Phase Diagrams
Quick Review of Linear Oscillator (Ch.3)
• Consider a 1d Linear Oscillator: Its state of motion
is completely specified if two quantities are given at
the initial time t0: x(t0), v(t0) = x(t0).
– 2 quantities because Newton’s 2nd Law is a 2nd order
differential equation in time!
 Its useful to consider x(t) & x(t) = v(t)  as
coordinates in a 2d phase space.
– For a 3d particle, the phase space would be 6 dimensional!
• At any time t, the 1d oscillator motion is completely
specified by specifying a point in this 2d phase space
(x - v or x - x plane).
• At time t, the oscillator motion is specified by
specifying point P = P(x,x). As t progresses, P
will move in this plane & trace out a
Phase Path / Phase Trajectory.
– Different initial conditions  Different phase paths
• The totality of all possible phase paths of the
particle  Phase Portrait / Phase Diagram of
the particle. Studying such diagrams gives insight
into the physics of the particle motion.
– This is not limited to oscillators, of course, but
clearly is a valid concept for any particle!
• Look in detail at the phase diagram for the 1d
simple harmonic oscillator:
x(t) = A sin(ω0t - δ) v(t) = x(t) = ω0A cos(ω0t - δ)
– Eliminating t from these 2 eqtns gives:
[x2/A2] + [x2/(A2ω02)] = 1
– This is a family of ellipses in the x - x plane!
– The phase diagram for the 1d oscillator = a family
of ellipses, each of which is a separate phase path,
for different initial conditions.
• The phase diagram for the 1d oscillator is a family of
ellipses, as in the Figure.
• Note: Oscillator total energy: E= (½)kA2.
Also, ω02 = (k/m)  The ellipse equation can be written:
[x2/(2E/k)] + [x2/(2E/m)] = 1
• Writing the elliptical phase path as
[x2/(2E/k)] + [x2/(2E/m)] = 1
 PHYSICS: Each phase path (ellipse)
corresponds to a definite total energy of the
oscillator (different initial conditions!).
No 2 phase paths of the oscillator can cross!
– If they could cross, this would mean that for given
initial conditions x(t0), v(t0), the motion could
proceed on different phase paths. This is
impossible, since the solutions to a linear 2nd Law
differential equation are unique!
• In constructing a phase diagram: Choose x as the
“x-axis” & x = v as the “y-axis”.
The motion of a typical point P(x,x) is always
clockwise! Because, for a harmonic oscillator, for x >
0, x = v decreases
& if x < 0, x = v increases!
• Earlier we obtained x(t) = A sin(ω0t - δ) &
v(t) = x(t) = ω0A cos(ω0t - δ) by integrating N’s 2nd
Law Eq. (a 2nd order differential equation):
(d2x/dt2)+ (ω0)2 x = 0
• But, we can get the phase path in a simpler way.
Phase Diagram with Damping
• Underdamped linear oscillator: A spiral phase path. A
continually decreasing magnitude of the radius vector to the
origin in the x-v plane is an indication of damped oscillations.
Phase Diagrams for Nonlinear Systems
Section 4.3
• At first, consider only conservative systems (so U(x)
is defined).
General Method to Construct a Phase Diagram
Use energy conservation!
E = (½)mv2 + U(x) = (½)mx2 + U(x)
(1)
 For fixed E, get the function x(x) or v(x) by
solving (1) for v = x:
v(x) = x(x) = ±[(2/m)(E -U(x))]½ (2)
(2) is especially useful for nonlinear systems! Given U(x), its
usually straightforward to numerically compute v(x).
• We can obtain a qualitative phase diagram for a
fixed energy E by noting the shape of U(x) & using:
v(x) = x(x) = ± [(2/m)(E -U(x))]½
• Consider a particle confined to the potential U(x) in the figure.
(Soft for x < 0. Hard for x > 0)
• Assuming no damping, &
using v(x) = ±[(2/m)(E -U(x))]½
we can reason that the phase
diagram must be as shown here:
3 oval phase paths,
corresponding to the 3 different
energies E.
• v(x) = ±[(2/m)(E -U(x))]½
For energy E only slightly above
the minimum of U(x), the ovals
approach ellipses (which, from the
linear oscillator discussion, is what they
would be for a parabola: Ulinear(x) = (½)kx2).
This corresponds to fact (already
discussed) that, near the minimum of
U(x), we can approximate U(x) as a
parabola (linear oscillator!).
• v(x) = ±[(2/m)(E -U(x))]½
• Recall: The damped linear oscillator
phase diagram is an inward spiraling path.
Similarly, in this nonlinear case, if we
add damping, the ovals will spiral inward
as the particle goes to the potential minimum
& eventually comes to rest at x = 0.
• The equilibrium point x = 0 is an example
of an attractor.
• ATTRACTOR  A set of points (or one
point) in x-v phase space towards which a system is
“attracted” when damping is present.
• For case just discussed, x = 0 is a position of stable
equilibrium, since (d2U/dx2)0 > 0 (x = 0 is a minimum).
 The motion will be bounded, no matter how
nonlinear F(x) is (no matter how non-parabolic U(x) is).
• By contrast, now consider a particle
moving near a point where U(x) has a
maximum (say x = 0, as in the figure).
In this case, x = 0 is a position of
unstable equilibrium (d2U/dx2)0 < 0
 For a particle at rest there, a slight
disturbance will cause it to “roll downhill”. That is, the
motion will be unbounded.
• Figure: Consider an asymmetric
U(x) with maximum at x = 0

