Nonlinear DynamicsSome exercises and solutions S. Strogatz – Nonlinear dynamics and chaos Dominik Zobel
[email protected] Please note: The following exercises should but mustn’t be correct. If you are convinced to have found an error, feel free to contact me. The Matlab codes below need some extra scripts which can be found at http://seriousjr.kyomu.43-1.org/notizen/. This work is licensed under the Creative Commons Attribution 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by/3.0/ version: 19 June 2013 correction(s): solution to exercise 3.1.4 Contents 2.1 2.2 2.4 2.7 3.1 3.2 A Geometric Way of Thinking . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Find all the fixed points of the flow. . . . . . . . . . . . . . . . . . . 2.1.2 At which points x does the flow have greatest velocity to the right? Fixed Points and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 x˙ = 4x2 − 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 x˙ = 1 − x14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 x˙ = x − x3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.4 x˙ = e−x sin (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.5 x˙ = 1 + 12 cos (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.6 x˙ = 1 − 2 cos (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.7 x˙ = ex − cos (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.10 Fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.13 Terminal velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear Stability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 x˙ = x(1 − x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.2 x˙ = x(1 − x)(2 − x) . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.3 x˙ = tan (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.4 x˙ = x2 (6 − x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2.4.5 x˙ = 1 − e−x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.6 x˙ = ln (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.7 x˙ = ax − x3 where a can be positive, negative, or zero. Discuss all three cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 x˙ = x(1 − x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2 x˙ = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.3 x˙ = sin (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.4 x˙ = 2 + sin (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.5 x˙ = − sinh (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.6 x˙ = r + x − x3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . Saddle–Node Bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 x˙ = 1 + rx + x2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 x˙ = r − cosh (x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 x˙ = r + x − ln (1 + x) . . . . . . . . . . . . . . . . . . . . . . . . x . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 x˙ = r + 12 x − (1+x) 3.1.5 (Unusual bifurcations) . . . . . . . . . . . . . . . . . . . . . . . . . Transcritical Bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 x˙ = rx + x2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 x˙ = rx − ln (1 + x) . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.3 x˙ = x − rx(1 − x) . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.4 x˙ = x(r − ex ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i 1 1 1 2 2 2 3 3 4 5 5 6 7 8 8 8 9 9 9 10 10 11 11 11 12 12 13 14 15 15 16 17 18 18 20 20 21 22 23 3.6 4.4 4.5 5.1 5.2 5.3 6.1 6.7 7.2 7.6 8.2 8.4 8.5 8.6 8.7 9.3 Imperfect Bifurcations and Catastrophes . . . . . . . . . . . . . . . . . . . 3.6.5 Mechanical example of imperfect bifurcation and catastrophe . . . . Overdamped Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.4 Torsional spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fireflies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Triangle wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Ellipses and energy conservation for the harmonic oscillator. . . . . 5.1.2 Consider the system x˙ = ax, y˙ = −y, where a < −1. . . . . . . . 5.1.3 x˙ = y, y˙ = −x . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.4 x˙ = 3x − 2y, y˙ = 2y − x . . . . . . . . . . . . . . . . . . . . . . 5.1.5 x˙ = 0, y˙ = x + y . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.6 x˙ = x, y˙ = 5x + y . . . . . . . . . . . . . . . . . . . . . . . . . . Classification of Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Consider the system x˙ = 4x − y, y˙ = 2x + y. . . . . . . . . . . . Love Affairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Consider the affair described by R˙ = J, J˙ = −R + J . . . . . . . Phase Portraits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.8 van der Pol oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.9 Dipole fixed point . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.10 Two–eyed monster . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.11 Parrot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.2 Pendulum driven by a constant torque . . . . . . . . . . . . . . . . Ruling Out Closed Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.10 Show that the system x˙ = y − x3 , y˙ = −x − y 3 has no closed orbits, by constructing a Liapunov function V = ax2 + by 2 with suitable a, b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Weakly Nonlinear Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.6 h(x, x) ˙ = xx˙ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.7 h(x, x) ˙ = (x4 − 1)x˙ . . . . . . . . . . . . . . . . . . . . . . . . . 7.6.8 h(x, x) ˙ = (|x| − 1)x˙ . . . . . . . . . . . . . . . . . . . . . . . . . Hopf Bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.12 Analytical criterion to decide if a Hopf bifurcation is subcritical or supercritical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Global Bifurcations of Cycles . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.3 Homoclinic bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . Hysteresis in the Driven Pendulum and Josephson Junction . . . . . . . . . 8.5.2 Consider the driven pendulum θ 00 + αθ 0 + sin (θ) = I. . . . . . . Coupled Oscillators and Quasiperiodicity . . . . . . . . . . . . . . . . . . . 8.6.7 Mechanical example of quasiperiodicity. . . . . . . . . . . . . . . . . Poincaré Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7.2 Consider the vector field on the cylinder given by θ˙ = 1, y˙ = ay. Chaos on a Strange Attractor . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 r = 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.3 r = 22 (transient chaos) . . . . . . . . . . . . . . . . . . . . . . . . 9.3.4 r = 24.5 (chaos and stable point co–exist) . . . . . . . . . . . . . . 9.3.5 r = 100 (surprise) . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.6 r = 126.52 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii 24 24 27 27 29 29 30 30 30 31 31 31 31 31 31 32 32 34 34 35 35 36 37 37 39 39 40 40 41 42 44 44 46 46 47 47 48 48 49 49 50 51 52 53 53 54 . . . .3 (intermittent chaos) . . . . . . . . .2 r = 212 (noisy periodicity) . . . . . . .3. . . 9. . . . . . 9. . . . . . .5. . . . . . . . . . .9. . . . . . . . . . 54 55 56 56 57 . . . . . . . . . . . .1 r = 166. . . . . .5 9.5. . . . . . . . . . . 9. . .8 Practice with the definition of an attractor Exploring Parameter Space . . . . . . . . . .7 r = 400 . iii . . . . .3. . . . . . . . . . . . . . . . . . . . 1. 1 . The flow has the greatest velocity to the right at all values x∗ . 2. 2. ! x˙ = 0 ⇔ sin (x) = 0 ⇒ x∗ = nπ ∀n ∈ N. At a fixed point. There are infinitely many fixed points. sin (x) = 1 ⇔ x∗ = π2 + n · 2π ∀n ∈ N.2 At which points x does the flow have greatest velocity to the right? The velocity and its direction are determined by the value of x. at the maximum positive value of the function ist the greatest velocity to the right.1 A Geometric Way of Thinking In the next three exercises. ˙ So.1 Find all the fixed points of the flow.1.Exercises for Chapter 2 2. the flow has to be zero. interpret x˙ = sin (x) as a flow on the line. 05:2. 2. x x˙ 100 0 ylim_extra=[1/6 -1/6]. 'MarkerEdgeColor'.2.5 % Gitter erzeugen [x y]=meshgrid([-1.2.Aufgabe 2.5 −5 % Differentialgleichung dx=1-x. sketch the vector field on the real line.startval).dx. .5 : 2).x_s]=ode23(inline('1-x. The fixed points are x∗1 = −1 (stable) and x∗2 = 1 (unstable).. ylim_extra).5).2 -.zeros(size(y)).%%'). ylim_extra)..1:4]. 'LineWidth'.5 [t_s.0). .[]. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x. .x_s.dx. hold on % Analytische Lösung t=0:0. 2. In each case. right: time– dependent behaviour x(t) with numerical solutions (start values x(0) = −4 : 0. 2. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x. 2 % Stabilitätsanalyse.2 x˙ = 1 − x14 No analytical solution found.5 −0. 50 −2 0 −50 −4 −2 0 x 2 −4 −0.25:1.[0 0.dx. right: time– dependent behaviour x(t) with numerical solutions (start values x(0) = −1 : 0.5.2..0025:1.2 Fixed Points and Stability Analyze the following equations graphically.'t'.'MarkerFaceColor'. classify their stability.2. for startval=-4:0.'Color'.'MarkerFaceColor'.5).55 0]) plot(t_s.'x').1: Left: Phase space of x˙ = 4x2 − 16.2: Left: Phase space of x˙ = 1 − x14 .55 0]) end . find all the fixed points.. 1 0 x x˙ 0.5. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.95 1/96].^14'.[0 0.[0../(1-exp(16*t)*C).5.5 t Fig. . . %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.55 0].Aufgabe 2.'LineWidth'.'o'.0). hold on % Numerische Lösung t=0:0. and sketch the graph of x(t)..75 0 0]) end 2. 150 2 % Differentialgleichung dx=x.'o'.1). 4 % Gitter erzeugen [x y]=meshgrid([-4:0. % Stabilitätsanalyse.startval.2.^14..5:2 C=(startval-2)/(startval+2).5 : 1.5 1 t Fig.1 -.zeros(size(y)). skizze_zeitverlauf(x.2*(1+exp(16*t)*C).y.5 4 0 0.[0 0.75 0 0]) plot(t.%%'). 0 ylim_extra=[-0.05:1.1 x˙ = 4x2 − 16 The analytical solution is: 2 x˙ = 4x − 16 ⇔ x=2 ⇔ 1 + C2 e16t 1 − C2 e16t Z 1 dx = 4 dt x2 − 4 x−2 C2 (t = 0) = x+2 Z ⇔ 1 x−2 ln = 4t + C1 4 x+2 There are two fixed points: x∗1 = −2 (which is stable) and x∗2 = 2 (unstable)..5 0 0. plot(0.75 0 0].y.[].. for startval=-1:0.2..t.startval.[].5:0.[0. 'MarkerEdgeColor'.dx.5 −10 −1 −1 0 x 1 −1.^4-16. 1.[].[0.5 1 1.5].2.'Color'. skizze_zeitverlauf(x. −0. plot(0. right: time– dependent behaviour x(t) with numerical solutions (start values x(0) = −2 : 0.[].[0.'o'.dx.01:5.4. 0 % Differentialgleichung dx=exp(-x). for i_count=2:4 set(handle(i_count).dx. t Fig.x_s.5.0).55 0]) plot(t_s.55 0]) end .2.x_s]=ode23s(inline('exp(-x).[].... Fixpunkte und zeitlicher Verlauf [substatusflag. t. 'LineWidth'.55 0].startval).[].*sin(x)'. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x. 3 hold on % Numerische Lösung t=0:0. ∀k ∈ N and the %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.handle]=stabilitaetsanalyse(x.y.*sin(x).4 x˙ = e−x sin (x) No analytical solution found..3 = ±1 (stable) and x∗2 = 0 (unstable).exp(t)*C.3 -. ...2. .startval.0).75 0 0]) plot(t. 0 0 x x˙ ylim_extra=[-1/3 -1/3].5*pi].2.'MarkerFaceColor'.'LineWidth'.'Color'. plot(0.4 -. right: time– dependent behaviour x(t) with numerical soluπ 9 tions (start values x(0) = − 13 4 π : 4 : 4 π)..25*pi:0.. skizze_zeitverlauf(x.2. 'MarkerEdgeColor'. 'MarkerEdgeColor'. 200 100 x x˙ 150 ylim_extra=[-87/512 -419/512]. zeros(size(y)). 50 −5 % Stabilitätsanalyse.5. The stable fixed points are x∗s = (2k − 1)π unstable fixed points are x∗u = 2kπ ∀k ∈ N.zeros(size(y)).startval.. .%%'). . ylim_extra).^3. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.'Color'.'o'. 0 −50 −10 −10 −5 0 x 5 0 2 4 t Fig.05*pi:2.75 0 0]..%%'). −1 −2 −2 −1 −2 −1 0 x 1 2 0 1 2 % Stabilitätsanalyse.'t'.25 : 2).5*pi [t_s.5)./sqrt(1+exp(2*t)*C^2). plot(0.[-2:2]) end skizze_zeitverlauf(x.Aufgabe 2.05:2.[0 0.2.05:2].4: Left: Phase space of x˙ = e−x sin (x).[0 0..75 0 0]) end 2.2.25:2 C=(startval)/sqrt(1-startval^2).[0 0.dx.[0.ylim_extra). for startval=-3. for startval=-2:0. 2.25*pi:0.[0.25*pi:2. .2.[]. 2 2 % Gitter erzeugen [x y]=meshgrid([-2:0.. 1 1 % Differentialgleichung dx=x-x.'x')..Aufgabe 2.3: Left: Phase space of x˙ = x − x3 . 5 % Gitter erzeugen [x y]=meshgrid([-3. hold on % Analytische Lösung t=0:0.3 x˙ = x − x3 The analytical solution is: x˙ = x − x ⇔ x = ±√ 3 Z ⇔ Cet 1 + C 2 e2t Z 1 1Z 1 1Z 1 1 dx = dx + dx − dx dt = 2 x(1 − x ) x 2 1−x 2 1+x x C(t = 0) = √ 1 − x2 Z There are three fixed points: x∗1.dx.'XTick'. 2.'MarkerFaceColor'.y.5). dx.Systems of the form x˙ = a + b cos (x). due to the definition of arctan (ϕ). However. 1 2 2.startval. this analytical solutions are restricted to the interval − π2 ≤ ϕ ≤ π2 .dx.. