Chapter 16

Examine the behavior of \(x_{t}\) for \(t=2,3, \cdots, 20\) of the difference equation $$ \begin{aligned} x_{t+1} &=\alpha_{0} x_{t}+\alpha_{1} x_{t-1} \\ x_{0} &=0.1, \quad x_{1}=0.2 \end{aligned} $$ for a. \(\alpha_{0}=\frac{5}{6}, \quad \alpha_{1}=-\frac{1}{2}\). b. \(\alpha_{0}=\frac{3}{2}, \quad \alpha_{1}=-\frac{1}{2}\). c. \(\alpha_{0}=\frac{5}{2}, \quad \alpha_{1}=-1\).

For the symbiosis system, $$ \begin{aligned} x_{n+1}-x_{n} &=\frac{5}{98} \cdot x_{n}\left(1+0.4 y_{n}-x_{n}\right) \\ y_{n+1}-y_{n} &=\frac{7}{120} \cdot y_{n}\left(1+\frac{5}{7} x_{n}-y_{n}\right) \end{aligned} $$ (1.76,2.0) is an equilibrium point. Choose the initial value, \(\left(x_{0}, y_{0}\right)=(1.76,2.1)\) and compute \(\left(x_{1}, y_{1}\right),\) \(\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right) .\)

For the dynamical system \(16.6 \mathrm{C}\), $$ \begin{array}{lll} x_{0} & =0.25 & x_{n+1} & =0.9 \times x_{n}+0.04 \times y_{n} \\ y_{0} & =0.5 & y_{n+1} & =0.1 \times x_{n}+0.86 \times y_{n} \end{array} $$ the characteristic roots are \(r_{1}=0.946\) and \(r_{2}=0.814 . x_{t}\) and \(y_{t}\) are given by $$ x_{t}=C_{1} \times r_{1}^{t}+C_{2} \times r_{2}^{t} $$ $$ y_{t}=D_{1} \times r_{1}^{t}+D_{2} \times r_{2}^{t} $$ Where \(C_{1}, C_{2}, D_{1}\) and \(D_{2}\) are computed from $$ \begin{array}{ll} x_{0}=C_{1}+C_{2} & y_{0}=D_{1}+D_{2} \\ x_{1}=C_{1} \times r_{1}+C_{2} \times r_{2} & y_{1}=D_{1} \times r_{1}+D_{2} \times r_{2} \end{array} $$ Compute \(C_{1}, C_{2}, D_{1}\) and \(D_{2}\) and use these values in Equations 16.12 and 16.13 to compute \(x_{2}, x_{3},\) and \(y_{2}, y_{3}\)

For the dynamical system \(16.6 \mathrm{D}\), $$ \begin{array}{lll} x_{0} & =0.25 & x_{n+1}= & 1.15 \times x_{n}-0.8 \times y_{n} \\ y_{0} & =0.5 & y_{n+1}= & 0.1 \times x_{n}+0.86 \times y_{n} \end{array} $$ the characteristic roots are $$ r_{1} \doteq 1.005+0.243 i, \quad r_{2} \doteq 1.005-0.243 i $$ \(x_{t}\) and \(y_{t}\) are given by $$ \begin{aligned} x_{t} &=C_{1} \rho^{t} \cos t \theta+C_{2} \rho^{t} \sin t \theta \\ y_{t} &=D_{1} \rho^{t} \cos t \theta+D_{2} \rho^{t} \sin t \theta \end{aligned} $$ where \(\rho=\sqrt{1.005^{2}+0.243^{2}}, \theta=\arccos (1.005 / \rho)\) and \(C_{1}, C_{2}, D_{1}\) and \(D_{2}\) are computed from $$ \begin{array}{ll} x_{0}=C_{1} & y_{0}=D_{1} \\ x_{1}=C_{1} \rho \cos \theta+C_{2} \rho \sin \theta & y_{1}=D_{1} \rho \cos \theta+D_{2} \rho \sin \theta \end{array} $$ Compute \(\rho, \theta\) and \(C_{1}, C_{2}\) and \(D_{1}, D_{2}\) and use these values in Equations 16.14 and 16.14 to compute \(x_{2},\) \(x_{3},\) and \(y_{2}, y_{3}\)

For the following systems, find all of the equilibrium points \(\left(x_{e}, y_{e}\right)\). a. For each equilibrium point with \(x_{e}>0\) and \(y_{e}>0,\) determine the stability of the system at that equilibrium point. b. Draw a phase plane and in each region of the phase plane bounded by null lines, draw a vector pointing from a point \(\left(x_{n}, y_{n}\right)\) toward \(\left(x_{n+1}, y_{n+1}\right)\). \(\begin{aligned} \text { a. } & x_{n+1}-x_{n}=0.1 * x_{n} *\left(1-0.5 y_{n}-x_{n}\right) \\ & y_{n+1}-y_{n}=0.05 * y_{n} *\left(1+0.2 x_{n}-y_{n}\right) \\ \text { b. } & x_{n+1}-x_{n}=0.05 * x_{n} *\left(1-0.2 y_{n}-x_{n}\right) \\ & y_{n+1}-y_{n}=0.1 * y_{n} *\left(1-0.5 x_{n}-y_{n}\right) \\ \text { c. } \quad & x_{n+1}-x_{n}=0.2 * x_{n} *\left(1+0.4 y_{n}-x_{n}\right) \\ & y_{n+1}-y_{n}=0.4 * y_{n} *\left(1+0.8 x_{n}-y_{n}\right) \\ \text { d. } \quad & x_{n+1}-x_{n}=0.05 * x_{n} *\left(1-0.2 y_{n}-x_{n}\right) \\ & y_{n+1}-y_{n}=0.1 * y_{n} *\left(1-y_{n}\right) \end{aligned}\)

