Chapter 3: Modeling with First-Order Differential Equations - PowerPoint PPT Presentation

Overview Linear Models Nonlinear Models Summary Chapter 3: Modeling with First-Order Differential Equations Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw September 26, 2013 DE Lecture 3 王奕翔 王奕翔

Overview Linear Models Nonlinear Models Summary 1 Overview 2 Linear Models Growth and Decay Cooling and Warming Mixtures Series Circuit 3 Nonlinear Models Population Dynamics and Logistic Equation Chemical Reactions 4 Summary DE Lecture 3 王奕翔

Overview Linear Models Nonlinear Models Summary 1 Overview 2 Linear Models Growth and Decay Cooling and Warming Mixtures Series Circuit 3 Nonlinear Models Population Dynamics and Logistic Equation Chemical Reactions 4 Summary DE Lecture 3 王奕翔

Overview Organization of Lectures in Chapter 2 and 3 Linear Models DE Lecture 3 Summary Nonlinear Models ��� Separable DE (2-2) ��� �� DE Linear (2-1) (2-3) Models (3-1) ���� Exact DE Nonlinear (2-6) (2-4) Models (3-2) ���� (2-5) 王奕翔

Overview Linear Models Nonlinear Models Summary What is covered in this lecture We will learn how to model a system by first-order differential equations, solve them by the methods that we have learned, and answer the questions in which we are interested. DE Lecture 3 解應用題基本步驟 : 1 寫下描述系統變化的微分方程式：定義自變數和應變數 2 用學過的方法來解寫下的微分方程式：得到描述系統特性的函數 3 解決一開始待解的問題 王奕翔

Overview Linear Models 4 Summary Chemical Reactions Population Dynamics and Logistic Equation 3 Nonlinear Models Series Circuit Mixtures Cooling and Warming Growth and Decay 2 Linear Models 1 Overview Series Circuit Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 王奕翔

Overview Series Circuit example, bacteria. This model can also be extended to other kinds of population, for Linear Models Population Growth 1 couple: 1 child per 10 months. 10 couples: 10 children per 10 months. Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 Thomas Malthus 人口論 (1798): 人口增加速率正比於人口總數。 Example ( 培養皿中的細菌數) Initially ( t = 0 ) there are P 0 number of bacteria. At time t = 1 hour, there are 2 P 0 number of bacteria. If the rate of growth is proportional to the total population, find the total population at time t , that is, P ( t ) , and the time at which there are 4 P 0 number of bacteria. 王奕翔

Overview Population Growth To determine k , we plug in the other condition: = Linear Models which can be solved as follows. dP dP Series Circuit Growth and Decay Nonlinear Models Summary DE Lecture 3 Mixtures Cooling and Warming A: First we write down the DE and initial conditions relating P ( t ) and t : dt = kP , P (0) = P 0 , k > 0 P = k dt , P ̸ = 0 = ⇒ ln | P | = kt + c 兩邊積分 ⇒ ln P 0 = c 代初始值 = ⇒ P ( t ) = P 0 e kt , t ≥ 0 . P (1) = 2 P 0 = P 0 e k = ⇒ k = ln 2 = ⇒ P ( t ) = P 0 2 t . Finally, the time it takes for the population to reach 4 P 0 is: log 2 4 = 2 hours. 王奕翔

Overview Series Circuit Estimate the age of the fossil given the half-life of C-14 is 5730 years. degenerate. Example: the half-life of C-14 is 5730 years. Linear Models Radioactive Decay For radioactive substances, the decay rate is proportional to the Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 concurrent total amount A ( t ) . dA dt = kA , A (0) = A 0 , k < 0 . Half-life : the time it takes for 1/2 of the atoms in a initial amount to Example ( 碳14定年) A piece of fossil is found to contain 0 . 1 % of the original amount of C-14. 王奕翔

Overview Series Circuit k = Linear Models Radioactive Decay Half-life = 5730 Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 A: We already know the function A ( t ) : A ( t )/ A 0 = e kt , A (0) = A 0 . ⇒ k = − ln 2 ⇒ 0 . 5 = exp (5730 k ) = 5730 . ⇒ t = − ln 1000 Now, 0 . 001 = e kt = = 5730 × 3 × log 2 10 ≈ 57104 . 王奕翔

Overview If the growth/decay rate is proportional to the current total To find k , we need to plug in another condition. X With experience, we know immediately that dX Linear Models amount/population, we have Growth and Decay Series Circuit Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 dt = kX , X (0) = X 0 = e kt . X 0 王奕翔

Overview Linear Models 4 Summary Chemical Reactions Population Dynamics and Logistic Equation 3 Nonlinear Models Series Circuit Mixtures Cooling and Warming Growth and Decay 2 Linear Models 1 Overview Series Circuit Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 王奕翔

