Nonlinear stability of stationary solutions to the Kuramoto-Sakaguchi equation with frustration

  • Received: 01 August 2019 Revised: 01 May 2020 Published: 09 September 2020
  • Primary: 92B25, 35Q92; Secondary: 35B40

  • We study measurable stationary solutions for the kinetic Kuramoto-Sakaguchi (in short K-S) equation with frustration and their stability analysis. In the presence of frustration, the total phase is not a conserved quantity anymore, but it is time-varying. Thus, we can not expect the genuinely stationary solutions for the K-S equation. To overcome this lack of conserved quantity, we introduce new variables whose total phase is conserved. In the transformed K-S equation in new variables, we derive all measurable stationary solution representing the incoherent state, complete and partial phase-locked states. We also provide several frameworks in which the complete phase-locked state is stable, whereas partial phase-locked state is semi-stable in the space of Radon measures. In particular, we show that the incoherent state is nonlinearly stable in a large frustration regime, whereas it can exhibit stable behavior or concentration phenomenon in a small frustration regime.

    Citation: Seung-Yeal Ha, Hansol Park, Yinglong Zhang. Nonlinear stability of stationary solutions to the Kuramoto-Sakaguchi equation with frustration[J]. Networks and Heterogeneous Media, 2020, 15(3): 427-461. doi: 10.3934/nhm.2020026

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  • We study measurable stationary solutions for the kinetic Kuramoto-Sakaguchi (in short K-S) equation with frustration and their stability analysis. In the presence of frustration, the total phase is not a conserved quantity anymore, but it is time-varying. Thus, we can not expect the genuinely stationary solutions for the K-S equation. To overcome this lack of conserved quantity, we introduce new variables whose total phase is conserved. In the transformed K-S equation in new variables, we derive all measurable stationary solution representing the incoherent state, complete and partial phase-locked states. We also provide several frameworks in which the complete phase-locked state is stable, whereas partial phase-locked state is semi-stable in the space of Radon measures. In particular, we show that the incoherent state is nonlinearly stable in a large frustration regime, whereas it can exhibit stable behavior or concentration phenomenon in a small frustration regime.



    Collective behaviors of oscillatory complex systems are ubiquitous in our nature, i.e., flashing of fireflies, beating of cardiac pacemaker cells, and arrays of Josephson junctions [1,10,27,30,32] etc. Recently, collective behaviors have received lots of attention from distinct scientific disciplines such as control theory, physics, neuroscience due to its applications in robot system, sensor network, and unmanned aerial vehicle. Among them, we are interested in the synchronization representing adjustment of rhythms of oscillators. The rigorous and systematic study for synchronization goes back to two pioneers Winfree and Kuramoto in a half century ago. In this paper, our focus lies in the Kuramoto model with frustration [29] (sometimes called the Kuramoto-Sakaguchi model). In order to fix the idea, let θi=θi(t)T and α be the phase of the i-th Kuramoto oscillator and the uniform frustration (phase shift) between oscillators. Then the phase dynamics of Kuramoto oscillators is governed by the following first-order system [20,21,28,29]:

    ˙θi=ωi+κNN=1sin(θθi+α),i=1,,N, (1.1)

    where ωi, κ and N are the natural frequency of the i-th Kuramoto oscillator, coupling strength, and the number of oscillators, respectively.

    Note that the K-S model (1.1) can be rewritten as

    ˙θi=ωi+κcosαNN=1sin(θθi)+κsinαNN=1cos(θθi). (1.2)

    The terms in the R.H.S. of (1.2) correspond to the natural frequency, synchronization enforcing force and integrable forcing term, respectively. When the system size N is sufficiently large, state of system (1.1) can be approximated by the continuity equation with nonlocal velocity field. More precisely, let F=F(t,θ,ω) be a one-particle distribution function, and f=f(t,θ,ω) be the conditional probability density function defined by the following relation:

    F(t,θ,ω)=f(t,θ,ω)g(ω),(t,θ,ω)R+×T×R,

    where g(ω) is the probability density for natural frequency. Then, for ωR, we define a measurable map ωfω:=f(,,ω) from R to P(R+×T) (set of all probability measures on T). Then the dynamics of conditional probability density fω satisfies

    {tfω+θ(V[fω]fω)=0,(t,θ,ω)R+×T×R,V[fω](θ,ω,t):=ω+κT×Rsin(θθ+α)g(ω)dfωdω,Tdfω=1, (1.3)

    where differential operator θ is the derivation on T interpreted in the sense of distribution: for any Ck function ϕ on the circle, the action of θ(V[fω]fω) is given by

    θ(V[fω]fω),ϕ=TϕV[fω]dfω.

    In the absence of frustration α=0, there have been lots of literatures on the Kuramoto model (1.1), to name a few [4,5,8,9,14,15,17,19,31] etc. As discussed in [6,7,22,25,26,33], frustration is needed for the realistic modeling of physical and biological oscillators. The mere change (1.1) of the Kuramoto model by adding frustration causes several analytical difficulties in the study of emergent dynamics. For example, the total phases:

    Nj=1θj:particle system,T×Rθg(ω)dfωdω:kinetic K-S equation.

    are not conserved quantitites, and gradient flow structure for (1.1) is also destroyed. Thus, we cannot use the useful machineries for the Kuramoto model in the study of emergent dynamics for (1.1) and (1.3). So far, there are only few works on the Kuramoto model with frustration [2,12,23]. Recently, the work [16] investigated the emergent property for a finite-N particle model with frustration |α|<π2, and Ha and his collaborators studied the stability and instability of incoherent solution in [11,18], if initial data are sufficient regular. However, As far as the authors know, there are no results concerning all measurable stationary solutions and their stability, if the initial datum is just a Radon measure for system (1.3). Thus, in this paper, we address the following questions:

    ● (Q1): Are there measurable stationary solutions for the K-S equation (1.3)?

    ● (Q2): If there exists a stationary solution, are they nonlinearly stable?

    The purpose of this paper is to answer the above questions. First, we discuss the first question (Q1). Mirollo and Strogatz [24] presented some special stationary solutions in the absence of frustration. In contrast, when frustration effect is present, many tricks employed in the Kuramoto model do not work mainly due to the non-conservation of the total phase (see Lemma 2.1 for details). Hence, we can not expect genuine stationary solutions. For this, we define a new variable ˜θ(θ,t):

    {t˜θ(t,θ)=T×RV[fω]g(ω)dfωdω=κR2(t)sinα,(t,θ)R+×T,˜θ(0,θ)=θ.

    By studying a new equation formulated in terms of ˜θ, we present all measurable stationary solutions for the reformulated equation. Second, we study a nonlinear stability of stationary solutions in more general case. Let M(T×R) be the set of non-negative Radon measures on T×R. Then, for a Radon measure μM(T×R), we use the standard duality relation:

    μ,h=T×Rh(θ,ω)μ(dθdω),for anyhC0(T×R),

    where C0 denotes the set of smooth functions vanishing at infinity.

    Consider the stability of stationary solutions arising from the Cauchy problem:

    {tμt+θ(V[μt]μt)=0,(t,θ,ω)R+×T×R,V[μt](θ,ω,t):=ω+κT×Rsin(θθ+α)μt(dθdω),μt|t=0=μ00,μ0(θ+2π)=μ0(θ),T×Rμt(dθdω)=1, (1.4)

    where μt(dθdω):=g(ω)dfωdω.

