On the convergence of triple Elzaki transform

Abstract

In this work, the convergence properties of triple Elzaki transform was examine and the results presented in the form of theorems on convergence, absolute convergence and uniform convergence of triple Elzaki transform. The triple Elzaki transform of triple integral examined for integral evaluation. Finally, Volterra integro-partial differential equation solved by using triple Elzaki transform.

1 Introduction

We have introduced the modified version of Sumudu and Laplace transforms [1, 2], namely Elzaki transform to solve some problems. In the recent years, this transform has become more accurate and efficient. Elzaki transformation [3,4,5,6,7,8,9,10,11], introduced by Tarig Elzaki in 2011. Elzaki transform solving differential equations with variable coefficients, which are not solved, by Sumudu transform [12,13,14,15,16].

In this paper, we have discussed the various convergence properties of triple Elzaki transformation. The triple Elzaki transform is a very useful technique to solve some differential equations [17,18,19,20], partial differential equations, and integral equations. This transform used as a very efficient tool in simplifying the calculations in many disciplines of engineering and mathematics.

Triple Elzaki transform:

Let ${\text{f }}\left( {{\text{x}}, {\text{y}}, {\text{t}}} \right)$ be a function that can be expressed as convergent infinite series, and let $\left( {{\text{x}}, {\text{y}}, {\text{t}}} \right) \in R_{3}^{ + }$, then, the triple Elzaki transform is denoted by:

$$ {\text{E}}_{3} \left[ {{\text{f}}\left( {{\text{x, y, t}}} \right):\left( {{\uprho} ,{\text{s}},\updelta} \right)} \right] ={\uprho} {\text{s}}\updelta\mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - {\frac{{\text{x}}}{{\uprho} }} - {\frac{{\text{y}}}{{\hbox{s}}}} - {\frac{{\text{t}}}{{\updelta}}}}} f\left( {x,y,t} \right)dxdydt, $$

where $x ,y,t > 0 \;and\; {\rho ,} {\text{s,}} \updelta$ are transform variables for $x ,y$ and $t$ respectively, whenever the improper integral is convergent.

2 Convergence theorems of triple Elzaki integral

In this section, we prove the convergence theorem of triple Elzaki integral.

Theorem 2.1

If $\emptyset \left( {x,y,t} \right)$ is a continuous function on the positive of the $x ,y ,t$-plane. If the integral converges at ${\uprho} ={\uprho}_{ \circ }$, ${\text{s}} = {\text{s}}_{ \circ } ,\;\updelta =\updelta_{ \circ }$ then, the integral:

$${\uprho} {\text{s}}\updelta\mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho} } - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} \emptyset \left( {x,y,t} \right)dxdydt $$

(1)

Converges for at ${\uprho} <{\uprho} _{ \circ } ,\;\text{s} < {\text{s}}_{ \circ } ,\;\updelta <\updelta_{ \circ }$For the proof, we will use the following lemmas.

Lemma 2.2

If the integral,

$$ {\text{s}}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy }} $$

(2)

converges at ${\text{s}} = {\text{s}}_{ \circ }$ then the integral converges for ${\text{s}} < {\text{s}}_{ \circ }$.

Proof

Consider the set,

$$ \alpha \left( {x,y,t} \right) = {\text{s}}_{\circ} \mathop \int \limits_{0}^{\text{y}}\,{\text{e}}^{{ -\frac{{\text{u}}}{{{\text{s}}_{ \circ } }}}} \emptyset \left({x,u,t} \right){\text{du }},\quad \left( {0 < y < \infty }\right) $$

(3)

Therefore $\alpha \left( {{\text{x}},0,{\text{t}}} \right) = 0$ and $\lim_{y \to \infty } \alpha \left( {x,y,t} \right)\;{\text{exist}}$ because, the integral ${\text{s}}\int_{0}^{\infty } {{\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy}}}$ converges at ${\text{s}} = {\text{s}}_{ \circ }$, by a fundamental theorem of calculus [17, 18] we have:

$$ \alpha_{y} \left( {x,y,t} \right) = {\text{s}}_{ \circ } \emptyset \left( {x,v,t} \right){\text{e}}^{{ - \frac{{\text{y}}}{{{\text{s}}_{ \circ } }}}} $$

$$ {\text{choose}}\;\epsilon_{1} \;{\text{and}}\;{\text{R}}_{1} \; such\;that\; \left( {0 < \epsilon_{1} < {\text{R}}_{1} } \right),\;{\text{then:}} $$

$$ \begin{aligned} {\text{s}}\mathop \int \limits_{{\epsilon_{1} }}^{{{\text{R}}_{1} }} {\text{e}}^{{ - \frac{{\text{y}}}{{\text{s}}}}} \emptyset \left( {x,y,t} \right){\text{dy}} & = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\mathop \int \limits_{{\epsilon_{1} }}^{{{\text{R}}_{1} }}\,{\text{e}}^{{ - \frac{{\text{y}}}{{\text{s}}}}} \alpha_{y} \left( {x,y,t} \right){\text{e}}^{{\frac{{\text{y}}}{{{\text{s}}_{ \circ } }}}} {\text{dy}} \\ & = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\mathop \int \limits_{{\epsilon_{1} }}^{{{\text{R}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \alpha_{y} \left( {x,y,t} \right) {\text{dy}}, \\ \end{aligned} $$

by using integration by parts

$$ \begin{aligned} & = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\left\{ {\left| {{\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \alpha \left( {x,y,t} \right)} \right|_{{\epsilon_{1} }}^{{{\text{R}}_{1} }} - \mathop \int \limits_{{\epsilon_{1} }}^{{{\text{R}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \left[ { - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)} \right]\alpha \left( {x,y,t} \right) {\text{dy}}} \right\} \\ & s\mathop \int \limits_{{\epsilon_{1} }}^{{{\text{R}}_{1} }} {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy}} = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\left\{ { {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{R}}_{1} }} \alpha \left( {x,{\text{R}}_{1} ,{\text{t}}} \right) - {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)\epsilon_{1} }} \alpha \left( {x,\epsilon_{1} ,{\text{t}}} \right) + \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)\mathop \int \limits_{{\epsilon_{1} }}^{{{\text{R}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \alpha \left( {x,y,t} \right) {\text{dy}}} \right\}, \\ \end{aligned} $$

