Rigorous mathematical proof of the equation of motion of a Sliding Chain with friction In most physics textbooks the sliding chain is considered a one dimensional problem, where every force applied to the chain in the x or y axes is immediately summed to calculate the net force (As if we were dealing with a pulley that connects to masses using a massless rope). I was wondering if there is a more rigorous explanation of why that is possible. I tried to work it out by myself wanted to know if my proof is sufficient.  
Let $\rho $ be the density of the chain and $l$ its constant length. Let
also $x$ be the part of the chain laying on the table and $y$ the part
of the chain hanging over the table.
The mass on top of the table and hanging from the table varies, so
$$\frac{d}{dt} ( \rho  x \dot{x} ) = - \mu g \rho x
\frac{\dot{x}}{|\dot{x}|} \tag{1}$$  (for motion with sliding friction)
for the $x$ direction with the unit vector pointing to the right
$ \frac{d}{dt} ( \rho  y \dot{y} ) =  g  \rho y \tag{2}$
for the y direction with the unit vector pointing downwards.
There is also
$ x + y = l \tag{3}$
and therefore
$\dot{x} = -\dot{y} \tag{4}$.
If we subtract  (2) from (1) we get
$\frac{d}{dt} ( x \dot{x} - y \dot{y} ) = - g y - \mu g x
\frac{\dot{x}}{|\dot{x}|} $.
After substituting x in (3) and (4) we get
$\ddot{y} = \frac{g}{l} y - \frac{\mu g}{l} (l - y)
\frac{\dot{y}}{|\dot{y}|}  \tag{5}$.
With $\frac{\dot{y}}{|\dot{y}|}=1$ we get
$\ddot{y} = \frac{g}{l} y - \frac{\mu g}{l} (l - y) \tag{6}$
which is the DE of the sliding chain with friction.

 A: Two forces that are relevant to potential acceleration act on the chain: gravity $\vec{G}$ on the $y$ part and a friction force $\vec{F}$ on the $x$ part. The latter arises from the Normal force. The vectors point in different directions but the corner acts as a frictionless pulley, 'transmitting' these forces and allowing us to use their scalars.
So we can write, as equation of motion:
$$ma=G-F$$
Filling in we get:
$$\rho l \ddot{y}=y\rho g-\mu \rho gx$$
And:
$$x+y=l \Rightarrow x=l-y$$
So:
$$ l \ddot{y}=gy-\mu gl+\mu gy$$
With minimal reworking this becomes OP's equation $(6)$:
$$\ddot{y} = \frac{g}{l} y - \frac{\mu g}{l} (l - y)\tag{6}$$
In turn this reworks easily to:
$$\ddot{y} = \frac{g(1+\mu)}{l}y-\mu g\tag{7}$$

Initial condition:
Only if $G-F>0$ does acceleration occur, so:
$$y\rho g-\mu \rho gx>0$$
$$y -\mu(l-y)>0$$
$$y>\frac{\mu}{1+\mu}l\tag{8}$$
So this is the minimum amount of chain that needs to dangle over the edge at $t=0$, to get acceleration.

Solving the DE:
This is outside the remit of the question but I was curious to see what form the solution of $(7)$ would take.

$$\ddot{y} = \frac{g(1+\mu)}{l}y-\mu g\tag{7}$$
Make a substitution:
$$\frac{g(1+\mu)}{l}y-\mu g=u$$
$$\frac{g(1+\mu)}{l}\dot{y}=\dot{u}$$
$$\frac{g(1+\mu)}{l}\ddot{y}=\ddot{u}$$
$$\ddot{y}=\frac{l}{g(1+\mu)}\ddot{u}$$
$$\frac{l}{g(1+\mu)}\ddot{u}-u=0$$
$$\ddot{u}-\frac{g(1+\mu)}{l}u=0$$
$$\ddot{u}-m^2u=0$$
Where:
$$m^2=\frac{g(1+\mu)}{l}$$
The solution is of the shape:
$$u=c_1 e^{mt}+c_2 e^{-mt}$$
Where the boundary conditions can be:
$$y=y_0, t=0$$
$$\dot{y}=0, t=0$$
So the acceleration is far from uniform, as is the case for free falling body (of constant mass).
