A look at an economic application of the Lotka-Volterra equations

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November 10, 2020

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Written by MichaelaMayow

The Lotka-Volterra model is mostly referred to as the predator-prey model. This model is used to describe lots of commonly encountered ecological processes. One of the more famous ones is the rabbit and the fox model. This is a cyclical relation between the amount of rabbits and foxes. More on this will be explained further on. In economics, we also encounter lots of cycles and that is why another application of this model could find a place in economics. Therefore, the Lotka-Volterra equations are potentially very useful in such an economic setting and may be able describe business or consumer cycles pretty accurately.

A quick introduction into the mathematics behind the model

We first start by looking a little bit more at the rabbits and foxes model. This model works as follows. First assume we start with a fixed number of rabbits and foxes. Then, if there is an excess amount of rabbits, the foxes have more than enough to eat and the number of foxes will increase. Because of the growing number of foxes more and more rabbits get eaten and the number of rabbits dwindles down. From this the number of foxes also starts to decrease due to a fall in food supply. Then finally the number of rabbits can start to increase again and we arrive back in the situation where we started. Hence this model describes a cyclical process. This cyclical process can also mathematically be explained. So let us now look at the mathematics. The Lotka-Volterra model consists of a system of the following 2 differential equations:

$\begin{equation*} \begin{split} \begin{cases} \frac{dx}{dt} & = (\alpha - \beta y) x \\ \frac{dy}{dt} & = (\delta x - \gamma) y \quad \text{With $\alpha, \beta, \delta, \gamma > 0$} \end{cases} \end{split} \end{equation*}$

To get a better feel of how this model works, we will first note that the only neutrally stable equilibrium solution is given by $(x,y) = (\frac{\gamma}{\delta},\frac{\alpha}{\beta})$ . This can be found by setting both derivatives equal to zero. We will come to why this solution is neutrally stable a little bit later. Now to find the general solution of this model, one can start by dividing the derivative of y by the derivative of x. This way, you join the two equations into one separable differential equation and solve for the general solution in the following way.

$\begin{equation*} \begin{split} \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = & \frac{dy}{dx} = \frac{-y}{x}\frac{\delta x - \gamma}{\beta y - \alpha} \iff \\ & \frac{\beta y- \alpha}{y} dy = \frac{\gamma - \delta x}{x} dx \iff \\ & \displaystyle\int (\beta - \frac{\alpha}{y})dy = \displaystyle\int (\frac{\gamma}{x} - \alpha)dx \iff \\ & \delta x + \beta y - \gamma ln|x| - \alpha ln|y| = C \qquad \text{with $C \in \mathbbm{R}$} \end{split} \end{equation*}$

To further analyze this model we will take a look at the Jacobian matrix of this system.

$\begin{equation*} J(x,y) = \begin{pmatrix} \alpha - \beta y & - \beta x \\ \delta y & \delta x - \gamma \end{pmatrix} \end{equation*}$

We are of course mostly interested in the Jacobian matrix evaluated at the equilibrium solution we found before.

$\begin{equation*} J\left(\frac{\gamma}{\delta},\frac{\alpha}{\beta}\right) = \begin{pmatrix} 0 & -\frac{\beta \gamma}{\delta} \\ \frac{\alpha \delta}{\beta} & 0 \end{pmatrix} \end{equation*}$

Then by looking at the eigenvalues of the Jacobian matrix evaluated at this stable equilibrium, which are given by $\lambda = \pm i \sqrt{\alpha \gamma}$ we can conclude that the solution curves are elliptic and periodic and from this it follows that the equilibrium is indeed neutrally stable. This all follows from the fact that both eigenvalues are purely imaginary and each other’s imaginary conjugates. The reason that the solution is neutrally stable stems from the fact that the real part of the eigenvalues is 0. This means that points in a neighbourhood of the equilibrium do not converge to the equilibrium nor diverge away from the equilibrium. We also know that these points in a neighbourhood are not equilibria and hence must move. The only way that this is possible is if the solution curves orbit around this neutrally stable equilibrium. Finally to get a better visual representation of this model, I wrote a program in R that sketches a few solution curves and the direction field of the system of differential equations in the situation where $\alpha = \beta = \delta = \gamma = 1$ . The values for the constant C are specified for each solution curve in the plot. And here we indeed observe the neutrally stable equilibrium at (1,1). The Rcode for this plot can be found here.

