# Linear kinetic PDE: Characteristics of the transport operator are given by the flow a Hamiltonian

I am trying to read and understand the article "Hypocoercivity for linear kinetic equations conserving mass." by Dolbeault, Mouhot, Schmeiser. doi: 10.1090/s0002-9947-2015-06012-7 (https://homepage.univie.ac.at/christian.schmeiser/DMS-I.pdf)

Already on first page we have the passage:

We consider linear kinetic equations which can be written as $$$$\partial_t f + \mathrm{T}f = \mathrm{L}f,$$$$ $$f=f(x,v,t)$$ with $$(x,v,t) \in \mathbb{T}^d \times \mathbb{R}^d \times (0, \infty)$$ and $$d \in \mathbb{N}.$$ The linear transport operator $$\mathrm{T} := v \cdot \nabla_x - \nabla_x V(x) \cdot \nabla_v$$ has characteristics given on the phase space $$\mathbb{T}^d \times \mathbb{R}^d$$ by the ﬂow of the Hamiltonian $$(x,v) \mapsto E(x,v) := \frac{1}{2} |v|^2 + V(x).$$
The external potential $$V(x)$$ is measurable function on $$\mathbb{R}^d.$$ I don't understand what is meant by "the characteristics of the transport operator are given by the flow the Hamiltonian"? And what is the connection of that Functional $$E$$ with the transport operator? I never heard the term "characteristics of operators", I just know this in context of the "method of characteristics" and Cauchy-Problems. Would be very grateful for further explanations and definitions.

This is related to the standard formulation of the method of characteristics in the following way. Consider the time-independent homogeneous version of this equation, i.e., an equation just involving the linear operator $$\mathrm{T}$$: $$\mathrm{T}u = 0\iff v\cdot\nabla_x u - \left[\nabla_x V\right]\cdot\nabla_v u=0,$$ where $$u$$ is a function of $$x$$ and $$v$$ only. Once can apply the method of characteristics to this system and obtain a system of $$2d+1$$ ODEs, although one will simply state that $$\dot{u}=0$$ along characteristic lines. Here I will let $$\dot{{}}$$ denote differentiation with respect to some real parameter $$s$$. The corresponding equations for $$x$$ and $$v$$ are then $$\dot{x} = v,\quad \dot{v}=-\nabla_x V.$$
Switching gears, given a smooth function $$H:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$$, the corresponding Hamiltonian system on $$\mathbb{R}^{2n}$$ is defined to be $$\dot{q} = \nabla_p H, \quad\dot{p}=-\nabla_q H,$$ where $$p,q\in\mathbb{R}^n$$. Importantly, $$H$$ is constant along trajectories of this system.
If we now make the identification $$(x,v) = (q,p)$$ and define $$H(q,p) = E(x,v)$$, then Hamilton's equations for this function read \begin{aligned} \dot{x} = \nabla_v E &= v \\ \dot{v} = -\nabla_x E &= -\nabla_x V, \end{aligned} which are our original characteristic equations for the operator $$\mathrm{T}$$.