Maxwell eigenproblems and compatible vector wave approximation

Let $n_1<n_2<\cdots$ and let ${\mu }_j>0$ be eigenvalues of $T_{n_j}$ converging to $\nu >0$, and choose unit eigenvectors $x_j$ with $T_{n_j}{x}_j={\mu }_j{x}_j$. Sequential collective compactness gives a subsequence for which $T_{n_j}{x}_j\to y$. Since ${\mu }_j\to \nu >0$, $x_j\to x:=y\mathrm{/}\nu$ and $\mathrm{\parallel }x\mathrm{\parallel }=1$. The uniform operator bound gives $\mathrm{\parallel }{T}_{n_j}(x_j-x)\mathrm{\parallel }\le ({\sup }_n\mathrm{\parallel }{T}_n\mathrm{\parallel })\mathrm{\parallel }{x}_j-x\mathrm{\parallel }\to 0$; strong convergence gives $T_{n_j}x\to Tx$. Hence $Tx=y=\nu x$, so $\nu$ is an eigenvalue of $T$.

By [F1], $E:=\ker (T-\mu I)$ is finite-dimensional; write $m=\dim E$. Choose $\delta >0$ so that $I=(\mu -\delta ,\mu +\delta )$ has $\mu -\delta >0$ and $\bar{I}\cap \sigma (T)=\{\mu \}$. Let $E_n$ be the orthogonal sum of the eigenspaces of $T_n$ whose eigenvalues lie in $I$. This sum is finite-dimensional because [F1] permits only finitely many eigenvalues outside a neighborhood of zero, and $I$ stays away from zero.

If $\dim E_n<m$ for infinitely many $n$, the orthogonal projection $P_n:H\to E_n$ restricts to a linear map from $E$ to a space of dimension less than $m$. By [F2] choose a unit $x_n\in E$ with $P_n{x}_n=0$. The compact self-adjoint spectral expansion of $T_n$ then gives $\mathrm{\parallel }(T_n-\mu I)x_n\mathrm{\parallel }\ge \delta$, since every spectral value in the orthogonal complement of $E_n$ is at distance at least $\delta$ from $\mu$. Strong convergence is uniform on the finite-dimensional unit sphere of $E$: for an orthonormal basis $e_1,\dots ,e_m$, write $x={\sum }_i{c}_i{e}_i$ and bound $\mathrm{\parallel }(T_n-T)x\mathrm{\parallel }\le {\sum }_i\mathrm{\mid }{c}_i\mathrm{\mid }\mathrm{\parallel }(T_n-T)e_i\mathrm{\parallel }$. Since $T{x}_n=\mu x_n$, this uniform convergence contradicts the lower bound. Therefore $\dim E_n\ge m$ for all sufficiently large $n$.

If $\dim E_n\ge m+1$ for infinitely many $n$, select $m+1$ orthonormal eigenvectors $x_{n,1},\dots ,x_{n,m+1}$ in $E_n$. Apply [F4] successively to these finitely many eigenvalue sequences to obtain one subsequence on which all converge in $\bar{I}$. By step 1.1 every such limit is in $\sigma (T)$, hence equals $\mu$. Sequential collective compactness, applied successively to the finite list of sequences $T_n{x}_{n,i}$, gives a subsequence on which all these images converge. Their eigenvalues stay bounded away from zero, so all $x_{n,i}$ converge in norm. Their limits remain orthonormal and, by step 1.1, lie in $E$, contradicting $\dim E=m$. Thus $\dim E_n\le m$ for all sufficiently large $n$, and together with step 2.1 the dimension equals $m$.

The eigenvalues of $T_n$ in $I$ all converge to $\mu$: otherwise a subsequence stays a positive distance from $\mu$ and has a convergent subsequence in $\bar{I}$ by [F4]; step 1.1 makes its limit a spectral point of $T$, while $\bar{I}\cap \sigma (T)=\{\mu \}$. Let $E_n$ continue to denote the $m$-dimensional cluster subspace. If unit vectors $x_n\in E_n$ had $\mathrm{dist}(x_n,E)\ge \epsilon >0$, the convergence just proved would give $\mathrm{\parallel }(T_n-\mu I)x_n\mathrm{\parallel }\to 0$. Collective compactness yields a subsequence with $T_n{x}_n$ convergent; then $x_n$ converges to a unit $x$ and the argument of step 1.1 gives $Tx=\mu x$, contradicting the distance bound. Thus ${\sup }_{x\in E_n,\mathrm{\parallel }x\mathrm{\parallel }=1}\mathrm{dist}(x,E)\to 0$.

