Darboux integrating factors of planar polynomial vector fields: - - PowerPoint PPT Presentation

Darboux integrating factors of planar polynomial vector fields: Inverse problems

(A Related Topic)

Mostly joint work with C. Christopher (Plymouth),

• J. Llibre (Barcelona), C. Pantazi (Barcelona)

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The problems

Consider planar complex polynomial vector field X = P ∂ ∂x + Q ∂ ∂y and associated ODE ˙ x = P(x, y) ˙ y = Q(x, y) Problem 1. Find all invariant algebraic curves for this system. Problem 2. Decide whether Darboux integrating factor exists. (Find it if answer is affirmative.)

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Some definitions and details

Given irreducible pairwise relatively prime polynomials f1, . . . , fr and nonzero d1, . . . , dr ∈ C, one says that X admits the Darboux integrating factor f −d1

1

· · · f −dr

r

if div f −d1

1

· · · f −dr

r

X = 0. Necessary for existence: Invariance of all complex zero sets Ci of fi for X. Equivalent: There are polynomials L1, . . . , Lr such that Xfi = Li · fi, 1 ≤ i ≤ r. ( fi is then called a semi-invariant of X, with cofactor Li. One also says: “X admits fi”.) Given invariance, X admits the Darboux integrating factor above iff d1 · L1 + · · · + dr · Lr = div X.

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Why?

• Classical problem (Poincaré, Darboux).
• Relation to Lie symmetries of first-order equations.
• Prelle and Singer (1983): Does elementary first integral exist

for planar ODE? “Missing link” in algorithmic decision: Darboux integrating factor (with rational di).

• Connection to qualitative properties.

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Some local results

Given local analytic (or formal) vector field X with stationary point 0, linearization not nilpotent, thus w.l.o.g. P(x, y) = λx + · · · Q(x, y) = µy + · · · with µ = 0. “Dicritical case” (problematic): λ/µ positive rational number. Local problem: Search for analytic (or formal) g and L, g(0) = 0 such that X(g) = L · g. Note that g may be multiplied by any invertible series. Proposition. In non-dicritical case there are at most two different semi-invariants (up to invertible factors) for X at 0. If λ/µ is not a rational number then there exists - up to constant factors - one and only one local integrating factor. It has the form (x + · · · )−1 · (y + · · · )−1

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Some global results

Theorem. Let X be a polynomial vector field of degree m such that all stationary points at infinity have non-nilpotent linearization and no stationary point at infinity is dicritical. (a) Then all irreducible semi-invariants have degree ≤ m + 1. (b) If one stationary point at infinity admits non-rational eigenvalue ratio then a Darboux integrating factor is necessarily of the form h−1, with h a polynomial of degree m + 1. Philosophically speaking: “Generically, things are simple.”

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Some sharper global results

• For curves (in projective setting):

Camaco and Sad, Cerveau and Lins Neto, Carnicer (1980s and 1990s)

• Algorithmic approach for curves (in affine setting):

Coutinho and Menasché Schechter (2009)

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Inverse problems

Given: Irreducible pairwise relatively prime polynomials f1, . . . , fr, and f := f1 · · · fr Inverse problem for curves: Find (characterize) the polynomial vector fields X = P ∂

∂x + Q ∂ ∂y

that admit all fi. (Equivalently: f is semi-invariant.) Inverse problem for integrating factors: Given, furthermore, nonzero complex constants d1, . . . , dr, find (characterize) the polynomial vector fields X with Darboux integrating factor f −d1

1

· · · f −dr

r

.

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Why inverse problems?

• Legitimate questions (e.g., what derivations leave given ideal

invariant?)

• Necessary for characterization (classification), in particular for

integrating factor case.

• Better understanding of obstacles to elementary integrability.
• Allows construction of vector fields with special properties.

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The inverse problem for curves - Orientation

X(f ) = L · f ⇔ X(fi) = Li · fi, 1 ≤ i ≤ r Vector fields admitting f form linear space V. Some elements: (i) Hamiltonian vector field of f , defined by Xf = −fy ∂ ∂x + fx ∂ ∂y . (ii) More generally vector fields of type X = a · Xf + f · X (arbitrary polynomial a, polynomial vector field X) lie in V. These form a subspace called V0 (trivial vector fields admitting f ). (iii) Refinement: All vector fields of type X =

• i

ai f fi · Xfi + f · X admit f . These vector fields form a subspace V1.

