Research Article Fractional Solutions of Bessel ...

Hindawi Publishing Corporation The Scientific World Journal Volume 2013, Article ID 685695, 8 pages http://dx.doi.org/10.1155/2013/685695

Research Article Fractional Solutions of Bessel Equation with 𝑁-Method Erdal Bas, Resat Yilmazer, and Etibar Panakhov Department of Mathematics, Firat University, 23119 Elazig, Turkey Correspondence should be addressed to Erdal Bas; [email protected] Received 30 April 2013; Accepted 16 July 2013 Academic Editors: R. M. Guedes, Z. Mukandavire, and R. Wangkeeree Copyright © 2013 Erdal Bas et al. 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. This paper deals with the design fractional solution of Bessel equation. We obtain explicit solutions of the equation with the help of fractional calculus techniques. Using the 𝑁-fractional calculus operator 𝑁] method, we derive the fractional solutions of the equation.

1. Introduction, Definitions, and Preliminaries Fractional calculus has an important place in the field of math. Firstly, L’Hospital and Leibniz were interested in the topic in 1695, [1]. Fractional calculus is an area of applied mathematics that deals with derivatives and integrals of arbitrary orders and their applications in science, engineering, mathematics, economics, and other fields. The seeds of fractional derivatives were planted over 300 years ago. Since then many efficient mathematicians of their times, such as N. H. Abel, M. Caputo, L. Euler, J. Fourier, A. K. Grunwald, J. Hadamard, G. H. Hardy, O. Heaviside, H. J. Holmgren, P. S. Laplace, G. W. Leibniz, A. V. Letnikov, J. Liouville, and B. Riemann, have contributed to this field; all these references can be seen in [1–5]. The mathematics involved appeared very different applications of this field. Fractional calculus has been applied to almost every field of science. They are viscoelasticity, electrical engineering, electrochemistry, biology, biophysics and bioengineering, signal and image processing, mechanics, mechatronics, physics, and control theory. During the last decade, Samko et al. [4], Nishimoto [6–12], and Podlubny [3] have been helpful in introducing the field to engineering, science, economics and finance, and pure and applied field. Furthermore, there were many studies in this field [5, 13–16]. Various scientists have studied that concept. The progress in this field continues [2–4, 6–12, 17– 20]. 𝑁-Fractional calculus is a very interesting method because this method is applied to singular equation. Note that fractional solutions can be obtained for kinds of singular

equation via this method [6–12, 17]. In this paper, our aim is to apply the same way for singular Sturm-Liouville equation with Bessel potential and find fractional solutions of this equation. Furthermore, we give some applications and their graphs of fractional solutions of the equation. Now, consider the following the Bessel equation:

𝑥

𝑝2 𝑑2 𝑧 𝑑𝑧 + + [𝜆𝑥 − ] 𝑧 = 0, 𝑑𝑥2 𝑑𝑥 𝑥

0 < 𝑥 ≤ 1,

(1)

where 𝜆 and 𝑝 are real numbers. By means of the substitution 𝑦 = √𝑥𝑧 (1) reduces to the form 𝑝2 − 1/4 𝑑2 𝑦 + (𝜆 − ) 𝑦 = 0. 2 𝑑𝑥 𝑥2

(2)

Bessel equation for having the analogous singularity is given in [21]. The differintegration operators and their generalizations [6–11, 17, 18] have been used to solve some classes of differential equations and fractional differential equations. Two of the most commonly encountered tools in the theory and applications of fractional calculus are provided by the Riemann-Liouville operator 𝑅𝑧𝜐 (𝜐 ∈ C) and the Weyl operator 𝑊𝑧𝜐 (𝜐 ∈ C), which are defined by [17, 19].

2

The Scientific World Journal Lemma 2 (linearity property). If the functions 𝑓(𝑧) and 𝑔(𝑧) are single-valued and analytic in some domain Ω ⊆ C, then

Consider 𝑅𝑧𝜐 𝑓 (𝑧) 𝑧 1 { ∫ (𝑧 − 𝑡)𝜐−1 𝑓 (𝑡) 𝑑𝑡 { { { Γ (𝜐) 0 ={ { { { 𝑑𝑛 𝜐+𝑛 𝑅 𝑓 (𝑧) { 𝑑𝑧𝑛 𝑧

(Re (𝜐) > 0) ,

(3)

(−𝑛 < Re (𝜐) ≤ 0; 𝑛 ∈ N) ,

(Re (𝜐) > 0) ,

(4)

(𝑓𝜎 (𝑧))] = 𝑓𝜎+] (𝑧) = (𝑓] (𝑧))𝜎

provided that the defining integrals in (3) and (4) exist, N being the set of positive integers. Definition 1 (cf. [6–10, 12, 20]). Let 𝐶 = {𝐶− , 𝐶+ } ,