now let $\epsilon_{1} \to 0$, both terms on the right which depend on $\epsilon_{1}$ approach a limit and,

$$ \mathop \int \limits_{ 0}^{{{\text{R}}_{1} }} \emptyset \left( {x,y,t} \right){\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} {\text{dy}} = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\left\{ { {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{R}}_{1} }} \alpha \left( {x,{\text{R}}_{1} ,{\text{t}}} \right) + \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)\mathop \int \limits_{0}^{{{\text{R}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \alpha \left( {x,y,t} \right) {\text{dy}}} \right\} $$

and then: let, ${\text{R}}_{1} \to \infty$. If, ${\text{s}} < {\text{s}}_{ \circ }$, the first term on the right approaches zero,

$$ {\text{s}}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right) {\text{dy}} = \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{s}}_{ \circ }^{2} }}} \right)\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \alpha \left( {x,y,t} \right) {\text{dy}} $$

(4)

the theorem proved if the integral on the right converges.

Directly by using the limit test for convergence (see [17]) we see,

$$ \begin{aligned} \mathop {\lim }\limits_{{{\text{y}} \to \infty }} {\text{y}}^{2} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} \alpha \left( {x,y,t} \right) & = \mathop {\lim }\limits_{{{\text{y}} \to \infty }} \frac{{{\text{y}}^{2} }}{{{\text{e}}^{{\left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} }}\left[ {\mathop { \lim }\limits_{{{\text{y}} \to \infty }} \alpha \left( {x,y,t} \right)} \right] \\ & = 0 *\left[ {\mathop { \lim }\limits_{{{\text{y}} \to \infty }} \alpha \left( {x,y,t} \right)} \right] = 0, \\ \end{aligned} $$

therefore, integral on the right of (4) converges for, ${\text{s}} < {\text{s}}_{ \circ }$, hence the integral,

$$ {\text{s}}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy,}}\;{\text{converges}}\;{\text{for,}}\;{\text{s}} < {\text{s}}_{ \circ } . $$

Lemma 2.3

If the integral,

$$ {\text{h}}\left( {{\text{x, s, t}}} \right) = {\text{s}}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right) {\text{dy}}, $$

(5)

converges for ${\text{s}} \le {\text{s}}_{ \circ }$ and integral,

$${\uprho} \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{x}{{\uprho} }}} {\text{h}}\left( {{\text{x, s, t}}} \right){\text{dx}}, $$

(6)

converges at ${\uprho} ={\uprho} _{ \circ }$ then the integral (6) converges for ${\uprho} <{\uprho} _{ \circ }$

Proof

$$ {\text{Let:}}\quad \beta \left( {x,s,t} \right) ={\uprho}_{\circ} \mathop \int\limits_{0}^{\text{x}}\,{\text{e}}^{{ -\frac{{\text{u}}}{{{\uprho} _{ \circ } }}}}{\text{h}}\left({{\text{u, s, t}}} \right){\text{du}}\quad 0 < x< \infty,$$

(7)

therefore, $\beta \left( {0,s,t} \right) = 0\; and\; \lim_{x \to \infty } \beta \left( {x,s,t} \right)$ exist, because ${\uprho} \int_{0}^{\infty } {{\text{e}}^{{ - \frac{{\text{x}}}{{\uprho} }}} } {\text{h}}\left( {{\text{x, s, t}}} \right){\text{dx}}$ converges at ${\uprho} ={\uprho} _{ \circ }$, by using fundamental theorem of calculus Eq. (7), becomes,

$$ \beta_{x} \left( {x,s,t} \right) ={\uprho} {\text{h}}\left( {{\text{x, s, t}}} \right){\text{e}}^{{ - \frac{{\text{x}}}{{{\uprho} _{ \circ } }}}} , $$

choose ∈₂ and R₂ and so that 0 < ∈₂ < R₂

$$ \begin{aligned}{\uprho} \mathop \int \limits_{{\epsilon_{2} }}^{{{\text{R}}_{2} }} {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho} }}} {\rm{h}}\left( {{\text{x, s, t}}} \right){\text{dx}} & = \frac{{\uprho} }{{{\uprho} _{ \circ } }}\mathop \int \limits_{{\epsilon_{2} }}^{{{\text{R}}_{2} }} {\text{e}}^{{ - \left( {\frac{{{\uprho} _{ \circ } -{\uprho} }}{{{\uprho}{\uprho} _{ \circ } }}} \right){\text{x}}}} \beta_{x} \left( {x,s,t} \right)dx \\ & = \frac{{\uprho} }{{{\uprho} _{ \circ } }}\left\{ {{\text{e}}^{{ - \left( {\frac{{{\uprho} _{ \circ } -{\uprho} }}{{{\uprho}{\uprho} _{ \circ } }}} \right)R_{2} }} \beta \left( {{\text{R}}_{2} ,s,t} \right) - {\text{e}}^{{ - \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right)\epsilon_{2} }} \beta \left( {\epsilon_{2} ,s,t} \right) + \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right)\mathop \int \limits_{{\epsilon_{2} }}^{{{\text{R}}_{2} }} {\text{e}}^{{ - \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right)x}} \beta \left( {x,s,t} \right)dx} \right\}, \\ \end{aligned} $$

now let ϵ₂ → 0. Both terms on the right which depend on ϵ₂ approach a limit and,

$${\uprho}\mathop \int \limits_{0}^{{{\text{R}}_{2} }} {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} {\text{h}}\left( {{\text{x, s, t}}} \right){\text{dx}} = \frac{{\uprho}}{{{\uprho}_{ \circ } }}\left\{ {{\text{e}}^{{ - \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right){\text{R}}_{2} }} \beta \left( {R_{2} ,s,t} \right) + \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right)\mathop \int \limits_{0}^{{{\text{R}}_{2} }} {\text{e}}^{{ - \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right)x}} \beta \left( {x,s,t} \right)dx } \right\} $$

(8)

now let R₂ → ∞, If ${\text{s}} < {\text{s}}_{ \circ }$, the first term on the right approaches zero.

$${\uprho}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} {\text{h}}\left( {{\text{x, s, t}}} \right){\text{dx}} = \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}_{ \circ }^{2} }}} \right)\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \left( {\frac{{{\uprho}_{ \circ } -{\uprho}}}{{{\uprho}{\uprho} _{ \circ } }}} \right)x}} \beta \left( {x,s,t} \right)dx ,\quad for\;{\uprho} <{\uprho}_{ \circ } , $$

(9)

converges.

