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Quantitatively analyzing a one-dimensional model (2 marks).

Consider the differential equation \(\frac{dx}{dt} = x(1-x)\) from last week’s assignment.

Find the fixed points (equilibrium values) of the system (analytically, on paper) (0.5 marks).

Use R to simulate this equation and plot \(x\) vs. \(t\) for \(0 \le t \le 10\). Plot trajectories for several different starting values of \(x\) (initial conditions) using

`ggplot`

and the`do`

function we used in class. Based on the long-time behaviour of the system, which fixed point(s) are stable and which are unstable? (1 mark)Analyze the stability of the fixed points on paper. Since this system is one-dimensional, we only need a single derivative: \(\frac{\partial f(x)}{\partial x}\), where \(f(x) = x(1-x)\). Calculate this derivative, evaluate it at each of the fixed points, and use the result to classify each point as stable or unstable (0.5 marks).

The Lotka-Volterra competition model (6 marks).

Consider the Lotka-Volterra competition model. Unlike the predator-prey model considered last week, in this model the growth rate of each species (\(x\) and \(y\)) depends on the per-capita effect the competing species have on each other, which can be thought of as the amount of overlap in resource use. The amount of overlap for each species is given by \(\alpha_1\) and \(\alpha_2\), respectively. \(\alpha_1\) and \(\alpha_2\) don’t need to be equal.

\[\frac{dx}{dt} = r_1 x \left(1-\frac{x+\alpha_1 y}{K_1}\right)\]

\[\frac{dy}{dt} = r_2 y \left(1-\frac{y+\alpha_2 x}{K_2}\right)\]

\(r\) and \(K\) are growth rates and carrying capacities, respectively.

Find the fixed points and nullclines of the system (analytically, on paper). The fixed points can be found by setting \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\) simultaneously, and the nullclines can be found by setting each derivative to zero individually and solving for \(y\) as a function of \(x\) in each case so that they can be plotted on a \(y\) vs. \(x\) phase portrait (0.75 marks).

Use R to simulate these equations and plot \(x\) and \(y\) vs. \(t\). Use the symmetric case: \(\alpha_1 = \alpha_2 = \alpha\), \(K_1 = K_2 = K\), and \(r_1 = r_2 = r\). Choose \(\alpha = 0.2, K = 50\), and \(r = 1\). Choose a starting value where \(x\) is different from \(y\) (0.75 marks).

- Create a phase portrait for this system in R using
`phaseR`

. Plot the following:- all the fixed points.
- the nullclines found in (a). Plot all the nullclines yourself instead of using
`nullcline`

from`phaseR`

—`nullcline`

doesn’t always plot all the nullclines correctly. - three to four trajectories (1 marks).

Analyze the stability of each fixed point using the same assumptions and parameter values as in part (b). Calculate the four partial derivatives needed for the Jacobian: \(\frac{\partial f(x,y)}{\partial x}\), \(\frac{\partial f(x,y)}{\partial y}\), \(\frac{\partial g(x,y)}{\partial x}\), \(\frac{\partial g(x,y)}{\partial y}\), where \(f(x) = r_1 x \left(1-\frac{x+\alpha_1 y}{K_1}\right)\) and \(f(y) = r_2 y \left(1-\frac{y+\alpha_2 x}{K_2}\right)\).

For example, for the fixed point \(x^* = 0\), \(y^* = K\) (\(x^*\) notation indicates fixed point):

And we can use our solutions for the partial derivatives above, plugging in the equilibrium values. For example, the partial derivative of \(f(x)\) with respect to x evaluated at the fixed point above is:

`# Partial derivatives / elements of the Jacobian df_dx <- r * (1 - (x_star + alpha * y_star) / K) - (1 / K) * r * x_star`

Now create the Jacobian matrix and then calculate the eigenvalues:

`J <- matrix(c(df_dx, df_dy, dg_dx, dg_dy), ncol = 2, nrow = 2, byrow = TRUE) e_vals <- eigen(J)$values`

For each point, give the eigenvalues and explain whether the fixed point is stable or unstable. Will it oscillate? Explain why or why not (1 marks).

Simplify the fixed points of the system on paper in the case when \(\alpha_1 = \alpha_2 = \alpha\), \(r_1 = r_2 = r\), and \(K_1 \neq K_2\). Create a plot in R showing \(x^*\) vs. \(\alpha\), \(y^*\) vs \(\alpha\), and \(x ^ * + y ^ *\) vs. \(\alpha\) for $ -1 < < 1.5$ for the

*non-trivial*fixed point where both \(x^*\) and \(y^*\) are not zero. Note that \(\alpha = -1\) and \(\alpha = 1\) will give infinity, so skip those two points when plotting. Label \(K_1\), \(K_2\), and \(K_1 + K_2\) on the y-axis. (2.5 marks total)Thinking of \(\alpha\) as the parameter that controls “overlap” between the two competing species, what is the biological meaning of \(\alpha = 0\)? (Hint: compare this fixed point to the non-trivial fixed point from the logistic model: \(N^* = K\).)

What is the biological meaning of \(0 < \alpha < 1\)? What is the biological meaning of \(\alpha = 1\), and of \(\alpha > 1\)? (Hint: what is the sum of the non-trivial fixed points \(x^*\) and \(y^*\) in each of these cases in terms of \(K_1\) and \(K_2\)?)

What would happen to the system if \(\alpha\) was negative? (Hint: is “competition” still an appropriate name for the model if \(\alpha < 0\)?)

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