(1 y) 2 = u 2 = 1 u = 1

Transcription

1 MATH 23 HOMEWORK #2 PART B SOLUTIONS Problem (Semistable Equilibrium Solutions). Sometimes a constant equilibrium solution has the property that solutions lying on one side of the equilibrium solution tend to approach it, whereas solutions lying on the other side depart from it. In this case the equilibrium solution is said to be semistable. (a) Consider the equation dy/ = k( y) 2, () where k is a positive constant. Show that y = is the only critical point, with the corresponding equilibrium solution y(t) =. (b) Sketch f (y) versus y. Show that y is increasing as a function of t for y < and also for y >. The phase line has upward-pointing arrows both below and above y =. Thus solutions below the equilibrium solution approach it, and those above it grow farther away. Therefore, y(t) = is semistable. (c) Solve Eq. () subject to the initial condition y(0) = y 0 and confirm the conclusions reached in part (b). Solution. (b) (a) If y is an equilibrium solution, then dy/ = 0 identically, so hence y(t) = identically. 0 = dy f (y) = dy = k( y)2 k 2 y Since dy/ = k( y) 2 > 0 for y =, then y is increasing for y < and y >. (c) Separating the variables, if y = we have dy du kt + C = k = ( y) 2 = u 2 = u = y

2 where we have made the substitution u = y, du = dy. Then y = kt + C = y = kt + C and the initial condition gives Thus y 0 = y(0) = C = C = y 0 = C = y 0. y = kt + y 0 = y 0 k( y 0 )t + or y = since we previously excluded this case. Since 0 = k( y 0 )t + = t = k( y 0 ) then y(t) has a vertical asymptote at t = and a horizontal asymptote k( y 0 ) at y =. If y >, then < y = kt + y 0 = 0 < kt + y = 0 kt + y 0 < 0. Then the numerator and denominator have different signs, so the denominator is negative: kt + < 0 = kt < = t < y 0 y 0 k( y 0 ) since k > 0. Then we are to the left of the vertical asymptote, so y(t) diverges to ± as t. On the other hand, if y >, then we are to the right of the k( y 0 ) vertical asymptote and y(t) as t since y(t) has a horizontal asymptote at y =. (See below for a plot of y(t) for the values k =, y 0 = 2.) 4 y t 4 Problem

3 (a) Solve the Gompertz equation dy/ = ry ln(k/y) subject to the initial condition y(0) = y 0. Hint: You may wish to let u = ln(y/k)l (b) For the data given in Example in the text (r = 0.7 per year, K = kg, y 0 /K = 0.25), use the Gompertz model to find the predicted value of y(2). (c) For the same data as in part (b), use the Gompertz model to find the time τ at which y(τ) = 0.75K. Solution. (a) If y = 0, we can make the suggested substitution u = ln(y/k). Then y = Ke u, so du = dy y/k = K dy y = K dy Ke u Then the Gompertz equation becomes = e u dy = dy = eu du. e u du = dy = ry ln(k/y) = rkeu ( u) = du = rku. Note that u = ln(y/k) = 0 y = K. Then for y = K we can separate variables: ln u = du u = rk = rkt + C 0 and exponentiating yields ln(y/k) = u = Ce rkt = y/k = e Ce rkt = y = Ke Ce rkt. The initial condition implies so y 0 = y(0) = Ke C = y 0 /K = e C = C = ln(y 0 /K) y = K exp (ln(y 0 /K) exp( rkt)) = K(y 0 /K) e rt. or the constant solution (which we previously excluded) y = K which clearly satisfies the Gompertz equation. (b) Using the provided data, we have y(2) = K exp (ln(y 0 /K) exp( 2rK)) kg. (c) We use Mathematica to find a numerical solution to the equation 0.75K = y(τ) = K exp (ln(y 0 /K) exp( rkτ)) which produces τ yrs. Problem Consider the cohort of individuals born in a given year (t = 0), and let n(t) be the number of these individuals surviving t years later. Let x(t) be the number of members of this cohort who have not had smallpox by year t and who are therefore still susceptible. Let β be the rate at which susceptibles contract smallpox, and let ν be the rate at which people who contract smallpox die from the disease. Finally, let µ(t) be the 3

4 death rate from all causes other than smallpox. Then dx/, the rate at which the number of susceptibles declines, is given by Also dx/ = [β + µ(t)]x. dn/ = νβx µ(t)n. (a) Let z = x/n, and show that z satisfies the initial value problem dz/ = βz( νz), z(0) =. (2) Observe that the initial value problem (2) does not depend on µ(t). (b) Find z(t) by solving (2). (c) Bernoulli estimated that ν = β = /8. Using these values, determine the proportion of 20-year-olds who have not had smallpox. Solution. (a) Since z = x/n, we use the quotient rule to find dz/: dz = nx xn n 2 = x n xn n 2 (β + µ)x = x ( νβx µn) n n2 = (β + µ)z + βν x2 n 2 + µ x n = βz µz + βνz2 + µz = βz + βνz 2 = βz( + νz) = βz( νz). Alternatively, we can use the multivariable chain rule: since z = x/n, then d z = z dx x + z dn n = ( n (β + µ(t))x + x ) n 2 ( νβx µ(t)n) = x x2 (β + µ(t)) + νβ n n 2 + µ(t) x n = z(β + µ) + νβz2 + µz = zβ + νβz 2 = βz( + νz) = βz( νz). Since x(0) = n(0) (no one is infected at time 0), then z(0) =. (b) If z = 0 and z = /ν, then we can separate variables as follows: dz z( νz) = β = βt + C 0. To integrate the lefthand side, we use partial fraction decomposition: z( νz) = A z + B = = A( νz) + Bz = (B Aν)z + A νz = A = and B Aν = 0 = A = and B = ν. So dz dz z( νz) = z + ν νz dz = ln z ln νz = ln z νz. 4

5 Exponentiating, then z νz = Ce βt = z = ( νz)ce βt = Ce βt νzce βt = z( + Cνe βt ) = z + νce βt z = Ce βt = z = Ce βt + Cνe βt = C eβt + ν. Using the initial condition z(0) =, then = z(0) = C + ν = C + ν = = C = ν = C = ν so we have z = C eβt + ν = ν + ( ν)e βt or z = 0 or z = /ν, since we excluded these cases and they clearly satisfy (2). (c) For β = ν = /8, we have z = (/8) + (7/8)e t/8 so z(20) = (/8) + (7/8)e20/8 5