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I have the following problem that I am dealing with, quite a long time, I must say. Let us assume that we have a predator-prey, Lotka-Volterra system given to us by: \begin{align} & \frac{dx}{dt}={\alpha}x-{\beta}xy \\ & \frac{dy}{dt}=-{\delta}y+{\gamma}xy \end{align} with all the parameters ${\alpha}, {\beta}, {\gamma}, {\delta}$ to be positive integers. A First Integral of this particular system would be given by: \begin{equation} F(x,y)={\gamma}x-{\delta}\ln(x)+{\beta}y-{\alpha}\ln(y) \end{equation} Next we consider the sets defined as: \begin{equation} {\Sigma}_c=\{(x,y)\in \mathbb{R}^2/ x>0, y>0, F(x,y)\leq c\}, \quad c\in \mathbb{R}^{+} \end{equation}

Edit: Reversed the Inequality

and we would like to prove that those ${\Sigma}_c$ sets are closed, bounded and convex, in order to use the Kakutani-Markov Fixed Point Theorem (not really sure if that is the theorem I must use, since the professor is not giving that away) and prove that, at the stable equilibrium of this particular Lotka Volterra the function $F(x,y)$ acquires its maximum value, that is $K_0=\max {F(x,y)}=F(\bar{x},\bar{y})$, where $(\bar{x},\bar{y})$ denotes the stable equilibrium. $$$$Any assistance would be much appreciated! Thank you all for your time!

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    $\begingroup$ Quick notation warning for readers: the usual notation uses $\frac{dy}{dt} = -\gamma y + \delta xy$ (note the swap of $\gamma$ and $\delta$). See for example the Wikipedia page. (Mathworld also has this switch, if you relate $C$ to $\gamma$ and $D$ to $\delta$.) $\endgroup$ – Mark Fischler Feb 9 '15 at 22:44
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    $\begingroup$ These sets are not bounded. For simplicity, consider the case for $c = 0$. For every $C > 0$, you can find an $x_0$ such that $Cx > \ln(x)$ for all $x > x_0$ (to see this, use l'Hopital to show that $\lim_{x \to \infty} \frac{\ln(x)}{x} = 0$). Setting $C:= \frac{\gamma}{\delta}$ or $C:=\frac{\beta}{\alpha}$ shows you that there exist $x_0, y_0$ such that $$F(x,y) \geq 0 \quad \forall (x,y) \in [x_0, \infty) \times [y_0, \infty)$$. Hence these sets are not bounded. $\endgroup$ – user159517 Feb 9 '15 at 22:52
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    $\begingroup$ So there must exist a mistake at the beginning of the exercise. At least that is what me and my fellow students ended up with. What if we consider the same sets but with the different inequality? Meaning: ${\Sigma}_c=\{(x,y)\in \mathbb{R}^2/ x>0, y>0, F(x,y)\leq c\}, \quad c\in \mathbb{R}^{+}$ $\endgroup$ – 010514 Feb 9 '15 at 22:57
  • $\begingroup$ In that case, the sets $\Sigma_c$ are indeed closed, bounded and convex (my comment above shows boundedness for this case). I have worked it all out in my head, so if you need help proving it, let me know. $\endgroup$ – user159517 Feb 10 '15 at 0:00
  • $\begingroup$ @user159517 I really do need that proof. Actually that is what I have been searching all along. I posted the inequality with $\geq$, since I had to know if that was right or not. Would you be so kind as to provide that proof for me? Thank you so much for your assistance! $\endgroup$ – 010514 Feb 10 '15 at 2:08
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  • Bounded: Prove boundedness by contradiction: $F(x,y) \to \infty$ if $x \to \infty$ or $y \to \infty$. It's intuitively clear that $\gamma x - \delta\ln(x) \to \infty$ as $x \to \infty$. For a formal argument show for an arbitrary $C > 0$ that $$\lim_{x\to\infty} \frac{\gamma x - C}{\delta \ln(x)} = \infty$$ using l'Hôpital. This implies the existence of an $x_0$ such that $$\frac{\gamma x - C}{\delta \ln(x)} \geq 1 \quad \forall x > x_0$$ which is equivalent to $$\gamma x - \delta \ln(x) \geq C \quad \forall x>x_0$$ You can show $\beta y - \alpha \ln(y) \to \infty$ as $y \to \infty$ in the same manner. Now assume that there is a set $\Sigma_c$ that contains an unbounded sequence $(x_n,y_n)$ and show that this contradicts the assumption that $F(x_n,y_n) \leq c$.
  • Closed: Let $(x_n,y_n)$ be a convergent sequence in $\mathbb{R}^2$ such that $(x_n,y_n) \in \Sigma_{c}$ for all $n \in \mathbb{N}$. Let $x := \lim_{n\to\infty} x_n$ and $y := \lim_{n\to\infty}y_n$. Clearly $x,y \geq 0$. If $x = 0$, then $\lim_{n\to\infty} F(x_n,y_n) = \infty$, a contradiction. So $x > 0$. You can show $y > 0$ in the same manner. The continuity of $F$ on $(0,\infty) \times (0, \infty)$ implies $$F(x,y) = F(\lim_{n\to\infty}(x_n,y_n)) = \lim_{n\to\infty}F(x_n,y_n) \leq c$$ Hence $(x,y) \in \Sigma_c$, so $\Sigma_c$ is closed.
  • Convex: Show that the function $-C\ln(x)$ for $C > 0$ is convex (consider its second derivative). Using that you can show for $(x,y), (a,b) \in \Sigma_c$ and $t \in [0,1]$ that $$ F\left(t\cdot \left( \begin{array}{c} x \\ y \end{array} \right) + (1-t)\cdot \left( \begin{array}{c} a \\ b \end{array} \right) \right) \leq t\cdot F \left( \begin{array}{c} x \\ y \end{array} \right) + (1-t)\cdot F\left( \begin{array}{c} a \\ b \end{array} \right) \leq t\cdot c + (1-t) \cdot c = c$$
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  • $\begingroup$ yeap, the whole deal! Thank you so much! And everyone who helped! Was not an easy task, I appreciate your time @user159517!! $\endgroup$ – 010514 Feb 10 '15 at 14:59
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I don't see why you need to use a fixed point theorem. The function $F$ is strictly convex. At the equilibrium point it has a minimum, not a maximum.

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  • $\begingroup$ yeah, I mentioned that I am not really sure yet if I have to use a Fixed point theorem.. But if I prove that these sets are closed, bounded and convex, which I am not able to, at the moment, I must then prove that there exists a $s \in \bigcap _{c>0} \Sigma_c$ which is an extrema for the $F$. $\endgroup$ – 010514 Feb 10 '15 at 3:07

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