# A Transformation Method for Delta Partial Difference Equations on Discrete Time Scale

Authors: Syed Sabyel Haider; Mujeeb Ur Rehman; Thabet Abdeljawad
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

The aim of this study is to develop a transform method for discrete calculus. We define the double Laplace transforms in a discrete setting and discuss its existence and uniqueness with some of its important properties. The delta double Laplace transforms have been presented for integer and noninteger order partial differences. For illustration, the delta double Laplace transforms are applied to solve partial difference equation.


---

## Body

## 1. Introduction

The origin of calculus of finite differences is found from Brook Taylor (1717), rather it was Jacob Stirling, who found the theory (1730) and introduce the deltaΔ symbol for the difference, which is in common use nowadays. The development on calculus of finite differences in the beginning of the nineteenth century by Lacroix and remarkable work of George Boole, Narlund, and Steffensen appeared later in the nineteenth century. Jordan discussed calculus of finite differences with the classical approach in [1]. In modern era, the focus of mathematician is to correlate the continuous and the discrete, to shape in comprehensive unified mathematics, and to eliminate ambiguity. The calculus of finite differences is applicable to both continuous and discrete functions. For difference equations, Bohner and Peterson treat the dynamic equations on time scales in [2] and get surprisingly different results from continuous counterpart. Some results can be found in [3–19] which has helped to construct the theory of discrete fractional calculus.Coon and Bernstein [20–22] defined the double Laplace transforms (continuous) and investigated many properties. Debnath [23] modified the properties and use the double Laplace transforms (continuous) to solve functional, integral, and partial differential equations. Dhunde and Waghmare [24] discussed convergence and absolute convergence of the double Laplace transforms (continuous) and, by application of double Laplace transforms, presented the solution of Volterra Integropartial differential equation. For applications of triple, quadruple, and n-dimensional Laplace transforms (continuous), we refer the readers to [25–27]. Goodrich and Peterson [10] developed discrete delta Laplace transforms analogous to Laplace transforms discussed by Bohner and Peterson [2] in the continuous case, to solve difference and summation equations with initial data by applying the delta Laplace transforms. The delta Laplace transforms is given for newly defined Hilfer difference operator [28] and substantial difference operator in [29]. Bohner et al. [30] generalized properties of the Laplace transforms to the delta Laplace transforms on arbitrary time scales and discussed translation theorems and transforms of periodic functions. Compatible discrete time Laplace transforms with Laplace transforms was introduced in [31]. Savoye [32] highlighted the importance of discrete time problems and relationship of Z transforms to Laplace transforms on time scale. Fractional double Laplace transform was introduced in [33]; during derivation of Corollary 1, authors neglected the violation of semigroup property of Mittage-Leffler functions, and a counter example for semigroup property of Mittage-Leffler functions is given in [34]. The qualitative analysis of delay partial difference equations is considered as discrete analog of delay partial differential equations by Zhang and Zhou [35]. For solving partial difference equations Ozpinar and Belgacem introduced discrete homotopy perturbation Sumudu transform method in [36]. For solving partial differential equations, double Laplace transform was applied in [37, 38].Here, we introduce the delta double Laplace transforms similar to the one presented by Bernstein [20] in such a way that properties and expressions bear a resemblance to that appearing in Debnath [23] for the continuous calculus. The double convolution product, we consider in this article, resemble with the convolution product defined for delta calculus in [2, 10], but it differs from the one defined by Atici in [8]. We consider the problem with constant coefficients in two independent variables and solve by applying the delta double Laplace transforms to partial difference equations with initial data.This paper is divided into five sections. In Section2, we shall present basic definitions and results from discrete calculus. Definition, existence, uniqueness, and series representation of the delta double Laplace transforms are given in Section 3. Some properties of the delta double Laplace transforms are proved in Section 4. In Section 5, we present the delta double Laplace transforms of partial differences.


