# On Multiple Convolutions and Time Scales

Authors: Hassan Eltayeb; Adem Kılıçman; Brian Fisher
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

The properties of the multiple Laplace transform and convolutions on a time scale are studied. Further, some related results are also obtained by utilizing the double Laplace transform. We also provide an example in order to illustrate the main result.


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## Body

## 1. Introduction

A time scale⊤ is a nonempty closed subset of the real numbers ℝ and it has the topology that it inherits from the real numbers with the standard topology. It is also well known that if the time scale is the set of real numbers, then the dynamic equation is a differential equation, whilst if the time scale is the set of integers, then the dynamic equation is reduced to a difference equation. Thus integral transform of differential and integral calculus on time scales allows us to develop a theory of differential equations. For the single Laplace transform on time scales, see [1–3]. Similarly, in [4–8], the authors defined the time scale to be any nonempty closed subset of the real numbers and provided the motivation and formulation of delta derivatives on a time scale as well as the properties of delta derivatives and integrals. In [2], Bohner and Peterson defined the Laplace transform of a time scale for function of single variable as follows.Definition 1.
Forf:⊤→ℂ, the time scale or generalized Laplace transform of f, denoted by L[f(t)] or F(z), is given by
(1)L[f(t)](z)=∫0∞‍e⊖zσ(t,0)f(t)Δt,forz∈D{f},
where D{f} consists of all z∈ℂ for which the improper integral exists and 1+μ1z≠0 for all t∈⊤.Definition 2.
The functionf:⊤→ℝ is said to be of exponential type I if there exist constants M,c>0 such that |f(t)|≤Mect. Furthermore, f  is said to be of exponential type II if there exist constants M,  c>0 such that |f(t)|≤Mec(t,0).The following two theorems were proved in [3].Theorem 3.
The integral∫0∞‍e⊖zσ(t,0)f(t)Δt converges absolutely for z∈D if f(t) is of  exponential type II with exponential constant c. For more details, see [3].Theorem 4.
Letα∈ℝ be regressive. Then
(I)(2)L{eα(t,0)}(z)=1z-α,
provided that limt→∞eα⊖z(t,0)=0,
(II)(3)L{cosα(t,0)}(z)=zz2+α2,
provided that limt→∞eiα⊖z(t,0)=limt→∞e-iα⊖z(t,0)=0,
(III)(4)L{sinα(t,0)}(z)=αz2+α2,
provided that limt→∞eiα⊖z(t,0)=limt→∞e-iα⊖z(t,0)=0.The following theorem was also studied by Bohner and Peterson in [2].Theorem 5.
Iff:⊤→ℝ is such that fΔ is regulated, then
(5)L[fΔ](z)=zL[f(t)](z)-f(0),
for all z∈D{f} such that limt→∞{e⊖z(t)f(t)}=0.Throughout the study, the following notation will be useful, where⊤1,⊤2 are time scales and
(6)ea⊕b(t1,t2)=eb(t1)eb(t2),e⊖z1⊖z2(t1,t2)=e⊖z1(t1)e⊖z2(t2).In [2], further it was proved that
(7)e⊖zσ=e⊖z1(1+μz)
was satisfied, where 1+μ1z≠0, by using the property
(8)ezσ=(1+μz)ez,⊖z=-z1+z,p⊖q=p-q(1+μq).Later in [9], Bohner and Guseinov defined the single convolution operation on time scale of two functions f,g:⊤→ℝ by the following formula:
(9)(f*g)(t)=∫t0t‍f⌢(t,σ(s))g(s)Δs.In this study, we introduce double Laplace transform on time scales and study some of the properties in solving partial differential equations. Further, we also extend theΔ-Laplace transform, which is given by Jackson in [10], to the multiple Δ-Laplace transform as follows.Definition 6.
