This is a preprint of: “Differential equations with a given set of solutions”, Jaume Llibre, Rafael Orlando Ramı́rez, Natalia Sadovskaia, Appl. Math. Comput., vol. 365, 124659:1–20, 2020. DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS JAUME LLIBRE1 , RAFAEL RAMÍREZ2 , AND NATALIA SADOVSKAIA3 Abstract. The aim of this paper is to study the following inverse problem of ordinary differential equations: For a given set of analytic functions ω = {z1 (t), . . . , zr (t)}, with zj (t) = xj (t) + iyj (t) and z̄j (t) = xj (t) − iyj (t) for j = 1, . . . , r, defined in the open interval I ⊆ R, we want to determine the differential equation ˙ . . . , z (n) , z̄ (n) ) = 0, F (t, z̄, z, ż, z̄, dj z where z (j) = for j = 1, . . . , n, in such a way that the given set of functions ω is a dtj set of solutions of this differential equation. 1. Introduction and statement of the main results In the theory of ordinary differential equations we can find two fundamental problems. The direct problem which consists in a broad sense in finding the solutions of a given ordinary differential equation, and the inverse problem. An inverse problem of ordinary differential equations is to find the more general differential equation satisfying a set of given properties (see [4, 10, 19, 20]. The inverse problem for determining the ordinary differential equations with given partial and first integrals was studied in [10]. The obtained results were applied in particular (i) to construct Lagrangian mechanical systems with a given set of linear constraints with respect to the velocity and to obtain Hamiltonian systems with a given set of first integrals (see [11, 20]), (ii) to solve the 16th Hilbert problem for algebraic limit cycles (see [11, 12, 13, 14, 21]), and (iii) to study the center-focus problem (see [15, 16, 17]). In the rest of this paper we assume that all the functions are analytic in their variables, but we remark that this condition is for simplicity, although most of the results remain valid for C r functions with a convenient r ≥ 1. We study the following two problems on the determination of ordinary differential equa- tions with a given set of solutions. Problem I. Determine the most general ordinary differential equations of order n of the form (1) F (t, z, ż, . . . , z (n) ) = 0, for which a given set of functions ω = {z1 (t), . . . , zr (t)}, with zj (t) = xj (t) + iyj (t) for j = 1, . . . , r, defined in the open interval I ⊆ R are solutions of the equation (1). We consider the following particular case of Problem I. Key words and phrases. planar differential system, inverse problem for ordinary differential equations, Ricatti equation, Abel equation, first integral. 1 2 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA Problem I.1 Determine the conditions on the set of functions ω = {z1 (t), . . . , zk (t)} in order that a differential equation of the form (2) ż = f (t, z), has first the functions zj (t), for j = 1, . . . , k, as solutions, and second an invariant of the form k Y (3) F = eΨ Λ(t) (z − zj (t))αj (z̄ − z̄j (t))γj , j=1 where Ψ = Ψ(z, z̄, t) and Λ = Λ(t) are functions, and αj and γj , are complex constants for j = 1, . . . , k. Problem I.1 is studied in sections 4 and 5. Problem II. Determine the most general ordinary differential equations of order n (4) ˙ . . . , z (n) , z̄ (n) ) = 0, F (t, z̄, z, ż, z̄, for which the given set of functions ω = {z1 (t), . . . , zr (t)}, with zj (t) = xj (t) + iyj (t) and z̄j (t) = xj (t) − iyj (t) for j = 1, . . . , r, defined in the open interval I ⊆ R are solutions of the equation (4). We shall study the following particular case of Problem II. Problem II.1 Let (5) ż = F (t, z, z̄), be a differential system, where F (t, z, z̄) is a polynomial of degree n in the variables z and z̄. (i) Determine the conditions of the set of functions ω = {z1 (t), . . . , zk (t)} for k ≥ 1, in order that there exists a differential equation (5) having the set of functions ω as solutions. (ii) Determine the maximum numbers of functions zj (t) for j = 1, . . . , k, which define a unique polynomial differential system (5). Problem II.1 is study in section 6. The main results are the following. Theorem 1. Let ω = {z1 (t), . . . , zn (t)} be a set of functions such that z1 (t) z2 (t) ... zn (t) ż1 (t) ż2 (t) ... żn (t) (6) det(A) := ∆(ω) = .. .. .. 6= 0. . .... . (n−1) (n−1) (n−1) z1 (t) z2 (t) . . . zn (t) Then the most general linear ordinary differential equations of order n of the form (7) a0 (t)z (n) + . . . + an−1 (t)ż + an (t)z = 0, a0 (t) 6= 0 having ω as a set of solutions is z z1 (t) z2 (t) ... zn (t) ż ż1 (t) ż2 (t) ... żn (t) W (z, z1 (t), . . . , zn (t)) = .. .. .. .. = 0. . . ... . . (n) (n) (n) z (n) z1 (t) z2 (t) . . . zn (t) n X Moreover z(t) = Cj zj (t) is a general solution of this differential equation, where Cj are j=1 arbitrary constants. DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 3 Corollary 2. If the set of functions ω = {z1 (t), . . . , zn (t)} satisfies (6), then the general solution of the differential equation a0 (t)z (n) + . . . + an (t)ż + an (t)z = f (t), a0 (t) 6= 0 is z = C1 (t)z1 (t) + C2 (t)z2 (t) + . . . + Cn (t)zn (t), where C1 (t), . . . , Cn (t) are functions such that AĊ = Λ, where C = (C1 , . . . , Cn )T , Λ = (0, . . . , 0, f (t))T , and the matrix A is defined in (6). Theorem 1 and Corollary 2 are proved in section 2. Theorem 3. Let ω = {z1 (t), . . . , zn+1 (t)} be a set of functions such that the Vandermonde determinant (8) z1n (t) z1n−1 (t) ... z1 (t) 1 z2n (t) z2n−1 (t) ... z2 (t) 1 Y 4(ω) = z3n (t) z3n−1 (t) ... z3 (t) 1 := det A = (zk (t) − zj (t)) 6= 0, . .. .. .. .. 