A Note on the Triple Laplace Transform and Its Applications to Some Kind of Third-Order Differential Equation

Abstract

We introduced a relatively new operator called the triple Laplace transform. We presented some properties and theorems about the relatively new operator. We examine the triple Laplace transform of some function of three variables. We make use of the operator to solve some kind of third-order differential equation called “Mboctara equations.”

1. Introduction

The topic of partial differential equations is one of the most important subjects in mathematics and other sciences. The behaviour of the solution very much depends essentially on the classification of PDEs therefore the problem of classification for partial differential equations is very natural and well known since the classification governs the sufficient number and the type of the conditions in order to determine whether the problem is well posed and has a unique solution. The Laplace transform has been intensively used to solve nonlinear and linear equations [1–7]. The Laplace transform is used frequently in engineering and physics; the output of a linear time invariant system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter is easier to solve because of its algebraic form. The Laplace transform can also be used to solve differential equations and is used extensively in electrical engineering [1–7]. The Laplace transform reduces a linear differential equation to an algebraic equation, which can then be solved by the formal rules of algebra. The original differential equation can then be solved by applying the inverse Laplace transform. The English electrical engineer Oliver Heaviside first proposed a similar scheme, although without using the Laplace transform, and the resulting operational calculus is credited as the Heaviside calculus. Recently Kılıçman et al. [8–11] extended the Laplace transform to the concept of double Laplace transform. This new operator has been intensively used to solve some kind of differential equation [11] and fractional differential equations. The aim of this work is to extend the Laplace transform to the triple Laplace transform. We will start with the definition of the triple Laplace transform.

2. Definitions and Theorems

Definition 1. Let 𝑓 be a continuous function of three variables; then, the triple Laplace transform of 𝑓(𝑥,𝑦,𝑡) is defined by 𝐿_{𝑥,𝑦,𝑡}𝑓𝑥,𝑦,𝑡=𝐹𝑝,𝑠,𝑘^∞₀[]𝑓exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡, (1) where, 𝑥,𝑦,𝑡>0 and 𝑝, 𝑠, 𝑘 are Laplace variables, and 𝑓=1𝑥,𝑦,𝑡2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥×12𝜋𝑖^𝛽+𝑖∞_{𝛽−𝑖∞}𝑒^𝑠𝑦×12𝜋𝑖^𝜇+𝑖∞_{𝜇−𝑖∞}𝑒^𝑘𝑡×𝐹𝑝,𝑠,𝑘𝑑𝑘𝑑𝑠𝑑𝑝 (2) is the inverse triple Laplace transform.

Property 2. Assuming that the continuous function 𝑓(𝑥,𝑦,𝑡) is triple Laplace transformable, then, 𝐿_{𝑡,𝑦,𝑥}𝜕³𝑓𝑥,𝑦,𝑡()()()(),𝐿𝜕𝑥𝜕𝑦𝜕𝑡=𝑝𝑠𝑘𝐹𝑝,𝑠,𝑘−𝑝𝑠𝐹𝑝,𝑠,0−𝑝𝑠𝐹𝑝,0,𝑘+𝑝𝐹𝑝,0,0−𝑠𝑘𝐹0,𝑠,𝑘+𝑠𝐹0,𝑠,0+𝑘𝐹0,0,𝑘−𝐹0,0,0_{𝑥,𝑥,𝑡}𝜕³𝑓𝑥,𝑦,𝑡𝜕𝑡𝜕𝑥²=𝑘𝑝²𝐹𝑝,𝑦,𝑘−𝑝𝑘𝐹0,𝑦,𝑘−𝜕𝐹0,𝑦,𝑘𝜕𝑥−𝑝²𝐹𝑝,𝑦,0+𝑝𝐹0,𝑦,0+𝜕𝐹0,𝑦,0,𝐿𝜕𝑥_𝑥𝑥𝑥𝜕³𝑓𝑥,𝑦,𝑡𝜕𝑥³=𝑝³𝐹𝑝,𝑦,𝑡−𝑝²𝐹0,𝑦,𝑡−𝑝𝜕𝐹0,𝑦,𝑡−𝜕𝜕𝑥²𝐹0,𝑦,𝑡𝜕𝑥². (3)

