Towards a new triple integral transform (Laplace–ARA–Sumudu) with applications

Abstract

The main objective of this work is to introduce a novel generalization of double transformations called the triple Laplace–ARA–Sumudu transform (TLARAST). This hybrid transformation extends the concepts of double Laplace–Sumudu, double Laplace–ARA and double ARA–Sumudu transforms into a triple hybrid transform. The article provides the definition of TLARAST and investigates its fundamental properties, including existence, inverses and related theorems. Furthermore, new results concerning TLARAST for partial derivatives and the theorem of multi-convolution are introduced and discussed. The practicality and efficacy of TLARAST are demonstrated by applying it to solve various types of partial differential equations with significant applications in physics and other scientific fields, such as the heat equation, Laplace equation, Poisson equation and wave equation. The solutions are illustrated through figures created using Mathematica software. Overall, this study underscores the usefulness and efficiency of TLARAST in solving partial differential equations involving multiple variables.

Keywords:

• ARA transform
• double transform
• integral transform
• Laplace transform
• partial differential equations
• Sumudu transform
• triple transform

1. Introduction

An integral transform is a specific type of mathematical operator (Debnath & Bhatta, 2014), functioning as a mapping or function that operates on elements in one space to produce different elements in another space through integration. This transformation allows certain characteristics of the original function to be discerned and manipulated more effectively than in its original space. By using the inverse of the integral transform (Akgül, Akgül, & Yavuz, 2021; Davies, 2002), the transformed function can be brought back to its original function space.

Partial differential equations (PDEs) hold significant interest for researchers, finding numerous real-life applications across various scientific fields, including physics (Evans, 2022; Mavi, Bekar, Haghighat, & Madenci, 2023), where they are utilized to solve problems like heat and Poisson equations. The heat equation describes the time-dependent diffusion of heat, with its solution representing the temperature of a body at time t and location x, considering specific initial or boundary conditions and a heat source. On the other hand, the Poisson equation (Abbas, 2020; Budhiraja, Chen, & Dupuis, 2013) addresses the stationary or equilibrium distribution of temperature, characterizing a situation where heat is no longer flowing, and it seeks to find a function of x alone that describes the temperature under these circumstances. As a result, mathematicians continually explore new techniques for solving a wide range of PDEs. In this study, we focus on integral transforms as a highly powerful method for solving differential equations, which will be thoroughly explored in this article.

The mechanism of an integral transform to produce a new function $F(y)$ is based on integrating the product of an original function $f(x)$ and a given kernel function $K(x, y)$ between appropriate limits $l_1$ and ${ l}_2.$ The process can be presented by the equation $$F\left(y\right)={\int }_{l_1}^{l_2}f\left(x\right)K\left(x,\mathrm{ }y\right)\mathrm{d}x.$$

Integral transforms are valuable for the simplicity that can be fulfilled, most often in dealing with differential equations subject to initial or boundary conditions (Cotta, Naveira-Cotta, & Knupp, 2016; Patil, 2018). It can be performed to a differential equation in a manner such that one of the dimensions is reduced from derivative operations to algebraic operations. This means that if we start with an ordinary differential equation (ODE), the integral transform will convert the equation to an algebraic equation. For a 2D problem, the integral transform maps the PDE to an ODE, and so on. Proper selection of the class of the integral transform facilitates not only the derivatives in a complicated differential equation but also the conditions values into terms of an equation that can be solved easily. The actual solution of the discussed problem can be gained using the inverse transformation.

There are many kinds of integral transforms with a wide variety of purposes, including engineering, physics, image and signal processing, mathematical statistics and functional analysis. Several single transforms have been developed in the literature. Laplace transform was proposed by Laplace in 1780. It is considered as one of the oldest and most widespread integral transforms. It transfers a function of a real variable (usually, in the time domain) to a function of a complex variable in the complex frequency domain, known as the s-domain. This approach has many advantages and applications that could utilize it in mechanical and electrical engineering problems (Wolf, 2013).

Another integral transform, called Sumudu transform, was introduced by Watugula (1993). Sumudu transform is applicable for both ODEs and PDEs, in particular in engineering control problems, see Belgacem and Karaballi (2006). One of the most strength points for Sumudu transform is that it has units preserving properties. Thus, it can be used to solve problems without resorting to the frequency domain.

A novel integral transformation, called ARA transform, was presented by Saadeh, Qazza, and Burqan (2020). It is a powerful transform that can be considered as a generalization for some variants of the classical transforms. It can also be utilized to handle ordinary and fractional differential equations, see Saadeh (2022a, 2022b). For more details on other single integral transforms, one may refer to the novel transform (Atangana & Kiliçman, 2013), Elzaki transform (Elzaki, 2011), natural transform (Khan & Khan, 2008), formable transform (Saadeh & Ghazal, 2021), Aboodh transform (Aboodh, Abdullahi, & Nuruddeen, 2017) and Laplace transform (Aghili & Parsa Moghaddam, 2011).

Double transformations have been developed to solve PDEs and obtain results outperforming numerical methods, see for example Dhunde, Bhondge, and Dhongle (2013). In fact, the extensions of double transforms are considered by many authors, such as double Laplace transform (Dhunde & Waghmare, 2017; Eltayeb & Kiliçman, 2013), double Sumudu transform (Ahmad Gani, Ahmad, & Jain, 2017; Eltayeb & Kiliçman, 2010; Tchuenche & Mbare, 2007), double Laplace–ARA transform (DLARAT; Sedeeg, Mahamoud, & Saadeh, 2022), double Laplace–Sumudu transform (DLST; Ahmed, 2021; Ahmed, Elzaki, Elbadri, & Mohamed, 2021; Ahmed, Elzaki, & Hassan, 2020), double ARA–Sumudu transform (DARAST; Kiliçman & Omran, 2017), double Elzaki (Hassan & Elzaki, 2020), double Shehu (Alfaqeih & Misirli, 2020), among others (Elzaki, Ahmed, Areshi, & Chamekh, 2022; Wang, Chiang, Liu, & Lee, 2001).

Recently, triple transformations are suggested. PDEs involving functions of three variables can be solved via triple integral transforms. Triple Laplace transform was proposed in 2013 (see Atangana, 2013; Khan, Shah, Khan, & Khan, 2017), and it was used to solve linear and non-linear third-order PDEs; more results on triple Laplace transform can be found in Bhanotar and Kaabar (2021). Also, in literature we have the triple Sumudu transform (Eltayeb & Mesloub, 2021; Kılıçman & Gadain, 2010), triple Elzaki transform (Elzaki & Mousa, 2019), triple natural (Neto, Quaresma, & Cotta, 2002) and lastly, in Saadeh (2023), Saadeh has presented a general formula for triple transforms of the Laplace kind.

Moreover, some researchers have investigated multi-dimensional transforms, such as multi-Wavlet transform (Rinoshika & Rinoshika, 2020), multi Laplace transform and others (Anjum, Suleman, Lu, He, & Ramzan, 2020; Fang, Nadeem, Islam, & Iambor, 2023; Nadeem, Islam, Karim, Mureşan, & Iambor, 2023; Tao, Anjum, & Yang, 2023).

The aim of the presented paper is to introduce a hybrid triple transform that combines Laplace, ARA and Sumudu transforms. We present the main properties of the transform such as conditions of existence, linearity and shifting properties. In addition, some of the main findings and results of this triple transform are used for transforming some popular functions. We also discuss the theory of convolution and the derivative characteristics of the new transform. Moreover, the new transform is applied to solve wide classes of PDEs. The importance of this study lies in developing one integral transform that combines the characteristics and advantages of the above three transforms, which leads to consider it as a generalization of many classes of double integral transform.

This study introduces a novel combination of three powerful transforms from the literature: Laplace, Sumudu and ARA transformations. While the last two transforms share similarities with Laplace, they possess additional properties and sometimes offer simpler solutions to problems. By incorporating these three transforms, the new approach proves effective in solving PDEs with multiple variables and higher orders. Additionally, it demonstrates applicability in solving various types of integral equations. Compared to alternative methods, the new technique requires less time and effort to obtain direct solutions without the need for linearization or homogenization. Moreover, utilizing triple Laplace–ARA–Sumudu transform (TLARAST) provides exact solutions, eliminating the necessity for comparison with other analytical methods.

Furthermore, it is worth mentioning that the results obtained in this article, as well as those from other triple transforms of the Laplace category, can all be considered as special cases of Elzaki et al. (2022). In that work, the author extensively studied the most popular relations of triple transforms of the Laplace kind. In this study, we focus on the strengths of the three proposed integral transforms, combining the advantages of each to create a triple transform that can be applied to tackle new problems.

The structure of the article is organized as follows: In Section 2, we provide the basic definitions and theorems for Laplace (Wang et al., 2001), ARA and Sumudu transforms. Section 3 presents fundamental characteristics of DLST, DARAST and DLARAT. The definition of the TLARAST and its main properties are discussed in Section 4. In Section 5, we apply the TLARAST to various types of PDEs. Finally, the conclusion section illustrates the obtained results.

2. Facts and properties of single transforms

This part of the study presents the basic properties of the single transforms: Laplace, Sumudu and ARA transforms.

2.1. Laplace transform

Herein, we introduce the definition and basic facts about Laplace transform.

