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 The Electrostatics v-formulation is derived based on the electrostatic approximation.
As a starting point, we have the conditions
∂ B ∂ t = 0 ∂ E ∂ t = 0. \begin{align}
\frac{\partial \boldsymbol{B}}{\partial t} &= 0 \\[10pt]
\frac{\partial \boldsymbol{E}}{\partial t} &= 0.
\end{align} ∂ t ∂ B ∂ t ∂ E = 0 = 0. Now with these conditions and the Maxwell’s equations, we have
∇ ⋅ ε E = ρ ∇ × E = 0 , \begin{align}
\nabla \cdot \varepsilon \boldsymbol{E} &= \rho \\[5pt]
\nabla \times \boldsymbol{E} &= 0,
\end{align} ∇ ⋅ ε E ∇ × E = ρ = 0 , where (3) is Gauss’s law and (4) is Faraday’s law.
From Faraday’s law we see that curl of E \boldsymbol{E} E E \boldsymbol{E} E v v v
E = − ∇ v , \begin{align}
\boldsymbol{E} = -\nabla v,
\end{align} E = − ∇ v , where v v v E \boldsymbol{E} E
∇ ⋅ ( − ε ∇ v ) = ρ ∇ ⋅ ( ε ∇ v ) + ρ = 0. \begin{align}
\nabla \cdot (-\varepsilon \nabla v) &= \rho \\[5pt]
\nabla \cdot (\varepsilon \nabla v) + \rho &= 0.
\end{align} ∇ ⋅ ( − ε ∇ v ) ∇ ⋅ ( ε ∇ v ) + ρ = ρ = 0. The partial differential equation is multiplied by the test function v ′ v^{\prime} v ′ Ω \Omega Ω
∫ Ω ( ∇ ⋅ ( ε ∇ v ) ) v ′ d Ω + ∫ Ω ρ v ′ d Ω = 0. \begin{align}
\int_{\Omega}\; (\nabla \cdot (\varepsilon \nabla v))\; v^{\prime}\; d \Omega
+ \int_{\Omega}\; \rho\; v^{\prime}\; d \Omega
= 0.
\end{align} ∫ Ω ( ∇ ⋅ ( ε ∇ v )) v ′ d Ω + ∫ Ω ρ v ′ d Ω = 0. Applying the Leibniz rule for nabla operator we get
∫ Ω ∇ ⋅ ( v ′ ε ∇ v ) d Ω − ∫ Ω ( ∇ v ′ ⋅ ε ∇ v ) d Ω + ∫ Ω ρ v ′ d Ω = 0. \begin{align}
\int_{\Omega}\; \nabla \cdot (v^{\prime}\; \varepsilon \nabla v)\; d \Omega
- \int_{\Omega}\; (\nabla v^{\prime} \cdot \varepsilon \nabla v)\; d \Omega
+ \int_{\Omega}\; \rho\; v^{\prime}\; d \Omega
= 0.
\end{align} ∫ Ω ∇ ⋅ ( v ′ ε ∇ v ) d Ω − ∫ Ω ( ∇ v ′ ⋅ ε ∇ v ) d Ω + ∫ Ω ρ v ′ d Ω = 0. For the first term we can use the Divergence theorem
∫ Γ n ⋅ ( v ′ ε ∇ v ) d Ω − ∫ Ω ( ∇ v ′ ⋅ ε ∇ v ) d Ω + ∫ Ω ρ v ′ d Ω = 0. \begin{align}
\int_{\Gamma}\; \boldsymbol{n} \cdot (v^{\prime}\; \varepsilon \nabla v)\; d \Omega
- \int_{\Omega}\; (\nabla v^{\prime} \cdot \varepsilon \nabla v)\; d \Omega
+ \int_{\Omega}\; \rho\; v^{\prime}\; d \Omega
= 0.
\end{align} ∫ Γ n ⋅ ( v ′ ε ∇ v ) d Ω − ∫ Ω ( ∇ v ′ ⋅ ε ∇ v ) d Ω + ∫ Ω ρ v ′ d Ω = 0. Rearranging the terms and using relation E = − ∇ v \boldsymbol{E} = - \nabla v E = − ∇ v
∫ Ω ( − ε ∇ v ⋅ ∇ v ′ ) d Ω + ∫ Γ ( − ε E ⋅ n ) v ′ d Γ + ∫ Ω ρ v ′ d Ω . \begin{align}
\int_{\Omega}\; (-\varepsilon \nabla v \cdot \nabla v^{\prime})\; d \Omega
+ \int_{\Gamma}\; (-\varepsilon \boldsymbol{E} \cdot \boldsymbol{n})\; v^{\prime} \; d \Gamma
+ \int_{\Omega}\; \rho\; v^{\prime}\; d \Omega.
\end{align} ∫ Ω ( − ε ∇ v ⋅ ∇ v ′ ) d Ω + ∫ Γ ( − ε E ⋅ n ) v ′ d Γ + ∫ Ω ρ v ′ d Ω. 