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
[1] | Seung-Yeal Ha, Hansol Park, Yinglong Zhang . Nonlinear stability of stationary solutions to the Kuramoto-Sakaguchi equation with frustration. Networks and Heterogeneous Media, 2020, 15(3): 427-461. doi: 10.3934/nhm.2020026 |
[2] | Hirotada Honda . Global-in-time solution and stability of Kuramoto-Sakaguchi equation under non-local Coupling. Networks and Heterogeneous Media, 2017, 12(1): 25-57. doi: 10.3934/nhm.2017002 |
[3] | Hirotada Honda . On Kuramoto-Sakaguchi-type Fokker-Planck equation with delay. Networks and Heterogeneous Media, 2024, 19(1): 1-23. doi: 10.3934/nhm.2024001 |
[4] | Seung-Yeal Ha, Yongduck Kim, Zhuchun Li . Asymptotic synchronous behavior of Kuramoto type models with frustrations. Networks and Heterogeneous Media, 2014, 9(1): 33-64. doi: 10.3934/nhm.2014.9.33 |
[5] | Tingting Zhu . Synchronization of the generalized Kuramoto model with time delay and frustration. Networks and Heterogeneous Media, 2023, 18(4): 1772-1798. doi: 10.3934/nhm.2023077 |
[6] | Xiaoxue Zhao, Zhuchun Li . Synchronization of a Kuramoto-like model for power grids with frustration. Networks and Heterogeneous Media, 2020, 15(3): 543-553. doi: 10.3934/nhm.2020030 |
[7] | Young-Pil Choi, Seung-Yeal Ha, Seok-Bae Yun . Global existence and asymptotic behavior of measure valued solutions to the kinetic Kuramoto--Daido model with inertia. Networks and Heterogeneous Media, 2013, 8(4): 943-968. doi: 10.3934/nhm.2013.8.943 |
[8] | Seung-Yeal Ha, Jeongho Kim, Jinyeong Park, Xiongtao Zhang . Uniform stability and mean-field limit for the augmented Kuramoto model. Networks and Heterogeneous Media, 2018, 13(2): 297-322. doi: 10.3934/nhm.2018013 |
[9] | Tingting Zhu . Emergence of synchronization in Kuramoto model with frustration under general network topology. Networks and Heterogeneous Media, 2022, 17(2): 255-291. doi: 10.3934/nhm.2022005 |
[10] | Seung-Yeal Ha, Shi Jin, Jinwook Jung . A local sensitivity analysis for the kinetic Kuramoto equation with random inputs. Networks and Heterogeneous Media, 2019, 14(2): 317-340. doi: 10.3934/nhm.2019013 |
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=ωi+κNN∑ℓ=1sin(θℓ−θi+α),i=1,⋯,N, | (1.1) |
where
Note that the K-S model (1.1) can be rewritten as
˙θi=ωi+κcosαNN∑ℓ=1sin(θℓ−θi)+κsinαNN∑ℓ=1cos(θℓ−θi). | (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
F(t,θ,ω)=f(t,θ,ω)g(ω),(t,θ,ω)∈R+×T×R, |
where
{∂∂tfω+∂θ(V[fω]fω)=0,(t,θ,ω)∈R+×T×R,V[fω](θ,ω,t):=ω+κ∫T×Rsin(θ∗−θ+α)g(ω∗)dfω∗dω∗,∫Tdfω=1, | (1.3) |
where differential operator
⟨∂θ(V[fω]fω),ϕ⟩=−∫Tϕ′V[fω]dfω. |
In the absence of frustration
N∑j=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
● (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
{∂∂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
⟨μ,h⟩=∫T×Rh(θ,ω)μ(dθdω),for anyh∈C∞0(T×R), |
where
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+κ∫T×Rsin(θ∗−θ+α)μt(dθ∗dω∗),μt|t=0=μ0≥0,μ0(θ+2π)=μ0(θ),∫T×Rμt(dθdω)=1, | (1.4) |
where
We first study the stability of the complete phase-locked state (stationary solution with
Wp(μt,νt)≤e−ctWp(μ0,ν0), |
where
{∂∂tΘ(t,θ,ω)=ω+κ∫T×Rsin(Θ∗−Θ+α)μ(dθ∗dω∗),Θ(0,θ,ω)=θ, |
and consider the dynamics of the following quantity:
D1θ(μt):=supΘ1,Θ2∈Bθ(t)(˙Θ1(t)−˙Θ2(t)), |
where
D1θ(μt)→0exponentially, as t→∞ |
which yields that
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
Reiψ:=∫Rg(ω)(∫Teiθdfω)dω. | (2.1) |
We divide both sides of relation (2.1) by
R=∫Rg(ω)(∫Tcos(θ−ψ)dfω)dω,0=∫Rg(ω)(∫Tsin(θ−ψ)dfω)dω. | (2.2) |
Moreover, we further divide both sides of relation (2.1) by
Rsin(ψ−θ+α)=∫Rg(ω)(∫Tsin(θ∗−θ+α)df∗ω)dω. | (2.3) |
Lemma 2.1. Suppose the natural frequency distribution
∫T×RV[fω]g(ω)dfωdω=κR2sinα. |
Proof. We use (2.3) to rewrite the nonlocal velocity
V[fω]=ω−κRsin(θ−ψ−α). |
Then we use (2.5) and (2.2) to get
∫Rg(ω)(∫T(ω−κRsin(θ−ψ−α))dfω)dω=∫Tωg(ω)dω−κR∫Rg(ω)(∫Tsin(θ−ψ−α)dfω)dω=−κRcosα∫Rg(ω)(∫Tsin(θ−ψ)dfω)dω+κRsinα∫Rg(ω)(∫Tcos(θ−ψ)dfω)dω=κR2sinα. |
By Lemma 2.1, we can not get equilibria for system (1.3) unless
d˜θdt=−κR2sinα,t>0,˜θ(0)=θ, |
and consider
{∂∂tfω+∂θ(V[fω]fω)=0,(t,θ,ω)∈R+×T×R,V[fω](θ,ω,t):=ω−κRsin(θ−ψ−α)−κR2sinα. | (2.4) |
Next, we recall several distinguished states in the following definition.
