Math 19. Lecture 29Fast and SlowT. JudsonFall 20041 Modeling t he Motion of the HeartSuppose that we have a system of equations with a periodic solution. Whatcan we say about the speed of the movement? For example, suppose thatwe look at a muscle controlling a heart valve. The muscle spends most of itstime moving slightly or very slowly near one of two positions (the value is inthe open position or the clo sed position). However, there is a rapid motionwhenever the valve is opened or closed. We can use the systemdxdt= −x33+ x + α (1)dαdt= −ǫx, (2)as a simple model, wherex(t) = the position of the muscle at time t,α(t) = the concentration of some chemical stimulusabove or below a fixed concentration at time t.Note that the x variable will be bounded. A heart valve can only move sofar. The inverse of the parameter ǫ > 0 will be used to estimate the amountof time that x spends at one or the other rest positions.11This equation is called van der Pol’s equation and occurs in circuit theory.12 Fast and Slow Subsyste msIn equations (1) and (2), we have a slow moving system interacting witha fast moving system. If ǫ > 0 is small and x is bounded, let us say that|x| < 10, then dα/dt is relatively small and cannot change quickly. Therefore,the slow moving system isdαdt= −ǫx.In this case, the speed at which α changes is never more than 10ǫ. On theother hand, the equationdxdt= −x33+ x + α,has no small number ǫ, and, t hus, its solutions can move relatively quickly.In fact,dxdt= xis exponential. In this case, the speed at which x can change can be relativelyhigh.3 What’s Going OnLet us see what happens to equationdxdt= −x33+ x + α,as we change α.• First, suppose that α = 2/3. Then x = 2 is a stable equilibrium fordxdt= −x33+ x +23.• However, α will not stay at 2/3, since it changes according todαdt= −ǫx.This equation tells us that α will decrease slowly.2• As α decreases (but stays above −2/3), the equa tiondxdt= −x33+ x + α (3)has a stable equilibrium point, say xα, fo r some positive x. This equi-librium point moves slowly to the left since α changes slowly.• At α = −2/3, the equilibrium point becomes unstable.• When α ( t) < −2/3, the equation (3) no longer even has a positiveequilibrium point.• As the positive equilibrium point becomes unstable and disappears, x(t)must travel from where it is nearly 1 to the remaining stable equilibriumpoint, where x < −2. The motion is governed bydxdt= −x33+ x + αand so it must occur at a speed governed by this equation. Since theequation does not include the parameter ǫ, the transition from wherex ≈ 1 to where x < −2 should occur much faster than it took for x todecrease from 2 to 1.• When x has switched to nearly −2, thendαdt= −ǫx.tells us that α must begin increasing again (dα/dt > 0 in this case).This will be a slow increase since ǫ is small. The increase will continueas long as α < 2/3.• As α r eaches 2/3, the point x = −1 is no longer a stable equilibriumpoint. Note that this is the mirror image of what happened before4 Existence of a Cyc l eWe can use the Poincar´e-Bendixson Theorem to show that we have a periodicsolution. The orig in is a repelling equilibrium point, and we ca n find a basin3of attraction bounded byα = x + 6α = x − 6x = 3x = −3α = 6α = −6.5 Fast and SlowWe wish to estimate the amount of time a perio dic solution stays in thevarious parts of it orbits in the (x, α)-plane. If ǫ > 0 is small, we can showthat it will take a very long time for α to move f r om 0 to −1/3 . On the otherhand, x moves from 1/2 to −3/2 relatively quickly when α < −1/2.Readings and References• C. Taubes. Modeling Di fferential Equations in Biology. Prentice Hall,Upper Saddle River, NJ, 2001. Chapter 24.• “Disparate Rates of Molecular Evolution in Cospeciating Hosts andParasites,” pp.
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