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

Authors: Abdon Atangana
Journal: Abstract and Applied Analysis
 (2013)

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## 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.”


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## Body

## 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.
Letf be a continuous function of three variables; then, the triple Laplace transform of  f(x,y,t) is defined by
(1)Lx,y,t[f(x,y,t)]=F(p,s,k)∭0∞exp[-px]exp[-sy]×exp[-kt]f(x,y,t)dxdydt,
where, x,y,t>0 and p, s, k are Laplace variables, and
(2)f(x,y,t)=12πi∫α-i∞α+i∞epx×[12πi∫β-i∞β+i∞esy×[12πi∫μ-i∞μ+i∞ekt×F(p,s,k)dk∫μ-i∞μ+i∞ekt12πi]ds∫β-i∞β+i∞esy12πi]dp
is the inverse triple Laplace transform.Property 2.
Assuming that the continuous functionf(x,y,t) is triple Laplace transformable, then,
(3)Lt,y,x[∂3f(x,y,t)∂x∂y∂t]=pskF(p,s,k)-psF(p,s,0)-psF(p,0,k)+pF(p,0,0)-skF(0,s,k)+sF(0,s,0)+kF(0,0,k)-F(0,0,0),Lx,x,t[∂3f(x,y,t)∂t∂x2]=kp2F(p,y,k)-pkF(0,y,k)-∂F(0,y,k)∂x-p2F(p,y,0)+pF(0,y,0)+∂F(0,y,0)∂x,Lxxx[∂3f(x,y,t)∂x3]=p3F(p,y,t)-p2F(0,y,t)-p∂F(0,y,t)∂x-∂2F(0,y,t)∂x2.


