Derive EM Field in a 1D PC Hill Equation from Maxwell's Eq's.

[zhillyz]

Homework Statement

Derive from Maxwell's equations these Hill equations for 's' and 'p' mode waves;

s\hspace{3mm} modes: E(r,t) = \Psi_{s}(z)e^{i(\beta x - \omega t)}y \\
\hspace{10mm}Hill\, Equation for\, \Psi_{s}(z)\\
\hspace{17mm} \dfrac{d^{2}\Psi_{s}(z)}{dz^{2}} + q^{2}(z)\Psi_{s}(z) = 0,\hspace{5mm}q(z+d) = q(z)\\

p\hspace{3mm} modes: B(r,t) = \Psi_{p}(z)e^{i(\beta x - \omega t)}y \\
\hspace{10mm}\mbox{General ODE with periodic coefficients for}\, \Psi_{p}(z)\\
\hspace{17mm} \dfrac{d^{2}\Psi_{p}(z)}{dz^{2}} -\dfrac{2}{n(z)}\dfrac{dn(z)}{dz}\dfrac{d\Psi_{p}(z)}{dz}+q^{2}(z)\Psi_{p}= 0

Homework Equations

Maxwell's equations, combining Faraday's and Ampere's Laws and a vector identity.

The Attempt at a Solution

I think I have the start down, I am just not sure on how to proceed from here;

Vector identity;

\nabla \times \nabla F = \nabla(\nabla \cdot F) - \nabla^{2}F

applied to \nabla\cdot E = 0

\nabla \times \nabla \times E = -\nabla^{2}E

Faraday's Law then says that the curl of E equals the negative partial derivative of B with respect to t, and B equals μH so therefore;

\nabla \times E = -\dfrac{\partial B}{\partial t} = -\mu \dfrac{\partial H}{\partial t}
\nabla \times(\nabla \times E) = -\mu \dfrac{\partial}{\partial t}(\nabla \times H)

and then as the curl of H equals the partial derivative of D with respect to t plus the current density which in this case will equal zero and D is equal to εE I can say;

\nabla \times H = (\dfrac{\partial D}{\partial t} + J(equal0)) \hspace{10mm} D = \epsilon E
\nabla \times \nabla \times E = -\mu \epsilon \dfrac{\partial}{\partial t}(\dfrac{\partial E}{\partial t})

\rightarrow \nabla^{2}E = \mu \epsilon \dfrac{\partial^{2}E}{\partial t^{2}}


Then because the question gives an equation for E I have tried to substitute this in and simplify but I can't make much sense of it.

Any help would be appreciated.

 

[zhillyz]

Forgot to say that;

q^{2}(z) = k^{2}n^{2}(z) - \beta^{2},\hspace{5mm} \beta = kn_{in}sin \theta_{in}, \hspace{5mm}k=\dfrac{2\pi}{\lambda}

and the second order differential of E with respect to t is;

\dfrac{\partial^{2} E}{\partial t^{2}} = -\omega^{2}\Psi(z)\exp^{i(\beta x-\omega t)}

or

\dfrac{\partial^{2} E}{\partial t^{2}} = \Psi(z)[-\omega^{2}cos(\beta x - \omega t) + isin(\beta x - \omega t)]

Nobody got any ideas to help me :/?