Overview Linear Models here is a shortcut. We can solve it by first finding an integrating factor (exercise). Instead, dT A: According to Newton, we have temperature? Example (Cooling of Hot Water) Cooling Series Circuit Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 I. Newton: 升(降)溫的速率與物體和周遭環境的溫差成正比 The initial temperature of a bottle of water is 100 ◦ . At t = 10 minutes, it becomes 40 ◦ . At t = 20 minutes, it becomes 28 ◦ . What is the room dt = k ( T − T m ) , T (0) = 100 , k < 0 . 王奕翔

Overview dt = = Linear Models Plug in the two conditions we immediately have DE Lecture 3 Cooling Series Circuit Mixtures Cooling and Warming Growth and Decay Nonlinear Models Summary Observe that d ( T − T m ) = k ( T − T m ) , T (0) − T m = 100 − T m . Hence, T − T m = e kt . 100 − T m 40 − T m 28 − T m = e 10 k , = e 20 k 100 − T m 100 − T m ( 40 − T m ) 2 = 28 − T m ⇒ 100 − T m 100 − T m ⇒ T m = 25 . 王奕翔

Overview Linear Models 4 Summary Chemical Reactions Population Dynamics and Logistic Equation 3 Nonlinear Models Series Circuit Mixtures Cooling and Warming Growth and Decay 2 Linear Models 1 Overview Series Circuit Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 王奕翔

Overview Example (Mixture of Salt Solutions) A A concentration changes with time. Let the total amount of salt We should do the calculation if we also want to know how the A: Immediately we see that the final answer is 2 lb/gal. (Why?) Linear Models In the figure, the concentration of salt of the incoming flow is 2 DE Lecture 3 Mixtures Nonlinear Models Summary Growth and Decay Mixture of Two Fluids Series Circuit Cooling and Warming input rate of brine 3 gal/min lb/gal. Initially there is 50 lb of salt in the tank. Find the concentration of salt in the tank as time t → ∞ . constant 300 gal at time t be A ( t ) lb. output rate of brine Incoming rate of salt: R in = 2 × 3 = 6 lb/min; 3 gal/min Outgoing: R out = ( current concentration ) × 3 = 300 × 3 = 100 . dt = 6 − A 100 , A (0) = 50 . ∴ dA 王奕翔

Overview dt t t e t t A Linear Models DE Lecture 3 Series Circuit Mixture of Two Fluids Nonlinear Models Summary Growth and Decay Cooling and Warming Mixtures dt = 6 − A 100 , A (0) = 50 . Solve dA 1 Derive Auxiliary DE: Let µ ( t ) := the integrating factor to be found. d ( µ A ) { } { d µ } dt + Ad µ + Ad µ µ = µ dA dt = µ 6 − A dt = 6 µ + dt − 100 100 2 Solve Auxiliary DE: d µ µ dt = 100 . One solution: µ ( t ) = e 100 . 3 Solve Original DE: Plug in µ ( t ) = e 100 and A (0) = 50 : 100 − 550 = − t 100 A = 600 e ⇒ A ( t ) = 600 − 550 e 100 王奕翔

Overview Linear Models 4 Summary Chemical Reactions Population Dynamics and Logistic Equation 3 Nonlinear Models Series Circuit Mixtures Cooling and Warming Growth and Decay 2 Linear Models 1 Overview Series Circuit Mixtures Cooling and Warming Growth and Decay Summary Nonlinear Models DE Lecture 3 王奕翔

Overview LRC Series Circuits C dt of charge: Current is the derivative Linear Models DE Lecture 3 Series Circuit Cooling and Warming Summary Mixtures Nonlinear Models Growth and Decay I n d u cto r ind u ctance L : henries (h) volta g e drop across: L di L i = dq dt R E ( t ) i C E ( t ) = total voltage drop: L E ( t ) = L d 2 q dt 2 + Rdq dt + q C a p a cito r Resisto r capacitance C : farads (f) resistance R : ohms ( Ω ) volta g e drop across: 1 q volta g e drop across: iR C i i R C 王奕翔

Overview Example: LR Series Circuit di Ldi Linear Models Consider the LR circuit shown on the left, Example DE Lecture 3 Series Circuit Growth and Decay Nonlinear Models Mixtures Summary Cooling and Warming where E ( t ) = 10 volts, R = 10 ohms, L = 0 . 5 L henry, and initial current i (0) = 0 . E Find i ( t ) . R A: Current i ( t ) satisfies the following DE: ⇒ 1 dt + Ri = E ( t ) = dt + 10 i = 10 = ⇒ di dt + 20 i = 20 2 王奕翔