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2.75 0 0]) end skizze_zeitverlauf(x.2. right: time–dependent behaviour x(t) with numerical solutions (start values x(0) = −3π : π2 : 4π). for startval=-3*pi:0.'x'). for startval=-3*pi:0. skizze_zeitverlauf(x..... % grafische Korrektur plot(0.'LineWidth'.x_s]=ode23(inline('1+0.'Color'.[0 0.%%').8).2.75 0 0]) plot(t.'t'. .75 0 0]. C(t = 0) = √ arctan √ tan .02:8. . 1 + s2 Inserting and integrating yields Z dt = t + C = Z 1 dx a + b cos (x) = Z 1 2 ds · 1−s2 a + b 1+s2 1 + s2 √ ! x 2 a−b .y.0).5: Left: Phase space of x˙ = 1 + 12 cos (x). we substitute s = tan 2 and get cos (x) = 1 − s2 1 + s2 and dx = 2 ds. . 'MarkerEdgeColor'. it is straightforward to show the analytical solutions of the following two integrals.5*cos(x).2*(atan(sqrt(3)*tan(sqrt(3)/4*(t+C)))+ . pi*floor((startval+pi)/2/pi))..55 0].5 −5 0 −5 0 5 x 10 0 2 4 6 8 t Fig.t.. 'LineWidth'.. ylim_extra=[3/4 1/6].5*pi:4*pi [t_s.8). 1 5 % Stabilitätsanalyse.55 0]) plot(t_s.4.startval. plot(0.5.x_s.5 -.[]. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x.5*cos(x)'.02:8..5*pi:4*pi % An der Stelle t=0 gilt C=4/sqrt(3)*(atan(tan(startval/2)/sqrt(3))+ .[0..dx.1*pi:4*pi]. 'MarkerEdgeColor'.55 0]) end . startval).. 2.[]. x = 2 arctan 4 2 3 3 Analyzing the phase portrait (or the formula) reveals: There are no fixed points. % Gitter erzeugen [x y]=meshgrid([-3*pi:0.5 0 0. 4 hold on % Analytische Lösung t=0:0.'o'..zeros(size(y)).[0 0.[0.2. .Aufgabe 2..[0 0. % Grafischer Korrekturterm pi*floor(((t+C)*sqrt(3)/2+pi)/2/pi)). x First... ylim_extra). . . hold on % Numerische Lösung t=0:0.5 x˙ = 1 + cos (x) Using the formula from above.[0. the analytical solution is √ !! ! √ 3 4 1 x 3 tan (t + C) . % Differentialgleichung dx=1+0. =√ 2 tan arctan √ 2 2 a+b a −b Having this form. The analytical solution of a a system x˙ = a + b cos (x) can be obtained with some tricks.'o'.'Color'.'MarkerFaceColor'.5. x 10 x˙ 1..'MarkerFaceColor'. .%%').05:5.'LineWidth'. C(t = 0) = √ arctan tan .55 0].55 0]) end . There is an unstable fixed point at zero and no fixed point for x > 0. −5 0 −20 −15 % Stabilitätsanalyse. the analytical solution is √ √ !! ! 3i 3 x i 2 x = 2 arctan √ tan (t + C) .2.Aufgabe 2. 0 % Differentialgleichung dx=exp(x)-cos(x). 'MarkerEdgeColor'..'Color'. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x.6 x˙ = 1 − 2 cos (x) Using the formula from above.5*pi:0. . startval). right: time–dependent behaviour x(t) with numerical solutions (start values x(0) = −3π : π2 : 4π).[].55 0]) end 2. % Stabilitätsanalyse..x_s]=ode23(inline('exp(x)-cos(x)'. .[0.55 0]) plot(t_s.y.'MarkerFaceColor'.'x').'t'.[0 0. hold on % Analytische Lösung t=0:0.'o'.[0 0.75 0 0]) plot(t.7 x˙ = ex − cos (x) No analytical solution found. for startval=-3*pi:0.7: Left: Phase space of x˙ = ex − cos (x).55 0]) plot(t_s. ylim_extra). skizze_zeitverlauf(x. 2.%%'). plot(0.dx.0). the space between stable and unstable fixed points is approaching a constant value (π) as x → −∞. plot(0.05:5.'LineWidth'.5*pi if (startval > 1) t=0:0.t.'t'.7 -.1*pi:4*pi]. .'MarkerFaceColor'.. . startval).. skizze_zeitverlauf(x...5). for startval=-5*pi:0.2.x_s.[]..'o'. end [t_s.6 -.6: Left: Phase space of x˙ = 1 − 2 cos (x). Fig.2. for startval=-3*pi:0.'x').[0. 20 x x˙ 10 ylim_extra=[1/24 -15/16].[0 0.2.startval.[0 0. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x.x_s]=ode23(inline('1-2*cos(x)'.5*pi:4*pi % An der Stelle t=0 gilt C=2/sqrt(3)/i*atan(tan(startval/2)/i*sqrt(3)).'o'. . 2 i 2 3 3i %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.startval. right: time–dependent behaviour x(t) with numerical solutions (start values x(0) = −5π : π2 : π2 ).75 0 0]) end 0 0 −5 −1 −5 0 5 10 % Differentialgleichung dx=1-2*cos(x)..2. 'LineWidth'..75 0 0].199..55 0]. x x˙ points are x∗u % Gitter erzeugen [x y]=meshgrid([-3*pi:0.Aufgabe 2. 0 2 4 t x skizze_zeitverlauf(x.[0. plot(0.4.2.dx. . In the left half plane..[0 0. 'MarkerEdgeColor'. 30 5 % Gitter erzeugen [x y]=meshgrid([-5*pi:0. ylim_extra).zeros(size(y)).[]. hold on % Numerische Lösung t=0:0.'Color'.1*pi:2*pi].0).5).dx. −10 −10 −15 −10 −5 x 0 5 0 2 4 t Fig.2*atan(i/sqrt(3)*tan(sqrt(3)/2*i*(t+C))).2.. .startval.5.[0 0. .2.t.y. 5 hold on % Numerische Lösung t=0:0.'Color'.[].. ylim_extra=[1/6 1/6]. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.5)..4.x_s.zeros(size(y)).5.dx. ∗ The stable fixed points are xs = 2kπ − 1 arccos 2 ∀k ∈ N and the unstbale fixed = 2kπ + arccos 3 10 2 5 1 1 2 ∀k ∈ N.'MarkerFaceColor'. 'MarkerEdgeColor'..01:0.05:5. 2.5.5*pi:4*pi [t_s.dx. One choice of an adjusted. At a fixed point. explain why not. e. If the flow should be zero for all values of x ⇔ x˙ = 0. one could use a polynomial with 100 zeros. c) There are precisely three fixed points. d) There are no fixed points. because of the mean value theorem at a smooth function.) a) Every real number is a fixed point. the flow has to be zero. g. which requires a (smooth) periodic function. A stable or unstable fixed point implies changing the sign of the function values locally. and all of them are stable. The flow must be zero et every integer. find an equation x˙ = f (x) with the stated properties or if there are no examples. unstable) must be a fixed point of the other type. Any function whose flow is never zero.2. Between any two fixed point of the same type (stable. periodic function is x˙ = sin (πx).10 Fixed points For each of (a)–(e). assume that f (x) is a smooth function. (In all cases. All constant functions x˙ = c ∀c ∈ R\{0} have this property. Thus. this property cannot be fulfilled. b) Every integer is a fixed point. e) There are precisely 100 fixed points.2. Without assembling functions or restricting periodic functions to intervals. k=1 6 . 100 Q (x − k). . and thereby re-derive a formula for the terminal velocity.'MarkerFaceColor'. . {''.. 'LineWidth'.0)..2. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x.zeros(size(y)).2. where m is the mass of the dkydiver.dx.[].0).. As can be seen.75 0 0]) plot(t.[].''.^2+2. As t → ∞... .. else t=0:0.2. k c) Give a graphical analysis of this problem.'Color'.[].4.%%'). {'$$-\sqrt{\frac{mg}{k}}$$'. ylim_extra).y.'$$t$$'.[]. end C=(startval-1)/(startval+1). '$$\sqrt{\frac{mg}{k}}$$'}.. v ∗ k is the terminal velocity.''.05:0.8: Left: Phase space of v˙ = g − m v . %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.5:0. v(t) → q mg .. assuming that v(0) = 0.[0. .. . Separate the variables and integrate using mv˙ = mg − kv 2 ⇔ − 1 2 s ⇔ − 1 R x2 −a2 ! √ v − mgk m ln =t+C √ gk v + mgk ⇒ Due to tanh (x) = C2 = −1 1−e−2x .0).4.Aufgabe 2.(1+exp(-4*t)*C).'Children').'0'.[0. % Gitter erzeugen [x y]=meshgrid([-1. 'MarkerEdgeColor'. % Differentialgleichung mg k v v˙ ylim_extra=[-1/3 1/6]. This limiting velocity is called the terminal velocity.. right: time–dependent behaviour v(t) with numerically obtained trajectories.5]./(1-exp(-4*t)*C)..''}.'$$v$$'. Therefore. 0 % Stabilitätsanalyse.{'0'.25:1.75 0 0]) end % Achsenbeschriftung anpassen ebenen=get(gcf. .26.. 0 t k 2 Fig.dx.4.''. {''.'$$g$$'}. b) Find the limit of v(t) as t → ∞. and k > 0 is a constant related to the amount of air resistance. 0 r mg − k r mg − k 0 v r mg k % Achsenbeschriftung anpassen ebenen=get(gcf.. k So the terminal velocity is v∞ = q mg . Z mZ 1 dx = dt g k v2 − m k r ⇔ r v(0) = 0 = ⇒ v(t) = v= √ gk −2 t m mg 1 + C2 e √ gk k 1 − C e−2 m t √ gk 2 mg 1 − e−2 m t √ gk k 1 + e−2 m t the result can also be written as v(t) = q mg k tanh q gk t m . g is the acceleration due to gravity.''.'0'. '$$\sqrt{\frac{mg}{k}}$$'. plot(0.5 if (startval < -1) t=0:0.''}. hold on % Analytische Lösung for startval=-1. 7 skizze_zeitverlauf(x.75 0 0].0625:1.25:0. 1+e−2x 1 2a ln x−a x+a + C.'Children').2).'0'.13 -. .startval.[0. renameaxis(ebenen(2)..'$$\dot{v}$$'. r g % Funktion und Parameter dx=-2*x.'$$v$$'.1:2. renameaxis(ebenen(2).'o'. .. physically meaningful solutions (v q > 0) approach the stable fixed ∗ point v = mg as t → ∞. .'$$-\sqrt{\frac{mg}{k}}$$'.26. 2. . a) Obtain the analytical solution for v(t).13 Terminal velocity The velocity v(t) of a skydiver falling to the ground is governed by mv˙ = mg − kv 2 .2. 4.[].05:2]. 3 Fig. 0 % Gitter erzeugen [x y]=meshgrid([-1:0.5 % Differentialgleichung dx=-x.9: Phase space of x˙ = x(1 − x). ylim_extra).%%'). Positive values indicate a positive slope and therefore an instable fixed point.5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. 2.1 x˙ = x(1 − x) The fixed points are x∗1 = 0 and x∗2 = 1. use graphical argument to decide the stability.1 -. 0.5. −2 −1 0 1 x 2 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.[]. x∗2 = 1 and x∗3 = 2. −1 ylim_extra=[-1/3 1/6].2 x˙ = x(1 − x)(2 − x) The fixed points are x∗1 = 0. 2.0). 0 x˙ % Gitter erzeugen [x y]=meshgrid([-1:0.05:3].. The derivative is x¨ = −2x + 1.zeros(size(y)). x∗2 is stable (¨ x(x∗2 ) = −1) and x∗3 is unstable (¨ x(x∗3 ) = 2) again. Inserting the x–values of the first fixed point yields x¨(x∗1 ) = 1.2. 2.[]. The second fixed point is stable due to x¨(x∗2 ) = −1. −1 0 1 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. . Therefore.%%')..^3-3*x.zeros(size(y)).[]. −0.4. % Differentialgleichung dx=x. 2.y. If linear stability analysis fails because f 0 (x∗ ) = 0.2 -. Linear stability analysis means calculating the derivative and evaluate it at the values of the fixed points.. x∗1 is unstable. 2 x Fig.. x∗1 is unstable (¨ x(x∗1 ) = 2). The derivative is x¨ = −3x2 − 6x + 2.dx.Aufgabe 2. graphical analysis is needed. If the derivative is zero at the fixed point. ylim_extra).^2+2*x.4.4 Linear Stability Analysis Use linear stability analysis to classify the fixed points of the following systems. . −1 ylim_extra=[-1/3 -1/3].Aufgabe 2. Negative values result in a stable fixed point with the same argumentation.5. 8 .0). x˙ 2 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.4.^2+x.10: Phase space of x˙ = x(1 − x)(2 − x).y.dx. %%').5 x˙ = 1 − e−x 2 The only fixed point is x∗ = 0 and the derivative is x¨ = 2xe−x .3 -.5 -. −50 ylim_extra=[-1/6 -1/6].dx. 1 Fig. the third fixed point is stable (¨ x(x∗3 ) = −36) and the stability of x∗1. 0. 2.5.4.%%').y.y.4 -..4. −10 −1 0 x % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. 2.[]. x˙ 10 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2 cannot be determined by linear stability analysis.4 % Differentialgleichung dx=1-exp(-x.%%').Aufgabe 2. Graphical analysis reveals that x∗1.[].y.49*pi]..0). 2.2 0 ylim_extra=[1/6 1/6]. −4 −2 0 2 x 4 6 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.. 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.Aufgabe 2.[].^3+6*x.4 x˙ = x2 (6 − x) The fixed points are x∗1. 10 x 2 Fig.2 = 0 and x∗3 = 6.4. 8 Fig. all fixed points are unstable (¨ x(x∗ ) = 1 ∀k ∈ N).^2. Graphical analysis can be used to classify the fixed point as semistable.1:8]. 0 % Differentialgleichung dx=-x.5. x˙ 0.. k π2 ) ∀k ∈ N is a fixed point x∗ = kπ. The derivative is x¨ = −3x2 + 12x.13: Phase space of x˙ = 1 − e−x .1*pi:4*pi].2 is semistable.Aufgabe 2. % Differentialgleichung dx=tan(x). 50 x˙ % Gitter erzeugen [x y]=meshgrid([-4:0.12: Phase space of x˙ = x2 (6 − x). 0.zeros(size(y)).[].5.^2).dx.zeros(size(y)).3 x˙ = tan (x) In any interval [(k − 1) π2 . 9 .0). ylim_extra).0). 0 % Gitter erzeugen [x y]=meshgrid([-0..2. −5 ylim_extra=[-1/3 -1/3].11: Phase space of x˙ = tan (x).[]. ..01*pi:0.4. ylim_extra). ylim_extra). 2.49*pi:0.8 % Gitter erzeugen [x y]=meshgrid([-3*pi:0.zeros(size(y)). Thus.4. . −5 0 5 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. Again.4. the stability cannot be determinded using linear stability analysis. As the derivative x¨ = 1 + tan (x)2 shows.[].dx.6 0. 2 2. . 100 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. . lower left: a > 0.y.4. So.'0'.0).05:4.26.%%'). Discuss all three cases The fixed points vary as the parameter a is varied.''. .'keep'.zeros(size(y)).''}.''. {''.'keep'. a) a < 0: There is only one fixed point x∗ = 0. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.[].4.'0'. {''.'0'. 4 Fig.'0'.15: Phase space of x˙ = ax − x3 .%%').0). x∗2 is unstable (¨ x(x∗2 ) = a) while the other ones are stable (¨ x(x∗1.Aufgabe 2.4. This fixed point is stable x¨(x∗ ) = a.. .'$$-\sqrt{a}$$'. 1 2 x 3 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.3 ) = −2a).7 x˙ = ax − x3 where a can be positive.'Children').'keep'.''}.'$$\sqrt{a}$$'. 0 x 0 √ − a 0 x √ a Fig.6 -.'XTick'. The derivative is x¨ = −3x2 + a.y.. x˙ 0 x % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.[].05:0.[]. Upper left: a < 0.2. {''.0). . 10 end % Achsenbeschriftung anpassen ebenen=get(gcf. .05:2].Aufgabe 2.''