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Analytical solutions for three and four diffraction orders interaction in Kerr media

N. Korneev

Open Access Open Access

Abstract

The analytical solution for the interaction of three diffraction orders in the Kerr medium is obtained by reducing the problem to the completely integrable Hamiltonian task. Intensities of all waves are periodic with propagation length and linearly related, the amplitudes are quasi-periodic and expressed in elliptic functions. Symmetrical four-order interaction also admits an analytical solution.

©2000 Optical Society of America

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Figures (3)
Fig. 1.
Fig. 1. The trajectory in the complex plane for S 1. It is seen that the amplitude gains a constant phase shift after one period of intensity. The numerical solution of equations (3–5) for K=1.8, κ=0.4, and initial conditions S -1=0.3, S 0=1, S 1=0.1+0.24i.

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Fig. 2.
Fig. 2. The maximal intensity transferred to the initially weak side order depending on the nonlinearity parameter q. The central beam intensity I 0=1.

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Fig. 3.
Fig. 3. Phase portraits of trajectories with different initial conditions on the Poincaré sphere for a four-beam interaction. The nonlinearity parameter is q=0.5.

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Equations (19)

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i ∂ z ψ + 1 2 ∂ xx ψ + κ ∣ ψ ∣ 2 ψ = 0 .
ψ ( x , z ) = ∑ k S k ( z ) exp ( i k x ) ,
i ∂ z S − 1 = 1 2 K 2 S − 1 − κ ( ∣ S − 1 ∣ 2 S − 1 + 2 ∣ S 0 ∣ 2 S − 1 + 2 ∣ S 1 ∣ 2 S − 1 + S 0 2 S 1 * ) ,
i ∂ z S 0 = − κ ( ∣ S 0 ∣ 2 S 0 + 2 ∣ S 1 ∣ 2 S 0 + 2 ∣ S − 1 ∣ 2 S 0 + 2 S 0 * S 1 S − 1 ) ,
i ∂ z S 1 = 1 2 K 2 S 1 − κ ( ∣ S 1 ∣ 2 S 1 + 2 ∣ S 0 ∣ 2 S 1 + 2 ∣ S − 1 ∣ 2 S 1 + S 0 2 S − 1 * ) .
H ( ψ , ψ * ) = 1 2 ∫ ( ∣ ∂ x ψ ∣ 2 − κ ∣ ψ ∣ 4 ) d x .
H ( S , S * ) = 1 2 ∑ k k 2 S k S k * − 1 2 κ ∑ a 1 + a 2 = a 3 + a 4 S a 1 S a 2 S a 3 * S a 4 * ,
i ∂ z S k = ∂ H ⁄ ∂ S k * .
i ∂ z U = { U , H } = ∑ k ∂ U ∂ S k ∂ H ∂ S k * − ∂ U ∂ S k * ∂ H ∂ S k
H = ( 1 2 K 2 − κ I 0 ) ( I − I 0 ) − 1 2 κ I 2 − 1 4 κ ( ( I − I 0 ) 2 − M 2 ⁄ K 2 ) − 2 κ Re ( S 0 * 2 S 1 S − 1 ) .
∂ z I 0 = − 4 κ I m ( S 0 * 2 S 1 S − 1 ) .
( Im ( S 0 * 2 S 1 S − 1 ) ) 2 + ( Re ( S 0 * 2 S 1 S − 1 ) ) 2 = I 0 2 I 1 I − 1 = 1 4 I 0 2 ( ( I − I 0 ) 2 − M 2 ⁄ K 2 ) ,
i ∂ z ln ( S 0 ) = − κ ( ∣ S 0 ∣ 2 + 2 ∣ S 1 ∣ 2 + 2 ∣ S − 1 ∣ 2 + 2 S 0 * 2 S 1 S − 1 ∣ S 0 ∣ − 2 ) .
i ∂ z S 1 = 1 8 K 2 S 1 − κ ( ( I + ∣ S 1 ∣ 2 + 2 Re ( S 1 S 2 * ) ) S 1 + ( ∣ S 1 ∣ 2 + 4 Re ( S 1 S 2 * ) ) S 2 ) ,
i ∂ z S 2 = 9 8 K 2 S 2 − κ ( ( I + ∣ S 2 ∣ 2 ) S 2 + ( ∣ S 1 ∣ 2 + 4 Re ( S 1 S 2 * ) ) S 1 ) .
H = 9 4 K 2 ∣ S 2 ∣ 2 + 1 4 K 2 ∣ S 1 ∣ 2 − 1 2 κ ( I 2 + 2 ∣ S 1 ∣ 4 + 2 ∣ S 2 ∣ 4 + 8 ∣ S 1 ∣ 2 Re ( S 1 S 2 * ) + 16 ( Re ( S 1 S 2 * ) ) 2 )
∂ z I 2 = − 2 κ ( ∣ S 1 ∣ 2 + 4 Re ( S 1 S 2 * ) ) Im ( S 1 S 2 * ) ,
A 2 + B 2 + C 2 = I 2 ⁄ 4 ,
2 K 2 A + κ ( A 2 + IC + 2 AC + 4 C 2 ) = const ,
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