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) Beijing; Sebastian Walcher Folie 1

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.) Beijing; Sebastian Walcher Folie 2

Some definitions and details Given irreducible pairwise relatively prime polynomials f 1 , . . . , f r and nonzero d 1 , . . . , d r ∈ C , one says that X admits the Darboux integrating factor f − d 1 · · · f − d r if 1 r div f − d 1 · · · f − d r X = 0 . 1 r Necessary for existence: Invariance of all complex zero sets C i of f i for X . Equivalent: There are polynomials L 1 , . . . , L r such that Xf i = L i · f i , 1 ≤ i ≤ r . ( f i is then called a semi-invariant of X , with cofactor L i . One also says: “ X admits f i ”.) Given invariance, X admits the Darboux integrating factor above iff d 1 · L 1 + · · · + d r · L r = div X . Beijing; Sebastian Walcher Folie 3

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 d i ). • Connection to qualitative properties. Beijing; Sebastian Walcher Folie 4

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 ) = λ x + · · · Q ( x , y ) = µ y + · · · 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 Beijing; Sebastian Walcher Folie 5

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.” Beijing; Sebastian Walcher Folie 6

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) Beijing; Sebastian Walcher Folie 7

Inverse problems Given: Irreducible pairwise relatively prime polynomials f 1 , . . . , f r , and f := f 1 · · · f r Inverse problem for curves: Find (characterize) the polynomial vector fields X = P ∂ ∂ x + Q ∂ ∂ y that admit all f i . (Equivalently: f is semi-invariant.) Inverse problem for integrating factors: Given, furthermore, nonzero complex constants d 1 , . . . , d r , find (characterize) the polynomial vector fields X with Darboux integrating factor f − d 1 · · · f − d r . 1 r Beijing; Sebastian Walcher Folie 8

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. Beijing; Sebastian Walcher Folie 9

The inverse problem for curves - Orientation X ( f ) = L · f ⇔ X ( f i ) = L i · f i , 1 ≤ i ≤ r Vector fields admitting f form linear space V . Some elements: (i) Hamiltonian vector field of f , defined by ∂ ∂ X f = − f y ∂ x + f x ∂ y . (ii) More generally vector fields of type X = a · X f + f · � X (arbitrary polynomial a , polynomial vector field � X ) lie in V . These form a subspace called V 0 ( trivial vector fields admitting f ). (iii) Refinement: All vector fields of type � f · X f i + f · � X = a i X f i i admit f . These vector fields form a subspace V 1 . Beijing; Sebastian Walcher Folie 10

Some commutative algebra P ∂ f ∂ x + Q ∂ f ∂ y = L · f By definition of ideal quotients: Necessary and sufficient condition L ∈ � f x , f y � : � f � (Quotient is related to singular points of curve.) Proposition. (a) A polynomial vector field X satisfies equation above with L ∈ � f x , f y � if and only if X ∈ V 0 . (b) The map sending vector field to cofactor induces an isomorphism of finite dimensional vector spaces V / V 0 ∼ = ( � f x , f y � : � f � ) / � f x , f y � . Beijing; Sebastian Walcher Folie 11

Important special case: Nondegenerate geometry Proposition. Assume that the zero sets C i of f i are smooth, that all pair intersections are transversal and there are no triple intersections. Then V = V 1 ; in other words, every vector field admitting f has the form � f · X f i + f · � X = a i X . f i i Beijing; Sebastian Walcher Folie 12

Algorithmic approach For simplicity’s sake: Consider only case with f x , f y relatively prime. • Find Gröbner basis G of � f x , f y � with respect to fixed monomial ordering. Only finitely many monomials m 1 , . . . , m d are not multiples of some leading monomial in G . The classes m i + � f x , f y � form a basis of C [ x , y ] / � f x , f y � . • Obtain cofactors from kernel of the map M f : g + � f x , f y � �→ f · g + � f x , f y � . • Vector fields obtained, in principle, from defining equation and Gröbner. (Nice shortcut: lift command in Singular .) Beijing; Sebastian Walcher Folie 13

An example f = ( y − x 2 )( y − x 3 ) V 1 has codimension one in V , only one more vector field Z = A ∂/∂ x + B ∂/∂ y needed. Computation with Singular : A = − 9 / 40 x 5 + 261 / 800 x 4 y − 1218891241 / 57600000 x 3 y 2 − 27 / 40 x 4 − 580752454969 / 4976640000 x 3 y − 1218891241 / 57600000 x 2 y 2 − 180423092156761 / 429981696000 x 3 − 580136595769 / 4976640000 x 2 y − 261 / 400 xy 2 + 1218891241 / 28800000 y 3 − 180444591241561 / 429981696000 x 2 + 9 / 10 xy + 582376083769 / 2488320000 y 2 + 1 / 2 x + 180229600393561 / 214990848000 y and B of similar size. Additional work (by hand) yields less involved expression 2 x − 3 x 4 + y � A = 2 x 3 + 3 x 5 + 4 y − 9 x 3 y � B = Beijing; Sebastian Walcher Folie 14

Sigma processes Alternative approach via blow-ups (seen as morphisms of affine plane). Prototype: � � � � x x Φ : C 2 → C 2 , �→ . y xy Lemma. Let 0 be a singular point for f and a polynomial vector field X = ( P , Q ) on C 2 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 xP ( x , xy ) � x · − 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 only simple intersections (Bendixson–Seidenberg). Beijing; Sebastian Walcher Folie 15

Inverse problem for integrating factors - Orientation Given f 1 , . . . , f r as above, f = f 1 · · · f r , and nonzero complex constants d 1 , . . . , d r . Vector fields with Darboux integrating factor � � − 1 f d 1 1 · · · f d r r form linear space F = F ( d 1 , . . . , d r ) (subspace of V ). Auxiliary construction: Given an arbitrary polynomial g , define � � Z g = Z ( d 1 ,..., d r ) f d 1 − 1 · · · f d r − 1 : Hamiltonian vector field of g / . g r 1 Then the polynomial vector field r � ( d i − 1 ) g f f d 1 1 · · · f d r · Z g = f · X g − · X f i r f i i = 1 admits the integrating factor ( f d 1 1 · · · f d r r ) − 1 . Vector fields of this type form a subspace F 0 = F 0 ( d 1 , . . . , d r ) of F . ( Trivial vector fields admitting given integrating factor .) Beijing; Sebastian Walcher Folie 16

A nontrivial example Theorem. (a) For all constants α i and every vector field � X with divergence zero, the vector field � f · X f i + f · � X = α i X f i i admits the integrating factor f − 1 . (X is not trivial if some α i � = 0 .) (b) In nondegenerate geometry setting, every vector field admitting the integrating factor f − 1 is of this type. Reminder. Nondegenerate geometry: All curves f i = 0 smooth, only pair intersections occur, intersections are transversal. Beijing; Sebastian Walcher Folie 17