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Polynomial Solutions of Equivariant Polynomial Abel Differential Equations

  • Jaume Llibre EMAIL logo and Clàudia Valls
Published/Copyright: December 16, 2017

Abstract

Let a(x) be non-constant and let bj(x), for j=0,1,2,3, be real or complex polynomials in the variable x. Then the real or complex equivariant polynomial Abel differential equation a(x)y˙=b1(x)y+b3(x)y3, with b3(x)0, and the real or complex polynomial equivariant polynomial Abel differential equation of the second kind a(x)yy˙=b0(x)+b2(x)y2, with b2(x)0, have at most 7 polynomial solutions. Moreover, there exist equations of this type having this maximum number of polynomial solutions.

MSC 2010: 34A05; 34C05; 37C10

1 Introduction and Statement of the Main Results

Abel differential equations of the first kind, namely,

(1) a ( x ) y ˙ = b 0 ( x ) + b 1 ( x ) y + b 2 ( x ) y 2 + b 3 ( x ) y 3

with b3(x)0, appear in many textbooks of ordinary differential equations as one of the first non-trivial examples of nonlinear differential equations, see, for instance, [11]. Here the dot denotes the derivative with respect to the independent variable x. If b3(x)=b0(x)=0 or b2(x)=b0(x)=0, the Abel differential equation reduces to a Bernoulli differential equation, while if b3(x)=0, the Abel differential equation reduces to a Riccati differential equation.

Abel differential equations of the form (1) have been studied intensively, either by calculating their solutions (see, for instance, [8, 12, 14, 15]), or by classifying their centers (see [2, 3, 4]), and recently in [7, 9, 10, 13], the polynomial solutions of the differential equation y=i=0nai(x)yi were studied as well.

The analysis of particular solutions (as polynomial or rational solutions) of differential equations is important for understanding the set of their solutions. In 1936 Rainville [16] characterized the Riccati differential equation y˙=b0(x)+b1(x)y+y2, with b0(x) and b1(x) being polynomials in the variable x, having polynomial solutions.

Campbell and Golomb [5] in 1954 provided an algorithm for determining the polynomial solutions of the Riccati differential equation a(x)y=b0(x)+b1(x)y+b2(x)y2, where a,b0,b1,b2 are polynomials in the variable x. Behloul and Cheng [1] in 2006 gave a different algorithm for finding the rational solutions of the differential equation a(x)y=i=0nbi(x)yi, where a,bi are polynomials in the variable x.

Here we consider the Abel differential equation (1), where a(x)𝔽[x]{0}, bi(x)𝔽[x], i=0,1,2,3, b3(x)0, and 𝔽=,, with 𝔽[x] being the ring of polynomials in the variable x with coefficients in 𝔽. We also assume that a(x) is not constant. The case where a(x) is constant has been studied in [10]. We say that the Abel differential equation (1) has degree η.

Equation (1) is reversible with respect to the change of variables (x,y)(x,-y) if the following equation coincides with equation (1):

- a ( x ) y ˙ = - ( b 0 ( x ) - b 1 ( x ) y + b 2 ( x ) y 2 - b 3 ( x ) y 3 ) .

In particular, this implies b1(x)=b3(x)=0, and since b3(x)=0, we do not consider these reversible differential equations.

The Abel differential equation (1) is equivariant with respect to the change of variables (x,y)(x,-y) if the following equation coincides with equation (1):

- a ( x ) y ˙ = b 0 ( x ) - b 1 ( x ) y + b 2 ( x ) y 2 - b 3 ( x ) y 3 .

This implies b0(x)=b2(x)=0. In this paper first we focus our study in these kind of equivariant polynomial Abel equations, i.e., in equations of the form

(2) a ( x ) y ˙ = b 1 ( x ) y + b 3 ( x ) y 3 .

Theorem 1.

The real or complex equivariant polynomial Abel differential equations, with b3(x)0 and a(x) being non-constant, have at most 7 polynomial solutions. Moreover, there exist equations of this type having this maximum number of polynomial solutions.

The proof of Theorem 1 is given in Section 2.

