Answers to P-Set # 02, 18.385j/2.036j MIT (Fall 2018) Rodolfo R. Rosales (MIT, Math. Dept., room 2-337, Cambridge, MA 02139) September 26, 2018 Contents 1 Problem 16.09.15 (Backwards: solutions to equation) 2 1.1 Problem 16.09.15 statement .......................................... 2 1.2 Problem 16.09.15 answer ............................................ 2 2 Problem 02.02.12 - Strogatz (A nonlinear resistor) 3 2.1 Problem 02.02.12 statement .......................................... 3 2.2 Problem 02.02.12 answer ............................................ 3 3 Problem 03.03.01 - Strogatz (Improved model of a laser) 3 3.1 Problem 03.03.01 statement .......................................... 3 3.2 Problem 03.03.01 answer ............................................ 5 4 Problem 03.04.06 - Strogatz (Find and classify bifurcations) 6 4.1 Problem 03.04.06 statement .......................................... 6 4.2 Problem 03.04.06 answer ............................................ 7 5 Problem 03.04.07 - Strogatz (Find and classify bifurcations) 8 5.1 Problem 03.04.07 statement .......................................... 8 5.2 Problem 03.04.07 answer ............................................ 8 6 Bifurcations in the circle problem #02 8 6.1 Statement: Bifurcations in the circle problem #02 ............................. 8 6.2 Answer: Bifurcations in the circle problem #02 ............................. 9 7 Excitable system with a refractory period 10 7.1 Statement: Excitable system with a refractory period ........................... 10 7.2 Answer: Excitable system with a refractory period ........................... 11 8 Perturbed pitchfork, with root preserved (bifurcation diagram) 12 8.1 Statement: Perturbed pitchfork, with root preserved (bifurcation diagram) ............... 12 8.2 Answer: Perturbed pitchfork, with root preserved (bifurcation diagram) ............... 12 9 Stability index for flows in the circle 13 9.1 Statement: Stability index for flows in the circle .............................. 13 9.2 Answer: Stability index for flows in the circle .............................. 13 List of Figures 1.1 Problem 16.09.15. Solutions to a one dimensional equation (on the line) ................ 2 3.1 Problem 03.03.01. Phase plane for improved laser model ......................... 7 4.1 Problem 03.04.06. Bifurcation diagram for dx/dt = ( r - (1 + x) -1 ) x .................. 7 1
Transcript
Answers to P-Set # 02, 18.385j/2.036j
MIT (Fall 2018)
Rodolfo R. Rosales (MIT, Math. Dept., room 2-337, Cambridge, MA 02139)
September 26, 2018
Contents
1 Problem 16.09.15 (Backwards: solutions to equation) 2
1 Problem 16.09.15 (Backwards: solutions to equation)
1.1 Statement for problem 16.09.15
(Working backwards, from solutions to equations). Find an equation x = f(x) whose solutions x = x(t) are
consistent with those shown in Figure 1.1. What is needed to make the consistency go beyond reproducing the topology
of the phase portrait, so as to capture the shapes 1 of the plots of the solutions as functions of time?
0 1 2 3
-1
0
1
t = time.
x =
sol
utio
n. Figure 1.1:
(Problem 16.09.15). Solutions to a one dimensional
equation (on the line): x = f(x).
Find an equation (i.e.: give f = f(x)) whose solu-
tions look like this, including the curves’ shapes.
Note that there is an infinite number of correct an-
swers — and wrong ones too.
Note: The figure shows you that there are two critical points on x ≥ −1, namely: x = 0 (stable) and x = 1 (unstable).
Thus you must produce x = f(x) where: f has zeros at x = 0 and x = 1; f(x) < 0 for 0 < x < 1; and f is positive
elsewhere. Actually, what f might do for x < −1 or x > 1.7 is not specified by the figure.
1.2 Answer for problem 16.09.15
A correct answer is provided by any (smooth 2) function f = f(x) with the following properties:
1. f(x) > 0 and f ′(x) < 0 for −1 ≤ x < 0.