1≤k≤j≤n . . ... . . n n−1 zn+1 (t) zn+1 (t) . . . zn+1 (t) 1 n+1 X and |żj |2 6= 0. j=1 Then any first order nonlinear ordinary differential equation of the form (9) a0 (t)z m ż + a1 (t)z n + . . . + an−1 (t)z 2 + an (t)z + an+1 (t) = 0, where n and m are non-negative integers with a0 (t) 6= 0 and having z1 (t), . . . , zn+1 (t) as solutions can be written as 1 1 1 ... 1 z z1 (t) z2 (t) ... zn+1 (t) z2 z12 (t) z22 (t) ... 2 zn+1 (t) (10) . .. . .. . .. .. := z m 4(ω)ż − Z(z, t) = 0, ... . n n n n z z1 (t) z2 (t) ... zn+1 (t) z m ż z1m (t)ż1 (t) z2m (t)ż2 (t) . . . m zn+1 (t)żn+1 (t) Moreover if zj = constant for j = 1, . . . , n then there exist the invariant   Z    z m dz(t) m zn+1 (t) dzn+1 (t)  (11)  − , Yn n Y    (z(t) − zj ) (zn+1 (t) − zj ) j=1 j=1 which is constant on the solutions of (9). Corollary 4. Differential equation (10) for m = 0 under the change 1 z = zn+1 + w can be written as (12) wn−2 ẇ = β1 (t) wn−1 + β2 (t) wn−2 + . . . + βn (t), ∂ j Z(z, t) where = −βj (t), for j = 1, . . . , n, where Z(z, t) is defined in (10). ∂z j z=zn+1 4 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA Theorem 3 and Corollary 4 are proved in section 3. The differential equation (13) z 0 = p(t)z 3 + q(t)z 2 + r(t)z + s(t) = Q(t, z), with p(t) 6= 0, is called the Abel equation of first kind and the differential equation (14) ww0 + Ψ(t)w = Φw2 + Υ(t), is called the Abel equation of a second kind. Note that Υ(t) 6= 0, otherwise we have a linear differential equation. The Abel equations in general has no invariants, but there exist some particular cases for which these equations have an invariant. We shall study some of these cases. Proposition 5. Let wj = wj (t) for j = 1, 2, 3 be solutions of the Abel differential equation of second kind (14). Then equation (14) has the invariant 3 X Z − mj Φ(t)dt 3 Y (15) F =e j=1 (w − wj (t))mj , j=1, if and only if X3 mj (16) = 0. w j=1 j (t) Here the mj are constants not all zero. Corollary 6. The Abel differential equation of the first kind has the invariant 3 X Z − mj Φ(t)dt 3  mj 3  mj Y z − zj Y z − zj F =e j=1 := Λ(t) , j=1, (z − z4 )(zj − z4 ) j=1, z − z4 if and only if 3 X (17) mj (zj − z4 )) = 0. j=1 3 X 3 X Moreover if mj = 0 then mj zj = 0 and the invariant F becomes j=1 j=1 3 Y  mj 3 Y z − zj mj (18) F = = Λ(t) (z − zj ) , zj − z4 k=1, k6=4 k=1, k6=4 Proposition 5 and Corollary 6 are proved in section 5. DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 5 Proposition 7. Let ω4 = {z1 (t), . . . , zκ (t)} for κ = (m + 1)(m + 2)/2 be a set of functions such that 1 1 ... 1 z1 z2 ... zκ z̄1 z̄2 ... z̄κ .. .. .. . . ... . ∆(ω) := 6= 0 with κ = (m + 1)(m + 2)/2 z1m z2m ... zκm .. .. .. . . ... . z1 z̄1m−1 z2 (t)z̄2m−1 ... zκ z̄κm−1 (t) z̄1m z̄2m ... z̄κm κ X |z˙j |2 6= 0, j=1 as solutions can be written as 1 1 1 ... 1 z z1 z2 ... zκ z̄ z̄1 z̄2 ... z̄κ .. .. .. .. . . . ... . (19) zm z1m z2m ... zκm = 0, .. .. .. .. . . . ... . z z̄ m−1 z1 z̄1m−1 z2 z̄2m−1 ... zκ z̄κm−1 z̄ m z̄1m z̄2m ... z̄κm ż ż1 ż2 ... żκ Moreover if zj = constant for j = 1, . . . , κ − 1 then there exist the invariant Z   dz dzκ (20) − = C = constant. Υ(z, z1 , . . . , zκ−1 ) Υ(zκ , z1 , . . . , zκ−1 ) where 1 1 1 ... 1 ξ z1 z2 ... zκ−1 ξ¯ z̄1 z̄2 ... z̄κ−1 .. .. .. .. . . . ... . Υ(ξ, z1 , . . . , zκ−1 ) = ξm z1m z2m ... m zκ−1 .. .. .. .. . . . ... . ξ ξ¯m−1 z1 z̄1m−1 z2 z̄2m−1 ... m−1 zκ−1 z̄κ−1 ξ¯m z̄1m z̄2m ... m z̄κ−1 Proposition 7 is proved in section 6. 6 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA Proposition 8. Consider the differential equation of the form m X X (21) ż = ajk (t)z j z̄ k , n=0 j+k=n with m + 1 complex solutions ω = (z1 (t), . . . , zm+1 (t)) such that (22) z̄j (t) = eiα zj (t), for j = 1, . . . , m + 1, where α ∈ R and the Vandermonde determinant 1 1 1 ... 1 z1 z2 z3 . . . zm+1 (23) ∆(ω) = z12 z22 z32 2 . . . zm+1 6= 0, .. .. .. .. . . . ... . z1m z1m z1m ... m zm+1 can be written as 1 1 1 1 ... 1 z z1 z2 z3 . . . zm+1 1 z2 z12 z22 z32 2 . . . zm+1 (24) .. .. .. .. .. = (z̄ − eiα z)P (z, z̄, t), ∆(ω) . . . . ... . zm z1m z1m z1m ... m zm+1 ż ż1 ż2 ż3 ... żm+1 where P (z, z̄, t) is a convenient polynomial of degree m − 1. Proposition 8 is proved in subsection 6.2. 2. Proofs of Theorem 1 and Corollary 2 Proof of Theorem 1. Clearly differential system (7) has the solutions given by the set of functions ω. Indeed, if z = zj (t) with j = 1, . . . , n, then in view of the properties of the determinant we have that W (zj (t), z1 (t), . . . , zn (t)) ≡ 0. The reciprocity is proved as follows. We assume that we have the linear differential equation of order n a0 (t)z (n) + . . . + an−1 (t)ż + an (t)z = 0, a0 (t) 6= 0, and let ω = {z1 (t), . . . , zn (t)} be the set of functions satisfying (6). Then, solving the system (n) a0 (t)zj + . . . + an−1 (t)żj + an (t)zj = 0, for j = 1, . . . , n. DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 7 with respect to a1 , . . . , an after some computations we get that z1 (t) z2 (t) ... zn (t) ż1 (t) ż2 (t) ... żn (t) (−1)n+2 a0 (t) .. .. .. (−1)n+2 a0 (t)∆1 (ω) a1 (t) = . . ... . := , ∆(ω) (n−2) (n−2) (n−2) ∆(ω) z1 (t) z2 (t) . . . zn (t) (n) (n) (n) z1 (t) z2 (t) ... zn (t) .. .. .. .. . . . . ż1 (t) ż2 (t) ... żn (t) −a0 (t) z̈1 (t) z̈2 (t) ... z̈n (t) −a0 (t)∆n (ω) an (t) = .. .. .. := , ∆(ω) . . ... . ∆(ω) (n) (n) (n) z1 (t) z2 (t) ... zn (t) Inserting a1 , . . . , an in (7) we obtain a0 (t)   ∆(ω) z (n) + (−1)n+2 ∆1 (ω)z (n−1) + . . . −∆n (ω)z = 0. ∆(ω) This last expression is the development of the determinant given in (8) using its first column. Hence the theorem is proved.  