## 2. Preliminaries

For convenience, this section comprises of some basic definitions and results from discrete calculus for later use in the following sections. The functions we consider usually are defined on the setℕa≔a,a+1,a+2,… and ℕab≔a,a+1,a+2,…,b, for fixed a,b∈ℝ.The following concepts are discussed in [10, 16].Falling function is defined for positive integern by(1)xn¯=xx−1x−2⋯x−n+1.The forward jump operator is defined forx∈ℕa by σx=x+1.The set of regressive functions is defined forx∈ℕa by ℛ=pi:1+pix≠0.The circle plus sum ofp1,p2∈ℛ is given by p1⊕p2=p1+p2+p1p2.The additive inverse ofp1∈ℛ is given by ⊖p1x=−p1x/1+p1x for x∈ℕa.Definition 1.
(see [10]). Assume p1∈ℛ and s∈ℕa. Then, the delta exponential function is given by(2)ep1x,s=∏t=sx−11+p1t,ifx∈ℕs,∏t=xs−11+p1t−1,ifx∈ℕas−1.
By the empty product convention,∏t=ss−1ht≔1 for any function h.Example 1.
Ifp1x=c is a constant such that c∈ℛ (that is c≠−1), then the delta exponential function for a constant is given by(3)ep1x,s=ecx,s=1+cx−s,forx∈ℕa.
For a particular choice ofs=a, that is, the initial point of the domain of definition,(4)ecx,a=1+cx−a,forx∈ℕa.Definition 2.
(see [10]). Assume f:ℕa⟶ℝ and b≤c are in ℕa; then, the delta definite integral is defined by(5)∫bcfxΔx=∑x=bc−1fx.
Note that the value of integral∫bcfxΔx, depending on the set b,b+1,…,c−1. Also, by the empty sum convention,(6)∑x=bb−kfx=0,wheneverk∈ℕ1.
The delta indefinite integral is defined by(7)∫b∞fxΔx=∑x=b∞fx.Definition 3.
(see [10]). Assume f:ℕa⟶ℝ. Then, the delta Laplace transform of f based at a is defined by(8)ℒxfp=∫a∞e⊖pσx,afxΔx,for all complex numbers p≠−1 such that this improper integral converges.
Note that throughout this article, we take the delta Laplace transform at the initial pointa of the set ℕa, unless stated otherwise.
The following concepts are also discussed in [10, 16].Definition 4.
(see [10]). A function f is of exponential order r1>0 if there exist a constant A1>0 and the following inequality:(9)fx≤A1r1x,holds for sufficiently largex∈ℕa.
Iff is of exponential order, then ℒxfp converges absolutely for p+1>r1, which ensures the existence of the Laplace transform. Even though the converse in not true, we restrict ourselves to only exponential order functions. For f:ℕa⟶ℝ, the following are useful expressions for the delta Laplace transform of f based at a:(10)ℒxfp=F˜p=∫0∞fa+jp+1j+1Δj=∑j=0∞fa+jp+1j+1,for all complex numbers p≠−1 such that this infinite series converges.Example 2.
Ifc≠−1, then for p+1>c+1, we have(11)ℒxecx,ap=1p−c.Definition 5.
(see [10]). Assume f,g:ℕa⟶ℝ. The convolution product is defined by(12)f∗gx=∑r=ax−1frgx−σr+a,forx∈ℕa.
Note that by the empty sum conventionf∗ga=0.Lemma 1.
(convolution theorem, see [10]). Assume f,g:ℕa⟶ℝ. If both ℒxfx and ℒxgx exist, then the delta Laplace transform of the convolution product is given by(13)ℒxf∗gx=ℒxfxℒxgx=ℒxg∗fx.Lemma 2.
(see [10]). Assume two functions v,w:ℕa⟶ℝ. Let b1,b2∈ℕa such that b1<b2, and we have the summation by parts formula:(14)∫b1b2vσtΔwtΔt=vtwtb1b2−∫b1b2wtΔvtΔt.Definition 6.
The generalized falling function is defined in term of gamma function by(15)tμ¯=ΓσtΓσt−μ,fort∈ℕa,μ∈ℝ,given that the expression in the above equation is justifiable. It is convenient to take tμ¯=0, whenever t+1 is natural number and t−μ+1 is a zero or negative integer.Definition 7.
The discrete Taylor monomial based ats=a is defined by(16)hnx,a=x−an¯n!,forx∈ℕa,and the μth order Taylor monomial is defined by(17)hμx,a=x−aμ¯Γμ+1,forx∈ℕa.Lemma 3.
(see [10]). The following hold for delta Laplace of Taylor monomial:(i)
ℒxhnx,ap=1/pn+1, for p+1>1, n∈ℕ0(ii)
ℒxx−an¯p=n!/pn+1, for p+1>1, n∈ℕ0(iii)
ℒxx−aμ¯p=Γμ+1p+1μ/pμ+1, for p+1>1, μ≥0.In the next definition, we consider only delta difference with increment 1, and do not bother the different operators that we will not be using here. One can find the details of Definition8 in [1, 39].Definition 8.
Assumeu:ℕa×ℕa⟶ℝ, a function of two independent variables. Then, the partial difference of ux,y with respect to x, regarding y as a constant is given by(18)Δxux,y=ux+1,y−ux,y.
The partial difference ofux,y with respect to y, regarding x as a constant, is given by(19)Δyux,y=ux,y+1−ux,y.
Partial difference equation is an equation containing partial differences.
Note thatΔxy=ΔyΔx=ΔxΔy=Δyx. Followed by the rule for integer order difference operator Δn=ΔΔn−1, we adopt the symbol for partial differences as follows: Δxn=ΔxΔxn−1 and Δym=ΔyΔym−1.


