The New Complex Integral Transform "Complex Sadik Transform" and It's Applications

In this work, we present a novel complex transform namely the "Complex Sadik Transform". The propositions of this transformation are investigated. The complex transform is used to convert the core problem to a simple algebraic equation. Then, the answer to this primary problem can be obtained to find the solution to this equation and apply the inverse of the complex Sadik transform. As well, the complex Sadik transform is applied and used to find the solution of linear higher order ordinary differential equations. As well, we present and discuss, some important real life problems such as pharmacokinetics problems, nuclear physics problems, and Beam problems.


Introduction
In (2018), researcher Sadik L. sheikh [11] presented a new integral transformation defined as follows : The Sadik integral transform of ( ) is defined as : Now, the complex Sadik transform is a new complex transform and it is applied and used to find the solution of ordinary differential equation and has applications in domains such as engineering, applied physics, and signed processing [3,4,7].
We analyze functions in the set defined by a novel complex transform defined for functions of exponential order: = { ( ): there exists , 1 and 2 are greater than zero such that where is a complex number. The constant must be a finite number for a particular function ( ) in the set , while 1 and 2 are may be finite or infinite.
The Complex Sadik Transform (CST) denoted by the operator {. }, the transform as follows: where ∈ ℂ , ∈ ℝ * , and ∈ ℝ . The aim of this work (complex transform) to find the solution of higher order ordinary differential equations.
Many researchers have proposed new integral transformations for the purpose of solving ordinary and partial differential equations and their applications [2,8,9,12].

A Novel Complex Transform "Complex Sadik Transform" of Important Functions
In this section, we present the complex Sadik transform of famous functions: Proof. Since after simple computations, we get: , > | | The result will be benefit in determining the complicated transform of:

The Sadik and Complex Sadik Integral Transforms for Some Basic Functions
In this section, we will present the Sadik transform and the novel complex transform for some basic functions in the following Table 1:
In this section, we present the inverse of Sadik complex integral transform of simple functions:

Complex Sadik Integral Transform of Derivatives:
Let ( ) be a continuous function and piecewise continuous on any interval, then the complex Sadik transform of first derivative of ( ) is given by: So theorem is true for ∈ ℕ.

Applications of Complex Sadik Integral Transform:
In this section, we introduce three real life problems: pharmacokinetics problem, nuclear physics and Beam problems. Example 4.1. For a physical explanation of the present scheme, we consider a problem from the field of "pharmacokinetics" for solving the concentration of the drug at any given time " " in the blood during continuous intravenous injection of drug and find its solution in this application. This
: is the constant velocity of elimination.
: the rate of infusion (in mg/min.) Volume: volume in which drug is distributed. Complex Sadik transform of both sides of equation (1) gives: Applying Theorem 3.1, we get: The use of the initial condition equation (2) in (4) gives: Applying inverse complex Sadik transform in equation (5 ), we get: By a fractional fraction, after simple computations, we get: Which is the required concentration of drug at any given time " " in the blood during continuous intravenous injection of a drug.

Example 4.2. "Complex Sadik Transform in Nuclear physics":
Consider the first order linear differential equation: This differential equation is the fundamental relationship describing radioactive decay, where ( ) represents the number of un decayed atoms remaining in a sample of radioactive isotope at the time " " and is the decay constant, [7,10].
We can apply the complex Sadik transform to find the solution to this differential equation. Rearranging the above differential equation, we obtain: ( ) + ( ) = 0 Taking complex Sadik transform on both sides, we have: Which is indeed the correct formula for radioactive decay.

Example 4.3. Problem to Beams:
A beam that is hinged at its ends, = 0 and = carries a uniform loud 0 per unit length. Find the deflection at any point .

Solutions:
The ordinary differential equation and boundary conditions are: