Page 160 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

150 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS

Let the sine transform of u be denoted by S n u(r,θ, z) = U n (r, n, z) with respect to θ
on the interval (0, π). Then

2 2 2
∂ U n 1 ∂U n n ∂ U n
+ − U n + + q(r)S n (1) = 0
∂r 2 r ∂r r 2 ∂z 2
where S n (1) is the sine transform of 1, and the boundary conditions for u(r,θ, z)on θ have
been used.
Note that the operator on in the r coordinate direction is

1 d d n 2

2
(r,µ j ) = r − =−µ
j
r dr dr r 2
With the boundary condition at r = 1 chosen as (1,µ j ) = 0 this gives the kernel function as
= rJ n (r,µ j ) with eigenvalues determined by J n (1,µ j ) = 0
We now apply the finite Hankel transform to the above partial differential equation and
denote the Hankel transform of U n by U jn .
After applying the boundary condition on r we find, after noting that

N β [U n (z, 1)] = S n (1)

N [ (1, z)] =−µ j J n+1 (µ j )

β
2
2 d U jn
−µ U jn + µ j J n+1 (µ j )S n (1) + 2 + Q j (µ j )S n (1) = 0. Here Q j (µ j ) is the Hankel trans-
j dz
form of q(r).
Solving the resulting ordinary differential equation and applying the boundary condition
at z = 0,

Q j (µ j ) + µ j J n+1 (µ j )
U jn (µ j , n, z) = S n (1) [1 − exp(−µ j z)]
µ 2 j

We now invert the transform for the sine and Hankel transforms according to Eq. (9.19)
and find that

∞
∞
4 U jn (µ j , n, z)
u(r,θ, z) = J n (µ j r)sin(nθ)
π [J n+1 (µ j )] 2
n=1 j=1
Note that

n
S n (1) = [1 − (−1) ]/n