For $x\in E$, let $P_n$ be the orthogonal projection onto $E_n$. On $E_n^{\perp }$, the spectrum of $T_n$ stays at distance at least $\delta$ from $\mu$. The spectral expansion therefore gives $\mathrm{\parallel }x-P_{n}x\mathrm{\parallel }\le {\delta }^{-1}\mathrm{\parallel }(T_n-\mu I)x\mathrm{\parallel }$. Strong convergence of $T_n$ to $T$ is uniform on the finite-dimensional space $E$, and $Tx=\mu x$ there, so the right side tends uniformly to zero for unit $x\in E$. Hence ${\sup }_{x\in E,\mathrm{\parallel }x\mathrm{\parallel }=1}\mathrm{dist}(x,E_n)\to 0$. Combined with the directed-gap limit proved in step 4.1, this is symmetric-gap convergence. The multiplicity and no-pollution conclusions follow.

The graph of distributional curl is closed in $L^2(\mathrm{\Omega };{\mathbb{R}}^3)\times L^2(\mathrm{\Omega };{\mathbb{R}}^3)$: if $v_j\to v$ and $\mathrm{curl}{v}_j\to w$ in $L^2$, [F3] gives ${\int }_{\mathrm{\Omega }}w\cdot \phi dx={\int }_{\mathrm{\Omega }}v\cdot \mathrm{curl}\phi dx$ for each test $\phi$, so $\mathrm{curl}v=w$. By [F1] the product is Hilbert, hence its closed graph is Hilbert with the graph inner product. Thus $H(\mathrm{curl};\mathrm{\Omega })$ is a Hilbert space.

The curls are bounded in the separable Hilbert space of [F1]. By [F2] choose a subsequence, still indexed by $n$, with $\mathrm{curl}{u}_n\rightharpoonup w$ weakly in $L^2$. For every test $\phi$, strong convergence of $u_n$ and [F3] give ${\int }_{\mathrm{\Omega }}w\cdot \phi dx={\lim }_n{\int }_{\mathrm{\Omega }}\mathrm{curl}{u}_n\cdot \phi dx={\lim }_n{\int }_{\mathrm{\Omega }}{u}_n\cdot \mathrm{curl}\phi dx={\int }_{\mathrm{\Omega }}u\cdot \mathrm{curl}\phi dx$. Therefore $\mathrm{curl}u=w$ distributionally and $u\in H(\mathrm{curl})$.

The strong convergence of $u_n$ and weak convergence of their curls show that $u_n\rightharpoonup u$ in the graph Hilbert space: for every $v\in H(\mathrm{curl})$, $⟨u_n,v{⟩}_{H(\mathrm{curl})}=(u_n,v{)}_{L^2}+(\mathrm{curl}{u}_n,\mathrm{curl}v{)}_{L^2}⟶(u,v{)}_{L^2}+(w,\mathrm{curl}v{)}_{L^2}=⟨u,v{⟩}_{H(\mathrm{curl})}$. For every $y\in H_0(\mathrm{curl}{)}^{\perp }$, each $⟨u_n,y{⟩}_{H(\mathrm{curl})}=0$; weak convergence gives $⟨u,y{⟩}_{H(\mathrm{curl})}=0$. By [F4], $u\in (H_0(\mathrm{curl}{)}^{\perp }{)}^{\perp }=H_0(\mathrm{curl})$.

Set $v=u$ in [F1]. Then ${\mu }^{-1}\mathrm{\parallel }\mathrm{curl}u{\mathrm{\parallel }}_{L^2}^2=0$. Since $\mu >0$, the squared norm vanishes and $\mathrm{curl}u=0$, so $u\in K$ by [F1].

Conversely, if $u\in K$, then [F1] gives $\mathrm{curl}u=0$. Hence ${\mu }^{-1}(\mathrm{curl}u,\mathrm{curl}v{)}_{L^2}=0$ for every $v\in V$, which is precisely the unquotiented eigenrelation at $\lambda =0$. The two inclusions identify the zero eigenspace with $K$.