    We first study the stability of the complete phase-locked state (stationary solution with R=1, see Definition 2.2 for details), and we show that the complete phase-locked state is nonlinearly stable in space of Radon measures in which the size of initial phase diameter is smaller than π2|α| (see Theorem 2.3 for more details). On the other hand, we discuss the stability of partial phase-locked state. Let μt,νtCw([0,T];M(T×R)) be measurable solutions to system (1.4) (see Definition 2.4 for details). One of difficulty that we confront with is the lack of exponential stability (see [3]):

    Wp(μt,νt)ectWp(μ0,ν0),

    where Wp(μt,νt) is the p-Wassertein distance between two measures μt and νt. In fact, let ϕi,i=1,2 be the pseudo-inverse functions associated to μt and νt, denote Ri by the order parameters of ϕi respectively, then by Lemma 2.1 and proof of Theorem 5.1 in [3], ddtWp(μt,νt) is bounded by |κR21(t)sinακR22(t)sinα|, so we have no knowledge whether Wp(μt,νt) will converge to zero. For this, we introduce the characteristic function for the K-S system:

    {tΘ(t,θ,ω)=ω+κT×Rsin(ΘΘ+α)μ(dθdω),Θ(0,θ,ω)=θ,

    and consider the dynamics of the following quantity:

    D1θ(μt):=supΘ1,Θ2Bθ(t)(˙Θ1(t)˙Θ2(t)),

    where Bθ(t) is the orthogonal θ-projection of supp(μt), and then show

    D1θ(μt)0exponentially, as t

    which yields that limt(Θ1Θ2) exists and finite, that is, phase-locked state emerge asymptotically. Moreover, we obtain the semi-stability of the partially phase-locked state in space of Radon measures under suitable conditions (see Theorem 2.6 for details). Finally, we study the stability of incoherent state to system (1.4) for identical oscillators, i.e., the natural frequency is a constant, and without loss of generality, we set ω=0 for simplicity. In a large frustration regime, the order parameter R tends to zero, as time tends to infinity, i.e., the incoherent solution is stable, whereas in a small frustration regime, the order parameter will tends to zero if there exists a constant M such that μt(dθ)Mμe(dθ) with μe a renormalized measure satisfying μe(T)=1, otherwise, the probability distribution concentrates and forms a Dirac mass.

    The rest of this paper is organized as follows. In Section 2, we briefly present main results of this paper. In Section 3, we discuss measurable stationary solutions. We first verify that the incoherent state is unique, then we find out all phase-locked states. In Sections 4 and 5, we study the stability of phase-locked states and incoherent state by providing suitable frameworks respectively.

    In this section, we present sufficient frameworks and main results on the existence of stationary solutions and their stabilities for the kinetic K-S equation.

    In this subsection, we introduce the order parameters and present stationary solutions for the K-S equation (1.3). We first define the order parameters (R,ψ) as follows:

    Reiψ:=Rg(ω)(Teiθdfω)dω. (2.1)

    We divide both sides of relation (2.1) by eiψ, and separate the real and imaginary part to obtain

    R=Rg(ω)(Tcos(θψ)dfω)dω,0=Rg(ω)(Tsin(θψ)dfω)dω. (2.2)

    Moreover, we further divide both sides of relation (2.1) by ei(θα), and compare the real and imaginary parts of the resulting relation to get

    Rsin(ψθ+α)=Rg(ω)(Tsin(θθ+α)dfω)dω. (2.3)

    Lemma 2.1. Suppose the natural frequency distribution g=g(ω) is integrable. Then, one has

    T×RV[fω]g(ω)dfωdω=κR2sinα.

    Proof. We use (2.3) to rewrite the nonlocal velocity V as

    V[fω]=ωκRsin(θψα).

    Then we use (2.5) and (2.2) to get

    Rg(ω)(T(ωκRsin(θψα))dfω)dω=Tωg(ω)dωκRRg(ω)(Tsin(θψα)dfω)dω=κRcosαRg(ω)(Tsin(θψ)dfω)dω+κRsinαRg(ω)(Tcos(θψ)dfω)dω=κR2sinα.

    By Lemma 2.1, we can not get equilibria for system (1.3) unless R0. Thus, we can get a relative equilibrium which is an equilibrium in a rotating frame with a constant velocity. For this, we define

    d˜θdt=κR2sinα,t>0,˜θ(0)=θ,

    and consider fω(t,˜θ,ω) instead of fω(t,θ,ω). For simplicity, we still write θ for ˜θ. Then our original K-S equation (1.3) can be rewritten as

    {tfω+θ(V[fω]fω)=0,(t,θ,ω)R+×T×R,V[fω](θ,ω,t):=ωκRsin(θψα)κR2sinα. (2.4)

    Next, we recall several distinguished states in the following definition.

    Definition 2.2. Let f be a measurable stationary solution to (2.4), and R=Rf is a corresponding order parameter associated with f in (2.1).

    1. If R0, then f is called an incoherent state.

    2. If 0<R<1, then f is called a partial phase-locked state.

    3. If R1, then f is called the complete phase-locked state.

    Now we are ready to present our first main theorem as follows.

    Theorem 2.3. Suppose that the natural frequency distribution g=g(ω)L1(R) and

    suppg=[L,L],g(ω)=g(ω),Rg(ω)dω=1. (2.5)

    Then, the following assertions hold.

    1. The incoherent state of (2.4) and (2.5) is unique, and fω=μe is the normalized Lebesgue measure on T with μe(T)=1.

    2. If the phase-locked state of (2.4) and (2.5) exists, then

    fω={CωκR2sinαωκR2sinακRsin(θα),for|ωκR2sinα|>κR,(1η(ω))δθω+η(ω)δθω,for|ωκR2sinα|κR,

    and the following relations must hold:

    {Rsinα=1κR+κR(2πCωω)(g(ω+κR2sinα)g(ωκR2sinα))dω1κRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω,Rcosα=|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω,

    where 0η(ω)1 is a constant, Cs is given by

    Cs=12πs2(κR)2or12πs2(κR)2for any constant s,

    with signs determined by TCssκRsin(θα)=1, phases θωα[π2,π2] and θωα=π(θωα) are the roots of equation:

    sinx=ωκR2sinακR.

    3. If g(ω) is non-increasing on [0,L], then

    LκRκR2sin|α|.

    Remark 1. From the proof of Theorem 2.3, we will see that the same results hold for identical case except the uniqueness of the incoherent state.

    In this subsection, we present main results on the stability of complete phase-locked state and partial phase-locked state. Denote by Cw([0,T);M(T×R)) the space of weakly continuous time-dependent measures. First, we present a definition of measure-valued solution for (1.4) as follows.

    Definition 2.4 (Measure-valued solution). For T(0,], we say μtCw([0,T);M(T×R)) be a measure-valued solution to (1.4) with an initial Radon measure μ0M(T×R) if μt satisfies the following conditions:

    (ⅰ) μt,h is continuous as a function of time t, for any hC0(T×R).

    (ⅱ) For any hC0([0,T]×T×R),

    μt,h(t,,)μ0,h(0,,)=t0μs,sh+V[μ]θhds, (2.6)

    where V[μ](s,θ,ω) is defined by

    V[μ](s,θ,ω):=ω+κT×Rsin(θθ+α)μ(dθdω).

    Remark 2. Below, we provide a brief comment on measure-valued solutions.

    1. Recall that supp(μ) (the support of a measure μ) is the closure of the set consisting of all points (x,v)R2d such that μ(Br((x,v)))>0,r>0. For a finite measure with a compact support, we can use hC1(R2d) as a test function in (2.6). Thus, we choose h=1,ω in (2.6) to get

    μt,1=μ0,1=1,μt,ω=μ0,ω.

    2. Let (θi(t),ωi(t)) be a solution of following ODE system:

    dθidt=ωi+κNNi=1sin(θjθi+α),dωidt=0,t>0.

    Then, the empirical measure

    μNt=1NNi=1δθi(t)δωi(t),

    is a measure-valued solution to system (1.4).

    Now let μtCw([0,T];M(T×R)) be a measurable solution to system (1.4). We define Bθ(t) and Bω(t) be the orthogonal θ and ω-projections of suppμt, respectively, i.e.,

    Bθ(t):=Pθsuppμt={θT|(θ,ω)suppμt},Bω(t):=Pωsuppμt={ωR|(θ,ω)suppμt},Dθ(μt):=diamBθ(t),Dω(μt):=diamBω(t),θc(t):=1M(t)μt,θ,ωc(t):=1M(t)μt,ω,

    where M and diamA are given as follows.

    M(t)=μt,1anddiamA:=supx,yA|xy|for AR.

    We use Remark 2 to see

    M(t)=μt,1=μ0,1=1,ωc(t)=μt,ω=μ0,ω=ωc(0).

    For identical oscillators with g(ω)=δωc, we set

    μ(dθdω):=δθc(t)×δωc(0).

    Our second main result deals with the stability of complete phase-locked state (i.e. for identical oscillators, g(ω)=δωc(0)) as follows.

