I am doing a course on PDEs and I am struggling to understand how to solve this:
$$\frac{\partial}{\partial s} (u(x+as, t+s) = f(x+as, t+s)$$
The notes say to integrate from $s = -t$ to $s= 0$ and they obtain $$u(x, t) = u(x-at, 0) + \int_{-t}^0 f(x+as, t+s) ds$$ but I don't understand how they have arrived here. This is probably fairly simple, but I'm looking for someone to help me fill in the details.
I am particularly struggling with the arguments on $u$, so how it has gone from $u(x+as, t+s)$ to $u(x, t)$ and how we have obtained the $u(x-at, 0)$ term
It is just the fundamental theorem of analysis. We have $$\int_a^b f(s)\,ds = F(b) - F(a) \tag{1}$$ with $F'(s) =f(s)$.
Now introduce the functions $$F_L(s) = u(x+as, t+s)$$ for the left-hand side and $$f_R(s) = f(x+as, t+s)$$ for the right-hand side.
From $f_L(s)= f_R(s)$, we obtain (from (1) with $a=-t$ and $b=0$) $$ F_L(0) - F_L(-t) = \int_{-t}^0f_R(s)\,ds = \int_{-t}^0 f(x+as, t+s)\,ds $$ or equivalently $$ u(x,t) - u(x- a t, 0) = \int_{-t}^0 f(x+as, t+s)\,ds\,.$$