Abstract

A new Integral Transform was introduced in this paper. Fundamental properties of this transform were derived and presented such as the convolution identity, and step Heaviside function. It is proven and tested to solve some basic linear-differential equations and had succesfully solved the Abel's Generalized equation and derived the Volterra Integral Equation of the second kind by means of Initial Value Problem. The Natural Logarithm (e.g log_ex=lnx) has been established and defined by means of modifying the Euler Definite Integral based on the Rangaig's fomulation. Hence, this transform may solve some different kind of integral and differential equations and it competes with other known transforms like Laplace, Sumudu and Elzaki Transform. Keywords: Rangaig Transform, Integral Transform, linear ordinary differential function, Integro-differential equation, Convolution Theorem.

1. Introduction

One of the most effective tools for solving problems in physics and engineering is using the transform method to obtain a solution for a given partial differential equations or ordinary differential equation by means of inverse transformation. Among these Transforms are the Fourier ¹, Laplace ², Hankel ³, Mellin ³ Sumudu ⁴, and Elzaki ⁵. These transforms play a versatile role in solving many problems of physical interests which can be described by integro-differential equations with a given appropriate initial and boundary conditions. In these past years, applicability of these transforms were shown and considered by many authors ^{6, 7, 8, 9, 10}.

In this paper, we investigate a new transform which is also based on the previous transforms but in another domain. This Transform takes another form in process of transforming an equation and obtains the solution of the equation wherein it follows the set of functions H, an exponential order defined as

(1)

For the set in (1), the arbitrary constant N must be a finite and the constants can be infinite or infinitely finite. Introducing a new transform which is defined in (1),

(2)

called Rangaig Transform. In this transform, the variable factorizes the variable of the function or in another sense, the function is mapped into the function of -space. Applicability and efficiency of this interesting transform will be shown in solving linear differential equations such as some problems on Classical Physics.

2. Rangaig Transform of Some Functions

For any function on such that the function exists. Then the Rangaig Transform (2) is satisfied if the condition for a function and is piecewise continuous with decreasing exponential order. Properties of this transform are presented in the subsequent sections.

Definition 2.1. Let denotes the Rangaig Transform of Then the following theorems holds:

Theorem 2.2. If h(t)=1, then its Rangaig Transform yields

and for yields

Now, the general form for we have

Proof: Now, from the definition of the Rangaig Transform, we have

Theorem 2.3. For exponential function of the form exp(at), then

Theorem 2.4. The Rangaig Transform of the trigonometric function such as cos(t) and sin(t) yields

and

respectively.

Proof: We can write the trigonometric function in terms of Taylor's series expansion

Then, obtaining the Rangaig Transform of sin(t) yields

and applying Theorem 2.2 and further simplification yields

Now, imposing the Binomial series expansion will result to

Similarly, for

Definition 2.5. If the function exists such that its n-th derivative also exists in the set (1), then let the Rangaig Transform of the n-th derivative of be denoted as holds the next theorem.

Theorem 2.6. (Rangaig Transform for derivatives) If then

Proof: (By Mathematical Induction). We verify for any satisfies the theorem. So, for For n=1, we have

applying integration by parts yields

Suppose, there exist an integer j such that holds for Letting so that Hence,

This finishes the proof.

Definition 2.7. Let the functions and Then the next Theorem holds the Rangaig Transform of Integrals denoted by

Theorem 2.8. (Rangaig Transform for Integrals) If

such that Then, the Rangaig Transform of is defined as

Proof: (By Mathematical Induction). We need to verify that Theorem 2.8 is satisfied for For we can obtain For we get

(see Theorem 2.6 and refer to it). To complete the proof, suppose there exists an integer such that for it is verified that this is true. Then for

Hence, the Proof has been done by Mathematical Induction.

Note that for multiple Integrals or antiderivative of a function h(t), we can write its equivalence as

Definition 2.9. Consider the functions and Then next Theorem holds the Rangaig Transform for the Convolution Identity.

Theorem 2.10. The Rangaig Transform for the Convolution Identity given by

is defined as

where and is the Rangaig Transform for and respectively.

Proof: Multiplying the Rangaig Transform of the functions and and to avoid complication, we assume that the limits are finite and thus resulting to

This formulation is possible since the integrands are considered to decrease exponentially. For the limit the integrands are very complicated to deal with, thus to reduce this complication, let us assume an integer t and T such that To verify these assumptions, we make use of the Jacobian's Transformation for differentials

or simply Substituting this expression to the above equation yields

Corollary 2.11. Let the functions and be in H. Then the Rangaig Transform of the nth antidifferential of the Convolution Identity is defined as

Furthermore, for any integer we have the more generalized Convolution Identity defined by its Rangaig Transform as

Proof: The same procedure in Theorem 2.8 for the expressions in this Corollary. The key to this proof is the associability and symmetric property of the convolution operator. Hence, Theorem 2.8 is the implication of Corollary 2.11.

Definition 2.12. The duality relation of the Rangaig Transform to Laplace Transform exists if the function exists such that the function is considered to change itself

Theorem 2.13. (Duality relation of Rangaig Transform and Laplace Transform) If and exist over then the relation of Rangaig Transform and Laplace Transform is

where is the Laplace Transform of

Proof: The Laplace Transform of the function is

following the given ranges of in (1). may exist and thus modifying the Laplace Transform into

Therefore the duality relation is satisfied.

Theorem 2.14. Let be a Heaviside step function. Then its Rangaig Transform takes the form

Proof: From the definition of the Rangaig Transform (2), we have

One proof of this can be done by the duality relation of the Rangaig Transform and Laplace Transform (Theorem 2.13).

