Let $H_n$ and $L_n$ express the amount of Hares and Lynxes the year $n$.
A basic model for the populations that satisfy the recursive functions:
$H_n = \frac{5}{2}H_{n-1} - \frac{3}{2}L_{n-1}$
$L_n = \frac{1}{2}H_{n-1} + \frac{1}{2}L_{n-1}$
One of the questions I am trying to answer is regarding my reasoning about the plausibility of the model, more specifically that it isn't entirely unreasonable.
I am arguing that this is due to the fact that for example the coefficient infront of $H_{n-1} >1$. Which represents the fact that the amount of Hares will increase each year. And the coefficient infront of $L_{n-1}<0$ which represents the fact that every year an amount of Hares will get eaten up by the Lynxes, so the coefficient is negative to indicate just that. A decrease in Hares each year.
And regarding the second equation my interpretation is that as long as there are enough Hares around each year, then the amount of Lynxes will increase this due to them being reliant on Hares for food. $L_n = \frac{H_{n-1}+L_{n-1}}{2}$. If the Hares die out then we'll see a big decrease in Lynxes each year.
Is my reasoning acceptable and am I missing any important detail?
My next question is how I can find explicit expressions (formulas) for the two sequences given $(H_{0},L_{0})=(3,1)$?
I have started by calculating the six first elements from each sequence:
$[(H_0, L_0)] = [(3.0,1.0),(6.0,2.0),(12.0,4.0),(24.0,8.0),(48.0,16.0),(96.0,32.0)]$
From this I can see the the quotient $\frac{H_n}{L_n}=3.0$ which seems to be constant, but I am not entirely sure what it represents or what I can do with this information. My interpretation of the quotient is that it tells us that we always seem to have three times more Hares than Lynxes each year. To my understanding this information should also help lead me to an explicit formula. I need help with this.
I managed to solve the simultaneous linear reccurence relation by substitution.
For anybody who is reading this and wants to know how to do so, here is how I did it.
Solve for $L_{n-1}$ out of the first relation and you get
$L_{n-1} = \frac{5}{3}H_{n-1} - \frac{2}{3}H_{n}$ of which follows
$L_{n} = \frac{5}{3}H_{n} - \frac{2}{3}H_{n+1}$
Sub these two expressions into the second relation and you'll end up with $H_{n+1} = 3H_{n} - 2H_{n-1}$
Now let $H_{n+1} = r^{n+1}$ which lets us rewrite the equation to $r^{n+1} = 3r^n - 2r^{n-1}$.
Now divide the whole equation with $r^{n-1}$ and we get $r^2-3r+2=0$ that has the solutions $r=2 \wedge r=1$.
The general solution then becomes $H_{n}=C_{1}2^n + C_{2}1^n$
Given the fact that $H_0 = 3 \wedge H_1 = 6$ we can find what $C_1 \wedge C_2$ are. $C_1=3 \wedge C_2=0$ giving us the final explicit expression $H_{n}=3*2^n$.
You can solve for an explicit expression for $L_n$ in the same way but by starting from (2) and subbing into (1).
For the system $$H_n = \frac{5}{2}H_{n-1} - \frac{3}{2}L_{n-1}\qquad \qquad L_n = \frac{1}{2}H_{n-1} + \frac{1}{2}L_{n-1}$$ study and use the material suggested by @Sil.
You should very quickly arrive at the result
$$H_n=\frac{1}{2} \left(3\times 2^n (H_0-L_0)+3 L_0-H_0\right)$$ $$L_n=\frac{1}{2} \left(1\times2^n (H_0-L_0)+3 L_0-H_0\right)$$ and quickly notice that if $\frac {H_0}{L_0}=3$, the system is stable in the sense that $\frac {H_n}{L_n}=3$ but this is the only case.
Let $H_0=x L_0$ to make, after simplifications, $$y_n=\frac {H_n}{L_n}=3+\frac{2 (x-3)}{2^n (x-1)+(3-x)}$$ So, $y_n$ is asymptotic to $3$. Looking what happens around $x=3$, using long division or Taylor series, we have $$y_n=3+2^{-n} (x-3)+O\left((x-3)^2\right)$$ This shows how the asymptote is approached (from above or below).