Solving a first order PDE Consider the PDE
$$u_t + uu_x +\alpha u = 0; \quad u(x,o)=u_0(x)$$
Whose characteristics are:
$$\begin{cases}dx/dt = u; \quad x(0) = \xi \\ du/dt = -\alpha u; \quad u(0)=u_0(\xi)\end{cases}$$
I can solve one of the equations to get
$$u = u_0(\xi)e^{-\alpha t}$$
and
$$x = ut+\xi$$
But the solution gives $x$ as
$$x = \xi +\frac{1}{\alpha}u_0(\xi)(1-e^{-\alpha t})$$
 A: I don't agree with your solution. One cannot directly integrate $\quad dx/dt=u\quad $ as $\quad x=ut+x(0)\quad $ because $u$ isn't constant wrt $t$.
$$u_t + u u_x =-\alpha u $$
The Charpit-Lagrange characteristic ODEs are :
$$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{-\alpha u}$$
A first characteristic equation coming from solving $\quad \frac{dx}{u}=\frac{du}{-\alpha u}\quad$ is :
$$u+\alpha x=c_1$$
A second characteristic equation comming from solving $\quad \frac{dt}{1}=\frac{du}{-\alpha u}\quad$ is :
$$e^{\alpha t}u=c_2$$
The general solution on the form of implicit equation $c_1=F(c_2)$ is:
$$\boxed{u+\alpha x=F\left(e^{\alpha t}u\right)}$$
$F$ is an arbitrary function, to be determined according to the initial condition.
Condition : $u(x,0)=u_0(x)$
$$u_0(x)+\alpha x=F\left(e^{0.\alpha}u_0(x)\right)=F(u_0(x))$$
Let $\quad \zeta=u_0(x)\quad\implies\quad x=u_0^{-1}(\zeta)$
$u_0^{-1}$ denotes the inverse function of the function $u_0(x)$.
$$F(\zeta)=\zeta+\alpha u_0^{-1}(\zeta)$$
Now the fonction $F$ is determined. We put it into the above general solution where the variable $\zeta$ is no longer  $=u_0(x)$ but is $=e^{\alpha t}u$ .
$$F(e^{\alpha t}u)=e^{\alpha t}u+\alpha u_0^{-1}(e^{\alpha t}u)$$
Thus the solution satisfying the initial condition is :
$$u+\alpha x=e^{\alpha t}u+\alpha u_0^{-1}(e^{\alpha t}u)$$
$$\boxed{x=\frac{1}{\alpha}(e^{\alpha t}-1)u+u_0^{-1}(e^{\alpha t}u)}$$
