1 Introduction

Integral transforms are very important tools in the different areas of science and technology. Many researchers have focused on the study of the theory and applications of integral transforms such as Laplace, Fourier, Mellin, Hankel, Sumudu, and gained a series of results. It is also known that the integral transform method is an effective method to solve some certain differential equations.

The Laplace transform is a wonderful tool for solving ordinary and partial differential equations and has enjoyed much success in this realm. Readers can refer to some classic literature [8, 19, 21, 23]. With respect to some more in-depth applications of the Laplace transform, readers may refer to the references [22, 25, 26]. In addition, some new type Laplace transforms are also worthy of our attention, see the reference [12, 24].

Compared to the single Laplace transformations, the study of the multiple Laplace transformations grew out of Jaeger’s work [15] in 1940 and Estrin and Higgins’s work [10] in 1951. In recent years, some researchers have focused on the study of the multiple Laplace transformations. Gadain [11] investigated that a modification of double Laplace decomposition method is proposed for the analytical approximation solution of a coupled system of pseudo-parabolic equation with initial conditions. Khan et al. [18] extended the concept of triple Laplace transform to the solution of fractional order partial differential equations by using Caputo fractional derivative.

In recent years, some mixed multiple integral transformations have been studied. In [20], the authors studied the properties of Sumudu transform, and studied the relationship between Laplace and Sumudu transforms and presented an example of the double Sumudu transform in order to solve the wave equation in one dimension. In [1], the authors combined two new methods, the conformable double Laplace-Sumudu transform and the modified decomposition technique, and solved some nonlinear fractional partial differential equations by using the new method. In [2, 3] and [9], the authors established an efficient new double Laplace-Sumudu transform method, and studied the properties of the double integral transform, and solved many types of partial differential equations by using the method.

The mixed multiple integral transformation is a powerful tool in handling the initial-boundary value problem of the partial differential equations. These work brings new ideas and methods for solving the partial differential equations.

However, to the best knowledge of the author, no work has been done on the triple mixed integral transformation and applications with respect to solving the initial-boundary value problem of the partial differential equations. Inspired by the above work, we develop and establish a novel triple mixed integral transformation method, and study the applications of this method for solving the initial-boundary value problems of the partial differential equations. The advantage of this method is to avoid repeatedly solving the related Sturm-Liouville problems, which are sometimes very difficult. The method brings great convenience for engineering and technical applications of the initial-boundary value problems of the partial differential equations.

Sumudu transform have many advantages [27, 28]. First, the Sumudu transform retains the consistency of units. For example, the unit-step function in the t-domain is transformed to unity in the u-domain by the Sumudu transform. Second, the Sumudu transform retains the consistency of scale. For example, scaling of the function f(t) in the t-domian is equivalent to scaling of F(u) in the u-domain by the same scale factor. It is easy to observe that it is consistent in units if the original variable t and the transformed variable u represent the same units in the Sumudu transform of partial derivatives. Additionally, the Sumudu transform of \(\frac{\partial f}{\partial t}\) can be visualized as the forward difference approximation of the partial derivative. These provide great convenience for the calculation of solutions for partial differential equations and systems. In control engineering, by using the scale and unit preserving properties, the Sumudu transform may be used to solve some problems without resorting to a new frequency domain. For more in-depth studies and applications of this integral transformation, readers may refer to references [4,5,6,7, 13, 14, 16, 17, 29].

By using such a special integral transform combination, triple Laplace-Sumudu transformation, on the one hand, the traditional advantages of Laplace transform can be retained, and on the other hand, the two advantages of Sumudu transform concerning the time variable t can be fully applied for solving some differential equations and engineering control problems. For the initial value and boundary value problems of the partial differential equation in one-dimensional time and two-dimensional space, this combination is more advantageous than other existing methods [1, 2, 9, 10, 13, 18, 28]. In general, the advantage of this method is to avoid repeatedly solving the related Sturm-Liouville problems, which are sometimes very difficult. It can easily be applied directly to solve a wide range of mathematical and physics equations by turning these equations into algebraic equations.

The organization of the paper is stated as follows: The second section is devoted to the fundamental definitions of the Laplace transform, the Sumudu transform and the triple Laplace-Sumudu transform as well as the Caputo fractional derivative. In Sect. 3, the basic properties of the triple Laplace-Sumudu transform are investigated, including the derivative properties, the convolution properties and the existence conditions for the triple Laplace-Sumudu transform. In Sect. 4, we give the triple Laplace-Sumudu transforms of some basic functions. In Sect. 5, we study the applications of the triple Laplace-Sumudu transform, the solutions of some heat flow equations and wave equations with the initial-boundary values are obtained by using the triple Laplace-Sumudu transform method. In Sect. 6, we summarize the idea of the whole paper and give the general steps of the application of this method.

2 Preliminary

In this section, some definitions and properties involving the Laplace transform and Sumudu transform are presented which are useful further for our main results in this paper.

Definition 1

The Laplace transform of the continuous function f(x) is defined by

$$\begin{aligned} L[f(x)]=\int _0^\infty e^{-\rho x}f(x){\mathrm d}x=\overline{f}(\rho ). \end{aligned}$$

Definition 2

The Sumudu transform of the function g(t) is defined by

$$\begin{aligned} S[g(t)]=\frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}g(t){\mathrm d}t=\overline{g}(\sigma ),\ \sigma \in (-\tau _1,\tau _2), \end{aligned}$$

over the set of the functions

$$\begin{aligned} \Omega =\left\{ g(t)\left| \right. \exists M,\tau _1,\tau _2>0, \left| g(t)\right| <Me^{\frac{|t|}{\tau _j}}\ \textrm{if}\ t\in (-1)^j\times [0,\infty )\right\} . \end{aligned}$$

Next, we give the main definition of the triple Laplace-Sumudu transformation in our work.

Definition 3

The triple Laplace-Sumudu transform of the function u(x, y, t) of three variables \(x>0,y>0\) and \(t>0\) is defined by

$$\begin{aligned} L_xL_yS_t[u(x,y,t)]=\frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}u(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t. \end{aligned}$$

The triple Laplace-Sumudu transform is denoted by \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\).

Clearly, the triple Laplace-Sumudu transform is a linear integral transformation as shown below. For any scalars \(\alpha , \beta\) and the functions u(x, y, t), v(x, y, t) of three variables, we have

$$\begin{aligned}{} & {} L_xL_yS_t[\alpha u(x,y,t)+\beta v(x,y,t)]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}[\alpha u(x,y,t)+\beta v(x,y,t)]{\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\alpha u(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t\\{} & {} \quad +\frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\beta v(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{\alpha }{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }} u(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t\\{} & {} \quad +\frac{\beta }{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }} v(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \alpha L_xL_yS_t[u(x,y,t)]+\beta L_xL_yS_t[v(x,y,t)]. \end{aligned}$$

The following is the definition of the inverse triple Laplace-Sumudu transformation.

Definition 4

The inverse of the triple Laplace-Sumudu transform \((L_xL_yS_t)^{-1}\) \(\overline{u}(\rho _1,\rho _2,\sigma )=u(x,y,t)\) is defined by

$$\begin{aligned}{} & {} (L_xL_yS_t)^{-1}\overline{u}(\rho _1,\rho _2,\sigma ) \\= & {} u(x,y,t) \\= & {} \left( \frac{1}{2\pi i}\right) \int _{\gamma _1-i\infty }^{\gamma _1+i\infty }e^{\rho _1x}{\mathrm d}\rho _1 \left( \frac{1}{2\pi i}\right) \int _{\gamma _2-i\infty }^{\gamma _2+i\infty }e^{\rho _2y}{\mathrm d}\rho _2 \left( \frac{1}{2\pi i}\right) \int _{\eta -i\infty }^{\eta +i\infty }\frac{1}{\sigma }e^{\frac{t}{\sigma }}\overline{u}(\rho _1,\rho _2,\sigma ){\mathrm d}\sigma . \end{aligned}$$

In order to discuss the derivative properties of the triple Laplace-Sumudu transform in a more general context, we give the definitions of the Caputo fractional derivatives with respect to different variables.

