In deriving the simple transport PDE,
Strauss "Introduction to Partial Differentials" shows
however, why is setting $h = 0$ necessary?
I provide my attempt at the derivation below, and I don't see where $h = 0$ is required to get equation (1).
$$ \begin{aligned} \frac{d}{db} \int_0^b u(x,t)dx &= \frac{d}{db} \int_{ch}^{b+ch} u(x, t+h)dx \\\\ \frac{d}{db} \left[U(b, t) - U(0, t) \right] &= \frac{d}{db} \left[ U(b+ch, t+h) - U(ch, t+h) \right] && \text{FTC} \\\\ u(b, t) &= u(b+ch, t+h) && \text{Keep upper lims of integrals because variable first argument to functions } U(\cdot) \\\\ \frac{d}{dh}[u(b, t)] &= \frac{d}{dh}[u(b+ch, t+h)] && \text{Differentiating w.r.t } h \\\\ 0 &= \frac{\partial u}{\partial x} \frac{\partial x}{\partial h} + \frac{\partial u}{\partial t} \frac{\partial t}{\partial h} && \text{Multivariable chain rule} \\\\ 0 &= u_x \frac{\partial}{\partial h}[b + ch] + u_t \frac{\partial}{\partial h}[t + h] \\\\ 0 &= u_x c + u_t 1 \\\\ 0 &= u_t + c u_x && \text{Rearranged} \end{aligned} $$