(5)

where 𝐶− is a curve along the cut joining two points 𝑧 and −∞+𝑖 Im(𝑧), 𝐶+ is a curve along the cut joining two points 𝑧 and ∞+𝑖 Im(𝑧), 𝐷− is a domain surrounded by 𝐶− , and 𝐷+ is a domain surrounded by 𝐶+ . (Here 𝐷 contains the points over the curve 𝐶.) 𝐷),

Moreover, let 𝑓 = 𝑓(𝑧) be a regular function in 𝐷 (𝑧 ∈ 𝑓] (𝑧) = (𝑓 (𝑧))] =

−

(] ∈ R \ Z ; Z = {. . . , −3, −2, −1}) ,

(6)

] → −𝑛

0 ≤ arg (𝑡 − 𝑧) ≤ 2𝜋 for 𝐶+ .

𝑑𝑡 Γ (] + 1) ) ∫ ]+1 2𝜋𝑖 𝐶 (𝑡 − 𝑧)

(] ∉ Z− )

(8)

with 𝑁

−𝑛

= lim 𝑁 ] → −𝑛

]

+

∞

] (𝑓 (𝑧) ⋅ 𝑔 (𝑧))] = ∑ ( ) 𝑓]−𝑛 (𝑧) ⋅ 𝑔𝑛 (𝑧) 𝑛 𝑛=0

(𝑛 ∈ Z ) .

(12)

(] ∈ R; 𝑧 ∈ Ω) , where 𝑔𝑛 (𝑧) is the ordinary derivative of 𝑔(𝑧) of order 𝑛 (𝑛 ∈ N0 := N ∪ {0}), being tacitly assumed (for simplicity) that 𝑔(𝑧) is the polynomial part (if any) of the product 𝑓(𝑧)𝑔(𝑧).

(𝑒𝜆𝑧 )] = 𝜆] 𝑒𝜆𝑧

(𝜆 ≠ 0; ] ∈ R; 𝑧 ∈ C) .

(13)

(𝜆 ≠ 0; ] ∈ R; 𝑧 ∈ C) .

(14)

Property 3. For a constant 𝜆, (7)

Then 𝑓] (𝑧) (] > 0) is said to be the fractional derivative of 𝑓(𝑧) of order ] and 𝑓] (𝑧) (] < 0) is said to be the fractional integral of 𝑓(𝑧) of order −], provided (in each case) that |𝑓] (𝑧)| < ∞ (] ∈ R). Finally, let the fractional calculus operator (Nishimoto’s operator) 𝑁] be defined by (cf. [6–10]) 𝑁] = (

Lemma 4 (generalized Leibniz rule). If the functions 𝑓(𝑧) and 𝑔(𝑧) are single-valued and analytic in some domain Ω ⊆ C, then

(𝑒−𝜆𝑧 )] = 𝑒−𝑖𝜋] 𝜆] 𝑒−𝜆𝑧

where 𝑡 ≠ 𝑧, −𝜋 ≤ arg (𝑡 − 𝑧) ≤ 𝜋 for 𝐶− ,

(11)

Property 2. For a constant 𝜆,

(𝑛 ∈ Z+ ) ,

𝑓−𝑛 (𝑧) = lim 𝑓] (𝑧)

(𝑓𝜎 (𝑧) ≠ 0; 𝑓] (𝑧) ≠ 0; 𝜎, ] ∈ R; 𝑧 ∈ Ω) .

Property 1. For a constant 𝜆,

𝑓 (𝑡) 𝑑𝑡 Γ (] + 1) ∫ ]+1 2𝜋𝑖 𝐶 (𝑡 − 𝑧) −

(10)

Lemma 3 (index law). If the function 𝑓(𝑧) is single-valued and analytic in some domain Ω ⊆ C, then

(−𝑛 < Re (𝜐) ≤ 0; 𝑛 ∈ N)

𝐷 = {𝐷− , 𝐷+ } ,

(] ∈ R; 𝑧 ∈ Ω) for any constants ℎ1 and ℎ2 .

𝑊𝑧𝜐 𝑓 (𝑧) ∞ 1 { ∫ (𝑡 − 𝑧)𝜐−1 𝑓 (𝑡) 𝑑𝑡 { { { Γ (𝜐) 𝑧 ={ { 𝑛 { {𝑑 𝑊𝜐+𝑛 𝑓 (𝑧) { 𝑑𝑧𝑛 𝑧

(ℎ1 𝑓 (𝑧) + ℎ2 𝑔 (𝑧))] = ℎ1 𝑓] (𝑧) + ℎ2 𝑔] (𝑧)

(9)

We find it to be worthwhile to recall here the following useful lemmas and properties associated with the fractional differintegration which was defined earlier (cf. e.g., [6–10, 12]).

