On the Double ARA-Sumudu Transform and Its Applications

Abstract

The main purpose of this work is to present a new double transform called the double ARA-Sumudu transform (DARA-ST). The application of the new double transform to some basic functions and the master properties are introduced. The convolution and existence theorems are also presented and proved. These new results are implemented to obtain the solution of partial differential equations (PDEs), integral equations (IEs) and functional equations. We obtain new formulas for solving families of PDEs. The latter ones are used to obtain exact solutions of some familiar PDEs such as the telegraph equation, the advection–diffusion equation, the Klein–Gordon equation and others. Moreover, a simple formula for solving a special kind of integral equations is presented and implemented in some applications. The outcomes show that DARA-ST is useful and efficient in handling such kinds of equations.

1. Introduction

Integral-transform methods are among the most important methods that have been implemented in solving PDEs. Since many phenomena in mathematical physics and science fields can be expressed mathematically in terms of PDEs [1,2,3,4,5,6,7,8,9,10,11,12], utilizing integral transforms enables us to transform these equations and obtain the exact solution of PDEs. Great efforts to improve these methods have been made by researchers and scientists, and they employ them to solve modern scientific problems. For instance, we have the Laplace transform [13], the Fourier transform [14], the Sumudu transform [15], the natural transform [16], the Elzaki transform [17], the Novel transform [18], the M-transform [19], the polynomial transform [20], the Aboodh transform [21] and the ARA transform [22,23,24]. Lately, the double Laplace transform has been used effectively and extensively to solve PDEs with unknown functions of two variables, obtaining strong results compared with numerical methods [25,26,27,28,29]. Furthermore, in the literature, there exist more extensions of the double Laplace transform, such as the double Shehu transform [30], the double Sumudu transform [31,32,33,34,35,36], the double Elzaki transform [37] and the Laplace–Sumudu transform [38,39].

The major objective of this paper is to introduce a new double transform, the DARA-ST. We present some main properties of the DARA-ST and compute it for some basic functions. Many simple theorems dealing with interesting properties of the DARA-ST are discussed. In addition, the convolution theorem of two functions and some properties with proofs are introduced.

In this article, we consider the nonhomogeneous linear PDE of the form

$$A{g}_{xx}+B{g}_{yy}+C{g}_x+D{g}_y+Eg\left(x,y\right)=u\left(x,y\right),$$

with the initial conditions (ICs)

$$g\left(x,0\right)=f_1\left(x\right), g_y\left(x,0\right)=f_2\left(x\right),$$

and the boundary conditions (BCs)

$$g\left(0,y\right)=h_1\left(y\right), g_x\left(0,y\right)=h_2\left(y\right),$$

where $g\left(x,y\right)$ is the unknown function; $A,B,C,D$ and $E$ are constants; and $u\left(x,y\right)$ is the source term. A simple formula for the solution of the above equation is established and employed in some applications.

Another achievement in this work, we consider the IE of the form

$$g\left(x,y\right)=f\left(x,y\right)+\lambda {{\displaystyle \int }}_0^x{{\displaystyle \int }}_0^{y}g\left(x-t,y-v\right)k\left(t,v\right)dtdv,$$

where $g\left(x,y\right)$ is the unknown function, λ is a real number, and $f\left(x,y\right)$ and $k\left(x,y\right)$ are known functions. This IE is considered, and a general formula is found to solve it.

This article is organized as follows: In Section 2, we introduce a new integral transform, the DARA-ST, that combines the ARA transform and the Sumudu transform and present some properties of this transform. In Section 3, we apply the DARA-ST to a family of PDEs, a special kind of IEs and a functional equation. In Section 4, some examples are presented and solved with the DARA-ST. In Section 5, we present some remarks and observations about the new approach, the DARA-ST. Finally, in the conclusion section, our results are summarized.

2. Double ARA-Sumudu Transform

In this section, a new integral transform, the DARA-ST, that combines the ARA transform and the Sumudu transform is introduced. We present basic properties concerning the existence conditions, linearity and the inverse of this transform. Moreover, some essential properties and results are used to compute the DARA-ST for some basic functions. We introduce the convolution theorem and the derivatives properties of the new transform.

Recall that the ARA transform of order n of a piecewise continuous function $g\left(x\right)$ on $\left[0,\infty \right)$ is given by

$${\mathcal{G}}_n\left[g\left(x\right)\right]=s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{x}^{n-1}{e}^{-sx}g\left(x\right)dx, s>0, n=1,2,\cdots .$$

The ARA transform of order one is given by

$${\mathcal{G}}_1\left[g\left(x\right)\right]=s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx}g\left(x\right)dx, s>0$$

For simplicity, let us denote ${\mathcal{G}}_1\left[g\left(x\right)\right]$ by $\mathcal{G}\left[g\left(x\right)\right]$.

Om the other hand, the Sumudu transform of function $g\left(x\right)$ is defined as

$$S\left[g\left(x\right)\right]=\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{t}{u}}g\left(x\right)dt, u>0$$

.

In the following, we introduce some basic properties of the ARA transform and Sumudu transform.

Let $g\left(x\right)$ and $h\left(x\right)$ be two continuous functions on $\left[0,\infty \right)$ in which the ARA transform exists. Then

$$\mathcal{G}\left[\alpha g\left(x\right)+\beta h\left(x\right)\right]=\alpha \mathcal{G}\left[g\left(x\right)\right]+\beta \mathcal{G}\left[h\left(x\right)\right]$$

(1)

where $\alpha$ and $\beta$ are nonzero constants.

$$\mathcal{G}\left[g^{\prime }\left(x\right)\right]=s\mathcal{G}\left[g\left(x\right)\right]-sg\left(0\right)$$

(2)

$$\mathcal{G}\left[x^{\alpha }\right]=\frac{\mathsf{\Gamma }\left(\alpha +1\right)}{s^{\alpha }}$$

(3)

$$\mathcal{G}\left[g\left(x+y\right)\right]=e^{sy} \mathcal{G}\left[g\left(x\right)\right], s>0$$

(4)

Let $g\left(x\right)$ and $h\left(x\right)$ be two continuous functions on $\left[0,\infty \right)$ in which the Sumudu transform exists. Then

$$S\left[\alpha g\left(x\right)+\beta h\left(x\right)\right]=\alpha S\left[g\left(x\right)\right]+\beta S\left[h\left(x\right)\right]$$

(5)

where $\alpha$ and $\beta$ are nonzero constants.

$$S\left[g^{\prime }\left(x\right)\right]=\frac{S\left[g\left(x\right)\right]}{u}-\frac{g\left(0\right)}{u},$$

(6)

$$S\left[x^{\alpha }\right]=\mathsf{\Gamma }\left(\alpha +1\right)u^{\alpha }$$

(7)

Now, we construct the definition of the DARA-ST and some properties.

