A.1 Preliminaries and assumptions

In the following, we consider a given $x$ -dependent incompressible MHD flow in the $y$ – $z$ plane,

(A 1a,b ) $$\begin{eqnarray}\boldsymbol{u}=u_{y}(x)\boldsymbol{e}_{y}+u_{z}(x)\boldsymbol{e}_{z},\quad \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0,\end{eqnarray}$$

such that it becomes constant at the boundaries,

(A 2a,b ) $$\begin{eqnarray}\lim _{x\rightarrow \pm \infty }u_{y}(x)=u_{0y}^{(\pm )},\quad \lim _{x\rightarrow \pm \infty }u_{z}(x)=u_{0z}^{(\pm )},\end{eqnarray}$$

i.e. we consider a localized velocity shear (the vorticity is vanishing at the boundaries, $\lim _{x\rightarrow \pm \infty }\unicode[STIX]{x1D735}\boldsymbol{\times }\boldsymbol{u}=0$ ). The magnetic field also lies on the $y$ – $z$ plane,

(A 3) $$\begin{eqnarray}\boldsymbol{B}(x)=B_{0y}H_{y}(x)\boldsymbol{e}_{y}+B_{0z}H_{z}(x)\boldsymbol{e}_{z}.\end{eqnarray}$$

The associated magnetic pressure is

(A 4a,b ) $$\begin{eqnarray}P_{B}(x)=\frac{B_{0}^{2}}{8\unicode[STIX]{x03C0}}{\mathcal{H}}(x),\quad {\mathcal{H}}(x)\equiv \frac{B_{0y}^{2}}{B_{0}^{2}}H_{y}^{2}(x)+\frac{B_{0z}^{2}}{B_{0}^{2}}H_{z}^{2}(x)\geqslant 0\quad \forall x,\end{eqnarray}$$

where we have defined $B_{0}$ as the (constant) value of $|\boldsymbol{B}|$ at the right boundary ( $x\rightarrow +\infty$ ):

(A 5) $$\begin{eqnarray}B_{0}\equiv \lim _{x\rightarrow +\infty }\sqrt{B_{0y}^{2}H_{y}^{2}(x)+B_{0z}^{2}H_{z}^{2}(x)},\quad \lim _{x\rightarrow +\infty }{\mathcal{H}}(x)=1.\end{eqnarray}$$

We further assume a polytropic relation for the thermal pressuresFootnote ⁸ :

(A 6a,b ) $$\begin{eqnarray}p_{\bot \text{p}}=p_{\bot \text{p},0}\,{\mathcal{F}}_{\bot }(x)=p_{\bot \text{p},0}\left(\frac{n(x)}{n_{0}}\right)^{\unicode[STIX]{x1D6FE}_{\bot \text{p}}},\quad p_{\bot \text{e}}=p_{\bot \text{e},0}{\mathcal{G}}_{\bot }(x)=p_{\bot \text{e},0}\left(\frac{n(x)}{n_{0}}\right)^{\unicode[STIX]{x1D6FE}_{\bot \text{e}}},\end{eqnarray}$$

and

(A 7a,b ) $$\begin{eqnarray}p_{\Vert \text{p}}=p_{\Vert \text{p},0}\,{\mathcal{F}}_{\Vert }(x)=p_{\Vert \text{p},0}\left(\frac{n(x)}{n_{0}}\right)^{\unicode[STIX]{x1D6FE}_{\Vert \text{p}}},\quad p_{\Vert \text{e}}=p_{\Vert \text{e},0}{\mathcal{G}}_{\Vert }(x)=p_{\Vert \text{e},0}\left(\frac{n(x)}{n_{0}}\right)^{\unicode[STIX]{x1D6FE}_{\Vert \text{e}}},\end{eqnarray}$$

where ${\mathcal{F}}_{\bot }$ , ${\mathcal{F}}_{\Vert }$ , ${\mathcal{G}}_{\bot }$ and ${\mathcal{G}}_{\Vert }$ are functions that reduce to unity for $x\rightarrow +\infty$ , as it is for  ${\mathcal{H}}$ .

A.2 General one-dimensional MHD solutions for incompressible flows

Within an (anisotropic) MHD model of plasma, the shear flow does not play a role in the equilibrium profile. In fact, when $\unicode[STIX]{x03C0}_{\text{p}}^{(1)}$ is neglected, the equilibrium condition for the above configuration simply consists of a balance between the magnetic pressure, $B^{2}(x)/8\unicode[STIX]{x03C0}$ , and total perpendicular thermal pressures, $P_{\bot }(x)$ :

(A 8) $$\begin{eqnarray}\frac{\text{d}}{\text{d}x}\left[p_{\bot \text{p}}(x)+p_{\bot \text{e}}(x)+\frac{B^{2}(x)}{2}\right]=0.\end{eqnarray}$$

In particular, the above condition allows also the widely adopted uniform and homogeneous plasma configuration: $p_{\bot \unicode[STIX]{x1D6FC}}=p_{\Vert \unicode[STIX]{x1D6FC},0}$ , $p_{\bot \unicode[STIX]{x1D6FC}}=p_{\bot \unicode[STIX]{x1D6FC},0}$ , $B_{y}=B_{0y}$ and $B_{z}=B_{0z}$ . Such homogeneous profiles are not an equilibrium solution when FLR corrections (or the full pressure-tensor equations) are included in the fluid description (Cerri Reference Cerri2012; Cerri et al. Reference Cerri, Henri, Califano, Del Sarto, Faganello and Pegoraro2013, Reference Cerri, Pegoraro, Califano, Del Sarto and Jenko2014), unless the velocity profile is a linear function of $x$ (see § A.3). In general, the solution of the MHD equilibrium condition in (A 8) is completely described by the magnetic pressure profile in (A 4), which determines all the other relevant functions, ${\mathcal{F}}_{\bot }(x)$ and ${\mathcal{G}}_{\bot }(x)$ . In fact, assuming $\unicode[STIX]{x1D6FE}_{\bot \text{e}}=\unicode[STIX]{x1D6FE}_{\bot \text{p}}$ for simplicity, quasi-neutrality reads as

(A 9) $$\begin{eqnarray}{\mathcal{G}}_{\bot }(x)={\mathcal{F}}_{\bot }(x)\end{eqnarray}$$

and the equilibrium condition finally gives ${\mathcal{F}}_{\bot }$ as function of ${\mathcal{H}}$ ,

(A 10) $$\begin{eqnarray}{\mathcal{F}}_{\bot }(x)=1+\frac{1}{\unicode[STIX]{x1D6FD}_{\bot ,0}}[1-{\mathcal{H}}(x)],\end{eqnarray}$$

where $\unicode[STIX]{x1D6FD}_{\bot ,0}=\unicode[STIX]{x1D6FD}_{\bot \text{p},0}+\unicode[STIX]{x1D6FD}_{\bot \text{e},0}$ (with $\unicode[STIX]{x1D6FD}_{\bot \unicode[STIX]{x1D6FC},0}\equiv 8\unicode[STIX]{x03C0}p_{\bot \unicode[STIX]{x1D6FC},0}/B_{0}^{2}$ ), and the constant is set to $1+\unicode[STIX]{x1D6FD}_{\bot ,0}$ by the boundary conditions at $x\rightarrow +\infty$ (the requirement ${\mathcal{F}}_{\bot }(x)\rightarrow 1$ for $x\rightarrow +\infty$ is then automatically satisfied due to (A 5)). Furthermore, since the function ${\mathcal{F}}_{\bot }(x)$ is related to the thermal pressure, it cannot assume negative values, which provides the additional condition

(A 11) $$\begin{eqnarray}{\mathcal{F}}_{\bot }(x)\geqslant 0\quad \forall x\quad \Longleftrightarrow \quad {\mathcal{H}}(x)\leqslant 1+\unicode[STIX]{x1D6FD}_{\bot ,0}.\end{eqnarray}$$

This states physically that any variation of the magnetic pressure, $\unicode[STIX]{x0394}B^{2}/8\unicode[STIX]{x03C0}=(B^{2}(x)-B_{0}^{2})/8\unicode[STIX]{x03C0}$ , cannot exceed the total thermal pressure, $P_{\bot ,0}=p_{\bot \text{p},0}+p_{\bot \text{e},0}$ , where $B_{0}$ , $p_{\bot \text{p},0}$ and $p_{\bot \text{e},0}$ are the values at $x\rightarrow +\infty$ . The parallel thermal pressures follow from the polytropic assumption, e.g.  ${\mathcal{F}}_{\Vert }(x)=[{\mathcal{F}}_{\bot }(x)]^{\unicode[STIX]{x1D6FE}_{\Vert \text{p}}/\unicode[STIX]{x1D6FE}_{\bot \text{p}}}$ . Analogously, the temperature profiles follow from $T_{\bot \unicode[STIX]{x1D6FC}}=p_{\bot \unicode[STIX]{x1D6FC}}/n$ and $T_{\Vert \unicode[STIX]{x1D6FC}}=p_{\Vert \unicode[STIX]{x1D6FC}}/n$ .

Starting from this MHD class of solutions, we self-consistently derive the corresponding equilibrium profiles with first-order FLR corrections.

A.3 General first-order FLR corrections to the one-dimensional MHD solutions

Let us now consider the changes to the MHD equilibrium profiles derived above that are induced by the velocity shear in (A 1) when first-order FLR corrections are taken into account. In this case, the only component of $\unicode[STIX]{x03C0}_{\text{p}}^{(1)}$ that is relevant to the equilibrium condition is

(A 12) $$\begin{eqnarray}\unicode[STIX]{x03C0}_{\text{p},xx}^{(1)}=-\frac{1}{2}\frac{m_{\text{p}}c}{e|\boldsymbol{B}|}\left(b_{z}\frac{\text{d}u_{y}}{\text{d}x}-b_{y}\frac{\text{d}u_{z}}{\text{d}x}\right)p_{\bot \text{p}}.\end{eqnarray}$$

The equilibrium condition in (A 8) now reads

(A 13) $$\begin{eqnarray}\frac{\text{d}}{\text{d}x}\left\{\!\left[1-\frac{1}{2}\frac{B_{z}(x)u_{y}^{\prime }(x)-B_{y}(x)u_{z}^{\prime }(x)}{eB^{2}(x)/m_{\text{p}}c}\right]p_{\bot \text{p}}(x)+p_{\bot \text{e}}(x)+\frac{B^{2}(x)}{8\unicode[STIX]{x03C0}}\!\right\}=0,\end{eqnarray}$$

where the prime denotes the $x$ -derivative. The above expressions can be explicitly written in terms of the fluid vorticity, $\unicode[STIX]{x1D74E}\equiv \unicode[STIX]{x1D735}\boldsymbol{\times }\boldsymbol{u}$ , and of the magnetic field direction, $\boldsymbol{b}$ :

(A 14) $$\begin{eqnarray}\unicode[STIX]{x03C0}_{\text{p},xx}^{(1)}=-\frac{1}{2}\frac{m_{\text{p}}c}{eB}(\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D74E})p_{\bot \text{p}}\quad \longrightarrow \quad \frac{\text{d}}{\text{d}x}\left[\left(1-\frac{m_{\text{p}}c}{eB}\frac{\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D74E}}{2}\right)p_{\bot \text{p}}+p_{\bot \text{e}}+\frac{B^{2}}{8\unicode[STIX]{x03C0}}\right]=0,\end{eqnarray}$$

where $\unicode[STIX]{x1D714}_{y}=-u_{z}^{\prime }$ and $\unicode[STIX]{x1D714}_{z}=u_{y}^{\prime }$ are the components of the fluid vorticity in our configuration. The dependence on $\boldsymbol{b}\boldsymbol{\cdot }\unicode[STIX]{x1D74E}$ highlights the intrinsic asymmetry in the system due to FLR corrections and related to the degree of alignment (or anti-alignment) between the flow vorticity and the magnetic field. We stress, however, that the simple dependence on the vorticity and magnetic-field direction in (A 12) is due to the one-dimensional character of the problem considered here.

We now seek FLR-corrected equilibrium profiles in the form $\widetilde{{\mathcal{F}}}_{\bot }(x)={\mathcal{F}}_{\bot }(x)f_{\bot }(x)$ , $\widetilde{{\mathcal{G}}}_{\bot }(x)={\mathcal{G}}_{\bot }(x)g_{\bot }(x)$ and $\widetilde{{\mathcal{H}}}(x)={\mathcal{H}}(x)h(x)$ , where $f_{\bot }$ , $g_{\bot }$ and $h$ are the ‘correction functions’. Due to the boundary conditions on the MHD flow, (A 2), the gyroviscous tensor vanishes at the boundaries, $\lim _{x\rightarrow \pm \infty }\unicode[STIX]{x03C0}_{\text{p}}^{(1)}=0$ , and thus the correction functions must reduce to unity accordingly, $\lim _{x\rightarrow \pm \infty }\{\,f_{\bot }(x),g_{\bot }(x),h(x)\}=1$ . Therefore, $\widetilde{{\mathcal{F}}}_{\bot }$ , $\widetilde{{\mathcal{G}}}_{\bot }$ and $\widetilde{{\mathcal{H}}}$ reduce to the corresponding MHD profiles away from the shear layer, where the vorticity vanishes (or, in general, where the vorticity becomes uniform and homogeneous). Moreover, since we want to preserve quasi-neutrality, $\widetilde{{\mathcal{F}}}(x)=\widetilde{{\mathcal{G}}}(x)$ must hold and therefore, using (A 9), we obtain the condition

(A 15) $$\begin{eqnarray}g_{\bot }(x)=f_{\bot }(x).\end{eqnarray}$$

In order to relate $h(x)$ and $f_{\bot }(x)$ , we actually need to impose a further constraint on the equilibrium. Such a condition cannot be derived from first principles and would rather be driven by a physical interpretation of the problem under study. Here we provide a viable option based on the plasma beta parameter (see, e.g. Cerri et al. Reference Cerri, Henri, Califano, Del Sarto, Faganello and Pegoraro2013, Reference Cerri, Pegoraro, Califano, Del Sarto and Jenko2014, for examples about different constraints). Since the (thermal) Larmor radius is sensitive to the (perpendicular) plasma beta, a very reasonable constraint is to require that the MHD profile $\unicode[STIX]{x1D6FD}_{\bot \text{p}}(x)$ does not change when passing to the corresponding FLR-corrected profile, i.e.