## 3. The Delta Double Laplace Transforms

In this section, we give abstract definition of the delta double Laplace transform. For convenience, we simplify definition to series representation followed by Goodrich and Peterson [10] simplification of the delta Laplace transform. Also, condition for existence, uniqueness, and linearity of the delta double Laplace transform has been revealed.Definition 9.
Assumef:ℕa×ℕa⟶ℝ. Then, the delta double Laplace transform of f based at (a,a) is the successive application of the delta Laplace transform on x and y in any order(20)ℒ2fx,yp,q=ℒxℒyfx,y;y⟶q;x⟶p=ℒyℒxfx,y;x⟶p;y⟶q=ℒyF˜p,y;y⟶q=F˜˜p,q,where ℒx and ℒy are the delta Laplace transforms (single) based at a with respect to x and y, respectively, and ℒ2 is the delta double Laplace transform based at a,a. The delta double Laplace transform of a function fx,y of two variables x and y is defined in p-q plane provided the following double sum converges:(21)ℒ2fp,q=∫a∞∫a∞e⊖pσx,ae⊖qσy,afx,yΔxΔy,for all complex numbers p≠−1 and q≠−1.
One can easily verify by using Lemma4 that ℒxℒy=ℒyℒx. Later, in Theorem 2, we will prove that the double infinite series is absolutely convergent. It is well known that absolutely convergent series behave nicely and change in the order of summation ∑k=0∞∑j=0∞ allowed. Therefore, we can operate in either way ℒxℒy=ℒyℒx.Lemma 4.
Assumef:ℕa×ℕa⟶ℝ. Then,(22)ℒ2fx,y=∑k=0∞∑j=0∞fa+j,a+kp+1j+1q+1k+1,for all complex numbers p≠−1 and q≠−1 such that the infinite series converges.Proof.
By using the definition of the delta double Laplace transform, consider the following:(23)ℒ2fp,q=∫a∞∫a∞e⊖pσx,ae⊖qσy,afx,yΔxΔy.
Now, by the definition of delta integral from discrete calculus, we obtain(24)=∑y=a∞∑x=a∞e⊖pσx,ae⊖qσy,afx,y=∑y=a∞∑x=a∞1⊖pσx−a1⊖qσy−afx,y=∑y=a∞∑x=a∞fx,yp+1x+1−aq+1y+1−a.
In preceding steps, we use the definition of delta exponential function and the fact that1⊖p=1/1+p and 1⊖q=1/1+q, since p and q are regressive functions. In the following step, we use x−a=j and y−a=k to reindex the sums as follows:(25)ℒ2fx,y=∑k=0∞∑j=0∞fa+j,a+kp+1j+1q+1k+1.Theorem 1.
Assume functionsfx,y:ℕa×ℕa⟶ℝ, gx:ℕa⟶ℝ, and hy:ℕa⟶ℝ such that the delta double Laplace transforms exist, then the following holds:(i)
ℒ2gxp,q=1/qℒxgxp(ii)
ℒ2hyp,q=1/pℒyhyq(iii)
Forfx,y=gxhy,ℒ2fx,yp,q=ℒxgxpℒyhyqProof.
Under the assumption stated above and by Lemma4,(i)
Forp≠−1 and q≠0,−1, we have(26)ℒ2gxp,q=∑k=0∞∑j=0∞ga+jp+1j+1q+1k+1=∑k=0∞1q+1k+1∑j=0∞ga+jp+1j+1=1qℒxgxp.(ii)
The proof is similar to part (i).(iii)
Forp≠−1 and q≠−1, we have(27)ℒ2fx,yp,q=∑k=0∞∑j=0∞ga+jha+kp+1j+1q+1k+1=∑j=0∞ga+jp+1j+1∑k=0∞ha+kq+1k+1=ℒxgxpℒyhyq.Example 3.
(i)
Iffx,y=1 for x,y∈ℕa, then ℒ21=1/pq,(ii)
Iffx,y=x−am¯y−an¯ for x,y∈ℕa, then ℒ2x−am¯y−an¯=m!n!/pm+1qn+1.(iii)
By Lemma4(28)ℒ21=∑k=0∞∑j=0∞1p+1j+1q+1k+1=∑k=0∞1q+1k+1∑j=0∞1p+1j+1=1pq,forp,q≠0,−1.(iv)
By using Theorem1 part (iii), we obtain(29)ℒ2x−am¯y−an¯=ℒxx−am¯ℒyy−an¯.
By using Lemma3,(30)=m!pm+1ℒyy−an¯=m!pm+1n!qn+1.
If we choose eitherm=0 or n=0, then as a special case of the above(31)ℒ2y−an¯=n!pqn+1,forp,q≠0,−1,ℒ2x−am¯=m!pm+1q,forp,q≠0,−1.
Coon and Bernstein [20, 21] defined the double Laplace transforms and discussed convergence and existence for the continuous case. We discuss discrete analogue of the double Laplace transforms.Definition 10.
A functionfx,y:ℕa×ℕa⟶ℝ is of exponential order r1,r2>0 with respect to x and y, respectively, if there exist a constant A>0 and m,n∈ℕ0 such that, for each x∈ℕa+m and y∈ℕa+n, the inequality fx,y≤Ar1xr2y holds, where A=maxA1,A2 for fx,a≤A1r1x and fa,y≤A2r2y.Theorem 2.
If a functionfx,y:ℕa×ℕa⟶ℝ is of exponential order r1,r2>0, then the delta double Laplace transform ℒ2fp,q converges absolutely for p and q provided p+1>r1, q+1>r2.Proof.
Assumefx,y:ℕa×ℕa⟶ℝ is of exponential order r1,r2>0. Then, there exists a constant A>0 and m,n∈ℕ0 such that, for each x∈ℕa+m and y∈ℕa+n, fx,y≤Ar1xr2y. Thus, for p+1>r1 and q+1>r2, we consider the following:(32)∑k=n∞∑j=m∞fa+j,a+kp+1j+1q+1k+1≤∑k=n∞∑j=m∞Ar1j+ar2k+ap+1j+1q+1k+1=Ar1ar2ap+1q+1∑k=n∞∑j=m∞r1p+1jr2q+1k=Ar1ar2ap+1q+1r1/p+1m1−r1/p+1r2/q+1n1−r2/q+1=Ar1a+mr2a+np+1mq+1np+1−r1q+1−r2<∞.
Sincep+1>r1 and q+1>r2, therefore p+1−r1>0 and q+1−r2>0. Hence, the delta double Laplace transform of f converges absolutely.
Theorem2 ensures the existence of the delta double Laplace transform. In general, the converse does not hold. We should consider functions f of some exponential order r>0, to ensure the delta double Laplace transform of f which does converge somewhere in the complex plane outside the both closed balls of radius r1,r2, centered at −1, that is, we can choose r=maxr1,r2 for p+1|>r1 and q+1|>r2.Theorem 3.
Supposef,g:ℕa×ℕa⟶ℝ. If the delta double Laplace transform of f,g converges for p+1>r1 and q+1>r2, where r1,r2>0, and let c1, c2∈ℂ, then the delta double Laplace transform of c1f+c2g converges for p+1>r1, q+1>r2, and ℒ2c1f+c2gp,q=c1ℒ2fp,q+c2ℒ2gp,q, converges for p+1>r1 and q+1>r2.Proof.
Sincef,g:ℕa×ℕa⟶ℝ and the delta double Laplace transform of f,g converges for p+1>r1 and q+1>r2, where r1,r2>0. We have that, for p+1>r1 and q+1>r2,(33)c1ℒ2fp,q+c2ℒ2gp,q=c1∑k=0∞∑j=0∞fa+j,a+kp+1j+1q+1k+1+c2∑k=0∞∑j=0∞ga+j,a+kp+1j+1q+1k+1=∑k=0∞∑j=0∞c1f+c2ga+j,a+kp+1j+1q+1k+1=ℒ2c1f+c2gp,q.
Theorem3 exposed the linearity property of the delta double Laplace transform, and Theorem 4 revealed the uniqueness.Theorem 4.
Supposef,g:ℕa×ℕa⟶ℝ and r1>0, r2>0. If ℒ2fp,q=ℒ2gp,q, provided p+1>r1, q+1>r2, and p,q≠0,−1, then fx,y=gx,y for all x,y∈ℕa.Proof.
By hypothesis, we have(34)ℒ2fp,q=ℒ2gp,q,for p+1>r1 and q+1>r2. This implies that(35)∑k=0∞∑j=0∞fa+j,a+kp+1j+1q+1k+1=∑k=0∞∑j=0∞ga+j,a+kp+1j+1q+1k+1,for p+1>r1 and q+1>r2. Since, by Theorem 2, the double infinite series is absolute convergent, therefore comparison of both sides implies that(36)fa+j,a+k=ga+j,a+k,for allj,k∈ℕ0.
For each fixj and for all y∈ℕa, this implies that(37)fa+j,y=ga+j,y.
For each fixk, we obtain(38)fx,y=gx,y,for allx,y∈ℕa.


