Abel Equations : Integrability and analytic properties of theirsolutions

Advances in Qualitative Theory of Differential Equations

Castro Urdiales, September 12-16, 2011

J.-P. Francoise

Universite Pierre-et-Marie Curie, Paris6,

Laboratoire J.-L. Lions, UMR 7598, CNRS

1

Hilbert’s 16th problem and Poincare’s center-focus problem for Abelequations

Poincare center–focus problem asks for finding homogeneous polynomialsf(x, y) and g(x, y) of degree d such that

x = y + f(x, y),y = −x+ g(x, y),

(1)

displays a neighborhood of the origin filled with periodic orbits. In such acase, the origin is said to be a center. Writing the system in polar coordinatesyields

drdt = rdA(θ)

dθdt = 1 + rd−1B(θ),

(2)

2

where A and B are homogeneous trigonometric polynomials. The associatedfoliation if defined by

dr

dθ=

rdA(θ)

1 + rd−1B(θ). (3)

Cherkas proposed to transform equation (3) by setting

ρ =rd−1

1 + rd−1B(θ),

and he obtained the following (now called trigonometric Abel equation) :

dρ

dθ= q(θ)ρ3 + p(θ)ρ2, (4)

where q(θ) = −(d− 1)f(θ)g(θ) and p(θ) = (d− 1)f(θ)− g′(θ).

3

These findings motivated to go over the subject polynomial Abel equa-tions (cf. [M. Briskin, Y. Yomdin, JPF, 1998])

dy

dx= q(x)y3 + p(x)y2, (5)

and consider the following problems. Given two points x0 = 0 and x1 = 1,(i) characterize the polynomials q, p ∈ R[x] such that for all initial

data (0, y0) the solution y(x, y0) of the equation satisfies y(0, y0) =y(1, y0) = y0 ; or

(ii) count and locate the number of isolated solutions (called limitcycles) of equation (5) such that y(0, y0) = y(1, y0) = y0.

In view of what precedes it is natural to call (i) the center problem for poly-nomial Abel equations, and (ii) the Hilbert’s 16th problem (sometimes alsocalled in this context the Pugh’s problem) for polynomial Abel equations.

4

In fact this is much more than a mere analogy as it can be understoodgoing to the complex. Consider indeed a complex Abel equation (now (x, y)are complex variables)

dy

dx= q(x)y3 + p(x)y2, (6)

where (q, p) ∈ C(x) are rational functions. Fix two points x0 and x1 in Cand a path γ connecting them. Assume that the path γ avoids the movablesingularities for an initial value y0 ∈ D(0, r), the disc centered at the originof radius r. Then equation defines a Poincare map Pγ : y0 7→ y1 using thesolution y(x, y0) such that y(0, y0) = y0 and y(1, y0) = y1. In this generalsetting we are interested in finding centers (equations for which Pγ = Id)

5

or for finding isolated fixed points of Pγ (the equivalent to limit cycles forthe equation).

In some sense, we can now forget about the initial motivation and studyAbel equations in their own with the hope that getting more informationsabout Abel equations we could still return on the initial questions. Is itrealistic ? What do we know already on these equations. Why are theycalled Abel equations ?

Abel went over these equations because in “some sense” they were themost natural extensions of Ricatti equations (which are of second orderin y). He asked himself what are the equations of this type which areintegrable ? There was a nice modern report on these classical contributionsand Abel’s followers (Liouville, Kamke, ...) in [Cheb-Terrab, Roche, 2000,2003, 2004, Computer Physics Comm., European Journal of Applied Maths]

6

Let K be a differential field, denotes Ω1K the K-module of the 1-forms

with coefficients in K.

A Godbillon-Vey sequence on K is a sequence of 1-forms ω =ω0, ω1, ω2, ..., ωn, ... such that :

dω = ω ∧ ω1,dω1 = ω ∧ ω2,dωn = ω ∧ ωn+1 + Σnk=1C

nkωk ∧ ωn−k+1,

(7)

7

An element H of a differential extension of K is said to be a first integralof ω iff

dH ∧ ω = 0.

