Math/Phys 594: Homework 7 Solutions

February 27, 2019

6.1.4 For x = y, y = x(1 + y) − 1, find the fixed points. Then sketch the nullclines, thevector field, and a plausible phase portrait.

The nullclines are given by x = 0 ⇒ y = 0 and y = 0 ⇒ y = 1x− 1. The fixed point(s)

are given by the points of intersection between the nullclines – plug y = 0 in y = 1x− 1 to

get x = 1. Thus, the only fixed point is (1, 0). The nullclines drawn in Figure 1(a) showthat the fixed point is a saddle. We can also guess the vector field away from the nullclinesand use that to sketch the phase portrait shown in Figure 1(b).

(a) Nullclines and vector field (b) The phase portrait

Figure 1: Problem 6.1.4

6.1.6 For x = x2 − y, y = x − y, find the fixed points. Then sketch the nullclines, thevector field, and a plausible phase portrait.

For finding the nullclines, we set x = 0 and y = 0 to get y = x2 and y = x respectively.The fixed points lie at the intersection points of these curves, namely (0, 0) and (1, 1). Thenullclines and a rough vector field are shown in Figure 2(a). The flow direction suggests thatthe system has a saddle point at (1, 1) and a stable spiral at (0, 0). That this is indeed so isconfirmed by the phase portrait in Figure 2(b).

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(a) Nullclines and vector field (b) The phase portrait

Figure 2: Problem 6.1.6

6.3.2 For x = sin(y), y = x−x3, find the fixed points, classify them, sketch the neighboringtrajectories, and try to fill in the rest of the phase portrait.

Setting x = 0 gives sin(y) = 0 ⇒ y = kπ for k ∈ Z. Similarly, setting y = 0 givesx − x3 = 0 ⇒ x = 0,±1. We conclude that the system has fixed points at (0, kπ), (1, kπ)and (−1, kπ).

In order to classify these fixed points, we compute the linearization. We have

A(x∗,y∗) =

(fx fygx gy

)=

(0 cos(y∗)

1− 3(x∗)2 0

).

Note that the trace of the above matrix is always zero so we only need to concern ourselves

with the determinant. At (0, kπ), we obtain A =

(0 (−1)k

1 0

)so that ∆ = (−1)k+1. Thus,

the linearization predicts centers for odd k and saddle points for even k.

Similarly, for (±1, kπ), we have A =

(0 (−1)k

−2 0

)so ∆ = 2(−1)k. In this case, the

linearization predicts centers for even k and saddle points for odd k.Recall that we are not supposed to trust the linearization when it predicts a center. Note,

however, that since

div(sin(y), x− x3) = 0

the system is conservative with

E(x, y) = − cos(y) +1

4x4 − 1

2x2

acting as a conserved quantity. We next show that the candidates for centers are strictmimima or maxima of E. We have

Ex = x3 − x, Ey = sin(y), Exx = 3x2 − 1, Eyy = cos(y), Exy = 0,

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so Ex(0, (2j + 1)π) = Ey(0, (2j + 1)π) = 0 and Ex(±1, 2jπ) = Ey(±1, 2jπ) = 0 for all j.Next, we compute the Hessian

D(0, (2j + 1)π) = (ExxEyy − E2xy)|(0,(2j+1)π) = 1 > 0

and, since Exx(0, (2j + 1)π) = −1 < 0, we conclude that (0, (2j + 1)π) is a strict maximumof E, for all j. Similarly, we have

D(±1, 2jπ) = (ExxEyy − E2xy)|(±1,2jπ) = 2 > 0

and, since Exx(±1, 2jπ) = 2 > 0, we conclude that (±1, 2jπ) are strict minima of E, for allj. This proves that the trajectories around these fixed points are indeed centers.

The corresponding phase portrait is shown in Figure 3. Note in particular the trajectoriesemanating out of the saddle point at the origin that eventually return to it. These areinstances of homoclinic orbits.