Definition 2.2. Let
1. If
2. If
3. If
Now we are ready to present our first main theorem as follows.
Theorem 2.3. Suppose that the natural frequency distribution
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
2. If the phase-locked state of (2.4) and (2.5) exists, then
fω={Cω−κR2sinαω−κR2sinα−κRsin(θ−α),for|ω−κR2sinα|>κR,(1−η(ω))δθω+η(ω)δθ∗ω,for|ω−κR2sinα|≤κR, |
and the following relations must hold:
{Rsinα=1κR∫+∞κR(2πCω−ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω−1κR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω,Rcosα=∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω, |
where
Cs=12π√s2−(κR)2or−12π√s2−(κR)2for any constant s, |
with signs determined by
sinx=ω−κR2sinακR. |
3. If
L≤κR−κR2sin|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
Definition 2.4 (Measure-valued solution). For
(ⅰ)
(ⅱ) For any
⟨μt,h(t,⋅,⋅)⟩−⟨μ0,h(0,⋅,⋅)⟩=∫t0⟨μs,∂sh+V[μ]∂θh⟩ds, | (2.6) |
where
V[μ](s,θ,ω):=ω+κ∫T×Rsin(θ∗−θ+α)μ(dθ∗dω∗). |
Remark 2. Below, we provide a brief comment on measure-valued solutions.
1. Recall that
⟨μt,1⟩=⟨μ0,1⟩=1,⟨μt,ω⟩=⟨μ0,ω⟩. |
2. Let
dθidt=ωi+κNN∑i=1sin(θj−θi+α),dωidt=0,t>0. |
Then, the empirical measure
μNt=1NN∑i=1δθi(t)⊗δωi(t), |
is a measure-valued solution to system (1.4).
Now let
Bθ(t):=Pθsuppμt={θ∈T|(θ,ω)∈suppμt},Bω(t):=Pωsuppμt={ω∈R|(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⟩anddiamA:=supx,y∈A|x−y|for A⊆R. |
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
μ∞(dθdω):=δθc(t)×δωc(0). |
Our second main result deals with the stability of complete phase-locked state (i.e. for identical oscillators,
Theorem 2.5. Suppose that the initial datum
Dθ(μ0)≤π−2|α|,Dω(μ0)=0. | (2.7) |
Then the measure-valued solution
Dθ(μt)≤Dθ(μ0)e−κΛ0t,t≥0,limt→∞d(μt,μ∞)=0exponential, |
where
For nonidentical oscillators, a difficulty we need to overcome is that the exponential decay (see [3]):
Wp(μt,νt)≤e−ctWp(μ0,ν0), |
where
{˙Θ(t,θ,ω)=ω+κ∫T×Rsin(Θ∗−Θ+α)μ(dθ∗dω∗),t>0,Θ(t,θ,ω)|t=0=θ, |
and define
D1θ(μt)=supΘ1,Θ2∈Bθ(t)(˙Θ1(t)−˙Θ2(t)). |
Our third main result deals with the exponential stability of
Theorem 2.6. Suppose the initial measure
Dθ(μ0)≤π−2|α|,0<Dω(μ0)<∞,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)−sin|α|, | (2.8) |
and let
Dθ(μt)<D∞,D1θ(μt)≤D1θ(μt0)e−12κcos(D∞+|α|)(t−t0),for allt≥t0, |
where
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)=κ∫Tsin(θ∗−θ+α)μt(dθ∗), | (2.9) |
where
Proposition 2.7. (Small frustration) Suppose frustration and initial datum satisfy
−π2<α<π2and∫T×Tln|sin(θ−θ∗2)|μ0(dθ)μ0(dθ∗)<∞, |
and let
1. If there exist a positive constant
limt→∞R(t)=0. |
2. If there does not exists a positive constant
limt→∞‖μt(dθ)μe(dθ)‖L∞=∞. |
Next, we consider a large frustration
∂∂tμt+∂θ(C[μt]μt)=0,(t,θ,ω)∈R+×T×RC[μt]=κ∫T×Rcos(θ∗−θ+ˆα)μt(dθ∗dω∗). | (2.10) |
Then, our last result is as follows.
