QUASI-PERIODIC SOLUTIONS TO THE RICCATI DIFFERENTIAL EQUATION

1. PRELIMINARY Let the Riccati differential equation ) 0 ) ( ( 0 ) ( ) ( ) ( ) ( ) ( ) ( 2 ≠ = + + + ′ x g x h x y x g x y x f x y (1) be given. We want to find QPS ) (x y y = for (1), i.e. to find the solution that satisfies the relation ) ( ) ( ) ( )) ( , ( ) ( x y x x y x x x y λ ω λ ω = = + , y D x x ∈ +ω , (2) where ) (x ω ω = is QP and ) (x λ λ = is QPC for the function ) (x y y = . The following theorem holds. Theorem 1.1. If DE (1) has QPS ) (x y y = with QP ) (x ω ω = and QPC ), (x λ then it is reduced to the algebraic equation with respect to the QPS y 56 M. Kujumdžieva Nikoloska, J. Mitevska Contributions, Sec. Math. Tech. Sci., XXVII–XXVIII, 1–2 (2006–2007), pp. 55–66 + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ′ − ′ ′ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ′ − ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 1 2 2 x y t f x x f t x t x x y t g x x g x t λ λ λ λ λ ) ( , 0 ) ( ) ( ) ( x x t t h x h t x ω λ + = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ′ − + . ) 3 ( Proof. Under the conditions of the theorem we have the system


PRELIMINARY
Let the Riccati differential equation be given.We want to find QPS ) (x y y = for (1), i.e. to find the solution that satisfies the relation where ) (x is QPC for the function ) (x y y = .
The following theorem holds.
then it is reduced to the algebraic equation with respect to the QPS y )

(
Proof.Under the conditions of the theorem we have the system Subtituting ( 5) and (6) in the second equation of the system (4), after short transfomations, we obtain (3).Remark 1.1.In general, solving equation ( 3) is not a simple problem and we can solve it only in some special cases.So, here we consider QPS for (1) with a constant QP and a constant QPC. and a constant QPC λ > 0 .Then it is reduced, with respect to QPS y, to the equation: we obtain eq. ( 7).
This equation has a particular solution and QPC 1, e respectively, according to the Theorem 2.2.every solution for the given equation is QPF.Indeed, its gen-   Proof.It can be proved in a similar manner as the Theorem 2.2.

Remark 2.3
The general solution for DE (1) is QPS to the DE ( 1) is also QPS to the equation where ( ) .Thus, from (11) we obtain (9).So, since the solution ( 9) is also the solution to DE (1), we obtain that the coefficients have to satisfy the relation (10).
Proof.From the relation (10) we have ( Since, under he conditions of the Theorem 2.5., QPS for DE (1) , respectively, and satisfy the condition (10).Thus, according to the Theorem 2.5, QPS for the given DE is ), or using (12): ( ) The last equation has solutions Since QPS to DE (1) is QPS for eq.( 14), it satisfies the equation 0 Thus we have: If we compare the obtained solution with the solution (16) or substituting it in the eq.( 14    where p and q are gien by (15).
Under the conditions of the Theorem 2.6., solving eq. ( 18) as a linear DE, we can find all QPS for DE (1).They are given by the formula Remark 2.5.Some examples in [4] are special cases from the obtained results in this paper.

eTheorem 2 . 5 .
respectively, according to the Theorem 2.4.every solution is QPF.Indeed, the general solution is Let the DE (1) have one QPS with QP c Then the QPS for DE (1) is

Corollary 2 . 1 .
Under the conditions of the Theorem 2.5.QPS for DE (1) is given by

Example 2 . 4 .
The Under the conditions of the theorem QPS for DE (1) and eq.(7) is also QPS to the equation 0

4 .
From eq. (14), after short transformations, we obtain that QPS for DE (1) is also a solution to the equation 0 It can be proved in a similar manner as the previous theorem.