The localized Newtonian potential and the ideal point-charge Coulomb field

By [F1], ${\omega }_2=\mathrm{\mid }{S}^2\mathrm{\mid }=4\pi$ and $\mathrm{\Phi }(z)=1\mathrm{/}(4\pi \mathrm{\mid }z\mathrm{\mid })$ for $z\mathrm{\ne }0$, so $\mathrm{\nabla }\mathrm{\Phi }(z)=-z\mathrm{/}(4\pi \mathrm{\mid }z{\mathrm{\mid }}^3)$ and $\mathrm{\mid }\mathrm{\nabla }\mathrm{\Phi }(z)\mathrm{\mid }=1\mathrm{/}(4\pi \mathrm{\mid }z{\mathrm{\mid }}^2)$ by the chain and power rules ([F4]); $\mathrm{\Phi }$ and $\mathrm{\nabla }\mathrm{\Phi }$ are locally integrable on ${\mathbb{R}}^3$, since ${\int }_{\mathrm{\mid }z\mathrm{\mid }<b}\mathrm{\mid }z{\mathrm{\mid }}^{-k}d{\lambda }_3=4\pi b^{3-k}\mathrm{/}(3-k)$ is finite for $k=1,2$, while away from the pole the kernels are continuous and hence bounded on compacta ([F8]). Let $\rho \in C_c^1$ with support in ${\bar{B}}_R$; by [F5] it is bounded, Lipschitz and has finite global $\alpha$-Hölder seminorm for $\alpha =1\mathrm{/}2$, and it vanishes off the open ball $B_R$ because it is continuous and zero on and beyond $S_R$.

The Hölder theorem [F2] applied with $\alpha =1\mathrm{/}2$ gives $N\rho \in C^2({\mathbb{R}}^3)$ with $-\mathrm{\Delta }(N\rho )=\rho$ pointwise; scaling $\phi ={\epsilon }_0^{-1}N\rho$ and using linearity of the second coordinate derivatives gives $-\mathrm{\Delta }\phi =\rho \mathrm{/}{\epsilon }_0$ pointwise, because ${\epsilon }_0>0$ is a constant ([F4]). With $E:=-\mathrm{\nabla }\phi$ one has $\mathrm{div}E=-\mathrm{div}\mathrm{\nabla }\phi =-\mathrm{\Delta }\phi =\rho \mathrm{/}{\epsilon }_0$ on ${\mathbb{R}}^3$ and $\mathrm{curl}E=-\mathrm{curl}\mathrm{\nabla }\phi =0$ by [F4], since $\phi$ is $C^2$.

Coulomb representation: since Lebesgue measure is translation invariant, $N\rho (x)=\int \mathrm{\Phi }(x-y)\rho (y)d{\lambda }_3(y)=\int \mathrm{\Phi }(z)\rho (x-z)d{\lambda }_3(z)$ ([F6]); fix $i\in \{1,2,3\}$, $x_0\in {\mathbb{R}}^3$, $\delta >0$ and the interval $J=(-\delta ,\delta )$ of coordinate increments, so $x_0+s{e}_i\in B(x_0,\delta )$ for $s\in J$. For such $x$ the integrand $\mathrm{\Phi }(z)\rho (x-z)$ is supported in the compact ball $K:={\bar{B}}_{\mathrm{\mid }{x}_0\mathrm{\mid }+\delta +R}(0)$, and its $s$-derivative $\mathrm{\Phi }(z){\mathrm{\partial }}_i\rho (x-z)$ is dominated on $J$ by $\mathrm{\parallel }\mathrm{\nabla }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mathrm{\Phi }(z)\mathrm{\mid }{\mathbf{1}}_K(z)$, which is integrable ([F5], [F8]); differentiation under the integral sign [F6] therefore gives ${\mathrm{\partial }}_{i}N\rho (x)=\int \mathrm{\Phi }(z){\mathrm{\partial }}_i\rho (x-z)d{\lambda }_3(z)$. For fixed $x$ put $h(z):=\rho (x-z)$, a $C^1$ compactly supported function; then ${\mathrm{\partial }}_i\rho (x-z)=-{\mathrm{\partial }}_{z_i}h(z)$, so ${\mathrm{\partial }}_{i}N\rho (x)=-\int \mathrm{\Phi }(z){\mathrm{\partial }}_{z_i}h(z)d{\lambda }_3(z)\mathrm{.}$ The kernel $\mathrm{\Phi }$ is locally integrable, continuous and smooth off the single point $0$, its classical partials are locally integrable, so The distributional gradient of a locally integrable function smooth off one point applies with $u=\mathrm{\Phi }$ and gives ${\mathrm{\partial }}_i{T}_{\mathrm{\Phi }}=T_{{\mathrm{\partial }}_i\mathrm{\Phi }}$; This identity is initially paired only with smooth tests: for $h_{\delta }$ of [F11], $\int \mathrm{\Phi }{\mathrm{\partial }}_i{h}_{\delta }=-\int {\mathrm{\partial }}_i\mathrm{\Phi }{h}_{\delta }$ by [F7]. Since $h\in C_c^1$, [F11] supplies uniform convergence of both $h_{\delta }$ and ${\mathrm{\partial }}_i{h}_{\delta }$ on a fixed compact support; the two locally integrable kernels therefore allow the limit $\int \mathrm{\Phi }{\mathrm{\partial }}_{i}h=-\int {\mathrm{\partial }}_i\mathrm{\Phi }h$, without pairing any distribution against a nonsmooth test. Substituting back and translating gives ${\mathrm{\partial }}_{i}N\rho (x)=\int {\mathrm{\partial }}_i\mathrm{\Phi }(x-y)\rho (y)d{\lambda }_3(y)=-\frac{1}{4\pi }\int \frac{x_i-y_i}{\mathrm{\mid }x-y{\mathrm{\mid }}^3}\rho (y)d{\lambda }_3(y)$ for every $x\in {\mathbb{R}}^3$, the integral being absolutely convergent because $\mathrm{supp}\rho \subseteq {\bar{B}}_R$ lies in the ball $B(x,\mathrm{\mid }x\mathrm{\mid }+2R)$ and ${\int }_{B(x,\mathrm{\mid }x\mathrm{\mid }+2R)}\mathrm{\mid }x-y{\mathrm{\mid }}^{-2}d{\lambda }_3(y)=4\pi (\mathrm{\mid }x\mathrm{\mid }+2R)<\mathrm{\infty }$, while $\mathrm{\mid }\rho \mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}$ on the support ([F1], [F5], [F8]). Multiplying by $-{\epsilon }_0^{-1}$ gives exactly the displayed Coulomb representation of $E=-\mathrm{\nabla }\phi$.

