Electric field integral equation

Example

Let $S$ be the smooth closed surface of a perfectly conducting body in $\R^3$, with outward unit normal $n$. Let a time-harmonic incident field $E^{\mathrm{inc}}$ with time factor $e^{-i\omega t}$ illuminate the body in a medium with permittivity $\varepsilon$ and permeability $\mu$, and let $k = \omega \sqrt{\varepsilon \mu}$. Let $G(x, y) = \frac{e^{ik\abs{x - y}}}{4\pi \abs{x - y}}$. The unknown is the tangential surface current density $J$ on $S$. The electric field integral equation is $$n(x) \times E^{\mathrm{inc}}(x) = -i\omega\mu \, n(x) \times \left[\int_S G(x, y) J(y) \, dS(y) + \frac{1}{k^2} \nabla_x \int_S G(x, y) \, (\nabla_S \cdot J)(y) \, dS(y)\right], \quad x \in S.$$ The unknown $J$ appears only under the integrals, so it is an integral equation of the first kind.

Check: the current $J$ radiates the scattered field $E^{\mathrm{s}} = i\omega A - \nabla \Phi$. Here $A = \mu \int_S G J \, dS$ is the vector potential and $\Phi = \frac{1}{\varepsilon} \int_S G \sigma \, dS$ the scalar potential. The charge density is $\sigma = \frac{1}{i\omega} \nabla_S \cdot J$ by the continuity equation. Since $\frac{1}{i\omega\varepsilon} = \frac{-i\omega\mu}{k^2}$, this gives $E^{\mathrm{s}} = i\omega\mu \left[\int_S G J \, dS + \frac{1}{k^2} \nabla \int_S G \, \nabla_S \cdot J \, dS\right]$. On a perfect conductor the tangential total field vanishes, $n \times (E^{\mathrm{inc}} + E^{\mathrm{s}}) = 0$ on $S$, which is the equation above.