Open Access

The continuous fractional Bessel wavelet transformation

  • Akhilesh Prasad1,
  • Ashutosh Mahato1,
  • Vishal Kumar Singh1 and
  • Madan Mohan Dixit2Email author
Boundary Value Problems20132013:40

DOI: 10.1186/1687-2770-2013-40

Received: 17 November 2012

Accepted: 25 January 2013

Published: 27 February 2013

Abstract

The main objective of this paper is to study the fractional Hankel transformation and the continuous fractional Bessel wavelet transformation and some of their basic properties. Applications of the fractional Hankel transformation (FrHT) in solving generalized n th order linear nonhomogeneous ordinary differential equations are given. The continuous fractional Bessel wavelet transformation, its inversion formula and Parseval’s relation for the continuous fractional Bessel wavelet transformation are also studied.

MSC:46F12, 26A33.

Keywords

Hankel transformation fractional Hankel transformation fractional Bessel wavelet transformation Bessel function

1 Introduction

Pathak and Dixit [1] introduced continuous and discrete Bessel wavelet transformations and studied their properties by exploiting the Hankel convolution of Haimo [2] and Hirschman [3]. Upadhyay et al. [4] studied the continuous Bessel wavelet transformation associated with the Hankel-Hausdorff operator.

Let L p ( R ) denote the class of measurable functions of ϕ on such that the integral R | ϕ ( x ) | p d t is finite. Also, let L ( R ) be a collection of almost everywhere bounded functions, hence endowed with the norm
ϕ L p = { ( R | ϕ ( x ) | p d x ) 1 / p , 1 p < , ess sup x R | ϕ ( x ) | , p = .
The Hankel transformation h μ , [5] of a conventional function φ L 1 ( R + ) , R + = ( 0 , ) is usually defined by
φ ˆ ( y ) = ( h μ φ ) ( y ) = 0 ( x y ) 1 2 J μ ( x y ) φ ( x ) d x , x R + , μ 1 / 2 ,
(1)
and its inversion formula is given by
φ ( x ) = ( h μ 1 φ ˆ ( y ) ) ( x ) = 0 ( x y ) 1 2 J μ ( x y ) φ ˆ ( y ) d y , y R + ,
(2)

where J μ is the Bessel function of the first kind of order μ.

The fractional Hankel transformation is the generalization of the conventional Hankel transformation in the fractional order with parameter θ and is effectively used in the design of lens, analysis of laser cavity study of wave propagation in quadratic refractive index medium when the system is axially symmetric. The earliest work on the fractional Hankel transformation was published by Namias in 1980 [6]. Recently, it has become of importance in various applications in optics [7, 8]. Kerr [9] has developed a theory of fractional power of Hankel transforms in Zemanian spaces. We define a one-dimensional fractional Hankel transformation (FrHT) with parameter θ of φ ( x ) for μ 1 / 2 and 0 < θ < π as follows:
φ ˆ μ θ ( y ) = ( h μ θ φ ) ( y ) = 0 K μ θ ( x , y ) φ ( x ) d x ,
(3)
where the kernel
K μ θ ( x , y ) = { c μ θ e i 2 ( x 2 + y 2 ) cot θ ( x y csc θ ) 1 2 J μ ( x y csc θ ) , θ n π , ( x y ) 1 2 J μ ( x y ) , θ = π 2 , δ ( x y ) , θ = n π , n Z ,
and
c μ θ = exp [ i ( 1 + μ ) ( π / 2 θ ) ] sin θ .
The inversion formula of (3) is given by
φ ( x ) = ( ( h μ θ ) 1 φ ˆ ) ( x ) = 0 K μ θ ( x , y ) ¯ ( h μ θ φ ) ( y ) d y ,
(4)
where
K μ θ ( x , y ) ¯ = exp [ i ( 1 + μ ) ( π / 2 θ ) ] e i 2 ( x 2 + y 2 ) cot θ ( x y csc θ ) 1 2 J μ ( x y csc θ ) = ( c μ θ ) ¯ sin θ e i 2 ( x 2 + y 2 ) cot θ ( x y csc θ ) 1 2 J μ ( x y csc θ ) ,
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Eque_HTML.gif

We assume that throughout this paper θ n π , n Z .

