Advances in Pure Mathematics, 2011, 1, 295-296
doi:10.4236/apm.2011.15053 Published Online September 2011 (http://www.SciRP.org/journal/apm)
A Closed Form Solution to a Special Normal Form of
Riccati Equation
Haiduke Sarafian
University College, The Pennsylvania State University, York, USA
Received May 18, 2011; revised June 28, 2011; accepted July 10, 2011
Abstract
We present the general solution to the Riccati differential equation, 2
d
d
n
wzw
z

, for arbitrary real number
n.
Keywords: Riccati Differential Equation, Bessel Functions
1. Introduction
We consider a special normal form of Riccati equation,
, where
2n
wzw
 ddww
z. We prove, for any real
number n, the equation has a closed analytic solution.
We show, the general solution for is a product
of
2n
n
z and a combination of Bessel functions of vari-
ous n—dependent orders and arguments. We discuss the
variable transformations leading to this general result.
Introducing another set of variable transformations con-
ducive to a closed analytic solution, we by-pass the sin-
gular behavior of the Bessel functions for n = –2. We
show, the general solution in the limit of is
identical to the solution of n = –2.
2n
2. Procedure
Riccati equation is given by,
 
2,uaz bzuczu
  (1)
where , and are analytic functions of z.
It is known, for ,

az

bz

cz

cz
0






2
1
22
bzc z
uw
czcz cz


transforms (1) into the normal form


2,wAz w

(2)
where

2
231
.
4242 2
bbcbc c
Az accc
 

 


It is the objective of this paper to solve (2) for a spe-
cial case where


n
A
zz

. We prove the solution for any
real value of n is analytic.
We begin with Euler variable transformation, [1, p.
112], namely


dd lnwzz yz . This linearizes (2)
0,
n
yzy


 (3)
Multiplying both sides of (3) by we compare the
result,
2
z
22
0
n
zyz y



 vs.


2222 22
14 0zyzv y

  


 of [1, p. 206] and
deduce the following identities







22, 1212,
and 122
nv n
n


 
 
Furthermore, according to the last reference, the solu-
tion of the given equation is,

yzfz


 where


f

is a solution of the Bessel equation of order v.
Therefore, we conclude the solution to, 22
0
n
zyz y




is


12
,yzcJ cY

 



 (4)
where Jv and Yv are the Bessel functions of the first and the
second kind of order ν, with c1 and c2 being two arbitrary
constants. Substituting (4) in Euler transformation and
applying the chain differentiation, 1
dd ddzz



 we
deduce






 
12 12
12
2,
2
vv vv
vv
cJcYz cJcY
wzcJ cY

 


 

 

(5)
c

with the prime notation indicating the derivative of the
Bessel functions with respect to variable

. Further-

H. SARAFIAN 296
more, applying two sets of recurrence relationships [2, p.
361],












11
11
1
2
,,
2
vv
vvv
JJ
YY
JJJJ
YYYY
v








 
 

 
 

 












  
 v



simplifies (5)
22
22
11 21
22
2
2
11 21
22
22
2,
22
22
nn
nn
nnn
n
nn
cJzcYz
nn
wz
cJz cYz
nn












 





2
2
2
n



(6)
And, (6) simplifies to its final form
22
2
11
22
2
22
11
22
22
22
,
22
22
nn
nn
nn n
n
nn
CJz Yz
nn
wz
CJz Yz
nn












 





2
2n




(7)
With 12
.Ccc
We observe that solution (2) is a one-parameter family
function.
The n = –2 is the pole of the order and the argument of
the Bessel functions and needs a special consideration.
For n = –2 Equation (2) takes the form
2
2
1,w
z
w
(8)
By inspection, 2
1
w  z solves (8). Substituting w
in (8) gives . We select
1
210
 

12 13i  
to further the analysis. The standard variable transforma-
tion [1, p. 50]

1
1
ww z

 , (9)
gives the general solution. Substituting (9) in (8) yields,
1
'21 0,vv
z

 (10)
The solution of (10) is,

1
2
1
21cz z

  , with C
being a constant. The solution to (8) yields,

3
11 1
13 1
2
3
i
wi
zCz i



 




Figure 1. The s olid curve is the graph of the solution given in
(12) for c1 = 1. The gray curve is the graph of the solution
given in (7) for n = –1.99 and C = 0.85.
Applying the identities, ln
z
zeand 1 tan
iz iz
iz iz
ee
zie e




and suitably selecting the value of 1
3
3
ci
i
Ce

 , (11)
becomes

1
13
13cot ln
22
wz
z,c





 

 




(12)
We show graphically, see Figure 1, that (7) in the limit
of is identical to (12).
2n
3. A comment and an Acknowledgement
By applying multiple step variable transformations (not
reported in this paper) the author devised a method of
solving (2) resulting in (7). The author appreciates the in
depth discussion with Prof. Rostamian.
4. References
[1] W. E. Boyce and R. C. DiPrima, “Elementary Differential
Equations and Boundary Value Problems,” 6th Edition,
Wiley, New York, 1997.
[2] M. Abramowitz and I. A. Stegun, “Handbook of Mathe-
matical Functions,” Dover Publications Inc., New York,
1970.
,
(11)