# Bessel Equation in the Semiunbounded Intervalx

Authors: Qing-Hua Zhang; Jian Ma; Yuanyuan Qu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2016)

---

## Abstract

This study expresses the solution of the Bessel equation in the neighbourhood ofx
=
∞ as the product of a known-form singular divisor and a specific nonsingular function, which satisfies the corresponding derived equation. Considering the failure of the traditional irregular solution constructed with the power series, we adopt the corrected Fourier series with only limited smooth degree to approximate the nonsingular function in the interval [
x
0
,
∞
]. In order to guarantee the series’ uniform convergence and uniform approximation to the derived equation, we introduce constraint and compatibility conditions and hence completely determine all undetermined coefficients of the corrected Fourier series. Thus, what we found is not an asymptotic solution at x
→
∞ (not to mention a so-called formal solution), but a solution in the interval [
x
0
,
∞
] with certain regularities of distribution. During the solution procedure, there is no limitation on the coefficient property of the equation; that is, the coefficients of the equation can be any complex constant, so that the solution method presented here is universal.


---

## Body

## 1. Introduction

The general form of the Bessel equation is(1)
x
2
y
′
′
x
+
λ
1
x
y
′
x
+
λ
2
x
2
+
λ
3
y
x
=
0
, where λ
j
(
j
=
1,2
,
3
) are arbitrary complex constants, with λ
1
≠
0 and λ
2
≠
0, and x
∈
[
0
,
∞
].The Bessel equation only has two singular points on the real axis:x
=
0 and x
=
∞. For convenience, set division point at x
0
>
0, to separate the entire solution interval into two parts, [
0
,
x
0
] and [
x
0
,
∞
], the latter of which is the semiunbounded interval that this study will solve the equation in. It is easy to verify that x
=
0 is a regular singular point, but x
=
∞ is irregular. Previous studies showed that the solution in the neighbourhood of the isolated singular point could be written as a combination of a known-form singular divisor and a specific nonsingular function. To find the solution in the neighbourhood of the regular singular point, one often uses the power series extension to reconstruct the nonsingular function. In the neighbourhood of the irregular singular point, attempts have been made with power series traditionally, but already existing efforts cannot find a uniformly convergent solution, which can only give an asymptotic solution at x
→
∞. This so-called regular formal solution can only indicate the variation tendency of the solution [1–4].However, the failure to obtain a solution with certain regularities of distribution using power series method does not necessarily mean inexistence of this kind of solution. As is known, the power series with uniform convergence in the complex number space converges to a complex analytic function, so that it is full differentiable. Yet, the Bessel equation is only a second-order differential equation, which demands limited smoothness (e.g., second-order derivative) only, but not full smoothness. Overdemanded smoothness could be the reason for the inability to find the solution with the uniformly convergent power series.This study adopts the corrected Fourier series [5–10] with limited smoothness to construct the nonsingular function and insert it into the derived equations from the Bessel equation. By introducing the constraint and compatibility conditions at the same time, all coefficients of the corrected Fourier series can be determined completely. Thus, one can obtain the solution with certain regularities of distribution in the interval x
∈
[
x
0
,
∞
], but not an asymptotic or formal solution. There is no limitation on the equation parameter λ
j in the whole solution procedure, which could be any arbitrary complex constant, so that the solution method given in this study is universal.


## 2. The Solution of the Bessel Equation in the Semiunbounded Interval[
x
0
,
∞
]

### 2.1. Obtaining the Derived Equation

For convenience, we introduce coordinate transformation(2)
y
x
=
y
x
0
ξ
-
1
=
y
e
ξ
,
ξ
=
x
0
x
,which transforms the semiunbounded interval x
∈
x
0
,
∞ to a bounded one ξ
∈
0,1. Equation (1) turns into the following form with the new coordinate (Appendix A):(3)
ξ
4
y
e
′
′
ξ
+
2
-
λ
1
ξ
3
y
e
′
ξ
+
λ
20
+
λ
3
ξ
2
y
e
ξ
=
0
,
ξ
∈
0,1
,
λ
20
=
λ
2
x
0
2
.It is easy to find that ξ
=
0 is an irregular singular point of the equation, and then the irregular solution can be written as(4)
y
e
ξ
=
R
ξ
V
ξ
,
R
ξ
=
ξ
ρ
e
σ
ξ
-
1
=
ξ
ρ
exp
⁡
σ
ξ
-
1
,where ρ, σ in R
(
ξ
) are constants to be determined.V
(
ξ
) is a nonsingular function with V
(
0
)
≠
0, which satisfies the following equation:(5)
ξ
4
V
′
′
ξ
+
P
1
ξ
V
′
ξ
+
P
2
ξ
V
ξ
=
0
,where(6)
P
1
ξ
=
2
ξ
4
R
′
R
+
2
-
λ
1
ξ
3
,
P
2
ξ
=
ξ
4
R
′
′
R
+
2
-
λ
1
ξ
3
R
′
R
+
λ
20
+
λ
3
ξ
2
.Using the differential relationship,(7)
R
′
R
=
ρ
ξ
-
1
-
σ
ξ
-
2
,
R
′
′
R
=
ρ
ρ
-
1
ξ
-
2
+
2
-
2
ρ
σ
ξ
-
3
+
σ
2
ξ
-
4
,one can obtain(8)
P
1
ξ
=
q
11
+
q
12
ξ
ξ
2
,
P
2
ξ
=
q
21
+
q
22
ξ
+
q
23
ξ
2
.Hence, (5) can be written as the following form:(9)
L
V
≡
ξ
4
V
′
′
+
q
11
+
q
12
ξ
ξ
2
V
′
+
q
21
+
q
22
ξ
+
q
23
ξ
2
V
=
0
,where(10)
q
11
=
-
2
σ
,
q
12
=
2
+
2
ρ
-
λ
1
,
q
21
=
σ
2
+
λ
20
,
q
22
=
λ
1
-
2
ρ
σ
,
q
23
=
ρ
ρ
+
1
-
λ
1
+
λ
3
.Let ξ
→
0, and, from(11)
L
V
=
0
,
L
V
′
=
0
,one can obtain(12)
q
21
=
q
22
=
0
;that is,(13)
σ
2
+
λ
20
=
0
,
λ
1
-
2
ρ
=
0
,which determines that(14a)
σ
=
σ
1
=
σ
0
,
σ
0
=
-
λ
20
=
-
λ
2
x
0
,
σ
2
=
-
σ
0
,
σ
0
=
-
λ
20
=
-
λ
2
x
0
,
(14b)
ρ
=
λ
1
2
,where(15)
q
11
=
-
2
σ
1
,
q
12
=
2
,
q
23
=
λ
1
2
1
-
λ
1
2
+
λ
3
.So, the derived equation is obtained:(16)
L
1
V
,
σ
≡
ξ
2
V
′
′
+
q
1
ξ
,
σ
V
′
+
q
0
V
=
0
,where(17)
q
1
ξ
,
σ
=
2
ξ
-
σ
,
q
0
=
q
23
.


