§ 4. Differential Manifolds
253
which (1) has a solution, which is, therefore, said to be maximal. I is clearly
open because of (a), but in general I = R : if X is an open subset of R
2 and L
is a vector field whose vectors can de deduced from each other by parallelism,
the trajectories are open line segments contained in X having as endpoints
border points of X.
If, for all x ∈ X, γ x (t) denotes the maximum trajectory such that γ x (0) =
x and I x is its interval of definition, we are led to introduce the set Ω ⊂ R×X
of (t, x) such that t ∈ I x and to set
γ(t, x) = γ x (t)
(15.3)
for (t, x) ∈ Ω. This give a map γ : Ω −→ X, the global flow of the vector
field L. The following additional result then holds :
(c) If L is of class C
k , then Ω is open in R × X and γ is of class C
k in Ω.
Similarly to (a) and (b), this statement is of a local nature: it amounts
to showing that the solution of (2) satisfying x(0) = ξ is defined and a C
k
function of (t, ξ) on the product of an interval centered at 0 and a ball with
given centre.
There are analogous statements for more general differential equations:
instead of a function L(x) of the only variable x ∈ R
p , L can be supposed to
depend on t, x and a parameter z varying in a Cartesian space. The purpose
is then to find a function x(t) satisfying
x
(t) = L [t, x(t), z] , x(0) = ξ
(15.4)
and to show that if L is C
k , then x(t) is a C
k function of t, ξ and z.
Everything on the subject and even more can be found in Dieudonn´ e,
vol. 1, X.4 ` a X.8, but as this reference is not very easy to read, here we will
substitute a low-powered proof to this type of high-powered one (expression
used by Spivak for Serge Lang’s Analysis II ) in the manner of the inventors
of the successive approximation method, for example ´
Emile Picard. We have
already used it in a particular case related to Bessel’s equation in Chap. VI,
n
◦ 10.
(i) Reduction to an integral equation. If L is assumed to be continuous
and if we consider x(0) = ξ, (4) amounts to solving
x(t) = ξ +
t
0
L [u, x(u), z] du ,
(15.5)
where this is an oriented integral. Setting
x(t) = y(t) + ξ , L (t, y + ξ, z) = L (t, y, ξ, z) ,
(15.6)
we are led to solve
y(t) =
t
0
L [u, y(u), ξ, z] du .
253
which (1) has a solution, which is, therefore, said to be maximal. I is clearly
open because of (a), but in general I = R : if X is an open subset of R
2 and L
is a vector field whose vectors can de deduced from each other by parallelism,
the trajectories are open line segments contained in X having as endpoints
border points of X.
If, for all x ∈ X, γ x (t) denotes the maximum trajectory such that γ x (0) =
x and I x is its interval of definition, we are led to introduce the set Ω ⊂ R×X
of (t, x) such that t ∈ I x and to set
γ(t, x) = γ x (t)
(15.3)
for (t, x) ∈ Ω. This give a map γ : Ω −→ X, the global flow of the vector
field L. The following additional result then holds :
(c) If L is of class C
k , then Ω is open in R × X and γ is of class C
k in Ω.
Similarly to (a) and (b), this statement is of a local nature: it amounts
to showing that the solution of (2) satisfying x(0) = ξ is defined and a C
k
function of (t, ξ) on the product of an interval centered at 0 and a ball with
given centre.
There are analogous statements for more general differential equations:
instead of a function L(x) of the only variable x ∈ R
p , L can be supposed to
depend on t, x and a parameter z varying in a Cartesian space. The purpose
is then to find a function x(t) satisfying
x
(t) = L [t, x(t), z] , x(0) = ξ
(15.4)
and to show that if L is C
k , then x(t) is a C
k function of t, ξ and z.
Everything on the subject and even more can be found in Dieudonn´ e,
vol. 1, X.4 ` a X.8, but as this reference is not very easy to read, here we will
substitute a low-powered proof to this type of high-powered one (expression
used by Spivak for Serge Lang’s Analysis II ) in the manner of the inventors
of the successive approximation method, for example ´
Emile Picard. We have
already used it in a particular case related to Bessel’s equation in Chap. VI,
n
◦ 10.
(i) Reduction to an integral equation. If L is assumed to be continuous
and if we consider x(0) = ξ, (4) amounts to solving
x(t) = ξ +
t
0
L [u, x(u), z] du ,
(15.5)
where this is an oriented integral. Setting
x(t) = y(t) + ξ , L (t, y + ξ, z) = L (t, y, ξ, z) ,
(15.6)
we are led to solve
y(t) =
t
0
L [u, y(u), ξ, z] du .