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Finite Element Embedded Library and Language in C++
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2D Maxwell Dimulation in a Diode
Author
Thomas Strub
Philippe Helluy
Christophe Prud'homme
Date
2011-06-01




Description

The Maxwell equations read:

\begin{eqnarray*} \frac{-1}{c^{2}}\frac{\partial \ensuremath{{\bm E}}\xspace}{\partial t}+\nabla\times \ensuremath{{\bm B}}\xspace & = & \mu_{0} \ensuremath{{\bm J}}\xspace\\ \ensuremath{{\bm B}}\xspace_{t}+\nabla\times \ensuremath{{\bm E}}\xspace & = & 0\\ \nabla \cdot \ensuremath{{\bm B}}\xspace & = & 0\\ \nabla \cdot \ensuremath{{\bm E}}\xspace & = & \frac{\rho}{\epsilon_{o}} \end{eqnarray*}

where $\ensuremath{{\bm E}}\xspace$ is the electric field, $\ensuremath{{\bm B}}\xspace$ the magnetic field, $\ensuremath{{\bm J}}\xspace$ the current density, $ c $ the speed of light, $ \ rho $ density of electric charge, $ \ mu_ {0} $ the vacuum permeability and $ \ epsilon_ {0} $ the vacuum permittivity.
In the midst industrial notament in aeronautics, systems Products must verify certain standards such as the receipt an electromagnetic wave emitted by a radar does not cause the inefficassité of part or all of the hardware in the system.
Thus, the simulation of such situations can develop when or during the certification of a new product to test its reaction to such attacks.
Also note that the last two equations are actually initial conditions, since if we assume they are true at the moment $ t = 0$ then it can be deduced from the first two.
At $t=0s$, we suppose that

\begin{eqnarray} \nabla \cdot \ensuremath{{\bm B}}\xspace & = & 0\\ \nabla \cdot \ensuremath{{\bm E}}\xspace & = & \frac{\rho}{\epsilon_{o}} \end{eqnarray}

Suppose that $\ensuremath{{\bm B}}\xspace = (B_x, B_y, B_z )^T$ and $\ensuremath{{\bm E}}\xspace=(E_x,E_y,E_z)^T$ i.e.

\begin{eqnarray} \frac{\partial B_{x}}{\partial x}(t=0)+\frac{\partial B_{y}}{\partial y}(t=0)+\frac{\partial B_{z}}{\partial z} & (t=0)= & 0\\ \frac{\partial E_{x}}{\partial x}(t=0)+\frac{\partial E_{y}}{\partial y}(t=0)+\frac{\partial E_{z}}{\partial z} & (t=0)= & \frac{\rho}{\epsilon_{o}} \end{eqnarray}

Differentiating the first of these two equations with respect to time, we get:

\begin{multline} \label{eq:6} \frac{\partial}{\partial t}\frac{\partial}{\partial x}B_{x}+\frac{\partial}{\partial t}\frac{\partial}{\partial y}B_{y}+\frac{\partial}{\partial t}\frac{\partial}{\partial z}B_{z} = \\ \frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}E_{z}-\frac{\partial}{\partial z}E_{y}\right)+\frac{\partial}{\partial y}\left(\frac{\partial}{\partial z}E_{x}-\frac{\partial}{\partial x}E_{z}\right)+\frac{\partial}{\partial z}\left(\frac{\partial}{\partial x}E_{y}-\frac{\partial}{\partial y}E_{x}\right)\\ = 0 \end{multline}

thanks to

\begin{equation} \label{eq:3} \ensuremath{{\bm B}}\xspace_{t}+\nabla\times \ensuremath{{\bm E}}\xspace =0 \end{equation}

So, for all $t\geq0$,

\begin{equation} \label{eq:4} \nabla \cdot \ensuremath{{\bm B}}\xspace(t)=\nabla\cdot\B(0)=0 \end{equation}

We deduce the same way the second equation, using the charge conservation equation :

\begin{equation} \label{eq:2} \frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \ensuremath{{\bm J}}\xspace) = 0 \end{equation}

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