Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36

Zero-Order Hankel Transform Method for Partial Differential Equations

Naol Tufa Negero Faculty of Natural and Computational Science, Department of , Wollega University, Nekemte,Ethiopia.

[email protected]

Abstract This paper deals with the solution of Partial Differential Equations (PDEs) by the method of Hankel transform. Firstly, Hankel transforms is derived using two dimensional , next its properties are presented because their results are used widely in solving partial differential equations in the polar and axisymmetric cylindrical configurations, finally it applied to partial differential equation with precise formulation of initial and boundary value problems. In this article, we use the Hankel transforms method, namely Hankel transforms of zero order and of order one are introduced, such that PDE reduced to an ODE, which can be subsequently solved using ODE techniques which involves the Bessel functions. Moreover the main result of this paper is the zero-order Hankel transform method for Partial differential equations in unbounded problems with radial symmetry that is appropriate in the polar and cylindrical coordinates, which is analyzed in some detail. Keywords: Radial Fourier transforms, Two Dimensional Fourier Transform, Polar, Cylindrical and spherical Coordinates, Zero-Order Hankel Transform, Partial Differential Equations

1. INTRODUCTION: There are several integral transforms which are frequently used as a tool for solving numerous scientific problems. In many real applications, Fourier transforms as well as Mellin transforms and Hankel transforms are all very useful. It is well known that the Fourier transform method is used to solve partial differential equations [12-21]. The transform is applied to PDEs on infinite and semi-infinite spatial domains [1]. Fourier transforms have been used to remove Cartesian coordinates from initial boundary value problems on infinite intervals; Fourier sine and cosine transforms are applicable to Cartesian coordinates on semi-infinite intervals[17]. To quote [23], “In two dimensions phenomena are richer than in one dimension.” True enough, working in two dimensions offers many new and rich possibilities. Contemporary applications of the Fourier transform are just as likely to come from problems in two, three, and even higher dimensions as they are in one Fourier transforms may be extend to functions of two or more variables[14]. We consider the case for functions of two variables primarily to understand an origin for a Hankel transform [5, 14]. Here, we show that radially sampled data can be processed directly using Fourier transforms in polar coordinates[6,19].In many practical problems, data are often acquired in such a form that is desirable to perform a two dimensional polar Fourier transform that is a Hankel transform rather than the Cartesian forms. So, we transform the Cartesian coordinates into the polar coordinates. The Hankel transform arises naturally in the discussion of problems posed in cylindrical coordinates [4, 19]. This transform is more appropriate in solving differential equations with boundary conditions in which there is an axial symmetry [7, 19]. The Hankel transform is frequently used as a tool for solving numerous scientific problems. It is widely used in several fields like, elasticity, optics, fluid mechanics, seismology, astronomy and image processing[4,23].The Hankel transform

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 24 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 becomes very useful in analysis of wave fields where it is used in mathematical handling of radiation, diffraction, and field projection. As written in Eq. (9), the transform is evaluated first with respect to 푟, and second with respect to 휃 [20]. We shall refer to this form of the polar Fourier Transform as a radial transform. This equation fits naturally with the interpretation of the 2-D Fourier Transform kernel as a superposition of 1-D cosine and sine waves in all directions [21]; here, each spoke of 푓 (푟, 휃) at fixed 휃 contributes the weighting coefficients for the cosine and sine waves in direction 휃 used in the synthesis of 퐹. 2. BASIC IDEA OF HANKEL TRANSFORM METHOD: 2.1 The Radial Fourier Transform For use in many applications, we’re going to consider one further aspects of the 2-dimensional case. A function on 푅2 is radial(also called radially symmetric or circularly symmetric) if it depends only on the distance from the origin [2]. In polar coordinates the distance from the origin is denoted by 푟, so to say that a function is radial is to say that it depends only on 푟 (and that it does not depend on 휃, writing the usual polar coordinates as (푟, 휃)).The definition of the Fourier transform is set up in Cartesian coordinates, and it’s clear that we’ll be better off writing it in polar coordinates if we work with radial functions. This is actually not so straightforward, or, at least, it involves introducing some special functions to write the formulas in a compact way. We have to convert

∞ ∞ 푒−2휋푖퐱.휿 푓 퐱 푑퐱 = 푒−2휋푖 푥푘1+푦푘2 푓 푥, 푦 푑푥푑푦 푅2 −∞ −∞ to polar coordinates. There are several steps: To say that 푓(퐱) is a radial function means that it becomes 푓(푟). To describe all of 푅2 in the limits of integration, we take 푟 going from 0 to ∞ and 휃 going from 0 to 2휋. The area element 푑푥푑푦 becomes 푟푑푟푑휃. Finally, the problem is the inner product

퐱. 휿 = 푥푘1 + 푦푘2 in the exponential and how to write it in polar coordinates.

If we identify 푥, 푦 = (푟, 휃) (varying over the 푥, 푦 − plane) and put 푘1, 푘2 = 휌, 휙 (fixed in the integral) then 퐱. 휿 = 퐱. 휿 푐표푠 휃 − 휙 = 푟휌푐표푠 휃 − 휙 .

∞ ∞ −2휋푖.휿 2휋 ∞ −2휋푖푟휌푐표푠 휃 − 휙 The Fourier transform of 푓 is thus −∞ −∞ 푒 푓 퐱 푑퐱 = 0 0 푓 r 푒 푟푑푟푑휃 [9].

There’s more to be done. First of all, because 푒−2휋푖푟휌푐표푠 휃 − 휙 , is periodic (in θ) of period 2휋,

2휋 −2휋푖푟휌푐표푠 휃 − 휙 the integral 0 푒 푑휃 does not depend on 휙. Consequently, 2휋 −2휋푖푟휌푐표푠 휃 − 휙 2휋 −2휋푖푟휌푐표푠휃 0 푒 푑휃 = 0 푒 푑휃. The next step is to define ourselves out of trouble. We 1 2휋 introduce the function 퐽 푎 = 푒−푖푎휌푐표푠휃 푑휃. We give this integral a name, 퐽 푎 , because, try 0 2휋 0 0 as you might, there is no simple closed form expression for it, so we take the integral as defining a new function. It is called the zero order of the first kind. Sorry, but Bessel functions, of whatever order and kind, always seem to come up in problems involving circular symmetry. 2휋 −2휋푖푟휌푐표푠휃 Incorporating 퐽0 into what we’ve done, 0 푒 푑휃 = 2휋퐽0 2휋푟휌 and the Fourier transform of ∞ 푓(푟) is 2휋 0 푓(푟)퐽0 2휋푟휌 푟푑푟. ∞ Generally, If 푓(퐱) is a radial function then its Fourier transform is 퐹 휌 = 2휋 0 푓(푟)퐽0 2휋푟휌 푟푑푟. In words, the important conclusion to take away from this is that the Fourier transform of a radial function is also radial. The formula for 퐹(휌) in terms of 푓(푟) is sometimes called the zero order Hankel transform of 푓(푟) but, again, we understand that it is nothing other than the Fourier transform of a radial function.

