In this section, we will derive phase diagrams that allow one to determine if and how two oscillators synchronize their activities. We hereby consider a system S of two oscillators A and B, both either inhibitorily or excitatorily coupled together with time delay τ .
To study the dynamics of S, we begin with assuming that the oscillator A just reaches the threshold and is reset to φA = 0 and φ = φB > 0. We further assume that if φB < τ , then φB must have fired φB time earlier. This assumption is not necessary in a mathematical sense, but makes the analysis easier by reducing the number of case distinctions. It should also be noted that since the speed of the oscillators is assumed to be one, the phase φ and the time φ is interchangeable. As the system S evolves, the phase positions of φA and φB at time t are to be denoted by φA(t) and φB(t), respectively. We next define a firemap and a return map for the system S of two oscillators A and B. Let t = tp,i denote the time when oscillator i has just fired its pth time and its phase is reset to zero. In this situation the system S is said to reach a ground state. The firemap and the return map are defined as follows, similarly as in [10]:
(i) Firemap h(φj(tp,i)) = φl(tq,k), where j 6= i, l 6= k and
Suppose the system S just reaches a ground state in oscillator i, then the firemap takes the phase position of the non-ground state oscillator into the phase position of the non-ground state oscillator when the system reaches the immediate next ground state. The return map takes the phase position of the non-ground state oscillator into the phase position of the non-ground state oscillator when the system reaches the next ground state in the same oscillators.
Before we begin the analysis of the different cases or configurations of the dynamics, we want to illustrate our motivation for our choice of intervals in the subspace of initial phase differences φ. Let us consider that S is a GS with oscillator A just being reset to φA= 0 such that φ = φB. In a first interval I1, both oscillators have fired, but their spikes did not reach their destination yet. Therefore, the consequences of the two pulses being received have to be evaluated. In a second interval I2, only the spike of oscillator A did not reach B and has to be taken into account. In a third interval I3, oscillator B will reach the threshold before the spike of A can be received. These considerations lead to the following definitions for I1, I2, and I3:
I1 : φB ∈ [0, τ ), I2 : φB ∈ [τ, 1 − τ ],
I3 : (Ex0, In0) : φB ∈ (1 − τ, 1].
On the one hand, the dynamics is very simple if we are looking at domain I3. After a time t = 1 − φ (from here on, we identify t with the time elapsed since S has been in a GS), S will be in a GS with φB = 0, which leads to a firemap h(φ) = 1 − φ. Additional case distinctions
5
are not required here, and I3 will be referenced as region Ex0 (for excitatory couplings) or In0 (for inhibitory couplings) for reasons of conformity.
On the other hand, the detailed analysis of I1 and I2following in the next section requires one to distinguish between excitatory and inhibitory coupling. In each case, we first heuristically describe the temporal development of S to motivate the mathematical notation of the dynamics that will follow afterwards. For brevity, we denote a spike originating from oscillator A as ”spike A” and the oscillator A simply as ”A”. Each domain will be partitioned into smaller regions, which we denote ExN or InN with successive numbering for excitatory and inhibitory coupling, respectively.
4. Excitatory Couplings, ǫ > 0 4.1. Construction of the FireMap.
Configuration I1
Ex1 : ǫ < 1 − f (τ + φ), ǫ < f (1 − φ) − f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, min{g(1 − ǫ) − τ, u1b, τ }), where u1b satisfies ǫ = f (1 − φ) − f (τ − φ).)
time t φA φB
0 0 φ
τ − φ τ − φ → F+(τ − φ, ǫ) τ
τ F+(τ − φ, ǫ) + φ < 1 τ + φ → F+(τ + φ, ǫ) < 1
h(φ) = 1 − |F+(τ + φ, ǫ) − F+(τ − φ, ǫ) − φ| (4.1) Using relation A3, we know that 0 ≤ F+(τ + φ, ǫ) − F+(τ − φ, ǫ) ≤ 2φ for all φ ∈ Ex1, and both the left and right equalities hold just as φ = 0. Then the firemap satisfies
h(φ) = 1 − |F+(τ + φ, ǫ) − F+(τ − φ, ǫ) − φ|
≥ 1 − |2φ − φ|
= 1 − φ > 1 − τ. (4.2)
Moreover, h(φ) = 1 − φ just as φ = 0.
Ex2a : 1 − f (φ + τ ) ≤ ǫ < f (1 − φ) − f (τ − φ), and φ ∈ I1 (For fixed ǫ, φ ∈ [g(1 − ǫ) − τ, min{u1b, τ }).)
time t φA φB
0 0 φ
τ − φ τ − φ → F+(τ − φ, ǫ) τ
τ F+(τ − φ, ǫ) + φ < 1 τ + φ → F+(τ + φ, ǫ) = 1
h(φ) = F+(τ − φ, ǫ) + φ (4.3)
Since F+(τ − φ, ǫ) > F+(τ + φ, ǫ) − 2φ ∀φ ∈ Ex2a, and F+(τ + φ, ǫ) = 1, h(φ) = F+(τ − φ, ǫ) + φ
> F+(τ + φ, ǫ) − 2φ + φ
≥ 1 − 2φ + φ = 1 − φ
≥ 1 − τ. (4.4)
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Ex2b : f (1 − φ) − f (τ − φ) ≤ ǫ < 1 − f (τ + φ), and φ ∈ I1 (For fixed ǫ, φ ∈ [u1b, min{g(1 − ǫ) − τ, τ }).)
