· abstract: fractional calculus theory includes definition of the derivatives and integrals of...

ABSTRACT: Fractional calculus theory includes definition of the derivatives and integrals of arbitrary order. This theory is used to solve some classes of singular differential equations and fractional order differential equations. One of these equations is the confluent hypergeometric equation. In this paper, we intend to solve this equation by applying

1

A Different Solution Method for the Confluent Hypergeometric Equation 1

2

ABSTRACT: Fractional calculus theory includes definition of the derivatives and 3

integrals of arbitrary order. This theory is used to solve some classes of singular 4

differential equations and fractional order differential equations. One of these equations 5

is the confluent hypergeometric equation. In this paper, we intend to solve this equation 6

by applying 𝑁𝑁𝜇𝜇 method as a different solution method. 7

Keywords: Fractional calculus theory, fractional solutions, Confluent hypergeometric 8

equation, 𝑁𝑁𝜇𝜇 method, generalized Leibniz rule 9

10

Konfluent Hipergeometrik Denklemi İçin Farklı Bir Çözüm Metodu 11

12

ÖZET: Kesirli hesap teorisi, keyfi mertebeden türev ve integral tanımını 13

kapsamaktadır. Diferansiyel denklemlerin ve kesirli diferansiyel denklemlerin bazı 14

sınıflarını çözmek için bu teori kullanılmaktadır. Bu denklemlerden birisi konfluent 15

hipergeometrik denklemidir. Bu makalede, farklı bir çözüm metodu olarak 𝑁𝑁𝜇𝜇 16

metodunun uygulanmasıyla bu denklemi çözmeyi hedeflemekteyiz. 17

Anahtar Kelimeler: Kesirli hesap teorisi, kesirli çözümler, Konfluent hipergeometrik 18

denklemi, 𝑁𝑁𝜇𝜇 metot, genelleştirilmiş Leibniz kuralı 19

INTRODUCTION 20

The fractional calculus theory enables a set of axioms and methods to generalize the 21

coordinate and corresponding derivative notions from integer 𝑘𝑘 to arbitrary order 𝜇𝜇, 22

{𝑥𝑥𝑘𝑘 ,𝜕𝜕𝑘𝑘 𝜕𝜕𝑥𝑥𝑘𝑘⁄ }→ {𝑥𝑥𝜇𝜇,𝜕𝜕𝜇𝜇 𝜕𝜕𝑥𝑥𝜇𝜇⁄ } in a good light. Fractional differential equations are 23

applied in a widespread manner in robot technology, PID control systems, Schrödinger 24

method as a different solution method.

Keywords: Fractional calculus theory, fractional solutions, Confluent hypergeometric equation,

1


2








10


12








INTRODUCTION 20





method, generalized Leibniz rule

ÖZET: Kesirli hesap teorisi, keyfi mertebeden türev ve integral tanımını kapsamaktadır. Diferansiyel denklemlerin ve kesirli diferansiyel denklemlerin bazı sınıflarını çözmek için bu teori kullanılmaktadır. Bu denklemlerden birisi konfluent hipergeometrik denklemidir. Bu makalede, farklı bir çözüm metodu olarak

Yapılması Gereken Değişiklikler (sy. 325-334)

2.) sy. 325/Türkçe özetin 3. satırında ‘‘farklı bir çözüm metodu olarak metodunun uygulanmasıyla’’ yerine ‘‘farklı bir çözüm metodu olarak metodunun uygulanmasıyla’’ yazılması,

-----------------------------------------------------------------------------------------------------------------

3.) sy. 326/Sol sütunun 4.-5. satırlarında ‘‘from integer to arbitrary order ,’’ yerine ‘‘from integer to arbitrary order ,’’ yazılması,

4.) sy. 326/Sol sütunun 6. satırında ‘‘a widespread manner’’ ifadesinin yazı karakteri metnin yazı karakterinden farklı olmuş. Bu ifadenin yazı karakterinin değiştirilmesi,

5.) sy. 326/Denklem (1)-(2) arasındaki ‘‘and’’ yerine ‘‘and,’’ yazılması ve ‘‘and,’’in sola dayalı yazılması,

6.) sy. 326/Denklem (2)’den sonraki satırın sola dayalı yazılması,

7.) sy. 326/Definition 2.1.’in 2. satırında ‘‘inside and on’’ yerine ‘‘inside and on ’’ yazılması,

8.) sy. 326/Sağ sütunun 2. satırında ‘‘realivity’’ yerine ‘‘relativity’’ yazılması,

9.) sy. 326/Sağ sütunun 13. satırındaki fazladan bir virgülün silinmesi,

10.) sy. 326/Sağ sütunun 14. satırında ‘‘and returns to the point at ,’’ yerine ‘‘and returns to the point at ,’’ yazılması,

-----------------------------------------------------------------------------------------------------------------

11.) sy. 327/1. satırdaki ‘‘where’’in sola dayalı yazılması ve bu ifadenin ‘‘where ,’’ olarak değiştirilmesi,

12.) sy. 327/Denklem (5)’ten sonraki ‘‘(Yilmazer and Ozturk, 2013).’’ün sola dayalı yazılması,

13.) sy. 327/Lemma 2.1.’de ‘‘ and fractional order derivatives exist,’’ yerine ‘‘fractional order derivatives and exist,’’ yazılması,

14.) sy. 327/Denklem (6)’dan sonraki satırın sola dayalı yazılması ve ‘‘and are constants and , ’’ yerine ‘‘ and are constants and’’ yazılması ve de enter yapılmadan bu ifadeden hemen sonra ‘‘ , .’’ ifadesinin gelmesi,

15.) sy. 327/Lemma 2.2.’de ‘‘( ) and ( ) fractional order derivatives exist,’’ yerine ‘‘fractional order derivatives ( ) and ( ) exist,’’ yazılması,

16.) sy. 327/Denklem (7)’den sonraki satırın sola dayalı yazılması ve ‘‘and . | ( ) ( ) ( )| .’’

yerine ‘‘and | ( ) ( ) ( )| .’’ yazılması,

17.) sy. 327/ Property 2.3’de ‘‘Let be a constant.’’ yerine ‘‘Let be a constant.’’ yazılması,

-----------------------------------------------------------------------------------------------------------------

18.) sy. 328/ Denklem (11)’den sonraki ‘‘and’’in sola dayalı yazılması ve bu ifadenin ‘‘and,’’ olarak değiştirilmesi,

19.) sy. 328/ Denklem (12)’den sonraki satırın sola dayalı yazılması,

metodunun uygulanmasıyla bu denklemi çözmeyi hedeflemekteyiz.

Anahtar Kelimeler: Kesirli hesap teorisi, kesirli çözümler, Konfluent hipergeometrik denklemi,

1


2








10


12








INTRODUCTION 20





metot, genelleştirilmiş Leibniz kuralı

A Different Solution Method for the Confluent Hypergeometric Equation

Konfluent Hipergeometrik Denklemi İçin Farklı Bir Çözüm Metodu

Ökkeş ÖZTÜRK1

Iğdı

r Ü

nive

rsit

esi F

en B

ilim

leri

Ens

titü

sü D

ergi

si

Iğdı

r U

nive

rsity

Jou

rnal

of

the

Inst

itute

of

Scie

nce

and

Tech

nolo

gy

Araştırma Makalesi / Research Article Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech. 7(2): 215-224, 2017

1 Bitlis Eren Üniversitesi, Fen-Edebiyat Fakültesi, Matematik Bölümü, Bitlis, Türkiye Sorumlu yazar/Corresponding Author: Ökkeş ÖZTÜRK, [email protected]

Geliş tarihi / Received: 28.07.2016Kabul tarihi / Accepted: 06.10.2016

Cilt

/Vol

ume:

7, S

ayı/I

ssue

: 2,

Say

fa/p

p: 2

15-2

24, 2

017

ISSN

: 214

6-05

74, e

-ISS

N: 2

536-

4618

D

OI:

10.

