Implementation of Ionospheric Generators in the Numerical Model of the Global Electric Circuit

Abstract

The consistent inclusion of global electric circuit sources of ionospheric and magnetospheric origin in distributed numerical models of the circuit is discussed. It is shown that the most natural approach to this inclusion is to introduce into the boundary conditions at the outer boundary of the model atmosphere the corresponding perturbation of the potential specified up to an unknown constant. As an example of the implementation of this approach, the solution of a model problem on a high-latitude magnetospheric convective generator with the use of a three-dimensional numerical model of the global electric circuit is demonstrated. It is shown that the specified potential perturbation in polar regions is projected into the lower layers of the atmosphere with preservation of their structure, which is a consequence of the quasi-one-dimensionality of the problem under the conditions of slow variation in all parameters with latitude and longitude at an approximately constant profile of conductivity.

Appendices

UNIQUENESS OF THE SOLUTION OF EQS. (4)–(6)

Let ${{\varphi }_{1}}\left( {\mathbf{r}} \right)$ and ${{\varphi }_{2}}\left( {\mathbf{r}} \right)$ be two solutions of problem (4)–(6) with constants ${{U}_{1}}$ and ${{U}_{2}}$, respectively. Then, for the difference $\delta \varphi \left( {\mathbf{r}} \right) = {{\varphi }_{1}}\left( {\mathbf{r}} \right) - {{\varphi }_{2}}\left( {\mathbf{r}} \right)$, we can write the system of equations

$${\text{div}}\left( {\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right)} \right) = 0,$$

(A.1)

$$\oint\limits_{{{{\Gamma }}_{1}}} {\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} = 0,$$

(A.2)

$${{\left. {\delta \varphi \left( {\mathbf{r}} \right)} \right|}_{{{{{\Gamma }}_{1}}}}} = 0,\,\,\,\,{{\left. {\delta \varphi \left( {\mathbf{r}} \right)} \right|}_{{{{{\Gamma }}_{2}}}}} = \delta U$$

(A.3)

with $\delta U = {{U}_{1}} - {{U}_{2}}.$ Integrating Eq. (A.1) over the region Ω, passing to the integral over the boundary, and using relationship (A.2), we obtain in addition

$$\oint\limits_{{{{\Gamma }}_{2}}} {\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} = 0.$$

(A.4)

Note that in the right-hand side of the identity (here and below, ${\text{d}}{\mathbf{r}}$ points to a volume integral)

$$\begin{gathered} \int\limits_{\Omega } {\sigma \left( {\mathbf{r}} \right){{{\left| {{\text{grad}}\delta \varphi \left( {\mathbf{r}} \right)} \right|}}^{2}}{\text{d}}{\mathbf{r}}} \\ = \int\limits_{\Omega } {{\text{div}}\left( {\delta \varphi \left( {\mathbf{r}} \right)~\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right)} \right){\text{d}}{\mathbf{r}}} \\ - \,\,\int\limits_{\Omega } {\delta \varphi \left( {\mathbf{r}} \right){\text{div}}\left( {\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right)} \right){\text{d}}{\mathbf{r}}} \\ \end{gathered} $$

the second summand equals zero by virtue of (A.1) and the first can be worked out to (here and below, we suppose that the vector ${\mathbf{n}}\left( {\mathbf{r}} \right)$ is directed outwards relative to the region Ω both on Γ₁ and on Γ₂)

$$\begin{gathered} \int\limits_{\Omega } {{\text{div}}\left( {\delta \varphi \left( {\mathbf{r}} \right)~\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right)} \right){\text{d}}{\mathbf{r}}} \\ = \oint\limits_{{{{\Gamma }}_{1}} \cup {{{\Gamma }}_{2}}} {\delta \varphi \left( {\mathbf{r}} \right)~\sigma \left( {\mathbf{r}} \right){\text{grad}}\delta \varphi \left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} , \\ \end{gathered} $$

which also equals zero by virtue of (A.3) and (A.4). Therefore,

$$\int\limits_{\Omega } {\sigma \left( {\mathbf{r}} \right){{{\left| {{\text{grad}}\delta \varphi \left( {\mathbf{r}} \right)} \right|}}^{2}}{\text{d}}{\mathbf{r}}} = 0,$$

which implies $\delta \varphi \left( {\mathbf{r}} \right) = {\text{const}}{\text{.}}$ However, together with the first equation from (A.3), this means that $\delta \varphi \left( {\mathbf{r}} \right) \equiv 0,$ i.e., ${{\varphi }_{1}}\left( {\mathbf{r}} \right) \equiv {{\varphi }_{2}}\left( {\mathbf{r}} \right)$ and ${{U}_{1}} = {{U}_{2}}.$ A rigorous proof of the existence and uniqueness of the solution for problem (4)–(6) in an appropriate functional space was presented by Kalinin and Slyunyaev (2017).

