Ordinary Differential Equations a Refresher Andreas Adelmann PSI November 12, 2018 CAS 2018 November 12, 2018 Page 1 / 43 Outline 1 What is a differential equation? 2 Initial Value Problems Linear first order differential equations Second order differential equations Recasting high order differential equations as a system of first order differential equations 3 Boundary Value Problems 4 Solution techniques for nonlinear differential equations Power series solutions 5 Stability Analysis 6 Numerical Solution with the Runge-Kutta Method CAS 2018 November 12, 2018 Page 2 / 43 Differential Equations: The BasicsI Ordinary differential equations are used to model change over an independent variable (for our purposes it will usually be t for time or x for a space like variable) without using partial derivatives. So we have equation involving the derivatives of an unknown function y of a single variable t over an interval t 2 (I). Differential equations contain three types of variables: an independent variable, at least one dependent variable (these will be functions of the independent variable), and the parameters. ODE's can contain multiple iterations of derivatives. They are named accordingly (i.e. if there are only first derivatives, then the ODE is called a first order ODE). CAS 2018 November 12, 2018 Page 3 / 43 Differential Equations: The BasicsII If the function F is linear in the variables a0; a1; : : : ; an the ODE is said to be linear. If, in addition, F is homogeneous then the ODE is said to be homogeneous. The general n-th order linear ODE can be written dny dn−1y dy a (x) + a (x) + ··· + a (x) + a (x)y = b(x): n dxn n−1 dxn−1 1 dx 0 CAS 2018 November 12, 2018 Page 4 / 43 General Solution of a Linear Differential Equation It represents the set of all solutions, i.e., the set of all functions which satisfy the equation in the interval (I). For example, the general solution of the differential equation y0 = 3x2 is y = x3 + C where C is an arbitrary constant. The constant C is the value of y at x = 0. This initial condition completely determines the solution. CAS 2018 November 12, 2018 Page 5 / 43 A System of ODE'sI 0 y1 = G1(x; y1; y2; : : : ; yn) (1) 0 y2 = G2(x; y1; y2; : : : ; yn) (2) . (3) 0 yn = Gn(x; y1; y2; : : : ; yn) (4) An n-th order ODE of the form y(n) = G(x; y; y0; : : : ; yn−1) can be transformed in the form of the system of first order DE's. If CAS 2018 November 12, 2018 Page 6 / 43 A System of ODE'sII 0 n−1 we introduce dependant variables y1 = y; y2 = y ; : : : ; yn = y we obtain the equivalent system of first order equations 0 y1 = y2; 0 y2 = y3; . 0 yn = G(x; y1; y2; : : : ; yn): For example, the ODE y00 = y is equivalent to the system 0 y1 = y2; 0 y2 = y1: CAS 2018 November 12, 2018 Page 7 / 43 A System of ODE's III In this way the study of n-th order equations can be reduced to the study of systems of first order equations. Some times, one called the latter as the normal form of the n-th order ODE. Systems of equations arise in the study of the motion of particles. For example, if P (x; y) is the position of a particle of mass m at time t, moving in a plane under the action of the force field (f(x; y); g(x; y)), we have d2x m = f(x; y); dt2 d2y m = g(x; y): dt2 CAS 2018 November 12, 2018 Page 8 / 43 A System of ODE'sIV The general first order ODE in normal form is y0 = F (x; y): @F If F and @y are continuous one can show that, given a; b, there is a unique solution with y(a) = b. CAS 2018 November 12, 2018 Page 9 / 43 A Simple Example: Population Modeling Population growth is commonly modelled with differential equations. In the following equation: t = time, P = population and k = proportionality constant. k represents the constant ratio between the growth rate of the population and the size of the population. dP = kP dt In this particular equation, the left hand side represents the growth rate of the population being proportional to the size of the population P . This is a very simple example of a first order, ordinary differential equation. CAS 2018 November 12, 2018 Page 10 / 43 Initial Value ProblemsI An initial value problem consists of a differential equation and an initial condition. So, going back to the population example, the