Hindawi Publishing Corporation ξ e Scientiο¬c World Journal Volume 2014, Article ID 401741, 8 pages http://dx.doi.org/10.1155/2014/401741
Research Article New Conditions for Obtaining the Exact Solutions of the General Riccati Equation Lazhar Bougoffa Department of Mathematics, Faculty of Science, Al Imam Mohammad Ibn Saud Islamic University (IMSIU), P.O. Box 90950, Riyadh 11623, Saudi Arabia Correspondence should be addressed to Lazhar Bougoffa; [email protected] Received 20 April 2014; Accepted 20 July 2014; Published 18 August 2014 Academic Editor: Kannan Krithivasan Copyright Β© 2014 Lazhar Bougoffa. 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. We propose a direct method for solving the general Riccati equation π¦σΈ = π(π₯)+π(π₯)π¦+β(π₯)π¦2 . We first reduce it into an equivalent equation, and then we formulate the relations between the coefficients functions π(π₯), π(π₯), and β(π₯) of the equation to obtain an equivalent separable equation from which the previous equation can be solved in closed form. Several examples are presented to demonstrate the efficiency of this method.
1. Introduction
A set of solutions to the Riccati equation is then given by
In realistic investigations of the dynamics of a physical system, nonlinearity may often be in the form of the wellknown Riccati equation: π¦σΈ = π (π₯) + π (π₯) π¦ + β (π₯) π¦2 .
(1)
The Riccati equation is used in different areas of physics, engineering, and mathematics such as quantum mechanics, thermodynamics, and control theory. It also appears in many engineering design simulations. The first well-known result in the analysis of the Riccati equation is that [1, 2] if one particular solution π¦1 can be found to (1), then the general solution is obtained as π¦ = π¦1 + π’,
(2)
where π’ satisfies the corresponding Bernoulli equation π’σΈ β (π (π₯) + 2β (π₯) π¦1 ) π’ = β (π₯) π’2 .
(3)
The substitution that is needed to solve this Bernoulli equation is π§=
1 . π’
(4)
1 π¦ = π¦1 + . π§
(5)
The second important result in the analysis of the solution of the Riccati equation is that the general equation can always be reduced to a second-order linear differential equation of the form [3, pages 23β25] π’σΈ σΈ β π (π₯) π’σΈ + π (π₯) π’ = 0,
(6)
βσΈ (π₯) , β (π₯) π (π₯) = π (π₯) β (π₯) .
(7)
where π (π₯) = π (π₯) +
A solution of this equation will lead to a solution π¦=β
1 π’σΈ β (π₯) π’
of the original Riccati equation.
(8)
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To our knowledge it is often very difficult, if not impossible, to find closed form solutions of such nonlinear differential equations. But a number of solutions of the Riccati equation can be obtained by assuming that the coefficients π(π₯), π(π₯), and β(π₯) satisfy certain constraints. Indeed, closed-form solutions are known if the following condition is satisfied [4]:
which yields the equation π’σΈ = π0 (π₯) + π1 (π₯) π’2 ,
(14)
where π0 (π₯) = πβ β« π(π₯)ππ₯ π(π₯) and π1 (π₯) = πβ« π(π₯)ππ₯ β(π₯). Now, let
π (π₯) + π (π₯) + β (π₯) = (log
πΌ(π₯) σΈ πΌ (π₯) β π½ (π₯) (πΌ (π₯) β (π₯) β π½ (π₯) π (π₯)) , ) β π½(π₯) πΌ (π₯) π½ (π₯) (9)
where πΌ(π₯) and π½(π₯) are arbitrary differentiable functions in πΌ β R with πΌ(π₯)π½(π₯) > 0. A new integrability condition for the Riccati equation was presented in [5] under the following constraint: π (π₯) =
πΉ (π₯) β π (π₯) β« πΉ (π₯) ππ₯ + πΆ0
σ΅¨σ΅¨ π (π₯) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ π’ = β σ΅¨σ΅¨σ΅¨ 0 σ΅¨V, σ΅¨σ΅¨ π1 (π₯) σ΅¨σ΅¨σ΅¨
and consider the case π0 (π₯)/π1 (π₯) > 0. Thus the transformation of (15) to (14) yields σΈ
(β β β (π₯) (β« πΉ (π₯) ππ₯ + πΆ0 ) ,
(10)
(π(π₯)β(π₯))
=Β±
σΈ
4π΄ βπ΅ (11)
holds, where π΄ and π΅ are either constants or π΄ = 0 and π΅ is an arbitrary function. In this paper, we present new solutions of the general Riccati equation. We first reduce it to an equivalent equation, and then we formulate the relations between the coefficients functions π(π₯), π(π₯), and β(π₯) of the equation to obtain the required relation πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯), π (π₯) β (π₯)
(16)
π (π₯) π (π₯) ) V. V = βπ0 (π₯) π1 (π₯) (1 + V ) β (β 0 ) (β 1 π1 (π₯) π0 (π₯) (17) σΈ
σΈ
3/2
π0 (π₯) σΈ π (π₯) ) V + (β 0 ) V = π0 (π₯) (1 + V2 ) . π1 (π₯) π1 (π₯)
Then (16) can be written in an equivalent form as
where πΉ(π₯) is an arbitrary differentiable function on πΌ. Also, the general solution was found in [6] when (π(π₯)β(π₯)) β 2 (π (π₯) + βσΈ (π₯) /β (π₯)) π (π₯) β (π₯)
(15)
(12)
where π(π₯) is an arbitrary function. Therefore the given Riccati equation can be transformed into a separable equation, which can be easily solved in two cases: π(π₯) equals a constant π, or certain functions. Several examples are studied in detail to illustrate the proposed technique.
2
An important remark can be made here. Equation (16) can be solved by assuming that the functions π0 (π₯) and π1 (π₯) satisfy the following condition: σΈ
π (π₯) π (π₯) ) = π1 (π₯) βπ0 (π₯) π1 (π₯), (β 0 ) (β 1 π1 (π₯) π0 (π₯)
(18)
where π1 (π₯) is an arbitrary function. This condition can be written in an equivalent form as π0σΈ (π₯) π1σΈ (π₯) β = π (π₯) βπ0 (π₯) π1 (π₯), π0 (π₯) π1 (π₯)
π (π₯) = 2π1 (π₯) , (19)
or equivalently
2. Exact Solutions of the Riccati Equation We begin our approach by converting the general Riccati equation (1) to a simpler form by the substitution
πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯). π (π₯) β (π₯)
π’ (π₯) = πβ β« π(π₯)ππ₯ π¦ (π₯) ,
This leads to the following important cases that solve (16).