## 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, letf(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
(4)supx,y,t>0|f(x,y,t)exp[ax+by+ct]|<0.
Under the previous condition, the triple Laplace transform,
(5)F(p,s,k)=∭0∞exp[-px]exp[-sy]×exp[-kt]f(x,y,t)dxdydt,
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.
Letf(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, ifα,β, and μare sufficiently large, then the integral expression, by
(6)f(x,y,t)=12πi∫α-i∞α+i∞epx×[12πi∫β-i∞β+i∞esy×[12πi∫μ-i∞μ+i∞ekt×F(p,s,k)dk∫μ-i∞μ+i∞ekt12πi]ds∫μ-i∞μ+i∞ekt12πi]dp,
for the triple inverse Laplace transform, can be used to obtain
(7)f(x,y,t)=12πi∫α-i∞α+i∞epx×[12πi∫β-i∞β+i∞esy×[12πi∫μ-i∞μ+i∞ekt×F(p,s,k)dk∫μ-i∞μ+i∞ekt12πi]ds∫μ-i∞μ+i∞ekt12πi]dp.
By hypothesis, we have that F(p,s,k)=G(p,s,k). then replacing this in the previous expression, we have the following:
(8)f(x,y,t)=12πi∫α-i∞α+i∞epx×[12πi∫β-i∞β+i∞esy×[12πi∫μ-i∞μ+i∞ekt×G(p,s,k)dk∫μ-i∞μ+i∞ekt12πi]ds∫μ-i∞μ+i∞ekt12πi]dp,
which boil down to
(9)f(x,y,t)=12πi∫α-i∞α+i∞epx×[12πi∫β-i∞β+i∞esy×[12πi∫μ-i∞μ+i∞ekt×G(p,s,k)dk∫μ-i∞μ+i∞ekt12πi]ds∫μ-i∞μ+i∞ekt12πi]dp,=g(x,y,t),
and this proves the uniqueness of the triple Laplace transform.Theorem 4.
If, at the point(p,s,k), the integrals
(10)F1(p,s,k)=∭0∞exp[-px]exp[-sy]×exp[-kt]f1(x,y,t)dxdydtF2(p,s,k)=∭0∞exp[-px]exp[-sy]×exp[-kt]f2(x,y,t)dxdydt
are convergent and in addition if
(11)F3(p,s,k)=∭0∞exp[-px]exp[-sy]×exp[-kt]f3(x,y,t)dxdydt
is absolutely convergent, then, the following expression:
(12)F(p,s,k)=F1(p,s,k)F2(p,s,k)F3(p,s,k)
is the Laplace transform of the function
(13)f(x,y,t)=∫0t∫0y∫0xf3((t1+τ)x-(x1+ρ),y-(y1+σ),t-(t1+τ))f2(x1-ρ,y1-σ,t1-τ)×f1(ρ,σ,τ)dρdσdτ,
and the integral
(14)F(p,s,k)=∭0∞exp[-px]exp[-sy]×exp[-kt]f(x,y,t)dxdydt
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 functionf(x,y,t) which is continuous on [0,∞) and satisfies the growth condition (4) can be recovered from only F(p,s,k)as
(15)f(x,y,t)=limn1→∞n2→∞n3→∞(-1)n1+n2+n3n1!n2!n3!(n1x)n1+1(n2y)n2+1×(n3t)n3+1Χn1+n2+n3[n1x,n2y,n3t].Evidently, the main difficulty in using Theorem5 for computing the inverse Laplace transform is the repeated symbolic differentiation of  F(p,s,k).Let us see how Theorem5 can be applicable. Let us consider the following functions:
(16)f(x,y,t)=exp[-ax-by-ct].
Naturally the triple Laplace transform of the previous function is given later as
(17)F(p,s,k)=1(p-a)(s-b)(k-c).
Now applying the high-order mixed derivative to the previous expression, we obtain the following:
(18)∂n1+n2+n3[F(p,s,k)]∂pn1∂sn2∂kn3=n1!n2!n3!(-1)n1+n2+n3×(a+P)-1-n1(s+b)-1-n2(c+k)-1-n3.
Applying Theorem 5 in the previous expression, we obtain the following result:
(19)f(x,y,t)=limn1→∞n2→∞n3→∞n11+n1n21+n2n31+n3xn1+1yn2+1tn3+1(a+n1x)-n1-1×(b+n2y)-n2-1(c+n3t)-n3-1.
Making a change of variable in the previous expression, we obtain the following simplified result:
(20)f(x,y,t)=limn1→∞n2→∞n3→∞(1+an1x)-n1-1(1+bn2y)-n2-1×(1+cn3t)-n3-1.
Using together, the application of logarithm and the L’Hôpital’s rule on the previous expression, we arrive at the following result:
(21)ln(f(x,y,t))=-ax-by-ct⟹f(x,y,t)=exp[-ax-by-ct].