.'0'. To determine its stability.'0'.''.. case 0 renameaxis(ebenen(2).''. for i_change_xtick=1:3 set(ebenen(i_change_xtick). 0 x˙ x˙ % Gitter erzeugen [x y]=meshgrid([-2:0. end .'keep'.4.''.''.6 x˙ = ln (x) The only fixed point is x∗ = 1.dx.[]. linear stability analysis cannot be used.''}. 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. √ √ c) a > 0: In this case. x∗2 = 0 and x∗3 = a). 0 ylim_extra=[-1/3 -1/3]. . upper right: a = 0. the fixed point is unstable (¨ x(x∗ ) = 1). 2. negative.^3..5.[].14: Phase space of x˙ = ln (x). three fixed points exist (x∗1 = − a.''}..[].''}.{''. case 1 renameaxis(ebenen(2).4. −3 ylim_extra=[1/6 1/6]..26. for a=-1:1 % Differentialgleichung dx=a*x-x.zeros(size(y)). The derivative is x¨ = x1 . 2. ylim_extra)..0).. or zero.7 -.0). there is only one fixed point x∗ = 0 with a multiplicity of three.'keep'. 'implementiert'). {''.[-2 -1 0 1 2]) end switch a case -1 renameaxis(ebenen(2)..26.''... ylim_extra). b) a = 0: Again.. −1 −2 % Differentialgleichung dx=log(x). 2.''}.'keep'.{''. . otherwise disp('Warnung: Keine Achsenanpassung '.[]. 1 x˙ 0 % Gitter erzeugen [x y]=meshgrid(0...dx. . right: potential function V (x) = −3x + C with C = 0. dx 2. 2.V'..5. ylim_extra=[1/6 1/6].'$$x$$'.[].1 x˙ = x(1 − x) 3 2 The potential of this function is V (x) = x3 − x2 + C. 2.'$$V$$').0). V x˙ −0..dx.Aufgabe 2. 2 −2 0 x 2 −2 0 x 2 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.'$$x$$'. 2. 5 3 V x˙ 3.y]=meshgrid(-2:0. ylim_extra=[-1/6 -1/3]. 11 % Potential darstellen customplot(x'.7 Potentials For each of the following vector fields. 1 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -..1:2.5 −2 0 −3 −2 2 ylim_extra=[-2/3 1/6].7. x x Fig.17: Left: Phase space of x˙ = 3.y]=meshgrid(-2:0. right: po3 2 tential function V (x) = x3 − x2 + C with C = 0. ylim_extra).dx.%%').5 % Differentialgleichung dx=3*ones(size(x)). 0 % Feld anlegen [x.%%').5 0 −1 % Differentialgleichung dx=x-x. 4 % Feld anlegen [x.ylim_extra.0). ylim_extra).7.[]. plot the potential function V (x) and identify all the equilibrium points and their stability The potential can be calculated with x˙ = − dV .1:3.5 −5 ylim_extra=[1/3 1/3].5 −1.[].^3/3-x. It can be seen.7. This function has no extremum (within finite values) and therefore no fixed points.[].'$$V$$'). −1 −2 % dazugehöriges Potential V=x.^2.2.[].1 -.. % Potential darstellen customplot(x'.[].16: Left: Phase space of x˙ = x(1 − x). 0 2 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. 0 % dazugehöriges Potential V=-3*x.[].^2/2.zeros(size(y)). .ylim_extra.Aufgabe 2.7. .5.[]. that the function has a local maximum at V (x∗u ) = 0 (indicating an unstable fixed point) and a local minimum at V (x∗s ) = 1 (stable fixed point).2 x˙ = 3 The potential of this function is V (x) = −3x + C. Fig. 0. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.y.2 -. 2.zeros(size(y)).y.V'. % Potential darstellen customplot(x'.[].'$$x$$'. ylim_extra). 0 % Differentialgleichung dx=2+sin(x).[]. ylim_extra).0).5 −0.5. 3 1. .dx. 12 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. There are no minima/maxima in V (x) and thus no fixed points.2..18: Left: Phase space of x˙ = sin (x).[].[]. 1 1 0. ylim_extra=[1/6 1/6]. 2.zeros(size(y)).Aufgabe 2.5 −1 −1 −5 0 x 5 % Feld anlegen [x.Aufgabe 2. 2.5 0 0 V x˙ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. % Potential darstellen customplot(x'.y]=meshgrid(-2*pi:0.zeros(size(y)). ylim_extra=[0 0].5. 2 V x˙ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.19: Left: Phase space of x˙ = 2 + sin (x). 10 2. −5 0 x 5 Fig.'$$x$$'.7. The minima of V (x) (stable fixed points) are V (x∗s ) = (2k − 1)π ∀k ∈ N and the maxima (unstable fixed points) are V (x∗u ) = 2kπ ∀k ∈ N. Fig..%%').y]=meshgrid(-2*pi:0.ylim_extra. % dazugehöriges Potential V=cos(x). .'$$V$$').05*pi:2*pi. right: potential function V (x) = cos (x) + C with C = 0.[].3 -.7. % Differentialgleichung dx=sin(x).. −5 1 −10 −5 0 x 5 ylim_extra=[1/6 1/6].V'.[]. −5 0 x 5 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.5 % dazugehöriges Potential V=-2*x+cos(x).dx. .%%'). 2.ylim_extra. ylim_extra=[1/6 1/6]. right: potential function V (x) = −2x + cos (x) + C with C = 0.[].3 x˙ = sin (x) The potential of this function is V (x) = cos (x) + C.y.[].5 0.0).7.7.05*pi:2*pi.4 x˙ = 2 + sin (x) The potential of this function is V (x) = −2x + cos (x) + C. −0.5 5 % Feld anlegen [x.V'..4 -.'$$V$$').y. 20: Left: Phase space of x˙ = − sinh (x). There is one global minimum at V (x∗ ) = 0 (stable fixed point).ylim_extra. −5 0 x 5 Fig. 13 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. % Potential darstellen customplot(x'. V x˙ 100 0 50 % dazugehöriges Potential V=cosh(x).'$$V$$').5 -. ylim_extra=[1/12 -1/3]. ylim_extra).y]=meshgrid(-2*pi:0.dx.7. .5. −100 0 −5 0 x 5 ylim_extra=[-1/6 -1/6].05*pi:2*pi.y. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. right: potential function V (x) = cos (x) + C with C = 0..zeros(size(y)).[]. 2.[]. .2.. 100 % Differentialgleichung dx=-sinh(x).5 x˙ = − sinh (x) The potential of this function is V (x) = cosh (x) + C.Aufgabe 2.V'.%%'). 150 % Feld anlegen [x.7.[].[].0).'$$x$$'. 5 1 1 0 V x˙ 4 The potential of this function is V (x) = x4 − x2 − rx + C. .5 −1 −1 −2 0 x 2 end 0 −2 0 x 2 Fig.25*x.2.^3.Aufgabe 2.5 0 % Differentialgleichung dx=r+x-x.'$$x$$'.0). % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. 0.dx..1/sqrt(3):4/sqrt(3). 1.6 -. 0.5 V x˙ 1 0 −0. .[]. r = 2 27 .y]=meshgrid(-4/sqrt(3):0. ylim_extra). The W–potential indicates the outer fixed points to be stable and q 4 the inner to be unstable. 1. 1 V x˙ 1 % Variation des Parameters for r=0:sqrt(4/27):2*sqrt(4/27) 0. 2 ylim_extra=[-2/5 -2/5].[]. r = 27 . For values of |r| < 27 .y.V'.^2-r*x. 4 right column: potential function V (x) = x4 − x2 C = 0.5 2 1 ylim_extra=[1/48 -5/8].[].zeros(size(y)).6 x˙ = r + x − x3 4 q 2 2 1.'$$V$$'). 14 % Potential darstellen customplot(x'. From top to bottom 2 − rx + C with q q 4 4 row: r = 0.5. there are three fixed points.%%').5 −1 0 −2 −2 0 x −2 2 0 x 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. 2.21: Left column: Phase space of x˙ = r + x − x3 .5*x.ylim_extra.. 0 −1 −0.^4-0.5 −2 0 x −2 2 0 x % dazugehöriges Potential V=0.5 2 % Feld anlegen [x.[].7.7. At |r| = 27 two fixed points annihilate each other and only a stable one remains. The curves approach f1 = −r and f2 = 0 as |r| → ∞.2) . 2 ylim_extra=[1/12 -1/3].'$$x$$'). . x˙ x˙ 15 10 5 % Variation des Bifurkationsparameters for r=0:2:4 5 0 0 −6 −4 −2 x 0 2 −6 −4 −2 x 0 % Differentialgleichung dx=1+r*x+x.. 2 x x˙ 4 0 0 −4 −6 % Bifurkationspunkte finden ch1=max(find(r_bf<=-2)). size(x_1(1:ch1).1:2. sketch the bifurcation diagram of fixed points x∗ versus r.Aufgabe 3. set x˙ = 1 + rx + qx = 0 2 to analyse the curve of the fixed points... . 15 % Plotte das Bifurkationsdiagramm customplot(.^2/4-1). to be determined.1 Saddle–Node Bifurcation For each of the following exercises. . .0).0 0 1 1]. ch2=min(find(r_bf>=2)). top left: r = 0. bottom right: bifurcation diagram of x˙ = 1 + rx + x2 . [min(r_bf) max(r_bf)].^2/4-1). '$$r$$'. 25 15 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.[].%%'). [x_1([1:ch1 ch2:end]) x_2([1:ch1 ch2:end])]'. bottom left: r = 4.1 x˙ = 1 + rx + x2 2 A stable and an unstable fixed point exist as |r| ≥ 2. . 3.1:4]...1. [r_bf([1:ch1 ch2:end]) r_bf([1:ch1 ch2:end])]'. . x_1=-r_bf/2-sqrt(r_bf. sketch all the qualitatively different vector fields that occur as r is varied. both functions describe the curve of the bifurcation diagram.2) size(x_1(ch2:end).[].^2. 3. The argument of the square root has to be nonnegative which is fulfilled for |r| ≥ 2. Show that a saddle–node bifurcation occurs at a critical value of r. Rearranging the terms yields x1.. end 6 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. [size(x_1(1:ch1). 20 10 % Feld anlegen [x.Exercises for Chapter 3 3.2 = − 2r ± r4 − 1.2) size(x_1(ch2:end).1.5.y. Finally. x_2=-r_bf/2+sqrt(r_bf... Finally.1 -... To see this. −2 −2 −4 −2 x 0 2 −4 −2 0 r 2 4 Fig. ylim_extra).[]..zeros(size(y)).2).1: All except bottom right: Phase space of x˙ = 1 + rx + x2 .y]=meshgrid(-6:0. top right: r = 2.dx.. 2 % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[-4:0... 2: All except bottom right: Phase space of x˙ = r − cosh (x).'$$r$$'.2 = ± arcosh (r).[size(x_1(ch1:end). [r_bf(ch1:end) r_bf(ch1:end)]'. . 0 1]. −6 −2 0 x 2 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. x2 = − arcosh (r) is the unstable fixed point.2 -. bottom right: bifurcation diagram of x˙ = r − cosh (x). top right: r = 1.%%').Aufgabe 3. . x_1=acosh(r_bf).1.1:3.[].. . While cosh (x) can never get smaller than 1. −2 −4 −4 % Feld anlegen [x. . [x_1([ch1:end]) x_2(ch1:end)]'.05:3].1.. top left: r = 0.y. [min(r_bf) max(r_bf)]..dx. 3. arcosh (r) must have an argument r ≥ 1. . .3. 0 0 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2 x˙ = r − cosh (x) A stable and an unstable fixed point exist as r ≥ 1. bottom left: r = 2..zeros(size(y)). 16 % Plotte das Bifurkationsdiagramm customplot(. −6 % Variation des Bifurkationsparameters for r=0:2 x˙ x˙ −2 −6 −8 −2 0 x 2 −2 0 x % Differentialgleichung dx=r-cosh(x). −1 0 1 r 2 3 Fig...2).. ylim_extra).2) size(x_1(ch1:end). x x˙ 0 −2 −4 −1 % Bifurkationspunkte finden ch1=min(find(r_bf>=1)). Rearranging the terms yields x1.[]. 2 ylim_extra=[-1/6 1/6]..y]=meshgrid(-3:0.'$$x$$')...0). x_2=-acosh(r_bf). We set x˙ = r − cosh (x) = 0 to analyse the curve of the fixed points.. []. 2 1 end 0 % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[-1:0.5.. So. the fixed point approaching x = −1 is stable.y.5].. ln (1 + x) has to have values x ≥ −1. r → −∞. solving to r results in r = ln (1 + x) + x.3 -.. [x_1([1:ch1]) x_1([ch1:end])]'.zeros(size(y)).1... bottom right: bifurcation diagram of x˙ = r + x − ln (1 + x).0). % Parametervariation und dazugehörige Fixpunktgleichungen x_1=[-1:0.2). If x → −1 or x → ∞. bottom left: r = 1. ylim_extra).'$$x$$'). the other one unstable..05:2.5.[size(x_1(1:ch1).05:3. there are no fixed points for r > 0. 0 1]. .[].[-1 1].dx. . −2 −3 −1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. top right: r = 0. % Feld anlegen [x.1.'$$r$$'. [r_bf(1:ch1) r_bf(ch1:end)]'.3: All except bottom right: Phase space of x˙ = r + x−ln (1 + x)..[]. −1 0 1 x 2 −2 −1 3 % Variation des Bifurkationsparameters for r=-1:1 0 1 x 2 % Differentialgleichung dx=r+x-log(1+x).. For r < 0.. 0 0 % Bifurkationspunkte finden ch1=max(find(x_1<=0)). top left: r = −1..3 x˙ = r + x − ln (1 + x) 0 1 −1 0 x˙ x˙ While solving x˙ = 0 for x is problematic. .%%').2) size(x_1(ch1:end). −1 −1 −1 0 1 x 2 3 −1 0 r 1 Fig... r_bf=log(1+x_1)-x_1.3. 3 2 2 x x˙ end 1 1 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. 3.y]=meshgrid(-1:0. 17 % Plotte das Bifurkationsdiagramm customplot(. 3 ylim_extra=[1 -1/6].Aufgabe 3. [1/6 1/6].. 4: All except bottom right: Phase space of x˙ = x r + 12 x − (1+x) .25).5-r_bf-sqrt(r_bf..2) size(x_1(ch2:end).. ch2=min(find(r_bf>=1. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.Aufgabe 3.. Due to the type of function (asymptotic behaviour for x → ±∞). bottom right: bifurcation x diagram of x˙ = r + 12 x − (1+x) .[size(x_1(1:ch1).1. 10 5 0 1 2 6 4 2 0 −2 −4 −6 −8 2 end % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x./(1+x)..5-3*sqrt(2):0. x_2=0.'$$r$$'.5+sqrt(2)))..5) delete(punkt(3)) else delete(punkt(1)) end % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[1..'Children'). −4 −3 −2 −1 x 10 −4 −3 −2 −1 x % Variation des Bifurkationsparameters for r=0:1. if (r < 1. top left: r = 0.2 = 12 −r ± r2 − 3r + 14 . 3. the fixed point farther away from −1 is always the unstable fixed point. bottom left: r = 2.4 x˙ = r + 12 x − x (1+x) q √ Solving for x yields x1.^2-3*r_bf+0.^2-3*r_bf+0.01*sqrt(2):1..y.. % Bifurkationspunkte finden ch1=max(find(r_bf<=1. punkt=get(ebenen(3).2) .25).5-r_bf+sqrt(r_bf. ylim_extra).'$$x$$').5. . 10 5 x˙ x˙ 0 −5 % Differentialgleichung dx=r+0.5*x-x.[]. size(x_1(1:ch1). As the root is nonnegative for |r − 32 | > 2. .dx. .y]=meshgrid(-4:0.4 -.1. the two fixed points cease to exist within this interval. top right: r = 1.5-sqrt(2))).0).5+3*sqrt(2)]. [x_1([1:ch1 ch2:end]) x_2([1:ch1 ch2:end])]'.. x_1=0. % Übergang von -Inf +Inf ist kein Fixpunkt-> Korrektur ebenen=get(gcf.'Children'). % Plotte das Bifurkationsdiagramm customplot(.. 1 0 0 1].. [r_bf([1:ch1 ch2:end]) r_bf([1:ch1 ch2:end])]'.[]. [].3.5 (Unusual bifurcations) In discussing the normal form of the saddle–node bifurcation.1.2).zeros(size(y))..1:2. . −2 −1 0 1 2 3 4 5 r Fig..%%'). 3.