2.1. Shear

To solve the problem analytically, the shear within the fluid layer will be treated in its simplest form, namely a Heaviside step function, with one part being at rest and the other one having a non-zero angular velocity (figure 1). To implement the remote effect of electrical conductivity (or magnetic permeability) anisotropy, we consider one rotor (case A), one stator (case B), or both one rotor and one stator (case C), located at a certain distance $e$ (case A and B), or $e_1$ and $e-e_1$ (case C), from the shear (figure 1).

In cases A and B, space is therefore divided into three cylindrical regions: $r \lt R$ , $R\lt r \lt R+e$ and $r \gt R+e$ , where $r$ is the radial coordinate, as illustrated in figures 1(a) and 1(b). Only one part, the rotor in case A and the stator in case B, is made of an anisotropic material. In case A the fluid inner part ( $R \lt r \lt R+e$ ) rotates at the same angular velocity $\varOmega$ as the rotor, while the fluid outer part ( $r \gt R+e$ ) is at rest. In case B the fluid outer part ( $R \lt r \lt R+e$ ) is at rest as the stator, while the inner part ( $r\lt R$ ) rotates at angular velocity $\varOmega$ . Therefore, in both cases, the shear occurs in the fluid layer at a distance $e$ from the anisotropic medium, namely at (a) $r=R+e$ or (b) $r=R$ .

In case C, space is divided into four cylindrical regions: $r \lt R, R\lt r \lt R+e_1$ , $R+e_1\lt r \lt R+e$ and $r \gt R+e$ , the fluid layer corresponding to the two intermediate parts $R\lt r \lt R+e_1$ and $R+e_1\lt r \lt R+e$ , as illustrated in figure 1(c). Both the rotor ( $r \lt R$ ) and the stator ( $r \gt R+e$ ) are made of the same anisotropic material. The inner part of the fluid layer ( $R\lt r\lt R+e_1$ ) rotates at the same angular velocity as the rotor, while the outer part ( $R+e_1\lt r\lt R+e$ ) is at rest as the stator. The shear then occurs at a distance $e_1$ from the rotor and $e-e_1$ from the stator, namely at $r=R+e_1$ .

In cases A and B, setting $e=0$ corresponds to a shear at the interface $r=R$ , between the fluid layer and the anisotropic material, which has already been studied in Plunian & Alboussière (Reference Plunian and Alboussière2023). In case C, setting $e_1=0$ , or $e-e_1=0$ , corresponds to the two cases studied in Plunian et al. (Reference Plunian, Gomez and Alboussière2025), with a shear at the interface between the fluid layer and either the rotor at $r=R$ or the stator at $r=R+e_1$ . In cases A and B, for $e\gg 1$ , the shear is pushed to an infinitely large distance from the anisotropic material, which makes the dynamo effect impossible. We therefore expect that an increase in $e$ will lead to an increase in the dynamo threshold. In case C, similarly, we expect an increase of the dynamo threshold for $e\gg e_1\gg 1$ . For the applications discussed in § 7, $e$ will be of the same order of magnitude as the rotor radius $e=O(R)$ .

The velocity field is given by $\boldsymbol{{U}}=r\tilde {\varOmega } \boldsymbol{{e}}_{\theta }$ , with

(2.1a) \begin{eqnarray} \tilde {\varOmega }^{({\rm A})}= \left \{ \begin{split} &\varOmega & \\ &\varOmega \\ &0 & \end{split}\!\!,\;\;\; \right . \tilde {\varOmega }^{({\rm B})}= \left \{ \begin{split} &\varOmega & \\ &0 & \\ &0 & \end{split}\!\!\!\!,\;\;\; \right . \text{for} \left \{\begin{split} &0\lt r\lt R \\ &R\lt r\lt R+e\\ &r\gt R+e \end{split}\,, \right . \end{eqnarray}

(2.1b) \begin{eqnarray} \tilde {\varOmega }^{({\rm C})}= \left \{ \begin{split} &\varOmega \\ &\varOmega \\ &0 \\ &0 \end{split}\,,\;\;\; \right . \text{for} \left \{\begin{split} &0\lt r\lt R \\ &R\lt r\lt R+e_1\\ &R+e_1\lt r\lt R+e\\ &r\gt R+e \end{split}\;\;\;, \right . \end{eqnarray}

where $(\boldsymbol{{e}}_r,\boldsymbol{{e}}_{\theta },\boldsymbol{{e}}_z)$ and $(r, \theta , z)$ are unit basis vectors and coordinates in cylindrical geometry, and where the superscripts $({\rm A}), ({\rm B})$ and $({\rm C})$ correspond to the three cases introduced above and depicted in figure 1.

2.2. Anisotropy

In each domain, we define two electrical conductivities and two magnetic permeabilities, $\sigma ^{\parallel }$ and $\mu ^{\parallel }$ in a given direction $\boldsymbol{{q}}$ , $\sigma ^{\perp }$ and $\mu ^{\perp }$ in the directions perpendicular to $\boldsymbol{{q}}$ , where $\boldsymbol{{q}}$ is a unit vector. In the direction parallel to $\boldsymbol{{q}}$ , Ohm’s law and the relation between $\boldsymbol{{H}}$ and $\boldsymbol{{B}}$ are written in the formFootnote ¹ $\boldsymbol{{J}}\boldsymbol{\cdot }\boldsymbol{{q}}=\sigma ^{\parallel } (\boldsymbol{{E}}\boldsymbol{\cdot }\boldsymbol{{q}})$ and $\boldsymbol{{B}}\boldsymbol{\cdot }\boldsymbol{{q}}=$ $\mu ^{\parallel }(\boldsymbol{{H}}\boldsymbol{\cdot }\boldsymbol{{q}})$ , while in the directions perpendicular to $\boldsymbol{{q}}$ they are written as $\boldsymbol{{J}}-(\boldsymbol{{J}}\boldsymbol{\cdot }\boldsymbol{{q}})$ $\boldsymbol{{q}}=\sigma ^{\perp } (\boldsymbol{{E}}-(\boldsymbol{{E}}\boldsymbol{\cdot }\boldsymbol{{q}})\boldsymbol{{q}})$ and $\boldsymbol{{B}}-(\boldsymbol{{B}}\boldsymbol{\cdot }\boldsymbol{{q}})\boldsymbol{{q}}=\mu ^{\perp }(\boldsymbol{{H}}-(\boldsymbol{{H}}\boldsymbol{\cdot }\boldsymbol{{q}})\boldsymbol{{q}})$ . This leads to two symmetric tensors, $\lbrack \sigma _{ij}\rbrack$ for the electrical conductivity and $\lbrack \mu _{ij}\rbrack$ for the magnetic permeability, defined by

(2.2a,b) \begin{align} \lbrack \sigma _{ij}\rbrack =\sigma ^{\perp } \delta _{ij} + (\sigma ^{\parallel }-\sigma ^{\perp })q_i q_j, \;\;\;\;\;\;\;\;\; \lbrack \mu _{ij}\rbrack =\mu ^{\perp } \delta _{ij} + (\mu ^{\parallel }-\mu ^{\perp })q_i q_j. \end{align}

For a fluid, with isotropic electrical conductivity $\sigma _F$ and magnetic permeability $\mu _F$ , the expressions (2.2a , b ) remain valid, with $\sigma ^{\perp }=\sigma ^{\parallel }=\sigma _F$ and $\mu ^{\perp }=\mu ^{\parallel }=\mu _F$ . Anticipating a renormalisation of the electrical conductivity and magnetic permeability by $\sigma ^{\perp }_{\rm }$ and $\mu ^{\perp }_{\rm }$ , it is then useful to rewrite (2.2a , b ) as

(2.3a,b) \begin{align} \lbrack \sigma _{ij}\rbrack =\sigma ^{\perp }_{\rm }\; \tilde {\varphi }_{\sigma } \left (\delta _{ij} -\frac {\tilde {\sigma }}{1+\tilde {\sigma }}q_i q_j\right ), \;\;\;\;\;\;\;\;\; \lbrack \mu _{ij}\rbrack =\mu ^{\perp }_{\rm }\; \tilde {\varphi }_{\mu } \left (\delta _{ij} -\frac {\tilde {\mu }}{1+\tilde {\mu }}q_i q_j\right ), \end{align}

with, in case A,

(2.4a–d) \begin{align} \tilde {\varphi }_{\sigma }^{({\rm A})}= \left \{ {\begin{matrix} &1 \\ &\varphi _{\sigma } \\ &\varphi _{\sigma } \end{matrix}} \right . , \quad \tilde {\sigma }^{({\rm A})}= \left \{ {\begin{matrix} &\sigma \\ &0\\ &0 \end{matrix}} \right . , \quad \tilde {\varphi }_{\mu }^{({\rm A})}= \left \{ {\begin{matrix} &1 \\ &\varphi _{\mu }\\ &\varphi _{\mu } \end{matrix}} \right . , \quad \tilde {\mu }^{({\rm A})}= \left \{ {\begin{matrix} &\mu \\ &0\\ &0 \end{matrix}} \right . , \quad \text{for} \left \{ {\begin{array}{l} r\lt R \\ R\lt r\lt R+e\\ r\gt R+e \end{array}}\right. ,\end{align}

in case B,

(2.5a–d) \begin{align} \tilde {\varphi }_{\sigma }^{({\rm B})}= \left \{ \begin{matrix} &\varphi _{\sigma } \\ &\varphi _{\sigma } \\ &1 \end{matrix} \right . ,\quad \tilde {\sigma }^{({\rm B})}= \left \{ \begin{matrix} &0 \\ &0\\ &\sigma \end{matrix} \right . , \quad \tilde {\varphi }_{\mu }^{({\rm B})}= \left \{ \begin{matrix} &\varphi _{\mu } \\ &\varphi _{\mu }\\ &1 \end{matrix} \right . , \quad \tilde {\mu }^{({\rm B})}= \left \{ \begin{matrix} &0 \\ &0\\ &\mu \end{matrix} \right . , \quad \text{for} \left \{ {\begin{array}{l} r\lt R \\ R\lt r\lt R+e\\ r\gt R+e \end{array}}\right. , \end{align}

