Compatible finite elements, constraints and coercive error control — Examples

Let u_0 be zero everywhere. Define a second measurable representative u_1 to equal t_e on e and zero on Omega minus e. A line segment has three-dimensional Lebesgue measure zero, so u_0 and u_1 represent the same zero L^2 class. Their distributional curls are both the zero distribution by the integral definition in [F1]; hence they represent the same element of H(curl;Omega).

Using the displayed representatives, the oriented edge integral of u_0 is zero, while that of u_1 is integral_e |t_e|^2 ds=length(e)>0. Therefore this integral is not a function of the H(curl) equivalence class, which refutes the universal raw edge-sampling claim. This does not affect the polynomial Whitney moments in [F2].

The counterexample concerns only unregularized line restriction. Under the distinct hypotheses in [F3], a smoothed relative commuting projection is available; its construction is not an edge-by-edge evaluation of the arbitrary H(curl) representative. Thus it supplies no missing edge trace to the refuted map and is consistent with steps 1.1-2.1.

Put k^2=k_mn^2. Since -Delta_t f_mn=k^2 f_mn and u=E_* f_mn e_z, direct coordinate differentiation gives curl u=E_(partial_y f_mn,-partial_x f_mn,0) and curl curl u=E_(0,0,-Delta_t f_mn)=k^2u. The field is nonzero because f_mn(a/(2m),b/(2n))=1 and E_*>0. The end-wall tangential trace is zero because u is normal to z=0,d; on x=0,a and y=0,b the whole field vanishes because the corresponding sine factor is zero. By [F1], B_z=0 for this TM mode, so n dot B=0 on the end walls; [F1] gives the same condition on the four side walls. Thus the field satisfies the PEC electric and magnetic wall conditions.

To verify the closure-defined condition in [F2], choose smooth one-dimensional cutoffs chi_delta^x,chi_delta^y,chi_delta^z that vanish within distance delta of their interval endpoints, equal one at distance at least 2delta, take values in [0,1], and satisfy |(chi_delta^.)'|<=C/delta. Put chi_delta=chi_delta^x chi_delta^y chi_delta^z and u_delta=chi_delta u; then u_delta is compactly supported and smooth in Omega. The L^2 norm of u_delta-u tends to zero because its support lies in boundary layers of volume O(delta). Also curl(u_delta-u)=(chi_delta-1)curl u+grad(chi_delta) x u. The first term tends to zero in L^2 by the same shrinking-support bound. In the second term, the z-cutoff derivative is parallel to u and has zero cross product; on the x- and y-boundary layers, |f_mn|<=C delta by the vanishing sine factor, so each remaining product |(chi_delta^.)' f_mn| is uniformly bounded on a set of volume O(delta), and its L^2 norm tends to zero. Hence u_delta->u in the H(curl) graph norm and u lies in H_0(curl;Omega).

For every v in C_c^\infty(Omega;R3), the distributional-curl definition and the smoothness of u give integral(curl u dot curl v)=integral(curl curl u dot v)=k^2 integral(u dot v). The identity extends to every v in H_0(curl;Omega) because C_c^\infty is graph-dense there and both pairings are continuous in that norm. Since omega_mn^2 epsilon=k^2/mu, the real-frequency form satisfies ${\int }_{\mathrm{\Omega }}({\mu }^{-1}\mathrm{curl}u\cdot \mathrm{curl}v-{\omega }_{mn}^2\epsilon u\cdot v)dx=0(v\in H_0(\mathrm{curl};\mathrm{\Omega }))$. Taking v=u gives a_omega(u,u)=0. Since u is nonzero, ||u||_{V,ell_0}>0 for every ell_0>0, so the form is not coercive and the zero-forcing weak problem has both the zero solution and a nonzero solution. This is an actual real-frequency cavity resonance and does not contradict the positive-shift theorem, which assumes real s>0 and a plus sign.