Chem Differential Eq HW Solutions Fall 2011 38

# Chem Differential Eq HW Solutions Fall 2011 38 - 38 Chapter...

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Unformatted text preview: 38 Chapter 3 Partial Diﬀerential Equations in Rectangular Coordinates Also g ∗(s + L) = −g ∗ (−s − L) = −g ∗ (−s − L + 2L) = −g ∗(L − s) = −g ∗(s), by the given symmetry property of g . So, using a change of variables, we have 1 2c x+ct+L g ∗(s) ds = x−ct−L x+ct 1 2c g ∗(s + L) ds = − x−ct 1 2c x+ct g ∗(s) ds. x−ct Putting these identities together, it follows that u(x, t + L ) = −u(x, t). c 17. (a) To prove that G is even, see Exercise 14(a). That G is 2L-periodic follows from the fact that g is 2L-periodic and its integral over one period is 0, because it is odd (see Section 2.1, Exercise 15). Since G is an antiderivative of g ∗, to obtain its Fourier series, we apply Exercise 33, Section 3.3, and get G(x) = A0 − L π ∞ n=1 bn (g ) nπ cos x, n L where bn (g ) is the nth Fourier sine coeﬃcient of g ∗, bn (g ) = L 2 L g (x) sin 0 nπ x dx L and ∞ bn (g ) L . π n=1 n A0 = In terms of b∗ , we have n L bn (g ) 2 = πn nπ L g (x) sin 0 nπ x dx = cb∗ , n L and so ∞ L π G(x) = n=1 ∞ bn (g ) L − n π cb∗ 1 − cos( n = n=1 ∞ n=1 bn (g ) nπ cos x n L nπ x) . L (b) From (a), it follows that ∞ cb∗ n G(x + ct) − G(x − ct) = 1 − cos( n=1 ∞ −cb∗ cos( n = n=1 nπ nπ (x + ct)) − 1 − cos( (x − ct)) L L nπ nπ (x + ct)) − cos( (x − ct)) L L ...
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## This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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