350 Chapter 5 Higher Dimensions and Other Coordinates
sin(λa)/λa=0, soλm=mπ/aform= 1 , 2 ,....Forothern’s, solutions of
Jn+ 1 / 2 (λa)=0 must be found numerically. For instance, forn=1theequation
is
sin(λa)−λacos(λa)= 0 ,with solutionsλa= 4 .493, 7.725, 10. 904 ,....(SeeHandbook of Mathematical
Functionsby Abramowitz and Stegun, listed in the Bibliography.)
Finally we can put together some product solutions. Clearly the factor func-
tionT(t)will be a sine or cosine ofλct. Thus our product solutions have the
form
ρ−^1 /^2 Jn+ 1 / 2 (λnmρ)Pn(
cos(φ))
sin(λnmct),
ρ−^1 /^2 Jn+ 1 / 2 (λnmρ)Pn(
cos(φ))
cos(λnmct).The solutionu(ρ, φ,t)will be an infinite series of constant multiples of these
functions. We will not write it out.
Let us summarize some of the information we have obtained. First, the fre-
quencies of vibration of a sphere areλmnc(radians per unit time), whereλmn
is themth positive solution of
Jn+ 1 / 2 (λa)= 0.Second, the nodal surfaces (loci of points where a product solution is 0 for all
time) are the values ofρandφfor which
Jn+ 1 / 2 (λmnρ)Pn(
cos(φ))
= 0.
One or the other factor must be 0, so these surfaces are either concentric
spheres,ρ=const., determined byJn+ 1 / 2 (λmnρ)=0, or else conesφ=const.,
determined byPn(cos(φ))=0.
Finally, let us observe that, becauseP 0 (cos(φ))≡1, the product solutions
withn=0 are precisely what we found as product solutions of the problem in
Section 5.8, Part B.
EXERCISES
1.Solve the potential equation in the sphere 0<ρ<1, 0<φ<πwith the
boundary conditionu( 1 ,φ)={
1 , 0 <φ<π/2,
0 ,π/ 2 <φ<π,together with appropriate boundedness conditions.