Green function for a randomly inhomogeneous medium

Mainly, the interest in the issue of wave propagation in randomly inhomogeneous media (which, for example, is the atmosphere ) can be explained by the rapid development of satellite technologies . In this case, it becomes important to calculate the characteristics (for example, amplitude ) of the wave transmitted through the medium and establish their relations with the parameter of the inhomogeneity of the medium. An important role here is played by the Green function for a randomly inhomogeneous medium , knowing which these characteristics can be determined. The passage of light through a medium with fluctuating dielectric constant is considered.

Content

1 Wave equation in a randomly inhomogeneous medium. Green function
2 Green function for a medium without fluctuations in dielectric constant
3 Green's function taking into account fluctuations of the dielectric constant
4 Literature

The wave equation in a randomly inhomogeneous medium. Green Function

The scattering of electromagnetic waves in such a medium is determined by the system of Maxwell equations . The main distinguishing features of scattering can be considered for a simplified model: a scalar field $u=u(\mathbf {r} ,\;t)$ ${\ displaystyle u = u (\ mathbf {r}, \; t)}$ . This scalar field replaces the electric and magnetic fields, then $u=u(\mathbf {r} ,\;t)$ ${\ displaystyle u = u (\ mathbf {r}, \; t)}$ satisfies the wave equation :

{\frac {\varepsilon (\mathbf {r} ,\;t)}{c^{2}}}{\frac {\partial ^{2}u(\mathbf {r} ,\;t)}{\partial t^{2}}}-\Delta u(\mathbf {r} ,\;t)=0,

{\ displaystyle {\ frac {\ varepsilon (\ mathbf {r}, \; t)} {c ^ {2}}} {\ frac {\ partial ^ {2} u (\ mathbf {r}, \; t )} {\ partial t ^ {2}}} - \ Delta u (\ mathbf {r}, \; t) = 0,}

\varepsilon (\mathbf {r} ,\;t)=\varepsilon _{0}+\delta \varepsilon (\mathbf {r} ,\;t),

{\ displaystyle \ varepsilon (\ mathbf {r}, \; t) = \ varepsilon _ {0} + \ delta \ varepsilon (\ mathbf {r}, \; t),}

Where $c$ ${\ displaystyle c}$ Is the speed of light in vacuum, $\varepsilon _{0}$ ${\ displaystyle \ varepsilon _ {0}}$ - the average value of the dielectric constant , $\delta \varepsilon (\mathbf {r} ,\;t)$ ${\ displaystyle \ delta \ varepsilon (\ mathbf {r}, \; t)}$ - fluctuations of the dielectric constant. Note that the average dielectric constant $\varepsilon _{0}$ ${\ displaystyle \ varepsilon _ {0}}$ It is assumed that it is independent of coordinates and time , that is, on average, the system is homogeneous and isotropic , and with good accuracy, as a first approximation, we can also assume that the unveraged dielectric constant $\varepsilon (\mathbf {r} ,\;t)=\varepsilon (\mathbf {r} )$ ${\ displaystyle \ varepsilon (\ mathbf {r}, \; t) = \ varepsilon (\ mathbf {r})}$ independent of time. This is explained by the fact that the characteristic times responsible for the molecular processes in the system are several orders of magnitude longer than the characteristic times of the electromagnetic field ; therefore, the medium “does not have time to react” to the change in the field.

A wave equation with such a dielectric constant is actually an example of a stochastic equation , since it contains a random parameter ${\delta \varepsilon (\mathbf {r} )}$ ${\ displaystyle {\ delta \ varepsilon (\ mathbf {r})}}$ . This parameter enters the equation by multiplication, that is, multiplicatively, and not by addition (additive), as in the well-known equation for Brownian motion .

