ODE jiangyushan

Pendulum As a example of a system that is nonlinear, consider the swinging pendulum shown above. When the mass of the pendulum is small in comparison with the mass m at the end of the pendulum, the equation of motion of the pendulum is as follows. Here is the angular position of the pendulum measured relative to vertical, m is mass at the end of the pendulum, L is the length of the pendulum, is the coefficient of viscous friction, and is the acceleration due to gravity. Again, this is a second-order equation in the form of (1.1)

Pendulum If we define the state vector to be, then from above (1.1) we get the following first-order system. This is an autonomous system because there is no explicit dependence on t. Furthermore, it is nonlinear due to the presence of the term.

Predator-prey Ecological System Dynamic systems occur in many fields of study. Consider, for example, the problem of modeling the population levels of a predator-prey pair of species. Let denote the population level of the prey, and let denote the population level of the predator. Suppose and are expressed in units of,say, thousands. The following simplified model of population growth is referred to as the Lotka-Volterra system.

Here the parameter denotes the normalized growth rat e of the prey when the predator is not present. Similarly, denotes the rate at which the predator popul ation decrease in the absence of prey. The term represents the decrease in the prey po pulation as a result of the actions by the predator, and the term represents the increase in the predat or population as a result of the availability of prey. Predator-prey Ecological System

This is a nonlinear system due to the presence of the product terms. As we shall see, it has a periodic solution in which the population levels of the predator and prey g o through ecological cycles. It can be shown that amplitu de of the cycle depends on the initial conditions, while th e period of the cycle is Predator-prey Ecological System

INITIAL AND BOUNDARY PROBLEM By itself, a differential equation does not uniquely determ ine a solution; additional side conditions must be impose d on the solution to make it unique. These side condition s prescribe values that the solution or its derivatives mus t have at some specified point or points. If all of the side conditions are specified at the same point, then we have an initial value problem, which we call it an Initial Value P roblem. If the side conditions are specified at more than one point, then we have a Boundary Value Problem.

Euler’s Method The example of dynamic systems introduced earlier repr esent special cases of the following general first-order no nlinear system where is an n×1 vector. we restrict our consideration to systems for which the rig ht-hand side function is sufficiently smooth that Equation (1.5) has a unique solution satisfying the initial condition, Sufficient conditions on to ensure the existence of a unique solution over can be foun d in (Vid78).

Euler’s Method We are interested in estimating for wh ere the are equality spaced over the interval. That is, where the step size is Suppose the value of is known. This is certainly tru e for because To find in te rms of we multiply both sides of Equation (1.5) by and then integrate from. This yields the follo wing reformulation of (1.5) as an integral equation.

Euler’s Method The problem with applying (1.7) directly is that we do not k now the value of for,and without it we c an not evaluate the integral. However, if the stepsize, is sufficiently small, we can approximate the integrand over t he interval,by it value at the start of t he interval.

Euler’s Method In this case, the integral in (1.7) simplifies to.If denotes the approximate solution obtained in this manner, this yields the following solution formula, which i s called Euler’s method. Euler’s method has a local truncation error of order and the global truncation error is of order

An Example for Euler’s method To illustrate the use of Euler’s method, consider the follo wing simple one-dimensional first-order system. Here the constant. This is a one-dimensional linea r system whose exact solution is. Appl ying Euler’s method in (1.8),we have

An Example for Euler’s method This difference equation is simple enough that we can write a closed-form expression for the solution. If then Recall that the exact solution is a decaying exponential t hat approaches zero in the steady state. The Euler estim ate of the solution will go to zero as approaches infin ity only if or

RUNGE-KUTTA METHODS The coefficients of the fourth-order Runge-Kutta method a re chosen to ensure that its local truncation error is of ord er, and its global truncation error is of order.

Consider the predator-prey equations discussion in (1.3). For convenience, suppose the parameters of the system are. Using the fourth-order Runge-Kutta method to solve this system from to using an initial condition of An Example for Runge-Kutta method

Objectives Know how to convert a higher-order differe ntial equation into an equivalent system of first-order equations. Understand the difference between initial and boundary conditions. Understand the relationship between local and global truncation error. Be able to apply the Runge-Kutta single-st ep solution methods.

Know how to adjust the step size to control the local truncation error. Understand how ordinary differential equat ion techniques can be used to solve practi cal engineering problems. Understand the relative strengths and wea knesses of each computational method an d know which are most applicable for a giv en problem.

Exercise: Chemical Reactor