18.034, Honors Differential Equations
Prof. Jason Starr
Notes from Lecture 36, 5/5/04
I. The winding number
Let R ⊂ IR2 be an open region, let
x' = F (x, y )
y ' = G (x, y )
be an autonomous differential system on R , and let C ⊂ R be an oriented, simpler
closed curve in R . In other words, C is the image of the circle under a 1-to-1 map
whose derivative vector is always nonzero, say h : [0,1] → R, h(1) = h(0) .
If C contains no equilibrium point of the system, the following function is welldefined and continuous:
f : C →
δ 1 ⊂ IR2 ,
f (q ) =
1
F (q ) + G (q )
2
2
⎡F (q ) ⎤
⎢
⎥.
⎣⎢G (q ) ⎦⎥
The composition foh : [0,1] → δ 1 is (essentially) a cts. map from the circle to the
circle. To such a map there is an associated integer n, the degree of the map. This
integer counts the number of times foh(t ) rotates counterclockwise around the circle
as t rotates once counterclockwise around the circle. If h , F and G are all
⎡g1 ⎤
continuously differentiable function, g = ⎢ ⎥ , and then the degree is simply
⎣g2 ⎦
1
2π
1
∫ [g
1
(t )g 2' (t ) − g1' (t )g 2 (t )] dt
0
This integer turns out to be independent of h(t ) (although it does depend on the
orientation of C ). it is called the winding number of (F, G ) about C .
Let p ∈ R be an equilibrium point. It is isolated if there exists ε > 0 such that p is
the only equilibrium point in the ε - ball about p . For any 0 < q < ε , consider the
circle C q of radius q centered at p . The winding number of (F ,6 ) about C q is
independent of q and is called the index of (F ,6 ) at p (or sometimes the Poincare
index).
Examples: (1) Let λ , µ > 0 and let F = λ x, G = µ y . Then p = (0,0) is an isolated
⎡q. cos (t )⎤
equilibrium point. Consider hq (t ) = ⎢
⎥, 0 < t
⎢⎣q. sin (t ) ⎥⎦
Then g(t ) =
⎡λ cos (t )⎤
⎢
⎥.
λ2 cos2 (t ) + µ 2 sin2 (t ) ⎣⎢ µ sin (t ) ⎦⎥
1
18.034, Honors Differential Equations
Prof. Jason Starr
Page 1 of 6
And g1 (t )g2' (t ) − g1' (t )g2' (t ) =
λµ
. This is closely related to the
λ cos (t ) + µ 2 sin 2 (t )
2
2
2π
Poisson kernel. It is nontrivial, but the integral
∫λ
0
2
λµ
dt
cos (t ) + µ 2 sin 2 (t )
2
can be
computed by elementary methods, and it equals 2π (consider the case that
λ = µ ).
So the index is +1.
(2)
λ, µ
< 0 . This is the same as above when
λ → - λ , µ → - µ . Notice the integral
does not change. So the index is +1.
(3)
λ <0, µ >0. now the integral is
2π
-
∫ (− λ )
2
0
(− λ )µ
dt .
cos 2 (t ) + µ 2 sin 2 (t )
This is -1 times the integral from (1). So the index is -1.
Theorem: Let C be a simple closed curve that contains no equ. pts. in R oriented so
the interior is always on the left. If the interior is contained in R , and if the interior
contains only finitely many equilibrium points, p1 ,....., pn , then the winding number
about C is index (p1 ) + .... + index (pn ) . (and 0 if there are no eq. pts).
Proof: This is proved, for instance in Theorem 3, § 11.9 on p.442 of Wilfred Kaplan,
Ordinary Differential Equations, Addison-Wesley, 1958.
Corollary:If R is simply-connected, then every cycle C contained in R contains an
equilibrium point in its interior.
Rmk: A region R in IR2 is simply-connected if for every simple closed curve C in R ,
the interior of C is contained in R . A cycle is a periodic orbit (that is necessarily a
simple closed curve).