## 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)}:
(22)(i)F(p+a,s+b,k+d)=Lx,y,t[e-ax-yb-ctf(x,y,t)](p,s,k).
We will present the proof
(23)Lx,y,t[e-ax-yb-ctf(x,y,t)](p,s,k)=∭0∞exp[-px]exp[-sy]exp[-kt]exp[-ax]×exp[-by]exp[-ct]f(x,y,t)dxdydt,∫0∞exp[-px]exp[-ax]×(∬0∞exp[-sy]exp[-kt]exp[-by]×exp[-ct]f(x,y,t)dtdy∬0∞)dt.
Note that the integral inside the bracket satisfies the properties of the double Laplace transform and is given as [11]
(24)(∬0∞exp[-sy]exp[-kt]exp[-by]exp[-ct]×f(x,y,t)dtdy∬)=F(x,s+b,k+d).
Thus
(25)∫0∞exp[-px]exp[-ax]F(x,s+b,k+d)dt=F(p+a,s+b,k+d),
and this completes the proof.(ii)   The following can also be observed:(26)1αβγF(pα,sβ,kγ)=Lx,y,t[f(αx,βy,γt)](p,s,k).We will present the proof(27)Lx,y,t[f(αx,βy,γt)](p,s,k)=∭0∞exp[-px]exp[-sy]exp[-kt]×f(αx,βy,γt)dxdydt,∫0∞exp[-px](∬0∞exp[-sy]exp[-kt]×f(αx,βy,γt)dydt∬)dx.
Note that the double integral inside the bracket satisfies the property of the double Laplace transform as [11]
(28)(∬0∞exp[-sy]exp[-kt]f(αx,βy,γt)dydt)=1βγF(αx,sβ,kγ).
Thus
(29)Lx,y,t[f(αx,βy,γt)](p,s,k)=∫0∞exp[-px]1βγF(αx,sβ,kγ)dx=1αβγF(pα,sβ,kγ),
and this completes the proof.(iii)   The following property can also be observed:(30)∂n+m+v[F(p,s,k)]∂pn∂sn∂kv=Lx,y,t[(-1)n+m+vxnymtvf(x,y,t)](p,s,k).
We will present the proof
(31)F(p,s,k)=∭0∞exp[-px]exp[-sy]exp[-kt]×f(x,y,t)dxdydt.
Then,
(32)∂n+m+v[F(p,s,k)]∂pn∂sn∂kv=∂n+m+v∂pn∂sn∂kv(∭0∞exp[-px]exp[-sy]×exp[-kt]f(x,y,t)dxdydt∭).
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:
(33)∂n+m+v[F(p,s,k)]∂pn∂sn∂kv=∂n∂pn∫0∞exp[-px]×(∂m+v∂sn∂kv∬0∞exp[-sy]exp[-kt]×f(x,y,t)dydt∂m+v∂sn∂kv)dx.
Note that the expression in the bracket satisfies the property of the double Laplace transform as [11]
(34)∂m+v∂sn∂kv∬0∞exp[-sy]exp[-kt]f(x,y,t)dydt=Ly,t[(-1)m+vymtvf(x,y,t)](s,k).
Thus
(35)∂n+m+v[F(p,s,k)]∂pn∂sn∂kv=∂n∂pn∫0∞exp[-px]×(Ly,t[(-1)m+vymtvf(x,y,t)](s,k))dx.
And finally, we obtain
(36)∂n+m+v[F(p,s,k)]∂pn∂sn∂kv=Lx,y,t[(-1)n+m+vxnymtvf(x,y,t)](p,s,k),
and this completes the proof.Now using the previous three properties, we will show the proof of Theorem5.Proof of Theorem5.
Let us define the set of functions depending on parametersm,  n, and v as
(37)hm,n,v(x,y,t)=mm+1nn+1vv+1m!n!v!xmyntve-mx-ny-vt.
