WITH GIVEN

. We determine all the C 1 planar vector ﬁelds with a given set of orbits of the form y − y ( x ) = 0 satisfying convenient assumptions. The case when these orbits are branches of an algebraic curve is also study. We show that if a quadratic vector ﬁeld admits a unique irreducible invariant algebraic curve g ( x, y ) = S X j =0 a j ( x ) y S − j = 0 with S branches with respect to the variable y , then the degree of the polynomial g is at most 4 S .


Introduction and statement of the main results
By definition an autonomous complex planar differential system is a system of the form where the dependent variables x = (x, y) are complex, and the independent variable (time t) is real. We assume that the vector field X = (P, Q) associated to the differential system (1) is C 1 in an open subset D of C 2 .
Let g = g(x) be a C 1 function. The curveg = 0 is an invariant curve of the vector field X if (2) X g| g=0 = 0, i.e. the curve g = 0 is formed by orbits of X . The vector field X is called a polynomial vector field of degree n if P and Q are polynomials such that the maximum of the degrees of P and Q is n. Let g be a complex polynomial in the variables x and y irreducible in the ring of polynomials C[x, y]. Suppose that g satisfies (2), then we say that g = 0 is an invariant algebraic curve of X . By the Hilbert's Nullstellensatz Theorem [6] if the polynomial X (g) vanishes when g = 0, there exist a non-negative integer m and a polynomial M = M (x, y) such that (X (g)) m = M g. Since g is irreducible, then there exist a polynomial K = K(x, y) such that X (g) = Kg, the polynomial K is called the cofactor of g = 0, and clearly the degree of K is at most n − 1.
We shall present briefly the contents of the paper.
In section 2 we prove our first main result (see Theorem 1 below) and determine a planar vector field from a given set of invariant orbits of the form y − y(x) = 0 where y(x) is an arbitrary C 1 function, and consequently the vector fields having such orbits are in general C 1 vector fields.
In section 3 we apply Theorem 1 to the set of orbits which are branches of an algebraic curve g(x, y) = 0 (see Proposition 10).
In section 4 we prove our second main result for the vector fields of degree two (see Theorem 3). More precisely for such vector fields having a unique irreducible invariant algebraic curve g = 0 we bound the degree of g by four times the number of its branches with respect to the variable x or y. The result is written with respect to the variable y.
In section 5 we determine the C 1 planar vector field with only one invariant curve of the form a 0 (x)y + a 1 (x) = 0 (see Proposition 14).
In section 6 as application of our previous results given in section 5 we determine the polynomial planar vector fields of degree 2, 3 or 4, with only one invariant algebraic curve of the type g = f (x)y + f (x) = 0 or g = f (x)y + f (x) = 0 where f is an orthogonal polynomial (see Proposition 17).
Finally in section 7 we analyze 16th Hilbert problem for limit cycles on a singular invariant algebraic curve, i.e. an invariant algebraic curve g = 0 having in the complex projective plane points such that the curve and its first derivatives are zero.
Our first main result is the following.
be a given set of orbits not formed by singular points of a complex planar differential system (S), where y j = y j (x) is a C 1 function for j = 1, . . . , S such that (y m − y j ) = 0, and there are at least two functions g 1 and g 2 for which Then the planar differential system (S) can be written as (y − y m ) and λ j = λ j (x, y) for j = 1, 2, . . . , S are arbitrary C 1 functions.
It is well known (see for instance [5]) that if g is a polynomial, then the polynomial differential system where λ, µ 1 and µ 2 are arbitrary polynomials on x and y, has g = 0 as an invariant algebraic curve.
Proposition 2. The polynomial system (7) can be written in the form (6).
We denote the degree of a polynomial g by deg g. For the definition of branches of an algebraic curve see the beginning of section 3. Our second main result is the following.
Theorem 3. Let X be the quadratic vector field associated to the quadratic system q jn x n , for j = 1, 2, and for which is the unique irreducible invariant algebraic curve. If the curve g = 0 has S > 1 branches with respect to the variable y (so a 0 = 0). Then deg g ≤ 4S.