Directly by means of the limit test for convergence (see [17]) we see,

$$ \begin{aligned} \mathop {\lim }\limits_{x \to \infty } {\text{x}}^{2} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)x}} \beta \left( {x,y,t} \right) & = \mathop {\lim }\limits_{{{\text{x}} \to \infty }} \frac{{{\text{x}}^{2} }}{{e^{{\left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{x}}}} }}\left[ {\mathop {\lim }\limits_{{{\text{x}} \to \infty }} \beta \left( {x,y,t} \right)} \right] \\ & = 0*\left[ {\mathop {\lim }\limits_{{{\text{x}} \to \infty }} \beta \left( {x,y,t} \right)} \right] = 0, \\ \end{aligned} $$

therefore, integral on the right of (6) converges for, ${\uprho} <{\uprho}_{ \circ }$.

Hence, the given integral ${\uprho}\int_{0}^{\infty } {{\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} } {\text{h}}\left( {{\text{x, t, s}}} \right){\text{dx}}$ converges for ${\uprho} <{\uprho}_{ \circ } .$

Lemma 2.4

If the integral,

$$ {\text{k}}\left( {{\uprho},{\text{y, t}}} \right) ={\uprho}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} {\text{h}}\left( {{\text{x, s, t}}} \right){\text{dx}}, $$

(10)

converges for ${\uprho} \le{\uprho}_{ \circ }$ and the integral

$$ \updelta\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{dt}}, $$

(11)

converges at $\updelta =\updelta_{ \circ }$ then the integral (11) converges for $\updelta <\updelta_{ \circ } .$

Proof

$${\text{Let:}}\quad \gamma \left( {{\uprho},y,{\text{t}}} \right)=\updelta_{ \circ } \mathop \int\limits_{0}^{\text{t}}\,{\text{e}}^{{ -\frac{{\text{w}}}{{\updelta_{\circ } }}}} {\text{k}}\left( {{\uprho},{\text{y, w}}}\right){\text{dw}},\quad 0 < t < \infty, $$

(12)

therefore, $\gamma \left( {{\uprho},y,0} \right) = 0 \;and\; \lim_{t \to \infty } \gamma \left( {x,y,t} \right)$ exist, because $\updelta\int_{0}^{\infty } {{\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{dt}}}$ converges at $\updelta =\updelta_{ \circ } ,$ by using the fundamental theorem of calculus, Eq. (12), develops as. $\gamma_{tx} \left( {{\uprho},y,{\text{t}}} \right) = {\text{k}}\left( {{\uprho},y,{\text{w}}} \right){\text{e}}^{{ - \frac{{\text{w}}}{{\updelta_{ \circ } }}}}$, choose ∈₃ and R₃ and so that 0 < ∈₃ < R₃

$$ \begin{aligned}\updelta\mathop \int \limits_{\epsilon 3 }^{{{\text{R}}3}} {\text{k}}\left( {{\uprho},y,{\text{t}}} \right){\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{dt}} & = \frac{\updelta}{{\updelta_{ \circ } }}\mathop \int \limits_{{\epsilon_{3} }}^{{{\text{R}}_{3} }} e^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)t}} \gamma_{t} \left( {{\uprho},y,{\text{t}}} \right)dt \\ & = \frac{\updelta}{{\updelta_{ \circ } }}\left\{ {\updelta^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right){\text{R}}_{3} }} \gamma \left( {{\uprho},y,{\text{R}}_{3} } \right) - {\text{e}}^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)\epsilon_{3} }} \gamma \left( {{\uprho},y,\epsilon_{3} } \right) + \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)\mathop \int \limits_{{\epsilon_{3} }}^{{{\text{R}}_{3} }} {\text{e}}^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)t}} \gamma \left( {{\uprho},y,{\text{t}}} \right)dt} \right\}, \\ \end{aligned} $$

now let ϵ₃ → 0. Both terms on the right which depend on ϵ₃ approach a limit and

$$ \updelta\mathop \int \limits_{0 }^{{{\text{R}}3}} {\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{dt}} = \frac{\updelta}{{\updelta_{ \circ } }}\left\{ {{\text{e}}^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)R_{3} }} \gamma \left( {{\uprho},y,{\text{R}}_{3} } \right) + \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)\mathop \int \limits_{0}^{{{\text{R}}_{3} }} {\text{e}}^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right){\text{t}}}} \gamma \left( {{\uprho},y,{\text{t}}} \right)dt} \right\}, $$

(13)

now let R₃ → ∞. If ${\text{s}} < {\text{s}}_{ \circ }$, the first term on the right approaches zero.

$$ \updelta\mathop \int \limits_{0 }^{\infty } {\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{dt}} = \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta_{ \circ }^{2} }}} \right)\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right)t}} \gamma \left( {{\uprho},y,{\text{t}}} \right)dt,\quad for \;\updelta <\updelta_{ \circ } , $$

(14)

the given theorem proved if the integral on the right converges.

