We are now going to deal with a model of **species competition**. Two species which don't predate on each other, but compete for a limited food source. Their respective populations are $N_1(t)$ and $N_2(t)$.

$\begin{align} \dot{N}_1(t) &= r_1 N_1(t) \left(1 - \frac{N_1(t)}{K_1} - b_{12} \frac{N_2}{K_1} \right) \\ \dot{N}_2(t) &= r_2 N_2(t) \left(1 - \frac{N_2(t)}{K_2} - b_{21} \frac{N_1}{K_2} \right) \end{align}$

Notice that the differential equation describing population change of each species $N_i$, is exactly the same as for the differential equation of the previous question, except it includes an additional term $b_{ij} \frac{N_j}{K_i}$. This term models inter-species competition. For instance, if $N_2(t)$ is large, this term exerts a negative pull on the population growth rate of $N_1(t)$.

We are now going to reduce the number of parameters in our model (this is called nondimensionalisation). We've done most of this for you. The equivalent,
nondimensionalised model is:

$\begin{align}
\frac{\d u_1}{\d \tau}(\tau) &= u_1(\tau) \left(1 - u_1(\tau) - a_{12}u_2(\tau)
\right) \\
\frac{\d u_2}{\d \tau}(\tau) &= \rho u_2(\tau) \left(1 - u_2(\tau) -
a_{21}u_1(\tau) \right)
\end{align}$ where the new parameters, expressed in terms of the original parameters, are: $u_1 = \frac{N_1}{K_1}, \quad u_2 = \frac{N_2}{K_2}, \quad \tau = r_1 t, \quad\rho = ?????, \ \ \ \\a_{12} = b_{12}\frac{K_2}{K_1}, \quad a_{21} = b_{21}\frac{K_1}{K_2}$

**a)** Figure out what $\rho$ is, in terms of the parameters of the original
equation. Describe what it represents biologically (Possible even if you couldn't
find its formula using common sense and after running the simulations below)

*Hint: for an arbitrary variable $x$, what is the relationship between $\dot{x}(t)
= \frac{\d x}{\d t}$ and $\frac{\d x}{\d \tau}$? E.g. $t$ could be in seconds and $r_1=60$ could then imply minutes.*

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**b)** We're now going to simulate the differential equation several times. So
write some code for simulating the differential equation. Plot a simulation, with
the following initial conditions and parameter values:
$u_1(0) = u_2(0) = 0.1$
$a_{12} = 0.9 \quad a_{21} = 1.1 \quad \rho = 1.6$

**Not a question:**
However, we will change the parameters repeatedly, to see how they influence the behaviour of the model. Simulate repeatedly with each of the following sets of parameter values (no need to write or plot anything for the report). The point of this is only to get a feel for how the model behaviour changes with the parameters.
$a_{12} = 0.9 \quad a_{21} = 1.1 \quad \rho = 1.6$
$a_{12} = 1.1 \quad a_{21} = 0.9 \quad \rho = 1.6$
$a_{12} = 1.1 \quad a_{21} = 0.9 \quad \rho = 0.6$
$a_{12} = 1.1 \quad a_{21} = 0.9 \quad \rho = 0.6$
$a_{12} = 0.6 \quad a_{21} = 0.4 \quad \rho = 5$
$a_{12} = 0.4 \quad a_{21} = 0.6 \quad \rho = 5$

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**c)** From simulating, with these parameter combinations, what seems to be more important for long term species survival in this model: its natural growth rate or the degree to which it is suppressed by its competitor? Plot ONE simulation from the parameter values above that supports your conclusion.

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**d)** Now calculate the four fixed points of the differential equation. You will
find three are easier, and one is harder. **For all four fixed points, briefly
describe (1/2 sentences) what each fixed point represents in terms of the
competition between the two species.** Although you don't need the previous
simulations for this question, they might be a useful sanity check for your
answers!

Our system of differential equations is in the general form:
$\dot{x}(t) = f\big(x(t) \big),$
where $x(t)$ is the vector $[u_1(\tau), u_2(\tau)]$. Note that $\tau$ is just time
with rescaled units (like minutes vs seconds), so you can treat it as time.
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**e)** Calculate the Jacobian matrix $J(x) = \frac{\d f}{\d x}(x)$ (2 marks)
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**f)** Evaluate this Jacobian matrix at the two fixed points you arrive at when
using the parameters:

$a_{12} = 0.9 \quad a_{21} = 1.1 \quad \rho = 1.6$
$a_{12} = 1.1 \quad a_{21} = 0.9 \quad \rho = 1.6$

**What are the requirements on the values of $a_{12}$ and $a_{21}$ for these fixed points to be stable?**

To do this, figure out conditions on when the Jacobian matrix has appropriate
eigenvalues. You don't necessarily need to calculate the individual eigenvalues.
Instead note that
1. The trace (sum of diagonal elements) of the matrix is equal to the sum of
eigenvalues.
2. The determinant of the matrix is equal to the product of the eigenvalues.