(13)
(20)
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2.1. Case 1: π(π₯) Equals a Constant. Substituting π(π₯) = π into (19) gives πV = βπ (π₯) β (π₯) (V2 β πV + 1) , ππ₯
V (π₯) (21)
πV = β« βπ (π₯) β (π₯)ππ₯. V2 β πV + 1
(22)
Based on the integral involving the rational algebraic functions of the form
β«
β4 β π { { { { { 2 { { { { { β4 β π 2 { { { Γ tan ( { { 2 { { { { { { π { { Γ [β« βπ (π₯) β (π₯)ππ₯ + πΆ] ) + , { { 2 { { { { { 2 { 4>π , { { { { { { { { { {(π + βπ2 β 4 + (βπ2 β 4 β π) { { { { { { { { 2 { { Γ exp [βπ β 4 (β« βπ (π₯) β (π₯)ππ₯ + πΆ)]) { { { { { β1 { { { { Γ(1 β exp [βπ2 β 4 (β« βπ(π₯)β(π₯)ππ₯ + πΆ)]) , { { { { { { { { π2 > 4, { { { { { { { { π + π exp [(π β π) (β« βπ (π₯) β (π₯)ππ₯ + πΆ)] { { { { { 1 β exp [(π β π) (β« βπ (π₯) β (π₯)ππ₯ + πΆ)] , ={ { { π2 > 4, { { { { { { { { where π and π are the roots of V2 β πV + 1, { { { { { { { 2 2 { { { βπ β 4 tanh (β βπ β 4 (β«βπ (π₯) β (π₯)ππ₯+ πΆ)) { { { 2 2 { { { { { π { { + , { { { 2 { { { { 2 2 2 { π > 4, (2V β π) < π β 4, { { { { { { { { βπ2 β 4 βπ2 β 4 { { { coth (β (β«βπ (π₯) β (π₯)ππ₯+ πΆ)) { { 2 2 { { { { { π { { + , { { { 2 { { { π2 > 4, (2V β π)2 > π2 β 4, { { { { { { { π 1 { { { + , β { { { β« βπ (π₯) β (π₯)ππ₯ + πΆ 2 { { { π2 = 4. { (24) 2
which is a first-order separable differential equation, and we can obtain its closed form solution from
β«
in view of this, the solution in a closed form is given by
πV πV2 + πV + π 2tanβ1 ((2πV + π) /β4ππ β π2 ) { { { , { { β4ππ β π2 { { { { { { { { { 4ππ > π2 , { { { { { { { { σ΅¨σ΅¨ σ΅¨ { 1 { σ΅¨ 2πV + π β βπ2 β 4ππ σ΅¨σ΅¨σ΅¨ { σ΅¨σ΅¨ , { log σ΅¨σ΅¨σ΅¨σ΅¨ { { βπ2 β 4ππ σ΅¨σ΅¨ 2πV + π + βπ2 β 4ππ σ΅¨σ΅¨σ΅¨ { { { { { { { { { π2 > 4ππ, { { { { { { { σ΅¨σ΅¨ V β π σ΅¨σ΅¨ { { 1 { σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ { log { σ΅¨σ΅¨ V β π σ΅¨σ΅¨σ΅¨ , { { π (π β π) σ΅¨ σ΅¨ { { { { { { { π2 > 4ππ, { { { 2 { = {where π and π are the roots of πV + πV + π = 0, { { { { { { { β2tanhβ1 ((2πV + π) /βπ2 β 4ππ) { { { , { { βπ2 β 4ππ { { { { { { { π2 > 4ππ, (2πV + π)2 < π2 β 4ππ, { { { { { { { { { { β2cothβ1 ((2πV + π) /βπ2 β 4ππ) { { { , { { { βπ2 β 4ππ { { { { { { { π2 > 4ππ, (2πV + π)2 > π2 β 4ππ, { { { { { { { { { 2 { { , β { { { 2πV +π { { { { { 2 { π = 4ππ, (23)
Once V is found then we can obtain π’ from (15). Finally, after finding π’, we can use π¦ = π’πβ« π(π₯)ππ₯ to return to the original variable. Our results can thus be summarized by the following lemma.
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Lemma 1. If the coefficients of the general Riccati equation (1) satisfy the condition πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = πβπ (π₯) β (π₯), π (π₯) β (π₯)
(25)
where π is a constant, then the general solutions of the Riccati equation can be exactly obtained as π¦ (π₯) π (π₯) β4 β π2 { { β { { { 2 { { β (π₯) { { β 4 β π2 { π { { Γ tan ( [β« βπ (π₯) β (π₯)ππ₯ + πΆ]) + , { { 2 2 { { { 2 { { , 4 > π { { { { { { { { π (π₯) { { β 2 β 2 { β { { β (π₯) (π + π β 4 + ( π β 4 β π) { { { { { { Γ exp [βπ2 β 4 (β« βπ (π₯) β (π₯)ππ₯ + πΆ)]) { { { { β1 { { { { Γ(1 β exp [βπ2 β 4 (β« βπ(π₯)β(π₯)ππ₯ + πΆ)]) , { { { { { { π2 > 4, { { { { { { { { { π (π₯) π + π exp [(π β π) (β« βπ (π₯) β (π₯)ππ₯ + πΆ)] { { β , { { β (π₯) 1 β exp [(π β π) (β« βπ (x) β (π₯)ππ₯ + πΆ)] { { { { { π2 > 4, ={ 2 {where π and π are the roots of V β πV + 1, { { { { { { { { { π (π₯) βπ2 β 4 { { β { { { β (π₯) 2 { { { { β π2 β 4 { π { { Γ tanh (β (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + , { { 2 2 { { { 2 2 2 { { > 4, β π) < π β 4, π (2V { { { { { { { { π (π₯) βπ2 β 4 { { { πβ« π(π₯)ππ₯ β { { β (π₯) 2 { { { { 2β4 β { π π { { Γ coth (β (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + , { { { 2 2 { { 2 2 2 { { > 4, β π) > π β 4, π (2V { { { { { π (π₯) 2 π { { { + , ββ { { β 2 (π₯) βπ β« β + πΆ (π₯) (π₯)ππ₯ { { { 2 π = 4, { (26) where πΆ is a constant. Remark 2. For the case π0 (π₯)/π1 (π₯) < 0, proceeding as before, we obtain the following condition: πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = πββπ (π₯) β (π₯) π (π₯) β (π₯)
(27)
and the first-order separable equation πV = ββπ (π₯) β (π₯) (βV2 β πV + 1) . ππ₯
(28)
Thus the general solutions can be similarly found. 2.2. Case 2: π(π₯) Equals an Arbitrary Function. We have πV = βπ (π₯) β (π₯) (V2 β π (π₯) V + 1) . ππ₯
(29)
This equation can be written in an equivalent form as π (π₯) 2 4 β π2 (π₯) πV = βπ (π₯) β (π₯) [(V β ) + ]. ππ₯ 2 4
(30)
The substitution of π€=Vβ
π (π₯) 2
(31)
into (30) yields ππ€ 4 β π2 (π₯) πσΈ (π₯) = βπ (π₯) β (π₯) [π€2 + ]β . ππ₯ 4 2
(32)
If we assume that the function π(π₯) satisfies the following condition: βπ (π₯) β (π₯) [
4 β π2 (π₯) πσΈ (π₯) ]β = πβπ (π₯) β (π₯), (33) 4 2
where π is a constant, then π2
βπ (π₯) β (π₯) πσΈ (π₯) , =β 2 (π₯) β 4 + 4π
(34)
which is a separable equation. Thus π (π₯) { (2π1 β1 β π exp [β2β1 β π β«βπ (π₯) β (π₯)ππ₯] { { { { { { { +2β1 β π) { { { { β1 { { { β1 β π β«βπ(π₯)β(π₯)ππ₯]) , { Γ (1 β π exp [β2 { 1 { { { { { 1 β π > 0, { { { { { { = {2βπ β 1 tan [ββπ β 1 β« βπ (π₯) β (π₯)ππ₯ { { { { { { +π1 βπ β 1] , { { { { { { 1 β π < 0, { { { { { { { 1 { { , { { { { { (1/2) β« βπ (π₯) β (π₯)ππ₯ + π1 { π = 1, {
(35)
where π1 is an arbitrary integration constant. Then it is easy to solve (32) in closed form. Thus ππ€ = βπ (π₯) β (π₯) [π€2 + π] , ππ₯
(36)
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or π€ = βπ tan (βπ β« βπ (π₯) β (π₯)ππ₯ + βππ2 ) ,
π =ΜΈ 0, (37)
π¦ (π₯)
where π2 is an arbitrary integration constant. Therefore, π (π₯) β V=π€+ = π tan (βπ β« βπ (π₯) β (π₯)ππ₯ + βππ2 ) 2 +
π (π₯) , 2
π =ΜΈ 0. (38)
Substituting the original expression for π¦, we obtain the final general solution as π¦ (π₯) = β
π (π₯) β [ π tan (βπ β« βπ (π₯) β (π₯)ππ₯ + βππ2 ) β (π₯) +
π (π₯) ], 2
π =ΜΈ 0. (39)
For π = 0, we have π€=
β1 . β« βπ (π₯) β (π₯)ππ₯ + π2
(40)
Thus π¦ (π₯) = β
π (π₯) β1 π (π₯) [ + ]. β (π₯) β« βπ (π₯) β (π₯)ππ₯ + π2 2
3.2. Special Case Where the Original Equation Has the Canonical Form
(41)
Lemma 3. If the coefficients of the general Riccati equation (1) satisfy the condition (42)
where π(π₯) is a function given by (35), then the general closed form solutions of the Riccati equation can be exactly obtained by (39) and (41).