Let0∈⊤1,⊤2,  sup{⊤1,⊤2}=∞, and t1∈⊤1,t2∈⊤2. The Δ-double Laplace transform of the function f(t1,t2)   (for  f∈C(⊤1×⊤2)) with respect to t1,t2 is given by
(10)Lt1Lt2[f(t1,t2)](z1,z2)=∬0∞e⊖z1⊖z2σ1σ2(t1,t2)f(t1,t2)Δ1t1Δ2t2
provided that the integral exists, further (1+μ1z1)≠0, and (1+μ2z2)≠0, for all (t1,t2)∈(⊤1×⊤2)+.Lemma 7.
Ifz1,z2∈ℂ are regressive, then
(11)e⊖z1⊖z2σ1σ2=e⊖z1⊖z2(1+μ1z1)(1+μ2z2).Proof.
By using the above analysis, we have(12)e⊖z1⊖z2σ1σ2=((1+μ1(⊖z1)))e⊖z1(1+μ2(⊖z2))e⊖z2=[1-μ1z11+μ1z1][1-μ2z21+μ2z2]e⊖z1e⊖z2=e⊖z1⊖z2(1+μ1z1)(1+μ2z2).As an example, we can easily computeLt2Lt1[ea⊕b(t1,t2,0,0)](z1,z2), where a,b are constants such that limt1→∞t2→∞ea⊕b⊖z1⊖z2(t1,t2,0,0)=0.We have(13)Lt2Lt1[ea⊕b(t1,t2,0,0)](z1,z2)=∬0∞ea⊕b(t1,t2,0,0)e⊖z1⊖z2σ1σ2(t1,t2)Δ1t1Δ2t2=∬0∞(1+μσ1(t1)⊖z1(t1))(1+μσ2(t2)⊖z2(t2))ea⊕b×(t1,t2,0,0)e⊖z1⊖z2(t1,t2)Δ1t1Δ2t2=∫0∞‍(1-μσ1(t1)z11+μσ1⊖z1)×[∫0∞‍(1-μσ1(t2)z21+μσ2⊖z2)eb⊖z2(t1,t2,0,0)Δ2t2]ea⊖z1×(t1,t2)Δ1t1=∫0∞‍(11+μσ1⊖z1)×[∫0∞‍(11+μσ2⊖z2)eb⊖z2(t1,t2,0,0)Δ2t2]×ea⊖z1(t1,t2,0,0)Δ1t1=1a-z1∫0∞‍[1b-z2∫0∞‍b⊖z2eb⊖z2(t1,t2,0,0)Δ2t2]×a⊖z1ea⊖z1(t1,t2,0,0)Δ1t1=1a-z1∫0∞‍[1b-z2∫0∞‍(eb⊖z2(t1,t2,0,0))Δ2t2Δ2t2]×(ea⊖z1(t1,t2,0,0))Δ1t1Δ1t1=1(z1-a)(z2-b).
For t1∈⊤1 and t2∈⊤2 with t1≥α and  t2≥β, we can easily see that
(14)cosh(α⊕β)(t1,t2)=e(αt1⊕βt2)+e-(αt1⊕βt2)2,sinh(α⊕β)(t1,t2)=e(αt1⊕βt2)-e-(αt1⊕βt2)2,cos(α⊕β)(t1,t2)=ei(αt1⊕βt2)+e-i(αt1⊕βt2)2,sin(α⊕β)(t1,t2)=ei(αt1⊕βt2)-e-i(αt1⊕βt2)2i,
and then the double Laplace transforms are given by
(15)Lt2Lt1[cosh(α⊕β)(t1,t2)]=1/((z1-α)(z2-β))+1/((z1+α)(z2+β))2=z1z2+αβ(z22-β2)(z12-α2),(16)Lt2Lt1[sinh(α⊕β)(t1,t2)]=1/((z1-α)(z2-β))-1/((z1+α)(z2+β))2=βz1+αz2(z22-β2)(z12-α2).Similarly, we have(17)Lt2Lt1[cos(α⊕β)(t1,t2)]=z1z2-αβ(z22+β2)(z12+α2),Lt2Lt1[sin(α⊕β)(t1,t2)]=z1z2+αβ(z22+β2)(z12+α2).Proposition 8.
Suppose⊤1,  ⊤2 are two time scales such that μ1<N1 and μ1<N2 for some N1,N2∈ℝ and all (t1,t2)∈⊤1×⊤2. Let f:(⊤1×⊤2)+→ℝ and
(18)limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(σ(t1),σ(t2),0,0)}=0.
