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Unformatted text preview: (θ) = λf (θ). (4.25) and The periodicity condition u(θ + π, t) = u(θ, t) becomes
f (θ + 2π )g (t) = f (θ)g (t),
and if g (t) is not identically zero, we must have
f (θ + 2π ) = f (θ).
Thus to ﬁnd the nontrivial solutions to the homogeneous linear part of the
problem requires us to ﬁnd the nontrivial solutions to the problem:
f (θ) = d2
(f (θ)) = λf (θ),
dθ2 f (θ + 2π ) = f (θ). (4.26) We will call (4.11) the eigenvalue problem for the diﬀerential operator
dθ2 acting on the space V of functions which are periodic of period 2π .
As before, we need to consider three cases.
Case 1: λ = 0. In this case, the eigenvalue problem (4.26) becomes
f (θ) = 0, f (θ + 2π ) = f (θ). The general solution to the diﬀerential equation is f (θ) = a + bθ, and
a + b(θ + 2π ) = a + b(θ) ⇒ b = 0.
Thus the only solution in this case is that where f is constant, and to be
consistent with our Fourier series conventions, we write f (θ) = a0 /2, where a0
is a constant.
Case 2: λ > 0. In this case, the diﬀerential equation
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- Winter '14