in case C,

(2.6a–d) \begin{align} \tilde {\varphi }_{\sigma }^{({\rm C})}= \left \{ \begin{matrix} &1\\ &\varphi _{\sigma } \\ &\varphi _{\sigma } \\ &1 \end{matrix} \right . , \quad\! \tilde {\sigma }^{({\rm C})}= \left \{ \begin{matrix} &\sigma \\ &0 \\ &0\\ &\sigma \end{matrix} \right . , \quad\! \tilde {\varphi }_{\mu }^{({\rm C})}= \left \{ \begin{matrix} &1\\ &\varphi _{\mu } \\ &\varphi _{\mu }\\ &1 \end{matrix} \right . , \quad\!\! \tilde {\mu }^{({\rm C})}= \left \{ \begin{matrix} &\mu \\ &0 \\ &0\\ &\mu \end{matrix} \right . , \quad\!\! \text{for} \; \left \{ {\begin{array}{l} r\lt R \\ R\lt r\lt R+e_1\\ R+e_1\lt r\lt R+e\\ r\gt R+e \end{array}}\right. , \end{align}

and

(2.7a–d) \begin{align} \varphi _{\sigma }=\frac {\sigma _F}{\sigma ^{\perp }}, \;\;\;\; \sigma =\frac {\sigma ^{\perp }}{\sigma ^{\parallel }}-1, \;\;\;\; \varphi _{\mu }=\frac {\mu _F}{\mu ^{\perp }}, \;\;\;\; \mu =\frac {\mu ^{\perp }}{\mu ^{\parallel }}-1.\end{align}

As in previous studies on the anisotropic dynamo (Plunian & Alboussière Reference Plunian and Alboussière2020, Reference Plunian and Alboussière2021, Reference Plunian and Alboussière2022, Reference Plunian and Alboussière2023; Gomez et al. Reference Gomez, Plunian and Alboussière2025; Plunian et al. Reference Plunian, Gomez and Alboussière2025), we choose $\boldsymbol{{q}}$ as a unit vector in the horizontal plane defined by

(2.8) \begin{equation} \boldsymbol{{q}}=c \; \boldsymbol{{e}}_r+s \; \boldsymbol{{e}}_{\theta }, \end{equation}

where $c=\cos \alpha _q$ , $s=\sin \alpha _q$ , with $\alpha _q$ a prescribed pitch angle. The vector $\boldsymbol{{q}}$ follows the direction of logarithmic spirals that are perpendicular to those depicted in figure 1 (see Ruderman & Ruzmaikin Reference Ruderman and Ruzmaikin1984; Alboussière et al. Reference Alboussière, Drif and Plunian2020 for an alternative expression in Cartesian geometry).

2.3. Induction equation

In the magnetohydrodynamic approximation, Maxwell’s equations and Ohm’s law take the forms

(2.9a) \begin{align} \boldsymbol{{H}}&=\lbrack \mu _{ij}\rbrack ^{-1}\boldsymbol{{B}},\\[-10pt]\nonumber \end{align}

(2.9b) \begin{align} \boldsymbol{\nabla }\boldsymbol{\cdot }\boldsymbol{{B}} &= 0,\\[-10pt]\nonumber \end{align}

(2.9c) \begin{align} \boldsymbol{{J}}&=\boldsymbol{\nabla }\times \boldsymbol{{H}},\\[-10pt]\nonumber \end{align}

(2.9d) \begin{align} \partial _t \boldsymbol{{B}} &= - \boldsymbol{\nabla }\times \boldsymbol{{E}}\\[-10pt]\nonumber \end{align}

(2.9e) \begin{align} \boldsymbol{{J}}&=\lbrack \sigma _{ij}\rbrack (\boldsymbol{{E}} + \boldsymbol{{U}} \times \boldsymbol{{B}}), \end{align}

leading to the induction equation,

(2.10) \begin{equation} \partial _t \boldsymbol{{B}} = \boldsymbol{\nabla }\times (\boldsymbol{{U}} \times \boldsymbol{{B}}) - \boldsymbol{\nabla }\times \left ( \lbrack \sigma _{ij}\rbrack ^{-1} \boldsymbol{\nabla }\times \left ( \lbrack \mu _{ij}\rbrack ^{-1}\boldsymbol{{B}} \right ) \right ), \end{equation}

where

(2.11a,b) \begin{align} \lbrack \sigma _{ij}\rbrack ^{-1}=\left (\tilde {\varphi }_{\sigma }\sigma ^{\perp }\right )^{-1} \left (\delta _{ij} + \tilde {\sigma } q_i q_j \right ), \;\;\;\;\;\;\;\;\; \lbrack \mu _{ij}\rbrack ^{-1}= \left (\tilde {\varphi }_{\mu }\mu ^{\perp }\right )^{-1} \left (\delta _{ij} + \tilde {\mu } q_i q_j \right ).\end{align}

Equations (2.11a,b ) have been derived from (2.3a,b ) knowing that, for any unit vector $\boldsymbol{{q}}$ and any scalar quantity $X \neq -1$ , we have

(2.12) \begin{equation} \left [\delta _{ij} -\frac {X}{1+X}q_i q_j\right ]^{-1} = \delta _{ij} + X q_i q_j. \end{equation}

With regard to the induction equation (2.10), and for an isotropic material such that $\lbrack \sigma _{ij}\rbrack$ and $\lbrack \mu _{ij}\rbrack$ are reduced to scalar quantities, it has been shown, using duality arguments, that swapping the electrical conductivity and magnetic permeability leads to the same dynamo threshold, provided that there is a change of sign in the velocity field and that the boundary conditions of the conducting domain can also be swapped (infinitely conducting with ferromagnetic) (Favier & Proctor Reference Favier and Proctor2013; Marcotte et al. Reference Marcotte, Gallet, Pétrélis and Gissinger2021). This demonstration can easily be generalised to the current case of an anisotropic material, and for the boundary conditions (3.1c–d ) given in § 3.1. Consequently, we expect to have the same dynamo threshold by swapping $\sigma$ and $\mu$ , $\varphi _{\sigma }$ and $\varphi _{\mu }$ , $\varOmega$ and $-\varOmega$ .

Since the velocity is stationary and independent of $z$ and $\theta$ , we look for a magnetic induction in the form

(2.13) \begin{equation} \boldsymbol{{B}}(r)\exp (\gamma t + \mathrm{i} kz+im\theta ), \end{equation}

where $\boldsymbol{{B}}(r)$ is the magnetic mode at vertical wavenumber $k$ and azimuthal wavenumber $m$ . In (2.13) a positive value of the real part of the magnetic growth rate $\gamma$ is the signature of dynamo action, the dynamo threshold corresponding to $\Re \{\gamma \}=0$ . The present work is restricted to the axisymmetric mode $m=0$ , motivated by the fact that it does not involve azimuthal gradients and is therefore the least dissipative candidate. This is also consistent with previous experimental studies of anisotropic dynamo (Alboussière et al. Reference Alboussière, Plunian and Moulin2022; Avalos-Zúñiga & Priede Reference Avalos-Zúñiga and Priede2023) in which the magnetic field is indeed axisymmetric. As mentioned in § 1.2, the generation of an axisymmetric mode $m=0$ is possible because in the anisotropic material the electric currents are deflected, implying that Cowling’s antidynamo theorem (Cowling Reference Cowling1934) does not hold (Plunian & Alboussière Reference Plunian and Alboussière2020).

Replacing (2.1a ), (2.1b ) and (2.13) in (2.10), and after some algebra (see, e.g. Plunian & Alboussière Reference Plunian and Alboussière2023), we obtain the following equations for $B_r(r)$ and $B_{\theta }(r)$ :

(2.14a) \begin{align} \tilde {\gamma } B_r &= -(\sigma ^{\perp }\mu ^{\perp })^{-1} \big[ \tilde {\mu } c^2 k^2 B_r + (1+\tilde {\sigma } s^2) D_k(B_r) - cs(\tilde {\sigma } -\tilde {\mu }) k^2 B_{\theta } \big],\\[-7pt]\nonumber \end{align}

(2.14b) \begin{align} \tilde {\gamma } B_{\theta } &= -(\sigma ^{\perp }\mu ^{\perp })^{-1} \left [ \tilde {\sigma } c^2 k^2 B_{\theta } + (1+\tilde {\mu } s^2)D_k (B_{\theta }) - cs(\tilde {\sigma } -\tilde {\mu }) D_k(B_r) \right ] . \end{align}

Here $\tilde {\gamma }=\tilde {\varphi }_{\sigma }\tilde {\varphi }_{\mu }\gamma$ is given by

(2.15a) \begin{eqnarray} \tilde {\gamma }^{({\rm A})}=\gamma \left \{ \begin{split} &1& \\ &\varphi _{\sigma }\varphi _{\mu } \\ &\varphi _{\sigma }\varphi _{\mu }& \end{split}\!\!\!, \right . \;\;\; \tilde {\gamma }^{({\rm B})}=\gamma \left \{ \begin{split} &\varphi _{\sigma }\varphi _{\mu } \\ &\varphi _{\sigma }\varphi _{\mu } \\ &1 \end{split} \right . , \;\;\;\text{for} \left \{ \begin{split} & r\lt R \\ & R\lt r\lt R+e \\ & r\gt R+e \end{split} \right .\! , \end{eqnarray}

(2.15b) \begin{eqnarray} \tilde {\gamma }^{({\rm C})}=\gamma \left \{ \begin{split} &1\\ &\varphi _{\sigma }\varphi _{\mu } \\ &\varphi _{\sigma }\varphi _{\mu } \\ &1 \end{split} \right . ,\;\;\; \text{for} \;\;\; \left \{ \begin{split} &r\lt R\\ &R\lt r\lt R+e_1 \\ &R+e_1\lt r\lt R+e \\ &r\gt R+e \end{split} \right . , \end{eqnarray}

and

(2.16) \begin{equation} D_{k}(X)=k^2X-\frac {\partial }{\partial r}\left (\frac {1}{r}\frac {\partial }{\partial r}(rX)\right ). \end{equation}