Definition 5

The \(\delta ,\xi ,\eta (\delta>0,\xi>0,\eta >0)\) order Caputo fractional derivative of the function u(x, y, t) with respect to x, y, t is defined respectively by

$$\begin{aligned}{} & {} \frac{\partial ^\delta u(x,y,t)}{\partial t^\delta }=\frac{1}{\Gamma (p-\delta )}\int _0^t(t-\tau )^{p-\delta -1}\frac{\partial ^p u(x,y,t)}{\partial \tau ^p}{\mathrm d}\tau ,\ p-1<\delta \leqslant p,\\{} & {} \frac{\partial ^\xi u(x,y,t)}{\partial x^\xi }=\frac{1}{\Gamma (q-\xi )}\int _0^x(x-\tau )^{q-\xi -1}\frac{\partial ^q u(x,y,t)}{\partial \tau ^q}{\mathrm d}\tau ,\ q-1<\xi \leqslant q,\\{} & {} \frac{\partial ^\eta u(x,y,t)}{\partial y^\eta }=\frac{1}{\Gamma (r-\eta )}\int _0^y(y-\tau )^{r-\eta -1}\frac{\partial ^r u(x,y,t)}{\partial \tau ^r}{\mathrm d}\tau ,\ r-1<\eta \leqslant r, \end{aligned}$$

where \(p,q,r\in \textbf{N}\).

3 Basic Properties of the Triple Laplace-Sumudu Transform

In this section, we investigate the basic properties of the triple Laplace-Sumudu transformation, which including the derivative properties, the convolution properties and the existence conditions for the triple Laplace-Sumudu transform.

3.1 Derivative Properties

The following result is useful for the derivative properties, which is from [19].

Theorem 1

([19]) Let \(\alpha >0, n-1<\alpha \leqslant n (n\in \textbf{N})\) be such that \(y(x)\in C^n(\textbf{R}^+), y^{(n)}\in L_1(0,b)\) for any \(b>0\), for any \(y^{(n)}(x)\) the estimate \(|y(x)|\leqslant Be^{q_0x} (x>b>0)\) holds for constants \(B>0\) and \(q_0>0\), the Laplace transforms (Ly)(p) and \(L[D^{n}y(t)]\) exist, and \(\lim _{x\rightarrow +\infty }(D^ky)(x)=0\) for \(k=0,1,\cdot \cdot \cdot ,n-1\). Then the following relation holds:

$$\begin{aligned} (L ^CD^\alpha _{0+}y)(s)=s^\alpha (Ly)(s)-\sum _{k=0}^{n-1}s^{\alpha -k-1}(D^ky)(0), \end{aligned}$$

where \(^CD^\alpha _{0+}y(s)\) is the Caputo fractional derivatives with respect to the function y(x).

Next, we prove some derivative properties of the triple Laplace-Sumudu transform.

If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\), then

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial ^\delta u(x,y,t)}{\partial t^\delta }\right] =\sigma ^{-\delta }\overline{u}(\rho _1,\rho _2,\sigma )-\sum \limits _{j=0}^{p-1}\sigma ^{-\delta +j}L_xL_y\left[ \frac{\partial ^j u(x,y,0)}{\partial t^j}\right] , \end{aligned}$$

where \(p-1<\delta \leqslant p,\ p\in \textbf{N}\).

Proof

By the definition of the triple Laplace-Sumudu transformation and Theorem 1, we get

$$\begin{aligned}{} & {} L_xL_yS_t\left[ \frac{\partial ^\delta u(x,y,t)}{\partial t^\delta }\right] \\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\left[ \frac{\partial ^\delta u(x,y,t)}{\partial t^\delta }\right] {\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \int _0^\infty e^{-\rho _2y} \int _0^\infty e^{-\rho _1x}\left[ \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}\frac{\partial ^\delta u(x,y,t)}{\partial t^\delta }{\mathrm d}t\right] {\mathrm d}x{\mathrm d}y\\= & {} \int _0^\infty e^{-\rho _2y} \int _0^\infty e^{-\rho _1x}\frac{1}{\sigma }\left[ \left( \frac{1}{\sigma }\right) ^\delta \left( S_t'u\right) \left( \frac{1}{\sigma }\right) -\sum \limits _{j=0}^{p-1}\left( \frac{1}{\sigma }\right) ^{\delta -j-1}\left( D^ju\right) (0)\right] {\mathrm d}x{\mathrm d}y\\= & {} L_yL_x\left[ \frac{1}{\sigma }\left( \frac{1}{\sigma }\right) ^\delta \left( S_t'u\right) \left( \frac{1}{\sigma }\right) -\sum \limits _{j=0}^{p-1}\sigma ^{-\delta +j}\frac{\partial ^j u(x,y,0)}{\partial t^j}\right] \\= & {} \sigma ^{-\delta }L_xL_yS_t[u(x,y,t)]-\sum \limits _{j=0}^{p-1}\sigma ^{-\delta +j}L_xL_y\left[ \frac{\partial ^j u(x,y,0)}{\partial t^j}\right] \\= & {} \sigma ^{-\delta }\overline{u}(\rho _1,\rho _2,\sigma )-\sum \limits _{j=0}^{p-1}\sigma ^{-\delta +j}L_xL_y\left[ \frac{\partial ^j u(x,y,0)}{\partial t^j}\right] , \end{aligned}$$

where \(p-1<\delta \leqslant p,\ p\in \textbf{N}\), the proof is complete. \(\square\)

When \(\delta =1\) or \(\delta =2\), we can get the following two special cases. If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\), then

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial u(x,y,t)}{\partial t}\right] =\frac{1}{\sigma }\overline{u}(\rho _1,\rho _2,\sigma )-\frac{1}{\sigma }L_xL_y[u(x,y,0)] \end{aligned}$$

and

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial ^2 u(x,y,t)}{\partial t^2}\right] =\frac{1}{\sigma ^2}\overline{u}(\rho _1,\rho _2,\sigma )-\frac{1}{\sigma ^2}L_xL_y\left[ u(x,y,0)\right] -\frac{1}{\sigma }L_xL_y\left[ u_t(x,y,0)\right] . \end{aligned}$$

Next, we calculate the triple Laplace-Sumudu transform of the \(\xi\) order Caputo fractional derivative with respect to the variable x.

If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\), then

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial ^\xi u(x,y,t)}{\partial x^\xi }\right] =\rho _1^\xi \overline{u}(\rho _1,\rho _2,\sigma )-\sum \limits _{k=0}^{q-1}\rho _1^{\xi -1-k}L_yS_t\left[ \frac{\partial ^k u(0,y,t)}{\partial x^k}\right] , \end{aligned}$$

where \(q-1<\xi \leqslant q,\ q\in \textbf{N}\).