### 2.2. Nonsingular Classic Solution of the Derived Solution

The form of the derived equation (see (16)) still seems to be a singular equation, but it can be guaranteed to have nonsingular classic solution after introducing certain constraint conditions (Appendix B). That is,(18a)
∂
∂
ξ
L
1
V
,
σ
ξ
=
0
=
0
;that is,(18b)
-
2
σ
V
′
′
0
,
σ
+
2
+
q
0
V
′
0
,
σ
=
0
.BecauseV
′
′
(
0
,
σ
)
<
∞, V
(
ξ
) can be expanded to the following corrected Fourier series:(19)
V
ξ
=
∑
n
≤
N
A
n
exp
⁡
i
α
n
ξ
+
∑
l
=
1
3
a
l
ξ
l
l
!
,
V
k
ξ
=
∑
n
≤
N
A
n
i
α
n
k
exp
⁡
i
α
n
ξ
+
∑
l
=
k
3
a
l
ξ
l
-
k
l
-
k
!
,
k
=
1,2
,which is uniformly convergent and where α
n
=
2
n
π. At the same time, we introduce the Fourier projection:(20)
F
-
1
n
=
∫
0
1
exp
⁡
-
i
α
n
ξ
d
ξ
.Inserting expansion equation (19) to the derived equation (see (16)), one can obtain (Appendix C)(21)
∑
n
A
n
G
R
n
,
ξ
,
σ
exp
⁡
i
α
n
ξ
+
∑
l
=
1
3
a
l
G
S
l
,
ξ
,
σ
=
0
.After the Fourier projection,(22)
∑
n
≤
N
A
n
R
n
,
n
0
,
σ
=
∑
l
=
1
3
a
l
S
l
,
n
0
,
σ
,
n
0
≤
N
,where(23)
R
n
,
n
0
,
σ
=
F
-
1
G
R
n
,
ξ
,
σ
exp
⁡
i
α
n
ξ
n
0
=
F
-
1
G
R
n
,
ξ
,
σ
n
0
-
n
,
S
l
,
n
0
,
σ
=
-
F
-
1
G
S
l
,
ξ
,
σ
n
0
.It can be derived that(24)
A
n
=
∑
l
=
1
3
a
l
A
e
l
,
n
,
σ
,where A
e
(
l
,
n
,
σ
) is the solution of the following equation:(25)
∑
n
A
e
l
,
n
,
σ
R
n
,
n
0
,
σ
=
S
l
,
n
0
,
σ
.Inserting solution (24) into extension (19), one can obtain (26)
V
ξ
,
σ
=
∑
l
=
1
3
a
l
Z
0
l
,
ξ
,
σ
,
V
k
ξ
,
σ
=
∑
l
=
1
3
a
l
Z
k
l
,
ξ
,
σ
,
k
=
1,2
,where(27)
Z
0
l
,
ξ
,
σ
=
∑
n
A
e
l
,
n
,
σ
exp
⁡
i
α
n
ξ
+
ξ
l
l
!
,
Z
k
l
,
ξ
,
σ
=
Z
0
k
l
,
ξ
,
σ
,
k
=
1,2
.The compatibility condition is required [5, 6] in Fourier projection of (16) (Appendix B):(28a)
L
1
V
ξ
,
σ
0
1
=
0
;that is,(28b)
V
′
′
1
,
σ
+
q
1
1
,
σ
V
′
1
,
σ
-
q
1
0
,
σ
V
′
0
,
σ
+
q
0
V
1
,
σ
-
V
0
,
σ
=
0
.Insert solution (26) into constraint (18a) and (18b) and consistency (28a) and (28b) conditions to obtain the equation determining a
l (Appendix D):(29)
∑
l
=
1
3
a
l
β
1
l
σ
=
0
,
∑
l
=
1
3
a
l
β
2
l
σ
=
0
.It can be solved that (30)
a
l
=
γ
l
,
σ
;hence, (31)
V
ξ
,
σ
=
∑
l
=
1
3
γ
l
,
σ
Z
0
l
,
ξ
,
σ
.Note that(32)
V
1
ξ
=
V
ξ
,
σ
1
=
V
ξ
,
σ
0
,
V
2
ξ
=
V
ξ
,
σ
2
=
V
ξ
,
-
σ
0
.Finally, the solution of (3) is obtained:(33)
y
e
1
ξ
=
ξ
λ
1
/
2
exp
⁡
σ
0
ξ
-
1
V
1
ξ
,
y
e
2
ξ
=
ξ
λ
1
/
2
exp
⁡
-
σ
0
ξ
-
1
V
2
ξ
.