Bounds and regime reading: for $\mathrm{\mid }x\mathrm{\mid }>2R$ the far-field lemma [F3] with $n=3$, ${\omega }_2=4\pi$ gives $\mathrm{\mid }\mathrm{\nabla }N\rho (x)\mathrm{\mid }\le 4\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(4\pi \mathrm{\mid }x{\mathrm{\mid }}^2)=\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(\pi \mathrm{\mid }x{\mathrm{\mid }}^2)$, hence $\mathrm{\mid }E(x)\mathrm{\mid }={\epsilon }_0^{-1}\mathrm{\mid }\mathrm{\nabla }N\rho (x)\mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(\pi {\epsilon }_0\mathrm{\mid }x{\mathrm{\mid }}^2)$. Since $\rho$ is continuous, compactly supported in ${\bar{B}}_R$ and zero off $B_R$, the decay theorem [F3] gives $\mathrm{\mid }N\rho (x)-Q\mathrm{/}(4\pi \mathrm{\mid }x\mathrm{\mid })\mathrm{\mid }\le R\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(\pi \mathrm{\mid }x{\mathrm{\mid }}^2)$ for $\mathrm{\mid }x\mathrm{\mid }>2R$, so for $\mathrm{\mid }x\mathrm{\mid }>2R$ we get $\mathrm{\mid }\phi (x)\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }\mathrm{/}(4\pi {\epsilon }_0\mathrm{\mid }x\mathrm{\mid })+R\mathrm{\parallel }\rho {\mathrm{\parallel }}_1\mathrm{/}(\pi {\epsilon }_0\mathrm{\mid }x{\mathrm{\mid }}^2)$ by the triangle inequality, as claimed. Both terms have units V: $Q$ and $\mathrm{\parallel }\rho {\mathrm{\parallel }}_1$ carry C, $R$ and $\mathrm{\mid }x\mathrm{\mid }$ carry m, and ${\epsilon }_0$ carries F m${}^{-1}$; the bound also gives $\phi (x)\to 0$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$. Finally the fields and density are time independent, and the pointwise identities of step 2.1 are exactly the reduced equations of the pure electrostatic regime, so $(E,0,\rho ,0)$ realises that regime with source $\rho$ in the sense of the fixed-time reading ([F9]).

Uniqueness (b): let $\psi \in C^2({\mathbb{R}}^3)$ with $-\mathrm{\Delta }\psi =\rho \mathrm{/}{\epsilon }_0$ pointwise and $\psi (x)\to 0$ as $\mathrm{\mid }x\mathrm{\mid }\to \mathrm{\infty }$, and put $h:=\psi -\phi$, where $\phi$ is the potential of clause (a). Then $h\in C^2({\mathbb{R}}^3)$, $\mathrm{\Delta }h=\mathrm{\Delta }\psi -\mathrm{\Delta }\phi =-\rho \mathrm{/}{\epsilon }_0+\rho \mathrm{/}{\epsilon }_0=0$ and $h(x)\to 0$ ([F4]). Given $\epsilon >0$ choose $R_{\epsilon }$ with $\mathrm{\mid }h(x)\mathrm{\mid }\le \epsilon$ for $\mathrm{\mid }x\mathrm{\mid }\ge R_{\epsilon }$; for $R>R_{\epsilon }$ the weak maximum principle [F10] applied on the bounded open ball $B_R$ to the harmonic $h$ gives ${\max }_{{\bar{B}}_R}h={\max }_{\mathrm{\partial }{B}_R}h\le \epsilon$, and applied to the harmonic $-h$ gives ${\min }_{{\bar{B}}_R}h\ge -\epsilon$; hence $\mathrm{\mid }h\mathrm{\mid }\le \epsilon$ on ${\bar{B}}_R$. Every point of ${\mathbb{R}}^3$ lies in $B_R$ for large $R$, so $\mathrm{\mid }h(x)\mathrm{\mid }\le \epsilon$ for all $x$ and all $\epsilon >0$, whence $h=0$ and $\psi =\phi$ ([F4]).