Describing scattering, the field characteristics are interesting. $u=u(\mathbf {r} ,\;t)$ ${\ displaystyle u = u (\ mathbf {r}, \; t)}$ averaged over fluctuations of the dielectric constant . These characteristics are: average field value $\langle u(\mathbf {r} ,\;t)\rangle$ ${\ displaystyle \ langle u (\ mathbf {r}, \; t) \ rangle}$ and intensity $I=I(\mathbf {r} ,\;t)$ ${\ displaystyle I = I (\ mathbf {r}, \; t)}$ determined by the average square of the field (averaging is also carried out over fluctuations of the dielectric constant ) $\langle {u(\mathbf {r} ,\;t)}^{2}\rangle$ ${\ displaystyle \ langle {u (\ mathbf {r}, \; t)} ^ {2} \ rangle}$ . We consider the fluctuation statistics to be given, and also take into account that the average deviation from the average permittivity is zero:

\langle \delta \varepsilon (\mathbf {r} )\rangle =0.

{\ displaystyle \ langle \ delta \ varepsilon (\ mathbf {r}) \ rangle = 0.}

The initial homogeneous wave equation always has a solution in the form $u(\mathbf {r} ,\;t)=0$ ${\ displaystyle u (\ mathbf {r}, \; t) = 0}$ . This is an obvious trivial decision. It is easy to show that in the absence of fluctuations, the nonzero solution is a plane monochromatic wave of the form:

u(\mathbf {r} ,\;t)=u_{0}\exp[i\mathbf {k_{0}} \mathbf {r} -iwt].

{\ displaystyle u (\ mathbf {r}, \; t) = u_ {0} \ exp [i \ mathbf {k_ {0}} \ mathbf {r} -iwt].}

Substitute this expression in the wave equation . We get:

-{\frac {w^{2}\varepsilon _{0}}{c^{2}}}u(\mathbf {r} ,\;t)+k_{0}^{2}u(\mathbf {r} ,\;t)=0.

{\ displaystyle - {\ frac {w ^ {2} \ varepsilon _ {0}} {c ^ {2}}} u (\ mathbf {r}, \; t) + k_ {0} ^ {2} u (\ mathbf {r}, \; t) = 0.}

It is clear from this that the proposed solution will satisfy the equation if the frequency of a plane wave $w$ ${\ displaystyle w}$ and wave vector $k_{0}$ ${\ displaystyle k_ {0}}$ are connected by the dispersion relation :

k_{0}^{2}={\frac {w^{2}\varepsilon _{0}}{c^{2}}}.

{\ displaystyle k_ {0} ^ {2} = {\ frac {w ^ {2} \ varepsilon _ {0}} {c ^ {2}}}.}

It is clear that any linear combination of waves corresponding to the dispersion relation is also a solution to the wave equation in the absence of fluctuations of the dielectric constant.

Define the Green's function $G(\mathbf {r} ,\;t)$ ${\ displaystyle G (\ mathbf {r}, \; t)}$ . Let this function be a solution to the initial wave equation , to the right of which is added a monochromatic source located at the origin (source frequency $w$ ${\ displaystyle w}$ ) We believe that the source “adiabatically turned on in the infinitely distant past”, for this we supplement the right side with a factor $\exp[\alpha t]$ ${\ displaystyle \ exp [\ alpha t]}$ where $\alpha$ ${\ displaystyle \ alpha}$ - small positive value. In the final expressions, we will tend it to zero. Total Green's function satisfies the equation:

{\frac {\varepsilon (\mathbf {r} )}{c^{2}}}{\frac {\partial ^{2}G(\mathbf {r} ,\;t)}{\partial t^{2}}}-\Delta G(\mathbf {r} ,\;t)=e^{-iwt+\alpha t}\delta (\mathbf {r} ).

{\ displaystyle {\ frac {\ varepsilon (\ mathbf {r})} {c ^ {2}}} {\ frac {\ partial ^ {2} G (\ mathbf {r}, \; t)} {\ partial t ^ {2}}} - \ Delta G (\ mathbf {r}, \; t) = e ^ {- iwt + \ alpha t} \ delta (\ mathbf {r}).}

It is convenient to look for a solution to this equation in the form $G(\mathbf {r} ,\;t)=e^{-iwt+\alpha t}G(\mathbf {r} )$ ${\ displaystyle G (\ mathbf {r}, \; t) = e ^ {- iwt + \ alpha t} G (\ mathbf {r})}$ . Substituting this expression into the equation for the Green's function, we obtain:

{\frac {\varepsilon (\mathbf {r} )G(\mathbf {r} )}{c^{2}}}{\frac {\partial ^{2}e^{-iwt+\alpha t}}{\partial t^{2}}}-e^{-iwt+\alpha t}\Delta G(\mathbf {r} )=e^{-iwt+\alpha t}\delta (\mathbf {r} ).

