5.7 Vibrations of a Circular Membrane 325
In order for the boundedness condition in Eq. (20) to be fulfilled,Dmust be
zero. Then we are left with
R(r)=Jm(λr).(Because any multiple of a solution is another solution, we can drop the con-
stantC.) The boundary condition of Eq. (20) becomes
R(a)=Jm(λa)= 0 ,implying thatλamust be a root of the equation
Jm(α)= 0.(See Table 1.) For each fixed integerm,αm 1 ,αm 2 ,αm 3 ,...are the first, second,
third,... solutions of the preceding equation. The values ofλfor whichJm(λr)
solves the differential equation and satisfies the boundary condition are
λmn=αmn
a , m=^0 ,^1 ,^2 ,..., n=^1 ,^2 ,^3 ,....
Now that the functionsRandQare determined, we can constructφ.For
m= 1 , 2 , 3 ,...andn= 1 , 2 , 3 ,..., both of the functions
Jm(λmnr)cos(mθ), Jm(λmnr)sin(mθ) (21)are solutions of the problem Eq. (18), both corresponding to the same eigen-
valueλ^2 mn.Form=0andn= 1 , 2 , 3 ,...,wehavethefunctions
J 0 (λ 0 nr), (22)which correspond to the eigenvaluesλ^20 n. (Compare with the simple case.) The
functionT(t)that is a solution of Eq. (17) is any combination of cos(λmnct)
and sin(λmnct).
Now the solutions of Eqs. (11)–(14) have any of the forms
Jm(λmnr)cos(mθ)cos(λmnct), Jm(λmnr)sin(mθ)cos(λmnct),
J (23)
m(λmnr)cos(mθ)sin(λmnct), Jm(λmnr)sin(mθ)sin(λmnct)form= 1 , 2 , 3 ,...andn= 1 , 2 , 3 ,.... In addition, there is the special case
m=0, for which solutions have the form
J 0 (λ 0 nr)cos(λ 0 nct), J 0 (λ 0 nr)sin(λ 0 nct). (24)The CD shows a few of these “standing waves” animated.
The general solution of the problem Eqs. (11)–(14) will thus have the form
of a linear combination of the solutions in Eqs. (23) and (24). We shall use