Planar DN comparison under disk enclosure and conductivity reciprocation

First obtain the weak theory on $D_1$ from [F1]. The map $T:D_1\to T(D_1)$, $T(z)=(1\mathrm{/}2,1\mathrm{/}2)+z\mathrm{/}4$, has image the radius-$1\mathrm{/}4$ disk contained in $(0,1{)}^2$. For a conductivity $\gamma$ on $D_1$, define $\tilde{\gamma }(Tz)=\gamma (z)$; the ellipticity bounds for the extended coefficients are $m_{\ast }:=\min \{m,1\}$ and $M_{\ast }:=\max \{M,1\}$. In two dimensions, pulling functions back by $T$ preserves the Dirichlet energy, while $\frac{1}{4}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(D_1)}\le \mathrm{\parallel }v\circ T^{-1}{\mathrm{\parallel }}_{H^1(T(D_1))}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(D_1)}$. Applying [F1] on $T(D_1)$ and pulling its weak solution and energy form back therefore gives existence, uniqueness, representative independence and bounded DN maps on $D_1$, with fixed norm-equivalence constants. The Alessandrini identity and double-annihilation norm equality are invariant under this transport. Given a conductivity ${\gamma }_B$ on $B=(1\mathrm{/}2,1\mathrm{/}2)+D_1$, set ${\gamma }_1(z)={\gamma }_B((1\mathrm{/}2,1\mathrm{/}2)+z)$ on $D_1$ and apply the result just proved to ${\gamma }_1$. Translation $z\mapsto (1\mathrm{/}2,1\mathrm{/}2)+z$ is an $H^1$ isometry from $D_1$ to $B$ and carries the solution, energy form and quotient DN map back to those on $B$. Thus the weak theory and norm comparison used on $B$ follow from the two successive transports $T(D_1)\leftrightarrow D_1\leftrightarrow B$.

For trace classes $f,g$ on $B$, let $u_1,u_2$ solve the two conductivity equations there. Their restrictions to $\mathrm{\Omega }$ are the respective weak solutions for their own restricted traces. Since the coefficient difference is zero on $B\setminus \mathrm{\Omega }$, applying the Alessandrini identity on the two domains gives $(({\mathrm{\Lambda }}_{{\gamma }_1}^B-{\mathrm{\Lambda }}_{{\gamma }_2}^B)f)(g)=(({\mathrm{\Lambda }}_{{\gamma }_1}^{\mathrm{\Omega }}-{\mathrm{\Lambda }}_{{\gamma }_2}^{\mathrm{\Omega }})[u_1{\mathrm{\mid }}_{\mathrm{\Omega }}])([u_2{\mathrm{\mid }}_{\mathrm{\Omega }}])$. In the sesquilinear convention this is exactly the common integral of $({\gamma }_1-{\gamma }_2)\mathrm{\nabla }{u}_1\cdot \overline{\mathrm{\nabla }{u}_2}$. Its absolute value is at most ${\delta }_{\mathrm{\Omega }}\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{H^1(B)}\mathrm{\parallel }{u}_2{\mathrm{\parallel }}_{H^1(B)}\le C{\delta }_{\mathrm{\Omega }}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathcal{D}(B)}\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathcal{D}(B)}$, using step 1.1. Hence the full disk operator norm is at most $C{\delta }_{\mathrm{\Omega }}$, and its constant-quotient norm is equal to that full norm by [F1]. This proves the first estimate.

On $D_1$ let $E=\mathcal{D}\mathrm{/}\mathbb{C}\mathbf{1}$ and let $D$ be the weak counterclockwise tangential derivative on its boundary. This is a bounded bijection $E\to E^{\ast }$ with bounded inverse. For completeness, the squared $E$ norm of a finite boundary Fourier sum ${\sum }_{n\mathrm{\ne }0}{a}_n{e}^{in\theta }$ is comparable to $2\pi \sum \mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }{a}_n{\mathrm{\mid }}^2$. The lower bound follows from the minimum Dirichlet energy of the harmonic lift $\sum a_n{r}^{\mathrm{\mid }n\mathrm{\mid }}{e}^{in\theta }$: subtracting that lift from any representative leaves a zero-trace function, whose gradient is orthogonal to the harmonic lift. Its energy is $2\pi \sum \mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }{a}_n{\mathrm{\mid }}^2$. The lift's squared $L^2$ norm is $\pi \sum \mathrm{\mid }{a}_n{\mathrm{\mid }}^2\mathrm{/}(\mathrm{\mid }n\mathrm{\mid }+1)$, which gives the upper bound for the quotient norm. Finite modes are dense in $E$: angular Fourier projections converge in $H^1(D_1)$ by polar Parseval first for smooth ambient restrictions, and then for general functions by [F3]; their quotient traces are the finite boundary modes. In these equivalent weighted sequence norms, $D$ multiplies the $n$th coefficient by $in$, and the dual norm uses weight $1\mathrm{/}\mathrm{\mid }n\mathrm{\mid }$. This proves its bounded invertibility.