3. Uniqueness and Existence of the Triple Laplace Transform

In this section, we will study the uniqueness and existence of triple Laplace transform. First of all, let 𝑓(𝑥,𝑦,𝑡) be a continuous function on the interval [0,∞) which is of exponential order, that is, for some 𝑎,𝑏,𝑐∈𝑅. Consider sup_{𝑥,𝑦,𝑡>0}𝑓𝑥,𝑦,𝑡exp𝑎𝑥+𝑏𝑦+𝑐𝑡<0. (4) Under the previous condition, the triple Laplace transform, 𝐹𝑝,𝑠,𝑘=^∞₀[]𝑓exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡, (5) exists for all 𝑝>𝑎, 𝑠>𝑏, and 𝑘>𝑐 and is in actuality infinitely differentiable with respect to 𝑝>𝑎, 𝑠>𝑏 and 𝑘>𝑐. All functions in this study are assumed to be of exponential order. The following theorem shows that 𝑓(𝑥,𝑦,𝑡) can be uniquely obtained from 𝐹(𝑝,𝑠,𝑡).

Theorem 3. Let 𝑓(𝑥,𝑦,𝑡) and 𝑔(𝑥,𝑦,𝑡) be continuous functions defined for 𝑥,𝑦,𝑡≥0and having Laplace transforms, 𝐹(𝑝,𝑠,𝑘) and𝐺(𝑝,𝑠,𝑘), respectively. If 𝐹(𝑝,𝑠,𝑘)=𝐺(𝑝,𝑠,𝑘), then 𝑓(𝑥,𝑦,𝑡)=𝑔(𝑥,𝑦,𝑡).

Proof. From the definition of the inverse Laplace transform, if 𝛼,𝛽, and 𝜇are sufficiently large, then the integral expression, by 𝑓=1𝑥,𝑦,𝑡2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥×12𝜋𝑖^𝛽+𝑖∞_{𝛽−𝑖∞}𝑒^𝑠𝑦×12𝜋𝑖^𝜇+𝑖∞_{𝜇−𝑖∞}𝑒^𝑘𝑡×𝐹𝑝,𝑠,𝑘𝑑𝑘𝑑𝑠𝑑𝑝, (6) for the triple inverse Laplace transform, can be used to obtain 𝑓=1𝑥,𝑦,𝑡2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥×12𝜋𝑖^𝛽+𝑖∞_{𝛽−𝑖∞}𝑒^𝑠𝑦×12𝜋𝑖^𝜇+𝑖∞_{𝜇−𝑖∞}𝑒^𝑘𝑡×𝐹𝑝,𝑠,𝑘𝑑𝑘𝑑𝑠𝑑𝑝. (7) By hypothesis, we have that 𝐹(𝑝,𝑠,𝑘)=𝐺(𝑝,𝑠,𝑘). then replacing this in the previous expression, we have the following: 𝑓=1𝑥,𝑦,𝑡2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥×12𝜋𝑖^𝛽+𝑖∞_{𝛽−𝑖∞}𝑒^𝑠𝑦×12𝜋𝑖^𝜇+𝑖∞_{𝜇−𝑖∞}𝑒^𝑘𝑡×𝐺𝑝,𝑠,𝑘𝑑𝑘𝑑𝑠𝑑𝑝, (8) which boil down to 𝑓=1𝑥,𝑦,𝑡2𝜋𝑖^𝛼+𝑖∞_{𝛼−𝑖∞}𝑒^𝑝𝑥×12𝜋𝑖^𝛽+𝑖∞_{𝛽−𝑖∞}𝑒^𝑠𝑦×12𝜋𝑖^𝜇+𝑖∞_{𝜇−𝑖∞}𝑒^𝑘𝑡,×𝐺𝑝,𝑠,𝑘𝑑𝑘𝑑𝑠𝑑𝑝,=𝑔𝑥,𝑦,𝑡 (9) and this proves the uniqueness of the triple Laplace transform.

Theorem 4. If, at the point (𝑝,𝑠,𝑘), the integrals 𝐹₁𝑝,𝑠,𝑘=^∞₀[]𝑓exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡₁𝐹𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡₂𝑝,𝑠,𝑘=^∞₀[]𝑓exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡₂𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡 (10) are convergent and in addition if 𝐹₃𝑝,𝑠,𝑘=^∞₀[]𝑓exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡₃𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡 (11) is absolutely convergent, then, the following expression: 𝐹𝑝,𝑠,𝑘=𝐹₁𝑝,𝑠,𝑘𝐹₂𝑝,𝑠,𝑘𝐹₃𝑝,𝑠,𝑘 (12) is the Laplace transform of the function 𝑓=𝑥,𝑦,𝑡_{𝑡0𝑦0𝑥0}𝑓₃𝑥−𝑥₁+𝜌,𝑦−𝑦₁,𝑡+𝜎𝑡−₁+𝜏𝑓₂𝑥₁−𝜌,𝑦₁−𝜎,𝑡₁−𝜏×𝑓₁𝜌,𝜎,𝜏𝑑𝜌𝑑𝜎𝑑𝜏, (13) and the integral 𝐹𝑝,𝑠,𝑘=^∞₀[]𝑓exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡 (14) is convergent at the point (𝑝,𝑠,𝑘); for the readers who are interested, they can see the proof in [11, 12].

