On Identifying Magnetized Anomalies Using Geomagnetic Monitoring Within a Magnetohydrodynamic model

Abstract

This paper is a continuation and an extension of our recent work (Deng et al. in Arch Ration Mech Anal 231(1):153–187, 2019) on the identification of magnetized anomalies using geomagnetic monitoring, which aims to establish a rigorous mathematical theory for the geomagnetic detection technology. Suppose a collection of magnetized anomalies is present in the shell of the Earth. By monitoring the variation of the magnetic field of the Earth due to the presence of the anomalies, we establish sufficient conditions for the unique recovery of these unknown anomalies. Deng et al. (2019), the geomagnetic model was described by a linear Maxwell system. In this paper, we consider a much more sophisticated and complicated magnetohydrodynamic model, which stems from the widely accepted dynamo theory of geomagnetics.

Appendices

Appendix A: Proof of Lemma 2.2

Define

$$\begin{aligned} N_t:=\Vert \Phi _s\Vert _{\mathrm{{TH}}(\mathrm {div}, \partial \Sigma )}+\Vert \Phi _c\Vert _{\mathrm{{TH}}(\mathrm {div}, \partial \Sigma _c)}+\Vert \nabla _{\partial \Sigma }\varphi _s\Vert _{\mathrm{{TH}}(\mathrm {div}, \partial \Sigma _c)}. \end{aligned}$$

To prove Lemma 2.2, we next make use of (2.14)–(2.16) to conduct the asymptotic analysis with respect to $k_s$. Denote by $\Delta _{\partial \Sigma }:=\nabla _{\partial \Sigma }\cdot \nabla _{\partial \Sigma }$. First, from (2.13) one can show that

$$\begin{aligned} \Vert \Delta _{\partial \Sigma } \varphi _s\Vert _{L^2(\partial \Sigma )}=\mathcal {O}(k_s^2N_t). \end{aligned}$$

(A.1)

Next, we derive the inequality

$$\begin{aligned} \Vert \nabla _{\partial \Sigma }\varphi _s\Vert _{\mathrm{TH}(\mathrm{div}, \partial \Sigma )}\leqq C \Vert \Delta _{\partial \Sigma }\varphi _s\Vert _{L^2(\partial \Sigma )}, \end{aligned}$$

(A.2)

where C is a constant depends only on $\Sigma$. In fact, for any $\Psi \in L_T^2(\partial \Sigma )$, by using the Helmholtz decomposition [2, 9], there exists a unique $u\in H^1(\partial \Sigma )$, $\int _{\partial \Sigma } u \mathrm{d}s=0$, such that

$$\begin{aligned} \Psi =\nabla _{\partial \Sigma } u + \nabla _{\partial \Sigma } u\times \nu . \end{aligned}$$

(A.3)

By integration by parts there holds

$$\begin{aligned} \begin{aligned} \Vert \nabla _{\partial \Sigma }\varphi _s\Vert _{L^2(\partial \Sigma )}&=\sup _{\Psi \in L_T^2(\partial \Sigma )}\frac{\langle \nabla _{\partial \Sigma } \varphi _s, \Psi \rangle }{\Vert \Psi \Vert _{\mathrm{TH}(\mathrm{div},\partial \Sigma )}}\\&=\sup _{\Psi \in L_T^2(\partial \Sigma )}\frac{\int _{\partial \Sigma } \nabla _{\partial \Sigma } \varphi _s \cdot \Big (\nabla _{\partial \Sigma } u + \nabla _{\partial \Sigma } u\times \nu \Big )\mathrm{d}s}{\Vert \nabla _{\partial \Sigma } u + \nabla _{\partial \Sigma } u\times \nu \Vert _{L^2(\partial \Sigma )}}\\&=\sup _{\Psi \in L_T^2(\partial \Sigma )}\frac{-\int _{\partial \Sigma } (\Delta _{\partial \Sigma } \varphi _s) u \mathrm{d}s}{\Vert \nabla _{\partial \Sigma } u\Vert _{L^2(\partial \Sigma )} + \Vert \nabla _{\partial \Sigma } u\times \nu \Vert _{L^2(\partial \Sigma )}}\\&\leqq \frac{1}{2}\sup _{\Psi \in L_T^2(\partial \Sigma )}\frac{\Vert \Delta _{\partial \Sigma } \varphi _s\Vert _{L^2(\partial \Sigma )} \Vert u\Vert _{L^2(\partial \Sigma )} }{\Vert \nabla _{\partial \Sigma } u\Vert _{L^2(\partial \Sigma )} } \leqq C \Vert \Delta _{\partial \Sigma } \varphi _s\Vert _{L^2(\partial \Sigma )}, \end{aligned} \end{aligned}$$