## 2.1. Obtaining the Derived Equation

For convenience, we introduce coordinate transformation(2)
y
x
=
y
x
0
ξ
-
1
=
y
e
ξ
,
ξ
=
x
0
x
,which transforms the semiunbounded interval x
∈
x
0
,
∞ to a bounded one ξ
∈
0,1. Equation (1) turns into the following form with the new coordinate (Appendix A):(3)
ξ
4
y
e
′
′
ξ
+
2
-
λ
1
ξ
3
y
e
′
ξ
+
λ
20
+
λ
3
ξ
2
y
e
ξ
=
0
,
ξ
∈
0,1
,
λ
20
=
λ
2
x
0
2
.It is easy to find that ξ
=
0 is an irregular singular point of the equation, and then the irregular solution can be written as(4)
y
e
ξ
=
R
ξ
V
ξ
,
R
ξ
=
ξ
ρ
e
σ
ξ
-
1
=
ξ
ρ
exp
⁡
σ
ξ
-
1
,where ρ, σ in R
(
ξ
) are constants to be determined.V
(
ξ
) is a nonsingular function with V
(
0
)
≠
0, which satisfies the following equation:(5)
ξ
4
V
′
′
ξ
+
P
1
ξ
V
′
ξ
+
P
2
ξ
V
ξ
=
0
,where(6)
P
1
ξ
=
2
ξ
4
R
′
R
+
2
-
λ
1
ξ
3
,
P
2
ξ
=
ξ
4
R
′
′
R
+
2
-
λ
1
ξ
3
R
′
R
+
λ
20
+
λ
3
ξ
2
.Using the differential relationship,(7)
R
′
R
=
ρ
ξ
-
1
-
σ
ξ
-
2
,
R
′
′
R
=
ρ
ρ
-
1
ξ
-
2
+
2
-
2
ρ
σ
ξ
-
3
+
σ
2
ξ
-
4
,one can obtain(8)
P
1
ξ
=
q
11
+
q
12
ξ
ξ
2
,
P
2
ξ
=
q
21
+
q
22
ξ
+
q
23
ξ
2
.Hence, (5) can be written as the following form:(9)
L
V
≡
ξ
4
V
′
′
+
q
11
+
q
12
ξ
ξ
2
V
′
+
q
21
+
q
22
ξ
+
q
23
ξ
2
V
=
0
,where(10)
q
11
=
-
2
σ
,
q
12
=
2
+
2
ρ
-
λ
1
,
q
21
=
σ
2
+
λ
20
,
q
22
=
λ
1
-
2
ρ
σ
,
q
23
=
ρ
ρ
+
1
-
λ
1
+
λ
3
.Let ξ
→
0, and, from(11)
L
V
=
0
,
L
V
′
=
0
,one can obtain(12)
q
21
=
q
22
=
0
;that is,(13)
σ
2
+
λ
20
=
0
,
λ
1
-
2
ρ
=
0
,which determines that(14a)
σ
=
σ
1
=
σ
0
,
σ
0
=
-
λ
20
=
-
λ
2
x
0
,
σ
2
=
-
σ
0
,
σ
0
=
-
λ
20
=
-
λ
2
x
0
,
(14b)
ρ
=
λ
1
2
,where(15)
q
11
=
-
2
σ
1
,
q
12
=
2
,
q
23
=
λ
1
2
1
-
λ
1
2
+
λ
3
.So, the derived equation is obtained:(16)
L
1
V
,
σ
≡
ξ
2
V
′
′
+
q
1
ξ
,
σ
V
′
+
q
0
V
=
0
,where(17)
q
1
ξ
,
σ
=
2
ξ
-
σ
,
q
0
=
q
23
.


## 2.2. Nonsingular Classic Solution of the Derived Solution

The form of the derived equation (see (16)) still seems to be a singular equation, but it can be guaranteed to have nonsingular classic solution after introducing certain constraint conditions (Appendix B). That is,(18a)
∂
∂
ξ
L
1
V
,
σ
ξ
=
0
=
0
;that is,(18b)
-
2
σ
V
′
′
0
,
σ
+
2
+
q
0
V
′
0
,
σ
=
0
.BecauseV
′
′
(
0
,
σ
)
<
∞, V
(
ξ
) can be expanded to the following corrected Fourier series:(19)
V
ξ
=
∑
n
≤
N
A
n
exp
⁡
i
α
n
ξ
+
∑
l
=
1
3
a
l
ξ
l
l
!
,
V
k
ξ
=
∑
n
≤
N
A
n
i
α
n
k
exp
⁡
i
α
n
ξ
+
∑
l
=
k
3
a
l
ξ
l
-
k
l
-
k
!
,
k
=
1,2
,which is uniformly convergent and where α
n
=
2
n
π. At the same time, we introduce the Fourier projection:(20)
F
-
1
n
=
∫
0
1
exp
⁡
-
i
α
n
ξ
d
ξ
.Inserting expansion equation (19) to the derived equation (see (16)), one can obtain (Appendix C)(21)
∑
n
A
n
G
R
n
,
ξ
,
σ
exp
⁡
i
α
n
ξ
+
∑
l
=
1
3
a
l
G
S
l
,
ξ
,
σ
=
0
.After the Fourier projection,(22)
∑
n
≤
N
A
n
R
n
,
n
0
,
σ
=
∑
l
=
1
3
a
l
S
l
,
n
0
,
σ
,
n
0
≤
N
,where(23)
R
n
,
n
0
,
σ
=
F
-
1
G
R
n
,
ξ
,
σ
exp
⁡
i
α
n
ξ
n
0
=
F
-
1
G
R
n
,
ξ
,
σ
n
0
-
n
,
S
l
,
n
0
,
σ
=
-
F
-
1
G
S
l
,
ξ
,
σ
n
0
.It can be derived that(24)
A
n
=
∑
l
=
1
3
a
l
A
e
l
,
n
,
σ
,where A
e
(
l
,
n
,
σ
) is the solution of the following equation:(25)
∑
n
A
e
l
,
n
,
σ
R
n
,
n
0
,
σ
=
S
l
,
n
0
,
σ
.Inserting solution (24) into extension (19), one can obtain (26)
V
ξ
,
σ
=
∑
l
=
1
3
a
l
Z
0
l
,
ξ
,
σ
,
V
k
ξ
,
σ
=
∑
l
=
1
3
a
l
Z
k
l
,
ξ
,
σ
,
k
=
1,2
,where(27)
Z
0
l
,
ξ
,
σ
=
∑
n
A
e
l
,
n
,
σ
exp
⁡
i
α
n
ξ
+
ξ
l
l
!
,
Z
k
l
,
ξ
,
σ
=
Z
0
k
l
,
ξ
,
σ
,
k
=
1,2
.The compatibility condition is required [5, 6] in Fourier projection of (16) (Appendix B):(28a)
L
1
V
ξ
,
σ
0
1
=
0
;that is,(28b)
V
′
′
1
,
σ
+
q
1
1
,
σ
V
′
1
,
σ
-
q
1
0
,
σ
V
′
0
,
σ
+
q
0
V
1
,
σ
-
V
0
,
σ
=
0
.Insert solution (26) into constraint (18a) and (18b) and consistency (28a) and (28b) conditions to obtain the equation determining a
l (Appendix D):(29)
∑
l
=
1
3
a
l
β
1
l
σ
=
0
,
∑
l
=
1
3
a
l
β
2
l
σ
=
0
.It can be solved that (30)
a
l
=
γ
l
,
σ
;hence, (31)
V
ξ
,
σ
=
∑
l
=
1
3
γ
l
,
σ
Z
0
l
,
ξ
,
σ
.Note that(32)
V
1
ξ
=
V
ξ
,
σ
1
=
V
ξ
,
σ
0
,
V
2
ξ
=
V
ξ
,
σ
2
=
V
ξ
,
-
σ
0
.Finally, the solution of (3) is obtained:(33)
y
e
1
ξ
=
ξ
λ
1
/
2
exp
⁡
σ
0
ξ
-
1
V
1
ξ
,
y
e
2
ξ
=
ξ
λ
1
/
2
exp
⁡
-
σ
0
ξ
-
1
V
2
ξ
.


