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What is the long-term behavior of solutions for dz/dt=4z−z3dz/dt = 4z - z^3 based on their initial conditions?

The long-term behavior is determined by the phase line. Any initial condition strictly above the unstable equilibrium z=0z = 0 will approach the stable equilibrium z=2z = 2. Any initial condition strictly below z=0z = 0 will approach the stable equilibrium z=−2z = -2. The unstable equilibrium z=0z = 0 acts as a separatrix between these two basins of attraction.

Conditions

  • The differential equation is dz/dt=4z−z3dz/dt = 4z - z^3.
  • Initial conditions are not exactly at the unstable equilibrium z=0z = 0.

Reasoning, step by step

  1. Identify the stable equilibria at z=−2z = -2 and z=2z = 2, and the unstable equilibrium at z=0z = 0.
  2. Observe the direction of the vector field (arrows) on the phase line.
  3. Note that for z>0z > 0, the flow is directed toward z=2z = 2.
  4. Note that for z<0z < 0, the flow is directed toward z=−2z = -2.
  5. Conclude that z(0)>0z(0) > 0 implies z(t)→2z(t) \to 2 as t→∞t \to \infty , and z(0)<0z(0) < 0 implies z(t)→−2z(t) \to -2 as t→∞t \to \infty .

Example

The speaker states, "initial conditions above zero would go to 2 and below zero would go to -2," summarizing the global behavior captured by the phase line diagram.

Common misconceptions

  • Assuming that all solutions converge to the same equilibrium.
  • Thinking that solutions starting exactly at z=0z = 0 will move away over time (they stay at 0).
  • Confusing the direction of time with the direction of the spatial axis.

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