
Piecewise Integral
I posted a related problem in Differential Equations, but I have realized that really all I need is to understand this integral and so I would more likely get help here (I know I only look at threads in areas that I know).
Let $\displaystyle f(t,x) = \begin{cases} 0 & x\le 0 \\ 1 & x > 0\end{cases}\,\quad f(0,x) = 0$
$\displaystyle u(x,t) = \int_0^t f(t,xa(ts))\, ds,\qquad a > 0$
So if $\displaystyle x \le a(ts),\; \int_0^t f(t,xa(ts))\, ds = \int_0^t\! 0\,ds$
$\displaystyle x > a(ts),\; \int_0^t f(t,xa(ts))\, ds = \int_0^t\! 1\,ds$
The solution is suppose to be
$\displaystyle u(x,t) = \begin{cases} 0 & x\le 0 \\ x/a & 0<x<at \\ t & x > at\end{cases}$
EDIT: Refer to other thread for solution.