ISRN Applied MathematicsVolume 2011 (2011), Article ID 852656, 44 pagesdoi:10.5402/2011/852656

Research Article

Rational Solutions of Riccati Differential Equation with Coefficients Rational

Nadhem Echi

Department of Mathematics, Tunis El Manar University, Tunis 1002, Tunisia

Received 28 March 2011; Accepted 16 April 2011

Academic Editors: H.-T. Hu and X. Meng

Copyright © 2011 Nadhem Echi. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

This paper presents a simple and efficient method for determining the rational solution of Riccati differential equation with coefficients rational. In case the differential Galois group of the differential equation , is reducible, we look for the rational solutions of Riccati differential equation , by reducing the number of checks to be made and by accelerating the search for the partial fraction decomposition of the solution reserved for the poles of which are false poles of . This partial fraction decomposition of solution can be used to code . The examples demonstrate the effectiveness of the method.

1. Introduction

The quadratic Riccati differential equation: where , and are in a differential field . The quadratic Riccati differential equation is first converted to a reduced Riccati differential equation: where , with and .

Furthermore, we put , reduced Riccati differential equation (1.2) is converted to a second-order linear ordinary differential equation If we have a particular solution nonzero of then general solution is where , constant (see [1–5]).

In the paper, we base ourselves mainly on the work of Kovacic [2] where the differential Galois group of the differential equation is reducible and we take .

The case where every solution of is Liouvillian corresponds to the case where reduced Riccati differential equation has algebraic solution over . The case where differential Galois group is reducible corresponds to the case where the Riccati differential equation has the rational solution solution of . The solution of is rational fraction if and only if is rational fraction with simples poles, integers residues and negative degree.

The field is differential extension of by exponential of an integral and if then two solutions of are linearly independent over field of constants . The ordinary extension is differential extension of , by an integral. is Picard-Vessiot extension of for the differential equation (see [2, 3, 6–8]). The existence of rational solution of Riccati differential equation and research primitive of give all solutions of .

This paper presents a simple and efficient method for determining the solution of Riccati differential equation with coefficients rational. In case the differential Galois group of the differential equation , is reducible, we look for the rational solutions of Riccati differential equation , by reducing the number of checks to be made and by accelerating the search for the partial fraction decomposition of the solution reserved for the poles of which are false poles of . This partial fraction decomposition of solution can be used to code . The examples demonstrate the effectiveness of the method.

2. Form of Rational Solution of Equation:

Let be rational fraction and let be the rational solution of Riccati differential equation .

2.1. Study in the Pole of Multiplicity of

We put We have Thus

Case 1 (). The function defines and is equal to at .Thus is pole of multiplicity of where

Case 2 (). We have The function defines and is equal to at .

Situation 1 c is double pole of and the residue of at have tow possibility values as follows: Thus, is double pole of and the residue of at simple pole equals where

Situation 2 is double pole of and the residue of at is .

Situation 3 is a simple pole of or is not pole of and the residue of at simple pole equals 1.

Proposition 2.1. Let such as . (1)The fraction: with and being polynomials relatively prime where , and are polynomials relatively prime pairwise. , and are with simple roots, and is without simple root (2)(a)Let . pole of multiplicity of .(b) simple pole of with residue , the residue of where equals where (c) simple pole of with residue .(d) simple pole of with residue or pole of and false pole of .

Corollary 2.2. One assumes that with and polynomials being relatively prime, where , and are polynomials relatively prime pairwise, , and are with simple roots, and is without simple root a rational fraction . Verify is the shape with being monic polynomial with simple roots and the roots are false poles of .

2.2. Study in the Infinity

Let rational fraction such as .

Case 1. We assume that .We have , and thus . We put We have Thus and .If: then the sum of residues of : .If: then the sum of residues of : where

Case 2. We assume that . constant . We put We have So constant equals :

Case 3. We assume that .We put with being rational fraction defined at 0. The scalar is the dominant coefficient of . We have Thus, So and the dominant coefficient of .

Proposition 2.3. Let such as . (1). Thus and (2). In the case is square root of : (3)We have . In the case(a); (b)the dominant coefficient of ) is square root of .

2.3. Determination of ;

We assume that is a rational fraction of degree and such as .

Let be the dominant coefficient of . Thus if tends to : The Taylor's expansion of order at 0 We have We have with rational fraction defined at 0, and Thus Imply

Proposition 2.4. Let be a rational fraction of degree , such as and the dominant coefficient of . If Then where

3. Determination of Partial Fraction Decomposition

Let be rational fraction with and polynomials being relatively prime, where , and polynomials being relatively prime pair-wise, , and with simple roots and without simple root Let be rational fraction. Verify .

3.1. Case and

We have .

This case corresponds to the fact that one pole of is simple or double with

Proposition 3.1. One assumes and . One has: with being positive integer of parity against that of with polynomial of degree

Proof. We have The sum of residues of equal with . In particular with .Remark 3.2. If then , , , and .

Proposition 3.3. One assumes and . (1) square root of such as is positive integer of same parity as .(2)One has with being polynomial of degree

Proof. We have We have Thus If is constant then and are constant.If is nonconstant then is not polynomial thus is a positive integer of same parity as .