From [10], wavelets as a family of functions constructed from translation and dilation of a single function ψ are called the mother wavelet defined by
ψ b , a ( x ) = 1 a ψ ( x b a ) , b R , a > 0 ,
where a is called the scaling parameter which measures the degree of compression or scale and b is a translation parameter which determines the time location of the wavelet. Shi et al. [11] defined the fractional mother wavelet as
ψ b , a , θ ( x ) = 1 a ψ ( x b a ) e i 2 ( x 2 b 2 ) cot θ ,

for all a, b and θ as above.

As per [2, 12], we defined the fractional Hankel convolution of functions φ , ψ L 1 ( R + ) as follows:
( φ # θ ψ ) ( x ) = 0 φ θ ( x , y ) ψ ( y ) d y = 0 ( τ x θ φ ) ( y ) ψ ( y ) d y = 0 φ ( y ) ( τ x θ ψ ) ( y ) d y ,
(5)
where the fractional Hankel translation of the function φ L 1 ( R + ) is defined by
( τ x θ φ ) ( y ) = φ θ ( x , y ) = e i 2 ( x 2 + y 2 ) cot θ 0 φ ( z ) D μ θ ( x , y , z ) d z ,
(6)
and
D μ θ ( x , y , z ) = c μ θ e i 2 ( x 2 + y 2 + z 2 ) cot θ 0 ξ ( μ 1 2 ) ( x ξ csc θ ) 1 2 J μ ( x ξ csc θ ) × ( y ξ csc θ ) 1 2 J μ ( y ξ csc θ ) ( z ξ csc θ ) 1 2 J μ ( z ξ csc θ ) d ξ = 2 μ 1 2 μ 1 c μ θ e i 2 ( x 2 + y 2 + z 2 ) cot θ ( x y z ) μ 1 / 2 π Γ ( μ + 1 / 2 ) ,
(7)

where ( x , y , z ) denotes the area of a triangle with sides x, y, z of such a triangle exists and zero otherwise. Clearly, | D μ θ ( x , y , z ) | 0 and is symmetric in x, y, z.

Now, setting ξ = 0 , we have
0 | D μ θ ( x , y , z ) | z μ + 1 / 2 d z ( x y ) μ + 1 / 2 2 μ Γ ( μ + 1 ) | ( sin θ ) μ + 1 / 2 |
(8)

for x , y , z R + .

Applying the inverse fractional Hankel transformation of D μ θ ( x , y , z ) , we obtain
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equh_HTML.gif
Lemma 1.1 If φ L 2 ( R + ) , then
x μ 1 / 2 ( τ x θ φ ) ( y ) L 2 1 | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) φ L 2 .

Proof

Since
( τ x θ φ ) ( y ) = φ θ ( x , y ) = e i 2 ( x 2 + y 2 ) cot θ 0 φ ( z ) D μ θ ( x , y , z ) d z ,
using (8), we have
| ( τ x θ φ ) ( y ) | 0 | φ ( z ) z 1 / 2 ( μ + 1 / 2 ) { D μ θ ( x , y , z ) } 1 / 2 z 1 / 2 ( μ + 1 / 2 ) { D μ θ ( x , y , z ) } 1 / 2 | d z ( 0 z ( μ + 1 / 2 ) | φ ( z ) | 2 | D μ θ ( x , y , z ) | d z ) 1 2 ( 0 z ( μ + 1 / 2 ) | D μ θ ( x , y , z ) | d z ) 1 2 ( ( x y ) μ + 1 / 2 | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ) 1 2 ( 0 z ( μ + 1 / 2 ) | φ ( z ) | 2 | D μ θ ( x , y , z ) | d z ) 1 2 ,
(9)
so that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equk_HTML.gif
Thus
x μ 1 / 2 ( τ x θ φ ) ( y ) L 2 1 | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) φ L 2 .

 □

Remark 1.1 If φ L 2 ( R + ) , then
0 | ( τ y θ φ ) ( x ) | 2 d x y 2 ( μ + 1 / 2 ) ( | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ) 2 0 | φ ( z ) | 2 d z ,
and
y μ 1 / 2 ( τ y θ φ ) ( x ) L 2 1 | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) φ L 2 .