Theorem 5. A function 𝑓(𝑥,𝑦,𝑡) which is continuous on [0,∞) and satisfies the growth condition (4) can be recovered from only 𝐹(𝑝,𝑠,𝑘)as 𝑓𝑥,𝑦,𝑡=lim_{𝑛1𝑛→∞2𝑛→∞3→∞}(−1)^{𝑛1+𝑛2+𝑛3}𝑛₁!𝑛₂!𝑛₃!𝑛₁𝑥^𝑛1+1𝑛₂𝑦^𝑛2+1×𝑛₃𝑡^𝑛3+1Χ^{𝑛1+𝑛2+𝑛3}𝑛₁𝑥,𝑛₂𝑦,𝑛₃𝑡. (15)

Evidently, the main difficulty in using Theorem 5 for computing the inverse Laplace transform is the repeated symbolic differentiation of 𝐹(𝑝,𝑠,𝑘).

Let us see how Theorem 5 can be applicable. Let us consider the following functions: 𝑓𝑥,𝑦,𝑡=exp−𝑎𝑥−𝑏𝑦−𝑐𝑡. (16) Naturally the triple Laplace transform of the previous function is given later as 𝐹=1𝑝,𝑠,𝑘().𝑝−𝑎𝑠−𝑏)(𝑘−𝑐 (17) Now applying the high-order mixed derivative to the previous expression, we obtain the following: 𝜕^{𝑛1+𝑛2+𝑛3}𝐹𝑝,𝑠,𝑘𝜕𝑝^𝑛1𝜕𝑠^𝑛2𝜕𝑘^𝑛3=𝑛₁!𝑛₂!𝑛₃!(−1)^{𝑛1+𝑛2+𝑛3}×(𝑎+𝑃)^−1−𝑛1(𝑠+𝑏)^−1−𝑛2(𝑐+𝑘)^−1−𝑛3. (18) Applying Theorem 5 in the previous expression, we obtain the following result: 𝑓𝑥,𝑦,𝑡=lim_{𝑛1𝑛→∞2𝑛→∞3→∞}𝑛₁^1+𝑛1𝑛₂^1+𝑛2𝑛₃^1+𝑛3𝑥^𝑛1+1𝑦^𝑛2+1𝑡^𝑛3+1𝑛𝑎+¹𝑥^−𝑛1−1×𝑛𝑏+²𝑦^−𝑛2−1𝑛𝑐+³𝑡^−𝑛3−1. (19) Making a change of variable in the previous expression, we obtain the following simplified result: 𝑓𝑥,𝑦,𝑡=lim_{𝑛1𝑛→∞2𝑛→∞3→∞}1+𝑎𝑛₁𝑥^−𝑛1−11+𝑏𝑛₂𝑦^−𝑛2−1×1+𝑐𝑛₃𝑡^−𝑛3−1. (20) Using together, the application of logarithm and the L’Hôpital’s rule on the previous expression, we arrive at the following result: 𝑓.ln𝑥,𝑦,𝑡=−𝑎𝑥−𝑏𝑦−𝑐𝑡⟹𝑓𝑥,𝑦,𝑡=exp−𝑎𝑥−𝑏𝑦−𝑐𝑡 (21)