3.1. Boundary conditions

The system of equations (2.14a,b ) is completed by the appropriate boundary conditions for $r=0$ and $r\rightarrow \infty$ ,

(3.1a–d) \begin{align} B_r(r=0)= B_{\theta }(r=0)= \lim _{r\rightarrow \infty }B_r=\lim _{r\rightarrow \infty }B_{\theta } = 0,\end{align}

and by the continuity across $r=R$ and $r=R+e$ of the normal component of $\boldsymbol{{B}}$ and of the tangential components of $\boldsymbol{{H}}$ and $\boldsymbol{{E}}$ ,

(3.2a–e) \begin{align} \left [B_r\right ]^{R^+}_{R^-} = \left [H_{\theta }\right ]^{R^+}_{R^-} =\left [H_z\right ]^{R^+}_{R^-} =\left [E_{\theta }\right ]^{R^+}_{R^-} =\left [E_z\right ]^{R^+}_{R^-} =0,\\[-10pt]\nonumber \end{align}

(3.3a–e) \begin{align} \left [B_r\right ]^{(R+e)^+}_{(R+e)^-} = \left [H_{\theta }\right ]^{(R+e)^+}_{(R+e)^-} =\left [H_z\right ]^{(R+e)^+}_{(R+e)^-} =\left [E_{\theta }\right ]^{(R+e)^+}_{(R+e)^-} =\left [E_z\right ]^{(R+e)^+}_{(R+e)^-} =0,\end{align}

and for case C, by the continuity across $r=R+e_1$ of the same quantities,

(3.4a–e) \begin{align} \left [B_r\right ]^{(R+e_1)^+}_{(R+e_1)^-} = \left [H_{\theta }\right ]^{(R+e_1)^+}_{(R+e_1)^-} =\left [H_z\right ]^{(R+e_1)^+}_{(R+e_1)^-} =\left [E_{\theta }\right ]^{(R+e_1)^+}_{(R+e_1)^-} =\left [E_z\right ]^{(R+e_1)^+}_{(R+e_1)^-} =0,\end{align}

where $[X]^{r_2}_{r_1} =X(r=r_2)-X(r=r_1)$ .

From (2.9a ) and (2.11b ), the magnetic field $\boldsymbol{{H}}$ takes the form

(3.5) \begin{equation} \boldsymbol{{H}}=\left (\tilde {\varphi }_{\mu }\mu ^{\perp }\right )^{-1}\begin{pmatrix} (1+\tilde {\mu } c^2)B_r+ \tilde {\mu } cs B_{\theta }\\[3pt] \tilde {\mu } cs B_r + (1+\tilde {\mu } s^2)B_{\theta }\\[3pt] B_z \end{pmatrix}, \end{equation}

with, from (2.9b ),

(3.6) \begin{equation} B_z = \mathrm{i} k^{-1} \left(\frac {B_r}{r}+\frac {\partial B_r}{\partial r}\right). \end{equation}

Then (3.2a – c ) and (3.3a – c ) can be rewritten, in case A, as

(3.7a) \begin{eqnarray} &B_r^{({\rm A})}(R^+)=B_r^{({\rm A})}(R^-),\end{eqnarray}

(3.7b) \begin{eqnarray} &\varphi _{\mu }^{-1}B_{\theta }^{({\rm A})}(R^+) =\mu cs B_r^{({\rm A})}(R^-)+(1+\mu s^2)B_{\theta }^{({\rm A})}(R^-), \end{eqnarray}

(3.7c) \begin{eqnarray} &\varphi _{\mu }^{-1}\left (\frac {B_r^{({\rm A})}(R^+)}{R} + \frac {\partial B_r^{({\rm A})}}{\partial r}(R^+)\right )=\frac {B_r^{({\rm A})}(R^-)}{R} + \frac {\partial B_r^{({\rm A})}}{\partial r}(R^-), \end{eqnarray}

(3.7d) \begin{eqnarray} &B_r^{({\rm A})}((R+e)^-)=B_r^{({\rm A})}((R+e)^+), \end{eqnarray}

(3.7e) \begin{eqnarray} &B_{\theta }^{({\rm A})}((R+e)^-) =B_{\theta }^{({\rm A})}((R+e)^+),\end{eqnarray}

(3.7f) \begin{align} \frac {B_r^{({\rm A})}((R+e)^-)}{(R+e)} + \frac {\partial B_r^{({\rm A})}}{\partial r}((R+e)^-)=\frac {B_r^{({\rm A})}((R+e)^+)}{(R+e)} + \frac {\partial B_r^{({\rm A})}}{\partial r}((R+e)^+), \end{align}

in case B, as

(3.8a) \begin{eqnarray} &B_r^{({\rm B})}(R^+)=B_r^{({\rm B})}(R^-), \end{eqnarray}

(3.8b) \begin{eqnarray} &B_{\theta }^{({\rm B})}(R^+) =B_{\theta }^{({\rm B})}(R^-), \end{eqnarray}

(3.8c) \begin{eqnarray} &\frac {B_r^{({\rm B})}(R^+)}{R} + \frac {\partial B_r^{({\rm B})}}{\partial r}(R^+)=\frac {B_r^{({\rm B})}(R^-)}{R} + \frac {\partial B_r^{({\rm B})}}{\partial r}(R^-) ,\end{eqnarray}

(3.8d) \begin{eqnarray} &B_r^{({\rm B})}((R+e)^-)=B_r^{({\rm B})}((R+e)^+) ,\end{eqnarray}

(3.8e) \begin{eqnarray} &\varphi _{\mu }^{-1}B_{\theta }^{({\rm B})}((R+e)^-) =\mu cs B_r^{({\rm B})}((R+e)^+) +(1+\mu s^2) B_{\theta }^{({\rm B})}((R+e)^+), \end{eqnarray}

(3.8f) \begin{align} \varphi _{\mu }^{-1}\left (\frac {B_r^{({\rm B})}((R+e)^-)}{(R+e)} + \frac {\partial B_r^{({\rm B})}}{\partial r}((R+e)^-)\right )=\frac {B_r^{({\rm B})}((R+e)^+)}{(R+e)} + \frac {\partial B_r^{({\rm B})}}{\partial r}((R+e)^+). \end{align}

In case C, (3.2a – c ), (3.3a – c ) and (3.4a – c ) can be rewritten as

(3.9a) \begin{eqnarray} &B_r^{({\rm C})}(R^+)=B_r^{({\rm C})}(R^-), \end{eqnarray}

(3.9b) \begin{eqnarray} &\varphi _{\mu }^{-1}B_{\theta }^{({\rm C})}(R^+) =\mu cs B_r^{({\rm C})}(R^-)+(1+\mu s^2)B_{\theta }^{({\rm C})}(R^-), \end{eqnarray}

(3.9c) \begin{eqnarray} &\varphi _{\mu }^{-1}\left (\frac {B_r^{({\rm C})}(R^+)}{R} + \frac {\partial B_r^{({\rm C})}}{\partial r}(R^+)\right )=\frac {B_r^{({\rm C})}(R^-)}{R} + \frac {\partial B_r^{({\rm C})}}{\partial r}(R^-) ,\end{eqnarray}

(3.9d) \begin{eqnarray} &B_r^{({\rm C})}((R+e_1)^+)=B_r^{({\rm C})}((R+e_1)^-) ,\end{eqnarray}

(3.9e) \begin{eqnarray} &B_{\theta }^{({\rm C})}((R+e_1)^+) =B_{\theta }^{({\rm C})}((R+e_1)^-) ,\end{eqnarray}

(3.9f) \begin{align} \frac {B_r^{({\rm C})}((R+e_1)^+)}{R+e_1} + \frac {\partial B_r^{({\rm C})}}{\partial r}((R+e_1)^+)=\frac {B_r^{({\rm C})}((R+e_1)^-)}{R+e_1} + \frac {\partial B_r^{({\rm C})}}{\partial r}((R+e_1)^-) ,\end{align}

(3.9g) \begin{eqnarray} &B_r^{({\rm C})}((R+e)^-)=B_r^{({\rm C})}((R+e)^+) ,\end{eqnarray}

(3.9h) \begin{eqnarray} &\varphi _{\mu }^{-1}B_{\theta }^{({\rm C})}((R+e)^-) =\mu cs B_r^{({\rm C})}((R+e)^+) +(1+\mu s^2)B_{\theta }^{({\rm C})}((R+e)^+), \end{eqnarray}

(3.9i) \begin{align} \varphi _{\mu }^{-1}\left (\frac {B_r^{({\rm C})}((R+e)^-)}{(R+e)} + \frac {\partial B_r^{({\rm C})}}{\partial r}((R+e)^-)\right )=\frac {B_r^{({\rm C})}((R+e)^+)}{(R+e)} + \frac {\partial B_r^{({\rm C})}}{\partial r}((R+e)^+). \end{align}

From (2.9d ) we have $E_{\theta }=-\mathrm{i} k^{-1}\gamma B_r$ , implying that the two conditions (3.2a ) and (3.2d ) are redundant, as are (3.3a ) and (3.3d ), and (3.4a ) and (3.4d ). As for the last boundary conditions (3.2e ), (3.3e ) and (3.4e ), using (2.9e ) we have

(3.10) \begin{equation} \tilde {\varphi }_{\sigma }^{-1}J_z + \sigma ^{\perp } r \tilde {\varOmega } B_r=\sigma ^{\perp }E_z \end{equation}

with $\tilde {\varOmega }$ defined in (2.1a ) and (2.1b ), leading to in case A,

(3.11a) \begin{align} \varphi _{\sigma }^{-1}J_z^{({\rm A})}(R^+) + \sigma ^{\perp }\varOmega R B_r^{({\rm A})}(R^{+})= J_z^{({\rm A})}(R^-)+ \sigma ^{\perp } \varOmega R B_r^{({\rm A})}(R^{-}), \end{align}

