252
IX – Multivariate Differential and Integral Calculus
Exercise 2. Let L 1 , . . . , L n be C
∞ vector fields on an open subset U of X ;
suppose that, for all x ∈ U , the L i (x) form a basis for X
(x). Show that all
differential operators on U can be written uniquely as a finite sum
pi≥0
a p1...pn (x)L
p1
1 . . . L
pn
n .
Note that as [L, M ] = N involves the derivatives of the components of L
and M , the value N (x) ∈ X
(x) of N at x does not only depend on the vectors
L(x) and M (x). The bracket [h, k] of two tangent vectors is not well-defined.
The existence of vector fields satisfying global conditions poses problems
related to the topology of manifolds: does there exist vector fields in X such
that L(x) = 0 for all x ? If X has dimension n, are there n vector fields L i
such that the L i (x) form a basis for X
(x) for all x ∈ X ? The answers are
already in the negative for the 2-dimensional sphere.
15 – Vector Fields and Differential Equations
Vector fields serve to generalize the theory of first order differential equations
to manifolds; we can start by searching for integral curves or trajectories
t → γ(t) from a vector field L onto a p-dimensional manifold X; they are
defined by the condition
γ
(t) = L[γ(t)] ,
(15.1)
where the left hand side is defined as in n
◦ 12, (ii). In a local chart (U, ϕ)
such that ϕ(U ) = R
p , this equivalent to looking for a function x(t) = ϕ[γ(t)]
satisfying
Dx(t) = L[x(t)] ,
(15.2)
where D = d/dt and where the functions x(t) and L(x) have values in R
p .
When L is C
1 , the following results hold:
(a) for all t 0 ∈ R and x 0 ∈ X, there exists a solution of (1) defined in the
neighbourhood of t 0 and such that γ(t 0 ) = x 0 ;
(b) two such solutions coincide on the interval on which they are simultaneously defined.
Statement (a) will follow from the analogous statement for equation (2).
Similarly for (b), because if two solutions of (1) defined on the same interval I
and equal at some point t ∈ I are known to be also equal on a neighbourhood
of t, then the set of t ∈ I where they are equal is both open and close
(continuity) in I, and so are equal to I.
Statement (b) shows that the solutions of (1) with a given value at a
given point t 0 are the restrictions to their intervals of definition of a unique
solution, defined on the union I of these intervals; I is the largest interval on
IX – Multivariate Differential and Integral Calculus
Exercise 2. Let L 1 , . . . , L n be C
∞ vector fields on an open subset U of X ;
suppose that, for all x ∈ U , the L i (x) form a basis for X
(x). Show that all
differential operators on U can be written uniquely as a finite sum
pi≥0
a p1...pn (x)L
p1
1 . . . L
pn
n .
Note that as [L, M ] = N involves the derivatives of the components of L
and M , the value N (x) ∈ X
(x) of N at x does not only depend on the vectors
L(x) and M (x). The bracket [h, k] of two tangent vectors is not well-defined.
The existence of vector fields satisfying global conditions poses problems
related to the topology of manifolds: does there exist vector fields in X such
that L(x) = 0 for all x ? If X has dimension n, are there n vector fields L i
such that the L i (x) form a basis for X
(x) for all x ∈ X ? The answers are
already in the negative for the 2-dimensional sphere.
15 – Vector Fields and Differential Equations
Vector fields serve to generalize the theory of first order differential equations
to manifolds; we can start by searching for integral curves or trajectories
t → γ(t) from a vector field L onto a p-dimensional manifold X; they are
defined by the condition
γ
(t) = L[γ(t)] ,
(15.1)
where the left hand side is defined as in n
◦ 12, (ii). In a local chart (U, ϕ)
such that ϕ(U ) = R
p , this equivalent to looking for a function x(t) = ϕ[γ(t)]
satisfying
Dx(t) = L[x(t)] ,
(15.2)
where D = d/dt and where the functions x(t) and L(x) have values in R
p .
When L is C
1 , the following results hold:
(a) for all t 0 ∈ R and x 0 ∈ X, there exists a solution of (1) defined in the
neighbourhood of t 0 and such that γ(t 0 ) = x 0 ;
(b) two such solutions coincide on the interval on which they are simultaneously defined.
Statement (a) will follow from the analogous statement for equation (2).
Similarly for (b), because if two solutions of (1) defined on the same interval I
and equal at some point t ∈ I are known to be also equal on a neighbourhood
of t, then the set of t ∈ I where they are equal is both open and close
(continuity) in I, and so are equal to I.
Statement (b) shows that the solutions of (1) with a given value at a
given point t 0 are the restrictions to their intervals of definition of a unique
solution, defined on the union I of these intervals; I is the largest interval on