Figure 3: Problem 6.3.2

6.3.4 For x = y+x−x3, y = −y, find the fixed points, classify them, sketch the neighboringtrajectories, and try to fill in the rest of the phase portrait.

Setting y = 0 gives y = 0. Plugged into x = y + x− x3 = 0, we obtain x(1− x2) = 0⇒x = 0,±1. Thus, the system has three fixed points at (±1, 0) and (0, 0).

The linearization is given by

A(x∗,y∗) =

(1− 3(x∗)2 1

0 −1

).

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Observe that since this is an upper-triangular matrix, the eigenvalues are simply given

by the diagonal entries. At (0, 0), we have A =

(1 10 −1

); the eigenvalues are λ1 = 1

and λ2 = −1 so the origin is a saddle point. Note that the corresponding eigenvectors

are v1 =

(10

)and v2 =

(−12

); these vectors indicate the unstable and stable manifolds

respectively close to the saddle.

At (±1, 0), we have A =

(−2 10 −1

); the eigenvalues are λ1 = −2 and λ2 = −1 so the

origin is a stable node. The corresponding eigenvectors are v1 =

(10

)(fast direction) and

v2 =

(11

)(slow direction). The resulting phase portrait is shown in Figure 4.

Figure 4: Problem 6.3.4: Observe that the trajectories become parallel to the slow directionsclose to the stable nodes (indicated by the dotted lines)

6.3.8 (Gravitational equilibrium) A particle moves along a line joining two stationarymasses, m1 and m2, and which are separated by a fixed distance a. Let x denote thedistance of the particle from m1.

(a) Show that x = Gm2

(x−a)2 −Gm1

x2, where G is the gravitational constant.

(b) Find the particle’s equilibrium position. Is it stable or unstable?

(a) Denote the mass of the particle by m. Then, the gravitational forces due to themasses m1 and m2 are −Gmm1

x2and Gmm2

(a−x)2 respectively. Adding these together and

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using Newton’s second law gives

mx =Gmm2

(x− a)2− Gmm1

x2⇒ x =

Gm2

(x− a)2− Gm1

x2.

(b) We first rewrite this as a first order system. Define y = x to obtain

x = y

y =Gm2

(x− a)2− Gm1

x2.

The fixed point must obey x = 0⇒ y = 0 and y = 0 which yields

Gm2

(x− a)2=Gm1

x2⇒(x− ax

)2

=m2

m1

⇒ x =a

1±√m2/m1

.

We shall ignore the solution corresponding to the negative sign above since that givesx < 0 or x > a. We conclude that the fixed point for the first order system is(

a

1+√m2/m1

, 0

)and the equilibrium position is x∗ = a

1+√m2/m1

.

To determine the stability of this fixed point, we linearize the system about it to get

A =

(0 1

2G(− m2

(x∗−a)3 + m1

(x∗)3

)0

).

This matrix is trace-less with determinant

∆ = −2G

(− m2

(x∗ − a)3+

m1

(x∗)3

).

Our fixed point obeys (x∗ − a) = −x∗√

m2

m1so we obtain

∆ = −2Gm1

(x∗)3

(√m1

m2

+ 1

)< 0.

We conclude that the fixed point is a saddle and hence unstable.

6.3.11 (Nonlinear terms can change a star into a spiral) Here’s another example that showsthat borderline fixed points are sensitive to nonlinear terms. Consider the system in polarcoordinates given by r = −r, θ = 1/ ln(r).

(a) Find r(t) and θ(t) explicitly, given an initial condition (r0, θ0).

(b) Show that r(t)→ 0 and |θ(t)| → ∞ as t→∞. Therefore the origin is a stable spiralfor the nonlinear system.

(c) Write the system in x, y coordinates.

(d) Show that the linearized system about the origin is x = −x, y = −y. Thus the originis a stable star for the linearized system.