    Theorem 2.5. Suppose that the initial datum μ0M(T×R) satisfies

    Dθ(μ0)π2|α|,Dω(μ0)=0. (2.7)

    Then the measure-valued solution μt to system (1.4) satisfies

    Dθ(μt)Dθ(μ0)eκΛ0t,t0,limtd(μt,μ)=0exponential,

    where Λ0:=2πcos(12Dθ(μ0)+|α|) is a positive constant, and μ(dθdω)=δθc(t)×δωc(0). In particular, we obtain the stability of the complete phase-locked state in space of Radon measure-valued solution whose initial data satisfy (2.7).

    For nonidentical oscillators, a difficulty we need to overcome is that the exponential decay (see [3]):

    Wp(μt,νt)ectWp(μ0,ν0),

    where Wp(μt,νt) is the p-Wasserstein distance between two measures μt and νt. As discussed in the introduction, ddtWp(μt,νt) is bounded by |κR21(t)sinακR22(t)sinα| with Ri,i=1,2 the order parameter of the pseudo-inverse functions associated to μt and νt, so we have no knowledge if Wp(μt,νt) will converge to zero. For this, we study the characteristic function for system (1.4):

    {˙Θ(t,θ,ω)=ω+κT×Rsin(ΘΘ+α)μ(dθdω),t>0,Θ(t,θ,ω)|t=0=θ,

    and define

    D1θ(μt)=supΘ1,Θ2Bθ(t)(˙Θ1(t)˙Θ2(t)).

    Our third main result deals with the exponential stability of D1θ.

    Theorem 2.6. Suppose the initial measure μ0M(T×R) satisfies

    Dθ(μ0)π2|α|,0<Dω(μ0)<,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)sin|α|, (2.8)

    and let μt be a measure-valued solution to (1.4). Then, there exists a positive time t0 such that

    Dθ(μt)<D,D1θ(μt)D1θ(μt0)e12κcos(D+|α|)(tt0),for alltt0,

    where D(0,π2|α|) is the solution of

    sin(x+|α|)=sin(Dθ(μ0)+|α|).

    In particular, we obtain the semi-stability of the partial phase-locked state in space of Radon measure-valued solution whose initial data satisfy (2.8).

    In this subsection, we provide stability and instability estimates for the incoherent state. First, we discuss the small frustration case:

    tμt+θ(V[μt]μt)=0,(θ,t)T×R+,V[μt](θ,t)=κTsin(θθ+α)μt(dθ), (2.9)

    where |α|<π2. Our main results are as follows.

    Proposition 2.7. (Small frustration) Suppose frustration and initial datum satisfy

    π2<α<π2andT×Tln|sin(θθ2)|μ0(dθ)μ0(dθ)<,

    and let μt be a measure-valued solution for system (2.9). Then, the following assertions hold.

    1. If there exist a positive constant M such that μt(dθ)Mμe(dθ), then we have

    limtR(t)=0.

    2. If there does not exists a positive constant M such that μt(dθ)Mμe(dθ), then we have

    limtμt(dθ)μe(dθ)L=.

    Next, we consider a large frustration |α|π2 case. For this, we set ˆα:=απ2. Then, the original identical system can be rewritten as

    tμt+θ(C[μt]μt)=0,(t,θ,ω)R+×T×RC[μt]=κT×Rcos(θθ+ˆα)μt(dθdω). (2.10)

    Then, our last result is as follows.

    Proposition 2.8. (Large frustration) Suppose frustration and initial datum satisfy

    0<ˆαπandT×Tln|sin(θθ2)|μ0(dθ)μ0(dθ)<,

    and let μ be a measure-valued solution for system (2.10). Then we have

    limtR(t)=0.

    In this section, we look for all measurable stationary solutions for the kinetic K-S equation (2.4). Without loss of generality, we may assume that phase order parameter ψ=0.

    Note that the stationary solution of (2.4) satisfies the equation:

    θ(V[fω]fω)=0.

    Recall that the distribution ξ on T satisfying θξ=0 is equal to a constant ˜Cω multiples of normalized Lebesgue measure μe on T. In other words, dμe=12πdθ. Hence, the stationary solution for (2.4) should satisfy

    V[fω]fω=˜Cωμe (3.1)

    In the following proposition, we show that a unique incoherent state is given by μe.

    Proposition 3.1. The incoherent state for (2.4) and (2.5) is unique, and

    fω=μe,

    where μe is the normalized Lebesgue measure on T.

    Proof. Note that the incoherent solutions for (2.4) satisfies

    V[fω]fω=˜Cωμe.

    Since R=0, we have

    V[fω]=ω.

    Thus, we have

    fω=˜Cωωμe.

    On the other hand, note that

    1=Tdfω=˜CωωTdμe.

    Thus,

    ˜Cωω=1,i.e.,fω=μe.

    Next, we classify all the phase-locked states.

    Proposition 3.2. The phase-locked state for (2.4) and (2.5) is

    fω={CωκR2sinαωκR2sinακRsin(θα),for|ωκR2sinα|>κR,(1η(ω))δθω+η(ω)δθω,for|ωκR2sinα|κR, (3.2)

    where η(ω) is a positive constant, and C±ωκR2sinα is given by

    CωκR2sinα=12π(ωκR2sinα)2(κR)2or12π(ωκR2sinα)2(κR)2,

    with signs determined by

    TCωκR2sinαωκR2sinακRsin(θα)dθ=1,

    phases θωα[π2,π2], θωα:=π(θωα) are the roots of equation:

    sinx=ωκR2sinακR.

    Proof. For a proof, we divide the domain of the natural frequency ω into two cases:

    ω[κR(1+Rsinα),κR(1+Rsinα)];ω[κR(1+Rsinα),κR(1+Rsinα)].

    Case A. Suppose that

    ω[κR(1+Rsinα),κR(1+Rsinα)],i.e.,|ωκR2sinα|>κR.

    Then we use relation (2.4) and assumption ψ=0 to get

    V[fω]0.

    Hence, we use relation (3.1) to obtain

    dfω=˜Cω/2πωκR2sinακRsin(θα)dθ.

    We use formula Tdfω=1 to get

    T˜Cω/2πωκR2sinακRsin(θα)dθ=1.

    By direct calculation, one has

    ˜Cω=±(ωκR2sinα)2(κR)2.

    We set

    CωκR2sinα=12π(ωκR2sinα)2(κR)2or12π(ωκR2sinα)2(κR)2,

    with signs of CωκR2sinα determined by

    TCωκR2sinαωκR2sinακRsin(θα)dθ=1. (3.3)

    Hence, we have

    fω=12π(ωκR2sinα)2(κR)2|ωκR2sinακRsin(θα)|=CωκR2sinαωκR2sinακRsin(θα).

    Case B. Suppose that

    ω[κR(1+Rsinα),κR(1+Rsinα)],that is,|ωκR2sinα|κR.

    In this case, we claim:

    ˜Cω=0. (3.4)

    Proof of claim (3.4). Suppose not, then for θ{ωκR2sinακRsin(θα)=0}

    dfω=˜Cω/2πωκR2sinακRsin(θα)dθ,

    Then, there exists θ{ωκR2sinακRsin(θα)=0} such that dfω<0. This gives a contradiction since dfω0, then our claim holds. Hence, we have

    V[fω]fω=0.

    We use Tdfω=1 to get

    V[fω]=0for|ωκR2sinα|κR,

    i.e.,

    ωκR2sinακRsin(θα)=0for|ωκR2sinα|κR.

    Thus, we can obtain

    fω=(1η(ω))δθω+η(ω)δθω,

    where θωα[π2,π2] is the solution of equation:

    sin(θωα)=ωκR2sinακR,

    and θω=πθω+2α. Now we summarize the value of fω as follows:

    fω={CωκR2sinαωκR2sinακRsin(θα),|ωκR2sinα|>κR,(1η(ω))δθω+η(ω)δθω,|ωκR2sinα|κR,

    In Proposition 3.2, we have determined the ansatz for fω in terms of R. It is clear that fω should satisfy the following relations with respect to order parameters:

    R=Rg(ω)(Tcosθdfω)dω,0=Rg(ω)(Tsinθdfω)dω. (3.5)

    Lemma 3.3. The phase-locked state in (3.2) satisfies the following relations:

    Rsinα=1κR+κR(2πCωω)(g(ω+κR2sinα)g(ωκR2sinα))dω1κRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω,Rcosα=|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω,

    where Cs=±12πs2(κR)2 for any variable s, η(ω) is a constant satisfying 0η(ω)1.