We introduce the following two conjectures which are commented in Remark 13 and in the paragraphs following this remark.
Conjecture 5. If a quadratic polynomial differential system (8) with a unique invariant irreducible algebraic curve g = 0 given in (9) does not admit a rational first integral, then deg g ≤ 12.
2. C 1 vector fields with at least two invariant curves of the form y = y(x) Proof of Theorem 1. Let X = (P, Q) be the vector field associated to differential system (6). Now we shall prove that the given orbits (3) are invariant curves of (6). Indeed the vector field X admits the equivalent representation (see also [13]).

PLANAR VECTOR FIELDS WITH A GIVEN SET OF ORBITS 5
Here we have used assumption (4), where . . . y S−2 S , for j = 1, 2, . . . S. Solving system (10) with respect to λ 1 and λ 2 we get that (12) (y − y m ), which appear in the denominator of λ 1 and λ 2 in (12), do not provide problems in the definition of λ 1 and λ 2 , because when we evaluate λ 1 (x, y) and λ 2 (x, y) in y = y j (x) for any j = 1, 2, . . . , S using the expression (11), then the factor y j − y j = 0 also appears in the numerator. So λ 1 and λ 2 are well defined. Substituting λ 1 and λ 2 in (6) we obtain the vector field Y. Hence Theorem 1 is proved. Remark 6. We recall that the determinant ∆ 0 is usually called a Vandermonde determinant.
Remark 7. The natural S ≥ 2 in Theorem 1 is arbitrary.
Proof. Developing by the last row the determinants of the statement of the corollary we get system (6).

Polynomial vector fields with invariant algebraic curves with at least two branches
In the rest of this paper we shall work with complex polynomial vector fields. First we shall study the planar polynomial vector field X of degree n having the invariant algebraic curve where a j = a j (x) for j = 0, . . . , S are polynomials. If a 0 (x) = 0, then it is well known where y j = y j (x), for j = 1, 2, . . . , S are algebraic functions. Moreover The functions g j = y − y j for j = 1, . . . , S are called the branches of the algebraic curve g = 0 with respect to the variable y.
Proposition 10. Let (13) be the product of all the invariant algebraic curves of a polynomial vector field X of degree n. Then the branches g j = y−y j (x) = 0 for j = 1, 2, . . . , S of g = 0 are invariant curves of the vector field X .
Proof. Using the branches g j = y − y j = 0 of g = 0 given by (14) we can write the vector field X in the form given by (6). ¿From Theorem 1 the proposition follows.
Remark 11. There are polynomial vector fields with an invariant algebraic curve having an arbitrary number of branches, this follows from Remark 3 and from the fact that the branches of an invariant algebraic curve are invariant curves of the vector field X (see Proposition 10).
Proof of Proposition 2. Choosing in (6) the arbitrary functions λ j as follows we obtain that system (6) becomeṡ We note that the curve g = 0 of system (7) is not necessarily irreducible.

Remark 12.
It is well known that if the invariant algebraic curve of a polynomial differential system of degree n is nonsingular in CP 2 , then deg g ≤ n + 1 (see for instance Corollary 4 of [4] and Theorem 2 of [16]). As a consequence if deg g > n + 1, then this curve is singular in CP 2

Quadratic system with a unique irreducible invariant algebraic curve
In this section we study the quadratic systems, i.e. polynomial differential systems of degree 2.
Proof of Theorem 3. From the relation Xg = (αy + βx + γ)g, taking the coefficients of the powers of y we obtain the following differential system (15) p where a = (a 0 , a 1 , ..., a S ) T , and a i = a i (x) for i = 0, 1, . . . , S, and A and B are the following (S + 1) × (S + 1) matrices It is known (see for instance [7]) that after a linear change of variables and a rescaling of the time any quadratic system (8) can be written aṡ where P (x, y) is one of the following ten polynomials 1 + xy, y + x 2 , y, 1, xy, −1 + x 2 , 1 + x 2 , x 2 , x, 0.