[].2) size(x_1(ch2:end). −10 0 1 2 x x˙ 5 0 −5 −10 0 1 ylim_extra=[1/6 1/6].5:3 0 −5 −10 −4 −3 −2 −1 x % Feld anlegen [x. we mentioned the as. . ∂f . To see what can happen if ∂f . 6= 0. sketch the sumption that a = ∂r . ∗ = 0. 2 = ±|r|.rc ) (x . Accordingly. So. 18 . and then plot the fixed points as a function of r. x1 = −r is stable for r < 0. x2 = r is unstable for r < 0 and stable otherwise. and for r > 0 unstable.rc ) vector fields for the follwing examples. ∗ ∂r (x . a) x˙ = r 2 − x2 : There is one stable and one unstable fixed point. Rearranging the terms gives x1. .5: All except bottom right: Phase space of x˙ = r2 − x2 . 3. [x_1([1:ch1 ch1:end]) x_2([1:ch1 ch1:end])]'. x_2=-sqrt(r_bf..y]=meshgrid(-3:0.2).'$$x$$'). x_1=sqrt(-r_bf... bottom right: bifurcation diagram of x˙ = r2 − x2 . bottom left: r = 1.y.1:3].zeros(size(y)).1.'$$x$$'). r = 0. 0 0 1 1]. .5 x x˙ 4 0 end 2 % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[-3:0.2) size(x_1(ch1:end). ylim_extra).. top right: r = 0. bottom right: bifurcation diagram of x˙ = r2 + x2 with only one halfstable fixed point at x = 0. x_1=sqrt(r_bf.1:3].'$$r$$'. 0 % Feld anlegen [x. Rearranging the terms gives x1.[].. x_2=-sqrt(-r_bf.2 = ±i|r|. −6 −2 0 x 2 −2 0 x 2 ylim_extra=[-1/3 1/6].[size(x_1(ch1). where all terms are purely imaginary except for r = 0. . .5b -. x˙ x˙ 4 2 % Variation des Bifurkationsparameters for r=-1:1 0 −2 0 x 2 −2 0 x 2 % Differentialgleichung dx=r^2+x. −3 −2 0 x 2 −2 0 r 2 Fig. −4 % Variation des Bifurkationsparameters for r=-1:1 −2 −4 % Differentialgleichung dx=r^2-x.[]. .^2.5. 6 6 4 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -..%%').y]=meshgrid(-3:0.. [r_bf([1:ch1 ch1:end]) r_bf([1:ch1 ch1:end])]'.[].1:3... ylim_extra).5. [].zeros(size(y)). 0 x˙ x˙ 0 −2 % Feld anlegen [x.dx. Fig. 19 % Plotte das Bifurkationsdiagramm customplot(.2) size(x_1(ch1:end). 3.'$$r$$'.. 0 r 1 % Bifurkationspunkte finden ch1=max(find(r_bf==0)). 0 −2 −1 −2 −4 % Bifurkationspunkte finden ch1=max(find(r_bf<=0))..[]. 0 1]..0).Aufgabe 3. bottom left: r = 1.^2).0).6: All except bottom right: Phase space of x˙ = r2 + x2 . 3 2 2 end x 1 x˙ 0 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. [x_1(ch1) x_2(ch1)]'.[]. −0.1.2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. % Plotte das Bifurkationsdiagramm customplot(. b) x˙ = r 2 + x2 : There is only one halfstable fixed point at x = 0 for r = 0.^2.[].%%').5a -.5 0 −2 0 x 2 −1 −1 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. size(x_1(1:ch1). 6 0.. % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[-3:0. 1 ylim_extra=[1/6 -1/3]..^2). top right: r = 0.[size(x_1(1:ch1)...1:3.dx. [r_bf(ch1) r_bf(ch1)]'... [].2).2) size(x_1(ch1).Aufgabe 3.2) .^2). top left: r = −1. .. top left: r = −1.^2).y.. .. 0).y. x_1=zeros(1.. 3.2)). 2 % Feld anlegen [x.2 Transcritical Bifurcation For each of the following exercises.'$$r$$'.. 0 2 −1 0 % Bifurkationspunkte finden ch1=min(find(r_bf>=0)).[]. ..%%').[].2). x1 represents the stable fixed point and x2 the unstable one. to be determined.^2.. x 4 x˙ % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. Show that a transcritical bifurcation occurs at a critical value of r. At r = 0 they change stability. .Aufgabe 3. r. ylim_extra).dx.. −2 −2 −3 −2 0 x 2 −2 0 r 2 Fig.. While r < 0. . size(x_1(1:ch1).1 x˙ = rx + x2 There are two fixed points. Finally.[size(x_1(1:ch1).3. 6 8 6 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. . 20 % Plotte das Bifurkationsdiagramm customplot(. top right: r = 0.2) . [x_1([1:ch1 ch1:end]) x_2([1:ch1 ch1:end])]'.2. 0 % Variation des Bifurkationsparameters for r=-2:2:2 x˙ x˙ 4 2 0 −2 −2 0 x 2 −2 0 x % Differentialgleichung dx=r*x+x.5.7: All except bottom right: Phase space of x˙ = rx + x2 .. described by x1 = 0 and x2 = −r. .1 -. bottom left: r = 2. 3 8 2 end 6 1 % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[-3:0. 2 ylim_extra=[1/6 -1/3].1:3]. 3. x_2=-r_bf.1:3.y]=meshgrid(-3:0. bottom right: bifurcation diagram of x˙ = rx + x2 . [r_bf([1:ch1 ch1:end]) r_bf([1:ch1 ch1:end])]'. 0 1 1 0].zeros(size(y))... sketch all the qualitatively different vector fields that occur as r is varied.. []. sketch the bifurcation diagram of fixed points x∗ vs..'$$x$$').2.2) size(x_1(ch1:end).[].2) size(x_1(ch1:end).size(r_bf. top left: r = −2.. 0.2 -.Aufgabe 3.2) ... % Plotte das Bifurkationsdiagramm customplot(.[size(x_1(1:ch1)..05:6].'$$r$$'.5.. 2 2 1.5 1 x˙ x˙ 1 0 0.5 1. x_2=zeros(1.8: All except bottom center: Phase space of x˙ = rx − ln (1 + x). It is stable while r < 1. middle right: r = 1. r_bf=log(1+x_1). . so r = ln (1+x) is used to describe the behaviour.. bottom center: bifurcation diagram of x˙ = rx − ln (1 + x). [x_1([1:ch1 ch1:end]) x_2([1:ch1 ch1 end])]'.y]=meshgrid(-1:0. 3. ylim_extra). % Variation des Bifurkationsparameters for r=0:0..2. one fixed point moves along x1 = 0.. 21 .dx. The x stable fixed points approaches x = −1 as r → ∞. Here./x_1. . middle left: r = 1. top left: r = 0. x˙ cannot be transformed to x = f (r).3..2.2) size(x_1([ch1 end]).5 −1 −2 −1 0 −0.'$$x$$'). 0 1 1 0]..5:1..size(x_1. .5 0 0 −1 0 1 x 2 3 −1 0 1 x 2 3 % Parametervariation und dazugehörige Fixpunktgleichungen x_1=[-1:0.2) size(x_1(ch1:end).2)..5.5. At r = 1 a second fixed point appears at x = ∞ changes its stability from unstable to stable at r = 1.[]. 2 1 0 −1 0 1 2 3 r Fig. 3 x end % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. size(x_1(1:ch1). [0 -1/6].2 x˙ = rx − ln (1 + x) Here..%%')..5 0 1 x 2 −1 3 0 1 x 2 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2)).[0 3].5 2 % Differentialgleichung dx=r*x-log(1+x).zeros(size(y)). [r_bf([1:ch1 ch1:end]) r_bf([1:ch1]) -1 1]'. % Feld anlegen [x. 4 % Bifurkationspunkte finden ch1=max(find(x_1<0)).y. x˙ x˙ 1 1 ylim_extra=[1/6 -1/6]. top right: r = 0.[].05:3.0). size(x_1(1:ch1)..5..5 −1 0 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. x_1=zeros(1.. 0 −4 −1 0 1 x 2 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x.zeros(size(y)).size(r_bf.[].. which yields in a different appearance of the fixed points around zero. −2 0 r 2 Fig.1:2.0). 22 . −3 −1 0 1 −1 2 0 1 2 % Variation des Bifurkationsparameters for r=[-1 0 0.. . else ylim_extra=[-1/6 -1/6].9: All except bottom right: Phase space of x˙ = x − rx(1 − x). middle left: r = 0. . ylim_extra).5.2) size(x_1(ch1:end).5:0. else if (r<0) ylim_extra=[-1/3 1/6]./r_bf.2) size(x_1(ch1:end). 2 1.3. x1 = 0 and x2 → 1.2)).[].'$$x$$'). x x˙ 2 2 0 −2 % Plotte das Bifurkationsdiagramm customplot(. x_2=(r_bf-1).[]. x1 = 0 is stable for x < 1 and x2 = r−1 for x > 0.3 x˙ = x − rx(1 − x) Two fixed points exist and interchange stability at r = 1.5 1 1 x˙ x˙ 0 0.Aufgabe 3.[size(x_1(1:ch1). [-1/6 -1/6]. bottom right: bifurcation diagram of x˙ = x − rx(1 − x). [x_1([1:ch1 ch1:end]) x_2([1:ch1 ch1:end])]'.. x2 → ∞ as r → 0 and x2 comes from r −∞ for r > 0. 6 4 4 % Bifurkationspunkte finden ch1=min(find(r_bf>=1)). top left: r = −1. . middle right: r = 1. end end x˙ x˙ 2 1 1 0 0 −1 −1 0 1 2 −1 x 0 1 2 x end 6 % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[-3:0.^2. 1 0 0 1].5 1 2] x x 2 4 % Differentialgleichung dx=x-r*x+r*x.. top right: r = 0.5.2). As |r| → ∞..%%'). [r_bf([1:ch1 ch1:end]) r_bf([1:ch1 ch1:end])]'.5. −2 −0.. Therefore. 3.2.dx.3 -. ..y]=meshgrid(-1.'$$r$$'. bottom left: r = 1. 3 if (r>0) ylim_extra=[1/6 -1/3].2)..y.5 % Feld anlegen [x.2. As can be seen..1:3]. −1 −2 0 1 2 3 r Fig.2) size(x_1(ch1:end). size(x_1(1:ch1).[]. % Variation des Bifurkationsparameters for r=-1:2 0 % Differentialgleichung dx=x..size(r_bf.. . x_2=log(r_bf). % Parametervariation und dazugehörige Fixpunktgleichungen r_bf=[0:0.2) size(x_1(ch1:end).2.dx. [r_bf([1:ch1 ch1:end]) r_bf([1:ch1 ch1:end])]'. . 0 1 1 0].2.%%'). x 0 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2).y.4 -. there is only one (stable) fixed point at x = 0. x˙ x˙ −5 −5 ylim_extra=[1/6 1/6].Aufgabe 3.2) .. So.10: All except bottom center: Phase space of x˙ = x(r − ex ). For r > 0 another fixed point emerges and merges with the stable fixed point at r = 1 to change its stability.y]=meshgrid(-4:0. % Bifurkationspunkte finden ch1=min(find(r_bf>=1)).1:3]. 23 . 5 0 0 x˙ x˙ −5 −5 −10 −10 −15 −15 −20 −4 −2 0 2 −4 −2 x 0 2 % Feld anlegen [x.5. top left: r = −1. As long as r < 0.zeros(size(y)).[size(x_1(1:ch1).. . 3.3. .[]. x_1=zeros(1. ylim_extra)..*(r-exp(x)). 0 x % Plotte das Bifurkationsdiagramm customplot(.4 x˙ = x(r − ex ) Two fixed points exist and interchange stability at r = 1... top right: r = 0. [x_1([1:ch1 ch1:end]) x_2([1:ch1 ch1:end])]'... middle left: r = 1.'$$r$$'. −10 −10 end −4 −2 0 2 −4 −2 x 0 2 x 1 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x..0). middle right: r = 2.1:2.. x1 = 0 is stable for x < 1 and x2 = ln (x) for x > 1.2)).. bottom center: bifurcation diagram of x˙ = x(r − ex ). [].'$$x$$').[]. 1+u2 . Now we have to modify the argument of the square root to get the dependence from u to x and we are done. g b) Show that the equilibrium equation can be written in dimensionless form as 1 − uh = √ R 2 for appropriate choices of R. Thus.6 Imperfect Bifurcations and Catastrophes 3. R = L0 a and h = 24 mg sin (θ) ak yields 1 − h u = √ R . we can restrict ourselves to w F mg sin (θ) forces in the direction of the wire. the sum of all forces acting on the bead must be zero.3. Dividing sin (θ) 0 by kx and rearranging yields 1 − mg kx = √xL2 +a 2 . choosing u = xa . While spring the gravitational force is simply mg sin (θ). coefficient k) is linearly dependent on the length of the spring.5 Mechanical example of imperfect bifurcation and catastrophe Consider the bead on a tilted wire discussed at the end of section 3. The spring force (relaxed length of spring L0 .6.proj = kx 1 − x2 +a2 which is equal to mg sin (θ). h and u. Therefore. a) Show that the equilibrium positions of the bead satisfy mg sin (θ) = kx 1 − √ L20 x +a2 . At an equilibrium position.6. The force projected on the √ x 2 + a2 yields direction of the wire is Fspring. AlFwire a x though we don’t know the normal force of the wire Fwire .proj = − L ). we need one term without x (which must be made 1). In short. θ the spring force requires some more calmg cos (θ) culation. a term with x1 and a term similiar to √1x2 . Fspring = k(w − L0 ). 1+u The variable in the dimensionless form is u (u ∼ x). Replacing w = x k(w 0 w L0 √ Fspring. −2 −2 −4 −5 0 u 5 −5 0 u % Gitter erzeugen [x y]=meshgrid([-8:0.[].^2)... else delete(elemente(3)). 8 6 6 f (u) f (u) 4 4 2 2 0 0 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. fu=inline('1-param. middle right: R = 3.[]. '$$h$$'. top left: R = −6.[size(hR..1:6. bottom left: R = 6.'param'. 1+u2 top right: R = −3. h = 1)./x-R.[0 10]). two (on curve) or three (above curve) fixed points.. −2 −4 −4 −5 0 u 5 −5 0 u 5 4 8 6 0 end R f (u) 2 −2 4 % Bifurkationspunkt linke Seite für R>1 verfolgen u=-12:0. bottom right: Dependence of h and R to have one (below curve).'Children').%%').2). depending on h.0:0. end renameaxis(bild(2).10e-9./sqrt(1+x.'Children').01:-0.. middle left: R = 0.1).10.1:8].'$$R$$').0].'$$f(u)$$')./u-((left+right)/2). which is in the right half plane (close to zero).[]. The location is determinded by the size of R and h (and close to 2 for R = 1.5 -.y. For R > 1 the situation is more involved. 6 6 for R=-6:3:6 % Differentialgleichung dx=1-h. again).. How many equilibria can exist in each case? The curve is approaching one as u → ∞.. all below in one and values on the curve in two.fu./sqrt(1+u. If R < 1. ylim_extra).dx. There are no oscillations but an overshoot on either side of the vertical axis is possible.11: All except bottom right: Plot of 1 − uh = √ R with h = 1. two more fixed points can exist (the location is depending on the size of both parameters. there is exactly one (unstable) fixed point..[0 1/32]. While the unstable fixed point is moving to infinity..1). 5 % Parameter festlegen h=1.c) Give a graphical analysis of the dimensionless equation for the cases R < 1 and R > 1. 3.'left'. h Fig.^2)'.0). elemente=get(bild(3). 'u'.6.'right'). −4 2 −6 −5 0 u 5 % Nachbearbeitung (kein Fixpunkt bei Null) bild=get(gcf. −2 % Stabilitätsanalyse und Fixpunkte stabilitaetsanalyse(x. if (R <0) delete(elemente(2)). .'$$u$$'. A numerical investigation for some values is shown below. 4 4 2 f (u) f (u) 2 0 0 ylim_extra=[-1/5-R/30 -1/5+R/30]. 25 % Grafische Ausgabe customplot(hR(:.zeros(size(y)). 