Then
(19)limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(t1,t2,0,0)}=0.Proof.
Since(20)limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(σ(t1),σ(t2),0,0)}=0
and on using the following relation:
(21)e⊖p1⊖p2(σ(t1),σ(t2),0,0)=1(1+μ1p1)(1+μ2p2)e⊖p1⊖p2(t1,t2,0,0)=⊖p1⊖p2p1p2e⊖p1⊖p2(t1,t2,0,0),
we have
(22)limt1→∞t2→∞{⊖z1(t1)⊖z2(t2)z1z2e⊖z1⊖z2(t1,t2)f(t1,t2,0,0)}=0,limt1→∞t2→∞{⊖z1(t1)⊖z2(t2)z1z2(1+μ1z1)(1+μ2z2)e⊖z1⊖z2×(t1,t2)f(t1,t2,0,0)⊖z1(t1)⊖z2(t2)z1z2(1+μ1z1)(1+μ2z2)}=0,limt1→∞t2→∞{1(1+μ1z1)(1+μ2z2)e⊖z1⊖z2×(t1,t2)f(t1,t2,0,0)1(1+μ1z1)(1+μ2z2)}=0.
Since μ1(t2)<N1 and μ1(t2)<N2, we get
(23)limt1→∞t2→∞|{e⊖z1⊖z2(t1,t2)f(t1,t2,0,0)(1+μ1z1)(1+μ2z2)}|>|limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(t1,t2,0,0)}(1+N1z1)(1+N2z2)|>0.
Thus we obtain that
(24)1(1+N1z1)(1+N2z2)×limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(t1,t2,0,0)}=0
and it follows that
(25)limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(t1,t2,0,0)}=0.Theorem 9.
Letf:(⊤1×⊤2)+→ℝ be regulated and let
(26)F(t1,t2)=∬0t1f(ζ,η)ΔζΔη
for (t1,t2)∈(⊤1×⊤2)+. Then
(27)Lt1Lt2[f(t1,t2)](z1,z2)=1z1z2Lt1Lt2[f(t1,t2)](z1,z2)-limt1→∞t2→∞e⊖z1⊖z2(t1,t2)f(t1,t2,0,0),
for z1,z2≠0.Proof.
By using the definition of the double Laplace transform on a time scale, we have(28)Lt1Lt2[f(t1,t2)](z1,z2)=∬0∞e⊖z1⊖z2σ1σ2(t1,t2)F(t1,t2,0,0)Δ1t1Δ2t2,
and on using (21), we have
(29)Lt1Lt2[f(t1,t2)](z1,z2)=1z1z2∬0∞e⊖z1⊖z2(t1,t2)(⊖z1⊖z2)×F(t1,t2,0,0)Δ1t1Δ2t2.
Now, applying integration by parts, and on using the fact that F(0,0)=0 together with the fundamental theorem of calculus, we obtain that
(30)Lt1Lt2[f(t1,t2)](z1,z2)=1z1z2Lt1Lt2[f(t1,t2)](z1,z2)-limt1→∞t2→∞e⊖z1⊖z2(t1,t2)f(t1,t2,0,0).Example 10.
Leta∈⊤1, b∈⊤2, a,b>0, and Ha,b(t1,t2)=0 for t1∈⊤1 and t2∈⊤2,t1<a, and t2<b whilst Ha,b(t1,t2)=1 for t1∈⊤1, t2∈⊤2 and t1≥a,  t2≥b, where Ha,b(t1,t2)=Ha,(t1)⊗Hb(t2) and the symbol ⊗ denotes the tensor product. Then we have
(31)Lt1Lt2[Ha,b(t1,t2)](z1,z2)=∬0∞e⊖z1⊖z2σ1σ2(t1,t2)Ha,b(t1,t2)Δ1t1Δ2t2=e⊖z1(a)e⊖z2(b)z1z2.Now assume thatf(t1,t2):⊤1×⊤2→ℂ such that  fΔ1 is continuous. Then the Δ-double Laplace transform
(32)Lt1Lt2[fΔ1(t1,t2)](z1,z2)=z1F=(z1,z2)-F-(0,z2)
holds for those regressive z1,z2∈ℂ with respect to t1 and t2 which further satisfies
(33)limt1→∞t2→∞{e⊖z1⊖z2(t1,t2)f(t1,t2)}=0.
