Thin closed current: divergence-free distribution and convergence of its mollified Biot-Savart fields

For $\psi \in \mathcal{D}({\mathbb{R}}^3;{\mathbb{C}}^3)$ the bound $\mathrm{\mid }j(\psi )\mathrm{\mid }\le \mathrm{\mid }I\mathrm{\mid }{\int }_0^L\mathrm{\mid }\psi (\gamma (s))\mathrm{\mid }ds\le \mathrm{\mid }I\mathrm{\mid }L{p}_0(\psi )$ holds, with $p_0$ the supremum seminorm over any compact set containing the bounded set $\gamma ([0,L])$; the constant does not depend on the compact set $K$ supporting $\psi$, so [F4] makes $j$ a vector distribution of order $\le 0$, supported in $\gamma ([0,L])$.

Put $A_k(x)={\mu }_{0}I{\int }_0^L{\gamma }_k^{\mathrm{\prime }}(s)\mathrm{\Phi }(x-\gamma (s))ds$ for $x\mathrm{\notin }\gamma$ and $A_k=0$ on $\gamma$. For a compact $Q\subseteq {\mathbb{R}}^3$ choose $R$ with $Q-\gamma ([0,L])\subseteq B_R(0)$; Tonelli [F5] and [F6] give ${\int }_Q\mathrm{\mid }{A}_k\mathrm{\mid }\le {\mu }_0\mathrm{\mid }I\mathrm{\mid }{\int }_0^L{\int }_{B_R}\mathrm{\mid }\mathrm{\Phi }\mathrm{\mid }=({\mu }_0\mathrm{\mid }I\mathrm{\mid }L{R}^2)\mathrm{/}2$ and ${\int }_Q\mathrm{\mid }{\mathrm{\partial }}_j{A}_k\mathrm{\mid }\le {\mu }_0\mathrm{\mid }I\mathrm{\mid }{\int }_0^L{\int }_{B_R}\mathrm{\mid }\mathrm{\nabla }\mathrm{\Phi }\mathrm{\mid }={\mu }_0\mathrm{\mid }I\mathrm{\mid }LR$, so $A_k$ and each first partial derivative ${\mathrm{\partial }}_j{A}_k$ belong to $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3)$ and define regular distributions by [F4].

For scalar $\phi \in \mathcal{D}({\mathbb{R}}^3)$ with support in $B_R(0)$ and $0<\delta <R$, the divergence theorem [F8] on the annulus $\{\delta <\mathrm{\mid }z\mathrm{\mid }<R\}$, presented by the six signed-maximal-coordinate regular graph faces of each boundary sphere whose seam arcs have zero surface measure, applied to the $C^1$ field $\mathrm{\Phi }\phi e_j$ gives ${\int }_{\delta <\mathrm{\mid }z\mathrm{\mid }<R}{\mathrm{\partial }}_j(\mathrm{\Phi }\phi )={\int }_{S_R}\mathrm{\Phi }\phi \frac{z_j}{R}dS-{\int }_{S_{\delta }}\mathrm{\Phi }\phi \frac{z_j}{\delta }dS$, where the outer term vanishes and the inner term is bounded by $\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\frac{4\pi {\delta }^2}{4\pi \delta }=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\delta \to 0$; expanding ${\mathrm{\partial }}_j(\mathrm{\Phi }\phi )={\mathrm{\partial }}_j\mathrm{\Phi }\phi +\mathrm{\Phi }{\mathrm{\partial }}_j\phi$ and letting $\delta \to 0$ with the absolutely integrable limits of [F6] yields $⟨{\mathrm{\partial }}_j{T}_{\mathrm{\Phi }},\phi ⟩=-\int \mathrm{\Phi }{\mathrm{\partial }}_j\phi =\int ({\mathrm{\partial }}_j\mathrm{\Phi })\phi$, that is ${\mathrm{\partial }}_j{T}_{\mathrm{\Phi }}=T_{{\mathrm{\partial }}_j\mathrm{\Phi }}$ with ${\mathrm{\partial }}_j\mathrm{\Phi }(z)=-z_j\mathrm{/}(4\pi \mathrm{\mid }z{\mathrm{\mid }}^3)$.

For scalar $\phi \in \mathcal{D}({\mathbb{R}}^3)$ one has $⟨\mathrm{div}j,\phi ⟩=-{\sum }_k{j}_k({\mathrm{\partial }}_k\phi )=-I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s)\cdot \mathrm{\nabla }\phi (\gamma (s))ds$ by [F4]; the chain rule [F1] gives ${\gamma }^{\mathrm{\prime }}(s)\cdot \mathrm{\nabla }\phi (\gamma (s))=(d\mathrm{/}ds)(\phi \circ \gamma )(s)$, and $s\mapsto (\phi \circ \gamma )(s)$ is $C^1$ with continuous derivative on $[0,L]$, hence integrable, so the second FTC [F2] and the periodicity $\phi (\gamma (L))=\phi (\gamma (0))$ give ${\int }_0^L(d\mathrm{/}ds)(\phi \circ \gamma )ds=0$. Therefore $\mathrm{div}j=0$.