(3.11b) \begin{align} \varphi _{\sigma }^{-1}J_z^{({\rm A})}((R+e)^+)= \varphi _{\sigma }^{-1} J_z^{({\rm A})}((R+e)^-)+\sigma ^{\perp } \varOmega (R+e) B_r^{({\rm A})}((R+e)^{-}), \end{align}

in case B,

(3.12a) \begin{align} \varphi _{\sigma }^{-1}J_z^{({\rm B})}(R^+) = \varphi _{\sigma }^{-1} J_z^{({\rm B})}(R^-)+ \sigma ^{\perp } \varOmega R B_r^{({\rm B})}(R^{-}), \end{align}

(3.12b) \begin{align} J_z^{({\rm B})}((R+e)^+)= \varphi _{\sigma }^{-1} J_z^{({\rm B})}((R+e)^-), \end{align}

and in case C,

(3.13a) \begin{align} \varphi _{\sigma }^{-1}J_z^{({\rm C})}(R^+) + \sigma ^{\perp }\varOmega R B_r^{({\rm C})}(R^{+})= J_z^{({\rm C})}(R^-)+ \sigma ^{\perp } \varOmega R B_r^{({\rm C})}(R^{-}), \end{align}

(3.13b) \begin{align} \varphi _{\sigma }^{-1}J_z^{({\rm C})}((R+e_1)^+)= \varphi _{\sigma }^{-1} J_z^{({\rm C})}((R+e_1)^-)+\sigma ^{\perp } \varOmega (R+e_1) B_r^{({\rm C})}((R+e_1)^{-}), \end{align}

(3.13c) \begin{align} J_z^{({\rm C})}((R+e)^+)= \varphi _{\sigma }^{-1} J_z^{({\rm C})}((R+e)^-). \end{align}

3.2. Magnetic Reynolds number and renormalisation

By defining the magnetic Reynolds number in terms of the radius $R$ , the angular velocity $\varOmega$ , the electrical conductivity and magnetic permeability, $\sigma ^{\perp }$ and $\mu ^{\perp }$ in cases A and C, and $\sigma ^{F}$ and $\mu ^{F}$ in case B, we obtain

(3.14a–c) \begin{align} Rm^{({\rm A})}=\sigma ^{\perp } \mu ^{\perp } R^2 |\varOmega ^{({\rm A})}|, \ Rm^{({\rm B})}=\sigma ^{F} \mu ^{F} R^2 |\varOmega ^{({\rm B})}|, \ Rm^{({\rm C})}=\sigma ^{\perp } \mu ^{\perp } R^2 |\varOmega ^{({\rm C})}|.\nonumber\\[4pt] \end{align}

Renormalising the distance and time by $R$ and $\sigma ^{\perp }\mu ^{\perp }R^2$ , and the current density $\boldsymbol{{J}}$ by $(\mu ^{\perp }R)^{-1}$ , corresponds to replacing $R$ , $\sigma ^{\perp }$ and $\mu ^{\perp }$ by unity in the system (2.14a , b ) and in the boundary conditions (3.7a )–(3.13c ). Finally, for the sake of clarity, in §§ 4, 5 and 6 the notations $\varOmega$ , $e$ and $e_1$ relate to dimensionless quantities. Then,

(3.15a–c) \begin{align} Rm^{({\rm A})}=|\varOmega ^{({\rm A})}|, \ Rm^{({\rm B})}=\varphi _{\sigma }\varphi _{\mu } |\varOmega ^{({\rm B})}|, \ Rm^{({\rm C})}=|\varOmega ^{({\rm C})}|. \end{align}

4.1. Derivation of the differential equations satisfied by $B_r$ and $B_{\theta }$

As we are interested in the dynamo threshold, the system (2.14a , b ) is solved for $\gamma =0$ , taking the following form:

(4.1a) \begin{align}\kern-18pt D_k(B_r) = -\left [\tilde {\mu } c^2k^2 B_r + \tilde {\sigma } s^2 D_k(B_r) - cs(\tilde {\sigma } -\tilde {\mu })k^2 B_{\theta } \right ] \end{align}

(4.1b) \begin{align} D_k (B_{\theta }) = - \left [\tilde {\sigma } c^2k^2 B_{\theta } + \tilde {\mu } s^2D_k (B_{\theta }) - cs(\tilde {\sigma } -\tilde {\mu }) D_k(B_r) \right ] . \end{align}

Considering the following linear combinations, $c$ (4.1a )+ $s$ (4.1b ) on the one hand, $s D_k$ (4.1a )- $ck^2$ (4.1b ) on the other hand, leads to

(4.2a) \begin{eqnarray} (1+\tilde {\mu } s^2) D_k(c B_r + s B_{\theta })&=& - \tilde {\mu } c^2 k^2 (c B_r + s B_{\theta }), \end{eqnarray}

(4.2b) \begin{eqnarray} (1+\tilde {\sigma } s^2) D_k(s D_k(B_r) -ck^2 B_{\theta })&=& - \tilde {\sigma } c^2 k^2 (s D_k(B_r) -ck^2 B_{\theta }). \end{eqnarray}

Introducing

(4.3a,b) \begin{align} k_{\tilde {\sigma }}=k\left (\frac {1+\tilde {\sigma }}{1+\tilde {\sigma } s^2}\right )^{1/2}, \qquad k_{\tilde {\mu }}=k\left (\frac {1+\tilde {\mu }}{1+\tilde {\mu } s^2}\right )^{1/2}, \end{align}

and with the help of the identity

(4.4) \begin{equation} D_{k_1}(X)=D_{k_2}(X) +(k_1^2-k_2^2)X, \end{equation}

we can show that

(4.5a) \begin{eqnarray} (1+\tilde {\sigma } s^2)D_{k_{\tilde {\sigma }}}(X) &=&(1+\tilde {\sigma } s^2) D_k(X)+ \tilde {\sigma } c^2 k^2 X, \end{eqnarray}

(4.5b) \begin{eqnarray} (1+\tilde {\mu } s^2)D_{k_{\tilde {\mu }}}(X)&=&(1+\tilde {\mu } s^2) D_k(X)+ \tilde {\mu } c^2 k^2 X . \end{eqnarray}

Then, (4.2a ) and (4.5b ) on the one hand, (4.2b ) and (4.5a ) on the other hand, lead to

(4.6a) \begin{eqnarray} \kern30pt D_{k_{\tilde {\mu }}}(c B_r + s B_{\theta })&=&0, \end{eqnarray}

(4.6b) \begin{eqnarray} D_{k_{\tilde {\sigma }}}(s D_k(B_r) -ck^2 B_{\theta })&=&0. \end{eqnarray}

Furthermore, from (4.1a ) and (4.5a ) on the one hand, from (4.1b ) and (4.5b ) on the other hand, we have

(4.7a) \begin{eqnarray}\kern-23pt (1+\tilde {\sigma } s^2) D_{k_{\tilde {\sigma }}}(B_r) &=& (\tilde {\sigma } -\tilde {\mu })ck^2 (cB_r + s B_{\theta }), \end{eqnarray}

(4.7b) \begin{eqnarray} (1+\tilde {\mu } s^2)D_{k_{\tilde {\mu }}} (B_{\theta }) &=&(\tilde {\sigma } -\tilde {\mu })c \left ( s D_k(B_r)-ck^2 B_{\theta } \right ) . \end{eqnarray}

Finally, using (4.6a , b ) and (4.7a , b ), leads to

(4.8a,b) \begin{align} \big(D_{k_{\tilde {\mu }}}\circ D_{k_{\tilde {\sigma }}}\big)(B_r)=\left (D_{k_{\tilde {\sigma }}}\circ D_{k_{\tilde {\mu }}}\right )(B_{\theta }) = 0. \end{align}

The two operators $D_{k_{\tilde {\sigma }}}$ and $D_{k_{\tilde {\mu }}}$ being commutative, $B_r$ and $B_{\theta }$ satisfy the same linear differential equation of fourth order.

4.2. General expressions of $B_r$ and $B_{\theta }$

As the solution of $D_{\nu }(X)=0$ is a linear combination of $I_1(\nu r)$ and $K_1(\nu r)$ , where $I_1$ and $K_1$ are first and second kind modified Bessel functions of order 1, the solutions of (4.8a,b ) are a linear combination of $I_1(k_{\tilde {\sigma }} r)$ , $K_1(k_{\tilde {\sigma }} r)$ , $I_1(k_{\tilde {\mu }} r)$ and $K_1(k_{\tilde {\mu }} r)$ . Applying the boundary conditions (3.1a – d ), $B_r$ and $B_{\theta }$ can be written in the following form for case A:

(4.9) \begin{align} B_r^{({\rm A})}= \left \{ \begin{array}{l@{\quad}l} r\lt 1, & -s\left (\lambda _{\sigma }^R \dfrac {I_1(k_{\sigma } r)}{I_1(k_{\sigma })}+ \lambda _{\mu }^R \dfrac {I_1(k_{\mu } r)}{I_1(k_{\mu })}\right ) \\[15pt] 1\lt r\lt 1+e & -s\left (\lambda _{r}^I \dfrac {I_1(k r)}{I_1(k)}+ \lambda _{r}^K \dfrac {K_1(k r)}{K_1(k)}\right ) \\[15pt] r\gt 1+e, & -s\lambda _{r}^S \dfrac {K_1(k r)}{K_1(k(1+e))}\end{array} \right . , \end{align}

(4.10) \begin{align} B_{\theta }^{({\rm A})}=&& \left \{ \begin{array}{ll} r\lt 1,&\,\,\,\, c \left (\lambda _{\sigma }^R \dfrac {I_1(k_{\sigma } r)}{I_1(k_{\sigma })}+ \dfrac {\mu s^2}{1+\mu s^2}\lambda _{\mu }^R \dfrac {I_1(k_{\mu } r)}{I_1(k_{\mu })}\right ) \\[15pt] 1\lt r\lt 1+e,& \,\,\,\, c \left (\lambda _{\theta }^I \dfrac {I_1(k r)}{I_1(k)}+ \lambda _{\theta }^K \dfrac {K_1(k r)}{K_1(k)}\right ) \\[15pt] r\gt 1+e,& \,\,\,\, c \lambda _{\theta }^S \dfrac {K_1(k r)}{K_1(k(1+e))} \end{array} \right ..\end{align}

Here (4.10a ) has been obtained by replacing (4.9a ) in (4.7a ), and where (4.9b , c ) and (4.10b , c ) result from $D_{k}(B_r)=D_{k}(B_{\theta })=0$ for $r\gt 1$ .