We next extend the result that was proved by Davis et al. in [3].Proposition 11.
Let(a,b)∈(⊤1×⊤2)+, a,b>0. Assume that f(t1,t2) is one of the following functions:
(34)ea⊕b(t1,t2,0,0),sinh(α⊕β)(t1+t2),cosh(α⊕β)(t1+t2),cos(α⊕β)(t1+t2),orsin(α⊕β)(t1+t2).
If z1,z2,α,andβ are regressive and satisfy
(35)limt1→∞t2→∞{eα⊖z1β⊖z2(t1,t2)f(t1,t2,a,b)}=limt1→∞t2→∞{eiα⊖z1iβ⊖z2(t1,t2)f(t1,t2,a,b)}=limt1→∞t2→∞{e-iα⊖z1-iβ⊖z2(t1,t2)f(t1,t2,a,b)},
then
(36)Lt1Lt2[Ha,b(t1,t2)f(t1,t2,a,b)](z1,z2)=e⊖z1⊖z2(a,b,0,0)Lt1Lt2[f(t1,t2,0,0)].Proof.
Let us study the casef(t1,t2,a,b)=eα⊕β(t1,t2,a,b). By using (21), we get
(37)eα⊕β(t1,t2,a,b)e⊖z1⊖z2(σ(t1),σ(t2),0,0)=eα⊕β(t1,t2,a,b)e⊖z1⊖z2(t1,t2,0,0)(1+μ1z1)(1+μ2z2)
and by using Theorem 4 and (21) the right hand side of equation (37) is
(38)=e⊖z1⊖z2(a,b,0,0)(α-z1)(β-z2)((α-z1)(β-z2)(1+μ1z1)(1+μ2z2))×eα⊖z1β⊖z2(t1,t2,a,b)=e⊖z1⊖z2(a,b,0,0)(α-z1)(β-z2)(α⊖z1)(β⊖z2)eα⊖z1β⊖z2(t1,t2,a,b).
Then
(39)Lt1Lt2[Ha,b(t1,t2)f(t1,t2,a,b)](z1,z2)=∬0∞e⊖z1⊖z2(σ(t1),σ(t2),0,0)Ha,b(t1,t2)eα⊕β×(t1,t2,a,b)Δ1t1Δ2t2=e⊖z1⊖z2(a,b,0,0)(α-z1)(β-z2)∬0∞Ha,b(t1,t2)×((α⊖z1)(β⊖z2))(t1,t2)×eα⊖z1β⊖z2(t1,t2,a,b)×Δ1t1Δ2t2=e⊖z1⊖z2(a,b,0,0)(α-z1)(β-z2)∫b∞‍∫a∞‍((α⊖z1)(β⊖z2))×eα⊖z1β⊖z2(t1,t2,a,b)×Δ1t1Δ2t2=e⊖z1⊖z2(a,b,0,0)(α-z1)(β-z2)=e⊖z1⊖z2(a,b,0,0)Lt1Lt2[eα⊕β(t1,t2,0,0)].Theorem 12.
Letf(t1,t2):⊤1×⊤2→ℂ and suppose that  fΔi and  fΔj are continuous for all i=1,2,…n and j=1,2,…,m, respectively; then the Δ-double Laplace transforms for fΔi and  fΔj are given by
(40)Lt1Lt2[fΔn1(t1,t2)](z1,z2)=z1nF̿(z1,z2)-z1n-1F-(0,z2)-⋯-F-Δ1n-1(0,z2),(41)Lt1Lt2[fΔm2(t1,t2)](z1,z2)=z2mF̿(z1,z2)-z2m-1F-(z1,0)-⋯-F-Δ2m-1(z1,0),
for all z1,z2∈ℂ with
(42)limt1→∞{e⊖z1⊖z2(t1,t2)f(t1,t2)}=limt1→∞{e⊖z1⊖z2(t1,t2)fΔn1(t1,t2)}⋯=limt1→∞{e⊖z1⊖z2(t1,t2)fΔn-11(t1,t2)}=0,limt2→∞{e⊖z1⊖z2(t1,t2)f(t1,t2)}=limt2→∞{e⊖z1⊖z2(t1,t2)fΔm2(t1,t2)}⋯=limt2→∞{e⊖z1⊖z2(t1,t2)fΔm-12(t1,t2)}=0,
respectively.Proof.
