Gauss’s law \text{Gauss's law} Gauss’s law ∇ ⋅ D = ρ \nabla \cdot \boldsymbol{D} = \rho ∇ ⋅ D = ρ Gauss’s law for magnetism \text{Gauss's law for magnetism} Gauss’s law for magnetism ∇ ⋅ B = 0 \nabla \cdot \boldsymbol{B} = 0 ∇ ⋅ B = 0 Faraday’s law \text{Faraday's law} Faraday’s law ∇ × E = − ∂ B ∂ t \nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t} ∇ × E = − ∂ t ∂ B Amp e ˋ re-Maxwell law \text{Ampère-Maxwell law} Amp e ˋ re-Maxwell law ∇ × H = J + ∂ D ∂ t \nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t} ∇ × H = J + ∂ t ∂ D We start from Maxwell’s equations along with the material relations
B = μ H + B r D = ε E + D r J = σ E + J r . \begin{align}
\boldsymbol{B}
&= \boldsymbol{\mu} \boldsymbol{H} + \boldsymbol{B_r} \\[5pt]
\boldsymbol{D}
&= \boldsymbol{\varepsilon} \boldsymbol{E}
+ \boldsymbol{D_r} \\[5pt]
\boldsymbol{J}
&= \boldsymbol{\sigma} \boldsymbol{E} + \boldsymbol{J_r}\ .
\end{align} B D J = μ H + B r = ε E + D r = σ E + J r . Where ε \boldsymbol{\varepsilon} ε μ \boldsymbol{\mu} μ σ \boldsymbol{\sigma} σ D r \boldsymbol{D_r} D r B r \boldsymbol{B_r} B r J r \boldsymbol{J_r} J r ϵ \boldsymbol{\epsilon} ϵ μ \boldsymbol{\mu} μ σ \boldsymbol{\sigma} σ D r \boldsymbol{D_r} D r B r \boldsymbol{B_r} B r J r \boldsymbol{J_r} J r
Using this assumptions we can rewrite Faraday’s law as
∇ × E = − μ ∂ H ∂ t μ − 1 ( ∇ × E ) = − ∂ H ∂ t . \begin{align}
\nabla \times \boldsymbol{E}
&= -\boldsymbol{\mu} \frac{\partial \boldsymbol{H}}{\partial t} \\[10pt]
\boldsymbol{\mu}^{-1} (\nabla \times \boldsymbol{E})
&= -\frac{\partial \boldsymbol{H}}{\partial t}.
\end{align} ∇ × E μ − 1 ( ∇ × E ) = − μ ∂ t ∂ H = − ∂ t ∂ H . We apply the curl operator to both sides and assume the curl and time derivative can be interchanged using the time derivative of a curl identity
∇ × ( μ − 1 ( ∇ × E ) ) = − ∂ ∂ t ( ∇ × H ) . \begin{align}
\nabla \times (\boldsymbol{\mu}^{-1} (\nabla \times \boldsymbol{E}))
&= -\frac{\partial}{\partial t} (\nabla \times \boldsymbol{H}).
\end{align} ∇ × ( μ − 1 ( ∇ × E )) = − ∂ t ∂ ( ∇ × H ) . Substituting in the Ampère-Maxwell, material relations (2) and (3) and by rearranging we obtain
∇ × ( μ − 1 ( ∇ × E ) ) + σ ∂ E ∂ t + ε ∂ 2 E ∂ t 2 = 0. \begin{align}
\nabla \times (\boldsymbol{\mu}^{-1} (\nabla \times \boldsymbol{E}))
+ \boldsymbol{\sigma} \frac{\partial \boldsymbol{E}}{\partial t} + \boldsymbol{\varepsilon} \frac{\partial^2 \boldsymbol{E}}{\partial t^2}
&= 0.
\end{align} ∇ × ( μ − 1 ( ∇ × E )) + σ ∂ t ∂ E + ε ∂ t 2 ∂ 2 E = 0. If we assume isotropic material, all tensor quantities become scalars, the equation results in
∇ × ( 1 μ ( ∇ × E ) ) + σ ∂ E ∂ t + ε ∂ 2 E ∂ t 2 = 0. \begin{align}
\nabla \times (\frac{1}{\mu} (\nabla \times \boldsymbol{E}))
+ \sigma \frac{\partial \boldsymbol{E}}{\partial t} + \varepsilon \frac{\partial^2 \boldsymbol{E}}{\partial t^2}
&= 0.
