Electrostatics (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

The formulation is derived based on the electrostatics approximation. Our starting points are Maxwell’s equations and conditions for

$$\begin{align} \frac{\partial \boldsymbol{B}}{\partial t} &= 0 \\[10pt] \frac{\partial \boldsymbol{E}}{\partial t} &= 0 \end{align}$$

Under these conditions Maxwell’s equations are reduced to

$$\begin{align} \nabla \cdot \varepsilon \boldsymbol{E} &= q_V \\[5pt] \nabla \times \boldsymbol{E} &= 0\\ \end{align}$$

where (3) is Gauss’s law and (4) is Faraday’s law. From Faraday’s law we can see that curl of $\boldsymbol{E}$ is zero, hence $\boldsymbol{E}$ is a conservative field. That means, that there exists a function $v: \mathbb{R}^3 \rightarrow \mathbb{R}$ such that:

$$\begin{align} \boldsymbol{E} = -\nabla v \end{align}$$

where $v$ is a scalar potential of the vector field $\boldsymbol{E}$. By substituting (6) into Gauss’s law we obtain a Poisson’s equation:

$$\begin{align} \nabla \cdot (-\varepsilon \nabla v) &= q_V \\[5pt] \nabla \cdot (\varepsilon \nabla v) + q_V &= 0 \\[5pt] \end{align}$$

Weak formulation

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

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

Applying the Leibniz rule for nabla operator we get

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

For the first term we can use the Divergence theorem

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

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

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