2. f(0) = 0.
3. f(x) < 0 and f has a single minimum for 0 < x < 1.
4. f(1) = 0.
5. f(x) > 0 and f ′(x) > 0 for 1 < x.
These properties are also needed for a correct answer. Note that no information about f is provided for x < −1, or
for x bigger than (roughly) x = 2. The critical points x = 0 and x = 1 are stable and unstable, respectively. One
(of the many possible) correct answer is: f(x) = x(x− 1) .
1Note that the middle plot involves a single inflection point, while the others have none.2Smoothness is not strictly needed. But, for f not smooth, one has to worry about details such as: Are there solutions? Are they unique?
3
2 Problem 02.02.12 - Strogatz (A nonlinear resistor)
2.1 Statement for problem 02.02.12
(A nonlinear resistor). Suppose the resistor in Example 2.2.2 (page 20) is replaced by a nonlinear resistor. In other
words, this resistor does not have a linear relation between voltage and current. Such a nonlinearity arises in certain
solid-state devices. Instead of IR = V/R, suppose that we have IR = g(V ), where g(V ) has the shape shown in
Figure 3 (page 38 of the book).
Redo Example 2.2.2 in this case. Derive the circuit equations, find all the fixed points, and analyze their stability.
What qualitative effects does the nonlinearity introduce (if any)?
2.2 Answer for problem 02.02.12
As in Example 2.2.2, we begin by writing the equation for the voltage balance across the circuit (shown in Figure 2.2.3,
page 20, of the book):
V0 = V +Q
C.
Here V0 is the applied voltage, V is the voltage drop across the resistor, Q/C is the voltage drop across the the
capacitor, Q is the charge in the capacitor, and C is the capacitance. We also have Q = I = g(V ), where I is the
current across the circuit. Thus, since the equation above yields V = V0 −Q/C, we can write:
dQ
dt= g(V0 −
Q
C) . (2.1)
This is our final equation. From Figure 3 in page 38 of the book, it is clear that the solution Q = Q(t) will still grow
up from Q(0) = 0, in a monotone fashion, to the final value Q∞ = CV0 — with the final approach exponential. In
fact, the curve Q = Q(t) will still be concave down, since equation (2.1) yields:
d2Q
dt2= − 1
C
dQ
dth(V0 −
Q
C) ,
where h > 0 is the derivative of g. SincedQ
dt> 0, it follows that
d2Q
dt2< 0.
Note. As far as I can see, there is no real difference introduced by the nonlinearity in this example. A difference
would arise if the plot of g(V ) as a function of V where to give g → constant as V → ±∞ (saturation). In this case
the circuit goes towards equilibrium at roughly a constant rate when at high voltages — in contrast to the linear
resistor circuit which goes towards equilibrium faster the farther away the circuit is from equilibrium. However, I
am not sure if this is what the plot intends to suggest.
3 Problem 03.03.01 - Strogatz (Improved model of a laser)
3.1 Statement for problem 03.03.01
In the simple laser model considered in Section 3.3, we wrote an algebraic equation relating N , the number of excited
atoms, to n, the number of laser photons. In more realistic models, this would be replaced by a differential equation.
For instance, Milonni and Eberly 3 show that after certain reasonable approximations, quantum mechanics leads to
3 Milonni, P. W., and Eberly, J. H. (1988) Lasers (Wiley, New York.)
4
the system
dn
dt= GnN − kn , (3.2)
dN
dt= −GnN − fN + p . (3.3)
Here G is the gain coefficient for stimulated emission, k is the decay rate due to loss of photons by mirror transmission,
scattering, etc., f is the decay rate for spontaneous emission, and p is the pump strength. All parameters are positive,
except p, which can have either sign.
This two dimensional system will be analyzed in Exercise 8.1.13. For now, let us convert it to a one dimensional
system, as follows.
a. Suppose that N relaxes much more rapidly than n. Then we may take the quasi-static approximation N = 0.
