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 Ampere-Maxwell law \text{Ampere-Maxwell law} Ampere-Maxwell law ∇ × H = J + ∂ D ∂ t \nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t} ∇ × H = J + ∂ t ∂ D Our starting points along with the material relation are
∇ ⋅ B = 0 ∇ × H = J B = μ H \begin{align}
\nabla \cdot \boldsymbol{B} &= 0 \\[5pt]
\nabla \times \boldsymbol{H} &= \boldsymbol{J} \\
\boldsymbol{B} &= \mu \boldsymbol{H}
\end{align} ∇ ⋅ B ∇ × H B = 0 = J = μ H Since the magnetic field is divergence-free, we can use the identity for divergence of a curl and write the magnetic field as a curl of some vector A \boldsymbol{A} A
B = ∇ × A \begin{align}
\boldsymbol{B} = \nabla \times \boldsymbol{A}
\end{align} B = ∇ × A Vector A \boldsymbol{A} A
H = 1 μ B = 1 μ ( ∇ × A ) \begin{align}
\boldsymbol{H}
= \frac{1}{\mu} \boldsymbol{B}
= \frac{1}{\mu} \left( \nabla \times \boldsymbol{A} \right)
\end{align} H = μ 1 B = μ 1 ( ∇ × A ) Substituting (5) in (2), the strong form becomes
∇ × H = 1 μ ( ∇ × A ) = J \begin{align}
\nabla \times \boldsymbol{H}
= \frac{1}{\mu} \left( \nabla \times \boldsymbol{A} \right)
= \boldsymbol{J}
\end{align} ∇ × H = μ 1 ( ∇ × A ) = J To obtain the weak form, we multiply the strong form with the test function A ′ \boldsymbol{A}^\prime A ′ Ω \Omega Ω
∫ Ω ( ∇ × H − J ) ⋅ A ′ d Ω = 0 ∫ Ω ( ∇ × H ) ⋅ A ′ d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 \begin{align}
\int_{\Omega}\ (\nabla \times \boldsymbol{H}
- \boldsymbol{J}) \cdot \boldsymbol{A^\prime}\ d \Omega
= 0 \\[5pt]
\int_{\Omega}\ (\nabla \times \boldsymbol{H}) \cdot \boldsymbol{A^\prime}\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0
\end{align} ∫ Ω ( ∇ × H − J ) ⋅ A ′ d Ω = 0 ∫ Ω ( ∇ × H ) ⋅ A ′ d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 Using the identity for divergence of a cross product on the first term we get
∫ Ω ∇ ⋅ ( H × A ′ ) d Ω + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 \begin{align}
\int_{\Omega}\ \nabla \cdot (\boldsymbol{H} \times \boldsymbol{A^\prime})\ d \Omega
+ \int_{\Omega}\ \boldsymbol{H} \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0
\end{align} ∫ Ω ∇ ⋅ ( H × A ′ ) d Ω + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 Applying the divergence theorem on the first term we obtain
∫ Γ ( H × A ′ ) ⋅ n d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 \begin{align}
\int_{\Gamma}\ (\boldsymbol{H} \times \boldsymbol{A^\prime}) \cdot \boldsymbol{n}\ d \Gamma
+ \int_{\Omega}\ \boldsymbol{H} \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0
\end{align} ∫ Γ ( H × A ′ ) ⋅ n d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 Using the scalar triple product rule on the boundary term we get
∫ Γ ( n × H ) ⋅ A ′ d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 \begin{align}
\int_{\Gamma}\ (\boldsymbol{n} \times \boldsymbol{H}) \cdot \boldsymbol{A^\prime}\ d \Gamma
+ \int_{\Omega}\ \boldsymbol{H} \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
- \int_{\Omega}\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
= 0
\end{align} ∫ Γ ( n × H ) ⋅ A ′ d Γ + ∫ Ω H ⋅ ( ∇ × A ′ ) d Ω − ∫ Ω J ⋅ A ′ d Ω = 0 Subsituting (5) in the above second term and by rearranging we derive we get the final weak form as:
∫ Ω 1 μ ( ∇ × A ) ⋅ ( ∇ × A ′ ) d Ω + ∫ Ω − J ⋅ A ′ d Ω + ∫ Γ ( n × H ) ⋅ A ′ d Γ = 0 \begin{align}
\int_{\Omega}\ \frac{1}{\mu}\ (\nabla \times \boldsymbol{A}) \cdot (\nabla \times \boldsymbol{A^\prime})\ d \Omega
+ \int_{\Omega}\ -\ \boldsymbol{J} \cdot \boldsymbol{A^\prime}\ d \Omega
+ \int_{\Gamma}\ (\boldsymbol{n} \times \boldsymbol{H}) \cdot \boldsymbol{A^\prime}\ d \Gamma
= 0
\end{align} ∫ Ω μ 1 ( ∇ × A ) ⋅ ( ∇ × A ′ ) d Ω + ∫ Ω − J ⋅ A ′ d Ω + ∫ Γ ( n × H ) ⋅ A ′ d Γ = 0 