Synchronization of Pulse- Coupled Biological Oscillators By Renato E. Mirollo and Steven H. Strogatz Presented by Ramil Berner Math 723 Spring Quarter 08

Overview Point of the Paper Model for 2 Oscillators Model for N Oscillators Main Theorem Conclusion

Synchronization of What coupled biological Who? Synchronization of Pulse-couple Biological Oscillators Examples of Biological Oscillators Fireflies Crickets Menstruating Women Pacemaker Pancreas The Idea Oscillators will evolve to synchronous firing. Shortest path to synchronization will be taken

Oscillators assume to interact by a simple form of pulse coupling –When a given oscillator fires, its pulls all the other oscillators up by an amount, or pulls them up to firing. –Whichever is less What are we talking about?

Peskin Conjectured –For arbitrary initial conditions, system approaches a state in which all oscillators are firing synchronously. –This remains true even when the oscillators are not quite identical. Proved the first for N=2 oscillators –Further assumptions of small coupling strength and small dissipation What are we talking about? cont…

Analysis of general version of Peskin’s Model for all N –Assume only the oscillators rise toward threshold with a time-course which is monotonic and concave down. –Retain two of Peskin’s assumptions Oscillators have identical dynamics Each is coupled to all the others. What are we talking about? cont…

Generalize the integrate and fire dynamics –Assume only that x evolves according to –ƒ:[0,1]  [0,1] is smooth, monotonically increasing and concave down –Here is a phase variable such that Ф=0 when the oscillator is at its lowest state x=0 Ф=1 at the end of the cycle when the oscillator reaches the threshold x=1. Two Oscillators – The Model ƒ satisfies ƒ(0)=0, ƒ(1)=1

g denotes the inverse function of ƒ. g maps states to their corresponding phases: g(x) = Ф. Because of the hypothesis on ƒ, the function g is increasing and concave up. The endpoint conditions are g(0)=0 and g(1)=1 Two Oscillators – The Model cont…

Example –Here – –Consider Two oscillators governed by ƒ Interact by pulse coupling rule Two Oscillators – The Model cont…

(a) Two Points on a fixed curve (b) Just before firing (c) Immediately After Firing Two Oscillators – The Model cont…

Strategy –To prove that the two oscillators always become synchronized, we first calculate the return map and then show the oscillators are driven closer together each time the map is iterated. –Perfect Synchrony when the oscillators have gotten so close together that the firing of one brings the other to threshold. –They remain synchronized thereafter because there dynamics are identical. Two Oscillators – The Model cont…

Calculating the Return Map Return Map –Two Oscillators A and B –Consider the system just after A has fired –Ф denotes the phase of B –The return map R(Ф) is defined to be the phase of B immediately after the next firing of A.

Calculating the Return Map cont… Observe that after a time 1-Ф oscillator B reaches threshold A moves from zero to an x-value given by x A = ƒ(1-Ф).

An instant later B fires and x A jumps to ε+ƒ(1-Ф) or 1. whichever is less. If x A =1, we are done, the oscillators have synchronized. Hence assume that x A =ε+ƒ(1-Ф)<1. Calculating the Return Map cont…

The corresponding phase of A is g(ε+f(1-Ф), where g = ƒ -1 h  (2.1)‏ Two iterations of h will give us R( Ф )‏ –R( Ф )=h(h( Ф )) (2.2)‏ Calculating the Return Map cont…

Domain of h and R: we assumed that ε+ƒ (1- Ф )<1. this assumption is satisfied for ε in [0,1) and Ф in (δ,1) where δ=1-g(1-ε). Thus the domain of h is the subinterval (δ,1)‏ Similarly the domain of R is the subinterval (δ,h -1 (δ))‏ This interval is non empty since δ<h -1 (δ) for ε<1 Calculating the Return Map cont…

Lemma 2.1

Proposition 2.2 By this we see R has simple dynamics. The system is always driven to synchrony.

Example We must construct an f function From the lemma we know h’(Ф)<-1 –Let h’(Ф)=- λ, λ>1 is independent of Ф –h(Ф)=-λ(Ф-Ф*)+Ф* –R(Ф)=λ 2 (Ф-Ф*)+Ф* –ƒ -1 must satisfy g’(ε+u)/g’(u)=λ for all u –Solution is a e bu –ƒ(Ф)=(1/b) (ln(1+e b-1 ) Ф)‏

As b goes to zero ƒ approaches the dashed line. For large b ƒ rises very rapidly and then levels off. –b measures how concave down the graph will be. –b is analogous to the conductance γ in the “leaky capacitor model Example cont…

Implications –Synchrony emerges more rapidly when the dissipation b or the pulse strength ε is large. –The time to synchronize is inversely proportional to the product εb Example cont…