Pf: By construction, (F ,6 ) is parallel to the tangent vector of C . Therefore the
winding number is +1. So, by the theorem, there is an equilibrium point in the
interior of C .
II. Lyapunov functions
Let R
()
⊂ IRn be an open region. Let x ’= F x be an autonomous system on R . Let
p ∈ R be a point.
Definition: A function V : R → IR is positive definite (resp. negative definite) if
(1) V (q ) ≥ 0 (resp. V (q ) ≤ 0 ) for all q ∈ R
(2) V (q ) = 0
iff q = p .
Let p be an equilibrium point.
18.034, Honors Differential Equations
Prof. Jason Starr
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Definition: A strong Lyapunov function is a continuously differentiable function
V : R → IR such that
(1) V is positive definite
(2) the function V ' :=
n
∑
i =1
dV(x )
Fi (x ) is negative definite.
dx i
Remark : It is often the case that there is no strong Lyapunov function on R , yet
there is an open subregion R ’ ⊂ R containing p and a strong Lyapunov function on
R ’. In this case, simply replace R by R ’ in what follows.
Hypothesis: Suppose a strong Lyapunov function exists. There is a minor issue that
your book does not deal with: long-time existence of solution curves. Let K ⊂ IRn be
a bounded closed region whose interior contains p and such that K ⊂ R . Define
r0 = minimum of V on the bounded closed set ∂K (a continuous function on a
bounded closed subset of IRn always attains a minimum). Because p ∈ interior of K ,
r0 > 0 . Define R ’ to be
R ’= KnV −1
( [0, r ] ) = { q ∈ K / V( ) < r }.
0
q
0
Observe this is an open region in R that contains p and is contained in the interior of
K.
Theorem:(1) For every x 0 ∈ R ’, the solution curve x (t ) is defined for all t > 0 .
(2) Moreover, lim x (t ) = p . Therefore p is an attractor and R ’ is in the basin of
t →∝
attraction of p .
Proof: For any x 0 ∈ R , if x (t ) is defined on the interval [0, t1 ) , consider V (x (t ))
defined on [0, t1 ) . By the Chain Rule, V (x (t )) is differentiable and
d
V (x (t )) =
dt
n
∂V
dx i
∑ ∂x (x (t )). dt
i =1
(t ) .
i
d
V (x (t )) = V ' (x (t )) .
dt
By hypothesis, this is nonpositive. Therefore V (x (t )) is a non-increasing function. In
particular, if x 0 ∈ R ’, then x (t ) is in R ’ for all t ∈ [0, t1 ) .
By hypothesis, x i' (t ) = Fi (x (t )) . Thus
(1) Let x 0 ∈ R ’. By way of contradiction, suppose that x (t ) is defined only on
[0, t1 ) where t1 is finite. By the theorem on maximally extended solutions,
lim x (t ) exists and is in ∂K . Therefore V ( lim x (t ) ) ≥ r0 . Since V is
t → t1
t → t1
continuous
V ( lim x (t ) ) = lim V ( x (t ) ). For all t ≥ 0 , V ( x (t ) ) ≤ V (x 0 ) < r0 .
t → t1
t → t1
So lim V ( x (t ) ) ≤ V (x 0 ) < r0 . This contradiction proves x (t ) is defined for all
t → t1
t>0.
18.034, Honors Differential Equations
Prof. Jason Starr
Page 3 of 6
(2) Let ε>0 and let Bε (p ) denote the open ball of radius ε centered at p . The set
difference K \ (KnBε (p )) is closed and bounded. Therefore V attains a
mimimum value r1 on this set. Since p is not in this set r1 > 0 . Also,
KnV −1 ([r1 , ∞ )) is a closed set contained in K . So it is closed and bounded ( K
is bounded). Therefore V ’ attains a maximum value −m1 on this set. Since p
is not in this set −m1 < 0 , i.e. m1 > 0 .
