Maxwell eigenproblems and compatible vector wave approximation — Examples

The function $-1\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ is smooth on the shell and $\mathrm{\nabla }(-1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })=x\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^3=u$. Therefore $\mathrm{curl}u=0$. Direct differentiation gives $\mathrm{div}u={\sum }_{i=1}^3(\mathrm{\mid }x{\mathrm{\mid }}^{-3}-3x_i^2\mathrm{\mid }x{\mathrm{\mid }}^{-5})=0$. For $0<\delta <1\mathrm{/}8$, choose a smooth radial cutoff ${\chi }_{\delta }(r)$ that vanishes near $r=1,2$ and equals one on $[1+2\delta ,2-2\delta ]$. The field $u_{\delta }(x)={\chi }_{\delta }(\mathrm{\mid }x\mathrm{\mid })u(x)$ is smooth and compactly supported in $\mathrm{\Omega }$. Both $\mathrm{\nabla }{\chi }_{\delta }$ and $u$ are radial, so $\mathrm{curl}{u}_{\delta }=0$. Since $\mathrm{\mid }u\mathrm{\mid }\le 1$ on the shell and $u-u_{\delta }$ is supported in a layer of volume $O(\delta )$, $\mathrm{\parallel }u-u_{\delta }{\mathrm{\parallel }}_{H(\mathrm{curl})}\to 0$. Thus [F1] gives $u\in H_0(\mathrm{curl})$ and $u\in K$.

If $q\in H_0^1(\mathrm{\Omega })$, choose $q_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ with $q_j\to q$ in $H^1$ by [F3]. For each $j$, integration by parts and $\mathrm{div}u=0$ give $(u,\mathrm{\nabla }{q}_j{)}_{L^2}=-(\mathrm{div}u,q_j{)}_{L^2}=0$. Passing to the limit shows $(u,\mathrm{\nabla }q{)}_{L^2}=0$. Hence this zero mode survives the zero-boundary gradient gauge.

Suppose for contradiction that $u=\mathrm{\nabla }q$ for some $q\in H_0^1(\mathrm{\Omega })$. For $1<r<2$ define the spherical mean $M(r):=\frac{1}{4\pi }{\int }_{S^2}q(r\omega )d\omega$. For $q_j$ from step 2.1, define $M_j$ by the same formula. Polar coordinates, Jensen's inequality and [F2] show $\mathrm{\parallel }{M}_j-M_k{\mathrm{\parallel }}_{H^1(1,2)}\le C\mathrm{\parallel }{q}_j-q_k{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$: the derivative is the spherical average of $\mathrm{\nabla }(q_j-q_k)(r\omega )\cdot \omega$, and $1\le r^2\le 4$. Thus $M_j\to M$ in $H^1(1,2)$. Each $M_j$ is smooth and compactly supported in $(1,2)$, so [F3] gives $M\in H_0^1(1,2)$. For a smooth $g$ on $[1,2]$, choose $s$ with $\mathrm{\mid }g(s)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^2(1,2)}$; then $\mathrm{\mid }g(r)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{L^2}+\mathrm{\parallel }{g}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2}\le \sqrt{2}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^1}$ for every $r$. Applying this estimate to $M_j-M_k$ makes $(M_j)$ uniformly Cauchy. Its continuous limit has value zero at both endpoints, and it represents $M$ because $M_j\to M$ in $L^2$. Thus $M(1)=M(2)=0$.

For the smooth $q_j$, differentiation under the spherical integral gives $M_j^{\mathrm{\prime }}(r)=(4\pi {)}^{-1}{\int }_{S^2}\mathrm{\nabla }{q}_j(r\omega )\cdot \omega d\omega$. Passing to the $H^1$ limit and using $\mathrm{\nabla }q=u$ yields $M^{\mathrm{\prime }}(r)=(4\pi {)}^{-1}{\int }_{S^2}{r}^{-2}d\omega =r^{-2}$ in distributions. Therefore $M(r)=c-r^{-1}$ for some constant $c$. The endpoint values from step 3.1 would force $c=1$ and $c=1\mathrm{/}2$ simultaneously, a contradiction. Thus $u\mathrm{\notin }\mathrm{\nabla }{H}_0^1(\mathrm{\Omega })$.

The field is a nonzero member of the full curl kernel by step 1.1, so it is excluded from $X=K^{\perp }$; by [F4] it is a zero eigenfield of the unquotiented pencil. Step 4.1 shows why the gradient-only gauge would fail on this shell. It is not a positive eigenvalue of the contractible-cavity approximation theorem.

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.

For positive integers $m,n$, set $E_{m,n}=(0,0,\sin (m\pi x)\sin (n\pi y))$. It is divergence-free and its tangential component vanishes on $x=0,1$ and $y=0,1$ because its third component vanishes there; on $z=0,1$ it is normal. A product cutoff ${\chi }_{\delta }(x,y,z)$ vanishing within distance $\delta$ of the cube faces, equal to one outside the $2\delta$ boundary layer, and with derivatives $O({\delta }^{-1})$ gives compactly supported smooth approximants ${\chi }_{\delta }{E}_{m,n}$. The $z$-derivative term crossed with $E_{m,n}$ is zero; on $x$ and $y$ layers the corresponding sine factor is $O(\delta )$, so $\mathrm{\parallel }\mathrm{\nabla }{\chi }_{\delta }\times E_{m,n}{\mathrm{\parallel }}_2=O(\sqrt{\delta })$. The remaining curl error is bounded on a layer of volume $O(\delta )$, hence tends to zero in $L^2$. Also $E_{m,n}-{\chi }_{\delta }{E}_{m,n}$ is bounded and supported in a layer of volume $O(\delta )$, so it tends to zero in $L^2$. Therefore $E_{m,n}\in H_0(\mathrm{curl})$.

Direct differentiation gives $\mathrm{curl}\mathrm{curl}{E}_{m,n}={\pi }^2(m^2+n^2)E_{m,n}$. Testing against compactly supported smooth fields and integrating by parts proves the weak eigenrelation; by the closure definition [F5] and continuity of both pairings it holds for every PEC test field. Testing with each $k$ in the curl kernel gives $0=\lambda (E_{m,n},k{)}_{ϵ}$, so the fields lie in the full-kernel gauge.

The eigenvalue is positive and isolated by [F3].

For $(m,n)=(1,2)$ and $(2,1)$ the eigenvalue is $\lambda ={\pi }^2(1^2+2^2)\mathrm{/}(ϵ\mu )=5{\pi }^2\mathrm{/}(ϵ\mu )$. Their mass inner product is zero because ${\int }_0^1\sin (\pi x)\sin (2\pi x)dx=0$; hence they are linearly independent and the eigenspace multiplicity is at least two.

Apply [F2] to the full eigenspace cluster: exactly its actual multiplicity of discrete eigenvalues converges to this value and the discrete eigenspaces converge in gap. The explicit two orthogonal modes show the multiplicity is at least two; no individual basis vector is selected by the theorem.