Final answer:
The Fitzhugh-Nagumo model has three equilibrium solutions for the electric potential: 0, a, and 1, corresponding to stable, unstable, and stable equilibria, respectively. A phase line would illustrate stable equilibria at 0 and 1 with attraction and an unstable equilibrium at a with repulsion.
Step-by-step explanation:
The equilibrium solutions of the differential equation dv/dt = -v(v^2 - (1+a)v + a) can be found by setting the right side of the equation to zero and solving for v. Since a is a constant such that 0 < a < 1, factoring gives us v(v - 1)(v - a) = 0. Thus, the equilibrium values of v are 0, 1, and a. Analyzing the sign of the derivative on intervals between these points gives us the phase line.
At v=0, any small positive v will make dv/dt negative, thus v decreases back to 0; it's a stable equilibrium. At the equilibrium v=a, the derivative changes sign at v, indicating that v=a is an unstable equilibrium. Finally, at v=1, any small deviation results in v returning to 1, making it a stable equilibrium. A phase line would show arrows pointing towards the stable equilibria (0 and 1) from the regions between the equilibria, and away from the unstable equilibrium (a).