Characteristic curves satisfy a straight line equation let $\Omega$ be an open set in $\mathbb{R}^{n}$
let $F: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \to \mathbb{R}$ be a smooth map defined as $ F(x,z,p) = \sum_{i=1}^{n} a_i(z) p_i$, where $a(z) = (a_1(z), ... , a_n(z)) \in \mathbb{R}^{n} $ is a smooth vector field.
Let $u$ be a smooth solution of $F(x, u(x), \nabla u(x)) =0$ for $ x \in \mathbb{R}^{n}$.
Let $x_0 \in \Omega$.
Let $(x_0, z_0, p_0) \in \Omega \times \mathbb{R} \times \mathbb{R}^{n}$.
Let $(X(t_0, x_0), z(t_0, x_0) ,p(t_0, x_0))$ be the associated bicharacteristic strip of $F$ passing  through the point $(x_0, z_0, p_0)$.
(The map $t \mapsto X(t, x_0)$ is called the characteristic curve)
Show that $ X(t, x_0)$ satisfies $ X(t, x_0) = x_0 + t a(u(x_0))$ in a straight line.
My attempt:
I am a beginner in PDE, having taken a course in ODE. I am stuck on this problem.
I have observed the following:
from hypothesis, $ \sum_{i=1}^{n} a_i(z) \frac{\partial u_i}{\partial x} = 0$.
Any inputs on how to proceed would be appreciated.
 A: I will show how to tackle the case $n=2$, see this Wikipedia article, "quasi-linear case" and "fully nonlinear case". So we have
$$
F(x, y, u, p, q) = a_1(u) p + a_2(u) q = 0
$$
with $p = u_x$ and $q = u_y$. To transform this PDE into differential equations, we introduce a parametrization of the solution of the form $u(x,y)$, $p(x,y)$, $q(x,y)$ with $x = x(t)$ and $y = y(t)$. Setting
$$
\dot x = F_p = a_1(u) , \qquad \dot y = F_q = a_2(u)
$$
gives us a system of ordinary differential equations. Now, let us compute the evolution of $u$ along these curves:
$$
\dot u = p\dot x + q\dot y = pF_p + qF_q = 0 \, ,
$$
according to the equation $F=0$.
Thus, $u = u(x_0, y_0)$ is constant along the integral curves $t\mapsto (x, y)$, as well as the derivatives $\dot x = a_1(u)$ and $\dot y = a_2(u)$ which are functions of $u$. In other words, the characteristic curves
$$
(x,y) = (x_0, y_0) + t\, a(u(x_0, y_0))
$$ are straight lines of the $x$-$y$ plane along which $u$ is constant. The generalisation for $n>2$ is straightforward.
