91Ó°ÊÓ

A very simple two-compartment model for gap dynamics in a forest assumes that gaps are created by disturbances (wind, fire, etc.) and that gaps revert to forest as trees grow in the gaps. We denote by \(x_{1}(t)\) the area occupied by gaps and by \(x_{2}(t)\) the area occupied by adult trees. We assume that the dynamics are given by $$ \begin{array}{l} \frac{d x_{1}}{d t}=-0.2 x_{1}+0.1 x_{2} \\ \frac{d x_{2}}{d t}=0.2 x_{1}-0.1 x_{2} \end{array} $$ (a) Find the corresponding compartment diagram. (b) Show that \(x_{1}(t)+x_{2}(t)\) is a constant. Denote the constant by \(A\) and give its meaning. [Hint: Show that \(\frac{d}{d t}\left(x_{1}+x_{2}\right)=0 .\) ] (c) Let \(x_{1}(0)+x_{2}(0)=20\). Use your answer in (b) to explain why this equation implies that \(x_{1}(t)+x_{2}(t)=20\) for all \(t>0\). (d) Use your result in (c) to replace \(x_{2}\) in (11.44) by \(20-x_{1}\), and show that doing so reduces the system \((11.44)\) and \((11.45)\) to $$ \frac{d x_{1}}{d t}=2-0.3 x_{1} $$ with \(x_{1}(t)+x_{2}(t)=20\) for all \(t \geq 0\). (e) Solve the system (11.44) and (11.45), and determine what fraction of the forest is occupied by adult trees at time \(t\) when \(x_{1}(0)=2\) and \(x_{2}(0)=18\). What happens as \(t \rightarrow \infty\) ?

Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{rr} -1 & 0 \\ 1 & -2 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=-1 \text { and } x_{2}(0)=-2 \text { . } $$

Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{cc} 4 & -7 \\ 2 & -5 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=13 \text { and } x_{2}(0)=3 \text { . } $$

23\. Solve $$ \frac{d^{2} x}{d t^{2}}=-4 x $$ with \(x(0)=0\) and \(\frac{d x(0)}{d t}=6\).

Let $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(2-x_{1}\right)-x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$ (a) Graph the zero isoclines. (b) Show that \((1,1)\) is an equilibrium. Use the graphical approach to determine its stability. 24\. Let $$ \begin{array}{l} \frac{d x_{1}}{d t}=x_{1}\left(2-x_{1}^{2}\right)-x_{1} x_{2} \\ \frac{d x_{2}}{d t}=x_{1} x_{2}-x_{2} \end{array} $$ (a) Graph the zero isoclines. (b) Show that \((1,1)\) is an equilibrium. Use the graphical approach to determine its stability.

Solve the given initial-value problem. $$ \left[\begin{array}{l} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{ll} -3 & 4 \\ -1 & 2 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=1 \text { and } x_{2}(0)=2 \text { . } $$

Solve $$ \frac{d^{2} x}{d t^{2}}=-9 x $$ with \(x(0)=0\) and \(\frac{d x(0)}{d t}=12\).

Solve the given initial-value problem. $$ \left[\begin{array}{c} \frac{d x_{1}}{d t} \\ \frac{d x_{2}}{d t} \end{array}\right]=\left[\begin{array}{rr} 4 & 7 \\ 1 & -2 \end{array}\right]\left[\begin{array}{l} x_{1}(t) \\ x_{2}(t) \end{array}\right] $$ $$ \text { with } x_{1}(0)=-3 \text { and } x_{2}(0)=1 \text { . } $$

\- Iranstorm the second-order dutferentral equation $$ \frac{d^{2} x}{d t^{2}}=3 x $$ into a system of first-order differential equations.

Transform the second-order differential equation $$ \frac{d^{2} x}{d t^{2}}=-\frac{1}{2} x $$ into a system of first-order differential equations.