Γ (] − 𝜆) 𝜆−] 𝑧 Γ (−𝜆) 󵄨󵄨 Γ (] − 𝜆) 󵄨󵄨 󵄨󵄨 󵄨 (] ∈ R; 𝑧 ∈ C; 󵄨󵄨󵄨 󵄨 < ∞) . 󵄨󵄨 Γ (−𝜆) 󵄨󵄨󵄨

(𝑧𝜆 )] = 𝑒−𝑖𝜋]

(15)

Now, let apply 𝑁-fractional method to nonhomogeneous Bessel equation.

2. The 𝑁] -Method Applied to Bessel Equation Theorem 5. Let 𝑦 ∈ {𝑦 : 0 ≠ |𝑦] | < ∞; ] ∈ R} and 𝑓 ∈ {𝑓 : 0 ≠ |𝑓] | < ∞; ] ∈ R}. We consider the nonhomogeneous Bessel equation: 𝐿 [𝑦, 𝑥, 𝜆, 𝑝] = 𝑦2 + 𝑦 [𝜆 −

𝑝2 − 1/4 ]=𝑓 𝑥2

(0 < 𝑥 ≤ 1) , (16)

The Scientific World Journal

3 Here, we choose 𝜂 such that

and it has particular solutions of the forms √𝜆𝑥

𝑦𝚤 = 𝑥𝑝+1/2 𝑒−𝑖

𝜂2 − 𝜂 + √𝜆𝑥

× {[(𝑓𝑥1/2−𝑝 𝑒𝑖 × 𝑒2𝑖

)

−𝑝−1/2

√𝜆𝑥 −𝑝−1/2

𝑥

𝑒−2𝑖

}

𝑝−1/2

√𝜆𝑥 𝑝−1/2

𝑥

]

−1

(17)

𝜂=

1/2−𝑝 −𝑖√𝜆𝑥

𝑒

)

2𝑖√𝜆𝑥 𝑝−1/2

−𝑝−1/2

−2𝑖√𝜆𝑥 −𝑝−1/2

×𝑒

𝑥

}

𝑒

𝑝−1/2

𝑥

]

−1

(18)

,

√𝜆𝑥

)

√𝜆𝑥 𝑝−1/2

𝑥

𝑝−1/2

}

𝑒−2𝑖

−𝑝−1/2

√𝜆𝑥 −𝑝−1/2

𝑥

]

(27)

𝜓2 𝑥 + 𝜓1 (2𝑝 + 1) + 𝜓𝜆𝑥 = 𝑓𝑥(1/2)−𝑝 .

(28)

𝜙 = 𝜙 (𝑥) .

−1

(19)

,

(𝑒𝜇𝑥 𝜙)2 𝑥 + (𝑒𝜇𝑥 𝜙)1 (2𝑝 + 1) + 𝑒𝜇𝑥 𝜙𝜆𝑥 = 𝑓𝑥(1/2)−𝑝 .

(𝑒𝜇𝑥 𝜙)1 = 𝑒𝜇𝑥 (𝜇𝜙 + 𝜙1 ) , √𝜆𝑥

)

√𝜆𝑥 𝑝−1/2

𝑥

𝑝−1/2

}

𝑒2𝑖

−𝑝−1/2

√𝜆𝑥 −𝑝−1/2

𝑥

]

(𝑒𝜇𝑥 𝜙)2 = 𝑒𝜇𝑥 (𝜇2 𝜙 + 2𝜇𝜙1 + 𝜙2 ) , −1

(20)

,

where 𝑦2 = 𝑑 𝑦/𝑑𝑥 , 𝑦 = 𝑦 (𝑧) (𝑧 ∈ C), 𝑓 = 𝑓(𝑧) (an arbitrary given function), and 𝑝, 𝜆 are given constants. Remark 6. The cases 𝑝 = 0 of (19) and (20) coincide with those (17) and (18). Proof. Set 𝜓 = 𝜓 (𝑥) .

(21)

Thus 𝑦1 = 𝜂𝑥𝜂−1 𝜓 + 𝑥𝜂 𝜓1 , 𝑦2 = 𝜂 (𝜂 − 1) 𝑥𝜂−2 𝜓 + 2𝜂𝑥𝜂−1 𝜓1 + 𝑥𝜂 𝜓2 .

(22)

𝑝2 − 1/4 ] = 𝑓. 𝑥2

With some rearrangement of the terms in (23), we have 𝜓2 𝑥 + 𝜓1 2𝜂 + 𝜓 [𝑥−1 (𝜂2 − 𝜂 +

𝜇2 + 𝜆 = 0.

(33)

𝜇 = ±√𝜆𝑖.