time t φA φB
0 0 φ
τ − φ τ − φ → F+(τ − φ, ǫ) < 1 τ
τ − φ + 1 − F+(τ − φ, ǫ) 1 → 0 τ + 1 − F+(τ − φ, ǫ) τ F+(τ − φ, ǫ) + φ − 1 ≥ 0 τ + φ → F+(τ + φ, ǫ) < 1
h(φ) = F+(τ − φ, ǫ) − F+(τ + φ, ǫ) + φ (4.5) Since F+(τ − φ, ǫ) − F+(τ + φ, ǫ) < 0 ∀φ ∈ Ex2b, the firemap satisfies
h(φ) = F+(τ − φ, ǫ) − F+(τ + φ, ǫ) + φ
< 0 + φ = φ. (4.6)
Ex3 : f (1 − φ) − f (τ − φ) ≤ ǫ < 1 − f (τ − φ), 1 − f (τ + φ) ≤ ǫ < 1 − f (τ − φ) and φ ∈ I1.
time t φA φB
0 0 φ
τ − φ τ − φ → F+(τ − φ, ǫ) < 1 τ
τ − φ + 1 − F+(τ − φ, ǫ) 1 → 0 τ + 1 − F+(τ − φ, ǫ) τ φ − 1 + F+(τ − φ, ǫ) ≥ 0 τ + φ → F+(τ + φ, ǫ) = 1
h(φ) = φ − 1 + F+(τ − φ, ǫ) (4.7)
Since F+(τ − φ, ǫ) < 1, the firemap satisfies
h(φ) = φ − 1 + F+(τ − φ, ǫ)
< φ − 1 + 1 = φ. (4.8)
Ex4 : ǫ ≥ 1 − f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, τ − g(1 − ǫ)).)
time t φA φB
0 0 φ
τ − φ τ − φ → F+(τ − φ, ǫ) = 1 τ
τ φ τ + φ → F+(τ + φ, ǫ) = 1
h(φ) = φ (4.9)
The firemap h is with the same initial conditions but A and B exchanged.
Configuration I2
Ex5 : ǫ < 1 − f (τ + φ) and φ ∈ I2 (For fixed ǫ, φ ∈ [τ, g(1 − ǫ) − τ ).)
time t φA φB
0 0 φ
τ τ τ + φ → F+(τ + φ, ǫ) < 1
h(φ) = 1 − F+(τ + φ, ǫ) + τ (4.10)
Since τ + φ < F+(τ + φ, ǫ) < 1 and τ ≤ φ, the firemap satisfies h(φ) = 1 − F+(τ + φ, ǫ) + τ
< 1 − τ − φ + φ = 1 − τ, (4.11)
and
h(φ) = 1 − F+(τ + φ, ǫ) + τ
> 1 − 1 + τ = τ. (4.12)
Ex6 : ǫ ≥ 1 − f (φ + τ ) and φ ∈ I2 (For fixed ǫ, φ ∈ [max{τ, g(1 − ǫ) − τ }, 1 − τ ].)
time t φA φB
0 0 φ
τ τ τ + φ → F+(τ + φ, ǫ) = 1
h(φ) = τ (4.13)
Configuration I3
Ex0 : φ ∈ I3
time t φA φB
0 0 φ
1 − φ 1 − φ 1 → 0
h(φ) = 1 − φ (4.14)
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4.2. Dynamics in Cases.
Configuration I1
Dynamics in Ex1 : ǫ < 1 − f (τ + φ), ǫ < f (1 − φ) − f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, min{g(1 − ǫ) − τ, u1b, τ }), where u1b satisfies ǫ = f (1 − φ) − f (τ − φ).)
h(φ) = 1 − |F+(τ + φ, ǫ) − F+(τ − φ, ǫ) − φ|
By Eq. (4.2), the firemap h : Ex1 7→ Ex0. Specifically, when φ satisfies F+(τ − φ, ǫ) + φ = F+(τ + φ, ǫ), h(φ) = 1, which means φA and φB synchronize. Moreover, the return map R satisfies
R(φ) = |F+(τ + φ, ǫ) − F+(τ − φ, ǫ) − φ| ≤ φ. (4.15) The equality holds just as φ = 0. Since φ ∈ Ex1 and R(φ) < φ, R(φ) ∈ Ex1.
Proposition 4.1. For each φ ∈ Ex1, Rk(φ) → 0, as k → ∞.
Proof. Suppose not, i.e., ∃φ ∈ Ex1, Rk(φ) 9 0, as k → ∞. Then Rk(φ) 6= 0 ∀k ∈ N. Otherwise, Rm(φ) = 0 for large enough m ∈ N. By Eq. (4.15), Rk(φ) ∈ Ex1 ∀k ∈ N and R(φ) < φ.
Inductively, φ > R(φ) > · · · > Rk−1(φ) > Rk(φ) > 0. Since {Rk(φ)} is a monotone and bounded sequence in Ex1, and R(φ) is continuous in Ex1. {Rk(φ)} is a convergent sequence, i.e., Rk(φ) → φ∗, for some φ∗ ∈ (0, τ ), and such φ∗ is a fixed point. It is a contradiction since no fixed point in Ex1\{0}, by Eq. (4.15).
Corollary 4.1. For each φ ∈ Ex1, φA and φB synchronize.
Dynamics in Ex2a: 1 − f (τ + φ) ≤ ǫ < f (1 − φ) − f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [g(1 − ǫ) − τ, min{u1b, τ }).)
h(φ) = F+(τ − φ, ǫ) + φ
Remark 4.1. φ = 0 6∈ Ex2a.
By Eq. (4.4), the firemap h : Ex2a7→ Ex0. Then the return map R satisfies R(φ) = h(h(φ)) = 1 − h(φ)
= 1 − F+(τ − φ, ǫ) − φ
= F+(τ + φ, ǫ) − F+(τ − φ, ǫ) − φ
Rk−1(φ) > Rk(φ) > 0. Thus, Rk(φ) → φ∗ for some φ∗ ∈ Ex2a, and such φ∗ is a fixed point. It makes a contradiction with Eq. (4.16).
Corollary 4.2. For each φ ∈ Ex2a, there is a k ∈ N, depending on φ, such that Rk(φ) ∈ Ex1.
