Jim Lambers
MAT 605
Fall Semester 2015-16
Lecture 22, 23 and 24 Notes
These notes correspond to Sections 6.1-6.4 in the text.
Stability of Fixed Points
We have previously studied the stability of fixed points through phase portraits. We now provide
a formal definition of this notion of stability.
Definition 1 (Stability of Fixed Points) Let X : D ⊆ Rn → Rn be a vector field with
~ and let x ∈ D. Let Ix = (ax , bx ) denote the maximum interval of existence for the
flow φ,
integral curve of x0 = X(x) passing through x at t = 0. A fixed point c of X is stable if
~ t (x) = c
lim φ
x→c
for each fixed t ∈ [0, bx ).
Because the flow is defined for all t > bx if it remains in a compact set, we have the following
proposition that extends the preceding definition somewhat.
Proposition 1 A fixed point c of X is stable if and only if
~ t (x) = c
lim φ
x→c
for all t ≥ 0.
If a fixed point c is stable, then that means the flow converges to c as the initial data x converges
to c. The following stronger notion of stability requires that the flow converges to c not just for x
that is arbitrarily close to c, but for all x within a neighborhood of c.
Definition 2 (Asymptotic Stability) A fixed point c of X is asymptotically stable
if it is stable and there exists δ > 0 such that
~ t (x) = c
lim φ
t→∞
for kx − ck < δ.
A key distinction between stability and asymptotic stability is that in stability, the size δ of the
ball around c depends on in the definition of a limit, whereas for asymptotic stability it does not.
1
Linear Stability of Fixed Points
For the case of linear systems, stability of fixed points can readily be determined from the fundamental matrix. To state results concerning stability, we use the following norms:
• For a vector x ∈ Rn , kxk is the Euclidean norm of x.
• For an n × n matrix A, kAk is the largest entry of A in absolute value.
We then have the following simple criterion for stability.
Theorem 1 (Linear Stability) If A is an n × n matrix that has eigenvalues with all
negative real parts, then there exist positive constants t0 , K, m such that
ketA k ≤ Ke−mt
for t ≥ t0 . It follows that
ketA xk ≤ nKe−mt kxk
for x ∈ Rn and t ≥ 0. It follows that 0 is an asymptotically stable fixed point of x0 = Ax.
Corollary 1 (Linear Instability) If A is an n × n matrix whose eigenvalues have
positive real parts, then there exist positive constants t0 , K, m such that
Kemt kxk ≤ ketA xk,
for x ∈ Rn and t ≥ t0 . It follows that
lim ketA xk = ∞
t→∞
for x 6= 0, and therefore 0 is an unstable fixed point of x0 = Ax.
Proof: Let −m be a strict upper bound on the real parts of all of the eigenvalues. Then, each
entry of emt etJ , where A = P JP −1 is the Jordan decomposition of A and J is the Jordan canonical
form of A, has a factor of the form e(m+a)t where a is the real part of an eigenvalue. Because these
exponentials all decay to zero as t → ∞, it follows that for some t0 > 0, all entries of emt etJ have
absolute value less than 1, and therefore
ketJ k ≤ e−mt
for t ≥ t0 . We then have
ketA k = kP etJ P −1 k
≤ n2 kP kkP −1 kketJ k
≤ Ke−mt ,
where we define K = n2 kP kkP −1 k. We also have

2

2
n
n
n
X
X
X
 [etA ]ij xj  ≤ e−2mt n2 K 2 
ketA xk2 =
xj  ≤ n2 K 2 e−2mt kxk2 .
i=1
j=1
j=1
2
To show that 0 is a stable fixed point of x0 = Ax, we let > 0 and let
δ=
n2 Kke−t0 A k
.
Then, for kxk < δ, we have
~ t (x) − 0k = ketA xk
kφ
≤ ke(−t0 +t+t0 )A kkxk
≤ nke−t0 A kke(t+t0 )A xk
≤ n2 Kke−t0 A ke−m(t+t0 ) δ
≤ e−m(t+t0 ) < and therefore 0 is a stable fixed point. The asymptotic stability of 0 follows from
lim etA x = 0
t→∞
for all x ∈ Rn . 2
It turns out that the assumptions in the preceding corollary can be significantly weakened.
Proposition 2 (Linear Instability) If A has at least one eigenvalue with positive real
part, then 0 is an unstable fixed point of x0 = Ax.
