# Note on Relation between Double Laplace Transform and Double Differential Transform

Authors: Hassan Eltayeb
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

Double differential transform method has been employed to compute double Laplace transform. To illustrate the method, four examples of different forms have been prepared.


---

## Body

## 1. Introduction

The concept of the DTM was first proposed by Zhou [1], who solved linear and nonlinear problems in electrical circuit problems. Chen and Ho [2] developed this method for partial differential equations and applied it to the system of differential equations. During the recent years, many authors have used this method for solving various types of equations. For example, this method has been used for differential algebraic equations [3], partial differential equations [4, 5], fractional differential equations [6], Volterra integral equations [7], and difference equations [8]. The main goal of this paper is to extend the study of single Laplace transform by using differential transform (see [9]) and to compute double Laplace transform by means of double differential transform method. As we know, the standard derivation of Laplace transforms inherits an improper integration which may, in certain cases, not be analytically tractable. However, in contrast, the proposed straightforward approach merely requires easy differentiations and algebraic operations. Three examples are proposed.The one-dimensional differential transform of the functionu(x) is defined by the following formula:
(1)U(k)=1k![dku(x)dxk]x=0,
where u(x) and U(k) are the original and transform functions, respectively. The inverse differential transform of U(k) is specified as follows:
(2)u(x)=∑k=0∞‍[U(k)xk].
Consider an analytical function u(x,t) of two variables; then this function can be represented as a series in (x0,t0)∈D using differential transform
(3)U(m,n)=1m!n![∂m+nu(x,t)∂xm∂tn]x=x0,t=t0
and inverse double differential transform
(4)u(x,t)=∑m=0∞‍∑n=0∞‍1m!n![∂m+nu(x,t)∂xm∂tn]x=0,t=0.
From the definition of double Laplace transform, we can write
(5)LxLt[f(x,t)]=F(p,s)=[∫x0x‍e-px[∫t0t‍e-stf(x,t)dt]t0=0t=∞dx]x0=0x=∞,
where x, t>0 and p, s is a complex value.On using double direct and inverse differential transform with respect tox and t for both sides of the previous,