3. Examples Listed below are some special cases where the above conditions are satisfied and the general solutions of the Riccati equation are found. 3.1. The Coefficients π(π₯), π(π₯), and β(π₯) Are Proportional. Example 1. Consider the following Riccati equation: π¦σΈ = π (π₯) (π + ππ¦ + ππ¦2 ) ,
σ΅¨σ΅¨ π σ΅¨σ΅¨ β4 β π2 { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { { σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 { { { { β 4 β π2 { π { { Γ tan ( [β|ππ| β« π (π₯) ππ₯ + πΆ]) + , { { 2 2 { { { { { |ππ| > π2 , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ (π + βπ2 β 4 + (βπ2 β 4 β π) { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ ={ Γ exp [βπ2 β 4 (β|ππ| β«π (π₯) ππ₯ + πΆ)]) { { { { β1 { { { βπ2 β 4 (β|ππ| β«π(π₯)ππ₯ + πΆ)]) , { Γ(1 β exp [ { { { { { { π2 > |ππ| , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { 2 π { σ΅¨ σ΅¨ { + , {ββ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { π 2 σ΅¨ σ΅¨ { β β« π ππ₯ + πΆ (π₯) |ππ| { { 2 π = |ππ| . { (44)
Example 2. Consider the following Riccati equation:
We have the following.
πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯), π (π₯) β (π₯)
The condition equation (25) holds if and only if π = β2π/β|ππ|. So the general solution to this equation can be found by
(43)
where π(π₯), π(π₯), and β(π₯) are proportional; that is, π(π₯) = ππ(π₯), π(π₯) = ππ(π₯), and β(π₯) = ππ(π₯).
π¦σΈ = π (π₯) + π¦2 ,
(45)
where π(π₯) = 1/(πΌπ₯ + π½)2 , π(π₯) = 0, and β(π₯) = 1. The condition equation (25) holds if and only if π = β2πΌ. Thus, the general solution is given as π¦ (π₯) β4 β π2 1 { { { { 2 πΌπ₯ + π½ { { { β 4 β π2 1 { { { Γ tan ( [ ln (πΌπ₯ + π½) + πΆ]) { { { 2 πΌ { { { π { { { + , { { 2 { { { 2 { 1 > πΌ , { { { { { { { { { 1 β 2 β 2 = { πΌπ₯ + π½ (π + π β 4 + ( π β 4 β π) { { 1 { { Γ exp [βπ2 β 4 ( ln (πΌπ₯ + π½) + πΆ)]) { { { πΌ { { β1 { 1 1 { 2 { β π β 4 ( , + πΆ)]) Γ(1 β exp [ { { { πΌ πΌπ₯ + π½ { { { { πΌ2 > 1, { { { { { { { 1 2 π { { { β + , { { πΌπ₯ + π½ 2 ln (πΌπ₯ + π½) + πΆ (1/πΌ) { { 2 πΌ = 1. { (46)
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3.3. Equations Containing Power Functions Example 3. Consider the following equation: π¦σΈ = ππ₯π + ππ₯π π¦2 , π
(47) π
where π(π₯) = ππ₯ , π(π₯) = 0, and β(π₯) = ππ₯ . The condition equation (25) holds if and only if π = (π β π)/β|ππ| with π + π = β2. So the general solution to this equation can be found for any values of π, π, π, and π by π¦ (π₯) σ΅¨σ΅¨ π σ΅¨σ΅¨ β4 β π2 (πβπ)/2 { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 π₯ { { { { { { { { β4 β π2 { { { Γ tan ( { { 2 { { { { { { { { 2 π { { β|ππ|π₯(π+π+2)/2 + πΆ] ) + , Γ[ { { π+π+2 2 { { { { { { { { 4 |ππ| > (π β π)2 , { { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π₯(πβπ)/2 (π + βπ2 β 4 + (βπ2 β 4 β π) { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { { { { { { { { Γ exp [βπ2 β 4 { { { { { { { { 2 { { { Γ( { { π + π+2 { { { { ={ { Γβ|ππ|π₯(π+π+2)/2 + πΆ)]) { { { { { { { { { { Γ (1 β exp [βπ2 β 4 { { { { { { { { { 2 { { { Γ( { { π + π+2 { { { { { { β1 { { (π+π+2)/2 { β Γ + πΆ)]) , { |ππ|π₯ { { { { { { { { { 4 |ππ| < (π β π)2 , { { { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ (πβπ)/2 { { { σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π₯ β β { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { { { { { π 2 { { + , Γ { { (π+π+2)/2 { +πΆ 2 (2/ (π + π + 2)) β|ππ|π₯ { { { { { 4 |ππ| = (π β π)2 . { (48)
π¦ (π₯) σ΅¨σ΅¨ π σ΅¨σ΅¨ β4 β π2 β(π+1) { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { { σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 π₯ { { { { β 4 β π2 { π { { [β|ππ| ln π₯ + πΆ]) + , { Γ tan ( { 2 2 { { { { { { |ππ| > (π + 1 + π)2 , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π₯β(π+1) (π + βπ2 β 4 + (βπ2 β 4 β π) { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { ={ Γ exp [βπ2 β 4 (β|ππ| ln π₯ + πΆ)]) { { { { { { β1 { { Γ(1 β exp [βπ2 β 4 (β|ππ| ln π₯ + πΆ)]) , { { { { { { { { |ππ| < (π + 1 + π)2 , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ β(π+1) { 2 π { σ΅¨σ΅¨ σ΅¨σ΅¨π₯ { β β + , { σ΅¨σ΅¨ σ΅¨σ΅¨ { { σ΅¨πσ΅¨ β|ππ| ln π₯ + πΆ 2 { { { |ππ| = (π + 1 + π)2 . { 3.4. Other Equations Example 5. Consider the following Riccati equation: π₯π¦σΈ = π₯2π + (π β π) π¦ + π₯2π π¦2 , 2πβ1
π π¦ + ππ₯π π¦2 , π₯
(49)
(51) 2πβ1
, π(π₯) = (π β π)/π₯, and β(π₯) = π₯ . We where π(π₯) = π₯ get π = 0, and the general solution to this equation is given by π¦=β
π (π₯) β4 β π2 β (π₯) 2
Γ tan (
β4 β π2 2
(52) π (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + . 2
Thus the exact closed form solution is π¦ = π₯πβπ tan (
π₯π+π + πΆ) , π+π
(53)
where πΆ is a constant of integration. Example 6. For the given equation 2π₯2 π¦σΈ = β2π2 π₯ + π₯π¦ + 2π¦2 , 2
(54) 2
where π(π₯) = βπ /π₯, π(π₯) = 1/2π₯, and β(π₯) = 1/π₯ , we get π = 0, and the general solution to this equation is given by π¦ = πβ« π(π₯)ππ₯ β
Example 4. For the equation π¦σΈ = ππ₯βπβ2 +
(50)
Γ tan (
π (π₯) β4 β π2 β (π₯) 2
β4 β π 2 2
π (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + . 2
(55)
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Thus the exact closed form solution is π¦ = |π| βπ₯ tan (
β2 |π| + πΆ) , π₯2
(56)
where πΆ is a constant of integration.