2 Properties of a fractional Hankel transformation

Zemanian [[5], p.129] introduced a function space H μ ( R + ) consisting of all complex-valued infinitely differentiable function φ defined on R + = ( 0 , ) , satisfying
Γ m , k μ ( φ ) = sup x R + | x m ( x 1 D ) k [ x μ 1 / 2 φ ( x ) ] | < , μ R , m , k N 0 .
(10)

Definition 2.1 (Test function space H μ , θ ( R + ) )

The space H μ , θ ( R + ) is defined as follows: φ is a member of H μ , θ ( R + ) if and only if it is a complex-valued C -function on R + and for every choice of m and k of non-negative integers, it satisfies
ϒ m , k θ ( φ ) = sup x R + | x m Δ μ , x k φ ( x ) | < , θ n π , n Z ,
(11)
where
Δ μ , x = [ d 2 d x 2 + 2 i x cot θ d d x + ( 1 4 μ 2 4 x 2 ) + i cot θ x 2 cot 2 θ ] ,
(12)
and
Δ μ , x k = x 2 k r = 0 2 k ( l = 0 2 k a l x 2 l ) ( x 1 d d x ) r ,

where the constants a l depend only on μ and parameter θ. On H μ , θ ( R + ) , we consider the topology generated by the family { ϒ m , k θ } m , k N 0 of seminorms.

Proposition 2.1 Let K μ θ ( x , y ) be the kernel of the fractional Hankel transformation. Then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equp_HTML.gif

where Δ μ , x = [ d 2 d x 2 2 i x cot θ d d x + ( 1 4 μ 2 4 x 2 ) i cot θ x 2 cot 2 θ ] and is known as a fractional Bessel operator with parameter θ.

Proof See [13]. □

Example 2.1 h μ θ [ ( Δ μ , x ) r δ ( x c ) ] ( y ) = ( y 2 csc 2 θ ) r K μ θ ( c , y ) , x , c R + .

The result can be easily shown by using Proposition 2.1(ii).

Proposition 2.2 Let φ L 1 ( R + ) . Then φ ˆ μ θ satisfies the following:
(i) φ ˆ μ θ L ( R + ) with φ ˆ μ θ L A μ , θ φ L 1 , (ii) φ ˆ μ θ ( y ) 0 as y + or , (iii) φ ˆ μ θ is continuous on R + ,

where A μ , θ is a positive constant depending on μ and θ.

Proof (i) Clearly, φ ˆ μ θ ( y ) = 0 K μ θ ( x , y ) φ ( x ) d x .

So, φ ˆ μ θ L A μ , θ φ L 1 .
  1. (ii)
    From Proposition 2.1(ii), where r = 1 , we have
    | φ ˆ μ θ ( y ) | = 1 | ( y 2 csc 2 θ ) | | ( h μ θ ( Δ μ , x φ ( x ) ) ) ( y ) | 0 as  y ± ,
     
although φ ˆ μ θ ( y ) 0 as y ± for every φ L 1 ( R + ) .
  1. (iii)
    Let h > 0 , consider
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equs_HTML.gif
     
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equt_HTML.gif
So,
sup y | φ ˆ μ θ ( y + h ) ϕ ˆ μ θ ( y ) | 0 as  h 0 .

This proves that φ ˆ μ θ ( y ) is continuous in R + . □

Proposition 2.3 (Parseval’s relation)

If Φ ( y ) = ( h μ θ φ ) ( y ) and Ψ ( y ) = ( h μ θ ψ ) ( y ) denote the fractional Hankel transformations of φ ( x ) and ψ ( x ) respectively, then
0 φ ( x ) ψ ( x ) ¯ d x = sin θ 0 ( h μ θ φ ) ( y ) ( h μ θ ψ ) ( y ) ¯ d y = sin θ 0 Φ ( y ) Ψ ( y ) ¯ d y ,
(13)
and
0 | φ ( x ) | 2 d x = sin θ 0 | ( h μ θ φ ) ( y ) | 2 d y .
(14)

