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World Academy of Science, Engineering and Technology
International Journal of Mathematical and Computational Sciences
Vol:11, No:3, 2017
An Analytical Method for Solving General Riccati
Equation
Y. Pala, M. O. Ertas
the equation, it cannot be considered a general method. Sugai
Abstract—In this paper, the general Riccati equation is transformed Riccati equation into a second order differential
analytically solved by a new transformation. By the method equation by suggesting a new transformation [11]. Since the
developed, looking at the transformed equation, whether or not an transformed equation is more complicated and unsolvable in
explicit solution can be obtained is readily determined. Since the most cases, it is also not general and applicable. Rao and
present method does not require a proper solution for the general Ukidave reduced Riccati equation into a separable form under
solution, it is especially suitable for equations whose proper solutions restricted condition [9]. It is of no importance in respect of
cannot be seen at first glance. Since the transformed second order engineering application. Siller also investigated a separability
linear equation obtained by the present transformation has the
simplest form that it can have, it is immediately seen whether or not condition of the equation [10].
the original equation can be solved analytically. The present method Integrability condition for the Riccati equation has been
is exemplified by several examples. studied by Mak and Harko, and a new method to generate
analytical solutions of the Riccati equation was presented [5].
Keywords—Riccati Equation, ordinary differential equation, Mortici gives a new method of the variation of constants
nonlinear differential equation, analytical solution, proper solution. which leads directly to an equation with separable variables.
I. INTRODUCTION This method also imposes several restrictions on the solution.
HE generalized Riccati equation is defined by Therefore, it cannot also be considered a general method [6].
T When all methods are investigated, we observe that no
݀ݕ ଶ method explains the question about whether or not the
ሺ ሻ ሺ ሻ ሺ ሻ (1)
݀ݔܲݔ ݕܳݔ ݕ െܴݔ ൌ0 analytical solution is obtained in explicit, implicit, or power
series form. In addition to finding the analytical solution to the
Here, ܲሺݔሻ, ܳሺݔሻ, and ܴሺݔሻ are arbitrary functions of ݔ. This problem for arbitrary values of ܲሺݔሻ, ܳሺݔሻ, and ܴሺݔሻ, the
equation is widely encountered in analytical mechanics, present method answers these important questions.
engineering and other fields. Therefore, depending on the
PECIAL TYPE RICCATI EQUATION
functions ܲሺݔሻ, ܳሺݔሻ and ܴሺݔሻ, several methods have been II. S
developed to solve various types of Riccati equations [3], [4]. In order to solve (1), consider a new transformation in the
⁄
If ሺݔሻ ൌ 0, then the transformation ݕൌ1 ݖ reduces (1) form
into a first order linear equation. For the case of ܴሺݔሻ ് 0,
ሺ ሻ ሺ ሻ
several methods are available in the literature. The equation in ሺ ሻ ௫ ௬ ௫ ௗ௫ (2)
this case is also known as Bernoulli equation. ݕത ൌ ݂ ݔ ݁
The classical method for solving Riccati equation makes where ሺݔሻ, ݃ሺݔሻ are functions to be determined in a
⁄ convenient manner. We differentiate both sides of (2) twice:
the transformation ݕതൌݕ 1 ݕ, where ݕ is a known
solution of the equation. However, it is not possible to see a ሺ ሻ ሺ ሻ
proper solution to the equation at every time. Therefore, this ݕത′ൌሺ݂ᇱ ݂݃ݕሻ݁ ௫ ௬ ௫ ௗ௫ (3)
method has a limited usage. Among the other methods, Rao and
transformation and Allen-Stein transformation can be ݕത′′ ൌ ሺ݂݃ݕᇱ 2݂ᇱ݃ݕ ݂݃ᇱݕ݂݃ଶݕଶ݂′′ሻ݁ሺ௫ሻ௬ሺ௫ሻௗ௫ (4)
mentioned. The basic idea here is to bring the main equation
into a separable form [1], [7], [8]. However, the Equation (4) can be written as
differentiability condition is a serious problem with these
International Science Index, Mathematical and Computational Sciences Vol:11, No:3, 2017 waset.org/Publication/10006683methods. ଶᇱᇱᇱᇱଵ ሺ ሻ ሺ ሻ
ᇱ ଶ ି ௫ ௬ ௫ ௗ௫
Harko et al. have investigated the Riccati Equation and ݕ ቂ ቃݕ݃ݕ ൌݕത′′݁ (5)
developed a restricted analytical solution [2]. Since the
method requires specific conditions among the coefficients in Recall that the left-hand side of (5) has the form of Riccati
equation. Comparing (1) and (5), we have
ଶᇱ ᇱ
Y. Pala, Prof. Dr., is with the Mechanical Engineering Department, ቂ ቃൌܲሺݔሻ (6a)
Uludag University, Bursa, Gorukle 16059 Turkey(phone: +90 533 7686390,
e-mail: mypala@uludag.edu.tr). ݃ሺݔሻ ൌܳሺݔሻ (6b)
M. O. Ertas, MSc, is with the Mechanical Engineering Department,
Uludag University, Bursa, Gorukle 16059 Turkey (phone: +90 542 5453387,
e-mail: moertas@gmail.com).