(6)F(p,s)=DTp-1DTs-1×{{[∫x0x‍e-px[∫t0t‍e-stf(x,t)dt]t0=0t=∞×dx∫x0x‍e-px[∫t0t‍e-stf(x,t)dt]t0=0t=∞]x0=0x=∞}DTxDTt×{[∫x0xe-px[∫t0te-stf(x,t)dt]t0=0t=∞dx]x0=0x=∞}},F(p,s)=[∫x0xe-px{DTt∫t0te-stf(x,t)dt}dxD1D447x∫x0x‍e-px{DTt∫t0t‍e-stf(x,t)dt}}]t0=0t=∞[{DTx∫x0x‍e-px{DTt∫t0t‍e-stf(x,t)dt}dx}DTp-1DTs-1×{{DTt∫t0t‍e-stf(x,t)dt}DTx∫x0xe-px{DTt∫t0te-stf(x,t)dt}dxD1D447x∫x0x‍e-px{DTt∫t0t‍e-stf(x,t)dt}}]t0=0t=∞{DTx∫x0x‍e-px{DTt∫t0t‍e-stf(x,t)dt}dx}]t0=0t=∞]x0=0x=∞.Lemma 1.
Letm and n be a finite positive integers such that m,  n≥1 and s, p>0, s, p∈R2; then
(7)limx→∞t→∞∑k=1∞∑r=1∞[kmrnk!r!(-st)k(-px)r]=0.Proof.
By using mathematical induction, lettingm, n=1, we have
(8)(∑k=1∞kk!(-st)k)(∑r=1∞rr!(-px)r)=(∑k=1∞1(k-1)!(-st)k)(∑r=1∞1(r-1)!(-px)r)=(∑k=0∞1k!(-st)k+1)(∑r=0∞1r!(-px)r+1)=((-st)∑k=0∞1k!(-st)k)((-px)∑r=0∞1r!(-px)r)=(-st)e-st(-px)e-px.
Then limx→∞t→∞∑k=1∞∑r=1∞[(kr/k!r!)(-st)k(-px)r]=limx→∞t→∞(-st)e-st(-px)e-px=0; since (∑k=1∞(k/k!)(-st)k)(∑r=1∞(r/r!)(-px)r)=(-st)e-st(-px)e-px we conclude that
(9)limx→∞t→∞(-st)(-px)(∑k=1∞kk!(-st)k)(∑r=1∞rr!(-px)r)=0,
for m, n=2,
(10)(∑k=1∞k2k!(-st)k)(∑r=1∞r2r!(-px)r)=(∑k=1∞k(k-1)!(-st)k)(∑r=1∞r(r-1)!(-px)r)=(∑k=0∞k+1k!(-st)k+1)(∑r=0∞r+1r!(-px)r+1)=(-st)(∑k=0∞kk!(-st)k+∑k=0∞(-st)kk!)×(-px)(∑r=0∞rr!(-px)r+∑r=0∞(-px)rr!)=(-st)[(-st)e-st+e-st]×(-px)[(-px)e-px+e-px].
Consequently
(11)limx→∞t→∞(∑k=1∞k2k!(-st)k)(∑r=1∞r2r!(-px)r)=limx→∞t→∞(-st)[(-st)e-st+e-st]×(-px)[(-px)e-px+e-px]=0.
Similarly
(12)limx→∞t→∞(-st)(-px)(∑k=1∞k2k!(-st)k)(∑r=1∞r2r!(-px)r)=limx→∞t→∞(-st)2[(-st)e-st+e-st]×(-px)2[(-px)e-px+e-px]=0.
Assume that, for m, n=1 to i, j, it holds that
(13)limx→∞t→∞∑k=1∞∑r=1∞[kmrnk!r!(-st)k(-px)r]=0,
and also
(14)limx→∞t→∞(-st)(-px)(∑k=1∞kmk!(-st)k)×(∑r=1∞rnr!(-px)r)=0.
Now, we are going to prove that
(15)limx→∞t→∞∑k=1∞∑r=1∞[km+1rn+1k!r!(-st)k(-px)r]=0,(16)∑k=1∞∑r=1∞[km+1rn+1k!r!(-st)k(-px)r]=∑k=1∞∑r=1∞[kmrn(k-1)!(r-1)!(-st)k(-px)r]=(-st)(-px)∑k=0∞∑r=0∞[(k+1)m(r+1)nk!r!(-st)k(-px)r].
By using the definition of polynomial, we have
(17)(-st)∑k=0∞[(k+1)mk!(-st)k](-px)∑r=0∞[(r+1)nr!(-px)r]=(-st)∑k=0∞[(1+(m1)k+(m2)k2+⋯+km)k!(-st)k]×(-px)∑r=0∞[(1+(n1)r+(n2)r2+⋯+rn)r!(-px)r]=((-st)∑k=0∞(-st)kk!+(m1)(-st)∑k=0∞k(-st)kk!+(m2)(-st)∑k=0∞k2(-st)kk!+⋯+(-st)∑k=0∞km(-st)kk!∑k=0∞(-st)kk!)×((-px)∑r=0∞(-px)rr!+(n1)(-px)∑r=0∞r(-px)rr!+(n2)(-px)∑r=0∞r2(-px)rr!+⋯+(-px)∑r=0∞rn(-px)rr!).
Thus
(18)limx→∞t→∞∑k=1∞∑r=1∞[km+1rn+1k!r!(-st)k(-px)r]=0.