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Some commutative algebra

P ∂f ∂x + Q ∂f ∂y = L · f By definition of ideal quotients: Necessary and sufficient condition L ∈ fx, fy : f (Quotient is related to singular points of curve.) Proposition. (a) A polynomial vector field X satisfies equation above with L ∈ fx, fy if and only if X ∈ V0. (b) The map sending vector field to cofactor induces an isomorphism of finite dimensional vector spaces V/V0 ∼ = (fx, fy : f ) / fx, fy .

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Important special case: Nondegenerate geometry

Proposition. Assume that the zero sets Ci of fi are smooth, that all pair intersections are transversal and there are no triple intersections. Then V = V1; in other words, every vector field admitting f has the form X =

• i

ai f fi · Xfi + f · X.

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Algorithmic approach

For simplicity’s sake: Consider only case with fx, fy relatively prime.

• Find Gröbner basis G of fx, fy with respect to fixed

monomial ordering. Only finitely many monomials m1, . . . , md are not multiples of some leading monomial in G. The classes mi + fx, fy form a basis of C[x, y]/fx, fy.

• Obtain cofactors from kernel of the map

Mf : g + fx, fy → f · g + fx, fy .

• Vector fields obtained, in principle, from defining equation and

Gröbner. (Nice shortcut: lift command in Singular.)

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An example

f = (y − x2)(y − x3) V1 has codimension one in V, only one more vector field Z = A∂/∂x + B∂/∂y needed. Computation with Singular: A = −9/40 x5 + 261/800 x4y − 1218891241/57600000 x3y2 − 27/40 x4 −580752454969/4976640000 x3y − 1218891241/57600000 x2y2 −180423092156761/429981696000 x3 − 580136595769/4976640000 x2y −261/400 xy2 + 1218891241/28800000 y3 −180444591241561/429981696000 x2 + 9/10 xy +582376083769/2488320000 y2 + 1/2x +180229600393561/214990848000 y and B of similar size. Additional work (by hand) yields less involved expression

• A

= 2x − 3x4 + y

• B

= 2x3 + 3x5 + 4y − 9x3y

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Sigma processes

Alternative approach via blow-ups (seen as morphisms of affine plane). Prototype: Φ : C2 → C2,

• x

y

• →
• x

xy

• .
• Lemma. Let 0 be a singular point for f and a polynomial vector

field X = (P, Q) on C2 be given. The following are equivalent: (i) The zero set of f is invariant for X with cofactor K. (ii) The vector field

• X = 1

x ·

• xP(x, xy)

−yP(x, xy) + Q(x, xy)

• is polynomial, and the zero set of ˆ

f := f (x, xy) is invariant for X with cofactor K(x, y) = K(x, xy). By finitely many sigma processes: Smooth irreducible curves with

• nly simple intersections (Bendixson–Seidenberg).

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Inverse problem for integrating factors - Orientation

Given f1, . . . , fr as above, f = f1 · · · fr, and nonzero complex constants d1, . . . , dr. Vector fields with Darboux integrating factor

• f d1

1 · · · f dr r

−1

form linear space F = F(d1, . . . , dr) (subspace of V). Auxiliary construction: Given an arbitrary polynomial g, define Zg = Z (d1,...,dr)

g

: Hamiltonian vector field of g/

• f d1−1

1

· · · f dr−1

r

• .

Then the polynomial vector field f d1

1 · · · f dr r

· Zg = f · Xg −

r

• i=1

(di − 1)g f fi · Xfi admits the integrating factor (f d1

1 · · · f dr r )−1. Vector fields of this

type form a subspace F0 = F0(d1, . . . , dr) of F. (Trivial vector fields admitting given integrating factor.)

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A nontrivial example

• Theorem. (a) For all constants αi and every vector field

X with divergence zero, the vector field X =

• i

αi f fi · Xfi + f · X admits the integrating factor f −1. (X is not trivial if some αi = 0.) (b) In nondegenerate geometry setting, every vector field admitting the integrating factor f −1 is of this type.

• Reminder. Nondegenerate geometry: All curves fi = 0 smooth,
• nly pair intersections occur, intersections are transversal.

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A reduction principle for nondegenerate geometry

Until further notice: Nondegenerate geometry is assumed.

• Lemma. Let e1, . . . , er ∈ C and k1, . . . , kr be nonnegative
• integers. Then

f k1

1 · · · f kr r

· F0(e1, . . . , er) ⊆ F0(e1 + k1, . . . , er + kr).

• Lemma. Let

X =

r

• i=1

ai f fi · Xfi + f · X ∈ V1 = V admit the given integrating factor, with dℓ = 1. Then X + 1 dℓ − 1 · f d1

1 · · · f dr r

· Zaℓ = fℓ · X ∗ and X ∗ ∈ F(d1, . . . , dℓ − 1, . . . , dr).