Since the last six possibilities for P (x, y) force that the quadratic system has an invariant straight line (real or complex) and by assumption our quadratic system has no invariant straight lines, the polynomial P (x, y) only can be 1 + xy, y + x 2 , y, 1. Case 1: Assume that P is either y +x 2 , or y. We consider the quadratic systeṁ with p 2 (x) = x 2 or p 2 (x) = 0. After the recursive integration of system (15), since the a j 's are polynomials, we deduce that where all the a ij are constants. Therefore deg g ≤ 2S.

PLANAR VECTOR FIELDS WITH A GIVEN SET OF ORBITS 9
Case 2: Assume that P = 1. Now we deal with the quadratic system We note that the previous differential system can be written as a Ricatti differential equation. Since this system has no singular points, the algebraic invariant curve g = 0 must be non-singular in the affine plane. If the curve is nonsingular in CP 2 then the degree of g is at most three (see remark 12). So if the algebraic curve g = 0 of (16) has degree larger than three, it is nonsingular in the affine plane and singular at infinity, i.e. in CP 2 . We shall determine the curve g = 0 solution of (15) with degree > 3. Assume that q 11 = 0. After the change of variables (q 11 x, y) −→ (y, x) and introducing the notations q 22 q 2
We study the case p 0 = 0. Therefore the differential system (15) is for j = 0, . . . , S, where a −1 = 0. Solving the first differential equation we get a 0 = C 0 x α , hence α must be a non-negative integer, and without loss of generality we can take C 0 = 1. Now substituting it into the differential equation of a 1 we obtain Solving this linear differential equation we have Since a 1 must be a polynomial we get that γ = αp 21 .
Solving the differential equation of a 2 we obtain Again since a 2 must be a polynomial we get that Doing similar arguments and considering that we can write for j ≥ 3, we can obtain solving the linear differential equation for a j that all a j for j ≥ 3 are polynomials choosing the arbitrary constant C j−1 conveniently. After the recursive integrations we finally deduce that where P m (x) is a polynomial of degree m in x and by definition P −1 (x) = 0. The invariant algebraic curve in this case admits the representation Hence deg g ≤ α + S. If α−S ≥ 0 then by considering that the curve is irreducible, we have α = S, and as a consequence deg g ≤ 2S. If α − S < 0 then deg g ≤ α + S < 2S. In short in case 2 and when q 11 = 0 we have that deg g ≤ 2S. Substituting a S and a S−1 in the last equation of (15) and taking the biggest coefficient in It is interesting the particular case 2 with q 11 = 0 when S > α = 1 and β = p 22 = 0, p 20 = γ = 0. The solutions of (15) are polynomial of degree one of the form a j = c j x + r j , for j = 0, 1, . . . , S where c j and r j are convenient constants satisfying the equations r j+1 = p 20 c j + (S + 1 − j)q 11 r j−1 , p 22 r j = q 11 (S + 1 − j)c j−1 , r −1 = c −1 = 0, for j = 0, 1, . . . , S. Hence we obtain that p 22 p 20 = Sq 11 , Consequently the curve g = 0 takes the form or equivalently where [x] is the integer part function of the real number x.
If we denote then we obtain the following representation for g g(x, y) = xH S (y) + p 20 S H S (y).
If α − Sq 0 = m we shall show that the functions a j (x) of (15) are polynomials of degree deg a j ≤ q 0 j + m if q 0 = 0, and of deg a j ≤ j + m if q 0 = 0, for j = 0, 1, . . . S.
