A specific 1st order PDE which looks almost like a linear PDE I have a PDE on the following form:
$$
\frac{\partial f}{\partial t}(t, x) + \mu \frac{\partial f}{\partial x}(t, x) + \lambda [f(t, x+1)-f(t, 1)]= 0
$$
where $\lambda$ and $\mu$ are positive constants and there is an end condition $f(T, x)=F(x)$, and I seek a solution on $x>0$. 
I am not sure how to approach this one. I have some experience with linear first-order PDEs but this seems to be of a slightly different form due to the $f(t, x+1)-f(t, 1)$ term. Is there a way to transform it into a regular linear PDE, perhaps?
I appreciate any pointers at how to deal with this sort of problem.
An attempt at using separation of variables: $f(t, x)=X(x)T(t)$. The equation now becomes (I use $\mu=\lambda=1$ for simplicity)
$$
X(x)T'(t) + X'(x)T(t) + T(t)X(x+1)-T(t)X(1)=0
$$
or
$$
\frac{T'(t)}{T(t)}=-\frac{X'(x)+X(x+1)-X(1)}{X(x)}=a
$$
for some constant $a$. The equation for $T$ is easy: $T(t)=ce^{at}$. The $X$ one is $X'(x)-aX(x)+X(x+1)-X(1)=0$. Not sure how to deal with the $X(x+1)$.
 A: NOTE: the text below is not perfect rigorous. If you want, you can suggest improvements in comments.
Ok, let's start. We have the equation
$$\frac{\partial f}{\partial t} + \mu\frac{\partial f}{\partial x} + \lambda[f(t, x + 1) - f(t, 1)]=0\tag1\label1$$
Following Fourier, we use separation of variables (in fact, this will lead us to Fourier transform). Let $f(t,x)=X(x)T(t)$:
$$
X(x)T'(t)+\mu X'(x)T(t) + \lambda [X(x+1)T(t) - X(1)T(t)] = 0
$$
Divide by $XT$:
$$
\frac{T'}{T} + \frac{\mu X' + \lambda X(x+1) - \lambda X(1)}{X} = 0\tag2
$$
Let $a$ be constant:
$$
\frac{T'}{T}=a\Longrightarrow T = e^{at}\tag3
$$
and
$$
\mu X' + aX + \lambda X(x+1) - \lambda X(1) = 0\tag4\label{4}
$$
We сould solve this equation by substitution $X = e^{px}$. But $X(1)$ doesn't allow it us. Here is some trick: let $X = X_0 + F$, where
$$
\mu \frac{dX_0}{dx} + aX_0 + \lambda X_0(x+1) = 0;\tag5
$$
therefore
$$
\mu F' + aF + \lambda F(x+1) - \lambda (X_0(1) + F(1)) = 0
$$
This equation have constant solutions. Let $F(x)=C$:
$$
 aC + \lambda C - \lambda (X_0(1) + C) = 0 \Longrightarrow C = \frac{\lambda}{a} X_0(1)
$$
$$
X = X_0 + \frac{\lambda}{a} X_0(1)\tag6
$$
And now back to $X_0$. Let $X_0 = e^{px}$:
$$
\mu p + a + \lambda e^p = 0
$$
So
$$
X_0(1)=e^p\Longrightarrow \frac{\lambda}{a}X_0(1)=\frac{\lambda}{a}e^p = -(1 + \frac{\mu}{a}p)
$$
and
$$
X = e^{px} - 1 - \frac{\mu}{a}p = e^{px} - \frac{\lambda e^p}{\lambda e^p + \mu p} = e^{px} - \frac{1}{1 + \frac{\mu}{\lambda}pe^{-p}}\tag7
$$
So, we have
$$
f(t,x) = e^{at} \left( e^{px} - \frac{1}{1 + \frac{\mu}{\lambda}pe^{-p}} \right) = e^{px + at} - \frac{e^{at}}{1 + \frac{\mu}{\lambda}pe^{-p}}\tag8\label{8}
$$
$p$ is an arbitrary number; thus, from linearity, we can write
$$
f(t,x) = \sum_pw(p) \left[e^{px + at} - \frac{e^{at}}{1 + \frac{\mu}{\lambda}pe^{-p}}\right]
$$
(it's not perfect rigorously part! But OP "completely unfamiliar with those techniques"). As Fourier himself, we can denote $p=i\omega$ and use integral instead of sum:
$$
\boxed{f(t,x) = \int\limits_{-\infty}^\infty \rho(\omega)e^{at} \left[e^{i\omega x} - \frac{1}{1 + \frac{\mu}{\lambda}i\omega e^{-i\omega}}\right] d\omega}\tag9
$$
(and $a = -(\mu i\omega + \lambda e^{i\omega})$ now). We made $p$ complex because we have theory for such (Fourier) integrals, actually :) Anyway, if I'm not mistaken, this is general solution of $\eqref{1}$.
(For rigorousity. We could apply Fourier transfrom w.r.t. $x$ directly to $\eqref{1}$:
$$
\frac{\partial\hat f}{\partial t} + \hat f(i\mu\omega + \lambda e^{i\omega}) = \lambda \mathcal F_x [f(t, 1)]\tag{10}
$$
What about RHS? Thus $f(t,1)=\lim_{x\to0} f(t,1+x)$, $\mathcal F_x [f(t, 1)] = e^{i\omega}\hat f(t,0)$. With trick for solving $\eqref{4}$, it leads to $\eqref{8}$. This is the end of excusable part :)
And now something completely different. For diff. eq. $\eqref{1}$ we should use initial condition $f(0, x) = g(x)$; so,
$$
\boxed{g(x) = \int\limits_{-\infty}^\infty \rho(\omega) \left[e^{i\omega x} - \frac{1}{1 + \frac{\mu}{\lambda}i\omega e^{-i\omega}}\right] d\omega}\tag{11}\label{11}
$$
or
$$
g(x) = \int\limits_{-\infty}^\infty \rho(\omega) e^{i\omega x}d\omega - \int\limits_{-\infty}^\infty \frac{\rho(\omega)d\omega}{1 + \frac{\mu}{\lambda}i\omega e^{-i\omega}}.\tag{12}\label{12}
$$
It's end of all hope :( I know nothing about solving of integral equations like $\eqref{11}$. And if first integral in $\eqref{12}$ is just inverse Fourier transform, then second scares me.
Maybe somebody save us?
