62. Maxwell equations

Maxwell equations describe electro-magnetic fields. We consider the special case of stationary magnetic fields. Maxwell equations are three-dimensional.

A magnetic field is caused by an electric current. We suppose that a current density

\[ j \in [L_2(\Omega)]^3 \]

is given. (Stationary) currents do not have sources, i.e., \(\operatorname{div} \, j = 0\).

The involved (unknown) fields are

• The magnetic flux \(B\) (in German: Induktion). The flux is free of sources, i.e.,

\[ \operatorname{div} \, B = 0. \]

• The magnetic field intensity \(H\) (in German: magnetische Feldstärke). The field is related to the current density by Henry’s law:

\[ \int_S j \cdot n \, ds = \int_{\partial S} H \cdot \tau \, ds \qquad \forall \; \mbox{Surfaces } S \]

By Stokes’ Theorem, one can derive Henry’s law in differential form:

\[ \operatorname{curl} \, H = j \]

The differential operator is \(\operatorname{curl} = \operatorname {rot} = \nabla \times\). Both fields are related by a material law. The coefficient \(\mu\) is called permeability:

\[ B = \mu H \]

The coefficient \(\mu\) is \(10^3\) to \(10^4\) times larger in iron (and other ferro-magnetic metals) as in most other media (air). In a larger range, the function \(B(H)\) is also highly non-linear.

Collecting the equations we have

\[ \operatorname{div} \, B = 0 \qquad B = \mu H \qquad \operatorname{curl} \, H = j \]

In principle, Maxwell equations are valid in the whole \(\mathbb{R}^3\). For simulation, we have to truncate the domain and have to introduce artificial boundary conditions.

Since \(\operatorname{div} B = 0\), and the domain \({\mathbb R}^3\) is simply connected, we may introduce a vector potential \(A\) such that

\[ B = \operatorname{curl} \, A. \]

Inserting the vector-potential into the system of equations, one obtains the second order equation

(62.1)\[\begin{equation} \operatorname{curl} \, \mu^{-1} \operatorname{curl} A = j. \end{equation}\]

The two original fields \(B\) and \(H\) can be obtained from the vector potential \(A\).

The vector-potential \(A\) is not uniquely defined. One may add a gradient field to \(A\), and the equation is still true. To obtain a unique solution, the so called Coloumb-Gauging can be applied:

(62.2)\[\begin{equation} \operatorname{div} \, A = 0. \end{equation}\]

As usual, we go over to the weak form. Both Equations together become: Find \(A\) such that

\[ \int_\Omega \mu^{-1} \operatorname{curl} A \, \operatorname{curl} v \, dx = \int_\Omega j \cdot v \, dx \qquad \forall \, v \in ? \]

and

\[ \int_\Omega A \cdot \nabla \psi \, dx = 0. \]

We want to choose the same space for \(A\) and the according test functions \(v\). But, then we have more equations than unknowns. The system is still solvable, since we have made the assumption \(\operatorname{div} \, j = 0\), and thus \(j\) is in the range of the \(\operatorname{curl}\)- operator. To obtain a symmetric system, we add a new scalar variable \(\varphi\). The problem is now: Find \(A \in V = ?\) and \(\varphi \in Q = H^1 / \mathbb{R}\) such that

(62.3)\[\begin{equation} \begin{array}{ccccll} \int \mu^{-1} \operatorname{curl} A \, \cdot \operatorname{curl} v \, dx & + & \int \nabla \varphi \cdot v \, dx & = & \int j \cdot v \, dx \qquad & \forall \, v \in V \\[0.5em] \int A \cdot \nabla \psi \, dx & & & = & 0 & \forall \, \psi \in Q \end{array} \end{equation}\]