Presentation is loading. Please wait.

Presentation is loading. Please wait.

Lecture 35 Numerical Analysis. Chapter 7 Ordinary Differential Equations.

Similar presentations


Presentation on theme: "Lecture 35 Numerical Analysis. Chapter 7 Ordinary Differential Equations."— Presentation transcript:

1 Lecture 35 Numerical Analysis

2 Chapter 7 Ordinary Differential Equations

3 Introduction Taylor Series Euler Method Runge-Kutta Method Predictor Corrector Method

4 INTRODUCTION Many problems in science and engineering when formulated mathematically are readily expressed in terms of ordinary differential equations (ODE) with initial and boundary condition.

5 Example The trajectory of a ballistic missile, the motion of an artificial satellite in its orbit, are governed by ordinary differential equations.

6 Theories concerning electrical networks, bending of beams, stability of aircraft, etc., are modeled by differential equations. To be more precise, the rate of change of any quantity with respect to another can be modeled by an ODE

7 Closed form solutions may not be possible to obtain, for every modeled problem, while numerical methods exist, to solve them using computers.

8 In general, a linear or non-linear ordinary differential equation can be written as

9 Here we shall focus on a system of first order differential equations of the form with the initial condition y (t 0 ) = y 0, which is called an initial value problem (IVP).

10 It is justified, in view of the fact that any higher order ODE can be reduced to a system of first order differential equations by substitution.

11 For example, consider a second order differential equation of the form

12 Introducing the substitution the above equation reduces to a system of two first order differential equations, such as the above equation reduces to a system of two first order differential equations, such as

13 Theorem Let f (t, y) be real and continuous in the strip R, defined by

14 Then for anyand for any y 1, y 2, there exists a constant L, satisfying the inequality so that for every Here, L is called Lipschitz constant.

15 If the above conditions are satisfied, then for any y 0, the IVP has a unique solution y ( t ), for In fact, we assume the existence and uniqueness of the solution to the above IVP

16 The function may be linear or non-linear. We also assume that the function f (t, y) is sufficiently differentiable with respect to either t or y.

17 TAYLOR’S SERIES METHOD

18 Consider an initial value problem described by Here, we assume that f (t, y) is sufficiently differentiable with respect to x and y.

19 If y (t) is the exact solution, we can expand y (t) by Taylor’s series about the point t = t 0 and obtain

20 Since, the solution is not known, the derivatives in the above expansion are not known explicitly. However, f is assumed to be sufficiently differentiable and therefore, the derivatives can be obtained directly from the given differential equation.

21 Noting that f is an implicit function of y, we have

22 Similarly

23 Continuing in this manner, we can express any derivative of y in terms of f (t, y) and its partial derivatives.

24 Example Using Taylor’s series method, find the solution of the initial value problem at t = 1.2, with h = 0.1 and compare the result with the closed form solution

25 Solution Let us compute the first few derivatives from the given differential equation as follows:

26 Prescribing the initial condition, that is, at t 0 =1, y 0 – y (t 0 ) = 0, we have

27 Now, using Taylor’s series method, we have

28 Substituting the above values of the derivatives, and the initial condition, we obtain

29 Therefore, Taking y 1 = 0.1103 at t = 1.1, the values of the derivatives are

30 Substituting the value of y 1 and its derivatives into Taylor’s series expansion we get, after retaining terms up to fifth derivative only……

31

32 To obtain the closed form solution, we rewrite the given IVP as or On integration, we get

33 Using the initial condition, we get or Therefore, the closed form solution is

34 When t = 1.2, the closed form solution becomes

35 Lecture 35 Numerical Analysis


Download ppt "Lecture 35 Numerical Analysis. Chapter 7 Ordinary Differential Equations."

Similar presentations


Ads by Google