π
3.5. Equations Containing Exponential Functions Example 7. Consider the following Riccati equation: π¦σΈ = ππππ₯ + ππ¦2 ,
(57)
where π(π₯) = ππππ₯ , π(π₯) = 0, and β(π₯) = π. The condition equation (42) holds if and only if π(π₯) = π/β|ππ|πβ(π/2)π₯ . Combining this value of π(π₯) with (35), we get π = 2. So π(π₯) = (2/β|ππ|)πβπ₯ . Thus the general solution to this equation can be found by (41); that is, σ΅¨σ΅¨ π σ΅¨σ΅¨ β1 2 βπ₯ σ΅¨ σ΅¨ π¦ (π₯) = β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ [ + π ]. σ΅¨σ΅¨ π σ΅¨σ΅¨ β|ππ|ππ₯ + π2 β|ππ|
(58)
4. Applications 4.1. The Central Potential Problem of the Power Law Type. Certain types of Newtons laws of motion are equivalent to the Riccati equation. For example, the equation for the energy conservation in the case of a central potential π(π) is given by the standard expression [7] π2 1 + π (π) . πΈ = ππ2Μ + 2 2ππ2
(59)
Under the influence of a power law central potential π(π) = πβ ππ and πΈ = 0, where πβ is the coupling constant and the exponent π can be either positive or negative, (59) can be transformed into the Riccati equation π€σΈ =
π+2 π+2 2 + π€. 2 2
4.2. Damped Harmonic Oscillators. Various problems in quantum optics, superconductivity, and nonrelativistic quantum mechanics can be described classically by [8]
(60)
ππ₯ ππ σ΅¨σ΅¨σ΅¨σ΅¨ π2 π₯ + π , = β σ΅¨ ππ‘2 ππ‘ ππ₯ σ΅¨σ΅¨σ΅¨π₯=π₯(π‘)
where π₯(π‘) is the particle coordinate, π is the mass of the particle, π is a damping constant, and π(π₯, π‘) is a potential energy that accounts for the interaction of the particle with its environment. If π(π₯, π‘) = βπ₯π(π‘), where π(π‘) is white or colored noise, then (60) becomes the well-known Langevin equation π
ππ₯ π2 π₯ +π β π (π‘) = 0. ππ‘2 ππ‘
ππ₯ π2 π₯ + 2π½ (π‘) + π2 (π‘) π₯ = 0, ππ‘2 ππ‘
of the original equation, where πσΈ = ππ/ππ and π(π‘) = 1/ππ2 . From the condition (25) of Lemma 1, we see that there exists a constant π = 0 for π =ΜΈ β2. Hence, the general solution of (60) is given by π€ (π) = tan (
|π + 2| π + πΆ1 ) . 2
(62)
Therefore, π (π) = πΆ2 [sec ( where πΆ2 is a constant.
2/|π+2| |π + 2| , π + πΆ1 )] 2
(63)
(66)
where π½ is the damping function and π is the the natural frequency of the undamped oscillator. The substitution π¦ = π₯σΈ /π₯ converts (66) to the Riccati equation π¦σΈ = βπ2 (π‘) β 2π½ (π‘) π¦ β π¦2 .
(67)
Here π(π‘) = βπ2 (π‘), π(π‘) = β2π½(π‘), and β(π‘) = β1. The condition (25) holds if and only if the coefficients π and π½ satisfy the following condition: πσΈ π + 2π½ = π, π 2
where π is a constant.
(68)
Then (68) can be written in an equivalent form as π πσΈ = β2π½ (π‘) π + π2 , 2
(69)
which is a Bernoulli equation and can be readily solved for π to obtain π (π‘) =
(61)
(65)
Also the equation of motion is
A solution of this equation will lead to a solution πσΈ =π€ π
(64)
1 β (π/2) + ππ2 β« π½(π‘)ππ‘
.
(70)
We conclude that if the coefficients π and π½ satisfy (70), then the general solution π¦(π‘) of (67) is given by (26). Returning to the original dependent variable by π¦ = π₯σΈ /π₯, we obtain the general solution to (66) as π₯ (π‘) = ππβ« π¦(π‘)ππ‘ ,
(71)
where π is a constant.
5. Conclusion We have converted the Riccati equation into an equivalent equation; then, by using the integrability condition for this equation, πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯), π (π₯) β (π₯)
(72)
8 we obtain a separable equation. The first case π(π₯) = π, where constant π is a constant, is considered. Thus, the general solutions of the Riccati equation can be exactly obtained. The second integrability case is obtained for the reduced Riccati equation with an arbitrary function π(π₯). We have considered several distinct examples to illustrate our new approach. The method is also applied to the Riccati equation arising in the solution of certain types of Newtons laws of motion and the damped harmonic oscillators equations.
Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.
References [1] A. D. Polyanin and V. F. Zaitsev, Handbook of Exact Solutions for Ordinary Differential Equations, Chapman & Hall/CRC, Boca Raton, Fla, USA, Second edition, 2003. [2] E. Hille, Ordinary Differential Equat ions in the Complex Domain, Dover Publications, Mineola, NY, USA, 1997. [3] E. L. Ince, Ordinary Differential Equations, Dover Publications, New York, NY, USA, 1956. [4] V. M. Strelchenya, βA new case of integrability of the general Riccati equation and its application to relaxation problems,β Journal of Physics A: Mathematical and General, vol. 24, no. 21, pp. 4965β4967, 1991. [5] M. K. Mak and T. Harko, βNew further integrability cases for the Riccati equation,β Applied Mathematics and Computation, vol. 219, no. 14, pp. 7465β7471, 2013. [6] A. Al Bastami, M. R. BeliΒ΄c, and N. Z. PetroviΒ΄c, βSpecial solutions of the Riccati equation with applications to the Gross-Pitaevskii nonlinear PDE,β Electronic Journal of Differential Equations, vol. 2010, no. 66, pp. 1β10, 2010. [7] M. Nowakowski and H. C. Rosu, βNewtonβs laws of motion in the form of a Riccati equation,β Physical Review E. Statistical, Nonlinear, and Soft Matter Physics, vol. 65, no. 4, Article ID 047602, 2002. [8] S. I. Denisov and W. Horsthemke, βAnomalous diffusion and stochastic localization of damped quantum particles,β Physics Letters A, vol. 282, no. 6, pp. 367β372, 2001.