(34)

That is,

(I) (i): For instance, taking 𝜇 = −√𝜆𝑖, we have 𝜙,

𝜙2 𝑥 + 𝜙1 [−√𝜆𝑖2𝑥 + 2𝑝 + 1] (23)

1 − 𝑝2 ) + 𝜆𝑥] = 𝑓𝑥1−𝜂 . 4 (24)

(32)

Choose 𝜇 such that

√𝜆𝑖𝑥

𝜓2 𝑥𝜂 + 𝜓1 2𝜂𝑥𝜂−1 + 𝜓𝜂 (𝜂 − 1) 𝑥𝜂−2 + 𝑥𝜂 𝜓 [𝜆 −

+ 𝜙 (2𝑝𝜇 + 𝜇 + 𝑥 (𝜇2 + 𝜆)) = 𝑓𝑥(1/2)−𝑝 𝑒−𝜇𝑥 .

𝜓 = 𝑒−

Putting (21) and (22) in (16), we obtained

(31)

and substituting from (29) and (31) in (30), we can express (30) as 𝜙2 𝑥 + 𝜙1 (2𝜇𝑥 + 2𝑝 + 1)

2

𝑦 = 𝑥𝜂 𝜓,

(30)

At this point, differentiating 𝑒𝜇𝑥 𝜙 two times,

√𝜆𝑥

× {[(𝑓𝑥1/2+𝑝 𝑒−𝑖

(29)

Rewrite (28) in the form

𝑦𝚤V = 𝑥−𝑝+1/2 𝑒𝑖

× 𝑒−2𝑖

𝑦 = 𝑥𝑝+(1/2) 𝜓,

𝜓 = 𝑒𝜇𝑥 𝜙,

× {[(𝑓𝑥1/2+𝑝 𝑒𝑖

2

(26)

Set

√𝜆𝑥

× 𝑒2𝑖

1 ± 𝑝. 2

(I) Let 𝜂 = 𝑝 + (1/2). From (21) and (24), we have

√𝜆𝑥

𝑦𝚤𝚤𝚤 = 𝑥−𝑝+1/2 𝑒−𝑖

(25)

That is,

,

𝑦𝚤𝚤 = 𝑥𝑝+1/2 𝑒𝑖

× {[(𝑓𝑥

1 − 𝑝2 = 0. 4

+ 𝜙 [−𝑖 (2𝑝 + 1) √𝜆] = 𝑓𝑥(1/2)−𝑝 𝑒−𝜇𝑥

(35)

(36)

from (29) and (32). Applying the operator 𝑁] to both members of (36), we find the following equality: [𝜙2 𝑥]] + {𝜙1 [−√𝜆𝑖2𝑥 + 2𝑝 + 1]}] + {𝜙 [−𝑖 (2𝑝 + 1) √𝜆]}] = [𝑓𝑥(1/2)−𝑝 𝑒−𝜇𝑥 ]] .

(37)

4

The Scientific World Journal Applying the operator 𝑁] to both members of (47), we have

Using (3)–(12), we have [𝜙2 𝑥]] = 𝑥𝜙2+] + ]𝜙1+] , {𝜙1 [−√𝜆𝑖2𝑥 + 2𝑝 + 1]}]

(38)

𝜙2+] 𝑥 + 𝜙1+] [2√𝜆𝑖𝑥 + 2𝑝 + 1 + ]] + 𝜙] [√𝜆𝑖 (2𝑝 + 1) 𝑖] = (𝑓𝑥(1/2)−𝑝 𝑒−𝑖

= 𝜙1+] [−√𝜆𝑖2𝑥 + 2𝑝 + 1] − √𝜆𝑖2]𝜙] . Making use of the relations (38), rewrite (37) in the following form:

√𝜆𝑥

− 𝜙] [√𝜆𝑖2] + 𝑖 (2𝑝 + 1) √𝜆] = [𝑓𝑥(1/2)−𝑝 𝑒𝑖

].

(39)

]

]

(40)

(41)

×𝑒

𝜙−𝑝+1/2 = 𝜔 = 𝜔 (𝑥) ,

(42)

we obtain the following equality from (41): 1 𝜔1 + 𝜔 [𝑥−1 (𝑝 + ) − 2√𝜆𝑖] 2 ]

−𝑝−1/2

(43)

𝑥−1 .

This is an ordinary differential equation of the first order which has a particular solution, 𝜔 = [[𝑓𝑥

(1/2)−𝑝 −𝑖√𝜆𝑥

𝑒

√𝜆𝑥

)

𝑒

]

−𝑝−1/2

2√𝜆𝑖𝑥 −𝑝−1/2

×𝑒

𝑥

−1 −2√𝜆𝑖𝑥 𝑝+1/2

𝑥 𝑒

𝑥

]

−1

𝑥

𝑥−1 𝑒2

(45)

satisfies (41). Therefore, (17) satisfies (16) because we have (27), (35), (44), and (45). (I) (ii): In the case when 𝜇 = √𝜆𝑖, we have √𝜆𝑖𝑥

𝜙,

(46)

𝜙2 𝑥 + 𝜙1 [2√𝜆𝑖𝑥 + 2𝑝 + 1] + 𝜙 [(2𝑝 + 1) √𝜆𝑖] = 𝑓𝑥(1/2)−𝑝 𝑒−𝑖

√𝜆𝑥

𝑥

𝑑𝑥]

−1

(52)

Thus, we have (18) from (52), (50), (46), and (27). (II) Let 𝜂 = −𝑝 + (1/2). With the help of the similar method in (I), replacing 𝑝 by −𝑝 in (I) (i) and (I) (ii), we have other solutions (19) and (20) different from (17) and (18), respectively, if 𝑝 ≠ 0.