Proof

We have
φ , ψ = 0 φ ( x ) ψ ( x ) ¯ d x = 0 φ ( x ) ( ( c μ θ ) ¯ sin θ 0 e i 2 ( x 2 + y 2 ) cot θ ( x y csc θ ) 1 2 J μ ( x y csc θ ) ( h μ θ ψ ) ( y ) d y ) ¯ d x = c μ θ sin θ 0 ( h μ θ ψ ) ¯ ( y ) ( 0 e i 2 ( x 2 + y 2 ) cot θ ( x y csc θ ) 1 2 J μ ( x y csc θ ) φ ( x ) d x ) d y = sin θ 0 ( h μ θ φ ) ( y ) ( h μ θ ψ ) ¯ ( y ) d y = sin θ 0 Φ ( y ) Ψ ( y ) ¯ d y .
If φ = ψ , then
0 | φ ( x ) | 2 d x = sin θ 0 | ( h μ θ φ ) ( y ) | 2 d y .

 □

3 Applications of the fractional Hankel transformation to generalized differential equations

We consider the generalized n th order linear nonhomogeneous ordinary differential equation
L φ ( x ) = f ( x ) ,
(15)
where L is the generalized n th order differential operator given by
L = a n ( Δ μ , x ) n + a n 1 ( Δ μ , x ) n 1 + + a 1 ( Δ μ , x ) + a 0 ,

where a n , a n 1 , , a 0 are constants and Δ μ , x is as given in Proposition 2.1.

Applying FrHT to both sides of equation (15), we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equy_HTML.gif
and equivalently,
P ( y 2 csc 2 θ ) ( h μ θ φ ) ( y ) = ( h μ θ f ) ( y ) , where  P ( z ) = r = 0 n a r z r .
Therefore,
( h μ θ φ ) ( y ) = ( h μ θ f ) ( y ) P ( y 2 csc 2 θ ) .
(16)
Now, an application of the inverse FrHT gives the solution
φ ( x ) = ( h μ θ ) 1 [ ( h μ θ f ) ( y ) P ( y 2 csc 2 θ ) ] .
Example 3.1 Let us consider ( 1 ( Δ μ , x ) 2 ) φ ( x ) = f ( x ) . Then we have
φ ( x ) = ( h μ θ ) 1 [ ( 1 y 4 csc 4 θ ) 1 ( h μ θ f ) ( y ) ] .
Example 3.2 Using the FrHT, we investigate the solution of the generalized differential equation
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equ17_HTML.gif
(17)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equ18_HTML.gif
(18)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equ19_HTML.gif
(19)
Let u ˆ 0 θ ( y , z ) be the FrHT of order zero of u ( x , z ) with respect to the variable x. Then, by definition,
u ˆ 0 θ ( y , z ) = 0 K 0 θ ( x , y ) u ( x , z ) d x ,
(20)

where K 0 θ ( x , y ) is the kernel of FrHT of order zero.

Taking the FrHT of order zero of (17), we get
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equac_HTML.gif
where D d d z , whose solution is
u ˆ 0 θ ( y , z ) = A e z y csc θ + B e z y csc θ .
(21)
Taking the FrHT of order zero of (18), we have
u ˆ 0 θ ( y , z ) 0 as  z .
(22)

Condition (22) is satisfied if we have A = 0 .

Therefore, from (21)
u ˆ 0 θ ( y , z ) = B e z y csc θ .
(23)
Taking the FrHT of order zero of (19), we have
0 u ( x , 0 ) K 0 θ ( x , y ) d x = 0 f ( x ) K 0 θ ( x , y ) d x , u ˆ 0 θ ( y , 0 ) = f ˆ 0 θ ( y ) ,
(24)

where f ˆ 0 θ ( y ) is the FrHT of zero order of f ( x ) .

Putting z = 0 in (23) and using (24), we get B = f ˆ 0 θ ( y ) .

Hence (23) reduces to
u ˆ 0 θ ( y , z ) = f ˆ 0 θ ( y ) e z y csc θ .
Applying the inversion formula, we have
u ( x , z ) = 0 f ˆ 0 θ ( y ) e z y csc θ K 0 θ ¯ ( x , y ) d y .