Our second objective in this paper is on the Abel differential equations of the second kind, i.e., on the equations of the form

(3) a ( x ) y y ˙ = b 0 ( x ) + b 1 ( x ) y + b 2 ( x ) y 2 ,

where a(x),bi(x)𝔽[x] for i=0,1,2, with a(x) and b2(x) being non-zero. We also consider the ones that are equivariant with respect to the change (x,y)(x,-y). Then we have that b1(x)=0, and so equation (3) becomes

(4) a ( x ) y y ˙ = b 0 ( x ) + b 2 ( x ) y 2 .

We also assume that b2(x)0 (otherwise the differential equation would be of separation of variables) and that a(x) is not constant, since the case where a(x) is constant has been studied in [7]. We say that equation (4) is the equivariant polynomial Abel differential equation of the second kind.

Theorem 2.

The real or complex equivariant polynomial Abel differential equations of the second kind, with b2(x)0 and a(x) being non-constant, have at most 7 polynomial solutions. Moreover, there exist equations of this type having this maximum number of polynomial solutions.

The proof of Theorem 2 is given in Section 3.

2 Proof of Theorem 1

First we recall that if y(x)0 is a solution of equation (2), then -y(x) is too, which is different from y(x).

Lemma 3.

Let y0(x)0, y1(x) and y2(x) be polynomial solutions of equation (2) such that y1(x)0, y2(x)0 and y2(x)-y1(x). Set y1(x)=g(x)y~1(x) and y2(x)=g(x)y~2(x), where g=gcd(y1,y2). Then, except the solution y=0, all the other polynomial solutions of equation (2) can be expressed as

(5) y 0 ( x ; c ) = ± y ~ 1 ( x ) y ~ 2 ( x ) g ( x ) ( c y ~ 1 2 ( x ) + ( 1 - c ) y ~ 2 2 ( x ) ) 1 / 2 ,

where c is a constant and (cy~12(x)+(1-c)y~22(x))1/2 is a polynomial.

Proof.

Let y be a non-zero polynomial solution of equation (2). The functions z0=1/y02, z1=1/y12 and z2=1/y22 are solutions of a linear differential equation and satisfy

- a ( x ) z ˙ i = 2 b 1 ( x ) z i + 2 b 3 ( x ) , i = 0 , 1 , 2 .

Therefore, we have

z ˙ 0 ( x ) - z ˙ 1 ( x ) z 0 ( x ) - z 1 ( x ) = z ˙ 2 ( x ) - z ˙ 1 ( x ) z 2 ( x ) - z 1 ( x ) .

Integrating this equality, we obtain

z 0 ( x ) = z 1 ( x ) + c ( z 2 ( x ) - z 1 ( x ) ) ,

with c being an arbitrary constant. So the general solution of equation (2) is

y 0 2 ( x ) = 1 z 0 ( x ) = 1 z 1 ( x ) + c ( z 2 ( x ) - z 1 ( x ) )
= y 1 2 ( x ) y 2 2 ( x ) c y 1 2 ( x ) + ( 1 - c ) y 2 2 ( x ) = y ~ 1 2 ( x ) y ~ 2 2 ( x ) g ( x ) 2 c y ~ 1 2 ( x ) + ( 1 - c ) y ~ 2 2 ( x ) ,

with c being an arbitrary constant. ∎

In view of Lemma 3, if y1(x) and y2(x) are polynomial solutions of equation (2) such that y1(x)0, y2(x)0 and y2(x)-y1(x), then any other polynomial solution different from them is of the form given in (5) for some appropriate constant c such that c{0,1}. In particular, cy~12(x)+(1-c)y~22(x) or y~12(x)+(1-c)y~22(x)/c is a square of a polynomial P, and P divides g. In view of Theorem 7 (see Appendix A), there is at most one constant c{0,1} such that cy~12+(1-c)y~22 is a square of a polynomial, meaning that equation (2) has at most seven different polynomial solutions 0, ±y1, ±y2 and y0, as in (5).

Example 4.

We consider the equivariant polynomial Abel differential equation (2) with

a ( x ) = - 2 x + 48 x 3 - 768 x 7 + 512 x 9 ,
b 1 ( x ) = 2 ( - 1 + 96 x 4 - 1536 x 6 + 768 x 8 ) ,
b 3 ( x ) = 64 .