Letn=1. On using the definition of the Δ-single Laplace transform and integrating parts of the function fΔ11(t1,t2), we have
(43)Lt1[fΔ11(t1,t2)](z1)=∫0∞‍e⊖z1σ1fΔ11(t1,t2)Δt1=[f(t1,t2)]0∞+z1∫0∞‍e⊖z1σ1f(t1,t2)Δt1=-f(0,t2)+z1F-(z1,t2).
Then the single Laplace transform with respect to t1 is given by
(44)Lt1[fΔ11(t1,t2)](z1,z2)=z1F-(z1,t2)-f(0,t2).
Similarly, Laplace transform for (44) with respect to t2 is given by
(45)Lt2Lt1[fΔ11(t1,t2)](z1)=z1F̿(z1,z2)-F-(0,z2).
Then (45) is called the Δ-double Laplace transform for the function fΔ11(t1,t2). Similarly, if m=1 in (41), the Δ-double Laplace transform for the function fΔ12(t1,t2) is given by
(46)Lt2Lt1[fΔ12(t1,t2)](z1,z2)=z2F̿(z1,z2)-F-(z1,0).
Now, in order to obtain the Δ-double Laplace transform for the function
(47)fΔ21(t1,t2)=∂2f(t1,t2)Δ1t12
with respect to t1,t2, we proceed as follows: first of all by taking the Δ-single Laplace transform with respect to the variable t1 and on using the fact that
(48)limt1→∞{e⊖z1(t1)fΔ11(t1,t2)}=0,limt1→∞{e⊖z1(t1)f(t1,t2)}=0,
we have
(49)Lt1[fΔ21(t1,t2)](z1)=[fΔ11(t1,t2)]0∞+z1∫0∞‍e⊖z1σ1fΔ11(t1,t2)Δt1=-fΔ11(0,t2)+z1∫0∞‍e⊖z1σ1fΔ11(t1,t2)Δt1.
Thus on using (44), we obtain that
(50)Lt1[fΔ21(t1,t2)](z1)=z12F-(z1,t2)-z1F-(0,t2)-fΔ1(0,t2).
In a similar way, the single Laplace transform with respect to t2 is given by
(51)Lt2Lt1[fΔ21(t1,t2)](z1,z2)=z12F̿(z1,z2)-z1F-(0,z2)-F-Δ1(0,z2).
Further, the Δ-double Laplace transform for the function fΔ22(t1,t2)=∂2f(t1,t2)/Δ2t22 with respect to t1,t2 is given by
(52)Lt2Lt1[fΔ22(t1,t2)](z1,z2)=z22F̿(z1,z2)-z2F-(z1,0)-F-Δ1(z1,0).
Finally, note that we can generalize the proof for n,m.Next, we define the double convolution of two time scale functions as follows.Definition 13.
The double convolution of the functionf,  g:T1×T2→ℝ is given by
(53)(f**g)(t1,t2)=∬0t1f(ζ,η)g(t1,t2,σ1(ζ),σ2(η))ΔζΔη.In the next theorem we prove the properties of the double convolution on a time scale as follows.Theorem 14.
Letf,g,h:T1×T2→ℝ be integrable functions on a time scale. Then the following properties are satisfied:
(54)(f**g)(t1,t2)=(g**f)(t1,t2),(f**g)**h=f**(g**h).Proof.
We can easily show that convolution is commutative by using the definition of the double convolution(55)(f**g)(t1,t2)=∬0t1f(ζ,η)g(t1,t2,σ1(ζ),σ2(η))ΔζΔη=∬0t1f(t1,t2,σ1(ζ),σ2(η))g(ζ,η)ΔζΔη.