Now by using the limit test for convergence (see [17]) we consider,

$$ \begin{aligned} \mathop {\lim }\limits_{{{\text{t}} \to \infty }} {\text{t}}^{2} {\text{e}}^{{ - \left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right){\text{t}}}} \gamma \left( {x,y,t} \right) & = \mathop {\lim }\limits_{{{\text{t}} \to \infty }} \frac{{{\text{t}}^{2} }}{{{\text{e}}^{{\left( {\frac{{\updelta_{ \circ } -\updelta}}{{\updelta \updelta _{ \circ } }}} \right){\text{t}}}} }}\left[ {\mathop {\lim }\limits_{{{\text{t}} \to \infty }} \gamma \left( {x,y,t} \right)} \right] \\ & = 0*\left[ {\mathop {\lim }\limits_{{{\text{t}} \to \infty }} \gamma \left( {x,y,t} \right)} \right] = 0 = {\text{finite}}, \\ \end{aligned} $$

therefore, integral on the right of (12) converges for, $\updelta <\updelta_{ \circ }$, hence, the given integral $\updelta\int_{0}^{\infty } {{\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{dt}}}$ converges for $\updelta <\updelta_{ \circ } .$

The proof of the Theorem 2.1 is as follows:

$$ \begin{aligned} &{\uprho}{\text{s}}\updelta\mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} \emptyset \left( {x,y,t} \right)dxdydt \\ & \quad =\updelta\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} \left\{ {{\uprho}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} \left\{ { {\text{s}}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy}}} \right\}dx} \right\}dt \\ & \quad =\updelta\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} \left\{ {{\uprho}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} h\left( {x,s,t} \right)dx} \right\}dt \\ & \quad =\updelta\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{dt}}, \\ \end{aligned} $$

(15)

where,

$$ h\left( {x,s,t} \right) = {\text{s}}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy}}, $$

$$ {\text{k}}\left( {{\uprho},{\text{y, t}}} \right) ={\uprho}\mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} h\left( {x,s,t} \right)dx, $$

by Lemma 2.2, integral ${\text{s}}\int_{0}^{\infty } {{\text{e}}^{{\frac{{\text{y}}}{\text{s}}}} \emptyset \left( {x,y,t} \right){\text{dy}}}$ converges for, ${\text{s}} < {\text{s}}_{ \circ }$, also by Lemma 2.3, integral ${\uprho}\int_{0}^{\infty } {{\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}}}} {\text{h}}\left( {{\text{x, s, t}}} \right) {\text{dx}}}$ converges for ${\uprho} <{\uprho}_{ \circ } ,$ and by Lemma 2.4, integral $\updelta\int_{0}^{\infty } {{\text{e}}^{{ - \frac{{\text{t}}}{\updelta}}} {\text{k}}\left( {{\uprho},{\text{y, t}}} \right){\text{dt}}}$ converges for $\updelta <\updelta_{ \circ } ,$ therefore, integral in the RHS (15) converges for ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } \;{\text{and}}\;\updelta <\updelta_{ \circ } ,$ hence the integral: ${\uprho}{\text{s}}\updelta\mathop \smallint \limits_{0}^{\infty } \mathop \smallint \limits_{0}^{\infty } \mathop \smallint \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} \emptyset \left( {x,y,t} \right)dxdydt$, converges for, ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } \;{\text{and}}\;\updelta <\updelta_{ \circ } ,$, then the proof of the Theorem 2.1 is completed.

Corollary 2.4

If the integral (1) diverges at ${\uprho} ={\uprho}_{ \circ } ,\;{\text{s}} = {\text{s}}_{ \circ } \;{\text{and}}\;\updelta =\updelta_{ \circ } ,$ then the integral, (1) diverges at

$${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } \;\updelta <\updelta_{ \circ } , $$

Corollary 2.5

The region of the convergence of the integral (1) is the positive quadrant of the ${\text{xyt - plane}}$. Now we prove absolute convergence of integral (1).

Theorem 2.6

If the integral (1) converges absolutely at ${{\uprho} ={\uprho} }_{ \circ } ,\;{\text{s}}\text{ = }{\text{s}}_{ \circ } \;{\text{and}}\;{\updelta = \updelta }_{ \circ } ,$ then integral (1) converges absolutely for ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } \;{\text{and}}\;\updelta <\updelta_{ \circ } ,$

Proof

We know that:

$$ {\text{e}}^{{ - \frac{{\text{x}}}{{{\uprho}_{ \circ } }} - \frac{{\text{y}}}{{{\text{s}}_{ \circ } }} - \frac{{\text{t}}}{{\updelta_{ \circ } }}}} \left| {\emptyset \left( {x,y,t} \right)} \right| \le {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} \quad {\text{for}} \left( {{\uprho} \le{\uprho}_{ \circ } < \infty ,{\text{s}} \le {\text{s}}_{ \circ } < \infty ,\updelta \le\updelta_{ \circ } < \infty } \right), $$

therefore,

$${\uprho}_{ \circ } {\text{s}}_{ \circ }\updelta_{ \circ } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{{\uprho}_{ \circ } }} - \frac{{\text{y}}}{{{\text{s}}_{ \circ } }} - \frac{{\text{t}}}{{\updelta_{ \circ } }}}} \left| {\emptyset \left( {x,y,t} \right)} \right|dxdydt \le{\uprho}{\text{s}}\updelta\mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} \left| {\emptyset \left( {x,y,t} \right)} \right|dxdydt, $$

form given hypothesis, ${\uprho}{\text{s}}\updelta\mathop \smallint \limits_{0}^{\infty } \mathop \smallint \limits_{0}^{\infty } \mathop \smallint \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} \left| {\emptyset \left( {x,y,t} \right)} \right|dxdydt,$ converge. Hence, we have: ${\uprho}_{ \circ } {\text{s}}_{ \circ }\updelta_{ \circ } \mathop \smallint \limits_{0}^{\infty } \mathop \smallint \limits_{0}^{\infty } \mathop \smallint \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{{\uprho}_{ \circ } }} - \frac{{\text{y}}}{{{\text{s}}_{ \circ } }} - \frac{{\text{t}}}{{\updelta_{ \circ } }}}} \left| {\emptyset \left( {x,y,t} \right)} \right|dxdydt$ converge for $\left( {{\uprho} \le{\uprho}_{ \circ } ,{\text{s}} \le {\text{s}}_{ \circ } ,\updelta \le\updelta_{ \circ } } \right).$

Therefore the integral (1) converges absolutely for, $\left( {{\uprho} \le{\uprho}_{ \circ } ,{\text{s}} \le {\text{s}}_{ \circ } ,\updelta \le\updelta_{ \circ } } \right)$.

