This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 12 Application of Finite Differences to ODE In this chapter we explore the application of finite differences in the simplest setting possible, namely where there is only one independent variable. The equations are then referred to as ordinary differential equations (ODE). We will use the setting of ODE’s to introduce several concepts and tools that will be useful in the numerical solution of partial differential equations. Furthermore, timemarching schemes are almost universally reliant on finite difference methods for discretization, and hence a study of ODE’s is time well spent. 12.1 Introduction Here we derive how an ODE may be obtained in the process of solving numerically a partial differential equations. Let us consider the problem of solving the following PDE: u t + cu x = νu xx , ≤ x ≤ L (12.1) subject to periodic boundary conditions. Equation (12.1) is an advection diffusion equation with c being the advecting velocity and ν the viscosity coeffficient. We will take c and ν to be positive constants. The two independent variables are t for time and x for space. Because of the periodicity, it is sensible to expand the unknown function in a Fourier series: u = ∞ X k =∞ ˆ u n ( t ) e ik n x (12.2) where ˆ u n are the complex amplitudes and depend only on the time variable, whereas e ikx are the Fourier functions with wavenumber k n . Because of the period icity requirement we have k n = 2 πn/L where n is an integer. The Fourier functions 93 94 CHAPTER 12. APPLICATION OF FINITE DIFFERENCES TO ODE form what is called an orthonormal basis, and can be determined as follows: mul tiply the two sides of equations (12.2) by e ik m x where m is integer and integrate over the interval [0 π ] to get: Z L ue ik m x d x = ∞ X k =∞ ˆ u n ( t ) Z L e i ( k n k m ) x d x (12.3) Now notice that the integral on the right hand side of equation (12.3) satisfies the orthogonality property: Z L e i ( k n k m ) x d x = e i ( k n k m ) L 1 i ( k n k m ) = e i 2 π ( n m ) L 1 i ( k n k m ) = 0 , n 6 = m L, n = m (12.4) The role of the integration is to pick out the m th Fourier component since all the other integrals are zero. We end up with the following expression for the Fourier coefficients: ˆ u m = 1 L Z L u ( x ) e ik m x d x (12.5) Equation (12.5) would allow us to calculate the Fourier coefficients for a known function u . Note that for a real function, the Fourier coefficients satisfy ˆ u n = ˆ u * n (12.6) where the * superscript stands for the complex conjugate. Thus, only the positive Fourier components need to be determined and the negative ones are simply the complex conjugates of the positive components....
View
Full Document
 Fall '08
 Staff
 Numerical Analysis, Ode, Tn, Partial differential equation, finite difference

Click to edit the document details