MTH 337. Differential Equations.

Problem Set 5

38 points

Due Monday, February 27, 2017 at 11:59 PM.

Michael Thomas Lynn

Problem 1 (8 points): Consider the following predator-prey system \begin{align*} \frac{dR}{dt} & = 2R - 1.2 RF \\ \frac{dF}{dt} & = -F + 0.9 RF, \end{align*} where $R = R(t)$ is the population of the prey at time $t$ and $F = F(t)$ is the population of the predator at time $t$.

a. How would you modify this system to include the effect of hunting the prey at a rate of $\alpha$ units of prey per unit time?

Solution:

b. How would you modify this system to include the effect of hunting of the predators at a rate proportional to the number of predators?

Solution:

c. Suppose the predators discover a second, unlimited source of food, but they still prefer to eat prey when they can catch them. How would you modify this system to include this assumption.

Solution:

d. Suppose the predators discover a second source of food that is limited in supply. How would you modify this system to include this assumption.

Solution:

Problem 2 (15 points): Consider the equation $$\frac{d^2 y}{dt^2} + \frac{k}{m} y = 0$$ for the motion of a simple harmonic oscillator.

a. Consider the function $y(t) = \cos \beta t$. Under what conditions on $\beta$ is $y(t)$ a solution to the differential equation?

Solution:

In order to find where $\beta$ is a solution to the differential equation, we must replace all the $y$ in the original differential equation with the function $cos \beta t$ from y(t).

The above action yields,$$- \beta ^2 cos(\beta t) + \frac{k}{m}cos(\beta t) = 0$$

Now factor out the $(\frac{k}{m} - \beta^2)$ to get,

now solve for $\beta$,

now find the initial conditions using what we know, $$c_1 = 1, c_2 = 0$$

b. Rewrite the initial value problem $y'' + y = 0$, $y(0) = 1$, and $y'(0) = 0$ as a system.

Hint: Let $v = y'$.

Solution:

The systems looks like the following.

and the initial conditions are $$y(0) = 1$$ $$v(0) = 0$$

c. What initial condition ($t = 0$) in the $yv$-plane corresponds to this solution?

Solution:

The inititial conditions $y(0) = 1$ and $v(0)=0$ correspond to the solution.

d. In terms of $k$ and $m$, what is the period of this solution?

Solution:

The period in terms of m and k is $$\frac{2\pi}{\sqrt{\frac{k}{m}}}$$

e. Use Sage to graph the solution curve in the $yv$-plane that corresponds to this solution when $m = 1$ and $k = 1$ (with initial conditions $y(0) = 1$ and $v(0) = 0$).

Solution:

Problem 3 (3 points): The spreading of cancer cells in the body is called {\em metastasis}. The Liotta-DeLisi model proposed in 1977 for the metastasis of malignant tumors in mice is given by the linear homogeneous system \begin{align*} x' & = - (\alpha + \beta)x \\ y' & = \beta x - \gamma y, \end{align*} where $x$ is the number of destroyed cancer cells, $y$ is the number of cells that invade the tissue, and $\alpha$, $\beta$, and $\gamma$ are positive constants depending on the type of cancer. Solve the system and give the physical interpretation for all possible choices of $\alpha$, $\beta$, and $\gamma$.

Solution: To solve the system we must first recognize that $x'$constitutes a seperated semi-autonomous differential equation. Knowing this, we can break $x'$, which is also $\frac{dx}{dt}$, apart onto both sides of the equals sign and solve.

We integrate to get,

now we apply some properties of logarithms and some fenagling of constants to get.

Next we rewrite the second equation as $$dy + \gamma y = \beta x$$.

Once rewritten in this form, we can see that an integration factor will be neccessary. Next we find an integration factor to solve the second, partially coupled, differnetial equation. $$\mu (x) = e^{-\gamma t}$$ Now that we have the I.F. we apply the method of the integration factor to find y(t). $$ye^{-\gamma t} = e^{-\gamma t} \beta x - e^{-\gamma t} \gamma y$$ with some further manipulation, and after plugging x(t) in for x in the y' equation we find. $$y(t) = ce^{-(\alpha + \beta)t} + ce^{-\gamma t}$$

Having found the solutions to the system for any values of $\alpha$, $\beta$ and $\gamma$ we can conclude that mice are dying any time those three terms are greater than 0.

Problem 4 (12 points):

Consider the system \begin{align*} x' & = y \\ y' & = x - x^3 -y \end{align*}

a. Find the equilibrium points of the system.

Solution:

To find the equilibrium points of the system, set both differential equations = 0. $$x' = y$$ So, $$y = 0$$ Now we can substitute 0 for y in the second equation. $$y' = x-x^{3}-0$$ Rearranging and simplifying at the same time yields. $$y' = x(1-x^{2})$$ Thus we find the equilibrium solutions to be $y = 0$, $x = 0$, $x = 1$ yielding points (0,0) and (0,1)

b. Determine the $x$ and $y$-nullclines.

Solution:

The nullclines are x = 0, x = 1 and y = 0

c. Use Sage to plot the phase plane for the system. Be sure to draw the $x$ and $y$-nullclines, the direction field, and a sample solution.