4. Some Properties of Triple Laplace Transform

In this section, we present some properties of the triple Laplace transform. Note that these properties follow from those of the double Laplace transform introduced by Kılıçman and Eltayeb [8]. The properties of the triple Laplace transform will enable us to find further transform pairs {𝑓(𝑥,𝑦,𝑡),𝐹(𝑝,𝑠,𝑘)}: (i)𝐹𝑝+𝑎,𝑠+𝑏,𝑘+𝑑=𝐿_{𝑥,𝑦,𝑡}𝑒^{−𝑎𝑥−𝑦𝑏−𝑐𝑡}𝑓𝑥,𝑦,𝑡𝑝,𝑠,𝑘. (22) We will present the proof 𝐿_{𝑥,𝑦,𝑡}𝑒^{−𝑎𝑥−𝑦𝑏−𝑐𝑡}𝑓𝑥,𝑦,𝑡𝑝,𝑠,𝑘=^∞₀[][][]𝑓exp−𝑝𝑥exp−𝑠𝑦exp−𝑘𝑡exp−𝑎𝑥×exp−𝑏𝑦exp−𝑐𝑡𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡,_∞0[]×exp−𝑝𝑥exp−𝑎𝑥_∞0[][]𝑓exp−𝑠𝑦exp−𝑘𝑡exp−𝑏𝑦×exp−𝑐𝑡𝑥,𝑦,𝑡𝑑𝑡𝑑𝑦𝑑𝑡. (23) Note that the integral inside the bracket satisfies the properties of the double Laplace transform and is given as [11] ^∞₀[][]().exp−𝑠𝑦exp−𝑘𝑡exp−𝑏𝑦exp−𝑐𝑡×𝑓𝑥,𝑦,𝑡𝑑𝑡𝑑𝑦=𝐹𝑥,𝑠+𝑏,𝑘+𝑑 (24) Thus ^∞₀[]𝐹(),exp−𝑝𝑥exp−𝑎𝑥𝑥,𝑠+𝑏,𝑘+𝑑𝑑𝑡=𝐹𝑝+𝑎,𝑠+𝑏,𝑘+𝑑 (25) and this completes the proof.

(ii)   The following can also be observed: 1𝐹𝑝𝛼𝛽𝛾𝛼,𝑠𝛽,𝑘𝛾=𝐿_{𝑥,𝑦,𝑡}𝑓𝛼𝑥,𝛽𝑦,𝛾𝑡𝑝,𝑠,𝑘. (26)

We will present the proof 𝐿_{𝑥,𝑦,𝑡}𝑓=𝛼𝑥,𝛽𝑦,𝛾𝑡𝑝,𝑠,𝑘_∞0exp−𝑝𝑥exp−𝑠𝑦exp[−𝑘𝑡]×𝑓𝛼𝑥,𝛽𝑦,𝛾𝑡𝑑𝑥𝑑𝑦𝑑𝑡,_∞0exp−𝑝𝑥^∞₀exp−𝑠𝑦exp[−𝑘𝑡]×𝑓𝛼𝑥,𝛽𝑦,𝛾𝑡𝑑𝑦𝑑𝑡𝑑𝑥. (27) Note that the double integral inside the bracket satisfies the property of the double Laplace transform as [11] ^∞₀[]𝑓=1exp−𝑠𝑦exp−𝑘𝑡𝛼𝑥,𝛽𝑦,𝛾𝑡𝑑𝑦𝑑𝑡𝐹𝑠𝛽𝛾𝛼𝑥,𝛽,𝑘𝛾. (28) Thus 𝐿_{𝑥,𝑦,𝑡}𝑓=𝛼𝑥,𝛽𝑦,𝛾𝑡𝑝,𝑠,𝑘_∞01exp−𝑝𝑥𝐹𝑠𝛽𝛾𝛼𝑥,𝛽,𝑘𝛾=1𝑑𝑥𝐹𝑝𝛼𝛽𝛾𝛼,𝑠𝛽,𝑘𝛾, (29) and this completes the proof.