## 4. Basic Properties of the Delta Double Laplace Transform

In this section, following Bohner et al. [30], we prove some properties of the delta double Laplace transform. We also define double convolution product of discrete functions followed by Goodrich and Peterson [10] convolution product (single) of discrete functions. We present, the delta double Laplace transform of double convolution product for later use to solve difference equations.Theorem 5.
Assumef:ℕa×ℕa⟶ℝ and ℒ2fx,y exists. If ℒ2fx,y=F˜˜p,q, then(39)ℒ2fx−α,y−βHx−α,y−β=e⊖pα,0e⊖qβ,0F˜˜p,q−∑s=0c−a−1∑τ=0c−a−1fa+τ,a+sp+1τ+1q+1s+1,where Hx,y is the Heaviside unit step function defined by(40)Hx−α,y−β=0,ifx−α,y−β∈ℕac−1,1,ifx−α,y−β∈ℕc.Proof.
We have, by Lemma4,(41)ℒ2fx−α,y−βHx−α,y−β=∑k=0∞∑j=0∞fa−α+j,a−β+kHa−α+j,a−β+kp+1j+1q+1k+1=∑k=β+c−a∞∑j=α+c−a∞fa−α+j,a−β+kp+1j+1q+1k+1.
Reindex byj−α=τ and k−β=s,(42)=∑s=c−a∞∑τ=c−a∞fa+τ,a+sp+1α+τ+1q+1β+s+1=1p+1αq+1β∑s=0∞∑τ=0∞fa+τ,a+sp+1τ+1q+1s+1−∑s=0c−a−1∑τ=0c−a−1fa+τ,a+sp+1τ+1q+1s+1,ℒ2fx−α,y−βHx−α,y−β=e⊖pα,0e⊖qβ,0F˜˜p,q−∑s=0c−a−1∑τ=0c−a−1fa+τ,a+sp+1τ+1q+1s+1.
In the last step, we use Lemma4 with the fact e⊖pα,0=1/p+1α and e⊖qβ,0=1/q+1β.
Theorem5 gives different results from its continuous counterpart stated in [23]. We state the useful shifting Theorem 6 for discrete setting.Theorem 6.
Assumef:ℕa×ℕa⟶ℝ and ℒ2fx,y exist. If ℒ2fx,y=F˜˜p,q, then(i)
ℒ2fx−c−a,y−c−aHx,y=1/p+1q+1c−aF˜˜p,q.(ii)
ℒ2fx+c−a,y+c−a=p+1q+1c−aF˜˜p,q−∑s=0c−a−1∑τ=0c−a−1fa+τ,a+s/p+1τ+1q+1s+1,whereHx,y is the Heaviside unit step function defined by(43)Hx,y=0,ifx,y∈ℕac−1,1,ifx,y∈ℕc.Proof.
(i)
We have by Lemma4, ℒ2fx−c−a,y−c−aHx,y(44)=∑k=0∞∑j=0∞fj+2a−c,k+2a−cHa+j,a+kp+1j+1q+1k+1=∑k=c−a∞∑j=c−a∞fj+2a−c,k+2a−cp+1j+1q+1k+1.Reindex byτ=j+a−c and s=k+a−c,(45)=∑s=0∞∑τ=0∞fa+τ,a+sp+1τ+c−a+1q+1s+c−a+1=1p+1q+1c−a∑s=0∞∑τ=0∞fa+τ,a+sp+1τ+1q+1s+1=1p+1q+1c−aF˜˜p,q.(ii)
By use of Lemma4 and reindex by τ=j+c−a and s=k+c−a,(46)ℒ2fx+c−a,y+c−a=∑k=0∞∑j=0∞fj+c,k+cp+1j+1q+1k+1=∑s=c−a∞∑τ=c−a∞fa+τ,a+sp+1τ+a−c+1q+1s+a−c+1=p+1q+1c−a∑s=c−a∞∑τ=c−a∞fa+τ,a+sp+1τ+1q+1s+1=p+1q+1c−aF˜˜p,q−∑s=0c−a−1∑τ=0c−a−1fa+τ,a+sp+1τ+1q+1s+1.Theorem 7.
Assumefx,y is periodic with T1,T2∈ℕ1 and ℒ2fx,y exist; then,(47)ℒ2fx,y=11−e⊖pT1,0e⊖qT2,0∑j=0T1−1∑k=0T2−1fa+j,a+kp+1j+1q+1k+1.Proof.
Under the assumption, we have, by Lemma4,(48)ℒ2fx,y=∑k=0∞∑j=0∞fa+j,a+kp+1j+1q+1k+1=∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1+∑k=T2∞∑j=T1∞fa+j,a+kp+1j+1q+1k+1=∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1+∑v=0∞∑u=0∞fa+u+T1,a+v+T2p+1T1+u+1q+1T2+v+1.
