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If the period is much bigger than the half-life, ie. the degradation rate is much bigger than the frequency : γ≫Ω ie. δ<<1, we can make some simplifications.
Let us recall the transcendental equation for the frequency :
Ω=ω+S2(δ2+1)14Im(ei2(arctan(ωγ)−arctan(δ))(11−exp(−(δ2+1)14λei2arctan(δ)d+iβ)+11−exp(−(δ2+1)14λei2arctan(δ)d−iβ)−1)) We do not know if Ωγ is negligible in comparison to ωγ so we will keep this term and write ei(arctan(ωγ)−arctan(Ωγ))=A+iB If Ω=ω+ϵ,ϵ<<1 then A=1 and B=ϵ In a simulation, we find that Ω−ω∼10−3 for example. It is useful to define the three following independant parameters :
λ=√Dγ (coupling range in meters) S=α√Dγ (coupling strength in 1/seconds)
δ=Ωγ
At first order in δ :
Ω=ω+S2Im((A+iB)(11−e−(1+iδλ+iβ)+11−e−(1+iδλ−iβ)−1))=ω+S2Im((A+iB)(11−e−(1λ+iβ)+11−e−(1λ−iβ)−1−iδ2λ(1e(1λ−iβ)−1+1e(1λ+iβ)−1)))
A common choice of parameters (alpha=50; omega=10; gamma=100.; D=1)would give δ∼0.1 and B∼0.001, so we will choose to take the zeroth order in B (consequently, A=1)and the first order in δ. Since the sum of two complex conjugates is real, taking the imaginary part leads to :
Ω=ω−S4δλ2Re(1e(1λ+iβ)−1)
(1e(1λ+iβ)−1)=e1λcos(β)−1−ie1λsin(β)|e(1λ+iβ)−1|2 so Ω=ω−Sδ2λe1λcos(β)−1|e(1λ+iβ)−1|2
Considering that λ=√Dγ, in many cases where γ is big, we have λ small, and we can approximate the previous expression taking the leading order in e1λ when β is not far from 0:
Ω=ω−Sδ2λe1λcos(β)−11−2e1λcos(β)+e2λ∼ω−Sδ2λe−1λcos(β)
We can finally solve the equation for Ω, recalling that δ=Ωγ :
Ω=ω(1+S2γλe−1λcos(β))−1∼ω(1−e−1λS2γλcos(β)), or putting the distance between the oscillators back in the formula : Ω=ω(1−e−dλSd2γλcos(β))
Conclusion : three hypotheses were necessary :
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