Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 88 / 3

Write $h=\log \psi$ and $q=2ϵh_{\theta }$, with $ϵ>0$. Direct differentiation givesThus the Cole-Hopf transformation with this negative-flux convention maps the heat equation to the stated viscous Burgers equation. Conversely its residual is independent of $\theta$ and can be removed by multiplying $\psi$ by a function of $Z$.

Convolving each initial exponential with the Gaussian heat kernel givesThe resulting finite-exponential Cole-Hopf solution isIt is a softmax function average of the constant states, so $q_1\le q\le q_N$ and $q_{\theta }={\mathrm{Var}}_w(q_j)/(2ϵ)\ge 0$.

For $N=2$, put $\Delta =q_2-q_1>0$ and $\overline{q}=(q_1+q_2)/2$. Equality of the two exponents occurs atThereforeThis is an increasing viscous shock from $q_1$ to $q_2$, moving at speed $-\overline{q}$ and with transition width of order $ϵ/\Delta$. That speed agrees with the Rankine-Hugoniot condition for flux $-q^2/2$; its concavity makes the increasing jump the entropy shock.

For general $N$, use the shock frame $\theta =-(q_1+q_N)Z/2+\xi$. The leading $Z$ coefficient in $A_j$ is $q_j(q_j-q_1-q_N)/(4ϵ)$. The endpoint coefficients coincide, and every interior coefficient is smaller by $(q_j-q_1)(q_N-q_j)/(4ϵ)>0$. The extremal-state asymptotics of exponential Burgers solutions therefore reduce the large-$Z$ profile to the same tanh shock with states $q_1,q_N$ and centreAt a fixed $\theta$, rather than in the moving frame, the endpoint of largest $|q_j|$ dominates. If $q_N=-q_1$, both endpoint rates coincide and the limiting shock is stationary. Intermediate plateaus disappear in the moving-frame limit.

Use time dependence $e^{i\omega t}$ and consider fields independent of $y$, with $\mathbf{E}=\hat{\mathbf{y}}{E}_y(x,z)$. The source-free Maxwell's equations are $\nabla \times \mathbf{E}=-i\omega \mu \mathbf{H}$ and $\nabla \times \mathbf{H}=i\omega \epsilon \mathbf{E}$. The first givesSubstitute these into the $y$ component of the second equation. The TE wave equation in a stratified magnetodielectric isBecause its coefficients depend only on $z$, ${\partial }_x{E}_y$ obeys the same equation. Since it is proportional to $\mu H_z$, substitution and division by $\mu$ giveFor a conserved transverse wavenumber $k_x$, replace ${\partial }_x^2$ by $-k_x^2$ to obtain the corresponding ordinary Helmholtz equations.

If “propagating along the $z$ axis” is read as strict normal incidence with no $x$ dependence, then $H_z=0$ identically. The nonzero magnetic component is $H_x$, and its equation isThe $H_z$ equation above applies to the oblique transverse electric polarization used in the next part and also includes the zero normal-incidence solution. This distinguishes the named component from the usual transverse magnetic field.

With constant permeability, the electric equation reduces to $({\partial }_x^2+{\partial }_z^2+k^2(z))E_y=0$. Insert the oblique-wave dependence $E_y=e_y(z)e^{-i{k}_0\sin {\theta }_{0}x}$ and set $s=z-z_0$. The inverse-square refractive-index profile givesSubstituting $e_y=\sqrt{|s|}w(\beta |s|)$ gives the Bessel equation of order $\nu$. On either side of the profile singularity, when ${\beta }^2>0$,Equivalently use Hankel functions for an incoming/outgoing wave basis. If ${\beta }^2<0$, write $\gamma =\sqrt{-{\beta }^2}$ and replace $J_{\nu },Y_{\nu }$ by modified Bessel functions $I_{\nu },K_{\nu }$ with argument $\gamma |s|$. For $\beta =0$, the solutions are $|s{|}^{1/2+\nu }$ and $|s{|}^{1/2-\nu }$, with a logarithmic second solution in the repeated-root case $\nu =0$. The constants on each side require incident-wave and boundary or matching data; the supplied singular profile alone does not choose them.