## 3. Examples and Discussion

As mentioned before, the solution method given here satisfies the equation with each parameter being any arbitrary complex constant, so that the first set of examples we will show are to provide the results of the functionsV
1
(
ξ
) and V
2
(
ξ
) according to several sets of different complex constants. Their accuracy of approximation to satisfy the derived equation will also be estimated. The second set of examples aim to solve V
1
(
ξ
), V
2
(
ξ
) and, in turn, the solution of the Bessel equation, y
e
1
(
ξ
) and y
e
2
(
ξ
), with several sets of certain common real constants. Noting their linear combination as y
1
(
x
) and y
2
(
x
), we will compare them with already known Bessel functions and modified Bessel functions.


### 3.1. First Set of Examples

Example 1.
Letλ
1
=
(
1
+
i
)
/
2,  λ
2
=
1,  λ
3
=
-
ν
2
(
ν
=
0.3
),  x
0
=
10.0, and N
=
50, and calculate V
1
(
ξ
) and V
2
(
ξ
). To examine the accuracy of the solution to approximate the derived equation, note(34)
g
1
ξ
,
j
=
ξ
2
V
j
′
′
ξ
,
g
2
ξ
,
j
=
q
1
ξ
,
σ
j
V
j
′
ξ
,
g
3
ξ
,
j
=
q
0
V
j
ξ
,
g
0
ξ
,
j
=
g
1
ξ
,
j
+
g
2
ξ
,
j
+
g
3
ξ
,
j
,with complex module (35)
E
l
ξ
,
j
=
g
l
ξ
,
j
,
l
=
0,1
,
2,3
.And then, we can define the relative accuracy of the solution V
j
(
ξ
) to satisfy the equation(36)
E
r
r
o
r
ξ
,
j
=
E
0
ξ
,
j
Max
⁡
E
1
ξ
,
j
,
E
2
ξ
,
j
,
E
3
ξ
,
j
.Values of the solution and relative error are shown in Table 1.Table 1
Solution and relative error for Example1.
ξ
V
1
(
ξ
)
V
2
(
ξ
)
E
r
r
o
r
(
ξ
,
1
)
E
r
r
o
r
(
ξ
,
2
)
0
(-
1.3512, -
2.7890)
(2.6727, 1.6044)
5.76
×
10
-
7
1.70
×
10
-
7
1/5
(-
1.3573, -
2.7902)
(2.6668, 1.6066)
1.47
×
10
-
4
1.47
×
10
-
4
2/5
(-
1.3635, -
2.7913)
(2.6608, 1.6086)
8.72
×
10
-
5
8.75
×
10
-
5
3/5
(-
1.3698, -
2.7923)
(2.6549, 1.6106)
1.19
×
10
-
4
1.19
×
10
-
4
4/5
(-
1.3760, -
2.7932)
(2.6488, 1.6124)
1.74
×
10
-
4
1.75
×
10
-
4
1
(-
1.3822, -
2.7939)
(2.6428, 1.6141)
6.43
×
10
-
7
3.80
×
10
-
7Example 2.
Letλ
1
=
1
,
λ
2
=
(
1
+
i
)
/
2, λ
3
=
-
ν
2
(
ν
=
0.3
), x
0
=
10.0, and N
=
50, and the results are presented in Table 2.Table 2
Solution and relative error for Example2.
ξ
V
1
(
ξ
)
V
2
(
ξ
)
E
r
r
o
r
(
ξ
,
1
)
E
r
r
o
r
(
ξ
,
2
)
0
(2.6352,-
0.3074)
(-
1.4742, 1.3684)
1.65
×
10
-
7
9.26
×
10
-
8
1/5
(2.6327,-
0.3118)
(-
1.4777, 1.3668)
1.92
×
10
-
4
2.83
×
10
-
4
2/5
(2.6301,-
0.3160)
(-
1.4812, 1.3650)
9.90
×
10
-
5
1.37
×
10
-
4
3/5
(2.6275,-
0.3202)
(
-
1.4846, 1.3632)
2.05
×
10
-
4
1.99
×
10
-
4
4/5
(2.6248,-
0.3242)
(-
1.4880, 1.3612)
2.82
×
10
-
4
3.28
×
10
-
4
1
(2.6220,-
0.3281)
(-
1.4914, 1.3592)
9.37
×
10
-
7
1.90
×
10
-
7Example 3.
Letλ
1
=
1
,
λ
2
=
1
,
λ
3
=
(
-
ν
2
+
i
)
/
2
(
ν
=
0.3
), x
0
=
10.0, and N
=
50, and the results are in Table 3.Table 3
Solution and relative error for Example3.
ξ
V
1
(
ξ
)
V
2
(
ξ
)
E
r
r
o
r
(
ξ
,
1
)
E
r
r
o
r
(
ξ
,
2
)
0
(
-
1.0161, -
0.4045)
(1.0966,-
0.2249)
7.67
×
10
-
7
1.11
×
10
-
6
1/5
(-
1.0220, -
0.4044)
(1.0916,-
0.2216)
1.57
×
10
-
4
1.56
×
10
-
4
2/5
(-
1.0280, -
0.4041)
(1.0865,-
0.2185)
9.41
×
10
-
5
9.44
×
10
-
5
3/5
(-
1.0339,  -
0.4037)
(1.0814,-
0.2154)
1.27
×
10
-
4
1.26
×
10
-
4
4/5
(-
1.0399, -
0.4032)
(1.0762,-
0.2125)
1.83
×
10
-
4
1.85
×
10
-
4
1
(-
1.0458, -
0.4025)
(1.0711,-
0.2097)
7.30
×
10
-
7
1.87
×
10
-
6From the above three examples, it is easy to see that functionsV
1
(
ξ
) and V
2
(
ξ
) vary very slowly with ξ; thus, it is hard to imagine to approximate them with uniform convergence by using the power series. On the other hand, when letting x
0
=
10
.
0 and N
=
50, they satisfy the derived equation at relative errors smaller than 4 × 10−4, meaning accuracy is significant enough.