0 2 4 6 % Nullstellen der Funktion finden [sf hR]=funczeros(u.hR(:.Aufgabe 3. All parameter values (points) above the R–h curve result in three fixed points. h) Interpret your results physically. A small value means small relative excitation. we have reduced the equilibrium equation to h + ru − 21 u3 ≈ 0. r can be seen as the length of the spring relative to its relaxed length. e) Find an approximate formula for the saddle–node bifurcation curves in the limit of small r. that the perpendiculat spring force has to be much higher than the force of the bead along the wire. in terms of the original dimensional variables. h and u. The saddle–node bifurcation occur at the local minimum/maximum of our equation h + ru − 12 u3 ≈ 0. |h| > |hc | in one. √ Using the approximation 1 + u2 ≈ 1 + 12 u2 for small values of u.d) Let r = R − 1. This can also be achieved by having a very small tilt angle. Values |h| < |hc | result in three fixed points. As the last part suggested. h is the ratio between the force of the bead along the wire and the spring force perpendicular to the wire. we obtain h r+1 = u 1 + 12 u2 1 ⇔ (u − h) 1 + u2 = ur + u 2 1 1 ⇔ h + u2 h + ru − u3 = 0 2 2 1− Ignoring 12 u2 h. 26 . changing the h less than hc results in one stable equilibrium point. Here. Show that the equilibrium equation reduces to h + ru − 12 u3 ≈ 0 for small r. We get the value where the bifurcation occurs with the help of the derivative d 1 3 ! h + ru − u3 = r − u2 = 0 du 2 2 s h(umin/max ) = ± s ⇔ umin/max = ± 2 r 3 8 3 r 27 The approximate formula q for the saddle–node bifurcation curves in the parameter 8 3 space h. h and u. a small value indicates. Otherwise the bead will have two stable equilibria (and an unstable one) on the wire. r is hc (r) = ± 27 r . as ξ is greater than one. As ξ resp. There are two fixed points on the circle (infinitely on the line) if |ζ| ≤ 1. Then the equation of motion becomes bθ˙ + mgL sin (θ) = Γ − kθ. On the other hand. the angular velocity is not uniquely defined and not on a circle. more fixed points emerge (always in pairs: a stable and an unstable one). if ξ > 0. What kind of bifurcations are they? If k = 0 (ξ = 0). only one (stable) fixed point exists. d) Show that many bifurcations occur as k is varied from 0 to ∞. and for the opposing force k ≥ 0. where at least one stable fixed point and up to n stable and n − 1 unstable fixed points may exist (n can be arbitrarily large). otherwise none. b) Nondimensionalize the equation Dividing by mgL and substituting τ = mgL t b with the definition θ0 = dθ dτ yields θ0 = ζ − ξθ − sin (θ). The most interesting intervall is 0 < ξ < 1. the pendulum will eventually approach a stable fixed point and therefore come to rest. 27 . mgL c) What does the pendulum do in the long run? A natural assumption is mgL > 0.4 Overdamped Pendulum 4. the overdamped pendulum would actually describe a well–defined vector field on the circle. the spring winds up and generates an opposing torque −kθ. we have exactly one (stable) fixed point where the system is driven to. if ξ ≥ 1. As the pendulum rotates.Exercises for Chapter 4 4. Thus. a) Does this equation give a well–defined vector field on the circle? ˙ ˙ + 2π) have to be the same (periodicity). because θ(θ) and θ(θ falsified by the existence of the term kθ. there are no fixed points possible for |ζ| > 1 and infinitely for |ζ| ≤ 1. This spontaneous emerging of two fixed points is typical for saddle–node bifurcations and can be seen in the plots below. As described in the previous part. If we allow k = 0 and therefore ξ = 0. To sum it up.4 Torsional spring Suppose that our overdamped pendulum is connected to a torsional spring. whereas ζ = Γ mgL and ξ = k .4. which can easily be No. k → 0. startval. t θ 10 15 % Differentialgleichung dx=zeta-xi*x-sin(x). skizze_zeitverlauf(x. −0. θ θ′ 5 0 0 −10 −5 −5 0 5 10 15 0 5 10 15 t θ 15 4 10 2 θ θ′ 5 end 0 0 −2 −5 −5 0 5 θ 10 15 0 5 10 15 t Fig.[].y.0).5 5 θ θ′ 2. ξ = 1.55 0]) bild=get(gcf. third row: ζ = π. ..4.t.zeros(size(y)). 28 hold on % Numerische Lösung t=0:0.2].55 0]. ξ = 0.'$$\theta$$').'MarkerEdgeColor'.'Children').. [0 0.8.'Color'.2. for startval=-3*pi:pi/2:5*pi [t_s..5 −5 0 5 10 15 0 5 10 15 for index=1:4 zeta=zeta_vec(index). right side: time dependence.x_s.[0 0.2.5 1 0 0.10). first row: ζ = 1. bild=get(gcf.Aufgabe 4. renameaxis(bild(2).5 −5 0 −5 0 5 10 15 0 5 10 15 t θ 15 1.x_s]=ode23(dgl_sys.10.'$$t$$'.'$$\theta^\prime$$').5 0 0 −5 zeta_vec=[1. % Stabilitätsanalyse.15 2 10 1. dgl_sys=@(t.dx. 4.'LineWidth'.[0 0.'MarkerFaceColor'.55 0]) plot(t_s.. ξ = 0.1: Left side: Phase portrait of θ0 = ζ − ξθ − sin (θ).'o'. θ θ′ 2 0.8 pi pi/2].[].'$$\theta$$'.2 0. ylim_extra).02:5*pi. 10 ylim_extra=[1/6 1/6]. renameaxis(bild(2). xi_vec=[0 0 1 0. fourth row: ζ = π2 .x)[zeta-xi*x-sin(x)].%%').5 10 1 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2.startval).. Fixpunkte und zeitlicher Verlauf stabilitaetsanalyse(x. ξ = 0. second row: ζ = 0.dx. plot(0. end .4 -.'Children'). % Gitter erzeugen [x y]=meshgrid([-3*pi:pi/48:5*pi]. xi=xi_vec(index). 4.5 Fireflies 4.5.1 Triangle wave In the firefly model, the sinusoidal form of the firefly’s response function was chosen ˙ = Ω, θ˙ = ω + Af (Θ − θ), somewhat arbitrarily. Consider the alternative model Θ where f is given now by a triangle wave, not a sine wave. Specifically, let ( f (φ) = φ, − π2 ≤ φ ≤ π2 π − φ, π2 ≤ φ ≤ 23 π on the interval − π2 ≤ φ ≤ 32 π, and extend f periodically outside this interval. a) Graph f (φ). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -- Aufgabe 4.5.1 -- %%'); π 2 theta1=-pi/2:0.1*pi/2:pi/2; theta2=pi/2:0.1*pi/2:3/2*pi; xval=[theta1,theta2,2*pi+theta1,2*pi+theta2]; yval=[theta1,pi-theta2,theta1,pi-theta2]; customplot(xval',yval',[],[1/6 1/6],[size(xval,2);0], ... '$$\theta$$','$$f(\theta)$$'); ebenen=get(gcf,'Children'); for i_correct=1:3 set(ebenen(i_correct),'XTick',[-pi/2:pi:7/2*pi], ... 'YTick',[-pi/2:pi/4:pi/2]) end f (θ) π 4 0 − π 4 − π 2 − π 2 π 2 3 π 2 5 π 2 renameaxis(ebenen(2),'keep','keep',[], ... {'$$-\frac{\pi}{2}$$';'$$\frac{\pi}{2}$$'; ... '$$\frac{3}{2}\pi$$';'$$\frac{5}{2}\pi$$'; ... '$$\frac{7}{2}\pi$$'},{'$$-\frac{\pi}{2}$$'; ... '$$-\frac{\pi}{4}$$';'0'; ... '$$\frac{\pi}{4}$$';'$$\frac{\pi}{2}$$'}); 7 π 2 θ Fig. 4.2: Triangle wave as defined above. b) Find the range of entrainment. In the range of entrainment, the firefly is able to synchronize (match frequency). ˙ − θ˙ to be zero and therefore This implies the difference φ˙ = Θ ˙ − θ˙ = 0 = Ω − ω − Af (Θ − θ) φ˙ = Θ ⇔ Ω = ω + Af (Θ − θ). f (θ) ranges from − π2 to π2 , so the range of entrainment is ω − A π2 ≤ Ω ≤ ω + A π2 . c) Assuming that the firefly is phase–locked to the stimulus, find a formula for the phase difference φ∗ . Being phase–locked, φ˙ = Ω − ω − Af (φ∗ ) = 0 which yields f (φ∗ ) = Ω−ω . As can be A seen from above, |f (φ∗ )| < π2 . d) Find a formula for Tdrift . Using the integration formula for Tdrift , inserting f (φ) and partitioning the integral in smooth intervals yields Tdrift = Z2π 0 1 = A 2π Z dt 1 dφ = dφ dφ Ω − ω − Af (φ) 0 π 2 Z − π2 3 1 1 dφ + Ω−ω A −φ A Z2 π π 2 29 Ω−ω A 1 2 dφ = ln A −π+φ Ω−ω A Ω−ω A + − π 2 π 2 ! . Exercises for Chapter 5 5.1 Definitions and Examples 5.1.1 Ellipses and energy conservation for the harmonic oscillator. Consider the harmonic oscillator x˙ = v, v˙ = −ω 2 x. a) Show that the orbits are given by ellipses ω 2 x2 + v 2 = C, where C is any nonnegative constant. (Hint: Divide the x˙ equation by the v˙ equation, separate the v’s from the x’s, and integrate the resulting separable equation.) v x˙ = v˙ −ω 2 x ⇔ ⇔ Z −ω 2 x dx = Z v dv ⇔ 1 2 2 1 2 ω x + v = C1 2 2 ω 2 x2 + v 2 = C Due to the addition of quadratic terms, the constant C must be nonnegative. b) Show that this conclusion is equivalent to conservation of energy. Multiplication with m yields 12 mω 2 x2 + 12 mv 2 = C2 . The second term on the left q side can be interpreted as the kinetic energy. Substituting ω = constant k in the first term gives the spring energy 12 kx2 . k m with the (spring) 5.1.2 Consider the system x˙ = ax, y˙ = −y, where a < −1. Show that all trajectories become parallel to the y–direction as t → ∞, and parallel to the x–direction as t → −∞. (Hint: Examine the slope dy dx = xy˙˙ .) Both equations can be observed independently. Separating the variables and integrating dx = ax yields x(t) = C1 eat C. For the second equation we obtain y(t) = C2 e−t . Inserting dt dy C2 t(−a−1) e both results in the slope yields dx = − aC . Since a < −1, the slope will go to zero 1 (parallel to x–axis) as t → −∞ or to infinity (parallel to y–axis) as t → ∞. 30 Write the following systems in matrix form. We make use of the fact, that two first order systems x ˙ and y˙ ! x˙ either or both variables can be written in the form =A y˙ with ! linear dependence in x . y 5.1.3 x˙ = y, y˙ = −x ! " # x y ! ! " # x y ! x˙ 0 −1 = y˙ −1 0 5.1.4 x˙ = 3x − 2y, y˙ = 2y − x x˙ 3 −2 = y˙ −1 2 5.1.5 x˙ = 0, y˙ = x + y ! " # x y ! ! " # x y ! x˙ 0 0 = y˙ 1 1 5.1.6 x˙ = x, y˙ = 5x + y x˙ 1 0 = y˙ 5 1 5.2 Classification of Linear Systems 5.2.1 Consider the system x˙ = 4x − y, y˙ = 2x + y. a) Write the system as x˙ = Ax. Show that the characteristic polynomial is λ2 − 5λ + 6, and find the eigenvalues and eigenvectors of A. ! " # x˙ 4 −1 = y˙ 2 1 ! x , y det(A) = (λ − 4)(λ − 1) + 2 = 0 ⇔ √ 5 ± 52 − 4 · 6 λ1,2 = ⇒ λ1 = 3, 2 " # ! −1 1 0 ∗ A(λ1 ) = v = ⇒ v1 = −2 2 1 0 " A∗(λ2 ) # ! −2 1 0 = v = −2 1 2 0 31 λ2 − 5λ + 6 = 0 λ2 = 2 ! 1 1 ! ⇒ 1 v2 = 2 ⇒ C1 = 2. b) Find the general solution of the system. d) Solve the system subject to the initial condition (x0 . our fixed point at the origin is unstable. C2 = 1 C1 + 2C2 = 4 So our solution is ! ! ! x(t) 2 3t 1 2t = e + e y(t) 2 2 5. a) Characterize the romantic styles of Romeo R and Juliet J . ! ! ! ! x(0) 1 3·0 1 2·0 3 = C1 e + C2 e = y(0) 1 2 4 ⇔ C1 + C2 = 3. We insert the initial condition in our general solution and determine the constants. Thus. the more Romeo loves Juliet. The general solution can be obtained by inserting the eigenvectors and eigenvalues in the fundamental solution z(t) = C1 v1 eλ1 t + C2 v2 eλ2 t with some constants C1 and C2 which yields ! ! ! x(t) 1 3t 1 2t = C1 e + C2 e y(t) 1 2 c) Classify the fixed point at the origin. Additionally.1 = 12 ± 23 i. y0 ) = (3.2 Consider the affair described by R˙ = J. What does this imply for the affair? The system can be described as R˙ 0 1 = ˙ −1 1 J ! " # ! R .3 Love Affairs 5. Calculating the trace (tr (A) = 5) and determinant (det (A) = 6) yields an unstable node. 32 . the faster his love for her grows and vice versa. Romeos affection grows or decays depending on Juliets state and size of affection. Juliet’s growth of affection is depending on her actual state of love. 4). The more Juliet loves him. the more Juliet’s love is decaying. they pull and push each other in infinitly growing (change of) love and hate. The eigenvalues are λ1. J √ which has the characteristic equation λ2 − λ + 1.3. Since both eigenvalues are positive. The eigenvalues are λ1 = 3 and λ2 = 2 and their eigenvectors are v1 = 1 1 v2 = 1 2 T T and . Two complex eigenvalues with a positive real part imply an unstable spiral as fixed point at the origin. b) Classify the fixed point at the origin. In contrary. J˙ = −R + J . Jt'. So our fundamental solution is ! X(t) = C1 1 2 ! √ 1√ ( 12 + 23 i)t e + C2 + 23 i 1 2 √ 1√ ( 12 + 23 i)t e . − 23 i Using R(0)√ = 1 and J(0) = 0.5*t).1*pi:4.'r--'. t'.. sin(sqrt(3)/2. To calculate the functions R(t) and J(t). For the second eigenvalue.%%'). 5.5*t).*t). . Jt=-exp(0..5*t).R you may find it useful to calculate the derivative of R(t) and examine it. hold(handles(3).1.-2/sqrt(3)*exp(0. '$$J(t)$$').*t)*2/sqrt(3).'r--'.. R(t) is zero for t = (n + 13 ) √23 π ∀n ∈ N.[].1 + ( + )v1. we let v2. √ left: exponential √ 1 3 1 R(t) = cos 2 t − √3 sin 23 t e 2 t . Further calculation yields 1 2 √ + 3 i 6 and √ ! √ !! 1 1 3 3 R(t) = cos t − √ sin t e2t 2 2 3 √ ! 1 2 3 t e 2 t. 0].9*pi]. 20 20 0 0 % Plotte R(t) customplot(. The limiting exponential functions can be obtained by finding the maximum amplitude which yields flim.[].2/sqrt(3)*exp(0.2). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. '$$t$$'. [-262/384 -112/384]..8)..1 = 1 and get v2. plot(t.8). assuming R(0) = 1.1 = 1 √results in v1.*sin(sqrt(3)/2...2 = 12 − 23 i. −20 −20 0 5 10 t 15 0 5 10 15 t Fig.[size(t. 0].1. J(t) R(t) % Zeitabhängige Gleichungen Rt=exp(0. % Laufkoordinate t=[0:0. one coordinate of the eigenvector can be chosen √ arbitrarily (second equation yields 0 = 0).1: Time behaviour of the solutions and the limiting functions..c) Sketch R(t) and J (t) as functions of t. To get flim.'on').'on').. plot(t.-(cos(11/6*pi)-sin(11/6*pi)/sqrt(3))* .2 = ( + i)v1. [-46/384 -323/384]. plot(t.'LineWidth'. .2 2 1 3 ⇔ −v1.1 = 0 4 4 With a compley pair of eigenvalues.J = ± √23 e 2 t ..2).. . √ 1 right: J(t) = √23 sin 23 t e 2 t .1 3 1 0 2 2 −1 − 2 i v1.1. t'.