    Proof. We split the proof into two cases:

    Either|ωκR2sinα|>κRor|ωκR2sinα|κR.

    Step A. We first estimate the integral:

    |ωκR2sinα|>κRg(ω)(Tcosθdfω)dω=|ωκR2sinα|>κRg(ω)CωκR2sinα(TcosθωκR2sinακRsin(θα)dθ)dω. (3.6)

    We use the relation (3.3) and

    Tcos(θα)ωκR2sinακRsin(θα)dθ =0

    to get

    CωκR2sinαTcosθωκR2sinακRsin(θα)dθ=CωκR2sinαTcos(θα)cosαsin(θα)sinαωκR2sinακRsin(θα)dθ=CωκR2sinαsinαTsin(θα)ωκR2sinακRsin(θα)dθ=CωκR2sinαsinα(ωκR2sinα)κR×T{1ωκR2sinακRsin(θα)1ωκR2sinα}dθ=2πCωκR2sinαsinακR(ωκR2sinα)sinακRTCωκR2sinαωκR2sinακRsin(θα)dθ=sinακR(2πCωκR2sinα(ωκR2sinα)), (3.7)

    where

    CωκR2sinα=12π(ωκR2sinα)2(κR)2or12π(ωκR2sinα)2(κR)2.

    In (3.6), we use the estimate (3.7) to see

    |ωκR2sinα|>κRg(ω)(Tcosθdfω)dω=|ωκR2sinα|>κRg(ω)sinακR[2πCωκR2sinα(ωκR2sinα)]dω=sinακR|ω|>κRg(ω+κR2sinα)(2πCωω)dω
    =sinακR+κRg(ω+κR2sinα)(2πCωω)dω+sinακRκRg(ω+κR2sinα)(2πCωω)dω=sinακR+κR(2πCωω)(g(ω+κR2sinα)g(ωκR2sinα))dω, (3.8)

    where we have used the relations:

    Cω=Cωandg(ω+κR2sinα)=g(ωκR2sinα).

    Similar to (3.8), we get

    |ωκR2sinα|>κRg(ω)(Tsinθdfω)dω=cosακR+κR(2πCω+ω)(g(ω+κR2sinα)g(ωκR2sinα))dω. (3.9)

    Step B. Next, we consider the integral:

    |ωκR2sinα|κRg(ω)(Tcosθdfω)dω=|ωκR2sinα|κRg(ω)((1η(ω))cosθω+η(ω)cosθω)dω. (3.10)

    By direct calculation, one has

    (1η(ω))cosθω+η(ω)cosθω=(1η(ω))(cos(θωα)cosαsin(θωα)sinα)+η(ω)(cos(θωα)cosαsin(θωα)sinα)=(1η(ω))(cosα1(ωκR2sinακR)2ωκR2sinακRsinα)+η(ω)(cosα1(ωκR2sinακR)2ωκR2sinακRsinα)=ωκR2sinακRsinα+(12η(ω))cosα1(ωκR2sinακR)2. (3.11)

    Note that

    sinακR|ωκR2sinα|κR(ωκR2sinα)g(ω)dω=sinακR|ω|κRωg(ω+κR2sinα)dω=sinακR{κR0ωg(ω+κR2sinα)dω+0κRωg(ω+κR2sinα)dω}=sinακRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω. (3.12)

    In (3.10), we combine estimates (3.11) and (3.12) to obtain

    |ωκR2sinα|κRg(ω)(Tcosθdfω)dω=sinακRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω+cosα|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω. (3.13)

    Similarly, we get

    |ωκR2sinα|κRg(ω)(Tsinθdfω)dω=cosακRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω+sinα|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω. (3.14)

    Now we substitute estimates (3.8), (3.9), (3.13) and (3.14) into (3.5) to obtain

    R=sinακR+κR(2πCωω)(g(ω+κR2sinα)g(ωκR2sinα))dωsinακRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω+cosα|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω, (3.15)

    and

    0=cosακR+κR(2πCω+ω)(g(ω+κR2sinα)g(ωκR2sinα))dω+cosακRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω+sinα|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω. (3.16)

    We multiply relation (3.15) by sinα, and multiply relation (3.16) by cosα, and then take the difference of the two resulting relations to derive

    Rsinα=1κR+κR(2πCωω)(g(ω+κR2sinα)g(ωκR2sinα))dω1κRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω. (3.17)

    Next, we substitute relation (3.17) into (3.15) to get

    R=Rsin2α+cosα|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω.

    Since α(π2,π2), we know cosα0. Thus, we have

    Rcosα=|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω.

    Lemma 3.4. If the phase-locked state for (2.4) and (2.5) exists, and g(ω) is non-increasing on [0,L], then the upper bound L of natural frequency should satisfy

    LκRκR2sin|α|.

    Proof. Recall that Lemma 3.3 yields

    Rsinα=2πκR+κRCω(g(ω+κR2sinα)g(ωκR2sinα))dω1κR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω.

    By direct calculation, we have

    +0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω=12+ω[g(ω+κR2sinα)g(ωκR2sinα)]dω=κR2sinα+g(ω)dω=κR2sinα.

    Thus, we have

    0=2πκR+κRCω(g(ω+κR2sinα)g(ωκR2sinα))dω. (3.18)

    Next, we will prove only α0 case. The other case α<0 can be treated similarly. Suppose that

    L>κRκR2sinα.

    Then, we will derive a contradiction by ruling out the following two cases:

    Case A (κR+κR2sinα). In this case, we analyze R.H.S term of relation (3.18) by using the non-increasing property of g on [0,+) to get

    +κRCω(g(ω+κR2sinα)g(ωκR2sinα))dω=κR2sinακRCω(g(ω+κR2sinα)g(ωκR2sinα))dω+κR2sinακR2sinαCωg(ωκR2sinα)dω+κR2sinακR2sinαCωg(ωκR2sinα)dω<0.

    This contradicts to relation (3.18).

    Case B (κRκR2sinα<<κR+κR2sinα). Note that

    +κRCω(g(ω+κR2sinα)g(ωκR2sinα))dω=+κR2sinακRCωg(ωκR2sinα)dω<0.

    This also contradicts to relation (3.18). Hence, we derived the desired estimate.

    Proof of Theorem 2.3. First, Proposition 3.1 gives (1) of Theorem 2.3 directly. Next, Proposition 3.2 shows if the phase-locked state exists, then

    fω={CωκR2sinαωκR2sinακRsin(θα),for|ωκR2sinα|>κR,(1η(ω))δθω+η(ω)δθω,for|ωκR2sinα|κR,

    where 0η(ω) is a constant, C±ωκR2sinα is given by

    CωκR2sinα=12π(ωκR2sinα)2(κR)2or12π(ωκR2sinα)2(κR)2,

    with signs determined by

    TCωκR2sinαωκR2sinακRsin(θα)dθ=1,

    and phases θωα[π2,π2], θωα:=π(θωα) are the roots of equation:

    sinx=ωκR2sinακR.

    Furthermore, Lemmas 3.3 and 3.4 show that the following relations must hold:

    {Rsinα=1κR+κR(2πCωω)(g(ω+κR2sinα)g(ωκR2sinα))dω,1κRκR0ω[g(ω+κR2sinα)g(ωκR2sinα)]dω,Rcosα=|ωκR2sinα|κR(12η(ω))g(ω)1(ωκR2sinακR)2dω,LκRκR2sin|α|.

    In this section, we study stability estimates of the phase-locked states for equation (1.4), i.e., stability of the complete phase-locked state and partial phase-locked state, respectively.

    We first derive a global existence of measure-valued solution for the corresponding mean-field model.

    Recall that the order parameters R and ψ can be redefined as follows.

    Reiψ:=T×Reiθμ(dθdω). (4.1)

    Then as in Section 2, one has

    R=μt,cos(θψ)=T×Rcos(θψ)μ(dθdω),0=μt,sin(θψ)=T×Rsin(θψ)μ(dθdω), (4.2)

    and

    V[μ](θ,ω,t)=ωκRsin(θψα).