(A 16) $$\begin{eqnarray}\unicode[STIX]{x1D6FD}_{\bot \text{p}}(x)|_{\text{MHD}}=\unicode[STIX]{x1D6FD}_{\bot \text{p}}(x)|_{\text{MHD}+\text{FLR}}\quad \Longrightarrow \quad h(x)=f_{\bot }(x).\end{eqnarray}$$

Then, using the above relations and the boundary conditions at $x\rightarrow +\infty$ to set the integration constant to $1+\unicode[STIX]{x1D6FD}_{\bot ,0}$ , from (A 13) we obtain the following equation for $f_{\bot }(x)$ :

(A 17) $$\begin{eqnarray}f_{\bot }(x)-\widetilde{U}^{\prime }(x)\sqrt{f_{\bot }(x)}-1=0,\end{eqnarray}$$

where we have defined

(A 18) $$\begin{eqnarray}\widetilde{U}^{\prime }(x)\equiv \frac{\widetilde{\unicode[STIX]{x1D6FD}}_{\bot \text{p},0}}{2}\frac{m_{\text{p}}c}{eB_{0}}\frac{{\mathcal{F}}_{\bot }(x)}{{\mathcal{H}}(x)}\left(\frac{B_{0z}}{B_{0}}H_{z}(x)u_{y}^{\prime }(x)-\frac{B_{0y}}{B_{0}}H_{y}(x)u_{z}^{\prime }(x)\right),\end{eqnarray}$$

with $\widetilde{\unicode[STIX]{x1D6FD}}_{\bot \text{p},0}\equiv \unicode[STIX]{x1D6FD}_{\bot \text{p},0}/(1+\unicode[STIX]{x1D6FD}_{\bot ,0})$ for brevity. Note that the above equation for $f_{\bot }(x)$ has been obtained taking into account the FLR corrections computed with the self-consistent equilibrium magnetic-field profile, $B(x)=B_{0}\sqrt{{\mathcal{H}}(x)f_{\bot }(x)}$ (we remind that $h(x)=f_{\bot }(x)$ holds). Finally, since $p_{\bot \text{p}}(x)$ must be a positive quantity, we require $f_{\bot }(x)\geqslant 0\,\forall x$ , so that the only physical solution of (A 17) is

(A 19) $$\begin{eqnarray}f_{\bot }(x)=\left\{\frac{\widetilde{U}^{\prime }(x)}{2}+\sqrt{1+\left(\frac{\widetilde{U}^{\prime }(x)}{2}\right)^{2}}\right\}^{2}.\end{eqnarray}$$

This correctly reduces to unity for vanishing FLR terms, $\widetilde{U}^{\prime }\rightarrow 0$ , recovering the MHD profiles. The resulting FLR-corrected profiles are therefore given by

(A 20a,b ) $$\begin{eqnarray}p_{\bot \text{p}}(x)=p_{\bot \text{p},0}\,{\mathcal{F}}_{\bot }(x)\,f_{\bot }(x),\quad p_{\Vert \text{p}}(x)=p_{\Vert \text{p},0}({\mathcal{F}}_{\bot }(x)\,f_{\bot }(x))^{\unicode[STIX]{x1D6FE}_{\Vert }/\unicode[STIX]{x1D6FE}_{\bot }},\end{eqnarray}$$

(A 21) $$\begin{eqnarray}n(x)=n_{0}({\mathcal{F}}_{\bot }(x)\,f_{\bot }(x))^{1/\unicode[STIX]{x1D6FE}_{\bot }},\end{eqnarray}$$

(A 22a,b ) $$\begin{eqnarray}B_{y}(x)=B_{0y}H_{y}(x)\sqrt{f_{\bot }(x)},\quad B_{z}(x)=B_{0z}H_{z}(x)\sqrt{f_{\bot }(x)},\end{eqnarray}$$

from which the current density, $\boldsymbol{J}=\unicode[STIX]{x1D735}\times \boldsymbol{B}$ , follows.

B.1 Perturbative expansion of the pressure-tensor equation

Let us consider the dynamic equation for the full pressure tensor,

(B 1) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6F1}_{ij}}{\unicode[STIX]{x2202}t}+\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x_{k}}(\unicode[STIX]{x1D6F1}_{ij}u_{k}+Q_{ijk})+\unicode[STIX]{x1D6F1}_{ik}\frac{\unicode[STIX]{x2202}u_{j}}{\unicode[STIX]{x2202}x_{k}}+\unicode[STIX]{x1D6F1}_{jk}\frac{\unicode[STIX]{x2202}u_{i}}{\unicode[STIX]{x2202}x_{k}}=\unicode[STIX]{x1D6FA}_{c\unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D716}_{ikl}\unicode[STIX]{x1D6F1}_{jk}+\unicode[STIX]{x1D716}_{jkl}\unicode[STIX]{x1D6F1}_{ik})b_{l},\end{eqnarray}$$

where $\unicode[STIX]{x1D716}_{ijk}$ is the Levi-Civita symbol, and perturbatively expand it with respect to the small parameter

(B 2) $$\begin{eqnarray}\unicode[STIX]{x1D700}\equiv \frac{\unicode[STIX]{x1D70C}}{L}\sim \frac{\unicode[STIX]{x1D714}}{\unicode[STIX]{x1D6FA}}\ll 1,\end{eqnarray}$$

where $\unicode[STIX]{x1D70C}$ is the Larmor radius, $L$ is the typical length scale of variation of fluid quantities and $\unicode[STIX]{x1D714}\sim u/L$ is the characteristic frequency of the fluid dynamics. Here we adopt the so-called ‘fast-dynamics ordering’, $u\sim v_{\text{th}}$ (Macmahon Reference Macmahon1965; Ramos Reference Ramos2005a ; Cerri et al. Reference Cerri, Henri, Califano, Del Sarto, Faganello and Pegoraro2013). Using dimensionless quantities denoted by a tildeFootnote ¹⁰ , equation (B 1) rewrites as

(B 3) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D716}_{ikl}\widetilde{\unicode[STIX]{x1D6F1}}_{jk}+\unicode[STIX]{x1D716}_{jkl}\widetilde{\unicode[STIX]{x1D6F1}}_{ik})\widetilde{b}_{l}\nonumber\\ \displaystyle & & \displaystyle \quad =\unicode[STIX]{x1D700}\frac{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{|\widetilde{\boldsymbol{B}}|}\left[\frac{\unicode[STIX]{x2202}\widetilde{\unicode[STIX]{x1D6F1}}_{ij}}{\unicode[STIX]{x2202}\widetilde{t}}+\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\widetilde{x}_{k}}(\widetilde{\unicode[STIX]{x1D6F1}}_{ij}\widetilde{u}_{k})+\widetilde{\unicode[STIX]{x1D6F1}}_{ik}\frac{\unicode[STIX]{x2202}\widetilde{u}_{j}}{\unicode[STIX]{x2202}\widetilde{x}_{k}}+\widetilde{\unicode[STIX]{x1D6F1}}_{jk}\frac{\unicode[STIX]{x2202}\widetilde{u}_{i}}{\unicode[STIX]{x2202}\widetilde{x}_{k}}+\frac{\unicode[STIX]{x2202}\widetilde{Q}_{ijk}}{\unicode[STIX]{x2202}\widetilde{x}_{k}}\right],\end{eqnarray}$$

where we have defined $\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}\equiv \text{sign}(e_{\unicode[STIX]{x1D6FC}})$ , i.e. the sign embedded in the cyclotron frequency, $\unicode[STIX]{x1D6FA}_{c\unicode[STIX]{x1D6FC}}=e_{\unicode[STIX]{x1D6FC}}B_{0}/m_{\unicode[STIX]{x1D6FC}}c=\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}|e_{\unicode[STIX]{x1D6FC}}|B_{0}/m_{\unicode[STIX]{x1D6FC}}c\equiv \unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}|\unicode[STIX]{x1D6FA}_{c\unicode[STIX]{x1D6FC}}|$ . We then expand the pressure tensor and heat-flux tensor in powers of $\unicode[STIX]{x1D700}$ , i.e.