Definition 1. Let $f(x)$ be a continuous function of $x$ specified for $x>0.$Then, Laplace transform of $f(x),$ denoted by $\mathcal{L}[f(x)],$ is defined by $$\mathcal{L}\left[f\left(x\right)\right]=F\left(r\right)={\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[f\left(x\right)\right]\mathrm{d}x,\mathrm{ }r>0.$$

The inverse Laplace transform can be presented as $${\mathcal{L}}^{-1}\left[F\left(r\right)\right]=\frac{1}{2\mathrm{\pi }i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{rx}F\left(r\right)\mathrm{d}r=f\left(x\right)\mathrm{ },\mathrm{ }x>0\mathrm{ }.$$

Theorem 1.

If $f\left(x\right)$ is a continuous function on a given interval $[0,\infty ),$ and is of exponential order $\alpha ,$ then $\mathcal{L}\left[f(x)\right]$ exists for $Re(r)>\alpha$ and satisfies $$\left|f\left(x\right)\right|\le M{e}^{\alpha x},$$ where $M>0,$ then, Laplace transform exists for $Re\left(r\right)>\alpha .$

Now, we present some properties of Laplace transform.

Suppose that $F(r)=\mathcal{L}\left[f\left(x\right)\right]$ and $G\left(r\right)=\mathcal{L}\left[g\left(x\right)\right]$ and $a,b\in \mathcal{R},$ then the following properties hold:

• $\mathcal{L}\left[a\mathrm{ }f\left(x\right)+b\mathrm{ }g(x)\right]=a\mathcal{ }\mathcal{L}\left[f\left(x\right)\right]+b\mathcal{ }\mathcal{L}\left[g(x)\right].$

• ${\mathcal{L}}^{-1}\left[a\mathrm{ }F\left(r\right)+b\mathrm{ }G\left(r\right)\right]=a\mathrm{ }{\mathcal{L}}^{-1}\left[F\left(r\right)\right]+b\mathrm{ }{\mathcal{L}}^{-1}\left[G\left(r\right)\right].$

• $\mathcal{L}\left[x^b\right]=\frac{\Gamma (b+1)}{r^{b+1}}\mathrm{ },b\ge 0,\text{ and Γ}$ is the regular gamma function.

• $\mathcal{L}\left[e^{bx}\right]=\frac{1}{r-b}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{L}\left[\sin \hspace{0.17em}bx\right]=\frac{b}{r^2+b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{L}\left[\cos \hspace{0.17em}bx\right]=\frac{s}{r^2+b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{L}\left[\sinh \hspace{0.17em}bx\right]=\frac{b}{r^2-b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{L}\left[\cosh \hspace{0.17em}bx\right]=\frac{s}{r^2\text{-}\mathrm{ }{b}^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{L}\left[f\prime \left(x\right)\right]=rF\left(r\right)-f\left(0\right).$

• $\mathcal{L}\left[f^{(n)}\left(x\right)\right]=r^{n}F\left(r\right)-{\sum }_{k=1}^n{r}^{n-k}{f}^{\left(k-1\right)}\left(0\right).$

2.2. ARA transform

In this section, we present the basic facts regarding the ARA transform.

Definition 2. The ARA transformation of order $n$ of a continuous function $f(y)$ on a given interval $(0, \infty )$ is given by $${\mathcal{G}}_n\left[f\left(y\right)\right]\left(p\right)=F\left(n,p\right)=p{\int }_0^{\mathrm{\infty }}{y}^{n-1}{e}^{-p\mathrm{y}}f\left(y\right)\mathrm{d}y,\mathrm{ }p>0.$$

In this research, we consider ${\mathcal{G}}_1\left[f\left(y\right)\right],$ ARA transformation of order 1 is given by $${\mathcal{G}}_1\left[f\left(y\right)\right]\left(p\right)=F\left(p\right)=p{\int }_0^{\mathrm{\infty }}{e}^{-py}f\left(y\right)\mathrm{d}y,p>0.$$

In the rest of the study, we denote ${\mathcal{G}}_1\left[f\left(y\right)\right]$ by $\mathcal{G}\left[f\left(y\right)\right].$

The inverse ARA transformation is expressed as $${\mathcal{G}}^{-1}\left[F\left(p\right)\right]=\frac{1}{2\mathrm{\pi }i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}\frac{e^{py}}{p}F\left(p\right)\mathrm{d}p=f\left(y\right).$$

Theorem 2.

(Existence conditions) If the function $f(y)$ is continuous in a given domain $0\le y\le \alpha$ and satisfying $$\left|y^{n-1}f\left(y\right)\right|\le \mathrm{M}{e}^{\alpha y},$$ where $\mathrm{M}$ is a positive constant, then, for all $p>\alpha$ the ARA transformation of order $n$ of the function $f(y)$ exists.

Suppose that $F(p)=\mathcal{G}\left[f\left(y\right)\right]$ and $G\left(p\right)=\mathcal{G}\left[g\left(y\right)\right]$ and $a,b\in \mathcal{R},$ then the following properties hold:

• $\mathcal{G}\left[a\mathrm{ }f\left(y\right)+b\mathrm{ }g(y)\right]=a\mathcal{ }\mathcal{G}\left[f\left(y\right)\right]+b\mathcal{ }\mathcal{G}\left[g(y)\right].$

• ${\mathcal{G}}^{-1}\left[a\mathrm{ }F\left(p\right)+b\mathrm{ }G\left(p\right)\right]=a\mathrm{ }{\mathcal{G}}^{-1}\left[F\left(p\right)\right]+b\mathrm{ }{\mathcal{G}}^{-1}\left[G\left(p\right)\right].$

• $\mathcal{G}\left[y^b\right]=\frac{\Gamma (b+1)}{p^b},\mathrm{ }b>0.$

• $\mathcal{G}\left[e^{by}\right]=\frac{p}{p-b}\mathrm{ },\mathrm{ }b\in \mathcal{R}.$

• $\mathcal{G}\left[\sin \hspace{0.17em}by\right]=\frac{bp}{p^2+b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{G}\left[\cos \hspace{0.17em}by\right]=\frac{p^2}{p^2+b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{G}\left[\sinh \hspace{0.17em}by\right]=\frac{bp}{p^2-b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{G}\left[\cosh \hspace{0.17em}by\right]=\frac{p^2}{p^2-b^2}\mathrm{ },b\in \mathcal{R}.$

• $\mathcal{G}\left[f\prime \left(y\right)\right]=p\mathcal{ }\mathcal{G}\left[f\left(y\right)\right]-pf(0).$

• $\mathcal{G}\left[f^{(n)}\left(y\right)\right]=p^n\mathcal{ }\mathcal{G}\left[f\left(y\right)\right]-{\sum }_{k=1}^n{p}^{n-k+1}{f}^{\left(k-1\right)}\left(0\right).$

2.3. Sumudu transform

Definition 3. If $f(t)$ is a function of $t$ defined over a positive domain (Davies, 2002; Debnath & Bhatta, 2014). Then, Sumudu transformation of $f(t),$ denoted by $S[f(t)],$ is given by (19) $$S\left[f\left(t\right)\right]=F\left(q\right)=\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{1}{q}t}\left[f\left(t\right)\right]\mathrm{d}t,\mathrm{ }q>0.$$(19)

The inverse Sumudu transformation is provided as (20) $$S^{-1}\left[F\left(q\right)\right]=\frac{1}{2\mathrm{\pi }i}{\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{\frac{1}{q}e}^{\frac{1}{q}t}F\left(q\right)\mathrm{d}q=f\left(t\right)\mathrm{ },\mathrm{ }t>0.$$(20)

Theorem 3.

If $f\left(t\right)$ is a continuous function defined for $t>0$ and of exponential order $\alpha .$ Then $S\left[f(t)\right]$ exists for $Re\left(\frac{1}{q}\right)>\alpha$ and satisfies $$\left|f\left(t\right)\right|\le M{e}^{\alpha t},$$ where $M>0,$ then Sumudu transformation exists for $Re\left(\frac{1}{q}\right)>\alpha .$

Suppose that $F(q)=S\left[f\left(t\right)\right]$ and $G\left(q\right)=S\left[g\left(t\right)\right]$ and $a,b\in \mathcal{R},$ then the following properties hold:

• $S\left[a\mathrm{ }f\left(t\right)+b\mathrm{ }g(t)\right]=a\mathrm{ }S\left[f\left(t\right)\right]+b\mathrm{ }S\left[g(t)\right].$

• $S^{-1}\left[a\mathrm{ }F\left(q\right)+b\mathrm{ }G\left(q\right)\right]=a\mathrm{ }{S}^{-1}\left[F\left(q\right)\right]+b\mathrm{ }{S}^{-1}\left[G\left(q\right)\right].$

• $S\left[t^b\right]=\mathrm{\Gamma }(\mathrm{b}+1)q^b\mathrm{ },b\ge 0.$

• $S\left[e^{bt}\right]=\frac{1}{1-bq}\mathrm{ },b\in \mathcal{R}.$

• $S\left[\sin \hspace{0.17em}bt\right]=\frac{bq}{1+b^2{q}^2}\mathrm{ },b\in \mathcal{R}.$

• $S\left[\cos \hspace{0.17em}bt\right]=\frac{1}{1+b^2{q}^2}\mathrm{ },b\in \mathcal{R}.$

• $S\left[\sinh \hspace{0.17em}bt\right]=\frac{aq}{1-b^2{q}^2}\mathrm{ },b\in \mathcal{R}.$

• $S\left[\cosh \hspace{0.17em}bt\right]=\frac{1}{1-b^2{q}^2}\mathrm{ },b\in \mathcal{R}.$

• $S\left[f\prime \left(t\right)\right]=\frac{1}{q}\left(F\left(q\right)-f\left(0\right)\right).$

• $S\left[f^{(n)}\left(t\right)\right]=q^{-n}\left(F\left(q\right)-{\sum }_{k=0}^{n-1}{q}^k{f}^{\left(k\right)}\left(0\right)\right).$

3. Double integral transforms

In this section, we present the main results and characteristics of the DLST, DLARAT and DARAST.

