Suppose that $M$ is a smooth manifold (without boundary), $S \subseteq M$ embedded hypersurface, $X$ smooth vector field on $M$ (nowhere vanishing on $M$ and nowhere tangent to $S$), and $f \in C^\infty(M)$. We would like to find a function $u \in C^\infty(M)$ satisfying equation $X(u) = f$ on $M$, with initial data $u|_S = \varphi$, where $\varphi \in C^\infty(S)$.
Local solvability of the Cauchy problem is treated, for example, in Theorem 9.51 in Lee: "Introduction to Smooth Manifolds" (2nd edition).
What are the additional conditions that might guarantee existence of the global solution to this problem? For example, is it enough to assume that $M$ is diffeomorphic to $S \times \mathbb{R}$ and $X$ is nowhere tangent to any copy of $S$?
Best Answer
Your condition is not sufficient. As a counterexample, let $M=\mathbb{R}^2$, where vertical lines are taken to be the copies of $S=\{0\}\times\mathbb{R}$, and consider the vector field $X(x,y)=((1+y^2)^{-1},1)$ with $f=1$ and $\varphi=0$. This solution will locally take the form $u(x,y)=y-\tan(\arctan(y)-x)$, which cannot be smoothly extended to $\mathbb{R}^2$.
This example suggests one thing that can go wrong: If the flowout of $S$ along $X$ is not surjective (i.e. not all integral curves interset $S$), there can be issues extending $u$ from thi image of the flowout. In fact a sufficient condition is that the flowout $\Theta:U\to M$, where $U$ is an open subset of $S\times\mathbb{R}$ is a diffeomorphism. In this case, you can pull the problem back to $U$, where it is easy to solved with integration. This condition is not necessary, of course, but in the case you outline in the question, it automatically holds when $S$ is compact.