The proof hinges on the fact that for f g v0 lf g

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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 L= d2 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 f...
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