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 25 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 Definition (Separability in Polar Coordinates) A function ), 푓(푟, 휃) is separable in polar coordinates if it can be written in the form 푓 푟, 휃 = 푓푟 푟 푓휃 휃 .Suppose such an f is circularly symmetric, with 푓휃 휃 = 1. As a specific example, we will consider a 2D circular (or cylinder) function:

1, 푟 < 1 1 푐푖푟푐 푟 = , 푟 = 1 2 0, 푟 > 1 ∞ The Fourier transform of a circularly symmetric function is 퐹(휌, 휙) = 2휋 0 푟퐽0 2휋휌푟 푓푟 푟 푑푟 This is also known as the Hankel transform of order zero and as the Fourier-Bessel transform. 1 2휋 The function 퐽 is the zero order Bessel function of the first kind defined as 퐽 푎 = 푒푖푎 휃−휙 푑휃 0 0 2휋 0 It oscillates like a damped cosine. Continuing with our specific example, the Fourier transform of 푐푖푟푐 푟 ∞ 1 is Ӻ 푐푖푟푐 푟 = 2휋 0 푟푐푖푟푐 푟 퐽0 2휋휌푟 푑푟 = 2휋 0 푟퐽0 2휋휌푟 푑푟. 1 2휋휌 퐽 2휋휌 Substitute 푟′ = 2휋휌푟, and 푑푟′ = 2휋휌푑푟 to find: Ӻ 푐푖푟푐 푟 = 푟′퐽 푟′ 푑푟′ = 1 , since 2휋 0 0 휌 훼 0 푥퐽0 푥 푑푥 = 훼퐽1 훼 where 퐽1 is the first order Bessel function of the first kind, similar to a damped sinusoid. The function 푠표푚푏 휌 or 푠표푚푏푟푒푟표 (also known as Mexican hat, Bessinc, and jinc) is defined as 푠표푚푏 휌 = 2퐽 2휋휌 1 . Thus, the Fourier transform of 푐푖푟푐 푟 is proportional to a 푠표푚푏푟푒푟표 function of 휌, the radial 2휋휌 coordinate in frequency space: Ӻ 푐푖푟푐 푟 = 휋푠표푚푏 휌 .The first result is that the radial Fourier transform is given by a Hankel transform. Suppose 푓 is a function on 푅푛. Its Fourier transform is −푖휿.퐱 푛 푓 휿 = 푒 푓(퐱)푑 퐱. Let 푟 = 퐱 and 푠 = 휿 . Write 푓 퐱 = 퐹(푟) and 푓 휿 = 퐹푛 푠 . Theorem: The radial Fourier transform in 푛 dimensions is given in terms of 푛−2 푛 ∞ 푛−2 2 2 푛 −2 2 the Hankel transform by 푠 퐹푛 푠 = 2휋 0 퐽 푠푟 푟 퐹(푟)푟푑푟 2 Here is the proof of the theorem. Introduce polar coordinates with the 푧 axis along 휿, so that 휿 · 퐱 = 푠푟 푐표푠(휃). Suppose that the function is radial, that is, 푓(푥) = 퐹(푟).

∞ 휋 −푖푠푟푐표푠 휃 푛−2 푛−1 푓 휿 = 퐹푛 푠 = 0 0 푒 퐹(푟)휔푛−2푠푖푛(휃) 푑휃푟 푑푟.

푛−2 푡 2 휋 −푖푡푐표푠(휃) 푛−2 푛 −2 Use 퐽 푡 = 푛 휔푛−2 0 푒 푠푖푛 휃 푑휃. For the case 푛 = 3 the Bessel function has order 2 2휋 2 ∞ sin 푠푟 1/2 and has the above expression in terms of elementary functions. So 퐹 푠 = 4휋 퐹(푟)푟2푑푟. 3 0 푠푟 ∞ ∞ For 푛 = 2 the Bessel function has order 0. We get 퐹2 푠 = 2휋 0 0 퐽0 푠푟 퐹(푟)푟푑푟.

We introduce the definition of the Hankel transform from the two dimensional Fourier transform and its inverse given by

