Chem Differential Eq HW Solutions Fall 2011 44

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online