An EM wave, amplitude $E_0$, frequency $\omega_0$, is incident upon a material with relative permittivity (dielectric function)
$$\varepsilon \left( z \right) = \left\{ \begin{gathered}{\varepsilon _0},z < 0 \\{\varepsilon _1},z \geqslant 0 \\ \end{gathered} \right.$$
$\varepsilon _0$ and $\varepsilon _1$ are constants, $\varepsilon _0 < \varepsilon _1$, z is the direction pependicular to material surface. If the incident angle is 0 and $\mathbf{E}$ is perpenducular to incident plane, then $E$ satisfies
$$\frac{{{\partial ^2}E\left( {t,z} \right)}}
{{\partial {t^2}}} – \frac{1}
{{{c^2}}}\frac{{{\partial ^2}E\left( {t,z} \right)}}
{{\partial {z^2}}} = 0~~~(1)$$
I start by guessing $E$ has the form
$$E\left( {t,z} \right) = \left\{ \begin{gathered} {E_0}\exp \left( { – i\omega t + ikz} \right) + R\exp \left( { – i\omega t – ikz} \right) , \ z<0\\ T\exp \left( { – i\omega t + ikz} \right), \ z\geqslant0\\ \end{gathered} \right.$$
then determine $R$ and $T$ with the Fresnel equations.
I want to know how to solve equation (1) with a bunch of boundary conditions, without "guessing" the solution form. What are the full form boundary conditions? Or maybe initial conditions too
Best Answer
This looks an incredibly difficult way of solving what is quite an easy problem.
If you use Faraday's law in integral form, constructing a small, rectangular loop that goes into and out of the interface, it is easy to show that the component of the E-field that is perpendicular to the normal surface vector (i.e. the E-field parallel to the interface plane) is the same immediately either side of the boundary.
Similarly, you can use the integral form of Ampere's law to show that the H-field parallel to the interface plane is continuous.
By equating the E-field and H-field of the (incident plus reflected) wave with the transmitted wave, you arrive at the Fresnel equations.
I don't think there is any way, given the problem that you have posed, that you can directly solve equation (1) to give an exact solution. For instance, any linear combination of functions of the form $f(\omega t \pm kz)$ is a solution to the wave equation.
I also don't think your wave equation is right. It is dimensionally incorrect and it should change as a function of the relative permittivity $\epsilon_r$. For non-magnetic, non-conductive media ($\mu_r=1$, $\sigma=0$) shouldn't it be $$\frac{{{\partial ^2}E\left( {t,z} \right)}} {{\partial {t^2}}} - \frac{c^2} {{{\epsilon_r}}}\frac{{{\partial ^2}E\left( {t,z} \right)}} {{\partial {z^2}}} = 0\ ?$$
Or if you prefer to leave $c$ as representing the speed of light in the medium $$\frac{{{\partial ^2}E\left( {t,z} \right)}} {{\partial {t^2}}} - c(z)^2 \frac{{{\partial ^2}E\left( {t,z} \right)}} {{\partial {z^2}}} = 0\ ?$$
A further problem is that your general solution is incorrect. Whilst it turns out to be right to assume that the frequency is the same either side of the interface, it is not correct to assume that $k$ is the same on each side.