With
$$
D=\frac{i}{2k_0}\partial_z^2,
\qquad
S=\frac{ik_0}{2}(n^2-1),
$$
the <parabolic wave equation> is $E_x=(D+S)E$. Freeze $S$ at $x_0$, or preferably at the step midpoint. Over a short distance $\Delta x$, <Lie-Trotter splitting> gives
$$
\boxed{
E(x_0+\Delta x)
\simeq e^{\Delta xD}e^{\Delta xS_0}E(x_0)}.
$$
The reversed ordering has the same first-order accuracy, while symmetric half-steps in $S$ give the more accurate Strang form.

The <commutator> can be displayed explicitly. If $q=n^2-1$, then
$$
[D,S]E
=-\frac14\left(q_{zz}E+2q_zE_z\right).
$$
The splitting assumption requires $\Delta x^2[D,S]E$ to be small relative to $E$. It is favored by a short range step, a transversely smooth refractive index, and a field without unresolved large transverse wavenumbers. Freezing $S$ also requires $\Delta x\,S_x$ to be small. These conditions supplement the one-way and <paraxial approximation> already used in part i.

The phase-screen substep is pointwise:
$$
e^{\Delta xS_0}E(x_0,z)
=\exp\left[
\frac{ik_0\Delta x}{2}
(n^2(x_0,z)-1)
\right]E(x_0,z).
$$
Define
$$
\boxed{
\widetilde E_0(x_0,z)
=e^{\Delta xS_0}E(x_0,z)}.
$$
Then solve the free-diffraction initial-value problem
$$
E_x=DE,
\qquad
E(x_0,z)=\widetilde E_0(x_0,z),
$$
to $x_0+\Delta x$. In transverse <Fourier transform> variables, this substep is simply
$$
\widehat E(x_0+\Delta x,k_z)
=e^{-ik_z^2\Delta x/(2k_0)}
\widehat{\widetilde E}_0(k_z).
$$
This is the <split-step Fourier method>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 335 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-335/1/ii/solution