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Research Article New Conditions for Obtaining the Exact Solutions of the General Riccati Equation Lazhar Bougoffa Department of Mathematics, Faculty of Science, Al Imam Mohammad Ibn Saud Islamic University (IMSIU), P.O. Box 90950, Riyadh 11623, Saudi Arabia Correspondence should be addressed to Lazhar Bougoffa; [email protected] Received 20 April 2014; Accepted 20 July 2014; Published 18 August 2014 Academic Editor: Kannan Krithivasan Copyright Β© 2014 Lazhar Bougoffa. 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. We propose a direct method for solving the general Riccati equation π¦σΈ = π(π₯)+π(π₯)π¦+β(π₯)π¦2 . We first reduce it into an equivalent equation, and then we formulate the relations between the coefficients functions π(π₯), π(π₯), and β(π₯) of the equation to obtain an equivalent separable equation from which the previous equation can be solved in closed form. Several examples are presented to demonstrate the efficiency of this method.
1. Introduction
A set of solutions to the Riccati equation is then given by
In realistic investigations of the dynamics of a physical system, nonlinearity may often be in the form of the wellknown Riccati equation: π¦σΈ = π (π₯) + π (π₯) π¦ + β (π₯) π¦2 .
(1)
The Riccati equation is used in different areas of physics, engineering, and mathematics such as quantum mechanics, thermodynamics, and control theory. It also appears in many engineering design simulations. The first well-known result in the analysis of the Riccati equation is that [1, 2] if one particular solution π¦1 can be found to (1), then the general solution is obtained as π¦ = π¦1 + π’,
(2)
where π’ satisfies the corresponding Bernoulli equation π’σΈ β (π (π₯) + 2β (π₯) π¦1 ) π’ = β (π₯) π’2 .
(3)
The substitution that is needed to solve this Bernoulli equation is π§=
1 . π’
(4)
1 π¦ = π¦1 + . π§
(5)
The second important result in the analysis of the solution of the Riccati equation is that the general equation can always be reduced to a second-order linear differential equation of the form [3, pages 23β25] π’σΈ σΈ β π (π₯) π’σΈ + π (π₯) π’ = 0,
(6)
βσΈ (π₯) , β (π₯) π (π₯) = π (π₯) β (π₯) .
(7)
where π (π₯) = π (π₯) +
A solution of this equation will lead to a solution π¦=β
1 π’σΈ β (π₯) π’
of the original Riccati equation.
(8)
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To our knowledge it is often very difficult, if not impossible, to find closed form solutions of such nonlinear differential equations. But a number of solutions of the Riccati equation can be obtained by assuming that the coefficients π(π₯), π(π₯), and β(π₯) satisfy certain constraints. Indeed, closed-form solutions are known if the following condition is satisfied [4]:
which yields the equation π’σΈ = π0 (π₯) + π1 (π₯) π’2 ,
(14)
where π0 (π₯) = πβ β« π(π₯)ππ₯ π(π₯) and π1 (π₯) = πβ« π(π₯)ππ₯ β(π₯). Now, let
π (π₯) + π (π₯) + β (π₯) = (log
πΌ(π₯) σΈ πΌ (π₯) β π½ (π₯) (πΌ (π₯) β (π₯) β π½ (π₯) π (π₯)) , ) β π½(π₯) πΌ (π₯) π½ (π₯) (9)
where πΌ(π₯) and π½(π₯) are arbitrary differentiable functions in πΌ β R with πΌ(π₯)π½(π₯) > 0. A new integrability condition for the Riccati equation was presented in [5] under the following constraint: π (π₯) =
πΉ (π₯) β π (π₯) β« πΉ (π₯) ππ₯ + πΆ0
σ΅¨σ΅¨ π (π₯) σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅¨ π’ = β σ΅¨σ΅¨σ΅¨ 0 σ΅¨V, σ΅¨σ΅¨ π1 (π₯) σ΅¨σ΅¨σ΅¨
and consider the case π0 (π₯)/π1 (π₯) > 0. Thus the transformation of (15) to (14) yields σΈ
(β β β (π₯) (β« πΉ (π₯) ππ₯ + πΆ0 ) ,
(10)
(π(π₯)β(π₯))
=Β±
σΈ
4π΄ βπ΅ (11)
holds, where π΄ and π΅ are either constants or π΄ = 0 and π΅ is an arbitrary function. In this paper, we present new solutions of the general Riccati equation. We first reduce it to an equivalent equation, and then we formulate the relations between the coefficients functions π(π₯), π(π₯), and β(π₯) of the equation to obtain the required relation πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯), π (π₯) β (π₯)
(16)
π (π₯) π (π₯) ) V. V = βπ0 (π₯) π1 (π₯) (1 + V ) β (β 0 ) (β 1 π1 (π₯) π0 (π₯) (17) σΈ
σΈ
3/2
π0 (π₯) σΈ π (π₯) ) V + (β 0 ) V = π0 (π₯) (1 + V2 ) . π1 (π₯) π1 (π₯)
Then (16) can be written in an equivalent form as
where πΉ(π₯) is an arbitrary differentiable function on πΌ. Also, the general solution was found in [6] when (π(π₯)β(π₯)) β 2 (π (π₯) + βσΈ (π₯) /β (π₯)) π (π₯) β (π₯)
(15)
(12)
where π(π₯) is an arbitrary function. Therefore the given Riccati equation can be transformed into a separable equation, which can be easily solved in two cases: π(π₯) equals a constant π, or certain functions. Several examples are studied in detail to illustrate the proposed technique.
2
An important remark can be made here. Equation (16) can be solved by assuming that the functions π0 (π₯) and π1 (π₯) satisfy the following condition: σΈ
π (π₯) π (π₯) ) = π1 (π₯) βπ0 (π₯) π1 (π₯), (β 0 ) (β 1 π1 (π₯) π0 (π₯)
(18)
where π1 (π₯) is an arbitrary function. This condition can be written in an equivalent form as π0σΈ (π₯) π1σΈ (π₯) β = π (π₯) βπ0 (π₯) π1 (π₯), π0 (π₯) π1 (π₯)
π (π₯) = 2π1 (π₯) , (19)
or equivalently
2. Exact Solutions of the Riccati Equation We begin our approach by converting the general Riccati equation (1) to a simpler form by the substitution
πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯). π (π₯) β (π₯)
π’ (π₯) = πβ β« π(π₯)ππ₯ π¦ (π₯) ,
This leads to the following important cases that solve (16).