A differential extension is said to be of Liouville type if it is obtained as asequence of extensions of type K ⊂ K(y) with y either algebraic over K ortranscendental such that dy = yγ1 + γ0, with γ0,1 ∈ Ω1

M . Michael Singerproved : (this formulation is due to J.-P. Rollin and F. Thouzet).

Theorem 1. A 1-form ω ∈ Ω1M displays a GV sequence of length at most

2 iff it admits a Liouville first integral.

8

A differential extension is said to be of Ricatti type if it is obtainedas successive extensions of type K ⊂ K(y)with y either algebraic ortranscendent so that

dy = y2γ2 + yγ1 + γ0,

with γ0,1,2 ∈ Ω1M . Guy Casale (2007) proved :

Theorem 2. A 1-form ω ∈ Ω1M displays a GV sequence of length at most

3 iff it displays a first integral of Ricatti type.

9

It is thus natural to extend with A differential extension is said to be ofAbel type if it is obtained as successive extensions of type K ⊂ K(y) withy either algebraic or transcendent so that

dy = y3γ3 + y2γ2 + yγ1 + γ0,

with γ0,1,2 ∈ Ω1M .

Theorem 3. A 1-form ω ∈ Ω1M displays a GV sequence of length at most

4 iff it displays a first integral of Abel type.

10

Are there examples of non-integrable Abel equations ? More precisely arethere examples of Abel equations non integrable by Liouville or Ricatti ?

Integrability of Abel equations is known to be closely related to integra-bility of Lienard equations.

dy

dx= q(x)y3 + p(x)y2, (8)

Replace y by 1/y, get

− dy

y2dx= q(x)y−3 + p(x)y−2, (9)

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which is the foliation associated to the vector fields :

dydt = −q(x)− p(x)ydxdt = y.

(10)

Consider for instance the case q(x) = x :

dydt = −x− p(x)ydxdt = y,

(11)

Lienard transform y = Y − P (x) yields

dxdt = Y − P (x)dYdt = −x. (12)

Surprisingly this type of transformation may have some variants.

12

Consider :

x = y − P (x2)y = −x. (13)

This is a Lienard equation whose orbits near the origin are all periodic(because symmetric if changing x into −x). By Poincare’s, this equationis locally integrable near the origin. In general, it is an open question tofind an explicit first integral. Right now, let us say that this equation isintegrable. Then changing y into Y = y − P (x2), still yields an integrableODE :

x = Y

Y = −x− 2xp(x2)Y.(14)

But then the 1-dimensional equation :

13

dY

xdx=−1− 2p(x2)Y

Y, (15)

is integrable. The same is true if we change x into X = x2

2 . This yieldsthat the following equation is integrable :

dY

dX= − 1

Y− 2p(2X), (16)

as well as the one obtained by changing Y into −1/Y , namely :

dY

dX= Y 3 + 2p(2X)Y 2. (17)

This tells nothing else than

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Theorem 4. Any Abel equation with constant leading term :

dY

dX= Y 3 + p(X)Y 2 (18)

is integrable on the set X > 0 or x < 0 provided that p is a differentiablefunction.

15

Any Abel equation

dy

dx= q(x)y3 + p(x)y2, (19)

writesdy

q(x)dx= y3 +

p(x)

q(x)y2, (20)

which after an ”algebraic change of variable” x 7→ Q(x) writes

dy

dx= y3 +

p

q(Q−1(x))y2. (21)

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Conjecture : An Abel equation is integrable iff pq(Q

−1(x)) is continuous andthe two sector first integrals matches. Note that this reduced form of Abelequation appeared already in Abel. This expression also appear in manyarticles related to the analytic extension of the generating function of themoments.

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We further discuss two examples.

Case without fixed singularities

Considerdy

dx= y3 + λxy2.

