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HARVARD MATH 19 - Lecture 18

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Math 19. Lecture 18No-Trawling ZonesT. JudsonFall 20041 No Trawling ZonesDeep trawling can have a devastating effect o n a fishery. If we were to imple-ment no trawling zones, what should be the minimum effective width of sucha strip? We will use lo bsters as an indicator species. Suppose that lobsterpopulations are destroyed outside of our no trawling zone by deep trawling.Inside our no trawling zone, the lobster population increases exponentially,saydudt= ru.We will designate infinitely long strips of width R as no trawl zones. Our goalis to estimate how wide the strip should be so that the lobster populationwill not decrease. The lobster population should obey the equation∂u∂t= µ∂2u∂x2+ ru.The first term on the right is the diffusion term accounting for lobsters ra n-domly roaming the botto m of the ocean. The second term on the right is thepopulation growth due to reproduction. We will assume that this growth isexponential. The constant µ is the diffusion constant. Both µ and r may beestimated by lab experiments or field observations.12 Boundary Conditio nsWe should look for solutions such thatu(t, 0) = u(t, R) = 0.We will show that the lobster population will grow with time providedR >µπ2r1/2.Hence, R depends on µ and r as we might expect.3 Separation of VariablesLet us assume that we can find a solution of the formu(t, x) = A(t)B(x).For u(t, x) = A(t)B(x) to be a solution to∂u∂t= µ∂2u∂x2+ ru,it must be the case thatB(x)dAdt= µA(t)d2Bdx2+ rA(t)B(x),where∂u∂t=dAdt(t)B(x),∂2u∂x2= A(t)d2Bdx2(x).Separating the variables in this last equation, we obtain1A(t)dAdt=µB(x)d2Bdx2+ r.2The variables x and t are independent. For a function of time t o be equal toa function of space, they must both be constant. Let1A(t)dAdt=µB(x)d2Bdx2+ r = λ.Thus, we obtain two ordinary differential equations,dAdt− λA = 0, (1)d2Bdx2−λ − rµB = 0. (2)The solution to the first equation isA(t) = A(0)eλt.Thus, A(t) grows if λ > 0 and decays if λ < 0. Since we are interested in ourlobster population surviving, we will assume that λ ≥ 0.We can rewrite equation (2) asd2Bdx2− cB = 0,wherec =λ − rµ.We will consider three cases, c > 0, c = 0, and c < 0. The only nontrivialcase occurs when c < 0. In this case,B(x) = α cos√−c x + β sin√−c x.The bo undary condition B(0) = 0 implies that α = 0. The second boundarycondition tells us thatβ sin√−c R= 0.Therefore, either β = 0 (no interesting solutions) orsin√−c R= 0.The sine function vanishes at multiples of π; hence, the later case is equivalentto√−c R = nπ,3where n is any integer. Since µ, r, a nd R are fixed, we are restricted on howwe may choose the value of λ. Our solution to∂u∂t= µ∂2u∂x2+ ru.isu(t, x) = A(t)B(x) = βeλtsinnπxR,where 0 < x < R and β > 0. Since our solution must be positive, n = 1.xun = 1xun = 2Therefore, our solution isu(t, x) = A(t)B(x) = βeλtsinπxR.Since c = (λ − r)/µ and√−c R = π,λ = r −µπ2R2.For the lobster population to g r ow,0 < λ = r −µπ2R2orR >µπ2r1/2.Readings and Referen ces• C. Taubes. Modeling Differe ntial Equations in Biology. Prentice Hall,Upper Saddle River, NJ, 2001. Chapter


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HARVARD MATH 19 - Lecture 18

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