1 ∞ ∞ Ӻ 푓 푥, 푦 = 퐹 푢, 푣 = 푒푥푝 −푖 휿. 풓 푓 푥, 푦 푑푥푑푦, (2.1) 2휋 −∞ −∞

1 ∞ ∞ Ӻ−1 퐹 푢, 푣 = 푓 푥, 푦 = 푒푥푝 −푖 휿. 풓 퐹 푢, 푣 푑푢푑푣, (2.2) 2휋 −∞ −∞

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 26 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 where 풓 = (푥, 푦) and 휿 = 푢, 푣 .Introducing polar coordinates (푥, 푦) = 푟(푐표푠 휃, 푠푖푛 휃) and 푢, 푣 = 휅 (푐표푠 휙, 푠푖푛 휙), we find 휿 · 풓 = 휅푟 푐표푠 (휃 − 휙) and then 1 ∞ 2휋 퐹 (휅, 휙) = 푟푑푟 푒푥푝 −푖휅푟푐표푠 휃 − 휙 푓 푟, 휃 푑휃, (2.3) 2휋 0 0 We next assume 푓(푟, 휃) = 푒푥푝 (푖푛휃) 푓(푟), which is not a very severe restriction, and make a π change of variable 휃 – 휙 = α − to reduce (2.3) to the form 2 1 ∞ 2휋+휙 π 퐹 (휅, 휙) = 푟푓(푟)푑푟 × 0 푒푥푝 푖푛푐표푠 휙 − + 푖 푛α − 휅푟푠푖푛α 푑α (2.4) 2휋 0 휙0 2 π where 휙 = − 휙. We use the integral representation of the Bessel function of order 푛 0 2 1 2휋+휙0 퐽푛 휅푟 = 푒푥푝 푖 푛α − 휅푟푠푖푛α 푑α (2.5) 2휋 휙0 so that integral (2.4) becomes π ∞ 퐹(휅, 휙) = 푒푥푝 푖푛 휙 − 푟퐽 휅푟 푓(푟)푑푟 2 0 푛 π = 푒푥푝 푖푛 휙 − 푓 휅 , (2.6) 2 푛 where 푓 푛 휅 is called the Hankel transform of 푓(푟) and is defined formally by ∞ 퐻푛 푓 (푟) = 푓푛 휅 = 0 푟퐽푛 휅푟 푓(푟)푑푟 (2.7) Similarly, in terms of the polar variables with the assumption 푓(푥, 푦) = 푓 (푟, 휃) = 푒푖푛휃 푓(푟) and with result (2.6), the inverse Fourier transform (2.2) becomes 1 ∞ 2휋 푒푖푛휃 푓 푟 = 휅푑휅 푒푥푝 푖휅푟푐표푠 휃 − 휙 퐹 휅, 휙 푑휙 2휋 0 0 1 ∞ 2휋 π = 휅푓 푛 휅 푑휅 푒푥푝 푖푛 휙 − + 푖휅푐표푠 휃 − 휙 푑휙 2휋 0 0 2 π π 휃 – 휙 = − α + and 휃 = − θ + , which is, by the change of variables 2 0 2 1 ∞ 2휋+휃0 = 휅푓 휅 푑휅 푒푥푝 푖푛 θ + α − 푖휅푟푠푖푛α 푑α 2휋 푛 0 휃0 푖푛휃 ∞ = 푒 0 휅퐽푛 휅푟 푓푛 휅 푑휅, by (2.5). (2.8) Thus, the inverse Hankel transform is defined by

−1 ∞ 퐻 푓푛 휅 = 푓 푟 = 0 휅퐽푛 휅푟 푓푛 휅 푑휅, (2.9)

Integrals (2.7) and (2.9) exist for certain large classes of functions, which usually occur in physical applications. 2.2 Hankel Transforms of Order Zero (풏 = ퟎ) In particular, the Hankel transforms of zero order (푛 = 0) and order (푛 = 1) are often useful for the solution of problems involving problems in an axisymmetric cylindrical geometry[10].The equation ∞ 1 푒−푘푧 퐽 푘푟 푑푘 = , 푧, 푟 > 0, (2.10) 0 0 푧2+푟 2 Shows that the zero-order Hankel transforms of 푓 푟, 푧 = 1/ 푧2 + 푟2 is given as −푘푧 퐻0 푘, 푧 = 1/푘 푒 . Using the transform formula (2.8), we obtain

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 27 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 ∞ 푟퐽 푘푟 1 0 푑푘 = 푒−푘푧 (2.11) 0 푧2+푟 2 푘

The result (2.9) can be verified by using the integral representation of the Bessel function 퐽0 푘푟 and interchanging the order of integration. A second result that can be obtained by using the series expansion of the Bessel function 퐽푛 푘푟 and integrating term by term is

∞ 2 1 푟 푛 푟2 푘퐽 푘푟 퐽 푘푠 푒−푘 푡푘푛 푑푘 = 푒푥푝 − . (2.12) 0 0 0 2푡 2푡 4푡 Example1. Obtain the zero-order Hankel transforms of 훿 푟 (a) 푟−1푒푥푝 −푎푟 (b) , (c) 퐻 푎 − 푟 , 푟 where 퐻 (푟) is the Heaviside unit step function. 1 ∞ 1 (a) 푓 휅 = 퐻 푒푥푝 −푎푟 = 푒푥푝 −푎푟 퐽 휅푟 푑푟 = 0 0 푟 0 0 휅 2+푎2 훿 푟 ∞ (b) 푓 휅 = 퐻 = 훿 푟 퐽 휅푟 푑푟 = 1 0 0 푟 0 0 ∞ 1 푎휅 푎 (c) 푓 휅 = 퐻 퐻 푎 − 푟 = 푟퐽 휅푟 푑푟 = 푝퐽 푝 푑푝 = 퐽 푎휅 . 0 0 0 0 휅 2 0 0 휅 1 Example 2.Find the 푛푡푕 -order Hankel transforms of (a) 푓 푟 = 푟푛 퐻 푎 − 푟 , (b) 푓 푟 = 푟푛 푒푥푝 −푎푟2 . ∞ −푎푟 −푎푟 휅 (b) 푓 휅 = 퐻1 푒 = 0 푟푒 퐽1 휅푟 푑푟 = 3 휅 2+푎 2 2

∞ 푎푛+1 (a) 푓 휅 = 퐻 푟푛퐻 푎 − 푟 = 푟푛+1퐽 휅푟 푑푟 = 퐽 푎휅 . 푛 0 푛 휅 푛+1 ∞ 휅 푛+1 푟2 (b) 푓 휅 = 퐻 푟푛푒푥푝 −푎푟2 = 푟푛+1퐽 푒푥푝 −푎푟2 푑푟 = 푒푥푝 − . 푛 0 푛 2푎 푛+1 4푎 A treatment of Hankel transform of order 푛 similar to Hankel transform of order zero is given in

[22].In order to associate a transform with 퐽푉 푟 , we must be aware of the behavior of Bessel functions for large 푟. It is shown in the theory of asymptotics that 퐽푉 푟 may be approximated for large 푟 by

2 휋 푣휋 퐽 푟 ≈ 푐표푠 푟 − − (2.13) 푣 휋푟 4 2 the approximation being better the larger the value of 푟. This means that for larger 푟, 퐽푉 푟 is oscillatory with an amplitude that decays at the same rate as 1/ 푟. Corresponding to interpretation we have the following Hankel integral formula. Theorem: I f 푓 푟 is absolutely integrable on 0 푟 ∞ and 푓 푟 is piecewise smooth 푟 ( ) < < , ( ) ∞ 푓 + +푓 − on every finite interval, then for 0 < 푟 < ∞, 푟 푟 = 휆퐴 휆 퐽푣 휆푟 푑휆 where 2 0 ∞ 퐴 휆 = 0 푟푓(푟) 퐽푣 휆푟 푑푟. In view of the asymptotic behavior of 퐽푣 휆푟 in expression (2.13), it is clear that absolute integrability of 푟 푓 (푟) guarantees convergence of the integral for 퐴(휆). Associated with this integral formula is the Hankel transform 푓 푣 휆 of a function푓(푟),

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 28 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 ∞ 푓푣 휆 = 0 푟푓(푟) 퐽푣 휆푟 푑푟 (2.14) ∞ (2.15) 푓 푟 = 0 휆푓푣 휆 퐽푣 휆푟 푑휆 where it is understood in (2.15) that 푓(푟) is defined as the average of left- and right-limits at points of discontinuity. We place a subscript 휈 on ourselves that the Hankel transform is dependent on the choice of 휈 in differential equation; changing 휈 changes the transform.