    Thus, equation (1.4) can be rewritten as

    ddtμt+θ(V[μt]μt)=0,V[μt](θ,ω,t):=ωκRsin(θψα). (4.3)

    In this subsection, we list the emergent estimates for the Kuramoto-Sakaguchi model for later use. Since the methodology for proofs is similar to the arguments given in [12,13,23], we leave detailed proofs in Appendix A. Consider the following N-particle Kuramoto model with frustration:

    ˙θi=ωi+κNNj=1sin(θjθi+α),t>0,|α|<π2, (4.4)

    and for a given phase vector Θ=(θ1,,θN), we define

    θM:=max1iNθi,θm:=min1iNθi,D(Θ):=θMθm,D(ω):=max1i,jN|ωiωj|. (4.5)

    Below, we state two asymptotic phase-locking for identical and non-identical ensembles.

    Proposition 4.1. The following assertions hold.

    1. Suppos natural frequencies, coupling strength and initial data satisfy

    ωi=0,1iN,κ>0,D(Θin)<π2|α|,

    and let θi be a solution to (A.1). Then, we have exponential synchronization:

    D(Θ(t))D(Θin)exp[2κπcos(12D(Θin)+|α|)t],fort0.

    2. Suppose natural frequencies, coupling strength and initial data satisfy

    0<D(Θin)<π2|α|,0<D(Ω)<,κ>κe:=D(Ω)sin(D(Θin)+|α|)sin|α|.

    Then, we have

    D(Θ(t))<D,for anyt>t0:=D(Θin)D(1κκe)D(Ω),

    where D(0,π2|α|) is the root of the following trigonometric equation:

    sin(x+|α|)=sin(D(Θin)+|α|).

    Proof. We leave its proof in Appendix A.

    Next, we briefly study a global existence of measure-valued solution for system (1.4) and its property.

    Theorem 4.2. For any μ0M(T×R), let μt be a unique measure-valued solution to system (1.4) with the initial data μ0. Then μt can be approximated by a sequence of empirical measures:

    μNt=1NNi=1δθi(t)δωi(t).

    Furthermore, one has

    d(μt,μNt)0,asN.

    Proof. Since the proof is nearly the same as in [3] using the N-particle theory in Section 4.1.1, we omit its proof.

    Lemma 4.3. Suppose the initial measure satisfies

    μ0,ω=0,

    and let μt be a measure-valued solution of system (1.4). Then, for t0, one has

    μt,θ=μ0,θ+t0κR2sinαds.

    Proof. We take h=θ in relation (2.6) and use system (4.3) to get

    μt,θ=μ0,θ+t0μs,V[μ]ds=μ0,θ+t0μs,ωκRsin(θψα)ds.

    Note that assumption on initial datum and Remark 2 (1) yield

    μt,ω=μ0,ω=0.

    Thus, we use relation (4.2) to obtain

    μt,θ=μ0,θ+t0κRμs,sin(θψα)ds=μ0,θt0κRcosαμs,sin(θψ)ds+t0κRsinαμs,cos(θψ)ds=μ0,θ+t0κR2sinαds.

    In this subsection, we study the stability estimate of the complete phase-locked state. Let μtC([0,T];M(T×R)) be a measurable weak solution to system (1.4).

    Recall that

    Bθ(t):=Pθsuppμt={θT|(θ,ω)suppμt},Bω(t):=Pωsuppμt={ωR|(θ,ω)suppμt},Dθ(μt):=diamBθ(t),Dω(μt):=diamBω(t),θc(t):=1M(t)μt,θ,ωc(t):=1M(t)μt,ω,

    where

    M(t)=μt,1=μ0,1=1,anddiamA:=supx,yA|xy|.

    Since Remark 2 (1) gives that μt,ω=μ0,ω, one has

    Bω(t)=Bω(0),t0.

    Note that for identical oscillators, without loss of generality, we may assume ωc(0)=0. Thus,

    ωc(t)=0,t0.

    Now we use Theorem 4.2 and Proposition 4.1 to get an exponential decay of Dθ(μt).

    Lemma 4.4. Suppose the initial datum μ0M(T×R) satisfies

    Dθ(μ0)π2|α|,Dω(μ0)=0,

    and let μt be a measure-valued solution to system (1.4). Then, there exists a positive constant Λ0:=2πcos(12Dθ(μ0)+|α|) such that

    Dθ(μt)Dθ(μ0)eκΛ0t,t0.

    Proof. We define μN0 as in reference [3]. Note that Proposition 4.1 gives that the approximate measure valued solution μNtM(T×R) satisfies

    Dθ(μNt)Dθ(μN0)eκΛ0t,t0.

    Now we use Theorem 4.2:

    d(μ,μNt)0as N

    to see

    Dθ(μNt)Dθ(μt)asN.

    Hence, our desired stability estimate is obtained.

    Now, we set

    μ(dθdω):=δθc(t)×δωc(0).

    Theorem 4.5. Suppose the initial datum μ0M(T×R) satisfies

    Dθ(μ0)π2|α|,Dω(μ0)=0,

    and let μt be a measure-valued solution to system (1.4). Then, one has

    limtd(μt,μ)=0exponentially.

    Proof. Let hC(T) be an arbitrary test function satisfying

    h1andhLip1.

    Then we have

    |T×Rh(θ)μt(dθ,dω)T×Rh(θ)μ(dθ,dω)|=|Th(θ)ˉμt(dθ)h(θc)|T|θθc|ˉμt(dθ)Dθ(μ0)ec0κt,

    where

    ˉμt(dθ):=Rμt(dθ,dω).

    We use Theorem 4.2 to conclude that

    d(μt,μ)0exponentially,ast.

    As a corollary of Theorem 4.5, we obtain the exponential stability of the complete phase-locked state in the space of Radon measure-valued solutions.

    Corollary 4.6. Suppose that the initial datum μ0M(T×R) satisfies

    Dθ(μ0)π2|α|,Dω(μ0)=0, (4.6)

    and let μt be a measure-valued solution to system (1.4). Then, μt is asymptotically phase-locked. In particular, we obtain the stability of the complete phase-locked state in the space of Radon measure solution with initial datum satisfying (4.6).

    Proof. Let Θ01 and Θ02 be initial measures satisfying

    |Θ01Θ02|π2|α|.

    Then, it follows from Theorem 4.5 that

    Θ1Θ20exponentially,

    where ˙Θ(t,θ,ω)=κT×Rsin(ΘΘ+α)μ(dθdω) is the characteristic function.

    In this subsection, we study the nonlinear stability of partial phase-locked state to the K-S equation (1.4).

    Consider the characteristic function defined by the following system:

    {˙Θ(t,θ,ω)=ω+κT×Rsin(ΘΘ+α)μ(dθdω),Θ(t,θ,ω)|t=0=θ.

    In this case, we set

    Φ(t,θ,ω):=˙Θ(t,θ,ω).

    Then, the new variable Φ(t) satisfies

    ˙Φ(t,θ,ω)=κT×Rcos(ΘΘ+α)(ΦΦ)μ(dθdω). (4.7)

    We also define

    D1θ(μt)=supΘ1,Θ2Bθ(t)(˙Θ1(t)˙Θ2(t)):=supΘ1,Θ2Bθ(t)(Φ1(t)Φ2(t)).

    Note that 0D1θ(μt)<. In fact, for any Θ(t)Bθ(t), we use the compactness of g in Lemma 3.4 to have

    |Φ|=|˙Θ|L+κ<.

    In the sequel, we will prove that the quantity D1θ(μt) tends to zero exponentially fast as t.

    Lemma 4.7. Let μ0M(T×R) be a given initial measure such that

    0<Dθ(μ0)π2|α|,0<Dω(μ0)<,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)sin|α|,

    and let μt be a measure-valued solution to equation (1.4) with an initial datum μ0. Then, there exists a time t0 such that

    Dθ(μt)<D,for allt>t0,

    where D(0,π2|α|) is the solution of

    sin(x+|α|)=sin(Dθ(μ0)+|α|).

    Proof. For given N>0, we consider the approximation μN0 for μ0:

    μN0=1NNi=1δθ0δω0.

    We solve the Cauchy problem for the following N-particle system:

    {dθidt=ωi+κNNj=1sin(θjθi+α),dωidt=0,

    with the initial data (θ0i,ω0i). We use Theorem 4.2 to obtain

    d(μt,μNt)0,asN.

    Furthermore, Proposition 4.1 shows that there exists time t0:

    t0:=Dθ(μN0)D,NDω(μN0)κ(sin(Dθ(μN0)+|α|)sin|α|)),

    with D,N+|α|=arcsin(Dθ(μN0)+|α|)(0,π2) such that

    Dθ(μNt)<D,N,for allt>tN0,

    for N large enough. Now let N tends to infinity to obtain the desired results.