(13)
(20)
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2.1. Case 1: π(π₯) Equals a Constant. Substituting π(π₯) = π into (19) gives πV = βπ (π₯) β (π₯) (V2 β πV + 1) , ππ₯
V (π₯) (21)
πV = β« βπ (π₯) β (π₯)ππ₯. V2 β πV + 1
(22)
Based on the integral involving the rational algebraic functions of the form
β«
β4 β π { { { { { 2 { { { { { β4 β π 2 { { { Γ tan ( { { 2 { { { { { { π { { Γ [β« βπ (π₯) β (π₯)ππ₯ + πΆ] ) + , { { 2 { { { { { 2 { 4>π , { { { { { { { { { {(π + βπ2 β 4 + (βπ2 β 4 β π) { { { { { { { { 2 { { Γ exp [βπ β 4 (β« βπ (π₯) β (π₯)ππ₯ + πΆ)]) { { { { { β1 { { { { Γ(1 β exp [βπ2 β 4 (β« βπ(π₯)β(π₯)ππ₯ + πΆ)]) , { { { { { { { { π2 > 4, { { { { { { { { π + π exp [(π β π) (β« βπ (π₯) β (π₯)ππ₯ + πΆ)] { { { { { 1 β exp [(π β π) (β« βπ (π₯) β (π₯)ππ₯ + πΆ)] , ={ { { π2 > 4, { { { { { { { { where π and π are the roots of V2 β πV + 1, { { { { { { { 2 2 { { { βπ β 4 tanh (β βπ β 4 (β«βπ (π₯) β (π₯)ππ₯+ πΆ)) { { { 2 2 { { { { { π { { + , { { { 2 { { { { 2 2 2 { π > 4, (2V β π) < π β 4, { { { { { { { { βπ2 β 4 βπ2 β 4 { { { coth (β (β«βπ (π₯) β (π₯)ππ₯+ πΆ)) { { 2 2 { { { { { π { { + , { { { 2 { { { π2 > 4, (2V β π)2 > π2 β 4, { { { { { { { π 1 { { { + , β { { { β« βπ (π₯) β (π₯)ππ₯ + πΆ 2 { { { π2 = 4. { (24) 2
which is a first-order separable differential equation, and we can obtain its closed form solution from
β«
in view of this, the solution in a closed form is given by
πV πV2 + πV + π 2tanβ1 ((2πV + π) /β4ππ β π2 ) { { { , { { β4ππ β π2 { { { { { { { { { 4ππ > π2 , { { { { { { { { σ΅¨σ΅¨ σ΅¨ { 1 { σ΅¨ 2πV + π β βπ2 β 4ππ σ΅¨σ΅¨σ΅¨ { σ΅¨σ΅¨ , { log σ΅¨σ΅¨σ΅¨σ΅¨ { { βπ2 β 4ππ σ΅¨σ΅¨ 2πV + π + βπ2 β 4ππ σ΅¨σ΅¨σ΅¨ { { { { { { { { { π2 > 4ππ, { { { { { { { σ΅¨σ΅¨ V β π σ΅¨σ΅¨ { { 1 { σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨ { log { σ΅¨σ΅¨ V β π σ΅¨σ΅¨σ΅¨ , { { π (π β π) σ΅¨ σ΅¨ { { { { { { { π2 > 4ππ, { { { 2 { = {where π and π are the roots of πV + πV + π = 0, { { { { { { { β2tanhβ1 ((2πV + π) /βπ2 β 4ππ) { { { , { { βπ2 β 4ππ { { { { { { { π2 > 4ππ, (2πV + π)2 < π2 β 4ππ, { { { { { { { { { { β2cothβ1 ((2πV + π) /βπ2 β 4ππ) { { { , { { { βπ2 β 4ππ { { { { { { { π2 > 4ππ, (2πV + π)2 > π2 β 4ππ, { { { { { { { { { 2 { { , β { { { 2πV +π { { { { { 2 { π = 4ππ, (23)
Once V is found then we can obtain π’ from (15). Finally, after finding π’, we can use π¦ = π’πβ« π(π₯)ππ₯ to return to the original variable. Our results can thus be summarized by the following lemma.
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Lemma 1. If the coefficients of the general Riccati equation (1) satisfy the condition πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = πβπ (π₯) β (π₯), π (π₯) β (π₯)
(25)
where π is a constant, then the general solutions of the Riccati equation can be exactly obtained as π¦ (π₯) π (π₯) β4 β π2 { { β { { { 2 { { β (π₯) { { β 4 β π2 { π { { Γ tan ( [β« βπ (π₯) β (π₯)ππ₯ + πΆ]) + , { { 2 2 { { { 2 { { , 4 > π { { { { { { { { π (π₯) { { β 2 β 2 { β { { β (π₯) (π + π β 4 + ( π β 4 β π) { { { { { { Γ exp [βπ2 β 4 (β« βπ (π₯) β (π₯)ππ₯ + πΆ)]) { { { { β1 { { { { Γ(1 β exp [βπ2 β 4 (β« βπ(π₯)β(π₯)ππ₯ + πΆ)]) , { { { { { { π2 > 4, { { { { { { { { { π (π₯) π + π exp [(π β π) (β« βπ (π₯) β (π₯)ππ₯ + πΆ)] { { β , { { β (π₯) 1 β exp [(π β π) (β« βπ (x) β (π₯)ππ₯ + πΆ)] { { { { { π2 > 4, ={ 2 {where π and π are the roots of V β πV + 1, { { { { { { { { { π (π₯) βπ2 β 4 { { β { { { β (π₯) 2 { { { { β π2 β 4 { π { { Γ tanh (β (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + , { { 2 2 { { { 2 2 2 { { > 4, β π) < π β 4, π (2V { { { { { { { { π (π₯) βπ2 β 4 { { { πβ« π(π₯)ππ₯ β { { β (π₯) 2 { { { { 2β4 β { π π { { Γ coth (β (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + , { { { 2 2 { { 2 2 2 { { > 4, β π) > π β 4, π (2V { { { { { π (π₯) 2 π { { { + , ββ { { β 2 (π₯) βπ β« β + πΆ (π₯) (π₯)ππ₯ { { { 2 π = 4, { (26) where πΆ is a constant. Remark 2. For the case π0 (π₯)/π1 (π₯) < 0, proceeding as before, we obtain the following condition: πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = πββπ (π₯) β (π₯) π (π₯) β (π₯)
(27)
and the first-order separable equation πV = ββπ (π₯) β (π₯) (βV2 β πV + 1) . ππ₯
(28)
Thus the general solutions can be similarly found. 2.2. Case 2: π(π₯) Equals an Arbitrary Function. We have πV = βπ (π₯) β (π₯) (V2 β π (π₯) V + 1) . ππ₯
(29)
This equation can be written in an equivalent form as π (π₯) 2 4 β π2 (π₯) πV = βπ (π₯) β (π₯) [(V β ) + ]. ππ₯ 2 4
(30)
The substitution of π€=Vβ
π (π₯) 2
(31)
into (30) yields ππ€ 4 β π2 (π₯) πσΈ (π₯) = βπ (π₯) β (π₯) [π€2 + ]β . ππ₯ 4 2
(32)
If we assume that the function π(π₯) satisfies the following condition: βπ (π₯) β (π₯) [
4 β π2 (π₯) πσΈ (π₯) ]β = πβπ (π₯) β (π₯), (33) 4 2
where π is a constant, then π2
βπ (π₯) β (π₯) πσΈ (π₯) , =β 2 (π₯) β 4 + 4π
(34)
which is a separable equation. Thus π (π₯) { (2π1 β1 β π exp [β2β1 β π β«βπ (π₯) β (π₯)ππ₯] { { { { { { { +2β1 β π) { { { { β1 { { { β1 β π β«βπ(π₯)β(π₯)ππ₯]) , { Γ (1 β π exp [β2 { 1 { { { { { 1 β π > 0, { { { { { { = {2βπ β 1 tan [ββπ β 1 β« βπ (π₯) β (π₯)ππ₯ { { { { { { +π1 βπ β 1] , { { { { { { 1 β π < 0, { { { { { { { 1 { { , { { { { { (1/2) β« βπ (π₯) β (π₯)ππ₯ + π1 { π = 1, {
(35)
where π1 is an arbitrary integration constant. Then it is easy to solve (32) in closed form. Thus ππ€ = βπ (π₯) β (π₯) [π€2 + π] , ππ₯
(36)
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or π€ = βπ tan (βπ β« βπ (π₯) β (π₯)ππ₯ + βππ2 ) ,
π =ΜΈ 0, (37)
π¦ (π₯)
where π2 is an arbitrary integration constant. Therefore, π (π₯) β V=π€+ = π tan (βπ β« βπ (π₯) β (π₯)ππ₯ + βππ2 ) 2 +
π (π₯) , 2
π =ΜΈ 0. (38)
Substituting the original expression for π¦, we obtain the final general solution as π¦ (π₯) = β
π (π₯) β [ π tan (βπ β« βπ (π₯) β (π₯)ππ₯ + βππ2 ) β (π₯) +
π (π₯) ], 2
π =ΜΈ 0. (39)
For π = 0, we have π€=
β1 . β« βπ (π₯) β (π₯)ππ₯ + π2
(40)
Thus π¦ (π₯) = β
π (π₯) β1 π (π₯) [ + ]. β (π₯) β« βπ (π₯) β (π₯)ππ₯ + π2 2
3.2. Special Case Where the Original Equation Has the Canonical Form
(41)
Lemma 3. If the coefficients of the general Riccati equation (1) satisfy the condition (42)
where π(π₯) is a function given by (35), then the general closed form solutions of the Riccati equation can be exactly obtained by (39) and (41).