In the last step, we usedj=T1+u and k=T2+v to reindex second double summation. In second double summation, periodicity of f implies that(49)ℒ2fx,y=∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1+∑v=0∞∑u=0∞fa+u,a+vp+1T1+u+1q+1T2+v+1=∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1+1p+1T11q+1T2∑v=0∞∑u=0∞fa+u,a+vp+1u+1q+1v+1=∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1+e⊖pT1,0e⊖qT2,0∑v=0∞∑u=0∞fa+u,a+vp+1u+1q+1v+1=∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1+e⊖pT1,0e⊖qT2,0ℒ2fx,y=11−e⊖pT1,0e⊖qT2,0∑k=0T2−1∑j=0T1−1fa+j,a+kp+1j+1q+1k+1.Definition 11.
Assumef,g:ℕa×ℕa⟶ℝ. The double convolution product is defined by(50)f∗∗gx,y=∑r=ax−1∑s=ay−1fr,sgx−σr+a,y−σs+aforx,y∈ℕa.
Note, by empty sum convention,f∗∗ga,a=0.Lemma 5.
Assumef,g:ℕa×ℕa⟶ℝ. The double convolution product is commutative:(51)f∗∗gx,y=g∗∗fx,yforx,y∈ℕa.Proof.
By Definition11 and the change of variables x−r−1+a=u and y−s−1+a=v, we have(52)g∗∗fx,y=∑r=ax−1∑s=ay−1gr,sfx−σr+a,y−σs+aforx,y∈ℕa,=∑u=ax−1∑v=ay−1gx−σu+a,y−σv+afu,v,=f∗∗gx,y,forx,y∈ℕa.Theorem 8.
(convolution theorem). Assumef,g:ℕa×ℕa⟶ℝ. If both ℒ2fx,y and ℒ2gx,y exist, then the delta double Laplace transform of double convolution product is(53)ℒ2f∗∗gx,y=ℒ2fx,yℒ2gx,y.Proof.
Under given assumption, we have, by Lemma4 and the fact f∗∗ga,a=0,(54)ℒ2f∗∗gx,y=∑k=0∞∑j=0∞f∗∗ga+j,a+kp+1j+1q+1k+1=∑k=1∞∑j=1∞f∗∗ga+j,a+kp+1j+1q+1k+1=∑k=1∞∑j=1∞1p+1j+1q+1k+1∑r=aa+k−1∑s=aa+j−1fr,sga+j−σr+a,a+k−σs+a.
In the last step, we used Definition11; next, making the change of variables r⟶a+r and s⟶a+s gives us that(55)=∑k=1∞∑j=1∞∑r=0k−1∑s=0j−1fa+r,a+sga+j−σr,a+k−σsp+1j+1q+1k+1=∑k=1∞∑r=0k−1∑j=1∞∑s=0j−1fa+r,a+sga+j−σr,a+k−σsp+1j+1q+1k+1=∑r=0∞∑k=1∞∑s=0∞∑j=1∞fa+r,a+sga+j−σr,a+k−σsp+1j+1q+1k+1=∑r=0∞∑τ2=0∞∑s=0∞∑τ1=0∞fa+r,a+sga+τ1,a+τ2p+1τ1+r+2q+1τ2+s+2=∑s=0∞∑r=0∞fa+r,a+sp+1r+1q+1s+1∑τ2=0∞∑τ1=0∞ga+τ1,a+τ2p+1τ1+1q+1τ2+1=ℒ2fx,yℒ2gx,y.
In the previous steps, we interchanged the order of first pairs and second pairs of summation and change variablesj−r−1=τ1 and k−s−1=τ2.Corollary 1.
Assumef,g:ℕa×ℕa⟶ℝ. If fx,y=u1xv1y and gx,y=u2xv2y and the delta Laplace transform exists, then(56)ℒ2f∗∗gx,y=ℒxu1∗u2xℒyv1∗v2y,where the product on right- and left-hand sides is given by Definitions 5 and 11, respectively.Proof.
By double convolution theorem, we have(57)ℒ2f∗∗gx,y=ℒ2fx,yℒ2gx,y.
Sinceℒ2fx,y=ℒ2u1xv1y=ℒxu1xℒyv1yandℒ2gx,y=ℒ2u2xv2y=ℒxu2xℒyv2y,(58)considerℒ2f∗∗gx,y=ℒxu1xℒyv1yℒxu2xℒyv2y=ℒxu1xℒxu2xℒyv1yℒyv2y=ℒxu1∗u2xℒyv1∗v2y.
The last step is followed from single convolution Lemma1.


