# Example we consider the operator lx d dx 109 x d dx

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Unformatted text preview: (θ) = λf (θ), or f (θ) − λf (θ) = 0 104 has the general solution f (θ) = ae( √ λθ ) + be−( √ λθ ) . Note that a = 0 ⇒ f (θ) → ±∞ as θ → ∞, while b = 0 ⇒ f (θ) → ±∞ as θ → −∞. Neither of these is consistent with the periodicity conditions f (θ + 2π ) = f (θ), so we conclude that a = b = 0, and we obtain no nontrivial solutions in this case. √ Case 3: λ < 0. In this case, we set ω = −λ, and rewrite the eigenvalue problem as f (θ) + ω 2 f (θ) = 0, f (θ + 2π ) = f (θ). We recognize once again our old friend, the diﬀerential equation of simple harmonic motion, which has the general solution f (θ) = a cos(ωθ) + b sin(ωθ) = A sin(ω (θ − θ0 )). The periodicity condition f (θ + 2π ) = f (θ) implies that ω = n, where n is an integer, which we can assume is positive, and f (θ) = an cos(nθ) + bn sin(nθ). Thus we see that the general solution to the eigenvalue problem (4.26) is λ=0 and f (θ) = a0 , 2 or λ = −n2 where n is a positive integer and f (θ) = an cos(nθ) + bn sin(nθ). Now we need to ﬁnd the corres...
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## This document was uploaded on 01/12/2014.

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