1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

318 Chapter 5 Higher Dimensions and Other Coordinates

The functionφ(r)=J 0 (λr)is a solution of Eq. (6), and we wish to chooseλ
so that Eq. (7) is satisfied. Then we must have

J 0 (λa)= 0

or

λn=

αn
a, n=^1 ,^2 ,...,
whereαnare the zeros of the functionJ 0. Thus the eigenfunctions and eigen-
values of Eqs. (6), (7), and (8) are

φn(r)=J 0 (λnr), λ^2 n=

(

αn

a

) 2

. (9)

TheseareshownontheCD.
Returning to Eq. (5), we determine that the time factorsTnare

Tn(t)=exp

(

−λ^2 nkt

)

.

We may now assemble the general solution of the partial differential equa-
tion (1), under the boundary condition (2) and boundedness condition (8),
as a general linear combination of our product solutions:

v(r,t)=

∑∞

n= 1

anJ 0 (λnr)exp

(

−λ^2 nkt

)

. (10)

It remains to determine the coefficientsanso as to satisfy the initial condi-
tion (3), which now takes the form

v(r, 0 )=

∑∞

n= 1

anJ 0 (λnr)=f(r), 0 <r<a. (11)

While this problem is not a routine exercise in Fourier series or even a reg-
ular Sturm–Liouville problem (see Section 2.7, especially Exercise 6 there), it
is nevertheless true that the eigenfunctions of Eqs. (6) and (7) are orthogonal,
as expressed by the relation
∫a

0

φn(r)φm(r)rdr= 0 (n=m)

or
∫a

0

J 0 (λnr)J 0 (λmr)rdr= 0 (n=m).

More importantly, the following theorem gives us justification for Eq. (11).