# n-Dimensional Fractional Frequency Laplace Transform by the Inverse Difference Operator

Authors: M. Meganathan; Thabet Abdeljawad; G. Britto Antony Xavier; Fahd Jarad
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

With the study of extensive literature on the Laplace transform with one and two variables and its properties, applications are available, but there is no work onn-dimensional Laplace transform. In this research article, we define n-dimensional fractional frequency Laplace transform with shift values. Several theorems are derived with properties of the Laplace transform. The results are numerically analyzed and discussed through MATLAB.


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## Body

## 1. Introduction

The fractional calculus is a branch of mathematics that focuses on arbitrary order integrals and derivatives. In spite of that, this type of calculus is as older as the conventional calculus, and it has attracted the interest of researchers for the last few decades. This is because of the results reported by these researchers as consequences of their attempts to model real-world phenomena using the fractional operators [1–4]. The discrete version of these operators fetched the attention of research studies as well. Many good results were reported when fractional sums and differences were used in studying related problems (see [5–17] and the references therein).The integral transforms such as Mellin, Laplace, and Fourier were applied to obtain the solution of differential equations. These transforms made effectively possible changes of a signal in the time domain into a frequencys-domain in the field of digital signal processing (DSP) [18]. The delta Laplace transform was first defined in a very general way by Bohner and Peterson [19]. In 2015, Ivic discussed the discrete Laplace transforms in the view of fast decay factor e−sx and obtained the Laplace transform of Px as ∫0∞Pxe−sxdx=πs−2∑n=1∞rne−π2/n. In practice, many applications of Laplace transform (LT), Lfx=∫0∞fxe−sxdx, and the forward discrete Laplace transform (DLT), Lfn=∑n=0∞fne−sn, are discussed and mentioned by several authors in [20–23]. For physical applications of Laplace transform, refer [24–27].In the existing Laplace transform, the shifting value of time domains is one. In 2016, Britto Antony Xavier et al. [28] defined the Laplace transform with shift value ℓ using the generalized difference operator and obtained the outcomes of polynomial and trigonometric functions. In this fractional frequency Laplace transform, the shift values νj′s,j=1,2,…,n lie in the interval [0, 1]. In [29], the author introduced the double Laplace transform and applied to solve initial and boundary value problems.In this research work, we extend the work of Laplace transform into ann-dimensional space in discrete case. We present several properties of the fractional transforms for functions such as polynomial factorial, exponential, and trigonometric functions. Also, we derive the relation between Laplace transform and Riemann zeta functions. Furthermore, we present the inverse Laplace transform to compare the results with the existing classical Laplace transform for the particular value of n.


## 2. Preliminaries

Here, we present some basic definitions and results which will be used further.Definition 1.
Letut¯ be the function with n-variables and h¯∈Rn be the shift values. Then, the n-dimensional partial difference operator is defined as(1)Δhiut¯=ut1,t2,…,ti+hi,…,tn−ut1,t2,…,tnhi,where t¯=t1,t2,…,tn and h¯=h1,h2,…,hn.Definition 2.
(see [30]). For h>0 and μ∈R, the rising h-polynomial factorial function is defined as(2)thμ=hμΓt/h+μΓt/h,where th0=1, Γ is the Euler gamma function, and t/h+μ,t/h∉0,−1,−2,−3,… as the division at a pole yields zero.Lemma 1.
(see [31]). Let h>0 and ut and wt be real-valued bounded functions. Then,(3)Δh−1utwt=utΔh−1wt−Δh−1Δh−1wt+hΔhut.Theorem 1.
(see [31]). Let t∈0,∞, h>0, and s>0; then,(4)Lh,νthμ=hμ+1μ!es1/vhes1/vh−1μ+1.