Definition 1.

Let $g\left(x,y\right)$be a continuous function of two variables  $x>0$ and  $y>0$. Then, the DARA-ST of the function  $g\left(x,y\right)$ is given by

$${\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]=G\left(s,u\right)=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy, s>0,u>0,$$

provided the integral exists.

Clearly, the DARA-ST is a linear transform, since

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[a g\left(x,y\right)+b h\left(x,y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}\left(a g\left(x,y\right)+b h\left(x,y\right)\right)dxdy \\ =\frac{a s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} g\left(x,y\right)dxdy+\frac{b s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} h\left(x,y\right)dxdy \\ =a {\mathcal{G}}_x{S}_y\left[ g\left(x,y\right)\right]+b {\mathcal{G}}_x{S}_y\left[ h\left(x,y\right)\right], \end{gathered}$$

where a and b are constants. The inverse DARA-ST is given by

$${\mathcal{G}}_x^{-1}{S}_y^{-1}\left[G\left(s,u\right)\right]=g\left(x,y\right)=\frac{1}{2\pi i}\underset{c-i\infty }{\overset{c+i\infty }{{\displaystyle \int }}}\frac{e^{sx}}{s}ds \frac{1}{2\pi i}\underset{\omega -i\infty }{\overset{\omega +i\infty }{{\displaystyle \int }}}\frac{e^{\frac{y}{u}}}{u}G\left(s,u\right)du.$$

2.1. DARA-ST for Some Basic Functions

a.

Let $g\left(x,y\right)=1, x>0, y>0$. Then

$${\mathcal{G}}_x{S}_y\left[1\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}dxdy=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx}dx\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}dy=1, \mathrm{Re}\left(s\right)>0.$$

b.

Let $g\left(x,y\right)=x^a{y}^b, x>0, y>0$. Then

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[x^a{y}^b\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{x}^a{y}^b{e}^{-sx-\frac{y}{u}}dxdy=s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{x}^a{e}^{-sx}dx ·\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{y}^b{e}^{-\frac{y}{u}}dy \\ =\mathcal{G}\left[x^a\right]·S\left[y^b\right]=\frac{u^b}{s^a}\mathsf{\Gamma }\left(a+1\right)\mathsf{\Gamma }\left(b+1\right), \mathrm{Re}\left(a\right)>-1, \mathrm{Re}\left(s\right)>0, \end{gathered}$$

where $a$ and $b$ are constants.

c.

Let $g\left(x,y\right)=e^{ax+by}$. Then

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[e^{ax+by}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{ax+by}{e}^{-sx-\frac{y}{u}}dxdy=s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{ax}{e}^{-sx}dx ·\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{by}{e}^{-\frac{y}{u}}dy \\ =\frac{s}{s-a}·\frac{1}{1-bu}=\frac{s}{\left(s-a\right)\left(1-bu\right)}, \mathrm{Re}\left(a\right)+ \mathrm{Re}\left(s\right)>0. \end{gathered}$$

Consequently

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[e^{i\left(ax+by\right)}\right]=\frac{s}{\left(s-ia\right)\left(1-ibu\right)}=\frac{s\left(s+ia\right)\left(1+ibu\right)}{\left(s^2+a^2\right)\left(1+b^2{u}^2\right)}=\frac{s\left(s+isbu+ia-abu\right)}{\left(s^2+a^2\right)\left(1+b^2{u}^2\right)} \\ =\frac{s\left(s-abu+i\left(sbu+a\right)\right)}{\left(s^2+a^2\right)\left(1+b^2{u}^2\right)}, \mathrm{Im}\left(a\right)< \mathrm{Re}\left(s\right). \end{gathered}$$

Using Euler’s formulas

$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}, \cos x=\frac{e^{ix}+e^{-ix}}{2},$$

and the formulas

$$\sinh x=\frac{e^x-e^{-x}}{2}, \cosh x=\frac{e^x+e^{-x}}{2},$$

we obtain the DARA-ST of the functions where $\mathrm{Im}\left(a\right)< \mathrm{Re}\left(s\right)\bigwedge \mathrm{Re}\left(s\right)>0$.

$${\mathcal{G}}_x{S}_y\left[\sin \left(ax+by\right)\right]=\frac{s \left(a+b s u\right)}{\left(a^2+s^2\right)\left(b^2 u^2+1\right)}$$

$${\mathcal{G}}_x{S}_y\left[\cos \left(ax+by\right)\right]=\frac{s \left(s-ab u\right)}{\left(a^2+s^2\right)\left(b^2 u^2+1\right)}.$$

$${\mathcal{G}}_x{S}_y\left[\sin \mathrm{h}(ax+by\right]=\frac{s \left(a+b s u\right)}{\left(a^2-s^2\right)\left(b^2 u^2-1\right)}$$

$${\mathcal{G}}_x{S}_y\left[\cosh \left(ax+by\right)\right]=\frac{s \left(s+ab u\right)}{\left(a^2-s^2\right)\left(b^2 u^2-1\right)}$$

d.

Let $g\left(x,y\right)=J_0\left(c\sqrt{xy} \right).$ Then

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[J_0\left(c\sqrt{xy} \right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} J_0\left(c\sqrt{xy} \right)dxdy \\ =\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}dy s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x}{J}_0\left(c\sqrt{xy} \right)dx=S\left[e^{-\frac{c^{2}y}{4s}}\right]=\frac{4s}{4s+c^{2}u}, \end{gathered}$$

where $J_0$ is the modified Bessel function of order zero.

e.

