What Quantity Is Conserved in the Lotka–Volterra System?

For the standard Lotka–Volterra predator–prey model, the populations change over time but a particular combination of them stays constant. Finding that combination gives a first integral (also called a conserved quantity) and helps describe the system’s trajectories.

Use the convention that $x$ is the prey population and $y$ is the predator population:

$$\begin{aligned} \frac{dx}{dt} &= x(a-by),\\ \frac{dy}{dt} &= y(dx-c), \end{aligned}$$

where $a,b,c,d$ are positive parameters. Then a conserved quantity is

$$H(x,y)=dx-c\ln x+by-a\ln y.$$

In other words, $H(x(t),y(t))$ has the same value at every time along a solution with $x,y>0$.

Derive the conserved quantity

The equations suggest using the relative growth rates, since dividing each equation by its population gives

$$\frac{1}{x}\frac{dx}{dt}=a-by, \qquad \frac{1}{y}\frac{dy}{dt}=dx-c.$$

Rearrange these to express the population terms in terms of rates:

$$by=a-\frac{1}{x}\frac{dx}{dt}, \qquad dx=c+\frac{1}{y}\frac{dy}{dt}.$$

A direct verification is to differentiate the proposed function. By the chain rule,

$$\frac{dH}{dt} =\left(d-\frac{c}{x}\right)\frac{dx}{dt} +\left(b-\frac{a}{y}\right)\frac{dy}{dt}.$$

Substitute the model equations:

$$\begin{aligned} \frac{dH}{dt} &=(dx-c)(a-by)+(by-a)(dx-c)\\ &=0. \end{aligned}$$

The two terms cancel, so $H$ is constant along each trajectory. The value of the constant depends on the initial populations. This verification is the quickest way to check the signs in the formula.

Worked example

Consider

$$\frac{dx}{dt}=x(2-y), \qquad \frac{dy}{dt}=y(x-3).$$

Here $a=2$, $b=1$, $c=3$, and $d=1$. Therefore,

$$H(x,y)=x-3\ln x+y-2\ln y.$$

Suppose the initial populations are $x(0)=1$ and $y(0)=1$. Since $\ln 1=0$, the conserved value is

$$H(1,1)=1+1=2.$$

Thus the solution remains on the level curve

$$x-3\ln x+y-2\ln y=2,$$

for as long as both populations are positive. You generally do not need to solve this equation for $y$ as a function of $x$; it describes the orbit implicitly.

For positive parameters, the coexistence equilibrium is $(x,y)=(c/d,a/b)$. The function $H$ is strictly convex and has a minimum there. Every level set $H=h$ with $h>H(c/d,a/b)$ is a closed curve in the positive quadrant. This explains the model’s recurring population cycles without claiming that the populations vary sinusoidally or with a fixed period.

Common mistakes

• Using the wrong signs. With the equations written as above, the logarithmic terms are $-c\ln x$ and $-a\ln y$. Changing the model’s sign convention changes the corresponding formula. Differentiate your proposed invariant to check it.
• Switching parameter roles. The coefficient $d$ multiplies $x$ in the invariant, while $b$ multiplies $y$. These pairings follow from the cross-interaction terms in the equations.
• Forgetting the domain. The logarithms require $x>0$ and $y>0$. The formula is an invariant for interior trajectories, not a finite-valued expression on the axes.
• Expecting one unique formula. If $H$ is conserved, then so is any function of $H$ wherever that function is defined. Adding a constant or multiplying by a constant also gives an equivalent invariant.
• Assuming every predator–prey model has this invariant. The result depends on the exact standard Lotka–Volterra equations. Logistic growth, harvesting, or other added terms usually change the dynamics and destroy this conserved quantity.

A common equivalent form

For a fully nondimensionalized model, exponentiating $H$ gives another way to write the same conservation law:

$$\frac{e^{dx+by}}{x^c y^a}=\text{constant}.$$

This follows because the expression on the left is $e^{H(x,y)}$. The logarithmic form is usually easier to differentiate; the exponential form may be useful when comparing level sets. If populations carry physical units, the logarithms should be written as $\ln(x/x_0)$ and $\ln(y/y_0)$, using fixed reference populations $x_0$ and $y_0$; this changes $H$ only by an additive constant and preserves its derivative. If time also carries units, $H$ has units of a rate, so form a dimensionless exponential by dividing by a fixed reference rate $H_0$ and using $e^{H/H_0}$. In a mathematical model, variables and parameters are often treated as dimensionless.