Dynamics in Ex2b : f (1 − φ) − f (τ − φ) ≤ ǫ < 1 − f (τ + φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [u1b, min{g(1 − ǫ) − τ, τ }).)
h(φ) = F+(τ − φ, ǫ) − F+(τ + φ, ǫ) + φ
Remark 4.2. φ = 0 6∈ Ex2b.
By Eq. (4.6), the firemap h : Ex2b 7→ Ex1 ∪ Ex2b.
Proposition 4.3. There is no φ ∈ Ex2b such that hk(φ) ∈ Ex2b for all k = 0, 1, 2, · · · .
Proof. Suppose not, i.e., ∃φ ∈ Ex2b such that hk(φ) ∈ Ex2b for all k = 0, 1, 2 · · · . Note that h(φ) is continuous in Ex2b. Moreover, h(φ) < φ, ∀φ ∈ Ex2b. Inductively φ > R(φ) > · · · >
Rk−1(φ) > Rk(φ) > 0. Thus, hk(φ) → φ∗ for some φ∗ ∈ Ex2b, and such φ∗ is a fixed point. It makes a contradiction with Eq. (4.6).
Corollary 4.3. For each φ ∈ Ex2b, there is a k ∈ N, depending on φ, such that hk(φ) ∈ Ex1.
Dynamics in Ex3 : 1 − f (τ + φ) ≤ ǫ < 1 − f (τ − φ), f (1 − φ) − f (τ − φ) ≤ ǫ < 1 − f (τ − φ) and φ ∈ I1.
h(φ) = φ − 1 + F+(τ − φ, ǫ)
Remark 4.3. φ = 0 6∈ Ex3.
By Eq.(4.8), the firemap h : Ex3 7→ I1. More precisely,
(i) If ǫ < 1 − f (τ ), h : Ex3 7→ Ex1 ∪ Ex2a∪ Ex2b∪ Ex3, since the slope of the lower bound of Ex3 is negative (−f′(τ + φ) < 0 and −f′(1 − φ) + f′(τ − φ) < 0.
(ii) If ǫ ≥ 1 − f (τ ), h : Ex3 7→ Ex3 ∪ Ex4 , since the slope of the upper bound of Ex3 is positive (f′(τ − φ) > 0).
Proposition 4.4. (i) If ǫ < 1 − f (τ ), there is no φ ∈ Ex3 such that hk(φ) ∈ Ex3 for all k = 0, 1, 2, · · · . (ii) If ǫ ≥ 1 − f (τ ), for each φ ∈ Ex3, hk(φ) ∈ Ex3 for all k ∈ N. Moreover, hk(φ) → τ − g(1 − ǫ) for all φ ∈ Ex3.
Proof.
(i) Suppose not, i.e., ∃φ ∈ Ex3 such that hk(φ) ∈ Ex3, for all k = 0, 1, 2, · · · . Since as ǫ < 1 − f (τ ), Ex3 = [g(1 − ǫ), τ ), or Ex3 = [u1b, τ ) and the sequence {hk(φ)} must satisfy hk(φ) → φ∗, where φ∗ is a fixed point for h in Ex3. It makes a contradiction with Eq. (4.8).
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(ii) If ǫ ≥ 1 − f (τ ), Ex3 = (τ − g(1 − ǫ), τ ), where τ − g(1 − ǫ) ≥ 0, and the equality holds just as ǫ = 1 − f (τ ). Because of
h(τ−) = τ − 1 + F+(τ − τ, ǫ)
= τ − 1 + g(ǫ) < τ, (4.17)
and
h((τ − g(1 − ǫ))+) = τ − g(1 − ǫ) − 1 + F+(τ − τ + g(1 − ǫ), ǫ)
= τ − g(1 − ǫ) − 1 + 1 = τ − g(1 − ǫ). (4.18) Also, for any φ ∈ Ex3,
h′(φ) = 1 + F+′(τ − φ, ǫ)
= 1 − g′(f (τ − φ) + ǫ)f′(τ − φ)
> 1 − g′(f (τ − φ))f′(τ − φ)
= 1 − 1 = 0. (4.19)
We obtain τ − g(1 − ǫ) = h((τ − g(1 − ǫ))+) < h(φ) < h(τ−) < τ . Therefore, for each φ ∈ Ex3, hk(φ) ∈ Ex3 for all k ∈ N. Moreover, by Eqs. (4.8), (4.17), (4.18), (4.19), the sequence {hk(φ)}
is decreasing. Hence, hk(φ) → φ∗. Since there is no fixed point in Ex3, φ∗∈ bd(Ex3). It follows that φ∗ = τ − g(1 − ǫ) which is also the boundary of Ex4.
Specifically, if ǫ = 1 − f (τ ), we obtain φ∗ = 0, i.e., φA and φB synchronize.
Corollary 4.4. (i) If ǫ < 1 − f (τ ), for each φ ∈ Ex3, hk(φ) ∈ Ex1 ∪ Ex2a∪ Ex2b for some k ∈ N. (ii) If ǫ ≥ 1 − f (τ ), for each φ ∈ Ex3, hk(φ) → τ − g(1 − ǫ) as k → ∞. Specifically, if ǫ = 1 − f (τ ), for each φ ∈ Ex3, φA and φB synchronize.
Dynamics in Ex4 : ǫ ≥ 1 − f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, τ − g(1 − ǫ)).) h(φ) = φ
Thus, h : Ex4 7→ Ex4. Specifically, if φ = 0 ∈ Ex4, hk(φ) = 0 ∈ Ex4, ∀k ∈ N.
Corollary 4.5. For each φ ∈ Ex4, φ is a marginal stable fixed point in Ex4.
Configuration I2 Proposition 4.5. There exists a unique fixed point for h in Ex5, and it is an attractor.
Proof. Define A(φ) = h(φ) − φ. Clearly, A′(φ) = h′(φ) − 1. From Lemma 4.2, we have h′(φ) < 0,
The last inequality follows directly from the definition of Ex5.