Solving the second nonzero differential equation of system (15) we obtain We assume that k ∈ C\{−1, 0, 1}. From the recursive integration of the system (15) we deduce that the vector a has the following components (20) Since a 0 = 0, we get that C 0 = 0. In order to simplify the proof and to avoid many cases, we assume that all the integration constants C j for j = 1, . . . , S are non-zero, otherwise working in a similar way we should get that some of the polynomials P l which appear in (20) would have lower degree, and this does not perturb the general bound for the deg g. Therefore, since the a j must be polynomials, k is an integer and as a consequence P j = P j (x) is a polynomial of degree j in x. Hence we obtain the deg a j ≤ kj + m, for j = 0, 1, 2, . . . S. Therefore deg g ≤ kS + m = q 0 S + α − q 0 S = α. On the other hand by considering that g = x m−j y S−j P j(k+1) = x m−S (xy) S−j P j(k+1) = 0, and in view that the curve must be irreducible we obtain that m = α−kS = S, therefore α = S(k + 1). Clearly if m − S < 0, then kS + m < S(k + 1) = α. So deg g ≤ (k + 1)S. We are interesting in determining the biggest finite upper bound of the degree of the polynomial g.
We shall study the last equation of system (15). We prove that if C S = 0 then the curve g = 0 has the cofactor K = αy. Indeed inserting a S and a S−1 in the last equation of system (15) we obtain that Hence β = γ = 0 and the cofactor is αy.
We prove that if C S C S−1 = 0 hence k = 3. Indeed, from the last equation of system (15) we obtain that the polynomials a S and a S−1 are such that After the integration we get On the other hand the polynomial a S has degree kS + m, therefore where r 0 is a real constant. Hence if C S C S−1 = 0, then k = q 0 = 3. Since deg g ≤ (k + 1)S = 4S.
In this case working as in the case that the constants C j were not zero with m = 0 we should get for the curve Now we assume that k = q 0 = 0. The recursive integration of system (15) produces the following polynomial solutions for j = 1, 2, . . . S, where r j are rational function in the variables q 11 , q 12 , q 21 , q 22 , q 20 , q 10 , α, β, and P m (x) is a polynomial of degree m in the variable x. Note that deg a j ≤ α + j. The polynomial g becomes where r 0 = 1 and P −1 (x) = 0. In view that the curve g = 0 is irreducible then α = S. If α − S < 0, then deg g ≤ α + S < 2S.
If k = 1, then system (15) becomes for j = 1, 2, . . . S, where a −1 = 0. Hence after integration it is easy to show that where P j = P j (x) is a polynomial of degree j in the variable x. Hence By considering that this curve must be irreducible, we have that m = α−S = 1. As a consequence S = α − 1 < α and deg g ≤ S + m = S + 1.
For the case when k = −1 system (15) takes the form After the recursive integrations we obtain that the polynomial solutions exist in particular if α + S = 0. In this case we obtain that deg a j ≤ j, and as a consequence deg g ≤ S.
In short Theorem 3 is proved.
be a polynomial of degree at most 4S + 1 in the variable x, where A j = A j (q 0 , q 11 , q 10 , q 22 , q 21 , q 20 , α, β, γ, C 0 , C 1 , . . . , C S ) , for j = 0, 1, . . . , 4S + 1. To determine the exact degree of the invariant curve g = 0 in all the cases studied in the proof of Theorem 3, it is necessary that the polynomial R(x) be zero. This holds if and only if all the coefficients are zero, i.e. A j = 0 for j = 0, 1, . . . , 4S + 1. The compatibility of all these equations is require. Working a little it is possible to reduce the system A j = 0 to a polynomial system in the variables q 0 , q 11 , q 10 , q 22 , q 21 , q 20 , α, β, γ, C 0 , C 1 , . . . , C S . These polynomials in general have high degree and it is not easy to work with them for proving that they do not have solution, and consequently the deg g (which from the proof of Theorem 3 must be a multiple of S smaller than or equal to 4S) seems that must be ≤ 3S.
In view of this remark and the comments later on we do the Conjecture 4 and Conjecture 5.
These conjectures are supported mainly by the following facts. First we are able to show that for S = 1, 2, . . . , 5 there are irreducible invariant algebraic curves g = 0 of degree 3S for convenient quadratic system (19). This curve has a cofactor K = 3Sy. On the other hand without loss of generality we can suppose that the given invariant curve has the form (22) for which the deg g ≤ 3S.
