# Dx e dz dz thus bessels equation 530 in the case

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

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 01/12/2014.

Ask a homework question - tutors are online