Hence h has a unique fixed point φ∗ ∈ Ex5, i.e., h(φ∗) = φ∗. Since R(φ∗) = h2(φ∗) and 0 < R′(φ) < 1, such φ∗ is also the unique fixed point for R in Ex5. Moreover, since
φ∗ < R(φ) < φ if φ > φ∗,
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and
φ∗ > R(φ) > φ if φ < φ∗, the fixed point φ∗ for R and hence for h is an attractor.
Corollary 4.6. For each φ ∈ Ex5, hk(φ) converges to the fixed point φ∗ ∈ Ex5, which is an attractor.
Dynamics in Ex6 : ǫ ≥ 1 − f (τ + φ) and φ ∈ I2 (For fixed ǫ, φ ∈ [max{τ, g(1 − ǫ) − τ }, 1 − τ ].)
h(φ) = τ Thus, h : Ex6 7→ I2.
Corollary 4.7. For each φ ∈ Ex6, the firemap maps φ to the line φ = τ . Configuration I3
Dynamics in Ex0 : φ ∈ I3
h(φ) = 1 − φ Thus, h : Ex0 7→ I1.
Corollary 4.8. For each φ ∈ Ex0, h(φ) ∈ I1.
4.3. Construction of Phase Diagrams.
Peskin modeled the pacemaker as a network of N ”integrate-and-fire” oscillators, each char-acterized by a voltagelike state variable xi, subject to the dynamics
dxi
dt = −rxi+ s, 0 ≤ xi≤ 1, i = 1, 2, · · · N. (4.20)
Mirollo and Strogatz [19] proposed a new model and they assume that x evolves according to x = f (φ), where f : [0, 1] → [0, 1] is smooth, monotonic increasing (f′ > 0), and concave down (f′′< 0). Here φ ∈ [0, 1] is a phase variable such that (i) dφ/dt = 1/T , where T is a cycle period. (ii) f (0) = 0. (iii) f (1) = 1. Then they generate the function f and its inverse function g which are given in the following, respectively.
f (φ) = s
In contrast, we assume there exists a function f which is smooth, monotonic increasing (f′ > 0), and concave up (f′′> 0). So, we switch the function f and g of Eq. (4.21) to be the model of our phase diagrams.
We fixed s = 1, and change the variable r, then we have three different kinds of figures in Figure 4.1:
(a) When r = 0.4 and τ = 0.2, f (1 − φ) − f (τ − φ) > 1 − f (τ + φ) ∀φ ∈ I1. (b) When r = 0.7 and τ = 0.2, f (1 − φ) − f (τ − φ) < 1 − f (τ + φ) ∀φ ∈ I1. (c) When r = 0.72 and τ = 0.35, f (1 − φ) − f (τ − φ) and 1 − f (τ + φ) have an
intersection in I1.
Moreover, we add Figure 4.2 which is a phase diagram we make from Ernst [10] as a contrast.
And it’s energy function f and inverse function g is exactly the Mirollo-Strogatz-type (Eq. 4.21) oscillators we just mentioned.
And there are some notations for the following figures:
l1 : ǫ = 1 − f (τ + φ)
l2 : ǫ = f (1 − φ) − f (τ − φ) l3 : ǫ = 1 − f (τ − φ)
In the following tables, we fixed an arbitrary ǫ > 0, and shift φ in [0,1]. Therefore, we clearly see which area (ExN ) does the oscillator begin and figure out where it ends.
15
(I) f (φ) = ln(s−rφs )
(II) f (φ) = s r −s
r(s − r
s )φ, f′ > 0, f′′ < 0
Ex4
l3 l2l1
1-f( )τ f(1- )τ
1-f(2 )τ
Ex1 Ex3 Ex2
Ex1 Ex1
Ex5 Ex6
Ex0
Figure 4.2. s := 1, r := 0.4, τ = 0.2, where s > r > 0
ǫ Dynamics Multistability
0 < ǫ < 1 − f (2τ ) φ → φ∗Ex5∗, if φ ∈ I1, Ex5∗, I3 Lag Syn. with Lag φ∗Ex5∗
φ →Period(τ, h(τ )), if φ ∈ I2\Ex5∗ Lag Syn. with Lag τ 1 − f (2τ ) ≤ ǫ < 1 − f (τ ) φ → φ∗Ex4 or τ , if φ ∈ I1, I3 Depends on the behavior of Ex2
φ → τ , if φ ∈ I2 Lag Syn. with Lag τ 1 − f (τ ) ≤ ǫ ≤ 1 φ → φ∗Ex4, if φ ∈ I1, I3 Lag Syn. with Lag φ∗Ex4
φ → τ , if φ ∈ I2 Lag Syn. with Lag τ
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5. Inhibitory Couplings, ǫ < 0 5.1. Construction of the FireMap.
Configuration I1
In1 : |ǫ| < f (τ − φ) and φ ∈ I1
⇒ ǫ > −f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, τ − g(−ǫ)).)
time t φA φB
0 0 φ
τ − φ τ − φ → F−(τ − φ, ǫ) > 0 τ
τ F−(τ − φ, ǫ) + φ τ + φ → F−(τ + φ, ǫ) > 0
h(φ) = 1 − F−(τ + φ, ǫ) + F−(τ − φ, ǫ) + φ (5.1) Using relation A4, F−(τ + φ, ǫ) ≥ F−(τ − φ, ǫ) + 2φ, we have φB(τ ) ≥ φA(τ ). Since F−(τ − φ, ǫ) > 0, the firemap satisfies
h(φ) = 1 − F−(τ + φ, ǫ) + F−(τ − φ, ǫ) + φ
> 1 − (τ + φ) + F−(τ − φ, ǫ) + φ
= 1 − τ + F−(τ − φ, ǫ)
> 1 − τ. (5.2)
In2 : f (τ − φ) ≤ |ǫ| < f (τ + φ) and φ ∈ I1
⇒ −f (τ − φ) ≥ ǫ > −f (τ + φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, τ ).)