Given this approximation, express N(t) in terms of n(t) and derive a first order system for n.
This procedure is often called adiabatic elimination, and one says that the evolution of N(t) is slaved to that
of n(t). See Haken;4 also remark 3.1 below.
b. Show that n∗ = 0 becomes unstable for p > pc, where pc is to be determined.
c. What type of bifurcation occurs at the laser threshold pc?
d. (Hard question). For what range of parameters is the approximation used in (a) valid?
Remark 3.1 The idea behind any adiabatic approximation is that, when the equations are appropriately non-
dimensionalized, one (or maybe more, in systems with many dimensions) of the equations takes the form:
dxndt
=1
τFn(x1, x2, . . . ) ,
where 0 < τ � 1 is a non–dimensional (fast) time scale — a measure of the speed at which xn attempts to reach
equilibrium, relative to the “main” time scale in the problem. Then, using the fact that τ is small, the differential
equation for xn can be replaced by the algebraic equation Fn = 0.
This adiabatic approximation idea is at the heart of most of the models in continuum mechanics (and many other
subjects). For example, in writing the equations for a (compressible) fluid one assumes that the internal energy, the
pressure, the density, and the temperature are related by equations of state, such as the ideal gas laws p = RρT and
e = cvT . However, thermodynamical concepts such as temperature and internal energy actually only have a strict
meaning at equilibrium; but, if the times scales for which this equilibrium is achieved are much shorter than the ones
involved in the dynamics of the gas, the type of arguments above justify their use. Finally, note that situations where
one cannot do this sort of stuff also occur. These cases, quite often, are open areas of research, where we do not have
a good description for the behavior of the system (e.g.: phase transitions in dynamical situations; say: triggered by a
strong compression wave.)
Note: Beware of typos in the answers at the back of the book.
• The solution given for part (a) cannot be right. It predicts N =∞ for n = f/G.
4 Haken, H. (1983) Synergetics, 3rd ed. (Springer, Berlin).
5
3.2 Answer for problem 03.03.01
We now solve the problem following the order outlined above.
a. The indicated quasi-static approximation yields
N =p
f +Gn,
as follows upon setting to zero the right hand side in equation (3.2). Using this to eliminate N in equation
(3.3), we obtain the first order system for n
dn
dt=
(Gp
f +Gn− k)n = k
(p
k− f
G− n
)Gn
f +Gn. (3.4)
b. Writing equation (3.4) in the formdn
dt= F (n), and calculating the derivative of the right hand side at n = 0,
we obtaindF
dn(0) = k
(p
k− f
G
)G
f=G
f(p− pc) ,
where pc = kf/G. Thus (given that G > 0, k > 0, and f > 0)
dF
dn(0) < 0 for p < pc , and
dF
dn(0) > 0 for p > pc .
It follows that n∗ = 0 is stable for p < pc and becomes unstable for p > pc.
c. Clearly a bifurcation occurs at pc (the laser threshold). From equation (3.4), we see that: at the the bifur-
cation point p = pc, the two roots (equilibrium points) of the right hand side of the equation 5 cross. Thus a
transcritical bifurcation occurs at p = pc.
d. When is the quasi-static approximation used to derive (3.4) from (3.2 – 3.3) valid? That is to say: for what
range of parameters can we use equation (3.4)?
The first thing we must do in order to answer this question (see remark 3.1) is to non-dimensionalize the
equations. The dimensions of the various parameters are as follows:
[k] = [f ] =1
time, [p] =
number
time, and [G] =
1
number× time.
Thus we have several possible choices of time and number units and we better think carefully to find the “right”
choice. We proceed to do this next.
First, it is clear that the critical points for equations (3.2 – 3.3) are:
(n,N) =
(0,p
f
), and (n,N) =
(p
k− f
G,k
G
).
