The source-free Maxwell equations in vacuum areTheir real and imaginary parts are exactlyPut $\theta =\omega t-K\cdot x$. Differentiating gives $\nabla \cdot W=(\Omega \cdot K)\sin \theta$ and $\nabla \times W=-(\Omega \times K)\sin \theta$, while $(i/c){\partial }_{t}W=-i\omega \Omega \sin \theta /c$. Thus the necessary and sufficient conditions areHere the complex dot product is bilinear, without complex conjugation. Taking the dot product of the second equation with $\Omega$ gives $\Omega \cdot \Omega =0$. Crossing it with $K$ givesso, since $\Omega \ne 0$, $|K{|}^2={\omega }^2/c^2$. Writing $\Omega =e+icb$, the real conditions areBoth physical fields are linearly polarized: $E=e\cos \theta$ and $B=b\cos \theta$ each oscillate along a fixed real direction. Combining two fields as $E+icB$ does not turn the electric field into a circularly polarized field.

For the regular Sturm-Liouville setting assume $p$ is positive and continuously differentiable, $q,w$ are continuous, and $w>0$. Apply a multiplier to $J[y]-\lambda ({\int }_a^{b}w{y}^2-1)$. Its first variation, divided by two, isEndpoint-vanishing variations and the allowed fundamental lemma giveIf an endpoint value is free, its variation is arbitrary, giving the natural condition $p{y}^{\prime }=0$ at that endpoint. At a prescribed endpoint value the variation vanishes there, so no additional natural condition is imposed.

For eigenfunctions $u,v$ with different eigenvalues, multiply their equations by $v,u$ and subtract after integration. The resulting boundary expression $[p(u{v}^{\prime }-u^{\prime }v){]}_a^b$ vanishes under common homogeneous Dirichlet or common natural boundary conditions. Hence $({\lambda }_u-{\lambda }_v){\int }_a^{b}wuv=0$, proving weighted orthogonality.

For the final Dirichlet problem the normalized stationary modes areSolving $y^{\prime \prime }+(\lambda +1)y=0$ with both endpoint values zero gives these and excludes nonpositive $\lambda +1$. Distinct sine modes have zero integral product on $[0,1]$. Their Dirichlet values also make the same boundary expression vanish, which explains orthogonality even though their endpoint derivatives do not obey the natural conditions.

For global minimality, the Fourier sine expansion of an endpoint-vanishing function and Parseval's identities give $\int y^{\prime 2}=\sum k^2{\pi }^2{a}_k^2\ge {\pi }^2\sum a_k^2={\pi }^2\int y^2$. Equality holds exactly in the first sine eigenspace. ThereforeDistinct modes are orthogonal; the two sign choices of the same minimum are not orthogonal to each other. This distinguishes the family of stationary modes from the two representatives of the minimizing eigenspace in the question's wording.