Proposition 2.8. (Large frustration) Suppose frustration and initial datum satisfy
0<ˆα≤πand∫T×Tln|sin(θ−θ∗2)|μ0(dθ)μ0(dθ∗)<∞, |
and let
limt→∞R(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
Note that the stationary solution of (2.4) satisfies the equation:
∂θ(V[fω]fω)=0. |
Recall that the distribution
V[fω]fω=˜Cωμe | (3.1) |
In the following proposition, we show that a unique incoherent state is given by
Proposition 3.1. The incoherent state for (2.4) and (2.5) is unique, and
fω=μe, |
where
Proof. Note that the incoherent solutions for (2.4) satisfies
V[fω]fω=˜Cωμe. |
Since
V[fω]=ω. |
Thus, we have
fω=˜Cωωμe. |
On the other hand, note that
1=∫Tdfω=˜Cωω∫Tdμe. |
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={Cω−κR2sinαω−κR2sinα−κRsin(θ−α),for|ω−κR2sinα|>κR,(1−η(ω))δθω+η(ω)δθ∗ω,for|ω−κR2sinα|≤κR, | (3.2) |
where
Cω−κR2sinα=12π√(ω−κR2sinα)2−(κR)2or−12π√(ω−κR2sinα)2−(κR)2, |
with signs determined by
∫TCω−κR2sinαω−κR2sinα−κRsin(θ−α)dθ=1, |
phases
sinx=ω−κR2sinακR. |
Proof. For a proof, we divide the domain of the natural frequency
ω∉[κR(−1+Rsinα),κR(1+Rsinα)];ω∈[κR(−1+Rsinα),κR(1+Rsinα)]. |
ω∉[κR(−1+Rsinα),κR(1+Rsinα)],i.e.,|ω−κR2sinα|>κR. |
Then we use relation (2.4) and assumption
V[fω]≠0. |
Hence, we use relation (3.1) to obtain
dfω=˜Cω/2πω−κR2sinα−κRsin(θ−α)dθ. |
We use formula
∫T˜Cω/2πω−κR2sinα−κRsin(θ−α)dθ=1. |
By direct calculation, one has
˜Cω=±√(ω−κR2sinα)2−(κR)2. |
We set
Cω−κR2sinα=12π√(ω−κR2sinα)2−(κR)2or−12π√(ω−κR2sinα)2−(κR)2, |
with signs of
∫TCω−κR2sinαω−κR2sinα−κRsin(θ−α)dθ=1. | (3.3) |
Hence, we have
fω=12π√(ω−κR2sinα)2−(κR)2|ω−κR2sinα−κRsin(θ−α)|=Cω−κR2sinαω−κR2sinα−κRsin(θ−α). |
ω∈[κR(−1+Rsinα),κR(1+Rsinα)],that is,|ω−κR2sinα|≤κR. |
In this case, we claim:
˜Cω=0. | (3.4) |
Proof of claim (3.4). Suppose not, then for
dfω=˜Cω/2πω−κR2sinα−κRsin(θ−α)dθ, |
Then, there exists
V[fω]fω=0. |
We use
V[fω]=0for|ω−κR2sinα|≤κR, |
i.e.,
ω−κR2sinα−κRsin(θ−α)=0for|ω−κR2sinα|≤κR. |
Thus, we can obtain
fω=(1−η(ω))δθω+η(ω)δθ∗ω, |
where
sin(θω−α)=ω−κR2sinακR, |
and
fω={Cω−κR2sinαω−κR2sinα−κRsin(θ−α),|ω−κR2sinα|>κR,(1−η(ω))δθω+η(ω)δθ∗ω,|ω−κR2sinα|≤κR, |
In Proposition 3.2, we have determined the ansatz for
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(2πCω−ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω−1κR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω,Rcosα=∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω, |
where
Proof. We split the proof into two cases:
Either|ω−κR2sinα|>κRor|ω−κR2sinα|≤κR. |
∫|ω−κR2sinα|>κRg(ω)(∫Tcosθdfω)dω=∫|ω−κR2sinα|>κRg(ω)Cω−κR2sinα(∫Tcosθω−κR2sinα−κRsin(θ−α)dθ)dω. | (3.6) |
We use the relation (3.3) and
∫Tcos(θ−α)ω−κR2sinα−κRsin(θ−α)dθ =0 |
to get
Cω−κR2sinα∫Tcosθω−κR2sinα−κRsin(θ−α)dθ=Cω−κR2sinα∫Tcos(θ−α)cosα−sin(θ−α)sinαω−κR2sinα−κRsin(θ−α)dθ=−Cω−κR2sinαsinα∫Tsin(θ−α)ω−κR2sinα−κRsin(θ−α)dθ=−Cω−κR2sinαsinα(ω−κR2sinα)κR×∫T{1ω−κR2sinα−κRsin(θ−α)−1ω−κR2sinα}dθ=2πCω−κR2sinαsinακR−(ω−κR2sinα)sinακR∫TCω−κR2sinαω−κR2sinα−κRsin(θ−α)dθ=sinακR(2πCω−κR2sinα−(ω−κR2sinα)), | (3.7) |
where
Cω−κR2sinα=12π√(ω−κR2sinα)2−(κR)2or−12π√(ω−κR2sinα)2−(κR)2. |
In (3.6), we use the estimate (3.7) to see
∫|ω−κR2sinα|>κRg(ω)(∫Tcosθdfω)dω=∫|ω−κR2sinα|>κRg(ω)sinακR[2πCω−κR2sinα−(ω−κR2sinα)]dω=sinακR∫|ω|>κRg(ω+κR2sinα)(2πCω−ω)dω |
=sinακR∫+∞κRg(ω+κR2sinα)(2πCω−ω)dω+sinακR∫−κR−∞g(ω+κR2sinα)(2πCω−ω)dω=sinακR∫+∞κR(2πCω−ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω, | (3.8) |