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Reduction at work

Starting point: X is congruent to some fℓ ·X ∗ mod F0(d1, . . . , dr). Note: Case closed if all di positive integers. First try: Reduce degree of vector field in every step. Does not work. Second try: Consider degree δy with respect to y, and start from assumption δy(fj) = δ(fj) for all j (no loss of generality).

• Proposition. There exist positive integers m1 and m2 (depending
• nly on f1, . . . , fr) such that “modulo F0” only vector fields

X = P ∂ ∂x + Q ∂ ∂y with δy(P) < m1 and δy(Q) < m2 remain to be investigated. (Moreover, one may assume all Re dj ≤ 1.)

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The reduced problem: Admitting invariant curves

• Proposition. Let m1 and m2 be positive integers, and consider the

C[x]-module of vector fields X = P ∂/∂x + Q ∂/∂y with δy(P) < m1 and δy(Q) < m2. Then there exist vector fields Yi = vi ∂/∂x + wi ∂/∂y, 1 ≤ i ≤ s, admitting f , with the following property: The vector field X admits f if and only if X = u1(x) · Y1 + · · · + us(x) · Ys with u1, . . . , us ∈ C[x]. Moreover the ui are uniquely determined.

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The reduced problem: Admitting integrating factors

• Theorem. Situation and notation as above. Given nonzero

d1, . . . , dr, there exist matrices V (x) and B(x) = Bd1,...,dr (x)

• ver C[x] with the following property:

X = u1(x) · Y1 + · · · + us(x) · Ys ∈ F(d1, . . . , dr) if and only if V (x) ·

  

u′

1

. . . u′

s

   = B(x) ·   

u1 . . . us

   .

The matrices V and B have s columns and at most max{m1, m2} − 1 rows. The entries of V do not depend on d1, . . . , dr, and V has maximal rank s.

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Finiteness, part one

• Theorem. In the nondegenerate geometry setting, one has

dim

• F(d1, . . . , dr)/F0(d1, . . . , dr)
• < ∞.

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Whence the linear differential equation?

Let Ki,j be the cofactor of fj with respect to Yi. For X as above, cofactor of fj: Lj =

• i

ui · Ki,j. Evaluation of the integrating factor condition div X = djLj:

• (ui · divYi + Yi(ui)) =
• (ui · divYi) +
• vi · u′

i =

• djLj,

hence

• i

vi · u′

i =

• i

ui

 

j

dj · Ki,j − div Yi

  .

Compare powers of y.

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Example: Irreducible cubic curves

Consider irreducible cubic polynomial f = y3 + p1(x) · y + p0(x) with smooth zero set C.

• Lemma. A vector field Y = b1 ∂/∂x + b2 ∂/∂y with δy(b1) ≤ 1

and δy(b2) ≤ 2 admits f if and only if Y = u1(x) · Y1 + u2(x) · Y2 with polynomials u1, u2 in one variable and Y1 :=

• 3p0 + 2p1y

p′

0y + p′ 1y2

• , Y2 :=
• −2p2

1 + 9p0y

2p′

0p1 − 3p0p′ 1 − p1p′ 1y + 3p′ 0y2

• .

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Example: Irreducible cubic curves (continued)

• Proposition. Assume that d is not a positive integer. The vector

field Y above admits the integrating factor f −d if and only if the polynomials u1 and u2 satisfy

• u′

1

u′

2

• = 1

∆ · B(x)

• u1

u2

• with the discriminant ∆ = 27p2

0 + 4p3 1 of f , and

B(x) =

• (3d − 4) ·

9p0p′

0 + 2p2 1p′ 1

• (3d − 5) · (−3p1) · (3p0p′

1 − 2p′ 0p1)

(3d − 4) · (3p0p′

1 − 2p′ 0p1)

(3d − 5) ·

9p0p′

0 + 2p2 1p′ 1

• .

Moreover (by Wronskian) dim F(d)/F0(d) ≤ 1.

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Example: Irreducible cubic curves (continued further)

Many classes of curves for which “F = F0”. For instance:

• Proposition. If δ(p0) ≥ 2δ(p1) then the linear meromorphic

system admits a first order pole at infinity, with the coefficient matrix of the lowest order term equal to A∞ =

• −3d−4

3

δ(p0) ∗ −3d−5

3

δ(p0)

• .

This implies F(d) = F0(d) whenever d is not a positive integer. Idea of proof: Assume there is a polynomial solution of meromorphic linear system. Then the negative of its degree is an eigenvalue of A∞.

• Note. Question F(d) = F0(d) still open in general.