We can fix arbitrarily λ, a convenient choice is λ = −2 This example isintegrable. Consider the equation :

dy

dx= y3 − 2xy2,

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and introduce the new variable

t = −1

y+ x2.

We obtain the Ricatti equation :

dx

dt= x2 − t.

It is then classical to associate to this Ricatti equation a second order

differential equation. Change x into −u′

u and we obtain

u′′ = xu.

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Hence the function u is a linear combination of the Airy functions :

u = αAi(t) + βBi(t).

Thus back to initial x, we get

x = −αAi′(t) + βBi′(t)

αAi(t) + βBi(t), (14)

which depends, of course, of only one arbitrary constant c = α/β. Weobtain in this way a first integral of the form :

c =xBi(t) + Bi′(t)

xAi(t) + Ai′(t)=xBi(−1

y + x2) + Bi′(−1y + x2)

xAi(−1y + x2) + Ai′(−1

y + x2).

By a theorem of [Gasull-Llibre, 1990] any polynomial Abel equation so thatq(x) never vanishes has at most three real limit cycles (the same property

20

was proved by N. Lloyd in 1979 for trigonometric Abel equations). Thisexample altough illustrate that the return map maybe still quite complicatedbecause if we write the conservation equation

x0Bi(− 1y0

+ x20) + Bi′(− 1y0

+ x20)

x0Ai(− 1y0

+ x20) + Ai′(− 1y0

+ x20)=x1Bi(− 1

y1+ x21) + Bi′(− 1

y1+ x21)

x1Ai(− 1y1

+ x21) + Ai′(− 1y1

+ x21)

, we obtain that for x0 = 1, x1 = 2 (for instance) there is infinitely manyreal solutions.

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Case of a single fixed singularity

This case has been investigated by N. Roytvarf, JPF, Y. Yomdin (JEMS2008).

It writes :dy

dx= cxy3 + y2.

The change of unknown function

y =v

x,

yields

− v

x2+v′

x=cv3

x2+v2

x2,

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and thus :

v′ =1

x[cv3 + v2 + v],

which obviously separates.

Write :1

cv3 + v2 + v=α

v+

β

v − v1+

γ

v − v2,

with :

v1 =−1 +

√1− 4c

2c, v2 =

−1−√

1− 4c

2c,

and

α = 1, β =1

cv1(v1 − v2), γ =

1

cv2(v2 − v1).

Integrating, we get for each its solution v(x)

v(v − v1)β(v − v2)γ = K · x,

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for a certain constant K. Equivalently

y(xy − v1)β(xy − v2)γ = K,

or

y(1− xyv1

)β(1− xyv2

)γ =K

vβ1 vγ2

= K ′. (M)

Notice that the only “fixed singularity” of the equation is x = 0. To startwith, let us take this point x = 0 as the initial point a. Now, the constantK ′ is evaluated by setting x = 0 and y = y0 and this yields K ′ = y0

Therefore, the solution y(y0, x) satisfying y(y0, 0) = y0 is given by

y(x)(1− xy(x)

v1)β(1− xy(x)

v2)γ = y0.

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Substituting the point x = b, we get the relation between yb = y(b) andy0 in the form :

y0 = yb(1−bybv1

)β(1− bybv2

)γ. (LCM)

These equations produce multivalued solutions.

We are interested in the “limit cycles” of the equation (M), i.e. inits solutions y(x) satisfying y(0) = y(b). This relation gives the followingequation for the limit cycles :

(1− by0v1

)β(1− by0v2

)γ = 1.

The accurate immediate interpretation of the equation (LCM) is that

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the algebraic curve Y = Yy0, defined by (LCM), passes through the points(0, y0) and (b, y0).

Certainly, this curve Y , parametrized as y = y(x), satisfies differentialequation (M). But a priori we do not even know whether Y is connected.So (LCM) by itself does not exclude a possibility that the points (0, y0) and(b, y0) belong to different leaves of the solutions of the differential equation(M). So we have to clarify the geometric interpretation of the “limit cycles”,specifying the continuation paths from a to b for the solutions y(x) satisfyingy(0) = y(b).

