Lecture Notes Ordinary and Partial Differential Equations

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Lecture Notes for Math 251: 1 Introduction to Ordinary and Partial Differential Equations Wen Shen Spring 2013 1 These notes are provided to students as a supplement to the textbook. They contain mainly examples that we cover in class. The explanation is not very “wordy”; that part you will get by attending the class.Contents 1 Introduction 3 1.1 Classification of Differential Equations . . . . . . . . . . . . . . . . . . . . . . . 3 1.2 Directional Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2 First Order Differential Equations 9 2.1 Linear equations; Method of integrating factors . . . . . . . . . . . . . . . . . . 9 2.2 Separable Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.3 Differences between linear and nonlinear equations . . . . . . . . . . . . . . . . 16 2.4 Modeling with first order equations . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.5 Autonomous equations and population dynamics . . . . . . . . . . . . . . . . . 23 2.6 Exact Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3 Second Order Linear Equations 32 3.1 Homogeneous equations with constant coefficients . . . . . . . . . . . . . . . . . 32 3.2 Solutions of Linear Homogeneous Equations; the Wronskian . . . . . . . . . . . 35 3.3 Complex Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 3.4 Repeated roots; reduction of order . . . . . . . . . . . . . . . . . . . . . . . . . 42 3.5 Non-homogeneous equations; method of undetermined coefficients . . . . . . . . 47 3.6 Mechanical Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 3.7 Forced Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 4 Higher Order Linear Equations 62 4.1 General Theory of n-th Order Linear Equations . . . . . . . . . . . . . . . . . . 62 4.2 Homogeneous Equations with Constant Coefficients. . . . . . . . . . . . . . . . 63 5 The Laplace Transform 66 5.1 Definition of the Laplace transform . . . . . . . . . . . . . . . . . . . . . . . . . 66 5.2 Solution of initial value problems . . . . . . . . . . . . . . . . . . . . . . . . . . 68 5.3 Step functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 5.4 Differential equations with discontinuous forcing functions . . . . . . . . . . . . 82 5.5 Impulse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 6 Systems of Two Linear Differential Equations 88 6.1 Introduction to systems of differential equations . . . . . . . . . . . . . . . . . . 88 6.2 Review of matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 6.3 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 16.4 Basic theory of systems of first order linear equation . . . . . . . . . . . . . . . 92 6.5 Homogeneous systems of two equations with constant coefficients. . . . . . . . . 93 6.6 Complex eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 6.7 Repeated eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 6.8 Stability of linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 7 Fourier Series 116 7.1 Introduction and Basic Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . 116 7.2 Even and Odd Functions; Fourier sine and Fourier cosine series. . . . . . . . . . 124 7.3 Properties of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 7.4 Two-Point Boundary Value Problems; Eigenvalue Problems . . . . . . . . . . . 130 8 Partial Differential Equations 137 8.