nonlinear oscillators. method of averaging

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1 Physics 4 Spring 7 nonlinear oscillators. method of averaging lecture notes, spring semester 7 rozman/courses/p4_7s/ Last modified: April, 7 Oscillator with nonlinear friction Let s consider the following second order non-linear differential equation with the initial conditions d ( 3 x dx + ε + x =, ε > ( x( =., ẋ( =. ( The equation describes a non-liner oscillator with the friction force that is proportional to the third power of the velocity. The parameter ε is a positive parameter that describes the rate of the energy loss in the system. Equation ( has no exact analytic solutions, therefore below we compare our analytics with the results of numerical calculations. Page of

2 Physics 4 Nonlinear oscillators Spring 7. Numerical integration To solve Eq. ( numerically, we introduce a new dependent variable, y = ẋ and rewrite Eq. ( as a system of two first order differential equations for two unknown x(t and y(t, dx dy = y, = εy 3 x, A typical result of the numerical integration of Eqs. (3 is presented in Fig.. (3 x(t t Figure : Typical solution of Eq. ( for small values of ε; ε =. (solid line. The approximation Eq. (35 is also shown (shed line.. Regular perturbation theory for nonlinear oscillator ẍ + x = εẋ 3. (4 A perturbative solution of this equation is obtained by expanding x(t as a power series in ε: x = x + εx + ε x +..., (5 Page of

3 Physics 4 Nonlinear oscillators Spring 7 where x ( =, ẋ ( =, and x n ( =, ẋ n ( = for n. Substituting Eq. (5 into Eq. (4 and equating coefficients of like powers of ε gives a sequence of linear differential equations of which all but the first are inhomogeneous: ẍ + x =, (6 ẍ + x = ẋ 3, ( ẍ n + x n = ẋ 3 n, ( The solution of Eq. (6 which satisfies x ( =, ẋ ( = is To solve Eq. (, recall that x (t = cos(t. (9 sin 3 t = 3 4 sin(t sin(3t. ( 4 ẍ + x = ẋ 3 = sin3 (t, x ( =, ẋ ( =. ( x (t = 9 3 sin(t + 3 sin(3t 3 t cos(t ( 8 The amplitude of oscillation of the solution Eq. ( grows unbounded as t. The term t cos(t, whose amplitude grows with t, is said to be a secular term. The secular term has appeared because sin 3 (t on the right of Eq. ( contains a component, sin(t, whose frequency equals the natural frequency of the unperturbed oscillator, i.e. because the inhomogeneity sin(t is itself a solution of the homogeneous equation associated with Eq. (: ẍ + x =. In general, secular terms always appear whenever the inhomogeneous term is itself a solution of the associated homogeneous constant-coefficient differential equation. A secular term always grows more rapidly than the corresponding solution of the homogeneous equation by at least a factor of t. However, the correct solution of Eq. (4, x(t, remains bounded for all t. Indeed, let s multiply Eq. (4 by ẋ. ẋẍ + ẋx = εẋ 4. (3 Rearraging terms in the left hand side, we obtain: ẋẍ + ẋx = d ( dx + d x = d ( dx + x = de, (4 Page 3 of

4 Physics 4 Nonlinear oscillators Spring 7 where E ( dx + x = ẋ + x (5 is the mechanical energy of the oscillator. The energy Eq. (5 is always non-negative. For the initial condition Eq. (, E( =. On the other hand the right hand side of Eq. (3 is always non-positive. Therefore, i.e. d E, (6 E(t E( (7 which means that that neither x(t nor ẋ can grow unbounded, in contradiction with the result Eq. (..3 The method of averaging To obtain an approximate analytic solution of Eq. (, we use a powerful method called the method of averaging. It is applicable to equations of the following general form: d ( x + x = εf x, dx, (8 where in our case We seek a solution to Eq. (8 in the form: F ( x, dx ( 3 dx =. (9 x = a(tcos(t + ψ(t, ( dx = a(tsin(t + ψ(t. ( The motivation for this ansatz is that when ε is zero, Eq. (8 has its solution of the form Eq. ( with a and ψ constants. For small values of ε we expect the same form of the solution to be approximately valid, but now a and ψ are expected to be slowly varying functions of t. Differentiating Eq. ( and requiring Eq. ( to hold, we obtain the following relation: ȧcos(t + ψ(t a ψ sin(t + ψ(t =. ( Page 4 of