For $j\ge 1$, set $c_j=x_0+r(1-2^{-j})e_1$ and $r_j=r{2}^{-j-4}$. The balls $B_j=B_{r_j}(c_j)$ lie in $B_r(x_0)$: $\mathrm{\mid }{c}_j-x_0\mathrm{\mid }+r_j=r(1-15\cdot 2^{-j-4})<r$. If $l>j$, $\mathrm{\mid }{c}_l-c_j\mathrm{\mid }=r(2^{-j}-2^{-l})\ge r{2}^{-j-1}$, while $r_j+r_l\le 3r{2}^{-j-5}<r{2}^{-j-1}$, so these balls are pairwise disjoint. By [F2], choose $\rho \in C_c^{\mathrm{\infty }}(B_1(0))$ equal to one near the origin. Set $\psi (\xi )={\xi }_1\rho (\xi )$, so ${\mathrm{\partial }}_1\psi (0)=1$, and put ${\psi }_j(x)=\psi ((x-c_j)\mathrm{/}{r}_j)$ and $A_j=(0,0,{\psi }_j)$. Then $v_j:=\mathrm{curl}{A}_j=({\mathrm{\partial }}_2{\psi }_j,-{\mathrm{\partial }}_1{\psi }_j,0)$ is a nonzero smooth vector field supported in $B_j$.

Each $v_j$ has compact support in $\mathrm{\Omega }$, hence $v_j\in V$. For every $k\in K$, [F3] gives $(v_j,k{)}_{L^2}={\int }_{\mathrm{\Omega }}k\cdot \mathrm{curl}{A}_{j}dx={\int }_{\mathrm{\Omega }}\mathrm{curl}k\cdot A_{j}dx=0$. Thus $v_j\in X$. Pairwise disjoint supports and $v_j\mathrm{\ne }0$ imply linear independence: restricting any finite vanishing linear combination to $B_j$ forces its $j$th coefficient to be zero. The gauge space therefore contains infinitely many linearly independent vectors and is infinite-dimensional.

On each cell $K$, [F2] defines the local interpolant $I_{K}w$ by its six edge moments. To verify constants belong to the local space, write ${\varphi }_{ij}={\lambda }_i\mathrm{\nabla }{\lambda }_j-{\lambda }_j\mathrm{\nabla }{\lambda }_i$ with ${\varphi }_{ji}=-{\varphi }_{ij}$. Then ${\sum }_{j\mathrm{\ne }i}{\varphi }_{ij}=-\mathrm{\nabla }{\lambda }_i$ because ${\sum }_j{\lambda }_j=1$. For each coordinate $\mathrm{\ell }$, the barycentric identity $x_{\mathrm{\ell }}={\sum }_i(x_i{)}_{\mathrm{\ell }}{\lambda }_i$ gives $e_{\mathrm{\ell }}=\mathrm{\nabla }{x}_{\mathrm{\ell }}={\sum }_i(x_i{)}_{\mathrm{\ell }}\mathrm{\nabla }{\lambda }_i$, so every constant vector lies in the local span. Unisolvence [F2] then makes $I_K$ reproduce constants. Fix $x_K\in K$ and put $r_K(x)=w(x)-w(x_K)$. Then $\mathrm{\parallel }{r}_K{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}\le h_K\mathrm{\parallel }Dw{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$, so every edge moment of $r_K$ has absolute value at most $h_K^2\mathrm{\parallel }Dw{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$. By [F1], summing the six moment coefficients against their basis functions gives $\mathrm{\parallel }{I}_K{r}_K{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }Dw{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$ and $\mathrm{\parallel }{r}_K{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }Dw{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$. Consequently $\mathrm{\parallel }{I}_{K}w-w{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }Dw{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$.