푣 Example: Find the Hankel transform of 푓 푟 = 푟 , 0 < 푟 < 푎 0, 푟 > 푎 푎 푣+1 Solution: By definition (2.14), 푓푣 휆 = 0 푟 퐽푣 휆푟 푑푟. 휆푎 푢 푣+1 푑푢 1 휆푎 If we set 푢 = 휆푟, then 푓 휆 = 퐽 푢 = 푢푣+1 퐽 푢 푑푢 푣 0 휆 푣 휆 휆푣+2 0 푣 1 휆푎 푑 1 = 푢푣+1퐽 푢 푑푢 = 푎푣+1퐽 휆푎 . 휆푣+2 0 푑푢 푣+1 휆 푣+1 The inverse Hankel transform then gives 푟푣, 0 < 푟 < 푎 ∞ 1 푣+1 푎푣 휆 푎 퐽푣+1 휆푎 퐽푣 휆푟 푑휆 = , 푟 = 푎 0 휆 2 0, 푟 > 푎, and from this we obtain the following useful integration formula 1 푣 ∞ 푟 , 0 < 푟 < 푎 푎 퐽푣+1 휆푎 퐽푣 휆푟 푑휆 = 푎 0 1/2푎, 푟 = 푎 0, 푟 > 푎, 2.3 Properties of Hankel Transforms Because of their increased generality over the Laplace and Fourier transforms, Hankel transforms do not have so many elementary properties as former [13]. We state the following properties of the Hankel transforms: (i) The Hankel transform operator, Hn is a linear integral operator, that is,

퐻푛 푎푓 푟 + 푏푔 푟 = 푎퐻푛 푓 푟 + 푏퐻푛 푔 푟 for any constants a and b. (ii) The Hankel transform satisfies the Parseval relation ∞ ∞ 0 푓 푟 푔 푟 푑푟 = 0 휅푓 휅 푔 휅 푑휅 (2.16) where 푓 휅 and 푔 휅 are Hankel transforms of 푓 (푟) and 푔 (푟) respectively. 푘 (iii) 퐻 푓′ 푟 = 푛 − 1 푓 푘 − 푛 + 1 푓 푘 푛 2푛 푛+1 푛−1 provided 푟푓 (푟) vanishes as 푟 → 0 and as 푟 → ∞. 1 푑 푑푓 푛2 (iv) 퐻 푟 − 푓 (푟) = −푘2푓 푘 (2.17) 푛 푟 푑푟 푑푟 푟2 푛 푑푓 provided both 푟 and 푟푓 (푟) vanishes as 푟 → 0 and as 푟 → ∞. 푑푟 (v) (Scaling). If 퐻푛 푓 푟 = 푓 푛 푘 , then 1 푘 퐻 푓 푎푟 = 푓 , 푎 > 0 (2.18) 푛 푎2 푛 푎 These results are used widely in solving partial differential equations in the axisymmetric cylindrical configurations. We illustrate this point by considering the following examples of applications.

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 29 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 3. RESULTS AND DISCUSSION : Example1: Obtain the solution of the free vibration of a large circular membrane governed by the initial-value problem 휕2푢 1 휕푢 1 휕2푢 + = , 0 < 푟 < ∞, 푡 > 0, 휕푟2 푟 휕푟 푐2 휕푡 2 푢 푟, 0 = 푓 푟 , 푢푡 푟, 0 = 푔 푟 , 0 ≤ 푟 < ∞, where 푐2 = (푇 /휌) = constant, 푇 is the tension in the membrane, and 휌 is the surface density of the membrane. Application of the Hankel transform of order zero ∞ 푢 푘, 푡 = 푟푢 푟, 푡 퐽0 푘푟 푑푟 0 푑2푢 to the vibration problem gives + 푘2푐2푢 = 0 푑푡 2

