Mathematical Models MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Fall 2018
Ordinary Differential Equations The topic of ordinary differential equations (ODEs) is a set of mathematical tools and notation for describing (or mathematically modeling) how quantities change with respect to an independent variable, usually time. An ODE which describes a physical quantity is sometimes called a mathematical model. Notation: We will denote the nth order derivative of y with respect to t as y (n) d n y dt n. As usual y = y (t) = y (1), y = y (t) = y (2), and so on.
ODEs and Solutions Definition An nth order ordinary differential equation is an equation of the form ( y (n) = f t, y, y, y,..., y (n 1)). A solution to the ODE on the interval α < t < β is a function φ(t) for which ( ) φ (n) (t) = f t, φ(t), φ (t), φ (t),..., φ (n 1) (t), for all α < t < β.
Initial Value Problems Definition An initial value problem is a system of equations containing an ordinary differential equation and one or more initial conditions of the form ( ) y (n) (t) = f t, y(t), y (t), y (t),..., y (n 1) (t) y(t 0 ) = y 0. y (n 1) (t 0 ) = y (n 1) 0
Solutions to Initial Value Problems Definition A solution to the IVP on the interval α < t < β with α < t 0 < β is a function φ(t) for which ( ) φ (n) (t) = f t, φ(t), φ (t), φ (t),..., φ (n 1) (t), for all α < t < β and φ(t 0 ) = y 0. φ (n 1) (t 0 ) = y (n 1) 0.
Linear and Nonlinear ODEs An ODE is linear if the function f is linear in the functions y, y, y,..., y (n 1). In this case the ODE can be written as y (n) = g(t) + a 0 (t)y + a 1 (t)y + a 2 (t)y + + a n 1 (t)y (n 1). Otherwise the ODE is said to be nonlinear. An ODE of the form y (n) = f is said to be autonomous. ( y, y, y,..., y (n 1)).
Spring/Mass System Later this semester we will develop from physical principles the mathematical model of a spring/mass system. Suppose an object of mass m is connected to a spring obeying Hooke s Law with spring constant k. The movement of the mass is subject to a viscous damping force proportional to the mass s velocity with proportionality constant γ and is also subject to an external force F(t). If the initial position of the mass at time t = 0 is y 0 and the initial velocity of the mass is v 0 then an mathematical model of the spring/mass system is the IVP: my (t) + γy (t) + ky(t) = F(t) y(0) = y 0 y (0) = v 0.
Comments my (t) + γy (t) + ky(t) = F(t) This is a 2nd order ODE. The ODE is linear. If F(t) = 0 for all t then the ODE is autonomous.
Example Suppose 10y (t) + y (t) + 8y(t) = cos t then the solution to the IVP is ( ( ) y(t) = e t/20 319 2233 cos 1595 20 t 2 5 cos t + 1 sin t. 5 y(0) = 1 y (0) = 0, 13 319 sin ( )) 319 20 t
Plot of Solution 1.0 y t 0.5 20 40 60 80 100 0.5
Frictionless Pendulum Consider an object of mass m connected to a rigid pendulum support of length L. The pendulum swings freely under the force of gravity only. A mathematical model of such a system can be written as mθ + mg ( π ) L sin 2 θ = 0 θ(0) = θ 0 θ (0) = θ 0.
Comments mθ + mg L sin ( π 2 θ ) = 0 The ODE is 2nd order. The ODE is nonlinear. The ODE is autonomous (therefore the choice of t 0 = 0 is arbitrary).
Example Suppose m = 10 kg and L = 2 m, then 10θ + 98 2 sin ( π 2 θ ) = 0 θ(0) = π 3 θ (0) = 0, a plot of the solution resembles the following. y t 1.5 1.0 0.5 2 1 1 2 x t 0.5 1.0 1.5 2.0
Autonomous ODEs and Equilibria For a first order autonomous ODE y (t) = f (y), a solution of the form y(t) = y e where y e is a constant, is called an equilibrium solution. Consider the logistic equation: ( y (t) = r y(t) 1 y(t) ). K The constant functions y(t) = 0 and y(t) = K are both equilibria.
Vector Fields (1 of 2) When not in equilibrium, the right-hand side of a first-order, autonomous ODE can indicate the direction of change of the solution to the ODE. Consider again the logistic equation: ( y (t) = r y(t) 1 y(t) ). K f y K y
Vector Fields (2 of 2) fhyl y K K y 0 t
Extended Example: Falling Body (1 of 3) Suppose an object of mass m falls under the influence of gravitational acceleration g and is also subject to atmospheric drag proportional to velocity. Let γ be the drag coefficient. According to Newton s Second Law of Motion: my (t) = mg γy (t) where positive y denotes the downward direction. For the sake of argument let m = 10 kg, γ = 2 kg/s, the initial velocity be v 0 = 0 m/s. We can use the techniques of elementary calculus to solve the equation.
Extended Example: Falling Body (2 of 3) 10y (t) = (10)(9.8) 2y (t) 10v (t) = 98 2v(t) 10 dv dt = 98 2v(t) 10 dv = dt 2v 98 v 5 t 0 u 49 du = ( 1) ds 0 5 ln v 49 5 ln 0 49 = t v(t) = 49 49e t/5 As t the mass approaches a terminal velocity of 49 m/s.
Extended Example: Falling Body (3 of 3) v t 50 40 30 20 10 If the initial height of the mass was 300 m, then 300 y y (t) = 49 49e t/5 1 du = t 0 49 49e s/5 ds y(t) = 545 49t 245e t/5.
Homework Read Section 1.2 Exercises: 1, 2, 4, 6, 10, 11