Angular Eigenvalue problem with Periodic Boundary Conditions
This page derives the allowed angular modes that arise from imposing periodicity in \( \theta \)
$$ g''(\theta) = \lambda g(\theta) \quad \text{with} \quad g(\theta) = g(\theta + 2\pi) $$Assume a solution of the form \( g(\theta) = e^{k\theta} \)
$$ k^2 e^{k \theta} = \lambda e^{k \theta} \quad \Rightarrow \quad k^2 = \lambda $$i) \( \lambda > 0 \)
$$ k = \pm \sqrt{\lambda} $$ $$ g(\theta) = c_1 e^{\sqrt{\lambda}\theta} + c_2 e^{-\sqrt{\lambda}\theta} $$ $$ g(\theta + 2\pi) = c_1 e^{\sqrt{\lambda}\theta}e^{2\sqrt{\lambda}\pi} + c_2 e^{-\sqrt{\lambda}\theta}e^{-2\sqrt{\lambda}\pi} $$Therefore, for \( \lambda > 0 \) \( \: g(\theta) \: \) is not periodic, since \( e^{\pm 2\sqrt{\lambda}\pi} \neq 1 \)
ii) \( \lambda = 0 \)
$$ k = 0 \quad \Rightarrow \quad g''(\theta) = 0 $$ $$ \therefore \, g(\theta) = c_1 \theta + c_2 $$ $$ g(\theta + 2\pi) = c_1(\theta + 2\pi) + c_2 $$This is periodic \(\Leftrightarrow \quad c_1 = 0\)
$$ g(\theta) = c_2 $$So \( \lambda = 0 \) corresponds to a constant angular solution.
iii) \( \lambda < 0 \)
$$ k = \pm i\sqrt{-\lambda} \quad |\lambda| = - \lambda $$Let \( \lambda = -b^2 \) where \( b \ge 0 \)
$$ g(\theta) = c_1 e^{i b \theta} + c_2 e^{-i b \theta} $$ $$ g(\theta + 2\pi) = c_1 e^{i b \theta}e^{2i b \pi} + c_2 e^{-i b \theta}e^{-2i b \pi} $$Therefore \( \, g(\theta) = g(\theta + 2\pi) \quad \Leftrightarrow \quad e^{\pm 2i b \pi} = 1 \)
$$ \Rightarrow \quad 2 b \pi = 2\pi n, \quad n = 0, 1, 2, ... $$ $$ b = n, \quad \lambda_n = -b^2 = -n^2 $$Therefore
$$ g_n(\theta) = c_{1,n}e^{in\theta} + c_{2,n}e^{-in\theta} \quad n = 0, 1, 2, ... $$Or equivalently
$$ g_n(\theta) = c_n e^{in\theta} \quad n \in \mathbb{Z} $$Equivalently, using trig functions
$$ g(\theta) = c_1 \sin(b \theta) + c_2 \cos(b \theta) $$ $$ g(\theta + 2\pi) = c_1 \sin(b (\theta + 2\pi) ) + c_2 \cos(b (\theta + 2\pi) ) $$Therefore \( \, g(\theta) = g(\theta + 2\pi) \quad \Leftrightarrow \quad 2\pi b = 2\pi n \) where \( n = 0, 1, 2, ... \)
$$ \lambda_n = -n^2 $$ $$ g_n(\theta) = c_{1,n} \cos(n\theta) + c_{2,n} \sin(n\theta) $$Negative values of \( n \) do not produce new angular modes since \( \cos(-x) = \cos x \) and \( \sin(-x) = - \sin x \), with the minus sign absorbed into the constants.