푢 푘, 0 = 푓 푘 , 푢 푡 푘, 0 = 푔 푘 . 푔 푘 The general solution of this transformed system is 푢 푘, 푡 = 푓 푘 푐표푠 푐푘푡 + 푠푖푛 푐푘푡 . 푐푘 The inverse Hankel transformation gives ∞ 1 ∞ 푢 푘, 푡 = 푘푓 푘 푐표푠 푐푘푡 퐽 푘푟 푑푘 + 푔 푘 푠푖푛 푐푘푡 퐽 푘푟 푑푟 (3.1) 0 0 푐 0 0 This is the desired solution. In particular, we consider the following initial conditions 퐴 푢 푟, 0 = 푓 푟 = 1 , 푢푡 푟, 0 = 푔 푡 = 0 푟2 1 + 2 푎2 ∞ 푟퐽 푘푟 푑푟 퐴푎 So that 푔 푘 = 0 and 푓 푘 = 퐴푎 0 = 푒−푎푘 by means of Example 1.Thus, solution (3.1) 0 1 푘 푎2+ 푟2 2 ∞ −푎푘 becomes 푢 푟, 푡 = 퐴푎 0 푒 퐽0 푘푟 푐표푠 푐푘푡 푑푘 1 ∞ − −푘(푎+푐푡) 2 2 2 = 퐴푎 푅푒 0 푒 퐽0 푘푟 푑푘 = 퐴푎 푅푒 푟 + 푎 + 푖푐푡 Example2: (Oscillation of an infinitely long hanging chain) The oscillation of a very long heavy chain can be modeled by assuming the chain is semi-infinite, with the fastened end sent to infinity. We will further assume that the transverse vibrations take place in one vertical plane. We place the chain on the 푥- axis, which we assume directed vertically and pointing upward. The boundary value problem governing the free motion of the chain becomes 휕2푢 휕2푢 휕푢 = 푔 푥 + , 0 < 푟 < ∞, 푡 > 0, 휕푡 2 휕푥2 휕푥 푢 푥, 0 = 푓 푥 , 푢푡 푥, 0 = 푣 푥 , 푥 > 0, Here 푔 is the gravitational acceleration, 푓(푥) is the initial displacement of the chain, and 푣(푥) is its initial velocity. To proceed with the solution, we make the change of variables 푧2 = 푥, 2푧푑푧 = 푑푥, and 휕2푢 푔 휕2푢 1 휕푢 transform the equations into = + , 휕푡 2 4 휕푧2 푧 휕푧 2 2 2 2 푢 푧 , 0 = 푓 푧 , 푢푡 푧 , 0 = 푣 푧 ., Let 푈 푠, 푡 denote then Hankel transform of order 0 of 푢 푧2, 푡 with respect to the variable t. Transforming the new set of equations and using (2.17) we get 푔 푢 푠, 푡 = −푠2 푈 푠, 푡 , 푡푡 4 2 2 푈 푠, 0 = 퐻0 푓 푧 푠 , 푈푡 푠, 0 = 퐻0 푣 푧 푠 . Solving the differential equation in 푡 and using the transformed initial conditions to determine the IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 30 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 푔 푔 arbitrary constants, we find 푈 푠, 푡 = 퐴 푠 푐표푠 푠푡 + 퐵 푠 푠푖푛 푠푡 , with 퐴 푠 = 퐻 푓 푧2 푠 , 2 2 0 2 2 and 퐵 푠 = 퐻0 푣 푧 푠 .The solution is now obtained by taking the inverse Hankel transform of 푈: 푔푠 ∞ 푔 푔 푈 푧2, 푡 = 퐴 푠 푐표푠 푠푡 + 퐵 푠 푠푖푛 푠푡 퐽 푧푠 푠푑푠 . 0 2 2 0 ∞ 푔 푔 Hence, the term 푥, we have 푢 푥, 푡 = 퐴 푠 푐표푠 푠푡 + 퐵 푠 푠푖푛 푠푡 퐽 푥푠 푠푑푠 0 2 2 0 The numerical solution for this problem is presented as[16], first when we describe the physical phenomenon corresponding to the data 푔 = 9.8 and 푓 푥 = 0, the chain starts to move from rest with an initial velocity of 푣 푥 = 푥.Second, to write the solution in the form of integral we have 퐴~0 and 2 1 2 2 ∞ 푔 1 퐵 푠 = 퐻0 푠 = . So 푢 푥, 푡 = 푠푖푛 푠푡 퐽0 푥푠 푑푠. 푔푠 푧 푔푠2 푔 0 2 푠 Example 3. Obtain the steady-state solution of the axisymmetric acoustic radiation problem governed by the wave equation in cylindrical polar coordinates (푟, 휃, 푧) 2 2 푐 ∇ 푢 = 푢푡푡 ,, 0 < 푟 < ∞, 푧 > 0,푡 > 0 푢푧 = 푓 (푟, 푡) on 푧 = 0, where 푓 (푟, 푡) is a given function and 푐 is a constant. We also assume that the solution is bounded and behaves as outgoing spherical waves. This is referred to as the Sommerfeld radiation condition. We seek a solution of the acoustic radiation potential 푢 = 푒푖휔푡 휙 (푟, 푧) so that 휙 satisfies the Helmholtz equation 1 휔 2 휙 + 휙 + 휙 + 휙 = 0, 0 < 푟 < ∞, 푧 > 0, 푟푟 푟 푟 푧푧 푐 2 with the boundary condition representing the normal velocity prescribed on the 푧 = 0 plane