3. Examples Listed below are some special cases where the above conditions are satisfied and the general solutions of the Riccati equation are found. 3.1. The Coefficients π(π₯), π(π₯), and β(π₯) Are Proportional. Example 1. Consider the following Riccati equation: π¦σΈ = π (π₯) (π + ππ¦ + ππ¦2 ) ,
σ΅¨σ΅¨ π σ΅¨σ΅¨ β4 β π2 { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { { σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 { { { { β 4 β π2 { π { { Γ tan ( [β|ππ| β« π (π₯) ππ₯ + πΆ]) + , { { 2 2 { { { { { |ππ| > π2 , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ (π + βπ2 β 4 + (βπ2 β 4 β π) { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ ={ Γ exp [βπ2 β 4 (β|ππ| β«π (π₯) ππ₯ + πΆ)]) { { { { β1 { { { βπ2 β 4 (β|ππ| β«π(π₯)ππ₯ + πΆ)]) , { Γ(1 β exp [ { { { { { { π2 > |ππ| , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { 2 π { σ΅¨ σ΅¨ { + , {ββ σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { π 2 σ΅¨ σ΅¨ { β β« π ππ₯ + πΆ (π₯) |ππ| { { 2 π = |ππ| . { (44)
Example 2. Consider the following Riccati equation:
We have the following.
πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯), π (π₯) β (π₯)
The condition equation (25) holds if and only if π = β2π/β|ππ|. So the general solution to this equation can be found by
(43)
where π(π₯), π(π₯), and β(π₯) are proportional; that is, π(π₯) = ππ(π₯), π(π₯) = ππ(π₯), and β(π₯) = ππ(π₯).
π¦σΈ = π (π₯) + π¦2 ,
(45)
where π(π₯) = 1/(πΌπ₯ + π½)2 , π(π₯) = 0, and β(π₯) = 1. The condition equation (25) holds if and only if π = β2πΌ. Thus, the general solution is given as π¦ (π₯) β4 β π2 1 { { { { 2 πΌπ₯ + π½ { { { β 4 β π2 1 { { { Γ tan ( [ ln (πΌπ₯ + π½) + πΆ]) { { { 2 πΌ { { { π { { { + , { { 2 { { { 2 { 1 > πΌ , { { { { { { { { { 1 β 2 β 2 = { πΌπ₯ + π½ (π + π β 4 + ( π β 4 β π) { { 1 { { Γ exp [βπ2 β 4 ( ln (πΌπ₯ + π½) + πΆ)]) { { { πΌ { { β1 { 1 1 { 2 { β π β 4 ( , + πΆ)]) Γ(1 β exp [ { { { πΌ πΌπ₯ + π½ { { { { πΌ2 > 1, { { { { { { { 1 2 π { { { β + , { { πΌπ₯ + π½ 2 ln (πΌπ₯ + π½) + πΆ (1/πΌ) { { 2 πΌ = 1. { (46)
6
The Scientific World Journal where π(π₯) = ππ₯βπβ2 , π(π₯) = π/π₯, and β(π₯) = ππ₯π , we get π = β2(π + 1 + π)/β|ππ|, and the general solution to this equation is given by
3.3. Equations Containing Power Functions Example 3. Consider the following equation: π¦σΈ = ππ₯π + ππ₯π π¦2 , π
(47) π
where π(π₯) = ππ₯ , π(π₯) = 0, and β(π₯) = ππ₯ . The condition equation (25) holds if and only if π = (π β π)/β|ππ| with π + π = β2. So the general solution to this equation can be found for any values of π, π, π, and π by π¦ (π₯) σ΅¨σ΅¨ π σ΅¨σ΅¨ β4 β π2 (πβπ)/2 { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 π₯ { { { { { { { { β4 β π2 { { { Γ tan ( { { 2 { { { { { { { { 2 π { { β|ππ|π₯(π+π+2)/2 + πΆ] ) + , Γ[ { { π+π+2 2 { { { { { { { { 4 |ππ| > (π β π)2 , { { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π₯(πβπ)/2 (π + βπ2 β 4 + (βπ2 β 4 β π) { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { { { { { { { { Γ exp [βπ2 β 4 { { { { { { { { 2 { { { Γ( { { π + π+2 { { { { ={ { Γβ|ππ|π₯(π+π+2)/2 + πΆ)]) { { { { { { { { { { Γ (1 β exp [βπ2 β 4 { { { { { { { { { 2 { { { Γ( { { π + π+2 { { { { { { β1 { { (π+π+2)/2 { β Γ + πΆ)]) , { |ππ|π₯ { { { { { { { { { 4 |ππ| < (π β π)2 , { { { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ (πβπ)/2 { { { σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π₯ β β { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { { { { { π 2 { { + , Γ { { (π+π+2)/2 { +πΆ 2 (2/ (π + π + 2)) β|ππ|π₯ { { { { { 4 |ππ| = (π β π)2 . { (48)
π¦ (π₯) σ΅¨σ΅¨ π σ΅¨σ΅¨ β4 β π2 β(π+1) { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ { { σ΅¨σ΅¨ π σ΅¨σ΅¨ 2 π₯ { { { { β 4 β π2 { π { { [β|ππ| ln π₯ + πΆ]) + , { Γ tan ( { 2 2 { { { { { { |ππ| > (π + 1 + π)2 , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨π₯β(π+1) (π + βπ2 β 4 + (βπ2 β 4 β π) { { σ΅¨σ΅¨ π σ΅¨σ΅¨ { { { ={ Γ exp [βπ2 β 4 (β|ππ| ln π₯ + πΆ)]) { { { { { { β1 { { Γ(1 β exp [βπ2 β 4 (β|ππ| ln π₯ + πΆ)]) , { { { { { { { { |ππ| < (π + 1 + π)2 , { { { { { { { σ΅¨σ΅¨ π σ΅¨σ΅¨ β(π+1) { 2 π { σ΅¨σ΅¨ σ΅¨σ΅¨π₯ { β β + , { σ΅¨σ΅¨ σ΅¨σ΅¨ { { σ΅¨πσ΅¨ β|ππ| ln π₯ + πΆ 2 { { { |ππ| = (π + 1 + π)2 . { 3.4. Other Equations Example 5. Consider the following Riccati equation: π₯π¦σΈ = π₯2π + (π β π) π¦ + π₯2π π¦2 , 2πβ1
π π¦ + ππ₯π π¦2 , π₯
(49)
(51) 2πβ1
, π(π₯) = (π β π)/π₯, and β(π₯) = π₯ . We where π(π₯) = π₯ get π = 0, and the general solution to this equation is given by π¦=β
π (π₯) β4 β π2 β (π₯) 2
Γ tan (
β4 β π2 2
(52) π (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + . 2
Thus the exact closed form solution is π¦ = π₯πβπ tan (
π₯π+π + πΆ) , π+π
(53)