2159

7/jis

t.201

7.15

3

Iğdır Üni. Fen Bilimleri Enst. Der. / Iğdır Univ. J. Inst. Sci. & Tech.216

Ökkeş ÖZTÜRK

INTRODUCTION

The fractional calculus theory enables a set of axioms and methods to generalize the coordinate and corresponding derivative notions from integer



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



to

arbitrary order μ,

1


2








10


12








INTRODUCTION 20





in a good light. Fractional differential equations are applied in a widespread manner in robot technology, PID

control systems, Schrödinger equation, KdV equations, heat transfer, relativity theory, economy, filtration, controller design, mechanics, optics, modelling and so on (Akgül, 2014; Akgül et al., 2015).

Riemann-Liouville fractional differentiation and fractional integration that are two most important definitions of fractional calculus are, respectively,

2

equation, KdV equations, heat transfer, realivity theory, economy, filtration, controller 25

design, mechanics, optics, modelling and so on (Akgül, 2014; Akgül et al., 2015). 26

Riemann-Liouville fractional differentiation and fractional integration that are two most 27

important definitions of fractional calculus are, respectively, 28

29

𝐷𝐷𝑎𝑎 𝑡𝑡𝜇𝜇𝑓𝑓(𝑡𝑡) = 1

Г(𝑘𝑘 − 𝜇𝜇)𝑑𝑑𝑘𝑘𝑑𝑑𝑡𝑡𝑘𝑘 ∫ 𝑓𝑓(𝜏𝜏)(𝑡𝑡− 𝜏𝜏)𝑘𝑘−𝜇𝜇−1

𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32

𝐷𝐷𝑎𝑎 𝑡𝑡−𝜇𝜇𝑓𝑓(𝑡𝑡) = 1

Г(𝜇𝜇)∫𝑓𝑓(𝜏𝜏)(𝑡𝑡 − 𝜏𝜏)𝜇𝜇−1𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33

where 𝑘𝑘 ∈ ℕ and Γ is Euler’s function gamma (Oldham and Spanier, 1974; Miller and 34

Ross, 1993; Podlubny, 1999; Yilmazer and Ozturk, 2013). 35

MATERIAL AND METHOD 36

Definition 2.1. If the function 𝑓𝑓(𝑧𝑧) is analytic (regular) inside and on 𝐶𝐶, where 37

𝐶𝐶 = {𝐶𝐶− ,𝐶𝐶+}, 𝐶𝐶− is a contour along the cut joining the points 𝑧𝑧 and −∞ + 𝑖𝑖Im(𝑧𝑧), 38

which starts from the point at −∞, encircles the point 𝑧𝑧 once counter-clockwise, and 39

returns to the point at −∞, and 𝐶𝐶+ is a contour along the cut joining the points 𝑧𝑧 and 40

∞ + 𝑖𝑖Im(𝑧𝑧), which starts from the point at ∞, encircles the point 𝑧𝑧 once counter-41

clockwise, and returns to the point at ∞, 42

43

(1)

and,

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

(2)

where

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

and

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

is Euler’s function gamma

(Oldham and Spanier, 1974; Miller and Ross, 1993;

Podlubny, 1999; Yilmazer and Ozturk, 2013).

MATERIAL AND METHOD

Definition 2.1. If the function

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

is analytic

(regular) inside and on

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

, where

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

is a contour along the cut joining the points

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

and

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

which starts from the point at

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

encircles the point

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

once counter-clockwise, and

returns to the point at

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

and

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

is a contour along

the cut joining the points

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

and

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

which

starts from the point at

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

encircles the point

2





29



𝑡𝑡

𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑘𝑘− 1 ≤ 𝜇𝜇 < 𝑘𝑘), (1)

30

and 31

32



𝑎𝑎

𝑑𝑑𝜏𝜏 (𝑡𝑡 > 𝑎𝑎, 𝜇𝜇 > 0), (2)

33










43

once

counter-clockwise, and returns to the point at



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



,

3

𝑓𝑓𝜇𝜇(𝑧𝑧) = [𝑓𝑓(𝑧𝑧)]𝜇𝜇 = Г(𝜇𝜇+ 1)2𝜋𝜋𝜋𝜋 ∫ 𝑓𝑓(𝜏𝜏)𝑑𝑑𝜏𝜏

(𝜏𝜏− 𝑧𝑧)𝜇𝜇+1𝐶𝐶

(𝜇𝜇 ∉ ℤ−),

𝑓𝑓−𝑘𝑘(𝑧𝑧) = lim𝜇𝜇→−𝑘𝑘

𝑓𝑓𝜇𝜇(𝑧𝑧) (𝑘𝑘 ∈ ℤ+),

(3)

44

where 𝜏𝜏 ≠ 𝑧𝑧, 45

46

−𝜋𝜋 ≤ arg(𝜏𝜏− 𝑧𝑧) ≤ 𝜋𝜋 for 𝐶𝐶− ,

0 ≤ arg(𝜏𝜏− 𝑧𝑧) ≤ 2𝜋𝜋 for 𝐶𝐶+ . (4)

47

In that case, 𝑓𝑓𝜇𝜇(𝑧𝑧) (𝜇𝜇 > 0) is the fractional derivative of 𝑓𝑓(𝑧𝑧) of order 𝜇𝜇 and 𝑓𝑓𝜇𝜇(𝑧𝑧) (𝜇𝜇 <48

0) is the fractional integral of 𝑓𝑓(𝑧𝑧) of order −𝜇𝜇, confirmed (in each case) that 49

|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)

(Yilmazer and Ozturk, 2013). 50

Lemma 2.1. (Linearity) When 𝑓𝑓𝜇𝜇 and 𝑔𝑔𝜇𝜇 fractional order derivatives exist, then 51

52

(𝐢𝐢) [𝑐𝑐1𝑓𝑓(𝑧𝑧)]𝜇𝜇 = 𝑐𝑐1[𝑓𝑓(𝑧𝑧)]𝜇𝜇, (6)

(𝐢𝐢𝐢𝐢) [𝑐𝑐1𝑓𝑓(𝑧𝑧) + 𝑐𝑐2𝑔𝑔(𝑧𝑧)]𝜇𝜇 = 𝑐𝑐1[𝑓𝑓(𝑧𝑧)]𝜇𝜇 + 𝑐𝑐2[𝑔𝑔(𝑧𝑧)]𝜇𝜇,

53

where 𝑓𝑓(𝑧𝑧), 𝑔𝑔(𝑧𝑧) are analytic and single-valued functions, 𝑐𝑐1 and 𝑐𝑐2 are constants and 54

𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55

Lemma 2.2. (Index law) If (𝑓𝑓𝜈𝜈)𝜇𝜇 and (𝑓𝑓𝜇𝜇)𝜈𝜈 fractional order derivatives exist, then 56

57

(3)

Cilt / Volume: 7, Sayı / Issue: 2, 2017 217


where



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



,

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

(4)

In that case,

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

is the fractional derivative of

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

of order

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

and

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

is the

fractional integral of

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

of order

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

confirmed (in each case) that

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

(5)

(Yilmazer and Ozturk, 2013).

