Hirsch M., Smale S., Devaney R., Differential Equations, Dynamical Systems, and an Introduction to Chaos

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266 Chapter 12 Applications in Circuit Theory

⫹

⫺

(1, 0)

a(y)

Figure 12.5 The

semi-Poincaré map.

about the origin. This implies that if (x(t ), y(t )) is a solution curve, then so is

(−x(t ), −y(t)).

Part of the graph of δ(y) is shown schematically in Figure 12.6. The interme-

diate value theorem and the proposition imply that there is a unique y

∈ v

with δ(y

) = 0. Consequently, α(y

) =−y

and it follows from the skew-

symmetry that the solution through (0, y

) is periodic. Since δ(y) = 0 except

at y

, we have α(y) = y for all other y-values, and so it follows that φ

(0, y

)

is the unique periodic solution.

We next show that all other solutions (except the equilibrium point) tend

to this periodic solution. Toward that end, we deﬁne a map β : v

−

→ v

sending each point of v

−

to the ﬁrst intersection of the solution (for t>0)

with v

. By symmetry we have

β(y) =−α(−y).

Note also that P(y) = β ◦α(y).

d(y)

Figure 12.6 The

graph of δ(y).

12.3 The van der Pol Equation 267

We identify the y-axis with the real numbers in the y coordinate. Thus if

, y

∈ v

∪ v

−

we write y

if y

is above y

. Note that α and β reverse

this ordering while P preserves it.

Now choose y ∈ v

with y>y

. Since α(y

) =−y

we have α(y) < −y

and P(y) >y

. On the other hand, δ(y) < 0, which implies that α(y) > −y.

Therefore P(y) = β(α(y)) <y. We have shown that y>y

implies y>P(y) >

. Similarly P(y) >P(P(y)) >y

and by induction P

(y) >P

n+1

(y) >y

for all n>0.

The decreasing sequence P

(y) has a limit y

≥ y

in v

. Note that y

is a

ﬁxed point of P, because, by continuity of P, we have

P(y

) − y

= lim

n→∞

P(P

(y)) − y

= y

− y

= 0.

Since P has only one ﬁxed point, we have y

= y

. This shows that the solution

through y spirals toward the periodic solution as t →∞. See Figure 12.7. The

same is true if y<y

; the details are left to the reader. Since every solution

except the equilibrium meets v

, the proof of the theorem is complete. 

Finally, we turn to the proof of the proposition. We adopt the following

notation. Let γ :[a, b]→

be a smooth curve in the plane and let F : R

→

R. We write γ (t ) = (x(t ), y(t)) and deﬁne



F(x, y) =



F(x(t ), y(t )) dt .

Figure 12.7 The phase

portrait of the van der Pol

equation.

268 Chapter 12 Applications in Circuit Theory

If it happens that x



(t) = 0 for a ≤ t ≤ b, then along γ , y is a function of x,

so we may write y = y(x). In this case we can change variables:



F(x(t ), y(t )) dt =



x(b)

x(a)

F(x, y(x))

dx.

Therefore



F(x, y) =



x(b)

x(a)

F(x, y(x))

dx/dt

dx.

We have a similar expression if y



(t) = 0.

Now recall the function

W (x, y) =



+ y



introduced in the previous section. Let p ∈ v

. Suppose α(p) = φ

(p). Let

γ (t ) = (x(t ), y(t )) for 0 ≤ t ≤ τ = τ (p) be the solution curve joining p ∈ v

to α(p) ∈ v

−

. By deﬁnition

δ(p) =



y(τ )

− y(0)



= W (x(τ ), y(τ )) − W (x(0), y(0)).

Thus

δ(p) =



W (x(t ), y(t)) dt .

Recall from Section 12.2 that we have

W =−xf (x) =−x(x

− x).

Thus we have

δ(p) =



−x(t )(x(t )

− x(t )) dt



x(t )

(1 − x(t)

) dt .

This immediately proves part (1) of the proposition because the integrand is

positive for 0 <x(t ) < 1.

12.3 The van der Pol Equation 269

⫹

Figure 12.8 The curves γ

, γ

and γ

depicted on the closed

orbit through y

We may rewrite the last equality as

δ(p) =





1 − x



We restrict attention to points p ∈ v

with p>y

∗

. We divide the correspond-

ing solution curve γ into three curves γ

, γ

as displayed in Figure 12.8.

The curves γ

and γ

are deﬁned for 0 ≤ x ≤ 1, while the curve γ

is deﬁned

for y

≤ y ≤ y

. Then

δ(p) = δ

(p) + δ

(p)

where

(p) =





1 − x



, i = 1, 2, 3.