Proof

By the definition of the triple Laplace-Sumudu transformation and Theorem 1, we have

$$\begin{aligned}{} & {} L_xL_yS_t\left[ \frac{\partial ^\xi u(x,y,t)}{\partial x^\xi }\right] \\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\frac{\partial ^\xi u(x,y,t)}{\partial x^\xi }{\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}\int _0^\infty e^{-\rho _2y} \int _0^\infty e^{-\rho _1x}\frac{\partial ^\xi u(x,y,t)}{\partial x^\xi }{\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}\int _0^\infty e^{-\rho _2y} \left[ \rho _1^\xi \left( L_xu\right) (\rho _1)-\sum \limits _{k=0}^{q-1}\rho _1^{\xi -1-k}\frac{\partial ^k u(0,y,t)}{\partial x^k}\right] {\mathrm d}y{\mathrm d}t\\= & {} \rho _1^\xi \overline{u}(\rho _1,\rho _2,\sigma )-\sum \limits _{k=0}^{q-1}\rho _1^{\xi -1-k}L_yS_t\left[ \frac{\partial ^k u(0,y,t)}{\partial x^k}\right] , \end{aligned}$$

where \(q-1<\xi \leqslant q,\ q\in \textbf{N}\). \(\square\)

When \(\xi =1\) or \(\xi =2\), we can obtain the following two special cases. If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\), then

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial u(x,y,t)}{\partial x}\right] =\rho _1\overline{u}(\rho _1,\rho _2,\sigma )-L_yS_t\left[ u(0,y,t)\right] \end{aligned}$$

and

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial ^2 u(x,y,t)}{\partial x^2}\right] =\rho _1^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _1L_yS_t\left[ u(0,y,t)\right] -L_yS_t\left[ u_x(0,y,t)\right] . \end{aligned}$$

For the convenience of the applications, we also calculate the triple Laplace-Sumudu transform of the \(\eta\) order Caputo fractional derivative with respect to the variable y.

If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\), then

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial ^\eta u(x,y,t)}{\partial y^\eta }\right] =\rho _2^\eta \overline{u}(\rho _1,\rho _2,\sigma )-\sum \limits _{k=0}^{r-1}\rho _2^{\eta -1-k}L_xS_t\left[ \frac{\partial ^k u(x,0,t)}{\partial y^k}\right] , \end{aligned}$$

where \(r-1< \eta \leqslant r, r\in \textbf{N}\).

Proof

By the definition of the triple Laplace-Sumudu transformation and Theorem 1, we obtain

$$\begin{aligned}{} & {} L_xL_yS_t\left[ \frac{\partial ^\eta u(x,y,t)}{\partial y^\eta }\right] \\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\frac{\partial ^\eta u(x,y,t)}{\partial y^\eta }{\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}\int _0^\infty e^{-\rho _1x} \int _0^\infty e^{-\rho _2y}\frac{\partial ^\eta u(x,y,t)}{\partial y^\eta }{\mathrm d}y{\mathrm d}x{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}\int _0^\infty e^{-\rho _1x} \left[ \rho _2^\eta \left( L_yu\right) (\rho _2)-\sum \limits _{k=0}^{r-1}\rho _2^{\eta -1-k}\frac{\partial ^k u(x,0,t)}{\partial y^k}\right] {\mathrm d}x{\mathrm d}t\\= & {} \rho _2^\eta \overline{u}(\rho _1,\rho _2,\sigma )-\sum \limits _{k=0}^{r-1}\rho _2^{\eta -1-k}L_xS_t\left[ \frac{\partial ^k u(x,0,t)}{\partial y^k}\right] , \end{aligned}$$

where \(r-1< \eta \le r\). \(\square\)

When \(\eta =1\) or \(\eta =2\), we have the following two special cases. If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\), then

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial u(x,y,t)}{\partial y}\right] =\rho _2\overline{u}(\rho _1,\rho _2,\sigma )-L_xS_t\left[ u(x,0,t)\right] \end{aligned}$$

and

$$\begin{aligned} L_xL_yS_t\left[ \frac{\partial ^2 u(x,y,t)}{\partial y^2}\right] =\rho _2^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _2L_xS_t\left[ u(x,0,t)\right] -L_xS_t\left[ u_y(x,0,t)\right] . \end{aligned}$$

3.2 Convolution Properties

In this subsection, we firstly prove a results which concerning the convolution properties of the triple Laplace-Sumudu transformation.

Theorem 2

If \(\overline{u}(\rho _1,\rho _2,\sigma )=L_xL_yS_t[u(x,y,t)]\),

$$\begin{aligned} H(x-\delta ,y-\eta ,t-\varepsilon )=\left\{ \begin{aligned}&1,&\quad&x>\delta ,y>\eta ,t>\varepsilon ,&\\&0,&\quad&\textrm{otherwise,}&\\ \end{aligned} \right. \end{aligned}$$

then

$$\begin{aligned} L_xL_yS_t[u(x-\delta ,y-\eta ,t-\varepsilon )H(x-\delta ,y-\eta ,t-\varepsilon )]=e^{-\rho _1\delta -\rho _2\eta -\frac{\varepsilon }{\sigma }}\overline{u}(\rho _1,\rho _2,\sigma ). \end{aligned}$$

Proof

By the definition of the triple Laplace-Sumudu transform, we have

$$\begin{aligned}{} & {} L_xL_yS_t[u(x-\delta ,y-\eta ,t-\varepsilon )H(x-\delta ,y-\eta ,t-\varepsilon )]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}u(x-\delta ,y-\eta ,t-\varepsilon )H(x-\delta ,y-\eta ,t-\varepsilon ){\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _\delta ^\infty \int _\eta ^\infty \int _\varepsilon ^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}u(x-\delta ,y-\eta ,t-\varepsilon ){\mathrm d}x{\mathrm d}y{\mathrm d}t. \end{aligned}$$

Let \(\alpha =x-\delta ,\beta =y-\eta ,\gamma =t-\varepsilon ,\) we have \(\alpha \geqslant 0,\beta \geqslant 0,\gamma \geqslant 0\), and

$$\begin{aligned}{} & {} L_xL_yS_t[u(x-\delta ,y-\eta ,t-\varepsilon )H(x-\delta ,y-\eta ,t-\varepsilon )]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1(\delta +\alpha )-\rho _2(\eta +\beta )-\frac{1}{\sigma }(\varepsilon +\gamma )}u(\alpha ,\beta ,\gamma ){\mathrm d}\alpha {\mathrm d}\beta {\mathrm d}\gamma \\= & {} e^{-\rho _1\delta -\rho _2\eta -\frac{\varepsilon }{\sigma }}\frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1\alpha -\rho _2\beta -\frac{\gamma }{\sigma }}u(\alpha ,\beta ,\gamma ){\mathrm d}\alpha {\mathrm d}\beta {\mathrm d}\gamma \\= & {} e^{-\rho _1\delta -\rho _2\eta -\frac{\varepsilon }{\sigma }}\overline{u}(\rho _1,\rho _2,\sigma ). \end{aligned}$$

The proof is complete. \(\square\)

Next, we introduce the definition of the convolution concerning two functions with three variables.

Definition 6

Let u(x, y, t), v(x, y, t) be integrable functions, then the convolution of u(x, y, t) and v(x, y, t) as

$$\begin{aligned} (u***v)(x,y,t)=\int _0^x \int _0^y \int _0^t u(x-\delta ,y-\eta ,t-\varepsilon )v(\delta ,\eta ,\varepsilon ){\mathrm d}\delta {\mathrm d}\eta {\mathrm d}\varepsilon , \end{aligned}$$

this is called the triple convolution and the symbol \(***\) denotes the triple convolution with respect to x, y, t.

The following theorem is the main result concerning the convolution properties of the triple Laplace-Sumudu transformation in the subsection.

