Chem Differential Eq HW Solutions Fall 2011 50

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

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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, ...
View Full Document

This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

Ask a homework question - tutors are online