The Goodwin model

Now that we are a little bit more familiar with this system of differential equations, we will take a look at an economic application of this model. Richard Goodwin was one of the first people to come up with such an application. Goodwin thought he could model the Marxist class struggle between the bourgeoisie and the proletariat. Here we have that the bourgeoisie controls all the capital and means of production and the proletariat forms the labour force and hence provides the labour. If we try thinking about why this specific proposition would go well with the Lotka-Volterra equations, it becomes apparent that we are dealing with so-called business cycles. The similarity here with the rabbits and foxes model stems from the fact that we are dealing with two groups in both cases. The assumption that every person falls into one of two groups is of course not very realistic, but models are often not realistic and never have the goal of perfectly explaining reality. Thus we can still use this assumption. As we saw, the Lotka-Volterra model also provides us with this cyclical behavior. So now let’s see what this cycle entails. Goodwin looked at the cyclical relation between the employment rate and the wage share in a working class economy. The macroeconomic effects that Goodwin modelled are explained by the circular flow chart below.

Now that the model is clear, we can dive into the mathematics behind the model. We start by creating the model from the two classes that form the basis of this model. These are of course the wage-earning workers and the profit-earning capitalists. Then by the macro-economic Labor Market model we have the total output Y is divided over these two groups such that $wL$ represents the total wages of the workers and $P = Y - wL$ the total profits of the capitalists. By denoting $\lambda = \frac{Y}{L}$ , we get that the total wage share is given by $\frac{wL}{Y} = \frac{w}{\lambda}$ and the profit share by $1 - \frac{w}{\lambda}$ . By the work of Kalecki, it seems a reasonable assumption that capitalists save everything and workers consume everything. And thus in our macro-economic labor market model we get $S = P = (1 - \frac{w}{\lambda})Y$ . Next to that, we obtain that $S = I$ , meaning all savings will be invested. Now we first need to determine the growth-rate $\delta$ of capital stock. We can do this in the following way by letting $v = \frac{K}{Y}$ be the capital output ratio.

$\begin{equation*} \delta = \frac{dK/dt}{K} = \frac{S}{K} = \frac{(1 - w/\lambda)Y}{K} = \frac{1 - w/\lambda}{v} \end{equation*}$

Now we let labor efficiency grow at a rate $\epsilon$ , that is, $\lambda = \lambda_0 e^{\epsilon t}$ . We can do so since employment is given by $L = \frac{Y}{\lambda}$ . Now to find the growth of labor we can do the following. Note from our equations that it follows that:

$\begin{equation*} \delta - \epsilon = \frac{dL/dt}{L} \end{equation*}$

Finally we still have to add the supply of workers to the model. Let N denote the amount of workers and let this grow with a rate n, that is, we can write $N = N_0 e^{n t}$ . And hence for the employment rate $\mu = \frac{L}{N}$ , we have that the growth rate of $\mu$ can be written as follows, when letting the labor share $u = \frac{w}{\lambda}$ .

$\begin{equation*} \frac{d \mu/dt}{\mu} = \frac{dL/dt}{L} - n = \delta - \epsilon - n = \frac{1 - u}{v} - \epsilon - n \end{equation*}$

This is the first equation in the Goodwin model and reflects the employment growth rate. Then for the second equation in the model we need to take a look at the growth of labor share, that is, using $u = \frac{w}{\lambda}$ we find

$\begin{equation*} \frac{du/dt}{u} = \frac{dw/dt}{w} - \epsilon \end{equation*}$

Now we only have to find $\frac{dw/dt}{w}$ and we can do this by assuming a Phillips curve relation that dominates the growth of wages. If we use the linear approximation of this given by $\frac{dw/dt}{w} = \rho \mu - \gamma$ for some $\rho, \gamma > 0$ , we can plug this in and hence find the two equations that form the Goodwin model.

$\begin{equation*} \begin{cases} \frac{d\mu}{dt} = (\frac{1}{v} - (\epsilon + n) - \frac{u}{v})\mu & \\ \frac{du}{dt} = (-(\gamma + \epsilon) + \rho \mu) u & \end{cases} \end{equation*}$

This model can be solved in the same way as the one described earlier in this article. In turn, this shows that the equations that are designed for something in nature also find a very useful application in economics. This may be a good incentive to look at nature for the answer when you encounter a difficult problem. It may not immediately become clear, but maybe similarities strike your thoughts and nature has already found a solution to the problem you are so desperately trying to solve.

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