Math 1270, Homework 3
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MATH 1270, HoMEwoRK 3
(1) For real numbers t1 and y1 , if φ(t) is a solution to the initial value problem y\ = f(t, y), y(t0 ) = y0
then the function φ 1 (t) defined by φ 1 (t) = φ(t - t1 + t0 ) + y1 - y0 solves the IVP
(*) y\ = f(t - t1 + t0 , y - y1 + y0 ), y(t1 ) = y1
We call the two IVPs equivalent because of the direct relationship between their solutions. (a) Solve the initial value problem y\ = 2ty , y(2) = 1, producing a function φ(t). (b) Now transform φ to a function φ 1 satisfying φ 1 (0) = 0 as above.
(c) Transform the IVP from part (a) to the equivalent one (in the sense of (*) above) “with initial point at the origin” – ie. with initial condition y(0) = 0 – then solve it explicitly. [Your solution should be identical to φ 1 from part (b).]
(2) Transform each initial value problem below into an equivalent one with initial point at the origin. [You are not asked to solve these!]
(a) y\ = 1 - y3 , y(1) = 2
(b) y\ = t2 + y2 , y(-1) = 3
(3) The Mean Value Theorem (MVT) states that if a function F (t) is differentiable on an interval (a, b), then for any a < t0 < t1 < b there is a real number s between t0 and t1 such that
F (t1 ) - F (t0 ) = F\ (s)(t1 - t0 )
(a) Use the MVT to show that if F (t) is differentiable and F\ (t) s 0 for each t e (a, b), then F is nonincreasing : for any t0 , t1 such that a < t0 < t1 < b, F (t0 ) > F (t1 ).
(b) Use the MVT to show that for any fixed t0 e (a, b) and y0 e R, a continuous function f(t) on (a, b) has a unique antiderivative F (t) such that F (t0 ) = y0 . (As discussed in class, by the Fundamental Theorem of Calculus there is at least one.)
(c) Use the MVT to show that for a two-variable function f(t, y) on a rectangle R = {(t, y) | a0 s t s a1 , b0 s y s b1 } in the (t, y)-plane, if ∂f/∂y is defined and continuous on R then for any fixed s e (a0 , a1 ) and b0 < y0 < y1 < b1 ,
|f(s, y1 ) - f(s, y0 )| s K|y1 - y0 |,
where K is the maximum of │ (t, y)│ over all (t, y) in R. [Since s is fixed, you can
apply the MVT to the one-variable function defined by g(y) = f(s, y).]
(4) This problem addresses sustainable harvesting (eg. of a fish population). We assume that in the absence of human interference, the population y would be governed by the logistic growth equation,
= r ╱ 1 - 、y,
where r > 0 is the intrinsic growth rate and K > 0 is the environmental carrying capacity. We also assume that the harvest rate (eg. the rate at which fish are caught) is proportional to the total population, ie. of the form Ey, where E is a constant measuring the effort expended in harvesting. The logistic equation is thus modified to:
= r ╱ 1 - 、y - Ey .
(This equation is called the Schaefer model.)
(a) If E > r, what is the limit of the solution y(t) as t → o, for any initial value y(0) = y0 > 0?
(b) Now assuming that E < r , show that there are two equilibrium solutions, y1 = 0 and y2 = K(1 - E/r) > 0.
(c) Show that y1 is unstable and y2 is asymptotically stable.
(d) A sustainable yield of the fishery is the product Y of E and the asymptotically stable population y2 . Write Y as a function of E .
(e) Determine the value of E that maximizes Y , and determine the maximum value Ym – the maximum sustainable yield.
(5) An alternate model for population growth with logistic constraints is the “Gompertz equation”, which is given (for positive constants r and K) by:
= ry ln ╱ 、 .
(a) Taking f(y) to be the right-hand side above, find its critical points and determine whether each is asymptotically stable or unstable. What is limy →0 f(y)?
(b) Determine the y between 0 and K at which solution curves are concave up and those at which they are concave down.
(c) For y between 0 and K, determine which is larger: ln(K/y) or 1 - y/K . Use this and the fundamental theorem of calculus to answer the following question. For solutions y1 (t), to the logistic equation, and y2 (t), to the Gompertz equation, with y1 (0) = y2 (0) = y0 e (0, K), is y1 (t) > y2 (t) or is y2 (t) > y1 (t)?
Note: you were not asked to solve the Gompertz equation explicitly. But you can! Do it as a bonus exercise in your copious free time. (Hint: rewrite ln(K/y) as - ln(y/K) and take this as u.)
(6) Consider the initial value problem for logistic growth with a threshold:
dy
where r > 0 and 0 < T < K are given constants.
(a) Solve the IVP in implicit form (ie. as an algebraic equation relating y and t). [Partial fractions are recommended for the ODE.]
(b) Without giving an explicit solution, use the direction field to characterize the limit- ing behavior limt→o y(t) of the different solutions y(t) for all possible values of y0 . [See Figure 2.5.8 in Boyce, DiPrima & Meade.]
2022-09-26