Maxwell equations PEC
Let \(\Omega \in \mathbb R^3\) denote a perfect electric conductor and \(\gamma_R \boldsymbol E^i = -\boldsymbol m\) the given Dirichlet trace of an incoming time-harmonic electromagnetic signal \(\boldsymbol E^i\). The scattered electromagnetic field with components \(\boldsymbol E\) and \(\boldsymbol H\) solves the following equations in the exterior domain \(\Omega^c\):
\[\begin{split} \left\{ \begin{array}{ccl} \mathbf{curl} \, \boldsymbol H &=& -i\omega\varepsilon \boldsymbol E\,, \\ \mathbf{curl} \, \boldsymbol E &=& i\omega\mu \boldsymbol H\,, \\ \gamma_R\, \boldsymbol E &=& \boldsymbol m \end{array} \right. \end{split}\]
From here we can derive two second order equations: one for the electric field \(\boldsymbol E\) and one for the magnetic field \(\boldsymbol H\).
Electric field boundary integral equations
The electric field \(\boldsymbol E\) solves the second order equation with Dirichlet boundary:
\[\begin{split}\left\{ \begin{array}{rcl l} \mathbf{curl} \, \mathbf{curl}\, \boldsymbol E - \kappa^2 \, \boldsymbol E &=& \boldsymbol 0, \quad &\textnormal{in } \Omega^c \subset \mathbb R^3\,,\\ \gamma_R \,\boldsymbol E &=& \boldsymbol m, \quad & \textnormal{on }\Gamma \\ \left\| \mathbf{curl} \, \boldsymbol E( x) - i\,\omega\,\epsilon \, \boldsymbol E( x)\right\| &=& \mathcal O\left( \displaystyle \frac{1}{\| x\|^2}\right), &\textnormal{for} \; \|x\| \to \infty\,.\end{array} \right. \end{split}\]
From here we can choose an direct or an indirect ansatz. We look first at the direct and then on the indirect ansatz.
Direct ansatz
Representation formula:
\[\begin{split} \begin{array}{rcl} \boldsymbol E(x) &=& \mathrm{SL}\left( \gamma_N \, \boldsymbol E\right)(x) + \mathrm{DL}\left( \gamma_D\,\boldsymbol E\right)(x) \\[1ex] &=&\underbrace{ \kappa\,\displaystyle \int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \boldsymbol j(y)\, \mathrm{d}\sigma_y + \frac{1}{\kappa} \nabla \displaystyle\int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi}\, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \mathrm{div}_\Gamma \boldsymbol j(y)\, \mathrm{d}\sigma_y }_{\displaystyle{ =\mathrm{SL}(\boldsymbol j) } } + \underbrace{ \nabla \times \displaystyle \int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \boldsymbol n(y) \times \boldsymbol{m}(y)\, \mathrm{d}\sigma_y }_{\displaystyle{ =\mathrm{DL}(\boldsymbol n\times \boldsymbol m)} } \,. \end{array}\end{split}\]
Variational formulation and discretisation:
\[ \forall \, \boldsymbol v\in H^{-\frac12}(\mathrm{div}_\Gamma, \Gamma): \quad \left\langle \gamma_R \, \mathrm{SL} (\boldsymbol j), \boldsymbol v \right\rangle_{-\frac12} = \left\langle \boldsymbol m, \boldsymbol v\right\rangle_{-\frac12} - \left\langle \gamma_R\,\mathrm{DL}(\boldsymbol n \times \boldsymbol{m}), \boldsymbol v\right\rangle_{-\frac12} \quad \stackrel{\textnormal{MoM}}{\Longrightarrow} \quad \mathrm{V} \, \mathbf{j} = \left( \frac12 \mathrm{M} - \mathrm{K}\right) \,\mathbf{m} \,, \]
Indirect ansatz
Representation formula:
\[\begin{split} \begin{array}{rcl} \boldsymbol E(x) &=& \mathrm{SL}\left( \gamma_N \, \boldsymbol E^t\right)(x) \\[1ex] &=&\underbrace{ \kappa\,\displaystyle \int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \boldsymbol j^t(y)\, \mathrm{d}\sigma_y + \frac{1}{\kappa} \nabla \displaystyle\int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi}\, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \mathrm{div}_\Gamma \boldsymbol j^t(y)\, \mathrm{d}\sigma_y }_{\displaystyle{ =\mathrm{SL}(\boldsymbol j^t) } } \,. \end{array}\end{split}\]
Variational formulation and discretisation:
\[ \forall \, \boldsymbol v\in H^{-\frac12}(\mathrm{div}_\Gamma, \Gamma): \quad \left\langle \gamma_R \, \mathrm{SL} (\boldsymbol j^t), \boldsymbol v \right\rangle_{-\frac12} = \left\langle \boldsymbol m, \boldsymbol v\right\rangle_{-\frac12} \quad \stackrel{\textnormal{MoM}}{\Longrightarrow} \quad \mathrm{V} \, \mathbf{j^t} = \mathrm{M} \,\mathbf{m} \,, \]
Notes
The bie from an indirect ansatz is often called EFIE (electric field integral equation).