### 3.2. Second Set of Examples

Example 1.
To compare with Bessel function of the first kind, letλ
1
=
λ
2
=
1
,
λ
3
=
-
ν
2
(
ν
=
0.3
,
2
ν
≠
i
n
t
e
g
e
r
), x
0
=
20.0, and N
=
200. For these parameters, the Bessel equation has solutions of the Bessel function of the first kind, J
ν
(
x
) and J
-
ν
(
x
), in the neighbourhood of x
=
0, which has infinite points of zero on the real axis. And when x
≫
1, they have the following Asymptotic Formula to determine the distribution of the zero points:(37)
x
1
m
=
m
+
ν
2
-
1
4
π
,
x
2
m
=
m
-
ν
2
-
1
4
π
.
The solutions obtained in this studyy
e
1
(
ξ
) and y
e
2
(
ξ
) are conjugate complex (∵
σ
0
=
i
λ
2
x
0; Appendix E); however, real solution compatible with J
ν
(
x
) and J
-
ν
(
x
) can be obtained by linear combination of certain form (Appendix F):(38)
y
1
x
=
b
11
y
e
1
ξ
+
b
12
y
e
2
ξ
,
y
2
x
=
b
21
y
e
1
ξ
+
b
22
y
e
2
ξ
,with the zero points noted as x
~
1
(
m
) and x
~
2
(
m
).Table4 compares the zero points x
1
(
m
) of J
ν
(
x
) and x
~
1
(
m
) of y
1
(
x
) and the zero points x
2
(
m
) of J
-
ν
(
x
) and x
~
2
(
m
) of y
2
(
x
). It is easy to find that the solution obtained here (after rearranging) has consistent distribution of zero points with Bessel function of the first kind, indicating the credibility of the solution method provided by the present study [1, 2].Table 4
Comparisons of the zero points of the Bessel function in Example1 between the asymptotic formula and the corrected Fourier series in this study.
m
x
1
(
m
)
x
~
1
(
m
)
x
2
(
m
)
x
~
2
(
m
)
7
21.6770
21.68
20.7345
20.73
8
24.8186
24.82
23.8761
23.87
9
27.9602
27.96
27.0177
27.01
10
31.1018
31.10
30.1593
30.16
11
34.2433
34.24
33.3009
33.30
12
37.3849
37.38
36.4425
36.44
13
40.5265
40.52
39.5841
39.58
14
43.6681
43.67
42.7257
42.72
15
46.8091
46.81
45.8672
45.86
16
49.9513
49.95
49.0088
49.00
17
53.0929
53.09
52.1504
52.15
18
56.2345
56.23
55.2920
55.29
19
59.3761
59.37
58.4336
58.43
20
62.5177
62.51
61.5752
61.57
Relative errors:
E
r
r
o
r
1
m
=
x
~
1
(
m
)
-
x
1
(
m
)
/
x
1
(
m
)
<
1.40
×
1
0
-
4,
E
r
r
o
r
2
m
=
x
~
2
(
m
)
-
x
2
(
m
)
/
x
2
(
m
)
<
2.55
×
1
0
-
4.Example 2.
Letλ
1
=
1, λ
2
=
-
1, λ
3
=
0, x
0
=
3.0, and N
=
200, which represent the Bessel equation with modified Bessel functions I
0
(
x
) and K
0
(
x
) in the neighbourhood of x
=
0. With this set of parameters, the solutions obtained by this paper are y
e
1
(
ξ
) and y
e
2
(
ξ
) in the interval x
∈
x
0
,
∞ or ξ
∈
0,1, with their linear combination also being solution of the equation(39)
y
1
x
=
b
11
y
e
1
ξ
+
b
12
y
e
2
ξ
,
y
2
x
=
b
21
y
e
1
ξ
+
b
22
y
e
2
ξ
.From the table of Bessel functions, one can look up the values of functions I
0
(
x
) and K
0
(
x
) in the interval x
∈
0,10. At the two arbitrarily selected points in the interval x
∈
3,10, we linearly fit y
1
(
x
) and I
0
(
x
) and y
2
(
x
) and K
0
(
x
), to determine constant b
i
j. The obtained distributions of functions y
1
(
x
) and y
2
(
x
) are compared with those of functions I
0
(
x
) and K
0
(
x
) in Table 5, which shows apparent consistency between the solution obtained with known modified Bessel functions and that calculated in this study.Table 5
Comparisons between the solution of Example2 obtained with known modified Bessel functions and that calculated in this study.
x
y
1
(
x
)
I
0
(
x
)
y
2
(
x
)
K
0
(
x
)
3.0
4.881
4.881
3.474
×
10
-
2
3.474
×
10
-
2
3.5
7.384
7.378
1.960
×
10
-
2
1.960
×
10
-
2
4.0
11.31
11.30
1.116
×
10
-
2
1.116
×
10
-
2
4.5
17.49
17.48
6.400
×
10
-
3
6.400
×
10
-
3
5.0
27.26
27.24
3.691
×
10
-
3
3.691
×
10
-
3
5.5
42.71
42.69
2.139
×
10
-
3
2.139
×
10
-
3
6.0
67.26
67.23
1.244
×
10
-
3
1.244
×
10
-
3
6.5
106.3
106.3
7.259
×
10
-
4
7.259
×
10
-
4
7.0
168.6
168.6
4.248
×
10
-
4
4.248
×
10
-
4
7.5
268.3
268.2
2.492
×
10
-
4
2.492
×
10
-
4
8.0
427.7
427.6
1.465
×
10
-
4
1.465
×
10
-
4
8.5
683.4
683.2
8.626
×
10
-
5
8.626
×
10
-
5
9.0
1094
1094
5.088
×
10
-
5
5.088
×
10
-
5
9.5
1754
1753
3.006
×
10
-
5
3.006
×
10
-
5
9.9
2562
2561
1.974
×
10
-
5
1.975
×
10
-
5
Relative errors:
E
r
r
o
r
1
x
=
y
1
(
x
)
-
I
0
(
x
)
/
I
0
(
x
)
<
0.88
×
1
0
-
3,
E
r
r
o
r
2
x
=
y
2
(
x
)
-
K
0
(
x
)
/
K
0
(
x
)
<
0.50
×
1
0
-
3.Example 3.
Letλ
1
=
1, λ
2
=
-
1, λ
3
=
-
1, x
0
=
3.0, and N
=
200, and this set of parameters corresponds to Bessel equation with modified Bessel functions I
1
(
x
) and K
1
(
x
) as the solution in the neighbourhood of x
=
0. The corresponding solution obtained here is y
e
1
(
ξ
) and y
e
2
(
ξ
), whose linear combination can construct new solution functions, after linear fit with functions I
1
(
x
) and K
1
(
x
):(40)
y
1
x
=
b
11
y
e
1
ξ
+
b
12
y
e
2
ξ
,
y
2
x
=
b
21
y
e
1
ξ
+
b
22
y
e
2
ξ
.Comparisons between their distributions and those of I
1
(
x
) and K
1
(
x
) are illustrated in Table 6. It can be easily seen that the solution obtained here is completely consistent with the known solution with modified Bessel functions [1, 2].Table 6
Comparisons between the solution of Example3 obtained with known modified Bessel functions and that calculated in this study.
x
y
1
(
x
)
I
1
(
x
)
y
2
(
x
)
K
1
(
x
)
3.0
3.953
3.953
4.016
×
10
-
2
4.016
×
10
-
2
3.5
6.203
6.206
2.224
×
10
-
2
2.224
×
10
-
2
4.0
9.755
9.759
1.248
×
10
-
2
1.248
×
10
-
2
4.5
15.39
15.39
7.079
×
10
-
3
7.078
×
10
-
3
5.0
24.33
24.34
4.045
×
10
-
3
4.045
×
10
-
3
5.5
38.59
38.59
2.326
×
10
-
3
2.326
×
10
-
3
6.0
61.35
61.34
1.344
×
10
-
3
1.344
×
10
-
3
6.5
97.75
97.74
7.799
×
10
-
4
7.799
×
10
-
4
7.0
156.1
156.0
4.542
×
10
-
4
4.542
×
10
-
4
7.5
249.6
249.6
2.653
×
10
-
4
2.653
×
10
-
4
8.0
399.9
399.9
1.554
×
10
-
4
1.554
×
10
-
4
8.5
641.7
641.6
9.120
×
10
-
5
9.120
×
10
-
5
9.0
1031
1031
5.364
×
10
-
5
5.364
×
10
-
5
9.5
1659
1658
3.160
×
10
-
5
3.160
×
10
-
5
9.9
2428
2428
2.072
×
10
-
5
2.072
×
10
-
5
Relative errors:
E
r
r
o
r
1
x
=
y
1
(
x
)
-
I
1
(
x
)
/
I
1
(
x
)
<
0.5
×
1
0
-
3,
E
r
r
o
r
2
x
=
y
2
(
x
)
-
K
1
(
x
)
/
K
1
(
x
)
<
1.5
×
1
0
-
4.


