## 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?