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(a) From r = −r, we deduce r(t) = r0e−t so that θ = 1/ ln(r0e

−t) = 1/(ln(r0) − t). Thisyields

θ(t) = θ0 + ln

∣∣∣∣ ln(r0)

ln(r0)− t

∣∣∣∣ .(b) As t→∞, we have r(t) = r0e

−t → 0. Similarly, as t→∞, we get ln(r0)/(ln(r0)−t)→0 so θ(t)→ −∞. (There is a singularity at t = ln(r0) if r0 > 1 where we start spinninginfinitely fast but this goes away when t goes past this value. It’s best to assume thatr0 < 1, in which case θ decays monotonically to −∞)

(c) Using x = r cos(θ) and y = r sin(θ), we have

x = r cos(θ)− r sin(θ)θ, y = r sin(θ) + r cos(θ)θ.

Plugging in the expressions for r and θ and using r = (x2 + y2)1/2 gives

x = −x− 2y

ln(x2 + y2), y = −y +

2x

ln(x2 + y2).

(d) Noting that x and y measure the deviations from the origin, we can linearize by omittingthe nonlinear terms in the system above to get

x = −x, y = −y

which predicts a star since the corresponding matrix for this problem is A = −I.

6.3.15 Consider the system r = r(1 − r2), θ = 1 − cos(θ), where r, θ represent polarcoordinates. Sketch the phase portrait and thereby show that the fixed point r∗ = 1,θ∗ = 0 is attracting but not Liapunov stable.

This system has fixed points at r = 0 and r = 1, θ = 0. From the phase portrait inFigure 5, it is clear that, while the fixed point at r = 1, θ = 0 is attractive, it is not Liapunovstable – a trajectory starting from r0 = 1 and θ0 small and positive will travel the unit circlebefore converging to the fixed point.

6.4.5 Now suppose that species #1 has a finite carrying capacity K1. Thus

N1 = r1N1(1−N1/K1)− b1N1N2

N2 = r2N2 − b2N1N2.

Nondimensionalize the model and analyze it. Show that there are two qualitatively differentkinds of phase portrait, depending on the size of K1. (Hint: Draw the nullclines.) Describethe long-term behavior in each case.

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Figure 5: Problem 6.3.15: The light blue trajectory is an example of a solution that startsclose to r = 1, θ = 0 but travels around the unit circle before converging.

Let T be the typical time-scale and length P1 and P2 be the typical population scales forspecies 1 and 2 respectively. Defining x1 = N1/P1, x2 = N2/P2 and τ = t/T then gives

P1

T

dx1dt

= x1(r1P1)

(1− P1

K1

x1

)− (b1P1P2)x1x2

P2

T

dx2dt

= (r2P2)x2 − (b2P1P2)x1x2.

⇒ dx1dt

= x1(r1T )

(1− P1

K1

x1

)− (b1P2T )x1x2

dx2dt

= (r2T )x2 − (b2P1T )x1x2.

Setting r2T = 1 and b2P1T = 1 gives T = r−12 and P1 = r2b−12 . Similarly, choosing

b1P1P2 = 1 gives P2 = r−12 b−11 b2. Finally, defining α = r1P1 = r1r2b−12 and γ = P1/K1 =

r2b−12 K−11 allows us to write the system as

dx1dt

= αx1 (1− γx1)− x1x2dx2dt

= x2 − x1x2.

We shall analyze this system by drawing its nullclines. There are two qualitativelydifferent cases, depending on the value of γ.

• For γ < 1, the system has three fixed points: saddles at (0, 0) and (1, α(1− γ)) and astable fixed point at (γ−1, 0) (see Figure 6). The phase portrait in Figure 6(b) indicates

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that if the initial size of species 2 is small enough, it goes extinct while the the firstpopulation survives. However, if the initial level is large, species 2 grows without boundand species 1 dies out.

(a) Nullclines and vector field (b) The phase portrait

Figure 6: Problem 6.4.5 with γ < 1 (we have set γ = 0.5 and α = 1 for these diagrams).Observe that the stable manifold of (1, α(1 − γ)) acts as a separatix; trajectories startingbelow it are attracted to (γ−1, 0) while those above it go to (0,∞).