## 5. The Delta Double Laplace Transforms of Partial Differences

In this section, we examine the action of the delta double Laplace transforms on first order partial differences. The results developed for first order partial differences are further used to establish properties of the delta double Laplace transforms of generalized order partial difference, similar to that appeared in [40] for fractional order partial derivatives. We usually consider functions ux,y:ℕa×ℕa⟶ℝ, of exponential order r1,r2>0 with respect to x and y, respectively, ensuring that delta Laplace and the delta double Laplace transforms of ux,y and its partial differences does exist.Lemma 6.
Assumeux,y:ℕa×ℕa⟶ℝ, such that the delta Laplace transforms exist for constants p≠−1 and q≠−1. Then,(59)ℒxΔxux,y=pℒxux,y−ua,y,(60)ℒyΔyux,y=qℒyux,y−ux,a,(61)ℒxΔyux,y=Δyℒxux,y,(62)ℒyΔxux,y=Δxℒyux,y.Proof.
By definition of the delta Laplace transforms onx,(63)ℒxΔxux,y=∫a∞e⊖pσx,aΔxux,yΔx.
Apply summation by parts (Lemma2) on x, and using the fact Δxe⊖pσx,a=⊖pe⊖px,a, we have that(64)=e⊖px,aux,yx=a∞−∫a∞ux,y⊖pe⊖px,aΔx.
Use the facte⊖px,a=1/p+1x−a and ⊖p=−p/p+1,(65)=1p+1x−aux,yx=a∞−∫a∞ux,y−pp+1e⊖px,aΔx.
Sincep+1e⊖pσx,a=e⊖px,a,(66)=0−ua,y+p∫a∞ux,ye⊖pσx,aΔx=−ua,y+pℒxux,y,ℒxΔxux,y=pℒxux,y−ua,y.
Letℒxux,y=u˜p,y. Consider the left-hand side of equation (61) and use the definition of delta difference:(67)ℒxΔyux,y=ℒxux,y+1−ux,y.
By using linearity property of the delta Laplace transforms, we obtain(68)=ℒxux,y+1−ℒxux,y=u˜p,y+1−u˜p,y.
Now, consider the right-hand side of equation (61) and use ℒxux,y=u˜p,y:(69)Δyℒxux,y=Δyu˜p,y.
By using the definition of delta difference, we obtain(70)=u˜p,y+1−u˜p,y.
Equality holds in equation (61) from equations (68) and (70). Proof of equations (60) and (62) is similar to proof of equations (59) and (61), respectively.Theorem 9.
Assumeux,y:ℕa×ℕa⟶ℝ, such that the delta double Laplace transforms exist for constants p≠−1 and q≠−1. Then,(i)
ℒ2Δxux,y=pℒ2ux,y−ℒyua,y(ii)
ℒ2Δyux,y=qℒ2ux,y−ℒxux,aProof.
Since, by definition, the delta double Laplace transforms is the successive application of the delta Laplace transforms onx and y in any order, therefore ℒ2=ℒxℒy=ℒyℒx.(i)
Consider(71)ℒ2Δxux,y=ℒyℒxΔxux,y.By using equation (59) of Lemma 6, we obtain(72)=ℒypℒxux,y−ua,y.Use linearity property of the delta Laplace transforms forℒy,(73)=pℒ2ux,y−ℒyua,y.(ii)
The proof is similar to part (i).
Note that, for constanta, Δxua,y=ua,y−ua,y=0. We adopt the following symbols in our result which are nonzero, in general, Δxua,y=Δxux,yx=a and Δyux,a=Δyux,yy=a, that is, first we take difference and then evaluate at a.Lemma 7.
Assumeux,y:ℕa×ℕa⟶ℝ, such that the delta Laplace transforms exist for constants p≠−1 and q≠−1. Then,(i)
ℒxΔxnux,y=pnℒxux,y−∑k=0n−1pn−1−kΔxkua,y(ii)
ℒyΔymux,y=qmℒyux,y−∑j=0m−1qm−1−jΔyjux,a(iii)
ℒxΔxynmux,y=pnℒxΔymux,y−∑k=0n−1pn−1−kΔxkΔymua,y(iv)
ℒyΔxynmux,y=qmℒyΔxnux,y−∑j=0m−1qm−1−jΔxnΔyjux,aProof.
(i)
