Figure 5: Bifurcation diagram for equation 4 as a function of K. n(t) << 1 then substituting into f(n) we get (using Taylor s theorem)

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1 Figure 5: Bifurcation diagram for equation 4 as a function of K n(t) << 1 then substituting into f(n) we get (using Taylor s theorem) dn = f(n + n) = f(n ) + f (N )n + higher order terms f (N )n Thus, as with the solution to equation 3 n(t) n 0 e f (N )t We call n 0 a small perturbation. If f (N ) > 0 the perturbation will grow, while if f (N ) < 0 the perturbation will shrink. Hence, the above definition of linear stability. Exponential growth, equation 3, has only one steady state, N = 0. This is (linearly) stable if d > b and unstable if b > d. Logistic growth, equation 4, has two steady states, N = 0 and N = K. f (0) = r and f (K) = r. Thus if r > 0, 0 is unstable and N is stable. Definition: A bifurcation diagram of dn = f b (N) for parameter b is a plot of the position and stability of the steady states of f b (N) as a function of b. Figure 5 shows the bifurcation diagram for equation 4 as a function of K. 8

2 2 Population dynamics: fishing policy Reading : Britton Harvesting a population with logistic growth In fishing, we want to find the maximum sustainable yield (MSY) that we can take from the population. For example, consider a population that when not fished exhibits logistic growth according to equation 2. We can add a death term to the fish population where each fish is caught at a constant rate E: dn = f(n) = rn(1 N/K) EN (5) where E is the effort put in to catching fish by the fishermen. We define Y (E) = EN to be the fishermen s yield. It is this they aim to maximise. The steady states of 5 are N = 0 and N = K(1 E/r), provided that E < r. The yield is Y (E) = EN = EK(1 E/r) To find the MSY we differentiate with respect to E and find that E M = r/2 is maximal. This gives yield Y M (r/2) = rk/4 This MSY gives a stable fish population, since N = K/2 and f (N ) = r 2rN /K r/2 = r/2 = E M r < 0 We should contrast the MSY with the maximum short term yield, which is simply to take all the fish away from the unfished population. This would give a short term yield of K but leave no fish! 2.2 Crashes in fish stock Consider the following model for a fish population N: dn = rn(1 N/K) EN b(n) 9

3 Figure 6: Sublinear predation b(n) where r, K and E are as before and b(n) is predation by seals. One possible form for b(n) is as follows b(n) = BN2 A 2 + N 2 where B and A are constants. This form of predation, shown in figure 6 is known as sublinear. For low densities the seals do not focus on the fish, but as the fish population increases the seals will focus on them more and more on catching them. We will return to predation functions in section 7.1 when we look at predator-prey models. The full model for the fish population is then dn = f(n) = rn(1 N/K) EN BN2 A 2 + N 2 (6) This model is much more complex than those we have looked at previously. It has five parameters (with units): r (time 1 ), K (biomass), E (time 1 ), A (biomass) and B (biomass.time 1 ). To simplify the model we can nondimensionalise it. Set, u = N/ ˆN and τ = t/t Note that, d = d dτ dτ = 1 d T dτ 10

4 Equation 6 becomes ( ˆN du = (r E) T ˆNu 1 r ˆNu ) B ˆN 2 u 2 K(r E) A 2 + ˆN 2 u 2 ( du = (r E)Tu 1 r ˆNu ) 2 B(T/ ˆN)u K(r E) (A 2 / ˆN 2 ) + u 2 Now comes the trick. We can choose ˆN and T, which are dimensionless quantities, to be any values we like. This choice can be motivated by some biological knowledge, or can be quite simply aimed at simplifying the model. Here we simplify the model by choosing ˆN = A, T = A/B. Giving du = su(1 u/q) u2 1 + u 2 (7) where (r E)A K(r E) s = and q = B ra It is best at this point to check that our non-dimensionalisation is correct! For more information on non-dimensionalisation see Edelstein-Keshet, Mathematical Models in Biology, pages We can now find the steady states of our simplified model, equation 7 u = 0 or s(1 u /q) = 1 + u 2 This last equation is best studied graphically. Figure 7 shows three possible solutions for the steady states. Still without solving equation 7 we can also determine the stability of the three steady states graphically. This is done in figure 8 by plotting equation 7. We now have enough information to sketch a bifurcation diagram for parameters q and s. We do this in figure 9. The bifurcation diagrams tell us something very important about the effects of fishing: either reducing s or q can result in a sudden crash in the fish population. In terms of our original parameters, an increase in E (fishing effort) corresponds to a decrease in both s and q. Thus a small increase in fishing effort can result in a large drop in fish populations. Furthermore, the reduction in effort required to recover the fish population must be much larger than the increase that caused the crash. 11 u

5 Figure 7: Three alternative possibilities (depending on the values of s and q) for the steady states of equation 7. 12

6 Figure 8: Three alternative possibilities (depending on the values of s and q) for the stability of the steady states of equation 7. 13

7 Figure 9: Sketch of a bifurcation diagram for equation 7. 14