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LECTURE 13: GEOMETRY AND DYNAMICS OF SMOOTH
VECTOR FIELDS
1. Geometry of Vector Fields: The Integral Curves
Suppose we have a smooth vector field defined on an Euclidian region. In calculus
and in ODE, we learned the conception of integral curves of such a vector field: they
are curves so that the given vector field is the tangent vector to the curves everywhere.
Here is an example of vector fields with many integral curves drawn:
The conception of integral curves above can be generalized to smooth manifolds
easily. To begin with, one need to explain the conceptions of “curve” and “tangent
vector to a curve” first.
Suppose M is a smooth manifold. A smooth curve in M is by definition a smooth
map γ : I → M, where I is an interval in R. 1 For any a ∈ I, the tangent vector of γ
at the point γ(a) is defined to be
γ˙ (a) = dγ(a) := dγ ( d ),
dt a dt
where d is the standard coordinate tangent vector of R.
dt
Definition 1.1. Let X ∈ Γ∞(TM) be a smooth vector field on M. A smooth curve
γ : I → M is called an integral curve of X if for any t ∈ I,
γ˙ (t) = Xγ(t).
1
Note: By a curve we really mean a “parametrized curve”. The parametrization is a part of the
definition. Different parametrizations of the “same geometric picture” represent different curves.
1
2 LECTURE 13: GEOMETRY AND DYNAMICS OF SMOOTH VECTOR FIELDS
∂ n
Example. Consider the coordinate vector field X = ∂x1 on R . Then the integral curves
of X are the straight lines parallel to the x1-axis, parametrized as
γ(t) = (c +t,c ,··· ,c ).
1 2 n
To check this, we note that for any smooth function f on Rn,
dγ( d )f = d (f ◦γ) = ∇f · dγ = ∂f .
dt dt dt ∂x1
Remark. Note that although the curve
γ˜(t) = (c + 2t,c ,··· ,c )
1 2 n
has the same picture (i.e. the same “horizontal line” passing the point (c ,··· ,c )) as
1 n
˙ ∂
γ, it is not an integral curve of X, since γ˜(t) = 2 1.
∂x
Example. Consider the vector field X = x ∂ −y ∂ on R2. Then if γ(t) = (x(t),y(t)) is
∂y ∂x
∞ 2
an integral curve of X, we must have for any f ∈ C (R ),
x′(t)∂f +y′(t)∂f = ∇f · dγ = X f = x(t)∂f −y(t)∂f,
∂x ∂y dt γ(t) ∂y ∂x
which is equivalent to the system
x′(t) = −y(t), y′(t) = x(t).
The solution to this system is
x(t) = acost−bsint, y(t) = asint+bcost.
These are circles centered at the origin in the plane parametrized by the angle (with
counterclockwise orientation).
Remark. In general, a re-parametrization of an integral curve is no longer an integral
curve. However, it is not hard to see that if γ : I → M is an integral curve of X, then
• Let I = {t | t + a ∈ I} and γ (t) := γ(t + a), then γ : I → M is an integral
a a a a
curve of X.
• Let Ia = {t | at ∈ I} and γa(t) := γ(at), then γa : Ia → M is an integral curve
for Xa = aX.
To study further properties of integral curves, we need to convert the equation
γ˙ (t) = Xγ(t) which is an equation relating tangent vectors on manifolds into equations
on functions defined on Euclidian region. To do so we first prove
Lemma 1.2. Let X be a smooth vector field on M, and suppose in a local chart
P i 1 n 1 n
(ϕ,U,V), X = X∂. Denote ϕ(p) = (x (p),··· ,x (p)) (so that x ,··· ,x are
i
smooth functions on U which represent the coordinates of p). Then Xi = X(xi).
Proof. Since ∂i(xj) = δj (Check This!), we have X(xj) = PXi∂i(xj) = Xj.
i
LECTURE 13: GEOMETRY AND DYNAMICS OF SMOOTH VECTOR FIELDS 3
Now let γ : I → M be an integral curve of X. To study the equation γ˙(t) = Xγ(t)
at a given point γ(t), WLOG we may assume γ(t) ∈ U, and (ϕ,U,V) is a coordinate
chart. By using the local chart map ϕ, one can convert the point γ(t) ∈ U to
1 n n
ϕ(γ(t)) = (x (γ(t)),··· ,x (γ(t))) ∈ R .