## 2. Relation between Double Laplace and Differential Transforms

In this section, we compute the double Laplace transform by means of double differential transform by proposing some examples as follows.Example 2.
Double Laplace transform of functioneax+bt; consider
(19)LxLt{eax+bt}=[{y0yeax-pxdxβ0βebt-stdt}{∫y0yeax-pxdx∫β0βebt-stdt}}DTp-1DTs-1×{{y0yeax-pxdxβ0βebt-stdt}{∫y0yeax-pxdx∫β0βebt-stdt}DTxDTt×{∫y0yeax-pxdx∫β0βebt-stdt}}]y0=0,β0=0y=∞,β=∞=[DTp-1{(-p+a)r-1r(r-1)!}]y0=0y=∞×[DTs-1{(-s+b)k-1k(k-1)!}]β0=0β=∞=[∑r=0∞{(-p+a)r-1yrr(r-1)!}]y0=0y=∞×[∑k=0∞{(-s+b)k-1βkk(k-1)!}]β0=0β=∞=1(a-p)[∑r=0∞{(-p+a)ryrr!}]y0=0y=∞×1(b-s)[∑k=0∞{(-s+b)kβkk!}]β0=0β=∞=1(a-p)[e(a-p)y]y0=0y=∞1(b-s)[e(b-s)β]β0=0β=∞=1(p-a)(s-b),p>a,s>b.
In the next example, we apply double differential transform to compute double Laplace transform as follows.Example 3.
If we consider the functionxmtn, then double Laplace transform is given by
(20)LxLt{xmtn;p,s}=[{DTxDTt{y0yxme-pxdxβ0βtne-stdt}}{DTxDTt{∫y0yxme-pxdx∫β0βtne-stdt}}DTp-1DTs-1×{DTxDTt{∫y0yxme-pxdx∫β0βtne-stdt}}]y0=0,β0=0y=∞,β=∞=[∑r=0∞{DTx{∫y0yxme-pxdx}yr}]y0=0y=∞×[∑k=0∞{DTt{∫β0βtne-stdt}βk}]β0=0β=∞.
From the properties of double differential transform, we have
(21)LxLt{xmtn}=0+[∑r=0∞{∑l=0r-1{δ(r-m-l-1)r(-p)ll!}yr}]y0=0y=∞×[∑k=0∞{∑c=0k-1{δ(k-n-c-1)k(-s)cc!}βk}]β0=0β=∞.
On using definition of Kronecker delta function forces, we have
(22)LxLt{xmtn}=[∑r=0∞{1r(-p)r-1-m(r-1-m)!}yr]y0=0y=∞×[∑k=0∞{1k(-s)k-1-n(k-1-n)!}βk]β0=0β=∞=[∑r=1∞{(r-1)(r-2)…(r-m)(-p)rr!yr}(-1)m+1pm+1×∑r=1∞{(r-1)(r-2)⋯(r-m)(-p)rr!yr}]y0=0y=∞×[(-1)n+1sn+1×∑k=0∞{(k-1)(k-2)⋯(k-n)(-s)kk!βk}]β0=0β=∞.
From the previous equation, we have
(23)LxLt{xmtn;p,s}=(-1)m+1pm+1([∑r=1∞{(-1)(-2)⋯(-m)(-p)rr!yr}]y0=0y=∞+∑i=1mailimy→∞∑r=1∞{rir!yr}-0)×(-1)n+1sn+1([∑k=1∞{(-1)(-2)⋯(-n)(-s)kk!βk}]β0=0β=∞+∑j=1nbjlimβ→∞∑k=1∞{kjk!βk}-0),
where ai and bj are constant coefficients of the polynomials generated by (r-1)(r-2)⋯(r-m) and (k-1)(k-2)⋯(k-n), respectively. According to the lemma, the last summations of (23), limy→∞∑r=1∞{(ri/r!)yr}, and limβ→∞∑k=1∞{(kj/k!)βk} are zeros, such that
(24)LxLt{xmtn}=(-1)m+1pm+1[∑r=1∞{(-1)(-2)⋯(-m)(-p)rr!yr}]y0=0y=∞×(-1)n+1sn+1[∑k=1∞{(-1)(-2)⋯(-n)(-s)kk!βk}]β0=0β=∞=(-1)m+1m!pm+1[∑r=1∞{(-p)rr!yr}]y0=0y=∞×(-1)n+1n!sn+1[∑k=1∞{(-s)kk!βk}]β0=0β=∞=-m!pm+1[∑r=1∞{(-p)rr!yr}]y0=0y=∞×-n!sn+1[∑k=1∞{(-s)kk!βk}]β0=0β=∞=-m!pm+1[(e-py-1)]y0=0y=∞×-n!sn+1[(e-sβ-1)]β0=0β=∞=m!n!pm+1sn+1.
