Uniqueness and flux compatibility for the electrostatic boundary value problem

Classical Dirichlet uniqueness. Let $U$ be a connected component of ${\mathrm{\Omega }}_v$. By [F3] $U$ is a bounded connected $C^1$ domain with $\bar{U}\subseteq \overline{{\mathrm{\Omega }}_v}$ and the restrictions ${\phi }_i{\mathrm{\mid }}_{\bar{U}}\in C^2(\bar{U})$ satisfy $-\mathrm{\Delta }({\phi }_i{\mathrm{\mid }}_U)=\rho \mathrm{/}{\epsilon }_0$ there, with equal traces on $\mathrm{\partial }U$ when ${\phi }_1={\phi }_2$ on $\mathrm{\partial }{\mathrm{\Omega }}_v$. By [F1] claim 1 applied on $U$, ${\phi }_1={\phi }_2$ on $\bar{U}$; as $U$ was an arbitrary component, ${\phi }_1={\phi }_2$ on all of $\overline{{\mathrm{\Omega }}_v}$, and then $E_1=-\mathrm{\nabla }{\phi }_1=-\mathrm{\nabla }{\phi }_2=E_2$ by [F3]. This is (a).

Classical Neumann uniqueness. Assume ${\mathrm{\partial }}_{\nu }{\phi }_1={\mathrm{\partial }}_{\nu }{\phi }_2$ on $\mathrm{\partial }{\mathrm{\Omega }}_v$. Applying [F2] to the bounded $C^1$ domain ${\mathrm{\Omega }}_v$ with its finitely many components gives that ${\phi }_1-{\phi }_2$ is constant on each component. Conversely, if $u$ solves the equation and $c_j$ is a constant on each component that is added to $u$, then $\mathrm{\Delta }(u+c_j)=\mathrm{\Delta }u$ and ${\mathrm{\partial }}_{\nu }(u+c_j)={\mathrm{\partial }}_{\nu }u$ because the gradient of a constant vanishes [F3], so the equation and the Neumann data are preserved; hence the field $E=-\mathrm{\nabla }u$ is unique while the potential is determined only up to componentwise constants. This is the uniqueness half of (b).

Weak variants. Clause (c)(i) is the uniqueness content of [F4]: two weak $H^1$ Dirichlet solutions with the same prescribed trace and the same right-hand side coincide. Clause (c)(ii) is the uniqueness-modulo-constants and normalisation clause of [F5], with the trace space of The fractional Sobolev space on a compact $C^1$ boundary; the componentwise statement is the same clause on each component. Clause (c)(iii) is the uniqueness clause of F6: outer Dirichlet data give a unique globally $H^1$ transmission solution, and compatible outer Neumann data give uniqueness modulo componentwise constants, fixed by zero means. This is (c).

Physics reading. On a star-shaped vacuum domain, [F7] supplies an interior $C^2$ potential for a curl-free $C^1$ field and nothing more: closure regularity, $H^1$ membership, a trace and boundary data are additional hypotheses, exactly as assumed in (a)-(c). For the bounded conductor model with its connected vacuum region [F10], fixed absolute conductor potentials and identical outer Dirichlet data are the Dirichlet data of (a) or (c)(i), so the difference of two admitted solutions has zero boundary data and is zero by the same uniqueness clause; grounding fixes the additive constant, while shifting both a relative-potential datum and every reference datum by a common constant leaves the difference of two solutions with the same data and hence with the same (fixed) uniqueness conclusion only for identical data — omitting outer data supplies no field uniqueness. This is (d).

Neumann flux compatibility. Let $U$ be a connected component of ${\mathrm{\Omega }}_v$, with outward normal $\nu$ and boundary $\mathrm{\partial }U$ a union of compact pieces of $\mathrm{\partial }{\mathrm{\Omega }}_v$ [F3]. Applying [F1] claim 3 (equivalently Necessary compatibility for the classical Neumann Poisson problem) to $u=\phi {\mathrm{\mid }}_U$ with $f=-\mathrm{\Delta }\phi =\rho \mathrm{/}{\epsilon }_0$ and $g={\mathrm{\partial }}_{\nu }\phi$ gives ${\int }_U\rho \mathrm{/}{\epsilon }_0=-{\int }_{\mathrm{\partial }U}{\mathrm{\partial }}_{\nu }\phi dS$, that is ${\int }_U\rho +{\epsilon }_0{\int }_{\mathrm{\partial }U}{\mathrm{\partial }}_{\nu }\phi dS=0$, the displayed identity for the component; no solvability is inferred from it. This completes (b).

Mixed conductor data. Suppose the vacuum region is connected, $m\ge 1$, $w$ is the difference of two admitted classical solutions with equal absolute conductor potentials and equal outer Neumann data. Then $w$ is harmonic, $w=0$ on the nonempty conductor boundary $\mathrm{\partial }D$ and ${\mathrm{\partial }}_{\nu }w=0$ on the outer boundary $\mathrm{\partial }\mathrm{\Omega }$. Green's first identity [F9] gives ${\int }_{{\mathrm{\Omega }}_v}\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2={\int }_{\mathrm{\partial }{\mathrm{\Omega }}_v}w{\mathrm{\partial }}_{\nu }wdS$, and each boundary piece carries one zero factor, so the integral vanishes; by [F9], $\mathrm{\nabla }w=0$ almost everywhere, and since $\mathrm{\nabla }w$ is continuous on the connected open ${\mathrm{\Omega }}_v$ it vanishes everywhere, so $w$ is constant by the vanishing-differential characterisation; its zero conductor trace forces the constant to be $0$. In the weak mixed branch, the difference of two admitted solutions has zero conductor trace, so it is an admissible test in the homogeneous difference equation and the same zero-energy argument applies, with the mean-zero coercivity inequality of [F10] converting $\mathrm{\nabla }w=0$ into $w=w_{{\mathrm{\Omega }}_v}$ as an $H^1$ class whose constant trace on the nonempty conductor boundary is zero. This sharpens (d); existence for mixed data remains unasserted.

The unbounded witness and scope. On ${\mathrm{\Omega }}_R=\{x:\mathrm{\mid }x\mathrm{\mid }>R\}$ put $u(x):=1-R\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, and take the comparison potential $\phi \equiv 0$. Then $u\in C^{\mathrm{\infty }}({\mathrm{\Omega }}_R)$; since $R\mathrm{/}\mathrm{\mid }x\mathrm{\mid }=4\pi R\mathrm{\Phi }(x)$ is a constant multiple of the $n=3$ fundamental solution, it is harmonic off the origin [F8], and the constant $1$ has zero Laplacian, so $\mathrm{\Delta }u=0$; moreover $u=0$ on $\mathrm{\mid }x\mathrm{\mid }=R$, $0<u<1$ on ${\mathrm{\Omega }}_R$ and $u(x)\to 1$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. Thus $0$ and $u$ are two distinct bounded harmonic functions with the same zero boundary trace and different limits at infinity, witnessing that the boundary and decay hypotheses of (a)-(c) cannot be dropped. No existence statement is inferred from uniqueness anywhere above; the classical branch requires $C^2$ up to the closure, the weak branches carry their own domain, coefficient and trace hypotheses, and no dielectric coefficient, empirical value or measurement is included. This is (e).