# Lotka-Volterra predator-prey model eigenvalue

I have a question about the eigenvalues of the prey-predator model called Lotka-Volterra. The model itself consists of 2 nonlinear differential equations of first order:

$$\begin{eqnarray} \frac{dx}{dt}&=&ax-bxy \\ \frac{dy}{dt}&=&cry-dy \end{eqnarray}$$

I found the stationary points to be both $(0,0)$ and $(d/c, a/b).$ Then I found the eigenvalues (e) to be:

\begin{eqnarray} e_1&=&a,\\ e_2&=&-d,\\ e_3&=&\pm i*\sqrt{ad}. \end{eqnarray}

I know the eigenvalues tells how the system behaves around the stationary points but my question is why are these eigenvalues important for the model? Can they tell something specific about the interaction between the species x and y in the model? I found the eigenvalues but I don't really know to do with them.

Regards Rasmus O.J (English is not my native language, so I hope its understandable)

As you said, the eigenvalues tell you about the behaviour near the stationary points. In particular, they say something about stability of those points.

The stationary point $(0,0)$ has one positive and one negative eigenvalue, so it is a saddle point and unstable. There is a curve (which happens to be the $y$ axis in this case) along which solutions approach $(0,0)$ as $t \to \infty$, but no other solutions will have this behaviour.

The stationary point $(d/c, a/b)$ has purely imaginary eigenvalues. This is actually a borderline case: in the linearized system the stationary point is a "centre", stable but not asymptotically stable, and the solutions near the stationary point form closed curves so as $t \to \infty$ they neither approach the stationary point nor move away from it. In general, in a nonlinear system a stationary point with imaginary eigenvalues could be asymptotically stable or unstable. In this case, it turns out that the nonlinear system is also a centre, but you can't tell that just from the eigenvalues.

As it happens, I have some on-line notes about this: Phase Portraits of Linear Systems, Phase Portraits of Nonlinear Systems, and Predator and Prey.

• I was looking at your notes, I was wondering on how you make those phase portrait with animations? I am interested in learning how to create them. – richitesenpai Dec 9 '15 at 19:13
• I used Maple. Each frame is constructed using some numerical solutions. The animation is then exported to GIF. – Robert Israel Dec 9 '15 at 20:02
• Here's a Maple worksheet that I used to make one of the animations. – Robert Israel Dec 9 '15 at 20:09
• thank you very much, I will try my best to understand it. – richitesenpai Dec 9 '15 at 20:24

The eigenvalues in this case, will tell you how the system behave as $t \to \infty$, for instance, in the case of this model $(0,0)$ is a saddle point. On the other hand, the positive equilibrium $e_3 = \pm i*\sqrt{ad}$ will predict that the system will oscillate. From an ecological point of view, it is important. Hope this picture helps.

• The fact that the eigenvalues are imaginary is not enough to ensure cycles. For example, the systems $\frac{dx}{dt} = y + (x^2 + y^2) x$, $\frac{dy}{dt} = -x + (x^2 + y^2) y$ and $\frac{dx}{dt} = y - (x^2 + y^2) x$, $\frac{dy}{dt} = -x - (x^2 + y^2) y$ also have imaginary eigenvalues for the equilibrium $(0,0)$, but the first is unstable while the second is asymptotically stable. – Robert Israel Dec 9 '15 at 18:49
• thanks, I wanted to say oscillation instead of cycles. – richitesenpai Dec 9 '15 at 19:10
• Thank you for the answer. This leads to another question about what you said: "From an ecological point of view, it is important." can you give me an example of this in connection with Lotka-Volterra prey-predator model? – rasmus ohrt Dec 10 '15 at 15:38
• One of the main reasons is that people that work in ecology, they are seeking for coexistence between species, in this case the oscillations will predict coexistence. However, this model is classified as pure consumer resource, so that in reality, a model will no behave as neat as this one. For instance, when there are no predators, the prey will grow exponentially, but in reality the exists some self- limitation. Most of the current models use a logistic term in the prey growth. – richitesenpai Dec 10 '15 at 19:25