3.1. Double Laplace–Sumudu transform

Definition 4. The DLST of a continuous function $z(x,t)$ on its domain of the positive variables $x$ and $t$ is defined by the double integral of the form (Dhunde et al., 2013; Dhunde & Waghmare, 2017; Eltayeb & Kiliçman, 2013): $$L_x{S}_t\left[z(x,t)\right]=Z(r,q)=\frac{1}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-rx-\frac{t}{q}}z(x,t)\mathrm{d}x\mathrm{ }\mathrm{d}t.$$

It can be shown that the double DLST is a linear integral transformation $$L_x{S}_t\left[\mathrm{a}\mathrm{ }z\left(x,t\right)+\mathrm{b}\mathrm{ }\psi (x,t)\right]=\mathrm{a}\mathrm{ }{L}_x{S}_t\left[z(x,t)\right]+\mathrm{b}\mathrm{ }{L}_x{S}_t\left[\psi (x,t)\right].$$ where $\mathrm{a}$ and $\mathrm{b}$ are constants, and $z\left(x,t\right),$ $\psi (x,t)$ are two continuous functions in which DLST exist.

Definition 5. The inverse DLST $L_x^{-1}{S}_t^{-1}$ is defined by the following form: $$L_x^{-1}{S}_t^{-1}\left[Z\left(r,q\right)\right]=z(x,t)=\frac{1}{2\mathrm{\pi }i}{\int }_{\gamma -i\mathrm{\infty }}^{\gamma +i\mathrm{\infty }}{e}^{rx}\mathrm{d}r\frac{1}{2\mathrm{\pi }i}{\int }_{\omega -i\mathrm{\infty }}^{\omega +i\mathrm{\infty }}\frac{1}{q}{e}^{\frac{t}{q}}Z(r,q)\mathrm{d}r\mathrm{ }\mathrm{d}q.$$

Now, the main properties and relations of DLST are provided in Table 1.

Table 1. Main properties of DLST.

$z(x,t)$	$L_x{S}_t\left[z\left(x,t\right)\right]=\mathrm{Z}\left(r,q\right)$
$1$	$\frac{1}{r}$
$x^{\alpha }{t}^{\beta }, \alpha \mathrm{and} \beta >-1$	$\frac{\Gamma (\alpha +1)\Gamma (\beta +1)q^{\beta }}{r^{\alpha +1}}$
$e^{\alpha x+\beta t}$	$\frac{1}{\left(r-\alpha \right)\left(1-\beta q\right)}, r>\alpha , q<\frac{1}{\beta }$
$\sin \hspace{0.17em}\left(\alpha x+\beta t\right)$	$\frac{\alpha +\beta qr}{\left(r^2+{\alpha }^2\right)\left(1+{\beta }^2{q}^2\right)}$
$\cos \hspace{0.17em}\left(\alpha x+\beta t\right)$	$\frac{r+\alpha \beta q}{\left(r^2+{\alpha }^2\right)\left(1+{\beta }^2{q}^2\right)}$

3.2. Double Laplace–ARA transform

Definition 6. The DLARAT is defined for a continuous function $z(x,y)$ on its domain of the positive variables $x,$ $y$ by the double integral as (Aghili & Parsa Moghaddam, 2011): $$L_x{\mathcal{G}}_y\left[z\left(x,y\right)\right]=Z\left(r,p\right)=p{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-\left(rx+py\right)}z\left(x,y\right)\mathrm{d}x\mathrm{ }\mathrm{d}y,\mathrm{ }r,p>0.$$

It is obvious that the DLARAT is a linear integral transformation: $$L_x{\mathcal{G}}_y\left[\mathrm{a}\mathrm{ }z\left(x,y\right)+\mathrm{b}\mathrm{ }\psi (x,y)\right]=\mathrm{a}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z(x,y)\right]+\mathrm{b} L_x{\mathcal{G}}_y\left[\psi (x,y)\right],$$ where $\mathrm{a}$ and $\mathrm{b}$ are constants, and $z\left(x,y\right),$ $\psi (x,y)$ are two continuous functions in which DLARAT exist.

Definition 7. The inverse DLARAT $L_x^{-1}{\mathcal{G}}_y^{-1}$ is formulated as: $$L_x^{-1}\left[{\mathcal{G}}_y^{-1}\left[Z\left(r,p\right)\right]\right]=\left(\frac{1}{2\pi i}\right){\int }_{c-i\mathrm{\infty }}^{c+i\mathrm{\infty }}{e}^{rx}\mathrm{d}r\left(\frac{1}{2\pi i}\right){\int }_{r-i\mathrm{\infty }}^{r+i\mathrm{\infty }}\frac{e^{py}}{p}Z\left(r,p\right)\mathrm{d}p=z\left(x,y\right).$$

Now, we give some properties and relations of DLARAT in Table 2.

Table 2. Basic properties of DLARAT.

$z(x,y)$	$L_x{\mathcal{G}}_y\left[z\left(x,y\right)\right]=\mathrm{Z}\left(r,p\right)$
$1$	$\frac{1}{r}$
$x^{\alpha }{y}^{\beta },$ $\alpha \mathrm{and} \beta >-1$	$\frac{\Gamma (\alpha +1)\Gamma (\beta +1)}{r^{\alpha +1}{ p}^{\beta }}$
$e^{\alpha x+\beta y}$	$\frac{p}{\left(r-\alpha \right)\left(p-\beta \right)}, r>\alpha , p>\beta$
$\sin \hspace{0.17em}\left(\alpha x+\beta y\right)$	$\frac{p(\alpha p+\beta r)}{\left(r^2+{\alpha }^2\right)\left(p^2+{\beta }^2\right)}$
$\cos \hspace{0.17em}\left(\alpha x+\beta y\right)$	$\frac{p(rp-\alpha \beta )}{\left(r^2+{\alpha }^2\right)\left(p^2+{\beta }^2\right)}$

3.3. Double ARA–Sumudu transform

Definition 8. The DARAST is defined for a continuous function $z(y,t)$ on its domain of the positive variables $y,$ $t$ by the double integral as (Ahmad Gani et al., 2017; Eltayeb & Kiliçman, 2010): $${\mathcal{G}}_y{S}_t\left[z\left(y,t\right)\right]=Z\left(p,q\right)=\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-\left(py+t/q\right)}z\left(y,t\right)\mathrm{d}y\mathrm{d}t,\mathrm{ }p,q>0.$$

It can be shown that the DARAST is a linear integral transformation: $${\mathcal{G}}_y{S}_t\left[\mathrm{a}\mathrm{ }z\left(y,t\right)+\mathrm{b}\mathrm{ }{\mathcal{G}}_y{S}_t\psi (y,t)\right]=\mathrm{a}\mathrm{ }{\mathcal{G}}_y{S}_t\left[z(y,t)\right]+\mathrm{b}\mathrm{ }{\mathcal{G}}_y{S}_t\left[\psi (y,t)\right].$$ where $\mathrm{a}$ and $\mathrm{b}$ are constants, and $z\left(y,t\right),$ $\psi (y,t)$ are two continuous functions in which DARAST exist.

Definition 9. The inverse DARAT ${\mathcal{G}}_y^{-1}{S}_t^{-1}$ is defined by the following form: $${\mathcal{G}}_y^{-1}\left[S_t^{-1}\left[Z\left(p,q\right)\right]\right]=\frac{1}{2\mathrm{\pi }i}\underset{c-i\mathrm{\infty }}{\overset{c+i\mathrm{\infty }}{\int }}\frac{e^{py}}{p}\mathrm{d}p\mathrm{ }\frac{1}{2\mathrm{\pi }i}\underset{\omega -i\mathrm{\infty }}{\overset{\omega +i\mathrm{\infty }}{\int }}\frac{e^{\frac{t}{q}}}{q}Z\left(p,q\right)\mathrm{d}q=z\left(y,t\right).$$

The basic properties and relations of DARAST are presented in Table 3.

Table 3. Basic properties of DARAST.

$z(y,t)$	${\mathcal{G}}_y{S}_t\left[z\left(y,t\right)\right]=\mathrm{Z}\left(p,q\right)$
$1$	$1$
$y^{\alpha }{t}^{\beta },$ $\alpha \mathrm{and} \beta >-1$	$\frac{\Gamma (\alpha +1)\Gamma (\beta +1){ p}^{\beta }}{r^{\alpha +1}}$
$e^{\alpha y+\beta t}$	$\frac{p}{\left(p-\alpha \right)\left(1-\beta q\right)}, p>\alpha , q<\frac{1}{\beta }$
$\sin \hspace{0.17em}\left(\alpha y+\beta t\right)$	$\frac{p \left(\alpha +\mathit{\text{β pq}}\right)}{\left({\alpha }^2+p^2\right)\left({\beta }^2 q^2+1\right)}$
$\cos \hspace{0.17em}\left(\alpha y+\beta t\right)$	$\frac{p \left(p-\mathit{\text{αβ q}}\right)}{\left({\alpha }^2+p^2\right)\left({\beta }^2 q^2+1\right)}$

4. Triple Laplace–ARA–Sumudu transform

Here, a novel triple integral transformation, TLARAST, is presented; it joins three interesting transformations, Laplace, ARA and Sumudu. The fundamental properties regarding the existence conditions, linearity and the inverse of this new triple transform are displayed. Furthermore, we establish new results related to partial derivatives and the triple convolution theorem. These results are implemented to compute the TLARAST for some basic functions.