time t φA φB
0 0 φ
τ − φ τ − φ → F−(τ − φ, ǫ) = 0 τ
τ φ τ + φ → F−(τ + φ, ǫ) > 0
h(φ) = 1 − |F−(τ + φ, ǫ) − φ| (5.3)
Since F−(τ + φ, ǫ) < τ + φ, the firemap satisfies
h(φ) = 1 − |F−(τ + φ, ǫ) − φ|
> 1 − |τ + φ − φ|
= 1 − τ. (5.4)
(i) If F−(τ + φ, ǫ) > φ,
(ii) If F−(τ + φ, ǫ) < φ,
Define In2c : f (φ) − f (τ + φ) > ǫ > −f (τ + φ) (For fixed ǫ, φ ∈ [g(−ǫ) − τ, min{l2b, τ }).) R(φ) = h(φ) = 1 + F−(τ + φ, ǫ) − φ
(iii) If F−(τ + φ, ǫ) = φ,
Define In2d: ǫ = f (φ) − f (τ + φ) (For fixed ǫ, φ = l2b.) h(φ) = 1
In3 : f (τ + φ) ≤ |ǫ| and φ ∈ I1
⇒ ǫ ≥ −f (τ + φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, min{g(−ǫ) − τ, τ }).)
time t φA φB
0 0 φ
τ − φ τ − φ → F−(τ − φ, ǫ) = 0 τ
τ φ τ + φ → F−(τ + φ, ǫ) = 0
h(φ) = R(φ) = 1 − φ (5.5)
Configuration I2
In4 : |ǫ| < f (τ + φ) − f (2τ ) and φ ∈ I2
⇒ ǫ > f (2τ ) − f (τ + φ) and φ ∈ I2 (For fixed ǫ, φ ∈ (l4, 1 − τ ], where l4 = g(f (2τ ) − ǫ) − τ .)
time t φA φB
0 0 φ
τ τ τ + φ → F−(τ + φ, ǫ) > 0
h(φ) = 1 − F−(τ + φ, ǫ) + τ (5.6)
We assume that |ǫ| is small enough such that F−(τ +φ, ǫ) > 2τ . Since 2τ < F−(τ +φ, ǫ) < τ +φ, the firemap satisfies
h(φ) = 1 − F−(τ + φ, ǫ) + τ
< 1 − 2τ + τ = 1 − τ, (5.7)
and
h(φ) = 1 − F−(τ + φ, ǫ) + τ
> 1 − τ − φ + τ
= 1 − φ ≥ τ. (5.8)
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In5 : f (τ + φ) − f (2τ ) ≤|ǫ| < f (τ + φ) and φ ∈ I2
⇒ f (2τ ) − f (τ + φ) ≥ ǫ > −f (τ + φ) and φ ∈ I2
time t φA φB
0 0 φ
τ τ τ + φ → F−(τ + φ, ǫ) > 0
h(φ) = 1 −|F−(τ + φ, ǫ) − τ | (5.9)
We assume that |ǫ| is bigger than that in In4 such that 0 < F−(τ + φ, ǫ) ≤ 2τ , then the firemap satisfies
h(φ) = 1 − |F−(τ + φ, ǫ) − τ |
≥ 1 − |2τ − τ |
= 1 − τ. (5.10)
In6 : f (τ + φ) ≤ |ǫ| and φ ∈ I2
⇒ −f (τ + φ) ≥ ǫ and φ ∈ I2 (For fixed ǫ, φ ∈ [τ, g(−ǫ) − τ ].)
time t φA φB
0 0 φ
τ τ τ + φ → F−(τ + φ, ǫ) = 0
h(φ) = R(φ) = 1 − τ (5.11)
Configuration I3
In0 : φ ∈ I3
time t φA φB
0 0 φ
1 − φ 1 − φ 1 → 0
h(φ) = 1 − φ (5.12)
5.2. Dynamics in Cases.
Configuration I1
Dynamics in In1 : ǫ > −f (τ − φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [0, τ − g(−ǫ)).) h(φ) = 1 − F−(τ + φ, ǫ) + F−(τ − φ, ǫ) + φ
By Eq. (5.2), thus h : In1 7→ In0. Then the return map satisfies R(φ) = h(h(φ)) = 1 − h(φ)
= F−(τ + φ, ǫ) − F−(τ − φ, ǫ) − φ (5.13)
≥ 2φ − φ = φ.
The equality holds just as φ = 0. Since φ ∈ In1 and R(φ) > φ, R(φ) ∈ In1 ∪ In2. Specifically, if φ = 0 ∈ In1, Rk(φ) = 0 ∈ In1.
Proposition 5.1. There is no φ ∈ In1\{0} such that Rk(φ) ∈ In1\{0} for all k = 0, 1, 2, · · · . Proof. Suppose not, i.e., ∃φ ∈ In1\{0} s.t. Rk(φ) ∈ In1\{0} for all k = 0, 1, 2, · · · . Note that R(φ) is continuous in (0, τ − g(−ǫ)], since
R(φ) =
( F−(τ + φ, ǫ) − F−(τ − φ, ǫ) − φ, φ ∈ In1;
F−(τ + φ, ǫ) − φ φ ∈ In2a∨ In2b. (5.14, 5.18) The return maps of In2a and In2b will be proved in Eqs. (5.14), (5.18).
⇒
( R((τ − g(−ǫ))−) = F−(2τ − g(−ǫ)) − (τ − g(−ǫ));
R(τ − g(−ǫ)) = F−(2τ − g(−ǫ)) − (τ − g(−ǫ)).