For case B,

(4.11) \begin{align} B_r^{({\rm B})}= \left \{ \begin{array}{rl} r\lt 1,& \,\,\,\, -s\lambda _{r}^R \dfrac {I_1(k r)}{I_1(k)} \\[16pt] 1\lt r\lt 1+e& \,\,\,\, -s\left (\lambda _{r}^I \dfrac {I_1(k r)}{I_1(k)}+ \lambda _{r}^K \dfrac {K_1(k r)}{K_1(k)}\right ) \\[16pt] r\gt 1+e,&\,\,\,\, -s\left (\lambda _{\sigma }^S \dfrac {K_1(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))}+ \lambda _{\mu }^S \dfrac {K_1(k_{\mu } r)}{K_1(k_{\mu }(1+e))}\right ) \end{array} \right . , \end{align}

(4.12) \begin{align} B_{\theta }^{({\rm B})}= \left \{ \begin{array}{rl} r\lt 1,&\,\,\,\, c \lambda _{\theta }^R \dfrac {I_1(k r)}{I_1(k)} \\[12pt] 1\lt r\lt 1+e,& \,\,\,\, c \left (\lambda _{\theta }^I \dfrac {I_1(k r)}{I_1(k)}+ \lambda _{\theta }^K \dfrac {K_1(k r)}{K_1(k)}\right ) \\[12pt] r\gt 1+e,& \,\,\,\,c \left (\lambda _{\sigma }^S \dfrac {K_1(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))}+ \dfrac {\mu s^2}{1+\mu s^2}\lambda _{\mu }^S \dfrac {K_1(k_{\mu } r)}{K_1(k_{\mu }(1+e))} \right ) \end{array} \right ., \end{align}

where (4.12c ) has been obtained by replacing (4.11c ) in (4.7a ), and where (4.11a , b ) and (4.12a , b ) result from $D_{k}(B_r)=D_{k}(B_{\theta })=0$ for $r\lt 1+e$ .

For case C,

(4.13) \begin{align} B_r^{({\rm C})}= \left \{ \begin{array}{rl} r\lt 1,&\,\,\,\, -s\left (\lambda _{\sigma }^R \dfrac {I_1(k_{\sigma } r)}{I_1(k_{\sigma })}+ \lambda _{\mu }^R \dfrac {I_1(k_{\mu } r)}{I_1(k_{\mu })}\right ) \\[15pt] 1\lt r\lt 1+e_1,&\,\,\,\, -s\left (\lambda _{r_1}^I \dfrac {I_1(k r)}{I_1(k)}+ \lambda _{r_1}^K \dfrac {K_1(k r)}{K_1(k)}\right ) \\[15pt] 1+e_1\lt r\lt 1+e,&\,\,\,\, -s\left (\lambda _{r_2}^I \dfrac {I_1(k r)}{I_1(k(1+e_1))}+ \lambda _{r_2}^K \dfrac {K_1(k r)}{K_1(k(1+e_1))}\right ) \\[15pt] r\gt 1+e,&\,\,\,\, -s\left (\lambda _{\sigma }^S \dfrac {K_1(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))}+ \lambda _{\mu }^S \dfrac {K_1(k_{\mu } r)}{K_1(k_{\mu }(1+e))}\right ) \end{array} \right ., \end{align}

(4.14) \begin{align} B_{\theta }^{({\rm C})}=& \left \{ \begin{array}{rl} r\lt 1,&\,\,\,\, c \left (\lambda _{\sigma }^R \dfrac {I_1(k_{\sigma } r)}{I_1(k_{\sigma })}+ \dfrac {\mu s^2}{1+\mu s^2}\lambda _{\mu }^R \dfrac {I_1(k_{\mu } r)}{I_1(k_{\mu })}\right ) \\[15pt] 1\lt r\lt 1+e_1,&\,\,\,\, c \left (\lambda _{\theta _1}^I \dfrac {I_1(k r)}{I_1(k)}+ \lambda _{\theta _1}^K \dfrac {K_1(k r)}{K_1(k)}\right ) \\[15pt] 1+e_1\lt r\lt 1+e,&\,\,\,\, c \left (\lambda _{\theta _2}^I \dfrac {I_1(k r)}{I_1(k(1+e_1))}+ \lambda _{\theta _2}^K \dfrac {K_1(k r)}{K_1(k(1+e_1))}\right ) \\[15pt] r\gt 1+e,&\,\,\,\, c \left (\lambda _{\sigma }^S \dfrac {K_1(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))}+ \dfrac {\mu s^2}{1+\mu s^2}\lambda _{\mu }^S \dfrac {K_1(k_{\mu } r)}{K_1(k_{\mu }(1+e))} \right ) \end{array} \right .,\nonumber\\[5pt] \end{align}

where (4.14a ) and (4.14d ) have been obtained by replacing (4.13a ) and (4.13d ) in (4.7a ), and where (4.13b , c ) and (4.14b , c ) result from $D_{k}(B_r)=D_{k}(B_{\theta })=0$ for $1\lt r\lt 1+e$ .

4.3. Application of other boundary conditions

From (4.9) and (4.10) in case A, and (4.11) and (4.12) in case B, there are eight unknowns, $(\lambda _{\sigma ,\mu }^R, \lambda _{r}^{I,K}, \lambda _{\theta }^{I,K}, \lambda _{r,\theta }^S)$ in case A and $(\lambda _{r,\theta }^R, \lambda _{r}^{I,K}, \lambda _{\theta }^{I,K}, \lambda _{\sigma ,\mu }^S)$ in case B. Therefore, each problem needs eight equations to be solved. In case C, from (4.13) and (4.14), there are twelve unknowns, $(\lambda _{\sigma ,\mu }^R, \lambda _{r_1}^{I,K}, \lambda _{\theta _1}^{I,K}, \lambda _{r_2}^{I,K}, \lambda _{\theta _2}^{I,K}, \lambda _{\sigma ,\mu }^S)$ , needing twelve equations to be solved.

In case A, the first six equations are given by the boundary conditions (3.7a – f ), leading to

(4.15a) \begin{eqnarray} &\lambda _{r}^I + \lambda _{r}^K = \lambda _{\sigma }^R + \lambda _{\mu }^R , \end{eqnarray}

(4.15b) \begin{eqnarray} &\varphi _{\mu }^{-1}\left (\lambda _{\theta }^I+\lambda _{\theta }^K\right )= \lambda _{\sigma }^R , \end{eqnarray}

(4.15c) \begin{eqnarray} &-k\varphi _{\mu }^{-1}\left (-\lambda _r^I\frac {I_0(k)}{I_1(k)}+\lambda _r^K\frac {K_0(k)}{K_1(k)} \right ) = k_{\sigma }\lambda _{\sigma }^R \frac {I_0(k_{\sigma })}{I_1(k_{\sigma })}+ k_{\mu }\lambda _{\mu }^R\frac {I_0(k_{\mu })}{I_1(k_{\mu })}, \end{eqnarray}

(4.15d) \begin{eqnarray} &\lambda _{r}^I \frac {I_1(k(1+e))}{I_1(k)} + \lambda _{r}^K \frac {K_1(k(1+e))}{K_1(k)} = \lambda _{r}^S , \end{eqnarray}

(4.15e) \begin{eqnarray} &\lambda _{\theta }^I\frac {I_1(k(1+e))}{I_1(k)}+\lambda _{\theta }^K\frac {K_1(k(1+e))}{K_1(k)}= \lambda _{\theta }^S , \end{eqnarray}

(4.15f) \begin{eqnarray} &-\lambda _r^I\frac {I_0(k(1+e))}{I_1(k)}+\lambda _r^K\frac {K_0(k(1+e))}{K_1(k)} =\lambda _{r}^S \frac {K_0(k(1+e))}{K_1(k(1+e))}. \end{eqnarray}

In case B, they are given by the boundary conditions (3.8a – f ), leading to

(4.16a) \begin{eqnarray} &\lambda _{r}^I + \lambda _{r}^K = \lambda _{r}^R , \end{eqnarray}

(4.16b) \begin{eqnarray} &\lambda _{\theta }^I+\lambda _{\theta }^K= \lambda _{\theta }^R , \end{eqnarray}

(4.16c) \begin{eqnarray} &\lambda _r^I\frac {I_0(k)}{I_1(k)}-\lambda _r^K\frac {K_0(k)}{K_1(k)} = \lambda _{r}^R \frac {I_0(k)}{I_1(k)}, \end{eqnarray}

(4.16d) \begin{eqnarray} &\lambda _{r}^I \frac {I_1(k(1+e))}{I_1(k)} + \lambda _{r}^K \frac {K_1(k(1+e))}{K_1(k)} = \lambda _{\sigma }^S + \lambda _{\mu }^S , \end{eqnarray}

(4.16e) \begin{eqnarray} &\varphi _{\mu }^{-1}\left (\lambda _{\theta }^I\frac {I_1(k(1+e))}{I_1(k)}+\lambda _{\theta }^K\frac {K_1(k(1+e))}{K_1(k)}\right )= \lambda _{\sigma }^S , \end{eqnarray}

(4.16f) \begin{align} k \varphi _{\mu }^{-1}\left (\!{-}\lambda _r^I\frac {I_0(k(1+e))}{I_1(k)}+\lambda _r^K\frac {K_0(k(1+e))}{K_1(k)} \right ) =k_{\sigma }\lambda _{\sigma }^S \frac {K_0(k_{\sigma }(1+e))}{K_1(k_{\sigma }(1+e))}+k_{\mu }\lambda _{\mu }^S \frac {K_0(k_{\mu }(1+e))}{K_1(k_{\mu }(1+e))}.\nonumber\\[6pt] \end{align}