3 Uniform convergence

In this section, we prove the uniform convergence of triple Elzaki Transform.

Theorem 3.1

If ${\text{f}}\left( {{\text{x, y, t}}} \right)$ is continuous on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$, and,

$$ {\text{H}}\left( {{\text{x, y, t}}} \right) ={\uprho}_{ \circ } {\text{s}}_{ \circ }\updelta_{ \circ } \mathop{\int} \limits_{0}^{x} \mathop{\int} \limits_{0}^{y} \mathop{\int} \limits_{0}^{t} {\text{e}}^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{{\text{v}}}{{{\hbox{s}}_{\circ }}} - \frac{{\text{w}}}{{\updelta_{\circ }}}}} f\left( {u,v,w} \right)dudvdw, $$

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is bounded on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$ then the triple of Elzaki transform of, f converges uniformly on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right) \times \left[ {\updelta, \infty } \right)$ if, ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ }$, $\updelta <\updelta_{ \circ }$. For the proof, we will use the following lemmas:

Lemma 3.2

If ${\text{g}}\left( {{\text{x, y, t}}} \right) = {\text{s}}_{ \circ } \int_{0}^{\text{y}} {{\text{e}}^{{ - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }}}} f\left( {x,v,t} \right) {\text{dv}}}$ is bounded on $\left[ {0, \infty } \right)$ then the Elzaki transform of, f with respect to s converges uniformly on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ .}$

Proof

If $0 \le {\text{r }} \le {\text{r}}_{1}$ then consider,

$$ {\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} f\left( {x,y,t} \right){\text{e}}^{{ - \frac{{\text{y}}}{{\text{s}}}}} {\text{dy}} = {\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} f\left( {x,y,t} \right){\text{e}}^{{\frac{{\text{y}}}{{{\text{s}}_{ \circ } }}}} {\text{dy}} = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} g_{y} \left( {x,y,t} \right)dy, $$

using integration by parts,

$$ = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\left\{ {{\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} g\left( {x,{\text{r}}_{1} ,{\text{t}}} \right) - {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} g\left( {x,{\text{r, t}}} \right) + \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} g\left( {x,y,t} \right)dy} \right\}, $$

therefore, if | (,y, ) | ≤  then:

$$ \begin{aligned} \left| {{\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} f\left( {x,y,t} \right) {\text{dy}}} \right| & \le {\text{M }}\left\{ {{\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} + {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} + \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right)\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} {\text{dy}}} \right\} \\ & = {\text{M}}\left\{ {{\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} + {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} - {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} + {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} } \right\} \\ & = 2{\text{M e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} \;{\text{for}}\;{\text{s}} \le {\text{s}}_{ \circ } , \\ \end{aligned} $$

by Cauchy criterion for uniform convergence [18], ${\text{s}}\int_{\text{r}}^{{{\text{r}}_{1} }} {f\left( {x,y,t} \right){\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} {\text{dy}}}$, converges uniformly on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ }$.

Hence, Elzaki transform of, f with respect to, s converges uniformly on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ } .$

Lemma 3.3

If $g\left( {{\text{x, y, t}}} \right) ={\uprho}_{ \circ } {\text{s}}_{ \circ } \mathop \int \nolimits_{0}^{x} \mathop \int \nolimits_{0}^{y} {\text{e}}^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }}}} f\left( {u,v,t} \right)dudv$ is bounded on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$ then the Elzaki transform of, f with respect to ${\uprho},s$ converges uniformly on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ }$ proof see double Elzaki transform [7].

Lemma 3.4

If the integral $g\left( {{\uprho}, {\text{s, t}}} \right) ={\uprho}{\text{s}}\mathop \int \nolimits_{0}^{\infty } \mathop \int \nolimits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}}}} f\left( {x,y,t} \right)dxdy$, ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ }$ and,$a\left( {{\uprho},{\text{s, t}}} \right) =\updelta_{ \circ } \int_{0}^{\text{t}} {{\text{e}}^{{ - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} {\text{g}}\left( {{\uprho},{\text{s, w}}} \right){\text{dw}}}$ is bounded on [0, ∞), then Elzaki transform of the function f with respect to ${\uprho},{\text{s}}$ converges uniformly on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right)$.

Proof

The proof of Theorem 3.1 is as follows,

$$ \begin{aligned} {\text{H}}\left( {{\text{x, y, t}}} \right) & ={\uprho}_{ \circ } {\text{s}}_{ \circ }\updelta_{ \circ } \mathop{\int} \limits_{0}^{x} \mathop \int \limits_{0}^{y} \mathop {\int} \limits_{0}^{t} {\text{e}}^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }} - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} f\left( {u,v,w} \right)dudvdw \\ & =\updelta_{ \circ } \mathop{\int} \limits_{0}^{t} {\text{e}}^{{ - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} \left\{ {{\uprho}_{ \circ } {\text{s}}_{ \circ } \mathop{\int} \limits_{0}^{x} \mathop{\int} \limits_{0}^{y} {\text{e}}^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }}}} f\left( {u,v,w} \right)dudv} \right\}dw =\updelta_{ \circ } \mathop{\int} \limits_{0}^{t} {\text{e}}^{{ - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} g\left( {x,y,w} \right)dw, \\ \end{aligned} $$

where $g\left( {{\text{x, y, w}}} \right) ={\uprho}_{ \circ } {\text{s}}_{ \circ } \mathop \smallint \limits_{0}^{x} \mathop \smallint \limits_{0}^{y} {\text{e}}^{{ - {\frac{{\text{u}}}{{{\uprho}_{ \circ } }}} - {\frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }}}}} f\left( {u,v,w} \right)dudv$ is bounded on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$, by Lemmas 3.2, and 3.3 double Elzaki transform of f with respect to ${\uprho},s$ converges uniformly on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right)$ if ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } .$

Also by Lemma 3.4, Elzaki transform of $g$ with respect to $\updelta$ converges uniformly on $\left[ {\updelta, \infty } \right)$ if $\updelta <\updelta_{ \circ }$.