Let $g\left(x,y\right)=f\left(x\right)h\left(y\right)$. Then

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]={\mathcal{G}}_x{S}_y\left[f\left(x\right)h\left(y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} f\left(x\right)h\left(y\right)dxdy \\ =\left(s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x}f\left(x\right)dx \right)\left( \frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}h\left(y\right)dy\right)=\mathcal{G}\left[f\left(x\right)\right]S\left[h\left(y\right)\right]. \end{gathered}$$

2.2. Existence Conditions of DARA-ST

Let $g\left(x,y\right)$ be a function of exponential orders $\lambda$ and $\gamma$ as $x\to \infty$ and $y\to \infty$.

If there exists a positive constant $M$ such that $\forall x>X$ and $y>Y$, we have

$$\left|g\left(x,y\right)\right|\le M{e}^{\lambda x+\gamma y}.$$

Hence, $g\left(x,y\right)=O\left(e^{\lambda x+\gamma y}\right)$ as $x\to \infty$ and $y\to \infty$, $s>\lambda$ and $\frac{1}{u}>\gamma$.

Theorem 1.

Let  $g\left(x,y\right)$ be a continuous function on the region  $\left[0,X\right)\times \left[0,Y\right)$ of exponential order  $\lambda$ and  $\gamma$. Then, the DARA-ST of  $g\left(x,y\right)$ exists for  $s$ and  $\frac{1}{u}$ provided  $Re\left[s\right]>\lambda$ and  $Re\left[\frac{1}{u}\right]>\gamma$.

Proof of Theorem 1.

The DARA-ST definition yields that

$$\begin{gathered} \left|G\left(s,u\right)\right|=\left|\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dx dy\right|\le \frac{s}{u} \underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}\left|g\left(x,y\right)\right|dx dy \\ \le \frac{M s}{u} \underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\left(s-\lambda \right)x}dx\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\left(\frac{1}{u}-\gamma \right)y}dy=\frac{Ms}{u\left(s-\lambda \right)\left(\frac{1}{u}-\gamma \right)} \\ =\frac{Ms}{\left(s-\lambda \right)\left(1-u\gamma \right)}, \mathrm{Re}\left[s\right]>\lambda \mathrm{and} \mathrm{Re}\left[\frac{1}{u}\right]>\gamma . \end{gathered}$$

The proof is completed. □

2.3. Basic Properties of DARA-ST

2.3.1. Shifting Property

See

$${\mathcal{G}}_x{S}_y\left[e^{cx+dy}g\left(x,y\right)\right]=\frac{s}{\left(s-c\right)\left(1-du\right)}G\left(s-c,\frac{u}{1-du}\right),$$

(8)

where $G\left(s,u\right)={\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]$.

Proof .

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[e^{cx+dy}g\left(x,y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\left(s-c\right)x-\left(\frac{1-du}{u}\right)y} g\left(x,y\right)dxdy \\ =\frac{s}{\left(s-c\right)\left(1-du\right)}\left(\frac{\left(s-c\right)\left(1-du\right)}{u}\right)\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\left(s-c\right)x-\left(\frac{1-du}{u}\right)y} g\left(x,y\right)dxdy \\ =\frac{s}{\left(s-c\right)\left(1-du\right)}G\left(s-c,\frac{u}{1-du}\right). \end{gathered}$$

□

2.3.2. Derivatives Properties

Let $G\left(s,u\right)={\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]$. Then, we have the following properties:

i.

$${\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial x}\right]=sG\left(s,u\right)-sS\left[g\left(0,y\right)\right]$$

(9)

ii.

$${\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial y}\right]=\frac{1}{u}G\left(s,u\right)-\frac{1}{u}\mathcal{G}\left[g\left(x,0\right)\right]$$

(10)

iii.

$${\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x^2}\right]=s^{2}G\left(s,u\right)-s^{2}S\left[g\left(0,y\right)\right]-sS\left[g_x\left(0,y\right)\right]$$

(11)

iv.

$${\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial y^2}\right]=\frac{1}{u^2}G\left(s,u\right)-\frac{1}{u^2}\mathcal{G}\left[g\left(x,0\right)\right]-\frac{1}{u}\mathcal{G}\left[g_y\left(x,0\right)\right]$$

(12)

v.

$${\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x\partial y}\right]=\frac{s}{u}\left(G\left(s,u\right)-S\left[g\left(0,y\right)\right]-\mathcal{G}\left[g\left(x,0\right)\right]+g\left(0,0\right)\right)$$

(13)

Here, we state the proofs of (i), (iii) and (v). We obtain the proofs of (ii) and (iv) in the same manner of the proofs of (i) and (iii), respectively.

Proof of (i).

$${\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial x}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} \frac{\partial g\left(x,y\right)}{\partial x}dxdy=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}dy \underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x}\frac{\partial g\left(x,y\right)}{\partial x}dx.$$

By integrating by parts, we obtain

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial x}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}dy \left(-g\left(0,y\right)+s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x}g\left(x,y\right)dx\right) \\ =\frac{-s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}g\left(0,y\right)dy+\frac{s^2}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x-\frac{y}{u}}g\left(x,y\right)dx \\ =sG\left(s,u\right)-sS\left[g\left(0,y\right)\right]. \end{gathered}$$

□

Proof of (iii).

$${\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x^2}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} \frac{{\partial }^{2}g\left(x,y\right)}{{\partial }^{2}x}dxdy=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}dy \underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x}\frac{{\partial }^{2}g\left(x,y\right)}{{\partial }^{2}x}dx.$$

By integrating by parts, we obtain

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x^2}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}dy \left(-g_x\left(0,y\right)-sg\left(0,y\right)+s^2\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x}g\left(x,y\right)dx\right) \\ =\frac{-s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}{g}_x\left(0,y\right)dy \frac{-s^2}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}}g\left(0,y\right)dy \\ +\frac{s^3}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s x-\frac{y}{u}}g\left(x,y\right)dx \\ =s^{2}G\left(s,u\right)-s^{2}S\left[g\left(0,y\right)\right]-sS\left[g_x\left(0,y\right)\right]. \end{gathered}$$

□

Proof of (v).