{\ displaystyle {\ frac {\ varepsilon (\ mathbf {r}) G (\ mathbf {r})} {c ^ {2}}} {\ frac {\ partial ^ {2} e ^ {- iwt + \ alpha t}} {\ partial t ^ {2}}} - e ^ {- iwt + \ alpha t} \ Delta G (\ mathbf {r}) = e ^ {- iwt + \ alpha t} \ delta (\ mathbf {r }).}

From the double differentiation of the exponent in time, a multiplier will appear $-(w+i\alpha )^{2}$ ${\ displaystyle - (w + i \ alpha) ^ {2}}$ , then we obtain the equation for the function $G(\mathbf {r} )$ ${\ displaystyle G (\ mathbf {r})}$ :

{\frac {-\varepsilon (\mathbf {r} )(w+i\alpha )^{2}}{c^{2}}}G(\mathbf {r} )-\Delta G(\mathbf {r} )=\delta (\mathbf {r} ).

{\ displaystyle {\ frac {- \ varepsilon (\ mathbf {r}) (w + i \ alpha) ^ {2}} {c ^ {2}}} G (\ mathbf {r}) - \ Delta G ( \ mathbf {r}) = \ delta (\ mathbf {r}).}

It is necessary to solve this equation for some permittivity $\varepsilon (\mathbf {r} ),$ ${\ displaystyle \ varepsilon (\ mathbf {r}),}$ and then average this decision over all sorts of deviations $\delta \varepsilon (\mathbf {r} )$ ${\ displaystyle \ delta \ varepsilon (\ mathbf {r})}$ . But it turns out that there is no way to obtain a solution to this equation for an arbitrary dielectric constant, so the solution is sought using perturbation theory , assuming the deviation $\delta \varepsilon (\mathbf {r} )$ ${\ displaystyle \ delta \ varepsilon (\ mathbf {r})}$ small size.

Green function for a medium without fluctuations in dielectric constant

First you need to find the Green's function $G(\mathbf {r} ,\;t)$ ${\ displaystyle G (\ mathbf {r}, \; t)}$ corresponding to the wave equation without dielectric permittivity deviations, i.e. $\varepsilon (\mathbf {r} )=\varepsilon _{0}$ ${\ displaystyle \ varepsilon (\ mathbf {r}) = \ varepsilon _ {0}}$ :

{\frac {\varepsilon _{0}}{c^{2}}}{\frac {\partial ^{2}G_{0}(\mathbf {r} ,\;t)}{\partial t^{2}}}-\Delta G_{0}(\mathbf {r} ,\;t)=e^{-iwt+\alpha t}\delta (\mathbf {r} ).

{\ displaystyle {\ frac {\ varepsilon _ {0}} {c ^ {2}}} {\ frac {\ partial ^ {2} G_ {0} (\ mathbf {r}, \; t)} {\ partial t ^ {2}}} - \ Delta G_ {0} (\ mathbf {r}, \; t) = e ^ {- iwt + \ alpha t} \ delta (\ mathbf {r}).}

(one)

Again we are looking for a solution in the form $G_{0}(\mathbf {r} ,\;t)=e^{-iwt+\alpha t}G_{0}(\mathbf {r} )$ ${\ displaystyle G_ {0} (\ mathbf {r}, \; t) = e ^ {- iwt + \ alpha t} G_ {0} (\ mathbf {r})}$ . Then $G_{0}(\mathbf {r} )$ ${\ displaystyle G_ {0} (\ mathbf {r})}$ satisfies the equation :