(iii)   The following property can also be observed: 𝜕^{𝑛+𝑚+𝑣}𝐹𝑝,𝑠,𝑘𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=𝐿_{𝑥,𝑦,𝑡}(−1)^{𝑛+𝑚+𝑣}𝑥^𝑛𝑦^𝑚𝑡^𝑣𝑓𝑥,𝑦,𝑡𝑝,𝑠,𝑘. (30) We will present the proof 𝐹𝑝,𝑠,𝑘=^∞₀exp−𝑝𝑥exp−𝑠𝑦exp[−𝑘𝑡]×𝑓𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡. (31) Then, 𝜕^{𝑛+𝑚+𝑣}𝐹𝑝,𝑠,𝑘𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=𝜕^{𝑛+𝑚+𝑣}𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣∞₀[]𝑓.exp−𝑝𝑥exp−𝑠𝑦×exp−𝑘𝑡𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡 (32) Now making use of the convergence properties of the improper integral involved, we can interchange the operation of differentiation and integration and differentiate with respect to 𝑝, 𝑠, and 𝑘 under the integral sign. Thus, we arrive at the following expression: 𝜕^{𝑛+𝑚+𝑣}𝐹𝑝,𝑠,𝑘𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=𝜕^𝑛𝜕𝑝^𝑛∞₀×𝜕exp−𝑝𝑥_𝑚+𝑣𝜕𝑠^𝑛𝜕𝑘^𝑣∞₀exp−𝑠𝑦exp[−𝑘𝑡]×𝑓𝑥,𝑦,𝑡𝑑𝑦𝑑𝑡𝑑𝑥. (33) Note that the expression in the bracket satisfies the property of the double Laplace transform as [11] 𝜕^𝑚+𝑣𝜕𝑠^𝑛𝜕𝑘^𝑣∞₀exp−𝑠𝑦exp[−𝑘𝑡]𝑓𝑥,𝑦,𝑡𝑑𝑦𝑑𝑡=𝐿_𝑦,𝑡(−1)^𝑚+𝑣𝑦^𝑚𝑡^𝑣𝑓𝑥,𝑦,𝑡(𝑠,𝑘). (34) Thus 𝜕^{𝑛+𝑚+𝑣}𝐹𝑝,𝑠,𝑘𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=𝜕^𝑛𝜕𝑝^𝑛∞₀×𝐿exp−𝑝𝑥_𝑦,𝑡(−1)^𝑚+𝑣𝑦^𝑚𝑡^𝑣𝑓𝑥,𝑦,𝑡(𝑠,𝑘)𝑑𝑥. (35) And finally, we obtain 𝜕^{𝑛+𝑚+𝑣}𝐹𝑝,𝑠,𝑘𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=𝐿_{𝑥,𝑦,𝑡}(−1)^{𝑛+𝑚+𝑣}𝑥^𝑛𝑦^𝑚𝑡^𝑣𝑓𝑥,𝑦,𝑡𝑝,𝑠,𝑘, (36) and this completes the proof.

Now using the previous three properties, we will show the proof of Theorem 5.

