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

[Dustinsfl]

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?

 

[chisigma]

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

$\chi$ $\sigma$

 

[Dustinsfl]

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.

 

[chisigma]

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

$\chi$ $\sigma$

 

[CaptainBlack]

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

 

[Dustinsfl]

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?