Clause (c): let $\rho \in L^1({\mathbb{R}}^3)$ be real, measurable and compactly supported. The distributional Poisson theorem [F2] gives that $N\rho$ is finite almost everywhere, belongs to $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$, is $C^{\mathrm{\infty }}$ and harmonic off the support of $\rho$, and satisfies $-\mathrm{\Delta }{T}_{N\rho }=T_{\rho }$ in ${\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^3)$. Multiplying the regular distribution by the positive constant ${\epsilon }_0^{-1}$ and using linearity of distributional differentiation, $-\mathrm{\Delta }{T}_{{\epsilon }_0^{-1}N\rho }={\epsilon }_0^{-1}(-\mathrm{\Delta }{T}_{N\rho })=T_{\rho \mathrm{/}{\epsilon }_0}$ ([F7]); this is the displayed identity, with the same smoothness and harmonicity assertions carried over.

Clause (d), first part: ${\phi }_q=(q\mathrm{/}{\epsilon }_0)\mathrm{\Phi }(\cdot -x_0)$, so ${\phi }_q\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3)$ and its classical partials ${\mathrm{\partial }}_i{\phi }_q$ are locally integrable, by the local integrability of $\mathrm{\mid }z{\mathrm{\mid }}^{-1}$ and $\mathrm{\mid }z{\mathrm{\mid }}^{-2}$ in dimension three ([F1], [F8]); the value at $x_0$ is arbitrary and immaterial ([F7]). By [F1] the translated kernel satisfies $-{\mathrm{\Delta }}_x{T}_{\mathrm{\Phi }(\cdot -x_0)}={\delta }_{x_0}$; multiplying by $q\mathrm{/}{\epsilon }_0$ and using linearity of distributional differentiation gives $-\mathrm{\Delta }{T}_{{\phi }_q}=(q\mathrm{/}{\epsilon }_0){\delta }_{x_0}$ ([F7]). Moreover $E_q=-\mathrm{\nabla }{\phi }_q$ on ${\mathbb{R}}^3\setminus \{x_0\}$: differentiating ${\phi }_q(x)=q\mathrm{/}(4\pi {\epsilon }_0\mathrm{\mid }x-x_0\mathrm{\mid })$ by the chain and power rules gives $\mathrm{\nabla }{\phi }_q(x)=-\frac{q(x-x_0)}{4\pi {\epsilon }_0\mathrm{\mid }x-x_0{\mathrm{\mid }}^3}$ for $x\mathrm{\ne }{x}_0$ ([F4]).

Clause (d), field identities and reading: by The distributional gradient of a locally integrable function smooth off one point applied to $u={\phi }_q$ (locally integrable, continuous and smooth off the single point $x_0$, classical partials locally integrable), ${\mathrm{\partial }}_i{T}_{{\phi }_q}=T_{{\mathrm{\partial }}_i{\phi }_q}$ ([F7]). Since $E_{q,i}=-{\mathrm{\partial }}_i{\phi }_q$ off $x_0$ and a singleton is null, the two functions agree almost everywhere, hence $T_{E_{q,i}}=-T_{{\mathrm{\partial }}_i{\phi }_q}=-{\mathrm{\partial }}_i{T}_{{\phi }_q}$ ([F7]); taking one more distributional derivative and summing over $i$ gives $\mathrm{div}{E}_q={\sum }_i{\mathrm{\partial }}_i{T}_{E_{q,i}}=-{\sum }_i{\mathrm{\partial }}_i{\mathrm{\partial }}_i{T}_{{\phi }_q}=-\mathrm{\Delta }{T}_{{\phi }_q}=\frac{q}{{\epsilon }_0}{\delta }_{x_0},$ and the curl components vanish because distributional partial derivatives commute: ${\mathrm{\partial }}_2{T}_{E_{q,3}}-{\mathrm{\partial }}_3{T}_{E_{q,2}}=-({\mathrm{\partial }}_2{\mathrm{\partial }}_3-{\mathrm{\partial }}_3{\mathrm{\partial }}_2)T_{{\phi }_q}=0,$ and likewise for the other two components ([F7]). By [F9] the distribution $q{\delta }_{x_0}$ is the fixed-time reading of the stationary point source with charge $q$ at $x_0$, so $(E_q,0)$ satisfies the distributional pure electrostatic equations with that source. No product of distributions was formed, no value of the field at $x_0$ was used, and no uniqueness, self-force, reaction or empirical claim is made; this is exactly the scope stated.