Moreover, R(φ) > φ, ∀φ ∈ In1\{0}. Thus, Rk(φ) → φ∗ for some φ∗ ∈ (0, τ − g(−ǫ)], and such φ∗ is a fixed point. That is R(φ∗) = φ∗. It follows that φ∗ = τ − g(−ǫ), or ǫ = −f (τ − φ∗), then F−(τ − φ∗, ǫ) = 0. Hence,
R(φ∗) = φ∗
⇒ F−(τ + φ∗, ǫ) − F−(τ − φ∗, ǫ) − φ∗ = φ∗
⇒ F−(τ + φ∗, ǫ) − φ∗ = φ∗
⇒ F−(τ + φ∗, ǫ) = 2φ∗
⇒ f (τ + φ∗) + ǫ = f (2φ∗)
⇒ f (τ + φ∗) − f (τ − φ∗) = f (2φ∗) − f (0) = f (φ∗+ φ∗) − f (φ∗− φ∗)
By relation A6, we know f (τ + φ∗) − f (τ − φ∗) > f (φ∗+ φ∗) − f (φ∗− φ∗), since τ > φ∗. →←
Corollary 5.1. For each φ ∈ In1\{0}, there is a k ∈ N, depending on φ, such that Rk(φ) ∈ In2.
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Dynamics in In2a : −f (τ − φ) ≥ ǫ ≥ −f (τ ) and φ ∈ I1 (For fixed ǫ, φ ∈ [l2a, τ ], where l2a = τ − g(−ǫ).)
h(φ) = 1 − F−(τ + φ, ǫ) + φ By Eq. (5.4), thus h : In2a7→ In0. Then the return map satisfies
R(φ) = h(h(φ)) = 1 − h(φ)
= F−(τ + φ, ǫ) − φ (5.14)
< τ + φ − φ = τ.
Thus R(φ) ∈ In1 ∪ In2.
Lemma 5.1. For each fixed ǫ, there exists a fixed point for R in In2a. Proof. Note that R(φ) is continuous in [τ − g(−ǫ), τ ].
If φ = l2a, we know that ǫ = −f (τ − l2a). The return map R(l2a) = F−(τ + l2a, ǫ) − l2a
= g(f (τ + l2a) − f (τ − l2a)) − l2a
> g(f (l2a+ l2a) − f (l2a − l2a)) − l2a (A6)
= 2l2a − l2a = l2a. (5.15)
If φ = τ , the return map
R(τ ) = F−(2τ, ǫ) − τ
< 2τ − τ = τ. (5.16)
By Eqs. (5.15), (5.16), there exists a fixed point for R in In2a.
Since R(φ) = φ,
R(φ) = φ
⇒ F−(τ + φ, ǫ) − φ = φ
⇒ F−(τ + φ, ǫ) = 2φ
⇒ g(f (τ + φ) + ǫ) = 2φ
⇒ ǫ = f (2φ) − f (τ + φ). (5.17)
Thus, the fixed point in In2a must satisfies ǫ(φ) = f (2φ) − f (τ + φ).
then the fixed point for R in In2a is unique.
For any φ ∈ In2a,
R′(φ) = F−′(τ + φ, ǫ) − 1
= g′(f (τ + φ) − |ǫ|)f′(τ + φ) − 1
> g′(f (τ + φ))f′(τ + φ) − 1
> 1 − 1 = 0.
Thus, l2a < R(l2a) < R(φ) < R(τ ) < τ . Hence, In2a is a trapping region.
Let φ∗ be the fixed point of R(φ), i.e., R(φ∗) = φ∗. (i)As φ > φ∗,
R(φ) < φ (Otherwise, there exists another fixed point in [φ, τ ].)
⇒ Rk(φ) < · · · < R2(φ) < R(φ) < φ, ∀k = 1, 2, · · ·
and Rk(φ) > Rk(φ∗) = φ∗, ∀k = 1, 2, · · · (∵ R is increasing.)
⇒ φ∗ < Rk(φ) < Rk−1(φ) < · · · < φ, ∀k = 1, 2, · · · Thus, the sequence {Rk(φ)} is decreasing, and Rk(φ) → φ∗. (ii)As φ < φ∗,
Similarly, the sequence {Rk(φ)} is increasing, and Rk(φ) → φ∗. The unique fixed point φ∗ is an attractor.
Corollary 5.2. For each φ ∈ In2a, Rk(φ) converges to the fixed point φ∗. Specifically, if φ = 0, then the fixed point φ∗ = 0, i.e., φA and φB synchronize.
Dynamics in In2b : −f (τ ) > ǫ > f (φ) − f (τ + φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [l2b, τ ), where l2b satisfies ǫ = f (φ) − f (τ + φ).)
h(φ) = 1 − F−(τ + φ, ǫ) + φ
Remark 5.1. φ = 0 6∈ In2b.
By Eq. (5.4), thus h : In2b7→ In0. Then the return map satisfies R(φ) = h(h(φ)) = 1 − h(φ)
= F−(τ + φ, ǫ) − φ (5.18)
< τ + φ − φ = τ.
Since −f (τ ) > ǫ, R(φ) 6∈ In1 ∪ In2a. It follows that R(φ) ∈ In2\In2a∪ In3.
23
Proposition 5.3. There is no φ ∈ In2b such that Rk(φ) ∈ In2b for all k = 0, 1, 2, · · · .
Proof. Note that R(φ) is continuous in [l2b, τ ]. Assume that there exists a fixed point for R in In2b. Since the return map of In2b is the same as that in In2a. It follows that
R(φ) = φ
⇔ ǫ = f (2φ) − f (τ + φ) > f (2φ − 2φ) − f (τ + φ − 2φ) (5.17, A5)
⇔ ǫ = f (2φ) − f (τ + φ) > −f (τ − φ) > −f (τ ) (For φ 6= 0 ∈ In2b.)
⇔ ǫ > −f (τ ).
This contradicts the region of the ǫ in In2b. Then there is no fixed point for R in In2b.
