1. Phase portrait for the damped pendulum

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PHY 380.03, Fall 2010 Homework 2 Solution
1. Phase portrait for the damped pendulum
The damped pendulum can be written as two first-order equations as follows:
q = w
w = -w 0 2 sin q - m w
where w 0 = gêL is the small angle oscillation frequency and m is the damping coefficient. This is a nontrivial
system and it takes some care in order to produce a good phase portrait. Let's begin by looking at some sample
phase space orbits, then we'll get more systematic in producing the final phase portrait.
Run for w 0 2 = 1 and m = 0.25 and plot some sample orbits with different initial
conditions
Let's do some experimentation in order to build our physical intuition for this system. Choose some
"typical" initial conditions and see what we get. FYI: although I suggested that you do this phase in
solving this problem, I do not necessarily recommend that you show all your experiments in the home
work writeup -- just the relevant ones that you end up using are sufficient.
In[1]:= m = 0.25; tend = 70.0;
In[2]:= q0 = 1.0; w0 = 0.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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In[5]:= q0 = 3; w0 = 0.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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In[13]:= q0 = 15.0; w0 = -2.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

What do you see here? For clarity, I left off the first orbit since it just adds confusion near the origin fixed point. I
hope what's happening is becoming clear: all orbits end up attracted to one of the many stable fixed points at even
multiples of p. Which one they end up attracted to depends on their initial conditions. It turns out that for any
attracting set, there is a region of initial conditions attracted to it, meaning that all initial conditions in that region
are attracted to the same attracting set. The attracting set for a particular attractor is called its basin of attraction.
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The attracting sets here are the attracting fixed points. So our goal now, in order to produce a meaningful phase
portrait, is to depict the basins of attraction for each attracting fixed point. The boundaries of the attracting basins
are orbits that separate basins - as it turns out, the basin boundaries will be what was the separatrix in the
undamped pendulum. Given this and the plot above it's a good guess that the basin boundaries are connected to
the unstable hyperbolic fixed points at odd multiples of p. This basin boundary is not easy to plot since, as in the
dissipationless pendulum, it represents an idealized orbit. We can explore near the hyperbolic fixed point at q ã p
to get a sense if what's happening. Let's see if we can determine these basin boundaries approximately by plotting
forward and backward in time.
Let's look at initial conditions very close to p: here's one that's equal to p to 6 significant digits, but slightly smaller
than p:
In[20]:= fp1 = 0.0; q0 = fp1 + 3.141592; w0 = 0.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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A couple things to notice: (1) the orbits never cross, but they intertwine, and (2) we get a hint of the old "football"
shaped separatrix from the undamped pendulum, but only in the upper left and lower right quadrants. One way to
remedy the second point is to integrate the equation of motion backwards in time -- in other words, use a negative
end time:
In[25]:= q0 = fp1 + 3.141592; w0 = 0.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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In[30]:= q0 = fp1 - 3.141592; w0 = 0.02;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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In[37]:= q0 = fp1 - 3.141592; w0 = 0.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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In[48]:= fp1 = -2 p; q0 = fp1 + 3.141592; w0 = 0.0;
soln = NDSolve@8q £ @tD ã w@tD, w £ @tD == -Sin@q@tDD - m w@tD, q@0D ã q0, w@0D ã w0

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In[60]:=
All3 = ShowAp7, p8, p9, p10, p11, p12, pl1,
pl2, pl3, pl4, pl5, pl6, plo1, plo2, plo3, plo4, plo5, plo6,
PlotRange Ø 88-6 p, 6 p

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Put it all together and you have a pretty good picture of the dynamics of a damped pendulum! Note that
I've left some whitespace in the upper right and lower left corners of the plot. We could have written
some code to repeat all the relevant orbits for the four remaining attracting fixed point on this plot but,
by symmetry, the results will be identical to the three basins show here shifted to each new fixed point.
In[96]:=
Show@All3, plo7, plo8, plo9, plo10, plo11, plo12,
PlotLabel Ø "Phase portrait for damped pendulum", ImageSize Ø LargeD
Phase portrait for damped pendulum
4
2
Out[96]=
q °
0
-2
-4
-6 p -4 p -2 p 0 2 p 4 p 6 p
q
This is a pretty good portrayal of the damped pendulum. You see 2 orbits within each of the three basins of
attraction, one starting at positive and one at negative angular velocity (green orbits). Depending on the initial
condition on each of these orbits they could represent: (1) a rotation that damps down into an oscillation and
eventually to no motion (high initial velocity), or (2) an oscillation that damps down to no motion (low initial
velocity).
In class we showed that there should be stable focus fixed points at even multiples of q = p and unstable hyperbolic
points at odd multiples of q = p (all at q ° = 0). That is clearly correct looking at the plot.
2. Fixed points of four nonlinear oscillators
We consider three nonlinear oscillators and find three fixed points. You were asked to choose one and determine
the type and stability of the fixed points. Here, I'll do all three.
A. The damped Duffing oscillator
The equation of motion is x – + m x ° + w 0 2 x + e x 3 = 0 and the fixed points are given by setting x ° = x – = 0,
leaving
xIw 0 2 + e x 2 M = 0,
which gives three values of x at the fixed points: x * = 0, ±Â w 0 í
e . The imaginary values cannot
represent physically meaningful positions so there is only real fixed point in phase space:
PHY 380.03, Spring 2013
Hx 1 * , v 1 * L = H0, 0L
© 2013 R. Martin