Theorem 3

If \(L_xL_yS_t[u(x,y,t)]=\overline{u}(\rho _1,\rho _2,\sigma ), L_xL_yS_t[v(x,y,t)]=\overline{v}(\rho _1,\rho _2,\sigma )\), then

$$\begin{aligned} L_xL_yS_t[(u***v)(x,y,t)]=\sigma \overline{u}(\rho _1,\rho _2,\sigma )\overline{v}(\rho _1,\rho _2,\sigma ). \end{aligned}$$

Proof

By the definition of the triple Laplace-Sumudu transform and Theorem 2, we have

$$\begin{aligned}{} & {} L_xL_yS_t[(u***v)(x,y,t)]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}(u***v)(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\left\{ \int _0^x \int _0^y \int _0^t u(x-\delta ,y-\eta ,t-\varepsilon )v(\delta ,\eta ,\varepsilon ){\mathrm d}\delta {\mathrm d}\eta {\mathrm d}\varepsilon \right\} {\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\left\{ \int _0^\infty \int _0^\infty \int _0^\infty u(x-\delta ,y-\eta ,t-\varepsilon )\right. \\ \quad{} & {} \times \left. H(x-\delta ,y-\eta ,t-\varepsilon )v(\delta ,\eta ,\varepsilon ){\mathrm d}\delta {\mathrm d}\eta {\mathrm d}\varepsilon \right\} {\mathrm d}x{\mathrm d}y{\mathrm d}t\\= & {} \int _0^\infty \int _0^\infty \int _0^\infty v(\delta ,\eta ,\varepsilon ){\mathrm d}\delta {\mathrm d}\eta {\mathrm d}\varepsilon \nonumber \\ \quad{} & {} \times \left\{ \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}u(x-\delta ,y-\eta ,t-\varepsilon )H(x-\delta ,y-\eta ,t-\varepsilon ){\mathrm d}x{\mathrm d}y{\mathrm d}t\right\} \\= & {} \int _0^\infty \int _0^\infty \int _0^\infty v(\delta ,\eta ,\varepsilon ){\mathrm d}\delta {\mathrm d}\eta {\mathrm d}\varepsilon \left\{ e^{-\rho _1\delta -\rho _2\eta -\frac{\varepsilon }{\sigma }}\overline{u}(\rho _1,\rho _2,\sigma )\right\} \\= & {} \overline{u}(\rho _1,\rho _2,\sigma )\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1\delta -\rho _2\eta -\frac{\varepsilon }{\sigma }}v(\delta ,\eta ,\varepsilon ){\mathrm d}\delta {\mathrm d}\eta {\mathrm d}\varepsilon \\= & {} \sigma \overline{u}(\rho _1,\rho _2,\sigma )\overline{v}(\rho _1,\rho _2,\sigma ). \end{aligned}$$

The proof is complete. \(\square\)

3.3 Sufficient Conditions for Existence of the Triple Laplace-Sumudu Transform

We say that u(x, y, t) is of exponential order if there exist positive numbers a, b, c and M such that

$$\begin{aligned} |u(x,y,t)|\leqslant Me^{ax+by+ct} \end{aligned}$$

holds for all sufficiently large x, y, t.

For example, the function \(u(x,y,t)=(\cos 2x)\cdot (\sin y)\cdot 5t\) is of exponential order.

Theorem 4

Suppose that a function u(x, y, t) is continuous on \([0,\infty )\times [0,\infty )\times [0,\infty )\) and of exponential order with

$$\begin{aligned} |u(x,y,t)|\leqslant Me^{ax+by+ct} \end{aligned}$$

for all \(x\geqslant 0, y\geqslant 0, t\geqslant 0\). Then \(L_xL_yS_t[u(x,y,t)]=\overline{u}(\rho _1,\rho _2,\sigma )\) exists for all \(\rho _1>a,\rho _2>b,\frac{1}{\sigma }>c.\)

Proof

By the definition of the triple Laplace-Sumudu transformation, we have

$$\begin{aligned} |\overline{u}(\rho _1,\rho _2,\sigma )|= & {} \left| \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}u(x,y,t){\mathrm d}x{\mathrm d}y{\mathrm d}t\right| \\\leqslant & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}|u(x,y,t)|{\mathrm d}x{\mathrm d}y{\mathrm d}t \\\leqslant & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}Me^{ax+by+ct}{\mathrm d}x{\mathrm d}y{\mathrm d}t \\\leqslant & {} \frac{M}{\sigma }\int _0^\infty e^{-(\rho _1-a)x}{\mathrm d}x \int _0^\infty e^{-(\rho _2-b)y}{\mathrm d}y\int _0^\infty e^{-\left( \frac{1}{\sigma }-c\right) t}{\mathrm d}t \\= & {} \frac{M}{(\rho _1-a)(\rho _2-b)(1-\sigma c)}<\infty . \end{aligned}$$

The proof is complete. \(\square\)

4 Triple Laplace-Sumudu Transforms of Some Basic Functions

In this section, in order to gain familiarity with the above fundamental properties of the transformation, let us first use it to obtain a few transforms of some basic functions.

(1) Let \(u(x,y,t)=1, x>0,y>0,t>0\), then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )= & {} L_xL_yS_t[1]=\frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}{\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \int _0^\infty e^{-\rho _1x}{\mathrm d}x \cdot \int _0^\infty e^{-\rho _2y}{\mathrm d}y \cdot \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}{\mathrm d}t=\frac{1}{\rho _1\rho _2}. \end{aligned}$$

   (2) Let \(u(x,y,t)=x^\alpha y^\beta t^\gamma , x>0,y>0,t>0\), then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )= & {} L_xL_yS_t[x^\alpha y^\beta t^\gamma ]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}x^\alpha y^\beta t^\gamma {\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \int _0^\infty e^{-\rho _1x}x^\alpha {\mathrm d}x \cdot \int _0^\infty e^{-\rho _2y}y^\beta {\mathrm d}y \cdot \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}t^\gamma {\mathrm d}t\\= & {} \frac{\sigma ^\gamma \Gamma (\alpha +1)\Gamma (\beta +1)\Gamma (\gamma +1)}{\rho _1^{\alpha +1}\rho _2^{\beta +1}}. \end{aligned}$$

It is easy to get the triple Laplace-Sumudu transforms of other basic functions of the same type. We present these basic results in the Table 1 below.

Table 1 Triple Laplace-Sumudu transformations of some basic functions
Full size table

(3) Let \(u(x,y,t)=e^{\alpha x+\beta y+\gamma t}, x>0,y>0,t>0\), then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )= & {} L_xL_yS_t[e^{\alpha x+\beta y+\gamma t}]=\frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\cdot e^{\alpha x+\beta y+\gamma t}{\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \frac{1}{\sigma }\int _0^\infty e^{(-\rho _1+\alpha )x}{\mathrm d}x \cdot \int _0^\infty e^{(-\rho _2+\beta )y}{\mathrm d}y \cdot \int _0^\infty e^{\left( -\frac{1}{\sigma }+\gamma \right) t}{\mathrm d}t\\= & {} \frac{1}{(\rho _1-\alpha )(\rho _2-\beta )(1-\sigma \gamma )}. \end{aligned}$$

It is also easy to obtain the triple Laplace-Sumudu transforms of other basic functions of the same type. We present these basic results in the Table 2 below.