Since we are (mostly) interested in studying the transition from non-lasing to lasing, we will begin by assuming
that these two critical points are (in some sense) in the “same neighborhood”. This is kind of a vague concept,
but we make it precise by assuming:
γ =Gp
kf= O(1) , (3.5)
where we note that γ is non-dimensional. In particular, this implies that the ratio of the N ’s for the critical
points is O(1), and that the value of n for the second critical point is the difference of two values of (roughly)
the same size.
5 Namely: n1 = 0 and n2 = pk− f
G.
6
With (3.5) in hand, we can now choose non-dimensional variables such that the critical points are both within
an O(1) distance of the origin. We do this by writing
n =f
Gn , N =
p
fN , and t =
1
kt , (3.6)
where the variables with bars are non-dimensional. As suggested by remark 3.1, the unit of time is selected
using a natural scale for the main phenomena of interest in the problem (that is: the number of laser photons,
with mean decay time — due to losses — given by (1/k)). The non-dimensional equations are thendn
dt= γnN − n , (3.7)
τdN
dt= −nN −N + 1 , (3.8)
where τ = k/f , and we have dropped the bars to simplify the notation. The critical points are now
(n,N) = (0, 1) and (n,N) = (γ − 1, 1/γ), both within O(1) distance to the origin (given (3.5)), as advertised.
Furthermore, the rate of change of n is O(1).
Equations (3.7 – 3.8) have the right form to implement an adiabatic approximation (as explained in remark 3.1),
provided that
τ =k
f� 1.
This is the condition (given (3.5)) needed for the validity of equation (3.4). That is: we need the decay rate
due to spontaneous emission, given by f , to be much larger than the decay rate due to mirror transmission,
scattering, etc., given by k. Said other way: the atoms fall from the excited state to the ground rate much
faster than the time scale over which photons can escape the system.
You can see what the solutions to the system of equations (3.7 – 3.8) look like in the phase plane (and how, except
for some rapid transient time, all the solutions end up on the line (1 + n)N = 1) in figure 3.1. This figure shows
two cases:
I. γ = 3/4 and τ = 0.1.
This is ”before” the bifurcation, that occurs for γ = 1.
The critical points are: (i) (n,N) = (0, 1), a node, and (ii) (n,N) = (−1/4, 4/3), a saddle.
II. γ = 4/3 and τ = 0.1.
This ”after” the bifurcation, that occurs for γ = 1.
The critical points are: (i) (n,N) = (0, 1), a saddle, and (ii) (n,N) = (1/3, 3/4), a node.
In general, the critical points of this system are (n,N) = (0, 1) and (n,N) = (γ − 1, 1/γ).
4 Problem 03.04.06 - Strogatz (Find and classify bifurcations)
4.1 Statement for problem 03.04.06
For the following equation, find the values of r at which bifurcations occur, and classify those as saddle node,
transcritical or pitchfork (supercritical or subcritical). Finally, sketch the bifurcation diagram of fixed points, x∗
versus r.dx
dt= r x− x
1 + x. (4.1)
Extra question: Notice that something “strange” happens for r = 0 in the bifurcation diagram. Is this a bifurcation?
If so, which type? Does the “principle of conservation of stability” apply? Hint: look at the equation satisfied by
y = 1/(1 + x).
7
-0.4 -0.2 0 0.2 0.40.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5 Laser equations, ! = 3/4 and " = 0.1
n
N
-0.2 0 0.2 0.4 0.6
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
Laser equations, ! = 4/3 and " = 0.1
n
N
Figure 3.1: (Problem 03.03.01). Phase plane portrait for the improved laser model non-dimensional equations (3.7–
3.8). Solutions converge rapidly to the curve N = 1/(n+ 1), because 0 < τ � 1. These figures are representative of
the situation on each side of the transcritical bifurcation at γ = 1.
4.2 Answer for problem 03.04.06
The equation is singular for x = −1, hence we should exclude this point from the phase space — which is then split
into two domains: x > −1 and x < −1. The critical points are given by:
1. x = 0, for any value of r and 2. x =1
r− 1, for r 6= 0.