where we have used the relations:
C−ω=−Cωandg(−ω+κR2sinα)=g(ω−κR2sinα). |
Similar to (3.8), we get
∫|ω−κR2sinα|>κRg(ω)(∫Tsinθdfω)dω=cosακR∫+∞κR(−2πCω+ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω. | (3.9) |
∫|ω−κR2sinα|≤κRg(ω)(∫Tcosθdfω)dω=∫|ω−κR2sinα|≤κRg(ω)((1−η(ω))cosθω+η(ω)cosθ∗ω)dω. | (3.10) |
By direct calculation, one has
(1−η(ω))cosθω+η(ω)cosθ∗ω=(1−η(ω))(cos(θω−α)cosα−sin(θω−α)sinα)+η(ω)(cos(θ∗ω−α)cosα−sin(θ∗ω−α)sinα)=(1−η(ω))(cosα√1−(ω−κR2sinακR)2−ω−κR2sinακRsinα)+η(ω)(−cosα√1−(ω−κR2sinακR)2−ω−κR2sinακRsinα)=−ω−κR2sinακRsinα+(1−2η(ω))cosα√1−(ω−κR2sinακR)2. | (3.11) |
Note that
−sinακR∫|ω−κR2sinα|≤κR(ω−κR2sinα)g(ω)dω=−sinακR∫|ω|≤κRωg(ω+κR2sinα)dω=−sinακR{∫κR0ωg(ω+κR2sinα)dω+∫0−κRωg(ω+κR2sinα)dω}=−sinακR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω. | (3.12) |
In (3.10), we combine estimates (3.11) and (3.12) to obtain
∫|ω−κR2sinα|≤κRg(ω)(∫Tcosθdfω)dω=−sinακR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω+cosα∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω. | (3.13) |
Similarly, we get
∫|ω−κR2sinα|≤κRg(ω)(∫Tsinθdfω)dω=cosακR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω+sinα∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω. | (3.14) |
Now we substitute estimates (3.8), (3.9), (3.13) and (3.14) into (3.5) to obtain
R=sinακR∫+∞κR(2πCω−ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω−sinακR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω+cosα∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω, | (3.15) |
and
0=cosακR∫+∞κR(−2πCω+ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω+cosακR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω+sinα∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω. | (3.16) |
We multiply relation (3.15) by
Rsinα=1κR∫+∞κR(2πCω−ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω−1κR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω. | (3.17) |
Next, we substitute relation (3.17) into (3.15) to get
R=Rsin2α+cosα∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω. |
Since
Rcosα=∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω. |
Lemma 3.4. If the phase-locked state for (2.4) and (2.5) exists, and
L≤κR−κR2sin|α|. |
Proof. Recall that Lemma 3.3 yields
Rsinα=2πκR∫+∞κRCω(g(ω+κR2sinα)−g(ω−κR2sinα))dω−1κR∫∞0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω. |
By direct calculation, we have
∫+∞0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω=12∫+∞−∞ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω=−κR2sinα∫+∞−∞g(ω)dω=−κR2sinα. |
Thus, we have
0=2πκR∫+∞κRCω(g(ω+κR2sinα)−g(ω−κR2sinα))dω. | (3.18) |
Next, we will prove only
L>κR−κR2sinα. |
Then, we will derive a contradiction by ruling out the following two cases:
∫+∞κRCω(g(ω+κR2sinα)−g(ω−κR2sinα))dω=∫ℓ−κR2sinακRCω(g(ω+κR2sinα)−g(ω−κR2sinα))dω−∫ℓ+κR2sinαℓ−κR2sinαCωg(ω−κR2sinα)dω≤−∫ℓ+κR2sinαℓ−κR2sinαCωg(ω−κR2sinα)dω<0. |
This contradicts to relation (3.18).
∫+∞κRCω(g(ω+κR2sinα)−g(ω−κR2sinα))dω=−∫ℓ+κR2sinακRCωg(ω−κR2sinα)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
fω={Cω−κR2sinαω−κR2sinα−κRsin(θ−α),for|ω−κR2sinα|>κR,(1−η(ω))δθω+η(ω)δθ∗ω,for|ω−κR2sinα|≤κR, |
where
Cω−κR2sinα=12π√(ω−κR2sinα)2−(κR)2or−12π√(ω−κR2sinα)2−(κR)2, |
with signs determined by
∫TCω−κR2sinαω−κR2sinα−κRsin(θ−α)dθ=1, |
and phases
sinx=ω−κR2sinακR. |
Furthermore, Lemmas 3.3 and 3.4 show that the following relations must hold:
{Rsinα=1κR∫+∞κR(2πCω−ω)(g(ω+κR2sinα)−g(ω−κR2sinα))dω,−1κR∫κR0ω[g(ω+κR2sinα)−g(ω−κR2sinα)]dω,Rcosα=∫|ω−κR2sinα|≤κR(1−2η(ω))g(ω)√1−(ω−κR2sinακR)2dω,L≤κR−κR2sin|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
Reiψ:=∫T×Reiθμ(dθdω). | (4.1) |
Then as in Section 2, one has
R=⟨μt,cos(θ−ψ)⟩=∫T×Rcos(θ−ψ)μ(dθdω),0=⟨μt,sin(θ−ψ)⟩=∫T×Rsin(θ−ψ)μ(dθdω), | (4.2) |
and
V[μ](θ,ω,t)=ω−κRsin(θ−ψ−α). |
Thus, equation (1.4) can be rewritten as
ddtμt+∂θ(V[μt]μt)=0,V[μt](θ,ω,t):=ω−κRsin(θ−ψ−α). | (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+κNN∑j=1sin(θj−θi+α),t>0,|α|<π2, | (4.4) |
and for a given phase vector
θM:=max1≤i≤Nθi,θm:=min1≤i≤Nθi,D(Θ):=θM−θm,D(ω):=max1≤i,j≤N|ωi−ωj|. | (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=0,1≤i≤N,κ>0,D(Θin)<π−2|α|, |
and let
D(Θ(t))≤D(Θin)exp[−2κπcos(12D(Θin)+|α|)t],fort≥0. |
2. Suppose natural frequencies, coupling strength and initial data satisfy
0<D(Θin)<π−2|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∞(1−κκe)D(Ω), |
where
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
μNt=1NN∑i=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,θ⟩=⟨μ0,θ⟩+∫t0κR2sinαds. |
Proof. We take
⟨μt,θ⟩=⟨μ0,θ⟩+∫t0⟨μs,V[μ]⟩ds=⟨μ0,θ⟩+∫t0⟨μs,ω−κRsin(θ−ψ−α)⟩ds. |
Note that assumption on initial datum and Remark 2
⟨μt,ω⟩=⟨μ0,ω⟩=0. |
Thus, we use relation (4.2) to obtain
⟨μt,θ⟩=⟨μ0,θ⟩+∫t0κR⟨μs,−sin(θ−ψ−α)⟩ds=⟨μ0,θ⟩−∫t0κRcosα⟨μs,sin(θ−ψ)⟩ds+∫t0κRsinα⟨μs,cos(θ−ψ)⟩ds=⟨μ0,θ⟩+∫t0κR2sinαds. |
In this subsection, we study the stability estimate of the complete phase-locked state. Let
Recall that
Bθ(t):=Pθsuppμt={θ∈T|(θ,ω)∈suppμt},Bω(t):=Pωsuppμt={ω∈R|(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,y∈A|x−y|. |
Since Remark 2
Bω(t)=Bω(0),t≥0. |
Note that for identical oscillators, without loss of generality, we may assume
ωc(t)=0,t≥0. |
Now we use Theorem 4.2 and Proposition 4.1 to get an exponential decay of
Lemma 4.4. Suppose the initial datum
Dθ(μ0)≤π−2|α|,Dω(μ0)=0, |
and let
Dθ(μt)≤Dθ(μ0)e−κΛ0t,t≥0. |
Proof. We define
Dθ(μNt)≤Dθ(μN0)e−κΛ0t,t≥0. |
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
Dθ(μ0)≤π−2|α|,Dω(μ0)=0, |
and let
limt→∞d(μt,μ∞)=0exponentially. |
Proof. Let
‖h‖∞≤1and‖h‖Lip≤1. |
Then we have
|∫T×Rh(θ)μt(dθ,dω)−∫T×Rh(θ)μ∞(dθ,dω)|=|∫Th(θ)ˉμt(dθ)−h(θc)|≤∫T|θ−θc|ˉμt(dθ)≤Dθ(μ0)e−c0κt, |
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
Dθ(μ0)≤π−2|α|,Dω(μ0)=0, | (4.6) |
and let
Proof. Let
|Θ01−Θ02|≤π−2|α|. |
Then, it follows from Theorem 4.5 that
Θ1−Θ2→0exponentially, |
where
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+κ∫T×Rsin(Θ∗−Θ+α)μ(dθ∗dω∗),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,θ,ω)=κ∫T×Rcos(Θ∗−Θ+α)(Φ∗−Φ)μ(dθ∗dω∗). | (4.7) |
We also define
D1θ(μt)=supΘ1,Θ2∈Bθ(t)(˙Θ1(t)−˙Θ2(t)):=supΘ1,Θ2∈Bθ(t)(Φ1(t)−Φ2(t)). |
Note that
|Φ|=|˙Θ|≤L+κ<∞. |
In the sequel, we will prove that the quantity
Lemma 4.7. Let
0<Dθ(μ0)≤π−2|α|,0<Dω(μ0)<∞,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)−sin|α|, |
and let
Dθ(μt)<D∞,for allt>t0, |
where
sin(x+|α|)=sin(Dθ(μ0)+|α|). |
Proof. For given
μN0=1NN∑i=1δθ0⊗δω0. |
We solve the Cauchy problem for the following N-particle system:
{dθidt=ωi+κNN∑j=1sin(θj−θi+α),dωidt=0, |
with the initial data
d(μt,μNt)→0,asN→∞. |
Furthermore, Proposition 4.1 shows that there exists time
t0:=Dθ(μN0)−D∞,NDω(μN0)−κ(sin(Dθ(μN0)+|α|)−sin|α|)), |
with
Dθ(μNt)<D∞,N,for allt>tN0, |
for
Theorem 4.8. Let