NUMERICAL SOLUTION OF PROBLEMS (7)–(8) AND (9)–(10)

Since problem (9)–(10) can be considered a particular case of problem (7)–(8), it is sufficient to implement the solve for problem (7)–(8). To solve this problem, we use the finite-element method.

First, let us formulate an equivalent problem in the form of an integral identity. It is easy to see that the problem of solving Eq. (7) is equivalent to the problem of searching for a function $\varphi \left( {\mathbf{r}} \right)$ that satisfies the integral relationship

$$\begin{gathered} \int\limits_{\Omega } {\psi \left( {\mathbf{r}} \right){\text{div}}\left( {\sigma \left( {\mathbf{r}} \right){\text{grad}}\varphi \left( {\mathbf{r}} \right)} \right){\text{d}}{\mathbf{r}}} \\ = \int\limits_{\Omega } {\psi \left( {\mathbf{r}} \right){\text{div }}{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} \\ \end{gathered} $$

(B.1)

for any function $\psi \left( {\mathbf{r}} \right)$ obeying the condition

$${{\left. {\psi \left( {\mathbf{r}} \right)} \right|}_{{{{{\Gamma }}_{1}}}}} = 0,\,\,\,\,{{\left. {\psi \left( {\mathbf{r}} \right)} \right|}_{{{{{\Gamma }}_{2}}}}} = 0.$$

(B.2)

Rewriting identity (B.1) in the form

$$\begin{gathered} \int\limits_{\Omega } {{\text{div}}\left( {\psi \left( {\mathbf{r}} \right)~\sigma \left( {\mathbf{r}} \right){\text{ grad}}\varphi \left( {\mathbf{r}} \right)} \right){\text{d}}{\mathbf{r}}} \\ - \,\,\int\limits_{\Omega } {\sigma \left( {\mathbf{r}} \right){\text{grad}}\varphi \left( {\mathbf{r}} \right) \cdot {\text{grad}}\psi \left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} \\ = \int\limits_{\Omega } {{\text{div}}\left( {\psi \left( {\mathbf{r}} \right)~{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right)} \right){\text{d}}{\mathbf{r}}} - \int\limits_{\Omega } {{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right) \cdot {\text{grad}}\psi \left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} \\ \end{gathered} $$

and passing to integrals over the boundary, we obtain

$$\begin{gathered} \oint\limits_{{{{\Gamma }}_{1}} \cup {{{\Gamma }}_{2}}} {\psi \left( {\mathbf{r}} \right)~\sigma \left( {\mathbf{r}} \right){\text{grad}}\varphi \left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} \\ - \,\,\int\limits_{\Omega } {\sigma \left( {\mathbf{r}} \right){\text{grad}}\varphi \left( {\mathbf{r}} \right) \cdot {\text{grad}}\psi \left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} \\ = \oint\limits_{{{{\Gamma }}_{1}} \cup {{{\Gamma }}_{2}}} {\psi \left( {\mathbf{r}} \right)~{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} \\ - \,\,\int\limits_{\Omega } {{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right) \cdot {\text{grad}}\psi \left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} , \\ \end{gathered} $$

which, with allowance for conditions (B.2), yields the final identity equivalent to (7),

$$\begin{gathered} \int\limits_{\Omega } {\sigma \left( {\mathbf{r}} \right){\text{grad}}\varphi \left( {\mathbf{r}} \right) \cdot {\text{grad}}\psi \left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} \\ = \int\limits_{\Omega } {{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right) \cdot {\text{grad}}\psi \left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}},} \\ \end{gathered} $$

(B.3)

for all functions $\psi \left( {\mathbf{r}} \right)$ satisfying (B.2). A more rigorous proof of the equivalence of relationship (B.3) to Eq. (7) was presented (e.g., Kalinin and Slyunyaev, 2017).