We prove this part by induction onn, and result for n=1 has been proved in Lemma 6. Assume the result is true for n≥1:(74)ℒxΔxnux,y=pnℒxux,y−∑k=0n−1pn−1−kΔxkua,y.We will try to establish result forn+1, beginning with the following:(75)ℒxΔxn+1ux,y=ℒxΔxΔxnux,y.Letwx,y=Δxnux,y, and we have that(76)=ℒxΔxwx,y.Again using equation (59) of Lemma 6,(77)=pℒxwx,y−wa,y=pℒxΔxnux,y−Δxnua,y.By using assumption forn,(78)=ppnℒxux,y−∑k=0n−1pn−1−kΔxkua,y−Δxnua,y=pn+1ℒxux,y−∑k=0n−1pn−kΔxkua,y−pn−nΔxnua,yℒxΔxn+1ux,y=pn+1ℒxux,y−∑k=0npn−kΔxkua,y.The result holds forn+1, whenever it holds for n. Hence, by induction, result in part (i) holds.(ii)(79)ℒxΔxynmux,y=ℒxΔxnΔymux,y.Letvx,y=Δymux,y, and use part (i) of the same Lemma:(80)=ℒxΔxnvx,y=pnℒxvx,y−∑k=0n−1pn−1−kΔxkva,y=pnℒxΔymux,y−∑k=0n−1pn−1−kΔxkΔymua,y.
Proof of (ii) and (iv) is similar as proof of part (i) and (iii), respectively.Theorem 10.
Assumeux,y:ℕa×ℕa⟶ℝ, such that the delta double Laplace transforms exist for constants p≠−1 and q≠−1. Then,(i)
ℒ2Δxnux,y=pnℒ2ux,y−∑k=0n−1pn−1−kℒyΔxkua,y(ii)
ℒ2Δymux,y=qmℒ2ux,y−∑j=0m−1qm−1−jℒxΔyjux,a(iii)
ℒ2Δxynmux,y=pnqmℒ2ux,y−∑k=0n−1p−1−kℒyΔxkua,y−∑j=0m−1q−1−jℒxΔyjux,a+∑j=0m−1∑k=0n−1p−1−kq−1−jΔxykjua,aProof.
Since by definition, the delta double Laplace transforms is the successive application of the delta Laplace transforms onx and y in any order; therefore, ℒ2=ℒxℒy=ℒyℒx.(i)
Using Lemma7 part (i) and linearity of Laplace, we consider the following:(81)ℒ2Δxnux,y=ℒyℒxΔxnux,y=ℒypnℒxux,y−∑k=0n−1pn−1−kΔxkua,y=pnℒyℒxux,y−ℒy∑k=0n−1pn−1−kΔxkua,y=pnℒ2ux,y−∑k=0n−1pn−1−kℒyΔxkua,y.(ii)
Proof is similar as in part (i).(iii)
Using Lemma7 part (iii) and linearity of Laplace, we consider the following:(82)ℒ2Δxynmux,y=ℒyℒxΔxynmux,y=ℒypnℒxΔymux,y−∑k=0n−1pn−1−kΔxkΔymua,y=pnℒyℒxΔymux,y−pn∑k=0n−1p−1−kℒyΔxkΔymua,y=pnℒ2Δymux,y−pn∑k=0n−1p−1−kqmℒyΔxkua,y−∑j=0m−1qm−1−jΔxkΔyjua,a.
In the previous step, we used Lemma7 part (iv). In the following step, using Theorem 10 part (ii),(83)=pnqmℒ2ux,y−∑j=0m−1qm−1−jℒyΔyjux,a−pn∑k=0n−1p−1−kqmℒyΔxkua,y−∑j=0m−1qm−1−jΔxkΔyjua,a=pnqmℒ2ux,y−∑k=0n−1p−1−kℒyΔxkua,y−∑j=0m−1q−1−jℒxΔyjux,a+∑j=0m−1∑k=0n−1p−1−kq−1−jΔxykjua,a.Theorem 11.
Assumefx,y:ℕa×ℕa⟶ℝ. If ℒ2fx,y=F˜˜p,q, then for constants p≠0,−1 and q≠0,−1, and we have(84)ℒ2∫ax∫ayft,τΔτΔt=F˜˜p,qpq.Proof.
Forx,y∈ℕa, let(85)ux,y=∫ax∫ayft,τΔτΔt.
Then, the differenceΔxy is(86)Δxyux,y=ΔyΔxux,y=Δy∫ax+1∫ayft,τΔτΔt−∫ax∫ayft,τΔτΔt=Δy∑t=ax∑τ=ay−1ft,τ−∑t=ax−1∑τ=ay−1ft,τ.
By separating last term fort=x, from the first double sum, we obtain(87)=Δy∑τ=ay−1fx,τ+∑t=ax−1∑τ=ay−1ft,τ−∑t=ax−1∑τ=ay−1ft,τ=Δy∑τ=ay−1fx,τ=∑τ=ayfx,τ−∑τ=ay−1fx,τ.
By separating last term forτ=y, from the first sum, we obtain(88)=fx,y+∑τ=ay−1fx,τ−∑τ=ay−1fx,τfx,y=Δxyux,y.
Now, for constantsp≠0,−1 and q≠0,−1, taking the delta double Laplace transforms on both sides,(89)ℒ2fx,y=ℒ2Δxyux,y.
By application of Theorem10 (iii) for m=1=n, ℒ2Δxyux,y=pqℒ2ux,y−qℒyua,y−pℒxux,a+ua,a. On right-hand side, we obtain(90)=pqℒ2ux,y−qℒyua,y−pℒxux,a+ua,aℒ2fx,y=pqℒ2ux,y.