It worth noting that the previous function is a kind of three-dimensional density of probability, and it therefore follows that
(38)∭0∞hm,n,v(x,y,t)dxdydt=1.
In addition of this, we will have that
(39)limm→∞n→∞v→∞∭0∞hm,n,v(x,y,t)ψ(x,y,t)dxdydt=ψ(1,1,1),
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:
(40)1αβγF(pα,sβ,kγ)=Lx,y,t[f(αx,βy,γt)](p,s,k).
Here if one applies the third property, in particular by replacing p=m/x,  s=n/y,  k=v/t  as follows:
(41)Lxyt(M(x,y,t))=1αβγF(pα,sβ,kγ),(42)∂n+m+v[Lxyt(M(x,y,t))]∂pn∂sn∂kv=∂n+m+v[(1/αβγ)F(p/α,s/β,k/γ)]∂pn∂sn∂kv=1αm+1βn+1γv+1×∂n+m+v[F(p/α,s/β,k/γ)]∂pn∂sn∂kv.
Now let us put ψ(x,y,t)=e-px-sy-ktM(x,y,t). Now if we make use of (38), we obtain the following
(43)ψ(1,1,1)=e-p-s-kM(1,1,1)=e-p-s-kf(α,β,γ)=limm→∞n→∞v→∞mm+1nn+1vv+1m!n!v!∭0∞xmyntve-px-sy-kt=limm→∞n→∞v→∞mm+1nn+1vv+1m!n!v!×e-mx-ny-vtΨ(x,y,t)dxdydt=limm→∞n→∞v→∞mm+1nn+1vv+1m!n!v!Lxyt[xmyntve-mx-ny-vtΨ(x,y,t)].
Now taking into account properties (i) and (ii), (42) together with the function M(x,y,t), we arrive at the following:
(44)Lxyt[xmyntve-mx-ny-vtΨ(x,y,t)]=(-1)m+n+v∂n+m+v[Lxyt(e-mx-ny-ktΨ(x,y,t))(p,s,k)]∂pn∂sn∂kv=(-1)m+n+v1αmβnγv×∂n+m+v[Lxyt(Ψ(x,y,t))(p+m,s+n,k+v)]∂pn∂sn∂kv=(-1)m+n+v1αmβnγv×(({k+vγ}∂n+m+v(∂pn∂sn∂kv)-1[{k+vγ}Lxyt(f(αx,βy,γt))×(p+mα,s+nβ,k+vγ)])(∂pn∂sn∂kv)-1)=(-1)m+n+v×1αmβnγv∂n+m+v[F((p+m)/α,(s+n)/β,(k+v)/γ)]∂pn∂sn∂kv.
Now observe that from (44) with the fact that f(α,β,t)=ψ(1,1,1)ep+s+k, we arrive at the following:
(45)f(α,β,t)=ep+s+klimm→∞n→∞v→∞mm+1nn+1vv+1m!n!v!×(mα)m+1(nβ)n+1(vγ)v+1×∂n+m+v[F((p+m)/α,(s+n)/β,(k+v)/γ)]∂pn∂sn∂kv.
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:(46)∂xytu(x,y,t)+u(x,y,t)=0.
The previous equation is called the Mboctara equation and is subjected to the following boundaries and initial conditions:
(47)u(x,y,0)=ex+y,u(x,0,t)=ex-t,u(0,y,t)=ey-t,u(x,y,1)=ex+y-1.
Now applying the triple Laplace transform on both sides of (46), we obtain the following:
(48)pskU(p,s,k)+U(p,s,k)=G(p,s,k).
Here
(49)G(p,s,k)=psU(p,s,0)+psU(p,0,k)-pU(p,0,0)+skU(0,s,k)-sU(0,s,0)-kU(0,0,k)+U(0,0,0).
Factorising the right side of equation (49), we obtain the following:
(50)U(p,s,k)=G(p,s,k)1+psk.
Now applying the inverse triple Laplace transform on the previous equation we obtain the following solution:
(51)u(x,y,t)=Lxyt-1[G(p,s,k)1+psk]=ex+y-t.
This is the exact solution for Mboctara equation.Example 2.
Let us consider the following nonhomogeneous Mboctara equation(52)∂xytu(x,y,t)+u(x,y,t)=-ex-2y+t
subjected to the following initial and boundaries conditions:
(53)u(x,0,0)=ex,∂tu(x,0,t)=ex+t,∂xu(x,0,t)=ex+t,u(0,0,0)=1,u(x,0.5,t)=ex+t-1.Now applying the triple Laplace transform on both sides of (52), we obtain the following:
(54)pskU(p,s,k)+U(p,s,k)=G(p,s,k)-1(1+p)(2+s)(1+k).
Factorising the right side of (54), we obtain the following:
(55)U(p,s,k)=G(p,s,k)-1/(1+p)(2+s)(1+k)1+psk.
Now applying the inverse triple Laplace transform on the previous equation, we obtain the following solution
(56)u(x,y,t)=Lxyt-1[G(p,s,k)-1/(1+p)(2+s)(1+k)1+psk]=ex-2y+t.
This is the exact solution for nonhomogeneous Mboctara equation.Example 3.
Let us consider the following nonhomogeneous Mboctara equation(57)∂xytu(x,y,t)+u(x,y,t)=cos(x)cos(y)cos(-t)-sin(x)sin(y)sin(-t),
subjected to the following initial and boundaries conditions:
(58)u(x,y,0)=cos(x)cos(y),∂tu(x,y,0)=∂xu(0,y,t)=∂yu(x,0,t)=0,u(x,π2,t)=u(x,y,π2)=u(π2,y,t)=0.
Now applying the triple Laplace transform on both sides of (57), we obtain the following:
(59)pskU(p,s,k)+U(p,s,k)=G(p,s,k)+ksp(1+p2)(1+s2)(-1+k2)-1(1+p2)(1+s2)(-1+k2).
Factorising the right side of (59), we obtain the following:
(60)U(p,s,k)=G(p,s,k)1+psk+ksp/(1+p2)(1+s2)(-1+k2)1+psk-1/(1+p2)(1+s2)(-1+k2)1+psk.
Now applying the inverse triple Laplace transform on the previous equation, we obtain the following solution:
(61)u(x,y,t)=Lxyt-1[G(p,s,k)1+psk+ksp/(1+p2)(1+s2)(-1+k2)1+psk-1/(1+p2)(1+s2)(-1+k2)1+psk]=cos(x)cos(y)cos(-t).
This is the exact solution for nonhomogeneous Mboctara equation.Example 4.
consider the following nonlinear  nonhomogeneous with variable coefficient Mboctara equation:(62)ex+y+t∂xytu(x,y,t)-3u2(x,y,t)+ex+y+tu(x,y,t)=e2x+2y+2t,ux(x,y,0)=ex+y,u(0,0,0)=1,u(1,0,0)=e,∂xytu(0,0,0)=1.
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:
(63)u(x,y,t)=ex+y+t.
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.Numerical simulation of the exact solutions of the Homogeneous and non-homogeneous Mboctara equations.
(a)
(b)
(c)
(d)