By checking the sign of x in each of the segments into which the critical points and x = −1 divide the phase space,
it is easy to see that the bifurcation diagram for this equation is as in figure 4.1.
-10 -5 0 5 10-4
-3
-2
-1
0
1
2Bifurcation diagram for dx/dt = (r - 1/(1+x)) x.
r
x
Figure 4.1: (Problem 03.04.06). Bifurcation diagram for equation (4.1). See remark 4.2 for details.
Remark 4.2 Bifurcation diagram explanation: The sign of x splits the bifurcation plane (x, r) into six regions,
separated by the critical point curves x = x∗(r) — where x = 0 — and by the singularity line x = −1. In figure 4.1
the sign of x in each of these regions is indicated by a vertical arrow (pointing up for x > 0, and down for x < 0).
The stable critical point curves are plotted with solid lines, and the unstable critical point curves with dashed lines.
The singularity line x = −1 is plotted by a think solid line. It should be obvious that a transcritical bifurcation
occurs for x = 0 and r = 1 — the bifurcation point is indicated by the small disk in the figure.
18.385 MIT, (Rosales) Bifurcations in the circle problem #02. 8
Finally, note that some kind of strange bifurcation occurs for r = 0. As r approaches r = 0, one of the critical
point curves diverges (x∗ → ±∞), the curve disappears for r = 0, and it switches stability across r = 0. In particular,
it would seem that the standard “conservation of stability” principle that holds for ‘regular” bifurcations, fails for r = 0.
But:
A. Does the conservation of stability principle actually fail?
B. What type of bifurcation occurs for r = 0?
The answers to these two questions are: A: no, it does not fail, and B: a transcritical bifurcation.
To see this, consider the equation satisfied by y = 1/(1 + x) — namely:
dy
dt= (r − y) (y − 1) y. (4.2)
This equation has no singularities and has the critical points y∗ = r, y∗ = 1, and y∗ = 0. Furthermore, it is easy to
see that it has two transcritical bifurcations: one for r = 1 and y = 1, and the other for r = 0 and y = 0. Translating
this result into the variable x, we see that: for equation (4.1), x∗ =∞ is also a critical point, unstable for
r < 0, and stable for r > 0. This branch of critical points crosses with the branch x∗ = 1r− 1 for
r = 0 and x∗ =∞, where a transcritical bifurcation, and an exchange of stability, occurs.
5 Problem 03.04.07 - Strogatz (Find and classify bifurcations)
5.1 Statement for problem 03.04.07
For equation (5.1) below, find the values of r at which a bifurcation occurs, and classify them as saddle-node,
transcritical, supercritical pitchfork, or subcritical pitchfork. Finally, sketch the bifurcation diagram of fixed points
x∗ versus r.dx
dt= 5− re−x
2
. (5.1)
5.2 Answer for problem 03.04.07
The critical points are given by r = 5 ex2
, for −∞ < x < ∞. Clearly, there is a saddle-node bifurcation at
(r, x) = (5, 0), with the branch of solutions r = 5 ex2
(for 0 < x <∞) being unstable, and the other branch being
stable. The bifurcation diagram for equation (5.1) is shown in figure 5.1.
6 Bifurcations in the circle problem #02
6.1 Statement: Bifurcations in the circle problem #02
For equation (6.1) find the values of r at which a bifurcation occurs, and classify them as saddle-node, transcritical,
supercritical pitchfork, or subcritical pitchfork. Finally, sketch the bifurcation diagram for the fixed points versus r,
including the flow direction and the stability of the various branches of solutions (solid lines for stable branches and
dashed ones for unstable ones).dθ
dt= (r − cos(2 θ)) sin(θ), (6.1)
where θ is an angle (in radians). Note that the bifurcation diagram — which is periodic in θ — should be for a 2π
range in θ, and a range of r that includes all the bifurcations.