Lemma 2.1. (Linearity) When fractional order derivatives

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

and

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

exist, then

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

(6)

where

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

are analytic and single-valued functions,



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



and



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



are constants and,

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

Lemma 2.2. (Index law) If fractional order derivatives

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

and

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

exist, then

4

{[𝑓𝑓(𝑧𝑧)]𝜈𝜈}𝜇𝜇 = [𝑓𝑓(𝑧𝑧)]𝜈𝜈+𝜇𝜇 = {[𝑓𝑓(𝑧𝑧)]𝜇𝜇}𝜈𝜈, (7)

58

where 𝑓𝑓(𝑧𝑧) is an analytic and single-valued function, 𝜈𝜈 and 𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ and 59

| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60

Property 2.1. Let 𝜆𝜆 be a constant. So, 61

62

(e𝜆𝜆z)𝜈𝜈 = 𝜆𝜆𝜈𝜈e𝜆𝜆𝑧𝑧 (𝜆𝜆 ≠ 0, 𝜈𝜈 ∈ ℝ, z ∈ ℂ). (8)

63


65

(e−𝜆𝜆𝑧𝑧)𝜈𝜈 = e−𝑖𝑖𝑖𝑖𝜈𝜈𝜆𝜆𝜈𝜈e−𝜆𝜆𝑧𝑧 (𝜆𝜆 ≠ 0,𝜈𝜈 ∈ ℝ, 𝑧𝑧 ∈ ℂ). (9)

66


68

(𝑧𝑧𝜆𝜆)𝜈𝜈 = e−𝑖𝑖𝑖𝑖𝜈𝜈𝑧𝑧𝜆𝜆−𝜈𝜈 Γ(𝜈𝜈− 𝜆𝜆)Γ(−𝜆𝜆) (𝜈𝜈 ∈ ℝ,𝑧𝑧 ∈ ℂ, |Γ

(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

(7)

where

3



(𝜇𝜇 ∉ ℤ−),



(3)

44


46



47



|𝑓𝑓𝜇𝜇(𝑧𝑧)| < ∞ (𝜇𝜇 ∈ ℝ). (5)



52



53


𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ. 55


57

is an analytic and single-valued function,

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

and

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

, and



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



Property 2.1. Let

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

be a constant. So,

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

(8)

Property 2.2. Let

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

be a constant. So,

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

(9)

Property 2.3. Let



-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------









-----------------------------------------------------------------------------------------------------------------



be a constant. So,

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

(10)


Ökkeş ÖZTÜRK

Property 2.4.

4


58


| Г(𝜈𝜈+𝜇𝜇+1)Г(𝜈𝜈+1)Г(𝜇𝜇+1)| < ∞. 60


62


63


65


66


68


(𝜈𝜈 − 𝜆𝜆)Γ(−𝜆𝜆) | < ∞). (10)

69

Property 2.4. 70

71

Г(𝑧𝑧+ 1) = 𝑧𝑧Г(𝑧𝑧+ 1) = 𝑧𝑧!, (11)

72

and 73

74

(11)

and,

5

Γ(𝜈𝜈− 𝑘𝑘) = (−1)𝑘𝑘 Γ(𝜈𝜈)Γ(1− 𝜈𝜈)Γ(𝑘𝑘+ 1− 𝜈𝜈), (12)

75

where 𝑘𝑘 ∈ ℤ0+ and 𝜈𝜈 ∈ ℝ. 76

Lemma 2.3. (𝑁𝑁𝜇𝜇 method) If 𝑓𝑓𝜇𝜇 and 𝑔𝑔𝜇𝜇 fractional order derivatives exist, then 77

generalized Leibniz rule is 78

79

𝑁𝑁𝜇𝜇(𝑓𝑓.𝑔𝑔) = (𝑓𝑓.𝑔𝑔)𝜇𝜇 = ∑ Г(𝜇𝜇+ 1)Г(𝜇𝜇+ 1− 𝑘𝑘)Г(𝑘𝑘+ 1)

∞

𝑘𝑘=0𝑓𝑓𝜇𝜇−𝑘𝑘 .𝑔𝑔𝑘𝑘 , (13)

80

where 𝑓𝑓(𝑧𝑧) and 𝑔𝑔(𝑧𝑧) are single-valued and analytic functions, 𝜇𝜇 ∈ ℝ, 𝑧𝑧 ∈ ℂ and 81

| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82

RESULTS AND DISCUSSION 83

The hypergeometric equation 84

85

𝑧𝑧(1− 𝑧𝑧)𝑑𝑑2𝑦𝑦(𝑧𝑧)𝑑𝑑𝑧𝑧2 + [𝑐𝑐 − (𝑎𝑎 + 𝑏𝑏 + 1)𝑧𝑧]𝑑𝑑𝑦𝑦(𝑧𝑧)

𝑑𝑑𝑧𝑧 − 𝑎𝑎𝑏𝑏𝑦𝑦(𝑧𝑧) = 0, (14)

86

has three regular singular points at 𝑧𝑧 = 0,1 and ∞ (𝑎𝑎, 𝑏𝑏 and 𝑐𝑐 are parameters). The 87

singularities can be merged at 𝑏𝑏 and infinity, where 𝑧𝑧 = 𝑥𝑥 𝑏𝑏⁄ and 𝑏𝑏 →∞. Thus, 88

confluent equation is 89

90

𝑥𝑥 𝑑𝑑2𝑦𝑦

𝑑𝑑𝑥𝑥2 + (𝑐𝑐 − 𝑥𝑥)𝑑𝑑𝑦𝑦𝑑𝑑𝑥𝑥 − 𝑎𝑎𝑦𝑦 = 0, (15)

91

(12)

where

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

Lemma 2.3.

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

fractional order derivatives

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

exist, then generalized Leibniz rule

is

(13)

where

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

are single-valued and analytic functions,

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

RESULTS AND DISCUSSION

The hypergeometric equation

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

(14)

has three regular singular points at

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

The

singularities can be merged at

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and infinity, where

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

and

5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

Thus, confluent equation is



5


75




79


∞


80


| Г(𝜇𝜇+1)Г(𝜇𝜇−𝑘𝑘+1)Г(𝑘𝑘+1)| < ∞. 82



85



86




90



91

(15)

solutions of which are the confluent hypergeometric

functions, which are defined as

6

solutions of which are the confluent hypergeometric functions, which are defined as 92

𝑀𝑀(𝑎𝑎, 𝑐𝑐; 𝑥𝑥). The confluent hypergeometric equation has a regular singular point at 𝑥𝑥 = 0 93

and an essential singularity at infinity. 𝐽𝐽𝑛𝑛(𝑥𝑥) (Bessel functions) and 𝐿𝐿𝑛𝑛(𝑥𝑥) (Laguerre 94

polynomials), can be formed in terms of the solutions of the confluent hypergeometric 95

equation as 96

97

𝐽𝐽𝑛𝑛(𝑥𝑥) = e−𝑖𝑖𝑖𝑖𝑛𝑛! (𝑥𝑥2)

𝑛𝑛𝑀𝑀(𝑛𝑛 + 1

2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),

𝐿𝐿𝑛𝑛(𝑥𝑥) = 𝑀𝑀(−𝑛𝑛, 1; 𝑥𝑥).

