Dx e dz dz thus bessels equation 530 in the case

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Unformatted text preview: r R is a Cauchy-Euler equidimensional equation and we can find a nontrivial solution by setting R(r) = rm . Then the equation becomes r d dr r dm (r ) dr = n2 r m , and carrying out the differentiation on the left-hand side yields the characteristic equation m2 − n2 = 0, which has the solutions m = ±n. In this case, the solution is R(r) = Arn + Br−n . Once again, in order for this equation to be well-behaved as r → 0 we must have B = 0, so R(r) is a constant multiple of rn , and un (r, θ) = an rn cos nθ + bn rn sin nθ. 139 The general solution to (5.25) which is well-behaved at r = 0 and satisfies the periodicity condition u(r, θ + 2π ) = u(r, θ) is therefore ∞ u(r, θ) = a0 (an rn cos nθ + bn rn sin nθ), + 2 n=1 where a0 , a1 , . . . , b1 , . . . are constants. To determine these constants we must apply the boundary condition: ∞ h(θ) = u(1, θ) = a0 + (an cos nθ + bn sin nθ). 2 n=1 We conclude that the constants a0 , a1 , . . . , b1 , . . . are simply the Fourier coefficients of h. Exercises: 5.6.1. Solve the following boundary value problem...
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This document was uploaded on 01/12/2014.

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