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Example: Graphs

Consider fi := y − pi(x); p1, . . . , pr polynomials in one variable. Convenient abbreviation: θi := di − 1; one may assume Re (θi) ≤ 0. Theorem. X ∗ :=

• i

ai(x) f fi · Xfi ∈ F(d1, . . . , dr) if and only if the polynomials ai satisfy the meromorphic linear system a′

i =

• j:j=i

(θjai − θiaj) · (pj − pi)′ pj − pi . The nonzero constant solution a1 = θ1, . . . , ar = θr yields a scalar multiple of f d1

1 · · · f dr r Z1. In particular

dim

• F/F0

≤ r − 1.

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Example: Graphs (continued)

Look at constant solutions. Assumption (w.l.o.g.): r > 1 and Re(θi) ≤ 0 for all i, but not all θi = 0. Define equivalence relation ∼ by i ∼ j if and only if pj − pi is constant.

• Proposition. Then there exist nontrivial constant solutions if and
• nly if there is an equivalence class I such that |I| > 1 and θj = 0

for all j ∈ I. In this case, the constant solutions are given by (α1, . . . , αr) with αj = 0 for all j ∈ I and αi arbitrary for i ∈ I. These correspond to the vector fields

• i∈I

αi f fi · Xfi ∈ F(d1, . . . , dr), and one has dim

• F(d1, . . . , dr)/F0(d1, . . . , dr)
• = |I| − 1.
• Note. Approach yields vector fields which are not integrable in

elementary manner.

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Example: Graphs (continued further)

• Theorem. Let d1, . . . , dr be arbitrary nonzero real constants, not

all of them positive integers. If every equivalence class contains just one element, then F(d1, . . . , dr) = F0(d1, . . . , dr). Main ingredient of proof: Let mij := δ(pi − pj) for i = j. Coefficient matrix of lowest order term for the system at infinity (first order pole): A∞ =

       

−

• j=1 m1jθj
• m12θ1

· · · m1rθ1 m12θ2 −

• j=2 m2jθj
• · · ·

m2rθ2 . . . . . . . . . m1rθr m2rθr · · · −

• j=r mrjθj
• 

      

. This is an M-matrix if all θi ≤ 0.

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Finiteness, part two

Drop nondegenerate geometry assumption now. Crucial result survives:

• Theorem. One always has

dim

• F(d1, . . . , dr)/F0(d1, . . . , dr)
• < ∞.

Proof via sigma processes. Problem: Control behavior of trivial vector fields.

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Some References

• C. Camacho, P. Sad: Invariant varieties through singularities of

holomorphic vector fields. Ann. Math. 15 (1982), 579–595.

• M. Carnicer: The Poincaré problem in the nondicritical case. Ann.
• Math. 140 (1994), 289–294.
• D. Cerveau, A. Lins Neto: Holomorphic foliations in CP(2) having

an invariant algebraic curve. Ann. Inst. Fourier 41 (1991), 883–903.

• C. Christopher, J. Llibre, C. Pantazi, S. Walcher: Inverse problems

for multiple invariant curves. Proc. Roy. Soc. Edinburgh 137A, 1197 - 1226 (2007).

• C. Christopher, J. Llibre, C. Pantazi, S. Walcher: Inverse problems

for invariant algebraic curves: Explicit computations. Proc. Roy.

• Soc. Edinburgh 139A, 1 - 16 (2009).
• C. Christopher, J. Llibre, C. Pantazi, S. Walcher: Darboux

integrating factors: Inverse problems. J. Differential Eqs. (to appear).

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• C. Christopher, J. Llibre, C. Pantazi, X. Zhang: Darboux

integrability and invariant algebraic curves for planar polynomial

• systems. J. Phys. A 35, 2457–2476 (2002).

S.C. Coutinho, L. Menasché Schechter: Algebraic solutions of plane vector fields. J. Pure Appl. Algebra 213, 144 - 153 (2009).

• J. Llibre, C. Pantazi, S. Walcher: Morphisms and inverse problems

for Darboux integrating factors. Preprint, 15 pp. (2008).

• M. J. Prelle and M. F. Singer: Elementary first integrals of

differential equations. Trans. Amer. Math. Soc. 279 (1983), 613–636. M.F. Singer: Liouvillian first integrals of differential equations.

• Trans. Amer. Math. Soc. 333 (1992), 673–688.
• S. Walcher: Plane polynomial vector fields with prescribed invariant
• curves. Proc. Royal Soc. Edinburgh 130A (2000), 633 - 649.
• S. Walcher: On the Poincaré problem. J. Differential Eqs. 166

(2000), 51–78.

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