Page 97 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS USING FOURIER SERIES AND INTEGRALS 87
Separation of variables as y = T(τ)R(η)S(θ), substituting into the equation and dividing by
TRS,
T 1 1 S
2
= (ηR ) + =−λ (5.79)
2
T ηR η S
The negative sign is because we anticipate sine and cosine solutions for T.
We also note that
η S
2 2 2
λ η + (ηR ) =− =±β (5.80)
R S
To avoid exponential solutions in the θ direction we must choose the positive sign. Thus we
have
2

T =−λ T
2
S =−β S (5.81)

2 2
2
η(ηR ) + (η λ − β )R = 0

The solutions of the first two of these are
T = A 1 cos(λτ) + A 2 sin(λτ)
(5.82)
S = B 1 cos(βθ) + B 2 sin(βθ)

The boundary condition on the initial velocity guarantees that A 2 = 0. β must be an integer so
that the solution comes around to the same place after θ goes from 0 to 2π.Either B 1 and B 2
can be chosen zero because it doesn’t matter where θ begins (we can adjust f (r,θ)).
T(τ)S(θ) = AB cos(λτ)sin(nθ) (5.83)

The differential equation for R should be recognized from our discussion of Bessel functions.
The solution with β = n is the Bessel function of the first kind order n. The Bessel function
of the second kind may be omitted because it is unbounded at r = 0. The condition that
R(1) = 0 means that λ is the mth root of J n (λ mn ) = 0. The solution can now be completed
using superposition and the orthogonality properties.
∞ ∞

y(τ, η, θ) = K nm J n (λ mn η)cos(λ mn τ)sin(nθ) (5.84)
n=0 m=1
Using the initial condition

∞ ∞

f (η, θ) = K nm J n (λ mn η)sin(nθ) (5.85)
n=0 m=1