98

Linearly independent solutions of ‘‘Eq. 15.’’ are defined as 99

100

𝑦𝑦1(𝑥𝑥) = 𝑀𝑀(𝑎𝑎, 𝑐𝑐; 𝑥𝑥) = 1 + 𝑎𝑎𝑐𝑐𝑥𝑥1! + 𝑎𝑎(𝑎𝑎+ 1)

𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103

𝑦𝑦2(𝑥𝑥) = 𝑥𝑥1−𝑐𝑐𝑀𝑀(𝑎𝑎+ 1 − 𝑐𝑐, 2− 𝑐𝑐; 𝑥𝑥) (𝑐𝑐 ≠ 2,3,4,… ),

104

Integral representation of the confluent hypergeometric functions 𝐹𝐹1 1(𝑎𝑎,𝑏𝑏; 𝑥𝑥) can be 105

defined as 106

107

𝑀𝑀(𝑎𝑎, 𝑐𝑐; 𝑥𝑥) = Г(𝑐𝑐)Г(𝑎𝑎)Г(𝑐𝑐 − 𝑎𝑎)∫ e𝑖𝑖𝑥𝑥𝑡𝑡𝑎𝑎−1(1− 𝑡𝑡)𝑐𝑐−𝑎𝑎−1𝑑𝑑𝑡𝑡

1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

The confluent hypergeometric equation has

a regular singular point at

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

and an

essential singularity at infinity.

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

(Bessel

functions) and

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

(Laguerre polynomials),

can be formed in terms of the solutions of the

confluent hypergeometric equation as

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

Linearly independent solutions of ‘‘Eq. 15.’’ are defined as

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

and,

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

Integral representation of the confluent hypergeometric functions

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107


1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

can be defined as

6





equation as 96

97



2 , 2𝑛𝑛 + 1; 2𝑖𝑖𝑥𝑥),


98


100


𝑐𝑐(𝑐𝑐 + 1)𝑥𝑥22! + 𝑎𝑎(𝑎𝑎 + 1)(𝑎𝑎+ 2)

𝑐𝑐(𝑐𝑐 + 1)(𝑐𝑐 + 2)𝑥𝑥33! +⋯,

(𝑐𝑐 ≠ 0,−1,−2, … ),

101

and 102

103


104


defined as 106

107

𝑀𝑀(𝑎𝑎, 𝑐𝑐; 𝑥𝑥) = Г(𝑐𝑐)Г(𝑎𝑎)Г(𝑐𝑐 − 𝑎𝑎) ∫ e𝑖𝑖𝑥𝑥𝑡𝑡𝑎𝑎−1(1− 𝑡𝑡)𝑐𝑐−𝑎𝑎−1𝑑𝑑𝑡𝑡

1

0

(𝑎𝑎, 𝑐𝑐 ∈ ℝ, 𝑐𝑐 > 𝑎𝑎 > 0).

(Bayın, 2006).

Remark 3.1. The familiar Bessel differential equation of general order

7

(Bayın, 2006). 108

Remark 3.1. The familiar Bessel differential equation of general order 𝑙𝑙: 109

𝑧𝑧2 𝑑𝑑2𝑓𝑓

𝑑𝑑𝑧𝑧2 + 𝑧𝑧𝑑𝑑𝑓𝑓𝑑𝑑𝑧𝑧 + (𝑧𝑧2 − 𝑙𝑙2)𝑓𝑓 = 0,

which is named after F. Wilheim Bessel. More precisely, just as in the earlier works 110

(Lin et al., 2005; Wang et al., 2006), we aim here at demonstrating how the underlying 111

simple fractional-calculus approach to the solutions of the classical differential 112

equation, which were considered in the earlier works (Lin et al., 2005; Wang et al., 113

2006; Akgül et al., 2013), would lead us analogously to several interesting 114

consequences including (for example) an alternative investigation of solutions of the 115

confluent hypergeometric equation. 116

Theorem 3.1. Let 𝑦𝑦 ∈ {𝑦𝑦: 0 ≠ ⎸𝑦𝑦𝜇𝜇⎸ < ∞, 𝜇𝜇 ∈ ℝ}. ‘‘Eq. 15.’’ can be written as 117

118

𝑥𝑥𝑦𝑦2 + (𝑐𝑐 − 𝑥𝑥)𝑦𝑦1 − 𝑎𝑎𝑦𝑦 = 0 (𝑥𝑥 ≠ 0). (16)

119

And ‘‘Eq. 16.’’ has particular solutions as follows 120

121

𝑦𝑦(I) = 𝐴𝐴[𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥]𝑎𝑎−1, (17)

122

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥−(𝑐𝑐+1)[𝑥𝑥1+𝑎𝑎e𝑥𝑥]−(2+𝑐𝑐−𝑎𝑎) , (18)

123

where 𝑦𝑦𝑘𝑘 = 𝑑𝑑𝑘𝑘𝑦𝑦 𝑑𝑑𝑥𝑥𝑘𝑘⁄ (𝑘𝑘 = 0,1,2, … ), 𝑦𝑦0 = 𝑦𝑦 = 𝑦𝑦(𝑥𝑥), 𝐴𝐴,𝐵𝐵 are constants. 124

Proof. (I) At first, by applying 𝑁𝑁𝜇𝜇 method to the both sides of ‘‘Eq. 16.’’, we obtain 125

126

:

7

(Bayın, 2006). 108












118


119


121


122


123



126

which is named after F. Wilheim Bessel. More

precisely, just as in the earlier works (Lin et al., 2005;

Wang et al., 2006), we aim here at demonstrating how

the underlying simple fractional-calculus approach

to the solutions of the classical differential equation,

which were considered in the earlier works (Lin et al.,

2005; Wang et al., 2006; Akgül et al., 2013), would

lead us analogously to several interesting consequences

including (for example) an alternative investigation of

solutions of the confluent hypergeometric equation.


Ökkeş ÖZTÜRK

Theorem 3.1. Let

7

(Bayın, 2006). 108












118


119


121


122


123



126

can be written as

7

(Bayın, 2006). 108












118


119


121


122


123



126

(16)

And ‘‘Eq. 16.’’ has particular solutions as follows

7

(Bayın, 2006). 108












118


119


121


122


123



126

(17)

7

(Bayın, 2006). 108












118


119


121


122


123



126

(18)

where

7

(Bayın, 2006). 108












118


119


121


122


123



126

are constants.

Proof. (I) At first, by applying

7

(Bayın, 2006). 108












118


119


121


122


123



126

method to the both sides of ‘‘Eq. 16.’’, we obtain

8

𝑥𝑥𝑦𝑦2+𝜇𝜇 + (𝜇𝜇 + 𝑐𝑐 − 𝑥𝑥)𝑦𝑦1+𝜇𝜇 − (𝜇𝜇+ 𝑎𝑎)𝑦𝑦𝜇𝜇 = 0. (19)

127

If we suppose that 128

129

𝜇𝜇 + 𝑎𝑎 = 0, (20)

130

then, we have 131

132

133

𝜇𝜇 = −𝑎𝑎. (21)

134

By substituting ‘‘Eq. 21.’’ into ‘‘Eq. 19.’’, we obtain 135

136

(𝑦𝑦1−𝑎𝑎)1 + [(𝑐𝑐 − 𝑎𝑎)𝑥𝑥−1 − 1]𝑦𝑦1−𝑎𝑎 = 0. (22)

137

Let 138

𝑦𝑦1−𝑎𝑎 = 𝑢𝑢 = 𝑢𝑢(𝑥𝑥) [𝑦𝑦(𝑥𝑥) = 𝑢𝑢𝑎𝑎−1]. (23)

139

So, we have differential equation as 140

141

𝑢𝑢1 + [(𝑐𝑐− 𝑎𝑎)𝑥𝑥−1 − 1]𝑢𝑢 = 0. (24)

142

Solution of the ‘‘Eq. 24.’’ is 143

144

(19)

If we suppose that

8


127


129

𝜇𝜇 + 𝑎𝑎 = 0, (20)

130

then, we have 131

132

133

𝜇𝜇 = −𝑎𝑎. (21)

134


136

(𝑦𝑦1−𝑎𝑎)1 + [(𝑐𝑐 − 𝑎𝑎)𝑥𝑥−1 − 1]𝑦𝑦1−𝑎𝑎 = 0. (22)