18.385 MIT, (Rosales) Bifurcations in the circle problem #02. 9
4.5 5 5.5 6 6.5-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1Bifurcation diagram for dx/dt = 5 - r e -x 2
.
r
x
Figure 5.1: (Problem 03.04.07). Bifurcation diagram for x = 5− re−x2
.
6.2 Answer: Bifurcations in the circle problem #02
The critical points for (6.1) are given by θ = nπ (n an integer), and the solutions to r = cos(2 θ). (6.2)
This has no solutions for |r| > 1 and four solutions (per period) for −1 < r < 1.
−1.5 −1 −0.5 0 0.5 1 1.5
Bifurcation diagram for d!/dt = (r − cos(2!)) sin(!).
r
!
−"/2
0
"/2
"
3"/2In each region (yellow or cyan), the black arrows
indicate the direction of the flow for the equation
θ = (r − cos(2 θ)) sin(θ).
Stable branches of critical points are plotted solid
blue, and unstable branches dashed red. Black dots
indicate the bifurcation points:
- Hard pitchforks at r = 1 and θ = 2nπ.
- Soft pitchforks at r = 1 and θ = 2nπ + π.
- Saddle nodes at r = −1 and θ = π2 + nπ.
Figure 6.1: Bifurcation diagram for equation 6.1.
At r = −1 the four solutions merge (by pairs) into the double roots θ = 12 π+nπ (n an integer), while at r = 1 they
merge (also by pairs) into the double roots θ = nπ.
The bifurcations occur for the values of r at which the number of solutions changes: r = −1 and r = 1. By looking
at the sign of θ in the regions into which the r-θ plane is divided by the curves r = cos(2 θ) and θ = nπ, it is easy to
ascertain the stability of the critical points, as well as the nature of the various bifurcations that occur. These are
as follows:
– Hard pitchfork bifurcations occur at r = 1 and θ = 2nπ, n integer.
– Soft pitchfork bifurcations occur at r = 1 and θ = 2nπ + π, n integer.
– Saddle node bifurcations at r = −1 and θ =π
2+ nπ, n integer.
18.385 MIT, (Rosales) Excitable system with a refractory period. 10
The results are summarized in figure 6.1.
7 Excitable system with a refractory period
7.1 Statement: Excitable system with a refractory period
The notion of an excitable system is introduced in the book by Strogatz, in problem 4.5.3. Quoting from there
Suppose you stimulate a neuron by injecting it with a pulse current. If the stimulus is small, nothing
dramatic happens: the neuron increases its membrane potential slightly, and then relaxes back to its
resting potential. However, if the stimulus exceeds a certain threshold, the neuron will ”fire” and produce
a large voltage spike before returning to rest. Surprisingly, the size of the spike does not depend much
on the size of the stimulus — anything above threshold will elicit essentially the same response.
Similar phenomena are found in other types of cells and even in some chemical reactions. These systems
are called excitable. The term is hard to define precisely, but roughly speaking, an excitable system is
characterized by two properties:
1. It has a unique, globally attracting rest state.
2. A large enough stimulus can send the system on a long excursion through phase space, before it
returns to the rest state.
Excitable systems can also exhibit an additional property: they have a refractory period: after going through an
excitation spike, there is a period of time during which it becomes harder to excite the system again.
This exercise deals with an extremely simple caricature of an excitable system with a refractory period. Let
dθ
dt= f(θ, r, z) (7.1)
where θ is an angle (in radians), and r, z are parameters satisfying 0 < r < 2 and r < z < 2π. The function f is
periodic of period 2π in θ, and it is defined as follows for 0 ≤ θ ≤ 2π
– f(θ) = θ (θ − r) for 0 ≤ x ≤ r,
– f(θ) = b (θ − r) (z − θ) for r ≤ x ≤ a,
– f(θ) = c (2π − θ) for a ≤ x ≤ 2π,
where
a =z + r
2+z − r
2
√1− r2
4, b =
4
(z − r)2, and c =
r2
8π − 4 a. (7.2)
The choice of b is so that f achieves a maximum value f
(z + r
2
)= 1.