In case C the first nine equations are given by the boundary conditions (3.9a – i ), leading to

(4.17a) \begin{align} \lambda _{r_1}^I + \lambda _{r_1}^K = \lambda _{\sigma }^R + \lambda _{\mu }^R , \end{align}

(4.17b) \begin{eqnarray} &\varphi _{\mu }^{-1}\left (\lambda _{\theta _1}^I+\lambda _{\theta _1}^K\right )= \lambda _{\sigma }^R , \end{eqnarray}

(4.17c) \begin{eqnarray} &-k\varphi _{\mu }^{-1}\left (-\lambda _{r_1}^I\frac {I_0(k)}{I_1(k)}+\lambda _{r_1}^K\frac {K_0(k)}{K_1(k)} \right ) = k_{\sigma }\lambda _{\sigma }^R \frac {I_0(k_{\sigma })}{I_1(k_{\sigma })}+ k_{\mu }\lambda _{\mu }^R\frac {I_0(k_{\mu })}{I_1(k_{\mu })}, \end{eqnarray}

(4.17d) \begin{eqnarray} &\lambda _{r_1}^I \frac {I_1(k(1+e_1))}{I_1(k)} + \lambda _{r_1}^K \frac {K_1(k(1+e_1))}{K_1(k)} = \lambda _{r_2}^I + \lambda _{r_2}^K , \end{eqnarray}

(4.17e) \begin{eqnarray} &\lambda _{\theta _1}^I\frac {I_1(k(1+e_1))}{I_1(k)}+\lambda _{\theta _1}^K\frac {K_1(k(1+e_1))}{K_1(k)}= \lambda _{\theta _2}^I+\lambda _{\theta _2}^K , \end{eqnarray}

(4.17f) \begin{align} &-\lambda _{r_1}^I\frac {I_0(k(1+e_1))}{I_1(k)}+\lambda _{r_1}^K\frac {K_0(k(1+e_1))}{K_1(k)} =-\lambda _{r_2}^I\frac {I_0(k(1+e_1))}{I_1(k(1+e_1))}+\lambda _{r_2}^K\frac {K_0(k(1+e_1))}{K_1(k(1+e_1))},\nonumber\\[5pt] \end{align}

(4.17g) \begin{eqnarray} &\lambda _{r_2}^I \frac {I_1(k(1+e))}{I_1(k(1+e_1))} + \lambda _{r_2}^K \frac {K_1(k(1+e))}{K_1(k(1+e_1))} = \lambda _{\sigma }^S + \lambda _{\mu }^S , \end{eqnarray}

(4.17h) \begin{eqnarray} &\varphi _{\mu }^{-1}\left (\lambda _{\theta _2}^I\frac {I_1(k(1+e))}{I_1(k(1+e_1))}+\lambda _{\theta _2}^K\frac {K_1(k(1+e))}{K_1(k(1+e_1))}\right )= \lambda _{\sigma }^S , \end{eqnarray}

(4.17i) \begin{align} & k \varphi _{\mu }^{-1}\left (-\lambda _{r_2}^I\frac {I_0(k(1+e))}{I_1(k(1+e_1))}+\lambda _{r_2}^K\frac {K_0(k(1+e))}{K_1(k(1+e_1))} \right )\nonumber\\[8pt]& \quad =k_{\sigma }\lambda _{\sigma }^S \frac {K_0(k_{\sigma }(1+e))}{K_1(k_{\sigma }(1+e))}+ k_{\mu }\lambda _{\mu }^S \frac {K_0(k_{\mu }(1+e))}{K_1(k_{\mu }(1+e))}. \end{align}

Finally, to apply the last boundary conditions (3.11a , b ) in case A, (3.12a , b ) in case B, (3.13a – c ) in case C, we need to calculate the $z$ component of the current density, that is given by

(4.18a,b) \begin{align} J_z= \tilde {\varphi }_{\mu }^{-1}\frac {1}{r}\frac {\partial }{\partial r}(r\tilde {\phi }), \quad \text{with} \; \tilde {\phi }=\tilde {\mu } cs B_r + (1+\tilde {\mu } s^2)B_{\theta }. \end{align}

In case A, replacing $B_r^{({\rm A})}$ and $B_{\theta }^{({\rm A})}$ given by (4.9) and (4.10) in (4.18b ) leads to

(4.19) \begin{eqnarray} \tilde {\phi }^{({\rm A})}=&& \left \{ \begin{split} r\lt 1, &\;\;\;c \lambda _{\sigma }^R\frac {I_1(k_{\sigma } r)}{I_1(k_{\sigma })} \\[5pt] 1\lt r\lt 1+e, &\;\;\;c \left (\lambda _{\theta }^I\frac {I_1(k r)}{I_1(k)}+ \lambda _{\theta }^K\frac {K_1(k r)}{K_1(k)}\right ) \\[5pt] r\gt 1+e, &\;\;\;c \lambda _{\theta }^S\frac {K_1(k r)}{K_1(k(1+e))} \end{split} \right . . \end{eqnarray}

In case B, replacing $B_r^{({\rm B})}$ and $B_{\theta }^{({\rm B})}$ given by (4.11) and (4.12) in (4.18b ) leads to

(4.20) \begin{eqnarray} \tilde {\phi }^{({\rm B})}=&& \left \{ \begin{split} r\lt 1, &\;\;\;c \lambda _{\theta }^R\frac {I_1(k r)}{I_1(k)} \\[5pt] 1\lt r\lt 1+e, &\;\;\;c \left (\lambda _{\theta }^I\frac {I_1(k r)}{I_1(k)}+ \lambda _{\theta }^K\frac {K_1(k r)}{K_1(k)}\right ) \\[5pt] r\gt 1+e, &\;\;\;c \lambda _{\sigma }^S\frac {K_1(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))} \end{split} \right . . \end{eqnarray}

In case C, replacing $B_r^{({\rm C})}$ and $B_{\theta }^{({\rm C})}$ given by (4.13) and (4.14) in (4.18b ) leads to

(4.21) \begin{eqnarray} \tilde {\phi }^{({\rm C})}=&& \left \{ \begin{split} r\lt 1, &\;\;\;c \lambda _{\sigma }^R\frac {I_1(k_{\sigma } r)}{I_1(k_{\sigma })} \\[5pt] 1\lt r\lt 1+e_1, &\;\;\;c \left (\lambda _{\theta _1}^I\frac {I_1(k r)}{I_1(k)}+ \lambda _{\theta _1}^K\frac {K_1(k r)}{K_1(k)}\right ) \\[5pt] 1+e_1\lt r\lt 1+e, &\;\;\;c \left (\lambda _{\theta _2}^I\frac {I_1(k r)}{I_1(k)}+ \lambda _{\theta _2}^K\frac {K_1(k r)}{K_1(k)}\right ) \\[5pt] r\gt 1+e, &\;\;\;c \lambda _{\sigma }^S\frac {K_1(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))} \end{split} \right . . \end{eqnarray}

Applying the relations $\partial _r(rI_1(\nu r))=\nu r I_0(\nu r)$ and $\partial _r(rK_1(\nu r))=-\nu r K_0(\nu r)$ , where $\nu$ is a dummy variable, to (4.18a ), leads to in case A,

(4.22) \begin{eqnarray} J_z^{({\rm A})}=&& \left \{ \begin{split} r\lt 1,&\;\;\;c k_{\sigma } \lambda _{\sigma }^R\frac {I_0(k_{\sigma } r)}{I_1(k_{\sigma })} \\[5pt] 1\lt r\lt 1+e,& \;\;\; ck\varphi _{\mu }^{-1}\left ( \lambda _{\theta }^I\frac {I_0(k r)}{I_1(k)}- \lambda _{\theta }^K\frac {K_0(k r)}{K_1(k)}\right ) \\[5pt] r\gt 1+e,&-c k \varphi _{\mu }^{-1} \lambda _{\theta }^S\frac {K_0(k r)}{K_1(k(1+e))} \end{split} \right . , \end{eqnarray}

in case B,

(4.23) \begin{eqnarray} J_z^{({\rm B})}=&& \left \{ \begin{split} r\lt 1,&\;\;\;c k\varphi _{\mu }^{-1} \lambda _{\theta }^R\frac {I_0(k r)}{I_1(k)} \\[4pt] 1\lt r\lt 1+e,& \;\;\; ck\varphi _{\mu }^{-1}\left ( \lambda _{\theta }^I\frac {I_0(k r)}{I_1(k)}- \lambda _{\theta }^K\frac {K_0(k r)}{K_1(k)}\right ) \\[4pt] r\gt 1+e,&-c k_{\sigma } \lambda _{\sigma }^S\frac {K_0(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))} \end{split} \right . , \end{eqnarray}

in case C,

(4.24) \begin{align} J_z^{({\rm C})}= \left \{ \begin{array}{r@{\quad}l} r\lt 1,& \!\!\! c k_{\sigma } \lambda _{\sigma }^R\dfrac {I_0(k_{\sigma } r)}{I_1(k_{\sigma })} \\[15pt] 1\lt r\lt 1+e_1,& \!\!\! ck\varphi _{\mu }^{-1}\left ( \lambda _{\theta _1}^I\dfrac {I_0(k r)}{I_1(k)}- \lambda _{\theta _1}^K\dfrac {K_0(k r)}{K_1(k)}\right ) \\[15pt] 1+e_1\lt r\lt 1+e,& \!\!\! ck\varphi _{\mu }^{-1}\left ( \lambda _{\theta _2}^I\dfrac {I_0(k r)}{I_1(k(1+e_1))}- \lambda _{\theta _2}^K\dfrac {K_0(k r)}{K_1(k(1+e_1))}\right ) \\[15pt] r\gt 1+e,& \!\!\! -c k_{\sigma } \lambda _{\sigma }^S\dfrac {K_0(k_{\sigma } r)}{K_1(k_{\sigma }(1+e))} \end{array} \right. \end{align}

