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Chapter 4: Problem 4

For each problem, determine equilibrium points, bifurcation points, and construct a bifurcation diagram. Discuss the different behaviors in each system. a. \(y^{\prime}=y-\mu y^{2}\) b. \(y^{\prime}=y(\mu-y)(\mu-2 y)\) c. \(x^{\prime}=\mu-x^{3}\) d. \(x^{\prime}=x-\frac{\mu x}{1+x^{2}}\).

Short Answer

Expert verified
a) Equilibrium points: \(y=0\), \(y=\mu\). Bifurcations occur at these points. b) Equilibrium points: \(y=0\), \(y=\mu\), \(y=\mu/2\). Bifurcations occur at these points. c) Equilibrium points: \(x=\sqrt[3]{\mu}\), \(x=-\sqrt[3]{\mu/2}\), \(x= \sqrt[3]{\mu/2}\). Bifurcations occur at these points. d) Equilibrium points: \(x=0\) and \(x=\sqrt{\mu-1}\),\(-\sqrt{\mu-1}\). Bifurcations occur at these points. Different behaviors observed in each system depend on the values of \(\mu\).

Step by step solution

01

Equilibrium Points for Equation a.

To find the equilibrium points, set the function equal to zero. Solve for y in the equation \(y^{\prime}=y-\mu y^{2}=0\). Doing this yields \(y=0\) and \(y= \mu\).
02

Bifurcation Points and Diagram for Equation a.

The system undergoes a bifurcation at \(y=0\) and \(y=\mu\). This is because these are the points where the stability of the system changes. Plot these points on a diagram with \(y\) on the y-axis and \(\mu\) on the x-axis.
03

Equilibrium Points for Equation b.

Solving the equation \(y^{\prime}=y(\mu-y)(\mu-2 y)=0\), gives three equilibrium points, \(y=0\), \(y=\mu\), and \(y=\mu/2\).
04

Bifurcation Points and Diagram for Equation b.

The system undergoes a bifurcation at \(y=0\), \(y=\mu\) and \(y=\mu/2\). Plot these points on a diagram with \(y\) on the y-axis and \(\mu\) on the x-axis.
05

Equilibrium Points for Equation c.

Solving the equation \(x^{\prime}=\mu-x^{3}=0\), we get the equilibrium points \(x=\sqrt[3]{\mu}\), \(x=-\sqrt[3]{\mu/2}\), and \(x= \sqrt[3]{\mu/2}\).
06

Bifurcation Points and Diagram for Equation c.

The system undergoes a bifurcation at \(x=\sqrt[3]{\mu}\), \(x=-\sqrt[3]{\mu/2}\), and \(x= \sqrt[3]{\mu/2}\). Plot these points on a diagram with \(x\) on the y-axis and \(\mu\) on the x-axis.
07

Equilibrium Points for Equation d.

Solving \(x^{\prime}=x-\frac{\mu x}{1+x^{2}}=0\) gives equilibrium points \(x=0\) and \(x=\sqrt{\mu-1}\),\(-\sqrt{\mu-1}\).
08

Bifurcation Points and Diagram for Equation d.

The system undergoes a bifurcation at \(x=0\), \(x=\sqrt{\mu-1}\), and \(x=-\sqrt{\mu-1}\). Plot these points on a diagram with \(x\) on the y-axis and \(\mu\) on the x-axis.
09

Discussion of system behavior

Each of the four system behaviors depends on the values of \(\mu\). For all, changes in stability occur at the bifurcation points which can be depicted in each bifurcation diagram. In the case of equations a and b, as \(\mu\) increases, the stable point transitions from one equilibrium point to another. For equations c and d, the systems show more complex bifurcations involving increases and decreases in stability at different equilibrium points.

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Most popular questions from this chapter

The Michaelis-Menten kinetics reaction is given by $$ E+S \frac{k_{3}}{k_{1}}+E S \underset{k_{2}}{ } E+P $$ The resulting system of equations for the chemical concentrations is $$ \begin{aligned} \frac{d[S]}{d t} &=-k_{1}[E][S]+k_{3}[E S] \\ \frac{d[E]}{d t} &=-k_{1}[E][S]+\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[E S]}{d t} &=k_{1}[E][S]-\left(k_{2}+k_{2}\right)[E S] \\ \frac{d[P]}{d t} &=k_{3}[E S] \end{aligned} $$ In chemical kinetics, one seeks to determine the rate of product formation \(\left(v=d[P] / d t=k_{3}[E S]\right)\). Assuming that \([E S]\) is a constant, find \(v\) as a function of \([S]\) and the total enzyme concentration \(\left[E_{T}\right]=[E]+[E S] .\) As a nonlinear dynamical system, what are the equilibrium points?

Consider a model of a food chain of three species. Assume that each population on its own can be modeled by logistic growth. Let the species be labeled by \(x(t), y(t)\), and \(z(t)\). Assume that population \(x\) is at the bottom of the chain. That population will be depleted by population \(y\). Population \(y\) is sustained by \(x^{\prime}\) s, but eaten by \(z^{\prime}\) s. A simple, but scaled, model for this system can be given by the system $$ \begin{aligned} x^{\prime} &=x(1-x)-x y \\ y^{\prime} &=y(1-y)+x y-y z \\ z^{\prime} &=z(1-z)+y z \end{aligned} $$ a. Find the equilibrium points of the system. b. Find the Jacobian matrix for the system and evaluate it at the equilibrium points. c. Find the eigenvalues and eigenvectors. d. Describe the solution behavior near each equilibrium point. e. Which of these equilibria are important in the study of the population model and describe the interactions of the species in the neighborhood of these point(s).

Solve the general logistic problem, $$ \frac{d y}{d t}=k y-c y^{2}, \quad y(0)=y_{0} $$ using separation of variables.

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.

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