Below we choose some specific values for the free parameter c in themodel equation (M). In particular, in all the considered cases we showthat in fact for y0 6= 0 the curve Y = Yy0 is connected. This allows usto give the following interpretation to the equation (LCM) : for each y0satisfying (LCM) there exists a path s from 0 to b such that the solu-tion y(y0, x) can be analytically continued along s, and this continuation

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satisfies y(y0, b) = y0.

On this base we produce an example of (M) with an infinite (or finite,but arbitrarily large) number of complex “limit cycles”.

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The case c = 14

We consider the case c = 14 and the equation

dy

dx=

1

4xy3 + y2.

With y = v/x, this equation yields :

dv

dx=

1

4x(v3 + 4v2 + 4v),

which separates and gives the solution y(x) corresponding to the initial datay0 as the solution to :

y

xy + 2e

2xy+2 =

e

2y0.

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Periodic orbits correspond to solutions of y(1) = y(0) = y to

2

y + 2e

2y+2 = e.

If we change y = 2ξ1−ξ, this yields

1− ξ = eξ.

We write ξ = x+ iy, and derive the two equations

1− x = excosy,

−y = exsiny.

Note that if (x, y) is a solution, then (x,−y) is also a solution. Then we canassume y > 0. Second equation implies sin(y) < 0 and we restrict ourselves

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now to y ∈](2n+ 1)π, (2n+ 2)π[. Now we plug x = −log(−sinyy ) into the

first equation. This displays :

F (y) = 1 + log(−siny

y) +

y

tany= 0.

Then we note that as y → (2n + 1)π, F (y) → +∞ and that as y →(2n+ 2)π, F (y)→ −∞. There is thus at least one solution (and in fact asingle one) in the interval. The Abel equation has thus infinitely many limitcycles : For c = 1

4 equation (M) has infinitely many different “limit cycles”,i.e. local solutions yj(x) at the origin, j = 1, . . . , n, and paths sj from 0to 1, such that each yj(x) being analytically continued along sj satisfiesy(0) = y(b). All these “limit cycles” are situated on different leaves of thePoincare mapping for (M). We add some comments on the geometry of the

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Riemann surface defined by the equation :

y

xy + 2e

2xy+2 =

e

2y0.

We proceed with coordinate Y defined by :

xy = 2(1/Y − 1),

and fix y0 = 1/e, to simplify. The equation of the Riemann surface becomes :

(1− Y )eY = x.

The different sheets of the Riemann surface, seen as a cover of the x-plane,can be explicitly parametrized. Write x = α+ iβ, and Y = u+ iv, then

u = −log(− v

αsinv + βcosv),

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and the sheets are parametrized by the different solutions v to the equation :

1 + log(− v

αsinv + βcosv) + v

αcosv − βsinv

αsinv + βcosv= 0

A more detailed analysis of this example can be done using properties ofthe so-called Lambert function (see for instance )

R. Corless, G. Gonnet, D. Hare, D. Jeffrey, D. Knuth, On the LambertW function, Advances in Computational Mathematics 5 (1996), 329-359.

Therefore, in these examples we see that the equation (M) may haveas many complex limit cycles as we wish, when the parameter c varies,although the degree of the coefficients of this equation remains bounded.This phenomenon reminds (in much simpler setting) the counterexample dueto Yu. Iliashenko of Petrowski-Landis claim. Note that Khovanski fewnomialstheory, or, rather, “additive complexity” arguments imply that the number

32

of real roots of equation (LCM) remains bounded independently of n. Sothat this example does not provide a counterexample to real Hilbert-Pughproblem.

Moreover, in all the cases considered we show that the above limitcycles “sit on different leaves” of the Poincare mapping. In other words,although the equality y(0) = y(b) is satisfied for a large number of theinitial values y0, it is realized on more and more complicated continuationpaths from a to b.

