18.4 Center Manifolds for Maps 259
or
N h(x) = h Ax + f x, h(x)) − Bh(x) − g x, h(x) = 0. (18.4.5)
(Note: the reader should compare (18.4.5) with (18.1.10).) The next theorem
justifies the approximate solution of (18.4.5) via power series expan-
sions.
Theorem 18.4.3 (Approximation) Let φ : Rc
→ Rs
be a C1
map with
φ(0) = 0, φ′
(0) = 0, and N φ(x) = O(|x|q
) as x → 0 for some q > 1.
Then
h(x) = φ(x) + O(|x|q
) as x → 0.
Proof: See Carr [1981]. ✷
We now give an example.
Example 18.4.1. Consider the map


u
v
w

 →


−1 0 0
0 −1
2
0
0 0 1
2




u
v
w

 +


vw
u2
−uv

 , (u, v, w) ∈ R3
. (18.4.6)
It should be clear that (u, v, w) = (0, 0, 0) is a fixed point of (18.4.6), and the
eigenvalues associated with the map linearized about this fixed point are −1, −1
2
,
1
2
. Thus, the linear approximation does not suffice to determine the stability or
instability. We will apply center manifold theory to this problem.
The center manifold can locally be represented as follows
Wc
(0) = (u, v, w) ∈ R3
| v = h1(u), w = h2(u), hi(0) = 0,
Dhi(0) = 0, i = 1, 2 (18.4.7)
for u sufficiently small. Recall that the center manifold must satisfy the following
equation
N h(x) = h Ax + f x, h(x) − Bh(x) − g x, h(x) = 0, (18.4.8)
where, in this example,
x = u, y ≡ (v, w), h = (h1, h2),
A = −1,
B =
−1
2
0
0 1
2
,
f(u, v, w) = vw,
g(u, v, w) =
u2
−uv
. (18.4.9)
We assume a center manifold of the form
h(u) =
h1(u)
h2(u)
=
a1u2
+ b1u3
+ O(u4
)
a2u2
+ b2u3
+ O(u4
)
. (18.4.10)
260 18. Center Manifolds
Substituting (18.4.10) into (18.4.8) and using (18.4.9) yields
N h(u) =
a1u2
− b1u3
+ O(u5
)
a2u2
− b2u3
+ O(u5
)
−
−1/2 0
0 1/2
a1u2
+ b1u3
+ · · ·
a2u2
+ b2u3
+ · · ·
−
u2
−uh1(u)
=
0
0
. (18.4.11)
Balancing powers of coefficients for each component gives
u2
:
a1 + 1
2
a1 − 1
a2 − 1
2
a2
=
0
0
⇒
a1 = 2
3
a2 = 0
, (18.4.12)
u3
:
−b1 + 1
2
b1
−b2 − 1
2
b2 + a1
=
0
0
⇒
b1 = 0
b2 = a1
2
3
= 4
9
;
hence, the center manifold is given by the graph of h1(u), h2(u) , where
h1(u) =
2
3
u2
+ O(u4
),
h2(u) =
4
9
u3
+ O(u4
). (18.4.13)
The map on the center manifold is given by
FIGURE 18.4.1.
u −→ −u +
8
27
u5
+ O(u6
); (18.4.14)
thus, the origin is attracting; see Figure 18.4.1.
End of Example 18.4.1
18.4 Center Manifolds for Maps 261
Example 18.4.2. Consider the map
x
y
→
0 1
−1
2
3
2
x
y
+
0
−y3 , (x, y) ∈ R2
. (18.4.15)
The origin is a fixed point of the map. Computing the eigenvalues of the map
linearized about the origin gives
λ1,2 = 1,
1
2
.
Therefore, there is a one-dimensional center manifold and a one-dimensional stable
manifold with the orbit structure in a neighborhood of (0, 0) determined by
the orbit structure on the center manifold.
We wish to compute the center manifold, but first we must put the linear part
in block diagonal form as given in (18.4.1). The matrix associated with the linear
transformation has columns consisting of the eigenvectors of the linearized map
and is easily calculated. It is given by
T =
1 2
1 1
with T−1
=
−1 2
1 −1
. (18.4.16)
Thus, letting
x
y
= T
u
v
,
our map becomes
u
v
→
1 0
0 1
2
u
v
+
−2(u + v)3
(u + v)3 . (18.4.17)
We seek a center manifold
Wc
(0) = { (u, v) | v = h(u); h(0) = Dh(0) = 0 } (18.4.18)
for u sufficiently small. The next step is to assume h(u) of the form
h(u) = au2
+ bu3
+ O(u4
) (18.4.19)
and substitute (18.4.19) into the center manifold equation
N h(u) = h Au + f u, h(u) − Bh(u) − g u, h(u) = 0, (18.4.20)
where, in this example, we have
A = 1,
B =
1
2
,
f(u, v) = −2(u + v)3
,
g(u, v) = (u + v)3
, (18.4.21)
262 18. Center Manifolds
and (18.4.20) becomes
a u − 2 u + au2
+ bu3
+ O(u4
)
3 2
+ b u − 2 u + au2
+ bu3
+ O(u4
)
3 3
+ · · · −
1
2
au2
+ bu3
+ O(u4
) − u + au2
+ bu3
+ O(u4
)
3
= 0.
(18.4.22)
or
au2
+ bu3
−
1
2
au2
−
1
2
bu3
− u3
+ O(u4
) = 0. (18.4.23)
FIGURE 18.4.2.
Equating coefficients of like powers to zero gives
u2
: a −
1
2
a = 0 ⇒ a = 0,
u3
: b −
1
2
b − 1 = 0 ⇒ b = 2. (18.4.24)
Thus, the center manifold is given by the graph of
h(u) = 2u3
+ O(u4
), (18.4.25)
and the map restricted to the center manifold is given by
u → u − 2 u + 2u3
+ O(u4
)
3
(18.4.26)
or
u → u − 2u3
+ O(u4
). (18.4.27)
Therefore, the orbit structure in the neighborhood of (0, 0) appears as in Figure
18.4.2 and (0, 0) is stable.
End of Example 18.4.2