### 2.1. Notations

(i)
Δnh¯ut¯=ΔhnΔhn−1⋯Δh2Δh1ut1,t2,…,tn(ii)
Δnh¯−1ut¯=Δhn−1Δhn−1−1⋯Δh2−1Δh1−1ut1,t2,…,tn(iii)
JDn denotes the set of all subsets of Dn=1,2,…,n(iv)
nDn−r¯ denotes the number of digits in Dn−r¯Definition 3.
(infinite inverse principle law). For the functionut¯, we define the infinite inverse principle law as follows:(5)Δnh¯−1a¯∞=∏j=1nhj∑r1=0∞∑r2=0∞⋯∑rn=0∞ua1+r1h1,a2+r2h2,…,an+r2h2,where a¯=a1,a2,…,an. In particular, if a¯=0,0,…,0, we obtain(6)Δnh¯−10∞=∏j=1nhj∑r1=0∞∑r2=0∞⋯∑rn=0∞ur1h1,r2h2,…,r2h2.Theorem 2.
Lett¯∈Rn,hj>0,j=1,2,…,n; then,(7)Δnh¯−1a−∑j=1nsj1/νjtj=∏j=1nhja−∑j=1nsj1/νjtj∏j=1na−sj1/νjhj−1.
Proof. Takingut¯=a−∑j=1nsj1/vjtj in (1) for the shift value h1, we obtain(8)Δh1a−∑j=1nsj1/vjtj=a−s11/ν1t1+h1−∑j=1nsj1/vjtj−a−∑j=1nsj1/vjtjh1=h1a−∑j=1nsj1/vjtja−s11/ν1h1−1.
In (8), applying Δh2−1 on both sides, we get(9)Δh2−1Δh1−1a−∑j=1nsj1/vjtj=h1a−s11/ν1h1−1Δh2−1a−∑j=1nsj1/vjtj=h1a−s11/ν1h1−1h2a−s21/ν2h2−1a−∑j=1nsj1/vjtj,Δh2−1Δh1−1a−∑j=1nsj1/vjtj=∏j=12hja−∑j=1nsj1/vjtj∏j=12a−sj1/νjhj−1.Proceeding like this for the induction onn, we get the proof of (7).Corollary 1.
Lett¯∈Rn,hj>0andj=1,2,…,n. Then, we have(10)Δnh¯−1e−∑j=1nsj1/vjtj=∏j=1nhje−∑j=1nsj1/vjtj∏j=1ne−sj1/νjhj−1.Proof. In the proof of Theorem2, replacing a by e, we get (10).Corollary 2.
In Theorem2, applying a=3, we get(11)Δnh¯−13−∑j=1nsj1/vjtj=∏j=1nhj3−∑j=1nsj1/vjtj∏j=1n3−sj1/νjhj−1.Example 1.
Letn=2 in (11); we get the result for the shift values h1 and h2 as(12)Δh2−1Δh1−13−s11/ν1t1+s21/ν2t2=h1h23−s11/ν1t1+s21/ν2t23−s11/ν1h1−13−s21/ν2h2−1.
Summing from 0 to∞ for t1 and t2 on both sides yields(13)Δh2−1Δh1−13−s11/v1t1+s21/v2t20∞0∞=h1h23−s11/v1h1−13−s21/v2h2−1.
Forn=2, the infinite principle law reads(14)Δh2−1Δh1−1ut1,t20∞0∞=h1h2∑r1=0∞∑r2=0∞ur1h1,r2h2.
Equating (13) and (14) for the function ut1,t2=3−s11/ν1t1+s21/ν2t2, we obtain(15)h1h2∑r1=0∞∑r2=0∞3−s11/ν1r1h1+s21/ν2r2h2=h1h23−s11/ν1h1−13−s21/ν2h2−1,which is verified for the particular values s1=2,s2=3,ν1=0.3,ν2=0.5,h1=0.4, and h2=0.7 by MATLAB coding as follows: 0.4.∗0.7.∗symsumsymsum3.∧−2.∧1./0.3.∗0.4.∗r1+3.∧1./0.5.∗0.7.∗r2,r1,0,∞,r2,0,∞=0.4.∗0.7/3.∧−2.∧1./0.3.∗0.4−1.∗3.∧−3.∧1./0.5.∗0.7−1=0.2837.