The pairing $j_{\epsilon }(x)=j({\eta }_{\epsilon }(x-\cdot ))$ uses the test function $y\mapsto {\eta }_{\epsilon }(x-y)\in \mathcal{D}({\mathbb{R}}^3)$, and expanding the definition of $j$ gives $j_{\epsilon }(x)=I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s){\eta }_{\epsilon }(x-\gamma (s))ds$ for every $x$; this is the mollified field of the Statement.

The translated identity ${\mathrm{\partial }}_j{T}_{\mathrm{\Phi }(\cdot -y)}=T_{{\mathrm{\partial }}_j\mathrm{\Phi }(\cdot -y)}$ follows from step 1.3 because the pairing is translation invariant [F5]; applying it under the absolutely integrable $s$-integral of step 1.2 gives ${\mathrm{\partial }}_j{T}_{A_k}=T_{{\mu }_{0}I{\int }_0^L{\gamma }_k^{\mathrm{\prime }}(s){\mathrm{\partial }}_j\mathrm{\Phi }(x-\gamma (s))ds}$, and with the componentwise distributional curl ${\mathrm{curl}}_{k}u={\sum }_{i,j}{\epsilon }_{kij}{\mathrm{\partial }}_i{u}_j$ of [F12] one obtains $\mathrm{curl}{T}_A=T_B$ for the stated $B$, since ${\epsilon }_{kij}{\gamma }_j^{\mathrm{\prime }}(s){\mathrm{\partial }}_i\mathrm{\Phi }(x-\gamma (s))={\epsilon }_{kij}{\gamma }_j^{\mathrm{\prime }}(s)\frac{-(x-\gamma (s){)}_i}{4\pi \mathrm{\mid }x-\gamma (s){\mathrm{\mid }}^3}=\frac{({\gamma }^{\mathrm{\prime }}(s)\times (x-\gamma (s)){)}_k}{4\pi \mathrm{\mid }x-\gamma (s){\mathrm{\mid }}^3}$ by [F12].

For scalar $\phi \in \mathcal{D}({\mathbb{R}}^3)$ set $H(y)={\int }_{{\mathbb{R}}^3}\mathrm{\Phi }(z)\phi (z+y)dz$, which is smooth near the compact curve $\gamma ([0,L])$ by differentiation under the integral sign [F3] against the locally integrable $\mathrm{\Phi }$; then $⟨\mathrm{div}{T}_A,\phi ⟩=-⟨T_A,\mathrm{\nabla }\phi ⟩=-{\mu }_{0}I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s)\cdot (\int \mathrm{\Phi }(z)\mathrm{\nabla }\phi (z+\gamma (s))dz)ds=-{\mu }_{0}I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s)\cdot \mathrm{\nabla }H(\gamma (s))ds$, the interchange being justified by the Tonelli bound of step 1.2, and the chain rule together with the FTC and periodicity of $s\mapsto H(\gamma (s))$ gives ${\int }_0^L{\gamma }^{\mathrm{\prime }}(s)\cdot \mathrm{\nabla }H(\gamma (s))ds=0$; hence $\mathrm{div}{T}_A=0$.

Componentwise, [F7] gives $-\mathrm{\Delta }{T}_{A_k}={\mu }_{0}I{\int }_0^L{\gamma }_k^{\mathrm{\prime }}(s){\delta }_{\gamma (s)}ds$: for a scalar test $\phi$, Fubini [F5] and the translated fundamental-solution identity of [F7] yield $⟨-\mathrm{\Delta }{T}_{A_k},\phi ⟩=-{\mu }_{0}I{\int }_0^L{\gamma }_k^{\mathrm{\prime }}(s)\int \mathrm{\Phi }(x-\gamma (s))\mathrm{\Delta }\phi (x)dxds={\mu }_{0}I{\int }_0^L{\gamma }_k^{\mathrm{\prime }}(s)\phi (\gamma (s))ds$; applying this to the components of a vector test $\psi$ gives $⟨-\mathrm{\Delta }{T}_A,\psi ⟩={\mu }_{0}I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s)\cdot \psi (\gamma (s))ds={\mu }_{0}j(\psi )$, that is $-\mathrm{\Delta }{T}_A={\mu }_{0}j$.