Proof of Corollary 2 . The proof of these Theorem is well known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.  3. Proofs of Theorem 3 and Corollary 4 Proof of Theorem 3. First we prove that the set of function ω = {z1 (t), . . . , zn+1 (t)} satisfies (10). Indeed, in view of the properties of the determinant we get that 1 1 1 ... 1 zj (t) z1 (t) z2 (t) . . . zn+1 (t) zj2 (t) z12 (t) z22 (t) 2 . . . zn+1 (t) .. .. .. .. = 0, . . . ... . zjn (t) z1n (t) z2n (t) n . . . zn+1 (t) żj (t) ż1 (t) ż2 (t) ... żn+1 (t) for j = 1, . . . , n + 1. The reciprocity is proved as follows. We assume that we have the nonlinear differential equation of first order (10), and let ω = {z1 (t), . . . , zn+1 (t)} be the set of functions satisfying (8). Then from the relations a1 (t)zjn + . . . + an−1 (t)zj2 + an (t)zj + an+1 (t) = −a0 (t)zjm żj . for j = 1, . . . , n + 1, Since this system is a linear system with respect to a1 (t), . . . , an+1 (t) it can be written in the matrix form  n     z1 (t) z1n−1 (t) . . . z1 (t) 1 a1 (t) z1m (t)ż1 (t)  z2n (t) z2n−1 (t) . . . z2 (t) 1     z2m (t)ż2 (t)   n   a2 (t)     z3 (t) z n−1 (t) . . . z3 (t)  1   a3 (t)  = −a0    z1m (t)ż3 (t)   3    .. .. .. ..   ..   ..   . . ... . .  .   .  n n−1 m zn+1 (t) zn+1 (t) . . . zn+1 (t) 1 an+1 (t) zn+1 (t)żn+1 (t) 8 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA In view of condition (8) and by applying the Cramer’s rule we obtain that z1m (t)ż1 (t) z1n−1 (t) ... z1 (t) 1 z2m (t)ż2 (t) z2n−1 (t) ... z2 (t) 1 a0 (t) z3m (t)ż3 (t) z3n−1 (t) ... z3 (t) 1 a0 (t)∆1 (ω) a1 (t) = − := − , ∆(ω) .. .. .. .. ∆(ω) . . ... . . m n−1 zn+1 (t)żn+1 (t) zn+1 (t) . . . zn+1 (t) 1 .. .. .. . . . , z1m (t)ż1 (t) z1n (t) z1n−1 (t) ... z1 (t) z2m (t)ż2 (t) z2n (t) z2n−1 (t) ... z2 (t) a0 (t) z3m (t)ż3 (t) z3n (t) z3n−1 (t) ... z3 (t) a0 (t)∆n+1 (ω) an+1 (t) = −(−)n := (−1)n+1 , ∆(ω) .. .. .. .. ∆(ω) . . ... . . m n n−1 zn+1 (t)żn+1 (t) zn+1 (t) zn+1 (t) . . . zn+1 (t) By inserting a1 (t), . . . , an+1 (t) into (9) after some computations we get (10). a0 (t)   ∆(ω)z m z (n) + (−1)n+2 ∆1 (ω)z (n−1) + . . . −∆n (ω)z = 0. ∆(ω) This last expression is the development of the determinant given in 10 using its first column. To prove the relation (11) it is easy to obtain by considering that żj = 0 for j = 1, . . . , n. n+1 X Clearly if |żj |2 = 0 then the all coefficients aj (t) for j = 1, . . . , n + 1 are equal to zero. j=1 Consequently differential system (9) becomes ż = 0. Thus the theorem is proved.  Proof of Corollary 4. By developing the polynomial Z(z, t) of degree n in the variable z (see formula (10)) in powers of z − zn+1 we get ∂Z(z, t) 1 ∂ 2 Z(z, t) Z(z, t) = Z(zn+1 , t) + (z − zn+1 ) + (z − zn+1 )2 ∂z z=zn+1 2! ∂z 2 z=zn+1 1 ∂ n Z(z, t) +... + (z − zn+1 )n n! ∂z n z=zn+1 = Z(zn+1 , t) − β1 (t)(z − zn+1 ) − β2 (t)(z − zn+1 )2 − . . . − βn (t)(z − zn+1 )n . Introducing the new variable 1/w = z −zn+1 and by considering that ∆(ω)żn+1 = Z(zn+1 , t) after some computations we obtain (12). Then the corollary is proved.  Example 1. For differential equation (10) for m = 0 and n = 2 ( Ricatti equation) z10 (z3 − z2 ) + z20 (z1 − z3 ) + z30 (z2 − z1 ) 2 z0 = z ∆(ω) z10 (z32 − z22 ) + z20 (z12 − z32 ) + z30 (z22 − z12 ) (25) − z ∆(ω) z10 (z3 − z2 )z3 z2 + z20 (z1 − z3 )z1 z3 + z30 (z2 − z1 )z2 z1 + . ∆(ω) differential equation (12) becomes 1 (26) ẇ = β1 (t) w + β2 (t), w= z − z3 DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 9 with (z2 − z3 )ż1 (z1 − z3 )ż2 (z1 − 2z3 + z2 )ż3 β1 (t) = − + (z1 − z3 )(z1 − z2 ) (z2 − z3 )(z1 − z2 ) (z2 − z3 )(z1 − z3 ) ż1 ż2 ż3 β2 (t) = − + − (z1 − z3 )(z1 − z2 ) (z2 − z3 )(z1 − z2 ) (z2 − z3 )(z1 − z3 ) Example 2. For differential equation (10) for m = 0 and n = 3 (Abel equation of first kind), having four solutions z1 , z2 , z3 , z4 satisfying (8) can be written as (10) or equivalently ∆1 (ω) 3 ∆2 (ω) 2 ∆3 (ω) ∆4 (ω) (27) z0 = z − z + z− := Z0 (z, t). ∆(ω) ∆(ω) ∆(ω) ∆(ω) equation (12) becomes 1 (28) wẇ = β1 (t) w2 + β2 (t)w + β3 (t), w= z − z4 with β1 (t) = λ(z1 , z2 , z3 , z4 )ż1 + λ(z2 , z1 , z3 , z4 )ż2 + λ(z1 , z3 , z2 , z4 )ż3 (z2 − z4 )(z1 − z4 ) + (z3 − z4 )(z1 − z2 − 2z4 ) + ż4 , (z1 − z4 )(z3 − z4 )(z2 − z4 ) β2 (t) = µ(z1 , z2 , z3 , z4 )ż1 + µ(z2 , z1 , z3 , z4 )ż2 + µ(z1 , z3 , z2 , z4 )ż3 + µ(z1 , z2 , z4 , z3 )ż4 , β3 (t) = ν(z1 , z2 , z3 , z4 )ż1 + ν(z2 , z1 , z3 , z4 )ż2 + ν(z1 , z3 , z2 , z4 )ż3 + ν(z1 , z2 , z4 , z3 )ż4 , (w − r)(v − u) v − 2r + w where λ(u, v, w, r) = − , µ(u, v, w, r) = , and (u − v)(u − w)(u − r) (u − v)(u − w)(u − r) 1 ν(u, v, w, r) = − . (u − v)(u − w)(u − r) We say that the set of functions ω1 = {z1 , z2 , . . . , zn+1 } which are solutions of equation (κ) (κ) (10) and the set of functions ωκ = {z1 , . . . , zn+1 } are equivalent and we write ω1 ' ωκ if ∆j (ω) ∆j (ωκ ) = , for j = 1, . . . , n + 1 κ ∈ N. ∆(ω) ∆(ωκ ) (κ) Clearly if ω1 ' ωκ then ω̃1 ' ω̃κ where ω̃1 = {z1 , z2 , zj−1 , zj , zj+1 , . . . , zn+1 } and ω̃κ = (κ) (κ) (κ) (κ) {z1 , . . . , zj−1 , zj , zj+1 , . . . , zn+1 , }, The set of equivalent solutions of (10) we shall denote by Ω. Example 3. For the Abel differential equations of the first kind 225 z 3 + 16(3t8 − 17t6 + 6t4 − 1)z ż = 4t(t4 − 1)(t4 − 4)(4t4 − 1) a set of equivalent solutions Ω is (see [6]) n 2 2 4 4 o z1 = −z2 = t(t2 + 1), z3 = −z4 = t(t2 − 1), z5 = −z6 = (t − 1), z7 = 0 5 3 15 4. Study of Problem I.1 4.1. Nonlinear differential equations with constant solutions. Problem I.1 has an easy solution when the functions zj (t), for j = 1, . . . , n + 1, are complex constant zj , because 10 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA n+1 Y then f (t, z) = λ(t) (z − zs )κs with κj ∈ N and λ(t) is a convenient function. Indeed, from s=1 (9) it follows that n+1 Y ż = λ(t) (z − zs )κs = λ(t)g(z), s=1 where for simplicity we assume that the function λ(t) is real. From the equations ż = λ(t)g(z) and z̄˙ = λ(t)g(z̄) it follows that dz dz̄ (29) − = 0. g(z) g(z̄) After the integration of (29) we get that Z   dz dz̄ F̃ = − f (z) f (z̄) is a first integral of the differential equation (29). It is possible to show that the function F = exp F̃ is an invariant, that in this case is a first integral n+1 Y A(1) (z − zs ) s s=1 F = eΨ n+1 Y Ā(1) (z̄ − z̄s ) s s=1 where ! (2) (k ) A1 1 A1 s Ψ := − + ... + + ... (z − z1 ) (1 − κ1 ) (z − z1 )−1+κ1 (2) (k ) ! n+1 Ān+1 1 Ān+1 − − + ... + , z̄ − z̄n+1 (1 − κn+1 ) (z̄ − z̄n+1 )−1+κn+1 (j) where Am is a convenient constants for j = 2, . . . , κn+1 and m = 1, . . . , n + 1. In short, Problem I.1 is solved when all the functions of ω are constant. 4.2. First order linear ordinary differential equations and Riccati equations. Here we see that Problem I.1 is easy to solve for first order linear ordinary differential equations and for the Riccati equations. Indeed, by Theorem 3 the first order linear differential equation with the two solutions z1 (t) and z2 (t) such that ∆(ω) = z1 (t) − z2 (t) 6= 0, can be written as (10) for n = 1 and m = 0 is z20 (t) − z10 (t)) z1 (t)z20 (t) − z2 (t)z10 (t) (30) z0 − z+ = 0. z2 (t) − z1 (t) z2 (t) − z1 (t) After some computations we get that z − z1 (t) (31) F = := Λ(t)(z − z1 (t)) = z2 (t) − z1 (t) is an invariant of the differential equation (5). By Theorem 3 the first order Riccati differential equation having the three solutions z1 (t), z2 (t) and z3 (t) satisfying (8) for n = 2 and m = 0 can be written as (25). By considering that Ricatti equation under the change w = 1/(z − z3 ) can be transformed to the linear differential equation (26), which has the invariant w − w1 (t) F = , w2 (t) − w1 (t) DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 11 (see formula (31)). Clearly this differential equation has the solution w1 = 1/(z1 −z3 ), w2 = 1/(z2 − z3 ), then in view of formula (31) we have that this linear equation has the invariant w − w3 1/(z − z3 ) − 1/(z1 − z3 ) (z − z1 )(z2 − z3 ) (z − z1 ) F =− =− = := Λ(t) . w1 − w2 1/(z1 − z3 ) − 1/(z2 − z3 ) (z − z3 )(z2 − z3 ) (z − z3 ) The invariant for the linear differential equations is trivial and for the Riccati case corre- sponds with the well known fact that the cross-ratio of four linearly independent solutions is a constant (see for instance [2]). These results are known but the interest of the proofs which we propose consists in showing that these proofs can obtained from the inverse approach: we have the explicit expression of these differential equations for the given solutions. 5. Abel differential equations having an invariant The main objective of this subsection is to study the problem of existence of the invariant (3) for the Abel differential equations . From Theorem 3 the Abel equation of the first kind (13) having four solutions z1 , z2 , z3 , z4 satisfying (8) for n = 3 and m = 0 can be written as (27) Clearly that in this case ∆1 (ω) z10 z20 p(t) = − = + ∆(ω) (z1 − z2 )(z1 − z3 )(z1 − z4 ) (z2 − z1 )(z2 − z3 )(z2 − z4 ) z30 z40 + + , (z3 − z2 )(z3 − z1 )(z3 − z4 ) (z4 − z2 )(z4 − z3 )(z4 − z1 ) ∆2 (ω) ∆3 (ω) ∆4 (ω) q(t) = , r(t) = − , s(t) = . ∆(ω) ∆(ω) ∆(ω) 1 As we prove in section 3, the Abel equation of first kind (27) under the change z = z4 (t)+ , w and introducing the respectively notations,becomes into the Abel equation of a second kind (28) or equivalent (14). Proof of Proposition 5. We consider   X3 X3 0 0 dF w − w j (t) (32) = − mj Φ(t) + mj  F. dt j=1 j=1 w − wj (t) Then in view of (14) and by considering that wj (t) is a solution of (14), i.e. wj wj0 + Ψ(t)wj = Φwj2 + Υ(t), then (32) becomes 3 X 3 X w0 − wj0 (t) − mj Φ(t) + mj j=1 j=1 w − wj (t) Υ(t) Υ(t) 3 X 3 X Φ(t)w − Ψ(t) + − Φ(t)wj + Ψ(t) − w wj = − mj Φ(t) + mj j=1 j=1 w − wj (t) 3 Υ(t) X mj = − = 0. z j=1 wj 12 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA   3 X dF Υ(t) m j Hence (32) can be written as = −  F. Since Υ(t) 6= 0, from this last dt u j=1 wj 3 X equality we have that dF/dt = 0 if and only if mj /wj (t) = 0. The proof of representation j=1 (15) it is easy to obtain. Hence this completes the proof of the proposition.  1 Proof of Corollary 6. It follows from Proposition 9 by considering that w = , and z − z4 3 X mj X3 1 wj = , for j = 1, 2, 3, consequently condition (16) becomes = mj (zj − zj − z4 j=1 wj j=1 z4 )) = 0.  Example 4. We illustrated Corollary 6 for the case when zj for j = 1, 2, 3 are constants. After some computations we can prove that differential equation (27) becomes (z − z1 )(z − z2 )(z − z3 )ż4 (t) ż = (z4 (t) − z1 )(z4 (t) − z2 )(z4 (t) − z3 ) The invariant F in this case is (see invariant (11))  m1  m2  m3 z − z1 z − z2 z − z3 fF = z4 (t) − z1 z4 (t) − z2 z4 (t) − z3 where 1 1 1 m1 = , m2 = , m3 = , (z1 − z2 )(z1 − z3 ) (z2 − z1 )(z2 − z3 ) (z3 − z1 )(z3 − z2 ) clearly that m1 + m2 + m3 = 0 and m1 z1 + m2 z2 + m3 z3 = 0. Z Φdt Clearly after the change of variables w 7−→ w e equation (14) becomes w (w0 + Ψ(t)) = Υ(t). Proposition 9. Let the sets of functions (κ) (κ) (κ) ω = {w1 , w2 , w3 }, ωκ = {w1 , w2 , w3 }, for 1 ≤ κ ≤ ς, be equivalent sets of solutions of the Abel differential equation of second kind (13). Denoting (1) (1) (ς) by Ω = {w1 , w2 , w3 , w4 = w1 , w5 = w2 , w6 = w3 , . . .