Energetics of stochastic limit-cycle oscillators: when does coupling reduce dissipation?

1. [1]

   tilted-washboard

   relative to the free system EPR σ (kij = 0) is given by σ (kij → ∞ ) σ (kij = 0) (β → 0) = N − 2/ 3 Γ[ 2 3 ]Γ[N +3 3 ] Γ[ 4 3 ]Γ[N +1 3 ]. (31) In the limit of infinitely many systems, we find for the relative EPR of coupled systems lim N →∞ σ (kij → ∞ ) σ (kij = 0) (β → 0) = lim N →∞ N − 2/ 3 Γ[ 2 3 ]Γ[N +3 3 ] Γ[ 4 3 ]Γ[N +1 3 ] = 3 − 2/ 3 Γ[ 2 3 ] Γ[ 4 3...

2. [2]

   Two coupled systems Coupling a single Cartesian coordinate breaks the sym- metry of the governing equations, making it intractable to find a closed solution. However, coupling both Carte- sian coordinates restores symmetry of the equations ˙xi = (1 − ri)xi +ωiyi +k(xj − xi) + √ 2Teffξxi ˙yi = (1 − ri)yi − ωixi +k(yj − yi) + √ 2Teffξyi, (53) which is most obv...

3. [3]

   Therefore, the EPR will always initially decrease with coupling strength, proportional to the variance of the square of the radii, which increases with effective tem- perature

   ≤ 0, (61) where equality holds at zero effective temperature. Therefore, the EPR will always initially decrease with coupling strength, proportional to the variance of the square of the radii, which increases with effective tem- perature. Although the total EPR decreases, the deriva- tive of the radial variance and average radius at k = 0 is given by d dk V...

4. [4]

   This EPR results from the center-of- mass motion

   but with the effective tem- perature halved. This EPR results from the center-of- mass motion. In units of the free system σ (k = 0) , the strong-coupling limit of the EPR is less than unity for all β , implying that, unlike in the radial case, strong Cartesian coupling always reduces the limiting EPR rel- ative to the free system, see Fig 5(d). The overal...

5. [5]

   N coupled systems ForN systems coupled through the Hookean interac- tion V = k 2 ∑N i<jcij|zj − zi|2, where cij = 1 if the i-th andj-th oscillators are coupled and cij = 0 otherwise, in Cartesian coordinates the equations read ˙xi = (1 − ri)xi +ωyi +k N∑ j=1 cij(xj − xi) + √ 2Teffξxi ˙yi = (1 − ri)yi − ωxi +k N∑ j=1 cij(yj − yi) + √ 2Teffξyi. (66) The proba...

6. [6]

   (74) That is, indefinitely increasing the number of coupled systems indefinitely increases the EPR reduction in the strong-coupling limit

   in units of the free system yields σ (kij → ∞ ) σ (kij = 0) (β → 0) = N − 2/ 3. (74) That is, indefinitely increasing the number of coupled systems indefinitely increases the EPR reduction in the strong-coupling limit. B. Single coordinate coupling When coupling only a single Cartesian coordinate with all-to-all coupling, without loss of generality, the Lan...

7. [7]

   (76) For the sake of argument, consider N = 2

   To illustrate the latter case, using ¯x = 1 N ∑N j=1xj , the coupling term for the i-th oscilla- tor can be written as k ∑ j̸=i (xj − xi) = kN (¯x − xi). (76) For the sake of argument, consider N = 2 . If the oscil- lator states are located on either side of the line x = 0 , such that ¯x ≈ 0, then Eq. (

8. [8]

   yields ˙xi ∼ (1 − ri)xi +ωyi − 2kxi + √ 2Teffξxi. (77) Suppose there is strong coupling and weak radial fluctu- ations, then the expression simplifies to ˙xi ∼ +ωyi − 2kxi + √ 2Teffξxi, (78) resulting in xi ∼ O ( ω 2k ) so that ri ∼ | yi| for large k. Consequently, ˙yi ∼ (1 − |yi|)yi + √ 2Teffξyi, (79) 14 FIG. 7. (a) The steady-state entropy production rate (E...

9. [9]

   8(a)(left), then ¯y ∼ ± 1 and ˙¯x ∼ ± ω and the angular driving ω of both oscillators work together to increase ¯x

   If both oscillator states are 15 positioned in the negative or positive y-plane, denoted in-phase in Fig. 8(a)(left), then ¯y ∼ ± 1 and ˙¯x ∼ ± ω and the angular driving ω of both oscillators work together to increase ¯x. However, if one oscillator is in the negative y-plane and the other in the positive, denoted anti-phase in Fig. 8(a)(right), then ¯y ∼ ...