Proof of Theorem 5. Let us define the set of functions depending on parameters 𝑚, 𝑛, and 𝑣 as ℎ_{𝑚,𝑛,𝑣}=𝑚𝑥,𝑦,𝑡^𝑚+1𝑛^𝑛+1𝑣^𝑣+1𝑥𝑚!𝑛!𝑣!^𝑚𝑦^𝑛𝑡^𝑣𝑒^{−𝑚𝑥−𝑛𝑦−𝑣𝑡}. (37) It worth noting that the previous function is a kind of three-dimensional density of probability, and it therefore follows that ^∞₀ℎ_{𝑚,𝑛,𝑣}𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡=1. (38) In addition of this, we will have that lim_{𝑚→∞𝑛→∞𝑣→∞}^∞₀ℎ_{𝑚,𝑛,𝑣}𝑥,𝑦,𝑡𝜓𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡=𝜓(1,1,1), (39) where 𝜓(𝑥,𝑦,𝑡) is any continuous function. Let Ψ(𝑝,𝑠,𝑘) denote the triple Laplace transform of the continuous function 𝜓(𝑥,𝑦,𝑡). However, if one defines the function 𝑀(𝑥,𝑦,𝑡)=𝑓(𝑥𝛼,𝑦𝛽,𝑡𝛾), making use of the second property established in (29), we arrive at the following: 1𝐹𝑝𝛼𝛽𝛾𝛼,𝑠𝛽,𝑘𝛾=𝐿_{𝑥,𝑦,𝑡}𝑓𝛼𝑥,𝛽𝑦,𝛾𝑡𝑝,𝑠,𝑘. (40) Here if one applies the third property, in particular by replacing 𝑝=𝑚/𝑥, 𝑠=𝑛/𝑦, 𝑘=𝑣/𝑡 as follows: 𝐿_𝑥𝑦𝑡𝑀=1𝑥,𝑦,𝑡𝐹𝑝𝛼𝛽𝛾𝛼,𝑠𝛽,𝑘𝛾,𝜕^{𝑛+𝑚+𝑣}𝐿_𝑥𝑦𝑡𝑀𝑥,𝑦,𝑡𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=𝜕^{𝑛+𝑚+𝑣}1/𝛼𝛽𝛾𝐹𝑝/𝛼,𝑠/𝛽,𝑘/𝛾𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=1𝛼^𝑚+1𝛽^𝑛+1𝛾^𝑣+1×𝜕^{𝑛+𝑚+𝑣}𝐹𝑝/𝛼,𝑠/𝛽,𝑘/𝛾𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣. (41)(42) Now let us put 𝜓(𝑥,𝑦,𝑡)=𝑒^{−𝑝𝑥−𝑠𝑦−𝑘𝑡}𝑀(𝑥,𝑦,𝑡). Now if we make use of (38), we obtain the following 𝜓(1,1,1)=𝑒^{−𝑝−𝑠−𝑘}𝑀(1,1,1)=𝑒^{−𝑝−𝑠−𝑘}𝑓𝛼,𝛽,𝛾=lim_{𝑚→∞𝑛→∞𝑣→∞}𝑚^𝑚+1𝑛^𝑛+1𝑣^𝑣+1𝑚!𝑛!𝑣!^∞₀𝑥^𝑚𝑦^𝑛𝑡^𝑣𝑒^{−𝑝𝑥−𝑠𝑦−𝑘𝑡}×𝑒^{−𝑚𝑥−𝑛𝑦−𝑣𝑡}Ψ𝑥,𝑦,𝑡𝑑𝑥𝑑𝑦𝑑𝑡=lim_{𝑚→∞𝑛→∞𝑣→∞}𝑚^𝑚+1𝑛^𝑛+1𝑣^𝑣+1𝐿𝑚!𝑛!𝑣!^𝑥𝑦𝑡𝑥^𝑚𝑦^𝑛𝑡^𝑣𝑒^{−𝑚𝑥−𝑛𝑦−𝑣𝑡}Ψ𝑥,𝑦,𝑡. (43) Now taking into account properties (i) and (ii), (42) together with the function 𝑀(𝑥,𝑦,𝑡), we arrive at the following: 𝐿_𝑥𝑦𝑡𝑥^𝑚𝑦^𝑛𝑡^𝑣𝑒^{−𝑚𝑥−𝑛𝑦−𝑣𝑡}Ψ=()𝑥,𝑦,𝑡−1^{𝑚+𝑛+𝑣}𝜕^{𝑛+𝑚+𝑣}𝐿_𝑥𝑦𝑡𝑒^{−𝑚𝑥−𝑛𝑦−𝑘𝑡}Ψ𝑥,𝑦,𝑡𝑝,𝑠,𝑘𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=(−1)^{𝑚+𝑛+𝑣}1𝛼^𝑚𝛽^𝑛𝛾^𝑣×𝜕^{𝑛+𝑚+𝑣}𝐿_𝑥𝑦𝑡Ψ𝑥,𝑦,𝑡𝑝+𝑚,𝑠+𝑛,𝑘+𝑣𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣=(−1)^{𝑚+𝑛+𝑣}1𝛼^𝑚𝛽^𝑛𝛾^𝑣×𝜕^{𝑛+𝑚+𝑣}𝐿_𝑥𝑦𝑡𝑓×𝛼𝑥,𝛽𝑦,𝛾𝑡𝑝+𝑚𝛼,𝑠+𝑛𝛽,𝑘+𝑣𝛾𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣−1=(−1)^{𝑚+𝑛+𝑣}×1𝛼^𝑚𝛽^𝑛𝛾^𝑣𝜕^{𝑛+𝑚+𝑣}𝐹𝑝+𝑚/𝛼,(𝑠+𝑛)/𝛽,(𝑘+𝑣)/𝛾𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣. (44) Now observe that from (44) with the fact that 𝑓(𝛼,𝛽,𝑡)=𝜓(1,1,1)𝑒^{𝑝+𝑠+𝑘}, we arrive at the following: 𝑓𝛼,𝛽,𝑡=𝑒^{𝑝+𝑠+𝑘}lim_{𝑚→∞𝑛→∞𝑣→∞}𝑚^𝑚+1𝑛^𝑛+1𝑣^𝑣+1×𝑚𝑚!𝑛!𝑣!𝛼^𝑚+1𝑛𝛽^𝑛+1𝑣𝛾^𝑣+1×𝜕^{𝑛+𝑚+𝑣}𝐹𝑝+𝑚/𝛼,(𝑠+𝑛)/𝛽,(𝑘+𝑣)/𝛾𝜕𝑝^𝑛𝜕𝑠^𝑛𝜕𝑘^𝑣. (45) The previously mentioned is true for any 𝑝, 𝑠, 𝑘 in the complete space, in particular, for 𝑝=0, 𝑠=0, 𝑘=0, and in this case Theorem 5 is covered.

5. Application to Third-Order Partial Differential Equation

In this section, we present the application of this operator for solving some kind of third-order partial differential equations.