Table 2 Triple Laplace-Sumudu transformations of some basic functions
Full size table

(4) Let \(u(x,y,t)=\sin (\alpha x+\beta y+ \gamma t),\) or \(u(x,y,t)=\cos (\alpha x+\beta y+ \gamma t), x>0,y>0,t>0\), then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )= & {} L_xL_yS_t[\sin (\alpha x+\beta y+ \gamma t)]=\frac{\rho _1\rho _2\sigma \gamma +\rho _1\beta +\rho _2\alpha -\sigma \gamma \alpha \beta }{(\rho _1^2+\alpha ^2)(\rho _2^2+\beta ^2)(1+\sigma ^2\gamma ^2)};\\ \overline{u}(\rho _1,\rho _2,\sigma )= & {} L_xL_yS_t[\cos (\alpha x+\beta y+ \gamma t)]=\frac{\rho _1\rho _2-\rho _1\beta \sigma \gamma -\rho _2\alpha \sigma \gamma -\alpha \beta }{(\rho _1^2+\alpha ^2)(\rho _2^2+\beta ^2)(1+\sigma ^2\gamma ^2)}. \end{aligned}$$

Proof

These transforms can be evaluated directly by using Definition 2.3. Our derivation will be based on Euler’s identity

$$\begin{aligned} \cos (\alpha x+\beta y+ \gamma t)+i\sin (\alpha x+\beta y+ \gamma t)=e^{i(\alpha x+\beta y+ \gamma t)}, \end{aligned}$$

we have

$$\begin{aligned}{} & {} L_xL_yS_t[\cos (\alpha x+\beta y+ \gamma t)]+iL_xL_yS_t[\sin (\alpha x+\beta y+ \gamma t)]\\= & {} L_xL_yS_t[e^{i(\alpha x+\beta y+\gamma t)}]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\cdot e^{i(\alpha x+\beta y+\gamma t)}{\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \int _0^\infty e^{(-\rho _1+i\alpha )x}{\mathrm d}x \cdot \int _0^\infty e^{(-\rho _2+i\beta )y}{\mathrm d}y \cdot \frac{1}{\sigma }\int _0^\infty e^{\left( -\frac{1}{\sigma }+i\gamma \right) t}{\mathrm d}t\\= & {} -\frac{1}{-\rho _1+i\alpha }\left( -\frac{1}{-\rho _2+i\beta }\right) \frac{1}{\sigma }\left( -\frac{1}{-\frac{1}{\sigma }+i\gamma }\right) \\= & {} \frac{(-\rho _1-i\alpha )(-\rho _2-i\beta )(1+i\sigma \gamma )}{(\rho _1^2+\alpha ^2)(\rho _2^2+\beta ^2)(1+\sigma ^2\gamma ^2)}\\= & {} \frac{\rho _1\rho _2-\rho _1\beta \sigma \gamma -\rho _2\alpha \sigma \gamma -\alpha \beta +i(\rho _1\rho _2\sigma \gamma +\rho _1\beta +\rho _2\alpha -\sigma \gamma \alpha \beta )}{(\rho _1^2+\alpha ^2)(\rho _2^2+\beta ^2)(1+\sigma ^2\gamma ^2)}, \end{aligned}$$

thus we get the result we wanted. \(\square\)

It is also easy to obtain the triple Laplace-Sumudu transforms of other basic functions of the same type. We present these basic results in the Tables 3 and 4 below.

Table 3 Triple Laplace-Sumudu transformations of some basic functions
Full size table
Table 4 Triple Laplace-Sumudu transformations of some basic functions
Full size table

(5) Let \(u(x,y,t)=\sinh (\alpha x+\beta y+\gamma t), x>0,y>0,t>0\), then

$$\begin{aligned}{} & {} \overline{u}(\rho _1,\rho _2,\sigma )\\= & {} L_xL_yS_t[\sinh (\alpha x+\beta y+\gamma t)]\\= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}\left[ \frac{e^{\alpha x+\beta y+\gamma t}-e^{-(\alpha x+\beta y+\gamma t)}}{2}\right] {\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \frac{1}{2\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}e^{\alpha x+\beta y+\gamma t}{\mathrm d}x{\mathrm d}y{\mathrm d}t\\ \quad{} & {} -\frac{1}{2\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}e^{-(\alpha x+\beta y+\gamma t)}{\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \frac{1}{2\sigma }\left[ \frac{1}{(\rho _1-\alpha )(\rho _2-\beta )(\frac{1}{\sigma }-\gamma )}-\frac{1}{(\rho _1+\alpha )(\rho _2+\beta )(\frac{1}{\sigma }+\gamma )}\right] \\= & {} \frac{\rho _1\rho _2\gamma \sigma +\rho _1\beta +\alpha \rho _2+\alpha \beta \gamma \sigma }{(\rho _1^2-\alpha ^2)(\rho _2^2-\beta ^2)(1-\sigma ^2\gamma ^2)}. \end{aligned}$$

Similarly, when \(u(x,y,t)=\cosh (\alpha x+\beta y+\gamma t), x>0,y>0,t>0\), we have

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )= & {} L_xL_yS_t[\cosh (\alpha x+\beta y+\gamma t)]\\= & {} \frac{\rho _1\rho _2+\rho _1\beta \gamma \sigma +\alpha \rho _2\gamma \sigma +\alpha \beta }{(\rho _1^2-\alpha ^2)(\rho _2^2-\beta ^2)(1-\sigma ^2\gamma ^2)}. \end{aligned}$$

It is also easy to obtain the triple Laplace-Sumudu transforms of other basic functions of the same type. We present these basic results in the Tables 5 and 6 below.

Table 5 Triple Laplace-Sumudu transformations of some basic functions
Full size table
Table 6 Triple Laplace-Sumudu transformations of some basic functions
Full size table

(6) Let \(u(x,y,t)=f(x)g(y)h(t)\), then

$$\begin{aligned} L_xL_yS_t[f(x)g(y)h(t)]= & {} \frac{1}{\sigma }\int _0^\infty \int _0^\infty \int _0^\infty e^{-\rho _1x-\rho _2y-\frac{t}{\sigma }}f(x)g(y)h(t){\mathrm d}x{\mathrm d}y{\mathrm d}t \\= & {} \int _0^\infty e^{-\rho _1x}f(x){\mathrm d}x \cdot \int _0^\infty e^{-\rho _2y}g(y){\mathrm d}y \cdot \frac{1}{\sigma }\int _0^\infty e^{-\frac{t}{\sigma }}h(t){\mathrm d}t\\= & {} L_x[f(x)]L_y[g(y)]S_t[h(t)]. \end{aligned}$$

   (7) Let \(u(x,y,t)=J_0(c\sqrt{xt})\), or \(u(x,y,t)=J_0(c\sqrt{yt})\), then

$$\begin{aligned}{} & {} L_xL_yS_t[J_0(c\sqrt{xt})]=L_yL_xS_t[J_0(c\sqrt{xt})]=L_y\left[ \frac{4}{4\rho _1+\sigma c^2}\right] =\frac{4}{\rho _2(4\rho _1+\sigma c^2)},\\{} & {} \quad L_xL_yS_t[J_0(c\sqrt{yt})]=L_x\left[ \frac{4}{4\rho _2+\sigma c^2}\right] =\frac{4}{\rho _1(4\rho _2+\sigma c^2)}, \end{aligned}$$

where \(J_0(c\sqrt{xt})\), \(J_0(c\sqrt{yt})\) are the modified Bessel function of order zero.

5 Applications of the Triple Laplace-Sumudu Transform

In this section, we investigate the applications of the triple Laplace-Sumudu transform. The solutions of some heat flow equations and wave equations with the initial-boundary values are obtained by using the triple Laplace-Sumudu transform method.