0<Dθ(μ0)≤π−2|α|,0<Dω(μ0)<∞,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)−sin|α|, |
and let
D1θ(μt)≤D1θ(μt0)e−12κcos(D∞+|α|)(t−t0),for allt≥t0. |
Proof. For
∫Φ≥ΦM,εμ(dθ∗dω∗)≤εand∫Φ≤Φm,εμ(dθ∗dω∗)≤ε. |
Then, we have
ddt(ΦM,ε−Φm,ε)=∫Φ∗≥ΦM,ε+∫Φ∗≤Φm,ε+∫Φm,ε≤Φ∗≤ΦM,ε=:J11+J12+J13. |
Below, we estimate
J11=κ∫Φ∗≥ΦM,ε[cos(Θ∗−ΘM,ε+α)(Φ∗−ΦM,ε)−cos(Θ∗−Θm,ε+α)(Φ∗−Φm,ε)]μ(dθ∗dω∗)≤κ∫Φ∗≥ΦM,εcos(Θ∗−ΘM,ε+α)(Φ∗−ΦM,ε)μ(dθ∗dω∗)≤κεD1θ(μt). |
J12=κ∬Φ∗≤Φm,ε[cos(Θ∗−ΘM,ε+α)(Φ∗−ΦM,ε)−cos(Θ∗−Θm,ε+α)(Φ∗−Φm,ε)]μ(dθ∗dω∗)≤κ∬Φ∗≥ΦM,ε−cos(Θ∗−Θm,ε+α)(Φ∗−Φm,ε)μ(dθ∗dω∗)≤κεD1θ(μt). |
J13=κ∫Φm,ε≤Φ∗≤ΦM,ε[cos(Θ∗−ΘM,ε+α)(Φ∗−ΦM,ε)−cos(Θ∗−Θm,ε+α)(Φ∗−Φm,ε)]μ(dθ∗dω∗)≤κcos(D∞+|α|)∫Φm,ε≤Φ∗≤ΦM,ε[(Φ∗−ΦM,ε)−(Φ∗−Φm,ε)]μ(dθ∗dω∗)=−κcos(D∞+|α|)∫Φm,ε≤Φ∗≤ΦM,ε(ΦM,ε−Φm,ε)μ(dθ∗dω∗)≤−κ(1−2ε)(1−ε)cos(D∞+|α|)D1θ(μt). |
Now, we combine all estimates to derive
ddt(ΦM,ε−Φm,ε)≤−κcosD∞+|α|[1−(3+2cos(D∞+|α|)−2ε)ε]D1θ(μt). |
Thus, we can derive
ddtD1θ(μt)≤supΦM,ε,Φm,εddt(ΦM,ε−Φm,ε)≤−12κcos(D∞+|α|)D1θ(μt). |
Finally, we use Gronwall's lemma to get
D1θ(μt)≤D1θ(μt0)e−12κ(cosD∞)(t−t0),for allt≥t0. |
Corollary 4.9. Let
0<Dθ(μ0)≤π−2|α|,0<Dω(μ0)<∞,κ>κe:=Dω(μ0)sin(Dθ(μ0)+|α|)−sin|α|. |
Then, the measure-valued solution
Proof. Note that Theorem 4.8 yields
limt→∞∫t0|∂sΘ1−∂sΘ2|ds<∞. |
Then we have
limt→∞(Θ1(t)−Θ2(t))=(Θ01−Θ02)+limt→∞∫t0(∂sΘ1−∂sΘ2)ds<∞. |
This means that for any
limt→∞(Θ1(t)−Θ2(t))exists and finite. |
Now we use Theorem 4.8 and definition of order parameters to get
ω−κRsin(Θ−ψ−α)−κR2sinα→0exponentially fast. |
We set
R∞:=limt→∞Randψ∞:=limt→∞ψ. |
Then, we can get
limt→∞sin(Θ∞−ψ∞−α)=ω−κ(R∞)2sinακR∞. |
Hence, for any
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(θ−θ∗2)|⟩=∫T×Tln|sin(θ−θ∗2)|μt(dθ)μt(dθ∗). |
Lemma 5.1. Let
μe(T)=1ordμe=12πdθ. |
Then we can have
∫T×Tln|sin(θ−θ∗2)|μt(dθ)μt(dθ∗)=−ln2. |
Proof. By direct estimate, we have
∫T×Tln|sin(θ−θ∗2)|μt(dθ)μt(dθ∗)=1(2π)2∫T×Tln|sin(θ−θ∗2)|μe(dθ)μe(dθ∗)=1(2π)2∫T×Tln|sin(θ2)|μe(dθ)μe(dθ∗)=12π∫Tln|sin(θ2)|μe(dθ)=1π∫π0ln(sinθ)μe(dθ)=−ln2. |
By Lemma 5.1, we can see that
Lemma 5.2. Suppose there exists positive constants
mμe(dθ)≤μt(dθ)≤Mμe(dθ)for all θ∈T. | (5.1) |
Then, the quantity
−4π2M2ln2≤I(t)≤−4π2m2ln2,t≥0. |
Proof. We use the assumptions (5.1) to get that for all
m2∫T×Tln|sin(θ−θ∗2)|μe(dθ)μe(dθ∗)≤I(t)≤M2∫T×Tln|sin(θ−θ∗2)|μe(dθ)μe(dθ∗). |
Note that
∫T×Tln|sin(θ−θ∗2)|μe(dθ)μe(dθ∗)=−4π2ln2. |
Thus, we have
−4π2M2ln2≤I(t)≤−4π2m2ln2,t≥0. |
In this subsection, we consider small frustration case with
ddtμt+∂θ{V[μt]μt(dθ)}=0,V[μt]=−κRsin(θ−ψ−α). | (5.2) |
We differentiate both sides of (4.1) with respect to
eiψ[˙R(t)+iR(t)˙ψ(t)]=∫Teiθ∂∂tμt(dθ)=∫Teiθ∂θ{κRsin(θ−ψ−α)μt(dθ)}=−i∫TeiθκRcos(θ−ψ−α)μ(dθ). | (5.3) |
Now we divide both sides of relation (5.3) by
˙R=κR∫Tsin(θ−ψ)sin(θ−ψ−α)μt(dθ),R˙ψ=−κR∫T×Rcos(θ−ψ)sin(θ−ψ−α)μt(dθ). | (5.4) |
Lemma 5.3. Let
ddtI(t)=−κR2cosαandd2dt2I(t)=−2κ2R2cosα∫Tsin(θ−ψ)sin(θ−ψ−α)μt(dθ). |
Proof.
ddtI(t)=∫T×Tln|sin(θ−θ∗2)|μt(dθ∗)ddtμt(dθ)+∫T×Tln|sin(θ−θ∗2)|μt(dθ)ddtμt(dθ∗)=κR∫T×Tln|sin(θ−θ∗2)|∂θ(sin(θ−ψ−α)μt(dθ))μt(dθ∗)+κR∫T×Tln|sin(θ−θ∗2)|∂θ∗(sin(θ∗−ψ−α)μt(dθ∗))μt(dθ). |
Integrations by parts yield
ddtI(t)=−κR∫T×T(sin(θ−ψ−α)∂θln|sin(θ−θ∗2)|+sin(θ∗−ψ−α)∂θ∗ln|sin(θ−θ∗2)|)μt(dθ)μt(dθ∗)=−κR∫T×T(sin(θ−ψ−α)−sin(θ∗−ψ−α))∂θln|sin(θ−θ∗2)|μt(dθ)μt(dθ∗). |
By direct calculation, one has
sin(θ−ψ−α)−sin(θ∗−ψ−α)=2cos(θ+θ∗−2ψ2−α)sin(θ∗−θ2). |
Now we use the above estimates and relation:
∂θln|sin(θ−θ∗2)|=cos(θ−θ∗2)2sin(θ−θ∗2) |