Notice that, along γ

, y(t ) may be regarded as a function of x. Hence we have

(p) =





1 − x



dx/dt





1 − x



y − f (x)

where f (x) = x

− x.Asp moves up the y-axis, y − f (x) increases [for (x, y)

on γ

]. Hence δ

(p) decreases as p increases. Similarly δ

(p) decreases as p

increases.

270 Chapter 12 Applications in Circuit Theory

On γ

, x(t ) may be regarded as a function of y, which is deﬁned for y ∈

, y

] and x ≥ 1. Therefore since dy/dt =−x, we have

(p) =



−x(y)



1 − x(y)





x(y)



1 − x(y)



so that δ

(p) is negative.

As p increases, the domain [y

, y

] of integration becomes steadily larger.

The function y → x(y) depends on p, so we write it as x

(y). As p

increases, the curves γ

move to the right; hence x

(y) increases and so

(y)(1 − x

(y)

) decreases. It follows that δ

(p) decreases as p increases;

and evidently lim

p→∞

(p) =−∞. Consequently, δ(p) also decreases and

tends to −∞ as p →∞. This completes the proof of the proposition.

12.4 A Hopf Bifurcation

We now describe a more general class of circuit equations where the resistor

characteristic depends on a parameter μ and is denoted by f

. (Perhaps μ is

the temperature of the resistor.) The physical behavior of the circuit is then

described by the system of differential equations on

= y − f

(x)

=−x.

Consider as an example the special case where f

is described by

(x) = x

− μx

and the parameter μ lies in the interval [−1, 1]. When μ = 1 we have the

van der Pol system from the previous section. As before, the only equilibrium

point lies at the origin. The linearized system is





μ 1

−10



Y ,

12.4 A Hopf Bifurcation 271

and the eigenvalues are



μ ±



− 4



Thus the origin is a spiral sink for −1 ≤ μ < 0 and a spiral source for

0 < μ ≤ 1. Indeed, when −1 ≤ μ ≤ 0, the resistor is passive as the graph of f

lies in the ﬁrst and third quadrants. Therefore all solutions tend to the origin in

this case. This holds even in the case where μ = 0 and the linearization yields

a center. Physically the circuit is dead in that, after a period of transition, all

currents and voltages stay at 0 (or as close to 0 as we want).

However, as μ becomes positive, the circuit becomes alive. It begins to

oscillate. This follows from the fact that the analysis of Section 12.3 applies

to this system for all μ in the interval (0, 1]. We therefore see the birth of

a (unique) periodic solution γ

as μ increases through 0 (see Exercise 4 at

the end of this chapter). Just as above, this solution attracts all other nonzero

solutions. As in Section 8.5, this is an example of a Hopf bifurcation. Further

elaboration of the ideas in Section 12.3 can be used to show that γ

→ 0as

μ → 0 with μ > 0. Figure 12.9 shows some phase portraits associated to this

bifurcation.

l⫽⫺0.5 l⫽⫺0.1

l⫽0.05 l⫽0.2

Figure 12.9 The Hopf bifurcation in the system x



= y −

+ μ x, y



= −x.

272 Chapter 12 Applications in Circuit Theory

12.5 Exploration: Neurodynamics

One of the most important developments in the study of the ﬁring of nerve

cells or neurons was the development of a model for this phenomenon in

giant squid in the 1950s by Hodgkin and Huxley [23]. They developed a four-

dimensional system of differential equations that described the electrochemical

transmission of neuronal signals along the cell membrane, a work for which

they later received the Nobel prize. Roughly speaking, this system is similar

to systems that arise in electrical circuits. The neuron consists of a cell body,

or soma, which receives electrical stimuli. This stimulus is then conducted

along the axon, which can be thought of as an electrical cable that connects to

other neurons via a collection of synapses. Of course, the motion is not really

electrical, because the current is not really made up of electrons, but rather

ions (predominantly sodium and potassium). See [15] or [34] for a primer on

the neurobiology behind these systems.

The four-dimensional Hodgkin-Huxley system is difﬁcult to deal with,

primarily because of the highly nonlinear nature of the equations. An impor-

tant breakthrough from a mathematical point of view was achieved by

Fitzhugh [18] and Nagumo et al. [35], who produced a simpler model of

the Hodgkin-Huxley model. Although this system is not as biologically accu-

rate as the original system, it nevertheless does capture the essential behavior

of nerve impulses, including the phenomenon of excitability alluded to below.