(B 4a,b ) $$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D6F1}}_{ij}=\mathop{\sum }_{n=0}^{\infty }\unicode[STIX]{x1D700}^{n}\widetilde{\unicode[STIX]{x1D6F1}}_{ij}^{(n)}\quad \text{and}\quad \widetilde{Q}_{ijk}=\mathop{\sum }_{n=0}^{\infty }\unicode[STIX]{x1D700}^{n}\widetilde{Q}_{ijk}^{(n)}.\end{eqnarray}$$

Hereafter, the tilde will be omitted for the sake of simplicity and all the quantities have to be understood as dimensionless. The $n$ th-order pressure-tensor equation then reads

(B 5) $$\begin{eqnarray}{\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D6F1}_{ij}^{(n)}]={\mathcal{R}}_{\boldsymbol{ u}}[\unicode[STIX]{x1D6F1}_{ij}^{(n-1)}]+{\mathcal{D}}[Q_{ij(k)}^{(n-1)}],\end{eqnarray}$$

where we have introduced the following linear operators:

(B 6) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D72B}]\equiv \{\unicode[STIX]{x1D72B}\times \boldsymbol{b}\}^{\text{(sym)}} & \displaystyle\end{eqnarray}$$

(B 7) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{R}}_{\boldsymbol{u}}[\unicode[STIX]{x1D72B}]\equiv \frac{\text{d}\unicode[STIX]{x1D72B}}{\text{d}t}+\unicode[STIX]{x1D72B}(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u})+\{\unicode[STIX]{x1D72B}\boldsymbol{ : }\unicode[STIX]{x1D735}\boldsymbol{u}\}^{\text{(sym)}} & \displaystyle\end{eqnarray}$$

(B 8) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{D}}[\unicode[STIX]{x1D64C}]\equiv \unicode[STIX]{x1D735}\boldsymbol{\cdot }\unicode[STIX]{x1D64C}, & \displaystyle\end{eqnarray}$$

which contribute to the evolution of the pressure tensor by involving only $\boldsymbol{B}$ , $\boldsymbol{u}$ and $\unicode[STIX]{x1D64C}$ , respectively ( $\unicode[STIX]{x2202}/\unicode[STIX]{x2202}t+\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}$ has been replaced by the Lagrangian time derivative $\text{d}/\text{d}t$ for shortness). The zero order, $n=0$ , gives

(B 9) $$\begin{eqnarray}(\unicode[STIX]{x1D716}_{ilm}\unicode[STIX]{x1D6F1}_{lj}^{(0)}+\unicode[STIX]{x1D716}_{jlm}\unicode[STIX]{x1D6F1}_{li}^{(0)})b_{m}=0,\end{eqnarray}$$

that means that $\unicode[STIX]{x1D6F1}_{ij}^{(0)}$ belongs to the kernel of the ${\mathcal{L}}_{\boldsymbol{B}}$ operator, whereas the first-order equation, $n=1$ , is

(B 10) $$\begin{eqnarray}(\unicode[STIX]{x1D716}_{ilm}\unicode[STIX]{x1D6F1}_{lj}^{(1)}+\unicode[STIX]{x1D716}_{jlm}\unicode[STIX]{x1D6F1}_{li}^{(1)})b_{m}=\frac{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\left[\frac{\text{d}\unicode[STIX]{x1D6F1}_{ij}^{(0)}}{\text{d}t}+\unicode[STIX]{x1D6F1}_{ij}^{(0)}\frac{\unicode[STIX]{x2202}u_{k}}{\unicode[STIX]{x2202}x_{k}}+\unicode[STIX]{x1D6F1}_{ik}^{(0)}\frac{\unicode[STIX]{x2202}u_{j}}{\unicode[STIX]{x2202}x_{k}}+\unicode[STIX]{x1D6F1}_{jk}^{(0)}\frac{\unicode[STIX]{x2202}u_{i}}{\unicode[STIX]{x2202}x_{k}}+\frac{\unicode[STIX]{x2202}Q_{ijk}^{(0)}}{\unicode[STIX]{x2202}x_{k}}\right].\end{eqnarray}$$

Before proceeding in the actual solution of the above equations, let us comment on their symmetry properties, in particular with respect to the magnetic-field direction.

B.2 Symmetry considerations on the perturbed equations

Let us consider the three operators, ${\mathcal{L}}_{\boldsymbol{B}}$ , ${\mathcal{R}}_{\boldsymbol{u}}$ and ${\mathcal{D}}$ . If we invert the direction of the magnetic field, $\boldsymbol{B}\rightarrow -\boldsymbol{B}$ , then such operators transform as

(B 11) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{L}}_{\boldsymbol{B}}[\bullet ]\quad \rightarrow \quad {\mathcal{L}}_{-\boldsymbol{B}}[\bullet ]=-{\mathcal{L}}_{\boldsymbol{B}}[\bullet ] & \displaystyle\end{eqnarray}$$

(B 12) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{R}}_{\boldsymbol{u}}[\bullet ]\quad \rightarrow \quad {\mathcal{R}}_{\boldsymbol{u}}[\bullet ] & \displaystyle\end{eqnarray}$$

(B 13) $$\begin{eqnarray}\displaystyle & \displaystyle {\mathcal{D}}[\bullet ]\quad \rightarrow \quad {\mathcal{D}}[\bullet ], & \displaystyle\end{eqnarray}$$

and this symmetry property has a direct consequence on the solutions.

Let us consider the zeroth-order equation, (B 9), and a possible solution $\unicode[STIX]{x1D6F1}_{+}^{(0)}$ . Then, if we reverse the direction of the magnetic field, the linear operator ${\mathcal{L}}_{\boldsymbol{B}}$ also changes sign, but the zeroth-order equation remains the same and $\unicode[STIX]{x1D72B}_{+}^{(0)}$ is still a solution (i.e. if $\unicode[STIX]{x1D72B}_{-}^{(0)}$ is the solution when the magnetic-field direction is reversed, then $\unicode[STIX]{x1D72B}_{-}^{(0)}=\unicode[STIX]{x1D72B}_{+}^{(0)}$ must hold in order to have a unique solution). Therefore, $\unicode[STIX]{x1D72B}^{(0)}$ is invariant under magnetic-field inversion and we can drop the ‘ $+$ ’ and ‘ $-$ ’ subscripts (see § B.3).

Let $\unicode[STIX]{x1D72B}_{+}^{(1)}$ be a solution of the first-order equation (B 10),

(B 14) $$\begin{eqnarray}{\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D72B}_{+}^{(1)}]={\mathcal{R}}_{\boldsymbol{ u}}[\unicode[STIX]{x1D72B}^{(0)}]+{\mathcal{D}}[\unicode[STIX]{x1D64C}^{(0)}].\end{eqnarray}$$

Now consider the same configuration, but with just the magnetic field in the opposite direction, i.e.  $\boldsymbol{b}\rightarrow -\boldsymbol{b}$ . Regardless of the actual behaviour of the gyrotropic heat-flux tensor, $\unicode[STIX]{x1D64C}^{(0)}$ , with respect to such inversionFootnote ¹¹ , if we assume that the first-order solution $\unicode[STIX]{x1D72B}_{+}^{(1)}$ is invariant with respect to $\boldsymbol{b}\rightarrow -\boldsymbol{b}$ , we then obtain a different equation:

(B 15) $$\begin{eqnarray}{\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D72B}_{+}^{(1)}]=-{\mathcal{R}}_{\boldsymbol{ u}}[\unicode[STIX]{x1D72B}^{(0)}]\mp {\mathcal{D}}[\unicode[STIX]{x1D64C}^{(0)}],\end{eqnarray}$$

where the $\mp$ sign in front of ${\mathcal{D}}[\unicode[STIX]{x1D64C}^{(0)}]$ takes into account for any possible behaviour of $\unicode[STIX]{x1D64C}^{(0)}$ with respect to such an inversion. Let us drop the heat-flux contribution for a moment and consider the two equations, ${\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D72B}_{+}^{(1)}]={\mathcal{R}}_{\boldsymbol{u}}[\unicode[STIX]{x1D72B}^{(0)}]$ and ${\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D72B}_{+}^{(1)}]=-{\mathcal{R}}_{\boldsymbol{u}}[\unicode[STIX]{x1D72B}^{(0)}]$ . Clearly, a non-zero solution $\unicode[STIX]{x1D72B}_{+}^{(1)}$ cannot satisfy simultaneously the two equations above, and so we must admit that there exists a different solution, $\unicode[STIX]{x1D72B}_{-}^{(1)}$ . Due to the linear nature of the operators, it is immediate to see that a relation $\unicode[STIX]{x1D72B}_{-}^{(1)}=-\unicode[STIX]{x1D72B}_{+}^{(1)}$ must hold. With the contribution of the heat flux the relation might not be straightforward as $\unicode[STIX]{x1D72B}_{-}^{(1)}=-\unicode[STIX]{x1D72B}_{+}^{(1)}$ , but, again, being ${\mathcal{L}}_{\boldsymbol{B}}$ , ${\mathcal{R}}_{\boldsymbol{u}}$ and ${\mathcal{D}}$ linear operators, there will be anyway a part of $\unicode[STIX]{x1D72B}^{(1)}$ that changes sign when $\boldsymbol{b}\rightarrow -\boldsymbol{b}$ . This is a feature deeply encoded in the governing equations of a plasma, but it first emerges only when the fluid hierarchy is retained up to the pressure-tensor equation (Cerri et al. Reference Cerri, Pegoraro, Califano, Del Sarto and Jenko2014; Del Sarto et al. Reference Del Sarto, Pegoraro and Califano2016) or first-order FLR corrections are included (Hazeltine, Kotschenreuther & Morrison Reference Hazeltine, Kotschenreuther and Morrison1985; Hsu, Hazeltine & Morrison Reference Hsu, Hazeltine and Morrison1986; Ramos Reference Ramos2005b ; Cerri et al. Reference Cerri, Henri, Califano, Del Sarto, Faganello and Pegoraro2013).

B.3 Zeroth-order solution: gyrotopic pressure tensor

At zero order, $\unicode[STIX]{x1D72B}^{(0)}$ must satisfy ${\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D72B}^{(0)}]=0$ , i.e. it will be a linear combination of the basis vector spanning the kernel of the (self-adjoint) linear operator ${\mathcal{L}}_{B}$ . Any linear combination of the identity, $\unicode[STIX]{x1D644}$ , and of the projector along the magnetic-field direction, $\boldsymbol{b}\boldsymbol{b}$ , i.e.  $\unicode[STIX]{x1D72B}^{(0)}=p_{1}\unicode[STIX]{x1D644}+p_{2}\boldsymbol{b}\boldsymbol{b}$ , is a zeroth-order solution. Defining the parallel and perpendicular pressures as $p_{\bot }=p_{1}$ and $p_{\Vert }=p_{1}+p_{2}$ , we recover the gyrotropic Chew-Goldberger-Low (CGL) pressure tensor (Chew et al. Reference Chew, Goldberger and Low1956):

(B 16) $$\begin{eqnarray}\unicode[STIX]{x1D72B}^{(0)}=p_{\bot }\unicode[STIX]{x1D749}+p_{\Vert }\boldsymbol{b}\boldsymbol{b}.\end{eqnarray}$$

The zeroth-order solution is insensitive to the operation $\boldsymbol{b}\rightarrow -\boldsymbol{b}$ , as anticipated. Note that the equation for $n=0$ , and thus its solution $\unicode[STIX]{x1D72B}_{\unicode[STIX]{x1D6FC}}^{(0)}$ , does not depend on the velocity field $\boldsymbol{u}$ or on the heat-flux tensor $\unicode[STIX]{x1D64C}$ , so the only information that we need is the direction of the magnetic field, $\boldsymbol{b}$ . Finally, note that there is an interesting consequence of this solution in an ordering for which $\unicode[STIX]{x1D714}/\unicode[STIX]{x1D6FA}_{c\unicode[STIX]{x1D6FC}}\ll 1$ : because the gyrofrequency is inversely proportional to the species’ mass, $\unicode[STIX]{x1D6FA}_{c\unicode[STIX]{x1D6FC}}\propto 1/m_{\unicode[STIX]{x1D6FC}}$ , within a low-frequency dynamics we expect the lighter species (e.g. the electrons) to be naturally found very close to a gyrotropic stateFootnote ¹² .

B.4 First-order solution: FLR corrections and dynamic equations for $p_{\Vert }$ and $p_{\bot }$

Before proceeding in the solution of the first-order equation in the perturbative expansion, (B 10), we recast it in a form that is invariant under the operation $\boldsymbol{b}\rightarrow -\boldsymbol{b}$ . In this way, we solve it only once for a solution $\unicode[STIX]{x1D72B}^{(1)}$ that encodes both $\unicode[STIX]{x1D72B}_{+}^{(1)}$ and $\unicode[STIX]{x1D72B}_{-}^{(1)}$ . At this stage, we need to take into account the fact that $\unicode[STIX]{x1D64C}^{(0)}$ changes sign when we reverse the direction of $\boldsymbol{B}$ (see e.g. Goswami et al. Reference Goswami, Passot and Sulem2005). Therefore, we introduce a coefficient that takes into account the relative orientation of the magnetic field with respect to the coordinate axes, $s_{m}\equiv \text{sign}[\boldsymbol{b}\boldsymbol{\cdot }\boldsymbol{e}_{m}]=\text{sign}[b_{m}]$ (such that $s_{m}^{-1}=s_{m}$ ), where $\boldsymbol{e}_{m}$ is the unit vector along the $m$ -axis of the reference system. The invariant equation now reads (Cerri et al. Reference Cerri, Henri, Califano, Del Sarto, Faganello and Pegoraro2013)

(B 17) $$\begin{eqnarray}\displaystyle & & \displaystyle (\unicode[STIX]{x1D716}_{ilm}\unicode[STIX]{x1D6F1}_{lj}^{(1)}+\unicode[STIX]{x1D716}_{jlm}\unicode[STIX]{x1D6F1}_{li}^{(1)})b_{m}\nonumber\\ \displaystyle & & \displaystyle \quad =\frac{s_{m}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\left[\frac{\text{d}\unicode[STIX]{x1D6F1}_{ij}^{(0)}}{\text{d}t}+\unicode[STIX]{x1D6F1}_{ij}^{(0)}\frac{\unicode[STIX]{x2202}u_{k}}{\unicode[STIX]{x2202}x_{k}}+\unicode[STIX]{x1D6F1}_{ik}^{(0)}\frac{\unicode[STIX]{x2202}u_{j}}{\unicode[STIX]{x2202}x_{k}}+\unicode[STIX]{x1D6F1}_{jk}^{(0)}\frac{\unicode[STIX]{x2202}u_{i}}{\unicode[STIX]{x2202}x_{k}}\right]+\frac{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\frac{\unicode[STIX]{x2202}Q_{ijk}^{(0)}}{\unicode[STIX]{x2202}x_{k}}.\end{eqnarray}$$

By evaluating every term in the above equation (see, e.g. Cerri et al. Reference Cerri, Henri, Califano, Del Sarto, Faganello and Pegoraro2013), one eventually gets the dynamic equations for the zeroth-order pressure components,