For every multi-index $\alpha$ the integrand ${\gamma }^{\mathrm{\prime }}(s)D^{\alpha }{\eta }_{\epsilon }(x-\gamma (s))$ is smooth in $x$ with $s$-independent bound $\sup \mathrm{\mid }{D}^{\alpha }{\eta }_{\epsilon }\mathrm{\mid }$ over the compact parameter interval $[0,L]$, so differentiation under the integral sign [F3] coordinatewise gives $D^{\alpha }{j}_{\epsilon }(x)=I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s)D^{\alpha }{\eta }_{\epsilon }(x-\gamma (s))ds$ and $j_{\epsilon }\in C^{\mathrm{\infty }}({\mathbb{R}}^3;{\mathbb{C}}^3)$; the integrand vanishes unless $x\in \gamma ([0,L])+\mathrm{supp}{\eta }_{\epsilon }\subseteq \gamma ([0,L])+{\bar{B}}_{\epsilon }(0)$, a closed bounded and hence compact set by [F11], so $\mathrm{supp}{j}_{\epsilon }$ is compact.

Since $T_B=\mathrm{curl}{T}_A$ by step 2.3 and constant-coefficient distributional derivatives commute [F4], the alternating-symbol cancellations give $\mathrm{div}{T}_B=\mathrm{div}\mathrm{curl}{T}_A=0$ and $\mathrm{curl}{T}_B=\mathrm{curl}\mathrm{curl}{T}_A=\mathrm{\nabla }\mathrm{div}{T}_A-\mathrm{\Delta }{T}_A=0+{\mu }_{0}j={\mu }_{0}j$; this proves the global distributional identities for $T_B$ without any limit through the singular support.

On ${\mathbb{R}}^3\setminus \gamma$ the field $B$ is given by the absolutely convergent integral of the Statement, with $\mathrm{\mid }B(x)\mathrm{\mid }\le \frac{{\mu }_0\mathrm{\mid }I\mathrm{\mid }}{4\pi }{\int }_0^L\mathrm{\mid }x-\gamma (s){\mathrm{\mid }}^{-2}ds$ because $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}(s)\times (x-\gamma (s))\mathrm{\mid }\le \mathrm{\mid }x-\gamma (s)\mathrm{\mid }$ by [F12]; the same Tonelli computation as step 1.2 gives ${\int }_Q\mathrm{\mid }B\mathrm{\mid }\le {\mu }_0\mathrm{\mid }I\mathrm{\mid }LR$ on every compact $Q$, so $B\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^3;{\mathbb{R}}^3)$ and $T_B$ is its regular distribution by [F4]; the curve is Lebesgue-null because for every $h>0$ the Lipschitz bound [F2] with $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }=1$ covers $\gamma ([0,L])$ by the $\lceil L\mathrm{/}h\rceil$ balls of radius $h$ about the equally spaced points $jL\mathrm{/}\lceil L\mathrm{/}h\rceil$, whose total volume $\lceil L\mathrm{/}h\rceil \frac{4}{3}\pi h^3$ tends to $0$ as $h\to 0$ by [F10], and arbitrary finite values assigned on that null set therefore give the same locally integrable class and a measurable representative; on a compact subset of the complement the distance gap of [F11] bounds the kernel and all its $x$-derivatives uniformly, so differentiation under the integral sign [F3] makes $B$ smooth off $\gamma$.

For every fixed $x$ one has $\mathrm{div}{j}_{\epsilon }(x)=I{\int }_0^L{\gamma }^{\mathrm{\prime }}(s)\cdot \mathrm{\nabla }{\eta }_{\epsilon }(x-\gamma (s))ds=-I{\int }_0^L(d\mathrm{/}ds)[{\eta }_{\epsilon }(x-\gamma (s))]ds=0$: the chain rule [F1] identifies the integrand with the negative $s$-derivative, the composite is $C^1$ on $[0,L]$ with integrable derivative [F2], and periodicity gives equal endpoint values.

If $\gamma ([0,L])\subseteq B_a(0)$ and $\mathrm{\mid }x\mathrm{\mid }>a$, then $\mathrm{\mid }x-\gamma (s)\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-a$ for every $s$, so step 3.3 gives $\mathrm{\mid }B(x)\mathrm{\mid }\le \frac{{\mu }_0\mathrm{\mid }I\mathrm{\mid }}{4\pi }\frac{L}{(\mathrm{\mid }x\mathrm{\mid }-a{)}^2}$, the stated decay at infinity.

