No, I do not see a problem.  I see an inherent negative feedback mechanism and the reason di is limited to and maintained at precisely .8A/s.

Consider what happens upon application of the 4 volt EMF.  Current begins to flow thru the inductor and...

1.  As di reaches a rate of change equal to .8A/s, the CEMF becomes equal to 4 volts.

2.  As per "B)" in the drawing, as the CEMF becomes equal to the EMF, di also reduces towards zero.

3.  But if di becomes less than .8A/s, the CEMF also becomes less than 4 volts which allows di to increase. 

4.  Return to number 1 above (loop forever)

Although described above in a rather step-wise fashion, it is, in reality, a smooth and continuous process and a classic example of a negative feedback loop.

Thanks for taking the time to post the drawings.  By all means, enjoy the 4th!

PW

Added:

Also note that I believe it more appropriate for L2 to be replaced with a simple conductor (wire).  Having an inductor inside a "model" of an inductor is redundant and could lead to confusion.

I would replace L2 with a wire and then place a dotted line box surrounding both the CEMF Vsource and the wire with di indicated.  The entire dotted line box would then be labeled as L2.

(I would also label the value for CEMF1 as X_v or X_v1 because the only time the CEMF=4volts is when di=.8A/s)

An ideal inductor sitting at rest, is essentially an ideal piece of wire. So when an ideal emf is placed across it, the current wants to instantly jump to infinity, right at t=0.

Due to the self-induced cemf however (the negative feedback), current is limited to rise at 0.8A/s (in our case with 4V and 5H).

There are two reasons the induced cemf = the applied emf:

1) The obvious reason is that the ideal voltage source is holding the inductor terminal voltage at 4V.

2) The not so obvious reason is due to the self-induction feedback process trying to be explained here.

To explain further, try thinking about the process right from the very beginning. As I said above, the moment Vin is applied, the current will try to instantly go to infinity. We know however that simultaneously, the inductor is self-inducing an opposing emf which is going to cause the current to increase at a set rate determined by Vin and L. This is the feedback mechanism that makes an inductor what it is. The inductor doesn't limit current, it limits the change in current. Any change in current is reflected in a corresponding opposing emf, which opposes the applied emf. In the case when R=0, the emf and cemf are always equal. The cemf settles or equalizes to the same value as the applied emf.

Yes, with a real coil at t=0 the cemf=emf, and yet, current flows. I've mentioned this a number of times now.
There is feedback as well with real coils. You can see the current is rising at 0.8A/s for the first 200ms when R=1 Ohm. After this point the voltage begins to trade off with the resistor until finally ALL the voltage is across the R. It started out with ALL the voltage across the L. Do you wish me to repost the sim result showing the 1 Ohm trace tracking the 1p Ohm trace for the first 200ms?
I'm not sure how you came to that conclusion, but it is wrong. What do you mean the current is not limited? Of course it is; in fact it is limited in two different ways: 1) a maximum of 0.8A/s current rate of rise, 2) a maximum current of Vin/R.

Here is the real coil (R=1 Ohm) voltage tradeoff traces.

At t=0 ALL the voltage is across the inductance, and 0V across the resistor. That is, at t=0, cemf=emf. And yet, current flows.

After 30s or so the inductance ends up with 0V, and the resistance 4V across them; they have traded places.

This model was an attempt to make sense of the Emf/Cemf feed back loop and D) represents said attempt. Do not be confused by your thought that I don't understand what you are saying because I understand completely, I just don't agree! I knew that my new model was not correct when I created the two Emf sources.

So, in another approach to resolve the matter, I propose we attempt to simulate the inductor that is, not use the internal inductor model(s) in any given simulator, but break it down into functional components as Webby suggests. 

LtSpice is my preference and it's toolbox includes arbitrary behavioral voltage and current sources plus voltage dependent current sources , current dependent voltage sources, etc, so all we need to do is basically establish the model and the math.

I see no reason why this should not be doable and it would be a great teaching tool.

I have attached a schematic that shows the general circuit we wish to simulate and what I believe to be your equivalent model, but I would like you to confirm or indicate any changes that should be made so we can proceed. I am still working on my own equivalent model and will post it as soon as it is complete enough for criticism.

pm

I was thinking of suggesting/doing this yesterday, but changed my mind. I've been down this road before and the work effort is not worth it. It will only be met with skepticism and will be readily dismissed as invalid.

Just thinking out loud...

If you can program an arbitrary current source so that it rises at a rate of .8A/s, it could be used as a reference current, Iref.  The applied EMF and Iref would be started simultaneously.  The current flowing thru the model would be detected, hereinafter, Idet.  You could then use the difference between Iref and Idet to generate an error signal to control the value of the CEMF Vsource using a comparator or the like.  In the real world of negative feedback, damping would also need to be dealt with...