Prey-Predators systems Resize


Orbits are closed

Our system is not explicitly solvable, that is, we can not determine the functions \(x\) and \(y\) of \(t\). However, a qualitative analysis provides sufficient information about the orbits \((x,y)\) that allows us to conclude that they are closed. The secret lies in the simple expression of the slope of the vectors of the field \(V\), as a function of the point \((x,y)\) given by \[\frac{dy}{dx}=\frac{(-C+Dx)y}{(A-By)x}\] in the complementar of \(y=\frac{A}{B}\) (where the field vectors are vertical) and \(x=0\) which biologically does not matter, as we are assuming the existence of pouts). This is a separable differential equation; let us begin by solving it: \[\begin{array}{ll} \frac{dy}{dx}=\frac{(-C+Dx)y}{(A-By)x}\Longleftrightarrow\frac{A-By}{y}\frac{dy}{dx}=\frac{-C+Dx}{x}\\ \Longleftrightarrow\left(\frac{A}{y}-B\right)\frac{dy}{dx}=-\frac{C}{x}+D\\ \Longleftrightarrow A\ln y-By=-C\ln x+Dx+k_{1} & k_{1}\mbox{ real constant}\\ \Longleftrightarrow\ln y^{A}+\ln x^{C}=By+Dx+k_{1}\\ \Longleftrightarrow\ln\left(y^{A}x^{C}\right)=By+Dx+k_{1}\\ \Longleftrightarrow y^{A}x^{C}=e^{By+Dx+k_{1}}\\ \Longleftrightarrow y^{A}x^{C}=e^{By}e^{Dx}k & k\in\mathbb{R}^{+}\\ \Longleftrightarrow\frac{y^{A}}{e^{By}}\frac{x^{C}}{e^{Dx}}=k & k\in\mathbb{R}^{+} \end{array}\]

Therefore, we conclude, that the orbits \((x,y)\) are contained in level curves of the function \[\begin{array}{cccc} H: & \mathbb{R}^{2} & \rightarrow & \mathbb{R}^{2}\\ & (x,y) & \rightarrow & \frac{y^{A}}{e^{By}}\frac{x^{C}}{e^{Dx}} \end{array}\]

The next step is then to study this function \(H\). For this, let us note that \(H(x,y)=f(y)g(x)\) being \(f(y)=\frac{y^{A}}{e^{By}}\) independent of \(x\) and \(g(x)=\frac{x^{C}}{e^{Dx}}\) independent of \(y\). We can thus write the previous equation as \(f(y)g(x)=k\), where \(k\) is a positive constant.

Let us study the behaviour of the function \(f\) (analogous conclusions for the function \(g\) that differs only from \(f\) in the constants \(C\) and \(D\)):

  • \(f(0)=0\)
  • \(f(y)>0\) for all \(y>0\)
  • \(lim_{y\rightarrow+\infty}f(y)=0\)
  • \(\begin{array}{cl} f'(y)=0 & \Longleftrightarrow\frac{y^{A-1}(A-By)}{e^{By}}=0\\ & \Longleftrightarrow y^{A-1}(A-By)=0\\ & \Longleftrightarrow y=0\vee y=\frac{A}{B} \end{array}\)
  • \(f\) reaches the absolute maximum value \(M_{f}\) only in \(y=\frac{A}{B}\)

Let \(M_{g}\) denote the absolute maximum value of \(g\), which is only reached at \(x=\frac{C}{D}\), and let us analyse the solutions of the equation \(f(y)g(x)=k\).

When \(k>M_{f}M_{g}\) , the \(H^{-1}(\left\{ k\right\})\) set is empty.

If \(k=M_{f}M_{g}\), taking into account that the maxima are only reached at \(y=\frac{A}{B}\) and \(x=\frac{C}{D}\), respectively, the equation \(f(y)g(x)=M_{f}M_{g}\) has a unique solution \((\frac{C}{D},\frac{A}{B})\).

It remains to analyse what happens when \(k<M_{f}M_{g}\). Now, such \(k\) can be written as the product \(\lambda M_{f}\), for some \(0<\lambda<M_{g}\). In this way, \[f(y)g(x)=k\Leftrightarrow f(y)g(x)=\lambda M_{f}\Leftrightarrow f(y)=\frac{\lambda}{g(x)}M_{f}\]

Taking into account the sketch of the plot of \(f\), it is important to distinguish the following cases where:

We note, first of all, that in the first case the set \(f^{-1}\left(\left\{ \frac{\lambda}{g(x)}M_{f}\right\} \right)\) has two elements, \(y_{1}\) and \(y_{2}\) , with \(y_{1}<\frac{A}{B}<y_{2}\); in the second case, it is a singular set, reducing to \(\left\{ \frac{A}{B}\right\}\) and the third case it is empty.

Let us now consider the equality \(g(x)=\lambda\). Now, for \(0<\lambda<M_{g}\), it has two solutions that we denote by \(x_{m}\) and \(x_{M}\), with \(x_{m}<\frac{C}{D}<x_{M}\). The inequality \(g(x)>\lambda\) is equivalent to \(x_{m}<x<x_{M}\); \(g(x)<\lambda\) corresponds to \(x<x_{m}\) or \(x>x_{M}\).

Putting together the information about \(x\) and \(y\) in each case, we conclude that the equation \(f(y)g(x)=\lambda M_{f}\) has solutions \((x,y)\) being

  • \(x_{m}<x<x_{M}\) and \(y=y_{1}\) or \(y=y_{2}\), with \(y_{1}<\frac{A}{B}<y_{2}\);
  • \(x=x_{m}\) or \(x=x_{M}\) and \(y=\frac{A}{B}\)

and has no solution for the values of \(x<x_{m}\) or \(x>x_{M}\).

To finally deduce that the solution curve is closed, it suffices now to show that the values of \(y_{1}\) and \(y_{2}\), ordinate of \(x^{*}\in]x_{m},x_{M}[\), approximate \(\frac{A}{B}\) when \(x^{*}\) tends to the the interval extrema. Now, when \(x^{*}\) approximates \(x_{m}\) (or \(x_{M}\)), \(g(x^{*})\) approximates \(\lambda\), and therefore \(\frac{\lambda}{g(x^{*})}\) tends to \(1\). Thus, \(f(y_{1})=f(y_{2})=\frac{\lambda}{g(x^{*})}M_{f}\) tends to \(M_{f}\), and therefore, \(y_{1}\) and \(y_{2}\) tend to \(\frac{A}{B}\).