Example 1

Consider the homogeneous initial-boundary value problem of the two dimensional heat flow equation

$$\begin{aligned} \left\{ \begin{aligned}&u_t=2(u_{xx}+u_{yy}), \ 0<x,y<\pi , \ t>0,\\&u(0,y,t)=u(\pi ,y,t)=0,\\&u(x,0,t)=u(x,\pi ,t)=0,\\&u_x(0,y,t)=e^{-4t}\sin y,\\&u_y(x,0,t)=e^{-4t}\sin x,\\&u(x,y,0)=\sin x\sin y. \end{aligned} \right. \end{aligned}$$
(1)

Firstly, taking the triple Laplace-Sumudu transform to the heat flow equation in (1), we get

$$\begin{aligned} L_xL_yS_t[u_t]=2L_xL_yS_t[u_{xx}]+2L_xL_yS_t[u_{yy}]. \end{aligned}$$

By the basic properties of the triple Laplace-Sumudu transform, we have

$$\begin{aligned}{} & {} \frac{1}{\sigma }\overline{u}(\rho _1,\rho _2,\sigma )-\frac{1}{\sigma }L_xL_y[u(x,y,0)]\\= & {} 2\left[ \rho _1^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _1L_yS_t[u(0,y,t)]-L_yS_t[u_x(0,y,t)]\right] \\{} & {} \quad +2\left[ \rho _2^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _2L_xS_t[u(x,0,t)]-L_xS_t[u_y(x,0,t)]\right] . \end{aligned}$$

By calculating, we obtain

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma }-2\rho _1^2-2\rho _2^2\right) \overline{u}(\rho _1,\rho _2,\sigma )\nonumber \\= & {} \frac{1}{\sigma }L_xL_y[u(x,y,0)]-2\rho _1L_yS_t[u(0,y,t)]-2L_yS_t[u_x(0,y,t)]\nonumber \\{} & {} -2\rho _2L_xS_t[u(x,0,t)]-2L_xS_t[u_y(x,0,t)]. \end{aligned}$$
(2)

By considering the transforms of the boundary conditions, we have

$$\begin{aligned} L_yS_t[u_x(0,y,t)]= & {} L_yS_t[e^{-4t}\sin y]=\frac{1}{1+4\sigma }\cdot \frac{1}{1+\rho _2^2}, \end{aligned}$$
(3)
$$\begin{aligned} L_xS_t[u_y(x,0,t)]= & {} L_xS_t[e^{-4t}\sin x]=\frac{1}{1+4\sigma }\cdot \frac{1}{1+\rho _1^2}. \end{aligned}$$
(4)

By substituting (3) and (4) into (2) and considering the transform of the initial condition, we obtain

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma }-2\rho _1^2-2\rho _2^2\right) \overline{u}(\rho _1,\rho _2,\sigma )\\= & {} \frac{1}{\sigma }\cdot \frac{1}{1+\rho _1^2}\cdot \frac{1}{1+\rho _2^2}-2\cdot \frac{1}{1+4\sigma }\cdot \frac{1}{1+\rho _2^2}-2\cdot \frac{1}{1+4\sigma }\cdot \frac{1}{1+\rho _1^2}\\= & {} \frac{1-2\sigma \rho _1^2-2\sigma \rho _2^2}{\sigma (1+4\sigma )(1+\rho _1^2)(1+\rho _2^2)}. \end{aligned}$$

Then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )=\frac{1}{(1+4\sigma )(1+\rho _1^2)(1+\rho _2^2)}. \end{aligned}$$

Taking the inverse triple Laplace-Sumudu transform of \(\overline{u}(\rho _1,\rho _2,\sigma )\), we get the solution of the homogeneous initial-boundary value problem of the two dimensional heat flow equation (1)

$$\begin{aligned} u(x,y,t)=(L_xL_yS_t)^{-1}\left[ \frac{1}{(1+4\sigma )(1+\rho _1^2)(1+\rho _2^2)}\right] =e^{-4t}\sin x\sin y. \end{aligned}$$

Example 2

Use the triple Laplace-Sumudu transformation method to solve the homogeneous initial-boundary value problem of the two dimensional wave equation

$$\begin{aligned} \left\{ \begin{aligned}&u_{tt}=2(u_{xx}+u_{yy}), \ 0<x,y<\pi , \ t>0,\\&u(0,y,t)=u(\pi ,y,t)=0,\\&u(x,0,t)=u(x,\pi ,t)=0,\\&u_x(0,y,t)=\sin y\sin (2t),\\&u_y(x,0,t)=\sin x\sin (2t),\\&u(x,y,0)=0,u_t(x,y,0)=2\sin x\sin y. \end{aligned} \right. \end{aligned}$$
(5)

Taking the triple Laplace-Sumudu transform to the wave equation in (5), we get

$$\begin{aligned} L_xL_yS_t[u_{tt}]=2L_xL_yS_t[u_{xx}]+2L_xL_yS_t[u_{yy}]. \end{aligned}$$

By the basic properties of the triple Laplace-Sumudu transform, we have

$$\begin{aligned}{} & {} \frac{1}{\sigma ^2}\overline{u}(\rho _1,\rho _2,\sigma )-\frac{1}{\sigma ^2}L_xL_y[u(x,y,0)]-\frac{1}{\sigma }L_xL_y[u_t(x,y,0)]\\= & {} 2\left[ \rho _1^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _1L_yS_t[u(0,y,t)]-L_yS_t[u_x(0,y,t)]\right] \\{} & {} \quad +2\left[ \rho _2^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _2L_xS_t[u(x,0,t)]-L_xS_t[u_y(x,0,t)]\right] . \end{aligned}$$

By calculating and considering the transform of the initial condition, we obtain

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma ^2}-2\rho _1^2-2\rho _2^2\right) \overline{u}(\rho _1,\rho _2,\sigma )\nonumber \\= & {} \frac{1}{\sigma }L_xL_y[u_t(x,y,0)]-2\rho _1L_yS_t[u(0,y,t)]-2L_yS_t[u_x(0,y,t)]\nonumber \\{} & {} -2\rho _2L_xS_t[u(x,0,t)]-2L_xS_t[u_y(x,0,t)]. \end{aligned}$$
(6)

Considering the transforms of the initial-boundary conditions, we have

$$\begin{aligned} L_xL_y[u_t(x,y,0)]= & {} L_xL_y[2\sin x\sin y]=\frac{2}{1+\rho _1^2}\cdot \frac{1}{1+\rho _2^2}, \end{aligned}$$
(7)
$$\begin{aligned} L_yS_t[u_x(0,y,t)]= & {} L_yS_t[\sin y\sin (2t)]=\frac{2\sigma }{1+4\sigma ^2}\cdot \frac{1}{1+\rho _2^2}, \end{aligned}$$
(8)
$$\begin{aligned} L_xS_t[u_y(x,0,t)]= & {} L_xS_t[\sin x\sin (2t)]=\frac{2\sigma }{1+4\sigma ^2}\cdot \frac{1}{1+\rho _1^2}. \end{aligned}$$
(9)

By substituting (7), (8) and (9) into (6), we obtain

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma ^2}-2\rho _1^2-2\rho _2^2\right) \overline{u}(\rho _1,\rho _2,\sigma )\\= & {} \frac{2}{\sigma }\cdot \frac{1}{1+\rho _1^2}\cdot \frac{1}{1+\rho _2^2}-2\cdot \frac{2\sigma }{1+4\sigma ^2}\cdot \frac{1}{1+\rho _2^2}-2\cdot \frac{2\sigma }{1+4\sigma ^2}\cdot \frac{1}{1+\rho _1^2}\\= & {} \frac{2-4\sigma ^2\rho _1^2-4\sigma ^2\rho _2^2}{\sigma (1+\rho _1^2)(1+\rho _2^2)(1+4\sigma ^2)}. \end{aligned}$$

Then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )=\frac{2\sigma }{(1+\rho _1^2)(1+\rho _2^2)(1+4\sigma ^2)}. \end{aligned}$$

Taking the inverse triple Laplace-Sumudu transform of \(\overline{u}(\rho _1,\rho _2,\sigma )\), we obtain the solution of the homogeneous initial-boundary value problem of the two dimensional wave equation (5)

$$\begin{aligned} u(x,y,t)=(L_xL_yS_t)^{-1}\left[ \frac{2\sigma }{(1+\rho _1^2)(1+\rho _2^2)(1+4\sigma ^2)}\right] =\sin x\sin y\sin (2t). \end{aligned}$$