Example 1. consider the following third-order partial differential equation: 𝜕_𝑥𝑦𝑡𝑢𝑥,𝑦,𝑡+𝑢𝑥,𝑦,𝑡=0. (46) The previous equation is called the Mboctara equation and is subjected to the following boundaries and initial conditions: 𝑢𝑥,𝑦,0=𝑒^𝑥+𝑦,𝑢(𝑥,0,𝑡)=𝑒^𝑥−𝑡,𝑢0,𝑦,𝑡=𝑒^𝑦−𝑡,𝑢𝑥,𝑦,1=𝑒^{𝑥+𝑦−1}. (47) Now applying the triple Laplace transform on both sides of (46), we obtain the following: 𝑝𝑠𝑘𝑈𝑝,𝑠,𝑘+𝑈𝑝,𝑠,𝑘=𝐺𝑝,𝑠,𝑘. (48) Here 𝐺()()()().𝑝,𝑠,𝑘=𝑝𝑠𝑈𝑝,𝑠,0+𝑝𝑠𝑈𝑝,0,𝑘−𝑝𝑈𝑝,0,0+𝑠𝑘𝑈0,𝑠,𝑘−𝑠𝑈0,𝑠,0−𝑘𝑈0,0,𝑘+𝑈0,0,0 (49) Factorising the right side of equation (49), we obtain the following: 𝑈=𝐺𝑝,𝑠,𝑘𝑝,𝑠,𝑘.1+𝑝𝑠𝑘 (50) Now applying the inverse triple Laplace transform on the previous equation we obtain the following solution: 𝑢𝑥,𝑦,𝑡=𝐿⁻¹_𝑥𝑦𝑡𝐺𝑝,𝑠,𝑘1+𝑝𝑠𝑘=𝑒^{𝑥+𝑦−𝑡}. (51) This is the exact solution for Mboctara equation.

Example 2. Let us consider the following nonhomogeneous Mboctara equation 𝜕_𝑥𝑦𝑡𝑢𝑥,𝑦,𝑡+𝑢𝑥,𝑦,𝑡=−𝑒^{𝑥−2𝑦+𝑡} (52) subjected to the following initial and boundaries conditions: 𝑢(𝑥,0,0)=𝑒^𝑥,𝜕_𝑡𝑢(𝑥,0,𝑡)=𝑒^𝑥+𝑡,𝜕_𝑥𝑢(𝑥,0,𝑡)=𝑒^𝑥+𝑡,𝑢(0,0,0)=1,𝑢(𝑥,0.5,𝑡)=𝑒^{𝑥+𝑡−1}. (53)

Now applying the triple Laplace transform on both sides of (52), we obtain the following: −1𝑝𝑠𝑘𝑈𝑝,𝑠,𝑘+𝑈𝑝,𝑠,𝑘=𝐺𝑝,𝑠,𝑘().1+𝑝2+𝑠)(1+𝑘 (54) Factorising the right side of (54), we obtain the following: 𝑈=𝐺()𝑝,𝑠,𝑘𝑝,𝑠,𝑘−1/1+𝑝2+𝑠)(1+𝑘.1+𝑝𝑠𝑘 (55) Now applying the inverse triple Laplace transform on the previous equation, we obtain the following solution 𝑢𝑥,𝑦,𝑡=𝐿⁻¹_𝑥𝑦𝑡𝐺𝑝,𝑠,𝑘−1/1+𝑝(2+𝑠)(1+𝑘)1+𝑝𝑠𝑘=𝑒^{𝑥−2𝑦+𝑡}. (56) This is the exact solution for nonhomogeneous Mboctara equation.