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 8.2 Heat Equation in 1D; Solution by Separation of Variable and Fourier series . . 138 8.3 Solutions of Wave Equation by Fourier Series . . . . . . . . . . . . . . . . . . . 144 8.4 Laplace Equation in 2D (probably skip) . . . . . . . . . . . . . . . . . . . . . . 147 8.5 D’Alembert’s Solution of Wave Equation; optional . . . . . . . . . . . . . . . . 150 2Chapter 1 Introduction 1.1 Classification of Differential Equations Definition: A differential equation is an equation which contains derivatives of the unknown. (Usually it is a mathematical model of some physical phenomenon.) Two classes of differential equations: • O.D.E. (ordinary differential equations): linear and non-linear; • P.D.E. (partial differential equations). (not covered in math250, but in math251) Some concepts related to differential equations: • system: a collection of several equations with several unknowns. • order of the equation: the highest order of derivatives. • linear or non-linear equations: Let y(t) be the unknown. Then, (n) (n−1) a (t)y +a (t)y +···+a (t)y =g(t), (∗) 0 1 n is a linear equations. If the equation can not be written as (∗), the it’s non-linear. Two things you must know: identify the linearity and the order of an equation. Example 1. Let y(t) be the unknown. Identify the order and linearity of the following equations. ′ (a). (y+t)y +y = 1, ′ 2 ′′ (b). 3y +(t+4)y =t +y , ′′′ (c). y = cos(2ty), √ (4) ′′′ y (d). y + ty +cost =e . Answer. Problem order linear? ′ (a). (y+t)y +y = 1 1 No ′ 2 ′′ (b). 3y +(t+4)y =t +y 2 Yes ′′′ (c). y = cos(2ty) 3 No √ (4) ′′′ y (d). y + ty +cost =e 4 No 3What is a solution? Solution is a function that satisfied the equation and the derivatives exist. at Example 2. Verify that y(t)=e is a solution of the IVP (initial value problem) ′ y =ay, y(0) = 1. Here y(0) = 1 is called the initial condition. Answer.Let’s check if y(t) satisfies the equation and the initial condition: ′ at 0 y =ae =ay, y(0) =e = 1. They are both OK. So it is a solution. −t ′ Example 3. Verify that y(t)= 10−ce with c a constant, is a solution to y +y = 10. Answer. ′ −t −t ′ −t −t y =−(−ce ) =ce , y +y =ce +10−ce = 10. OK. Let’s try to solve one equation. Example 4. Consider the equation ′ 2 (t+1)y =t We can rewrite it as (for t = 6 −1) 2 2 t t −1+1 (t+1)(t−1)+1 1 ′ y = = = = (t−1)+ t+1 t+1 t+1 t+1 ′ To findy, we need to integrate y : Z Z   2 1 t ′ y = y (t)dt = (t−1)+ dt = −t+lnt+1+c t+1 2 where c is an integration constant which is arbitrary. This means there are infinitely many solutions. Additional condition: initial condition y(0) = 1. (meaning: y = 1 whent = 0) Then 2 t y(0) = 0+ln1+c =c = 1, so y(t) = −t+lnt+1+1. 2 R ′ So for equation like y =f(t), we can solve it by integration: y = f(t)dt. 4Review on integration: Z 1 n n+1 x dx = x +c, (n6= 1) n+1 Z 1 dx = lnx+c x Z sinxdx = −cosx+c Z cosxdx = sinx+c Z x x e dx = e +c Z x a x a dx = +c lna Integration by parts: Z Z udv =uv− vdu Chain rule: d ′ ′ (f(g(t)) =f (g(t))·g (t) dt 1.2 Directional Fields ′ Directional field: for first order equations y =f(t,y). ′ Interpret y as the slope of the tangent to the solution y(t) at point (t,y) in the y−t plane. ′ • If y = 0, the tangent line is horizontal; ′ • If y 0, the tangent line goes up; ′ • If y 0, the tangent line goes down; ′ • The value ofy determines the steepness. 