Chem Differential Eq HW Solutions Fall 2011 44

Chem Differential Eq HW Solutions Fall 2011 44 - 44 Chapter...

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Unformatted text preview: 44 Chapter 3 Partial Differential Equations in Rectangular Coordinates as in Example 1: ∞ Bk sin (2k + 1) u(x, 0) = f (x) ⇒ f (x) = k =0 ⇒ f (x) sin (2n + 1) π x 2L ∞ = Bk sin (2k + 1) k =0 L f (x) sin (2n + 1) ⇒ 0 ∞ Bk π x dx 2L sin (2k + 1) 0 k =0 L ⇒ π π x sin (2n + 1) x 2L 2L L = π x; 2L π π x sin (2n + 1) x dx 2L 2L π x dx 2L π (2n + 1) x dx, 2L f (x) sin (2n + 1) 0 L sin2 = Bn 0 where we have integrated the series term by term and used the orthogonality of the functions sin (2k + 1) 2π x on the interval [0, L]. The orthogonality L can be checked directly by verifying that L sin (2k + 1) 0 π π x sin (2n + 1) x dx = 0 2L 2L if n = k. Solving for Bn and using that L sin2 (2n + 1) 0 π L x dx = 2L 2 (check this using a half-angle formula), we find that Bn = L 2 L f (x) sin (2n + 1) 0 π x dx. 2L 9. This is a straightforward application of Exercise 7. For Exercise 1 the average is 100. For Exercise 2 the average is a0 = 0. 13. The solution is given by (8), where cn is given by (11). We have 1 sin2 µn x dx = 0 = 1 2 1 (1 − cos(2µn x) dx 0 1 1 sin(2µn x) x− 2 2µn 1 = 0 1 2 1− 1 sin(2µn ) . 2µn Since µn is a solution of tan µ = −µ, we have sin µn = −µn cos µn , so sin 2µn = 2 sin µn cos µn = −2µn cos2 µn , and hence 1 sin2 µn x dx = 0 1 1 + cos2 µn . 2 ...
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