(A.4)

where the last inequality is obtained by using the Poincaré inequality (see, for example, [4]). By using (A.1) and (A.2), one thus derives that

$$\begin{aligned} \Vert \nabla _{\partial \Sigma }\varphi _s\Vert _{\mathrm{TH}(\mathrm{div}, \partial \Sigma )}\sim \Vert \Delta _{\partial \Sigma }\varphi _s\Vert _{L^2(\partial \Sigma )}=\mathcal {O}(k_s^2N_t). \end{aligned}$$

(A.5)

Furthermore, from (2.15) one obtains

$$\begin{aligned} \Vert \Phi _s\Vert _{\mathrm{{TH}}(\mathrm {div}, \partial \Sigma )}=\mathcal {O}(k_s^2N_t). \end{aligned}$$

(A.6)

By substituting (A.5)–(A.6) into (2.14), it holds that

$$\begin{aligned} \begin{aligned} \Big (-\frac{I}{2}+\mathcal {M}^{0}_{\Sigma _c}\Big )[\Phi _c] =\nu \times \mathbf {H}_0|_{\partial \Sigma _c}^++\mathcal {O}(k_s^2). \end{aligned} \end{aligned}$$

(A.7)

Since $-\frac{I}{2} + \mathcal {M}^{0}_{\Sigma _c}$ is invertible on $\mathrm{TH}(\mathrm{div}, \partial \Sigma _c)$, by combining (A.5)–(A.7), one can obtain (2.18), which completes the proof.

Appendix B: Proof of Lemma 2.3

By the continuity of $\nu \times \mathbf {H}$ across $\partial \Sigma$ and $\partial D_{l}$, and using the same notation as that in Appendix A and the jump formulas (2.4) and (2.6), one can show that

$$\begin{aligned} \begin{aligned}&\mathcal {M}_{\Sigma ,\Sigma _c}^0[\Psi _0]+\Big (-\frac{I}{2}+\mathcal {M}_{\Sigma }^0\Big )[\Phi _0]+ \mathcal {L}_{\Sigma }^0[\nabla _{\partial \Sigma }\varphi _0]\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {M}_{\Sigma , D_{l'}}^0[\Phi _{l'}]+\mathcal {L}_{\Sigma , D_{l'}}^0[\nabla _{\partial D_{l'}}\varphi _{l'}]\big )\\&\quad =\mathcal {M}_{\Sigma ,\Sigma _c}^{k_s}[\Psi _0]+\Big (\frac{I}{2}+\mathcal {M}_{\Sigma }^{k_s}\Big )[\Phi _0]+ \mathcal {L}_{\Sigma }^{k_s}[\nabla _{\partial \Sigma }\varphi _0]\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {M}_{\Sigma , D_{l'}}^{k_s}[\Phi _{l'}]+\mathcal {L}_{\Sigma , D_{l'}}^{k_s}[\nabla _{\partial D_{l'}}\varphi _{l'}]\big ) \end{aligned} \end{aligned}$$

