Uniform DN-to-scattering stability for planar conductivities

For each conductivity ${\gamma }_j$ on $D_1$, define its transport ${\tilde{\gamma }}_j$ on $T(D_1)$ by ${\tilde{\gamma }}_j(Tz)={\gamma }_j(z)$, where $T(z)=(1\mathrm{/}2,1\mathrm{/}2)+z\mathrm{/}4$. The image is the radius-$1\mathrm{/}4$ disk in $(0,1{)}^2$, so [F2] applies there; pulling the weak solutions and energy forms back by $T$ gives the weak theory on $D_1$. In two dimensions the Dirichlet energy is invariant, and $\frac{1}{4}\mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(D_1)}\le \mathrm{\parallel }v\circ T^{-1}{\mathrm{\parallel }}_{H^1(T(D_1))}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H^1(D_1)}$, so the trace quotients and bounded DN maps transport with fixed norm-equivalence constants. The equality between the full and constant-quotient norms follows from the double-annihilation argument in [F2]. Put $f_1=[u_1(\cdot ,-k)]$ and $f_2=[u_2(\cdot ,k)]$, and set $U=u_2(z,k)$, $V=\overline{u_1(z,-k)}$. By the weak Alessandrini identity from [F2], $(({\mathrm{\Lambda }}_{{\gamma }_1}-{\mathrm{\Lambda }}_{{\gamma }_2})f_1)(f_2)={\int }_{D_1}({\gamma }_1-{\gamma }_2)\mathrm{\nabla }{u}_1(z,-k)\cdot \overline{\mathrm{\nabla }{u}_2(z,k)}dA$. Since the conductivities are real and equal to $1$ near $\mathrm{\partial }{D}_1$, Green's identity on that collar identifies the complex conjugate of this pairing with $W:={\int }_{\mathrm{\partial }{D}_1}(U{\mathrm{\partial }}_{\nu }V-V{\mathrm{\partial }}_{\nu }U)ds$. This uses the trace of $u_1(\cdot ,-k)$ in the first slot and the trace of $u_2(\cdot ,k)$ in the second slot; it does not treat either solution as a solution for the other conductivity. The identity is valid for weak interior solutions because only their smooth harmonic boundary collars are integrated by parts. On every exterior annulus $1<\mathrm{\mid }z\mathrm{\mid }<R$, both $U$ and $V$ are harmonic. Applying the second Green identity to their real and imaginary parts shows that the Wronskian with counterclockwise orientation has the same value on every circle $\mathrm{\mid }z\mathrm{\mid }=R>1$.

On a fixed circle $\mathrm{\mid }z\mathrm{\mid }=R>1$, write the holomorphic pieces as $H(z)=e^{ikz}{A}_2(z,k)$ for $U$ and $G(z)=e^{-ikz}{J}_1(z,-k)$ for $V$; the other pieces are antiholomorphic. A holomorphic–antiholomorphic cross term integrates to zero because its Wronskian density is a constant multiple of the tangential derivative of the product. For the holomorphic pair, $H{G}^{\mathrm{\prime }}-G{H}^{\mathrm{\prime }}=A_2{\mathrm{\partial }}_z{J}_1-J_1{\mathrm{\partial }}_z{A}_2-2ik{A}_2{J}_1$. Its $z^{-1}$ Laurent coefficient is $-2ik\overline{{\tau }_1(-k)}$, since $A_2=1+O(z^{-1})$, ${\mathrm{\partial }}_z{A}_2=O(z^{-2})$, and $J_1(z,-k)=\overline{{\tau }_1(-k)}{z}^{-1}+O(z^{-2})$. If $L(z)={\sum }_{m\in \mathbb{Z}}{c}_m{z}^m$ is any of these exterior Laurent series, direct parametrization $z=R{e}^{i\theta }$ gives ${\int }_0^{2\pi }{e}^{i\theta }L(R{e}^{i\theta })Rd\theta =2\pi c_{-1}$; all other powers integrate to zero by angular orthogonality. Since ${\mathrm{\partial }}_{\nu }H=e^{i\theta }{H}^{\mathrm{\prime }}$, this proves that the holomorphic–holomorphic Wronskian contribution is $-4\pi ik\overline{{\tau }_1(-k)}$ without invoking a general residue theorem. For the antiholomorphic pair $\tilde{H}=e^{-i\bar{k}\bar{z}}\overline{J_2(z,k)}$ and $\tilde{G}=e^{i\bar{k}\bar{z}}\overline{A_1(z,-k)}$, the coefficient of ${\bar{z}}^{-1}$ in $\tilde{H}{\mathrm{\partial }}_{\bar{z}}\tilde{G}-\tilde{G}{\mathrm{\partial }}_{\bar{z}}\tilde{H}$ is $2i\bar{k}{\tau }_2(k)$. Direct parametrization, with $\bar{z}=R{e}^{-i\theta }$, gives its contribution $4\pi i\bar{k}{\tau }_2(k)$. Hence $W=4\pi i[\bar{k}{\tau }_2(k)-k\overline{{\tau }_1(-k)}]$. These calculations use only the exterior Laurent expansions in [F1].

Apply the same calculation with the two conductivities equal to ${\gamma }_1$. The boundary pairing in step 1.1 is then zero, so $k\overline{{\tau }_1(-k)}=\bar{k}{\tau }_1(k)$. Substituting this relation in step 2.1 gives $W=4\pi i\bar{k}({\tau }_2(k)-{\tau }_1(k))$. By [F2], the absolute DN pairing is at most $\delta$ times the product of the two trace-class norms modulo constants. Subtract $1$ from each solution representative to bound these norms by the $H^1$ norms of $u_j(\cdot ,\pm k)-1$. The uniform bound in [F1] therefore gives $4\pi \mathrm{\mid }k\mathrm{\mid }\mathrm{\mid }{\tau }_2(k)-{\tau }_1(k)\mathrm{\mid }\le C\delta \mathrm{\mid }k{\mathrm{\mid }}^2{e}^{C(1+\mathrm{\mid }k\mathrm{\mid })}$. Divide by $4\pi \mathrm{\mid }k\mathrm{\mid }$ and enlarge $C$. At $k=0$, both amplitudes are zero by their definition. Finally $\mathrm{\mid }k\mathrm{\mid }\le e^{\mathrm{\mid }k\mathrm{\mid }}$ gives the second stated bound.