Lecture 5. Exponential of vector fields and solutions of differential equations

Let x\in C^{1-var} ([0,T], \mathbb{R}^d) and let V : \mathbb{R}^e \to \mathbb{R}^{e\times d} be a Lipschitz continuous map. In order to analyse the solution of the differential equation,
y(t)=y_0+\int_0^t V(y(s)) dx(s),
and make the geometry enter into the scene, it is convenient to see V as a collection of vector fields V=(V_1, \cdots, V_d), where the V_i‘s are the columns of the matrix V. The differential equation then of course writes
y(t)=y_0+\sum_{i=1}^d \int_0^t V_i (y(s)) dx^i(s),

Generally speaking, a vector field V on \mathbb{R}^{e} is a map
\begin{array}{llll}  V: & \mathbb{R}^{e}& \rightarrow  & \mathbb{R}^{e} \\  & x & \rightarrow  & (v_{1}(x),...,v_{e}(x)).  \end{array}
A vector field V can be seen as a differential operator acting on differentiable functions f: \mathbb{R}^{e} \rightarrow \mathbb{R} as follows:
Vf(x)=\langle V(x), \nabla f (x) \rangle= \sum_{i=1}^e v_i (x) \frac{\partial f}{\partial x_i}.
We note that V is a derivation, that is for f,g \in \mathcal{C}^{1} (\mathbb{R}^e , \mathbb{R} ),
V(fg)=(Vf)g +f (Vg).
For this reason we often use the differential notation for vector fields and write:
V=\sum_{i=1}^d v_i(x) \frac{\partial }{\partial x_i}.
Using this action of vector fields on functions, the change of variable formula for solutions of differential equations takes a particularly concise form:

Proposition: Let y be a solution of a differential equation that writes
y(t)=y_0+\sum_{i=1}^d \int_0^t V_i (y(s)) dx^i(s),
then for any C^1 function f: \mathbb{R}^{e} \rightarrow \mathbb{R},
f(y(t))=f(y_0)+\sum_{i=1}^d \int_0^t V_i f (y(s)) dx^i(s).

Let V be a Lipschitz vector field on \mathbb{R}^e. For any y_0 \in \mathbb{R}^e, the differential equation
y(t)=y_0+\int_0^t V(y(s)) ds
has a unique solution y: \mathbb{R} \to \mathbb{R}^e. By time homogeneity of the equation, the flow of this equation satisfies
\pi ( t_1 , \pi( t_2 ,y_0 ) )=\pi (t_1 +t_2,y_0).
and therefore \{ \pi( t, \cdot), t \in \mathbb{R}\} is a one parameter group of diffeomorphisms \mathbb{R}^e \to \mathbb{R}^e. This group is generated by V in the sense that for every y_0 \in \mathbb{R}^e,
\lim_{t\to 0} \frac{\pi(t,y_0) -y_0}{t}=V(y_0).
For these reasons, we write \pi(t,y_0)=e^{tV}(y_0). Let us now assume that V is a C^1 Lipschitz vector field on \mathbb{R}^e. If \phi :\mathbb{R}^e \to \mathbb{R}^e is a diffeomorphism, the pull-back \phi^{\ast}V of the vector field V by the map \phi is the vector field defined by the chain rule,
\phi^{\ast}V (x)=(d \phi^{-1} )_{\phi (x) } \left( V (\phi(x)) \right). In particular, if V' is another C^1 Lipschitz vector field on \mathbb{R}^e, then for every t \in \mathbb{R}, we have a vector field (e^{tV})^{\ast} V'. The Lie bracket [V,V'] between V and V' is then defined as
[V,V']=\left( \frac{d}{dt} \right)_{t=0} (e^{tV})^{\ast}V'.
It is computed that
[ V, V' ](x)=\sum_{i=1}^e \left( \sum_{j=1}^e v_j (x) \frac{\partial v'_i}{\partial x_j}(x)- v'_j (x) \frac{\partial v_i}{\partial x_j}(x)\right)\frac{\partial}{\partial x_i}.
Observe that the Lie bracket obviously satisfies [V,V']=-[V',V] and the so-called Jacobi identity that is:
[V,[V',V'']]+[V',[V'',V]]+[V'',[V,V']]=0.
What the Lie bracket [V,V'] really quantifies is the lack of commutativity of the respective flows generated by V and V'.

Lemma: Let V,V' be two C^1 Lipschitz vector fields on \mathbb{R}^e. Then, [V,V']=0 if and only if for every s,t \in \mathbb{R},
e^{sV} e^{t V'}=e^{sV+tV'}=e^{t V'} e^{sV}.

Proof: This is a classical result in differential geometry, so we only give one part the proof. From the very definition of the Lie bracket and the multiplicativity of the flow, we see that [V,V']=0 if and only if for every s \in \mathbb{R}, (e^{sV})^{\ast}V'=V'. Now, suppose that [V,V']=0. Let y be the solution of the equation
y(t)=y_0+\int_0^t V'(y(s)) ds.
Since (e^{sV})^{\ast}V'=V', we obtain that e^{sV} (y(t)) is also a solution of the equation. By uniqueness of solutions, we obtain that
e^{sV}(y(t))=e^{tV'} ( e^{sV}(y_0)).
As a conclusion,
e^{sV} e^{t V'}=e^{t V'} e^{sV}
\square

If we consider a differential equation
y(t)=y_0+\sum_{i=1}^d \int_0^t V_i (y(s)) dx^i(s),
as we will see it throughout this class, the Lie brackets [V_i,V_j] play an important role in understanding the geometry of the set of solutions. The easiest result in that direction is the following:

Proposition: Let x\in C^{1-var} ([0,T], \mathbb{R}^d) and let V_1,\cdots, V_d be C^1 Lipschitz vector fields on \mathbb{R}^e. Assume that for every 1 \le i,j \le d , [V_i,V_j]=0, then the solution of the differential equation
y(t)=y_0+\sum_{i=1}^d \int_0^t V_i (y(s)) dx^i(s), \quad 0 \le t \le T,
can be represented as
y(t)= \exp \left( \sum_{i=1}^d x^i(t) V_i \right) (y_0).

Proof: Let
F(x_1,\cdots,x_n)= \exp \left( \sum_{i=1}^d x_i V_i \right) (y_0).
Since the flows generated by the V_i‘s are commuting, we get that
\frac{\partial F}{\partial x_i}(x)=V_i (F(x)).
The change of variable formula for bounded variation paths implies then that F(x^1(t),\cdots,x^n(t)) is a solution and we conclude by uniqueness \square

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5 Responses to Lecture 5. Exponential of vector fields and solutions of differential equations

  1. taramata says:

    It seems I can not follow your arugments when you define the pull-back of a vector field. On the RHS you have a differential and some function in the index. Is this well-known notation?

  2. The notation (d \phi^{-1} )_{\phi (x) } \left( V (\phi(x)) \right) means the following: (d \phi^{-1} )_{\phi (x) } is the differential of \phi^{-1} at the point \phi(x). This is a linear map that we apply then to the vector V (\phi(x)).

  3. taramata says:

    Thank you for the response. As far as I understand $(d \phi^{-1})_{\phi(x)}(V(\phi(x)))=D^{-1}\phi(x) V(x)$, here $D^{-1} \phi(x)$ stands for the inverse of the derivative of $\phi$, in our case this is some matrix. Right?

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