Accordingly, we may ask to what extent this property remains valid forthe Poincare mapping of a general polynomial Abel equation.

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Proof that the number of real limit cycles of the first equation isless than 3

Consider the solution y(x, y0) of the Abel equation

dy

dx= S(x, y),

at time x with initial value y(0, y0) = y0. We obtain

∂

∂y0

(dy(x, y0)

dx

)= S′y(x, y(x, y0))

dy(x, y0)

dx.

34

Since

dy(0, y0)

dx= Id,

we have

dy(x, y0)

dy0= exp

∫ x

0

S′y(u, y(u, y0)) du.

So that if we denote y0 7→ L(y0) the Poincare mapping which associates tothe initial data y0 the point y(1, y0), we obtain

L′(y0) = exp

∫ 1

0

S′y(u, y(u, y0))du.

35

This expression easily yields

L′′(y0) =

(∫ 1

0

S′′y (u, y(u, y0))

[exp

∫ u

0

S′y(v, y(v, y0))dv

]du

)·

exp

∫ 1

0

S′y(u, y(u, y0))du.

To lighten the notations we write

E(u, y0) = exp

∫ u

0

S′y(v, y(v, y0))dv,

and

D(u, y0) = E(u, y0)S′′y (u, y(u, y0)).

36

With these notations we can write

L′(y0) = E(1, y0), L′′(y0) = E(1, y0)

∫ 1

0

D(u, y0)du.

The third derivative is

L′′′(y0) = E(1, y0)

[(

∫ 1

0

D(u, y0)du)2 +

∫ 1

0

D(u, y0)

∫ u

0

D(v, y0)dv+∫ 1

0

E(u, y0)2S′′′y (u, y(u, y0))du

].

The integration by part of the term in between yields finally to

L′′′(y0) = E(1, y0)

[3

2(

∫ 1

0

D(u, y0)du)2 +

∫ 1

0

E(u, y0)2S′′′y (u, y(u, y0))du

].

37

Applying this third derivative to our equation, we have S′′′y = 6. So thethird derivative never vanishes. Then the Rolle’s lemma shows that thereare at most three limit cycles (i.e. solutions of L(y0) = y0).

Airy functions are easy to handle, for instance using Mathematica. Thismakes quite possible to study the equation of the limit cycles of equation(7). Limit cycles are solutions of the equation (7) such that y1 = y0 with

Bi(1− 1/y1) + Bi′(1− 1/y1)

Ai(1− 1/y1) + Ai′(1− 1/y1)=

Bi′(−1/y0)

Ai′(−1/y0).

Hence these solutions are contained in the set of zeros of the function

Bi(1− 1/y) + Bi′(1− 1/y)

Ai(1− 1/y) + Ai′(1− 1/y)− Bi′(−1/y)

Ai′(−1/y).

Working with Mathematica we find 3 limit cycles, which correspond to the

38

following 3 zerosy1 = 0.0238374105198454,y2 = 0.0814786531950691,y3 = 0.8551647091959168.

of the previous function.

39

The two functions E(u, y0) and D(u, y0) allows to compute all thederivatives of the return map. One can observe that they are solutions of aPDE system.

∫ u

0

D(v, y0)dv =∂E

∂y0(u, y0)/E(u, y0), (22)

then∂

∂u[∂E

∂y0(u, y0)/E(u, y0)] = D(u, y0), (23)

and then∂2

∂y0∂u[∂E∂y0(u, y0)/E(u, y0)] =

∂∂u[∂E∂y0(u, y0)/E(u, y0)]

∂E∂y0

(u, y0)/E(u, y0)

+6q(u)E(u, y0)2.

(24)

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Alternatively, this PDE can be displaid as a partial differential system :

∂E∂y0

= E.D∂D∂u = F

∂F∂y0

= F.D + 6q(u)E2.(25)

Note that this PDE system is with constant coefficients in case q(u) isconstant.

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C’est fini...Merci de votre attention

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