} the set of all equivalent solutions where κ ∈ N and µ = 3(ς + 1). Then equation (14) has the invariant Xµ Z − mj Φ(t)dt µ Y F = e j=1 (w − wj )mj , with wj ∈ Ω, j=1, if and only if µ X mj = 0. j=1 wj Here the mj are constants not all zero. Moreover, under the changes Z Z Φdt Φdt w = We , wj = Wj e , DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 13 the invariant F becomes µ Y F = (W − Wj (t))mj , j=1 Corollary 10. Let the sets of functions (κ) (κ) (κ) (κ) ω1 = {z1 , z2 , z3 , z4 }, ωκ = (z1 , z2 , z3 , z4 }, for 1 ≤ κ ≤ ς, be equivalent sets of solutions of the Abel differential equation of first kind (13). Denoting (1) (1) (ς) (ς) by Ω = {z1 , z2 , z3 , z4 , z5 = z1 , z6 = z2 , z7 = z3 , z8 = z4 , . . .} the set of all equivalent solutions. Then there exists the invariant µ Y  mj z − zj F = Λ(t) , with zj ∈ Ω, z − z4 k=1, k6=4 if and only if µ X mj (zj − z4 ) = 0. j=1, j6=4 µ X µ X Moreover if mj = 0 then mj zj = 0 and the invariant F becomes j=1 j=1 µ Y  mj µ Y z − zj mj F = = Λ(t) (z − zj ) . zj − z4 k=1, k6=4 k=1, k6=4 Proof. It is analogous to the proof of Proposition 9.  There are a lot of results on the existence of invariants F for the equation (17), or for its canonical form (see for instance [2]). Z Introducing the new independent variable x = − Ψ(t)dt we get the canonical form of the Abel equation of the second kind becomes   du −Υ (33) u − 1 = R(x) with R(x) = . dx Ψ Clearly that this differential equation with two solutions ω = {u1 , u2 } where u1 = u1 (x) and u2 = u2 (x), can be written as ∆1 (ω) ∆2 (ω) (34) uu0 + u+ = 0, ∆(ω) ∆(ω) d ∆1 (ω1 ) ∆2 (ω1 ) where 0 := , and = −1, = R(x) dx ∆(ω1 ) ∆(ω1 ) Proposition 11. Let the sets of functions (κ) (κ) ω1 = {u1 , u2 }, ωκ = {u1 , u2 }, for 1 ≤ κ ≤ ς, be equivalent sets of solutions of the Abel differential equation of second kind (13). Denoting (1) (1) by Ω = {u1 , u2 , u3 = w1 , u4 = w2 , . . .} the set of all equivalent solutions of the Abel differential equation of second kind (34). Then there exists the invariant µ Y mj F = (u − uj ) , with uj ∈ Ω, k=1, k 14 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA if and only if Xµ mj (35) = 0. u j=1, j Proof. It is analogous to the proof of Proposition 9 with Φ = 0 and Ψ = −1.  Example 5. We study equation (34) in the particular case   du 3x A B (36) u −1 = − √ − √ . dx 16 3 x 3 x5 In [2] the authors showed that the set of function ω1 = {u1 , u2 } where r √ 3 3√ √ −3B u1 = x + 3A + −3B x + √ 3 , 4 2 3 x (37) r √ 3 3√ √ −3B u2 = x − 3A + −3B x + √ 3 , 4 2 3 x are solutions of equation (36). The following set of functions ω2 = {u3 , u4 } r √ 3 3√ √ −3B u3 = x + 3A − −3B x − √ 3 , 4 2 3 x r √ 3 3√ √ −3B u4 = x − 3A − −3B x − √ 3 , 4 2 3 x is equivalent to (37). Indeed it is easy to show that ∆1 (ω1 ) ∆1 (ω2 ) = = −1, ∆(ω1 ) ∆(ω2 ) ∆2 (ω1 ) ∆2 (ω2 ) 3x A B = = − √ − √ ∆(ω1 ) ∆(ω2 ) 16 3 x 3 x5 By considering that m1 /u1 + m2 /u2 + m3 /u3 + m4 /u4 = 0, where p √ p √ m1 = 12A − 6 −3B, m2 = − 12A − 6 −3B, p √ p √ m3 = − 12A + 6 −3B, m4 = 12A + 6 −3B. Then in view of Proposition 11 we obtain that equation (36) has the invariant F = (u − u1 )m1 (u − u2 )m2 (u − u3 )m3 ((u − u4 )m4 ,  m1  m4 u − u1 u − u4 = . u − u2 u − u3 According to the result given by Alexeeva, Zaitsev and Zhvets [2], any Abel equation of the second kind in the canonical form (33) has the invariant Ym (38) F = (u − uj (s))mj j=1, where uj (s) is a particular solution of (33) and mj is a convenient constant for j = 1, . . . , m if the following relation holds   Xm   k6Y=j Xm   (39)  mj ( du − duj (s) (u − uk (s)) = du Φj (s) + Ψj (s)uj , j=1 ds ds j=1 ds k=1,..,n where Φν (s) = −Ψν (s) Ψν−1 (s) = −R(s)Φν (s), and equate the other Φk and Ψk with zero. DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 15 Selecting different values ν = 1, . . . , n − 1, we obtain n − 1 systems of different-algebraic equations; only one of the systems, corresponding to mk 6= 0 for all k = 1, . . . , n and ui 6= uj for i 6= j, leads to a non degenerate solution of the form (38). Clearly condition (39) is complicated to compute unlike conditions (17). In the monograph [2] there is an example of solutions of (33) obtained from (39) for a convenient R(x). All these solutions satisfy (35) for a convenient constants mj . The examples given in [2] are the following and as we show all of them satisfy the condition (35). Example 6. For n = 2 differential equation (33) with R(x) = Ax + b where a and B are constants has the following solutions √ √ 1 + 1 + 4A 1 + 1 + 4A u1 == (Ax + b) , u2 == − √ (Ax + b) , 2A 2A + 1 + 1 + 4A √ and m1 = 2A + 1 + 1 + 4A, m2 = 2A. Clearly that m1 /u1 + m2 /u2 = 0. √ For n = 3 and for R(x) = −2/9t x + A + B/ x the obtained solutions in [2] are 2 √ 3B uj = ( x + λj x) + , 3 λj 2A mj = for j = 1, 2, 3, 3(2λ2j − 3A) where λ is a root of λ3 − 9/2Aλ − 9/2B = 0. It is easy to show that m1 /u1 + m2 /u2 + m3 /u3 = 0. We can prove that Ω = {u1 , u2 , u3 } is a set of equivalent solutions. Indeed if ω1 = {u1 , u2 }, ω2 = {u1 , u3 } and ω3 = {u2 , u3 } then ∆1 (ω1 ) ∆1 (ω2 ) ∆1 (ω3 ) = = = −1, ∆(ω1 ) ∆(ω2 ) ∆(ω3 ) ∆2 (ω1 ) ∆2 (ω2 ) ∆1 (ω3 ) √ = = == −2/9 x + A + B/ x. ∆(ω1 ) ∆(ω2 ) ∆(ω3 ) 3x A B For m = 4 and − √ − √ the authors proposed the solutions given in example 5 16 3 x 3 x5 √ √ with wrong constants m1 = −m2 = −2A + −3B, and m3 = −m4 = 4A2 + 3B, because m1 /u1 + m2 /u2 + m3 /u3 + m4 /u4 6= 0. 