(B.1)

holds on $\partial \Sigma$ and

$$\begin{aligned} \begin{aligned}&\mathcal {M}_{D_l,\Sigma _c}^{k_s}[\Psi _0]+\mathcal {M}_{D_l,\Sigma }^{k_s}[\Phi _0]+ \mathcal {L}_{D_l, \Sigma }^{k_s}[\nabla _{\partial \Sigma }\varphi _0]-\frac{\Phi _l}{2}\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {M}_{D_l, D_{l'}}^{k_s}[\Phi _{l'}]+\mathcal {L}_{D_l, D_{l'}}^{k_s}[\nabla _{\partial D_{l'}}\varphi _{l'}]\big )\\&\quad =\mathcal {M}_{D_l,\Sigma _c}^{k_l}[\Psi _0]+\mathcal {M}_{D_l,\Sigma }^{k_l}[\Phi _0]+ \mathcal {L}_{D_l, \Sigma }^{k_l}[\nabla _{\partial \Sigma }\varphi _0]+\frac{\Phi _l}{2}\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {M}_{D_l, D_{l'}}^{k_l}[\Phi _{l'}]+\mathcal {L}_{D_l, D_{l'}}^{k_l}[\nabla _{\partial D_{l'}}\varphi _{l'}]\big ) \end{aligned} \end{aligned}$$

(B.2)

holds on $\partial D_l$, $l=1, 2, \ldots , l_0$. On the other hand, by using the continuity of $\nu \cdot \mu \mathbf {H}$ across $\partial \Sigma$ and $\partial D_{l}$ (see Lemma 3.3 in [16]), one can further show that

$$\begin{aligned} \begin{aligned}&\mathcal {N}_{\Sigma ,\Sigma _c}^0[\Psi _0]+\mathcal {N}_{\Sigma }^0[\Phi _0]+ \Big (\frac{I}{2}+(\mathcal {K}_{\Sigma }^0)^*\Big )[\Delta _{\partial \Sigma }\varphi _0]\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {N}_{\Sigma , D_{l'}}^0[\Phi _{l'}]+\mathcal {K}_{\Sigma , D_{l'}}^0[\Delta _{\partial D_{l'}}\varphi _{l'}]\big )\\&\quad =\mathcal {N}_{\Sigma ,\Sigma _c}^{k_s}[\Psi _0]+\mathcal {N}_{\Sigma }^{k_s}[\Phi _0]+ \Big (-\frac{I}{2}+(\mathcal {K}_{\Sigma }^{k_s})^*\Big )[\Delta _{\partial \Sigma }\varphi _0]+k_s^2\nu \cdot \mathcal {A}_{\Sigma }^{k_s}[\nabla _{\partial \Sigma }\varphi _0]\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {N}_{\Sigma , D_{l'}}^{k_s}[\Phi _{l'}] +\mathcal {K}_{\Sigma , D_{l'}}^{k_s}[\Delta _{\partial D_{l'}}\varphi _{l'}]\big ) \end{aligned} \end{aligned}$$

(B.3)

holds on $\partial \Sigma$ and

$$\begin{aligned} \begin{aligned}&\mu _0\Big (\mathcal {N}_{D_l,\Sigma _c}^{k_s}[\Psi _0]+\mathcal {N}_{D_l,\Sigma }^{k_s}[\Phi _0]+ \mathcal {K}_{D_l,\Sigma }^{k_s}[\Delta _{\partial \Sigma }\varphi _0]+\frac{\varphi _l}{2}\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {N}_{D_l, D_{l'}}^{k_s}[\Phi _{l'}]+k_s^2\nu \cdot \mathcal {A}_{D_l, D_{l'}}^{k_s}[\nabla _{\partial D_{l'}}\varphi _{l'}]+\mathcal {K}_{D_l, D_{l'}}^{k_s}[\Delta _{\partial D_{l'}}\varphi _{l'}]\big )\Big )\\&\quad =\mu _l\Big (\mathcal {N}_{D_l,\Sigma _c}^{k_l}[\Psi _0]+\mathcal {N}_{D_l,\Sigma }^{k_l}[\Phi _0]+ \mathcal {K}_{D_l,\Sigma }^{k_l}[\Delta _{\partial \Sigma }\varphi _0]-\frac{\varphi _l}{2}\\&\qquad +\,\sum _{l'=1}^{l_0}\big (\mathcal {N}_{D_l, D_{l'}}^{k_l}[\Phi _{l'}]+k_l^2\nu \cdot \mathcal {A}_{D_l, D_{l'}}^{k_l}[\nabla _{\partial D_{l'}}\varphi _{l'}]+\mathcal {K}_{D_l, D_{l'}}^{k_l}[\Delta _{D_{l'}}\varphi _{l'}]\big )\Big ) \end{aligned} \end{aligned}$$