4 The continuous fractional Bessel wavelet transformation

The continuous fractional Bessel wavelet transformation (CFrBWT) is a generalization of the ordinary continuous Bessel wavelet transformation (CBWT) with parameter θ, that is, CBWT is a special case of CFrBWT with parameter θ = π 2 . In this section, we define the continuous fractional Bessel wavelet transformation and study some of its properties using the theory of fractional Hankel convolution (5) corresponding to [10].

A fractional Bessel wavelet is a function ψ L 2 ( R + ) which satisfies the condition
C μ , ψ , θ = 0 x 2 μ 2 | ( h μ θ ψ ) ( x ) | 2 d x < , μ 1 / 2 ,
where C μ , ψ , θ is called the admissibility condition of the fractional Bessel wavelet and ( h μ θ ψ ) is the fractional Hankel transformation of ψ. The fractional Bessel wavelets ψ b , a θ are generated from one single function ψ L 2 ( R + ) by dilation and translation with parameters a > 0 and b 0 respectively by
ψ b , a θ ( x ) = 1 a D a τ b θ ψ ( x ) = 1 a D a ψ θ ( b , x ) = 1 a ψ θ ( b a , x a ) = 1 a e i 2 ( b 2 a 2 + x 2 a 2 ) cot θ 0 ψ ( z ) D μ θ ( b a , x a , z ) d z .
Lemma 4.1 If ψ L 2 ( R + ) , then
ψ b , a θ L 2 b ( μ + 1 / 2 ) a ( μ + 1 / 2 ) | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ψ L 2 .

Proof

We have
ψ b , a θ ( x ) = 1 a e i 2 ( b 2 a 2 + x 2 a 2 ) cot θ 0 ψ ( z ) D μ θ ( b a , x a , z ) d z .
Now,
| ψ b , a θ ( x ) | 1 a 0 | ψ ( z ) | z 1 / 2 ( μ + 1 / 2 ) | { D μ θ ( b a , x a , z ) } 1 / 2 | z 1 / 2 ( μ + 1 / 2 ) × | { D μ θ ( b a , x a , z ) } 1 / 2 | d z 1 a ( 0 z ( μ + 1 / 2 ) | ψ ( z ) | 2 | D μ θ ( b a , x a , z ) | d z ) 1 / 2 × ( 0 | D μ θ ( b a , x a , z ) | z ( μ + 1 / 2 ) d z ) 1 / 2 1 a ( ( b x ) μ + 1 / 2 a 2 ( μ + 1 / 2 ) | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ) 1 / 2 × ( 0 z ( μ + 1 / 2 ) | ψ ( z ) | 2 | D μ θ ( b a , x a , z ) | d z ) 1 / 2 .
Therefore,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equak_HTML.gif
Thus,
ψ b , a θ L 2 b ( μ + 1 / 2 ) a ( μ + 1 / 2 ) | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ψ L 2 .

 □

Theorem 4.1 Let f , ψ L 2 ( R + ) . Then the continuous fractional Bessel wavelet transformation B ψ θ is defined on f by
( B ψ θ f ) ( b , a ) = a μ sin θ c μ θ ¯ 0 e i 2 ( 1 a 2 1 ) ( a x ) 2 cot θ x μ 1 2 ( b x csc θ ) 1 2 × J μ ( b x csc θ ) ( h μ θ e i 2 ( ) 2 cot θ f ) ( x ) ( h μ θ ψ ) ¯ ( a x ) d x .

Proof

We have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equan_HTML.gif
by putting ξ a = x , then the continuous fractional Bessel wavelet transformation can be written as
( B ψ θ f ) ( b , a ) = a μ sin θ c μ θ ¯ 0 e i 2 ( 1 a 2 1 ) ( a x ) 2 cot θ x μ 1 2 ( b x csc θ ) 1 2 × J μ ( b x csc θ ) ( h μ θ e i 2 ( ) 2 cot θ f ) ( x ) ( h μ θ ψ ) ¯ ( a x ) d x .
This means that
h μ θ { e i 2 b 2 cot θ ( B ψ θ f ) ( b , a ) } = a μ sin θ ( x μ 1 2 e i 2 a 2 x 2 cot θ ( h μ θ e i 2 ( ) 2 cot θ f ) ( x ) ( h μ θ ψ ) ¯ ( a x ) ) .