Hence triple Elzaki transform of, f converges uniformly on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right) \times \left[ {\updelta, \infty } \right)$ if ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } \;{\text{and}}\;\updelta <\updelta_{ \circ }$.

We now prove the differentiability of triple Elzaki transform.

Theorem 3.5

If $f\left( {x,y,t} \right)$ is continuous on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$ and,

$$ {\text{H}}\left( {{\text{x, y, t}}} \right) ={\uprho}_{ \circ } {\text{s}}_{ \circ }\updelta_{ \circ } \mathop \int \limits_{0}^{x} \mathop \int \limits_{0}^{y} \mathop \int \limits_{0}^{t} e^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }} - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} f\left( {u,v,w} \right)dudvdw, $$

is bounded on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$, then the triple Elzaki transform of the fuction f is infinitely differentiable with respect to ${\uprho}, s$ and $\updelta$ on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right) \times \left[ {\updelta, \infty } \right)$ ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ }$, ${\text{and}}\;\updelta <\updelta_{ \circ }$ with

$$ \frac{{\partial^{m + n} }}{{\partial \rho^{m} \partial s^{n} }} \bar{f}\left( {{\uprho},s,\updelta} \right) = \left( { - 1} \right)^{{{\text{m}} + {\text{n}} + {\text{k}}}}{\uprho}{\text{s}}\updelta\mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}} - \frac{{\text{t}}}{\updelta}}} x^{m} y^{n} t^{k} f\left( {x,y,t} \right)dxdydt, $$

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for the proof, we will use the following lemmas.

Lemma 3.6

If ${\text{g}}\left( {{\text{x, y, t}}} \right) = \int_{0}^{y} {{\text{e}}^{{ - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }}}} f\left( {x,v,t} \right) {\text{dv}}}$ is bounded on $\left[ {0, \infty } \right)$ then the Elzaki transform of f is infinitely differentiable with respect to s on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ }$ with,

$$ \frac{{\partial^{n} }}{{\partial s^{n} }} \bar{f}\left( {{\text{x, s}}} \right) = \left( { - 1} \right)^{\text{n}} \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} y^{n} f\left( {x,y,t} \right)dy $$

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Proof

Foremost we will prove that the integrals,

$$ I_{n} \left( {{\text{x, s, t}}} \right) = \left( { - 1} \right)^{\text{n}} \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} y^{n} f\left( {x,y,t} \right)dy , \quad n = 0,1,2,3, \ldots , $$

all converge uniformly on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ }$ and if $0 \le {\text{r }} \le {\text{r}}_{1}$, then,

$$ \begin{aligned} {\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \frac{{\text{y}}}{{\text{s}}}}} y^{n} f\left( {x,y,t} \right){\text{dy}} & = {\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} y^{n} g_{y} \left( {x,y,t} \right)dy \\ & = \frac{{\text{s}}}{{{\text{s}}_{ \circ } }}\left\{ { {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} {\text{r}}_{1}^{n} g\left( {x,{\text{r}}_{1} ,{\text{t}}} \right) - {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} r^{n} g\left( {x,{\text{r, t}}} \right) - \mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} [\frac{{\text{d}}}{{\text{dt}}}{\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{y}}}} y^{n} ]g\left( {x,y,t} \right)dy} \right\}, \\ \end{aligned} $$

therefore, if | (,y, ) | ≤  $< \infty$ on $\left[ {0, \infty } \right)$ then,

$$ \begin{aligned} & \left| {{\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} y^{n} f\left( {x,y,t} \right){\text{dy}}} \right| \le {\text{M}}\left\{ {{\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} {\text{r}}_{1}^{n} + {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} r^{n} - {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}_{1} }} {\text{r}}_{1}^{n} + {\text{e}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} r^{n} } \right\} \\ & \left| {{\text{s}}\mathop \int \limits_{\text{r}}^{{{\text{r}}_{1} }} {\text{e}}^{{ - \frac{{\text{y}}}{\text{s}}}} y^{n} f\left( {x,y,t} \right){\text{dy}}} \right| \le 2{\text{Me}}^{{ - \left( {\frac{{{\text{s}}_{ \circ } - {\text{s}}}}{{{\text{ss}}_{ \circ } }}} \right){\text{r}}}} r^{n} \;for\;0 \le {\text{r }} \le {\text{r}}_{1} . \\ \end{aligned} $$

By Cauchy criterion for uniform convergence [18] ${\text{I}}_{\text{n}} \left( {{\text{x, s, t}}} \right)$ converges uniformly on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ }$. Now, using Trench [18] and induction proof, we have (17). That is Elzaki transform of f is infinitely differentiable with respect to ${\text{s}}$, on $\left[ {{\text{s}}, \infty } \right)$ if ${\text{s}} < {\text{s}}_{ \circ } .$

Lemma 3.7

If $g\left( {{\text{x, y, t}}} \right) ={\uprho}_{ \circ } {\text{s}}_{ \circ } \mathop \int \nolimits_{0}^{x} \mathop \int \nolimits_{0}^{y} {\text{e}}^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{v}{{{\text{s}}_{ \circ } }}}} f\left( {u,v,w} \right)dudv$ is bounded on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$ then double Elzaki transform of f is infinitely differentiable with respect to ${\uprho},s$ on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right)$ if ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ }$ with,

$$ \frac{{\partial^{m + n} }}{{\partial \rho^{m} \partial s^{n} }} \bar{f}\left( {{\uprho},{\text{s, t}}} \right) = \left( { - 1} \right)^{{{\text{m}} + {\text{n}}}}{\uprho}{\text{s}}\mathop \int \limits_{0}^{\infty } \mathop \int \limits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}}}} x^{m} y^{n} f\left( {x,y,t} \right)dxdy, $$

for the proof see double Elzaki transform [7].

