Solve $u_t+bu_{x}=s(x,t)$ with $u(x,0)=u_0(x)$ Consider the inhomogeneous transport equation $u_t+bu_{x}=s(x,t)$ with initial condition $u(x,0)=u_0(x)$. I'm trying to write the solution to this PDE using the method of characteristics.
$\frac{d}{dt}u(\xi(t),t)=u_x\xi'(t)+u_t=u_xb+u_t=s(\xi(t),t)$ and so $u(\xi(t),t)=\int_0^ts(\xi(r),r)dr+u(\xi(0),0)$ 
and since the characteristics are given by $\xi(t)=bt+x_0$ we have that $x=\xi(t)$ iff $x_0=x-bt$ and so the solution is given by $$u(x,t)=\int_0^ts(x,r)dr+u_0(x-bt)$$
Is this correct?
 A: You are very close, but slightly off, and I think the confusion comes in this variable of integration $r$. 
I'll present the exposition that has always made most sense to me. Characteristics $(x(\tau),t(\tau))$ originating from a point $(x(0),t(0)) = (\xi,0)$ are given by \begin{align*}
\dot x &= b, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x(0) = \xi, \\
\dot t &= 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, t(0) = 0, \\
\dot z &= s(x,t), \,\,\, z(0) = u_0(\xi),
\end{align*} where $z(\tau;\xi) = u(x(\tau;\xi),t(\tau;\xi))$. Solving these gives \begin{align*}
x(\tau;\xi) &= b\tau+\xi, \\
t(\tau;\xi) &= \tau, \\
z(\tau;\xi) &= u_0(\xi) + \int^\tau_{0} s(x(r;\xi),t(r;\xi))dr = u_0(\xi) + \int^\tau_{0} s(br+\xi,r)dr.
\end{align*} Now if we can invert the relationships $(x(\tau;\xi),t(\tau;\xi))$ to arrive at $(\tau(x,t),\xi(x,t))$, then we have the solution $u(x,t) = z(\tau(x,t);\xi(x,t))$. Note that the integration variable $r$ will stay! Only $\tau$ and $\xi$ will be replaced. As you've noted $\xi = x-bt$, and we also have $\tau = t$. Thus \begin{align*} u(x,t) &= u_0(x-bt)+\int^t_0 s(br+x-bt,r)dr \\
&= u_0(x-bt)+\int^t_0 s(x-b(t-r),r)dr.
\end{align*} We can check that this is indeed a solution. Note that when you differentiate the integral term with respect to $t$, you need to use the Leibniz rule. We see: \begin{align*} u_t(x,t) &= -bu'_0(x-bt) + s(x,t) -b \int^t_0 s_x(x-b(t-r),r)dr,\\
u_x(x,t) &= u'_0(x-bt) + \int^t_0 s_x(x-b(t-r),r)dr.
\end{align*} Thus we do indeed have $$u_t(x,t)+bu_x(x,t) = s(x,t),$$ and clearly $u(x,0) = u_0(x)$. 