   2020

10. [10]

    tilted washboard

    Since in these coordinates the degrees of freedom decouple, they each contribute to the entropy production independently. Let us first consider the relative phase ϕ . The corresponding Langevin equation corresponds to the well-known “tilted washboard” problem for Brownian motion in the potential V (ϕ ) = ∆ ωϕ − 2k cosϕ , where ∆ ω plays the role of a drivi...

11. [11]

    Hence, the EPR is given by σϕ = ∫ 2π 0 J 2 ϕ 2Dps dϕ = Jϕ 2D ∫ 2π 0 Jϕ ps dϕ = − π ∆ ω D Jϕ

    Entropy production rate The relative current obeys Jϕ = (− ∆ ω − 2k sinϕ )ps − 2D∂ϕps ⇔ Jϕ ps = − ∆ ω − 2k sinϕ − 2D∂ϕ lnps ⇒ ∫ 2π 0 Jϕ ps dϕ = − 2π ∆ ω − 2k ∫ 2π 0 sinϕ dϕ − 2D ∫ 2π 0 ∂ϕ lnps dϕ = − 2π ∆ ω, (91) where the last two terms vanish due to periodicity. Hence, the EPR is given by σϕ = ∫ 2π 0 J 2 ϕ 2Dps dϕ = Jϕ 2D ∫ 2π 0 Jϕ ps dϕ = − π ∆ ω D Jϕ....

12. [12]

    Moments of the free system are then given by ⟨rn⟩0 = In/ I0

    is given by In(β ) := ∫ ∞ 0 rdre β ( r2 2 − r3 3 )rn = 1 8 3 − 1+n 3 β − 2 3 − n 3 ( 8 Γ ( 2+n 3 ) 3F2 ( { 1 3 + n 6, 5 6 + n 6 }; { 1 3, 2 3 }; β 6 ) + 4 3 2 3β 1 3 Γ ( 4+n 3 ) 3F2 ( { 2 3 + n 6, 7 6 + n 6 }; { 2 3, 4 3 }; β 6 ) + 3 3 1 3β 2 3 Γ ( 2 + n 3 ) 3F2 ( {1 + n 6, 3 2 + n 6 }; { 4 3, 5 3 }; β 6 ) ) , (98) where 3F2 denotes the generalized hyperg...

13. [13]

    Entropy production rate With steady-state distribution in Eq. ( 100) resulting from the addition of a conservative interaction V , the steady-state probability current is given by J = 1 Z J0e− βV , (104) where J0 is the probability current of the free system, where J0,θ i = −ωip× N 0 . The EPR is then σ = ∫ J⊤ D− 1J p dx = ∫ J⊤ 0D− 1J0 p× N 0 e− βV Z dx =...

14. [14]

    Expansion of EPR The total EPR for two coupled oscillators can be ex- panded into moments of the free system as follows σ = 2∑ i=1 ∫ ∞ 0 J 2 θi Dθiprjdrjridri = ω 2 Teff 2∑ i=1 ∫ ∞ 0 r2 ip(ri,r j )rjdrjridri = ω 2 Teff 2∑ i=1 ∫ ∞ 0 r2 i ∫ ∞ 0 1 Zpi 0pj 0 ∞∑ n=0 1 n! [ − k 2Teff (ri − rj )2 ]n rjdrjridri = ω 2 Teff 1 Z 2∑ i=1 ∞∑ n=0 1 n! ( − k 2Teff ) n ∫ ∞ 0 ∫...

15. [15]

    (110) This expression greatly simplifies at k = 0 since any covariance terms between oscillator coordinates vanish, i.e., Cov0(rn i,r m j ) = 0 for all n,m and i ̸= j

    EPR gradient For N oscillators coupled through the interaction V ({rj}) = k 2 ∑N i<jcij(ri − rj)2, the EPR gradient reads d dkσ =ω 2β N∑ i=0 d dk Vark(ri) + 2 ⟨ri⟩k d dk ⟨ri⟩k = − ω 2β 2 2 N∑ i=0 [Covk((ri − ⟨ri⟩k)2, N∑ l<j clj(rl − rj )2) + 2 ⟨ri⟩k Covk(ri, N∑ l<j clj(rl − rj)2)]. (110) This expression greatly simplifies at k = 0 since any covariance term...