Lemma 3.8

If the integral $\emptyset \left( {{\uprho}, {\text{s, t}}} \right) ={\uprho}{\text{s}}\mathop \int \nolimits_{0}^{\infty } \mathop \int \nolimits_{0}^{\infty } {\text{e}}^{{ - \frac{{\text{x}}}{{\uprho}} - \frac{{\text{y}}}{\text{s}}}} x^{m} y^{n} f\left( {x,y,t} \right)dxdy$ converges uniformly on $\left[ {{\uprho}, \infty } \right),\left[ {{\text{s}}, \infty } \right)$ if ${\uprho} <{\uprho}_{ \circ } , {\text{s}} < {\text{s}}_{ \circ }$ and ${\text{h}}\left( {{\uprho}, {\text{s, t}}} \right) =\updelta\int_{0}^{\text{t}} {{\text{e}}^{{ - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} \emptyset \left( {{\uprho}, {\text{s, t}}} \right){\text{dt}}}$ is bounded on [0, ∞) then the Elzaki transform of $\emptyset$is infinitely differentiable with respect to $\updelta$ on $\left[ {\updelta, \infty } \right)$ if $\updelta <\updelta_{ \circ }$, with the,

$$ \frac{{\partial^{k} }}{{\partial\updelta^{k} }} \emptyset \left( {{\uprho}, {\text{s, t}}} \right) = \left( { - 1} \right)^{\text{k}} \mathop \int \limits_{0}^{\infty } {\rm{e}}^{{ - \frac{{\text{t}}}{\updelta}}} t^{k} \emptyset \left( {{\uprho}, {\text{s, t}}} \right)dt $$

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Proof

The validation is a similar to Lemmas 3.6 and 3.7.

The validation of the Theorem 3.5 is as follows:

$$ {\text{H}}\left( {{\text{x, y, t}}} \right) ={\uprho}_{ \circ } {\text{s}}_{ \circ }\updelta_{ \circ } \mathop \int \limits_{0}^{x} \mathop \int \limits_{0}^{y} \mathop \int \limits_{0}^{t} {\text{e}}^{{ - \frac{{\text{u}}}{{{\uprho}_{ \circ } }} - \frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }} - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} f\left( {u,v,w} \right)dudvdw =\updelta_{ \circ } \mathop \int \limits_{0}^{\text{t}}\,{\text{e}}^{{ - \frac{{\text{w}}}{{\updelta_{ \circ } }}}} g\left( {{\text{x, y, w}}} \right){\text{dw}}, $$

where $g\left( {{\text{x, y, w}}} \right) ={\uprho}_{ \circ } {\text{s}}_{\circ } \mathop{\int} \limits_{0}^{{\rm x}} \mathop{\int} \limits_{0}^{{\rm y}} {\text{e}}^{{ - {\frac{{\text{u}}}{{{\uprho}_{ \circ } }}} - {\frac{{\text{v}}}{{{\hbox{s}}_{ \circ } }}}}} f\left( {{\text{u, v,w}}} \right){\text{dudv}}$, is bounded on $\left[ {0, \infty } \right) \times \left[ {0, \infty } \right)$. By Lemma 3.7, Elzaki transform of f is infinitely differentiable with respect to ${\uprho},s$ on $\left[ {{\uprho}, \infty } \right),\left[ {{\text{s}}, \infty } \right)$ if ${\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } ,$ also by Lemma 3.8, Elzaki transform of $g$ is infinitely differentiable with respect to $\updelta$ on $\left[ {\updelta, \infty } \right)$ if $\updelta <\updelta_{ \circ }$.

Hence triple Elzaki transform of f is infinitely differentiable with respect to ${\uprho},s$ and $\updelta$ on $\left[ {{\uprho}, \infty } \right) \times \left[ {{\text{s}}, \infty } \right) \times \left[ {\updelta, \infty } \right)\;{\uprho} <{\uprho}_{ \circ } ,\;{\text{s}} < {\text{s}}_{ \circ } ,\;{\text{and}}\;\updelta <\updelta_{ \circ } .$

4 Triple Elzaki transform of triple integral

Now we find the triple Elzaki transform of triple integral.

Theorem 4.1

If ${\text{E}}_{\rm {x}}\,{\text{E}}_{\rm {y}}\,{\text{E}}_{\rm {t}}$ $f\left( {x,y,t} \right) = \bar{f}(p,s,\updelta)$ and,

$$ g\left( {{\text{x, y, t}}} \right) = \mathop \int \limits_{0}^{\text{x}} \mathop \int \limits_{0}^{\text{y}} \mathop \int \limits_{0}^{\text{t}} f\left( {u,v,w} \right)dudvdw, $$

(20)

$$ {\text{then:}}\quad {\text{E}}_{\rm {x}}\,{\text{E}}_{\rm {y}} {\text{E}}_{\rm {t}} \left\{ {\mathop \int \limits_{0}^{\text{x}} \mathop \int \limits_{0}^{\text{y}} \mathop \int \limits_{0}^{\text{t}} f\left( {u,v,w} \right)dudvdw} \right\} ={\uprho}{\text{s}}\updelta{\bar{\text{f}}}\left( {{\text{p, s}},\updelta} \right). $$

(21)

Proof

If we denote, ${\text{h}}\left( {{\text{x, y, t}}} \right) = \int_{0}^{\text{y}} {f\left( {{\text{x, v, t}}} \right){\text{dv}}}$, by using fundamental theorem of calculous,

$$ h_{y} \left( {x,y,t} \right) = f\left( {x,y,t} \right), $$

(22)

since,

$$ h\left( {x,0,t} \right) = 0, $$

(23)

taking double Elzaki transform of Eq. (22), we get:

$$ {\bar{\text{h}}}\left( {{\uprho},{\text{s}},\updelta} \right) = s \bar{f}\left( {{\uprho},{\text{s}},\updelta} \right), $$