(B 18) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}p_{\Vert \unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x2202}t}+\unicode[STIX]{x1D735}\boldsymbol{\cdot }(p_{\Vert \unicode[STIX]{x1D6FC}}\boldsymbol{u}_{\unicode[STIX]{x1D6FC}})+2p_{\Vert \unicode[STIX]{x1D6FC}}(\boldsymbol{b}\boldsymbol{b}\boldsymbol{ : }\unicode[STIX]{x1D735}\boldsymbol{u}_{\unicode[STIX]{x1D6FC}})+\unicode[STIX]{x1D735}\boldsymbol{\cdot }(q_{\Vert \unicode[STIX]{x1D6FC}}\boldsymbol{b})-2q_{\bot \unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{b})=0, & \displaystyle\end{eqnarray}$$

(B 19) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\unicode[STIX]{x2202}p_{\bot \unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x2202}t}+\unicode[STIX]{x1D735}\boldsymbol{\cdot }(p_{\bot \unicode[STIX]{x1D6FC}}\boldsymbol{u}_{\unicode[STIX]{x1D6FC}})+p_{\bot \unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D749}\boldsymbol{ : }\unicode[STIX]{x1D735}\boldsymbol{u}_{\unicode[STIX]{x1D6FC}})+\unicode[STIX]{x1D735}\boldsymbol{\cdot }(q_{\bot \unicode[STIX]{x1D6FC}}\boldsymbol{b})+q_{\bot \unicode[STIX]{x1D6FC}}(\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{b})=0, & \displaystyle\end{eqnarray}$$

and the expressions for the components of $\unicode[STIX]{x1D72B}_{\unicode[STIX]{x1D6FC}}^{(1)}$ ,

(B 20) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xx}^{(1)}=-\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yy}^{(1)}=-\frac{s_{3}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{2}\frac{p_{\bot \unicode[STIX]{x1D6FC}}}{B}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},x}}{\unicode[STIX]{x2202}y}+\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},y}}{\unicode[STIX]{x2202}x}\right) & \displaystyle\end{eqnarray}$$

(B 21) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xy}^{(1)}=\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yx}^{(1)}=-\frac{s_{3}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{2}\frac{p_{\bot \unicode[STIX]{x1D6FC}}}{B}\left(\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},y}}{\unicode[STIX]{x2202}y}-\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},x}}{\unicode[STIX]{x2202}x}\right) & \displaystyle\end{eqnarray}$$

(B 22) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xz}^{(1)}=\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zx}^{(1)}=-\frac{s_{3}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\left[(2p_{\Vert \unicode[STIX]{x1D6FC}}-p_{\bot \unicode[STIX]{x1D6FC}})\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},y}}{\unicode[STIX]{x2202}z}+p_{\bot \unicode[STIX]{x1D6FC}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},z}}{\unicode[STIX]{x2202}y}\right]-\frac{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\frac{\unicode[STIX]{x2202}q_{\bot \unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x2202}y} & \displaystyle\end{eqnarray}$$

(B 23) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yz}^{(1)}=\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zy}^{(1)}=\frac{s_{3}\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\left[(2p_{\Vert \unicode[STIX]{x1D6FC}}-p_{\bot \unicode[STIX]{x1D6FC}})\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},x}}{\unicode[STIX]{x2202}z}+p_{\bot \unicode[STIX]{x1D6FC}}\frac{\unicode[STIX]{x2202}u_{\unicode[STIX]{x1D6FC},z}}{\unicode[STIX]{x2202}x}\right]+\frac{\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}}{B}\frac{\unicode[STIX]{x2202}q_{\bot \unicode[STIX]{x1D6FC}}}{\unicode[STIX]{x2202}x} & \displaystyle\end{eqnarray}$$

(B 24) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zz}^{(1)}=0. & \displaystyle\end{eqnarray}$$

By neglecting the parallel heat fluxes, $q_{\Vert }$ and $q_{\bot }$ , the above expressions can be compared with the classical results given in Braginskii (Reference Braginskii1965) for the collisional case by setting $\unicode[STIX]{x1D702}_{0}=\unicode[STIX]{x1D702}_{1}=\unicode[STIX]{x1D702}_{2}=0$ , $\unicode[STIX]{x1D702}_{3}=p_{\bot }/2\unicode[STIX]{x1D6FA}$ and $\unicode[STIX]{x1D702}_{4}=p_{\bot }/\unicode[STIX]{x1D6FA}$ in the Braginskii’s gyro-viscous coefficients. Moreover, in our expressions there is a contribution to $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xz}^{(1)}$ and to $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yz}^{(1)}$ that is due to the pressure anisotropy, $[2(p_{\Vert }-p_{\bot })/\unicode[STIX]{x1D6FA}]\unicode[STIX]{x2202}_{z}u_{\unicode[STIX]{x1D6FC},x}$ and $[2(p_{\Vert }-p_{\bot })/\unicode[STIX]{x1D6FA}]\unicode[STIX]{x2202}_{z}u_{\unicode[STIX]{x1D6FC},y}$ , respectively, which is missing in Braginskii (Reference Braginskii1965) because of the assumed isotropic temperature, $T_{\Vert }=T_{\bot }=T$ . The above expressions for the FLR corrections explicitly account for the orientation of the magnetic field with respect to the $z$ -axis through the $s_{3}$ coefficient.

C.1 Assumptions and general $n$ th-order solution

In order to find a solution of (C 2) to all orders, we first need to make four assumption on the configuration, on the energy and on the closure. The first is to (i) neglect the heat-flux tensor. The second is that (ii) the inhomogeneity direction, the flow direction and the magnetic-field direction form a right-handed basisFootnote ¹³ , e.g. $\boldsymbol{u}=u_{y}(x)\boldsymbol{e}_{y}$ and $\boldsymbol{B}=B_{z}(x)\boldsymbol{e}_{z}$ . The third assumption is (iii) stationarity, i.e. no time dependence. Finally, (iv) we assume that any contribution to the pressure tensor beyond the gyrotropic pressure is traceless, which means that we are considering corrections at constant thermal energy. So, summarizing the hypothesis under which we find the solution:

1. (i) $\unicode[STIX]{x1D64C}^{(n)}=0\,\forall n$ ;

2. (ii) $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0$ and $\boldsymbol{B}\boldsymbol{\times }(\unicode[STIX]{x1D735}\boldsymbol{\times }\boldsymbol{u})=0$ ;

3. (iii) $\unicode[STIX]{x2202}/\unicode[STIX]{x2202}t=0$ ;

4. (iv) $\text{Tr}[\unicode[STIX]{x1D72B}_{\unicode[STIX]{x1D6FC}}^{(n)}]=0\,\forall n\geqslant 1$ .

