Why does a homogeneous first-order linear PDE have $n-1$ functionally independent solutions? Why does a homogeneous first-order linear PDE$$\sum_{i=0}^n\xi_i(\mathbf{x})\frac{\partial u}{\partial x_i}=0,$$where $\mathbf{x} = (x_1, x_2, …, x_n)$, have $n-1$ functionally independent solutions for $u(\mathbf{x})$?
 A: This is a local result that is a consequence of the Inverse Function Theorem and the ODE existence theorems.
We assume $\xi$ is a smooth vector field.
Suppose $x_0$ is a point where $\xi\ne0$. Then there are vectors
$v_1,\ldots,v_{n-1}$  such that the list
$v_1,\ldots,v_{n-1},\xi(x_0)$ is linearly independent. Points $x$ near
$x_0$ can be written in the form
$$
  x = x_0+\sum_{k=1}^{n-1}c_kv_k+t\xi(x_0)
$$
with $(c_1,\ldots, c_{n-1},t)$ near the zero vector. The rough idea is that
each of $c_1$ to $c_{n-1}$, thought of as a function of $x$,
almost qualifies as a solution of the type we seek. To correct this
idea and make it work, we do the following.
The plane through $x_0$ that is orthogonal to $\xi(x_0)$ consists of
the points
$$
 y(0) = x_0+\sum_{k=0}^{n-1} c_kv_k.
$$
There is a solution to the ODE
$$
 y'(t) = \xi(y(t))
$$
for each initial value $y(0)$. Let $f(c_1,\ldots,c_{n-1},t)$ denote
the value of the solution $y(t)$. The existence theorems for ODEs
will give that the solutions  $y$ exist for at least a short time
for all the initial values in a small compact set, and that $y$
is continuously differentiable as a function of the $c_k$ as well as
of $t$. Thus $f$ is a $C^1$ function.
The derivative matrix of $f$ is
$$
 Df = \big[y_{c_1} \ y_{c_2} \ \cdots \ y_{c_{n-1}} \ \xi\big]
$$
and differentiating the initial values with respect to each $c_k$ we get
$$
 Df(0,0) = \big[v_1 \ v_2 \ \cdots \ v_{n-1} \ \xi(x_0)\big]
$$
which is invertible.
By the Inverse Function Theorem then $f^{-1}$
exists in a  neighborhood of $x_0$. That means that $c_k$ can be viewed
as a function $u_k$  of points near $x_0$, rather than just a coefficient
of points on the plane of initial values.
Since $u_k$ is constant on the surface formed by varying the
$c$'s other than $c_k$, and since this surface consists of  curves
tangent to $\xi$, it follows that:
$$
 \xi\cdot\nabla u_k = 0, \qquad\qquad k=1,\ldots,n-1.
$$
The $u_k$ are functionally independent because $Df$ is invertible.