Proposition 3.4. We assume and . Let be the dominant coefficient of and one considers the Taylor's expansion at infinity: (1) where is a positive integer, , and same parity of .(2)One has with being polynomial of degree and the dominant coefficient of where

Proof. We have Thus with positive integer. Thus , and have the same parity.

3.2. Case: and

This case corresponds to the fact that a pole of is simple or double with only double pole such as:

Proposition 3.5. Consider (3.21) and let be a rational fraction. Verify . One assumes . Accordingly, in view of (2.11) and (2.16) one has with being polynomial of degree where one half positive integer of same parity as .

Proof. We have Thus

Proposition 3.6. Consider (3.21) and let be a rational fraction. Verify: . One assumes . One has: with and being polynomial of degree where is positive integer of parity against that of .

Proof. We have Thus

Proposition 3.7. Consider (3.21) and let be a rational fraction. Verify . One assumes . Let be the dominant coefficient of , let be the dominant coefficient of , and Taylors expansion: One has with being polynomial of degree:

Proof. We have Thus Thus

3.3. Case: ( and ) or ( and )

and are polynomials being relatively prime. Thus there are only two polynomials and such as Thus We have Because does not have simple roots verify or . Thus . Thus We consider the rational fraction where

Proposition 3.8. One assumes or . (1)If then: and .(2)If then: where .(3)For all root of one has (4)For all root of of multiplicity one has

Proof. (1) is from negative degree.(2) .(3) Let be root of (4) Let be root of

Lemma 3.9. Let be nonzero rational fraction. is a finished set such as (1)There exists open connected set on which one has a square root holomorphic of , containing for every a half-right closed by origin . (2)If, besides, is from even degree then one can choose of complementary compact.

Proof. We fix is not an element of . Let be a finished set containing roots, poles of and . They consider all rights linked to the different pairs of points of . They choose a point not being on these rights. For all , straight line which joins right and does not contain any other point of .Case 1 ( even). Replacing by rational fraction .Assume . We put . is compact-connected.We put . is open at infinity such as for all reaching right in private of is contained in .If is a shoelace of then is included in one connected component of . Thus: Thus there exist the primitive of and the determination of logarithm of and the square root of in .Case 2 ( odd). They use the previous case in replacing by rational fraction .

Notation 3.3. We choose a square root of the polynomial of even degree on a connected open at infinity, roots of and root of . We put

Proposition 3.10. One assumes or . One has where .

Proof. For root of multiplicity of Thus Partial fraction of , associated in root of , minus , is in sign meadows that of .For root of Thus Partial fraction of , associated in root of , minus , is in sign meadows that of . Thus

Proposition 3.11. One assumes and or . One has where being a polynomial of degree

Proof. We have Thus

Proposition 3.12. One assumes . One has where and being a polynomial of degree:

Proof. We have Thus

Proposition 3.13. One assumes . One has where being a polynomial of degree:

Proof. Let be dominant coefficient of , and we have sum of residues of , Thus Thus

3.4. Case and

and are polynomials being relatively prime. Thus there are only two polynomials and such as We have We consider the rational fraction:

Proposition 3.14. One assumes . (1)If then: and .(2)If then: where . (3)For all root of of multiplicity one has

Proof. (1) is from negative degree.(2) (3) Let be root of :

Notation 3.4. We choose a square root of the polynomial of even degree: on a connected open at infinity, root of . We put

Proposition 3.15. One assumes . One has where and being a polynomial of degree:

Proof. Thus

Proposition 3.16. One assumes . One has where and being a polynomial of degree

Proof. We have Thus

Proposition 3.17. One assumes . One has where and being a polynomial of degree

Proof. Let dominant coefficient of . . , where . Thus Thus Thus

4. Recurrent Method at Infinity

4.1. Presentation of as Determinant

Proposition 4.1. Let be complex constants. We put For , we put and (1)One has (2)One has

Proof. are polynomials at with real coefficients. Thus, and are polynomials at with real coefficients. Thus, to have both identities we can assume are reals. Furthermore, we can restrict to the open of Zariski distinct real nonzero.We have for all .We put We put The scalar product The Gram matrix of is Thus We put We notice The th column of Gram of matrix which defines is Thus where We obtain The cofactor of is . The adjoint of is nonzero. We prove that adjoint of of rank 1. . Because: and for all . Thus, the th column of is is a linear combination of other column.We consider the matrix It is the determinant of Gram of Thus But Thus . Or is scalar nonzero, is invertible matrix.With symmetric matrix, Com symmetric matrix, we have Thus Or is invertible and Com, Com of rank 1. Thus For , we put: , Thus: Or is monic polynomial and We obtain

4.2. Taylor Expansion of

Proposition 4.2. Let be pairwise distinct constants. (1)We put, for : and (2)We put . Taylor expansion on the neighborhood of infinity: (a)(b)

Proof. We have .(1)(2)We put . We have as Thus

4.3. Research of

Let be a rational fraction and where .

We put where and monic polynomial of degree . We have with and polynomials as .