3. The Operator 𝑁] -Method to a Homogeneous Bessel Equation 𝑜

𝐿 [𝑦, 𝑥, 𝑝] = 𝑦2 + 𝑦 [𝜆 −

𝜓=𝑒

𝑥

.

(44)

Making use of the reverse process to obtain 𝑦𝚤 , we finally obtain the solution (17) from (44), (42), (35), and (27). Inversely, (44) satisfies (43); then

from (29) and (32).

−𝑝−1/2

Theorem 7. If 𝑦 ∈ ℘, just as in Theorem 5, then the homogeneous Bessel equation

.

𝜙 = 𝜔𝑝−1/2

)

(51)

−1

√𝜆𝑖𝑥 𝑝+1/2

−𝑝−1/2

−2√𝜆𝑖𝑥 −𝑝−1/2

(1/2)−𝑝 𝑖√𝜆𝑥

(50)

from (48). A particular solution of (51) is given by 𝜗 = [(𝑓𝑥1/2−𝑝 𝑒−𝑖

from (39). Next, writing

√𝜆𝑥

(49)

1 𝜗1 + 𝜗 [2√𝜆𝑖 + (𝑝 + ) 𝑥−1 ] 2 = (𝑓𝑥

1 − 2√𝜆𝑖𝑥] 2

−𝑝−1/2

= [𝑓𝑥(1/2)−𝑝 𝑒𝑖

1 2

𝜙−𝑝+1/2 = 𝜗 = 𝜗 (𝑥) ,

We then have

√𝜆𝑥

]

we then obtain

1 ] = −𝑝 − . 2

= [𝑓𝑥(1/2)−𝑝 𝑒𝑖

(48)

and replacing

Choose ] such that

𝜙−𝑝+3/2 𝑥 + 𝜙−𝑝+1/2 [𝑝 +

).

Choosing ] such that ] = −𝑝 −

𝜙2+] 𝑥 + 𝜙1+] [−√𝜆𝑖2𝑥 + 2𝑝 + 1 + ]]

√𝜆𝑥

(47)

𝑝2 − 1/4 ] = 0, 𝑥2

(53)

(0 < 𝑥 ≤ 1) , has solutions of the forms 𝑦𝚤 = 𝑘𝑥𝑝+1/2 𝑒−𝑖

√𝜆𝑥

√𝜆𝑥

𝑦𝚤𝚤 = 𝑘𝑥𝑝+1/2 𝑒𝑖

𝑦𝚤𝚤𝚤 = 𝑘𝑥−𝑝+1/2 𝑒−𝑖 𝑦𝚤V = 𝑘𝑥−𝑝+1/2 𝑒𝑖

{𝑒2𝑖

√𝜆𝑥 −𝑝−1/2

{𝑒−2𝑖

√𝜆𝑥

√𝜆𝑥

𝑥

}

√𝜆𝑥 −𝑝−1/2

𝑥

{𝑒2𝑖

{𝑒−2𝑖

√𝜆𝑥 𝑝−1/2

𝑝−1/2

}

𝑥

}

√𝜆𝑥 𝑝−1/2

}

𝑥

for 𝑝 ≠ 0, where 𝑘 is an arbitrary constant.

,

(54)

,

(55)

𝑝−1/2

,

(56)

,

(57)

−𝑝−1/2

−𝑝−1/2


5

Remark 8. In the case when 𝑝 = 0, (56) and (57) coincide with (54) and (55). Proof. When 𝑓 = 0 in Section 2, we have

0.4

0.6

0.8

1.0

−0.5

1 𝜔1 + 𝜔 [(𝑝 + ) 𝑥−1 − 2√𝜆𝑖] = 0, 2

(58)

1 𝜗1 + 𝜗 [(𝑝 + ) 𝑥−1 + 2√𝜆𝑖] = 0, 2

(59)

−1.0

for 𝜇 = −𝑖√𝜆 and 𝜇 = 𝑖√𝜆, instead of (43) and (51). Therefore, we get (54) for (58) and (55) for (59). And, for 𝜂 = −𝑝 + (1/2), replacing 𝑝 by −𝑝 in (58) and (59), we have (56) and (57). 𝑜

0.2

−1.5

Figure 1: Rey.