Define t1 =
r0 − r1
.
m1
The claim is that for all x 0 ∈ R ’, V ( x (t ) ) < r1 for all. In particular, since
x (t ) ∈ R' & V (x (t )) < r1 , x (t ) is in R ’ n Bε (p ) . By way of contradiction, suppose
V ( x (t ) ) ≥ r1 . By the mean value theorem, there exists t ’ with 0 < t ' < t such
( ( ))
()
that V (x 0 ) − V (x (t )) = −V ' x t ' ⋅ t . Since V ( x (t ) ) ≥ r1 , also V ( x t ' ) ≥ r1 .
()
'
Therefore x t & KnV
−1
[[r1, ∞ )] .
Thus - V ’( x (t ) ) ≥ m1 . So V (x 0 ) − V (x (t )) ≥ m1t > m1t1 = r0 − r1 . But V (x 0 ) < r0
and V ( x (t ) ) ≥ r1 . This is a contradiction, proving V ( x (t ) ) < r1 for all t > t1 .
The definition of a weak Lyapunov function as well as the statements of
Lyapunov’s second and third theorems are in the textbook.
III. A criterion for asymptotic stability.
Let V be a real vector space of dimension n , eg. IRn. Let R
()
⊂ V be an open region,
and let x ’= F x be an autonomous system on R . Let p ∈ R be an equilibrium
point.
⎡ ∂F ⎤
Theorem: If F is differentiable at p , and if every eigenvalue of ⎢ i ⎥ has negative
⎣⎢ ∂R j ⎦⎥ p
real part, then there is an open region R ’ ⊂ R contains p and a story Lyapunov
function on R ’.
Proof:There is a beautiful proof in the first edition of the textbook, which is stapled at
the end. Here we give a closely related, but different argument.
The Jacobian of F at p is a linear transformation T : V → V with the property
that, for my norm − on V , for every ε > 0 , ∃ ∂ 2 >0 such that if V < ∂ 2 , then
F(p + v ) − F(p ) − Tv ≤ ε . V . Notice this is independent of the system of coordinates on
V . Without loss of generality, translate so p =0.
As we have alluded to earlier in the semester, for each real vector space V there is
an associated complex vector space Vc defined as a set to be V × V with elements
(v, w ) written v + iw . The addition is defined component-by-component. And for
18.034, Honors Differential Equations
Prof. Jason Starr
Page 4 of 6
each complex number α + iβ , (α + iβ ) . (v + iw ) is defined to be (αv − βw ) + i (βv + αw ) .
The original vector space V is a subset by v 6 v + i ⋅ 0. And T : V → V extends to a
CI-linear transformation T⊄ : V ⊄ → V⊄ by T⊄ (v + iw ) = T (v ) + iT (w ) .
By the Jordan normal form theorem, there exists a direct sum decomposition
V⊄ = V1 , ⊕ ….. ⊕ V n and for each i = 1,......, n an ordered basis Bi for Vi s.t.
(1) for each i = 1,......, n ,
T ( Vi )
⊂ Vi
(2) the corresponding linear transformation Ti : V i → V i has matrix
[Ti ] B ,B
i
i
⎡λ ⎤
⎣0 ⎦
=⎢ ⎥
for some
For any nonzero
α Є CI, , there is also a basis
λ.
Bi ,α s.t.
[Ti ]Bi ,α , Bi ,α = [ ] .
(
)
Indeed, if Bi = (v1 ,....., v m ) , then Bi ,α = v1 , α v 2 , α 2v 3 ,...., α m −1v m .
For each ordered basis B for V i , there is a “dual basis of coordinates” x1 ,....., x m :
Vi → C
I, s.t.
V = x1 (v )v1 + ...... + x m (v ) ⋅ v m for every v Є V : (B = (v1 ,...., v m )) . There is a
corresponding Hermition inner product,
<·,·> B : V i × V i → C
I,
m
< v, w > B =
∑ x (v ) x (w) .
i
i =1
i
In particular, this is bilinear, positive definite and < w, v > B = (< v , w > B ).