휙푧 = 푓 (푟) on 푧 = 0, where 푓 (푟) is a known function of 푟.We solve the problem by means ∞ of the zero-order Hankel transformation 휙 푘, 푧 = 0 푟휙 푟, 푧 퐽0 푘푟 푑푟 so that the given differential system becomes 1 휔 2 2 휙 = 푘2휙 , 푧 > 0, 휙 = 푓 푘 on 푧 = 0 where 푘 = 푘2 − . 푧푧 푧 푐2 The solution of this system is 휙 푘, 푧 = −푘−1푓 푘 푒−푘푧 , where 휅 is real and positive for 휔 휔 푘 > , and purely imaginary for 푘 < .The inverse transformation yields the solution 푐 푐 ∞ −1 −푘푧 휙 푟, 푧 = − 0 푘 푓 푘 푘퐽0 푘푟 푒 푑푘 . Since the exact evaluation of this integral is difficult, we choose a simple form of 푓 (푟) as 푓 (푟) = 퐴퐻 (푎 − 푟) , where 퐴 is a constant and 퐻 (푥) is the 푎 푎 Heaviside unit step function so that 푓 푘 = 푘 퐽 푎푘 푑푘 = 퐽 푎푘 . Then the solution for this 0 0 푘 1 ∞ −1 −푘푧 special case is given by 휙 푟, 푧 = −퐴푎 0 푘 퐽1 푎푘 퐽0 푘푟 푒 푑푘. For an asymptotic evaluation of this integral, we express it in terms of the spherical polar coordinates (푅, 휃, 휙), (푥 = 푅 푠푖푛 휃 푐표푠 휙, 푦 = 푅 푠푖푛 휃 푠푖푛 휙, 푧 = 푅 푐표푠 휃), combined with the 1 2 휋 asymptotic result 퐽 푘푟 ~ 2 푐표푠 푘푟 − as 푟 → ∞ so that the acoustic potential 푢 = 0 휋푘푟 4 퐴푎 2푒푖휔푡 ∞ 휋 푒푖휔푡 휙 is 푢~ − 푘−1퐽 푎푘 푐표푠 푘푅푠푖푛휃 − 푒−푘푧 푑푘, where 푧 = 푅 푐표푠 휃.This integral 휋푅푠푖푛 휃 0 1 4 can be evaluated asymptotically for 푅 → ∞ by using the stationary phase approximation formula 휔푅 퐴푎푐 푖 휔푡 − to obtain 푢~ − 퐽 푎푘 푒 푐 ,where 푘 = 휔/푐 푠푖푛 휃 is the stationary point. This 휔푅푠푖푛 휃 1 1 1 solution represents the outgoing spherical waves with constant velocity 푐 and decaying amplitude IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 31 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 as 푅 → ∞. Example 4: (Laplace Equation on an Axial Source) We consider Laplace’s equation in three dimensions with a source concentrated on the 푧-axis. Introducing cylindrical coordinates (푟, 휃, 푧) noting the axial symmetry, we ask for a 휃 independent solution 푢(푟, 푧) of Laplace’s equation, 1 ∇2푢 푟, 푧 = 푢 푟, 푧 + 푢 푟, 푧 + 푢 푟, 푧 = 0, 푟 > 0, −∞ < 푥 < ∞, (3.2) 푟푟 푟 푟 푧푧 With the conditions on the 푧-axis (i.e., 푟 = 0) given as 2 휕푢 푟,푧 lim푟→0 푟 푢 푟, 푧 = 0, lim2휋푟 = −푓 푧 , −∞ < 푥 < ∞, (3.3) 푟→0 휕푟 So that 푓 푧 is a measure of the strength of the source. Applying the zero- order Hankel transform to (3.2), we multiply across by 푟퐽0 푘푟 and integrate from 0 to ∞. [The zero-order transform is used because the term 푛2/푟2푢 푟, 푧 does not occur in equation (3.2).]Defining the zero-order Hankel transform 푈0(푘, 푧) of 푢 푟, 푧 as ∞ 푈0 푘, 푧 = 0 푟푢 푟, 푧 퐽0 푘푟 푑푟 (3.4) ∞ 1 0 = 푟 푢푟푟 푟, 푧 + 푢푟 푟, 푧 + 푢푧푧 푟, 푧 퐽0 푘푟 푑푟 0 푟 휕2푈 푘,푧 푟=∞ = −푘2푈 푘, 푧 + 0 + 푟푢 푟, 푧 퐽 푘푟 − 푘푟푢 푟, 푧 퐽′ 푘푟 (3.5) 0 휕푧2 푟 0 0 푟=0 푘푟 Now 퐽 0 = 1퐽′ 푘푟 = −퐽 푘푟 and 퐽 푘푟 ≈ as 푟 → 0. Assuming that the contribution from the 0 0 1 1 2 푘푟2 limit at infinite vanish in (3.5) and noting that 푟푢퐽′ 푘푟 ≈ − 푢 as 푟 → 0, we obtain on using (3.5), 0 2 휕2푈 푘,푧 1 0 − 푘2푈 푘, 푧 = − 푓 푧 , −∞ < 푥 < ∞. 휕푧2 0 2휋 Requiring that 푈0 푘, 푧 → 0 as 푧 → ∞, we apply Fourier transform method to this differential equation 1 ∞ and the solution is 푈 푘, 푧 = 푒−푘 푧−푠 푓 푠 푑푠, 0 4휋푘 −∞ Inverting the transform 푈0 푘, 푧 gives 1 ∞ ∞ −푘 푧−푠 1 ∞ 푓 푠 푑푠 푢 푟, 푧 = 푒 퐽0 푘푟 푓 푠 푑푘푑푠 = , (3.6) 4휋 −∞ 0 4휋 −∞ 푟2+ 푧−푠 2 1 1 On interchanging the order of integration and using (2.10).The term is the source function or 4휋 푟2+ 푧−푠 2 free –space Green’s function for Laplace’s Equation, with the source point at 푥, 푦, 푧 = 0,0, 푠 . Thus, the solution (3.6) represents a superposition of point source functions over the 푧-axis with density 푓 푧 . Example 5:(The Diffusion Equation with Axially Symmetric Data) We consider the three dimensional diffusion equation in cylindrical coordinates where the concentration 푢 푥, 푦, 푧, 푡 has an initial distribution that depends only on 푟 = 푥2 + 푦2. We introduce cylindrical coordinates (푟, 휃, 푧) and looks for a solution 푢 푟, 푡 of the initial value problem 1 푢 푟, 푡 = 퐷 푢 푟, 푡 + 푢 푟, 푧 , 푡 > 0, 푢 푟, 0 = 푓 푟 , 푟 > 0 (3.7) 푡 푟푟 푟 푟 Where 퐷 > 0 is the diffusion constant, Applying the zero-order Hankel transform to (3.7), we obtain, 휕2푈 푘,푡 0 − 퐷푘2푈 푘, 푧 = 0, 푡 > 0, 푈 푘, 0 = 퐹 (푘). 휕푡2 0 0 0 −퐷푘2푡 Thus, 푈0 푘, 푡 = 퐹0(푘)푒 1 ∞ ∞ 2 And 푢 푟, 푡 = 푘푠퐽 푘푟 퐽 푘푠 푒−퐷푘 푡푓 푠 푑푘푑푠 , (3.9) 4휋 0 0 0 0