    Theorem 4.8. Let μ0M(T×R) be a given initial measure such that

    0<Dθ(μ0)π2|α|,0<Dω(μ0)<,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)sin|α|,

    and let μt be a measure valued solution to equation (1.4). Then there exists a positive time t0 such that

    D1θ(μt)D1θ(μt0)e12κcos(D+|α|)(tt0),for alltt0.

    Proof. For t>t0 and any ε(0,18), we take any ΘM,εBM,ε and Θm,εBm,ε such that

    ΦΦM,εμ(dθdω)εandΦΦm,εμ(dθdω)ε.

    Then, we have

    ddt(ΦM,εΦm,ε)=ΦΦM,ε+ΦΦm,ε+Φm,εΦΦM,ε=:J11+J12+J13.

    Below, we estimate J1i,i=1,2,3 one by one.

    Case A (Estimate on J11). It follows from relation (4.7) and Lemma 4.7 that

    J11=κΦΦM,ε[cos(ΘΘM,ε+α)(ΦΦM,ε)cos(ΘΘm,ε+α)(ΦΦm,ε)]μ(dθdω)κΦΦM,εcos(ΘΘM,ε+α)(ΦΦM,ε)μ(dθdω)κεD1θ(μt).

    Case B (Estimate on J12). Similar to Case A, we get

    J12=κΦΦm,ε[cos(ΘΘM,ε+α)(ΦΦM,ε)cos(ΘΘm,ε+α)(ΦΦm,ε)]μ(dθdω)κΦΦM,εcos(ΘΘm,ε+α)(ΦΦm,ε)μ(dθdω)κεD1θ(μt).

    Case C (Estimate on J13). We use Lemma 4.7 to obtain

    J13=κΦm,εΦΦM,ε[cos(ΘΘM,ε+α)(ΦΦM,ε)cos(ΘΘm,ε+α)(ΦΦm,ε)]μ(dθdω)κcos(D+|α|)Φm,εΦΦM,ε[(ΦΦM,ε)(ΦΦm,ε)]μ(dθdω)=κcos(D+|α|)Φm,εΦΦM,ε(ΦM,εΦm,ε)μ(dθdω)κ(12ε)(1ε)cos(D+|α|)D1θ(μt).

    Now, we combine all estimates to derive

    ddt(ΦM,εΦm,ε)κcosD+|α|[1(3+2cos(D+|α|)2ε)ε]D1θ(μt).

    Thus, we can derive

    ddtD1θ(μt)supΦM,ε,Φm,εddt(ΦM,εΦm,ε)12κcos(D+|α|)D1θ(μt).

    Finally, we use Gronwall's lemma to get

    D1θ(μt)D1θ(μt0)e12κ(cosD)(tt0),for alltt0.

    Corollary 4.9. Let μ0M(T×R) be an initial measure satisfying

    0<Dθ(μ0)π2|α|,0<Dω(μ0)<,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)sin|α|.

    Then, the measure-valued solution μt to equation (1.4) is asymptotically phase-locked. In particular, we obtain the semi-stability of the partially phase-locked state in space of Radon measure solution whose initial data satisfy (2.8).

    Proof. Note that Theorem 4.8 yields

    limtt0|sΘ1sΘ2|ds<.

    Then we have

    limt(Θ1(t)Θ2(t))=(Θ01Θ02)+limtt0(sΘ1sΘ2)ds<.

    This means that for any Θ1 and Θ2,

    limt(Θ1(t)Θ2(t))exists and finite.

    Now we use Theorem 4.8 and definition of order parameters to get

    ωκRsin(Θψα)κR2sinα0exponentially fast.

    We set

    R:=limtRandψ:=limtψ.

    Then, we can get

    limtsin(Θψα)=ωκ(R)2sinακR.

    Hence, for any Θ(0) satisfy assumption (2.8), Θ(t) approaches to an equilibrium described in (2) of Theorem 2.3. In particular, the partially phase-locked state we obtained in Section 3 is semi-stable.

    In this section, we study the stability of incoherent solution to the K-S equation with a frustration for identical oscillators. For this, we define a new quantity:

    I(t):=μt,ln|sin(θθ2)|=T×Tln|sin(θθ2)|μt(dθ)μt(dθ).

    Lemma 5.1. Let μt:=μe2π be a uniform distribution on S1 such that

    μe(T)=1ordμe=12πdθ.

    Then we can have

    T×Tln|sin(θθ2)|μt(dθ)μt(dθ)=ln2.

    Proof. By direct estimate, we have

    T×Tln|sin(θθ2)|μt(dθ)μt(dθ)=1(2π)2T×Tln|sin(θθ2)|μe(dθ)μe(dθ)=1(2π)2T×Tln|sin(θ2)|μe(dθ)μe(dθ)=12πTln|sin(θ2)|μe(dθ)=1ππ0ln(sinθ)μe(dθ)=ln2.

    By Lemma 5.1, we can see that I(t) works well (at least for the uniform distribution on S1). As a corollary of Lemma 5.1, we have upper and lower bound estimates for I.

    Lemma 5.2. Suppose there exists positive constants m and M such that

    mμe(dθ)μt(dθ)Mμe(dθ)for all θT. (5.1)

    Then, the quantity I(t) satisfies

    4π2M2ln2I(t)4π2m2ln2,t0.

    Proof. We use the assumptions (5.1) to get that for all t0,

    m2T×Tln|sin(θθ2)|μe(dθ)μe(dθ)I(t)M2T×Tln|sin(θθ2)|μe(dθ)μe(dθ).

    Note that

    T×Tln|sin(θθ2)|μe(dθ)μe(dθ)=4π2ln2.

    Thus, we have

    4π2M2ln2I(t)4π2m2ln2,t0.

    In this subsection, we consider small frustration case with α(π2,π2). Similar to system (4.3), we use order parameters to rewrite equation (2.9) as

    ddtμt+θ{V[μt]μt(dθ)}=0,V[μt]=κRsin(θψα). (5.2)

    We differentiate both sides of (4.1) with respect to t and use equation (5.2) to get

    eiψ[˙R(t)+iR(t)˙ψ(t)]=Teiθtμt(dθ)=Teiθθ{κRsin(θψα)μt(dθ)}=iTeiθκRcos(θψα)μ(dθ). (5.3)

    Now we divide both sides of relation (5.3) by eiψ, and compare the real and imaginary parts to obtain

    ˙R=κRTsin(θψ)sin(θψα)μt(dθ),R˙ψ=κRT×Rcos(θψ)sin(θψα)μt(dθ). (5.4)

    Lemma 5.3. Let μtM(T×R) be a measure-valued solution for equation (2.9). Then, for all t0,

    ddtI(t)=κR2cosαandd2dt2I(t)=2κ2R2cosαTsin(θψ)sin(θψα)μt(dθ).

    Proof. (i) We use equation (5.2) to obtain

    ddtI(t)=T×Tln|sin(θθ2)|μt(dθ)ddtμt(dθ)+T×Tln|sin(θθ2)|μt(dθ)ddtμt(dθ)=κRT×Tln|sin(θθ2)|θ(sin(θψα)μt(dθ))μt(dθ)+κRT×Tln|sin(θθ2)|θ(sin(θψα)μt(dθ))μt(dθ).

    Integrations by parts yield

    ddtI(t)=κRT×T(sin(θψα)θln|sin(θθ2)|+sin(θψα)θln|sin(θθ2)|)μt(dθ)μt(dθ)=κRT×T(sin(θψα)sin(θψα))θln|sin(θθ2)|μt(dθ)μt(dθ).

    By direct calculation, one has

    sin(θψα)sin(θψα)=2cos(θ+θ2ψ2α)sin(θθ2).

    Now we use the above estimates and relation:

    θln|sin(θθ2)|=cos(θθ2)2sin(θθ2)

    to obtain

    ddtI(t)=κRT×Tcos(θ+θ2ψ2α)sin(θθ2)cos(θθ2)sin(θθ2)μt(dθ)μt(dθ)=κRcosαT×Tcos(θ+θ2ψ2)cos(θθ2)μt(dθ)μt(dθ)κRsinαT×Tsin(θ+θ2ψ2)cos(θθ2)μt(dθ)μt(dθ)=:J21+J22.