*t)/sqrt(3)). 33 % Und die begrenzenden Exponentialfunktionen handles=get(gcf.'r--'.1.'LineWidth'.8). J(t) = √ sin 2 3 It can easily be seen.Aufgabe 5.'Children').. Choosing v1. the constants can be calculated as C1 = C2 = 12 − 63 i. '$$t$$'..'LineWidth'.3..2c -. plot(t.5*t).2 = 12 + 23 i.5*t). % Und die begrenzenden Exponentialfunktionen handles=get(gcf.5*t). % Plotte J(t) customplot(. we first have to obtain the eigenvectors.*(cos(sqrt(3)/2.(cos(11/6*pi)-sin(11/6*pi)/sqrt(3))* .'r--'.'Children'). exp(0.8). hold(handles(3). '$$R(t)$$'). that J(t) is zero for t = n √23 π ∀n ∈ N. exp(0.[size(t.. J (0) = 0. Rt'.R = ± cos 11 π 6 − sin 11 π √13 6 1 e 2 t and 1 flim.'LineWidth'.1 0 ∗ A(λ1 ) = = ⇔ v1. √ √ " # ! ! 3 1 − 21 − 23 i 1√ v1. 'o'. 2 for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.2). e.8 van der Pol oscillator x˙ = y. -v(1)+v(2)*(1-v(1)^2)]. y˙ = −x + y(1 − x2 ) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.[0.'LineWidth'.ts.2.'MarkerEdgeColor'.g.3 0..[2. .1 Phase Portraits Computer work: Plot computer–generated phase portraits of the following systems. 6.3 0.res(:.3 0.1). [0.3]) . x˙ 4 % Feld anlegen lval=[-4 4]. x2 ) with numerical results.. dy=-x+y.1).2).3]) plot(res(:.res(1. As always.[].^2).res]=ode23(dgl_sys.'MarkerFaceColor'. vectorfield(x. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)). 2 % van der Pol Oscillator dx=y.1.3 0. dgl_sys=@(t.%%').. .3 0.1: van der Pol oscillator x˙ = y. MacMath (Hubbard and West 1992).y.1..[]. 0 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.Exercises for Chapter 6 6.Aufgabe 6.3]. .dy).-1]).dx. [0.3 0. hold on −2 % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0..1:4. . [lval(1) lval(2)]'...*(1-x.'Color'. −4 −4 −2 0 x Fig.8 -. 6. you may write your own computer programs or use any readymade software. [startx starty]).. 4 y˙ = −x + y(1 − end 34 end plot(res(1.v)[v(2). %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. 4 % Dipole fixed point dx=2*x.[0. end [t_s.3: Two–eyed monster x˙ = y + y 2 ..'Color'.10 Two–eyed monster x˙ = y + y 2 . .3]) 6.ts.[startx starty]).dx.3 0. .'LineWidth'. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)).^2.3].y... y˙ = − 12 x + 15 y − xy + 56 y 2 (from Borelli and Coleman 1987.^2. dgl_sys=@(t.1).6. . [lval(1) lval(2)]'.v)[v(2)+v(2)^2.dy). [0.-1]).2*v(2)-v(1)*v(2)+1.. .5*10^(-starty):8*10^(-starty).2). numerical results.'Color'.res(:.[]..[].3]) plot(res(:. 385) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.5*x+0.9 Dipole fixed point y˙ = y 2 − x2 x˙ = 2xy.%%').3 0. % Feld anlegen lval=[-4.'LineWidth'. [0.[].ts.3 0. -0.3 0.dx.5 3]. .3]) plot(res(:.[2.res(1. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)). vectorfield(x.3 0.-1]). .'MarkerEdgeColor'.. v(2)^2-v(1)^2].1).'MarkerEdgeColor'. −4 −4 −2 0 for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.v)[2*v(1)*v(2).5*v(1)+0.[startx starty]).[0.*y.9 -..3]..2).^2-x. [0. % Feld anlegen lval=[-4 4].3]) .2).3 0.1:10. 2 x Fig. 35 end end plot(res(1.y.^2.res]=ode23(dgl_sys.'o'. 4 y˙ = y 2 − x2 with end end plot(res(1. x˙ 2 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.1. p. dy=y.1.3 0.Aufgabe 6. else ts=0:0..'MarkerFaceColor'. hold on −2 % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.res(:. . 6. [0. dy=-0.2: Dipole fixed point x˙ = 2xy. x˙ 0 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.03:5.2.. 6.[2.%%').3 0.1. dgl_sys=@(t.Aufgabe 6. y˙ = − 12 x + 1 6 2 5 y − xy + 5 y with numerical results.2*y.3 0.2. dass Werte oben mitte divergieren if ((startx == 0) && (starty > 0)) ts=0:0. % Berücksichtige.1).dy).'MarkerFaceColor'.1).[]. [lval(1) lval(2)]'. hold on 0 % Numerische Lösungen bestimmen und dazu zeichnen for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) −2 −4 −4 −2 0 x 2 Fig. 2 % Two--eyed monster dx=y+y..2).'o'..3 0. vectorfield(x.2*v(2)^2].res(1.10 -.3 0.*y+1.2*y-x.res]=ode23(dgl_sys...1.3 0. ^2...'o'.3 0. . 5 Fig. x˙ 5 % Parrot dx=y+y.'MarkerFaceColor'.%%'). dy=-x+0. [lval(1) lval(2)]'. % Feld anlegen lval=[-8 8]. 6.2)...'MarkerEdgeColor'.'Color'. . 36 end end plot(res(1.3]) .3 0...Aufgabe 6.*y+1.11 -.2*y.-1]). 0 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'. . [0.^2. y˙ = −x + 15 y − xy + 65 y 2 (from Borelli and Coleman 1987..1.6.dx. 384) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.res(1.2*v(2)-v(1)*v(2)+1. y˙ = −x + 15 y − xy + 65 y 2 with numerical results.3 0.4: Parrot x˙ = y + y 2 .3 0. p.[startx starty]).res]=ode23s(dgl_sys.[0.2).1). [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)). hold on −5 % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.3]) plot(res(:.[].res(:.2*v(2)^2]. dgl_sys=@(t.11 Parrot x˙ = y + y 2 .dy). -v(1)+0.[].2*y-x.[2. vectorfield(x..y.v)[v(2)+v(2)^2.'LineWidth'. . −5 0 x for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.1).03:10. [0.3 0.ts.2.1.3 0.3]. we calculate the Jacobian insert the values and evaluate its determinant. b) Sketch the vector field. First. J = cos (arcsin (γ)) yields infinitely many positive (centers) and negative values (saddles) for one β. we can rewrite the system with two first–order systems θ˙ = v v˙ = − sin (θ) + γ. see d) c) Is the system conservative? If so. Is the system reversible? d Multiplying with θ˙ suggests a time derivative. arcsin (γ) has infinitely many points for one γ (periodicity in vertical direction). − cos (θ) 0 " # 0 1 J = . than 2nπ + β and (2n − 1)π − β are also. To classify them.6. Rearraning to dt yields ˙ =0 θ˙θ¨ + θ˙ sin (θ) − γ θ) d 1 2 d 1 ˙2 2 2 ⇔ θ − cos (θ) − γθ = 0 ⇔ mv − mR cos (θ) − mR γθ = 0 dt 2 dt 2 ˙ In this equation.7. 1 mv 2 + V (θ) represents a constant (energy) and where v = θR. find a conserved quantity. but there are infinitely many values for each defined coordinate (due to its vertical periodicity). saddle nodes for det (J ∗ ) < 0 or non–isolated fixed points otherwise. In the cosine function cos (arcsin (γ)). 2 therefore a conservative quantity. we have centers for det (J ∗ ) > 0. or an undamped Josephson junction driven by a constant bias current. a) Find all the equilibrium points and classify them as γ varies. det (J ∗ ) = 0 results in non–isolated fixed points and other γ are not allowed. T T The equilibrium points are θ v = arcsin (γ) 0 . choosing γ = 1 or γ = −1. Since tr (J ∗ ) = 0. More precisely. In short.7 Pendulum 6. − cos (arcsin (γ)) 0 ∗ det (J ) = cos (arcsin (γ)).2 Pendulum driven by a constant torque The equation θ¨ + sin (θ) = γ describes the dynamics of an undamped pendulum driven by a constant torque. the second set of periodic solutions will produce results with an opposing sign due to the periodicity shift of both functions. if β is a solution of arcsin (γ). 37 . arcsin (γ) is only defined for −1 ≤ γ ≤ 1. " ˙ ∂θ J= ∂θ ∂ v˙ ∂θ ∗ ∂ θ˙ ∂v ∂ v˙ ∂v # " # 0 1 = . the system is reversible. −1 < γ < 1 yields infinitely many centers and saddle nodes. Since θ¨ + sin (θ) = γ is invariant to t → −t (second derivative will have the same sign). As a result. −5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.. [lval(1) lval(2)]'.3 0. To associate the eigenvalues λ with the frequency ω. vectorfield(x.y. for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)). As our frequency can’t get larger than one intervall of√the arcsin (γ) function. hold on −5 −5 0 5 −5 θ˙ 0 5 θ˙ % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.'LineWidth'..2.5:1 % DGL dx=y.3 0.. middle left: γ = 0. e) Find the approximate frequency of small oscillations about any centers in the phase portrait. one can for example calculate the q fundamental equation and observe the time behaviour to be λ = iω and thus ω = cos (arcsin (γ)).'$$\dot{v}$$').2).3 0.v)[v(2).7..2).2 = ±i cos (arcsin (γ)). v˙ = − sin (θ) + γ. bottom left: γ = 1.[2..res(1.. dgl_sys=@(t. v˙ 5 0 −5 −5 0 end 5 end end plot(res(1.qDetermining its eigenvalues from the characteristic equation yields λ1.[startx starty]).5: Numerical solutions and vector field to θ˙ = v.1). % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.3]) plot(res(:.dx. 6..3 0..[]. ..5 5 0 0 v˙ v˙ d) Sketch the phase portrait on the plane as γ varies.res]=ode23s(dgl_sys.3 0. [0.-1]. the origin is on a center (det (J ∗ ) > 0).'o'..3].1). [0.2:8.Aufgabe 6.. As we know from our observation of the Jacobian.'MarkerFaceColor'.'Color'.res(:..'MarkerEdgeColor'.2 -. . middle right: γ = − 12 . we can also use the trigonometric identity and write ω = 4 1 − γ 2 38 .3]) θ˙ Fig. for gamma=-1:0. top right: γ = − 12 . -sin(v(1))+gamma]. Top left: γ = −1.[0. dy=-sin(x)+gamma. .%%'). '$$\dot{\theta}$$'.[]. −5 −5 0 5 −5 0 5 θ˙ 5 5 0 0 v˙ v˙ θ˙ −5 % Feld anlegen lval=[-8 8].dy).3 0.ts. -1]. by constructing a Liapunov function V = ax2 + by 2 with suitable a.2.2).. .5:10 15:1:30 35:5:70 80:10:200].dy). '$$\dot{\theta}$$'. end 39 end plot(res(1. -v(1)-v(2)^3].res(:..[startx starty]).2).10 Show that the system x˙ = y − x3 ..1). our Liapunov function is V = ax2 + ay 2 with a > 0..'o'.5:10]. −0.5 −1 0 1 end θ˙ Fig..'LineWidth'.3 0.'$$\dot{v}$$').3 0.1). y˙ = −x − y 3 with % Spirale weiter verfolgen ts=[0:0.ts..5 1 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.[0.2.^3.. the first two terms are always negative with positive a.5 −1 −1. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. 1..res]=ode23(dgl_sys.Aufgabe 7.3 0.'LineWidth'.3 0. for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.3 0.'MarkerFaceColor'. b.'MarkerFaceColor'.5]. Our Liapunov function must be zero at our equilibrium point 0 0 points.3 0. .3]) end plot(res(1.res]=ode23(dgl_sys. dgl_sys=@(t.'MarkerEdgeColor'.3 0. [0. % DGL dx=y-x.^3.res(1..%%').5 1.ts.1: Phase space of x˙ = y − x3 .[].3].[0.3]) plot(res(:.3 0. [lval(1) lval(2)]'.3 0.3]) plot(res(:.[]..3 0.1).v)[v(2)-v(1)^3. Calculating the derivative and inserting yields T and greater at other V˙ = 2axx˙ + 2by y˙ = 2ax(y − x3 ) + 2by(−x − y 3 ) = −2ax4 − 2by 4 + 2axy − 2bxy.. [0.'Color'. 7.3 0. So. y˙ = −x − y 3 has no closed orbits.y.1).'MarkerEdgeColor'.5 0 % Numerische Lösungen bestimmen und dazu zeichnen ts=[0:0.2 Ruling Out Closed Orbits 7.'o'. [0. dy=-x-y. for startx=lval(1)/9:(lval(2)-lval(1))/27:lval(2)/9 for starty=lval(1)/9:(lval(2)-lval(1))/27:lval(2)/9 [t_s. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)). vectorfield(x... hold on v˙ 0. % Feld anlegen lval=[-1. . For our derivative to be negative for all values.res(1..2.2).Exercises for Chapter 7 7. b and the second two terms must vanish by choosing a = b. [0..'Color'.[startx starty]).3 0.2).dx.1:5 6:0..res(:.3]) .1:5 6:0.2.10 -.3].[2.. numerical solutions.. the the system doesn’t change in the long–term and the amplitude of a limit cycles is only depending on the initial condition.[2. . Thus. [0.. dgl_sys=@(t. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. nor our slow–varying phase φ(x0 ) changes (both derivatives are zero).[].[]. x) ˙ = 0 with 0 < ε 1. dy=-epsilon*x. [0. E E 1 · 3 · 5 .v)[v(2). -epsilon*v(1)*v(2)-v(1)].01.'MarkerFaceColor'. which are defined by 2π 1 Z r = h(θ) sin (θ) dθ ≡ hh sin (θ)i 2π 0 0 2π 1 Z rφ = h(θ) cos (θ) dθ ≡ hh cos (θ)i .01.2: Phase space of x ¨ + x + εxx. epsilon=0..3 0. x) ˙ = xx˙ Inserting yields h = r cos (θ)(−r sin (θ)). h(x.. % Feld anlegen lval=[-50 50].%%'). −r sin (θ)) We can make our life easier with the help of some relations: hsin (ϕ)i = hcos (ϕ)i = 0 E D sin (ϕ)2n = cos (ϕ)2n = D sin (ϕ)2n+1 = cos (ϕ)2n+1 = 0. end 40 end plot(res(1.res(1.. ˙ with ε = 0.1). % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.1:8.ts. x) ˙ = h(r cos (θ).Aufgabe 7.3 0. (2n) D D E n≥1 n≥1 7.3]) plot(res(:.*y-x.2).'Color'. Neither our slow–varying amplitude r(x0 ). Find the amplitude of any limit cycles for the original system. for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.'MarkerEdgeColor'. hold on 0 x 50 Fig. r and φ are the slowly–varying amplitude and phase of the approximate (averaged) solution x0 = r cos (τ + φ) of x. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)).y.6 -. 7. [lval(1) lval(2)]'.'LineWidth'.6 h(x. calculate the averaged equations. vectorfield(x.3]) . (2n − 1) .. 2 · 4 · 6 .dy).-1]). 0 −50 −50 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.3].1).3 0...'o'.3 0. Therefore D E D E D E D E r0 = −r2 sin (θ)2 cos (θ) = −r2 hcos (θ)i + r2 cos (θ)3 = 0 rφ0 = −r2 sin (θ) cos (θ)2 = −r2 hsin (θ)i + r2 sin (θ)3 = 0.dx.[0.6.[startx starty]).6 Weakly Nonlinear Oscillators For each of the following systems x ¨ + x + εh(x. . . 2π 0 0 Here..3 0.2).7. Then analyze the long–term behavior of the averaged system.3 0. .res]=ode23(dgl_sys. 50 x˙ % DGL dx=y.6. .2.res(:. .