Determine whether the following systems are stable. a. \(\quad \begin{aligned} x_{t+1} &=0.48 x_{t}+0.2 y_{t} \\ y_{t+1} &=0.128 x_{t}+0.72 y_{t} \end{aligned}\) b. \(\quad \begin{aligned} x_{t+1} &=0.8 x_{t}-0.5 y_{t} \\ y_{t+1} &=0.2 x_{t}+0.9 y_{t} \end{aligned}\) c. \(\quad \begin{aligned} x_{t+1} &=1.1 x_{t}-0.5 y_{t} \\ y_{t+1} &=0.2 x_{t}+0.9 y_{t} \end{aligned}\) d. \(\quad \begin{aligned} x_{t+1} &=0.8 x_{t}-0.1 y_{t} \\ y_{t+1} &=0.1 x_{t}+0.6 y_{t} \end{aligned}\) e. \(\quad \begin{aligned} x_{t+1} &=1.2 x_{t}-0.1 y_{t} \\ y_{t+1} &=0.9 x_{t}+0.6 y_{t} \end{aligned}\) f. \(\quad \begin{aligned} x_{t+1} &=0.56 x_{t}+0.4 y_{t} \\ y_{t+1} &=0.256 x_{t}+1.04 y_{t} \end{aligned}\)

In Equation 16.11 , we claim that for $$ x_{t}=C_{1} r_{1}^{t}+C_{2} \times t \times r_{1}^{t}, \quad \lim _{t \rightarrow \infty} A_{t}=0 \quad \text { if } \quad\left|r_{1}\right|<1 $$ Use L'Hospital's Theorem 14.5.1 to show that \(\lim _{t \rightarrow \infty} t \times r^{t}=0\) if \(|r|<1\).

Show that the characteristic equation of $$ \begin{aligned} x_{n+1} &=p \cdot x_{n}-q \cdot y_{n} \\ y_{n+1} &=x_{n} \end{aligned} $$ is Exercise 16.2 .12 Show that the characteristic equation of $$ \begin{array}{l} x_{n+1}=p \cdot x_{n}-q \cdot y_{n} \\ y_{n+1}=x_{n} \end{array} $$ is $$ x_{n+2}-p x_{n+1}+q x_{n}=0 $$

Compute the characteristic values of $$ \begin{array}{l} \xi_{n+1}=0.7647 \xi_{n}+-0.04706 \eta_{n} \\ \eta_{n+1}=0.02157 \xi_{n}+0.7843 \eta_{n} \end{array} $$ and determine whether both of them are of magnitude less than \(1 .\)

For each of the following systems, find an equilibrium point which has both coordinates positive, if there is one, and determine whether it is stable. $$ \begin{array}{l} \text { a. } \quad x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.5 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.3 y_{n}\left(1+0.2 x_{n}-y_{n}\right) \\ \text { b. } x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.8 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.3 y_{n}\left(1+0.5 x_{n}-y_{n}\right) \end{array} $$ c. \(x_{n+1}=x_{n}+1.2 x_{n}\left(1-0.9 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.8 y_{n}\left(1+0.2 x_{n}-y_{n}\right)\) d. \(x_{n+1}=x_{n}+1.8 x_{n}\left(1-1.2 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+1.2 y_{n}\left(1+0.5 x_{n}-y_{n}\right)\) e. \(\quad x_{n+1}=x_{n}+1.4 x_{n}\left(1-1.0 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.8 y_{n}\left(1+0.5 x_{n}-y_{n}\right)\) f. \(x_{n+1}=x_{n}+1.4 x_{n}\left(1-1.1 y_{n}-x_{n}\right) \quad y_{n+1}=y_{n}+0.8 y_{n}\left(1+0.5 x_{n}-y_{n}\right)\)

For each of the following systems, find an equilibrium point which has both coordinates positive, if there is one, and determine whether it is stable. $$ \begin{array}{lll} \text { a. } & x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.5 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.3 y_{n}\left(1-0.2 x_{n}-y_{n}\right) \\ \text { b. } & x_{n+1}=x_{n}+0.7 x_{n}\left(1-0.8 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.3 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \\ \text { c. } & x_{n+1}=x_{n}+1.2 x_{n}\left(1-0.9 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.8 y_{n}\left(1-0.2 x_{n}-y_{n}\right) \\ \text { d. } & x_{n+1}=x_{n}+1.8 x_{n}\left(1-1.2 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+1.2 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \\ \text { e. } & x_{n+1}=x_{n}+1.4 x_{n}\left(1-1.0 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.8 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \\ \text { f. } & x_{n+1}=x_{n}+1.4 x_{n}\left(1-0.9 y_{n}-x_{n}\right) & y_{n+1}=y_{n}+0.8 y_{n}\left(1-0.5 x_{n}-y_{n}\right) \end{array} $$