Maxwell’s equations
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
In the case of a non-zero constant current flow Ampere-Maxwell law states, that
∇ × H = J + ∂ D ∂ t . \begin{align}
\nabla \times \boldsymbol{H}
= \boldsymbol{J}
+ \frac{\partial \boldsymbol{D}}{\partial t}.
\end{align} ∇ × H = J + ∂ t ∂ D .
Taking the divergence of both sides we get
∇ ⋅ ( ∇ × H ) = ∇ ⋅ ( J + ∂ D ∂ t ) \begin{align}
\nabla \cdot (\nabla \times \boldsymbol{H})
&= \nabla \cdot (\boldsymbol{J}
+ \frac{\partial \boldsymbol{D}}{\partial t})
\end{align} ∇ ⋅ ( ∇ × H ) = ∇ ⋅ ( J + ∂ t ∂ D )
and
0 = ∇ ⋅ ( J + ∂ D ∂ t ) , \begin{align}
0
&= \nabla \cdot (\boldsymbol{J}
+ \frac{\partial \boldsymbol{D}}{\partial t}),
\end{align} 0 = ∇ ⋅ ( J + ∂ t ∂ D ) ,
since the divergence of a curl is zero.
Under the assumption of constant current flow (or approximating a slowly varying current), the time derivative of the electric displacement field ∂ D ∂ t \frac{\partial \boldsymbol{D}}{\partial t} ∂ t ∂ D equals zero.
∇ ⋅ J = 0 \begin{align}
\nabla \cdot \boldsymbol{J} = 0
\end{align} ∇ ⋅ J = 0
This indicates that the current flow is non-divergent.
We can write a current density using Ohm’s law in differential form as
J = σ E , \begin{align}
\boldsymbol{J} &= \sigma \boldsymbol{E},
\end{align} J = σ E ,
where σ \sigma σ is the conductivity of the material.
Thus we have
∇ ⋅ ( σ E ) = 0 ∇ ⋅ ( − σ ∇ v ) = 0 ∇ ⋅ ( σ ∇ v ) = 0 \begin{align}
\nabla \cdot (\sigma \boldsymbol{E}) &= 0 \\[5pt]
\nabla \cdot (- \sigma \nabla v) &= 0 \\[5pt]
\nabla \cdot (\sigma \nabla v) &= 0 \\[5pt]
\end{align} ∇ ⋅ ( σ E ) ∇ ⋅ ( − σ ∇ v ) ∇ ⋅ ( σ ∇ v ) = 0 = 0 = 0
resulting in Laplace’s equation.
The partial differential equation is multiplied by the test function v ′ v^{\prime} v ′ and integrated over the whole domain Ω \Omega Ω .
∫ Ω ( ∇ ⋅ ( σ ∇ v ) ) v ′ d Ω = 0 \begin{align}
\int_{\Omega}\; (\nabla \cdot (\sigma \nabla v))\; v^{\prime}\; d \Omega = 0
\end{align} ∫ Ω ( ∇ ⋅ ( σ ∇ v )) v ′ d Ω = 0
Applying the Leibniz rule for nabla operator we get
∫ Ω ∇ ⋅ ( v ′ σ ∇ v ) d Ω + ∫ Ω − ( ∇ v ′ ⋅ σ ∇ v ) d Ω = 0 \begin{align}
\int_{\Omega}\; \nabla \cdot (v^{\prime}\; \sigma \nabla v)\; d \Omega
+ \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega
= 0
\end{align} ∫ Ω ∇ ⋅ ( v ′ σ ∇ v ) d Ω + ∫ Ω − ( ∇ v ′ ⋅ σ ∇ v ) d Ω = 0
For the first term we can use the Divergence theorem
∫ Γ ( v ′ σ ∇ v ) ⋅ n d Γ + ∫ Ω − ( ∇ v ′ ⋅ σ ∇ v ) d Ω = 0 \begin{align}
\int_{\Gamma}\; (v^{\prime}\; \sigma \nabla v) \cdot \boldsymbol{n}\; d \Gamma
+ \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega
= 0
\end{align} ∫ Γ ( v ′ σ ∇ v ) ⋅ n d Γ + ∫ Ω − ( ∇ v ′ ⋅ σ ∇ v ) d Ω = 0
Rearranging the terms and using relation E = − ∇ v \boldsymbol{E} = - \nabla v E = − ∇ v we obtain
∫ Ω ( − σ ∇ v ⋅ ∇ v ′ ) d Ω + ∫ Γ − σ E ⋅ n v ′ d Γ = 0 \begin{align}
\int_{\Omega}\; (-\sigma \nabla v \cdot \nabla v^{\prime})\; d \Omega
+ \int_{\Gamma}\; -\sigma \boldsymbol{E} \cdot \boldsymbol{n}\; v^{\prime}\; d \Gamma
= 0
\end{align} ∫ Ω ( − σ ∇ v ⋅ ∇ v ′ ) d Ω + ∫ Γ − σ E ⋅ n v ′ d Γ = 0
Usign relation J = σ E \boldsymbol{J} = \sigma \boldsymbol{E} J = σ E on the Neumann term results in
∫ Ω ( − σ ∇ v ⋅ ∇ v ′ ) d Ω + ∫ Γ ( − J ⋅ n ) v ′ d Γ = 0 \begin{align}
\int_{\Omega}\; (-\sigma \nabla v \cdot \nabla v^{\prime})\; d \Omega
+ \int_{\Gamma}\; (-\boldsymbol{J} \cdot \boldsymbol{n})\; v^{\prime}\; d \Gamma
= 0
\end{align} ∫ Ω ( − σ ∇ v ⋅ ∇ v ′ ) d Ω + ∫ Γ ( − J ⋅ n ) v ′ d Γ = 0
Applies a fixed value to the scalar potential v v v . Use this when you need to fix
scalar potential at a node or within a region. This can be used to drive potential difference for two ends of a conductor.
Applies a prescribed scalar or vector current density to a selected region. Use this interaction to drive the system with a current source instead of a voltage (potential difference).
Lumped voltage or current source applied between terminals.
Lumped voltage or current source applied across an internal cut.
Imposes periodic boundary conditions for the v v v .
Reduces computational domain size for symmetric problems.