Definition 10. Let $z\left(x,y,t\right)$ be a continuous function of three variables $x,y \mathrm{and} t>0.$ TLARAST of $z\left(x,y,t\right)$ is defined and denoted by (1) $$L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\mathrm{\Phi }\left(r,p,q\right)=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}z\left(x,y,t\right)\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t,r,p,q>0.$$(1)

The inverse TLARAST is defined by (2) $$L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}\left[\mathrm{\Phi }\left(r,p,q\right)\right]=z\left(x,y,t\right)=\frac{1}{2\pi i}\underset{\mathrm{a}-i\mathrm{\infty }}{\overset{\mathrm{a}+i\mathrm{\infty }}{\int }}{e}^{rx}\mathrm{d}r\mathrm{ }\frac{1}{2\pi i}\underset{\mathrm{b}-i\mathrm{\infty }}{\overset{\mathrm{b}+i\mathrm{\infty }}{\int }}\frac{e^{px}}{p}\mathrm{d}p\mathrm{ }\frac{1}{2\pi i}\underset{\mathrm{c}-i\mathrm{\infty }}{\overset{\mathrm{c}+i\mathrm{\infty }}{\int }}\frac{e^{\frac{t}{q}}}{q}\mathrm{\Phi }\left(r,p,q\right)\mathrm{d}q.$$(2) where $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are real constants.

Property 1. (Linearity) If $L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\mathrm{\Phi }\left(r,p,q\right)$ and $L_x{\mathcal{G}}_y{S}_t\left[\psi (x,y,t)\right]=\mathrm{\Psi }\left(r,p,q\right),$ then for any constants $\mathrm{A}$ and $\mathrm{B},$ we have (3) $$L_x{\mathcal{G}}_y{S}_t\left[\mathrm{A}\mathrm{ }z\left(x,y,t\right)+\mathrm{B}\mathrm{ }\psi (x,y,t)\right]=\mathrm{A}\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]+\mathrm{B}\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\psi (x,y,t)\right].$$(3)

Proof

of Property 1. $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[\mathrm{A}\mathrm{ }z\left(x,y,t\right)+\mathrm{B}\mathrm{ }\psi (x,y,t)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[\mathrm{A}\mathrm{ }z\left(x,y,t\right)+\mathrm{B}\mathrm{ }\psi (x,y,t)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\\ =\mathrm{A}\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-rx-py- \frac{t}{q}}\left[z\left(x,y,t\right)\right]\mathrm{d}x \mathrm{d}y\mathrm{ }\mathrm{d}t+\mathrm{B}\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[\psi (x,y,t)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }\mathrm{ }\\ =\mathrm{A}\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]+\mathrm{B}\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\psi (x,y,t)\right].\end{array}$$

Thus, TLARAST is a linear integral transformation. Similarly, we can show the inverse TLARAST is also linear.

Property 2. (Shifting property) If $L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\mathrm{\Phi }\left(r,p,q\right),$ then for any real constants $\mathrm{a},\mathrm{b}$ and $\mathrm{c},$ we have (4) $$L_x{\mathcal{G}}_y{S}_t\left[e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t}\mathrm{ }z\left(x,y,t\right)\right]=\frac{p}{\left(p\text{-}\mathrm{b}\right)\left(1\text{-}\mathrm{c}q\right)}\Phi \left(r\text{-}\mathrm{a},p\text{-}\mathrm{b},\frac{q}{1\text{-}\mathrm{c}q}\right).$$(4)

Proof

of Property 2. $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t}\mathrm{ }z\left(x,y,t\right)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t}\mathrm{ }z\left(x,y,t\right)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\mathrm{ }\\ =\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\left(r\text{-}\mathrm{a}\right)x-\left(p\text{-}\mathrm{b}\right)y-\left(\frac{1\text{-}\mathrm{c}q}{q}\right)t}\left[z\left(x,y,t\right)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\mathrm{ }\\ \begin{array}{c}=\frac{p}{\left(p\text{-}\mathrm{b}\right)\left(1\text{-}\mathrm{c}q\right)}\left(\frac{\left(p\text{-}\mathrm{b}\right)\left(1\text{-}\mathrm{c}q\right)}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\left(r\text{-}\mathrm{a}\right)x-\left(p\text{-}\mathrm{b}\right)y-\left(\frac{1\text{-}\mathrm{c}q}{q}\right)t}\left[z\left(x,y,t\right)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\mathrm{ }\right)\\ \mathrm{ }=\frac{p}{\left(p\text{-}\mathrm{b}\right)\left(1\text{-}\mathrm{c}q\right)}\mathrm{\Phi }\left(r\text{-}\mathrm{a},p\text{-}\mathrm{b},\frac{q}{1\text{-}\mathrm{c}q}\right).\end{array}\end{array}$$

Property 3. Let $z\left(x,y,t\right)=f\left(x\right)h\left(y\right)g\left(t\right)\mathrm{ },x>0,y>0,t>0.$ Then (5) $$L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=L_x\left[f\left(x\right)\right]{{\mathcal{G}}_y\left[h\right(y\left)\right]S}_t\left[g\left(t\right)\right].$$(5)

Proof

of Property 3. $$\begin{array}{c}{L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=L}_x{\mathcal{G}}_y{S}_t\left[f\left(x\right)h\left(y\right)g\left(t\right)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[f\left(x\right)h\left(y\right)g\left(t\right)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }\\ ={\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[f\left(x\right)\right]\mathrm{d}x·p{\int }_0^{\mathrm{\infty }}{e}^{-py}\left[h\left(y\right)\right]\mathrm{d}y·\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\left[g\left(t\right)\right]\mathrm{d}t\\ =L_x\left[f\left(x\right)\right]{{\mathcal{G}}_y\left[h\right(y\left)\right]S}_t\left[g\left(t\right)\right].\end{array}$$

Now, we compute the TLARAST for some essential functions.

1. Let $z\left(x,y,t\right)=1.$ Then $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[1\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }={\int }_0^{\mathrm{\infty }}{e}^{-rx}\mathrm{d}x\mathrm{ }p{\int }_0^{\mathrm{\infty }}{e}^{-py}\mathrm{d}y\mathrm{ }\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\mathrm{d}t\\ =L_x\left[1\right]{{\mathcal{G}}_y\left[1\right]S}_t\left[1\right]=\frac{1}{r}\mathrm{ }.\end{array}$$

2. Let $z\left(x,y,t\right)=x^{\alpha }{y}^{\lambda }{t}^{\beta },x>0,y>0$ and $t>0$ and $\mathrm{\alpha },\mathrm{\lambda }$ and $\mathrm{\beta }$ are positive constants. Then $$L_x{\mathcal{G}}_y{S}_t\left[x^{\mathrm{\alpha }}{y}^{\mathrm{\lambda }}{t}^{\mathrm{\beta }}\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[x^{\mathrm{\alpha }}{y}^{\mathrm{\lambda }}{t}^{\mathrm{\beta }}\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\mathrm{ }={\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[x^{\mathrm{\alpha }}\right]\mathrm{d}x\mathrm{ }p{\int }_0^{\mathrm{\infty }}{e}^{-py}\left[y^{\mathrm{\lambda }}\right]\mathrm{d}y\mathrm{ }\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\left[t^{\mathrm{\beta }}\right]\mathrm{d}t.$$

Thus, we get using the properties of single transforms $$L_x{\mathcal{G}}_y{S}_t\left[x^{\alpha }{y}^{\lambda }{t}^{\beta }\right]=\frac{\Gamma (\mathrm{\alpha }+1)}{r^{\mathrm{\alpha }+1}}\frac{\Gamma (\mathrm{\lambda }+1)}{p^{\mathrm{\lambda }}}\Gamma (\mathrm{\beta }+1)q^{\mathrm{\beta }}.$$

• iii. Let $z\left(x,y,t\right)=e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t},x>0,y>0$ and $t>0$ and $\mathrm{a},\mathrm{b}$ and $\mathrm{c}$ are constants. Then, $$L_x{\mathcal{G}}_y{S}_t\left[e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t}\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t}\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }={\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[e^{\mathrm{a}x}\right]\mathrm{d}x\mathrm{ }p{\int }_0^{\mathrm{\infty }}{e}^{-py}\left[e^{\mathrm{b}y}\right]\mathrm{d}y\mathrm{ }\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\left[e^{\mathrm{c}t}\right]\mathrm{d}t.$$

Thus, we get using the properties of single transforms $$L_{}{\mathcal{G}}_y{S}_t\left[e^{\mathrm{a}x+\mathrm{b}y+\mathrm{c}t}\right]=\frac{p}{\left(r\text{-}\mathrm{a}\right)\left(p\text{-}\mathrm{b}\right)\left(1\text{-}\mathrm{c}q\right)}.$$

Similarly, $$L_x{\mathcal{G}}_y{S}_t\left[e^{i\left(\mathrm{a}x+\mathrm{b}y+\mathrm{c}t\right)}\right]=\frac{p}{\left(r-i\mathrm{a}\right)\left(p-i\mathrm{b}\right)\left(1-i\mathrm{c}q\right)}.$$

Thus, one can obtain $$L_x{\mathcal{G}}_y{S}_t\left[e^{i\left(\mathrm{a}x+\mathrm{b}y+\mathrm{c}t\right)}\right]=\frac{p\left(rp\text{-}\mathrm{ab}\text{-}\mathrm{c}qr\mathrm{b}\text{-}\mathrm{c}q\mathrm{a}p\right)+iv(r\mathrm{b}+\mathrm{a}p+rp\mathrm{c}q\text{-}\mathrm{a}b\mathrm{c}q)}{\left(r^2+{\mathrm{a}}^2\right)\left(p^2+{\mathrm{b}}^2\right)\left(1+{\mathrm{c}}^2{q}^2\right)}$$

