Spring 2017, MATH 481, Project 6
Linearization and the analysis of equilibria

Due Friday May 12 Friday May 19

Outline

• The solution of the initial value problem for a linear system of differential equation $dx/dt = Ax$, $x(0)=x_0$, where $A$ is an $n\times n$ constant matrix, may be expressed explicitly in terms of the eigenvalues and eigenvectors of $A$.
• The origin is an equilibrium of the differential equation $dx/dt = Ax$. The stability of the origin is determined completely by the eigenvalues of $A$. Thus, if you do have the eigenvalues of $A$, you should be able to tell immediately which of the following pictures depicts the phase portrait.

  stable node	unstable node	saddle	stable spiral	unstable spiral	center
  Click image to download PDF

• If you don't have the eigenvalues of $A$, you can still tell to which case your system belongs by evaluating the trace, determinant, and discriminant of $A$, and appealing to the following flowchart.

  Click image to download PDF

• The Taylor series expansion of a function $f : R^n \to R^n$ (that is, a vector field) introduces the concept of the Jacobian matrix, $Df$.
• Consider the nonlinear system of differential equations $dx/dt = f(x)$ associated with a vector field $f : R^n \to R^n$. Let $\bar{x}$ be an equilibrium. The Hartman-Grobman theorem asserts that the structure of the orbits in a sufficiently small neighborhood of $\bar{x}$ is homeomorphic to that of the linear system $d\xi/dt = A \xi$ near the origin, where $A = Df\big|_{\bar{x}}$, provided that no eigenvalues of $A$ lie on the imaginary axis in the complex plane.
• The predator-prey system \begin{align*} \frac{dx}{dt} &= ( a_1 - b_{11} x - b_{12} y)x, \\ \frac{dy}{dt} &= (-a_2 + b_{21} x - b_{22} y)y \end{align*} has equilibria at $E_1 = (0,0)$, $E_2 = (a_1/b_{11},0)$, $E_3 = (0,-a_2/b_{22})$, and $E_4 = (\bar{x}, \bar{y})$.

  Assume $a_1/b_{11} > a_2/b_{21}$ and note that this implies that $E_4$ lies in the first quadrant. Examine the type/stability of each of the four equilibria.

• In Project #5 you plotted the system's phase portrait after picking \[ a_1 = 4\quad b_{11} = 1\quad b_{12} = 1\quad a_2 = 1\quad b_{21} = 1\quad b_{22} = 1. \] Does $E_4$ lie in the first quadrant with these choices? Is $E_4$ a node, spiral, or a saddle?
• It is possible to have a stable node (not a spiral) $E_4$ in the first quadrant. To demonstrate that, search for suitable coefficients $a_i$ and $b_{ij}$ selected from the set of numbers assigned to you in the following table:

  Bolton [1, 2, 4]	Burchette [1, 2, 5]	Funk [1, 2, 6]	Gandhi [1, 3, 4]	Gerstman [1, 3, 5]
  Gomes [1, 3, 6]	Gwanvoma [1, 4, 5]	Hayo [1, 4, 6]	Holmes [1, 5, 6]	Holtschneider [2, 3, 4]
  Jordan [2, 3, 5]	Lee [2, 3, 6]	Mcaulife [2, 4, 5]	Mcelhattan [2, 4, 6]	Sharma [2, 5, 6]
  Tawes [3, 4, 5]	Uchendu [3, 4, 6]	Whigham [3, 5, 6]	Willis [4, 5, 6]

  Plot the system's phase portrait within the first quadrant. Please note that the plotting ranges xmax and ymax from the previous project will almost certainly be wrong choices here. Apply your judgment to choose a plotting region in the $x$-$y$ plane which best shows the behavior of the orbits near $E_4$. Plot the nullclines first to get an idea where to focus your attention.

Author: Rouben Rostamian