## 3.1. First Set of Examples

Example 1.
Letλ
1
=
(
1
+
i
)
/
2,  λ
2
=
1,  λ
3
=
-
ν
2
(
ν
=
0.3
),  x
0
=
10.0, and N
=
50, and calculate V
1
(
ξ
) and V
2
(
ξ
). To examine the accuracy of the solution to approximate the derived equation, note(34)
g
1
ξ
,
j
=
ξ
2
V
j
′
′
ξ
,
g
2
ξ
,
j
=
q
1
ξ
,
σ
j
V
j
′
ξ
,
g
3
ξ
,
j
=
q
0
V
j
ξ
,
g
0
ξ
,
j
=
g
1
ξ
,
j
+
g
2
ξ
,
j
+
g
3
ξ
,
j
,with complex module (35)
E
l
ξ
,
j
=
g
l
ξ
,
j
,
l
=
0,1
,
2,3
.And then, we can define the relative accuracy of the solution V
j
(
ξ
) to satisfy the equation(36)
E
r
r
o
r
ξ
,
j
=
E
0
ξ
,
j
Max
⁡
E
1
ξ
,
j
,
E
2
ξ
,
j
,
E
3
ξ
,
j
.Values of the solution and relative error are shown in Table 1.Table 1
Solution and relative error for Example1.
ξ
V
1
(
ξ
)
V
2
(
ξ
)
E
r
r
o
r
(
ξ
,
1
)
E
r
r
o
r
(
ξ
,
2
)
0
(-
1.3512, -
2.7890)
(2.6727, 1.6044)
5.76
×
10
-
7
1.70
×
10
-
7
1/5
(-
1.3573, -
2.7902)
(2.6668, 1.6066)
1.47
×
10
-
4
1.47
×
10
-
4
2/5
(-
1.3635, -
2.7913)
(2.6608, 1.6086)
8.72
×
10
-
5
8.75
×
10
-
5
3/5
(-
1.3698, -
2.7923)
(2.6549, 1.6106)
1.19
×
10
-
4
1.19
×
10
-
4
4/5
(-
1.3760, -
2.7932)
(2.6488, 1.6124)
1.74
×
10
-
4
1.75
×
10
-
4
1
(-
1.3822, -
2.7939)
(2.6428, 1.6141)
6.43
×
10
-
7
3.80
×
10
-
7Example 2.
Letλ
1
=
1
,
λ
2
=
(
1
+
i
)
/
2, λ
3
=
-
ν
2
(
ν
=
0.3
), x
0
=
10.0, and N
=
50, and the results are presented in Table 2.Table 2
Solution and relative error for Example2.
ξ
V
1
(
ξ
)
V
2
(
ξ
)
E
r
r
o
r
(
ξ
,
1
)
E
r
r
o
r
(
ξ
,
2
)
0
(2.6352,-
0.3074)
(-
1.4742, 1.3684)
1.65
×
10
-
7
9.26
×
10
-
8
1/5
(2.6327,-
0.3118)
(-
1.4777, 1.3668)
1.92
×
10
-
4
2.83
×
10
-
4
2/5
(2.6301,-
0.3160)
(-
1.4812, 1.3650)
9.90
×
10
-
5
1.37
×
10
-
4
3/5
(2.6275,-
0.3202)
(
-
1.4846, 1.3632)
2.05
×
10
-
4
1.99
×
10
-
4
4/5
(2.6248,-
0.3242)
(-
1.4880, 1.3612)
2.82
×
10
-
4
3.28
×
10
-
4
1
(2.6220,-
0.3281)
(-
1.4914, 1.3592)
9.37
×
10
-
7
1.90
×
10
-
7Example 3.
Letλ
1
=
1
,
λ
2
=
1
,
λ
3
=
(
-
ν
2
+
i
)
/
2
(
ν
=
0.3
), x
0
=
10.0, and N
=
50, and the results are in Table 3.Table 3
Solution and relative error for Example3.
ξ
V
1
(
ξ
)
V
2
(
ξ
)
E
r
r
o
r
(
ξ
,
1
)
E
r
r
o
r
(
ξ
,
2
)
0
(
-
1.0161, -
0.4045)
(1.0966,-
0.2249)
7.67
×
10
-
7
1.11
×
10
-
6
1/5
(-
1.0220, -
0.4044)
(1.0916,-
0.2216)
1.57
×
10
-
4
1.56
×
10
-
4
2/5
(-
1.0280, -
0.4041)
(1.0865,-
0.2185)
9.41
×
10
-
5
9.44
×
10
-
5
3/5
(-
1.0339,  -
0.4037)
(1.0814,-
0.2154)
1.27
×
10
-
4
1.26
×
10
-
4
4/5
(-
1.0399, -
0.4032)
(1.0762,-
0.2125)
1.83
×
10
-
4
1.85
×
10
-
4
1
(-
1.0458, -
0.4025)
(1.0711,-
0.2097)
7.30
×
10
-
7
1.87
×
10
-
6From the above three examples, it is easy to see that functionsV
1
(
ξ
) and V
2
(
ξ
) vary very slowly with ξ; thus, it is hard to imagine to approximate them with uniform convergence by using the power series. On the other hand, when letting x
0
=
10
.
0 and N
=
50, they satisfy the derived equation at relative errors smaller than 4 × 10−4, meaning accuracy is significant enough.