to obtain
ddtI(t)=−κR∫T×Tcos(θ+θ∗−2ψ2−α)sin(θ∗−θ2)cos(θ−θ∗2)sin(θ−θ∗2)μt(dθ)μt(dθ∗)=−κRcosα∫T×Tcos(θ+θ∗−2ψ2)cos(θ−θ∗2)μt(dθ)μt(dθ∗)−κRsinα∫T×Tsin(θ+θ∗−2ψ2)cos(θ−θ∗2)μt(dθ)μt(dθ∗)=:J21+J22. |
We use relation (4.2) to derive
∫T×Tcos(θ+θ∗−2ψ2)cos(θ−θ∗2)μt(dθ)μt(dθ∗)=∫T×T(cos(θ−ψ)+cos(θ∗−ψ))μt(dθ)μt(dθ∗)=R,∫T×Tsin(θ+θ∗−2ψ2)cos(θ−θ∗2)μt(dθ)μt(dθ∗)=12∫T×T(sin(θ−ψ)+sin(θ∗−ψ))μt(dθ)μt(dθ∗)=0. |
Thus, we have
J21=−κR2cosα,J22=0. | (5.5) |
This yields
ddtI(t)=−κR2cosα,t>0. | (5.6) |
d2dt2I(t)=−2κcosαR˙R,t>0. |
Hence, we use relation (5.4) to obtain
d2dt2I(t)=−2κ2R2cosα∫Tsin(θ−ψ)sin(θ−ψ−α)μt(dθ),t>0. |
Proposition 5.4. Suppose the frustration and initial datum satisfy
α∈(−π2,π2)and∫T×Tln|sin(θ−θ∗2)|μ0(dθ)μ0(dθ∗)<∞, |
and let
1. If there exist constant
limt→∞R(t)=0. |
2. If not, we have
limt→∞‖μt(dθ)μe(dθ)‖L∞=∞. |
Proof. (1) We use Lemma 5.3 to see that
I(t)≥−4π2M2ln2. |
Hence, we deduce
limt→∞I(t)exists. |
Together with initial assumption
I(0):=∫T×Tln|sin(θ−θ∗2)|μ0(dθ)μ0(dθ∗)<∞, |
we obtain
limt→∞∫t0ddsI(s)ds=limt→∞I(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
limt→∞R(t)=0. |
(2) If we can not find a positive constant
limt→∞I(t)=−∞. |
Furthermore, we can see
I(t)=∫T×Tln|sin(θ−θ∗2)|μt(dθ)μt(dθ∗)≥‖μt(dθ)μe(dθ)‖2L∞∫T×Tln|sin(θ−θ∗2)|μe(dθ)μe(dθ∗)=−ln2‖μt(dθ)μe(dθ)‖2L∞. |
Thus, we can deduce
‖μt(dθ)μe(dθ)‖L∞→∞,ast→∞. |
In this subsection, we consider a large frustration case (
∂∂tμt+∂θ(C[μt]μt)=0,(t,θ,ω)∈R+×T×R,C[μt]=κ∫T×Rcos(θ∗−θ+ˆα)μt(dθ∗dω∗). | (5.7) |
Note that the assumption on
ˆα∈(−3π2,−π]=(π2,π]. |
We divide both sides of relation (4.1) by
Rcos(θ−ψ−ˆα)=∫T×Rcos(θ∗−θ+ˆα)μt(dθ∗dω∗). | (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(θ−ψ−ˆα). | (5.9) |
We differentiate both sides of (4.1) with respect to
eiψ[˙R(t)+iR(t)˙ψ(t)]=∫T×Reiθddtμt(dθdω)=−∫T×Reiθ∂θ{κRcos(θ−ψ−ˆα)μt(dθdω)}=i∫T×ReiθκRcos(θ−ψ−ˆα)μt(dθdω). |
Now, we divide both sides of above relation by
˙R(t)=−κR∫T×Rsin(θ−ψ)cos(θ−ψ−ˆα)μt(dθdω),R(t)˙ψ(t)=κR∫T×Rcos(θ−ψ)cos(θ−ψ−ˆα)μt(dθdω). | (5.10) |
Now, we define the first phase moment
m1(t):=∫Tθμt(dθ). |
Lemma 5.5. Let
(i)ddtI(t)=κR2sinˆα,ddtm1(t)=κR2cosˆα.(ii)d2dt2I(t)=−2κ2R2sinˆα∫Tsin(θ−ψ)cos(θ−ψ−ˆα)μt(dθ). |
Proof.
ddtI(t)=∫T×Tln|sin(θ−θ∗2)|μt(dθ∗)ddtμt(dθ)+∫T×Tln|sin(θ−θ∗2)|μt(dθ)ddtμt(dθ∗)=−κR∫T×Tln|sin(θ−θ∗2)|∂θ(cos(θ−ψ−ˆα)μt(dθ))μt(dθ∗)−κR∫T×Tln|sin(θ−θ∗2)|∂θ∗(cos(θ∗−ψ−ˆα)μt(dθ∗))μt(dθ). |
We use integration by parts to get
ddtI(t)=κR∫T×T(cos(θ−ψ−ˆα)∂θln|sin(θ−θ∗2)|+cos(θ∗−ψ−ˆα)∂θ∗ln|sin(θ−θ∗2)|)μt(dθ)μt(dθ∗)=κR∫T×T(cos(θ−ψ−ˆα)−cos(θ∗−ψ−ˆα))∂θln|sin(θ−θ∗2)|μt(dθ)μt(dθ∗). | (5.11) |
By direct calculation, one has
cos(θ−ψ−ˆα)−cos(θ∗−ψ−ˆα)=−2sin(θ+θ∗−2ψ2−ˆα)sin(θ∗−θ2). | (5.12) |
Now, we use (5.11), (5.12) and relation
∂θln|sin(θ−θ∗2)|=cos(θ−θ∗2)2sin(θ−θ∗2) |
to obtain
ddtI(t)=−κR∫T×Tsin(θ+θ∗−2ψ2−ˆα)sin(θ∗−θ2)cos(θ−θ∗2)sin(θ−θ∗2)μt(dθ)μt(dθ∗)=−κRcosˆα∫T×Tsin(θ+θ∗−2ψ2)cos(θ−θ∗2)μt(dθ)μt(dθ∗)+κRsinˆα∫T×Tcos(θ+θ∗−2ψ2)cos(θ−θ∗2)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)=−κR∫Tθ∂θ(cos(θ−ψ−ˆα)μt)=κR∫Tcos(θ−ψ−ˆα)μt(dθ)=κRcosˆα∫Tcos(θ−ψ)μt(dθ)+κRsinˆα∫Tsin(θ−ψ)μt(dθ)=κR2cosˆα. |
d2dt2I(t)=2κR˙Rsinˆα,t>0. |
We use relation (5.10) to deduce
d2dt2I(t)=−2κ2R2sinˆα∫Tsin(θ−ψ)cos(θ−ψ−ˆα)μt(dθ),t>0. |
Corollary 5.6. Let
cosˆαI(t)−sinˆαm1(t)=cosˆαI(0)−sinˆαm1(0),for allt≥0. |
Proof. We combine estimate
cosˆαddtm1(t)I(t)−sinˆαddtm1(t)=0,t>0. |
This yields our desired estimate.
