Chapter 8

The equation \(x^{\prime \prime}=a x^{\prime}-b\left(x^{\prime}\right)^{3}-k x\) was devised by Lord Rayleigh to model the motion of the reed in a clarinet. With \(a=5, b=4\), and \(k=5\), solve this equation numerically with initial conditions \(x(0)=A\), and \(x^{\prime}(0)=0\) over the interval \([0,10]\) for the three choices \(A=0.5,1\), and 2. Use MATLAB's hold command to prepare both time plots and phase plane plots containing the solutions to all three initial value problems superimposed. Describe the relationship that you see between the three solutions.

Harmonic motion. An unforced, damped oscillator is modeled by the equation $$ m y^{\prime \prime}+c y^{\prime}+k y=0, $$ where \(m\) is the mass, \(c\) is the damping constant, and \(k\) is the spring constant. Write the equation as a system of first order ODEs and create a function ODE file that allows the passing of parameters \(m, c\), and \(k\). In each of the following cases compute the solution with initial conditions \(y(0)=1\), and \(y^{\prime}(0)=0\) over the interval \([0,20]\). Prepare both a plot of \(y\) versus \(t\) and a phase plane plot of \(y^{\prime}\) versus \(y\). a) (No damping) \(m=1, c=0\), and \(k=16\). b) (Under damping) \(m=1, c=2\), and \(k=16\). c) (Critical damping) \(m=1, c=8\), and \(k=16\). d) (Over damping) \(m=1, c=10\), and \(k=16\).

The system $$ \begin{aligned} \varepsilon \frac{d x}{d t} &=x(1-x)-\frac{(x-q)}{(q+x)} f z \\ \frac{d z}{d t} &=x-z \end{aligned} $$ models a chemical reaction called an oregonator. Suppose that \(\varepsilon=10^{-2}\) and \(q=9 \times 10^{-5}\). Put the system into normal form and write an ODE function M-file for the system that passes \(f\) as a parameter. The idea is to vary the parameter \(f\) and note its affect on the solution of the oregonator model. We will use the initial conditions \(x(0)=0.2\) and \(y(0)=0.2\) and the solution interval \([0,50]\). a) Use ode 45 to solve the system with \(f=1 / 4\). This should provide no difficulties. b) Use ode 45 to solve the system with \(f=1\). This should set off all kinds of warning messages. c) Try to improve the accuracy with options = odeset ('RelTol ', 1e-6) and using options as the options parameter in ode45. You should find that this slows computation to a crawl as the system is very stiff. d) The secret is to use ode15s instead of ode45. Then you can note the oscillations in the reaction.

The system $$ \begin{aligned} &m_{1} x^{\prime \prime}=-k_{1} x+k_{2}(y-x) \\ &m_{2} y^{\prime \prime}=-k_{2}(y-x) \end{aligned} $$ models a coupled oscillator. Imagine a spring (with spring constant \(k_{1}\) ) attached to a hook in the ceiling. Mass \(m_{1}\) is attached to the spring, and a second spring (with spring constant \(k_{2}\) ) is attached to the bottom of mass \(m_{1}\). If a second mass, \(m_{2}\), is attached to the second spring, you have a coupled oscillator, where \(x\) and \(y\) represent the displacements of masses \(m_{1}\) and \(m_{2}\) from their respective equilibrium positions. If you set \(x_{1}=x, x_{2}=x^{\prime}, x_{3}=y\), and \(x_{4}=y^{\prime}\), you can show that $$ \begin{aligned} x_{1}^{\prime} &=x_{2}, \\ x_{2}^{\prime} &=-\frac{k_{1}}{m_{1}} x_{1}+\frac{k_{2}}{m_{1}}\left(x_{3}-x_{1}\right) \\ x_{3}^{\prime} &=x_{4}, \\ x_{4}^{\prime} &=-\frac{k_{2}}{m_{2}}\left(x_{3}-x_{1}\right) \end{aligned} $$ Assume \(k_{1}=k_{2}=2\) and \(m_{1}=m_{2}=1\). Create an ODE file and name it couple.m. Suppose that the first mass is displaced upward two units, the second downward two units, and both masses are released from rest. Plot the position of each mass versus time.

In the 1920 's, the Italian mathematician Umberto Volterra proposed the following mathematical model of a predator-prey situation to explain why, during the first World War, a larger percentage of the catch of Italian fishermen consisted of sharks and other fish eating fish than was true both before and after the war. Let \(x(t)\) denote the population of the prey, and let \(y(t)\) denote the population of the predators. In the absence of the predators, the prey population would have a birth rate greater than its death rate, and consequently would grow according to the exponential model of population growth, i.e. the growth rate of the population would be proportional to the population itself. The presence of the predator population has the effect of reducing the growth rate, and this reduction depends on the number of encounters between individuals of the two species. Since it is reasonable to assume that the number of such encounters is proportional to the number of individuals of each population, the reduction in the growth rate is also proportional to the product of the two populations, i.e., there are constants \(a\) and \(b\) such that $$ x^{\prime}=a x-b x y $$ Since the predator population depends on the prey population for its food supply it is natural to assume that in the absence of the prey population, the predator population would actually decrease, i.e. the growth rate would be negative. Furthermore the (negative) growth rate is proportional to the population. The presence of the prey population would provide a source of food, so it would increase the growth rate of the predator species. By the same reasoning used for the prey species, this increase would be proportional to the product of the two populations. Thus, there are constants \(c\) and \(d\) such that $$ y^{\prime}=-c y+d x y . $$ a) A typical example would be with the constants given by \(a=0.4, b=0.01, c=0.3\), and \(d=0.005\). Start with initial conditions \(x_{1}(0)=50\) and \(x_{2}(0)=30\), and compute the solution to (8.22) and (8.23) over the interval \([0,100]\). Prepare both a time plot and a phase plane plot. After Volterra had obtained his model of the predator-prey populations, he improved it to include the effect of "fishing," or more generally of a removal of individuals of the two populations which does not discriminate between the two species. The effect would be a reduction in the growth rate for each of the populations by an amount which is proportional to the individual populations. Furthermore, if the removal is truly indiscriminate, the proportionality constant will be the same in each case. Thus, the model in equations \((8.22)\) and (8.23) must be changed to $$ \begin{aligned} &x^{\prime}=a x-b x y-e x \\ &y^{\prime}=-c y+d x y-e y \end{aligned} $$ where \(e\) is another constant. b) To see the effect of indiscriminate reduction, compute the solutions to the system in \((8.24)\) when \(e=0\), \(0.01,0.02,0.03\), and \(0.04\), and the other constants are the same as they were in part a). Plot the five solutions on the same phase plane, and label them properly. c) Can you use the plot you constructed in part b) to explain why the fishermen caught more sharks during World War I? You can assume that because of the war they did less fishing.