𝑜

Theorem 9. Let 𝑦 ∈ ℘ and 𝑓 ∈ ℘ just as in Theorem 5. Then the nonhomogeneous modified Sturm-Liouville equation (16) is satisfied by the fractional differintegrated functions 𝑦 = 𝑦𝚤 + 𝑦𝚤 .

(60)

0.8 0.6

Proof. It is clear by Theorems 5 and 7.

0.4

Application 1. If we substitute 𝑝 = 0, 𝜆 = 1/4, and 𝑓 = 𝑖𝑥−1/2 𝑒−(𝑖/2)𝑥 in (16), then we obtain the following equation:

0.2

1 1 𝑦2 + 𝑦 ( + 2 ) = 𝑖𝑥−1/2 𝑒−(𝑖/2)𝑥 , 4 4𝑥

(61)

0.2

0.4

0.6

0.8

1.0

Figure 2: Imy.

and its solution is 𝑦 = 𝑥1/2 𝑒(−𝑖/2)𝑥 {[(𝑖𝑥−1/2 𝑒−(𝑖/2)𝑥 𝑥1/2 𝑒(𝑖/2)𝑥 )−1/2 × 𝑒−𝑖𝑥 𝑥−1/2 ]−1 𝑒𝑖𝑥 𝑥−1/2 }

−1/2

.

(62)

By performing the necessary operations in (62), we get 𝑦 = 𝑥1/2 𝑒(−𝑖/2)𝑥 {[

2𝑖√𝑥 −𝑖𝑥 −1/2 𝑒 𝑥 ] 𝑒𝑖𝑥 𝑥−1/2 } , √𝜋 −1 −1/2

(63)

𝑥 𝑖 1 2𝑖√𝑥 , 𝑑𝑡 = ∫ √𝜋 Γ (1/2) 0 √𝑥 − 𝑡

𝑦=𝑥

1/2 (−𝑖/2)𝑥

𝑒

2𝑥−1/2 ) , (− √𝜋 −1/2

𝑦 = −2𝑥1/2 𝑒−(𝑖/2)𝑥 .

(64)

(65) (66)

Now, let us show that the last equality is the solution of (61): 1 1 𝑦2 = 𝑒−(𝑖/2)𝑥 [ 𝑥1/2 + 𝑥−3/2 + 𝑖𝑥−1/2 ] . 2 2

𝑦2 −

3 𝑦=0 4𝑥2

(0 < 𝑥 ≤ 1) ,

𝑦 = 𝑘𝑥−1/2 {𝑥1/2 }−3/2 .

and using the definitions of Riemann Liouville operator again, we obtain the following solution: 𝑥 2𝑡−1/2 1 2𝑥−1/2 ) =− 𝑑𝑡 = −2, (− ∫ √𝜋 −1/2 Γ (1/2) 0 √𝜋√𝑥 − 𝑡

Application 2. If we substitute 𝑝 = −1 and 𝜆 = 0 in (53), then we obtain the following equation: (68)

and its solution is

where Riemann Liouville operator is [𝑖]−1/2 =

Obviously, if (66) and (67) are put in (61), it is satisfied. The graph of the solution of (61) is given in Figures 1 and 2.

(67)

(69)

We prove that 𝑦𝚤 is the solution of (67). With the help of Riemann Liouville operator, [𝑥−1/2 ]−1/2 =

𝑥 √𝜋𝑥2 𝑡1/2 1 𝑑𝑡 = , ∫ Γ (3/2) 0 √𝑥 − 𝑡 4

𝑦=

𝑘𝑥3/2 √𝜋 . 4

(70) (71)

Now, let us show that the last equality is the solution of (67), 𝑦2 =

3𝑘𝑥−1/2 √𝜋 . 16

(72)

Obviously, if (71) and (72) are put into (68), it is satisfied. The graph of the solution of (68) is given in Figure 3.

6

The Scientific World Journal 𝑦(𝚤𝚤𝚤) = 𝑥−𝑝𝑖+1/2 𝑒−𝑖

2.0

√𝜆𝑥 √𝜆𝑥

× {[(𝑓𝑥1/2+𝑝𝑖 𝑒𝑖

)

𝑝𝑖−1/2

1.5

× 𝑒2𝑖

1.0

√𝜆𝑥 𝑝𝑖−1/2

𝑥

}

𝑒−2𝑖

−𝑝𝑖−1/2

√𝜆𝑥 −𝑝𝑖−1/2

𝑥

]

−1

,

√𝜆𝑥

𝑦(𝚤V) = 𝑥−𝑝𝑖+1/2 𝑒𝑖

0.5

× {[(𝑓𝑥1/2+𝑝𝑖 𝑒−𝑖 0.2

0.4

0.6

0.8

1.0

√𝜆𝑥

)