$${\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x\partial y}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} \frac{{\partial }^{2}g\left(x,y\right)}{\partial x\partial y}dxdy=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}} e^{-sx}dx \underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}} \frac{{\partial }^{2}g\left(x,y\right)}{\partial x\partial y}dy.$$

By integrating by parts, we obtain

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x\partial y}\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx}dx \left(-g_x\left(x,0\right)+\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}} \frac{\partial g\left(x,y\right)}{\partial x}dy\right) \\ =\frac{-s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx}{g}_x\left(x,0\right) dx+\frac{s}{u^2}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}\frac{\partial g\left(x,y\right)}{\partial x}dxdy \\ =-\frac{1}{u}\mathcal{G}\left[g_x\left(x,0\right)\right]+\frac{1}{u}{\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial x}\right]. \end{gathered}$$

Using the facts in Equations (2) and (9), we obtain

$${\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x\partial y}\right]=\frac{s}{u}\left(G\left(s,u\right)-S\left[g\left(0,y\right)\right]-\mathcal{G}\left[g\left(x,0\right)\right]+g\left(0,0\right)\right).$$

□

2.3.3. DARA-ST of a Periodic Function

Theorem 2.

If the DARA-ST of  $g\left(x,y\right)$ exists, where  $g\left(x,y\right)$ is a periodic function of periods  $c$ and  $d$ such that  $g\left(x+c,y+d\right)=g\left(x,y\right)$ for all  $x,y$, then

$${\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]={\left(1-e^{-sc-\frac{d}{u}}\right)}^{-1}\left(\frac{s}{u}\underset{0}{\overset{c}{{\displaystyle \int }}}\underset{0}{\overset{d}{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy\right)$$

(14)

Proof of Theorem 2.

From the definition of the DARA-ST,

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy \\ =\frac{s}{u}\underset{0}{\overset{c}{{\displaystyle \int }}}\underset{0}{\overset{d}{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy+\frac{s}{u}\underset{c}{\overset{\infty }{{\displaystyle \int }}}\underset{d}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy. \end{gathered}$$

(15)

Let $x=c+p \mathrm{and} y=d+q$ on both integrals in Equation (15). We obtain

$$\begin{gathered} G\left(s,u\right)={\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right] \\ =\frac{s}{u}\underset{0}{\overset{c}{{\displaystyle \int }}}\underset{0}{\overset{d}{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy \\ +\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s\left(c+p\right)-\frac{d+q}{u}}g\left(c+p,d+q\right)dpdq. \end{gathered}$$

(16)

Using the periodicity of the function $g\left(x,y\right)$, Equation (16) becomes

$$G\left(s,u\right)=\frac{s}{u}\underset{0}{\overset{c}{{\displaystyle \int }}}\underset{0}{\overset{d}{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy+e^{-sc-\frac{d}{u}}\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{e^{-sp-\frac{q}{u}}}g\left(p,q\right)dpdq.$$

(17)

From the definition of the DARA-ST, we obtain

$$G\left(s,u\right)=\frac{s}{u}\underset{0}{\overset{c}{{\displaystyle \int }}}\underset{0}{\overset{d}{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy+e^{-sc-\frac{d}{u}}G\left(s,u\right).$$

(18)

Equation (18) can be simplified into

$$G\left(s,u\right)={\left(1-e^{-sc-\frac{d}{u}}\right)}^{-1}\left(\frac{s}{u}\underset{0}{\overset{c}{{\displaystyle \int }}}\underset{0}{\overset{d}{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x,y\right)dxdy\right).$$

(19)

□

2.3.4. Convolution Theorem of DARA-ST

Theorem 3.

Let  $G\left(s,u\right)={\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]$. Then

$${\mathcal{G}}_x{S}_y\left[g\left(x-\delta ,y-ϵ\right)H\left(x-\delta ,y-ϵ\right)\right]=e^{-s\delta -\frac{ϵ}{y}}G\left(s,u\right),$$

(20)

where  $H\left(x,y\right)$ is the Heaviside unit step function given by

$$H\left(x-\delta ,y-ϵ\right)=\left\{\begin{array}{c}1, x>\delta , y>ϵ\\ 0, otherwise.\end{array}\right.$$

Proof of Theorem 3.

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[g\left(x-\delta ,y-ϵ\right)H\left(x-\delta ,y-ϵ\right)\right] \\ =\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} \left(g\left(x-\delta ,y-ϵ\right)H\left(x-\delta ,y-ϵ\right)\right)dxdy \\ =\frac{s}{u}\underset{\delta }{\overset{\infty }{{\displaystyle \int }}}\underset{ϵ}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}} g\left(x-\delta ,y-ϵ\right)dxdy. \end{gathered}$$

(21)

Letting $x-\delta =p \mathrm{and} y-ϵ=q$ in Equation (21), we obtain

$${\mathcal{G}}_x{S}_y\left[g\left(x-\delta ,y-ϵ\right)H\left(x-\delta ,y-ϵ\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-s\left(\delta +p\right)-\frac{ϵ+q}{u}} g\left(p,q\right)dpdq.$$

(22)

Equation (22) can be simplified into

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[g\left(x-\delta ,y-ϵ\right)H\left(x-\delta ,y-ϵ\right)\right]=e^{-s\delta -\frac{ϵ}{u}}\left(\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sp-\frac{q}{u}} g\left(p,q\right)dpdq\right) \\ =e^{-s\delta -\frac{ϵ}{y}}G\left(s,u\right). \end{gathered}$$

□

Theorem 4

(Convolution Theorem). If ${\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]=G\left(s,u\right)$ and${\mathcal{G}}_x{S}_y\left[k\left(x,y\right)\right]=K\left(s,u\right)$, then

$${\mathcal{G}}_x{S}_y\left[\left(g\ast \ast k\right)\left(x,y\right)\right]=\left(\frac{u}{s}\right)G\left(s,u\right)K\left(s,u\right)$$

(23)

where

$$\left(g\ast \ast k\right)\left(x,y\right)=\underset{0}{\overset{x}{{\displaystyle \int }}}\underset{0}{\overset{y}{{\displaystyle \int }}}g\left(x-\delta ,y-ϵ\right)k\left(\delta ,ϵ\right)d\delta dϵ.$$

Proof of Theorem 4.