This equation has the following 7 polynomial solutions:

y 1 ( x ) = 0 ,
y 2 , 3 ( x ) = ± ( 2 2 x 3 + x 2 ) ,
y 4 , 5 ( x ) = ± ( x - 4 x 3 ) ,
y 6 , 7 ( x ) = ± ( 1 4 2 - 2 2 x 4 ) .

3 Proof of Theorem 2

First we recall that if y(x)0 is a solution of equation (4), then -y(x) is too, which is different from y(x).

Lemma 5.

Let y0(x)0, y1(x) and y2(x) be polynomial solutions of equation (4) such that y1(x)0, y2(x)0 and y2(x)-y1(x). Set y1(x)=g(x)y~1(x) and y2(x)=g(x)y~2(x), where g=gcd(y1,y2). Then, except the solution y=0, all the other polynomial solutions of equation (4) can be expressed as

(6) y 0 ( x ; c ) = ± g ( x ) ( c y ~ 1 2 ( x ) + ( 1 - c ) y ~ 2 2 ( x ) ) 1 / 2 ,

where c is a constant.

Proof.

Let y be a non-zero polynomial solution of equation (2). The functions z0=y02, z1=y12 and z2=y22 are solutions of a linear differential equation and satisfy

a ( x ) z ˙ i = 2 b 0 ( x ) + 2 b 2 ( x ) z i , i = 0 , 1 , 2 .

Therefore, we have

z ˙ 0 ( x ) - z ˙ 1 ( x ) z 0 ( x ) - z 1 ( x ) = z ˙ 2 ( x ) - z ˙ 1 ( x ) z 2 ( x ) - z 1 ( x ) .

Integrating this equality, we obtain

z 0 ( x ) = z 1 ( x ) + c ( z 2 ( x ) - z 1 ( x ) ) ,

with c being an arbitrary constant. So the general solution of equation (2) is

y 0 2 ( x ) = z 0 ( x ) = z 1 ( x ) + c ( z 2 ( x ) - z 1 ( x ) )
= ( 1 - c ) y 1 2 ( x ) + c y 2 2 ( x )
= g 2 ( x ) ( ( 1 - c ) y ~ 1 2 + c y ~ 2 2 ) ,

with c being an arbitrary constant. ∎

In view of Lemma 3, if y1(x) and y2(x) are polynomial solutions of equation (4) such that y1(x)0, y2(x)0 and y2(x)-y1(x), then any other polynomial solution is of the form (6) for some appropriate constant c. In particular, cy~12(x)+(1-c)y~22(x) is a square of a polynomial P. Again, in view of Theorem 7, this c is unique, and we conclude that equation (4) has at most seven different polynomial solutions 0, ±y1, ±y2 and y0, as in (6).

Example 6.

We consider the equivariant polynomial Abel differential equation of the second kind (4) with

a ( x ) = 2 x 4 - 3 x 2 + 1 8 ,
b 0 ( x ) = x 2 - 8 x 5 ,
b 2 ( x ) = 4 x 3 - 3 x .

This equation has the following 7 polynomial solutions:

y 1 ( x ) = 0 ,
y 2 , 3 ( x ) = ± ( 2 x 2 - 1 2 ) ,
y 4 , 5 ( x ) = ± ( 2 x 2 + 1 2 2 ) ,
y 6 , 7 ( x ) = ± 2 x .


Communicated by Yiming Long


Funding statement: The first author is partially supported by a FEDER-MINECO grant MTM2016-77278-P, a MINECO grant MTM2013-40998-P, and an AGAUR grant number 2014SGR-568. The second author is partially supported by FCT/Portugal through UID/MAT/04459/2013.

A Theorem 7

The following theorem is proved in [6]. For completeness we reproduce their proof here.

Theorem 7.

Let p,qC[x] be coprime polynomials (i.e., gcd(p,q)=1) such that

(7) p 2 + q 2 = r 2 𝑎𝑛𝑑 p 2 + α 2 q 2 = s 2 ,

with r,sC[x] and αC. Then either α=0 or α2=1.

Proof.