137

Let 138


139


141

𝑢𝑢1 + [(𝑐𝑐− 𝑎𝑎)𝑥𝑥−1 − 1]𝑢𝑢 = 0. (24)

142


144

(20)

then, we have

8


127


129

𝜇𝜇 + 𝑎𝑎 = 0, (20)

130

then, we have 131

132

133

𝜇𝜇 = −𝑎𝑎. (21)

134


136

(𝑦𝑦1−𝑎𝑎)1 + [(𝑐𝑐 − 𝑎𝑎)𝑥𝑥−1 − 1]𝑦𝑦1−𝑎𝑎 = 0. (22)

137

Let 138


139


141

𝑢𝑢1 + [(𝑐𝑐− 𝑎𝑎)𝑥𝑥−1 − 1]𝑢𝑢 = 0. (24)

142


144

(21)

By substituting ‘‘Eq. 21.’’ into ‘‘Eq. 19.’’, we obtain

8


127


129

𝜇𝜇 + 𝑎𝑎 = 0, (20)

130

then, we have 131

132

133

𝜇𝜇 = −𝑎𝑎. (21)

134


136

(𝑦𝑦1−𝑎𝑎)1 + [(𝑐𝑐 − 𝑎𝑎)𝑥𝑥−1 − 1]𝑦𝑦1−𝑎𝑎 = 0. (22)

137

Let 138


139


141

𝑢𝑢1 + [(𝑐𝑐− 𝑎𝑎)𝑥𝑥−1 − 1]𝑢𝑢 = 0. (24)

142


144

(22)

Let

8


127


129

𝜇𝜇 + 𝑎𝑎 = 0, (20)

130

then, we have 131

132

133

𝜇𝜇 = −𝑎𝑎. (21)

134


136

(𝑦𝑦1−𝑎𝑎)1 + [(𝑐𝑐 − 𝑎𝑎)𝑥𝑥−1 − 1]𝑦𝑦1−𝑎𝑎 = 0. (22)

137

Let 138


139


141

𝑢𝑢1 + [(𝑐𝑐− 𝑎𝑎)𝑥𝑥−1 − 1]𝑢𝑢 = 0. (24)

142


144

(23)

So, we have differential equation as

8


127


129

𝜇𝜇 + 𝑎𝑎 = 0, (20)

130

then, we have 131

132

133

𝜇𝜇 = −𝑎𝑎. (21)

134


136

(𝑦𝑦1−𝑎𝑎)1 + [(𝑐𝑐 − 𝑎𝑎)𝑥𝑥−1 − 1]𝑦𝑦1−𝑎𝑎 = 0. (22)

137

Let 138


139


141

𝑢𝑢1 + [(𝑐𝑐− 𝑎𝑎)𝑥𝑥−1 − 1]𝑢𝑢 = 0. (24)

142


144

(24)

Solution of the ‘‘Eq. 24.’’ is

9

𝑢𝑢 = 𝐴𝐴𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥 , (25)

145

and we obtain a fractional solution of the ‘‘Eq. 16.’’ as 146

147

𝑦𝑦(I) = 𝐴𝐴[𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥]𝑎𝑎−1. (26)

(II) We suppose that 148

149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150

where ℎ = ℎ(𝑥𝑥). Then, 151

152

𝑦𝑦1 = 𝑡𝑡𝑥𝑥𝑡𝑡−1ℎ + 𝑥𝑥𝑡𝑡ℎ1, (28)

153

and 154

155

𝑦𝑦2 = 𝑡𝑡(𝑡𝑡 − 1)𝑥𝑥𝑡𝑡−2ℎ + 2𝑡𝑡𝑥𝑥𝑡𝑡−1ℎ1 + 𝑥𝑥𝑡𝑡ℎ2. (29)

156

By substituting ‘‘Eq. 27.’’, ‘‘Eq. 28.’’ and ‘‘Eq. 29.’’ into ‘‘Eq. 16.’’, we have 157

158

𝑥𝑥ℎ2 + (2𝑡𝑡 + 𝑐𝑐 − 𝑥𝑥)ℎ1 + [𝑡𝑡(𝑡𝑡 + 𝑐𝑐 + 1)𝑥𝑥−1 − (𝑡𝑡 + 𝑎𝑎)]ℎ = 0. (30)

159

We suppose that 160

161

(25)



and we obtain a fractional solution of the ‘‘Eq. 16.’’ as

9


145


147



149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150


152


153

and 154

155


156


158


159

We suppose that 160

161

(26)

(II) We suppose that

9


145


147



149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150


152


153

and 154

155


156


158


159

We suppose that 160

161

(27)

where

9


145


147



149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150


152


153

and 154

155


156


158


159

We suppose that 160

161

Then,

9


145


147



149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150


152


153

and 154

155


156


158


159

We suppose that 160

161

(28)

and,

9


145


147



149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150


152


153

and 154

155


156


158


159

We suppose that 160

161

(29)

By substituting ‘‘Eq. 27.’’, ‘‘Eq. 28.’’ and ‘‘Eq. 29.’’ into ‘‘Eq. 16.’’, we have

9


145


147



149

𝑦𝑦 = 𝑥𝑥𝑡𝑡ℎ (𝑥𝑥 ≠ 0), (27)

150


152


153

and 154

155


156


158


159

We suppose that 160

161

(30)

We suppose that

10

𝑡𝑡(𝑡𝑡+ 𝑐𝑐 + 1) = 0. (31)

162

So, 163

164

𝑡𝑡 = 0 or 𝑡𝑡 = −(𝑐𝑐 + 1). (32)

165

(II-i) We have ‘‘Eq. 16.’’ from ‘‘Eq. 30.’’, where 𝑡𝑡 = 0. 166

(II-ii) We write equation as 167

168

𝑥𝑥ℎ2 − (𝑐𝑐 + 2 + 𝑥𝑥)ℎ1 + (𝑐𝑐 + 1− 𝑎𝑎)ℎ = 0. (33)

169

where 𝑡𝑡 = −(𝑐𝑐 + 1). Now, by applying 𝑁𝑁𝜇𝜇 method to the both sides of ‘‘Eq. 33.’’, we 170

obtain 171

172

𝑥𝑥ℎ2+𝜇𝜇 + [𝜇𝜇 − (𝑐𝑐 + 2 + 𝑥𝑥)]ℎ1+𝜇𝜇 + (−𝜇𝜇+ 𝑐𝑐 + 1−𝑎𝑎)ℎ𝜇𝜇 = 0 (34)

173

If we choose that 174

175

−𝜇𝜇 + 𝑐𝑐 + 1−𝑎𝑎 = 0, (35)

176

then, we find 177

178

𝜇𝜇 = 𝑐𝑐 + 1 −𝑎𝑎 . (36)

179

(31)

So,

10

𝑡𝑡(𝑡𝑡+ 𝑐𝑐 + 1) = 0. (31)

162

So, 163

164

𝑡𝑡 = 0 or 𝑡𝑡 = −(𝑐𝑐 + 1). (32)

165



168

𝑥𝑥ℎ2 − (𝑐𝑐 + 2 + 𝑥𝑥)ℎ1 + (𝑐𝑐 + 1− 𝑎𝑎)ℎ = 0. (33)

169


obtain 171

172


173


175

−𝜇𝜇 + 𝑐𝑐 + 1−𝑎𝑎 = 0, (35)

176

then, we find 177

178

𝜇𝜇 = 𝑐𝑐 + 1 −𝑎𝑎 . (36)

179

(32)

