# Fortunately computers can do the calculations for us

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: found that for each choice of n, Bessel’s equation has a one-dimensional space of well-behaved solutions, which are constant multiples of the Bessel function of the ﬁrst kind Jn (x). Here is an important fact regarding these Bessel functions: Theorem. For each nonnegative integer n, Jn (x) has inﬁnitely many positive zeros. Graphs of the functions J0 (x) and J1 (x) suggest that this theorem might well be true, but it takes some eﬀort to prove rigorously. For completeness, we sketch the proof for the case of J0 (x) at the end of the section. The zeros of the Bessel functions are used to determine the eigenvalues of the Laplace operator on the disk. To see how, note ﬁrst that the boundary condition, √ y ( −λa) = 0, √ requires that −λa be one of the zeros of Jn (x). Let αn,k denote the k -th positive root of the equation Jn (x) = 0. Then √ −λa = αn,k ⇒ λ=− 2 αn,k , a2 and R(r) = Jn (αn,k r/a) will be a solution to (5.29) vanishing at r = a. Hence, in the case where n = 0, λ0,k = − 2 α0,k , a2 and f0,k (r, θ) = J0 (α0,k r/a) will be a solution to the eigenvalue problem 1∂ r ∂r r ∂f ∂r + 1 ∂2f = λf, r2 ∂θ2 f |∂D = 0. Similarly, in the case of general n, λn,k = − 2 αn,k , a2 and fn,k (r, θ) = Jn (αn,k r/a) cos(nθ) or gn,k = Jn (αn,k r/a) sin(nθ), for n = 1, 2, . . . , will be solutions to this e...
View Full Document

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

Ask a homework question - tutors are online