where πΆ is a constant of integration. Example 6. For the given equation 2π₯2 π¦σΈ = β2π2 π₯ + π₯π¦ + 2π¦2 , 2
(54) 2
where π(π₯) = βπ /π₯, π(π₯) = 1/2π₯, and β(π₯) = 1/π₯ , we get π = 0, and the general solution to this equation is given by π¦ = πβ« π(π₯)ππ₯ β
Example 4. For the equation π¦σΈ = ππ₯βπβ2 +
(50)
Γ tan (
π (π₯) β4 β π2 β (π₯) 2
β4 β π 2 2
π (β« βπ (π₯) β (π₯)ππ₯ + πΆ)) + . 2
(55)
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7
Thus the exact closed form solution is π¦ = |π| βπ₯ tan (
β2 |π| + πΆ) , π₯2
(56)
where πΆ is a constant of integration.
π
3.5. Equations Containing Exponential Functions Example 7. Consider the following Riccati equation: π¦σΈ = ππππ₯ + ππ¦2 ,
(57)
where π(π₯) = ππππ₯ , π(π₯) = 0, and β(π₯) = π. The condition equation (42) holds if and only if π(π₯) = π/β|ππ|πβ(π/2)π₯ . Combining this value of π(π₯) with (35), we get π = 2. So π(π₯) = (2/β|ππ|)πβπ₯ . Thus the general solution to this equation can be found by (41); that is, σ΅¨σ΅¨ π σ΅¨σ΅¨ β1 2 βπ₯ σ΅¨ σ΅¨ π¦ (π₯) = β σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨ [ + π ]. σ΅¨σ΅¨ π σ΅¨σ΅¨ β|ππ|ππ₯ + π2 β|ππ|
(58)
4. Applications 4.1. The Central Potential Problem of the Power Law Type. Certain types of Newtons laws of motion are equivalent to the Riccati equation. For example, the equation for the energy conservation in the case of a central potential π(π) is given by the standard expression [7] π2 1 + π (π) . πΈ = ππ2Μ + 2 2ππ2
(59)
Under the influence of a power law central potential π(π) = πβ ππ and πΈ = 0, where πβ is the coupling constant and the exponent π can be either positive or negative, (59) can be transformed into the Riccati equation π€σΈ =
π+2 π+2 2 + π€. 2 2
4.2. Damped Harmonic Oscillators. Various problems in quantum optics, superconductivity, and nonrelativistic quantum mechanics can be described classically by [8]
(60)
ππ₯ ππ σ΅¨σ΅¨σ΅¨σ΅¨ π2 π₯ + π , = β σ΅¨ ππ‘2 ππ‘ ππ₯ σ΅¨σ΅¨σ΅¨π₯=π₯(π‘)
where π₯(π‘) is the particle coordinate, π is the mass of the particle, π is a damping constant, and π(π₯, π‘) is a potential energy that accounts for the interaction of the particle with its environment. If π(π₯, π‘) = βπ₯π(π‘), where π(π‘) is white or colored noise, then (60) becomes the well-known Langevin equation π
ππ₯ π2 π₯ +π β π (π‘) = 0. ππ‘2 ππ‘
ππ₯ π2 π₯ + 2π½ (π‘) + π2 (π‘) π₯ = 0, ππ‘2 ππ‘
of the original equation, where πσΈ = ππ/ππ and π(π‘) = 1/ππ2 . From the condition (25) of Lemma 1, we see that there exists a constant π = 0 for π =ΜΈ β2. Hence, the general solution of (60) is given by π€ (π) = tan (
|π + 2| π + πΆ1 ) . 2
(62)
Therefore, π (π) = πΆ2 [sec ( where πΆ2 is a constant.
2/|π+2| |π + 2| , π + πΆ1 )] 2
(63)
(66)
where π½ is the damping function and π is the the natural frequency of the undamped oscillator. The substitution π¦ = π₯σΈ /π₯ converts (66) to the Riccati equation π¦σΈ = βπ2 (π‘) β 2π½ (π‘) π¦ β π¦2 .
(67)
Here π(π‘) = βπ2 (π‘), π(π‘) = β2π½(π‘), and β(π‘) = β1. The condition (25) holds if and only if the coefficients π and π½ satisfy the following condition: πσΈ π + 2π½ = π, π 2
where π is a constant.
(68)
Then (68) can be written in an equivalent form as π πσΈ = β2π½ (π‘) π + π2 , 2
(69)
which is a Bernoulli equation and can be readily solved for π to obtain π (π‘) =
(61)
(65)
Also the equation of motion is
A solution of this equation will lead to a solution πσΈ =π€ π
(64)
1 β (π/2) + ππ2 β« π½(π‘)ππ‘
.
(70)
We conclude that if the coefficients π and π½ satisfy (70), then the general solution π¦(π‘) of (67) is given by (26). Returning to the original dependent variable by π¦ = π₯σΈ /π₯, we obtain the general solution to (66) as π₯ (π‘) = ππβ« π¦(π‘)ππ‘ ,
(71)
where π is a constant.
5. Conclusion We have converted the Riccati equation into an equivalent equation; then, by using the integrability condition for this equation, πσΈ (π₯) βσΈ (π₯) β β 2π (π₯) = π (π₯) βπ (π₯) β (π₯), π (π₯) β (π₯)
(72)
8 we obtain a separable equation. The first case π(π₯) = π, where constant π is a constant, is considered. Thus, the general solutions of the Riccati equation can be exactly obtained. The second integrability case is obtained for the reduced Riccati equation with an arbitrary function π(π₯). We have considered several distinct examples to illustrate our new approach. The method is also applied to the Riccati equation arising in the solution of certain types of Newtons laws of motion and the damped harmonic oscillators equations.
Conflict of Interests The author declares that there is no conflict of interests regarding the publication of this paper.
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