(II-i) We have ‘‘Eq. 16.’’ from ‘‘Eq. 30.’’, where


21.) sy. 328/ Denklem (13)’ün

( ) ( ) ∑ ( ) ( ) ( )

olarak değiştirilmesi,




-----------------------------------------------------------------------------------------------------------------


25.) sy. 329/ ( ) ve ( ) denklemleri arasındaki ‘‘and’’in sola dayalı yazılması ve bu ifadenin ‘‘and,’’ olarak değiştirilmesi,

26.) sy. 329/ ( ) denkleminden sonraki ‘‘(Bayın, 2006).’’nın sola dayalı yazılması,

27.) sy. 329/Remark 3.1.’de ‘‘of general order :’’ yerine ‘‘of general order :’’ yazılması ve burdaki denklemden sonraki satırın sola dayalı yazılması,

-----------------------------------------------------------------------------------------------------------------

28.) sy. 330/Theorem 3.1.’de ‘‘Let.’’ yerine ‘‘Let’’ yazılması,

-----------------------------------------------------------------------------------------------------------------

29) sy. 331/Denklem (28)-(29) arasındaki ‘‘and’’ yerine ‘‘and,’’ yazılması,

30.) sy. 331/(II-i)’de ‘‘where .’’ yerine ‘‘where .’’ yazılması,

31.) sy. 331/(II-ii)’de ‘‘equation as .’’ yerine ‘‘equation as’’ yazılması,

32.) sy. 331/Denklem (33)’den sonra ‘‘where .’’ yerine ‘‘where ( ).’’ yazılması ve ‘‘by applying method’’ yerine ‘‘by applying method’’ yazılması,

33.) sy. 331/Denklem (34)’ün ortalanarak

[ ( )] ( )

şeklinde yazılması,

-----------------------------------------------------------------------------------------------------------------

34.) sy. 332/Denklem (37)’nin

( ) [( ) ]



= 0 .

(II-ii) We write equation as

10

𝑡𝑡(𝑡𝑡+ 𝑐𝑐 + 1) = 0. (31)

162

So, 163

164

𝑡𝑡 = 0 or 𝑡𝑡 = −(𝑐𝑐 + 1). (32)

165



168

𝑥𝑥ℎ2 − (𝑐𝑐 + 2 + 𝑥𝑥)ℎ1 + (𝑐𝑐 + 1− 𝑎𝑎)ℎ = 0. (33)

169


obtain 171

172


173


175

−𝜇𝜇 + 𝑐𝑐 + 1−𝑎𝑎 = 0, (35)

176

then, we find 177

178

𝜇𝜇 = 𝑐𝑐 + 1 −𝑎𝑎 . (36)

179

(33)

where


21.) sy. 328/ Denklem (13)’ün

( ) ( ) ∑ ( ) ( ) ( )





-----------------------------------------------------------------------------------------------------------------





-----------------------------------------------------------------------------------------------------------------


-----------------------------------------------------------------------------------------------------------------






[ ( )] ( )


-----------------------------------------------------------------------------------------------------------------


( ) [( ) ]



. Now, by applying N method to the both sides of ‘‘Eq. 33.’’, we obtain


21.) sy. 328/ Denklem (13)’ün

( ) ( ) ∑ ( ) ( ) ( )





-----------------------------------------------------------------------------------------------------------------





-----------------------------------------------------------------------------------------------------------------


-----------------------------------------------------------------------------------------------------------------






[ ( )] ( )


-----------------------------------------------------------------------------------------------------------------


( ) [( ) ]



(34)


Ökkeş ÖZTÜRK

If we choose that

10

𝑡𝑡(𝑡𝑡+ 𝑐𝑐 + 1) = 0. (31)

162

So, 163

164

𝑡𝑡 = 0 or 𝑡𝑡 = −(𝑐𝑐 + 1). (32)

165



168

𝑥𝑥ℎ2 − (𝑐𝑐 + 2 + 𝑥𝑥)ℎ1 + (𝑐𝑐 + 1− 𝑎𝑎)ℎ = 0. (33)

169


obtain 171

172


173


175

−𝜇𝜇 + 𝑐𝑐 + 1−𝑎𝑎 = 0, (35)

176

then, we find 177

178

𝜇𝜇 = 𝑐𝑐 + 1 −𝑎𝑎 . (36)

179

(35)

then, we find

10

𝑡𝑡(𝑡𝑡+ 𝑐𝑐 + 1) = 0. (31)

162

So, 163

164

𝑡𝑡 = 0 or 𝑡𝑡 = −(𝑐𝑐 + 1). (32)

165



168

𝑥𝑥ℎ2 − (𝑐𝑐 + 2 + 𝑥𝑥)ℎ1 + (𝑐𝑐 + 1− 𝑎𝑎)ℎ = 0. (33)

169


obtain 171

172


173


175

−𝜇𝜇 + 𝑐𝑐 + 1−𝑎𝑎 = 0, (35)

176

then, we find 177

178

𝜇𝜇 = 𝑐𝑐 + 1 −𝑎𝑎 . (36)

179

(36)

By substituting ‘‘Eq. 36.’’ into ‘‘Eq. 34.’’, we have

11

By substituting ‘‘Eq. 36.’’ into ‘‘Eq. 34.’’, we have 180

181

(ℎ2+𝑐𝑐−𝑎𝑎)1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]ℎ2+𝑐𝑐−𝑎𝑎 = 0 (37)

182

Let 183

184

ℎ2+𝑐𝑐−𝑎𝑎 = 𝑤𝑤 = 𝑤𝑤(𝑥𝑥), [ℎ(𝑥𝑥) = 𝑤𝑤−(2+𝑐𝑐−𝑎𝑎)]. (38)

185

Then, we have another differential equation as 186

187

𝑤𝑤1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]𝑤𝑤 = 0. (39)

188

The solution of ‘‘Eq. 39’’ is 189

190

𝑤𝑤 = 𝐵𝐵𝑥𝑥1+𝑎𝑎e𝑥𝑥 , (40)

191

and 192

193

ℎ = 𝐵𝐵(𝑥𝑥1+𝑎𝑎e𝑥𝑥)−(2+𝑐𝑐−𝑎𝑎) , (41)

194

and finally, we find another fractional solution of the ‘‘Eq. 16.’’ as 195

196

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥−(𝑐𝑐+1)(𝑥𝑥1+𝑎𝑎e𝑥𝑥)−(2+𝑐𝑐−𝑎𝑎) . (42)

Example 3.1. Let 𝑎𝑎 = 2 and 𝑐𝑐 = 1 for ‘‘Eq. 16.’’. So, we have equation as 197

. (37)

Let

11


181

(ℎ2+𝑐𝑐−𝑎𝑎)1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]ℎ2+𝑐𝑐−𝑎𝑎 = 0 (37)

182

Let 183

184


185


187

𝑤𝑤1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]𝑤𝑤 = 0. (39)

188


190


191

and 192

193


194


196



(38)

Then, we have another differential equation as

11


181

(ℎ2+𝑐𝑐−𝑎𝑎)1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]ℎ2+𝑐𝑐−𝑎𝑎 = 0 (37)

182

Let 183

184


185


187

𝑤𝑤1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]𝑤𝑤 = 0. (39)

188


190


191

and 192

193


194


196



(39)

The solution of ‘‘Eq. 39’’ is

11


181

(ℎ2+𝑐𝑐−𝑎𝑎)1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]ℎ2+𝑐𝑐−𝑎𝑎 = 0 (37)

182

Let 183

184


185


187

𝑤𝑤1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]𝑤𝑤 = 0. (39)

188


190


191

and 192

193


194


196



(40)

and,

11


181

(ℎ2+𝑐𝑐−𝑎𝑎)1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]ℎ2+𝑐𝑐−𝑎𝑎 = 0 (37)

182

Let 183

184


185


187

𝑤𝑤1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]𝑤𝑤 = 0. (39)

188


190


191

and 192

193


194


196



(41)

and finally, we find another fractional solution of the ‘‘Eq. 16.’’ as

11


181

(ℎ2+𝑐𝑐−𝑎𝑎)1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]ℎ2+𝑐𝑐−𝑎𝑎 = 0 (37)

182

Let 183

184


185


187

𝑤𝑤1 − [(1 + 𝑎𝑎)𝑥𝑥−1 + 1]𝑤𝑤 = 0. (39)

188


190


191

and 192

193


194


196



(42)

Example 3.1. Let 36.) sy. 332/Example 3.1.’de ‘‘Let and for ‘‘Eq. 16.’’.’’ yerine ‘‘Let and for ‘‘Eq. 16.’’.’’ yazılması,

37.) sy. 332/Example 3.1.’de

( ) ( )

and,

yerine

( ) ( )

and,

( ) ( )

yazılması,

-----------------------------------------------------------------------------------------------------------------

38.) sy. 333/Theorem 3.2. Proof’da ‘‘(46)’’ denklem numarasının

[ ]

denklemine ait olması,

-----------------------------------------------------------------------------------------------------------------


[ ]


40.) sy. 334/CONCLUSION bölümünün ilk satırında ‘‘we used method’’ yerine ‘‘we used method’’ yazılması gerekmektedir.

NOT: Genel olarak, metindeki bütün denklemlerden önce ve sonra birer boşluk bırakılması gerekmektedir.

Teşekkürler, iyi çalışmalar…

Yazar: Ökkeş Öztürk

Makale: Konfluent Hipergeometrik Denklemi İçin Farklı Bir Çözüm Metodu

for ‘‘Eq. 16.’’. So, we have equation as

12

198

𝑥𝑥𝑦𝑦2 + (1− 𝑥𝑥)𝑦𝑦1 − 2𝑦𝑦 = 0 (𝑥𝑥 ≠ 0). (43)

199

Therefore, we can write solutions of the ‘‘Eq. 43.’’ as follows 200

201

𝑦𝑦(I) = 𝐴𝐴(𝑥𝑥e𝑥𝑥)1,

202

and, 203

204

205

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥−2(𝑥𝑥3e𝑥𝑥)−1.

206

So, we obtain solutions by calculating as 207

208

𝑦𝑦(I) = 𝐴𝐴(𝑥𝑥e𝑥𝑥)1 = 𝐴𝐴𝑑𝑑(𝑥𝑥e𝑥𝑥)𝑑𝑑𝑥𝑥 ,

209

= 𝐴𝐴e𝑥𝑥(𝑥𝑥 + 1),

210

and, 211

212

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥−2(𝑥𝑥3e𝑥𝑥)−1 = 𝐵𝐵𝑥𝑥−2 ∫ 𝑥𝑥3e𝑥𝑥 𝑑𝑑𝑥𝑥,

213

(43)

Therefore, we can write solutions of the ‘‘Eq. 43.’’ as follows

36.) sy. 332/Example 3.1.’de ‘‘Let and for ‘‘Eq. 16.’’.’’ yerine ‘‘Let and for ‘‘Eq. 16.’’.’’ yazılması,

37.) sy. 332/Example 3.1.’de

( ) ( )

and,

yerine

( ) ( )

and,

( ) ( )

yazılması,

-----------------------------------------------------------------------------------------------------------------


[ ]


-----------------------------------------------------------------------------------------------------------------


[ ]







36.) sy. 332/Example 3.1.’de ‘‘Let and for ‘‘Eq. 16.’’.’’ yerine ‘‘Let and for ‘‘Eq. 16.’’.’’ yazılması,

37.) sy. 332/Example 3.1.’de

( ) ( )

and,

yerine

( ) ( )

and,

( ) ( )

yazılması,

-----------------------------------------------------------------------------------------------------------------


[ ]


-----------------------------------------------------------------------------------------------------------------


[ ]







So, we obtain solutions by calculating as

12

198

𝑥𝑥𝑦𝑦2 + (1− 𝑥𝑥)𝑦𝑦1 − 2𝑦𝑦 = 0 (𝑥𝑥 ≠ 0). (43)

199


201


202

and, 203

204

205


206


208


209


210

and, 211

212


213

12

198

𝑥𝑥𝑦𝑦2 + (1− 𝑥𝑥)𝑦𝑦1 − 2𝑦𝑦 = 0 (𝑥𝑥 ≠ 0). (43)

199


201


202

and, 203

204

205


206


208


209


210

and, 211

212


213

and,



12

198

𝑥𝑥𝑦𝑦2 + (1− 𝑥𝑥)𝑦𝑦1 − 2𝑦𝑦 = 0 (𝑥𝑥 ≠ 0). (43)

199


201


202

and, 203

204

205


206


208


209


210

and, 211

212


213

13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214

Theorem 3.2. Let |(𝑥𝑥𝑎𝑎−𝑐𝑐)𝑘𝑘| < ∞ (𝑘𝑘 ∈ ℤ+ ∪ {0}), 𝑥𝑥 ≠ 0, and |1𝑥𝑥| < 1. The solution of 215

‘‘Eq. 26.’’ can be written as follows 216

217

𝑦𝑦(I) = 𝐴𝐴𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥 𝐹𝐹02 [1− 𝑎𝑎, 𝑐𝑐 − 𝑎𝑎; 1𝑥𝑥] . (44)

218

Proof. By means of ‘‘Eq. 13.’’, we have 219

220

𝑦𝑦(I) = 𝐴𝐴∑ Γ(𝑎𝑎)Γ(𝑎𝑎− 𝑘𝑘)Γ(𝑘𝑘+ 1)

∞

𝑘𝑘=0

(𝑥𝑥𝑎𝑎−𝑐𝑐)𝑘𝑘(e𝑥𝑥)𝑎𝑎−1−𝑘𝑘 . (45)

221

By using ‘‘Eq. 8.’’, ‘‘Eq. 10.’’, ‘‘Eq. 11.’’ and ‘‘Eq. 12.’’, we can rewrite the ‘‘Eq. 45.’’ 222

as follows 223

224

𝑦𝑦(I) = 𝐴𝐴∑ Γ(𝑘𝑘+ 1− 𝑎𝑎)(−1)𝑘𝑘Γ(1− 𝑎𝑎)

1𝑘𝑘!

∞

𝑘𝑘=0

(−1)𝑘𝑘𝑥𝑥𝑎𝑎−𝑐𝑐−𝑘𝑘 Γ(𝑘𝑘+ 𝑐𝑐 − 𝑎𝑎)Γ(𝑐𝑐− 𝑎𝑎) e𝑥𝑥 ,

225

= 𝐴𝐴𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥∑[1−𝑎𝑎]𝑘𝑘∞

𝑘𝑘=0

[𝑐𝑐 − 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

226

= 𝐴𝐴𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥 𝐹𝐹02 [1 −𝑎𝑎 ,𝑐𝑐 − 𝑎𝑎; 1𝑥𝑥]. (46)

227

Theorem 3.2. Let

13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214



217


218


220


∞

𝑘𝑘=0


221


as follows 223

224


1𝑘𝑘!

∞

𝑘𝑘=0


225


𝑘𝑘=0


1𝑥𝑥)

𝑘𝑘,

226


227

The solution of ‘‘Eq. 26.’’ can

be written as follows

13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214



217


218


220


∞

𝑘𝑘=0


221


as follows 223

224


1𝑘𝑘!

