Elasticity_ Barber

384 24 Spherical harmonics
The attentive reader will recognize a similarity between the logarithmic
terms in this series and the functions φ −1 , φ −2 , . . . of equation (23.23), obtained
from the source solution by successive partial integrations. However,
the two sets of functions are not identical. In effect, the functions in (23.23)
correspond to distributions of sources along the negative z-axis, whereas those
in (24.34) involve distributions of sources along both the positive and negative
z-axes. The functions (23.23) are actually more useful, since they are
harmonic throughout the region z > 0 and hence can be applied to problems
of the half space with concentrated loading, as we discovered in §23.2.1.
The functions Q n (x) can all be written in the form
( ) 1 + x
Q n (x) = P n (x) ln − W n−1 (x) , (24.35)
1 − x
where
W n−1 (x) =
n∑
k=1
1
k P k−1(x)P n−k (x) (24.36)
is a finite polynomial of degree of degree (n−1). In other words, the multiplier
on the logarithmic term is the Legendre polynomial P n (x) of the same
order. This is also true for the functions of equation (23.23), as can be seen
by comparing them with equations (24.34). A convenient way to extend the
sequence of functions (23.23) is therefore to assume a solution of the form
φ −n−1 = R n P n (z/R) ln(R + z) + R n (R, z) , (24.37)
where R n is a general polynomial of degree n in R, z, substitute into the
Laplace equation
∇ 2 φ = ∂2 φ
∂r 2 + 1 ∂φ
r ∂r + 1 ∂ 2 φ
r 2 ∂θ 2 + ∂2 φ
∂z 2 = 0 (24.38)
and use the resulting equations to determine the coefficients in R n . This
method is used in the Maple and Mathematica files ‘sing’.
24.6.1 Logarithmic functions for cylinder problems
The function ln(R+z) corresponds to a uniform distribution of sources along
the negative z-axis. A similar distribution along the positive z-axis would lead
to the related harmonic function ln(R−z). Alternatively, we could add these
two functions, obtaining the solution for a uniform distribution of sources
along the entire z-axis (−∞<z <∞) with the result
ln(R + z) + ln(R − z) = ln(R 2 − z 2 ) = 2 ln(r) . (24.39)
Not surprisingly, this function is independent of z. It is actually the source
solution for the two-dimensional Laplace equation