Predator-prey systems with differential equations
How to identify cooperative, competitive, and predator-prey systems
When it comes to a system of two populations, we can classify all systems as one of these:
Cooperative
Competitive
Predator-prey
We’ll always have an equation for ???dx/dt???, which models population ???x??? over time ???t???; and another equation for ???dy/dt???, which models population ???y??? over time ???t???.
The sign on the term in each equation that contains both ???x??? and ???y??? is what tells you whether the system is cooperative, competitive, or predator-prey.
If both signs are positive, the system is cooperative.
???\frac{dx}{dt}=ax+bxy???
???\frac{dy}{dt}=cy+dxy???
If both signs are negative, the system is competitive.
???\frac{dx}{dt}=ax-bxy???
???\frac{dy}{dt}=cy-dxy???
If one sign is positive and the other is negative, the system is predator-prey.
???\frac{dx}{dt}=ax+bxy???
???\frac{dy}{dt}=cy-dxy???
or
???\frac{dx}{dt}=ax-bxy???
???\frac{dy}{dt}=cy+dxy???
In a predator-prey system, the equation whose ???xy???-term is positive represents the predator population; the equation whose ???xy???-term is negative represents the prey population.
What it means when the system contains higher-degree terms
Sometimes one or both equations will contain higher-degree terms. In the system
???\frac{dx}{dt}=ax+bx^2+cxy???
???\frac{dy}{dt}=fy-gxy???
???x??? is the predator population because ???cxy??? is positive, and ???y??? is the prey population because ???-gxy??? is negative. Because there are no higher-degree terms in the equation for ???dy/dt???, the prey population is only effected by the predators. On the other hand, because the equation for ???dx/dt??? contains the higher-degree term ???bx^2???, it means that the predator population is effected by the prey population, as well as another factor, like carrying capacity.
How to interpret population stability from the equilibrium solutions
Solving the system for an equilibrium solution ???(x,y)??? will help you draw conclusions about both populations.
An equilibrium solution ???(0,0)??? means that both populations are at ???0???.
An equilibrium solution ???(a,b)??? means that the system is stable. The size ???a??? of population ???x??? supports in balance the size ???b??? of population ???y???, and vice versa.
An equilibrium solution ???(a,0)??? means that population ???x??? is stable at size ???a???, but population ???y??? is at ???0???.
An equilibrium solution ???(0,b)??? means that population ???y??? is stable at size ???b???, but population ???x??? is at ???0???.
How to solve predator-prey systems
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How to determine whether the system is cooperative, competitive, or predator-prey
Example
Does the system of lions and zebras represent a cooperative, competitive, or predator-prey system? What are the equilibrium solutions of the system and what do they mean?
???\frac{dL}{dt}=-0.5L+0.0001LZ???
???\frac{dZ}{dt}=2Z-0.0002Z^2-0.01LZ???
The equation for lions ???dL/dt??? has a positive ???LZ???-term, but the equation for zebras ???dZ/dt??? has a negative ???LZ???-term, which means this is a predator-prey system in which the lions are the predators and the zebras are the prey.
To find equilibrium solutions, we’ll factor both equations.
???\frac{dL}{dt}=-0.5L+0.0001LZ???
???\frac{dL}{dt}=0.0001L(-5,000+Z)???
???\frac{dL}{dt}=0.0001L(Z-5,000)???
and
???\frac{dZ}{dt}=2Z-0.0002Z^2-0.01LZ???
???\frac{dZ}{dt}=0.0002Z(10,000-Z-50L)???
Setting both equations equal to ???0??? gives
???0.0001L(Z-5,000)=0???
???0.0001L=0??? and ???Z-5,000=0???
???L=0??? and ???Z=5,000???
and
???0.0002Z(10,000-Z-50L)=0???
???0.0002Z=0??? and ???10,000-Z-50L=0???
???Z=0??? and ???Z+50L=10,000???
We need to test both solutions from the first equation with both solutions from the second equation. So if we pair ???L=0??? with both ???Z=0??? and ???Z+50L=10,000???, and then in addition we pair ???Z=5,000??? with both ???Z=0??? and ???Z+50L=10,000???, we get the following combinations:
???L=0???
???Z=0???
or
???L=0???
???Z+50L=10,000???
or
???Z=5,000???
???Z=0???
or
???Z=5,000???
???Z+50L=10,000???
We’re looking for each pair to generate an equilibrium solution ???(L,Z)???. The third pair, ???Z=5,000??? and ???Z=0???, doesn’t include a value for ???L???, so we can eliminate that pair completely and focus on just the other three:
???L=0???
???Z=0???
or
???L=0???
???Z+50L=10,000???
or
???Z=5,000???
???Z+50L=10,000???
The first pair, ???L=0??? and ???Z=0???, is the solution ???(0,0)???, so the pairs become
???(0,0)???
or
???L=0???
???Z+50L=10,000???
or
???Z=5,000???
???Z+50L=10,000???
To turn the remaining two pairs into equilibrium solutions, we’ll solve each pair as a system of linear equations. So
???L=0???
???Z+50L=10,000???
gives
???L=0???
???Z+50(0)=10,000???
which becomes
???L=0???
???Z=10,000???
So the pairs are now
???(0,0)???
or
???(0,10,000)???
or
???Z=5,000???
???Z+50L=10,000???
Solving the last pair,
???Z=5,000???
???Z+50L=10,000???
gives
???Z=5,000???
???5,000+50L=10,000???
which becomes
???Z=5,000???
???50L=5,000???
or
???Z=5,000???
???L=100???
And so the pairs are now
???(0,0)???
or
???(0,10,000)???
or
???(100,5,000)???
These are the three equilibrium solutions of the system, and if we analyze them we can say:
At ???(L,Z)=(0,0)??? both populations are at ???(0,0)???.
At ???(L,Z)=(0,10,000)??? the zebra population is stable at ???10,000???, but the lion population is at ???0???.
At ???(L,Z)=(100,5,000)??? the system is stable. ???100??? lions supports in balance ???5,000??? zebras and vice versa.