(B.4)

holds on $\partial D_l$, where $\mathcal {K}_{\Sigma ', \Sigma ''}^{k}: L^2(\partial \Sigma ')\rightarrow L^2(\partial \Sigma '')$, $\Sigma ', \Sigma ''\in \{\Sigma , \Sigma _c, D_1, D_2, \ldots , D_{l_0}\}$, $k\in \{0, k_s, k_1, k_2, \ldots , k_{l_0}\}$, are defined by

$$\begin{aligned} \mathcal {K}_{\Sigma ', \Sigma ''}^k[\varphi ]:=\nu \cdot \nabla \mathcal {S}^{k}_{\Sigma ''}[\varphi ]|_{\partial \Sigma '}^{-}. \end{aligned}$$

(B.5)

If $\Sigma '=\Sigma ''$, then $\mathcal {K}_{\Sigma ', \Sigma '}^k[\varphi ]:=(\mathcal {K}_{\Sigma '}^k)^*$. Similarly, $\mathcal {A}_{\Sigma ', \Sigma ''}^{k}: \mathrm{TH}(\mathrm{div}, \partial \Sigma ')\rightarrow \mathrm{TH}(\mathrm{div}, \partial \Sigma '')$, is defined by

$$\begin{aligned} \mathcal {A}_{\Sigma ', \Sigma ''}^k[\Phi ]:=\mathcal {A}^{k}_{\Sigma ''}[\Phi ]|_{\partial \Sigma '}^{-}. \end{aligned}$$

(B.6)

Furthermore, by using the boundary magnetic field $\nu \times \mathbf {H}$ on $\partial \Sigma _c$, one also has

$$\begin{aligned} \begin{aligned}&\Big (-\frac{I}{2}+\mathcal {M}_{\Sigma _c}^{k_s}\Big )[\Psi _0]+\mathcal {M}_{\Sigma _c,\Sigma }^{k_s}[\Phi _0]+ \mathcal {L}_{\Sigma _c,\Sigma }^{k_s}[\nabla _{\partial \Sigma }\varphi _0]\\&\quad +\,\sum _{l'=1}^{l_0}\big (\mathcal {M}_{\Sigma _c, D_{l'}}^{k_s}[\Phi _{l'}]+\mathcal {L}_{\Sigma _c, D_{l'}}^{k_s}[\nabla _{\partial D_{l'}}\varphi _{l'}]\big )=\nu \times \mathbf {H}.\ \ \ \text{ on }\ \partial \Sigma _c. \end{aligned} \end{aligned}$$

(B.7)

By combining (2.19)–(B.7), along with straightforward calculations and the similar inequality that is deduced in “Appendix A”, one can derive the following estimates sequentially:

$$\begin{aligned} \Vert \Phi _0\Vert _{\mathrm{TH}(\mathrm{div}, \partial \Sigma )}= & {} \mathcal {O}(\omega N_f),\nonumber \\ \Vert \nabla _{\partial \Sigma }\varphi _0\Vert _{\mathrm{TH}(\mathrm{div}, \partial \Sigma )}= & {} \mathcal {O}(\omega N_f),\quad \Vert \Phi _l\Vert _{\mathrm{TH}(\mathrm{div}, \partial D_l)}=\mathcal {O}(\omega N_f), \end{aligned}$$

(B.8)

where $l=1, 2, \ldots , l_0$ and $N_f:=\mathcal {O}(\Vert \Psi _0\Vert _{\mathrm{TH}(\mathrm{div}, \partial \Sigma _c)}+\sum _{l'=1}^{l_0}\Vert \nabla _{\partial D_{l'}}\varphi _{l'}\Vert _{\mathrm{TH}(\mathrm{div}, (\partial D_l)})$, and

$$\begin{aligned} \begin{aligned} \Psi _0&=\Big (-\frac{I}{2}+\mathcal {M}_{\Sigma _c}^0\Big )^{-1}[\nu \times \mathbf {H}]-\sum _{l'=1}^{l_0}\Big (-\frac{I}{2}+\mathcal {M}_{\Sigma _c}^0\Big )^{-1}\mathcal {L}_{\Sigma _c, D_{l'}}^{0}[\Delta _{D_{l'}}\varphi _{l'}]\\&\quad +\,\mathcal {O}\Big (\omega \sum _{l'=1}^{l_0}\Vert \Delta _{D_{l'}}\varphi _{l'}\Vert _{L^2(\partial D_l)}\Big ). \end{aligned} \end{aligned}$$