## 3.2. Second Set of Examples

Example 1.
To compare with Bessel function of the first kind, letλ
1
=
λ
2
=
1
,
λ
3
=
-
ν
2
(
ν
=
0.3
,
2
ν
≠
i
n
t
e
g
e
r
), x
0
=
20.0, and N
=
200. For these parameters, the Bessel equation has solutions of the Bessel function of the first kind, J
ν
(
x
) and J
-
ν
(
x
), in the neighbourhood of x
=
0, which has infinite points of zero on the real axis. And when x
≫
1, they have the following Asymptotic Formula to determine the distribution of the zero points:(37)
x
1
m
=
m
+
ν
2
-
1
4
π
,
x
2
m
=
m
-
ν
2
-
1
4
π
.
The solutions obtained in this studyy
e
1
(
ξ
) and y
e
2
(
ξ
) are conjugate complex (∵
σ
0
=
i
λ
2
x
0; Appendix E); however, real solution compatible with J
ν
(
x
) and J
-
ν
(
x
) can be obtained by linear combination of certain form (Appendix F):(38)
y
1
x
=
b
11
y
e
1
ξ
+
b
12
y
e
2
ξ
,
y
2
x
=
b
21
y
e
1
ξ
+
b
22
y
e
2
ξ
,with the zero points noted as x
~
1
(
m
) and x
~
2
(
m
).Table4 compares the zero points x
1
(
m
) of J
ν
(
x
) and x
~
1
(
m
) of y
1
(
x
) and the zero points x
2
(
m
) of J
-
ν
(
x
) and x
~
2
(
m
) of y
2
(
x
). It is easy to find that the solution obtained here (after rearranging) has consistent distribution of zero points with Bessel function of the first kind, indicating the credibility of the solution method provided by the present study [1, 2].Table 4
Comparisons of the zero points of the Bessel function in Example1 between the asymptotic formula and the corrected Fourier series in this study.
m
x
1
(
m
)
x
~
1
(
m
)
x
2
(
m
)
x
~
2
(
m
)
7
21.6770
21.68
20.7345
20.73
8
24.8186
24.82
23.8761
23.87
9
27.9602
27.96
27.0177
27.01
10
31.1018
31.10
30.1593
30.16
11
34.2433
34.24
33.3009
33.30
12
37.3849
37.38
36.4425
36.44
13
40.5265
40.52
39.5841
39.58
14
43.6681
43.67
42.7257
42.72
15
46.8091
46.81
45.8672
45.86
16
49.9513
49.95
49.0088
49.00
17
53.0929
53.09
52.1504
52.15
18
56.2345
56.23
55.2920
55.29
19
59.3761
59.37
58.4336
58.43
20
62.5177
62.51
61.5752
61.57
Relative errors:
E
r
r
o
r
1
m
=
x
~
1
(
m
)
-
x
1
(
m
)
/
x
1
(
m
)
<
1.40
×
1
0
-
4,
E
r
r
o
r
2
m
=
x
~
2
(
m
)
-
x
2
(
m
)
/
x
2
(
m
)
<
2.55
×
1
0
-
4.Example 2.
Letλ
1
=
1, λ
2
=
-
1, λ
3
=
0, x
0
=
3.0, and N
=
200, which represent the Bessel equation with modified Bessel functions I
0
(
x
) and K
0
(
x
) in the neighbourhood of x
=
0. With this set of parameters, the solutions obtained by this paper are y
e
1
(
ξ
) and y
e
2
(
ξ
) in the interval x
∈
x
0
,
∞ or ξ
∈
0,1, with their linear combination also being solution of the equation(39)
y
1
x
=
b
11
y
e
1
ξ
+
b
12
y
e
2
ξ
,
y
2
x
=
b
21
y
e
1
ξ
+
b
22
y
e
2
ξ
.From the table of Bessel functions, one can look up the values of functions I
0
(
x
) and K
0
(
x
) in the interval x
∈
0,10. At the two arbitrarily selected points in the interval x
∈
3,10, we linearly fit y
1
(
x
) and I
0
(
x
) and y
2
(
x
) and K
0
(
x
), to determine constant b
i
j. The obtained distributions of functions y
1
(
x
) and y
2
(
x
) are compared with those of functions I
0
(
x
) and K
0
(
x
) in Table 5, which shows apparent consistency between the solution obtained with known modified Bessel functions and that calculated in this study.Table 5
Comparisons between the solution of Example2 obtained with known modified Bessel functions and that calculated in this study.
x
y
1
(
x
)
I
0
(
x
)
y
2
(
x
)
K
0
(
x
)
3.0
4.881
4.881
3.474
×
10
-
2
3.474
×
10
-
2
3.5
7.384
7.378
