How does changing the parameter C from 0 to 1 affect the equilibria of $dz/dt = 4z - z^3 + C$?

The long-term behavior is determined by the phase line. Any initial condition strictly above the unstable equilibrium $z = 0$ will approach the stable equilibrium $z = 2$.

Conditions: The differential equation is $dz/dt = 4z - z^3$.; Initial conditions are not exactly at the unstable equilibrium $z = 0$.

A bifurcation diagram is a plot that shows how the equilibrium values of a dynamical system change as a parameter varies. It is constructed by plotting the equilibrium values (z) on the vertical axis against the parameter (C) on the horizontal axis.

Conditions: The system is described by a family of ODEs parameterized by C.; Equilibria are determined for a range of C values.

Equilibria are found by setting the time derivative to zero, which means finding the roots of the right-hand side function. By plotting $f(z) = 4z - z^3$, the equilibria correspond exactly to the points where the graph intersects the horizontal z-axis.

Conditions: The differential equation is autonomous, meaning the right-hand side depends only on z.; Equilibria are defined as constant solutions where $dz/dt = 0$.

Equilibria correspond to the roots of the right-hand side function $f(z) = 4z - z^3 + C$. As C increases, the graph of $f(z)$ shifts vertically upward.

Conditions: The differential equation is $dz/dt = 4z - z^3 + C$.; $C = 4$, which is greater than the critical value C* ≈ 3.079.

For $C = 4$, the system has only one stable equilibrium. Because there are no other equilibria to act as separatrices, all initial conditions, regardless of their starting value, will converge to this single positive equilibrium as time progresses.

Conditions: The differential equation is $dz/dt = 4z - z^3 + 4$.; $C = 4$ is greater than the bifurcation threshold C* ≈ 3.079.