(B.9)

Hence

$$\begin{aligned} \begin{aligned}&\left( \varsigma _l I -(\mathcal {K}_{D_l}^0)^*+\mathcal {P}_{D_l, \Sigma _c}\mathcal {L}_{\Sigma _c, D_l}^0\right) [\Delta _{\partial D_l}\varphi _l]\\&\qquad -\,\sum _{l'\ne l}^{l_0} \Big (\mathcal {K}_{D_l, D_{l'}}^0-\mathcal {P}_{D_l, \Sigma _c}\mathcal {L}_{\Sigma _c, D_{l'}}^0\Big )[\Delta _{\partial D_{l'}}\varphi _{l'}]\\&\quad =\mathcal {P}_{D_l, \Sigma _c}[\nu \times \mathbf {H}|_{\partial \Sigma _c}]+\mathcal {O}(\omega ), \end{aligned} \end{aligned}$$

(B.10)

where the operators $\mathcal {P}_{D_l, \Sigma _c}: \mathrm{TH}(\mathrm{div}, \partial \Sigma _c)\rightarrow \mathrm{TH}(\mathrm{div}, \partial D_l)$, $l=1, 2, \ldots , l_0$, are defined by

$$\begin{aligned} \mathcal {P}_{D_l, \Sigma _c}[\Phi ]:=\mathcal {N}_{D_l, \Sigma _c}^0\Big (-\frac{I}{2}+\mathcal {M}_{\Sigma _c}^0\Big )^{-1}[\Phi ]. \end{aligned}$$

(B.11)

Noting that $(\mathcal {K}_{D_l}^0)^*$ and $\mathcal {P}_{D_l, \Sigma _c}\mathcal {L}_{\Sigma _c, D_l}^0$ are compact operators on $L^2(\partial D_l)$, one can prove the invertibility of $\varsigma _l I -(\mathcal {K}_{D_l}^0)^*+\mathcal {P}_{D_l, \Sigma _c}\mathcal {L}_{\Sigma _c, D_l}^0$ on $L^2(\partial D_l)$ by following a similar proof of Lemma 2.2 in [14]. In fact, by the Fredholm theory, it suffices to show the uniqueness of a trivial solution to the following integral equation:

$$\begin{aligned} \left( \varsigma _l I -(\mathcal {K}_{D_l}^0)^*+\mathcal {P}_{D_l, \Sigma _c}\mathcal {L}_{\Sigma _c, D_l}^0\right) [\Delta _{\partial D_l}\varphi _l]=0. \end{aligned}$$

(B.12)

Note that there exists only a trial solution to the following system (see Appendix A in [14]):

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle {\nabla \times \mathbf {H}=0, \quad \nabla \cdot \mathbf {H}=0}, &{} \text{ in } \,\ (\mathbb {R}^3{\setminus }\overline{D_l \cup \Sigma _c})\cup D_l,\\ \displaystyle {\nu _l\times \mathbf {H}|_+=\nu _l\times \mathbf {H}|_-,} &{} \text{ on } \, \ \partial D_l, \\ \displaystyle {\mu _0\nu _l\cdot \mathbf {H}|_+=\mu _l\nu _l\cdot \mathbf {H}|_-, }&{} \text{ on } \,\ \partial D_l, \\ \displaystyle {\nu \times \mathbf {H}|_+=0, \quad \int _{\partial \Sigma _c}\nu \cdot \mathbf {H}|_+ =0,} &{}\text{ on } \,\ \partial \Sigma _c, \\ \displaystyle {\mathbf {H}(\mathbf {x})=\mathcal {O}(\Vert \mathbf {x}\Vert ^{-2}), \quad \Vert \mathbf {x}\Vert \rightarrow \infty .} \end{array} \right. \end{aligned}$$

(B.13)