Under the assumptions (i)–(iv), considering the inhomogeneity to be in $x$ -direction for simplicity, the solution of (C 2) $\forall n\geqslant 1$ is:

(C 5) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(n)}=0\quad \text{if }i\neq j,\\ \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xx}^{(n)}=-\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yy}^{(n)}=(\widetilde{\unicode[STIX]{x1D712}}_{\unicode[STIX]{x1D6FC}}(x))^{n}p_{\bot \unicode[STIX]{x1D6FC}},\\ \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zz}^{(n)}=0,\end{array}\right\}\end{eqnarray}$$

where we have defined the function $\widetilde{\unicode[STIX]{x1D712}}_{\unicode[STIX]{x1D6FC}}(x)$ as

(C 6) $$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D712}}_{\unicode[STIX]{x1D6FC}}(x)\equiv -\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}\frac{\unicode[STIX]{x1D74E}_{\unicode[STIX]{x1D6FC}}\boldsymbol{\cdot }\boldsymbol{b}}{2\unicode[STIX]{x1D6FA}_{c\unicode[STIX]{x1D6FC}}}=-\unicode[STIX]{x1D70E}_{\unicode[STIX]{x1D6FC}}\frac{s_{z}}{2|\boldsymbol{B}|}\frac{\text{d}u_{\unicode[STIX]{x1D6FC},y}}{\text{d}x}.\end{eqnarray}$$

Note that, in general, $\unicode[STIX]{x1D6F1}_{zz}^{(n)}$ is undetermined at each order, so we make the reasonable choice to take it non-zero only for $n=0$ , i.e. $\unicode[STIX]{x1D6F1}_{zz}^{(n)}=p_{\Vert }\unicode[STIX]{x1D6FF}_{n0}$ , which then, together with the traceless condition (iv), gives us the relation $\unicode[STIX]{x1D6F1}_{xx}^{(n)}+\unicode[STIX]{x1D6F1}_{yy}^{(n)}=2p_{\bot }\unicode[STIX]{x1D6FF}_{n0}$ .

C.2 General $n$ th-order solution: proof

We now proceed to prove that (C 5) is the solution of (C 2), for all $n$ . In order to do that, we are going to use the so-called mathematical induction method. Later on, we will omit the $\unicode[STIX]{x1D6FC}$ index for the species for shortness.

1. (i) $n=1$ : For $n=1$ , (C 2) is ${\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(1)}]=\hat{{\mathcal{R}}}_{\boldsymbol{u}}[\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(0)}]$ , or written in matrix form

   (C 7) $$\begin{eqnarray}\left(\begin{array}{@{}ccc@{}}2\unicode[STIX]{x1D6F1}_{xy}^{(1)} & \unicode[STIX]{x1D6F1}_{yy}^{(1)}-\unicode[STIX]{x1D6F1}_{xx}^{(1)} & \unicode[STIX]{x1D6F1}_{yz}^{(1)}\\ \unicode[STIX]{x1D6F1}_{yy}^{(1)}-\unicode[STIX]{x1D6F1}_{xx}^{(1)} & -2\unicode[STIX]{x1D6F1}_{xy}^{(1)} & -\unicode[STIX]{x1D6F1}_{xz}^{(1)}\\ \unicode[STIX]{x1D6F1}_{yz}^{(1)} & -\unicode[STIX]{x1D6F1}_{xz}^{(1)} & 0\end{array}\right)=\frac{s_{3}\unicode[STIX]{x1D70E}}{|\boldsymbol{B}|}\left(\begin{array}{@{}ccc@{}}\displaystyle \frac{\text{d}}{\text{d}t}\unicode[STIX]{x1D6F1}_{xx}^{(0)} & \displaystyle \unicode[STIX]{x1D6F1}_{xx}^{(0)}\frac{\text{d}u_{y}}{\text{d}x} & 0\\ \displaystyle \unicode[STIX]{x1D6F1}_{xx}^{(0)}\frac{\text{d}u_{y}}{\text{d}x} & \displaystyle -\frac{\text{d}}{\text{d}t}\unicode[STIX]{x1D6F1}_{xx}^{(0)} & 0\\ 0 & 0 & 0\end{array}\right)\end{eqnarray}$$
   whose solution under our assumptions is:
   (C 8) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(1)}=0\quad \text{if }i\neq j,\\ \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xx}^{(1)}=-\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yy}^{(1)}=-\frac{s_{z}\unicode[STIX]{x1D70E}}{2|\boldsymbol{B}|}\frac{\text{d}u_{y}}{\text{d}x}p_{\bot }\equiv \widetilde{\unicode[STIX]{x1D712}}(x)p_{\bot },\\ \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zz}^{(1)}=0,\end{array}\right\}\end{eqnarray}$$
   where we have used the assumptions (ii) and (iii) in order to have $\text{d}\unicode[STIX]{x1D6F1}_{xx}^{(0)}/\text{d}t=0$ : since every quantity can be function only of $x$ and the flow is along the $y$ -direction due to assumption (ii), we get $\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D735}\unicode[STIX]{x1D6F1}_{xx}^{(0)}=u_{y}\unicode[STIX]{x2202}\unicode[STIX]{x1D6F1}_{xx}^{(0)}/\unicode[STIX]{x2202}y=0$ and thus, due also to the stationarity assumption (iii), $(\text{d}/\text{d}t)\unicode[STIX]{x1D6F1}_{xx}^{(0)}=(\unicode[STIX]{x2202}/\unicode[STIX]{x2202}t+u_{y}\unicode[STIX]{x2202}/\unicode[STIX]{x2202}y)\unicode[STIX]{x1D6F1}_{xx}^{(0)}=0$ .

2. (ii) $n=2$ : For $n=2$ , (C 2) is ${\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(2)}]=\hat{{\mathcal{R}}}_{\boldsymbol{u}}[\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(1)}]$ , with $\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(1)}$ given in (C 8). Such an equation, written in matrix form reads

   (C 9) $$\begin{eqnarray}\displaystyle \left(\begin{array}{@{}ccc@{}}2\unicode[STIX]{x1D6F1}_{xy}^{(2)} & \unicode[STIX]{x1D6F1}_{yy}^{(2)}-\unicode[STIX]{x1D6F1}_{xx}^{(2)} & \unicode[STIX]{x1D6F1}_{yz}^{(2)}\\ \unicode[STIX]{x1D6F1}_{yy}^{(2)}-\unicode[STIX]{x1D6F1}_{xx}^{(2)} & -2\unicode[STIX]{x1D6F1}_{xy}^{(2)} & -\unicode[STIX]{x1D6F1}_{xz}^{(2)}\\ \unicode[STIX]{x1D6F1}_{yz}^{(2)} & -\unicode[STIX]{x1D6F1}_{xz}^{(2)} & 0\end{array}\right)=\frac{s_{z}\unicode[STIX]{x1D70E}}{|\boldsymbol{B}|}\left(\begin{array}{@{}ccc@{}}\displaystyle \frac{\text{d}}{\text{d}t}\unicode[STIX]{x1D6F1}_{xx}^{(1)} & \displaystyle \unicode[STIX]{x1D6F1}_{xx}^{(1)}\frac{\text{d}u_{y}}{\text{d}x} & 0\\ \displaystyle \unicode[STIX]{x1D6F1}_{xx}^{(1)}\frac{\text{d}u_{y}}{\text{d}x} & \displaystyle -\frac{\text{d}}{\text{d}t}\unicode[STIX]{x1D6F1}_{xx}^{(1)} & 0\\ 0 & 0 & 0\end{array}\right) & & \displaystyle\end{eqnarray}$$
   whose solution, using again the fact that $\text{d}\unicode[STIX]{x1D6F1}_{xx}^{(1)}/\text{d}t=0$ , is:
   (C 10) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(2)}=0\quad \text{if }i\neq j,\\ \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xx}^{(2)}=-\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yy}^{(2)}=\frac{1}{4|\boldsymbol{B}|^{2}}\left(\frac{\text{d}u_{y}}{\text{d}x}\right)^{2}p_{\bot }\equiv (\widetilde{\unicode[STIX]{x1D712}}(x))^{2}p_{\bot },\\ \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zz}^{(2)}=0,\end{array}\right\}\end{eqnarray}$$
   where we used the fact that $s_{z}^{2}=1$ and $\unicode[STIX]{x1D70E}^{2}=1$ .

