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
17
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)
19
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.
21
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.
27
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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