# T2 514 on the other hand the potential energy stored

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Unformatted text preview: + B2 sin(2πx/a) sinh(2πb/a)+ . . . + Bk sin(2πk/a) sinh(kπb/a) + . . . . We see that Bk sinh(kπb/a) is the k -th coeﬃcient in the Fourier sine series for f (x). For example, if a = b = π , and f (x) = 3 sin x + 7 sin 2x, then we must have f (x) = B1 sin(x) sinh(π ) + B2 sin(2x) sinh(2π ), and hence B1 = 3 , sinh(π ) B2 = 7 , sinh(2π ) Bk = 0 for k = 3, . . . . Thus the solution to Dirichlet’s problem in this case is u(x, y ) = 3 7 sin x sinh y + sin 2x sinh 2y. sinh(π ) sinh(2π ) Exercises: 5.2.1. Which of the following functions are harmonic? a. f (x, y ) = x2 + y 2 . b. f (x, y ) = x2 − y 2 . c. f (x, y ) = ex cos y . d. f (x, y ) = x3 − 3xy 2 . 5.2.2. a. Solve the following Dirichlet problem for Laplace’s equation in a square region: Find u(x, y ), 0 ≤ x ≤ π, 0 ≤ y ≤ π , such that ∂2u ∂2u + 2 = 0, ∂x2 ∂y u(x, 0) = 0, u(0, y ) = u(π, y ) = 0, u(x, π ) = sin x − 2 sin 2x + 3 sin 3x. 122 10 5 3 0 -5 -10 0 2 1 1 2 3 Figure 5.1: Graph of u(x, y ) = 3 sinh(π ) 0 sin...
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## This document was uploaded on 01/12/2014.

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