ATRACTOR

To give an answer to our initial problem,

Fishing decreases each species, the greater the fishing intensity \(P\) and the number of fish of each species are, the bigger is the decrease of species. Then, the variations \(x^{'}\) and \(y^{'}\) must also be affected by a negative term proportional to \(Px\) and \(Py\), respectively:

We thus obtain a new system that models the evolution of pouts and sharks:

\[\begin{cases} \begin{array}{ccc} x' & = & Ax-Bxy-EPx\\ y' & = & -Cy+Dxy-FPy \end{array}\end{cases},\] that is, \[\begin{cases} \begin{array}{ccc} x' & = & (A-EP)x-Bxy\\ y' & = & -(C+FP)y+Dxy \end{array}\end{cases}\]

This system is similar to the one we have analysed previously as long as the new constants \(A–EP\) and \(C+FP\) are positive. We have that \(C+FP\) is a sum of positive constants; as for \(A–EP\) the condition \(EP<A\) means that we require that fishing is moderate, ensuring that preys have the capacity to recover their losses by fishing.

(Click on the figure to observe the effect of decreasing fishing in the average values.)

According to the study we have already done, we can say that the mean values of this new system are: \[\begin{array}{cc} \bar{x}=\frac{C+FP}{D} & \bar{y}=\frac{A-EP}{B}\end{array}\]

Now, by decreasing fishing intensity \(P\), we see that the mean number of sharks \(\bar{y}\) increases, but the average number of pouts \(\bar{x}\) decreases. Therefore, the reduction of fishing favours sharks when compared to pouts which explains the percentual increase in the D'Ancona values.

Try a joint manipulation of the parameters involved in the problem and observe the changes in the vector fields. Relate the results to the study we did and to the we answers we got.