Euler’s formulas implies that $\sin \hspace{0.17em}x=\frac{e^{ix}-e^{-ix}}{2i}\text{ and }\cos \hspace{0.17em}x=\frac{e^{ix}+e^{-ix}}{2},$ and: $\sinh \hspace{0.17em}x=\frac{e^x-e^{-x}}{2i}\mathrm{ }\&\mathrm{ }\cosh \hspace{0.17em}x=\frac{e^x+e^{-x}}{2}.$

Consequently, the TLARAST of some essential functions can be obtained as: (6) $$L_x{\mathcal{G}}_y{S}_t\left[\sin \hspace{0.17em}\left(\mathrm{a}x+\mathrm{b}y+\mathrm{c}t\right)\right]=\frac{v(r\mathrm{b}+\mathrm{a}p+rp\mathrm{c}q\text{-}\mathrm{abc}q)}{\left(r^2+{\mathrm{a}}^2\right)\left(p^2+{\mathrm{b}}^2\right)\left(1+{\mathrm{c}}^2{q}^2\right)},$$(6) (7) $$L_x{\mathcal{G}}_y{S}_t\left[\cos \hspace{0.17em}\left(\mathrm{a}x+\mathrm{b}y+\mathrm{c}t\right)\right]=\frac{p\left(rp\text{-}\mathrm{ab}\text{-}\mathrm{c}qr\mathrm{b}\text{-}\mathrm{c}q\mathrm{a}p\right)}{\left(r^2+{\mathrm{a}}^2\right)\left(p^2+{\mathrm{b}}^2\right)\left(1+{\mathrm{c}}^2{q}^2\right)},$$(7) (8) $$L_x{\mathcal{G}}_y{S}_t\left[\sinh \hspace{0.17em}\left(\mathrm{a}x+\mathrm{b}y+\mathrm{c}t\right)\right]=\frac{v(r\mathrm{b}+\mathrm{a}p+rp\mathrm{c}q\text{-}\mathrm{abc}q)}{\left(r^2-{\mathrm{a}}^2\right)\left(p^2-{\mathrm{b}}^2\right)\left(1-{\mathrm{c}}^2{q}^2\right)},$$(8) (9) $$L_x{\mathcal{G}}_y{S}_t\left[\cosh \hspace{0.17em}\left(\mathrm{a}x+\mathrm{b}y+\mathrm{c}t\right)\right]=\frac{p\left(rp+\mathrm{ab}+\mathrm{c}qr\mathrm{b}+\mathrm{c}q\mathrm{a}p\right)}{\left(r^2-{\mathrm{a}}^2\right)\left(p^2-{\mathrm{b}}^2\right)\left(1-{\mathrm{c}}^2{q}^2\right)}.$$(9)

• v.Let $z\left(x,y,t\right)=\sin \hspace{0.17em}\mathrm{a}x\sin \hspace{0.17em}\mathrm{b}y\sin \hspace{0.17em}\mathrm{c}t,x>0,y>0$ and $t>0$ and $\mathrm{a},\mathrm{b}$ and $\mathrm{ }\mathrm{c}$ are constants. Then

$$L_x{\mathcal{G}}_y{S}_t\left[\sin \hspace{0.17em}\mathrm{a}x\sin \hspace{0.17em}\mathrm{b}y\sin \hspace{0.17em}\mathrm{c}t\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[\sin \hspace{0.17em}\mathrm{a}x\sin \hspace{0.17em}\mathrm{b}y\sin \hspace{0.17em}\mathrm{c}t\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }={\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[\sin \hspace{0.17em}\mathrm{a}x\right]\mathrm{d}x\mathrm{ }p{\int }_0^{\mathrm{\infty }}{e}^{-py}\left[\sin \hspace{0.17em}\mathrm{b}y\right]\mathrm{d}y\mathrm{ }\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\left[\sin \hspace{0.17em}\mathrm{c}t\right]\mathrm{d}t.$$

Thus, we get using the properties of single transforms: $$L_x{\mathcal{G}}_y{S}_t\left[\sin \hspace{0.17em}\mathrm{a}x\sin \hspace{0.17em}\mathrm{b}y\sin \hspace{0.17em}\mathrm{c}t\right]=\frac{\mathrm{a}p\mathrm{c}q}{\left(r^2+{\mathrm{a}}^2\right)\left(p^2+{\mathrm{b}}^2\right)\left(1+{\mathrm{c}}^2{q}^2\right)}.$$

Definition 11. If $z\left(x,y,t\right)$ is defined on $\left[0,X\right]\times \left[0,Y\right]\times [0,T],$ and satisfies the condition $$\left|z\left(,y,t\right)\right|\le R{e}^{\alpha x+\beta y+\gamma t},\mathrm{ }\exists R>0,\mathrm{ }\forall x>X,\mathrm{ }\forall y>Y\text{ and }t>T.$$

Then, $z\left(x,y,t\right)$ is said to be a function of exponential orders $\alpha ,$ $\beta$ and $\gamma$ as $x,y,t\to \infty$.

Theorem 4.

The existence condition of TLARAST of the continuous function $z\left(x,y,t\right)$ defined on $\left[0,X\right]\times \left[0,Y\right]\times [0,T]$ is to be of exponential orders $\alpha ,$ $\beta$ and $\gamma$ , for $Re\left[r\right]>\alpha$ , $Re\left[p\right]>\beta$ and $Re\left[\frac{1}{q}\right]>\gamma .$

Proof

of Theorem 4. The definition of TLARAST implies that $$\left|\mathrm{\Phi }\left(r,p,q\right)\right|=\left|\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}z\left(x,y,t\right)\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\right|\le \frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left|z\left(x,y,t\right)\right|\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\le \mathrm{ }\frac{R\mathrm{ }p}{q}\mathrm{ }\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\left(r-\alpha \right)x}\mathrm{d}x\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\left(p-\beta \right)y}\mathrm{d}y\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\left(\frac{1}{q}-\gamma \right)t}\mathrm{d}t=\frac{R\mathrm{ }p}{\left(r-\alpha \right)\left(p-\beta \right)\left(1-\gamma q\right)},\mathrm{ }Re\left[r\right]>\alpha \mathrm{ },Re\left[p\right]>\beta \mathrm{ and}\mathrm{ }Re\left[\frac{1}{q}\right]>\gamma .$$

Definition 12. The convolution of $z\left(x,y,t\right)$ and $\psi \left(x,y,t\right)$ is denoted by $\left(\mathit{\text{z***ψ}}\right)\left(x,y,t\right)$ and defined by (10) $$\left(z\text{***}\psi \right)\left(x,y,t\right)=\underset{0}{\overset{x}{\int }}\underset{0}{\overset{y}{\int }}\underset{0}{\overset{t}{\int }}z\left(x-\delta ,y-\epsilon ,t-\sigma \right)\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma .$$(10)

Theorem 5.

Let $L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\Phi \left(r,p,q\right).$ Then, $$L_x{\mathcal{G}}_y{S}_t\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]=e^{-\delta r-ϵp\text{-}\mathrm{ }\frac{\sigma }{q}}\mathrm{\Phi }\left(r,p,q\right),$$ where $H(x,y,t)$ denotes the Heaviside function defined by $$H\left(x-\delta ,y-\epsilon ,t-\sigma \right)=\{\begin{array}{l}1, x>\delta , y>ϵ,t>\sigma \\ 0, \mathit{otherwise}. \end{array}$$

Proof of Theorem 5.

From the definition of TLARAST, we have (11) $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]=\\ \frac{p}{q}{\int }_0^{\mathrm{\infty }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\int }_0^{\mathrm{\infty }}{e}^{-rx-py-\frac{t}{q}}\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\\ =\frac{p}{q}{\int }_0^{\mathrm{\infty }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\int }_0^{\mathrm{\infty }}{e}^{-rx-py-\frac{t}{q}}\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]\mathrm{d}x \mathrm{d}y \mathrm{d}t.\end{array}$$(11)

Putting $-\delta =\rho$, $y-ϵ=\beta$ and $t-\sigma =\lambda$ in Equation (11)(11) $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]=\\ \frac{p}{q}{\int }_0^{\mathrm{\infty }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\int }_0^{\mathrm{\infty }}{e}^{-rx-py-\frac{t}{q}}\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\\ =\frac{p}{q}{\int }_0^{\mathrm{\infty }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\int }_0^{\mathrm{\infty }}{e}^{-rx-py-\frac{t}{q}}\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]\mathrm{d}x \mathrm{d}y \mathrm{d}t.\end{array}$$(11) , we obtain (12) $$L_x{\mathcal{G}}_y{S}_t\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]=\frac{p}{q}{\int }_0^{\mathrm{\infty }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\int }_0^{\mathrm{\infty }}{e}^{-r(\rho +\delta )-p(\beta +ϵ)-\frac{1}{q}\left(\lambda +\sigma \right)}\left[Z\left(\rho ,\beta ,\lambda \right)\right]\mathrm{d}\rho \mathrm{d}\beta \mathrm{d}\lambda .$$(12)

Thus, Equation (12)(12) $$L_x{\mathcal{G}}_y{S}_t\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]=\frac{p}{q}{\int }_0^{\mathrm{\infty }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{\int }_0^{\mathrm{\infty }}{e}^{-r(\rho +\delta )-p(\beta +ϵ)-\frac{1}{q}\left(\lambda +\sigma \right)}\left[Z\left(\rho ,\beta ,\lambda \right)\right]\mathrm{d}\rho \mathrm{d}\beta \mathrm{d}\lambda .$$(12) can be simplified into $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\right]\\ =e^{-\delta r-ϵp\text{-}\mathrm{ }\frac{\sigma }{q}}\left(\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-r\rho -p\beta \text{-}\mathrm{ }\frac{\lambda }{q}}\left[z\left(\rho ,\beta ,\lambda \right)\right]\mathrm{d}\rho \mathrm{d}\beta \mathrm{d}\lambda \right)=e^{-\delta r-ϵp\text{-}\mathrm{ }\frac{\sigma }{q}}\mathrm{ }\mathrm{\Phi }\left(r,p,q\right).\end{array}$$

Theorem 6.