−2𝑖√𝜆𝑥 𝑝𝑖−1/2

×𝑒

𝑥

𝑝𝑖−1/2

𝑒2𝑖

}

−𝑝𝑖−1/2

Figure 3

√𝜆𝑥 −𝑝𝑖−1/2

𝑥

]

−1

. (76)

(ii) In the same way, for the solutions for (75), substituting the relations (21), and (22) into (75), we have

4. Two Further Cases of Modified Bessel Equation Theorem 10. In the similar way as in the previous sections, we can solve the following nonhomogeneous modified Bessel equation:

𝜙2 𝑥 + 𝜙1 2] + 𝜙 [(]2 − ] + Choose ] as follows: ]2 − ] +

(1/4) + 𝑝2 ] = 𝑓, 𝑦2 + 𝑦 [𝜆 + 𝑥2

(73)

(1/4) + 𝑝2 ] = 𝑓, 𝑦2 + 𝑦 [−𝜆 + 𝑥2

(74)

(75)

(i) Therefore, the solutions for (74) are given by replacing 𝑝 by 𝑖𝑝 in (17), (18), (19), and (20) as follows:

(𝚤𝚤)

𝑦

=𝑥

𝑦 = 𝑥𝑝𝑖+(1/2) 𝜙,

1/2−𝑝𝑖 𝑖√𝜆𝑥

𝑒

)

−2𝑖√𝜆𝑥 𝑝𝑖−1/2

−𝑝𝑖−1/2

√𝜆𝑥 −𝑝𝑖−1/2

.

(81)

Next, set (29); then (81) is rewritten in the form

Substituting the relations (29) and (31) into (82), we have 𝜓2 𝑥 + 𝜓1 (2𝜇𝑥 + 2𝑝𝑖 + 1) + 𝜓 [(𝜇2 − 𝜆) 𝑥 + (2𝑝𝑖 + 1) 𝜇] = 𝑓𝑥(1/2)−𝑖𝑝 𝑒−𝜇𝑥 .

𝑥

}

𝑒

𝑝𝑖−1/2

𝑥

]

−1

(83)

𝜇2 − 𝜆 = 0.

(84)

𝜇 = ±√𝜆.

(85)

That is,

,

(ii. 1) In the case when 𝜇 = −√𝜆, we have

𝑝𝑖+1/2 𝑖√𝜆𝑥

𝑒

√𝜆𝑥

𝜙 = 𝑒−

1/2−𝑝𝑖 −𝑖√𝜆𝑥

× 𝑒−2𝑖

(80) (1/2)−𝑝𝑖

Choose 𝜇 as follows:

√𝜆𝑥

× {[(𝑓𝑥

(79)

(𝑒𝜇𝑥 𝜓)2 𝑥 + (𝑒𝜇𝑥 𝜓)1 (2𝑖𝑝 + 1) − 𝑒𝜇𝑥 𝜓𝜆𝑥 = 𝑓𝑥(1/2)−𝑝𝑖 . (82)

2

× 𝑒2𝑖

1 ± 𝑝𝑖. 2

𝜙2 𝑥 + 𝜙1 (2𝑝𝑖 + 1) − 𝜙𝜆𝑥 = 𝑓𝑥

(1/4) − (𝑖𝑝) 𝑦2 + 𝑦 [−𝜆 + ] = 𝑓. 𝑥2

× {[(𝑓𝑥

(78)

Let ] = 𝑝𝑖 + (1/2). From (21) and (77), we have

2

(1/4) − (𝑖𝑝) ] = 𝑓, 𝑦2 + 𝑦 [𝜆 + 𝑥2

𝑦(𝚤) = 𝑥𝑝𝑖+1/2 𝑒−𝑖

1 + 𝑝2 = 0. 4

That is ]=

which are obtained by replacing 𝑝 by 𝑖𝑝 (−𝜆 instead of 𝜆) in (16); that is,

1 2 − (𝑝𝑖) ) 𝑥−1 − 𝜆𝑥] = 𝑓𝑥1−] . 4 (77)

𝑒

)

−𝑝𝑖−1/2

√𝜆𝑥 −𝑝𝑖−1/2

𝑥

2𝑖√𝜆𝑥 𝑝𝑖−1/2

}

𝑒

𝑝𝑖−1/2

𝑥

,

]

−1

𝜓,

(86)

𝜓2 𝑥 + 𝜓1 (−2√𝜆𝑥 + 2𝑝𝑖 + 1) √𝜆𝑥

− 𝜓 [√𝜆 (2𝑝𝑖 + 1)] = 𝑓𝑥(1/2)−𝑝𝑖 𝑒 from (29) and (83).

(87)


7

Applying the operator 𝑁] to both members of (87), we then obtain (𝜓2 𝑥)] + [𝜓1 (−2√𝜆𝑥 + 2𝑝𝑖 + 1)]] √𝜆𝑥

+ {𝜓 [−√𝜆 (2𝑝𝑖 + 1)]}] = (𝑓𝑥(1/2)−𝑝𝑖 𝑒

).