International Scholarly and Scientific Research & Innovation 11(3) 2017 125 scholar.waset.org/1307-6892/10006683
World Academy of Science, Engineering and Technology
International Journal of Mathematical and Computational Sciences
Vol:11, No:3, 2017
ᇱᇱ ൌെܴሺݔሻ (6c) ݃ൌ1 (16b)
ᇱᇱ ଶହ ଶ ହ
If we wish to solve an equation of the form ൌ ݔ (16c)
ᇱ ଶᇱ ᇱ ଶ ᇱᇱ ସ ଶ
ݕ ቂ ቃݕ݃ݕ ൌ0 (7) The function ݂ can be obtained via (16a) and (16b)
ఱ௫మ (17)
ర
then, by (5), in the first place, we can have ݂ ൌܿ݁
ᇱᇱ Using (9), we find
ݕത ൌ0 → ݕതൌܽݔܾ (8)
ത
The functions ݂ and ݃ are to be obtained such that (6) are ଵ ௗ ௬ ଵ ௗ ௫ା (18)
ݕൌ ቀln ቁൌ ቆln ఱమቇ
satisfied. Now, using the inverse transformation from (2), we ௗ௫ ଵௗ௫ రೣ
obtain
ത or ଵ ହ
ݕൌଵ ௗ ቀln௬ቁ (9) ݕൌ ̅ െ ݔ, ܿ̅ൌܾ/ܽ (19)
ௗ௫ ሺ௫ାሻ ଶ
We recall that, for some simple types of Riccati equations, Again (19) satisfies (15).
the solution is obtained without having any difficulty in Example 3: (20) is required to be solved.
solving the second order equation ((8)). In the other words, ᇱ ଶ ଶ ଵ ଶ
the solution is found without solving a complex second order ݕ 2ݔݕݔݕ െ௫మ௫ర1ൌ0 (20)
differential equation with variable coefficients. The examples
that are numbered from 1 to 4 illustrate the method. Comparison of (20) with (7) yields
Example 1: We first try to solve the equation ଶᇱ ᇱ
ᇱ ଶ ቂ ቃൌ2ݔ (21a)
ݕ 2ݕݕ 1ൌ0 (10)
ଶ (21b)
whose solution can also be found by integration. Comparing ݃ൌݔ
(10) with (7) gives ᇱᇱ ଵ ଶ
ൌെమ ర1 (21c)
ଶᇱ ᇱ ௫ ௫
ቂ ቃൌ2 (11a)
݃ൌ1 (11b) The function ݂ can readily be obtained
మ
ଵ ೣ (22)
ᇱᇱ ݂ ൌܿ. .݁మ
ൌ1 (11c) ௫
Inserting ݃ൌ1 into (11a) and solving the equation yields Using (9), we obtain
ത
௫ ݕൌଵ ௗ ቀln௬ቁൌଵ ௗ ቆln ௫ା మቇ (23)
݂ ൌܿ݁ (12) ௗ௫ ௫మௗ௫ షభ ೣ
Equation (11c) is automatically satisfied. Now, using (9), .௫ . మ
we have or ݕൌ ଵ െଵଵ ܿ̅ൌܾ/ܽ
మ ̅ య (24)
ത ௫ ሺ௫ାሻ ௫ ௫
ଵ ௗ ௬ ଵ ௗ ௫ା
ݕൌ ቀln ቁൌ ቀln ೣቁ (13)
ௗ௫ ଵௗ௫ In the examples ever seen, the examples satisfying the
തሺ ሻ ⁄
or condition ܴ ݔ ൌ݂′′ ݂݃ automatically have been chosen.