• For γ > 1, we only have two fixed points at (0, 0) and (γ−1, 0) and both are saddles(Figure 7). As a result, if we start with non-zero populations for both species, wealways approach (0,∞). This corresponds to species 1 dying out and species 2 takingover the ecosystem.

(a) Nullclines and vector field (b) The phase portrait

Figure 7: Problem 6.4.5 with γ > 1 (actually γ = 2 and α = 1). Observe that all thetrajectories approach (0,∞).

Note that the behavior is controlled by the ratio γ = r2b2K1

. When the carrying capacity

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for species 1 is large, we see that it may survive. However, if it is too low, it will invariablydie out.

6.5.7 (General relativity and planetary orbits) The relativistic equation for the orbit of aplanet around the sun is

d2u

dθ2+ u = α + εu2

where u = 1/r and r, θ are the polar coordinates of the planet in its plane of motion. Theparameter α is positive and can be found explicitly from classical Newtonian mechanics;the term εu2 is Einstein’s correction. Here ε is a very small positive parameter.

(a) Rewrite the equation as a system in the (u, v) phase plane, where v = du/dθ.

(b) Find all the equilibrium points of the system.

(c) Show that one of the equilibria is a center in the (u, v) phase plane, according to thelinearization. Is it a nonlinear center?

(d) Show that the equilibrium point found in (c) corresponds to a circular planetary orbit.

(a) Setting v = du/dθ allows us to write the system as

du

dθ= v,

dv

dθ= α− u+ εu2.

(b) At equilibrium, we must have v = 0 and

α− u+ εu2 = 0⇒ u =1±√

1− 4αε

2ε.

The two equilibria of this system are therefore u∗± =(

1±√1−4αε2ε

, 0)

.

(c) The linearization of this system is

A(u∗, v∗) =

(0 1

−1 + 2εu∗ 0

).

This matrix has τ = 0 and ∆ = 1 − 2εu∗. Note that for u∗− =(

1−√1−4αε2ε

, 0)

, we get

∆ =√

1− 4αε > 0 so this fixed point is a center according to the linearization. Notethat the system is conservative with

E(u, v) =1

2v2 +

1

2u2 − αu− ε

3u3

acting as the conserved quantity. We have

Eu = u− α− εu2, Ev = v, Euu = 1− 2εu, Evv = 1, Euv = 0

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so Eu(u∗−) = Ev(u

∗−) = 0 with Hessian

D(u∗−) = (EuuEvv − E2uv)|u∗

−=√

1− 4αε > 0.

Since Evv(u∗−) = 1 > 0 in addition, we infer that u∗− is a strict minimum of E so it is

in fact a nonlinear center. (The other equilibrium u∗+ is a saddle.)

(d) Using u = 1/r, we get

du

dθ=

d

dθ

(1

r

)= − 1

r2dr

dθ.

Since du/dθ = 0 at the equilibrium, we obtain dr/dθ = 0. As a result, the equilibriumcorresponds to a constant value of r (i.e., independent of θ) which gives a circular orbit.

6.5.6 (Bonus 1) (Epidemic model revisited) In Exercise 3.7.6, you analyzed the Kermack-McKendrick model of an epidemic by reducing it to a certain first-order system. In thisproblem you’ll see how much easier the analysis becomes in the phase plane. As before, letx(t) ≥ 0 denote the size of the healthy population and y(t) ≥ 0 denote the size of the sickpopulation. Then the model is

x = −kxy, y = kxy − ly

where k, l > 0. (The equation for z(t) the number of deaths, plays no role in the x, ydynamics so we omit it.)

(a) Find and classify all the fixed points.

(b) Sketch the nullclines and the vector field.

(c) Find a conserved quantity for the system. (Hint: Form a differential equation fordy/dx. Separate the variables and integrate both sides.)

(d) Plot the phase portrait. What happens as t→∞?