-k_{0}^{2}G_{0}(\mathbf {r} )-\Delta G_{0}(\mathbf {r} )=\delta (\mathbf {r} ),

{\ displaystyle -k_ {0} ^ {2} G_ {0} (\ mathbf {r}) - \ Delta G_ {0} (\ mathbf {r}) = \ delta (\ mathbf {r}),}

(2)

where the magnitude $k_{0}={\frac {{\sqrt {\varepsilon _{0}}}(w+i\alpha )}{c}}$ ${\ displaystyle k_ {0} = {\ frac {{\ sqrt {\ varepsilon _ {0}}} (w + i \ alpha)} {c}}}$ . It can be seen that $k_{0}$ ${\ displaystyle k_ {0}}$ there is an imaginary positive part, then we need it. The equation $(2)$ ${\ displaystyle (2)}$ it is convenient to solve using the Fourier transform of the form:

F(\mathbf {k} )=\int \limits _{-\infty }^{\infty }F(\mathbf {r} )e^{-i\mathbf {k} \mathbf {r} }\,d\mathbf {r} ,

{\ displaystyle F (\ mathbf {k}) = \ int \ limits _ {- \ infty} ^ {\ infty} F (\ mathbf {r}) e ^ {- i \ mathbf {k} \ mathbf {r} } \, d \ mathbf {r},}

(3)

F(\mathbf {r} )={\frac {1}{(2\pi )^{3}}}\int \limits _{-\infty }^{\infty }F(\mathbf {k} )e^{i\mathbf {k} \mathbf {r} }\,d\mathbf {k} ,

{\ displaystyle F (\ mathbf {r}) = {\ frac {1} {(2 \ pi) ^ {3}}} \ int \ limits _ {- \ infty} ^ {\ infty} F (\ mathbf { k}) e ^ {i \ mathbf {k} \ mathbf {r}} \, d \ mathbf {k},}

(four)

Expression $(3)$ ${\ displaystyle (3)}$ - direct Fourier transform , $F(\mathbf {k} )$ ${\ displaystyle F (\ mathbf {k})}$ - Fourier transform of the function $F(\mathbf {r} )$ ${\ displaystyle F (\ mathbf {r})}$ , expression $(four)$ ${\ displaystyle (4)}$ Is the inverse Fourier transform . Green function image $G_{0}(\mathbf {r} )$ ${\ displaystyle G_ {0} (\ mathbf {r})}$ will be denoted by $G_{0}(\mathbf {k} )$ ${\ displaystyle G_ {0} (\ mathbf {k})}$ . Applying Fourier transforms to an equation $(2)$ ${\ displaystyle (2)}$ and given that $\delta$ ${\ displaystyle \ delta}$ -function is the Fourier transform of unity, we get:

(k^{2}-k_{0}^{2})G_{0}(\mathbf {k} )=1,

{\ displaystyle (k ^ {2} -k_ {0} ^ {2}) G_ {0} (\ mathbf {k}) = 1,}

(5)

G_{0}(\mathbf {k} )={\frac {1}{k^{2}-k_{0}^{2}}}.

{\ displaystyle G_ {0} (\ mathbf {k}) = {\ frac {1} {k ^ {2} -k_ {0} ^ {2}}}.}

(6)

To get a function $G_{0}(\mathbf {r} )$ ${\ displaystyle G_ {0} (\ mathbf {r})}$ doing the inverse Fourier transform $G_{0}(\mathbf {k} )$ ${\ displaystyle G_ {0} (\ mathbf {k})}$ :

G_{0}(\mathbf {r} )={\frac {1}{(2\pi )^{3}}}\int \limits _{-\infty }^{\infty }{\frac {e^{i\mathbf {k} \mathbf {r} }}{k^{2}-k_{0}^{2}}}\,d\mathbf {k} .