Nonlinear Stability
We now seek to use stability theory for fixed points of linear systems to determine the stability of
fixed points for nonlinear systems. This can be accomplished using topological equivalence.
Proposition 3 (Invariance of Stability) Let f be a homeomorphism such that x0 =
X(x) and y0 = Y(y) are topologically equivalent. Let c be a fixed point of X, and let
b = f (c). Then c is stable if and only if b is stable, and c is asymptotically stable if and
only if b is asymptotically stable.
~ X and φ
~ Y are the flows for X and Y, respectively, then they are
Proof: First, we show that if φ
t
t
related by
~Y = f ◦ φ
~ X ◦ f −1 .
φ
t
t
~
Let α
~ (t) be an integral curve of X, and let x = α
~ (0). Then β(t)
= f (~
α(t)) is an integral curve of
Y, and we let y = f (x). Then
~ Y (y) = β(t)
~
φ
t
= f (~
α(t))
~ X (x))
= f (φ
t
~ X (f −1 (y)).
= f (φ
t
3
We assume c is stable and prove that b is stable; the converse is similar. Let > 0. Furthermore,
because f is a homeomorphism, there exists 0 > 0 such that
f (B(c, 0 )) ⊂ B(b, ).
Because c is stable, there exists δ0 > 0 such that
~ X (B(c, δ0 )) ⊂ B(c, 0 ).
φ
t
Then, again because f (by extension f −1 ) is a homeomorphism, there exists δ > 0 such that
f −1 (B(b, δ)) ⊂ B(c, δ0 ).
We then have, for all t ≥ 0,
~ Y (B(b, δ)) = f (φ
~ X (f −1 (B(b, δ))))
φ
t
t
~ X (B(c, δ0 )))
⊂ f (φ
t
⊂ f (B(c, 0 ))
⊂ B(b, ).
Therefore b is stable.
Now, suppose that c is also asymptotically stable. Then there exists δ0 > 0 such that
~ X (x) = c
lim φ
t
t→∞
for all x ∈ B(c, δ0 ). Because f −1 is a homeomorphism, there exists δ > 0 such that
f −1 (B(b, δ)) ⊂ B(c, δ0 ).
By the continuity of f and f −1 , we have, for all y ∈ B(b, δ), x = f −1 (y) ∈ B(c, δ0 ) and
~ X (f −1 (y)))
lim f (φ
t
~ X (f −1 (y))
= f lim φ
t
t→∞
~ X (x)
= f lim φ
t
~ Y (y) =
lim φ
t
t→∞
t→∞
t→∞
= f (c)
= b.
We conclude that b is asymptotically stable. 2
We can now use linear stability theory to determine the stability of fixed points of nonlinear
systems.
Corollary 2 (Nonlinear Stability) Let c be a simple fixed point of X. If all eigenvalues
of X0 (c) have negative real parts, then c is asymptotically stable. If any eigenvalue of
X0 (c) has a positive real part, then c is unstable.
Proof: Let A = X0 (c). Because c is a simple fixed point, by the Linearization Theorem,
x0 = X(x) and y0 = Ay are topologically equivalent for x near c and y near 0. By the preceding
proposition, c and 0 have the same stability type. The Linear Stability Theorem can be used to
determine the stability type of 0. 2
4
Liapunov Functions
We now consider the case where a fixed point c of X is not hyperbolic, so the Linearization Theorem
does not apply. First, we need the following definitions.
Definition 3 (Covariant Derivative) Let F : D ⊆ Rn → R be C 1 . The covariant
derivative of F along X, denoted by ∇X F , is defined by
∇X F (x) ≡ ∇F (x) · X(x) =
n
X
X j (x)
j=1
∂F
(x).
∂xj
Definition 4 (Liapunov Functions) Let c be a fixed point of X. A Liapunov function for c is a C 1 function Λ defined on a neighborhood U of c such that
1. Λ(c) < Λ(x) for x ∈ U \{c}, and
2. ∇X Λ(x) ≤ 0 for x ∈ U \{c}.
If, in addition, Λ satisfies the strict inequality ∇X Λ(x) < 0 for x ∈ U \{c}, then Λ is
called a strict Liapunov function for c.
Now, we can easily characterize the stability of fixed points.
Theorem 2 (Liapunov Stability Theorem) Let c be a fixed point of X. If there
exists a Liapunov function Λ for c, then c is a stable fixed point. If Λ is a strict Liapunov
function for c, then c is an asymptotically stable fixed point.