Let $K$ be nonempty compact with $d=\mathrm{dist}(K,\gamma ([0,L]))>0$ by [F11], let $0<\epsilon <d\mathrm{/}2$, and let $A_{\epsilon }(x)=\frac{{\mu }_0}{4\pi }\int \frac{j_{\epsilon }(y)}{\mathrm{\mid }x-y\mathrm{\mid }}dy$ be the Newtonian potential of the smooth compactly supported $j_{\epsilon }$; differentiating under the integral sign [F3] on $K$, where $\mathrm{\mid }x-y\mathrm{\mid }\ge d\mathrm{/}2$ for $y\in \mathrm{supp}{j}_{\epsilon }$ by step 3.1, gives the Biot-Savart field $B_{\epsilon }(x)=\mathrm{curl}{A}_{\epsilon }(x)=\frac{{\mu }_0}{4\pi }\int \frac{j_{\epsilon }(y)\times (x-y)}{\mathrm{\mid }x-y{\mathrm{\mid }}^3}dy$; substituting the formula of step 2.2 for $j_{\epsilon }$, applying Fubini [F5] and translating $y=\gamma (s)+u$ with the measure-preserving change of variables [F5] yields $B_{\epsilon }(x)=\frac{{\mu }_{0}I}{4\pi }{\int }_0^L{\int }_{{\bar{B}}_{\epsilon }(0)}\frac{{\gamma }^{\mathrm{\prime }}(s)\times (x-\gamma (s)-u)}{\mathrm{\mid }x-\gamma (s)-u{\mathrm{\mid }}^3}{\eta }_{\epsilon }(u)duds$, all integrals absolutely convergent because $\mathrm{\mid }x-\gamma (s)-u\mathrm{\mid }\ge d-\epsilon >0$.

For every multi-index $\alpha$ and $x\in K$, differentiate the absolutely convergent formulas of steps 3.3 and 4.3 under the integral signs [F3] and write $F_s(w)={\gamma }^{\mathrm{\prime }}(s)\times w\mathrm{/}\mathrm{\mid }w{\mathrm{\mid }}^3$: $D^{\alpha }{B}_{\epsilon }(x)=\frac{{\mu }_{0}I}{4\pi }{\int }_0^L\int D^{\alpha }{F}_s(x-\gamma (s)-u){\eta }_{\epsilon }(u)duds$ and $D^{\alpha }B(x)=\frac{{\mu }_{0}I}{4\pi }{\int }_0^L{D}^{\alpha }{F}_s(x-\gamma (s))ds$; on the annulus $\mathrm{\mid }w\mathrm{\mid }\ge d\mathrm{/}2$ one has $\mathrm{\mid }{D}_w{D}^{\alpha }{F}_s(w)\mathrm{\mid }\le C_{\alpha }\mathrm{\mid }w{\mathrm{\mid }}^{-3-\mathrm{\mid }\alpha \mathrm{\mid }}$ uniformly in $s$ because $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}(s)\mathrm{\mid }=1$ and the derivatives of $w\mathrm{/}\mathrm{\mid }w{\mathrm{\mid }}^3$ are homogeneous of the corresponding degrees, so the Lipschitz bound [F2] applied along the segment from $x-\gamma (s)$ to $x-\gamma (s)-u$ gives $\mathrm{\mid }{D}^{\alpha }{F}_s(x-\gamma (s)-u)-D^{\alpha }{F}_s(x-\gamma (s))\mathrm{\mid }\le C_{\alpha }(d\mathrm{/}2{)}^{-3-\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\mid }u\mathrm{\mid }$; integrating against the unit-mass kernel supported in ${\bar{B}}_{\epsilon }(0)$ gives $\mathrm{\mid }{D}^{\alpha }{B}_{\epsilon }(x)-D^{\alpha }B(x)\mathrm{\mid }\le \frac{{\mu }_0\mathrm{\mid }I\mathrm{\mid }L}{4\pi }{C}_{\alpha }(d\mathrm{/}2{)}^{-3-\mathrm{\mid }\alpha \mathrm{\mid }}\epsilon$, which tends to $0$ uniformly over $x\in K$ as $\epsilon \to 0$: the Biot-Savart fields converge with every derivative uniformly on $K$.

For $\gamma (s)=a(\cos (s\mathrm{/}a),\sin (s\mathrm{/}a),0)$ and $x=z{e}_3$, step 3.3's formula has numerator ${\gamma }^{\mathrm{\prime }}(s)\times (x-\gamma (s))=(z\cos (s\mathrm{/}a),z\sin (s\mathrm{/}a),a)$ and denominator $(a^2+z^2{)}^{3\mathrm{/}2}$, so $B(x)=\frac{{\mu }_{0}I}{4\pi (a^2+z^2{)}^{3\mathrm{/}2}}{\int }_0^{2\pi a}(z\cos (s\mathrm{/}a),z\sin (s\mathrm{/}a),a)ds$; with $ds=ad\theta$, the transverse entries integrate to $0$ and the third entry integrates to $2\pi a^2$, because ${\int }_0^{2\pi }\cos ={\int }_0^{2\pi }\sin =0$ by the FTC and the fundamental period of [F13]; hence $B_z={\mu }_{0}I{a}^2\mathrm{/}(2(a^2+z^2{)}^{3\mathrm{/}2})$, while no value of the singular line integral on the curve and no self-energy is asserted.