The choice of a is so that f(a) =r2
4— note that the minimum value achieved by f is f
(r2
)= −r
2
4.
The choice of c guarantees continuity at θ = a.
Note that f is Lipschitz continuous — in fact, it is piece-wise smooth.
Let the “signal” put out by equation (7.1) be σ = θ, and assume that r is small (0 < r � 1) and z is somewhere
in the center of the interval 0 < θ < 2π, say 6 z ≈ π. Then show that (7.1) models an excitable system with a
refractory period. That is, show that:6 The exact value of z does not matter, as long as it is not close to 0 or to 2π.
18.385 MIT, (Rosales) Excitable system with a refractory period. 11
A. The system has a unique, globally attracting, 7 critical point. Note that, while there, σ ≡ 0.
B. A small perturbation (to the attracting critical point) results in a very small signal: σ never exceeds r2/2.
C. Above a certain threshold, an O(1) signal can result from a perturbation to the globally attracting critical
point. The signal then reaches a maximum value σ = 1, (almost) independently of the size of the perturbation
beyond the threshold. What is the threshold?
D. In case C, after the signal becomes small again (σ = O(r2)), there is a time period (in fact, a long time period
for this example), where a small perturbation cannot trigger another spike — the refractory period.
Hint: investigate the properties of f when 0 < r � 1. It is OK to do this graphically: that is, plot the function using,
for example, MatLab.
7.2 Answer: Excitable system with a refractory period
Figure 7.1 illustrates the behavior of the function f in (7.2) for r small and z away from both θ = 0 and θ = 2π.
Notice that:
A. The system has two critical points, θ = 0 (stable), and θ = r (unstable). In fact θ = 0 is a global attractor.
B. Perturbations smaller than r (the threshold value) cause the system to return to equilibrium, with |θ| never
exceeding r. Hence σ never exceeds r2/4, since σ = f , and |f | does not exceed r2/4 except on the “bump”
from θ = r to θ = a.
0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
!
y
Excitable with refractory period.
Figure 7.1: Excitable system with a refractory period. Plot of the function defined in (7.2) for r = 0.5 and z = 3.
C. Perturbations that cause the value of θ to exceed r, result in a solution that traverses all the way along the
circle (back to θ = 2π ≡ 0). Along the way σ reaches the value σ = 1. Only very large perturbations, which
send θ beyond the position of the maximum of f at θ = 12 (z + r), fail to produce σ = 1.
D. In case C, the signal becomes small again after θ reaches θ = a, and stays small thereafter. However, till θ
gets close to the attractor, small perturbations cannot reset the solution so that it (again) traverses the region
r < θ < a, which is associated with σ = O(1).
7 Globally attracting means that all the solutions approach the critical point as t→∞.
8 Perturbed pitchfork, with root preserved (bifurcation diagram)
8.1 Statement: Perturbed pitchfork, with root preserved (bifurcation diagram)
Consider the structural stability for a (soft) pitchfork bifurcation, with the restriction that the “main” solution
branch is preserved across the bifurcation. Specifically, consider the situation where:
dx
dt= g(x, r) (g odd in x), (8.1)
has a (soft) pitchfork bifurcation at (x, r) = (0, 0). Assume that the
problem depends on a hidden parameter h — i.e. let g(x, r) = f(x, r, h)h=0
,
where you only know that h is small (but it may not be zero). Assume
also that you know that f(0, r, h) = 0, though f may not be odd for h 6= 0. Provided that f is reasonably smooth,
and f is generic, it can be shown that the canonical equation 8 describing this situation is
dx
dt= r x+ hx2 − x3. (8.2)
Tasks: Assume h 6= 0 small (say, h = 0.05), and draw the bifurcation diagram for (8.2), including the flow lines
— recall that the bifurcation diagram is, basically, all the phase portraits (one for each r) stacked in one single 2-D
plot. What happens to the pitchfork? Furthermore: estimate the level of noise (in x) under which the distinction
between the pitchfork and the new behavior will be hidden — do this in terms of h.