## 2.1. Notations

(i)
Δnh¯ut¯=ΔhnΔhn−1⋯Δh2Δh1ut1,t2,…,tn(ii)
Δnh¯−1ut¯=Δhn−1Δhn−1−1⋯Δh2−1Δh1−1ut1,t2,…,tn(iii)
JDn denotes the set of all subsets of Dn=1,2,…,n(iv)
nDn−r¯ denotes the number of digits in Dn−r¯Definition 3.
(infinite inverse principle law). For the functionut¯, we define the infinite inverse principle law as follows:(5)Δnh¯−1a¯∞=∏j=1nhj∑r1=0∞∑r2=0∞⋯∑rn=0∞ua1+r1h1,a2+r2h2,…,an+r2h2,where a¯=a1,a2,…,an. In particular, if a¯=0,0,…,0, we obtain(6)Δnh¯−10∞=∏j=1nhj∑r1=0∞∑r2=0∞⋯∑rn=0∞ur1h1,r2h2,…,r2h2.Theorem 2.
Lett¯∈Rn,hj>0,j=1,2,…,n; then,(7)Δnh¯−1a−∑j=1nsj1/νjtj=∏j=1nhja−∑j=1nsj1/νjtj∏j=1na−sj1/νjhj−1.
Proof. Takingut¯=a−∑j=1nsj1/vjtj in (1) for the shift value h1, we obtain(8)Δh1a−∑j=1nsj1/vjtj=a−s11/ν1t1+h1−∑j=1nsj1/vjtj−a−∑j=1nsj1/vjtjh1=h1a−∑j=1nsj1/vjtja−s11/ν1h1−1.
In (8), applying Δh2−1 on both sides, we get(9)Δh2−1Δh1−1a−∑j=1nsj1/vjtj=h1a−s11/ν1h1−1Δh2−1a−∑j=1nsj1/vjtj=h1a−s11/ν1h1−1h2a−s21/ν2h2−1a−∑j=1nsj1/vjtj,Δh2−1Δh1−1a−∑j=1nsj1/vjtj=∏j=12hja−∑j=1nsj1/vjtj∏j=12a−sj1/νjhj−1.Proceeding like this for the induction onn, we get the proof of (7).Corollary 1.
Lett¯∈Rn,hj>0andj=1,2,…,n. Then, we have(10)Δnh¯−1e−∑j=1nsj1/vjtj=∏j=1nhje−∑j=1nsj1/vjtj∏j=1ne−sj1/νjhj−1.Proof. In the proof of Theorem2, replacing a by e, we get (10).Corollary 2.
In Theorem2, applying a=3, we get(11)Δnh¯−13−∑j=1nsj1/vjtj=∏j=1nhj3−∑j=1nsj1/vjtj∏j=1n3−sj1/νjhj−1.Example 1.
Letn=2 in (11); we get the result for the shift values h1 and h2 as(12)Δh2−1Δh1−13−s11/ν1t1+s21/ν2t2=h1h23−s11/ν1t1+s21/ν2t23−s11/ν1h1−13−s21/ν2h2−1.
Summing from 0 to∞ for t1 and t2 on both sides yields(13)Δh2−1Δh1−13−s11/v1t1+s21/v2t20∞0∞=h1h23−s11/v1h1−13−s21/v2h2−1.
Forn=2, the infinite principle law reads(14)Δh2−1Δh1−1ut1,t20∞0∞=h1h2∑r1=0∞∑r2=0∞ur1h1,r2h2.
Equating (13) and (14) for the function ut1,t2=3−s11/ν1t1+s21/ν2t2, we obtain(15)h1h2∑r1=0∞∑r2=0∞3−s11/ν1r1h1+s21/ν2r2h2=h1h23−s11/ν1h1−13−s21/ν2h2−1,which is verified for the particular values s1=2,s2=3,ν1=0.3,ν2=0.5,h1=0.4, and h2=0.7 by MATLAB coding as follows: 0.4.∗0.7.∗symsumsymsum3.∧−2.∧1./0.3.∗0.4.∗r1+3.∧1./0.5.∗0.7.∗r2,r1,0,∞,r2,0,∞=0.4.∗0.7/3.∧−2.∧1./0.3.∗0.4−1.∗3.∧−3.∧1./0.5.∗0.7−1=0.2837.