{\ displaystyle G_ {0} (\ mathbf {r}) = {\ frac {1} {(2 \ pi) ^ {3}}} \ int \ limits _ {- \ infty} ^ {\ infty} {\ frac {e ^ {i \ mathbf {k} \ mathbf {r}}} {k ^ {2} -k_ {0} ^ {2}}} \, d \ mathbf {k}.}

(7)

We will calculate this integral in a spherical coordinate system by choosing the polar axis along the vector $\mathbf {r}$ ${\ displaystyle \ mathbf {r}}$ (by the polar axis, we mean the axis from which the angle is measured $\theta$ ${\ displaystyle \ theta}$ ):

G_{0}(\mathbf {r} )={\frac {1}{(2\pi )^{3}}}\int \limits _{-\infty }^{\infty }{\frac {e^{i\mathbf {k} \mathbf {r} }}{k^{2}-k_{0}^{2}}}\,d\mathbf {k} ={\frac {1}{(2\pi )^{3}}}\int \limits _{0}^{2\pi }\,d\varphi \int \limits _{0}^{\pi }\sin \theta \,d\theta \int \limits _{0}^{\infty }k^{2}{\frac {e^{ikr\cos \theta }}{k^{2}-k_{0}^{2}}}\,dk=

{\ displaystyle G_ {0} (\ mathbf {r}) = {\ frac {1} {(2 \ pi) ^ {3}}} \ int \ limits _ {- \ infty} ^ {\ infty} {\ frac {e ^ {i \ mathbf {k} \ mathbf {r}}} {k ^ {2} -k_ {0} ^ {2}}} \, d \ mathbf {k} = {\ frac {1} {(2 \ pi) ^ {3}}} \ int \ limits _ {0} ^ {2 \ pi} \, d \ varphi \ int \ limits _ {0} ^ {\ pi} \ sin \ theta \, d \ theta \ int \ limits _ {0} ^ {\ infty} k ^ {2} {\ frac {e ^ {ikr \ cos \ theta}} {k ^ {2} -k_ {0} ^ {2} }} \, dk =}

={\frac {2\pi }{8\pi ^{3}}}\int \limits _{\pi }^{0}\,d\cos \theta \int \limits _{0}^{\infty }k^{2}{\frac {e^{ikr\cos \theta }}{k^{2}-k_{0}^{2}}}\,dk={\frac {1}{4\pi ^{2}}}\int \limits _{0}^{\infty }{\frac {k^{2}}{k^{2}-k_{0}^{2}}}{\Bigl [}{\frac {e^{ikr\cos \theta }}{ikr\cos \theta }}{\Bigr ]}_{\pi }^{0}\,dk=

{\ displaystyle = {\ frac {2 \ pi} {8 \ pi ^ {3}}} \ int \ limits _ {\ pi} ^ {0} \, d \ cos \ theta \ int \ limits _ {0} ^ {\ infty} k ^ {2} {\ frac {e ^ {ikr \ cos \ theta}} {k ^ {2} -k_ {0} ^ {2}}}, dk = {\ frac {1 } {4 \ pi ^ {2}}} \ int \ limits _ {0} ^ {\ infty} {\ frac {k ^ {2}} {k ^ {2} -k_ {0} ^ {2}} } {\ Bigl [} {\ frac {e ^ {ikr \ cos \ theta}} {ikr \ cos \ theta}} {\ Bigr]} _ {\ pi} ^ {0} \, dk =}

={\frac {1}{4\pi ^{2}}}\int \limits _{0}^{\infty }{\frac {k^{2}}{k^{2}-k_{0}^{2}}}{\Bigl (}{\frac {e^{ikr}}{ikr}}-{\frac {e^{-ikr}}{ikr}}{\Bigr )}\,dk={\frac {1}{4\pi ^{2}ir}}{\frac {1}{2}}\int \limits _{-\infty }^{\infty }{\frac {k(e^{ikr}-e^{-ikr})}{k^{2}-k_{0}^{2}}}\,dk=

{\ displaystyle = {\ frac {1} {4 \ pi ^ {2}}} \ int \ limits _ {0} ^ {\ infty} {\ frac {k ^ {2}} {k ^ {2} - k_ {0} ^ {2}}} {\ Bigl (} {\ frac {e ^ {ikr}} {ikr}} - {\ frac {e ^ {- ikr}} {ikr}} {\ Bigr)} \, dk = {\ frac {1} {4 \ pi ^ {2} ir}} {\ frac {1} {2}} \ int \ limits _ {- \ infty} ^ {\ infty} {\ frac { k (e ^ {ikr} -e ^ {- ikr})} {k ^ {2} -k_ {0} ^ {2}}} \, dk =}

