Download Definition1 , f is and is an equilibrium point

Note: f(x) ~ o(x) 
f (x)
.
x
Theorem: Let x  W be a sink of x = f (x) . Suppose every eigenvalue of Df (x )
f1
f1
 f1

(x)
(x) 
(x)

x2
xn
 x1

 
 



 has negative real part less than  c , c  0, then  a




 f n ( x ) f n ( x )  f n ( x ) 
 x

x2
xn
 1

neighborhood u  W of x such that
(a) t (x) is defined and in u  x  u , t >0 .
(b)  a Euclidean norm on R n such that t ( x)  x  e  tc x  x ,  x  u , t  0.
(c) For any norm on R n ,  B > 0,  t ( x)  x  B e  tc x  x ,  x  u , t  0.
Proof:
Assume x  0, A  Df (0) , Re   b  c  0 where  is an eigenvalue of A.
f(x)=f(0)+Df(0)(x-0)+o(x) ( by Taylor’s series expansion )
=Ax+o(x)
2
By Lemma 2, we know that Ax , x   b x
x  R n .
f ( x )  Ax
 0 as x  0 .
x
f ( x )  Ax , x
 lim
 0.
2
x 0
x
Hence
So pick a small   0   0  x   , then x  W
and
f ( x ), x
x

2

Ax , x
x
2
 .
2
2
f ( x ), x  Ax , x   x  (b  ) x  c x

2

Let u  x  R n x   and x(t), 0  t  t o be a solution curve in u, and x(t)  0 .
x , x
f ( x ), x
cx
d
x


 c x
dt
x
x
x
2
since x ( t ) is decreasing , then x(t )  u t  0,  ( u is compact) ----(a)
 t (x)  x(t )  ect x(0) t  0 ----(b)
(c) is clear from (b) since norms are equivalent in finite dimension.
Next, let us consider the stability issue of a nonlinear dynamical system.
Definition1:
x  f ( x ) , f is C1 and x is an equilibrium point. Then x is stable if 
neighborhood V of x  a neighborhood V1 of x such that every
solution x(t) with x(0) in V1 is defined and in V  t > 0.
x
V1
x (0)
V
Definition2:
In addition, lim x ( t )  x , then x is asymptotically stable.
t 
Example1:
A sink is asymptotically stable and stable.
Example2:
  AX has imaginary eigenvalues, then the origin is stable but not
If A in X
asymptotically stable.
V
Definition3:
x  f ( x ) , f is C1 and x is an equilibrium point. Then x is unstable if 
neighborhood V of x ,  neighborhood V1 of x such that  at least one
solution x(t) with x(0) in V1 is not entirely in V  t > 0.
Motivation:
If all eigenvalues are on the imaginary axis, then the origin is not a sink, but is it
asymptotically stable?
Theorem (Lyapunov)
Let x  W be an equilibrium point for x  f ( x ) . Let u  W be an open set and
V : u  R be a differentiable function on u  x and continuous on u .
If (a) V( x ) =0 and V( x ) >0 if x  x ,
  0 in U  x ,
(b) V
then x is stable.
 < 0 in u  x , then x is asymptotically stable.
Moreover, if (c) V
Proof:
Let  > 0 be small enough such that B (x)  x  u | x  x |   u . Let  be the
minimum value of V on S (x)  x  u | x  x | . Then   0 by (a).
Let u1  x  B (x) V(x)   be an open set, then no solution starting in u1 can
meet S ( x ) since V is nonincreasing on solution curves. Hence, every solution starting
in u1 stays in B ( x ) .  x is stable.
Moreover, if V is strictly decreasing in u  x , then x ( t n )  zo  B ( x ) for some
sequence t n   ( B ( x ) is compact ).
We contend that zo  x .
 V( x ( t ))  V(zo ) t  0 . Hence, V(x(t n ))  V(zo ) ( x(t n )  zo and V is
continuous ).
If zo  x , let z( t ) be the solution of initial z o , we have V(z(s))  V(zo ) s  0.
Hence, for any solution y (s ) starting very close to z o , we have V( y(s))  V(zo ) .
Let y(0)  x(t n ) for sufficiently large n. Hence, V( x( t n s ))  V(zo )  .
Therefore, zo  x .