And < Ti v, v > Biα = λ x1
2
+ α x1 x2 + λ x2
2
2
+ α x 2 x 3 + ...... + αx m −1 x m + λ x m .
Lemma1: For each n , the function on C
I n,
(x1 ,...., x n ) 6
x1
2
− x1 x2 + x2
∑x
+ ... + x k
2
n −1
=
2
k
− x k x k +1 + x n
2
− x k x k + 1 + x k +1
2
+ ... + − x n −1 x n + x n
2
2
k =1
is positive definite.
n −1
∑(
1
1
2
Proof: it is simply
x1 +
x k − x k +1
2
2 k =1
18.034, Honors Differential Equations
Prof. Jason Starr
2
)
+
1
2
xn .
2
Page 5 of 6
Since it is a sum of squares, it is nonnegative. It is zero iff x1 = 0 ,
x 2 − x1 = 0,...., x n − x n −1 = 0 and x n = 0, i.e. x1 = .... = x n = 0 .
Lemma 2:If Re (λ ) < 0 and if α < −Re (λ ) , then 2 Re < Ti v , v > βi ,α is negative definite.
(
)
Moreover, 2 Re < Ti v , v > βi ,α ≤ 2 Re (λ ) + α ⋅ ⎛⎜ x1
⎝
2
2
+ ..... + x n ⎞⎟ .
⎠
(
)
2
2
Proof: 2 Re < Ti v , v > βi ,α = 2 Re (λ ) ⎛⎜ x1 + ... + x n ⎞⎟ + 2Re α x1 x2 + ... + α x n −1 x n
⎝
⎠
2
2
≤ 2Re (λ ) ⎛⎜ x1 + ... + x n ⎞⎟ +2 α ⎛⎜ x1 x2 + ... + x n −1 x n ⎞⎟
⎝
⎠
⎝
⎠
2
2
2
2
2
2
=2 Re (λ ) + α ⎛⎜ x1 + ... + x n ⎞⎟ − α ⎛⎜ x1 − x1 x2 + x2 + ... + x n −1 − x n −1 x n + x n ⎞⎟
⎠
⎠
⎝
⎝
(
)
2
2
2
2
By Lemma 1, - α ⎛⎜ x1 − x1 x2 + x2 + ... + x n −1 − x n −1 x n + x n ⎞⎟ is negative
⎝
⎠
2
2⎞
⎛
definite. Because Re (λ ) + α < 0 , also 2 (Re (λ ) + α ) ⎜ x1 + ... + x n ⎟ is negative
⎝
⎠
definite.
For each i = 1,....., n, let Re (λ ) > ε i > 0 . Let α i + Re (λ ) < −ε i , i.e. 0< α i < Re (λ ) − ε i .
Define the function ⋅
2
by v1 + .... + v n
n
2
∑ < v ,v
=
i
i
This is a positive definite function. Moreover, ⋅ :=
.
where v i Є V i
>
i =1
Bi ,αi
2
is a norm. Therefore there is
a δ > 0 s.t. if v < δ , then F (v ) − Tv ≤ min (ε 1 ,..., ε n ) v .
Now
d
2
x (t ) = 2Re < F (x ), x (t ) >=
dt
So 2Re < F (x ), x > ≤
n
∑ 2R
e
< Ti x, x > +2Re < F(x )Tv , v >
i =1
n
∑ 2R
e
< Ti x, x > + F(x ) − Tx . x .
i =1
By Lemma 2, this is ≤ -2 min (ε 1 ,..., ε n ) x
2
If x < δ , this is ≤ -2 min (ε 1 ,..., ε n ) x
+ min(ε 1 ,..., ε n ) x
= -min (ε 1 ,..., ε n ) x
2
+ F(x ) − Tx ⋅ x
2
2
So this is negative semidefinite. Therefore ⋅
2
is a strong Lyapunov function on the
ball of radius R centered at p.
18.034, Honors Differential Equations
Prof. Jason Starr
Page 6 of 6