 □

Remark 4.1 If f L 2 ( R + ) is a homogeneous function of degree n, then
( B ψ θ f ) ( λ b , λ a ) = λ n + 1 2 ( B ψ θ f ) ( b , a ) .
Theorem 4.2 If ψ 1 and ψ 2 are two wavelets and ( B ψ 1 θ f ) ( b , a ) and ( B ψ 2 θ g ) ( b , a ) denote the continuous fractional Bessel wavelet transformations of f , g L 2 ( R + ) respectively, then
0 0 ( B ψ 1 θ f ) ( b , a ) ( B ψ 2 θ g ) ¯ ( b , a ) d b d a a 2 = sin 2 θ C μ , ψ 1 , ψ 2 , θ f , g ,
where
C μ , ψ 1 , ψ 2 , θ = 0 a 2 μ 2 ( h μ θ ψ 1 ) ¯ ( a ) ( h μ θ ψ 2 ) ( a ) d a < .

Proof

We have
( B ψ 1 θ f ) ( b , a ) = a μ sin θ c μ θ ¯ 0 e i 2 ( 1 a 2 1 ) ( a x ) 2 cot θ x μ 1 2 ( b x csc θ ) 1 2 × J μ ( b x csc θ ) ( h μ θ e i 2 ( ) 2 cot θ f ) ( x ) ( h μ θ ψ 1 ) ¯ ( a x ) d x .
Now,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equau_HTML.gif

 □

Theorem 4.3 If ψ is a wavelet and ( B ψ θ f ) ( b , a ) and ( B ψ θ g ) ( b , a ) are the continuous fractional Bessel wavelet transformations of f , g L 2 ( R + ) respectively, then
0 0 ( B ψ θ f ) ( b , a ) ( B ψ θ g ) ¯ ( b , a ) d b d a a 2 = sin 2 θ C μ , ψ , θ f , g .

Proof The proof of Theorem 4.3 can be easily deduced by setting ψ 1 = ψ 2 = ψ in Theorem 4.2. □

Remark 4.2 If f = g and ψ 1 = ψ 2 = ψ , then from Theorem 4.3, we have
0 0 | ( B ψ θ f ) ( b , a ) | 2 d b d a a 2 = sin 2 θ C μ , ψ , θ f 2 2 .
Theorem 4.4 Let f L 2 ( R + ) . Then f can be reconstructed by the formula
f ( t ) = 1 sin 2 θ C μ , ψ , θ 0 0 ( B ψ θ f ) ( b , a ) ψ b , a θ ( t ) d b d a a 2 , a > 0 .
Proof For any g L 2 ( R + ) , we have
sin 2 θ C μ , ψ , θ f , g = 0 0 ( B ψ θ f ) ( b , a ) ( B ψ θ g ) ¯ ( b , a ) d b d a a 2 = 0 0 ( B ψ θ f ) ( b , a ) ( 0 g ( t ) ψ b , a θ ( t ) ¯ d t ¯ ) d b d a a 2 = 0 [ 0 0 ( B ψ θ f ) ( b , a ) ψ b , a θ ( t ) d b d a a 2 ] g ( t ) ¯ d t = 0 0 ( B ψ θ f ) ( b , a ) ψ b , a θ ( t ) d b d a a 2 , g ( t ) .
Therefore,
f ( t ) = 1 sin 2 θ C μ , ψ , θ 0 0 ( B ψ θ f ) ( b , a ) ψ b , a θ ( t ) d b d a a 2 .

 □

Theorem 4.5 If ψ L 2 ( R + ) , then
0 [ ( B ψ θ f ) ( b , a ) ( B ψ θ g ) ( b , a ) ¯ ] d b = a 2 μ sin 2 θ F , G ,
where
{ F ( x ) : = e i 2 ( ( 2 a 2 ) x 2 ) cot θ x μ 1 2 ( h μ θ e i 2 ( ) 2 cot θ f ) ( x ) ( h μ θ ψ ) ¯ ( a x ) , G ( x ) : = e i 2 ( ( 2 a 2 ) x 2 ) cot θ x μ 1 2 ( h μ θ e i 2 ( ) 2 cot θ g ¯ ) ( x ) ( h μ θ ψ ) ¯ ( a x ) .