## 3.n-Dimensional Fractional Frequency Laplace Transform

Definition 4.
For the functionut¯ with n-variables t1,t2,…,tn, the n-dimensional fractional frequency Laplace transform is defined as(16)ℒnhut¯=Uns¯=Δnh¯−1ut¯e−∑j=1nsj1/νjtjtj=0,j=1,2,…,n∞=∏j=1nhj∑r1=0∞∑r2=0∞⋯∑rn=0∞ur1h1,r2h2,…,rnhne−∑j=1nsj1/νjrjhj.Remark 1.
(i)
Then-dimensional fractional frequency Laplace transform satisfies the linear property.(ii)
In the aforesaid equation (16), we represent the Laplace transform of the functions in two ways: one in the closed-form solution and another one in the summation form solution. In this paper, we numerically verified and analyzed with MATLAB that both solutions are equal.Theorem 3.
Lett¯∈Rn,h¯>0, νj be a fraction, and sj>0,j=1,2,…,n. Then, we have(17)ℒnh1=∏j=1nhj∏j=1ne−sj1/νjhj−1.
Proof. Takingut¯=1 in (16) yields(18)ℒnh1=Δnh¯−1e−∑∑j=1nsj1/vjtjtj=0,j=1,2,…,n∞=Δhn−1e−sn1/vntn0∞⋯Δh2−1e−s21/ν2t20∞Δh1−1e−s11/v1t10∞=hne−sn1/vnhn−1⋯h2e−s21/v2h2−1h1e−s11/v1h1−1,(using Corollary 1), which completes the proof.Theorem 4.
Lett¯∈Rn,h¯>0, νj be a fraction, and sj>0,j=1,2,…,n; then,(19)ℒnhe∑j=1najtj=∏j=1nhj∏j=1ne−sj1/νj−ajhj−1.
Proof. Takingut¯=e∑j=1najtj in (16), we have(20)ℒnhe∑j=1najtj=Δnh¯−1e∑j=1najtje−∑j=1nsj1/vjtjtj=0,j=1,2,…,n∞=Δhn−1e−sn1/vn−antn0∞⋯Δh2−1e−s21/v2−a2t20∞Δh1−1e−s11/v1−a1t10∞=hne−sn1/vn−anhn−1⋯h2e−s21/v2−a2h2−1h1e−s11/v1−a1h1−1,which gives (19).Example 2.
Forn=2, the summation solution of the exponential function given by the infinite inverse principle law and the closed form of the solution given as(21)ℒnhea1t1+a2t2=h1h2∑r2=0∞∑r1=0∞e−s11/ν1−a1r1h1+s21/ν2−a2r2h2=h2e−s21/ν2−a2h2−1h1e−s11/ν1−a1h1−1,is numerically verified for the particular values h1=7,h2=3,a1=5,a2=9,ν1=0.1,ν2=0.3,s1=11, and s2=13 by MATLAB coding as follows: 21.∗symsumsymsumexp−11.∧1./1.0−5.∗7.∗r1+13.∧1./0.3−9.∗3.∗r2,r1,0,10,r2,0,10=21./exp−11.∧1./0.1−5.∗7−1.∗exp−13.∧1./0.3−9.∗3−1.
The following are the graphical representations of the exponential function in time and frequency domains. Figure1 is the graphical representation of the input function e5t1+9t2 in the time domain, and Figure 2 is the graphical representation of the output in the frequency domain for the particular values of ν1=0.0001,ν2=0.0003. One can easily choose the values of fraction νi′s to get the output in the frequency domain.Figure 1
Time signal (function)e5t1+9t2.Figure 2
Frequency signal forν1=0.0001andν2=0.0003.Theorem 5.
Lett¯∈Rn,h¯>0, νj be a fraction, and sj>0,j=1,2,…,n; then,(22)ℒnhei∑j=1najtj=∏j=1nhj∑r¯∈JDn−1nDn−r¯e−iar¯hr¯esDn−r¯hDn−r¯2n∏j=1ncoshsj1/νjhj−cosajhj,where e−iar¯hr¯=e−ia1h1+a2h2+⋯+anhn for r¯=1,2,…,n, esDn−r¯hDn−r¯=es11/ν1h1+s21/ν2h2+⋯+sn1/νnhn for Dn−r¯=1,2,…,n, and e−ia0h0=es0h0=1.