^5.'LineWidth'.v)[v(2).[0 0.[].r_s]=ode23(inline('1/2*r-1/16*r^5. ts.1).2:4. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/16:lval(2)).6. dr=0.. ylim_extra=[1/6 1/6].'r').-1]). [lval(1) lval(2)]'..res]=ode23(dgl_sys.[2. hold on 4 5 % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.dx. dy=-epsilon*(x. -epsilon*(v(1)^4-1)*v(2)-v(1)]. end plot(res(1. ˙ with ε = 0.^4-1).ts.2.7. vectorfield(x. % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.3 0. . Therefore D E D E D E r0 = −r5 cos (θ)4 sin (θ)2 + r sin (θ)2 = r5 cos (θ)6 − r5 cos (θ)4 + r hsin (θ)i 3 1 1 1 5 = r5 − r5 + r = r − r5 8 2 2 16 D16 E D E 5 0 5 rφ = −r cos (θ) sin (θ) + r sin (θ) cos (θ) = −r5 cos (θ)5 sin (θ) + r hsin (θ) cos (θ)i = 0 + 0. [0.01.'Children').'MarkerEdgeColor'.2. 41 .'o'.3 0. .Aufgabe 7. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.'Color'. 'MarkerEdgeColor'.55 0]) plot(t_s.3 0.[startx starty])..3 0. 7. dgl_sys=@(t.. % Zeitlicher Verlauf skizze_zeitverlauf(r.'MarkerFaceColor'.[0 0. .1).[0 0.. hold on % Numerische Lösung ts=0:0..dy). epsilon=0. right: time dependency of the averaged amplitude r(t). % Feld anlegen lval=[-8 8].01. 0 r x˙ 2 0 −2 −5 −4 −5 0 x 5 0 2 4 end t 4 Fig.'$$t$$'.'.startval). [0. 1 5 r to get our amplitude.dr.55 0]) end % Achsenbeschriftung anpassen ebenen=get(gcf.'LineWidth'.3]) % DGL für r r=-4:0.'MarkerFaceColor'. % DGL dx=y.55 0].2). renameaxis(ebenen(2).startval.res(1.. We need to solve r0 = 12 r − 16 Z 2 dr = dt r(1 − 18 r4 ) Z r4 ln 8 − r4 ⇔ ⇔ r= v u u 4 t e−t C2 ! =t+C 8 +1 There is a stable limit cycle with amplitude lim r = t→∞ √ 4 8.3 0. x) ˙ = (x4 − 1)x˙ Inserting yields h = −r5 cos (θ)4 sin (θ) + r sin (θ).r_s..5). plot(0.7 -.3]) plot(res(:.res(:.. for startval=-4:1:4 [t_s.1:5.6.'$$r$$').1:5.y.'t'.'Color'.3 0. There is no long–term phase change.'o'.[0.5*r-1/16*r..3].2). for startx=lval(1):(lval(2)-lval(1))/8:lval(2) for starty=lval(1):(lval(2)-lval(1))/8:lval(2) [t_s.%%').*y-x.3: Left: Phase space of x ¨ +x+ε(x −1)x.7 h(x.[].. 8 h(x. Before we start. x) ˙ = (|x| − 1)x˙ Inserting yields h = −r2 | cos (θ)| sin (θ)+r sin (θ). we consider the following: sin (θ)3 3 Z cos (θ)3 2 sin (θ) cos (θ) dθ = − .7. 3 Z sin (θ)2 cos (θ) dθ = Inserting the values yields D r0 = −r2 | cos (θ)| sin (θ)2 + r sin (θ)2 E.6. . D = −r2 sin (θ)2 cos (θ) . E D 0≤θ< π2 . 3π ≤θ<2π 2 E. . + r2 sin (θ)2 cos (θ) . π 2 π 2 D ≤θ< 3π 2 + r sin (θ)2 E 3π 2 2π r2 Z r2 Z r2 Z 2 2 =− sin (θ) cos (θ) dθ − sin (θ) cos (θ) dθ + sin (θ)2 cos (θ) dθ 2π 2π 3π 2π π 0 2 2 2π r Z sin (θ)2 dθ + 2π 0 r2 r2 r r 2r2 r − − + = − 6π 3π 2 2 3π D 6π E 0 2 rφ = −r | cos (θ)| sin (θ) cos (θ) + r sin (θ) cos (θ) 2 =− . D E = −r2 sin (θ) cos (θ)2 . . . 3π ≤θ<2π 2 D E + r2 sin (θ) cos (θ)2 . 0≤θ< π2 . . 3π 3π 2 2 There is no long–term phase change. π 2 π ≤θ< 3π 2 + r hsin (θ) cos (θ)i 3π 2π 2 2 r2 Z r2 Z r2 Z 2 =− sin (θ) cos (θ)2 dθ + 0 sin (θ) cos (θ) dθ − sin (θ) cos (θ)2 dθ + 2π 2π 3π 2π π 0 2 =− 2 2 2 r r + + 0 + 0 = 0. We need to solve r0 = 12 r − 3π r to get our amplitude. t→∞ 42 . Z Z 2 dr + r Z 2 dr = dt 3 π − r 4 ! r4 ⇔ 2 ln =t+C 8 − r4 3 1 ⇔ r = π −1t 4 e 2 +1 C2 There is a stable limit cycle with amplitude lim r = 34 π. renameaxis(ebenen(2).'MarkerFaceColor'.1:8..*y-x.3 0. [0. [0.[].'.[0 0.55 0]) % Achsenbeschriftung anpassen ebenen=get(gcf..3 0..[]. % Zeitlicher Verlauf skizze_zeitverlauf(r.ts.4: Left: Phase space of x ¨ +x+ε(|x|−1)x.1). dy=-epsilon*(abs(x)-1).Aufgabe 7. % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.3 0.8 -.%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.55 0]. t Fig..dr.'o'.3].'t'. [lval(1) lval(2)]'.[2. 100 r x˙ 2 0 0 −100 −200 −200 end −2 0 x 200 0 2 4 end plot(res(1.v)[v(2). dgl_sys=@(t. 7.ts.'Children').[startx starty]).2.[0 0..'Color'.6.3 0. end end [t_s. % Feld anlegen lval=[-200 200].startval). -epsilon*(abs(v(1))-1)*v(2)-v(1)].01.-1]).1:10. 'MarkerEdgeColor'.3]) % DGL für r r=-2:0.. 200 4 for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.2...startval. ˙ with ε = 0.01.'LineWidth'. right: time dependency of the averaged amplitude r(t). vectorfield(x.'o'. dr=0. ylim_extra=[1/6 1/6]..'MarkerEdgeColor'.res(:.2).55 0]) plot(t_s.'MarkerFaceColor'.1).5).5*r-2/(3*pi)*r.'$$t$$'.y. 43 .'$$r$$')..^2.dx.[0 0. % DGL dx=y. .res]=ode23s(dgl_sys.[0 0 0]) plot(res(:. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2))...2:5.'Color'.2). hold on % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.dy).r_s]=ode23(inline('1/2*r-2/(3*pi)*r^2..r_s.res(1. . 'r').'LineWidth'. plot(0. hold on % Numerische Lösung for startval=-2:1:5 if (startval < 0) ts=0:1/(10*(1-startval)):1. epsilon=0. else ts=0:0.%%'). if a > 0 the bifurcation is subcritical.Exercises for Chapter 8 8. y) = −x2 . y) = xy 2 and g(x. f (x. Inserting yields ! 1 16a = 0 + 2 + 0 + 0 + 2y(0 + 2x) − 0(2 + 0) − 0 · 2 + 2x · 0 1 = 2 + 4xy Evaluating at (0. y). y˙ = ωx + g(x. a) Calculate a for the system x˙ = −y + xy 2 .2. 0). y˙ = x + µy − x2 at µ = 0. Here we have ω = 1. b) Use part (a) to decide which type of Hopf bifurcation occurs for x˙ = −y + µx + xy 2 .2 Hopf Bifurcations 8. As shown by Guckenheimer and Holmes (1983. where f and g contain only higher–order nonlinear terms that vanish at the origin. the Hopf bifurctaion is also subcritical. the bifurcation is supercritical. y˙ = x − x2 . the hopf bifurcation is subcritical. Thus. y). pp. 152–156). 0) yields a = 18 . Since the system at µ = 0 is identical to that of part (a). 44 .12 Analytical criterion to decide if a Hopf bifurcation is subcritical or supercritical Any system at a Hopf bifurcation can be put into the following form by suitable changes of variables: x˙ = −ωy + f (x. The criterion is: If a < 0. one can decide whether the bifurcation is subcritical or supercritical by calculating the sign of the following quantity: 1 fxy (fxx + fyy ) − gxy (gxx + gyy ) − fxx gxx + fyy gyy 16a = fxxx + fxyy + gxxy + gyyy + ω ! where the subscripts denote partial derivatives evaluated at (0. 2.[2.2).3 0.3 0. else ts=0:0.Aufgabe 8. 45 end end end plot(res(1...ts.5 % DGL dx=-y+mu*x+x.res(1.1).. .2).'Color'.3 0.'o'.^2.-1]).'MarkerFaceColor'.'MarkerEdgeColor'.2. bottom: µ = 21 .. 1 1 0 0 x˙ x˙ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. [lval(1) lval(2)]'.6]. 8. [0.^2.dy).dx. hold on x˙ 1 % Numerische Lösungen bestimmen und dazu zeichnen for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) if ((abs(startx) > 1) || (abs(starty) > 1)) ts=0:0.. vectorfield(x.1).'LineWidth'. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2))..3 0. for mu=-0. [0.y. dy=x+mu*y-x. end 0 −1 −1 0 x 1 [t_s..c) Verify your results by plotting phase portraits on the computer.*y. v(1)+mu*v(2)-v(1)^2]..res]=ode23(dgl_sys. top left: µ = − 21 .3]) .1: Phase space of x˙ = −y + µx + xy . dgl_sys=@(t.[]..[].05:5.3].%%'). top right: µ = 0.6 1. −1 % Feld anlegen lval=[-1.v)[-v(2)+mu*v(1)+v(1)*v(2)^2.5:0..5:0.[startx starty]).3 0.12 -.[0. 2 Fig. −1 −1 0 x 1 −1 0 x 1 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.res(:.3 0. .3]) plot(res(:.1:20. y˙ = x + µy − x2 . res(1.res(:.. for mu=[-0.'o'.'o'.^2..res]=ode23(dgl_sys..2: Phase space of x˙ = µx + y − x2 .3]) .dx.06626 bis 10000 getestet Fig. %für mu=0.1] [t_s. .1).4 0 x 0.5 0 0 x˙ x˙ % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.. vectorfield(x. " # FP1 = 0 0 FP2 = where FP1 is a spiral and FP2 a saddle node. Sketch the phase portrait just above and below the bifurcation.3 0. Our system has two fixed points.-1]). But if µ approaches the critical value.3]) % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'. 0 −0.5 0 x 0.[2.4 0.2).dy).3 0..3]...[0.1:8..res]=ode23(dgl_sys. bottom left: µ = 0. hold on 1 1 0.3]) plot(res(:.. in the bottom row only the trajectories of four values near to FP1 are shown..5 0 x end plot(res(1..^2.6 0.'Color'. y˙ = −x + µy + 2x2 undergoes a homoclinic bifurcation.v)[mu*v(1)+v(2)-v(1)^2. . dgl_sys=@(t.6 0.[2.06626 as shown below.'LineWidth'. [0.2 −0.3 Homoclinic bifurcation Using numerical integration.3 0.2).1).ts.5 1 end % Feld anlegen lval=[-0..3 0. % Feld anlegen lval=[-1 1]. -v(1)+mu*v(2)+2*v(1)^2]. hold on −0.3 0. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/16:lval(2)).2..1).ts.'LineWidth'.v)[mu*v(1)+v(2)-v(1)^2..y..dx.06627] % DGL dx=mu*x+y-x. before a homoclinic bifurcation takes place. [startx starty]).06626 before the homoclinic bifurcation (at least within the first 10000 s).1:360. dgl_sys=@(t.1). 8.4 −0. . [startx starty]). [0..3 0. the points near FP1 approach the homoclinic orbit of the saddle node.6 0. [lval(1) lval(2)]'... −0.4.'MarkerFaceColor'. Here.-1]).3]) plot(res(:.%%'). find the value of µ at which the system x˙ = µx + y − x2 . Numerical investigation reveals µcrit ≈ 0.1 0. If µ passed the critical value.4 0.2] % DGL dx=mu*x+y-x.2). -v(1)+mu*v(2)+2*v(1)^2]. .y.[].res(1.2.^2.3 0.5 % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0. 1+µ2 2+µ 1−2µ+µ2 −2µ3 2 4+4µ+µ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.dy). end end end plot(res(1.5 0.2 −0.res(:. . [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2)). vectorfield(x. points near FP1 are attracted by the spiral or repelled to a limit cycle.8.1 0.'MarkerEdgeColor'.'MarkerFaceColor'.. −0.06626 0.[0.[].[].6].'MarkerEdgeColor'. % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'.[].1. depending on µ. ...'Color'.4.3 0.5 end 0 x −1 −1 1 0.3 0. top left: µ = −0. dy=-x+mu*y+2*x.06627 after the bifurcation. [0. values near FP1 can escape the local domain. y˙ = −x + µy + 2x2 . % Und auch den Übergang im Detail zeichnen for mu=[0.2 0 x˙ x˙ −1 −1 −0.2.Aufgabe 8.3 0. 46 for startx=[-0. [0.3 0.1 0. top right: µ = 0. dy=-x+mu*y+2*x. [lval(1) lval(2)]'.2 0..3 -..1] for starty=[-0.3].5 for startx=lval(1):(lval(2)-lval(1))/6:lval(2) for starty=lval(1):(lval(2)-lval(1))/6:lval(2) [t_s.4 Global Bifurcations of Cycles 8..^2. bottom right: µ = 0.2).3 0. 1). v = I − αv − sin (θ).res(1.. while for α < 1 we have a homoclinic bifurcation.5.-1]). . 1 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.2). [lval(1)/3 lval(2)/3]'.1). 'MarkerFaceColor'. hold on θ ′′ 2 θ ′′ 2 % DGL dx=y.[].'String'. '$$\theta^\prime$$'). lval(1)/3:(lval(2)-lval(1))/32:lval(2)/3).4 75 90. First.0.'XLabel')..3] for I=[1. −5 0 θ′ 5 for alpha=[0.. set(get(ebenen(2).[].3 0.2 0. 5 θ′ 1 1 plot(res(1.. vectorfield(x.ebenen(2).6 0.'Children').2.3: Phase space of θ = v. dgl_sys=@(t.. . verify that if α is fixed and sufficiently small..2). right column: α = 1.[0. I-alpha*v(2)-sin(v(1))]. the system’s stable limit cycle is destroyed in a homoclinic bifurcation as I decreases. dy=I-alpha*y-sin(x).res(:.Aufgabe 8. 0 0 −1 −1 −5 0 5 −5 0 θ′ 5 θ′ 0 end 0 Fig. 8.85 0.3]) θ ′′ 2 θ ′′ 2 for startx=lval(1):(lval(2)-lval(1))/24:lval(2) for starty=[0].v)[v(2)..'o'.3..4. [0.dx.dy. As I is passing 1. top row: I = 1.5 4]). the bifurcation is an infinite–period bifurcation instead. ... If α > 1.6*pi].[2. we rewrite the second–order system as two first–order systems θ0 = v and v 0 = I − αv − sin (θ). . depending on the value of I (which must also be between zero and one).2 0. By numerical computation of the phase portrait.3 0.6. .3 0.%%'). 'Color'.5 1..6*pi 2.5 Hysteresis in the Driven Pendulum and Josephson Junction 8.0 0.. the bifurcations appear.2.'YLabel').'LineWidth'. . .res]=ode23(dgl_sys.3 0. 0 0 −1 −1 −5 0 5 −5 0 θ′ % Numerische Lösungen bestimmen und dazu zeichnen ts=0:0.8..[startx starty]).8.[0.5...2 -.y. bottom row: I = 0.[0. we have an infinite– period bifurcation as I passes 1.ebenen(1).2 Consider the driven pendulum θ 00 + αθ 0 + sin (θ) = I.1:18. 47 end end end % Achsenbeschriftungen anpassen ebenen=get(gcf. '$$\theta^{\prime\prime}$$').8 0.4] θ′ 1 1 % Zeichenebene vorbereiten und Vektorfeld zeichnen customplot([lval(1) lval(2)]'. θ ′′ 2 θ ′′ 2 0 0 −1 −1 −5 0 5 % Feld anlegen lval=[-1. .'Children').5. 'MarkerEdgeColor'. [t_s. Left column: α = 0.3]) plot(res(:.3]. [x y]=meshgrid(lval(1):(lval(2)-lval(1))/12:lval(2). .3 0. Show that if α is too large.'String'. set(get(ebenen(2).6 0. middle row: I = 0.ts. ebenen=get(gcf..3 0. ) e) Solve the equations on a computer. Here r. Taking into account small radial perturbations r = r0 + δr yields m d2 h (r0 + δr) = −k 2 dt m(r0 + δr)3 c) Show that these small radial oscillations correspond to quasiperiodic motion by calculating the winding number ωr /ωθ . We can write s h2 −k m¨ r=0= mr03 ⇔ r0 = 3 h2 . The equations m¨ r= h − k. 48 . (To say it in a more interesting way. hm b) Find the frequency ωr for small radial oscillations about the circular orbit. mr3 h θ˙ = mr2 govern the motion of a mass m subject to a central force of constant strength k > 0. θ. mk Afterwards.8. corresponding to uniform circular motion of radius r0 and frequency ωθ . Find formulas for r0 and ωθ .6. θ˙ = ωθ .6 Coupled Oscillators and Quasiperiodicity 8. a) Show that the system has a solution of the form r = r0 . the second equation can be rearranged as follows h θ˙ = ωθ = mr02 s ⇔ ωθ = 3 k2 . the motion is never chaotic. and plot the particle’s path in the plane with polar coordinates r. Having circular motion.7 Mechanical example of quasiperiodicity. d) Show by a geometric argument that the motion is either periodic or quasiperiodic for any amplitude of radial oscillation. θ are polar coordinates and h > 0 is a constant (the angular momentum of the particle). there is no change of the radius (¨ r = 0). 