Example 3. Let us consider the following nonhomogeneous Mboctara equation 𝜕_𝑥𝑦𝑡𝑢(𝑥)𝑦()(𝑥)𝑦(),𝑥,𝑦,𝑡+𝑢𝑥,𝑦,𝑡=coscoscos−𝑡−sinsinsin−𝑡 (57) subjected to the following initial and boundaries conditions: 𝑢(𝑥)𝑦,𝜕𝑥,𝑦,0=coscos_𝑡𝑢𝑥,𝑦,0=𝜕_𝑥𝑢0,𝑦,𝑡=𝜕_𝑦𝑢()𝑢𝜋𝑥,0,𝑡=0,𝑥,2𝜋,𝑡=𝑢𝑥,𝑦,2𝜋=𝑢2,𝑦,𝑡=0. (58) Now applying the triple Laplace transform on both sides of (57), we obtain the following: +𝑝𝑠𝑘𝑈𝑝,𝑠,𝑘+𝑈𝑝,𝑠,𝑘=𝐺𝑝,𝑠,𝑘𝑘𝑠𝑝1+𝑝²1+𝑠²−1+𝑘²−11+𝑝²1+𝑠²−1+𝑘². (59) Factorising the right side of (59), we obtain the following: 𝑈=𝐺𝑝,𝑠,𝑘𝑝,𝑠,𝑘+1+𝑝𝑠𝑘𝑘𝑠𝑝/1+𝑝²1+𝑠²−1+𝑘²−1+𝑝𝑠𝑘1/1+𝑝²1+𝑠²−1+𝑘².1+𝑝𝑠𝑘 (60) Now applying the inverse triple Laplace transform on the previous equation, we obtain the following solution: 𝑢𝑥,𝑦,𝑡=𝐿⁻¹_𝑥𝑦𝑡𝐺𝑝,𝑠,𝑘+1+𝑝𝑠𝑘𝑘𝑠𝑝/1+𝑝²1+𝑠²−1+𝑘²−1+𝑝𝑠𝑘1/1+𝑝²1+𝑠²−1+𝑘²(𝑥)𝑦().1+𝑝𝑠𝑘=coscoscos−𝑡 (61) This is the exact solution for nonhomogeneous Mboctara equation.

Example 4. consider the following nonlinear nonhomogeneous with variable coefficient Mboctara equation: 𝑒^{𝑥+𝑦+𝑡}𝜕_𝑥𝑦𝑡𝑢𝑥,𝑦,𝑡−3𝑢²𝑥,𝑦,𝑡+𝑒^{𝑥+𝑦+𝑡}𝑢𝑥,𝑦,𝑡=𝑒^{2𝑥+2𝑦+2𝑡},𝑢_𝑥𝑥,𝑦,0=𝑒^𝑥+𝑦()𝑢(),𝑢0,0,0=1,1,0,0=𝑒,𝜕_𝑥𝑦𝑡𝑢(0,0,0)=1. (62) Now applying the triple Laplace transform on both sides of (62) and then using the properties of the triple Laplace transform and after factorising as in the previous examples and taking the inverse triple Laplace transform, we obtain the following as an exact solution of this type of Mboctara equation: 𝑢𝑥,𝑦,𝑡=𝑒^{𝑥+𝑦+𝑡}. (63) The numerical simulations of the exact solutions of the Mboctara equation are depicted in Figure 1(a) (4.1), Figure 1(b) (4.6), Figure 1(c) (4.6) and Figure 1(d) (4.11), respectively.

(a)

(b)

(c)

(d)

Figure 1 

Numerical simulation of the exact solutions of the Homogeneous and non-homogeneous Mboctara equations.

6. Triple Laplace Transform of Some Functions of Three Variables

In this section, we examine the triple Laplace transform of some functions in Table 1: 𝐿_𝑥𝑦𝑡𝑌_𝑛(𝑥)𝑌_𝑛𝑦𝑌_𝑛(𝑡)=8^−𝑛𝑘𝑠𝑝^−1−𝑛𝐶𝑠𝑐³[]×−𝑛𝜋4𝑘^2𝑛×Hypergeometric2𝐹11+𝑛2,2+𝑛21,1−𝑛,−𝑘²()×+cos𝑛𝜋Hypergeometric2𝐹11+𝑛2,2+𝑛21,1+𝑛,−𝑘²×−4𝑠^2𝑛×Hypergeometric2𝐹11+𝑛2,2+𝑛21,1−𝑛,−𝑠²()×+cos𝑛𝜋Hypergeometric2𝐹11+𝑛2,2+𝑛21,1+𝑛,−𝑠²×−4𝑝^2𝑛×Hypergeometric2𝐹11+𝑛2,2+𝑛21,1−𝑛,−𝑝²()×+cos𝑛𝜋Hypergeometric2𝐹11+𝑛2,2+𝑛21,1+𝑛,−𝑝². (64)

Table 1 

Table of triple Laplace transform for some function of three variables.

7. Conclusion

This work presents the definition of the triple Laplace transform. Some triple Laplace transform is presented in Table 1. Some theorems and properties of this new relatively new operator are presented. Applications of the new operator, for solving some kind of third-order partial differential equations called Mboctara equation, are presented. Numerical solutions of the Mboctara equation are given.