5 Physics 4 Nonlinear oscillators Spring 7 where we introduced the notation: ȧ, ψ dψ. (3 Differentiation of Eq. ( and substitution the result into Eq. (8 gives ȧsin(t + ψ a ψ cos(t + ψ = εa 3 sin 3 (t + ψ. (4 Solving Eqs. ( and (4 for ȧ and ψ, we obtain the following system of two differential equations: dψ So far our treatment has been exact. = εa 3 sin 4 (t + ψ (5 = εa sin 3 (t + ψcos(t + ψ. (6 Now we introduce the following approximation: since ε is small, dψ and are also small. Hence a(t and ψ(t are slowly varying functions of t. Thus over one cycle of oscillations the quantities a(t and ψ(t on the right hand sides of Eqs. (5 and (6 can be treated as nearly constant, and thus these right hand sides may be replaced by their averages: Eqs. (5 and (6 become π dφ... (7 π dψ = ε π π = ε π π dφa 3 sin 4 (φ (8 dφa sin 3 (φcos(φ (9 The right hand side of Eq. (9 is zero. The averaging in Eq. (8 can be done using the following trigonometric identities: sin (φ = ( cos(φ, π dφcos (nφ = π dφsin (nφ = π π, n =,,... Page 5 of

6 Physics 4 Nonlinear oscillators Spring 7 π dφcos(nφ = π dφsin(nφ =, n =,,... π π π dφsin 4 (φ = π dφ( π π ( cos(φ = = π dφ ( cos(φ + cos (φ = 4 π = ( + = (3 The averaged equations of motion are as following: The solution of Eq. (3 is We can chose the constant to be. dψ Eq. (3 can be solved by separating the variables: = ε 3 8 a3 (3 = (3 ψ = const. (33 a 3 = 3 8 ε a (t = 3 4 εt + a ( a(t = 3 4 εt +, (34 a ( where a( is the amplitude of oscillations at t =. Finally, cos(t x(t = 3 4 εt + a ( (35 Van der Pol oscillator The second order non-linear autonomous differential equation d x + ε ( x dx + x =, ε > (36 Page 6 of

7 Physics 4 Nonlinear oscillators Spring 7 is called van der Pol equation. The parameter ε is positive and indicates the nonlinearity and the strength of the mping. The equation models a non-conservative system in which energy is added to and subtracted from the system, resulting in a periodic motion called a limit cycle. The sign of the coefficient in the mping term in Eq. (36, ( x changes, depending whether x is larger or smaller than one, describing the inflow and outflow of the energy. The equation was originally proposed in the late 9-th to describe stable oscillations in electrical circuits employing vacuum tubes. a C d + V L I I b Figure : The Fitzhugh-Nagumo circuit used to model the nerve membrane. With a cubically nonlinear tunnel diode, I = I + F(V V, F(V = ( α 3 V 3 βv, it is described by van der Pol equation. V Figure 3: I(V for a tunnel diode. Note the negative resistance, di/dv <, for V V. Van der Pol oscillator is the example of a system that exibits the so called limit cycle. A limit cycle is an isolated closed trajectory. Isolated means that neighboring trajectories are not closed; they spiral either toward or away from the limit cycle. If all neighboring trajectories approach the limit cycle, we say the limit cycle is stable or attracting. Otherwise the limit cycle is in general unstable. Stable limit cycles model systems, e.g. the beating of a heart, that exhibit self-sustained oscillations. These systems oscillate even in the absence of external periodic forcing. There is a stanrd oscillation of some preferred period, waveform, and amplitude. If the system is perturbed slightly, it returns to the stanrd cycle. Limit cycles are inherently nonlinear phenomena. They can t occur in linear systerns. Of course, a linear system, such as a linear differential equation, can have closed orbits periodic solutions, but they won t be isolated. If x(t is a periodic solution, then so Page 7 of

