Radiation asymptotics and controlled multipole expansion — Examples

Choose $0<r_0<R_0<a$ and apply [F1] to obtain a dimensionless $f\in C^{\mathrm{\infty }}({\mathbb{R}}^3)$ with $0\le f\le 1$, $f=1$ on ${\bar{B}}_{r_0}(0)$ and $\mathrm{supp}f\subset B_{R_0}(0)$. By [F5], $F:={\int }_{{\mathbb{R}}^3}f(y)dy\ge {\lambda }_3(B_{r_0})>0$. By [F2], choose dimensionless ${\chi }_G,{\chi }_H\in C_c^{\mathrm{\infty }}(\mathbb{R})$ with ${\chi }_G=1$ near $u$ and ${\chi }_H=1$ on a neighborhood of $[u-a\mathrm{/}c,u+a\mathrm{/}c]$. Define $G(t):=\alpha {\chi }_G(t)(t-u{)}^2$ and $H(t):=\beta {\chi }_H(t)(t-u{)}^3$, with units $\mathrm{C}{\mathrm{m}}^{-2}$ and $\mathrm{C}{\mathrm{m}}^{-1}$, respectively. Then $G^{\mathrm{\prime }\mathrm{\prime }}(u)=2\alpha$ and $H^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(s)=6\beta$ for every $s\in [u-a\mathrm{/}c,u+a\mathrm{/}c]$.

Let $e_1=(1,0,0)$ and $e_3=(0,0,1)$, and define ${\displaystyle \rho (y,t):=-G(t){\mathrm{\partial }}_{3}f(y),J(y,t):=G^{\mathrm{\prime }}(t)f(y)e_3+H^{\mathrm{\prime }}(t)(-{\mathrm{\partial }}_{3}f(y)e_1+{\mathrm{\partial }}_{1}f(y)e_3)\mathrm{.}}$ These maps $\rho :{\mathbb{R}}^3\times \mathbb{R}\to \mathbb{R}$ and $J:{\mathbb{R}}^3\times \mathbb{R}\to {\mathbb{R}}^3$ have units $\mathrm{C}{\mathrm{m}}^{-3}$ and $\mathrm{A}{\mathrm{m}}^{-2}$: both $G^{\mathrm{\prime }}f$ and $H^{\mathrm{\prime }}{\mathrm{\partial }}_{i}f$ have the latter units. They are jointly smooth and supported in $\mathrm{supp}f\times (\mathrm{supp}G\cup \mathrm{supp}{G}^{\mathrm{\prime }}\cup \mathrm{supp}{H}^{\mathrm{\prime }})$, a compact subset of $B_a\times \mathbb{R}$. Their derivatives satisfy ${\displaystyle {\mathrm{\partial }}_t\rho +{\mathrm{div}}_{y}J=-G^{\mathrm{\prime }}(t){\mathrm{\partial }}_{3}f+G^{\mathrm{\prime }}(t){\mathrm{\partial }}_{3}f+H^{\mathrm{\prime }}(t)(-{\mathrm{\partial }}_1{\mathrm{\partial }}_{3}f+{\mathrm{\partial }}_3{\mathrm{\partial }}_{1}f)=0}$ by [F4], so the source is admissible for the controlled multipole theorem on any nondegenerate compact interval containing $u$.

The current derivative is $J_t=G^{\mathrm{\prime }\mathrm{\prime }}f{e}_3+H^{\mathrm{\prime }\mathrm{\prime }}(-{\mathrm{\partial }}_{3}f{e}_1+{\mathrm{\partial }}_{1}f{e}_3)$. Each integral ${\int }_{{\mathbb{R}}^3}{\mathrm{\partial }}_{i}fdy$ is zero by the divergence theorem applied to $f{e}_i$ on a ball containing $\mathrm{supp}f$, since its boundary flux vanishes. Thus [F3] gives $p^{\mathrm{\prime }\mathrm{\prime }}(u)=\int J_t(y,u)dy=G^{\mathrm{\prime }\mathrm{\prime }}(u)F{e}_3=2\alpha F{e}_3$, which is nonzero and parallel to $n=e_3$. Therefore $n\times (n\times p^{\mathrm{\prime }\mathrm{\prime }}(u))=0$.