Example 3

Consider the initial-boundary value problem of the heat equation with a lateral heat loss

$$\begin{aligned} \left\{ \begin{aligned}&u_{t}=3(u_{xx}+u_{yy})-2u, \ 0<x,y<\pi , \ t>0,\\&u(0,y,t)=-u(\pi ,y,t)=e^{-8t}\sin y,\\&u(x,0,t)=-u(x,\pi ,t)=e^{-8t}\sin x,\\&u_x(0,y,t)=e^{-8t}\cos y,\\&u_y(x,0,t)=e^{-8t}\cos x,\\&u(x,y,0)=\sin (x+y). \end{aligned} \right. \end{aligned}$$
(10)

By taking the triple Laplace-Sumudu transform to the heat equation in (10), we get

$$\begin{aligned} L_xL_yS_t[u_{t}]=3L_xL_yS_t[u_{xx}]+3L_xL_yS_t[u_{yy}]-2L_xL_yS_t[u]. \end{aligned}$$

By using the basic properties of the triple Laplace-Sumudu transform, we obtain

$$\begin{aligned}{} & {} \frac{1}{\sigma }\overline{u}(\rho _1,\rho _2,\sigma )-\frac{1}{\sigma }L_xL_y[u(x,y,0)]\\= & {} 3\left[ \rho _1^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _1L_yS_t[u(0,y,t)]-L_yS_t[u_x(0,y,t)]\right] \\{} & {} +3\left[ \rho _2^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _2L_xS_t[u(x,0,t)]-L_xS_t[u_y(x,0,t)]\right] -2\overline{u}(\rho _1,\rho _2,\sigma ). \end{aligned}$$

By calculating, we have

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma }-3\rho _1^2-3\rho _2^2+2\right) \overline{u}(\rho _1,\rho _2,\sigma )\nonumber \\= & {} \frac{1}{\sigma }L_xL_y[u(x,y,0)]-3\rho _1L_yS_t[u(0,y,t)]-3L_yS_t[u_x(0,y,t)]\nonumber \\{} & {} -3\rho _2L_xS_t[u(x,0,t)]-3L_xS_t[u_y(x,0,t)]. \end{aligned}$$
(11)

Considering the transforms of the initial-boundary conditions, we have

$$\begin{aligned} L_xL_y[u(x,y,0)]= & {} L_xL_y[\sin (x+y)]=\frac{\rho _1+\rho _2}{(1+\rho _1^2)(1+\rho _2^2)}, \end{aligned}$$
(12)
$$\begin{aligned} L_yS_t[u(0,y,t)]= & {} L_yS_t[e^{-8t}\sin y]=\frac{1}{1+8\sigma }\cdot \frac{1}{1+\rho _2^2}, \end{aligned}$$
(13)
$$\begin{aligned} L_yS_t[u_x(0,y,t)]= & {} L_yS_t[e^{-8t}\cos y]=\frac{1}{1+8\sigma }\cdot \frac{\rho _2}{1+\rho _2^2}, \end{aligned}$$
(14)
$$\begin{aligned} L_xS_t[u(x,0,t)]= & {} L_xS_t[e^{-8t}\sin x]=\frac{1}{1+8\sigma }\cdot \frac{1}{1+\rho _1^2}. \end{aligned}$$
(15)
$$\begin{aligned} L_xS_t[u_y(x,0,t)]= & {} L_xS_t[e^{-8t}\cos x]=\frac{1}{1+8\sigma }\cdot \frac{\rho _1}{1+\rho _1^2}. \end{aligned}$$
(16)

By substituting (12), (13), (14), (15) and (16) into (11), we get

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma }-3\rho _1^2-3\rho _2^2+2\right) \overline{u}(\rho _1,\rho _2,\sigma )\\= & {} \frac{1}{\sigma }\cdot \frac{\rho _1+\rho _2}{(1+\rho _1^2)(1+\rho _2^2)}-3\rho _1\cdot \frac{1}{(1+8\sigma )(1+\rho _2^2)} -3\cdot \frac{1}{1+8\sigma }\cdot \frac{\rho _2}{1+\rho _2^2}\\{} & {} -3\rho _2\cdot \frac{1}{(1+8\sigma )(1+\rho _1^2)}-3\cdot \frac{1}{1+8\sigma }\cdot \frac{\rho _1}{1+\rho _1^2}\\= & {} \frac{(\rho _1+\rho _2)(1+2\sigma -3\rho _1^2\sigma -3\rho _2^2\sigma )}{\sigma (1+\rho _1^2)(1+\rho _2^2)(1+8\sigma )}. \end{aligned}$$

Thus

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )=\frac{\rho _1+\rho _2}{(1+\rho _1^2)(1+\rho _2^2)(1+8\sigma )}. \end{aligned}$$

By computing the inverse triple Laplace-Sumudu transform of \(\overline{u}(\rho _1,\rho _2,\sigma )\), we obtain the solution of the initial-boundary value problem of two dimensional heat flow equation with the lateral heat loss (10)

$$\begin{aligned} u(x,y,t)=(L_xL_yS_t)^{-1}\left[ \frac{\rho _1+\rho _2}{(1+\rho _1^2)(1+\rho _2^2)(1+8\sigma )}\right] =e^{-8t}\sin (x+y). \end{aligned}$$

Example 4

Solve the following inhomogeneous initial-boundary value problem of the heat equation

$$\begin{aligned} \left\{ \begin{aligned}&u_{t}=u_{xx}+u_{yy}+2\cos (x+y), \ 0<x,y<\pi , \ t>0,\\&u(0,y,t)=e^{-2t}\sin y+\cos y,\\&u(\pi ,y,t)=-(e^{-2t}\sin y+\cos y),\\&u(x,0,t)=e^{-2t}\sin x+\cos x,\\&u(x,\pi ,t)=-(e^{-2t}\sin x+\cos x),\\&u_x(0,y,t)=e^{-2t}\cos y-\sin y,\\&u_y(x,0,t)=e^{-2t}\cos x-\sin x,\\&u(x,y,0)=\sin (x+y)+\cos (x+y). \end{aligned} \right. \end{aligned}$$
(17)

We take the triple Laplace-Sumudu transform to the inhomogeneous heat equation in (17) and get

$$\begin{aligned} L_xL_yS_t[u_{t}]=L_xL_yS_t[u_{xx}]+L_xL_yS_t[u_{yy}]+L_xL_yS_t[2\cos (x+y)]. \end{aligned}$$

By using the basic properties of the triple Laplace-Sumudu transform, we have

$$\begin{aligned}{} & {} \frac{1}{\sigma }\overline{u}(\rho _1,\rho _2,\sigma )-\frac{1}{\sigma }L_xL_y[u(x,y,0)]\\= & {} \rho _1^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _1L_yS_t[u(0,y,t)]-L_yS_t[u_x(0,y,t)]\\{} & {} +\rho _2^2\overline{u}(\rho _1,\rho _2,\sigma )-\rho _2L_xS_t[u(x,0,t)]-L_xS_t[u_y(x,0,t)]+\frac{2(\rho _1\rho _2-1)}{(1+\rho _1^2)(1+\rho _2^2)}. \end{aligned}$$

By calculating, we obtain

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma }-\rho _1^2-\rho _2^2\right) \overline{u}(\rho _1,\rho _2,\sigma )\nonumber \\= & {} \frac{1}{\sigma }L_xL_y[u(x,y,0)]-\rho _1L_yS_t[u(0,y,t)]-L_yS_t[u_x(0,y,t)]\nonumber \\{} & {} -\rho _2L_xS_t[u(x,0,t)]-L_xS_t[u_y(x,0,t)]+\frac{2(\rho _1\rho _2-1)}{(1+\rho _1^2)(1+\rho _2^2)}. \end{aligned}$$
(18)

