= Solution
Interpret the longitudinal delta <covariance> as a white-noise limit. It is a generalized random field, not a pointwise unit-variance <function>. Let $B_x(\mathbf r)$ have independent longitudinal increments with
$$
\langle dB_x(\mathbf r)dB_x(\mathbf r')\rangle=A(\mathbf r-\mathbf r')dx.
$$
The smooth-medium limit gives the Stratonovich equation $dE=DE\,dx+iaE\circ dB$. Conversion to Itô form contributes the local <variance> drift:
$$
dE=\left(DE-\frac{a^2A(0)}2E\right)dx+iaE\,dB.
$$
The Itô <stochastic integral> has zero mean, so <coherent attenuation in a white-noise random medium> satisfies
$$
\boxed{m_x=\frac{i}{2k}\Delta_\perp m-\gamma m,\qquad\gamma=\frac{k^2\mu^2A(0)}2,\qquad
m(x,\mathbf r)=e^{-\gamma x}(U_xE_0)(\mathbf r).}
$$
Here $U_x=e^{ix\Delta_\perp/(2k)}$ is the <Fresnel propagator>. Equivalently,
$$
m(x,\mathbf r)=e^{-\gamma x}\frac{k}{2\pi ix}\int_{\mathbb R^2}\exp\left(\frac{ik|\mathbf r-\mathbf r'|^2}{2x}\right)E_0(\mathbf r')d\mathbf r'.
$$
For a constant entrance envelope this is simply its constant value times $e^{-\gamma x}$.
Put $B_n(r)=\mu^2A(r)$, the transverse <covariance> coefficient of the refractive-index fluctuations. The spectrum required here is the transverse spectrum of that longitudinal <covariance> coefficient, or equivalently the longitudinal-zero-frequency slice of a prelimit stationary medium. Radial symmetry refers to the transverse plane; exact <white noise> in one direction is not full three-dimensional isotropy. With the Cartesian Fourier convention,
$$
S_F(\boldsymbol\nu)=\int_{\mathbb R^2}B_n(\mathbf r)e^{-i\boldsymbol\nu\cdot\mathbf r}d\mathbf r,
$$
angular integration gives $S_F(\nu)=2\pi\int_0^\infty B_n(r)J_0(\nu r)r\,dr$. Its inverse gives
$$
B_n(0)=\frac1{2\pi}\int_0^\infty S_F(\nu)\nu\,d\nu,\qquad
\boxed{m(x)=\exp\left[-\frac{k^2x}{4\pi}\int_0^\infty S_F(\nu)\nu\,d\nu\right]U_xE_0.}
$$
The printed transform hint mixes two different normalizations. Its radial transform and inverse constitute the self-reciprocal Hankel pair, whose spectrum is $S_H=S_F/(2\pi)$, rather than the unnormalized Cartesian transform also displayed there. Using that Hankel convention consistently yields instead
$$
\boxed{m(x)=\exp\left[-\frac{k^2x}{2}\int_0^\infty S_H(\nu)\nu\,d\nu\right]U_xE_0.}
$$
These are the same physical solution after converting spectra. The distinction is <Fourier-Hankel normalization for an isotropic spectrum>; $J_0(0)=1$ fixes the zero-separation <covariance> in either convention.
Back to article page