5.1. Julia invariant. Assume that the functions u1 (s), . . . , uµ (s) satisfy the equation uµ + a1 (s)uµ−1 + . . . + aµ−2 (s)u2 + aµ (s) = 0, and are solutions of the canonic Abel second kind differential equation (33), then there exists an invariant µ Y (40) F = (u − uj (s)). j=1 Indeed, under the given assumption the functions u1 (s), . . . , um (s) satisfy the relation µ Y µ Y µ−1 Y uj (s) + uj (s) + . . . + uj (s) = 0, j=2 j=1, j6=2 j=1 µ X 1 or what is equivalent = 0, consequently, in view of Proposition 11 with mj = 1 for j=1 uj (s) j = 1, . . . , µ there exists the invariant F. 16 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA 1 1 In particular if µ = 2 we have that if + = 0 or, what is equivalent u1 + u2 = 0, u1 u2 then the invariant becomes F = (u − u1 )(u + u1 ). The French mathematician Gaston Julia proved 1933 in [5] (6, p. 82f) that the equation dy (Dy + E) + Ay 2 + By + C = 0, dt for A, B, C, D and E functions of t, has an implicit solution if the condition L := E(2A − D0 ) − D(B − E 0 ) = 0, with D 6= 0 is satisfied. Then the invariant is Z Z 2A   Z 2A dt y 2 E C dt F =e D + y + e D = const. 2 D D Doing two changes of variables the previous differential equation yields the general Abel E equation of first kind. First we do the change of variables y = w(t) − and we get D A L E 2 A − EDB + CD2 (41) ww0 + w2 + 2 w + = 0. D D D3 The invariant under the given change becomes Z Z 2A   Z 2A dt E 2 C dt F =e D 2 w − 2 + e D . D D Z A dt By putting L = 0 and after the change w −→ we D we obtain that the differential equation (41) becomes Z −2A  dt E 2 A − EDB + CD2  0 ww + e D = 0. D3 This last differential equation has the invariant F = (u − u1 )(u + u1 ), where v Z Z u 2A 2A u 2 dt Z dt tE C u1 = −u2 = e D + e D , D2 D which corresponds to the invariant (40) with µ = 2. Finally we observe that there exits a big numbers of papers devoted to study the existence of exactly solutions and applications of the Abel equation (see for instance [1, 3, 7, 8, 9, 18]). 6. Planar polynomial differential equations with given solutions In this section we study the problem II.1. We determine the planar differential systems (42) ẋ = P (x, y, t) ẏ = Q(x, y, t), having the given set of solutions ω = {z1 (t), . . . , zk (t)} where k ≥ 1. After the change z = x + iy and z̄ = x − iy differential system (42) becomes (43) ż = R(z, z̄, t) where R = R(z, z̄, t) = (P + iQ)|x=(z+z̄)/2, y=(z−z̄)/(2i) . DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 17 First we determine the simplest case when ω = {z1 (t)}. Proposition 12. The most general planar analytic differential system with the given solu- tion z1 (t) can be written as (44) ż = ż1 + Φ(z, z̄, t), or equivalently ẋ = ẋ1 (t) + ReΦ(z, z̄, t)), ẏ = ẏ1 (t) + ImΦ(z, z̄, t)), where z1 (t) = x1 (t) + iy1 (t) and Φ is any analytic function (Erugyn function) such that (45) Φ(z, z̄, t))|z=z1 (t), z̄=z̄1 (t) = 0. Proof. Indeed z = z1 (t) is a solution of (44) because (45) holds. Assume that differential system ż = L(z, z̄, t) is another differential equation having z = z1 (t) as a solution, i.e. L(z1 (t), z1 (t), t) = ż1 (t). taking Φ(z, z̄, t) = L(z, z̄, t) − ż1 (t) we get that this differential equation is a particular case of (44).  Proposition 13. The most general planar polynomial differential system of degree m with coefficients functions depending on t and with the given solution z1 (t), can be written as (44) with either ˜ M (t)ξ T >, Φ = α1 (t)H11 + α2 (t)H12 + α3 (t)H21 + α4 (t)H22 =< ξ(t), where αj (t) are arbitrary functions, Hjk is a component of the matrix H = G(t, z, z̄)G(t, z1 (t), z̄1 (t)) − G(t, z1 (t), z̄1 (t))G(t, z, z̄) := [G(t, z, z̄), G(t, z1 (t), z̄1 (t))], with the matrix   1 1 G(t, z, z̄) = , p1 (t, z, z̄) p2 (t, z, z̄) and m X m X p1 (t, z, z̄) = anj (t)z n z̄ j , p2 (t, z, z̄) = bnj (t)z n z̄ j , n+j=0 n+j=0 ξ= 2 2 2 m m (1, z, z̄, z , z z̄ , z̄ , . . . , z , . . . , z̄ ), ˜ = ξ| ξ(t) z=z1 (t), z̄=z̄1 (t) , and M (t) is a convenient (m + 1)(m + 2)/2 × (m + 1)(m + 2)/2 antisymmetric matrix with components functions of t. In particular, the most general differential equation with a given singular point z1 (t) = α + iβ ∈ C can be written as (44) with ż1 (t) = 0. Proof. The proof of this statement require a huge number of computations. We prove this proposition only for linear and quadratic differential equations. First we study the linear case, when p1 (t, z, z̄) = z, p2 (t, z, z̄) = z̄ Hence       1 1 1 1 1 1 1 1 H= − z z̄ z1 (t) z̄( t) z1 (t) z̄( t) z z̄   z1 (t) − z z̄1 (t) − z̄ = z̄z1 (t) − z̄1 z + z − z1 (t) z − z1 (t) Consequently Φ = α̃1 (t)(z − z1 (t)) − α2 (t)(z̄ − z̄1 (t)) + α3 (t)(z̄z1 (t) − z z̄1 (t))     0 α̃1 (t) −α2 (t) 1 = 1, z1 (t), z̄1 (t)  −α̃1 (t) 0 α3 (t)   z  , α2 (t) −α3 (t) 0 z̄ 18 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA where α̃1 (t) = −α1 (t) + α3 (t) + α4 (t). Thus from (44) we obtain that the most general planar linear differential system of first order with the given solution z1 (t) is ż = ż1 + α̃1 (t)(z − z1 (t)) − α2 (t)(z̄ − z̄1 (t)) + α3 (t)(z̄z1 (t) − z z̄1 (t)), where α̃1 (t), α2 (t) and α3 (t) are arbitrary functions. So the proposition is proved for the linear differential equations of first order with the given solution z1 (t). For the quadratic differential systems we have that ˜ ξ(t)M (t)ξ(t) = −(a46 z12 + a56 z1 w1 + a26 z1 + a36 z̄1 + a16 )z̄ 2 (a56 w12 − a45 z12 − a25 z1 − a35 z̄1 − a15 )z z̄ +(a45 z1 z̄1 + a46 z̄12 − a24 z1 − a34 z̄1 − a14 )z 2 (a34 z12 + a35 z1 z̄1 + a36 z̄12 − a23 z1 − a13 )z̄ +(a24 z12 + a25 z1 z̄1 + a26 z̄12 + a23 z1 − a12 )z a14 z12 + a15 z1 z̄1 + a16 z12 + a12 z1 + a13 z̄1 , ˜ where Mjk = ajk (t). From the equation Φ = σ1 z 2 + σ2 z̄ 2 + σ3 z z̄ + σ4 z + σ5 z̄ = ξ(t)M (t)ξ(t) we get that σ1 = a16 + a46 z12 + a56 z1 w1 + a26 z1 + a36 z̄1 , σ2 = −a15 + a56 w12 − a45 z12 − a25 z1 − a35 z̄1 , σ3 = −a14 + a45 z1 z̄1 + a46 z̄12 − a24 z1 − a34 z̄1 , σ4 = −a13 + a34 z12 + a35 z1 z̄1 + a36 z̄12 , σ5 = −a12 + a24 z12 + a25 z1 z̄1 + a26 z̄12 + a23 z1 , Φ|z=z1 , z̄=z̄1 = σ1 z12 + σ2 z̄12 + σ3 z1 z̄1 + σ4 z1 + σ5 z̄1 = 0. This completes the proof of the proposition for the quadratic polynomial differential systems with a given solution z1 (t). In similar way we could obtain the result of Proposition 13 for polynomial differential equations of degree m > 2 with a given solution z1 (t), but the computations increases strongly.  