Planar BV logarithmic Calderón stability

Extend both coefficients by $1$ and use the fixed translation in [F2]. Let $K=\max \{2,M,m^{-1}\}$, $\kappa =(K-1)\mathrm{/}(K+1)\in (0,1)$, and ${\mu }_j=(1-{\gamma }_j)\mathrm{/}(1+{\gamma }_j)$. Then $\mathrm{\mid }{\mu }_j\mathrm{\mid }\le \kappa$ and their support lies in $D_a$. The scalar function $x\mapsto (1-x)\mathrm{/}(1+x)$ is Lipschitz on $[\min \{m,1\},\max \{M,1\}]$, so [F1] gives ${\omega }_p({\mu }_j,t)\le C_p{t}^{1\mathrm{/}p}$ for all sufficiently large finite $p$. Choose $2<r\le p_{\kappa }$ sufficiently close to $2$ that $q=2r\mathrm{/}(r-2)$ and $p=2q=4r\mathrm{/}(r-2)$ both lie in this range. These are fixed class-dependent exponents. The same bounds apply to $-{\mu }_j$, which correspond to the reciprocal conductivities. Thus all hypotheses of [F3]–[F6] hold uniformly for both conductivity pairs, and [F2] bounds both corresponding disk measurement errors by $C\epsilon$.

For either of these two pairs, write the conductivity phases as ${\delta }_j$ and amplitudes as ${\tau }_j$. Fix $z\in D_2$ and any $w\in \mathbb{C}$, and put $g(k)={\delta }_1(z,k)-{\delta }_2(w,k)$. Then $g(0)=0$ and $g(k)=i(z-w)k+O(\mathrm{\mid }k{\mathrm{\mid }}^{1-b})$ by [F3], with a constant independent of $z,w$. The phase equation in [F4] gives ${\bar{\mathrm{\partial }}}_{k}g=\sigma g+E$, where, when $g\mathrm{\ne }0$, $\sigma =-i{\tau }_2[e^{{\bar{\delta }}_1-{\delta }_1}-e^{{\bar{\delta }}_2-{\delta }_2}]\mathrm{/}g$, and set $\sigma =0$ where $g=0$; also $E=-i({\tau }_1-{\tau }_2)e^{{\bar{\delta }}_1-{\delta }_1}$. Since these exponentials have modulus one and the exponential is Lipschitz on the imaginary axis, $\mathrm{\mid }\sigma \mathrm{\mid }\le 2\mathrm{\mid }{\tau }_2\mathrm{\mid }\le 2$. By [F2]–[F4], $\mathrm{\mid }E\mathrm{\mid }\le C\epsilon e^{C\mathrm{\mid }k\mathrm{\mid }}$ after increasing $C$, and $\mathrm{\mid }{\mathrm{\nabla }}_{k}E\mathrm{\mid }\le C{e}^{C(1+\mathrm{\mid }k\mathrm{\mid })}$: differentiate the amplitude difference and the unit-modulus exponential, using only the parameter derivative bound for ${\delta }_1(z,k)$ with $z\in D_2$. This last estimate is independent of $w$. For $w\mathrm{\ne }z$, these are the hypotheses of [F5] with nonzero $\mathrm{\ell }=i(z-w)$; its positive-error conclusion is available when the stated threshold for $\mathrm{\mid }z-w\mathrm{\mid }$ holds. No application of [F5] is made when $w=z$. If necessary reduce $b$ below $1$; this only weakens the already proved phase bound.

At the fixed parameter $k=1$, surjectivity of ${\delta }_2(\cdot ,1)$ in [F3] supplies a point $w$ with ${\delta }_2(w,1)={\delta }_1(z,1)$. If $w=z$, the distance and solution-difference bounds below hold immediately. If $w\mathrm{\ne }z$, then $g(1)=0$, so for sufficiently small $0<\epsilon <t_0$, step 2.1 and [F5] force $\mathrm{\mid }z-w\mathrm{\mid }<D[\log (1\mathrm{/}\epsilon ){]}^{-b}$: the opposite inequality would exclude this nonzero-parameter zero. Reduce $t_0$ so this distance is below one; then $w\in D_3$. The uniform Hölder continuity from [F3], with exponent ${\theta }_0=1-2\mathrm{/}r>0$, gives $\mathrm{\mid }{u}_1(z,1)-u_2(z,1)\mathrm{\mid }=\mathrm{\mid }{u}_2(w,1)-u_2(z,1)\mathrm{\mid }\le C\mathrm{\mid }w-z{\mathrm{\mid }}^{{\theta }_0}\le C[\log (1\mathrm{/}\epsilon ){]}^{-b{\theta }_0}$. The bound holds uniformly for $z\in D_2$, and for both the original and reciprocal pairs. If $\epsilon =0$, the amplitude differences in [F4] vanish, so $E=0$; if $w\mathrm{\ne }z$, the zero-error clause of [F5] contradicts $g(1)=0$. Hence $w=z$. In that case both conductivity CGO pairs agree exactly.

By the definition of the two conductivity combinations, $f_{{\mu }_j}(z,1)=\mathrm{Re}{u}_{{\gamma }_j}(z,1)+i\mathrm{Im}{u}_{{\gamma }_j^{-1}}(z,1)$. Applying step 3.1 to the two pairs therefore gives $\mathrm{\parallel }{f}_{{\mu }_1}(\cdot ,1)-f_{{\mu }_2}(\cdot ,1){\mathrm{\parallel }}_{L^{\mathrm{\infty }}(D_2)}\le C[\log (1\mathrm{/}\epsilon ){]}^{-b{\theta }_0}$ for small positive $\epsilon$, and zero difference when $\epsilon =0$. At this fixed parameter, [F3] gives the uniform $W^{1,r}(D_3)$ bound and the inverse-gradient integral with $\sigma =r-2>0$. The coefficient modulus needed in [F6] is ${\omega }_q({\mu }_j,t)\le C_q{t}^{1\mathrm{/}q}$ from step 1.1. Reduce $t_0$ so the displayed solution difference is at most one, as required by [F6].

Apply [F6] with its modulus exponent $\theta =1\mathrm{/}q$ and inverse-gradient exponent $\sigma =r-2$. Its positive exponent is $\eta =\theta \sigma \mathrm{/}[(1+\theta )(2+\sigma )]$. Step 4.1 yields $\mathrm{\parallel }{\gamma }_1-{\gamma }_2{\mathrm{\parallel }}_{L^2(D_1)}\le C[\log (1\mathrm{/}\epsilon ){]}^{-\rho }$, with $\rho =b{\theta }_0\eta >0$, for $0<\epsilon <t_0$. The translated coefficient difference vanishes outside the original $\mathrm{\Omega }$, so this is exactly its $L^2(\mathrm{\Omega })$ norm. If $\epsilon =0$, step 4.1 and the zero-error clause of [F6] give equality of the coefficients and the required value $\omega (0)=0$. Finally, for $\epsilon \ge t_0$, use $\mathrm{\parallel }{\gamma }_1-{\gamma }_2{\mathrm{\parallel }}_2\le 2M\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}$ and enlarge $C$ to at least $2M\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}[\log (1\mathrm{/}{t}_0){]}^{\rho }$. Taking $t_0<1$ yields precisely the saturated modulus in the statement. All exponents and constants were fixed using only the listed class and geometry parameters.