In the last step,ux,a,ua,y,ua,a are zero by empty sum convention, and on further simplification, we obtain(91)ℒ2∫ax∫ayft,τΔτΔt=F˜˜p,qpq.Example 4.
(a)
Solve the partial difference equation:(92)Δxux,y−Δyux,y=0withux,a=x−a1¯,ua,y=y−a1¯.Application of the delta Laplace transforms to initial conditions by Lemma3,(93)ℒxux,a=ℒxx−a1¯=1p2,ℒyua,y=ℒyy−a1¯=1q2.Apply the delta double Laplace transforms to difference equation and then use linearity property:(94)ℒ2Δxux,y−Δyux,y=0,ℒ2Δxux,y−ℒ2Δyux,y=0.Using Theorem9,(95)pℒ2ux,y−ℒyua,y−qℒ2ux,y−ℒxux,a=0,p−qℒ2ux,y=1q2−1p2,ℒ2ux,y=1pq2+1p2q.Inverting the delta Laplace transforms pairs(96)ux,y=x−a1¯+y−a1¯.(b)
Solve the same partial difference equation with slightly different initial conditions:(97)ux,a=x−a2¯,ua,y=y−a2¯.Application of the delta Laplace transforms to initial conditions by Lemma3:(98)ℒxux,a=ℒxx−a2¯=2p3,ℒyua,y=ℒyy−a2¯=2q3.From equation7p−qℒ2ux,y=2q3−2p3,ℒ2ux,y=2pq3+2p2q2+2p3q.Inverting delta Laplace transforms pairs,(99)ux,y=x−a2¯+2x−a1¯y−a1¯+y−a2¯.
Assumef:ℕa⟶ℝ, then the Riemann–Liouville fractional difference of order N−1<α≤N, for N∈ℕ1 is given by Δaαfx=ΔNΔa−N−αfx, for x∈ℕa+N−α. By using the discussion and results, from Theorem 2.65 to Theorem 2.70, in [10], we take the starting point of the double Laplace a+α−N and a+N−α, respectively, for sum and difference operator.Corollary 2.
Assumeux,y:ℕa×ℕa⟶ℝ, such that the delta double Laplace transforms exists for constants p≠−1 and q≠−1 and denote ℒ2ux,y=u˜˜p,q. Then, for ⌈N−α⌉=M and ⌈k+α−N⌉=L, the delta double Laplace transforms of fractional order operators is given by(100)iℒ2Δa−αux,yp,q=p+1α−Nq+1α−Npαqαu˜˜p,q,whereN−1<α<N,(101)iiℒ2Δxαux,yp,q=pαqα−Np+1N−α−Mq+1N−α−Mu˜˜p,q−∑k=0N−1pN−1−kqk+α−Nq+1L−k+α−Nu˜a,q−∑j=0L−1qjΔxk+α−N−1ua,a+L−k+α−N,(102)iiiℒ2Δyαux,yp,q=pα−Nqαp+1N−α−Mq+1N−α−Mu˜˜p,q−∑k=0N−1qN−1−kpk+α−Np+1L−k+α−Nu˜p,a−∑j=0L−1pjΔyk+α−N−1ua+L−k+α−N,a.Proof.
(i)
Proof is an implication of Definition 3.1 and Theorem 2.67 of [10](ii)
Result is obtained by application of Theorem 5.4 part (i) and Theorem 2.70 of [10](iii)
Result is obtained by application of Theorem 5.4 part (ii) and Theorem 2.70 of [10]Example 5.
Solve the fractional difference equation for0<α<1:(103)Δxαux,y=y−a1¯,withua,y=0.
Apply the delta Laplace transforms to initial conditionℒyua,y=ℒy0=0. For 0<α<1, we have N=1 which implies k=0 and therefore ⌈k+α−1⌉=L=0, also ⌈1−α⌉=M=1. Application of the delta double Laplace transforms on both sides of fractional difference equation (103) and making use of equation (101) on left-hand side, and on the right-hand side we used Example 3 to obtain(104)pαqα−1p+1αq+1αu˜˜p,q−p0qα−1q+11−αu˜a,q=1pq2.
Usingu˜a,q=0 and simplifying the above,(105)u˜˜p,q=p+1αq+1αpα+1qα+1.
Inverting the delta Laplace transforms pairs by making use of Theorem1 (iii), together with Lemma 3 (iii),(106)ux,y=x−aα¯Γα+1y−aα¯Γα+1.



---

*Source: 3902931-2020-07-10.xml*