 A particle at energies E0, E1, E2
would have unbounded motion. To
understand the phase diagram (below): Imagine
first the case where, instead of the one in the
figure, U(x) is an “upside down” parabola:
U(x) = - (½)kx2. In that case, the phase paths
with energy E0 would be straight lines & those
with energies E1, E2 would be hyperbolas. The
phase paths in the figure approach these
in the limit as the nonlinearity in F(x)
(the non-parabolicity of U(x)) goes away.
• By looking at the phase diagrams for the
2 potentials we’ve
discussed (“valley” &
“hill”)  One can
(qualitatively) construct
the phase diagram for an
arbitrary U(x), with
both “hills” & “valleys”.
It would be some
superposition or
combination of the two phase diagrams. Near a “valley”, it
looks like the one on the left, near a “hill”, it looks like the one on the right!
Van der Pol Equation
• An interesting example of a non-linear equation of motion:
 “The van der Pol Equation”
(d2x/dt2) + μ(x2 - a2)(dx/dt) + ω02x = 0 (1)
μ > 0 (usually small). Find x(t)!
• The physics behind this equation? The corresponding U(x)?
– Electrical oscillators in vacuum tube circuits. See footnote, p. 153.
• Compare & contrast (1) with the equation of motion for the
linear oscillator with damping:
(d2x/dt2) + 2β(dx/dt) + ω02x = 0 (2)
 Comparing (1) & (2): 2β  μ(x2 -a2)  The Van der Pol
Equation is  linear oscillator with damping, but with an x
dependent damping coefficient (the amount of damping depends
on x = x(t), which is the unknown solution to the equation (1)!)
Van
der
Pol
Phase
Diagrams
2
2
2
2
2
(d x/dt ) + μ(x - a )(dx/dt) + ω0 x = 0
• Note: If the amplitude |x(t)| exceeds a critical value |a|, then
μ(x2 - a2) > 0 & the system is damped.  The x-v phase diagram is
Schematic
an inward spiral (towards x = a).
of a phase
• Note: If the amplitude |x(t)| is
diagram
less than |a|, then μ(x2 - a2) < 0 &
the system is “negatively” damped!
OR: x(t) increases exponentially
a
-a
without limit!  The x-v phase
diagram is an outward spiral (towards x = a).
• Note: If |x(t)| = |a|, then μ(x2 - a2) = 0 & the system is an
undamped linear oscillator. (No increase or decrease in x(t) with time).
 The x-v phase diagram is an ellipse.
(d2x/dt2) + μ(x2 - a2)(dx/dt) + ω02x = 0 (1)
• For (1), the curve |x(t)| = |a| is an ellipse in the x-v
phase plane. This curve is an example of a general
concept in the x-v phase plane:
“Limit Cycle”  A curve such that all paths
outside it spiral inward & all paths inside it
spiral outward.
The limit cycle is also the attractor for this system.
• The Limit Cycle defines locally bounded
motion  Therefore it also represents a stable
state for this non-linear oscillator.
Van der Pol Equation
(d2x/dt2) + μ(x2 - a2)(dx/dt) + ω02x = 0 (1)
• A system which obeys the Van der Pol Equation is an
example of a system which is SELF-LIMITING.
• Once the oscillator is set into motion under initial
conditions which lead to increasing amplitude |x(t)| < |a|
so that μ(x2 -a2) < 0, the amplitude is prevented from
growing without bound.
– In this case, there is a maximum (limiting) amplitude = |a|
• Once it is set into motion under initial conditions which
lead to decreasing amplitude |x(t)| > |a| so that μ(x2 -a2) >
0 the amplitude is prevented from decreasing to zero.
– In this case, there is a minimum (limiting) amplitude = |a|
Numerical Solution of the Van der Pol Eqtn
(d2x/dt2) + μ(x2 - a2)(dx/dt) + ω02x = 0 (1)
• Note: The following differs from your text, which I think has
the wrong explanation to go with the figures they generated!
To make the numerics easier, choose a system of units where
a = 2, ω02 = 1 (text says a = 1?)  (1) becomes:
(d2x/dt2) + μ(x2 - 4)(dx/dt) + x = 0 (2)
• The limit cycle is an ellipse semimajor axis 2. The
authors solve this on a computer (using Mathcad) using
μ = 0.05 (small damping)  It will take a long time for
x to reach the limit cycle |x| = |a| = 2. Look at it for 2
different initial conditions: x(0) = 1.0 (< a = 2) &
x(0) = 3.0 (> a = 2). In both cases, x(0) = v(0) = 0.
(d2x/dt2) + μ(x2 - 4)(dx/dt) + x = 0 (2)
• Limit cycle = ellipse
semimajor axis = 2. μ = 0.05
2 initial conditions:
x(0) = 1.0 (< a = 2)
x(0) = 3.0 (> a = 2)
v(0) = 0. Solutions
in the figure

Both cases: Clearly the
solution spirals towards
the limit cycle for long
times.
In this small damping
case, x(t) & v(t) are both
sinusoidal in time for long times.
(d2x/dt2) + μ(x2 - 4)(dx/dt) + x = 0 (2)
• Limit cycle = ellipse semimajor axis = 2. μ = 0.5 (large damping)
2 initial conditions:
x(0) = 1.0 (< a = 2)
x(0) = 3.0 (> a = 2)
v(0) = 0. Solutions
in the figure

Both cases: The solution
spirals towards the limit
cycle much more quickly
than in the other case. In
this large damping case,
x(t) & v(t) are skewed
from being sinusoidal in
time for long times.