={\frac {1}{8\pi ^{2}ir}}\int \limits _{-\infty }^{\infty }{\frac {ke^{ikr}}{k^{2}-k_{0}^{2}}}\,dk-{\frac {1}{8\pi ^{2}ir}}\int \limits _{-\infty }^{\infty }{\frac {ke^{-ikr}}{k^{2}-k_{0}^{2}}}\,dk={\frac {2\pi ie^{ik_{0}r}}{8\pi ^{2}ir2}}+{\frac {2\pi ie^{ik_{0}r}}{8{\pi }^{2}ir2}}={\frac {e^{i{k_{0}}r}}{4\pi r}}.

{\ displaystyle = {\ frac {1} {8 \ pi ^ {2} ir}} \ int \ limits _ {- \ infty} ^ {\ infty} {\ frac {ke ^ {ikr}} {k ^ { 2} -k_ {0} ^ {2}}}, dk - {\ frac {1} {8 \ pi ^ {2} ir}} \ int \ limits _ {- \ infty} ^ {\ infty} { \ frac {ke ^ {- ikr}} {k ^ {2} -k_ {0} ^ {2}}}, dk = {\ frac {2 \ pi ie ^ {ik_ {0} r}} {8 \ pi ^ {2} ir2}} + {\ frac {2 \ pi ie ^ {ik_ {0} r}} {8 {\ pi} ^ {2} ir2}} = {\ frac {e ^ {i { k_ {0}} r}} {4 \ pi r}}.}

To calculate the integral over spherical coordinates , we used the parity of the function ${\frac {k(e^{ikr}-e^{-ikr})}{k^{2}-k_{0}^{2}}}$ ${\ displaystyle {\ frac {k (e ^ {ikr} -e ^ {- ikr})} {k ^ {2} -k_ {0} ^ {2}}}}$ , as well as the last integrals were taken from the residues . For the first term, the integration loop is closed from above; in this half-plane, it attenuates $e^{ikr}$ ${\ displaystyle e ^ {ikr}}$ then the deduction is taken in $k=k_{0}={\frac {{\sqrt {\varepsilon _{0}}}(w+i\alpha )}{c}}$ ${\ displaystyle k = k_ {0} = {\ frac {{\ sqrt {\ varepsilon _ {0}}} (w + i \ alpha)} {c}}}$ . For the second term, they closed the contour in the lower half-plane, and then the residue at the point $k=-k_{0}$ ${\ displaystyle k = -k_ {0}}$ , it must be remembered that the circuit is circumvented clockwise, while the residue theorem uses counterclockwise circumvention. The bypass direction can be easily changed by adding a factor in the second term $(-1)$ ${\ displaystyle (-1)}$ .

The final expression for the Green's function will be:

G_{0}(\mathbf {r} ,\;t)={\frac {e^{ik_{0}r-iwt+\alpha t}}{4\pi r}}.

{\ displaystyle G_ {0} (\ mathbf {r}, \; t) = {\ frac {e ^ {ik_ {0} r-iwt + \ alpha t}} {4 \ pi r}}.}

This is a diverging spherical wave . The amplitude of this wave decreases as ${\frac {1}{r}}$ ${\ displaystyle {\ frac {1} {r}}}$ as you move away from the source.

Green function taking into account fluctuations of the dielectric constant

Rewrite the equation

-{\frac {\varepsilon (\mathbf {r} )(w+i\alpha )^{2}}{c^{2}}}G(\mathbf {r} )-\Delta G(\mathbf {r} )=\delta (\mathbf {r} )

{\ displaystyle - {\ frac {\ varepsilon (\ mathbf {r}) (w + i \ alpha) ^ {2}} {c ^ {2}}} G (\ mathbf {r}) - \ Delta G ( \ mathbf {r}) = \ delta (\ mathbf {r})}

as

(-k_{0}^{2}-\Delta )G(\mathbf {r} )={\frac {\delta \varepsilon (\mathbf {r} )}{\varepsilon _{0}}}k_{0}^{2}G(\mathbf {r} )+\delta (\mathbf {r} ).

