gpt 6 astra ultra attempt on Math-318

Calderón uniqueness for bounded measurable real isotropic conductivities

Definitions and mathematical statement

Let $n\ge3$ and let $\Omega\subset{\mathbb{R}}^n$ be a bounded connected Lipschitz domain. For a real scalar conductivity $\gamma\in L^\infty(\Omega)$ with $0<c\le\gamma\le C<\infty$, the weak solution with boundary value $f\in H^{1/2}(\partial\Omega)$ satisfies $\nabla\cdot(\gamma\nabla u)=0$. Its Dirichlet-to-Neumann map is the bounded operator determined by \[ \langle\Lambda_\gamma f,g\rangle =\int_\Omega\gamma\nabla u\cdot\nabla v{\,\mathrm{d}} x, \qquad v|_{\partial\Omega}=g. \] The value is independent of the chosen $H^1$ extension $v$. The uniqueness question is whether $\Lambda_{\gamma_1}=\Lambda_{\gamma_2}$ implies $\gamma_1=\gamma_2$ almost everywhere. The hypotheses impose no derivative regularity, smallness, or prior agreement near the boundary. This scalar question is distinct from anisotropic uniqueness modulo boundary-fixing changes of coordinates.

Formulation and concept references: [S1].

Short English statement

Do all boundary voltage-to-current measurements uniquely determine a bounded measurable positive scalar conductivity in dimension at least three?

Research attempt

Endpoint and source audit. The coefficients here are real scalar $L^\infty$ functions with a fixed positive lower bound. There is no Sobolev derivative available and no assumption that two coefficients agree in a collar of the boundary. The primary note [S1] proposes a Clifford-algebra/Beltrami strategy as a possible first step toward this endpoint; its abstract does not claim the full measurable-conductivity uniqueness theorem. Results for Lipschitz or other derivative-controlled conductivities and two-dimensional measurable results do not remove these distinctions. We give exact identities, two rough-coefficient subcases, and injectivity of the linearization at a constant coefficient. The final nonlinear completeness step remains unproved.

The variational formulation needs no coefficient derivatives. Fix a bounded extension operator from $H^{1/2}(\partial\Omega)$ to $H^1(\Omega)$, available for a Lipschitz domain. If $F$ extends $f$, seek $u=F+w$, $w\in H_0^1(\Omega)$, with \[ \int_\Omega\gamma\nabla w\cdot\nabla v =-\int_\Omega\gamma\nabla F\cdot\nabla v \quad(v\in H_0^1(\Omega)). \] Boundedness, uniform ellipticity, and Poincaré's inequality give existence and uniqueness by Lax--Milgram. Testing with $w$ gives a bound for $\|u\|_{H^1}$ in terms of $f$, the domain, and the ellipticity constants. Changing an extension of the second boundary argument by an element of $H_0^1$ changes the DN pairing by zero. This proves the asserted well-definedness and boundedness of $\Lambda_\gamma$ without assigning a classical normal derivative to $u$.

Let $u_1^f$ solve for $\gamma_1$ with trace $f$, and $u_2^g$ solve for $\gamma_2$ with trace $g$. Symmetry of each real-coefficient DN form and testing the weak equations with differences having zero trace give \[ \langle(\Lambda_{\gamma_1}-\Lambda_{\gamma_2})f,g\rangle =\int_\Omega(\gamma_1-\gamma_2) \nabla u_1^f\cdot\nabla u_2^g\,dx. \] For example the first term can use $u_2^g$ as its extension; the second can use $u_1^f$ after symmetry and the zero-trace replacement of its own solution. This is the basic orthogonality identity. Equality of the DN maps supplies orthogonality to these products, not orthogonality to every $L^1$ function. The density or special-solution theorem needed to bridge that gap is the substantive issue.

Monotonicity and an ordered constant-background subcase. For a fixed real boundary value $f$, the solution uniquely minimizes \[ E_\gamma(v)=\int_\Omega\gamma|\nabla v|^2 \] among all $v$ with trace $f$. Indeed writing $v=u+w$ makes the cross term zero and leaves $E_\gamma(v)=E_\gamma(u)+E_\gamma(w)$. For solutions $u_1,u_2$ with that common trace, comparison with the other minimizer gives \[ \int(\gamma_1-\gamma_2)|\nabla u_1|^2 \leq\langle(\Lambda_{\gamma_1}-\Lambda_{\gamma_2})f,f\rangle \leq\int(\gamma_1-\gamma_2)|\nabla u_2|^2. \] In particular pointwise ordering of conductivities implies ordering of their DN quadratic forms.

Suppose now that $\gamma\geq c_0>0$ a.e., where $c_0$ is constant, and $\Lambda_\gamma=\Lambda_{c_0}$. Use $f=x_j|_{\partial\Omega}$. The constant-coefficient solution is $x_j$. If $u$ is the $\gamma$ solution, \[ E_{c_0}(x_j)\leq E_{c_0}(u)\leq E_\gamma(u) =\langle\Lambda_\gamma f,f\rangle=E_{c_0}(x_j). \] Equality and uniqueness of the constant-coefficient minimizer force $u=x_j$. It follows that $\int_\Omega(\gamma-c_0)=0$, hence $\gamma=c_0$ a.e. If instead $\gamma\leq c_0$, use \[ E_\gamma(u)\leq E_\gamma(x_j)=\int\gamma \leq c_0|\Omega|=E_{c_0}(x_j)=E_\gamma(u) \] to reach the same conclusion. This proves uniqueness against a constant background under a one-sided ordering, with no regularity assumption on $\gamma$. It does not prove uniqueness for two unordered coefficients. For a general variable comparison coefficient, the corresponding gradients are unknown; declaring them nonzero on every possible discrepancy set would insert an additional unique-continuation or localization assertion.