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[\left(g\ast \ast k\right)\left(x,y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}\left(g\ast \ast k\right)\left(x,y\right)dxdy \\ =\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}\left(\underset{0}{\overset{x}{{\displaystyle \int }}}\underset{0}{\overset{y}{{\displaystyle \int }}}g\left(x-\delta ,y-ϵ\right) k\left(\delta ,ϵ\right)d\delta dϵ\right)dxdy. \end{gathered}$$

(24)

Using the Heaviside unit step function, Equation (24) can be written as

$${\mathcal{G}}_x{S}_y\left[\left(g\ast \ast k\right)\left(x,y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}\left(\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}g\left(x-\delta ,y-ϵ\right)K\left(x-\delta ,y-ϵ\right) k\left(\delta ,ϵ\right)d\delta dϵ\right)dxdy.$$

Thus

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[\left(g\ast \ast k\right)\left(x,y\right)\right] \\ =\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}k\left(\delta ,ϵ\right)d\delta dϵ\left(\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x-\delta ,y-ϵ\right)K\left(x-\delta ,y \\ -ϵ\right) dxdy\right)=\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}k\left(\delta ,ϵ\right)d\delta dϵ\left(e^{-sx-\frac{y}{u}}G\left(s,u\right)\right) \\ =G\left(s,u\right)\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}k\left(\delta ,ϵ\right) e^{-sx-\frac{y}{u}} d\delta dϵ=\frac{u}{s}G\left(s,u\right)K\left(s,u\right). \end{gathered}$$

□

Now, we summarize the previous results in

Table 1. The DARA-STs of some basic functions.

$\mathit{g}\left(\mathit{x},\mathit{y}\right)$	${\mathcal{G}}_{\mathit{x}}{\mathit{S}}_{\mathit{y}}\left[\mathit{g}\left(\mathit{x},\mathit{y}\right)\right]=\mathit{G}\left(\mathit{s},\mathit{u}\right)$
$1$	$1$
$x^a{y}^b$	$s^{-a}\mathsf{\Gamma }\left(a+1\right)u^b\mathsf{\Gamma }\left(b+1\right)$
$e^{ax+yb}$	$\frac{s}{\left(s-a\right)\left(1-bu\right)}$
$e^{i \left(ax+yb\right)}$	$\frac{i s}{\left(s-i a\right)\left(bu+i\right)}$
$\sin \left(ax+yb\right)$	$\frac{s \left(a+b s u\right)}{\left(a^2+s^2\right)\left(b^2 u^2+1\right)}$
$\cos \left(ax+yb\right)$	$\frac{s \left(s-abu\right)}{\left(a^2+s^2\right)\left(b^2 u^2+1\right)}$
$\sinh \left(ax+yb\right)$	$\frac{s \left(a+b s u\right)}{\left(a^2-s^2\right)\left(b^2 u^2-1\right)}$
$\cosh \left(ax+yb\right)$	$\frac{s \left(s+abu\right)}{\left(a^2-s^2\right)\left(b^2 u^2-1\right)}$
$J_0\left(c\sqrt{xy} \right)$	$\frac{4s}{4s+c^{2}u}$
$g\left(x-\delta ,y-ϵ\right)$ $H\left(x-\delta ,y-ϵ\right)$	$e^{-s\delta -\frac{ϵ}{y}}G\left(s,u\right)$
$\left(g\ast \ast k\right)\left(x,y\right)$	$\left(\frac{u}{s}\right)G\left(s,u\right)K\left(s,u\right)$

.

3. Applications of DARA-ST

This section contains three main parts. In the first one, we apply the DARA-ST to a family of PDEs and obtain a simple formula for the general solution. In the second part, the DARA-ST is implemented to solve IEs of a special kind. Finally, the DARA-ST is used to obtain a solution of a functional equation.

3.1. DARA-ST for Solving PDEs

Consider the nonhomogeneous PDE of the form

$$A{g}_{xx}+B{g}_{yy}+C{g}_x+D{g}_y+Eg\left(x,y\right)=u\left(x,y\right)$$

(25)

with the initial conditions (ICs)

$$g\left(x,0\right)=f_1\left(x\right), g_y\left(x,0\right)=f_2\left(x\right)$$

(26)

and the boundary conditions (BCs)

$$g\left(0,y\right)=h_1\left(y\right), g_x\left(0,y\right)=h_2\left(y\right),$$

(27)

where $g\left(x,y\right)$ is the unknown function; $A,B,C,D$ and $E$ are constants; and $u\left(x,y\right)$ is the source term.

The main idea of this technique is to apply the DARA-ST to Equation (25) and simplify the obtained one. The single ARA transform is applied to the conditions in Equation (26)

$$\mathcal{G}\left[f_1\left(x\right)\right]=F_1\left(s\right), \mathcal{G}\left[f_2\left(x\right)\right]=F_2\left(s\right),$$

and the single Sumudu transform is applied to the conditions in Equation (27)

$$\mathcal{G}\left[h_1\left(y\right)\right]=H_1\left(u\right), \mathcal{G}\left[h_2\left(y\right)\right]=H_2\left(u\right).$$

By applying the DARA-ST to Equation (25), we have

$$\begin{gathered} A {\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial x^2}\right]+B {\mathcal{G}}_x{S}_y\left[\frac{{\partial }^{2}g\left(x,y\right)}{\partial y^2}\right]+C {\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial x}\right] \\ +D {\mathcal{G}}_x{S}_y\left[\frac{\partial g\left(x,y\right)}{\partial y}\right]+E {\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]={\mathcal{G}}_x{S}_y\left[u\left(x,y\right)\right]. \end{gathered}$$

(28)

The properties of the derivatives in Equations (9)–(15), ICs (26) and BCs (27) yield

$$\begin{gathered} A\left( s^{2}G\left(s,u\right)-s^2{H}_1\left(u\right)-s{H}_2\left(u\right)\right)+B\left(\frac{1}{u^2}G\left(s,u\right)-\frac{1}{u^2}{F}_1\left(s\right)-\frac{1}{u}{F}_2\left(s\right)\right)+C \left(sG\left(s,u\right)-s{H}_1\left(u\right)\right) \\ +D\left(\frac{1}{u}G\left(s,u\right)-\frac{1}{u}{F}_1\left(s\right)\right)+E G\left(x,y\right)=U\left(x,y\right). \end{gathered}$$