Where is the fixed point? –Proposition 2.2 F(Ф)= Ф-h(Ф)=0 Rewrite as ε= ƒ (Ф*)- ƒ (1- Ф*)‏ –ƒ( Ф)=(1/b) (ln(1+e b -1) Ф)‏ –Ф*=(e b(1+ ε) -1)/((e b -1)(e bε +1))‏ –Graph of Ф* Example cont…

As ε  0 the fixed point  ½ –Holds for general ƒ –In the limit of small coupling; repelling fixed point always occurs with the oscillators in antiphase. Example cont…

What determines the Stability Type –The Eigenvalues λ=e bε Determines the stability ε>0, b>0 In this case λ>1 and Ф* is a repeller. –Note, if either ε<0 or b<0; Ф* is automatically stabilized –If both ε<0 & b<0 then the fixed point is again a repeller Example cont…

As System evolves, –oscillators clump together –Groups fire at the same time As Groups get bigger they create a larger pulse when they fire –Absorbs Other oscillators Ultimately 1 group remains –The population is then synchronized Population of Oscillators

Computer Simulation and the Two Main Proofs State of the system is characterized by the phases Ф 1,… Ф n of the remaining n=N-1 Oscillators –The possible states are S={(Ф 1,… Ф n ) ε R n s.t. 0< Ф 1 < Ф 2 <…< Ф n <1} Ф 0 =0 –The flow preserves cyclic ordering of the oscillators Because Oscillators are assumed to have identical dynamics and coupling is all to all Same frequencies so no order change between firings Monotonicity ensures the order is maintained –Ф n i s the next to fire. Then Ф n-1 and so on. After Ф n fires we re-label it to Ф 0 and increase the indices by one when the next fires. If the oscillators had different frequencies this indexing scheme would fail. –Oscillators could pass another and the dynamics would be more difficult to analyze

The firing map Let Ф=(Ф 1,… Ф n ) be the vector of phases immediately after firing. –Want to find the firing map h that transforms Ф to the vector phases right after the next firing. –Note: next firing occurs after a time 1-Ф n Oscillator I has drifted to the phase Ф i +1-Ф n, where I = 1,2,3,…,n-1 –Let σ=(Ф 1,… Ф n ) )=(1- Ф n ),Ф 1 +1-Ф n,.. Ф n-1 +1-Ф n ) ) =(σ 1,… σ n ) After firing, new phases are given by the map: –τ(σ 1,… σ n )=(g(f(σ 1 ) + ε),…, g(f(σ n ) + ε))‏ –h(Ф)= τ(σ(Ф))‏ Describes the new phases of the oscillators after one firing.

Absorption Set S is invariant under the affine map σ –Not invariant under the map τ Since f(σ n )+ ε ≧ 1 –When this happens Firing of oscillator n has also brought oscillator n-1 to threshold with it. –Both oscillators now act as one. –This is what is meant by absorption

Two complications of absorption –Problem 1 Domain of h is not all of S, instead we have –S ε ={(Ф 1,… Ф n ) ε S s.t f(Ф n-1 +1-Ф n ) + ε <1} –Or S ε ={(Ф 1,… Ф n ) ε S s.t Ф n -Ф n-1 >1-g(1-ε)} –Problem 2 Groups are created that will have enhanced pulse strength. –We must now allow for the possibility of different pulse strengths in the population. Absorption cont…

Main Theorem Two parts –Part I Rules out the possibility that elements of a set will not all be absorbed –Part II Rules out the possibility that there might exist sets which never get absorbed.

Main Theorem Consider two sets A which is the set of all oscillators within S that did not absorb B which is another set in S that did not absorb If B or A are not empty we have a problem.

Theorem 3.1 and Lebesgue measure The set A has Lebesgue Measure zero. For measure zero, set is empty

Theorem 3.1 and Lebesgue measure cont…

Assume another set B exists which is in S n but never achieves synchrony The set B has Lebesgue Measure zero. Theorem 3.2

Theorem 3.2 cont…

Numerical Results Computer Simulation –Let N=100 –S 0 = 2 –γ = 1 –ε = 0.3

Number of Oscillators over time –Little coherence in the beginning Organization is rather slow. –Down the road the synchrony builds up By t=9T – Perfect synchronization Numerical Results cont…

The slow beginning makes sense –Real world example Asian Fireflies at dawn –Buildup to synchrony is slow, due to stimulus of light from many sources. –Conflicting pulses in the incoherent initial stage. Numerical Results cont…

Evolution of the system in state space. –System strobed after each firing of oscillator i=1 –In the beginning some shallow parts –By seventh firing parts have become completely flat. Numerical Results cont…

Meaning the corresponding oscillators are firing in unison –Dominant group has emerged by the tenth firing. Numerical Results cont…

Conclusion Point of the Paper Model for 2 Oscillators Model for N Oscillators Main Theorem Conclusion

References “Synchronization of Pulse-Coupled Biological Oscillators”, Renato E. Mirollo and Steven H. Strogatz, Siam Journal of Applied Mathematics, Vol. 50, No. 6 (Dec 1990), PP

Questions?