16. [16]

    (116) In the strong-coupling limit kij → ∞ for all i,j p({ri}) = 1 Z ′( N∏ i=1 pi 0) N∏ i<j δ(ri − rj)

    Strong-coupling limit The probability distribution function for N -radially coupled oscillators is given by p({ri}) = 1 Z ( N∏ i=1 pi 0) N∏ i<j e− 1 2Teff kij (ri− rj )2 . (116) In the strong-coupling limit kij → ∞ for all i,j p({ri}) = 1 Z ′( N∏ i=1 pi 0) N∏ i<j δ(ri − rj). (117) where the coefficients emerging from the δ-function limit cancel with the sam...

17. [17]

    For β ≳ 1

    Distribution under coupling Figure 10(a) shows the gradient of the average ra- dius ⟨r⟩k versus coupling strength k and inverse effec- tive temperature β . For β ≳ 1. 8, ⟨r⟩k increases with coupling strength until plateau. For β ≲ 1. 8, ⟨r⟩k ini- tially decreases then increases with coupling strength un- til plateau. Figure 10(b) shows the gradient of the ...

18. [18]

    (121) Now, D− 1 = 1 Teff      1 0 0 0 0 1 0 0 0 0 r2 1 +r2 2 1 2 (r2 2 − r2 1) 0 0 1 2 (r2 2 − r2

    Entropy production rate The total EPR is given by σ = ∫ dX J⊤D− 1J p . (121) Now, D− 1 = 1 Teff      1 0 0 0 0 1 0 0 0 0 r2 1 +r2 2 1 2 (r2 2 − r2 1) 0 0 1 2 (r2 2 − r2

19. [19]

    (122) 22 FIG

    1 4 (r2 1 +r2 2)      = 1 Teff      1 0 0 0 0 1 0 0 0 0 a b 0 0 b 1 4a      . (122) 22 FIG. 10. Radial distribution deformation under radial cou- pling. (a) Heat map of the derivative of the average radius ⟨r⟩k (in units of ⟨r⟩0) with respect to coupling strength k ver- sus k and the inverse effective temperature β . (b) Heat map of the deriv...

20. [20]

    The Fisher information is defined as J (θ) = ∫ (∂θ lnp(x;θ))2p(x;θ)dx, (140) where the integral is performed over the x-domain

    Fisher Information Consider a probability distribution p(x;θ), where θ parametrizesp. The Fisher information is defined as J (θ) = ∫ (∂θ lnp(x;θ))2p(x;θ)dx, (140) where the integral is performed over the x-domain. In Eq. ( 42) we straightforwardly identify the EPR contri- bution σriϕ as a weighted radial average of a Fisher in- formation term since ⟨(∂ri l...

21. [21]

    Approximation The expression for the EPR in Eq. (

22. [22]

    can be approx- imated with Eq. ( 44). To this end, the reduced current is given by jϕ = − Dϕϕ 2πρ ∞∑ n=−∞ In(α )In(− α ) n2+ρ 2 , (148) whereα := − 2k/D ϕϕ andρ := ∆ ω/D ϕϕ . With Eq. ( 44), the expression for ⟨sin2ϕ ⟩ϕ is found to be ⟨sin2ϕ ⟩ϕ = ∞∑ n=−∞ In(α ) n2+ρ 2 [ In(−α ) − 1 2 (In− 2(−α ) − In+2(−α )) ] 2 ∞∑ n=−∞ In(α )In(− α ) n2+ρ 2 . (149) The r...

23. [23]

    (154) 25 To further evaluate the foregoing expression, we note that |z1 − z2|2 = |r1eiθ1 − r2eiθ2|2 =r2 1|1 − r2 r1 e− iϕ |2 =r2 1(1 − r2 r1 e− iϕ )(1 − r2 r1 eiϕ )

    Expansion of EPR The normalization constant is given by Z = ∫ 2π 0 dθ1dθ2 ∫ ∞ 0 r1dr1r2dr2p1 0p2 0e− k 2Teff |z1− z2|2 = ∫ 2π 0 dθ1dθ2 ∫ ∞ 0 r1dr1r2dr2p1 0p2 0 ∞∑ n=0 1 n! ( − k 2Teff ) n |z1 − z2|2n. (154) 25 To further evaluate the foregoing expression, we note that |z1 − z2|2 = |r1eiθ1 − r2eiθ2|2 =r2 1|1 − r2 r1 e− iϕ |2 =r2 1(1 − r2 r1 e− iϕ )(1 − r2 r...

24. [24]

    (161) At k = 0 , the radii and phases are independent, so ⟨r2 i cos(θl − θj)⟩0 = ⟨r2 i ⟩⟨cos(θl − θj)⟩0 = 0

    EPR gradient For N oscillators coupled through the interaction V ({rj}) = k 2 ∑N i<jcij[r2 i +r2 j − 2rirj cos(θi − θj)], the gradient of the second radial moment reads d dk ⟨r2 i ⟩k = − β 2 Covk(r2 i, N∑ l<j clj[r2 l +r2 j − 2rlrj cos(θl − θj)]) = − β 2 Covk(r2 i, N∑ l<j clj[r2 l +r2 j ]) − β 2 Covk(r2 i, N∑ l<j clj[− 2rlrj cos(θl − θj)]). (161) At k = 0...