(29)

After simple computations, Equation (29) can be simplified as

$$G\left(s,u\right)=\frac{U\left(s,u\right)+\left(A s^2+Cs\right)H_1\left(u\right)+As H_2\left(u\right)+\left(\frac{B}{u^2}+\frac{D}{u}\right)F_1\left(s\right)+\frac{B}{u}{F}_2\left(s\right)}{A{s}^2+\frac{B}{u^2}+Cs+\frac{D}{u}+E}.$$

(30)

To find the solution of Equation (25) in the original space, take the inverse DARA-ST to both sides of Equation (30) to obtain

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{U\left(s,u\right)+\left(A s^2+Cs\right)H_1\left(u\right)+As H_2\left(u\right)+\left(\frac{B}{u^2}+\frac{D}{u}\right)F_1\left(s\right)+\frac{B}{u}{F}_2\left(s\right)}{A{s}^2+\frac{B}{u^2}+Cs+\frac{D}{u}+E}\right].$$

(31)

3.2. DARA-ST for Solving IEs

Consider the IE of the form

$$g\left(x,y\right)=f\left(x,y\right)+\lambda \underset{0}{\overset{x}{{\displaystyle \int }}}\underset{0}{\overset{y}{{\displaystyle \int }}}g\left(x-t,y-v\right)k\left(t,v\right)dtdv,$$

(32)

where $g\left(x,y\right)$ is the unknown function, $\lambda$ is a real number, and $f\left(x,y\right)$ and $k\left(x,y\right)$ are known functions.

To obtain the solution of the IE (32), we apply the DARA-ST to Equation (32) to obtain

$$G\left(s,u\right)=F\left(s,u\right)+\lambda {\mathcal{G}}_x{S}_y\left[\left(g\ast \ast k\right)\left(x,y\right)\right].$$

(33)

Using Theorem 4 of the convolution theorem, Equation (33) can be simplified as

$$G\left(s,u\right)=F\left(s,u\right)+\lambda \left(\frac{u}{s}\right)G\left(s,u\right)K\left(s,u\right).$$

where ${\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\right]=G\left(s,u\right)$, ${\mathcal{G}}_x{S}_y\left[k\left(x,y\right)\right]=K\left(s,u\right)$ and ${\mathcal{G}}_x{S}_y\left[f\left(x,y\right)\right]=F\left(s,u\right)$.

Consequently

$$G\left(s,u\right)=\frac{F\left(s,u\right)}{1-\lambda \frac{u}{s}K\left(s,u\right)}.$$

(34)

Hence, by running the inverse DARA-ST on Equation (35), we obtain

$$\begin{gathered} g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{F\left(s,u\right)}{1-\lambda \frac{u}{s}K\left(s,u\right)}\right]={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{1}{\frac{u}{s}-\lambda \frac{u^2}{s^2}K\left(s,u\right)}\frac{u}{s}F\left(s,u\right)\right] \\ ={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{u}{s}F\left(s,u\right)M\left(s,u\right)\right]=\left(g\ast \ast m\right)\left(x,y\right) \\ =\underset{0}{\overset{x}{{\displaystyle \int }}}\underset{0}{\overset{y}{{\displaystyle \int }}}f\left(x-t,y-v\right)m\left(t,v\right)dtdv, \end{gathered}$$

(35)

where $M\left(s,u\right)=\frac{1}{\frac{u}{s}-\lambda \frac{u^2}{s^2}K\left(s,u\right)}$ and $m\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[M\left(s,u\right)\right].$

Thus, we obtain the general solution of the IE (32).

3.3. DARA-ST for Solving Functional Equations

Consider the Cauchy functional equation of the form

$$g\left(x+y\right)=g\left(x\right)+g\left(y\right),$$

(36)

where $g$ is an unknown function.

By applying the DARA-ST to the left-hand side of Equation (36), to obtain

$$\begin{gathered} {\mathcal{G}}_x{S}_y\left[g\left(x+y\right)\right]=\frac{s}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx-\frac{y}{u}}g\left(x+y\right)dxdy \\ =\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}} dy s\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-sx}g\left(x+y\right)dx=\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}} \mathcal{G}\left[g\left(x+y\right)\right]dy. \end{gathered}$$

(37)

Using the fact in Equation (4), Equation (37) becomes

$${\mathcal{G}}_x{S}_y\left[g\left(x+y\right)\right]=\frac{1}{u}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{y}{u}} e^{sy}G\left(s\right)dy=G\left(s\right) S\left[e^y\right]=\frac{G\left(s\right)}{1-su},$$

where $G\left(s\right)=\mathcal{G}\left[g\left(x\right)\right]$.

Now, we apply the DARA-ST to Equation (36), to obtain

$$G\left(s\right)\frac{1}{1-su}=G\left(s\right)+G\left(u\right),$$

(38)

where $\mathcal{G}\left[g\left(x\right)\right]=G\left(s\right)$ and $S\left[g\left(y\right)\right]=G\left(u\right).$

By simplifying Equation (38), we have

$$sG\left(s\right)=\frac{G\left(u\right)}{u}.$$

Two functions of two variables are equal, which implies that

$$sG\left(s\right)=\frac{G\left(u\right)}{u}=k, k\in ℝ.$$

Hence, one can obtain

$$G\left(s\right)=\frac{k}{s}.$$

By applying the inverse ARA transform, we have

$$g\left(x\right)={\mathcal{G}}^{-1}\left[\frac{k}{s}\right]=kx$$

The solution of functional Equation (36) is $g\left(x\right)=kx$, for some constant $k$.

4. Illustrative Examples

In this section, some examples are presented and solved by applying the DARA-ST. The efficiency and simplicity of the proposed technique are illustrated in solving these examples and obtaining the exact solutions.

Example 1.

Consider the wave equation

$$g_{yy}\left(x,y\right)=g_{xx}\left(x,y\right), x\ge 0, y\ge 0,$$

(39)

with the ICs

$$g\left(x,0\right)=\sin x, g_y\left(x,0\right)=2,$$

(40)

and the BCs

$$g\left(0,y\right)=2y, g_x\left(0,y\right)=\cos y.$$

(41)

Solution. 