(Convolution Theorem) If $L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\mathrm{\Phi }\left(r,p,q\right)$ and $L_x{\mathcal{G}}_y{S}_t\left[\psi (x,y,t)\right]=\mathrm{\Psi }\left(r,p,q\right),$ then (13) $$L_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\frac{q}{p}\mathrm{\Phi }\left(r,p,q\right)\mathrm{\Psi }\left(r,p,q\right).$$(13)

Proof of Theorem 6.

From the definition of TLARAST, we have (14) $$L_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[\underset{0}{\overset{x}{\int }}\underset{0}{\overset{y}{\int }}\underset{0}{\overset{t}{\int }}z\left(x-\delta ,y-\epsilon ,t-\sigma \right)\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma \right]\mathrm{d}x\mathrm{d}y\mathrm{d}t$$(14)

The definition of Heaviside function, Equation (14)(14) $$L_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[\underset{0}{\overset{x}{\int }}\underset{0}{\overset{y}{\int }}\underset{0}{\overset{t}{\int }}z\left(x-\delta ,y-\epsilon ,t-\sigma \right)\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma \right]\mathrm{d}x\mathrm{d}y\mathrm{d}t$$(14) can be written as (15) $$L_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}[\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma ]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }$$(15)

Thus, Equation (15)(15) $$L_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}[\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma ]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\mathrm{ }$$(15) can be written as $$L_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{ }\mathrm{d}\epsilon \mathrm{ }\mathrm{d}\sigma [\frac{p}{q}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\mathrm{ }z\left(x-\delta ,y-\epsilon ,t-\sigma \right)H\left(x-\delta ,y-\epsilon ,t-\sigma \right)]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t.$$

Using Theorem 5, we have $$\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[\left(z\text{***}\psi \right)\left(x,y,t\right)\right]=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma \mathrm{ }{e}^{-\delta r-ϵp\text{-}\mathrm{ }\frac{\sigma }{q}}\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)\\ =\mathrm{\Phi }\left(r,p,q\right)\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\delta r-ϵp\text{-}\mathrm{ }\frac{\sigma }{q}}\psi \left(\delta ,\epsilon ,\sigma \right)\mathrm{d}\delta \mathrm{d}\epsilon \mathrm{d}\sigma \mathrm{ }\\ =\frac{q}{p}\mathrm{\Phi }\left(r,p,q\right)\mathrm{\Psi }\left(r,p,q\right).\end{array}$$

The following theorem presents TLARAST for the partial derivatives of orders 1 and 2.

Theorem 7.

(Derivative properties) If $L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\mathrm{\Phi }\left(r,p,q\right),$ then

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial x}\right]=r\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial y}\right]=p\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-p\mathrm{ }{L}_x{S}_t\left[z\left(x,0,t\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial t}\right]=\frac{1}{q}\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-\frac{1}{q}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial x^2}\right]=r^2\mathrm{\Phi }\left(r,p,q\right)-r\mathrm{ }{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial y^2}\right]=p^2\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]-p\mathrm{ }{L}_x{S}_t\left[z_y\left(x,0,t\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial t^2}\right]=\frac{1}{q^2}\mathrm{\Phi }\left(r,p,q\right)-\frac{1}{q^2}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right].$

Proof

of Theorem 7. Based on the definition of TLARAST, we get

1. $\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial x}\right]=\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-rx-py\text{-}\mathrm{ }\frac{t}{q}}\left[\frac{\partial z\left(x,y,t\right)}{\partial x}\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{d}t\\ =\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-py\text{-}\mathrm{ }\frac{t}{q}}\mathrm{ }\mathrm{d}y\mathrm{d}t\mathrm{ }{\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[\frac{\partial z\left(x,y,t\right)}{\partial x}\right]\mathrm{d}x\mathrm{ }\end{array}$

   Appling integrating by parts with: $$u=e^{-rx}\mathrm{ }⟹\mathrm{ }\mathrm{d}u=\mathrm{ }-r{e}^{-rx}\mathrm{d}x,$$ $$\mathrm{d}v=\frac{\partial z\left(x,y,t\right)}{\partial x}\mathrm{d}x\mathrm{ }⟹\mathrm{ }v=z(x,y,t),$$ we obtain (16) $$\begin{array}{c}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-rx}\left[\frac{\partial z\left(x,y,t\right)}{\partial x}\right]\mathrm{d}x\mathrm{ }=\left(-z\left(0,y,t\right)+r{\int }_0^{\mathrm{\infty }}{e}^{-rx}\left[z\left(x,y,t\right)\right]\mathrm{d}x\right)\\ \therefore \mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial x}\right]=r\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right].\end{array}$$(16)

2. $\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial y}\right]=\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\infty }{\int }_0^{\mathrm{\infty }}{e}^{-rx-py-\frac{t}{q}}\left[\frac{\partial z\left(x,y,t\right)}{\partial y}\right]\mathrm{d}x\mathrm{ }\mathrm{d}y\mathrm{ }\mathrm{d}t\\ =\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-rx\text{-}\mathrm{ }\frac{t}{q}}\mathrm{ }\mathrm{d}x\mathrm{d}t\mathrm{ }{\int }_0^{\mathrm{\infty }}{e}^{-py}\left[\frac{\partial z\left(x,y,t\right)}{\partial y}\right]\mathrm{d}y\mathrm{ }\end{array}$

   Using integrating by parts, we obtain:

   Let $u=e^{-py}\mathrm{ }⟹\mathrm{ }\mathrm{d}u=\mathrm{ }-p{e}^{-py}\mathrm{d}x,$ $$\mathrm{d}v=\frac{\partial z\left(x,y,t\right)}{\partial y}\mathrm{d}y\mathrm{ }⟹\mathrm{ }v=z(x,y,t).$$

Thus (17) $$\begin{array}{c}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-py}\left[\frac{\partial z\left(x,y,t\right)}{\partial y}\right]\mathrm{d}y\mathrm{ }=\left(-z\left(x,0,t\right)+p{\int }_0^{\mathrm{\infty }}{e}^{-py}\left[z\left(x,y,t\right)\right]\mathrm{d}y\right)\\ \therefore \mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial y}\right]=p\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-p\mathrm{ }{L}_x{S}_t\left[z\left(x,0,t\right)\right].\end{array}$$(17)

• iii. $\begin{array}{c}{L}_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial t}\right]=\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\infty }{\int }_0^{\mathrm{\infty }}{e}^{-rx-py- \frac{t}{q}}\left[\frac{\partial z\left(x,y,t\right)}{\partial t}\right]\mathrm{d}x \mathrm{d}y\mathrm{d}t\\ =\frac{p}{q}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-rx\text{-}\mathrm{ }py}\mathrm{ }\mathrm{d}x\mathrm{d}y\mathrm{ }{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\left[\frac{\partial z\left(x,y,t\right)}{\partial t}\right]\mathrm{d}t\mathrm{ }\end{array}$

  Appling integrating by parts with: $$u=e^{-\frac{t}{q}}\mathrm{ }⟹\mathrm{ }\mathrm{d}u=-\frac{1}{q}{e}^{-\frac{t}{q}}\mathrm{ }\mathrm{d}t,$$ $$\mathrm{d}v=\frac{\partial z\left(x,y,t\right)}{\partial t}\mathrm{d}t\mathrm{ }⟹\mathrm{ }v=z(x,y,t),$$

we get (18) $$\begin{array}{c}\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-\frac{t}{q}}\mathrm{ }\left[\frac{\partial z\left(x,y,t\right)}{\partial t}\right]\mathrm{d}t\mathrm{ }=\left(-z\left(x,y,0\right)+\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{t}{q}}\left[z\left(x,y,t\right)\right]\mathrm{d}t\right)\\ \therefore \mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\frac{\partial z\left(x,y,t\right)}{\partial t}\right]=\frac{1}{q}\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-\frac{1}{q}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right].\end{array}$$(18)

Similarly, we can prove that:

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial x^2}\right]=r^2\mathrm{\Phi }\left(r,p,q\right)-r\mathrm{ }{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial y^2}\right]=p^2\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]-p\mathrm{ }{L}_x{S}_t\left[z_y\left(x,0,t\right)\right].$

• $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial t^2}\right]=\frac{1}{q^2}\mathrm{\Phi }\left(r,p,q\right)-\frac{1}{q^2}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right].$

Corollary 1.