1 ′ Example 5. Consider the equation y = (3−y). We know the following: 2 ′ • If y = 3, then y = 0, flat slope, ′ • If y 3, then y 0, down slope, ′ • If y 3, then y 0, up slope. See the directional field below (with some solutions sketched in red): 55 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 6 We note that, if y(0) =3, then y(t)= 3 is the solution. Asymptotic behavior: As t→∞, we have y→ 3. Remarks: ′ (1). For equation y (t) = a(b−y) with a 0, it will have similar behavior as Example 5, 1 where b = 3 and a = . Solution will approach y =b as t→ +∞. 2 ′ ′ (2). Now considery (t)=a(b−y), but witha 0. This changes the sign ofy . We now have – If y(0) =b, then y(t) =b; – If y(0)b, then y→ +∞ as t→ +∞; – If y(0)b, then y→−∞ as t→ +∞. ′ Example 6: Let y (t) = (y−1)(y−5). Then, ′ • If y = 1 or y = 5, then y = 0. ′ • If y 1, then y 0; ′ • If 1y 5, then y 0; ′ • If y 5, then y 0. Directional field looks like: 68 7 6 5 4 3 2 1 0 −1 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 What can we say about the solutions? • If y(0) = 1, then y(t) = 1; • If y(0) = 5, then y(t) = 5; • If y(0) 1, then y→ 1 as t→ +∞; • If 1y(0) 5, then y→ 1 as t→ +∞; • If y(0) 5, then y→ +∞ as t→ +∞. ′ Remark: If we have y (t) =f(y), and for some y we have f(y ) = 0, then, y(t) =y is a 0 0 0 solution. Example 7: Given the plot of a directional field, 76 5 4 3 2 1 0 −1 −2 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 which of the following ODE could have generate it? ′ (a). y (t) = (y−2)(y−4) ′ 2 (b). y (t) = (y−1) (y−3) ′ 2 (c). y (t) = (y−1)(y−3) ′ 2 (d). y (t) =−(y−1)(y−3) We first check the constant solution, y = 1 and y = 3. Then (a) can not be. Then, we ′ check the sign of y on the intervals: y 1, 1 y 3, and y 3, to match the directional field. We found that (c) could be the equation. ′ Example 8. Consider a more complicated situation where y depends on both t and y. ′ Consider y =t+y. To generate the directional field, we see that: ′ • We have y = 0 wheny =−t, ′ • We have y 0 wheny−t, ′ • We have y 0 wheny−t. One can sketch the directional field along lines of y =−t+c for various values of c. ′ • If y =−t , then y = 0; ′ • If y =−t−1 , then y =−1; ′ • If y =−t−2 , then y =−2; ′ • If y =−t+1 , then y = 1; ′ • If y =−t+2 , then y = 2; 8Below is the graph of the directional field, with some solutions plotted in red. 2 1 0 −1 −2 −3 −4 −5 −1 0 1 2 3 4 5 What can we say about the solutions? The solution depends on the initial condition y(0). • We see first that if y(0) =−1, the solution is y(t)=−t−1; • If y(0)−1, then y→ +∞ as t→ +∞; • If y(0)−1, then y→∓∞ as t→ +∞. We can also discuss the asymptotic behavior as t→−∞: • If y(0)−1, then y→ +∞ as t→−∞. • If y(0)−1, then y→ +∞ as t→−∞. 9Chapter 2 First Order Differential Equations We consider the equation dy =f(t,y) dt Overview: • Two special types of equations: linear, and separable; • Linear vs. nonlinear; • modeling; • autonomous equations. 2.1 Linear equations; Method of integrating factors The function f(t,y) is a linear function in y, i.e, we can write f(t,y)=−p(t)y+g(t). So we will study the equation ′ y +p(t)y =g(t). (A) We introduce the method of integrating factors (due to Leibniz): We multiply equation (A) by a function μ(t) on both sides ′ μ(t)y +μ(t)p(t)y =μ(t)g(t) The function μ is chosen such that the equation is integrable, meaning the LHS (Left Hand Side) is the derivative of something. In particular, we require: ′ ′ ′ ′ ′ μ(t)y +μ(t)p(t)y = (μ(t)y), ⇒ μ(t)y +μ(t)p(t)y =μ(t)y +μ(t)y which requires dμ dμ ′ μ(t) = =μ(t)p(t), ⇒ =p(t)dt dt μ 10Integrating both sides Z lnμ(t) = p(t)dt which gives a formula to compute μ   Z μ(t) = exp p(t)dy . Therefore, this μ is called the integrating factor. Note that μ is not unique. In fact, adding an integration constant, we will get a different μ. But we don’t need to be bothered, since any such μ will work. We can simply choose one that is convenient. Putting back into equation (A), we get Z d (μ(t)y) =μ(t)g(t), μ(t)y = μ(t)g(t)dt+c dt which give the formula for the solution     Z Z 1 y(t) = μ(t)g(t)dt+c , where μ(t)= exp p(t)dt . μ(t) ′ Example 1. Solve y +ay =b (a6= 0). Answer.We have p(t) =a and g(t) =b. So Z  at μ = exp adt =e so Z   b b −at at −at at −at y =e e bdt =e e +c = +ce , a a wherec is an arbitrary constant. Pay attention to where one adds this integration constant ′ 2t Example 2. Solve y +y =e . 2t Answer.We have p(t) = 1 andg(t) =e . So   Z t μ(t) = exp 1dt =e and   Z Z 1 1 −t t 2t t 3t −t 3t 2t −t y(t) =e e e dt =e e dt =e e +c = e +ce . 3 3 Example 3. Solve  −2 2 ′ 2 (1+t )y +4ty = 1+t , y(0) = 1. 11Answer.First, let’s rewrite the equation into the normal form  4t −3 ′ 2 y + y = 1+t , 2 1+t so  4t −3 2 p(t) = , g(t) 1+t . 2 1+t Then     Z Z 4t μ(t) = exp p(t)dt = exp dt 2 1+t  2 2 2 2 2 = exp(2ln(1+t )) = exp(ln(1+t ) ) = 1+t . Then Z Z arctant+c 2 −2 2 2 2 −3 2 −2 2 −1 y = (1+t ) (1+t ) (1+t ) dt = (1+t ) (1+t ) dt = . 2 2 (1+t ) By the IC y(0) = 1: 0+c arctant+1 y(0) = =c = 1, ⇒ y(t) = . 2 2 1 (1+t ) ′ 2 −t Example 4. Solve ty −y =t e , (t0). Answer.Rewrite it into normal form 1 ′ −t y − y =te t so −t p(t) =−1/t, g(t) =te . We have Z  1 μ(t)= exp (−1/t)dt = exp(−lnt)= t and Z Z  1 −t −t −t −t y(t) =t te dt =t e dt =t −e +c =−te +ct. t 1 −t Example 5. Solvey− y =e , withy(0) =a, anddiscussionthebehaviorofy ast→∞, 3 as one chooses different initial value a. Answer.Let’s solve it first. We have 1 − t 3 μ =e so Z Z   1 1 1 4 1 4 3 t − t −t t − t t − t 3 3 3 3 3 3 y(t) =e e e dt =e e dt =e − e +c . 4 12Plug in the IC to findc 3 3 0 y(0) =e (− +c) =a, c=a+ 4 4 so     1 4 3 3 3 3 t − t −t t/3 3 3 y(t) =e − e +a+ =− e + a+ e . 4 4 4 4 −t To see the behavior of the solution, we see that it contains two terms. The first term e t/3 3 goes to 0 as t grows. The second term e goes to ∞ as t grows, but the constant a+ is 4 multiplied on it. So we have 3 3 • If a+ = 0, i.e., if a =− , we have y→ 0 as t→∞; 4 4 3 3 • If a+ 0, i.e., if a− , we have y→∞ as t→∞; 4 4 3 3 • If a+ 0, i.e., if a− , we have y→−∞ as t→∞; 4 4 −t On the other hand, as t → −∞, the term e will blow up to −∞, and will dominate. Therefore, y→−∞ as t→−∞ for any values of a. See plot below: ′ 2 Example 6. Solve ty +2y = 4t , y(1) = 2. Answer.Rewrite the equation first 2 ′ y + y = 4t, (t = 6 0) t Sop(t) = 2/t and g(t) = 4t. We have Z  2 μ(t) = exp 2/tdt = exp(2lnt)=t 13and Z  −2 2 −2 4 y(t)=t 4t·t dy =t t +c By ICy(1) = 2, y(1) = 1+c =2, c = 1 we get the solution: 1 2 y(t)=t + , t 0. 2 t Note the condition t 0 comes from the fact that the initial condition is given at t = 1, and we require t6= 0. In the graph below we plot several solutions in the t−y plan, depending on initial data. The one for our solution is plotted with dashed line where the initial point is marked with a ‘×’. 