Remark 3. For all
I(t)=I(0), |
and for
Proposition 5.7. Suppose frustration and initial datum satisfy
ˆα∈(0,π]and∫T×Tln|sin(θ−θ∗2)|μ0(dθ)dμ0(dθ∗)<∞, |
and let
limt→∞R(t)=0. |
Proof. It follows from Lemma 5.5 that
ddtI(t)=κR2sinˆα,d2dt2I(t)=−2κ2R2sinˆα∫Tsin(θ−ψ)cos(θ−ψ−ˆα)μt(dθ). |
This yields that
limt→∞∫t0ddsI(s)ds=limt→∞I(t)−I(0)≤−I(0)<∞. |
Thus, we apply Barbalat's Lemma to conclude
limt→∞ddtI(t)=0. | (5.14) |
Now we use the relation
limt→∞R(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=κNN∑j=1sin(θj−θi+α),t>0,|α|<π2. | (A.1) |
Lemma A.1. (Phase coherence) Suppose the coupling strength and initial data satisfy
κ>0,D(Θin)<π−2|α|, |
and let
sup0≤t<∞D(Θ(t))≤π−2|α|,sup0≤t<∞D(Θ(t))≤D(Θin). |
Proof. (ⅰ) The first relation can be obtained from Lemma
(ⅱ) Note that the phase diameter
ddtD(Θ)=ddt(θM−θm)=κN{N∑j=1sin(θj−θM+α)−N∑j=1sin(θj−θm+α)}≤0, |
where we have used the following relations:
|θj−θM+α|≤D(Θ)+|α|≤π−|α|,|θj−θm+α|≤D(Θ)+|α|≤π−|α|. |
Now, we are ready to provide the first estimate in Proposition 4.1.
Let
ddtD(θ)=κNN∑j=1{sin(θj−θM+α)−sin(θj−θm+α)}=−2κNN∑j=1cos(θj−θM2+θj−θm2+α)sin(θM−θm2). | (A.2) |
Since
θj−θM2+α≤θj−θM2+θj−θm2+α≤θj−θm2+α, |
one has
θj−θM2+α∈(−12D(Θin)−|α|,|α|)⊆(−π2,π2),θj−θm2+α∈(|α|,12D(Θin)+|α|)⊆(−π2,π2). | (A.3) |
Then, we use (A.3) to get
θj−θM2+θj−θm2+α∈(−12D(Θin)−|α|,12D(Θin)+|α|)⊆(−π2,π2). |
This yields
cos(θj−θM2+θj−θm2+α)≥cos(12D(Θin)+|α|). | (A.4) |
Now we substitute (A.4) into (A.2) to derive a differential inequality:
dD(Θ)dt≤−2κNcos(12D(Θin)+|α|)N∑j=1sin(θM−θm2). | (A.5) |
Since
sin(θM−θm2)≥2πD(Θ)2=D(Θ)π. | (A.6) |
Now we combine (A.5) and (A.6) to obtain
dD(Θ)dt≤−2κπcos(12D(Θin)+|α|)D(Θ). | (A.7) |
We integrate the differential inequality (A.7) with respect to time
D(Θ(t))≤D(Θin)exp{−2κπcos(12D(Θin)+|α|)t},fort≥0. |
In this subsection, we consider the non-identical Kuramoto-Sakaguchi equation (4.4). We find the exact time
Lemma A.2. Suppose initial data, natural frequencies and coupling strength satisfy
0<D(Θin)<π−2|α|,0<D(Ω)<∞,κ>κe:=D(Ω)sin(D(Θin)+|α|)−sin|α|. |
Then we have
sup0≤t<∞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.,
D(Θ(t))<D(Θin),for allt∈(0,t0)andD(Θ)(t0)=D(Θin). | (A.9) |
Then we use continuity of
ddt|t=t0D(Θ)>0. | (A.10) |
On the other hand, note that the system (4.4) can be rewritten as
˙θi=ωi+κcosαNN∑j=1sin(θj−θi)+κsinαNN∑j=1cos(θj−θi),t>0,|α|<π2. |
Hence, we use definition of
ddtD(Θ)≤D(Ω)+κcosαNN∑j=1(sin(θj−θM)−sin(θj−θm))+κsinαNN∑j=1(cos(θj−θM)−cos(θj−θm)). | (A.11) |
We use (A.9) to see that for
cos(θj−θM)−cos(θj−θm)≤1−cosD(Θ), | (A.12) |
and
sin(θM−θj)θM−θj>sinD(Θ)D(Θ)andsin(θj−θm)θj−θm>sinD(Θ)D(Θ),t∈(0,t0). |
Hence, for
sin(θj−θM)−sin(θj−θm)<sinD(Θ)D(Θ)((θj−θM)−(θj−θm))=−sinD(Θ). | (A.13) |
Now we substitute (A.12) and (A.13) into (A.11) to obtain
dD(Θ)dt≤D(Ω)−κcosαsinD(Θ)+κsin|α|(1−cosD(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
Now, we are ready to provide a proof of the second part.
Suppose initial data, natural frequencies and coupling strength satisfy
0<D(Θin)<π−2|α|,0<D(Ω)<∞,κ>κe:=D(Ω)sin(D(Θin)+|α|)−sin|α|. |
Then we claim: for all
D(Θ(t))<D∞, | (A.15) |
where
sin(x+|α|)=sin(D(Θin)+|α|). |
We split the proof of claim (A.15) into two steps.
D(Θ(t0))<D∞. |
For this, we consider two cases.
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).
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
sin(D(Θ)+|α|)≥sin(D∞+α)=sin(D(Θin)+|α|). | (A.17) |
We use relation (A.11), relation (A.17) and assumption of
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)=(1−κκe)D(Ω). |
Hence, we have
D(Θ(t))<D(Θin)+(1−κκe)D(Ω)t,t>0. |
Thus, in order to have
t>te:=D(Θin)−D∞(1−κκe)D(Ω). |
D(Θ(t))<D∞,for allt>t0. |
Suppose that there exists a time
D(Θ(t))<D∞fort∈(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.
1. | Seung-Yeal Ha, Javier Morales, Yinglong Zhang, Kuramoto order parameters and phase concentration for the Kuramoto-Sakaguchi equation with frustration, 2021, 20, 1553-5258, 2579, 10.3934/cpaa.2021013 |