An explicit PEC standing wave in the unit cube

Put $f(x,y)=\sin (\pi x)\sin (\pi y)$. On the faces $x=0,1$ and $y=0,1$, $E=0$, so its tangential component vanishes there; on $z=0,1$, $E$ is normal. To verify the closure convention in [F2], choose a smooth one-variable cutoff ${\theta }_{\delta }$ which is zero within distance $\delta$ of $0,1$, one beyond distance $2\delta$, and has derivative bounded by $C\mathrm{/}\delta$. Let ${\chi }_{\delta }(x,y,z)={\theta }_{\delta }(x){\theta }_{\delta }(y){\theta }_{\delta }(z)$. Then $E_{\delta }={\chi }_{\delta }E\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^3)$ and $\mathrm{curl}(E_{\delta }-E)=\mathrm{\nabla }{\chi }_{\delta }\times E+({\chi }_{\delta }-1)\mathrm{curl}E$. The $z$-derivative term in the cross product is zero because it is parallel to $E$; on an $x$- or $y$-boundary layer, the corresponding sine factor is $O(\delta )$ and cancels the $O({\delta }^{-1})$ cutoff derivative. The remaining terms are bounded and supported in a layer of volume $O(\delta )$, so their $L^2$ norms tend to zero. Also $E_{\delta }\to E$ in $L^2$. Thus $E\in H_0(\mathrm{curl})$.

Direct differentiation gives $\mathrm{div}E=0$, $\mathrm{curl}E=(\pi \sin (\pi x)\cos (\pi y),-\pi \cos (\pi x)\sin (\pi y),0)$, and $\mathrm{curl}\mathrm{curl}E=2{\pi }^{2}E$. For every compactly supported smooth test field $v$, distributional integration by parts yields $(\mathrm{curl}E,\mathrm{curl}v{)}_{L^2}=(\mathrm{curl}\mathrm{curl}E,v{)}_{L^2}$. By [F2] and continuity of both pairings, the identity extends to all $v\in H_0(\mathrm{curl})$. Hence [F1] holds with $\lambda =2{\pi }^2\mathrm{/}(ϵ\mu )$. Testing that identity with any $k\in K$ gives $0=\lambda (E,k{)}_{ϵ}$, so $E\perp K$ and $E\in X$.

The eigenvalue is positive and the field is nonzero. By [F3], its full continuous eigenspace cluster is approximated with its actual multiplicity; the argument makes no selection of an individual limiting discrete eigenvector.