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 32 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 on interchanging the order of integration in the last integral. We cite a known result to evaluate the inner integral. That is, ∞ 2 1 푟2+푠2 푟푠 푘퐽 푘푟 퐽 푘푠 푒−퐷푘 푡푑푘 = 푒푥푝 − 퐼 , (3.10) 0 0 0 2퐷푡 4퐷푡 0 2퐷푡 where 퐼0 푧 is the modified Zero-order Bessel function. We note the property of 퐼0 푧 that 퐼0 0 = 1, so that for 푠 = 0 in (3.10) [Since 퐽0 0 = 1]. We find that (3.10) reduces to [2.12] with 푛 = 0 and 퐷 = 0. Inserting (3.10) into (3.9) gives 1 ∞ 푟2+푠2 푟푠 푢 푟, 푡 = 푒푥푝 − 퐼 푓 푠 푠푑푠 , 2퐷푡 0 4퐷푡 0 2퐷푡 It is of interest to show this solution reduces to the fundamental solution for the two-dimensional heat or diffusion equation if we let 푓 푟 represent a concentrated source at 푟 = 0. We require that with 푓 푟 ≥ 0, 휖 lim휖→0 2휋 0 푓 푟 푟푑푟 = 1, (3.11) Where we assume that the source is concentrated in circle of radius 휖 and let the radius tends to zero but keep the source strength fixed at unity. It may be assumed that 푓 푟 vanishes for 푟 > 휖. Using the generalized mean value theorem for integrals, we have [since 푓 푟 ≥ 0]. 1 푟2+푠 2 푟푠 휖 푢 푟, 푡 = 푒푥푝 − 퐼 푓 푠 푠푑푠 , where 푠 = 푠 (휖) → 0 as 휖 → 0.Then as 휖 → 0, we 2퐷푡 4퐷푡 0 2퐷푡 0 1 푟2 1 푥 2+푦 2 obtain 푢 푟, 푡 = 푒푥푝 − = 푒푥푝 − . 4휋퐷푡 4퐷푡 4휋퐷푡 4퐷푡 On using the result (3.11) and 퐼0 0 = 1. This is the fundamental solution of the diffusion equation in the two dimensional case. Example 6: (Poisson’s Equation) Let us solve the Poisson equation 1 휕 휕푢 휕2푢 푟 + = −4휋훿 푟 훿 푧 , 0 ≤ 푟 < ∞, −푎 < 푧 < 푎, (3.12) 푟 휕푟 휕푟 휕푧2 Subject to the boundary conditions lim푟→0 푢 푟, 푧 < ∞, lim푟→∞ 푢 푟, 푧 → 0, −푎 < 푧 < 푎 and 푢 푟, −푎 = 푢 푟, 푎 = 0, 0 ≤ 푟 < ∞ Mathematically we are finding the Green’s function for Poisson equation for the given domain. Physically we are computing electrostatic potential in the free space between two grounded planes at 푧 = ±푎 when a unit point charge is located at the origin. Again, we begin by taking the Hankel transform of equation (3.12) and obtain ∞ 휕2푢 ∞ 휕 휕푢 ∞ 퐽 푘푟 푟푑푟 + 푟 퐽 푘푟 푑푟 = −4휋훿 푧 훿 푟 퐽 푘푟 푟푑푟 (3.13) 0 휕푧2 0 0 휕푟 휕푟 0 0 0 ∞ 휕 휕푢 ∞ Because 푟 퐽 푘푟 푑푟 = −푘2 푢 푟, 푧 퐽 푘푟 푟푑푟 0 휕푟 휕푟 0 0 0 ∞ 휕2푢 푑2 ∞ and 퐽 푘푟 푟푑푟 = 푢 푟, 푧 퐽 푘푟 푟푑푟 . Equation (3.13) becomes 0 휕푧2 0 푑푧2 0 0 휕2푈 푘,푧 − 푘2푈 푘, 푧 = −2훿 푧 , − 푎 < 푥 < 푎 (3.14) 휕푧2 ∞ With the boundary conditions 푈 푘, −푎 = 푈 푘, 푎 = 0, where 푈 푘, 푧 = 0 푢 푟, 푧 퐽0 푘푟 푟푑푟. To solve equation (3.14), we divide the region −푎 < 푥 < 푎 into two sub regions, −푎 < 푥 < 0 and 푑2푈 푘,푧 0 < 푥 < 푎. Within each region, ± − 푘2푈 푘, 푧 = 0.From the boundary conditions, we find that 푑푧2 ± 푈+ 푘, 푎 = 푈− 푘, −푎 = 0. The corresponding solutions are 푈+ 푘, 푧 = 퐴푠푖푛푕 푘 푧 − 푎 , 0 < 푥 < 푎, (3.15) 푈− 푘, 푧 = 퐵푠푖푛푕 푘 푧 + 푎 , − 푎 < 푥 < 0, (3.16) We now must evaluate A and B. Of course, 푈 푘, 푧 must be continuous and

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 33 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 푈+ 푘, 0 = 푈− 푘, 0 . For the second condition, we integrate equation (3.14) over the infinitesimal 0+ 0+ interval 0−, 0+ and find that 푈′′ 푘, 푧 푑푧 = −2 훿 푧 푑푧, or 0 0 푈′′ 푘, 0+ − 푈′′ 푘, 0− = −2 (3.17) Up on substituting equation (3.15) and equation (3.16) into equation (3.17), we obtain 푒 −푘푧 푐표푠푕 푘푧 푒−푘푎 푒푘푧 푐표푠푕 푘푧 푒−푘푎 푈 푘, 푧 = − , 푈 푘, 푧 = − , + 푘 푐표푠푕 푘푎 푘 − 푘 푐표푠푕 푘푎 푘 푒−푘 푧 푐표푠푕 푘푧 푒−푘푎 Therefore, 푈 푘, 푧 = − . 푘 푐표푠푕 푘푎 푘 Finally, taking the inverse Hankel transform, ∞ ∞ 푐표푠푕 푘푧 1 ∞ 푐표푠푕 푘푧 푢 푘, 푧 = 푒−푘 푧 퐽 푘푟 푑푘 − 푒−푘푎 퐽 푘푟 푑푘 = − 푒−푘푎 퐽 푘푟 푑푘 0 0 0 푐표푠푕 푘푎 0 푟2+푧2 0 푐표푠푕 푘푎 0 1 ∞ since = 푒−푘푧 퐽 푘푟 푑푘, 0 < 푧 푟2+푧2 0 0 EXAMPLE 7: Water Waves Consider the surface waves generated within an incompressible ocean of infinite depth due to an explosion above it that produces the pressure field 푓 푟, 푡 . Assuming potential flow, the continuity equation is 1 휕 휕푢 휕2푢 푟 + , 0 < 푟 < ∞, 푧 < 푎 (3.18) 푟 휕푟 휕푟 휕푧2 This Velocity potential 푢 푟, 푧, 푡 must satisfy the boundary condition that 휕2푢 휕푢 1 휕푓 + 푔 = 퐻 푟 − 퐻 푟 − 푟 (푡) (3.19) 휕푧2 휕푧 휌 휕푡 0

At 푧 = 0, where 휌 is the density of the liquid and 푟0(푡) is the extent of the blast. If the fluid is initially at rest, a general solution to equation (3.18) is clearly ∞ 푘푧 푢 푟, 푧, 푡 = 0 퐴 푘, 푡 푒 퐽0 푘푟 푘푑푘 (3.20) Because the Hankel transform representation of 푓 푟, 푡 is ∞ 푟0(푡) 푓 푟, 푡 = 0 0 퐹 훼, 푡 퐽0 푘훼 훼푑훼 퐽0 푘푟 푘푑푘, Equation (3.19) becomes 휕2퐴 푘,푡 1 휕 푟 (푡) + 푔푘퐴 푘, 푡 = 0 퐹 훼, 푡 퐽 푘훼 훼푑훼 (3.21) 휕푡 2 휌 휕푡 0 0 The general solution to equation (3.21) is