Proposition 14. The most general planar polynomial differential system of degree m with the given solutions ω = {z1 (t), 0}, can be written as ż1 (t) (46) ż = z + Φ(z, z̄, t), z1 (t) where ˜ M (t)ξ T >, Φ(z, z̄, t) =< ξ(t), ˜ = ξ| and ξ = (z, z̄, z 2 , z z̄ 2 , z̄ 2 , . . . , z m , . . . , z̄ m ), ξ(t) z=z1 (t), z̄=z̄1 (t) and M = M (t) is a m(m+ 3)/2 × m(m + 3)/2 antisymmetric matrix with components arbitrary functions of t. In particular, the most general differential equation with a given singular points z1 (t) = α+iβ ∈ C and z2 = 0 can be written as (46) with ż1 (t) = 0. Proof. It is analogous to the proof of Proposition 13.  DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 19 6.1. Construction the quadratic differential system with a given set of solutions. Proposition 15. Let ω = {z1 (t), z2 (t), z3 (t), z4 (t), z5 (t), z6 (t)} be a set of functions such that 1 1 1 1 1 1 z1 z2 z3 z4 z5 z6 z̄1 z̄2 z̄3 z̄4 z̄5 z̄6 (47) ∆(ω) := 6= 0 z12 z22 z32 z42 z52 z62 z̄12 z̄22 z̄32 z̄42 z̄52 z̄62 z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 z6 z̄6 κ X where |z˙j |2 6= 0. Then the most general quadratic differential equations of the first order j=1 (48) ż = a0 (t) + a1 (t)z + a2 (t)z̄ + a3 (t)z 2 + a4 (t)z z̄ + a5 (t)z̄ 2 , for which ω is a given set of solutions can be rewritten as 1 1 1 1 1 1 1 z z1 z2 z3 z4 z5 z6 z̄ z̄1 z̄2 z̄3 z̄4 z̄5 z̄6 (49) z2 z12 z22 z32 z54 z52 z62 = 0. 2 z̄ z̄12 z̄22 z̄32 z̄42 z̄52 z̄62 z z̄ z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 z6 z̄6 ż ż1 ż2 ż3 ż4 ż5 ż6 Moreover the quadratic vector field with the set of solutions ω = {z1 , z2 , z3 , z4 , z5 , z6 } where zj for j = 1, 2, 3, 4, 5 are constants solutions such than (47) holds has the invariant Z   dz dz6 (50) − = C = contstant. Υ(z, z1 , z2 , z3 , z4 , z5 ) Υ(z6 , z1 , z2 , z3 , z4 , z5 ) where 1 1 1 1 1 1 ξ z1 z2 z3 z4 z5 ξ¯ z̄1 z̄2 z̄3 z̄4 z̄5 Υ(ξ, z1 , z2 , z3 , z4 , z5 ) = 2 . ξ z12 z22 z32 z54 z52 ξ¯2 z̄12 z̄22 z̄32 z̄42 z̄52 ξ ξ¯ z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 Proof. By applying the property of the determinant we easily prove that z = zj (t) are solutions of (51). The reciprocity we prove as follows. From the equations żj = a0 (t) + a1 (t)zj + a2 (t)z̄j + a3 (t)zj2 + a4 (t)zj z̄j + a5 (t)z̄j2 , for j = 1, . . . , 6, 20 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA or equivalently      1 z1 z̄1 ... z̄12 a0 ż1  1 z2 z̄2 ... z̄22  a1   ż2        1 z3 z̄3 ... z̄32  a3   ż3    =   .. .. .. ..  ..   ..   . . . ... .  .   .  1 z6 z̄6 ... z̄62 a6 ż6 we obtain that z1 z2 z3 z4 z5 z6 z̄1 z̄2 z̄3 z̄4 z̄5 z̄6 1 z12 z22 z32 z42 z52 z62 ∆0 (ω) a0 = := , ∆(ω) z̄12 z̄22 z̄32 z̄42 z̄52 z̄62 ∆(ω) z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 z6 z̄6 ż1 ż2 ż3 ż4 ż5 ż6 1 1 1 1 1 1 z̄1 z̄2 z̄3 z̄4 z̄5 z̄6 1 z12 z22 z32 z42 z52 z62 ∆1 (ω) a1 (t) = − := , ∆(ω) z̄12 z̄22 z̄32 z̄42 z̄52 z̄62 ∆(ω) z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 z6 z̄6 ż1 ż2 ż3 ż4 ż5 ż6 1 1 1 1 1 1 z1 z2 z3 z4 z5 z6 1 z12 z22 z32 z42 z52 z62 ∆2 (ω) a2 (t) = := , ∆(ω) z̄12 z̄22 z̄32 z̄42 z̄52 z̄62 ∆(ω) z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 z6 z̄6 ż1 ż2 ż3 ż4 ż5 ż6 .. .. .. .. . . . . By inserting into (48) we get ∆0 (ω) ∆1 (ω) ∆2 (ω) ∆3 (ω) 2 ∆4 (ω) ∆5 (ω) 2 ż = + z+ z̄ + z + z z̄ + z̄ , ∆(ω) ∆(ω) ∆(ω) ∆(ω) ∆(ω) ∆(ω) after some computations we obtain (51). The proof of (50) it is easy to obtain. Indeed from the equation DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 21 1 1 1 1 1 1 1 z z1 z2 z3 z4 z5 z6 z̄ z̄1 z̄2 z̄3 z̄4 z̄5 z̄6 (51) z2 z12 z22 z32 z54 z52 z62 = 0. z̄ 2 z̄12 z̄22 z̄32 z̄42 z̄52 z̄62 z z̄ z1 z̄1 z2 z̄2 z3 z̄3 z4 z̄4 z5 z̄5 z6 z̄6 ż 0 0 0 0 0 ż6 and developing this determinant with respect to the last row we get (20). Clearly if Xκ |z˙j |2 = 0, then differential equation (51) becomes ż = 0.  j=1 Proof of Proposition 7. The problem on the construction the polynomial planar vector field (48) of degree m from a given set of (m + 1)(m + 2)/2 solutions is solved in Proposition 7. The proof of this result is analogous to the proof of Proposition 15.  (κ) (κ) The set of functions ωκ = {z1 , . . . , zκ ) for 1 ≤ κ ≤ µ we call equivalent to the set ω1 = {(z1 (t), . . . , zκ (t))} if ∆j (ω1 ) ∆j (ωκ ) = . ∆(ω1 ) ∆(Υκ ) for j = 2, . . . , κ. Another interesting problem will be to determine the set of solutions ω such that ∆j (ω1 ) = aj = constants, for j = 1, . . . , 6. ∆(ω1 ) Remark 16. By considering that (m+1)(m+2)/2 < m2 for m = 2, 3 then the most general quadratic and cubic polynomial differential systems can be obtained from (19) taking zj = constants for m = 2, 3. 6.2. Complex polynomial differential equations of degree m with m + 1 real solu- tions. In this section we study the problem on the determination the complex polynomial differential system of degree m of the form (21) with m + 1 complex solution satisfying (22) and m + 1 real solutions. Proof of Proposition 8. Assume that (21) has m + 1 solutions such that (22) holds. Then from (21) it follows that (52) żj = λ0 + λ1 zj + . . . + λm zjm , for j = 1, . . . , m + 1 22 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA k X where λk = einα ak−n,n (t), , for k = 0, . . . , m. By solving (52) with respect to λk for n=0 k = 0, . . . , m, we get that ż1 z1 z12 ... z1m ż2 z2 z22 ... z2m 1 ż3 z3 z32 ... z3m λ0 = , ∆(ω) .. .. .. .. . . . ... . 