Mangetism 𝜑

Maxwell’s equations

$$\text{Gauss's law}$$$$\nabla \cdot \boldsymbol{D} = \rho$$$$\text{Gauss's law for magnetism}$$$$\nabla \cdot \boldsymbol{B} = 0$$$$\text{Faraday's law}$$$$\nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t}$$$$\text{Ampere-Maxwell law}$$$$\nabla \times \boldsymbol{H} = \boldsymbol{J} + \frac{\partial \boldsymbol{D}}{\partial t}$$

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Strong formulation

Our starting points along with the material relation are

$$\begin{align} \nabla \cdot \boldsymbol{B} &= 0 \\[5pt] \nabla \times \boldsymbol{H} &= \boldsymbol{J} \\ \boldsymbol{B} &= \mu\boldsymbol{H} \end{align}$$

Under the assumption of no current flow ($\boldsymbol{J} = 0$) Gauss’s law for magnetism and Ampere’s law state

$$\begin{align} \nabla \cdot \boldsymbol{B} &= 0 \\[5pt] \nabla \times \boldsymbol{H} &= 0 \end{align}$$

The curl of $\boldsymbol{H}$ is zero, hence $\boldsymbol{H}$ is a conservative field. Meaning there exists a scalar function $\varphi$ such that

$$\begin{align} \boldsymbol{H} = -\nabla \varphi \end{align}$$

Here, $\varphi$ is called the magnetic scalar potential. Substituting (5) and the constitutive relation $\boldsymbol{B} = \mu \boldsymbol{H}$ in Gauss’s law for magnetism we get

$$\begin{align} \nabla \cdot \mu \boldsymbol{H} &= 0 \\[5pt] \nabla \cdot \mu \nabla \varphi &= 0 \end{align}$$

Resulting in Laplace’s equation.

Weak formulation

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

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

Using the Leibniz rule for nabla operator on the divergence term we get

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

Applying the divergence theorem on the divergence term we get

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

Rearranging the terms and using relation $\boldsymbol{B} = \mu \boldsymbol{H} = -\mu \nabla \varphi$ on the Neumann term, we obtain

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