ex14_032708 - 3/27/08 EP 471 - Engineering Problem Solving...

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3/27/08 EP 471 -- Engineering Problem Solving II Exercise 14: Numerical Integration: Gaussian Quadrature In CS310/NEEP271 you probably discussed methods of numerical integration that involved breaking up a one-dimensional spatial domain into uniform intervals of width Δ x , subsequently using the trapezoidal rule, Simpson’s rule, or something similar, to evaluate the integral. The problem with this method is that it is not easily extendable to 2D or 3D problems and it provides relatively low order accuracy. The purpose of this and the subsequent exercise is to introduce a method that is more accurate and extendable to 2D or 3D problems, and that is also the foundation of integration in finite element analysis (FEA). Along the way, we’ll also answer a question that has occasionally proved vexing during your construction of Matlab scripts: What is a Jacobian determinant, and why is it mucking up my programming? Gaussian integration or “quadrature” differs from other methods in that it doesn’t use uniform intervals as the basis of integration. The idea is that by permitting non-uniform spacing in the evaluation of the integrand, we can introduce another degree-of-freedom that permits higher order accuracy with fewer terms. To begin, we consider a “second order integration” over the interval from 1 to +1. We want to approximate the integral below with a weighted sum of the integrand evaluated at two locations. We want to choose the “best” a , b , t 1 and t 2 so that the approximation provides the highest order accuracy of the integral. The non-uniform spacing is evident in that, at this point, t 1 and t 2 can be any locations within the domain from 1 to +1. The weighting coefficients a and b are also not yet known. These four “degrees-of-freedom” should, in principle, allow us to get an exact representation of the integration of a cubic polynomial, which also produces four terms. Our goal then is to represent the integral as the weighted sum below: 2 1 2 1 1 1 , , , ; ) ( ) ( ) ( t t b a t bf t af dt t f + + to be determined The conditions that allow us to pick values for the four degrees-of-freedom involve an exact match between the left and right hand sides for certain f ( t ). Suppose we let f ( t ) be t 3 , t 2 , t and 1 respectively. These generate four equations: (1) 3 2 3 1 1 1 3 3 0 : ) ( bt at dt t t t f + = = = + 2 2 2 1 1 1 2 2 3 2 : ) ( bt at dt t t t f + = = = + (2) 2 1 1 1 0 : ) ( bt at dt t t t f + = = = + (3) b a dt t f + = = = + 2 : 1 ) ( 1 1 (4)
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3/27/08 Multiplying the third equation by , and subtracting from the first, we have: 2 1 t ) )( ( ) ( 0 0 1 2 1 2 2 2 1 2 3 2 t t t t bt t t t b + = + = We can satisfy this equation with either b = 0, t 2 = 0, t 1 = t 2 or t 2 = t 1 . Only the last of these options is acceptable, so we choose t 2 = t 1 . Equation (3) then tells us that a = b , and Equation (4) that a = b = 1. Equation (1) is now identically satisfied, so that leaves Equation (2), which tells us that: 5773 . 0
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This note was uploaded on 01/30/2012 for the course ENGINEERIN 471 taught by Professor Witt during the Spring '08 term at Wisconsin.

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ex14_032708 - 3/27/08 EP 471 - Engineering Problem Solving...

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