8 Physics 4 Nonlinear oscillators Spring 7 is αx(t for any constant α. Hence x(t is surrounded by a family of closed orbits. Consequently, the amplitude of a linear oscillation is set entirely by its initial conditions. Any slight disturbance to the amplitude will persist forever. In contrast, limit cycle oscillations are determined by the structure of the system itself. Limit cycles are only possible in systems with dissipation. System that conserve energy do not have isolated closed trajectories.... Numerical integration Let s write Eq. (36 as a first order system of differential equations, dx dy = y, = ε ( x dx x, The results of numerical integration of Eqs. (37 are presented in Figs Numerical integration of Eq. (37 shows that every initial condition (except x =, ẋ = approaches a unique periodic motion. The nature of this limit cycle is dependent on the value of ε. For small values of ε the motion is nearly harmonic. Numerical integration shows that the limit cycle is a closed curve enclosing the origin in the x-y phase plane. From the fact that Eqs. (37 are invariant under the transformation x x, y y, we may conclude that the curve representing the limit cycle is point symmetric about the origin. (37. Averaging In order to obtain information regarding the approach to the limit cycle, we use the method of averaging. The method is applicable to equations of the following general form: d ( x + x = εf x, dx, (38 where in our case F ( x, dx = ( x dx. (39 Page 8 of

9 Physics 4 Nonlinear oscillators Spring 7 x(t t ẋ(t t Figure 4: Typical solution of van der Pol equation for small values of ε; top graph x(t, bottom graph ẋ(t; ε =. (solid line. The approximations Eq. (7, (7 shown as shed line. We seek a solution to Eq. (38 in the form: x = a(tcos(t + ψ(t, (4 Page 9 of

10 Physics 4 Nonlinear oscillators Spring 7 ẋ(t x Figure 5: Typical phase space trajectory of van der Pol equation for small values of ε. dx = a(tsin(t + ψ(t. (4 Our motivation for this ansatz is, as in the example before, that when ε is zero, Eq. (38 has its solution of the form Eq. (4 with a and ψ constants. For small values of ε we expect the same form of the solution to be approximately valid, but now a and ψ are expected to be slowly varying functions of t. Differentiating Eq. (4 and requiring Eq. (4 to hold, we obtain: cos(t + ψ(t adψ sin(t + ψ(t =. (4 Differentiating Eq. (4 and substituting the result into Eq. (38 gives sin(t + ψ adψ cos(t + ψ = εf (a(tcos(t + ψ, a(tsin(t + ψ. (43 Solving Eqs. (4 and (43 for where dψ dψ and, we obtain: = εf (a(tcos(t + ψ, a(tsin(t + ψsin(t + ψ (44 = ε F (acos(t + ψ, asin(t + ψcos(t + ψ, (45 a F (... = a ( a cos (t + ψ sin(t + ψ. (46 Page of

11 Physics 4 Nonlinear oscillators Spring 7 dψ So far our treatment has been exact. = εa ( a cos (t + ψ sin (t + ψ (47 = ε ( a cos (t + ψ sin(t + ψcos(t + ψ (48 Now we introduce the following approximation: since ε is small, dψ and are also small. Hence a(t and ψ(t are slowly varying functions of t. Thus over one cycle of oscillations the quantities a(t and ψ(t on the right hand sides of Eqs. (47 and (48 can be treated as nearly constant, and thus these right hand sides may be replaced by their averages: Eqs. (47 and (48 become π π... dφ (49 dψ As shown in the Appendix, = εa 3 cos (φsin (φ + εa sin (φ (5 = εa cos 3 (φsin(φ + εcos(φsin(φ (5 cos 3 (φsin(φ π I 3, =, (5 cos(φsin(φ π I, =, (53 thus the right hand side of Eq. (5 is zero. Therefore, i.e. ψ = C, where C is an integration constant. We can chose that dψ =, (54 ψ =. (55 The averaged terms in Eq. (47 are as following: cos (φsin (φ π I, = 8, (56 Page of

