Showing a function $u(x,t)$ solves a partial differential equation. I'm trying to show that the function $$u(x,t) = \int^t_0 s(x + b(\tau - t), \tau) d\tau$$ satisfies the partial differential equation $$u_t + bu_x = s(x,t).$$
I start by finding $$u_t(x,t) = \frac{\partial}{\partial t}\int^t_0 s(x + b(\tau - t), \tau) \, d\tau =s(x,t)$$ and then $$u_x(x,t) = \frac{\partial}{\partial x}\int^t_0 s(x + b(\tau - t), \tau) \,  d\tau$$ $$= \int^t_0 \frac{\partial}{\partial x}s(x + b(\tau - t),\tau) \, d\tau$$ and this is where I get stuck.
Am I on the right track? 
 A: Your calculation of $u_t(x,t)$ is incorrect. Should be
$u_t(x,t) = \frac{\partial}{\partial t}\int^t_0 s(x + b(\tau - t), \tau) \, d\tau =s(x,t) - b\int^t_0 s_x(x + b(\tau - t),\tau) \, d\tau$
as the integrand also depends on t in the first argument of $s$. See the Leibniz rule: http://en.wikipedia.org/wiki/Leibniz_integral_rule
A: As @PaulCarter mentions, your trouble is in the calculation of $u_t$. 
In gory detail, according to Leibniz's integral rule, 
$$\begin{eqnarray*}
u_t(x,t) &=& \frac{\partial}{\partial t} \int_0^t s(x+b(\tau-t),\tau) d\tau \\
&=& \frac{\partial t}{\partial t} s(x+b(\tau-t),\tau)|_{\tau=t} + 
\int_0^t \frac{\partial}{\partial t}s(x+b(\tau-t),\tau) d\tau \\
&=& s(x,t) -b \int_0^t s^{(1,0)}(x+b(\tau-t),\tau) d\tau,
\end{eqnarray*}$$
where $s^{(1,0)}(x+b(\tau-t),\tau) = \frac{\partial}{\partial X} s(X,\tau)|_{X=x+b(\tau-t)}$. 
Chain rule brings out the factor $-b$.
Notice that the derivative with respect to $t$ acts on the first argument and that $s^{(1,0)}(x+b(\tau-t),\tau) = s_x(x+b(\tau-t),\tau)$.
Likewise 
$$\begin{eqnarray*}
u_x(x,t) &=& \int_0^t s^{(1,0)}(x+b(\tau-t),\tau) d\tau.
\end{eqnarray*}$$
Therefore, 
$u_t + b u_x = s(x,t)$, as claimed. 