Proof

Using Theorem 4.1 and Theorem 4.2, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equbc_HTML.gif

This completes the proof of the theorem. □

Theorem 4.6 If ψ L 2 ( R + ) is a Bessel wavelet and f is a bounded integrable function, then the convolution ( ψ θ f ) is a fractional Bessel wavelet, where
( ψ θ f ) ( x ) = 0 ( τ x θ ψ ) ( y ) y ( μ + 1 / 2 ) f ( y ) d y .

Proof

We have
( ψ θ f ) ( x ) = 0 ( τ x θ ψ ) ( y ) y ( μ + 1 / 2 ) f ( y ) d y .
Therefore,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equbf_HTML.gif
This implies that
( ψ θ f ) ( x ) L 2 1 | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ψ L 2 f L 1 < .
We have ( ψ θ f ) L 2 ( R + ) . Moreover,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equbh_HTML.gif

Thus, the convolution function ( ψ θ f ) is a fractional Bessel wavelet. □

Theorem 4.7 If f , ψ L 2 ( R + ) and ( B ψ θ f ) ( b , a ) is the continuous fractional Bessel wavelet transformation, then
(i) ( B ψ θ f ) ( b , a ) is continuous on R + × R + , (ii) ( B ψ θ f ) ( b , a ) L b ( μ + 1 / 2 ) a ( μ + 1 / 2 ) | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) f L 2 ψ L 2 .
Proof (i) Let ( b 0 , a 0 ) be an arbitrary but fixed point in R + × R + . Then, by the Hölder inequality,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equbj_HTML.gif
Since
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equbk_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-40/MediaObjects/13661_2012_Article_294_Equbl_HTML.gif
by the dominated convergence theorem and the continuity of D μ θ ( b a , x a , z ) in the variable b and a, we have
lim b b 0 lim a a 0 | ( B ψ θ f ) ( b , a ) ( B ψ θ f ) ( b 0 , a 0 ) | = 0 .
This proves that ( B ψ θ f ) ( b , a ) is continuous on R + × R + .
  1. (ii)
    We have
    ( B ψ θ f ) ( b , a ) = 0 f ( x ) ( 1 a e i 2 ( b 2 a 2 + x 2 a 2 ) cot θ 0 ψ ( z ) D μ θ ( b a , x a , z ) d z ) ¯ d x .
     
Therefore, by the Hölder inequality, we have
| ( B ψ θ f ) ( b , a ) | 1 a ( 0 0 x ( μ + 1 / 2 ) | f ( x ) | 2 z ( μ + 1 / 2 ) | D μ θ ( b a , x a , z ) | d x d z ) 1 / 2 × ( 0 0 z ( μ + 1 / 2 ) | ψ ( z ) | 2 x ( μ + 1 / 2 ) | D μ θ ( b a , x a , z ) | d x d z ) 1 / 2 = 1 a ( 0 x ( μ + 1 / 2 ) | f ( x ) | 2 d x 0 | D μ θ ( b a , x a , z ) | z ( μ + 1 / 2 ) d z ) 1 / 2 × ( 0 z ( μ + 1 / 2 ) | ψ ( z ) | 2 d z 0 | D μ θ ( b a , x a , z ) | x ( μ + 1 / 2 ) d x ) 1 / 2 1 a ( b μ + 1 / 2 a 2 ( μ + 1 / 2 ) | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ) 1 / 2 ( 0 | f ( x ) | 2 d x ) 1 / 2 × ( b μ + 1 / 2 a 1 | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) ) 1 / 2 ( 0 | ψ ( z ) | 2 d z ) 1 / 2 = b ( μ + 1 / 2 ) a ( μ + 1 / 2 ) | ( sin θ ) μ + 1 / 2 | 2 μ Γ ( μ + 1 ) f L 2 ψ L 2 .

 □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Authors’ Affiliations

(1)
Department of Applied Mathematics, Indian School of Mines
(2)
Department of Mathematics, NERIST

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