On the other hand, one can verify that

$$\begin{aligned} \mathbf {H}=\Big (-\nabla \mathcal {S}_{D_l}^{0}+\nabla \times \mathcal {A}_{\Sigma _c}^0\left( -\frac{I}{2}+\mathcal {M}_{\Sigma _c}^{0}\right) ^{-1}\nu \times \nabla \mathcal {S}_{D_l}^{0}\Big )[\Delta _{\partial D_l}\varphi _l] \end{aligned}$$

(B.14)

is also the solution to (B.13). Therefore, one has

$$\begin{aligned} \Big (-\nabla \mathcal {S}_{D_l}^{0}+\nabla \times \mathcal {A}_{\Sigma _c}^0\left( -\frac{I}{2}+\mathcal {M}_{\Sigma _c}^{0}\right) ^{-1}\nu \times \nabla \mathcal {S}_{D_l}^{0}\Big )[\Delta _{\partial D_l}\varphi _l]=0 \quad \text{ in } \, \ \mathbb {R}^3{\setminus }\overline{\Sigma _c}. \end{aligned}$$

(B.15)

Hence $\Delta _{\partial D_l}\varphi _l=0$, which proves the unique trial solution to (B.12). Note that $D_l$, $l=1, 2, \ldots , l_0$ are small inclusions which are disjoint from each, one can prove the unique solvability of (B.10) (see Appendix B in [14]).

Appendix C: Harmonic Representation of Vectorial Spherical Polynomials

In this appendix, we shall represent $\mathbf {A}_n^m({\hat{\mathbf {x}}})\xi$, where $\mathbf {A}_n^m({\hat{\mathbf {x}}})$ is defined in (3.10) and $\xi \in \mathbb {R}^3$, in terms of vectorial spherical harmonic functions. Recall that the vectorial spherical harmonic functions of degree n are composed of $\mathbf {M}_{n+1}^m({\hat{\mathbf {x}}})$, $\mathbf {Q}_{n-1}^m({\hat{\mathbf {x}}})$ and $\mathbf {T}_n^m({\hat{\mathbf {x}}})$, which are defined in (3.2) and (3.3). From (3.10) one has

$$\begin{aligned} \begin{aligned} \mathbf {A}_n^m({\hat{\mathbf {x}}})\xi&=(n+1){\hat{\mathbf {x}}}\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi +(n+1)Y_n^m({\hat{\mathbf {x}}})\xi -(n+1)(n+3)Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}{\hat{\mathbf {x}}}^T\xi \\&\quad -\,\nabla _S(\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi )+(n+2)\nabla _SY_n^m({\hat{\mathbf {x}}}) {\hat{\mathbf {x}}}^T\xi . \end{aligned} \end{aligned}$$

(C.1)

By vector calculus identity and integration by parts, we have that

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {T}_n^m({\hat{\mathbf {x}}})}\cdot (\mathbf {A}_n^m({\hat{\mathbf {x}}})\xi ) \mathrm{d}s=(n+1)\int _{\mathbb {S}}Y_n^m({\hat{\mathbf {x}}})\overline{\mathbf {T}_n^m({\hat{\mathbf {x}}})}\cdot \xi \mathrm{d}s\\&\quad =(n+1)\int _{\mathbb {S}}\Vert \mathbf {x}\Vert ^nY_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}\cdot (\xi \times \nabla (\Vert \mathbf {x}\Vert ^n \overline{Y_n^m({\hat{\mathbf {x}}})}))\mathrm{d}s\\&\quad =(n+1)\int _{B_1}\nabla \cdot \big (\Vert \mathbf {x}\Vert ^nY_n^m({\hat{\mathbf {x}}})\xi \times \nabla (\Vert \mathbf {x}\Vert ^n \overline{Y_n^m({\hat{\mathbf {x}}})})\big )\mathrm{d}\mathbf {x}\\&\quad =(n+1)\int _{B_1}\nabla (\Vert \mathbf {x}\Vert ^n \overline{Y_n^m({\hat{\mathbf {x}}})})\cdot \big (\nabla (\Vert \mathbf {x}\Vert ^nY_n^m({\hat{\mathbf {x}}}))\times \xi \big )\mathrm{d}\mathbf {x}=0. \end{aligned} \end{aligned}$$