For the proof of the associative law, we also use the definition as follows:
(56)(f**g)**h=∬0t1(f**g)(t1,t2,σ1(ζ),σ2(η))h(ζ,η)ΔζΔη=∬0t1∫σ2(η)t1‍∫σ1(ζ)t1‍f(t1,t2,σ1(α),σ2(β))×g(α,β,σ1(ζ),σ2(η))×h(ζ,η)ΔζΔηΔαΔβ=∬0t1f(t1,t2,σ1(α),σ2(β))(f**g)(α,β)ΔαΔβ×f**(g**h)(t1,t2).Theorem 15 (convolution theorem).
Letf,  g:T1×T2→ℂ be regulated functions. Then the double Laplace transform of the double convolution is given by
(57)Lt2Lt1[(f**g)(t1,t2);z1,z2]=F(z1,z2)G(z1,z2).Proof.
By using the definition of the double Laplace transform on a time scale, we obtain that(58)Lt2Lt1[(f**g)(t1,t2);z1,z2]=∬0∞e⊖z1⊖z2σ1σ2(t1,t2,0,0)[(f**g)(t1,t2)]Δ1t1Δ2t2=∬0∞[∬0t1f(ζ,η)g(t1,t2,σ1(ζ),σ2(η))×ΔζΔηe⊖z1⊖z2σ1σ2(t1,t2,0,0)∬t1]Δ1t1Δ2t2=∬0∞f(ζ,η)[∫σ1(ζ)∞‍∫σ2(η)∞‍g(t1,t2,σ1(ζ),σ2(η))⊖z1⊖z2σ1σ2×(t1,t2,0,0)Δ1t1Δ2t2∫σ2(η)∞‍]ΔζΔη=∬0∞f(ζ,η)[Lt2Lt1[Hσ1(ζ),σ2(η)(t1,t2)×g(t1,t2,σ1(ζ),σ2(η))Hσ1(ζ),σ2(η)]Lt2Lt1]ΔζΔη,
where Hσ1(ζ),σ2(η)(t1,t2)=Hσ1(ζ),(t1)⊗Hσ2(ζ),(t2)  and the symbol ⊗ denotes the tensor product. Thus we have
(59)Lt2Lt1[(f**g)(t1,t2);z1,z2]=∬0∞f(ζ,η)[G(z1,z2)⊖z1⊖z2σ1σ2(ζ,η,0,0)]ΔζΔη=G(z1,z2)∬0∞f(ζ,η)⊖z1⊖z2σ1σ2(ζ,η,0,0)ΔζΔη=F(z1,z2)G(z1,z2).Finally, we give the solution to the wave equation in one dimension as follows. Consider the wave equation in the form of(60)uΔ1Δ1-uΔ2Δ2=0,
under the conditions
(61)u(x,0)=e1(x),u(0,t)=e1(t),uΔ2(0,t)=e1(t),uΔ1(x,0)=e1(x).
Then by formally taking the double Laplace transform, we get
(62)Lt2Lt1[u(t,x);z1,z2]=z1+1(z2-1)(z12-z22)-z2+1(z1-1)(z12-z22).
By using partial fractions, we have
(63)Lt2Lt1[u(t,x);z1,z2]=12(z2-1)(z1-z2)+12(z2-1)(z1+z2)+12z2(z2-1)(z1-z2)-12z2(z2-1)(z1+z2)-12(z1-1)(z1-z2)+12(z1-1)(z1+z2)-12z1(z1-1)(z1-z2)-12z1(z1-1)(z1+z2).
Thus the solution to the above equation is given by
(64)u(t,x)=e1⊕1(t,x,0,0).
In particular, if we consider the case T1×T2=ℝ×ℝ, σ(t)=t,σ(x)=x, for all (t,x)∈T1×T2, then ea⊕b(t,x,0,0)=eax+bt for any constant a,b∈ℝ and so the solution is given by
(65)u(t,x)=et+x.



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*Source: 217656-2013-12-18.xml*