Do linearly independent vector fields with vanishing Lie bracket always have integral manifolds which are level sets? Suppose I have a smooth distribution on an open subset of $M = R^n_{++}$ consisting of $n-1$ vector fields $X^i$. I know three things about this distribution:

*

*At each point $p \in M$ the set $(X^1_p,\ldots,X^{n-1}_p)$ is linearly independent and so spans an $(n-1)$-dimensional subspace of $R^n$.


*For each pair $i,j, \; 1 \leq i < j \leq n-1, \; [X^i,X^j] = 0$ (so the system is in involution).


*At each point of $M$ the normal to the subspace spanned by this system is non-negative, with the last coordinate strictly positive.
By 2, Frobenius tells us this system is integrable with its integral manifolds forming a foliation of $M$. I understand this also means that there are local coordinates $x^i$ such that $X^i = \partial/\partial x^i$. I have two questions:

*

*In this particular case does there always exist a single function $f: M \rightarrow R$ (smooth or not) such that the integral manifolds (hypersurfaces) are level sets of $f$; i.e. each leaf $L$ of the foliation is of the form $f^{-1}(x)$ for some $x \in R$?


*Is each leaf $L$ of the foliation globally the graph of some function $h:D \rightarrow R$ (i.e. $L = (x_1,\ldots,x_{n-1},h(x_1,\ldots,x_{n-1}))$ where $D$ is some domain in $R^{n-1}$?
My sense is that the first statement is not true but (because of the third assumption) that the second is. As to 1., it has been suggested to me that if I looked at the distribution from the perspective of the normal field (i.e. 1-form approach) that my assumption that the Lie bracket vanishes is equivalent to the "well-known" integrability conditions insuring there exists a function f with the required property, but I'm not sure this is right.
I would appreciate any insight into this (surely) simple issue that I'm just not seeing.
Edit: In case it helps I can give a concrete example of the vector fields I'm working with when $n=3$. $X^1 = (-m(x_1,x_2,x_3),1, 0), \; X^2 = (-n(x_1,x_2,x_3),0,1)$. Here the smooth functions $m$ and $n$ are strictly positive everywhere on $M$ and bounded away from zero and $\infty$ on every compact subset of $M$. I also simply impose the condition $[X^1,X^2] = mn_1 - nm_1 -n_2 +m_3 = 0$. Note that the normal field is $N_{(x_1,x_2,x_3)} = (1,m(x_1,x_2,x_3),n(x_1,x_2,x_3)) \gg 0$.
Further clarification: Assumptions on m are, for each fixed $x_2,x_3$ that $\lim_{x \rightarrow \infty}m(x,x_2,x_3) = 0, \lim_{x \rightarrow 0}m(x,x_2,x_3) = \infty$. A similar assumption holds for $n$; i.e. for each fixed $x_1,x_3$ $\lim_{x \rightarrow \infty}n(x_1,x,x_3) = 0, \lim_{x \rightarrow 0}n(x_1,x,x_3) = \infty$.
 A: I will try the case $n=2$. Consider a field of directions on $(0,\infty)^2$, with the field of normals
$n(x,y) = (n_1(x,y), n_2(x,y))\ $, $n_1\ge 0$, $n_2>0$.
The leafs of the foliations will be the graphs of functions $y = \phi(x)$ solutions of the differential equation
$$a_1 dx + a_2 dy = 0$$
or
$$\frac{dy}{dx} = -\frac{a_1(x,y)}{a_2(x,y)}$$
(in the case $n=2$, we get an ordinary differential equation, so there are no integrability conditions)
The graphs of different solutions will be the leaves of the foliation.
Note that to get a solution we fix $x_0$, $y_0>0$ and consider the solution that satisfies furthermore $\phi(x_0) = y_0$.  Now, there are in fact $\infty^1$, and not $\infty^2$ such solutions, since for all pairs $(x^*, y^*)$ on the graph of one, we get the same one.
In other words, the family of solutions ( of leaves if you want) depends on $1$ parameter ( something that they tell you in ODE class while hand-waving). However, globally labelling such solutions could be tricky.
To label the solutions( leaves), find a transversal curve $t\mapsto (\psi_1(t), \psi_2(t))$ that intersects each leaf in exactly one point. This has to work globally. (Foliations are  tricky...). Now I think we could consider the transversal $t \mapsto (t, t)$. In this way we'll get a function $f\colon (0, \infty)^2$ with the leaves the level curves. The description of the function is this:
for every $(x_0, y_0)$, the solution of the Cauchy problem
$$\frac{dy}{dx} = - \frac{n_1(x,y)}{n_2(x,y)}, \  y(x_0) = x_0 $$
has a unique fixed point $t = t(x_0, y_0)$. We take $f(x_0, y_0) = t$.
Notes:

*

*We see that in this case every level set is a leaf so connected. This is a nice situation, where leaves can be labeled "linearly". However, you can start with a submersion $f$, but whose level sets are not connected. Now, look only at the distribution of hyperplanes, and so at the foliation. All is good, integrable, but trying to write each leaf separately as a level set would be a problem.


*The case of general $n$ : the Frobenius integrability condition ensure that the solution of the Cauchy-Frobenius problem
$$\frac{\partial y}{\partial x_i} = - \frac{a_i(x', y)}{a_n(x',y)}$$
exists locally. Extend it all the way and get the graphs as the leaves of the foliations. Again, we have to label them. Each of the leaves will intersect the diagonal $(t, t, \ldots, t)$ in exactly one point. Now the function $f(x_{01}, \ldots, x_{0n-1}, x_{0n})$ is as follows: the solution $y_{(x'_0, x_{0,n})}$ of
$$\frac{\partial y}{\partial x_i} = -\frac{a_i(x',y)}{a_n(x',y)} \ \ ,  y(x'_0) = x_{0n}$$ has a unique "fixed point" $(t,t\ldots, t)$. Define $f((x'_0, x_{0n})=t$.
