POPULATION DYNAMICS WITH SYMMETRIC AND ASYMMETRIC HARVESTING

We study the positive solutions to steady state reaction diffusion equations with Dirichlet boundary conditions of the forms: u ′′ =

x ∈ (0, L) ∪ (1 − L, 1), (A) u(0) = 0 = u(1), and x ∈ ( 1 2 , 1), (B) Here λ, a, b, c and L are positive constants with 0 < L < 1 2 .Such steady state equations arise in population dynamics with logistic type growth and constant yield harvesting.Here u is the population density, 1  λ is the diffusion coefficient and c is the harvesting effort.In particular, model A corresponds to a symmetric harvesting case and model B to an asymmetric harvesting case.Our objective is to study the existence of positive solutions and also discuss the effects of harvesting.We will develop appropriate quadrature methods via which we will establish our results.

Introduction
In [9] the authors study the nonlinear boundary value problem arising in population dynamics.
Here Ω is a smooth bounded region with ∂Ω ∈ C 2 , u is the population density, au − bu 2 represents logistic growth where a > 0, b > 0 are constants.Here h is assumed to be a smooth function representing the intrinsic properties of the region while c ≥ 0 represents the harvesting effort.The assumptions in the development of this steady state model are: (i) the species disperses randomly in the bounded environment; (ii) the reproduction of the species follows logistic growth; (iii) the boundary is hostile to the species; and (iv) the environment is homogeneous (i.e., the diffusion coefficient is independent of spatial variable x).
Here we extend this study to cases where h(x) is a non-smooth function.However we restrict our analysis to the one-dimensional case with Ω = (0, 1) and study the following two models: There has been a raging controversy over the decline of western Atlantic bluefin tuna since the 1970s (see [8]).The U.S. fishing industries demand that the bluefin tuna movements between the western and eastern Atlantic be considered for a better assessments of the effect of 'overfishing' of bluefin by fishermen in central and eastern Atlantic, including the Mediterranean Sea.The model B is a simplified representation of the bluefin tuna population in the Atlantic Ocean in which we focus on the diffusion effects (the convection effects are not considered).It tries to capture the scenario where fishing is restricted to one half of the ocean.The relevance of the study is to see if such regulations can improve the depleting stock of fish in the Atlantic (see also [3] and [10]).
The classes of such models in which the reaction term is negative at the origin are known as semipositone problems.It is well documented in the literature that the study of positive solutions to such classes of problems are mathematically challenging (see [1] and [7]).Also see [4], [5] and references therein.We analyze these problems by modifying the quadrature method discussed in [6] to accommodate such models with discontinuous reaction terms.For further information on the quadrature method see [2] and [4].
In Section 2 the quadrature method that was developed in [6] is briefly outlined.In Section 3 and Section 4 we extend the quadrature method for model A and model B respectively, and study the existence of a positive solution.Though in both models (A) and (B), in one of the harvesting regions the harvesting effort is zero, our methods easily extend to models with unequal harvesting rates in different regions.In Section 5, we provide various computational results on the bifurcation diagrams of positive solutions.

Preliminaries
Here we describe the quadrature method that has been used to prove our results.Consider the boundary value problem where f : [0, ∞) → (0, ∞) is a C 1 function and λ is a positive parameter.Suppose u is a solution of (2) in (0, 1) such that u ′ (x 0 ) = 0 for x 0 ∈ (0, 1).Then u(x 0 +x) = u(x 0 −x) for all x ∈ [0, c) where c = min{x 0 , 1 − x 0 }.This follows by noting the fact that both v(x) := u(x 0 + x) and w(x) := u(x 0 − x) satisfy the initial value problem Hence if u is a positive solution of (2) then u must be symmetric about x = 1 2 , increasing on (0, 1  2 ) and decreasing on ( where F (s) := s 0 f (t)dt, and integrating from 0 to 1 2 we get Here we have used u EJQTDE Spec.Ed.I, 2009 No. 2 and integrate from u(0) = 0 to u(x) after making the substitution z = u(x).This gives Further since u( 12 ) = ρ we obtain Thus, if there exists a positive solution u of (2) with u( 12 ) = ρ, then ρ must be such that G(ρ) exists and satisfy Now suppose there exists ρ > 0 such that √ λ = G(ρ).Define u(x) for x ∈ [0, 1  2 ] using (6).It can be shown that u is a C 2 function that satisfies (2).Thus (2) has a positive solution u with u( 12 Hence by analyzing G(ρ) for ρ ∈ (0, ∞) we can precisely discuss the existence, non-existence and multiplicity of positive solutions as λ varies.Further, when the solution exists, (6) defines the solution.