12 Physics 4 Nonlinear oscillators Spring 7 sin (φ π I, =, (57 where Eq. (6 and (7 have been used. Thus, the averaged Eq. (5 is = ε 8 a(4 a. (58 Eq. (58 can be solved separating variables: Decomposing the left hand side into partial fractions, we obtain Integrating both sides a( a( + a = ε. (59 8 a( a( + a = 4 a + 8 a a 8 + a, (6 a + a = ε, (6 + a d( a a d( + a + a = ε, (6 dlog ( a dlog a dlog( + a = ε. (63 ( a log = ε(t t (a +. (64 a We can choose an integration constant t to be. Exponentiating, which is a transcendental equation for a(t. a (a + a = eεt, (65 Since we are primary interested in the limit cycle solution of the van der Pol equation, we do not need the complete solution of Eq. (65 but only its limit as t. Inverting Eq. (65, we get which is possible if a( =. (a + a a = e εt as t, (66 Page of

13 Physics 4 Nonlinear oscillators Spring 7 To find the rate of approach to the limit cycle, we substitute the solution in the form a(t = δ(t (67 into Eq. (66 and keep only the term linear in δ. We assume that the initial conditions are such that a <, so that δ(t >. (a + a (4 + δ(tδ(t a = 4 + δ(t + δ(t 4δ(t 4 + δ(t = δ(t + δ(t δ(t ( δ(t δ(t. (68 We obtain that Thus, Finally, for t ε, δ(t = e εt. (69 a(t e εt as t. (7 x(t = a(tcos(t + ψ(t = ( e εt cost, (7 ẋ(t = a(tsin(t + ψ(t = ( e εt sint (7 are the parametric equations of the limit cycle in the phase plane. 3 Oscillator with the slowly changing frequency where the new dependence is highligted in bold. d ( x + x = εf x, dx, εt, (73 d x + ω (εtx =, (74 where ω(εt. (75 τ = f (t (76 dx = dx dτ dτ = df dx dτ, (77 Page 3 of

14 Physics 4 Nonlinear oscillators Spring 7 d x = d ( dx = d ( df = d f dx dτ + df d dτ dx = d f dx dτ dτ + df df = d f dx dτ + ( dx dτ Substituting Eq. (78 into Eq. (74 and introducing the notations ( d dx dτ ( df d x dτ (78 ẋ = dx dτ, ẍ = d x dτ, (79 where ( df ẍ + d f ẋ + ω (εtx =. (8 ( df = ω (εt df = ω(εt, (8 τ = t ω(εudu, dτ = ω(εt (8 d f = dω(εt = εω (T, (83 T = εt. (84 d x dτ + ε ω (εt dx ω + x =. (85 (εt dτ Eq. (85 is in the form Eq. (73. Using the method of averaging we obtain the following equations for a(t and ψ(t: dτ dψ dτ = ε ω (εt ω (εt a sin (τ + ψ, (86 = ε ω (εt ω a sin(τ + ψcos(τ + ψ. (87 (εt Averaging Eq. (86, (87 we obtain: dτ dψ dτ = ε ω (εt ω a, (88 (εt =. (89 Page 4 of

15 Physics 4 Nonlinear oscillators Spring 7 Eq. (89 telss us that ψ = const, and we can chose Eq. (88 can be solved separating variables a = ε ω (εt ω (εt dτ = ε x(t = a(tcos(τ = ψ =. (9 ω (εt ω (εt ω(εt = ω (εt d(εt. (9 ω(εt dlog(a = dlog(ω(εt, (9 log(a = log + C, (93 ω(εt a = C ω(εt (94 ( C t cos ω(εt ω(εt. (95 ẋ(t = C ( t ω(εt sin ω(εt. (96 E(t = ẋ + ω(εt x = C ω, (97 E(t = const. ω(εt (98 4 Problems Problem. Find (a the time dependence of the amplitude and (b the frequency of the Duffing oscillator: ẍ + x + ɛx 3 =, (99 where ɛ is a small parameter (ɛ ; x( =, ẋ( =. Compare your analytic approximation with the numerical solution of the differential equation. Page 5 of

