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I have read that Lotka-Volterra equations have periodic solutions. I would like to find a proof of this fact, but I haven't found one that is satisfactory.

We can compute to find the solution that satisfies the equation $$\log(x_1(t)) + 2\log(x_2(t)) - x_1(t)/50 - x_2(t)/25 = C$$ for some constant $C$, (where $\begin{bmatrix} x_1(t)\\ x_2(t) \end{bmatrix}$ is a solution to the IVP of the Lotka-Volterra a-equation), which is an equation for a closed curve. All the explanations I have found simply say that lying on this curve implies that the solutions to the Lotka-Volterra equation are periodic. This is not satisfactory to me, because I feel I need to show the solution travels along the entire curve, so that since the curve is closed, the solution must pass a location a second time. It is the statement that the solution travels around the curve (and not just a part of the curve) that I don't know how to prove. Furthermore, even if the solution travels around the curve at least once, does this guarantee the periodicity? This is not obvious to me.

I appreciate you pointing out what I am missing here.

user628623
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1 Answers1

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Just a sketch

Let's consider the Lotka-Volterra equations in the form $$x'(t)= x(t)[\alpha-\beta y(t)]\qquad y'(t)=-y(t)[\gamma-\delta x(t)]$$ where the solutions must lie on the curve given by the implicit equation $$C=\delta x-\gamma \ln(x)+\beta y - \alpha \ln(y)=:F(x,y).\qquad (1)$$ Now proceed as follows:

  1. Check that $F:\mathbb{R}^+\times \mathbb{R}^+ \to \mathbb{R}$ is a convex function, by checking that its Hessian matrix is a (diagonal) positive-definite matrix. Check that $F$ diverges to $+\infty$ in the vicinity of the $x-$ and $y-$axes and that the minimum of $F$ lies at $(\gamma/\delta,\alpha/\beta)=:(x^*,y^*)$

  2. Remember that for all $C\in \mathbb{R}$, $F^{-1}[(-\infty,C)]$ is a convex set (since $F$ is convex)... so the contour $F^{-1}(C)$ is a kind of oval enclosing the point $(x^*,y^*)$.

  3. In particular, when we turn to the polar coordinates $$x-x^*=r\cos(\theta)\qquad y-y^*=r\sin(\theta)$$ the implicit equation (1) yields only one solution $r(\theta)$ for every $\theta \in [0,2\pi)$.

  4. Now compute an expression for $\theta'(t)$ (given the Lotka-Volterra equations) and show that it is bounded away from $0$ (I don't know by heart if it is positive or negative). Now you should be able to conclude, at least as far as the "revisiting of each point" is concerned.

With the revisiting of every point concluded, periodicity of the solution is more a matter of uniqueness of solutions to differential equations (Gronwall lemma etc.) but I'll refer you to any standard book for that part.

Hecatonchires
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  • Could you please explain a little for step 3 how the implicit equation (1) gives a solution $(x(t),y(t))$ to the IVP, if that is the idea in your answer? Did you imply that there is a parametrization $(x(t),y(t))$ of $F(x,y) = C$ and that this parametrization is a solution of the Lotka-Volterra equation, and so the polar coordinates parameterization $(r(t),\theta(t))$ is a reparameterization of $(x(t), y(t))$ mentioned above? If so, I would try to fill in the details of how exactly this coordinate change is done in the parameterization of $F(x,y)=C$ and how to find $\theta'(t)$. – user628623 Jul 27 '24 at 21:58
  • Yes, I think you are roughly on the right path. The gist of my method is to transform to a type of polar coordinates... recurrence is then about the angle growing from $0$ to $2\pi$. But to complete the logic behind that reasoning, we have to be sure that every angle corresponds to only one possible radius... Steps 1, 2 and 3 were all about completing that final part. – Hecatonchires Jul 27 '24 at 23:25
  • Let me add that with a bit more abstract thinking we wouldn't need to invoke the convexity of the curve $F(x,y)=C$ per se (and the "explanations" you referred to in your post probably implicitly invoke the more general principles), but I wanted to avoid more general abstract thinking and give you a more ad hoc proof. – Hecatonchires Jul 27 '24 at 23:31