My Evans textbook presents a nonlinear first-order PDE \begin{cases}F(Du,u,x)=0 & \text{in }U \\ u=g & \text{on } \Gamma \subseteq \partial U\end{cases} where $x \in U \subset \mathbb{R}^n$. Also, $u : \bar{U} \longrightarrow \mathbb{R}$ is the unknown, $u=u(x)$.
Finally, $F=F(p,z,x)=F(p_1,\ldots,p_n,z,x_1,\ldots,x_n)$ and $g$ are assumed to be smooth functions, for $p \in \mathbb{R}^n, z \in \mathbb{R}, x \in U$.
We also set \begin{cases}D_pF = (F_{p_1},\ldots,F_{p_n}) \\ D_z F=F_z \\ D_xF=(F_{x_1},\ldots,F_{x_n}) \end{cases}
The textbook eventually goes on to differentiate this wtih respect to $x_i$: $$\sum_{j=1}^n F_{p_j}(Du,u,x)u_{x_j x_i} + F_z(Du,u,x)u_{x_i}+F_{x_i}(Du,u,x)=0$$
I have two questions about this differentiated equation.
- Is the chain rule for several variables is used, which is why we have $\sum_{j=1}^n F_{p_j}+F_z+F_{x_i}$?
- Why does the first term $F_{p_j}$ have a summation but the last term $F_{x_i}$ does not?
$\bullet\,$ Of course, yes. For instance, taking $i=1$ you get $$ \begin{align} \bigl(F(Du,u,x)\bigr)_{x_1}=\frac{\partial\,}{\partial x_1}\Bigl(F\bigl(u_{x_1}(x),\dots,u_{x_n}(x)\,,u(x),x_1\,,\dots,x_n\bigr)\Bigr)\\ =\sum_{j=1}^n F_{p_j}(p_1\,,\dots,p_n,u,x)\frac{\partial p_j}{\partial x_1} + F_z(Du,z,x)\frac{\partial z}{\partial x_1}+F_{x_1}(Du,u,x_1\,,\dots,x_n) \end{align} $$ where $p_j=u_{x_j}(x_1\,,\dots,x_n)$ while $z=u(x_1\,,\dots,x_n)$.
$\bullet\,$ Strictly according to the multivariate chain rule. The reason is the same why $$ \Bigl(f\bigl(g_1(x),\dots,g_n(x)\bigr)\Bigr)_{x_1}=\sum_{j=1}^n f_{g_j}(g_1\,,\dots,g_n)\cdot g_{x_1}(x) $$ involves summation, while $\bigl(h(x_1\,,\dots,x_n)\bigr)_{x_1}=h_{x_1}(x)$ does not.