By [F3] the local fields assemble to a global $H(\mathrm{curl})$ field. If $w=0$, then $I_{h}w=0\in H_0(\mathrm{curl};\mathrm{\Omega })$. Otherwise let $d=\mathrm{dist}(\mathrm{supp}w,\mathrm{\partial }\mathrm{\Omega })>0$ and take $h<d\mathrm{/}2$. Any cell with a nonzero edge moment lies within distance $h$ of $\mathrm{supp}w$, so the assembled interpolant has compact support in $\mathrm{\Omega }$. Its zero extension $U$ is in $H(\mathrm{curl};{\mathbb{R}}^3)$: cellwise integration by parts cancels the interior face terms by [F3], and there is no boundary jump because $U$ vanishes near $\mathrm{\partial }\mathrm{\Omega }$. Choose $\rho$ as in [F5] with $r=1\mathrm{/}2$ and $R=1$, and put $c=({\int }_{{\mathbb{R}}^3}\rho {)}^{-1}$. Then $0<c<\mathrm{\infty }$ by [F6]. For $\eta >0$, set ${\kappa }_{\eta }(x)=c{\eta }^{-3}\rho (x\mathrm{/}\eta )$. This family is nonnegative, has unit integral and $L^1$ norm one, and is supported in $B_{\eta }(0)$; hence it is an $L^1$ approximate identity by An $L^1$ approximate identity on ${\mathbb{R}}^n$. Take ${\eta }_j=(d-h)\mathrm{/}(j+2)$, so ${\eta }_j<d-h$ and ${\eta }_j\to 0$. By [F8], ${\kappa }_{{\eta }_j}\ast U$ is smooth. The support of $U$ lies within distance $h$ of $\mathrm{supp}w$, so $\mathrm{dist}(\mathrm{supp}U,\mathrm{\partial }\mathrm{\Omega })\ge d-h$. Since $\mathrm{supp}{\kappa }_{{\eta }_j}\subset B_{{\eta }_j}(0)$, the convolution vanishes when $\mathrm{dist}(x,\mathrm{supp}U)>{\eta }_j$; its support is therefore in the closed ${\eta }_j$-neighborhood of $\mathrm{supp}U$, which is compactly contained in $\mathrm{\Omega }$ because ${\eta }_j<d-h$. Thus it is in $C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^3)$. By [F7], its curl is ${\kappa }_{{\eta }_j}\ast (\mathrm{curl}U)$. The approximate-identity convergence [F4] applied to both $U$ and $\mathrm{curl}U$ gives convergence in the $H(\mathrm{curl})$ graph norm. Thus $I_{h}w\in H_0(\mathrm{curl};\mathrm{\Omega })$ by the closure definition.

By [F1], $\mathrm{curl}{I}_{K}w$ is a constant vector $c_K$ on $K$. For each oriented face $F\subset \mathrm{\partial }K$, [F3] and the defining edge moments give ${\int }_F{c}_K\cdot n_{F}dS={\int }_F\mathrm{curl}w\cdot n_{F}dS$. With $c_0=\mathrm{curl}w(x_K)$, smoothness bounds the difference between the right side and $\mathrm{\mid }F\mathrm{\mid }{c}_0\cdot n_F$ by $\mathrm{\mid }F\mathrm{\mid }{h}_K\mathrm{\parallel }D(\mathrm{curl}w){\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$. Three face normals span ${\mathbb{R}}^3$. More precisely, choose three reference faces whose unit normals form a basis. Their physical normals are normalized cofactor images under $B_K^{-T}$; shape regularity bounds the condition number of $B_K\mathrm{/}{h}_K$, so the inverse of this three-normal matrix is uniformly bounded. Hence $\mathrm{\mid }{c}_K-c_0\mathrm{\mid }\le C{h}_K\mathrm{\parallel }D(\mathrm{curl}w){\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$ and $\mathrm{\parallel }\mathrm{curl}{I}_{K}w-\mathrm{curl}w{\mathrm{\parallel }}_{L^2(K)}\le C{h}_K^{5\mathrm{/}2}\mathrm{\parallel }D(\mathrm{curl}w){\mathrm{\parallel }}_{L^{\mathrm{\infty }}(K)}$.

Shape regularity gives $\mathrm{\mid }K\mathrm{\mid }\ge c{h}_K^3$, so ${\sum }_K{h}_K^5\le h^2{\sum }_K{h}_K^3\le C{h}^2\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }$. Squaring the local bounds in steps 1.1 and 2.1, summing over the mesh, and taking square roots gives $\mathrm{\parallel }{I}_{h}w-w{\mathrm{\parallel }}_{H(\mathrm{curl};\mathrm{\Omega })}\le C_{w}h⟶0$. The constant depends on the fixed smooth field, the domain and the mesh shape-regularity bound, but not on $h$.

By [F1] and Countable Choice [A2], choose for each $n$ $z_n\in H_0^1(\mathrm{\Omega };{\mathbb{R}}^3)$ and $p_n\in H^1(\mathrm{\Omega })$ with $u_n=z_n+\mathrm{\nabla }{p}_n$ and $\mathrm{\parallel }{z}_n{\mathrm{\parallel }}_{H^1}\le C_{\mathrm{\Omega }}\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{H(\mathrm{curl})}$. Hence $(z_n)$ is bounded in $W_0^{1,2}(\mathrm{\Omega };{\mathbb{R}}^3)$.

