Lotka-Voltera predator prey model isocline?

Question

In lotka voltera predator prey model, why do we get a straight isocline with variable prey/predator population for single value of corresponding predator/prey?

I understand that we get the prey isocline by the equation

N1=prey population ; n2=predator population

What I don't understand is why we are getting variable prey population on isocline for same value of predator population. Shouldn't the prey population be also constant? PS: I picked the graph image from the internet so the symbols are different. Sorry.

Answer 1

I think your issue is mainly with the interpretation of the isoclines.

If you assume this standard predator-prey model:

$$\frac{dN_1}{dt} = rN_1-pN_1N_2$$ $$\frac{dN_2}{dt} = apN_1N_2-mN_2$$

you get the isoclines: $$ N_2 = \frac{r}{p} $$ $$ N_1 = \frac{m}{ap}$$

with the first isocline referring to prey and the second to predators. These isoclines mean that the population growth rate of prey and predators (respectively) will be zero exactly at these isoclines (as well as at the point N₁=N₂=0). Since the isoclines, for this partuicular model, are fixed values and not functions of N₁ and/or N₂ they will be strait lines. Also, in a sense, you should view the isoclines as the borderline case where population growth rate shifts between a positive and a negative value (ie a growing or a decreasing population). The isoclines are not giving you the population size of the prey at a certain predator population size (which you seem to be implying with "...why we are getting variable prey population on isocline for same value of predator population"), but the locations in the plane where the population growth rate (for either prey or predators) is zero.

A graphical view of only the prey isocline (with some color coding) can be:

This means that the prey population (blue) is growing (arrows to the right) below the blue prey isocline, so there if a "flow" to the right there, while it is declining at predator population sizes above the isocline (arrows to the left). So basically, for all predator population sizes (N₂) below isocline the prey population will grow, and for all N₂ above the isocline the prey population will decline. In reality the rate of increase/decrease differs depending on how far from the isocline you are (so arrows should be of different lenghts), but to simplify the issue and to show the general principle I'm just showing the direction of flow here.

The same picture for the predator population is:

Here the predator population (orange) is declining to the left of the orange predator isocline (arrows down) because there is too little prey to sustain the population, and increasing to the right of the isocline (arrows up).

If you combine these you get this, with the black arrows showing the overall direction of the combined population trajectories:

I hope this clarifies the issue a bit. Otherwise, please comment and hightlight what parts you find unclear.

Also, the dynamics of this Lotka-Volterra model also means that populations that start exactly at the intersection of the isoclines (i.e. both N₁ and N₂ will lie at that point) will stay there, since that is an (unstable) equilibrium point. Populations that are started anywhere else in this plane (except for cases where N₁ and/or N₂ is zero), will follow an oscillating population trajectory, that will not change if there isn't any external pertubations (environmental changes, input/output through immigration/emmigration etc). Therefore, the combined trajectories of N₁ and N₂ will form an ellipse in the plane, until they are perturbed (by something outside the current model), when they will then settle at a new ellipsis. For examples of the ellipses, see this phase-plane graph from Wikipedia, with initial values as points:

However, also note that changing the model, i.e. to include a logistic component to the prey population, will completely change the population dynamics.

Answer 2

I am going to try and cover the math with a few more words, to show that your intuition is in fact, correct (which with math, is not always given).

In the differential equation you wrote $$ \frac{d N_1}{dt} = r_1 N_1 - p_1 N_1N_2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1) $$ the numbers $r_1$ and $p_1$ are simply constants that give a strength of coupling, between the change of a variable (the left-hand side) and how this change is computed (the right hand side).
Would you set $p_1=0$, then the resulting solution of (1) would be an unhindered exponential growth. If we go and increase $p_1$ to $p_1>0$, then we increase the "efficiency of conversion from $N_1N_2$ into negative $N_1$", or more simply put "the predators become more efficient at hunting".

Hence, intuitively, as the predators become more efficient, by increasing $p_1$, the prey population cannot grow exponentially grow to infinity any more. It must now level off (as the solution of (1) will show you, if you keep $N_2$ constant, see also the logistic function "In ecology: modeling population growth").
And the stable value towards which it is leveling off, is given by the requirement $\frac{d N_1}{dt} =0$, which leads us to the isocline $0=r_1-p_1N_2$.

Now, coming back to your question, it is important to notice that by requiring $\frac{d N_1}{dt} =0$, we have already set $N_1=const.$, we just have no means yet to compute this value.
So the prey population is in fact constant! And the predator population that can live off this unknown value is given by the ratio of prey multiplication vs. hunting efficiency, which is the isocline.

As a last comment, it is strange that an equation for $N_1$ turns out to give us the most information about $N_2$, but that's how it often is with coupled differential equations.