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Current flow (v-formulation)

In the case of a non-zero constant current flow Ampere-Maxwell law states, that

×H=J+Dt.\begin{align} \nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}. \end{align}

Taking the divergence of both sides we get

(×H)=(J+Dt)\begin{align} \nabla \cdot (\nabla \times \boldsymbol{H}) &= \nabla \cdot (\boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}) \end{align}

and

0=(J+Dt),\begin{align} 0 &= \nabla \cdot (\boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}), \end{align}

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 Dt\frac{\partial \boldsymbol{D}}{\partial t} equals zero.

J=0\begin{align} \nabla \cdot \boldsymbol{J} = 0 \end{align}

This indicates that the current flow is non-divergent. We can write a current density using Ohm’s law in differential form as

J=σE,\begin{align} \boldsymbol{J} &= \sigma \boldsymbol{E}, \end{align}

where σ\sigma is the conductivity of the material.

Thus we have

(σE)=0(σv)=0(σv)=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}

resulting in Laplace’s equation.

The partial differential equation is multiplied by the test function vv^{\prime} and integrated over the whole domain Ω\Omega.

Ω  ((σv))  v  dΩ=0\begin{align} \int_{\Omega}\; (\nabla \cdot (\sigma \nabla v))\; v^{\prime}\; d \Omega = 0 \end{align}

Applying the Leibniz rule for nabla operator we get

Ω  (v  σv)  dΩ+Ω  (vσv)  dΩ=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}

For the first term we can use the Divergence theorem

Γ  (v  σv)n  dΓ+Ω  (vσv)  dΩ=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}

Rearranging the terms and using relation E=v\boldsymbol{E} = - \nabla v we obtain

Ω  (σvv)  dΩ+Γ  σEn  v  dΓ=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}

Usign relation J=σE\boldsymbol{J} = \sigma \boldsymbol{E} on the Neumann term results in

Ω  (σvv)  dΩ+Γ  (Jn)  v  dΓ=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}

Applies a fixed value to the scalar potential vv. 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 vv. Reduces computational domain size for symmetric problems.