Proof. From the previous theorem, we obtain the Laplace transform for the trigonometric functionei∑j=1najtj as(23)ℒnhei∑j=1najtj=∏j=1nhj∏j=1ne−sj1/vj−iajhj−1.The proof then can be continued by making use of the conjugate and the product of each term inn-variables.Theorem 6.
Lett¯∈Rn,h¯>0, νj be a fraction, and sj>0,j=1,2,…,n; then,(24)ℒ2hsin∑j=12ajtj=h1h2es21/ν2h2sina1h1+es11/ν1h1sina2h2−sina1h1+a2h24∏j=12coshsj1/νjhj−cosajhj,(25)ℒ2hcos∑j=12ajtj=h1h2cos∑j=12ajtj−es21/v2h2cosa1h1−es11/v1h1cosa2h2+e∑j=12sj1/vjhj4∏j=12coshsj1/vjhj−cosajhj.Proof. The proof follows by takingn=2 in Theorem 5, making the product by its conjugate terms and separating the real and imaginary parts to get the double Laplace transform for the sine and cosine terms.Example 3.
Equation (24) is the closed-form solution of the sine function. Now, for n=2, the summation solution of the sine function given by the infinite inverse principle law is(26)ℒ2hsin∑j=12ajtj=h1h2∑r2=0∞∑r1=0∞sina1r1h1+a2r2h2e−s11/ν1r1h1+s21/ν2r2h2.
Now, (24) and (26) are numerically verified for the particular values h1=2,h2=3,a1=1,a2=4,ν1=0.4,ν2=0.6,s1=5, and s2=6 by MATLAB coding as follows: 6.∗symsumsymsumsin2.∗r1+12.∗r2.∗exp−5.∧1./0.4.∗2.∗r1+6.∧1./0.6.∗3.∗r2,r1,0,10,r2,0,∞=6.∗exp6.∧1./0.6.∗3.∗sin2−exp5.∧1./0.4.∗2.∗sin12−sin14/4.∗cosh6.∧1./0.6.∗3−cos12.∗cosh5.∧1./0.4.∗2−cos2.
The following are the graphical representations of the sine function in time and frequency domains. Figure3 represents the input time-domain signal (function) for the sine function. Figure 4 represents the output in the frequency domain for the particular values of ν1=0.4andν2=0.6. Figure 5 represents the output in the frequency domain for the particular values of ν1=0.3andν2=0.5. Figure 6 represents the output in the frequency domain for the particular values of ν1=0.1andν2=0.7. Similarly, one can analyze the solution in the frequency domain by choosing diverse values of fraction νi′s.Figure 3
Time signal forsint1+4t2.Figure 4
Frequency signal forν1=0.4andν2=0.6.Figure 5
Frequency signal forν1=0.3andν2=0.5.Figure 6
Frequency signal forν1=0.1andν2=0.7.Theorem 7.
Lett¯∈Rn,h¯>0, νj be a fraction, and sj,μj>0,j=1,2,…,n; then,(27)ℒnh∏j=1ntjhjμj=∏j=1nhjμj+1μj!esj1/νjhje−sj1/νj−1.
Proof. Takingut¯=∏j=1ntjhjμj in (16), we have(28)ℒnh∏j=1ntjhjμj=Δnh¯−1∏j=1ntjhjμje−∑j=1nsj1/νjtjtj=0,j=1,2,…,n∞=Δhn−1tnhnμne−sn1/νntn0∞⋯Δh2−1t2h2μ2e−s21/ν2t20∞Δh1−1t1h1μ1e−s11/ν1t10∞=hnμn+1μn!esn1/νnhne−sn1/νn−1⋯h2μ2+1μ2!es21/ν2h2e−s21/ν2−1h1μ1+1μ1!es11/ν1h1e−s11/ν1−1,by Lemma 1, which gives (4).