The solution \(\boldsymbol j^t\) of the EFIE is the Neumann trace of the total electric field \(\boldsymbol E^t = \boldsymbol E + \boldsymbol E^i\).
The density \(\boldsymbol j\) from the direct ansatz is not the same as the density \(\boldsymbol j^t\) from the indirect ansatz. It holds
\[\begin{split} \begin{array}{rcl cl l} \boldsymbol j &=& \gamma_N \, \boldsymbol E &=& \dfrac{1}{\kappa} \, \boldsymbol n \times \mathbf{curl}\,\boldsymbol E \quad &\textnormal{Neumann trace of scattered field} \\ \boldsymbol j^t &=& \gamma_N \, \boldsymbol E^t &=& \dfrac{1}{\kappa} \, \boldsymbol n \times \mathbf{curl}\,\boldsymbol E^t \quad & \textnormal{Neumann trace of total field}\,. \end{array} \end{split}\]
\[ \boldsymbol j^t = \boldsymbol j + \boldsymbol j^i\]
Magnetic field boundary integral equations
The magnetic field solves a second order equation with Neumann boundary conditions:
\[\begin{split} \left\{ \begin{array}{rcl l} \mathbf{curl} \, \mathbf{curl}\, \boldsymbol H - \kappa^2 \, \boldsymbol H &=& \boldsymbol 0, \quad &\textnormal{in } \Omega^c \subset \mathbb R^3\,,\\ \gamma_N \,\boldsymbol H &=& -\dfrac{i\omega\varepsilon}{\kappa} \, \boldsymbol n\times \boldsymbol m, \quad & \textnormal{on }\Gamma \\[1ex] \left\| \mathbf{curl} \, \boldsymbol H( x) + i\,\omega\,\mu \, \boldsymbol H( x)\right\| &=& \mathcal O\left( \displaystyle \frac{1}{\| x\|^2}\right), &\textnormal{for} \; \|x\| \to \infty\,.\end{array} \right. \end{split}\]
Again, we can choose an direct or an indirect ansatz. We look first at the direct and then on the indirect ansatz.
Direct ansatz
Representation formula:
\[\begin{split}\begin{array}{rcl} \boldsymbol H(x) &=& \mathrm{SL}\left( \gamma_N \, \boldsymbol H\right)(x) +\mathrm{DL}\left( \gamma_R\,\boldsymbol H\right)(x) \\[1ex] &=& -\dfrac{i\omega\varepsilon}{\kappa} \, \Big(\underbrace{ \kappa\, \displaystyle\int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \boldsymbol n(y)\times \boldsymbol m(y)\, \mathrm{d}\sigma_y + \frac{1}{\kappa} \nabla \displaystyle\int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi}\, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \mathrm{curl}_\Gamma \,\boldsymbol m(y)\, \mathrm{d}\sigma_y }_{\displaystyle{\mathrm{SL}(\boldsymbol n \times \boldsymbol m)} } \Big) + \dfrac{ \kappa }{ i\omega\mu} & \underbrace{ \nabla \times \displaystyle\int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \boldsymbol{j}(y) \, \mathrm{d}\sigma_y }_{\displaystyle{ \mathrm{DL} (\boldsymbol{j}) } }\,.\end{array}\end{split}\]
Apply rotated Dirichlet trace:
\[\begin{split}\begin{array}{c rcl} & \gamma_D \,\boldsymbol H &=& -\dfrac{i\omega\varepsilon}{\kappa} \gamma_D \,\mathrm{SL}(\boldsymbol n\times \boldsymbol m) + \dfrac{\kappa}{i\omega\mu} \gamma_D \,\mathrm{DL}( \boldsymbol j) \\[1ex] \Rightarrow & \dfrac{\kappa}{i\omega\mu}\boldsymbol j &=& -\dfrac{i\omega\varepsilon}{\kappa} \gamma_D \, \mathrm{SL}(\boldsymbol n\times \boldsymbol m) + \dfrac{\kappa}{i\omega\mu} \gamma_D \,\mathrm{DL}( \boldsymbol j) \quad \Rightarrow \quad \boldsymbol j = \gamma_D \,\mathrm{SL}( \boldsymbol n\times \boldsymbol m) + \gamma_D\, \mathrm{DL}(\boldsymbol j) \end{array}\end{split}\]
Variational formulation and discretisation:
\[ \forall \, \boldsymbol v\in H^{-\frac12}(\mathrm{curl}_\Gamma, \Gamma): \quad \left\langle \boldsymbol v, \boldsymbol j\right\rangle_{-\frac12} - \left\langle \boldsymbol v, \gamma_D \,\mathrm{DL} (\boldsymbol j) \right\rangle_{-\frac12} = \left\langle \boldsymbol v, \gamma_D\, \mathrm{SL}(\boldsymbol n \times \boldsymbol{m}) \right\rangle_{-\frac12} \quad \stackrel{\textnormal{MoM}}{\Longrightarrow} \quad \left( \frac12 \mathrm{M}^\intercal + \mathrm{K}^\intercal\right) \,\mathbf{j} = -\mathrm D \, \mathbf m \,, \]
Indirect ansatz
Representation formula:
\[ \boldsymbol H(x) = \mathrm{DL}\left( \gamma_R\,\boldsymbol H^t\right)(x) = \dfrac{ \kappa }{ i\omega\mu} \underbrace{ \nabla \times \displaystyle\int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \boldsymbol{j}^t(y) \, \mathrm{d}\sigma_y }_{\displaystyle{ \mathrm{DL} (\boldsymbol{j}^t) } }\,.\]
Apply rotated Dirichlet trace and use \(\boldsymbol j = \boldsymbol j^t - \boldsymbol j^i\):
\[\begin{split}\begin{array}{ l c rcl} \gamma_D \,\boldsymbol H = \dfrac{\kappa}{i\omega\mu} \gamma_D \,\mathrm{DL}( \boldsymbol j^t) \quad &\Rightarrow & \dfrac{\kappa}{i\omega\mu}\boldsymbol j &=& \dfrac{\kappa}{i\omega\mu} \gamma_D \,\mathrm{DL}( \boldsymbol j^t) \\[2ex] &\Rightarrow &\boldsymbol j^t &=& \gamma_D \,\mathrm{DL}(\boldsymbol j^t) + \boldsymbol j^i \end{array}\end{split}\]
Variational formulation and discretisation:
\[ \forall \, \boldsymbol v\in H^{-\frac12}(\mathrm{curl}_\Gamma, \Gamma): \quad \left\langle \boldsymbol v, \boldsymbol j^t\right\rangle_{-\frac12} - \left\langle \boldsymbol v, \gamma_D \,\mathrm{DL} (\boldsymbol j^t) \right\rangle_{-\frac12} = \left\langle \boldsymbol v, \boldsymbol{j}^i \right\rangle_{-\frac12} \quad \stackrel{\textnormal{MoM}}{\Longrightarrow} \quad \left( \frac12 \mathrm{M}^\intercal + \mathrm{K}^\intercal\right) \,\mathbf{j} = \mathrm M \, \mathbf j^i \,, \]
Notes
Also the magnetic field leads to boundary integral equations \(\boldsymbol j\) and \(\boldsymbol j^t\). The boundary integral equation is are integral equations of second type.
consider \(H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\) conforming finite elements for test and trial space. Here is the hypersingular operator entry \(lk\)
\[ \mathrm{D}_{lk} = \int\limits_\Gamma \int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \langle \boldsymbol n(y)\times \boldsymbol v_l(y), \boldsymbol n(x) \times \boldsymbol \varphi_k(x)\rangle\, \mathrm{d}\sigma_y \, \mathrm{d}\sigma_x - \int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi}\, \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \, \mathrm{curl}_\Gamma \,\boldsymbol v_l(y)\, \mathrm{curl}_\Gamma\,\boldsymbol \varphi_k(x) \mathrm{d}\sigma_y \mathrm{d}\sigma_x \]
consider a trial function \(\boldsymbol \psi_k \in H^{-\frac12}(\mathrm{div}_\Gamma,\Gamma)\) and a test function \(\boldsymbol v_l \in H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\). Here is the adjoint double layer potential operator entry \(lk\)
\[ \mathrm{K}'_{lk} = \int\limits_\Gamma \int\limits_\Gamma \displaystyle{ \frac{1}{4\,\pi} \, \big\langle \boldsymbol n(y)\times \boldsymbol v_l(y), \nabla_{x} \frac{e^{i\,\kappa\,\|x-y\|}}{\| x-y\|} } \times \boldsymbol \psi_k(y) \big\rangle\, \mathrm{d}\sigma_y \, \mathrm{d}\sigma_x \]
Conclusion
Operator Name |
Symbol |
trial space |
test space |
rotation |
|---|
single layer operator |
\(\mathrm V \) |
\(H^{-\frac12}(\mathrm{div}_\Gamma,\Gamma)\) |
\(H^{-\frac12}(\mathrm{div}_\Gamma,\Gamma)\) |
- |
double layer operator |
\(\mathrm K \) |
\(H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\) |
\(H^{-\frac12}(\mathrm{div}_\Gamma,\Gamma)\) |
trial space |
hypersingular operator |
\(\mathrm D\) |
\(H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\) |
\(H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\) |
- |
adjoint double layer operator |
\(\mathrm K' \) |
\(H^{-\frac12}(\mathrm{div}_\Gamma,\Gamma)\) |
\(H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\) |
test space |
mass matrix |
\(\mathrm M \) |
\(H^{-\frac12}(\mathrm{div}_\Gamma,\Gamma)\) |
\(H^{-\frac12}(\mathrm{curl}_\Gamma,\Gamma)\) |
- |
NG-BEM Python Functions
Operator |
Python Function |
|---|
\(\mathrm V \) |
MaxwellSingleLayerPotentialOperator |
\(\mathrm K \) |
MaxwellDoubleLayerPotentialOperator |
\(\mathrm D \) |
MaxwellSingleLayerPotentialOperatorCurl |