It is important to note that when possible, it is best to use the Linearization Theorem to determine
the stability of simple fixed points, because it can be very difficult to determine whether c has a
Liapunov function. Liapunov functions should therefore only be sought when c is not simple.
The geometric intuition behind Theorem 2 is that because ∇X Λ ≤ 0, either X is tangent to
a level curve of Λ (if it is orthogonal to ∇Λ), or X makes an obtuse angle with the direction of
steepest ascent of Λ, and therefore points toward c. It follows that as t increases, Λ decreases along
integral curves near c, and because c is a local minimum of Λ, we conclude that integral curves near
c converge to c as t → ∞, which implies stability. This is the main idea of the proof of Theorem 2.
Example 1 Consider the vector field
X(x, y, z) = (y(z − 1), x(z + 1), −2xy) .
The origin c = (0, 0, 0) is a fixed point of X, but from


0
z−1 y
0
x 
X0 (x, y, z) =  z + 1
−2y −2x 0
5
we obtain


0 −1 0
X0 (0, 0, 0) =  1 0 0  .
0 0 0
This matrix is not invertible, so c is not a simple fixed point. Therefore, the Linearization Theorem
does not apply. Instead, we show that
1
Λ(x, y, z) = (x2 + y 2 + z 2 )
2
is a Liapunov function for c.
First, if (x, y, z) 6= c, then Λ(x, y, z) > 0 = Λ(c), so the first condition in the definition of
Liapunov functions is satisfied. To check the second condition, we again assume (x, y, z) 6= c and
compute
∇X Λ(x, y, z) = ∇Λ(x, y, z) · X(x, y, z) = (x, y, z) · (y(z − 1), x(z + 1), −2xy) = 0.
We conclude that Λ is a (non-strict) Liapunov function for c, and therefore c is a stable fixed point.
2
Example 2 We consider the vector field
1 2
1 2
3
2
X(x, y) = −y + xy − x − xy , −3y + xy + x y − xy .
2
2
This vector field has a fixed point c = (0, 0). From
y − 3x2 − 12 y 2 −1 + x − xy
0
X (x, y) =
y + 2xy − 12 y 2 −3 + x − xy
we obtain
0
X (0, 0) =
0 −1
0 −3
.
This matrix is not invertible, so c is not a simple fixed point and the Linearization Theorem does
not apply.
We consider whether
1
Λ(x, y) = (3x2 − 2xy + y 2 )
2
is a Liapunov function. From
1
∇Λ(x, y) = (6x − 2y, −2x + 2y)
2
we easily see that c is a critical point of Λ. We then compute the Hessian
6 −2
HΛ (x, y) =
.
−2 2
Let λ1 and λ2 be the eigenvalues of HΛ . From tr(HΛ ) = λ1 + λ2 = 8 and det(HΛ ) = λ1 + λ2 = 16,
we conclude that HΛ is symmetric positive definite, due to its positive eigenvalues λ1 = λ2 = 4.
Because this is the case on all of Rn , we conclude that c is a strict global minimizer of Λ. That is,
Λ(x) > Λ(c) for all x 6= c.
6
To check the second condition in the definition of a Liapunov function, we assume (x, y, z) 6=
(0, 0, 0) and compute
∇X Λ(x, y) = ∇Λ(x, y) · X(x, y)
1
1
1
=
(6x − 2y, −2x + 2y) · −y + xy − x3 − xy 2 , −3y + xy + x2 y − xy 2
2
2
2
3
1
= −3xy + 3x2 y − 3x4 − x2 y 2 + y 2 − xy 2 + x3 y + xy 3 +
2
2
1
1
3xy − x2 y − x3 y + x2 y 2 − 3y 2 + xy 2 + x2 y 2 − xy 3
2
2
= 2x2 y − 3x4 − 2y 2
= −x4 + 2x2 y − y 2 − 3x4 − 2y 2 + x4 + y 2
= −(x2 − y)2 − 2x4 − y 2
< 0.
It follows that Λ is a strict Liapunov function for c.
It is worth noting that we could have tried a similar function as in the previous example,
1
F (x, y) = (x2 + y 2 ),
2
to see if it is a Liapunov function, but in this case it is not, because c is a saddle point of ∇X F ,
rather than a strict global maximizer as it was for Λ. 2
Exercises
Section 6.2: Exercises 1, 2, 3
Section 6.4: Exercise 2(i)ac
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