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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 enter image description here

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? enter image description here PS: I picked the graph image from the internet so the symbols are different. Sorry.

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    $\begingroup$ Welcome to Bio.SE! What attempts to answer this question have you already taken? We ask that all question posters here attempt to search for an answer to their own question and explicitly indicate what research they've already done, what they learned, and what is still confusing or unknown to them. Our goal is not to simply be an answer site, but rather a site that promotes self-learning with some expert help along the way :). Please take a moment to edit your post with this additional detail, and it will likely be received more positively by our community. Thanks! $\endgroup$ – theforestecologist Dec 1 '19 at 19:08
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    $\begingroup$ @theforestecologist Sorry for the trouble, It was my first time posting a question. $\endgroup$ – helppp Dec 2 '19 at 7:48
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    $\begingroup$ I dont understand exactly what you are asking. Are you wondering how the prey and predator populations can fluctuate with a "fixed" isocline? Or if the prey population is stable while the predator population is stable ("...getting variable prey population on isocline for same value of predator population"), which isn't the case? Can you please clarify. $\endgroup$ – fileunderwater Dec 2 '19 at 14:29
  • $\begingroup$ @helppp thanks for the updates! You've improved your post :). However, I, too, am confused regarding exactly what you're asking about. Please use fileunderwater's comment as a guide for further improving your question. You're getting closer to an answer -- just a few more clarifications and we'll see if we can help you out! $\endgroup$ – theforestecologist Dec 2 '19 at 19:09
  • $\begingroup$ @fileunderwater Yes I don't understand why prey or predator population fluctuates with a "fixed " isocline? If we are saying that rate of change of prey population is 0 for the isocline, then how come we are getting a line parallel to prey axis? $\endgroup$ – helppp Dec 3 '19 at 9:07
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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:

enter image description here

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:

enter image description here

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:

enter image description here

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:

enter image description here

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

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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.

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  • $\begingroup$ Okay, that does clear up a few things. But shouldn't a fixed predator population of N2=r1/p1 correspond to a fixed value of prey population, so instead of a line we get a point ? $\endgroup$ – helppp Dec 3 '19 at 9:14
  • $\begingroup$ @helppp Could you post the link where you've got the image from? Then I could check with their variable namings. The way I see it the line means exactly that: For one particular N2, there is one constant N1. However, now if we change N2 to some other number N2', then N1 has also to change, in a correspondingly linear fashion, along the isocline. $\endgroup$ – AtmosphericPrisonEscape Dec 3 '19 at 10:41
  • $\begingroup$ here you go: nre509.wikidot.com/lecture-15-nov-4 $\endgroup$ – helppp Dec 3 '19 at 13:10
  • $\begingroup$ Why are you bringing the logistic function into this? The prey model in the Q is not based on the logistic, and the levelling off in the logistic model has nothing to do with the isocline in this Q. $\endgroup$ – fileunderwater Dec 4 '19 at 9:14
  • $\begingroup$ @fileunderwater: Because when keeping $N_2$ constant, this is a logistic DEQ. It helps develop what you expect from a solution. The first term is an enxponential growth, the second one an exponential decline. If the logistic level-off is actually reached, depends on the ratio of the coefficients. $\endgroup$ – AtmosphericPrisonEscape Dec 4 '19 at 11:11

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