Symmetric Constant Yield Harvesting
In this section we consider the model A with harvesting allowed only in the interior.Quadrature methods are developed for the regions with and without harvesting separately to prove the existence of a positive solution.We call such classes of positive symmetric solutions as 'Type I' solutions.In this section we consider Type I solutions.Let f (u) := au − bu 2 and f (u) := au − bu 2 − c.We first consider the interval (L, 1  2 ).On the interval (L, 1 2 ) we have Using similar calculations as in Section 2, we have where F (s) := s 0 f(t)dt and ρ = u( 12 ).Hence on (L, 1 2 ), solutions of the model A will satisfy Letting σ := u(L) and evaluating ( 8) at x = L, we have Simplifying ( 9) we get We next consider the interval (0, L).On the interval (0, L) we have As proceeding in the previous case we observe that the solution u of model A on (0, L) satisfies where F (s) := s 0 f (t)dt and m = u ′ (0).Letting σ = u(L), and evaluating (11 In order for u to be a C 1 solution of model A, it is necessary that Substituting in (12) and simplifying we get .If ρ ∈ S then G 1 , G 2 are well-defined.
Proof.We note here that α is the first non-zero root of F (u) = au 2 2 − bu 3 3 − cu and β is the first largest positive root of F ′ (u) = f (u).

Asymmetric Harvesting
In this section we analyze existence of solutions to model B using the quadrature method.We call such classes of positive asymmetric solutions as 'Type II' solutions.In this section we consider Type II solutions.Let f (u and L be the point at which u is maximum.We consider the following two cases: We develop the quadrature method for the interval 0 ≤ x ≤ 1 2 first; the intervals and solving for m2 2 and then substituting x = L we get Similarly for L ≤ x ≤ 1 we get Note here that u is decreasing and thus u ′ is negative on Lemma 4.1.Let T := {ρ : α < ρ < β} where α is as defined in Lemma 3.2 and β = a b .If ρ ∈ T then G 1 , G 2 are well-defined.Proof.We have is positive for all v ∈ (0, ρ) and f (ρ) > 0, if ρ ∈ (0, β).So G 2 is also well-defined for all ρ ∈ T .Lemma 4.2.Let ρ ∈ T be fixed, then there exists a unique σ * (ρ) such that G 1 (σ * (ρ), ρ) = G 2 (σ * (ρ), ρ).Moreover, if u is a positive solution of (B) of Type II with L > 1 2 , then u( 12 ) = σ * (ρ).

Computational Results
In this section we discuss the results obtained using Mathematica computation of bifurcation curves of positive solutions developed in Sections Here again the the set S is non-empty only for ρ ∈ (1.484, 399.259).In Figure 8, we notice that for higher birth rates (i.e.large a such as a = 100) there exists two solutions for λ ∈ (0.2401, 0.3787) .

Figure 1 :
Figure 1: Typical Type I solution

1 2 ≤
x ≤ L and L ≤ x ≤ 1 are considered separately.As explained in Section 2 in the interval 0 ≤ x ≤ 1 2 t)dt and m = u ′ (0).To solve for m 2 2 we set u ′ ( 1

Figure 9 :Figure 10 :Figure 11 :
Figure 9: λ vs ρ when a = 100,b = 1 4 and c = 74 -(2 solutions) Figure 10 is the bifurcation diagram of the asymmetric harvesting problem (B) with parameter values a = 10, b = 1 4 and c = 74.Here ρ has to lie in the interval (26.536, 40) for the set T in Theorem 4.2 to be non-empty.Figures 11 and 12 are the bifurcation diagrams of the asymmetric harvesting problem (B) with parameter values a = 100, b = 1 4 and c = 74.As in the symmetric harvesting problem we get two solutions for higher birth rates when λ ∈ (0.1053, 0.142).The T in Theorem 4.2 is non-empty if ρ ∈ (1.48367, 400).