Set $g_n:=\mathrm{\nabla }{p}_n=u_n-z_n$. By [F3], $z_n\in H_0(\mathrm{curl})$; since $u_n\in H_0(\mathrm{curl})$, their difference is in that space. Its distributional curl is zero: for every compactly supported smooth vector test $\phi$, integration by parts in the weak-derivative definition gives $⟨\mathrm{curl}\mathrm{\nabla }{p}_n,\phi ⟩=-{\int }_{\mathrm{\Omega }}{p}_n\mathrm{div}(\mathrm{curl}\phi )dx=0$, because $\mathrm{div}(\mathrm{curl}\phi )=0$. Thus $g_n\in K$ for the kernel $K$ defining $X$.

Apply [F2] componentwise and take the finite successive subsequence to obtain indices $n_j$ for which $z_{n_j}$ converges strongly, and therefore is Cauchy, in $L^2(\mathrm{\Omega };{\mathbb{R}}^3)$. The finite product of the three component $L^2$ norms is the vector $L^2$ norm.

For any $j,k$, $u_{n_j}-u_{n_k}\in X=K^{\perp }$ and $g_{n_j}-g_{n_k}\in K$, so their $L^2$ inner product is zero. Since $u_{n_j}-u_{n_k}=(z_{n_j}-z_{n_k})+(g_{n_j}-g_{n_k})$, it follows that $\mathrm{\parallel }{u}_{n_j}-u_{n_k}{\mathrm{\parallel }}_2^2=⟨u_{n_j}-u_{n_k},z_{n_j}-z_{n_k}{⟩}_{L^2}$. Cauchy–Schwarz implies $\mathrm{\parallel }{u}_{n_j}-u_{n_k}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{z}_{n_j}-z_{n_k}{\mathrm{\parallel }}_2$ (with the zero case immediate). Step 2.1 therefore makes $(u_{n_j})$ Cauchy in $L^2$; [F4] gives its strong $L^2$ limit. This proves compactness.

If the curl Poincare inequality failed, Countable Choice [A2] selects for each $n$ $v_n\in X$ with $\mathrm{\parallel }{v}_n{\mathrm{\parallel }}_2=1$ and $\mathrm{\parallel }\mathrm{curl}{v}_n{\mathrm{\parallel }}_2<1\mathrm{/}n$. The graph norms are bounded, so compactness proved in steps 1.1–3.1 gives a subsequence converging strongly in $L^2$ to some $v$. By [F5], after a further subsequence $v\in H_0(\mathrm{curl})$ and $\mathrm{curl}v=0$. For each $k\in K$, strong convergence and $v_n\perp K$ give $(v,k{)}_{L^2}=0$, so $v\in X\cap K=\{0\}$. But strong convergence preserves $\mathrm{\parallel }v{\mathrm{\parallel }}_2=1$, a contradiction. Therefore a uniform constant $C_{\mathrm{\Omega }}$ exists.

The estimate in step 4.1 makes $\mathrm{\parallel }u{\mathrm{\parallel }}_{H(\mathrm{curl})}^2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\mathrm{curl}u{\mathrm{\parallel }}_2^2$ bounded by a constant times $\mathrm{\parallel }\mathrm{curl}u{\mathrm{\parallel }}_2^2$ on $X$. Hence the curl form is coercive in the graph norm.

By [F1], for every $j\ge 1$ there is ${\phi }_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^3)$ with $\mathrm{\parallel }u-{\phi }_j{\mathrm{\parallel }}_{H(\mathrm{curl})}<1\mathrm{/}j$. Countable Choice selects one such field for each $j$. By [F2], for each fixed $j$ there is an index $N_j$ such that for every $n\ge N_j$, $I_{h_n}{\phi }_j\in V_{h_n}$ and $\mathrm{\parallel }{I}_{h_n}{\phi }_j-{\phi }_j{\mathrm{\parallel }}_{H(\mathrm{curl})}<1\mathrm{/}j$; choose the least such $N_j$, and replace the sequence by increasing thresholds $M_j:=\max \{j,N_1,\dots ,N_j,M_{j-1}+1\}$, with $M_0=0$.

Each $K_{h_n}$ is a finite-dimensional subspace of the real $L^2$ Hilbert space. By Countable Choice [A1], choose an orthonormal basis for each $K_{h_n}$ using [F3]. If a sequence in $K_{h_n}$ converges in $L^2$, its coordinates against this finite basis converge, and the limit is the corresponding finite linear combination; thus $K_{h_n}$ is closed. By [F3] it has an orthogonal projection $P_n$ satisfying $\mathrm{\parallel }{P}_{n}y{\mathrm{\parallel }}_2\le \mathrm{\parallel }y{\mathrm{\parallel }}_2$, including $P_n=0$ when $K_{h_n}=\{0\}$.