10 5 0 −5 −4 −3 −2 −1 0 1 2 3 4 2.2 Separable Equations We study first order equations that can be written as dy M(x) =f(x,y) = dx N(y) where M(x) and N(y) are suitable functions of x and y only. Then we have Z Z N(y)dy =M(x)dx, ⇒ N(y)dy = M(x)dx 14and we get implicitly defined solutions of y(x). Example 1. Consider dy sinx = . 2 dx 1−y We can separate the variables: Z Z 1 2 3 (1−y )dy = sinxdx, ⇒ y− y =−cosx+c. 3 If one has IC as y(π) = 2, then 1 5 3 2− ·2 =−cosπ+c, ⇒ c =− , 3 3 so the solution y(x) is implicitly given as 1 5 3 y− y +cosx+ = 0. 3 3 Example 2. Find the solution in explicit form for the equation 2 dy 3x +4x+2 = , y(0) =−1. dx 2(y−1) Answer.Separate the variables Z Z 2 2 3 2 2(y−1)dy = (3x +4x+2)dx, ⇒ (y−1) =x +2x +2x+c Set in the IC y(0) =−1, i.e., y =−1 whenx = 0, we get 2 2 3 2 (−1−1) = 0+c, c = 4, (y−1) =x +2x +2x+4. In explicitly form, one has two choices: p 3 2 y(t)= 1± x +2x +2x+4. To determine which sign is the correct one, we check again by the initial condition: √ y(0) = 1± 4 = 1±2, must have y(0) =−1. We see we must choose the ‘-’ sign. The solution in explicitly form is: p 3 2 y(x) = 1− x +2x +2x+4. On which interval will this solution be defined? 3 2 2 x +2x +2x+4≥ 0, ⇒ x (x+2)+2(x+2)≥ 0 2 ⇒ (x +2)(x+2)≥ 0, ⇒ x≥−2. 15We can also argue that when x =−2, we have y = 1. At this point dy/dx→∞, therefore solution can not be defined at this point. The plot of the solution is given below, where the initial data is marked with ‘x’. We also include the solution with the ‘+’ sign, using dotted line. 10 8 6 4 2 0 −2 −4 −6 −8 −3 −2 −1 0 1 2 3 ′ 2 2 2 Example 3. Solve y = 3x +3x y , y(0) = 0, and find the interval where the solution is defined. Answer.Let’s first separate the variables. Z Z dy 1 2 2 2 3 = 3x (1+y ), ⇒ dy = 3x dx, ⇒ arctany =x +c. 2 dx 1+y Set in the IC: arctan0 = 0+c, ⇒ c = 0 we get the solution 3 3 arctany =x , ⇒ y = tan(x ). 3 π π Since the initial data is given atx = 0, i.e.,x = 0, and tan is defined on the interval (− , ), 2 2 we have h i h i 1/3 1/3 π π π π 3 − x , ⇒ − x . 2 2 2 2 Example 4. Solve 2 1+3x ′ y = , y(0) = 1 2 3y −6y and identify the interval where solution is valid. Answer.Separate the variables Z Z 2 2 3 2 3 (3y −6y)dy = (1+3x )dx y −3y =x+x +c. 16Set in the IC: x = 0,y = 1, we get 1−3 =c, ⇒ c =−2, Then, 3 2 3 y −3y =x −x−2. Note that solution is given in implicitly form. ′ 2 To find the valid interval of this solution, we note that y is not defined if 3y −6y = 0, i.e., wheny = 0 or y = 2. These are the two so-called “bad points” where you can not define the solution. To find the corresponding values of x, we use the solution expression: 3 y = 0: x +x−2 = 0, 2 ⇒ (x +x+2)(x−1) = 0, ⇒ x = 1 and 3 3 y = 2 : x +x−2 =−4, ⇒ x +x+2= 0, 2 ⇒ (x −x+2)(x+1) = 0, ⇒ x =−1 2 2 (Note that we used the facts x +x+2= 6 0 and x −x+2 = 6 0 for all x.) Draw the real line and work on it as following:  - - × × x −2 −1 0 1 2 Therefore the interval is−1x 1. 2.3 Differences between linear and nonlinear equations We will take this chapter before the modeling (ch. 2.3). For a linear equation ′ y +p(t)y =g(t), y(t ) =y , 0 0 we have the following existence and uniqueness theorem. Theorem . If p(t) and g(t) are continuous and bounded on an open interval containing t , then it has an unique solution on that interval. 0 Example 1. Find the largest interval where the solution can be defined for the following problems. ′ 3 (A). ty +y =t , y(−1) = 3. ′ 1 2 Answer.Rewrite: y + y =t , sot6= 0. Since t =−1, the interval is t 0. 