A measurable layered subcase with full boundary data. Take $\Omega=(0,1)^n$ and $\gamma(x)=a(x_1)$, where $a$ is merely measurable and $0<c\leq a\leq C$. For the boundary voltage $f=x_2$, the function $u=x_2$ is a weak solution because $\operatorname{div}(a(x_1)e_2)=0$ in distributions. For a smooth boundary test $g$ compactly supported in the relative interior of the face $\{x_2=0\}$, the variational DN pairing is \[ \langle\Lambda_a f,g\rangle =-\int_{(0,1)^{n-1}}a(x_1)g(x_1,0,x_3,\ldots,x_n) \,dx_1dx_3\cdots dx_n. \] To justify the formula at this regularity, take a smooth extension supported near that face and integrate $a(x_1)\partial_2v$ in the $x_2$ variable using Fubini. No derivative or boundary trace of $a$ in its own variable is taken.

If $a_1,a_2$ have equal full DN maps, choose product tests $g=\psi(x_1)\chi(x_3,\ldots,x_n)$ with $\chi$ of integral one. Then $\int_0^1(a_1-a_2)\psi=0$ for every compactly supported smooth $\psi$, so $a_1=a_2$ a.e. This subcase genuinely permits jumps and arbitrarily oscillatory measurable layers. It uses accessible side faces and the a priori layered geometry, not just measurements on the two faces perpendicular to the layers. Those more limited measurements can collapse to an effective conductivity and would not justify this argument.

Injectivity of the derivative at the identity conductivity. Let $\gamma_\varepsilon=1+\varepsilon h$ with real $h\in L^\infty(\Omega)$ and $|\varepsilon|\|h\|_\infty\leq1/2$. Write $u_\varepsilon^f=u_0^f+w_\varepsilon$, where $u_0^f$ is harmonic. The weak equation gives \[ \int(1+\varepsilon h)\nabla w_\varepsilon\cdot\nabla v =-\varepsilon\int h\nabla u_0^f\cdot\nabla v, \quad \|\nabla w_\varepsilon\|_2 \leq2|\varepsilon|\|h\|_\infty\|\nabla u_0^f\|_2. \] The preceding integral identity therefore yields the Fréchet derivative \[ \langle D\Lambda_1[h]f,g\rangle =\int_\Omega h\nabla u_0^f\cdot\nabla u_0^g, \] with a remainder bounded by a constant times $\varepsilon^2\|h\|_\infty^2\|f\|_{H^{1/2}}\|g\|_{H^{1/2}}$. Complexify this real bilinear identity. For $\xi\ne0$, choose a real unit vector $e\perp\xi$ and put \[ \zeta=\frac{i\xi}{2}+\frac{|\xi|e}{2},\qquad \eta=\frac{i\xi}{2}-\frac{|\xi|e}{2}. \] Then $\zeta\cdot\zeta=\eta\cdot\eta=0$, $\zeta+\eta=i\xi$, and $\zeta\cdot\eta=-|\xi|^2/2$. The functions $e^{\zeta\cdot x}$ and $e^{\eta\cdot x}$ are smooth harmonic functions on the bounded domain. If $D\Lambda_1[h]=0$, substitution gives \[ 0=-\frac{|\xi|^2}{2}\int_\Omega h(x)e^{i\xi\cdot x}\,dx. \] The zero extension of $h$ is integrable and compactly supported. Its Fourier transform vanishes at every nonzero frequency and, by continuity, at zero. Fourier uniqueness gives $h=0$ a.e. Thus the linearized inverse problem is injective throughout the requested $L^\infty$ class of perturbation directions.

The nonlinear gap is not removed by this derivative. An injective derivative in infinite-dimensional spaces need not have a bounded inverse in the relevant norms and does not itself imply nonlinear global injectivity. At a variable coefficient, the exponentials used above are not conductivity solutions. Replacing them by complex geometric optics solutions with errors small enough to recover Fourier data is a new analytic construction, not a formal repetition of the constant-coefficient proof. Likewise the formal substitution $v=\sqrt\gamma\,u$ introduces $q=\gamma^{-1/2}\Delta\sqrt\gamma$. At mere $L^\infty$ regularity this expression cannot automatically be treated as an ordinary potential or multiplied with arbitrary rough solutions. Smooth approximation may produce unbounded derivative norms: $1+\delta\sin(jx_1)$ stays uniformly elliptic for fixed $0<\delta<1$ while its gradient grows with $j$.

Finally, a boundary-fixing change of variables transforms a scalar coefficient into a tensor proportional to $DF\,DF^T/|\det DF|$, composed with the inverse map. That tensor is generally not scalar. The usual anisotropic gauge therefore does not directly provide a counterexample here, and singular cloaking coefficients violate the stipulated bounded positive scalar class.

Verification and outcome. The diagnostic checks the complex-vector identities numerically at specified frequencies and checks layered boundary pairings exactly for a piecewise constant rational example. The infinite-dimensional variational and Fourier arguments are supplied above, not inferred from that finite sample. The attempt proves ordered constant-background uniqueness, layered uniqueness, and injectivity of the constant-background linearization. It supplies no complete family of products of solutions for two arbitrary measurable coefficients, which is the first missing step toward the full target.

Sources

• S1 Matteo Santacesaria, Note on Calderón's inverse problem for measurable conductivities, Inverse Problems and Imaging13$1$$2019$,149–157; arXiv1803.06931v2. https://arxiv.org/pdf/1803.06931v2. Formulation source.