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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 &gt; 0. In this case, the diﬀerential equation f...
View Full Document

Ask a homework question - tutors are online