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Write $m=⟨E⟩$ and $L_0=i{\partial }_z^2/(2k)$. Substituting $n=1+\mu W$ into part (a), without prematurely replacing $n^2$ by its linear part, givesTaking the expectation yields the exact mean equation within the paraxial approximation:The correlation term cannot be discarded just because $⟨W⟩=0$: the field depends on the same random medium as $W$. Nor can $⟨W^{2}E⟩$ be factored exactly from the given variance alone. The printed statistics do not specify the autocorrelation function of a random field, so they do not determine a unique local closed equation for $m$.

A useful covariance-dependent equation follows to second order in $\mu$. Defineand letbe the Fresnel propagator kernel for one transverse coordinate. The Duhamel principle applied to the linear fluctuation term giveswhere $E^{(0)}=E_{\mathrm{free}}$. Multiply by $W(x,z)$ and average. Also $⟨W(x,z{)}^{2}E(x,z)⟩=E^{(0)}(x,z)+O(\mu )$. Since $m=E^{(0)}+O({\mu }^2)$, replacing $E^{(0)}$ by $m$ in terms already multiplied by ${\mu }^2$ preserves second-order accuracy. We obtain the weak-fluctuation memory equation for the coherent fieldThis perturbation statement is for fixed propagation intervals with sufficient covariance regularity and moment bounds. The kernel is understood as an oscillatory integral. If the field is jointly Gaussian, odd perturbative contributions vanish and the remainder can be improved under the corresponding expansion hypotheses. Joint Gaussianity itself still leaves the covariance unspecified. The term $ik{\mu }^{2}m/2$ is the quadratic refractive-index shift in the coherent field; it comes from the mean of $W^2$.

For the usual local answer, impose the additional Markov approximation for a random medium. Its longitudinal correlation length must be much shorter than the distance on which the envelope varies, and diffraction across that correlation distance must not resolve substantial transverse covariance variation. Away from the entry layer, replace the slowly varying factors in the memory integral by their local values and extend the longitudinal integral to infinity. Define the integrated covarianceFor an integrable real stationary covariance, $C(s,0)$ is even, and $B(0)\ge 0$ is its zero-longitudinal-frequency spectral density. The conventional leading weak-index model keeps only the linear random potential $ik\mu W$; in that model the local first-moment equation isThe factor $1/2$ can be checked independently in the white-noise model. If ${\mathcal{B}}_x(z)$ is a Brownian field with $⟨d{\mathcal{B}}_x(z)d{\mathcal{B}}_x(z^{\prime })⟩=B(z-z^{\prime })dx$, writeConverting the Stratonovich integral to an Itô integral adds $-k^2{\mu }^{2}B(0)Edx/2$. The remaining stochastic integral has zero mean, giving the boxed equation andIf the finite-correlation model retains $n^2$ consistently through order ${\mu }^2$, the local second-order expansion also has $+ik{\mu }^{2}m/2$, adding the phase factor $e^{ik{\mu }^2(x-x_0)/2}$. This phase is computed before the white-noise idealization: squaring ideal Gaussian white noise is not the finite-variance operation $W^2$.

The need for a correlation assumption can be seen without any closure argument. The stationary Gaussian random field $W(x,z)=Z$, with a single standard normal variable $Z$, has all the printed one-point statistics. In the linear weak-index model it givesIts attenuation is quadratic in propagation distance, whereas the Markov model gives linear-distance exponential attenuation. The local attenuation equation requires the additional correlation/Markov assumption; it does not follow from statistical homogeneity and unit variance alone.

Interpret the longitudinal Dirac delta function as the Markov approximation for a random medium. The two arguments on the left of the printed covariance must be $x_1$ and $x_2$. The transverse covariance kernel is $B(|{\mathbf{z}}_1-{\mathbf{z}}_2|)$. Here statistical isotropy means transverse isotropy: a medium with a distinguished longitudinal white noise direction and a smooth transverse covariance is not literally isotropic in all three directions. Also, ideal Gaussian white noise replaces the original finite-variance field; it cannot simultaneously satisfy a pointwise normalization $\mathbb{E}{W}^2=1$.

Write $Q(r)={\mu }^{2}B(r)$ for the covariance strength of the refractive index fluctuations. Define the transverse power spectrum using the unnormalized forward Fourier transform printed in equation 3:This is the spectrum of the fluctuations, excluding the deterministic mean index. Equivalently it is the zero-longitudinal-frequency slice of the three-dimensional fluctuation spectrum before the Markov approximation. The angular integral is $2\pi J_0(\nu r)$, where $J_0$ is the Bessel function of the first kind. Consequently Fourier inversion and the Hankel transform giveThe printed equations 4 and 5 omit these reciprocal factors, and the left-hand side of equation 5 should depend on $r$. They are instead a consistent order-zero Hankel transform pair if their $F$ denotes $S_H=S_F/(2\pi )$.