1.960
×
10
-
2
1.960
×
10
-
2
4.0
11.31
11.30
1.116
×
10
-
2
1.116
×
10
-
2
4.5
17.49
17.48
6.400
×
10
-
3
6.400
×
10
-
3
5.0
27.26
27.24
3.691
×
10
-
3
3.691
×
10
-
3
5.5
42.71
42.69
2.139
×
10
-
3
2.139
×
10
-
3
6.0
67.26
67.23
1.244
×
10
-
3
1.244
×
10
-
3
6.5
106.3
106.3
7.259
×
10
-
4
7.259
×
10
-
4
7.0
168.6
168.6
4.248
×
10
-
4
4.248
×
10
-
4
7.5
268.3
268.2
2.492
×
10
-
4
2.492
×
10
-
4
8.0
427.7
427.6
1.465
×
10
-
4
1.465
×
10
-
4
8.5
683.4
683.2
8.626
×
10
-
5
8.626
×
10
-
5
9.0
1094
1094
5.088
×
10
-
5
5.088
×
10
-
5
9.5
1754
1753
3.006
×
10
-
5
3.006
×
10
-
5
9.9
2562
2561
1.974
×
10
-
5
1.975
×
10
-
5
Relative errors:
E
r
r
o
r
1
x
=
y
1
(
x
)
-
I
0
(
x
)
/
I
0
(
x
)
<
0.88
×
1
0
-
3,
E
r
r
o
r
2
x
=
y
2
(
x
)
-
K
0
(
x
)
/
K
0
(
x
)
<
0.50
×
1
0
-
3.Example 3.
Letλ
1
=
1, λ
2
=
-
1, λ
3
=
-
1, x
0
=
3.0, and N
=
200, and this set of parameters corresponds to Bessel equation with modified Bessel functions I
1
(
x
) and K
1
(
x
) as the solution in the neighbourhood of x
=
0. The corresponding solution obtained here is y
e
1
(
ξ
) and y
e
2
(
ξ
), whose linear combination can construct new solution functions, after linear fit with functions I
1
(
x
) and K
1
(
x
):(40)
y
1
x
=
b
11
y
e
1
ξ
+
b
12
y
e
2
ξ
,
y
2
x
=
b
21
y
e
1
ξ
+
b
22
y
e
2
ξ
.Comparisons between their distributions and those of I
1
(
x
) and K
1
(
x
) are illustrated in Table 6. It can be easily seen that the solution obtained here is completely consistent with the known solution with modified Bessel functions [1, 2].Table 6
Comparisons between the solution of Example3 obtained with known modified Bessel functions and that calculated in this study.
x
y
1
(
x
)
I
1
(
x
)
y
2
(
x
)
K
1
(
x
)
3.0
3.953
3.953
4.016
×
10
-
2
4.016
×
10
-
2
3.5
6.203
6.206
2.224
×
10
-
2
2.224
×
10
-
2
4.0
9.755
9.759
1.248
×
10
-
2
1.248
×
10
-
2
4.5
15.39
15.39
7.079
×
10
-
3
7.078
×
10
-
3
5.0
24.33
24.34
4.045
×
10
-
3
4.045
×
10
-
3
5.5
38.59
38.59
2.326
×
10
-
3
2.326
×
10
-
3
6.0
61.35
61.34
1.344
×
10
-
3
1.344
×
10
-
3
6.5
97.75
97.74
7.799
×
10
-
4
7.799
×
10
-
4
7.0
156.1
156.0
4.542
×
10
-
4
4.542
×
10
-
4
7.5
249.6
249.6
2.653
×
10
-
4
2.653
×
10
-
4
8.0
399.9
399.9
1.554
×
10
-
4
1.554
×
10
-
4
8.5
641.7
641.6
9.120
×
10
-
5
9.120
×
10
-
5
9.0
1031
1031
5.364
×
10
-
5
5.364
×
10
-
5
9.5
1659
1658
3.160
×
10
-
5
3.160
×
10
-
5
9.9
2428
2428
2.072
×
10
-
5
2.072
×
10
-
5
Relative errors:
E
r
r
o
r
1
x
=
y
1
(
x
)
-
I
1
(
x
)
/
I
1
(
x
)
<
0.5
×
1
0
-
3,
E
r
r
o
r
2
x
=
y
2
(
x
)
-
K
1
(
x
)
/
K
1
(
x
)
<
1.5
×
1
0
-
4.


## 4. Conclusions

The present study proves that the solution of the Bessel equation in the interval with irregular singular point can be expressed as the product of a singular divisor and a nonsingular function, the latter of which can be approximated with uniform convergence by corrected Fourier series with limited smoothness in the interval. Thus, we are able to obtain the solution with certain regularities of distribution of the Bessel equation in the semiunbounded interval with irregular singular point, but no more an asymptotic solution atx
→
∞, or a regular formal solution. This indicates that it is incorrect to conclude existence of only asymptotic or regular formal solution in the neighbourhood of an irregular singular point only because of the inadequacy of certain solution method.



---

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