If we denote y = xi ◦ γ : I → R, then we can convert the (vector!) equation defining
i
integral curves into equations on these one-variable functions y ’s. More precisely,
i
according to the previous lemma, we have
d X d i X i ′ X′
γ˙ (t) = dγ ( ) = dγ ( )(x )∂ = (x ◦γ)(t)∂ = y (t)∂ .
t dt t dt i i i i
i i i
So the integral curve equation γ˙(t) = Xγ(t) becomes
X′ X i X i −1
y (t)∂ = X(γ(t))∂ = X ◦ϕ (y (t),··· ,y (t)).
i i i 1 n
i i i
for all t ∈ I. In conclusion, we convert the integral curve equation into the following
system of ODEs on the one-variable functions yi’s:
′ i −1
y (t) = X ◦ϕ (y ,··· ,y ), ∀t ∈ I, ∀1 ≤ i ≤ n.
i 1 n
Recall:
Theorem1.3(TheFundamentalTheoremforSystemsofFirstOrderODEs). Suppose
n 1 n n
V ⊂ R is open, and F = (F ,··· ,F ) : V → R a smooth vector-valued function.
Consider the initial value problem
(1) y˙i(t) = Fi(y1(t),··· ,yn(t)), i = 1,··· ,n
yi(t ) = ci, i = 1,··· ,n
0
1 n
for arbitrary t0 ∈ R and c0 = (c ,··· ,c ) ∈ V .
(1) Existence : For any t0 ∈ R and any c0 ∈ V, there exist an open interval I0
containing t0 and an open subset V0 containing c0 so that for any c ∈ V0, the
system (1) has a smooth solution y (t) = (y1(t),··· ,yn(t)) for t ∈ I .
c 0
(2) Uniqueness : If y is a solution to the system (1) for t ∈ I and y is a solution
1 0 2
to the system (1) for t ∈ J , then y = y for t ∈ I ∩ J .
0 1 2 0 0
(3) Smoothness : The solution function Y(c,t) := y (t) in part (1) is smooth on
c
(c,t) ∈ V0 × I0.
Wewill refer to Lee’s book, Appendix D (Page 663-671) for a proof. According to
the fundamental theorem of systems of ODEs, we conclude
Theorem 1.4 (Local Existence, Uniqueness and Smoothness). Suppose X is a smooth
vector field on M. Then for any point p ∈ M, there exists a neighborhood U of p, an
p
εp > 0 and a smooth map
Γ:(−ε ,ε )×U →M
p p p
so that for any q ∈ U, the curve γ : (−ε,ε) → M defined by
q
γ (t) := Γ(t,q)
q
4 LECTURE 13: GEOMETRY AND DYNAMICS OF SMOOTH VECTOR FIELDS
is an integral curve of X with γ(0) = q. Moreover, this integral curve is unique in the
sense that if σ : I → M is another integral curve of X with σ(0) = q, then σ(t) = γq(t)
for t ∈ I ∩ (−ε ,ε ).
p p
2. Complete vector fields
As a consequence of the uniqueness, any integral curve has a maximal defining
interval. We are interested in those vector fields whose maximal defining interval is R.
Definition 2.1. A vector field X on M is complete if for any p ∈ M, there is an
integral curve γ : R → M such that γ(0) = p.
Note every smooth vector field is complete.
2 d
Example. Consider the vector field X = t dt on R. Let γ(t) = (x(t)) be its integral
curve. Then according to the integral curve equation,
′ d 2 d ′ 2
x(t)dt = Xγ(t) = x(t) dt =⇒x(t)=x(t) .
The solution to this ODE is with initial condition x(0) = c is
x (t) = 1 for c 6= 0
c −t+1/c
and
x0(t) = 0 for c = 0.
Note that the maximal interval of x (t) is
c
Ic = (−∞,1/c) for c > 0
and
Ic = (1/c,+∞) for c < 0.
Since the integral curves starting at any c 6= 0 is not defined for all t ∈ R, we conclude
that X is not complete.
Wewill use complete vector fields to construct global flows next time. We will end
this lecture with a sufficient condition for a vector field to be complete. As in the case
of functions, we can define the support of a vector field by
supp(X) = {p ∈ M | X(p) 6= 0}.
Our criteria is
Theorem 2.2. If X is a compactly supported vector field on M, then it is complete.
Proof. Let C = supp(X). First suppose q ∈ M \C, i.e. Xq = 0. We define a “constant
curve” γ on M by letting γ (t) = q for all t ∈ R, then we see
q q
γ˙ (t) = 0 = X = X .
q q γq(t)
In other words, the constant curve γq (whose domain is R) is the unique integral curve
of X passing q.
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