Since R(l2b) = 0 and R(τ ) = F−(2τ, ǫ) − τ < τ , it follows that R(φ) < φ. Otherwise, there exists fixed points in [l2b, τ ]. Thus, φ is decreasing in In2b.
Suppose not, i.e., ∃φ ∈ In2b, s.t. Rk(φ) ∈ In2b, for all k = 0, 1, 2, · · · . Since φ is decreasing, Rk(φ) → φ∗, for some φ∗ ∈ [l2b, τ ]. It follows that φ∗ = l2b is a fixed point in In2a. It makes contradiction.
Corollary 5.3. For each φ ∈ In2b, there is a k ∈ N, depending on φ, such that Rk(φ) ∈ In2c∪ In2d∪ In3.
Dynamics in In2c : f (φ) − f (τ + φ) > ǫ > −f (τ + φ) and φ ∈ I1 (For fixed ǫ, φ ∈ [g(−ǫ) − τ, min{l2b, τ }).)
R(φ) = h(φ) = 1 + F−(τ + φ, ǫ) − φ
Remark 5.2. φ = 0 6∈ In2c.
By Eq. (5.4), thus R, h : In2c 7→ In0. We iterate the return map and derive h(R(φ)) = 1 − R(φ) = 1 − h(φ)
= φ − F−(τ + φ, ǫ) (5.19)
< φ.
Since R(φ) ∈ In0, h(R(φ)) < φ, h(R(φ)) ∈ In2c∪ In3.
Note that h(R(φ)) is continuous in In2c, and h(R(g(−ǫ) − τ )) = g(−ǫ) − τ . Clearly, we know that h′(R(φ)) = 1 − F−′(τ + φ, ǫ) < 0. For each φ ∈ In2c, since g(−ǫ) − τ < φ, we have
h(R(g(−ǫ) − τ )) > h(R(φ))
⇒ g(−ǫ) − τ > h(R(φ))
Dynamics in In2d: ǫ = f (φ) − f (τ + φ) and φ ∈ I1 (For fixed ǫ, φ = l2b.) Lemma 5.2. There exists a unique fixed point for h in In4.
Proof. From Eq. (5.20), we know that h(l4) = 1 − τ > l4. It follows that
Now, we claim that the fixed point for h in In4 is unique.
Suppose not, i.e., ∃φ1< φ2 ∈ In4 s.t. h(φ1) = φ1 and h(φ2) = φ2. By Mean Value Theorem, h′(φ′) = h(φ2) − h(φ1)
φ2− φ1
for some φ′ ∈ (φ1, φ2)
⇒ h′(φ′) = 1 for some φ′ ∈ (φ1, φ2)
This is a contradiction with Eq. (5.21). Thus, there exists a unique fixed point φ∗ for h in In4.
25
Since there exists a fixed point for h, h(φ) = φ
⇒ 1 − F−(τ + φ, ǫ) + τ = φ
⇒ 1 + τ − φ = g(f (τ + φ) + ǫ)
⇒ ǫ = f (1 + τ − φ) − f (τ + φ). (5.23)
Thus, the fixed point in In4 must satisfy ǫ(φ) = f (1 + τ − φ) − f (τ + φ).
Proposition 5.4. Except for φ = φ∗, there is no φ ∈ In4 such that hk(φ) ∈ In4 for all k = 0, 1, 2, · · · . i.e., The unique fixed point φ∗ for h in In4 is a repellor.
Proof. Suppose not, i.e., ∃φ ∈ In4 s.t. hk(φ) ∈ In4, for all k = 0, 1, 2, · · · . Let
S1:= {φ : h1(φ) ∈ In4} ⇒ S1 is an interval, S1 6= ∅. (∵ φ∗ ∈ S1 and h′(φ) < −1) S2:= {φ : h2(φ) ∈ In4} ⇒ S2 is an interval, S2 6= ∅. (∵ φ∗ ∈ S2 ⊆ S1 and d
dφh2(φ) > 1)
... ...
⇒ S := {φ : hk(φ) ∈ In4 ∀k} = ∞T
k=1
Sk is an interval (noempty).
By Eq. (5.21), R′(φ) = h′(h(φ))h′(φ) > 1. Let H(φ) = R(φ) − φ, then we have H′(φ) = R′(φ) − 1 > 0.
1◦
(i) If φ < φ∗,
H(φ) < H(φ∗) = 0 ⇒ R(φ) < φ.
(ii) If φ > φ∗,
H(φ) > H(φ∗) = 0 ⇒ φ < R(φ).
Then (i) Rk(φ) < · · · < R2(φ) < R(φ) < φ < φ∗, ∀k ∈ N.
and (ii) Rk(φ) > · · · > R2(φ) > R(φ) > φ > φ∗, ∀k ∈ N.
2◦
(i) If φ < φ∗,
H(R(φ)) < H(φ)
⇒ R(R(φ)) − R(φ) < R(φ) − φ
⇒ |R(φ) − φ| < |R2(φ) − R(φ)|
(ii) If φ > φ∗,
H(φ) < H(R(φ))
⇒ |R(φ) − φ| < |R2(φ) − R(φ)|
Corollary 5.7. For each φ ∈ In4, there exists a unique fixed point φ∗ which is a repellor. Also there is a k ∈ N, depending on φ, such that hk(φ) ∈ In5 ∪ In6.
Dynamics in In5 : f (2τ ) − f (τ + φ) ≥ ǫ > −f (τ + φ) and φ ∈ I2
h(φ) = 1 − |F−(τ + φ, ǫ) − τ |
By Eq. (5.10), thus h : In5 7→ In0 ∪ l1−τ, where l1−τ is the line φ = 1 − τ . The firemap h(φ) = 1 − τ ∈ l1−τ just as φ = l4. Also, if φA(τ ) > φB(τ ), R(φ) = h(φ).
