# Math Help - Piecewise Integral

1. ## 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 $f(t,x) = \begin{cases} 0 & x\le 0 \\ 1 & x > 0\end{cases}\,\quad f(0,x) = 0$

$u(x,t) = \int_0^t f(t,x-a(t-s))\, ds,\qquad a > 0$

So if $x \le a(t-s),\; \int_0^t f(t,x-a(t-s))\, ds = \int_0^t\! 0\,ds$
$x > a(t-s),\; \int_0^t f(t,x-a(t-s))\, ds = \int_0^t\! 1\,ds$

The solution is suppose to be
$u(x,t) = \begin{cases} 0 & x\le 0 \\ x/a & 0 at\end{cases}$

EDIT: Refer to other thread for solution.