3. Some Application of Rangaig Transform

In this section, we will present an application of Rangaig Transform to some differential equations. The following examples are governed by linear differential equations with initial condition wherein Rangaig Transform is used in solving some initial value problems described by ordinary differential equations.

Example 3.1.

Consider the first-order linear differential equation of the form

where g(t) is an input function.

Sol'n: Let the initial condition be f(0) = 1 so that we can write the Rangaig Transform for the first-order derivative as

Obtaining Rangaig transform will yield

Further manipulation, we can obtain the general solution as

and taking the inverse Rangaig Transform leads to the solution. For , we have the solution

Example 3.2.

For an RC-circuit, the potential on the capacitor for is After a long time the potential on the capacitor follows the first-order linear differential equation of the form

from Theorem 2.6, we have

and evaluating yields

Inversing the Rangaig Transform (using Theorem 2.3) we have the potential for the capacitor for long time

Example 3.3.

A bullet moving horizontally and experiences a resistive force during the motion. Then consider the equation of motion with constant k ¹⁰

then find the velocity at any time t and given that

Sol'n: This problem is almost the same with the previous Example 3.2. Following similar step, we get the Rangaig Transformed equation of motion as

thus inversing, we obtain the solution

Example 3.4.

Solve the differential equation

using the Rangaig transform if the initial condition is

Sol'n: Again, Using the Theorem 2.6, we get the Rangaig Transform of the first derivative of the function as

thus the transformed differential equation takes the form

Further manipulation, we get

Using the Previous theorems, we get the solution as

Example 3.5. (Application of Rangaig Transform on Abel's Generalized equation)

The Generalized Abel's equation is given by

Where is assumed to be known and is unknown.

In this example, we will show the effectiveness of the Rangaig Transform in this kind of equation. Particularly, the validity of the Rangaig Transform on Convolution Theorem will be shown.

Taking the Rangaig Transform for both side of the equation yields

Notice that the right-hand side of this expression demonstrates the Convolution Identity given in Theorem 2.10 if Then we can write the Rangaig Transform in accordance to the Convolution Theorem as

Rearranging this expression yields

and multiplying the left side of the expression by we have

By using Theorem 2.10 and 2.8 we get

then from ³, the factorial term of this expression can be written as

Substituting this expression to the previous equation and then differentiating. Hence, we have obtained the unknown term of Abel's equation as

Example 3.6. (Derivation of Volterra Integral Equation of the Second Kind Using Rangaig Transform)

Let us consider first an Initial Value Problem of the form

with the condition Now, let then getting its Rangaig Transform yields

Simplifying, we get

Using Theorem 2.8 with n = 2, obtaining the inverse Rangaig Transform of this expression gives

Therefore, we get the Volterra Integral Equation of the second kind as

4. Uniqueness of the Rangaig Transform

The Gamma Function is typically well defined for if n < 0, It is hard to define since it diverges for The Gamma function can be defined by its second deffnition known as Euler's Definite Integral ¹³

Logically, if the Gamma function happens to have a negative value of n, of course it is difficult to deal, but what we do is to create a counter part of the Gamma function just in case for n to take values from negative. If we let be the counter form of the Gamma function if n < 0. Another problem arises, that is, how can we make an equivalence form of and in order to do this, we need to modify the Euler Definite Integral based on the formulation of the Rangaig Transform. We have the proposed modification of the Euler Definite Integral based on the form of Rangaig Transform as

The reason of this formulation is restraining the values of n to become negative. Since this formulation is based on Rangaig Transform, it is required that this result is applicable only to Rangaig Transform for n < 0. Consider the Function where n < 0, then we have its Rangaig Transform as

For which the Laplace Transform of is Hence, this gives us the property of a Natural Logarithm (Napierian).

Proposition 4.1. For a function Then we have obtained the Rangaig Transform for Natural Logarithm defined as

Proof: From the Integral definition of natural logarithm, we have

Recall, from Theorem 2.8 the Rangaig transform of a given integral

for we get

Actually, using the Uniqueness of Rangaig Transform, we have Therefore, the Proposition is satisfied.

Lemma 4.2. For the function

and

Then, thier Rangaig Transform takes the form

respectively.

Proof: (By Mathematical Induction). We will verify that for is true. For we have

which is true (can be verified by integrating). Suppose there exist an integer such that for it is indeed verified. Now, we will show that for all is true, that is

This finishes the proof.

Similar proof can be done for

Example 4.3. Effeciency of Uniqueness of the Rangaig Transform

In this following examples, we wil going to show the validity of the Uniqueness of the Rangaig Transform to some separable differential equation.

Consider a differential equation of the form

where So, the equation can be immediately written as

Using Theorem 2.8 we have

Now, Using the Uniqueness we get

and inversing both sides of the equation, we can obtain its solution as

For general solution, we have

Example 4.4. Solve this differential equation

by applying the Rangaig Transform.

Sol'n: Since the equation is separable we can write it as

Applying the Rangaig Transform for both sides of the equation we have

and further manipulations, we have the solution as

5. Conclusion

The deffnition of Rangaig Transform and its application to the solution of ordinary differential equation has been presented. It was shown that the Rangaig Transform has a clearer and deeper connection to Laplace Transform. However, there are cases that Laplace Transform and Elzaki Transform cannot solve integro-differential equations but can be solved in Rangaig Transform. Therefore, the Rangaig Transform can be used as an effective tool in solving integro-differential equations.

Acknowledgements

The authors would like to thank the Department of Physics for the support and ideas in developing this paper. The authors would also like to thank the Mindanao State University-Main Campus for the support extended on this work.

References