∞

𝑘𝑘=0


225


𝑘𝑘=0


1𝑥𝑥)

𝑘𝑘,

226


227

(44)

Proof. By means of ‘‘Eq. 13.’’, we have

13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214



217


218


220


∞

𝑘𝑘=0


221


as follows 223

224


1𝑘𝑘!

∞

𝑘𝑘=0


225


𝑘𝑘=0


1𝑥𝑥)

𝑘𝑘,

226


227

(45)

By using ‘‘Eq. 8.’’, ‘‘Eq. 10.’’, ‘‘Eq. 11.’’ and ‘‘Eq. 12.’’, we can rewrite the ‘‘Eq. 45.’’ as follows

13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214



217


218


220


∞

𝑘𝑘=0


221


as follows 223

224


1𝑘𝑘!

∞

𝑘𝑘=0


225


𝑘𝑘=0


1𝑥𝑥)

𝑘𝑘,

226


227 13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214



217


218


220


∞

𝑘𝑘=0


221


as follows 223

224


1𝑘𝑘!

∞

𝑘𝑘=0


225


𝑘𝑘=0


1𝑥𝑥)

𝑘𝑘,

226


227 13

= 𝐵𝐵𝑥𝑥−2[e𝑥𝑥(𝑥𝑥3 − 3𝑥𝑥2 + 6𝑥𝑥 − 6)].

214



217


218


220


∞

𝑘𝑘=0


221


as follows 223

224


1𝑘𝑘!

∞

𝑘𝑘=0


225


𝑘𝑘=0


1𝑥𝑥)

𝑘𝑘,

226


227

(46)

Theorem 3.3. Let

14

Theorem 3.3. Let |(𝑥𝑥1+𝑎𝑎)𝑘𝑘| < ∞ (𝑘𝑘 ∈ ℤ+ ∪ {0}), 𝑥𝑥 ≠ 0, and |1𝑥𝑥| < 1. The solution of 228


230

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥 𝐹𝐹02 [2 −𝑎𝑎 + 𝑐𝑐,−1−𝑎𝑎; 1𝑥𝑥] . (47)

231


233

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥−(𝑐𝑐+1) ∑ Γ(𝑎𝑎 − 𝑐𝑐 − 1)Γ(𝑎𝑎− 𝑐𝑐 − 1− 𝑘𝑘)

∞

𝑘𝑘=0

1𝑘𝑘! (𝑥𝑥1+𝑎𝑎)𝑘𝑘(e𝑥𝑥)−(2+𝑐𝑐−𝑎𝑎+𝑘𝑘) . (48)

234

By using ‘‘Eq. 8.’’, ‘‘Eq. 10.’’, ‘‘Eq. 11.’’ and ‘‘Eq. 12.’’, we rewrite the ‘‘Eq. 48.’’ as 235

follows 236

𝑦𝑦(II) = 𝐵𝐵𝑥𝑥−(𝑐𝑐+1) ∑ Γ(𝑘𝑘+ 2−𝑎𝑎 + 𝑐𝑐)(−1)𝑘𝑘Γ(2−𝑎𝑎 + 𝑐𝑐)

∞

𝑘𝑘=0

1𝑘𝑘! (−1)𝑘𝑘𝑥𝑥1+𝑎𝑎−𝑘𝑘 Γ

(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237

=𝐵𝐵𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥 ∑[2−𝑎𝑎 + 𝑐𝑐]𝑘𝑘∞

𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238

=𝐵𝐵𝑥𝑥𝑎𝑎−𝑐𝑐e𝑥𝑥 𝐹𝐹02 [2− 𝑎𝑎 + 𝑐𝑐,−1− 𝑎𝑎; 1𝑥𝑥]. (49)

CONCLUSION 239

In this paper, we used 𝑁𝑁𝜇𝜇 method for the confluent hypergeometric equation. We also 240

obtained hypergeometric forms of the fractional solutions. The most important 241

advantage of this method is that can be applied for singular equations. 242

, and

14



230


231


233


∞

𝑘𝑘=0


234


follows 236


∞

𝑘𝑘=0


(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237


𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238


CONCLUSION 239




The solution of ‘‘Eq. 42.’’

can be written as follows

14



230


231


233


∞

𝑘𝑘=0


234


follows 236


∞

𝑘𝑘=0


(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237


𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238


CONCLUSION 239




(47)

Proof. By means of ‘‘Eq. 13.’’, we have


Ökkeş ÖZTÜRK

14



230


231


233


∞

𝑘𝑘=0


234


follows 236


∞

𝑘𝑘=0


(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237


𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238


CONCLUSION 239




(48)

By using ‘‘Eq. 8.’’, ‘‘Eq. 10.’’, ‘‘Eq. 11.’’ and ‘‘Eq. 12.’’, we rewrite the ‘‘Eq. 48.’’ as follows

14



230


231


233


∞

𝑘𝑘=0


234


follows 236


∞

𝑘𝑘=0


(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237


𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238


CONCLUSION 239



advantage of this method is that can be applied for singular equations. 242 14



230


231


233


∞

𝑘𝑘=0


234


follows 236


∞

𝑘𝑘=0


(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237


𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238


CONCLUSION 239



advantage of this method is that can be applied for singular equations. 242 14



230


231


233


∞

𝑘𝑘=0


234


follows 236


∞

𝑘𝑘=0


(𝑘𝑘 − 1− 𝑎𝑎)Γ(−1− 𝑎𝑎) e𝑥𝑥 ,

237


𝑘𝑘=0

[−1− 𝑎𝑎]𝑘𝑘1𝑘𝑘! (

1𝑥𝑥)

𝑘𝑘,

238


CONCLUSION 239




(49)

CONCLUSION

In this paper, we used N method for the

confluent hypergeometric equation. We also obtained

hypergeometric forms of the fractional solutions. The

most important advantage of this method is that can be

applied for singular equations.

REFERENCESAkgül A, 2014. A new method for approximate solutions of

fractional order boundary value problems. Neural Parallel and Scientific Computations 22(1-2): 223-237.

Akgül A, Inc M, Karatas E, Baleanu D, 2015. Numerical solutions of fractional differential equations of Lane-Emden type by an accurate technique. Advances in Difference Equations, 220: 12 pages.

Akgül A, Kılıçman A, Inc M, 2013. Improved (G'/G)-expansion method for the space and time fractional foam drainage and KdV equations. Abstract and Applied Analysis, 2013: 7 pages.

Bayın S, 2006. Mathematical Methods in Science and Engineering. John Wiley & Sons, USA, 709p.

Lin SD, Ling WC, Nishimoto K, Srivastava HM, 2005. A simple fractional-calculus approach to the solutions of the Bessel differential equation of general order and some of its applications. Computers & Mathematics with Applications, 49: 1487-1498.

Miller K, Ross B, 1993. An Introduction to the Fractional Calculus and Fractional Differential Equations. John Wiley & Sons, USA, 376p.

Oldham K, Spanier J, 1974. The Fractional Calculus: Theory and Applications of Differentiation and Integration to Arbitrary Order. Academic Press, USA, 240p.

Podlubny I, 1999. Fractional Differential Equations: An Introduction to Fractional Derivatives, Fractional Differential Equations, Methods of Their Solution and Some of Their Applications. Academic Press, USA, 365p.

Wang PY, Lin SD, Srivastava HM, 2006. Remarks on a simple fractional-calculus approach to the solutions of the Bessel differential equation of general order and some of its applications. Computers & Mathematics with Applications, 51: 105-114.

Yilmazer R, Ozturk O, 2013. Explicit Solutions of Singular Differential Equation by means of Fractional Calculus Operators. Abstract and Applied Analysis, 2013: 6 pages.

· abstract: fractional calculus theory includes definition of the derivatives and integrals of...

Documents