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Unformatted text preview: neous initial
condition u(r, θ, 0) = h(r, θ) is also satisﬁed. If we set t = 0 in (5.33), this
∞ ∞ ∞ a0,k f0,k +
k=1 [an,k fn,k + bn,k gn,k ] = h. (5.34) n=1 k=1 Thus we need to express an arbitrary initial displacement h(r, θ) as a superposition of these eigenfunctions.
It is here that the inner product , on V comes to the rescue. The lemma
implies that eigenfunctions corresponding to distinct eigenvalues are perpendicular. This implies, for example, that
f0,j , f0,k = 0,
152 unless j = k. Similarly, for arbitrary n ≥ 1,
fn,j , fn,k = fn,j , gn,k = gn,j , gn,k = 0, unless j = k. The lemma does not immediately imply that
fn,j , gn,j = 0.
To obtain this relation, recall that in terms of polar coordinates, the inner
product , on V takes the form
1 2π f, g = f (r, θ)g (r, θ)rdθ dr.
0 0 If we perform integration with respect to θ ﬁrst, we can conclude from the
familiar integral formula
2π sin nθ cos nθdθ = 0
fn,j , gn,j = 0,
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This document was uploaded on 01/12/2014.
- Winter '14