PDEs - Nonlinear Characteristics - On computation In general, for the (Nonlinear) Method of Characteristics set up, we have a parameterized curve:
$$x(s) = (x^1(s),...,x^n(s))$$ and some related variables , for an unknown (smooth enough) function, $u:\mathbb{R}^n\to\mathbb{R}$:
$$z(s):= u(x(s))$$
$$p^i(s):= \partial_{x^i}u(x(s)),$$
for $i\in\{1,...,n\}.$

With these definitions, one proceeds to a certain system of ODE's that I won't list here.
In a special case relating to Conservation Laws, we have time as a separate domain variable, which changes the above to:
$$y(s) := (x(s),t(s)) = \big(y^1(s),...,y^n(s),y^{n+1}(s)\big)$$
$$z(s) := u(y(s))$$
$$q(s):= \big(q^1(s),...,q^n(s),q^{n+1}(s)\big)$$
where 
$\forall i\neq n+1,\text{ }\text{ }y^i(s):= x^i(s)$; 
$y^{n+1}(s) := t(s)$; 
$\forall i\neq n+1,\text{ }\text{ }q^i(s):= p^i(s)$; and 
$q^{n+1}(s):= \partial_{n+1}u(y(s)) = \partial_tu(x(s),t(s)).$ 
This time the unknown function is $u:\mathbb{R}^{n}\times [0,\infty)\to\mathbb{R}$.

Consider now the PDE, $F((x,t),u,Du) = 0$, instantiated as:
$$u_t + H(x,Du) = 0.$$
If we restrict to the curve, $x(s)$, this looks like $F(y,z,q) = 0$:
$$\bigg[\text{ }q^{n+1}(s)+H(x(s),p(s)) = 0\text{ }\bigg],$$
with $x$ and $p$ understood as the first $n$ coordinates of $y$ and $q$ respectively.

$\underline{Question:}$
When computing the RHS for the system of ODE's (not listed), the author states: $\partial_{z}F = 0$. At first glance, you don't see $z$ in the equation, but recall we defined $q^{i}(s) = \partial_{y^i}z(s)$. So there are contributions from both terms $\big(q^{n+1},H(x,p)\big)$ that require:
$$\partial_z(\partial_{y^i}z).$$
How does one further evaluate?

The reference for this is p.113 of Evan's PDE text.
 A: See below comment.

If we write out the difference quotient for $z$ (using function variables), we get:
$$\partial_z(\partial_{y^i}z(s)) := \lim\limits_{h(s)\to 0(s)}\frac{(\partial_{y^i})\big(z(s)+h(s)\big) - (\partial_{y^i})\big(z(s)\big)}{h(s)}$$
$$= \lim\limits_{h(s)\to 0(s)}\frac{(\partial_{y^i})\big(z(s)\big)+(\partial_{y^i})\big(h(s)\big) - (\partial_{y^i})\big(z(s)\big)}{h(s)}$$
$$\lim\limits_{h(s)\to 0(s)}\frac{(\partial_{y^i})\big(h(s)\big)}{h(s)}\text{ }\text{ }\text{ }\text{ }\text{ }(\star)$$
by Linearity of the partial operators, $\partial_{y^i}(\cdot)$, in their function variable. But $$\partial_{y^i}h(s) = 0(s)$$ since $h(s)$ doesn't contain $y^i$ as a variable (technicality here). So the desired limit $(\star)$ is zero.

This gives:
$$\partial_zF(y,z,q) = \partial_{z}q^{n+1} + \partial_zH(x,p) = \partial_z(\partial_{y^{n+1}}z) + \sum\limits_{j=1}^n\partial_{p^j}H(x,p)\cdot\partial_z(\partial_{y^j}z) = 0.$$

** I found a potential solution above. So this kind of changed to a solution verification/alternate input post. (Back to my future post that points here).