It follows from (7) that

p 2 = r 2 - q 2 = ( r - q ) ( r + q ) and p 2 = s 2 - α 2 q 2 = ( s + α q ) ( s - α q ) .

Let

A = r + q , B = r - q , C = s + α q and D = s - α q .

We get that gcd(A,B)=gcd(C,D)=1 because gcd(p,q)=1. Moreover, α(A-B)=C-D.

We assume that α0 and we will show that α2=1. The proof that α2=1 will be given by induction in the degree of the polynomial p.

First note that the degree of p must be greater than or equal to 1 because if the degree of p is zero, then p2=r2-q2=(r+q)(r-q) has degree 0 which will imply that also r and q have degree zero contradicting the hypotheses.

If the degree of p is one, then we must have p2=AB=λp2λ for some non-zero λ and p2=CD=μp2μ for some non-zero μ. From (7), we conclude that

α ( λ - p 2 λ ) = ± ( μ - p 2 μ ) α λ μ = p 2 ( α λ 1 μ ) .

Hence, αλμ=0, which implies μ=±αλ, and also αλ1μ=0, which implies α2=1. So, the proof follows when the degree of p is one.

Now we assume that the result is true for all p with degree less than or equal to n-1, and we will prove the result for any p with degree n. Note that

degree ( A ) + degree ( B ) = 2 d e g r e e ( p ) , degree ( C ) + degree ( D ) = 2 d e g r e e ( p ) .

We consider four different cases. Case 1: At least one of the polynomials A,B,C or D has degree strictly between 0 and n. Assume that 0<degree(A)<n (the other cases can be done exactly in the same way). In this case, using that AB=CD, the irreducible components of A have to be in C or D, and gcd(A,C) and gcd(A,D) have degree less than n and at least one of them has positive degree. Let c1=gcd(A,C) and assume that 0<degree(c1)<n. Also let c2=gcd(B,D). Note that gcd(c1,c2)=1. From AB=CD, we get that there exist a,b[x] and a non-zero λ such that

A = c 1 a , B = b c 2 , C = λ b c 1 and D = 1 λ a c 2 .

It follows from (7) that

(8) ( λ b - α a ) c 1 = ( a λ - α b ) c 2 .

Note that gcd(a,b)=1. So, (λb-αa,aλ-αb)=1 and it follows from (8) that there exists β such that

λ b - α a = β c 2 and a λ - α b = β c 1 .

Assuming that α21, we will reach a contradiction. If α21, we get that

a = ( λ c 1 + α c 2 ) β 1 - α 2 and b = ( α c 1 + 1 λ c 2 ) β 1 - α 2 .

Since p2=c1c2ab and any pair among c1, c2, a and b has no common divisors, we conclude that c1,c2 and ab are perfect squares. Hence, c1+αλc2 and c1+1αλc2 are also perfect squares. But since 0<degree(c1)<n and gcd(c1,c2)=1, by the induction hypotheses, we get α2=1, which is not possible. Case 2: The degrees of A and C are both 2n, and the degrees of B and D are both 0. This case can be done in a similar way as the case in which the degree of p is one, with decompositions p2=AB=p2λλ=CD=p2μμ for some non-zero λ,μ. Again, using (7), we conclude that μ=λα and α2=1. Case 3: The degree of A is 2n, the degree of B is 0, and the degrees of C and D are both n. In this case we have that degree(A-B)=2n and degree(C-D)n, which is impossible in view of (7). Case 4: The degrees of A, B, C, D are all n. Let c1=gcd(A,C) and c2=gcd(B,D). If degree(c1)=degree(c2)=n, then A=λC and B=1λD for some non-zero λC, and then, using (7), we readily conclude that α2=1. So, we can assume that degee(c1)<n. If degree(c1)>0, then the proof follows as in case 1. If degree(c1)=0, then gcd(A,D)=A and gcd(B,C)=B, and the proof follows as the subcase above degree(c1)=degree(c2)=n. ∎

Acknowledgements

The authors wish to thank to the reviewer and also to the authors of [6] for pointing out an error in a previous version of this paper.

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Received: 2017-04-28
Revised: 2017-11-10
Accepted: 2017-11-27
Published Online: 2017-12-16
Published in Print: 2018-08-01

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