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Good morning everybody. My question is inspired from the following fact:

Consider $\mathbb R^3$ endowed with coordinates $(x,y,z)$. Of course if we were to solve the second order pde $\partial_x^2 g(x,y,z)=0$ we would find $g(x,y,z)=x p_0(y,z)+p_1(y,z)$. Then it will be possible, playing with $p_0$ and $p_1$, to prescribe for example both the behaviour on the hyperplanes $x=0$ and $x=1$ (Am I right?)

The question then is as follows: Suppose $X,Y,Z$ to be three arbitrary linearly independent vector fields (still in $\mathbb R^3$ with standard coordinates) and assume we want to solve the equation $X^2 g(x,y,z)=0$.

Assume $\pi_0$ and $\pi_1$ to be two hypersurfaces transverse to $X$ (they play the same roles of the previous hyperplanes).

Is it possible as above to find $g$ that satisfies ALSO prescribed conditions both on $\pi_0$ and $\pi_1$?

Intuitively this seems to be right to me, however I am not in possess of a rigorous proof, nor I know enough general theory to convince myself (therefore references are welcomed).

Thanks in advance for your kindness.

Regards,

-Guido-

Edit In response to prof. Bryant comment, let's say that my case of interest concerns with $X=\partial_x-\frac y2\partial_z$, but still I have no particular constraints on $\pi_0$ and $\pi_1$, except for them being different and both transverse to $X$.

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    $\begingroup$ You need to specify what you mean by 'play the same roles'. Just requiring $X$ to be transverse to both $\pi_0$ and $\pi_1$ is not sufficient in general. You need to know more about the dynamics of the vector field $X$ before you can assert that there exists a solution with specified values on $\pi_0$ and $\pi_1$. $\endgroup$
    – Robert Bryant
    Commented Dec 9, 2015 at 10:06
  • $\begingroup$ @RobertBryant I've specified which vector field $X$ I am concerned with. However your answer already contains an interesting point about existence of solution. May I ask for any particular reference that you have in mind? Best regards. $\endgroup$
    – guido giuliani
    Commented Dec 9, 2015 at 10:29
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    $\begingroup$ In the case of this specific vector field, the differentiable solutions of your equation are of the form $g(x,y,z) = p_0(y,2z{+}xy)+xp_1(y,2z{+}xy)$ where $p_0(u,v)$ and $p_1(u,v)$ are differentiable functions of their arguments, and if $\pi_0$ and $\pi_1$ are disjoint surfaces transverse to $X$ such that each of them meets each integral curve of $X$ exactly once, then knowing $g$ on each of $\pi_0$ and $\pi_1$ will uniquely determine $p_0$ and $p_1$. I don't have a reference in mind for the counterexample in the general case, just a counterexample. $\endgroup$
    – Robert Bryant
    Commented Dec 9, 2015 at 14:12
  • $\begingroup$ Do $Y$ and $Z$ play any role here? $\endgroup$
    – Deane Yang
    Commented Dec 9, 2015 at 22:33
  • $\begingroup$ @DeaneYang Nope... they do not play any particular role. $\endgroup$
    – guido giuliani
    Commented Dec 10, 2015 at 9:06

1 Answer 1

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Let us look at the local problem: taking $X$ a non-zero smooth vector field in a neighborhood of 0 in $\mathbb R^3$, you may choose local coordinates such that $X=\partial_z$. If $π_1, π_2$ are smooth hypersurfaces such that $X$ is transverse to both of them, you may assume that they are given locally by $$ π_j=\{(x,y,z), z=f_j(x,y)\}, j=1,2. $$ The matter is now to solve $$ \partial_z^2u=0,\quad u(x,y,f_j(x,y))=w_j(x,y), $$ i.e. $ u(x,y,z)=g_0(x,y) +zg_1(x,y),$ $$ g_0(x,y) +f_1(x,y)g_1(x,y)=w_1(x,y),\quad g_0(x,y) +f_2(x,y)g_1(x,y)=w_2(x,y). $$ Since the $f_j, w_j$ are given, it means that we want to solve a $2\times 2$ linear system with unknowns $g_0,g_1$: a simple sufficient condition for this to be possible is that $$ f_1-f_2\not=0, $$ which is true geometrically when $π_0, π_1$ are disjoint transverse hypersurfaces to whom $X $ is transverse. I guess that the global problem could be more complicated.

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