In case A, replacing (4.9) and (4.22) in (3.11a , b ) leads to

(4.25a) \begin{align} \varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta }^I\frac {I_0(k)}{I_1(k)}- \lambda _{\theta }^K\frac {K_0(k)}{K_1(k)}\right ) - \varOmega s (\lambda _r^I+\lambda _r^K) = c k_{\sigma } \lambda _{\sigma }^R \frac {I_0(k_{\sigma })}{I_1(k_{\sigma })} -\varOmega s \left (\lambda _{\sigma }^R+\lambda _{\mu }^R\right ), \end{align}

(4.25b) \begin{align} &\varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta }^I\frac {I_0(k(1+e))}{I_1(k)}- \lambda _{\theta }^K\frac {K_0(k(1+e))}{K_1(k)}\right )\nonumber\\[5pt]& \quad\qquad - \varOmega s (1+e) \left (\lambda _r^I\frac {I_1(k(1+e))}{I_1(k)}+\lambda _r^K\frac {K_1(k(1+e))}{K_1(k)}\right )\nonumber \\[5pt] & \qquad = -\varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}c k \lambda _{\theta }^S \frac {K_0(k(1+e))}{K_1(k(1+e))}. \end{align}

In case B, replacing (4.11) and (4.23) in (3.12a , b ) leads to

(4.26a) \begin{align} \varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta }^I\frac {I_0(k)}{I_1(k)}- \lambda _{\theta }^K\frac {K_0(k)}{K_1(k)}\right ) = \varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}c k \lambda _{\theta }^R \frac {I_0(k)}{I_1(k)} -\varOmega s \lambda _{r}^R ,\end{align}

(4.26b) \begin{align} \varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta }^I\frac {I_0(k(1+e))}{I_1(k)}- \lambda _{\theta }^K\frac {K_0(k(1+e))}{K_1(k)}\right ) = -c k_{\sigma } \lambda _{\sigma }^S \frac {K_0(k_{\sigma }(1+e))}{K_1(k_{\sigma }(1+e))}. \end{align}

In case C, replacing (4.13) and (4.24) in (3.13a , c ) leads to

(4.27a) \begin{align} &\varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta _1}^I\frac {I_0(k)}{I_1(k)}- \lambda _{\theta _1}^K\frac {K_0(k)}{K_1(k)}\right ) - \varOmega s \big(\lambda _{r_1}^I+\lambda _{r_1}^K\big)\nonumber \\[5pt] & \quad = c k_{\sigma } \lambda _{\sigma }^R \frac {I_0(k_{\sigma })}{I_1(k_{\sigma })} -\varOmega s \left (\lambda _{\sigma }^R+\lambda _{\mu }^R\right ),\end{align}

(4.27b) \begin{align} &\varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta _1}^I\frac {I_0(k(1+e_1))}{I_1(k)}- \lambda _{\theta _1}^K\frac {K_0(k(1+e_1))}{K_1(k)}\right ) \nonumber \\[5pt] & \qquad - \varOmega s (1+e_1) \left (\lambda _{r_1}^I\frac {I_1(k(1+e_1))}{I_1(k)}+\lambda _{r_1}^K\frac {K_1(k(1+e_1))}{K_1(k)}\right ) \nonumber \\[5pt] & \quad = \varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta _2}^I\frac {I_0(k(1+e_1))}{I_1(k(1+e_1))}- \lambda _{\theta _2}^K\frac {K_0(k(1+e_1))}{K_1(k(1+e_1))}\right ) ,\end{align}

(4.27c) \begin{align} \varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}ck\left (\lambda _{\theta _2}^I\frac {I_0(k(1+e))}{I_1(k(1+e_1))}- \lambda _{\theta _2}^K\frac {K_0(k(1+e))}{K_1(k(1+e_1))}\right ) = -c k_{\sigma } \lambda _{\sigma }^S \frac {K_0(k_{\sigma }(1+e))}{K_1(k_{\sigma }(1+e))}. \end{align}

6.1. Cases A and B for $e=0$

Taking $e=0$ implies that $(k', k'_\sigma , k'_\mu )=(k, k_\sigma , k_\mu )$ and $\alpha '=\beta '=1$ . Then, the dispersion relations (5.1) and (5.2) take the form

(6.1) \begin{align} \varOmega ^{({\rm A})}_{e=0}=-\frac {c}{s}\frac {(\varphi _{\sigma }^{-1}\varPsi (k)+\varPhi (k_{\sigma }))(\varphi _{\mu }^{-1}\varPsi (k)+\varPhi (k_{\mu }))}{\varPhi (k_{\sigma })-\varPhi (k_{\mu })} \end{align}

and

(6.2) \begin{align} \varOmega ^{({\rm B})}_{e=0}=-\frac {c}{s}\frac {(\varphi _{\sigma }^{-1}\varPhi (k)+\varPsi (k_{\sigma }))(\varphi _{\mu }^{-1}\varPhi (k)+\varPsi (k_{\mu }))}{\varPsi (k_{\sigma })-\varPsi (k_{\mu })}. \end{align}

In Plunian & Alboussière (Reference Plunian and Alboussière2023) the dynamo threshold has been derived in the form

(6.3) \begin{align} \varOmega _{2023} = - \frac {c}{s}\frac {\left (\varPhi (k_{\sigma _{\textrm {in}}})+\overline {\varphi }_{\sigma }^{-1}\varPsi (k_{\sigma _{\textrm {out}}})\right )\left (\varPhi (k_{\mu _{\textrm {in}}})+\overline {\varphi }_{\mu }^{-1}\varPsi (k_{\mu _{\textrm {out}}})\right )}{\varPhi (k_{\sigma _{\textrm {in}}})+\varPsi (k_{\sigma _{\textrm {out}}})-\varPhi (k_{\mu _{\textrm {in}}})-\varPsi (k_{\mu _{\textrm {out}}})}, \end{align}

where the subscripts `in’ and `out’ correspond to $r\lt R$ and $r\gt R$ , respectively; $\varOmega _{2023}$ denotes the rotor velocity normalised by $R^2\sigma _{\textrm {in}}^{\perp }\mu _{\textrm {in}}^{\perp }$ , $\overline {\varphi }_{\sigma }=\sigma _{\textrm {out}}^{\perp }/\sigma _{\textrm {in}}^{\perp }$ and $\overline {\varphi }_{\mu }=\mu _{\textrm {out}}^{\perp }/\mu _{\textrm {in}}^{\perp }$ .

In case A, $(\sigma _{\textrm {in}}, \mu _{\textrm {in}})=(\sigma ,\mu )$ and $(\sigma _{\textrm {out}}, \mu _{\textrm {out}})=(0,0)$ , implying that $\overline {\varphi }_{\sigma }=\varphi _{\sigma }$ and $\overline {\varphi }_{\mu }=\varphi _{\mu }$ . Then, $\varOmega ^{({\rm A})}_{e=0}$ , that is given in (6.1), corresponds directly to $\varOmega _{2023}$ in (6.3), with $k_{\sigma _{\textrm {in}}}=k_{\sigma }$ , $k_{\mu _{\textrm {in}}}=k_{\mu }$ and $k_{\sigma _{\textrm {out}}}=k_{\mu _{\textrm {out}}}=k$ .

In case B, as already mentioned in (3.15b ), $|\varOmega ^{({\rm B})}_{e=0}|$ must be multiplied by $\varphi _{\sigma }\varphi _{\mu }$ to obtain the magnetic Reynolds number $Rm^{({\rm B})}$ . In addition, $(\sigma _{\textrm {in}}, \mu _{\textrm {in}})=(0,0)$ and $(\sigma _{\textrm {out}}, \mu _{\textrm {out}})=(\sigma ,\mu )$ , implying that $\overline {\varphi }_{\sigma }=\varphi _{\sigma }^{-1}$ and $\overline {\varphi }_{\mu }=\varphi _{\mu }^{-1}$ . Therefore, the expression of $\varOmega ^{({\rm B})}_{e=0}$ given in (6.2), corresponds to $\varphi _{\sigma }^{-1}\varphi _{\mu }^{-1}\varOmega _{2023}$ in (6.3), with $k_{\sigma _{\textrm {in}}}=k_{\mu _{\textrm {in}}}=k$ , $k_{\sigma _{\textrm {out}}}=k_{\sigma }$ and $k_{\mu _{\textrm {out}}}=k_{\mu }$ .

6.2. Case C for $e\gg 1$

In the asymptotic limit $e\gg 1$ , we have $(k', k'_\mu , k'_\sigma ) =O(e)$ and $(k,k_1)=O(1)$ . In addition, we know that, for $x\gg 1$ , $(I_0(x), I_1(x)) \propto \exp (x)/\sqrt {x}$ and $(K_0(x), K_1(x)) \propto \exp (-x)/\sqrt {x}$ (Abramowitz & Stegun Reference Abramowitz and Stegun1968). Then, we have $(\alpha , \alpha ') \propto \exp (e)/\sqrt {e}$ , $(\beta , \beta ') \propto \exp (-e)/\sqrt {e}$ and $(\alpha _1, \beta _1) =O(1)$ , implying that $(A(\nu ), B(\nu ), E_{\sigma \mu })=O(1)$ and $(C(\nu ), D(\nu ), F_{\sigma \mu })=O(e)$ . Then $P(\nu )\approx - ({\alpha }/{\beta _1})B(\nu )C(\nu )$ , $Q(\nu ) \approx \alpha C(\nu )$ and $E_{\sigma \mu }Q(\sigma )Q(\mu ) \gg F_{\sigma \mu }R(\sigma )R(\mu )$ . Replacing in (5.3) then leads to (5.1) in which $e$ must be replaced by $e_1$ . Therefore, case C with $e\gg 1$ corresponds to case A.