### 3.1.n-Kind Riemann Zeta Function in the Discrete Case

In Theorem7, when νj=1 and hj⟶0 for j=1,2,…,n., we get(29)ℒnht1μ1−1t2μ2−1…tnμn−1=Γμn…Γμ2Γμ1snμn…s2μ2s1μ1.We know that the Riemann zeta function is defined as(30)ζμ=∑s=1∞1sμ.Equation (29) can be written as(31)Γμn…Γμ2Γμ1snμn…s2μ2s1μ1=Δhn−1tnμn−1e−sntn0∞⋯Δh2−1t2μ2−1e−s2t20∞Δh1−1t1μ1−1e−s1t10∞.Taking summation onsj,j=1,2,…,∞, on both sides, we get(32)∏j=1nΓμj∑sn=1∞1snμn⋯∑s2=1∞1s2μ2∑s1=1∞1s1μ1=Δhn−1tnμn−1∑sn=1∞e−sntn|0∞⋯Δh2−1t2μ2−1∑s2=1∞e−s2t2|0∞Δh1−1t1μ1−1∑s1=1∞e−s1t1|0∞,∏j=1nΓμjζμn⋯ζμ2ζμ1=Δhn−1tnμn−11etn−10∞⋯Δh2−1t2μ2−11et2−10∞Δh1−1t1μ1−11et1−10∞,ζμn⋯ζμ2ζμ1=1ΓμnΔhn−1tnμn−1etn−10∞⋯1Γμ2Δh2−1t2μ2−1et2−10∞1Γμ1Δh1−1t1μ1−1et1−10∞,which is the product of nth-kind Riemann zeta function in the discrete case.


### 3.2. One-Dimensional Laplace Transform on the Fractional Difference Equation

Letut and vt be the two functions. The Leibniz rule of noninteger order is Δνutvt=∑r=0∞νrΔν−rutΔrvt+ν−r. Here, we present the product formula on the fractional difference operator as Δhνutvt=∑r=0∞νrΔhν−rutΔhrvt+ν−rh.The following theorem plays an important role in solving the fractional difference equation by one-dimensional Laplace transform.Theorem 8.
Letut1 be a real-valued function and s1,h1,ν1>0. Then, we have(33)Lh1Δh1νut=1−e−s11/ν1h1ν1h1ν1Lh1ut1+ν1h1−∑r1=1∞1−e−s11/ν1h1ν1−r1h1ν1−r1Δh1r1−1uν1−r1h1.Proof. Takingut1=Δh1ut1 in (16), we get Lh1Δh1ut1=Δh1−1Δh1ut1e−s11/ν1t10∞. Now, applying (3) and solving, we get(34)Lh1Δh1ut1=1−e−s11/ν1h1h1Lh1ut1+h1−u0.Again takingut1=Δh12ut1, using (3) and (16), and applying (34) give(35)Lh1Δh12ut1=1−e−s11/ν1h12h12Lh1ut1+2h1−1−e−s11/ν1h1h1uh1−Δh1u0.Continuing this process for integern, we arrive at(36)Lh1Δh1nut1=1−e−s11/ν1h1nh1nLh1ut1+nh1−∑r1=1n1−e−s11/ν1h1n−r1h1n−r1Δh1r1−1un−r1h1.Since the order is a fraction, we consider (36) for fraction ν as mentioned in (33).


## 3.1.n-Kind Riemann Zeta Function in the Discrete Case

In Theorem7, when νj=1 and hj⟶0 for j=1,2,…,n., we get(29)ℒnht1μ1−1t2μ2−1…tnμn−1=Γμn…Γμ2Γμ1snμn…s2μ2s1μ1.We know that the Riemann zeta function is defined as(30)ζμ=∑s=1∞1sμ.Equation (29) can be written as(31)Γμn…Γμ2Γμ1snμn…s2μ2s1μ1=Δhn−1tnμn−1e−sntn0∞⋯Δh2−1t2μ2−1e−s2t20∞Δh1−1t1μ1−1e−s1t10∞.Taking summation onsj,j=1,2,…,∞, on both sides, we get(32)∏j=1nΓμj∑sn=1∞1snμn⋯∑s2=1∞1s2μ2∑s1=1∞1s1μ1=Δhn−1tnμn−1∑sn=1∞e−sntn|0∞⋯Δh2−1t2μ2−1∑s2=1∞e−s2t2|0∞Δh1−1t1μ1−1∑s1=1∞e−s1t1|0∞,∏j=1nΓμjζμn⋯ζμ2ζμ1=Δhn−1tnμn−11etn−10∞⋯Δh2−1t2μ2−11et2−10∞Δh1−1t1μ1−11et1−10∞,ζμn⋯ζμ2ζμ1=1ΓμnΔhn−1tnμn−1etn−10∞⋯1Γμ2Δh2−1t2μ2−1et2−10∞1Γμ1Δh1−1t1μ1−1et1−10∞,which is the product of nth-kind Riemann zeta function in the discrete case.