The Fitzhugh-Nagumo system of equations is given by



= y + x −

+ I



=−x + a − by

where a and b are constants satisfying

0 <

(1 − a) <b<1

and I is a parameter. In these equations x is similar to the voltage and rep-

resents the excitability of the system; the variable y represents a combination

of other forces that tend to return the system to rest. The parameter I is a

stimulus parameter that leads to excitation of the system; I is like an applied

current. Note the similarity of these equations with the van der Pol equation

of Section 12.3.

1. First assume that I = 0. Prove that this system has a unique equilibrium

point (x

, y

). Hint: Use the geometry of the nullclines for this rather than

Exercises 273

explicitly solving the equations. Also remember the restrictions placed on

a and b.

2. Prove that this equilibrium point is always a sink.

3. Now suppose that I = 0. Prove that there is still a unique equilibrium

point (x

, y

) and that x

varies monotonically with I .

4. Determine values of x

for which the equilibrium point is a source and

show that there must be a stable limit cycle in this case.

5. When I = 0, the point (x

, y

) is no longer an equilibrium point.

Nonetheless we can still consider the solution through this point. Describe

the qualitative nature of this solution as I moves away from 0. Explain

in mathematical terms why biologists consider this phenomenon the

“excitement” of the neuron.

6. Consider the special case where a = I = 0. Describe the phase plane for

each b>0 (no longer restrict to b<1) as completely as possible. Describe

any bifurcations that occur.

7. Now let I vary as well and again describe any bifurcations that occur.

Describe in as much detail as possible the phase portraits that occur in the

I, b–plane, with b>0.

8. Extend the analysis of the previous problem to the case b ≤ 0.

9. Now ﬁx b = 0 and let a and I vary. Sketch the bifurcation plane (the

I, a–plane) in this case.

EXERCISES

1. Find the phase portrait for the differential equation



= y − f (x), f (x) = x



=−x.

Hint: Exploit the symmetry about the y-axis.

2. Let

f (x) =

⎧

⎨

⎩

2x − 3ifx>1

−x if −1 ≤ x ≤ 1

2x + 3ifx<−1.

Consider the system



= y − f (x)



=−x.

(a) Sketch the phase plane for this system.

(b) Prove that this system has a unique closed orbit.

274 Chapter 12 Applications in Circuit Theory

3. Let

(x) =

⎧

⎨

⎩

2x + a − 2ifx>1

ax if −1 ≤ x ≤ 1

2x − a + 2ifx<−1.

Consider the system



= y − f

(x)



=−x.

(a) Sketch the phase plane for this system for various values of a.

(b) Describe the bifurcation that occurs when a = 0.

4. Consider the system described in Section 12.4



= y − (x

− μx)



=−x

where the parameter μ satisﬁes 0 ≤ μ < 1. Fill in the details of the proof

that a Hopf bifurcation occurs at μ = 0.

5. Consider the system



= μ(y − (x

− x)), μ > 0



=−x.

Prove that this system has a unique nontrivial periodic solution γ

. Show

that as μ →∞, γ

tends to the closed curve consisting of two horizontal

line segments and two arcs on y = x

− x as in Figure 12.10. This type

of solution is called a relaxation oscillation. When μ is large, there are

two quite different timescales along the periodic solution. When moving

horizontally, we have x



very large, and so the solution makes this transit

Figure 12.10

Exercises 275

Figure 12.11

very quickly. On the other hand, near the cubic nullcline, x



= 0 while y



is bounded. Hence this transit is comparatively much slower.

6. Find the differential equations for the network in Figure 12.11, where the

resistor is voltage controlled; that is, the resistor characteristic is the graph

of a function g :

R → R, i

= g (v

7. Show that the LC circuit consisting of one inductor and one capacitor

wired in a closed loop oscillates.

8. Determine the phase portrait of the following differential equation and in

particular show there is a unique nontrivial periodic solution:



= y − f (x)



=−g (x)

where all of the following are assumed:

(a) g (−x) =−g (x) and xg (x) > 0 for all x = 0;

(b) f (−x) =−f (x) and f (x) < 0 for 0 <x<a;

(d) f (x) →∞as x →∞.

9. Consider the system



= y



= a(1 − x

)y − x

(a) Find all equilibrium points and classify them.

(b) Sketch the phase plane.

(d) Prove that there exists a unique closed orbit for this system when

a>0.

(e) Show that all nonzero solutions of the system tend to this closed orbit

when a>0.