Let us now use the finite-element method. We search for solution of problem (7)–(8) in the form

$$\varphi \left( {\mathbf{r}} \right) \approx \sum\limits_{i = 1}^N {{{\varphi }_{i}}{{\psi }_{i}}\left( {\mathbf{r}} \right)} = \sum\limits_{i = 1}^M {{{\varphi }_{i}}{{\psi }_{i}}\left( {\mathbf{r}} \right)} + \sum\limits_{i = M + 1}^N {{{\varphi }_{i}}{{\psi }_{i}}\left( {\mathbf{r}} \right)} ,$$

where ${{\psi }_{1}}\left( {\mathbf{r}} \right),{{\psi }_{2}}\left( {\mathbf{r}} \right), \ldots ,{{\psi }_{N}}\left( {\mathbf{r}} \right)$ are linear Lagrangian finite elements in the region Ω (they are enumerated so that only first M of them are set to zero at the boundary) and ${{\varphi }_{1}},{{\varphi }_{2}}, \ldots ,{{\varphi }_{N}}$ are unknown coefficients (degrees of freedom). Then, identity (B.3) can be approximated with the equations ($j = 1,{{\;}}2, \ldots ,M$)

$$\begin{gathered} \int\limits_{\Omega } {\sigma \left( {\mathbf{r}} \right)} \left( {\sum\limits_{i = 1}^N {{{\varphi }_{i}}{\text{grad}}{{\psi }_{i}}\left( {\mathbf{r}} \right)} } \right) \cdot {\text{grad}}{{\psi }_{j}}\left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}} \\ = \int\limits_{\Omega } {{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right) \cdot {\text{grad}}{{\psi }_{j}}\left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} \\ \end{gathered} $$

—in other words, the problem is reduced to a system of linear equations ($j = 1,{{\;}}2, \ldots ,M$)

$$\sum\limits_{i = 1}^N {{{A}_{{ji}}}{{\varphi }_{i}}} = {{b}_{j}},$$

where

$${{A}_{{ji}}} = \sum\limits_{C \in {{T}_{{\Omega }}}} {\int\limits_C {\sigma \left( {\mathbf{r}} \right){\text{grad}}{{\psi }_{i}}\left( {\mathbf{r}} \right) \cdot {\text{grad}}{{\psi }_{j}}\left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} } $$

(B.4)

and

$${{b}_{j}} = \sum\limits_{C \in {{T}_{{\Omega }}}} {\int\limits_C {{{{\mathbf{j}}}^{{{\text{source}}}}}\left( {\mathbf{r}} \right) \cdot {\text{grad}}{{\psi }_{j}}\left( {\mathbf{r}} \right){\text{d}}{\mathbf{r}}} } ;$$

(B.5)

here, ${{T}_{{\Omega }}}$ denotes triangulation (i.e., a set of cells) of the whole Ω region and the variable C enumerates the corresponding cells. Boundary conditions (8) are taken into account by adding $N - M$ equations of the form ${{\varphi }_{i}} = 0$ or ${{\varphi }_{i}} = {{\varphi }_{{{\text{source}}}}}\left( {{{{\mathbf{r}}}_{i}}} \right)$ into the system (depending on whether the degree of freedom corresponds to the component of the boundary Γ₁ or to the component of the boundary Γ₂; ${{{\mathbf{r}}}_{i}}$ are coordinates of the triangulation node corresponding to ${{\varphi }_{i}}$) for $i = M + 1,M + 2, \ldots ,N.$

Integrals over cells in (B.4) and (B.5) are calculated with the five-point Gaussian quadrature rule with respect to each coordinate. The integrals over the outer boundary of the atmosphere that enter into relationship (11) are reduced with the finite-element method to integrals over cells F of the triangulation of this boundary ${{T}_{{{{{\Gamma }}_{2}}}}}{\text{:}}$

$$\begin{gathered} \oint\limits_{{{{\Gamma }}_{2}}} {\sigma \left( {\mathbf{r}} \right){\text{grad}}\varphi \left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} \\ = \oint\limits_{{{{\Gamma }}_{2}}} {\sigma \left( {\mathbf{r}} \right)\left( {\sum\limits_{i = 1}^N {{{\varphi }_{i}}{\text{grad}}{{\psi }_{i}}\left( {\mathbf{r}} \right)} } \right)} \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right) = \sum\limits_{i = 1}^N {{{k}_{i}}{{\varphi }_{i}}} , \\ \end{gathered} $$

where

$${{k}_{i}} = \sum\limits_{F \in {{T}_{{{{{\Gamma }}_{2}}}}}} {\oint\limits_F {\sigma \left( {\mathbf{r}} \right){\text{grad}}{{\psi }_{i}}\left( {\mathbf{r}} \right) \cdot {\mathbf{n}}\left( {\mathbf{r}} \right){\text{ds}}\left( {\mathbf{r}} \right)} } ;$$

the integrals over cells in the last expression are also calculated with the five-point Gaussian quadrature rule with respect to each coordinate.