If $L_x{\mathcal{G}}_y{S}_t\left[z\left(x,y,t\right)\right]=\mathrm{\Phi }\left(r,p,q\right),$ then

1. $L_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial x\partial y}\right]=pr\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-p\mathrm{ }{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-p\mathrm{ }{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{S}_t\left[z\left(0,0,t\right)\right]$

2. $\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial x\partial t}\right]=\frac{r}{q}\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-\frac{1}{q}\mathrm{ }{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-\frac{1}{q}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]+\frac{1}{q}\mathrm{ }{\mathcal{G}}_y\left[z\left(0,y,0\right)\right].$

3. $\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\frac{{\partial }^{2}z\left(x,y,t\right)}{\partial y\partial t}\right]=\frac{p}{q}\mathrm{ }\mathrm{\Phi }\left(r,p,q\right)-\frac{p}{q}\mathrm{ }{L}_x{S}_t\left[z\left(x,0,t\right)\right]-\frac{p}{q}\mathrm{ }{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]+\frac{p}{q}\mathrm{ }{L}_x\left[z\left(x,0,0\right)\right].$

The proof can be obtained by direct applications of partial derivatives in Theorem 7.

5. Applications

In this section, we use the proposed triple transform TLARAST for solving some types of linear PDEs.

Example 1.

Consider the following nonhomogeneous heat equation (19) $$\mathrm{ }{z}_t\left(x,y,t\right)=z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+2\cos \hspace{0.17em}\left(x+y\right),\left(x,y\right)\in {\mathcal{R}}_{+}^2,t>0.$$(19)

Subject to the boundary and initial conditions (20) $$\{\begin{array}{ll}z\left(0,y,t\right)=e^{-2t}\sin \hspace{0.17em}y+\cos \hspace{0.17em}y,\mathrm{ }& \mathrm{ }{z}_x\left(0,y,t\right)=e^{-2t}\cos \hspace{0.17em}y-\sin \hspace{0.17em}y,\mathrm{ }\\ z\left(x,0,t\right)=e^{-2t}\sin \hspace{0.17em}x+\cos \hspace{0.17em}x,\mathrm{ }& \mathrm{ }{z}_y\left(x,0,t\right)=e^{-2t}\cos \hspace{0.17em}x-\sin \hspace{0.17em}x,\\ z\left(x,y,0\right)=\sin (x+y)+\cos \hspace{0.17em}\left(x+y\right).\mathrm{ }& .\end{array}$$(20)

Applying TLARAST on both sides of Equation (19)(19) $$S\left[f\left(t\right)\right]=F\left(q\right)=\frac{1}{q}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{1}{q}t}\left[f\left(t\right)\right]\mathrm{d}t,\mathrm{ }q>0.$$(19) , we have (21) $$L_x{\mathcal{G}}_y{S}_t\mathrm{ }\left[z_t\left(x,y,t\right)\right]=\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t[z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+2\cos \hspace{0.17em}\left(x+y\right)].\mathrm{ }$$(21)

Using the linearity and partial derivative properties of TLARAST, we get (22) $$\frac{1}{q}\Phi \left(r,p,q\right)-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]=r^2\Phi \left(r,p,q\right)-r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+\mathrm{ }{p}^2\Phi \left(r,p,q\right)-p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]-p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{2p(rp-1)}{\left(r^2+1\right)\left(p^2+1\right)}.\mathrm{ }$$(22)

Rearranging the terms, we have (23) $$\Phi \left(r,p,q\right)=\frac{q}{r^{2}q+p^{2}q-1}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{2p(rp-1)}{\left(r^2+1\right)\left(p^2+1\right)}).$$(23)

Substituting the transformed values: $$\begin{array}{c}{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]=\frac{p}{\left(p^2+1\right)\left(1+2q\right)}+\frac{p^2}{p^2+1},\mathrm{ }{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]=\frac{p^2}{\left(p^2+1\right)\left(1+2q\right)}-\frac{p}{p^2+1},\\ L_x{S}_t\left[z\right(x,0,t)=\frac{1}{\left(r^2+1\right)\left(1+2q\right)}+\frac{r}{r^2+1},\mathrm{ }{L}_x{S}_t[z_y\left(x,0,t\right)]=\frac{r}{\left(r^2+1\right)\left(1+2q\right)}-\frac{1}{r^2+1},\\ {\mathrm{ }L}_x{\mathcal{G}}_y\left[z\right(x,y,0)=\frac{p\left(p+r\right)}{\left(r^2+1\right)\left(p^2+1\right)}+\frac{p\left(pr-1\right)}{\left(r^2+1\right)\left(p^2+1\right)}.\end{array}$$ in Equation (23)(23) $$\Phi \left(r,p,q\right)=\frac{q}{r^{2}q+p^{2}q-1}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{2p(rp-1)}{\left(r^2+1\right)\left(p^2+1\right)}).$$(23) and simplifying, we obtain (24) $$\Phi \left(r,p,q\right)=\frac{q}{r^{2}q+p^{2}q-1}\left(\frac{\left[p^2+pr+\left(p^{2}r-p\right)(1+2q)\right]\left(r^{2}q+p^{2}q-1\right)}{q\left(r^2+1\right)\left(p^2+1\right)\left(1+2q\right)}\right)$$(24)

Taking the inverse transform $L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}$ for Equation (24)(24) $$\Phi \left(r,p,q\right)=\frac{q}{r^{2}q+p^{2}q-1}\left(\frac{\left[p^2+pr+\left(p^{2}r-p\right)(1+2q)\right]\left(r^{2}q+p^{2}q-1\right)}{q\left(r^2+1\right)\left(p^2+1\right)\left(1+2q\right)}\right)$$(24) , we get $$z\left(x,y,t\right)=L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}\left[\frac{p\left(p+r\right)}{\left(r^2+1\right)\left(p^2+1\right)(1+2q)}+\frac{p\left(pr-1\right)}{\left(r^2+1\right)\left(p^2+1\right)}\right]=e^{-2t}\sin \hspace{0.17em}(x+y)+\cos \hspace{0.17em}(x+y).$$

Figure 1 shows the contour graph of different surfaces of the solution $z(x,y,t)$ of Example 1.

Figure 1. The contour graph of the solution of the heat equation in Example 1, with different surfaces.

Example 2.

Consider the following Laplace equation (25) $$z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+z_{tt}\left(x,y,t\right)=0\mathrm{ },\mathrm{ }\left(x,y,t\right)\in {\mathcal{R}}_{+}^3.$$(25) subjected to the conditions (26) $$\{\begin{array}{ll}z\left(0,y,t\right)=0,\mathrm{ }& \mathrm{ }{z}_x\left(0,y,t\right)=\sin \hspace{0.17em}y\sinh \hspace{0.17em}\sqrt{2}t,\\ z\left(x,0,t\right)=0,& \mathrm{ }{z}_y\left(x,0,t\right)=\sin \hspace{0.17em}x\sinh \hspace{0.17em}\sqrt{2}t,\\ z\left(x,y,0\right)=0,& \mathrm{ }{z}_t\left(x,y,0\right)=\sqrt{2}\sin \hspace{0.17em}x\sin \hspace{0.17em}y.\end{array}$$(26)

Applying TLARAST on both sides of Equation (25)(25) $$z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+z_{tt}\left(x,y,t\right)=0\mathrm{ },\mathrm{ }\left(x,y,t\right)\in {\mathcal{R}}_{+}^3.$$(25) , we have (27) $$L_x{\mathcal{G}}_y{S}_t\left[z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+z_{tt}\left(x,y,t\right)\right]=0.$$(27)

By linearity and partial derivative properties of TLARAST, we obtain (28) $$\mathrm{ }{r}^2\Phi \left(r,p,q\right)-r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+\mathrm{ }{p}^2\Phi \left(r,p,q\right)-p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]-p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{1}{q^2}\Phi \left(r,p,q\right)-\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]=0.$$(28)

Rearranging the terms, we have (29) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2+1}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]+\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]).$$(29)

Substituting $${\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]=0,\mathrm{ }{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]=\frac{\sqrt{2}pq}{\left(p^2+1\right)\left(1-2q^2\right)},$$ $$L_x{S}_t\left[z\right(x,0,t\left)\right]=0,\mathrm{ }{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]=\frac{\sqrt{2}q}{\left(r^2+1\right)\left(1-2q^2\right)},$$ $$L_{\mathrm{x}}{\mathcal{G}}_y\left[z\right(x,y,0\left)\right]=0,\mathrm{ }{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]=\frac{\sqrt{2}p}{\left(r^2+1\right)\left(p^2+1\right)}.$$ in Equation (29)(29) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2+1}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]+\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]).$$(29) and simplifying, we obtain (30) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2+1}\left(\frac{\sqrt{2}p\left(r^2{q}^2+p^2{q}^2+1\right)}{q\left(r^2+1\right)\left(p^2+1\right)\left(1-2q^2\right)}\right).\mathrm{ }$$(30)

Taking the inverse transform $L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}$ for Equation (30), we get $$z\left(x,y,t\right)=L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}\left[\frac{\sqrt{2}pq}{\left(r^2+1\right)\left(p^2+1\right)\left(1-2q^2\right)}\right]=\sin \hspace{0.17em}x\sin \hspace{0.17em}y\sinh \hspace{0.17em}\sqrt{2}t.$$

Figure 2 below, shows the contour graph of different surfaces of the solution $z(x,y,t)$ of Example 2.

Figure 2. The contour graph of the solution of Example 2, with different surfaces.

Example 3.

