Working out the analysis of a circuit with non-linear resistor I am working on a problem from the Strogatz book, specifically problem 2.2.12. I am having trouble evaluating the fixed point that I found. I have summarized the problem statement here, and show my work below:
Suppose a series circuit with a voltage source $V_0$, a non linear resistor and a capacitor. Assume that the the non linear nature of the resistor is such that $I = g(V)$, which is graphed in the book and I have tried to reproduce:

So, to model the equation:
$$V_0 = V_r + Q/C \Leftrightarrow V_r = V_0 - Q/C$$
Where $V_r$ is the voltage drop across the non-linear resistor. Coupling this with the fact (given in the text) that $\dot{Q} = I$, the function $g(V)$ can be rewritten as 
$$\dot{Q} = g(V_0 - \frac{Q}{C})$$
and we see that $\dot{Q} = 0$ When $Q = \frac{V_0}{C}$
I thought to use the linearization technique mentioned in this chapter, namely to find out the sign of $f'(Q, t) = \ddot{Q} = -\frac{1}{C}g'(V_0 - \frac{Q}{C})$ at the fixed point.
So here is where I get a little stumped. I see that $g'(V_0 - \frac{Q}{C})$ is positive and non-zero at (0,0) and therefore the overall value is negative (stable) due to the factor of $-\frac{1}{C}$, but I don't know how to determine end behavior. The task was to determine the qualitative differences between this model and the one where voltage drop is simply given by $V_r = I_rR$. 
That means, in my mind, that $Q(t) \rightarrow Q^* = V_0 / C$ as $t \rightarrow \infty$, which is different than the fixed point in the example of a linear resistor. Is that all that the question is asking? But the stability remains the same... except that I recall stability analysis for non-linear equations is supposed to be local only, so maybe there is something else I am missing?
 A: I got bugged by this exercise too. I must say that you put me in the right direction for a better understanding of it, and therefore I will put my thoughts here for the sake of completeness.
Once you have the flow equation:
$$
\dot{Q} = g(V_0 - \frac{Q}{C})
$$
The rest is to be solved and explained qualitatively (the linear stability analysis comes later in the chapter so I don't think it should be necessary to answer here).
A small mistake in your post though, from the equation above, and given that $g(0) = 0$, we find $Q^* = CV_0$, exactly like in the linear resistor case (when $I = V_r/R$). That fixed point is indeed stable, but we can draw the phase portrait from the sketch of $g(V)$: you just need to flip around the graph w.r.t the origin and translate of $V_0$ along the x-axis (this is a qualitative sketch, there is also a scaling by $1/C$ to be applied to be exact, and we don't even know $g$ despite the sketch of its graph). Also in the plot below, I probably did not reproduce exactly the intended $g$, I don't think there should be an infinite derivative at the fixed point.

The only qualitative difference that I conclude from that is that for the non-linear case the approach to the stable fixed point is much faster in its vicinity, while slower when starting far away from it...
