Applications of New Double Integral Transform (Laplace– Sumudu Transform) in Mathematical Physics

The primary purpose of this research is to demonstrate an efficient replacement of double transform called the double Laplace– Sumudu transform (DLST) and prove some related theorems of the new double transform. Also, we will discuss the fundamental properties of the double Laplace–Sumudu transform of some basic functions. Then, by utilizing those outcomes, we will apply it to the partial differential equations to show its simplicity, efficiency, and high accuracy.


Introduction
Partial differential equations are used to describe many realworld problems arising in all the fields of applied science and issues associated with engineering. Due to the rapid advancement in science and engineering, various integral transforms have been used to solve the differential and integral equations, for instance, the Laplace transform [1], the Sumudu transform [2], the Elzaki transform [3], the natural transform [4], and many other double integral transforms [5,6]. However, most of the existing integral transforms have some limitations and cannot be used directly to solve nonlinear problems or many complex mathematical models. As a result, some researchers have combined these integral transforms with other methods such as the differential transform method, homotophy perturbation method, Adomian decomposition method and variational iteration method [7][8][9][10][11][12][13] for solving many nonlinear differential equations.
The usefulness of these equations has attracted the attention of many scholars throughout the history of applied sciences. One of the problems with these equations is that they are very difficult to solve equations with unknown functions of two variables sometimes. For this reason, this problem was solved by combining the Laplace transforms and Sumudu transform to give another double transform which is called the double Laplace-Sumudu transform.
In the present research, we consider linear, one-dimensional, time-dependent partial differential equation (PDE) of the form with the initial conditions: and boundary conditions: where c n , 0 ≤ n ≤ N; d m , 1 ≤ m ≤ M are given coefficients and N, M are positive integers and gðx, tÞ is the source term.
The main objective of this paper is to develop new applications of the double Laplace-Sumudu transform for solving linear partial differential equations of the type (1) subject to the initial conditions (2) and boundary conditions (3). The proposed integral transform is successfully applied to a wide range of linear partial differential equations in mathematical physics.

Preliminaries
Definition 1. The double Laplace-Sumudu transform of the function ϕðx, tÞ of two variables x > 0 and t > 0 is denoted by L x S t ½ϕðx, tÞ = ϕðρ, σÞ and defined as whenever that integral exists. Here, ρ and σ are complex numbers.
Clearly, the double Laplace-Sumudu transform is a linear integral transformation as shown below: where γ and η are constants.
Definition 2. The inverse double Laplace-Sumudu transform L −1 x S −1 t ½ϕðρ, σÞ = ϕðx, tÞ is defined by the following form: Definition 3. The double Laplace-Sumudu transform formulas for the partial derivatives of an arbitrary integer order are given by (see the proof in Section 3.2)

Existence Condition for the Double Laplace-Sumudu
Transform. If ϕðx, tÞ is an exponential order c and d as and we write Or, equivalently, The function ϕðx, tÞ is called an exponential order as x → ∞, t → ∞, and clearly, it does not grow faster than K e c x+d t as x → ∞, t → ∞. Proof. From Definition 1, we have Then, from Equation (11) Proof.

Elucidative Examples
In this section, the applications of the proposed transform are presented. The simplicity, efficiency, and high accuracy of the double Laplace-Sumudu transform are clearly illustrated.

Abstract and Applied Analysis
Substituting in (22) and simplifying, we get a solution of (23)

The Reaction-Diffusion Equation.
By substituting N = 1, M = 2, g = 0, c 1 = 1, d 1 = 0 , and d 2 > 0 in (1), we have the reaction-diffusion equation: with ICs and BCs: then (19) gives the solution of (27) as with ICs and BCs: then (29) gives the solution of (30) as Example 6. Putting d 2 = 1 in (30) to yield with the conditions, Substituting in (32) and simplifying, we get a solution of (33) Abstract and Applied Analysis = g = 0 in (1), we have got the linear telegraph equation with ICs and BCs: then (19) gives the solution of (37) as Taking c 0 = c 1 = 0, c 2 = 1 in (37), we obtain the linear wave equation: with ICs and BCs then (39) gives the solution of (40) as