Why do two equilibria disappear when C is increased to 4 for $dz/dt = 4z - z^3 + C$?

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 $dz/dt = 0$.

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.

Increasing C from 0 to 1 translates the graph of the right-hand side function vertically upward by one unit. This shift causes the three equilibria to move: the two stable equilibria shift to the right, and the unstable equilibrium shifts to the left.

Conditions: The differential equation is $dz/dt = 4z - z^3 + C$.; C is changed from 0 to 1.