퐴 푘, 푡 = 퐴0 푘 푐표푠 휍푡 + 퐵0 푘 푠푖푛 휍푡 1 ∞ 휕 푟 (푠) + 0 퐹 훼, 푠 퐽 푘훼 훼푑훼 푠푖푛 휍(푡 − 푠) 푑푠, (3.22) 휌휍 0 휕푠 0 0 2 Where 휍 = 푔푘. From the initial conditions, 퐴0 푘 = 퐵0 푘 = 0. After integration by parts, equation (3.22) becomes ∞ 푡 푟0(푠) 1 푘푧 푢 푟, 푧, 푡 = 푒 퐽0 푘푟 퐹 훼, 푠 퐽0 푘훼 훼푑훼 푥 푐표푠 휍(푡 − 푠) 푑푠 푘푑푘 휌 0 0 0 4. CONCLUSION: It becomes desirable to apply Fourier transforms as widely as possible but it is defined only in terms of Cartesian coordinates and so may not make great use of any symmetry inherent in the problem. For problems with other than Cartesian geometry, there are yet other transforms that are more natural and therefore more useful. Frequently one encounters problems in two or three dimensional space that symmetry renders expressible by equations in less than two or three variables. Perhaps the most common examples are spherical symmetry and axisymmetry.

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 34 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 Writing the equations for such systems in Cartesian coordinates fails to take advantage of the symmetry. In two dimensions with spherical symmetry when we define the Hankel transform in full momentarily, attention should be paid to its relationship with Fourier transform. The author of this paper extends Fourier transform to the two-dimensional Fourier transform of a function which shows a circular symmetry. For problems with other than Cartesian geometry, there are yet other transforms that are more natural and therefore more useful. This result shows that, the two-dimensional Fourier transform of a circularly symmetric function is, in fact, a Hankel transform of order zero. For example, in unbounded problems with radial symmetry in either the plane or the space, so that the appropriate coordinate, are polar, cylindrical, or spherical, the natural transform for the radial variable involves Bessel functions. For problems in polar and cylindrical coordinates where in the radial coordinate has range r ≥ 0, the Hankel transform is prominent. In particular, Hankel transforms of order zero and order one are often useful for the solution of problems involving vibration of a large circular membrane, Oscilla-tion of an infinitely long hanging chain, acoustic radiation, Laplace’s equation, Diffusion Equation, Poisson’s equation and water waves in polar and cylindrical geometry. 5. REFERENCE: [1] Naol T., Fourier Transform Methods for Partial Differential Equations. International Journal of Partial Differential Equations and Applications, 3(2014)44-57 [2] Grafakos L., and Teschl G., On Fourier Transforms of Radial Functions and Distributions, J. Fourier Anal. Appl., 19(2013) 167–179 [3] J.Berian James., Integral Transforms for You & Me. Royal Observatory Edinburgh Institute for Astronomy, 2008. [4] Irfan N., and Siddik A., A Wavelet Algorithm for Fourier-Bessel Transform Arising in Optics. International Journal of Engineering Mathematics, 2015 [5] J. V. Cornacchio and R. P. Soni., On a Relation between Two-Dimensional Fourier Integrals and Series of Hankel Transforms. Journal of Research of the National Bureau of Standards-B. Mathematics and Mathematical Physics Vol. 69B, 1965 [6] Amidro I., The Fourier-spectrum of circular sine and cosine gratings with arbitrary radial phases. Optics Communications 149 (1998) 127–134 [7] Garg M., Rao A., and S.L. Kalla,On a generalized Finite Hankel transform. Applied Mathematics and Computation 190 (2007) 705–711 [8] I. Ali, S.L. Kalla, A generalized Hankel transform and its use for solving certain partial differential equations, J. Aust. Math. Soc. 41B (1999) 105–117. [9] Duffy D., Transform methods for solving partial differential equations, London New York Washington, D.C,2004 [10]Myint-U T., and Debnath L., Linear Partial Differential Equations for Scientists and Engineers. New York, USA, 2007 [11] Piessens., The Hankel Transform.Boca Raton:CRC Press LLC, 2000 [12] Zauderer E., Partial Differential Equations of Applied Mathematics. John Wiley &Sons,Inc, New Jersey, 2006 [13] Davies B., Integral Transforms and their applications. Springer Verlag New York, 2002 [14] Hanna J., and Rowland J., , Transforms, and Boundary Value Problems. John Wiley &Sons, USA, 1990 [15] Hayek S., Advanced Mathematical Methods in Science and Engineering. Marcel Dekker,Inc, New York, 2001 [16] Asmara N.H., Partial Differential Equations with Foureir Series and Boundary Value Problems. 2nd edition, Pearson Education, Inc, USA,2005 [17] Donald W., Applied Partial Differential Equations, 2013, P427-473 [18] Cox B., and Joseph P., A Comparison of Hankel Transform Algorithms’ Performance for Use in Shallow Water Applications. Institute of sound and vibration research, Southampton University Southampton, UK, SO17 IBJ [19] E. B. POSTNIKOV., About Calculation of the Hankel Transform Using Preliminary Wavelet Transform. Journal of Applied Mathematics, 6 (2003) 319–325 [20] E. Coggins, P. Zhou, Polar Fourier transforms of radially sampled NMR data. Journal of Magnetic Resonance 182 (2006) 84–95 [21] R.N. Bracewell, The Fourier Transform and Its Applications. McGraw-Hill, Boston, 2000 IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 35 Naol Tufa Negero / International Journal of Modern Sciences and Engineering Technology (IJMSET) ISSN 2349-3755; Available at https://www.ijmset.com Volume 3, Issue 10, 2016, pp.24-36 [22] I.N. Sneddon, The Use of Integral Transforms, McGraw-Hill. New York, 1993 [23] R.N. Bracewell, Two-Dimensional Imaging. Prentice-Hall, Englewood Cliffs, NJ, 1995 [24] Vineet K. Singh, Rajesh K. Pandey , Saurabh Singh, A stable algorithm for Hankel transforms using hybrid of Block- pulse and Legendre polynomials. Journal of Computer Physics Communications 181(2010)1–10

AUTHOR’S BRIEF BIOGRAPHY:

Mr. Naol Tufa: He is currently working as Lecturer, Department of Mathematics, College of Natural & Computational Sciences, Wollega University,Ethiopia. He has obtained M.Sc in Mathematics from Haramaya University, in 2012. He has published one article in International Journals. His interested area is Methods and Partial Differential equations. .

IJMSET-Advanced Scientific Research Forum (ASRF), All Rights Reserved “ASRF promotes research nature, Research nature enriches the world’s future” 36