For $n=e_3$, the $e_1$ component of the exact shifted current moment is ${\displaystyle V_1(e_3,u)=-{\int }_{B_a}{H}^{\mathrm{\prime }\mathrm{\prime }}(u+y_3\mathrm{/}c){\mathrm{\partial }}_{3}f(y)dy=\frac{1}{c}{\int }_{B_a}f(y)H^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(u+y_3\mathrm{/}c)dy=\frac{6\beta F}{c}>0.}$ For the middle equality, apply [F4] to $f(y)H^{\mathrm{\prime }\mathrm{\prime }}(u+y_3\mathrm{/}c)e_3$ on a ball containing its support. Its boundary flux is zero and its divergence is $H^{\mathrm{\prime }\mathrm{\prime }}(u+y_3\mathrm{/}c){\mathrm{\partial }}_{3}f+f{H}^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(u+y_3\mathrm{/}c)\mathrm{/}c$. Also $\mathrm{\mid }{y}_3\mathrm{\mid }\le a$ on the support of $f$, so $H^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=6\beta$ there by step 1.1. The term $G^{\mathrm{\prime }\mathrm{\prime }}f{e}_3$ contributes no $e_1$ component. Hence $V(e_3,u)$ has a nonzero component perpendicular to $e_3$.

By [F6], $E_{\mathrm{r}\mathrm{a}\mathrm{d}}(e_3,u)=\frac{{\mu }_0}{4\pi }{e}_3\times (e_3\times V(e_3,u))\mathrm{\ne }0$, whereas the dipole coefficient $\frac{{\mu }_0}{4\pi }{e}_3\times (e_3\times p^{\mathrm{\prime }\mathrm{\prime }}(u))$ is zero by step 3.1. Therefore the absolute multipole remainder is nonzero at a direction where the dipole reference amplitude vanishes; no finite uniform relative-error bound with that dipole amplitude as denominator can hold.

The data in the Example exist: by [F1] one may choose a dimensionless $f\in C_c^{\mathrm{\infty }}(B_a(0))$ with $0\le f\le 1$ and $f=1$ on some ball $B_{r_0}(0)$ with $0<r_0<a$. Then $\int fdy\ge {\lambda }_3(B_{r_0})>0$ by [F3]. For any $f$ admitted in the Example write $F:=\int fdy>0$. Choose $G$ by applying the smooth cutoff supplier to a compact interval slightly larger than $[u-a\mathrm{/}c,u+a\mathrm{/}c]$ and multiply that dimensionless cutoff by $\alpha (t-u{)}^2$. The resulting $G$ is compactly supported, has units $\mathrm{C}{\mathrm{m}}^{-2}$, equals $\alpha (t-u{)}^2$ on a neighborhood of the delay interval, and satisfies $G^{\mathrm{\prime }\mathrm{\prime }}=2\alpha$ there.

The maps $\rho ,J$ in the Example are jointly smooth and compactly supported in $B_a$ in space and in $\mathrm{supp}G\cup \mathrm{supp}{G}^{\mathrm{\prime }}$ in time. Their derivatives obey ${\rho }_t=-G^{\mathrm{\prime }}{\mathrm{\partial }}_{3}f$ and $\mathrm{div}J=G^{\mathrm{\prime }}{\mathrm{\partial }}_{3}f$, so ${\rho }_t+\mathrm{div}J=0$. Thus the retarded-history theorem applies.

For $n=e_1$, every source point in $B_a$ has $\mathrm{\mid }{y}_1\mathrm{\mid }<a$, so $G^{\mathrm{\prime }\mathrm{\prime }}(u+y_1\mathrm{/}c)=2\alpha$. The shifted-current moment of [F2] is therefore ${\displaystyle V(e_1,u)={\int }_{B_a}{J}_t(y,u+y_1\mathrm{/}c)dy=2\alpha F{e}_3\mathrm{.}}$ Since $e_1\cdot e_3=0$, $e_1\times (e_1\times V(e_1,u))=-2\alpha F{e}_3\mathrm{\ne }0$. Apply [F2] on any nondegenerate compact interval containing $u$ to obtain ${\displaystyle E_{\mathrm{r}\mathrm{a}\mathrm{d}}(e_1,u)=-\frac{{\mu }_0\alpha F}{2\pi }{e}_3\mathrm{\ne }0,}$ so $rE(r{e}_1,u+r\mathrm{/}c)$ converges to a nonzero vector.

Let $t_f\in \mathbb{R}$ be a fixed observation time and choose $t_{-}<\inf (\mathrm{supp}G\cup \mathrm{supp}{G}^{\mathrm{\prime }})$. For $y\in B_a$ and $x=rn$, $\mathrm{\mid }x-y\mathrm{\mid }\ge r-a$, so the source retarded time satisfies $t_f-\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}c\le t_f-(r-a)\mathrm{/}c$. If $r>a+c(t_f-t_{-})$, this is less than $t_{-}$ for every $y\in B_a$, with a uniform strict margin. The source and all its derivatives vanish on a time neighborhood of every such argument; because the inequalities remain strict on a neighborhood of $(x,t_f)$, the retarded potentials are identically zero there and their derivatives vanish. Hence $E(rn,t_f)=B(rn,t_f)=0$ for all such $r$. This fixed-time limit and the nonzero scaled retarded-time limit of step 3.1 are different paths to infinity, not contradictory field values.