We study the particular systems of (15) satisfying (23) with S = 4, and we obtain the family of quadratic systems where a is a nonzero parameter, which admits the following family of invariant algebraic curves of degree 12 The singular points of system (14) are foci, hence it has no rational first integrals. From this example we show that the degree of the invariant algebraic curve of the studied quadratic systems without rational first integral is greater than or equal to 12.
5. C 1 vector fields with only one invariant curve of the form y = y(x) Now we determine the differential system which admits a unique invariant curve Proposition 14. A differential system having the orbit g = y − y 1 (x) = 0, where y 1 = y 1 (x) is a C 1 function, can be written as where λ, µ and ν are arbitrary C 1 functions.
Proof. We set X = (P, Q). First we prove that the curve g = 0 is invariant of the vector field X . Indeed X (g) = (−µy 1 + ν)g. Hence g = 0 is an invariant curve of the differential system associated to the vector field X . Let Y = (Y 1 (x, y), Y 2 (x, y)) = (Y 1 , Y 2 ) be another vector field with the given invariant curve, i.e.
and inserting them into (26) we obtain the vector field Y.
Then system (26) takes the the forṁ which is the most general system for which the curve G = 0 is invariant.

Quadratic system with a unique invariant algebraic curve with one branch
Now we consider the vector field X associated to the quadratic system (27)ẋ = p 2 ,ẏ = q 0 y 2 + q 1 y + q 2 , where p 2 = 0 and q j are polynomials of degree j in the variable x. We assume that X has the curve G = 0, given in (25) with a 0 and a 1 polynomials, as an invariant algebraic curve. System (15) is valid in this case.
Hence a 0 is an orthogonal polynomial because the degrees of p 2 = p 2 (x), p 2 − (βx + γ) − r and κ are 2, 1, 0 respectively (see for instance [1]) For a precise definition of a family of orthogonal polynomials see [1]. A very important class of orthogonal polynomials f 0 , f 1 , . . . f n , . . . are the ones satisfying the differential equation where p = p(x) is a polynomial of degree at most two, q = q(x) is a linear polynomial, and r is nonzero constant.
(i) There exists a quadratic polynomial differential system having the invariant algebraic curveG = f (x)y + f (x) = 0. (i) There exists a polynomial differential system of degree 2,3 or 4 having the invariant algebraic curve G = f (x)y + f (x).
The cofactor in this case is y. This system admits three, two or one invariant algebraic curve depending of degree of p(x). Hence statement (a) is proved.
This system has degree two, three or four and admits one, two or three invariant algebraic curves depending on the degree of polynomial p(x). Moreover the cofactor of G = 0 is p(x)y − q(x).
From Proposition 17 it follows that there exist polynomial differential systems with invariant algebraic curves of arbitrary high degree.
7. On the 16th Hilbert problem for limit cycles on singular invariant algebraic curve One of the motivations of this paper was to study the 16th Hilbert problem for limit cycles on singular invariant algebraic curves. As it follows from the results exposed in [10] and [11] for solving this problem it is necessary to determine the maximum degree of the invariant algebraic curves (Poincaré's problem). It is well known that if the invariant algebraic curve g = 0 of a polynomial vector field of degree n is non-singular in CP 2 then deg g ≤ n + 1 (see for instance [4]). Additionally in that paper the authors gave the following result: if all the singularities on the invariant algebraic curve g = 0 are double and ordinary, then deg g ≤ 2n.
If the algebraic curve is of nodal type, i.e. it is singular and all its singularities are normal crossing type (that is at any singularity of the curve there are exactly two branches of g = 0 which intersect transversally), then deg g ≤ n+2 (see for more deatails [2]).
To determine an upper bound for the degree of a singular invariant algebraic curve is in general an open problem. In this direction there are the following two results. Theorem 18 ([16]). If there exists an integer κ such that for all polynomial vector fields of degree n having the singular invariant algebraic curve g = 0, we have that (g, ∂ y g) ≤ κ, where (g, ∂ y g) is the intersection number of the curves g = 0 and ∂ y g = 0 at a point of g ∩ ∂ y g, then deg g ≤ 4 + 2n + κ + (4 + 2n + κ) 2 + 16κn 2 4 .