2 m żm zm+1 zm+1 ... zm+1 (53) .. .. .. . . . 1 z1 z12 ... ż1 1 z2 z22 ... ż2 1 1 z3 z32 ... ż3 λm = , ∆(ω) .. .. .. .. . . . ... . 2 1 zm+1 zm+1 ... żm By considering that am,0 = λm − am−1,1 eiα − am−2,2 e2iα − . . . − a0,m eimα , am−1,0 = λm−1 − am−2,1 eiα − . . . − a0,m−1 ei(m−1)α , .. .. .. . . . a00 = λ0 , and inserting them (21) we get m X ż − λj z j = am−1,1 (z̄z m−1 − eiα z m ) j=0 +am−2,2 (z̄ 2 z m−2 − e2iα z m ) + . . . + a0,m (z̄ m − eimα z m )  + . . . + a0,1 (z̄ − eiα z) = (z̄ − eiα z) am−1,1 z m−1 +am−2,2 z m−2 (z̄ + eiα z) + . . . + a0,m (z̄ m−1 + z̄ m−2 eiα z  + . . . + ei(m−1)α z m−1 ) + . . . + a0,1 := (z̄ − eiα z)P (z, z, t). Hence in view of (53) we get that 1 1 1 1 ... 1 z z1 z2 z3 . . . zm+1 m X 1 z2 z12 z22 z32 2 . . . zm+1 ż − λj z j = .. .. .. .. .. . j=0 ∆(ω) . . . . ... . zm z1m z1m z1m ... m zm+1 ż ż1 ż2 ż3 ... żm+1 Therefore we obtain the proof of the proposition.  Corollary 17. The most general complex polynomial differential equations of degree m with m + 1 real solutions ω = (z1 (t), . . . , zm+1 (t)) such that the Vandermonde determinant (23) DIFFERENTIAL EQUATIONS WITH A GIVEN SET OF SOLUTIONS 23 is non-zero can be written as 1 1 1 1 ... 1 z z1 z2 z3 . . . zm+1 1 z2 z12 z22 z32 2 . . . zm+1 (54) .. .. .. .. .. = (z̄ − z)P (z, z̄, t), ∆(ω) . . . . ... . zm z1m z1m z1m ... m zm+1 ż ż1 ż2 ż3 ... żm+1 where P (z, z̄, t) is a convenient polynomial of degree m − 1. Proof. It follows from Proposition 8 with α = 0.  Remark 18. From Proposition 17 it follows the next remarks. (i) If a complex polynomial differential equation of degree m has m + 1 singular points such that z̄j = eiα zj , for j = 1, . . . , m + 1 , then it has the singular straight line z̄ − eiα z = 0, i.e. (1 − cos α)x + sin α y = 0. The proof of this result follows trivially from (24). Indeed, in this case differential equation (54) becomes ż = (z̄ − eiα z)P̃ (z, z, t). where P̃ (z, z, t) is a convenient polynomial of degree m − 1. (ii) If a complex polynomial differential equation of degree m has m + 1 real singular points, then it has the singular straight line z̄ − z = 0. The proof of this well known result follows trivially from (21). Indeed, in this case differential equation (54) becomes ż = (z̄ − z)P (z, z, t). (iii) If a complex polynomial differential equation of degree m has m + 2 real solutions, then we have 1 1 1 1 ... 1 zm+2 z1 z2 z3 . . . zm+1 2 zm+2 z12 z22 z32 2 . . . zm+1 (55) .. .. .. .. .. = 0. . . . . ... . m zm+2 z1m z1m z1m ... m zm+1 żm+2 ż1 ż2 ż3 ... żm+1 Indeed if z = zm+2 (t) is another real solutions of a complex polynomial differential equation of degree m, then from (54) it follows (55). Consequently zm+2 (t) is a solution of the generalized real Abel equation m X ẋ = aj (t)xj . j=0 6.3. Polynomial differential system of degree m > 3 with m2 singular points. We shall determine the most general polynomial differential system of degree m for which ω = {(z1 , z2 , . . . , zm2 )} where zj is constant for j = 1, . . . , m2 such that zj 6= zk for j, k = 1, . . . , m with j 6= k. Proposition 19. The most general differential polynomial differential system of degree m with m2 singular points can be represented as follows (56) ż = α11 (t)H11 (z, z̄, t) + α21 (t)H21 (z, z̄, t) + α22 (t)H22 (z, z̄, t) + . . . + ακκ (t)Hκκ (z, z̄, t), 24 J. LLIBRE, R. RAMÍREZ, AND N. SADOVSKAIA where ajn (t) is an arbitrary function for j, n = 1, . . . , κ > m, and Hjn (z, z̄, t) for j, n = 1, . . . , κ are polynomials of degree m elements of the κ × κ matrix H = Hm2 where h i Hr = Hr−1 (z, z̄, t), Hr−1 (zr , z̄r , t) , where [H, G] = HG − GH is the Lie bracket, r = 1, . . . , m2 and H0 = G(z, z̄, t) is an arbitrary κ × κ matrix whose are elements are arbitrary polynomials of degree m. h i Proof. Clearly that all elements of the matrix H1 = H0 (z, z̄, t), H0 (z1 , z̄1 , t) are vanishing when z = z1 , and all elements of the matrix hh i h ii H2 = H0 (z, z̄, t), H0 (z1 , z̄1 , t) , H0 (z2 , z̄2 , H0 (z1 , z̄1 , t) are vanishing when z = z1 , and z = z2 . By continuing this process we obtain that all elements of the matrix h i Hm2 = Hm2 −1 (z, z̄, t), Hm2 −1 (zm2 , z̄m2 , t) , are vanishing when z = z1 , . . . , z = zm2 . Consequently the function Φ = α11 (t)H11 + α12 (t)H12 + α21 (t)H21 + α22 (t)H22 + . . . + ακκ (t)Hκκ , is a polynomial of degree m for which the given m2 complex constants are it zeros. In short the proposition is proved.  Corollary 20. The most general differential polynomial differential system of degree m with m2 singular points can be represented as follows ˜ M (t)ξ T >, ż =< ξ(t), ˜ = ξ| where ξ = (1, z, z̄, z 2 , z z̄ 2 , z̄ 2 , . . . , z m , . . . , z̄ m ), ξ(t) , and M (t) is an z=z m2 (t), z̄=z̄m2 (t) ˜ M (t)ξ T (t) >= arbitrary (m+1)(m+2)/2×(m+1)(m+2)/2 antisymmetric matrix such that < ξ(t), j 0 with ξj = (1, z, z̄j , zj2 , z z̄j2 , z̄j2 , . . . , zjm , . . . , z̄jm ), for j = 1, . . . , m2 . Proof. 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Zhang, The 16th Hilbert problem on algebraic limit cycles, J. Differential Equations 251 (2011), 1778–1789. 1 Departament de Matemàtiques, Universitat Autònoma de Barcelona, 08193 Bellaterra, Barcelona, Catalonia, Spain. Email address: jllibre@mat.uab.cat 2 Departament d’Enginyeria Informàtica i Matemàtiques, Universitat Rovira i Virgili, Avin- guda dels Paı̈sos Catalans 26, 43007 Tarragona, Catalonia, Spain. Email address: rafaelorlando.ramirez@urv.cat 3Departament de Matemàtica Aplicada II, Universitat Politècnica de Catalunya, C. Pau Gar- gallo 5, 08028 Barcelona, Catalonia, Spain. Email address: natalia.sadovskaia@upc.edu