leaving
xIw 0 2 + e x 2 M = 0,
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which gives three values of x at the fixed points: x * = 0, ±Â w 0 í e . The imaginary values cannot
represent physically meaningful positions so there is only real fixed point in phase space:
Hx 1 * , v 1 * L = H0, 0L
B. The Lotka-Volterra equations
The equations of motion are already in standard form:
x ° = Hb - p yL x = F x Hx, yL
y ° = Hr x - dL y = F y Hx, yL
and the fixed points are given by setting x ° = y ° = 0, leaving
xHb - p yL = 0 and yHr x - dL = 0,
From the first equation we have either x = 0 or y = bêp and from the second equation we have either
y = 0 or x = d êr. Combining, we have two possible fixed points:
Hx 1 * , v 1 * L = H0, 0L
Hx 2 * , v 2 * L = Hd êr, bêpL
If x and y represent the numbers of predator and prey species, the first fixed point represents both
populations dying out, while the second fixed point represents an equilibrium value of the populations.
C. The Rayleigh Oscillator
The equation of motion is x – + eJx ° 2 - 1N x ° + x = 0 and the fixed points are given by setting x ° = x – = 0 the
first two terms vanish, leaving
x = 0,
which gives one value of x at the fixed points, so there is one fixed point at the origin in phase space:
Hx 1 * , v 1 * L = H0, 0L
D. Van der Pol oscillator
This equation of motion is x – - eI1 - x 2 M x ° + x = 0 and the fixed points are given by setting x ° = x – = 0, again
leaves only
x = 0,
which is the same as the Rayleigh oscillator and gives one value of x at the fixed points, so there is one
fixed point at the origin:
PHY 380.03, Spring 2013
Hx 1 * , v 1 * L = H0, 0L
© 2013 R. Martin

This equation of motion is x - eI1 - x M x + x = 0 and the fixed points are given by setting x = x = 0, again
leaves only
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x = 0,
which is the same as the Rayleigh oscillator and gives one value of x at the fixed points, so there is one
fixed point at the origin:
Hx 1 * , v 1 * L = H0, 0L
Problem 3: Stability of Rayleigh fixed points
I'll apply the methods discussed in class, either the stability matrix or, for Newtonian systems, all you
have to do is compute the two coefficients a and b of the linearized equation(s) of motion.
The equation of motion is x – + eJx ° 2 - 1N x ° + x = 0 and the only fixed point is:
Hx 1 * , v 1 * L = H0, 0L
This is a Newtonian system, with FHx, vL = -eIv 2 - 1M v - x so we use the standard method from class.
The linear coefficients are:
a = ∂F ∂F
= -1, and b = = ∂x H0,0L ∂v H0,0L -IeIv2 - 1M + 2 e v 2 M = e
H0,0L
Here the coefficient a is always negative, while b depends on the value of e. The discriminant is
b 2 + 4 a = e 2 - 4. If e 2 > 4, both eigenvalues will be real and positive, while if the opposite is true the
eigenvalues will be complex with a positive real part. By the results of our general 2-D fixed point
derivation in class, this means the fixed point is either a focus or a node. Since e > 0, the fixed point is
unstable, so this fixed point is:
an unstable node if e 2 > 4 and e > 0
an unstable star/improper node if e 2 = 4 and e > 0
an unstable focus if e 2 < 4 and e > 0
Here, if we increase e from less than 2 to greater than 2, there will be a bifurcation at e = 2 as the fixed
point changes from a focus to a node.
Note that you can also use the stability matrix:
M = 0 1
a b
and the eigenvalues are given by
det
-a 1
a
b - a
= a 2 - b a - a = 0
with a = -1 and b = e the solution is
PHY 380.03, Spring 2013
a = e± e2 -4
2
which yields to same fixed point character as the calculation above (of course).
© 2013 R. Martin

-a 1
det
= a 2 - b a - a = 0
a b - a
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with a = -1 and b = e the solution is
a = e± e2 -4
2
which yields to same fixed point character as the calculation above (of course).
PHY 380.03, Spring 2013
© 2013 R. Martin