Darboux coordinate for contact geometry

Question

I'm reading Geiges' notes. (https://arxiv.org/pdf/math/0307242.pdf) In the proof of Theorem 2.44 on page 17, the existence of the contact version Darboux coordinate is reduced to solving $H_t$ for each $t$, the PDE near the origin of $\mathbb{R}^{2n+1}$ $$\dot{\alpha}_t (R_{\alpha_t})+dH_t(R_{\alpha_t} )= 0$$ where $\alpha_t$ is a $1$-parameter family of contact forms and $R_{\alpha_t}$ is the corresponding reeb vector field. And he said that this equation always has a solution by integration if the neighborhood is small enough so that $R_{\alpha_t}$ has no closed orbit.

My question is why this is obvious? What I know is that this equation is a quasilinear first order PDE and can possibly be solved by the method of characteristics. But I can't find a reference that contains a clear statement when this kind of equation can be solved.

Thank you.

Answer 1

My answer is probably late... But I would like to know it there is any error in what I did.

I use the following theorem: Let $M$ be a smooth manifold. Suppose we are given an embedded hypersurface $S\subseteq M$, a smooth vector field $V$ on $M$, that is nowhere tangent to $S$, smooth functions $b,g:M\to\mathbb{R}$ and $\varphi:S\to\mathbb{R}$. Then for some neighbourhood $U$ of $S$ in $M$, there exists a unique smooth solution $u:U\to\mathbb{R}$ to the Cauchy problem: $$Vu+bu=g$$ $$u\vert_S=\varphi$$ The proof of this statement can be found in Introduction to Manifolds by the professor Lee.

I've also read the same notes so I've tried to avoid mentioning those steps that are not part of your doubt ...

Let $p_0\in M$ and $\varphi=(x^1,\cdots,x^n,y^1,\cdots,y^n,z)$ the smooth coordinates mentioned in the beginning of the proof.

1. The manifold is $J\times U_1$, here $J$ is a bounded and open interval containing $[0,1]$, $U_1$ a neighbourhood of $p_0$, recall that $U_1$ is chosen in such a way $\alpha_s$ is a contact form on $U_1$, thus the smoothness of $\alpha_t$ and the condition at $p_0$ given in the notes, implies that $R(t,p)=R_{\alpha_t}\vert_{p}$ is a smooth time dependent vector field, we can add the condition $Rz\neq 0$ to the problem of finding $J$ and $U_1$.

2. $g(t,p)= \alpha_0\vert_{p}(R(t,p))-\alpha_1\vert_{p}(R(t,p))$ , $b,\varphi = 0$ and $V=\hat{R}$, where $\hat{R}$ is the smooth vector field defined by just changing $\vert_{p}$ to $\vert_{(t,p)}$ in the expression of $R(t,p)$, i.e. if $R(t,p)=a(t,p)\partial_z\vert_{p}$ then $\hat{R}(t,p)=a(t,p)\partial_z\vert_{(t,p)}$. Moreover, $S=z^{-1}(0)$ is a hypersurface in $J\times U_1$ s.t. $\hat{R}$ is not tangent in any point.

By the theorem, we know there exists $u$ defined on some neighbourhood of $S$ s.t. $\hat{R}u=g$ and $u\vert_S=0$. So, by the openness of the neighbourhood of $S$ and compactness of any bounded interval, we can assume that $u$ is defined on $I\times U_2$, here $[0,1]\subseteq I\subseteq J$ and $U_2\subseteq U_1$, thus we define $H_t:U_2\to\mathbb{R}$ s.t. $H_t(p)=u(t,p)$, and for obvious reasons $\{H_t\}_{t\in I}$ is a smooth family of smooth functions.

We also got $H_t(p_0)=0$($(t,p_0)\in S$) and $dH_t\vert_{p_0}=0$($du_{(t,p_0)}=0$ since $g(t,p_0)=0$).

The reason I use that theorem it's because we need to prove the smoothness of the family $\{H_t\}$, since $\{X_t=H_tR_{\alpha_t}+Y_t\}$ needs to be a smooth family in order to define a smooth flow.