Consider $\Omega = \Omega_1 \cup \Omega_2$, where $\Omega _1$ and $\Omega_2$ are two different media with conductivity and permeability

\begin{equation} \sigma= \begin{cases} \sigma _1 & \text{in } \Omega _1\\ \sigma _2 & \text{in } \Omega_2 \end{cases}, \ \ \ \mu= \begin{cases} \mu _1 & \text{in } \Omega _1\\ \mu _2 & \text{in } \Omega_2 \end{cases} \end{equation}

respectively. From the Maxwell equations we know the interface conditions

\begin{eqnarray} n \times (E_1 - E_2) = 0 \tag{1} \\ n \cdot (B_1 - B_2) = 0 \tag{2} \\ n \times (H_2 - H_1) = j \tag{3} \end{eqnarray}

for the electric field $E$, the magnetic induction $B$, the magnetic field $H$ and the electric current $j$, where $B$ and $H$ are connected by

\begin{equation} B=\mu H. \end{equation}

Further we have Faraday's law

\begin{equation} \nabla \times E = - \frac{\partial B}{\partial t} \tag{4} \end{equation}

and Ohm's law

\begin{equation} E = B \times u + \frac{1}{\sigma} \nabla \times \left(\frac{B}{\mu} \right), \end{equation}

yielding the induction equation

\begin{equation} \partial _t B + \nabla \times (B\times u) + \nabla \times \left(\frac{1}{\sigma}\nabla \times \left( \frac{1}{\mu}B \right)\right) = 0 \tag{5} \end{equation}

Now I would like to deduce the variational form of this induction equation,

\begin{equation} \int _0^T \int _\Omega - B \cdot \partial _t \phi + (B \times u) \cdot (\nabla \times \phi) - \frac{1}{\sigma} \left( \nabla \times \frac{B}{\mu} \right) \cdot (\nabla \times \phi)\ dx\ dt \tag{6} \end{equation}

for any smooth and compactly supported function $\phi \in C_c^\infty ((0,T)\times \Omega)$. Using (1) and (4) this follows immediately, but I would like to get this directly from (5). My problem is the last term in the integral, for which I obtain some strange boundary terms,

\begin{eqnarray} &&\int _0^T \int _\Omega \nabla \times \left(\frac{1}{\sigma}\nabla \times \left( \frac{1}{\mu}B \right)\right) \cdot \phi \ dx \ dt \\ = &&\int _0^T \int _{\Omega_1} \nabla \times \left(\frac{1}{\sigma _1}\nabla \times \left( \frac{1}{\mu _1}B_1 \right)\right) \cdot \phi \ dx \ dt + \int _0^T \int _{\Omega_2} \nabla \times \left(\frac{1}{\sigma _2}\nabla \times \left( \frac{1}{\mu _2}B_2 \right)\right) \cdot \phi \ dx \ dt \\ = && \int _0^T \int _{\Omega_1} \left(\frac{1}{\sigma _1}\nabla \times \left( \frac{1}{\mu _1}B_1 \right)\right) \cdot (\nabla \times \phi) \ dx \ dt \\ + && \int _0^T\int_{\partial \Omega _1} \left( \frac{1}{\sigma _1} \left( \nabla \times \left( \frac{1}{\mu _1} B_1 \right) \right) \times \phi \right) \cdot n \ ds \ dt \\ + && \int _0^T \int _{\Omega_2} \left(\frac{1}{\sigma _2}\nabla \times \left( \frac{1}{\mu _2}B_2 \right)\right) \cdot (\nabla \times \phi) \ dx \ dt \\ - && \int _0^T\int_{\partial \Omega _2} \left( \frac{1}{\sigma _2} \left( \nabla \times \left( \frac{1}{\mu _2} B_2 \right) \right) \times \phi \right) \cdot n \ ds \ dt. \end{eqnarray}

Now, if the boundary integrals cancel each other we will again end up with (6). I hope to get this from (3), but I can't see how. Or are the interface conditions wrong?


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