Assignment 1
Logistic equation and Lotka
Volterra two species competition
model
Edvin Listo Zec
[email protected]
920625-2976
August 28, 2014
Co-workers: Emma Ekberg, Sofia Toivonen
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Question 4 - Logistic equation for large time
intervals
The logistic equation is described as:
x0i = ri xi (1 −
xi
)
Ki
Limited resources lead to competition within the population of a species. Below we solved the differential equation with ODE45 in Matlab and plotted the
result for different values on r, the intrinsic growth rate. Ki > 0 is the carrying
capacity.
The values used were x(0) = {1, 2, 3, 4} , x(0) = {6, 7, 8, 9} for top and bottom pictures respectively in figures 1, 2 and 3 and K = 5.
Figure 1: r = 1
Figure 2: r = 0.1
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Figure 3: r = 0.01
We see that when t → ∞, x(t) → K, no matter what values r has. We see
though that the larger r is the faster the solution converges towards K.
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Question 5 - Lotka-Volterra model for large
time intervals
If we instead look at a system where two species compete against eachother we
arrive at the Lotka-Volterra two species competition model:
(
x1
x1 = r1 x1 (1 − K
) − αx1 x2
1
x2
) − βx1 x2
x2 = r2 x2 (1 − K
2
We see that for large t, both species can reach an equilibrium (see figure 4)
lower that their respective carrying capacity and survive. However, if α or β
is much larger than the other (as seen in figure 5), one species out-survive the
other and reaches its carrying capacity (while the other goes towards zero).
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Figure 4: Both species survive and converge towards an equilibrium.
Figure 5: When β is much larger than α one species survives and the other dies.
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Question 7 - Logistic equation theory
Assuming that xi (0) > 0, we want to show that xi (t) > 0 for the logistic
equation. Assume further that Ki and ri are postive constants and that t > 0.
The ordinary differential equation
x0i = ri xi (1 −
xi
)
Ki
has the solution
Kec1 K+rt
.
ec1 K+rt − 1
According to the assumption we have that
x(t) =
x(0) =
Kec1 K
>0
ec1 K − 1
Which implies that ec1 K > 1. Also we know that ert > 1 because r and t are
both > 0. This therefore implies that x(t) > 0 because ec1 K ert > 1.
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Question 8 - Carrying capacity
We see that the solution goes upwards if x(0) < K and that the solution goes
downwards if x(0) > K. This can be explained by realising that the derivative
is postive and negative in both cases respectively. If x(0) = K the derivative is
0 (K is a fixed point) and we get a straight line x(t) = K as solution.
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Question 9 & 10 - Lotka-Volterra two species
competition model theory
Assuming that x1 (0) > 0 and x2 (0) > 0 we want to show that x1 (t) > 0 and
x2 (t) > 0 and that in this case
x01 ≤ r1 x1 (1 −
x1
x2
); x02 ≤ r2 x2 (1 −
)
K1
K2
Assume further that r1 , r2 , K1 , K2 , α, β are positive constants and that t > 0.
(
x1
x01 = r1 x1 (1 − K
) − αx1 x2
1
(1)
x2
0
) − βx1 x2
x2 = r2 x2 (1 − K
2
We also assume that Ki > xi (0) which yields positive evolution for the first
x1
equation if r1 x1 (1 − K
) > αx1 x2 (if not it will decline towards zero, as in the
1
example with α much larger than β). Furtherly we see that if x1 approaches
0 it implies that x01 does so too, resulting in that the population never will get
negative. When t goes towards ∞ we see that x2 approaches the logistic growth
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(limit towards K2 ).
Consider the case with
r1 x1 (1 −
x1
x2
) > αx1 x2 ; r2 x2 (1 −
) > βx1 x2
K1
K2
Numerically it was shown that this converges towards an equilibrium with
x0i = 0. Setting the both equations in (1) equal to zero gives one trivial solution x1 = 0 and x2 = 0. More importantly we also get the solution x1 = 0
which gives us the logistic equation with x2 = K2 . A third solution is at the
points of equilibrium.
Checking for stability for the first solution gives us the matrix:
r1 0
0 r2
Since r1 and r2 are positive, we have positive eqigenvalues implying that the
solution is unstable. For the other solution we get:
r1 − αK2
0
(2)
−r2
−βK2
It is obvious that the stability now is dependent on α , β , r1 , r2 , K1 and K2 .
Briefly explained, this solution is stable if β (or α) is large enough. However,
if α and β both are small the equilibrium is stable (Hsu, pp. 88-91). If the
opposite is true, i.e. for large α , β, it is not known what behaviour the equation
will take and it’s called bistability.
We see that if α and β are small the difference between the solutions for the
logistic equation and the Lotka-Volterra two species competition model is small.
The larger α and β are the larger the difference becomes. Also, if α and β are
large the solution goes faster towards K. One can realise this fact by looking
at the two different differential equations and noting that when α and β are
small, the Lotka-Volterra two species competition model is similiar to the logistic equation. In figures 6 and 7 we plotted the solutions to the Lotka-Volterra
(upper figures) and to the logistic equation (bottom figures).
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Figure 6: For Lotka r1 , r2 = 1,
α = 0.6, β = 0.8, for logistic r = 1.
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Figure 7: For Lotka r1 , r2 = 1, α, β = 0,
for logistic r = 1.
Question 11 - Distance between solutions
We now want to estimate the difference between the equations theoretically by
using the Grönwall inequality:
x0i (t) ≤ ri (1 −
This gives us
xi (t) ≤ xi (0)eri t e
xi
)xi (t)
Ki
r
− Ki
i
Rt
0
xi (s)ds
As mentioned previously, the solutions to the two different systems are very
similiar when α and β are small (the inequality is an equality for α = β = 0).
However, for large α and β we get a bound from the Grönwall inequality. In
conclusion, we have that for small α and β the solutions are close to eachother:
r1
x1 (t) − x2 (t) ≤ x1 (0)er1 t e− K1
Rt
0
x1 (s)ds
r2
− x2 (0)er2 t e− K2
Rt
0
x2 (s)ds
One can however still see that the difference is very much dependent on ri , Ki ,
and the initial values xi (0).
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