Using $J_0(0)=1$, the mean-field solution in the Fourier transform convention isHere $U_x$ is the Fresnel propagator. Finiteness of the integral ensures a finite screen wave phase variance. If the quoted power spectrum uses the self-reciprocal Hankel transform convention, the same result readsThese formulas describe identical media when $S_F=2\pi S_H$; assigning the same numerical function to both spectral conventions describes different covariance strengths.

Use the time-harmonic wave convention $\psi e^{-i\omega t}$ and write $\mathbf{z}=(y,z)$, $E(0,\mathbf{z})=E_0(\mathbf{z})$, with a deterministic incident wave envelope. The printed speed ratio is inconsistent with $k=k_{0}n$: the refractive index used below is $n=c_0/c$. Substituting $\psi =e^{i{k}_{0}x}E$ in the Helmholtz equation givesThe paraxial approximation discards ${\partial }_x^{2}E$. The usual weak-fluctuation model also linearizes $n^2-1=2\mu W+O({\mu }^2{W}^2)$, givingDropping the quadratic contrast is an additional weak-fluctuation assumption, not a consequence of small propagation angles. Although the linearized random potential generates attenuation of order ${\mu }^2$, it does not retain every effect of that order in the literal finite-correlation index: the discarded quadratic contrast can also produce a mean wave phase shift. This linearization must precede a Gaussian white noise limit: the square of ideal white noise has no ordinary pointwise meaning.

The split-step Fourier method alternates free-space diffraction, $U_h=e^{hD}$, with a random phase screen,For jointly Gaussian random fields, Gaussian phase averaging gives $\mathbb{E}{e}^{i{\varphi }_h}=e^{-\mathrm{Var}({\varphi }_h)/2}$. More generally, a product of $p$ fields and $q$ conjugate fields picks up $\exp (i{\sum }_a{ϵ}_a{\varphi }_h({\mathbf{z}}_a))$, with signs ${ϵ}_a=+1$ or $-1$. Its screen average is determined entirely by the wave phase covariance matrix. The deterministic diffraction step acts on each coordinate, with opposite signs on factors formed by complex conjugation. This is the basis of the field-moment equations.

There is an important closure qualification. A stationary Gaussian random field need not have independent longitudinal increments. For finite-correlation fluctuations, the unlinearized parabolic wave equation instead givesIn the weak-fluctuation model, the exact first-moment equation isThe last term cannot in general be replaced by a constant times $m$. An exact generic solution of the linearized model iswhere $\mathcal{T}$ denotes time ordering of the propagation operators. For the unlinearized model, add $i{k}_0{\mu }^2{W}^2/2$ inside the propagation generator. The covariance function of the medium is needed to evaluate this expression; One-point Gaussian distributions alone would not even determine the joint wave phase statistics.

For example, omit diffraction and take a longitudinal autocovariance function $R(s)=\mathbb{E}[W(s,\mathbf{z})W(0,\mathbf{z})]$. Direct Gaussian phase averaging givesEven here, unit-variance stationary Gaussian random fields with different $R$ give different answers. In the general problem the diffraction and multiplication operators do not commute, so this scalar attenuation cannot simply be multiplied by $U_x$ without an additional approximation.

The standard closed answer uses the Markov approximation for a random medium, made explicit in part ii. Let the longitudinally integrated covariance kernel beReplace the medium by longitudinal Gaussian white noise with this strength. A screen of thickness $h$ then has $\mathbb{E}[{\varphi }_h({\mathbf{z}}_a){\varphi }_h({\mathbf{z}}_b)]=k_0^2{\mu }^{2}hB({\mathbf{z}}_a-{\mathbf{z}}_b)$ and is independent of the incoming field. For the moment $M=\mathbb{E}{\prod }_a{E}_a^{({ϵ}_a)}$, expanding both steps to order $h$ givesIn particular, the coherent attenuation in a white-noise random medium and its Fresnel propagator solution areEquivalently, write $Wdx=d{\beta }_x$, where ${\beta }_x(\mathbf{z})$ is Brownian motion in $x$ with transverse covariance kernel $\mathbb{E}[d{\beta }_x(\mathbf{z})d{\beta }_x({\mathbf{z}}^{\prime })]=B(\mathbf{z}-{\mathbf{z}}^{\prime })dx$. The Stratonovich integral formulation is $dE=DEdx+i{k}_0\mu E\circ d{\beta }_x$. Its Itô integral form isThe mean of the Itô integral vanishes, independently confirming the attenuation drift. For $x>0$ in two transverse dimensions,Thus a unit plane wave has $m=e^{-\gamma x}$. For a general incident wave envelope, the Fresnel propagator supplies its spreading. The attenuation is redistribution between the coherent and diffuse wave fields, rather than wave absorption: for a field and its conjugate at the same point, the screen contribution in the second-moment equation cancels.