3. (iii) Inductive step: We now assume that (C 5) is the correct $n$ th-order solution and we want to solve (C 2) for the ( $n+1$ )th order. That is, ${\mathcal{L}}_{\boldsymbol{B}}[\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(n+1)}]=\hat{{\mathcal{R}}}_{\boldsymbol{u}}[\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(n)}]$ , which in matrix form reads

   (C 11) $$\begin{eqnarray}\displaystyle \!\left(\begin{array}{@{}ccc@{}}2\unicode[STIX]{x1D6F1}_{xy}^{(n+1)} & \unicode[STIX]{x1D6F1}_{yy}^{(n+1)}-\unicode[STIX]{x1D6F1}_{xx}^{(n+1)} & \unicode[STIX]{x1D6F1}_{yz}^{(n+1)}\\ \unicode[STIX]{x1D6F1}_{yy}^{(n+1)}-\unicode[STIX]{x1D6F1}_{xx}^{(n+1)} & -2\unicode[STIX]{x1D6F1}_{xy}^{(n+1)} & -\unicode[STIX]{x1D6F1}_{xz}^{(n+1)}\\ \unicode[STIX]{x1D6F1}_{yz}^{(n+1)} & -\unicode[STIX]{x1D6F1}_{xz}^{(n+1)} & 0\end{array}\right)=\frac{s_{z}\unicode[STIX]{x1D70E}}{|\boldsymbol{B}|}\left(\begin{array}{@{}ccc@{}}\displaystyle \frac{\text{d}}{\text{d}t}\unicode[STIX]{x1D6F1}_{xx}^{(n)} & \displaystyle \unicode[STIX]{x1D6F1}_{xx}^{(n)}\frac{\text{d}u_{y}}{\text{d}x} & 0\\ \displaystyle \unicode[STIX]{x1D6F1}_{xx}^{(n)}\frac{\text{d}u_{y}}{\text{d}x} & \displaystyle -\frac{\text{d}}{\text{d}t}\unicode[STIX]{x1D6F1}_{xx}^{(n)} & 0\\ 0 & 0 & 0\end{array}\right)\hspace{-13.20007pt} & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$
   whose solution, using again the fact that our assumptions are such that $\text{d}\unicode[STIX]{x1D6F1}_{xx}^{(n)}/\text{d}t=0$ , is:
   (C 12) $$\begin{eqnarray}\left.\begin{array}{@{}c@{}}\displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},ij}^{(n+1)}=0\quad \text{if }i\neq j,\\ \displaystyle \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},xx}^{(n+1)}=-\unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},yy}^{(n+1)}=\left(-\frac{s_{z}\unicode[STIX]{x1D70E}}{2|\boldsymbol{B}|}\frac{\text{d}u_{y}}{\text{d}x}\right)^{n+1}p_{\bot }\equiv (\widetilde{\unicode[STIX]{x1D712}}(x))^{n+1}p_{\bot },\\ \unicode[STIX]{x1D6F1}_{\unicode[STIX]{x1D6FC},zz}^{(n+1)}=0,\end{array}\right\}\end{eqnarray}$$
   which finally proves the thesis. ◻

C.3 Summability, convergence and stability of the complete pressure tensor

Now that we have proved the expression for the general $n$ th-order solution of (C 2), we want to go back from the FLR expansion to the full pressure tensor, (C 1). In order to be able to do that, the series must be summable and it should converge.

If we put all the FLR contributions together, the full pressure-tensor components are:

(C 13) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{ij}=0\quad \text{if }i\neq j, & \displaystyle\end{eqnarray}$$

(C 14) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{zz}=p_{\Vert }, & \displaystyle\end{eqnarray}$$

(C 15) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{xx}=[1+\widetilde{\unicode[STIX]{x1D712}}+\widetilde{\unicode[STIX]{x1D712}}^{2}+\cdots \,]p_{\bot }=\left[1+\widetilde{\unicode[STIX]{x1D712}}\mathop{\sum }_{n=0}^{\infty }(\widetilde{\unicode[STIX]{x1D712}})^{n}\right]p_{\bot }, & \displaystyle\end{eqnarray}$$

(C 16) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{yy}=[1-\widetilde{\unicode[STIX]{x1D712}}-\widetilde{\unicode[STIX]{x1D712}}^{2}-\cdots \,]p_{\bot }=\left[1-\widetilde{\unicode[STIX]{x1D712}}\mathop{\sum }_{n=0}^{\infty }(\widetilde{\unicode[STIX]{x1D712}})^{n}\right]p_{\bot }, & \displaystyle\end{eqnarray}$$

so the main request for absolute convergence is that the geometric series $\sum _{n}|\widetilde{\unicode[STIX]{x1D712}}|^{n}$ converge, which is true if and only if

(C 17) $$\begin{eqnarray}|\widetilde{\unicode[STIX]{x1D712}}(x)|<1\quad \Longleftrightarrow \quad |\unicode[STIX]{x1D74E}\boldsymbol{\cdot }\boldsymbol{b}|<2\unicode[STIX]{x1D6FA}_{c}\quad \forall x,\end{eqnarray}$$

which is the absolute convergence condition from the mathematical point of view and represents a limit on the shear strength. If the condition (C 17) holds, then the resulting diagonal components of the pressure tensor are

(C 18) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{xx}=\left(1+\frac{\widetilde{\unicode[STIX]{x1D712}}(x)}{1-\widetilde{\unicode[STIX]{x1D712}}(x)}\right)p_{\bot }, & \displaystyle\end{eqnarray}$$

(C 19) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{yy}=\left(1-\frac{\widetilde{\unicode[STIX]{x1D712}}(x)}{1-\widetilde{\unicode[STIX]{x1D712}}(x)}\right)p_{\bot }, & \displaystyle\end{eqnarray}$$

(C 20) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D6F1}_{zz}=p_{\Vert }. & \displaystyle\end{eqnarray}$$

However, since the components of the (diagonal) pressure tensor cannot be negative in order to have a physical meaning, the function $\widetilde{\unicode[STIX]{x1D712}}(x)$ – and thus the shear strength $\text{d}u_{y}/\text{d}x$ – has to fulfil the positivity condition. This request gives a physical condition on the shear strength which reads

(C 21) $$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D712}}(x)\leqslant {\textstyle \frac{1}{2}}\quad \Longleftrightarrow \quad \unicode[STIX]{x1D74E}\boldsymbol{\cdot }\boldsymbol{b}\geqslant -\unicode[STIX]{x1D6FA}_{c},\end{eqnarray}$$

where now, in principle, the shear can be as negative as one wishes, without no limitations. If we put together the physical condition (C 21) and the mathematical condition (C 17), gives the asymmetric condition

(C 22) $$\begin{eqnarray}-1<\widetilde{\unicode[STIX]{x1D712}}(x)\leqslant {\textstyle \frac{1}{2}}\quad \Longleftrightarrow \quad -\unicode[STIX]{x1D6FA}_{c}\leqslant \unicode[STIX]{x1D74E}\boldsymbol{\cdot }\boldsymbol{b}<2\unicode[STIX]{x1D6FA}_{c}\quad \forall x,\end{eqnarray}$$

The condition above is also a stability condition for the shear-flow configuration. In fact, that is in agreement with Del Sarto et al. (Reference Del Sarto, Pegoraro and Califano2016), where the stability condition is found to be $\unicode[STIX]{x1D6FA}^{\prime }\equiv \unicode[STIX]{x1D6FA}+\unicode[STIX]{x2202}_{x}u_{y}\geqslant 0$ , which translated in our notation correspond to $\widetilde{\unicode[STIX]{x1D712}}(x)\leqslant 1/2$ Footnote ¹⁴ .