## 3.2. One-Dimensional Laplace Transform on the Fractional Difference Equation

Letut and vt be the two functions. The Leibniz rule of noninteger order is Δνutvt=∑r=0∞νrΔν−rutΔrvt+ν−r. Here, we present the product formula on the fractional difference operator as Δhνutvt=∑r=0∞νrΔhν−rutΔhrvt+ν−rh.The following theorem plays an important role in solving the fractional difference equation by one-dimensional Laplace transform.Theorem 8.
Letut1 be a real-valued function and s1,h1,ν1>0. Then, we have(33)Lh1Δh1νut=1−e−s11/ν1h1ν1h1ν1Lh1ut1+ν1h1−∑r1=1∞1−e−s11/ν1h1ν1−r1h1ν1−r1Δh1r1−1uν1−r1h1.Proof. Takingut1=Δh1ut1 in (16), we get Lh1Δh1ut1=Δh1−1Δh1ut1e−s11/ν1t10∞. Now, applying (3) and solving, we get(34)Lh1Δh1ut1=1−e−s11/ν1h1h1Lh1ut1+h1−u0.Again takingut1=Δh12ut1, using (3) and (16), and applying (34) give(35)Lh1Δh12ut1=1−e−s11/ν1h12h12Lh1ut1+2h1−1−e−s11/ν1h1h1uh1−Δh1u0.Continuing this process for integern, we arrive at(36)Lh1Δh1nut1=1−e−s11/ν1h1nh1nLh1ut1+nh1−∑r1=1n1−e−s11/ν1h1n−r1h1n−r1Δh1r1−1un−r1h1.Since the order is a fraction, we consider (36) for fraction ν as mentioned in (33).


## 4.n-Dimensional Inverse Laplace Transform

Then-dimensional inverse Laplace transform is defined by(37)ℒnh−1Uns¯=ut¯=Δnh¯−1Uns¯e∑j=1nsj1/vjtjcj−i∞cj+i∞,j=1,2,…,n.Since we can easily represent then-dimensional Laplace transform of the functions mentioned, we can present some results listed as follows:(38)ℒnh−1∏j=1nhj∏j=1ne−sj1/vjhj−1=1,ℒnh−1∏j=1nhj∏j=1ne−sj1/vj−ajhj−1=e∑j=1najtj,ℒnh−1∏j=1nhj∑r¯∈JDn−1nDn−r¯e−iar¯hr¯esDn−r¯hDn−r¯2n∏j=1ncoshsj1/vjhj−cosajhj=ei∑j=1najtj,ℒnh−1∏j=1nhjμj+1μj!esj1/vjhje−sj1/vj−1=∏j=1ntjhjμj,ℒnh−1Γμn…Γμ2Γμ1snμn…s2μ2s1μ1=t1μ1−1t2μ2−1…tnμn−1.


## 5. Results and Discussion

Whenn=2, ν1=ν2=1, and h1,h2⟶0, in all the above results, we have(i)
ℒ21=1/s1s2.(ii)
ℒ2ea1t1+a2t2=1/s1−a1s2−a2.(iii)
ℒ2sina1t1+a2t2=a1a2/s12+a12s22+a22.(iv)
ℒ2cosa1t1+a2t2=s1s2/s12+a12s22+a22.Similarly, the following result for the hyperbolic functions can be obtained:(i)
ℒ2t1μ1t2μ2=μ1!μ2!/s1μ1+1s2μ2+1.These results match the formulas of the double Laplace transform of functions available in the literature.


## 6. Conclusion

The fractional frequency is used to derive then-dimensional Laplace transform with shift values hj,j=1,2,…,n, that presents more accuracy outputs of the input functions such as exponential, polynomial factorial, polynomial, and trigonometric functions. Also, the numerical results and the solutions are analyzed graphically by MATLAB. The major application of this research work is also provided by considering the classical Laplace transform according to particular values of n which are νj=1 and hj⟶0,j=1,2,…,n.



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