Considering the transforms of the initial-boundary conditions, we have

$$\begin{aligned} L_xL_y[u(x,y,0)]= & {} L_xL_y[\sin (x+y)+\cos (x+y)]\nonumber \\= & {} \frac{\rho _1+\rho _2}{(1+\rho _1^2)(1+\rho _2^2)}+\frac{\rho _1\rho _2-1}{(1+\rho _1^2)(1+\rho _2^2)}, \end{aligned}$$
(19)
$$\begin{aligned} L_yS_t[u(0,y,t)]= & {} L_yS_t[e^{-2t}\sin y+\cos y]=\frac{1}{1+2\sigma }\cdot \frac{1}{1+\rho _2^2}+\frac{\rho _2}{1+\rho _2^2}, \end{aligned}$$
(20)
$$\begin{aligned} L_yS_t[u_x(0,y,t)]= & {} L_yS_t[e^{-2t}\cos y-\sin y]=\frac{1}{1+2\sigma }\cdot \frac{\rho _2}{1+\rho _2^2}-\frac{1}{1+\rho _2^2}, \end{aligned}$$
(21)
$$\begin{aligned} L_xS_t[u(x,0,t)]= & {} L_xS_t[e^{-2t}\sin x+\cos x]=\frac{1}{1+2\sigma }\cdot \frac{1}{1+\rho _1^2}+\frac{\rho _1}{1+\rho _1^2}, \end{aligned}$$
(22)
$$\begin{aligned} L_xS_t[u_y(x,0,t)]= & {} L_xS_t[e^{-2t}\cos x-\sin x]=\frac{1}{1+2\sigma }\cdot \frac{\rho _1}{1+\rho _1^2}-\frac{1}{1+\rho _1^2}. \end{aligned}$$
(23)

By substituting (19), (20), (21), (22) and (23) into (18), we get

$$\begin{aligned}{} & {} \left( \frac{1}{\sigma }-\rho _1^2-\rho _2^2\right) \overline{u}(\rho _1,\rho _2,\sigma )\\= & {} \frac{1}{\sigma }\left( \frac{\rho _1+\rho _2}{(1+\rho _1^2)(1+\rho _2^2)}+\frac{\rho _1\rho _2-1}{(1+\rho _1^2)(1+\rho _2^2)}\right) -\rho _1\left( \frac{1}{1+2\sigma }\cdot \frac{1}{1+\rho _2^2}+\frac{\rho _2}{1+\rho _2^2}\right) \\{} & {} \quad -\frac{1}{1+2\sigma }\cdot \frac{\rho _2}{1+\rho _2^2}+\frac{1}{1+\rho _2^2}-\rho _2\left( \frac{1}{1+2\sigma }\cdot \frac{1}{1+\rho _1^2}+\frac{\rho _1}{1+\rho _1^2}\right) \\ \quad{} & {} -\frac{1}{(1+2\sigma )}\cdot \frac{\rho _1}{(1+\rho _1^2)}+\frac{1}{(1+\rho _1^2)}+\frac{2(\rho _1\rho _2-1)}{(1+\rho _1^2)(1+\rho _2^2)}\\= & {} \frac{A}{\sigma (1+\rho _1^2)(1+\rho _2^2)(1+2\sigma )}, \end{aligned}$$

where

$$\begin{aligned} A= & {} (\rho _1+\rho _2+\rho _1\rho _2-1)(1+2\sigma )-\rho _1(1+(1+2\sigma )\rho _2)(1+\rho _1^2)\sigma \\{} & {} -\sigma (1+\rho _1^2)(\rho _2-(1+2\sigma )) \\{} & {} -\rho _2(1+(1+2\sigma )\rho _1)\sigma (1+\rho _2^2)-\sigma (1+\rho _2^2)(\rho _1-(1+2\sigma ))+2\sigma (1+2\sigma )(\rho _1\rho _2-1)\\= & {} \rho _1+\rho _2+\rho _1\rho _2-1-2\sigma -\rho _1^3\sigma -\rho _1^3\sigma \rho _2-2\sigma ^2\rho _1^3\rho _2-\rho _2\rho _1^2\sigma +\sigma \rho _1^2+2\sigma ^2\rho _1^2\\{} & {} -\rho _2^3\sigma -\sigma \rho _1\rho _2^3-2\sigma ^2\rho _1\rho _2^3-\sigma \rho _1\rho _2^2+\sigma \rho _2^2+2\sigma ^2\rho _2^2+2\sigma \rho _1\rho _2\\= & {} \left[ \rho _1+\rho _2+(1+2\sigma )(\rho _1\rho _2-1)\right] \left[ 1-\sigma (\rho _1^2+\rho _2^2)\right] . \end{aligned}$$

Then

$$\begin{aligned} \overline{u}(\rho _1,\rho _2,\sigma )= & {} \frac{\rho _1+\rho _2+(1+2\sigma )(\rho _1\rho _2-1)}{(1+\rho _1^2)(1+\rho _2^2)(1+2\sigma )} \\= & {} \frac{\rho _1+\rho _2}{(1+2\sigma )(1+\rho _1^2)(1+\rho _2^2)}+\frac{\rho _1\rho _2-1}{(1+\rho _1^2)(1+\rho _2^2)}. \end{aligned}$$

By taking the inverse Laplace-Sumudu transform of \(\overline{u}(\rho _1,\rho _2,\sigma )\), we obtain the solution of the initial-boundary value problem of the inhomogeneous heat flow equation (17)

$$\begin{aligned} u(x,y,t)= & {} (L_xL_yS_t)^{-1}\left[ \frac{\rho _1+\rho _2}{(1+2\sigma )(1+\rho _1^2)(1+\rho _2^2)}+\frac{\rho _1\rho _2-1}{(1+\rho _1^2)(1+\rho _2^2)}\right] \\= & {} e^{-2t}\sin (x+y)+\cos (x+y). \end{aligned}$$

Remark 1

For two-dimensional, time-dependent initial-boundary value problems of the partial differential equations, the efficiency of this method is higher than that of other methods. It can directly transform the initial-boundary value problem of the partial differential equation into an algebraic equation and obtain the solution of the problem. Other methods require the assistance of other techniques. For example, solving a related ordinary differential equation at the same time as the integral transform, or iterating at the same time as the integral transform. Examples from the references [10, 11, 28] are available for comparisons.

6 Conclusion

In this study, we have introduced a new mixed integral transformation method for solving the initial-boundary value problems of the partial differential equations. The basic properties of the triple Laplace-Sumudu transform have been studied and the triple Laplace-Sumudu transforms of some basic functions have been given. Some important results have been described and have been tested with the help of some examples. The results show that the triple mixed Laplace-Sumudu transform is more efficient and useful to deal with some initial-boundary value problems.

The general procedure to solve the initial-boundary value problems with 2-dimensional space by using the triple mixed Laplace-Sumudu transform is as follows:

Step 1: The two sides of the partial differential equation are transformed to obtain an equation \((\sharp )\) by using the triple mixed Laplace-Sumudu transform. The resulting equation \((\sharp )\) includes the double integral transformations of initial-boundary conditions and the function \(\overline{u}(\rho _1,\rho _2,\sigma )\);

Step 2: We calculate the double integral transformations of the initial-boundary conditions by using the given initial-boundary conditions;

Step 3: By substituting the results of Step 2 into the equation \((\sharp )\) and some basic algebraic computation, we obtain an algebraic expression \((\star )\) about \(\overline{u}(\rho _1,\rho _2,\sigma )\);

Step 4: The solution u(x, y, t) of the initial boundary value problem is obtained by the inverse triple mixed Laplace-Sumudu transformation of the expression \(\overline{u}(\rho _1,\rho _2,\sigma )\).

Step 5: We need to check if the result satisfies the partial differential equation and initial-boundary conditions.