(88)

]

Using (4), (10), and (11), we have

√𝜆𝑥

+ 𝜓] [−√𝜆 (2𝑝𝑖 + 1 + 2V)] = (𝑓𝑥(1/2)−𝑝𝑖 𝑒

).

(89)

]

Choose ] such that (90)

We then have

from (91). This is an ordinary differential equation of the first order which has a particular solution

2√𝜆𝑥 −𝑝𝑖−1/2

×𝑒

𝑥


𝑥

]

−1

(94)

.

We finally obtain the solution √𝜆𝑥

𝑦(𝚤) = 𝑥𝑝𝑖+1/2 𝑒−

√𝜆𝑥

× {[(𝑓𝑥1/2−𝑝𝑖 𝑒 √𝜆𝑥

× 𝑒2

)


−𝑝𝑖−1/2

𝑥−𝑝𝑖−1/2 }

𝑒−2

𝑝𝑖−1/2

𝑥

]

−1

(95)

.

from (94), (92), (86) and (80). (ii. 2) Similarly, in the case when 𝜇 = √𝜆, we obtain √𝜆𝑥

𝑦(𝚤𝚤) = 𝑟𝑝𝑖+1/2 𝑒

√𝜆𝑥

× {[(𝑓𝑥1/2−𝑝𝑖 𝑒−

)


−𝑝𝑖−1/2

√𝜆𝑥 −𝑝𝑖−1/2

× 𝑒−2

𝑥

}

𝑒2

𝑝𝑖−1/2

.

𝑥

√𝜆𝑥 −𝑝𝑖−1/2

𝑥

(𝑒2𝑖

(𝑒−2𝑖


𝑥


𝑥

𝑝𝑖−1/2

)

)

)

,

𝑝𝑖−1/2

, (97)

−𝑝𝑖−1/2

−𝑝𝑖−1/2

,

,

(92)

1 √ 𝑢1 + 𝑢 [−2√𝜆 + (𝑝𝑖 + ) 𝑥−1 ] = (𝑓𝑥(1/2)−𝑝𝑖 𝑒 𝜆𝑥 ) 𝑥−1 −𝑝𝑖−1/2 2 (93)

𝑒−2

)

The 𝑁-fractional calculus operator 𝑁] -method is applied to the nonhomogeneous and homogeneous Bessel equation. Explicit fractional solutions of Bessel equations are obtained. Furthermore, similar solutions were obtained for the modified same equation by using the method.

we obtain

−𝑝𝑖−1/2

√𝜆𝑥

𝑥

5. Conclusion

−𝑝𝑖−1/2

𝜓1/2−𝑝𝑖 = 𝑢 = 𝑢 (𝑥) ,

)

(𝑒−2𝑖

√𝜆𝑥

𝑦(𝚤V) = 𝛼𝑥−𝑝𝑖+1/2 𝑒𝑖

√𝜆𝑥 −𝑝𝑖−1/2

(91)

)

from (89). Next, writing

√𝜆𝑥

√𝜆𝑥

𝑦(𝚤𝚤) = 𝛼𝑥𝑝𝑖+1/2 𝑒𝑖

(𝑒2𝑖

for 𝑝 ≠ 0, where 𝛼 is an arbitrary constant.

1 𝜓2−𝑝𝑖−1/2 𝑥 + 𝜓1−𝑝𝑖−1/2 [−2√𝜆𝑥 + 𝑝𝑖 + ] 2

𝑢 = [(𝑓𝑥(1/2)−𝑝𝑖 𝑒

√𝜆𝑥

𝑦(𝚤𝚤𝚤) = 𝛼𝑥−𝑝𝑖+1/2 𝑒−𝑖

1 ] = −𝑝𝑖 − . 2

√𝜆𝑥

Theorem 11. In the homogeneous case for (74) with 𝑓 = 0, using the solutions (54), (55), (56), and (57) and replacing 𝑝 by 𝑝𝑖, we obtain 𝑦(𝚤) = 𝛼𝑥𝑝𝑖+1/2 𝑒−𝑖

𝜓2+] 𝑥 + 𝜓1+] (−2√𝜆𝑥 + 2𝑝𝑖 + 1 + ])

= (𝑓𝑥(1/2)−𝑝𝑖 𝑒

Let ] = −𝑖𝑝 + (1/2). In the same way as in the procedure in (ii), replacing 𝑖𝑝 by –𝑖𝑝 (ii. 1) and (ii. 2), we can obtain 𝑦(𝚤𝚤𝚤) and 𝑦(𝚤V) .

]

−1

(96)

Acknowledgment The authors sincerely thank the reviewers for their valuable suggestions and useful comments.

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