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The medium being lossless, the total energy must be constant jQ1 k; t j2 jQ2 k; t j2 d3 k 0: This implies that o1 k and o2 k are real and
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We write now the Bourret s equation (see Section 3.4) for the mean propagator of Equation (3.181)
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As there are only two modes, it is more convenient to have a scalar operator instead of a matrix one. For the unperturbed propagator being a diagonal matrix, it is suf cient to introduce, over the solid line, a superscript indicating the wave mode. Using this convention we write (3.187) as
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As there are no self-coupling terms, hG12 iB and hG21 iB vanish. The consequence of the absence of self-coupling terms is thus the absence of coupling between the mean propagators (this is due to the fact that any Bourret diagram has an even number of vertices). But this will not prevent energy transfer between the wave modes, because the energy densities are calculated from the mean double propagator. Let us nd the behavior of hG11 iB for t ) tint. It is determined by the scalar terms of the Bourret series and thus by lim
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In explicit from Equation (3.188a) becomes iz io1 k ib1 k hG11 iB 1: To derive this equation we have replaced we obtain 3:190
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The frequency o1 k has thus been renormalized. For the second wave mode from (3.188b) we nd a similar result with b2 k b k . 1 Let us calculate b1 k . Writing hB12 k; k0 B21 k0 ; k00 i G k k0 d k k00 we obtain b1 k lim e G k; k0 d3 k 0 : o1 k o2 k iZ 3:193 3:192
G k; k0 is a positive measure (FT of a covariance function), accordingly the imaginary part of b1 k is positive and hG11 k; t i is damped. It is possible to evaluate this imaginary part by a dimensional analysis of (3.188a)   Imb1 k  2 2 3    o k  $ e K ( 1: 1 3:194
Now we turn our attention to the mean double propagator and calculate the mean spectral energy densities hjQ1 k; t j2 i, hjQ2 k; t j2 i. They do not satisfy any propagation ~ ~ ~ ~ equation but are deducible from hQ1 k; z Q k; z0 i and hQ2 k; z Q k; z0 i, which 1 2 satisfy the following equations of the Bourret s approximation
~ ~* Q1 (k ; z )Q1 (k ; z' ) =
~ ~* Q1 (k ;0 )Q1 (k ;0 )+
~ ~* Q2 (k' ; z )Q2 (k' ; z' )
~ ~* Q2 (k ; z ) Q2 (k ; z' ) =
~* ~ Q2 (k ;0)Q2 (k ;0)+
~* ~ Q 1 (k ' ; z ) Q 1 (k ' ; z' )
3:195b ~ ~ ~ ~ hQ1 k; z Q k; z0 i is the FT of hQ1 k; t Q k; t0 i with respect to t and t0 . It is 1 1 interesting to solve (2.195a) with the following initial conditions: ~ ~ Q1 k; 0 Q k; 0 E0 d k k0 1 ~ ~ Q2 k; 0 Q k; 0 0
for which the total initial energy of the wave is concentrated in the rst wave mode ~ ~ with a single spectral line. Eliminating hQ1 k; z Q k; z0 i between (3.195a) and 1 (3.195b) we obtain:
~ ~* Q2 (k ; z )Q2 (k ; z' ) = 2 1 2 1 E0d (k' k0 ) + 2 1 2 1 ~ ~* Q2 (k' ; z ) Q2 (k' ; z' ) .