Joint regularity of the radial vector and scalar potentials on a star-shaped domain

Extension and radial-admissibility. By the joint $C^k$ convention there are an open neighbourhood $N\subseteq {\mathbb{R}}^4$ of $U\times I$ and a $C^k$ map $\tilde{B}:N\to {\mathbb{R}}^3$ with $\tilde{B}{\mathrm{\mid }}_{U\times I}=B$; for the scalar witness take $N$ to be the intersection of the two neighbourhoods and let $\tilde{C}$ be the corresponding $C^k$ extension, which exists on that intersection. Put $P:=\{(x,t)\in N:(a+s(x-a),t)\in N\text{ for every }s\in [0,1]\}$; then $U\times I\subseteq P$ by [F1], and $P$ is open: if $(x,t)\in P$, the radial image $R:=\{(a+s(x-a),t):0\le s\le 1\}$ is compact and contained in the open set $N$, use the indexed cover of all balls $B(q,\epsilon )$ with $q\in R$, $\epsilon >0$ and $B(q,2\epsilon )\subseteq N$. Openness supplies at least one at each point, without selecting a radius function. Take a finite subcover and put $\rho$ equal to the minimum of its finitely many positive radii. Every point within distance $\rho$ of a point of $R$ then belongs to $N$. The map $\mathrm{\Phi }(x^{\mathrm{\prime }},t^{\mathrm{\prime }},s)=(a+s(x^{\mathrm{\prime }}-a),t^{\mathrm{\prime }})$ satisfies $\parallel \mathrm{\Phi }(x^{\mathrm{\prime }},t^{\mathrm{\prime }},s)-\mathrm{\Phi }(x,t,s){\parallel }_2\le \parallel x^{\mathrm{\prime }}-x{\parallel }_2+\mid t^{\mathrm{\prime }}-t\mid$ uniformly in $s$, so for $\parallel x^{\mathrm{\prime }}-x{\parallel }_2+\mid t^{\mathrm{\prime }}-t\mid <\rho$ every point of the radial image of $(x^{\mathrm{\prime }},t^{\mathrm{\prime }})$ lies within distance $\rho$ of $R$, hence in $N$, and $(x^{\mathrm{\prime }},t^{\mathrm{\prime }})\in P$.

Joint regularity of the radial integrals on $P$: for every compact box $Q\times J\subseteq P$ and every multi-index $\alpha$ in the variables $(x,t)$ with $\mid \alpha \mid \le k$, the partial derivative ${\mathrm{\partial }}^{\alpha }$ of the integral $\tilde{A}(x,t):={\int }_0^{1}s\tilde{B}(a+s(x-a),t)\times (x-a)ds$ exists on $Q\times J$ and equals ${\int }_0^1{\mathrm{\partial }}^{\alpha }[s\tilde{B}(a+s(x-a),t)\times (x-a)]ds$, and this is jointly continuous on $Q\times J$; the same holds for $\tilde{\mathrm{\Psi }}(x,t):={\int }_0^1⟨\tilde{C}(a+s(x-a),t),x-a⟩ds$. Indeed, on the compact set $(Q\times J)\times [0,1]$ every radial argument $(a+s(x-a),t)$ lies in a compact subset of $N$, and each integrand is a sum of products of the $C^k$ functions $\tilde{B}$, $\tilde{C}$ composed with the smooth map $(x,t,s)\mapsto (a+s(x-a),t)$, the polynomial factor $s$ and the polynomial vector $x-a$, hence is $C^k$ jointly in $(x,t,s)$ by [F4]. Induction on $\mid \alpha \mid$: for $\mid \alpha \mid =0$ joint continuity of the $s$-integral follows from uniform continuity of the integrand on the compact product and [F6]; if the claim holds for $\alpha$, then differentiating the integral in one further spatial or time variable under the integral sign is legitimate by [F5] applied with parameter $s\in [0,1]$ and the other variables frozen at values in $Q\times J$, and the resulting integral is again jointly continuous by the same uniform-continuity argument; at an included endpoint of a nondegenerate $J$ this gives the relative derivative, as stated there.

Restriction to $U\times I$: since $U\times I\subseteq P$ and the extensions agree with $B$ and $C$ there, the maps $A=\tilde{A}{\mathrm{\mid }}_{U\times I}$ and $-\tilde{\mathrm{\Psi }}{\mathrm{\mid }}_{U\times I}$ are the restrictions of the $C^k$ maps $\tilde{A}$ and $-\tilde{\mathrm{\Psi }}$ on the open set $P$, hence are jointly $C^k$ on $U\times I$; the restricted integrals are exactly the formulas displayed in the Statement because $\tilde{B}{\mathrm{\mid }}_{U\times I}=B$ and $\tilde{C}{\mathrm{\mid }}_{U\times I}=C$.

Field identities at each fixed time. Fix $t\in I$. The slice $B(\cdot ,t)$ is $C^1$ on $U$ (as $k\ge 1$) and divergence-free, and $A(\cdot ,t)$ coincides with the radial witness of [F2] for that slice; hence $\mathrm{curl}A(\cdot ,t)=B(\cdot ,t)$ by [F2]. For the scalar witness put ${\mathrm{\Psi }}_C(x,t):={\int }_0^1⟨C(a+s(x-a),t),x-a⟩ds$, so that $\varphi =-{\mathrm{\Psi }}_C$; the slice $C(\cdot ,t)$ is $C^1$ and closed on $U$. Differentiating ${\mathrm{\Psi }}_C$ in $x_j$ under the integral sign, legitimate by step 2.1 with $\mid \alpha \mid =1$, gives ${\mathrm{\partial }}_j{\mathrm{\Psi }}_C(x,t)={\int }_0^1[C_j(a+s(x-a),t)+s{\sum }_i(x-a{)}_i{\mathrm{\partial }}_j{C}_i(a+s(x-a),t)]ds$; closedness of the slice replaces ${\mathrm{\partial }}_j{C}_i$ by ${\mathrm{\partial }}_i{C}_j$, so the integrand is $\frac{d}{ds}[s{C}_j(a+s(x-a),t)]$ by the chain and product rules [F4], and the fundamental theorem [F7] gives ${\mathrm{\partial }}_j{\mathrm{\Psi }}_C(x,t)=C_j(x,t)-0\cdot C_j(a,t)=C_j(x,t)$; hence $\mathrm{\nabla }\varphi (\cdot ,t)=-C(\cdot ,t)$.

Jointly $C^{\mathrm{\infty }}$ data: if $B$ and $C$ are jointly $C^{\mathrm{\infty }}$, choose one smooth extension of each on a common neighbourhood, repeat steps 1.1 through 4.1 for every $k$, and conclude that the same radial integrals are jointly $C^k$ on $U\times I$ for every $k$, hence jointly $C^{\mathrm{\infty }}$, with the same field identities at every $t$.

Degenerate and unbounded time domains: for $I=\mathrm{\varnothing }$ the statements over $U\times I$ are vacuous; for $I=\{t_0\}$ the constant-in-time extensions $\tilde{B}(x,t)=B(x,t_0)$ and $\tilde{C}(x,t)=C(x,t_0)$ are $C^k$ on $U\times \mathbb{R}$ and the construction of steps 1.1 through 4.1 produces the two potentials with the stated spatial identities at $t_0$, while the values of the extension away from $t_0$ are not determined by the original data, so no unique time jets are asserted; for unbounded $I$ the set $P$ is constructed from the given neighbourhood $N$ of $U\times I$ without any fixed uniform enlargement, and the argument applies verbatim. Hence both witnesses are jointly $C^k$ with the stated identities, which is the Statement.