An initial value problem for $p_t+ap_x = f(x,t)$

Question

Consider the portion of this solution where $p(x,t)= \displaystyle \frac{2}{a}$.

For this case, why doesn't $ p(x,0) = \displaystyle \frac{2}{a}$? If it did, this would clearly contradict the initial value condition because $a > 0 \Rightarrow \displaystyle \frac{2}{a} > 0$.

Answer 1

Well, if $t=0$ then you will never hit that case, since $x$ could not simultaneously be both $\geq 1$ and $\leq -1$.