1. yn+1 5 % Anzahl der Iterationen steps=4. % Startwerte für die Cobweb-Trajektorien startval=[-2 2].7.'String'. top right: a = 0. 8. It has a stable fixed point at zero for a < 0 (and therefore the original system θ˙ = 1.'String'.7 Poincaré Maps 8.'Children'). y˙ = ay. Classify its stability for all real values of a.'YLabel'). 0 % Zeichne das Cobweb cobwebplot(P. bottom center: a = 0.1:0. −4 −2 0 yn 2 4 % Variiere a for a=-0. 2 4 1 2 yn+1 yn+1 So we have a Poincaré map P (y) = ye2πa .2 -.7.Aufgabe 8. Since the time of flight can be seen with θ˙ = 1 to be 2π. 49 .startval. Show that the system has a periodic orbit.1 % Poincaré Map der Differentialgleichung P=inline(['exp('.%%').'$$y_{n+1}$$').steps. y˙ = ay has a stable limit cycle at y = 0). y˙ = ay.2 Consider the vector field on the cylinder given by θ˙ = 1. set(get(ebenen(2). −5 % Achsenbeschriftungen anpassen ebenen=get(gcf. set(get(ebenen(2).interval.[1/12 1/12]).8.1:0.1. we can take 0 and 2π as our integration limits for time and evaluate the integral dy = ay dt y1 ⇔ 2π Z 1Z 1 dy = dt ay y 0 0 ⇔ 2πa y1 = y0 e . top left: a = −0.'var').1. a > 0 yields an unstable fixed point and therefore no stable limit cycle. % Zu betrachtendes Intervall interval=[-4 4].4: Poincaré map P (y) = ye2πa of the system θ˙ = 1.'*2*pi)*var'].num2str(a). Define an appropriate Poincaré map and find a formula for it.'XLabel').'$$y_n$$'). −4 −2 0 yn 2 4 end Fig. The case a = 0 yields neutrally stable limit cycles. 0 0 −1 −2 −2 −4 −4 −2 0 yn 2 4 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. 0 0 α α = 1. use a computer to explore the dynamics of the Lorenz system. 20. z. r−1 . 10.z = ±1 q q − b(r − 1) and − b(r − 1). assuming σ = 10 and b = 8/3 as usual.y. Also in some cases you may want to ignore the transient behavior. The following initial values were used for the numerical investigation of the Lorenz system: α 0 0 0 . and plot only the sustained long–term behavior. and x vs. In each case plot x(t). 0 . γx. y(t).Exercises for Chapter 9 9.3 Chaos on a Strange Attractor (Numerical experiments) For each of the values of r given below. γz q 0 qb(r − 1) 0. The system has three fixed points. 100. α . b(r − 1) 0 r−1 50 β = 1. γx β γy . You should investigate the consequences of choosing different initial conditions and lengths of integration. ..x)[sigma*(x(2)-x(1))..3.3. '): x_e='.0 10 0. -1 -1 1. .:)).res]=ode23(dgl_sys. 0 0 1.3. % 1 -1 -1.[]...'$$y$$').8]. 1 0.res(:. %startvalmat=20*startvalmat.res(:...res(:.2: Phase space of Lorenz system (σ = 10. '.num2str(res(end.2).[].. .'$$t$$'.6 0.1). 3.[size(ts. either 24 24 9 or − 24 − 24 9 .. y_e='.2). top left: time dependency x(t).2)). . −10 −20 −10 0 100 200 0 100 t dgl_sys=@(t. 20 aufgabe=2.[]. -1 1 1.. .. 1 1 -1. 9. %startvalmat=[1 1 1. 1 1 0. 1 -1 1.100 0 0.0 0 100].2).8]. All other initial values approach a √ √ T T √ √ stable fixed point.005:10 10. 200 t %ts=[0:0.. . top right: time dependency y(t)..2 0 0 100 200 0 100 t t 200 −1 0 x 1 Fig. .1) [t_s.[1/6 1/6]. z_e='.'$$x$$'.. r*x(1)-x(2)-x(1)*x(3)..num2str(res(end. r=r_vec(aufgabe).10 0 0.:)).01:0.3).startvalmat(count.0 0 10. 3. % Lorenz DGL sigma=10.1).02:0. 20 x Fig.[size(ts.. z 20 0 for count=1:size(startvalmat. 40 startvalmat=[1 0 0.2 r = 10 0 0 α T T converges to the fixed point 0 0 0 .02:200].9.. 20 10 30 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -.8 z y x 0.num2str(startvalmat(count. -1 -1 -1].. ts=[0:0..%%'). 9. bottom: phase space of z(x). -1 1 -1. '.'$$z$$'). customplot(t_s. bottom: phase space of z(x).01:120].5 100 126. [size(ts.005:10 10. top left: time dependency x(t). 0 1 0. y x 10 0 0 r_vec=[0 10 22 24. b=8/3.1: Phase space of Lorenz system (σ = 10.'$$x$$'). b = 83 ) with r = 10 for initial values (20 − 20 − 20)T .52 400]. customplot(res(:.5 −0.Aufgabe 9. top right: time dependency y(t). −20 0 10 disp(['start: ('.'$$t$$'. .[1/6 1/6].0 100 0.4 −0.2).5 −1 −1 0. x(1)*x(2)-b*x(3)]. b = 83 ) with r = 10 for initial values (0 1 0)T .ts.1))..[1/6 1/6].8]. .x -.5 0 0 end customplot(t_s.5 0..01:0.num2str(res(end.3))]). . 51 .01:40 40. 9. bottom: phase space of z(x). bottom: phase space of z(x). b = 83 ) with r = 22 for initial values (0 1 0)T .3. 52 . top left: time dependency x(t). 0 0 α √ T fixed point 0 0 0 .3: Phase space of Lorenz system (σ = 10.4: Phase space of Lorenz system (σ = 10. top right: time dependency y(t). 9. top right: time dependency y(t). b = 83 ) with r = 22 for initial values (0 100 0)T . Again. 40 100 20 50 T converges to the T √ 56 56 21 or z y x 100 0 0 50 −50 −20 0 0 100 0 200 100 t 200 −20 0 t 20 x Fig.3 r = 22 (transient chaos) Some initial values produce a transient chaos (as title suggested) but for all initial val ues convergence to a fixed point can be observed.9. 20 40 20 30 10 0 z y x 10 0 10 −10 −10 0 100 200 20 0 0 100 t t 200 −10 0 10 x Fig. top left: time dependency x(t). while the other values eventually approach √ T √ − 56 − 56 21 . top right: time dependency y(t). 0 0 α T T converges to 0 0 0 .7: Phase space of Lorenz system (σ = 10. b = 83 ) with r = 24. bottom: phase space of z(x). top left: time dependency x(t). 9. b = 83 ) with r = 100 for initial values (0 100 0)T . bottom: phase space of z(x).4 r = 24.5: Phase space of Lorenz system (σ = 10. 20 20 40 10 30 0 z 0 y x 10 20 −10 −10 10 −20 −20 0 −30 0 100 200 0 100 t 200 −20 −10 0 t 10 x Fig. but other initial values seem to trace out a strange attractor. top right: time dependency y(t). some values spiral down to one fixed point q or − 188 3 q − T 188 3 23.5 but others don’t (at least in the observed time interval).5 for initial values (20 − 20 − 20)T . − 264 − 264 99 250 40 100 200 20 50 150 y z 150 x 60 100 0 0 −20 −50 50 −100 0 −40 0 50 100 0 50 t 100 t −20 0 20 40 x Fig. top right: time dependency y(t). 9. 30 60 40 20 40 20 z y x 10 0 20 0 0 −10 −20 −20 −20 0 100 200 0 100 t 200 −10 0 t 10 20 x Fig.5 (chaos and stable point co–exist) Still. Eventually. 9.5 r = 100 (surprise) T T As before. some values don’t exhibit the typical ring–like structure but seem to √ T √ T √ √ fill out the space around the fixed points . top left: time dependency x(t).9. top left: time dependency x(t). 0 0 α converges to 0 0 0 . bottom: phase space of z(x).5 q 188 3 q 188 3 T 23.3. 9.6: Phase space of Lorenz system (σ = 10. 53 .3. but all other initial values seem to be on a strange attractor.5 for initial values (−20 − 20 20)T . Here. b = 83 ) with r = 24. 264 264 99 . top left: time dependency x(t). top right: time dependency y(t).8: Phase space of Lorenz system (σ = 10.9.3. 54 .52 q . Two fixed points are at q 8368 25 q 8368 25 .7 r = 400 T T 200 600 50 0 400 0 −50 −200 −100 −400 0 50 100 z 400 800 100 y x Again.6 r = 126. bottom: phase space of z(x). 9. This indicates an attracting limit cycle.52 .9: Phase space of Lorenz system (σ = 10. z y x 100 0 0 100 −20 −50 50 −40 −100 0 50 T 8368 25 250 40 0 q − 100 0 50 t 100 −40 −20 0 t 20 40 x Fig. 0 0 α converges to 0 0 0 . b = 83 ) with r = 400 for initial values (0 10 0)T . but all other initial values are after some transient behavior on a narrow. 1064 1064 399 . 0 0 α converges to 0 0 0 strange attractor. 9. periodic band.3. − 150 60 8368 25 200 20 50 150 125. T √ T √ √ √ Two fixed points are at . top left: time dependency x(t).52 for initial values (1 0 0)T . b = 83 ) with r = 126. but all other initial values seem to be on a T 125. 9. top right: time dependency y(t). − 1064 − 1064 399 200 0 0 50 t 100 t −50 0 50 100 x Fig. bottom: phase space of z(x).52 T T Again. 9. it is no attractor. It means.. 'MarkerFaceColor'.3 0.24) end d) Repeat part (c) for the circle x2 + y 2 = 1. that they are also attracted.75:3 for starttheta=-pi/2:pi/8:pi/2 [t_s.[startr starttheta]).3].2). set(textfelder(i_resize).3 0.'Marker'.[0. a) Is D an invariant set? Since D : r2 ≤ 1. 1 180 for startr=-3:0.2).[0. .g. D is not minimal.1:2]. c) Is D an attractor? If not. why not? If so.res(1.3 0.1)).r)[r(1)*(1-r(1)^2).[0.%%'). so it is an invariant set. r = 1. 60 ts=[0:0..8 -. 120 90 3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. all initial values in D will stay in D.. Let D be the disk x2 + y 2 ≤ 1.11 < 1) shows. for i_resize=1:length(textfelder).res]=ode23(dgl_sys.2.'text'). t2=polar(res(:.3.'o'.3 0.'LineWidth'. D0 attracts an open set of initial conditions but D0 is minimal. 55 . 1]. 'MarkerEdgeColor'. R \ {0. D0 : r2 = 1 is an invariant set because initial values on the circle stay on it (r = 1 ∈ D ⇒ r˙ = 0).3 0. end set(t1.1)). find its basin of attraction. Since all values without the origin are attracted. 2 150 30 dgl_sys=@(t. So D attracts an open set of initial conditions. b) Does D attract an open set of initial conditions? The condition can be stated as r˙ = r(1 − r2 ) ≤ 1. the length of the radii isn’t allowed to grow for the value to be attracted. 0 210 t1=polar(res(1.8 Practice with the definition of an attractor Consider the following familiar system in polar coordinates: r˙ = r(1 − r 2 ).1 ⇒ r˙ = 0. Testing values outside the disk (e. hold on.3 0. Therefore.res(:.3]) set(t2.'FontSize'. 330 240 300 270 end Fig. 0} is the basin of attraction.9. The circles are lines of the same radius and the values are shown in degrees.3. since it has a smaller attractor within it. .'Type'.ts..Aufgabe 9.10: Phase space of D with numerical solutions. θ˙ = 1.3]) clear t1 t2 textfelder=findall(gcf.'Color'. As for D in part (b). .startvalmat(count. 0 1 0.01:120].res(:. 0 0 r_vec=[0 10 22 24.8].0 10 0.'$$t$$'.. '): x_e='. y_e='.02:200]. . 1 1 -1. 9. -1 1 1. q Two fixed points are at 2204 5 q 2204 5 T 165. .. '. z.100 0 0. z_e='..0 0 100].[1/6 1/6].3.3 q ..'$$t$$'.10 0 0. -1 -1 -1]..num2str(res(end. %startvalmat=[1 1 1.x -..[size(ts...'$$y$$'). b = 83 ) with r = 166. bottom: phase space of z(x). In each case..res(:. -1 1 -1. all seem to be on an attracting limit cycle. 3. 300 z 200 100 for count=1:size(startvalmat.. '.. r*x(1)-x(2)-x(1)*x(3). x(1)*x(2)-b*x(3)]. 60 x Fig.1 r = 166.0 0 10..num2str(res(end.1) [t_s.[1/6 1/6].2). startvalmat=[1 0 0.5.Aufgabe 9. 0 0 1.ts. .res]=ode23(dgl_sys. 0 0 α converges to 0 0 0 . The other values exhibit transient chaos on a strange attractor.2). %startvalmat=20*startvalmat.%%'). 50 y x 100 % Lorenz DGL sigma=10. .'$$z$$')..005:10 10.5 Exploring Parameter Space (Numerical experiments) For each of the values of r fiven below.02:0. .9.2).3 (intermittent chaos) T T As in previous parts concerning the Lorenz system. b=8/3.1)).2).1).res(:.. 100 t %ts=[0:0.11: Phase space of Lorenz system (σ = 10.01:0.3). ts=[0:0.num2str(startvalmat(count.01:0. 9.3))]). [size(ts. 56 end customplot(t_s.8].[size(ts.. 3.3 . 1 -1 1.2))..52 400].. .:)). plot x(t). assuming σ = 10 and b = 8/3 as usual. 0 −40 −20 0 20 40 disp(['start: ('.1).3 for initial values (10 0 0)T .x)[sigma*(x(2)-x(1)). −100 −50 0 50 100 0 50 t dgl_sys=@(t.[].5 100 126. but after time.num2str(res(end. % 1 -1 -1. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% disp('%% -. y(t) and x vs.8]. use a computer to explore the dynamics of the Lorenz system.. r=r_vec(aufgabe).[]. 200 aufgabe=2.[].005:10 10.. top right: time dependency y(t).01:40 40.[1/6 1/6]. . . − 2204 5 q − 2204 5 T 165.'$$x$$').3..'$$x$$'.0 100 0.:)). -1 -1 1. . top left: time dependency x(t). customplot(res(:. customplot(t_s. . 5. bottom: phase space of z(x). the band on the strange attractor is broader but still represents a limit cycle (hence the name noisy periodicity). b = 83 ) with r = 212 for initial values (0 10 0)T . 57 . 9.9.2 r = 212 (noisy periodicity) T T Again 0 0 α converges to 0 0 0 . This time. Two fixed points are at q 1688 3 q 1688 3 T 211 q . top right: time dependency y(t). − 1688 3 400 100 300 0 200 z y x −50 100 .12: Phase space of Lorenz system (σ = 10. top left: time dependency x(t). 211 0 −200 50 T 1688 3 100 −100 0 − 200 50 0 q 0 50 t 100 t −50 0 50 x Fig.