Hi I have following assignment and I have not really a clue on how to solve it, because it is completely different then what we did at class.
The simple fishery model reads $\dot{N} = rN(1-N/K)-H$, where $N$ is the fish population, $K$ is a carrying capacity, $r$ is the growth rate, and $H$ is the term considering the effect of fishing. In a refined version, the model is improved to $\dot{N}=rN(1-N/K)-H\frac{N}{A+N}$, with parameters $H$ and $A$ being larger than zero.
(a) Why is the refined model more realistic than the original, simple one?
(b) Give a biological interpretation of parameter $A$; what does it measure?
(c) Show that the refined system can be rewritten in dimensionless form as $$ \frac{dx}{d\tau} = x(1-x) - \frac{hx}{a+x} $$
(d) Show that the refined system can have one, two, or three fixed points, depending on the numerical values of parameters $a$ and $h$. Classify the stability of the fixed points.
(e) Analyze the dynamics of the system near $x=0$, and show that a bifurcation occurs if $h=a$. Which kind of bifurcation is it?
(f) Plot the stability diagram of the system in the $a$, $q$ parameter space. Can hysteresis occur in any of the stability regions?
d and e I have some idea but with the others i am clueless :( Greetings Max
a) Fishers cannot reduce the fish population at a constant rate $H$. If no fish is in the water, they fish nothing, i.e. the fishing term must equal zero if $N=0$. In the refined version, one can observe that the fishing rate increases with the amount of fish, but is bounded by $H$.
b) If $A=0$ the simple model is recovered. If $A\to \infty$, no fishing occurs at all. In between, the parameter $A$ sets the slope of the fishing term at $N=0$ (decreasing with $A$). Therefore, $A$ measures the inefficiency of the fishing community.
c) Let us set $t = \alpha\tau$ and $x = \beta N$. We get $$ \frac{\text d x}{\text d \tau} = \alpha r {x} \left(1-\frac{x}{\beta K}\right) - \alpha H \frac {x}{\beta A+x} \, . $$ Setting $\alpha =1/r$ and $\beta =1/K$, the desired equation is obtained, where $h = \alpha H$ and $a = \beta A$.
d) This question amounts to solve ${\text d x}/{\text d \tau}=0$, i.e. $x=0$ or $x^2 + (a-1) x + h -a =0$. The solutions are $x=0$ or $$ \begin{aligned} x & = x_{\pm} \\ & = \frac{1}{2}\left(1-a \pm \sqrt{(1+a)^2 -4h}\right) . \end{aligned} $$ Thus, we have one, two or three equilibria.
e) This question amounts to evaluate the derivative $J (x)$ of ${\text d x}/{\text d \tau}$ at the origin $x=0$. Here, $$ J (x) = 1-2x - h \frac{a}{(a+x)^2} , $$ and $J (0) = 1 - h/a$. If $J (0)<0$, i.e. $h > a$, the state $x=0$ is an asymptotically stable equilibrium. If $h < a$ , the state $x=0$ is an unstable equilibrium. Looking back to the quadratic equation above, a pitchfork bifurcation may occur at $a = h$.
f) Now let us determine the stability of the other equilibrium points $x_{\pm}$, which exist provided that $h \leq \frac{1}{4} (1+a)^2$. The sign of $J(x_+)$ tells us that $x_+$ is asymptotically stable if $a<1$, or if $a>1$ and $h < a$. The sign of $J(x_-)$ tells us that $x_-$ is asymptotically stable if $a>1$, or if $a<1$ and $h < a$. Here are the stability diagrams in $hy, x$ space for the two cases $a<1$ and $a>1$ are provided, where $x$ is restricted to positive values:
The diagrams suggest that if $a>1$ (the fishing community is not efficient), then one has only one attractive equilibrium value of $x$ simultaneously: no hysteresis occurs. However, if $a<1$ (the fishing community is more efficient), then less fish is available at equilibrium. Moreover, one has simultaneously two attractive equilibrium values for $a<h<\frac{1}{4}(1+a)^2$: hysteresis can occur.
Note that similar diagrams can be drawn in $a, x$ space.