quiz3solutions

quiz3solutions - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 18.085 Quiz 3 December 7, 2007 Professor Strang Your PRINTED name is: Grading 1 2 3 ** NOTE AT NOON A BIG CHEMISTRY CLASS IS COMING !!! 1) (30 pts.) (a) Solve by a Fourier sine series u ( x ) = b k sin kx : ⎧ ⎨ 1 < x < π − u + 4 u ( x ) = f ( x ) = ⎩ with u ( − π ) = u ( π ) = 0 . − 1 − π < x < That right side f ( x ) is the square wave SW( x ) on page 318. (b) What is the decay rate of the coefficients b k ? What is the smoothness of u ( x ) — which derivative jumps ? 1 We’re told that the solution u ( x ) has the form of a Fourier sine series, u ( x ) = ∞ k =1 b k sin kx . The problem is to find the coefficients b k . I think the easiest way to do this 1 is to take the given differential equation and substitute in (1) the Fourier series for u ( x ) and (2) the Fourier series for the right hand side given on page 318 of the textbook. d 2 ∞ ∞ 4 1 − dx 2 b k sin kx + 4 b k sin kx = π k sin kx k =1 k =1 k odd ∞ 4 1 ( k 2 + 4) b k sin kx = sin kx π k k =1 k odd Setting the coefficients of sin kx on the left and right sides equal to each other gives the following equations for the b k ’s. ⎧ ⎫ ⎨ 4 k = 1 , 3 , 5 , ... ⎬ ( k 2 + 4) b k = πk ⎩ k = 2 , 4 , 6 , ... ⎭ So, for odd k , 4 b k = , πk ( k 2 + 4) and, for even k , b k = 0. The solution is 4 u ( x ) = sin kx. πk ( k 2 + 4) k odd Since the denominators of the coefficients contain a k 3 , the decay rate is 1 /k 3 . By comparison with the chart on page 321, this means the first derivative u ( x ) is continuous but the second derivative u ( x ) jumps....
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quiz3solutions - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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