6.3. Case C for $e_1\gg 1$ and $e=O(e_1)$

In the asymptotic limit $e_1\gg 1$ and $e=O(e_1)$ , we have $(k_1, k', k'_\mu , k'_\sigma ) =O(e_1)$ and $k=O(1)$ . As written above, we know that, for $x\gg 1$ , $(I_0(x), I_1(x)) \propto \exp (x)/\sqrt {x}$ and $(K_0(x), K_1(x)) \propto \exp (-x)/\sqrt {x}$ (Abramowitz & Stegun Reference Abramowitz and Stegun1968). Then, we have $(\alpha _1, \alpha ') \propto \exp (e_1)/\sqrt {e_1}$ , $(\beta _1, \beta ') \propto \exp (-e_1)/\sqrt {e_1}$ and $(\alpha , \beta ) =O(1)$ . Then we have $(A(\nu ), B(\nu ), E_{\sigma \mu })=O(1)$ and $(C(\nu ), D(\nu ), F_{\sigma \mu })=O(e_1)$ , implying that $P(\nu )\approx - ({\alpha }/{\beta _1})B(\nu )C(\nu )$ , $Q(\nu ) =O(e_1)$ , $R(\nu )\approx -B(\nu )/\beta _1$ and $F_{\sigma \mu }R(\sigma )R(\mu )\gg E_{\sigma \mu }Q(\sigma )Q(\mu )$ . Replacing in (5.3) then leads to

(6.4) \begin{align} \varOmega ^{(c)}\approx -\frac {c}{s}\frac {\alpha ^2}{(1+e_1)^2}\frac {C(\sigma )C(\mu )}{F_{\sigma \mu }}. \end{align}

In order to compare (6.4) to the dynamo threshold (5.2) of case B, we must take into account that the rotor of dimensionless radius equal to unity in case B corresponds in case C to the rotating part of radius $1+e_1$ . Therefore, we have

(6.5) \begin{align} \frac {1+e^{({\rm C})}}{1+e_1}=1+e^{({\rm B})}, \end{align}

where $e^{({\rm B})}$ and $e^{({\rm C})}$ denote the parameter $e$ in cases B and C. As in both cases, time has been normalised by $R^2$ (where $R$ is the dimensional radius of the rotor), in order to retrieve $\varOmega ^{({\rm B})}$ from $\varOmega ^{({\rm C})}$ we have to multiply (6.4) by $(1+e_1)^2$ . In addition, $k_1$ in case C is now equivalent to $k$ in case B. Furthermore, $k'$ in cases B and C are equal (see Appendix E for a detailed proof). This implies that $\alpha$ in case C becomes $\alpha '$ in case B, and that $C(\nu )$ and $F_{\sigma \mu }$ in case B are the same as those in case C. This demonstrates that (6.4) and (5.2) are equivalent, and that case C with $e_1\gg 1$ and $e=O(e_1)$ corresponds to case B.

6.4. Case C for $e=0$

Taking $e=0$ implies that $e_1=0$ , and then $k'=k_1=k$ , $k'_\sigma =k_\sigma$ , $k'_\mu =k_\mu$ and $\alpha _1=\beta _1=\alpha '=\beta '=\alpha =\beta$ . Then, $P(\nu )=-\varphi _{\nu }^{-1}(\varPhi (k_{\nu })+\varPsi (k_{\nu }))(\varPhi (k)+\varPsi (k))$ and $Q(\nu )=-R(\nu )=\varphi _{\nu }^{-1}(\varPhi (k)+\varPsi (k))$ , leading to

(6.6) \begin{align} \varOmega ^{({\rm C})}_{e=0}=-\frac {c}{s}\frac {(\varPhi (k_\sigma )+\varPsi (k_\sigma ))(\varPhi (k_\mu )+\varPsi (k_\mu ))}{\varPhi (k_\sigma )+\varPsi (k_\sigma )-\varPhi (k_\mu )-\varPsi (k_\mu )} .\end{align}

Applying the Wronskian (Abramowitz & Stegun Reference Abramowitz and Stegun1968) given in (A1) of Appendix A, leads to

(6.7) \begin{align} \varOmega ^{({\rm C})}_{e=0}=\frac {c}{s}(I_1(k_\sigma )K_1(k_\sigma )-I_1(k_\mu )K_1(k_\mu ))^{-1}, \end{align}

which is the dynamo threshold found in Plunian & Alboussière (Reference Plunian and Alboussière2021).

6.5. Case C for $e_1=0$

Taking $e_1=0$ implies that $k_1=k$ , $\alpha _1=\beta _1=1$ , $\alpha =\alpha '$ and $\beta =\beta '$ . Then $P(\nu )=\beta 'A(\nu )D(\nu )-\alpha 'B(\nu )C(\nu )$ , $Q(\nu ) = \alpha ' C(\nu ) - \beta ' D(\nu )$ , $R(\nu ) = A(\nu )-B(\nu )$ . Replacing in (5.3) leads to

(6.8) \begin{align} \varOmega ^{({\rm C})}_{e_1=0}=-\frac {c}{s} \frac {L(\sigma )L(\mu )}{E_{\sigma \mu }G(\sigma )G(\mu )+F_{\sigma \mu }H(\sigma )H(\mu )}, \end{align}

with

(6.9a) \begin{eqnarray} &L(\nu )=\beta 'A(\nu )D(\nu )-\alpha 'B(\nu )C(\nu ), \end{eqnarray}

(6.9b) \begin{eqnarray} & G(\nu )=\alpha 'C(\nu )-\beta 'D(\nu ), H(\nu )=A(\nu )-B(\nu ). \end{eqnarray}

Now, the dispersion relation given in Plunian et al. (Reference Plunian, Gomez and Alboussière2025) can be rewritten as (see Appendix F)

(6.10) \begin{align} \varOmega _{2025} = \omega -\frac {c}{s} \frac {L(\sigma )L(\mu )+ ({s}/{c})\omega (1+e)^2\left (F_{\sigma \mu }M(\sigma )M(\mu )+\alpha '^2\beta '^2E_{\sigma \mu }N(\sigma )N(\mu )\right )}{E_{\sigma \mu }G(\sigma )G(\mu )+F_{\sigma \mu }H(\sigma )H(\mu )+ ({s}/{c})\omega (1+e)^2(\alpha '-\beta ')^2F_{\sigma \mu }E_{\sigma \mu }},\nonumber\\[5pt] \end{align}

with

(6.11) \begin{eqnarray} &M(\nu )=-\alpha 'B(\nu )+\beta 'A(\nu ), N(\nu )=C(\nu )-D(\nu ). \end{eqnarray}

We note that $H(\nu )$ and $N(\nu )$ are related to each other by (A5) given in Appendix A. This case corresponds to a rotating rotor at velocity $\varOmega _{2025}$ , a stator at rest, both made of anisotropic material, and between both, a fluid layer rotating at velocity $\omega$ . Then, it is straightforward to show that (6.8) is equivalent to (6.10) for $\omega =0$ . Therefore, the case C for $e_1=0$ is equivalent to Plunian et al. (Reference Plunian, Gomez and Alboussière2025) with $\omega =0$ .

6.6. Case C for $e_1=e$

For $e_1=e$ , we have $k_1=k'$ , $\alpha =\beta =1$ , $\alpha _1=\alpha '$ and $\beta _1=\beta '$ . Then $P(\nu )=$ $( {1}/{\alpha '})A(\nu )D(\nu )- ({1}/{\beta '})B(\nu )C(\nu )$ , $Q(\nu ) = C(\nu ) - D(\nu )$ , $R(\nu ) = ({1}/{\alpha '})A(\nu )- ({1}/{\beta '})B(\nu )$ . Replacing in (5.3) leads to

(6.12) \begin{align} \varOmega ^{({\rm C})}_{e_1=e}=-\frac {c}{s}\frac {1}{(1+e)^2} \frac {L(\sigma )L(\mu )}{F_{\sigma \mu }M(\sigma )M(\mu )+\alpha '^2\beta '^2 E_{\sigma \mu }N(\sigma )N(\mu )}. \end{align}

Now taking $\omega =\varOmega _{2025}$ in (6.10) leads to (6.12). Therefore, the case C for $e_1=e$ is equivalent to Plunian et al. (Reference Plunian, Gomez and Alboussière2025) with $\omega =\varOmega _{2025}$ .

Table 1.

Values of the electromagnetic properties taken for the four applications I–IV. The electrical conductivities $\sigma ^{\perp }$ , $\sigma ^{\parallel }$ , $\sigma ^{\perp }$ and $\sigma ^{\parallel }$ are given in units of $10^6 \; \varOmega ^{-1}\,\rm m^{-1}$ . The magnetic permeabilities $\mu ^{\perp }$ , $\mu ^{\parallel }$ , $\mu ^{\perp }$ and $\mu ^{\parallel }$ are given in units of $4\pi \times 10^{-7}\,\rm Hm^{-1}$ . Their values have been taken from appropriate references (Jiles Reference Jiles1991; Gale & Totemeir Reference Gale and Totemeir2004; Wu & Chung Reference Wu and Chung2008; DuPont $^{ \text{TM}}$ , 2006; Driscoll & Du Reference Driscoll and Du2019). The dimensionless parameters $\sigma$ , $\mu$ , $\varphi _{\sigma }$ and $\varphi _{\mu }$ are calculated from (2.7a–d ).

Table 1. Long description

A table comparing electromagnetic properties for different cases and materials. The table has 4 rows and 12 columns. Column headers are: Case, T (degrees C), sigma perpendicular, mu perpendicular, sigma parallel, mu parallel, sigma, mu, Mat., sigma F, mu F, phi sigma, phi mu. Row 1: Case I, 20 degrees C, 63, 1, 5 x 10^-22, 1, 13 x 10^22, 0, Ga, 3.86, 1, 0.06, 1. Row 2: Cases IIa, IIb, 100 degrees C, 47, 1, 2.5 x 10^-19, 1, 19 x 10^19, 0, Na, 10.6, 1, 0.23, 1. Row 3: Case III, 100 degrees C, 6.9, 5 x 10^3, 10.6, 1, -0.35, 5 x 10^3, Na, 10.6, 1, 1.53, 2 x 10^-4. Row 4: Case IV, 4400 degrees C, 1, 1, (sigma + 1)^-1, 1, sigma, 0, Fe, 1, 1, 1, 1.