Solving the bessel functions using the sadik transform

Introduction

Bessel function are now essential for solving problems in the fields of engineering, Heat transfer, vibrations, hydrodynamics, fluid mechanics, nuclear physics, mathematical physics, atomic physics, acoustics, radio physics, and stress analysis, and nuclear reactor flux distribution. The nonnegative integer is represented by the Bessel function of order [2,5,7,8].It is also clear how crucial integral transformations are for precisely solving differential equations. Numerous integral transformations have been proposed by researchers, each of which addresses a problem in life sometimes social and economic, sometimes engineering, physical, medical, or astronomical with the goal of finding solutions using less complicated, more technical, and simpler approaches. It is thought that the authentic transformation is easier to use and more universal [9-13] .

For 𝑛𝑛 = 0 ,The zero order Bessel function is represented by. 𝐽𝐽 0(𝑡𝑡) The following infinite power series is used to define.

For 𝑛𝑛 = 1, The following infinite power series defines Bessel function of order one, symbolizes by 𝐽𝐽 1(𝑡𝑡).

Some useful properties of sadik transform: Sadik Transform (𝑺𝑺 -Transform)

Integral transforms methods (𝐿𝐿 -Transform, 𝑀𝑀𝑀𝑀 -Transform, 𝑀𝑀 -Transform, 𝑆𝑆 -Transform ,……… etc ) are practical mathematical tools that can be used to solve complex science and engineering problems that can be is expressed in terms of integral and the differential equations.

Definition: [3,8] The Sadik Transform of the function Ꝼ(ɤ) , ɤ ≥ 0 can be is defined

when 𝑣𝑣 is a complex variable and 𝛼𝛼 ≠ 0 , 𝛽𝛽 are any actual numbers. The Sadik transform operator is denoted by the letter S.

It should be noted that his Sadik transform of then functions Ꝼ(ɤ) since ɤ≥0 exists if Ꝼ(ɤ)is The only prerequisites for the existence of Sadik transforms the function are piecewise continuity and exponential order Ꝼ(ɤ).

Convolution Theorem the Sadik Transform

To solve integral equation in any transform we should use the convolution theorem it as we have observed in solving the integral equations by previous transform. Let the sadik Transform of the functions Ꝼ 1 (ɤ) and Ꝼ 2 (ɤ) is 𝑆𝑆 �Ꝼ 1 (ɤ)� and 𝑆𝑆 �Ꝼ 2 (ɤ)� respectively, then: Inverse of S -Transform [3].

If 𝑆𝑆 {Ꝼ(ɤ)} is the Sadik Transforms of the function Ꝼ(ɤ), the Ꝼ(ɤ) is known as the inverse of the Sadik Transform and has the following mathematical definition.

Such that operator 𝑆𝑆 -1 is called the inverse S -Transform operator and for sure it have that linearity property:

Table 1: (Sadik -Transform) for some elementary functions

S.N Ꝼ (ɤ) SꝻ (ɤ)

5

𝑒𝑒 𝑎𝑎ɤ 1 𝑣𝑣 𝛽𝛽 (𝑣𝑣 𝛼𝛼 -𝑎𝑎)

Dualities between transforms

To illustrate the significance of these linkages for the integral transforms mentioned, we give examples using φ as a symbol for the other transform result. The dualities between the L-Transform and a number of popular transforms, including the M-Transform, S-Transform, Mo-Transform, and γ-Transform, will be covered in this section.

Mahgoub -Sadik Duality

𝑀𝑀ahgoubTransform of then is function Ꝼ(ɤ) , since ɤ ≥ 0 such that define as:

and 𝑆𝑆adik Transforms of then function Ꝼ(ɤ) , since ɤ ≥ 0 symbolizes by 𝑆𝑆{Ꝼ(ɤ)} and can be defined as:

On the other hand: Let 𝑟𝑟 = 𝑣𝑣 𝛼𝛼 , then

Sadik -mohand duality 𝑆𝑆adik Transforms of then function Ꝼ(ɤ) , since 𝑥𝑥 ≥ 0 denoted by 𝑆𝑆 {Ꝼ(ɤ)} and can be they know:

. Now let 𝑣𝑣 𝛼𝛼 = 𝑢𝑢 , then we have:

𝜑𝜑(𝑢𝑢) . On the other hand:

Let 𝑣𝑣 = 𝑟𝑟 𝛼𝛼 , then: By applying the second derivative Sadik transform property, we arrive at:

Example 1 Solve the following integral equation using sadik integral transforms

As a result, the Sadik transform of from 𝐽𝐽 0(𝑡𝑡) equation ( 7) is:

(ℎ) ``𝑡𝑡 = 𝐽𝐽 0(𝑡𝑡) and h(0)=0 and h'(0)=0

Consequently, by applying the transform relation linked to the Bessel function of order one

and taking the inverse Sadik transform, we ultimately obtain:

Equation ( 7) thus gives the transform of{𝐽𝐽 0(𝑡𝑡) }, and the inverse transform step that links the answer to {𝐽𝐽 0(𝑡𝑡) } yields equation (8).

Practical example: Transient response of a cylindrical system

The transient response of a cylindrical system in physics and engineering frequently requires the solution of differential equations with boundary conditions determined by the object's geometry. system, like a thin circular plate or a cylindrical heat conductor, responds to a time-dependent stimulus by using the Sadik Transform. the Problem since by equation (7,8,10) we option: Examine a mechanical system in which the following secondorder differential equation controls the displacement h(t):

With Initial conditions:h(0)=0 and h'(0)=0

We use the following property for second-order derivatives by applying the Sadik Transform S{h(t)} = h(v) to both sides of the equation:

Substituting the known transform for the zero-order Bessel function,

Using the inverse Sadik transform property for Bessel functions, the solution in the time domain is:

References

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5. Luchko Y. Some schemata for applications of the integral transforms of mathematical physics. Mathematics. 2019;7(3):254. doi:10.3390/math7030254.
6. Wazwaz AM. Linear and Nonlinear Integral Equations. Berlin: Springer; 2011.
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