0 t ′ 3 (B). ty +y =t , y(1) =−3. Answer.The equation is same as (A), so t = 6 0. t = 1, the interval is t 0. 0 ′ (C). (t−3)y +(lnt)y = 2t, y(1) = 2 17lnt 2t ′ Answer.Rewrite: y + y = , so t6= 3 and t 0 for the ln function. Since t = 1, 0 t−3 t−3 the interval is then 0t 3. ′ (D). y +(tant)y = sint, y(π) = 100. 2k+1 Answer.Sincet =π, andfortanttobedefinedwemusthavet = 6 π,k =±1,±2,···. 0 2 π 3π So the interval is t . 2 2 For non-linear equation ′ y =f(t,y), y(t ) =y , 0 0 we have the following theorem: ∂f Theorem . If f(t,y), (t,y) are continuous and bounded on an rectangle (α t ∂y β,a y b) containing (t ,y ), then there exists an open interval around t , contained in 0 0 0 (α,β), where the solution exists and is unique. We note that the statement of this theorem is not as strong as the one for linear equation. Below we give two counter examples. Example 1. Loss of uniqueness. Consider dy t =f(t,y)=− , y(−2)= 0. dt y ′ We first note that at y = 0, which is the initial value of y, we have y =f(t,y)→∞. So the conditions of the Theorem are not satisfied, and we expect something to go wrong. Solve the equation as an separable equation, we get Z Z 2 2 ydy =− tdt, y +t =c, 2 2 2 and by IC we getc = (−2) +0= 4, soy +t = 4. In they−t plan, this is the equation for a circle, centered at the origin, with radius 2. The initial condition is given at t =−2,y = 0, 0 0 √ 2 wherethetangentlineisvertical (i.e., withinfiniteslope). Wehavetwosolutions: y = 4−t √ 2 and y =− 4−t . We lose uniqueness of solutions. Consider a simple non-linear equation: Example 2. Blow-up of solution. ′ 2 y =y , y(0) = 1. 2 Note that f(t,y) = y , which is defined for all t and y. But, due to the non-linearity of f, solution can not be defined for all t. This equation can be easily solved as a separable equation. Z Z 1 1 −1 dy = dt, − =t+c, y(t) = . 2 y y t+c By ICy(0) = 1, we get 1 =−1/(0+c), and so c=−1, and −1 y(t) = . t−1 We see that the solution blows up as t→ 1, and can not be defined beyond that point. This kind of blow-up phenomenon is well-known for nonlinear equations. 182.4 Modeling with first order equations General modeling concept: derivatives describe “rates of change”. Model I: Exponential growth/decay. Q(t) = amount of quantity at time t Assume the rate of change ofQ(t) is proportional to the quantity at timet. We can write dQ (t) =r·Q(t), r : rate of growth/decay dt If r 0: exponential growth If r 0: exponential decay Differential equation: ′ Q =rQ, Q(0) =Q . 0 Solve it: separable equation. Z Z 1 rt+c rt dQ = rdt, ⇒ lnQ =rt+c, ⇒ Q(t) =e =ce Q Here r is called the growth rate. By IC, we get Q(0) =C =Q . The solution is 0 rt Q(t) =Q e . 0 Two concepts: • For r 0, we define Doubling time T , as the time such that Q(T )= 2Q . D D 0 ln2 rT rT D D Q(T ) =Q e = 2Q , e = 2, rT = ln2, T = . D 0 0 D D r 1 • For r 0, we define Half life (or half time) T , as the time such that Q(T )= Q . H H 0 2 1 1 1 ln2 rT rT H H Q(T ) =Q e = Q , e = , rT = ln =−ln2, T = . H 0 0 H H 2 2 2 −r Note here that T 0 since r 0. H NB T , T do not depend on Q . They only depend on r. D H 0 Example 1. If interest rate is 8%, compounded continuously, find doubling time. ln2 Answer.Since r = 0.08, we have T = . D 0.08 Example 2. A radio active material is reduced to 1/3 after 10 years. Find its half life. dQ rt Answer.Model: =rQ,r isratewhichisunknown. We havethesolutionQ(t) =Q e . 0 dt So 1 1 −ln3 10r Q(10) = Q , Q e = Q , r = . 0 0 0 3 3 10 19