Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 88 / 4

Write $\psi =e^{ikx}E(x,\mathbf{s})$, $\mathbf{s}=(y,z)$. In the weak-contrast approximation $n^2\simeq 1+2\mu W$, neglecting $E_{xx}$ relative to $k{E}_x$ gives the parabolic wave equationFor the mutual coherence $\Gamma (x;{\mathbf{s}}_1,{\mathbf{s}}_2)=\mathbb{E}[E_1{E}_2^{\ast }]$, differentiating gives the exact moment equation within this wave approximation,This is an evolution equation but not a closed one. Even a jointly Gaussian random field does not make $W$ independent of $E$, which depends on its history. Transverse statistical homogeneity alone therefore does not close the second moment. A covariance or longitudinal Markov approximation is additional information.

Under the white-in-$x$ assumption introduced in the next part, interpret $Wdx=d{\mathcal{B}}_x$, a longitudinal Brownian field with transverse covariance $C$, and use the Stratonovich integral as the smooth-medium limit. Conversion to the Itô integral yieldsThe product rule with quadratic variation gives the closed mutual coherence in a white-noise parabolic medium equation,The two attenuation drifts contribute $-k^2{\mu }^{2}C(0)$ and the cross variation contributes $+k^2{\mu }^{2}C({\mathbf{s}}_1-{\mathbf{s}}_2)$. The coefficient $C$ is the white-noise covariance intensity; it replaces, rather than literally retains, the original finite pointwise variance of $W$.

For a general colored medium and unit plane-wave input $E(0,\mathbf{s})=1$, let $U_W(x,0)$ be the propagator of ${\partial }_x=i{\Delta }_{\perp }/(2k)+ik\mu W(x,\mathbf{s})$. Then the solution isA time-ordered operator exponential defines $U_W$. Its expectation depends on the full longitudinal covariance; the initial Gaussian and transverse-stationarity assumptions do not specify a universal closed expression.

With the stated longitudinal Gaussian white noise and transverse statistical homogeneity, the moment depends only on $\mathbf{\rho }={\mathbf{s}}_1-{\mathbf{s}}_2$. Hence $({\Delta }_1-{\Delta }_2)\Gamma =0$. The closed equation in the previous part and initial value $\Gamma (0)=1$ giveIn particular $\Gamma (x,0)=1$: the ensemble mean intensity remains constant for plane-wave input, even though coherent attenuation in a white-noise random medium reduces the mean complex field.

Use a consistent transverse power spectrum convention,For an isotropic spectrum, angular integration gives the isotropic transverse covariance spectrum, $C(\rho )=(2\pi {)}^{-1}{\int }_0^{\infty }\Phi (\kappa )J_0(\kappa \rho )\kappa d\kappa$. ThereforeThe supplied two-dimensional Fourier transform definition has angular factor $2\pi$, whereas the displayed radial transform pair omits it. The latter is a normalized Hankel transform pair. If its radial spectrum is denoted ${\Phi }_H=\Phi /(2\pi )$, use ${\Phi }_H$ in the integral and omit the prefactor $1/(2\pi )$. These are equivalent conventions; mixing them changes the attenuation rate.

Set $\psi =e^{ikx+\mu {\chi }_1+O({\mu }^2)}$. Substitution into the Helmholtz equation gives the first Rytov approximation equationFor the forward parabolic approximation, omit ${\partial }_x^2{\chi }_1$. With zero perturbation at $x=0$, the Fresnel propagator then givesA jointly Gaussian random field makes ${\chi }_1$ a centered complex Gaussian variable. The complex Gaussian exponential moment therefore gives the first-Rytov coherent meanThe second moment here has no complex conjugation. It cannot be replaced by the positive intensity variance $\mathbb{E}[|{\chi }_1{|}^2]$.

For example, let $\hat{C}(u,v;\mathbf{\kappa })$ be the transverse spectrum of the covariance between depths $u$ and $v$. The Fourier multiplier of the Fresnel propagator is $e^{-i(x-u)|\mathbf{\kappa }{|}^2/(2k)}$, soThis supplies the requested mean in terms of medium statistics. For the longitudinal white-noise model, $\hat{C}=\delta (u-v)\Phi$, which reduces the double longitudinal integral to one.

The exponential is the mean of the first-order logarithmic model. It is not a complete second-order perturbation expansion of the true mean: the omitted logarithmic term ${\mu }^2{\chi }_2$ can itself have a nonzero mean. Strictly at first order, $\mathbb{E}[\psi ]=e^{ikx}+O({\mu }^2)$. As an independent check, the closed white-noise parabolic model gives exactlyfor plane-wave input. A truncated first-Rytov covariance expression with transverse diffraction need not equal this exact mean; omitted logarithmic terms account for the difference.