ݕൌଵ ̅െ1, ܿ̅ ൌܾ/ܽ (14) Otherwise, the unknown function ݂ሺݔሻ must be found such
International Science Index, Mathematical and Computational Sciences Vol:11, No:3, 2017 waset.org/Publication/10006683 ௫ାthat the equations must be simultaneously satisfied.
It can be verified that (14) satisfies (10). തሺ ሻ ଶᇱ ᇱ
Example 2: We now solve the equation ܲ ݔ ൌቂ ቃ (25a)
ଶହ ହ തሺ ሻ (25b)
ᇱ ଶ ଶ ܳ ݔ ൌ݃
ݕ 5ݔݕݕ ସ ݔ ଶൌ0 (15) തሺ ሻ ᇱᇱ
Comparison of (15) with (7) yields ܴ ݔ ൌ (25c)
ଶᇱ ᇱ However, it is not always possible to obtain a unique ݂ሺݔሻ
ቂ ቃൌ5ݔ (16a) that satisfies both (25a) and (25c). To show this case, we
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World Academy of Science, Engineering and Technology
International Journal of Mathematical and Computational Sciences
Vol:11, No:3, 2017
finally consider fourth example in this section. In order to remove the restriction in Section II, we now
Example 4: We try to solve the equation assume that we seek the solution of the equation
ᇱ ଶ ଶ തሺ ሻ ᇱ ଶᇱ ᇱ ଶ ᇱᇱ
ݕ ߙݔݕߚݔݕ ܴݔ ൌ0 (26) ݕ ቂ ቃݕ݃ݕ ൌܵሺݔሻ (37)
Using (25a) and (25b), we can write where ܵሺݔሻ is a function to be determined. Thus, (36) gives
ଶᇱ ᇱ ଶ
ቂ ቃൌߙݔ (27a) ሺ ሻ ሺ ሻ
ᇱᇱ ሺ ሻ ௫ ௬ ௫ ௗ௫ (38)
݃ൌߚݔ (27b) ݕത ൌ݂.݃.ܵݔ .݁
However, by (33), we can also write
The function ݂ can be obtained as ᇱᇱ ሺ ሻ
ݕത ൌ݃ܵݔ ሺݕതሻ (39)
ିଵ ఈ௫య Replacing ݕത by ݑሺݔሻ, we obtain
݂ ൌܿ.ݔଶ.݁ (28)
ݑᇱᇱ െ݃ܵݑൌ0 (40)
We then have
തሺ ሻ ᇱᇱ ఈమ ଷ ଷ ିଷ ఈ Equation (40) is a linear second order ordinary differential
ܴ ݔ ൌൌସఉݔ ସఉݔ ଶఉ (29) equation whose general solution is generally obtained in
Thus, the present method gives a solution as long as the power series. However, if ݃ܵ is a constant or something that
leads to the analytical solution of (40), then the explicit form
condition given in (29) is satisfied. The new form of (24) of ݑ or ݕ can always be obtained.
becomes Now, an attention should be paid to (40). According to this
ᇱ ଶ ଶ ఈమ ଷ ଷ ିଷ ఈ result, we are led to conclude that one of the main advantage
ݕ ߙݔݕߚݔݕ ସఉݔ ସఉݔ ଶఉൌ0 (30) of the transformation given by (33) is that it converts Riccati
equation directly into the linear form of (40). Equation (40)
The solution can be obtained in the following form has the simplest form of a second order differential equation.
This form also provides us the knowledge of whether we can
ത obtain the explicit solution of (37) or not. After (40) is solved
ݕൌଵ ௗ ቀln௬ቁൌଵ ௗ ቆln ௫ା యቇ (31) via classical methods, the inverse transformation yields the
ௗ௫ ఉ௫ௗ௫ భ ഀೣ
ష ల
.௫ మ. solution as
or ଵ ఈ௫ ଵ ଵ ௗ ௨ሺ௫ሻ
ݕൌ െ ܿ̅ ൌܾ/ܽ ݕൌ ቀln ቁ (41)
̅ మ (32) ሺ௫ሻௗ௫ ሺ௫ሻ
ఉ௫ሺ௫ାሻ ଶఉ ଶఉ௫ Since (41) does not involve any expression to be integrated,
III. THE RICCATI EQUATION OF GENERAL TYPE there exists no difficulty in obtaining the explicit form of
The transformation in (2) prevents the free choice of the ሺ ሻ
term ܳሺݔሻ in (1). So, the method can be valid only for special ݕ ݔ .