(24)

from Eq. (20) we get,

$$ r\left( {{\text{x, y, t}}} \right) = \mathop \int \limits_{0}^{\text{t}} {\text{h}}\left( {{\text{x, y, w}}} \right){\text{dw}}, $$

$$ r_{t} \left( {x.y,t} \right) = {\text{h}}\left( {{\text{x}}.{\text{y, t}}} \right)\;{\text{and}}\;{\text{r}}\left( {{\text{x, y}},0} \right) = 0, $$

$$ {\bar{\rm{r}}}\left( {{\uprho},{\text{s}},\updelta} \right) =\updelta{\bar{\text{h}}}\left( {{\uprho},{\text{s}},\updelta} \right), $$

(25)

$g\left( {{\text{x, y, t}}} \right) = \int\limits_{0}^{\text{x}} {{\text{r}}\left( {{\text{u, y, t}}} \right){\text{du}},}$ and $g\left( {0,{\text{y, t}}} \right) = 0,$

$$ \bar{g}\left( {{\uprho},{\text{s}},\updelta} \right) ={\uprho}{\bar{\text{r}}}\left( {{\uprho},{\text{s}},\updelta} \right), $$

(26)

now by using (24), (25) and (26), we obtain,

$$ {\text{E}}_{\rm {x}}\,{\text{E}}_{\rm {y}}\,{\text{E}}_{\rm {t}} \left\{ {\mathop \int \limits_{0}^{\text{x}} \mathop \int \limits_{0}^{\text{y}} \mathop \int \limits_{0}^{\text{t}} f\left( {u,v,w} \right)dudvdw} \right\} ={\uprho}{\text{s}}\updelta{\bar{\text{f}}}\left( {{\text{p}}, {\text{s}},\updelta} \right). $$

5 Application of triple Elzaki transform in Volterra integro-partial differential equation

In this section, we use the triple Elzaki transform to solve the Volterra integro-partial differential equation, which already solved in [19] by means of differential transform method.

Example 5.1

Consider the Volterra integro-partial differential equation,

$$ \frac{{\partial {\text{u}}\left( {{\text{x, y, t}}} \right)}}{{\partial {\text{x}}}} + \frac{{\partial {\text{u}}\left( {{\text{x, y, t}}} \right)}}{{\partial {\text{y}}}} + \frac{{\partial {\text{u}}\left( {{\text{x, y, t}}} \right)}}{{\partial {\text{t}}}} = - {\text{x}}^{2} {\text{y}}^{2} {\text{t}}^{2} + 4{\text{xy}} + 4{\text{xt}} + 4{\text{yt}} + 2\mathop \int \limits_{0}^{\text{x}} \mathop \int \limits_{0}^{\text{y}} \mathop \int \limits_{0}^{\text{t}} {\text{u}}\left( {{\text{k, r, s}}} \right){\text{dkdrds}}, $$

(27)

with,

$$ {\text{u}}\left( {{\text{x}},y,0} \right) = 0,{\text{u}}\left( {{\text{x}},0,t} \right) = 0,{\text{u}}\left( {0,y,t} \right) = 0, $$

(28)

applying triple Elzaki transform of Eq. (27) to get,

$$ \begin{aligned} & \frac{1}{\upsigma}{\bar{\text{u}}}\left( {{\uprho},s,\updelta} \right) -\upsigma{\bar{\text{u}}}\left( {0,s,\updelta} \right) + \frac{1}{{\uprho}}{\bar{\text{u}}}\left( {{\uprho},s,\updelta} \right) -{\uprho}{\bar{\text{u}}}\left( {{\uprho},0,\updelta} \right) + \frac{1}{\updelta}{\bar{\text{u}}}\left( {{\uprho},{\text{s,}}\updelta} \right) -\updelta{\bar{\text{u}}}\left( {{\uprho},s,0} \right) \\ & \quad = - 8{\uprho}^{4} s^{4}\updelta^{4} + 4{\uprho}^{3} s^{3}\updelta^{2} + 4{\uprho}^{3} s^{2}\updelta^{3} + 4{\uprho}^{2} s^{3}\updelta^{3} +{\uprho}s\updelta{\bar{\text{u}}}\left( {{\uprho},s,\updelta} \right), \\ \end{aligned} $$

(29)

the double Elzaki transforms of Eq. (28)

$$ {\bar{\text{u}}}\left( {0,s,\updelta} \right) = 0,\; {\bar{\text{u}}}\left( {{\uprho},0,\updelta} \right) = 0, \;{\bar{\text{u}}}\left( {{\uprho},s,0} \right) = 0, $$

(30)

substituting (30) in (29), simplifying, and we obtain,

$$ \frac{1}{\upsigma}{\bar{\text{u}}}\left( {{\uprho},{\text{s}},\updelta} \right) + \frac{1}{{\uprho}}{\bar{\text{u}}}\left( {{\uprho},{\text{s}},\updelta} \right) + \frac{1}{\updelta}{\bar{\text{u}}}\left( {{\uprho},{\text{s}},\updelta} \right) = - 8{\uprho}^{4} {\text{s}}^{4}\updelta^{4} + 4{\uprho}^{3} s^{3}\updelta^{2} + 4{\uprho}^{3} s^{2}\updelta^{3} + 4{\uprho}^{2} s^{3}\updelta^{3} +{\uprho}s\updelta{\bar{\text{u}}}\left( {{\uprho},{\text{s}},\updelta} \right) $$

So ${\bar{\text{u}}}\left( {{\uprho},s,\updelta} \right) = 4{\uprho}^{3} s^{3}\updelta^{3} ,$ applying triple inverse Elzaki transform, to ascertain the exact solution in the shape,

$$ {\text{u}}\left( {{\text{x, y, t}}} \right) = 4{\text{xyt}} . $$

Which is the same result that obtained by using differential transform method, see [19].

To compere this method with other methods you can refer to papers [1, 19], because the main objectives of this paper is to study the convergence, absolute convergence and uniform convergence of triple Elzaki transform.

6 Conclusion

We have demonstrated the convergence, absolute convergence and uniform convergence of triple Elzaki transform they have been shown. Other than these, we got triple Elzaki transform of triple integral and use to illuminate Volterra integro-partial differential equation.