Current flow (v-formulation)

Maxwell’s equations

$$\nabla \cdot \boldsymbol{D} = \rho$$$$\nabla \cdot \boldsymbol{B} = 0$$$$\nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t}$$$$\nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}$$

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Strong formulation

In case of non-zero constant current flow Ampere-Maxwell law states

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

Taking the divergence of both sides we get

$$\begin{align} \nabla \cdot (\nabla \times \boldsymbol{H}) &= \nabla \cdot (\boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}) \\[10pt] 0 &= \nabla \cdot (\boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}) \\[5pt] \end{align}$$

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

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

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

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

Where $\sigma$ is the conductivity of the material.

$$\begin{align} \nabla \cdot (\sigma \boldsymbol{E}) &= 0 \\[5pt] \nabla \cdot (- \sigma \nabla v) &= 0 \\[5pt] \nabla \cdot (\sigma \nabla v) &= 0 \\[5pt] \end{align}$$

Resulting in Laplace’s equation.

Weak formulation

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

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

Applying the Leibniz rule for nabla operator we get

$$\begin{align} \int_{\Omega}\; \nabla \cdot (v^{\prime}\; \sigma \nabla v)\; d \Omega + \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega = 0 \end{align}$$

For the first term we can use the Divergence theorem

$$\begin{align} \int_{\Gamma}\; (v^{\prime}\; \sigma \nabla v) \cdot \boldsymbol{n}\; d \Gamma + \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega = 0 \end{align}$$

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

$$\begin{align} \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega + \int_{\Gamma}\; -\sigma \boldsymbol{E} \cdot \boldsymbol{n}\; v^{\prime}\; d \Gamma = 0 \end{align}$$

Usign relation $\boldsymbol{J} = \sigma \boldsymbol{E}$ results in

$$\begin{align} \int_{\Omega}\; -(\nabla v^{\prime} \cdot \sigma \nabla v)\; d \Omega + \int_{\Gamma}\; -\boldsymbol{J} \cdot \boldsymbol{n}\; v^{\prime}\; d \Gamma = 0 \end{align}$$