    We use relation (4.2) to derive

    T×Tcos(θ+θ2ψ2)cos(θθ2)μt(dθ)μt(dθ)=T×T(cos(θψ)+cos(θψ))μt(dθ)μt(dθ)=R,T×Tsin(θ+θ2ψ2)cos(θθ2)μt(dθ)μt(dθ)=12T×T(sin(θψ)+sin(θψ))μt(dθ)μt(dθ)=0.

    Thus, we have

    J21=κR2cosα,J22=0. (5.5)

    This yields

    ddtI(t)=κR2cosα,t>0. (5.6)

    (ii) Now we differentiate relation (5.6) with respect to time t to get

    d2dt2I(t)=2κcosαR˙R,t>0.

    Hence, we use relation (5.4) to obtain

    d2dt2I(t)=2κ2R2cosαTsin(θψ)sin(θψα)μt(dθ),t>0.

    Proposition 5.4. Suppose the frustration and initial datum satisfy

    α(π2,π2)andT×Tln|sin(θθ2)|μ0(dθ)μ0(dθ)<,

    and let μ be a measure-valued solution for equation (2.9).

    1. If there exist constant M such that μ(dθ)Mμe(dθ), then we have

    limtR(t)=0.

    2. If not, we have

    limtμt(dθ)μe(dθ)L=.

    Proof. (1) We use Lemma 5.3 to see that ddtI(t) is uniformly continuous and I(t) is decreasing. Now if there exist a constant M such that μ(dθ)Mμe(dθ), we can use Lemma 5.2 to get

    I(t)4π2M2ln2.

    Hence, we deduce

    limtI(t)exists.

    Together with initial assumption

    I(0):=T×Tln|sin(θθ2)|μ0(dθ)μ0(dθ)<,

    we obtain

    limtt0ddsI(s)ds=limtI(t)I(0)exists.

    Thus, we can use Barbalat's Lemma to conclude

    ddtI(t)0,ast.

    Now we use

    ddtI(t)=κR2cosαand|α|<π2,

    to see

    limtR(t)=0.

    (2) If we can not find a positive constant M such that μ(dθ)Mμe(dθ), then we obtain

    limtI(t)=.

    Furthermore, we can see

    I(t)=T×Tln|sin(θθ2)|μt(dθ)μt(dθ)μt(dθ)μe(dθ)2LT×Tln|sin(θθ2)|μe(dθ)μe(dθ)=ln2μt(dθ)μe(dθ)2L.

    Thus, we can deduce

    μt(dθ)μe(dθ)L,ast.

    In this subsection, we consider a large frustration case (|α|π2). For this, we define ˆα=απ2. Then, the original K-S equation becomes

    tμt+θ(C[μt]μt)=0,(t,θ,ω)R+×T×R,C[μt]=κT×Rcos(θθ+ˆα)μt(dθdω). (5.7)

    Note that the assumption on α and definition of ˆα, we have ˆα[0,π], and since α[π2,π], it is easy to see that ˆα[0,π2]. As α(π,π2], we have

    ˆα(3π2,π]=(π2,π].

    We divide both sides of relation (4.1) by ei(θˆα) and take the real part of the resulting relation to have

    Rcos(θψˆα)=T×Rcos(θθ+ˆα)μt(dθdω). (5.8)

    We use relation (5.8) to rewrite the equation (5.7) to rewrite as

    tμt+θ(C[μt]μt)=0,C[μt]=κRcos(θψˆα). (5.9)

    We differentiate both sides of (4.1) with respect to t and use system (5.9) to get

    eiψ[˙R(t)+iR(t)˙ψ(t)]=T×Reiθddtμt(dθdω)=T×Reiθθ{κRcos(θψˆα)μt(dθdω)}=iT×ReiθκRcos(θψˆα)μt(dθdω).

    Now, we divide both sides of above relation by eiψ and separate the real and imaginary parts to obtain

    ˙R(t)=κRT×Rsin(θψ)cos(θψˆα)μt(dθdω),R(t)˙ψ(t)=κRT×Rcos(θψ)cos(θψˆα)μt(dθdω). (5.10)

    Now, we define the first phase moment m1(t) as follows.

    m1(t):=Tθμt(dθ).

    Lemma 5.5. Let μtM(T×R) be a measure-valued solution for equation (2.10). Then one has

    (i)ddtI(t)=κR2sinˆα,ddtm1(t)=κR2cosˆα.(ii)d2dt2I(t)=2κ2R2sinˆαTsin(θψ)cos(θψˆα)μt(dθ).

    Proof. (i) We use relation (5.9) to obtain that

    ddtI(t)=T×Tln|sin(θθ2)|μt(dθ)ddtμt(dθ)+T×Tln|sin(θθ2)|μt(dθ)ddtμt(dθ)=κRT×Tln|sin(θθ2)|θ(cos(θψˆα)μt(dθ))μt(dθ)κRT×Tln|sin(θθ2)|θ(cos(θψˆα)μt(dθ))μt(dθ).

    We use integration by parts to get

    ddtI(t)=κRT×T(cos(θψˆα)θln|sin(θθ2)|+cos(θψˆα)θln|sin(θθ2)|)μt(dθ)μt(dθ)=κRT×T(cos(θψˆα)cos(θψˆα))θln|sin(θθ2)|μt(dθ)μt(dθ). (5.11)

    By direct calculation, one has

    cos(θψˆα)cos(θψˆα)=2sin(θ+θ2ψ2ˆα)sin(θθ2). (5.12)

    Now, we use (5.11), (5.12) and relation

    θln|sin(θθ2)|=cos(θθ2)2sin(θθ2)

    to obtain

    ddtI(t)=κRT×Tsin(θ+θ2ψ2ˆα)sin(θθ2)cos(θθ2)sin(θθ2)μt(dθ)μt(dθ)=κRcosˆαT×Tsin(θ+θ2ψ2)cos(θθ2)μt(dθ)μt(dθ)+κRsinˆαT×Tcos(θ+θ2ψ2)cos(θθ2)μt(dθ)μt(dθ)=:J31+J32.

    Similar to the derivation of (5.5) in Lemma 4.7, we have

    J31=0,J32=κR2sinˆα.

    Hence, we conclude

    ddtI(t)=κR2sinˆα,t>0. (5.13)

    We use relation (5.9) to obtain

    ddtm1(t)=Tθddtμ(dθ)=κRTθθ(cos(θψˆα)μt)=κRTcos(θψˆα)μt(dθ)=κRcosˆαTcos(θψ)μt(dθ)+κRsinˆαTsin(θψ)μt(dθ)=κR2cosˆα.

    (ii) Now we differentiate relation (5.13) with respect to time t to get

    d2dt2I(t)=2κR˙Rsinˆα,t>0.

    We use relation (5.10) to deduce

    d2dt2I(t)=2κ2R2sinˆαTsin(θψ)cos(θψˆα)μt(dθ),t>0.

    Corollary 5.6. Let μtM(T×R) be a measure-valued solution for equation (2.10). Then, the quantity cosˆαI(t)sinˆαm1(t) is invariant with respect to time t:

    cosˆαI(t)sinˆαm1(t)=cosˆαI(0)sinˆαm1(0),for allt0.

    Proof. We combine estimate (i) of Lemma 5.5 (i) to get

    cosˆαddtm1(t)I(t)sinˆαddtm1(t)=0,t>0.

    This yields our desired estimate.

    Remark 3. For all t0, for ˆα=0, Lemma 5.5 gives

    I(t)=I(0),

    and for ˆα=π2, Lemma 5.5 gives m1(t)=m1(0).

    Proposition 5.7. Suppose frustration and initial datum satisfy

    ˆα(0,π]andT×Tln|sin(θθ2)|μ0(dθ)dμ0(dθ)<,

    and let μtM(T×R) be a measure-valued solution for equation (2.10). Then we have

    limtR(t)=0.

    Proof. It follows from Lemma 5.5 that

    ddtI(t)=κR2sinˆα,d2dt2I(t)=2κ2R2sinˆαTsin(θψ)cos(θψˆα)μt(dθ).

    This yields that ddtI(t) is uniformly continuous. Furthermore, we use the assumption on μ0 to get

    limtt0ddsI(s)ds=limtI(t)I(0)I(0)<.