25. [25]

    (164) To find the strong-coupling limit, it is convenient to transform the system into the relative and center-of-mass coordinates which diagonalize the interaction term

    Strong-coupling limit The steady-state probability distribution of N cou- pled oscillators with interaction V = k 2 ∑ i<jcij|zi − zj|2, where cij = 1 if the i-th and j-th oscillators are coupled and cij = 0 otherwise, reads p({zi}) = 1 Ze β N∑ i=1 ( 1 2 |zi|2− 1 3 |zi|3)− βk 2 ∑ i<j cij|zi− zj |2 . (164) To find the strong-coupling limit, it is convenient ...

26. [26]

    (168) Now lim k→∞ exp [ − βk 2 N − 1∑ α =1 λα |qα |2 ] ∝ N − 1∏ α =1 δ(2)(qα ), (169) where δ(2) denotes the 2d Dirac delta function

    becomes p(q0, {qα }) = 1 Ze β N∑ i=1 ( 1 2 |(Uq )i|2− 1 3 |(Uq )i|3)− βk 2 N − 1∑ α =1 λ α |qα |2 . (168) Now lim k→∞ exp [ − βk 2 N − 1∑ α =1 λα |qα |2 ] ∝ N − 1∏ α =1 δ(2)(qα ), (169) where δ(2) denotes the 2d Dirac delta function. Since λα > 0 for all α ≥ 1, an integration over the delta func- tions sets qα → 0. Then (Uq )i = (q0e0)i = ¯z. That is, in ...

27. [27]

    For β ≳ 0

    Distribution under coupling Figure 11(a) shows the gradient of the average ra- dius ⟨r⟩k versus coupling strength k and inverse effec- tive temperature β . For β ≳ 0. 5, ⟨r⟩k initially decreases then increases with coupling strength until plateau. For β ≲ 0. 5, ⟨r⟩k decreases with coupling strength until plateau. Figure 11(b) shows the gradient of the radi...

28. [28]

    The EPR is then [15, 33] σ = ∫ 2π 0 dφ ∫ ∞ 0 r2J 2 φ Dp0 rdr = ∫ ∞ 0 r2(α − ζr2)2 D p0(r)rdr = ∫ ∞ 0 r2α 2 +ζ2r6 − 2αζr 4 D p0(r)rdr

    Entropy production rate The non-zero angular current reads Jφ = (α − ζr2)p0 = 1 2π (α − ζr2)p0(r). The EPR is then [15, 33] σ = ∫ 2π 0 dφ ∫ ∞ 0 r2J 2 φ Dp0 rdr = ∫ ∞ 0 r2(α − ζr2)2 D p0(r)rdr = ∫ ∞ 0 r2α 2 +ζ2r6 − 2αζr 4 D p0(r)rdr. (178) Using the moments of p0, this becomes σ = α 2 D ⟨r2⟩0 + ζ2 D ⟨r6⟩0 − 2αζ D ⟨r4⟩0. (179) where the mth moment is given ...

29. [29]

    ( 173) yields com- pletely analogous results presented in Eq

    Radial coupling Introducing radial coupling to Eqs. ( 173) yields com- pletely analogous results presented in Eq. ( 19). Specifi- cally, Eq. ( 178) generalizes to non-zero k σ = ∑ i=1 α 2 D ⟨r2 i ⟩k + ζ2 D ⟨r6 i ⟩k − 2αζ D ⟨r4 i ⟩k = ∑ i=1 1 D ⟨r2 i ˜ω 2(ri)⟩k, (181) whereω 2(ri) describes the radial dependent angular fre- quency and the non-zero k-expecta...

30. [30]

    In this case, approximations could be made, for example, in Ref

    Cartesian Coupling Unlike with radial coupling, introducing Cartesian cou- pling is not analytically solvable for ζ ̸= 0 . In this case, approximations could be made, for example, in Ref. [34]. When ζ = 0 , similarly to above, analogous expressions to Eq. (

31. [31]

    V for the EPR can be found with a modified free particle distribution p0

    in Sec. V for the EPR can be found with a modified free particle distribution p0. H. Phase coupling Likewise to Cartesian coupling, when ζ = 0 , the anal- ysis in Sec. IV is completely analogous. 28

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