(e) Let (x0, y0) be the initial condition. An epidemic is said to occur if y(t) increasesinitially. Under what condition does an epidemic occur?

(a) Setting x = 0 gives −kxy = 0 ⇒ x = 0 or y = 0. Similarly, setting y = 0 yieldsy(kx − l) = 0 ⇒ x = l/k or y = 0. Observe that y = 0 is sufficient and necessary forthe flow to be zero; we conclude that all points of the form (a, 0), for any a ≥ 0, arefixed points of this system. This gives a line of fixed points (so these are non-isolated).

In order to identify the stability of these fixed points, we linearize to obtain

A(a,0) =

(0 −ka0 ka− l

).

The linearization predicts unstable non-isolated fixed points for a > l/k and stablenon-isolated fixed points for a < l/k. While we should take this conclusion with apinch of salt, the nullclines and phase portrait shown in Figure 8 validate this.

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(b) The vertical nullcline is x = 0, the horizontal nullcline is x = l/k while y = 0 is a setof fixed points. See Figure 8(a).

(c) We have

dy

dx=y

x=kxy − ly−kxy

= −1 +l

kx⇒ y = −x+

l

kln(x) + C

so we always have x+y− lk

ln(x) = C along any trajectory. We conclude that E(x, y) =x+ y − l

kln(x) is a conserved quantity for this system.

(a) Nullclines and vector field (b) The phase portrait

Figure 8: Problem 6.5.6 (with k = l = 1).

(d) See Figure 8(b) for the phase portrait. Observe that if y0 > 0, the system evolves toan infection-less state where there are no infecteds and a non-zero healthy population.If y0 = 0, there would be no change since the disease does not spread.

(e) We require y|(x0,y0) > 0. Since y|(x0,y0) = y0(kx0 − l), we need y0 > 0 and x0 > l/kfor an epidemic to occur. This translates into having an infection to begin with and alarge enough susceptible population. This is also borne out by the vector field shownin Figure 8(a): for x0 < l/k, the number of infecteds is on a downward trend from thestart so an epidemic cannot occur.

6.5.13 (Bonus 2) (Nonlinear centers)

(a) Show that the Duffing equation x + x + εx3 = 0 has a nonlinear center at the originfor all ε > 0.

(b) If ε < 0, show that all trajectories near the origin are closed. What about trajectoriesthat are far from the origin?

(a) Defining y = x allows us to write

x = y, y = −x− εx3.

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This system is conservative with the energy function

E(x, y) =1

2y2 +

1

2x2 +

ε

4x4.

Next, we find the maxima and minima of E. We have

∇E(x, y) = (Ex, Ey) = (x+ εx3, y).

Setting ∇E = 0 gives y = 0 and x = 0, x = ±√−1ε

(which is valid only for ε < 0).

For ε > 0, E has a single critical point at (0, 0). Note further that Exx = 1 + 3εx2,Exy = 0 and Eyy = 1 so

D(0, 0) = (ExxEyy − (Exy)2)|(0,0) = 1 > 0.

Along with Eyy > 0, this shows that E has a strict minimum at (0, 0) so all trajectoriesare closed curves around it. We conclude that there is a nonlinear center at the origin.

(b) For ε < 0, we still have a strict minimum at the origin. However, the additional critical

points(±√−1ε, 0)

come into play now. The Hessian at these points is

D

(±√−1

ε, 0

)= −2 < 0

so these are saddle points. We can deduce again that all trajectories close to the originare closed but the origin is no longer a global center. Since the trajectories lie on curvesof the form

y2 = E − x2 − ε

2x4

for different values of E, away from the origin the trajectories look like parabolasy ≈ ±

√− ε

2x2. See Figure 9 for an illustration of the trajectories near and far from

the origin.

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(a) Closed orbits near the origin; note the het-erclinic orbits connecting the saddle nodes at(±√− 1

ε , 0)

.

(b) Parabolas away from the origin

Figure 9: Problem 6.5.13 (with ε = −2).

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