{\ displaystyle (-k_ {0} ^ {2} - \ Delta) G (\ mathbf {r}) = {\ frac {\ delta \ varepsilon (\ mathbf {r})} {\ varepsilon _ {0}} } k_ {0} ^ {2} G (\ mathbf {r}) + \ delta (\ mathbf {r}).}

To use the perturbation theory , in which we will consider $\delta \varepsilon (\mathbf {r} )$ ${\ displaystyle \ delta \ varepsilon (\ mathbf {r})}$ small value, it is more convenient to go to the integral analogue of the previous equation:

G(\mathbf {r} )=G_{0}(\mathbf {r} )+{\frac {{k_{0}}^{2}}{\varepsilon _{0}}}\int G_{0}(\mathbf {r} -\mathbf {r_{1}} )\delta \varepsilon (\mathbf {r_{1}} )G(\mathbf {r_{1}} )\,dr_{1}.

{\ displaystyle G (\ mathbf {r}) = G_ {0} (\ mathbf {r}) + {\ frac {{k_ {0}} ^ {2}} {\ varepsilon _ {0}}} \ int G_ {0} (\ mathbf {r} - \ mathbf {r_ {1}}) \ delta \ varepsilon (\ mathbf {r_ {1}}) G (\ mathbf {r_ {1}}) \, dr_ {1 }.}

Then you can easily write an iterative solution in the form of a series:

G(\mathbf {r} )=G_{0}(\mathbf {r} )+{\frac {k_{0}^{2}}{\varepsilon _{0}}}\int G_{0}(\mathbf {r} -\mathbf {r_{1}} )\delta \varepsilon (\mathbf {r_{1}} )G_{0}(\mathbf {r_{1}} )\,d\mathbf {r_{1}} +

{\ displaystyle G (\ mathbf {r}) = G_ {0} (\ mathbf {r}) + {\ frac {k_ {0} ^ {2}} {\ varepsilon _ {0}}} \ int G_ { 0} (\ mathbf {r} - \ mathbf {r_ {1}}) \ delta \ varepsilon (\ mathbf {r_ {1}}) G_ {0} (\ mathbf {r_ {1}}) \, d \ mathbf {r_ {1}} +}

+{\frac {k_{0}^{4}}{\varepsilon _{0}^{2}}}\int G_{0}(\mathbf {r} -\mathbf {r_{1}} )\delta \varepsilon (\mathbf {r_{1}} )\int G_{0}(\mathbf {r_{1}} -\mathbf {r_{2}} )\delta \varepsilon (\mathbf {r_{2}} )G_{0}(\mathbf {r_{2}} )\,d\mathbf {r_{1}} \,d\mathbf {r_{2}} +\ldots

{\ displaystyle + {\ frac {k_ {0} ^ {4}} {\ varepsilon _ {0} ^ {2}}} \ int G_ {0} (\ mathbf {r} - \ mathbf {r_ {1} }) \ delta \ varepsilon (\ mathbf {r_ {1}}) \ int G_ {0} (\ mathbf {r_ {1}} - \ mathbf {r_ {2}}) \ delta \ varepsilon (\ mathbf {r_ {2}}) G_ {0} (\ mathbf {r_ {2}}) \, d \ mathbf {r_ {1}} \, d \ mathbf {r_ {2}} + \ ldots}

Value $G(\mathbf {r} )$ ${\ displaystyle G (\ mathbf {r})}$ - random value. In the future, it must be averaged over all possible fluctuations of the dielectric constant. This represents the next laborious step.

Literature

Rytov S. M., Kravtsov Yu. A., Tatarsky V. I. Introduction to statistical radiophysics. - Part 2. Random fields. - M.: Science, 1978.
Ishimaru A. Propagation and scattering of waves in randomly inhomogeneous media. - T. 1, 2. - M .: Mir, 1981.