Consider the following nonhomogeneous wave equation (31) $${2z}_{tt}\left(x,y,t\right)=z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+24t^2+4y\mathrm{ },\mathrm{ }\left(x,y\right)\in {\mathcal{R}}_{+}^2,t>0.\mathrm{ }$$(31) subjected to the conditions (32) $$\{\begin{array}{ll}z\left(0,y,t\right)=t^4+y{t}^2,\mathrm{ }\mathrm{ }& \mathrm{ }{z}_x\left(0,y,t\right)=\sin \hspace{0.17em}y\sin \hspace{0.17em}t,\\ z\left(x,0,t\right)=t^4,& \mathrm{ }{z}_y\left(x,0,t\right)=t^2+\sin \hspace{0.17em}x\sin \hspace{0.17em}t,\\ z\left(x,y,0\right)=0,& \mathrm{ }{z}_t\left(x,y,0\right)=\sin \hspace{0.17em}x\sin \hspace{0.17em}y.\end{array}$$(32)

Applying TLARAST on both sides of Equation (31)(31) $${2z}_{tt}\left(x,y,t\right)=z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+24t^2+4y\mathrm{ },\mathrm{ }\left(x,y\right)\in {\mathcal{R}}_{+}^2,t>0.\mathrm{ }$$(31) , we have (33) $${2L}_x{\mathcal{G}}_y{S}_t\left[z_{tt}\left(x,y,t\right)\right]=\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t[z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+24t^2+4y].$$(33)

By linearity and partial derivative properties of TLARAST, we get (34) $$2\left(\frac{1}{q^2}\Phi \left(r,p,q\right)-\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]\right)=\mathrm{ }{r}^2\Phi \left(r,p,q\right)-r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+\mathrm{ }{p}^2\Phi \left(r,p,q\right)-p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]-p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{48q^2}{r}+\mathrm{ }\frac{4}{rp}.\mathrm{ }$$(34)

Rearranging the terms, we have (35) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2-2}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]-\frac{2}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{2}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]-\frac{48q^2}{r}-\frac{4}{rp}.).$$(35)

Substituting the transformed values $${\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]=24q^4+\frac{2q^2}{p},\mathrm{ }{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]=\frac{pq}{\left(p^2+1\right)\left(1+q^2\right)},$$ $$L_x{S}_t\left[z\right(x,0,t)=\frac{24q^4}{r},\mathrm{ }{L}_x{S}_t[z_y\left(x,0,t\right)]=\frac{2q^2}{r}+\frac{q}{\left(r^2+1\right)\left(1+q^2\right)},$$ $$L_x{\mathcal{G}}_y\left[z\right(x,y,0)=0,\mathrm{ }{L}_x{\mathcal{G}}_y[z_t\left(x,y,0\right)]=\frac{p}{\left(r^2+1\right)\left(p^2+1\right)}.$$ in Equation (35)(35) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2-2}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]-\frac{2}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{2}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]-\frac{48q^2}{r}-\frac{4}{rp}.).$$(35) and simplifying, we obtain (36) $$\Phi \left(r,p,q\right)=\left(\frac{24q^4}{r}+\frac{2q^2}{rp}+\frac{pq}{\left(r^2+1\right)\left(p^2+1\right)\left(1+q^2\right)}\right).\mathrm{ }$$(36)

Taking the inverse transform $L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}$ for Equation (36)(36) $$\Phi \left(r,p,q\right)=\left(\frac{24q^4}{r}+\frac{2q^2}{rp}+\frac{pq}{\left(r^2+1\right)\left(p^2+1\right)\left(1+q^2\right)}\right).\mathrm{ }$$(36) , we get $$z\left(x,y,t\right)=L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}\left[\frac{24q^4}{r}+\frac{2q^2}{rp}+\frac{pq}{\left(r^2+1\right)\left(p^2+1\right)\left(1+q^2\right)}\right]=t^4+y{t}^2+\sin \hspace{0.17em}x\sin \hspace{0.17em}y\sin \hspace{0.17em}t.$$

Figure 3 shows the contour graph of different surfaces of the solution $z(x,y,t)$ of Example 3.

Figure 3. The contour graph of the solution of Example 3, with different surfaces.

Example 4.

Consider the following Poisson PDE (37) $$z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+z_{tt}\left(x,y,t\right)=2\sin \hspace{0.17em}x\cos \hspace{0.17em}y\sinh \hspace{0.17em}2t\mathrm{ },\mathrm{ }\left(x,y,t\right)\in {\mathfrak{R}}_{+}^3.$$(37) subjected to the conditions (38) $$\{\begin{array}{ll}z\left(0,y,\mathrm{t}\right)=0,\mathrm{ }& \mathrm{ }{z}_x\left(0,y,t\right)=\cos \hspace{0.17em}y\sinh \hspace{0.17em}2t,\\ z\left(x,0,t\right)=\sin \hspace{0.17em}x\sinh \hspace{0.17em}2t,& \mathrm{ }{z}_y\left(x,0,t\right)=0,\\ z\left(x,y,0\right)=0,& \mathrm{ }{z}_t\left(x,y,0\right)=2\sin \hspace{0.17em}x\cos \hspace{0.17em}y.\end{array}$$(38)

Applying TLARAST on both sides of Equation (38), we have (39) $$L_x{\mathcal{G}}_y{S}_t\left[z_{xx}\left(x,y,t\right)+z_{yy}\left(x,y,t\right)+z_{tt}\left(x,y,t\right)\right]=2\mathrm{ }{L}_x{\mathcal{G}}_y{S}_t\left[\sin \hspace{0.17em}x\cos \hspace{0.17em}x\sinh \hspace{0.17em}2t\right].$$(39)

By linearity and partial derivative properties of TLARAST, we obtain (40) $$r^2\Phi \left(r,p,q\right)-r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]-{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+\mathrm{ }{p}^2\Phi \left(r,p,q\right)-p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]-p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{1}{q^2}\Phi \left(r,p,q\right)-\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]-\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]=\frac{4q{p}^2}{\left(r^2+1\right)\left(p^2+1\right)\left(1-4q^2\right)}.\mathrm{ }$$(40)

Rearranging the terms, we have (41) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2+1}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]+\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]+\frac{4q{p}^2}{\left(r^2+1\right)\left(p^2+1\right)\left(1-4q^2\right)}).$$(41)

Substituting the transformed values $${\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]=0,\mathrm{ }{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]=\frac{2q{p}^2}{\left(p^2+1\right)\left(1-4q^2\right)},$$ $$L_x{S}_t\left[z\right(x,0,t)=\frac{2q}{\left(r^2+1\right)\left(1-4q^2\right)},\mathrm{ }{L}_x{S}_t[z_y\left(x,0,t\right)]=0,$$ $$L_x{\mathcal{G}}_y\left[z\right(x,y,0)=0,\mathrm{ }{L}_x{\mathcal{G}}_y[z_t\left(x,y,0\right)]=\frac{2p^2}{\left(r^2+1\right)\left(p^2+1\right)}.$$ in Equation (41)(41) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2+1}(r{\mathcal{G}}_y{S}_t\left[z\left(0,y,t\right)\right]+{\mathcal{G}}_y{S}_t\left[z_x\left(0,y,t\right)\right]+p^2{L}_x{S}_t\left[z\left(x,0,t\right)\right]+p{L}_x{S}_t\left[z_y\left(x,0,t\right)\right]+\frac{1}{q^2}{L}_x{\mathcal{G}}_y\left[z\left(x,y,0\right)\right]+\frac{1}{q}{L}_x{\mathcal{G}}_y\left[z_t\left(x,y,0\right)\right]+\frac{4q{p}^2}{\left(r^2+1\right)\left(p^2+1\right)\left(1-4q^2\right)}).$$(41) and simplifying, we obtain (42) $$\Phi \left(r,p,q\right)=\frac{q^2}{r^2{q}^2+p^2{q}^2+1}\left(\frac{2p^2\left(r^2{q}^2+p^2{q}^2+1\right)}{q\left(r^2+1\right)\left(p^2+1\right)\left(1-4q^2\right)}\right)$$(42)

Taking the inverse transform $L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}$ for Equation (42), we get $$z\left(x,y,t\right)=L_x^{-1}{\mathcal{G}}_y^{-1}{S}_t^{-1}\left[\frac{2q{p}^2}{\left(r^2+1\right)\left(p^2+1\right)\left(1-4q^2\right)}\right]=\sin \hspace{0.17em}x\cos \hspace{0.17em}y\sinh \hspace{0.17em}2t.$$

Figure 4 shows the contour graph of different surfaces of the solution $z(x,y,t)$ of Example 4.

Figure 4. The contour graph of the solution of Example 4, with different surfaces.

6. Conclusion

This article introduced a new triple integral transform called the TLARAST. The definition of TLARAST and its fundamental properties, including linearity, existence, inverse and specific values for basic functions were discussed and proved. Moreover, new results concerning the triple convolution theorem and partial derivatives were presented. The effectiveness of the new transform was shown by applying it to solve various types of PDEs. The results obtained in this study highlighted the simplicity and practicality of the TLARAST approach, achieving the goal of providing new techniques to solve PDEs with significant scientific applications. Moving forward, our aim is to extend the application of TLARAST to handle non-linear PDEs and systems of differential equations. To ensure comprehensive analysis, we plan in the future to compare the results with other analytical methods, such as the He Laplace method and related approaches (Abdoon, Saadeh, Berir, EL Guma, & Ali, 2023; Anjum et al., 2020; Anjum & He, 2019; Nadeem et al., 2023; Suleman, Lu, Yue, Ul Rahman, & Anjum, 2019; Tao et al., 2023). Additionally, we encourage readers to explore the solutions of fractional differential equations and non-linear problems. For the latter, we recognize the potential for a fruitful combination of TLARAST with other analytical methods such as the Adomian decomposition method. Such investigations could lead to further advancements and broader applications of TLARAST in the realm of mathematical problem-solving.

Disclosure statement

No potential conflict of interest was reported by the author(s).