3-6 - 174 Finite Element Analysis and Design 6 Consider the following differential equation d 2u u x 0 dx 2 u(0 0 du dx 0x 1 1 x 1 The solution is

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174 Finite Element Analysis and Design 6. Consider the following differential equation: 2 2 1 0, 0 1 (0) 0 1 x du u x x dx u du dx The solution is approximated as 2 12 () u x c x c x  Calculate the unknown coefficients using Galerkin method. Compare u ( x ) and du ( x )/ dx with the exact solution: u ( x ) = 3.7 sin x x by plotting the solution. Solution: Substitute the approximate solution and make the weighted residuals equal to zero 1 , 0 ( ) 1,2 xx i u u x dx i  Use integration by parts for the first term: 1 1 1 1 , , , 0 0 0 0 ( ) ( ) ( ) 0 x i x i x i i u u dx u dx x dx From the boundary term, we can recognize the essential and natural BCs. The essential boundary specifies u or i , while the natural boundary specifies u , x . Implementing the BCs and also substituting the approximate solution, we obtain the Galerkin equations:
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This note was uploaded on 08/22/2011 for the course EML 4500 taught by Professor Staff during the Fall '08 term at University of Florida.

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3-6 - 174 Finite Element Analysis and Design 6 Consider the following differential equation d 2u u x 0 dx 2 u(0 0 du dx 0x 1 1 x 1 The solution is

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