For any disk conductivity $m_{\ast }\le \gamma \le M_{\ast }$, its induced map ${\dot{\mathrm{\Lambda }}}_{\gamma }:E\to E^{\ast }$ is bounded and coercive uniformly. Indeed, if $u$ solves its boundary class $f$, then $\mathrm{\parallel }f{\mathrm{\parallel }}_E\le \mathrm{\parallel }u-u_{D_1}{\mathrm{\parallel }}_{H^1}\le C\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$ by [F2], while $⟨{\dot{\mathrm{\Lambda }}}_{\gamma }f,f⟩=\int \gamma \mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2\ge m_{\ast }\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2^2$. The quotient of a Hilbert space by its closed constant subspace is Hilbert, so Lax–Milgram [F2] gives a bounded inverse with a uniform norm. This argument also applies to ${\gamma }^{-1}$ with reciprocal ellipticity bounds.

We prove the conductivity-conjugate relation for rough coefficients. If $u\in H^1(D_1)$ is $\gamma$-harmonic, the $L^2$ vector field $F=J\gamma \mathrm{\nabla }u$, where $J(x,y)=(-y,x)$, has distributional curl zero. It has an $H^1$ potential $v$ with $\mathrm{\nabla }v=F$: mollify $F$ inside shrinking concentric disks, and for each smooth curl-free field use $v_{ϵ}(x)={\int }_0^1{F}_{ϵ}(tx)\cdot xdt$, whose derivative is $F_{ϵ}$ by the curl identity. Subtract its mean on $D_{1\mathrm{/}2}$. Poincaré–Wirtinger on each disk of radius between $1\mathrm{/}2$ and $1$ gives a uniform $L^2$ bound by the gradient norm; the normalization on $D_{1\mathrm{/}2}$ controls the disk mean by Cauchy–Schwarz. Since the gradients converge in $L^2$ on every fixed interior disk by [F4], these normalized potentials are Cauchy in $H^1$ there. Their limits have a uniform global $H^1$ bound by exhaustion, giving the required $v$. Then $\mathrm{div}({\gamma }^{-1}\mathrm{\nabla }v)=\mathrm{div}(J\mathrm{\nabla }u)=0$. The weak boundary identities are $D[v]={\dot{\mathrm{\Lambda }}}_{\gamma }[u]$ and ${\dot{\mathrm{\Lambda }}}_{{\gamma }^{-1}}[v]=-D[u]$. For smooth functions these are ${\mathrm{\partial }}_{T}v=\gamma {\mathrm{\partial }}_{\nu }u$ and ${\gamma }^{-1}{\mathrm{\partial }}_{\nu }v=-{\mathrm{\partial }}_{T}u$. For the rough solutions they follow by applying the sesquilinear identity $\int (v_y\overline{w_x}-v_x\overline{w_y})=⟨D[v],[w]⟩$ first to smooth approximations and then in $H^1$, using [F3]. Thus no boundary regularity of $\gamma$ or interior second derivatives are assumed.

The two identities in step 5.1 imply ${\dot{\mathrm{\Lambda }}}_{{\gamma }^{-1}}{D}^{-1}{\dot{\mathrm{\Lambda }}}_{\gamma }=-D$, hence ${\dot{\mathrm{\Lambda }}}_{{\gamma }^{-1}}=-D{\dot{\mathrm{\Lambda }}}_{\gamma }^{-1}D$. Subtract this formula for the two coefficients and use $A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1}$. The bounded norms of $D$ from step 3.1 and of the inverses from step 4.1 give ${\delta }_B({\gamma }_1^{-1},{\gamma }_2^{-1})\le C{\delta }_B({\gamma }_1,{\gamma }_2)$. Together with step 2.1 this proves both norm comparisons. Finally the translated unit square lies in the disk of radius $1\mathrm{/}\sqrt{2}<1$ about zero, so the supports of both original and reciprocal Beltrami coefficients lie in $D_a$ for any fixed $1\mathrm{/}\sqrt{2}<a<1$.