Lotka-Volterra First Integral and Fixed Point

Question

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!

Answer 1

• 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 $$ F\left(t\cdot \left( \begin{array}{c} x \\ y \end{array} \right) + (1-t)\cdot \left( \begin{array}{c} a \\ b \end{array} \right) \right) \leq t\cdot F \left( \begin{array}{c} x \\ y \end{array} \right) + (1-t)\cdot F\left( \begin{array}{c} a \\ b \end{array} \right) \leq t\cdot c + (1-t) \cdot c = c$$

Answer 2

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.