Furthermore,
h(φ) = 1
⇔ |F−(τ + φ, ǫ) − τ | = 0
⇔ g(f (τ + φ) + ǫ) = τ
⇔ ǫ = f (τ ) − f (τ + φ).
Therefore, φA and φB synchronize just as ǫ(φ) = f (τ ) − f (τ + φ).
Corollary 5.8. For each φ ∈ In5, h : In5 7→ In0 ∪ l1−τ. Specifically, when ǫ(φ) = f (τ ) − f (τ + φ), φA and φB synchronize.
Dynamics in In6 : −f (τ + φ) ≥ ǫ and φ ∈ I2 (For fixed ǫ, φ ∈ [τ, g(−ǫ) − τ ].) h(φ) = R(φ) = 1 − τ
Thus h : In6 7→ In4 ∪ In5.
Corollary 5.9. For each φ ∈ In6, h : In6 7→ In4 ∪ In5.
Configuration I3
Dynamics in In0 : φ ∈ I3
h(φ) = 1 − φ Thus, h : In0 7→ I1.
Corollary 5.10. In0 is the same as Ex0 and we have h : In0 7→ I1.
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5.3. Construction of Phase Diagrams.
Since for all φ ∈ In6, h(φ) = 1 − τ , and the line 1 − τ belongs to In4, In5, In6. Let’s find out how the oscillators run when φ = 1 − τ . We find that different τ leads to three different cases.
Here K′(ǫ) = 0 as ǫ = −1
2). Moreover, the collection of such ǫ is an interval.
Thus, there is a ǫ such that h(1 − τ ) ∈ In6 with 1 − τ ∈ In4 if and only if τ ≤ g(1
(i) The proof is the same as Case 1-(ii).
(ii)The proof is the same as Case 1-(iii).
Case 3:
If τ > 1 2g(1
2), thenf (2τ ) − 1 > −f (2τ ).
(i) The proof is the same as Case 1-(ii).
(ii)The proof is the same as Case 1-(iii).
In the following Figure 5.1 ∼ 5.3, we use the particular energy function f and inverse function g which are used in Section 4.3 (The inverse of Eq. (4.21)). Also, we fix s := 1, r := 0.9, and three different τ in Figure 5.1 ∼ 5.3.
Moreover, we add Figure 5.4 which is the phase diagram we make from Ernst [10] as a contrast.
And it’s energy function f and inverse function g is exactly the Mirollo-Strogatz-type (Eq. 4.21) oscillators we mentioned before. Also, we fixed s := 1, r := 0.4, and τ = 0.2 in Figure 5.4.
And there are some notations for the following figures:
l1 : ǫ = −f (τ − φ) l2 : ǫ = −f (τ + φ) l3 : ǫ = f (2τ ) − f (τ + φ) D1 : ∀φ ∈ D1, φ is an attractor.
D2 : ∀φ ∈ D2, the complete synchronization occurs.
D3 : ∀φ ∈ D3, φ is a repellor.
D4 : ∀φ ∈ D4, the complete synchronization occurs.
In the following tables, we fixed an arbitrary ǫ < 0, and shift φ in [0,1]. Therefore, we clearly see which area (InN ) does the oscillator begin and figure out where it ends.
29
(I) f (φ) = ln(s−rφs )
Case 2: g(1
Case 3: τ > 1 2g(1
2)
In1
In0 In3
In2
aIn6 In4
In5 In2
cIn2
b-f( ) τ
-f(2 )+ τ f( ) τ
-f(2 ) τ f(2 )-1 τ
l
1l
2l
3D
3= In2
dD
2D
4D
1Figure 5.3. s := 1, r := 0.9, τ = 0.4, where s > r > 0
ǫ Dynamics Multistability
−f (τ ) ≤ ǫ < 0 φ→ φ∗I n2a, if φ ∈ I1, I2\D3, I3 Lag Syn. with Lag φ∗I n2a
φ→ φ, if φ ∈ D3 Lag Syn. with Lag φ
f(2τ ) − 1 < ǫ < −f (τ ),
−1 ≤ ǫ < f (2τ ) − 1
φ→Period(φ∗I n3,1 − φ∗I n3), if φ ∈ I1\D2, I2\{D3∪ D4}, I3 Lag Syn. with Lag φ∗I n3
φ→ 0, if φ ∈ D2, D4 Complete Syn.
φ→ φ, if φ ∈ D3 Lag Syn. with Lag φ
(II) f (φ) = s r −s
r(s − r
s )φ, f′ > 0, f′′< 0
l3
In1
In3
In5 -f( )τ
f(2 )-1τ -f(2 )τ
In4
In0
In6
l2
In2 l1
Figure 5.4. s := 1, r := 0.4, τ = 0.2, where s > r > 0
ǫ Dynamics Multistability
−f (τ ) ≤ ǫ < 0 φ → 0, if φ ∈ I1, In5, I3 Complete Syn.
φ → φ∗I n4, if φ ∈ In4 Lag Syn. with Lag φ∗I n4
f (2τ ) − 1 < ǫ < −f (τ ) φ → φ∗I n3, if φ ∈ I1, In5, I3 Lag Syn. with Lag φ∗I n3
φ → φ∗I n4, if φ ∈ In4 Lag Syn. with Lag φ∗I n4
−1 ≤ ǫ < f (2τ ) − 1 φ → φ∗I n3, if φ ∈ I1, I2, I3 Lag Syn. with Lag φ∗I n3
33
6. Conclusion
It was numerically demonstrated in [10] that with N ≫ 2 convex oscillators, the system reveal multistable phase clustering for inhibitory couplings. For N ≫ 2 concave oscillators the corresponding system also has multistable phase clustering for excitatory couplings. Since for N = 2, the corresponding system has stable in-phase synchronization. It will be interesting to treat N as a parameter so as to see how the system evolves from synchronization to clustering synchronization as N increases.
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