Mathematics 22: Lecture 5

Transcription

1 Mathematics 22: Lecture 5 Autonomous Equations Dan Sloughter Furman University January 11, 2008 Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

2 Solving the logistics model Recall the logistics population model: where r > 0 is a constant. dp (1 dt = r p ) p, p(0) = p 0, K Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

3 Solving the logistics model Recall the logistics population model: where r > 0 is a constant. dp (1 dt = r p ) p, p(0) = p 0, K To simplify the equation, we will define new variables: Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

4 Solving the logistics model Recall the logistics population model: where r > 0 is a constant. dp (1 dt = r p ) p, p(0) = p 0, K To simplify the equation, we will define new variables: P = p, the fraction of the maximum sustainable population, and K Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

5 Solving the logistics model Recall the logistics population model: where r > 0 is a constant. dp (1 dt = r p ) p, p(0) = p 0, K To simplify the equation, we will define new variables: P = p, the fraction of the maximum sustainable population, and K τ = rt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

6 Solving the logistics model Recall the logistics population model: where r > 0 is a constant. dp (1 dt = r p ) p, p(0) = p 0, K To simplify the equation, we will define new variables: P = p, the fraction of the maximum sustainable population, and K τ = rt. There is a sense in which both P and τ are dimensionless. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

7 Solving the logistic model (cont d) Now dp dt = d dp dτ (KP) = K dt dτ dt = rk dp dτ. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

8 Solving the logistic model (cont d) Now dp dt = d dp dτ (KP) = K dt dτ dt Hence the logistic equation becomes that is, rk dp dτ dp dτ = r(1 P)PK, = P(1 P), with initial condition P(0) = p 0 K = α. = rk dp dτ. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

9 Solving the logistic model (cont d) Hence 1 dp P(1 P) dτ dτ = dτ = τ + k 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

10 Solving the logistic model (cont d) Hence Now 1 dp P(1 P) dτ dτ = 1 dp P(1 P) dτ dτ = dτ = τ + k 1. 1 P(1 P) dp. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

11 Solving the logistic model (cont d) Hence Now 1 dp P(1 P) dτ dτ = 1 dp P(1 P) dτ dτ = Note: Using partial fractions, dτ = τ + k 1. 1 P(1 P) dp. 1 P(1 P) = 1 P P. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

12 Solving the logistic model (cont d) Hence Now 1 dp P(1 P) dτ dτ = 1 dp P(1 P) dτ dτ = Note: Using partial fractions, And so dτ = τ + k 1. 1 P(1 P) dp. 1 P(1 P) = 1 P P. 1 P(1 P) dp = log(p) log(1 P) + k 2 = log ( ) P + k 2. 1 P Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

13 Solving the logistic model (cont d) Hence ( ) P log = τ + c. 1 P Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

14 Solving the logistic model (cont d) Hence ( ) P log = τ + c. 1 P ( ) α Evaluating at τ = 0, we find c = log. 1 α Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

15 Solving the logistic model (cont d) Hence ( ) P log = τ + c. 1 P ( α Evaluating at τ = 0, we find c = log It follows that 1 α P 1 P = αeτ 1 α. ). Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

16 Solving the logistic model (cont d) Hence ( ) P log = τ + c. 1 P ( α Evaluating at τ = 0, we find c = log It follows that Solving for P, P = 1 α P 1 P = αeτ 1 α. ). αe τ 1 + α(e τ 1) = α α + (1 α)e τ. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

17 Solving the logistic model (cont d) In terms of the original variables, p(t) = p 0 K p 0 + (K p 0 )e rt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

18 Solving the logistic model (cont d) P P = α α + (1 α)e τ α τ Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

19 Separation of variables An autonomous equation du dt = f (u) may be written as 1 du f (u) dt = 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

20 Separation of variables An autonomous equation du dt = f (u) may be written as 1 du f (u) dt = 1. Integrating both sides yields 1 du f (u) dt dt = dt = t + c. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