\end{align} ∇ × ( μ 1 ( ∇ × E )) + σ ∂ t ∂ E + ε ∂ t 2 ∂ 2 E = 0. To obtain the weak formulation of (7) we multiply the equation by a test function E ′ \boldsymbol{E}' E ′ Ω \Omega Ω
∫ Ω ∇ × ( μ − 1 ∇ × E ) ⋅ E ′ d Ω + ∫ Ω σ ∂ E ∂ t ⋅ E ′ d Ω + ∫ Ω ϵ ∂ 2 E ∂ t 2 ⋅ E ′ d Ω = 0. \begin{align}
\int_\Omega\ \nabla \times (\boldsymbol{\mu}^{-1} \, \nabla \times \boldsymbol{E}) \cdot \boldsymbol{E}' \, d\Omega
+ \int_\Omega\ \boldsymbol{\sigma} \frac{\partial \, \boldsymbol{E}}{\partial t} \cdot \boldsymbol{E}' \, d\Omega
+ \int_\Omega\ \boldsymbol{\epsilon} \frac{\partial^2 \, \boldsymbol{E}}{\partial t^2} \cdot \boldsymbol{E}' \, d\Omega
&= 0.
\end{align} ∫ Ω ∇ × ( μ − 1 ∇ × E ) ⋅ E ′ d Ω + ∫ Ω σ ∂ t ∂ E ⋅ E ′ d Ω + ∫ Ω ϵ ∂ t 2 ∂ 2 E ⋅ E ′ d Ω = 0. We can rewrite the first term using the divergence of cross product identity
∫ Ω ∇ ⋅ ( ( μ − 1 ∇ × E ) × E ′ ) d Ω + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) , \begin{align}
\int_{\Omega}\ \nabla \cdot ((\boldsymbol{\mu}^{-1} \nabla \times \boldsymbol{E}) \times \boldsymbol{E}^{\prime})\ d\Omega
+ \int_{\Omega}\ (\boldsymbol{\mu}^{-1} \nabla \times \boldsymbol{E}) \cdot \ (\nabla \times \boldsymbol{E}^{\prime}),
\end{align} ∫ Ω ∇ ⋅ (( μ − 1 ∇ × E ) × E ′ ) d Ω + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) , where the divergence theorem can rewritten using the divergence theorem
∫ Γ ( ( μ − 1 ∇ × E ) × E ′ ) ⋅ n d Γ + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) d Ω . \begin{align}
\int_{\Gamma}\ ((\boldsymbol{\mu}^{-1} \nabla \times \boldsymbol{E}) \times \boldsymbol{E}^{\prime}) \cdot \boldsymbol{n}\ d\Gamma
+ \int_{\Omega}\ (\boldsymbol{\mu}^{-1} \nabla \times \boldsymbol{E}) \cdot \ (\nabla \times \boldsymbol{E}^{\prime})\ d\Omega.
\end{align} ∫ Γ (( μ − 1 ∇ × E ) × E ′ ) ⋅ n d Γ + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) d Ω. Substituting in the Faraday’s law and using the material relation (1) we obtain
− ∫ Γ ( ∂ H ∂ t × E ′ ) ⋅ n d Γ + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) d Ω . \begin{align}
-\int_{\Gamma}\ (\frac{\partial \boldsymbol{H}}{\partial t} \times \boldsymbol{E}^{\prime}) \cdot \boldsymbol{n}\ d\Gamma
+ \int_{\Omega}\ (\boldsymbol{\mu}^{-1} \nabla \times \boldsymbol{E}) \cdot \ (\nabla \times \boldsymbol{E}^{\prime})\ d\Omega.
\end{align} − ∫ Γ ( ∂ t ∂ H × E ′ ) ⋅ n d Γ + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) d Ω. Finally, by using the scalar triple product identity we derive the weak formulation
− ∫ Γ ( n × ∂ H ∂ t ) ⋅ E ′ d Γ + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) d Ω + ∫ Ω σ ∂ E ∂ t ⋅ E ′ d Ω + ∫ Ω ϵ ∂ 2 E ∂ t 2 ⋅ E ′ d Ω = 0. \begin{align}
-\int_{\Gamma}\ (\boldsymbol{n} \times \frac{\partial \boldsymbol{H}}{\partial t}) \cdot \boldsymbol{E}^{\prime}\ d\Gamma
+ \int_{\Omega}\ (\boldsymbol{\mu}^{-1} \nabla \times \boldsymbol{E}) \cdot \ (\nabla \times \boldsymbol{E}^{\prime})\ d\Omega \\[10pt]
+ \int_{\Omega}\ \boldsymbol{\sigma} \frac{\partial \boldsymbol{E}}{\partial t} \cdot \boldsymbol{E}^{\prime}\ d\Omega
+ \int_{\Omega}\ \boldsymbol{\epsilon} \frac{\partial^2 \boldsymbol{E}}{\partial t^2} \cdot \boldsymbol{E}^{\prime}\ d\Omega
&= 0.
\end{align} − ∫ Γ ( n × ∂ t ∂ H ) ⋅ E ′ d Γ + ∫ Ω ( μ − 1 ∇ × E ) ⋅ ( ∇ × E ′ ) d Ω + ∫ Ω σ ∂ t ∂ E ⋅ E ′ d Ω + ∫ Ω ϵ ∂ t 2 ∂ 2 E ⋅ E ′ d Ω = 0. 