
Theorem 3.2
The origin is exponentially stable if and only if Re[i ] < 0
for all eigenvalues of A
The origin is unstable if Re[i ] > 0 for some i
Linearization fails when Re[i ] 0 for all i, with Re[i ] = 0
for some i
Example 3.3
3
x = ax ,
f
2
=
3ax
=0
= max V (x)
dkxkr
Z t
V (x(t) = V (x(0) +
V (x( ) d V (x(0) t
0
This inequality contradicts the assumption c > 0
The origin is asymptotically stable
The condition kxk V (x) implies that the set
c = cfw_x Rn | V (x) c is compact for every c > 0. This is
Variable Gradient Method
V
f (x) = g T (x)f (x)
V (x) =
x
g(x) = V = (V /x)T
Choose g(x) as the gradient of a positive definite function
V (x) that would make V (x) negative definite
g(x) is the gradient of a scalar function if and only if
gj
gi
=
, i, j
Proof
V (x) 0 in
V (x(t) is a decreasing
V (x) is continuous in V (x) b = min V (x)
x
lim V (x(t) = a
t
x(t) x(t) is bounded L+ exists
Moreover, L+ and x(t) approaches L+ as t
For any p L+ , there is cfw_tn with limn tn = such that
x(tn ) p as n
V (x)
Theorem 3.1
The equilibrium point x = 0 of x = Ax is stable if and only if
all eigenvalues of A satisfy Re[i ] 0 and for every eigenvalue
with Re[i ] = 0 and algebraic multiplicity qi 2,
rank(A i I) = n qi , where n is the dimension of x. The
equilibrium
f (x) =
Z
1
J(x) d x
0
Set A = J(0) and add and subtract Ax
Z 1
f (x) = [A + G(x)]x, where G(x) =
[J(x) J(0)] d
0
G(x) 0 as x 0
This suggests that in a small neighborhood of the origin we
can approximate the nonlinear system x = f (x) by its
linearization
Linear Systems
x = Ax
V (x) = xT P x,
P = PT > 0
def
V (x) = xT P x + x T P x = xT (P A + AT P )x = xT Qx
If Q > 0, then A is Hurwitz
Or choose Q > 0 and solve the Lyapunov equation
P A + AT P = Q
If P > 0, then A is Hurwitz
MATLAB: P = lyap(A , Q)
Nonlin
Furthermore, if V (x) > 0, x 6= 0,
kxk V (x)
and V (x) < 0, x 6= 0, then the origin is globally
asymptotically stable
Nonlinear Control Lecture # 2 Stability of Equilibrium Points
Theorem 4.5
Suppose
c1 kxk2 V (x) c2 kxk2
V
f (t, x) c3 kxk2 , x D with kxk , t 0
x
p
for some positive constants c1 to c3 , and < c/c2 . Then,
c = cfw_V (x) c is positively invariant and x(t0 ) c
V (x(t) max V (x(t0 )e(c3 /c2 )(tt0 ) , c2 2 , t t0
kx(t
V
V
V
Ax =
f (x)
G(x)x
x
x
x
c3 kxk2 + c4 Lkxk2
= (c3 c4 L)kxk2
def
Take L < c3 /c4 , = (c3 c4 L) > 0
V
Ax kxk2 , kxk < mincfw_r0 , r1
x
The origin of x = Ax is exponentially stable
Nonlinear Control Lecture # 3 Stability of Equilibrium Points
A continuously differentiable function V (x) satisfying the
conditions for stability is called a Lyapunov function. The
surface V (x) = c, for some c > 0, is called a Lyapunov surface
or a level surface
c3
c2
V (x) = c 1
c 1 <c 2 <c 3
Nonlinear Control Le
The Invariance Principle
Definitions
Let x(t) be a solution of x = f (x)
A point p is a positive limit point of x(t) if there is a sequence
cfw_tn , with limn tn = , such that x(tn ) p as n
The set of all positive limit points of x(t) is called the posit