In the next example, we apply double integral transform as follows.Example 4.
If we consider the functionsin(ax)sin(bt), then double Laplace transform is given by
(25)LxLt{sin(ax)sin(bt);p,s}=[{y0yβ0βsin(ax)sin(bt)e-pxe-stdtdx}}DTp-1DTs-1×{{y0yβ0βsin(ax)sin(bt)e-pxe-stdtdx}DTxDTt×{∫y0y∫β0βsin(ax)sin(bt)e-pxe-stdtdx}}]y0=0,β0=0y=∞,β=∞,LxLt{sin(ax)sin(bt);p,s}=[∑l=0r-1{ajj!sin(jπ2)(-p)r-j-1/((r-j-1)!)r}DTp-1×{∑l=0r-1{ajj!sin(jπ2)(-p)r-j-1/(r-j-1)!r}}]y0=0y=∞×[∑i=0k-1{bii!sin(iπ2)(-s)k-i-1/((k-i-1)!)k}DTs-1×{∑i=0k-1{bii!sin(iπ2)(-s)k-i-1/(k-i-1)!k}}]β0=0β=∞,LxLt{sin(ax)sin(bt);p,s}=[∑r=0∞{∑l=1,3,5r-1{ajj!(r-1)(r-2)⋯(r-j)pj+1×(-1)(j-1)/2-j-1}yr(-p)rr!}]y0=0y=∞×[∑k=0∞{∑i=1,3,5k-1{bii!(k-1)(k-2)⋯(k-i)si+1×(-1)(i-1)/2-i-1bii!(k-1)(k-2)⋯(k-i)si+1}βk(-s)kk!}]β0=0β=∞,LxLt{sin(ax)sin(bt);p,s}=[∑r=0∞{∑l=1,3,5r-1{ajj!(r-1)(r-2)⋯(r-j)pj+1×(-1)(j+3)/2ajj!(r-1)(r-2)⋯(r-j)pj+1}yr(-p)rr!}]y0=0y=∞×[∑k=0∞{∑i=1,3,5k-1{bii!(k-1)(k-2)⋯(k-i)si+1×(-1)(i+3)/2}βk(-s)kk!}]β0=0β=∞.
By calculating the summation inside the bracket we have
(26)LxLt{sin(ax)sin(bt);p,s}=[∑r=0∞{ap2(r-1)(-py)rr!-a3(r-1)(r-2)(r-3)p4(-py)rr!+a5(r-1)(r-2)(r-3)p6(-py)rr!+⋯}∑r=0∞{ap2(r-1)(-py)rr!]y0=0y=∞×[∑k=0∞{bs2(k-1)(-sβ)kk!-b3(k-1)(k-2)(k-3)s4(-sβ)kk!+b5(k-1)(k-2)(k-3)s6(-sβ)kk!+⋯}∑k=0∞{bs2(k-1)(-sβ)kk!]β0=0β=∞.
From the previous lemma, we know that, for m, n≥1,  limy→∞∑r=1∞{(rm/r!)(-py)r}=0, and limβ→∞∑k=1∞{(km/k!)(-sβ)k}=0, we have the following form:
(27)LxLt{sin(ax)sin(bt);p,s}=[∑r=0∞{-ap2(-py)rr!+a3p4(-py)rr!-a5p6(-py)rr!+⋯}]y0=0y=∞×[∑k=0∞{-bs2(-sβ)kk!+b3s4(-sβ)kk!-b5s6(-sβ)kk!+⋯}]β0=0β=∞=[(-ap2+a3p4-a5p6+⋯)e-py]y0=0y=∞×[(-bs2+b3s4-b5s6+⋯)e-sβ]β0=0β=∞=(ap2-a3p4+a5p6+⋯)×(bs2-b3s4+b5s6+⋯).
From the definition of infinite geometric series and the summation of the previous terms, we have
(28)LxLt{sin(ax)sin(bt);p,s}=limm→∞ap2(1-(-a2/p2)m+1)1-(-a2/p2)×limn→∞bs2(1-(-b2/s2)n+1)1-(-b2/s2)=(a/p21+a2/p2)(b/s21+b2/s2)=ab(p2+a2)(s2+b2).In the next example, we apply double differential transform to find double Laplace transform of the function(sin(x)/x)(sin(t)/t) as follows.Example 5.
The double Laplace transform of the function(sin(x)/x)(sin(t)/t), as follows:
(29)LxLt{sin(x)xsin(t)t;p,s}=[DTp-1DTs-1{DTxDTt{∫y0y∫β0βsin(x)xsin(t)t×e-pxe-stdtdx∫y0y∫β0βsin(x)xsin(t)t}DTxDTt{∫y0y∫β0βsin(x)xsin(t)t}DTp-1DTs-1{DTxDTt{∫y0y∫β0βsin(x)xsin(t)t]y0=0,β0=0y=∞,β=∞.
By using sin series, we have
(30)sin(x)xsin(t)t=(∑i=0∞((-1)ix2i+1/(2i+1)!)x)×(∑j=0∞((-1)jt2j+1/(2j+1)!)t)=(∑i=0∞(-1)ix2i(2i+1)!)(∑j=0∞(-1)jt2j(2j+1)!).
By applying double differential transform, we have
(31)DTxDTt{sin(x)xsin(t)t}={∑i=0∞(-1)i(2i+1)!DTx{x2i}}×{∑j=0∞(-1)j(2j+1)!DTt{t2j}}.
On using double inverse differential transform, we have
(32)LxLt{sin(x)xsin(t)t;p,s}=[DTp-1{(-1)m/2(m+1)!(-p)k-m-1/(k-m-1)!k}]y0=0y=∞×[DTs-1{(-1)n/2(n+1)!(-s)β-n-1/(β-n-1)!β}]β0=0β=∞=[∑k=0∞{∑m=0,2,4k-1{(-1)m/2(k-1)(k-2)⋯(k-m)(m+1)!(-p)m+1×(-1)m+1(-1)m/2(k-1)(k-2)⋯(k-m)(m+1)!(-p)m+1}(-px)kk!∑m=0,2,4k-1{(-1)m/2(k-1)(k-2)⋯(k-m)(m+1)!(-p)m+1}]y0=0y=∞×[∑β=0∞{∑n=0,2,4β-1{(-1)n/2(β-1)(β-2)⋯(β-n)(n+1)!(-s)n+1×(-1)n+1(-1)n/2(β-1)(β-2)⋯(β-n)(n+1)!(-s)n+1}(-st)ββ!∑n=0,2,4β-1{(-1)n/2(β-1)(β-2)⋯(β-n)(n+1)!(-s)n+1}∑β=0∞{∑n=0,2,4β-1{(-1)n/2(β-1)(β-2)⋯(β-n)(n+1)!(-s)n+1]β0=0β=∞,(33)LxLt{sin(x)xsin(t)t;p,s}=[∑k=0∞{(-1p+(k-1)(k-2)3!p3-(k-1)(k-2)(k-3)(k-4)5!p5+(7!p7)-1((k-1)(k-2)(k-3)(k-1)×(k-5)(k-6))×(7!p7)-1+⋯(k-1)(k-2)3!p3)×(-px)kk!(-1p+(k-1)(k-2)3!p3}∑k=0∞{(-1p+(k-1)(k-2)3!p3]y0=0y=∞×[∑β=0∞{(-1s+(β-1)(β-2)3!s3-(β-1)(β-2)(β-3)(β-4)5!s5+((7!s7)-1(β-1)(β-2)(β-3)×(β-1)(β-5)(β-6))×(7!s7)-1+⋯(β-1)(β-2)3!s3)×(-st)ββ!(-1s+(β-1)(β-2)3!s3}∑β=0∞{(-1s+(β-1)(β-2)3!s3]β0=0β=∞.
According to the previous lemma, we have
(34)LxLt{sin(x)xsin(t)t;p,s}=[∑k=0∞{(-px)kk!(-1p+2!3!p3-4!5!p5+6!7!p7+⋯)×(-px)kk!}∑k=0∞{(-1p+2!3!p3-4!5!p5+6!7!p7+⋯)]y0=0y=∞×[∑β=0∞{(-st)ββ!(-1s+2!3!s3-4!5!s5+6!7!s7+⋯)×(-st)ββ!}∑β=0∞{(-1s+2!3!s3-4!5!s5+6!7!s7+⋯)]β0=0β=∞,(35)LxLt{sin(x)xsin(t)t;p,s}=(-1p+2!3!p3-4!5!p5+6!7!p7+⋯)[e-px]y0=0y=∞×(-1s+2!3!s3-4!5!s5+6!7!s7+⋯)[e-st]β0=0β=∞,(36)LxLt{sin(x)xsin(t)t;p,s}=(1p-13p3+15p5-17p7+⋯)×(1s-13s3+15s5-17s7+⋯).
By using the definition of
(37)∫dxx2+a2=1aarctan(xa),
we have
(38)arctan(1p)arctan(1s)=(1p-13p3+15p5-17p7+⋯)×(1s-13s3+15s5-17s7+⋯).
So that
(39)LxLt{sin(x)xsin(t)t;p,s}=arctan(1p)arctan(1s).Also, we can use the same idea to compute double Laplace transform for convolution function, single or double.



---

*Source: 535020-2013-06-24.xml*