Chem Differential Eq HW Solutions Fall 2011 50

# Chem Differential Eq HW Solutions Fall 2011 50 - 50 Chapter...

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Unformatted text preview: 50 Chapter 3 Partial Diﬀerential Equations in Rectangular Coordinates Solutions to Exercises 3.9 1. We apply (2), with a = b = 1: ∞ ∞ u(x, y ) = Emn sin mπx sin nπy, n=1 m=1 where −4 π 2 (m2 + n2 ) Emn = 1 1 x sin mπx sin nπy dx dy 0 0 = 1−(−1)n nπ 1 1 −4 x sin mπx dx sin nπy dy π 2 (m2 + n2 ) 0 0 −4 1 − (−1)n x cos(m π x) sin(m x) − + 4 (m2 + n2 ) π n m m2 π 4 1 − (−1)n (−1)m . π 4 (m2 + n2 ) n m = = = 1 0 Thus u(x, y ) = 8 π4 ∞ ∞ k =0 m=1 (−1)m sin mπx sin((2k + 1)πy ). (m2 + (2k + 1)2)m(2k + 1) 5. We will use an eigenfunction expansion based on the eigenfunctions φ(x, y ) = sin mπx sin nπy , where ∆π (x, y ) = −π 2 m2 + n2 sin mπx sin nπy . So plug ∞ ∞ u(x, y ) = Emn sin mπx sin nπy n=1 m=1 into the equation ∆u = 3u − 1, proceed formally, and get ∞ ∞ ∞ ∞ n=1 m=1 Emn sin mπx sin nπy ) = 3 n=1 m=1 Emn sin mπx sin nπy − 1 ∞ ∞ ∞ ∞ n=1 m=1 Emn ∆ (sin mπx sin nπy ) = 3 n=1 m=1 Emn sin mπx sin nπy − 1 ∞ ∞ 2 2 2 n=1 m=1 −Emn π m + n sin mπx sin nπy ∞ =3 ∞ n=1 m=1 Emn sin mπx sin nπy − 1 ∞ ∞ 2 2 2 Emn sin mπx sin nπy = 1. n=1 m=1 3 + π m + n ∆( Thinking of this as the double sine series expansion of the function identically 1, it follows that 3 + π 2 m2 + n2 Emn are double Fourier sine coeﬃcients, given by (see (8), Section 3.7) 1 1 3 + π 2 m2 + n2 Emn = 4 sin mπx sin nπy dx dy 0 =4 = 0 1 − (−1)m 1 − (−1)n mπ nπ 0 if either m or n is even, . 16 if both m and n are even. π2 m n Thus 0 Emn = 16 π 2 m n 3+π 2 m2 +n2 if either m or n is even, if both m and n are even, ...
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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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