type Riccati equations. Now, we wish to remove this Example 5: As a first example in this section, we solve the
restriction to have the analytical solution of general type equation
Riccati equation involving arbitrary ܲሺݔሻ, ܳሺݔሻ, and ܴሺݔሻ. ᇱ ଶ ଶ
ݕ 2ݔݕെݕ െሺ1ݔሻൌ0 (42)
Again, we consider the same transformation
ሺ ሻ ሺ ሻ Comparing (42) and (37), we have
ሺ ሻ ௫ ௬ ௫ ௗ௫ (33)
ݕതൌ݂ ݔ ݁ ݃ൌെ1 (43a)
Differentiating (33) twice yields ଶᇱ ᇱ
ሺ ሻ ሺ ሻ ቂ ቃൌ2x (43b)
International Science Index, Mathematical and Computational Sciences Vol:11, No:3, 2017 waset.org/Publication/10006683 ݕത′ ൌ ሺ݂ᇱ ݂݃ݕሻ݁ ௫ ௬ ௫ ௗ௫ (34) ᇱᇱሺଶሻሺ ሻ
and ݕത′′ ൌ ሺ݂݃ݕᇱ 2݂ᇱ݃ݕ ݂݃ᇱݕ݂݃ଶݕଶ ൌെ1ݔ , ܵ ݔ ൌ0 (43c)
(35) మ⁄
ሺ ሻ ሺ ሻ ௫ ଶ
݂′′ሻ݁ ௫ ௬ ௫ ௗ௫ Solving (43a) and (43b) gives ݂ൌܿ݁ (ܿ=constant), and
⁄
we can see immediately that ݂′′ ݂݃ directly equals ሺݔሻ.
Equation (35) can be written as Hence, ܵሺݔሻ can be taken as zero. This means that ݕതᇱᇱ ൌ0.
ଶᇱ ᇱ ᇱᇱ ଵ ሺ ሻ ሺ ሻ The solution of ݕത is given by ݕതൌܽݔܾ. Substituting this
ᇱ ଶ ି ௫ ௬ ௫ ௗ௫ result into the inverse transformation, after some operations,
ݕ ቂ ቃݕ݃ݕ ൌݕത′′݁ (36) we obtain
International Scholarly and Scientific Research & Innovation 11(3) 2017 127 scholar.waset.org/1307-6892/10006683
World Academy of Science, Engineering and Technology
International Journal of Mathematical and Computational Sciences
Vol:11, No:3, 2017
ݕൌݔെଵ ̅ ሺܿ̅ ൌܿ݊ݏݐܽ݊ݐሻ (44) is obvious that (55a) and (55b) are valid:
௫ା ଵ
It can be checked that (44) satisfies (42). ݃ൌ (55a)
Example 6: We seek the solution of the equation ଶᇱ ᇱ ௫ ଵ
ቂ ቃൌ4െ (55b)
ᇱ ଶ ଶ (45) ௫
ݕ 8ݔݕ4ݕ 4ݔ െ3ൌ0 ଶ௫
Comparing (37) and (45), we have Solving (55a) and (55b) gives ݂ൌܿ݁ (ܿ=constant). Now,
noting that
݃ൌ4 (46a) ᇱᇱ ൌ4ݔ
ଶᇱ ᇱ (56)
ቂ ቃൌ8ݔ (46b)
మ from (37), we obtain ܵሺݔሻ as
ଶ௫
Solving (46a) and (46b) gives ݂ൌܿ݁ (ܿ=constant). The ᇱᇱ ሺ ሻ ଶ ሺ ሻ ଶ
other condition yields െܵݔ ൌെݔ 4ݔ⟹ 4ݔെܵݔ ൌെݔ (57)
ሺ ሻ ଶ
ᇱᇱ ଶ ሺ ሻ 4ݔ ⟹ ܵ ݔ ൌݔ
ൌ4ݔ 1, ܵ ݔ ൌ4 (47) Hence, (40) reads
The transformed equation in this case becomes ݑᇱᇱ െݔݑൌ0 (58)
ᇱᇱ
