Q:

Method of Characteristics(Advection equation with initial and boundary condition)

Solve, using the Method of Characteristics, the equation
$\frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}=-\mu\rho$
for $x,t>0$ with the conditions $\rho(x,0)=f(x)$ and $\rho(0,t)=g(t)$. 
I'm currently trying to solve above problem using method of characteristics.  I used initial condition $\rho(x,0)=f(x)$ and obtained $|\rho(x,t)|=|f(x-t)|e^{- \mu t}$. But I need to find $\rho(x,t)$ and also I don't know how to use the boundary condition $\rho(0,t)=g(t)$. I'm posting this problem here to get a hint. Thanks in advance.     

A:

$$\frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}=-\mu\rho$$
System of ODEs for the characteristics equations : $\quad \frac{dt}{1}=\frac{dx}{1}=\frac{d\rho}{-\mu\rho}$
First family of characteristic curves, from $\quad \frac{dt}{1}=\frac{dx}{1} \quad\to\quad t-x=c_1$ 
Second family of characteristic curves, from $\quad \frac{dt}{1}=\frac{d\rho}{-\mu\rho}\quad\to\quad \rho e^{\mu t}=c_2$
General solution on form of implicit equation : $\quad\Phi(t-x\:,\:\rho e^{\mu t})=0$
$\Phi$ is any differentiable function of two variables. 
Or equivalently, on explicit form : $\quad\rho e^{\mu t} =F(t-x)$
$F(X)$ is any differentiable function where $X=t-x$.
$$\rho(x,t)=e^{-\mu t}F(t-x)$$
Case of condition : $\rho(0,t)=g(t)$ :
$\rho(0,t)=g(t)=e^{-\mu t}F(t-0)\quad$ which determines the function $F$ :
$F(X)=g(X)e^{\mu X}\quad$ that we put into the above general solution with $X=t-x$
$$\rho(x,t)=e^{-\mu t}g(t-x)e^{\mu(t-x)} = g(t-x)e^{-\mu x} $$
Case of condition : $\rho(x,0)=f(x)$ :
$\rho(x,0)=f(x)=e^{-\mu 0}F(-x)\quad$ which determines the function $F$ :
$F(X)=f(-x)\quad$ that we put into the above general solution with $X=t-x$
$$\rho(x,t)=e^{-\mu t}f(-(t-x))=f(x-t)e^{-\mu t} $$
If the two conditions are specified together and without information about the functions $f$ and $g$, it seems doubtful that further calculus be possible about the respective ranges of validity of the two above different functions $\rho(x,t)$ and the boundaries between them. 
Comment :
If we don't know what are the functions $f$ and $g$ ( smooth or piecewise, related functions or not), in the general case $f(0)\neq g(0)$. Then there would be a contradiction between the conditions $\rho(0,0)=e^{-\mu 0}f(0-0)=f(0)$ and $\rho(0,0)=e^{-\mu 0}g(0-0)=g(0)$.
This implies that $\rho(x,t)$ would be a piecewise function like this, for example : 
$$\rho(x,t)=H(x-t)f(x-t)e^{-\mu t}+H(t-x)g(t-x)e^{-\mu x}$$
$H$ is the Heaviside step function. 
In writing such a form of solution, we accept that the PDE be not valid on the line $x=t$ where the partial derivatives would be not continuous.