MHB Why does the population go to extinction if the solution is real?

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The discussion centers on a mathematical model describing population dynamics under predation, represented by the difference equation u_{t+1} = (au_t^2)/(b^2 + u_t^2). It identifies equilibria at u_* = 0 and u_* = (a ± √(a² - 4b²))/2, indicating that if a² > 4b², extinction is possible if the population falls below a critical size. The analysis reveals two attractive fixed points and one repulsive fixed point, suggesting that populations starting below a certain threshold will decline to extinction. The recursive nature of the model shows that if the initial population is less than the critical size, it will converge to zero, leading to extinction. Thus, even with real solutions, the population can go extinct due to the dynamics of the model.
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The population of a certain species subjected to a specific kind of predation is modeled
by the difference equation

$$
u_{t+1}=\frac{au_t^2}{b^2+u_t^2}, \quad a>0.
$$

Determine the equilibria and show that if $a^2 > 4b^2$ it is possible for the populationto be driven to extinction if it becomes less than a critical size which you should find.

So the steady states are

$u_*=0$

$u_*=\frac{a\pm\sqrt{a^2-4b^2}}{2}$

So why if the solution is real, does the population go to extinction?
 
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dwsmith said:
The population of a certain species subjected to a specific kind of predation is modeled
by the difference equation

$$
u_{t+1}=\frac{au_t^2}{b^2+u_t^2}, \quad a>0.
$$

Determine the equilibria and show that if $a^2 > 4b^2$ it is possible for the populationto be driven to extinction if it becomes less than a critical size which you should find.

So the steady states are

$u_*=0$

$u_*=\frac{a\pm\sqrt{a^2-4b^2}}{2}$

So why if the solution is real, does the population go to extinction?

The recursive relation can be written in the form...

$\displaystyle \Delta_{n}=u_{n+1}-u_{n}= - \frac{b^{2}\ u_{n} - a\ u^{2}_{n} - u^{3}_{n}}{b^{2}+u^{2}_{n}}= f(u_{n})\ ,\ a>0$ (1)

Under the hypothesis that is $a^{2}>4\ b^{2}$ and that is $u_{n} \ge 0\ \forall n$ the function f(x) has two 'attractive fixed points' in $x_{0}=0$ and $\displaystyle x_{+}= \frac{a+ \sqrt{a^{2}-4 b^{2}}}{2}$ and a 'repulsive fixed point' in $\displaystyle x_{-}= \frac{a- \sqrt{a^{2}-4 b^{2}}}{2}$. The solution of (1) depends from the 'initial state' $u_{0}$ and precisely...

a) if $0 \le u_{0} < x_{-}$ the sequence has limit $x_{0}=0$...

b) if $u_{0}=x_{-}$ the sequence has limit $x_{-}$...

c) if $u_{0}>x_{-}$ the sequence has limit $x_{+}$...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
The recursive relation can be written in the form...

$\displaystyle \Delta_{n}=u_{n+1}-u_{n}= - \frac{b^{2}\ u_{n} - a\ u^{2}_{n} - u^{3}_{n}}{b^{2}+u^{2}_{n}}= f(u_{n})\ ,\ a>0$ (1)

Under the hypothesis that is $a^{2}>4\ b^{2}$ and that is $u_{n} \ge 0\ \forall n$ the function f(x) has two 'attractive fixed points' in $x_{0}=0$ and $\displaystyle x_{+}= \frac{a+ \sqrt{a^{2}-4 b^{2}}}{2}$ and a 'repulsive fixed point' in $\displaystyle x_{-}= \frac{a- \sqrt{a^{2}-4 b^{2}}}{2}$. The solution of (1) depends from the 'initial state' $u_{0}$ and precisely...

a) if $0 \le u_{0} < x_{-}$ the sequence has limit $x_{0}=0$...

b) if $u_{0}=x_{-}$ the sequence has limit $x_{-}$...

c) if $u_{0}>x_{-}$ the sequence has limit $x_{+}$...

Kind regards

$\chi$ $\sigma$

Unfortunately, I just don't understand what you are doing to help in this post (or the others). I am not saying you aren't helping. I just don't understand your procedures and methods.
 
dwsmith said:
Unfortunately, I just don't understand what you are doing to help in this post (or the others). I am not saying you aren't helping. I just don't understand your procedures and methods.

On MHF I wrote a tutorial section where the general procedure for solving this type of recursive relations was illustrated ... 'unfortunately' MHF collapsed and now, with the approval of Administrators, I can try to rewrite the same tutorial section on MHB...

Kind regards

$\chi$ $\sigma$
 
dwsmith said:
The population of a certain species subjected to a specific kind of predation is modeled
by the difference equation

$$
u_{t+1}=\frac{au_t^2}{b^2+u_t^2}, \quad a>0.
$$

Determine the equilibria and show that if $a^2 > 4b^2$ it is possible for the populationto be driven to extinction if it becomes less than a critical size which you should find.

So the steady states are

$u_*=0$

$u_*=\frac{a\pm\sqrt{a^2-4b^2}}{2}$

So why if the solution is real, does the population go to extinction?

Consider:

\[ \frac{u_{t+1}}{u_t}=\frac{au_t}{b^2+u_t^2}\]

\( \{u_t\} \) is a decreasing sequence if the above is less than \(1\). So solve:

\[ \frac{au_t}{b^2+u_t^2}<1\]

and the result will follow (you will need to justify it going to zero rather than decreasing to a non-zero limit but that is easily done).

CB
 
CaptainBlack said:
Consider:

\[ \frac{u_{t+1}}{u_t}=\frac{au_t}{b^2+u_t^2}\]

\( \{u_t\} \) is a decreasing sequence if the above is less than \(1\). So solve:

\[ \frac{au_t}{b^2+u_t^2}<1\]

and the result will follow (you will need to justify it going to zero rather than decreasing to a non-zero limit but that is easily done).

CB

I am not sure how or why the result follows? That leads me to the same quadratic to solve but with an inequality.

From the cob web plot, I see there is only the zero steady state when a^2<4b^2

How could there be a critical extinction point if the only steady state is 0?
 
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