    Thus, we apply Barbalat's Lemma to conclude

    limtddtI(t)=0. (5.14)

    Now we use the relation ddtI(t)=κR2sinˆα, ˆα(0,π) and (5.14) to obtain

    limtR(t)=0.

    In this appendix, we provide proofs for Proposition 4.1 on the emergent dynamics of the Kuramoto-Sakaguchi equation.

    Consider an ensemble of identical oscillators. Without loss of generality, we may assume

    ωi=0,i=1,,N.

    In this case, the Kuramoto model becomes

    ˙θi=κNNj=1sin(θjθi+α),t>0,|α|<π2. (A.1)

    Lemma A.1. (Phase coherence) Suppose the coupling strength and initial data satisfy

    κ>0,D(Θin)<π2|α|,

    and let {θi} be a solution to (A.1). Then, the following relations hold:

    sup0t<D(Θ(t))π2|α|,sup0t<D(Θ(t))D(Θin).

    Proof. (ⅰ) The first relation can be obtained from Lemma 3.1 in [12].

    (ⅱ) Note that the phase diameter D(Θ) satisfies

    ddtD(Θ)=ddt(θMθm)=κN{Nj=1sin(θjθM+α)Nj=1sin(θjθm+α)}0,

    where we have used the following relations:

    |θjθM+α|D(Θ)+|α|π|α|,|θjθm+α|D(Θ)+|α|π|α|.

    Now, we are ready to provide the first estimate in Proposition 4.1.

    Let Θ be a solution to system (A.1). By definition of D(Θ), one has

    ddtD(θ)=κNNj=1{sin(θjθM+α)sin(θjθm+α)}=2κNNj=1cos(θjθM2+θjθm2+α)sin(θMθm2). (A.2)

    Since

    θjθM2+αθjθM2+θjθm2+αθjθm2+α,

    one has

    θjθM2+α(12D(Θin)|α|,|α|)(π2,π2),θjθm2+α(|α|,12D(Θin)+|α|)(π2,π2). (A.3)

    Then, we use (A.3) to get

    θjθM2+θjθm2+α(12D(Θin)|α|,12D(Θin)+|α|)(π2,π2).

    This yields

    cos(θjθM2+θjθm2+α)cos(12D(Θin)+|α|). (A.4)

    Now we substitute (A.4) into (A.2) to derive a differential inequality:

    dD(Θ)dt2κNcos(12D(Θin)+|α|)Nj=1sin(θMθm2). (A.5)

    Since θMθm2(0,π), one has

    sin(θMθm2)2πD(Θ)2=D(Θ)π. (A.6)

    Now we combine (A.5) and (A.6) to obtain

    dD(Θ)dt2κπcos(12D(Θin)+|α|)D(Θ). (A.7)

    We integrate the differential inequality (A.7) with respect to time t to get

    D(Θ(t))D(Θin)exp{2κπcos(12D(Θin)+|α|)t},fort0.

    In this subsection, we consider the non-identical Kuramoto-Sakaguchi equation (4.4). We find the exact time t0 such that all the particles trapped in half circle after time t0, which is important for asymptotic phase-locking of non-identical oscillators in Section 4.3.

    Lemma A.2. Suppose initial data, natural frequencies and coupling strength satisfy

    0<D(Θin)<π2|α|,0<D(Ω)<,κ>κe:=D(Ω)sin(D(Θin)+|α|)sin|α|.

    Then we have

    sup0t<D(Θ(t))<D(Θin).

    Proof. We will use the continuity argument. For this, we define

    T:=sup{T|D(Θ(t))<D(Θin),for allt(0,T]}.

    We claim:

    T=+. (A.8)

    Suppose not, i.e., T<+. Then there exists t0 such that

    D(Θ(t))<D(Θin),for allt(0,t0)andD(Θ)(t0)=D(Θin). (A.9)

    Then we use continuity of D(Θ)(t) to see

    ddt|t=t0D(Θ)>0. (A.10)

    On the other hand, note that the system (4.4) can be rewritten as

    ˙θi=ωi+κcosαNNj=1sin(θjθi)+κsinαNNj=1cos(θjθi),t>0,|α|<π2.

    Hence, we use definition of D(Θ) in (4.5) to obtain

    ddtD(Θ)D(Ω)+κcosαNNj=1(sin(θjθM)sin(θjθm))+κsinαNNj=1(cos(θjθM)cos(θjθm)). (A.11)

    We use (A.9) to see that for t(0,t0)

    cos(θjθM)cos(θjθm)1cosD(Θ), (A.12)

    and

    sin(θMθj)θMθj>sinD(Θ)D(Θ)andsin(θjθm)θjθm>sinD(Θ)D(Θ),t(0,t0).

    Hence, for t(0,t0), one has

    sin(θjθM)sin(θjθm)<sinD(Θ)D(Θ)((θjθM)(θjθm))=sinD(Θ). (A.13)

    Now we substitute (A.12) and (A.13) into (A.11) to obtain

    dD(Θ)dtD(Ω)κcosαsinD(Θ)+κsin|α|(1cosD(Θ))=D(Ω)κ(sin(D(Θ)+|α|)sin|α|),t(0,t0). (A.14)

    Thus, we use (A.14) to get

    ddt|t=t0D(Θ)D(Ω)κ(sin(D(Θ)(t0)+|α|)sin|α|)<D(Ω)D(Ω)sin(D(Θin)+|α|)sin|α|(sin(D(Θ)(t0)+|α|)sin|α|)=0.

    This contradicts (A.10). Thus, we deduce that T=+, and our assertion holds.

    Now, we are ready to provide a proof of the second part.

    Suppose initial data, natural frequencies and coupling strength satisfy

    0<D(Θin)<π2|α|,0<D(Ω)<,κ>κe:=D(Ω)sin(D(Θin)+|α|)sin|α|.

    Then we claim: for all t>t0=D(Θin)D(1κκe)D(Ω),

    D(Θ(t))<D, (A.15)

    where D(0,π2|α|) is a root of the equation:

    sin(x+|α|)=sin(D(Θin)+|α|).

    We split the proof of claim (A.15) into two steps.

    Step A. We need to find time t0 such that

    D(Θ(t0))<D.

    For this, we consider two cases.

    Case I. Suppose

    0<D(Θin)π2.

    Then, one has

    D=D(Θin).

    Thus, it follows from Lemma A.2 that we have the desired estimate (A.15).

    Case II. Suppose

    D(Θin)(π2,π).

    Then, it follows from Lemma A.2 that

    D(Θ(t))<D(Θin)for all t>0.

    Now we claim:

    ddtD(Θ)<0,for a.e. t such thatD(Θ(t))(D,D(Θin)). (A.16)

    Proof of Claim (A.16). Since (D+|α|,D(θ0)+|α|)(|α|,π|α|), we have

    sin(D(Θ)+|α|)sin(D+α)=sin(D(Θin)+|α|). (A.17)

    We use relation (A.11), relation (A.17) and assumption of κ to get

    ddtD(Θ)D(Ω)κ(sin(D(Θ)+|α|)sin|α|)D(Ω)κ(sin(D(Θin)+|α|)sin|α|)<D(Ω)D(Ω)sin(D(Θ0)+|α|)sin|α|(sin(D(Θin)+|α|)sin|α|)=0.

    In fact, we can get

    ddtD(Θ)D(Ω)κ(sin(D(Θin)+|α|)sin|α|)=D(Ω)κκeD(Ω)=(1κκe)D(Ω).

    Hence, we have

    D(Θ(t))<D(Θin)+(1κκe)D(Ω)t,t>0.

    Thus, in order to have D(Θ(t))<D, we need

    t>te:=D(Θin)D(1κκe)D(Ω).

    Step B. We need to verify

    D(Θ(t))<D,for allt>t0.

    Suppose that there exists a time t1>t0 such that

    D(Θ(t))<Dfort(t0,t1)D(Θ(t1))=D.

    This yields

    ddt|t=t1D(Θ)>0. (A.18)

    However, we use relation (A.11) again to obtain that

    ddt|t=t1D(Θ)D(Ω)κ(sin(D+|α|)sin|α|)=D(Ω)κ(sin(D(Θin)+|α|)sin|α|)<D(Ω)D(Ω)sin(D(Θin)+|α|)sin|α|(sin(D(Θin)+|α|)sin|α|)=0.

    This contradicts to the relation (A.18). Thus, our assertion holds.



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