Chapter 4: Problem 5

Consider the family of differential equations \(x^{\prime}=x^{3}+\delta x^{2}-\mu x\) a. Sketch a bifurcation diagram in the \(x \mu\)-plane for \(\delta=0\). b. Sketch a bifurcation diagram in the \(x \mu\)-plane for \(\delta>0\). Hint: Pick a few values of \(\delta\) and \(\mu\) in order to get a feel for how this system behaves.

Short Answer

Expert verified

The bifurcation diagrams in the \(x - \mu\) plane would show how the equilibrium solutions vary with \(\mu\) for \(\delta = 0\) and \(\delta > 0\). The equilibrium solutions for \(\delta = 0\) are \(x = 0\), \(x = \mu\), and \(x = -\mu\). When \(\delta > 0\), the equilibrium solutions will differ, and will depend on the specific values of \(\delta\) and \(\mu\).

Step by step solution

Table of Values for \(\delta = 0\)

First, sketch the bifurcation diagram for \(\delta = 0\) in the \(x - \mu\) plane. In this case, the equation is \(x' = x^3 - \mu x\). Set the equation equal to zero and solve for \(x\), \(0 = x^3 - \mu x = x(x-\mu)(x+\mu)\), to find the equilibrium solutions. This gives \(x = 0\), \(x = \mu\), and \(x = -\mu\). These are the values we will plot on the bifurcation diagram.

Graph for \(\delta = 0\)

Plot these equilibrium solutions on the \(x - \mu\) plane, which yields two lines \(x = \mu\) and \(x = -\mu\), and a line at \(x = 0\). These represent equilibriums changing with \(\mu\). If \(\mu > 0\), then \(x = \mu\) is unstable and \(x = 0\) and \(x = -\mu\) are stable. For \(\mu < 0\), \(x = -\mu\) is unstable, while \(x = \mu\) and \(x = 0\) are stable.

Table of Values for \(\delta > 0\)

Next, sketch the bifurcation diagram for \(\delta > 0\). Now the equation is \(x' = x^3 + \delta x^2 - \mu x\). Again, find the equilibrium points by setting the equation to zero.

Graph for \(\delta > 0\)

Plot the equilibrium solutions on the \(x - \mu\) plane. Because of the inclusion of the term \(\delta x^2\), the bifurcation diagram will look different than for \(\delta = 0\). Different values of \(\delta\) and \(\mu\) will shift and alter the equilibriums.