Dynamical Systems in Neuroscience:

486 Synchronization (see www.izhikevich.com)H - H - H - H - H -H+ H+ H+ H+ H+ H+ H+nH - H -traveling waveFigure 10.32: Traveling wave solutions in chains of oscillators (10.18) describe undulatorylocomotion and central pattern generation.Frequency differencesSuppose the connections in (10.18) look qualitatively similar to those in Fig. 10.26, inparticular, H + (0) = H − (0) = 0. If the frequencies are all equal, then the in-phasesynchronized solution ϕ 1 = · · · = ϕ n exists and is stable. A traveling wave exists whenthe frequencies are not all equal.Let us seek the conditions for the existence of a traveling wave with a constantphase shift, say χ = φ i+1 − φ i , along the chain. Subtracting each equation from thesecond one, we find that0 = ω 2 − ω 1 + H − (−χ) , 0 = ω 2 − ω i , 0 = ω 2 − ω n + H + (χ) ,and ω 0 = ω 1 +ω n −2ω 2 . In particular, if ω 1 ≤ ω 2 = · · · = ω n−1 ≤ ω n , which correspondsto the first oscillator being tuned up and the last oscillator being tuned down, thenχ < 0 and the traveling wave moves up, as in Fig. 10.32, i.e., from the fastest to theslowest oscillator. Interestingly, such an ascending wave exists even when H − = 0, i.e.,even when the coupling is only in the opposite, descending direction.When there is a linear gradient of frequencies (ω 1 > ω 2 > · · · > ω n or vice versa),as in the cases of the smooth muscle of intestines or leech CPG for swimming, onemay still observe a traveling wave but with a non-constant phase difference along thechain. When the gradient is large enough, the synchronized solution corresponding toa single traveling wave disappears, and frequency plateaus may appear (Ermentroutand Kopell 1984). That is, solutions occur in which the first k < n oscillators arephase locked and the last n − k oscillators are phase locked as well, but the two poolsoscillate with different frequencies. There may be many frequency plateaus.Coupling functionsA traveling wave solution may exist even when all the frequencies are equal, if eitherH + (0) ≠ 0 or H − (0) ≠ 0. As an example, consider the case of descending coupling(H − = 0)ϕ ′ i = ω + H + (ϕ i+1 − ϕ i ) , i = 1, . . . , n − 1 .