Case 1 (). We have and . Taylors expansion of , and at order .We obtain the Taylors expansion of at order .The Taylors expansion equals the Taylors expansion of . Let . In Taylors expansion of (see Proposition 4.2) the coefficient of is: . For , coefficient of use: .Besides, we have Thus, Taylors expansion of and the coefficient of is determine by recurrence at .

Case 2 (). The coefficient of in Taylors expansion of is By recurrence at condition We have with .For , to be at function of . If then, the coefficient of is nonzero implied that there is no solution, or the coeffcient of is zero implied that is arbitrary and ; depend in a unique way. is polynomial determined by by determinant formulae; the problem of nonunique suite is put if is positive integer equal to or less than , then is square of integer and

Example 4.3. In this example we consider the Riccati differential equation (1.2) where We have , , and .We can assume , . Thus .Thus , and . is a solution of Riccati equation if and only if verifies Thus where Taylors expansion of at As . Thus For , equivalent to .For all, , . For example Thus, the polynomial is partner to Thus

5. Method of Last Minor

Let be a rational fraction and as: .

solution of (1.2) if and only if verifies We choose a complex number not pole of and we use the expression of polynomials following the powers of .

For the sake of simplicity, in the following we assume that . Constant coefficient constant of is nonzero.

Denote by , and coefficients of , in , and , respectively, . If then .

5.1. Case

We put .

If then are pole unique, it is simple or double. If simple pole of then of degree −1 and (1.2) has no rational solution.

If double pole of then , and .

Thus,(5.1) The solutions where nonzero constant.

We have, an infinity of the other rational solution when , with the same parity of .

We assume and we consider a polynomial of degree that verifies (5.1).

Proposition 5.1. One assumes and verifies (5.1). (1)One has (2)One has .(3)One has

Proof. (1) . If then If then (2) We have and coefficients of in (5.1).(3) Accordingly, in view of .

We put The th element of is of degree equal to or less than and of coefficient For , the coefficient of is zero if and only if of degree equal to or less than .

For the coefficient of more low degree of is coefficient of equal .

Let matrix of row and th row is row of coefficients of in basis of .

is linearly independent system.

Equation (5.1) gives obtained Thus, the matrix is rank or . Accordingly, existence of correspondent is linearly dependent in row of .

Let roots of distinct.

We put, for : and The column vector of column is linearly independent.

The equation of hyperplane engendered by vectors is The coefficient of is and as equation where scalar satisfying The derivative of is derived from last column. Equation (5.1) is equivalent to: Thus, for all , . With application of Taylor, this is equivalent to columns of matrix in hyperplane , The polynomial is partner to Thus For , we put where .

The matrix partner to is triangular where th entry diagonal equal . The coefficient is nonzero except possibly for a single value of . Thus, rank of is or .

For rank, (5.1) as solution is equivalent to which is linearly independent.

Hence, we assume rank. In that case and are linear combinations of . One of both combinations has to express . Thus we have two situations.

Situation 1 linear combination of where coefficient of is nonzero. By replacement of , we obtain a linear combination of .Thus, is linear combination of . libre system where for 0. Thus: .If where is constant, then As , thus for all , equivalent to . Thus .

First Way of Having : has to be the sum of simple elements of the shape where is greater or equal to 1.Thus and .For all root of where .Thus Thus is primitive nonzero in roots of .

Second Way of Having : where is constant . The relations which are linearly dependent between and are Thus we have an infinity of solutions: where is arbitrarily constant.

Situation 2 is linear combination of where coefficient of is nonzero and is linear combination of . Thus, is linear combination of . Thus, . is libre system.If then As . Thus, for all , equivalent to and .

First Way of Having : We put . , Thus with ; rational function.The coefficient of in is zero. (; ).

Second Way of Having thus is arbitrarily constant.

5.2. Case

We put and . We have Thus Thus, and where is dominant coefficient of .

We put The th element of is is of degree equal to or less than where coefficient of equal . For , the th element of is of degree: . The th element of is of degree equal to or less than .

Proposition 5.2. One puts where . (1)(2) libre system and where constants. (3) is solution if and only if

Example 5.3. In this example we consider the Riccati differential equation (1.2) where . We have

Study at (−1)Laurent series development at −1:

Study at InfinityWe have where , Thus and , where , Case and are to be rejected because we obtain negative values of . (i)If and then: .(ii)If and then: .

Case 1 ( and ). We have the following where . Research of coefficients of , , where and . Thus ,

Case 2 ( and ). We obtain the rational fraction It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.

Example 5.4. In this example we consider the Riccati differential equation (1.2) where and . We have Laurent series development at 1: Laurent series development at −2: We have Thus where Cases and are to be rejected because we obtain negative values of .

Case 1. If then , Research of coefficients of , , where , , . Thus, ,

Case 2. If , , then: , It is not a solution because the sum of this fraction with the solution already found is not the logarithmic derivative of a rational fraction.

Case 3. If , , then , Research of coefficients of , We consider the minor obtained by column vectors , and 9. In . Accordingly, row 6 is not linear combination of other row and So no solution .

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