Assignment 1 Logistic equation and Lotka Volterra two species competition model

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1 Assignment 1 Logistic equation and Lotka Volterra two species competition model Edvin Listo Zec edvinli@student.chalmers.se August 28, 2014 Co-workers: Emma Ekberg, Sofia Toivonen

2 1 Question 4 - Logistic equation for large time intervals The logistic equation is described as: x i = r i x i (1 x i 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. > 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 1

3 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. 2 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: { x 1 = r 1 x 1 (1 x1 αx 1 x 2 x 2 = r 2 x 2 (1 x2 βx 1 x 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. 2

4 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. 3

5 3 Question 7 - Logistic equation theory Assuming that x i (0 > 0, we want to show that x i (t > 0 for the logistic equation. Assume further that and r i are postive constants and that t > 0. The ordinary differential equation has the solution x i = r i x i (1 x i x(t = According to the assumption we have that x(0 = Kec1K+rt e c1k+rt 1. Kec1K e c1k 1 > 0 Which implies that e c1k > 1. Also we know that e rt > 1 because r and t are both > 0. This therefore implies that x(t > 0 because e c1k e rt > 1. 4 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 (s a fixed point and we get a straight line x(t = K as solution. 5 Question 9 & 10 - Lotka-Volterra two species competition model theory Assuming that x 1 (0 > 0 and x 2 (0 > 0 we want to show that x 1 (t > 0 and x 2 (t > 0 and that in this case x 1 r 1 x 1 (1 x 1 ; x 2 r 2 x 2 (1 x 2 Assume further that r 1, r 2,,, α, β are positive constants and that t > 0. { x 1 = r 1 x 1 (1 x1 αx 1 x 2 x (1 2 = r 2 x 2 (1 x2 βx 1 x 2 We also assume that > x i (0 which yields positive evolution for the first equation if r 1 x 1 (1 x1 > αx 1 x 2 (if not it will decline towards zero, as in the example with α much larger than β. Furtherly we see that if x 1 approaches 0 it implies that x 1 does so too, resulting in that the population never will get negative. When t goes towards we see that x 2 approaches the logistic growth 4

6 (limit towards. Consider the case with r 1 x 1 (1 x 1 > αx 1 x 2 ; r 2 x 2 (1 x 2 > βx 1 x 2 Numerically it was shown that this converges towards an equilibrium with x i = 0. Setting the both equations in (1 equal to zero gives one trivial solution x 1 = 0 and x 2 = 0. More importantly we also get the solution x 1 = 0 which gives us the logistic equation with x 2 =. A third solution is at the points of equilibrium. Checking for stability for the first solution gives us the matrix: ( r1 0 0 r 2 Since r 1 and r 2 are positive, we have positive eqigenvalues implying that the solution is unstable. For the other solution we get: ( r1 α 0 (2 r 2 β It is obvious that the stability now is dependent on α, β, r 1, r 2, and. 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 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. 5

7 Figure 6: For Lotka r 1, r 2 = 1, α = 0.6, β = 0.8, for logistic r = 1. Figure 7: For Lotka r 1, r 2 = 1, α, β = 0, for logistic r = 1. 6 Question 11 - Distance between solutions We now want to estimate the difference between the equations theoretically by using the Grönwall inequality: This gives us x i(t r i (1 x i x i (t x i (t x i (0e rit e r i t 0 xi(sds 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: x 1 (t x 2 (t x 1 (0e r1t e r 1 t 0 x1(sds x 2 (0e r2t e r 2 t 0 x2(sds One can however still see that the difference is very much dependent on r i,, and the initial values x i (0. 6