ex01_012208 - 01/21/08 NEEP 602 - Engineering Problem...

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01/21/08 NEEP 602 -- Engineering Problem Solving II Exercise 1 A large fraction of the problems we’ll examine this semester will be differential equations, so we’ll begin there for illustration and, in the process, remind ourselves of some of the features of Matlab. In discussing ordinary differential equations (ODEs), we can divide these into two broad categories that depend on how boundary conditions are prescribed. First we note that the number of boundary conditions depends on the order of the equation: a first order ODE requires only one boundary condition, a second order ODE requires two, and so forth. If the conditions are all prescribed at the same point, the information about the solution has to be originated at that point and marched or propagated to different parts of the solution domain. This results in one kind of solution approach. If the conditions are prescribed at different points (say, at the beginning and end of an interval), then information about the solution in the rest of the domain has to span the whole interval with the boundary condition information incorporated into the solution. This latter condition results in a second kind of approach. We’ll refer to problems having the first kind of boundary condition information as Initial Value Problems (IVPs) and the second as Boundary Value Problems (BVPs). We’ll begin our treatment of differential equations by considering the simplest (first order) IVPs. Matlab provides several utilities (among them, ode45 ) for solving IVPs, but what goes on inside the utility is hidden from us, so we’ll spend some time today talking about how we would advance the solution ourselves, and in the process introduce some terms that will be helpful in characterizing computational approaches to problem-solving. Numerical Solution of First Order IVP: ) , ( ' y x f y = To provide some motivation for the Runge-Kutta method embedded in, for instance, Matlab’s ode45 utility, we begin with simpler ways of advancing the solution and work towards more accurate ones. For reference, consider the first order IVP: 1 ) 0 ( ; = + = y y x dx dy Since you’ve all taken Math 319 (or some other course on the solution of ODEs), you’ll know that this first order, linear, non-homogeneous ODE has a solution given by: 1 2 ) ( = x e x y x As we’ll do often over the course of this semester, we’ll use the analytical solution to compare the accuracy of various numerical methods. Instead of beginning with a numerical solution of the differential equation itself, let’s begin with a Taylor series expansion of the function in the neighborhood of x = 0. We know that a general expression for the Taylor series in the neighborhood of a point x = x o , is: ... ) ( ! 3 ) ( ' ' ' ) ( ! 2 ) ( ' ' ) )( ( ' ) ( ) ( 3 2 + + + + = o o o o o o o x x x y x x x y x x x y x y x y
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ex01_012208 - 01/21/08 NEEP 602 - Engineering Problem...

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