(C.2)

Hence, $\mathbf {A}_n^m({\hat{\mathbf {x}}})\xi$ is a linear combination of the spherical harmonics $\{\mathbf {N}_{n+1}^m({\hat{\mathbf {x}}})\}$ and $\{\mathbf {Q}_{n-1}^m({\hat{\mathbf {x}}})\}$. By straightforward computations, one can obtain that

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {N}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot ({\hat{\mathbf {x}}}\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi ) \mathrm{d}s=\frac{n'+1}{n'} \int _{\mathbb {S}} \overline{\mathbf {Q}_{n'-1}^{m'}({\hat{\mathbf {x}}})}\cdot ({\hat{\mathbf {x}}}\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi ) \mathrm{d}s\\&\quad =(n'+1)\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi \mathrm{d}s. \end{aligned} \end{aligned}$$

(C.3)

Similarly, one can show that

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {N}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot (Y_n^m({\hat{\mathbf {x}}})\xi ) \mathrm{d}s\\&\quad =-\int _{\mathbb {S}}Y_n^m({\hat{\mathbf {x}}})\nabla _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}^T\xi \mathrm{d}s+(n'+1)\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}^T\xi \mathrm{d}s, \end{aligned} \end{aligned}$$

(C.4)

and

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {Q}_{n'-1}^{m'}({\hat{\mathbf {x}}})}\cdot (Y_n^m({\hat{\mathbf {x}}})\xi ) \mathrm{d}s\\&\quad =\int _{\mathbb {S}}Y_n^m({\hat{\mathbf {x}}})\nabla _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}^T\xi \mathrm{d}s+n'\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}^T\xi \mathrm{d}s, \end{aligned} \end{aligned}$$

(C.5)

and

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {N}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot (Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}{\hat{\mathbf {x}}}^T\xi ) \mathrm{d}s=\frac{n'+1}{n'} \int _{\mathbb {S}} \overline{\mathbf {Q}_{n'-1}^{m'}({\hat{\mathbf {x}}})}\cdot (Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}{\hat{\mathbf {x}}}^T\xi ) \mathrm{d}s\\&\quad =(n'+1)\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}}^T\xi \mathrm{d}s. \end{aligned} \end{aligned}$$

(C.6)

Next, for the last two terms in (C.1), using integration by parts, we have

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {N}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot \nabla _S(\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi ) \mathrm{d}s=-\int _{\mathbb {S}} \overline{\mathbf {Q}_{n'-1}^{m'}({\hat{\mathbf {x}}})}\cdot \nabla _S(\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi ) \mathrm{d}s\\&\quad =-\int _{\mathbb {S}} \nabla _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\cdot \nabla _S(\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi )\mathrm{d}s\\&\quad =\int _{\mathbb {S}} \Delta _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi \mathrm{d}s=-n'(n'+1)\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\nabla _SY_n^m({\hat{\mathbf {x}}})^T\xi \mathrm{d}s. \end{aligned} \end{aligned}$$

(C.7)

Furthermore, it holds that

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {N}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot (\nabla _SY_n^m({\hat{\mathbf {x}}}) {\hat{\mathbf {x}}}^T\xi ) \mathrm{d}s=-\int _{\mathbb {S}} \overline{\mathbf {Q}_{n'-1}^{m'}({\hat{\mathbf {x}}})}\cdot (\nabla _SY_n^m({\hat{\mathbf {x}}}) {\hat{\mathbf {x}}}^T\xi ) \mathrm{d}s\\&\quad =-\int _{\mathbb {S}} \nabla _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\cdot (\nabla _SY_n^m({\hat{\mathbf {x}}}) {\hat{\mathbf {x}}}^T\xi )\mathrm{d}s\\&\quad =-\int _{\mathbb {S}} \nabla _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\cdot \big (\nabla (Y_n^m({\hat{\mathbf {x}}}) {\hat{\mathbf {x}}}^T\xi )-Y_n^m({\hat{\mathbf {x}}})\xi \big )\mathrm{d}s\\&\quad =-n'(n'+1)\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}Y_n^m({\hat{\mathbf {x}}}) {\hat{\mathbf {x}}}^T\xi \mathrm{d}s+\int _{\mathbb {S}} Y_n^m({\hat{\mathbf {x}}})\nabla _S\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}^T\xi \mathrm{d}s. \end{aligned} \end{aligned}$$