following is an example of an initial value problem: dP = kP; P (0) = P dt 0 The solution to this set of equations is a function, call it P (t), that satisfies both equations. CAS 2018 November 12, 2018 Page 11 / 43 Initial Value ProblemsII Ansatz: P (t) = Cekt CAS 2018 November 12, 2018 Page 12 / 43 General Solutions to a Differential Equation Let's look at a simple example and walk through the steps of finding a general solution to the following equation dy = (ty)2 dt We will simply \separate" the variables then integrate the both sides of the equation to find the general solution. dy = t2y2 dt 1 dy = t2 dt y2 Z 1 Z dy = t2 dt y2 CAS 2018 November 12, 2018 Page 13 / 43 t3 −y−1 = + c 3 1 t3 − = + c y 3 1 ) y(t) = − t3 3 + c where c 2 < is any real number. CAS 2018 November 12, 2018 Page 14 / 43 Linear First Order Differential EquationsI Initial value problems consist of a differential equation and an initial value. We will work through the example below: dx 1 = −xt; x(0) = p dt π dx First we will need to find the general solution to dt = −xt, then p1 use the initial value x(0) = π to solve for c. Since we do not CAS 2018 November 12, 2018 Page 15 / 43 Linear First Order Differential EquationsII know what x(t) is, we will need to "separate" the equation before integrating. dx = −xt dt 1 − dx = t dt x Z 1 Z − dx = t dt x CAS 2018 November 12, 2018 Page 16 / 43 Linear First Order Differential Equations Continued t2 − ln x = + c 2 2 −( t +c) x = e 2 2 −( t ) −c x = e 2 e 2 − t x = ke 2 dx The above function of t is the general solution to dt = −xt where k is some constant. Since we have the initial value p1 x(0) = π , we can solve for k. CAS 2018 November 12, 2018 Page 17 / 43 Solving Initial Value Problems Thus we can see that the solution to the initial value problem dx 1 = −xt; x(0) = p dt π is 2 1 − 0 x(0) = p = ke 2 π 2 1 − t x(t) = p e 2 π CAS 2018 November 12, 2018 Page 18 / 43 Second Order Differential EquationsI Second order differential equations simply have a second derivative of the dependent variable. The following is a common example that models a simple harmonic oscillator: d2y k + y = 0 dt2 m where m and k are determined by the mass and spring involved. This second order differential equation can be rewritten as the following first order differential equation: dv k = − y dt m CAS 2018 November 12, 2018 Page 19 / 43 Second Order Differential EquationsII where v denotes velocity. dy dv If v(t) is velocity, then v = dt . Thus, we can substitute in dt into our second order differential equation and essentially turn it into a first order differential equation. d2y k dv k = − y , = − y dt2 m dt m Now we have the following system of first order differential equations to describe the original second order differential equation: CAS 2018 November 12, 2018 Page 20 / 43 Second Order Differential Equations III dy = v dt dv k = − y dt m With k=m = 1 consider the following initial value problem: d2y + y = 0 dt2 with y(0) = 0 and y0(0) = v(0) = 1. Let's show that y(t) = sin(t) is a solution. CAS 2018 November 12, 2018 Page 21 / 43 Second Order Differential EquationsIV dy d = sin(t) = cos(t) = v dt dt dv = − sin(t) = −y dt d2y ) = − sin(t) dt2 d2y d2(sin(t)) ) + y = + sin(t) dt2 dt2 = − sin(t) + sin(t) = 0 CAS 2018 November 12, 2018 Page 22 / 43 Reminder: Lie transformationsI A Lie transformation is written as: M = e−t:H: (5) where the Lie operator :H : is defined by: @H @ @H @ :H := − : (6) @~q @~p @~p @~q ~q are the coordinates and ~p the conjugate momenta; h is a function of ~q and ~p. The exponential operator is defined in terms of its series expansion: t2 t3 e−t:H: = 1 − t:H :+ :H :2 − :H :3 + ··· (7) 2 3! CAS 2018 November 12, 2018 Page 23 / 43 Reminder: Lie transformationsII If H is the Hamiltonian of the system, then the evolution of any function of the phase space variables is given by: df = −:H :f; f(t) = e−t:H:f(0): (8) dt CAS 2018 November 12, 2018 Page 24 / 43 Lie transformation The operator e: g : is called a Lie transformation. To see how this works, consider the example of a familiar system: a simple harmonic oscillator in one degree of freedom. The Hamiltonian is: 1 1 H = p2 + !2q2: (9) 2 2 Suppose we want to find the coordinate q as a function of time t.