Solving the PDE $\frac{\partial u}{\partial t}+f(x)\frac{\partial u}{\partial x}+u=0$ subject to $u(x,0)=e^{-b x}$. I am hoping to obtain a closed-form solution to the PDE
$$\frac{\partial u}{\partial t}+f(x)\frac{\partial u}{\partial x}+u=0$$
subject to the initial condition $u(x,0)=e^{-b x}$.
If this is not possible, the special case with $b=0$ and/or $f(x)=(1-a)+a(1+cx)^{-1}$ would be of interest. Any information about the long-term behavior of the solution would also be useful.
 A: $$\frac{\partial u}{\partial t}+f(x)\frac{\partial u}{\partial x}=-u \tag 1$$
$$\frac{dt}{1}=\frac{dx}{f(x)}=\frac{du}{-u}$$
A first characteristic equation comes from $\frac{dt}{1}=\frac{dx}{f(x)}$ which solution is :
$$-t+\int{\frac{dx}{f(x)}}=c_1$$
A second characteristic equation comes from $\frac{dt}{1}=\frac{du}{-u}$ which solution is :
$$u\:e^t=c_2$$
The general solution of the PDE $(1)$ expressed on the form of implicit equation is :
$$\Phi\left(c_1\:,\:c_2 \right)=\Phi\left(-t+\int{\frac{dx}{f(x)}}\:,\:u\:e^t \right)=0$$
where $\Phi$ is an arbitrary function of two variables.
Or equivalently on explicit form :
$\quad u\:e^t=F\left(-t+\int{\frac{dx}{f(x)}} \right)$
$$u(x,t)=e^{-t}F\left(-t+\int{\frac{dx}{f(x)}} \right)\tag 2$$
where $F$ is an arbitrary function.
The function $F$ has to be determined according to the initial condition.
$$u(x,0)=e^{-bx}=e^{-0}F\left(-0+\int{\frac{dx}{f(x)}} \right)$$
$$F\left(\int{\frac{dx}{f(x)}} \right)=e^{-bx}$$
Let $\quad \int{\frac{dx}{f(x)}}=g(x) $
Since $f(x)$is a given function, $g(x)$ can be considered as a known function.
Let $x=h(g(x))$ the inverse function of $g(x)$.
Since $g(x)$ is a known function the inverse function $h$ can be considered as a known function.
$$F(g)=e^{-bh(g)}$$
Since $h$ is a known function, now the function $F$ is determined. 
Putting it into the general solution $(2)$ leads to the solution of the problem (which agrees to the PDE and to the initial condition) :
$$u(x,t)=e^{-t}e^{-b\:h\left(g(x)-t\right)}$$
Of course, this equation is difficult to understand, due to the presence of an inverse function. This becomes simpler if the function $f(x)$ is explicitly given.
CASE $\quad b=0$ :
The solution is trivial 
$$u(x,t)=e^{-t}$$
Putting it into $(1)$ shows that the PDE is satisfied any $f(x)$ because $\frac{\partial u}{\partial x}=0.$ 
Also, the initial condition $u(x,0)=e^{-0\:x}=1$ is satisfied.
CASE $\quad b\neq 0\quad$ and $\quad f(x)=(1-a)+\frac{a}{1+cx}$
$g(x)=\int{\frac{dx}{(1-a)+\frac{a}{1+cx}}}=\int{\frac{(1+cx)dx}{1+(1-a)cx}}=\frac{x}{1-a}-\frac{a}{(1-a)^2c}\ln|1+(1-a)cx|$
At this step, we have to compute $h$ the inverse function of $g(x)$. This is rather complicated because it cannot be expressed in terms of a finite number of elementary functions. $h$ involves a special function , namely the Lambert W function.
Theoretically this is possible. Sorry, I am too lazy and I let you continue. Hang in there !