Here, we have

$$f_1\left(x\right)=\sin x, f_2\left(x\right)=2$$

(42)

$$h_1\left(y\right)=2y, h_2\left(y\right)=\cos y.$$

(43)

By applying the single ARA transform to $f_1\left(x\right)$ and $f_2\left(x\right)$ in Equation (42), we obtain

$$F_1\left(s\right)=\frac{s}{s^2+1}, F_2\left(s\right)=2.$$

By applying the single Sumudu transform to $h_1\left(y\right)$ and $h_2\left(y\right)$ in Equation (43), we obtain

$$H_1\left(s\right)=2u, H_2\left(s\right)=\frac{1}{u^2+1}.$$

By substituting the values of functions $F_1,F_2,H_1$ and $H_2$ in the general formula in Equation (31), we obtain

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[2u+\frac{s}{\left(s^2+1\right)\left(u^2+1 \right)}\right]=2y+\sin x\cos y.$$

Figure 1 below shows the solution of wave Equation (39) with ICs (40) and BCs (41).

Example 2.

Consider the heat equation

$$g_y\left(x,y\right)=g_{xx}\left(x,y\right)-3g\left(x,y\right)+3, x\ge 0, y\ge 0,$$

(44)

with the IC

$$g\left(x,0\right)=1+\sin x,$$

(45)

and the BCs

$$g\left(0,y\right)=1, g_x\left(0,y\right)=e^{-4y}.$$

(46)

Solution. 

Here, we have

$$f_1\left(x\right)=1+\sin x,$$

(47)

$$h_1\left(y\right)=2y, h_2\left(y\right)=\cos y.$$

(48)

By applying the single ARA transform to $f_1\left(x\right)$ in Equation (47), we obtain

$$F_1\left(s\right)=1+\frac{s}{s^2+1}.$$

By applying the single Sumudu transform to $h_1\left(y\right)$ and $h_2\left(y\right)$ in Equation (48), we obtain

$$H_1\left(s\right)=1, H_2\left(s\right)=\frac{1}{1+4u}.$$

Substituting the values of functions $F_1,H_1$ and $H_2$ in the general formula in Equation (31), to obtain

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[1+\frac{s}{\left(s^2+1\right)\left(1+4u \right)}\right]=1+e^{-4y}\sin x.$$

Figure 2 below shows the solution of heat Equation (44) with the IC (45) and the BCs (46).

Example 3.

Consider the telegraph equation

$$g_{xx}\left(x,y\right)=g_{yy}\left(x,y\right)+g_y\left(x,y\right)-g\left(x,y\right), x\ge 0, y\ge 0,$$

(49)

with the ICs

$$g\left(x,0\right)=e^x, g_y\left(x,0\right)=-2e^x,$$

(50)

and the BCs

$$g\left(0,y\right)=e^{-2y}, g_x\left(0,y\right)=e^{-2y}.$$

(51)

Solution. 

Here, we have

$$f_1\left(x\right)=e^x, f_2\left(x\right)=-2e^x$$

(52)

$$h_1\left(y\right)=e^{-2y}, h_2\left(y\right)=e^{-2y}.$$

(53)

By applying the single ARA transform to $f_1\left(x\right)$ and $f_2\left(x\right)$ in Equation (52), we obtain

$$F_1\left(s\right)=\frac{s}{s-1}, F_2\left(s\right)=\frac{-2s}{s-1}.$$

By applying the single Sumudu transform to $h_1\left(y\right)$ and $h_2\left(y\right)$ in Equation (53), we obtain

$$H_1\left(s\right)=\frac{1}{1+2u}, H_2\left(s\right)=\frac{1}{1+2u}.$$

By substituting the values of functions $F_1, F_2, H_1$ and $H_2$ in the general formula in Equation (31), to obtain

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{s}{\left(s-1\right)\left(1+2u\right)}\right]=e^{x-2y}.$$

Figure 3 below shows the solution of telegraph Equation (49) with ICs (50) and BCs (51).

Example 4.

Consider the advection-diffusion equation

$$g_y\left(x,y\right)=A{g}_{xx}\left(x,y\right)+B{g}_x\left(x,y\right), x\ge 0,y\ge 0,$$

(54)

with the IC

$$g\left(x,0\right)=f_1\left(x\right),$$

(55)

and the BCs

$$g\left(0,y\right)=h_1\left(y\right), g_x\left(0,y\right)=h_2\left(y\right).$$

(56)

Solution. 

By applying the single ARA transform to $f_1\left(x\right)$ in Equation (55), we obtain

$$F_1\left(s\right)=\mathcal{G}\left[f_1\left(x\right)\right].$$

By applying the single Sumudu transform to $h_1\left(y\right)$ and $h_2\left(y\right)$ in Equation (56), we obtain

$$H_1\left(s\right)=S\left[h_1\left(y\right)\right], H_2\left(s\right)=S\left[h_2\left(y\right)\right].$$

By substituting the values of functions $F_1,H_1$ and $H_2$ in the general formula in Equation (31), to obtain

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{F_1\left(s\right)-A{s}^{2}u{H}_1\left(u\right)-Bsu{H}_1\left(u\right)-Asu{H}_2\left(u\right)}{1-A{s}^{2}u-Bsu}\right].$$

(57)

By putting $A=1, B=-1, f_1\left(x\right)=e^x-x, h_1\left(y\right)=1+y$ and $h_2\left(y\right)=0$ in Equations (54)–(56), we obtain the solution

$$\begin{gathered} g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{\frac{s}{s-1}-\frac{1}{s}-s^{2}u\left(1+u\right)+su\left(1+u\right)}{1-s^{2}u+su}\right] \\ ={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{s}{s-1}-\frac{1}{s}+u\right]=e^x-x+y. \end{gathered}$$

Figure 4 below shows the solution of advection–diffusion Equation (54) with IC $g\left(x,0\right)=e^x-x$ and BCs $g\left(0,y\right)=1+y, g_x\left(0,y\right)=0$.

Example 5.