21 Separation of variables An autonomous equation du dt = f (u) may be written as 1 du f (u) dt = 1. Integrating both sides yields 1 du f (u) dt dt = Moreover, we also have 1 du f (u) dt dt = dt = t + c. 1 f (u) du. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

22 Separation of variables An autonomous equation du dt = f (u) may be written as 1 du f (u) dt = 1. Integrating both sides yields 1 du f (u) dt dt = Moreover, we also have 1 du f (u) dt dt = dt = t + c. 1 f (u) du. Hence the general solution is 1 du = t + c, f (u) provided we can evaluate the integral on the left. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

23 Example Consider the equation du dt = ru, with u(0) = u 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

24 Example Consider the equation du dt = ru, with u(0) = u 0. Then 1 u du = rdt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

25 Example Consider the equation du dt = ru, with u(0) = u 0. Then 1 u du = rdt. And so log u = rt + c. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

26 Example Consider the equation du dt = ru, with u(0) = u 0. Then 1 u du = rdt. And so log u = rt + c. Hence u = e c e rt, and so for some constant k. u = ke rt Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

27 Example Consider the equation du dt = ru, with u(0) = u 0. Then 1 u du = rdt. And so Hence u = e c e rt, and so for some constant k. log u = rt + c. u = ke rt Now u 0 = u(0) = k, and so the solution to the initial value problem is u(t) = u 0 e rt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

28 Radioactive decay If u(t) is the amount of a radioactive substance remaining at time t, starting with an initial amount u 0 at time t = 0, then du = rt for dt some constant r > 0. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

29 Radioactive decay If u(t) is the amount of a radioactive substance remaining at time t, starting with an initial amount u 0 at time t = 0, then du = rt for dt some constant r > 0. Hence u(t) = u 0 e rt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

30 Radioactive decay If u(t) is the amount of a radioactive substance remaining at time t, starting with an initial amount u 0 at time t = 0, then du = rt for dt some constant r > 0. Hence u(t) = u 0 e rt. Suppose the substance has a half-life of T years. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

31 Radioactive decay If u(t) is the amount of a radioactive substance remaining at time t, starting with an initial amount u 0 at time t = 0, then du = rt for dt some constant r > 0. Hence u(t) = u 0 e rt. Suppose the substance has a half-life of T years. Then 1 2 u 0 = u 0 e rt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

32 Radioactive decay If u(t) is the amount of a radioactive substance remaining at time t, starting with an initial amount u 0 at time t = 0, then du = rt for dt some constant r > 0. Hence u(t) = u 0 e rt. Suppose the substance has a half-life of T years. Then Hence 1 2 u 0 = u 0 e rt. r = log(2) T. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

33 Carbon-14 dating Carbon-14 has a half-life of 5730 years. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

34 Carbon-14 dating Carbon-14 has a half-life of 5730 years. If u(t) is the amount of Carbon-14 after t years, then u(t) = u 0 e rt, where r = log(2) Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

35 Carbon-14 dating Carbon-14 has a half-life of 5730 years. If u(t) is the amount of Carbon-14 after t years, then u(t) = u 0 e rt, where r = log(2) Suppose an object has 10% of its original Carbon-14. Let T be the age of the object. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

36 Carbon-14 dating Carbon-14 has a half-life of 5730 years. If u(t) is the amount of Carbon-14 after t years, then u(t) = u 0 e rt, where r = log(2) Suppose an object has 10% of its original Carbon-14. Let T be the age of the object. Then 0.10u 0 = u 0 e rt. Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11

37 Carbon-14 dating Carbon-14 has a half-life of 5730 years. If u(t) is the amount of Carbon-14 after t years, then u(t) = u 0 e rt, where r = log(2) Suppose an object has 10% of its original Carbon-14. Let T be the age of the object. Then And so 0.10u 0 = u 0 e rt. T = log(0.10) r 5730 log(0.10) = 19, 035 years. log(2) Dan Sloughter (Furman University) Mathematics 22: Lecture 5 January 11, / 11