16 Physics 4 Nonlinear oscillators Spring 7. Figure 6: Typical solution of the Duffing equation.5 Eq. (99, ε =. (solid line. The approximation. obtained by the method -.5 of averaging is also shown (shed line. -. Problem. friction: x(t t Find the time dependence of the amplitude of an oscillator with dry where γ is a small parameter (γ ; x( =, ẋ( =,, α >, sign(α =, α =, α <. Determine the time until the full stop. ẍ + γ sign(ẋ + x =, ( Compare your analytic approximation with the numerical solution of the differential equation. Figure 7: Typical solution of the dry friction oscillator Eq. ( for small values of γ; γ =.3 (solid line. The approximation obtained by the method of averaging is also shown (shed line. x(t t Problem 3. Find the solution of the following nonlinear differential equation: ẍ + ɛẋ 5 + x =, x( = x, ẋ( =, ( where ɛ is a small positive parameter. Compare your analytic approximation with the numerical solution of the differential equation. Page 6 of

17 Physics 4 Nonlinear oscillators Spring 7 Figure 8: Typical solution of the nonlinear friction oscillator Eq. ( for small values of ɛ; γ =.3 (solid line. The approximation obtained by the method of averaging is also shown (shed line. x(t t Appendix A. Integrals for the method of averaging The method of averaging requires the evaluation of integrals of the form where p and q are positive integers. I p,q = π cos p x sin q x dx, ( First, notice that the integration in Eq. ( is over the period of the integrand, thus for arbitrary u. π cos p x sin q x dx = π+u u cos p x sin q x dx (3 I p,q is zero if at least one of p or q is odd. Indeed, consider separately the three possible cases:. If p is even and q is odd, i.e. if p = m and q = n +, then I m,n+ = π cos m (xsin n+ (xdx = since the integrand is an odd function.. If both p and q are odd, i.e. p = m + and q = n +, then I m+,n+ = π cos m+ (xsin n+ (xdx = π π π π Page 7 of cos m (xsin n+ (xdx = (4 cos m (xsin n (xsin(xdx = (5

18 Physics 4 Nonlinear oscillators Spring 7 since the integrand is again an odd function; here we used the identity cos(xsin(x = sin(x. 3. If p is odd and q is even, i.e. if p = m + and q = n, then, using the identities sin(x = cos ( x π and cos(x = sin ( x π, I m+,n = π cos m+ (xsin n (xdx = π sin m+ ( x π ( cos n x π dx = 3 π π sin π m+ (ucos n (udu = since the last integral is from an odd function. To evaluate I pq when both p and q are even, let s proceed as following. π sin m+ (ucos n (udu = (6 I m,n = π ( cos (x m ( sin (x π n ( dx = cos (x m ( cos (x n dx π = ( cos (x m ( cos (x π n dx + ( cos (x m ( cos (x n dx. (7 π Let s introduce the new integration variable, u = cos x, u, du = cosx sinx dx. (8 In the first integral in Eq. (8, x π, thus both cos(x and sin(x are positive, therefore cos(x = u and sin(x = ( u. So, du = u ( u dx, (9 i.e. du dx =. ( u ( u Page 8 of

19 Physics 4 Nonlinear oscillators Spring 7 In the second integral in Eq. (8, π x π, thus cos(x is negative and sin(x is positive, therefore cos(x = u and sin(x = ( u. So, i.e. Substituting Eqs. ( ( into Eq. (7, we obtain I m,n = u m ( u n du + du = u ( u dx, ( du dx =. ( u ( u u m ( u n du = u m+ ( u n+ du. (3 The last integral is B ( m +,n +, therefore ( I m,n = B m +,n + In particular, Γ ( = π and Γ ( =, thus I, = Γ ( Γ ( Γ ( m + = Γ ( m + Γ (m + n +. (4 = π. (5 This trivial by itself result (obviously I, = π dx = π confirms the correctness of Eq. (4. Furthermore, Γ ( + = Γ ( = π, Γ ( =, Γ (3 = Γ ( =, thus I, = Γ ( ( + Γ ( = Γ = π (6 Γ ( and Finally, Γ ( + = 3 Γ ( 3 = 3 4 π, and I, = Γ ( + Γ (3 I 4, = Γ ( + Γ (3 Γ ( Page 9 of = π 4. (7 = 3π 4. (8

20 Physics 4 Nonlinear oscillators Spring 7 References [] S. H. Strogatz, Nonlinear Dynamics and Chaos: with Applications to Physics, Biology, Chemistry, and Engineering. Westview Press,. [] C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers. Springer Verlag, 999. Page of