Consider the Klein-Gordon equation

$$g_{yy}\left(x,y\right)-Ag\left(x,y\right)=B{g}_{xx}\left(x,y\right)+u\left(x,y\right),$$

(58)

with the ICs

$$g\left(x,0\right)=f_1\left(x\right), g_2\left(x,0\right)=f_2\left(x\right),$$

(59)

and the BCs

$$g\left(0,y\right)=h_1\left(y\right), g_x\left(0,y\right)=h_2\left(y\right).$$

(60)

Solution. 

By applying the single ARA transform to $f_1\left(x\right)$ and $f_2\left(x\right)$ in Equation (59), we obtain

$$F_1\left(s\right)=\mathcal{G}\left[f_1\left(x\right)\right], F_2\left(s\right)=\mathcal{G}\left[f_2\left(x\right)\right].$$

By applying the single Sumudu transform to $h_1\left(y\right)$ and $h_2\left(y\right)$ in Equation (60), we obtain

$$H_1\left(s\right)=S\left[h_1\left(y\right)\right], H_2\left(s\right)=S\left[h_2\left(y\right)\right].$$

By substituting the values of functions $F_1,F_2,H_1$ and $H_2$ in the general formula in Equation (31), to obtain

$$\begin{gathered} g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{-U\left(s,u\right)+B{s}^2{H}_1\left(u\right)+Bs{H}_2\left(u\right)-\frac{1}{u^2}{F}_1\left(s\right)-\frac{1}{u}{F}_2\left(s\right)}{B{s}^2-\frac{1}{u}+Au}\right] \\ ={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{F_1\left(s\right)+u{F}_2\left(s\right)-B{s}^2{u}^2{H}_1\left(u\right)-Bs{u}^2{H}_2\left(u\right)+u^{2}U\left(s,u\right)}{1-B{s}^2{u}^2-A{u}^2}\right]. \end{gathered}$$

(61)

By putting $A=2, B=1, u\left(x,y\right)=2\sin x\sin y, f_1\left(x\right)=0, f_2\left(x\right)=\sin x, h_1\left(y\right)=$0 and $h_2\left(y\right)=\sin y$, in Equations (58)–(60), we obtain the solution

$$\begin{gathered} g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{\frac{us}{s^2+1}+0-\frac{s{u}^3}{u^2+1}-\frac{2u^{3}s}{\left(s^2+1\right)\left(u^2+1\right)}}{1-2u^2-s^2{u}^2}\right] \\ ={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{su}{\left(s^2+1\right)\left(u^2+1\right)}\right]=\sin x\sin y. \end{gathered}$$

Figure 5 below shows the solution of Klein–Gordon Equation (58) with ICs (59) and BCs (60).

Example 6. Consider the following IE

$$g\left(x,y\right)=a-\lambda \underset{0}{\overset{x}{{\displaystyle \int }}}\underset{0}{\overset{y}{{\displaystyle \int }}}g\left(t,v\right)dt dv,$$

(62)

where  $a$  and  $\lambda$  are two constants,$g\left(x,y\right)$is the unknown function.

Solution. By applying the DARA-ST to Equation (62) and using the convolution theorem, we obtain

$$G\left(s,u\right)={\mathcal{G}}_x{S}_y\left[a\right]-\lambda {\mathcal{G}}_x{S}_y\left[1\ast \ast g\left(x,y\right)\right]=a-\lambda \frac{u}{s}G\left(s,u\right).$$

(63)

By simplifying Equation (63), we obtain

$$G\left(s,u\right)=\frac{as}{s+\lambda u}.$$

(64)

Now, running the inverse DARA-ST on Equation (64), we obtain the solution of Equation (62) as

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[\frac{as}{s+\lambda u}\right]=a J_0\left(2\sqrt{\lambda xy}\right).$$

Figure 6 below shows the solution of IE (62).

Example 7.

Consider the following IE

$$\underset{0}{\overset{x}{{\displaystyle \int }}}\underset{0}{\overset{y}{{\displaystyle \int }}}g\left(x-t,y-v\right)g\left(t,v\right)dt dv=a^2,$$

(65)

where$a$is any constant.

Solution. 

By applying the DARA-ST to Equation (65), we obtain

$${\mathcal{G}}_x{S}_y\left[g\left(x,y\right)\ast \ast g\left(x,y\right)\right]={\mathcal{G}}_x{S}_y\left[a^2\right].$$

(66)

Using the convolution theorem, Equation (66) becomes

$$\frac{u}{s}G\left(s,u\right)G\left(s,u\right)=a^2$$

$$G\left(s,u\right)=\sqrt{\frac{a^{2}s}{u}}=a\sqrt{\frac{s}{u}}.$$

By running the inverse DARA-ST, we obtain the solution of Equation (65) as

$$g\left(x,y\right)={\mathcal{G}}_x^{-1}{S}_y^{-1}\left[a\sqrt{\frac{s}{u}}.\right]=\frac{a}{\pi }\frac{1}{\sqrt{xy}}.$$

Figure 7 below shows the solution of IE (65).

5. Observations about the New Scheme

In this part of the article, we present some observations about the new approach, the DARA-ST:

• The main purpose of this research study is to obtain a new technique for solving PDEs by turning them into algebraic ones using the new double transform and then obtain the solution in a new space. After that, running the inverse transform, we obtain the solutions of the target equations in the original space. Moreover, solutions of a special kind of IEs are presented;
• The usage of the single ARA and the Sumudu transforms for solving PDEs with an unknown function of two variables are useless and hard sometimes. Hence, the new DARA-ST can help researchers to overcome such problems;
• DARA-ST produces a quick convergence of the exact solution without any complex computation compared with other techniques;
• For solving nonhomogeneous equations, the source terms must satisfy the existence conditions of the DARA-ST;
• A disadvantage of this transform is that it cannot be applied directly to solve nonlinear PDEs. It needs to be combined with other iterative methods that are often used, such as the homotopy method or the differential transform method.

6. Conclusions

In this research study, a new transform approach called the DARA-ST is presented. Basic properties of the proposed double transform are introduced and implemented to obtain solutions of PDEs, IEs and functional equations. The usage of the new double transform is illustrated by solving some interesting examples and obtaining the exact solutions of them. New results of the DARA-ST will be discussed in the future and implemented for solving fractional DEs and FPDEs with variable coefficients.