8.2 Answer: Perturbed pitchfork, with root preserved (bifurcation diagram)
The critical points for (8.2) are given by x = 0 for all values of (r, h), and
xu =h
2+
√r +
h2
4, xd =
h
2−√r +
h2
4, for r ≥ −h
2
4. (8.3)
Alternatively, instead of (8.3), we can write
r = x2 − hx = (x− 12 h)2 − 1
4 h2, (8.4)
which parameterizes the nonzero critical points by giving r as a function of x. Before drawing the bifurcation
diagram, we notice that we can scale-out h from equation (8.2) by the transformation
x = hX, r = h2R, and t = T/h2, (8.5)
as long as h 6= 0. This reduces (8.2) todX
dT= RX +X2 −X3. (8.6)
The bifurcation diagram for this equation can be found in figure 8.1. Note that the pitchfork is broken into a
transcritical and a saddle node (separated by a distance which is O(h) in x — this follows from the scaling in (8.5).
Remark The scaling in (8.5) indicates that the changes to the pitchfork bifurcation diagram occur within a neigh-
borhood of the critical point at x, whose size is O(h). Outside this neighborhood the diagram looks the same as that
8 That is, near the bifurcation, the full problem can be mapped into equation (8.2).
18.385 MIT, (Rosales) Stability index for flows in the circle. 13
0.5 0.25 0 0.25 0.5 0.750.5
0
0.5
1
1.5 Bifurcation diagram for dX/dT = RX + X2 X3.
R
X
Figure 8.1: Scaled perturbation diagram for a
perturbed pitchfork, with perturbation preserving
the “main” solution. The canonical equation for
the situation is (8.2), which (for h 6= 0) can be
transformed into: dX/dT = RX +X2 −X3.
The diagram on the left corresponds to this equa-
tion. The stable branches of solutions are indi-
cated by solid blue lines, while the unstable ones
are in dashed red. The black dots indicate the
location of the bifurcations.
of the pitchfork. It follows that if the system is noisy, with a level of noise above O(h), it becomes impossible to
detect the difference between the diagram in figure 8.1, and that of the unperturbed system. ♣
9 Stability index for flows in the circle
9.1 Statement: Stability index for flows in the circle
Show that the stability index S for any flow in the circle vanishes. To be precise, consider an equation of the form
dθ
dt= f(θ), (9.1)
where θ is an angle (in radians), and f is periodic of period 2π and Lipschitz continuous. Assume also that the
equation has a finite number of critical points: 9 θ1 < θ2 < · · · < θN < θ1 + 2π. Now assign a weight w = 1 to each
stable critical point, a weight w = −1 to each unstable critical point, and a weight w = 0 to each semi-stable critical
point. Then show that
S =
N∑n=1
wn = 0. (9.2)
Hint 9.1 Consider the intervals In, 1 ≤ n ≤ N , where In is the interval θn < θ < θn+1 — here θN+1 = θ1 + 2π,
which is the same point as θ1 because we are in the circle. Then in each such interval either 10 f > 0 or f < 0.
Define σn = 1 if f > 0 in In, and σn = −1 if f < 0 in In. Then relate the wn to the σn to show (9.2). What
information do the σn capture? ♣
9.2 Answer: Stability index for flows in the circle
The σn characterize the direction of the flow given by (9.1). If σn = 1, then the flow is from θn towards θn+1. If
σn = −1, the flow is in the opposite direction. It is then easy to see that
wn =1
2(σn−1 − σn) for 1 ≤ n ≤ N, (9.3)
9 The critical points are the zeros of f .10 If f were to switch sign in In, then (since it is continuous) it would have a zero in In. This zero would no be one of the θn, which are
supposed to be all the zeros.
18.385 MIT, (Rosales) Stability index for flows in the circle. 14
where σ0 = σN (again, we are in a circle, so that I0 is the same as IN ). It follows that