ݑ െ16ݑൌ0 (48) Equation (58) is Airy equation, and its solution is given by
The characteristic equation of (48) is ݕത ൌݑൌܿܣ݅ݎݕܣ݅ሺݔሻܿܣ݅ݎݕܤ݅ሺݔሻ (59)
݉ଶെ16ൌ0 (49) ଵ ଶ
Here, the functions ܣ݅ݎݕܣ݅ሺݔሻ and ܣ݅ݎݕܤ݅ሺݔሻ are given as
whose roots are ݉ ൌ4 and ݉ ൌെ4. Thus, the solution of
ଵ ଶ ሺ ሻ ஶ ௫య
∑
(48) has the form ܣ݅ݎݕܣ݅ ݔ ൌ1 ୀଵ ሺ ሻሺ ሻ
ସ௫ ିସ௫ ଶ.ଷ.ହ.… ଷିଷ . ଷିଵ .ଷ (60)
ݑൌܿ݁ ܿ݁ (50) ௫యశభ
ଵ ଶ ሺ ሻ ஶ
∑
ܣ݅ݎݕܤ݅ ݔ ൌݔ ୀଵ
ሺ ሻ ሺ ሻ
Using (41), we obtain ݕሺݔሻ as ଷ.ସ..… ଷିଶ .ଷ. ଷାଵ
రೣ షరೣ Their derivatives are given by
ଵ ௗ ௨ሺ௫ሻ ଵ ௗ ା
ݕൌ ቀln ቁൌ ቀln భ మ ቁ (51) యషభ
మ ᇱ ௫
ሺ௫ሻௗ௫ ሺ௫ሻ ସௗ௫ మೣ ሺ ሻ ஶ
൫ܣ݅ݎݕܣ݅ ݔ ൯ ൌ ∑
ୀଵ ሺ ሻሺ ሻ
రೣି షరೣ ଶ.ଷ.ହ.… ଷିଷ . ଷିଵ (61)
ݕൌభ రೣ మ షరೣെݔ (52) ௫య
ା ஶ
భ మ ሺ ሻ ∑
ሺܣ݅ݎݕܤ݅ ݔ ሻ′ ൌ 1 ୀଵ ሺ ሻ
It can be verified that ଷ.ସ.ହ..… ଷିଶ .ଷ
ఴೣ Hence, using the inverse transformation, we obtain
ି̅ ⁄ (53)
ݕൌ െݔ,ሺܿ̅ ൌܿ ܿ ሻ ᇲ ᇲ
ఴೣ ̅ ଶ ଵ ሺ ሻ ሺ ሻ
௬ ௫ ା. ௬ ௫
ା ൫ ൯ ൫ ൯ (62)
ݕൌݔሺ ሺ ሻ ሺ ሻ െ2ሻ
Note that if we take ܿ ݁ସ௫ as the solution of ݑ, then we ௬ ௫ ା.௬ ௫
ଵ
obtain the proper solution ݕ ൌ1െݔ. In addition, if we take where ܿ is a constant.
ିସ௫ Example 8: We try to solve the equation which has been
ܿ ݁ only, then we find the other proper solution of (44) as
ଶ studied by Mortici [6]
ݕ ൌെ1െݔ. Indeed, we can verify that these are proper
International Science Index, Mathematical and Computational Sciences Vol:11, No:3, 2017 waset.org/Publication/10006683solutions of (45). Thus, the method presented also gives the ᇱఉଶ ఊ
proper solutions of Riccati Equation. Then, when desired, one ݕ െ௫ݕെߙݕ െ௫మൌ0 (63)
̅ሺ ሻ ⁄
can use the famous transformation ݕൌܵݔ 1 ݔ to find the
general solution. Here, ܵ̅ሺݔሻ is a proper solution. Comparing (37) and (63), we have
Example 7: We now try to solve the equation ݃ൌെߙ (64a)
ᇱ ଵ ଵ ଶ ଶ
ݕ ቀ4െ ቁݕ ݕ െݔ 4ݔൌ0 (54) ଶᇱ ᇱ ఉ
௫ ௫ ቂ ቃൌെ (64b)
We recall that a proper solution of (54) is not readily seen. ௫ ିఉ⁄ଶ
Therefore, the well-known classical method cannot be used. It Solving (64a) and (64b) gives ݂ൌܿݔ . (ܿ=constant).
Noting that
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