## 6. Triple Laplace Transform of Some Functions of Three Variables

In this section, we examine the triple Laplace transform of some functions in Table1:
(64)Lxyt(Yn(x)Yn(y)Yn(t))=8-n(ksp)-1-nCsc3[nπ]×(2+n2-(4k2)nHypergeometric2F1×(1+n2,2+n2,1-n,-1k2)+cos(nπ)Hypergeometric2F1×(1+n2,2+n2,1+n,-1k2))×(2+n2-(4s2)nHypergeometric2F1×(1+n2,2+n2,1-n,-1s2)+cos(nπ)Hypergeometric2F1×(1+n2,2+n2,1+n,-1s2))×(2+n2-(4p2)nHypergeometric2F1×(1+n2,2+n2,1-n,-1p2)+cos(nπ)Hypergeometric2F1×(1+n2,2+n2,1+n,-1p2)).Table 1
Table of triple Laplace transform for some function of three variables.
Functionsf(x,y,t)
Triple laplace transformF(p,s,k)
a
b
c
a
b
c
p
s
k
x
y
t
1
p
2
s
2
k
2
x
n
y
m
t
v
,
n
,
m
,
v  are natural numbers
k
-
1
-
v
s
-
1
-
m
p
-
n
-
1
Γ
(
1
+
n
)
Γ
(
1
+
m
)
Γ
(
1
+
v
)
x
n
y
m
t
v
e
-
a
x
-
b
y
-
c
t
(
k
+
c
)
-
1
-
v
(
s
+
b
)
-
1
-
m
(
p
+
a
)
-
n
-
1
Γ
(
1
+
n
)
Γ
(
1
+
m
)
Γ
(
1
+
v
)
e
-
a
x
-
b
y
-
c
t
1
(
a
+
p
)
(
b
+
s
)
(
c
+
k
)
cos
(
x
)
cos
(
y
)
cos
(
t
)
k
s
p
(
1
+
p
2
)
(
1
+
s
2
)
(
1
+
k
2
)
sin
(
x
)
sin
(
y
)
sin
(
t
)
1
(
1
+
p
2
)
(
1
+
s
2
)
(
1
+
k
2
)
sin
(
x
+
y
+
t
)
-
1
+
p
s
+
k
(
p
+
s
)
(
1
+
p
2
)
(
1
+
s
2
)
(
1
+
k
2
)
cos
(
x
+
y
+
t
)
-
k
+
p
+
s
-
k
p
s
(
1
+
p
2
)
(
1
+
s
2
)
(
1
+
k
2
)
x
y
t
π
π
8
3
k
s
p
e
a
x
+
y
b
+
c
t
sinh
(
a
x
)
sinh
(
b
y
)
sinh
(
c
t
)
(
b
)
(
c
)
(
a
)
(
-
2
a
p
+
p
2
)
(
-
2
b
s
+
s
2
)
(
-
2
c
k
+
k
2
)
e
a
x
+
y
b
+
c
t
cosh
(
a
x
)
cosh
(
b
y
)
cosh
(
c
t
)
(
b
-
s
)
(
c
-
k
)
(
a
-
p
)
(
-
2
a
p
+
p
2
)
(
-
2
b
s
+
s
2
)
(
-
2
c
k
+
k
2
)
Erf
[
a
2
x
]
Erf
[
b
2
y
]
Erf
[
c
2
t
]
e
-
c
2
k
-
b
2
s
k
p
s
(
-
1
+
e
-
c
2
k
)
(
1
-
e
-
a
2
p
)
(
-
1
+
e
-
b
2
s
)
sin
(
a
x
)
x
sin
(
b
y
)
y
sin
(
c
t
)
t
arctan
(
a
2
p
)
arctan
(
b
2
s
)
arctan
(
c
2
k
)
cos
(
a
x
)
x
n
cos
(
b
y
)
y
m
cos
(
c
t
)
t
v
k
-
1
+
v
(
1
+
b
2
s
2
)
1
/
2
(
-
1
+
m
)
s
-
1
+
m
cos
[
c
t
]
cos
(
(
-
1
+
m
)
arctan
[
b
2
s
]
)
Γ
(
1
-
m
)
Γ
(
1
-
v
)  ×(1+a2p2)1/2(-1+n)p-1+ncos((n-1)arctan(|a|p))Γ(1-n)
sin
(
a
x
)
x
n
sin
(
b
y
)
y
m
sin
(
c
t
)
t
v
k
-
1
+
v
(
1
+
b
2
s
2
)
1
/
2
(
-
1
+
m
)
s
-
1
+
m
Γ
(
1
-
m
)
Γ
(
1
-
v
)
(
1
+
a
2
p
2
)
1
/
2
(
-
1
+
n
) ×p-1+nΓ(1-n)sign(a)sin((n-1)arctan(|a|p))Γ(1-n)
J
n
(
x
)
J
n
(
y
)
J
n
(
t
)
8
-
n
(
k
s
p
)
-
1
-
n
Hypergeometric
2
F
1
(
1
+
n
2
,
2
+
n
2
,
-
1
k
2
)
Hypergeometric
2
F
1
(
1
+
n
2
,
2
+
n
2
,
-
1
s
2
)
Hypergeometric
2
F
1
(
1
+
n
2
,
2
+
n
2
,
-
1
p
2
)
I
n
(
x
)
I
n
(
y
)
I
n
(
t
)
8
-
n
(
k
s
p
)
-
1
-
n
Hypergeometric
2
F
1
(
1
+
n
2
,
2
+
n
2
,
1
+
n
,
1
k
2
)
Hypergeometric
2
F
1
(
1
+
n
2
,
2
+
n
2
,
1
+
n
,
1
s
2
)
Hypergeometric
2
F
1
(
1
+
n
2
,
2
+
n
2
,
1
+
n
,
1
p
2
)


## 7. Conclusion

This work presents the definition of the triple Laplace transform. Some triple Laplace transform is presented in Table1. 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.



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*Source: 769102-2013-06-24.xml*