(C.8)

In what follows, we define

$$\begin{aligned} \mathbf {a}_{n',n}^{m',m}:=\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}\nabla _SY_n^m({\hat{\mathbf {x}}}) \mathrm{d}s,\quad \mathbf {b}_{n',n}^{m',m}:=\int _{\mathbb {S}}\overline{Y_{n'}^{m'}({\hat{\mathbf {x}}})}Y_n^m({\hat{\mathbf {x}}}){\hat{\mathbf {x}}} \mathrm{d}s. \end{aligned}$$

(C.9)

By combining (C.3)–(C.8) and using (C.1), one thus has

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {S}} \overline{\mathbf {N}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot (\mathbf {A}_n^m({\hat{\mathbf {x}}})\xi ) \mathrm{d}s\\&\quad =\left( (n'+1)(n'+n+1)\mathbf {a}_{n',n}^{m',m}-(n'+1)(n'+n+1)(n+2)\mathbf {b}_{n',n}^{m',m}+\overline{\mathbf {a}_{n,n'}^{m,m'}}\right) ^T\xi , \end{aligned} \end{aligned}$$

(C.10)

and

$$\begin{aligned} \begin{aligned} \int _{\mathbb {S}} \overline{\mathbf {Q}_{n'+1}^{m'}({\hat{\mathbf {x}}})}\cdot (\mathbf {A}_n^m({\hat{\mathbf {x}}})\xi ) \mathrm{d}s=\left( n'(n-n')\mathbf {a}_{n',n}^{m',m}-n'(n-n')(n+2)\mathbf {b}_{n',n}^{m',m}-\overline{\mathbf {a}_{n,n'}^{m,m'}}\right) ^T\xi . \end{aligned} \end{aligned}$$

(C.11)

By using the elementary result (cf. [23])

$$\begin{aligned} \mathbf {a}_{n',n}^{m',m}=\mathbf {b}_{n',n}^{m',m}=0, \quad \text{ for } \text{ any } \, n'\ne n-1, \, n+1, \text{ and } \, m'\ne m-1, m, m+1, \end{aligned}$$

(C.12)

one finally obtains

$$\begin{aligned} \begin{aligned} \mathbf {A}_n^m({\hat{\mathbf {x}}})\xi&=\sum _{m'=m-1}^{m+1}\Big ((\mathbf {c}_{n-1,n}^{m',m})^T\xi \mathbf {N}_{n}^{m'}({\hat{\mathbf {x}}})+(\mathbf {c}_{n+1,n}^{m',m})^T\xi \mathbf {N}_{n+2}^{m'}({\hat{\mathbf {x}}})\\&\quad +\,(\mathbf {d}_{n-1,n}^{m',m})^T\xi \mathbf {Q}_{n-2}^{m'}({\hat{\mathbf {x}}})+(\mathbf {d}_{n+1,n}^{m',m})^T\xi \mathbf {Q}_{n}^{m'}({\hat{\mathbf {x}}})\Big ), \end{aligned} \end{aligned}$$

(C.13)

where

$$\begin{aligned} \mathbf {c}_{n',n}^{m',m}:=\frac{(n'+1)(n'+n+1)\mathbf {a}_{n',n}^{m',m}-(n'+1)(n'+n+1)(n+2)\mathbf {b}_{n',n}^{m',m}+\overline{\mathbf {a}_{n,n'}^{m,m'}}}{(n'+1)(2n'+1)}, \end{aligned}$$

(C.14)

and

$$\begin{aligned} \mathbf {d}_{n',n}^{m',m}:=\frac{n'(n-n')\mathbf {a}_{n',n}^{m',m}-n'(n-n')(n+2)\mathbf {b}_{n',n}^{m',m}-\overline{\mathbf {a}_{n,n'}^{m,m'}}}{n'(2n'+1)}. \end{aligned}$$

(C.15)